Calculus Single Multivariable Solutions Manual
User Manual:
Open the PDF directly: View PDF 
.
Page Count: 1886
| Download | |
| Open PDF In Browser | View PDF |
Calculus
Single & Multivariable
6th Edition
INSTRUCTOR
SOLUTIONS
MANUAL
Hughes-Hallett
Gleason
McCallum
et al.
1.1 SOLUTIONS
1
CHAPTER ONE
Solutions for Section 1.1
Exercises
1. Since t represents the number of years since 1970, we see that f (35) represents the population of the city in 2005. In
2005, the city’s population was 12 million.
2. Since T = f (P ), we see that f (200) is the value of T when P = 200; that is, the thickness of pelican eggs when the
concentration of PCBs is 200 ppm.
3. If there are no workers, there is no productivity, so the graph goes through the origin. At first, as the number of workers
increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest
level, after which it goes downward; in other words, as the number of workers increases beyond that point, productivity
decreases. This might, for example, be due either to the inefficiency inherent in large organizations or simply to workers
getting in each other’s way as too many are crammed on the same line. Many other reasons are possible.
4. The slope is (1 − 0)/(1 − 0) = 1. So the equation of the line is y = x.
5. The slope is (3 − 2)/(2 − 0) = 1/2. So the equation of the line is y = (1/2)x + 2.
6. Using the points (−2, 1) and (2, 3), we have
Slope =
2
1
3−1
= = .
2 − (−2)
4
2
Now we know that y = (1/2)x + b. Using the point (−2, 1), we have 1 = −2/2 + b, which yields b = 2. Thus, the
equation of the line is y = (1/2)x + 2.
6−0
7. Slope =
= 2 so the equation is y − 6 = 2(x − 2) or y = 2x + 2.
2 − (−1)
5
5
8. Rewriting the equation as y = − x + 4 shows that the slope is − and the vertical intercept is 4.
2
2
9. Rewriting the equation as
2
12
y =− x+
7
7
shows that the line has slope −12/7 and vertical intercept 2/7.
10. Rewriting the equation of the line as
−2
x−2
4
1
y = x + 2,
2
−y =
we see the line has slope 1/2 and vertical intercept 2.
11. Rewriting the equation of the line as
4
12
x−
6
6
2
y = 2x − ,
3
y=
12. (a)
(b)
(c)
(d)
(e)
(f)
we see that the line has slope 2 and vertical intercept −2/3.
is (V), because slope is positive, vertical intercept is negative
is (IV), because slope is negative, vertical intercept is positive
is (I), because slope is 0, vertical intercept is positive
is (VI), because slope and vertical intercept are both negative
is (II), because slope and vertical intercept are both positive
is (III), because slope is positive, vertical intercept is 0
2
Chapter One /SOLUTIONS
13. (a)
(b)
(c)
(d)
(e)
(f)
is (V), because slope is negative, vertical intercept is 0
is (VI), because slope and vertical intercept are both positive
is (I), because slope is negative, vertical intercept is positive
is (IV), because slope is positive, vertical intercept is negative
is (III), because slope and vertical intercept are both negative
is (II), because slope is positive, vertical intercept is 0
14. The intercepts appear to be (0, 3) and (7.5, 0), giving
6
2
−3
=−
=− .
7.5
15
5
Slope =
The y-intercept is at (0, 3), so a possible equation for the line is
2
y = − x + 3.
5
(Answers may vary.)
15. y − c = m(x − a)
16. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then
Slope =
29.2 − 27.8
1.4
=
= 14.
5.3 − 5.2
0.1
Using the point-slope formula, with the point (5.2, 27.8), we get the equation
y − 27.8 = 14(x − 5.2)
which is equivalent to
y = 14x − 45.
17. y = 5x − 3. Since the slope of this line is 5, we want a line with slope − 51 passing through the point (2, 1). The equation
is (y − 1) = − 51 (x − 2), or y = − 51 x + 57 .
18. The line y + 4x = 7 has slope −4. Therefore the parallel line has slope −4 and equation y − 5 = −4(x − 1) or
−1
= 41 and equation y − 5 = 41 (x − 1) or y = 0.25x + 4.75.
y = −4x + 9. The perpendicular line has slope (−4)
19. The line parallel to y = mx + c also has slope m, so its equation is
y = m(x − a) + b.
The line perpendicular to y = mx + c has slope −1/m, so its equation will be
y=−
1
(x − a) + b.
m
20. Since the function goes from x = 0 to x = 4 and between y = 0 and y = 2, the domain is 0 ≤ x ≤ 4 and the range is
0 ≤ y ≤ 2.
21. Since x goes from 1 to 5 and y goes from 1 to 6, the domain is 1 ≤ x ≤ 5 and the range is 1 ≤ y ≤ 6.
22. Since the function goes from x = −2 to x = 2 and from y = −2 to y = 2, the domain is −2 ≤ x ≤ 2 and the range is
−2 ≤ y ≤ 2.
23. Since the function goes from x = 0 to x = 5 and between y = 0 and y = 4, the domain is 0 ≤ x ≤ 5 and the range is
0 ≤ y ≤ 4.
24. The domain is all numbers. The range is all numbers ≥ 2, since x2 ≥ 0 for all x.
1
.
2
26. The value of f (t) is real provided t2 − 16 ≥ 0 or t2 ≥ 16. This occurs when either t ≥ 4, or t ≤ −4. Solving f (t) = 3,
we have
25. The domain is all x-values, as the denominator is never zero. The range is 0 < y ≤
p
t2 − 16 = 3
t2 − 16 = 9
t2 = 25
1.1 SOLUTIONS
3
so
t = ±5.
27. We have V = kr 3 . You may know that V =
4 3
πr .
3
d
.
t
29. For some constant k, we have S = kh2 .
28. If distance is d, then v =
30. We know that E is proportional to v 3 , so E = kv 3 , for some constant k.
31. We know that N is proportional to 1/l2 , so
N=
k
,
l2
for some constant k.
Problems
32. The year 1983 was 25 years before 2008 so 1983 corresponds to t = 25. Thus, an expression that represents the statement
is:
f (25) = 7.019
33. The year 2008 was 0 years before 2008 so 2008 corresponds to t = 0. Thus, an expression that represents the statement
is:
f (0) meters.
34. The year 1965 was 2008 − 1865 = 143 years before 2008 so 1965 corresponds to t = 143. Similarly, we see that the
year 1911 corresponds to t = 97. Thus, an expression that represents the statement is:
f (143) = f (97)
35. Since t = 1 means one year before 2008, then t = 1 corresponds to the year 2007. Similarly, t = 0 corresponds to
the year 2008. Thus, f (1) and f (0) are the average annual sea level values, in meters, in 2007 and 2008, respectively.
Because 1 millimeter is the same as 0.001 meters, an expression that represents the statement is:
f (0) = f (1) + 0.001.
Note that there are other possible equivalent expressions, such as: f (1) − f (0) = 0.001.
36. (a)
(b)
(c)
(d)
Each date, t, has a unique daily snowfall, S, associated with it. So snowfall is a function of date.
On December 12, the snowfall was approximately 5 inches.
On December 11, the snowfall was above 10 inches.
Looking at the graph we see that the largest increase in the snowfall was between December 10 to December 11.
37. (a) When the car is 5 years old, it is worth $6000.
(b) Since the value of the car decreases as the car gets older, this is a decreasing function. A possible graph is in Figure 1.1:
V (thousand dollars)
(5, 6)
a (years
Figure 1.1
(c) The vertical intercept is the value of V when a = 0, or the value of the car when it is new. The horizontal intercept
is the value of a when V = 0, or the age of the car when it is worth nothing.
4
Chapter One /SOLUTIONS
38. (a) The story in (a) matches Graph (IV), in which the person forgot her books and had to return home.
(b) The story in (b) matches Graph (II), the flat tire story. Note the long period of time during which the distance from
home did not change (the horizontal part).
(c) The story in (c) matches Graph (III), in which the person started calmly but sped up later.
The first graph (I) does not match any of the given stories. In this picture, the person keeps going away from home,
but his speed decreases as time passes. So a story for this might be: I started walking to school at a good pace, but since I
stayed up all night studying calculus, I got more and more tired the farther I walked.
39. (a) f (30) = 10 means that the value of f at t = 30 was 10. In other words, the temperature at time t = 30 minutes was
10◦ C. So, 30 minutes after the object was placed outside, it had cooled to 10 ◦ C.
(b) The intercept a measures the value of f (t) when t = 0. In other words, when the object was initially put outside,
it had a temperature of a◦ C. The intercept b measures the value of t when f (t) = 0. In other words, at time b the
object’s temperature is 0 ◦ C.
40. (a) The height of the rock decreases as time passes, so the graph falls as you move from left to right. One possibility is
shown in Figure 1.2.
s (meters)
t (sec)
Figure 1.2
(b) The statement f (7) = 12 tells us that 7 seconds after the rock is dropped, it is 12 meters above the ground.
(c) The vertical intercept is the value of s when t = 0; that is, the height from which the rock is dropped. The horizontal
intercept is the value of t when s = 0; that is, the time it takes for the rock to hit the ground.
41. (a) We find the slope m and intercept b in the linear equation C = b + mw. To find the slope m, we use
m=
We substitute to find b:
∆C
12.32 − 8
=
= 0.12 dollars per gallon.
∆w
68 − 32
C = b + mw
8 = b + (0.12)(32)
b = 4.16 dollars.
The linear formula is C = 4.16 + 0.12w.
(b) The slope is 0.12 dollars per gallon. Each additional gallon of waste collected costs 12 cents.
(c) The intercept is $4.16. The flat monthly fee to subscribe to the waste collection service is $4.16. This is the amount
charged even if there is no waste.
42. We are looking for a linear function y = f (x) that, given a time x in years, gives a value y in dollars for the value of the
refrigerator. We know that when x = 0, that is, when the refrigerator is new, y = 950, and when x = 7, the refrigerator
is worthless, so y = 0. Thus (0, 950) and (7, 0) are on the line that we are looking for. The slope is then given by
m=
950
−7
It is negative, indicating that the value decreases as time passes. Having found the slope, we can take the point (7, 0) and
use the point-slope formula:
y − y1 = m(x − x1 ).
So,
950
(x − 7)
7
950
x + 950.
y=−
7
y−0 = −
1.1 SOLUTIONS
5
43. (a) The first company’s price for a day’s rental with m miles on it is C1 (m) = 40 + 0.15m. Its competitor’s price for a
day’s rental with m miles on it is C2 (m) = 50 + 0.10m.
(b) See Figure 1.3.
C (cost in dollars)
C1 (m) = 40 + 0.15m
150
100
C2 (m) = 50 + 0.10m
50
0
200
400
600
800
m (miles)
Figure 1.3
(c) To find which company is cheaper, we need to determine where the two lines intersect. We let C1 = C2 , and thus
40 + 0.15m = 50 + 0.10m
0.05m = 10
m = 200.
If you are going more than 200 miles a day, the competitor is cheaper. If you are going less than 200 miles a day, the
first company is cheaper.
∆$
55 − 40
44. (a) Charge per cubic foot =
=
= $0.025/cu. ft.
∆ cu. ft.
1600 − 1000
Alternatively, if we let c = cost, w = cubic feet of water, b = fixed charge, and m = cost/cubic feet, we obtain
c = b + mw. Substituting the information given in the problem, we have
40 = b + 1000m
55 = b + 1600m.
Subtracting the first equation from the second yields 15 = 600m, so m = 0.025.
(b) The equation is c = b + 0.025w, so 40 = b + 0.025(1000), which yields b = 15. Thus the equation is c =
15 + 0.025w.
(c) We need to solve the equation 100 = 15 + 0.025w, which yields w = 3400. It costs $100 to use 3400 cubic feet of
water.
45. See Figure 1.4.
driving speed
time
Figure 1.4
46. See Figure 1.5.
distance driven
time
Figure 1.5
6
Chapter One /SOLUTIONS
47. See Figure 1.6.
distance from exit
time
Figure 1.6
48. See Figure 1.7.
distance between cars
distance driven
Figure 1.7
(i) f (1985) = 13
(ii) f (1990) = 99
(b) The average yearly increase is the rate of change.
49. (a)
Yearly increase =
f (1990) − f (1985)
99 − 13
=
= 17.2 billionaires per year.
1990 − 1985
5
(c) Since we assume the rate of increase remains constant, we use a linear function with slope 17.2 billionaires per year.
The equation is
f (t) = b + 17.2t
where f (1985) = 13, so
13 = b + 17.2(1985)
Thus, f (t) = 17.2t − 34,129.
b = −34,129.
50. (a) The largest time interval was 2008–2009 since the percentage growth rate increased from −11.7 to 7.3 from 2008 to
2009. This means the US consumption of biofuels grew relatively more from 2008 to 2009 than from 2007 to 2008.
(Note that the percentage growth rate was a decreasing function of time over 2005–2007.)
(b) The largest time interval was 2005–2007 since the percentage growth rates were positive for each of these three
consecutive years. This means that the amount of biofuels consumed in the US steadily increased during the three
year span from 2005 to 2007, then decreased in 2008.
51. (a) The largest time interval was 2005–2007 since the percentage growth rate decreased from −1.9 in 2005 to −45.4 in
2007. This means that from 2005 to 2007 the US consumption of hydroelectric power shrunk relatively more with
each successive year.
(b) The largest time interval was 2004–2007 since the percentage growth rates were negative for each of these four
consecutive years. This means that the amount of hydroelectric power consumed by the US industrial sector steadily
decreased during the four year span from 2004 to 2007, then increased in 2008.
52. (a) The largest time interval was 2004–2006 since the percentage growth rate increased from −5.7 in 2004 to 9.7 in
2006. This means that from 2004 to 2006 the US price per watt of a solar panel grew relatively more with each
successive year.
(b) The largest time interval was 2005–2006 since the percentage growth rates were positive for each of these two
consecutive years. This means that the US price per watt of a solar panel steadily increased during the two year span
from 2005 to 2006, then decreased in 2007.
1.1 SOLUTIONS
7
53. (a) Since 2008 corresponds to t = 0, the average annual sea level in Aberdeen in 2008 was 7.094 meters.
(b) Looking at the table, we see that the average annual sea level was 7.019 fifty years before 2008, or in the year 1958.
Similar reasoning shows that the average sea level was 6.957 meters 125 years before 2008, or in 1883.
(c) Because 125 years before 2008 the year was 1883, we see that the sea level value corresponding to the year 1883 is
6.957 (this is the sea level value corresponding to t = 125). Similar reasoning yields the table:
Year
1883
1908
1933
1958
1983
2008
S
6.957
6.938
6.965
6.992
7.019
7.094
54. (a) We find the slope m and intercept b in the linear equation S = b + mt. To find the slope m, we use
m=
66 − 113
∆S
=
= −0.94.
∆t
50 − 0
When t = 0, we have S = 113, so the intercept b is 113. The linear formula is
S = 113 − 0.94t.
(b) We use the formula S = 113 − 0.94t. When S = 20, we have 20 = 113 − 0.94t and so t = 98.9. If this linear
model were correct, the average male sperm count would drop below the fertility level during the year 2038.
55. (a) This could be a linear function because w increases by 5 as h increases by 1.
(b) We find the slope m and the intercept b in the linear equation w = b + mh. We first find the slope m using the first
two points in the table. Since we want w to be a function of h, we take
m=
∆w
171 − 166
=
= 5.
∆h
69 − 68
Substituting the first point and the slope m = 5 into the linear equation w = b + mh, we have 166 = b + (5)(68),
so b = −174. The linear function is
w = 5h − 174.
The slope, m = 5, is in units of pounds per inch.
(c) We find the slope and intercept in the linear function h = b + mw using m = ∆h/∆w to obtain the linear function
h = 0.2w + 34.8.
Alternatively, we could solve the linear equation found in part (b) for h. The slope, m = 0.2, has units inches per
pound.
56. We will let
T = amount of fuel for take-off,
L = amount of fuel for landing,
P = amount of fuel per mile in the air,
m = the length of the trip in miles.
Then Q, the total amount of fuel needed, is given by
Q(m) = T + L + P m.
57. (a) The variable costs for x acres are $200x, or 0.2x thousand dollars. The total cost, C (again in thousands of dollars),
of planting x acres is:
C = f (x) = 10 + 0.2x.
This is a linear function. See Figure 1.8. Since C = f (x) increases with x, f is an increasing function of x. Look
at the values of C shown in the table; you will see that each time x increases by 1, C increases by 0.2. Because C
increases at a constant rate as x increases, the graph of C against x is a line.
8
Chapter One /SOLUTIONS
(b) See Figure 1.8 and Table 1.1.
Table 1.1
Cost of
planting
seed
x
C (thosand $)
10
C
0
10
2
10.4
3
10.6
4
10.8
5
11
6
11.2
C = 10 + 0.2x
5
2
4
6
x (acres)
Figure 1.8
(c) The vertical intercept of 10 corresponds to the fixed costs. For C = f (x) = 10 + 0.2x, the intercept on the vertical
axis is 10 because C = f (0) = 10 + 0.2(0) = 10. Since 10 is the value of C when x = 0, we recognize it as the
initial outlay for equipment, or the fixed cost.
The slope 0.2 corresponds to the variable costs. The slope is telling us that for every additional acre planted, the
costs go up by 0.2 thousand dollars. The rate at which the cost is increasing is 0.2 thousand dollars per acre. Thus the
variable costs are represented by the slope of the line f (x) = 10 + 0.2x.
58. See Figure 1.9.
Distance from
Kalamazoo
155
120
Time
start in
Chicago
arrive in
Kalamazoo
arrive in
Detroit
Figure 1.9
59. (a) The line given by (0, 2) and (1, 1) has slope m =
2−1
−1
= −1 and y-intercept 2, so its equation is
y = −x + 2.
The points of intersection of this line with the parabola y = x2 are given by
x2 = −x + 2
x2 + x − 2 = 0
(x + 2)(x − 1) = 0.
The solution x = 1 corresponds to the point we are already given, so the other solution, x = −2, gives the xcoordinate of C. When we substitute back into either equation to get y, we get the coordinates for C, (−2, 4).
(b) The line given by (0, b) and (1, 1) has slope m = b−1
= 1 − b, and y-intercept at (0, b), so we can write the equation
−1
for the line as we did in part (a):
y = (1 − b)x + b.
We then solve for the points of intersection with y = x2 the same way:
2
x2 = (1 − b)x + b
x − (1 − b)x − b = 0
x2 + (b − 1)x − b = 0
(x + b)(x − 1) = 0
Again, we have the solution at the given point (1, 1), and a new solution at x = −b, corresponding to the other point
of intersection C. Substituting back into either equation, we can find the y-coordinate for C is b2 , and thus C is given
by (−b, b2 ). This result agrees with the particular case of part (a) where b = 2.
1.2 SOLUTIONS
9
60. Looking at the given data, it seems that Galileo’s hypothesis was incorrect. The first table suggests that velocity is not
a linear function of distance, since the increases in velocity for each foot of distance are themselves getting smaller.
Moreover, the second table suggests that velocity is instead proportional to time, since for each second of time, the
velocity increases by 32 ft/sec.
Strengthen Your Understanding
61. The line y = 0.5 − 3x has a negative slope and is therefore a decreasing function.
62. If y is directly proportional to x we have y = kx. Adding the constant 1 to give y = 2x + 1 means that y is not
proportional to x.
63. One possible answer is f (x) = 2x + 3.
8
64. One possible answer is q = 1/3 .
p
65. False. A line can be put through any two points in the plane. However, if the line is vertical, it is not the graph of a
function.
66. True. Suppose we start at x = x1 and increase x by 1 unit to x1 + 1. If y = b + mx, the corresponding values of y are
b + mx1 and b + m(x1 + 1). Thus y increases by
∆y = b + m(x1 + 1) − (b + mx1 ) = m.
67. False. For example, let y = x + 1. Then the points (1, 2) and (2, 3) are on the line. However the ratios
2
=2
1
and
3
= 1.5
2
are different. The ratio y/x is constant for linear functions of the form y = mx, but not in general. (Other examples are
possible.)
68. False. For example, if y = 4x + 1 (so m = 4) and x = 1, then y = 5. Increasing x by 2 units gives 3, so y = 4(3) + 1 =
13. Thus, y has increased by 8 units, not 4 + 2 = 6. (Other examples are possible.)
√
69. (b) and (c). For g(x) = x, the domain and range are all nonnegative numbers, and for h(x) = x3 , the domain and range
are all real numbers.
Solutions for Section 1.2
Exercises
1. The graph shows a concave up function.
2. The graph shows a concave down function.
3. This graph is neither concave up or down.
4. The graph is concave up.
5. Initial quantity = 5; growth rate = 0.07 = 7%.
6. Initial quantity = 7.7; growth rate = −0.08 = −8% (decay).
7. Initial quantity = 3.2; growth rate = 0.03 = 3% (continuous).
8. Initial quantity = 15; growth rate = −0.06 = −6% (continuous decay).
t
9. Since e0.25t = e0.25 ≈ (1.2840)t , we have P = 15(1.2840)t . This is exponential growth since 0.25 is positive. We
can also see that this is growth because 1.2840 > 1.
10. Since e−0.5t = (e−0.5 )t ≈ (0.6065)t , we have P = 2(0.6065)t . This is exponential decay since −0.5 is negative. We
can also see that this is decay because 0.6065 < 1.
11. P = P0 (e0.2 )t = P0 (1.2214)t . Exponential growth because 0.2 > 0 or 1.2214 > 1.
12. P = 7(e−π )t = 7(0.0432)t . Exponential decay because −π < 0 or 0.0432 < 1.
10
Chapter One /SOLUTIONS
13. (a) Let Q = Q0 at . Then Q0 a5 = 75.94 and Q0 a7 = 170.86. So
170.86
Q 0 a7
=
= 2.25 = a2 .
Q 0 a5
75.94
So a = 1.5.
(b) Since a = 1.5, the growth rate is r = 0.5 = 50%.
14. (a) Let Q = Q0 at . Then Q0 a0.02 = 25.02 and Q0 a0.05 = 25.06. So
Q0 a0.05
25.06
=
= 1.001 = a0.03 .
Q0 a0.02
25.02
So
a = (1.001)
(b) Since a = 1.05, the growth rate is r = 0.05 = 5%.
100
3
= 1.05.
15. (a) The function is linear with initial population of 1000 and slope of 50, so P = 1000 + 50t.
(b) This function is exponential with initial population of 1000 and growth rate of 5%, so P = 1000(1.05)t .
16. (a) This is a linear function with slope −2 grams per day and intercept 30 grams. The function is Q = 30 − 2t, and the
graph is shown in Figure 1.10.
Q (grams)
Q (grams)
30
30
Q = 30 − 2t
15
Figure 1.10
Q = 30(0.88)t
t (days)
15
t (days)
Figure 1.11
(b) Since the quantity is decreasing by a constant percent change, this is an exponential function with base 1 − 0.12 =
0.88. The function is Q = 30(0.88)t , and the graph is shown in Figure 1.11.
17. The function is increasing and concave up between D and E, and between H and I. It is increasing and concave down
between A and B, and between E and F . It is decreasing and concave up between C and D, and between G and H.
Finally, it is decreasing and concave down between B and C, and between F and G
18. (a) It was decreasing from March 2 to March 5 and increasing from March 5 to March 9.
(b) From March 5 to 8, the average temperature increased, but the rate of increase went down, from 12◦ between March
5 and 6 to 4◦ between March 6 and 7 to 2◦ between March 7 and 8.
From March 7 to 9, the average temperature increased, and the rate of increase went up, from 2◦ between March
7 and 8 to 9◦ between March 8 and 9.
Problems
19. (a) A linear function must change by exactly the same amount whenever x changes by some fixed quantity. While h(x)
decreases by 3 whenever x increases by 1, f (x) and g(x) fail this test, since both change by different amounts
between x = −2 and x = −1 and between x = −1 and x = 0. So the only possible linear function is h(x), so it
will be given by a formula of the type: h(x) = mx + b. As noted, m = −3. Since the y-intercept of h is 31, the
formula for h(x) is h(x) = 31 − 3x.
(b) An exponential function must grow by exactly the same factor whenever x changes by some fixed quantity. Here,
g(x) increases by a factor of 1.5 whenever x increases by 1. Since the y-intercept of g(x) is 36, g(x) has the formula
g(x) = 36(1.5)x . The other two functions are not exponential; h(x) is not because it is a linear function, and f (x)
is not because it both increases and decreases.
20. Table A and Table B could represent linear functions of x. Table A could represent the constant linear function y = 2.2
because all y values are the same. Table B could represent a linear function of x with slope equal to 11/4. This is because
x values that differ by 4 have corresponding y values that differ by 11, and x values that differ by 8 have corresponding y
values that differ by 22. In Table C, y decreases and then increases as x increases, so the table cannot represent a linear
function. Table D does not show a constant rate of change, so it cannot represent a linear function.
1.2 SOLUTIONS
11
21. Table D is the only table that could represent an exponential function of x. This is because, in Table D, the ratio of y
values is the same for all equally spaced x values. Thus, the y values in the table have a constant percent rate of decrease:
4.5
2.25
9
=
=
= 0.5.
18
9
4.5
Table A represents a constant function of x, so it cannot represent an exponential function. In Table B, the ratio between y
values corresponding to equally spaced x values is not the same. In Table C, y decreases and then increases as x increases.
So neither Table B nor Table C can represent exponential functions.
22. (a) Let P represent the population of the world, and let t represent the number of years since 2010. Then we have
P = 6.91(1.011)t .
(b) According to this formula, the population of the world in the year 2020 (at t = 10) will be P = 6.9(1.011)10 = 7.71
billion people.
(c) The graph is shown in Figure 1.12. The population of the world has doubled when P = 13.82; we see on the graph
that this occurs at approximately t = 63.4. Under these assumptions, the doubling time of the world’s population is
about 63.4 years.
P (billion people)
13.82
6.91
63.4
t (years)
Figure 1.12
23. (a) We have P0 = 1 million, and k = 0.02, so P = (1,000,000)(e0.02t ).
(b)
P
1,000,000
t
24. The doubling time t depends only on the growth rate; it is the solution to
2 = (1.02)t ,
since 1.02t represents the factor by which the population has grown after time t. Trial and error shows that (1.02)35 ≈
1.9999 and (1.02)36 ≈ 2.0399, so that the doubling time is about 35 years.
25. (a) We have
Reduced size = (0.80) · Original size
or
1
Reduced size = (1.25) Reduced size,
(0.80)
so the copy must be enlarged by a factor of 1.25, which means it is enlarged to 125% of the reduced size.
(b) If a page is copied n times, then
New size = (0.80)n · Original.
Original size =
We want to solve for n so that
(0.80)n = 0.15.
8
By trial and error, we find (0.80) = 0.168 and (0.80)9 = 0.134. So the page needs to be copied 9 times.
12
Chapter One /SOLUTIONS
26. (a) See Figure 1.13.
customers
time
Figure 1.13
(b) “The rate at which new people try it” is the rate of change of the total number of people who have tried the product.
Thus, the statement of the problem is telling you that the graph is concave down—the slope is positive but decreasing,
as the graph shows.
27. (a) Advertising is generally cheaper in bulk; spending more money will give better and better marginal results initially,
(Spending $5,000 could give you a big newspaper ad reaching 200,000 people; spending $100,000 could give you a
series of TV spots reaching 50,000,000 people.) See Figure 1.14.
(b) The temperature of a hot object decreases at a rate proportional to the difference between its temperature and the
temperature of the air around it. Thus, the temperature of a very hot object decreases more quickly than a cooler
object. The graph is decreasing and concave up. See Figure 1.15 (We are assuming that the coffee is all at the same
temperature.)
temperature
revenue
advertising
Figure 1.14
time
Figure 1.15
28. (a) This is the graph of a linear function, which increases at a constant rate, and thus corresponds to k(t), which increases
by 0.3 over each interval of 1.
(b) This graph is concave down, so it corresponds to a function whose increases are getting smaller, as is the case with
h(t), whose increases are 10, 9, 8, 7, and 6.
(c) This graph is concave up, so it corresponds to a function whose increases are getting bigger, as is the case with g(t),
whose increases are 1, 2, 3, 4, and 5.
29. (a) This is a linear function, corresponding to g(x), whose rate of decrease is constant, 0.6.
(b) This graph is concave down, so it corresponds to a function whose rate of decrease is increasing, like h(x). (The rates
are −0.2, −0.3, −0.4, −0.5, −0.6.)
(c) This graph is concave up, so it corresponds to a function whose rate of decrease is decreasing, like f (x). (The rates
are −10, −9, −8, −7, −6.)
30. Since we are told that the rate of decay is continuous, we use the function Q(t) = Q0 ert to model the decay, where Q(t)
is the amount of strontium-90 which remains at time t, and Q0 is the original amount. Then
Q(t) = Q0 e−0.0247t .
So after 100 years,
Q(100) = Q0 e−0.0247·100
and
so about 8.46% of the strontium-90 remains.
Q(100)
= e−2.47 ≈ 0.0846
Q0
1.2 SOLUTIONS
13
31. We look for an equation of the form y = y0 ax since the graph looks exponential. The points (0, 3) and (2, 12) are on the
graph, so
3 = y0 a 0 = y0
and
x
12 = y0 · a2 = 3 · a2 ,
giving
Since a > 0, our equation is y = 3(2 ).
a = ±2.
32. We look for an equation of the form y = y0 ax since the graph looks exponential. The points (−1, 8) and (1, 2) are on the
graph, so
8 = y0 a−1 and 2 = y0 a1
−1
y0 a
1
8
= 2 , giving a = 21 , and so 2 = y0 a1 = y0 · 21 , so y0 = 4.
Therefore =
2
y
a
a
x0
Hence y = 4 12 = 4(2−x ).
33. We look for an equation of the form y = y0 ax since the graph looks exponential. The points (1, 6) and (2, 18) are on the
graph, so
6 = y0 a1 and 18 = y0 a2
2
18
= 3, and so 6 = y0 a = y0 · 3; thus, y0 = 2. Hence y = 2(3x ).
Therefore a = yy00aa =
6
34. The difference, D, between the horizontal asymptote and the graph appears to decrease exponentially, so we look for an
equation of the form
D = D0 ax
where D0 = 4 = difference when x = 0. Since D = 4 − y, we have
4 − y = 4ax or y = 4 − 4ax = 4(1 − ax )
The point (1, 2) is on the graph, so 2 = 4(1 − a1 ), giving a = 21 .
Therefore y = 4(1 − ( 12 )x ) = 4(1 − 2−x ).
35. Since f is linear, its slope is a constant
m=
Thus f increases 5 units for unit increase in x, so
f (1) = 15,
20 − 10
= 5.
2−0
f (3) = 25,
f (4) = 30.
Since g is exponential, its growth factor is constant. Writing g(x) = abx , we have g(0) = a = 10, so
g(x) = 10 · bx .
Since g(2) = 10 · b2 = 20, we have b2 = 2 and since b > 0, we have
√
b = 2.
√
Thus g increases by a factor of 2 for unit increase in x, so
√
√
√
g(1) = 10 2, g(3) = 10( 2)3 = 20 2,
√
g(4) = 10( 2)4 = 40.
Notice that the value of g(x) doubles between x = 0 and x = 2 (from g(0) = 10 to g(2) = 20), so the doubling time of
g(x) is 2. Thus, g(x) doubles again between x = 2 and x = 4, confirming that g(4) = 40.
36. We see that 1.09
≈ 1.03, and therefore h(s) = c(1.03)s ; c must be 1. Similarly
1.06
3.65
a = 2. Lastly, 3.47 ≈ 1.05, so g(s) = b(1.05)s ; b ≈ 3.
2.42
2.20
= 1.1, and so f (s) = a(1.1)s ;
37. (a) Because the population is growing exponentially, the time it takes to double is the same, regardless of the population
levels we are considering. For example, the population is 20,000 at time 3.7, and 40,000 at time 6.0. This represents
a doubling of the population in a span of 6.0 − 3.7 = 2.3 years.
How long does it take the population to double a second time, from 40,000 to 80,000? Looking at the graph once
again, we see that the population reaches 80,000 at time t = 8.3. This second doubling has taken 8.3 − 6.0 = 2.3
years, the same amount of time as the first doubling.
Further comparison of any two populations on this graph that differ by a factor of two will show that the time
that separates them is 2.3 years. Similarly, during any 2.3 year period, the population will double. Thus, the doubling
time is 2.3 years.
(b) Suppose P = P0 at doubles from time t to time t + d. We now have P0 at+d = 2P0 at , so P0 at ad = 2P0 at . Thus,
canceling P0 and at , d must be the number such that ad = 2, no matter what t is.
14
Chapter One /SOLUTIONS
38. (a) After 50 years, the amount of money is
P = 2P0 .
After 100 years, the amount of money is
P = 2(2P0 ) = 4P0 .
After 150 years, the amount of money is
P = 2(4P0 ) = 8P0 .
(b) The amount of money in the account doubles every 50 years. Thus in t years, the balance doubles t/50 times, so
P = P0 2t/50 .
39. (a) Since 162.5 = 325/2, there are 162.5 mg remaining after H hours.
Since 81.25 = 162.5/2, there are 81.25 mg remaining H hours after there were 162.5 mg, so 2H hours after there
were 325 mg.
Since 40.625 = 81.25/2, there are 41.625 mg remaining H hours after there were 81.25 mg, so 3H hours after there
were 325 mg.
(b) Each additional H hours, the quantity is halved. Thus in t hours, the quantity was halved t/H times, so
A = 325
1 t/H
2
.
40. (a) The quantity of radium decays exponentially, so we know Q = Q0 at . When t = 1620, we have Q = Q0 /2 so
Q0
= Q0 a1620 .
2
Thus, canceling Q0 , we have
a1620 =
a=
Thus the formula is Q = Q0
(b) After 500 years,
1 1/1620
2
t
= Q0
Fraction remaining =
1
2
1 1/1620
2
.
1 t/1620
.
2
1
· Q0
Q0
so 80.740% is left.
500/1620
1
2
= 0.80740.
41. Let Q0 be the initial quantity absorbed in 1960. Then the quantity, Q, of strontium-90 left after t years is
Q = Q0
Since 2010 − 1960 = 50 years, in 2010
Fraction remaining =
1
· Q0
Q0
t/29
1
2
1 50/29
2
=
.
1 50/29
2
= 0.30268 = 30.268%.
42. Direct calculation reveals that each 1000 foot increase in altitude results in a longer takeoff roll by a factor of about 1.096.
Since the value of d when h = 0 (sea level) is d = 670, we are led to the formula
d = 670(1.096)h/1000 ,
where d is the takeoff roll, in feet, and h is the airport’s elevation, in feet.
Alternatively, we can write
d = d0 a h ,
where d0 is the sea level value of d, d0 = 670. In addition, when h = 1000, d = 734, so
734 = 670a1000 .
Solving for a gives
a=
so
734
670
1/1000
= 1.00009124,
d = 670(1.00009124)h .
1.2 SOLUTIONS
15
43. (a) Since the annual growth factor from 2005 to 2006 was 1 + 1.866 = 2.866 and 91(1 + 1.866) = 260.806, the US
consumed approximately 261 million gallons of biodiesel in 2006. Since the annual growth factor from 2006 to 2007
was 1 + 0.372 = 1.372 and 261(1 + 0.372) = 358.092, the US consumed about 358 million gallons of biodiesel in
2007.
(b) Completing the table of annual consumption of biodiesel and plotting the data gives Figure 1.16.
Year
2005
2006
2007
2008
2009
Consumption of biodiesel (mn gal)
91
261
358
316
339
consumption of
biodiesel (mn gal)
400
200
year
2006
2008
Figure 1.16
44. (a) False, because the annual percent growth is not constant over this interval.
(b) The US consumption of biodiesel more than doubled in 2005 and more than doubled again in 2006. This is because
the annual percent growth was larger than 100% for both of these years.
(c) The US consumption of biodiesel more than tripled in 2005, since the annual percent growth in 2005 was over 200%
45. (a) Since the annual growth factor from 2006 to 2007 was 1 − 0.454 = 0.546 and 29(1 − 0.454) = 15.834, the US
consumed approximately 16 trillion BTUs of hydroelectric power in 2007. Since the annual growth factor from 2005
29
= 32.222, the US consumed about 32 trillion BTUs of hydroelectric
to 2006 was 1 − 0.10 = 0.90 and
(1 − 0.10)
power in 2005.
(b) Completing the table of annual consumption of hydroelectric power and plotting the data gives Figure 1.17.
Year
2004
2005
2006
2007
2008
2009
Consumption of hydro. power (trillion BTU)
33
32
29
16
17
19
(c) The largest decrease in the US consumption of hydroelectric power occurred in 2007. In this year, the US consumption of hydroelectric power dropped by about 13 trillion BTUs to 16 trillion BTUs, down from 29 trillion BTUs in
2006.
consumption of hydro.
power (trillion BTU)
35
25
15
5
year
2005
2007
Figure 1.17
2009
16
Chapter One /SOLUTIONS
46. (a) From the figure we can read-off the approximate percent growth for each year over the previous year:
Year
2005
2006
2007
2008
2009
% growth over previous yr
25
50
30
60
29
Since the annual growth factor from 2006 to 2007 was 1 + 0.30 = 1.30 and
341
= 262.31,
(1 + 0.30)
the US consumed approximately 262 trillion BTUs of wind power energy in 2006. Since the annual growth factor
from 2007 to 2008 was 1 + 0.60 = 1.60 and 341(1 + 0.60) = 545.6, the US consumed about 546 trillion BTUs of
wind power energy in 2008.
(b) Completing the table of annual consumption of wind power and plotting the data gives Figure 1.18.
Year
2005
2006
2007
2008
2009
Consumption of wind power (trillion BTU)
175
262
341
546
704
(c) The largest increase in the US consumption of wind power energy occurred in 2008. In this year the US consumption
of wind power energy rose by about 205 trillion BTUs to 546 trillion BTUs, up from 341 trillion BTUs in 2007.
consumption of wind power
energy (trillion BTU)
700
500
300
100
year
2007
2009
Figure 1.18
47. (a) The US consumption of wind power energy increased by at least 40% in 2006 and in 2008, relative to the previous year. In 2006 consumption increased by just under 50% over consumption in 2005, and in 2008 consumption
increased by about 60% over consumption in 2007. Consumption did not decrease during the time period shown
because all the annual percent growth values are positive, indicating a steady increase in the US consumption of wind
power energy between 2005 and 2009.
(b) Yes. From 2006 to 2007 consumption increased by about 30%, which means x(1 + 0.30) units of wind power energy
were consumed in 2007 if x had been consumed in 2006. Similarly,
(x(1 + 0.30))(1 + 0.60)
units of wind power energy were consumed in 2008 if x had been consumed in 2006 (because consumption increased
by about 60% from 2007 to 2008). Since
(x(1 + 0.30))(1 + 0.60) = x(2.08) = x(1 + 1.08),
the percent growth in wind power consumption was about 108%, or just over 100%, in 2008 relative to consumption
in 2006.
Strengthen Your Understanding
48. The function y = e−0.25x is decreasing but its graph is concave up.
49. The graph of y = 2x is a straight line and is neither concave up or concave down.
1.3 SOLUTIONS
17
50. One possible answer is q = 2.2(0.97)t .
51. One possible answer is f (x) = 2(1.1)x .
52. One possibility is y = e−x − 5.
53. False. The y-intercept is y = 2 + 3e−0 = 5.
54. True, since, as t → ∞, we know e−4t → 0, so y = 5 − 3e−4t → 5.
55. False. Suppose y = 5x . Then increasing x by 1 increases y by a factor of 5. However increasing x by 2 increases y by a
factor of 25, not 10, since
y = 5x+2 = 5x · 52 = 25 · 5x .
(Other examples are possible.)
56. True. Suppose y = Abx and we start at the point (x1 , y1 ), so y1 = Abx1 . Then increasing x1 by 1 gives x1 + 1, so the
new y-value, y2 , is given by
y2 = Abx1 +1 = Abx1 b = (Abx1 )b,
so
y2 = by1 .
Thus, y has increased by a factor of b, so b = 3, and the function is y = A3x .
However, if x1 is increased by 2, giving x1 + 2, then the new y-value, y3 , is given by
y3 = A3x1 +2 = A3x1 32 = 9A3x1 = 9y1 .
Thus, y has increased by a factor of 9.
57. True. For example, f (x) = (0.5)x is an exponential function which decreases. (Other examples are possible.)
58. True. If b > 1, then abx → 0 as x → −∞. If 0 < b < 1, then abx → 0 as x → ∞. In either case, the function
y = a + abx has y = a as the horizontal asymptote.
59. True, since e−kt → 0 as t → ∞, so y → 20 as t → ∞.
Solutions for Section 1.3
Exercises
1.
y
(a)
y
(b)
4
4
4
x
−2
−2
−4
4
x
−4
y
(f)
4
2
x
−4
y
(e)
4
2
2
−4
y
−2
−2
x
2
(d)
y
(c)
x
−2
2
−4
x
−2
2
−4
18
Chapter One /SOLUTIONS
2.
y
(a)
y
(b)
4
4
4
x
−2
x
2
−2
−4
(e)
4
x
−2
−4
y
(b)
2
y
(c)
4
4
x
2
−2
−4
x
2
−2
2
−4
y
−4
y
(e)
4
y
(f)
4
4
x
2
−4
−2
−4
x
−2
−2
x
2
−4
4
(d)
y
(f)
x
2
y
2
−4
4
−2
(a)
−2
y
4
3.
x
2
−4
y
(d)
y
(c)
x
−2
2
−4
x
−2
2
−4
1.3 SOLUTIONS
4. This graph is the graph of m(t) shifted upward by two units. See Figure 1.19.
4
n(t)
−4
4
6
t
−4
Figure 1.19
5. This graph is the graph of m(t) shifted to the right by one unit. See Figure 1.20.
y
4
p(t)
−4
4
6
t
−4
Figure 1.20
6. This graph is the graph of m(t) shifted to the left by 1.5 units. See Figure 1.21.
4
k(t)
−4
4
6
t
−4
Figure 1.21
7. This graph is the graph of m(t) shifted to the right by 0.5 units and downward by 2.5 units. See Figure 1.22.
4
−4
6
w(t)
−4
Figure 1.22
t
19
20
Chapter One /SOLUTIONS
8. (a)
(b)
(c)
(d)
(e)
9. (a)
(b)
(c)
(d)
(e)
10. (a)
(b)
(c)
(d)
(e)
11. (a)
(b)
(c)
(d)
(e)
12. (a)
(b)
(c)
13. (a)
(b)
(c)
(d)
(e)
f (g(1)) = f (1 + 1) = f (2) = 22 = 4
g(f (1)) = g(12 ) = g(1) = 1 + 1 = 2
f (g(x)) = f (x + 1) = (x + 1)2
g(f (x)) = g(x2 ) = x2 + 1
f (t)g(t) = t2 (t + 1)
√
√
2
= 5
f (g(1)) = f (1
√ ) = f (1) =√ 1 + 4 √
2
g(f (1)) = g( 1 + 4)
√= g( 5) = ( 5) = 5
2
2
f (g(x)) = f (x
√ ) = x +√4
g(f (x)) = g(√ x + 4) = ( √
x + 4)2 = x + 4
f (t)g(t) = ( t + 4)t2 = t2 t + 4
f (g(1)) = f (12 ) = f (1) = e1 = e
g(f (1)) = g(e1 ) = g(e) = e2
2
f (g(x)) = f (x2 ) = ex
g(f (x)) = g(ex ) = (ex )2 = e2x
f (t)g(t) = et t2
1
f (g(1)) = f (3 · 1 + 4) = f (7) =
7
g(f (1)) = g(1/1) = g(1) = 7
1
f (g(x)) = f (3x + 4) =
3x
+ 4
3
1
1
=3
+4 = +4
g(f (x)) = g
x
x
x
4
1
f (t)g(t) = (3t + 4) = 3 +
t
t
g(2 + h) = (2 + h)2 + 2(2 + h) + 3 = 4 + 4h + h2 + 4 + 2h + 3 = h2 + 6h + 11.
g(2) = 22 + 2(2) + 3 = 4 + 4 + 3 = 11, which agrees with what we get by substituting h = 0 into (a).
g(2 + h) − g(2) = (h2 + 6h + 11) − (11) = h2 + 6h.
f (t + 1) = (t + 1)2 + 1 = t2 + 2t + 1 + 1 = t2 + 2t + 2.
f (t2 + 1) = (t2 + 1)2 + 1 = t4 + 2t2 + 1 + 1 = t4 + 2t2 + 2.
f (2) = 22 + 1 = 5.
2f (t) = 2(t2 + 1) = 2t2 + 2.
2
(f (t))2 + 1 = t2 + 1 + 1 = t4 + 2t2 + 1 + 1 = t4 + 2t2 + 2.
14. m(z + 1) − m(z) = (z + 1)2 − z 2 = 2z + 1.
15. m(z + h) − m(z) = (z + h)2 − z 2 = 2zh + h2 .
16. m(z) − m(z − h) = z 2 − (z − h)2 = 2zh − h2 .
17. m(z + h) − m(z − h) = (z + h)2 − (z − h)2 = z 2 + 2hz + h2 − (z 2 − 2hz + h2 ) = 4hz.
18. (a) f (25) is q corresponding to p = 25, or, in other words, the number of items sold when the price is 25.
(b) f −1 (30) is p corresponding to q = 30, or the price at which 30 units will be sold.
19. (a) f (10,000) represents the value of C corresponding to A = 10,000, or in other words the cost of building a 10,000
square-foot store.
(b) f −1 (20,000) represents the value of A corresponding to C = 20,000, or the area in square feet of a store which
would cost $20,000 to build.
20. f −1 (75) is the length of the column of mercury in the thermometer when the temperature is 75◦ F.
21. (a) The equation is y = 2x2 + 1. Note that its graph is narrower than the graph of y = x2 which appears in gray. See
Figure 1.23.
8
y = 2x2 + 1
6
4
y = x2
2
Figure 1.23
8
7
6
5
4
3
2
1
y = 2(x2 + 1)
y = x2
Figure 1.24
1.3 SOLUTIONS
21
(b) y = 2(x2 + 1) moves the graph up one unit and then stretches it by a factor of two. See Figure 1.24.
(c) No, the graphs are not the same. Since 2(x2 + 1) = (2x2 + 1) + 1, the second graph is always one unit higher than
the first.
22. Figure 1.25 shows the appropriate graphs. Note that asymptotes are shown as dashed lines and x- or y-intercepts are
shown as filled circles.
y
(a)
y
(b)
5
4
y=3
2
2
x
−5 −3
x
−5 −3
2
2
5
−2
y
(c)
5
y
(d)
5
5
y=4
2
2
1
x
−10
−6
x
2
−1
−5
−1
2
5
Figure 1.25
23. The function is not invertible since there are many horizontal lines which hit the function twice.
24. The function is not invertible since there are horizontal lines which hit the function more than once.
25. Since a horizontal line cuts the graph of f (x) = x2 + 3x + 2 two times, f is not invertible. See Figure 1.26.
y
4
f (x) = x2 + 3x + 2
y = 2.5
3
2
x
−3
Figure 1.26
22
Chapter One /SOLUTIONS
26. Since a horizontal line cuts the graph of f (x) = x3 − 5x + 10 three times, f is not invertible. See Figure 1.27.
y
f (x) = x3 − 5x + 10
10
y=8
x
-2
2
Figure 1.27
27. Since any horizontal line cuts the graph once, f is invertible. See Figure 1.28.
y
f (x) = x3 + 5x + 10
40
y = 30
20
-2
2
x
y = −19
Figure 1.28
28.
f (−x) = (−x)6 + (−x)3 + 1 = x6 − x3 + 1.
29.
Since f (−x) 6= f (x) and f (−x) 6= −f (x), this function is neither even nor odd.
f (−x) = (−x)3 + (−x)2 + (−x) = −x3 + x2 − x.
Since f (−x) 6= f (x) and f (−x) 6= −f (x), this function is neither even nor odd.
30. Since
we see f is even
f (−x) = (−x)4 − (−x)2 + 3 = x4 − x2 + 3 = f (x),
31. Since
f (−x) = (−x)3 + 1 = −x3 + 1,
we see f (−x) 6= f (x) and f (−x) 6= −f (x), so f is neither even nor odd
32. Since
we see f is odd.
f (−x) = 2(−x) = −2x = −f (x),
1.3 SOLUTIONS
33. Since
f (−x) = e(−x)
2
−1
= ex
2
−1
23
= f (x),
we see f is even.
34. Since
we see f is odd
f (−x) = (−x)((−x)2 − 1) = −x(x2 − 1) = −f (x),
35. Since
f (−x) = e−x + x,
we see f (−x) 6= f (x) and f (−x) 6= −f (x), so f is neither even nor odd
Problems
36. f (x) = x3 ,
g(x) = x + 1.
37. f (x) = x + 1, g(x) = x3 .
√
38. f (x) = x, g(x) = x2 + 4
39. f (x) = ex ,
g(x) = 2x
40. This looks like a shift of the graph y = −x2 . The graph is shifted to the left 1 unit and up 3 units, so a possible formula
is y = −(x + 1)2 + 3.
41. This looks like a shift of the graph y = x3 . The graph is shifted to the right 2 units and down 1 unit, so a possible formula
is y = (x − 2)3 − 1.
42. (a) We find f −1 (2) by finding the x value corresponding to f (x) = 2. Looking at the graph, we see that f −1 (2) = −1.
(b) We construct the graph of f −1 (x) by reflecting the graph of f (x) over the line y = x. The graphs of f −1 (x) and
f (x) are shown together in Figure 1.29.
y
4
f (x)
x
−4
6
f −1 (x)
−4
Figure 1.29
43. Values of f −1 are as follows
x
3
f −1 (x)
1
−7
2
19
4
178
2
1
3
4
5
6
7
The domain of f −1 is the set consisting of the integers {3, −7, 19, 4, 178, 2, 1}.
44. f is an increasing function since the amount of fuel used increases as flight time increases. Therefore f is invertible.
45. Not invertible. Given a certain number of customers, say f (t) = 1500, there could be many times, t, during the day at
which that many people were in the store. So we don’t know which time instant is the right one.
46. Probably not invertible. Since your calculus class probably has less than 363 students, there will be at least two days in
the year, say a and b, with f (a) = f (b) = 0. Hence we don’t know what to choose for f −1 (0).
47. Not invertible, since it costs the same to mail a 50-gram letter as it does to mail a 51-gram letter.
48. The volume of the balloon t minutes after inflation began is: g(f (t)) ft3 .
49. The volume of the balloon if its radius were twice as big is: g(2r) ft3 .
24
Chapter One /SOLUTIONS
50. The time elapsed is: f −1 (30) min.
51. The time elapsed is: f −1 (g −1 (10,000)) min.
52. We have v(10) = 65 but the graph of u only enables us to evaluate u(x) for 0 ≤ x ≤ 50. There is not enough information
to evaluate u(v(10)).
53. We have approximately v(40) = 15 and u(15) = 18 so u(v(40)) = 18.
54. We have approximately u(10) = 13 and v(13) = 60 so v(u(10)) = 60.
55. We have u(40) = 60 but the graph of v only enables us to evaluate v(x) for 0 ≤ x ≤ 50. There is not enough information
to evaluate v(u(40)).
56. (a) Yes, f is invertible, since f is increasing everywhere.
(b) The number f −1 (400) is the year in which 400 million motor vehicles were registered in the world. From the picture,
we see that f −1 (400) is around 1979.
(c) Since the graph of f −1 is the reflection of the graph of f over the line y = x, we get Figure 1.30.
(year)
’10
’05
’00
’95
’90
’85
’80
’75
’70
’65
200
400
600
800
1000
(millions)
Figure 1.30: Graph of f −1
57. f (g(1)) = f (2) ≈ 0.4.
58. g(f (2)) ≈ g(0.4) ≈ 1.1.
59. f (f (1)) ≈ f (−0.4) ≈ −0.9.
60. Computing f (g(x)) as in Problem 57, we get Table 1.2. From it we graph f (g(x)) in Figure 1.31.
Table 1.2
x
g(x)
f (g(x))
−3
0.6
−0.5
−2.5
−1.1
−1.5
−1.9
−2
−1
−0.5
0
0.5
−1.9
−1.4
−0.5
0.5
1.4
−1.3
−1.2
3
−1.2
−1.3
−1
−0.6
−0.2
1
2
1.5
2.2
2
1.6
0
2.5
0.1
3
−2.5
−0.7
f (g(x))
x
3
−3
0.4
0.5
0.1
−3
Figure 1.31
61. Using the same way to compute g(f (x)) as in Problem 58, we get Table 1.3. Then we can plot the graph of g(f (x)) in
Figure 1.32.
1.3 SOLUTIONS
25
Table 1.3
x
−3
f (x)
g(f (x))
3
−2.6
−2.5
0.1
−1.5
−1.3
−2
−1
−0.5
0
0.5
1
1.5
0.8
−1
−1.4
−1.2
−1.7
−1
−0.8
−1
−0.6
−0.1
0.3
−3
3
x
−0.3
0.3
1.1
2.5
0.9
2
3
1.6
2.2
2
g(f (x))
−1.4
−0.6
−0.4
3
−1.8
−3
Figure 1.32
62. Using the same way to compute f (f (x)) as in Problem 59, we get Table 1.4. Then we can plot the graph of f (f (x)) in
Figure 1.33.
Table 1.4
x
f (x)
−3
3
1.6
−2.5
0.1
−0.7
−1.5
−1.3
−0.5
−1
−2
−1
0
0.5
1
1.5
2
−1
−1.2
−0.8
−0.6
f (f (x))
−1.2
−1.3
−1.1
−1
−0.9
0.3
−0.6
2.5
0.9
3
1.6
f (f (x))
−1.2
−0.4
−0.1
3
−1.3
−3
3
x
−0.8
−0.4
0
−3
Figure 1.33
63. (a) The graph shows that f (15) is approximately 48. So, the place to find find 15 million-year-old rock is about 48
meters below the Atlantic sea floor.
(b) Since f is increasing (not decreasing, since the depth axis is reversed!), f is invertible. To confirm, notice that the
graph of f is cut by a horizontal line at most once.
(c) Look at where the horizontal line through 120 intersects the graph of f and read downward: f −1 (120) is about 35.
In practical terms, this means that at a depth of 120 meters down, the rock is 35 million years old.
(d) First, we standardize the graph of f so that time and depth are increasing from left to right and bottom to top. Points
(t, d) on the graph of f correspond to points (d, t) on the graph of f −1 . We can graph f −1 by taking points from
the original graph of f , reversing their coordinates, and connecting them. This amounts to interchanging the t and d
axes, thereby reflecting the graph of f about the line bisecting the 90◦ angle at the origin. Figure 1.34 is the graph of
f −1 . (Note that we cannot find the graph of f −1 by flipping the graph of f about the line t = d in because t and d
have different scales in this instance.)
26
Chapter One /SOLUTIONS
✒
✒
f −1
30
100
Time
Depth
f (standardized)
10
20
✠
10
Time
30
✠
Figure 1.34: Graph of f , reflected to give that of f
20
Depth
100
−1
64. The tree has B = y − 1 branches on average and each branch has n = 2B 2 − B = 2(y − 1)2 − (y − 1) leaves on
average. Therefore
Average number of leaves = Bn = (y − 1)(2(y − 1)2 − (y − 1)) = 2(y − 1)3 − (y − 1)2 .
65. The volume, V , of the balloon is V = 43 πr 3 . When t = 3, the radius is 10 cm. The volume is then
V =
4
4000π
π(103 ) =
cm3 .
3
3
66. (a) The function f tells us C in terms of q. To get its inverse, we want q in terms of C, which we find by solving for q:
C = 100 + 2q,
C − 100 = 2q,
C − 100
= f −1 (C).
q=
2
(b) The inverse function tells us the number of articles that can be produced for a given cost.
67. Since Q = S − Se−kt , the graph of Q is the reflection of y about the t-axis moved up by S units.
68.
x
f (x)
g(x)
−3
h(x)
0
0
0
−2
2
2
−1
2
2
−2
0
0
0
1
2
2
2
−2
−2
3
0
0
0
−2
−2
0
−2
Strengthen Your Understanding
69. The graph of f (x) = −(x + 1)3 is the graph of g(x) = −x3 shifted left by 1 unit.
70. Since f (g(x)) = 3(−3x − 5) + 5 = −9x − 10, we see that f and g are not inverse functions.
71. While y = 1/x is sometimes referred to as the multiplicative inverse of x, the inverse of f is f −1 (x) = x.
72. One possible answer is g(x) = 3 + x. (There are many answers.)
73. One possibility is f (x) = x2 + 2.
74. Let f (x) = 3x, then f −1 (x) = x/3. Then for x > 0, we have f (x) > f −1 (x).
1.4 SOLUTIONS
27
75. We have
g(x) = f (x + 2)
because the graph of g is obtained by moving the graph of f to the left by 2 units. We also have
g(x) = f (x) + 3
because the graph of g is obtained by moving the graph of f up by 3 units. Thus, we have f (x + 2) = f (x) + 3. The graph
of f climbs 3 units whenever x increases by 2. The simplest choice for f is a linear function of slope 3/2, for example
f (x) = 1.5x, so g(x) = 1.5x + 3.
76. True. The graph of y = 10x is moved horizontally by h units if we replace x by x − h for some number h. Writing
100 = 102 , we have f (x) = 100(10x ) = 102 · 10x = 10x+2 . The graph of f (x) = 10x+2 is the graph of g(x) = 10x
shifted two units to the left.
77. True. If f is increasing then its reflection about the line y = x is also increasing. An example is shown in Figure 1.35.
The statement is true.
y
x=y
4
f (x)
x
−4
6
f −1 (x)
−4
Figure 1.35
78. True. If f (x) is even, we have f (x) = f (−x) for all x. For example, f (−2) = f (2). This means that the graph of f (x)
intersects the horizontal line y = f (2) at two points, x = 2 and x = −2. Thus, f has no inverse function.
79. False. For example, f (x) = x and g(x) = x3 are both odd. Their inverses are f −1 (x) = x and g −1 (x) = x1/3 .
2
80. False. For x < 0, as x increases, x2 decreases, so e−x increases.
81. True. We have g(−x) = g(x) since g is even, and therefore f (g(−x)) = f (g(x)).
82. False. A counterexample is given by f (x) = x2 and g(x) = x + 1. The function f (g(x)) = (x + 1)2 is not even because
f (g(1)) = 4 and f (g(−1)) = 0 6= 4.
83. True. The constant function f (x) = 0 is the only function that is both even and odd. This follows, since if f is both even
and odd, then, for all x, f (−x) = f (x) (if f is even) and f (−x) = −f (x) (if f is odd). Thus, for all x, f (x) = −f (x)
i.e. f (x) = 0, for all x. So f (x) = 0 is both even and odd and is the only such function.
84. Let f (x) = x and g(x) = −2x. Then f (x) + g(x) = −x, which is decreasing. Note f is increasing since it has positive
slope, and g is decreasing since it has negative slope.
85. This is impossible. If a < b, then f (a) < f (b), since f is increasing, and g(a) > g(b), since g is decreasing, so
−g(a) < −g(b). Therefore, if a < b, then f (a) − g(a) < f (b) − g(b), which means that f (x) + g(x) is increasing.
86. Let f (x) = ex and let g(x) = e−2x . Note f is increasing since it is an exponential growth function, and g is decreasing
since it is an exponential decay function. Then f (x)g(x) = e−x , which is decreasing.
87. This is impossible. As x increases, g(x) decreases. As g(x) decreases, so does f (g(x)) because f is increasing (an
increasing function increases as its variable increases, so it decreases as its variable decreases).
Solutions for Section 1.4
Exercises
1. Using the identity eln x = x, we have eln(1/2) = 21 .
28
Chapter One /SOLUTIONS
2. Using the identity 10log x = x, we have
10log(AB) = AB
3. Using the identity eln x = x, we have 5A2 .
4. Using the identity ln (ex ) = x, we have 2AB.
5. Using the rules for ln, we have
ln
1
e
+ ln AB = ln 1 − ln e + ln A + ln B
= 0 − 1 + ln A + ln B
= −1 + ln A + ln B.
6. Using the rules for ln, we have 2A + 3e ln B.
7. Taking logs of both sides
log 3x = x log 3 = log 11
log 11
= 2.2.
x=
log 3
8. Taking logs of both sides
log 17x = log 2
x log 17 = log 2
log 2
≈ 0.24.
x=
log 17
9. Isolating the exponential term
20 = 50(1.04)x
20
= (1.04)x .
50
Taking logs of both sides
2
= log(1.04)x
5
2
log = x log(1.04)
5
log(2/5)
= −23.4.
x=
log(1.04)
log
10.
5x
4
= x
7
3 x
4
5
=
7
3
Taking logs of both sides
log
4
7
= x log
5
3
log(4/7)
x=
≈ −1.1.
log(5/3)
1.4 SOLUTIONS
11. To solve for x, we first divide both sides by 5 and then take the natural logarithm of both sides.
7
= e0.2x
5
ln(7/5) = 0.2x
ln(7/5)
≈ 1.68.
x=
0.2
12. ln(2x ) = ln(ex+1 )
x ln 2 = (x + 1) ln e
x ln 2 = x + 1
0.693x = x + 1
1
x=
≈ −3.26
0.693 − 1
13. To solve for x, we first divide both sides by 600 and then take the natural logarithm of both sides.
50
= e−0.4x
600
ln(50/600) = −0.4x
ln(50/600)
≈ 6.212.
x=
−0.4
14.
ln(2e3x ) = ln(4e5x )
ln 2 + ln(e3x ) = ln 4 + ln(e5x )
0.693 + 3x = 1.386 + 5x
x = −0.347
15. Using the rules for ln, we get
ln 7x+2 = ln e17x
(x + 2) ln 7 = 17x
x(ln 7 − 17) = −2 ln 7
−2 ln 7
x=
≈ 0.26.
ln 7 − 17
16.
ln(10x+3 ) = ln(5e7−x )
(x + 3) ln 10 = ln 5 + (7 − x) ln e
2.303(x + 3) = 1.609 + (7 − x)
3.303x = 1.609 + 7 − 2.303(3)
x = 0.515
17. Using the rules for ln, we have
2
2x − 1 = x2
x − 2x + 1 = 0
(x − 1)2 = 0
x = 1.
18.
4e2x−3 = e + 5
ln 4 + ln(e2x−3 ) = ln(e + 5)
1.3863 + 2x − 3 = 2.0436
x = 1.839.
log a
.
19. t =
log b
29
30
Chapter One /SOLUTIONS
log
P
P0
log P − log P0
=
.
log a
log a
21. Taking logs of both sides yields
20. t =
nt =
Hence
t=
Q
Q0
log
n log a
Q
Q0
log
log a
.
log Q − log Q0
.
n log a
=
22. Collecting similar terms yields
t
a
b
Hence
t=
=
log
log
Q0
.
P0
Q0
P0
a
b
.
a
23. t = ln .
b
ln PP0
P
= kt, so t =
.
P0
k
25. Since we want (1.5)t = ekt = (ek )t , so 1.5 = ek , and k = ln 1.5 = 0.4055. Thus, P = 15e0.4055t . Since 0.4055 is
positive, this is exponential growth.
24. ln
26. We want 1.7t = ekt so 1.7 = ek and k = ln 1.7 = 0.5306. Thus P = 10e0.5306t .
27. We want 0.9t = ekt so 0.9 = ek and k = ln 0.9 = −0.1054. Thus P = 174e−0.1054t .
28. Since we want (0.55)t = ekt = (ek )t , so 0.55 = ek , and k = ln 0.55 = −0.5978. Thus P = 4e−0.5978t . Since
−0.5978 is negative, this represents exponential decay.
29. If p(t) = (1.04)t , then, for p−1 the inverse of p, we should have
(1.04)p
p
−1
−1
(t)
= t,
(t) log(1.04) = log t,
log t
p−1 (t) =
≈ 58.708 log t.
log(1.04)
30. Since f is increasing, f has an inverse. To find the inverse of f (t) = 50e0.1t , we replace t with f −1 (t), and, since
f (f −1 (t)) = t, we have
−1
t = 50e0.1f (t) .
We then solve for f −1 (t):
t = 50e0.1f
−1
(t)
−1
t
= e0.1f (t)
50
t
= 0.1f −1 (t)
ln
50
t
t
1
f −1 (t) =
ln
= 10 ln
.
0.1
50
50
31. Using f (f −1 (t)) = t, we see
f (f −1 (t)) = 1 + ln f −1 (t) = t.
So
ln f −1 (t) = t − 1
f −1 (t) = et−1 .
1.4 SOLUTIONS
31
Problems
32. The population has increased by a factor of 48,000,000/40,000,000 = 1.2 in 10 years. Thus we have the formula
P = 40,000,000(1.2)t/10 ,
and t/10 gives the number of 10-year periods that have passed since 2000.
In 2000, t/10 = 0, so we have P = 40,000,000.
In 2010, t/10 = 1, so P = 40,000,000(1.2) = 48,000,000.
In 2020, t/10 = 2, so P = 40,000,000(1.2)2 = 57,600,000.
To find the doubling time, solve 80,000,000 = 40,000,000(1.2)t/10 , to get t = 38.02 years.
33. In ten years, the substance has decayed to 40% of its original mass. In another ten years, it will decay by an additional
factor of 40%, so the amount remaining after 20 years will be 100 · 40% · 40% = 16 kg.
34. We can solve for the growth rate k of the bacteria using the formula P = P0 ekt :
1500 = 500ek(2)
ln(1500/500)
k=
.
2
Knowing the growth rate, we can find the population P at time t = 6:
P = 500e(
ln 3 )6
2
≈ 13,500 bacteria.
35. (a) Assuming the US population grows exponentially, we have population P (t) = 281.4ekt at time t years after 2000.
Using the 2010 population, we have
308.7 = 281.4e10k
ln(308.7/281.4)
k=
= 0.00926.
10
We want to find the time t in which
350 = 281.4e0.00926t
ln(350/281.4)
= 23.56 years.
t=
0.00926
This model predicts the population to go over 350 million 23.56 years after 2000, in the year 2023.
(b) Evaluate P = 281.4e0.00926t for t = 20 to find P = 338.65 million people.
36. If C0 is the concentration of NO2 on the road, then the concentration x meters from the road is
C = C0 e−0.0254x .
We want to find the value of x making C = C0 /2, that is,
C0 e−0.0254x =
C0
.
2
Dividing by C0 and then taking natural logs yields
ln e−0.254x = −0.0254x = ln
so
1
2
= −0.6931,
x = 27 meters.
At 27 meters from the road the concentration of NO2 in the air is half the concentration on the road.
32
Chapter One /SOLUTIONS
37. (a) Since the percent increase in deaths during a year is constant for constant increase in pollution, the number of deaths
per year is an exponential function of the quantity of pollution. If Q0 is the number of deaths per year without
pollution, then the number of deaths per year, Q, when the quantity of pollution is x micrograms per cu meter of air
is
Q = Q0 (1.0033)x .
(b) We want to find the value of x making Q = 2Q0 , that is,
Q0 (1.0033)x = 2Q0 .
Dividing by Q0 and then taking natural logs yields
ln ((1.0033)x ) = x ln 1.0033 = ln 2,
so
ln 2
= 210.391.
ln 1.0033
When there are 210.391 micrograms of pollutants per cu meter of air, respiratory deaths per year are double what
they would be in the absence of air pollution.
x=
38. (a) Since there are 4 years between 2004 and 2008 we let t be the number of years since 2004 and get:
450,327 = 211,800er4 .
Solving for r, we get
450,327
= er4
211,800
ln
Substituting t = 1, 2, 3 into
450,327
211,800
= 4r
r = 0.188583
211,800 e(0.188583)t ,
we find the three remaining table values:
Year
2004
2005
2006
2007
2008
Number of E85 vehicles
211,800
255,756
308,835
372,930
450,327
(b) If N is the number of E85-powered vehicles in 2003, then
211,800 = N e0.188583
or
211,800
= 175,398 vehicles.
e0.188583
(c) From the table, we can see that the number of E85 vehicles slightly more than doubled from 2004 to 2008, so the
percent growth between these years should be slightly over 100%:
N=
Percent growth from
= 100
2004 to 2008
39. (a)
(b)
(c)
(d)
450,327
−1
211,800
= 1.12619 = 112.619%.
The initial dose is 10 mg.
Since 0.82 = 1 − 0.18, the decay rate is 0.18, so 18% leaves the body each hour.
When t = 6, we have A = 10(0.82)6 = 3.04. The amount in the body after 6 hours is 3.04 mg.
We want to find the value of t when A = 1. Using logarithms:
1 = 10(0.82)t
0.1 = (0.82)t
ln(0.1) = t ln(0.82)
t = 11.60 hours.
After 11.60 hours, the amount is 1 mg.
1.4 SOLUTIONS
33
40. (a) Since the initial amount of caffeine is 100 mg and the exponential decay rate is −0.17, we have A = 100e−0.17t .
(b) See Figure 1.36. We estimate the half-life by estimating t when the caffeine is reduced by half (so A = 50); this
occurs at approximately t = 4 hours.
A (mg)
100
50
t (hours)
4
Figure 1.36
(c) We want to find the value of t when A = 50:
50 = 100e−0.17t
0.5 = e−0.17t
ln 0.5 = −0.17t
t = 4.077.
The half-life of caffeine is about 4.077 hours. This agrees with what we saw in Figure 1.36.
41. Since y(0) = Ce0 = C we have that C = 2. Similarly, substituting x = 1 gives y(1) = 2eα so
2eα = 1.
Rearranging gives eα = 1/2. Taking logarithms we get α = ln(1/2) = − ln 2 = −0.693. Finally,
y(2) = 2e2(− ln 2) = 2e−2 ln 2 =
1
.
2
42. The function ex has a vertical intercept of 1, so must be A. The function ln x has an x-intercept of 1, so must be D. The
graphs of x2 and x1/2 go through the origin. The graph of x1/2 is concave down so it corresponds to graph C and the
graph of x2 is concave up so it corresponds to graph B.
43. (a) B(t) = B0 e0.067t
(b) P (t) = P0 e0.033t
(c) If the initial price is $50, then
B(t) = 50e0.067t
P (t) = 50e0.033t .
We want the value of t such that
B(t) = 2P (t)
0.067t
50e
= 2 · 50e0.033t
e0.067t
= e0.034t = 2
e0.033t
ln 2
t=
= 20.387 years .
0.034
Thus, when t = 20.387 the price of the textbook was predicted to be double what it would have been had the price
risen by inflation only. This occurred in the year 2000.
44. (a) We assume f (t) = Ae−kt , where A is the initial population, so A = 100,000. When t = 110, there were 3200
tigers, so
3200 = 100,000e−k·110
Solving for k gives
e−k·110 =
so
k=−
3200
= 0.0132
100,000
1
ln(0.0132) = 0.0313 = 3.13%
110
f (t) = 100,000e−0.0313t .
34
Chapter One /SOLUTIONS
(b) In 2000, the predicted number of tigers was
f (100) = 100,000e−0.0313(100) = 4372.
In 2010, we know the number of tigers was 3200. The predicted percent reduction is
3200 − 4372
= −0.268 = −26.8%.
4372
Thus the actual decrease is larger than the predicted decrease.
45. The population of China, C, in billions, is given by
C = 1.34(1.004)t
where t is time measured from 2011, and the population of India, I, in billions, is given by
I = 1.19(1.0137)t .
The two populations will be equal when C = I, thus, we must solve the equation:
1.34(1.004)t = 1.19(1.0137)t
for t, which leads to
(1.0137)t
1.34
=
=
1.19
(1.0004)t
Taking logs on both sides, we get
t log
so
1.0137
1.004
t
.
1.0137
1.34
= log
,
1.004
1.19
log (1.34/1.19)
= 12.35 years.
log (1.0137/1.004)
This model predicts the population of India will exceed that of China in 2023.
t=
46. Let A represent the revenue (in billions of dollars) at Apple t years since 2005. Since A = 3.68 when t = 0 and we want
the continuous growth rate, we write A = 3.68ekt . We use the information from 2010, that A = 15.68 when t = 5, to
find k:
15.68 = 3.68ek·5
4.26 = e5k
ln(4.26) = 5k
k = 0.2899.
We have A = 3.68e0.2899t , which represents a continuous growth rate of 28.99% per year.
47. Let P (t) be the world population in billions t years after 2010.
(a) Assuming exponential growth, we have
P (t) = 6.9ekt .
In 2050, we have t = 40 and we expect the population then to be 9 billion, so
9 = 6.9ek·40 .
Solving for k, we have
ek·40 =
9
1
ln
40
6.9
(b) The “Day of 7 Billion” should occur when
k=
9
6.9
= 0.00664 = 0.664% per year.
7 = 6.9e0.00664t .
Solving for t gives
7
6.9
ln(7/6.9)
t=
= 2.167 years.
0.00664
So the “Day of 7 Billion” should be 2.167 years after the end of 2010. This is 2 years and 0.167 · 365 = 61 days; so
61 days into 2013. That is, March 2, 2013.
e0.00664t =
1.4 SOLUTIONS
35
48. If r was the average yearly inflation rate, in decimals, then 41 (1 + r)3 = 2,400,000, so r = 211.53, i.e. r = 21,153%.
49. To find a half-life, we want to find at what t value Q = 12 Q0 . Plugging this into the equation of the decay of plutonium240, we have
1
= e−0.00011t
2
ln(1/2)
≈ 6,301 years.
t=
−0.00011
The only difference in the case of plutonium-242 is that the constant −0.00011 in the exponent is now −0.0000018.
Thus, following the same procedure, the solution for t is
t=
ln(1/2)
≈ 385,081 years.
−0.0000018
50. Given the doubling time of 5 hours, we can solve for the bacteria’s growth rate;
2P0 = P0 ek5
ln 2
k=
.
5
So the growth of the bacteria population is given by:
P = P0 eln(2)t/5 .
We want to find t such that
3P0 = P0 eln(2)t/5 .
Therefore we cancel P0 and apply ln. We get
t=
5 ln(3)
= 7.925 hours.
ln(2)
51. (a) The pressure P at 6194 meters is given in terms of the pressure P0 at sea level to be
P = P0 e−0.00012h
= P0 e(−0.00012)6194
= P0 e−0.74328
≈ 0.4756P0
or about 47.6% of sea level pressure.
(b) At h = 12,000 meters, we have
P = P0 e−0.00012h
= P0 e(−0.00012)12,000
= P0 e−1.44
≈ 0.2369P0
or about 23.7% of sea level pressure.
52. We know that the y-intercept of the line is at (0,1), so we need one other point to determine the equation of the line. We
observe that it intersects the graph of f (x) = 10x at the point x = log 2. The y-coordinate of this point is then
y = 10x = 10log 2 = 2,
so (log 2, 2) is the point of intersection. We can now find the slope of the line:
m=
1
2−1
=
.
log 2 − 0
log 2
Plugging this into the point-slope formula for a line, we have
y − y1 = m(x − x1 )
1
(x − 0)
y−1 =
log 2
1
y=
x + 1 ≈ 3.3219x + 1.
log 2
36
Chapter One /SOLUTIONS
53. If t is time in decades, then the number of vehicles, V, in millions, is given by
V = 246(1.155)t .
For time t in decades, the number of people, P, in millions, is given by
P = 308.7(1.097)t .
V
= 1, or V = P . Thus, we solve for t in the equation:
P
There is an average of one vehicle per person when
246(1.155)t = 308.7(1.097)t ,
which leads to
Taking logs on both sides, we get
1.155
1.097
t
t log
so
=
(1.155)t
308.7
=
(1.097)t
246
308.7
1.155
= log
,
1.097
246
log (308.7/246)
= 4.41 decades.
log (1.155/1.097)
This model predicts one vehicle per person in 2054
t=
54. We assume exponential decay and solve for k using the half-life:
e−k(5730) = 0.5
so
k = 1.21 · 10−4 .
t
t=
ln 0.995
= 41.43 years.
−1.21 · 10−4
Now find t, the age of the painting:
−4
e−1.21·10
= 0.995,
so
Since Vermeer died in 1675, the painting is a fake.
55. Yes, ln(ln(x)) means take the ln of the value of the function ln x. On the other hand, ln2 (x) means take the function
ln x and square it. For example, consider each of these functions evaluated at e. Since ln e = 1, ln2 e = 12 = 1, but
ln(ln(e)) = ln(1) = 0. See the graphs in Figure 1.37. (Note that ln(ln(x)) is only defined for x > 1.)
−1
2
1 f (x) = ln(ln x)
2
1
f (x) = ln x
1
2
3
4
5
1
−1
−2
1
2
3
4
f (x) = ln2 x
5
1
2
3
4
5
Figure 1.37
56. (a) The y-intercept of h(x) = ln(x + a) is h(0) = ln a. Thus increasing a increases the y-intercept.
(b) The x-intercept of h(x) = ln(x + a) is where h(x) = 0. Since this occurs where x + a = 1, or x = 1 − a, increasing
a moves the x-intercept to the left.
57. The vertical asymptote is where x + a = 0, or x = −a. Thus increasing a moves the vertical asymptote to the left.
58. (a) The y-intercept of g(x) = ln(ax + 2) is g(0) = ln 2. Thus increasing a does not effect the y-intercept.
(b) The x-intercept of g(x) = ln(ax + 2) is where g(x) = 0. Since this occurs where ax + 2 = 1, or x = −1/a,
increasing a moves the x-intercept toward the origin. (The intercept is to the left of the origin if a > 0 and to the
right if a < 0.)
59. The vertical asymptote is where x + 2 = 0, or x = −2, so increasing a does not effect the vertical asymptote.
60. The vertical asymptote is where ax + 2 = 0, or x = −2/a. Thus increasing a moves the vertical asymptote toward the
origin. (The asymptote is to the left of the origin for a > 0 and to the right of the origin for a < 0.)
Strengthen Your Understanding
61. The function − log |x| is even, since | − x| = |x|, which means − log | − x| = − log |x|.
1.5 SOLUTIONS
62. We have
ln(100x) = ln(100) + ln x.
In general, ln(100x) 6= 100 · ln x.
63. One possibility is f (x) = −x, because ln(−x) is only defined if −x > 0.
64. One possibility is f (x) = ln(x − 3).
65. True, as seen from the graph.
66. False, since log(x − 1) = 0 if x − 1 = 1, so x = 2.
67. False. The inverse function is y = 10x .
68. False, since ax + b = 0 if x = −b/a. Thus y = ln(ax + b) has a vertical asymptote at x = −b/a.
Solutions for Section 1.5
Exercises
1. See Figure 1.38.
3π
= −1 is negative.
2
3π
=0
cos
2
3π
tan
is undefined.
2
sin
✲
Figure 1.38
2. See Figure 1.39.
sin(2π) = 0
cos(2π) = 1 is positive.
tan(2π) = 0
✻
Figure 1.39
37
38
Chapter One /SOLUTIONS
3. See Figure 1.40.
π
4
π
cos
4
π
tan
4
sin
is positive
is positive
is positive
■
Figure 1.40
4. See Figure 1.41.
sin 3π = 0
cos 3π = −1 is negative
tan 3π = 0
❄
Figure 1.41
5. See Figure 1.42.
π
is positive.
6
π
cos
is positive.
6
π
is positive.
tan
6
sin
▼
Figure 1.42
1.5 SOLUTIONS
6. See Figure 1.43.
4π
3
4π
cos
3
4π
tan
3
sin
is negative
is negative
is positive
❥
Figure 1.43
7. See Figure 1.44.
sin
−4π
3
is positive.
−4π
is negative.
3
−4π
is negative.
tan
3
cos
✯
Figure 1.44
8. 4 radians ·
180◦
=
π radians
720
π
◦
≈ 240◦ . See Figure 1.45.
sin 4
is negative
cos 4
is negative
tan 4
is positive.
❥
Figure 1.45
39
40
Chapter One /SOLUTIONS
9. −1 radian ·
180◦
π radians
=−
180◦
π
≈ −60◦ . See Figure 1.46.
sin (−1)
is negative
cos (−1)
is positive
tan (−1)
is negative.
✙
Figure 1.46
10. The period is 2π/3, because when t varies from 0 to 2π/3, the quantity 3t varies from 0 to 2π. The amplitude is 7, since
the value of the function oscillates between −7 and 7.
11. The period is 2π/(1/4) = 8π, because when u varies from 0 to 8π, the quantity u/4 varies from 0 to 2π. The amplitude
is 3, since the function oscillates between 2 and 8.
12. The period is 2π/2 = π, because as x varies from −π/2 to π/2, the quantity 2x + π varies from 0 to 2π. The amplitude
is 4, since the function oscillates between 4 and 12.
13. The period is 2π/π = 2, since when t increases from 0 to 2, the value of πt increases from 0 to 2π. The amplitude is 0.1,
since the function oscillates between 1.9 and 2.1.
x
.
14. This graph is a sine curve with period 8π and amplitude 2, so it is given by f (x) = 2 sin
4
x
15. This graph is a cosine curve with period 6π and amplitude 5, so it is given by f (x) = 5 cos
.
3
16. This graph is an inverted sine curve with amplitude 4 and period π, so it is given by f (x) = −4 sin(2x).
x
.
17. This graph is an inverted cosine curve with amplitude 8 and period 20π, so it is given by f (x) = −8 cos
10
18. This graph has period 6, amplitude 5 and no vertical or horizontal shift, so it is given by
f (x) = 5 sin
2π
π
x = 5 sin
x .
6
3
19. The graph is a cosine curve with period 2π/5 and amplitude 2, so it is given by f (x) = 2 cos(5x).
20. The graph is an inverted sine curve with amplitude 1 and period 2π, shifted up by 2, so it is given by f (x) = 2 − sin x.
21. This can be represented by a sine function of amplitude 3 and period 18. Thus,
f (x) = 3 sin
π
x .
9
x
+ 2.
4
23. This graph has period 8, amplitude 3, and a vertical shift of 3 with no horizontal shift. It is given by
22. This graph is the same as in Problem 14 but shifted up by 2, so it is given by f (x) = 2 sin
f (x) = 3 + 3 sin
π
2π
x = 3 + 3 sin
x .
8
4
41
1.5 SOLUTIONS
24.
π
5
▼✌
cos −
π
5
π
5
π
(by picture)
5
= 0.809.
= cos
25.
(0,1)
π
5
1
sin
cos
26.
π
5
π
5
sin
π
5
(1,0)
cos
π
5
By the Pythagorean Theorem, (cos π5 )2 + (sin π5 )2 = 12 ;
p
p
so (sin π5 )2 = 1 − (cos π5 )2 , and sin π5 = 1 − (cos π5 )2 = 1 − (0.809)2 ≈ 0.588.
We take the positive square root since by the picture we know that sin π5 is positive.
(0,1)
cos
sin
cos
π
12
π
12
π
12
sin
π
12
π
12
1
(1,0)
By the Pythagorean Theorem, (cos
p
1−
(0.259)2
π 2
π 2
) +(sin 12
)
12
= 12 ; so (cos
π 2
)
12
= 1−(sin
π 2
)
12
and cos
≈ 0.966. We take the positive square root since by the picture we know that cos
27. We first divide by 5 and then use inverse sine:
2
= sin(3x)
5
sin−1 (2/5) = 3x
x=
sin−1 (2/5)
≈ 0.1372.
3
There are infinitely many other possible solutions since the sine is periodic.
π
12
π
12
=
p
1 − (sin
is positive.
π 2
)
12
=
42
Chapter One /SOLUTIONS
28. We first isolate cos(2x + 1) and then use inverse cosine:
1 = 8 cos(2x + 1) − 3
4 = 8 cos(2x + 1)
0.5 = cos(2x + 1)
cos−1 (0.5) = 2x + 1
x=
cos−1 (0.5) − 1
≈ 0.0236.
2
There are infinitely many other possible solutions since the cosine is periodic.
29. We first isolate tan(5x) and then use inverse tangent:
8 = 4 tan(5x)
2 = tan(5x)
tan
−1
2 = 5x
tan−1 2
x=
= 0.221.
5
There are infinitely many other possible solutions since the tangent is periodic.
30. We first isolate (2x + 1) and then use inverse tangent:
1 = 8 tan(2x + 1) − 3
4 = 8 tan(2x + 1)
0.5 = tan(2x + 1)
arctan(0.5) = 2x + 1
arctan(0.5) − 1
= −0.268.
x=
2
There are infinitely many other possible solutions since the tangent is periodic.
31. We first isolate sin(5x) and then use inverse sine:
8 = 4 sin(5x)
2 = sin(5x).
But this equation has no solution since −1 ≤ sin(5x) ≤ 1.
Problems
32. (a) h(t) = 2 cos (t − π/2)
(b) f (t) = 2 cos t
(c) g(t) = 2 cos (t + π/2)
33. sin x2 is by convention sin(x2 ), which means you square the x first and then take the sine.
sin2 x = (sin x)2 means find sin x and then square it.
sin(sin x) means find sin x and then take the sine of that.
Expressing each as a composition: If f (x) = sin x and g(x) = x2 , then
sin x2 = f (g(x))
sin2 x = g(f (x))
sin(sin x) = f (f (x)).
34. Suppose P is at the point (3π/2, −1) and Q is at the point (5π/2, 1). Then
Slope =
2
1 − (−1)
= .
5π/2 − 3π/2
π
If P had been picked to the right of Q, the slope would have been −2/π.
1.5 SOLUTIONS
43
35. (a) See Figure 1.47.
900
P
800
700
100
t (months)
Jan. 1
Jul. 1
Jan. 1
Figure 1.47
= 800, amplitude = 900−700
= 100, and period = 12 months, so B =
(b) Average value of population = 700+900
2
2
2π/12 = π/6. Since the population is at its minimum when t = 0, we use a negative cosine:
P = 800 − 100 cos
36. We use a cosine of the form
πt
.
6
H = A cos(Bt) + C
and choose B so that the period is 24 hours, so 2π/B = 24 giving B = π/12.
The temperature oscillates around an average value of 60◦ F, so C = 60. The amplitude of the oscillation is 20◦ F.
To arrange that the temperature be at its lowest when t = 0, we take A negative, so A = −20. Thus
A = 60 − 20 cos
π
t .
12
37. (a) f (t) = −0.5 + sin t, g(t) = 1.5 + sin t, h(t) = −1.5 + sin t, k(t) = 0.5 + sin t.
(b) The values of g(t) are one more than the values of k(t), so g(t) = 1+k(t). This happens because g(t) = 1.5+sin t =
1 + 0.5 + sin t = 1 + k(t).
(c) Since −1 ≤ sin t ≤ 1, adding 1.5 everywhere we get 0.5 ≤ 1.5 + sin t ≤ 2.5 and since 1.5 + sin t = g(t), we get
0.5 ≤ g(t) ≤ 2.5. Similarly, −2.5 ≤ −1.5 + sin t = h(t) ≤ −0.5.
π
38. Depth = 7 + 1.5 sin
t
3
39. (a) Beginning at time t = 0, the voltage will have oscillated through a complete cycle when cos(120πt) = cos(2π),
1
1
second. The period is 60
second.
hence when t = 60
(b) V0 represents the amplitude of the oscillation.
(c) See Figure 1.48.
V
V0
1
120
1
60
t
Figure 1.48
40. (a) When the time is t hours after 6 am, the solar panel outputs f (t) = P (θ(t)) watts. So,
f (t) = 10 sin
where 0 ≤ t ≤ 14 is the number of hours after 6 am.
π
t
14
44
Chapter One /SOLUTIONS
(b) The graph of f (t) is in Figure 1.49:
P (watts)
10
7
14
t (hrs after 6 am)
Figure 1.49
(c) The power output is greatest when sin(πt/14) = 1. Since 0 ≤ πt/14 ≤ π, the only point in the domain of f at
which sin(πt/14) = 1 is when πt/14 = π/2. Therefore, the power output is greatest when t = 7, that is, at 1 pm.
The output at this time will be f (7) = 10 watts.
(d) On a typical winter day, there are 9 hours of sun instead of the 14 hours of sun. So, if t is the number of hours since
8 am, the angle between a solar panel and the sun is
π
14
θ= t
9
9
The solar panel outputs g(t) = P (φ(t)) watts:
where 0 ≤ t ≤ 9.
φ=
g(t) = 10 sin
where 0 ≤ t ≤ 9 is the number of hours after 8 am.
π
t
9
41. The function R has period of π, so its graph is as shown in Figure 1.50. The maximum value of the range is v02 /g and
occurs when θ = π/4.
2
v0
g
R
R=
π
4
π
2
2
v0
g
sin 2θ
θ
Figure 1.50
42. Over the one-year period, the average value is about 75◦ and the amplitude of the variation is about 90−60
= 15◦ . The
2
function assumes its minimum value right at the beginning of the year, so we want a negative cosine function. Thus, for t
in years, we have the function
2π
t .
f (t) = 75 − 15 cos
12
(Many other answers are possible, depending on how you read the chart.)
43. (a)
(b)
(c)
(d)
D = the average depth of the water.
A = the amplitude = 15/2 = 7.5.
Period = 12.4 hours. Thus (B)(12.4) = 2π so B = 2π/12.4 ≈ 0.507.
C is the time of a high tide.
44. Using the fact that 1 revolution = 2π radians and 1 minute = 60 seconds, we have
200
rad
1 rad
rev
= (200) · 2π
= 200 · 2π
min
min
60 sec
(200)(6.283)
≈
60
≈ 20.94 radians per second.
Similarly, 500 rpm is equivalent to 52.36 radians per second.
1.5 SOLUTIONS
45. 200 revolutions per minute is
1
200
minutes per revolution, so the period is
1
200
45
minutes, or 0.3 seconds.
46. The earth makes one revolution around the sun in one year, so its period is one year.
47. The moon makes one revolution around the earth in about 27.3 days, so its period is 27.3 days ≈ one month.
48. (a) The period of the tides is 2π/0.5 = 4π = 12.566 hours.
(b) The boat is afloat provided the water is deeper than 2.5 meters, so we need
d(t) = 5 + 4.6 sin(0.5t) > 2.5.
Figure 1.51 is a graph of d(t), with time t in hours since midnight, 0 ≤ t ≤ 24. The boat leaves at t = 12 (midday).
To find the latest time the boat can return, we need to solve the equation d(t) = 5 + 4.6 sin(0.5t) = 2.5.
A quick way to estimate the solution is to trace along the line y = 2.5 in Figure 1.51 until we get to the first
point of intersection to the right of t = 12. The value we want is about t = 20. Thus the water remains deep enough
until about 8 pm.
To find t analytically, we solve
5 + 4.6 sin(0.5t) = 2.5
2.5
= −0.5435
sin(0.5t) = −
4.6
1
t=
arcsin(−0.5435) = −1.149.
0.5
This is the value of t immediately to the left of the vertical axis. The water is also 2.5 meters deep one period later at
t = −1.149 + 12.566 = 11.417. This is shortly before the boat leaves, while the water is rising. We want the next
time the time the water is this depth.
The water was at its deepest (that is, d(t) was a maximum) when t = 12.566/4 = 3.142. From the figure, the
time between when the water was 2.5 meters and when it was deepest was 3.142 + 1.149 = 4.291 hours. Thus, the
value of t that we want is
t = 11.417 + 2 · 4.291 = 19.999.
depth in meters
d(t)
10
7.5
5
2.5
5
10
15
20
25
t, hours since midnight
Figure 1.51
49. Since b is a positive constant, f is a vertical shift of sin t where the midline lies above the t-axis. So f matches Graph C.
Function g is the sum of sin t plus a linear function at + b. We suspect then that the graph of g might periodically
oscillate about a line at + b, just like the graph of sin t oscillates about its midline. When adding at + b to sin t, we note
that every zero of sin t, (t, 0), gets displaced to a corresponding point (t, at + b) that lies both on the graph of g, and on
the line at + b. See Figure 1.52. So g matches Graph B.
Function h is the sum of sin t plus an increasing exponential function ect + d. We suspect then that the graph of
g might periodically oscillate about the graph of ect + d, just like the graph of sin t oscillates about its midline. When
adding ect + d to sin t, we note that every zero of sin t, (t, 0), gets displaced to a corresponding point (t, ect + d) that lies
both on the graph of h, and on the graph of ect + d. See Figure 1.52. So h matches Graph A. Note that the oscillations on
the graph of h may not be visible for all t values.
Function r is the sum of sin t plus an decreasing exponential function −ect + b. We suspect then that the graph of
r might periodically oscillate about the graph of −ect + b, just like the graph of sin t oscillates about its midline. When
adding −ect + b to sin t, we note that every zero of sin t, (t, 0), gets displaced to a corresponding point (t, −ect + b)
that lies both on the graph of r, and on the graph of −ect + b. See Figure 1.52. So r matches Graph D. Note that the
oscillations on the graph of r may not be visible for all t values.
46
Chapter One /SOLUTIONS
A
B
(t, ect + b)
(t, at + b)
C
(t, −ect + b)
D
(t, 0)
t
Figure 1.52
50. (a) The monthly mean CO2 increased about 10 ppm between December 2005 and December 2010. This is because the
black curve shows that the December 2005 monthly mean was about 381 ppm, while the December 2010 monthly
mean was about 391 ppm. The difference between these two values, 391 − 381 = 10, gives the overall increase.
(b) The average rate of increase is given by
Average monthly increase of monthly mean =
1
391 − 381
= ppm/month.
60 − 0
6
This tells us that the slope of a linear equation approximating the black curve is 1/6. Since the vertical intercept is
about 381, a possible equation for the approximately linear black curve is
y=
1
t + 381,
6
where t is measured in months since December 2005.
(c) The period of the seasonal CO2 variation is about 12 months since this is approximately the time it takes for the
function given by the blue curve to complete a full cycle. The amplitude is about 3.5 since, looking at the blue curve,
the average distance between consecutive maximum and minimum values is about 7 ppm. So a possible sinusoidal
function for the seasonal CO2 cycle is
π
t .
y = 3.5 sin
6
π
1
t and g(t) = t + 381, we have
(d) Taking f (t) = 3.5 sin
6
6
π
1
t + t + 381.
h(t) = 3.5 sin
6
6
See Figure 1.53.
ppm
395
h(t)
390
385
380
375
12
24
36
48
Figure 1.53
60
t (months since Dec 2005)
1.5 SOLUTIONS
47
51. (a) The period is 2π.
(b) After π, the values of cos 2θ repeat, but the values of 2 sin θ do not (in fact, they repeat but flipped over the x-axis).
After another π, that is after a total of 2π, the values of cos 2θ repeat again, and now the values of 2 sin θ repeat also,
so the function 2 sin θ + 3 cos 2θ repeats at that point.
52. Figure 1.54 shows that the cross-sectional area is one rectangle of area hw and two triangles. Each triangle has height h
and base x, where
h
h
= tan θ so x =
.
x
tan θ
2
1
h
Area of triangle = xh =
2
2 tan θ
Total area = Area of rectangle + 2(Area of triangle)
h2
h2
= hw +
.
2 tan θ
tan θ
= hw + 2 ·
✛x✲
✛x✲
✻θ
✻
θ
h
θ
h
❄
w
❄θ
Figure 1.54
53. (a)
(b)
(c)
(d)
Two solutions: 0.4 and 2.7. See Figure 1.55.
arcsin(0.4) is the first solution approximated above; the second is an approximation to π − arcsin(0.4).
By symmetry, there are two solutions: −0.4 and −2.7.
−0.4 ≈ − arcsin(0.4) and −2.7 ≈ −(π − arcsin(0.4)) = arcsin(0.4) − π.
y
1
0.4
−2.7
−0.4
x
0.4
2.7
−0.4
−1
Figure 1.55
54. The ramp in Figure 1.56 rises 1 ft over a horizontal distance of x ft.
(a) For a 1 ft rise over 12 ft, the angle in radians is θ = arctan(1/12) = 0.0831. To find the angle in degrees, multiply
by 180/π. Hence
1
180
arctan
= 4.76◦ .
θ=
π
12
(b) We have
180
1
θ=
arctan = 7.13◦ .
π
8
(c) We have
1
180
arctan
= 2.86◦ .
θ=
π
20
1 ft
θ
x ft
Figure 1.56
48
Chapter One /SOLUTIONS
55. (a) A table of values for g(x) is:
x
arccos x
−1
3.14
−0.8
2.50
−0.6
2.21
−0.4
−0.2
1.98
1.77
0
0.2
0.4
0.6
0.8
1
1.57
1.37
1.16
0.93
0.64
0
(b) See Figure 1.57.
π
y = arccos x
y
✻
Range
π
2
−1
1
✛
✲
Domain
❄
x
Figure 1.57
y
1
y = cos x
−π/2
π
π/2
x
−1
Figure 1.58
(c) The domain of arccos is −1 ≤ x ≤ 1, because its inverse, cosine, takes all values from -1 to 1. The domains of
arccos and arcsin are the same because their inverses, sine and cosine, have the same range.
(d) Figure 1.57 shows that the range of y = arccos x is 0 ≤ θ ≤ π.
(e) The range of an inverse function is the domain of the original function. The arcsine is the inverse function to the piece
of the sine having domain [− π2 , π2 ]. Hence, the range of the arcsine is [− π2 , π2 ]. But the piece of the cosine having
domain [− π2 , π2 ] does not have an inverse, because there are horizontal lines that intersect its graph twice. Instead,
we define arccosine to be the inverse of the piece of cosine having domain [0, π], so the range of arccosine is [0, π],
which is different from the range of arcsine. See Figure 1.58.
Strengthen Your Understanding
56. Increasing the value of B decreases the period. For example, f (x) = sin x has period 2π, whereas g(x) = sin(2x) has
period π.
57. The maximum value of A sin(Bx) is A, so the maximum value of y = A sin(Bx) + C is y = A + C.
58. For B > 0, the period of y = sin(Bx) is 2π/B. Thus, we want
2π
= 23
B
so
B=
2π
.
23
The function is f (x) = sin(2πx/23)
59. The midline is y = (1200 + 2000)/2 = 1600 and the amplitude is y = (2000 − 1200)/2 = 400, so a possible function
is
f (x) = 400(cos x) + 1600.
60. False, since cos θ is decreasing and sin θ is increasing.
61. False. The period is 2π/(0.05π) = 40
1.6 SOLUTIONS
49
62. True. The period is 2π/(200π) = 1/100 seconds. Thus, the function executes 100 cycles in 1 second.
63. False. If θ = π/2, 3π/2, 5π/2 . . ., then θ − π/2 = 0, π, 2π . . ., and the tangent is defined (it is zero) at these values.
64. False: When x < 0, we have sin |x| = sin(−x) = − sin x 6= sin x.
65. False: When π < x < 2π, we have sin |x| = sin x < 0 but | sin x| > 0.
66. False: When π/2 < x < 3π/2, we have cos |x| = cos x < 0 but | cos x| > 0.
67. True: Since cos(−x) = cos x, cos |x| = cos x.
68. False. For example, sin(0) 6= sin((2π)2 ), since sin(0) = 0 but sin((2π)2 ) = 0.98.
69. True. Since sin(θ + 2π) = sin θ for all θ, we have g(θ + 2π) = esin(θ+2π) = esin θ = g(θ) for all θ.
70. False. A counterexample is given by f (x) = sin x, which has period 2π, and g(x) = x2 . The graph of f (g(x)) = sin(x2 )
in Figure 1.59 is not periodic with period 2π.
y
f (x)
1
−2π
−π
−1
π
x
2π
Figure 1.59
71. True. If g(x) has period k, then g(x + k) = g(x). Thus we have
f (g(x + k)) = f (g(x))
which shows that f (g(x)) is periodic with period k.
72. True, since | sin(−x)| = | − sin x| = sin x.
Solutions for Section 1.6
Exercises
1. As x → ∞, y → ∞.
As x → −∞, y → −∞.
2. As x → ∞, y → ∞.
As x → −∞, y → 0.
3. Since f (x) is an even power function with a negative leading coefficient, it follows that f (x) → −∞ as x → +∞ and
f (x) → −∞ as x → −∞.
4. Since f (x) is an odd power function with a positive leading coefficient, it follows that f (x) → +∞ as x → +∞ and
f (x) → −∞ as x → −∞.
5. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest
degree term of 5x4 . Thus, as x → ±∞, we see that f (x) → +∞.
6. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest
degree term of −5x3 . Thus, as x → +∞, we see that f (x) → −∞ and as x → −∞, we see that f (x) → +∞.
7. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest
3x2
degree terms in the numerator and denominator. Thus, as x → ±∞, we see that f (x) behaves like 2 = 3. We have
x
f (x) → 3 as x → ±∞.
8. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest
−3x3
= −3/2. We
degree terms in the numerator and denominator. Thus, as x → ±∞, we see that f (x) behaves like
2x3
have f (x) → −3/2 as x → ±∞.
9. As x → ±∞, we see that 3x−4 gets closer and closer to 0, so f (x) → 0 as x → ±∞.
50
Chapter One /SOLUTIONS
10. As x → +∞, we have f (x) → +∞. As x → −∞, we have f (x) → 0.
11. The power function with the higher power dominates as x → ∞, so 0.2x5 is larger.
12. An exponential growth function always dominates a power function as x → ∞, so 10e0.1x is larger.
13. An exponential growth function always dominates a power function as x → ∞, so 1.05x is larger.
14. The lower-power terms in a polynomial become insignificant as x → ∞, so we are comparing 2x4 to the leading term
10x3 . In comparing two power functions, the higher power dominates as x → ∞, so 2x4 is larger.
15. The lower-power terms in a polynomial become insignificant as x → ∞, so we are comparing the leading term 20x4 to
the leading term 3x5 . In comparing two power functions, the higher power dominates as x → ∞, so the polynomial with
leading term 3x5 is larger. As x → ∞, we see that 25 − 40x2 + x3 + 3x5 is larger.
√
16. A power function with positive exponent dominates a log function, so as x → ∞, we see that x is larger.
17. (I) (a)
(b)
(II) (a)
(b)
(III) (a)
(b)
(IV) (a)
(b)
(V) (a)
(b)
Minimum degree is 3 because graph turns around twice.
Leading coefficient is negative because y → −∞ as x → ∞.
Minimum degree is 4 because graph turns around three times.
Leading coefficient is positive because y → ∞ as x → ∞.
Minimum degree is 4 because graph turns around three times.
Leading coefficient is negative because y → −∞ as x → ∞.
Minimum degree is 5 because graph turns around four times.
Leading coefficient is negative because y → −∞ as x → ∞.
Minimum degree is 5 because graph turns around four times.
Leading coefficient is positive because y → ∞ as x → ∞.
18. (a) From the x-intercepts, we know the equation has the form
y = k(x + 2)(x − 1)(x − 5).
Since y = 2 when x = 0,
2 = k(2)(−1)(−5) = k · 10
1
k= .
5
Thus we have
y=
1
(x + 2)(x − 1)(x − 5).
5
19. (a) Because our cubic has a root at 2 and a double root at −2, it has the form
y = k(x + 2)(x + 2)(x − 2).
Since y = 4 when x = 0,
4 = k(2)(2)(−2) = −8k,
1
k=− .
2
Thus our equation is
1
y = − (x + 2)2 (x − 2).
2
20. f (x) = k(x + 3)(x − 1)(x − 4) = k(x3 − 2x2 − 11x + 12), where k < 0. (k ≈ − 61 if the horizontal and vertical scales
are equal; otherwise one can’t tell how large k is.)
21. f (x) = kx(x + 3)(x − 4) = k(x3 − x2 − 12x), where k < 0. (k ≈ − 29 if the horizontal and vertical scales are equal;
otherwise one can’t tell how large k is.)
22. f (x) = k(x + 2)(x − 1)(x − 3)(x − 5) = k(x4 − 7x3 + 5x2 + 31x − 30), where k > 0. (k ≈
vertical scales are equal; otherwise one can’t tell how large k is.)
1
15
if the horizontal and
1
23. f (x) = k(x + 2)(x − 2)2 (x − 5) = k(x4 − 7x3 + 6x2 + 28x − 40), where k < 0. (k ≈ − 15
if the scales are equal;
otherwise one can’t tell how large k is.)
1.6 SOLUTIONS
51
24. There are only two functions, h and p, which can be put in the form y = Cbx , where C and b are constants:
p(x) =
a 3 bx
= (a3 /c)bx ,
c
where C = a3 /c since a, c are constants.
−1
= −5−(x−2) = −5−x+2 = (−25)5−x .
5x−2
Thus, h and p are the only exponential functions.
h(x) =
25. There is only one function, r, who can be put in the form y = Ax2 + Bx + C:
√
√
√
r(x) = −x + b − cx4 = − c x2 − x + b, where A = − c, since c is a constant.
Thus, r is the only quadratic function.
26. There is only one function, q, which is linear. Function q is a constant linear function whose vertical intercept is the
constant ab2 /c, since a, b and c are constants.
Problems
27. Consider the end behavior of the graph; that is, as x → +∞ and x → −∞. The ends of a degree 5 polynomial are in
Quadrants I and III if the leading coefficient is positive or in Quadrants II and IV if the leading coefficient is negative.
Thus, there must be at least one root. Since the degree is 5, there can be no more than 5 roots. Thus, there may be 1, 2, 3,
4, or 5 roots. Graphs showing these five possibilities are shown in Figure 1.60.
(a) 5 roots
(b) 4 roots
(c) 3 roots
x
x
(d) 2 roots
(e) 1 root
x
x
x
Figure 1.60
28. g(x) = 2x2 , h(x) = x2 + k for any k > 0. Notice that the graph is symmetric about the y-axis and limx→∞ f (x) = 2.
29. The graphs of both these functions will resemble that of x3 on a large enough window. One way to tackle the problem is to
graph them both (along with x3 if you like) in successively larger windows until the graphs come together. In Figure 1.61,
f , g and x3 are graphed in four windows. In the largest of the four windows the graphs are indistinguishable, as required.
Answers may vary.
y
y
106
109
x
−100
x
100
−106
−1000
1000
−109
y
1012
y
1015
x
−104
x
104
−105
−1012
105
−1015
Figure 1.61
52
Chapter One /SOLUTIONS
30. (a) A polynomial has the same end behavior as its leading term, so this polynomial behaves as −5x4 globally. Thus we
have:
f (x) → −∞ as x → −∞, and f (x) → −∞ as x → +∞.
(b) Polynomials behave globally as their leading term, so this rational function behaves globally as (3x2 )/(2x2 ), or 3/2.
Thus we have:
f (x) → 3/2 as x → −∞, and f (x) → 3/2 as x → +∞.
(c) We see from a graph of y = ex that
f (x) → 0 as x → −∞, and f (x) → +∞ as x → +∞.
31. Substituting w = 65 and h = 160, we have
(a)
s = 0.01(650.25 )(1600.75 ) = 1.3 m2 .
(b) We substitute s = 1.5 and h = 180 and solve for w:
1.5 = 0.01w0.25 (1800.75 ).
We have
1.5
= 3.05.
0.01(1800.75 )
w0.25 =
Since w0.25 = w1/4 , we take the fourth power of both sides, giving
w = 86.8 kg.
(c) We substitute w = 70 and solve for h in terms of s:
s = 0.01(700.25 )h0.75 ,
so
h0.75 =
s
.
0.01(700.25 )
Since h0.75 = h3/4 , we take the 4/3 power of each side, giving
h=
so
s
0.01(700.25 )
4/3
=
s4/3
(0.014/3 )(701/3 )
h = 112.6s4/3 .
32. Let D(v) be the stopping distance required by an Alpha Romeo as a function of its velocity. The assumption that stopping
distance is proportional to the square of velocity is equivalent to the equation
D(v) = kv 2
where k is a constant of proportionality. To determine the value of k, we use the fact that D(70) = 177.
D(70) = k(70)2 = 177.
Thus,
k=
It follows that
D(35) =
and
177
≈ 0.0361.
702
177
177
(35)2 =
= 44.25 ft
702
4
177
(140)2 = 708 ft.
702
Thus, at half the speed it requires one fourth the distance, whereas at twice the speed it requires four times the distance,
as we would expect from the equation. (We could in fact have figured it out that way, without solving for k explicitly.)
D(140) =
33. (a) Since the rate R varies directly with the fourth power of the radius r, we have the formula
R = kr 4
where k is a constant.
1.6 SOLUTIONS
(b) Given R = 400 for r = 3, we can determine the constant k.
400 = k(3)4
400 = k(81)
400
≈ 4.938.
k=
81
So the formula is
R = 4.938r 4
(c) Evaluating the formula above at r = 5 yields
R = 4.928(5)4 = 3086.42
cm 3
.
sec
34. Let us represent the height by h. Since the volume is V , we have
x2 h = V.
Solving for h gives
V
.
x2
The graph is in Figure 1.62. We are assuming V is a positive constant.
h=
h
h = V /x2
x
Figure 1.62
35. (a) Let the height of the can be h. Then
V = πr 2 h.
The surface area consists of the area of the ends (each is πr 2 ) and the curved sides (area 2πrh), so
S = 2πr 2 + 2πrh.
Solving for h from the formula for V , we have
h=
V
.
πr 2
Substituting into the formula for S, we get
S = 2πr 2 + 2πr ·
2V
V
= 2πr 2 +
.
πr 2
r
(b) For large r, the 2V /r term becomes negligible, meaning S ≈ 2πr 2 , and thus S → ∞ as r → ∞.
(c) The graph is in Figure 1.63.
S
100
75
S = 2πr 2 +
V = 10
50
25
0
r
1
2
3
4
Figure 1.63
5
2V
r
53
54
Chapter One /SOLUTIONS
36. To find the horizontal asymptote, we look at end behavior. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest degree term of the numerator and denominator. Thus, as x → ±∞,
we see that
5
5x
= .
f (x) →
2x
2
There is a horizontal asymptote at y = 5/2.
To find the vertical asymptotes, we set the denominator equal to zero. When 2x + 3 = 0, we have x = −3/2 so there is
a vertical asymptote at x = −3/2.
37. To find the horizontal asymptote, we look at end behavior. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest degree term of the numerator and denominator. Thus, as x → ±∞,
we see that
x2
f (x) → 2 = 1.
x
There is a horizontal asymptote at y = 1.
To find the vertical asymptotes, we set the denominator equal to zero. When x2 − 4 = 0, we have x = ±2 so there are
vertical asymptotes at x = −2 and at x = 2.
38. To find the horizontal asymptote, we look at end behavior. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest degree term of the numerator and denominator. Thus, as x → ±∞,
we see that
5x3
f (x) → 3 = 5.
x
There is a horizontal asymptote at y = 5.
To find the vertical asymptotes, we set the denominator equal to zero. When x3 − 27 = 0, we have x = 3 so there is a
vertical asymptote at x = 3.
39. (a) The object starts at t = 0, when s = v0 (0) − g(0)2 /2 = 0. Thus it starts on the ground, with zero height.
(b) The object hits the ground when s = 0. This is satisfied at t = 0, before it has left the ground, and at some later time
t that we must solve for.
0 = v0 t − gt2 /2 = t (v0 − gt/2)
Thus s = 0 when t = 0 and when v0 − gt/2 = 0, i.e., when t = 2v0 /g. The starting time is t = 0, so it must hit the
ground at time t = 2v0 /g.
(c) The object reaches its maximum height halfway between when it is released and when it hits the ground, or at
t = (2v0 /g)/2 = v0 /g.
(d) Since we know the time at which the object reaches its maximum height, to find the height it actually reaches we just
use the given formula, which tells us s at any given t. Substituting t = v0 /g,
s = v0
=
v0
g
1
− g
2
v2
2v02 − v02
= 0.
2g
2g
v02
g2
=
v02
v2
− 0
g
2g
40. The pomegranate is at ground level when f (t) = −16t2 + 64t = −16t(t − 4) = 0, so when t = 0 or t = 4.
At time t = 0 it is thrown, so it must hit the ground at t = 4 seconds. The symmetry of its path with respect to
time may convince you that it reaches its maximum height after 2 seconds. Alternatively, we can think of the graph of
f (t) = −16t2 + 64t = −16(t − 2)2 + 64, which is a downward parabola with vertex (i.e., highest point) at (2, 64). The
maximum height is f (2) = 64 feet.
41. (a)
(i) If (1, 1) is on the graph, we know that
1 = a(1)2 + b(1) + c = a + b + c.
(ii) If (1, 1) is the vertex, then the axis of symmetry is x = 1, so
−
and thus
b
= 1,
2a
b
a = − , so b = −2a.
2
But to be the vertex, (1, 1) must also be on the graph, so we know that a + b + c = 1. Substituting b = −2a,
we get −a + c = 1, which we can rewrite as a = c − 1, or c = 1 + a.
55
1.6 SOLUTIONS
(iii) For (0, 6) to be on the graph, we must have f (0) = 6. But f (0) = a(02 ) + b(0) + c = c, so c = 6.
(b) To satisfy all the conditions, we must first, from (a)(iii), have c = 6. From (a)(ii), a = c − 1 so a = 5. Also from
(a)(ii), b = −2a, so b = −10. Thus the completed equation is
y = f (x) = 5x2 − 10x + 6,
which satisfies all the given conditions.
42. The function is a cubic polynomial with positive leading coefficient. Since the figure given in the text shows that the
function turns around once, we know that the function has the shape shown in Figure 1.64. The function is below the
x-axis for x = 5 in the given graph, and we know that it goes to +∞ as x → +∞ because the leading coefficient is
positive. Therefore, there are exactly three zeros. Two zeros are shown, and occur at approximately x = −1 and x = 3.
The third zero must be to the right of x = 10 and so occurs for some x > 10.
Figure 1.64
43. We use the fact that at a constant speed, Time = Distance/Speed. Thus,
Horizontal asymptote: x-axis.
Vertical asymptote: x = 0 and x = 2.
Total time = Time running + Time walking
6
3
.
= +
x
x−2
44. (a) II and III because in both cases, the numerator and denominator each have x2 as the highest power, with coefficient
= 1. Therefore,
x2
y → 2 = 1 as x → ±∞.
x
(b) I, since
x
y → 2 = 0 as x → ±∞.
x
(c) II and III, since replacing x by −x leaves the graph of the function unchanged.
(d) None
(e) III, since the denominator is zero and f (x) tends to ±∞ when x = ±1.
45. h(t) cannot be of the form ct2 or kt3 since h(0.0) = 2.04. Therefore h(t) must be the exponential, and we see that
the ratio of successive values of h is approximately 1.5. Therefore h(t) = 2.04(1.5)t . If g(t) = ct2 , then c = 3 since
g(1.0) = 3.00. However, g(2.0) = 24.00 6= 3 · 22 . Therefore g(t) = kt3 , and using g(1.0) = 3.00, we obtain
g(t) = 3t3 . Thus f (t) = ct2 , and since f (2.0) = 4.40, we have f (t) = 1.1t2 .
46. The graphs are shown in Figure 1.65.
y
(a)
9
y
(b)
x4
3x
y
(c)
625
50000
x4
3x
x
3x
x4
x
5
5
Figure 1.65
x
10
56
Chapter One /SOLUTIONS
47. (a) R(P ) = kP (L − P ), where k is a positive constant.
(b) A possible graph is in Figure 1.66.
R
P
L
Figure 1.66
48. Since the parabola opens upward, we must have a > 0. To determine a relationship between x and y at the point of
intersection P , we eliminate a from the parabola and circle equations. Since y = x2 /a, we have a = x2 /y. Putting this
into the circle equation gives x2 + y 2 = 2x4 /y 2 . Rewrite this as
x2 y 2 + y 4 = 2x4
2
y 4 + x2 y 2 − 2x4 = 0
(y + 2x2 )(y 2 − x2 ) = 0.
This means x2 = y 2 (since y 2 cannot equal −2x2 ). Thus x = y since P is in the first quadrant. So P moves out along
the line y = x through the origin.
49. (a) The graph is shown in Figure 1.67. The graph represented by the exact formula has a vertical asymptote where the
denominator is undefined. This happens when
v2
= 0, or at v 2 = c2 .
c2
1−
Since v > 0, the graph of the exact formula has a vertical asymptote where
v = c = 3 · 108 m/sec.
E (1017 joules)
5
E = mc2
4
3
2
E=
1
1
2
3
4
√
1
1−v 2 /c2
−1
1
mv2
2
5
v (108 m/sec)
Figure 1.67
(b) The first formula does not give a good approximation to the exact formula when the graphs are not close together.
This happens for v > 1.5 · 108 m/sec. For v < 1.5 · 108 m/sec, the graphs look close together. However, the vertical
scale we are using is so large and the graphs are so close to the v-axis that a more careful analysis should be made.
We should zoom in and redraw the graph.
Strengthen Your Understanding
50. The graph of a polynomial of degree 5 cuts the horizontal axis at most five times, but it could be fewer. For example,
f (x) = x5 cuts the x-axis only once.
1.7 SOLUTIONS
57
51. The rational function f (x) = (x3 + 1)/x has no horizontal asymptotes. To see this, observe that
y = f (x) =
x3
x3 + 1
≈
= x2
x
x
for large x. Thus, y → ∞ as x → ±∞.
52. One possibility is p(x) = (x − 1)(x − 2)(x − 3) = x3 − 6x2 + 11x − 6.
53. A possible function is
f (x) =
3x
.
x − 10
f (x) =
1
.
x2 + 1
54. One possibility is
1
. Other answers are possible.
x + 7π
1
56. Let f (x) =
. This function has an asymptote corresponding to every factor
(x − 1)(x − 2)(x − 3) · · · (x − 16)(x − 17)
in the denominator. Other answers are possible.
x−1
has y = 1 as the horizontal asymptote and x = 2 as the vertical asymptote. These lines cross
57. The function f (x) =
x−2
at the point (2, 1). Other answers are possible.
55. Let f (x) =
58. False. The polynomial f (x) = x2 + 1, with degree 2, has no real zeros.
59. True. If the degree of the polynomial, p(x), is n, then the leading term is an xn with an 6= 0.
If n is odd and an is positive, p(x) tends toward ∞ as x → ∞ and p(x) tends toward −∞ as x → −∞. Since the
graph of p(x) has no breaks in it, the graph must cross the x-axis at least once.
If n is odd and an is negative, a similar argument applies, with the signs reversed, but leading to the same conclusion.
60. (a), (c), (d), (e), (b). Notice that f (x) and h(x) are decreasing functions, with f (x) being negative. Power functions
grow slower than exponential growth functions, so k(x) is next. Now order the remaining exponential functions, where
functions with larger bases grow faster.
Solutions for Section 1.7
Exercises
1. Yes, because x − 2 is not zero on this interval.
2. No, because x − 2 = 0 at x = 2.
3. Yes, because 2x − 5 is positive for 3 ≤ x ≤ 4.
4. Yes, because the denominator is never zero.
5. Yes, because 2x + x2/3 is defined for all x.
6. No, because 2x + x−1 is undefined at x = 0.
7. No, because cos(π/2) = 0.
8. No, because sin 0 = 0.
9. No, because ex − 1 = 0 at x = 0.
10. Yes, because cos θ is not zero on this interval.
11. We have that f (0) = −1 < 0 and f (1) = 1 > 0 and that f is continuous. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f (c) = k = 0.
12. We have that f (0) = 1 > 0 and f (1) = e − 3 < 0 and that f is continuous. Thus, by the Intermediate Value Theorem
applied to k = 0, there is a number c in [0, 1] such that f (c) = k = 0.
13. We have that f (0) = −1 < 0 and f (1) = 1 − cos 1 > 0 and that f is continuous. Thus, by the Intermediate Value
Theorem applied to k = 0, there is a number c in [0, 1] such that f (c) = k = 0.
58
Chapter One /SOLUTIONS
14. Since f is not continuous at x = 0, we consider instead the smaller interval [0.01, 1]. We have that f (0.01) = 20.01 −
100 < 0 and f (1) = 2 − 1/1 = 1 > 0 and that f is continuous. Thus, by the Intermediate Value Theorem applied to
k = 0, there is a number c in [0.01, 1], and hence in [0, 1], such that f (c) = k = 0.
15. (a) At x = 1, on the line y = x, we have y = 1. At x = 1, on the parabola y = x2 , we have y = 1. Thus, f (x) is
continuous. See Figure 1.68.
(b) At x = 3, on the line y = x, we have y = 3. At x = 3, on the parabola y = x2 , we have y = 9. Thus, g(x) is not
continuous. See Figure 1.69.
y
y
g(x)
f (x)
x
x
1
3
Figure 1.68
Figure 1.69
Problems
16. (a) Even if the car stops to refuel, the amount of fuel in the tank changes smoothly, so the fuel in the tank is a continuous
function; the quantity of fuel cannot suddenly change from one value to another.
(b) Whenever a student joins or leaves the class the number jumps up or down immediately by 1 so this is not a continuous
function, unless the enrollment does not change at all.
(c) Whenever the oldest person dies the value of the function jumps down to the age of the next oldest person, so this is
not a continuous function.
17. Two possible graphs are shown in Figures 1.70 and 1.71.
velocity
distance
time
time
Figure 1.70: Velocity of the car
Figure 1.71: Distance
The distance moved by the car is continuous. (Figure 1.71 has no breaks in it.) In actual fact, the velocity of the car
is also continuous; however, in this case, it is well-approximated by the function in Figure 1.70, which is not continuous
on any interval containing the moment of impact.
18. The voltage f (t) is graphed in Figure 1.72.
12
6
time (seconds)
7
Figure 1.72: Voltage change from 6V to 12V
1.7 SOLUTIONS
59
Using formulas, the voltage, f (t), is represented by
(
f (t) =
6, 0 < t ≤ 7
12,
7 0, we have |x| = x, so f (x) = 1. For x < 0, we have |x| = −x, so f (x) = −1. Thus, the function is given by
f (x) =
(
1
0
−1
x>0
x=0
x<1 ,
so f is not continuous on any interval containing x = 0.
32. The graph of g suggests that g is not continuous on any interval containing θ = 0, since g(0) = 1/2.
33. The drug first increases linearly for half a second, at the end of which time there is 0.6 ml in the body. Thus, for 0 ≤ t ≤
0.5, the function is linear with slope 0.6/0.5 = 1.2:
Q = 1.2t
for
0 ≤ t ≤ 0.5.
At t = 0.5, we have Q = 0.6. For t > 0.5, the quantity decays exponentially at a continuous rate of 0.002, so Q has the
form
Q = Ae−0.002t
0.5 < t.
We choose A so that Q = 0.6 when t = 0.5:
0.6 = Ae−0.002(0.5) = Ae−0.001
A = 0.6e0.001 .
Thus
Q=
(
1.2t
0 ≤ t ≤ 0.5
0.6e0.001 e−.002t
0.5 < t.
34.
0.8
0.2
0.4
0.6
1
0.8
0.2
0.4
0.6
1
0.8
0.2
0.4
0.6
1
62
Chapter One /SOLUTIONS
35. Since polynomials are continuous, and since p(5) < 0 and p(10) > 0 and p(12) < 0, there are two zeros, one between
x = 5 and x = 10, and another between x = 10 and x = 12. Thus, p(x) is a cubic with at least two zeros.
If p(x) has only two zeros, one would be a double zero (corresponding to a repeated factor). However, since a
polynomial does not change sign at a repeated zero, p(x) cannot have a double zero and have the signs it does.
Thus, p(x) has three zeros. The third zero can be greater than 12 or less than 5. See Figures 1.77 and 1.78.
y
y
p(x)
x
x
p(x)
Figure 1.77
Figure 1.78
36. (a) The graphs of y = ex and y = 4 − x2 cross twice in Figure 1.79. This tells us that the equation ex = 4 − x2 has two
solutions.
Since y = ex increases for all x and y = 4 − x2 increases for x < 0 and decreases for x > 0, these are only
the two crossing points.
y
y = ex
−2
2
y = x2 − 4
x
Figure 1.79
(b) Values of f (x) are in Table 1.5. One solution is between x = −2 and x = −1; the second solution is between x = 1
and x = 2.
Table 1.5
x
f (x)
−4
12.0
−3
5.0
−2
0.1
−1
−2.6
0
1
2
3
4
−3
−0.3
7.4
25.1
66.6
37. (a) Figure 1.80 shows a possible graph of f (x), yours may be different.
9
2
x
3
5
Figure 1.80
(b) In order for f to approach the horizontal asymptote at 9 from above it is necessary that f eventually become concave
up. It is therefore not possible for f to be concave down for all x > 6.
1.7 SOLUTIONS
63
38. (a) Since f (x) is not continuous at x = 1, it does not satisfy the conditions of the Intermediate Value Theorem.
(b) We see that f (0) = e0 = 1 and f (2) = 4 + (2 − 1)2 = 5. Since ex is increasing between x = 0 and x = 1, and
since 4 + (x − 1)2 is increasing between x = 1 and x = 2, any value of k between e1 = e and 4 + (1 − 1)2 = 4,
such as k = 3, is a value such that f (x) = k has no solution.
Strengthen Your Understanding
39. The Intermediate Value theorem only makes this guarantee for a continuous function, not for any function.
40. The Intermediate Value Theorem guarantees that for at least one value of x between 0 and 2, we have f (x) = 5, but it
does not tell us which value(s) of x give f (x) = 5.
41. We want a function which has a value at every point but where the graph has a break at x = 15. One possibility is
f (x) =
(
1
x ≥ 15
−1
x < 15
42. One example is f (x) = 1/x, which is not continuous at x = 0. The Intermediate Value Theorem does not apply on an
interval that contains a point where a function is not continuous.
43. Let f (x) =
(
44. Let f (x) =
(
1
x≤2
x x>2
x
2x
. Then f (x) is continuous at every point in [0, 3] except at x = 2. Other answers are possible.
x≤3
x>3
. Then f (x) is increasing for all x but f (x) is not continuous at x = 3. Other answers are
possible.
45. False. For example, let f (x) =
(
1 x≤3
2 x>3
, then f (x) is defined at x = 3 but it is not continuous at x = 3. (Other
examples are possible.)
46. False. A counterexample is graphed in Figure 1.81, in which f (5) < 0.
f (x)
100
x
5
10
Figure 1.81
47. False. A counterexample is graphed in Figure 1.82.
f (a)
f (x)
f (b)
a
Figure 1.82
x
b
64
Chapter One /SOLUTIONS
Solutions for Section 1.8
Exercises
1. (a) As x approaches −2 from either side, the values of f (x) get closer and closer to 3, so the limit appears to be about 3.
(b) As x approaches 0 from either side, the values of f (x) get closer and closer to 7. (Recall that to find a limit, we are
interested in what happens to the function near x but not at x.) The limit appears to be about 7.
(c) As x approaches 2 from either side, the values of f (x) get closer and closer to 3 on one side of x = 2 and get closer
and closer to 2 on the other side of x = 2. Thus the limit does not exist.
(d) As x approaches 4 from either side, the values of f (x) get closer and closer to 8. (Again, recall that we don’t care
what happens right at x = 4.) The limit appears to be about 8.
lim f (x) = 1.
2. (a)
x→1−
(b) lim f (x) does not exist.
x→1+
(c) lim f (x) does not exist.
x→1
lim f (x) = 1.
(d)
x→2−
(e) lim f (x) = 1.
x→2+
(f) lim f (x) = 1.
x→2
3. From the graphs of f and g, we estimate lim f (x) = 3, lim g(x) = 5,
x→1−
x→1−
lim f (x) = 4, lim g(x) = 1.
x→1+
x→1+
(a)
lim (f (x) + g(x)) = 3 + 5 = 8
x→1−
(b) lim (f (x) + 2g(x)) = lim f (x) + 2 lim g(x) = 4 + 2(1) = 6
x→1+
(c)
x→1+
x→1+
lim (f (x)g(x)) = ( lim f (x))( lim g(x)) = (3)(5) = 15
x→1−
(d) lim (f (x)/g(x)) =
x→1+
x→1−
−
x→1
lim f (x) /
x→1+
lim g(x) = 4/1 = 4
x→1+
4. We see that f (x) goes to −∞ on both ends, so one possible graph is shown in Figure 1.83. Other answers are possible.
y
x
Figure 1.83
5. We see that f (x) goes to +∞ on the left and to −∞ on the right. One possible graph is shown in Figure 1.84. Other
answers are possible.
y
x
Figure 1.84
1.8 SOLUTIONS
65
6. We see that f (x) goes to +∞ on the left and approaches a y-value of 1 on the right. One possible graph is shown in
Figure 1.85. Other answers are possible.
y
1
x
Figure 1.85
7. We see that f (x) approaches a y-value of 3 on the left and goes to −∞ on the right. One possible graph is shown in
Figure 1.86. Other answers are possible.
y
3
x
Figure 1.86
8. We see that f (x) goes to +∞ on the right and that it also passes through the point (−1, 2). (Notice that this must be a
point on the graph since the instructions require that f (x) be defined and continuous.) One possible graph is shown in
Figure 1.87. Other answers are possible.
y
2
x
−1
Figure 1.87
9. We see that f (x) goes to +∞ on the left and that it also passes through the point (3, 5). (Notice that this must be a
point on the graph since the instructions require that f (x) be defined and continuous.) One possible graph is shown in
Figure 1.88. Other answers are possible.
y
5
x
3
Figure 1.88
66
Chapter One /SOLUTIONS
10. We see that lim f (x) = −∞ and lim f (x) = −∞.
x→−∞
x→+∞
11. As x → ±∞, we know that f (x) behaves like its leading term −2x3 . Thus, we have
lim f (x) = +∞ and
x→−∞
lim f (x) = −∞.
x→+∞
12. As x → ±∞, we know that f (x) behaves like its leading term x5 . Thus, we have lim f (x) = −∞ and lim f (x) =
x→−∞
+∞.
x→+∞
13. As x → ±∞, we know that f (x) behaves like the quotient of the leading terms of its numerator and denominator. Since
f (x) →
3
3x3
= ,
5x3
5
we have lim f (x) = 3/5 and lim f (x) = 3/5.
x→−∞
x→+∞
14. As x → ±∞, we know that x−3 gets closer and closer to zero, so we have lim f (x) = 0 and lim f (x) = 0.
x→−∞
x→+∞
15. We have lim f (x) = 0 and lim f (x) = +∞.
x→−∞
x→+∞
16. The break in the graph at x = 0 suggests that lim
x→0
|x|
does not exist. See Figure 1.89.
x
1
|x|
x
−1
x
1
−1
Figure 1.89
17. For −1 ≤ x ≤ 1, − 1 ≤ y ≤ 1, the graph of y = x ln |x| is in Figure 1.90. The graph suggests that
lim x ln |x| = 0.
x→0
y
1
y = x ln |x|
x
−1
1
−1
Figure 1.90
|x|
|x|
|x|
= 1 and lim
= −1, we say that lim
does not exist. In addition f (x) is not defined at 0.
x
x→0 x
x→0− x
Therefore, f (x) is not continuous on any interval containing 0.
18. Since lim
x→0+
1.8 SOLUTIONS
19. For −0.5 ≤ θ ≤ 0.5, 0 ≤ y ≤ 3, the graph of y =
67
sin (2 θ)
sin (2 θ)
is shown in Figure 1.91. Therefore, lim
= 2.
θ→0
θ
θ
y=
2
sin(2θ)
θ
1
−0.5
0
0.5
θ
Figure 1.91
20. For −1 ≤ θ ≤ 1, −1 ≤ y ≤ 1, the graph of y =
cos θ − 1
cos θ − 1
is shown in Figure 1.92. Therefore, lim
= 0.
θ→0
θ
θ
1
y=
cos θ−1
θ
−1
θ
1
−1
Figure 1.92
21. For −90◦ ≤ θ ≤ 90◦ , 0 ≤ y ≤ 0.02, the graph of y =
sin θ
curve, we see that in degrees, lim
= 0.01745 . . ..
θ→0
θ
sin θ
is shown in Figure 1.93. Therefore, by tracing along the
θ
sin θ
θ
0.02
0.01
−90
0
90
θ (degree)
Figure 1.93
22. For −0.5 ≤ θ ≤ 0.5, 0 ≤ y ≤ 0.5, the graph of y =
curve, we see that lim
θ→0
θ
is shown in Figure 1.94. Therefore, by tracing along the
tan(3θ)
θ
= 0.3333 . . ..
tan(3θ)
0.5
−0.5
θ
tan(3θ)
0
0.5
θ
Figure 1.94
eh − 1
in a window such as −0.5 ≤ h ≤ 0.5 and 0 ≤ y ≤ 3 appears to indicate y → 1 as h → 0.
h
eh − 1
Therefore, we estimate that lim
= 1.
h→0
h
23. A graph of y =
68
Chapter One /SOLUTIONS
e5h − 1
in a window such as −0.5 ≤ h ≤ 0.5 and 0 ≤ y ≤ 6 appears to indicate y → 5 as h → 0.
h
e5h − 1
= 5.
Therefore, we estimate that lim
h→0
h
2h − 1
in a window such as −0.5 ≤ h ≤ 0.5 and 0 ≤ y ≤ 1 appears to indicate y → 0.7 as h → 0. By
25. A graph of y =
h
zooming in on the graph, we can estimate the limit more accurately. Therefore, we estimate that
24. A graph of y =
lim
h→0
2h − 1
= 0.693.
h
3h − 1
in a window such as −0.5 ≤ h ≤ 0.5 and 0 ≤ y ≤ 1.5 appears to indicate y → 1.1 as h → 0.
26. A graph of y =
h
By zooming in on the graph, we can estimate the limit more accurately. Therefore, we estimate that
lim
h→0
3h − 1
= 1.098.
h
cos(3h) − 1
in a window such as −0.5 ≤ h ≤ 0.5 and −1 ≤ y ≤ 1 appears to indicate y → 0 as
27. A graph of y =
h
h → 0. Therefore, we estimate that
cos(3h) − 1
lim
= 0.
h→0
h
sin(3h)
in a window such as −0.5 ≤ h ≤ 0.5 and 0 ≤ y ≤ 4 appears to indicate y → 3 as h → 0.
28. A graph of y =
h
Therefore, we estimate that
sin(3h)
lim
= 3.
h→0
h
(
x−4
x>4
|x − 4|
1
x>4
x
−
4
=
=
29. f (x) =
x−4
−x − 4 x < 4
−1 x < 4
x−4
Figure 1.95 confirms that lim f (x) = 1, lim f (x) = −1 so lim f (x) does not exist.
x→4+
x→4
x→4−
f (x)
1
x
0
2
4
6
8
−1
Figure 1.95
30. f (x) =
x−2
, x>2
x
|x − 2|
=
x
−x − 2, x < 2
x
Figure 1.96 confirms that lim f (x) = lim f (x) = lim f (x) = 0.
x→2+
x→2
x→2−
1
f (x)
0
x
1
2
3
Figure 1.96
4
1.8 SOLUTIONS
31. f (x) =
69
2
x − 2 0 < x < 3
2
x=3
2x + 1
3 0, there is a δ > 0 such that
|g(h) − K| < ǫ for all 0 < |h − a| < δ.
Problems
34. At x = 0, the function is not defined. In addition, limx→0 f (x) does not exist. Thus, f (x) is not continuous at x = 0.
35. Since limx→0+ f (x) = 1 and limx→0− f (x) = −1, we see that limx→0 f (x) does not exist. Thus, f (x) is not continuous at x = 0
36. Since x/x = 1 for x 6= 0, this function f (x) = 1 for all x. Thus, f (x) is continuous for all x.
37. Since 2x/x = 2 for x 6= 0, we have limx→0 f (x) = 2, so
lim f (x) 6= f (0) = 3.
x→0
Thus, f (x) is not continuous at x = 0.
70
Chapter One /SOLUTIONS
38. The answer (see the graph in Figure 1.99) appears to be about 2.7; if we zoom in further, it appears to be about 2.72,
which is close to the value of e ≈ 2.71828.
3.5
3
2.5
2
1.5
1
0.5
(1 + x)1/x
x
−0.1
0.1
Figure 1.99
39. We use values of h approaching, but not equal to, zero. If we let h = 0.01, 0.001, 0.0001, 0.00001, we calculate the values
2.7048, 2.7169, 2.7181, and 2.7183. If we let h = −0.01 −0.001, −0.0001, −0.00001, we get values 2.7320, 2.7196,
2.7184, and 2.7183. These numbers suggest that the limit is the number e = 2.71828 . . .. However, these calculations
cannot tell us that the limit is exactly e; for that a proof is needed.
40. When x = 0.1, we find xe1/x ≈ 2203. When x = 0.01, we find xe1/x ≈ 3 × 1041 . When x = 0.001, the value of xe1/x
is too big for a calculator to compute. This suggests that lim xe1/x does not exist (and in fact it does not).
x→0+
41. If x > 1 and x approaches 1, then p(x) = 55. If x < 1 and x approaches 1, then p(x) = 34. There is not a single number
that p(x) approaches as x approaches 1, so we say that lim p(x) does not exist.
x→1
42. The limit appears to be 1; a graph and table of values is shown below.
y
1.4
1.2
1
0.8
0.6
0.4
0.2
0
x
0.02 0.04 0.06 0.08 0.1
x
xx
0.1
0.7943
0.01
0.9550
0.001
0.9931
0.0001
0.9990
0.00001
0.9999
43. The only change is that, instead of considering all x near c, we only consider x near to and greater than c. Thus the phrase
“|x − c| < δ” must be replaced by “c < x < c + δ.” Thus, we define
lim f (x) = L
x→c+
to mean that for any ǫ > 0 (as small as we want), there is a δ > 0 (sufficiently small) such that if c < x < c + δ, then
|f (x) − L| < ǫ.
44. The only change is that, instead of considering all x near c, we only consider x near to and less than c. Thus the phrase
“|x − c| < δ” must be replaced by “c − δ < x < c.” Thus, we define
lim f (x) = L
x→c−
to mean that for any ǫ > 0 (as small as we want), there is a δ > 0 (sufficiently small) such that if c − δ < x < c, then
|f (x) − L| < ǫ.
45. Instead of being “sufficiently close to c,” we want x to be “sufficiently large.” Using N to measure how large x must be,
we define
lim f (x) = L
x→∞
to mean that for any ǫ > 0 (as small as we want), there is a N > 0 (sufficiently large) such that if x > N , then
|f (x) − L| < ǫ.
1.8 SOLUTIONS
71
46. From Table 1.6, it appears the limit is 1. This is confirmed by Figure 1.100. An appropriate window is −0.0033 < x <
0.0033, 0.99 < y < 1.01.
1.01
Table 1.6
x
f (x)
x
f (x)
0.1
1.3
−0.0001
0.9997
−0.01
0.97
0.01
1.03
0.001
1.003
0.0001
1.0003
−0.001
0.997
−0.1
0.7
0.99
−0.0033
0.0033
Figure 1.100
47. From Table 1.7, it appears the limit is −1. This is confirmed by Figure 1.101. An appropriate window is −0.099 < x <
0.099, −1.01 < y < −0.99.
−0.99
Table 1.7
x
f (x)
x
f (x)
0.1
−0.99
−0.0001
−0.99999999
−0.999999
−0.01
−0.9999
−0.9999
0.01
0.001
0.0001
−0.99999999
−0.001
−0.1
−0.999999
−1.01
−0.099
−0.99
0.099
Figure 1.101
48. From Table 1.8, it appears the limit is 0. This is confirmed by Figure 1.102. An appropriate window is −0.005 < x <
0.005, −0.01 < y < 0.01.
0.01
Table 1.8
x
f (x)
x
f (x)
0.1
0.1987
−0.0001
−0.0002
−0.01
−0.0200
0.01
0.0200
0.001
0.0020
0.0001
0.0002
−0.001
−0.1
−0.0020
−0.01
−0.005
−0.1987
0.005
Figure 1.102
49. From Table 1.9, it appears the limit is 0. This is confirmed by Figure 1.103. An appropriate window is −0.0033 < x <
0.0033, −0.01 < y < 0.01.
0.01
Table 1.9
x
f (x)
x
f (x)
0.1
0.2955
−0.0001
−0.0003
−0.01
−0.0300
0.01
0.0300
0.001
0.0030
0.0001
0.0003
−0.001
−0.1
−0.0030
−0.2955
−0.01
−0.0033
Figure 1.103
0.0033
72
Chapter One /SOLUTIONS
50. From Table 1.10, it appears the limit is 2. This is confirmed by Figure 1.104. An appropriate window is −0.0865 < x <
0.0865, 1.99 < y < 2.01.
2.01
Table 1.10
x
f (x)
x
f (x)
0.1
1.9867
−0.0001
2.0000
−0.01
1.9999
0.01
1.9999
0.001
2.0000
0.0001
2.0000
−0.001
2.0000
−0.1
1.9867
1.99
−0.0865
0.0865
Figure 1.104
51. From Table 1.11, it appears the limit is 3. This is confirmed by Figure 1.105. An appropriate window is −0.047 < x <
0.047, 2.99 < y < 3.01.
3.01
Table 1.11
x
f (x)
x
f (x)
0.1
2.9552
−0.0001
3.0000
−0.01
2.9996
0.01
2.9996
0.001
3.0000
0.0001
3.0000
−0.001
3.0000
−0.1
2.9552
2.99
−0.047
0.047
Figure 1.105
52. From Table 1.12, it appears the limit is 1. This is confirmed by Figure 1.106. An appropriate window is −0.0198 < x <
0.0198, 0.99 < y < 1.01.
1.01
Table 1.12
x
f (x)
x
f (x)
0.1
1.0517
−0.0001
1.0000
−0.01
0.9950
0.01
1.0050
0.001
1.0005
0.0001
1.0001
−0.001
0.9995
−0.1
0.9516
0.99
−0.0198
0.0198
Figure 1.106
53. From Table 1.13, it appears the limit is 2. This is confirmed by Figure 1.107. An appropriate window is −0.0049 < x <
0.0049, 1.99 < y < 2.01.
2.01
Table 1.13
x
f (x)
x
f (x)
0.1
2.2140
−0.0001
1.9998
−0.01
1.9801
0.01
2.0201
0.001
2.0020
0.0001
2.0002
−0.001
1.9980
−0.1
1.8127
1.99
−0.0049
Figure 1.107
0.0049
1.8 SOLUTIONS
54. Divide numerator and denominator by x:
f (x) =
so
lim f (x) = lim
x→∞
x→∞
x+3
1 + 3/x
=
,
2−x
2/x − 1
limx→∞ (1 + 3/x)
1
1 + 3/x
=
=
= −1.
2/x − 1
limx→∞ (2/x − 1)
−1
55. Divide numerator and denominator by x:
f (x) =
so
lim f (x) = lim
x→∞
x→∞
π + 3x
(π + 3x)/x
=
,
πx − 3
(πx − 3)/x
limx→∞ (π/x + 3)
π/x + 3
3
=
= .
π − 3/x
limx→∞ (π − 3/x)
π
56. Divide numerator and denominator by x2 :
f (x) =
so
x−5
(1/x) − (5/x2 )
=
,
2
5 + 2x
(5/x2 ) + 2
limx→∞ ((1/x) − (5/x2 ))
(1/x) − (5/x2 )
0
=
= = 0.
2
x→∞
(5/x ) + 2
limx→∞ ((5/x2 ) + 2)
2
lim f (x) = lim
x→∞
57. Divide numerator and denominator by x2 , giving
f (x) =
so
lim f (x) = lim
x→∞
x→∞
1 + 2/x − 1/x2
x2 + 2x − 1
=
,
3 + 3x2
3/x2 + 3
limx→∞ (1 + 2/x − 1/x2 )
1
1 + 2/x − 1/x2
=
= .
3/x2 + 3
limx→∞ (3/x2 + 3)
3
58. Divide numerator and denominator by x, giving
f (x) =
x2 + 4
x + 4/x
=
,
x+3
1 + 3/x
so
lim f (x) = +∞.
x→∞
59. Divide numerator and denominator by x3 , giving
f (x) =
so
lim f (x) = lim
x→∞
x→∞
2 − 16/x
2x3 − 16x2
=
,
4x2 + 3x3
4/x + 3
2 − 16/x
limx→∞ (2 − 16/x)
2
=
= .
4/x + 3
limx→∞ (4/x + 3)
3
60. Divide numerator and denominator by x5 , giving
f (x) =
so
lim f (x) =
x→∞
1/x + 3/x4
x4 + 3x
=
,
4
5
x + 2x
1/x + 2
0
limx→∞ (1/x + 3/x4 )
= = 0.
limx→∞ (1/x + 2)
2
73
74
Chapter One /SOLUTIONS
61. Divide numerator and denominator by ex , giving
f (x) =
so
lim f (x) =
x→∞
3 + 2e−x
3ex + 2
=
,
2ex + 3
2 + 3e−x
limx→∞ (3 + 2e−x )
3
= .
limx→∞ (2 + 3e−x )
2
62. Since limx→∞ 2−x = limx→∞ 3−x = 0, we have
lim f (x) = lim
x→∞
63. f (x) =
x→∞
limx→∞ (2−x + 5)
5
2−x + 5
=
= .
−x
3 +7
limx→∞ (3−x + 7)
7
limx→∞ (2e−x + 3)
2e−x + 3
3
, so lim f (x) =
= .
−x
x→∞
3e + 2
limx→∞ (3e−x + 2)
2
64. Because the denominator equals 0 when x = 4, so must the numerator. This means k2 = 16 and the choices for k are 4
or −4.
65. Because the denominator equals 0 when x = 1, so must the numerator. So 1 − k + 4 = 0. The only possible value of k
is 5.
66. Because the denominator equals 0 when x = −2, so must the numerator. So 4 − 8 + k = 0 and the only possible value
of k is 4.
67. Division of numerator and denominator by x2 yields
1 + 3/x + 5/x2
x2 + 3x + 5
=
.
4x + 1 + xk
4/x + 1/x2 + xk−2
As x → ∞, the limit of the numerator is 1. The limit of the denominator depends upon k. If k > 2, the denominator
approaches ∞ as x → ∞, so the limit of the quotient is 0. If k = 2, the denominator approaches 1 as x → ∞, so the
limit of the quotient is 1. If k < 2 the denominator approaches 0+ as x → ∞, so the limit of the quotient is ∞. Therefore
the values of k we are looking for are k ≥ 2.
68. For the numerator,
If k = 0,
lim
x→−∞
lim
x→−∞
e2x − 5 = −5. If k > 0,
lim
x→−∞
ekx + 3 = 3, so the quotient has a limit of −5/3.
ekx + 3 = 4, so the quotient has limit of −5/4. If k < 0, the limit of the quotient is given by
lim (e2x − 5)/(ekx + 3) = 0.
x→−∞
69. Division of numerator and denominator by x3 yields
1 − 6/x3
x3 − 6
=
.
xk + 3
xk−3 + 3/x3
As x → ∞, the limit of the numerator is 1. The limit of the denominator depends upon k. If k > 3, the denominator
approaches ∞ as x → ∞, so the limit of the quotient is 0. If k = 3, the denominator approaches 1 as x → ∞, so the
limit of the quotient is 1. If k < 3 the denominator approaches 0+ as x → ∞, so the limit of the quotient is ∞. Therefore
the values of k we are looking for are k ≥ 3.
70. We divide both the numerator and denominator by 32x , giving
3(k−2)x + 6/32x
3kx + 6
=
.
2x
+4
1 + 4/32x
x→∞ 3
lim
In the denominator, lim 1 + 4/32x = 1. In the numerator, if k < 2, we have lim 3(k−2)x + 6/32x = 0, so the quotient
x→∞
x→∞
has a limit of 0. If k = 2, we have lim 3(k−2)x + 6/32x = 1, so the quotient has a limit of 1. If k > 2, we have
x→∞
lim 3(k−2)x + 6/32x = ∞, so the quotient has a limit of ∞.
x→∞
71. In the denominator, we have lim 32x + 4 = 4. In the numerator, if k < 0, we have lim 3kx + 6 = ∞, so the
x→−∞
x→−∞
quotient has a limit of ∞. If k = 0, we have lim 3kx + 6 = 7, so the quotient has a limit of 7/4. If k > 0, we have
x→−∞
lim 3kx + 6 = 6, so the quotient has a limit of 6/4.
x→−∞
1.8 SOLUTIONS
75
72. By tracing on a calculator or solving equations, we find the following values of δ:
For ǫ = 0.2, δ ≤ 0.1.
For ǫ = 0.1, δ ≤ 0.05.
For ǫ = 0.02, δ ≤ 0.01.
For ǫ = 0.01, δ ≤ 0.005.
For ǫ = 0.002, δ ≤ 0.001.
For ǫ = 0.001, δ ≤ 0.0005.
73. By tracing on a calculator or solving equations, we find the following values of δ:
For ǫ = 0.1, δ ≤ 0.46.
For ǫ = 0.01, δ ≤ 0.21.
For ǫ = 0.001, δ < 0.1. Thus, we can take δ ≤ 0.09.
74. The results of Problem 72 suggest that we can choose δ = ǫ/2. For any ǫ > 0, we want to find the δ such that
|f (x) − 3| = |−2x + 3 − 3| = |2x| < ǫ.
Then if |x| < δ = ǫ/2, it follows that |f (x) − 3| = |2x| < ǫ. So limx→0 (−2x + 3) = 3.
75. (a) Since sin(nπ) = 0 for n = 1, 2, 3, . . . the sequence of x-values
1 1 1
,
,
,...
π 2π 3π
works. These x-values → 0 and are zeroes of f (x).
(b) Since sin(nπ/2) = 1 for n = 1, 5, 9 . . . the sequence of x-values
2 2 2
,
,
,...
π 5π 9π
works.
(c) Since sin(nπ)/2 = −1 for n = 3, 7, 11, . . . the sequence of x-values
2 2
2
,
,
...
3π 7π 11π
works.
(d) Any two of these sequences of x-values show that if the limit were to exist, then it would have to have two (different)
values: 0 and 1, or 0 and −1, or 1 and −1. Hence, the limit can not exist.
76. From Table 1.14, it appears the limit is 0. This is confirmed by Figure 1.108. An appropriate window is −0.015 < x <
0.015, −0.01 < y < 0.01.
0.01
Table 1.14
x
f (x)
x
f (x)
−0.0001
−0.0001
−0.01
−0.0067
0.1
0.0666
0.01
0.0067
0.001
0.0007
0.0001
0.0001
−0.001
−0.1
−0.0007
−0.01
−0.015
−0.0666
0.015
Figure 1.108
77. From Table 1.15, it appears the limit is 0. This is confirmed by Figure 1.109. An appropriate window is −0.0029 < x <
0.0029, −0.01 < y < 0.01.
0.01
Table 1.15
x
f (x)
x
f (x)
0.1
0.3365
−0.0001
−0.0004
−0.01
−0.0337
0.01
0.0337
0.001
0.0034
0.0001
0.0004
−0.001
−0.1
−0.0034
−0.3365
−0.01
−0.0029
Figure 1.109
0.0029
76
Chapter One /SOLUTIONS
78. (a) If b = 0, then the property says limx→c 0 = 0, which is easy to see is true.
ǫ
, then multiplying by |b| gives
(b) If |f (x) − L| < |b|
|b||f (x) − L| < ǫ.
Since
|b||f (x) − L| = |b(f (x) − L)| = |bf (x) − bL|,
we have
|bf (x) − bL| < ǫ.
(c) Suppose that lim f (x) = L. We want to show that lim bf (x) = bL. If we are to have
x→c
x→c
|bf (x) − bL| < ǫ,
then we will need
|f (x) − L| <
We choose δ small enough that
|x − c| < δ
implies
ǫ
.
|b|
|f (x) − L| <
By part (b), this ensures that
ǫ
.
|b|
|bf (x) − bL| < ǫ,
as we wanted.
79. Suppose lim f (x) = L1 and lim g(x) = L2 . Then we need to show that
x→c
x→c
lim (f (x) + g(x)) = L1 + L2 .
x→c
Let ǫ > 0 be given. We need to show that we can choose δ > 0 so that whenever |x − c| < δ, we will have
|(f (x) + g(x)) − (L1 + L2 )| < ǫ. First choose δ1 > 0 so that |x − c| < δ1 implies |f (x) − L1 | < 2ǫ ; we can do this
since lim f (x) = L1 . Similarly, choose δ2 > 0 so that |x − c| < δ2 implies |g(x) − L2 | < 2ǫ . Now, set δ equal to the
x→c
smaller of δ1 and δ2 . Thus |x − c| < δ will make both |x − c| < δ1 and |x − c| < δ2 . Then, for |x − c| < δ, we have
|f (x) + g(x) − (L1 + L2 )| = |(f (x) − L1 ) + (g(x) − L2 )|
≤ |(f (x) − L1 )| + |(g(x) − L2 )|
ǫ
ǫ
≤ + = ǫ.
2
2
This proves limx→c (f (x) + g(x)) = limx→c f (x) + limx→c g(x), which is the result we wanted to prove.
80. (a) We need to show that for any given ǫ > 0, there is a δ > 0 so that √
|x − c| < δ implies |f (x)g(x)| < ǫ. If ǫ > 0
is given, choose δ1 so that when |x − c| < δ1 , we have |f (x)| < ǫ. This can be done since limx→0 f (x) = 0.
√
Similarly, choose δ2 so that when |x − c| < δ2 , we have |g(x)| < ǫ. Then, if we take δ to be the smaller of δ1
√
√
and δ2 , we’ll have that |x − c|√< δ√implies both |f (x)| < ǫ and |g(x)| < ǫ. So when |x − c| < δ, we have
|f (x)g(x)| = |f (x)| |g(x)| < ǫ · ǫ = ǫ. Thus lim f (x)g(x) = 0.
x→c
(b) (f (x) − L1 ) (g(x) − L2 ) + L1 g(x) + L2 f (x) − L1 L2
= f (x)g(x) − L1 g(x) − L2 f (x) + L1 L2 + L1 g(x) + L2 f (x) − L1 L2 = f (x)g(x).
(c) lim (f (x) − L1 ) = lim f (x)− lim L1 = L1 −L1 = 0, using the second limit property. Similarly, lim (g(x) − L2 ) =
x→c
x→c
x→c
x→c
0.
(d) Since lim (f (x) − L1 ) = lim (g(x) − L2 ) = 0, we have that lim (f (x) − L1 ) (g(x) − L2 ) = 0 by part (a).
x→c
x→c
x→c
(e) From part (b), we have
lim f (x)g(x) = lim ((f (x) − L1 ) (g(x) − L2 ) + L1 g(x) + L2 f (x) − L1 L2 )
x→c
x→c
= lim (f (x) − L1 ) (g(x) − L2 ) + lim L1 g(x) + lim L2 f (x) + lim (−L1 L2 )
x→c
x→c
(using limit property 2)
= 0 + L1 lim g(x) + L2 lim f (x) − L1 L2
x→c
x→c
(using limit property 1 and part (d))
= L1 L2 + L2 L1 − L1 L2 = L1 L2 .
x→c
x→c
1.8 SOLUTIONS
77
81. We will show f (x) = x is continuous at x = c. Since f (c) = c, we need to show that
lim f (x) = c
x→c
that is, since f (x) = x, we need to show
lim x = c.
x→c
Pick any ǫ > 0, then take δ = ǫ. Thus,
|f (x) − c| = |x − c| < ǫ
for all
|x − c| < δ = ǫ.
82. Since f (x) = x is continuous, Theorem 1.3 on page 64 shows that products of the form f (x) · f (x) = x2 and f (x) · x2 =
x3 , etc., are continuous. By a similar argument, xn is continuous for any n > 0.
83. If c is in the interval, we know lim f (x) = f (c) and limx→c g(x) = g(c). Then,
x→c
lim (f (x) + g(x)) = lim f (x) + lim g(x) by limit property 2
x→c
x→c
x→c
= f (c) + g(c),
so f + g is continuous at x = c.
Also,
lim (f (x)g(x)) = lim f (x) lim g(x) by limit property 3
x→c
x→c
x→c
= f (c)g(c) so f g is continuous at x = c.
Finally,
lim
x→c
limx→c f (x)
f (x)
=
by limit property 4
g(x)
limx→c g(x)
f (c)
f
=
, so is continuous at x = c.
g(c)
g
Strengthen Your Understanding
84. Though P (x) and Q(x) are both continuous for all x, it is possible for Q(x) to be equal to zero for some x. For any such
value of x, where Q(x) = 0, we see that P (x)/Q(x) is undefined, and thus not continuous. For example,
P (x)
x
=
Q(x)
x−1
is not defined or continuous at x = 1.
85. The left- and right-hand limits are not the same:
lim
x→1−
x−1
= −1,
|x − 1|
but
x−1
= 1.
|x − 1|
Since the left- and right-hand limits are not the same, the limit does not exist and thus is not equal to 1.
lim
x→1+
86. For f to be continuous at x = c, we need limx→c f (x) to exist and to be equal to f (c).
87. One possibility is
f (x) =
We have limx→1 f (x) = 4 but f (1) is undefined.
(x + 3)(x − 1)
.
x−1
78
Chapter One /SOLUTIONS
88. One example is
f (x) =
2|x|
.
x
89. True, by Property 3 of limits in Theorem 1.2, since limx→3 x = 3.
90. False. If limx→3 g(x) does not exist, then limx→3 f (x)g(x) may not even exist. For example, let f (x) = 2x + 1 and
define g by:
n
g(x) = 1/(x − 3) if x 6= 3
4
if x = 3
Then limx→3 f (x) = 7 and g(3) = 4, but limx→3 f (x)g(x) 6= 28, since limx→3 (2x + 1)/(x − 3) does not exist.
91. True, by Property 2 of limits in Theorem 1.2.
92. True, by Properties 2 and 3 of limits in Theorem 1.2.
lim g(x) = lim (f (x) + g(x) + (−1)f (x)) = lim (f (x) + g(x)) + (−1) lim f (x) = 12 + (−1)7 = 5.
x→3
x→3
x→3
x→3
93. False. For example, define f as follows:
f (x) =
n
2x + 1 if x 6= 2.99
1000
if x = 2.99.
n
2x + 1 if x 6= 3.01
−1000 if x = 3.01.
Then f (2.9) = 2(2.9) + 1 = 6.8, whereas f (2.99) = 1000.
94. False. For example, define f as follows:
f (x) =
Then f (3.1) = 2(3.1) + 1 = 7.2, whereas f (3.01) = −1000.
95. True. Suppose instead that limx→3 g(x) does not exist but limx→3 (f (x)g(x)) did exist. Since limx→3 f (x) exists and is
not zero, then limx→3 ((f (x)g(x))/f (x)) exists, by Property 4 of limits in Theorem 1.2. Furthermore, f (x) 6= 0 for all x
in some interval about 3, so (f (x)g(x))/f (x) = g(x) for all x in that interval. Thus limx→3 g(x) exists. This contradicts
our assumption that limx→3 g(x) does not exist.
96. False. For some functions we need to pick smaller values of δ. For example, if f (x) = x1/3 + 2 and c = 0 and L = 2,
then f (x) is within 10−3 of 2 if |x1/3 | < 10−3 . This only happens if x is within (10−3 )3 = 10−9 of 0. If x = 10−3 then
x1/3 = (10−3 )1/3 = 10−1 , which is too large.
97. False. The definition of a limit guarantees that, for any positive ǫ, there is a δ. This statement, which guarantees an ǫ for
a specific δ = 10−3 , is not equivalent to limx→c f (x) = L. For example, consider a function with a vertical asymptote
within 10−3 of 0, such as c = 0, L = 0, f (x) = x/(x − 10−4 ).
98. True. This is equivalent to the definition of a limit.
99. False. Although x may be far from c, the value of f (x) could be close to L. For example, suppose f (x) = L, the constant
function.
100. False. The definition of the limit says that if x is within δ of c, then f (x) is within ǫ of L, not the other way round.
101. (a) This statement follows: if we interchange the roles of f and g in the original statement, we get this statement.
(b) This statement is true, but it does not follow directly from the original statement, which says nothing about the case
g(a) = 0.
(c) This follows, since if g(a) 6= 0 the original statement would imply f /g is continuous at x = a, but we are told it is
not.
(d) This does not follow. Given that f is continuous at x = a and g(a) 6= 0, then the original statement says g continuous
implies f /g continuous, not the other way around. In fact, statement (d) is not true: if f (x) = 0 for all x, then g could
be any discontinuous, non-zero function, and f /g would be zero, and therefore continuous. Thus the conditions of
the statement would be satisfied, but not the conclusion.
SOLUTIONS to Review Problems for Chapter One
79
Solutions for Chapter 1 Review
Exercises
1. The line of slope m through the point (x0 , y0 ) has equation
y − y0 = m(x − x0 ),
so the line we want is
y − 0 = 2(x − 5)
y = 2x − 10.
2. We want a function of the form y = a(x − h)2 + k, with a < 0 because the parabola opens downward. Since (h, k) is the
vertex, we must take h = 2, k = 5, but we can take any negative value of a. Figure 1.110 shows the graph with a = −1,
namely y = −(x − 2)2 + 5.
y
(2, 5)
x
Figure 1.110: Graph of y = −(x − 2)2 + 5
3. A parabola with x-intercepts ±1 has an equation of the form
y = k(x − 1)(x + 1).
Substituting the point x = 0, y = 3 gives
3 = k(−1)(1)
so
k = −3.
Thus, the equation we want is
y = −3(x − 1)(x + 1)
y = −3x2 + 3.
4. The equation of the whole circle is
√
x2 + y 2 = ( 2)2 ,
so the bottom half is
y=−
p
2 − x2 .
5. A circle with center (h, k) and radius r has equation (x − h)2 + (y − k)2 = r 2 . Thus h = −1, k = 2, and r = 3, giving
(x + 1)2 + (y − 2)2 = 9.
Solving for y, and taking the positive square root gives the top half, so
(y − 2)2 = 9 − (x + 1)2
y = 2+
See Figure 1.111.
p
9 − (x + 1)2 .
80
Chapter One /SOLUTIONS
y
✛
✲
3
(−1, 2)
x
Figure 1.111: Graph of y = 2 +
p
9 − (x + 1)2
6. A cubic polynomial of the form y = a(x−1)(x−5)(x−7) has the correct intercepts for any value of a 6= 0. Figure 1.112
shows the graph with a = 1, namely y = (x − 1)(x − 5)(x − 7).
y
1
5
7
x
Figure 1.112: Graph of y = (x − 1)(x − 5)(x − 7)
7. Since the vertical asymptote is x = 2, we have b = −2. The fact that the horizontal asymptote is y = −5 gives a = −5.
So
−5x
.
x−2
8. The amplitude of this function is 5, and its period is 2π, so y = 5 cos x.
y=
9. See Figure 1.113.
heart
rate
time
administration
of drug
Figure 1.113
10. Factoring gives
(2 − x)(2 + x)
.
x(x + 1)
The values of x which make g(x) undefined are x = 0 and x = −1, when the denominator is 0. So the domain is all
x 6= 0, −1. Solving g(x) = 0 means one of the numerator’s factors is 0, so x = ±2.
g(x) =
11. (a) The domain of f is the set of values of x for which the function is defined. Since the function is defined by the graph
and the graph goes from x = 0 to x = 7, the domain of f is [0, 7].
(b) The range of f is the set of values of y attainable over the domain. Looking at the graph, we can see that y gets as
high as 5 and as low as −2, so the range is [−2, 5].
(c) Only at x = 5 does f (x) = 0. So 5 is the only zero of f (x).
(d) Looking at the graph, we can see that f (x) is decreasing on (1, 7).
(e) The graph indicates that f (x) is concave up at x = 6.
(f) The value f (4) is the y-value that corresponds to x = 4. From the graph, we can see that f (4) is approximately 1.
(g) This function is not invertible, since it fails the horizontal-line test. A horizontal line at y = 3 would cut the graph of
f (x) in two places, instead of the required one.
SOLUTIONS to Review Problems for Chapter One
12. (a)
(b)
(c)
(d)
(e)
f (n) + g(n) = (3n2 − 2) + (n + 1) = 3n2 + n − 1.
f (n)g(n) = (3n2 − 2)(n + 1) = 3n3 + 3n2 − 2n − 2.
The domain of f (n)/g(n) is defined everywhere where g(n) 6= 0, i.e. for all n 6= −1.
f (g(n)) = 3(n + 1)2 − 2 = 3n2 + 6n + 1.
g(f (n)) = (3n2 − 2) + 1 = 3n2 − 1.
81
13. (a) Since m = f (A), we see that f (100) represents the value of m when A = 100. Thus f (100) is the minimum annual
gross income needed (in thousands) to take out a 30-year mortgage loan of $100,000 at an interest rate of 6%.
(b) Since m = f (A), we have A = f −1 (m). We see that f −1 (75) represents the value of A when m = 75, or the size
of a mortgage loan that could be obtained on an income of $75,000.
log 7
= 1.209.
14. Taking logs of both sides yields t log 5 = log 7, so t =
log 5
log 2
≈ 35.003.
15. t =
log 1.02
3 t
2
16. Collecting similar factors yields
= 75 , so
t=
1.04 t
1.03
17. Collecting similar factors yields
=
12.01
.
5.02
5
7
3
2
log
log
= −0.830.
Solving for t yields
t=
log
log
= 90.283.
12.01
5.02
1.04
1.03
18. We want 2t = ekt so 2 = ek and k = ln 2 = 0.693. Thus P = P0 e0.693t .
19. We want 0.2t = ekt so 0.2 = ek and k = ln 0.2 = −1.6094. Thus P = 5.23e−1.6094t .
20. f (x) = ln x,
3
21. f (x) = x ,
g(x) = x3 . (Another possibility: f (x) = 3x,
g(x) = ln x.)
g(x) = ln x.
22. The amplitude is 5. The period is 6π. See Figure 1.114.
y
5
x
−6π − 9π −3π− 3π
2
3π
2
2
3π
6π
y = 5 sin
9π
2
−5
1
3x
Figure 1.114
23. The amplitude is 2. The period is 2π/5. See Figure 1.115.
y
6
4
2
y = 4 − 2 cos(5x)
x
− 4π
5
− 2π
5
2π
5
Figure 1.115
4π
5
82
Chapter One /SOLUTIONS
24. (a) We determine the amplitude of y by looking at the coefficient of the cosine term. Here, the coefficient is 1, so the
amplitude of y is 1. Note that the constant term does not affect the amplitude.
(b) We know that the cosine function cos x repeats itself at x = 2π, so the function cos(3x) must repeat itself when
3x = 2π, or at x = 2π/3. So the period of y is 2π/3. Here as well the constant term has no effect.
(c) The graph of y is shown in the figure below.
y
6
5
4
x
2π
3
25. (a) Since f (x) is an odd polynomial with a positive leading coefficient, it follows that f (x) → ∞ as x → ∞ and
f (x) → −∞ as x → −∞.
(b) Since f (x) is an even polynomial with negative leading coefficient, it follows that f (x) → −∞ as x → ±∞.
(c) As x → ±∞, x4 → ∞, so x−4 = 1/x4 → 0.
(d) As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest
degree terms in its numerator and denominator. So as x → ±∞, f (x) → 6.
26. Exponential growth dominates power growth as x → ∞, so 10 · 2x is larger.
27. As x → ∞, 0.25x1/2 is larger than 25,000x−3 .
28. This is a line with slope −3/7 and y-intercept 3, so a possible formula is
3
y = − x + 3.
7
29. Starting with the general exponential equation y = Aekx , we first find that for (0, 1) to be on the graph, we must have
A = 1. Then to make (3, 4) lie on the graph, we require
4 = e3k
ln 4 = 3k
ln 4
≈ 0.4621.
k=
3
Thus the equation is
y = e0.4621x .
Alternatively, we can use the form y = ax , in which case we find y = (1.5874)x .
30. This looks like an exponential function. The y-intercept is 3 and we use the form y = 3ekt . We substitute the point (5, 9)
to solve for k:
9 = 3ek5
3 = e5k
ln 3 = 5k
k = 0.2197.
A possible formula is
y = 3e0.2197t .
Alternatively, we can use the form y = 3at , in which case we find y = 3(1.2457)t .
SOLUTIONS to Review Problems for Chapter One
83
31. y = −kx(x + 5) = −k(x2 + 5x), where k > 0 is any constant.
32. Since this function has a y-intercept at (0, 2), we expect it to have the form y = 2ekx . Again, we find k by forcing the
other point to lie on the graph:
1 = 2e2k
1
= e2k
2
ln
1
2
= 2k
k=
ln( 21 )
≈ −0.34657.
2
This value is negative, which makes sense since the graph shows exponential decay. The final equation, then, is
y = 2e−0.34657x .
Alternatively, we can use the form y = 2ax , in which case we find y = 2(0.707)x .
33. z = 1 − cos θ
34. y = k(x + 2)(x + 1)(x − 1) = k(x3 + 2x2 − x − 2), where k > 0 is any constant.
35. x = ky(y − 4) = k(y 2 − 4y), where k > 0 is any constant.
36. y = 5 sin
πt
20
37. This looks like a fourth degree polynomial with roots at −5 and −1 and a double root at 3. The leading coefficient is
negative, and so a possible formula is
y = −(x + 5)(x + 1)(x − 3)2 .
38. This looks like a rational function. There are vertical asymptotes at x = −2 and x = 2 and so one possibility for the
denominator is x2 − 4. There is a horizontal asymptote at y = 3 and so the numerator might be 3x2 . In addition, y(0) = 0
which is the case with the numerator of 3x2 . A possible formula is
y=
3x2
.
−4
x2
39. There are many solutions for a graph like this one. The simplest is y = 1 − e−x , which gives the graph of y = ex , flipped
over the x-axis and moved up by 1. The resulting graph passes through the origin and approaches y = 1 as an upper
bound, the two features of the given graph.
40. The graph is a sine curve which has been shifted up by 2, so f (x) = (sin x) + 2.
41. This graph has period 5, amplitude 1 and no vertical shift or horizontal shift from sin x, so it is given by
f (x) = sin
2π
x .
5
42. Since the denominator, x2 + 1, is continuous and never zero, g(x) is continuous on [−1, 1].
43. Since
h(x) =
1
1
=
,
1 − x2
(1 − x)(1 + x)
we see that h(x) is not defined at x = −1 or at x = 1, so h(x) is not continuous on [−1, 1].
44. (a) lim f (x) = 1.
x→0
(b) lim f (x) does not exist.
x→1
(c) lim f (x) = 1.
x→2
(d)
lim f (x) = 0.
x→3−
84
Chapter One /SOLUTIONS
3
x (2x − 6) = 2x3 ,
x>3
x3 |2x − 6|
x−3
=
3
x−3
x (−2x + 6) = −2x3 , x < 3
x−3
Figure 1.116 confirms that lim f (x) = 54 while lim f (x) = −54; thus lim f (x) does not exist.
45. f (x) =
x→3+
x→3
x→3−
100
f (x)
50
1
0
2
x
3
4
−50
−100
Figure 1.116
46. f (x) =
x
e
1
−1 < x < 0
x=0
cos x
0 a so, from the graph, F is negative, indicating an attractive force,
which pulls the atoms back together.
(ii) If the atoms are moved closer together, then r < a so, from the graph, F is positive, indicating an attractive
force, which pushes the atoms apart again.
(b) At r = a, the force is zero. The answer to part (a)(i) tells us that if the atoms are pulled apart slightly, so r > a,
the force tends to pull them back together; the answer to part (a)(ii) tells us that if the atoms are pushed together, so
r < a, the force tends to push them back apart. Thus, r = a is a stable equilibrium.
51. (a)
52. If the pressure at sea level is P0 , the pressure P at altitude h is given by
P = P0 1 −
0.4
100
h/100
,
since we want the pressure to be multiplied by a factor of (1 − 0.4/100) = 0.996 for each 100 feet we go up to make it
decrease by 0.4% over that interval. At Mexico City h = 7340, so the pressure is
P = P0 (0.996)7340/100 ≈ 0.745P0 .
So the pressure is reduced from P0 to approximately 0.745P0 , a decrease of 25.5%.
53. Assuming the population of Ukraine is declining exponentially, we have population P (t) = 45.7ekt at time t years after
2009. Using the 2010 population, we have
45.42 = 45.7e−k·1
45.42
= 0.0061.
k = − ln
45.7
SOLUTIONS to Review Problems for Chapter One
87
We want to find the time t at which
45 = 45.7e−0.0061t
ln(45/45.7)
= 2.53 years.
t=−
0.0061
This model predicts the population to go below 45 million 2.53 years after 2009, in the year 2011.
54. (a) We compound the daily inflation rate 30 times to get the desired monthly rate r:
1+
r
100
1
= 1+
0.67
100
30
= 1.2218.
Solving for r, we get r = 22.18, so the inflation rate for April was 22.18%.
(b) We compound the daily inflation rate 365 times to get a yearly rate R for 2006:
1+
R
100
1
= 1+
0.67
100
365
= 11.4426.
Solving for R, we get R = 1044.26, so the yearly rate was 1044.26% during 2006. We could have obtained approximately the same result by compounding the monthly rate 12 times. Computing the annual rate from the monthly
gives a lower result, because 12 months of 30 days each is only 360 days.
55. (a) The US consumption of hydroelectric power increased by at least 10% in 2009 and decreased by at least 10% in 2006
and in 2007, relative to each corresponding previous year. In 2009 consumption increased by 11% over consumption
in 2008. In 2006 consumption decreased by 10% over consumption in 2005, and in 2007 consumption decreased by
about 45% over consumption in 2006.
(b) False. In 2009 hydroelectric power consumption increased only by 11% over consumption in 2008.
(c) True. From 2006 to 2007 consumption decreased by 45.4%, which means x(1 − 0.454) units of hydroelectric power
were consumed in 2007 if x had been consumed in 2006. Similarly,
(x(1 − 0.454))(1 + 0.051)
units of hydroelectric power were consumed in 2008 if x had been consumed in 2006, and
(x(1 − 0.454)(1 + 0.051))(1 + 0.11)
units of hydroelectric power were consumed in 2009 if x had been consumed in 2006. Since
x(1 − 0.454)(1 + 0.051)(1 + 0.11) = x(0.637) = x(1 − 0.363),
the percent growth in hydroelectric power consumption was −36.3%, in 2009 relative to consumption in 2006. This
amounts to about 36% decrease in hydroelectric power consumption from 2006 to 2009.
56. (a) For each 2.2 pounds of weight the object has, it has 1 kilogram of mass, so the conversion formula is
k = f (p) =
1
p.
2.2
(b) The inverse function is
p = 2.2k,
and it gives the weight of an object in pounds as a function of its mass in kilograms.
57. Since f (x) is a parabola that opens upward, we have f (x) = ax2 + bx + c with a > 0. Since g(x) is a line with negative
slope, g(x) = b + mx, with slope m < 0. Therefore
g(f (x)) = b + m(ax2 + bx + c) = max2 + mbx + mc + b.
The coefficient of x2 is ma, which is negative. Thus, the graph is a parabola opening downward.
58. (a) is g(x) since it is linear. (b) is f (x) since it has decreasing slope; the slope starts out about 1 and then decreases to
1
1
. (c) is h(x) since it has increasing slope; the slope starts out about 10
and then increases to about 1.
about 10
59. Given the doubling time of 2 hours, 200 = 100ek(2) , we can solve for the growth rate k using the equation:
2P0 = P0 e2k
ln 2 = 2k
ln 2
.
k=
2
88
Chapter One /SOLUTIONS
Using the growth rate, we wish to solve for the time t in the formula
P = 100e
ln 2 t
2
where P = 3,200, so
3,200 = 100e
ln 2 t
2
t = 10 hours.
60. (a) The y-intercept of f (x) = a ln(x + 2) is f (0) = a ln 2. Since ln 2 is positive, increasing a increases the y-intercept.
(b) The x-intercept of f (x) = a ln(x + 2) is where f (x) = 0. Since this occurs where x + 2 = 1, so x = −1, increasing
a does not affect the x-intercept.
61. Since the factor by which the prices have increased after time t is given by (1.05)t , the time after which the prices have
doubled solves
2 = (1.05)t
log 2 = log(1.05t ) = t log(1.05)
log 2
≈ 14.21 years.
t=
log 1.05
62. Using the exponential decay equation P = P0 e−kt , we can solve for the substance’s decay constant k:
(P0 − 0.3P0 ) = P0 e−20k
ln(0.7)
.
k=
−20
Knowing this decay constant, we can solve for the half-life t using the formula
0.5P0 = P0 eln(0.7)t/20
20 ln(0.5)
≈ 38.87 hours.
t=
ln(0.7)
63. (a) We know the decay follows the equation
P = P0 e−kt ,
and that 10% of the pollution is removed after 5 hours (meaning that 90% is left). Therefore,
0.90P0 = P0 e−5k
1
k = − ln(0.90).
5
Thus, after 10 hours:
P = P0 e−10((−0.2) ln 0.90)
P = P0 (0.9)2 = 0.81P0
so 81% of the original amount is left.
(b) We want to solve for the time when P = 0.50P0 :
0.50P0 = P0 et((0.2) ln 0.90)
0.2t )
0.50 = eln(0.90
0.50 = 0.900.2t
5 ln(0.50)
≈ 32.9 hours.
t=
ln(0.90)
SOLUTIONS to Review Problems for Chapter One
(c)
89
P
P0
5 10
t
32.9
(d) When highly polluted air is filtered, there is more pollutant per liter of air to remove. If a fixed amount of air is
cleaned every day, there is a higher amount of pollutant removed earlier in the process.
64. Since the amount of strontium-90 remaining halves every 29 years, we can solve for the decay constant;
0.5P0 = P0 e−29k
ln(1/2)
.
k=
−29
Knowing this, we can look for the time t in which P = 0.10P0 , or
0.10P0 = P0 eln(0.5)t/29
29 ln(0.10)
= 96.336 years.
t=
ln(0.5)
65. One hour.
66. (a) V0 represents the maximum voltage.
(b) The period is 2π/(120π) = 1/60 second.
(c) Since each oscillation takes 1/60 second, in 1 second there are 60 complete oscillations.
67. The US voltage has a maximum value of 156 volts and has a period of 1/60 of a second, so it executes 60 cycles a second.
The European voltage has a higher maximum of 339 volts, and a slightly longer period of 1/50 seconds, so it
oscillates at 50 cycles per second.
68. (a) The amplitude of the sine curve is |A|. Thus, increasing |A| stretches the curve vertically. See Figure 1.121.
(b) The period of the wave is 2π/|B|. Thus, increasing |B| makes the curve oscillate more rapidly—in other words, the
function executes one complete oscillation in a smaller interval. See Figure 1.122.
A=3
A=2
A=1
✲
✲
✲
B=1
B=2
y
❘
3
1
❄
B=3
✠
2
1
−2π
2π
2π
x
x
−2π
−1
−3
Figure 1.121
69. (a)
y
Figure 1.122
(i) The water that has flowed out of the pipe in 1 second is a cylinder of radius r and length 3 cm. Its volume is
V = πr 2 (3) = 3πr 2 .
(ii) If the rate of flow is k cm/sec instead of 3 cm/sec, the volume is given by
V = πr 2 (k) = πr 2 k.
90
Chapter One /SOLUTIONS
(i) The graph of V as a function of r is a quadratic. See Figure 1.123.
(b)
V
V
r
k
Figure 1.123
Figure 1.124
(ii) The graph of V as a function of k is a line. See Figure 1.124.
70. Looking at g, we see that the ratio of the values is:
3.74
4.49
5.39
6.47
3.12
≈
≈
≈
≈
≈ 0.83.
3.74
4.49
5.39
6.47
7.76
Thus g is an exponential function, and so f and k are the power functions. Each is of the form ax2 or ax3 , and since
k(1.0) = 9.01 we see that for k, the constant coefficient is 9.01. Trial and error gives
k(x) = 9.01x2 ,
since k(2.2) = 43.61 ≈ 9.01(4.84) = 9.01(2.2)2 . Thus f (x) = ax3 and we find a by noting that f (9) = 7.29 = a(93 )
so
7.29
a = 3 = 0.01
9
and f (x) = 0.01x3 .
71. (a) See Figure 1.125.
(b) The graph is made of straight line segments, rising from the x-axis at the origin to height a at x = 1, b at x = 2, and
c at x = 3 and then returning to the x-axis at x = 4. See Figure 1.126.
y
y
a
4
3
2
c
1
b
x
−1
0
1
2
3
4
5
x
−1
0
1
Figure 1.125
2
3
4
5
Figure 1.126
72. (a) Reading the graph of θ against t shows that θ ≈ 5.2 when t = 1.5. Since the coordinates of P are x = 5 cos θ,
y = 5 sin θ, when t = 1.5 the coordinates are
(x, y) ≈ (5 cos 5.2, 5 sin 5.2) = (2.3, −4.4).
(b) As t increases from 0 to 5, the angle θ increases from 0 to about 6.3 and then decreases to 0 again. Since 6.3 ≈ 2π,
this means that P starts on the x-axis at the point (5, 0), moves counterclockwise the whole way around the circle (at
which time θ ≈ 2π), and then moves back clockwise to its starting point.
73. (a)
(b)
(c)
(d)
III
IV
I
II
SOLUTIONS to Review Problems for Chapter One
91
74. The functions y(x) = sin x and zk (x) = ke−x for k = 1, 2, 4, 6, 8, 10 are shown in Figure 1.127. The values of f (k) for
k = 1, 2, 4, 6, 8, 10 are given in Table 1.16. These values can be obtained using either tracing or a numerical root finder
on a calculator or computer.
From Figure 1.127 it is clear that the smallest solution of sin x = ke−x for k = 1, 2, 4, 6 occurs on the first period
of the sine curve. For small changes in k, there are correspondingly small changes in the intersection point. For k = 8 and
k = 10, the solution jumps to the second period because sin x < 0 between π and 2π, but ke−x is uniformly positive.
Somewhere in the interval 6 ≤ k ≤ 8, f (k) has a discontinuity.
2
✙
zk (x) = ke−x
✰
1
0
6
2
4
Table 1.16
y(x) = sin x
10
12
x
8
−1
Figure 1.127
k
f (k)
1
0.588
2
0.921
4
1.401
6
1.824
8
6.298
10
6.302
75. By tracing on a calculator or solving equations, we find the following values of δ:
For ǫ = 0.1, δ ≤ 0.1
For ǫ = 0.05, δ ≤ 0.05.
For ǫ = 0.0007, δ ≤ 0.00007.
76. By tracing on a calculator or solving equations, we find the following values of δ:
For ǫ = 0.1, δ ≤ 0.45.
For ǫ = 0.001, δ ≤ 0.0447.
For ǫ = 0.00001, δ ≤ 0.00447.
77. For any values of k, the function is continuous on any interval that does not contain x = 2.
Since 5x3 − 10x2 = 5x2 (x − 2), we can cancel (x − 2) provided x 6= 2, giving
f (x) =
5x3 − 10x2
= 5x2
x−2
x 6= 2.
Thus, if we pick k = 5(2)2 = 20, the function is continuous.
78. At x = 0, the curve y = k cos x has y = k cos 0 = k. At x = 0, the curve y = ex − k has y = e0 − k = 1 − k. If j(x)
is continuous, we need
1
k = 1 − k, so k = .
2
CAS Challenge Problems
79. (a) A CAS gives f (x) = (x − a)(x + a)(x + b)(x − c).
(b) The graph of f (x) crosses the x-axis at x = a, x = −a, x = −b, x = c; it crosses the y-axis at a2 bc. Since the
coefficient of x4 (namely 1) is positive, the graph of f looks like that shown in Figure 1.128.
y
a2 bc
−b
−a a
c
x
Figure 1.128: Graph of
f (x) =
(x−a)(x+a)(x+b)(x−c)
92
Chapter One /SOLUTIONS
80. (a) A CAS gives f (x) = −(x − 1)2 (x − 3)3 .
(b) For large |x|, the graph of f (x) looks like the graph of y = −x5 , so f (x) → ∞ as x → −∞ and f (x) → −∞ as
x → ∞. The answer to part (a) shows that f has a double root at x = 1, so near x = 1, the graph of f looks like
a parabola touching the x-axis at x = 1. Similarly, f has a triple root at x = 3. Near x = 3, the graph of f looks
like the graph of y = x3 , flipped over the x-axis and shifted to the right by 3, so that the “seat” is at x = 3. See
Figure 1.129.
x
1
3
Figure 1.129: Graph of
f (x) = −(x − 1)2 (x − 3)3
81. (a) As x → ∞, the term e6x dominates and tends to ∞. Thus, f (x) → ∞ as x → ∞.
As x → −∞, the terms of the form ekx , where k = 6, 5, 4, 3, 2, 1, all tend to zero. Thus, f (x) → 16 as x → −∞.
(b) A CAS gives
f (x) = (ex + 1)(e2x − 2)(ex − 2)(e2x + 2ex + 4).
Since ex is always positive, the factors (ex + 1) and (e2x + 2ex + 4) are never zero. The other factors each lead to a
zero, so there are two zeros.
(c) The zeros are given by
e2x = 2
so
ex = 2
so
ln 2
2
x = ln 2.
x=
Thus, one zero is twice the size of the other.
82. (a) Since f (x) = x2 − x,
f (f (x)) = (f (x))2 − f (x) = (x2 − x)2 − (x2 − x) = x − 2 x3 + x4 .
Using the CAS to define the function f (x), and then asking it to expand f (f (f (x))), we get
f (f (f (x))) = −x + x2 + 2 x3 − 5 x4 + 2 x5 + 4 x6 − 4 x7 + x8 .
(b) The degree of f (f (x)) (that is, f composed with itself 2 times) is 4 = 22 . The degree of f (f (f (x))) (that is, f
composed with itself 3 times), is 8 = 23 . Each time you substitute f into itself, the degree is multiplied by 2, because
you are substituting in a degree 2 polynomial. So we expect the degree of f (f (f (f (f (f (x)))))) (that is, f composed
with itself 6 times) to be 64 = 26 .
83. (a) A CAS or division gives
3
x3 − 30
= x2 + 3x + 9 −
,
x−3
x−3
so p(x) = x2 + 3x + 9, and r(x) = −3, and q(x) = x − 3.
(b) The vertical asymptote is x = 3. Near x = 3, the values of p(x) are much smaller than the values of r(x)/q(x).
Thus
−3
f (x) ≈
for x near 3.
x−3
(c) For large x, the values of p(x) are much larger than the value of r(x)/q(x). Thus
f (x) =
f (x) ≈ x2 + 3x + 9
as x → ∞, x → −∞.
(d) Figure 1.130 shows f (x) and y = −3/(x − 3) for x near 3. Figure 1.131 shows f (x) and y = x2 + 3x + 9 for
−20 ≤ x ≤ 20. Note that in each case the graphs of f and the approximating function are close.
PROJECTS FOR CHAPTER ONE
y
f (x)
50
y=
93
y
f (x)
500
−3
x−3
y = x2 + 3x + 9
f (x)
❄
5
x
y=
x
−3
x−3
−20
20
−50
−500
Figure 1.130: Close-up view of f (x) and
y = −3/(x − 3)
Figure 1.131: Far-away view of f (x) and
y = x2 + 3x + 9
84. Using the trigonometric expansion capabilities of your CAS, you get something like
sin(5x) = 5 cos4 (x) sin(x) − 10 cos2 (x) sin3 (x) + sin5 (x).
Answers may vary. To get rid of the powers of cosine, use the identity cos2 (x) = 1 − sin2 (x). This gives
2
sin(5x) = 5 sin(x) 1 − sin2 (x)
Finally, using the CAS to simplify,
− 10 sin3 (x) 1 − sin2 (x) + sin5 (x).
sin(5x) = 5 sin(x) − 20 sin3 (x) + 16 sin5 (x).
85. Using the trigonometric expansion capabilities of your computer algebra system, you get something like
cos(4x) = cos4 (x) − 6 cos2 (x) sin2 (x) + sin4 (x).
Answers may vary.
(a) To get rid of the powers of cosine, use the identity cos2 (x) = 1 − sin2 (x). This gives
2
cos(4x) = cos4 (x) − 6 cos2 (x) 1 − cos2 (x) + 1 − cos2 (x) .
Finally, using the CAS to simplify,
cos(4x) = 1 − 8 cos2 (x) + 8 cos4 (x).
(b) This time we use sin2 (x) = 1 − cos2 (x) to get rid of powers of sine. We get
2
cos(4x) = 1 − sin2 (x)
− 6 sin2 (x) 1 − sin2 (x) + sin4 (x) = 1 − 8 sin2 (x) + 8 sin4 (x).
PROJECTS FOR CHAPTER ONE
1. Notice that whenever x increases by 0.5, f (x) increases by 1, indicating that f (x) is linear. By inspection, we
see that f (x) = 2x.
Similarly, g(x) decreases by 1 each time x increases by 0.5. We know, therefore, that g(x) is a linear
−1
= −2. The y-intercept is 10, so g(x) = 10 − 2x.
function with slope 0.5
h(x) is an even function which is always positive. Comparing the values of x and h(x), it appears that
h(x) = x2 .
F (x) is an odd function that seems to vary between −1 and 1. We guess that F (x) = sin x and check
with a calculator.
G(x) is also an odd function that varies between −1 and 1. Notice that G(x) = F (2x), and thus G(x) =
sin 2x.
Notice also that H(x) is exactly 2 more than F (x) for all x, so H(x) = 2 + sin x.
94
Chapter One /SOLUTIONS
2. (a) Begin by finding a table of correspondences between the mathematicians’ and meteorologists’ angles.
θmet (in degrees)
0
45
90
135
180
225
270
315
θmath (in degrees)
270
225
180
135
90
45
0
315
The table is linear for 0 ≤ θmet ≤ 270, with θmath decreasing by 45 every time θmet increases by 45,
giving slope ∆θmet /∆θmath = 45/(−45) = −1.
The interval 270 < θmet < 360 needs a closer look. We have the following more detailed table for
that interval:
θmet
280
290
300
310
320
330
340
350
θmath
350
340
330
320
310
300
290
280
Again the table is linear, this time with θmath decreasing by 10 every time θmet increases by 10, again
giving slope −1. The graph of θmath against θmet contains two straight line sections, both of slope −1.
See Figure 1.132.
(b) See Figure 1.132.
270 − θmet if 0 ≤ θmet ≤ 270
.
θmath =
630 − θmet if 270 < θmet < 360
θmath
360◦
270◦
180◦
90◦
90◦
180◦ 270◦ 360◦
Figure 1.132
θmet
2.1 SOLUTIONS
95
CHAPTER TWO
Solutions for Section 2.1
Exercises
1. For t between 2 and 5, we have
Average velocity =
∆s
400 − 135
265
=
=
km/hr.
∆t
5−2
3
The average velocity on this part of the trip was 265/3 km/hr.
2. The average velocity over a time period is the change in position divided by the change in time. Since the function x(t)
gives the position of the particle, we find the values of x(0) = −2 and x(4) = −6. Using these values, we find
Average velocity =
x(4) − x(0)
−6 − (−2)
∆x(t)
=
=
= −1 meters/sec.
∆t
4−0
4
3. The average velocity over a time period is the change in position divided by the change in time. Since the function x(t)
gives the position of the particle, we find the values of x(2) = 14 and x(8) = −4. Using these values, we find
Average velocity =
x(8) − x(2)
−4 − 14
∆x(t)
=
=
= −3 angstroms/sec.
∆t
8−2
6
4. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the distance of the particle from a point, we read off the graph that s(0) = 1 and s(3) = 4. Thus,
Average velocity =
s(3) − s(0)
∆s(t)
4−1
=
=
= 1 meter/sec.
∆t
3−0
3
5. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the distance of the particle from a point, we read off the graph that s(1) = 2 and s(3) = 6. Thus,
Average velocity =
s(3) − s(1)
∆s(t)
6−2
=
=
= 2 meters/sec.
∆t
3−1
2
6. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the distance of the particle from a point, we find the values of s(2) = e2 − 1 = 6.389 and s(4) = e4 − 1 = 53.598.
Using these values, we find
Average velocity =
s(4) − s(2)
∆s(t)
53.598 − 6.389
=
=
= 23.605 µm/sec.
∆t
4−2
2
7. The average velocity over a time period is the change in the distance divided by the change
√ s(t)
√ in time. Since the function
gives the distance of the particle from a point, we find the values of s(π/3) = 4 + 3 3/2 and s(7π/3) = 4 + 3 3/2.
Using these values, we find
√
√
∆s(t)
s(7π/3) − s(π/3)
4 + 3 3/2 − (4 + 3 3/2)
Average velocity =
=
=
= 0 cm/sec.
∆t
7π/3 − π/3
2π
Though the particle moves, its average velocity is zero, since it is at the same position at t = π/3 and t = 7π/3.
96
Chapter Two /SOLUTIONS
8. (a) Let s = f (t).
(i) We wish to find the average velocity between t = 1 and t = 1.1. We have
Average velocity =
f (1.1) − f (1)
3.63 − 3
=
= 6.3 m/sec.
1.1 − 1
0.1
(ii) We have
Average velocity =
(iii) We have
f (1.01) − f (1)
3.0603 − 3
=
= 6.03 m/sec.
1.01 − 1
0.01
f (1.001) − f (1)
3.006003 − 3
=
= 6.003 m/sec.
1.001 − 1
0.001
(b) We see in part (a) that as we choose a smaller and smaller interval around t = 1 the average velocity appears to be
getting closer and closer to 6, so we estimate the instantaneous velocity at t = 1 to be 6 m/sec.
Average velocity =
9. (a) Let s = f (t).
(i) We wish to find the average velocity between t = 0 and t = 0.1. We have
Average velocity =
0.004 − 0
f (0.1) − f (0)
=
= 0.04 m/sec.
0.1 − 0
0.1
(ii) We have
Average velocity =
(iii) We have
0.000004
f (0.01) − f (0)
=
= 0.0004 m/sec.
0.01 − 0
0.01
f (0.001) − f (0)
4 × 10−9 − 0
=
= 4 × 10−6 m/sec.
1.001 − 1
0.001
(b) We see in part (a) that as we choose a smaller and smaller interval around t = 0 the average velocity appears to be
getting closer and closer to 0, so we estimate the instantaneous velocity at t = 0 to be 0 m/sec.
Looking at a graph of s = f (t) we see that a line tangent to the graph at t = 0 is horizontal, confirming our
result.
Average velocity =
10. (a) Let s = f (t).
(i) We wish to find the average velocity between t = 1 and t = 1.1. We have
Average velocity =
0.808496 − 0.909297
f (1.1) − f (1)
=
= −1.00801 m/sec.
1.1 − 1
0.1
Average velocity =
0.900793 − 0.909297
f (1.01) − f (1)
=
= −0.8504 m/sec.
1.01 − 1
0.01
(ii) We have
(iii) We have
f (1.001) − f (1)
0.908463 − 0.909297
=
= −0.834 m/sec.
1.001 − 1
0.001
(b) We see in part (a) that as we choose a smaller and smaller interval around t = 1 the average velocity appears to be
getting closer and closer to −0.83, so we estimate the instantaneous velocity at t = 1 to be −0.83 m/sec. In this case,
more estimates with smaller values of h would be very helpful in making a better estimate.
Average velocity =
11. See Figure 2.1.
distance
time
Figure 2.1
2.1 SOLUTIONS
12. See Figure 2.2.
distance
time
Figure 2.2
13. See Figure 2.3.
distance
time
Figure 2.3
Problems
14. Using h = 0.1, 0.01, 0.001, we see
(3 + 0.1)3 − 27
= 27.91
0.1
3
(3 + 0.01) − 27
= 27.09
0.01
3
(3 + 0.001) − 27
= 27.009.
0.001
These calculations suggest that lim
h→0
(3 + h)3 − 27
= 27.
h
15. Using radians,
h
0.01
0.001
0.0001
These values suggest that lim
h→0
(cos h − 1)/h
−0.005
−0.0005
−0.00005
cos h − 1
= 0.
h
16. Using h = 0.1, 0.01, 0.001, we see
70.1 − 1
= 2.148
0.1
70.01 − 1
= 1.965
0.01
70.001 − 1
= 1.948
0.001
70.0001 − 1
= 1.946.
0.0001
This suggests that lim
h→0
7h − 1
≈ 1.9.
h
97
98
Chapter Two /SOLUTIONS
17. Using h = 0.1, 0.01, 0.001, we see
(e1+h − e)/h
h
0.01
These values suggest that lim
h→0
2.7319
0.001
2.7196
0.0001
2.7184
e1+h − e
= 2.7. In fact, this limit is e.
h
18.
Slope
Point
−3
−1
F
C
0
1/2
1
2
E
A
B
D
19. The slope is positive at A and D; negative at C and F . The slope is most positive at A; most negative at F .
20. 0 < slope at C < slope at B < slope of AB < 1 < slope at A. (Note that the line y = x, has slope 1.)
21. Since f (t) is concave down between t = 1 and t = 3, the average velocity between the two times should be less than the
instantaneous velocity at t = 1 but greater than the instantaneous velocity at time t = 3, so D < A < C. For analogous
reasons, F < B < E. Finally, note that f is decreasing at t = 5 so E < 0, but increasing at t = 0, so D > 0. Therefore,
the ordering from smallest to greatest of the given quantities is
F < B < E < 0 < D < A < C.
22.
Average velocity
0 < t < 0.2
Average velocity
0.2 < t < 0.4
!
=
s(0.2) − s(0)
0.5
=
= 2.5 ft/sec.
0.2 − 0
0.2
!
=
s(0.4) − s(0.2)
1.3
=
= 6.5 ft/sec.
0.4 − 0.2
0.2
A reasonable estimate of the velocity at t = 0.2 is the average: 12 (6.5 + 2.5) = 4.5 ft/sec.
23. One possibility is shown in Figure 2.4.
f (t)
t
Figure 2.4
24. (a) When t = 0, the ball is on the bridge and its height is f (0) = 36, so the bridge is 36 feet above the ground.
(b) After 1 second, the ball’s height is f (1) = −16 + 50 + 36 = 70 feet, so it traveled 70 − 36 = 34 feet in 1 second,
and its average velocity was 34 ft/sec.
(c) At t = 1.001, the ball’s height is f (1.001) = 70.017984 feet, and its velocity about 70.017984−70
= 17.984 ≈ 18
1.001−1
ft/sec.
2.1 SOLUTIONS
99
(d) We complete the square:
f (t) = −16t2 + 50t + 36
25
t + 36
= −16 t2 −
8
625
625
25
2
t+
+ 36 + 16
= −16 t −
8
256
256
2
1201
= −16(t − 25
)
+
16
16
so the graph of f is a downward parabola with vertex at the point (25/16, 1201/16) = (1.6, 75.1). We see from
Figure 2.5 that the ball reaches a maximum height of about 75 feet. The velocity of the ball is zero when it is at the
peak, since the tangent is horizontal there.
= 1.6.
(e) The ball reaches its maximum height when t = 25
16
y
(1.6, 75.1)
80
60
40
20
1
2
3
t
Figure 2.5
4 + 4h + h2 − 4
(2 + h)2 − 4
= lim
= lim (4 + h) = 4
h
h→0
h
h→0
h→0
(1 + h)3 − 1
h(3 + 3h + h2 )
1 + 3h + 3h2 + h3 − 1
26. lim
= lim
= lim
= lim 3 + 3h + h2 = 3.
h→0
h→0
h→0
h→0
h
h
h
3(2 + h)2 − 12
h(12 + 3h)
12 + 12h + 3h2 − 12
27. lim
= lim
= lim
= lim 12 + 3h = 12.
h→0
h→0
h→0
h→0
h
h
h
2
2
2
2
(3 + h) − (3 − h)
9 + 6h + h − 9 + 6h − h
12h
= lim
= lim
= lim 6 = 6.
28. lim
h→0
h→0 2h
h→0
h→0
2h
2h
25. lim
Strengthen Your Understanding
29. Speed is the magnitude of velocity, so it is always positive or zero; velocity has both magnitude and direction.
30. We expand and simplify first
lim
h→0
(4 + 4h + h2 ) − 4
(2 + h)2 − 22
4h + h2
= lim
= lim
= lim (4 + h) = 4.
h→0
h→0
h→0
h
h
h
31. Since the tangent line to the curve at t = 4 is almost horizontal, the instantaneous velocity is almost zero. At t = 2 the
slope of the tangent line, and hence the instantaneous velocity, is relatively large and positive.
32. f (t) = t2 . The slope of the graph of y = f (t) is negative for t < 0 and positive for t > 0.
Many other answers are possible.
33. One possibility is the position function s(t) = t2 . Any function that is symmetric about the line t = 0 works.
For s(t) = t2 , the slope of a tangent line (representing the velocity) is negative at t = −1 and positive at t = 1, and
that the magnitude of the slopes (the speeds) are the same.
34. False. For example, the car could slow down or even stop at one minute after 2 pm, and then speed back up to 60 mph at
one minute before 3 pm. In this case the car would travel only a few miles during the hour, much less than 50 miles.
35. False. Its average velocity for the time between 2 pm and 4 pm is 40 mph, but the car could change its speed a lot during
that time period. For example, the car might be motionless for an hour then go 80 mph for the second hour. In that case
the velocity at 2 pm would be 0 mph.
100
Chapter Two /SOLUTIONS
36. True. During a short enough time interval the car can not change its velocity very much, and so it velocity will be nearly
constant. It will be nearly equal to the average velocity over the interval.
37. True. The instantaneous velocity is a limit of the average velocities. The limit of a constant equals that constant.
38. True. By definition, Average velocity = Distance traveled/Time.
39. False. Instantaneous velocity equals a limit of difference quotients.
Solutions for Section 2.2
Exercises
1. The derivative, f ′ (2), is the rate of change of x3 at x = 2. Notice that each time x changes by 0.001 in the table, the value
of x3 changes by 0.012. Therefore, we estimate
Rate of change
0.012
≈
= 12.
of f at x = 2
0.001
f ′ (2) =
The function values in the table look exactly linear because they have been rounded. For example, the exact value of
x3 when x = 2.001 is 8.012006001, not 8.012. Thus, the table can tell us only that the derivative is approximately 12.
Example 5 on page 95 shows how to compute the derivative of f (x) exactly.
2. With h = 0.01 and h = −0.01, we have the difference quotients
f (1.01) − f (1)
= 3.0301
0.01
and
f (0.99) − f (1)
= 2.9701.
−0.01
and
f (0.999) − f (1)
= 2.997001.
−0.001
With h = 0.001 and h = −0.001,
f (1.001) − f (1)
= 3.003001
0.001
The values of these difference quotients suggest that the limit is about 3.0. We say
f ′ (1) =
Instantaneous rate of change of f (x) = x3
with respect to x at x = 1
≈ 3.0.
3. (a) Using the formula for the average rate of change gives
Average rate of change
of revenue for 1 ≤ q ≤ 2
Average rate of change
of revenue for 2 ≤ q ≤ 3
=
R(2) − R(1)
160 − 90
=
= 70 dollars/kg.
1
1
=
R(3) − R(2)
210 − 160
=
= 50 dollars/kg.
1
1
So we see that the average rate decreases as the quantity sold in kilograms increases.
(b) With h = 0.01 and h = −0.01, we have the difference quotients
R(2.01) − R(2)
= 59.9 dollars/kg
0.01
and
R(1.99) − R(2)
= 60.1 dollars/kg.
−0.01
and
R(1.999) − R(2)
= 60.01 dollars/kg.
−0.001
With h = 0.001 and h = −0.001,
R(2.001) − R(2)
= 59.99 dollars/kg
0.001
The values of these difference quotients suggest that the instantaneous rate of change is about 60 dollars/kg. To
confirm that the value is exactly 60, that is, that R′ (2) = 60, we would need to take the limit as h → 0.
2.2 SOLUTIONS
101
4. (a) Using a calculator we obtain the values found in the table below:
x
1
1.5
2
2.5
3
ex
2.72
4.48
7.39
12.18
20.09
(b) The average rate of change of f (x) = ex between x = 1 and x = 3 is
Average rate of change =
e3 − e
20.09 − 2.72
f (3) − f (1)
=
≈
= 8.69.
3−1
3−1
2
(c) First we find the average rates of change of f (x) = ex between x = 1.5 and x = 2, and between x = 2 and x = 2.5:
f (2) − f (1.5)
e2 − e1.5
7.39 − 4.48
=
≈
= 5.82
2 − 1.5
2 − 1.5
0.5
f (2.5) − f (2)
e2.5 − e2
12.18 − 7.39
=
≈
= 9.58.
Average rate of change =
2.5 − 2
2.5 − 2
0.5
Average rate of change =
Now we approximate the instantaneous rate of change at x = 2 by averaging these two rates:
Instantaneous rate of change ≈
5.82 + 9.58
= 7.7.
2
5. (a)
Table 2.1
x
1
1.5
2
2.5
3
log x
0
0.18
0.30
0.40
0.48
(b) The average rate of change of f (x) = log x between x = 1 and x = 3 is
f (3) − f (1)
log 3 − log 1
0.48 − 0
=
≈
= 0.24
3−1
3−1
2
(c) First we find the average rates of change of f (x) = log x between x = 1.5 and x = 2, and between x = 2 and
x = 2.5.
0.30 − 0.18
log 2 − log 1.5
=
≈ 0.24
2 − 1.5
0.5
log 2.5 − log 2
0.40 − 0.30
=
≈ 0.20
2.5 − 2
0.5
Now we approximate the instantaneous rate of change at x = 2 by finding the average of the above rates, i.e.
the instantaneous rate of change
of f (x) = log x at x = 2
≈
0.24 + 0.20
= 0.22.
2
6. In Table 2.2, each x increase of 0.001 leads to an increase in f (x) by about 0.031, so
f ′ (3) ≈
0.031
= 31.
0.001
Table 2.2
x
2.998
2.999
3.000
3.001
3.002
x3 + 4x
38.938
38.969
39.000
39.031
39.062
102
Chapter Two /SOLUTIONS
y
7.
1
y = sin x
π
2π
3π
4π
x
−1
Since sin x is decreasing for values near x = 3π, its derivative at x = 3π is negative.
log(1 + h) − log 1
log(1 + h)
8. f ′ (1) = lim
= lim
h→0
h→0
h
h
Evaluating log(1+h)
for
h
=
0.01,
0.001,
and
0.0001, we get 0.43214, 0.43408, 0.43427, so f ′ (1) ≈ 0.43427. The
h
corresponding secant lines are getting steeper, because the graph of log x is concave down. We thus expect the limit to be
more than 0.43427 . If we consider negative values of h, the estimates are too large. We can also see this from the graph
below:
y
log(1+h)
x
h
for h < 0
f ′ (1)x
✛
✛
❘
log(1+h)
x
h
for h > 0
x
1
9. We estimate f ′ (2) using the average rate of change formula on a small interval around 2. We use the interval x = 2 to
x = 2.001. (Any small interval around 2 gives a reasonable answer.) We have
f ′ (2) ≈
f (2.001) − f (2)
32.001 − 32
9.00989 − 9
=
=
= 9.89.
2.001 − 2
2.001 − 2
0.001
10. (a) The average rate of change from x = a to x = b is the slope of the line between the points on the curve with x = a
and x = b. Since the curve is concave down, the line from x = 1 to x = 3 has a greater slope than the line from
x = 3 to x = 5, and so the average rate of change between x = 1 and x = 3 is greater than that between x = 3 and
x = 5.
(b) Since f is increasing, f (5) is the greater.
(c) As in part (a), f is concave down and f ′ is decreasing throughout so f ′ (1) is the greater.
11. Since f ′ (x) = 0 where the graph is horizontal, f ′ (x) = 0 at x = d. The derivative is positive at points b and c, but the
graph is steeper at x = c. Thus f ′ (x) = 0.5 at x = b and f ′ (x) = 2 at x = c. Finally, the derivative is negative at points
a and e but the graph is steeper at x = e. Thus, f ′ (x) = −0.5 at x = a and f ′ (x) = −2 at x = e. See Table 2.3.
Thus, we have f ′ (d) = 0, f ′ (b) = 0.5, f ′ (c) = 2, f ′ (a) = −0.5, f ′ (e) = −2.
Table 2.3
x
f ′ (x)
d
0
b
0.5
c
2
a
−0.5
e
−2
2.2 SOLUTIONS
103
12. One possible choice of points is shown below.
y
F
A
E
C
x
D
B
Problems
13. The statements f (100) = 35 and f ′ (100) = 3 tell us that at x = 100, the value of the function is 35 and the function is
increasing at a rate of 3 units for a unit increase in x. Since we increase x by 2 units in going from 100 to 102, the value
of the function goes up by approximately 2 · 3 = 6 units, so
f (102) ≈ 35 + 2 · 3 = 35 + 6 = 41.
14. The answers to parts (a)–(d) are shown in Figure 2.6.
Slope= f ′ (3)
❄
✻
✻
✻
✛
❄
f (4) − f (2)
f (x)
Slope =
f (5)−f (2)
5−2
f (4)
❄
1
2
3
4
x
5
Figure 2.6
15. (a) Since f is increasing, f (4) > f (3).
(b) From Figure 2.7, it appears that f (2) − f (1) > f (3) − f (2).
f (2) − f (1)
(c) The quantity
represents the slope of the secant line connecting the points on the graph at x = 1
2−1
and x = 2. This is greater than the slope of the secant line connecting the points at x = 1 and x = 3 which is
f (3) − f (1)
.
3−1
(d) The function is steeper at x = 1 than at x = 4 so f ′ (1) > f ′ (4).
104
Chapter Two /SOLUTIONS
f (x)
f (3) − f (2)
✻
❄
✻
f (2) − f (1)
✻✻ ❄
f (3)−f (1)
3−1
slope =
slope =
f (2)−f (1)
2−1
x
1
2
3
4
5
Figure 2.7
16. Figure 2.8 shows the quantities in which we are interested.
Slope = f ′ (2)
Slope = f ′ (3)
f (x)
✻
f (x)
f (3)−f (2)
Slope =
3−2
= f (3) − f (2)
x
2
3
Figure 2.8
The quantities f ′ (2), f ′ (3) and f (3) − f (2) have the following interpretations:
• f ′ (2) = slope of the tangent line at x = 2
• f ′ (3) = slope of the tangent line at x = 3
(2)
= slope of the secant line from f (2) to f (3).
• f (3) − f (2) = f (3)−f
3−2
From Figure 2.8, it is clear that 0 < f (3) − f (2) < f ′ (2). By extending the secant line past the point (3, f (3)), we can
see that it lies above the tangent line at x = 3.
Thus
0 < f ′ (3) < f (3) − f (2) < f ′ (2).
17. The coordinates of A are (4, 25). See Figure 2.9. The coordinates of B and C are obtained using the slope of the tangent
line. Since f ′ (4) = 1.5, the slope is 1.5
From A to B, ∆x = 0.2, so ∆y = 1.5(0.2) = 0.3. Thus, at C we have y = 25 + 0.3 = 25.3. The coordinates of
B are (4.2, 25.3).
From A to C, ∆x = −0.1, so ∆y = 1.5(−0.1) = −0.15. Thus, at C we have y = 25 − 0.15 = 24.85. The
coordinates of C are (3.9, 24.85).
2.2 SOLUTIONS
Tangent line
B
1.5(0.2) = 0.3
0.2
A = (4, 25)
0.15
C
0.1
Figure 2.9
18. (a) Since the point B = (2, 5) is on the graph of g, we have g(2) = 5.
(b) The slope of the tangent line touching the graph at x = 2 is given by
Slope =
Thus, g ′ (2) = −0.4.
5 − 5.02
−0.02
Rise
=
=
= −0.4.
Run
2 − 1.95
0.05
19. See Figure 2.10.
y
y = f (x)
✻✻
(c) f (x + h) − f (x)
❨
❄
✻
(e) Slope =
f (x+h)−f (x)
h
(b) f (x + h)
(a) f (x)
❄
❄
x
✛
(d) h
x
x+h
✲
Figure 2.10
20. See Figure 2.11.
y
(e) Slope =
✻
✻
✠
(c) f (x + h) − f (x) (which is negative)
(a) f (x)
(b) f (x + h)
❄
x
✛
(d) h
f (x+h)−f (x)
h
❄
✻
❄
y = f (x)
x+h
✲
Figure 2.11
x
105
106
Chapter Two /SOLUTIONS
21. (a) For the line from A to B,
f (b) − f (a)
.
b−a
(b) The tangent line at point C appears to be parallel to the line from A to B. Assuming this to be the case, the lines have
the same slope.
(c) There is only one other point, labeled D in Figure 2.12, at which the tangent line is parallel to the line joining A and
B.
Slope =
B
C
D
A
Figure 2.12
22. (a) Figure 2.13 shows the graph of an even function. We see that since f is symmetric about the y-axis, the tangent line
at x = −10 is just the tangent line at x = 10 flipped about the y-axis, so the slope of one tangent is the negative of
that of the other. Therefore, f ′ (−10) = −f ′ (10) = −6.
(b) From part (a) we can see that if f is even, then for any x, we have f ′ (−x) = −f ′ (x). Thus f ′ (−0) = −f ′ (0), so
f ′ (0) = 0.
f (x)
60
x
−10
10
−60
Figure 2.13
23. Figure 2.14 shows the graph of an odd function. We see that since g is symmetric about the origin, its tangent line at
x = −4 is just the tangent line at x = 4 flipped about the origin, so they have the same slope. Thus, g ′ (−4) = 5.
15
10
5
g(x)
x
−4
4
−5
−10
−15
Figure 2.14
24. (a)
h in degrees 0
z}|{
z}|{
sin h − sin 0
sin h
f (0) = lim
=
.
h→0
h
h
′
To four decimal places,
sin 0.2
sin 0.1
sin 0.01
sin 0.001
≈
≈
≈
≈ 0.01745
0.2
0.1
0.01
0.001
so f ′ (0) ≈ 0.01745.
(b) Consider the ratio sinh h . As we approach 0, the numerator, sin h, will be much smaller in magnitude if h is in degrees
than it would be if h were in radians. For example, if h = 1◦ radian, sin h = 0.8415, but if h = 1 degree,
sin h = 0.01745. Thus, since the numerator is smaller for h measured in degrees while the denominator is the same,
we expect the ratio sinh h to be smaller.
2.2 SOLUTIONS
107
25. We find the derivative using a difference quotient:
(3 + h)2 + 3 + h − (32 + 3)
f (3 + h) − f (3)
= lim
h→0
h→0
h
h
7h + h2
9 + 6h + h2 + 3 + h − 9 − 3
= lim
= lim (7 + h) = 7.
= lim
h
h→0
h
h→0
h→0
f ′ (3) = lim
Thus at x = 3, the slope of the tangent line is 7. Since f (3) = 32 + 3 = 12, the line goes through the point (3, 12), and
therefore its equation is
y − 12 = 7(x − 3) or y = 7x − 9.
The graph is in Figure 2.15.
y
y = x2 + x
y = 7x − 9
12
x
3
Figure 2.15
26. Using a difference quotient with h = 0.001, say, we find
1.001 ln(1.001) − 1 ln(1)
= 1.0005
1.001 − 1
2.001 ln(2.001) − 2 ln(2)
= 1.6934
f ′ (2) ≈
2.001 − 2
f ′ (1) ≈
The fact that f ′ is larger at x = 2 than at x = 1 suggests that f is concave up between x = 1 and x = 2.
27. We want f ′ (2). The exact answer is
f ′ (2) = lim
h→0
(2 + h)2+h − 4
f (2 + h) − f (2)
= lim
,
h→0
h
h
but we can approximate this. If h = 0.001, then
(2.001)2.001 − 4
≈ 6.779
0.001
and if h = 0.0001 then
(2.0001)2.0001 − 4
≈ 6.773,
0.0001
so f ′ (2) ≈ 6.77.
28. Notice that we can’t get all the information we want just from the graph of f for 0 ≤ x ≤ 2, shown on the left in
Figure 2.16. Looking at this graph, it looks as if the slope at x = 0 is 0. But if we zoom in on the graph near x = 0, we
get the graph of f for 0 ≤ x ≤ 0.05, shown on the right in Figure 2.16. We see that f does dip down quite a bit between
x = 0 and x ≈ 0.11. In fact, it now looks like f ′ (0) is around −1. Note that since f (x) is undefined for x < 0, this
derivative only makes sense as we approach zero from the right.
y
y
x
6
5
4
−0.0025
−0.005
f (x) = 3x3/2 − x
3
2
1
x
0.5
1
1.5
2
−0.0075
−0.01
−0.0125
−0.015
−0.0175
Figure 2.16
0.01 0.02 0.03 0.04 0.05
f (x) = 3x3/2 − x
108
Chapter Two /SOLUTIONS
We zoom in on the graph of f near x = 1 to get a more accurate picture from which to estimate f ′ (1). A graph of
f for 0.7 ≤ x ≤ 1.3 is shown in Figure 2.17. [Keep in mind that the axes shown in this graph don’t cross at the origin!]
Here we see that f ′ (1) ≈ 3.5.
y
f (x) = 3x3/2 − x
3
2.5
2
1.5
x
0.7 0.8 0.9
1.1 1.2 1.3
Figure 2.17
29.
f (1 + h) − f (1)
ln(cos(1 + h)) − ln(cos 1)
= lim
h
h→0
h
For h = 0.001, the difference quotient = −1.55912; for h = 0.0001, the difference quotient = −1.55758.
The instantaneous rate of change of f therefore appears to be about −1.558 at x = 1.
At x = π4 , if we try h = 0.0001, then
f ′ (1) = lim
h→0
difference quotient =
ln[cos( π4 + 0.0001)] − ln(cos π4 )
≈ −1.0001.
0.0001
The instantaneous rate of change of f appears to be about −1 at x =
π
.
4
30. The quantity f (0) represents the population on October 17, 2006, so f (0) = 300 million.
The quantity f ′ (0) represents the rate of change of the population (in millions per year). Since
1/106 million people
1 person
=
= 2.867 million people/year,
11 seconds
11/(60 · 60 · 24 · 365) years
so we have f ′ (0) = 2.867.
31. We want to approximate P ′ (0) and P ′ (7). Since for small h
P ′ (0) ≈
P (h) − P (0)
,
h
if we take h = 0.01, we get
P ′ (0) ≈
P ′ (7) ≈
1.267(1.007)0.01 − 1.267
= 0.00884 billion/year
0.01
= 8.84 million people/year in 2000,
1.267(1.007)7.01 − 1.267(1.007)7
= 0.00928 billion/year
0.01
= 9.28 million people/year in 2007
2.2 SOLUTIONS
109
32. (a) From Figure 2.18, it appears that the slopes of the tangent lines to the two graphs are the same at each x. For x = 0,
the slopes of the tangents to the graphs of f (x) and g(x) at 0 are
f (0 + h) − f (0)
h
f (h) − 0
= lim
h→0
h
1 2
h
= lim 2
h→0 h
1
= lim h
h→0 2
= 0,
g ′ (0) = lim
f ′ (0) = lim
h→0
h→0
= lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
g(0 + h) − g(0)
h
g(h) − g(0)
h
1 2
h +3−3
2
h
1 2
h
2
h
1
h
2
= 0.
For x = 2, the slopes of the tangents to the graphs of f (x) and g(x) are
f (2 + h) − f (2)
h
1
(2 + h)2 − 21 (2)2
2
lim
h
h→0
1
(4
+
4h
+ h2 ) − 2
lim 2
h→0
h
2 + 2h + 12 h2 − 2
lim
h→0
h
2h + 12 h2
lim
h→0
h
1
lim 2 + h
h→0
2
2,
f ′ (2) = lim
h→0
=
=
=
=
=
=
g(2 + h) − g(2)
h
1
(2 + h)2 + 3 − ( 21 (2)2 + 3)
2
lim
h
h→0
2
1
1
(2
+
h)
−
(2)2
2
lim 2
h→0
h
1
(4
+
4h
+ h2 ) − 2
lim 2
h→0
h
2 + 2h + 12 (h2 ) − 2
lim
h→0
h
2h + 12 (h2 )
lim
h→0
h
1
lim 2 + h
2
h→0
2.
g ′ (2) = lim
h→0
=
=
=
=
=
=
=
g(x) = f (x) + 3
f (x) =
1 2
x
2
Figure 2.18
For x = x0 , the slopes of the tangents to the graphs of f (x) and g(x) are
110
Chapter Two /SOLUTIONS
f (x0 + h) − f (x0 )
h
2
1
(x
+
h)
− 21 x20
0
lim 2
h→0
h
2
2
1 2
1
(x
+
2x
0 h + h ) − 2 x0
0
lim 2
h
h→0
x0 h + 12 h2
lim
h→0
h
1
lim x0 + h
h→0
2
x0 ,
f ′ (x0 ) = lim
h→0
=
=
=
=
=
g(x0 + h) − g(x0 )
h
2
1
(x
+
h)
+ 3 − ( 21 (x0 )2 + 3)
0
lim 2
h→0
h
2
1
1
(x
+
h)
−
(x0 )2
0
2
lim 2
h
h→0
2
2
1
1 2
(x
+
2x
0 h + h ) − 2 x0
0
lim 2
h→0
h
x0 h + 21 h2
lim
h→0
h
1
lim x0 + h
h→0
2
x0 .
g ′ (x0 ) = lim
h→0
=
=
=
=
=
=
(b)
g(x + h) − g(x)
h
f (x + h) + C − (f (x) + C)
= lim
h→0
h
f (x + h) − f (x)
= lim
h→0
h
= f ′ (x).
g ′ (x) = lim
h→0
33. As h gets smaller, round-off error becomes important. When h = 10−12 , the quantity 2h − 1 is so close to 0 that the
calculator rounds off the difference to 0, making the difference quotient 0. The same thing will happen when h = 10−20 .
34. (a) Table 2.4 shows that near x = 1, every time the value of x increases by 0.001, the value of x2 increases by approximately 0.002. This suggests that
0.002
f ′ (1) ≈
= 2.
0.001
Table 2.4
Values of f (x) = x2 near x = 1
Difference in
x
x2
0.998
0.996004
0.999
0.998001
1.000
1.000000
1.001
1.002001
1.002
1.004004
↑
successive x2 values
0.001997
0.001999
0.002001
0.002003
↑
x increments
All approximately
of 0.001
0.002
(b) The derivative is the limit of the difference quotient, so we look at
f ′ (1) = lim
h→0
f (1 + h) − f (1)
.
h
Using the formula for f , we have
f ′ (1) = lim
h→0
(1 + 2h + h2 ) − 1
2h + h2
(1 + h)2 − 12
= lim
= lim
.
h→0
h→0
h
h
h
2.2 SOLUTIONS
111
Since the limit only examines values of h close to, but not equal to zero, we can cancel h in the expression (2h +
h2 )/h. We get
h(2 + h)
= lim (2 + h).
f ′ (1) = lim
h→0
h→0
h
′
2
This limit is 2, so f (1) = 2. At x = 1 the rate of change of x is 2.
(c) Since the derivative is the rate of change, f ′ (1) = 2 means that for small changes in x near x = 1, the change in
f (x) = x2 is about twice as big as the change in x. As an example, if x changes from 1 to 1.1, a net change of 0.1,
then f (x) changes by about 0.2. Figure 2.19 shows this geometrically. Near x = 1 the function is approximately
linear with slope of 2.
✛
f (x) = x2
✻
1.21
Zooming
1
✲
✛ 0.1 ✲
f (x) = x2
Slope ≈ 2
0.21
❄
x
1 1.1
Figure 2.19: Graph of f (x) = x2 near x = 1 has slope ≈ 2
(−3 + h)2 − 9
h(−6 + h)
9 − 6h + h2 − 9
= lim
= lim
= lim −6 + h = −6.
h→0
h→0
h→0
h
h
h
2
3
3
h(−12 + 6h − h2 )
(2 − h) − 8
8 − 12h + 6h − h − 8
= lim
= lim
= lim −12 + 6h − h2 = −12.
36. lim
h→0
h→0
h→0
h→0
h
h
h
1 − (1 + h)
1
1
−1
37. lim
− 1 = lim
= lim
= −1
h→0 h
h→0 (1 + h)h
h→0 1 + h
1+h
35. lim
h→0
1
1 − (1 + 2h + h2 )
−2 − h
− 1 = lim
= lim
= −2
2
(1 + h)
h→0
h(1 + h)2
h→0 (1 + h)2
√
√
√
( 4 + h − 2)( 4 + h + 2)
4+h−4
h
√
= √
= √
.
39. 4 + h − 2 =
4
+
h
+
2
4
+
h
+
2
4
+
h+2
√
4+h−2
1
1
Therefore lim
= lim √
=
h→0
h→0
h
4
4+h+2
√
√
√
(2 − 4 + h)(2 + 4 + h)
4 − (4 + h)
2− 4+h
1
1
√
√
√
√
=
− =
= √
.
40. √
2
2 4 + h(2 + 4 + h)
4+h
2 4 + h
2 4 + h(2 + 4 + h)
1
−1
1
1
1
√
√
= lim √
Therefore lim
−
=−
h→0 2 4 + h(2 +
h→0 h
2
16
4+h
4 + h)
1
38. lim
h→0 h
41. Using the definition of the derivative, we have
f (10 + h) − f (10)
h
5(10 + h)2 − 5(10)2
lim
h→0
h
500 + 100h + 5h2 − 500
lim
h→0
h
100h + 5h2
lim
h→0
h
h(100 + 5h)
lim
h
h→0
lim 100 + 5h
f ′ (10) = lim
h→0
=
=
=
=
=
h→0
= 100.
112
Chapter Two /SOLUTIONS
42. Using the definition of the derivative, we have
f (−2 + h) − f (−2)
h
3
(−2 + h) − (−2)3
lim
h→0
h
(−8 + 12h − 6h2 + h3 ) − (−8)
lim
h→0
h
12h − 6h2 + h3
lim
h→0
h
h(12 − 6h + h2 )
lim
h→0
h
lim (12 − 6h + h2 ),
f ′ (−2) = lim
h→0
=
=
=
=
=
h→0
which goes to 12 as h → 0. So f ′ (−2) = 12.
43. Using the definition of the derivative
g(−1 + h) − g(−1)
h
((−1 + h)2 + (−1 + h)) − ((−1)2 + (−1))
= lim
h→0
h
(1 − 2h + h2 − 1 + h) − (0)
= lim
h→0
h
−h + h2
= lim (−1 + h) = −1.
= lim
h
h→0
h→0
g ′ (−1) = lim
h→0
44.
((1 + h)3 + 5) − (13 + 5)
f (1 + h) − f (1)
= lim
h→0
h→0
h
h
1 + 3h + 3h2 + h3 + 5 − 1 − 5
3h + 3h2 + h3
= lim
= lim
h→0
h
h→0
h
= lim (3 + 3h + h2 ) = 3.
f ′ (1) = lim
h→0
45.
1
−
g(2 + h) − g(2)
= lim 2+h
h→0
h→0
h
h
2 − (2 + h)
−h
= lim
= lim
h→0 h(2 + h)2
h→0 h(2 + h)2
−1
1
= lim
=−
h→0 (2 + h)2
4
g ′ (2) = lim
1
2
46.
1
− 212
g(2 + h) − g(2)
(2+h)2
= lim
h→0
h→0
h
h
22 − (2 + h)2
4 − 4 − 4h − h2
= lim 2
= lim
h→0 2 (2 + h)2 h
h→0
4h(2 + h)2
g ′ (2) = lim
−4 − h
−4h − h2
= lim
h→0 4(2 + h)2
h→0 4h(2 + h)2
−4
1
=
=− .
4(2)2
4
= lim
2.3 SOLUTIONS
113
47. As we saw in the answer to Problem 41, the slope of the tangent line to f (x) = 5x2 at x = 10 is 100. When x = 10,
f (x) = 500 so (10, 500) is a point on the tangent line. Thus y = 100(x − 10) + 500 = 100x − 500.
48. As we saw in the answer to Problem 42, the slope of the tangent line to f (x) = x3 at x = −2 is 12. When x = −2,
f (x) = −8 so we know the point (−2, −8) is on the tangent line. Thus the equation of the tangent line is y = 12(x +
2) − 8 = 12x + 16.
49. We know that the slope of the tangent line to f (x) = x when x = 20 is 1. When x = 20, f (x) = 20 so (20, 20) is on
the tangent line. Thus the equation of the tangent line is y = 1(x − 20) + 20 = x.
50. First find the derivative of f (x) = 1/x2 at x = 1.
1
− 112
f (1 + h) − f (1)
(1+h)2
= lim
h→0
h→0
h
h
12 − (1 + h)2
1 − (1 + 2h + h2 )
= lim
= lim
h→0
h→0
h(1 + h)2
h(1 + h)2
f ′ (1) = lim
= lim
h→0
−2 − h
−2h − h2
= lim
= −2
h→0 (1 + h)2
h(1 + h)2
Thus the tangent line has a slope of −2 and goes through the point (1, 1), and so its equation is
y − 1 = −2(x − 1)
or
y = −2x + 3.
Strengthen Your Understanding
51. The graph of f (x) = log x is increasing, so f ′ (0.5) > 0.
52. The derivative of a function at a point is the slope of the tangent line, not the tangent line itself.
53. f (x) = ex .
Many other answers are possible.
54. A linear function is of the form f (x) = ax + b. The derivative of this function is the slope of the line y = ax + b, so
f ′ (x) = a, so a = 2. One such function is f (x) = 2x + 1.
55. True. The derivative of a function is the limit of difference quotients. A few difference quotients can be computed from
the table, but the limit can not be computed from the table.
56. True. The derivative f ′ (10) is the slope of the tangent line to the graph of y = f (x) at the point where x = 10. When
you zoom in on y = f (x) close enough it is not possible to see the difference between the tangent line and the graph of f
on the calculator screen. The line you see on the calculator is a little piece of the tangent line, so its slope is the derivative
f ′ (10).
57. True. This is seen graphically. The derivative f ′ (a) is the slope of the line tangent to the graph of f at the point P where
x = a. The difference quotient (f (b) − f (a))/(b − a) is the slope of the secant line with endpoints on the graph of f
at the points where x = a and x = b. The tangent and secant lines cross at the point P . The secant line goes above the
tangent line for x > a because f is concave up, and so the secant line has higher slope.
58. (a). This is best observed graphically.
Solutions for Section 2.3
Exercises
1. (a) We use the interval to the right of x = 2 to estimate the derivative. (Alternately, we could use the interval to the left
of 2, or we could use both and average the results.) We have
f ′ (2) ≈
24 − 18
6
f (4) − f (2)
=
= = 3.
4−2
4−2
2
We estimate f ′ (2) ≈ 3.
(b) We know that f ′ (x) is positive when f (x) is increasing and negative when f (x) is decreasing, so it appears that
f ′ (x) is positive for 0 < x < 4 and is negative for 4 < x < 12.
114
Chapter Two /SOLUTIONS
2. For x = 0, 5, 10, and 15, we use the interval to the right to estimate the derivative. For x = 20, we use the interval to the
left. For x = 0, we have
f (5) − f (0)
70 − 100
−30
=
=
= −6.
f ′ (0) ≈
5−0
5−0
5
Similarly, we find the other estimates in Table 2.5.
Table 2.5
x
f ′ (x)
0
5
10
15
20
−6
−3
−1.8
−1.2
−1.2
3. The graph is that of the line y = −2x + 2. The slope, and hence the derivative, is −2. See Figure 2.20.
4
x
−4
4
−4
Figure 2.20
4. See Figure 2.21.
4
x
−4
4
−4
Figure 2.21
5. See Figure 2.22.
4
−4
4
−4
Figure 2.22
x
2.3 SOLUTIONS
115
6. See Figure 2.23.
4
x
−4
4
−4
Figure 2.23
7. The slope of this curve is approximately −1 at x = −4 and at x = 4, approximately 0 at x = −2.5 and x = 1.5, and
approximately 1 at x = 0. See Figure 2.24.
4
x
−4
4
−4
Figure 2.24
8. See Figure 2.25.
15
x
−3 − 1
2
3
Figure 2.25
9. See Figure 2.26.
4
−4
4
−4
Figure 2.26
x
116
Chapter Two /SOLUTIONS
10. See Figure 2.27.
4
x
−4
4
−4
Figure 2.27
11. See Figure 2.28.
4
x
−4
4
−4
Figure 2.28
12. See Figure 2.29.
4
−4
4
x
−4
Figure 2.29
13. See Figures 2.30 and 2.31.
f (x) = 5x
10
f ′ (x)
5
6
2
x
−2 −1
1
2
−6
x
−10
−2
Figure 2.30
−1
1
Figure 2.31
2
2.3 SOLUTIONS
14. See Figures 2.32 and 2.33.
4
4
f (x)
f ′ (x)
2
2
x
−2
−1
1
2
−2
x
−2
2
−4
Figure 2.32
Figure 2.33
15. See Figures 2.34 and 2.35.
f ′ (x)
4
3
2
2
x
f (x)
−1
1
1
−2
x
−1
1
2
2
−4
Figure 2.34
Figure 2.35
16. The graph of f (x) and its derivative look the same, as in Figures 2.36 and 2.37.
10
10
8
8
6
6
f (x)
4
f ′ (x)
4
2
2
x
−3
−2
−1
1
2
3
x
−3
−2
−1
Figure 2.36
1
2
3
Figure 2.37
17. See Figures 2.38 and 2.39.
f (x)
1
1
f ′ (x)
π
2
π
−1
x
3π
2
π
2
2π
π
−1
Figure 2.38
Figure 2.39
x
3π
2
2π
117
118
Chapter Two /SOLUTIONS
18. See Figures 2.40 and 2.41.
f (x)
x
−1
1
f ′ (x)
x
1
1
Figure 2.40
Figure 2.41
19. Since 1/x = x−1 , using the power rule gives
k′ (x) = (−1)x−2 = −
1
.
x2
Using the definition of the derivative, we have
1
− x1
x − (x + h)
k(x + h) − k(x)
= lim x+h
= lim
h
h→0
h
h→0 h(x + h)x
h→0
−1
1
−h
= lim
= − 2.
= lim
h→0 (x + h)x
h→0 h(x + h)x
x
k′ (x) = lim
20. Since 1/x2 = x−2 , using the power rule gives
l′ (x) = −2x−3 = −
2
.
x3
Using the definition of the derivative, we have
l′ (x) = lim
1
(x+h)2
h→0
h
−
1
x2
= lim
h→0
x2 − (x + h)2
h(x + h)2 x2
x2 − (x2 + 2xh + h2 )
−2xh − h2
= lim
= lim
2
2
h→0 h(x + h)2 x2
h→0
h(x + h) x
−2x
2
−2x − h
= 2 2 = − 3.
= lim
x x
x
h→0 (x + h)2 x2
21. Using the definition of the derivative,
2(x + h)2 − 3 − (2x2 − 3)
g(x + h) − g(x)
= lim
h→0
h→0
h
h
2(x2 + 2xh + h2 ) − 3 − 2x2 + 3
4xh + 2h2
= lim
= lim
h→0
h→0
h
h
= lim (4x + 2h) = 4x.
g ′ (x) = lim
h→0
22. Using the definition of the derivative, we have
m(x + h) − m(x)
1
1
1
= lim
−
h→0 h
h→0
h
x+h+1
x+1
1 x+1−x−h−1
−h
= lim
= lim
h→0 h
(x + 1)(x + h + 1)
h→0 h(x + 1)(x + h + 1)
−1
= lim
h→0 (x + 1)(x + h + 1)
−1
=
.
(x + 1)2
m′ (x) = lim
2.3 SOLUTIONS
119
Problems
23.
y
(a)
(b)
y
x
x
y
(c)
(d)
y
x
x
24. Since f ′ (x) > 0 for x < −1, f (x) is increasing on this interval.
Since f ′ (x) < 0 for x > −1, f (x) is decreasing on this interval.
Since f ′ (x) = 0 at x = −1, the tangent to f (x) is horizontal at x = −1.
One possible shape for y = f (x) is shown in Figure 2.42.
x
−1
Figure 2.42
25.
x
ln x
x
ln x
x
ln x
x
ln x
0.998
−0.0020
1.998
0.6921
4.998
1.6090
9.998
2.3024
−0.0010
1.999
0.6926
4.999
1.6092
9.999
2.3025
0.0000
2.000
0.6931
5.000
1.6094
10.000
2.3026
1.001
0.0010
2.001
0.6936
5.001
1.6096
10.001
2.3027
1.002
0.0020
2.002
0.6941
5.002
1.6098
10.002
2.3028
0.999
1.000
At x = 1, the values of ln x are increasing by 0.001 for each increase in x of 0.001, so the derivative appears to be 1.
At x = 2, the increase is 0.0005 for each increase of 0.001, so the derivative appears to be 0.5. At x = 5, ln x increases
by 0.0002 for each increase of 0.001 in x, so the derivative appears to be 0.2. And at x = 10, the increase is 0.0001 over
intervals of 0.001, so the derivative appears to be 0.1. These values suggest an inverse relationship between x and f ′ (x),
namely f ′ (x) = x1 .
f (x + h) − f (x)
. For this problem, we’ll take the average of the values obtained for h = 1
h
f (x + 1) − f (x − 1)
and h = −1; that’s the average of f (x + 1) − f (x) and f (x) − f (x − 1) which equals
. Thus,
2
f ′ (0) ≈ f (1) − f (0) = 13 − 18 = −5.
f ′ (1) ≈ (f (2) − f (0))/2 = (10 − 18)/2 = −4.
f ′ (2) ≈ (f (3) − f (1))/2 = (9 − 13)/2 = −2.
f ′ (3) ≈ (f (4) − f (2))/2 = (9 − 10)/2 = −0.5.
f ′ (4) ≈ (f (5) − f (3))/2 = (11 − 9)/2 = 1.
f ′ (5) ≈ (f (6) − f (4))/2 = (15 − 9)/2 = 3.
f ′ (6) ≈ (f (7) − f (5))/2 = (21 − 11)/2 = 5.
26. We know that f ′ (x) ≈
120
Chapter Two /SOLUTIONS
f ′ (7) ≈ (f (8) − f (6))/2 = (30 − 15)/2 = 7.5.
f ′ (8) ≈ f (8) − f (7) = 30 − 21 = 9.
The rate of change of f (x) is positive for 4 ≤ x ≤ 8, negative for 0 ≤ x ≤ 3. The rate of change is greatest at about
x = 8.
27. The value of g(x) is increasing at a decreasing rate for 2.7 < x < 4.2 and increasing at an increasing rate for x > 4.2.
7.4 − 6.0
∆y
=
= 2.8
∆x
5.2 − 4.7
∆y
9.0 − 7.4
=
= 3.2
∆x
5.7 − 5.2
between x = 4.7 and x = 5.2
between x = 5.2 and x = 5.7
Thus g ′ (x) should be close to 3 near x = 5.2.
28. (a) x3
(b) x4
(c) x5
(d) x3
29. This is a line with slope 1, so the derivative is the constant function f ′ (x) = 1. The graph is the horizontal line y = 1.
See Figure 2.43.
f ′ (x)
1
x
−3
3
Figure 2.43
30. This is a line with slope −2, so the derivative is the constant function f ′ (x) = −2. The graph is a horizontal line at
y = −2. See Figure 2.44.
1
x
1
2
−1
f ′ (x)
−2
Figure 2.44
31. See Figure 2.45.
f ′ (x)
x
2
4
Figure 2.45
32. See Figure 2.46.
x
1
2
f ′ (x)
Figure 2.46
2.3 SOLUTIONS
121
33. See Figure 2.47.
f ′ (x)
4
x
Figure 2.47
34. See Figure 2.48.
f ′ (x)
x
Figure 2.48
35. See Figure 2.49.
20
f ′ (x)
x
−3
3
−20
Figure 2.49
36. One possible graph is shown in Figure 2.50. Notice that as x gets large, the graph of f (x) gets more and more horizontal.
Thus, as x gets large, f ′ (x) gets closer and closer to 0.
f ′ (x)
x
−1
1
2
3
Figure 2.50
37. See Figure 2.51.
f ′ (x)
−1
1
2 3
x
4
Figure 2.51
5
6
122
Chapter Two /SOLUTIONS
38. See Figure 2.52.
f ′ (x)
x
Figure 2.52
39. See Figure 2.53.
f ′ (2) = 1
f ′ (3) = 0
(1, 3)
f ′ (0) = 3
1
3
5
Figure 2.53
40. (a) Graph II
(b) Graph I
(c) Graph III
41. (a) t = 3
(b) t = 9
(c) t = 14
(d)
V ′ (t)
1
15
3
6
9
18
12
t
−2
42. The derivative is zero whenever the graph of the original function is horizontal. Since the current is proportional to
the derivative of the voltage, segments where the current is zero alternate with positive segments where the voltage is
increasing and negative segments where the voltage is decreasing. See Figure 2.54. Note that the derivative does not exist
where the graph has a corner.
current
time
Figure 2.54
2.3 SOLUTIONS
123
43. (a) The function f is increasing where f ′ is positive, so for x1 < x < x3 .
(b) The function f is decreasing where f ′ is negative, so for 0 < x < x1 or x3 < x < x5 .
44. On intervals where f ′ = 0, f is not changing at all, and is therefore constant. On the small interval where f ′ > 0, f is
increasing; at the point where f ′ hits the top of its spike, f is increasing quite sharply. So f should be constant for a while,
have a sudden increase, and then be constant again. A possible graph for f is shown in Figure 2.55.
f (t)
t
Figure 2.55: Step function
45. (a) The population varies periodically with a period of 1 year. See below.
4500
4000
3500
P (t)
t (in months)
J
t=0
F M A M J
J A S O N D J
t=1
(b) The population is at a maximum on July 1st . At this time sin(2πt − π2 ) = 1, so the actual maximum population is
4000 + 500(1) = 4500. Similarly, the population is at a minimum on January 1st . At this time, sin(2πt − π2 ) = −1,
so the minimum population is 4000 + 500(−1) = 3500.
(c) The rate of change is most positive about April 1st and most negative around October 1st .
(d) Since the population is at its maximum around July 1st , its rate of change is about 0 then.
46. The derivative of the accumulated federal debt with respect to time is shown in Figure 2.56. The derivative represents the
rate of change of the federal debt with respect to time and is measured in trillions of dollars per year.
rate of change in debt
(trillions of $/year)
0.75
0.5
0.25
year
1975
1985
1995
Figure 2.56
2005
124
Chapter Two /SOLUTIONS
47. From the given information we know that f is increasing for values of x less than −2, is decreasing between x = −2 and
x = 2, and is constant for x > 2. Figure 2.57 shows a possible graph—yours may be different.
y
x
−4
−2
2
4
Figure 2.57
48. Since f ′ (x) > 0 for 1 < x < 3, we see that f (x) is increasing on this interval.
Since f ′ (x) < 0 for x < 1 and for x > 3, we see that f (x) is decreasing on these intervals.
Since f ′ (x) = 0 for x = 1 and x = 3, the tangent to f (x) will be horizontal at these x’s.
One of many possible shapes of y = f (x) is shown in Figure 2.58.
y
x
1
2
3
4
Figure 2.58
49. If lim f (x) = 50 and f ′ (x) is positive for all x, then f (x) increases to 50, but never rises above it. A possible graph of
x→∞
f (x) is shown in Figure 2.59. If lim f ′ (x) exists, it must be zero, since f looks more and more like a horizontal line.
x→∞
If f ′ (x) approached another positive value c, then f would look more and more like a line with positive slope c, which
would eventually go above y = 50.
50
40
f (x)
30
20
10
x
Figure 2.59
50. If f (x) is even, its graph is symmetric about the y-axis. So the tangent line to f at x = x0 is the same as that at x = −x0
reflected about the y-axis.
2.4 SOLUTIONS
y
125
y
y = f (x)
y = f ′ (x)
x
x
So the slopes of these two tangent lines are opposite in sign, so f ′ (x0 ) = −f ′ (−x0 ), and f ′ is odd.
51. If g(x) is odd, its graph remains the same if you rotate it 180◦ about the origin. So the tangent line to g at x = x0 is the
tangent line to g at x = −x0 , rotated 180◦ .
y
y
y = g(x)
x
y = g ′ (x)
x
But the slope of a line stays constant if you rotate it 180◦ . So g ′ (x0 ) = g ′ (−x0 ); g ′ is even.
Strengthen Your Understanding
52. Since f (x) = cos x is decreasing on some intervals, its derivative f ′ (x) is negative on those intervals, and the graph of
f ′ (x) is below the x-axis where cos x is decreasing.
53. In order for f ′ (x) to be greater than zero, the slope of f (x) has to be greater than zero. For example, f (x) = e−x is
positive for all x but since the graph is decreasing everywhere, f (x) has negative derivative for all x.
54. Two different functions can have the same rate of change. For example, f (x) = 1, g(x) = 2 both are constant, so
f ′ (x) = g ′ (x) = 0 but f (x) 6= g(x).
55. f (t) = t(1 − t). We have f (t) = t − t2 , so f ′ (t) = 1 − 2t so the velocity is positive for 0 < t < 0.5 and negative for
0.5 < t < 1.
Many other answers are possible.
56. Every linear function is of the form f (x) = b + mx and has derivative f ′ (x) = m. One family of functions with the
same derivative is f (x) = b + 2x.
57. True. The graph of a linear function f (x) = mx + b is a straight line with the same slope m at every point. Thus
f ′ (x) = m for all x.
58. True. Shifting a graph vertically does not change the shape of the graph and so it does not change the slopes of the tangent
lines to the graph.
59. False. If f ′ (x) is increasing then f (x) is concave up. However, f (x) may be either increasing or decreasing. For example,
the exponential decay function f (x) = e−x is decreasing but f ′ (x) is increasing because the graph of f is concave up.
60. False. A counterexample is given by f (x) = 5 and g(x) = 10, two different functions with the same derivatives:
f ′ (x) = g ′ (x) = 0.
Solutions for Section 2.4
Exercises
1. (a) The statement f (200) = 1300 means that it costs $1300 to produce 200 gallons of the chemical.
126
Chapter Two /SOLUTIONS
(b) The statement f ′ (200) = 6 means that when the number of gallons produced is 200, costs are increasing at a rate of
$6 per gallon. In other words, it costs about $6 to produce the next (the 201st ) gallon of the chemical.
2. (a) The statement f (5) = 18 means that when 5 milliliters of catalyst are present, the reaction will take 18 minutes.
Thus, the units for 5 are ml while the units for 18 are minutes.
(b) As in part (a), 5 is measured in ml. Since f ′ tells how fast T changes per unit a, we have f ′ measured in minutes/ml.
If the amount of catalyst increases by 1 ml (from 5 to 6 ml), the reaction time decreases by about 3 minutes.
3. (Note that we are considering the average temperature of the yam, since its temperature is different at different points
inside it.)
(a) It is positive, because the temperature of the yam increases the longer it sits in the oven.
(b) The units of f ′ (20) are ◦ F/min. The statement f ′ (20) = 2 means that at time t = 20 minutes, the temperature T
would increase by approximately 2◦ F if the yam is in the oven an additional minute.
4. (a) As the cup of coffee cools, the temperature decreases, so f ′ (t) is negative.
(b) Since f ′ (t) = dH/dt, the units are degrees Celsius per minute. The quantity f ′ (20) represents the rate at which the
coffee is cooling, in degrees per minute, 20 minutes after the cup is put on the counter.
5. (a) The function f takes quarts of ice cream to cost in dollars, so 200 is the amount of ice cream, in quarts, and $600 is
the corresponding cost, in dollars. It costs $600 to produce 200 quarts of ice cream.
(b) Here, 200 is in quarts, but the 2 is in dollars/quart. After producing 200 quarts of ice cream, the cost to produce one
additional quart is about $2.
6. (a) If the price is $150, then 2000 items will be sold.
(b) If the price goes up from $150 by $1 per item, about 25 fewer items will be sold. Equivalently, if the price is decreased
from $150 by $1 per item, about 25 more items will be sold.
7. Units of C ′ (r) are dollars/percent. Approximately, C ′ (r) means the additional amount needed to pay off the loan when
the interest rate is increased by 1%. The sign of C ′ (r) is positive, because increasing the interest rate will increase the
amount it costs to pay off a loan.
8. The units of f ′ (x) are feet/mile. The derivative, f ′ (x), represents the rate of change of elevation with distance from the
source, so if the river is flowing downhill everywhere, the elevation is always decreasing and f ′ (x) is always negative. (In
fact, there may be some stretches where the elevation is more or less constant, so f ′ (x) = 0.)
9. Units of P ′ (t) are dollars/year. The practical meaning of P ′ (t) is the rate at which the monthly payments change as the
duration of the mortgage increases. Approximately, P ′ (t) represents the change in the monthly payment if the duration is
increased by one year. P ′ (t) is negative because increasing the duration of a mortgage decreases the monthly payments.
10. Since B is measured in dollars and t is measured in years, dB/dt is measured in dollars per year. We can interpret dB
as the extra money added to your balance in dt years. Therefore dB/dt represents how fast your balance is growing, in
units of dollars/year.
11. (a) This means that investing the $1000 at 5% would yield $1649 after 10 years.
(b) Writing g ′ (r) as dB/dr, we see that the units of dB/dr are dollars per percent (interest). We can interpret dB as
|
≈ 165 means that
the extra money earned if interest rate is increased by dr percent. Therefore g ′ (5) = dB
dr r=5
the balance, at 5% interest, would increase by about $165 if the interest rate were increased by 1%. In other words,
g(6) ≈ g(5) + 165 = 1649 + 165 = 1814.
12. (a) The units of lapse rate are the same as for the derivative dT /dz, namely (units of T )/(units of z) = ◦ C/km.
(b) Since the lapse rate is 6.5, the derivative of T with respect to z is dT /dz = −6.5◦ C/km. The air temperature drops
about 6.5◦ for every kilometer you go up.
Problems
13. (a) Since W = f (c) where W is weight in pounds and c is the number of Calories consumed per day:
f (1800) = 155
means that
consuming 1800 Calories per day
results in a weight of 155 pounds.
f ′ (2000) = 0
means that
consuming 2000 Calories per day causes
neither weight gain nor loss.
f −1 (162) = 2200
means that
a weight of 162 pounds is caused by
a consumption of 2200 Calories per day.
(b) The units of dW/dc are pounds/(Calories/day).
2.4 SOLUTIONS
127
14. (a) Let f (t) be the volume, in cubic km, of the Greenland Ice Sheet t years since 2011 (Alternatively, in year t). We are
given information about f ′ (t), which has unit km3 per year.
(b) If t is in years since 2011, we know f ′ (0) is between −224 and −82 cubic km/year. (Alternatively, f ′ (2011) is
between −224 and −82.)
15. The graph is increasing for 0 < t < 15 and is decreasing for 15 < t < 30. One possible graph is shown in Figure 2.60.
The units on the horizontal axis are years and the units on the vertical axis are people.
people
f (t)
years
15
30
Figure 2.60
The derivative is positive for 0 < t < 15 and negative for 15 < t < 30. Two possible graphs are shown in
Figure 2.61. The units on the horizontal axes are years and the units on the vertical axes are people per year.
people/year
people/year
f ′ (t)
f ′ (t)
years
15
years
30
15
30
Figure 2.61
16. Since f (t) = 1.34(1.004)t , we have
f (9) = 1.34(1.004)9 = 1.389.
To estimate f ′ (9), we use a small interval around 9:
f ′ (9) ≈
1.34(1.004)9.001 − 1.34(1.004)9
f (9.001) − f (9)
=
= 0.0055.
9.001 − 9
0.001
We see that f (9) = 1.389 billion people and f ′ (9) = 0.0055 billion (that is, 5.5 million) people per year. Since t = 9 in
2020, this model predicts that the population of China will be about 1,389,000,000 people in 2009 and growing at a rate
of about 5,500,000 people per year at that time.
17. f (10) = 240,000 means that if the commodity costs $10, then 240,000 units of it will be sold. f ′ (10) = −29,000 means
that if the commodity costs $10 now, each $1 increase in price will cause a decline in sales of 29,000 units.
18. Let p be the rating points earned by the CBS Evening News, let R be the revenue earned in millions of dollars, and let
R = f (p). When p = 4.3,
Rate of change of revenue ≈
$5.5 million
= 55 million dollars/point.
0.1 point
Thus
f ′ (4.3) ≈ 55.
128
Chapter Two /SOLUTIONS
19. (a) The units of P are millions of people, the units of t are years, so the units of f ′ (t) are millions of people per year.
Therefore the statement f ′ (6) = 2 tells us that at t = 6 (that is, in 1986), the population of Mexico was increasing at
2 million people per year.
(b) The statement f −1 (95.5) = 16 tells us that the year when the population was 95.5 million was t = 16 (that is, in
1996).
(c) The units of (f −1 )′ (P ) are years per million of population. The statement (f −1 )′ (95.5) = 0.46 tells us that when
the population was 95.5 million, it took about 0.46 years for the population to increase by 1 million.
20. (a)
(b)
(c)
(d)
When t = 10, that is, at 10 am, 3.1 cm of rain has fallen.
We are told that when 5 cm of rain has fallen, 16 hours have passed (t = 16); that is, 5 cm of rain has fallen by 4 pm.
The rate at which rain is falling is 0.4 cm/hr at t = 10, that is, at 10 am.
The units of (f −1 )′ (5) are hours/cm. Thus, we are being told that when 5 cm of rain has fallen, rain is falling at a
rate such that it will take 2 additional hours for another centimeter to fall.
21. (a)
(b)
(c)
(d)
The depth of the water is 3 feet at time t = 5 hours.
The depth of the water is increasing at 0.7 feet/hour at time t = 5 hours.
When the depth of the water is 5 feet, the time is t = 7 hours.
Since 5 is the depth in feet and h−1 (5) is time in hours, the units of (h−1 )′ are hours/feet. Thus, (h−1 )′ (5) = 1.2
tells us that when the water depth is 5 feet, the rate of change of time with depth is 1.2 hours per foot. In other words,
when the depth is 5 feet, water is entering at a rate such that it takes 1.2 hours to add an extra foot of water.
The pressure in dynes/cm2 at a depth of 100 meters.
The depth of water in meters giving a pressure of 1.2 · 106 dynes/cm2 .
The pressure at a depth of h meters plus a pressure of 20 dynes/cm2 .
The pressure at a depth of 20 meters below the diver.
The rate of increase of pressure with respect to depth, at 100 meters, in units of dynes/cm2 per meter. Approximately,
p′ (100) represents the increase in pressure in going from 100 meters to 101 meters.
(f) The depth, in meters, at which the rate of change of pressure with respect to depth is 100,000 dynes/cm2 per meter.
22. (a)
(b)
(c)
(d)
(e)
23. The units of g ′ (t) are inches/year. The quantity g ′ (10) represents how fast Amelia Earhart was growing at age 10, so we
expect g ′ (10) > 0. The quantity g ′ (30) represents how fast she was growing at age 30, so we expect g ′ (30) = 0 because
she was probably not growing taller at that age.
24. Units of g ′ (55) are mpg/mph. The statement g ′ (55) = −0.54 means that at 55 miles per hour the fuel efficiency (in miles
per gallon, or mpg) of the car decreases at a rate of approximately one half mpg as the velocity increases by one mph.
25. Units of dP/dt are barrels/year. dP/dt is the change in quantity of petroleum per change in time (a year). This is negative.
We could estimate it by finding the amount of petroleum used worldwide over a short period of time.
26. (a)
velocity
terminal
velocity
t
(b) The graph should be concave down because air resistance decreases your acceleration as you speed up, and so the
slope of the graph of velocity is decreasing.
(c) The slope represents the acceleration due to gravity.
27. (a) The derivative, dW/dt, measures the rate of change of water in the bathtub in gallons per minute.
(b) (i) The interval t0 < t < tp represents the time before the plug is pulled. At that time, the rate of change of W is
0 since the amount of water in the tub is not changing.
(ii) Since dW/dt represents the rate at which the amount of water in the tub is changing, after the plug is pulled and
water is leaving the tub, the sign of dW/dt is negative.
(iii) Once all the water has drained from the tub, the amount of water in the tub is not changing, so dW/dt = 0.
28. (a) The company hopes that increased advertising always brings in more customers instead of turning them away. Therefore, it hopes f ′ (a) is always positive.
(b) If f ′ (100) = 2, it means that if the advertising budget is $100,000, each extra dollar spent on advertising will bring
in about $2 worth of sales. If f ′ (100) = 0.5, each dollar above $100 thousand spent on advertising will bring in
about $0.50 worth of sales.
2.4 SOLUTIONS
129
(c) If f ′ (100) = 2, then as we saw in part (b), spending slightly more than $100,000 will increase revenue by an amount
greater than the additional expense, and thus more should be spent on advertising. If f ′ (100) = 0.5, then the increase
in revenue is less than the additional expense, hence too much is being spent on advertising. The optimum amount
to spend is an amount that makes f ′ (a) = 1. At this point, the increases in advertising expenditures just pay for
themselves. If f ′ (a) < 1, too much is being spent; if f ′ (a) > 1, more should be spent.
29. (a) The derivative has units of people/second, so we find the rate of births, deaths, and migrations per second and combine
them.
1
people per second
8
1
Death rate =
people per second
13
1
Migration rate =
people per second
27
Birth rate =
Thus
1
1
1
−
+
= 0.0851 people/second.
8
13
27
In other words, the population is increasing at 0.0851 people per second.
(b) From the answer to part (a), we see that it took 1/0.0851 = 11.75 ≈ 12 seconds to add one person.
f ′ (0) = Rate of change of population =
30. Since O′ (2000) = −1, 25, we know the ODGI is decreasing at 1.25 units per year. To reduce the ODGI from 95 to 0 will
take 95/1.25 = 76 years. Thus the ozone hole is predicted to recover by 2076.
31. Since
P (67) − P (66)
≈ P ′ (66),
67 − 66
we may think of P ′ (66) as an estimate of P (67) − P (66), and the latter is the number of people between 66 and 67
inches tall. Alternatively, since
P (66.5) − P (65.5)
is a better estimate of P ′ (66),
66.5 − 65.5
we may regard P ′ (66) as an estimate of the number of people of height between 65.5 and 66.5 inches. The units for
P ′ (x) are people per inch. Since there are about 300 million people in the US, we guess that there are about 250 million
full-grown persons in the US whose heights are distributed between 60 inches (5 ft) and 75 inches (6 ft 3 in). There are
probably quite a few people of height 66 inches—between one and two times what we would expect from an even, or
uniform, distribution— because 66 inches is nearly average. An even distribution would yield
P ′ (66) =
250 million
≈ 17 million people per inch,
15 ins
so we expect P ′ (66) to be between 17 and 34 million people per inch.
The value of P ′ (x) is never negative because P (x) is never decreasing. To see this, let’s look at an example involving
a particular value of x, say x = 70. The value P (70) represents the number of people whose height is less than or equal
to 70 inches, and P (71) represents the number of people whose height is less than or equal to 71 inches. Since everyone
shorter than 70 inches is also shorter than 71 inches, P (70) ≤ P (71). In general, P (x) is 0 for small x, and increases as
x increases, and is eventually constant (for large enough x).
32. (a) The units of compliance are units of volume per units of pressure, or liters per centimeter of water.
(b) The increase in volume for a 5 cm reduction in pressure is largest between 10 and 15 cm. Thus, the compliance
appears maximum between 10 and 15 cm of pressure reduction. The derivative is given by the slope, so
Compliance ≈
0.70 − 0.49
= 0.042 liters per centimeter.
15 − 10
(c) When the lung is nearly full, it cannot expand much more to accommodate more air.
33. Solving for dp/dδ, we get
dp
=
dδ
p
δ + (p/c2 )
γ.
(a) For δ ≈ 10 g/cm3 , we have log δ ≈ 1, so, from Figure 2.38 in the text, we have γ ≈ 2.6 and log p ≈ 13.
130
Chapter Two /SOLUTIONS
Thus p ≈ 1013 , so p/c2 ≈ 1013 /(9 · 1020 ) ≈ 10−8 , and
1013
dp
≈
2.6 ≈ 2.6 · 1012 .
dδ
10 + 10−8
The derivative can be interpreted as the ratio between a change in pressure and the corresponding change in density.
The fact that it is so large says that a very large change in pressure brings about a very small change in density. This
says that cold iron is not a very compressible material.
(b) For δ ≈ 106 , we have log δ ≈ 6, so, from Figure 2.38 in the text, γ ≈ 1.5 and log p ≈ 23.
Thus p ≈ 1023 , so p/c2 ≈ 1023 /(9 · 1020 ) ≈ 102 , and
1023
dp
≈ 6
1.5 ≈ 1.5 · 1017 .
dδ
10 + 102
This tells us that the matter in a white dwarf is even less compressible than cold iron.
Strengthen Your Understanding
34. Since we are not given the units of either t or s we cannot conclude that the units of the derivative are meters/second.
35. Since air is leaking from the balloon, the radius of the balloon must be decreasing, so r ′ (t) < 0.
36. Since T has units of minutes, its derivative with respect to P will have units of minutes/page.
37. Let T (P ) be the time, in years, to repay a loan of P dollars, then the derivative dT /dP is given in years/dollar.
There are many other possible answers.
38. Let m = f (t) be the total distance, in miles, driven in a car, t days since it was purchased. Then the derivative dm/dt is
given in miles/day.
There are many other possible answers.
39. True. The two sides of the equation are different frequently used notations for the very same quantity, the derivative of f
at the point a.
40. True. The derivatives f ′ (t) and g ′ (t) measure the same thing, the rate of chemical production at the same time t, but they
measure it in different units. The units of f ′ (t) are grams per minute, and the units of g ′ (t) are kilograms per minute. To
convert from kg/min to g/min, multiply by 1000.
41. False. The derivatives f ′ (t) and g ′ (t) measure different things because they measure the rate of chemical production at
different times. There is no conversion possible from one to the other.
42. (b) and (e) (b), (e)
43. (b) and (d) are equivalent, with (d) containing the most information. Notice that (a) and (c) are wrong.
Solutions for Section 2.5
Exercises
1. (a) Increasing, concave up
(b) Decreasing, concave down
2. (a) Since the graph is below the x-axis at x = 2, the value of f (2) is negative.
(b) Since f (x) is decreasing at x = 2, the value of f ′ (2) is negative.
(c) Since f (x) is concave up at x = 2, the value of f ′′ (2) is positive.
3. At B both dy/dx and d2 y/dx2 could be positive because y is increasing and the graph is concave up there. At all the
other points one or both of the derivatives could not be positive.
4. The two points at which f ′ = 0 are A and B. Since f ′ is nonzero at C and D and f ′′ is nonzero at all four points, we get
the completed Table 2.6:
2.5 SOLUTIONS
131
Table 2.6
Point
f
f′
f ′′
A
−
0
+
+
0
C
+
−
D
−
−
B
+
−
+
5.
(a)
(b)
f (x)
f (x)
x
(c)
x
(d)
f (x)
f (x)
x
x
6. The graph must be everywhere decreasing and concave up on some intervals and concave down on other intervals. One
possibility is shown in Figure 2.62.
x
Figure 2.62
7. Since velocity is positive and acceleration is negative, we have f ′ > 0 and f ′′ < 0, and so the graph is increasing and
concave down. See Figure 2.63.
height
time
Figure 2.63
132
Chapter Two /SOLUTIONS
8. f ′ (x) = 0
f ′′ (x) = 0
9. f ′ (x) < 0
f ′′ (x) = 0
10. f ′ (x) > 0
f ′′ (x) > 0
11. f ′ (x) < 0
f ′′ (x) > 0
12. f ′ (x) > 0
f ′′ (x) < 0
13. f ′ (x) < 0
f ′′ (x) < 0
14. The velocity is the derivative of the distance, that is, v(t) = s′ (t). Therefore, we have
s(t + h) − s(t)
h
(5(t + h)2 + 3) − (5t2 + 3)
= lim
h→0
h
2
10th + 5h
= lim
h→0
h
h(10t + 5h)
= lim
= lim (10t + 5h) = 10t
h→0
h→0
h
v(t) = lim
h→0
The acceleration is the derivative of velocity, so a(t) = v ′ (t):
10(t + h) − 10t
h
10h
= lim
= 10.
h→0 h
a(t) = lim
h→0
Problems
15. (a) The derivative, f ′ (t), appears to be positive since the number of cars is increasing. The second derivative, f ′′ (t),
appears to be negative during the period 1975–1990 because the rate of change is increasing. For example, between
1975 and 1980, the rate of change is (121.6 − 106.7)/5 = 2.98 million cars per year, while between 1985 and 1990,
the rate of change is 1.16 million cars per year.
(b) The derivative, f ′ (t), appears to be negative between 1990 and 1995 since the number of cars is decreasing, but
increasing between 1995 and 2000. The second derivative, f ′′ (t), appears to be positive during the period 1990–
2000 because the rate of change is increasing. For example, between 1990 and 1995, the rate of change is (128.4 −
133.7)/5 = −1.06 million cars per year, while between 1995 and 2000, the rate of change is 1.04 million cars per
year.
(c) To estimate f ′ (2000) we consider the interval 2000–2005
f ′ (2005) ≈
f (2005) − f (2000)
136.6 − 133.6
3
≈
= = 0.6.
2005 − 2000
5
5
We estimate that f ′ (2005) ≈ 0.6 million cars per year. The number of passenger cars in the US was increasing at a
rate of about 600,000 cars per year in 2005.
16. To measure the average acceleration over an interval, we calculate the average rate of change of velocity over the interval.
The units of acceleration are ft/sec per second, or (ft/sec)/sec, written ft/sec2 .
Average acceleration
for 0 ≤ t ≤ 1
=
v(1) − v(0)
30 − 0
Change in velocity
=
=
= 30 ft/sec2
Time
1
1
Average acceleration
for 1 ≤ t ≤ 2
=
52 − 30
= 22 ft/sec2
2−1
2.5 SOLUTIONS
17.
(a)
(b)
(c)
18. Since the graph of this function is a line, the second derivative (of any linear function) is 0. See Figure 2.64.
y
4
x
−4
4
−4
Figure 2.64
19. See Figure 2.65.
y
4
x
−4
4
−4
Figure 2.65
20. See Figure 2.66.
y
x
−4
4
Figure 2.66
133
134
Chapter Two /SOLUTIONS
21. See Figure 2.67.
y
x
−4
4
Figure 2.67
22. See Figure 2.68.
y
x
−4
4
Figure 2.68
23. See Figure 2.69.
y
x
−4
4
Figure 2.69
24. (a) dP/dt > 0 and d2 P/dt2 > 0.
(b) dP/dt < 0 and d2 P/dt2 > 0 (but dP/dt is close to zero).
25. (a)
utility
quantity
2.5 SOLUTIONS
135
(b) As a function of quantity, utility is increasing but at a decreasing rate; the graph is increasing but concave down. So
the derivative of utility is positive, but the second derivative of utility is negative.
26. (a) Let N (t) be the number of people below the poverty line. See Figure 2.70.
N (t)
t
Figure 2.70
(b) dN/dt is positive, since people are still slipping below the poverty line. d2 N/dt2 is negative, since the rate at which
people are slipping below the poverty line, dN/dt, is decreasing.
27. (a) The EPA will say that the rate of discharge is still rising. The industry will say that the rate of discharge is increasing
less quickly, and may soon level off or even start to fall.
(b) The EPA will say that the rate at which pollutants are being discharged is leveling off, but not to zero—so pollutants
will continue to be dumped in the lake. The industry will say that the rate of discharge has decreased significantly.
28. (a)
(b)
(c)
(d)
(e)
(f)
At x4 and x5 , because the graph is below the x-axis there.
At x3 and x4 , because the graph is sloping down there.
At x3 and x4 , because the graph is sloping down there. This is the same condition as part (b).
At x2 and x3 , because the graph is bending downward there.
At x1 , x2 , and x5 , because the graph is sloping upward there.
At x1 , x4 , and x5 , because the graph is bending upward there.
29. (a)
(b)
(c)
(d)
(e)
At t3 , t4 , and t5 , because the graph is above the t-axis there.
At t2 and t3 , because the graph is sloping up there.
At t1 , t2 , and t5 , because the graph is concave up there
At t1 , t4 , and t5 , because the graph is sloping down there.
At t3 and t4 , because the graph is concave down there.
30. Since f ′ is everywhere positive, f is everywhere increasing. Hence the greatest value of f is at x6 and the least value of
f is at x1 . Directly from the graph, we see that f ′ is greatest at x3 and least at x2 . Since f ′′ gives the slope of the graph
of f ′ , f ′′ is greatest where f ′ is rising most rapidly, namely at x6 , and f ′′ is least where f ′ is falling most rapidly, namely
at x1 .
31. To the right of x = 5, the function starts by increasing, since f ′ (5) = 2 > 0 (though f may subsequently decrease) and
is concave down, so its graph looks like the graph shown in Figure 2.71. Also, the tangent line to the curve at x = 5 has
slope 2 and lies above the curve for x > 5. If we follow the tangent line until x = 7, we reach a height of 24. Therefore,
f (7) must be smaller than 24, meaning 22 is the only possible value for f (7) from among the choices given.
y
T
24
f (x)
20
x
5
Figure 2.71
7
136
Chapter Two /SOLUTIONS
32. (a) From the information given, C(1994) = 3200 ppt and C(2010) = 2750 ppt.
(b) Since the change has been approximately linear, the rate of change is constant:
C ′ (1994) = C ′ (2010) =
2750 − 3200
= −28.125 ppt per year .
2010 − 1994
(c) The slope is −28.125 and C(1994) = 3200.
If t is the year, we have
C(t) = 3200 − 28.125(t − 1994).
(d) We solve
1850 = 3200 − 28.125(t − 1994)
28.125(t − 1994) = 3200 − 1850
3200 − 1850
t = 1994 +
= 2042.
28.125
The CFC level in the atmosphere above the US is predicted to return to the original level in 2042.
(e) Since C ′′ (t) > 0, the graph bends upward, so the answer to part (d) is too early. The CFCs are expected to reach
their original level later than 2042.
Strengthen Your Understanding
33. A linear function is neither concave up nor concave down.
34. When the acceleration of a car is zero, the car is not speeding up or slowing down. This happens whenever the velocity is
constant. The car does not have to be stationary for this to happen.
35. One possibility is f (x) = b + ax, a 6= 0.
36. One possibility is f (x) = x2 . We have f ′ (x) = 2x, which is zero at x = 0 but f ′′ (x) = 2.
There are many other possible answers.
37. True. The second derivative f ′′ (x) is the derivative of f ′ (x). Thus the derivative of f ′ (x) is positive, and so f ′ (x) is
increasing.
38. True. Instantaneous acceleration is a derivative, and all derivatives are limits of difference quotients. More precisely,
instantaneous acceleration a(t) is the derivative of the velocity v(t), so
a(t) = lim
h→0
v(t + h) − v(t)
.
h
39. True.
40. True; f (x) = x3 is increasing over any interval.
41. False; f (x) = x2 is monotonic on intervals which do not contain the origin (unless the origin is an endpoint).
Solutions for Section 2.6
Exercises
1. (a) Function f is not continuous at x = 1.
(b) Function f appears not differentiable at x = 1, 2, 3.
2. (a) Function g appears continuous at all x-values shown.
(b) Function g appears not differentiable at x = 2, 4. At x = 2, the curve is vertical, so the derivative does not exist. At
x = 4, the graph has a corner, so the derivative does not exist.
3. No, there are sharp turning points.
4. Yes.
2.6 SOLUTIONS
137
Problems
5. Yes, f is differentiable at x = 0, since its graph does not have a “corner” at x = 0. See below.
1.64
−0.4
0.4
Another way to see this is by computing:
lim
h→0
2
(h + |h|)2
h2 + 2h|h| + |h|2
f (h) − f (0)
= lim
= lim
.
h→0
h→0
h
h
h
2
Since |h| = h , we have:
lim
h→0
f (h) − f (0)
2h2 + 2h|h|
= lim
= lim 2(h + |h|) = 0.
h→0
h→0
h
h
So f is differentiable at 0 and f ′ (0) = 0.
6. As we can see in Figure 2.72, f oscillates infinitely often between the x-axis and the line y = 2x near the origin. This
means a line from (0, 0) to a point (h, f (h)) on the graph of f alternates between slope 0 (when f (h) = 0) and slope 2
(when f (h) = 2h) infinitely often as h tends to zero. Therefore, there is no limit of the slope of this line as h tends to
zero, and thus there is no derivative at the origin. Another way to see this is by noting that
lim
h→0
does not exist, since
sin( h1 )
h sin( h1 ) + h
f (h) − f (0)
1
= lim
= lim sin
+1
h→0
h→0
h
h
h
does not have a limit as h tends to zero. Thus, f is not differentiable at x = 0.
y
y = f (x)
x
−2
3π
2
5π
y = 2x
Figure 2.72
7. We can see from Figure 2.73 that the graph of f oscillates infinitely often between the curves y = x2 and y = −x2 near
= h)
the origin. Thus the slope of the line from (0, 0) to (h, f (h)) oscillates between h (when f (h) = h2 and f (h)−0
h−0
= −h) as h tends to zero. So, the limit of the slope as h tends to zero is 0, which
and −h (when f (h) = −h2 and f (h)−0
h−0
is the derivative of f at the origin. Another way to see this is to observe that
lim
h→0
h2 sin( h1 )
h
1
= lim h sin( )
h→0
h
= 0,
f (h) − f (0)
= lim
h
h→0
since lim h = 0 and −1 ≤ sin( h1 ) ≤ 1 for any h. Thus f is differentiable at x = 0, and f ′ (0) = 0.
h→0
138
Chapter Two /SOLUTIONS
2
3π
2
− 3π
Figure 2.73
8. (a) The graph is concave up everywhere, except at x = 2 where the derivative is undefined. This is the case if the graph
has a corner at x = 2. One possible graph is shown in Figure 2.74.
(b) The graph is concave up for x < 2 and concave down for x > 2, and the derivative is undefined at x = 2. This is the
case if the graph is vertical at x = 2. One possible graph is shown in Figure 2.75.
f (x)
f (x)
x
x
2
2
Figure 2.74
Figure 2.75
9. We want to look at
(h2 + 0.0001)1/2 − (0.0001)1/2
.
h→0
h
As h → 0 from positive or negative numbers, the difference quotient approaches 0. (Try evaluating it for h = 0.001,
0.0001, etc.) So it appears there is a derivative at x = 0 and that this derivative is zero. How can this be if f has a corner
at x = 0?
The answer lies in the fact that what appears to be a corner is in fact smooth—when you zoom in, the graph of f
looks like a straight line with slope 0! See Figure 2.76.
lim
f (x)
f (x)
2
0.2
1
0.1
x
−2
−1
0
1
x
2
−0.2
−0.1
0
0.1
Figure 2.76: Close-ups of f (x) = (x2 + 0.0001)1/2 showing differentiability at x = 0
10. (a)
g
GM
R2
r
R
Figure 2.77
0.2
2.6 SOLUTIONS
139
(b) The graph certainly looks continuous. The only point in question is r = R. Using the second formula with r = R
gives
GM
.
g=
R2
Then, using the first formula with r approaching R from below, we see that as we get close to the surface of the earth
GM R
GM
=
.
R3
R2
g≈
Since we get the same value for g from both formulas, g is continuous.
GM
. For r > R, the graph of g looks like 1/x2 , and
(c) For r < R, the graph of g is a line with a positive slope of =
R3
so has a negative slope. Therefore the graph has a “corner” at r = R and so is not differentiable there.
11. (a) The graph of Q against t does not have a break at t = 0, so Q appears to be continuous at t = 0. See Figure 2.78.
Q
1
−2
−1
1
2
t
Figure 2.78
(b) The slope dQ/dt is zero for t < 0, and negative for all t > 0. At t = 0, there appears to be a corner, which does not
disappear as you zoom in, suggesting that I is defined for all times t except t = 0.
12. (a) Notice that B is a linear function of r for r ≤ r0 and a reciprocal for r > r0 . The constant B0 is the value of B at
r = r0 and the maximum value of B. See Figure 2.79.
B
B0
r
r0
Figure 2.79
(b) B is continuous at r = r0 because there is no break in the graph there. Using the formula for B, we have
lim B =
−
r→r0
r0
B0 = B0
r0
and
lim B =
+
r→r0
r0
B0 = B0 .
r0
(c) The function B is not differentiable at r = r0 because the graph has a corner there. The slope is positive for r < r0
and the slope is negative for r > r0 .
13. (a) Since
lim E = kr0
−
r→r0
and
lim E =
+
r→r0
kr02
= kr0
r0
and
E(r0 ) = kr0 ,
we see that E is continuous at r0 .
140
Chapter Two /SOLUTIONS
(b) The function E is not differentiable at r = r0 because the graph has a corner there. The slope is positive for r < r0
and the slope is negative for r > r0 .
(c) See Figure 2.80.
E
kr0
r
r0
Figure 2.80
14. (a) The graph of g(r) does not have a break or jump at r = 2, and so g(r) is continuous there. See Figure 2.81. This is
confirmed by the fact that
g(2) = 1 + cos(π2/2) = 1 + (−1) = 0
so the value of g(r) as you approach r = 2 from the left is the same as the value when you approach r = 2 from the
right.
2
g(r)
r
−2
2
Figure 2.81
(b) The graph of g(r) does not have a corner at r = 2, even after zooming in, so g(r) appears to be differentiable at
r = 0. This is confirmed by the fact that cos(πr/2) is at the bottom of a trough at r = 2, and so its slope is 0 there.
Thus the slope to the left of r = 2 is the same as the slope to the right of r = 2.
15. (a) The graph of φ does not have a break at y = 0, and so φ appears to be continuous there. See figure Figure 2.82.
φ
1
y
−2
−1
1
2
Figure 2.82
(b) The graph of φ has a corner at y = 0 which does not disappear as you zoom in. Therefore φ appears not be
differentiable at y = 0.
16. (a) The graph of
f (x) =
y
(
0 if x < 0.
x2 if x ≥ 0.
is shown to the right. The graph is continuous and has no vertical segments or corners,
so f (x) is differentiable everywhere.
y = f (x)
x
2.6 SOLUTIONS
141
By Example 4 on page 94,
′
f (x) =
(
y
0 if x < 0
2x if x ≥ 0
y = f ′ (x)
So its graph is shown to the right.
x
(b) The graph of the derivative has a corner at x = 0 so f ′ (x) is
not differentiable at x = 0. The graph of
f ′′ (x) =
(
y
y = f ′′ (x)
0 if x < 0
2
2 if x > 0
looks like:
x
The second derivative is not defined at x = 0. So it is
certainly neither differentiable nor continuous at x = 0.
Strengthen Your Understanding
17. There are several ways in which a function can fail to be differentiable at a point, one of which is because the graph has
a sharp corner at the point. Other cases are when the function is not continuous at a point or if the graph has a vertical
tangent line.
18. The converse of this statement is true. However, a function can be continuous and not differentiable at a point; for example,
f (x) = |x| is continuous but not differentiable at x = 0.
19. f (x) = |x − 2|. This is continuous but not differentiable at x = 2.
√
√
20. f (x) = x, x ≥ 0. This is invertible but f ′ (x) = 1/(2 x), which is not defined at x = 0.
21. Let
x2 − 1
.
x2 − 4
Since x2 − 1 = (x − 1)(x + 1), this function has zeros at x = ±1. However, at x = ±2, the denominator x2 − 4 = 0,
so f (x) is undefined and not differentiable.
f (x) =
22. True. Let f (x) = |x − 3|. Then f (x) is continuous for all x but not differentiable at x = 3 because its graph has a corner
there. Other answers are possible.
23. True. If a function is differentiable at a point, then it is continuous at that point. For example, f (x) = x2 is both differentiable and continuous on any interval. However, one example does not establish the truth of this statement; it merely
illustrates the statement.
24. False. Being continuous does not imply differentiability. For example, f (x) = |x| is continuous but not differentiable at
x = 0.
25. True. If a(function were differentiable, then it would be continuous. For example,
1 x≥0
is neither differentiable nor continuous at x = 0. However, one example does not establish the
−1 x < 0
truth of this statement; it merely illustrates the statement.
f (x) =
26. False. For example, f (x) = |x| is not differentiable at x = 0, but it is continuous at x = 0.
27. (a) This is not a counterexample, since it does not satisfy the conditions of the statement, and therefore does not have the
142
Chapter Two /SOLUTIONS
potential to contradict the statement.
(b) This contradicts the statement, because it satisfies its conditions but not its conclusion. Hence it is a counterexample.
Notice that this counterexample could not actually exist, since the statement is true.
(c) This is an example illustrating the statement; it is not a counterexample.
(d) This is not a counterexample, for the same reason as in part (a).
Solutions for Chapter 2 Review
Exercises
1. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the distance of the particle from a point, we find the values of s(3) = 72 and s(10) = 144. Using these values, we
find
∆s(t)
s(10) − s(3)
144 − 72
72
Average velocity =
=
=
=
= 10.286 cm/sec.
∆t
10 − 3
7
7
2. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the position of the particle, we find the values of s(3) = 12 · 3 − 32 = 27 and s(1) = 12 · 1 − 12 = 11. Using
these values, we find
Average velocity =
s(3) − s(1)
27 − 11
∆s(t)
=
=
= 8 mm/sec.
∆t
3−1
2
3. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the position of the particle, we find the values of s(3) = ln 3 and s(1) = ln 1. Using these values, we find
Average velocity =
s(3) − s(1)
∆s(t)
ln 3 − ln 1
ln 3
=
=
=
= 0.549 mm/sec.
∆t
3−1
2
2
4. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the position of the particle, we find the values of s(3) = 11 and s(1) = 3. Using these values, we find
∆s(t)
s(3) − s(1)
11 − 3
=
=
= 4 mm/sec.
∆t
3−1
2
Average velocity =
5. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the position of the particle, we find the values of s(3) = 4 and s(1) = 4. Using these values, we find
Average velocity =
s(3) − s(1)
4−4
∆s(t)
=
=
= 0 mm/sec.
∆t
3−1
2
Though the particle moves, its average velocity over the interval is zero, since it is at the same position at t = 1 and t = 3.
6. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the position of the particle, we find the values on the graph of s(3) = 2 and s(1) = 3. Using these values, we find
Average velocity =
s(3) − s(1)
∆s(t)
2−3
1
=
=
= − mm/sec.
∆t
3−1
2
2
7. The average velocity over a time period is the change in position divided by the change in time. Since the function s(t)
gives the position of the particle, we find the values on the graph of s(3) = 2 and s(1) = 2. Using these values, we find
Average velocity =
s(3) − s(1)
∆s(t)
2−2
=
=
= 0 mm/sec.
∆t
3−1
2
Though the particle moves, its average velocity over the interval is zero, since it is at the same position at t = 1 and t = 3.
SOLUTIONS to Review Problems for Chapter Two
143
8. (a) Let s = f (t).
(i) We wish to find the average velocity between t = 1 and t = 1.1. We have
Average velocity =
(ii) We have
Average velocity =
(iii) We have
f (1.1) − f (1)
7.84 − 7
=
= 8.4 m/sec.
1.1 − 1
0.1
f (1.01) − f (1)
7.0804 − 7
=
= 8.04 m/sec.
1.01 − 1
0.01
f (1.001) − f (1)
7.008004 − 7
=
= 8.004 m/sec.
1.001 − 1
0.001
(b) We see in part (a) that as we choose a smaller and smaller interval around t = 1 the average velocity appears to be
getting closer and closer to 8, so we estimate the instantaneous velocity at t = 1 to be 8 m/sec.
Average velocity =
9. See Figure 2.83.
distance
time
Figure 2.83
10. See Figure 2.84.
distance
t
Figure 2.84
11. (a) Figure 2.85 shows a graph of f (x) = x sin x.
y
y = x sin x
8
4
x
−10
−5
5
−4
Figure 2.85
10
144
Chapter Two /SOLUTIONS
(b) Seven, since x sin x = 0 at x = 0, ±π, ±2π, ±3π.
(c) From the graph, we see x sin x is increasing at x = 1, decreasing at x = 4.
(d) We calculate both average rates of change
f (2) − f (0)
2 sin 2 − 0
=
= sin 2 ≈ 0.91
(2 − 0)
2
8 sin 8 − 6 sin 6
f (8) − f (6)
=
≈ 4.80.
(8 − 6)
2
So the average rate of change over 6 ≤ x ≤ 8 is greater.
(e) From the graph, we see the slope is greater at x = −9.
12. (a) Using the difference quotient
f (0.8) − f (0.4)
0.5
=
= 1.25.
0.8 − 0.4
0.4
Substituting x = 0.6, we have y = 3.9, so the tangent line is y − 3.9 = 1.25(x − 0.6), that is y = 1.25x + 3.15.
(b) The equation from part (a) gives
f ′ (0.6) ≈
f (0.7) ≈ 1.25(0.7) + 3.15 = 4.025
f (1.2) ≈ 1.25(1.2) + 3.15 = 4.65
f (1.4) ≈ 1.25(1.4) + 3.15 = 4.9
The estimate for f (0.7) is likely to be reliable as 0.7 is close to 0.6 (and f (0.8) = 4, which is not too far off). The
estimate for f (1.2) is less reliable as 1.2 is outside the given data (from 0 to 1.0). The estimate for f (1.4) less reliable
still.
13. See Figure 2.86.
f ′ (x)
x
Figure 2.86
14. See Figure 2.87.
x
−1.5
f ′ (x)
Figure 2.87
15. See Figure 2.88.
1
−1
x
2
3
4
f ′ (x)
Figure 2.88
SOLUTIONS to Review Problems for Chapter Two
16. See Figure 2.89.
x
f ′ (x)
Figure 2.89
17. See Figure 2.90.
f ′ (x)
x
Figure 2.90
18. See Figure 2.91.
f ′ (x)
x
Figure 2.91
19. See Figure 2.92.
y
5
4
3
2
−1
1
x
−1
2
Figure 2.92
4
6
145
146
Chapter Two /SOLUTIONS
20. See Figure 2.93.
y
x
−4
4
Figure 2.93
21. See Figure 2.94.
y
x
−4
4
Figure 2.94
22. Using the definition of the derivative
f (x + h) − f (x)
h
5(x + h)2 + x + h − (5x2 + x)
lim
h→0
h
5(x2 + 2xh + h2 ) + x + h − 5x2 − x
lim
h→0
h
10xh + 5h2 + h
lim
h→0
h
lim (10x + 5h + 1) = 10x + 1
f ′ (x) = lim
h→0
=
=
=
=
h→0
23. Using the definition of the derivative, we have
n′ (x) = lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
n(x + h) − n(x)
h
h
i
1
1
1
+1 −
+1
h
x+h
x
1
1
1
−
h x+h
x
x − (x + h)
hx(x + h)
−h
hx(x + h)
−1
−1
= 2.
x(x + h)
x
SOLUTIONS to Review Problems for Chapter Two
147
24. We need to look at the difference quotient and take the limit as h approaches zero. The difference quotient is
[(3 + h)2 + 1] − 10
9 + 6h + h2 + 1 − 10
6h + h2
h(6 + h)
f (3 + h) − f (3)
=
=
=
=
.
h
h
h
h
h
Since h 6= 0, we can divide by h in the last expression to get 6 + h. Now the limit as h goes to 0 of 6 + h is 6, so
f ′ (3) = lim
h→0
h(6 + h)
= lim (6 + h) = 6.
h→0
h
So at x = 3, the slope of the tangent line is 6. Since f (3) = 32 + 1 = 10, the tangent line passes through (3, 10), so its
equation is
y − 10 = 6(x − 3), or y = 6x − 8.
25. By joining consecutive points we get a line whose slope is the average rate of change. The steeper this line, the greater
the average rate of change. See Figure 2.95.
(i) C and D. Steepest slope.
(ii) B and C. Slope closest to 0.
(b) A and B, and C and D. The two slopes are closest to each other.
(a)
D
E
C
k(x)
B
A
x
Figure 2.95
26. Using the definition of the derivative,
f (x + h) − f (x)
h
(3(x + h) − 1) − (3x − 1)
lim
h→0
h
3x + 3h − 1 − 3x + 1
lim
h→0
h
3h
lim
h→0 h
lim 3
f ′ (x) = lim
h→0
=
=
=
=
h→0
= 3.
27. Using the definition of the derivative,
f (x + h) − f (x)
h
(5(x + h)2 ) − (5x2 )
= lim
h→0
h
5(x2 + 2xh + h2 ) − 5x2
= lim
h→0
h
f ′ (x) = lim
h→0
148
Chapter Two /SOLUTIONS
5x2 + 10xh + 5h2 − 5x2
h→0
h
2
10xh + 5h
= lim
h→0
h
= lim (10x + 5h)
= lim
h→0
= 10x.
28. Using the definition of the derivative,
f (x + h) − f (x)
h
((x + h)2 + 4) − (x2 + 4)
lim
h→0
h
x2 + 2xh + h2 + 4 − x2 − 4
lim
h→0
h
2xh + h2
lim
h→0
h
lim (2x + h)
f ′ (x) = lim
h→0
=
=
=
=
h→0
= 2x.
29. Using the definition of the derivative,
f (x + h) − f (x)
h
(3(x + h)2 − 7) − (3x2 − 7)
lim
h→0
h
(3(x2 + 2xh + h2 ) − 7) − (3x2 − 7)
lim
h→0
h
3x2 + 6xh + 3h2 − 7 − 3x2 + 7
lim
h
h→0
6xh + 3h2
lim
h→0
h
lim (6x + 3h)
f ′ (x) = lim
h→0
=
=
=
=
=
h→0
= 6x.
30. Using the definition of the derivative,
f (x + h) − f (x)
h
(x + h)3 − x3
lim
h→0
h
x3 + 3x2 h + 3xh2 + h3 − x3
lim
h→0
h
3x2 h + 3xh2 + h3
lim
h→0
h
2
lim (3x + 3xh + h2 )
f ′ (x) = lim
h→0
=
=
=
=
h→0
= 3x2 .
(a + h)2 − a2
a2 + 2ah + h2 − a2
= lim
= lim (2a + h) = 2a
h→0
h→0
h→0
h
h
a − (a + h)
1
−1
−1
1
1
−
= lim
= 2
32. lim
= lim
h→0 (a + h)a
h→0 h
h→0 (a + h)ah
a+h
a
a
31. lim
33. lim
h→0
1
h
1
1
− 2
(a + h)2
a
= lim
h→0
(−2a − h)
−2
a2 − (a2 + 2ah + h2 )
= lim
= 3
(a + h)2 a2 h
h→0 (a + h)2 a2
a
SOLUTIONS to Review Problems for Chapter Two
√
√
√
√
√
149
√
( a + h − a)( a + h + a)
a+h−a
h
√
a=
= √
√
√ = √
√ .
a
+
h
+
a
a
+
h
+
a
a
+
h
+ a
√
√
1
1
a+h− a
= lim √
Therefore lim
√ = √
h→0
h→0
h
2 a
a+h+ a
√
√
35. We combine terms in the numerator and multiply top and bottom by a + a + h.
√
√
√
√
√
√
( a − a + h)( a + a + h)
a− a+h
1
1
√
√
−√ = √
= √
√
√ √
a
a+h
a+h a
a + h a( a + a + h)
a − (a + h)
√
= √
√ √
a + h a( a + a + h)
34.
a+h−
Therefore lim
h→0
1
h
√
1
1
−√
a
a+h
= lim √
h→0
−1
−1
√
= √ 3
√ √
2( a)
a + h a( a + a + h)
Problems
36. The function is everywhere increasing and concave up. One possible graph is shown in Figure 2.96.
x
Figure 2.96
37. First note that the line y = t has slope 1. From the graph, we see that
0 < Slope at C < Slope at B < Slope between A and B < 1 < Slope at A.
Since instantaneous velocity is represented by the slope at a point and average velocity is represented by the slope
between two points, we have
0 < Inst. vel. at C < Inst. vel. at B < Av. vel. between A and B < 1 < Inst. vel. at A.
38. (a)
(b)
(c)
(d)
The only graph in which the slope is 1 for all x is Graph (III).
The only graph in which the slope is positive for all x is Graph (III).
Graphs where the slope is 1 at x = 2 are Graphs (III) and (IV).
Graphs where the slope is 2 at x = 1 are Graphs (II) and (IV).
39. (a) Velocity is zero at points A, C, F , and H.
(b) These are points where the acceleration is zero, at which the particle switches from speeding up to slowing down or
vice versa.
40. (a) The derivative, f ′ (t), appears to be positive between 2003–2005 and 2006–2007, since the number of cars increased
in these intervals. The derivative, f ′ (t), appears to be negative from 2005–2006, since the number of cars decreased
then.
(b) We use the average rate of change formula on the interval 2005 to 2007 to estimate f ′ (2006):
f ′ (2006) ≈
−0.7
135.9 − 136.6
=
= −0.35.
2007 − 2005
2
We see that f ′ (2006) ≈ −0.35 million cars per year. The number of passenger cars in the US was decreasing at a
rate of about 0.35 million, or 350,000, cars per year in 2006.
41. (a) If f ′ (t) > 0, the depth of the water is increasing. If f ′ (t) < 0, the depth of the water is decreasing.
(b) The depth of the water is increasing at 20 cm/min when t = 30 minutes.
(c) We use 1 meter = 100 cm, 1 hour = 60 min. At time t = 30 minutes
Rate of change of depth = 20
cm 60 min
1m
cm
= 20
·
·
= 12 meters/hour.
min
min
1 hr
100 cm
150
Chapter Two /SOLUTIONS
42. Since f (t) = 45.7e−0.0061t , we have
f (6) = 45.7e−0.0061·6 = 44.058.
To estimate f (6), we use a small interval around 6:
′
f ′ (6) ≈
45.7e−0.0061·6.001 − 45.7e−0.0061·6
f (6.001) − f (6)
=
= −0.269.
6.001 − 6
0.001
We see that f (6) = 44.058 million people and f ′ (6) = −0.269 million (that is, −269,000) people per year. Since t = 6
in 2015, this model predicts that the population of Ukraine will be about 44,058,000 people in 2015 and declining at a
rate of about 269,000 people per year at that time.
43. (a) The units of R′ (3) are thousands of dollars per (dollar per gallon).
The derivative R′ (3) tells us the rate of change of revenue with price. That is, R′ (3) gives approximately how much
the revenue changes if the gas price increases by $1 per gallon from $3 per gallon.
(b) The units of R−1 (5) are dollars per gallon. Thus, the units of (R−1 )′ (5) are dollars/gallon per thousand dollars.
The derivative (R−1 )′ (5) tells us the rate of change of price with revenue. That is, (R−1 )′ (5) gives approximately
how much the price of gas changes if the revenue increases by $1000 from $5000 to $6000.
44. (a) A possible example is f (x) = 1/|x − 2| as lim 1/|x − 2| = ∞.
x→2
(b) A possible example is f (x) = −1/(x − 2)2 as lim −1/(x − 2)2 = −∞.
x→2
45. For x < −2, f is increasing and concave up. For −2 < x < 1, f is increasing and concave down. At x = 1, f has a
maximum. For x > 1, f is decreasing and concave down. One such possible f is in Figure 2.97.
y
x
−2 −1
1
Figure 2.97
46. Since f (2) = 3 and f ′ (2) = 1, near x = 2 the graph looks like the segment shown in Figure 2.98.
Slope = 1
3
x
2
Figure 2.98
(a) If f (x) is even, then the graph of f (x) near x = 2 and x = −2 looks like Figure 2.99. Thus f (−2) = 3 and
f ′ (−2) = −1.
(b) If f (x) is odd, then the graph of f (x) near x = 2 and x = −2 looks like Figure 2.100. Thus f (−2) = −3 and
f ′ (−2) = 1.
3
3
x
−2
x
−2
2
Figure 2.99: For f even
2
−3
Figure 2.100: For f odd
SOLUTIONS to Review Problems for Chapter Two
151
47. The slopes of the lines drawn through successive pairs of points are negative but increasing, suggesting that f ′′ (x) > 0
for 1 ≤ x ≤ 3.3 and that the graph of f (x) is concave up.
48. Using the approximation ∆y ≈ f ′ (x)∆x with ∆x = 2, we have ∆y ≈ f ′ (20) · 2 = 6 · 2, so
f (22) ≈ f (20) + f ′ (20) · 2 = 345 + 6 · 2 = 357.
49. (a)
6
Student B’s
5
✛ answer
= slope
Student C’s answer
=slope of this line
❘
of this line
✛ Student A’s answer
=slope of this line
4
x
1
2
3
4
5
6
(b) The slope of f appears to be somewhere between student A’s answer and student B’s, so student C’s answer, halfway
in between, is probably the most accurate.
(x)
(c) Student A’s estimate is f ′ (x) ≈ f (x+h)−f
, while student B’s estimate is f ′ (x) ≈ f (x)−fh(x−h) . Student C’s
h
estimate is the average of these two, or
f (x) − f (x − h)
1 f (x + h) − f (x)
+
f (x) ≈
2
h
h
′
=
f (x + h) − f (x − h)
.
2h
This estimate is the slope of the chord connecting (x − h, f (x − h)) to (x + h, f (x + h)). Thus, we estimate that
the tangent to a curve is nearly parallel to a chord connecting points h units to the right and left, as shown below.
50. (a) Since the point A = (7, 3) is on the graph of f , we have f (7) = 3.
(b) The slope of the tangent line touching the curve at x = 7 is given by
Slope =
3.8 − 3
0.8
Rise
=
=
= 4.
Run
7.2 − 7
0.2
Thus, f ′ (7) = 4.
51. At point A, we are told that x = 1 and f (1) = 3. Since A = (x2 , y2 ), we have x2 = 1 and y2 = 3. Since h = 0.1, we
know x1 = 1 − 0.1 = 0.9 and x3 = 1 + 0.1 = 1.1.
Now consider Figure 2.101. Since f ′ (1) = 2, the slope of the tangent line AD is 2. Since AB = 0.1,
BD
Rise
=
= 2,
Run
0.1
so BD = 2(0.1) = 0.2. Therefore y1 = 3 − 0.2 = 2.8 and y3 = 3 + 0.2 = 3.2.
152
Chapter Two /SOLUTIONS
y
3.2
D
A
3
✛
✲
2.8
B
✻0.2
❄
0.1
0.1 0.1
✛
✲
✛
✲
x
0.9 1 1.1
Figure 2.101
52. A possible graph of y = f (x) is shown in Figure 2.102.
y
−3
−5
5
−3
2.5
x
7
Figure 2.102
53. (a) Negative.
(b) dw/dt = 0 for t bigger than some t0 (the time when the fire stops burning).
(c) |dw/dt| increases, so dw/dt decreases since it is negative.
54. (a) The yam is cooling off so T is decreasing and f ′ (t) is negative.
(b) Since f (t) is measured in degrees Fahrenheit and t is measured in minutes, df /dt must be measured in units of
◦
F/min.
55. (a) The statement f (140) = 120 means that a patient weighing 140 pounds should receive a dose of 120 mg of the
painkiller. The statement f ′ (140) = 3 tells us that if the weight of a patient increases by one pound (from 140
pounds), the dose should be increased by about 3 mg.
(b) Since the dose for a weight of 140 lbs is 120 mg and at this weight the dose goes up by about 3 mg for one pound, a 145
lb patient should get about an additional 3(5) = 15 mg. Thus, for a 145 lb patient, the correct dose is approximately
f (145) ≈ 120 + 3(5) = 135 mg.
56. Suppose p(t) is the average price level at time t. Then, if t0 = April 1991,
“Prices are still rising” means p′ (t0 ) > 0.
“Prices rising less fast than they were” means p′′ (t0 ) < 0.
“Prices rising not as much less fast as everybody had hoped” means H < p′′ (t0 ), where H is the rate of change in rate of
change of prices that people had hoped for.
57. The rate of change of the US population is P ′ (t), so
P ′ (t) = 0.8% · Current population = 0.008P (t).
SOLUTIONS to Review Problems for Chapter Two
153
58. (a) See Figure 2.103.
2
x
1
5
−2
−4
Figure 2.103
(b) Exactly one. There can’t be more than one zero because f is increasing everywhere. There does have to be one zero
because f stays below its tangent line (dotted line in above graph), and therefore f must cross the x-axis.
(c) The equation of the (dotted) tangent line is y = 21 x − 21 , and so it crosses the x-axis at x = 1. Therefore the zero of
f must be between x = 1 and x = 5.
(d) lim f (x) = −∞, because f is increasing and concave down. Thus, as x → −∞, f (x) decreases, at a faster and
x→−∞
faster rate.
(e) Yes.
(f) No. The slope is decreasing since f is concave down, so f ′ (1) > f ′ (5), i.e. f ′ (1) > 21 .
f (0.8) − f (0.6)
4.0 − 3.9
=
= 0.5.
0.8 − 0.6
0.2
f (0.6) − f (0.4)
0.4
=
= 2.
0.6 − 0.4
0.2
′
′
f (0.6) − f (0.5)
0.5 − 2
−1.5
(b) Using the values of f ′ from part (a), we get f ′′ (0.6) ≈
=
=
= −15.
0.6 − 0.5
0.1
0.1
(c) The maximum value of f is probably near x = 0.8. The minimum value of f is probably near x = 0.3.
59. (a) f ′ (0.6) ≈
60. (a) Slope of tangent line = limh→0
tangent line is about 0.25.
(b)
√
√
4+h− 4
.
h
f ′ (0.5) ≈
√
Using h = 0.001,
√
4.001− 4
0.001
= 0.249984. Hence the slope of the
y − y1 = m(x − x1 )
y − 2 = 0.25(x − 4)
y − 2 = 0.25x − 1
y = 0.25x + 1
(c) f (x) = kx2
If (4, 2) is on the graph of f , then f (4) = 2, so k · 42 = 2. Thus k = 18 , and f (x) = 81 x2 .
(d) To find where the graph of f crosses then line y = 0.25x + 1, we solve:
1 2
x = 0.25x + 1
8
x2 = 2x + 8
x2 − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4 or x = −2
1
f (−2) = (4) = 0.5
8
Therefore, (−2, 0.5) is the other point of intersection. (Of course, (4, 2) is a point of intersection; we know that from
the start.)
√
61. (a) The slope of the tangent line at (0,
√ 19) is zero: it is horizontal.
The slope of the tangent line at ( 19, 0) is undefined: it is vertical.
(b) The slope appears to be about 12 . (Note that when x is 2, y is about −4, but when x is 4, y is approximately −3.)
154
Chapter Two /SOLUTIONS
4
2
−4
−2
2
4
−2
−4
(c) Using symmetry we can determine: Slope at (−2,
√
(2, 15): about − 21 .
√
15): about
1
.
2
√
Slope at (−2, − 15): about − 12 . Slope at
62. (a) IV, (b) III, (c) II, (d) I, (e) IV, (f) II
63. (a) The population varies periodically with a period of 12 months (i.e. one year).
5000
4000
April
(b)
(c)
(d)
(e)
3
6
9
12
15
18
21
24
July
Oct
Jan
April
July
Oct
Jan
April
The herd is largest about June 1st when there are about 4500 deer.
The herd is smallest about February 1st when there are about 3500 deer.
The herd grows the fastest about April 1st . The herd shrinks the fastest about July 15 and again about December 15.
It grows the fastest about April 1st when the rate of growth is about 400 deer/month, i.e about 13 new fawns per day.
64. (a) The graph looks straight because the graph shows only a small part of the curve magnified greatly.
(b) The month is March: We see that about the 21st of the month there are twelve hours of daylight and hence twelve
hours of night. This phenomenon (the length of the day equaling the length of the night) occurs at the equinox, midway
between winter and summer. Since the length of the days is increasing, and Madrid is in the northern hemisphere, we
are looking at March, not September.
(c) The slope of the curve is found from the graph to be about 0.04 (the rise is about 0.8 hours in 20 days or 0.04
hours/day). This means that the amount of daylight is increasing by about 0.04 hours (about 2 21 minutes) per calendar
day, or that each day is 2 12 minutes longer than its predecessor.
65. (a) A possible graph is shown in Figure 2.104. At first, the yam heats up very quickly, since the difference in temperature
between it and its surroundings is so large. As time goes by, the yam gets hotter and hotter, its rate of temperature
increase slows down, and its temperature approaches the temperature of the oven as an asymptote. The graph is thus
concave down. (We are considering the average temperature of the yam, since the temperature in its center and on its
surface will vary in different ways.)
temperature
200◦ C
20◦ C
time
Figure 2.104
SOLUTIONS to Review Problems for Chapter Two
155
(b) If the rate of temperature increase were to remain 2◦ /min, in ten minutes the yam’s temperature would increase 20◦ ,
from 120◦ to 140◦ . Since we know the graph is not linear, but concave down, the actual temperature is between 120◦
and 140◦ .
(c) In 30 minutes, we know the yam increases in temperature by 45◦ at an average rate of 45/30 = 1.5◦ /min. Since the
graph is concave down, the temperature at t = 40 is therefore between 120 + 1.5(10) = 135◦ and 140◦ .
(d) If the temperature increases at 2◦ /minute, it reaches 150◦ after 15 minutes, at t = 45. If the temperature increases at
1.5◦ /minute, it reaches 150◦ after 20 minutes, at t = 50. So t is between 45 and 50 mins.
66. (a) We construct the difference quotient using erf(0) and each of the other given values:
erf(1) − erf(0)
= 0.84270079
1−0
erf(0.1) − erf(0)
erf ′ (0) ≈
= 1.1246292
0.1 − 0
erf(0.01) − erf(0)
= 1.128342.
erf ′ (0) ≈
0.01 − 0
erf ′ (0) ≈
Based on these estimates, the best estimate is erf ′ (0) ≈ 1.12; the subsequent digits have not yet stabilized.
(b) Using erf(0.001), we have
erf(0.001) − erf(0)
erf ′ (0) ≈
= 1.12838
0.001 − 0
and so the best estimate is now 1.1283.
67. (a)
Table 2.7
x
sinh(x+0.001)−sinh(x)
0.001
sinh(x+0.0001)−sinh(x)
0.0001
so f ′ (0) ≈
cosh(x)
0
1.00000
1.00000
1.00000
1.00000
0.3
1.04549
1.04535
1.04535
1.04534
0.7
1.25555
1.25521
1.25521
1.25517
1
1.54367
1.54314
1.54314
1.54308
(b) It seems that they are approximately the same, i.e. the derivative of sinh(x) = cosh(x) for x = 0, 0.3, 0.7, and 1.
68. (a) Since the sea level is rising, we know that a′ (t) > 0 and m′ (t) > 0. Since the rate is accelerating, we know that
a′′ (t) > 0 and m′′ (t) > 0.
(b) The rate of change of sea level for the mid-Atlantic states is between 2 and 4, we know 2 < a′ (t) < 4. (Possibly also
a′ (t) = 2 or a′ (t) = 4.)
Similarly, 2 < m′ (t) < 10. (Possibly also m′ (t) = 2 or m′ (t) = 10.)
(c) (i) If a′ (t) = 2, then sea level rise = 2 · 100 = 200 mm.
If a′ (t) = 4, then sea level rise = 4 · 100 = 400 mm.
So sea level rise is between 200 mm and 400 mm.
(ii) The shortest amount of time for the sea level in the Gulf of Mexico to rise 1 meter occurs when the rate is largest,
10 mm per year. Since 1 meter = 1000 mm,
shortest time to rise 1 meter = 1000/10 = 100 years.
CAS Challenge Problems
69. The CAS says the derivative is zero. This can be explained by the fact that f (x) = sin2 x + cos2 x = 1, so f ′ (x) is the
derivative of the constant function 1. The derivative of a constant function is zero.
70. (a) The CAS gives f ′ (x) = 2 cos2 x − 2 sin2 x. Form of answers may vary.
(b) Using the double angle formulas for sine and cosine, we have
f (x) = 2 sin x cos x = sin(2x)
f ′ (x) = 2 cos2 x − 2 sin2 x = 2(cos2 x − sin2 x) = 2 cos(2x).
Thus we get
d
sin(2x) = 2 cos(2x).
dx
156
Chapter Two /SOLUTIONS
2
71. (a) The first derivative is g ′ (x) = −2axe−ax , so the second derivative is
g ′′ (x) =
4a2 x2
−2 a
d2 −ax2
= ax2 + ax2 .
e
2
dx
e
e
Form of answers may vary.
(b) Both graphs get narrow as a gets larger; the graph of g ′′ is below the x-axis along the interval where g is concave
down, and is above the x-axis where g is concave up. See Figure 2.105.
y
g ′′ (x)
y
y
3
2
1 g(x)
g ′′ (x)
3
2
1
g ′′ (x) 3
2
1
g(x)
✠
x
−2
2
x
−2
−2
2
−2
−2
−4
−4
−6
−6
a=1
a=2
g(x)
✠
x
2
−2
−4
−6
a=3
Figure 2.105
(c) The second derivative of a function is positive when the graph of the function is concave up and negative when it is
concave down.
72. (a) The CAS gives the same derivative, 1/x, in all three cases.
(b) From the properties of logarithms, g(x) = ln(2x) = ln 2 + ln x = f (x) + ln 2. So the graph of g is the same
shape as the graph of f , only shifted up by ln 2. So the graphs have the same slope everywhere, and therefore the two
functions have the same derivative. By the same reasoning, h(x) = f (x) + ln 3, so h and f have the same derivative
as well.
73. (a) The computer algebra system gives
d
(x2 + 1)2 = 4x(x2 + 1)
dx
d
(x2 + 1)3 = 6x(x2 + 1)2
dx
d
(x2 + 1)4 = 8x(x2 + 1)3
dx
(b) The pattern suggests that
d
(x2 + 1)n = 2nx(x2 + 1)n−1 .
dx
Taking the derivative of (x2 + 1)n with a CAS confirms this.
74. (a) Using a CAS, we find
d
sin x = cos x
dx
d
cos x = − sin x
dx
d
(sin x cos x) = cos2 x − sin2 x = 2 cos2 x − 1.
dx
(b) The product of the derivatives of sin x and cos x is cos x(− sin x) = − cos x sin x. On the other hand, the derivative
of the product is cos2 x − sin2 x, which is not the same. So no, the derivative of a product is not always equal to the
product of the derivatives.
PROJECTS FOR CHAPTER TWO
157
PROJECTS FOR CHAPTER TWO
1. (a) S(0) = 12 since the days are always 12 hours long at the equator.
(b) Since S(0) = 12 from part (a) and the formula gives S(0) = a, we have a = 12.
Since S(x) must be
π
=
12
+
b
continuous at x = x0 , and
the
formula
gives
S(x
)
=
a
+
b
arcsin(1)
0
2 and also S(x0 ) = 24,
π
π
24
we must have 12 + b 2 = 24 so b 2 = 12 and b = π ≈ 7.64.
(c) S(32◦ 13′ ) ≈ 14.12 and S(46◦ 4′ ) ≈ 15.58.
(d)
hours of sunlight
S(x)
24
18
12
6
30
60
90
x (◦ )
Figure 2.106
(e) The graph in Figure 2.106 appears to have a corner at x0 = 66◦ 30′ . We compare the slope to the right of
x0 and to the left of x0 . To the right of S0 , the function is constant, so S ′ (x) = 0 for x > 66◦ 30′ .
We estimate the slope immediately to the left of x0 . We want to calculate the following:
lim
h→0−
S(x0 + h) − S(x0 )
.
h
We approximate it by taking x0 = 66.5 and h = −0.1, − 0.01, − 0.001:
S(66.49) − S(66.5)
22.3633 − 24
≈
= 16.38,
−0.1
−0.1
S(66.499) − S(66.5)
23.4826 − 24
≈
= 51.83,
−0.01
−0.01
23.8370 − 24
S(66.4999) − S(66.5)
≈
= 163.9.
−0.001
−0.001
These approximations suggest that, for x0 = 66.5,
lim−
h→0
S(x0 + h) − S(x0 )
h
does not exist.
This evidence suggests that S(x) is not differentiable at x0 . A proof requires the techniques found in
Chapter 3.
2. (a) (i) Estimating derivatives using difference quotients (but other answers are possible):
92.0 − 76.0
P (1910) − P (1900)
=
= 1.6 million people per year
10
10
150.7 − 131.7
P (1950) − P (1940)
P ′ (1945) ≈
=
= 1.9 million people per year
10
10
281.4 − 248.7
P (2000) − P (1990)
=
= 3.27 million people per year
P ′ (2000) ≈
10
10
P ′ (1900) ≈
(ii) The population growth rate was at its greatest at some time between 1950 and 1960.
P (1960) − P (1950)
179.0 − 150.7
(iii) P ′ (1950) ≈
=
= 2.83 million people per year,
10
10
′
so P (1956) ≈ P (1950) + P (1950)(1956 − 1950) = 150.7 + 2.83(6) ≈ 167.7 million people.
158
Chapter Two /SOLUTIONS
(iv) If the growth rate between 2000 and 2010 was the same as the growth rate from 1990 to 2000, then
the total population should be about 314 million people in 2010.
(b) (i) f −1 (100) is the point in time when the population of the US was 100 million people (somewhere
between 1910 and 1920).
(ii) The derivative of f −1 (P ) at P = 100 represents the ratio of change in time to change in population,
and its units are years per million people. In other words, this derivative represents about how long it
took for the population to increase by 1 million, when the population was 100 million.
(iii) Since the population increased by 105.7 − 92.0 = 13.7 million people in 10 years, the average rate
of increase is 1.37 million people per year. If the rate is fairly constant in that period, the amount of
time it would take for an increase of 8 million people (100 million − 92.0 million) would be
8 million people
≈ 5.8 years ≈ 6 years
1.37 million people/year
Adding this to our starting point of 1910, we estimate that the population of the US reached 100
million around 1916, i.e. f −1 (100) ≈ 1916.
(iv) Since it took 10 years between 1910 and 1920 for the population to increase by 105.7 − 92.0 = 13.7
million people, the derivative of f −1 (P ) at P = 100 is approximately
10 years
= 0.73 years/million people
13.7 million people
(c) (i) Clearly the population of the US at any instant is an integer that varies up and down every few seconds
as a child is born, a person dies, or a new immigrant arrives. So f (t) has “jumps;” it is not a smooth
function. But these jumps are small relative to the values of f , so f appears smooth unless we zoom
in very closely on its graph (to within a few seconds).
Major land acquisitions such as the Louisiana Purchase caused larger jumps in the population,
but since the census is taken only every ten years and the territories acquired were rather sparsely
populated, we cannot see these jumps in the census data.
(ii) We can regard rate of change of the population for a particular time t as representing an estimate of
how much the population will increase during the year after time t.
(iii) Many economic indicators are treated as smooth, such as the Gross National Product, the Dow Jones
Industrial Average, volumes of trading, and the price of commodities like gold. But these figures only
change in increments, not continuously.
3.1 SOLUTIONS
CHAPTER THREE
Solutions for Section 3.1
Exercises
1. The derivative, f ′ (x), is defined as
f ′ (x) = lim
h→0
f (x + h) − f (x)
.
h
If f (x) = 7, then
f ′ (x) = lim
h→0
0
7−7
= lim = 0.
h→0 h
h
2. The definition of the derivative says that
f ′ (x) = lim
h→0
f (x + h) − f (x)
.
h
Therefore,
f ′ (x) = lim
h→0
[17(x + h) + 11] − [17x + 11]
17h
= lim
= 17.
h→0 h
h
3. The x is in the exponent and we haven’t learned how to handle that yet.
4. y ′ = 3x2 .
′
5. y = πx
π−1
(power rule)
.
(power rule)
6. y ′ = 12x11 .
7. y ′ = 11x10 .
8. y ′ = 11x−12 .
9. y ′ = −12x−13 .
10. y ′ = 3.2x2.2 .
11. y ′ = − 43 x−7/4 .
12. y ′ = 43 x1/3 .
13. y ′ = 34 x−1/4 .
dy
= 2x + 5.
14.
dx
15. f ′ (t) = 3t2 − 6t + 8.
16. f ′ (x) = −4x−5 .
1
17. Since g(t) = 5 = t−5 , we have g ′ (t) = −5t−6 .
t
1
18. Since f (z) = − 6.1 = −z −6.1 , we have f ′ (z) = −(−6.1)z −7.1 = 6.1z −7.1 .
z
1
dy
7
19. Since y = 7/2 = r −7/2 , we have
= − r −9/2 .
dx
2
r
√
1
dy
= x−1/2 .
20. Since y = x = x1/2 , we have
dx
2
21. f ′ (x) = 14 x−3/4 .
1
1
22. Since h(θ) = √
= θ−1/3 , we have h′ (θ) = − θ−4/3 .
3
3
θ
23. Since f (x) =
r
1
1
3
= 3/2 = x−3/2 , we have f ′ (x) = − x−5/2 .
x3
2
x
159
160
Chapter Three /SOLUTIONS
24. h(x) = ax · ln e = ax, so h′ (x) = a.
25. y ′ = 6x1/2 − 52 x−1/2 .
26. f ′ (t) = 6t − 4.
27. y ′ = 17 + 12x−1/2 .
28. y ′ = 2z −
1
.
2z 2
29. The power rule gives f ′ (x) = 20x3 −
2
.
x3
3 −1/2
w
2
31. y ′ = −12x3 − 12x2 − 6.
30. h′ (w) = 6w−4 +
32. y ′ = 15t4 − 52 t−1/2 −
33. y ′ = 6t −
36.
37.
38.
39. f (z) =
42.
43.
44.
45.
46.
2
.
t3
√
1
z
1
1
1
+ z −1 =
z + z −1 , so f ′ (z) =
1 − z −2 =
3
3
3
3
3
z2 − 1
.
z2
1
3
1 1
3 5
d
x 2 + x−1 + x− 2 = x− 2 − x−2 − x− 2 .
dx
2
2
√
y = √θθ − √1θ = θ − √1θ
1
y ′ = 2√
+ 2θ13/2 .
θ
√
t(1 + t)
t1/2 · 1 + t1/2 t
t1/2
t3/2
3
1
Since g(t) =
=
= 2 + 2 = t−3/2 + t−1/2 , we have g ′ (t) = − t−5/2 − t−3/2 .
2
2
t
t
t
t
2
2
2ax
3x2
′
+
−c
j (x) =
a
b
ax + b
ax
b
Since f (x) =
=
+ = a + bx−1 , we have f ′ (x) = −bx−2 .
x
x
x
a
b
a
ax + b
= x + , we have h′ (x) = .
Since h(x) =
c
c
c
c
Since 4/3, π, and b are all constants, we have
40. g ′ (x) =
41.
+
dy
3
1
x(x + 1) = x1/2 x + x1/2 · 1 = x3/2 + x1/2 , we have
= x1/2 + x−1/2 .
dx
2
2
√
dy
Since y = t3/2 (2 + t) = 2t3/2 + t3/2 t1/2 = 2t3/2 + t2 , we have dx
= 3t1/2 + 2t.
4
3
Since h(t) = + 2 = 3t−1 + 4t−2 , we have h′ (t) = −3t−2 − 8t−3 .
t
t
1
Since h(θ) = θ(θ−1/2 − θ−2 ) = θθ−1/2 − θθ−2 = θ1/2 − θ−1 , we have h′ (θ) = θ−1/2 + θ−2 .
2
y = x + x1 , so y ′ = 1 − x12 .
34. Since y =
35.
6
t3/2
7
.
t2
dV
4
8
= π(2r)b = πrb.
dr
3
3
47. Since w is a constant times q, we have dw/dq = 3ab2 .
48. Since a, b, and c are all constants, we have
dy
= a(2x) + b(1) + 0 = 2ax + b.
dx
49. Since a and b are constants, we have
dP
1
b
= 0 + b t−1/2 = √ .
dt
2
2 t
Problems
50. So far,√we can only take the derivative of powers of x and the sums of constant multiples of powers of x. Since we cannot
write x + 3 in this form, we cannot yet take its derivative.
3.1 SOLUTIONS
161
51. The x is in the exponent and we have not learned how to handle that yet.
52. g ′ (x) = πx(π−1) + πx−(π+1) , by the power and sum rules.
53. y ′ = 6x.
(power rule and sum rule)
54. We cannot write
′
3
55. y = −2/3z .
1
3x2 +4
as the sum of powers of x multiplied by constants.
(power rule and sum rule)
56.
y ′ = 3x2 − 18x − 16
5 = 3x2 − 18x − 16
0 = 3x2 − 18x − 21
0 = x2 − 6x − 7
0 = (x + 1)(x − 7)
x = −1 or x = 7.
When x = −1, y = 7; when x = 7, y = −209.
Thus, the two points are (−1, 7) and (7, −209).
57. Differentiating gives
f ′ (x) = 6x2 − 4x so f ′ (1) = 6 − 4 = 2.
Thus the equation of the tangent line is (y − 1) = 2(x − 1) or y = 2x − 1.
58. (a) We have f (2) = 8, so a point on the tangent line is (2, 8). Since f ′ (x) = 3x2 , the slope of the tangent is given by
m = f ′ (2) = 3(2)2 = 12.
Thus, the equation is
y − 8 = 12(x − 2)
y = 12x − 16.
or
(b) See Figure 3.1. The tangent line lies below the function f (x) = x3 , so estimates made using the tangent line are
underestimates.
30
y = x3
y = 12x − 16
20
10
x
−2
2
4
−10
Figure 3.1
59. To calculate the equation of the tangent line to y = f (x) = x2 + 3x − 5 at x = 2, we need to find the y-coordinate and
the slope at x = 2. The y-coordinate is
y = f (2) = 22 + 3(2) − 5 = 5,
so a point on the line is (2, 5). The slope is found using the derivative: f ′ (x) = 2x + 3. At the point x = 2, we have
Slope = f ′ (2) = 2(2) + 3 = 7.
The equation of the line is
y − 5 = 7(x − 2)
y = 7x − 9.
When we graph the function and the line together, the line y = 7x − 9 appears to lie tangent to the curve y = x2 + 3x − 5
at the point x = 2 as we expect.
162
Chapter Three /SOLUTIONS
60. The slope of the tangent line is the value of the first derivative at x = 2. Differentiating gives
d
dx
x3
4
−
2
3x
For x = 2,
f ′ (2) =
d 1 3 4 −1
x − x
dx 2
3
1
4
= · 3x2 − (−1)x−2
2
3
4
3
= x2 + 2 .
2
3x
=
4
1
3 2
(2) +
= 6 + = 6.333
2
3(2)2
3
and
4
2
23
−
= 4 − = 3.333.
2
3(2)
3
To find the y-intercept for the tangent line equation at the point (2, 3.333), we substitute in the general equation, y =
b + mx, and solve for b.
f (2) =
3.333 = b + 6.333(2)
−9.333 = b.
The tangent line has the equation
2
y = −9.333 + 6.333x.
61. The slopes of the tangent lines to y = x − 2x + 4 are given by y ′ = 2x − 2. A line through the origin has equation
y = mx. So, at the tangent point, x2 − 2x + 4 = mx where m = y ′ = 2x − 2.
x2 − 2x + 4 = (2x − 2)x
x2 − 2x + 4 = 2x2 − 2x
−x2 + 4 = 0
−(x + 2)(x − 2) = 0
x = 2, −2.
Thus, the points of tangency are (2, 4) and (−2, 12). The lines through these points and the origin are y = 2x and
y = −6x, respectively. Graphically, this can be seen in Figure 3.2.
y
y = x2 − 2x + 4
(−2, 12)
y = 2x
y = −6x
(2, 4)
x
Figure 3.2
62. Decreasing means f ′ (x) < 0:
′
f ′ (x) = 4x3 − 12x2 = 4x2 (x − 3),
so f (x) < 0 when x < 3 and x 6= 0. Concave up means f ′′ (x) > 0:
so f ′′ (x) > 0 when
f ′′ (x) = 12x2 − 24x = 12x(x − 2)
12x(x − 2) > 0
x < 0 or
So, both conditions hold for x < 0 or 2 < x < 3.
x > 2.
3.1 SOLUTIONS
163
63. The graph increases when dy/dx > 0:
dy
= 5x4 − 5 > 0
dx
5(x4 − 1) > 0 so
x4 > 1 so x > 1 or x < −1.
The graph is concave up when d2 y/dx2 > 0:
d2 y
= 20x3 > 0 so x > 0.
dx2
We need values of x where {x > 1 or x < −1} AND {x > 0}, which implies x > 1. Thus, both conditions hold for all
values of x larger than 1.
64.
f ′ (x) = 12x2 + 12x − 23 ≥ 1
12x2 + 12x − 24 ≥ 0
12(x2 + x − 2) ≥ 0
12(x + 2)(x − 1) ≥ 0.
65.
Hence x ≥ 1 or
x ≤ −2.
f ′ (x) = 3(2x − 5) + 2(3x + 8) = 12x + 1
f ′′ (x) = 12.
66. (a) We have p(x) = x2 − x. We see that p′ (x) = 2x − 1 < 0 when x < 12 . So p is decreasing when x < 21 .
(b) We have p(x) = x1/2 − x, so
p′ (x) =
1 −1/2
x
−1 < 0
2
1 −1/2
x
<1
2
−1/2
x
<2
1
1/2
x
>
2
1
x> .
4
Thus p(x) is decreasing when x > 14 .
(c) We have p(x) = x−1 − x, so
p′ (x) = −1x−2 − 1 < 0
−x−2 < 1
x−2 > −1,
which is always true where x−2 is defined since x−2 = 1/x2 is always positive. Thus p(x) is decreasing for x < 0
and for x > 0.
67. Since W is proportional to r 3 , we have W = kr 3 for some constant k. Thus, dW/dr = k(3r 2 ) = 3kr 2 . Thus, dW/dr
is proportional to r 2 .
68. Since f (t) = 700 − 3t2 , we have f (5) = 700 − 3(25) = 625 cm. Since f ′ (t) = −6t, we have f ′ (5) = −30 cm/year.
In the year 2010, the sand dune will be 625 cm high and eroding at a rate of 30 centimeters per year.
d
= dt
(1250 − 16t2 ) = −32t.
69. (a) Velocity v(t) = dy
dt
Since t ≥ 0, the ball’s velocity is negative. This is reasonable, since its height y is decreasing.
d
(b) Acceleration a(t) = dv
= dt
(−32t) = −32.
dt
So its acceleration is the negative constant −32.
164
Chapter Three /SOLUTIONS
(c) The ball hits the ground when its height y = 0. This gives
1250 − 16t2 = 0
t = ±8.84 seconds
We discard t = −8.84 because time t is nonnegative. So the ball hits the ground 8.84 seconds after its release, at
which time its velocity is
v(8.84) = −32(8.84) = −282.88 feet/sec = −192.84 mph.
70. (a) The average velocity between t = 0 and t = 2 is given by
Average velocity =
−4.9(22 ) + 25(2) + 3 − 3
33.4 − 3
f (2) − f (0)
=
=
= 15.2 m/sec.
2−0
2−0
2
(b) Since f ′ (t) = −9.8t + 25, we have
Instantaneous velocity = f ′ (2) = −9.8(2) + 25 = 5.4 m/sec.
(c) Acceleration is given f ′′ (t) = −9.8. The acceleration at t = 2 (and all other times) is the acceleration due to gravity,
which is −9.8 m/sec2 .
(d) We can use a graph of height against time to estimate the maximum height of the tomato. See Figure 3.3. Alternately,
we can find the answer analytically. The maximum height occurs when the velocity is zero and v(t) = −9.8t+25 = 0
when t = 2.6 sec. At this time the tomato is at a height of f (2.6) = 34.9. The maximum height is 34.9 meters.
height (m)
34.9
2.6
5.2
t (sec)
Figure 3.3
(e) We see in Figure 3.3 that the tomato hits ground at about t = 5.2 seconds. Alternately, we can find the answer
analytically. The tomato hits the ground when
f (t) = −4.9t2 + 25t + 3 = 0.
We solve for t using the quadratic formula:
t=
−25 ±
p
(25)2 − 4(−4.9)(3)
2(−4.9)
√
−25 ± 683.8
−9.8
t = −0.12 and t = 5.2.
t=
We use the positive values, so the tomato hits the ground at t = 5.2 seconds.
71. Recall that v = dx/dt. We want to find the acceleration, dv/dt, when x = 2. Differentiating the expression for v with
respect to t using the chain rule and substituting for v gives
dv
d 2
dx
=
(x + 3x − 2) ·
= (2x + 3)v = (2x + 3)(x2 + 3x − 2).
dt
dx
dt
Substituting x = 2 gives
Acceleration =
dv
dt
x=2
= (2(2) + 3)(22 + 3 · 2 − 2) = 56 cm/sec2 .
3.1 SOLUTIONS
165
72. (a) Given h(t) = f (t) − g(t), this means h′ (t) = f ′ (t) − g ′ (t), so
h(0) = f (0) − g(0) = 2000 − 1500 = 500
h′ (0) = f ′ (0) − g ′ (0) = 11 − 13.5
= −2.5.
This tells us that initially there are 500 more acre-feet of water in the first reservoir than the second, but that this
difference drops at an initial rate of 2.5 acre-feet per day.
(b) Assume h′ is constant, this means h′ (t) = −2.5. It also means that h is a linear function:
h(t) = Starting value + Rate of change × Time
|{z}
{z
|
h(0)
= 500 − 2.5t.
}
|
{z
}
h′ (0)
t
To find zeros of h, we write:
h(t) = 0
500 − 2.5t = 0
t = 200.
This tells us the difference in water level will be zero—that is, the water levels will be equal—after 200 days.
73. (a) We have
d
dg
1
d
2GM
= GM
= GM
r −2 = GM (−2)r −3 = − 3 .
2
dr
dr r
dr
r
(b) The derivative dg/dr is the rate of change of acceleration due to the pull of gravity with respect to distance. The
further away from the center of the earth, the weaker the pull of gravity is. So g is decreasing and therefore its
derivative, dg/dr, is negative.
(c) By part (a),
dg
dr
r=6400
=−
2GM
r3
r=6400
=−
2(6.67 × 10−20 )(6 × 1024 )
≈ −3.05 × 10−6 .
(6400)3
(d) Since the magnitude of dg/dr is small, the value of g is not changing much near r = 6400. It is reasonable to assume
that g is a constant near the surface of the earth.
74. (a) T = 2π
r
1
l
2π
dT
2π
= √ l 2 , so
= √
g
g
dl
g
1
2
1
l− 2
π
= p .
gl
dT
is positive, the period T increases as the length l increases.
dl
2
75. (a) A = πr
dA
= 2πr.
dr
(b) This is the formula for the circumference of a circle.
for small h. When h > 0, the numerator of the difference quotient denotes the area of the
(c) A′ (r) ≈ A(r+h)−A(r)
h
region contained between the inner circle (radius r) and the outer circle (radius r + h). See figure below. As h
approaches 0, this area can be approximated by the product of the circumference of the inner circle and the “width”
of the region, i.e., h. Dividing this by the denominator, h, we get A′ = the circumference of the circle with radius r.
(b) Since
✛
h
r
We can also think about the derivative of A as the rate of change of area for a small change in radius. If the radius
increases by a tiny amount, the area will increase by a thin ring whose area is simply the circumference at that radius
times the small amount. To get the rate of change, we divide by the small amount and obtain the circumference.
166
Chapter Three /SOLUTIONS
76. V = 43 πr 3 . Differentiating gives dV
= 4πr 2 = surface area of a sphere.
dr
V (r+h)−V (r)
is the volume between two spheres divided by the change in radius. FurtherThe difference quotient
h
more, when h is very small, the difference between volumes, V (r + h) − V (r), is like a coating of paint of depth h
applied to the surface of the sphere. The volume of the paint is about h · (Surface Area) for small h: dividing by h gives
back the surface area.
Thinking about the derivative as the rate of change of the function for a small change in the variable gives another
way of seeing the result. If you increase the radius of a sphere a small amount, the volume increases by a very thin layer
whose volume is the surface area at that radius multiplied by that small amount.
77. If f (x) = xn , then f ′ (x) = nxn−1 . This means f ′ (1) = n · 1n−1 = n · 1 = n, because any power of 1 equals 1.
78. Since f (x) = axn , f ′ (x) = anxn−1 . We know that f ′ (2) = (an)2n−1 = 3, and f ′ (4) = (an)4n−1 = 24. Therefore,
24
f ′ (4)
=
f ′ (2)
3
(an)4n−1
=
(an)2n−1
n−1
4
2
=8
2n−1 = 8, and thus n = 4.
Substituting n = 4 into the expression for f ′ (2), we get 3 = a(4)(8), or a = 3/32.
dy
d n
dy
= y. We first calculate
=
(x ) = nxn−1 . The
79. Yes. To see why, we substitute y = xn into the equation 13x
dx
dx
dx
differential equation becomes
13x(nxn−1 ) = xn
But 13x(nxn−1 ) = 13n(x · xn−1 ) = 13nxn , so we have
13n(xn ) = xn
This equality must hold for all x, so we get 13n = 1, so n = 1/13. Thus, y = x1/13 is a solution.
80. (a)
(x + h)−1 − x−1
d(x−1 )
1
1
1
= lim
= lim
−
h→0
h→0 h x + h
dx
h
x
1 x − (x + h)
1
−h
= lim
= lim
h→0 h
h→0 h
x(x + h)
x(x + h)
−1
−1
= lim
= 2 = −1x−2 .
h→0 x(x + h)
x
h
d(x−3 )
(x + h)−3 − x−3
= lim
h→0
dx
h
1
1
1
−
= lim
h→0 h
(x + h)3
x3
x3 − (x + h)3
x3 (x + h)3
−3hx2 − 3xh2 − h3
x3 (x + h)3
= lim
1
h
= lim
1
h
= lim
1
h
= lim
−3x2 − 3xh − h2
x3 (x + h)3
h→0
h→0
h→0
h→0
=
x3 − (x3 + 3hx2 + 3h2 x + h3 )
x3 (x + h)3
−3x2
= −3x−4 .
x6
(b) For clarity, let n = −k, where k is a positive integer. So xn = x−k .
d(x−k )
(x + h)−k − x−k
= lim
h→0
dx
h
i
3.1 SOLUTIONS
= lim
1
h
= lim
1
h
h→0
h→0
1
1
− k
(x + h)k
x
xk − (x + h)k
xk (x + h)k
167
terms involving h2 and higher powers of h
z }| {
1 xk − xk − khxk−1 − . . . − hk
= lim
h→0 h
xk (x + h)k
=
−k
−kxk−1
= k+1 = −kx−(k+1) = −kx−k−1 .
xk (x)k
x
81. (a) We see that f (1) = a · 1 = a, and the graph of x2 + 3 goes through the point (1, 4), so f (x) is continuous when
a = 4.
(b) No, f (x) does not have a derivative at (1, 4). There is a corner there. We can see this without a graph by noticing that
d 2
(x + 3) = 2x
dx
has a value of 2 at x = 1, but
d
(4x) = 4
dx
has a value of 4 at x = 1.
82. Since ax + b and x2 + 3 are each differentiable for all x, the function is differentiable on 0 < x < 2 except at x = 1. For
continuity, we must have a · 1 + b = 12 + 3.
For differentiability, the derivatives at x = 1 of the two pieces must be the same. Since
d
(ax + b) = a
dx
and
d 2
(x + 3) = 2x,
dx
at x = 1, we have a = 2 · 1 = 2. We can then solve for b:
2 · 1 + b = 4, so b = 2.
Strengthen Your Understanding
d 2
(x + a) = 2x for any constant a.
dx
84. The function can be written as f (x) = x−2 so the power rule gives f ′ (x) = −2x−3 = −2/x3 .
83. Since the derivative of a constant is zero,
85. One possible example is f (x) = x2 and g(x) = 3x. More generally, f (x) = x2 + c and g(x) = 3x + k work for any c
and k.
86. Any function of the form g(x) = x2 + c, where c is a positive constant works. One possibility is g(x) = x2 + 1.
87. If f (x) = 3x2 , we have f ′ (x) = 6x and f ′′ (x) = 6. Other answers are possible.
88. True. Since d(xn )/dx = nxn−1 , the derivative of a power function is a power function, so the derivative of a polynomial
is a polynomial.
89. False, since
d
dx
π
x2
=
d
−2π
πx−2 = −2πx−3 = 3 .
dx
x
90. True. The slope of f (x) + g(x) at x = 2 is the sum of the derivatives, f ′ (2) + g ′ (2) = 3.1 + 7.3 = 10.4.
91. True. Since f ′′ (x) > 0 and g ′′ (x) > 0 for all x, we have f ′′ (x) + g ′′ (x) > 0 for all x, which means that f (x) + g(x) is
concave up.
92. False. Let f (x) = 2x2 and g(x) = x2 . Then f (x) − g(x) = x2 , which is concave up for all x.
168
Chapter Three /SOLUTIONS
Solutions for Section 3.2
Exercises
1. f ′ (x) = 2ex + 2x.
2. y ′ = 10t + 4et .
3. Using the chain rule gives f ′ (x) = 5 ln(a)a5x .
4. f ′ (x) = 12ex + (ln 11)11x .
5. y ′ = 10x + (ln 2)2x .
6. f ′ (x) = (ln 2)2x + 2(ln 3)3x .
dy
= 4(ln 10)10x − 3x2 .
7.
dx
8. z ′ = (ln 4)ex .
1
33 − 32
dy
= (ln 3)3x −
(x ).
9.
dx
3
2
dy
2
= (ln 2)2x − 6x−4 .
10. Since y = 2x + 3 = 2x + 2x−3 , we have
x
dx
11. z ′ = (ln 4)2 4x .
12. f ′ (t) = (ln(ln 3))(ln 3)t .
dy
= 5 · 5t ln 5 + 6 · 6t ln 6
13.
dx
14. h′ (z) = (ln(ln 2))(ln 2)z .
15. f ′ (x) = exe−1 .
dy
16.
= π x ln π
dx
17. f ′ (x) = (ln π)π x .
18. This is the sum of an exponential function and a power function, so f ′ (x) = ln(π)π x + πxπ−1 .
19. Since e and k are constants, ek is constant, so we have f ′ (x) = (ln k)kx .
20. f (x) = e1+x = e1 · ex . Then, since e1 is just a constant,
f ′ (x) = e · ex = e1+x .
21. f (t) = et · e2 . Then, since e2 is just a constant, f ′ (t) =
′
22. f (θ) = ke
d
(et e2 )
dt
d t
= e2 dt
e = e2 et = et+2 .
kθ
23. y ′ (x) = ax ln a + axa−1
2
24. f ′ (x) = π 2 x(π −1) + (π 2 )x ln(π 2 )
1
d
x
(2x − x−1/3 + 3x − e) = 2 +
25. g ′ (x) =
4 + 3 ln 3.
dx
3
3x
26. f ′ (x) = 6x(ex − 4) + (3x2 + π)ex = 6xex − 24x + 3x2 ex + πex .
Problems
27. y ′ = 2x + (ln 2)2x .
1
28. y ′ = 12 x− 2 − ln 12 ( 21 )x =
1
√
2 x
+ ln 2( 21 )x .
29. We can take the derivative of the sum x2 + 2x , but not the product.
30. f (s) = 5s es = (5e)s , so f ′ (s) = ln(5e) · (5e)s = (1 + ln 5)5s es .
31. Since y = e5 ex , y ′ = e5 ex = ex+5 .
32. y = e5x = (e5 )x , so y ′ = ln(e5 ) · (e5 )x = 5e5x .
33. The exponent is x2 , and we haven’t learned what to do about that yet.
√ √
34. f ′ (z) = (ln 4)( 4)z = (ln 2)2z .
3.2 SOLUTIONS
35. We can’t use our rules if the exponent is
√
169
θ.
36. This is the composition of two functions each of which we can take the derivative of, but we don’t know how to take the
derivative of the composition.
37. Once again, this is a product of two functions, 2x and x1 , each of which we can take the derivative of; but we don’t know
how to take the derivative of the product.
38. The derivative is
P ′ (t) = 300(ln 1.044)(1.044)t
so
P ′ (5) = 300(ln 1.044)(1.044)5 = 16.021.
The value
P ′ (5) = 16.021
means that when t = 5, the population is increasing by approximately 16 animals per year.
39. Since P = 1 · (1.05)t ,
dP
dt
= ln(1.05)1.05t . When t = 10,
dP
= (ln 1.05)(1.05)10 ≈ $0.07947/year ≈ 7.95c//year.
dt
40. (a) Substituting t = 4 gives V (4) = 25(0.85)4 = 25(0.522) = 13.050. Thus the value of the car after 4 years is
$13,050.
(b) We have a function of the form f (t) = Cat . We know that such functions have a derivative of the form (C ln a) · at .
Thus, V ′ (t) = (25 ln 0.85) · (0.85)t = −4.063(0.85)t . The units are the change in value (in thousands of dollars)
with respect to time (in years), or thousands of dollars/year.
(c) Substituting t = 4 gives V ′ (4) = −4.063(0.85)4 = −4.063(0.522) = −2.121. This means that at the end of the
fourth year, the value of the car is decreasing by $2121 per year.
(d) The function V (t) is positive and decreasing, so that the value of the automobile is positive and decreasing. The
function V ′ (t) is negative, and its magnitude is decreasing, meaning the value of the automobile is always dropping,
but the yearly loss of value decreases as time goes on. The graphs of V (t) and V ′ (t) confirm that the value of the car
decreases with time. What they do not take into account are the costs associated with owning the vehicle. At some
time, t, it is likely that the yearly costs of owning the vehicle will outweigh its value. At that time, it may no longer
be worthwhile to keep the car.
41. For t in years since 2009, the population of Mexico is given by the formula
M = 111(1 + 0.0113)t = 111(1.0113)t million
and that of the US by
U = 307(1 + 0.00975)t = 307(1.00975)t million,
The rate of change of each population, in people/year is given by
dM
dt
dU
dt
and
Since
dU
dt
>
t=0
= 111
t=0
= 307
t=0
dM
dt
d
(1.0113)t
dt
= 111(1.0113)t ln(1.0113)
t=0
d
(1.00975)t
dt
= 1.247 million people per year
t=0
= 307(1.00975)t ln(1.00975)
t=0
= 2.979 million people per year.
t=0
, the population of the US was growing faster in 2009.
t=0
42. Differentiating gives
Rate of change of price =
dV
= 75(1.35)t ln 1.35 ≈ 22.5(1.35)t dollar/yr.
dt
43. (a) f (x) = 1 − ex crosses the x-axis where 0 = 1 − ex , which happens when ex = 1, so x = 0. Since f ′ (x) = −ex ,
f ′ (0) = −e0 = −1.
(b) y = −x
(c) The negative of the reciprocal of −1 is 1, so the equation of the normal line is y = x.
170
Chapter Three /SOLUTIONS
44. Since y = 2x , y ′ = (ln 2)2x . At (0, 1), the tangent line has slope ln 2 so its equation is y = (ln 2)x + 1. At c, y = 0, so
0 = (ln 2)c + 1, thus c = − ln12 .
45.
g(x) = ax2 + bx + c
f (x) = ex
′
g (x) = 2ax + b
f ′ (x) = ex
g ′′ (x) = 2a
f ′′ (x) = ex
So, using g ′′ (0) = f ′′ (0), etc., we have 2a = 1, b = 1, and c = 1, and thus g(x) =
Figure 3.4.
1 2
x
2
+ x + 1, as shown in
ex
1 2
x
2
+x+1
x
Figure 3.4
The two functions do look very much alike near x = 0. They both increase for large values of x, but ex increases
much more quickly. For very negative values of x, the quadratic goes to ∞ whereas the exponential goes to 0. By choosing
a function whose first few derivatives agreed with the exponential when x = 0, we got a function which looks like the
exponential for x-values near 0.
46. The first and second derivatives of ex are ex . Thus, the graph of y = ex is concave up. The tangent line at x = 0 has
slope e0 = 1 and equation y = x + 1. A graph that is always concave up is always above any of its tangent lines. Thus
ex ≥ x + 1 for all x, as shown in Figure 3.5.
y y = ex
y = x+1
(0, 1)
x
Figure 3.5
47. For x = 0, we have y = a0 = 1 and y = 1 + 0 = 1, so both curves go through the point (0, 1) for all values of a.
Differentiating gives
d(ax )
dx
d(1 + x)
dx
= ax ln a|x=0 = a0 ln a = ln a
x=0
= 1.
x=0
The graphs are tangent at x = 0 if
ln a = 1
so
a = e.
d(ax )
is positive and when it is negative. The quantity ax is always positive.
48. We are interested in when the derivative
dx
However ln a > 0 for a > 1 and ln a < 0 for 0 < a < 1. Thus the function ax is increasing for a > 1 and decreasing
for a < 1.
Strengthen Your Understanding
49. The function is an exponential function so the power rule does not apply. The derivative of f is f ′ (x) = (ln 2)2x .
3.3 SOLUTIONS
171
50. Since π and e are constants, f is a constant function, so f ′ (x) = 0.
51. The derivative of f (x) = ax is f ′ (x) = (ln a)ax which is negative when ln a < 0. Since ln a < 0 when 0 < a < 1, we
see that 0.5x is such a function. There are many other possible examples.
52. A possibility is f (x) = ex . Then f ′ (x) = ex , f ′′ (x) = ex , and f ′′′ (x) = ex , so f ′′′ (x) = f (x).
53. False. If f (x) = ln x, then f ′ (x) = 1/x, which is decreasing for x > 0.
54. False, since f ′ (x) = f (x) = ex for all x.
55. False. If f (x) = |x|, then f (x) is not differentiable at x = 0 and f ′ (x) does not exist at x = 0.
Solutions for Section 3.3
Exercises
1. By the product rule, f ′ (x) = 2x(x3 + 5) + x2 (3x2 ) = 2x4 + 3x4 + 10x = 5x4 + 10x. Alternatively, f ′ (x) =
(x5 + 5x2 )′ = 5x4 + 10x. The two answers should, and do, match.
2. Using the product rule,
f ′ (x) = (ln 2)2x 3x + (ln 3)2x 3x = (ln 2 + ln 3)(2x · 3x ) = ln(2 · 3)(2 · 3)x = (ln 6)6x
or, since 2x · 3x = (2 · 3)x = 6x ,
f ′ (x) = (6x )′ = (ln 6)(6x ).
The two answers should, and do, match.
3. f ′ (x) = x · ex + ex · 1 = ex (x + 1).
4. y ′ = 2x + x(ln 2)2x = 2x (1 + x ln 2).
√
1
5. y ′ = 2√
2x + x(ln 2)2x .
x
dy
= 2tet + (t2 + 3)et = et (t2 + 2t + 3).
dt
1
1
1 − 12
1
′
2
x
x
x
2
2
2
√
7. f (x) = (x − x ) · 3 (ln 3) + 3 2x − x
= 3 (ln 3)(x − x ) + 2x −
.
2
2 x
6.
8. y ′ = (3t2 − 14t)et + (t3 − 7t2 + 1)et = (t3 − 4t2 − 14t + 1)et .
ex (1 − x)
1−x
ex · 1 − x · ex
=
=
.
9. f ′ (x) =
x
2
(e )
(ex )2
ex
50xex − 25x2 ex
50x − 25x2
=
.
2x
e
ex
t
t
t
1 · 2 − (t + 1)(ln 2)2
2 (1 − (t + 1) ln 2)
1 − (t + 1) ln 2
dy
=
=
=
11.
dx
(2t )2
(2t )2
2t
10. g ′ (x) =
3.2w2.2 − w3.2 (ln 5)
3.2w2.2 (5w ) − (ln 5)(w3.2 )5w
=
.
2w
5
5w
15r + 6 − 15r
6
3(5r + 2) − 3r(5)
=
=
13. q ′ (r) =
(5r + 2)2
(5r + 2)2
(5r + 2)2
(t + 4) − (t − 4)
8
14. g ′ (t) =
=
.
(t + 4)2
(t + 4)2
3(5t + 2) − (3t + 1)5
dz
15t + 6 − 15t − 5
1
15.
=
=
=
.
dt
(5t + 2)2
(5t + 2)2
(5t + 2)2
12. g ′ (w) =
16. z ′ =
(2t + 5)(t + 3) − (t2 + 5t + 2)
t2 + 6t + 13
=
.
(t + 3)2
(t + 3)2
d
17. f (t) =
dt
′
1
2te − √
t
t
= 2et + 2tet +
1
.
2t3/2
172
Chapter Three /SOLUTIONS
18. Divide and then differentiate
3
x
3
f ′ (x) = 1 − 2 .
x
f (x) = x +
19. w = y 2 − 6y + 7. w′ = 2y − 6, y 6= 0.
1
√
−2
√
t−1/2 = √
20. g ′ (t) = −4(3 + t)−2
2
t(3 + t)2
21.
d
dz
22.
z2 + 1
√
z
=
√
1
d 23
3 1
1 3
z
(z + z − 2 ) = z 2 − z − 2 =
(3 − z −2 ).
dz
2
2
2
(−3)(5 + 3z) − (5 − 3z)(3)
dw
=
dz
(5 + 3z)2
−30
−15 − 9z − 15 + 9z
=
=
(5 + 3z)2
(5 + 3z)2
d
23. h (r) =
dr
′
r2
2r + 1
=
2r(r + 1)
(2r)(2r + 1) − 2r 2
=
.
(2r + 1)2
(2r + 1)2
24. Notice that you can cancel a z out of the numerator and denominator to get
f (z) =
3z
,
5z + 7
z 6= 0
Then
(5z + 7)3 − 3z(5)
(5z + 7)2
15z + 21 − 15z
=
(5z + 7)2
21
=
, z 6= 0.
(5z + 7)2
f ′ (z) =
[If you used the quotient rule correctly without canceling the z out first, your answer should simplify to this one, but
it is usually a good idea to simplify as much as possible before differentiating.]
17ex (2x )(1 − ln 2)
17ex (1 − ln 2)
17ex (2x ) − (ln 2)(17ex )2x
=
=
.
25. w′ (x) =
2x
2x
2
2
2x
2
2
2p(3 + 2p ) − 4p(1 + p )
6p + 4p3 − 4p − 4p3
2p
26. h′ (p) =
=
=
.
(3 + 2p2 )2
(3 + 2p2 )2
(3 + 2p2 )2
(x + 2)(x + 1)
x2 + 3x + 2
can be written as f (x) =
which reduces to f (x) = x + 2,
x+1
x+1
′
giving f (x) = 1, or use the quotient rule which gives
27. Either notice that f (x) =
f ′ (x) =
(x + 1)(2x + 3) − (x2 + 3x + 2)
(x + 1)2
=
2x2 + 5x + 3 − x2 − 3x − 2
(x + 1)2
=
x2 + 2x + 1
(x + 1)2
(x + 1)2
(x + 1)2
= 1.
=
3.3 SOLUTIONS
173
28. We use the quotient rule. We have
f ′ (x) =
(cx + k)(a) − (ax + b)(c)
acx + ak − acx − bc
ak − bc
=
=
.
(cx + k)2
(cx + k)2
(cx + k)2
29. Using the product and chain rules, we have
dy
= 3(x2 + 5)2 (2x)(3x3 − 2)2 + (x2 + 5)3 [2(3x3 − 2)(9x2 )]
dx
= 3(2x)(x2 + 5)2 (3x3 − 2)[(3x3 − 2) + (x2 + 5)(3x)]
= 6x(x2 + 5)2 (3x3 − 2)[6x3 + 15x − 2].
30. f ′ (x) =
d
(2 − 4x − 3x2 )(6xe − 3π) = (−4 − 6x)(6xe − 3π) + (2 − 4x − 3x2 )(6exe−1 ).
dx
Problems
31. Using the product rule, we know that h′ (x) = f ′ (x) · g(x) + f (x) · g ′(x). We use slope to compute the derivatives. Since
f (x) is linear on the interval 0 < x < 2, we compute the slope of the line to see that f ′ (x) = 2 on this interval. Similarly,
we compute the slope on the interval 2 < x < 4 to see that f ′ (x) = −2 on the interval 2 < x < 4. Since f (x) has a
corner at x = 2, we know that f ′ (2) does not exist.
Similarly, g(x) is linear on the interval shown, and we see that the slope of g(x) on this interval is −1 so we have
g ′ (x) = −1 on this interval.
(a) We have h′ (1) = f ′ (1) · g(1) + f (1) · g ′ (1) = 2 · 3 + 2(−1) = 6 − 2 = 4.
(b) Since f (x) has a corner at x = 2, we know that f ′ (2) does not exist. Therefore, we cannot use the product rule (and,
in this case, h′ (2) does not exist).
(c) We have h′ (3) = f ′ (3) · g(3) + f (3) · g ′ (3) = (−2)1 + 2(−1) = −2 − 2 = −4.
32. Using the quotient rule, we know that k′ (x) = (f ′ (x) · g(x) − f (x) · g ′ (x))/(g(x))2 , when f and g are differentiable.
We use slope to compute the derivatives. Since f (x) is linear on the interval 0 < x < 2, we compute the slope of the line
to see that f ′ (x) = 2 on this interval. Similarly, we compute the slope on the interval 2 < x < 4 to see that f ′ (x) = −2
on the interval 2 < x < 4. Since f (x) has a corner at x = 2, we know that f ′ (2) does not exist.
Similarly, g(x) is linear on the interval shown, and we see that the slope of g(x) on this interval is −1 so we have
g ′ (x) = −1 on this interval.
(a) We have
2 · 3 − 2(−1)
6+2
8
f ′ (1) · g(1) − f (1) · g ′ (1)
=
=
= .
(g(1))2
32
9
9
(b) Since f (x) has a corner at x = 2, we know that the quotient rule does not apply. We know that f ′ (2) does not exist,
and k is not differentiable at x = 2.
(c) We have
f ′ (3) · g(3) − f (3) · g ′ (3)
(−2)1 − 2(−1)
−2 + 2
k′ (3) =
=
=
= 0.
(g(3))2
12
1
k′ (1) =
33. Using the quotient rule, we know that j ′ (x) = (g ′ (x) · f (x) − g(x) · f ′ (x))/(f (x))2 . We use slope to compute the
derivatives. Since f (x) is linear on the interval 0 < x < 2, we compute the slope of the line to see that f ′ (x) = 2 on this
interval. Similarly, we compute the slope on the interval 2 < x < 4 to see that f ′ (x) = −2 on the interval 2 < x < 4.
Since f (x) has a corner at x = 2, we know that f ′ (2) does not exist.
Similarly, g(x) is linear on the interval shown, and we see that the slope of g(x) on this interval is −1 so we have
g ′ (x) = −1 on this interval.
(a) We have
(−1)2 − 3 · 2
g ′ (1) · f (1) − g(1) · f ′ (1)
−2 − 6
−8
=
=
=
= −2.
(f (1))2
22
4
4
(b) Since f (x) has a corner at x = 2, so the quotient rule does not apply. We know that f ′ (2) does not exist, so j ′ (2)
does not exist.
(c) We have
(−1)2 − 1(−2)
g ′ (3) · f (3) − g(3) · f ′ (3)
−2 + 2
=
=
= 0.
j ′ (3) =
(f (3))2
22
4
j ′ (1) =
174
Chapter Three /SOLUTIONS
34. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ′ (1) ≈ 0.5, g(1) ≈ 2, and g ′ (1) ≈ 1. By the product
rule,
h′ (1) = f ′ (1) · g(1) + f (1) · g ′ (1) ≈ (0.5)2 + (−0.4)1 = 0.6.
35. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ′ (1) ≈ 0.5, g(1) ≈ 2, and g ′ (1) ≈ 1. By the quotient
rule, to one decimal place
k′ (1) =
(0.5)2 − (−0.4)1
f ′ (1) · g(1) − f (1) · g ′ (1)
≈
= 0.4.
(g(1))2
22
36. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ′ (2) ≈ 1.1, g(2) ≈ 1.6, and g ′ (2) ≈ −0.5. By the
product rule, to one decimal place
h′ (2) = f ′ (2) · g(2) + f (2) · g ′ (2) ≈ 1.1(1.6) + 0.3(−0.5) = 1.6.
37. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ′ (2) ≈ 1.1, g(2) ≈ 1.6, and g ′ (2) ≈ −0.5. By the
quotient rule, to one decimal place
k′ (2) =
1.1(1.6) − 0.3(−0.5)
f ′ (2) · g(2) − f (2) · g ′ (2)
≈
= 0.7.
(g(2))2
(1.6)2
38. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ′ (1) ≈ 0.5, g(1) ≈ 2, and g ′ (1) ≈ 1. By the quotient
rule, to one decimal place
l′ (1) =
g ′ (1) · f (1) − g(1) · f ′ (1)
1(−0.4) − 2(0.5)
≈
= −8.8.
(f (1))2
(−0.4)2
39. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ′ (2) ≈ 1.1, g(2) ≈ 1.6, and g ′ (2) ≈ −0.5. By the
quotient rule, to one decimal place
l′ (2) =
(−0.5)0.3 − 1.6(1.1)
g ′ (2) · f (2) − g(2) · f ′ (2)
≈
= −21.2.
(f (2))2
(0.3)2
40.
1
et
et · 0 − et · 1
f ′ (t) =
(et )2
−1
= t = −e−t .
e
f (t) =
41. f (x) = ex · ex
f ′ (x) = ex · ex + ex · ex = 2e2x .
42.
f (x) = ex e2x
f ′ (x) = ex (e2x )′ + (ex )′ e2x
= 2ex e2x + ex e2x (from Problem 41)
= 3e3x .
43. We have
f ′ (x) = ex + xex
f ′′ (x) = ex + ex + xex = (2 + x)ex .
Since f (x) is concave up when f ′′ (x) > 0, we see that f (x) is concave up when x > −2.
3.3 SOLUTIONS
175
44. Using the quotient rule, we have
g ′ (x) =
−2x
0 − 1(2x)
= 2
(x2 + 1)2
(x + 1)2
g ′′ (x) =
−2(x2 + 1)2 + 2x(4x3 + 4x)
(x2 + 1)4
=
−2(x2 + 1)2 + 8x2 (x2 + 1)
(x2 + 1)4
=
−2(x2 + 1) + 8x2
(x2 + 1)3
=
2(3x2 − 1)
.
(x2 + 1)3
Since (x2 + 1)3 > 0 for all x, we have g ′′ (x) < 0 if (3x2 − 1) < 0, or when
3x2 < 1
1
1
−√ < x < √ .
3
3
45. Since f (0) = −5/1 = −5, the tangent line passes through the point (0, −5), so its vertical intercept is −5. To find the
slope of the tangent line, we find the derivative of f (x) using the quotient rule:
f ′ (x) =
7
(x + 1) · 2 − (2x − 5) · 1
=
.
(x + 1)2
(x + 1)2
At x = 0, the slope of the tangent line is m = f ′ (0) = 7. The equation of the tangent line is y = 7x − 5.
46. This is the same function we were asked to differentiate in Problem 24, so we know that, if x 6= 0,
f ′ (x) =
So at x = 1,
21
.
(5x + 7)2
3
1
= ,
12
4
7
21
′
=
.
y =
144
48
y = f (1) =
So,
1
7
=
(x − 1).
4
48
7
5
x+
.
y=
48
48
y−
47. (a) Although the answer you would get by using the quotient rule is equivalent, the answer looks simpler in this case if
you just use the product rule:
d
dx
d
dx
d
dx
ex
x
ex
x2
ex
x3
=
d
1
ex ·
dx
x
=
d
1
ex · 2
dx
x
=
d
1
ex · 3
dx
x
=
ex
ex
− 2
x
x
=
2ex
ex
−
x2
x3
=
3ex
ex
− 4 .
3
x
x
d ex
ex
nex
= n − n+1
n
dx x
x
x
.
48. By the product rule, we have
(b)
d 2
d 2
(x f (x)) =
(x ) · f (x) + x2 · f ′ (x) = 2xf (x) + x2 f ′ (x).
dx
dx
176
Chapter Three /SOLUTIONS
49. By the product rule, we have
d x
d
d x
(4 (f (x) + g(x))) =
(4 ) · (f (x) + g(x)) + 4x ·
(f (x) + g(x))
dx
dx
dx
x
x
′
= (ln 4 · 4 )(f (x) + g(x)) + 4 (f (x) + g ′ (x))
= 4x (ln 4 · f (x) + ln 4 · g(x) + f ′ (x) + g ′ (x)).
50. By the quotient rule, we have
d
dx
f (x)
g(x) + 1
=
d
f ′ (x)(g(x) + 1) − f (x) · dx
(g(x) + 1)
f ′ (x)g(x) + f ′ (x) − f (x)g ′ (x)
=
2
(g(x) + 1)
(g(x) + 1)2 .
51. By the quotient rule, we have
d
dx
f (x)g(x)
h(x)
=
d
(f (x)g(x))
dx
· h(x) − (f (x)g(x)) · h′ (x)
(h(x))2
(f ′ (x)g(x) + f (x)g ′ (x)) · h(x) − f (x)g(x)h′ (x)
(h(x))2
′
f (x)g(x)h(x) + f (x)g ′ (x)h(x) − f (x)g(x)h′ (x)
=
(h(x))2 .
=
52. (a) We have h′ (2) = f ′ (2) + g ′ (2) = 5 − 2 = 3.
(b) We have h′ (2) = f ′ (2)g(2) + f (2)g ′ (2) = 5(4) + 3(−2) = 14.
f ′ (2)g(2) − f (2)g ′ (2)
5(4) − 3(−2)
26
13
(c) We have h′ (2) =
=
=
=
.
(g(2))2
42
16
8
′
′
′
53. (a) G (z) = F (z)H(z) + H (z)F (z), so
G′ (3) = F ′ (3)H(3) + H ′ (3)F (3) = 4 · 1 + 3 · 5 = 19.
4(1) − 3(5)
F ′ (w)H(w) − H ′ (w)F (w)
, so G′ (3) =
= −11.
(b) G′ (w) =
[H(w)]2
12
54. Use the product rule and rewrite the expression as
f ′ (x)g(x) + f (x)g ′ (x) − g(x) + 4f ′ (x).
At x = 3, we have
Using the given values,
f ′ (3)g(3) + f (3)g ′ (3) − g(3) + 4f ′ (3).
4
1
1
· 12 + 6 · − 12 + 4 · = 6 + 8 − 12 + 2 = 4.
2
3
2
55. f ′ (x) = 10x9 ex + x10 ex is of the form g ′ h + h′ g, where
g(x) = x10 , g ′ (x) = 10x9
and
h(x) = ex , h′ (x) = ex .
Therefore, using the product rule, let f = g · h, with g(x) = x10 and h(x) = ex . Thus
f (x) = x10 ex .
56. (a) f (140) = 15,000 says that 15,000 skateboards are sold when the cost is $140 per board.
f ′ (140) = −100 means that if the price is increased from $140, roughly speaking, every dollar of increase will
decrease the total sales by 100 boards.
3.3 SOLUTIONS
(b)
177
dR
d
d
=
(p · q) =
(p · f (p)) = f (p) + pf ′ (p).
dp
dp
dp
So,
dR
dp
= f (140) + 140f ′ (140)
p=140
= 15,000 + 140(−100) = 1000.
(c) From (b) we see that
dR
dp
= 1000 > 0. This means that the revenue will increase by about $1000 if the price
p=140
is raised by $1.
57. We want dR/dr1 . Solving for R:
1
1
1
r2 + r1
r1 r2
=
+
=
, which gives R =
.
R
r1
r2
r1 r2
r2 + r1
So, thinking of r2 as a constant and using the quotient rule,
r2 (r2 + r1 ) − r1 r2 (1)
r22
dR
=
=
.
dr1
(r2 + r1 )2
(r1 + r2 )2
58. (a) If the museum sells the painting and invests the proceeds P (t) at time t, then t years have elapsed since 2000, and
the time span up to 2020 is 20 − t. This is how long the proceeds P (t) are earning interest in the bank. Each year the
money is in the bank it earns 5% interest, which means the amount in the bank is multiplied by a factor of 1.05. So,
at the end of (20 − t) years, the balance is given by
B(t) = P (t)(1 + 0.05)20−t = P (t)(1.05)20−t .
(b)
B(t) = P (t)(1.05)20 (1.05)−t = (1.05)20
P (t)
.
(1.05)t
(c) By the quotient rule,
B ′ (t) = (1.05)20
So,
B ′ (10) = (1.05)20
P ′ (t)(1.05)t − P (t)(1.05)t ln 1.05
.
(1.05)2t
5000(1.05)10 − 150,000(1.05)10 ln 1.05
(1.05)20
= (1.05)10 (5000 − 150,000 ln 1.05)
≈ −3776.63.
59. Note first that f (v) is in
(a) g(v) =
1
.
f (v)
(This
km
liters
, and v is in hour
.
km
km
is in liter .) Differentiating
gives
g ′ (v) =
So,
g(80) =
g ′ (80) =
(b) h(v) = v · f (v). (This is in
km
hour
·
−f ′ (v)
.
(f (v))2
1
km
.
= 20 liter
0.05
−0.0005
1 km
= − liter
for each 1 km
increase in speed.
hr
(0.05)2
5
liters
km
=
liters
.)
hour
Differentiating gives
h′ (v) = f (v) + v · f ′ (v),
so
.
h(80) = 80(0.05) = 4 liters
hr
for each 1 km
increase in speed.
h′ (80) = 0.05 + 80(0.0005) = 0.09 liters
hr
hr
178
Chapter Three /SOLUTIONS
(c) Part (a) tells us that at 80 km/hr, the car can go 20 km on 1 liter. Since the first derivative evaluated at this velocity
is negative, this implies that as velocity increases, fuel efficiency decreases, i.e., at higher velocities the car will not
go as far on 1 liter of gas. Part (b) tells us that at 80 km/hr, the car uses 4 liters in an hour. Since the first derivative
evaluated at this velocity is positive, this means that at higher velocities, the car will use more gas per hour.
60. Assume for g(x) 6= f (x), g ′ (x) = g(x) and g(0) = 1. Then for
h(x) =
h′ (x) =
g(x)
ex
ex (g ′ (x) − g(x))
g ′ (x) − g(x)
g ′ (x)ex − g(x)ex
=
=
.
(ex )2
(ex )2
ex
But, since g(x) = g ′ (x), h′ (x) = 0, so h(x) is constant. Thus, the ratio of g(x) to ex is constant. Since
g(0)
1
= = 1,
e0
1
g(x)
must equal 1 for all x. Thus g(x) = ex = f (x) for all x, so f and g are the same function.
ex
61. (a) f ′ (x) = (x − 2) + (x − 1).
(b) Think of f as the product of two factors, with the first as (x − 1)(x − 2). (The reason for this is that we have already
differentiated (x − 1)(x − 2)).
f (x) = [(x − 1)(x − 2)](x − 3).
Now f ′ (x) = [(x − 1)(x − 2)]′ (x − 3) + [(x − 1)(x − 2)](x − 3)′
Using the result of a):
f ′ (x) = [(x − 2) + (x − 1)](x − 3) + [(x − 1)(x − 2)] · 1
= (x − 2)(x − 3) + (x − 1)(x − 3) + (x − 1)(x − 2).
(c) Because we have already differentiated (x − 1)(x − 2)(x − 3), rewrite f as the product of two factors, the first being
(x − 1)(x − 2)(x − 3):
f (x) = [(x − 1)(x − 2)(x − 3)](x − 4)
Now f ′ (x) = [(x − 1)(x − 2)(x − 3)]′ (x − 4) + [(x − 1)(x − 2)(x − 3)](x − 4)′ .
f ′ (x) = [(x − 2)(x − 3) + (x − 1)(x − 3) + (x − 1)(x − 2)](x − 4)
+[(x − 1)(x − 2)(x − 3)] · 1
= (x − 2)(x − 3)(x − 4) + (x − 1)(x − 3)(x − 4)
+(x − 1)(x − 2)(x − 4) + (x − 1)(x − 2)(x − 3).
From the solutions above, we can observe that when f is a product, its derivative is obtained by differentiating each
factor in turn (leaving the other factors alone), and adding the results.
62. From the answer to Problem 61, we find that
f ′ (x) = (x − r1 )(x − r2 ) · · · (x − rn−1 ) · 1
+(x − r1 )(x − r2 ) · · · (x − rn−2 ) · 1 · (x − rn )
+(x − r1 )(x − r2 ) · · · (x − rn−3 ) · 1 · (x − rn−1 )(x − rn )
+ · · · + 1 · (x − r2 )(x − r3 ) · · · (x − rn )
1
1
1
.
+
+··· +
= f (x)
x − r1
x − r2
x − rn
63. (a) We can approximate
d
[F (x)G(x)H(x)]
dx
using the large rectangular solids by which our original cube is increased:
Volume of whole − volume of original solid = change in volume.
F (x + h)G(x + h)H(x + h) − F (x)G(x)H(x) = change in volume.
3.3 SOLUTIONS
179
The volume of this slab is F ′ (x)G(x)H(x)h
G(x)
G′ (x)h
F ′ (x)h
F (x)
H ′ (x)h
H(x)
As in the book, we will ignore the smaller regions which are added (the long, thin rectangular boxes and the
small cube in the corner.) This can be justified by recognizing that as h → 0, these volumes will shrink much faster
than the volumes of the big slabs and will therefore be insignificant. (Note that these smaller regions have an h2 or
h3 in the formulas of their volumes.) Then we can approximate the change in volume above by:
F (x + h)G(x + h)H(x + h) − F (x)G(x)H(x) ≈ F ′ (x)G(x)H(x)h (top slab)
+ F (x)G′ (x)H(x)h (front slab)
+ F (x)G(x)H ′(x)h
(other slab).
Dividing by h gives
F (x + h)G(x + h)H(x + h) − F (x)G(x)H(x)
h
≈ F ′ (x)G(x)H(x) + F (x)G′ (x)H(x) + F (x)G(x)H ′(x).
Letting h → 0
(F GH)′ = F ′ GH + F G′ H + F GH ′ .
(b) Verifying,
d
[(F (x) · G(x)) · H(x)] = (F · G)′ (H) + (F · G)(H)′
dx
= [F ′ G + F G′ ]H + F GH ′
= F ′ GH + F G′ H + F GH ′
as before.
(c) From the answer to (b), we observe that the derivative of a product is obtained by differentiating each factor in turn
(leaving the other factors alone), and adding the results. So, in general,
(f1 · f2 · f3 · . . . · fn )′ = f1′ f2 f3 · · · fn + f1 f2′ f3 · · · fn + · · · + f1 · · · fn−1 fn′ .
180
Chapter Three /SOLUTIONS
64. (a) Since x = a is a double zero of a polynomial P (x), we can write P (x) = (x − a)2 Q(x), so P (a) = 0. Using the
product rule, we have
P ′ (x) = 2(x − a)Q(x) + (x − a)2 Q′ (x).
Substituting in x = a, we see P ′ (a) = 0 also.
(b) Since P (a) = 0, we know x = a is a zero of P , so that x − a is a factor of P and we can write
P (x) = (x − a)Q(x),
where Q is some polynomial. Differentiating this expression for P using the product rule, we get
P ′ (x) = Q(x) + (x − a)Q′ (x).
Since we are told that P ′ (a) = 0, we have
P ′ (a) = Q(a) + (a − a)Q′ (a) = 0
and so Q(a) = 0. Therefore x = a is a zero of Q, so again we can write
Q(x) = (x − a)R(x),
where R is some other polynomial. As a result,
P (x) = (x − a)Q(x) = (x − a)2 R(x),
so that x = a is a double zero of P .
65. Using the product rule
d
d
d
d2
(f (x)g(x)) =
(f (x)g(x)) =
f ′ (x)g(x) + f (x)g ′ (x)
2
dx
dx dx
dx
= f ′′ (x)g(x) + f ′ (x)g ′ (x) + f ′ (x)g ′ (x) + f (x)g ′′ (x)
= f ′′ (x)g(x) + 2f ′ (x)g ′ (x) + f (x)g ′′ (x).
Strengthen Your Understanding
66. The derivative of x2 ex is not the product of the derivatives of the factors. Applying the product rule gives
df
d 2 x
d
=
(x )e + x2 (ex ) = (2x + x2 )ex .
dx
dx
dx
67. The terms in the numerator of f ′ (x) are transposed. Applying the quotient rule with u(x) = x and v(x) = x + 1 we get
d
dx
so
f ′ (x) =
u
v
=
du
dx
·v−u·
v2
dv
dx
d
d
(x + 1) dx
(x) − x dx
(x + 1)
.
2
(x + 1)
68. Rewrite f (x) as f (x) = ex (x + 1) then the product rule gives f ′ (x) = ex (x + 1) + ex · 1 = (x + 2)ex .
69. One possibility is f (x) = ex sin x. More complicated examples include g(x) = 10x sin(3x).
70. Functions like f (x) = x2 can be differentiated using the power rule. However, we can also view it as f (x) = x · x and
apply the product rule.
71. False. The product rule gives
(f g)′ = f g ′ + f ′ g.
Differentiating this and using the product rule again, we get
(f g)′′ = f ′ g ′ + f g ′′ + f ′ g ′ + f ′′ g = f g ′′ + 2f ′ g ′ + f ′′ g.
Thus, the right hand side is not equal to f g ′′ + f ′′ g in general.
3.4 SOLUTIONS
181
72. True; looking at the statement from the other direction, if both f (x) and g(x) are differentiable at x = 1, then so is their
quotient, f (x)/g(x), as long as it is defined there, which requires that g(1) 6= 0. So the only way in which f (x)/g(x)
can be defined but not differentiable at x = 1 is if either f (x) or g(x), or both, is not differentiable there.
73. False. Let f (x) = x2 and g(x) = x2 − 1. Let h(x) = f (x)g(x). Then h′′ (x) = 12x2 − 2. Since h′′ (0) < 0, clearly h
is not concave up for all x.
74. (a) This is not a counterexample. Although the product rule says that (f g)′ = f ′ g + f g ′ , that does not rule out the
possibility that also (f g)′ = f ′ g ′ . In fact, if f and g are both constant functions, then both f ′ g + f g ′ and f ′ g ′ are
zero, so they are equal to each other.
(b) This is not a counterexample. In fact, it agrees with the product rule:
d
(xf (x)) =
dx
d
d
(x) f (x) + x f (x) = f (x) + xf ′ (x) = xf ′ (x) + f (x).
dx
dx
(c) This is not a counterexample. Although the product rule says that
d
d
(f (x)2 ) =
f (x) · f (x) = f ′ (x)f (x) + f (x)f ′ (x) = 2f (x)f ′ (x),
dx
dx
it could be true that f ′ (x) = 1, so that the derivative is also just 2f (x). In fact, f (x) = x is an example where this
happens.
(d) This would be a counterexample. If f ′ (a) = g ′ (a) = 0, then
d
(f (x)g(x))
dx
= f ′ (a)g(a) + f (a)g ′ (a) = 0.
x=a
So f g cannot have positive slope at x = a. Of course such a counterexample could not exist, since the product rule
is true.
Solutions for Section 3.4
Exercises
1. f ′ (x) = 99(x + 1)98 · 1 = 99(x + 1)98 .
2. w′ = 100(t3 + 1)99 (3t2 ) = 300t2 (t3 + 1)99 .
d
d
(4x2 + 1)7 = 7(4x2 + 1)6 (4x2 + 1) = 7(4x2 + 1)6 · 8x = 56x(4x2 + 1)6 .
3. g ′ (x) =
dx
dx
−x
1
1
.
4. f ′ (x) = (1 − x2 )− 2 (−2x) = p
2
1 − x2
d √ x
d x
1
d
ex
dy
=
( e + 1) =
(e + 1)1/2 = (ex + 1)−1/2 (ex + 1) = √ x
.
dx
dx
dx
2
dx
2 e +1
√
√
1
50
= √
( t + 1)99 .
6. w′ = 100( t + 1)99 2√
t
t
5.
7. h′ (w) = 5(w4 − 2w)4 (4w3 − 2)
8. s′ (t) = 3(3t2 + 4t + 1)2 · (6t + 4) = 3(6t + 4)(3t2 + 4t + 1)2 .
9. We can write w(r) = (r 4 + 1)1/2 , so
2r 3
1
.
w′ (r) = (r 4 + 1)−1/2 (4r 3 ) = √
2
r4 + 1
10. k′ (x) = 4(x3 + ex )3 (3x2 + ex ).
11. f ′ (x) = 2e2x [x2 + 5x ] + e2x [2x + (ln 5)5x ] = e2x [2x2 + 2x + (ln 5 + 2)5x ].
3
12. y ′ = 32 e 2 w .
13. g(x) = πeπx .
dB
14.
= 15e0.20t · 0.20 = 3e0.20t .
dt
182
Chapter Three /SOLUTIONS
2
2
dw
= 100e−x · (−2x) = −200xe−x .
dx
16. f (θ) = (2−1 )θ = ( 12 )θ so f ′ (θ) = (ln 21 )2−θ .
15.
17. y ′ = (ln π)π (x+2) .
18. g ′ (x) = 2(ln 3)3(2x+7) .
19. f ′ (t) = 1 · e5−2t + te5−2t (−2) = e5−2t (1 − 2t).
20. p′ (t) = 4e4t+2 .
21. Using the product rule gives v ′ (t) = 2te−ct − ce−ct t2 = (2t − ct2 )e−ct .
2 d
2
2
d (1+3t)2
= e(1+3t)
22.
e
(1 + 3t)2 = e(1+3t) · 2(1 + 3t) · 3 = 6(1 + 3t)e(1+3t) .
dt
dt
1 √
23. w′ = √ e s .
2 s
24. y ′ = −4e−4t .
3s2
.
25. y ′ = √
2 s3 + 1
2
2
26. y ′ = 1 · e−t + te−t (−2t)
√
1
27. f ′ (z) = √ e−z − ze−z .
2 z
(ln 2)2x
.
28. z ′ (x) = p
3
3 (2x + 5)2
29. z ′ = 5 · ln 2 · 25t−3 .
3 √
3√ 2 x
x · 5 [2x(5x ) + (ln 5)(x2 )(5x )] = x2 53x (2 + x ln 5).
30. w′ =
2
2
12
5
1
= 10 2 − 2 y
5 1
31. f (y) = 10(5−y)
f ′ (y) = (ln 10) 10 2 − 2 y
−
32. We can write this as f (z) =
′
33. y =
′
z
2√
2 z
√
√
− ( z)(ln 2)(2z )
22z
34. y = 2
x2 + 2
3
2x
3
35. We can write h(x) =
=
1
h (x) =
2
5
1
1
= − (ln 10)(10 2 − 2 y ).
2
√
1
ze−z , in which case it is the same as problem 27. So f ′ (z) = √ e−z − ze−z .
2 z
=
1 − 2z ln 2
√ .
2z+1 z
4
x x2 + 2
9
x2 + 9
x+3
′
1
2
1/2
x2 + 9
x+3
, so
−1/2
2x(x + 3) − (x2 + 9)
(x + 3)2
1
=
2
r
x+3
x2 + 9
36.
(ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x )
dy
=
dx
(ex + e−x )2
(ex + e−x )2 − (ex − e−x )2
(e2x + 2 + e−2x ) − (e2x − 2 + e−2x )
=
x
−x
2
(e + e )
(ex + e−x )2
4
= x
(e + e−x )2
=
37. y ′ =
−(3e3x + 2x)
.
(e3x + x2 )2
38. h′ (z) =
−8b4 z
(a + z 2 )5
x2 + 6x − 9
.
(x + 3)2
3.4 SOLUTIONS
3
1
39. f ′ (x) = − (x3 + 1)− 2 · (3x2 ) = −1.5x2 (x3 + 1)−1.5 .
2
−2ez
.
40. f ′ (z) = −2(ez + 1)−3 · ez = z
(e + 1)3
41. w′ = (2t + 3)(1 − e−2t ) + (t2 + 3t)(2e−2t ).
3x
42. h′ (x) = (ln 2)(3e3x )2e
′
5x
43. f (x) = 6(e )(5) + (e
3x
= 3e3x 2e
−x2
ln 2.
2
)(−2x) = 30e5x − 2xe−x .
2
44. f ′ (x) = e−(x−1) · (−2)(x − 1).
45.
2
2
f ′ (w) = (ew )(10w) + (5w2 + 3)(ew )(2w)
2
= 2wew (5 + 5w2 + 3)
2
= 2wew (5w2 + 8).
46. The power and chain rules give
f ′ (θ) = −1(eθ + e−θ )−2 ·
d θ
(e + e−θ ) = −(eθ + e−θ )−2 (eθ + e−θ (−1)) = −
dθ
eθ − e−θ
(eθ + e−θ )2
2
47. We write y = (e−3t + 5)1/2 , so
2
2
2
2
d
1
d
1
dy
= (e−3t + 5)−1/2 · (e−3t + 5) = (e−3t + 5)−1/2 · e−3t · (−3t2 )
dt
2
dt
2
dt
2
=
48. Using the product and chain rules, we have
2
1 −3t2
3te−3t
(e
+ 5)−1/2 · e−3t · (−6t) = − p
.
2
e−3t2 + 5
d
dz
= 9(te3t + e5t )8 · (te3t + e5t ) = 9(te3t + e5t )8 (1 · e3t + t · e3t · 3 + e5t · 5)
dt
dt
= 9(te3t + e5t )8 (e3t + 3te3t + 5e5t ).
(y2 )
49. f ′ (y) = ee
h
2
(y2 )
i
+y 2 ]
(ey )(2y) = 2ye[e
2t
2t
50. f ′ (t) = 2(e−2e )(−2e2t )2 = −8(e−2e
.
+2t
).
′
51. Since a and b are constants, we have f (x) = 3(ax2 + b)2 (2ax) = 6ax(ax2 + b)2 .
52. Since a and b are constants, we have f ′ (t) = aebt (b) = abebt .
53. We use the product rule. We have
f ′ (x) = (ax)(e−bx (−b)) + (a)(e−bx ) = −abxe−bx + ae−bx .
54. Using the product and chain rules, we have
−2α
g ′ (α) = eαe
−2α
= eαe
·
−2α
d
(αe−2α ) = eαe
(1 · e−2α + αe−2α (−2))
dx
(e−2α − 2αe−2α )
−2α
= (1 − 2α)e−2α eαe
.
55. We have
−cx ′
dy
= ae−be
dx
′ −be−cx
ae
chain rule
= −be−cx
.
183
184
Chapter Three /SOLUTIONS
−cx
= −ab −ce−cx e−be
−cx
= abce−cx e−be
.
−cx
dy
= abce−be −cx .
dx
56. One approach is to expand this at the outset:
This can also be written
2
−x
y = ex − e−x
= ex − e
ex − e−x
x
−x
e−x} − e|−x
= (ex )2 − e|x{z
{ze} + e
| {z }
e2x
2x
so
1
1
−2x
= e +e
− 2,
′
dy
2x
−2x
= e +e
−2
dx
= 2e2x − 2e−2x .
2
| {z }
e−2x
Another approach is to use the Chain Rule:
2 ′
dy
= ex − e−x
dx
′
= 2 ex − e−x · ex − e−x
= 2 ex − e−x · ex + e−x
2
= 2 (ex )2 + ex e−x − e−x ex − e−x
= 2 e2x − e−2x ,
which is equivalent to our first answer.
Problems
57. When f and g are differentiable, the chain rule gives h′ (x) = f ′ (g(x)) · g ′ (x). We use slope to compute the derivatives.
Since f (x) is linear on the interval 0 < x < 2, we compute the slope of the line to see that f ′ (x) = 2 on this interval.
Similarly, we compute the slope on the interval 2 < x < 4 to see that f ′ (x) = −2 on the interval 2 < x < 4. Since f (x)
has a corner at x = 2, we know that f ′ (2) does not exist.
Similarly, g(x) is linear on the interval shown, and we see that the slope of g(x) on this interval is −1 so we have
g ′ (x) = −1 on this interval.
(a) We have h′ (1) = f ′ (g(1)) · g ′ (1) = (f ′ (3))(−1) = (−2)(−1) = 2.
(b) Since f (x) has a corner at x = 2, we know that f ′ (2) does not exist. Therefore, the chain rule does not apply.
(c) We have h′ (3) = f ′ (g(3)) · g ′ (3) = (f ′ (1))(−1) = 2(−1) = −2.
58. When f and g are differentiable, the chain rule gives u′ (x) = g ′ (f (x)) · f ′ (x). We use slope to compute the derivatives.
Since f (x) is linear on the interval 0 < x < 2, we compute the slope of the line to see that f ′ (x) = 2 on this interval.
Similarly, we compute the slope on the interval 2 < x < 4 to see that f ′ (x) = −2 on the interval 2 < x < 4. Since f (x)
has a corner at x = 2, we know that f ′ (2) does not exist.
Similarly, g(x) is linear on the interval shown, and we see that the slope of g(x) on this interval is −1 so we have
g ′ (x) = −1 on this interval.
(a) We have u′ (1) = g ′ (f (1)) · f ′ (1) = (g ′ (2))2 = (−1)2 = −2.
(b) Since f (x) has a corner at x = 2, the chain rule is not applicable.
(c) We have u′ (3) = g ′ (f (3)) · f ′ (3) = (g ′ (2))(−2) = (−1)(−2) = 2.
59. When f and g are differentiable, the chain rule gives v ′ (x) = f ′ (f (x)) · f ′ (x). We use slope to compute the derivatives.
Since f (x) is linear on the interval 0 < x < 2, we compute the slope of the line to see that f ′ (x) = 2 on this interval.
Similarly, we compute the slope on the interval 2 < x < 4 to see that f ′ (x) = −2 on the interval 2 < x < 4. Since f (x)
has a corner at x = 2, we know that f ′ (2) does not exist.
(a) To use the chain rule, we need f ′ (f (1)) = f ′ (2). Since f (x) has a corner at x = 2, we know that f ′ (2) does not
exist. Therefore, the chain rule does not apply.
(b) Since f (x) has a corner at x = 2, we know that f ′ (2) does not exist. Therefore, the chain rule does not apply.
3.4 SOLUTIONS
185
(c) To use the chain rule, we need f ′ (f (3)) = f ′ (2). Since f (x) has a corner at x = 2, we know that f ′ (2) does not
exist. Therefore, the chain rule does not apply.
60. When f and g are differentiable, the chain rule gives w′ (x) = g ′ (g(x)) · g ′ (x). We use slope to compute the derivatives.
Since g(x) is linear on the interval shown, with slope equal to −1, we have g ′ (x) = −1 on this interval.
(a) We have w′ (1) = g ′ (g(1)) · g ′ (1) = (g ′ (3))(−1) = (−1)(−1) = 1.
(b) We have w′ (2) = g ′ (g(2)) · g ′ (2) = (g ′ (2))(−1) = (−1)(−1) = 1.
(c) We have w′ (3) = g ′ (g(3)) · g ′ (3) = (g ′ (1))(−1) = (−1)(−1) = 1.
61. The chain rule gives
d
f (g(x))
dx
1
1
= f ′ (g(30))g ′ (30) = f ′ (55)g ′ (30) = (1)( ) = .
2
2
x=30
62. The chain rule gives
d
f (g(x))
dx
= f ′ (g(70))g ′ (70) = f ′ (60)g ′ (70) = (1)(0) = 0.
x=70
63. The chain rule gives
d
g(f (x))
dx
x=30
= g ′ (f (30))f ′ (30) = g ′ (20)f ′ (30) = (1/2)(−2) = −1.
64. The chain rule gives
d
g(f (x))
dx
x=70
1
1
= g ′ (f (70))f ′ (70) = g ′ (30)f ′ (70) = (1)( ) = .
2
2
3
65. We have f (2) = (2 − 1) = 1, so (2, 1) is a point on the tangent line. Since f ′ (x) = 3(x − 1)2 , the slope of the tangent
line is
m = f ′ (2) = 3(2 − 1)2 = 3.
The equation of the line is
y − 1 = 3(x − 2) or y = 3x − 5.
66.
f (x) = 6e5x + e−x
f (1) = 6e5 + e−1
2
f ′ (x) = 30e5x − 2xe−x
2
f ′ (1) = 30e5 − 2(1)e−1
y − y1 = m(x − x1 )
y − (6e5 + e−1 ) = (30e5 − 2e−1 )(x − 1)
y − (6e5 + e−1 ) = (30e5 − 2e−1 )x − (30e5 − 2e−1 )
y = (30e5 − 2e−1 )x − 30e5 + 2e−1 + 6e5 + e−1
≈ 4451.66x − 3560.81.
67. To calculate the equation of the tangent line, we need to find the function value and the slope at t = 2. The function value
is
f (2) = 100e−0.3(2) = 54.8812,
so a point on the line is (2, 54.8812).
The slope is found using the derivative: f ′ (t) = 100e−0.3t (−0.3). At the point t = 2, we have
Slope = f ′ (2) = 100e−0.3(2) (−0.3) = −16.4643.
The equation of the line is
y − 54.8812 = −16.4643(t − 2)
y = −16.464t + 87.810.
186
Chapter Three /SOLUTIONS
68. The graph is concave down when f ′′ (x) < 0.
2
f ′ (x) = e−x (−2x)
h
i
2
2
f ′′ (x) = e−x (−2x) (−2x) + e−x (−2)
4x2
2
− x2
ex2
e
4x2 − 2
<0
=
ex2
=
The graph is concave down when 4x2 < 2. This occurs when x2 < 12 , or − √12 < x <
1
√
.
2
69. We rewrite e−x = 1/ex so that we can use the quotient rule, then
x
,
ex
(1 − x)ex
1−x
1 · ex − x · ex
=
=
,
f ′ (x) =
(ex )2
(ex )2
ex
−ex − ex + xex
(−2 + x)ex
x−2
−1 · ex − (1 − x)ex
=
=
=
.
f ′′ (x) =
x
2
x
2
(e )
(e )
(ex )2
ex
f (x) =
70.
Since e−x > 0, for all x, we have f ′′ (x) < 0 if x − 2 < 0, that is, x < 2.
f ′ (x) = [10(2x + 1)9 (2)][(3x − 1)7 ] + [(2x + 1)10 ][7(3x − 1)6 (3)]
= (2x + 1)9 (3x − 1)6 [20(3x − 1) + 21(2x + 1)]
= [(2x + 1)9 (3x − 1)6 ](102x + 1)
f ′′ (x) = [9(2x + 1)8 (2)(3x − 1)6 + (2x + 1)9 (6)(3x − 1)5 (3)](102x + 1)
+(2x + 1)9 (3x − 1)6 (102).
71. (a) P (12) = 10e0.6(12) = 10e7.2 ≈ 13,394 fish. There are 13,394 fish in the area after 12 months.
(b) We differentiate to find P ′ (t), and then substitute in to find P ′ (12):
P ′ (t) = 10(e0.6t )(0.6) = 6e0.6t
P ′ (12) = 6e0.6(12) ≈ 8037 fish/month.
The population is growing at a rate of approximately 8037 fish per month.
72. (a) With µ and σ constant, differentiating m(t) = eµt+σ
′
m (t) = e
ut+σ 2 t2 /2
·
2 2
t /2
with respect to t gives
2σ 2 t
µ+
2
= eµt+σ
2 2
t /2
(µ + σ 2 t).
Thus,
Mean = m′ (0) = e0 (µ + 0) = µ.
(b) Differentiating m′ (t) = eµt+σ
2 2
t /2
(µ + σ 2 t), we have
m′′ (t) = eµt+σ
2 2
t /2
(µ + σ 2 t)2 + eµt+σ
2 2
t /2
σ2.
Thus
Variance = m′′ (0) − (m′ (0))2 = e0 µ2 + e0 σ 2 − µ2 = σ 2 .
73. (a) If
p(x) = k(2x),
then
p′ (x) = k′ (2x) · 2.
When x = 12 ,
p′
1
2
= k′ 2 ·
1
(2) = 2 · 2 = 4.
2
3.4 SOLUTIONS
187
(b) If
q(x) = k(x + 1),
then
q ′ (x) = k′ (x + 1) · 1.
When x = 0,
q ′ (0) = k′ (0 + 1)(1) = 2 · 1 = 2.
(c) If
r(x) = k
then
r ′ (x) = k′
When x = 4,
r ′ (4) = k′
1
x ,
4
1
1
x · .
4
4
1
1
1
1
4
=2· = .
4
4
4
2
√
dx
dx
and evaluate it. To do this, first we calculate
.
74. Yes. To see why, simply plug x = 3 2t + 5 into the expression 3x2
dt
dt
By the chain rule,
1
2
1
dx
d
2
2
= (2t + 5) 3 = (2t + 5)− 3 = [(2t + 5) 3 ]−2 .
dt
dt
3
3
1
But since x = (2t + 5) 3 , we have (by substitution)
2
dx
= x−2 .
dt
3
dx
2 −2
= 3x2
x
= 2.
dt
3
75. We see that m′ (x) is nearly of the form f ′ (g(x)) · g ′ (x) where
It follows that 3x2
f (g) = eg
and
g(x) = x6 ,
but g ′ (x) is off by a multiple of 6. Therefore, using the chain rule, let
6
m(x) =
e(x )
f (g(x))
=
.
6
6
76. (a) H(x) = F (G(x))
H(4) = F (G(4)) = F (2) = 1
(b) H(x) = F (G(x))
H ′ (x) = F ′ (G(x)) · G′ (x)
H ′ (4) = F ′ (G(4)) · G′ (4) = F ′ (2) · 6 = 5 · 6 = 30
(c) H(x) = G(F (x))
H(4) = G(F (4)) = G(3) = 4
(d) H(x) = G(F (x))
H ′ (x) = G′ (F (x)) · F ′ (x)
H ′ (4) = G′ (F (4)) · F ′ (4) = G′ (3) · 7 = 8 · 7 = 56
F (x)
(e) H(x) = G(x)
H ′ (x) =
H ′ (4) =
G(x)·F ′ (x)−F (x)·G′ (x)
[G(x)]2
G(4)·F ′ (4)−F (4)·G′ (4)
= 2·7−3·6
[G(4)]2
22
77. (a) Differentiating g(x) =
p
f (x) = (f (x))
=
1/2
14−18
4
=
−4
4
= −1
, we have
g ′ (x) =
f ′ (x)
1
(f (x))−1/2 · f ′ (x) = p
2
2 f (x)
f ′ (1)
3
3
g ′ (1) = p
= √ = .
4
2 4
2 f (1)
188
Chapter Three /SOLUTIONS
√
(b) Differentiating h(x) = f ( x), we have
√
1
h′ (x) = f ′ ( x) · √
2 x
√
f ′ (1)
1
3
h′ (1) = f ′ ( 1) · √ =
= .
2
2
2 1
78. (a) Using the chain rule, h′ (x) = ef (x) · f ′ (x). Since ef (x) 6= 0 for all values of x, the derivative h′ (x) is zero when
f ′ (x) = 0 and only then.
From the graph we see that f ′ (x) = 0 occurs when x ≈ −2.5 and x ≈ 1.5.
(b) Taking the derivative of p, we have p′ (x) = f ′ (ex ) · ex . Since ex 6= 0 for any value of x, the derivative p′ (x) is zero
only when f ′ (ex ) = 0.
From the graph we see that f ′ (y) = 0 for y ≈ 1.5 and y ≈ −2.5. Thus, f ′ (ex ) equals zero when ex = 1.5 or
x
e = −2.5. Since ex never equals −2.5, the only solution to f ′ (ex ) = 0 is x = ln(1.5).
79. We have h(0) = f (g(0)) = f (d) = d. From the chain rule, h′ (0) = f ′ (g(0))g ′ (0). From the graph of g, we see that
g ′ (0) = 0, so h′ (0) = f ′ (g(0)) · 0 = 0.
80. We have h(−c) = f (g(−c)) = f (−b) = 0. From the chain rule,
h′ (−c) = f ′ (g(−c))g ′ (−c).
Since g is increasing at x = −c, we know that g ′ (−c) > 0. We have
f ′ (g(−c)) = f ′ (−b),
and since f is decreasing at x = −b, we have f ′ (g(−c)) < 0. Thus,
h′ (−c) = f ′ (g(−c)) · g ′ (−c) < 0,
so h is decreasing at x = −c.
|
{z
} | {z }
−
+
81. We have
h′ (a) = f ′ (g(a))g ′(a).
From the graph of g, we see that g is decreasing at x = a, so g ′ (a) < 0. We have
f ′ (g(a)) = f ′ (b),
and from the graph of f , we see that f is increasing at x = b, so f ′ (b) > 0. Thus,
h′ (a) = f ′ (g(a)) · g ′ (a) < 0,
| {z } |{z}
+
so h is decreasing at x = a.
−
82. We have h(d) = f (g(d)) = f (−d) = d so h(d) is positive. From the chain rule,
h′ (d) = f ′ (g(d))g ′(d).
We have
f ′ (g(d)) = f ′ (−d).
From the graph of f , we see that f ′ (−d) < 0, and from the graph of g, we see that g ′ (d) < 0. This means the sign of
h′ (d) is the product of two negative numbers, so h′ (d) > 0.
83. On the interval −d < x < −b, we see that the value of g(x) increases from −d to 0. On the interval −d < x < 0, the
value of f (x) decreases from d to −d. Thus, the value of h(x) = f (g(x)) decreases on the interval −d < x < −b from
h(−d) = f (g(−d)) = f (−d) = d
to
h(−b) = f (g(−b)) = f (0) − d.
Confirming this using derivatives and the chain rule, we see
h′ (x) = f ′ (g(x)) · g ′ (x),
and since g ′ (x) is negative on −d < x < −b and f ′ (g(x)) is positive on this interval, the value of h(x) is decreasing.
3.4 SOLUTIONS
189
84. We have f (0) = 6.91 and f (10) = 6.91e0.011(10) = 7.713. The derivative of f (t) is
f ′ (t) = 6.91e0.011t · 0.011 = 0.0760e0.011t ,
and so f ′ (0) = 0.0760 and f ′ (10) = 0.0848.
These values tell us that in 2010 (at t = 0), the population of the world was 6.91 billion people and the population
was growing at a rate of 0.0760 billion people per year. In the year 2020 (at t = 10), this model predicts that the population
of the world will be 7.71 billion people and growing at a rate of 0.0848 billion people per year.
85. Since the population is 308.75 million on April 1, 2010 and growing exponentially, P = 308.75ekt , where P is the
population in millions and t is time in years since the census in 2010. Then
dP
= 308.75kekt ,
dt
so, since the population is growing at 2.85 million/year on April 1, 2010,
dP
dt
= 308.75kek·0 = 308.75k = 2.85
t=0
k=
2.85
= 0.00923,
308.75
so P = 308.75e0.00923t .
86. (a) The function f (t) is linear; g(t) is exponential; h(t) is quadratic (polynomial of degree 2.)
(b) In 2010, we have t = 60. For f (t), the rate of change is
f ′ (t) = 1.3,
f ′ (60) = 1.3 ppm per year.
For g(t), the rate of change is
g ′ (t) = 304(0.0038)e0.0038t = 1.1552e0.0038t ,
g ′ (60) = 1.1552e0.0038(60) = 1.451 ppm per year .
For h(t), the rate of change is
h′ (t) = 0.0135 · 2t + 0.5133 = 0.027t + 0.5135,
h′ (60) = 0.027(60) + 0.5133 = 2.133 ppm per year .
(c) Since 1.3 < 1.451 < 2.133, we see
Linear prediction < Exponential < Quadratic.
(d) The linear growth rate remains constant and will always be the smallest. The other two growth rates increase with
time, and eventually the exponential growth rate will overtake the quadratic growth rate.
For example, if t = 1000, we have
f ′ (1000) = 1.3
g ′ (1000) = 1.1552e0.0038(1000) = 51.639
′
h (1000) = 0.027(1000) + 0.5133 = 27.513.
Thus, at t = 1000, the exponential growth rate is largest.
87. We have f (5) = 5.1e0.043(5) = 6.3 billion dollars. Since f ′ (t) = 5.1e0.043t · 0.043, we have
f ′ (5) = 5.1e0.043(5) (0.043) = 0.272 billion dollars per year.
In 2013, net sales at Hershey are predicted to be 6.3 billion dollars and to be increasing at a rate of 0.272 billion dollars
per year.
190
Chapter Three /SOLUTIONS
88. Since we’re given that the instantaneous rate of change of T at t = 30 is 2, we want to choose a and b so that the derivative
of T agrees with this value. Differentiating, T ′ (t) = ab · e−bt . Then we have
2 = T ′ (30) = abe−30b or e−30b =
2
.
ab
We also know that at t = 30, T = 120, so
120 = T (30) = 200 − ae−30b or e−30b =
Thus
80
.
a
80
2
= e−30b =
,
a
ab
so
b=
1
= 0.025
40
a = 169.36.
and
dB
r t
r
dB
. The expression
= P 1+
ln 1 +
tells us how fast the amount of money in the bank is
dt
100
100
dt
changing with respect to time for fixed initial investment P and interest rate r.
dB
r t−1 1
dB
= Pt 1 +
. The expression
indicates how fast the amount of money changes with respect
(b)
dr
100
100
dr
to the interest rate r , assuming fixed initial investment P and time t.
89. (a)
90. (a) We see from the formula that $1000 was deposited initially, and that the money is earning interest at 8% compounded
continuously.
(b) We have f (10) = 1000e0.08(10) = 2225.54 dollars. Since f ′ (t) = 1000e0.08t · 0.08, we have
f ′ (10) = 1000e0.08(10) (0.08) = 178.04 dollars per year.
Ten years after the money was deposited, the balance is $2225.54 and is growing at a rate of $178.04 per year.
91. (a)
"
dm
d
m0
=
dv
dv
v2
1− 2
c
1
1−
= m0 −
2
m0 v
= 2 q
c
1
1−
v2
c2
−1/2 #
v2
c2
3 .
−3/2
−
2v
c2
dm
represents the rate of change of mass with respect to the speed v.
dv
dQ
= 0.
92. (a) For t < 0, I =
dt
dQ
Q0 −t/RC
For t > 0, I =
=−
e
.
dt
RC
(b) For t > 0, t → 0 (that is, as t → 0+ ),
(b)
I=−
Q0
Q0 −t/RC
e
→−
.
RC
RC
Since I = 0 just to the left of t = 0 and I = −Q0 /RC just to the right of t = 0, it is not possible to define I at
t = 0.
(c) Q is not differentiable at t = 0 because there is no tangent line at t = 0.
93. Let f have a zero of multiplicity m at x = a so that
f (x) = (x − a)m h(x),
h(a) 6= 0.
Differentiating this expression gives
f ′ (x) = (x − a)m h′ (x) + m(x − a)(m−1) h(x)
3.4 SOLUTIONS
191
and both terms in the sum are zero when x = a so f ′ (a) = 0. Taking another derivative gives
f ′′ (x) = (x − a)m h′′ (x) + 2m(x − a)(m−1) h′ (x) + m(m − 1)(x − a)(m−2) h(x).
Again, each term in the sum contains a factor of (x − a) to some positive power, so at x = a this will evaluate to 0.
Differentiating repeatedly, all derivatives will have positive integer powers of (x − a) until the mth and will therefore
vanish. However,
f (m) (a) = m!h(a) 6= 0.
94. Using the chain and product rule:
d
d
d
d2
(f (g(x))) =
(f (g(x))) =
f ′ (g(x)) · g ′ (x)
2
dx
dx dx
dx
= f ′′ (g(x)) · g ′ (x) · g ′ (x) + f ′ (g(x)) · g ′′ (x)
= f ′′ (g(x)) · g ′ (x)
2
+ f ′ (g(x)) · g ′′ (x).
95. Using the chain and product rules:
d2
dx2
f (x)
g(x)
=
d
dx
d
dx
f (x)(g(x))−1
=
d
f ′ (x)(g(x))−1 − f (x)(g(x))−2 g ′ (x)
dx
= f (x)(g(x))−1 − f ′ (x)(g(x))−2g ′ (x) − f ′ (x)(g(x))−2g ′ (x)
′′
′′
= f (x)(g(x))
−1
′
− 2f (x)(g(x))
−2 ′
+2f (x)(g(x))−3 (g ′ (x))2 − f (x)(g(x))−2g ′′ (x)
g (x) + 2f (x)(g(x))−3 (g ′ (x))2 − f (x)(g(x))−2 g ′′ (x).
Strengthen Your Understanding
96. The derivative of the inside function, ex , is missing.
Let z = h(x) = ex + 2, so g(z) = z 5 and g ′ (z) = 5z 4 · h′ (x). Taking the derivative of h, we have h′ (x) = ex , so
g ′ (x) = 5(ex + 2)4 ex = 5ex (ex + 2)4 .
2
97. To calculate the derivative of ex we need to apply the chain rule. Take z = g(x) = x2 as the inside function and
f (z) = ez as the outside function. Since g ′ (x) = 2x and f ′ (z) = ez , the chain rule gives
2
w′ (x) = ez · 2x = 2xex .
98. Two possibilities are
f (x) = sin (ex )
and
g(x) = esin x .
More complicated examples include
f (x) = sin(3 · 10x )
and
g(x) = 5 · 73 sin 2x .
99. One possibility is f (x) = (x2 + 1)2 , which can be differentiated using the chain rule with x2 + 1 as the inside function
and x2 as the outside function:
f ′ (x) = 2(x2 + 1)1 · 2x.
In addition, we can expand the function into a polynomial: f (x) = x4 + 2x2 + 1, and now differentiate term-by-term:
f ′ (x) = 4x3 + 4x.
100. False; for example, if both f (x) and g(x) are constant functions, such as f (x) = 6, g(x) = 10, then (f g)′ (x) = 0, and
f ′ (x) = 0 and g ′ (x) = 0.
192
Chapter Three /SOLUTIONS
101. False; for example, if both f and g are constant functions, then the derivative of f (g(x)) is zero, as is the derivative of
f (x). Another example is f (x) = 5x + 7 and g(x) = x + 2.
2
2
102. False. Let f (x) = e−x and g(x) = x2 . Let h(x) = f (g(x)) = e−x . Then h′ (x) = −2xe−x and h′′ (x) = (−2 +
2
4x2 )e−x . Since h′′ (0) < 0, clearly h is not concave up for all x.
Solutions for Section 3.5
Exercises
1.
Table 3.1
x
cos x
0
1.0
0.1
0.995
0.2
0.98007
0.3
0.95534
0.4
0.92106
0.5
0.87758
0.6
0.82534
Difference Quotient
−0.0005
−0.099833
−0.296
−0.29552
−0.19916
−0.38988
−0.47986
−0.56506
3. s′ (θ) = − sin θ sin θ + cos θ cos θ = cos2 θ − sin2 θ = cos 2θ.
4. z ′ = −4 sin(4θ).
5. f ′ (x) = cos(3x) · 3 = 3 cos(3x).
dy
= 5 cos(3t) · 3 = 15 cos(3t).
6.
dt
dP
7.
= 4(− sin(2t)) · 2 = −8 sin(2t).
dt
d
d
8.
sin(2 − 3x) = cos(2 − 3x) (2 − 3x) = −3 cos(2 − 3x).
dx
dx
9. Using the chain rule gives R′ (x) = 3π sin(πx).
10. g ′ (θ) = 2 sin(2θ) cos(2θ) · 2 − π = 4 sin(2θ) cos(2θ) − π
11. g ′ (t) = 3(2 + sin(πt))2 · (cos(πt) · π) = 3π cos(πt)(2 + sin(πt))2 .
12. f ′ (x) = (2x)(cos x) + x2 (− sin x) = 2x cos x − x2 sin x.
14. f ′ (x) = (ecos x )(− sin x) = − sin xecos x .
15. f ′ (y) = (cos y)esin y .
16. z ′ = ecos θ − θ(sin θ)ecos θ .
17. Using the chain rule gives R′ (θ) = 3 cos(3θ)esin(3θ) .
cos(tan θ)
18. g ′ (θ) =
cos2 θ
2x
19. w′ (x) =
cos2 (x2 )
20.
1
f (x) = (1 − cos x) 2
1
1
f ′ (x) = (1 − cos x)− 2 (−(− sin x))
2
sin x
.
= √
2 1 − cos x
0.0
−0.10033
2. r ′ (θ) = cos θ − sin θ.
13. w′ = et cos(et ).
− sin x
−0.19867
−0.38942
−0.47943
−0.56464
3.5 SOLUTIONS
1
1
(3 + sin(8x))− 2 · (cos(8x) · 8) = 4 cos(8x)(3 + sin(8x))−0.5 .
2
22. f ′ (x) = [− sin(sin x)](cos x).
cos x
.
23. f ′ (x) =
cos2 (sin x)
21. f ′ (x) =
24. k′ (x) =
′
3
2
p
sin(2x)(2 cos(2x)) = 3 cos(2x)
p
sin(2x).
25. f (x) = 2 · [sin(3x)] + 2x[cos(3x)] · 3 = 2 sin(3x) + 6x cos(3x)
26. y ′ = eθ sin(2θ) + 2eθ cos(2θ).
27. f ′ (x) = (e−2x )(−2)(sin x) + (e−2x )(cos x) = −2 sin x(e−2x ) + (e−2x )(cos x) = e−2x [cos x − 2 sin x].
cos t
.
28. z ′ = √
2 sin t
29. y ′ = 5 sin4 θ cos θ.
ez
30. g ′ (z) =
.
cos2 (ez )
31. z ′ =
−3e−3θ
.
cos2 (e−3θ )
32. w′ = (− cos θ)e− sin θ .
dQ
= − sin(e2x ) · (e2x · 2) = −2e2x sin(e2x ).
33.
dx
34. h′ (t) = 1 · (cos t) + t(− sin t) + cos12 t = cos t − t sin t +
1
.
cos2 t
′
35. f (α) = − sin α + 3 cos α
36. k′ (α) = (5 sin4 α cos α) cos3 α + sin5 α(3 cos2 α(− sin α)) = 5 sin4 α cos4 α − 3 sin6 α cos2 α
37. f ′ (θ) = 3θ2 cos θ − θ3 sin θ.
38. y ′ = −2 cos w sin w − sin(w2 )(2w) = −2(cos w sin w + w sin(w2 ))
39. y ′ = cos(cos x + sin x)(cos x − sin x)
40. y ′ = 2 cos(2x) sin(3x) + 3 sin(2x) cos(3x).
41. We use the quotient rule. We have
(− sin t) · t3 − (3t2 ) · cos t
−t3 sin t − 3t2 cos t
−t sin t − 3 cos t
dP
=
=
=
.
3
2
dt
(t )
t6
t4
(sin2 θ + cos2 θ)
1
− sin θ sin θ − cos θ cos θ
=
−
=− 2 .
sin2 θ
sin2 θ
sin θ
43. Using the power and quotient rules gives
42. t′ (θ) =
f ′ (x) =
1
2
1
=
2
1
=
2
44.
d
dy
y
cos y + a
=
1 − sin x
1 − cos x
r
r
−1/2 − cos x(1 − cos x) − (1 − sin x) sin x
1 − cos x
1 − sin x
1 − cos x
1 − sin x
(1 − cos x)2
− cos x(1 − cos x) − (1 − sin x) sin x
(1 − cos x)2
1 − cos x − sin x
.
(1 − cos x)2
cos y + a − y(− sin y)
cos y + a + y sin y
=
.
(cos y + a)2
(cos y + a)2
2 sin x cos x(cos2 x + 1) + 2 sin x cos x(sin2 x + 1)
(cos2 x + 1)2
or, using sin2 x + cos2 x = 1,
6 sin x cos x
.
G′ (x) =
(cos2 x + 1)2
45. The quotient rule gives G′ (x) =
46.
dy
= a cos(bt) · b = ab cos(bt).
dt
193
194
47.
Chapter Three /SOLUTIONS
dP
= a(− sin(bt + c) · b) = −ab sin(bt + c).
dt
3.5 SOLUTIONS
195
Problems
48. We begin by taking the derivative of y = sin(x4 ) and evaluating at x = 10:
dy
= cos(x4 ) · 4x3 .
dx
Evaluating cos(10,000) on a calculator (in radians) we see cos(10,000) < 0, so we know that dy/dx < 0, and therefore
the function is decreasing.
Next, we take the second derivative and evaluate it at x = 10, giving sin(10,000) < 0:
d2 y
= cos(x4 ) · (12x2 ) + 4x3 · (− sin(x4 ))(4x3 ) .
dx2
|
{z
} |
{z
}
negative
positive, but much
larger in magnitude
From this we can see that d2 y/dx2 > 0, thus the graph is concave up.
49. To calculate the equation of the tangent line, we need to find the y-coordinate and the slope at t = π. The y-coordinate is
y = f (π) = 3 sin(2π) + 5 = 5,
so a point on the line is (π, 5).
The slope is found using the derivative: f ′ (t) = 3 cos(2t) · 2 = 6 cos(2t). At the point t = π, we have
Slope = f ′ (π) = 6 cos(2π) = 6.
The equation of the line is
y − 5 = 6(t − π)
y = 6t − 13.850.
50. The pattern in the table below allows us to generalize and say that the (4n)th derivative of cos x is cos x, i.e.,
d8 y
d4n y
d4 y
=
=
·
·
·
=
= cos x.
dx4
dx8
dx4n
Thus we can say that d48 y/dx48 = cos x. From there we differentiate twice more to obtain d50 y/dx50 = − cos x.
n
nth
derivative
1
2
3
4
− sin x
− cos x
sin x
cos x
···
48
49
50
cos x
− sin x
− cos x
51. Differentiating with respect to t using the chain rule and substituting for dx/dt gives
d
d2 x
=
dt2
dt
52. We see that q ′ (x) is of the form
dx
dt
=
dx
d
(x sin x) ·
= (sin x + x cos x)x sin x.
dx
dt
g(x) · f ′ (x) − f (x) · g ′ (x)
,
(g(x))2
with f (x) = ex and g(x) = sin x. Therefore, using the quotient rule, let
q(x) =
f (x)
ex
=
.
g(x)
sin x
196
Chapter Three /SOLUTIONS
53. Since F ′ (x) is of the form sin u, we can make an initial guess that
F (x) = cos(4x),
then
F ′ (x) = −4 sin(4x)
so we’re off by a factor of −4. To fix this problem, we modify our guess by a factor of −4, so the next try is
F (x) = −(1/4) cos(4x),
which has
F ′ (x) = sin(4x).
54. (a) We have
f ′ (x) = 2(sin x)1 (cos x) + 2(cos x)1 (− sin x) = 2 sin x cos x − 2 sin x cos x = 0.
(b) Since sin2 x + cos2 x = 1, we see that f (x) = 1 so f ′ (x) = 0.
55. We have
′
′
= f′
x2 x2
2 2
h′ (x) = f x2
= sin
x
Chain rule
since f ′ (x) = sin x2
· 2x
= 2x sin x4
′
so h′′ (x) = 2x sin x4
′
= (2x)′ sin x4 + 2x sin x4
Product rule
4 ′
4
4
= 2 sin x + 2x cos x
x
= 2 sin x4 + 2x cos x4 4x3
4
4
4
+ 8x cos x
= 2 sin x
.
56. (a) Differentiating gives
dy
4.9π
π
=−
sin
t .
dt
6
6
The derivative represents the rate of change of the depth of the water in feet/hour.
(b) The derivative, dy/dt, is zero where the tangent line to the curve y is horizontal. This occurs when dy/dt =
sin( π6 t) = 0, or at t = 6, 12, 18 and 24 (6 am, noon, 6 pm, and midnight). When dy/dt = 0, the depth of the
water is no longer changing. Therefore, it has either just finished rising or just finished falling, and we know that the
harbor’s level is at a maximum or a minimum.
dy
d
57. (a) v(t) =
= (15 + sin(2πt)) = 2π cos(2πt).
dt
dt
y
(b)
16
v
15
14
2π
v = 2π cos 2πt
y = 15 + sin 2πt
1
1
2
3
t
2
3
t
−2π
58. (a) Differentiating, we find
Rate of change of voltage
dV
=
= −120π · 156 sin(120πt)
with time
dt
= −18720π sin(120πt) volts per second.
(b) The rate of change of voltage with time is zero when sin(120πt) = 0. This occurs when 120πt equals any multiple
of π. For example, sin(120πt) = 0 when 120πt = π, or at t = 1/120 seconds. Since there are an infinite number of
multiples of π, there are many times when the rate of change dV /dt is zero.
(c) The maximum value of the rate of change is 18720π = 58810.6 volts/sec.
197
3.5 SOLUTIONS
59. (a) When
pk
v=A
a=
p mk
t=
cos
m
k
−A m
(b) T = √2π
sin
k/m
π
the spring is farthest from the equilibrium position. This occurs at time t =
2 p
p
= 2π
k
t
m
k
t
m
, so the maximum velocity occurs when t = 0
, so the maximum acceleration occurs when
pm
k
pk
m
t=
3π
,
2
pm
π
2
k
which is at time t =
3π
2
pm
k
1
2π 1
π
dT
= √ · m− 2 = √
(c)
dm
2
k
km
dT
Since
> 0, an increase in the mass causes the period to increase.
dm
60. (a) The population varies periodically with a period of 1 year. See Figure 3.6.
P (t)
4500
4000
3500
t
J
t=0
F
M A M
J
J
A S
O
N
D
J
t=1
Figure 3.6
(b) The population is at a maximum on July 1st . At this time sin(2πt − π2 ) = 1, so the actual maximum population is
4000 + 500(1) = 4500. Similarly, the population is at a minimum on January 1st . At this time, sin(2πt − π2 ) = −1,
so the minimum population is 4000 + 500(−1) = 3500.
(c) The rate of change is most positive about April 1st and most negative around October 1st .
(d) Since the population is at its maximum around July 1st , its rate of change is about 0 then.
61. (a) The function d(t) is increasing at a constant rate for the period 0 ≤ t ≤ 2, when the derivative of d(t) is k.
(b) The functions d(t) must be continuous, since the depth of water cannot shift suddenly and instantly (even the fastest
change takes some amount of time), so we know that
2k = 50 + sin(0.2),
so
k = 25.099.
This means that the derivative for 0 < t < 2 is 25.099, whereas the derivative for t > 2 is 0.1 cos(0.1t). In other
words
n 25.099t
0≤t≤2
d(t) =
50 + sin(0.1t)
t > 2,
so
n 25.099
0≤t<2
d′ (t) =
0.1 cos(0.1t)
t > 2,
At t = 2, the derivative is undefined, since 0.1 cos(0.1 · 2) 6= 25.099.
62. (a) Using triangle OP D in Figure 3.7, we see
OD
= cos θ
a
PD
= sin θ
a
so
OD = a cos θ
so
P D = a sin θ.
P
l
a
θ
O
D
d
Figure 3.7
Q
198
Chapter Three /SOLUTIONS
Using triangle P QD, we have
(P D)2 + d2 = l2
so
a2 sin2 θ + d2 = l2 ,
d=
Thus,
p
l2 − a2 sin2 θ.
x = OD + DQ
= a cos θ +
(b) Differentiating, regarding a and l as constants,
p
l2 − a2 sin2 θ.
1 (−2a2 sin θ cos θ)
dx
= −a sin θ +
p
dθ
2
l2 − a2 sin2 θ
a2 sin θ cos θ
.
= −a sin θ − p
l2 − a2 sin2 θ
We want to find dx/dt. Using the chain rule and the fact that dθ/dt = 2, we have
dx dθ
dx
dx
=
·
=2 .
dt
dθ dt
dθ
(i) Substituting θ = π/2, we have
dx
dx
=2
dt
dθ
θ=π/2
= −2a sin
π
a2 sin( π2 ) cos( π2 )
− 2p
l2 − a2 sin2 ( π2 )
π
a2 sin( π4 ) cos( π4 )
− 2p
l2 − a2 sin2 ( π4 )
2
= −2a cm/sec.
(ii) Substituting θ = π/4, we have
dx
dx
=2
dt
dθ
θ=π/4
= −2a sin
4
√
a2
= −a 2 − p
cm/sec.
l2 − a2 /2
d
(sin x) = cos x. The tangent line at x = 0 has slope f ′ (0) = cos 0 = 1
dx
and goes through the point (0, 0). Consequently, its equation is y = g(x) = x. The approximate value of sin(π/6) given
by this equation is g(π/6) = π/6 ≈ 0.524.
Similarly, the tangent line at x = π3 has slope
63. The tangent lines to f (x) = sin x have slope
f′
and goes through the point (π/3,
√
π
3
= cos
1
π
=
3
2
3/2). Consequently, its equation is
√
3 3−π
1
.
y = h(x) = x +
2
6
The approximate value of sin(π/6) given by this equation is then
√
π
6 3−π
≈ 0.604.
h
=
6
12
The actual value of sin(π/6) is 12 , so the approximation from 0 is better than that from π/3. This is because the slope
of the function changes less between x = 0 and x = π/6 than it does between x = π/6 and x = π/3. This is illustrated
by the following figure.
3.5 SOLUTIONS
199
y
y = g(x)
y = sin x
1
y = h(x)
π
6
x
π
3
64. If the graphs of y = sin x and y = ke−x are tangent, then the y-values and the derivatives,
−ke−x , are equal at that point, so
sin x = ke−x
dy
dy
= cos x and
=
dx
dx
cos x = −ke−x .
and
Thus sin x = − cos x so tan x = −1. The smallest x-value is x = 3π/4, which leads to the smallest k value
k=
3π
3π
, we have y = sin
When x =
4
4
sin(3π/4)
= 7.46.
e−3π/4
1
= √ so the point is
2
3π 1
,√ .
4
2
65. For f (x) and g(x) to be tangent at x = 0 we must have f (0) = g(0) and f ′ (0) = g ′ (0).
From f (0) = g(0) we have
(1 + R) cos(−wt) = T cos(−wt)
and thus
1 + R = T.
Since
f ′ (x) = −k1 sin(k1 x − wt) + k1 R sin(−k1 x − wt)
g ′ (x) = −k2 T sin(k2 x − wt)
the condition f ′ (0) = g ′ (0) gives
−k1 sin(−wt) + k1 R sin(−wt) = −k2 T sin(−wt)
and thus
−k1 + k1 R = −k2 T.
Substituting T = 1 + R and solving for R yields
−k1 + k1 R = −k2 (1 + R)
(k1 + k2 )R = k1 − k2
k1 − k2
.
R=
k1 + k2
Finally
T =1+R =1+
k1 − k2
2k1
=
.
k1 + k2
k1 + k2
66. (a) If f (x) = sin x, then
sin(x + h) − sin x
h
(sin x cos h + sin h cos x) − sin x
= lim
h→0
h
sin x(cos h − 1) + sin h cos x
= lim
h→0
h
sin h
cos h − 1
+ cos x lim
.
= sin x lim
h→0
h→0
h
h
f ′ (x) = lim
h→0
200
Chapter Three /SOLUTIONS
(b) cos hh−1 → 0 and
(c) Similarly,
sin h
h
→ 1, as h → 0. Thus, f ′ (x) = sin x · 0 + cos x · 1 = cos x.
cos(x + h) − cos x
h
(cos x cos h − sin x sin h) − cos x
lim
h→0
h
cos x(cos h − 1) − sin x sin h
lim
h→0
h
cos h − 1
sin h
cos x lim
− sin x lim
h→0
h→0
h
h
− sin x.
g ′ (x) = lim
h→0
=
=
=
=
67. (a) Sector OAQ is a sector of a circle with radius cos1 θ and angle ∆θ. Thus its area is the left side of the inequality.
Similarly, the area of Sector OBR is the right side of the equality. The area of the triangle OQR is 21 ∆ tan θ since it
is a triangle with base ∆ tan θ (the segment QR) and height 1 (if you turn it sideways, it is easier to see this). Thus,
using the given fact about areas (which is also clear from looking at the picture), we have
∆θ
·π
2π
1
cos θ
(b) Dividing the inequality through by
∆θ
2
2
≤
1
∆θ
· ∆(tan θ) ≤
·π
2
2π
1
cos(θ + ∆θ)
2
.
and canceling the π’s gives:
1
cos θ
2
∆ tan θ
≤
≤
∆θ
1
cos(θ + ∆θ)
2
2
Then as ∆θ → 0, the right and left sides both tend toward cos1 θ while the middle (which is the difference quotient
for tangent) tends to (tan θ)′ . Thus, the derivative of tangent is “squeezed” between two values heading toward the
2
same thing and must, itself, also tend to that value. Therefore, (tan θ)′ = cos1 θ .
(c) Take the identity sin2 θ + cos2 θ = 1 and divide through by cos2 θ to get (tan θ)2 + 1 = ( cos1 θ )2 . Differentiating
with respect to θ yields:
2(tan θ) · (tan θ)′ = 2
2
sin θ
cos θ
·
1 1 ′
·
cos θ
cos θ
2
1
1
1
=2
(cos θ)′
· (−1)
cos θ
cos θ
cos θ
sin θ
1
2
= (−1)2
(cos θ)′
cos3 θ
cos3 θ
− sin θ = (cos θ)′ .
2
(d)
d
d
sin2 θ + cos2 θ =
(1)
dθ
dθ
2 sin θ · (sin θ)′ + 2 cos θ · (cos θ)′ = 0
2 sin θ · (sin θ)′ + 2 cos θ · (− sin θ) = 0
(sin θ)′ − cos θ = 0
(sin θ)′ = cos θ.
Strengthen Your Understanding
68. The function sin(cos x) is a composition of two functions and requires the chain rule to differentiate. The correct derivative
computation is
d
sin(cos x) = cos(cos x) · (− sin x).
dx
3.6 SOLUTIONS
201
69. The function f is not a product; it is a composition. We need to use the chain rule. Let z = g(x) = sin x, so f (z) = sin z.
Then, using the chain rule, f ′ (x) = cos z · g ′ (x) = cos(sin x) cos x.
70. One possibility is f (x) = cos(x2 ) whose derivative is f ′ (x) = − sin(x2 ) · 2x = −2x sin(x2 ).
71. The function f (x) = sin x satisfies this condition because
d
sin x = cos x
dx
d
cos x = − sin x = −f (x).
f ′′ (x) =
dx
f ′ (x) =
There are many other possibilities, including f (x) = cos x.
72. True, since cos θ and therefore cos2 θ are periodic, and
1
d
(tan θ) =
.
dθ
cos2 θ
73. True. If f (x) is periodic with period c, then f (x + c) = f (x) for all x. By the definition of the derivative, we have
f ′ (x) = lim
h→0
and
f ′ (x + c) = lim
h→0
f (x + h) − f (x)
h
f (x + c + h) − f (x + c)
.
h
Since f is periodic, for any h 6= 0, we have
f (x + h) − f (x)
f (x + c + h) − f (x + c)
=
.
h
h
′
′
′
Taking the limit as h → 0, we get that f (x) = f (x + c), so f is periodic with the same period as f (x).
74. False; the fourth derivative of cos t+C, where C is any constant, is indeed cos t. But any function of the form cos t+p(t),
where p(t) is a polynomial of degree less than or equal to 3, also has its fourth derivative equal to cos t. So cos t + t2 will
work.
Solutions for Section 3.6
Exercises
2t
.
t2 + 1
1
−1
=
.
f ′ (x) =
1−x
x−1
10x
1
· (10x) =
.
f ′ (x) =
5x2 + 3
5x2 + 3
1
3
f ′ (x) = 4x + 3 = 4x + .
x
x
1
dy
.
= p
dx
1 − (x + 1)2
1. f ′ (t) =
2.
3.
4.
5.
1
3
·3=
.
1 + (3x)2
1 + 9x2
1
6x + 15
dP
=3 2
· (2x + 5) = 2
.
7.
dt
x + 5x + 3
x + 5x + 3
dQ
1
ab
8.
=a
·b=
.
dx
bx + c
bx + c
2x
9. Since ln(e ) = 2x, the derivative f ′ (x) = 2.
6. f ′ (x) =
2x2 +3
10. Since eln(e
)
= e2x
2
+3
, the derivative f ′ (x) = 4xe2x
2
+3
.
202
Chapter Three /SOLUTIONS
1
e−x
−x
·
(−e
)(−1)
=
.
1 − e−x
1 − e−x
cos α
1
· cos α =
.
f ′ (α) =
sin α
sin α
1
f ′ (x) = x
· ex .
e +1
1
dy
= ln x + x
− 1 = ln x
dx
x
ax
ae
j ′ (x) = ax
(e + b)
We use the product rule. We have
11. f ′ (x) =
12.
13.
14.
15.
16.
dy
1
= 3x2 · ln x + x3 · = 3x2 ln x + x2 .
dx
x
17. Using the product and chain rules gives h′ (w) = 3w2 ln(10w) + w3
10
= 3w2 ln(10w) + w2 .
10w
1
· (e7x )7 = 7.
e7x
(Note also that ln(e7x ) = 7x implies f ′ (x) = 7.)
18. f ′ (x) =
19. Note that f (x) = eln x · e1 = x · e = ex. So f ′ (x) = e. (Remember, e is just a constant.) You might also use the chain
rule to get:
f ′ (x) = e(ln x)+1 · x1 .
[Are the two answers the same? Of course they are, since
e(ln x)+1
20. f ′ (θ) =
− sin θ
cos θ
1
x
= eln x · e
1
x
= xe
1
x
= e.]
= − tan θ.
21. f (t) = ln t (because ln ex = x or because eln t = t), so f ′ (t) = 1t .
2y
.
22. f ′ (y) = p
1 − y4
d
−1
23. s′ (x) =
(arctan(2 − x)) =
.
dx
1 + (2 − x)2
24. g(α) = α, so g ′ (α) = 1.
25. g ′ (t) = earctan(3t
2
)
1
1 + (3t2 )2
− sin(ln t)
.
t
27. h′ (z) = (ln 2)z (ln 2−1) .
(6t) = earctan(3t
2
26. g ′ (t) =
)
6t
.
1 + 9t4
w
.
1 − w2
′
29. Note that f (x) = kx so, f (x) = k.
28. h′ (w) = arcsin w + √
30. Using the chain rule gives r ′ (t) = √
′
31. j (x) = − sin sin
−1
2
.
1 − 4t2
1
x · √
1 − x2
1
32. f (x) = − sin(arctan 3x)
1 + (3x)2
−1
1
.
33. f ′ (z) = −1(ln z)−2 · =
z
z(ln z)2
34. Using the quotient rule gives
′
g ′ (t) =
k
kt
x
= −√
1 − x2
(3) =
−3 sin(arctan 3x)
.
1 + 9x2
+ 1 (ln(kt) − t) − (ln(kt) + t)
(ln(kt) − t)2
k
kt
−1
3.6 SOLUTIONS
+ 1 (ln(kt) − t) − (ln(kt) + t) 1t − 1
(ln(kt) − t)2
ln(kt)/t − 1 + ln(kt) − t − ln(kt)/t − 1 + ln(kt) + t
g ′ (t) =
(ln(kt) − t)2
2
ln(kt)
−
2
g ′ (t) =
.
(ln(kt) − t)2
1
t
g ′ (t) =
35. f ′ (w) = 3w−1/2 − 2w−3 + 5
36.
2
3
5
1
= √ − 3 + .
w
w
w
w
1
dy
= 2(ln x + ln 2) + 2x
dx
x
− 2 = 2(ln x + ln 2) = 2 ln(2x)
37. Using the chain rule gives f ′ (x) =
cos x − sin x
.
sin x + cos x
1
1 1
· =
ln t t
t ln t
39. Using the chain rule gives
38. f ′ (t) =
"
′
T (u) =
1
1+
2
u
1+u
#
(1 + u) − u
(1 + u)2
1
(1 + u)2
=
(1 + u)2 + u2 (1 + u)2
1
=
.
1 + 2u + 2u2
40. Since ln
1 − cos t
1 + cos t
4
= 4 ln
h 1 − cos t i
1 + cos t
a′ (t) = 4
=
h
we have
1 + cos t sin t(1 + cos t) + sin t(1 − cos t)
(1 + cos t)2
1 − cos t
1 + cos t
1 − cos t
8 sin t
1 − cos2 t
8
.
=
sin t
i
8 sin t
(1 + cos t)2
=
−(x + 1)
1
41. f ′ (x) = − sin(arcsin(x + 1))( p
)= p
.
1 − (x + 1)2
1 − (x + 1)2
Problems
42. (a) We have
3
1
1
·3=
= .
3x
3x
x
(b) Using properties of logs, we have f (x) = ln 3 + ln x.
(c) Since ln 3 is a constant, the derivative of f (x) = ln 3 + ln x is
f ′ (x) =
f ′ (x) = 0 +
1
1
= .
x
x
Yes, the answer is the same as that obtained in part (a), as it should be.
43. Differentiating
f ′ (x) =
1
· 2x = 2x(x2 + 1)−1
x2 + 1
203
204
Chapter Three /SOLUTIONS
f ′′ (x) = 2(x2 + 1)−1 − 2x(x2 + 1)−2 · 2x
=
4x2
2x2 + 2
4x2
2
− 2
= 2
− 2
(x2 + 1)
(x + 1)2
(x + 1)2
(x + 1)2
=
2(1 − x2 )
.
(x2 + 1)2
Since (x2 + 1)2 > 0 for all x, we see that f ′′ (0) > 0 for 1 − x2 > 0 or x2 < 1. That is, ln(x2 + 1) is concave up on the
interval −1 < x < 1.
44. Let
g(x) = arcsin x
so
sin[g(x)] = x.
Differentiating,
cos[g(x)] · g ′ (x) = 1
1
g ′ (x) =
cos[g(x)]
Using the fact that sin2 θ + cos2 θ = 1, and cos[g(x)] ≥ 0, since − π2 ≤ g(x) ≤
cos[g(x)] =
Therefore,
Since sin[g(x)] = x, we have
π
,
2
we get
p
1 − (sin[g(x)])2 .
1
g ′ (x) = p
1 − (sin[g(x)])2
g ′ (x) = √
1
, −1 < x < 1.
1 − x2
45. Let
g(x) = log x.
Then
10g(x) = x.
Differentiating,
(ln 10)[10g(x) ]g ′ (x) = 1
1
(ln 10)[10g(x) ]
1
.
g ′ (x) =
(ln 10)x
g ′ (x) =
46. pH = 2 = − log x means log x = −2 so x = 10−2 . Rate of change of pH with hydrogen ion concentration is
d
d
−1
1
pH = − (log x) =
=−
= −43.4
dx
dx
x(ln 10)
(10−2 ) ln 10
47. We have g(5000) = 20 ln(0.001(5000)) = 32.189 years. Since
g ′ (F ) = 20 ·
20
1
· 0.001 =
,
0.001F
F
we have
20
= 0.004
5000
years per dollar. It takes about 32.189 years for the investment to grow to $5000, and it takes about 0.004 years (or about
1.5 days) for the investment to grow by one more dollar.
g ′ (5000) =
3.6 SOLUTIONS
205
48. The marginal revenue, M R, is obtained by differentiating the total revenue function, R. We use the chain rule so
MR =
When q = 10,
1
d
1
dR
=
·
· 2000q.
1000q 2 =
dq
1 + 1000q 2 dq
1 + 1000q 2
Marginal revenue =
2000 · 10
= $0.2/unit.
1 + 1000 · 102
49. (a) For y = ln x, we have y ′ = 1/x, so the slope of the tangent line is f ′ (1) = 1/1 = 1. The equation of the tangent
line is y − 0 = 1(x − 1), so, on the tangent line, y = g(x) = x − 1.
(b) Using a value on the tangent line to approximate ln(1, 1), we have
ln(1.1) ≈ g(1.1) = 1.1 − 1 = 0.1.
Similarly, ln(2) is approximated by
ln(2) ≈ g(2) = 2 − 1 = 1.
(c) From Figure 3.8, we see that f (1.1) and f (2) are below g(x) = x − 1. Similarly, f (0.9) and f (0.5) are also below
g(x). This is true for any approximation of this function by a tangent line since f is concave down (f ′′ (x) = − x12 <
0 for all x > 0). Thus, for a given x-value, the y-value given by the function is always below the value given by the
tangent line.
g(x) = x − 1
4
f (x) = ln x
x
2
4
6
−4
Figure 3.8
50. (a) Let g(x) = ax2 + bx + c be our quadratic and f (x) = ln x. For the best approximation, we want to find a
quadratic with the same value as ln x at x = 1 and the same first and second derivatives as ln x at x = 1. g ′ (x) =
2ax + b, g ′′ (x) = 2a, f ′ (x) = x1 , f ′′ (x) = − x12 .
g(1) = a(1)2 + b(1) + c f (1) = 0
g ′ (1) = 2a(1) + b f ′ (1) = 1
g ′′ (1) = 2a f ′′ (1) = −1
Thus, we obtain the equations
a+b+c = 0
2a + b = 1
2a = −1
We find a = − 21 , b = 2 and c = − 23 . Thus our approximation is:
3
1
g(x) = − x2 + 2x −
2
2
(b) From the graph below, we notice that around x = 1, the value of f (x) = ln x and the value of g(x) = − 21 x2 +2x− 23
are very close.
206
Chapter Three /SOLUTIONS
y
f (x) = ln x
x
g(x) = − 21 x2 + 2x −
3
2
(c) g(1.1) = 0.095 g(2) = 0.5
Compare with f (1.1) = 0.0953, f (2) = 0.693.
51. (a)
1
1
1
+
· (− 2 )
1 + x2
x
1 + x12
1
1
+ − 2
=
2
1+x
x +1
1
1
=
−
1 + x2
1 + x2
=0
f ′ (x) =
(b) f is a constant function. Checking at a few values of x,
Table 3.2
arctan x
arctan x−1
f (x) = arctan x + arctan x−1
1
0.785392
0.7853982
1.5707963
2
1.1071487
0.4636476
1.5707963
3
1.2490458
0.3217506
1.5707963
x
52. The closer you look at the function, the more it begins to look like a line with slope equal to the derivative of the function
at x = 0. Hence, functions whose derivatives at x = 0 are equal will look the same there.
y
y
y
y
The following functions look like the line y = x since, in all cases, y ′ = 1 at x = 0.
=x
y′ = 1
= sin x
y ′ = cos x
= tan x
y ′ = cos12 x
1
= ln(x + 1)
y ′ = x+1
The following functions look like the line y = 0 since, in all cases, y ′ = 0 at x = 0.
y = x2
y ′ = 2x
y = x sin x
y ′ = x cos x + sin x
3
y=x
y ′ = 3x2
y ′ = 2x · 21 · x21+1 = x2x+1
y = 12 ln (x2 + 1)
y = 1 − cos x
y ′ = sin x
The following
functions look like the line x = 0 since, in all cases, as x → 0+ , the slope y ′ → ∞.
√
y ′ = 2√1 x
y= x
p x+1
p x
1
y ′ = (x+1)−x
y=
· 21 · √ 1x = 2(x+1)
2 ·
x+1
x
(x+1)2
x+1
√
y = 2x − x2
= √1−x
y ′ = (2 − 2x) 21 · √ 1
2x−x2
2x−x2
3.6 SOLUTIONS
207
53. Since the chain rule gives h′ (x) = n′ (m(x))m′(x) = −2 we must find values a and x such that a = m(x) and
n′ (a)m′ (x) = −2.
Calculating slopes from the graph of n gives
′
n (a) =
(
1
if 0 < a < 50
1/2 if 50 < a < 100.
Calculating slopes from the graph of m gives
m′ (x) =
(
−2 if 0 < x < 50
2
if 50 < x < 100.
The only values of the derivative n′ are 1 and 1/2 and the only values of the derivative m′ are 2 and −2. In order to
have n′ (a)m′ (x) = −2 we must therefore have n′ (a) = 1 and m′ (x) = −2. Thus 0 < a < 50 and 0 < x < 50.
Now a = m(x) and from the graph of m we see that 0 < m(x) < 50 for 25 < x < 75.
The two conditions on x we have found are both satisfied when 25 < x < 50. Thus h′ (x) = −2 for all x in the
interval 25 < x < 50. The question asks for just one of these x values, for example x = 40.
54. Since the chain rule gives h′ (x) = n′ (m(x))m′ (x) = 2 we must find values a and x such that a = m(x) and
n′ (a)m′ (x) = 2.
Calculating slopes from the graph of n gives
′
n (a) =
(
1
if 0 < a < 50
1/2 if 50 < a < 100.
Calculating slopes from the graph of m gives
m′ (x) =
(
−2 if 0 < x < 50
2
if 50 < x < 100.
The only values of the derivative n′ are 1 and 1/2 and the only values of the derivative m′ are 2 and −2. In order to
have n′ (a)m′ (x) = 2 we must therefore have n′ (a) = 1 and m′ (x) = 2. Thus 0 < a < 50 and 50 < x < 100.
Now a = m(x) and from the graph of m we see that 0 < m(x) < 50 for 25 < x < 75.
The two conditions on x we have found are both satisfied when 50 < x < 75. Thus h′ (x) = 2 for all x in the
interval 50 < x < 75. The question asks for just one of these x values, for example x = 60.
55. Since the chain rule gives h′ (x) = n′ (m(x))m′ (x) = 1 we must find values a and x such that a = m(x) and
n′ (a)m′ (x) = 1.
Calculating slopes from the graph of n gives
′
n (a) =
(
1
if 0 < a < 50
1/2 if 50 < a < 100.
Calculating slopes from the graph of m gives
m′ (x) =
(
−2 if 0 < x < 50
2
if 50 < x < 100.
The only values of the derivative n′ are 1 and 1/2 and the only values of the derivative m′ are 2 and −2. In order to
have n′ (a)m′ (x) = 1 we must therefore have n′ (a) = 1/2 and m′ (x) = 2. Thus 50 < a < 100 and 50 < x < 100.
Now a = m(x) and from the graph of m we see that 50 < m(x) < 100 for 0 < x < 25 or 75 < x < 100.
The two conditions on x we have found are both satisfied when 75 < x < 100. Thus h′ (x) = 1 for all x in the
interval 75 < x < 100. The question asks for just one of these x values, for example x = 80.
56. Since the chain rule gives h′ (x) = n′ (m(x))m′(x) = −1 we must find values a and x such that a = m(x) and
n′ (a)m′ (x) = −1.
Calculating slopes from the graph of n gives
′
n (a) =
(
1
if 0 < a < 50
1/2 if 50 < a < 100.
208
Chapter Three /SOLUTIONS
Calculating slopes from the graph of m gives
′
m (x) =
(
−2 if 0 < x < 50
2
if 50 < x < 100.
The only values of the derivative n′ are 1 and 1/2 and the only values of the derivative m′ are 2 and −2. In order to
have n′ (a)m′ (x) = −1 we must therefore have n′ (a) = 1/2 and m′ (x) = −2. Thus 50 < a < 100 and 0 < x < 50.
Now a = m(x) and from the graph of m we see that 50 < m(x) < 100 for 0 < x < 25 or 75 < x < 100.
The two conditions on x we have found are both satisfied when 0 < x < 25. Thus h′ (x) = −1 for all x in the
interval 0 < x < 25. The question asks for just one of these x values, for example x = 10.
57. We have
(f −1 )′ (5) =
1
.
f ′ (f −1 (5))
From the graph of f (x) we see that f −1 (5) = 13. From the graph of f ′ (x) we see that f ′ (13) = 0.36. Thus (f −1 )′ (5) =
1/0.36 = 2.8.
58. We have
(f −1 )′ (10) =
1
.
f ′ (f −1 (10))
From the graph of f (x) we see that f −1 (10) = 23. From the graph of f ′ (x) we see that f ′ (23) = 0.62. Thus
(f −1 )′ (10) = 1/0.62 = 1.6.
59. We have
(f −1 )′ (15) =
1
.
f ′ (f −1 (15))
From the graph of f (x) we see that f −1 (15) = 30. From the graph of f ′ (x) we see that f ′ (30) = 0.73. Thus
(f −1 )′ (15) = 1/0.73 = 1.4.
60. Since the point (2, 5) is on the curve, we know f (2) = 5. The point (2.1, 5.3) is on the tangent line, so
Slope tangent =
0.3
5.3 − 5
=
= 3.
2.1 − 2
0.1
Thus, f ′ (2) = 3.
By the chain rule
h′ (2) = 3(f (2))2 · f ′ (2) = 3 · 52 · 3 = 225.
61. Since the point (2, 5) is on the curve, we know f (2) = 5. The point (2.1, 5.3) is on the tangent line, so
Slope tangent =
5.3 − 5
0.3
=
= 3.
2.1 − 2
0.1
Thus, f ′ (2) = 3.
By the chain rule
k′ (2) = −(f (2))−2 · f ′ (2) = −5−2 · 3 = −0.12.
62. Since the point (2, 5) is on the curve, we know f (2) = 5. The point (2.1, 5.3) is on the tangent line, so
Slope tangent =
0.3
5.3 − 5
=
= 3.
2.1 − 2
0.1
Thus, f ′ (2) = 3. Since g is the inverse function of f and f (2) = 5, we know f −1 (5) = 2, so g(5) = 2.
Differentiating, we have
1
1
1
g ′ (2) = ′
= ′
= .
f (g(5))
f (2)
3
63. (a) Since f (x) = x3 , we have f ′ (x) = 3x2 . Thus, f ′ (2) = 3(2)2 = 12.
(b) To find f −1 (x), we switch xs and ys and solve for y.
Since y = x3 , we get x = y 3 .
√
Solving for y gives
y = 3 x.
√
−1
Thus, f (x) = 3 x.
3.6 SOLUTIONS
(c) To find (f −1 )′ (x), we differentiate. Since f −1 (x) =
√
3
209
x = x1/3 , we get
(f −1 )′ (x) =
1 −2/3
x
.
3
Thus,
1 −2/3
1
1
1
(8)
=
=
=
.
3
3·4
12
3 · 82/3
−1
(d) The point (2, 8) is on the graph of f . Thus the point (8, 2) is on the graph of f , so f −1 (8) = 2. Therefore,
(f −1 )′ (8) =
(f −1 )′ (8) =
1
1
1
= ′
=
.
f ′ (f −1 (8))
f (2)
12
64. (a) Since f (x) = 2x5 + 3x3 + x, we differentiate to get f ′ (x) = 10x4 + 9x2 + 1.
(b) Because f ′ (x) is always positive, we know that f (x) is increasing everywhere. Thus, f (x) is a one-to-one function
and is invertible.
(c) To find f (1), substitute 1 for x into f (x). We get f (1) = 2(1)5 + 3(1)3 + 1 = 2 + 3 + 1 = 6.
(d) To find f ′ (1), substitute 1 for x into f ′ (x). We get f ′ (1) = 10(1)4 + 9(1)2 + 1 = 20.
(e) Since f (1) = 6, we have f −1 (6) = 1, so
(f −1 )′ (6) =
1
1
1
= ′
=
.
f ′ (f −1 (6))
f (1)
20
65. To find (f −1 )′ (3), we first look in the table to find that 3 = f (9), so f −1 (3) = 9. Thus,
(f −1 )′ (3) =
1
1
1
= ′
= .
f ′ (f −1 (3))
f (9)
5
66. (a) The statement f (2) = 4023 tells us that when the price is $2 per gallon, 4023 gallons of gas are sold.
(b) Since f (2) = 4023, we have f −1 (4023) = 2. Thus, 4023 gallons are sold when the price is $2 per gallon.
(c) The statement f ′ (2) = −1250 tells us that if the price increases from $2 per gallon, the sales decrease at a rate of
1250 gallons per $1 increase in price.
(d) The units of (f −1 )′ (4023) are dollars per gallon. We have
(f −1 )′ (4023) =
1
1
1
= ′
=−
= −0.0008.
f ′ (f −1 (4023))
f (2)
1250
Thus, when 4023 gallons are already sold, sales decrease at the rate of one gallon per price increase of 0.0008 dollars.
In others words, an additional gallon is sold if the price drops by 0.0008 dollars.
67. (a) Knowing f (2005) = 296 tells us that the US population was 296 million in the year 2005.
(b) Since f (2005) = 296, we have f −1 (296) = 2005. This tells us that the year in which the US population was 296
million was 2005.
(c) Knowing f ′ (2005) = 2.65 tells us that in the year 2005, the US population was growing at a rate of 2.65 million
people per year.
(d) Using parts (b) and (c), we have
(f −1 )′ (296) =
1
1
1
= ′
=
= 0.377.
f ′ (f −1 (296))
f (2005)
2.65
The units of the derivative of f −1 are years per million people (the reciprocal of the units of f ′ ). The statement
(f −1 )′ (296) = 0.377 tells us that when the US population was 296 million, it took 0.377 of a year (between 4 and
5 months) for the population to increase by another million.
68. Each grid mark on the horizontal axis represents 5 years and each grid mark on the vertical axis represents 100 million
vehicles.
(a) When t = 20, the year is 1985. Reading from the graph, we find that in 1985
f (20) ≈ 500 million vehicles.
This tells us that 20 years after 1965, in 1985, there were 500 million registered vehicles.
210
Chapter Three /SOLUTIONS
(b) Drawing a tangent line to the curve at t = 20, we have
Slope = f ′ (20) ≈
175
= 17.5 million vehicles/year.
10
Thus, 20 years after 1965, in 1985, the number of registered vehicles was increasing at 17.5 million vehicles per year.
(c) From the graph or part (a)
f −1 (500) = 20 years.
Thus, there were 500 million cars registered when t = 20, that is, in 1985.
(d) We have
1
1
1
= ′
=
= 0.0571 years/million vehicles.
(f −1 )′ (500) = ′ −1
f (f (500))
f (20)
17.5
Thus, when 500 million vehicles were already registered, it took 0.0571 year, or about 21 days, for another million
to be registered.
69. We have (f −1 )′ (8) = 1/f ′ (f −1 (8)). From the graph we see f −1 (8) = 4. Thus (f −1 )′ (8) =
1
1
=
.
f ′ (4)
3.0
1
1
= ′
. Therefore (f −1 )′ (10)f ′ (20) =
f ′ (f −1 (10))
f (20)
70. Since f (20) = 10, we have f −1 (10) = 20, so (f −1 )′ (10) =
1.
Option (b) is wrong.
71. All three values equal 1.
(a) We have f −1 (A) = a, so (f −1 )′ (A) =
(b) We have f
−1
(c) We have f
−1
(B) = b, so (f
−1 ′
(C) = c, so (f
−1 ′
) (B) =
) (C) =
1
f ′ (f −1 (A))
1
f ′ (f −1 (B))
1
f ′ (f −1 (C))
=
=
=
1
. Thus f ′ (a)(f −1 )′ (A) = 1.
f ′ (a)
1
. Thus f ′ (b)(f −1 )′ (B) = 1.
f ′ (b)
1
. Thus f ′ (c)(f −1 )′ (C) = 1.
f ′ (c)
72. A continuous invertible function f (x) cannot be increasing on one interval and decreasing on another because it would fail
the horizontal line test. The same is true of the inverse function f −1 (x). Either f −1 (x) is increasing and (f −1 )′ (x) ≥ 0
for all x, or f −1 (x) is decreasing and (f −1 )′ (x) ≤ 0 for all x. We can not have both (f −1 )′ (10) = 8 and (f −1 )′ (20) =
−6.
73. (a) The definition of the derivative of ln(1 + x) at x = 0 is
lim
h→0
ln(1 + h)
ln(1 + h) − ln 1
1
= lim
=
h→0
h
h
1+x
= 1.
x=0
(b) The rules of logarithms give
lim
h→0
ln(1 + h)
1
= lim ln(1 + h) = lim ln(1 + h)1/h = 1.
h→0 h
h→0
h
Thus, taking e to both sides and using the fact that eln A = A, we have
elimh→0 ln(1+h)
1/h
= lim eln(1+h)
h→0
1/h
= e1
lim (1 + h)1/h = e.
h→0
This limit is sometimes used as the definition of e.
(c) Let n = 1/h. Then as h → 0+ , we have n → ∞. Since
lim (1 + h)1/h = lim (1 + h)1/h = e,
h→0
h→0+
we have
lim
n→∞
1+
This limit is also sometimes used as the definition of e.
1
n
n
= e.
3.7 SOLUTIONS
211
Strengthen Your Understanding
74. To calculate the derivative of ln(1 + x4 ) we need to apply the chain rule. Take z = g(x) = 1 + x4 as the inside function
and f (z) = ln z as the outside function. Since g ′ (x) = 4x3 and f ′ (z) = 1/z, the chain rule gives
1
4x3
dw
= · 4x3 =
.
dx
z
1 + x4
1
1
1
· =
.
ln x x
x ln x
−1 ′
′
−1
−1
76. The formula for (f ) (2) is wrong. We need f (f (2)), and we are not given f (2).
75. The function f is not a product. We need the chain rule, so f ′ (x) =
77. Any exponential function of the form c · ax with c 6= 0, a > 0 and a 6= e is a constant multiple of its derivative but it is
d
not equal to its derivative: dx
(c · ax ) = (ln a)(c · ax ). One example is f (x) = 2x .
78. Since
1
d
ln(x) = ,
dx
x
any function of the form y = c ln x for some constant c will have derivative c/x. A particular example is y = 2 ln x.
79. f (x) = ln x is a possible answer since
d
1
ln x =
dx
x
and
d
d
c
1
f (cx) =
ln(cx) =
=
dx
dx
cx
x
by the chain rule.
80. For the statement to be true, we need f ′ (x) = 1, so f (x) = x is a function to try. Then f −1 (x) = x, so
(f −1 )′ (x) =
1
1
1
= ′
= = 1.
f ′ (f −1 (x))
f (x)
1
81. False. Since
2
1
2
d2
d 2
d
= − 2,
ln(x2 ) = 2 · 2x =
and
ln(x2 ) =
dx
x
x
dx2
dx x
x
we see that the second derivative of ln(x2 ) is negative for x > 0. Thus, the graph is concave down.
82. False; For example, the inverse function of f (x) = x3 is x1/3 , and the derivative of x1/3 is (1/3)x−2/3 , which is not
1/f ′ (x) = 1/(3x2 ).
Solutions for Section 3.7
Exercises
1. We differentiate implicitly both sides of the equation with respect to x.
2x + 2y
dy
=0,
dx
2x
x
dy
=−
=− .
dx
2y
y
2. We differentiate implicitly both sides of the equation with respect to x.
dy
=0
dx
dy
= −2x
3y 2
dx
−2x
dy
=
.
dx
3y 2
2x + 3y 2
212
Chapter Three /SOLUTIONS
3. We differentiate implicitly both sides of the equation with respect to x.
2x + y + x
x
dy
dx
− 3y 2
dy
dy
= y 2 + x(2y)
,
dx
dx
dy
dy
dy
− 3y 2
− 2xy
= y 2 − y − 2x ,
dx
dx
dx
dy
y 2 − y − 2x
=
.
dx
x − 3y 2 − 2xy
4. We differentiate implicitly both sides of the equation with respect to x.
2x + 2y
dy
dy
+3−5
dx
dx
dy
dy
2y
−5
dx
dx
dy
(2y − 5)
dx
dy
dx
=0
= −2x − 3
= −2x − 3
=
−2x − 3
.
2y − 5
5. Implicit differentiation gives
1·y+x·
dy
dy
+1+
= 0.
dx
dx
Solving for dy/dx, we have
dy
1+y
=−
.
dx
1+x
6.
dy
dy
−2
=0
dx
dx
dy
(x2 − 2)
= −2xy
dx
−2xy
dy
= 2
dx
(x − 2)
2xy + x2
7. We differentiate implicitly both sides of the equation with respect to x.
(x2 · 3y 2
dy
dy
+ 2x · y 3 ) − (x ·
+ 1 · y) =
dx
dx
dy
dy
3x2 y 2
+ 2xy 3 − x
−y =
dx
dx
dy
dy
3x2 y 2
−x
=
dx
dx
dy
=
(3x2 y 2 − x)
dx
dy
=
dx
0
0
y − 2xy 3
y − 2xy 3
y − 2xy 3
.
3x2 y 2 − x
8. We differentiate implicitly both sides of the equation with respect to x.
x1/2 = 5y 1/2
5
dy
1 −1/2
x
= y −1/2
2
2
dx
q
1 −1/2
x
dy
1 y
1
= 25 −1/2 =
=
.
dx
5
x
25
y
2
We can also obtain this answer by realizing that the original equation represents part of the line x = 25y which has slope 1/25.
3.7 SOLUTIONS
9. We differentiate implicitly both sides of the equation with respect to x.
1
1
x 2 + y 2 = 25 ,
1 1 dy
1 − 12
x
=0,
+ y− 2
2
2
dx
1
√
1
q
1 −2
x
y
dy
x− 2
y
= − 2 1 = − 1 = −√ = −
.
1 −2
−2
dx
x
x
y
y
2
10. We differentiate implicitly with respect to x.
y+x
3dy
dy
−1−
=
dx
dx
dy
=
(x − 3)
dx
dy
=
dx
0
1−y
1−y
x−3
11.
12x + 8y
dy
=0
dx
−12x
−3x
dy
=
=
dx
8y
2y
12.
2ax − 2by
dy
=0
dx
−2ax
ax
dy
=
=
dx
−2by
by
13. We differentiate implicitly both sides of the equation with respect to x.
ln x + ln(y 2 ) = 3
1
dy
1
+ 2 (2y)
=0
x
y
dx
dy
−1/x
y
=
=− .
dx
2y/y 2
2x
14. We differentiate implicitly both sides of the equation with respect to x.
ln y + x
1 dy
dy
1
+ 3y 2
=
y dx
dx
x
dy
1
x dy
+ 3y 2
= − ln y
y dx
dx
x
dy
dx
dy
dx
x
+ 3y 2
y
x + 3y 3
y
=
1 − x ln y
x
=
1 − x ln y
x
(1 − x ln y)
dy
y
=
·
dx
x
(x + 3y 3 )
213
214
Chapter Three /SOLUTIONS
15. We differentiate implicitly both sides of the equation with respect to x.
dy
=2
dx
dy
=2
y cos(xy) + x cos(xy)
dx
2 − y cos(xy)
dy
=
.
dx
x cos(xy)
cos(xy) y + x
16. We differentiate implicitly both sides of the equation with respect to x.
ecos y (− sin y)
dy
dx
−e
cos y
1 dy
dy
= 3x2 arctan y + x3
dx
1 + y 2 dx
x3
sin y −
1 + y2
= 3x2 arctan y
3x2 arctan y
dy
=
.
dx
−ecos y sin y − x3 (1 + y 2 )−1
17. We differentiate implicitly both sides of the equation with respect to x.
arctan(x2 y) = xy 2
dy
dy
1
(2xy + x2 ) = y 2 + 2xy
1 + x4 y 2
dx
dx
dy
dy
2xy + x2
= [1 + x4 y 2 ][y 2 + 2xy ]
dx
dx
dy 2
[x − (1 + x4 y 2 )(2xy)] = (1 + x4 y 2 )y 2 − 2xy
dx
dy
y 2 + x4 y 4 − 2xy
= 2
.
dx
x − 2xy − 2x5 y 3
18. We differentiate implicitly both sides of the equation with respect to x.
2
ex + ln y = 0
2
1 dy
2xex +
=0
y dx
2
dy
= −2xyex .
dx
19. We differentiate implicitly both sides of the equation with respect to x.
(x − a)2 + y 2 = a2
dy
=0
dx
dy
2y
= 2a − 2x
dx
dy
2a − 2x
a−x
=
=
.
dx
2y
y
2(x − a) + 2y
20.
dy
x−1/3
y 1/3
2 −1/3 2 −1/3 dy
x
+ y
·
= 0,
= − −1/3 = − 1/3 .
3
3
dx
dx
y
x
3.7 SOLUTIONS
215
21. Differentiating implicitly on both sides with respect to x,
dy
dy
− b sin(bx) = y + x
dx
dx
dy
(a cos(ay) − x)
= y + b sin(bx)
dx
dy
y + b sin(bx)
=
.
dx
a cos(ay) − x
a cos(ay)
22. Differentiating x2 + y 2 = 1 with respect to x gives
2x + 2yy ′ = 0
so that
y′ = −
x
y
At the point (0, 1) the slope is 0.
23. Differentiating sin(xy) = x with respect to x gives
(y + xy ′ ) cos(xy) = 1
or
xy ′ cos(xy) = 1 − y cos(xy)
so that
y′ =
1 − y cos(xy)
.
x cos(xy)
As we move along the curve to the point (1, π2 ), the value of dy/dx → ∞, which tells us the tangent to the curve at (1, π2 )
has infinite slope; the tangent is the vertical line x = 1.
24. Differentiating with respect to x gives
3x2 + 2xy ′ + 2y + 2yy ′ = 0
so that
y′ = −
3x2 + 2y
2x + 2y
At the point (1, 1) the slope is − 54 .
25. The slope is given by dy/dx, which we find using implicit differentiation. Notice that the product rule is needed for the
second term. We differentiate to obtain:
3x2 + 5x2
dy
dy
dy
+ 10xy + 4y
=4
dx
dx
dx
dy
= −3x2 − 10xy
(5x2 + 4y − 4)
dx
dy
−3x2 − 10xy
=
.
dx
5x2 + 4y − 4
At the point (1, 2), we have dy/dx = (−3 − 20)/(5 + 8 − 4) = −23/9. The slope of this curve at the point (1, 2) is
−23/9.
26. First, we must find the slope of the tangent, i.e.
dy
dx
. Differentiating implicitly, we have:
(1,−1)
dy
= 0,
dx
2
dy
y
y
=−
=− .
dx
2xy
2x
y 2 + x(2y)
Substitution yields
dy
dx
(1,−1)
=−
1
−1
= . Using the point-slope formula for a line, we have that the equation for the
2
2
tangent line is y + 1 = 12 (x − 1) or y = 21 x − 32 .
216
Chapter Three /SOLUTIONS
dy
, at (1, e2 ). Differentiating implicitly, we have:
dx
27. First we must find the slope of the tangent,
1
xy
dy
+y = 2
dx
2xy − y
dy
=
.
dx
x
x
Evaluating dy/dx at (1, e2 ) yields (2(1)e2 − e2 )/1 = e2 . Using the point-slope formula for the equation of the line, we
have:
y − e2 = e2 (x − 1),
or
y = e2 x.
28. First, we must find the slope of the tangent,
dy
dx
. Implicit differentiation yields:
(4,2)
dy
2x(xy − 4) − x2 x dx
+y
dy
=
.
2y
dx
(xy − 4)2
Given the complexity of the above equation, we first want to substitute 4 for x and 2 for y (the coordinates of the point
dy
. Substitution yields:
where we are constructing our tangent line), then solve for
dx
dy
dy
(2 · 4)(4 · 2 − 4) − 42 4 dx
+2
8(4) − 16(4 dx
+ 2)
dy
dy
=
=
= −4 .
2·2
2
dx
(4 · 2 − 4)
16
dx
4
Solving for
dy
dy
= −4 ,
dx
dx
dy
, we have:
dx
dy
= 0.
dx
The tangent is a horizontal line through (4, 2), hence its equation is y = 2.
x
dy
. Rewriting y =
as y(y + a) = x so that
29. First, we must find the slope of the tangent at the origin, that is
dx (0,0)
y+a
we have
y 2 + ay = x
and differentiating implicitly gives
dy
dy
+a
=1
dx
dx
dy
(2y + a) = 1
dx
dy
1
=
.
dx
2y + a
2y
Substituting x = 0, y = 0 yields
dy
dx
=
(0,0)
1
. Using the point-slope formula for a line, we have that the equation for
a
the tangent line is
y−0 =
30. First, we must find the slope of the tangent,
dy
dx
1
(x − 0)
a
or
y=
x
.
a
. We differentiate implicitly, obtaining:
(a,0)
2 − 13
2 1 dy
x
=0,
+ y− 3
3
3
dx
3.7 SOLUTIONS
217
1
√
2 −3
3 y
x
dy
=−3 1 =−√
.
3
2
−
dx
x
y 3
3
dy
Substitution yields,
dx
(a,0)
√
3
0
= 0. The tangent is a horizontal line through (a, 0), hence its equation is y = 0.
= √
3
a
Problems
31. (a) By implicit differentiation, we have:
2x + 2y
dy
dy
−4+7
=0
dx
dx
dy
= 4 − 2x
(2y + 7)
dx
dy
4 − 2x
=
.
dx
2y + 7
(b) The curve has a horizontal tangent line when dy/dx = 0, which occurs when 4 − 2x = 0 or x = 2. The curve has a
horizontal tangent line at all points where x = 2.
The curve has a vertical tangent line when dy/dx is undefined, which occurs when 2y + 7 = 0 or when y = −7/2.
The curve has a vertical tangent line at all points where y = −7/2.
32. (a) Taking derivatives implicitly, we get
2 dy
2
x+ y
=0
25
9 dx
dy
−9x
=
.
dx
25y
(b) The slope is not defined anywhere along the line y = 0. This ellipse intersects that line in two places, (−5, 0) and
(5, 0). (These are the “ends” of the ellipse where the tangent is vertical.)
33. Differentiating implicitly with respect to x, gives
d
dx
(x − 2)2
y2
+
16
4
=
d
(1) .
dx
Treating y as a function of x and using the chain rule, we get
y dy
x−2
+ ·
=0
8
2 dx
dy
=
dx
−
x − 2
8
y
2
=
2−x
.
4y
At x = 2, the value of dy/dx is 0 and the tangent lines are horizontal. The corresponding y-values satisfy
y2
02
+
=1
2
16
4
y2 = 4
y = ±2.
Thus the tangent line equations at (2, 2) and (2, −2) are the horizontal lines y = 2 and y = −2, respectively.
dy
implicitly:
34. (a) If x = 4 then 16 + y 2 = 25, so y = ±3. We find
dx
2x + 2y
dy
=0
dx
dy
x
=−
dx
y
218
Chapter Three /SOLUTIONS
So the slope at (4, 3) is − 34 and at (4, −3) is 34 . The tangent lines are:
4
(y − 3) = − (x − 4)
3
and
(y + 3) =
4
(x − 4)
3
(b) The normal lines have slopes that are the negative of the reciprocal of the slopes of the tangent lines. Thus,
and
3
x
4
3
(x − 4)
4
so
y=
3
(y + 3) = − (x − 4)
4
so
3
y=− x
4
(y − 3) =
are the normal lines.
(c) These lines meet at the origin, which is the center of the circle.
35. (a) Solving for
dy
dx
by implicit differentiation yields
3x2 + 3y 2
dy
dy
− y 2 − 2xy
=0
dx
dx
dy
y 2 − 3x2
=
.
dx
3y 2 − 2xy
(b) We can approximate the curve near x = 1, y = 2 by its tangent line. The tangent line will have slope
1
= 0.125. Thus its equation is
8
y = 0.125x + 1.875
(2)2 −3(1)2
3(2)2 −2(1)(2)
=
Using the y-values of the tangent line to approximate the y-values of the curve, we get:
x
0.96
0.98
1
1.02
1.04
approximate y
1.995
1.9975
2.000
2.0025
2.005
(c) When x = 0.96, we get the equation 0.963 + y 3 − 0.96y 2 = 5, whose solution by numerical methods is 1.9945,
which is close to the one above.
dy
dy
(d) The tangent line is horizontal when dx
is zero and vertical when dx
is undefined. These will occur when the numerator
is zero and when the denominator is zero, respectively.
√
Thus, we know that the tangent is horizontal when y 2 − 3x2 = 0 ⇒ y = ± 3x. To find the points that satisfy
this condition, we substitute back into the original equation for the curve:
x3 + y 3 − xy 2 = 5
√
x ± 3 3x3 − 3x3 = 5
3
5
√
±3 3 − 2
So x ≈ 1.1609 or x ≈ −0.8857.
x3 =
Substituting,
√
y = ± 3x so y ≈ 2.0107
or
y ≈ 1.5341.
Thus, the tangent line is horizontal at (1.1609, 2.0107) and (−0.8857, 1.5341).
Also, we know that the tangent is vertical whenever 3y 2 − 2xy = 0, that is, when y = 23 x or y = 0. Substituting
into the original equation for the curve gives us x3 + ( 32 x)3 − ( 32 )2 x3 = 5. This means x3 ≈ 5.8696, so x ≈ 1.8039,
√
y ≈ 1.2026. The other vertical tangent is at y = 0, x = 3 5.
36. The slope of the tangent to the curve y = x2 at x = 1 is 2 so the equation of such a tangent will be of the form y = 2x +c.
As the tangent must pass through (1, 1), c = −1 and so the required tangent is y = 2x − 1.
Any circle centered at (8, 0) will be of the form
(x − 8)2 + y 2 = R2 .
The slope of this curve at (x, y) is given by implicit differentiation:
2(x − 8) + 2yy ′ = 0
3.7 SOLUTIONS
or
219
8−x
y
For the tangent to the parabola to be tangential to the circle we need
y′ =
8−x
=2
y
so that at the point of contact of the circle and the line the coordinates are given by (x, y) when y = 4 − x/2. Substituting
into the equation of the tangent line gives x = 2 and y = 3. From this we conclude that R2 = 45 so that the equation of
the circle is
(x − 8)2 + y 2 = 45.
37. (a) Differentiating both sides of the equation with respect to P gives
d
dP
4f 2 P
1 − f2
=
dK
= 0.
dP
By the product rule
d
dP
4f 2 P
1 − f2
=
d
dP
=
=
4f 2
1 − f2
P+
4f 2
1 − f2
·1
4f 2
1 − f2
(1 − f 2 )(8f ) − 4f 2 (−2f )
(1 − f 2 )2
8f
(1 − f 2 )2
df
P+
dP
df
P+
dP
4f 2
1 − f2
= 0.
So
−4f 2 /(1 − f 2 )
−1
df
=
=
f (1 − f 2 ).
dP
8f P/(1 − f 2 )2
2P
(b) Since f is a fraction of a gas, 0 ≤ f ≤ 1. Also, in the equation relating f and P we can’t have f = 0, since that
would imply K = 0, and we can’t have f = 1, since the left side is undefined there. So 0 < f < 1. Thus 1 − f 2 > 0.
Also, pressure can’t be negative, and from the equation relating f and P , we see that P can’t be zero either, so P > 0.
Therefore df /dP = −(1/2P )f (1 − f 2 ) < 0 always. This means that at larger pressures less of the gas decomposes.
38. Let the point of intersection of the tangent line with the smaller circle be (x1 , y1 ) and the point of intersection with the
larger be (x2 , y2 ). Let the tangent line be y = mx + c. Then at (x1 , y1 ) and (x2 , y2 ) the slopes of x2 + y 2 = 1 and
y 2 +(x−3)2 = 4 are also m. The slope of x2 +y 2 = 1 is found by implicit differentiation: 2x+2yy ′ = 0 so y ′ = −x/y.
Similarly, the slope of y 2 + (x − 3)2 = 4 is y ′ = −(x − 3)/y. Thus,
m=
p
where y1 = 1 − x21 and y2 =
our choice of m > 0. We obtain
p
(x2 − 3)
x1
y2 − y1
=−
=−
,
x2 − x1
y1
y2
4 − (x2 − 3)2 . The positive values for y1 and y2 follow from Figure 3.9 and from
x1
p
1
1
− x21
x21
− x21
2
x2 − 3
= p
4 − (x2 − 3)2
=
(x2 − 3)2
4 − (x2 − 3)2
x21 [4 − (x2 − 3) ] = (1 − x21 )(x2 − 3)2
4x21 − (x21 )(x2 − 3)2 = (x2 − 3)2 − x21 (x2 − 3)2
4x21 = (x2 − 3)2
2|x1 | = |x2 − 3|.
From the picture x1 < 0 and x2 < 3. This gives x2 = 2x1 + 3 and y2 = 2y1 . From
y2 − y1
x1
=− ,
x2 − x1
y1
220
Chapter Three /SOLUTIONS
substituting y1 =
p
1 − x21 , y2 = 2y1 and x2 = 2x1 + 3 gives
1
x1 = − .
3
From x2 = 2x1 + 3 we get x2 = 7/3. In addition, y1 =
√
y2 = 4 2/3.
√
1 − x21 gives y1 = 2 2/3, and finally y2 = 2y1 gives
p
y
2
1
−1
1 2
x
3
4
5
−1
−2
Figure 3.9
39. Using implicit differentiation we have
1 = (cos y)
Therefore,
dy
.
dx
1
dy
=
.
dx
cos y
√
Solving the trig identity sin2 y + cos2 y = 1 for cos y and substituting x = sin y gives cos y = ± 1 − x2 . However,
because y = arcsin x, we have −π/2 ≤ y ≤ π/2, so cos y ≥ 0 and thus we take the positive root:
1
1
1
dy
.
=
= p
= √
dx
cos y
1 − x2
1 − sin2 y
m
m
40. y = x n . Taking nth powers of both sides of this expression yields (y)n = (x n )n , or y n = xm .
d m
d n
(y ) =
(x )
dx
dx
dy
= mxm−1
ny n−1
dx
dy
m xm−1
=
dx
n y n−1
m xm−1
=
n (xm/n )n−1
m xm−1
m
n xm− n
m
m
m m
= x(m−1)−(m− n ) = x n −1 .
n
n
=
41. Implicit differentiation gives
1−
2n2 a dV
V 3 dP
dV
dP
−
(V − nb) +
2n2 a
(V − nb) +
V3
dV
dP
P−
P+
n2 a
V2
P+
n2 a
V2
dV
=0
dP
n2 a
2n3 ab
+
2
V
V3
= nb − V
= nb − V
dV
nb − V
=
.
dP
P − n2 a/V 2 + 2n3 ab/V 3
3.8 SOLUTIONS
221
Strengthen Your Understanding
42. Since we cannot solve for y in terms of x, we need to differentiate implicitly. This gives
d
d
(y) =
(sin(xy)) ,
dx
dx
so
dy
dy
= cos(xy) 1 · y + x ·
.
dx
dx
Solving for dy/dx gives
y cos(xy)
dy
=
.
dx
1 − x cos(xy)
43. The formula applies only when the point (x, y) is on the circle.
44. One possible format for the slope is
x2 − 4
dy
=
.
dx
y−2
To confirm that the tangents exist, we would need to have the formula for the implicit function and check that there are
points satisfying it with y = 2 and x = ±2.
45. A hyperbola opening up and down works here.
x
dy
= . Although the denominator is zero when y = 0, the original
dx
y
equation is √
not satisfied by y = 0. Thus, there are no vertical tangents to the curve. Horizontal tangents occur when x = 0
and y = ± 1.
For example, −x2 + y 2 = 1. Note that
46. True; differentiating the equation with respect to x, we get
2y
dy
dy
+y+x
= 0.
dx
dx
Solving for dy/dx, we get that
dy
−y
=
.
dx
2y + x
Thus dy/dx exists where 2y + x 6= 0. Now if 2y + x = 0, then x = −2y. Substituting for x in the original equation,
y 2 + xy − 1 = 0, we get
y 2 − 2y 2 − 1 = 0.
This simplifies to y 2 + 1 = 0, which has no solutions. Thus dy/dx exists everywhere.
Solutions for Section 3.8
Exercises
d
(sinh(3z + 5)) = cosh(3z + 5) · 3 = 3 cosh(3z + 5).
dz
d
2. Using the chain rule,
(cosh(2x)) = (sinh(2x)) · 2 = 2 sinh(2x).
dx
3. Using the chain rule,
d
cosh2 t = 2 cosh t · sinh t.
dt
.
1. Using the chain rule,
4. Using the chain rule,
.
d
(cosh(sinh t)) = sinh(sinh t) · cosh t
dt
222
Chapter Three /SOLUTIONS
5. Using the product rule,
d 3
t sinh t = 3t2 sinh t + t3 cosh t.
dt
d
(cosh(3t) sinh(4t)) = 3 sinh(3t) sinh(4t) + 4 cosh(3t) cosh(4t).
dt
d
1
7. Using the chain rule,
(tanh(3 + sinh x)) =
· cosh x.
dx
cosh2 (3 + sinh x)
2
2
2
2
2
d
cosh(et ) = sinh(et ) · et · 2t = 2tet sinh(et ).
8. Using the chain rule twice,
dt
9. Using the chain rule,
6. Using the product and chain rules,
sinh(1 + θ)
d
1
(ln (cosh(1 + θ))) =
· sinh(1 + θ) =
= tanh(1 + θ).
dθ
cosh(1 + θ)
cosh(1 + θ)
10. Using the chain rule twice,
d
(sinh (sinh(3y))) = cosh (sinh(3y)) · cosh(3y) · 3
dy
= 3 cosh(3y) · cosh (sinh(3y)) .
11. Using the chain rule, f ′ (t) = 2 cosh t sinh t − 2 sinh t cosh t = 0. This is to be expected since cosh2 t − sinh2 t = 1.
12. Using the formula for sinh x and the fact that d(e−x )/dx = −e−x , we see that
d
dx
ex − e−x
2
=
ex + e−x
= cosh x.
2
13. Substitute x = 0 into the formula for sinh x. This yields
sinh 0 =
e0 − e−0
1−1
=
= 0.
2
2
14. Substituting −x for x in the formula for sinh x gives
sinh(−x) =
e−x − ex
ex − e−x
e−x − e−(−x)
=
=−
= − sinh x.
2
2
2
15. By definition cosh x = (ex + e−x )/2 so, since eln t = t and e− ln t = 1/eln t = 1/t, we have
cosh(ln t) =
t + 1/t
t2 + 1
eln t + e− ln t
=
=
.
2
2
2t
16. By definition sinh x = (ex − e−x )/2 so, since eln t = t and e− ln t = 1/eln t = 1/t, we have
sinh(ln t) =
t − 1/t
t2 − 1
eln t − e− ln t
=
=
.
2
2
2t
Problems
17. The graph of sinh x in the text suggests that
As x → ∞,
As x → −∞,
1 x
e .
2
1
sinh x → − e−x .
2
sinh x →
3.8 SOLUTIONS
Using the facts that
As x → ∞,
As x → −∞,
e−x → 0,
ex → 0,
we can obtain the same results analytically:
As x → ∞,
1
ex − e−x
→ ex .
2
2
1
ex − e−x
→ − e−x .
sinh x =
2
2
sinh x =
As x → −∞,
18. Using the identity
cosh2 t − sinh2 t = 1,
we see
x2 − y 2 = 1.
19. First we observe that
sinh(2x) =
e2x − e−2x
.
2
Now let’s calculate
(sinh x)(cosh x) =
ex − e−x
2
ex + e−x
2
(ex )2 − (e−x )2
4
e2x − e−2x
=
4
1
= sinh(2x).
2
=
Thus, we see that
sinh(2x) = 2 sinh x cosh x.
20. First, we observe that
cosh(2x) =
e2x + e−2x
.
2
Now let’s use the fact that ex · e−x = 1 to calculate
2
cosh x =
ex + e−x
2
ex − e−x
2
2
(ex )2 + 2ex · e−x + (e−x )2
4
e2x + 2 + e−2x
=
.
4
=
Similarly, we have
sinh2 x =
2
(ex )2 − 2ex · e−x + (e−x )2
4
e2x − 2 + e−2x
.
=
4
=
223
224
Chapter Three /SOLUTIONS
Thus, to obtain cosh(2x), we need to add (rather than subtract) cosh2 x and sinh2 x, giving
e2x + 2 + e−2x + e2x − 2 + e−2x
4
2e2x + 2e−2x
=
4
e2x + e−2x
=
2
= cosh(2x).
cosh2 x + sinh2 x =
Thus, we see that the identity relating cosh(2x) to cosh x and sinh x is
cosh(2x) = cosh2 x + sinh2 x.
21. Recall that
1 A
(e − e−A )
2
Now substitute, expand and collect terms:
sinh A =
and
cosh A =
1 A
(e + e−A ).
2
1 A
1
1
1
(e + e−A ) · (eB + e−B ) + (eA − e−A ) · (eB − e−B )
2
2
2
2
1 A+B
=
e
+ eA−B + e−A+B + e−(A+B)
4
+eB+A − eB−A − e−B+A + e−A−B
1 A+B
=
e
+ e−(A+B)
2
= cosh(A + B).
cosh A cosh B + sinh A sinh B =
22. Recall that
1 A
(e − e−A )
2
Now substitute, expand and collect terms:
sinh A =
and
cosh A =
1 A
(e + e−A ).
2
1
1
1
1 A
(e − e−A ) · (eB + e−B ) + (eA + e−A ) · (eB − e−B )
2
2
2
2
1 A+B
e
+ eA−B − e−A+B − e−(A+B)
=
4
+eB+A + eB−A − e−B+A − e−A−B
sinh A cosh B + cosh A sinh B =
1 A+B
e
− e−(A+B)
2
= sinh(A + B).
=
23. limx→∞ sinh(2x)/ cosh(3x) = limx→∞ (e2x − e−2x )/(e3x + e−3x ) = limx→∞ (1 − e−4x )/(ex + e−5x ) = 0.
24. Using the definition of sinh x, we have sinh 2x =
lim
x→∞
e2x − e−2x
. Therefore
2
e2x
2e2x
= lim 2x
x→∞ e
sinh(2x)
− e−2x
2
= lim
x→∞ 1 − e−4x
= 2.
3.8 SOLUTIONS
2
25. Using the definition of cosh x and sinh x, we have cosh x2 =
2
2
225
2
ex − e−x
ex + e−x
and sinh x2 =
. Therefore
2
2
2
2
sinh(x2 )
ex − e−x
= lim x2
2
x→∞ e
x→∞ cosh(x )
+ e−x2
lim
2
2
ex (1 − e−2x )
= lim x2
2
x→∞ e (1 + e−2x )
2
1 − e−2x
2
x→∞ 1 + e−2x
= 1.
= lim
26. Using the definition of cosh x and sinh x, we have cosh 2x =
e3x − e−3x
e2x + e−2x
and sinh 3x =
. Therefore
2
2
cosh(2x)
e2x + e−2x
= lim 3x
x→∞ sinh(3x)
x→∞ e
− e−3x
lim
e2x (1 + e−4x )
x→∞ e2x (ex − e−5x )
= lim
1 + e−4x
x→∞ ex − e−5x
= 0.
= lim
27. Note that
ekx + e−kx
2
e(k−3)x + e−(k+3)x
=
.
2
e−3x cosh kx = e−3x
If |k| = 3, then the limit as x → ∞ is 1/2.
If |k| > 3, then the limit as x → ∞ does not exist.
If |k| < 3, then the limit as x → ∞ is 0.
28. Note that
ekx − e−kx
sinh kx
= 2x
cosh 2x
e + e−2x
e2x (e(k−2)x − e−(k+2)x )
=
e2x (1 + e−4x )
=
e(k−2)x − e−(k+2)x
.
1 + e−4x
If k = 2, then the limit as x → ∞ is 1.
If |k| > 2, then the limit as x → ∞ does not exist.
If |k| < 2, then the limit as x → ∞ is 0.
29. (a) Since the cosh function is even, the height, y, is the same at x = −T /w and x = T /w. The height at these endpoints
is
w T
T
T e1 + e−1
T
·
= cosh 1 =
.
y = cosh
w
T w
w
w
2
At the lowest point, x = 0, and the height is
y=
T
T
cosh 0 = .
w
w
T
T
=
w
w
Thus the “sag” in the cable is given by
Sag =
T
w
e + e−1
2
−
e + e−1
−1
2
≈ 0.54
T
.
w
226
Chapter Three /SOLUTIONS
(b) To show that the differential equation is satisfied, take derivatives
T w
wx
dy
=
· sinh
dx
w T
T
w
d2 y
wx
=
cosh
.
dx2
T
T
= sinh
wx
T
Therefore, using the fact that 1 + sinh2 a = cosh2 a and that cosh is always positive, we have:
w
T
r
1+
dy
dx
So
2
r
w
wx
=
1 + sinh2
T
T
wx
w
cosh
.
=
T
T
w
T
r
1+
y
30.
dy
dx
2
=
w
=
T
r
cosh2
wx
T
d2 y
.
dx2
(0, 615)
x
(265, 0)
We know x = 0 and y = 615 at the top of the arch, so
615 = b − a cosh(0/a) = b − a.
This means b = a + 615. We also know that x = 265 and y = 0 where the arch hits the ground, so
0 = b − a cosh(265/a) = a + 615 − a cosh(265/a).
We can solve this equation numerically on a calculator and get a ≈ 100, which means b ≈ 715. This results in the
equation
x
.
y ≈ 715 − 100 cosh
100
31. (a) The graph in Figure 3.10 looks like the graph of y = cosh x, with the minimum at about (0.5, 6.3).
y
y = 2ex + 5e−x
40
20
(0.5, 6.3)
x
−3
3
Figure 3.10
3.8 SOLUTIONS
227
(b) We want to write
A x−c A −(x−c)
e
+ e
2
2
A x −c A −x c
= e e + e e
2
2 c
−c
Ae
Ae
ex +
e−x .
=
2
2
y = 2ex + 5e−x = A cosh(x − c) =
Thus, we need to choose A and c so that
Ae−c
=2
2
and
Aec
= 5.
2
Dividing gives
5
Aec
=
Ae−c
2
e2c = 2.5
1
c = ln 2.5 ≈ 0.458.
2
Solving for A gives
A=
4
= 4ec ≈ 6.325.
e−c
Thus,
y = 6.325 cosh(x − 0.458).
Rewriting the function in this way shows that the graph in part (a) is the graph of cosh x shifted to the right by 0.458
and stretched vertically by a factor of 6.325.
32. We want to show that for any A, B with A > 0, B > 0, we can find K and c such that
Ke(x−c) + Ke−(x−c)
2
K x −c K −x c
=
e e + e e
2 −c 2 c
Ke
Ke
ex +
e−x .
=
2
2
y = Aex + Be−x =
Thus, we want to find K and c such that
Ke−c
=A
2
and
Kec
= B.
2
Dividing, we have
B
Kec
=
Ke−c
A
B
2c
e =
A
B
1
.
c = ln
2
A
If A > 0, B > 0, then there is a solution for c. Substituting to find K, we have
Ke−c
=A
2
K = 2Aec = 2Ae(ln(B/A))/2
r
√
√
B
ln B/A
= 2A
= 2Ae
= 2 AB.
A
Thus, if A > 0, B > 0, there is a solution for K also.
The fact that y = Aex + Be−x can be rewritten in this way shows that the graph of y = Aex + Be−x is the graph
of cosh x, shifted over by c and stretched (or shrunk) vertically by a factor of K.
228
Chapter Three /SOLUTIONS
33. (a) Substituting x = 0 gives
tanh 0 =
1−1
e0 − e−0
=
= 0.
e0 + e−0
2
ex − e−x
and ex + e−x is always positive, tanh x has the same sign as ex − e−x . For x > 0, we
ex + e−x
have ex > 1 and e−x < 1, so ex − e−x > 0. For x < 0, we have ex < 1 and e−x > 1, so ex − e−x < 0. For x = 0,
we have ex = 1 and e−x = 1, so ex − e−x = 0. Thus, tanh x is positive for x > 0, negative for x < 0, and zero for
x = 0.
(c) Taking the derivative, we have
d
1
.
(tanh x) =
dx
cosh2 x
Thus, for all x,
d
(tanh x) > 0.
dx
Thus, tanh x is increasing everywhere.
(d) As x → ∞ we have e−x → 0; as x → −∞, we have ex → 0. Thus
(b) Since tanh x =
lim tanh x = lim
x→∞
x→∞
lim tanh x = lim
x→−∞
x→−∞
ex − e−x
ex + e−x
ex − e−x
ex + e−x
= 1,
= −1.
Thus, y = 1 and y = −1 are horizontal asymptotes to the graph of tanh x. See Figure 3.11.
y
1
x
−5
5
−1
Figure 3.11: Graph of y = tanh x
(e) The graph of tanh x suggests that tanh x is increasing everywhere; the fact that the derivative of tanh x is positive
for all x confirms this. Since tanh x is increasing for all x, different values of x lead to different values of y, and
therefore tanh x does have an inverse.
Strengthen Your Understanding
34. Since f (x) = cosh x = (ex + e−x )/2, the function is not periodic.
35. Since f (x) = cosh x = (ex + e−x )/2 we have
d
f (x) =
dx
′
ex + e−x
2
=
ex − e−x
= sinh x.
2
36. The required identity for hyperbolic functions is cosh2 x − sinh2 x = 1.
37. Since cosh x and sinh x behave like ex /2 as x → ∞ then tanh x = sinh x/ cosh x → 1 as x → ∞.
38. f (x) = cosh x is concave up. Other answers are possible.
39. We have
lim ekx cosh x = lim ekx
x→∞
x→∞
ex + e−x
2
This limit does not exist for any value k > −1: for example, k = 0.
= lim
x→∞
e(k+1)x + e(k−1)x
.
2
40. Since cosh 0 = 1, we can shift cosh x to the right by 1 and up 2. Thus, f (x) = cosh(x − 1) + 2 is a possibility.
3.9 SOLUTIONS
229
41. True. We have tanh x = (sinh x) / cosh x = (ex − e−x )/(ex + e−x ). Replacing x by −x in this expression gives
(e−x − ex )/(e−x + ex ) = − tanh x.
42. False. The second, fourth and all even derivatives of sinh x are all sinh x.
43. True. The definitions of sinh x and cosh x give
sinh x + cosh x =
ex + e−x
2ex
ex − e−x
+
=
= ex .
2
2
2
44. False. Since (sinh x)′ = cosh x > 0, the function sinh x is increasing everywhere so can never repeat any of its values.
45. False. Since (sinh2 x)′ = 2 sinh x cosh x and (2 sinh x cosh x)′ = 2 sinh2 x + 2 cosh2 x > 0, the function sinh2 x is
concave up everywhere.
Solutions for Section 3.9
Exercises
√
√
1. With f (x) = 1 + x, the chain rule gives f ′ (x) = 1/(2 1 + x), so f (0) = 1 and f ′ (0) = 1/2. Therefore the tangent
line approximation of f near x = 0,
f (x) ≈ f (0) + f ′ (0)(x − 0),
becomes
√
x
.
2
√
This means that, near x = 0, the function 1 + x can be approximated by its tangent line y = 1 + x/2. (See Figure 3.12.)
1+x≈1+
y
y = 1 + x/2
2
y=
√
1+x
1
x
−2
−1
0
1
2
Figure 3.12
2. With f (x) = ex , the tangent line approximation to f near x = 0 is f (x) ≈ f (0) + f ′ (0)(x − 0) which becomes
ex ≈ e0 + e0 x = 1 + 1x = 1 + x. Thus, our local linearization of ex near x = 0 is ex ≈ 1 + x.
3. With f (x) = 1/x, we see that the tangent line approximation to f near x = 1 is
f (x) ≈ f (1) + f ′ (1)(x − 1),
which becomes
1
≈ 1 + f ′ (1)(x − 1).
x
Since f ′ (x) = −1/x2 , f ′ (1) = −1. Thus our formula reduces to
1
≈ 1 − (x − 1) = 2 − x.
x
This is the local linearization of 1/x near x = 1.
230
Chapter Three /SOLUTIONS
4. Since f (1) = 1 and we showed that f ′ (1) = 2, the local linearization is
f (x) ≈ 1 + 2(x − 1) = 2x − 1.
2
2
5. With f (x) = ex , we get a tangent line approximation of f (x) ≈ f (1) + f ′ (1)(x − 1) which becomes ex ≈ e +
2xex
2
2
x=1
2
(x − 1) = e + 2e(x − 1) = 2ex − e. Thus, our local linearization of ex near x = 1 is ex ≈ 2ex − e.
√
6. With f (x) = 1/( 1 + x), we see that the tangent line approximation to f near x = 0 is
f (x) ≈ f (0) + f ′ (0)(x − 0),
which becomes
√
1
≈ 1 + f ′ (0)x.
1+x
Since f ′ (x) = (−1/2)(1 + x)−3/2 , f ′ (0) = −1/2. Thus our formula reduces to
√
1
≈ 1 − x/2.
1+x
1
near x = 0.
1+x
7. Let f (x) = e−x . Then f ′ (x) = −e−x . So f (0) = 1, f ′ (0) = −e0 = −1. Therefore, e−x ≈ f (0) + f ′ (0)x = 1 − x.
This is the local linearization of √
8. The graph of x2 is concave
√up and lies above its tangent line; therefore, the linearization will always be too small. See
Figure 3.13. The graph of x is concave down and lies below its tangent line, and therefore the linearization will be too
large. See Figure 3.14.
y
2
1
y
y = x2
y=
√
x
x
−1
x
1
2
Figure 3.13
4
Figure 3.14
9. From Figure 3.15, we see that the error has its maximum magnitude at the end points of the interval, x = ±1. The
magnitude of the error can be read off the graph as less than 0.2 or estimated as
|Error| ≤ |1 − sin 1| = 0.159 < 0.2.
The approximation is an overestimate for x > 0 and an underestimate for x < 0.
y
❄
x
1
Error
✻
sin x
x
−1
1
−1
Figure 3.15
3.9 SOLUTIONS
231
10. Figure 3.16 shows that 1 + x is an underestimate of ex for −1 ≤ x ≤ 1. On this interval, the error has the largest
magnitude at x = 1. Its magnitude can be estimated from the graph as less than 0.8, or estimated as
|Error| = e − 1 − 1 = 0.718 < 0.8.
y
3
ex
✻
❄
Error
2
1+x
1
x
−1
1
Figure 3.16
Problems
11. (a) If f (x) = ex , then f ′ (x) = ex , so f ′ (0) = e0 = 1. Thus, the local linearization near x = 0 is
L(x) = f (0) + f ′ (0)(x − 0)
L(x) = 1 + x.
x
(b) Since the curve f (x) = e is above the line L(x) = 1 + x (see Figure 3.17), the error is positive everywhere except
at x = 0, where it is 0.
(c) The true value of the function at x = 1 is f (1) = e1 = e = 2.718. The approximation is L(1) = 1 + 1 = 2. The
error is E(1) = f (1) − L(1) = e − 2 = 0.718. See Figure 3.17.
(d) We expect E(1) to be larger than E(0.1), because 0.1 is closer to x = 0, the point at which the linear approximation,
L(x), is exactly equal to the function, f (x).
(e) We have
E(0.1) = f (0.1) − L(0.1) = e0.1 − (1 + 0.1) = 1.105 − 1.1 = 0.005.
✛
✲
✻
✛
E(1)
✻
f (1)
L(1)
1
f (x) = ex
L(x) = 1 + x
❄ ❄x
−1
1
Figure 3.17
12. (a) Since
d
(cos x) = − sin x,
dx√
the slope√
of the tangent line is − sin(π/4) = −1/ 2. Since the tangent line passes through the point (π/4, cos(π/4)) =
(π/4, 1/ 2), its equation is
1
1
π
y − √ = −√ x −
4
2
2
π
1
1
y = −√ x + √
+1 .
2
2 4
232
Chapter Three /SOLUTIONS
Thus, the tangent line approximation to cos x is
1
1
cos x ≈ − √ x + √
2
2
π
+1 .
4
(b) From Figure 3.18, we see that the tangent line approximation is an overestimate.
(c) From Figure 3.18, we see that the maximum error for 0 ≤ x ≤ π/2 is either at x = 0 or at x = π/2. The error can
either be estimated from the graph, or as follows. At x = 0,
1
|Error| = cos 0 − √
2
At x = π/2,
|Error| = cos
π
+1
4
1 π
1
π
+√
−√
2
2
2
2
Thus, for 0 ≤ x ≤ π/2, we have
Error
= 0.262 < 0.3.
π
+1
4
= 0.152 < 0.2.
|Error| < 0.3.
✻
❄
√
1/ 2
Tangent Line
1
1
−√ x + √
2
2
cos x
π/4
x
π
+1
4
π/2
Figure 3.18
13. (a) See Figure 3.19.
(b) Since f (2) = 23 − 3 · 22 + 3 · 2 + 1 = 3, we have f ′ (x) = 3x2 − 6x + 3, so f ′ (2) = 3 · 22 − 6 · 2 + 3 = 3, and
the local linearization is y = 3 + 3(x − 2) = 3x − 3.
(c) See Figure 3.19.
y
f (x) = x3 − 3x2 + 3x + 1
3
✛
True value
Approximation
Error
y = 3x − 3
x
2
−3
Figure 3.19
14. (a) Let f (x) = (1 + x)k . Then f ′ (x) = k(1 + x)k−1 . Since
f (x) ≈ f (0) + f ′ (0)(x − 0)
is the tangent line approximation, and f (0) = 1, f ′ (0) = k, for small x we get
f (x) ≈ 1 + kx.
3.9 SOLUTIONS
233
√
(b) Since 1.1 = (1 + 0.1)1/2 ≈ 1 + (1/2)0.1 = 1.05 by the above method, this estimate is about right.
(c) The real answer is less than 1.05. Since (1.05)2 = (1 + 0.05)2 = 1 + 2(1)(0.05) + (0.05)2 = 1.1 + (0.05)2 > 1.1,
we have (1.05)2 > 1.1 Therefore
√
1.1 < 1.05.
√
Graphically,
this because the graph of 1 + x is concave down, so it bends below its tangent line. Therefore the true
√
value ( 1.1) which is on the curve is below the approximate value (1.05) which is on the tangent line.
15. Since the line meets the curve at x = 1, we have a = 1. Since the point with x = 1 lies on both the line and the curve,
we have
f (a) = f (1) = 2 · 1 − 1 = 1.
The approximation is an underestimate because the line lies under the curve. Since the linear function approximates f (x),
we have
f (1.2) ≈ 2(1.2) − 1 = 1.4.
16. We have f (x) = ex + x, so f ′ (x) = ex + 1. Thus f ′ (0) = 2, so
Local linearization near x = 0
f (x) ≈ f (0) + f ′ (0)x = 1 + 2x.
is
We get an approximate solution using the local linearization instead of f (x), so the equation becomes
1 + 2x = 2,
x=
with solution
1
.
2
A computer or calculator gives the actual solution as x = 0.443.
17. We have f (x) = x + ln(1 + x), so f ′ (x) = 1 + 1/(1 + x). Thus f ′ (0) = 2 so
Local linearization near x = 0
is
f (x) ≈ f (0) + f ′ (0)x = 2x.
We get an approximate solution using the local linearization instead of f (x), so the equation becomes
2x = 0.2,
with solution
x = 0.1.
A computer or calculator gives the actual value as x = 0.102.
18. (a) The line tangent to the graph of f at x = 7 is given by
y = f (7) + f ′ (7)(x − 7)
= 13 − 0.38(x − 7).
We can use the tangent line to approximate the value of f (7.1):
f (7.1) ≈ 13 − 0.38(7.1 − 7)
= 13 − 0.38(0.1)
= 12.962.
′′
(b) If f (x) < 0, then the graph of f is everywhere concave down, so it lies below its tangent line. Thus, our tangent
line approximation is an overestimate.
19. (a) Zooming in on the graphs of y = et and y = 0.02t + 1.098 shows they cross just to the right of the origin.
Numerically, we see that at t = 0
e0 = 1 < 0.02 · 0 + 1.098.
At t = 0.2, we have
e0.2 = 1.221 > 0.02 · 2 + 1.098 = 1.102.
Therefore, somewhere between t = 0 and t = 0.2 the equation et = 0.02t + 1.098 has a solution.
(b) The linearization of et near 0 is 1 + t, so the new equation is
1 + t = 0.02t + 1.098
0.98t = 0.098
0.098
t=
= 0.1.
0.98
234
Chapter Three /SOLUTIONS
20. We have
f (0) = 331.3
T
1
· 331.3 1 +
2
273.15
f ′ (0) = 0.606.
f ′ (T ) =
−1/2
1
273.15
Thus, for temperatures, T , near zero, we have
Speed of sound = f (T ) ≈ f (0) + f ′ (0)T = 331.3 + 0.606T meters/second.
21. (a) See Figure 3.20.
P
30
−5
P = 30e−3.23×10
h
h
Figure 3.20
(b) Using the chain rule, we have
−5
dP
= 30e−3.23×10 h (−3.23 × 10−5 )
dh
so
dP
dh
h=0
= −30(3.23 × 10−5 ) = −9.69 × 10−4
Hence, at h = 0, the slope of the tangent line is −9.69 × 10−4 , so the equation of the tangent line is
y − 30 = (−9.69 × 10−4 )(h − 0)
y = (−9.69 × 10−4 )h + 30 = 30 − 0.000969h.
(c) The rule of thumb says
Drop in pressure from
h
=
sea level to height h
1000
But since the pressure at sea level is 30 inches of mercury, this drop in pressure is also (30 − P ), so
30 − P =
h
1000
giving
P = 30 − 0.001h.
(d) The equations in (b) and (c) are almost the same: both have P intercepts of 30, and the slopes are almost the same
(9.69 × 10−4 ≈ 0.001). The rule of thumb calculates values of P which are very close to the tangent lines, and
therefore yields values very close to the curve.
(e) The tangent line is slightly below the curve, and the rule of thumb line, having a slightly more negative slope, is
slightly below the tangent line (for h > 0). Thus, the rule of thumb values are slightly smaller.
22. (a) The derivative gives the rate of change of the number of Android users. We estimate A′ (0) as the rate of change over
the year:
10.9 − 0.866
= 10.034 million users/year.
A′ (0) =
1
(b) Similarly,
13.5 − 7.8
= 5.7 million users/year.
P ′ (0) =
1
3.9 SOLUTIONS
235
(c) Since 866,000 = 0.866 million, the tangent line approximation to A(t) near t = 0 is
A(t) ≈ 0.866 + 10.034t.
The tangent line approximation for P (t) is
P (t) ≈ 7.8 + 5.7t.
The two groups of users are predicted to be the same size when
0.866 + 10.034t = 7.8 + 5.7t
10.034t − 5.7t = 7.8 − 0.866
7.8 − 0.866
= 1.600 years.
t=
10.034 − 5.7
Thus, Android and iPhones were predicted to have the same number of users 1.6 years after 2009; that is, in mid
2011.
(d) We are assuming that the rates of change of each group of users remain the same in the future.
23. (a) Suppose g is a constant and
T = f (l) = 2π
Then
r
l
.
g
2π 1
π
f ′ (l) = √ l−1/2 = √ .
g2
gl
Thus, local linearity tells us that
π
f (l + ∆l) ≈ f (l) + √ ∆l.
gl
Now T = f (l) and ∆T = f (l + ∆l) − f (l), so
π
∆T ≈ √ ∆l = 2π
gl
r
l 1 ∆l
T ∆l
·
=
.
g 2 l
2 l
(b) Knowing that the length of the pendulum increases by 2% tells us that
∆l
= 0.02.
l
Thus,
∆T ≈
So
T
(0.02) = 0.01T.
2
∆T
≈ 0.01.
T
Thus, T increases by 1%.
24. (a) Considering l as a constant, we have
T = f (g) = 2π
Then,
Thus, local linearity gives
r
l
.
g
√ 1
f ′ (g) = 2π l − g −3/2 = −π
2
f (g + ∆g) ≈ f (g) − π
Since T = f (g) and ∆T = f (g + ∆g) − f (g), we have
r
l
∆g = −2π
∆T ≈ −π
g3
−T ∆g
∆T ≈
.
2 g
r
r
r
l
.
g3
l
(∆g).
g3
l ∆g
−T ∆g
=
.
g 2g
2 g
236
Chapter Three /SOLUTIONS
(b) If g increases by 1%, we know
∆g
= 0.01.
g
Thus,
∆T
1 ∆g
1
≈−
= − (0.01) = −0.005,
T
2 g
2
So, T decreases by 0.5%.
25. Since f has a positive second derivative, its graph is concave up, as in Figure 3.21 or 3.22. This means that the graph of
f (x) is above its tangent line. We see that in both cases
f (1 + ∆x) ≥ f (1) + f ′ (1)∆x.
(The diagrams show ∆x positive, but the result is also true if ∆x is negative.)
y
y
f (x)
f (1 + ∆x)
f (1) + f ′ (1)∆x
Tangent line
Slope = f ′ (1)
f (1)
f (1)
f (x)
f (1 + ∆x)
f (1) +
f ′ (1)∆x
Tangent line
Slope = f ′ (1)
x
x
1
1
1 + ∆x
Figure 3.21
1 + ∆x
Figure 3.22
26. (a) Since f ′ is decreasing, f ′ (5) is larger.
(b) Since f ′ is decreasing, its derivative, f ′′ , is negative. Thus, f ′′ (5) is negative, so 0 is larger.
(c) Since f ′′ (x) is negative for all x, the graph of f is concave down. Thus the graph of f (x) is below its tangent line.
From Figure 3.23, we see that f (5 + ∆x) is below f (5) + f ′ (5)∆x. Thus, f (5) + f ′ (5)∆x is larger.
✛
f ′ (5)∆x
∆x
✲
✻
Tangent line
Slope = f ′ (5)
✻
❄
✻
f (x)
f (5 + ∆x)
f (5)
❄
5
❄
x
5 + ∆x
Figure 3.23
27. (a) The range is f (20) = 16,398 meters.
(b) We have
π
πθ
25510 cos
90
90
f ′ (20) = 682.
f ′ (θ) =
Thus, for angles, θ, near 20◦ , we have
Range = f (θ) ≈ f (20) + f ′ (20)(θ − 20) = 16398 + 682(θ − 20) meters.
(c) The true range for 21◦ is f (21) = 17,070 meters. The linear approximation gives
Approximate range = 16398 + 682(21 − 20) = 17080 meters
which is a little too high.
3.9 SOLUTIONS
237
28. (a) The time in the air is t(20) = 34.9 seconds
(b) We have
πθ
π
102 cos
180
180
t′ (20) = 1.67
t′ (θ) =
Thus, for angles, θ, near 20◦ , we have
Time in air = t(θ) ≈ t(20) + t′ (20)(θ − 20) = 34.9 + 1.67(θ − 20) seconds.
(c) The true air time for 21◦ is t(21) = 36.6 seconds. The linear approximation gives
Approximate peak altitude = 34.9 + 1.67(21 − 20) = 36.6 seconds
which is the correct value to three significant figures.
29. (a) The peak altitude is h(20) = 1492 meters
(b) We have
πθ
πθ
π
12755 sin
cos
180
180
180
h′ (20) = 143.
h′ (θ) = 2 ·
Thus, for angles, θ, near 20◦ , we have
Peak altitude = h(θ) ≈ h(20) + h′ (20)(θ − 20) = 1490 + 143(θ − 20) meters.
(c) The true peak altitude for 21◦ is h(21) = 1638 meters. The linear approximation gives
Approximate peak altitude = 1492 + 143(21 − 20) = 1635 meters
which is a little too low.
30. We have f (x) = (1 + x)r , so f ′ (x) = r(1 + x)r−1 . Thus f ′ (0) = r so the local linearization near x = 0 is
f (x) ≈ f (0) + f ′ (0)x = 1 + rx.
Thus
(1 + x)r ≈ 1 + rx
for small values of x.
Using the linearization with r = 3/5 and x = 0.2, we have
1.23/5 = 1 +
3
0.2 = 1 + 0.12 = 1.12.
5
The actual value is 1.23/5 = 1.117.
31. We have f (x) = ekx , so f ′ (x) = kekx . Thus f ′ (0) = k so the local linearization near x = 0 is
f (x) ≈ f (0) + f ′ (0)x = 1 + kx.
Thus
ekx ≈ 1 + kx
for small values of x.
Using the linearization with k = 0.3 and x = 1, we have
e0.3 ≈ 1 + 0.3 = 1.3
The actual value is e0.3 = 1.350.
32. We have f (x) = (b2 + x)1/2 , so f ′ (x) = (1/2)(b2 + x)−1/2 . Thus f ′ (0) = 1/(2b) so the local linearization near x = 0
is
1
f (x) ≈ f (0) + f ′ (0)x = b + x.
2b
Thus
p
1
b2 + x ≈ b + x for small values of x.
2b
Using the linearization with b = 5 and x = 1, we have
√
1
26 ≈ 5 +
1 = 5.1.
10
√
The actual value is 26 = 5.099.
238
Chapter Three /SOLUTIONS
33. We have f (1) = 1 and f ′ (1) = 4. Thus
E(x) = x4 − (1 + 4(x − 1)).
Values of E(x)/(x − 1) near x = 1 are in Table 3.3.
Table 3.3
x
1.1
1.01
1.001
E(x)/(x − 1)
0.641
0.060401
0.006004
From the table, we can see that
E(x)
≈ 6(x − 1),
(x − 1)
so k = 6 and
E(x) ≈ 6(x − 1)2 .
In addition, f ′′ (1) = 12, so
f ′′ (1)
(x − 1)2 .
2
The same result can be obtained by rewriting the function x4 using x = 1 + (x − 1) and expanding:
E(x) ≈ 6(x − 1)2 =
x4 = (1 + (x − 1))4 = 1 + 4(x − 1) + 6(x − 1)2 + 4(x − 1)3 + (x − 1)4 .
Thus,
E(x) = x4 − (1 + 4(x − 1)) = 6(x − 1)2 + 4(x − 1)3 + (x − 1)4 .
For x near 1, the value of x − 1 is small, so we ignore powers of x − 1 higher than the first, giving
E(x) ≈ 6(x − 1)2 .
34. We have f (0) = 1 and f ′ (0) = 0. Thus
E(x) = cos x − 1.
Values for E(x)/(x − 0) near x = 0 are in Table 3.4.
Table 3.4
x
0.1
0.01
0.001
E(x)/(x − 0)
−0.050
−0.0050
−0.00050
From the table, we can see that
so k = −1/2 and
In addition, f ′′ (0) = −1, so
E(x)
≈ −0.5(x − 0),
(x − 0)
1
1
E(x) ≈ − (x − 0)2 = − x2 .
2
2
f ′′ (0) 2
1
E(x) ≈ − x2 =
x .
2
2
3.9 SOLUTIONS
35. We have f (0) = 1 and f ′ (0) = 1. Thus
E(x) = ex − (1 + x).
Values of E(x)/(x − 0) near x = 0 are in Table 3.5.
Table 3.5
x
0.1
0.01
0.001
E(x)/(x − 0)
0.052
0.0050
0.00050
From the table, we can see that
E(x)
≈ 0.5(x − 0)
(x − 0)
so k = 1/2 and
E(x) ≈
1
1
(x − 0)2 = x2 .
2
2
In addition, f ′′ (0) = 1, so
E(x) ≈
f ′′ (0) 2
1 2
x =
x
2
2
36. We have f (1) = 1 and f ′ (1) = 1/2. Thus
E(x) =
√
x − (1 +
1
(x − 1)).
2
Values of E(x)/(x − 1) near x = 1 are in Table 3.6.
Table 3.6
x
1.1
1.01
1.001
E(x)/(x − 1)
−0.0119
−0.00124
−0.000125
From the table, we can see that
E(x)
≈ −0.125(x − 1)
(x − 1)
so k = −1/8 and
In addition, f ′′ (1) = −1/4, so
1
E(x) ≈ − (x − 1)2 .
8
f ′′ (1)
1
E(x) ≈ − (x − 1)2 =
(x − 1)2 .
8
2
37. We have f (1) = 0 and f ′ (1) = 1. Thus
E(x) = ln x − (x − 1).
Values of E(x)/(x − 1) near x = 1 are in Table 3.7.
Table 3.7
x
1.1
1.01
1.001
E(x)/(x − 1)
−0.047
−0.0050
−0.00050
From the table, we see that
so k = −1/2 and
In addition, f ′′ (1) = −1, so
E(x)
≈ −0.5(x − 1),
(x − 1)
1
E(x) ≈ − (x − 1)2 .
2
1
f ′′ (1)
E(x) ≈ − (x − 1)2 =
(x − 1)2 .
2
2
239
240
Chapter Three /SOLUTIONS
38. The local linearization of ex near x = 0 is 1 + 1x so
ex ≈ 1 + x.
Squaring this yields, for small x,
e2x = (ex )2 ≈ (1 + x)2 = 1 + 2x + x2 .
Local linearization of e2x directly yields
e2x ≈ 1 + 2x
for small x. The two approximations are consistent because they agree: the tangent line approximation to 1 + 2x + x2 is
just 1 + 2x.
The first approximation is more accurate. One can see this numerically or by noting that the approximation for e2x
given by 1 + 2x is really the same as approximating ey at y = 2x. Since the other approximation approximates ey at
y = x, which is twice as close to 0 and therefore a better general estimate, it’s more likely to be correct.
39. (a) Let f (x) = 1/(1 + x). Then f ′ (x) = −1/(1 + x)2 by the chain rule. So f (0) = 1, and f ′ (0) = −1. Therefore, for
x near 0, 1/(1 + x) ≈ f (0) + f ′ (0)x = 1 − x.
(b) We know that for small y, 1/(1 + y) ≈ 1 − y. Let y = x2 ; when x is small, so is y = x2 . Hence, for small x,
1/(1 + x2 ) ≈ 1 − x2 .
(c) Since the linearization of 1/(1 + x2 ) is the line y = 1, and this line has a slope of 0, the derivative of 1/(1 + x2 ) is
zero at x = 0.
40. The local linearizations of f (x) = ex and g(x) = sin x near x = 0 are
f (x) = ex ≈ 1 + x
and
g(x) = sin x ≈ x.
Thus, the local linearization of ex sin x is the local linearization of the product:
ex sin x ≈ (1 + x)x = x + x2 ≈ x.
We therefore know that the derivative of ex sin x at x = 0 must be 1. Similarly, using the local linearization of 1/(1 + x)
near x = 0, 1/(1 + x) ≈ 1 − x, we have
1
ex sin x
= (ex )(sin x)
1+x
1+x
so the local linearization of the triple product
x = 0 is 1.
≈ (1 + x)(x)(1 − x) = x − x3
ex sin x
ex sin x
at x = 0 is simply x. And therefore the derivative of
at
1+x
1+x
41. Note that
f (x + h)g(x + h) − f (x)g(x)
.
h
We use the hint: For small h, f (x + h) ≈ f (x) + f ′ (x)h, and g(x + h) ≈ g(x) + g ′ (x)h. Therefore
[f (x)g(x)]′ = lim
h→0
f (x + h)g(x + h) − f (x)g(x) ≈ [f (x) + hf ′ (x)][g(x) + hg ′ (x)] − f (x)g(x)
= f (x)g(x) + hf ′ (x)g(x) + hf (x)g ′ (x)
+h2 f ′ (x)g ′ (x) − f (x)g(x)
= hf ′ (x)g(x) + hf (x)g ′ (x) + h2 f ′ (x)g ′ (x).
Therefore
lim
h→0
f (x + h)g(x + h) − f (x)g(x)
hf ′ (x)g(x) + hf (x)g ′ (x) + h2 f ′ (x)g ′ (x)
= lim
h→0
h
h
h (f ′ (x)g(x) + f (x)g ′ (x) + hf ′ (x)g ′ (x))
= lim
h→0
h
= lim f ′ (x)g(x) + f (x)g ′ (x) + hf ′ (x)g ′ (x)
h→0
′
= f (x)g(x) + f (x)g ′ (x).
3.9 SOLUTIONS
241
A more complete derivation can be given using the error term discussed in the section on Differentiability and Linear
Approximation in Chapter 2. Adapting the notation of that section to this problem, we write
f (x + h) = f (x) + f ′ (x)h + Ef (h)
and
g(x + h) = g(x) + g ′ (x)h + Eg (h),
Eg (h)
Ef (h)
= lim
= 0. (This implies that lim Ef (h) = lim Eg (h) = 0.)
h→0
h→0
h→0
h
h
We have
where lim
h→0
f (x + h)g(x + h) − f (x)g(x)
f (x)g(x)
Eg (h)
Ef (h)
=
+ f (x)g ′ (x) + f ′ (x)g(x) + f (x)
+ g(x)
h
h
h
h
f (x)g(x)
Ef (h)Eg (h)
′
′
′
′
−
+f (x)g (x)h + f (x)Eg (h) + g (x)Ef (h) +
h
h
The terms f (x)g(x)/h and −f (x)g(x)/h cancel out. All the remaining terms on the right, with the exception of the
second and third terms, go to zero as h → 0. Thus, we have
[f (x)g(x)]′ = lim
h→0
f (x + h)g(x + h) − f (x)g(x)
= f (x)g ′ (x) + f ′ (x)g(x).
h
42. Note that
[f (g(x))]′ = lim
h→0
f (g(x + h)) − f (g(x))
.
h
Using the local linearizations of f and g, we get that
f (g(x + h)) − f (g(x)) ≈ f g(x) + g ′ (x)h − f (g(x))
≈ f (g(x)) + f ′ (g(x))g ′(x)h − f (g(x))
= f ′ (g(x))g ′(x)h.
Therefore,
f (g(x + h)) − f (g(x))
h
f ′ (g(x))g ′(x)h
= lim
h→0
h
= lim f ′ (g(x))g ′(x) = f ′ (g(x))g ′(x).
[f (g(x))]′ = lim
h→0
h→0
A more complete derivation can be given using the error term discussed in the section on Differentiability and Linear
Approximation in Chapter 2. Adapting the notation of that section to this problem, we write
f (z + k) = f (z) + f ′ (z)k + Ef (k)
where lim
h→0
and
g(x + h) = g(x) + g ′ (x)h + Eg (h),
Ef (k)
Eg (h)
= lim
= 0.
k→0
h
k
Now we let z = g(x) and k = g(x + h) − g(x). Then we have k = g ′ (x)h + Eg (h). Thus,
f (z + k) − f (z)
f (g(x + h)) − f (g(x))
=
h
h
f (z) + f ′ (z)k + Ef (k) − f (z)
f ′ (z)k + Ef (k)
=
=
h
h
k
Ef (k)
f ′ (z)g ′ (x)h + f ′ (z)Eg (h)
+
·
=
h
k
h
f ′ (z)Eg (h)
Ef (k) g ′ (x)h + Eg (h)
′
′
= f (z)g (x) +
+
h
k
h
f ′ (z)Eg (h)
g ′ (x)Ef (k)
Eg (h) · Ef (k)
+
+
h
k
h·k
Now, if h → 0 then k → 0 as well, and all the terms on the right except the first go to zero, leaving us with the term
f ′ (z)g ′ (x). Substituting g(x) for z, we obtain
= f ′ (z)g ′ (x) +
[f (g(x))]′ = lim
h→0
f (g(x + h)) − f (g(x))
= f ′ (g(x))g ′(x).
h
242
Chapter Three /SOLUTIONS
43. We want to show that
lim
x→a
Substituting for f (x) we have
lim
x→a
f (x) − f (a)
= L.
x−a
f (a) + L(x − a) + EL (x) − f (a)
f (x) − f (a)
= lim
x→a
x−a
x−a
EL (x)
EL (x)
= lim L +
= L + lim
= L.
x→a
x→0 x − a
x−a
Thus, we have shown that f is differentiable at x = a and that its derivative is L, that is, f ′ (a) = L.
44. We know that the local linearization is y = 2x − 1. The fact that f (x) is differentiable tells us that
lim
x→0
x2 − (2x − 1)
E(x)
= lim
= 0.
x→0
x−1
x−1
Suppose we take ǫ = 1/10, then there is a δ such that
x2 − (2x − 1)
1
<
x−1
10
for all
|x − 1| < δ.
Thus
1
1
(x − 1) < x2 − (2x − 1) <
(x − 1) for all
10
10
To find this δ, we observe that
−
1 − δ < x < 1 + δ.
(x − 1)2
x2 − (2x − 1)
x2 − 2x + 1
=
=
= |x − 1|.
x−1
x−1
x−1
Therefore we can take δ = 1/10. Then
x2 − (2x − 1)
1
<
x−1
10
so
−
for all |x − 1| <
1
10
1
1
(x − 1) < |x2 − (2x − 1)| <
(x − 1).
10
10
Strengthen Your Understanding
45. The line y = x + 1 is the linear approximation for f (x) = ex near x = 0. If we move far from x = 0, the approximation
is useless.
For example, for x = 1, the approximation gives e1 ≈ 2 (instead of 2.718). For x = 2, our estimate of 3 is not a
good approximation for e2 = 7.389.
This linear approximation is only useful near x = 0.
46. The graph of F is concave down to the left of x = 0 and concave up to the right. The tangent at x = 0 lies beneath the
graph for x > 0 and above the graph for x < 0. Thus, the linear approximation near x = 0 is an overestimate for values
less than zero and an underestimate for values greater than zero.
47. In order to have the same linear approximation, two functions f and g must go through the same point at x = 0 and have
the same slope at x = 0. Thus, f (0) = g(0) and f ′ (0) = g ′ (0). While there are many answers, let f (x) = x3 + 1 and
g(x) = x4 + 1. Then the linear approximation for both functions near x = 0 is y = 1.
48. One possible answer is g(x) = cos x.
49. The linear approximation of a function f, for values of x near a, is given by f (x) ≈ f (a) + f ′ (a)(x − a). Since
f (x) = |x + 1| does not have a derivative at x = −1 this function does not have a linear approximation for x near −1.
Other answers are possible.
50. (a) False. Only if k = f ′ (a) is L the local linearization of f .
(b) False. Since f (a) = L(a) for any k, we have limx→a (f (x) − L(x)) = f (a) − L(a) = 0, but only if k = f ′ (a) is
L the local linearization of f .
3.10 SOLUTIONS
243
Solutions for Section 3.10
Exercises
1. False. The derivative, f ′ (x), is not equal to zero everywhere, because the function is not continuous at integral values of
x, so f ′ (x) does not exist there. Thus, the Constant Function Theorem does not apply.
2. True. If f ′ is positive on [a, b], then f is continuous and the Increasing Function Theorem applies. Thus, f is increasing
on [a, b], so f (a) < f (b).
3. False. Let f (x) = x3 on [−1, 1]. Then f (x) is increasing but f ′ (x) = 0 for x = 0.
4. False. The horse that wins the race may have been moving faster for some, but not all, of the race. The Racetrack Principle
guarantees the converse—that if the horses start at the same time and one moves faster throughout the race, then that horse
wins.
5. True. If g(x) is the position of the slower horse at time x and h(x) is the position of the faster, then g ′ (x) ≤ h′ (x) for
a < x < b. Since the horses start at the same time, g(a) = h(a), so, by the Racetrack Principle, g(x) ≤ h(x) for
a ≤ x ≤ b. Therefore, g(b) ≤ h(b), so the slower horse loses the race.
6. Yes, it satisfies the hypotheses and the conclusion. This function has two points, c, at which the tangent to the curve is
parallel to the secant joining (a, f (a)) to (b, f (b)), but this does not contradict the Mean Value Theorem. The function is
continuous and differentiable on the interval [a, b].
7. No, it does not satisfy the hypotheses. The function does not appear to be differentiable. There appears to be no tangent
line, and hence no derivative, at the “corner.”
No, it does not satisfy the conclusion as there is no horizontal tangent.
8. No. This function does not satisfy the hypotheses of the Mean Value Theorem, as it is not continuous.
However, the function has a point c such that
f ′ (c) =
f (b) − f (a)
.
b−a
Thus, this satisfies the conclusion of the theorem.
9. No, it does not satisfy the hypotheses. This function does not appear to be continuous.
No, it does not satisfy the conclusion as there is no horizontal tangent.
Problems
10. The Mean Value Theorem tells us that
f ′ (4) =
f (b) − f (a)
9−5
4
=
= .
b−a
7−2
5
Thus, the slope of the tangent line is 4/5. Its equation is
y =b+
4
x.
5
8 = b+
4
(4)
5
Substituting x = 4, y = 8 gives
24
= b.
5
Thus, the tangent line is
y=
4
24
+ x.
5
5
11. The Mean Value Theorem tells us that
f ′ (c) =
f (b) − f (a)
7 − 12
=
= −0.5.
b−a
13 − 3
244
Chapter Three /SOLUTIONS
Since the slope at x1 is negative but less steep than at c, we have
f ′ (x1 ) > −0.5.
The slope at x2 is negative and steeper than at c. Thus
f ′ (x2 ) < −0.5.
12. We notice that f (x) = p′ (x). We have
p(1) = 15 + 8 · 14 − 30 · 13 + 30 · 12 − 31 · 1 + 22 = 0.
p(2) = 25 + 8 · 24 − 30 · 23 + 30 · 22 − 31 · 2 + 22 = 0.
Thus, by Rolle’s Theorem, f (x) has a zero between x = 1 and x = 2.
13. A polynomial p(x) satisfies the conditions of Rolle’s Theorem for all intervals a ≤ x ≤ b.
Suppose a1 , a2 , a3 , a4 , a5 , a6 , a7 are the seven distinct zeros of p(x) in increasing order. Thus p(a1 ) = p(a2 ) = 0,
so by Rolle’s Theorem, p′ (x) has a zero, c1 , between a1 and a2 .
Similarly, p′ (x) has 6 distinct zeros, c1 , c2 , c3 , c4 , c5 , c6 , where
a1 < c1 < a2
a2 < c2 < a3
a3 < c3 < a4
a4 < c4 < a5
a5 < c5 < a6
a6 < c6 < a7 .
′
′
The polynomial p (x) is of degree 6, so p (x) cannot have more than 6 zeros.
14. Let f (x) = sin x and g(x) = x. Then f (0) = 0 and g(0) = 0. Also f ′ (x) = cos x and g ′ (x) = 1, so for all x ≥ 0
we have f ′ (x) ≤ g ′ (x). So the graphs of f and g both go through the origin and the graph of f climbs slower than the
graph of g. Thus the graph of f is below the graph of g for x ≥ 0 by the Racetrack Principle. In other words, sin x ≤ x
for x ≥ 0.
15. Let g(x) = ln x and h(x) = x − 1. For x ≥ 1, we have g ′ (x) = 1/x ≤ 1 = h′ (x). Since g(1) = h(1), the
Racetrack Principle with a = 1 says that g(x) ≤ h(x) for x ≥ 1, that is, ln x ≤ x − 1 for x ≥ 1. For 0 < x ≤ 1,
we have h′ (x) = 1 ≤ 1/x = g ′ (x). Since g(1) = h(1), the Racetrack Principle with b = 1 says that g(x) ≤ h(x) for
0 < x ≤ 1, that is, ln x ≤ x − 1 for 0 < x ≤ 1.
y
16.
y = ex
y=x
y = x−1
y = ln x
1
x
1
y =x+1
Graphical solution: If f and g are inverse functions then the graph of g is just the graph of f reflected through the
line y = x. But ex and ln x are inverse functions, and so are the functions x + 1 and x − 1. Thus the equivalence is clear
from the figure.
Algebraic solution: If x > 0 and
x + 1 ≤ ex ,
then, replacing x by x − 1, we have
x ≤ ex−1 .
Taking logarithms, and using the fact that ln is an increasing function, gives
ln x ≤ x − 1.
We can also go in the opposite direction, which establishes the equivalence.
3.10 SOLUTIONS
245
17. The Decreasing Function Theorem is: Suppose that f is continuous on [a, b] and differentiable on (a, b). If f ′ (x) < 0 on
(a, b), then f is decreasing on [a, b]. If f ′ (x) ≤ 0 on (a, b), then f is nonincreasing on [a, b].
To prove the theorem, we note that if f is decreasing then −f is increasing and vice-versa. Similarly, if f is nonincreasing, then −f is nondecreasing. Thus if f ′ (x) < 0, then −f ′ (x) > 0, so −f is increasing, which means f is
decreasing. And if f ′ (x) ≤ 0, then −f ′ (x) ≥ 0, so −f is nondecreasing, which means f is nonincreasing.
18. Dominic’s trip lasted 88 minutes, so his average velocity for the trip was 116/88 miles per minute, or 116/88 · 60 ≈ 79.1
miles per hour. By the Mean Value Theorem, there must have been some time at which Dominic’s instantaneous velocity
was 79.1 miles per hour. Since the speed limit on I-10 between Phoenix and Tucson is never more than 75 miles per hour,
Dominic must have been speeding at that time.
19. Use the Racetrack Principle, Theorem 3.10, with g(x) = x. Since f ′ (x) ≤ g ′ (x) for all x and f (0) = g(0), then
f (x) ≤ g(x) = x for all x ≥ 0.
20. First apply the Racetrack Principle, Theorem 3.10, to f ′ (t) and g(t) = 3t. Since f ′′ (t) ≤ g ′ (t) for all t and f ′ (0) = 0 =
g(0), then f ′ (t) ≤ 3t for all t ≥ 0. Next apply the Racetrack Principle again to f (t) and h(t) = 23 t2 . Since f ′ (t) ≤ h′ (t)
for all t ≥ 0 and f (0) = 0 = h(0), then f (t) ≤ h(t) = 32 t2 for all t ≥ 0.
21. Apply the Constant Function Theorem, Theorem 3.9, to h(x) = f (x) − g(x). Then h′ (x) = 0 for all x, so h(x) is
constant for all x. Since h(5) = f (5) − g(5) = 0, we have h(x) = 0 for all x. Therefore f (x) − g(x) = 0 for all x, so
f (x) = g(x) for all x.
22. By the Mean Value Theorem, Theorem 3.7, there is a number c, with 0 < c < 1, such that
f ′ (c) =
f (1) − f (0)
.
1−0
Since f (1) − f (0) > 0, we have f ′ (c) > 0.
Alternatively if f ′ (c) ≤ 0 for all c in (0, 1), then by the Increasing Function Theorem, f (0) ≥ f (1).
23. Since f ′′ (t) ≤ 7 for 0 ≤ t ≤ 2, if we apply the Racetrack Principle with a = 0 to the functions f ′ (t) − f ′ (0) and 7t,
both of which go through the origin, we get
f ′ (t) − f ′ (0) ≤ 7t
for 0 ≤ t ≤ 2.
The left side of this inequality is the derivative of f (t) − f ′ (0)t, so if we apply the Racetrack Principle with a = 0 again,
this time to the functions f (t) − f ′ (0)t and (7/2)t2 + 3, both of which have the value 3 at t = 0, we get
f (t) − f ′ (0)t ≤
7 2
t + 3 for 0 ≤ t ≤ 2.
2
That is,
7 2
t for 0 ≤ t ≤ 2.
2
In the same way, we can show that the lower bound on the acceleration, 5 ≤ f ′′ (t) leads to:
f (t) ≤ 3 + 4t +
f (t) ≥ 3 + 4t +
5 2
t
2
for 0 ≤ t ≤ 2.
If we substitute t = 2 into these two inequalities, we get bounds on the position at time 2:
21 ≤ f (2) ≤ 25.
24. Consider the function f (x) = h(x) − g(x). Since f ′ (x) = h′ (x) − g ′ (x) ≥ 0, we know that f is nondecreasing by the
Increasing Function Theorem. This means f (x) ≤ f (b) for a ≤ x ≤ b. However, f (b) = h(b) − g(b) = 0, so f (x) ≤ 0,
which means h(x) ≤ g(x).
25. If f ′ (x) = 0, then both f ′ (x) ≥ 0 and f ′ (x) ≤ 0. By the Increasing and Decreasing Function Theorems, f is both
nondecreasing and nonincreasing, so f is constant.
26. Let h(x) = f (x) − g(x). Then h′ (x) = f ′ (x) − g ′(x) = 0 for all x in (a, b). Hence, by the Constant Function Theorem,
there is a constant C such that h(x) = C on (a, b). Thus f (x) = g(x) + C.
246
Chapter Three /SOLUTIONS
27. We will show f (x) = Cex by deducing that f (x)/ex is a constant. By the Constant Function Theorem, we need only
show the derivative of g(x) = f (x)/ex is zero. By the quotient rule (since ex 6= 0), we have
f ′ (x)ex − ex f (x)
.
(ex )2
g ′ (x) =
Since f ′ (x) = f (x), we simplify and obtain
g ′ (x) =
0
f (x)ex − ex f (x)
= 2x = 0,
(ex )2
e
which is what we needed to show.
28. Apply the Racetrack Principle to the functions f (x) − f (a) and M (x − a); we can do this since f (a) − f (a) = M (a − a)
and f ′ (x) ≤ M . We conclude that f (x) − f (a) ≤ M (x − a). Similarly, apply the Racetrack Principle to the functions
m(x − a) and f (x) − f (a) to obtain m(x − a) ≤ f (x) − f (a). If we substitute x = b into these inequalities we get
m(b − a) ≤ f (b) − f (a) ≤ M (b − a).
Now, divide by b − a.
29. (a) Since f ′′ (x) ≥ 0, f ′ (x) is nondecreasing on (a, b). Thus f ′ (c) ≤ f ′ (x) for c ≤ x < b and f ′ (x) ≤ f ′ (c) for
a < x ≤ c.
(b) Let g(x) = f (c)+f ′ (c)(x−c) and h(x) = f (x). Then g(c) = f (c) = h(c), and g ′ (x) = f ′ (c) and h′ (x) = f ′ (x).
If c ≤ x < b, then g ′ (x) ≤ h′ (x), and if a < x ≤ c, then g ′ (x) ≥ h′ (x), by (a). By the Racetrack Principle,
g(x) ≤ h′ (x) for c ≤ x < b and for a < x ≤ c, as we wanted.
Strengthen Your Understanding
30. The function f is not differentiable at x = 0, so the Mean value Theorem does not apply.
31. This function does not satisfy the conclusion of the Mean Value Theorem because it is not continuous at x = 0.
32. To apply the Constant Function Theorem, we need f to be continuous on a ≤ x ≤ b and differentiable on a < x < b.
For example, the function
(
1 if 0 ≤ x < 1
f (x) =
0 if x = 1
has a zero derivative on 0 < x < 1 but is not constant on 0 ≤ x ≤ 1.
33. The function f must be continuous on a ≤ x ≤ b and differentiable on a < x < b. Any interval avoiding x ≤ 0 will
suffice, so, for example 1 ≤ x ≤ 2.
34. The function f must be continuous on a ≤ x ≤ b and differentiable on a < x < b. Any interval including x = 0 will
suffice, so, for example −1 ≤ x ≤ 1.
35. The function f (x) = |x| is continuous on [−1, 1], but there is no number c, with −1 < c < 1, such that
f ′ (c) =
|1| − | − 1|
= 0;
1 − (−1)
that is, the slope of f (x) = |x| is never 0.
36. Let f be defined by
f (x) =
(
x
19
if 0 ≤ x < 2
if x = 2
′
Then f is differentiable on (0, 2) and f (x) = 1 for all x in (0, 2). Thus there is no c in (0, 2) such that
f ′ (c) =
f (2) − f (0)
19
=
.
2−0
2
The reason that this function does not satisfy the conclusion of the Mean Value Theorem is that it is not continuous
at x = 2.
SOLUTIONS to Review Problems for Chapter Three
247
37. Let f be defined by
f (x) =
(
x2
1/2
if 0 ≤ x < 1
if x = 1.
Then f is not continuous at x = 1, but f is differentiable on (0, 1) and f ′ (x) = 2x for 0 < x < 1. Thus, c = 1/4
satisfies
f (1) − f (0)
1
1
1
1
f ′ (c) =
= , since f ′
=2· = .
1−0
2
4
4
2
38. True, by the Increasing Function Theorem, Theorem 3.8.
39. False. For example, let f (x) = x + 5, and g(x) = 2x − 3. Then f ′ (x) ≤ g ′ (x) for all x, but f (0) > g(0).
40. False. For example, let f (x) = 3x + 1 and g(x) = 3x + 7.
41. False. For example, if f (x) = −x, then f ′ (x) ≤ 1 for all x, but f (−2) = 2, so f (−2) > −2.
Solutions for Chapter 3 Review
Exercises
1. w′ = 100(t2 + 1)99 (2t) = 200t(t2 + 1)99 .
2. f ′ (t) = e3t · 3 = 3e3t .
(2t + 3)(t + 1) − (t2 + 3t + 1)
dz
dz
t2 + 2t + 2
3. Using the quotient rule gives
=
or
=
.
2
dt
(t + 1)
dt
(t + 1)2
√
1
√
(t2 + 1) − t(2t)
4. y ′ = 2 t
.
(t2 + 1)2
5. Using the quotient rule,
8
(−1)(4 + t) − (4 − t)
=−
.
h′ (t) =
(4 + t)2
(4 + t)2
6. f ′ (x) = exe−1 .
7. f ′ (x) =
8.
f ′ (x) =
x3
3x2
(3 ln x − 1) +
9
9
3
x
= x2 ln x −
x2
x2
+
= x2 ln x
3
3
(2 + 3x + 4x2 )(1) − (1 + x)(3 + 8x)
(2 + 3x + 4x2 )2
=
2 + 3x + 4x2 − 3 − 11x − 8x2
(2 + 3x + 4x2 )2
=
−4x2 − 8x − 1
.
(2 + 3x + 4x2 )2
9. Using the chain rule, g ′ (θ) = (cos θ)esin θ .
√
√ √
1
θ
dy
θ+ √
= θ1/2 θ1/2 + √ = θ + 1, we have dx
= 1.
10. Since y = θ
θ
θ
1
11. f ′ (w) =
[− sin(w − 1)] = − tan(w − 1).
cos(w − 1)
[This could be done easily using the answer from Problem 20 and the chain rule.]
1
1
3
d
ln ln(2y 3 ) =
6y 2 =
.
12.
dy
ln(2y 3 ) 2y 3
y ln(2y 3 )
d
13. g ′ (x) =
xk + kx = kxk−1 + kx ln k.
dx
14. y ′ = 0
dz
= 3 sin2 θ cos θ
15.
dθ
248
Chapter Three /SOLUTIONS
16.
f ′ (t) = 2 cos(3t + 5) · (− sin(3t + 5))3
= −6 cos(3t + 5) · sin(3t + 5)
17.
M ′ (α) = 2 tan(2 + 3α) ·
= 6·
1
·3
cos2 (2 + 3α)
tan(2 + 3α)
cos2 (2 + 3α)
d
sin2 (3θ − π) = 6 cos(3θ − π) sin(3θ − π).
dθ
1
19. h′ (t) = −t
−e−t − 1 .
e −t
20.
sin(5 − θ)
cos(5 − θ)(−1)θ2 − sin(5 − θ)(2θ)
d
=
2
dθ
θ
θ4
18. s′ (θ) =
=−
21. w′ (θ) =
θ cos(5 − θ) + 2 sin(5 − θ)
.
θ3
2θ cos θ
1
−
sin2 θ
sin3 θ
e−θ
.
(1 + e−θ )2
1
2w ln 2 + ew
d
=−
.
23. g ′ (w) =
w
w
dw 2 + e
(2w + ew )2
1
1
1
24. f (t) = 2 + − 4 = t−2 + t−1 − t−4
t
t
t
f ′ (t) = −2t−3 − t−2 + 4t−5 .
22. f ′ (θ) = −1(1 + e−θ )−2 (e−θ )(−1) =
25. Using the quotient rule and the chain rule,
1
h (z) =
2
′
=
=
sin(2z)
cos(2z)
cos(2z)
sin(2z)
−1/2
1/2
2 cos(2z) cos(2z) − sin(2z)(−2 sin(2z))
cos2 (2z)
cos2 (2z) + sin2 (2z)
cos2 (2z)
1
(cos(2z))1/2
p
= p
.
(sin(2z))1/2 cos2 (2z)
sin(2z) cos3 (2z)
26. Using the chain rule and simplifying,
q ′ (θ) =
4θ − 2 sin(2θ) cos(2θ)
1
(4θ2 − sin2 (2θ))−1/2 (8θ − 2 sin(2θ)(2 cos(2θ))) = p
.
2
4θ2 − sin2 (2θ)
27. Using the product rule and factoring gives
dw
= 2−4z [−4 ln(2) sin(πz) + π cos(πz)] .
dz
3
.
(3t − 4)2 + 1
θ −θ
θ
−θ
d
e(e +e ) = e(e +e ) eθ − e−θ .
29. r ′ (θ) =
dθ
30. Using the chain rule, we get:
m′ (n) = cos(en ) · (en )
28. g ′ (t) =
SOLUTIONS to Review Problems for Chapter Three
31. Using the chain rule we get:
G′ (α) = etan(sin α) (tan(sin α))′ = etan(sin α) ·
1
· cos α.
cos2 (sin α)
32. Here we use the product rule, and then the chain rule, and then the product rule.
√
√
√
√
√
g ′ (t) = cos( tet ) + t(cos tet )′ = cos( tet ) + t(− sin( tet ) · ( tet )′ )
√
√
√ t
1
= cos( tet ) − t sin( tet ) ·
te + √ et
2 t
33. f ′ (r) = e(tan 2 + tan r)e−1 (tan 2 + tan r)′ = e(tan 2 + tan r)e−1
34.
35.
36.
37.
38.
39.
40.
1
cos2 r
1
d
xetan x = etan x + xetan x
.
dx
cos2 x
dy
= 2e2x sin2 (3x) + e2x (2 sin(3x) cos(3x)3) = 2e2x sin(3x)(sin(3x) + 3 cos(3x))
dx
6x
6x
=
g ′ (x) =
2
4 + 6x2 + 2
2
9x
1 + (3x + 1)
dy
= (ln 2)2sin x cos x · cos x + 2sin x (− sin x) = 2sin x (ln 2) cos2 x − sin x
dx
Simplifying first gives F (x) = ax ln e + b = ax + b. Thus F ′ (x) = a.
The same result is obtained by differentiating first and then simplifying.
d
d θ −1
(e e ) = e−1 eθ = eθ e−1 = eθ−1 .
y = eθ e−1 y ′ =
dθ
dθ
Using the product rule and factoring gives f ′ (t) = e−4kt (cos t − 4k sin t).
41. Using the product rule gives
H ′ (t) = 2ate−ct − c(at2 + b)e−ct
= (−cat2 + 2at − bc)e−ct .
1
sin θ cos θ
dp 2
a − sin2 θ = p
(−2 sin θ cos θ) = − p
.
dθ
2 a2 − sin2 θ
a2 − sin2 θ
43. y ′ = (ln 5)5x .
42.
44. Using the quotient rule gives
f ′ (x) =
(−2x)(a2 + x2 ) − (2x)(a2 − x2 )
−4a2 x
= 2
.
2
2
2
(a + x )
(a + x2 )2
45. Using the quotient rule gives
w′ (r) =
=
46. Using the quotient rule gives
′
f (s) =
2ar(b + r 3 ) − 3r 2 (ar 2 )
(b + r 3 )2
2abr − ar 4
.
(b + r 3 )2
√
−2s a2 + s2 − √
s
a2 +s2
(a2 − s2 )
(a2 + s2 )
2
2
=
−2s(a + s ) − s(a2 − s2 )
(a2 + s2 )3/2
=
−2a2 s − 2s3 − a2 s + s3
(a2 + s2 )3/2
=
−3a2 s − s3
.
(a2 + s2 )3/2
249
250
47.
Chapter Three /SOLUTIONS
1
dy
=
2
dx
1 + x2
−2
x2
−2
x2 + 4
=
cos( kt ) 1
.
sin( kt ) k
48. Using the chain rule gives r ′ (t) =
49. Since g(w) = 5(a2 − w2 )−2 , g ′ (w) = −10(a2 − w2 )−3 (−2w) =
2e2x (x2 + 1) − e2x (2x)
2e2x (x2 + 1 − x)
dy
=
=
2
2
dx
(x + 1)
(x2 + 1)2
au
ae
51. g ′ (u) = 2
a + b2
52. Using the quotient and chain rules, we have
20w
(a2 − w2 )3
50.
dy
(aeax + ae−ax )(eax + e−ax ) − (eax − e−ax )(aeax − ae−ax )
=
dx
(eax + e−ax )2
=
a(eax + e−ax )2 − a(eax − e−ax )2
(eax + e−ax )2
a[(e2ax + 2 + e−2ax ) − (e2ax − 2 + e−2ax )]
(eax + e−ax )2
4a
= ax
(e + e−ax )2
=
53. Using the quotient rule gives
1 + ln x − x( x1 )
(1 + ln x)2
ln x
.
=
(1 + ln x)2
f ′ (x) =
54. Using the quotient and chain rules
dz
=
dt
=
2
d
(et
dt
2
et ·
2
d
+ t) · sin(2t) − (et + t) dt
(sin(2t))
(sin(2t))2
d
(t2 )
dt
2
d
+ 1 sin(2t) − (et + t) cos(2t) dt
(2t)
sin2 (2t)
2
=
2
(2tet + 1) sin(2t) − (et + t)2 cos(2t)
.
sin2 (2t)
55. Using the chain rule twice:
′
f (t) = cos
√
et
√
√
√
d t
d √ t
1
1
et + 1
t
t cos
t
t
√
· (e + 1) = cos e + 1 √
e =e
.
+1
e + 1 = cos e + 1 √
dt
2 et + 1 dt
2 et + 1
2 et + 1
56. Using the chain rule twice:
(y3 )
g ′ (y) = e2e
57. y ′ = 18x2 + 8x − 2.
58. g(z) = z 5 + 5z 4 − z
g ′ (z) = 5z 4 + 20z 3 − 1.
3
d
2e(y )
dy
(y3 )
= 2e2e
e(y
3
)
3
(y3 )
d 3
(y ) = 6y 2 e(y ) e2e
.
dy
SOLUTIONS to Review Problems for Chapter Three
251
59. f ′ (z) = (2 ln 3)z + (ln 4)ez .
dy
60.
= 3 − 2(ln 4)4x .
dx
61. f ′ (x) = 3x2 + 3x ln 3
62. f ′ (θ) = 2θ sin θ + θ2 cos θ + 2 cos θ − 2θ sin θ − 2 cos θ = θ2 cos θ.
63.
1
1
dy
= (cos(5θ))− 2 (− sin(5θ) · 5) + 2 sin(6θ) cos(6θ) · 6
dθ
2
5 sin(5θ)
+ 12 sin(6θ) cos(6θ)
=− p
2 cos(5θ)
d
sin[(3θ − π)2 ] = cos[(3θ − π)2 ] · 2(3θ − π) · 3 = 6(3θ − π) cos[(3θ − π)2 ].
dθ
65. It is easier to do this by multiplying it out first, rather than using the product rule first: z = s4 − s,
64. r ′ (θ) =
′
z ′ = 4s3 − 1.
66. Since tan(arctan(kθ)) = kθ, because tangent and arctangent are inverse functions, we have N (θ) = k.
67. Using the product rule gives h′ (t) = kekt (sin at + cos bt) + ekt (a cos at − b sin bt).
68. f ′ (y) = (ln 4)4y (2 − y 2 ) + 4y (−2y) = 4y ((ln 4)(2 − y 2 ) − 2y).
69. f ′ (t) = 4(sin(2t) − cos(3t))3 [2 cos(2t) + 3 sin(3t)]
√
√
3
70. Since cos2 y + sin2 y = 1, we have s(y) = 3 1 + 3 = 4. Thus s′ (y) = 0.
71.
f ′ (x) = (−2x + 6x2 )(6 − 4x + x7 ) + (4 − x2 + 2x3 )(−4 + 7x6 )
= (−12x + 44x2 − 24x3 − 2x8 + 6x9 ) + (−16 + 4x2 − 8x3 + 28x6 − 7x8 + 14x9 )
= −16 − 12x + 48x2 − 32x3 + 28x6 − 9x8 + 20x9
72.
h′ (x) = −
2
1
+ 3
x2
x
2x3 + 4 +
1
1
− 2
x
x
8
4
+ 3 + 6x − 6
x2
x
= 4x − 2 − 4x−2 + 8x−3
= −2x + 4 −
73. Note: f (z) = (5z)1/2 + 5z 1/2 + 5z −1/2 −
√
5z −1/2 +
√
5, so f ′ (z) =
6x2
√
5
5
5
5 −3/2
(5z)−1/2 + z −1/2 − z −3/2 +
z
.
2
2
2
2
74.
3x2 + 3y 2
dy
dy
− 8xy − 4x2
=0
dx
dx
dy
(3y 2 − 4x2 )
= 8xy − 3x2
dx
8xy − 3x2
dy
=
dx
3y 2 − 4x2
75. Using the relation cos2 y + sin2 y = 1, the equation becomes:
dy
1 = y + 2 or y = −1. Hence,
= 0.
dx
76. We wish to find the slope m = dy/dx. To do this, we can implicitly differentiate the given formula in terms of x:
x2 + 3y 2 = 7
d
dy
=
(7) = 0
2x + 6y
dx
dx
dy
−2x
−x
=
=
.
dx
6y
3y
Thus, at (2, −1), m = −(2)/3(−1) = 2/3.
252
Chapter Three /SOLUTIONS
77. Taking derivatives implicitly, we find
dy
dy
+ cos y
+ 2x = 0
dx
dx
−2x
dy
=
dx
1 + cos y
So, at the point x = 3, y = 0,
dy
(−2)(3)
−6
=
=
= −3.
dx
1 + cos 0
2
78. First, we differentiate with respect to x:
x·
dy
dy
+ y · 1 + 2y
=0
dx
dx
dy
(x + 2y) = −y
dx
dy
−y
=
.
dx
x + 2y
At x = 3, we have
3y + y 2 = 4
y 2 + 3y − 4 = 0
(y − 1)(y + 4) = 0.
Our two points, then, are (3, 1) and (3, −4).
−1
1
1
dy
=
= − ; Tangent line: (y − 1) = − (x − 3).
dx
3 + 2(1)
5
5
−(−4)
4
4
dy
=
= − ; Tangent line: (y + 4) = − (x − 3).
At (3, −4),
dx
3 + 2(−4)
5
5
At (3, 1),
Problems
79. f ′ (t) = 6t2 − 8t + 3
80.
and
f ′′ (t) = 12t − 8.
√
f ′ (x) = −8 + 2 2x
√
f ′ (r) = −8 + 2 2r = 4
√
12
r = √ = 3 2.
2 2
81. Since f (x) = x3 − 6x2 − 15x + 20, we have f ′ (x) = 3x2 − 12x − 15. To find the points at which f ′ (x) = 0, we solve
3x2 − 12x − 15 = 0
3(x2 − 4x − 5) = 0
′
3(x + 1)(x − 5) = 0.
We see that f (x) = 0 at x = −1 and at x = 5. The graph of f (x) in Figure 3.24 appears to be horizontal at x = −1 and
at x = 5, confirming what we found analytically.
f (x)
5
−1
Figure 3.24
x
SOLUTIONS to Review Problems for Chapter Three
253
82. (a) Since the power of x will go down by one every time you take a derivative (until the exponent is zero after which the
derivative will be zero), we can see immediately that f (8) (x) = 0.
(b) f (7) (x) = 7 · 6 · 5 · 4 · 3 · 2 · 1 · x0 = 5040.
83. (a) Applying the product rule to h(x) we get h′ (1) = t′ (1)s(1) + t(1)s′ (1) ≈ (−2) · 3 + 0 · 0 = −6.
(b) Applying the product rule to h(x) we get h′ (0) = t′ (0)s(0) + t(0)s′ (0) ≈ (−2) · 2 + 2 · 2 = 0.
(−2) · 2 − 2 · 2
t′ (0)s(0) − t(0)s′ (0)
≈
= −2.
(c) Applying the quotient rule to p(x) we get p′ (0) =
(s(0))2
22
′
Note that since t(x) is a linear function whose slope looks like −2 from the graph, t (x) ≈ −2 everywhere. To find
s′ (1), draw a line tangent to the curve at the point (1, s(1)), and estimate the slope.
84. Since r(x) = s(t(x)), the chain rule gives r ′ (x) = s′ (t(x)) · t′ (x). Thus,
r ′ (0) = s′ (t(0)) · t′ (0) ≈ s′ (2) · (−2) ≈ (−2)(−2) = 4.
Note that since t(x) is a linear function whose slope looks like −2 from the graph, t′ (x) ≈ −2 everywhere. To find
s′ (2), draw a line tangent to the curve at the point (2, s(2)), and estimate the slope.
85. (a) Applying the chain rule we get h′ (1) = s′ (s(1)) · s′ (1) ≈ s′ (3) · 0 = 0.
(b) Applying the chain rule we get h′ (2) = s′ (s(2)) · s′ (2) ≈ s′ (2) · s′ (2) = (−2)2 = 4.
To find s′ (2), draw a line tangent to the curve at the point (2, s(2)), and estimate the slope.
86. We need to find all values for x such that
dy
= s′ (s(x)) · s′ (x) = 0.
dx
This is the case when either s′ (s(x)) = 0 or s′ (x) = 0. From the graph we see that s′ (x) = 0 when x ≈ 1. Also,
s′ (s(x)) = 0 when s(x) ≈ 1, which happens when x ≈ −0.4 or x ≈ 2.4.
To find s′ (a), for any a, draw a line tangent to the curve at the point (a, s(a)), and estimate the slope.
87. (a) Applying the product rule we get h′ (−1) = 2 · (−1) · t(−1) + (−1)2 · t′ (−1) ≈ (−2) · 4 + 1 · (−2) = −10.
(b) Applying the chain rule we get p′ (−1) = t′ ((−1)2 ) · 2 · (−1) = −2 · t′ (1) ≈ (−2) · (−2) = 4.
Note that since t(x) is a linear function whose slope looks like −2 from the graph, t′ (x) ≈ −2 everywhere.
88. We have r(1) = s(t(1)) ≈ s(0) ≈ 2. By the chain rule, r ′ (x) = s′ (t(x)) · t′ (x), so
r ′ (1) = s′ (t(1)) · t′ (1) ≈ s′ (0) · (−2) ≈ 2(−2) = −4.
Thus the equation of the tangent line is
y − 2 = −4(x − 1)
y = −4x + 6.
Note that since t(x) is a linear function whose slope looks like −2 from the graph, t′ (x) ≈ −2 everywhere. To find
′
s (0), draw a line tangent to the curve at the point (0, s(0)), and estimate the slope.
89. Estimates may vary. From the graphs, we estimate g(1) ≈ 2, g ′ (1) ≈ 1, and f ′ (2) ≈ 0.8. Thus, by the chain rule,
h′ (1) = f ′ (g(1)) · g ′ (1) ≈ f ′ (2) · g ′ (1) ≈ 0.8 · 1 = 0.8.
90. Estimates may vary. From the graphs, we estimate f (1) ≈ −0.4, f ′ (1) ≈ 0.5, and g ′ (−0.4) ≈ 2. Thus, by the chain
rule,
k′ (1) = g ′ (f (1)) · f ′ (1) ≈ g ′ (−0.4) · 0.5 ≈ 2 · 0.5 = 1.
91. Estimates may vary. From the graphs, we estimate g(2) ≈ 1.6, g ′ (2) ≈ −0.5, and f ′ (1.6) ≈ 0.8. Thus, by the chain
rule,
h′ (2) = f ′ (g(2)) · g ′ (2) ≈ f ′ (1.6) · g ′ (2) ≈ 0.8(−0.5) = −0.4.
92. Estimates may vary. From the graphs, we estimate f (2) ≈ 0.3, f ′ (2) ≈ 1.1, and g ′ (0.3) ≈ 1.7. Thus, by the chain rule,
k′ (2) = g ′ (f (2)) · f ′ (2) ≈ g ′ (0.3) · f ′ (2) ≈ 1.7 · 1.1 ≈ 1.9.
93. Taking the values of f , f ′ , g, and g ′ from the table we get:
(a)
(b)
(c)
(d)
(e)
(f)
h(4) = f (g(4)) = f (3) = 1.
h′ (4) = f ′ (g(4))g ′ (4) = f ′ (3) · 1 = 2.
h(4) = g(f (4)) = g(4) = 3.
h′ (4) = g ′ (f (4))f ′ (4) = g ′ (4) · 3 = 3.
h′ (4) = (f (4)g ′ (4) − g(4)f ′ (4)) /f 2 (4) = −5/16.
h′ (4) = f (4)g ′ (4) + g(4)f ′ (4) = 13.
254
Chapter Three /SOLUTIONS
94. (a) H ′ (2) = r ′ (2)s(2) + r(2)s′ (2) = −1 · 1 + 4 · 3 = 11.
1
r ′ (2)
−1
(b) H ′ (2) = p
= √ =− .
4
2 4
2 r(2)
(c) H ′ (2) = r ′ (s(2))s′ (2) = r ′ (1) · 3, but we don’t know r ′ (1).
(d) H ′ (2) = s′ (r(2))r ′ (2) = s′ (4)r ′ (2) = −3.
95. (a) f (x) = x2 − 4g(x)
f ′ (x) = 2x − 4g ′ (x)
f ′ (2) = 2(2) − 4(−4) = 4 + 16 = 20
x
(b) f (x) = g(x)
f ′ (x) =
g(x)−xg ′ (x)
(g(x))2
g(2)−2g ′ (2)
= 3−2(−4)
(g(2))2
(3)2
2
f ′ (2) =
= 11
9
(c) f (x) = x g(x)
f ′ (x) = 2xg(x) + x2 g ′ (x)
f ′ (2) = 2(2)(3) + (2)2 (−4) = 12 − 16 = −4
(d) f (x) = (g(x))2
f ′ (x) = 2g(x) · g ′ (x)
f ′ (2) = 2(3)(−4) = −24
(e) f (x) = x sin(g(x))
f ′ (x) = sin(g(x)) + x cos(g(x)) · g ′ (x)
f ′ (2) = sin(g(2)) + 2 cos(g(2)) · g ′ (2)
= sin 3 + 2 cos(3) · (−4)
= sin 3 − 8 cos 3
(f) f (x) = x2 ln(g(x))
′
(x)
)
f ′ (x) = 2x ln(g(x)) + x2 ( gg(x)
)
f ′ (2) = 2(2) ln 3 + (2)2 ( −4
3
= 4 ln 3 −
16
3
96. (a) f (x) = x2 − 4g(x)
f (2) = 4 − 4(3) = −8
f ′ (2) = 20
Thus, we have a point (2, −8) and slope m = 20. This gives
−8 = 2(20) + b
b = −48,
x
(b) f (x) =
g(x)
2
f (2) =
3
11
′
f (2) =
9
Thus, we have point (2, 32 ) and slope m =
so
y = 20x − 48.
11
.
9
This gives
11
2
= ( )(2) + b
3
9
2
22
−16
b= −
=
,
3
9
9
11
16
y=
x−
.
9
9
(c) f (x) = x2 g(x)
f (2) = 4 · g(2) = 4(3) = 12
f ′ (2) = −4
Thus, we have point (2, 12) and slope m = −4. This gives
12 = 2(−4) + b
so
SOLUTIONS to Review Problems for Chapter Three
b = 20,
255
so
y = −4x + 20.
(d) f (x) = (g(x))2
f (2) = (g(2))2 = (3)2 = 9
f ′ (2) = −24
Thus, we have point (2, 9) and slope m = −24. This gives
9 = 2(−24) + b
b = 57,
so
y = −24x + 57.
(e) f (x) = x sin(g(x))
f (2) = 2 sin(g(2)) = 2 sin 3
f ′ (2) = sin 3 − 8 cos 3
We will use a decimal approximation for f (2) and f ′ (2), so the point (2, 2 sin 3) ≈ (2, 0.28) and m ≈ 8.06. Thus,
0.28 = 2(8.06) + b
b = −15.84,
so
y = 8.06x − 15.84.
(f) f (x) = x2 ln g(x)
f (2) = 4 ln g(2) = 4 ln 3 ≈ 4.39
16
≈ −0.94.
f ′ (2) = 4 ln 3 −
3
Thus, we have point (2, 4.39) and slope m = −0.94. This gives
4.39 = 2(−0.94) + b
b = 6.27,
so
y = −0.94x + 6.27.
97. When we zoom in on the origin, we find that two functions are not defined there. The other functions all look like straight
lines through the origin. The only way we can tell them apart is their slope.
The following functions all have slope 0 and are therefore indistinguishable:
2
x
.
sin x − tan x, x2x+1 , x − sin x, and 1−cos
cos x
These functions all have slope 1 at the origin, and are thus indistinguishable:
sin x
x
arcsin x, 1+sin
, arctan x, ex − 1, x+1
, and x2x+1 .
x
sin x
Now, x − 1 and −x ln x both are undefined at the origin, so they are distinguishable from the other functions. In
addition, while sinx x − 1 has a slope that approaches zero near the origin, −x ln x becomes vertical near the origin, so
they are distinguishable from each other.
√
Finally, x10 + 10 x is the only function defined at the origin and with a vertical tangent there, so it is distinguishable
from the others.
98. It makes sense to define the angle between two curves to be the angle between their tangent lines. (The tangent lines are
the best linear approximations to the curves). See Figure 3.25. The functions sin x and cos x are equal at x = π4 .
√
2
π
′ π
For f1 (x) = sin x, f1 ( ) = cos( ) =
4
4
2
√
π
2
′ π
For f2 (x) = cos x, f2 ( ) = − sin( ) = −
.
4
4
2
Using the point ( π4 ,
√
2
)
2
√
for each tangent line we get y =
√
2
x + 22 (1 − π4 )
2
√
and y = −
√
2
x + 22 (1 + π4 ),
2
respectively.
256
Chapter Three /SOLUTIONS
y
√
2
2
1+
π
4
√
2
2 x
y=
√
2
2
+
1−
π
4
y = sin x
√
2
2
y
1+
π
4
✻
√
2π
8
β
1
2α
❄
α
π
4
√
2
2
1−
π
4
√
y = cos x
y=−
2
2 x
+
√
2
2
1+
π
4
x
π
4
√
2
2
Figure 3.25
1−
π
4
Figure 3.26
There are two possibilities of how to define the angle between the tangent lines, indicated by α and β above. The
choice is arbitrary, so we will solve for both. To find the angle, α, we consider the triangle formed by these two lines and
the y-axis. See Figure 3.26.
√2π/8 √
2
1
α =
=
tan
2
π/4
2
1
α = 0.61548 radians
2
α = 1.231 radians, or 70.5◦ .
Now let us solve for β, the other possible measure of the angle between the two tangent lines. Since α and β are
supplementary, β = π − 1.231 = 1.909 radians, or 109.4◦ .
99. The curves meet when 1 + x − x2 = 1 − x + x2 , that is when 2x(1 − x) = 0 so that x = 1 or x = 0. Let
y1 (x) = 1 + x − x2
and y2 (x) = 1 − x + x2 .
Then
y1 ′ = 1 − 2x
and y2 ′ = −1 + 2x.
At x = 0, y1 ′ = 1, y2 ′ = −1 so that y1 ′ · y2 ′ = −1 and the curves are perpendicular. At x = 1, y1 ′ = −1, y2 ′ = 1
so that y1 ′ · y2 ′ = −1 and the curves are perpendicular.
100. The curves meet when 1 − x3 /3 = x − 1, that is when x3 + 3x − 6 = 0. So the roots of this equation give us the xcoordinates of the intersection point. By numerical methods, we see there is one solution near x = 1.3. See Figure 3.27.
Let
x3
and y2 (x) = x − 1.
y1 (x) = 1 −
3
So we have
y1 ′ = −x2 and y2 ′ = 1.
However, y2 ′ (x) = +1, so if the curves are to be perpendicular when they cross, then y1 ′ must be −1. Since y1 ′ = −x2 ,
y1 ′ = −1 only at x = ±1 which is not the point of intersection. The curves are therefore not perpendicular when they
cross.
y
5
1
−2
y = x3 + 3x − 6
−1
x
2
−5
−10
−15
−20
Figure 3.27
SOLUTIONS to Review Problems for Chapter Three
dy
101. Differentiating gives
= ln x + 1 − b.
dx
To find the point at which the graph crosses the x-axis, set y = 0 and solve for x:
0 = x ln x − bx
0 = x(ln x − b).
Since x > 0, we have
ln x − b = 0
x = eb .
At the point (eb , 0), the slope is
Thus the equation of the tangent line is
dy
= ln eb + 1 − b = b + 1 − b = 1.
dx
y − 0 = 1(x − eb )
y = x − eb .
102. Using the definition of cosh x and sinh x, we have cosh 2x =
e3x − e−3x
e2x + e−2x
and sinh 3x =
. Therefore
2
2
cosh(2x)
e2x + e−2x
= lim 3x
x→−∞ sinh(2x)
x→−∞ e
− e−3x
lim
e−2x (e4x + 1)
− e−x )
= lim
x→−∞ e−2x (e5x
e4x + 1
− e−x
= lim
x→−∞ e5x
= 0.
103. Using the definition of sinh x we have sinh 2x =
e2x − e−2x
. Therefore
2
2e−2x
e−2x
= lim 2x
x→−∞ e
x→−∞ sinh(2x)
− e−2x
2
= lim 4x
x→−∞ e
−1
= −2.
lim
2
104. Using the definition of cosh x and sinh x, we have cosh x2 =
2
2
2
ex + e−x
ex − e−x
and sinh x2 =
. Therefore
2
2
2
2
sinh(x2 )
ex − e−x
= lim x2
2
x→−∞ e
x→−∞ cosh(x )
+ e−x2
lim
2
2
ex (1 − e−2x )
2
2
x→−∞ ex (1 + e−2x )
= lim
2
1 − e−2x
2
x→−∞ 1 + e−2x
= 1.
= lim
257
258
Chapter Three /SOLUTIONS
√
√
105. (a) Since f (x) = x, we have f ′ (x) = (1/2)x−1/2 . So f (4) = 4 = 2 and f ′ (4) = (1/2)4−1/2 = 1/4, and the
tangent line approximation is
f (x) ≈ f (4) + f ′ (4)(x − 4)
1
≈ 2 + (x − 4).
4
See Figure 3.28.
(b) For x = 4.1, the true value is
f (4.1) =
√
4.1 = 2.02485...,
whereas the approximation is
1
(4.1 − 4) = 2.025.
4
Thus, the approximation differs from the true value by about 0.00015.
(c) For x = 16 the true value is:
√
f (16) = 16 = 4
f (4.1) ≈ 2 +
whereas the approximation is
f (16) ≈ 2 +
1
(16 − 4) = 5.
4
Thus, the approximation differs from the true value by 1.
(d) The tangent line is a good approximation to the graph near x = 4, but not necessarily far away. Of course, there’s
no reason to expect that the curve will look like the tangent line if we go too far away, and usually it does not. (See
Figure 3.28.) The problem is that we have traveled too far from the place where the curve looks like a line with slope
1/4.
Tangent line
at x = 4
5
4
f (x) =
√
x
x
4
16
Figure 3.28: Local linearization: Approximating
√
f (x) = x by its tangent line at x = 4
106. (a) From the figure, we see a = 2. The point with x = 2 lies on both the line and the curve. Since
y = −3 · 2 + 7 = 1,
we have
f (a) = 1.
Since the slope of the line is −3, we have
(b) We use the line to approximate the function, so
f ′ (a) = −3.
f (2.1) ≈ −3(2.1) + 7 = 0.7.
This is an underestimate, because the line is beneath the curve for x > 2. Similarly,
f (1.98) ≈ −3(1.98) + 7 = 1.06.
This is an overestimate because the line is above the curve for x < 2.
The approximation f (1.98) ≈ 1.06 is likely to be more accurate because 1.98 is closer to 2 than 2.1 is. Since
the graph of f (x) appears to bend away from the line at approximately the same rate on either side of x = 2, in this
example, the error is larger for points farther from x = 2.
SOLUTIONS to Review Problems for Chapter Three
259
107. (a) The function f (t) is linear; g(t) is quadratic (polynomial of degree 2); and h(t) is exponential.
(b) In 2010, we have t = 130. For f (t), the rate of change is
f ′ (t) = 0.006
f ′ (130) = 0.006◦ C per year.
For g(t)
g ′ (t) = 0.00012t − 0.0017
g ′ (130) = 0.00012 · 130 − 0.0017 = 0.0139◦ C per year.
For h(t)
h′ (t) = 13.63(0.0004)e0.0004t = 0.00545e0.0004t
h′ (130) = 0.00545e0.0004(130) = 0.00574◦ C per year.
(c) For f (t) : Change = 0.006 · 130 = 0.78◦ C.
For g(t) : Change = 0.0139 · 130 = 1.807◦ C.
For h(t) : Change = 0.00574 · 130 = 0.746◦ C.
(d) For f (t) : Predicted change = f (130) − f (0) = (13.625 + 0.006 · 130) − 13.625 = 0.78◦ C.
For g(t) : Predicted change = g(130) − g(0) = (0.00006(1302 ) − 0.0017(130) + 13.788) − 13.788 = 0.793◦ C
For h(t) : Predicted change = h(130) − h(0) = 13.63e0.0004(130) − 13.63 = 0.728◦ C.
(e) For the linear model, the answers in parts (c) and (d) are equal.
(f) For the quadratic model, the discrepancy is largest.
108. (a) We have P = 9.906(0.997)11 = 9.584 million.
(b) Differentiating, we have
dP
= 9.906(ln 0.997)(0.997)t
dt
so
dP
dt
t=11
= 9.906(ln 0.997)(0.997)11 = −0.0288 million/year.
Thus in 2020, Hungary’s population will be decreasing by about 28,800 people per year.
dF
2GM m
109.
=−
.
dr
r3
t
110. (a) If the distance s(t) = 20e 2 , then the velocity, v(t), is given by
t
v(t) = s′ (t) = 20e 2
(b) Observing the differentiation in (a), we note that
s′ (t) = v(t) =
t
Substituting s(t) for 20e 2 , we obtain s′ (t) = 21 s(t).
′
=
1
2
t
1
20e 2
2
t
20e 2
=
t
= 10e 2 .
1
s(t).
2
111. (a) The rate of change of the population is P ′ (t). If P ′ (t) is proportional to P (t), we have
P ′ (t) = kP (t).
(b) If P (t) = Aekt , then P ′ (t) = kAekt = kP (t).
112. (a) Differentiating, we see
dy
= −2πωy0 sin(2πωt)
dt
dv
a=
= −4π 2 ω 2 y0 cos(2πωt).
dt
v=
(b) We have
y = y0 cos(2πωt)
v = −2πωy0 sin(2πωt)
a = −4π 2 ω 2 y0 cos(2πωt).
260
Chapter Three /SOLUTIONS
So
Amplitude of y is |y0 |,
Amplitude of v is |2πωy0 | = 2πω|y0 |,
Amplitude of a is |4π 2 ω 2 y0 | = 4π 2 ω 2 |y0 |.
The amplitudes are different (provided 2πω 6= 1). The periods of the three functions are all the same, namely 1/ω.
(c) Looking at the answer to part (a), we see
d2 y
= a = −4π 2 ω 2 (y0 cos(2πωt))
dt2
= −4π 2 ω 2 y.
So we see that
d2 y
+ 4π 2 ω 2 y = 0.
dt2
1000000
= 1000000. Thus, in the long run, close to 1,000,000
1 + 5000e−0.1t
people will have had the disease. This can be seen in Figure 3.29.
(b) The rate at which people fall sick is given by the first derivative N ′ (t).
N ′ (t) ≈ ∆N
, where ∆t = 1 day.
∆t
113. (a) Since lim e−0.1t = 0, we see that lim
t→∞
t→∞
N ′ (t) =
500,000,000
500,000,000
= 0.1t
e0.1t (1 + 5000e−0.1t )2
e
+ 25,000,000e−0.1t + 104
In Figure 3.30, we see that the maximum value of N ′ (t) is approximately 25,000. Therefore the maximum
number of people to fall sick on any given day is 25,000. Thus there are no days on which a quarter million or more
get sick.
dN
dt
25,000
N
20,000
1,000,000
N (t)
15,000
10,000
N ′ (t)
5,000
50
t
Figure 3.29
100
Figure 3.30
114. (a) We solve for t to find the time it takes for the population to reach 10 billion.
P (t) = 10
6.7ekt = 10
10
ekt =
6.7
ln(10/6.7)
0.40048
t=
=
years
k
k
Thus the time is
0.40048
years.
k
(b) The time to reach 10 billion with a growth rate of 1.2% is
f (k) =
f (0.012) =
0.40048
= 33.4 years.
0.012
150
200
t
SOLUTIONS to Review Problems for Chapter Three
261
(c) We have
−0.40048
k2
f ′ (0.012) = −2781.
f ′ (k) =
Thus, for growth rates, k, near 1.2% we have
Time for world population to reach 10 billion = f (k) ≈ f (0.012) + f ′ (0.012)(k − 0.012)
f (k) = 33.4 − 2781(k − 0.012) years
(d) True time to reach 10 billion when the growth rate is 1.0% is f (0.01) = 0.40048/0.01 = 40.0 years. The linear
approximation gives
Approximate time = 33.4 − 2781(0.01 − 0.012) = 39.0 years.
115. (a) Suppose
g = f (r) =
Then
f ′ (r) =
GM
.
r2
−2GM
.
r3
So
f (r + ∆r) ≈ f (r) −
2GM
(∆r).
r3
Since f (r + ∆r) − f (r) = ∆g, and g = GM/r 2 , we have
∆g ≈ −2
GM
∆r
(∆r) = −2g
.
r3
r
(b) The negative sign tells us that the acceleration due to gravity decreases as the distance from the center of the earth
increases.
(c) The fractional change in g is given by
∆r
∆g
≈ −2
.
g
r
So, since ∆r = 4.315 km and r = 6400 km, we have
4.315
∆g
≈ −2
g
6400
= −0.00135 = −0.135%.
116. Since g is the inverse of f , we know that g(4) = f −1 (4) = 3, so
g ′ (4) =
117. We must have
(f −1 )′ (5) =
1
1
1
= ′
= .
f ′ (g(4))
f (3)
6
1
1
1
= ′
= .
f ′ (f −1 (5))
f (10)
8
118. We know that the velocity is given by
dx
= v(x).
dt
By the chain rule,
Acceleration =
dv
dv dx
=
·
= v ′ (x)v(x).
dt
dx dt
119. Since f (x) is decreasing, its inverse function f −1 (x) is also decreasing. Thus (f −1 )′ (x) ≤ 0 for all x. Option (b) is
incorrect.
262
Chapter Three /SOLUTIONS
120. (a) If y = ln x, then
y′ =
1
x
y ′′ = −
1
x2
2
x3
3·2
=− 4
x
y ′′′ =
y ′′′′
and so
y (n) = (−1)n+1 (n − 1)!x−n .
x
(b) If y = xe , then
y ′ = xex + ex
y ′′ = xex + 2ex
y ′′′ = xex + 3ex
so that
y (n) = xex + nex .
(c) If y = ex cos x, then
y ′ = ex (cos x − sin x)
y ′′ = −2ex sin x
y ′′′ = ex (−2 cos x − 2 sin x)
y (4) = −4ex cos x
y (5) = ex (−4 cos x + 4 sin x)
y (6) = 8ex sin x.
Combining these results we get
y (n) = (−4)(n−1)/4 ex (cos x − sin x),
n = 4m + 1,
m = 0, 1, 2, 3, . . .
y (n) = −2(−4)(n−2)/4 ex sin x,
n = 4m + 2,
m = 0, 1, 2, 3, . . .
n = 4m + 3,
m = 0, 1, 2, 3, . . .
y (n) = (−4)(n/4) ex cos x,
n = 4m,
m = 1, 2, 3, . . . .
y
(n)
= −2(−4)
(n−3)/4 x
e (cos x + sin x),
121. (a) We multiply through by h = f · g and cancel as follows:
g′
h′
f′
+
=
f
g
h
g′
f′
+
f
g
· fg =
h′
· fg
h
f′
g′
h′
· fg +
· fg =
·h
f
g
h
f ′ · g + g ′ · f = h′ ,
which is the product rule.
(b) We start with the product rule, multiply through by 1/(f g) and cancel as follows:
f ′ · g + g′ · f
1
(f ′ · g + g ′ · f ) ·
fg
1
1
+ (g ′ · f ) ·
(f ′ · g) ·
fg
fg
g′
f′
+
f
g
which is the additive rule shown in part (a).
= h′
1
fg
1
= h′ ·
fg
h′
= ,
h
= h′ ·
SOLUTIONS to Review Problems for Chapter Three
263
122. This problem can be solved by using either the quotient rule or the fact that
f′
d
=
(ln f )
f
dx
and
g′
d
=
(ln g).
g
dx
We use the second method. The relative rate of change of f /g is (f /g)′ /(f /g), so
(f /g)′
d
=
ln
f /g
dx
f
g
=
d
d
d
f′
g′
(ln f − ln g) =
(ln f ) −
(ln g) =
− .
dx
dx
dx
f
g
Thus, the relative rate of change of f /g is the difference between the relative rates of change of f and of g.
CAS Challenge Problems
123. (a) Answers from different computer algebra systems may be in different forms. One form is:
d
(x + 1)x = x(x + 1)x−1 + (x + 1)x ln(x + 1)
dx
d
(sin x)x = x cos x(sin x)x−1 + (sin x)x ln(sin x)
dx
(b) Both the answers in part (a) follow the general rule:
d
f (x)x = xf ′ (x) (f (x))x−1 + (f (x))x ln(f (x)).
dx
(c) Applying this rule to g(x), we get
d
(ln x)x = x(1/x)(ln x)x−1 + (ln x)x ln(ln x) = (ln x)x−1 + (ln x)x ln(ln x).
dx
This agrees with the answer given by the computer algebra system.
(d) We can write f (x) = eln(f (x)) . So
(f (x))x = (eln(f (x)) )x = ex ln(f (x)) .
Therefore, using the chain rule and the product rule,
d
d
d
(f (x))x =
(x ln(f (x))) · ex ln(f (x)) = ln(f (x)) + x
ln(f (x)) ex ln(f (x))
dx
dx
dx
f ′ (x)
(f (x))x = ln(f (x)) (f (x))x + xf ′ (x) (f (x))x−1
= ln(f (x)) + x
f (x)
= xf ′ (x) (f (x))x−1 + (f (x))x ln(f (x)).
124. (a) A CAS gives f ′ (x) = 1.
(b) By the chain rule,
f ′ (x) = cos(arcsin x) · √
p
1
.
1 − x2
Now cos t = ± p1 − sin2 t. Furthermore, if −π/2 ≤ t ≤ π/2 then cos t ≥ 0, so we take the positive square root
and get cos t = 1 − sin2 t. Since −π/2 ≤ arcsin x ≤ π/2 for all x in the domain of arcsin, we have
cos(arcsin x) =
so
p
1 − (sin(arcsin x))2 =
p
1 − x2 ,
p
1
d
sin(arcsin(x)) = 1 − x2 · √
= 1.
dx
1 − x2
(c) Since sin(arcsin(x)) = x, its derivative is 1.
125. (a) A CAS gives g ′ (r) = 0.
(b) Using the product rule,
d
d −2r
(2
) · 4r + 2−2r (4r ) = −2 ln 2 · 2−2r 4r + 2−2r ln 4 · 4r
dr
dr
= − ln 4 · 2−2r 4r + ln 4 · 2−2r 4r = (− ln 4 + ln 4)2−2r 4r = 0 · 2−2r 4r = 0.
g ′ (r) =
(c) By the laws of exponents, 4r = (22 )r = 22r , so 2−2r 4r = 2−2r 22r = 20 = 1. Therefore, its derivative is zero.
264
Chapter Three /SOLUTIONS
126. (a) A CAS gives h′ (t) = 0
(b) By the chain rule
′
h (t) =
=
d
dt
1−
1−
1
t
1
t
+
d
dt
t
t−1
t
t−1
=
1
t2
t−1
t
+
1
t−1
t
(t−1)2
t
t−1
−
(t − 1) − t
1
1
−1
+
= 2
+ 2
= 0.
t2 − t
t2 − t
t −t
t −t
(c) The expression inside the first logarithm is 1 − (1/t) = (t − 1)/t. Using the property log A + log B = log(AB),
we get
ln 1 −
1
t
+ ln
t
t−1
Thus h(t) = 0, so h′ (t) = 0 also.
t−1
t
+ ln
t
t−1
t − 1
t
·
= ln 1 = 0.
= ln
t
1−t
= ln
PROJECTS FOR CHAPTER THREE
1. Let r = i/100. (For example if i = 5%, r = 0.05.) Then the balance, $B, after t years is given by
B = P (1 + r)t ,
where $P is the original deposit. If we are doubling our money, then B = 2P , so we wish to solve for t in the
equation 2P = P (1 + r)t . This is equivalent to
2 = (1 + r)t .
Taking natural logarithms of both sides and solving for t yields
ln 2 = t ln(1 + r),
ln 2
t=
.
ln(1 + r)
We now approximate ln(1 + r) near r = 0. Let f (r) = ln(1 + r). Then f ′ (r) = 1/(1 + r). Thus, f (0) = 0
and f ′ (0) = 1, so
f (r) ≈ f (0) + f ′ (0)r
becomes
ln(1 + r) ≈ r.
Therefore,
t=
ln 2
100 ln 2
70
ln 2
≈
=
≈
,
ln(1 + r)
r
i
i
as claimed. We expect this approximation to hold for small values of i; it turns out that values of i up to 10
give good enough answers for most everyday purposes.
2. (a) (i) Set f (x) = sin x, so f ′ (x) = cos x. Guess x0 = 3. Then
sin 3
≈ 3.1425
cos 3
sin x1
x2 = x1 −
≈ 3.1415926533,
cos x1
x1 = 3 −
which is correct to one billionth!
PROJECTS FOR CHAPTER THREE
265
(ii) Newton’s method uses the tangent line at x = 3, i.e. y − sin 3 = cos(3)(x − 3). Around x = 3,
however, sin x is almost linear, since the second derivative sin′′ (π) = 0. Thus using the tangent line
to get an approximate value for the root gives us a very good approximation.
✙
f (x) = sin x
✛ tangent line
3
x
(iii) For f (x) = sin x, we have
f (3) = 0.14112
f (4) = −0.7568,
so there is a root in [3, 4]. We now continue bisecting:
[3, 3.5] : f (3.5) = −0.35078 (bisection 1)
[3, 3.25] : f (3.25) = −0.10819 (bisection 2)
[3.125, 3.25] : f (3.125) =
0.01659 (bisection 3)
[3.125, 3.1875] : f (3.1875) = −0.04584 (bisection 4)
We continue this process; after 11 bisections, we know the root lies between 3.1411 and 3.1416, which
still is not as good an approximation as what we get from Newton’s method in just two steps.
(b) (i) We have f (x) = sin x − 23 x and f ′ (x) = cos x − 32 .
Using x0 = 0.904,
x1 = 0.904 −
x2 = 4.704 −
sin(0.904) − 32 (0.904)
≈ 4.704,
cos(0.904) − 23
sin(4.704) − 32 (4.704)
≈ −1.423,
cos(4.704) − 23
x3 = −1.433 −
sin(−1.423) − 32 (−1.423)
≈ −1.501,
cos(−1.423) − 32
x4 = −1.499 −
sin(−1.501) − 32 (−1.501)
≈ −1.496,
cos(−1.501) − 32
x5 = −1.496 −
sin(−1.496) − 32 (−1.496)
≈ −1.496.
cos(−1.496) − 32
Using x0 = 0.905,
x1 = 0.905 −
x2 = 4.643 −
sin(0.905) − 32 (0.905)
≈ 4.643,
cos(0.905) − 23
sin(4.643) − 32 (4.643)
≈ −0.918,
cos(4.643) − 23
266
Chapter Three /SOLUTIONS
x3 = −0.918 −
sin(−0.918) − 32 (−0.918)
≈ −3.996,
cos(−0.918) − 32
x4 = −3.996 −
sin(−3.996) − 32 (−3.996)
≈ −1.413,
cos(−3.996) − 32
x5 = −1.413 −
sin(−1.413) − 32 (−1.413)
≈ −1.502,
cos(−1.413) − 32
x6 = −1.502 −
sin(−1.502) − 32 (−1.502)
≈ −1.496.
cos(−1.502) − 32
Now using x0 = 0.906,
x1 = 0.906 −
x2 = 4.584 −
x3 = −0.510 −
sin(0.906) − 32 (0.906)
≈ 4.584,
cos(0.906) − 23
sin(4.584) − 32 (4.584)
≈ −0.509,
cos(4.584) − 23
sin(−0.509) − 32 (−0.509)
≈ .207,
cos(−0.509) − 32
x4 = −1.300 −
sin(.207) − 32 (.207)
≈ −0.009,
cos(.207) − 23
x5 = −1.543 −
sin(−0.009) − 32 (−0.009)
≈ 0,
cos(−0.009) − 32
(ii) Starting with 0.904 and 0.905 yields the same value, but the two paths to get to the root are very
different. Starting with 0.906 leads to a different root. Our starting points were near the maximum
value of f . Consequently, a small change in x0 makes a large change in x1 .
4.1 SOLUTIONS
CHAPTER FOUR
Solutions for Section 4.1
Exercises
1. See Figure 4.1.
local max
✻
❥
local max
✻
local min
Figure 4.1
2. There are many possible answers. One possible graph is shown in Figure 4.2.
Critical point
Inflection point
x
2
4
6
8
Figure 4.2
3. We sketch a graph which is horizontal at the two critical points. One possibility is shown in Figure 4.3.
Critical point
Not local max or min
Local min
x
Figure 4.3
267
268
Chapter Four /SOLUTIONS
4. To find the critical points, we set f ′ (x) = 0. Since f ′ (x) = 3x2 − 18x + 24, we have
3x2 − 18x + 24 = 0
3(x2 − 6x + 8) = 0
3(x − 2)(x − 4) = 0
x = 2, 4.
There are two critical points: at x = 2, x = 4.
To find the inflection points, we look for points where f ′′ is zero or undefined. Since f ′′ (x) = 6x − 18, it is defined
everywhere, and setting f ′′ (x) = 0 we get
6x − 18 = 0
x = 3.
Furthermore, 6x − 18 is positive when x > 3 and negative when x < 3, so f ′′ changes sign at x = 3. Thus, there is one
inflection point: x = 3.
5. To find the critical points, we set f ′ (x) = 0. Since f ′ (x) = 5x4 − 30x2 , we have
5x4 − 30x2 = 0
5x2 (x2 − 6) = 0
√ √
x = 0, − 6, 6.
√
√
There are three critical points: x = 0, x = − 6, and x = 6.
To find the inflection points, we look for points where f ′′ is undefined or zero. Since f ′′ (x) = 20x3 − 60x, it is defined
everywhere. Setting it equal to zero, we get
20x3 − 60x = 0
20x(x2 − 3) = 0
√ √
x = 0, − 3, 3.
√
√
√
√
Furthermore, 20x3 − 60x is positive when − 3 < x < 0 or x > 3 and negative when x <√− 3 or 0 <√x < 3, so
′′
f changes sign at each of the solutions. Thus, there are three inflection points: x = 0, x = − 3, and x = 3.
6. To find the critical points, we set f ′ (x) = 0. Since f ′ (x) = 5x4 + 60x3 , we have
5x4 + 60x3 = 0
5x3 (x + 12) = 0
x = 0, −12.
There are two critical points: x = 0, x = −12.
To find the inflection points, look for points where f ′′ is undefined or zero. Since f ′′ (x) = 20x3 + 180x2 , it is defined
everywhere. Setting it equal to zero, we get
20x3 + 180x2 = 0
20x2 (x + 9) = 0
x = 0, −9.
Furthermore, 20x2 (x + 9) is negative when x < −9, and positive or zero when x > −9. So f ′′ does change sign at
x = −9, but not at x = 0. Thus, there is one inflection point: x = −9.
7. To find the critical points, we set f ′ (x) = 0. Since f ′ (x) = 5 − 3(1/x), we have
3
=0
x
3
5=
x
5x = 3
5−
x = 3/5.
There is one critical point, at x = 3/5.
To find the inflection points, we look for points where f ′′ (x) is undefined or zero. Since f ′′ (x) = 3x−2 , it is undefined
at x = 0. However, f (x) is also undefined at x = 0, so this is not an inflection point. Also, 3/x2 is never equal to zero,
so there are no inflection points.
4.1 SOLUTIONS
269
8. To find the critical points, we set f ′ (x) = 0. Using the product rule, we have f ′ (x) = 4x · (e3x · 3) + 4 · e3x .
12xe3x + 4e3x = 0
4e3x (3x + 1) = 0
x = −1/3.
There is one critical point, at x = −1/3.
To find the inflection points, we look for points where f ′′ is zero or undefined. Since f ′′ (x) = 12x · (e3x · 3) + 12 · e3x +
4e3x · 3 = 12e3x (3x + 2), it is defined everywhere. Setting it equal to zero, we get
12e3x (3x + 2) = 0
x = −2/3.
Since 12e3x is always positive, 12e3x (3x + 2) is negative when x < −2/3 and positive when x > −2/3, so f ′′ changes
sign at x = −2/3. Thus, there is one inflection point, at x = −2/3.
9. f ′ (x) = 12x3 − 12x2 . To find critical points, we set f ′ (x) = 0. This implies 12x2 (x − 1) = 0. So the critical points of
f are x = 0 and x = 1. To the left of x = 0, f ′ (x) < 0. Between x = 0 and x = 1, f ′ (x) < 0. To the right of x = 1,
f ′ (x) > 0. Therefore, f (1) is a local minimum, but f (0) is not a local extremum. See Figure 4.4.
y
f (x) = 3x4 − 4x3 + 6
x
1
Figure 4.4
10. f ′ (x) = 7(x2 − 4)6 2x = 14x(x − 2)6 (x + 2)6 . The critical points of f are x = 0, x = ±2. To the left of x = −2,
f ′ (x) < 0. Between x = −2 and x = 0, f ′ (x) < 0. Between x = 0 and x = 2, f ′ (x) > 0. To the right of x = 2,
f ′ (x) > 0. Thus, f (0) is a local minimum, whereas f (−2) and f (2) are not local extrema. See Figure 4.5.
y
x
−2
2
f (x) = (x2 − 4)7
Figure 4.5
270
Chapter Four /SOLUTIONS
11. f ′ (x) = 4(x3 − 8)3 3x2
= 12x2 (x − 2)3 (x2 + 2x + 4)3 .
So the critical points are x = 0 and x = 2.
To the left of x = 0, f ′ (x) < 0.
Between x = 0 and x = 2, f ′ (x) < 0.
To the right of x = 2, f ′ (x) > 0.
Thus, f (2) is a local minimum, whereas f (0) is not a local extremum. See Figure 4.6.
y
f (x) = (x3 − 8)4
x
2
Figure 4.6
12.
f ′ (x) =
1 − x2
(1 − x)(1 + x)
x2 + 1 − x · 2x
= 2
=
.
2
2
(x + 1)
(x + 1)2
(x2 + 1)2
Critical points are x = ±1. To the left of x = −1, f ′ (x) < 0.
Between x = −1 and x = 1, f ′ (x) > 0.
To the right of x = 1, f ′ (x) < 0.
So, f (−1) is a local minimum, f (1) a local maximum. See Figure 4.7.
y
x
−1 1
f (x) =
x
x2 + 1
Figure 4.7
13. We have
g ′ (x) = e−3x − 3xe−3x = (1 − 3x)e−3x .
To find critical points, we set g ′ (x) = 0. Then
(1 − 3x)e−3x = 0.
4.1 SOLUTIONS
271
Therefore, the critical point of g is x = 1/3. To the left of x = 1/3, we have g ′ (x) > 0. To the right of x = 1/3, we have
g ′ (x) < 0. Thus g(1/3) is a local maximum. See Figure 4.8.
y
g(x) = xe−3x
x
1/3
Figure 4.8
14. We have
h′ (x) = 1 − 1/x2 .
′
To find critical points, we set h (x) = 0. Then
1−
1
=0
x2
1
1= 2
x
x = ±1.
Therefore, the critical points of h are x = −1 and x = 1. For 0 < x < 1, we have h′ (x) < 0, and for x > 1, we have
h′ (x) > 0. Thus we have a local minimum at x = 1. For x < −1, we have h′ (x) > 0 and for −1 < x < 0, we have
h′ (x) < 0. Thus x = −1 is a local maximum. See Figure 4.9.
y
x
h(x) = x +
1
x
−1
1
Figure 4.9
2
15. (a) A graph of f (x) = e−x is shown in Figure 4.10. It appears to have one critical point, at x = 0, and two inflection
points, one between 0 and 1 and the other between 0 and −1.
2
Critical
1 point
Inflection
point
Inflection
point
x
−2
−1
1
Figure 4.10
2
272
Chapter Four /SOLUTIONS
2
(b) To find the critical points, we set f ′ (x) = 0. Since f ′ (x) = −2xe−x = 0, there is one solution, x = 0. The only
critical point is at x = 0.
To find the inflection points, we first use the product rule to find f ′′ (x). We have
2
2
2
2
f ′′ (x) = (−2x)(e−x (−2x)) + (−2)(e−x ) = 4x2 e−x − 2e−x .
We set f ′′ (x) = 0 and solve for x by factoring:
2
2
4x2 e−x − 2e−x = 0
2
(4x2 − 2)e−x = 0.
2
Since e−x is never zero, we have
4x2 − 2 = 0
1
x2 =
2
√
x = ±1/ 2.
√
√
There are exactly two inflection points, at x = 1/ 2 ≈ 0.707 and x = −1/ 2 ≈ −0.707.
16. (a) Increasing for x > 0, decreasing for x < 0.
(b) f (0) is a local and global minimum, and f has no global maximum.
17. (a) Increasing for all x.
(b) No maxima or minima.
18. (a) Decreasing for x < 0, increasing for 0 < x < 4, and decreasing for x > 4.
(b) f (0) is a local minimum, and f (4) is a local maximum.
19. (a) Decreasing for x < −1, increasing for −1 < x < 0, decreasing for 0 < x < 1, and increasing for x > 1.
(b) f (−1) and f (1) are local minima, f (0) is a local maximum.
Problems
20. (a) The function f (x) is defined for x ≥ 0. We set the derivative equal to zero to find critical points:
f ′ (x) = 1 +
1 −1/2
ax
=0
2
a
1 + √ = 0.
2 x
√
Since a > 0 and x > 0 for x > 0, we have f ′ (x) > 0 for x > 0, so there is no critical point with x > 0. The only
critical point is at x = 0 where f ′ (x) is undefined.
(b) We see in part (a) that the derivative is positive for x > 0 so the function is increasing for all x > 0. The second
derivative is
1
f ′′ (x) = − ax−3/2 .
4
Since a and x−3/2 are positive for x > 0, the second derivative is negative for all x > 0. Thus the graph of f is
concave down for all x > 0.
21. (a) This function is defined for x > 0. We set the derivative equal to zero and solve for x to find critical points:
f ′ (x) = 1 − b
1
=0
x
b
1=
x
x = b.
The only critical point is at x = b.
(b) Since f ′ (x) = 1 − bx−1 , the second derivative is
f ′′ (x) = bx−2 =
b
.
x2
Since b > 0, the second derivative is always positive. Thus, the function is concave up everywhere and f has a local
minimum at x = b.
4.1 SOLUTIONS
273
22. (a) To find the critical points, we set the derivative of f (x) = ax−2 + x equal to zero and solve for x.
f ′ (x) = −2ax−3 + 1 = 0
−2a
+1 = 0
x3
x3 = 2a
√
3
x = 2a.
There is one critical point, at x =
in the domain of f .
(b) The second derivative is
√
3
2a. Although f ′ is undefined at x = 0, this is not a critical point since it is not
6a
.
x4
4
′′
If a and x are positive,
f (x) is positive and the graph is concave up everywhere, so the function has a local
√
minimum at x = 3 2a. Similarly, if a is negative,
then f ′′ (x) is negative and the graph is concave down everywhere,
√
3
so the function has a local maximum at x = 2a.
f ′′ (x) = 6ax−4 =
23. To find the critical points, we set the derivative equal to zero and solve for t.
F ′ (t) = U et + V e−t (−1) = 0
V
U et − t = 0
e
V
U et = t
e
U e2t = V
V
e2t =
U
2t = ln(V /U )
ln(V /U )
.
t=
2
The derivative F ′ (t) is never undefined, so the only critical point is t = 0.5 ln(V /U ).
24. The critical points of f are zeros of f ′ . Just to the left of the first critical point f ′ > 0, so f is increasing. Immediately
to the right of the first critical point f ′ < 0, so f is decreasing. Thus, the first point must be a maximum. To the left of
the second critical point, f ′ < 0, and to its right, f ′ > 0; hence it is a minimum. On either side of the last critical point,
f ′ > 0, so it is neither a maximum nor a minimum. See the figure below. See Figure 4.11.
f ′ (x)
local min
❄
✻
x
✻
local max
neither max
nor min
Figure 4.11
25. To find inflection points of the function f we must find points where f ′′ changes sign. However, because f ′′ is the derivative of f ′ , any point where f ′′ changes sign will be a local maximum or minimum on the graph of f ′ . See Figure 4.12.
✙
x-values of
these points give
inflection points of f
❄
f ′ (x)
❯
❄
Figure 4.12
x
274
Chapter Four /SOLUTIONS
26. The inflection points of f are the points where f ′′ changes sign. See Figure 4.13.
f ′′ (x)
inflection points of f
☛
❯
x
Figure 4.13
27. See Figure 4.14.
y
y′ = 0
y ′′ = 0
y ′′ = 0 y ′ = 0
y = f (x)
y′ > 0
y ′′ < 0
y′ > 0
y ′′ > 0
y′ > 0
y′′ < 0
x1
x2
y′ < 0
y ′′ < 0
x3
x
Figure 4.14
28. See Figure 4.15.
y ′ < 0 everywhere
y
y
′′
=0
y ′′ = 0 y ′′ = 0
y ′′ > 0 y ′′ < 0
y ′′ > 0 y ′′ < 0
y = f (x)
x1
x2
Figure 4.15
x3
x
4.1 SOLUTIONS
275
29. See Figure 4.16.
y
y ′ , y ′′ undefined
y′ = 0
y′ < 0
y ′′ > 0
y′ > 0
y ′′ > 0
y′ > 0
y ′′ > 0
y = f (x)
x1
x
x2
Figure 4.16
30. See Figure 4.17.
y
y = f (x)
y′ > 0
y ′′ > 0
y′ = 2
y ′′ = 0
x1
x
Figure 4.17
31. (a) A critical point occurs when f ′ (x) = 0. Since f ′ (x) changes sign between x = 2 and x = 3, between x = 6 and
x = 7, and between x = 9 and x = 10, we expect critical points at around x = 2.5, x = 6.5, and x = 9.5.
(b) Since f ′ (x) goes from positive to negative at x ≈ 2.5, a local maximum should occur there. Similarly, x ≈ 6.5 is a
local minimum and x ≈ 9.5 a local maximum.
32. (a) It appears that this function has a local maximum at about x = 1, a local minimum at about x = 4, and a local
maximum at about x = 8.
(b) The table now gives values of the derivative, so critical points occur where f ′ (x) = 0. Since f ′ is continuous, this
occurs between 2 and 3, so there is a critical point somewhere around 2.5. Since f ′ is positive for values less than
2.5 and negative for values greater than 2.5, it appears that f has a local maximum at about x = 2.5. Similarly, it
appears that f has a local minimum at about x = 6.5 and another local maximum at about x = 9.5.
33. See Figure 4.18.
depth of water
Time at which water
reaches corner of vase
✠
Figure 4.18
time
276
Chapter Four /SOLUTIONS
34. See Figure 4.19.
depth of water
time at which water reaches
widest part of urn
✠
time
Figure 4.19
35. Differentiating using the product rule gives
f ′ (x) = 3x2 (1 − x)4 − 4x3 (1 − x)3 = x2 (1 − x)3 (3(1 − x) − 4x) = x2 (1 − x)3 (3 − 7x).
The critical points are the solutions to
f ′ (x) = x2 (1 − x)3 (3 − 7x) = 0
3
x = 0, 1, .
7
For x < 0, since 1 − x > 0 and 3 − 7x > 0, we have f ′ (x) > 0.
For 0 < x < 37 , since 1 − x > 0 and 3 − 7x > 0, we have f ′ (x) > 0.
For 37 < x < 1, since 1 − x > 0 and 3 − 7x < 0, we have f ′ (x) < 0.
For 1 < x, since 1 − x < 0 and 3 − 7x < 0, we have f ′ (x) > 0.
Thus, x = 0 is neither a local maximum nor a local minimum; x = 3/7 is a local maximum; x = 1 is a local
minimum.
36. By the product rule
f ′ (x) =
dxm (1 − x)n
= mxm−1 (1 − x)n − nxm (1 − x)n−1
dt
= xm−1 (1 − x)n−1 (m(1 − x) − nx)
= xm−1 (1 − x)n−1 (m − (m + n)x).
We have f ′ (x) = 0 at x = 0, x = 1, and x = m/(m + n), so these are the three critical points of f .
We can classify the critical points by determining the sign of f ′ (x).
If x < 0, then f ′ (x) has the same sign as (−1)m−1 : negative if m is even, positive if m is odd.
If 0 < x < m/(m + n), then f ′ (x) is positive.
If m/(m + n) < x < 1, then f ′ (x) is negative.
If 1 < x, then f ′ (x) has the same sign as (−1)n : positive if n is even, negative if n is odd.
If m is even, then f ′ (x) changes from negative to positive at x = 0, so f has a local minimum at x = 0.
If m is odd, then f ′ (x) is positive to both the left and right of 0, so x = 0 is an inflection point of f .
At x = m/(m + n), the derivative f ′ (x) changes from positive to negative, so x = m/(m + n) is a local maximum
of f .
If n is even, then f ′ (x) changes from negative to positive at x = 1, so f has a local minimum at x = 1. If n is odd,
then f ′ (x) is negative to both the left and right of 1, so x = 1 is an inflection point of f .
4.1 SOLUTIONS
37. (a) From the graph of P =
277
2000
in Figure 4.20, we see that the population levels off at about 2000 rabbits.
1 + e(5.3−0.4t)
population of rabbits
2000
1500
1000
500
years since 1774
10
20
30
40
Figure 4.20
(b) The population appears to have been growing fastest when there were about 1000 rabbits, about 13 years after Captain
Cook left them there.
(c) The rabbits reproduce quickly, so their population initially grew very rapidly. Limited food and space availability and
perhaps predators on the island probably account for the population being unable to grow past 2000.
38. First, we wish to have f ′ (6) = 0, since f (6) should be a local minimum:
f ′ (x) = 2x + a = 0
a
x=− =6
2
a = −12.
Next, we need to have f (6) = −5, since the point (6, −5) is on the graph of f (x). We can substitute a = −12 into our
equation for f (x) and solve for b:
f (x) = x2 − 12x + b
f (6) = 36 − 72 + b = −5
b = 31.
Thus, f (x) = x2 − 12x + 31.
39. We wish to have f ′ (3) = 0. Differentiating to find f ′ (x) and then solving f ′ (3) = 0 for a gives:
f ′ (x) = x(aeax ) + 1(eax ) = eax (ax + 1)
f ′ (3) = e3a (3a + 1) = 0
3a + 1 = 0
1
a=− .
3
Thus, f (x) = xe−x/3 .
40. Using the product rule on the function f (x) = axebx , we have f ′ (x) = aebx + abxebx = aebx (1 + bx). We want
f ( 13 ) = 1, and since this is to be a maximum, we require f ′ ( 31 ) = 0. These conditions give
f (1/3) = a(1/3)eb/3 = 1,
f ′ (1/3) = aeb/3 (1 + b/3) = 0.
Since ae(1/3)b is non-zero, we can divide both sides of the second equation by ae(1/3)b to obtain 0 = 1 + 3b . This
implies b = −3. Plugging b = −3 into the first equation gives us a( 13 )e−1 = 1, or a = 3e. How do we know we have
a maximum at x = 13 and not a minimum? Since f ′ (x) = aebx (1 + bx) = (3e)e−3x (1 − 3x), and (3e)e−3x is always
positive, it follows that f ′ (x) > 0 when x < 31 and f ′ (x) < 0 when x > 31 . Since f ′ is positive to the left of x = 31 and
negative to the right of x = 13 , f ( 13 ) is a local maximum.
278
Chapter Four /SOLUTIONS
41. The graph of f (x) = x + sin x is in Figure 4.21 and the graph of f ′ (x) = 1 + cos x is in Figure 4.22.
Where is f increasing most rapidly? At the points x = . . . , −2π, 0, 2π, 4π, 6π, . . . , because these points are
local maxima for f ′ , and f ′ has the same value at each of them. Likewise f is increasing least rapidly at the points
x = . . . , −3π, −π, π, 3π, 5π, . . . , since these points are local minima for f ′ . Notice that the points where f is increasing
most rapidly and the points where it is increasing least rapidly are inflection points of f .
Point of inflection
slope minimum
f
✛
❄
−2π
f′
x
π
−π
2
Point of inflection
slope maximum
2π
3π
−3π
−2π
−π
π
0
x
2π
3π
Figure 4.22: Graph of f ′ (x) = 1 + cos x
Figure 4.21: Graph of f (x) = x + sin x
42. Figure 4.23 contains the graph of f (x) = x2 + cos x.
y
15
8
g(x) = x2
f (x) = x2 + cos x
10
◆
✛
4
5
f (x) = x2 + cos x
x
−2
x
2
−4 −2
Figure 4.23
2
4
Figure 4.24
The graph looks like a parabola with no waves because f ′′ (x) = 2 − cos x, which is always positive. Thus, the graph
of f is concave up everywhere; there are no waves. If you plot the graph of f (x) together with the graph of g(x) = x2 ,
you see that the graph of f does wave back and forth across the graph of g, but never enough to change the concavity of
f . See Figure 4.24.
43. Neither B nor C is 0 where A has its maxima and minimum. Therefore neither B nor C is the derivative of A, so A = f ′′ .
We can see B could be the derivative of C because where C has a maximum, B is 0. However, C is not the derivative of
B because B is decreasing for some x-values and C is never negative. Thus, C = f , B = f ′ , and A = f ′′ .
44. A has zeros where B has maxima and minima, so A could be a derivative of B. This is confirmed by comparing intervals
on which B is increasing and A is positive. (They are the same.) So, C is either the derivative of A or the derivative of C is
B. However, B does not have a zero at the point where C has a minimum, so B cannot be the derivative of C. Therefore,
C is the derivative of A. So B = f , A = f ′ , and C = f ′′ .
45. Since the derivative of an even function is odd and the derivative of an odd function is even, f and f ′′ are either both odd
or both even, and f ′ is the opposite. Graphs I and III represent even functions; II represents an odd function, so II is f ′ .
Since the maxima and minima of II occur where I crosses the x-axis, I must be the derivative of f ′ , that is, f ′′ . In addition,
the maxima and minima of III occur where II crosses the x-axis, so III is f .
46. Since the derivative of an even function is odd and the derivative of an odd function is even, f and f ′′ are either both
odd or both even, and f ′ is the opposite. Graphs I and II represent odd functions; III represents an even function, so III
is f ′ . Since the maxima and minima of III occur where I crosses the x-axis, I must be the derivative of f ′ , that is, f ′′ . In
addition, the maxima and minima of II occur where III crosses the x-axis, so II is f .
47. Using the quotient rule, since we have
f (x) =
x + 50
,
x2 + 525
4.1 SOLUTIONS
f ′ (x) =
2
279
2
x + 525 − 2x(x + 50)
−x − 100x + 525
=
.
(x2 + 525)2
(x2 + 525)2
Since the denominator is positive for all x, setting f ′ (x) = 0 gives:
x2 + 100x − 525 = 0
(x − 5)(x + 105) = 0
x = 5, −105.
These are the only two critical points. Checking signs using the formula for f ′ (x), we find f (x) is increasing for −105 <
x < 5 and f (x) is decreasing for x < −105 and x > 5.
It is difficult to solve this problem with a calculator for two reasons. First, the critical points are at −105 and 5, which
are very far apart. Second, the range of values around the critical points is very small. For example,
f (−100) = −0.0047506
f (−105) = −0.0047619
f (−110) = −0.0047525
Thus, although x = −105 gives a local minimum, the graph of f is extremely flat in the neighborhood of −105. It is hard
to see these small differences on a graph of f .
48. (a) When a number grows larger, its reciprocal grows smaller. Therefore, since f is increasing near x0 , we know that
g (its reciprocal) must be decreasing. Another argument can be made using derivatives. We know that (since f is
′
(x)
increasing) f ′ (x) > 0 near x0 . We also know (by the chain rule) that g ′ (x) = (f (x)−1 )′ = − ff(x)
2 . Since both
′
2
′
f (x) and f (x) are positive, this means g (x) is negative, which in turn means g(x) is decreasing near x = x0 .
(b) Since f has a local maximum near x1 , f (x) increases as x nears x1 , and then f (x) decreases as x exceeds x1 . Thus
the reciprocal of f , g, decreases as x nears x1 and then increases as x exceeds x1 . Thus g has a local minimum at
′
(x)
x = x1 . To put it another way, since f has a local maximum at x = x1 , we know f ′ (x1 ) = 0. Since g ′ (x) = − ff(x)
2,
′
′
′
′
′
g (x1 ) = 0. To the left of x1 , f (x1 ) is positive, so g (x) is negative. To the right of x1 , f (x1 ) is negative, so g (x)
is positive. Therefore, g has a local minimum at x1 .
(c) Since f is concave down at x2 , we know f ′′ (x2 ) < 0. We also know (from above) that
f ′′ (x2 )
2f ′ (x2 )2
1
−
=
g (x2 ) =
f (x2 )3
f (x2 )2
f (x2 )2
′′
2f ′ (x2 )2
− f ′′ (x2 ) .
f (x2 )
Since f (x12 )2 > 0, 2f ′ (x2 )2 > 0, and f (x2 ) > 0 (as f is assumed to be everywhere positive), we see that
g (x2 ) is positive. Thus g is concave up at x2 .
Note that for the first two parts of the problem, we did not need to require f to be positive (only non-zero).
However, it was necessary here.
′′
49. We know that x = 1 is the only solution to f (x) = 0, and that x = 2 is the only solution to f ′ (x) = 0.
(a) To find the zeros of this function, we solve the equation y = 0:
f x2 − 3 = 0
x2 − 3 = 1
because f (1) = 0
x = ±2.
Thus, the zeros are x = ±2,
(b) To find the critical points of this function, we solve the equation y ′ = 0:
f
′
′
f x2 − 3
2
2
=0
′
x − 3 (x − 3) = 0
f
′
2
x − 3 · 2x = 0
so either
or
2x = 0
x=0
f ′ x2 − 3 = 0
Chain rule
280
Chapter Four /SOLUTIONS
x2 − 3 = 2
√
because f ′ (2) = 0
x = ± 5.
√
Thus, the critical points are x = 0, ± 5.
50. We know that x = 1 is the only solution to f (x) = 0, and that x = 2 is the only solution to f ′ (x) = 0.
(a) To find the zeros of this function, we solve the equation y = 0:
(f (x))2 + 3 = 0
(f (x))2 = −3. no solution
Thus, this function has no zeros.
(b) To find the critical points of this function, we solve the equation y ′ = 0:
′
(f (x))2 + 3
′
=0
2f (x)f (x) = 0
so either
f (x) = 0
or
f ′ (x) = 0
Chain rule
x=1
because f (1) = 0
x = 2. because f ′ (2) = 0
Thus, this function has critical points at x = 1, 2.
51. We know f ′ (x) < 0 for all x. Since the graph of f lies entirely above the x-axis, we know f (x) > 0 for all x.
For y = (f (x))2 , we have:
dy
d
=
(f (x))2
dx
dx
= 2f (x)f ′ (x)
= 2 (A positive number) (A negative number)
|
{z
f (x)>0
= A negative number.
}|
{z
f ′ (x)<0
}
We see that dy/dx is everywhere negative.
(a) Since the derivative is defined everywhere and has no zeros, there are no critical points.
(b) Since the derivative is everywhere negative, this is a decreasing function.
52. We know f ′ (x) < 0 for all x. Since the graph of f lies entirely above the x-axis, we know f (x) > 0 for all x.
For y = f (x2 ), we have:
d
dy
=
f x2
dx
dx
= 2xf ′ x2
= 2x · (A negative number)
|
{z
f ′ (x2 )<0
}
= 2x · (A negative number) .
We see that:
• dy/dx = 0 at x = 0
• dy/dx < 0 for x > 0
• dy/dx > 0 for x < 0
(a) Since dy/dx = 0 at x = 0, this means there is a critical point at x = 0.
(b) Since dy/dx < 0 for x > 0, the function decreases for x > 0. And since dy/dx > 0 for x < 0, the function
increases for x < 0.
Strengthen Your Understanding
53. The function f (x) = x3 is increasing and has an inflection point at x = 0.
4.1 SOLUTIONS
281
54. Consider f (x) = x4 . We have f ′′ (x) = 12x2 so f ′′ (0) = 0. However, f ′′ is positive for all x 6= 0. Thus, the graph of f
does not change concavity at x = 0.
55. One possible example is f (x) = x, for which f ′ (x) = 1, so there are no critical points. There are many other examples.
56. One possible example is f (x) = |x − 1|. The function is defined at x = 1 since f (1) = 0, but f is not differentiable at
x = 1. Thus, f has a critical point at x = 1.
57. One possible answer is f (x) = cos x which has local maxima at 0, ±2π, ±4π, . . . (that is, at x = 2nπ for all integers
n). The value at all the local maxima is 1. The local minima are at ±π, ±3π, . . . (that is, at x = (2n + 1)π), all with
value −1.
The function f (x) = sin x is another possibility, and there are many more.
58. True. Since the domain of f is all real numbers, all local minima occur at critical points.
59. True. Since the domain of f is all real numbers, all local maxima occur at critical points. Thus, if x = p is a local
maximum, x = p must be a critical point.
60. False. A local maximum of f might occur at a point where f ′ does not exist. For example, f (x) = −|x| has a local
maximum at x = 0, but the derivative is not 0 (or defined) there.
61. False, because x = p could be a local minimum of f . For example, if f (x) = x2 , then f ′ (0) = 0, so x = 0 is a critical
point, but x = 0 is not a local maximum of f .
62. False. For example, if f (x) = x3 , then f ′ (0) = 0, but f (x) does not have either a local maximum or a local minimum at
x = 0.
63. True. Suppose f is increasing at some points and decreasing at others. Then f ′ (x) takes both positive and negative values.
Since f ′ (x) is continuous, by the Intermediate Value Theorem, there would be some point where f ′ (x) is zero, so that
there would be a critical point. Since we are told there are no critical points, f must be increasing everywhere or decreasing
everywhere.
64. True. Since f ′′ changes sign at the inflection point x = p, by the Intermediate Value Theorem, f ′′ (p) = 0.
65. False. For example, if f (x) = x3 , then f ′ (0) = 0, so x = 0 is a critical point, but x = 0 is neither a local maximum nor
a local minimum.
66. True. A cubic polynomial is a function of the form f (x) = Ax3 + Bx2 + Cx + D with A 6= 0. We have f ′′ (x) =
6Ax + 2B. There is an inflection point where f ′′ (x) = 0, at x = −B/(3A).
67. f (x) = x2 + 1 is positive for all x and concave up.
68. This is impossible. If f (a) > 0, then the downward concavity forces the graph of f to cross the x-axis to the right or left
of x = a, which means f (x) cannot be positive for all values of x. More precisely, suppose that f (x) is positive for all
x and f is concave down. Thus there must be some value x = a where f (a) > 0 and f ′ (a) is not zero, since a constant
function is not concave down. The tangent line at x = a has nonzero slope and hence must cross the x-axis somewhere to
the right or left of x = a. Since the graph of f must lie below this tangent line, it must also cross the x-axis, contradicting
the assumption that f (x) is positive for all x.
69. f (x) = −x2 − 1 is negative for all x and concave down.
70. This is impossible. If f (a) < 0, then the upward concavity forces the graph of f to cross the x-axis to the right or left
of x = a, which means f (x) cannot be negative for all values of x. More precisely, suppose that f (x) is negative for all
x and f is concave up. Thus there must be some value x = a where f (a) < 0 and f ′ (a) is not zero, since a constant
function is not concave up. The tangent line at x = a has nonzero slope and hence must cross the x-axis somewhere to
the right or left of x = a. Since the graph of f must lie above this tangent line, it must also cross the x-axis, contradicting
the assumption that f (x) is negative for all x.
71. This is impossible. Since f ′′ exists, so must f ′ , which means that f is differentiable and hence continuous. If f (x) were
positive for some values of x and negative for other values, then by the Intermediate Value Theorem, f (x) would have
to be zero somewhere, but this is impossible since f (x)f ′′ (x) < 0 for all x. Thus either f (x) > 0 for all values of x,
in which case f ′′ (x) < 0 for all values of x, that is f is concave down. But this is impossible by Problem 68. Or else
f (x) < 0 for all x, in which case f ′′ (x) > 0 for all x, that is f is concave up. But this is impossible by Problem 70.
72. This is impossible. Since f ′′′ exists, f ′′ must be continuous. By the Intermediate Value Theorem, f ′′ (x) cannot change
sign, since f ′′ (x) cannot be zero. In the same way, we can show that f ′ (x) and f (x) cannot change sign. Since the
product of these three with f ′′′ (x) cannot change sign, f ′′′ (x) cannot change sign. Thus f (x)f ′′ (x) and f ′ (x)f ′′′ (x)
cannot change sign. Since their product is negative for all x, one or the other must be negative for all x. By Problem 71,
this is impossible.
73. (a) and (c). a is a critical point and f (a) is a local minimum.
282
Chapter Four /SOLUTIONS
Solutions for Section 4.2
Exercises
1. See Figure 4.25.
y
Global and local max
8
6
Local min
4
2
Global and local min
x
1
2
3
4
5
Figure 4.25
2. The global maximum is achieved at the two local maxima, which are at the same height. See Figure 4.26.
Local and
global max
y
Local and
global max
4
Local min
3
2
Local min
1
Local and global min
x
2
4
6
8
10
Figure 4.26
3. (a) We see in Figure 4.27 that the maximum occurs at about x = 4 and the maximum value of y is about y = 60.
(b) The maximum value occurs at a critical point, so we find all critical points of y:
dy
= 12x − 3x2 = 0
dx
3x(4 − x) = 0
x = 0, x = 4.
Since y is a cubic polynomial with negative leading coefficient, the critical point x = 0 gives a local minimum and
the critical point x = 4 gives a local maximum. Since we require x > 0, the maximum value of the function occurs
at x = 4. We find the maximum value of y by substituting x = 4:
Maximum value of y = 25 + 6 · 42 − 43 = 57.
y
57
50
x
4
Figure 4.27
4.2 SOLUTIONS
283
4. Since f (x) = x3 − 3x2 + 20 is continuous and the interval −1 ≤ x ≤ 3 is closed, there must be a global maximum
and minimum. The candidates are critical points in the interval and endpoints. Since there are no points where f ′ (x) is
undefined, we solve f ′ (x) = 0 to find all the critical points:
f ′ (x) = 3x2 − 6x = 3x(x − 2) = 0,
so x = 0 and x = 2 are the critical points; both are in the interval. We then compare the values of f at the critical points
and the endpoints:
f (−1) = 16, f (0) = 20, f (2) = 16, f (3) = 20.
Thus the global maximum is 20 at x = 0 and x = 3, and the global minimum is 16 at x = −1 and x = 2.
5. Since f (x) = x4 − 8x2 is continuous and the interval −3 ≤ x ≤ 1 is closed, there must be a global maximum and
minimum. The possible candidates are critical points in the interval and endpoints. Since there are no points where f ′ (x)
is undefined, we solve f ′ (x) = 0 to find all the critical points:
f ′ (x) = 4x3 − 16x = 4x(x2 − 4) = 0,
so x = −2, 0, 2 are the critical points; only x = −2, 0 are in the interval. We then compare the values of f at the critical
points and the endpoints:
f (−3) = 9, f (−2) = −16, f (0) = 0, f (1) = −7.
Thus the global maximum is 9 at x = −3 and the global minimum is −16 at x = −2.
2
6. Since f (x) = xe−x /2 is continuous and the interval −2 ≤ x ≤ 2 is closed, there must be a global maximum and
minimum. The possible candidates are critical points in the interval and endpoints. Since there are no points where f ′ (x)
is undefined, we solve f ′ (x) = 0 to find all the critical points:
f ′ (x) = e−x
2
/2
+ xe−x
2
/2
(−x) = e−x
2
/2
(1 − x2 ).
2
Since e−x /2 6= 0, the critical points are x = −1, 1; both are in the interval. We then compare the values of f at the
critical points and the endpoints:
f (2) = 2e−2 = 0.271,
f (1) = e−1/2 = 0.607,
f (−1) = −f (1) = −0.607
f (−2) = −f (2) = −0.271.
Thus, the global maximum is 0.607 at x = 1 and the global minimum is −0.607 at x = −1
7. Since f (x) = 3x1/3 − x is continuous and the interval −1 ≤ x ≤ 8 is closed, there must be a global maximum and
minimum. The possible candidates are critical points in the interval and endpoints. One critical point is x = 0, since f ′ is
undefined there. We solve f ′ (x) = 0 to find the other critical points:
f ′ (x) = x−2/3 − 1 = 0,
so x = −1 and x = 1. Thus the critical points are x = −1, 0, 1; all are in the interval. We then compare the values of f
at the critical points and the endpoints:
f (−1) = −2,
f (0) = 0,
f (1) = 2,
f (8) = −2.
Thus, the global maximum is 2 at x = 1, and the global minimum is −2 at x = −1 and x = 8.
8. Since f (x) = x − 2 ln(x + 2) is continuous on the closed interval 0 ≤ x ≤ 2, there must be a global maximum and
minimum. The possible candidates are critical points in the interval and endpoints. Since there are no points in the interval
where f ′ (x) is undefined, we solve f ′ (x) = 0 to find the critical points:
f ′ (x) = 1 −
2
= 0,
x+1
so x + 1 − 2 = 0. Thus the only critical point is x = 1, which is in the interval. We then compare the values of f at the
critical point and the endpoints:
f (0) = 0,
f (1) = 1 − 2 ln(2) = −0.386,
f (2) = 2 − 2 ln(3) = −0.197.
Thus, the global maximum is 0 at x = 0, and the global minimum is −0.386 at x = 1.
284
Chapter Four /SOLUTIONS
9. Since f (x) = x2 − 2|x| is continuous and the interval −3 ≤ x ≤ 4 is closed, there must be a global maximum and
minimum. The possible candidates are critical points in the interval and endpoints. The derivative f ′ is not defined at
x = 0. To find the other critical points we solve f ′ (x) = 0. For x > 0 we have
f (x) = x2 − 2x, so f ′ (x) = 2x − 2 = 0.
Thus, x = 1 is the only critical point for 0 < x < 4. For x < 0, we have:
f (x) = x2 + 2x, so f ′ (x) = 2x + 2 = 0.
Thus x = −1 is the only critical point for −3 < x < 0. We then compare the values of f at the critical points and the
endpoints:
f (−3) = 3, f (−1) = −1, f (0) = 0, f (1) = −1. f (4) = 8.
Thus the global maximum is 8 at x = 4 and the global minimum is −1 at x = −1 and x = 1.
10. Since the denominator is never 0, we have that f (x) = (x + 1)/(x2 + 3) is continuous. As the interval −1 ≤ x ≤ 2 is
closed, there must be a global maximum and minimum. The candidates are critical points in the interval and endpoints.
Since there are no points where f ′ (x) is undefined, we solve f ′ (x) = 0 to find all the critical points:
f ′ (x) =
(x2 + 3) − (x + 1)(2x)
−x2 − 2x + 3
=
= 0.
(x2 + 3)2
(x2 + 3)2
Thus −x2 − 2x + 3 = −(x + 3)(x − 1) = 0, so x = −3 and x = 1 are the critical points; only x = 1 is in the interval.
We then compare the values of f at the critical points and the endpoints:
f (−1) = 0,
f (1) =
1
,
2
f (2) =
3
.
7
Thus the global maximum is 1/2 at x = 1, and the global minimum is 0 at x = −1.
11. (a) We have f ′ (x) = 10x9 − 10 = 10(x9 − 1). This is zero when x = 1, so x = 1 is a critical point of f . For values of
x less than 1, x9 is less than 1, and thus f ′ (x) is negative when x < 1. Similarly, f ′ (x) is positive for x > 1. Thus
f (1) = −9 is a local minimum.
We also consider the endpoints f (0) = 0 and f (2) = 1004. Since f ′ (0) < 0 and f ′ (2) > 0, we see x = 0 and
x = 2 are local maxima.
(b) Comparing values of f shows that the global minimum is at x = 1, and the global maximum is at x = 2.
12. (a) f ′ (x) = 1 − 1/x. This is zero only when x = 1. Now f ′ (x) is positive when 1 < x ≤ 2, and negative when
0.1 < x < 1. Thus f (1) = 1 is a local minimum. The endpoints f (0.1) ≈ 2.4026 and f (2) ≈ 1.3069 are local
maxima.
(b) Comparing values of f shows that x = 0.1 gives the global maximum and x = 1 gives the global minimum.
13. (a) Differentiating gives
f (x) = sin2 x − cos x for 0 ≤ x ≤ π
f ′ (x) = 2 sin x cos x + sin x = (sin x)(2 cos x + 1),
so f ′ (x) = 0 when sin x = 0 or when 2 cos x + 1 = 0. Now, sin x = 0 when x = 0 or when x = π. On the
other hand, 2 cos x + 1 = 0 when cos x = −1/2, which happens when x = 2π/3. So the critical points are x = 0,
x = 2π/3, and x = π.
Note that sin x > 0 for 0 < x < π. Also, 2 cos x + 1 < 0 if 2π/3 < x ≤ π and 2 cos x + 1 > 0 if
0 < x < 2π/3. Therefore,
f ′ (x) < 0
for
f ′ (x) > 0
for
2π
0 for x < 2 and g (x) < 0 for x > 2, the critical point at x = 2 is a local maximum.
As x → ±∞, the value of g(x) → −∞. Thus, the local maximum at x = 2 is a global maximum of g(2) =
4 · 2 − 22 − 5 = −1. There is no global minimum. See Figure 4.28.
2
x
−1
g(x) = 4x − x2 − 5
Figure 4.28
15. Differentiating gives
f ′ (x) = 1 −
1
,
x2
so the critical points satisfy
1−
1
=0
x2
x2 = 1
x=1
(We want x > 0).
Since f ′ is negative for 0 < x < 1 and f ′ is positive for x > 1, there is a local minimum at x = 1.
Since f (x) → ∞ as x → 0+ and as x → ∞, the local minimum at x = 1 is a global minimum; there is no global
maximum. See Figure 4.29. The the global minimum is f (1) = 2.
2
f (x) = x + 1/x
x
1
Figure 4.29
16. Differentiating using the product rule gives
g ′ (t) = 1 · e−t − te−t = (1 − t)e−t ,
so the critical point is t = 1.
Since g ′ (t) > 0 for 0 < t < 1 and g ′ (t) < 0 for t > 1, the critical point is a local maximum.
As t → ∞, the value of g(t) → 0, and as t → 0+ , the value of g(t) → 0. Thus, the local maximum at x = 1 is
a global maximum of g(1) = 1e−1 = 1/e. In addition, the value of g(t) is positive for all t > 0; it tends to 0 but never
reaches 0. Thus, there is no global minimum. See Figure 4.30.
1/e
g(t) = te−t
t
1
Figure 4.30
286
Chapter Four /SOLUTIONS
17. Differentiating gives
f ′ (x) = 1 −
1
,
x
so the critical points satisfy
1−
1
=0
x
1
=1
x
x = 1.
Since f ′ is negative for 0 < x < 1 and f ′ is positive for x > 1, there is a local minimum at x = 1.
Since f (x) → ∞ as x → 0+ and as x → ∞, the local minimum at x = 1 is a global minimum; there is no global
maximum. See Figure 4.31. Thus, the global minimum is f (1) = 1.
f (x) = x − ln x
1
x
1
Figure 4.31
18. Differentiating using the quotient rule gives
f ′ (t) =
1(1 + t2 ) − t(2t)
1 − t2
=
.
(1 + t2 )2
(1 + t2 )2
The critical points are the solutions to
1 − t2
=0
(1 + t2 )2
t2 = 1
t = ±1.
Since f ′ (t) > 0 for −1 < t < 1 and f ′ (t) < 0 otherwise, there is a local minimum at t = −1 and a local maximum at
t = 1.
As t → ±∞, we have f (t) → 0. Thus, the local maximum at t = 1 is a global maximum of f (1) = 1/2, and the
local minimum at t = −1 is a global minimum of f (−1) = −1/2. See Figure 4.32.
1/2
−1
f (t) =
1
−1/2
t
1+t2
t
Figure 4.32
19. Differentiating using the product rule gives
f ′ (t) = 2 sin t cos t · cos t − (sin2 t + 2) sin t = 0
sin t(2 cos2 t − sin2 t − 2) = 0
sin t(2(1 − sin2 t) − sin2 t − 2) = 0
sin t(−3 sin2 t) = −3 sin3 t = 0.
4.2 SOLUTIONS
287
Thus, the critical points are where sin t = 0, so
t = 0, ±π, ±2π, ±3π, . . . .
′
3
Since f (t) = −3 sin t is negative for −π < t < 0, positive for 0 < t < π, negative for π < t < 2π, and so on, we find
that t = 0, ±2π, . . . give local minima, while t = ±π, ±3π, . . . give local maxima. Evaluating gives
f (0) = f (±2π) = (0 + 2)1 = 2
f (±π) = f (±3π) = (0 + 2)(−1) = −2.
Thus, the global maximum of f (t) is 2, occurring at t = 0, ±2π, . . ., and the global minimum of f (t) is −2, occurring at
t = ±π, ±3π, . . . . See Figure 4.33.
2
−3π−2π−π
π 2π 3π
t
−2
Figure 4.33
20. Let y = x3 − 4x2 + 4x. To locate the critical points, we solve y ′ = 0. Since y ′ = 3x2 − 8x + 4 = (3x − 2)(x − 2), the
critical points are x = 2/3 and x = 2. To find the global minimum and maximum on 0 ≤ x ≤ 4, we check the critical
points and the endpoints: y(0) = 0; y(2/3) = 32/27; y(2) = 0; y(4) = 16. Thus, the global minimum is at x = 0 and
x = 2, the global maximum is at x = 4, and 0 ≤ y ≤ 16.
2
2
21. Let y = e−x . Since y ′ = −2xe−x , y is increasing for x < 0 and decreasing for x > 0. Hence y = e0 = 1 is a global
maximum.
When x = ±0.3, y = e−0.09 ≈ 0.9139, which is a global minimum on the given interval. Thus e−0.09 ≤ y ≤ 1 for
|x| ≤ 0.3.
22. Examination of the graph suggests that 0 ≤ x3 e−x ≤ 2. The lower bound of 0 is the best possible lower bound since
f (0) = (0)3 e−0 = 0.
To find the best possible upper bound, we find the critical points. Differentiating, using the product rule, yields
f ′ (x) = 3x2 e−x − x3 e−x
Setting f ′ (x) = 0 and factoring gives
3x2 e−x − x3 e−x = 0
x2 e−x (3 − x) = 0
So the critical points are x = 0 and x = 3. Note that f ′ (x) < 0 for x > 3 and f ′ (x) > 0 for x < 3, so f (x) has a local
maximum at x = 3. Examination of the graph tells us that this is the global maximum. So 0 ≤ x3 e−x ≤ f (3).
f (3) = 33 e−3 ≈ 1.34425
So 0 ≤ x3 e−x ≤ 33 e−3 ≈ 1.34425 are the best possible bounds for the function.
2
33 e−3 ≈ 1.34425
1
f (x) = x3 e−x
Figure 4.34
288
Chapter Four /SOLUTIONS
23. The graph of y = x + sin x in Figure 4.35 suggests that the function is nondecreasing over the entire interval. You can
confirm this by looking at the derivative:
y ′ = 1 + cos x
y
2π
π
y = x + sin x
x
π
−π
2π
−π
Figure 4.35: Graph of y = x + sin x
Since cos x ≥ −1, we have y ′ ≥ 0 everywhere, so y never decreases. This means that a lower bound for y is 0 (its
value at the left endpoint of the interval) and an upper bound is 2π (its value at the right endpoint). That is, if 0 ≤ x ≤ 2π:
0 ≤ y ≤ 2π.
These are the best bounds for y over the interval.
24. Let y = ln(1 + x). Since y ′ = 1/(1 + x), y is increasing for all x ≥ 0. The lower bound is at x = 0, so, ln(1) = 0 ≤ y.
There is no upper bound.
25. Let y = ln(1 + x2 ). Then y ′ = 2x/(1 + x2 ). Since the denominator is always positive, the sign of y ′ is determined by
the numerator 2x. Thus y ′ > 0 when x > 0, and y ′ < 0 when x < 0, and we have a local (and global) minimum for y
at x = 0. Since y(−1) = ln 2 and y(2) = ln 5, the global maximum is at x = 2. Thus 0 ≤ y ≤ ln 5, or (in decimals)
0 ≤ y < 1.61. (Note that our upper bound has been rounded up from 1.6094.)
Problems
26. We want to maximize the height, y, of the grapefruit above the ground, as shown in the figure below. Using the derivative
we can find exactly when the grapefruit is at the highest point. We can think of this in two ways. By common sense, at the
peak of the grapefruit’s flight, the velocity, dy/dt, must be zero. Alternately, we are looking for a global maximum of y,
so we look for critical points where dy/dt = 0. We have
dy
−50
= −32t + 50 = 0
and so
t=
≈ 1.56 sec.
dt
−32
Thus, we have the time at which the height is a maximum; the maximum value of y is then
y ≈ −16(1.56)2 + 50(1.56) + 5 = 44.1 feet.
y
60
40
20
1
2
3
t
27. We find all critical points:
dy
= 2ax + b = 0
dx
b
.
2a
Since y is a quadratic polynomial, its graph is a parabola which opens up if a > 0 and down if a < 0. The critical value
is a maximum if a < 0 and a minimum if a > 0.
x=−
4.2 SOLUTIONS
289
28. To find the value of w that minimizes S, we set dS/dw equal to zero and solve for w. To find dS/dw, we first solve for
S:
S − 5pw = 3qw2 − 6pq
S = 5pw + 3qw2 − 6pq.
We now find the critical points:
dS
= 5p + 6qw = 0
dw
w=−
5p
.
6q
There is one critical point. Since S is a quadratic function of w with a positive leading coefficient, the function has a
minimum at this critical point.
29. To find the value of t that maximizes y, we set dy/dt equal to zero and solve for t. We use the product rule to find dy/dt.
dy
= at2 e−bt (−b) + 2ate−bt = 0
dt
dy
= ate−bt (−bt + 2) = 0
dt
2
t = 0, t = .
b
Since a, t are nonnegative, ate−bt is nonnegative. Thus dy/dt is positive for −bt + 2 > 0 and negative for −bt + 2 < 0.
Also, t ≥ 0, so
dy/dt
= 0 for
t = 0, 2/b
dy/dt is positive for 0 < t < 2/b
dy/dt is negative for t > 2/b.
Thus, y has a local maximum, and also a global maximum, at t = 2/b. The graph of y = at2 e−bt , using a = 1 and b = 1,
is shown in Figure 4.36. We see that the function has a minimum at the critical point t = 0 and a maximum at t = 2/b.
y
t
Figure 4.36
30. (a) We have
T (D) =
and
CD2
C
D
D3
D2 =
−
−
,
2
3
2
3
dT
= CD − D2 = D(C − D).
dD
Since, by this formula, dT /dD is zero when D = 0 or D = C, negative when D > C, and positive when D < C,
we have (by the first derivative test) that the temperature change is maximized when D = C.
(b) The sensitivity is dT /dD = CD − D2 ; its derivative is d2 T /dD2 = C − 2D, which is zero if D = C/2, negative
if D > C/2, and positive if D < C/2. Thus by the first derivative test the sensitivity is maximized at D = C/2.
290
Chapter Four /SOLUTIONS
31. (a) Since a/q decreases with q, this term represents the ordering cost. Since bq increases with q, this term represents the
storage cost.
(b) At the minimum,
−a
dC
= 2 +b=0
dq
q
giving
q
a
a
so q =
.
q2 =
b
b
Since
2a
d2 C
= 3 > 0 for q > 0,
dq 2
q
we know that q =
minimum.
p
a/b gives a local minimum. Since q =
32. We look for critical points of M :
p
a/b is the only critical point, this must be the global
1
dM
= wL − wx.
dx
2
Now dM/dx = 0 when x = L/2. At this point d2 M/dx2 = −w so this point is a local maximum. The graph of M (x)
is a parabola opening downward, so the local maximum is also the global maximum.
33. (a) If we expect the rate to be nonnegative, then we must have 0 ≤ y ≤ a. See Figure 4.37.
rate (gm/sec)
Max rate
a/2
y (gm)
a
Figure 4.37
(b) The maximum value of the rate occurs at y = a/2, as can be seen from Figure 4.37, or by setting
d
(rate) = 0
dy
d
d
(rate) =
(kay − ky 2 ) = ka − 2ky = 0
dy
dy
a
y= .
2
From the graph, we see that y = a/2 gives the global maximum.
34. We set dU /dx = 0 to find the critical points:
b
−2a2
a
+ 2
x3
x
=0
−2a2 + ax = 0
x = 2a.
The only critical point is at x = 2a. When x < 2a we have dU/dx < 0, and when x > 2a we have dU/dx > 0. The
potential energy, U , is minimized at x = 2a.
4.2 SOLUTIONS
291
35. We set f ′ (r) = 0 to find the critical points:
3B
2A
− 4 =0
r3
r
2Ar − 3B
=0
r4
2Ar − 3B = 0
3B
r=
.
2A
The only critical point is at r = 3B/(2A). If r > 3B/(2A), we have f ′ > 0 and if r < 3B/(2A), we have f ′ < 0.
Thus, the force between the atoms is minimized at r = 3B/(2A).
36. (a) To show that R is an increasing function of r1 , we show that dR/dr1 > 0 for all values of r1 . We first solve for R:
1
1
1
=
+
R
r1
r2
r2 + r1
1
=
R
r1 r2
r1 r2
R=
.
r2 + r1
We use the quotient rule (and remember that r2 is a constant) to find dR/dr1 :
(r2 + r1 )(r2 ) − (r1 r2 )(1)
(r2 )2
dR
=
=
.
2
dr1
(r2 + r1 )
(r2 + r1 )2
Since dR/dr1 is the square of a number, we have dR/dr1 > 0 for all values of r1 , and thus R is increasing for all
r1 .
(b) Since R is increasing on any interval a ≤ r1 ≤ b, the maximum value of R occurs at the right endpoint r1 = b.
37. (a) We want the maximum value of I. Using the properties of logarithms, we rewrite the expression for I as
I = k(ln S − ln S0 ) − S + S0 + I0 .
Since k and S0 are constant, differentiating with respect to S gives
dI
k
= − 1.
dS
S
Thus, the critical point is at S = k. Since dI/dS is positive for S < k and dI/dS is negative for S > k, we see that
S = k is a local maximum.
We only consider positive values of S. Since S = k is the only critical point, it gives the global maximum value
for I, which is
I = k(ln k − ln S0 ) − k + S0 + I0 .
(b) Since both k and S0 are in the expression for the maximum value of I, both the particular disease and how it starts
influence the maximum.
38. Suppose the points are given by x and −x, where x ≥ 0. The function is odd, since
y=
(−x)3
x3
=−
,
4
1 + (−x)
1 + x4
so the corresponding y-coordinates are also opposite. See Figure 4.38. For x > 0, we have
m=
x3
1+x4
3
x
− − 1+x
4
x − (−x)
=
1
2x3
x2
·
=
.
4
2x 1 + x
1 + x4
For the maximum slope,
x2 (4x3 )
2x
dm
=
−
=0
dx
1 + x4
(1 + x4 )2
2x(1 + x4 ) − 4x5
=0
(1 + x4 )2
292
Chapter Four /SOLUTIONS
2x(1 − x4 )
=0
(1 + x4 )2
x 1 − x4 = 0
x = 0, ±1.
For x > 0, there is one critical point, x = 1. Since m tends to 0 when x → 0 and when x → ∞, the critical point x = 1
gives the maximum slope. Thus, the maximum slope occurs when the line has endpoints
−1, −
1
2
and
1
.
2
1,
y
x,
−x
x
−x,
−x3
1 + x4
x3
1 + x4
x
Figure 4.38
39. Since the function is positive, the graph lies above the x-axis. If there is a global maximum at x = 3, t′ (x) must be
positive, then negative. Since t′ (x) and t′′ (x) have the same sign for x < 3, they must both be positive, and thus the
graph must be increasing and concave up. Since t′ (x) and t′′ (x) have opposite signs for x > 3 and t′ (x) is negative,
t′′ (x) must again be positive and the graph must be decreasing and concave up. A possible sketch of y = t(x) is shown
in Figure 4.39.
y
(3, 3)
y = t(x)
x
Figure 4.39
40. One possible graph of g is in Figure 4.40.
(−2, g(−2))
g(x)
(0, g(0))
(2, g(2))
Figure 4.40
(a) From left to right, the graph of g(x) starts “flat”, decreases slowly at first then more rapidly, most rapidly at x = 0.
The graph then continues to decrease but less and less rapidly until flat again at x = 2. The graph should exhibit
symmetry about the point (0, g(0)).
4.2 SOLUTIONS
293
(b) The graph has an inflection point at (0, g(0)) where the slope changes from negative and decreasing to negative and
increasing.
(c) The function has a global maximum at x = −2 and a global minimum at x = 2.
(d) Since the function is decreasing over the interval −2 ≤ x ≤ 2
g(−2) = 5 > g(0) > g(2).
Since the function appears symmetric about (0, g(0)), we have
g(−2) − g(0) = g(0) − g(2).
41. (a) We know that h′′ (x) < 0 for −2 ≤ x < −1, h′′ (−1) = 0, and h′′ (x) > 0 for x > −1. Thus, h′ (x) decreases to its
minimum value at x = −1, which we know to be zero, and then increases; it is never negative.
(b) Since h′ (x) is non-negative for −2 ≤ x ≤ 1, we know that h(x) is never decreasing on [−2, 1]. So a global
maximum must occur at the right hand endpoint of the interval.
(c) The graph below shows a function that is increasing on the interval −2 ≤ x ≤ 1 with a horizontal tangent and an
inflection point at (−1, 2).
y
h(x)
(−1, 2)
x
−2
−1
1
42. Suppose f has critical points x = a and x = b. Suppose a < b. By the Extreme Value Theorem, we know that
the derivative function, f ′ (x), has global extrema on [a, b]. If both the maximum and minimum of f ′ (x) occur at the
endpoints of [a, b], then f ′ (a) = 0 = f ′ (b), so f ′ (x) = 0 for all x in [a, b]. In this case, f would have more than two
critical points. Since f has only two critical points, there is a local maximum or minimum of f ′ inside the interval [a, b].
43. To find the critical points, set dD/dx = 0:
dD
= 2(x − a1 ) + 2(x − a2 ) + 2(x − a3 ) + · · · + 2(x − an ) = 0.
dx
Dividing by 2 and solving for x gives
x + x + x + · · · + x = a1 + a2 + a3 + · · · + an .
Since there are n terms on the left,
nx = a1 + a2 + a3 + · · · + an
x=
n
1X
a1 + a2 + a3 + · · · + an
=
ai .
n
n
i=1
The expression on the right is the average of a1 , a2 , a3 , · · · , an .
Since D is a quadratic with positive leading coefficient, this critical point is a minimum.
44. (a) If both the global minimum and the global maximum are at the endpoints, then f (x) = 0 everywhere in [a, b], since
f (a) = f (b) = 0. In that case f ′ (x) = 0 everywhere as well, so any point in (a, b) will do for c.
(b) Suppose that either the global maximum or the global minimum occurs at an interior point of the interval. Let c be
that point. Then c must be a local extremum of f , so, by the theorem concerning local extrema on page 188, we have
f ′ (c) = 0, as required.
45. (a) The equation of the secant line between x = a and x = b is
y = f (a) +
f (b) − f (a)
(x − a)
b−a
294
Chapter Four /SOLUTIONS
and
f (b) − f (a)
(x − a),
b−a
so g(x) is the difference, or distance, between the graph of f (x) and the secant line. See Figure 4.41.
g(x) = f (x) − f (a) −
y
y = f (x)
y = f (a) +
Secant
line
❄
a
✻
❄
■
f (b) − f (a)
b−a
(x − a)
g(x)
x
b
Figure 4.41: Value of g(x) is the difference between the secant line
and the graph of f (x)
(b) Figure 4.41 shows that g(a) = g(b) = 0. You can also easily check this from the formula for g(x). By Rolle’s
Theorem, there must be a point c in (a, b) where g ′ (c) = 0.
(c) Differentiating the formula for g(x), we have
g ′ (x) = f ′ (x) −
So from g ′ (c) = 0, we get
f ′ (c) =
as required.
f (b) − f (a)
.
b−a
f (b) − f (a)
,
b−a
Strengthen Your Understanding
46. The function f has a critical point at x = 1 where f ′′ < 0, so f has a local maximum at (1, 0). However, checking the
endpoints we find that f takes its largest value at the endpoint x = 3, where f (3) = 4.
47. Consider the interval 1 ≤ x ≤ 2. Since f is increasing on this interval, its global minimum occurs at the left end-point of
the interval, x = 1. If x = 0 is inside the interval a ≤ x ≤ b, then the global minimum of f occurs at x = 0. Otherwise,
the global minimum of f will occur at an endpoint.
48. The function has a vertical asymptote at x = 1 where it is not defined. There is no global minimum nor maximum and
therefore no upper or lower bound for f (x).
49. One possible answer is f (x) = 1 − x for which f ′ (x) = −1 so there are no critical points. Also f (0) = 1 is a global
maximum and f (1) = 0 is a global minimum. There are many other possible answers.
50. The function must be a constant. One possible answer is f (x) = 1. Any constant function’s global maximum is equal to
its global minimum.
√
√
√
51. Solving
x2 = 5 and x2 = 2, we see that one such interval is 2 ≤ x ≤ 5. A second possible interval is − 5 ≤ x ≤
√
− 2.
52. The function f (x) = sin x is differentiable and has 1 as its global maximum and −1 as its global minimum on any
interval containing −π/2 ≤ x ≤ π/2 such as the interval −4 ≤ x ≤ 4. The function f (x) = cos x is another such
function. There are many other possibilities.
53. True, f (x) ≤ 4 on the interval (0, 2).
54. False. The values of f (x) get arbitrarily close to 4, but f (x) < 4 for all x in the interval (0, 2).
55. True. The values of f (x) get arbitrarily close to 0, but f (x) > 0 for all x in the interval (0, 2).
56. False. On the interval (−1, 1), the global minimum is 0.
57. True, by the Extreme Value Theorem, Theorem 4.2.
58. (a) This is not implied; just because a function satisfies the conclusions of the statement, that does not mean it has to
satisfy the conditions.
4.3 SOLUTIONS
295
(b) This is not implied; if a function fails to satisfy the conditions of the statement, then the statement does not tell us
anything about it.
(c) This is implied; if a function fails to satisfy the conclusions of the statement, then it could not satisfy the conditions
of the statement, because if it did the statement would imply it also satisfied the conclusions.
59. False, since f (x) = 1/x takes on arbitrarily large values as x → 0+ . The Extreme Value Theorem requires the interval
to be closed as well as bounded.
60. False. The Extreme Value Theorem says that continuous functions have global maxima and minima on every closed,
bounded interval. It does not say that only continuous functions have such maxima and minima.
61. True. If the maximum is not at an endpoint, then it must be at critical point of f . But x = 0 is the only critical point of
f (x) = x2 and it gives a minimum, not a maximum.
62. True. For example, A = 1 and A = 2 are both upper bounds for f (x) = sin x.
63. True. If f ′ (0) > 0, then f would be increasing at 0 and so f (0) < f (x) for x just to the right of 0. Then f (0) would not
be a maximum for f on the interval 0 ≤ x ≤ 10.
Solutions for Section 4.3
Exercises
1. Let the numbers be x and y. Then x + y = 100, so y = 100 − x.
Since both numbers are nonnegative, we restrict to 0 ≤ x ≤ 100.
The product is
P = xy = x(100 − x) = 100x − x2 .
Differentiating to find the maximum,
dP
= 100 − 2x = 0
dx
100
= 50.
x=
2
So there is a critical point at x = 50; the end points are at x = 0, 100.
Evaluating gives
At x = 0, we have P = 0.
At x = 50, we have P = 50(100 − 50) = 2500.
At x = 100, we have P = 100(100 − 100) = 0.
Thus the maximum value is 2500.
2. Let the numbers be x and y. Then xy = 784, so y = 784/x.
Since both numbers are positive, we restrict to x > 0.
The sum is
784
.
S =x+y =x+
x
Differentiating to find the minimum,
dS
784
= 1− 2 = 0
dx
x
x2 = 784
√
so x = ± 784 = ±28.
For x > 0, there is only one critical point at x = 28. We find
d2 S
784
=2 3 .
dx2
x
Since d2 S/dx2 > 0 for x > 0, there is a local minimum at x = 28. The derivative dS/dx is negative for 0 < x < 28
and dS/dx is positive for x > 28. Thus, x = 28 gives the global minimum for x > 0.
The minimum value of the sum is
784
= 56.
S = 28 +
28
296
Chapter Four /SOLUTIONS
3. Let the numbers be x, y, z and let y = 2x. Then
x + y + z = 3x + z = 36,
so z = 36 − 3x.
Since all the numbers are nonnegative, we restrict to 0 ≤ x ≤ 12.
The product is
P = xyz = x · 2x · (36 − 3x) = 72x2 − 6x3 .
Differentiating to find the maximum,
dP
= 144x − 18x2 = 0
dx
−18x(x − 8) = 0
x = 0, 8.
So there are critical points at x = 0 and x = 8; the end points are at x = 0, 12.
Evaluating gives:
At x = 0, we have P = 0.
At x = 8, we have P = 8 · 16 · (36 − 3 · 8) = 1536.
At x = 12, we have P = 12 · 24 · (36 − 3 · 12) = 0.
Thus, the maximum value of the product is 1536.
4. Let the sides be x and y cm. Then 2x + 2y = 64, so y = 32 − x.
Since both sides are nonnegative, we restrict to 0 ≤ x ≤ 32.
The area in cm2 is
A = xy = x(32 − x) = 32x − x2 .
Differentiating to find the maximum,
dA
= 32 − 2x = 0
dx
32
= 16.
x=
2
So there is a critical point at x = 16 cm; the end points are at x = 0 and x = 32 cm.
Evaluating gives:
At x = 0, we have A = 0 cm2 .
At x = 16, we have A = 16(32 − 16) = 256 cm2 .
At x = 32, we have A = 32(32 − 32) = 0 cm2 .
Thus, the maximum area occurs in a square of side 16 cm.
5. Let w and l be the width and length, respectively, of the rectangular area you wish to enclose. Then
w + w + l = 100 feet
l = 100 − 2w
Area = w · l = w(100 − 2w) = 100w − 2w2
WALL
w
w
l
To maximize area, we solve A′ = 0 to find critical points. This gives A′ = 100 − 4w = 0, so w = 25, l = 50. So
the area is 25 · 50 = 1250 square feet. This is a local maximum by the second derivative test because A′′ = −4 < 0.
Since the graph of A is a parabola, the local maximum is in fact a global maximum.
6. We want to minimize the surface area S of the box, shown in Figure 4.42. The box has 6 faces: the top and bottom, each
of which has area x2 and the four sides, each of which has area xh. Thus we want to minimize
S = 2x2 + 4xh.
4.3 SOLUTIONS
297
The volume of the box is 8 = x2 h, so h = 8/x2 . Substituting this expression in for h in the formula for S gives
S = 2x2 + 4x ·
32
8
= 2x2 +
.
x2
x
Differentiating gives
dS
32
= 4x − 2 .
dx
x
To minimize S we look for critical points, so we solve 0 = 4x − 32/x2 . Multiplying by x2 gives
0 = 4x3 − 32,
so x = 81/3 . Then we can find
h=
8
8
= 2/3 = 81/3 .
x2
8
Thus x = h = 81/3 cm (the box is a cube).
We can check that this critical point is a minimum of S by checking the sign of
64
d2 S
=4+ 3
dx2
x
which is positive when x > 0. So S is concave up at the critical point and therefore x = 81/3 gives a minimum value
of S.
h
x
x
Figure 4.42
7. We want to minimize the surface area S of the box, shown in Figure 4.43. The box has 5 faces: the bottom which has area
x2 and the four sides, each of which has area xh. Thus we want to minimize
S = x2 + 4xh.
The volume of the box is 8 = x2 h, so h = 8/x2 . Substituting this expression in for h in the formula for S gives
S = x2 + 4x ·
32
8
= x2 +
.
x2
x
Differentiating gives
dS
32
= 2x − 2 .
dx
x
To minimize S we look for critical points, so we solve 0 = 2x − 32/x2 . Multiplying by x2 gives
0 = 2x3 − 32,
so x = 161/3 cm. Then we can find
h=
cm.
16
161/3
8
8
=
= 2/3 =
2
2/3
x
2
16
2 · 16
298
Chapter Four /SOLUTIONS
We can check that this critical point is a minimum of S by checking the sign of
64
d2 S
=2+ 3
dx2
x
which is positive when x > 0. So S is concave up at the critical point and therefore x = 161/3 gives a minimum value
of S.
h
x
x
Figure 4.43
8. We want to minimize the surface area S of the cylinder, shown in Figure 4.44. The cylinder has 3 pieces: the top and
bottom disks, each of which has area πr 2 and the tube, which has area 2πrh. Thus we want to minimize
S = 2πr 2 + 2πrh.
The volume of the cylinder is 8 = πr 2 h, so h = 8/πr 2 . Substituting this expression for h in the formula for S gives
S = 2πr 2 + 2πr ·
16
8
= 2πr 2 +
.
πr 2
r
Differentiating gives
16
dS
= 4πr − 2 .
dr
r
To minimize S we look for critical points, so we solve 0 = 4πr − 16/r 2 . Multiplying by r 2 gives
0 = 4πr 3 − 16,
so r = (4/π)1/3 cm. Then we can find
h=
8
=
πr 2
π
8
4 2/3
=
π
4
π
4 2/3
π
2
=2
1/3
4
π
.
We can check that this critical point is a minimum of S by checking the sign of
d2 S
32
= 4π + 3
dr 2
r
which is positive when r > 0. So S is concave up at the critical point and therefore r = (4/π)1/3 cm is a minimum.
r
✻
h
❄
Figure 4.44
✲
4.3 SOLUTIONS
299
9. We want to minimize the surface area S of the cylinder, shown in Figure 4.45. The cylinder has two pieces: one disk (the
base) which has area πr 2 and the tube (the sides), which has area 2πrh. Thus we want to minimize
S = πr 2 + 2πrh.
The volume of the cylinder is 8 = πr 2 h, so h = 8/πr 2 . Substituting this expression in for h in the formula for S gives
S = πr 2 + 2πr ·
16
8
= πr 2 +
.
πr 2
r
Differentiating gives
16
dS
= 2πr − 2 .
dr
r
To minimize S we look for critical points, so we solve 0 = 2πr − 16/r 2 . Multiplying by r 2 gives
0 = 2πr 3 − 16,
so r = (8/π)1/3 cm. Then we can find
h=
8
=
πr 2
π
8
=
8 2/3
π
8
π
1/3
8
=
.
2/3
8
π
π
We can check that this critical point is a minimum of S by checking the sign of
32
d2 S
= 2π + 3
dr 2
r
which is positive when r > 0. So S is concave up at the critical point and therefore r = (8/π)1/3 cm is a minimum.
r
✻
✲
h
❄
Figure 4.45
10. We want to maximize the area function
A(x) = Base × Height
= xf (x)
1
= x3 − 50x2 + 1000x.
3
Critical points of A are solutions of
A′ (x) = x2 − 100x + 1000 = 0.
√
√
There are two critical points, x = 50 − 10 15 = 11.27 and x = 50 + 10 15 = 88.73, but only the first is in the interval
0 ≤ x ≤ 20.
The maximum occurs at the critical point in the interval or at one of the endpoints. Since
A(0) = 0
A(11.27) = 5396.5
√
the maximum area occurs with x = 50 − 10 15 = 11.270.
A(20) = 2666.67
300
Chapter Four /SOLUTIONS
11. We want to maximize the area function
1
× Base × Height
2
1
= · 20 · f (x)
2
10x2
=
− 500x + 10000.
3
A(x) =
Critical points of A are solutions of
20x
− 500 = 0.
3
There is one critical point, x = 75, but it is outside the interval 0 ≤ x ≤ 20.
The maximum of A in the interval occurs at one of the endpoints. Since
A′ (x) =
A(0) = 10000
A(20) = 4000/3 = 1333.3
the maximum area occurs with x = 0.
We can also see that x = 0 gives te maximum area without calculus, since the base of the triangle is always 20 and
the height is largest when x = 0.
12. The rectangle in Figure 4.46 has area, A, given by
A = 2xy =
2x
1 + x2
for x ≥ 0.
At a critical point,
2
dA
=
+ 2x
dx
1 + x2
−2x
(1 + x2 )2
=0
2(1 + x2 − 2x2 )
=0
(1 + x2 )2
1 − x2 = 0
x = ±1.
Since A = 0 for x = 0 and A → 0 as x → ∞, the critical point x = 1 is a local and global maximum for the area. Then
y = 1/2, so the vertices are
1
1
, −1,
.
(−1, 0) , (1, 0) , 1,
2
2
y
y=
1
1 + x2
✻
y
✛x✲
❄
Figure 4.46
13. The triangle in Figure 4.47 has area, A, given by
A=
1
x
xy = e−x/3 .
2
2
At a critical point,
1
x
dA
= e−x/3 − e−x/3 = 0
dx
2
6
1 −x/3
e
(3 − x) = 0
6
x = 3.
x
4.3 SOLUTIONS
301
Substituting the critical point and the endpoints into the formula for the area gives:
For x = 1, we have A = 21 e−1/3 = 0.358
For x = 3, we have A = 32 e−1 = 0.552
For x = 5, we have A = 52 e−5/3 = 0.472
Thus, the maximum area is 0.552 and the minimum area is 0.358.
y
(x, e−x/3 )
y✻
✛
x
✲
❄
x
Figure 4.47
14. (a) The rectangle in Figure 4.48 has area, A, given by
A = xy = xe−2x .
At a critical point, we have
dA
= 1 · e−2x − 2xe−2x = 0
dx
e−2x (1 − 2x) = 0
1
x= .
2
Since A = 0 when x = 0 and A → 0 as x → ∞, the critical point x = 1/2 is a local and global maximum. Thus
the maximum area is
1
1
.
A = e−2(1/2) =
2
2e
(b) The rectangle in Figure 4.48 has perimeter, P , given by
P = 2x + 2y = 2x + 2e−2x .
At a critical point, we have
dP
= 2 − 4e−2x = 0
dx
1
e−2x =
2
1
2
1 1
1
x = − ln = ln 2.
2 2
2
−2x = ln
To see if this critical point gives a maximum or minimum, we find
d2 P
= 8e−2x .
dx2
Since d2 P/dx2 > 0 for all x, including x =
minimum perimeter is
P =2
1
2
ln 2, the critical point is a local and global minimum. Thus, the
1
1
1
ln 2 + 2e−2( 2 ln 2) = ln 2 + 2e− ln 2 = ln 2 + 2 · = ln 2 + 1.
2
2
302
Chapter Four /SOLUTIONS
y
✻
y = e−x/2
y
✛
x
❄
✲
x
Figure 4.48
Problems
15. We take the derivative, set it equal to 0, and solve for x:
1
dt
=
dx
6
2
(2000 − x) =
3
4
2
(2000 − x) =
9
5
4
2
(2000 − x) =
9
9
−
−1/2
1 1
·
(2000 − x)2 + 6002
· 2(2000 − x) = 0
4 2
(2000 − x)2 + 6002
(2000 − x)2 + 6002
· 6002
1/2
r
4
1200
· 6002 = √
5
5
1200
x = 2000 − √ feet.
5
2000 − x =
√
Note that 2000 − (1200/ 5) ≈ 1463 feet, as given in the example.
16. Let the sides of the rectangle have lengths a and b. We shall look for the minimum of the square s of the length of either
diagonal, i.e. s = a2 + b2 . The area is A = ab, so b = A/a. This gives
s(a) = a2 +
A2
.
a2
To find the minimum squared length we need to find the critical points of s. Differentiating s with respect to a gives
A2
ds
= 2a + (−2)A2 a−3 = 2a 1 − 4
da
a
The derivative ds/da = 0 when a =
2 1+
3A2
a4
√
A, that is when a = b and so the rectangle is a square. Because
> 0, this is a minimum.
17. From the triangle shown in Figure 4.49, we see that
2
w
2
2
h
= 302
2
w2 + h2 = 4(30)2 = 3600.
+
30
w/2
Figure 4.49
h/2
d2 s
=
da2
4.3 SOLUTIONS
The strength, S, of the beam is given by
303
S = kwh2 ,
for some constant k. To make S a function of only one variable, substitute for h2 , giving
S = kw(3600 − w2 ) = k(3600w − w3 ).
Differentiating and setting dS/dw = 0,
dS
= k(3600 − 3w2 ) = 0.
dw
Solving for w gives
w=
√
1200 = 34.64 cm,
so
h2 = 3600 − w2 = 3600 − 1200 = 2400
√
h = 2400 = 48.99 cm.
Thus, w = 34.64 cm and h = 48.99 cm give a critical point. To check that this is a local maximum, we compute
d2 S
= −6w < 0 for
dw2
w > 0.
Since d2 S/dw2 < 0, we see that w = 34.64 cm is a local maximum. It is the only critical point, so it is a global
maximum.
18. The area of the rectangle on the left is 2x. The entire rectangle has area 2, so the area of the rectangle on the right is
2 − 2x. We are to maximize the product
f (x) = 2x(2 − 2x) = 4x − 4x2
of the two areas where 0 ≤ x ≤ 1.
The critical points of f occur where
f ′ (x) = 4 − 8x = 0
at x = 1/2.
The maximum of f on the interval 0 ≤ x ≤ 1 is at the critical point x = 1/2 or one of the endpoints x = 0 or
x = 1. We have
1
f (0) = f (1) = 0 and f
= 1.
2
The product of the areas is a maximum when x = 1/2.
19. First find the length h of the vertical segment at x. Since the top edge of the triangle has slope h/x = 2/1, we have
h = 2x. Thus
1
1
Area of smaller triangle = × Base × Height = · x · 2x = x2 .
2
2
The entire triangle has area 1, so the area of the trapezoid on the right is 1 − x2 . We are to maximize the product
f (x) = x2 (1 − x2 ) = x2 − x4
of the two areas where 0 ≤ x ≤ 1
The critical points of f occur where
f ′ (x) = 2x − 4x3 = 2x(1 − 2x2 ) = 0
√
√
√
at x = 0, x = 1/ 2 or x = −1/ 2. The only critical point in the interval 0 < x < 1 is x = 1/ 2.
The
√ maximum of f on the interval 0 ≤ x ≤ 1 is at one of the endpoints x = 0 or x = 1 or at the critical point
x = 1/ 2. We have
1
1
= .
f (0) = f (1) = 0 and f √
4
2
√
The product of the areas is a maximum when x = 1/ 2.
304
Chapter Four /SOLUTIONS
20. (a) The rectangle on the left has area xy, and the semicircle on the right with radius y/2 has area (1/2)π(y/2)2 = πy 2 /8.
We have
π
Area of entire region = xy + y 2 .
8
(b) The perimeter of the figure is made of two horizontal line segments of length x each, one vertical segment of length
y, and a semicircle of radius y/2 of length πy/2. We have
Perimeter of entire region = 2x + y +
π
π
y = 2x + (1 + )y.
2
2
(c) We want to maximize the area xy + πy 2 /8, with the perimeter condition 2x + (1 + π/2)y = 100. Substituting
x = 50 −
into the area formula, we must maximize
A(y) = 50 −
on the interval
π
1
+
2
4
1
π
+
2
4
y = 50 −
y y+
2+π
y
4
π 2
y = 50y −
8
1
π
+
2
8
y2
0 ≤ y ≤ 200/(2 + π) = 38.8985
where y ≥ 0 because y is a length and y ≤ 200/(2 + π) because x ≥ 0 is a length.
The critical point of A occurs where
A′ (y) = 50 − 1 +
at
π
4
y=0
200
= 28.005.
4+π
A maximum for A must occur at the critical point y = 200/(4 + π) or at one of the endpoints y = 0 or y =
200/(2 + π). Since
200
200
A(0) = 0
A
= 700.124
A
= 594.2
4+π
2+π
the maximum is at
200
y=
.
4+π
Hence
2 + π 200
100
2+π
y = 50 −
=
.
x = 50 −
4
4 4+π
4+π
The dimensions giving maximum area with perimeter 100 are
y=
x=
100
= 14.0
4+π
y=
200
= 28.0.
4+π
The length y is twice as great as x.
21. (a) The rectangle has area xy. The two semicircles together form a circle of radius y/2 and area π(y/2)2 = πy 2 /4. We
have
π
Area of entire region = xy + y 2 .
4
(b) The perimeter of the figure is the sum 2x of the lengths of the two straight sections and the circumference πy of a
circle of diameter y. Thus
Perimeter of entire region = 2x + πy.
(c) We want to maximize the area xy + πy 2 /4, with the perimeter condition
2x + πy = 100.
Substituting
x = 50 −
π
y
2
into the area formula, we must maximize
A(y) = 50 −
π
π
π
y y + y 2 = 50y − y 2
2
4
4
4.3 SOLUTIONS
305
on the interval
100
π
where y ≥ 0 because y is a length and y ≤ 100/π because x ≥ 0 is a length.
The critical point of A occurs where
π
A′ (y) = 50 − y = 0
2
at y = 100/π, which is an endpoint. A maximum for A occurs at one of the endpoints y = 0 or y = 100/π. Since
0≤y≤
A(0) = 0
A
100
=
π
2500
π
the maximum is at y = 100/π. Hence x = 0.
The dimensions giving maximum area with perimeter 100 are x = 0, y = 100/π. The figure is a circle.
22. (a) The rectangle has area xy, the two semicircles with radius x/2 together have area π(x/2)2 = πx2 /4, and the two
semicircles with radius y/2 together have area πy 2 /4. We have
Area of entire figure = xy +
π 2 π 2
x + y .
4
4
(b) The perimeter of the figure is the sum of the circumference of a circle of diameter x and a circle of diameter y. Thus
Perimeter of entire figure = πx + πy.
(c) We want to maximize the area xy + πx2 /4 + πy 2 /4, with the perimeter condition
πx + πy = 100.
Substituting
x=
100
−y
π
into the area formula, we must maximize
A(y) =
on the interval
π
100
−y y+
π
4
100
−y
π
2
+
π 2
y
4
0 ≤ y ≤ 100/π = 31.831
where y ≥ 0 because y is a length and y ≤ 100/π because x ≥ 0 is a length.
The critical point of A occurs where
A′ (y) =
100
− 50 + (π − 2)y = 0
π
at y = 50/π = 15.916. A maximum for A must occur at one of the endpoints y = 0 or y = 100/π or at the critical
point y = 50/π. Since
A(0) = A
100
π
=
2500
= 795.8
π
A
50
π
= 651.2,
the maximum is at y = 0, giving x = 100/π, or at y = 100/π, when x = 0.
The dimensions giving maximum area with perimeter 100 are x = 0, y = 100/π and x = 100/π, y = 0. The
figure is a circle.
23. (a) The length of the piece of wire made into a circle is x cm, so the length of the piece made into a square is (L − x) cm.
See Figure 4.50.
L−x
x
Wire
s
Circle:
Perimeter x
r
Square:
Perimeter L − x
Figure 4.50
306
Chapter Four /SOLUTIONS
The circumference of the circle is x, so its radius, r cm, is given by
r=
x
cm.
2π
The perimeter of the square is (L − x), so the side length, s cm, is given by
s=
L−x
cm.
4
Thus, the sum of areas is given by
A = πr 2 + s2 = π
x
2π
2
+
L−x
4
Setting dA/dx = 0 to find the critical points gives
2
=
(L − x)2
x2
+
,
4π
16
for 0 ≤ x ≤ L.
(L − x)
dA
x
=
−
=0
dx
2π
8
8x = 2πL − 2πx
(8 + 2π)x = 2πL
πL
2πL
=
≈ 0.44L.
x=
8 + 2π
4+π
To find the maxima and minima, we substitute the critical point and the endpoints, x = 0 and x = L, into the area
function.
L2
.
For x = 0, we have A =
16
πL
πL
4L
For x =
, we have L − x = L −
=
. Then
4+π
4+π
4+π
A=
=
For x = L, we have A =
Thus, x =
1
π 2 L2
+
4π(4 + π)2
16
4L
4+π
2
=
πL2
L2
+
2
4(4 + π)
(4 + π)2
L2
L2
πL2 + 4L2
=
=
.
2
4(4 + π)
4(4 + π)
16 + 4π
L2
.
4π
L2
πL
gives the minimum value of A =
.
4+π
16 + 4π
L2
Since 4π < 16, we see that x = L gives the maximum value of A =
.
4π
This corresponds to the situation in which we do not cut the wire at all and use the single piece to make a circle.
(b) At the maximum, x = L, so
0
Length of wire in square
=
= 0.
Length of wire in circle
L
0
Area of square
= 2
= 0.
Area of circle
L /4π
At the minimum, x =
πL
πL
4L
, so L − x = L −
=
.
4+π
4+π
4+π
4L/(4 + π)
4
Length of wire in square
=
= .
Length of wire in circle
πL/(4 + π)
π
L2 /(4 + π)2
4
Area of square
=
= .
Area of circle
πL2 /(4(4 + π)2 )
π
(c) For a general value of x,
Length of wire in square
L−x
=
.
Length of wire in circle
x
(L − x)2 /16
Area of square
π (L − x)2
=
= ·
.
2
Area of circle
x /(4π)
4
x2
4.3 SOLUTIONS
307
If the ratios are equal, we have
π (L − x)2
L−x
= ·
.
x
4
x2
So either L − x = 0, giving x = L, or we can cancel (L − x) and multiply through by 4x2 , giving
4x = π(L − x)
πL
x=
.
4+π
Thus, the two values of x found in part (a) are the only values of x in 0 ≤ x ≤ L making the ratios in part (b) equal.
(The ratios are not defined if x = 0.)
24. The minimum value of x in the interval 0 ≤ x ≤ 10 is x = 0 and the maximum is x = 10.
25. The geometry shows that as x increases
from 0√to 10 the length y decreases. The minimum value of y is 5 when x = 10,
√
and the maximum value of y is 102 + 52 = 125 when x = 0.
We can also use calculus. By the Pythagorean theorem we have
y=
The extreme values of
p
52 + (10 − x)2 .
y = f (x) =
p
52 + (10 − x)2
for 0 ≤ x ≤ 10 occur at the endpoints x = 0 or x = 10 or at a critical point of f . The only solution of the equation
x − 10
=0
f ′ (x) = p
52 + (10 − x)2
is x = 10, which is the only critical point of f . We have
√
f (0) = 125 = 11.18
√
Therefore the minimum value of y is 5 and the maximum value is
f (10) = 5.
√
125 = 5 5.
26. By the Pythagorean Theorem we have
y=
The extreme values of
p
52 + (10 − x)2 .
p
f (x) = x + 2y = x + 2
52 + (10 − x)2
for 0 ≤ x ≤ 10 occur at the endpoints x = 0 or x = 10 or at a critical point of f in the interval.
To find critical points, solve the equation
f ′ (x) = 0
2(10 − x)
=0
1− p
2
5 + (10 − x)2
2(10 − x) =
2
p
2
3(10 − x) = 5
52 + (10 − x)2
5
x = 10 ± √ .
3
√
The only critical point of f in the interval 0 ≤ x ≤ 10 is at x = 10 − 5/ 3. Comparing values at the endpoints and
critical points we have
√
f (0) = 10 5 = 22.36
√
5
f (10 − √ ) = 10 + 5 3 = 18.66.
3
√
√
Therefore the minimum value of x + 2y for 0 ≤ x ≤ 10 is 10 + 5 3 and the maximum value is 10 5.
f (10) = 20
308
Chapter Four /SOLUTIONS
27. By the Pythagorean theorem we have
y=
The extreme values of
p
52 + (10 − x)2 .
f (x) = 2x + y = 2x +
p
52 + (10 − x)2
for 0 ≤ x ≤ 10 occur at the endpoints x = 0 or x = 10 or at a critical point of f in the interval.
To find critical points, solve the equation
f ′ (x) = 0
10 − x
=0
2− p
52 + (10 − x)2
p
10 − x = 2
2
52 + (10 − x)2
3(10 − x) = −100.
The function f has no critical points. Comparing values at the endpoints we have
√
f (10) = 25.
f (0) = 5 5 = 11.18
√
The minimum value of 2x + y for 0 ≤ x ≤ 10 is 5 5 and the maximum value is 25.
√
28. The distance d(x) from the point (x, 1 − x) on the curve to the origin is given by
d(x) =
Since x is in the domain of y =
√
p
√
x2 + ( 1 − x)2 = x2 + (1 − x).
q
1 − x, we have −∞ < x ≤ 1. Differentiating gives
2x − 1
d′ (x) = √
,
2 x2 − x + 1
′
′
so x = 1/2 is the only critical point.
√ Since d (x) < 0 for x < 1/2 and d (x) > 0 for x > 1/2, the point x = 1/2 is a
minimum for x. The point (1/2, 1/ 2) is the closest point on the curve to the origin.
29. (a) For points (x, y) on the ellipse, we have y 2 = 1 − x2 /9 and −3 ≤ x ≤ 3. We wish to minimize the distance
D=
p
(x − 2)2 + (y − 0)2 =
r
(x − 2)2 + 1 −
x2
.
9
To do so, we find the value of x minimizing d = D2 for −3 ≤ x ≤ 3. This x also minimizes D. Since d =
(x − 2)2 + 1 − x2 /9, we have
2x
16x
d′ (x) = 2(x − 2) −
=
− 4,
9
9
which is 0 when x = 9/4. Since d′′ (9/4) = 16/9 > 0, we see d has a local minimum at x = 9/4. Since the graph
of d is a parabola, the local minimum is in fact a global minimum. Solving for y, we have
y2 = 1 −
x2
=1−
9
2
9
4
·
1
7
=
,
9
16
√
√
so y = ± 7/4. Therefore, the points on the ellipse closest to (2, 0) are 9/4, ± 7/4 .
(b) This time, we wish to minimize
r
√
x2
.
D = (x − 8)2 + 1 −
9
√
Again, let d = D2 and minimize d(x) for −3 ≤ x ≤ 3. Since d = (x − 8)2 + 1 − x2 /9,
√
√
16x
2x
d′ (x) = 2(x − 2 2) −
=
− 4 2.
9
9
√
√
′
Therefore, d (x) = 0 when x = 9 2/4. But 9 2/4 > 3, so there are not any critical points on the interval
′
−3 ≤ x ≤ 3. The minimum distance must be attained at an endpoint. Since
√ d (x) < 0 for all x between −3 and 3,
the minimum is at x = 3. So (3, 0) is the point on the ellipse closest to ( 8, 0).
4.3 SOLUTIONS
309
30. Let the radius of the can be r and let its height be h. The surface area, S, and volume, V , of the can are given by
S = Area of the sides of can + Area of top and bottom
S = 2πrh + 2πr 2
V = πr 2 h.
Since S = 280, we have 2πrh + 2πr 2 = 280, we have
140 − πr 2
πr
h=
and
140 − πr 2
= 140r − πr 3 .
πr
p
Since h ≥ 0, we have πr 2 ≤ 140 and thus 0 < r ≤ 140/π = 6.676.
We have
V ′ (r) = 140 − 3πr 2 ,
V = πr 2
p
so the only critical point of V with r > 0 is r = 140/3π = 3.854. The critical point is in the domain 0 < r ≤ 6.676.
Since V ′′ (r) = −6πr is negative for all r > 0, the critical point r = 3.854 cm gives the maximum value of the
volume. For this value of r, we have h = 7.708 cm and V = 359.721 cm3 .
√
31. Figure 4.51 shows the vertical cross section through the cylinder and sphere. The circle has equation y = 1 − x2 , so if
the cylinder has radius x and height y, its volume, V , is given by
V = πx2 y = πx2
At a critical point, dV /dx = 0, so
p
p
dV
= 2πx 1 − x2 + πx2
dx
1 − x2
1
2
2πx
πx
√
1 − x2
for 0 ≤ x ≤ 1.
(1 − x2 )−1/2 (−2x) = 0
p
1 − x2 − √
p
2
1−
x2
2
πx3
=0
1 − x2
−x
2
=0
x(2 − 3x2 ) = 0
x = 0, ±
r
2
.
3
p
Since V = 0 at the endpoints x = 0 and x = 1, and V is positive at the only critical point, x = 2/3, in the interval,
p
the critical point x = 2/3 is a local and global maximum. Thus, the cylinder with maximum volume has
Radius = x =
r
2
3
v
u
r !2 r
u
2
1
t
Height = y = 1 −
=
.
3
y
1
✻y = √1 − x2
y
−1
✛
x
✲
Figure 4.51
❄
1
x
3
310
Chapter Four /SOLUTIONS
32. (a) If we expect the rate to be nonnegative, we must have 0 ≤ y ≤ a and 0 ≤ y ≤ b. Since we assume a < b, we restrict
y to 0 ≤ y ≤ a.
In fact, the expression for the rate is nonnegative for y greater than b, but these values of y are not meaningful
for the reaction. See Figure 4.52.
rate (gm/sec)
kab
a
y (gm)
b
Figure 4.52
(b) From the graph, we see that the maximum rate occurs when y = 0; that is, at the start of the reaction.
33. Call the stacks A and B. (See below.) Assume that A corresponds to k1 , and B corresponds to k2 .
20 miles
A
x
B
Suppose the point where the concentration of deposit is a minimum occurs at a distance of x miles from stack A. We
want to find x such that
k2
1
7
k1
=
k
+
S= 2 +
2
x
(20 − x)2
x2
(20 − x)2
is a minimum, which is the same thing as minimizing f (x) = 7x−2 + (20 − x)−2 since k2 is nonnegative.
We have
f ′ (x) = −14x−3 − 2(20 − x)−3 (−1) =
−14(20 − x)3 + 2x3
2
−14
+
=
.
3
3
x
(20 − x)
x3 (20 − x)3
Thus we want to find x such that −14(20 − x)3 + 2x3 = 0, which implies 2x3 = 14(20 − x)3 . That’s equivalent to x3 =
= (1/7)1/3 ≈ 0.523. Solving for x, we have 20 − x = 0.523x, whence x = 20/1.523 ≈ 13.13.
7(20 − x)3 , or 20−x
x
To verify that this minimizes f , we take the second derivative:
42
6
+
>0
x4
(20 − x)4
f ′′ (x) = 42x−4 + 6(20 − x)−4 =
for any 0 < x < 20, so by the second derivative test the concentration is minimized 13.13 miles from A.
34. We only consider λ > 0. For such λ, the value of v → ∞ as λ → ∞ and as λ → 0+ . Thus, v does not have a maximum
velocity. It will have a minimum velocity. To find it, we set dv/dλ = 0:
dv
1
=k
dλ
2
Solving, and remembering that λ > 0, we obtain
λ
c
+
c
λ
−1/2
1
c
− 2
c
λ
c
1
− 2 =0
c
λ
1
c
= 2
c
λ
λ2 = c2 ,
= 0.
4.3 SOLUTIONS
311
so
λ = c.
Thus, we have one critical point. Since
dv
<0
dλ
for λ < c
and
dv
> 0 for λ > c,
dλ
the first derivative test tells us that we have a local minimum of v at x = c. Since λ = c is the only critical point, it gives
the global minimum. Thus the minimum value of v is
v=k
q
c √
c
+ = 2k.
c
c
35. The domain for E is all real x. Note E → 0 as x → ±∞. The critical points occur where dE/dx = 0. The derivative is
kx(2x)
k
3
dE
=
− ·
2 3/2
2
2
dx
2
(x + r0 )
(x + r02 )5/2
=
k x2 + r02 − 3x2
=
k r02 − 2x2
(x2 + r02 )5/2
(x2 + r02 )
5/2
.
So dE/dx = 0 where
r02 − 2x2 = 0
r0
x = ±√ .
2
Looking at the formula for dE/dx shows
r0
dE
r0
> 0 for − √ < x < √
dx
2
2
dE
r0
< 0 for x < − √
dx
2
r0
dE
< 0 for x > √ .
dx
2
√
√
Therefore, x = −r0 / 2 gives the minimum value of E and x = r0 / 2 gives the maximum value of E.
E
r0
−√
2
r
√0
2
x
36. A graph of F against θ is shown in Figure 4.53.
Taking the derivative:
mgµ(cos θ − µ sin θ)
dF
=−
.
dθ
(sin θ + µ cos θ)2
At a critical point, dF/dθ = 0, so
cos θ − µ sin θ = 0
1
tan θ =
µ
θ = arctan
1
µ
.
312
Chapter Four /SOLUTIONS
If µ = 0.15, then θ = arctan(1/0.15) = 1.422 ≈ 81.5◦ . To calculate the maximum and minimum values of F , we
evaluate at this critical point and the endpoints:
At θ = 0,
F =
0.15mg
= 1.0mg newtons.
sin 0 + 0.15 cos 0
At θ = 1.422, F =
0.15mg
= 0.148mg newtons.
sin(1.422) + 0.15 cos(1.422)
At θ = π/2,
0.15mg
= 0.15mg newtons.
sin( π2 ) + 0.15 cos( π2 )
F =
Thus, the maximum value of F is 1.0mg newtons when θ = 0 (her arm is vertical) and the minimum value of F is
0.148mg newtons is when θ = 1.422 (her arm is close to horizontal). See Figure 4.54.
F (newtons)
F
1.0mg
1.0mg
F =
0.15mg
sin θ+0.15 cos θ
F =
0.15mg
sin θ+0.15 cos θ
0.148mg
π
2
θ
1.422
Figure 4.53
θ
Figure 4.54
37. (a) We must locate the critical point(s) of A(r) — i.e. the places where its derivative is zero. The derivative is
A′ (r) =
T
2r
− 2,
T
r
so A′ (r) = 0 when
A′ (r) =
2r
T
− 2 =0
T
r
T
2r
= 2
T
r
2r 3 = T 2
T2
r3 =
2
T 2/3
r = 1/3 .
2
The only critical point of A(r) is thus r = (T 2/3 )/(21/3 ). To check that this is a local (and, since it is the only
critical point, global) minimum, we employ the second derivative test. The second derivative A′′ (r) is
A′′ (r) =
2T
2
+ 3,
T
r
and at the critical point we have
4
6
2
T 2/3
+
=
> 0,
A
=
T
T
T
21/3
which implies A is concave up at this critical point. It is therefore a local and global minimum.
(b) The new period is 2T . To find the radius for this new period, substitute 2T for T in the expression for r:
′′
(2T )2/3
21/3
T 2/3
= 22/3 1/3 .
2
r=
Notice this is 22/3 times the original radius. To find the percentage increase, compute 22/3 ≈ 1.5874. The new
radius therefore represents an increase of 58.74% over the original.
4.3 SOLUTIONS
313
38. Let x equal the number of chairs ordered in excess of 300, so 0 ≤ x ≤ 100.
Revenue = R = (90 − 0.25x)(300 + x)
= 27, 000 − 75x + 90x − 0.25x2 = 27, 000 + 15x − 0.25x2
At a critical point dR/dx = 0. Since dR/dx = 15 − 0.5x, we have x = 30, and the maximum revenue is $27, 225 since
the graph of R is a parabola which opens downward. The minimum is $0 (when no chairs are sold).
39. If v is the speed of the boat in miles per hour, then
Cost of fuel per hour (in $/hour) = kv 3 ,
where k is the constant of proportionality. To find k, use the information that the boat uses $100 worth of fuel per hour
when cruising at 10 miles per hour: 100 = k103 , so k = 100/103 = 0.1. Thus,
Cost of fuel per hour (in $/hour) = 0.1v 3 .
From the given information, we also have
Cost of other operations (labor, maintenance, etc.) per hour (in $/hour) = 675.
So
Total Cost per hour (in $/hour) = Cost of fuel (in $/hour) + Cost of other (in $/hour)
= 0.1v 3 + 675.
However, we want to find the Cost per mile, which is the Total Cost per hour divided by the number of miles that the ferry
travels in one hour. Since v is the speed in miles/hour at which the ferry travels, the number of miles that the ferry travels
in one hour is simply v miles. Let C = Cost per mile. Then
Cost per mile (in $/mile) =
C=
Total Cost per hour (in $/hour)
Distance traveled per hour (in miles/hour)
675
0.1v 3 + 675
= 0.1v 2 +
.
v
v
We also know that 0 < v < ∞. To find the speed at which Cost per mile is minimized, set
dC
675
= 2(0.1)v − 2 = 0
dv
v
so
675
v2
675
3
= 3375
v =
2(0.1)
v = 15 miles/hour.
2(0.1)v =
Since
2(675)
d2 C
= 0.2 +
>0
dv 2
v3
for v > 0, v = 15 gives a local minimum for C by the second-derivative test. Since this is the only critical point for
0 < v < ∞, it must give a global minimum.
40. (a) The business must reorder often enough to keep pace with sales. If reordering is done every t months, then,
Quantity sold in t months = Quantity reordered in each batch
rt = q
q
t = months.
r
(b) The amount spent on each order is a + bq, which is spent every q/r months. To find the monthly expenditures, divide
by q/r. Thus, on average,
Amount spent on ordering per month =
a + bq
ra
=
+ rb dollars.
q/r
q
314
Chapter Four /SOLUTIONS
(c) The monthly cost of storage is kq/2 dollars, so
C = Ordering costs + Storage costs
kq
ra
+ rb +
dollars.
C=
q
2
(d) The optimal batch size minimizes C, so
dC
−ra
k
= 2 + =0
dq
q
2
k
ra
=
q2
2
2ra
q2 =
k
so
q=
r
2ra
items per order.
k
41. (a) The line in the left-hand figure has slope equal to the rate worms arrive. To understand why, see line (1) in the
right-hand figure. (This is the same line.) For any point Q on the loading curve, the line P Q has slope
QT
load
QT
=
=
.
PT
P O + OT
traveling time + searching time
(b) The slope of the line P Q is maximized when the line is tangent to the loading curve, which happens with line (2).
The load is then approximately 7 worms.
(c) If the traveling time is increased, the point P moves to the left, to point P ′ , say. If line (3) is tangent to the curve, it
will be tangent to the curve further to the right than line (2), so the optimal load is larger. This makes sense: if the
bird has to fly further, you’d expect it to bring back more worms each time.
load
(number of worms)
load
(number of worms)
8
(2)
8
Number of worms
Number of worms
(3)
4
Q
(1)
time
P
P′
O
time
P
O T
Traveling time
Searching time
42. Let x be asp
indicated in the figure in the text. Then the distance from S to Town 1 is
√
Town 2 is (4 − x)2 + 42 = x2 − 8x + 32.
Total length of pipe = f (x) =
p
1 + x2 +
√
1 + x2 and the distance from S to
p
x2 − 8x + 32.
We want to look for critical points of f . The easiest way is to graph f and see that it has a local minimum at about x = 0.8
miles. Alternatively, we can use the formula:
2x − 8
2x
+ √
f ′ (x) = √
2 1 + x2
2 x2 − 8x + 32
x−4
x
+√
= √
1 + x2
x2 − 8x + 32
√
√
x x2 − 8x + 32 + (x − 4) 1 + x2
√
√
= 0.
=
1 + x2 x2 − 8x + 32
f ′ (x) is equal to zero when the numerator is equal to zero.
x
p
x2 − 8x + 32 + (x − 4)
x
p
p
1 + x2 = 0
x2 − 8x + 32 = (4 − x)
p
1 + x2 .
4.3 SOLUTIONS
315
Squaring both sides and simplifying, we get
x2 (x2 − 8x + 32) = (x2 − 8x + 16)(14x2 )
x4 − 8x3 + 32x2 = x4 − 8x3 + 17x2 − 8x + 16
15x2 + 8x − 16 = 0,
(3x + 4)(5x − 4) = 0.
So x = 4/5. (Discard x = −4/3 since we are only interested in x between 0 and 4, between the two towns.) Using the
second derivative test, we can verify that x = 4/5 is a local minimum.
43. (a) The distance the pigeon flies over water is
BP =
AB
500
=
,
sin θ
sin θ
and over land is
P L = AL − AP = 2000 −
500
500 cos θ
= 2000 −
.
tan θ
sin θ
Therefore the energy required is
500 cos θ
500
+ e 2000 −
sin θ
sin θ
2 − cos θ
500
π
= 500e
+ 2000e, for arctan
≤θ≤ .
sin θ
2000
2
2 − cos θ
must have the same critical points since the graph of E is just a
(b) Notice that E and the function f (θ) =
sin θ
2 − cos θ
500
stretch and a vertical shift of the graph of f . The graph of
for arctan( 2000
) ≤ θ ≤ π2 in Figure 4.55
sin θ
shows that E has precisely one critical point, and that a minimum for E occurs at this point.
E = 2e
arctan
π
2
π
3
500
2000
Figure 4.55: Graph of f (θ) =
2−cos θ
sin θ
θ
500
for arctan( 2000
)≤θ≤
π
2
To find the critical point θ, we solve f ′ (θ) = 0 or
sin θ · sin θ − (2 − cos θ) · cos θ
E = 0 = 500e
sin2 θ
1 − 2 cos θ
= 500e
.
sin2 θ
Therefore 1 − 2 cos θ = 0 and so θ = π/3.
(c) Letting a = AB and b = AL, our formula for E becomes
′
a
a cos θ
+e b−
sin θ
sin θ
2 − cos θ
a
π
+ eb, for arctan
≤θ≤ .
= ea
sin θ
b
2
2 − cos θ
. Thus, the critical point θ = π/3
Again, the graph of E is just a stretch and a vertical shift of the graph of
sin θ
is independent of e, a, and b. But the maximum of E on the domain arctan(a/b) ≤ θ ≤ π2 is dependent on the ratio
AB
a/b =
. In other words, the optimal angle is θ = π/3 provided arctan(a/b) ≤ π3 ; otherwise, the optimal angle
AL
is arctan(a/b), which means the pigeon should fly over the lake for the entire trip—this occurs when a/b > 1.733.
E = 2e
316
Chapter Four /SOLUTIONS
44. We want to maximize the viewing angle, which is θ = θ1 − θ2 . See Figure 4.56. Now
92
x
46
.
so θ2 = arctan
x
92
x
46
tan(θ2 ) =
x
so θ1 = arctan
tan(θ1 ) =
Then
θ = arctan
92
− arctan
x
We look for critical points of the function by computing dθ/dx:
46
x
for
x > 0.
1
−92
1
dθ
=
−
dx
1 + (92/x)2
x2
1 + (46/x)2
−46
−92
− 2
= 2
x + 922
x + 462
−92(x2 + 462 ) + 46(x2 + 922 )
=
(x2 + 922 ) · (x2 + 462 )
=
−46
x2
46(4232 − x2 )
.
(x2 + 922 ) · (x2 + 462 )
Setting dθ/dx = 0 gives
x2 = 4232
√
x = ± 4232.
√
that this is indeed where θ attains a maximum,
Since x > 0, the critical point is x = √4232 ≈ 65.1 meters. To verify√
we note that dθ/dx
√ > 0 for 0 < x < 4232 and dθ/dx < 0 for x > 4232. By the First Derivative Test, θ attains a
maximum at x = 4232 ≈ 65.1.
✻
92
✻
θ = θ1 − θ2
46
θ1
θ2
❄
❄
x
c Wesley Hitt/Getty Images
Figure 4.56
45. (a) Since the speed of light is a constant, the time of travel is minimized when the distance of travel is minimized. From
Figure 4.57,
Thus,
p
−
−
→ p
Distance OP = x2 + 12 = x2 + 1
p
−
−
→ p
Distance P Q = (2 − x)2 + 12 = (2 − x)2 + 1
Total distance traveled = s =
p
x2 + 1 +
p
(2 − x)2 + 1.
4.3 SOLUTIONS
317
The total distance is a minimum if
ds
1
1
= (x2 + 1)−1/2 · 2x + ((2 − x)2 + 1)−1/2 · 2(2 − x)(−1) = 0,
dx
2
2
giving
√
Squaring both sides gives
2−x
x
−p
=0
x2 + 1
(2 − x)2 + 1
2−x
x
√
= p
x2 + 1
(2 − x)2 + 1
(2 − x)2
x2
=
.
+1
(2 − x)2 + 1
x2
Cross multiplying gives
x2 ((2 − x)2 + 1) = (2 − x)2 (x2 + 1).
Multiplying out
x2 (4 − 4x + x2 + 1) = (4 − 4x + x2 )(x2 + 1)
4x2 − 4x3 + x4 + x2 = 4x2 − 4x3 + x4 + 4 − 4x + x2 .
Collecting terms and canceling gives
0 = 4 − 4x
x = 1.
We can see that this value of x gives a minimum by comparing the value of s at this point and at the endpoints,
x = 0, x = 2.
At x = 1,
p
p
s = 12 + 1 + (2 − 1)2 + 1 = 2.83.
At x = 0,
s=
At x = 2,
s=
p
02 + 1 +
p
22 + 1 +
p
(2 − 0)2 + 1 = 3.24.
p
(2 − 2)2 + 1 = 3.24.
Thus the shortest travel time occurs when x = 1; that is, when P is at the point (1, 1).
y
P = (x, 1)
1
A
O
x
(2 − x)
x
2
Figure 4.57
(b) Since x = 1 is halfway between x = 0 and x = 2, the angles θ1 and θ2 are equal.
318
Chapter Four /SOLUTIONS
46. (a) We have
x1/x = eln (x
1/x
)
= e(1/x) ln x .
Thus
d( 1 ln x) (1/x) ln x
d(e(1/x) ln x )
d(x1/x )
=
= x
e
dx
dx
dx
ln x
1
= − 2 + 2 x1/x
x
x
(
= 0 when x = e
x1/x
= 2 (1 − ln x) < 0 when x > e
x
> 0 when x < e.
(b)
(c)
47. (a)
(b)
Hence e1/e is the global maximum for x1/x , by the first derivative test.
Since x1/x is increasing for 0 < x < e and decreasing for x > e, and 2 and 3 are the closest integers to e, either
21/2 or 31/3 is the maximum for n1/n . We have 21/2 ≈ 1.414 and 31/3 ≈ 1.442, so 31/3 is the maximum.
Since e < 3 < π, and x1/x is decreasing for x > e, 31/3 > π 1/π .
√
If, following the hint, we set f (x) = (a + x)/2 − ax, then f (x) represents the difference between the arithmetic
and geometric means for some fixed a and any x > 0. We can find where this difference is minimized by solving
√
√
√
f ′ (x) = 0. Since f ′ (x) = 12 − 12 ax−1/2 , if f ′ (x) = 0 then 21 ax−1/2 = 21 , or x = a. Since f ′′ (x) = 41 ax−3/2
is positive for all positive
at x = a, and f (a) = 0. Thus
√
√ x, by the second derivative test f (x) has a minimum
f (x) = (a + x)/2 − ax ≥ 0 for all x > 0, which means (a + x)/2 ≥ ax. This means that the arithmetic mean
is greater than the geometric mean unless a = x, in which case the two means are equal.
Alternatively, and without using calculus, we obtain
√
a − 2 ab + b
a+b √
− ab =
2
2
√
√
( a − b)2
≥ 0,
=
2
√
and again we have (a + b)/2 ≥ ab.
√
Following the hint, set f (x) = a+b+x
− 3 abx. Then f (x) represents the difference between the arithmetic and
3
geometric means for some fixed a, b√and any x > 0. We can find where this
√ difference is minimized√by solving
f ′ (x) = 0. Since f ′ (x) = 13 − 31 3 abx−2/3 , f ′ (x) = 0 implies that 31 3 abx−2/3 = 31 , or x = ab. Since
√
√
3
f ′′ (x) = 29 abx−5/3 is positive for all positive x, by the second derivative test f (x) has a minimum at x = ab.
But
√
√
√
p
√
√
3
a + b + ab √
a + b − 2 ab
a + b + ab
f ( ab) =
− ab ab =
− ab =
.
3
3
3
√
√
− ab ≥ 0, which implies that a + b − 2 ab ≥ 0. Thus
By the first part of this problem, we know that a+b
2
√
√
√
ab
f ( ab) = a+b−2
≥ 0. Since f has a maximum at x = ab, f (x) is always nonnegative. Thus f (x) =
3
√
√
3
a+b+c
a+b+x
−
abx
≥
0,
so
≥ 3 abc. Note that equality holds only when a = b = c. (Part (b) may also be done
3
3
without calculus, but it’s harder than (a).)
48. For x > 0, the line in Figure 4.58 has
Slope =
y
x2 e−3x
=
= xe−3x .
x
x
If the slope has a maximum, it occurs where
d
(Slope) = 1 · e−3x − 3xe−3x = 0
dx
e−3x (1 − 3x) = 0
1
x= .
3
For this x-value,
1
1
1 −3(1/3)
e
= e−1 =
.
3
3
3e
Figure 4.58 shows that the slope tends toward 0 as x → ∞; the formula for the slope shows that the slope tends toward 0
as x → 0. Thus the only critical point, x = 1/3, must give a local and global maximum.
Slope =
4.3 SOLUTIONS
319
y
(x, x2 e−3x )
y✻
✛
x
✲
❄
x
Figure 4.58
49. (a) For a point (t, s), the line from the origin has rise = s and run = t; See Figure 4.59. Thus, the slope of the line OP
is s/t.
(b) Sketching several lines from the origin to points on the curve (see Figure 4.60), we see that the maximum slope occurs
at the point P , where the line to the origin is tangent to the graph. Reading from the graph, we see t ≈ 2 hours at this
point.
s (km)
s (km)
(t, s)
P
✻
P
s
O
✛
t
✲
❄
t (hours)
t (hours)
Figure 4.59
Figure 4.60
(c) The instantaneous speed of the cyclist at any time is given by the slope of the corresponding point on the curve. At
the point P , the line from the origin is tangent to the curve, so the quantity s/t equals the cyclist’s speed at the point
P.
50. (a) To maximize benefit (surviving young), we pick 10, because that’s the highest point of the benefit graph.
(b) To optimize the vertical distance between the curves, we can either do it by inspection or note that the slopes of the
two curves will be the same where the difference is maximized. Either way, one gets approximately 9.
51. (a) At higher speeds, the bird uses more energy so the graph rises to the right. The initial drop is due to the fact that the
energy it takes a bird to fly at very low speeds is greater than that needed to fly at a slightly higher speeds. (This is
analogous to our swimming in a pool).
(b) The value of f (v) measures energy per second; the value of a(v) measures energy per meter. In one second, a bird
traveling at rate v travels v meters and consumes v · a(v) joules. Thus v · a(v) represents the energy consumption per
second, so f (v) = v · a(v).
(c) Since v · a(v) = f (v), we have
f (v)
.
a(v) =
v
This ratio is the slope of a line passing from the origin through the point (v, f (v)) on the curve. (See Figure 4.61.)
Thus a(v) is minimal when the slope of this line is minimal. This occurs where the line is tangent to the curve.
To find the value of v minimizing a(v) symbolically, we solve a′ (v) = 0. By the quotient rule,
a′ (v) =
vf ′ (v) − f (v)
.
v2
Thus a′ (v) = 0 when vf ′ (v) = f (v), or when f ′ (v) = f (v)/v = a(v). Thus a(v) is minimized when a(v) =
f ′ (v).
(d) Assuming the bird wants to go from one particular point to another, that is, when the distance is fixed, the bird should
minimize a(v). Then minimizing a(v) minimizes the total energy used for the flight.
320
Chapter Four /SOLUTIONS
energy
f (v)
a(v)
v
Figure 4.61
52. (a) Figure 4.62 contains the graph of total drag, plotted on the same coordinate system with induced and parasite drag. It
was drawn by adding the vertical coordinates of Induced and Parasite drag.
drag
(thousands
of pounds)
3
2
1
Total
Drag
Induced✲
Drag
❄
✛
Parasite
Drag
speed
200
400
600 (mph)
Figure 4.62
(b) Air speeds of approximately 160 mph and 320 mph each result in a total drag of 1000 pounds. Since two distinct
air speeds are associated with a single total drag value, the total drag function does not have an inverse. The parasite
and induced drag functions do have inverses, because they are strictly increasing and strictly decreasing functions,
respectively.
(c) To conserve fuel, fly the at the air speed which minimizes total drag. This is the air speed corresponding to the lowest
point on the total drag curve in part (a): that is, approximately 220 mph.
53. (a) To obtain g(v), which is in gallons per mile, we need to divide f (v) (in gallons per hour) by v (in miles per hour).
Thus, g(v) = f (v)/v.
(b) By inspecting the graph, we see that f (v) is minimized at approximately 220 mph.
(c) Note that a point on the graph of f (v) has the coordinates (v, f (v)). The line passing through this point and the
origin (0, 0) has
f (v)
f (v) − 0
=
= g(v).
Slope =
v−0
v
So minimizing g(v) corresponds to finding the line of minimum slope from the family of lines which pass through
the origin (0, 0) and the point (v, f (v)) on the graph of f (v). This line is the unique member of the family which is
tangent to the graph of f (v). The value of v corresponding to the point of tangency will minimize g(v). This value
of v will satisfy f (v)/v = f ′ (v). From the graph in Figure 4.63, we see that v ≈ 300 mph.
4.4 SOLUTIONS
321
f (v) (gallons/hour)
100
75
f (v)
50
25
100
200
300 400
500
600
v (miles/hour)
Figure 4.63
(d) The pilot’s goal with regard to f (v) and g(v) would depend on the purpose of the flight, and might even vary within
a given flight. For example, if the mission involved aerial surveillance or banner-towing over some limited area, or
if the plane was flying a holding pattern, then the pilot would want to minimize f (v) so as to remain aloft as long
as possible. In a more normal situation where the purpose was economical travel between two fixed points, then the
minimum net fuel expenditure for the trip would result from minimizing g(v).
Strengthen Your Understanding
54. Since A = 2x2 , the area increases with x, and the maximum A occurs at the endpoint x = 10.
55. The function V = h(20 − 2h)2 has domain 0 ≤ h ≤ 10 for this model, or we cannot make the box from the cardboard.
We maximize V on 0 ≤ h ≤ 10.
56. The solution of an optimization problem depends on both the modeling function and the interval over which that function
is optimized. The solution can occur either at a critical point of the function or at an endpoint of the interval. For example,
if the modeling function is f (x) = −(x − 2)2 + 6 on the interval 5 ≤ x ≤ 10, the absolute minimum occurs at x = 10
and the maximum occurs at x = 5; neither occurs at the vertex x = 2.
57. For example, if sides are 9 cm and 1 cm, the perimeter is 20 cm and the area is 9 cm2 .
58. If we interpret xy as the area of a rectangular field and 2x + 6y as a weighted perimeter of the field, one possible problem
would be: A farmer would like to fence in a rectangular field of area 120 square feet. Fencing material along the north
and south sides of the field costs $1 per foot and along the east and west sides it costs $3 per foot. What dimensions of
the field minimize the cost for the fencing material?
59. The only additional information we have to provide is the cost of the material of the can. One possible choice is: The
material for the bottom and the top of the can costs 2 cents per square centimeter and the material for the side of the can
costs 1 cent per square centimeter.
Solutions for Section 4.4
Exercises
1. (a) See Figure 4.64.
large a
small a
❄
❘
x
Figure 4.64
322
Chapter Four /SOLUTIONS
(b) We see in Figure 4.64 that in each case the graph of f is a parabola with one critical point, its vertex, on the positive
x-axis. The critical point moves to the right along the x-axis as a increases.
(c) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = 2(x − a) = 0
x = a.
The only critical point is at x = a. As we saw in the graph, and as a increases, the critical point moves to the right.
2. (a) See Figure 4.65.
x
■
✻
large a
small a
Figure 4.65
(b) We see in Figure 4.65 that in each case f has two critical points placed symmetrically about the origin, one in each
of quadrants II and IV. As a increases, they appear to move farther apart, the one in quadrant II up and to the left, the
one in quadrant IV down and to the right.
(c) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = 3x2 − a = 0
a
x2 =
3
x=±
p
p
q
a
.
3
There are two critical points, at x = a/3 and x = − a/3. (Since the parameter a is positive, the critical points
exist.) As we saw in the graph, and as a increases, the critical points both move away from the vertical axis.
3. (a) See Figure 4.66.
large a
❄
small a
x
❄
Figure 4.66
(b) We see in Figure 4.66 that in each case f has two critical points placed symmetrically about the origin, one in each
of quadrants II and IV. As a increases, the critical points appear to move closer together, the one in quadrant II down
and to the right, the one in quadrant IV up and to the left.
(c) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = 3ax2 − 1 = 0
1
x2 =
3a
x=±
p
p
r
1
.
3a
There are two critical points, at x = 1/(3a) and x = − 1/(3a). (Since the parameter a is positive, the critical
points exist.) As we saw in the graph, and as a increases, the critical points both move toward the vertical axis.
4.4 SOLUTIONS
323
4. (a) See Figure 4.67.
small a
❄
x
✻
large a
Figure 4.67
(b) The domain of f is x ≥ 0 since the formula for f (x) contains the square root of x. We see in Figure 4.67 that in each
case f appears to have one critical point in quadrant IV where the function has a local minimum and possibly a second
critical point at the origin where the graph appears to have a vertical tangent line. As the parameter a increases, the
critical point in quadrant IV appears to move down and to the right.
(c) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = 1 −
1 −1/2
ax
2
a
1− √
2 x
√
2 x
√
x
=0
=0
=a
a
=
2
a2
.
x=
4
There is a critical point at x = a2 /4. Since f ′ is undefined at x = 0, there is also a critical point at x = 0. Increasing
a has no effect on the critical point at x = 0. As we saw in the graph, as a increases the x-value of the other critical
point increases and the critical point moves to the right.
5. (a) See Figure 4.68.
small a
❄
large a
❄
x
Figure 4.68
(b) We see in Figure 4.68 that in each case f appears to have two critical points. One critical point is a local minimum at
the origin and the other is a local maximum in quadrant I. As the parameter a increases, the critical point in quadrant I
appears to move down and to the left, closer to the origin.
(c) To find the critical points, we set the derivative equal to zero and solve for x. Using the product rule, we have:
f ′ (x) = x2 · e−ax (−a) + 2x · e−ax = 0
xe−ax (−ax + 2) = 0
2
x = 0 and x = .
a
There are two critical points, at x = 0 and x = 2/a. As we saw in the graph, as a increases the nonzero critical point
moves to the left.
324
Chapter Four /SOLUTIONS
6. (a) See Figure 4.69.
large a
❄
small a
x
Figure 4.69
(b) We see in Figure 4.69 that in each case f appears to have one critical point, a local minimum in quadrant I. As the
parameter a increases, the critical point appears to move up and to the right.
(c) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = −2ax−3 + 1 = 0
2a
1= 3
x
x3 = 2a
√
3
x = 2a.
There is one critical point, at x =
√
3
2a. As we saw in the graph, as a increases the critical point moves to the right.
7. (a) The larger the value of |A|, the steeper the graph (for the same x-value).
(b) The graph is shifted horizontally by B. The shift is to the left for positive B, to the right for negative B. There is a
vertical asymptote at x = −B. See Figure 4.70.
y
(c)
A = 2, B = 5
10
A = 20, B = 0
✛
A = 2, B = 0
−10
x
10
−10
Figure 4.70
8. To find the critical points, we set the derivative of f (w) = Aw−2 − Bw−1 equal to zero and solve for w.
f ′ (w) = −2Aw−3 − (−1)Bw−2
−2A
B
+ 2
w3
w
−2A + Bw
w3
−2A + Bw
=0
=0
=0
=0
2A
w=
.
B
Although f ′ is undefined at w = 0, this is not a critical point since it is not in the domain of f . The only critical point is
w = 2A/B.
9. We have f ′ (x) = −ae−ax , so that f ′ (0) = −a. We see that as a increases the slopes of the curves at the origin become
more and more negative. Thus, A corresponds to a = 1, B to a = 2, and C to a = 5.
10. We have f ′ (x) = (1 − ax)e−ax , so that the critical points occur when 1 − ax = 0, that is, when x = 1/a. We see that as
a increases the x-value of the extrema moves closer to the origin. Looking at the figure we see that A’s local maximum is
the farthest from the origin, while C’s is the closest. Thus, A corresponds to a = 1, B to a = 2, and C to a = 3.
325
4.4 SOLUTIONS
11. (a) Figure 4.71 shows the effect of varying a with b = 1.
(b) Figure 4.72 shows the effect of varying b with a = 1.
large b
❄
✒
✻
small a
✒
large a
small b
x
x
Figure 4.71
Figure 4.72
(c) From Figure 4.71 it appears that increasing a shifts the graph and its critical point to the right. From Figure 4.72 it
appears that increasing b moves the graph and its critical point up.
(d) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = 2(x − a) = 0
x = a.
There is one critical point, at x = a. As we saw in the graph, as a increases the critical point moves to the right. As b
increases the critical point does not move horizontally.
12. (a) Figure 4.73 shows the effect of varying a with b = 1.
(b) Figure 4.74 shows the effect of varying b with a = 1.
small a
large b
❄
❄
x
✛
✻
small b
x
large a
Figure 4.73
Figure 4.74
(c) From the graphs it appears that there are two critical points, a local maximum at x = 0 and a local minimum in the
region x > 0. From Figure 4.73 it appears that increasing a moves the local minimum down and to the right, and
does not move the local maximum. From Figure 4.74 it appears that increasing b moves both critical points up.
(d) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = 3x2 − 2ax = 0
x(3x − 2a) = 0
2a
x = 0 or x =
.
3
There are two critical points, at x = 0 and at x = 2a/3. As a increases the critical point to the right moves farther
right. As b increases, neither critical point moves horizontally. This matches what we saw in the graphs.
13. (a) Figure 4.75 shows the effect of varying a with b = 1.
(b) Figure 4.76 shows the effect of varying b with a = 1.
326
Chapter Four /SOLUTIONS
large a
large b
❄
❄
✻
x
small a
small b
x
✲
Figure 4.75
Figure 4.76
(c) In each case f appears to have two critical points, a local maximum in quadrant I and a local minimum on the
positive x-axis. From Figure 4.75 it appears that increasing a moves the local maximum up and does not move the
local minimum. From Figure 4.76 it appears that increasing b moves the local maximum up and to the right and
moves the local minimum to the right along the x-axis.
(d) To find the critical points, we set the derivative equal to zero and solve for x. Using the product rule, we have:
f ′ (x) = ax · 2(x − b) + a · (x − b)2 = 0
a(x − b)(2x + (x − b)) = 0
a(x − b)(3x − b) = 0
b
x = b or x = .
3
There are two critical points, at x = b and at x = b/3. Increasing a does not move either critical point horizontally.
Increasing b moves both critical points to the right. This confirms what we saw in the graphs.
14. (a) Figure 4.77 shows the effect of varying a with b = 1.
(b) Figure 4.78 shows the effect of varying b with a = 1.
✻
✻
✻
x
x
large b
✻
small a
large a
small b
Figure 4.77
Figure 4.78
(c) From the graphs it appears that in all cases f has two critical points symmetrically placed about the origin, a local
maximum in quadrant I and a local minimum in quadrant III. From Figure 4.77 it appears that increasing a moves
the local maximum up and the local minimum down. From Figure 4.78 it appears that increasing b moves the local
maximum down and to the right and the local minimum up and to the left. Thus increasing b appears to move both
critical points closer to the horizontal axis and farther from the vertical axis.
(d) To find the critical points, we set the derivative equal to zero and solve for x. Using the quotient rule, we have:
f ′ (x) =
(x2 + b) · a − ax · 2x
=0
(x2 + b)2
ax2 + ab − 2ax2
=0
(x2 + b)2
ab − ax2
=0
(x2 + b)2
ab − ax2 = 0
a(b − x2 ) = 0
√
x = ± b.
√
√
There are two critical points, at x = b and x = − b. Increasing a does not move either critical point horizontally.
Increasing b moves both critical points farther from the vertical axis. This confirms what we saw in the graphs.
327
4.4 SOLUTIONS
15. (a) Figure 4.79 shows the effect of varying a with b = 1.
(b) Figure 4.80 shows the effect of varying b with a = 1.
small a
large a
small b
❄
❄
❄
large b
✠
x
x
Figure 4.79
Figure 4.80
p
√
√
(c) The domain of f (x) = b − (x − a)2 is limited by the square root to b−(x−a)2 ≥ 0, so it is a− b < x < a+ b.
From the graphs in parts (a) and (b) it appears that in all cases f has three critical points, one a local maximum
and the other two on the x-axis at the endpoints of the domain of f where the graph of the function has a vertical
tangent line. From Figure 4.79 it appears that increasing a moves all three critical points to the right. From Figure 4.80
it appears that increasing b moves the local maximum up and moves the other two critical points away from each other
horizontally along the x-axis.
(d) To find the critical points, we set the derivative equal to zero and solve for x. Using the chain rule, we have:
f ′ (x) =
1
(b − (x − a)2 )−1/2 (−2(x − a)) = 0
2
−(x − a)
p
=0
b − (x − a)2
−(x − a) = 0
x = a.
There is a critical point of f at x = a. As a increases this critical point moves to the right. As b increases, it does not
move horizontally.
p
The derivative of f is undefined when b − (x − a)2 = 0. Solving for x, we have:
p
b − (x − a)2 = 0
b − (x − a)2 = 0
(x − a)2 = b
√
x−a= ± b
√
x = a ± b.
√
√
The function f has two additional critical points, at x = a + b and x = a − b, where the derivative f ′ is not
defined and the graph of f has a vertical tangent. As a increases both these critical points move to the right, and as b
increases they move in opposite directions away from the line x = a containing the other critical point. We saw this
in the graphs.
16. (a) Figure 4.81 shows the effect of varying a with b = 1.
(b) Figure 4.82 shows the effect of varying b with a = 1.
✛
large a
large b
❄
❄
✛
small a
small b
x
Figure 4.81
x
Figure 4.82
(c) It appears from the graphs that in all cases f has one critical point, a local minimum in quadrant I. From Figure 4.81
it appears that increasing a moves the critical point up and to the right. From Figure 4.82 it appears that increasing b
moves the critical point up and to the left.
328
Chapter Four /SOLUTIONS
(d) To find the critical points, we set the derivative equal to zero and solve for x.
f ′ (x) = −ax−2 + b = 0
−a
+b = 0
x2
a
b= 2
x
a
x2 =
b
x=±
q
a
.
b
Sincep
the domain of f is x > 0 we are only interested in positive values for x, so there is one critical point at
x = a/b. As a increases, this value gets larger and the critical point moves to the right. As b increases, this value
gets smaller and the critical point moves to the left. This confirms what we see in the graphs.
Problems
17. We have f ′ (x) = 1 − a2 /x2 , so the critical points occur at 1 − a2 /x2 = 0, that is x = ±a. We see that A has a minimum
at x = 3, B at x = 2, and C at x = 1. Thus, C corresponds to a = 1, B to a = 2, and A to a = 3. The third value is 3.
18. For the curves to intersect in the same point for any positive value of a, say a and b 6= a, we need x+a sin x = x+b sin x,
that is (a − b) sin x = 0. Because a − b 6= 0, we find that sin x = 0, which occurs at x = 0, ±π, ±2π . . .. For any integer
n, the graphs go through the points (nπ, nπ) for all a.
19. (a) Graphs of y = xe−bx for b = 1, 2, 3, 4 are in Figure 4.83. All the graphs rise at first, passing through the origin,
reach a maximum and then decay toward 0. If b is small, the graph rises longer and to a higher maximum before the
decay begins.
(b) Since
dy
= (1 − bx)e−bx ,
dx
we see
dy
1
= 0 at x = .
dx
b
The critical point has coordinates (1/b, 1/(be)). If b is small, the x and y-coordinates of the critical point are both
large, indicating a higher maximum further to the right. See Figure 4.84.
y
0.5
b=1
✠
y
b=2
✠
✻
b=4
1
be
b=3
y = xe−bx
✠
x
1
b
x
1
3
Figure 4.83
Figure 4.84
20. (a) Let f (x) = axe−bx . To find the local maxima and local minima of f , we solve
′
f (x) = ae
−bx
− abxe
−bx
= ae
−bx
(1 − bx)
(
= 0 if x = 1/b
< 0 if x > 1/b
> 0 if x < 1/b.
Therefore, f is increasing (f ′ > 0) for x < 1/b and decreasing (f ′ > 0) for x > 1/b. A local maximum occurs at
x = 1/b. There are no local minima. To find the points of inflection, we write
4.4 SOLUTIONS
329
f ′′ (x) = −abe−bx + ab2 xe−bx − abe−bx
= −2abe−bx + ab2 xe−bx
= ab(bx − 2)e−bx ,
so f ′′ = 0 at x = 2/b. Therefore, f is concave up for x < 2/b and concave down for x > 2/b, and the inflection
point is x = 2/b.
(b) Varying a stretches or flattens the graph but does not affect the critical point x = 1/b and the inflection point x = 2/b.
Since the critical and inflection points are depend on b, varying b will change these points, as well as the maximum
f (1/b) = a/be. For example, an increase in b will shift the critical and inflection points to the left, and also lower
the maximum value of f .
(c)
y
Varying a
Varying b
a =4
4
be
3
be
2
be
1
be
a =3
b=1
a =2
a =1
b=2
x
1
1
2
2
x
b=3
21. Cubic polynomials are all of the form f (x) = Ax3 + Bx2 + Cx + D. There is an inflection point at the origin (0, 0)
if f ′′ (0) = 0 and f (0) = 0. Since f (0) = D, we must have D = 0. Since f ′′ (x) = 6Ax + 2B, giving f ′′ (0) = 2B,
we must have B = 0. The family of cubic polynomials with inflection point at the origin is the two parameter family
f (x) = Ax3 + Cx.
22. (a) Writing y = L(1 + Ae−kt )−1 , we find the first derivative by the chain rule
LAke−kt
dy
= −L(1 + Ae−kt )−2 (−Ake−kt ) =
.
dt
(1 + Ae−kt )2
Using the quotient rule to calculate the second derivative gives
−LAk2 e−kt (1 + Ae−kt )2 − 2LAke−kt (1 + Ae−kt )(−Ake−kt )
d2 y
=
dt2
(1 + Ae−kt )4
=
LAk2 e−kt (−1 + Ae−kt )
.
(1 + Ae−kt )3
(b) Since L, A > 0 and e−kt > 0 for all t, the factor LAk2 e−kt and the denominator are never zero. Thus, possible
inflection points occur where
−1 + Ae−kt = 0.
Solving for t gives
ln A
.
k
(c) The second derivative is positive to the left of t = ln(A)/k and negative to the right, so the function changes from
concave up to concave down at t = ln(A)/k.
t=
330
Chapter Four /SOLUTIONS
23. (a) See Figure 4.85.
a = 0.5
10
a=3
x
−10
10
−10
Figure 4.85
(b) The function f (x) = x + a sin x is increasing for all x if f ′ (x) > 0 for all x. We have f ′ (x) = 1 + a cos x. Because
cos x varies between −1 and 1, we have 1 + a cos x > 0 for all x if −1 < a < 1 but not otherwise. When a = 1,
the function f (x) = x + sin x is increasing for all x, as is f (x) = x − sin x, obtained when a = −1. Thus f (x) is
increasing for all x if −1 ≤ a ≤ 1.
24. (a) See Figure 4.86.
a=1
a = 20
100
x
−10
10
−100
Figure 4.86
(b) The function f (x) = x2 + a sin x is concave up for all x if f ′′ (x) > 0 for all x. We have f ′′ (x) = 2 − a sin x.
Because sin x varies between −1 and 1, we have 2 − a sin x > 0 for all x if −2 < a < 2 but not otherwise. Thus
f (x) is concave up for all x if −2 < a < 2.
25. For −5 ≤ x ≤ 5, we have the graphs of y = a cosh(x/a) shown below.
y
a=1
10
a=2
a=3
1
x
−5
5
Increasing the value of a makes the graph flatten out and raises the minimum value. The minimum value of y occurs
at x = 0 and is given by
y = a cosh
0
a
=a
e0/a + e−0/a
2
= a.
4.4 SOLUTIONS
331
26. Graphs of y = e−ax sin(bx) for b = 1 and various values of a are shown in Figure 4.87. The parameter a controls the
amplitude of the oscillations.
y
1
a=
a=
a=
a=
✠
✠
✠
✠
0.2
0.4
0.6
0.8
0
x
π
2π
−1
Figure 4.87
y
27.
1
b=1
❄
0
✻
✻ ✻b = 2
−1
x
π
2π
b=3
b=4
The larger the value of b, the narrower the humps and more humps per given region there are in the graph.
28. Since f ′ (x) = abe−bx , we have f ′ (x) > 0 for all x. Therefore, f is increasing for all x. Since f ′′ (x) = −ab2 e−bx , we
have f ′′ (x) < 0 for all x. Therefore, f is concave down for all x.
29. (a) The critical point will occur where f ′ (x) = 0. But by the product rule,
f ′ (x) = be1+bx + b2 xe1+bx = 0
be1+bx (1 + bx) = 0
1 + bx = 0
1
x=− .
b
Thus the critical point of f is located at x = − 1b .
(b) To determine whether this critical point is a local minimum or local maximum, we can use the first derivative test.
Since both b and e1+bx are positive for all x, the sign of f ′ (x) = be1+bx (1 + bx) depends on the sign of 1 + bx.
for x < − 1b ,
f ′ (x) is negative
f ′ (x) = 0
for x = 1b ,
for x >
− 1b ,
′
f (x) is positive
Therefore, f goes from decreasing to increasing at x = − 1b , making this point a local minimum.
(c) We only need to substitute the critical point x = − 1b into the original function f :
f −
1
b
=b −
1 1+b(− 1b )
e
b
= −1e1−1
= −1.
This answer does not depend on the value of b. Though the x-coordinate of the critical point depends on b, the
y-coordinate does not.
332
Chapter Four /SOLUTIONS
30. We study h′ (x) = k − e−x . Notice that the critical point of h, if it exists, is located where h′ (x) = 0:
h′ (x) = k − e−x = 0
e−x = k
−x = ln k
x = − ln k.
(a) Since the domain of the natural logarithm is all positive real numbers, x = − ln k does not exist (and hence there
will be no critical point of h) if k ≤ 0.
(b) In order for the critical point x = − ln k to exist, we must have k > 0.
(c) There is a horizontal asymptote for h only when lim h(x) or lim h(x) exists and is finite. But
x→−∞
x→+∞
lim e−x = +∞ and
lim e−x = 0,
x→−∞
x→+∞
so
lim h(x) = +∞ and
x→−∞
lim h(x) = lim kx.
x→+∞
x→+∞
The first limit is never finite and the only way to make the second one finite is to set k = 0.
31. The function g will have a critical point when g ′ (x) = 0. Solving this equation gives
g ′ (x) = 1 − kex = 0
kex = 1
1
ex =
k
x = ln
1
= − ln k.
k
Since the natural logarithm has a domain of all positive real numbers, such a value for x may only exist for k > 0.
32. Since the horizontal asymptote is y = 5, we know a = 5. The value of b can be any number. Thus y = 5(1 − e−bx ) for
any b > 0.
2
33. Since the maximum is on the y-axis, a = 0. At that point, y = be−0
/2
= b, so b = 3.
2
−(x−a) /b
34. The maximum of y = e
occurs at x = a. (This is because the exponent −(x − a)2 /b is zero when x = a and
negative for all other x-values. The same result can be obtained by taking derivatives.) Thus we know that a = 2.
Points of inflection occur where d2 y/dx2 changes sign, that is, where d2 y/dx2 = 0. Differentiating gives
Since e−(x−2)
2
2(x − 2) −(x−2)2 /b
dy
=−
e
dx
b
2
2
4(x − 2)2 −(x−2)2 /b
2
d2 y
2
2
= e−(x−2) /b −1 + (x − 2)2 .
= − e−(x−2) /b +
e
2
2
dx
b
b
b
b
/b
is never zero, d2 y/dx2 = 0 where
−1 +
2
(x − 2)2 = 0.
b
We know d2 y/dx2 = 0 at x = 1, so substituting x = 1 gives
−1 +
2
(1 − 2)2 = 0.
b
Solving for b gives
−1 +
2
=0
b
b = 2.
Since a = 2, the function is
y = e−(x−2)
2
/2
.
4.4 SOLUTIONS
333
You can check that at x = 2, we have
d2 y
2
= e−0 (−1 + 0) < 0
dx2
2
so the point x = 2 does indeed give a maximum. See Figure 4.88.
y
Point of inflection
at x = 1
✛
✲
Max at x = 2
✛
Point of inflection
at x = 3
x
1
2
3
Figure 4.88: Graph of y = e−(x−2)
2
/2
35. We want a function of the form
L
.
1 + Ae−kt
Since the carrying capacity is 12, we have L = 12. The y-intercept is 4, so
y=
12
L
=
=4
1+A
1+A
A = 2.
The point of inflection is at (0.5, 6), so
ln(2)
ln(A)
=
k
k
k = 1.386.
0.5 =
Thus, the function is
y=
12
.
1 + 2e−1.386x
36. Since y(0) = a/(1 + b) = 2, we have a = 2 + 2b. To find a point of inflection, we calculate
abe−t
dy
=
,
dt
(1 + be−t )2
and using the quotient rule,
−abe−t (1 + be−t )2 − abe−t 2(1 + be−t )(−be−t )
d2 y
=
2
dx
(1 + be−t )4
=
abe−t (−1 + be−t )
.
(1 + be−t )3
The second derivative is equal to 0 when be−t = 1, or b = et . When t = 1, we have b = e. The second derivative
changes sign at this point, so we have an inflection point. Thus
y=
2 + 2e
.
1 + e1−t
37. Since the x3 term has coefficient 1, the polynomial is of the form
y = x3 + ax2 + bx + c.
Differentiating gives
dy
= 3x2 + 2ax + b.
dx
334
Chapter Four /SOLUTIONS
There is a critical point at x = 2, so dy/dx = 0 at x = 2. Thus
dy
dx
x=2
= 3(22 ) + 2a(2) + b = 12 + 4a + b = 0, so 4a + b = −12.
We take the second derivative to look for the inflection point. We find
d2 y
= 6x + 2a,
dx2
and for an inflection point at x = 1, we have 6 + 2a = 0, so a = −3. We now use a = −3 and the relationship
4a + b = −12, which gives 4(−3) + b = −12, so b = 0.
We now have
y = x3 − 3x2 + c,
and using the point (1,4) gives
1 − 3 + c = 4,
c = 6.
Thus, y = x3 − 3x2 + 6.
38. Since the graph is symmetric about the y-axis, the polynomial must have only even powers. Also, since the y-intercept is
0, the constant term must be zero. Thus, the polynomial is of the form
y = ax4 + bx2 .
Differentiating gives
dy
= 4ax3 + 2bx = x(4ax2 + 2b).
dx
Thus dy/dx = 0 at x = 0 and when 4ax2 + 2b = 0. Since the maxima occur where x = ±1, we have 4a + 2b = 0, so
b = −2a.
We are given y = 2 when x = ±1, so using y = ax4 + bx2 gives
a + b = 2,
a − 2a = −a = 2,
which gives a = −2 and b = 4. Thus
y = −2x4 + 4x2 .
To see if the points (1, 2) and (−1, 2) are local maxima, we take the second derivative,
d2 y
= −24x2 + 8,
dx2
which is negative if x = ±1. The critical points at x = ±1 are local maxima. Since the leading coefficient of this
polynomial is negative, we know the graph decreases without bound as |x| approaches ±∞, so these critical points are
also global maxima. Note that at x = 0 the second derivative is positive, so the point (0, 0) is a local minimum. There is
no global minimum.
39. Differentiating gives
dy
= 2abt cos(bt2 ).
dt
Since the first critical point for positive t occurs at t = 1, we have b = π/2. Then at t = 2, we have
dy
dt
so a =
= πa(2) cos(2π) = 2aπ = 3,
t=2
3
, and
2π
3
y=
sin
2π
πt2
2
.
4.4 SOLUTIONS
335
40. Differentiating gives
dy
= −2abt sin(bt2 ).
dx
√
Since the first critical point for positive t occurs at t = 1, we have b = π. Then at t = 1/ 2, we have
dy
dt
so a =
t= √1
2
π
2πa
= − √ sin
2
2
√
2
, and
π
y=
2aπ
= − √ = −2,
2
√
2
cos(πt2 ).
π
41. Differentiating y = ae−x + bx gives
dy
= −ae−x + b.
dx
Since the global minimum occurs for x = 1, we have −a/e + b = 0, so b = a/e.
The value of the function at x = 1 is 2, so we have 2 = a/e + b, which gives
2=
so a = e and b = 1. Thus
a
2a
a
+ =
,
e
e
e
y = e1−x + x.
We compute d2 y/dx2 = e1−x , which is always positive, so this confirms that x = 1 is a local minimum. Because
the value of e1−x + x as x → ±∞ grows without bound, this local minimum is a global minimum.
42. Differentiating y = bxe−ax gives
dy
= be−ax − abxe−ax = be−ax (1 − ax).
dx
Since we have a critical point at x = 3, we know that 1 − 3a = 0, so a = 1/3.
If b > 0, the first derivative goes from positive values to the left of x = 3 to negative values on the right of x = 3,
so we know this critical point is a local maximum. Since the function value at this local maximum is 6, we have
6 = 3be−3/3 =
so b = 2e and
3b
,
e
y = 2xe1−x/3 .
43. Notice that since y is an odd function, choosing a and b so there is a minimum at (3, 12) automatically results in a local
maximum at (−3, −12).
Differentiating y = at + b/t, we have
d2 y
dy
= a − bt−2 , and 2 = 2bt−3 .
dt
dt
The critical points require dy/dt = 0 at t = ±3. This gives a − b/9 = 0 so b = 9a.
Substituting the point (3, 12) in the original equation, we have
12 = 3a + 9a/3 = 6a.
Thus a = 2, and b = 18, and
y = 2t + 18/t.
The second derivative test affirms that (3, 12) is a local minimum, as the second derivative is positive at t = 3.
44. We begin by finding the intercepts, which occur where f (x) = 0, that is
√
x−k x = 0
√ √
x( x − k) = 0
336
Chapter Four /SOLUTIONS
√
so x = 0 or
x = k, x = k2 .
2
So 0 and k are the x-intercepts. Now we find the location of the critical points by setting f ′ (x) equal to 0:
f ′ (x) = 1 − k
This means
k
1= √ ,
2 x
1 −(1/2)
x
2
√
so
x=
k
= 1 − √ = 0.
2 x
1
k,
2
and x =
1 2
k .
4
2
We can use the second derivative to verify that x = k4 is a local minimum. f ′′ (x) = 1 + 4xk3/2 is positive for all x > 0.
So the critical point, x = 14 k2 , is 1/4 of the way between the x-intercepts, x = 0 and x = k2 . Since f ′′ (x) = 41 kx−3/2 ,
f ′′ ( 14 k2 ) = 2/k2 > 0, this critical point is a minimum.
45. (a) The x-intercept occurs where f (x) = 0, so
ax − x ln x = 0
x(a − ln x) = 0.
Since x > 0, we must have
a − ln x = 0
ln x = a
x = ea .
(b) See Figures 4.89 and 4.90.
1
1
x
1
2
x
3
1
−1
2
3
−1
Figure 4.89: Graph of f (x) with
a = −1
Figure 4.90: Graph of f (x) with a = 1
(c) Differentiating gives f ′ (x) = a − ln x − 1. Critical points are obtained by solving
a − ln x − 1 = 0
ln x = a − 1
x = ea−1 .
Since ea−1 > 0 for all a, there is no restriction on a. Now,
f (ea−1 ) = aea−1 − ea−1 ln(ea−1 ) = aea−1 − (a − 1)ea−1 = ea−1 ,
so the coordinates of the critical point are (ea−1 , ea−1 ). From the graphs, we see that this critical point is a local
maximum; this can be confirmed using the second derivative:
f ′′ (x) = −
1
<0
x
for x = ea−1 .
46. (a) Figures 4.91- 4.94 show graphs of f (x) = x2 + cos(kx) for various values of k. For k = 0.5 and k = 1, the graphs
look like parabolas. For k = 3, there is some waving in the parabola, which becomes more noticeable if k = 5. The
waving begins to happen at about k = 1.5.
x
Figure 4.91: k = 0.5
x
Figure 4.92: k = 1
x
Figure 4.93: k = 3
x
Figure 4.94: k = 5
4.4 SOLUTIONS
337
(b) Differentiating, we have
f ′ (x) = 2x − k sin(kx)
f ′′ (x) = 2 − k2 cos(kx).
√
If k2 ≤ 2, then f ′′ (x) ≥ 2 − 2 cos(kx) ≥ 0, since cos(kx) ≤ 1. Thus, the graph is always concave up if k ≤ 2.
2
′′
2
If k > 2, then f (x) changes sign whenever cos(kx) = 2/k , which occurs for infinitely many values of x, since
0 < 2/k2 < 1.
(c) Since f ′ (x) = 2x − k sin(kx), we want to find all points where
2x − k sin(kx) = 0.
Since
−1 ≤ sin(kx) ≤ 1,
f ′ (x) 6= 0 if x > k/2 or x < −k/2. Thus, all the roots of f ′ (x) must be in the interval −k/2 ≤ x ≤ k/2. The roots
occur where the line y = 2x intersects the curve y = k sin(kx), and there are only a finite number of such points for
−k/2 ≤ x ≤ k/2.
47. (a) Figure 4.95 suggests that each graph decreases to a local minimum and then increases sharply. The local minimum
appears to move to the right as k increases. It appears to move up until k = 1, and then to move back down.
k=2 k=4
k=1
k = 1/2
k = 1/4
x
Figure 4.95
(b) f ′ (x) = ex − k = 0 when x = ln k. Since f ′ (x) < 0 for x < ln k and f ′ (x) > 0 for x > ln k, f is decreasing to
the left of x = ln k and increasing to the right, so f reaches a local minimum at x = ln k.
(c) The minimum value of f is
f (ln k) = eln k − k(ln k) = k − k ln k.
Since we want to maximize the expression k − k ln k, we can imagine a function g(k) = k − k ln k. To maximize
this function we simply take its derivative and find the critical points. Differentiating, we obtain
g ′ (k) = 1 − ln k − k(1/k) = − ln k.
Thus g ′ (k) = 0 when k = 1, g ′ (k) > 0 for k < 1, and g ′ (k) < 0 for k > 1. Thus k = 1 is a local maximum for
g(k). That is, the largest global minimum for f occurs when k = 1.
48. (a) The graph of r has a vertical asymptote if the denominator is zero. Since (x − b)2 is nonnegative, the denominator
can only be zero if a ≤ 0. Then
a + (x − b)2 = 0
(x − b)2 = −a
√
x − b = ± −a
√
x = b ± −a.
In order for there to be a vertical asymptote, a must be less than or equal to zero. There are no restrictions on b.
(b) Differentiating gives
−1
r ′ (x) =
· 2(x − b),
(a + (x − b)2 )2
so r ′ = 0 when x = b. If a ≤ 0, then r ′ is undefined at the same points at which r is undefined. Thus the only critical
point is x = b. Since we want r(x) to have a maximum at x = 3, we choose b = 3. Also, since r(3) = 5, we have
r(3) =
1
1
= =5
a + (3 − 3)2
a
so
a=
1
.
5
338
Chapter Four /SOLUTIONS
p
p
49. (a) f ′ (x) = 4x3 + 2ax = 2x(2x2 + a); so x = 0 and x = ± −a/2 (if ± −a/2 is real, i.e. if −a/2 ≥ 0) are
critical points.
(b) x = 0 is a critical point for any value of a. In order to guarantee that x = 0 is the only critical point, the factor
2x2 + a should not have a root other than possibly x = 0. This means a ≥ 0, since 2x2 + a has only one root (x = 0)
for a = 0, and no roots for a > 0. There is no restriction on the constant b.
Now f ′′ (x) = 12x2 + 2a and f ′′ (0) = 2a.
If a > 0, then by the second derivative test, f (0) is a local minimum.
If a = 0, then f (x) = x4 + b, which has a local minimum at x = 0.
So x = 0 is a local minimum when a ≥ 0.
(c) Again, b will have no effect on the location of the critical points. In order for f ′ (x) = 2x(2x2 + a) to have three
different roots, the constant a has to be negative. Let a = −2c2 , for some c > 0. Then
f ′ (x) = 4x(x2 − c2 ) = 4x(x − c)(x + c).
p
The critical points of f are x = 0 and x = ±c = ± −a/2.
To the left of x = −c, f ′ (x) < 0.
Between x = −c and x = 0, f ′ (x) > 0.
Between x = 0 and x = c, f ′ (x) < 0.
To the right of x = c, f ′ (x) > 0.
So, f (−c) and f (c) are local minima and f (0) is a local maximum.
(d) For a ≥ 0, there is exactly one critical point, x = 0. For a < 0 there are exactly three different critical points. These
exhaust all the possibilities. (Notice that the value of b is irrelevant here.)
50. (a) The graphs are shown in Figures 4.96–4.101.
y
y
A=B=1
y
x
Figure 4.96: A > 0, B > 0
y
A=2
B=1
A = −B = 1
x
x
Figure 4.97: A > 0, B < 0
y
Figure 4.98: A > 0, B > 0
y
A=2
B = −1
x
x
A = −2
B = −1
Figure 4.99: A > 0, B < 0
Figure 4.100: A < 0, B < 0
x
A = −2
B=1
Figure 4.101: A < 0, B > 0
(b) If A and B have the same sign, the graph is U -shaped. If A and B are both positive, the graph opens upward. If A
and B are both negative, the graph opens downward.
(c) If A and B have different signs, the graph appears to be everywhere increasing (if A > 0, B < 0) or decreasing (if
A < 0, B > 0).
(d) The function appears to have a local maximum if A < 0 and B < 0, and a local minimum if A > 0 and B > 0.
To justify this, calculate the derivative
dy
= Aex − Be−x .
dx
Setting dy/dx = 0 gives
Aex − Be−x = 0
4.4 SOLUTIONS
339
Aex = Be−x
B
e2x = .
A
This equation has a solution only if B/A is positive, that is, if A and B have the same sign. In that case,
B
A
2x = ln
1
B
ln
.
2
A
x=
This value of x gives the only critical point.
To determine whether the critical point is a local maximum or minimum, we use the first derivative test. Since
dy
= Aex − Be−x ,
dx
we see that:
If A > 0, B > 0, we have dy/dx > 0 for large positive x and dy/dx < 0 for large negative x, so there is a
local minimum.
If A < 0, B < 0, we have dy/dx < 0 for large positive x and dy/dx > 0 for large negative x, so there is a
local maximum.
51. T (t) = the temperature at time t = a(1 − e−kt ) + b.
(a) Since at time t = 0 the yam is at 20◦ C, we have
T (0) = 20◦ = a 1 − e0 + b = a(1 − 1) + b = b.
Thus b = 20◦ C. Now, common sense tells us that after a period of time, the yam will heat up to about 200◦ , or oven
temperature. Thus the temperature T should approach 200◦ as the time t grows large:
lim T (t) = 200◦ C = a(1 − 0) + b = a + b.
t→∞
Since a + b = 200◦ , and b = 20◦ C, this means a = 180◦ C.
(b) Since we’re talking about how quickly the yam is heating up, we need to look at the derivative, T ′ (t) = ake−kt :
T ′ (t) = (180)ke−kt .
We know T ′ (0) = 2◦ C/min, so
2 = (180)ke−k(0) = (180)(k).
So k = (2◦ C/min)/180◦ C =
1
min−1 .
90
52. (a) Since
a2 − ax
= 0 when x = a,
x2
the x-intercept is x = a. There is a vertical asymptote at x = 0 and a horizontal asymptote at U = 0.
(b) Setting dU/dx = 0, we have
U =b
2a2
dU
a
=b − 3 + 2
dx
x
x
=b
−2a2 + ax
x3
= 0.
So the critical point is
x = 2a.
When x = 2a,
U =b
The second derivative of U is
a
a2
−
4a2
2a
d2 U
=b
dx2
When we evaluate this at x = 2a, we get
d2 U
=b
dx2
b
=− .
4
.
=
b
> 0.
8a2
2a
6a2
− 3
x4
x
6a2
2a
−
(2a)4
(2a)3
Since d2 U/dx2 > 0 at x = 2a, we see that the point (2a, −b/4) is a local minimum.
340
Chapter Four /SOLUTIONS
(c)
U
a
2a
x
(2a, −b/4)
53. Both U and F have asymptotes at x = 0 and the x-axis. In Problem 52 we saw that U has intercept (a, 0) and local
minimum (2a, −b/4). Differentiating U gives
F =b
Since
F =b
2a2
a
− 2
x3
x
2a2 − ax
x3
.
= 0 for x = 2a,
F has one intercept: (2a, 0). Differentiating again to find the critical points:
6a2
2a
dF
=b − 4 + 3
dx
x
x
so x = 3a. When x = 3a,
F =b
=b
a
2a2
− 2
27a3
9a
−6a2 + 2ax
x4
=−
= 0,
b
.
27a
By the first or second derivative test, x = 3a is a local minimum of F . See figure below.
U
F
a
2a 3a
(2a, −b/4)
x
(3a, −b/(27a))
54. (a) To find limr→0+ V (r), first rewrite V (r) with a common denominator:
A
B
− 6
r 12
r
A − Br 6
= lim
r 12
r→0+
A
→ + → +∞.
0
lim V (r) = lim
r→0+
r→0+
As the distance between the two atoms becomes small, the potential energy diverges to +∞.
(b) The critical point of V (r) will occur where V ′ (r) = 0:
6B
12A
+ 7 =0
r 13
r
−12A + 6Br 6
=0
r 13
−12A + 6Br 6 = 0
2A
r6 =
B
2A 1/6
r=
B
V ′ (r) = −
4.4 SOLUTIONS
341
To determine whether this is a local maximum or minimum, i we can use the first derivative test. Since r is positive,
the sign of V ′ (r) is determined by the sign of −12A + 6Br 6 . Notice that this is an increasing function of r for r > 0,
so V ′ (r) changes sign from − to + at r = (2A/B)1/6 . The first derivative test yields
r
V ′ (r)
2A 1/6
B
←
neg.
zero
→
pos.
Thus V (r) goes from decreasing to increasing at the critical point r = (2A/B)1/6 , so this is a local minimum.
(c) Since F (r) = −V ′ (r), the force is zero exactly where V ′ (r) = 0, i.e. at the critical points of V . The only critical
point was the one found in part (b), so the only such point is r = (2A/B)1/6 .
(d) Since the numerator in r = (2A/B)1/6 is proportional to A1/6 , the equilibrium size of the molecule increases when
the parameter A is increased. Conversely, since B is in the denominator, when B is increased the equilibrium size of
the molecules decrease.
55. (a) The force is zero where
f (r) = −
B
A
+ 3 =0
r2
r
Ar 3 = Br 2
B
r= .
A
The vertical asymptote is r = 0 and the horizontal asymptote is the r-axis.
(b) To find critical points, we differentiate and set f ′ (r) = 0:
3B
2A
− 4 =0
r3
r
2Ar 4 = 3Br 3
3B
r=
.
2A
f ′ (r) =
Thus, r = 3B/(2A) is the only critical point. Since f ′ (r) < 0 for r < 3B/(2A) and f ′ (r) > 0 for r > 3B/(2A),
we see that r = 3B/(2A) is a local minimum. At that point,
f
Differentiating again, we have
3B
2A
=−
A
9B 2 /4A2
+
B
27B 3 /8A3
=−
4A3
.
27B 2
12B
6
6A
+ 5 = − 5 (Ar − 2B).
r4
r
r
So f ′′ (r) < 0 where r > 2B/A and f ′′ (r) > 0 when r < 2B/A. Thus, r = 2B/A is the only point of inflection.
At that point
B
A3
A
2B
= − 2.
f
=− 2 2 +
3
3
A
4B /A
8B /A
8B
f ′′ (r) = −
(c)
f (r)
r
B
A
( 3B
,
2A
(d)
2B −A3
−4A3 ( A , 8B 2 )
)
2
27B
(i) Increasing B means that the r-values of the zero, the minimum, and the inflection point increase, while the f (r)
values of the minimum and the point of inflection decrease in magnitude. See Figure 4.102.
342
Chapter Four /SOLUTIONS
(ii) Increasing A means that the r-values of the zero, the minimum, and the point of inflection decrease, while the
f (r) values of the minimum and the point of inflection increase in magnitude. See Figure 4.103.
✛
✛
Small B
✛
Large A
✛
Large B
Small A
r
Figure 4.102: Increasing B
r
Figure 4.103: Increasing A
b
b−c·0
56. (a) The vertical intercept is W = Ae−e
= Ae−e . There is no horizontal intercept since the exponential function
is always positive. There is a horizontal asymptote. As t → ∞, we see that eb−ct = eb /ect → 0, since t is positive.
Therefore W → Ae0 = A, so there is a horizontal asymptote at W = A.
(b) The derivative is
b−ct
b−ct
dW
= Ae−e
(−eb−ct )(−c) = Ace−e
eb−ct .
dt
Thus, dW/dt is always positive, so W is always increasing and has no critical points. The second derivative is
b−ct
b−ct d
d
d2 W
= (Ace−e
(eb−ct )
)eb−ct + Ace−e
dt2
dt
dt
b−ct
b−ct
eb−ct eb−ct + Ace−e
b−ct
eb−ct (eb−ct − 1).
= Ac2 e−e
= Ac2 e−e
(−c)eb−ct
Now eb−ct decreases from eb > 1 when t = 0 toward 0 as t → ∞. The second derivative changes sign from positive
to negative when eb−ct = 1, i.e., when b − ct = 0, or t = b/c. Thus the curve has an inflection point at t = b/c,
b−(b/c)c
where W = Ae−e
= Ae−1 .
(c) See Figure 4.104.
W
A = 50, b = 2, c = 5
A = 50, b = 2, c = 1
A = 20, b = 2, c = 1
t
Figure 4.104
(d) The final size of the organism is given by the horizontal asymptote W = A. The curve is steepest at its inflection
point, which occurs at t = b/c, W = Ae−1 . Since e = 2.71828 . . . ≈ 3, the size the organism when it is growing
fastest is about A/3, one third its final size. So yes, the Gompertz growth function is useful in modeling such growth.
Strengthen Your Understanding
57. As a counterexample, f (x) = x2 + 1 has no zeros.
4.5 SOLUTIONS
343
p
58. Differentiating f we see that f ′ (x) = −a/x2 + b. Solving f ′ (x) = 0 we see that x = ± a/b are critical points
provided a and b have the same sign. So for example, f (x) = 1/x + x has two critical points. But if a and b have opposite
sign, f has no critical points. The function f (x) = 1/x − x has no critical points, for example.
59. The family of functions f (x) = kx(x − b) are all quadratics and satisfy f (0) = f (b) = 0, as required.
60. The condition for critical points is f ′ (x) = 3ax2 − b = 0 or
x2 =
b
.
3a
So there will be no critical points if
b
< 0.
3a
As long as a and b have opposite signs, we have b/(3a) < 0. For example, we can take a = 1, b = −1 to get f (x) =
x3 + x.
61. Let f (x) = ax2 , with a 6= 0. Then f ′ (x) = 2ax, so f has a critical point only at x = 0.
62. Let g(x) = ax3 + bx2 , where neither a nor b are allowed to be zero. Then
g ′ (x) = 3ax2 + 2bx = x(3ax + 2b).
Then g(x) has two distinct critical points, at x = 0 and at x = −2b/3a. Since
g ′′ (x) = 6ax + 2b,
there is exactly one point of inflection, x = −2b/6a = −b/3a.
p
p
p
p
b/a, then as b increases ± b/a increases. Therefore (a).
p
p
√
√
√
b
= ± ab ± ab = ±2 ab. Therefore (c) is also
The critical values of f (x) are f (± b/a) = a(± b/a) + p
± b/a
true.
63. (a) and (c). Since the critical points of f (x) are x = ±
64. (b) and (c). Since the critical points of f (x) are x = ±
b/a, then as a increases ± b/a decreases. Therefore (b).
p
p
√
√
√
b
The critical values of f (x) are f (± b/a) = a(± b/a) + p
= ± ab ± ab = ±2 ab. Therefore (c) is also
± b/a
true.
Solutions for Section 4.5
Exercises
1. The profit function is positive when R(q) > C(q), and negative when C(q) > R(q). It’s positive for 5.5 < q < 12.5,
and negative for 0 < q < 5.5 and 12.5 < q. Profit is maximized when R(q) > C(q) and R′ (q) = C ′ (q) which occurs
at about q = 9.5. See Figure 4.105.
$ (thousands)
C(q)
400
Profit function
is positive
✛
✲
R(q)
300
200
100
5
10
Figure 4.105
15
q (thousands)
344
Chapter Four /SOLUTIONS
2. Since fixed costs are represented by the vertical intercept, they are $1.1 million. The quantity that maximizes profit is
about q = 70, and the profit achieved is $(3.7 − 2.5) = $1.2 million
3. The fixed costs are $5000, the marginal cost per item is $2.40, and the price per item is $4.
4. The cost function C(q) = b + mq satisfies C(0) = 500, so b = 500, and M C = m = 6. So
C(q) = 500 + 6q.
The revenue function is R(q) = 12q, so the profit function is
π(q) = R(q) − C(q) = 12q − 500 − 6q = 6q − 500.
5. The cost function C(q) = b + mq satisfies C(0) = 35,000, so b = 35,000, and M C = m = 10. So
C(q) = 35,000 + 10q.
The revenue function is R(q) = 15q, so the profit function is
π(q) = R(q) − C(q) = 15q − 35,000 − 10q = 5q − 35,000.
6. The cost function C(q) = b + mq satisfies C(0) = 5000, so b = 5000, and M C = m = 15. So
C(q) = 5000 + 15q.
The revenue function is R(q) = 60q, so the profit function is
π(q) = R(q) − C(q) = 60q − 5000 − 15q = 45q − 5000.
7. The cost function C(q) = b + mq satisfies C(0) = 0, so b = 0. There are 32 ounces in a quart, so 160 in 5 quarts, or 20
cups at 8 ounces per cup. Thus each cup costs the operator 20 cents = 0.20 dollars, so M C = m = 0.20. So
C(q) = 0.20q.
The revenue function is R(q) = 0.25q. So the profit function is
π(q) = R(q) − C(q) = 0.25q − 0.20q = 0.05q.
8. The profit π(q) is given by
π(q) = R(q) − C(q) = 500q − q 2 − (150 + 10q) = 490q − q 2 − 150.
The maximum profit occurs when
π ′ (q) = 490 − 2q = 0
so
q = 245 items.
′′
Since π (q) = −2, this critical point is a maximum. Alternatively, we obtain the same result from the fact that the graph
of π is a parabola opening downward.
9. First find marginal revenue and marginal cost.
M R = R′ (q) = 450
M C = C ′ (q) = 6q
Setting M R = M C yields 6q = 450, so marginal cost is equal to marginal revenue when
q=
450
= 75 units.
6
Is profit maximized at q = 75? Profit = R(q) − C(q);
R(75) − C(75) = 450(75) − (10,000 + 3(75)2 )
= 33,750 − 26,875 = $6875.
4.5 SOLUTIONS
345
Testing q = 74 and q = 76:
R(74) − C(74) = 450(74) − (10,000 + 3(74)2 )
= 33,300 − 26,428 = $6872.
R(76) − C(76) = 450(76) − (10,000 + 3(76)2 )
= 34,200 − 27,328 = $6872.
Since profit at q = 75 is more than profit at q = 74 and q = 76, we conclude that profit is maximized locally at q = 75.
The only endpoint we need to check is q = 0.
R(0) − C(0) = 450(0) − (10,000 + 3(0)2 )
= −$10,000.
This is clearly not a maximum, so we conclude that the profit is maximized globally at q = 75, and the total profit at this
production level is $6,875.
10. The marginal revenue, M R, is given by differentiating the total revenue function, R. We use the chain rule so
MR =
When q = 10,
1
d
1
dR
=
·
· 2000q.
1000q 2 =
dq
1 + 1000q 2 dq
1 + 1000q 2
2000 · 10
= $0.20/item.
1 + 1000 · 102
When 10 items are produced, each additional item produced gives approximately $0.20 in additional revenue.
Marginal revenue =
11. (a) Profit is maximized when R(q)−C(q) is as large as possible. This occurs at q = 2500, where profit = 7500−5500 =
$2000.
(b) We see that R(q) = 3q and so the price is p = 3, or $3 per unit.
(c) Since C(0) = 3000, the fixed costs are $3000.
12. (a) At q = 5000, M R > M C, so the marginal revenue to produce the next item is greater than the marginal cost. This
means that the company will make money by producing additional units, and production should be increased.
(b) Profit is maximized where M R = M C, and where the profit function is going from increasing (M R > M C) to
decreasing (M R < M C). This occurs at q = 8000.
Problems
13. (a) The value of C(0) represents the fixed costs before production, that is, the cost of producing zero units, incurred for
initial investments in equipment, and so on.
(b) The marginal cost decreases slowly, and then increases as quantity produced increases. See Problem 74, graph (b).
(c) Concave down implies decreasing marginal cost, while concave up implies increasing marginal cost.
(d) An inflection point of the cost function is (locally) the point of maximum or minimum marginal cost.
(e) One would think that the more of an item you produce, the less it would cost to produce extra items. In economic
terms, one would expect the marginal cost of production to decrease, so we would expect the cost curve to be concave down. In practice, though, it eventually becomes more expensive to produce more items, because workers and
resources may become scarce as you increase production. Hence after a certain point, the marginal cost may rise
again. This happens in oil production, for example.
14. Since marginal revenue is larger than marginal cost around q = 2000, as you produce more of the product your revenue
increases faster than your costs, so profit goes up, and maximal profit will occur at a production level above 2000.
15. Since for q = 500, we have M C(500) = C ′ (500) = 75 and M R(500) = R′ (500) = 100, so M R(500) > M C(500).
Thus, increasing production from q = 500 increases profit.
16. Her marginal cost is at a minimum of $25 when the quantity sold is 100, which is still below the marginal revenue of $35.
Thus she will continue to make a profit as the quantity increases, although her profit will start to decrease if the marginal
cost goes above $35. So the quantity that maximizes profit is greater than 100.
17. (a) We know that Profit = Revenue − Cost, so differentiating with respect to q gives:
Marginal Profit = Marginal Revenue − Marginal Cost.
346
Chapter Four /SOLUTIONS
We see from the figure in the problem that just to the left of q = a, marginal revenue is less than marginal cost, so
marginal profit is negative there. To the right of q = a marginal revenue is greater than marginal cost, so marginal
profit is positive there. At q = a marginal profit changes from negative to positive. This means that profit is decreasing
to the left of a and increasing to the right. The point q = a corresponds to a local minimum of profit, and does not
maximize profit. It would be a terrible idea for the company to set its production level at q = a.
(b) We see from the figure in the problem that just to the left of q = b marginal revenue is greater than marginal cost, so
marginal profit is positive there. Just to the right of q = b marginal revenue is less than marginal cost, so marginal
profit is negative there. At q = b marginal profit changes from positive to negative. This means that profit is increasing
to the left of b and decreasing to the right. The point q = b corresponds to a local maximum of profit. In fact, since
the area between the M C and M R curves in the figure in the text between q = a and q = b is bigger than the area
between q = 0 and q = a, q = b is in fact a global maximum.
18. (a) The fixed cost is 0 because C(0) = 0.
(b) Profit, π(q), is equal to money from sales, 7q, minus total cost to produce those items, C(q).
π = 7q − 0.01q 3 + 0.6q 2 − 13q
π ′ = −0.03q 2 + 1.2q − 6
p
(1.2)2 − 4(0.03)(6)
≈ 5.9 or 34.1.
−0.06
Now π ′′ = −0.06q + 1.2, so π ′′ (5.9) > 0 and π ′′ (34.1) < 0. This means q = 5.9 is a local min and q = 34.1 a
local max. We now evaluate the endpoint, π(0) = 0, and the points nearest q = 34.1 with integer q-values:
π ′ = 0 if q =
−1.2 ±
π(35) = 7(35) − 0.01(35)3 + 0.6(35)2 − 13(35) = 245 − 148.75 = 96.25,
π(34) = 7(34) − 0.01(34)3 + 0.6(34)2 − 13(34) = 238 − 141.44 = 96.56.
So the (global) maximum profit is π(34) = 96.56. The money from sales is $238, the cost to produce the items is
$141.44, resulting in a profit of $96.56.
(c) The money from sales is equal to price×quantity sold. If the price is raised from $7 by $x to $(7 + x), the result is a
reduction in sales from 34 items to (34 − 2x) items. So the result of raising the price by $x is to change the money
from sales from (7)(34) to (7 + x)(34 − 2x) dollars. If the production level is fixed at 34, then the production costs
are fixed at $141.44, as found in part (b), and the profit is given by:
π(x) = (7 + x)(34 − 2x) − 141.44
This expression gives the profit as a function of change in price x, rather than as a function of quantity as in part (b).
We set the derivative of π with respect to x equal to zero to find the change in price that maximizes the profit:
dπ
= (1)(34 − 2x) + (7 + x)(−2) = 20 − 4x = 0
dx
So x = 5, and this must give a maximum for π(x) since the graph of π is a parabola which opens downward. The
profit when the price is $12 (= 7 + x = 7 + 5) is thus π(5) = (7 + 5)(34 − 2(5)) − 141.44 = $146.56. This is
indeed higher than the profit when the price is $7, so the smart thing to do is to raise the price by $5.
19. For each month,
Profit = Revenue − Cost
π = pq − wL = pcK α Lβ − wL
The variable on the right is L, so at the maximum
dπ
= βpcK α Lβ−1 − w = 0
dL
Now β − 1 is negative, since 0 < β < 1, so 1 − β is positive and we can write
βpcK α
=w
L1−β
giving
1
βpcK α 1−β
w
Since β − 1 is negative, when L is just above 0, the quantity Lβ−1 is huge and positive, so dπ/dL > 0. When L is large,
Lβ−1 is small, so dπ/dL < 0. Thus the value of L we have found gives a global maximum, since it is the only critical
point.
L=
4.5 SOLUTIONS
347
20. (a) We have N (x) = 100 + 20x, graphed in Figure 4.106.
(b) (i) Differentiating gives N ′ (x) = 20 and the graph of N ′ is a horizontal line. This means that rate of increase of
the number of bees with acres of clover is constant—each acre of clover brings 20 more bees.
(ii) On the other hand, N (x)/x = 100/x + 20 means that the average number of bees per acre of clover approaches
20 as more acres are put under clover. See Figure 4.107. As x increases, 100/x decreases to 0, so N (x)/x
approaches 20 (i.e. N (x)/x → 20). Since the total number of bees is 20 per acre plus the original 100, the
average number of bees per acre is 20 plus the 100 shared out over x acres. As x increases, the 100 are shared
out over more acres, and so its contribution to the average becomes less. Thus the average number of bees per
acre approaches 20 for large x.
bees/acre
number of bees
N (x) = 100 + 20x
2100
✠
✐
20
100
50
100
x (acres)
50
Figure 4.106
N(x)
x
=
100
x
+ 20
N ′ (x) = 20
x (acres)
100
Figure 4.107
21. This question implies that the line from the origin to the point (x, R(x)) has some relationship to r(x). The slope of this
line is R(x)/x, which is r(x). So the point x0 at which r(x) is maximal will also be the point at which the slope of this
line is maximal. The question claims that the line from the origin to (x0 , R(x0 )) will be tangent to the graph of R(x). We
can understand this by trying to see what would happen if it were otherwise.
If the line from the origin to (x0 , R(x0 )) intersects the graph of R(x), but is not tangent to the graph of R(x) at x0 ,
then there are points of this graph on both sides of the line — and, in particular, there is some point x1 such that the line
from the origin to (x1 , R(x1 )) has larger slope than the line to (x0 , R(x0 )). (See the graph below.) But we picked x0 so
that no other line had larger slope, and therefore no such x1 exists. So the original supposition is false, and the line from
the origin to (x0 , R(x0 )) is tangent to the graph of R(x).
(a) See (b).
(b)
Line through origin
is tangent here
R(x)
❄
Optimal point on r(x)
❄
r(x)
x
(c)
R(x)
x
xR′ (x) − R(x)
′
r (x) =
x2
r(x) =
348
Chapter Four /SOLUTIONS
So when r(x) is maximized 0 = xR′ (x) − R(x), the numerator of r ′ (x), or R′ (x) = R(x)/x = r(x). i.e.
when r(x) is maximized, r(x) = R′ (x).
Let us call the x-value at which the maximum of r occurs xm . Then the line passing through R(xm ) and the
origin is y = x · R(xm )/xm . Its slope is R(xm )/xm , which also happens to be r(xm ). In the previous paragraph,
we showed that at xm , this is also equal to the slope of the tangent to R(x). So, the line through the origin is the
tangent line.
22. (a) The value of M C is the slope of the tangent to the curve at q0 . See Figure 4.108.
(b) The line from the curve to the origin joins (0, 0) and (q0 , C(q0 )), so its slope is C(q0 )/q0 = a(q0 ).
(c) Figure 4.109 shows that the line whose slope is the minimum a(q) is tangent to the curve C(q). This line, therefore,
also has slope M C, so a(q) = M C at the q making a(q) minimum.
$
C(q)
Slope= M C
$
C(q)
❄
■
✛
Slope=
C(q0 )
q0
Slope of
this line is
minimum a(q)
= a(q0 )
q
q0
q
Figure 4.108
Figure 4.109
23. (a) The average cost is a(q) = C(q)/q, so the total cost is C(q) = 0.01q 3 − 0.6q 2 + 13q.
(b) Taking the derivative of C(q) gives an expression for the marginal cost:
C ′ (q) = M C(q) = 0.03q 2 − 1.2q + 13.
To find the smallest M C we take its derivative and find the value of q that makes it zero. So: M C ′ (q) = 0.06q−1.2 =
0 when q = 1.2/0.06 = 20. This value of q must give a minimum because the graph of M C(q) is a parabola
opening upward. Therefore the minimum marginal cost is M C(20) = 1. So the marginal cost is at a minimum when
the additional cost per item is $1.
(c) Differentiating gives a′ (q) = 0.02q − 0.6.
Setting a′ (q) = 0 and solving for q gives q = 30 as the quantity at which the average is minimized, since the graph
of a is a parabola which opens upward. The minimum average cost is a(30) = 4 dollars per item.
(d) The marginal cost at q = 30 is M C(30) = 0.03(30)2 − 1.2(30) + 13 = 4. This is the same as the average cost at
this quantity. Note that since a(q) = C(q)/q, we have a′ (q) = (qC ′ (q) − C(q))/q 2 . At a critical point, q0 , of a(q),
we have
q0 C ′ (q0 ) − C(q0 )
,
0 = a′ (q0 ) =
q02
so C ′ (q0 ) = C(q0 )/q0 = a(q0 ). Therefore C ′ (30) = a(30) = 4 dollars per item.
Another way to see why the marginal cost at q = 30 must equal the minimum average cost a(30) = 4 is to view
C ′ (30) as the approximate cost of producing the 30th or 31st good. If C ′ (30) < a(30), then producing the 31st
good would lower the average cost, i.e. a(31) < a(30). If C ′ (30) > a(30), then producing the 30th good would
raise the average cost, i.e. a(30) > a(29). Since a(30) is the global minimum, we must have C ′ (30) = a(30).
24. (a) Since the company can produce more goods if it has more raw materials to use, the function f (x) is increasing. Thus,
we expect the derivative f ′ (x) to be positive.
(b) The cost to the company of acquiring x units of raw material is wx, and the revenue from the sale of f (x) units of
the product is pf (x). The company’s profit is π(x) = Revenue − Cost = pf (x) − wx.
(c) Since profit π(x) is maximized at x = x∗ , we have π ′ (x∗ ) = 0. From π ′ (x) = pf ′ (x) − w, we have pf ′ (x∗ ) − w =
0. Thus f ′ (x∗ ) = w/p.
(d) Computing the second derivative of π(x) gives π ′′ (x) = pf ′′ (x). Since π(x) has a maximum at x = x∗ , the second
derivative π ′′ (x∗ ) = pf ′′ (x∗ ) is negative. Thus f ′′ (x∗ ) is negative.
(e) Differentiate both sides of pf ′ (x∗ ) − w = 0 with respect to w. The chain rule gives
p
d ′ ∗
f (x ) − 1 = 0
dw
4.5 SOLUTIONS
pf ′′ (x∗ )
349
∗
dx
−1 = 0
dw
1
dx∗
=
.
dw
pf ′′ (x∗ )
Since f ′′ (x∗ ) < 0, we see dx∗ /dw is negative.
(f) Since dx∗ /dw < 0, the quantity x∗ is a decreasing function of w. If the price w of the raw material goes up, the
company should buy less.
25. (a) See Figure 4.110.
x2 + y 2 = 9
x2 + y 2 = 4
x2 + y 2 = 1
y
❘
❘
❘
4
x+y =4
x
4
Figure 4.110
(b) As C increases in the equation x2 + y 2 = C, the circle expands outward. For C = 4, the circle does not intersect the
line. As C increases from 4, the circle expands until it touches the line. At C = 9, the circle cuts the line twice.
The minimum value of C = x2 + y 2 occurs where a circle tangent to the line. For larger C-values, x2 + y 2 = C
cuts the line twice, for smaller C-values, the circle does not touch the line.
(c) At the point at when the circle touches the line, the slope of the circle equals the slope of the line, namely −1. Implicit
differentiation gives the slope of x2 + y 2 = C:
2x + 2y · y ′ = 0
−x
.
y′ =
y
Thus, at the point where a circle touches the line, we have
−
x
= −1
y
x = y.
Substitution into x + y = 4 gives 2x = 4, so x = 2 and y = 2. Thus, the minimum is x2 + y 2 = 22 + 22 = 8.
26. (a) See Figure 4.111.
y
Q = 100
✠
5
✠
Q = 200
Q = 300
✠ C = x + 2y = 10
✠
x
10
Figure 4.111
350
Chapter Four /SOLUTIONS
(b) Comparing the curves Q = 100, Q = 200, and Q = 300, we see that production increases as we move away from
the origin. The curve Q = 100 cuts the line C = x + 2y = 10 twice while the curves Q = 200 and Q = 300 do not
cut the line.
The maximum possible Q occurs where a curve touches the line. At this point, the slope of the production curve
equals the slope of the budget line, namely −1/2.
(c) Using implicit differentiation, the slope of the curve 10xy = C is given by
10y + 10xy ′ = 0
y
y′ = − .
x
Thus, at the point where the curve touches the line, whose slope is −1/2, we have
1
y
=−
x
2
x
y= .
2
Substituting into C = x + 2y = 10 gives 2x = 10, so x = 5 and y = 5/2. Thus, the maximum production is
5
Q = 10 · 5 · = 125.
2
−
27. (a) See Figure 4.112.
y
4
C=4
✠
C=3
2
✠
C=2
✠
Q = x1/2 y 1/2
x
2
4
Figure 4.112
(b) Comparing the lines C = 2, C = 3, C = 4, we see that the cost increases as we move away from the origin. The
line C = 2 does not cut the curve Q = 1; the lines C = 3 and C = 4 cut twice.
The minimum cost occurs where a cost line is tangent to the production curve.
(c) Using implicit differentiation, the slope of x1/2 y 1/2 = 1 is given by
1 −1/2 1/2 1 1/2 −1/2 ′
x
y
+ x y
y =0
2
2
y
−x−1/2 y 1/2
y ′ = 1/2 −1/2 = − .
x
x y
The cost lines all have slope −2. Thus, if the curve is tangent to a line, we have
y
− = −2
x
y = 2x.
Substituting into Q = x1/2 y 1/2 = 1 gives
x1/2 (2x)1/2 = 1
√
2x = 1
1
x= √
2
√
1
y = 2 · √ = 2.
2
Thus the minimum cost is
√
√
1
C = 2 √ + 2 = 2 2.
2
4.6 SOLUTIONS
351
Strengthen Your Understanding
28. We are only given the cost and revenue when q = 100. Since we do not have any information about the marginal revenue
and the marginal cost at q = 100, we do not know how an increase in production will affect either the revenue or the cost.
Thus we cannot determine whether the profit will increase or decrease if production is increased from 100. We only know
that when q = 100 the profit will be R(100) − C(100) = $60.
29. The profit, π is given by
π(q) = R(q) − C(q).
Thus we have
π ′ (q) = R′ (q) − C ′ (q),
and the critical points of π(q) occur where R′ (q) = C ′ (q). In Figure 4.82, there are two such points, at approximately
q = 3.5 and q = 13. To the left of q = 10, the graph shows that the costs exceed revenue, so the company is losing
money. Profit is positive between q = 10 and q = 15. So the critical point at which profit is a maximum is q = 13. The
maximum profit occurs when the quantity produced is about 13,000 units.
30. We want a quantity where the slope of the cost curve, C, is greater than the slope of the revenue curve, R. For example,
small quantities, such as q = 2, or large quantities, such as q = 15.
31. The cost curve, C, lies above the revenue curve, R, for all quantities, q. See Figure 4.113.
y
C
R
x
Figure 4.113
32. (e). A company should maximize profit, that is, it should maximize revenue minus cost.
33. (a) If M R − M C > 0, so M R > M C, then revenue is increasing faster than cost with production, so profit increases.
Solutions for Section 4.6
Exercises
1. The rate of change of temperature is
dH
= 16(−0.02)e−0.02t = −0.32e−0.02t .
dt
When t = 0,
When t = 10,
dH
= −0.32e0 = −0.32◦ C/min.
dt
dH
= −0.32e−0.02(10) = −0.262◦ C/min.
dt
2. The rate of growth, in billions of people per year, was
dP
= 6.7(0.011)e0.011t .
dt
On January 1, 2007, we have t = 0, so
dP
= 6.7(0.011)e0 = 0.0737 billion/year = 73.7 million people/year.
dt
352
Chapter Four /SOLUTIONS
3. The rate of change of the power dissipated is given by
dP
81
= − 2.
dR
R
4. (a) The rate of change of the period is given by
π
1
1
2π
2π d √
π
dT
· l−1/2 = √
·√ = √
.
= √
( l) = √
dl
9.8 dl
9.8 2
9.8
l
9.8l
√
(b) The rate decreases since l is in the denominator.
5. Let y(t) be the height of the aircraft, in feet, as a function of time, t, in minutes, and H(y) be the outside temperature in
◦
C as a function of height, y, in feet. We have
dy
= 500 feet/minute,
dt
dH
−2 ◦
=
C/foot,
dy
1000
so we have
dH dy
dH
=
·
dt
dy dt
−2
=
· 500
1000
◦
= −1 C/minute
6. (a) Since P = 25, we have
−5
25 = 30e−3.23×10 h
ln(25/30)
= 5644 ft.
h=
−3.23 × 10−5
(b) Both P and h are changing over time, and we differentiate with respect to time t, in minutes:
−5
dP
dh
= 30e−3.23×10 h −3.23 × 10−5
.
dt
dt
We see in part (a) that h = 5644 ft when P = 25 inches of mercury, and we know that dP/dt = 0.1. Substituting
gives:
−5
0.1 = 30e−3.23×10
×5644.630
dh
= −124 ft/minute.
dt
−3.23 × 10−5
dh
dt
The glider’s altitude is decreasing at a rate of about 124 feet per minute.
7. We differentiate F = k/r 2 with respect to t using the chain rule to give
2k dr
dF
=− 3 ·
.
dt
r
dt
We know that k = 1013 newton · km2 and that the rocket is moving at 0.2 km/sec when r = 104 km. In other words,
dr/dt = 0.2 km/sec when r = 104 . Substituting gives
2 · 1013
dF
=−
· 0.2 = −4 newtons/sec.
dt
(104 )3
8. We know dR/dt = 0.2 when R = 5 and V = 9 and we want to know dI/dt. Differentiating I = V /R with V constant
gives
1 dR
dI
=V − 2
,
dt
R dt
so substituting gives
1
dI
= 9 − 2 · 0.2 = −0.072 amps per second.
dt
5
4.6 SOLUTIONS
9. We have
So
353
9
dθ
dA
=
[4 − 4 cos(4θ)] .
dt
16
dt
dA
dt
=
θ=π/4
π
9
4 − 4 cos 4 ·
16
4
0.2 =
9
(4 + 4)0.2 = 0.9 cm2 /min.
16
10. Differentiating with respect to time t, in seconds, we have
dφ
= 2π
dt
1 2
dx
(x + 4)−1/2 2x
2
dt
−
dx
.
dt
Since the point is moving to the left, dx/dt = −0.2. Substituting x = 3 and dx/dt = −0.2, we have
dφ
= 2π
dt
1
2
(32 + 4)−1/2 (2 · 3(−0.2)) − (−0.2) = 0.211.
The potential is changing at a rate of 0.211 units per second.
11. (a) The rate of change of average cost as quantity increases is
dC
a
= − 2 dollars/cell phone.
dq
q
(b) We are told that dq/dt = 100, and we want dC/dt. The chain rule gives
dC dq
a
100a
dC
=
·
= − 2 · 100 = − 2 dollars/week.
dt
dq dt
q
q
Since a is positive, dC/dt is negative, so C is decreasing.
12. When we differentiate x2 + y 2 = 25 with respect to time, we have
2x
dx
dy
+ 2y
= 0.
dt
dt
(a) We use the fact that x2 + y 2 = 25 to solve for y when x = 0. Since 02 + y 2 = 25, we have y = 5 since y is positive.
Substituting, we have
dy
dx
+ 2y
=0
dt
dt
dy
=0
2(0)(6) + 2(5)
dt
dy
= 0.
dt
2x
(b) We use the fact that x2 + y 2 = 25 to solve for y when x = 3. Since 32 + y 2 = 25, we have y = 4 since y is positive.
Substituting, we have
dy
dx
+ 2y
=0
dt
dt
dy
=0
2(3)(6) + 2(4)
dt
dy
= −4.5.
dt
2x
(c) We use the fact that x2 + y 2 = 25 to solve for y when x = 4. Since 42 + y 2 = 25, we have y = 3 since y is positive.
Substituting, we have
dy
dx
+ 2y
=0
dt
dt
dy
2(4)(6) + 2(3)
=0
dt
dy
= −8.
dt
2x
354
Chapter Four /SOLUTIONS
13. Since h is constant, we have
2 dx
dV
= xh .
dt
3
dt
Thus for h = 120, x = 150, and dx/dt = −0.002, we have
dV
2
= (150)(120)(−0.002) = −24 meters3 /yr.
dt
3
The volume is decreasing at 24 meters3 per year.
14. (a) Treating a and l as constants, by the chain and quotient rules, we have
K(2a + l) da
K(35)
dF
=−
=−
2 = −K(7.78 × 10−4 ) newtons/min.
dt
[a(a + l)]2 dt
[15(20)]2
(b) We have
dF
Ka
dl
K(15)
=−
=−
(−2) = K(3.33 × 10−4 ) newtons/min.
dt
[a(a + l)]2 dt
[15(20)]2
15. (a) We have
Surface area = 0.01(600.25 )(1500.75 ) = 1.19 meters2 .
(b) Since h = 150 is constant, we have
s = 0.01w0.25 1500.75 .
Differentiating with respect to time t, in years,we have
dw
ds
= 0.01 0.25w−0.75
1500.75 .
dt
dt
Substituting dw/dt = 0.5 and w = 62, we have
ds
= 0.01(0.25(62−0.75 )(0.5))(1500.75 ) = 0.0024 meter2 /year.
dt
The surface area is increasing at a rate of 0.0024 meter2 per year, or 24 cm2 per year.
Problems
16. Let the other side of the rectangle be x cm. Then the area is A = 10x, so differentiating with respect to time gives
dA
dx
= 10 .
dt
dt
We are interested in the instant when x = 12 and dx/dt = 3, giving
dx
dA
= 10
= 10 · 3 = 30 cm2 per minute.
dt
dt
17. Let the other side of the rectangle be x cm. Then the diagonal is
D=
Differentiating with respect to time gives
p
82 + x2 =
p
64 + x2 .
dx
dx
2x
x
dD
= √
= √
.
dt
2 64 + x2 dt
64 + x2 dt
We are interested in the instant when x = 6 and dx/dt = 3, giving
6
6
dD
= √
· 3 = 1.8 cm per minute.
·3=
dt
10
64 + 62
4.6 SOLUTIONS
18. Let the other leg be x cm. Then the area is
A=
355
1
7
7x = x.
2
2
Differentiating with respect to time gives
7 dx
dA
=
.
dt
2 dt
We are interested in the instant when x = 10 and dx/dt = −2, where the negative sign reflects the fact that x is
decreasing. Thus
dA
7
= (−2) = −7 cm2 per second.
dt
2
The area is decreasing at 7 cm2 per second.
√
19. The side length is s = A cm. Differentiating with respect to time gives
ds
1 1 dA
= √
.
dt
2 A dt
We are interested in the instant when A = 576 and dA/dt = 3, giving
1 1
1
ds
= √
3=
.
dt
2 576
16
Thus, the side is increasing at 1/16 cm per minute.
20. Since R1 is constant, we know dR1 /dt = 0. We know that dR2 /dt = 2 ohms/min when R2 = 20 ohms. At the moment
we are interested in, R1 = 10 and R2 = 20, so
1
1
1
3
=
+
=
,
R
10
20
20
Differentiating with respect to time t,
so R =
20
ohms.
3
1
1
1
=
+
,
R
R1
R2
gives
−
1 dR
1 dR1
1 dR2
=− 2
−
.
R2 dt
R1 dt
R2 2 dt
Substituting gives
−
dR
1
1
1
= − 2 ·0− 2 ·2
(20/3)2 dt
10
20
20 2 1
dR
2
· 2 · 2 = ohms/min.
=
dt
3
20
9
21. (a) The rate of change of temperature change is
d
dT
=
dD
dD
CD2
D3
−
2
3
=
2CD
3D2
−
= CD − D2 .
2
3
(b) We want to know for what values of D the value of dT /dD is positive. This occurs when
dT
= (C − D) D > 0.
dD
We only consider positive values of D, since a zero dosage obviously has no effect and a negative dosage does not
make sense. If D > 0, then (C − D)D > 0 when C − D > 0, or D < C. So the rate of change of temperature
change is positive for doses less than C.
22. (a)
(i) Differentiating thinking of r as a constant gives
r
dP
= 500ert/100 ·
= 5rert/100 .
dt
100
Substituting t = 0 gives
dP
= 5rer·0/100 = 5r dollars/yr.
dt
356
Chapter Four /SOLUTIONS
(ii) Substituting t = 2 gives
dP
= 5rer·2/100 = 5re0.02r dollars/yr.
dt
(b) To differentiate thinking of r as variable, think of the function as
P = 500er(t)·t/100 ,
and use the chain rule (for ert/100 ) and the product rule (for r(t) · t):
1
d
dP
= 500er(t)·t/100 ·
· (r(t) · t) = 5er(t)·t/100 r(t) · 1 + r ′ (t) · t .
dt
100 dt
Substituting t = 2, r = 4, and r ′ (2) = 0.3 gives
dP
= 5e4·2/100 (4 + 0.3 · 2) = 24.916 dollars/year
dt
Thus, the price is increasing by about $25 per year at that time.
23. (a) The rate of change of force with respect to distance is
2A
3B
dF
= 3 − 4.
dr
r
r
The units are units of force per units of distance.
(b) We are told that dr/dt = k and we want dF/dt. By the chain rule
dF
dF dr
=
·
=
dt
dr dt
The units are units of force per unit time.
2A
3B
− 4
r3
r
k.
24. The rate of change of velocity is given by
dv
mg
k
=−
− e−kt/m = ge−kt/m .
dt
k
m
When t = 0,
dv
dt
= g.
t=0
When t = 1,
dv
= ge−k/m .
dt t=1
These answers give the acceleration at t = 0 and t = 1. The acceleration at t = 0 is g, the acceleration due to gravity,
and at t = 1, the acceleration is ge−k/m , a smaller value.
25. We let x represent the distance from the base of the wall to the base of the ladder and y represent the distance from the
top of the ladder to the base of the wall. Notice that x and y are changing over time, but that the length of the ladder is
fixed at 10 m. Using the Pythagorean Theorem, we have
x2 + y 2 = 102 .
Differentiating with respect to time, we have
2x
dy
dx
+ 2y
= 0.
dt
dt
We have dx/dt = 0.5 and we are finding dy/dt.
(a) When x = 4, we have 42 + y 2 = 102 so y = 9.165. Substituting, we have
2(4)(0.5) + 2(9.165)
dy
=0
dt
dy
= −0.218 m/sec.
dt
The top of the ladder is sliding down at 0.218 m/sec.
(b) When x = 8, we have 82 + y 2 = 102 so y = 6. Substituting, we have
dy
=0
dt
dy
= −0.667 m/sec.
dt
The top of the ladder is sliding down at 0.667 m/sec. Notice that the top slides down faster and faster as the bottom
moves away at a constant speed.
2(8)(0.5) + 2(6)
4.6 SOLUTIONS
357
26. Since the radius is 3 feet, the volume of the gas when the depth is h is given by
V = π32 h = 9πh.
We want to find dV /dt when h = 4 and dh/dt = 0.2. Differentiating gives
dh
dV
= 9π
= 9π(0.2) = 5.655 ft3 /sec.
dt
dt
Notice that the value H = 4 is not used. This is because V is proportional to h so dV /dt does not depend on h.
27. The rate of 3 meters3 /min is a derivative and the units tell us it is dV /dt where V is volume in cubic meters and t is time
in minutes. The rate at which the water level is rising is dh/dt where h is the height of the water in meters. We must relate
V and h.
When the water level is h, the volume, V , of water in the cylinder is
V = πr 2 h
where r = 5 meters is the radius, so
V = 25πh.
Differentiating with respect to time gives
dh
dV
= 25π .
dt
dt
We substitute dV /dt = 3 and solve for dh/dt:
3 = 25π
dh
dt
dh
3
=
= 0.038 meter/min.
dt
25π
The water level is rising 0.038 meters per minute. Notice that this rate does not depend on the fact that the cylinder is half
full. We have dh/dt = 0.038 for all water levels h.
28. When the radius is r, the volume V of the snowball is
V =
4 3
πr .
3
We know that dr/dt = −0.2 when r = 15 and we want to know dV /dt at that time. Differentiating, we have
4
dr
dr
dV
= π3r 2
= 4πr 2 .
dt
3
dt
dt
Substituting dr/dt = −0.2 gives
dV
dt
r=15
= 4π(15)2 (−0.2) = −180π = −565.487 cm3 /hr.
Thus, the volume is decreasing at 565.487 cm3 per hour.
29. If V is the volume of the balloon and r is its radius, then
V =
4 3
πr .
3
We want to know the rate at which air is being blown into the balloon, which is the rate at which the volume is increasing,
dV /dt. We are told that
dr
= 2 cm/sec when r = 10 cm.
dt
Using the chain rule, we have
dV dr
dr
dV
=
·
= 4πr 2 .
dt
dr dt
dt
Substituting gives
dV
= 4π(10)2 2 = 800π = 2513.3 cm3 /sec.
dt
358
Chapter Four /SOLUTIONS
30. If C represents circulation time in seconds and B represent body mass in kilograms, we have
C = 17.40B 1/4 .
As the child grows, both the body mass and the circulation time change. Differentiating with respect to time t, we have
1 −3/4 dB
dC
= 17.40
B
dt
4
dt
Substituting B = 45 and dB/dt = 0.1, we have
.
dC
1
= 17.40
(45)−3/4 (0.1) = 0.025.
dt
4
The circulation time is increasing at a rate of 0.025 seconds per month.
31. (a) Since P = 1 when V = 20, we have
Thus, we have
k = 1 · (201.4 ) = 66.29.
P = 66.29V −1.4 .
Differentiating gives
dP
= 66.29(−1.4V −2.4 ) = −92.8V −2.4 atmospheres/cm3 .
dV
(b) We are given that dV /dt = 2 cm3 /min when V = 30 cm3 . Using the chain rule, we have
atm
cm3
dP
dP dV
2
=
·
= −92.8V −2.4 3
dt
dV dt
cm
min
atm
−2.4
= −92.8 30
2
min
= −0.0529 atmospheres/min
Thus, the pressure is decreasing at 0.0529 atmospheres per minute.
32. If the length of a horizontal rod is x and the length of a vertical rod is h, the volume, V , is given by
V = x2 h.
Taking derivatives gives
dx
dh
dV
= 2xh
+ x2 .
dt
dt
dt
When the area of the square base is 9 cm2 , we have x = 3 cm. Since the volume is then 180 cm3 , we know h = 180/9 =
20 cm. So
dV
= 2 · 3 · 20(0.001) + 32 (0.002) = 0.138 cm3 /hr.
dt
33. (a) Since the slick is circular, if its radius is r meters, its area, A, is A = πr 2 . Differentiating with respect to time using
the chain rule gives
dr
dA
= 2πr .
dt
dt
We know dr/dt = 0.1 when r = 150, so
dA
= 2π150(0.1) = 30π = 94.248 m2 /min.
dt
(b) If the thickness of the slick is h, its volume, V , is given by
V = Ah.
Differentiating with respect to time using the product rule gives
dA
dh
dV
=
h+A .
dt
dt
dt
4.6 SOLUTIONS
359
We know h = 0.02 and A = π(150)2 and dA/dt = 30π. Since V is fixed, dV /dt = 0, so
0 = 0.02(30π) + π(150)2
dh
.
dt
Thus
dh
0.02(30π)
=−
= −0.0000267 m/min,
dt
π(150)2
so the thickness is decreasing at 0.0000267 meters per minute.
34. Let the volume of clay be V. The clay is in the shape of a cylinder, so V = πr 2 L. We know dL/dt = 0.1 cm/sec and we
want to know dr/dt when r = 1 cm and L = 5 cm. Differentiating with respect to time t gives
dr
dL
dV
= π2rL
+ πr 2
.
dt
dt
dt
However, the amount of clay is unchanged, so dV /dt = 0 and
2rL
dL
dr
= −r 2
,
dt
dt
therefore
r dL
dr
=−
.
dt
2L dt
When the radius is 1 cm and the length is 5 cm, and the length is increasing at 0.1 cm per second, the rate at which
the radius is changing is
dr
1
=−
· 0.1 = −0.01 cm/sec.
dt
2·5
Thus, the radius is decreasing at 0.01 cm/sec.
35. (a) When full
Volume of water in filter =
1
1 2
πr h = π62 · 10 = 120π.
3
3
Water flows out at a rate of 1.5 cm3 per second, so
Time to empty =
120π
= 80π = 251.327 secs.
1.5
The time taken is 251.327 sec or just over 4 minutes.
(b) Let the radius of the surface of the water be r cm when the depth is h cm. See Figure 4.114. Then by similar triangles
r
6
=
10
h
3
r = h.
5
Thus, when the depth of the water is h,
Volume of water = V =
1
1 2
πr h = π
3
3
2
3
h
5
h=
3
πh3 cm3 .
25
(c) We know that water is flowing out at 1.5 cm3 per second, so dV /dt = −1.5. We want to know dh/dt when h = 8.
Differentiating the answer to part (b), we have
3
dh
9
dh
dV
=
π3h2
=
πh2 .
dt
25
dt
25
dt
Substituting dV /dt = −1.5 and h = 8 gives
dh
9
π82 ·
25
dt
1.5 · 25
dh
=−
= −0.0207 cm/sec.
dt
9π82
−1.5 =
Thus, the water level is dropping by 0.0207 cm per second.
360
Chapter Four /SOLUTIONS
✛6 cm ✲
✻
r
✛✲
10 cm
✻
h
❄
❄
Figure 4.114
36. The surface area of an open cone is given by√A = πrs where s is the length of the side of the cone. Thus, if the height of
the cone is h and the radius r, the side s = r 2 + h2 . Since the cone has radius 4 in and depth 4 in, when the depth is h,
the radius of the surface of the water is also h, so r = h. Thus
p
p
√
A = πr r 2 + h2 = πr r 2 + r 2 = 2πr 2 .
We want dA/dt when r = 2.5 and dr/dt = 3 in/min. Differentiating gives
√
dr
dA
= 2 2πr
dt
dt
√
dA
= 2 2π · 2.5 · 3 = 66.643 in2 /min.
dt
37. Since the slant side of the cone makes an angle of 45◦ with the vertical, we have r = h, so the volume of the cone is
V =
1 3
πh .
3
Differentiating with respect to time gives
dV
dh
= πh2 .
dt
dt
We know dV /dt = 2 cubic meters/min, and we want to know dh/dt when h = 0.5 meters. Substituting gives
2 = π(0.5)2
dh
,
dt
dh
2
8
=
= meters/min.
dt
π(0.5)2
π
38. (a) Let x, y be the distances, in miles, of the car and truck respectively, from the gas station. See Figure 4.115. If the car
and truck are h miles apart, Pythagoras’ Theorem gives
h2 = x2 + y 2 .
We know that when x = 3, dx/dt = −100 (the negative sign represents the fact that the distance from the gas
station is decreasing), and y = 4, dy/dt = 80. Thus
h2 = 32 + 42 = 25 so h = 5 miles.
We want to find dh/dt. Differentiating h2 = x2 + y 2 gives
2h
dx
dy
dh
= 2x
+ 2y
dt
dt
dt
dh
= 3(−100) + 4(80)
dt
−300 + 320
dh
=
= 4 mph.
dt
5
5
Thus, the distance is increasing at 4 mph.
4.6 SOLUTIONS
361
(b) If dy/dt = 70, we have
dh
= 3(−100) + 4(70)
dt
−300 + 280
dh
=
= −4 mph.
dt
5
5
Thus, the distance is decreasing at 4 mph.
Truck
h
y
Gas station
x
Car
Figure 4.115
39. Let the origin be at the center of the wheel and (x, y) be the coordinates of a point on the wheel. Then x2 + y 2 = R2 ,
where R = 67.5 meters is the radius of the wheel. One minute into the ride, we know the passenger is rising at 0.06
meters per second, so dy/dt = 0.06. We want to know dx/dt. Differentiating with respect to t gives
2x
dx
dy
+ 2y
= 0,
dt
dt
so
y dy
dx
=−
.
dt
x dt
Suppose we are looking at the wheel in such a way that it appears to be rotating counter clockwise. In one minute,
the wheel travels through 360◦ /27 = 13.33◦ . From Figure 4.116, we see that at this time the coordinates of the passenger
are x = R sin 13.33◦ and y = −R cos 13.33◦ . Since the vertical speed of the cabin is dy/dt = 0.06 meters per second,
the horizontal speed of the wheel, dx/dt, is
−R cos 13.33◦
dx
=−
0.06 = 0.253 meters/second.
dt
R sin 13.33◦
y
x
13.33◦
R
Start
•
(x, y)
Figure 4.116
40. (a) From the second figure in the problem, we see that θ ≈ 3.3 when t = 2. The coordinates of P are given by x = cos θ,
y = sin θ. When t = 2, the coordinates of P are
(x, y) ≈ (cos 3.3, sin 3.3) = (−0.99, −0.16).
362
Chapter Four /SOLUTIONS
(b) Using the chain rule, the velocity in the x-direction is given by
vx =
dx
dx dθ
dθ
=
·
= − sin θ ·
.
dt
dθ dt
dt
From Figure 4.117, we estimate that when t = 2,
dθ
dt
t=2
≈ 2.
So
dx
≈ −(−0.16) · (2) = 0.32.
dt
Similarly, the velocity in the y-direction is given by
vx =
vy =
When t = 2
vy =
dy
dy dθ
dθ
=
·
= cos θ ·
.
dt
dθ dt
dt
dy
≈ (−0.99) · (2) = −1.98.
dt
θ
3.3
t
2
Figure 4.117
41. (a) On the interval 0 < M < 70, we have
Slope =
2.8
∆G
=
= 0.04 gallons per mile.
∆M
70
On the interval 70 < M < 100, we have
Slope =
∆G
4.6 − 2.8
1.8
=
=
= 0.06 gallons per mile.
∆M
100 − 70
30
(b) Gas consumption, in miles per gallon, is the reciprocal of the slope, in gallons per mile. On the interval 0 < M < 70,
gas consumption is 1/(0.04) = 25 miles per gallon. On the interval 70 < M < 100, gas consumption is 1/(0.06) =
16.667 miles per gallon.
(c) In Figure 4.90 in the text, we see that the velocity for the first hour of this trip is 70 mph and the velocity for the
second hour is 30 mph. The first hour may have been spent driving on an interstate highway and the second hour may
have been spent driving in a city. The answers to part (b) would then tell us that this car gets 25 miles to the gallon
on the highway and about 16 miles to the gallon in the city.
(d) Since M = h(t), we have G = f (M ) = f (h(t)) = k(t). The function k gives the total number of gallons of gas
used t hours into the trip. We have
G = k(0.5) = f (h(0.5)) = f (35) = 1.4 gallons.
The car consumes 1.4 gallons of gas during the first half hour of the trip.
(e) Since k(t) = f (h(t)), by the chain rule, we have
dG
= k′ (t) = f ′ (h(t)) · h′ (t).
dt
Therefore:
dG
dt
t=0.5
= k′ (0.5) = f ′ (h(0.5)) · h′ (0.5) = f ′ (35) · 70 = 0.04 · 70 = 2.8 gallons per hour,
4.6 SOLUTIONS
363
and
dG
dt
t=1.5
= k′ (1.5) = f ′ (h(1.5)) · h′ (1.5) = f ′ (85) · 30 = 0.06 · 30 = 1.8 gallons per hour.
Gas is being consumed at a rate of 2.8 gallons per hour at time t = 0.5 and is being consumed at a rate of 1.8
gallons per hour at time t = 1.5. Notice that gas is being consumed more quickly on the highway, even though the
gas mileage is significantly better there.
42. The rate of change of temperature with distance, dH/dy, at altitude 4000 ft approximated by
∆H
38 − 52
dH
≈
=
= −7◦ F/thousand ft.
dy
∆y
6−4
A speed of 3000 ft/min tells us dy/dt = 3000, so
Rate of change of temperature with time =
◦
dH dy
F
3 thousand ft
·
≈ −7
·
= −21◦ F/min.
dy dt
thousand ft
min
Other estimates can be obtained by estimating the derivative as
dH
∆H
52 − 60
≈
=
= −4◦ F/thousand ft
dy
∆y
4−2
or by averaging the two estimates
−7 − 4
dH
≈
= −5.5◦ F/thousand ft.
dy
2
If the rate of change of temperature with distance is −4◦ /thousand ft, then
Rate of change of temperature with time =
◦
dH dy
F
3 thousand ft
·
≈ −4
·
= −12◦ F/min.
dy dt
thousand ft
min
Thus, estimates for the rate at which temperature was decreasing range from 12◦ F/min to 21◦ F/min.
43. (a) Assuming that T (1) = 98.6 − 2 = 96.6, we get
96.6 = 68 + 30.6e−k·1
28.6 = 30.6e−k
0.935 = e−k .
So
k = − ln(0.935) ≈ 0.067.
(b) We’re looking for a value of t which gives T ′ (t) = −1. First we find T ′ (t):
T (t) = 68 + 30.6e−0.067t
T ′ (t) = (30.6)(−0.067)e−0.067t ≈ −2e−0.067t .
Setting this equal to −1 per hour gives
−1 = −2e−0.067t
ln(0.5) = −0.067t
ln(0.5)
≈ 10.3.
t=−
0.067
Thus, when t ≈ 10.3 hours, we have T ′ (t) ≈ −1◦ F per hour.
(c) The coroner’s rule of thumb predicts that in 24 hours the body temperature will decrease 25◦ F, to about 73.6◦ F. The
formula predicts a temperature of
T (24) = 68 + 30.6e−0.067·24 ≈ 74.1◦ F.
364
Chapter Four /SOLUTIONS
44. (a) Using Pythagoras’ theorem, we see
z 2 = 0.52 + x2
so
p
z = 0.25 + x2 .
(b) We want to calculate dz/dt. Using the chain rule, we have
dz
dz dx
2x
dx
=
·
= √
.
dt
dx dt
2 0.25 + x2 dt
Because the train is moving at 0.8 km/hr, we know that
dx
= 0.8 km/hr.
dt
At the moment we are interested in z = 1 km so
12 = 0.25 + x2
giving
x=
√
0.75 = 0.866 km.
Therefore
2(0.866)
dz
· 0.8 = 0.866 · 0.8 = 0.693 km/min.
= √
dt
2 0.25 + 0.75
(c) We want to know dθ/dt, where θ is as shown in Figure 4.118. Since
x
= tan θ
0.5
we know
x
θ = arctan
,
0.5
so
dθ
1
1 dx
=
·
.
dt
1 + (x/0.5)2 0.5 dt
√
We know that dx/dt = 0.8 km/min and, at the moment we are interested in, x = 0.75. Substituting gives
dθ
1
1
=
·
· 0.8 = 0.4 radians/min.
dt
1 + 0.75/0.25 0.5
0
x km
0.5
,
Train
✲
z km
θ
Camera
Figure 4.118
45. Using the triangle OSL in Figure 4.119, we label the distance x.
L
θ
2
O
x
S
Figure 4.119
We want to calculate dx/dθ. First we must find x as a function of θ. From the triangle, we see
x
= tan θ so x = 2 tan θ.
2
Thus,
dx
2
=
.
dθ
cos2 θ
4.6 SOLUTIONS
46. The radius r is related to the volume by the formula V = 34 πr 3 . By implicit differentiation, we have
4
dr
dr
dV
= π3r 2
= 4πr 2 .
dt
3
dt
dt
The surface area of a sphere is 4πr 2 , so we have
dV
dr
= s·
,
dt
dt
but since
1
dV
= s was given, we have
dt
3
dr
1
= .
dt
3
47. The volume of a cube is V = x3 . So
dV
dx
= 3x2 ,
dt
dt
and
3 dx
1 dV
=
.
V dt
x dt
The surface area of a cube is A = 6x2 . So
dx
dA
= 12x ,
dt
dt
and
2 dx
1 dA
=
.
A dt
x dt
1 dV
Thus the percentage rate of change of the volume of the cube,
, is larger.
V dt
48. (a) The end of the pipe sweeps out a circle of circumference 2π · 20 = 40π meters in 5 minutes, so
Speed =
40π
= 8π = 25.133 meters/min.
5
(b) The distance, h, between P and Q is given by the Law of Cosines:
h2 = 502 + 202 − 2 · 50 · 20 cos θ.
When θ = π/2, we have
h2 = 502 + 202 − 2 · 50 · 20 · 0.
√
h = 2900 = 53.852 m.
When θ = 0, we have h = 30 m.
Since the pipe makes one rotation of 2π radians every 5 minutes, we know
2π
dθ
=
radians/minute.
dt
5
Differentiating the relationship h2 = 502 + 202 − 2 · 50 · 20 cos θ gives
2h
dh
dθ
= 2 · 50 · 20 sin θ .
dt
dt
When θ = π/2, we have
√
2π
dh
= 2 · 50 · 20 · 1 ·
2 2900
dt
5
dh
50 · 20 2π
= √
·
= 23.335 meters/min.
dt
2900 5
When θ = 0, we have
2 · 30
dh
2π
= 2 · 50 · 20 · 0 ·
dt
5
dh
= 0 meters/min.
dt
365
366
Chapter Four /SOLUTIONS
49. The volume, V, of a cone of radius r and height h is
V =
1 2
πr h.
3
Figure 4.120 shows that h/r = 10/8, thus r = 8h/10, so
V =
1
π
3
Differentiating V with respect to time, t, gives
2
8
h
10
h=
64
πh3 .
300
dV
64
dh
=
πh2 .
dt
100
dt
Since water is flowing into the tank at 0.1 cubic meters/min but leaking out at a rate of 0.004h2 cubic meters/min, we also
have
dV
= 0.1 − 0.004h2 .
dt
Equating the two expressions for dV /dt, we have
dh
64
πh2
= 0.1 − 0.004h2 .
100
dt
Solving for dh/dt gives
0.1(100 − 4h2 )
dh
=
.
dt
64πh2
Notice dh/dt is positive when
100 − 4h2 > 0 that is, when 4h2 < 100.
We conclude that if h < 5 then dh/dt > 0. Therefore, the depth increases until h = 5. For h > 5, we have dh/dt < 0,
so the depth decreases whenever h > 5. Since the tank is more than 5 meters deep, it does not overflow.
✛8 cm ✲
✻
r
✛✲
10 cm
✻
h
❄
❄
Figure 4.120
50. (a) Since the elevator is descending at 30 ft/sec, its height from the ground is given by h(t) = 300 − 30t,
t ≤ 10.
(b) From the triangle in the figure,
tan θ =
0≤
h(t) − 100
300 − 30t − 100
200 − 30t
=
=
.
150
150
150
Therefore
θ = arctan
and
dθ
=
dt
1+
for
1
·
200−30t 2
150
−30
150
200 − 30t
150
1
=−
5
1502
1502 + (200 − 30t)2
.
is always negative, which is reasonable since θ decreases as the elevator descends.
Notice that dθ
dt
(c) If we want to know when θ changes (decreases) the fastest, we want to find out when dθ/dt has the largest magnitude.
This will occur when the denominator, 1502 + (200 − 30t)2 , in the expression for dθ/dt is the smallest, or when
200 − 30t = 0. This occurs when t = 200
seconds, and so h( 200
) = 100 feet, i.e., when the elevator is at the level
30
30
of the observer.
4.6 SOLUTIONS
367
51. (a) We differentiate a2 (t) + b2 (t) = c with respect to t to find
d 2
d
(a (t) + b2 (t)) = c,
dt
dt
or
giving
2a(t) · a′ (t) − 2b(t) · b′ (t) = 0,
a(t) · a′ (t) = −b(t) · b′ (t).
(i) If Angela likes Brian, then a(t) > 0, so b′ (t) < 0. This means that b(t) is decreasing, so Brian’s affection
decreases when Angela likes him.
(ii) If Angela dislikes Brian, then a(t) < 0, so b′ (t) > 0. This means that b(t) is increasing, so Brian’s affection
increases when Angela dislikes him.
(c) Substituting b′ (t) = −a(t) into a(t) · a′ (t) = −b(t) · b′ (t) gives
(b)
a(t) · a′ (t) = −b(t) · b′ (t) = −b(t)(−a(t)),
so
a′ (t) = b(t).
(i) If Brian likes Angela, then b(t) > 0, so a′ (t) > 0. This means that a(t) is increasing, so Angela’s affection
increases when Brian likes her.
(ii) If Brian dislikes Angela, then b(t) < 0, so a′ (t) < 0. This means that a(t) is decreasing, so Angela’s affection
decreases when Brian dislikes her.
(d) When t = 0, they both like each other. This means that Angela’s affection increases, while Brian’s decreases.
Eventually b(t) < 0, when he dislikes her.
52. (a) We have either x(0) = 50 and y(0) = 40, or y(0) = 50 and x(0) = 40. In the first case c = x2 (0) − y 2 (0) =
502 − 402 = 900 whereas in the second c = x2 (0) − y 2 (0) = 402 − 502 = −900. But c > 0, so c = 900 and we
have x2 (t) − y 2 (t) = 900.
(b) Because√
x2 (t) − y 2 (t) = 900 we have x2 (3) − y 2 (3) = 900 so x2 (3) = y 2 (3) + 900 = 162 + 900 = 1156, giving
x(3) = 1156 = 34. After 3 hours, y has 16 ships, and x has 34 ships.
(c) The condition y(T ) = 0 means that there are no more ships on that side, so the battle ends at time T hours.
(d) We have x2 (T ) − y 2 (T ) = 900 with y(T ) = 0 so x(T ) = 30 ships.
(e) The rate per hour at which y loses ships is y ′ (t), so y ′ (t) = kx. Because y is decreasing, k is negative.
(f) We differentiate x2 (t) − y 2 (t) = 900 with respect to t to find
d
d 2
(x (t) − y 2 (t)) = 900,
dt
dt
or
giving
2x(t) · x′ (t) − 2y(t) · y ′ (t) = 0,
x′ (t) =
y(t) ′
y (t).
x(t)
But y ′ (t) = kx(t) so
x′ (t) =
y(t)
kx(t) = ky(t).
x(t)
(g) From part (b), we know that when t = 3 we have x(3) = 34, y(3) = 16; we are now given that x′ (3) = 32. But
x′ (t) = ky(t) so 32 = ky(3) = 16k giving k = 2. In this case y ′ (3) = kx(3) = 2 · 34 = 68 ships/hour.
Strengthen Your Understanding
dD dR
dR
dD
=
=2
.
53. The radius of a circle and the circle’s diameter are related by the linear function D = 2R. Thus,
dt
dR dt
dt
The rate of change of the diameter is twice the rate of change of the radius. For example, if the radius is increasing at a
constant rate of 5 feet per second, then the circle’s diameter is increasing at a constant rate of 10 feet per second.
54. Differentiating y = 1 − x2 with respect to t gives
dx
dy
= −2x .
dt
dt
368
Chapter Four /SOLUTIONS
dy
dy dx
=
, in
dt
dx dt
order for dy/dt to be constant, we must also have dy/dx a constant. Choosing, for example, f (x) = 2x + 1 we have:
55. We want dx/dt = g ′ (t) to be constant, so g can be any linear function of t, such as g(t) = 5t. Since
dx
=5
dt
and
dy
dx
=2·
= 2(5) = 10.
dt
dt
There are many other possible solutions.
56. We want dx/dt = g ′ (t) to be constant, so g can be any linear function of t, such as g(t) = 5t. If we choose f to be
a linear function, then dy/dt will be constant. So we choose a function f which is not linear, such as f (x) = x2 . With
y = x2 and x = 5t, we have
dx
=5
dt
There are other possible examples.
dy
dy dx
=
·
= 2x(5) = 10t.
dt
dx dt
and
57. True. The circumference C and radius r are related by C = 2πr, so dC/dt = 2πdr/dt. Thus if dr/dt is constant, so is
dC/dt.
58. False. The circumference A and radius r are related by A = πr 2 , so dA/dt = 2πrdr/dt. Thus dA/dt depends on r and
since r is not constant, neither is dA/dt.
59. (c). Differentiating x = 5 tan θ gives
5 dθ
dx
=
.
dt
cos2 θ dt
Since the light rotates at 2 revolutions per minute = 4π radians per minute, we know dθ/dt = 4π. Thus, we can calculate
dx/dt, the speed at which the spot is moving, for any angle θ.
Differentiating any of the other relationships introduces dr/dt, whose values we cannot find as easily as we can find
dθ/dt.
Solutions for Section 4.7
Exercises
1. Since lim (x − 2) = lim (x2 − 4) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→2
x→2
lim
x→2
1
1
x−2
= lim
= .
x2 − 4
x→2 2x
4
2. Since lim (x2 + 3x − 4) = lim (x − 1) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→1
x→1
lim
x→1
x2 + 3x − 4
2x + 3
= lim
= 5.
x→1
x−1
1
3. Since lim (x6 − 1) = lim (x4 − 1) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→1
x→1
lim
x→1
6x5
6
x6 − 1
= lim
= = 1.5.
4
x→1 4x3
x −1
4
4. Since lim (ex − 1) = lim (sin x) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→0
x→0
lim
x→0
ex
1
ex − 1
= lim
= = 1.
x→0 cos x
sin x
1
5. Since the limit is not in the form 0/0 or ∞/∞, l’Hopital’s rule does not apply in this case. We have
lim
x→0
sin x
0
= = 0.
ex
1
4.7 SOLUTIONS
369
6. Since lim (ln x) = lim (x − 1) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→1
x→1
lim
x→1
(1/x)
1
ln x
= lim
= = 1.
x−1
x→1
1
1
7. Since lim (ln x) = lim (x) = ∞, this is an ∞/∞ form, and l’Hopital’s rule applies:
x→∞
x→∞
lim
x→∞
(1/x)
ln x
= lim
= 0.
x→∞
x
1
8. Since lim ((ln x)3 ) = lim (x2 ) = ∞, this is an ∞/∞ form, and l’Hopital’s rule applies. In fact, after applying
x→∞
x→∞
l’Hopital’s rule, we again obtain the ∞/∞ form. We apply l’Hopital’s rule repeatedly, simplifying algebraically at each
step:
lim
x→∞
3(ln x)2 (1/x)
3(ln x)2
6 ln x(1/x)
6/x
(ln x)3
6 ln x
6
= lim
= lim
= lim
= lim
= lim
= lim
.
2
x→∞
x→∞
x→∞
x→∞ 4x2
x→∞ 8x
x→∞ 8x2
x
2x
2x2
4x
The last limit has the form 6/∞ so the limit is 0.
9. Since lim (e4x − 1) = 0 and lim (cos x) = 1, l’Hopital’s rule does not apply in this case. We have
x→0
x→0
lim
x→0
e4x − 1
0
= = 0.
cos x
1
10. Since lim (xa − 1) = lim (xb − 1) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→1
x→1
lim
x→1
axa−1
a
xa − 1
= lim
= .
b
x −1
x→1 bxb−1
b
√
√
11. Since lim ( 3 x − 3 a) = lim (x − a) = 0, this is a 0/0 form, and l’Hopital’s rule applies:
x→a
x→a
lim
√
3
x→a
√
(1/3)x−2/3
1
x− 3 a
= lim
= a−2/3 .
x−a
x→a
1
3
12. The larger power dominates. Using l’Hopital’s rule
5x4
20x3
x5
= lim
= lim
7
6
x→∞ 0.7x
x→∞ 4.2x5
x→∞ 0.1x
2
120x
120
60x
= lim
= lim
=0
= lim
x→∞ 84x3
x→∞ 252x2
x→∞ 21x4
lim
so 0.1x7 dominates.
13. We apply l’Hopital’s rule twice to the ratio 50x2 /0.01x3 :
50x2
100x
100
= lim
= lim
= 0.
x→∞ 0.01x3
x→∞ 0.03x2
x→∞ 0.06x
lim
Since the limit is 0, we see that 0.01x3 is much larger than 50x2 as x → ∞.
14. The power function dominates. Using l’Hopital’s rule
1
x0.8
ln(x + 3)
(x+3)
=
lim
=
lim
.
x0.2
x→∞ 0.2x−0.8
x→∞ 0.2(x + 3)
x→∞
lim
Using l’Hopital’s rule again gives
x0.8
0.8x−0.2
= lim
= 0,
x→∞ 0.2(x + 3)
x→∞
0.2
lim
so x0.2 dominates.
370
Chapter Four /SOLUTIONS
15. The exponential dominates. After 10 applications of l’Hopital’s rule
lim
x→∞
10x9
10!
x10
= lim
= · · · = lim
= 0,
x→∞ 0.1e0.1x
x→∞ (0.1)10 e0.1x
e0.1x
so e0.1x dominates.
Problems
16. Observe that both f (4) and g(4) are zero. Also, f ′ (4) = 1.4 and g ′ (4) = −0.7, so by l’Hopital’s rule,
lim
x→4
f ′ (4)
1.4
f (x)
= ′
=
= −2.
g(x)
g (4)
−0.7
17. Since f ′ (a) > 0 and g ′ (a) < 0, l’Hopital’s rule tells us that
lim
x→a
f ′ (a)
f (x)
= ′
< 0.
g(x)
g (a)
18. Since f ′ (a) < 0 and g ′ (a) < 0, l’Hopital’s rule tells us that
lim
x→a
f (x)
f ′ (a)
= ′
> 0.
g(x)
g (a)
19. Here f (a) = g(a) = f ′ (a) = g ′ (a) = 0, and f ′′ (a) > 0 and g ′′ (a) < 0.
lim
x→a
f (x)
f ′ (x)
f ′′ (a)
= lim ′
= ′′
<0
x→a g (x)
g(x)
g (a)
20. Note that f (0) = g(0) = 0 and f ′ (0) = g ′ (0). Since x = 0 looks like a point of inflection for each curve, f ′′ (0) =
g ′′ (0) = 0. Therefore, applying l’Hopital’s rule successively gives us
lim
x→0
f ′ (x)
f ′′ (x)
f ′′′ (x)
f (x)
= lim ′
= lim ′′
= lim ′′′
.
x→0 g (x)
x→0 g (x)
x→0 g (x)
g(x)
Now notice how the concavity of f changes: for x < 0, it is concave up, so f ′′ (x) > 0, and for x > 0 it is concave
down, so f ′′ (x) < 0. Thus f ′′ (x) is a decreasing function at 0 and so f ′′′ (0) is negative. Similarly, for x < 0, we see g
is concave down and for x > 0 it is concave up, so g ′′ (x) is increasing at 0 and so g ′′′ (0) is positive. Consequently,
f ′′′ (0)
f (x)
= lim ′′′
< 0.
x→0 g (0)
x→0 g(x)
lim
21. The denominator approaches zero as x goes to zero and the numerator goes to zero even faster, so you should expect that
the limit to be 0. You can check this by substituting several values of x close to zero. Alternatively, using l’Hopital’s rule,
we have
2x
x2
= lim
= 0.
lim
x→0 cos x
x→0 sin x
22. The numerator goes to zero faster than the denominator, so you should expect the limit to be zero. Using l’Hopital’s rule,
we have
sin2 x
2 sin x cos x
lim
= lim
= 0.
x→0
x→0
x
1
23. The denominator goes to zero more slowly than x does, so the numerator goes to zero faster than the denominator, so you
should expect the limit to be zero. With l’Hopital’s rule,
lim
x→0
cos x
sin x
= lim 1 −2/3 = lim 3x2/3 cos x = 0.
x→0 x
x→0
x1/3
3
4.7 SOLUTIONS
371
24. The denominator goes to zero more slowly than x. Therefore, you should expect that the limit to be 0. Using l’Hopital’s
rule,
1
3(sin x)2/3
x
=
lim
=
lim
= 0,
lim
x→0 1 (sin x)−2/3 cos x
x→0
x→0 (sin x)1/3
cos x
3
since sin 0 = 0 and cos 0 = 1.
25. Since limx→∞ x = limx→∞ ex = ∞, this is an ∞/∞ form, and l’Hopital’s rule applies directly.
26. We have limx→1 x = 1 and limx→1 (x − 1) = 0, so l’Hopital’s rule does not apply.
27. We have limt→∞ (1/t) − (2/t2 ) = 0 − 0 = 0. This is not an indeterminate form and l’Hopital’s rule does not apply.
28. We have limt→0+ 1/t = limt→0+ 1/(et − 1) = ∞, so this is an ∞ − ∞ form. Adding the fractions we get
et − 1 − t
,
t(et − 1)
lim
t→0+
which is a 0/0 form to which l’Hopital’s rule can be applied.
29. We have limx→0 (1 + x)x = (1 + 0)0 = 1. This is not an indeterminate form and l’Hopital’s rule does not apply.
30. This is an ∞0 form. With y = limx→∞ (1 + x)1/x , we take logarithms to get
ln y = lim
x→∞
This limit is a 0 · ∞ form,
lim
x→∞
1
ln(1 + x),
x
which can be rewritten as the ∞/∞ form
lim
1
ln(1 + x).
x
x→∞
ln(1 + x)
,
x
to which l’Hopital’s rule applies.
31. Let f (x) = ln x and g(x) = x2 − 1, so f (1) = 0 and g(1) = 0 and l’Hopital’s rule can be used. To apply l’Hopital’s
rule, we first find f ′ (x) = 1/x and g ′ (x) = 2x, then
lim
x→1
1/x
1
1
ln x
= lim
= lim
= .
x→1 2x2
x2 − 1 x→1 2x
2
32. Let f (t) = sin2 t and g(t) = t − π, then f (π) = 0 and g(π) = 0 but f ′ (t) = 2 sin t cos t and g ′ (t) = 1, so f ′ (π) = 0
and g ′ (π) = 1. L’Hopital’s rule can be used, giving
lim
t→π
0
sin2 t
= = 0.
t−π
1
33. Rewriting this as lim n1/n , we see that it is an ∞0 form. The logarithm of the limit is
n→∞
lim
n→∞
1/n
ln n
1
ln n = lim
= lim
= 0.
n→∞ n
n→∞ 1
n
Thus the original limit is e0 = 1.
34. Let f (x) = ln x and g(x) = 1/x so f ′ (x) = 1/x and g ′ (x) = −1/x2 and
lim
x→0+
1/x
x
ln x
= lim
= lim
= 0.
2
1/x
x→0+ −1/x
x→0+ −1
35. If f (x) = sinh(2x) and g(x) = x, then f (0) = g(0) = 0, so we use l’Hopital’s Rule:
lim
x→0
sinh 2x
2 cosh 2x
= lim
= 2.
x→0
x
1
372
Chapter Four /SOLUTIONS
36. If f (x) = 1 − cosh(3x) and g(x) = x, then f (0) = g(0) = 0, so we use l’Hopital’s Rule:
−3 sinh 3x
1 − cosh 3x
= lim
= 0.
x→0
x
1
lim
x→0
37. Since cos 0 = 1, we have cos−1 (1) = 0 and lim cos−1 x = 0. Therefore, both cos−1 x and (x − 1) tend to 0 as
x→1−1
x →√1− , so l’Hopital’s rule can be applied. Let f (x) = cos−1 x and g(x) = x − 1, and differentiate to get f ′ (x) =
−1/ 1 − x2 and g ′ (x) = 1. Applying l’Hopital’s rule gives
√
−1/ 1 − x2
cos−1 x
lim
= lim
1
x→1− x − 1
x→1−
−1
.
= lim √
x→1−
1 − x2
However,
√
1 − x2 → 0 as x → 1− , so the limit does not exist.
38. To get this expression in a form in which l’Hopital’s rule applies, we combine the fractions:
1
1
sin x − x
−
=
.
x
sin x
x sin x
Letting f (x) = sin x − x and g(x) = x sin x, we have f (0) = 0 and g(0) = 0 so l’Hopital’s rule can be used.
Differentiating gives f ′ (x) = cos x − 1 and g ′ (x) = x cos x + sin x, so f ′ (0) = 0 and g ′ (0) = 0, so f ′ (0)/g ′ (0) is
undefined. Therefore, to apply l’Hopital’s rule we differentiate again to obtain f ′′ (x) = − sin x and g ′′ (x) = 2 cos x −
x sin x, for which f ′′ (0) = 0 and g ′′ (0) = 2 6= 0. Then
lim
x→0
1
1
−
x
sin x
sin x − x
x→0
x sin x
cos x − 1
= lim
x→0 x cos x + sin x
− sin x
= lim
x→0 2 cos x − x sin x
0
= = 0.
2
= lim
39. Adding the fractions, we get
lim
t→0+
1
2
− t
t
e −1
= lim
t→0+
1
2et − 2 − t
2et − 1
= lim t
= = ∞.
t
t
+
t(e − 1)
0
t→0 e + te − 1
Note that the limit is ∞, not −∞ because et > 1 for t > 0, so the denominator is always positive.
40. Adding the fractions and applying l’Hopital’s rule, we get
lim
t→0
1
1
− t
t
e −1
= lim
t→0
et − 1 − t
et − 1
= lim t
.
t
t→0 e + tet − 1
t(e − 1)
This is again a 0/0 form, so we apply l’Hopital’s rule again to get
lim
t→0 et
et
1
et − 1
= lim t
=
t
t→0 2e + tet
+ te − 1
2
41. Let y = (1 + sin 3/x)x . Taking logs gives
ln y = x ln 1 + sin
To use l’Hopital’s rule, we rewrite ln y as a fraction:
3
.
x
4.7 SOLUTIONS
lim ln y = lim x ln 1 + sin
373
3
x
ln(1 + sin(3/x))
.
= lim
x→∞
1/x
x→∞
x→∞
Let f (x) = ln(1 + sin(3/x)) and g(x) = 1/x then
cos(3/x)(−3/x2 )
(1 + sin(3/x))
f ′ (x) =
and
g ′ (x) = −
1
.
x2
Now apply l’Hopital’s rule to get
f (x)
g(x)
f ′ (x)
= lim ′
x→∞ g (x)
lim ln y = lim
x→∞
x→∞
cos(3/x)(−3/x2 )/(1 + sin(3/x))
x→∞
−1/x2
3 cos(3/x)
3 cos 0
= lim
=
x→∞ 1 + sin(3/x)
1 + sin 0
= 3.
= lim
Since limx→∞ ln y = 3, we have
lim y = e3 .
x→∞
Thus,
lim
x→∞
1 + sin
3
x
x
= e3 .
42. Since limt→0 sin2 At = 0 and limt→0 cos At − 1 = 1 − 1 = 0, this is a 0/0 form. Applying l’Hopital’s rule we get
lim
t→0
2A sin At cos At
sin2 At
= lim
= lim −2 cos At = −2.
t→0
cos At − 1 t→0
−A sin At
43. We rewrite this in the form
tn
.
t→∞
t→∞
et
From Example 5 on page 244 of the text, with k = A = 1 and p = n, we know that et dominates tn , so
lim et − tn = lim et 1 −
lim et 1 −
t→∞
tn
et
= ∞ · (1 − 0) = ∞.
44. To get this expression in a form in which l’Hopital’s rule applies, we rewrite it as a fraction:
xa ln x =
ln x
.
x−a
Letting f (x) = ln x and g(x) = x−a , we have
lim f (x) = lim ln x = −∞,
x→0+
x→0+
and
lim g(x) = lim
x→0+
x→0+
1
= ∞.
xa
So l’Hopital’s rule can be used. To apply l’Hopital’s rule we differentiate to get f ′ (x) = 1/x and g ′ (x) = −ax−a−1 .
Then
374
Chapter Four /SOLUTIONS
ln x
x−a
1/x
= lim
−a−1
x→0+ −ax
1
= − lim xa
a x→0+
= 0.
lim xa ln x = lim
x→0+
x→0+
45. Let f (x) = sin(2x) and g(x) = x. Observe that f (1) = sin 2 6= 0 and g(1) = 1 6= 0. Therefore l’Hopital’s rule does
not apply. However,
sin 2
sin 2x
=
= 0.909297.
lim
x→1
x
1
46. Let f (x) = cos x and g(x) = x. Observe that since f (0) = 1, l’Hopital’s rule does not apply. But since g(0) = 0,
lim
x→0
cos x
x
does not exist.
47. Let f (x) = e−x and g(x) = sin x. Observe that as x increases, f (x) approaches 0 but g(x) oscillates between −1 and
1. Since g(x) does not approach 0 in the limit, l’Hopital’s rule does not apply. Because g(x) is in the denominator and
oscillates through 0 forever, the limit does not exist.
48. Let n = 1/x, so n → ∞ as x → 0+ . Thus
lim (1 + x)1/x = lim
n→∞
x→0+
1+
1
n
n
= e.
49. Let k = n/2, so k → ∞ as n → ∞. Thus,
lim
n→∞
1+
2
n
n
= lim
k→∞
2k
1+
1
k
1+
1
n
nkt
1
n
= lim
k→∞
1+
1
k
k 2
= e2 .
50. Let n = 1/(kx), so n → ∞ as x → 0+ . Thus
lim (1 + kx)t/x = lim
n→∞
x→0+
51. Let
y = 1−
Then
= lim
n→∞
n
1+
1
n
n kt
= ekt .
.
ln(1 − 1/n)
1
1 n
= n ln 1 −
=
.
n
n
1/n
As n → ∞, both the numerator and the denominator of the last fraction tend to 0. Thus, applying L’Hopital’s rule, we
have
1
· n12
ln(1 − 1/n)
1
1−1/n
= lim
= lim −
= −1.
lim
n→∞
n→∞
n→∞
1/n
−1/n2
1 − 1/n
Thus,
ln y = ln 1 −
lim ln y = −1
n→∞
lim y = e−1 ,
n→∞
so
lim
n→∞
1−
1
n
= e−1 .
4.7 SOLUTIONS
52. Let k = n/λ, so k → ∞ as n → ∞. Thus
lim
n→∞
1−
λ
n
n
= lim
k→∞
1−
1
k
kλ
= lim
k→∞
1−
1
k
k λ
= (e−1 )λ = e−λ .
53. This limit is of the form 00 so we apply l’Hopital’s rule to
ln (3t + 5t )/2
.
ln f (t) =
t
We have
lim ln f (t) =
t→−∞
=
=
=
(ln 3)3t + (ln 5)5t / 3t + 5t
lim
t→−∞
1
t
(ln 3)3 + (ln 5)5t
lim
t→−∞
3t + 5t
ln 3 + (ln 5)(5/3)t
lim
t→−∞
1 + (5/3)t
ln 3 + 0
= ln 3.
1+0
Thus
lim f (t) = lim eln f (t) = elimt→−∞ ln f (t) = eln 3 = 3.
t→−∞
t→−∞
54. This limit is of the form ∞0 so we apply l’Hopital’s rule to
ln (3t + 5t )/2
.
ln f (t) =
t
We have
lim ln f (t) =
t→+∞
=
=
=
(ln 3)3t + (ln 5)5t / 3t + 5t
lim
t→+∞
1
(ln 3)3t + (ln 5)5t
lim
t→+∞
3t + 5t
(ln 3)(3/5)t + ln 5
lim
t→+∞
(3/5)t + 1
0 + ln 5
lim
= ln 5.
t→+∞ 0 + 1
Thus
lim f (t) = lim eln f (t) = elimt→−∞ ln f (t) = eln 5 = 5.
t→−∞
t→−∞
55. This limit is of the form 1∞ so we apply l’Hopital’s rule to
ln (3t + 5t )/2
.
ln f (t) =
t
We have
(ln 3)3t + (ln 5)5t / 3t + 5t
lim ln f (t) = lim
1
t→0
t→0
(ln 3)3t + (ln 5)5t
= lim
t→0
3t + 5t
ln 3 + ln 5
ln 15
=
=
.
1+1
2
Thus
lim f (t) = lim eln f (t) = elimt→−∞ ln f (t) = e(ln 15)/2 =
t→−∞
t→−∞
√
15.
375
376
Chapter Four /SOLUTIONS
56. To evaluate, we use l’Hopital’s Rule:
lim
x→0
sinh 2x
2 cosh 2x
= lim
= 2.
x→0
x
1
57. To evaluate, we use l’Hopital’s Rule:
lim
x→0
−3 sinh 3x
1 − cosh 3x
= lim
= 0.
x→0
x
1
58. To evaluate, we use l’Hopital’s Rule:
lim
x→0
−5 sinh 5x
−25 cosh 5x
1 − cosh 5x
= lim
= lim
= −25/2.
x2
x→0
2x
x→0
2
59. To evaluate, we use l’Hopital’s Rule:
lim
x→0
x − sinh x
1 − cosh x
− sinh x
− cosh x
= lim
= lim
= lim
= −1/6.
x→0
x→0
x→0
x3
3x2
6x
6
60. Since the limit is of the form 0/0, we can apply l’Hopital’s rule. Doing it twice we have
lim
x→0
ex − 1 − ln(1 + x)
ex − 1/(1 + x)
= lim
2
x→0
x
2x
ex + 1/(1 + x)2
1+1
=
= 1.
= lim
x→0
2
2
61. Since the limit is of the form 0/0, we can apply l’Hopital’s rule. We have
lim
x→π/2
− cos x − sin x
−1
1 − sin x + cos x
= lim
=
= 1.
sin x + cos x − 1
−1
x→π/2 cos x − sin x
62. Since the limit is of the form 0/0, we can apply l’Hopital’s rule. First note that
1
dxx
dex ln x
=
= ex ln x x · + 1 · ln x = xx + xx ln x.
dx
dx
x
Applying l’Hopital’s rule twice we have
lim
x→1
xx + xx ln x − 1
xx − x
= lim
x→1
1 − x + ln x
−1 + 1/x
xx + xx ln x + (xx + xx ln x) ln x + xx (1/x)
1+0+0+1
= lim
=
= −2.
x→1
−1/x2
−1
Strengthen Your Understanding
63. There is no such n because the function ex dominates xn for every n, no matter how large. This means that for any n, the
values of ex are much larger than the values of xn for large x. Thus, xn cannot dominate ex .
To see why ex dominates xn for all positive integers n, let f (x) = xn and g(x) = ex . Then limx→∞ f (x) = ∞
and limx→∞ g(x) = ∞ so we apply l’Hopital’s rule:
lim
x→∞
f (x)
f ′ (x)
xn
= lim
= lim ′
x
x→∞ g(x)
x→∞ g (x)
e
nxn−1
ex
n · (n − 1)xn−2
= lim
x→∞
ex
:
n!
= lim x = 0
x→∞ e
= lim
x→∞
so ex dominates xn for all positive integers n.
4.8 SOLUTIONS
377
64. If f (x) = 5x + cos x and g(x) = x, we have limx→∞ f (x) = ∞ and limx→∞ g(x) = ∞, so we try to use l’Hopital’s
rule:
f (x)
f ′ (x)
5 − sin x
5x + cos x
= lim
= lim ′
= lim
.
lim
x→∞ g(x)
x→∞ g (x)
x→∞
x→∞
x
1
However, limx→∞ 5 − sin x does not exist, so l’Hopital’s rule is not applicable.
65. When finding limx→a f (x)/g(x), we can only apply l’Hopital’s rule if f (a) = g(a) = 0 or limx→a f (x) = ±∞ and
limx→a g(x) = ±∞. Thus, there are many examples of limits of rational functions to which l’Hopital’s rule cannot be
.
applied. One example is limx→0 x+1
x+2
66. Since limx→∞ ln x = ∞, if limx→∞ f (x) = ±∞ we could apply L’Hopital’s rule to try to find limx→∞
possible example is f (x) = x.
f (x)
.
ln x
One
67. False. To use l’Hopital’s rule, we need f (a) = g(a) = 0. For example, if f (x) = 3 and g(x) = x, then g(1) = 1 and
f ′ (1)/g ′ (1) = 0/1 = 0, but limx→1 (f (x)/g(x)) = 3/1 = 3.
68. (b)
Solutions for Section 4.8
Exercises
1. Between times t = 0 and t = 1, x goes at a constant rate from 0 to 1 and y goes at a constant rate from 1 to 0. So the
particle moves in a straight line from (0, 1) to (1, 0). Similarly, between times t = 1 and t = 2, it goes in a straight line
to (0, −1), then to (−1, 0), then back to (0, 1). So it traces out the diamond shown in Figure 4.121.
y
1
t = 0, t = 4
t=1
x
t=3
−1
1
−1
t=2
Figure 4.121
2. This is like Example 2, except that the x-coordinate goes all the way to 2 and back. So the particle traces out the rectangle
shown in Figure 4.122.
y
t=3
t=2
t = 0, t = 4
t=1
1
Figure 4.122
x
2
378
Chapter Four /SOLUTIONS
3. As the x-coordinate goes at a constant rate from 2 to 0, the y-coordinate goes from 0 to 1, then down to −1, then back
to 0. So the particle zigs and zags from (2, 0) to (1.5, 1) to (1, 0) to (.5, −1) to (0, 0). Then it zigs and zags back again,
forming the shape in Figure 4.123.
y
1
t = 2.5
t=1
t=3
t=2
−1
t = .5
t=0
t=4
1
t = 1.5
x
2
t = 3.5
Figure 4.123
4. Between times t = 0 and t = 1, x goes from −1 to 1, while y stays fixed at 1. So the particle goes in a straight line from
(−1, 1) to (1, 1). Then both the x- and y-coordinates decrease at a constant rate from 1 to −1. So the particle goes in a
straight line from (1, 1) to (−1, −1). Then it moves across to (1, −1), then back diagonally to (−1, 1). See Figure 4.124.
y
t=0
t=4
1
t=1
x
−1
t=2
1
−1
t=3
Figure 4.124
5. One possible answer is x = 3 cos t, y = −3 sin t, 0 ≤ t ≤ 2π.
6. One possible answer is x = −2, y = t.
7. One possible answer is x = 2 + 5 cos t, y = 1 + 5 sin t, 0 ≤ t ≤ 2π.
8. The parameterization x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ 2π, is a circle of radius 2 traced out counterclockwise starting at
the point (2, 0). To start at (−2, 0), put a negative in front of the first coordinate
x = −2 cos t
y = 2 sin t,
0 ≤ t ≤ 2π.
Now we must check whether this parameterization traces out the circle clockwise or counterclockwise. Since when t
increases from 0, sin t is positive, the point (x, y) moves from (−2, 0) into the second quadrant. Thus, the circle is traced
out clockwise and so this is one possible parameterization.
4.8 SOLUTIONS
379
9. The slope of the line is
3 − (−1)
= −4.
1−2
The equation of the line with slope −4 through the point (2, −1) is y − (−1) = (−4)(x − 2), so one possible parameterization is x = t and y = −4t + 8 − 1 = −4t + 7.
m=
10. The ellipse x2 /25 + y 2 /49 = 1 can be parameterized by x = 5 cos t, y = 7 sin t, 0 ≤ t ≤ 2π.
11. The parameterization x = −3 cos t, y = 7 sin t, 0 ≤ t ≤ 2π, starts at the right point but sweeps out the ellipse
in the wrong direction (the y-coordinate becomes positive as t increases). Thus, a possible parameterization is x =
−3 cos(−t) = −3 cos t, y = 7 sin(−t) = −7 sin t, 0 ≤ t ≤ 2π.
12. For 0 ≤ t ≤ π2 , we have x = sin t increasing and y = cos t decreasing, so the motion is clockwise for 0 ≤ t ≤
, and 3π
≤ t ≤ 2π.
Similarly, we see that the motion is clockwise for the time intervals π2 ≤ t ≤ π, π ≤ t ≤ 3π
2
2
π
.
2
13. The particle moves clockwise: For 0 ≤ t ≤ π2 , we have x = cos t decreasing and y = − sin t decreasing. Similarly, for
, and 3π
≤ t ≤ 2π, we see that the particle moves clockwise.
the time intervals π2 ≤ t ≤ π, π ≤ t ≤ 3π
2
2
14. Let f (t) = t2 . The particle is moving clockwise when f (t) is decreasing, that is, when f ′ (t) = 2t < 0, so when t < 0.
The particle is moving counterclockwise when f ′ (t) = 2t > 0, so when t > 0.
′
2
15. Let f (t) = t3 − t. The particle is moving clockwise when f (t) is decreasing, that is, when
p 1− 1 < 0,
p 1 f (t) = 3t
′
2
and counterclockwise when f (t) = 3t − 1 > 0. That is, it moves clockwise when − 3 < t <
, between
3
p
p1
p1 3 p1
p1 3 p1
p1 3 p1
),
sin((−
)
+
))
and
(cos((
)
−
),
sin((
)
−
),
and
counterclockwise
(cos((− 31 )3 +
3 p
3
3
3
3
3
3
p
1
.
when t < − 13 or t >
3
16. Let f (t) = ln t. Then f ′ (t) = 1t . The particle is moving counterclockwise when f ′ (t) > 0, that is, when t > 0. Any
other time, when t ≤ 0, the position is not defined.
17. Let f (t) = cos t. Then f ′ (t) = − sin t. The particle is moving clockwise when f ′ (t) < 0, or − sin t < 0, that is, when
2kπ < t < (2k + 1)π,
where k is an integer. The particle is otherwise moving counterclockwise, that is, when
(2k − 1)π < t < 2kπ,
where k is an integer. Actually, the particle does not fully trace out a circle. The range of f (t) is [−1, 1] so the particle
oscillates between the points (cos(−1), sin(−1)) and (cos 1, sin 1).
18. The circle (x − 2)2 + (y − 2)2 = 1.
19. The line segment y + x = 4 , for 1 ≤ x ≤ 3.
20. The parabola y = (x − 2)2 , for 1 ≤ x ≤ 3.
21. We see from the parametric equations that the particle moves along a line. It suffices to plot two points: at t = 0, the
particle is at point (1, −4), and at t = 1, the particle is at point (4, −3). Since x increases as t increases, the motion is
left to right on the line as shown in Figure 4.125.
y
x
−2
2
4
6
−2
−4
t=0
Figure 4.125
Alternately, we can solve the first equation for t, giving t = (x − 1)/3, and substitute this into the second equation
to get
1
13
x−1
−4= x−
.
y=
3
3
3
The line is y = 13 x − 13
.
3
380
Chapter Four /SOLUTIONS
22. If we eliminate the parameter t, we see that
y = (x − 3) − 2 = x − 5.
Thus, the particle moves along a line. Notice, however, that t2 + 3 ≥ 3 and so x ≥ 3. Similarly, y ≥ −2. The graph is
the ray, or half-line, shown in Figure 4.126.
y
4
x
5
−2
t=0
Figure 4.126
Notice that at t = 0, the particle is at point (3, −2), at t = 1, the particle is at point (4, −1), and at t = −1, the
particle is at point (4, −1). The particle is at the endpoint of the ray when t = 0. As t increases through negative t values,
the particle moves down the ray toward the point (3, −2). At t = 0, it changes direction. As t increases through positive
t values, the particle moves up the ray again.
23. The Cartesian equation of the motion is obtained by eliminating the parameter t, giving
y = (x − 4)2 − 3 = x2 − 8x + 13.
The graph of this parabola is shown in Figure 4.127. At t = 0, the particle is at the vertex of the parabola (4, −3), and the
motion is left to right since x increases steadily as t increases.
y
8
4
x
4
−3
8
t=0
Figure 4.127
24. The graph is a circle centered at the origin with radius 1. The equation is
x2 + y 2 = (cos 3t)2 + (sin 3t)2 = 1.
The particle is at the point (1, 0) when t = 0, and motion is counterclockwise. See Figure 4.128.
y
1
t=0
x
1
−1
−1
Figure 4.128
4.8 SOLUTIONS
381
25. The graph is a circle centered at the origin with radius 3. The equation is
x2 + y 2 = (3 cos t)2 + (3 sin t)2 = 9.
The particle is at the point (3, 0) when t = 0, and motion is counterclockwise. See Figure 4.129.
y
3
−3
3
t=0
x
−3
Figure 4.129
26. The graph is a circle centered at the point (2, 7) with radius 5. The equation is
(x − 2)2 + (y − 7)2 = (5 cos t)2 + (5 sin t)2 = 25.
The particle is at the point (7, 7) when t = 0, and motion is counterclockwise. See Figure 4.130.
y
10
(2, 7)
t=0
(7, 7)
5
x
5
Figure 4.130
27. We have
dy/dt
dy
2t
=
= 2
.
dx
dx/dt
3t − 1
Thus when t = 2, the slope of the tangent line is 4/11. Also when t = 2, we have
x = 23 − 2 = 6,
Therefore the equation of the tangent line is
(y − 4) =
28. We have
y = 22 = 4.
4
(x − 6).
11
dy/dt
2t + 2
dy
=
=
.
dx
dx/dt
2t − 2
When t = 1, the denominator is zero and the numerator is nonzero, so the tangent line is vertical. Since x = −1 when
t = 1, the equation of the tangent line is x = −1.
382
Chapter Four /SOLUTIONS
29. We have
dy
dy/dt
4 cos(4t)
=
=
.
dx
dx/dt
3 cos(3t)
Thus when t = π, the slope of the tangent line is −4/3. Since x = 0 and y = 0 when t = π, the equation of the tangent
line is y = −(4/3)x.
30. We have dx/dt = 2t and dy/dt = 3t2 . Therefore, the speed of the particle is
v=
r
dx
dt
2
+
2
dy
dt
=
p
((2t)2 + (3t2 )2 ) = |t| ·
p
(4 + 9t2 ).
√
The particle comes to a complete stop when its speed is 0, that is, if t 4 + 9t2 = 0, and so when t = 0.
31. We have dx/dt = −2t sin(t2 ) and dy/dt = 2t cos(t2 ). Therefore, the speed of the particle is given by
v=
=
p
(−2t sin(t2 ))2 + (2t cos(t2 ))2
p
4t2 (sin(t2 ))2 + 4t2 (cos(t2 ))2
p
= 2|t|
sin2 (t2 ) + cos2 (t2 )
= 2|t|.
The particle comes to a complete stop when speed is 0, that is, if 2|t| = 0, and so when t = 0 .
32. We have
dy
dx
= −2 sin 2t,
= cos t.
dt
dt
The speed is
p
4 sin2 (2t) + cos2 t.
v=
Thus, v = 0 when sin(2t) = cos t = 0, and so the particle stops when t = ±π/2, ±3π/2, . . . or t = (2n + 1) π2 , for any
integer n.
33. We have
dx
= 2t − 4,
dt
dy
= 3t2 − 12.
dt
The speed is given by:
v=
2
p
(2t − 4)2 + (3t2 − 12)2 .
The particle stops when 2t − 4 = 0 and 3t − 12 = 0. Since these are both satisfied only by t = 2, this is the only time
that the particle stops.
34. At t = 2, the position is (22 , 23 ) = (4, 8), the velocity in the x-direction is 2 · 2 = 4, and the velocity in the y-direction
is 3 · 22 = 12. So we want the line going through the point (4, 8), with the given x- and y-velocities:
x = 4 + 4t,
y = 8 + 12t.
Problems
35. The graphs are in Figure 4.131.
y
x
1
0
1
f (t)
8
16
24
32
t
0
Figure 4.131
g(t)
8
16
24
32
t
383
4.8 SOLUTIONS
36. The graphs are in Figure 4.132.
y
x
1
0
f (t)
8
1
16
24
32
t
0
g(t)
8
16
24
32
t
Figure 4.132
37. (a) We get the part of the line with x < 10 and y < 0.
(b) We get the part of the line between the points (10, 0) and (11, 2).
38. (a) If t ≥ 0, we have x ≥ 2, y ≥ 4, so we get the part of the line to the right of and above the point (2, 4).
(b) When t = 0, (x, y) = (2, 4). When t = −1, (x, y) = (−1, −3). Restricting t to the interval −1 ≤ t ≤ 0 gives the
part of the line between these two points.
(c) If x < 0, giving 2 + 3t < 0 or t < −2/3. Thus t < −2/3 gives the points on the line to the left of the y-axis.
39. (a) Eliminating t between
x = 2 + t,
y = 4 + 3t
gives
y − 4 = 3(x − 2),
y = 3x − 2.
Eliminating t between
gives
x = 1 − 2t,
y = 1 − 6t
y − 1 = 3(x − 1),
y = 3x − 2.
Since both parametric equations give rise to the same equation in x and y, they both parameterize the same line.
(b) Slope = 3, y-intercept = −2.
40. In all three cases, y = x2 , so that the motion takes place on the parabola y = x2 .
In case (a), the x-coordinate always increases at a constant rate of one unit distance per unit time, so the equations
describe a particle moving to the right on the parabola at constant horizontal speed.
In case (b), the x-coordinate is never negative, so the particle is confined to the right half of the parabola. As t moves
from −∞ to +∞, x = t2 goes from ∞ to 0 to ∞. Thus the particle first comes down the right half of the parabola,
reaching the origin (0, 0) at time t = 0, where it reverses direction and goes back up the right half of the parabola.
In case (c), as in case (a), the particle traces out the entire parabola y = x2 from left to right. The difference is that
the horizontal speed is not constant. This is because a unit change in t causes larger and larger changes in x = t3 as t
approaches −∞ or ∞. The horizontal motion of the particle is faster when it is farther from the origin.
41. (a) C1 has center at the origin and radius 5, so a = b = 0, k = 5 or −5.
(b) C2 has center at (0, 5) and radius 5, so a = 0, b = 5, k = 5 or −5.
p
√
√
(c) C3 has center at (10, −10), so a = 10, b = −10. The radius of C3 is 102 + (−10)2 = 200, so k = 200 or
√
k = − 200.
42. (I) has a positive slope and so must be l1 or l2 . Since its y-intercept is negative, these equations must describe l2 . (II)
has a negative slope and positive x-intercept, so these equations must describe l3 .
43. It is a straight line through the point (3, 5) with slope −1. A linear parameterization of the same line is x = 3 + t,
y = 5 − t.
44. (a) The curve is a spiral as shown in Figure 4.133.
384
Chapter Four /SOLUTIONS
(b) At t = 2, the position is (2 cos 2, 2 sin 2) = (−0.8323, 1.8186), and at t = 2.01 the position is (2.01 cos 2.01, 2.01 sin 2.01) =
(−0.8546, 1.8192). The distance between these points is
p
(−0.8546 − (−0.8323))2 + (1.8192 − 1.8186)2 ≈ 0.022.
Thus the speed is approximately 0.022/0.01 ≈ 2.2. See Figure 4.134.
y
y
t=2
t=6
x
x
t=4
Figure 4.133: The spiral
x = t cos t, y = t sin t for 0 ≤ t ≤ 4π
Figure 4.134: The spiral x = t cos t, y = t sin t and three velocity
vectors
(c) Evaluating the exact formula
v=
gives :
p
v(2) =
45. (a) The chain rule gives
(cos t − t sin t)2 + (sin t + t cos t)2
p
(−2.235)2 + (0.077)2 = 2.2363.
dy/dt
4e2t
dy
=
= t = 4et .
dx
dx/dt
e
2t
t 2
t
(b) We are given y = 2e so y = 2(e ) . Since x = e , we can substitute x for et . Thus y = 2x2 .
(c) Differentiating y = 2x2 with respect to x, we get dy/dx = 4x. Notice that, since x = et , this is equivalent to the
answer that we obtained in part (a).
46. (a) In order for the particle to stop, its velocity both dx/dt and dy/dt must be zero,
dx
= 3t2 − 3 = 3(t − 1)(t + 1) = 0,
dt
dy
= 2t − 2 = 2(t − 1) = 0.
dt
The value t = 1 is the only solution. Therefore, the particle stops when t = 1 at the point (t3 − 3t, t2 − 2t)|t=1 =
(−2, −1).
(b) In order for the particle to be traveling straight up or down, the velocity in the x-direction must be 0. Thus, we
solve dx/dt = 3t2 − 3 = 0 and obtain t = ±1. However, at t = 1 the particle has no vertical motion, as we
saw in part (a). Thus, the particle is moving straight up or down only when t = −1. The position at that time is
(t3 − 3t, t2 − 2t)|t=−1 = (2, 3).
(c) For horizontal motion we need dy/dt = 0. That happens when dy/dt = 2t − 2 = 0, and so t = 1. But from part (a)
we also have dx/dt = 0 also at t = 1, so the particle is not moving at all when t = 1. Thus, there is no time when
the motion is horizontal.
47. (a)
(i) A horizontal tangent occurs when dy/dt = 0 and dx/dt 6= 0. Thus,
dy
= 6e2t − 2e−2t = 0
dt
6e2t = 2e−2t
1
e4t =
3
1
4t = ln
3
1 1
t = ln = −0.25 ln 3 = −0.275.
4 3
4.8 SOLUTIONS
385
We need to check that dx/dt 6= 0 when t = −0.25 ln 3. Since dx/dt = 2e2t + 2e−2t is always positive, dx/dt
is never zero.
(ii) A vertical tangent occurs when dx/dt = 0 and dy/dt 6= 0. Since dx/dt = 2e2t + 2e−2t is always positive,
there is no vertical tangent.
(b) The chain rule gives
dy/dt
6e2t − 2e−2t
3e2t − e−2t
dy
=
= 2t
= 2t
.
−2t
dx
dx/dt
2e + 2e
e + e−2t
(c) As t → ∞, we have e−2t → 0. Thus,
lim
t→∞
3e2t − e−2t
3e2t
dy
= lim 2t
= lim 2t = 3.
t→∞ e
t→∞ e
dx
+ e−2t
As t → ∞, the fraction gets closer and closer to 3.
48. (a) Since y = f ′ (t) > 0 for all points on the curve, f is increasing.
(b) Since x = f (t) is increasing by part (a), the motion is to the right from P to Q.
(c) As t increases and the curve is traced from P to Q, the values of the derivative y = f ′ (t) decrease, so f is concave
down.
49. (a) Figure 4.135 shows the path and the clockwise direction of motion. (The curve is an ellipse.)
4
y
−5
5
x
−4
Figure 4.135
(b) At t = π/4, the position is given by
x(π/4) = 5 sin
π
2π
= 5 sin = 5 and
4
2
y(π/4) = 4 cos
2π
π
= 4 cos = 0.
4
2
Differentiating, we get x′ (t) = 10 cos(2t) and y ′ (t) = −8 sin(2t). At t = π/4, the velocity is given by
2π
π
2π
π
= 10 cos = 0 and y ′ (π/4) = −8 sin
= −8 sin = −8.
4
2
4
2
x′ (π/4) = 10 cos
(c) As t increases from 0 to 2π, the ellipse is traced out twice. Thus, the particle passes through the point (5, 0) twice.
(d) Since x′ (π/4) = 0 and y ′ (π/4) = −8, when t = π/4, the particle is moving in the negative y-direction, parallel to
the y-axis.
(e) At time t,
p
p
Speed = (x′ (t))2 + (y ′ (t))2 = (10 cos(2t))2 + (−8 sin(2t))2 .
When t = π,
Speed =
p
(10 cos(2π))2 + (−8 sin(2π))2 =
p
(10 · 1)2 + (−8 · 0)2 = 10.
50. (a) Substituting α = 36◦ = π/5 and v0 = 60 into x(t) = (v0 cos α)t and y(t) = (v0 sin α)t − 21 gt2 , we get
π
π
t and y(t) = 60 sin
5
5
(b) Figure 4.136 shows the path and the direction of motion.
x(t) = 60 cos
t−
1
π
(32)t2 = 60 sin
2
5
y
x
Figure 4.136
t − 16t2 .
386
Chapter Four /SOLUTIONS
(c) When the football hits the ground, y(t) = 0, so
60 sin
π
5
t − 16t2 = 0
π
− 4t = 0
5
15 sin(π/5)
= 2.204 seconds.
t = 0 or t =
4
The ball hits the ground in approximately 2.204 seconds. The ball’s distance from the spot where it was kicked is
x(2.204) = 106.994 feet.
(d) At its highest point, the football is moving neither upward nor downward, so y ′ (t) is zero. To find the time when the
football reaches its maximum height, we set y ′ (t) = 0, giving
4t 15 sin
π
− 32t = 0
5
60 sin(π/5)
t=
= 1.102 seconds.
32
This makes sense since this is half the time it took the football to reach the ground. The maximum height is
y(1.102) = 19.434 feet.Thus the football reaches
19.434 feet.
(e) Since x(t) = 60 cos π5 t and y(t) = 60 sin π5 t − 16t2 , we have
y ′ (t) = 60 sin
x′ (t) = 60 cos
π
5
y ′ (t) = 60 sin
and
π
− 32t.
5
Thus,
Speed =
At t = 1,
p
(x′ (t))2 + (y ′ (t))2 =
Speed =
r
60 cos
π
5
2
r
60 cos
+ 60 sin
π
5
2
π
− 32t
5
+ 60 sin
2
π
− 32
5
2
.
= 48.651 feet/sec.
51. (a) To determine if the particles collide, we check whether they are ever at the same point at the same time. We first set
the two x-coordinates equal to each other:
4t − 4 = 3t
t = 4.
When t = 4, both x-coordinates are 12. Now we check whether the y-coordinates are also equal at t = 4:
yA (4) = 2 · 4 − 5 = 3
yB (4) = 42 − 2 · 4 − 1 = 7.
Thus, the particles do not collide since they are not at the same point at the same time.
(b) For the particles to collide, we need both x- and y-coordinates to be equal. Since the x-coordinates are equal at t = 4,
we find the k value making yA (4) = yB (4).
Substituting t = 4 into yA (t) = 2t − k and yB (t) = t2 − 2t − 1, we have
8 − k = 16 − 8 − 1
k = 1.
(c) To find the speed of the particles, we differentiate.
For particle A,
x(t) = 4t − 4, so x′ (t) = 4, and x′ (4) = 4
y(t) = 2t − 1, so y ′ (t) = 2, and y ′ (4) = 2
Speed A =
p
p
p
p
(x′ (t))2 + (y ′ (t))2 =
For particle B,
x(t) = 3t, so x′ (t) = 3, and x′ (4) = 3
y(t) = t2 − 2t − 1, so y ′ (t) = 2t − 2, and y ′ (4) = 6
Speed B = (x′ (t))2 + (y ′ (t))2 =
Thus, when t = 4, particle B is moving faster.
42 + 22 =
32 + 62 =
√
20.
√
45.
4.8 SOLUTIONS
387
52. (a) Since x = t3 + t and y = t2 , we have
w=
dy/dt
2t
dy
=
= 2
.
dx
dx/dt
3t + 1
Differentiating w with respect to t, we get
(3t2 + 1)2 − (2t)(6t)
−6t2 + 2
dw
=
=
,
2
2
dt
(3t + 1)
(3t2 + 1)2
so
dw/dt
dw
−6t2 + 2
d2 y
=
=
=
.
2
dx
dx
dx/dt
(3t2 + 1)3
(b) When t = 1, we have d2 y/dx2 = −1/16 < 0, so the curve is concave down.
53. (a) The x and y-coordinates of the point on the graph when t = π/3 are given by
x=3·
π
=π
3
and
y = cos
2π
3
1
=− .
2
Thus when t = π/3, the particle is at the point (π, −1/2).
To find the slope, we find dy/dx
dy
dy/dt
−2 sin(2t)
=
=
.
dx
dx/dt
3
When t = π/3,
√
−2 sin(2π/3)
3
dy
=
=−
.
dx
3
3
The equation of the tangent line when t = π/3 is:
√
1
3
(x − π).
y+ =−
2
3
(b) To find the smallest positive value of t for which the y-coordinate is a local maximum, we set dy/dt = 0. We have
dy
= −2 sin(2t) = 0
dt
2t = π or 2t = 2π
π
or t = π.
t=
2
There is a minimum of y = cos(2t) at t = π/2, and a maximum at t = π.
(c) To find d2 y/dx2 when t = 2, we use the formula:
dw/dt
d2 y
=
dx2
dx/dt
where
w=
dy
.
dx
Since w = −2 sin(2t)/3 from part (a), we have
−4 cos(2t)/3
d2 y
=
.
dx2
3
When t = 2, we have
−4 cos(4)/3
d2 y
=
= 0.291.
dx2
3
Since the second derivative is positive, the graph is concave up when t = 2.
54. (a) We differentiate for both x and y in terms of t, giving us:
dy/dt
2e2t + 6et
dy
=
=
= 2et + 6.
dx
dx/dt
et
388
Chapter Four /SOLUTIONS
(b) To find d2 y/dx2 , we use the formula:
dw/dt
d2 y
=
dx2
dx/dt
w=
where
dy
dx
2et
d2 y
= t = 2.
2
dx
e
Since the second derivative is always positive, the graph is concave up everywhere.
(c) We are given y = e2t + 6et + 9. We can factor this to y = (et + 3)2 . Since x = et + 3, we can substitute x for
et + 3. Thus, y = x2 for x > 3.
(d) Since y = x2 , dy/dx = 2x and d2 y/dx2 = 2.
From part (a), we have dy/dx = 2et + 6. We can factor this to dy/dx = 2(et + 3) = 2x.
Part (b) tells us that d2 y/dx2 = 2, which is what we have just determined.
Our graph is a parabola that is concave up everywhere.
55. (a) The particle touches the x-axis when y = 0. Since y = cos(2t) = 0 for the first time when 2t = π/2, we have
t = π/4. To find the speed of the particle at that time, we use the formula
Speed =
When t = π/4,
Speed =
r
dx
dt
2
+
p
dy
dt
2
=
p
(cos t)2 + (−2 sin(2t))2 .
(cos(π/4))2 + (−2 sin(π/2))2 =
(b) The particle is at rest when its speed is zero. Since
cos t = 0
p
q√
( 2/2)2 + (−2 · 1)2 =
p
9/2.
(cos t)2 + (−2 sin(2t))2 ≥ 0, the speed is zero when
and
− 2 sin(2t) = 0.
Now cos t = 0 when t = π/2 or t = 3π/2. Since −2 sin(2t) = −4 sin t cos t, we see that this expression also
equals zero when t = π/2 or t = 3π/2.
(c) We need to find d2 y/dx2 . First, we must determine dy/dx. We know
dy/dt
−2 sin 2t
−4 sin t cos t
dy
=
=
=
= −4 sin t.
dx
dx/dt
cos t
cos t
Since dy/dx = −4 sin t, we can now use the formula:
dw/dt
d2 y
=
dx2
dx/dt
where
w=
dy
dx
−4 cos t
d2 y
=
= −4.
dx2
cos t
56. Let
Since d2 y/dx2 is always negative, our graph is concave down everywhere.
Using the identity y = cos(2t) = 1 − 2 sin2 t, we can eliminate the parameter and write the original equation
as y = 1 − 2x2 , which is a parabola that is concave down everywhere.
w=
We want to find
dy/dt
dy
=
.
dx
dx/dt
dw
dw/dt
d2 y
=
=
dt.
dx2
dx
dx
To find dw/dt, we use the quotient rule:
(dx/dt)(d2 y/dt2 ) − (dy/dt)(d2 x/dt2 )
dw
=
.
dt
(dx/dt)2
We then divide this by dx/dt again to get the required formula, since
dw/dt
dw
d2 y
=
=
.
dx2
dx
dx/dt
4.8 SOLUTIONS
57. For 0 ≤ t ≤ 2π, we get Figure 4.137.
y
x
−1
1
Figure 4.137
58. For 0 ≤ t ≤ 2π, we get Figure 4.138.
y
1
x
1
Figure 4.138
59. For 0 ≤ t ≤ 2π, we get Figure 4.139.
y
x
1
Figure 4.139
389
390
Chapter Four /SOLUTIONS
60. This curve never closes on itself. The plot for 0 ≤ t ≤ 8π is in Figure 4.140.
y
1
x
−1
1
−1
Figure 4.140
61. (a) To find the equations of the moon’s motion relative to the star, you must first calculate the equation of the planet’s
motion relative to the star, and then the moon’s motion relative to the planet, and then add the two together.
The distance from the planet to the star is R, and the time to make one revolution is one unit, so the parametric
equations for the planet relative to the star are x = R cos t, y = R sin t.
The distance from the moon to the planet is 1, and the time to make one revolution is twelve units, therefore, the
parametric equations for the moon relative to the planet are x = cos 12t, y = sin 12t.
Adding these together, we get:
x = R cos t + cos 12t,
y = R sin t + sin 12t.
(b) For the moon to stop completely at time t, the velocity of the moon must be equal to zero. Therefore,
dx
= −R sin t − 12 sin 12t = 0,
dt
dy
= R cos t + 12 cos 12t = 0.
dt
There are many possible values to choose for R and t that make both of these equations equal to zero. We choose
t = π, and R = 12.
(c) The graph with R = 12 is shown in Figure 4.141.
Figure 4.141
Strengthen Your Understanding
62. The parameterization does give the required line segment but traversed in the wrong direction.
SOLUTIONS to Review Problems for Chapter Four
391
63. The given parametric equations give a circle of radius 2 but centered at (0, 0). We need x = 2 cos πt, y = 1 + 2 sin πt,
0 ≤ t ≤ 2. There are many other possibilities.
64. One possible choice is x = 2 cos t, y = 2 sin t, 0 ≤ t ≤
π
.
2
There are many other possibilities.
65. One possible choice is x = t, y = 2t, 0 ≤ t ≤ 1. There are many other possibilities.
66. False. If the particle tracing out the curve comes to a complete stop, it can then head off in a completely new direction.
For example, the curve given parametrically by x = t3 and y = t2 is the same as the graph of y = x2/3 which has a cusp
at x = 0.
67. False. The slope is given by
dy/dt
2t cos(t2 )
cos(t2 )
dy
=
=
=−
.
2
dx
dx/dt
−2t sin(t )
sin(t2 )
Solutions for Chapter 4 Review
Exercises
1. See Figure 4.142.
Local and global max
50
Local max
f (x)
40
Local max
30
20
Local min
10
Local and global min
x
1
2
3
4
5
6
Figure 4.142
2. See Figure 4.143.
Local and global max
8
Local max
f (x)
6
Critical point
(not max or min)
4
2
Local and global min
x
1
2
3
4
5
Figure 4.143
3. (a) We wish to investigate the behavior of f (x) = x3 − 3x2 on the interval −1 ≤ x ≤ 3. We find:
f ′ (x) = 3x2 − 6x = 3x(x − 2)
f ′′ (x) = 6x − 6 = 6(x − 1)
(b) The critical points of f are x = 2 and x = 0 since f ′ (x) = 0 at those points. Using the second derivative test, we
find that x = 0 is a local maximum since f ′ (0) = 0 and f ′′ (0) = −6 < 0, and that x = 2 is a local minimum since
f ′ (2) = 0 and f ′′ (2) = 6 > 0.
(c) There is an inflection point at x = 1 since f ′′ changes sign at x = 1.
392
Chapter Four /SOLUTIONS
(d) At the critical points, f (0) = 0 and f (2) = −4.
At the endpoints: f (−1) = −4, f (3) = 0.
So the global maxima are f (0) = 0 and f (3) = 0, while the global minima are f (−1) = −4 and f (2) = −4.
(e) See Figure 4.144.
| incr. | decreasing | incr. |
| concave down | concave up |
−1
1
3
−4
Figure 4.144
First we find f ′ and f ′′ ; f ′ (x) = 1 + cos x and f ′′ (x) = − sin x.
The critical point of f is x = π, since f ′ (π) = 0.
Since f ′′ changes sign at x = π, it means that x = π is an inflection point.
Evaluating f at the critical point and endpoints, we find f (0) = 0, f (π) = π, f (2π) = 2π,. Therefore, the global
maximum is f (2π) = 2π, and the global minimum is f (0) = 0. Note that x = π is not a local maximum or
minimum of f , and that the second derivative test is inconclusive here.
(e) See Figure 4.145.
4. (a)
(b)
(c)
(d)
π
|
increasing
| concave down | concave up
2π
|
|
Figure 4.145
5. (a) First we find f ′ and f ′′ :
f ′ (x) = −e−x sin x + e−x cos x
f ′′ (x) = e−x sin x − e−x cos x
−e−x cos x − e−x sin x
= −2e−x cos x
(b) The critical points are x = π/4, 5π/4, since f ′ (x) = 0 here.
(c) The inflection points are x = π/2, 3π/2, since f ′′ changes sign at these points.
√
(d) At the endpoints,
f (0) = 0, f (2π) = 0. So we have f (π/4) = (e−π/4 )( 2/2) as the global maximum; f (5π/4) =
√
−e−5π/4 ( 2/2) as the global minimum.
(e) See Figure 4.146.
SOLUTIONS to Review Problems for Chapter Four
π
2
| conc. down |
| incr. |
3π
2
π
x
| conc. down |
|
increasing
concave up
|
decreasing
2π
393
Figure 4.146
6. (a) We first find f ′ and f ′′ :
2 5
1 2
1 5
f ′ (x) = − x− 3 + x− 3 = x− 3 (x − 2)
3
3
3
10 − 83
2 5
2 8
f ′′ (x) =
x
− x− 3 = − x− 3 (x − 5)
9
9
9
(b) Critical point: x = 2.
(c) There are no inflection points, since f ′′ does not change sign on the interval 1.2 ≤ x ≤ 3.5.
(d) At the endpoints, f (1.2) ≈ 1.94821 and f (3.5) ≈ 1.95209. So, the global minimum is f (2) ≈ 1.88988 and the
global maximum is f (3.5) ≈ 1.95209.
(e) See Figure 4.147.
2
1.9
1.8
1.2
|
|
2
decr
|
3.5
x
incr
|
concave up
|
Figure 4.147
7. The polynomial f (x) behaves like 2x3 as x goes to ∞. Therefore, lim f (x) = ∞ and lim f (x) = −∞.
x→∞
x→−∞
We have f ′ (x) = 6x2 − 18x + 12 = 6(x − 2)(x − 1), which is zero when x = 1 or x = 2.
Also, f ′′ (x) = 12x − 18 = 6(2x − 3), which is zero when x = 3/2. For x < 3/2, f ′′ (x) < 0; for x > 3/2,
f ′′ (x) > 0. Thus x = 3/2 is an inflection point.
The critical points are x = 1 and x = 2, and f (1) = 6, f (2) = 5. By the second derivative test, f ′′ (1) = −6 < 0,
so x = 1 is a local maximum; f ′′ (2) = 6 > 0, so x = 2 is a local minimum.
Now we can draw the diagrams below.
y′ > 0
y′ < 0
y′ > 0
increasing x = 1 decreasing x = 2 increasing
y ′′ < 0
concave down
y ′′ > 0
x = 3/2
concave up
394
Chapter Four /SOLUTIONS
The graph of f (x) = 2x3 − 9x2 + 12x + 1 is shown in Figure 4.148. It has no global maximum or minimum.
f (x) = 2x3 − 9x2 + 12x + 1
6
5
x
1
2
Figure 4.148
8. If we divide the denominator and numerator of f (x) by x2 we have
lim
x→±∞
4
4x2
= lim
=4
x2 + 1 x→±∞ 1 + x12
since
lim
x→±∞
1
= 0.
x2
Using the quotient rule we get
f ′ (x) =
(x2 + 1)8x − 4x2 (2x)
8x
= 2
,
(x2 + 1)2
(x + 1)2
which is zero when x = 0, positive when x > 0, and negative when x < 0. Thus f (x) has a local minimum when x = 0,
with f (0) = 0.
Because f ′ (x) = 8x/(x2 + 1)2 , the quotient rule implies that
f ′′ (x) =
=
(x2 + 1)2 8 − 8x[2(x2 + 1)2x]
(x2 + 1)4
8(1 − 3x2 )
8x2 + 8 − 32x2
=
.
(x2 + 1)3
(x2 + 1)3
The denominator is always positive, so f ′′ (x) = 0 when x = ±
p
p
negative when x > 1/3 or x < − 1/3. This gives the diagram
y′ < 0
decreasing
x=−
and the graph of f looks Figure 4.149.
1/3, positive when −
1/3
1/3 < x <
p
1/3, and
y′ > 0
y ′′ > 0
concave up
p
p
increasing
x=0
y ′′ < 0
concave down
p
x=
4
y ′′ < 0
concave down
p
1/3
f (x) =
4x2
x2 +1
x
Figure 4.149
with inflection points x = ±
equals 4).
p
1/3, a global minimum at x = 0, and no local or global maxima (since f (x) never
SOLUTIONS to Review Problems for Chapter Four
395
9. As x → −∞, e−x → ∞, so xe−x → −∞. Thus limx→−∞ xe−x = −∞.
As x → ∞, exx → 0, since ex grows much more quickly than x. Thus limx→∞ xe−x = 0.
Using the product rule,
f ′ (x) = e−x − xe−x = (1 − x)e−x ,
which is zero when x = 1, negative when x > 1, and positive when x < 1. Thus f (1) = 1/e1 = 1/e is a local
maximum.
Again, using the product rule,
f ′′ (x) = −e−x − e−x + xe−x
= xe−x − 2e−x
= (x − 2)e−x ,
which is zero when x = 2, positive when x > 2, and negative when x < 2, giving an inflection point at (2,
above, we have the following diagram:
y′ > 0
2
).
e2
With the
y′ < 0
increasing
decreasing
x=1
y ′′ < 0
y ′′ > 0
concave down
x=2
concave up
The graph of f is shown in Figure 4.150.
f (x) = xe−x
x
1
2
Figure 4.150
and f (x) has one global maximum at 1/e and no local or global minima.
10. Since f (x) = e−x sin x is continuous and the interval 0 ≤ x ≤ 2π is closed, there must be a global maximum and
minimum. The possible candidates are critical points in the interval and endpoints. Since there are no points where f ′ is
undefined, we solve f ′ (x) = 0 to find all critical points:
f ′ (x) = −e−x sin x + e−x cos x = e−x (− sin x + cos x) = 0.
Since e−x 6= 0, the critical points are when sin x = cos x; the only solutions in the given interval are x = π/4 and
x = 5π/4. We then compare the values of f at the critical points and the endpoints:
√
√
2
− 2
−5π/4
−π/4
= 0.322, f (5π/4) = e
= −0.0139, f (2π) = 0.
f (0) = 0, f (π/4) = e
2
2
Thus the global maximum is 0.322 at x = π/4, and the global minimum is −0.0139 at x = 5π/4.
396
Chapter Four /SOLUTIONS
11. Since f (x) = ex + cos x is continuous on the closed interval 0 ≤ x ≤ π, there must be a global maximum and minimum.
The possible candidates are critical points in the interval and endpoints. Since there are no points in the interval where
f ′ (x) is undefined, we solve f ′ (x) = 0 to find the critical points:
f ′ (x) = ex − sin x = 0.
Since ex > 1 for all x > 0 and sin x ≤ 1 for all x, the only possibility is x = 0, but e0 − sin 0 = 1. Thus there are no
critical points in the interval. We then compare the values of f at the endpoints:
f (0) = 2,
f (π) = eπ + cos(π) = 22.141.
Thus, the global maximum is 22.141 at x = π, and the global minimum is 2 at x = 0.
12. Since f (x) = x2 + 2x + 1 is continuous and the interval 0 ≤ x ≤ 3 is closed, there must be a global maximum
and minimum. The candidates are critical points in the interval and endpoints. Since there are no points where f ′ (x) is
undefined, we solve f ′ (x) = 0 to find all the critical points:
f ′ (x) = 2x + 2 = 0,
so the only critical point, x = −1, is not in the interval 0 ≤ x ≤ 3. At the endpoints, we have
f (0) = 1,
f (3) = 16.
Thus, the global maximum is 16 at x = 3 , and the global minimum is 1 at x = 0.
2
13. Since f (x) = e−x is continuous and the interval 0 ≤ x ≤ 10 is closed, there must be a global maximum and minimum.
The candidates are critical points in the interval and endpoints. Since there are no points where f ′ (x) is undefined, we
solve f ′ (x) = 0 to find all the critical points:
2
f ′ (x) = −2xe−x = 0,
so the only critical point, x = 0, is at an endpoint of the interval. At the endpoints, we have
f (0) = 1,
f (10) = e−100 ≈ 0.
Thus, the global maximum is 1 at x = 0 , and the global minimum is near 0 at x = 10.
14. We rewrite h(z) as h(z) = z −1 + 4z 2 .
Differentiating gives
h′ (z) = −z −2 + 8z,
so the critical points satisfy
−z −2 + 8z = 0
z −2 = 8z
8z 3 = 1
1
z3 =
8
1
z= .
2
Since h′ is negative for 0 < z < 1/2 and h′ is positive for z > 1/2, there is a local minimum at z = 1/2.
Since h(z) → ∞ as z → 0+ and as z → ∞, the local minimum at z = 1/2 is a global minimum; there is no global
maximum. See Figure 4.151. Thus, the global minimum is h(1/2) = 3.
h(z) =
3
1
z
+ 4z 2
z
1/2
Figure 4.151
SOLUTIONS to Review Problems for Chapter Four
397
15. Since g(t) is always decreasing for t ≥ 0, we expect it to a global maximum at t = 0 but no global minimum. At t = 0,
we have g(0) = 1, and as t → ∞, we have g(t) → 0.
Alternatively, rewriting as g(t) = (t3 + 1)−1 and differentiating using the chain rule gives
g ′ (t) = −(t3 + 1)−2 · 3t2 .
Since 3t2 = 0 when t = 0, there is a critical point at t = 0, and g decreases for all t > 0. See Figure 4.152.
1
g(t) =
1
t3 + 1
x
Figure 4.152
16. We begin by rewriting f (x):
f (x) =
1
= ((x − 1)2 + 2)−1 = (x2 − 2x + 3)−1 .
(x − 1)2 + 2
Differentiating using the chain rule gives
f ′ (x) = −(x2 − 2x + 3)−2 (2x − 2) =
2 − 2x
,
(x2 − 2x + 3)2
so the critical points satisfy
2 − 2x
=0
(x2 − 2x + 3)2
2 − 2x = 0
2x = 2
x = 1.
Since f ′ is positive for x < 1 and f ′ is negative for x > 1, there is a local maximum at x = 1.
Since f (x) → 0 as x → ∞ and as x → −∞, the local maximum at x = 1 is a global maximum; there is no global
minimum. See Figure 4.153. Thus, the global maximum is f (1) = 1/2.
1/2
f (x) =
1
(x − 1)2 + 2
x
1
Figure 4.153
17. lim f (x) = +∞, and lim f (x) = −∞.
x→∞
x→−∞
There are no asymptotes.
f ′ (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1). Critical points are x = −3, x = 1.
f ′′ (x) = 6(x + 1).
x
f′
+
f ′′
−
f
ր⌢
−3
0
−
−
−
ց⌢
−1
−
0
1
−
+
ց⌣
0
+
+
+
ր⌣
398
Chapter Four /SOLUTIONS
Thus, x = −1 is an inflection point. f (−3) = 12 is a local maximum; f (1) = −20 is a local minimum. There are
no global maxima or minima. See Figure 4.154.
f (x) = x3 + 3x2 − 9x − 15
12
x
−3
1
−20
Figure 4.154
18.
lim f (x) = +∞, and lim f (x) = −∞.
x→+∞
x→−∞
There are no asymptotes.
f ′ (x) = 5x4 − 45x2 = 5x2 (x2 − 9) = 5x2 (x + 3)(x − 3).
The critical points are x = 0, x = ±3. f ′ changes sign at 3 and −3√
but not at 0.
f ′′ (x) = 20x3 − 90x = 10x(2x2 − 9). f ′′√changes sign at 0, ±3/ 2.
So, inflection points are at x = 0, x = ±3/ 2.
x
f′
+
f ′′
−
f
ր⌢
−3
−
0
−
√
−3/ 2
−
0
−
0
+
ց⌢
√
3/ 2
0
0
ց⌣
−
−
3
−
0
+
ց⌢
ց⌣
0
+
+
+
ր⌣
Thus, f (−3) is a local maximum; f (3) is a local minimum. There are no global maxima or minima.
172
10
x
−3
3
−152
19.
f (x) = x5 − 15x3 + 10
lim f (x) = +∞, and lim f (x) = +∞.
x→+∞
x→0+
Hence, x = 0 is a vertical asymptote.
2
x−2
f ′ (x) = 1 − =
, so x = 2 is the only critical point.
x
x
2
f ′′ (x) = 2 , which can never be zero. So there are no inflection points.
x
x
f′
f ′′
f
f (x) = x − ln x
2
−
+
ց⌣
0
+
+
+
ր⌣
Thus, f (2) is a local and global minimum.
x
2
SOLUTIONS to Review Problems for Chapter Four
20. Since lim f (x) = lim f (x) = 0, y = 0 is a horizontal asymptote.
x→−∞
x→+∞
−x2
′
. So, x = 0 is the only critical point.
f (x) = −2xe
√
2 √
2
2
−x2
f ′′ (x) = −2(e−x
) = 2e−x (2x2 − 1) = 2e−x ( 2x − 1)( 2x + 1).
√ + x(−2x)e
Thus, x = ±1/ 2 are inflection points.
Table 4.1
√
−1/ 2
x
f′
f ′′
f
+
+
+
0
ր⌣
√
1/ 2
0
+
0
−
−
ր⌢
−
−
−
−
0
+
ց⌢
ց⌣
Thus, f (0) = 1 is a local and global maximum.
f (x) = e−x
2
x
21.
lim f (x) = +∞, lim f (x) = 0.
x→+∞
x→−∞
y = 0 is the horizontal asymptote.
f ′ (x) = 2xe5x + 5x2 e5x = xe5x (5x + 2).
Thus, x = − 25 and x = 0 are the critical points.
f ′′ (x) = 2e5x + 2xe5x · 5 + 10xe5x + 25x2 e5x
= e5x (25x2 + 20x + 2).
So, x =
−2 ±
5
√
2
are inflection points.
−2−
5
x
f′
f ′′
f
+
+
+
0
ր⌣
√
2
+
−
ր⌢
−2+
5
− 25
−
0
−
−
√
2
0
−
−
0
ց⌢
+
ց⌣
f (x) = x2 e5x
x
−0.4
So, f (− 25 ) is a local maximum; f (0) is a local and global minimum.
0
+
+
+
ր⌣
399
400
22.
Chapter Four /SOLUTIONS
lim f (x) = lim f (x) = 1.
x→+∞
x→−∞
Thus, y = 1 is a horizontal asymptote. Since x2 + 1 is never 0, there are no vertical asymptotes.
f ′ (x) =
2x(x2 + 1) − x2 (2x)
2x
= 2
.
(x2 + 1)2
(x + 1)2
So, x = 0 is the only critical point.
2(x2 + 1)2 − 2x · 2(x2 + 1) · 2x
(x2 + 1)4
f ′′ (x) =
=
2(x2 + 1 − 4x2 )
(x2 + 1)3
=
2(1 − 3x2 )
.
(x2 + 1)3
So, x = ± √13 are inflection points.
Table 4.2
−1
√
3
x
f′
f ′′
f
−
−
−
√1
3
0
−
0
+
ց⌢
0
+
+
+
+
+
0
−
ց⌣
ր⌣
ր⌢
Thus, f (0) = 0 is a local and global minimum. A graph of f (x) can be found in Figure 4.155.
y=1
f (x) =
x2
x2 +1
x
Figure 4.155
23. For 0 ≤ r ≤ a, the speed at which air is expelled is given by
v(r) = k(a − r)r 2 = kar 2 − kr 3 .
Thus, the derivative is defined for all r and given by
v ′ (r) = 2kar − 3kr 2 = 2kr a −
3
r .
2
The derivative is zero if r = 23 a or r = 0. These are the critical points of v. To decide if the critical points give global
maxima or minima, we evaluate v at the critical point:
v
2
2
a = k a− a
3
3
2
a 4a
=k
3
9
4ka3
,
=
27
2
a
3
2
and we evaluate v at the endpoints:
v(0) = v(a) = 0.
Thus, v has a global maximum at r =
2
a.
3
The global minimum of v = 0 occurs at both endpoints r = 0 and r = a.
SOLUTIONS to Review Problems for Chapter Four
401
24. The slope of the curve, dy/dt, is given by
−2
−2
dy
= −50 1 + 6e−2t
−12e−2t = 600e−2t 1 + 6e−2t
.
dt
If the slope has a maximum, it occurs at a critical point of dy/dt or at the endpoint t = 0. We have
−3
−2
d2 y
−12e−2t
− 1200e−2t 1 + 6e−2t
= 600 −2e−2t 1 + 6e−2t
2
dt
1200e−2t
6e−2t − 1 .
=
−2t
3
(1 + 6e )
At a critical point of dy/dt, we have d2 y/dt2 = 0, so
6e−2t − 1 = 0
1
t = ln 6 = 0.896.
2
Since
dy
dt
=
t=0
600
= 12.25
49
and
dy
dt
= 25,
t=(1/2) ln 6
the maximum slope occurs at t = 12 ln 6.
We see in Figure 4.156 that the slope increases as t increases from 0 and tends to 0 as t → ∞, so the only critical
1
1
point of slope,
t = 2 ln 6, is a local and global maximum for the slope. At t = 2 ln 6, we have y = 25. The point
1
ln 6, 25 is the point where slope is maximum.
2
y
50
y=
50
1 + 6e−2t
25
1
2
t
ln 6
Figure 4.156
25. We have f ′ (x) = (a2 − x2 )/(x2 + a2 )2 , so that the critical points occur when a2 − x2 = 0, that is, when x = ±a. We
see that as a increases the x-value of the extrema moves away the origin.
Looking at the figure we see that A’s local maximum has the smallest x-value, while C’s has the largest x-value.
Thus, A corresponds to a = 1, B to a = 2, and C to a = 3.
26. We have f ′ (x) = ae−ax , so that f ′ (0) = a. We see that as a increases the slopes of the curves at the origin become more
and more positive. Thus, C corresponds to a = 1, B to a = 2, and A to a = 3.
27. (a) We set the derivative equal to zero and solve for x to find critical points:
f ′ (x) = 4x3 − 4ax = 0
4x(x2 − a) = 0.
We see that there are three critical points:
Critical points:
x = 0,
x=
√
a,
√
x = − a.
To find possible inflection points, we set the second derivative equal to zero and solve for x:
f ′′ (x) = 12x2 − 4a = 0.
There are two possible inflection points:
Possible inflection points:
x=
q
a
,
3
x=−
q
a
.
3
402
Chapter Four /SOLUTIONS
To see if these are inflection points, we determine whether concavity changes by evaluating f ′′ at values on either
side of each of the potential inflection points. We see that
q
f ′′ (−2
so f is concave up to the left of x = −
so f is concave down between x = −
p
a/3. Also,
f ′′ (0) = −4a < 0,
p
a/3 and x =
q
f ′′ (2
a
a
) = 12(4 ) − 4a = 16a − 4a = 12a > 0,
3
3
p
a/3. Finally, we see that
a
a
) = 12(4 ) − 4a = 16a − 4a = 12a > 0,
3
3
p
p
so f is concave up to the right of x = a/3. Since f (x) changes concavity at x = a/3 and x = −
points are inflection points.
√
(b) The only positive critical point is at x = a, so to have a critical point at x = 2, we substitute:
√
x= a
√
2= a
p
a/3, both
a = 4.
Since the critical point is at the point (2, 5), we have
f (2) = 5
4
2
2 − 2(4)2 + b = 5
16 − 32 + b = 5
b = 21.
The function is f (x) = x4 − 8x2 + 21.
p
p
(c) We have seen that a = 4, so the inflection points are at x = 4/3 and x = − 4/3.
28. (a) The function f (x) is defined for x ≥ 0.
We set the derivative equal to zero and solve for x to find critical points:
f ′ (x) = 1 −
1 −1/2
ax
=0
2
a
1− √ = 0
2 x
√
2 x=a
a2
.
x=
4
Notice that f ′ is undefined at x = 0 so there are two critical points: x = 0 and x = a2 /4.
(b) We want the critical point x = a2 /4 to occur at x = 5, so we have:
a2
4
20 = a2
√
a = ± 20.
5=
Since a is positive, we use the positive square root. The second derivative,
f ′′ (x) =
1√
1 −3/2
ax
=
20x−3/2
4
4
is positive for all x > 0, so the function is concave up and x = 5 gives a local minimum. See Figure 4.157.
SOLUTIONS to Review Problems for Chapter Four
403
y
5
x
5
Figure 4.157
29. The domain is all real numbers except x = b. The function is undefined at x = b and has a vertical asymptote there. To
find the critical points, we set the derivative equal to zero and solve for x. Using the quotient rule, we have:
f ′ (x) =
(x − b)2ax − (ax2 )1
=0
(x − b)2
2ax2 − 2abx − ax2
=0
(x − b)2
ax2 − 2abx
= 0.
(x − b)2
The first derivative is equal to zero if
ax2 − 2abx = 0
ax(x − 2b) = 0
x = 0 or
x = 2b.
There are two critical points: at x = 0 and x = 2b.
30. Let the numbers be x and y. Then
Average =
x+y
= 180,
2
so
y = 360 − x.
Since both numbers are nonnegative, we restrict to 0 ≤ x ≤ 360.
The product is
P = xy = x(360 − x) = 360x − x2 .
Differentiating to find the maximum,
dP
= 360 − 2x = 0
dx
360
= 180.
x=
2
So there is a critical point at x = 180; the end points are at x = 0, 360.
Evaluating gives
At x = 0, we have P = 0.
At x = 180, we have P = 180(360 − 180) = 32,400.
At x = 360, we have P = 360(360 − 360) = 0.
Thus, the maximum value is 32,400.
31. Let the numbers be x, y, z and let y = 2x. Then
xyz = 2x2 z = 192,
so z =
192
96
= 2.
2x2
x
Since all the numbers are positive, we restrict to x > 0.
The sum is
96
96
S = x + y + z = x + 2x + 2 = 3x + 2 .
x
x
404
Chapter Four /SOLUTIONS
Differentiating to find the minimum,
96
dS
= 3−2 3 = 0
dx
x
192
3= 3
x
192
3
= 64
x =
3
so x = 4.
There is only one critical point at x = 4. We find
192
d2 S
=3 4 .
dx2
x
Since d2 S/dx2 > 0 for all x, there is a local minimum at x = 4. The derivative dS/dx is negative for 0 < x < 4 and
dS/dx is positive for x > 4. Thus, x = 4 gives the global minimum for x > 0.
The minimum value of the sum obtained from the three numbers 4, 8, and 6 is
S =3·4+
96
= 18.
42
32. Let x be the larger number and y be the smaller number. Then x − y = 24, so y = x − 24.
Since both numbers are 100 or larger, we restrict to x ≥ 124.
The product is
P = xy = x(x − 24) = x2 − 24x.
Differentiating to find the minimum,
dP
= 2x − 24 = 0
dx
24
= 12.
x=
2
So there is a critical point at x = 12. There is an end point at x = 124; the domain x ≥ 124 has no critical points.
Since the derivative, dP/dx, is positive for all x greater than 12, the minimum value of P occurs at the left end of
the domain, x = 124.
The minimum value of the product is
P = 124(124 − 24) = 12,400.
33. (a) Total cost, in millions of dollars, C(q) = 3 + 0.4q.
(b) Revenue, in millions of dollars, R(q) = 0.5q.
(c) Profit, in millions of dollars, π(q) = R(q) − C(q) = 0.5q − (3 + 0.4q) = 0.1q − 3.
34. The square is S = x2 . Differentiating with respect to time gives
dx
dS
= 2x .
dt
dt
We are interested in the instant when x = 10 and dS/dt = 5, giving
5 = 2 · 10
so
35. We have
If x = 5, then
dx
,
dt
dx
5
1
=
= unit per second.
dt
20
4
dx
dM
= (3x2 + 0.4x3 ) .
dt
dt
dM
= [3(52 ) + 0.4(53 )](0.02) = 2.5 gm/hr.
dt
SOLUTIONS to Review Problems for Chapter Four
36. We have
405
dθ
dA
= (3 + 4 cos θ + cos(2θ)) .
dt
dt
So
dA
dt
= 3 + 4 cos
θ=π/2
π
2
+ cos π 0.3 = 0.6 cm2 /min.
37. We see from the parametric equations that the particle moves along a line. It suffices to plot two points: at t = 0, the
particle is at point (4, 1), and at t = 1, the particle is at point (2, 5). Since x decreases as t increases, the motion is right
to left and the curve is shown in Figure 4.158.
Alternately, we can solve the first equation for t, giving t = −(x − 4)/2, and substitute this into the second equation
to get
−(x − 4)
+ 1 = −2x + 9.
y=4
2
The line is y = −2x + 9.
y
2 t=0
y
6
4
2
−2
t=0
2
x
x
−2
2
4
6
−2
−2
Figure 4.158
Figure 4.159
38. The graph is a circle centered at the origin with radius 2. The equation is
x2 + y 2 = (2 sin t)2 + (2 cos t)2 = 4.
The particle is at the point (0, 1) when t = 0, and motion is clockwise. See Figure 4.159.
Problems
39. (a) The function f is a local maximum where f ′ (x) = 0 and f ′ > 0 to the left, f ′ < 0 to the right. This occurs at the
point x3 .
(b) The function f is a local minimum where f ′ (x) = 0 and f ′ < 0 to the left, f ′ > 0 to the right. This occurs at the
points x1 and x5 .
(c) The graph of f is climbing fastest where f ′ is a maximum, which is at the point x2 .
(d) The graph of f is falling most steeply where f ′ is the most negative, which is at the point 0.
40. The function f has critical points at x = 1, x = 3, x = 5.
By the first derivative test, since f ′ is positive to the left of x = 1 and negative to the right, x = 1 is a local maximum.
Since f ′ is negative to the left of x = 3 and positive to the right, x = 3 is a local minimum.
Since f ′ does not change sign at x = 5, this point is neither a local maximum nor a local minimum.
41. The critical points of f occur where f ′ is zero. These two points are indicated in the figure below.
f ′ (x)
f has a
local min.
f has crit. pt.
Neither max or min
Note that the point labeled as a local minimum of f is not a critical point of f ′ .
406
Chapter Four /SOLUTIONS
42. From the first condition, we get that x = 2 is a local minimum for f . From the second condition, it follows that x = 4 is
an inflection point. A possible graph is shown in Figure 4.160.
Point of inflection
Local min
−2 −1
1
2
3
4
x
5
6
Figure 4.160
43. Since the x3 term has coefficient of 1, the cubic polynomial is of the form y = x3 + ax2 + bx + c. We now find a, b, and
c. Differentiating gives
dy
= 3x2 + 2ax + b.
dx
The derivative is 0 at local maxima and minima, so
dy
dx
= 3(1)2 + 2a(1) + b = 3 + 2a + b = 0
x=1
dy
dx
= 3(3)2 + 2a(3) + b = 27 + 6a + b = 0
x=3
Subtracting the first equation from the second and solving for a and b gives
24 + 4a = 0
so
a = −6
b = −3 − 2(−6) = 9.
Since the y-intercept is 5, the cubic is
y = x3 − 6x2 + 9x + 5.
Since the coefficient of x3 is positive, x = 1 is the maximum and x = 3 is the minimum. See Figure 4.161. To confirm
that x = 1 gives a maximum and x = 3 gives a minimum, we calculate
d2 y
= 6x + 2a = 6x − 12.
dx2
d2 y
= −6 < 0, so we have a maximum.
dx2
2
d y
At x = 3,
= 6 > 0, so we have a minimum.
dx2
At x = 1,
y
5
x
1
3
Figure 4.161: Graph of y = x3 − 6x2 + 9x + 5
SOLUTIONS to Review Problems for Chapter Four
407
44. Since the graph of the quartic polynomial is symmetric about the y-axis, the quartic must have only even powers and be
of the form
y = ax4 + bx2 + c.
The y-intercept is 3, so c = 3. Differentiating gives
dy
= 4ax3 + 2bx.
dx
Since there is a maximum at (1, 4), we have dy/dx = 0 if x = 1, so
4a(1)3 + 2b(1) = 4a + 2b = 0
b = −2a.
so
The fact that dy/dx = 0 if x = −1 gives us the same relationship
−4a − 2b = 0
so
b = −2a.
We also know that y = 4 if x = ±1, so
a(1)4 + b(1)2 + 3 = a + b + 3 = 4
so
a + b = 1.
Solving for a and b gives
2
a − 2a = 1
2
so
a = −1 and b = 2.
Finding d y/dx so that we can check that x = ±1 are maxima, not minima, we see
d2 y
= 12ax2 + 2b = −12x2 + 4.
dx2
Thus
d2 y
= −8 < 0 for x = ±1, so x = ±1 are maxima. See Figure 4.162.
dx2
y
(−1, 4)
(1, 4)
3
x
Figure 4.162: Graph of y = −x4 + 2x2 + 3
45. Differentiating y = axb ln x, we have
1
dy
= abxb−1 ln x + axb · = axb−1 (b ln x + 1).
dx
x
Since the maximum occurs at x = e2 , we know that
a(e2 )b−1 (b ln(e2 ) + 1) = 0.
Since a 6= 0 and (e2 )b−1 6= 0 for all b, we have
b ln(e2 ) + 1 = 0.
Since ln(e2 ) = 2, the equation becomes
2b + 1 = 0
1
b=− .
2
408
Chapter Four /SOLUTIONS
Thus y = ax−1/2 ln x. When x = e2 , we know y = 6e−1 , so
y = a(e2 )−1/2 ln e2 = ae−1 (2) = 6e−1
a = 3.
Thus y = 3x−1/2 ln x. To check that x = e2 gives a local maximum, we differentiate twice
3
1
3
dy
= − x−3/2 ln x + 3x−1/2 · = − x−3/2 ln x + 3x−3/2 ,
dx
2
x
2
9 −5/2
3 −3/2 1
3
d2 y
= x
ln x − x
· − · 3x−5/2
dx2
4
2
x
2
9 −5/2
3 −5/2
−5/2
= x
ln x − 6x
= x
(3 ln x − 8).
4
4
At x = e2 , since ln(e2 ) = 2, we have a maximum because
3
d2 y
3
= (e2 )−5/2 3 ln(e2 ) − 8 = e−5 (3 · 2 − 8) < 0.
2
dx
4
4
See Figure 4.163.
y
y
(5, 2)
y = 1.75
(e2 , 6e−1 )
(15, 1.5)
x
20
60
x
100
5
10
15
Figure 4.164: Graph of
y = 0.25 sin(πx/10) + 1.75
Figure 4.163: Graph of y = 3x−1/2 ln x
46. Since the maximum is y = 2 and the minimum is y = 1.5, the amplitude is A = (2 − 1.5)/2 = 0.25. Between
the maximum and the minimum, the x-value changes by 10. There is half a period between a maximum and the next
minimum, so the period is 20. Thus
π
2π
= 20
so
B=
.
B
10
The mid-line is y = C = (2 + 1.5)/2 = 1.75. Figure 4.164 shows a graph of the function
y = 0.25 sin
πx
10
+ 1.75.
47. First notice that since this function approaches 0 as x approaches either plus or minus infinity, any local extrema that we
find are also global extrema.
2
Differentiating y = axe−bx gives
2
2
2
dy
= ae−bx − 2abx2 e−bx = ae−bx (1 − 2bx2 ).
dx
Since we have a critical points at x = 1 and x = −1, we know 1 − 2b = 0, so b = 1/2.
The global maximum is 2 at x = 1, so we have 2 = ae−1/2 which gives a = 2e1/2 . Notice that this value of a also
gives the global minimum at x = −1.
Thus,
y = 2xe(
1−x2
2
)
.
SOLUTIONS to Review Problems for Chapter Four
409
48. We want to maximize the volume V = x2 h of the box, shown in Figure 4.165. The box has 6 faces: the top and bottom,
each of which has area x2 and the four sides, each of which has area xh. Thus 8 = 2x2 + 4xh, so
h=
8 − 2x2
.
4x
Substituting this expression in for h in the formula for V gives
V = x2 ·
8 − 2x2
1
= (8x − 2x3 ).
4x
4
Differentiating gives
1
dV
= (8 − 6x2 ).
dx
4
p
To maximize V we look for critical points, so we solve 0 = (8 − 6x2 )/4, getting x = ± 4/3. We discard the negative
solution, since x is a positive length. Then we can find
8 − 2 34
8 − 2x2
p
h=
=
4x
4 43
p
16
3
4
3
= p = p =
4
4 43
3
r
4
.
3
Thus x = h = 4/3 cm (the box is a cube).
We can check that this critical point is a maximum of V by checking the sign of
d2 V
= −3x2
dx2
which is negative when x 6= 0. So V is concave down at the critical point and therefore x =
value of V .
h
p
4/3 gives a maximum
h
x
x
x
x
Figure 4.165
Figure 4.166
49. We want to maximize the volume V = x2 h of the box, shown in Figure 4.166. The box has 5 faces: the bottom, which
has area x2 and the four sides, each of which has area xh. Thus 8 = x2 + 4xh, so
h=
8 − x2
.
4x
Substituting this expression in for h in the formula for V gives
V = x2 ·
1
8 − x2
= (8x − x3 ).
4x
4
Differentiating gives
dV
1
= (8 − 3x2 ).
dx
4
p
To maximize V we look for critical points, so we solve 0 = (8 − 3x2 )/4, getting x = ± 8/3. We discard the negative
solution, since x is a positive length. Then we can find
16
4
8− 8
8 − x2
3
3
h=
= p3 = p
= p
=
4x
4 83
4 83
2 23
r
2
.
3
410
Chapter Four /SOLUTIONS
p
p
Thus x = 8/3 cm and h = 2/3 cm.
We can check that this critical point is a maximum of V by checking the sign of
3
d2 V
= − x,
dx2
2
which is negative when x > 0. So V is concave down at the critical point and therefore x =
value of V .
p
8/3 gives a maximum
50. We want to maximize the volume V = πr 2 h of the cylinder, shown in Figure 4.167. The cylinder has 3 pieces: the top
and bottom disks, each of which has area πr 2 and the tube, which has area 2πrh. Thus 8 = 2πr 2 + 2πrh. Solving for h
gives
8 − 2πr 2
.
h=
2πr
Substituting this expression in for h in the formula for V gives
V = πr 2 ·
8 − 2πr 2
= 4r − πr 3 .
2πr
Differentiating gives
dV
= 4 − 3πr 2 .
dr
√
To maximize V we look for critical points, so we solve 0 = 4 − 3πr 2 , thus r = ±2/ 3π. We discard the negative
solution, since r is a positive length. Substituting this value in for r in the formula for h gives
h=
8 − 2π
4
3π
2π √23π
=
16
3
2π √23π
=
8
3π
√2
3π
4
= √ .
3π
We can check that this critical point is a maximum of V by checking the sign of
d2 V
= −6πr
dr 2
√
which is negative when r > 0. So V is concave down at the critical point and therefore r = 2/ 3π is a maximum.
r
✻
r
✲
✻
h
✲
h
❄
❄
Figure 4.167
Figure 4.168
51. We want to maximize the volume V = πr 2 h of the cylinder, shown in Figure 4.168. The cylinder has 2 pieces: the end
disk, of area πr 2 and the tube, which has area 2πrh. Thus 8 = πr 2 + 2πrh. Solving for h gives
h=
8 − πr 2
.
2πr
Substituting this expression in for h in the formula for V gives
V = πr 2 ·
1
8 − πr 2
= (8r − πr 3 ).
2πr
2
Differentiating gives
dV
1
= (8 − 3πr 2 ).
dr
2
SOLUTIONS to Review Problems for Chapter Four
411
p
To maximize V we look for critical points, so we solve 0 = (8 − 3πr 2 )/2, thus r = ± 8/(3π). We discard the negative
solution, since r is a positive length. Substituting this value in for r in the formula for h gives
h=
8−π
2π
8
p 3π
=
8
3π
2π
16
8
3
p
= p3π8 =
8
3π
3π
r
8
.
3π
We can check that this critical point is a maximum of V by checking the sign of
d2 V
= −3πr
dr 2
which is negative when r > 0. So V is concave down at the critical point and therefore r =
y
52.
p
8/(3π) is a maximum.
5π/4
❄
x
✻
π/4
Letting f (x) = e−x sin x, we have
f ′ (x) = −e−x sin x + e−x cos x.
Solving f ′ (x) = 0, we get sin x = cos x. This means x = arctan(1) = π/4, and π/4 plus multiples of π, are the critical
points of f (x). By evaluating f (x) at the points kπ + π/4, where k is an integer, we can find:
e−5π/4 sin(5π/4) ≤ e−x sin x ≤ e−π/4 sin(π/4),
since f (0) = 0 at the endpoint. So
−0.014 ≤ e−x sin x ≤ 0.322.
53. Let f (x) = x sin x. Then f ′ (x) = x cos x + sin x.
f ′ (x) = 0 when x = 0, x ≈ 2, and x ≈ 5. The latter two estimates we can get from the graph of f ′ (x).
Zooming in (or using some other approximation method), we can find the zeros of f ′ (x) with more precision. They
are (approximately) 0, 2.029, and 4.913. We check the endpoints and critical points for the global maximum and minimum.
f (0) = 0,
f (2.029) ≈ 1.8197,
f (2π) = 0,
f (4.914) ≈ −4.814.
Thus for 0 ≤ x ≤ 2π, −4.81 ≤ f (x) ≤ 1.82.
54. To find the best possible bounds for f (x) = x3 − 6x2 + 9x + 5 on 0 ≤ x ≤ 5, we find the global maximum and minimum
for the function on the interval. First, we find the critical points. Differentiating yields
f ′ (x) = 3x2 − 12x + 9
Letting f ′ (x) = 0 and factoring yields
3x2 − 12x + 9 = 0
3(x2 − 4x + 3) = 0
3(x − 3)(x − 1) = 0
So x = 1 and x = 3 are critical points for the function on 0 ≤ x ≤ 5. Evaluating the function at the critical points and
endpoints gives us
f (0) = (0)3 − 6(0)2 + 9(0) + 5 = 5
f (1) = (1)3 − 6(1)2 + 9(1) + 5 = 9
f (3) = (3)3 − 6(3)2 + 9(3) + 5 = 5
f (5) = (5)3 − 6(5)2 + 9(5) + 5 = 25
412
Chapter Four /SOLUTIONS
So the global minimum on this interval is f (0) = f (3) = 5 and the global maximum is f (5) = 25. From this we
conclude
5 ≤ x3 − 6x2 + 9x + 5 ≤ 25
are the best possible bounds for the function on the interval 0 ≤ x ≤ 5.
55. We first solve for P
and find the derivative
P = −6jm2 + 4jk − 5km,
dP
= −12jm − 5k.
dm
Since the derivative is defined for all m, we find the critical points by solving dP/dm = 0:
dP
= −12jm − 5k = 0,
dm
m=−
5k
.
12j
There is one critical point at m = −5k/(12j). Since P is a quadratic function of m with a negative leading coefficient
−6j, the critical point gives the global maximum of P . There is no global minimum because P → −∞ as m → ±∞.
56. Differentiating gives
dy
= a(e−bx − bxe−bx ) = ae−bx (1 − bx).
dx
Thus, dy/dx = 0 when x = 1/b. Then
1
a
y = a e−b·1/b = e−1 .
b
b
Differentiating again gives
d2 y
= −abe−bx (1 − bx) − abe−bx
dx2
= −abe−bx (2 − bx)
When x = 1/b,
1
d2 y
= −abe−b·1/b 2 − b ·
= −abe−1 .
dx2
b
Therefore the point ( 1b , ab e−1 ) is a maximum if a and b are positive. We can make (2, 10) a maximum by setting
1
1
= 2 so b =
b
2
and
a −1
a −1
e =
e = 2ae−1 = 10 so
b
1/2
a = 5e.
Thus a = 5e, b = 1/2.
57. (a) We set the derivative equal to zero and solve for t to find critical points. Using the product rule, we have:
f ′ (t) = (at2 )(e−bt (−b)) + (2at)e−bt = 0
ate−bt (−bt + 2) = 0
2
t = 0 or t = .
b
There are two critical points: t = 0 and t = 2/b.
(b) Since we want a critical point at t = 5, we substitute and solve for b:
5 = 2/b
2
b = = 0.4.
5
To find the value of a, we use the fact that f (5) = 12, so we have:
a(52 )e−0.4(5) = 12
a · 25e−2 = 12
12e2
= 3.547.
a=
25
SOLUTIONS to Review Problems for Chapter Four
413
(c) To show that f (t) has a local minimum at t = 0 and a local maximum at t = 5, we can use the first derivative test or
the second derivative test. Using the first derivative test, we evaluate f ′ at values on either side of t = 0 and t = 5.
Since f ′ (t) = 3.547te−0.4t (−0.4t + 2), we have
f ′ (−1) = −3.547e0.4 (2.4) = −12.700 < 0
and
f ′ (1) = 3.547e−0.4 (1.6) = 3.804 > 0,
and
f ′ (6) = 3.547(6)e−2.4 (−0.4) = −0.772 < 0.
The function f is decreasing to the left of t = 0, increasing between t = 0 and t = 5, and decreasing to the right of
t = 5. Therefore, f (t) has a local minimum at t = 0 and a local maximum at t = 5. See Figure 4.169.
f (t)
12
t
5
Figure 4.169
58. We have f (x) = x2 + 2ax = x(x + 2a) = 0 when x = 0 or x = −2a.
′
f (x) = 2x + 2a = 2(x + a)
(
= 0 when x = −a
> 0 when x > −a
< 0 when x < −a.
See Figure 4.170. Furthermore, f ′′ (x) = 2, so that f (−a) = −a2 is a global minimum, and the graph is always concave
up.
x
−a
−2a
−a2
Figure 4.170
Increasing |a| stretches the graph horizontally. Also, the critical value (the value of f at the critical point) drops
further beneath the x-axis. Letting a < 0 would reflect the graph shown through the y-axis.
59.
a = −2
a=0
a=2
a=4
a=6
✲
✲
✲
✲
✲
✛
y = − 21 x3
414
Chapter Four /SOLUTIONS
dy
d
To solve for the critical points, we set dx
= 0. Since dx
x3 − ax2 = 3x2 − 2ax, we want 3x2 − 2ax = 0, so
2
x = 0 or x = 3 a. At x = 0, we have y = 0. This first critical point is independent of a and lies on the curve y = − 21 x3 .
3
2
a .
3
−bt
4 3
a = − 21
At x = 23 a, we calculate y = − 27
60. We want the maximum value of r(t) = ate
Thus the second critical point also lies on the curve y = − 21 x3 .
to be 0.3 ml/sec and to occur at t = 0.5 sec. Differentiating gives
r ′ (t) = ae−bt − abte−bt ,
so r ′ (t) = 0 when
ae−bt (1 − bt) = 0
t=
or
1
.
b
Since the maximum occurs at t = 0.5, we have
1
= 0.5
b
so
b = 2.
Thus, r(t) = ate−2t . The maximum value of r is given by
r(0.5) = a(0.5)e−2(0.5) = 0.5ae−1 .
Since the maximum value of r is 0.3, we have
0.5ae−1 = 0.3
a=
so
0.3e
= 1.63.
0.5
Thus, r(t) = 1.63te−2t ml/sec.
61. We know that dp/dt = −3 mm/sec when p = 35 mm and we want to know q and dq/dt at that time. We also know that
f = 15 mm. Since
1
1
1
+ = ,
p
q
f
substituting p = 35 and f = 15 gives
1
1
4
1
=
−
=
,
q
15
35
105
so q =
105
= 26.25 mm.
4
Differentiating with respect to time t gives,
−
1 dp
1 dq
− 2
= 0.
p2 dt
q dt
Substituting gives
−
1
dq
1
(−3) −
= 0,
352
(105/4)2 dt
so
(105/4)2 3
27
dq
=
=
= 1.688 mm/sec.
dt
352
16
Thus, the image is moving away from the lens at 1.688 mm per second.
62.
r(λ) = a(λ)−5 (eb/λ − 1)−1
r ′ (λ) = a(−5λ−6 )(eb/λ − 1)−1 + a(λ−5 )
b b/λ
e
(eb/λ − 1)−2
λ2
(0.96, 3.13) is a maximum, so r ′ (0.96) = 0 implies that the following holds, with λ = 0.96:
5λ−6 (eb/λ − 1)−1 = λ−5
5λ(eb/λ − 1) = beb/λ
5λeb/λ − 5λ = beb/λ
5λeb/λ − beb/λ = 5λ
5λ − b
eb/λ = 1
5λ
4.8 − b b/0.96
e
− 1 = 0.
4.8
b b/λ
e
(eb/λ − 1)−2
λ2
SOLUTIONS to Review Problems for Chapter Four
415
Using Newton’s method, or some other approximation method, we search for a root. The root should be near 4.8. Using
our initial guess, we get b ≈ 4.7665. At λ = 0.96, r = 3.13, so
a
or
3.13 =
0.965 (eb/0.96 − 1)
a = 3.13(0.96)5 (eb/0.96 − 1)
≈ 363.23.
As a check, we try r(4) ≈ 0.155, which looks about right on the given graph.
63. Since I(t) is a periodic function with period 2π/w, it is enough to consider I(t) for 0 ≤ wt ≤ 2π. Differentiating, we
find
√
dI
= −w sin(wt) + 3w cos(wt).
dt
At a critical point
√
−w sin(wt) + 3w cos(wt) = 0
√
sin(wt) = 3 cos(wt)
√
tan(wt) = 3.
So wt = π/3 or 4π/3, or these values plus multiples of 2π. Substituting into I, we see
√
√
π
π
π
1 √
3
At wt = : I = cos
+ 3 sin
= + 3·
= 2.
3
3
3
2
2
√
√
4π
4π
4π
1 √
3
: I = cos
= −2.
At wt =
+ 3 sin
=− − 3·
3
3
3
2
2
64.
Thus, the maximum value is 2 amps and the minimum is −2 amps.
(µ + θ)(1 − 2µθ) − (θ − µθ2 )
µ(1 − 2µθ − θ2 )
dE
=
=
.
2
dθ
(µ + θ)
(µ + θ)2
p
p
Now dE/dθ = 0 when θ = −µ ± 1 + µ2 . Since θ > 0, the only possible critical point is when θ = −µ + µ2 + 1.
Differentiating again gives E ′′ < 0 at this point and so it is a local maximum. Since E(θ) is continuous for θ > 0 and
E(θ) has only one critical point, the local maximum is the global maximum.
√
65. The top half of the circle has equation y = 1 − x2 . The rectangle in Figure 4.171 has area, A, given by
A = 2xy = 2x
At a critical point,
p
1 − x2 ,
for 0 ≤ x ≤ 1.
p
−1/2
dA
1
= 2 1 − x2 + 2x
(−2x)
1 − x2
dt
2
p
2x2
2 1 − x2 − √
1 − x2
2
√
2
1 − x2 − 2x2
√
1 − x2
2(1 − x2 − x2 )
√
1 − x2
2(1 − 2x2 )
=0
=0
=0
=0
=0
1
x = ±√ .
2
√
Since A = 0 at the endpoints x = 0 and x = 1, and since A is positive at the only critical point, x = 1/ 2,
in
p the interval 0 ≤√x ≤ 1, the critical point is a local and global maximum. The vertices on the circle have y =
1 − (1/2)2 = 1/ 2. Thus the coordinates of the rectangle with maximum area are
1
√ , 0 ;
2
1
1
√ , √
2
2
;
1
−√ , 0 ;
2
1
1
−√ , √
2
2
416
Chapter Four /SOLUTIONS
and the maximum area is
1
1
A = 2 √ · √ = 1.
2
2
y
1
✻y = √1 − x2
y
✛
−1
✲
x
❄
x
1
Figure 4.171
66. The triangle in Figure 4.172 has area, A, given by
1
1
x · y = x3 e−3x .
2
2
A=
If the area has a maximum, it occurs where
3
3
dA
= x2 e−3x − x3 e−3x = 0
dx
2
2
3 2
x (1 − x) e−3x = 0
2
x = 0, 1.
The value x = 0 gives the minimum area, A = 0, for x ≥ 0. Since
3
dA
= x2 (1 − x)e−3x ,
dx
2
we see that
dA
> 0 for 0 < x < 1
dx
Thus, x = 1 gives the local and global maximum of
A=
and
dA
< 0 for x > 1.
dx
1
1 3 −3·1
1 e
= 3.
2
2e
y
(x, x2 e−3x )
✛
✲
x
✻y
❄
x
Figure 4.172
67. The distance from a given point on the parabola (x, x2 ) to (1, 0) is given by
D=
p
(x − 1)2 + (x2 − 0)2 .
Minimizing this is equivalent to minimizing d = (x − 1)2 + x4 . (We can ignore the square root if we are only
interested in minimizing because the square root is smallest when the thing it is the square root of is smallest.) To minimize
d, we find its critical points by solving d′ = 0. Since d = (x − 1)2 + x4 = x2 − 2x + 1 + x4 ,
d′ = 2x − 2 + 4x3 = 2(2x3 + x − 1).
By graphing d′ = 2(2x3 + 2x − 1) on a calculator, we see that it has only 1 root, x ≈ 0.59. This must give a minimum
because d → ∞ as x → −∞ and as x → +∞, and d has only one critical point. This is confirmed by the second
derivative test: d′′ = 12x2 + 2 = 2(6x2 + 1), which is always positive. Thus the point (0.59, 0.592 ) ≈ (0.59, 0.35) is
approximately the closest point of y = x2 to (1, 0).
417
SOLUTIONS to Review Problems for Chapter Four
68. Any point on the curve can be written (x, x2 ). The distance between such a point and (3, 0) is given by
s(x) =
p
(3 − x)2 + (0 − x2 )2 =
p
(3 − x)2 + x4 .
Plotting this function in Figure 4.173, we see that there is a minimum near x = 1.
To find the value of x that minimizes the distance we can instead minimize the function Q = s2 (the derivative is
simpler). Then we have
Q(x) = (3 − x)2 + x4 .
Differentiating Q(x) gives
dQ
= −6 + 2x + 4x3 .
dx
Plotting the function 4x3 + 2x − 6 shows that there is one real solution at x = 1, which can be verified by substitution; the
required coordinates are therefore (1, 1). Because Q′′ (x) = 2 + 12x2 is always positive, x = 1 is indeed the minimum.
See Figure 4.174.
y
y
17.5
100
15
12.5
50
10
7.5
x
−4
5
x
−2
2
4
−50
2.5
−4
−2
0
2
−100
4
Figure 4.173
Figure 4.174
69. We see that the width of the tunnel is 2r. The area of the rectangle is then (2r)h. The area of the semicircle is (πr 2 )/2.
The cross-sectional area, A, is then
1
A = 2rh + πr 2
2
and the perimeter, P , is
P = 2h + 2r + πr.
From A = 2rh + (πr 2 )/2 we get
h=
πr
A
−
.
2r
4
Thus,
A
πr
πr
A
+ 2r + πr =
−
+ 2r +
.
2r
4
r
2
We now have the perimeter in terms of r and the constant A. Differentiating, we obtain
P =2
dP
A
π
= − 2 +2+ .
dr
r
2
To find the critical points we set P ′ = 0:
−
π
A
+ +2 = 0
r2
2
r2
2
=
A
4+π
r=
r
2A
.
4+π
Substituting this back into our expression for h, we have
√
√
A
4+π
π
2A
h=
− ·√
.
· √
2
4
4+π
2A
418
Chapter Four /SOLUTIONS
Since P → ∞ as r → 0+ and as r → ∞, this critical point must be a global minimum. Notice that the h-value simplifies
to
r
2A
= r.
h=
4+π
70. Consider the rectangle of sides x and y shown in Figure 4.175.
y
x
Figure 4.175
The total area is xy = 3000, so y = 3000/x. Suppose the left and right edges and the lower edge have the shrubs
and the top edge has the fencing. The total cost is
C = 45(x + 2y) + 20(x)
= 65x + 90y.
Since y = 3000/x, this reduces to
C(x) = 65x + 90(3000/x) = 65x + 270,000/x.
Therefore, C ′ (x) = 65 − 270,000/x2 . We set this to 0 to find the critical points:
65 −
270,000
=0
x2
270,000
= 65
x2
2
x = 4153.85
x = 64.450 ft
so that
y = 3000/x = 46.548 ft.
+
Since C(x) → ∞ as x → 0 and x → ∞, we see that x = 64.450 is a minimum. The minimum total cost is then
C(64.450) ≈ $8378.54.
71. Figure 4.176 shows the the pool has dimensions x by y and the deck extends 5 feet at either side and 10 feet at the ends
of the pool.
10
5
y
5
x
10
Figure 4.176
The dimensions of the plot of land containing the pool are then (x + 5 + 5) by (y + 10 + 10). The area of the land
is then
A = (x + 10)(y + 20),
SOLUTIONS to Review Problems for Chapter Four
419
which is to be minimized. We also are told that the area of the pool is xy = 1800, so
y = 1800/x
and
1800
+ 20
x
18000
+ 200.
= 1800 + 20x +
x
A = (x + 10)
We find dA/dx and set it to zero to get
18000
dA
= 20 −
=0
dx
x2
20x2 = 18000
x2 = 900
x = 30 feet.
Since A → ∞ as x → 0+ and as x → ∞, this critical point must be a global minimum. Also, y = 1800/30 = 60 feet.
The plot of land is therefore (30 + 10) = 40 by (60 + 20) = 80 feet.
72. (a) Suppose n passengers sign up for the cruise. If n ≤ 100, then the cruise’s revenue is R = 2000n, so the maximum
revenue is
R = 2000 · 100 = 200,000.
If n > 100, then the price is
and hence the revenue is
p = 2000 − 10(n − 100)
R = n(2000 − 10(n − 100)) = 3000n − 10n2 .
To find the maximum revenue, we set dR/dn = 0, giving 20n = 3000 or n = 150. Then the revenue is
R = (2000 − 10 · 50) · 150 = 225,000.
Since this is more than the maximum revenue when n ≤ 100, the boat maximizes its revenue with 150 passengers,
each paying $1500.
(b) We approach this problem in a similar way to part (a), except now we are dealing with the profit function π. If
n ≤ 100, we have
π = 2000n − 80,000 − 400n,
so π is maximized with 100 passengers yielding a profit of
π = 1600 · 100 − 80,000 = $80,000.
If n > 100, we have
π = n(2000 − 10(n − 100)) − (80,000 + 400n).
We again set dπ/dn = 0, giving 2600 = 20n, so n = 130. The profit is then $89,000. So the boat maximizes profit
by boarding 130 passengers, each paying $1700. This gives the boat $89,000 in profit.
73. (a) π(q) is maximized when R(q) > C(q) and they are as far apart as possible. See Figure 4.177.
(b) π ′ (q0 ) = R′ (q0 ) − C ′ (q0 ) = 0 implies that C ′ (q0 ) = R′ (q0 ) = p.
Graphically, the slopes of the two curves at q0 are equal. This is plausible because if C ′ (q0 ) were greater than p
or less than p, the maximum of π(q) would be to the left or right of q0 , respectively. In economic terms, if the cost
were rising more quickly than revenues, the profit would be maximized at a lower quantity (and if the cost were rising
more slowly, at a higher quantity).
(c) See Figure 4.178.
420
Chapter Four /SOLUTIONS
$
$
C(q)
C ′ (q)
R(q)
■
p
maximum π(q)
q
q0
Figure 4.177
Figure 4.178
74.
marginal
cost
marginal
revenue
profit
q1
q
q0
q
q2
q1
q
q2
75.
q1
q2
q
D (km)
15 knots
3
I
T
2
12 knots
3 km
1
S
45◦
0.05
Figure 4.179: Position of the tanker
and ship
√
3 2
2
0.1
Figure 4.180: Distance between the
ship at S and the tanker at T
Suppose t is the time, in hours, since the ships were 3 km apart. Then T I =
− (12)(1.85)t. So the distance, D(t), in km, between the ships at time t is
D(t) =
s
√
t (hr)
3 2
− 27.75t
2
2
+
√
3 2
− 22.2t
2
2
√
3 2
2
− (15)(1.85)t and SI =
.
Differentiating gives
dD
=
dt
−55.5
2
r
√3
2
− 27.75 t − 44.4
√3
2
− 27.75 t
2
+
√3
2
√3
2
− 22.2 t
− 22.2 t
2
.
Solving dD/dt = 0 gives a critical point at t = 0.0839 hours when the ships will be approximately 331 meters apart. So
the ships do not need to change course. Alternatively, tracing along the curve in Figure 4.180 gives the same result. Note
that this is after the eastbound ship crosses the path of the northbound ship.
SOLUTIONS to Review Problems for Chapter Four
421
76. Since the volume is fixed at 200 ml (i.e. 200 cm3 ), we can solve the volume expression for h in terms of r to get (with h
and r in centimeters)
200 · 3
.
h=
7πr 2
Using this expression in the surface area formula we arrive at
S = 3πr
r
r2 +
600 2
7πr 2
By plotting S(r) we see that there is a minimum value near r = 2.7 cm.
t
t
= 1−ln
, which is zero if t = e, negative if t > e, and positive if t < e, since ln t is
77. (a) We have g ′ (t) = t(1/t)−ln
t2
t2
1
increasing. Thus g(e) = e is a global maximum for g. Since t = e was the only point at which g ′ (t) = 0, there is no
minimum.
(b) Now ln t/t is increasing for 0 < t < e, ln 1/1 = 0, and ln 5/5 ≈ 0.322 < ln(e)/e. Thus, for 1 < t < e, ln t/t
increases from 0 to above ln 5/5, so there must be a t between 1 and e such that ln t/t = ln 5/5. For t > e, there
is only one solution to ln t/t = ln 5/5, namely t = 5, since ln t/t is decreasing for t > e. For 0 < t < 1, ln t/t is
negative and so cannot equal ln 5/5. Thus ln x/x = ln t/t has exactly two solutions.
(c) The graph of ln t/t intersects the horizontal line y = ln 5/5, at x = 5 and x ≈ 1.75.
78. (a) x-intercept: (a, 0), y-intercept: (0,
(b) Area = 12 (a)( a21+1 ) = 2(a2a+1)
(c)
1
)
a2 +1
A=
a
2(a2 + 1)
A′ =
2(a2 + 1) − a(4a)
4(a2 + 1)2
=
2(1 − a2 )
4(a2 + 1)2
=
(1 − a2 )
.
2(a2 + 1)2
If A′ = 0, then a = ±1. We only consider positive values of a, and we note that A′ changes sign from positive to
negative at a = 1. Hence a = 1 is a local maximum of A which is a global maximum because A′ < 0 for all a > 1
and A′ > 0 for 0 < a < 1.
(d) A = 12 (1)( 12 ) = 41
(e) Set 2(a2a+a) = 31 and solve for a:
5a = 2a2 + 2
2a2 − 5a + 2 = 0
(2a − 1)(a − 2) = 0.
79. (a) Since the volume of water in the container is proportional to its depth, and the volume is increasing at a constant rate,
d(t) = Depth at time t = Kt,
where K is some positive constant. So the graph is linear, as shown in Figure 4.181. Since initially no water is in the
container, we have d(0) = 0, and the graph starts from the origin.
depth of water
depth of water
d(t)
d(t)
slope = K
t
t
Figure 4.181
Figure 4.182
422
Chapter Four /SOLUTIONS
(b) As time increases, the additional volume needed to raise the water level by a fixed amount increases. Thus, although
the depth, d(t), of water in the cone at time t, continues to increase, it does so more and more slowly. This means
d′ (t) is positive but decreasing, i.e., d(t) is concave down. See Figure 4.182.
80. (a) The concavity changes at t1 and t3 , as shown in Figure 4.183.
f (t)
y3
y2
y1
t1
t2
t3
t
Figure 4.183
(b) f (t) grows most quickly where the vase is skinniest (at y3 ) and most slowly where the vase is widest (at y1 ). The
diameter of the widest part of the vase looks to be about 4 times as large as the diameter at the skinniest part. Since
the area of a cross section is given by πr 2 , where r is the radius, the ratio between areas of cross sections at these
two places is about 42 , so the growth rates are in a ratio of about 1 to 16 (the wide part being 16 times slower).
81. The volume, V, of a cone of radius r and height h is
V =
1 2
πr h.
3
However, Figure 4.184 shows that h/r = 12/5, thus r = 5h/12, so
V =
1
π
3
Differentiating with respect to time, t, gives
2
5
h
12
h=
25
πh3 .
432
25
dh
dV
=
πh2 .
dt
144
dt
When the depth of chemical in the tank is 1 meter, the level is falling at 0.1 meter/min so h = 1 and dh/dt = −0.1. Thus
dV
25
=−
· π · 12 · 0.1 = −0.0545 m3 /min.
dt
144
✛5m✲
✻
r
✛✲
12 m
✻
h
❄
❄
Figure 4.184
82. Evaluating the limits in the numerator and the denominator we get 0/e0 = 0/1 = 0, so this is not an indeterminate form.
l’Hopital’s rule does not apply.
83. We have limx→1 sin πx = sin π = 0, and limx→1 x − 1 = 0, so this is a 0/0 form and l’Hopital’s rule applies directly.
SOLUTIONS to Review Problems for Chapter Four
423
84. This is a 0/0 form. Applying l’Hopital’s rule twice, we get
lim
t→0
et − 0 − 1
et
1
et − 1 − t
= lim
= lim
= .
2
t→0
t→0 2
t
2t
2
85. Let f (t) = 3 sin t − sin 3t and g(t) = 3 tan t − tan 3t, then f (0) = 0 and g(0) = 0. Similarly,
f ′ (t) = 3 cos t − 3 cos 3t
f ′ (0) = 0
f ′′′ (t) = −3 cos t + 27 cos 3t
f ′′′ (0) = 24
′′
f ′′ (0) = 0
f (t) = −3 sin t + 9 sin 3t
′
2
2
g ′ (0) = 0
g (t) = 3sec t − 3sec 3t
g ′′ (t) = 6 sec2 t tan t − 18 sec2 3t tan 3t
′′′
4
g ′′ (0) = 0
g (t) = −54 sec 3t − 108 sec 37 tan 3t + 6 sec t + 12 sec t tan t g ′′′ (0) = −48
Since the first and second derivatives of f and g are both 0 at t = 0, we have to go as far as the third derivative to use
l’Hopital’s rule. Applying l’Hopital’s rule gives
lim
t→0+
2
2
4
2
2
f ′ (t)
f ′′ (t)
f ′′′ (t)
24
1
3 sin t − sin 3t
= lim ′
= lim ′′
= lim ′′′
=
=− .
+
+
+
3 tan t − tan 3t
g
(t)
g
(t)
−48
2
t→0
t→0
t→0 g (t)
86. If f (x) = 1 − cosh(5x) and g(x) = x2 , then f (0) = g(0) = 0, so we use l’Hopital’s Rule:
lim
x→0
−5 sinh 5x
−25 cosh 5x
25
1 − cosh 5x
= lim
= lim
=− .
x2
x→0
2x
x→0
2
2
87. If f (x) = x − sinh x and g(x) = x3 , then f (0) = g(0) = 0. However, f ′ (0) = g ′ (0) = f ′′ (0) = g ′′ (0) = 0 also, so
we use l’Hopital’s Rule three times. Since f ′′′ (x) = − cosh x and g ′′′ (x) = 6:
lim
x→0
1 − cosh x
− sinh x
− cosh x
1
x − sinh x
= lim
= lim
= lim
=− .
x→0
x→0
x→0
x3
3x2
6x
6
6
88. (a) The population is increasing if dP/dt > 0, that is, if
kP (L − P ) > 0.
Since P ≥ 0 and k, L > 0, we must have P > 0 and L − P > 0 for this to be true. Thus, the population is increasing
if 0 < P < L.
The population is decreasing if dP/dt < 0, that is, if P > L.
The population remains constant if dP/dt = 0, so P = 0 or P = L.
(b) Differentiating with respect to t using the chain rule gives
d2 P
d
d
dP
=
(kP (L − P )) =
(kLP − kP 2 ) ·
= (kL − 2kP )(kP (L − P ))
dt2
dt
dP
dt
= k2 P (L − 2P )(L − P ).
89. We are given that the volume is increasing at a constant rate
formula V = 43 πr 3 . By implicit differentiation, we have
dV
dt
= 400. The radius r is related to the volume by the
dV
4
dr
dr
= π3r 2
= 4πr 2
dt
3
dt
dt
Plugging in
dV
dt
= 400 and r = 10, we have
400 = 400π
so
dr
dt
=
1
π
dr
dt
≈ 0.32µm/day.
90. Let r be the radius of the raindrop. Then its volume V = 34 πr 3 cm3 and its surface area is S = 4πr 2 cm2 . It is given that
dV
= 2S = 8πr 2 .
dt
424
Chapter Four /SOLUTIONS
Furthermore,
dV
= 4πr 2 ,
dr
so from the chain rule,
dV dr
dV
=
·
dt
dr dt
dr
dV /dt
=
= 2.
dt
dV /dr
and thus
Since dr/dt is a constant, dr/dt = 2, the radius is increasing at a constant rate of 2 cm/sec.
91. (a) Since dθ/dt represents the rate of change of θ with time, dθ/dt represents the angular velocity of the disk.
(b) Suppose P is the point on the rim shown in Figure 4.185.
P
■
a
s
θ
❑
x
Figure 4.185
Any other point on the rim is moving at the same speed, though in a different direction. We know that since θ is
in radians,
s = aθ.
Since a is a constant, we know
dθ
ds
=a .
dt
dt
But ds/dt = v, the speed of the point on the rim, so
v=a
92. We have
93. We have
dθ
.
dt
dD
dr
dr
= K[e−r − (r + 1)e−r ]
= −Kre−r .
dt
dt
dt
1
dM
=K 1−
dt
1+r
dr
.
dt
94. Let V be the volume of the ice, so that V = 3π(r 2 − 12 ). Now,
dr
dV
= 6πr .
dt
dt
Thus for r = 1.5, we have
dV
= 6π(1.5)(0.03) = 0.848 cm3 /hr.
dt
95. The volume, V , of a cone of height h and radius r is
1 2
πr h.
3
√
Since the angle of the cone is π/6, so r = h tan(π/6) = h/ 3
V =
V =
1
π
3
h
√
3
2
h=
1 3
πh .
9
SOLUTIONS to Review Problems for Chapter Four
425
Differentiating gives
dV
1
= πh2 .
dh
3
To find dh/dt, use the chain rule to obtain
dV dh
dV
=
.
dt
dh dt
So,
dV /dt
0.1meters/hour
0.3
dh
=
=
=
meters/hour.
dt
dV /dh
πh2 /3
πh2
√
Since r = h tan(π/6) = h/ 3, we have
1 0.3
dr
dh 1
√ = √
=
meters/hour.
dt
dt 3
3 πh2
96. (a) The surface of the water is circular with radius r cm. Applying Pythagoras’ Theorem to the triangle in Figure 4.186
shows that
(10 − h)2 + r 2 = 102
so
r=
p
102 − (10 − h)2 =
p
20h − h2 cm.
(b) We know dh/dt = −0.1 cm/hr and we want to know dr/dt when h = 5 cm. Differentiating
r=
gives
p
20h − h2
1
dh
dr
dh
= (20h − h2 )−1/2 20
− 2h
dt
2
dt
dt
Substituting dh/dt = −0.1 and h = 5 gives
dr
dt
h=5
= √
= √
10 − h
dh
.
·
2
dt
20h − h
5
1
· (−0.1) = − √ = −0.0577 cm/hr.
20 · 5 − 52
2 75
Thus, the radius is decreasing at 0.0577 cm per hour.
✻
(10 − h)
❄
✻
10
■
h
❄
r
Figure 4.186
97. We have
K(a2 − 2y 2 ) dy
dF
= 2
.
dt
(a + y 2 )5/2 dt
(a) When y = 0, we have
dy
dF
= Ka−3 .
dt
dt
So, dF /dt is positive
and F is increasing.
√
(b) When y = a/ 2, we have
K(a2 − a2 ) dy
dF
= 2
= 0.
dt
(a + (a2 /2))5/2 dt
So, dF /dt = 0 and F is not changing.
(c) When y = 2a, we have
K(a2 − 8a2 ) dy
dF
7Ka2 dy
= 2
=
−
.
dt
(a + 4a2 )5/2 dt
(5a2 )5/2 dt
So, dF /dt is negative and F is decreasing.
426
Chapter Four /SOLUTIONS
98. The rate at which the voltage, V, is changing is obtained by differentiating V = IR to get
dR
dI
dV
=I
+R .
dt
dt
dt
Since the voltage remains constant, dV /dt = 0. Thus
R dI
dR
=−
,
dt
I dt
and the rate at which the resistance is changing is
1000
dR
=−
(0.001) = −10 ohms/min.
dt
0.1
We conclude that the resistance is falling by 10 ohms/min.
99. From Figure 4.187, Pythagoras’ Theorem shows that the ground distance, d, between the train and the point, B, vertically
below the plane is given by
d2 = x 2 + y 2 .
Figure 4.188 shows that
z 2 = d2 + 42
so
z 2 = x2 + y 2 + 42 .
We know that when x = 1, dx/dt = 80, y = 5, dy/dt = 500, and we want to know dz/dt. First, we find z:
√
z 2 = 12 + 52 + 42 = 42, so z = 42.
Differentiating z 2 = x2 + y 2 + 42 gives
2z
Canceling 2s and substituting gives
√
dz
dx
dy
= 2x
+ 2y .
dt
dt
dt
dz
= 1(80) + 5(500)
dt
dz
2580
= 398.103 mph.
= √
dt
42
42
B Plane: 4 miles above
✻
d
this point
y
Plane
z
Train
A
✛
✲
4 miles
❄
x
Figure 4.187: View from air
Train
A
d
B
Figure 4.188: Vertical view
SOLUTIONS to Review Problems for Chapter Four
427
100. We want to find dP/dV . Solving P V = k for P gives
P = k/V
so,
k
dP
= − 2.
dV
V
101. (a) Since V = k/P , the volume decreases.
(b) Since P V = k and P = 2 when V = 10, we have k = 20, so
V =
20
.
P
We think of both P and V as functions of time, so by the chain rule
dV
dV dP
=
,
dt
dP dt
20 dP
dV
=− 2
.
dt
P dt
We know that dP/dt = 0.05 atm/min when P = 2 atm, so
20
dV
= − 2 · (0.05) = −0.25 cm3 /min.
dt
2
CAS Challenge Problems
102. (a) Since k > 0, we have lim e−kt = 0. Thus
t→∞
lim P = lim
t→∞
t→∞
L
L
=
= L.
1 + Ce−kt
1+C ·0
The constant L is called the carrying capacity of the environment because it represents the long-run population in the
environment.
(b) Using a CAS, we find
LCk2 e−kt (1 − Ce−kt )
d2 P
=−
.
2
dt
(1 + Ce−kt )3
Thus, d2 P/dt2 = 0 when
1 − Ce−kt = 0
t=−
ln(1/C)
.
k
Since e−kt and (1 + Ce−kt ) are both always positive, the sign of d2 P/dt2 is negative when (1 − Ce−kt ) > 0,
that is, for t > − ln(1/C)/k. Similarly, the sign of d2 P/dt2 is positive when (1 − Ce−kt ) < 0, that is, for
t < − ln(1/C)/k. Thus, there is an inflection point at t = − ln(1/C)/k.
For t = − ln(1/C)/k,
L
L
L
=
= .
P =
1 + C(1/C)
2
1 + Celn(1/C)
Thus, the inflection point occurs where P = L/2.
103. (a) The graph has a jump discontinuity whose position depends on a. The function is increasing, and the slope at a given
x-value seems to be the same for all values of a. See Figure 4.189.
428
Chapter Four /SOLUTIONS
y
y
2
y
2
2
x
x
1/a
−2
x
1/a
−2
a = 0.5
1/a
−2
a=1
a=2
Figure 4.189
(b) Most computer algebra systems will give a fairly complicated answer for the derivative. Here is one example; others
may be different.
√
√ √
dy
x + a ax
=
√ .
√ √
dx
2x 1 + a + 2 a x + x + ax − 2 ax
When we graph the derivative, it appears that we get the same graph for all values of a. See Figure 4.190.
dy/dx
x
Figure 4.190
(c) Since a and x are positive, we have
derivative:
√
ax =
√ √
a x. We can use this to simplify the expression we found for the
√
√ √
dy
x + a ax
=
√ √
√
dx
2x 1 + a + 2 a x + x + ax − 2 ax
√ √ √
√
x+ a a x
=
√ √
√ √
2x 1 + a + 2 a x + x + ax − 2 a x
√
√
√
√
x+a x
x
(1 + a) x
=
=
=
2x (1 + a + x + ax)
2x(1 + a)(1 + x)
2x(1 + x)
Since a has canceled out, the derivative is independent of a. This explains why all the graphs look the same in part
(b). (In fact they are not exactly the same, because f ′ (x) is undefined where f (x) has its jump discontinuity. The
point at which this happens changes with a.)
104. (a) A CAS gives
1
d
arcsinh x = √
dx
1 + x2
(b) Differentiating both sides of sinh(arcsinh x) = x, we get
cosh(arcsinh x)
d
(arcsinh x) = 1
dx
1
d
(arcsinh x) =
.
dx
cosh(arcsinh x)
SOLUTIONS to Review Problems for Chapter Four
429
p
Since cosh2 x − sinh2 x = 1, cosh xp= ± 1 + sinh2 x. Furthermore, since cosh x > 0 for all x, we take
p
1 + sinh2 x. Therefore, cosh(arcsinh x) =
1 + (sinh(arcsinh x))2 =
the positive square root, so cosh x =
√
1 + x2 . Thus
1
d
arcsinh x = √
.
dx
1 + x2
105. (a) A CAS gives
1
d
arccosh x = √
,
dx
x2 − 1
(b) Differentiating both sides of cosh(arccosh x) = x, we get
sinh(arccosh x)
x ≥ 1.
d
(arccosh x) = 1
dx
d
1
(arccosh x) =
.
dx
sinh(arccosh x)
p
2
2
sinh x ≥ 0, so we take the positive square
Since cosh2 x − sinh
p x = 1, sinh x = ± cosh x − 1. If x ≥ 0, thenp
√
2
root. So sinh x = cosh x − 1, x ≥ 0. Therefore, sinh(arccosh x) = (cosh(arccosh x))2 − 1 = x2 − 1, for
x ≥ 1. Thus
d
1
.
arccosh x = √
dx
x2 − 1
106. (a) Using a computer algebra system or differentiating by hand, we get
√
a+x
1
√
√
√
−
.
f ′ (x) = √
√
√
2 x( a + x)2
2 a + x( a + x)
Simplifying gives
f ′ (x) =
2
√
√ √
−a + a x
.
√ 2 √ √
a+ x
x a+x
The denominator
derivative is always positive if x > 0, and the numerator is zero when x = a. Writing the
√ the √
√ of
numerator as a( x − a), we see that the derivative changes from negative to positive at x = a. Thus, by the first
derivative test, the function has a local minimum at x = a.
(b) As a increases, the local minimum moves to the right. See Figure 4.191. This is consistent with what we found in
part (a), since the local minimum is at x = a.
y
1
a=1
a=3
a=5
x
10
Figure 4.191
(c) Using a computer algebra system to find the second derivative when a = 2, we get
√
√
√
4 2 + 12 x + 6 x3/2 − 3 2 x2
′′
.
f (x) =
√
√ 3
4
2 + x x3/2 (2 + x)3/2
Using the computer algebra system again to solve f ′′ (x) = 0, we find that it has one zero at x = 4.6477. Graphing
the second derivative, we see that it goes from positive to negative at x = 4.6477, so this is an inflection point.
430
Chapter Four /SOLUTIONS
107. (a) Different CASs give different answers. (In fact, their answers could be more complicated than what you get by hand.)
One possible answer is
tan x
dy
= q 2 .
dx
x
2 1−cos
1+cos x
(b) The graph in Figure 4.192 is a step function:
f (x) =
(
1/2
2nπ < x < (2n + 1)π
−1/2
(2n + 1)π < x < (2n + 2)π.
y
0.5
x
−4π
−2π
2π
4π
−0.5
Figure 4.192
Figure 4.192, which shows the graph in disconnected line segments, is correct. However, unless you select certain
graphing options in your CAS, it may join up the segments. Use the double angle formula cos(x) = cos2 (x/2) −
sin2 (x/2) to simplify the answer in part (a). We find
tan(x/2)
tan(x/2)
tan(x/2)
dy
= q
= q
= q
dx
1−cos(2·(x/2))
1−cos2 (x/2)+sin2 (x/2)
1−cos x
2 1+cos x
2 1+cos(2·(x/2))
2 1+cos2 (x/2)−sin2 (x/2)
tan(x/2)
tan(x/2)
tan(x/2)
= q
= p
=
2
2 |tan(x/2)|
2 sin2 (x/2)
2
tan
(x/2)
2 2 cos2 (x/2)
Thus, dy/dx = 1/2 when tan(x/2) > 0, i.e. when 0 < x < π (more generally, when 2nπ < x < (2n + 1)π), and
dy/dx = −1/2 when tan(x/2) < 0, i.e., when π < x < 2π (more generally, when (2n + 1)π < x < (2n + 2)π,
where n is any integer).
108. (a)
A
✛
x
E
✲✛
❑
√
3−x
1m
❯
C
D
Figure 4.193
B
✲
PROJECTS FOR CHAPTER FOUR
431
We want to maximize the sum of the lengths EC and CD in Figure 4.193. Let x be the distance AE. Then x
can be between 0 and 1, the length of the left rope. By the Pythagorean theorem,
EC =
p
1 − x2 .
The length of the rope from B to C can also be found by the Pythagorean theorem:
BC =
p
EC 2 + EB 2 =
p
√
√
1 − x2 + ( 3 − x)2 = 4 − 2 3x.
q
Since the entire rope from B to D has length 3 m, the length from C to D is
CD = 3 −
The distance we want to maximize is
f (x) = EC + CD =
Differentiating gives
p
√
4 − 2 3x.
p
1 − x2 + 3 −
√
4 − 2 3x,
p
for
0 ≤ x ≤ 1.
√
−2 3
−2x
− p
f ′ (x) = √
√ .
2 1 − x2
2 4 − 2 3x
Setting f ′ (x) = 0 gives the cubic equation
√
2 3x3 − 7x2 + 3 = 0.
√
√
√
Using a computer algebra system to solve the equation gives three roots: x = −1/ 3, x = 3/2, x = 3. We
discard the negative
√ root. Since x cannot be larger than 1 meter (the length of the left rope), the only critical point of
interest is x = 3/2, that is, halfway between A and B.
To find the global maximum, we calculate the distance of the weight from the ceiling at the critical point and at
the endpoints:
√
√
f (0) = 1 + 3 − 4 = 2
r
√ r
√
√
3
3
3
= 2.5
f
= 1− +3− 4−2 3·
2
4
2
p
√
√
√
f (1) = 0 + 3 − 4 − 2 3 = 4 − 3 = 2.27.
√
Thus, the weight is at the maximum distance from the ceiling when x = 3/2; that is, the weight comes to rest at a
point halfway between points A and B.
(b) No, the equilibrium position depends on the length of the rope. For example, suppose that the left-hand rope was 1
cm long. Then there is no way for the pulley at its end to move to a point halfway between the anchor points.
PROJECTS FOR CHAPTER FOUR
1.
(a) (i) The slope, dV /dt, is the rate of change of the volume of air that has been exhaled over time. Thus,
it is the rate of flow of the air out of the lungs, measured in liters per second. The greater the slope,
the faster air is being expelled from the lungs. Notice that dV /dt is a volume flow; that is, the rate at
which the volume of air exhaled is changing. It is not the speed at which air particles are leaving the
lungs, which would be measured in meters per second.
(ii) As the horizontal asymptote on the graph, the vital capacity VC is the volume of air expelled at the
end of the exhalation. It is the maximum volume of air the patient can exhale in one breath.
(b) (i) The slope of the flow-volume curve tells you how much the flow rate of air out of the lungs changes
in response to exhalation of an additional volume of air.
432
Chapter Four /SOLUTIONS
(ii) For V near 0, when little air has been exhaled, the slope of the flow-volume curve is steep and positive.
This means that the flow rate dV /dt is rapidly increasing as air is exhaled. The patient is expelling
air faster and faster. The flow volume curve has zero slope when approximately 1 liter has been
exhaled. This corresponds to the maximum flow rate dV /dt; air is expelled most rapidly at this stage.
Thereafter, the slope of the flow-volume curve is negative, so the flow rate decreases and air is expelled
slower and slower until approximately 5.5 liters of air have been exhaled. At that point, the patient
has no more air to exhale and must take a breath.
(iii) Suppose the flow rate, dV /dt, is given as function of V by the function f , so dV /dt = f (V ); the
flow-volume curve we were given is a graph of f . Now we are asked to plot f ′ (V ), whose graph is in
Figure 4.194.
flow rate/liter
20
15
10
5
f ′ (V )
1
2
3
4
5
6
V , liters
Figure 4.194
(iv) The maximal rate is identified as the highest point on the flow-volume curve, which occurs at about
V = 1 liter. Thus, the patient’s peak expiratory flow occurs when 1 liter of air has been exhaled. The
peak expiratory flow is about 11.5 liters per second.
On the slope graph in Figure 4.194, a critical point occurs where the slope is 0; that is, where
the slope graph intersects the horizontal axis. This occurs at V = 1. The first derivative test tells us
that this critical point gives a maximum flow rate since the slope is positive to the left of V = 1 and
negative to the right of V = 1.
(c) The increased resistance to airflow means that the patient exhales more slowly. This makes the volumetime curve rise more slowly. In fact, sometimes the patient exhales so slowly that they “run out of time”
and cannot exhale all the volume they want to, leading to air trapping. This corresponds to a lowered vital
capacity, VC. See Figure 4.195.
The peak expiratory flow will be lower. This means that the flow-volume curve does not rise as high.
See Figure 4.196.
V , liters
6
5
4
3 VC
2
1
t, sec
1 2 3 4 5 6 7 8 9 10
Figure 4.195: Volume-time curve for an
asthmatic
dV /dt, liters/sec
12
10
8
6
4
2
V , liters
1 2 3 4 5 6
Figure 4.196: Flow-volume curve for an
asthmatic
PROJECTS FOR CHAPTER FOUR
2.
✛
Remaining wall
✛
A
433
Glass
✲
✻✠
D
h
❄
✛x✲✛
θ
y
✛
Removed
wall
✲
Figure 4.197: A Cross-section of the Projected Greenhouse
Suppose that the glass is at an angle θ (as shown in Figure 4.197), that the length of the wall is l, and that
the glass has dimensions D ft by l ft. Since your parents will spend a fixed amount, the area of the glass, say
k ft2 , is fixed:
Dl = k.
The width of the extension is D cos θ. If h is the height of your tallest parent, he or she can walk in a distance
of x, and
h
h
= tan θ, so y =
.
y
tan θ
Thus,
x = D cos θ − y = D cos θ −
h
tan θ
for 0 < θ <
π
.
2
We maximize x since doing so maximizes the usable area:
h
1
dx
= −D sin θ +
·
=0
dθ
(tan θ)2 (cos θ)2
h
sin3 θ =
D
1/3 !
h
θ = arcsin
.
D
This is the only critical point, and x → 0 when θ → 0 and when θ → π/2. Thus, the critical point is a global
maximum. Since
s
2/3
p
h
2
,
cos θ = 1 − sin θ = 1 −
D
the maximum value of x is
h cos θ
h
= D cos θ −
tan θ
sin θ
2/3 !1/2
h
h
h
cos θ = D −
= D−
· 1−
sin θ
D
(h/D)1/3
1/2
h2/3
= (D − h2/3 D1/3 ) · 1 − 2/3
D
1/2
3/2
2/3
h2/3
h2/3
h
= D 1 − 2/3
.
= D 1 − 2/3 · 1 − 2/3
D
D
D
x = D cos θ −
434
Chapter Four /SOLUTIONS
This means
Maximum Usable Area = lx
3/2
h2/3
= lD 1 − 2/3
D
2/3 !3/2
hl
=k 1−
k
3. (a) The point on the line y = mx corresponding to the point (2, 3.5) has y-coordinate given by y = m(2) =
2m. Thus, for the point (2, 3.5)
Vertical distance to the line = |2m − 3.5|.
We calculate the distance similarly for the other two points. We want to minimize the sum, S, of the
squares of these vertical distances
S = (2m − 3.5)2 + (3m − 6.8)2 + (5m − 9.1)2 .
Differentiating with respect to m gives
dS
= 2(2m − 3.5) · 2 + 2(3m − 6.8) · 3 + 2(5m − 9.1) · 5.
dm
Setting dS/dm = 0 gives
2 · 2(2m − 3.5) + 2 · 3(3m − 6.8) + 2 · 5(5m − 9.1) = 0.
Canceling a 2 and multiplying out gives
4m − 7 + 9m − 20.4 + 25m − 45.5 = 0
38m = 72.9
m = 1.92.
Thus, the best fitting line has equation y = 1.92x.
(b) To fit a line of the form y = mx to the data, we take y = V and x = r3 . Then k will be the slope m. So
we make the following table of data:
r
2
5
7
8
x = r3
8
125
343
512
y=V
8.7
140.3
355.8
539.2
To find the best fitting line of the form y = mx, we minimize the sums of the squares of the vertical
distances from the line. For the point (8, 8.7) the corresponding point on the line has y = 8m, so
Vertical distance = |8m − 8.7|.
We find distances from the other points similarly. Thus we want to minimize
S = (8m − 8.7)2 + (125m − 140.3)2 + (343m − 355.8)2 + (512m − 539.2)2 .
Differentiating with respect to m, which is the variable, and setting the derivative to zero:
dS
= 2(8m − 8.7) · 8 + 2(125m − 140.3) · 125 + 2(343m − 355.8) · 343 + 2(512m − 539.2) · 512 = 0.
dm
PROJECTS FOR CHAPTER FOUR
435
After canceling a 2, solving for m leads to the equation
82 m + 1252 m + 3432 m + 5122m = 8 · 8.7 + 125 · 140.3 + 343 · 355.8 + 512 · 539.2
m = 1.051.
Thus, k = 1.051 and the relationship between V and r is
V = 1.051r3 .
(In fact, the correct relationship is V = πr3 /3, so the exact value of k is π/3 = 1.047.)
(c) The best fitting line minimizes the sum of the squares of the vertical distances from points to the line.
Since the point on the line y = mx corresponding to (x1 , y1 ) is the point with y = mx1 ; for this point we
have
Vertical distance = |mx1 − y1 |.
We calculate the distance from the other points similarly. Thus we want to minimize
S = (mx1 − y1 )2 + (mx2 − y2 )2 · · · + (mxn − yn )2 .
The variable is m (the xi s and yi s are all constants), so
dS
= 2(mx1 − y1 )x1 + 2(mx2 − y2 )x2 + · · · + 2(mxn − yn )xn = 0
dm
2(m(x21 + x22 + · · · + x2n ) − (x1 y1 + x2 y2 + · · · + xn yn )) = 0.
Solving for m gives
m=
n
P
xi yi
x1 y1 + x2 y2 + · · · + xn yn
i=1
.
=
n
P
x21 + x22 + · · · + x2n
x2i
i=1
4. (a) (i) We want to minimize A, the total area lost to the forest, which is made up of n firebreaks and 1 stand
of trees lying between firebreaks. The area of each firebreak is (50 km)(0.01 km) = 0.5 km2 , so
the total area lost to the firebreaks is 0.5n km2 . There are n total stands of trees between firebreaks.
The area of a single stand of trees can be found by subtracting the firebreak area from the forest and
dividing by n, so
2500 − 0.5n
.
Area of one stand of trees =
n
Thus, the total area lost is
A = Area of one stand + Area lost to firebreaks
2500
2500 − 0.5n
+ 0.5n =
− 0.5 + 0.5n.
=
n
n
We assume that A is a differentiable function of a continuous variable, n. Differentiating this function
yields
dA
2500
= − 2 + 0.5.
dn
n
p
2
At critical points, dA/dn = 0, so 0.5 = 2500/n or n = 2500/0.5 ≈ 70.7. Since n must be an
integer, we check that when n = 71, A = 70.211 and when n = 70, A = 70.214. Thus, n = 71 gives
a smaller area lost.
We can check that this is a local minimum since the second derivative is positive everywhere
5000
d2 A
= 3 > 0.
dn2
n
Finally, we check the endpoints: n = 1 yields the entire forest lost after a fire, since there is only
one stand of trees in this case and it all burns. The largest n is 5000, and in this case the firebreaks
remove the entire forest. Both of these cases maximize the area of forest lost. Thus, n = 71 is a global
minimum. So 71 firebreaks minimizes the area of forest lost.
436
Chapter Four /SOLUTIONS
(ii) Repeating the calculation using b for the width gives
A=
2500
− 50b + 50bn,
n
and
dA
−2500
=
+ 50b,
dn
n2
p
with a critical point when b = 50/n2 so n = 50/b. So, for example, if we make the width b four
times as large we need half as many firebreaks.
(b) We want to minimize A, the total area lost to the forest, which is made up of n firebreaks in one direction,
n firebreaks in the other, and one square of trees surrounded by firebreaks. The area of each firebreak is
0.5 km2 , and there are 2n of them, giving a total of 0.5 · 2n. But this is larger than the total area covered
by the firebreaks, since it counts the small intersection squares, of size (0.01)2 , twice. Since there are n2
intersections, we must subtract (0.01)2 n2 from the total area of the 2n firebreaks. Thus,
Area covered by the firebreaks = 0.5 · 2n − (0.01)2 n2 .
To this we must add the area of one square patch of trees lost in a fire. These are squares of side (50 −
0.01n)/n = 50/n − 0.01. Thus the total area lost is
A = n − 0.0001n2 + (50/n − 0.01)2
Treating n as a continuous variable and differentiating this function yields
dA
50
−50
.
= 1 − 0.0002n + 2
− 0.01
dn
n
n2
Using a computer algebra system to find critical points we find that dA/dn = 0 when n ≈ 17 and
n = 5000. Thus n = 17 gives a minimum lost area, since the endpoints of n = 1 and n = 5000 both yield
A = 2500 or the entire forest lost. So we use 17 firebreaks in each direction.
5.1 SOLUTIONS
437
CHAPTER FIVE
Solutions for Section 5.1
Exercises
1. (a)
(b)
(c)
(d)
(e)
2. (a)
Left sum
Upper estimate
6
∆t = 2
Upper estimate is approximately 4 · 2 + 2.9 · 2 + 2 · 2 + 1.5 · 2 + 1 · 2 + 0.8 · 2 = 24.4.
(i) Since the velocity is increasing, for an upper estimate we use a right sum. Using n = 4, we have ∆t = 3, so
Upper estimate = (37)(3) + (38)(3) + (40)(3) + (45)(3) = 480.
(ii) Using n = 2, we have ∆t = 6, so
Upper estimate = (38)(6) + (45)(6) = 498.
(b) The answer using n = 4 is more accurate as it uses the values of v(t) when t = 3 and t = 9.
(c) Since the velocity is increasing, for a lower estimate we use a left sum. Using n = 4, we have ∆t = 3, so
Lower estimate = (34)(3) + (37)(3) + (38)(3) + (40)(3) = 447.
3. (a) Since the velocity is decreasing, for an upper estimate, we use a left sum. With n = 5,we have ∆t = 2. Then
Upper estimate = (44)(2) + (42)(2) + (41)(2) + (40)(2) + (37)(2) = 408.
(b) For a lower estimate, we use a right sum, so
Lower estimate = (42)(2) + (41)(2) + (40)(2) + (37)(2) + (35)(2) = 390.
4. (a) Lower estimate = 60 + 40 + 25 + 10 + 0 = 135 feet. Upper estimate = 88 + 60 + 40 + 25 + 10 = 223 feet.
(b) See Figure 5.1.
velcoity
88
60
40
25
10
time
1
2
3
4
5
Figure 5.1
(c) The difference between the estimates = 223 − 135 = 88 feet. This is the sum of the lightly shaded areas in the graph,
namely (88 − 0) · 1 = 88 feet.
438
Chapter Five /SOLUTIONS
5. The distance traveled is represented by area under the velocity curve. We can approximate the area using left- and righthand sums. Alternatively, counting the squares (each of which has area 10), and allowing for the broken squares, we can
see that the area under the curve from 0 to 6 is between 140 and 150. Hence the distance traveled is between 140 and 150
meters.
6. Using ∆t = 2,
Lower estimate = v(0) · 2 + v(2) · 2 + v(4) · 2
= 1(2) + 5(2) + 17(2)
= 46
Upper estimate = v(2) · 2 + v(4) · 2 + v(6) · 2
= 5(2) + 17(2) + 37(2)
= 118
46 + 118
= 82
Average =
2
Distance traveled ≈ 82 meters.
7. (a) The velocity is always positive, so the particle is moving in the same direction throughout. However, the particle is
speeding up until shortly before t = 0, and slowing down thereafter.
(b) The distance traveled is represented by the area under the curve. We can calculate over and understimates for the area
using a combination of left- and right-hand sums. Alternatively, using whole grid squares, we can overestimate the
area as 3 + 3 + 3 + 3 + 2 + 1 = 15 cm, and we can underestimate the area as 1 + 2 + 2 + 1 + 0 + 0 = 6 cm.
8. Using ∆t = 0.2, our upper estimate is
1
1
1
1
1
(0.2) +
(0.2) +
(0.2) +
(0.2) +
(0.2) ≈ 0.75.
1+0
1 + 0.2
1 + 0.4
1 + 0.6
1 + 0.8
The lower estimate is
1
1
1
1
1
(0.2) +
(0.2) +
(0.2) +
(0.2)
(0.2) ≈ 0.65.
1 + 0.2
1 + 0.4
1 + 0.6
1 + 0.8
1+1
Since v is a decreasing function, the bug has crawled more than 0.65 meters, but less than 0.75 meters. We average the
two to get a better estimate:
0.65 + 0.75
= 0.70 meters.
2
9. From t = 0 to t = 3 the velocity is constant and positive, so the change in position is 2 · 3 cm, that is 6 cm to the right.
From t = 3 to t = 5, the velocity is negative and constant, so the change in position is −3 · 2 cm, that is 6 cm to the left.
Thus the total change in position is 0. The particle moves 6 cm to the right, followed by 6 cm to the left, and returns to
where it started.
10. From t = 0 to t = 5 the velocity is positive so the change in position is to the right. The area under the velocity graph
gives the distance traveled. The region is a triangle, and so has area (1/2)bh = (1/2)5 · 10 = 25. Thus the change in
position is 25 cm to the right.
11. The velocity is constant and negative, so the change in position is −3 · 5 cm, that is 15 cm to the left.
12. From t = 0 to t = 4 the velocity is positive so the change in position is to the right. The area under the velocity graph
gives the distance traveled. The region is a triangle, and so has area (1/2)bh = (1/2)4 · 8 = 16. Thus the change in
position is 16 cm to the right for t = 0 to t = 4. From t = 4 to t = 5, the velocity is negative so the change in position is
to the left. The distance traveled to the left is given by the area of the triangle, (1/2)bh = (1/2)1 · 2 = 1. Thus the total
change in position is 16 − 1 = 15 cm to the right.
13. (a) With n = 4, we have ∆t = 2. Then
t0 = 15, t1 = 17, t2 = 19, t3 = 21, t4 = 23
and f (t0 ) = 10, f (t1 ) = 13, f (t2 ) = 18, f (t3 ) = 20, f (t4 ) = 30
(b)
Left sum = (10)(2) + (13)(2) + (18)(2) + (20)(2) = 122
Right sum = (13)(2) + (18)(2) + (20)(2) + (30)(2) = 162.
5.1 SOLUTIONS
439
(c) With n = 2,we have ∆t = 4. Then
t0 = 15, t1 = 19, t2 = 23
f (t0 ) = 10, f (t1 ) = 18, f (t2 ) = 30
and
(d)
Left sum = (10)(4) + (18)(4) = 112
Right sum = (18)(4) + (30)(4) = 192.
14. (a) With n = 4, we have ∆t = 4. Then
t0 = 0, t1 = 4, t2 = 8, t3 = 12, t4 = 16 and
f (t0 ) = 25, f (t1 ) = 23, f (t2 ) = 22, f (t3 ) = 20, f (t4 ) = 17
(b)
Left sum = (25)(4) + (23)(4) + (22)(4) + (20)(4) = 360
Right sum = (23)(4) + (22)(4) + (20)(4) + (17)(4) = 328.
(c) With n = 2,we have ∆t = 8. Then
t0 = 0, t1 = 8, t2 = 16
and f (t0 ) = 25, f (t1 ) = 22, f (t2 ) = 17
(d)
Left sum = (25)(8) + (22)(8) = 376
Right sum = (22)(8) + (17)(8) = 312.
Problems
15. (a) Note that 15 minutes equals 0.25 hours. Lower estimate = 11(0.25) + 10(0.25) = 5.25 miles. Upper estimate
= 12(0.25) + 11(0.25) = 5.75 miles.
(b) Lower estimate = 11(0.25) + 10(0.25) + 10(0.25) + 8(0.25) + 7(0.25) + 0(0.25) = 11.5 miles. Upper estimate
= 12(0.25) + 11(0.25) + 10(0.25) + 10(0.25) + 8(0.25) + 7(0.25) = 14.5 miles.
(c) The difference between Roger’s pace at the beginning and the end of his run is 12 mph. If the time between the
measurements is h, then the difference between the upper and lower estimates is 12h. We want 12h < 0.1, so
0.1
≈ 0.0083 hours = 30 seconds
12
Thus Jeff would have to measure Roger’s pace every 30 seconds.
h<
16. The change in position is calculated from the area between the velocity graph and the t-axis, with the region below the
axis corresponding to negatives velocities and counting negatively.
Figure 5.2 shows the graph of f (t). From t = 0 to t = 3 the velocity is positive. The region under the graph of f (t)
is a triangle with height 6 cm/sec and base 3 seconds. Thus, from t = 0 to t = 3, the particle moves
1
· 3 · 6 = 9 centimeters.
2
From t = 3 to t = 4, the velocity is negative. The region between the graph of f (t) and the t-axis is a triangle with height
2 cm/sec and base 1 second, so in this interval the particle moves
Distance moved to right =
1
· 1 · 2 = 1 centimeter.
2
Thus, the total change in position is 9 − 1 = 8 centimeters to the right.
Distance moved to left =
y
6
Motion to right: 9 cm
✠
4
3
−2
Motion to left: 1 cm
Figure 5.2
✒
t
f (t)
440
Chapter Five /SOLUTIONS
17. Since f is increasing, the right-hand sum is the upper estimate and the left-hand sum is the lower estimate. We have
f (a) = 13, f (b) = 23 and ∆t = (b − a)/n = 2/100. Thus,
|Difference in estimates| = |f (b) − f (a)|∆t
1
1
= .
= |23 − 13|
50
5
18. Since f is decreasing, the right-hand sum is the lower estimate and the left-hand sum is the upper estimate. We have
f (a) = 24, f (b) = 9 and ∆t = (b − a)/n = 3/500. Thus,
|Difference in estimates| = |f (b) − f (a)|∆t
3
= 0.09.
= |9 − 24|
500
19. Since f is increasing, the right-hand sum is the upper estimate and the left-hand sum is the lower estimate. We have
f (0) = 0, f (π/2) = 1 and ∆t = (b − a)/n = π/200. Thus,
|Difference in estimates| = |f (b) − f (a)|∆t
π
= 0.0157.
= |1 − 0|
200
20. Since f is decreasing, the right-hand sum is the lower estimate and the left-hand sum is the upper estimate. We have
f (0) = 1, f (2) = e−2 and ∆t = (b − a)/n = 2/20 = 1/10. Thus,
|Difference in estimates| = |f (b) − f (a)|∆t
1
= 0.086.
= |e−2 − 1|
10
21. (a) See Figure 5.3.
(b) The peak of the flight is when the velocity is 0, namely t = 3. The height at t = 3 is given by the area under the
graph of the velocity from t = 0 to t = 3; see Figure 5.3. The region is a triangle of base 3 seconds and altitude 96
ft/sec, so the height is (1/2)3 · 96 = 144 feet.
(c) The velocity is negative from t = 3 to t = 5, so the motion is downward then. The distance traveled downward can
be calculated by the area of the triangular region which has base of 2 seconds and altitude of −64 ft/sec. Thus, the
baseball travels (1/2)2 · 64 = 64 feet downward from its peak height of 144 feet at t = 3. Thus, the height at time
t = 5 is the total change in position, 144 − 64 = 80 feet.
96
3
6
t
v(t)
−96
Figure 5.3
22. From t = 0 to t = 3, you are moving away from home (v > 0); thereafter you move back toward home. So you are the
farthest from home at t = 3. To find how far you are then, we can measure the area under the v curve as about 9 squares,
or 9 · 10 km/hr · 1 hr = 90 km. To find how far away from home you are at t = 5, we measure the area from t = 3 to
t = 5 as about 25 km, except that this distance is directed toward home, giving a total distance from home during the trip
of 90 − 25 = 65 km.
5.1 SOLUTIONS
441
23. (a) When the aircraft is climbing at v ft/min, it takes 1/v minutes to climb 1 foot. Therefore
1 min
1 min
1 min
(1000 ft) +
(1000 ft) + · · · +
(1000 ft)
925 ft
875 ft
490 ft
≈ 14.73 minutes.
1 min
1 min
1 min
(1000 ft) +
(1000 ft) + · · · +
(1000 ft)
Upper estimate =
875 ft
830 ft
440 ft
≈ 15.93 minutes.
Lower estimate =
Note: The Pilot Operating Manual for this aircraft gives 16 minutes as the estimated time required to climb to 10,000
ft.
(b) The difference between upper and lower sums with ∆x = 500 ft would be
Difference =
1 min
440 ft
−
1 min
(500 ft) = 0.60 minutes.
925 ft
24. (a) At t = 20 minutes, she stops moving toward the lake (with v > 0) and starts to move away from the lake (with
v < 0). So at t = 20 minutes the cyclist turns around.
(b) The cyclist is going the fastest when v has the greatest magnitude, either positive or negative. Looking at the graph,
we can see that this occurs at t = 40 minutes, when v = −25 and the cyclist is pedaling at 25 km/hr away from the
lake.
(c) From t = 0 to t = 20 minutes, the cyclist comes closer to the lake, since v > 0; thereafter, v < 0 so the cyclist
moves away from the lake. So at t = 20 minutes, the cyclist comes the closest to the lake. To find out how close she
is, note that between t = 0 and t = 20 minutes the distance she has come closer is equal to the area under the graph
of v. Each box represents 5/6 of a kilometer, and there are about 2.5 boxes under the graph, giving a distance of about
2 km. Since she was originally 5 km away, she then is about 5 − 2 = 3 km from the lake.
(d) At t = 20 minutes she turns around , since v changes sign then. Since the area below the t-axis is greater than the
area above, the farthest she is from the lake is at t = 60 minutes. Between t = 20 and t = 60 minutes, the area under
the graph is about 10.8 km. (Since 13 boxes · 5/6 = 10.8.) So at t = 60 she will be about 3 + 10.8 = 13.8 km from
the lake.
25. (a) Since car B starts at t = 2, the tick marks on the horizontal axis (which we assume are equally spaced) are 2 hours
apart. Thus car B stops at t = 6 and travels for 4 hours.
Car A starts at t = 0 and stops at t = 8, so it travels for 8 hours.
(b) Car A’s maximum velocity is approximately twice that of car B, that is 100 km/hr.
(c) The distance traveled is given by the area of under the velocity graph. Using the formula for the area of a triangle, the
distances are given approximately by
1
· Base · Height =
2
1
Car B travels = · Base · Height =
2
Car A travels =
1
· 8 · 100 = 400 km
2
1
· 4 · 50 = 100 km.
2
26. (a) Car A has the largest maximum velocity because the peak of car A’s velocity curve is higher than the peak of B’s.
(b) Car A stops first because the curve representing its velocity hits zero (on the t-axis) first.
(c) Car B travels farther because the area under car B’s velocity curve is the larger.
27. (a) See Figure 5.4.
(b) The distance traveled is the area under the graph of the velocity in Figure 5.4. The region is a triangle of base 5
seconds and altitude 50 ft/sec, so the distance traveled is (1/2)5 · 50 = 125 feet.
(c) The slope of the graph of the velocity function is the same, so the triangular region under it has twice the altitude and
twice the base (it takes twice as long to stop). See Figure 5.5. Thus, the area is 4 times as large and the car travels 4
times as far.
442
Chapter Five /SOLUTIONS
100
50
50
v(t)
v(t)
5
t
5
Figure 5.4
10
t
Figure 5.5
28. The graph of her velocity against time is a straight line from 0 mph to 60 mph; see Figure 5.6. Since the distance traveled
is the area under the curve, we have
1
Shaded area = · t · 60 = 10 miles
2
Solving for t gives
1
t = hr = 20 minutes .
3
velocity (mph)
60
✛
t
Area = 10
time (hr)
Figure 5.6
29. Since acceleration is the derivative of velocity (a(t) = v ′ (t)), and a(t) is a constant, v(t) must be a linear function. Its
v-intercept is 0 since the object has zero initial velocity. Thus the graph of v is a line through the origin with slope 32 and
equation v(t) = 32t ft/sec. See Figure 5.7.
Using ∆t = 1, the right sum, 32 + 64 + 96 + 128 = 320 feet, is an upper bound on the distance traveled. The left
sum, 0 + 32 + 64 + 96 = 192 feet, is a lower bound.
To find the actual distance, we find the exact area of the region below the line—it is the triangle in Figure 5.7 with
height 32 · 4 = 128 and base 4, so its area is 256 ft.
128
96
64
v(t) = 32t
32
1
2
3
4
t
Figure 5.7
30. No, the 2010 Prius Plug-in Prototype cannot travel half a mile in EV-only mode in under 25 seconds, because the upper
estimate for the distance traveled in 25 seconds is only about 0.3 miles. To see this, we convert seconds to hours (1 sec
= 1/3600 hours) so that the upper distance estimate is in miles:
Total distance traveled
between t = 0 and t = 25
(Upper estimate)
= 20 ·
5
5
5
5
5
+ 33 ·
+ 45 ·
+ 50 ·
+ 59 ·
< 0.3 miles
3600
3600
3600
3600
3600
5.1 SOLUTIONS
443
31. To estimate the distance between the two cars at t = 15, we calculate upper and lower bound estimates for this distance
and then average these estimates. Note that because the speed of the Prius is in miles per hour, we need to convert seconds
to hours (1 sec = 1/3600 hours) to calculate the distance estimates in miles. We may then convert miles to feet (1 mile
= 5280 feet).
Distance between cars at t = 15
(Upper Estimate)
≈ (Difference in speeds at t = 5) · (Travel time)
+ (Difference in speeds at t = 10) · (Travel time)
+ (Difference in speeds at t = 15) · (Travel time)
5
5
5
+ (53 − 33) ·
+ (70 − 45) ·
= 0.081 miles
= (33 − 20) ·
3600
3600
3600
Distance between cars at t = 15
(Lower Estimate)
≈ (Difference in speeds at t = 0) · (Travel time)
+ (Difference in speeds at t = 5) · (Travel time)
+ (Difference in speeds at t = 10) · (Travel time)
5
5
5
+ (33 − 20) ·
+ (53 − 33) ·
= 0.046 miles
= (0 − 0) ·
3600
3600
3600
Since:
0.081 + 0.046
Average of Lower and Upper Estimates of
=
= 0.0635 miles = 335 feet,
2
distance between cars at t = 15
the distance between the two cars is about 335 feet, 15 seconds after leaving the stoplight.
Strengthen Your Understanding
32. This is only true if the car accelerates at a constant rate, that is, if the graph of the velocity is a linear function increasing
from 0 to 50 in 10 seconds. If the graph of velocity is above that line, the car travels further, and if the graph of velocity
is below that line the car travels less far.
33. Recording the velocity every 0.1 seconds means there are 10 subdivisions, so if the velocity of the car in ft/sec is f (t)
after t seconds, and if f is increasing during the second, then the difference between the lower and upper estimates is
(f (b) − f (a))
b−a
,
10
where [a, b] is the interval during which the velocity is recorded. Since the interval is one second long, b − a = 1, so the
difference is (f (b) − f (a))/10. This could be much bigger than 0.1 feet if the car is accelerating quickly. For example, if
f (a) = 0 and f (b) = 10, then the difference between the estimates is 1 foot.
34. If f is decreasing on the interval then the right sum is less than the left sum, so f (x) = x2 , [a, b] = [−2, −1] is an
example.
35. If f (t) is increasing on the interval [a, b], then the difference between the upper and lower estimates is |f (b)−f (a)|·∆t =
|f (b) − f (a)|(b − a)/n, where n is the number of subdivisions. So if, for example, f (b) − f (a) = 10 and b − a = 1,
then
10
b−a
=
,
|f (b) − f (a)|
n
n
and n would have to be ≥ 100 to make this ≤ 0.1. A velocity function and interval that satisfy these conditions are
f (x) = 10x and [a, b] = [0, 1].
36. True. Since the velocity is increasing, each term in the left-hand sum is less than the corresponding term in the right-hand
sum.
37. True. If the velocity is f (t) and the interval is a ≤ t ≤ b, then
Difference between left and right sums = (f (a) − f (b)) · ∆t.
Since ∆t is halved when the number of subdivisions is doubled, the difference is halved also.
444
Chapter Five /SOLUTIONS
38. False. For example, for a constant function, the difference does not get smaller, since it is always 0. Another example is
the velocity function f (x) = x2 on the interval −1 ≤ x ≤ 1. By the symmetry of the graph, for an even number of
subdivisions the difference between the left and right sums is always 0.
39. (a)
(b)
(c)
(d)
On [3, 5] and on [9, 10], since v = 0 there.
3600 feet to the east, since this is the area under the velocity curve between t = 0 and t = 3.
At t = 8 minutes, since the areas above and below the curve between t = 0 and t = 8 are equal.
It will take her 30 seconds longer. By calculating areas, we see that at t = 11,
Distance from home = 2 · 30 · 60 − 3 · 30 · 60 + 0.5 · 30 · 60 = −900 feet.
Thus, at t = 11, she is 900 feet west of home. At a velocity of 30 ft/sec eastward, it takes 900/30 = 30 seconds to
get home.
Solutions for Section 5.2
Exercises
1. (a)
(b)
(c)
(d)
Right sum
Upper estimate
3
∆x = 2
2. (a)
(b)
(c)
(d)
Left sum
Lower estimate
3
∆x = 2
3. (a) Left-hand sum. Right-hand sum would be smaller.
(b) We have a = 0, b = 2, n = 6, ∆x = 62 = 13 .
4.
32
32
24
24
f (t)
16
f (t)
16
8
8
2
4
6
t
8
Figure 5.8: Left Sum, ∆t = 4
2
4
6
8
t
Figure 5.9: Right Sum, ∆t = 4
(a) Left-hand sum = 32 · 4 + 24 · 4 = 224.
(b) Right-hand sum = 24 · 4 + 0 · 4 = 96.
32
32
24
24
16
f (t)
16
f (t)
8
8
2
4
6
8
t
Figure 5.10: Left Sum, ∆t = 2
(c) Left-hand sum = 32 · 2 + 30 · 2 + 24 · 2 + 14 · 2 = 200.
(d) Right-hand sum = 30 · 2 + 24 · 2 + 14 · 2 + 0 · 2 = 136.
2
4
6
8
Figure 5.11: Right Sum, ∆t = 2
t
5.2 SOLUTIONS
5. We use a calculator or computer to see that
4
Z
(x2 + x) dx = 28.5.
1
6. We use a calculator or computer to see that
3
Z
2x dx = 10.0989.
0
7. We use a calculator or computer to see that
1
Z
2
e−x dx = 1.4936.
−1
3
8. We use a calculator or computer to see that
Z
ln(y 2 + 1) dy = 3.406.
0
9. We use a calculator or computer to see that
1
Z
sin(t2 )dt = 0.3103.
0
10. We use a calculator or computer to see that
4
Z
√
ez + z dz = 6.111.
3
11. We estimate
R 40
0
f (x)dx using left- and right-hand sums:
Left sum = 350 · 10 + 410 · 10 + 435 · 10 + 450 · 10 = 16,450.
Right sum = 410 · 10 + 435 · 10 + 450 · 10 + 460 · 10 = 17,550.
We estimate that
Z
40
f (x)dx ≈
0
16450 + 17550
= 17,000.
2
In this estimate, we used n = 4 and ∆x = 10.
12. With ∆x = 3, we have
Left-hand sum = 3(32 + 22 + 15 + 11) = 240,
Right-hand sum = 3(22 + 15 + 11 + 9) = 171.
The average of these two sums is our best guess for the value of the integral;
Z
12
0
f (x) dx ≈
240 + 171
= 205.5.
2
13. We take ∆x = 3. Then:
Left-hand sum = 50(3) + 48(3) + 44(3) + 36(3) + 24(3)
= 606
Right-hand sum = 48(3) + 44(3) + 36(3) + 24(3) + 8(3)
= 480
606 + 480
= 543.
Average =
2
So,
Z
0
15
f (x) dx ≈ 543.
14. Since we have 5 subdivisions,
b−a
7−3
=
= 0.8.
n
5
The interval begins at x = 3 and ends at x = 7. Table 5.1 gives the value of f (x) at the pertinent points.
∆x =
Table 5.1
x
3.0
3.8
4.6
5.4
6.2
7.0
f (x)
1
1 + 3.0
1
1 + 3.8
1
1 + 4.6
1
1 + 5.4
1
1 + 6.2
1
1 + 7.0
So a right-hand sum is
1
1
1
(0.8) +
(0.8) + · · · +
(0.8).
1 + 3.8
1 + 4.6
1 + 7.0
445
446
15.
Chapter Five /SOLUTIONS
Z
20
f (x) dx is equal to the area shaded in Figure 5.12. We estimate the area by counting boxes. There are about 15 boxes
0
and each box represents 4 square units, so the area shaded is about 60. We have
20
Z
f (x) dx ≈ 60.
0
5
f (x)
4
3
2
1
x
4
8 12 16 20
Figure 5.12
16. We know that
Z
15
−10
f (x)dx = Area under f (x) between x = −10 and x = 15.
The area under the curve consists of approximately 14 boxes, and each box has area (5)(5) = 25. Thus, the area under
the curve is about (14)(25) = 350, so
15
Z
−10
17. We know that
Z
f (x)dx ≈ 350.
5
f (x)dx = Area above the axis − Area below the axis.
−3
The area above the axis is about 3 boxes. Since each box has area (1)(5) = 5, the area above the axis is about (3)(5) = 15.
The area below the axis is about 11 boxes, giving an area of about (11)(5) = 55. We have
Z
5
−3
f (x)dx ≈ 15 − 55 = −40.
Problems
18. The graph given shows that f is positive for 0 ≤ t ≤ 1. Since the graph is contained within Ra rectangle of height 100 and
1
length 1, the answers −98.35 and 100.12 are both either too small or too large to represent 0 f (t)dt. Since the graph of
f is above the horizontal line y = 80 for 0 ≤ t ≤ 0.95, the best estimate is IV, 93.47 and not 71.84.
19. (a) The total area between f (x) and the x-axis is the sum of the two given areas, so
Area = 7 + 6 = 13.
(b) To find the integral, we note that from x = 3 to x = 5, the function lies below the x-axis, and hence makes a negative
contribution to the integral. So
Z
0
5
f (x) dx =
Z
3
f (x)dx +
0
Z
3
5
f (x)dx = 7 − 6 = 1.
5.2 SOLUTIONS
447
20. The graph of y = 4 − x2 crosses the x-axis at x = 2 since solving y = 4 − x2 = 0 gives x = ±2. See Figure 5.13. To
find the total area, we find the area above the axis and the area below the axis separately. We have
Z
0
2
(4 − x2 )dx = 5.3333
Z
and
2
3
(4 − x2 )dx = −2.3333.
As expected, the integral from 2 to 3 is negative. The area above the axis is 5.3333 and the area below the axis is 2.3333,
so
Total area = 5.3333 + 2.3333 = 7.6666.
Thus the area is 7.667.
y
4
x
2
y = 4 − x2
3
Figure 5.13
21. (a) The graph of f (x) = x3 − x crosses the x-axis at x = 1 since solving f (x) = x3 − x = 0 gives x = 0 and x = ±1.
See Figure 5.14. To find the total area, we find the area above the axis and the area below the axis separately. We have
Z
0
1
(x3 − x)dx = −0.25
and
Z
3
1
(x3 − x)dx = 16.
As expected, the integral from 0 to 1 is negative. The area above the axis is 16 and the area below the axis is 0.25 so
Total area = 16.25.
(b) We have
3
Z
(x3 − x)dx = 15.75.
0
Notice that the integral is equal to the area above the axis minus the area below the axis, as expected.
(c) No, they are not the same, since the integral counts area below the axis negatively while total area counts all area as
positive. The two answers are not expected to be the same unless all the area lies above the horizontal axis.
y
f (x) = x3 − x
x
1
2
Figure 5.14
22. A graph of y = 6x3 − 2 shows that this function is nonnegative on the interval x = 5 to x = 10. Thus,
Area =
Z
5
The integral was evaluated on a calculator.
10
(6x3 − 2) dx = 14,052.5.
448
Chapter Five /SOLUTIONS
23. Since cos t ≥ 0 for 0 ≤ t ≤ π/2, the area is given by
π/2
Z
Area =
cos t dt = 1.
0
The integral was evaluated on a calculator.
24. A graph of y = ln x shows that this function is non-negative on the interval x = 1 to x = 4. Thus,
4
Z
Area =
ln x dx = 2.545.
1
The integral was evaluated on a calculator.
25. A graph of y = 2 cos(t/10) shows that this function is nonnegative on the interval t = 1 to t = 2. Thus,
Area =
Z
2
Z
2
2 cos
1
The integral was evaluated on a calculator.
√
26. Since cos x > 0 for 0 ≤ x ≤ 2, the area is given by
Area =
cos
t
dt = 1.977.
10
√
x dx = 1.106.
0
The integral was evaluated on a calculator.
√
27. The graph of y = 7 − x2 has intercepts x = ± 7. See Figure 5.15. Therefore we have
√
Area =
The integral was evaluated on a calculator.
Z
7
√
− 7
(7 − x2 ) dx = 24.694.
y
7
y = 7 − x2
√
− 7
√
x
7
Figure 5.15
√
4
28. The graph of y = x4 − 8 has intercepts x = ± 8. See Figure 5.16. Since the region is below the x-axis, the integral is
negative, so
√
Area = −
The integral was evaluated on a calculator.
Z
−
4
8
√
4
8
(x4 − 8) dx = 21.527.
y
y = x4 − 8
−
√
4
8
√
4
8
Figure 5.16
x
5.2 SOLUTIONS
449
29. (a) The area between the graph of f and the x-axis between x = a and x = b is 13, so
Z
b
Z
c
f (x) dx = 13.
a
(b) Since the graph of f (x) is below the x-axis for b < x < c,
f (x) dx = −2.
b
(c) Since the graph of f (x) is above the x-axis for a < x < b and below for b < x < c,
c
Z
f (x) dx = 13 − 2 = 11.
a
(d) The graph of |f (x)| is the same as the graph of f (x) except that the part below the x-axis is reflected to be above it.
See Figure 5.17. Thus
Z
c
|f (x)| dx = 13 + 2 = 15.
a
|f (x)|
Area = 13
✠
Area = 2
✠
a
c
b
x
Figure 5.17
30. The region shaded between x = 0 and x = 2 appears
R 0to have approximately the same area as the region shaded between
x = −2 and x = 0, but it lies below the axis. Since −2 f (x)dx = 4, we have the following results:
(a)
R0
R2
f (x)dx ≈ − −2 f (x)dx = −4.
0
R2
(b) −2 f (x)dx ≈ 4 − 4 = 0.
(c) The total area shaded is approximately 4 + 4 = 8.
31. (a)
(b)
Z
0
Z−34
f (x) dx = −2.
f (x) dx =
Z
0
−3
−3
f (x) dx +
Z
3
f (x) dx +
0
Z
3
4
f (x) dx = −2 + 2 −
A
A
=− .
2
2
32. We can compute each integral in this problem by finding the difference between the area that lies above the x-axis and the
area that lies below the x-axis on the given interval.
(a) For
R2
0
f (x) dx, on 0 ≤ x ≤ 1 the area under the graph is 1; on 1 ≤ x ≤ 2 the areas above and below the x-axis are
R2
equal and cancel each other out. Therefore, 0 f (x) dx = 1.
(b) On 3 ≤ x ≤ 7 the graph of f (x) is the upper half circle of radius 2 centered at (5, 0). The integral is equal to the
area between the graph of f (x) and the x-axis, which is the area of a semicircle of radius 2. This area is 2π, and so
7
Z
f (x) dx =
3
π22
= 2π.
2
(c) On 2 ≤ x ≤ 7 we are looking at two areas: We already know that the area of the semicircle on 3 ≤ x ≤ 7 is 2π. On
2 ≤ x ≤ 3, the graph lies below the x-axis and the area of the triangle is 21 . Therefore,
Z
2
7
f (x) dx = −
1
+ 2π.
2
(d) For this portion of the problem, we can split the region between the graph and the x-axis into a quarter circle on
5 ≤ x ≤ 7 and a trapezoid on 7 ≤ x ≤ 8 below the x-axis. The semicircle has area π, the trapezoid has area 3/2.
Therefore,
Z 8
3
f (x) dx = π − .
2
5
450
Chapter Five /SOLUTIONS
y
33. (a)
y = f (x)
2
A1
(b) A1 =
Z
x
0
f (x) dx = 2.667.
−2
A2 = −
1
A2
−2
Z
1
f (x) dx = 0.417.
0
So total area = A1 + A2 ≈ 3.084. Note that while A1 and A2 are accurate to 3 decimal places, the quoted value for
A
Z 1 + A2 is accurate only to 2 decimal places.
1
(c)
−2
34.
Z
4
cos
f (x) dx = A1 − A2 = 2.250.
√
x dx = 0.80 = Area A1 – Area A2 . See Figure 5.18.
0
1
cos
√
x
A2
A1
4
x
❄
( π2 )2
−1
Figure 5.18
35. Looking at the graph of e−x sin x for 0 ≤ x ≤ 2π in Figure 5.19, we see that the area, A1 , below the curve for 0 ≤ x ≤ π
is much greater than the area, A2 , above the curve for π ≤ x ≤ 2π. Thus, the integral is
Z
2π
0
e−x sin x dx = A1 − A2 > 0.
e−x sin x
A1
A2
π
❄
x
2π
Figure 5.19
2
36. Since e−x is decreasing between x = 0 and x = 1, the left sum is an overestimate and the right sum is an underestimate
2
of the integral. Letting f (x) = e−x , we divide the interval 0 ≤ x ≤ 1 into n = 5 sub-intervals to create Table 5.2.
451
5.2 SOLUTIONS
Table 5.2
x
0.0
0.2
0.4
0.6
0.8
1.0
f (x)
1.000
0.961
0.852
0.698
0.527
0.368
(a) Letting ∆x = (1 − 0)/5 = 0.2, we have:
Left-hand sum = f (0)∆x + f (0.2)∆x + f (0.4)∆x + f (0.6)∆x + f (0.8)∆x
= 1(0.2) + 0.961(0.2) + 0.852(0.2) + 0.698(0.2) + 0.527(0.2)
= 0.808.
(b) Again letting ∆x = (1 − 0)/5 = 0.2, we have:
Right-hand sum = f (0.2)∆x + f (0.4)∆x + f (0.6)∆x + f (0.8)∆x + f (1)∆x
= 0.961(0.2) + 0.852(0.2) + 0.698(0.2) + 0.527(0.2) + 0.368(0.2)
= 0.681.
37. (a) See Figure 5.20.
Left sum = f (1)∆x + f (1.5)∆x = (ln1)0.5 + ln(1.5)0.5 = (ln1.5)0.5.
(b) See Figure 5.21.
Right sum = f (1.5)∆x + f (2)∆x = (ln1.5)0.5 + (ln2)0.5.
(c) Right sum is an overestimate, left sum is an underestimate.
ln x
1
1.5
ln x
x
2
1
Figure 5.20: Left sum
1.5
x
2
Figure 5.21: Right sum
38. (a) See Figure 5.22.
(b) See Figure 5.23.
(c) Since the rectangle above the curve in Figure 5.22 and the rectangle below the Rcurve in Figure 5.23 look approximately
π
equal in area (in fact, they are exactly equal), the left hand approximation to −π sin x dx is 0.
sin x
sin x
π
−π
x
π
−π
Figure 5.22
Figure 5.23
39. (a) The integral gives the shaded area in Figure 5.24. We find
y
Z
6
(x2 + 1) dx = 78.
0
40
y = x2 + 1
20
x
3
Figure 5.24
6
x
452
Chapter Five /SOLUTIONS
(b) Using n = 3, we have
Left-hand sum = f (0) · 2 + f (2) · 2 + f (4) · 2 = 1 · 2 + 5 · 2 + 17 · 2 = 46.
This sum is an underestimate. See Figure 5.25.
y
y
40
40
y = x2 + 1
y = x2 + 1
20
20
x
3
x
6
3
Figure 5.25
6
Figure 5.26
(c) Using n = 3 gives
Right-hand sum = f (2) · 2 + f (4) · 2 + f (6) · 2 = 5 · 2 + 17 · 2 + 37 · 2 = 118.
This sum is an overestimate. See Figure 5.26.
40. (a) See Figure 5.27.
(b) Since each of the triangular regions in Figure 5.27 have area 1/2, we have
Z
0
2
f (x) dx =
1
1
+ = 1.
2
2
(c) Using ∆x = 1/2 in the 4-term Riemann sum shown in Figure 5.28, we have
Left hand sum = f (0)∆x + f (0.5)∆x + f (1)∆x + f (1.5)∆x
1 1
1
1 1
1
+
+0
+
= 1.
=1
2
2 2
2
2 2
We notice that in this case the approximation is exactly equal to the exact value of the integral.
1
1
f (x)
f (x)
x
1
2
Figure 5.27
x
1
1
2
3
2
2
Figure 5.28
41. Left-hand sum gives: 12 (1/4) + (1.25)2 (1/4) + (1.5)2 (1/4) + (1.75)2 (1/4) = 1.96875.
Right-hand sum gives: (1.25)2 (1/4) + (1.5)2 (1/4) + (1.75)2 (1/4) + (2)2 (1/4) = 2.71875.
We estimate the value of the integral by taking the average of these two sums, which is 2.34375. Since x2 is monotonic on 1 ≤ x ≤ 2, the true value of the integral lies between 1.96875 and 2.71875. Thus the most our estimate could
be off is 0.375. We expect it to be much closer. (And it is—the true value of the integral is 7/3 ≈ 2.333.)
42. We have ∆x = 2/500 = 1/250. The formulas for the left- and right-hand Riemann sums give us that
Left = ∆x[f (−1) + f (−1 + ∆x) + ... + f (1 − 2∆x) + f (1 − ∆x)]
Right = ∆x[f (−1 + ∆x) + f (−1 + 2∆x) + ... + f (1 − ∆x) + f (1)].
Subtracting these yields
Right − Left = ∆x[f (1) − f (−1)] =
4
2
1
[6 − 2] =
=
.
250
250
125
5.2 SOLUTIONS
453
43. See Figure 5.29.
y
x
a
b
Figure 5.29: Integral vs. Left- and
Right-Hand Sums
44. The statement is rarely true. The graph of almost any non-linear monotonic function, such as x10 for 0 < x < 1, should
provide convincing geometric evidence. Furthermore, if the statement were true, then (LHS+RHS)/2 would always give
the exact value of the definite integral. This is not true.
45. (a) If the interval 1 ≤ t ≤ 2 is divided into n equal subintervals of length ∆t = 1/n, the subintervals are given by
1 ≤t≤1+
1
2
n−1
1
, 1 + ≤ t ≤ 1 + , ..., 1 +
≤ t ≤ 2.
n
n
n
n
The left-hand sum is given by
Left sum =
n−1
X
r=0
r
f 1+
n
n−1
n−1
r=0
r=0
X
X 1
1
1
1
=
· =
n
1 + r/n n
n+r
and the right-hand sum is given by
Right sum =
n
X
f 1+
r=1
r
n
n
X 1
1
=
.
n
n+r
r=1
Since f (t) = 1/t is decreasing in the interval 1 ≤ t ≤ 2, we know that the right-hand sum is less than
the left-hand sum is larger than this integral. Thus we have
n
X
r=1
1
n+r
Z
<
1
t
1
dt
t
<
n−1
X
r=0
R2
1
1/t dt and
1
.
n+r
(b) Subtracting the sums gives
n−1
X
r=0
(c) Here we need to find n such that
n
X 1
1
1
1
1
−
= −
=
.
n+r
n+r
n
2n
2n
r=1
1
≤ 5 × 10−6 ,
2n
so n ≥
1
× 106 = 105 .
10
Strengthen Your Understanding
46. This does not say that f (x) ≥ 0. If f (x) < 0 on [1, 3] then the integral is negative, but an area cannot be negative.
47. There are too many terms in the sum; last term should be sin(1.9). So the correct left-hand sum is
0.1 (sin(1) + sin(1.1) + · · · + sin(1.9)) .
48. Any function which is negative on the whole interval will do, for example f (x) = −1 and [a, b] = [0, 1]. There are also
examples like f (x) = −1 + x with 0 ≤ x ≤ 1.1.
454
Chapter Five /SOLUTIONS
49. This is true if the function is negative on the interval [2, 3]; for example f (x) = 2 − x.
50. False. The integral is the change in position from t = a to t = b. If the velocity changes sign in the interval, the total
distance traveled and the change in position will not be the same.
51. False. A counterexample is given by f (x) = 1 on [0, 2].Then
Z
2
f (x) dx = 2.
0
On the other hand, using ∆x = 1/2 in the 4-term Riemann sum we have
Left hand sum = f (0)∆x + f (0.5)∆x + f (1)∆x + f (1.5)∆x
1
1
1
1
+1
+1
+1
= 2.
=1
2
2
2
2
52. False. A counterexample is given by the functions f and g in Figure 5.30. The function f is decreasing, g is increasing,
and we have
Z 2
Z 2
g(x) dx,
f (x) dx =
1
1
because both integrals equal 1/2, the area of of the same sized triangle.
1
1
f (x)
g(x)
x
1
x
2
1
2
Figure 5.30
53. False. Any function f (x) that is negative between x = 2 and x = 3 has
R2
R3
2
f (x) dx < 0, so
R2
0
f (x) dx >
R3
0
f (x) dx.
54. True. Since 0 f (x) dx is a number, if we use the variable t instead of the variable x in the function f , we get the same
number for the definite integral.
55. False. Let f (x) = x and g(x) = 5. Then
f (x) > g(x) for 5 < x < 6.
56. An example is graphed in Figure 5.31.
R6
2
f (x) dx = 16 and
R6
2
g(x) dx = 20, so
1
f (x)
x
10
Figure 5.31
57. An example is graphed in Figure 5.32.
5
f (x)
x
0.4
10
Figure 5.32
R6
2
f (x) dx ≤
R6
2
g(x) dx, but
5.3 SOLUTIONS
455
Solutions for Section 5.3
Exercises
1. The units of measurement are dollars.
2. The units of measurement are meters per second (which are units of velocity).
3. The units of measurement are foot-pounds (which are units of work).
4. The integral
5. The integral
R3
1
v(t) dt represents the change in position between time t = 1 and t = 3 seconds; it is measured in meters.
R6
a(t) dt represents the change in velocity between times t = 0 and t = 6 seconds; it is measured in km/hr.
R5
s(x) dx represents the change in salinity (salt concentration) in the first 5 cm of water; it is measured in
0
R 2011
6. The integral 2005 f (t) dt represents the change in the world’s population between the years 2005 and 2011. It is measured
in billions of people.
7. The integral
gm/liter.
0
8. (a) One small box on the graph corresponds to moving at 750 ft/min for 15 seconds, which corresponds to a distance of
187.5 ft. Estimating the area beneath the velocity curves, we find:
Distance traveled by car 1 ≈ 5.5 boxes = 1031.25 ft.
Distance traveled by car 2 ≈ 3 boxes = 562.5 ft.
(b) The two cars will have gone the same distance when the areas beneath their velocity curves are equal. Since the two
areas overlap, they are equal when the two shaded regions have equal areas, at t ≈ 1.6 minutes. See Figure 5.33.
v ft/min
Car 2
Car 1
1.6
t
Figure 5.33
9. We have f (t) = F ′ (t) = 2t, so by the Fundamental Theorem of Calculus,
Z
1
3
2t dt = F (3) − F (1) = 9 − 1 = 8.
10. We have f (t) = F ′ (t) = 6t + 4, so by the Fundamental Theorem of Calculus,
5
Z
(6t + 4) dt = F (5) − F (2) = 95 − 20 = 75.
2
11. We have f (t) = F ′ (t) = 1/t, so by the Fundamental Theorem of Calculus,
Z
1
5
1
dt = F (5) − F (1) = ln 5 − ln 1 = ln 5.
t
12. We have f (t) = F ′ (t) = cos t, so by the Fundamental Theorem of Calculus,
Z
0
π/2
cos t dt = F (π/2) − F (0) = 1 − 0 = 1.
456
Chapter Five /SOLUTIONS
13. We have f (t) = F ′ (t) = 7 ln(4) · 4t , so by the Fundamental Theorem of Calculus,
Z
3
2
7 ln(4) · 4t dt = F (3) − F (2) = 448 − 112 = 336.
14. We have f (t) = F ′ (t) = 1/(cos2 t), so by the Fundamental Theorem of Calculus,
Z
π
0
1/(cos2 t) dt = F (π) − F (0) = 0 − 0 = 0.
Problems
15. (a) We have
d 3
(x + x) = 3x2 + 1.
dx
(b) By the Fundamental Theorem of Calculus with F ′ (x) = 3x2 + 1 and F (x) = x3 + x, we have
2
Z
(3x2 + 1) dx = F (2) − F (0) = (23 + 2) − (03 + 0) = 10.
0
16. (a) We have
d
(sin t) = cos t.
dt
(b) Since v(t) is the derivative of distance traveled, by the Fundamental Theorem of Calculus with F ′ (t) = v(t) = cos t
and F (t) = sin t, we have
Distance =
Z
π/2
v(t) dt =
0
Z
π/2
cos t dt = F
0
π
2
− F (0) = sin
π
2
− sin 0 = 1.
17. (a) By the chain rule,
d
dx
R 0.4
1
1
sin2 t = · 2 sin t cos t = sin t cos t.
2
2
(i) Using a calculator, 0.2 sin t cos t dt = 0.056
(ii) The Fundamental Theorem of Calculus tells us that the integral is
Z
0.4
0.2
sin t cos t dt = F (0.4) − F (0.2) =
18. (a) By the chain rule,
F ′ (x) =
R1
2
2
d
ex
dx
1
sin2 (0.4) − sin2 (0.2) = 0.05609.
2
2
= 2xex .
(i) Using a calculator, 0 2xex dx = 1.718.
(ii) The Fundamental Theorem of Calculus says we can get the exact value of the integral by looking at
Z
0
1
2
2
2
2xex dx = F (1) − F (0) = e1 − e0 = e − 1 = 1.71828.
19. (a) On day 12 pollution is removed from the lake at a rate of 500 kg/day.
(b) The limits of the integral are t = 5 and t = 15. Since t is time in days, the units of the 5 and 15 are days. The units
of the integral are obtained by multiplying the units of f (t), kg/day, by the units of dt, day. Thus the units of the
integral are
kg
× day = kg.
day
The 4000 has units of kilograms.
(c) The integral of a rate gives the total change. Here f (t) is the rate of change of the quantity of pollution that has been
removed from the lake. The integral gives the change in the quantity of pollution that has been removed during the
time interval; in other words, the total quantity removed during that time period. During the 10 days from day 5 to
day 15, a total of 4000 kg were removed from the lake.
5.3 SOLUTIONS
457
20. For any t, consider the interval [t, t + ∆t]. During this interval, oil is leaking out at an approximately constant rate of f (t)
gallons/minute. Thus, the amount of oil which has leaked out during this interval can be expressed as
Amount of oil leaked = Rate × Time = f (t) ∆t
and the units of f (t) ∆t are gallons/minute × minutes = gallons. The total amount of oil leaked is obtained by adding
all these amounts between t = 0 and t = 60. (An hour is 60 minutes.) The sum of all these infinitesimal amounts is the
integral
Total amount of
=
oil leaked, in gallons
21. (a) The amount leaked between t = 0 and t = 2 is
Z
Z
60
f (t) dt.
0
2
R(t) dt.
0
(b)
(0, 2)
2
(1, 1.6)
(2, 1)
R(t)
1
1
t
2
1
. Of these 45 are wholly beneath the curve, hence the area
(c) The rectangular boxes on the diagram each have area 16
under the curve is certainly more than 45
>
2.81.
There
are 9 more partially beneath the curve, and so the desired
16
54
area is completely covered by 54 boxes. Therefore the area is less than 16
< 3.38.
These are very safe estimates but far apart. We can do much better by estimating what fractions of the broken
boxes are beneath the curve. Using this method, we can estimate the area to be about 3.2, which corresponds to 3.2
gallons leaking over two hours.
22. (a) Using rectangles under the curve, we get
Acres defaced ≈ (1)(0.2 + 0.4 + 1 + 2) = 3.6 acres.
(b) Using rectangles above the curve, we get
Acres defaced ≈ (1)(0.4 + 1 + 2 + 3.5) = 6.9 acres.
(c) The number of acres defaced is between 3.6 and 6.9, so we estimate the average, 5.25 acres.
R5
23. (a) Quantity used = 0 f (t) dt.
(b) Using a left sum, our approximation is
32e0.05(0) + 32e0.05(1) + 32e0.05(2) + 32e0.05(3) + 32e0.05(4) = 177.270.
Since f is an increasing function, this represents an underestimate.
(c) Each term is a lower estimate of one year’s consumption of oil.
24. (a) Solving v(t) = −4t2 + 16t = 0 gives t = 0 and t = 4. At t = 0 the jump has just started; at t = 4 the jumper
is momentarily stopped before starting back up. In Figure 5.34, we see the velocity is positive from t = 0 to t = 4
and negative from t = 4 to t = 5. The total number of meters traveled is given by the total area between the velocity
curve and the t-axis. We find the total area by finding the areas above and below the axis separately. We have
Z
0
4
(−4t2 + 16t) dt = 42.6667
and
Z
4
5
(−4t2 + 16t) dt = −9.3333.
The jumper traveled 42.6667 meters downward and then 9.3333 meters upward, so
Total number of meters traveled = 42.6667 + 9.3333 = 52 meters.
(b) Since the jumper fell 42.6667 meters downward and came back up 9.3333 meters,
Ending position = 42.6667 − 9.3333 = 33.333 meters below the starting point.
458
Chapter Five /SOLUTIONS
(c) We find that
Z
5
(−4t2 + 16t) dt = 33.333 meters.
0
Thus, the net change in position is a positive 33.333 meters, or 33.333 meters below the starting point. The net change
is given by the integral, while the total number of meters traveled is given by the total area.
v(t) = −4t2 + 16t
15
10
5
x
4
−5
−10
−15
Figure 5.34
25. (a) Note that the rate r(t) sometimes increases and sometimes decreases in the interval. We can calculate an upper
estimate of the volume by choosing ∆t = 5 and then choosing the highest value of r(t) on each interval, and
similarly a lower estimate by choosing the lowest value of r(t) on each interval:
Upper estimate = 5[20 + 24 + 24] = 340 liters.
Lower estimate = 5[12 + 20 + 16] = 240 liters.
(b) A graph of r(t) along with the areas represented by the choices of r(t) in calculating the lower estimate is shown in
Figure 5.35.
24
20
16
12
r(t)
5
10
15
t
Figure 5.35
26. We use left- and right-hand sums to estimate the total amount of coal produced during this period:
Left sum = (1.090)2 + (1.094)2 + (1.121)2 + (1.072)2 + (1.132)2 + (1.147)2 = 13.312.
Right sum = (1.094)2 + (1.121)2 + (1.072)2 + (1.132)2 + (1.147)2 + (1.073)2 = 13.278.
We see that
13.312 + 13.278
= 13.295 billion tons.
2
The total amount of coal produced is the definite integral of the rate of coal production r = f (t) given in the table. Since
t is in years since 1997, the limits of integration are t = 0 and t = 12. We have
Total amount of coal produced ≈
Total amount of coal produced =
Z
0
12
f (t) dt billion tons.
5.3 SOLUTIONS
27. Since W is in tons per week and t is in weeks since January 1, 2005, the integral
tons, produced during the year 2005.
0
459
W dt gives the amount of waste, in
52
Z
Total waste during the year =
R 52
3.75e−0.008t dt = 159.5249 tons.
0
Since waste removal costs $15/ton, the cost of waste removal for the company is 159.5249 · 15 = $2392.87.
28. The total number of “worker-hours” is equal to the area under the curve. The total area is about 14.5 boxes. Since each
box represents (10 workers)(8 hours) = 80 worker-hours, the total area is 1160 worker-hours. At $10 per hour, the total
cost is $11,600.
29. The time period 9am to 5pm is represented by the time t = 0 to t = 8 and t = 24 to t = 32. The area under the curve,
or total number of worker-hours for these times, is about 9 boxes or 9(80) = 720 worker-hours. The total cost for 9am to
5pm is (720)(10) = $7200. The area under the rest of the curve is about 5.5 boxes, or 5.5(80) = 440 worker-hours. The
total cost for this time period is (440)(15) = $6600. The total cost is about 7200 + 6600 = $13,800.
30. The area under the curve represents the number of cubic feet of storage times the number of days the storage was used.
This area is given by
Area under graph = Area of rectangle + Area of triangle
1
= 30 · 10,000 + · 30(30,000 − 10,000)
2
= 600,000.
Since the warehouse charges $5 for every 10 cubic feet of storage used for a day, the company will have to pay
(5)(60,000) = $300,000.
31. If H(t) is the temperature of the coffee at time t, by the Fundamental Theorem of Calculus
Change in temperature = H(10) − H(0) =
Therefore,
H(10) = H(0) +
Z
Z
10
H ′ (t) dt =
Z
0
0
10
−7e−0.1t dt.
10
0
−7(0.9t ) dt ≈ 90 − 44.2 = 45.8◦ C.
32. The change in the amount of water is the integral of rate of change, so we have
Number of liters pumped out =
Z
60
(5 − 5e−0.12t )dt = 258.4 liters.
0
Since the tank contained 1000 liters of water initially, we see that
Amount in tank after one hour = 1000 − 258.4 = 741.6 liters.
33. By the Fundamental Theorem,
f (1) − f (0) =
Z
1
f ′ (x) dx,
0
Since f ′ (x) is negative for 0 ≤ x ≤ 1, this integral must be negative and so f (1) < f (0).
34. First rewrite each of the quantities in terms of f ′ , since we have the graph of f ′ . If A1 and A2 are the positive areas shown
in Figure 5.36:
f (3) − f (2) =
f (4) − f (3) =
3
Z
f ′ (t) dt = −A1
2
Z
4
f ′ (t) dt = −A2
3
f (4) − f (2)
1
=
2
2
Z
2
4
f ′ (t) dt = −
A1 + A2
2
460
Chapter Five /SOLUTIONS
Since Area A1 > Area A2 ,
A1 + A2
< A1
2
A2 <
so
−A1 < −
and therefore
f (3) − f (2) <
A1 + A2
< −A2
2
f (4) − f (2)
< f (4) − f (3).
2
y
3
2
4
x
1
A1
A2
y = f ′ (x)
Figure 5.36
35. We know that the the integral of F , and therefore the work, can be obtained by computing the areas in Figure 5.37.
F (newtons)
2
1
A1
A2
A3
4
−1
8
14
10
16
A4
x (meters)
−2
Figure 5.37
W =
Z
16
0
F (x) dx = Area above x-axis − Area below x-axis
= A1 + A2 + A3 − A4
1
1
1
·4·2+4·2+ ·2·2− ·2·2
2
2
2
= 12 newton · meters.
=
36. According to the Fundamental Theorem,
f (2) − f (1) =
|{z}
7
Z
2
f ′ (t) dt
1
f (2) = 7 +
Z
2
2
2
e−t dt, since f ′ (t) = e−t .
1
2
We estimate the integral using left and right sums. Since f ′ (t) = e−t is decreasing between t = 1 and t = 2, the
left sum overestimates and the right sum underestimates the integral.
To find left- and right-hand sums of 5 rectangles, we let ∆t = (2 − 1)/5 = 0.2. The table gives values of f ′ (t).
t
1.0
1.2
1.4
1.6
1.8
2.0
f ′ (t)
0.368
0.237
0.141
0.077
0.039
0.018
5.3 SOLUTIONS
461
We have the following estimates:
LHS = ∆t f ′ (1.0) + f ′ (1.2) + f ′ (1.4) + f ′ (1.6) + f ′ (1.8) = 0.2 (0.862) = 0.1724
RHS = ∆t f ′ (1.2) + f ′ (1.4) + f ′ (1.6) + f ′ (1.8) + f ′ (2.0) = 0.2(0.512) = 0.1024.
So
0.1024 <
Z
2
2
e−t dt < 0.1734.
1
Adding 7, we estimate that
7.1024 < f (2) < 7.1724.
37. All the integrals have positive values, since f ≥ 0. The integral in (ii) is about one-half the integral in (i), due to the
apparent symmetry of f . The integral in (iv) will be much larger than the integral in (i), since the two peaks of f 2 rise to
10,000. The integral in (iii) will be smaller than half of the integral in (i), since the peaks in f 1/2 will only rise to 10. So
Z
2
(f (x))
1/2
dx <
Z
1
f (x) dx <
0
0
Z
0
2
f (x) dx <
Z
2
(f (x))2 dx.
0
38. (a)
(b)
(c)
(d)
(e)
V, since the slope is constant.
IV, since the net area under this curve is the most negative.
III, since the area under the curve is largest.
II, since the steepest ascent at t = 0 occurs on this curve.
III, since average velocity is (totalZdistance)/5, and III moves the largest total distance.
5
1
1
v ′ (t) dt = (v(5) − v(0)), and in I, the velocity increases the most from start
(f) I, since average acceleration is
5 0
5
(t = 0) to finish (t = 5).
39. To make this estimate, we first observe that, for constant speed,
Fuel used =
Miles traveled
Speed · Time
=
.
Miles/gallon
Miles/gallon
We make estimates at the start and end of each of the six intervals given. The 60 mph increase in speed (from 10 mph
to 70 mph) takes half an hour, or 30 minutes. Since the speed increases at a constant rate, each 10 mph increase takes 5
minutes.
At the start of the first 5 minutes, the speed is 10 mph and at the end, the speed is 20 mph. Since 5min = 5/60 hour,
during the first 5 minutes the distance traveled is between
5
5
mile
and
20 ·
mile.
60
60
During this first five minute period, the speed is between 10 and 20 mph, so the fuel efficiency is between 15 mpg and
18 mpg. So the fuel used during this period is between
10 ·
5
1
· 10 ·
gallons
18
60
Thus, an underestimate of the fuel used is
Fuel =
and
1
5
· 20 ·
gallons.
15
60
1
5
1
1
1
1
1
· 10 +
· 20 +
· 30 +
· 40 +
· 50 +
· 60
= 0.732 gallons.
18
21
23
24
25
26
60
An overestimate of the fuel used is
5
1
1
1
1
1
1
· 20 +
· 30 +
· 40 +
· 50 +
· 60 +
· 70
= 1.032 gallons.
Fuel =
15
18
21
23
24
25
60
We can get better bounds by using the actual distance traveled during each five minute period. For example, in the
first five minutes the average speed is 15 mph (halfway between 10 and 20 mph because the speed is increasing at a
constant rate). Since 5 minutes is 5/60 of an hour, in the first five minutes the car travels 15(5/60)=5/4 miles. Thus the fuel
used during this period was between
5 1
5 1
·
and
·
.
4 18
4 15
Using this method for each five minute period gives a lower estimate of 0.843 gallons and an upper estimate of 0.909
gallons.
40. The value of this integral tells us how much oil is pumped from the well between day t = 0 and day t = t0 .
462
Chapter Five /SOLUTIONS
41. The time period from t0 to 2t0 is shorter (and
Z contained within) the time period from 0 to 2t0 . Thus, the amount of oil
2t0
r(t) dt, is less than the amount of oil pumped out in the longer timer
pumped out during the shorter time period,
period,
Z
t0
2t0
r(t) dt. This means
0
Z
2t0
Z
3t0
r(t) dt <
t0
Z
2t0
Z
2t0
r(t) dt.
0
The length of the time period from 2t0 to 3t0 is the same as the length from t0 to 2t0 : both are t0 days. But the rate
at which oil is pumped is going down, since r ′ (t) < 0. Thus, less oil is pumped out during the later time period, so
r(t) dt <
We conclude that
Z
3t0
r(t) dt <
2t0
Z
2t0
r(t) dt <
t0
r(t) dt.
t0
2t0
Z
2t0
r(t) dt.
0
42. (a) The black curve is for boys, the colored one for girls. The area under each curve represents the change in growth in
centimeters. Since men are generally taller than women, the curve with the larger area under it is the height velocity
of the boys.
(b) Each square below the height velocity curve has area 1 cm/yr · 1 yr = 1 cm. Counting squares lying below the black
curve gives about 43 cm. Thus, on average, boys grow about 43 cm between ages 3 and 10.
(c) Counting squares lying below the black curve gives about 23 cm growth for boys during their growth spurt. Counting
squares lying below the colored curve gives about 18 cm for girls during their growth spurt.
(d) We can measure the difference in growth by counting squares that lie between the two curves. Between ages 2 and
12.5, the average girl grows faster than the average boy. Counting squares yields about 5 cm between the colored and
black curves for 2 ≤ x ≤ 12.5. Counting squares between the curves for 12.5 ≤ x ≤ 18 gives about 18 squares.
Thus, there is a net increase of boys over girls by about 18 − 5 = 13 cm.
43. Using a left-hand Riemann sum with ∆t = 0.1, we have:
Z
0
0.5
f (t) dt ≈ f (0)∆t + f (0.1)∆t + · · · + f (0.4)∆t
= 0.3(0.1) + 0.2(0.1) + 0.2(0.1) + 0.3(0.1) + 0.4(0.1)
using values from the table
= 0.14.
44. From the Fundamental Theorem of Calculus, we have:
Z
0.5
g ′ (t) dt = g(0.5) − g(0.2)
0.2
= 0.8 − 5.1
since g is an antiderivative of g ′
using values from the table
= −4.3.
45. Referring to the table, we see that:
g (f (0.0)) = g(0.3) = 5.1
g (f (0.1)) = g(0.2) = 5.1
g (f (0.2)) = g(0.2) = 5.1
Using a left-hand Riemann sum with ∆t = 0.1, we have:
Z
0
0.3
g (f (t)) dt ≈ g (f (0)) ∆t + g (f (0.1)) ∆t + g (f (0.2)) ∆t.
= 5.1(0.1) + 5.1(0.1) + 5.1(0.1)
= 1.53.
5.3 SOLUTIONS
463
46. Since C is an antiderivative of c, we know by the Fundamental Theorem that
Z
24
15
c(n) dn = C(24) − C(15) = 13 − 8 = 5.
Since C(24) is the cost to plow a 24 inch snowfall, and C(15) is the cost to plow a 15 inch snowfall, this tells us that it
costs $5 million more to plow a 24 inch snowfall than a 15 inch snowfall.
47. Since C is an antiderivative of c, we know from the Fundamental Theorem that
15
Z
c(n) dn = C(15) −C(0)
|0
{z
7.5
so
| {z }
8
}
C(0) = 8 − 7.5 = 0.5.
This tells us the cost of preparing for a storm, even if no snow falls, is $0.5 million, or $500,000.
48. By the Fundamental Theorem, we have
c(15) +
24
Z
c′ (n) dn = c(15) + c(24) − c(15)
15
= c(24) = 0.4.
Since c(n) = C ′ (n), we have C ′ (24) = 0.4, which tells us the cost of plowing increases at the instantaneous rate of $0.4
million/inch, or by $400,000 per each additional inch, after 24 inches have already fallen.
49. The expression
R2
0
r(t) dt represents the amount of water that leaked from the ruptured pipe during the first two hours.
R4
Likewise, the expression 2 r(t) dt represents the amount of water that leaked from the ruptured pipe during the next two
hours. Since the leak “worsened throughout the afternoon,” we know that r is an increasing function, which means that
more water leaked out during the second two hours than during the first two hours. Therefore,
Z
2
r(t) dt <
Z
4
r(t) dt.
2
0
R4
50. The expression 0 r(t) dt represents the amount of water that leaked from the ruptured pipe during the first four hours.
The expression r(4) represents the rate water was leaking from the pipe at time t = 4, or four hours after the leak
began. Had water leaked at this rate during the whole four-hour time period,
Total amount of water leaked
at a constant rate of r(4)
= |{z}
Rate × Time elapsed = 4r(4) gallons.
r(4)
|
{z
4
}
According to the article, the leak “worsened throughout the afternoon,” so the rate that water leaked out initially was
less than r(4). Thus, 4r(4) overestimates the total amount of water leaked during the first four hours.
We conclude that
Z
4
r(t) dt < 4r(4).
0
51. According to the article, the leak “worsened throughout the afternoon,” eventually reaching (before being shut off) 8
million gallons/hour. Thus, 0 < r(t) < 8 million gal/hr. The leak began around 10 am and ended after 6 pm. The
R8
expression 0 r(t) dt represents the total amount of water leaked during the first 8 hours (or almost the total amount of
water leaked altogether). Since r(t) < 8 million gal/hr,
Total amount of water leaked
< Rate (million gal/hr) × Time elapsed = 64 million gallons.
during first 8 hours
Thus,
Z
0
8
|
{z
8
}
|
r(t) dt < 64 million gallons.
{z
8
}
464
Chapter Five /SOLUTIONS
Strengthen Your Understanding
R4
52. Since f (t) represents the rate of change of the weight of the dog, 0 f (t)dt represents the total change in the weight of
the dog from the time the dog is born to its fourth birthday. This is not the same as the weight of the dog when it is four
years old, because the dog has a nonzero weight when it is born.
53. In the Fundamental√Theorem of Calculus, the integrand
is the derivative of the function that is evaluated at the limits of
√
the integral. Since x is not the derivative of x, the statement is not correct. One way to write a correct statement is
R9
√
√
√
1/(2 x) dx = 9 − 4.
4
54. Since
d x
e
dx
= ex , f (x) = ex with a = 2 and b = 4 is one example.
55. If the car travels at a constant velocity of 50 miles per hour, it travels 200 miles in 4 hours as shown in Figure 5.38.
velocity
(miles/hour)
50
1
2
3
t (hours)
4
Figure 5.38
56. False. The units of the integral are the product of the units for f (x) times the units for x.
Solutions for Section 5.4
Exercises
1. Note that
Rb
a
g(x) dx =
Rb
a
Z
g(t) dt. Thus, we have
b
(f (x) + g(x)) dx =
a
2. Note that
Rb
a
f (z) dz =
Rb
a
b
Z
f (x) dx +
a
Rb
a
(g(x))2 dx =
Z
a
b
b
g(x) dx = 8 + 2 = 10.
a
f (x) dx. Thus, we have
Z
b
cf (z) dz = c
a
3. Note that
Z
Rb
a
Z
b
f (z) dz = 8c.
a
(g(t))2 dt. Thus, we have
Z
b
Z
b
(f (x))2 − (g(x))2 dx =
a
(f (x))2 dx −
4. We have
Z
a
b
(f (x))2 dx −
f (x) dx
a
2
Z
a
b
(g(x))2 dx = 12 − 3 = 9.
= 12 − 82 = −52.
5.4 SOLUTIONS
465
5. We write
b
Z
c1 g(x) + (c2 f (x))2 dx =
a
Z
b
c1 g(x) + c22 (f (x))2 dx
a
=
Z
b
c1 g(x) dx +
b
c22 (f (x))2 dx
a
a
= c1
Z
b
Z
g(x) dx + c22
Z
b
(f (x))2 dx
a
a
= c1 (2) + c22 (12) = 2c1 + 12c22 .
6. The graph of y = f (x − 5) is the graph of y = f (x) shifted to the right by 5. Since the limits of integration have also
Rb
R b+5
shifted by 5 (to a + 5 and b + 5), the areas corresponding to a+5 f (x − 5) dx and a f (x) dx are the same. Thus,
b+5
Z
a+5
7. Average value =
1
2−0
Z
1
8. Average value =
10 − 0
9. We have
2
(1 + t) dt =
0
Z
10
et dt =
0
f (x − 5) dx =
f (x) dx = 8.
a
1
(22025) = 2202.5
10
Z
1
=
b−a
Z
=
b
1
(4) = 2.
2
1
b−a
Average value =
Z
b
f (x) dx
a
a
b
1
2 dx =
·
b−a
Area of rectangle
of height 2 and base b − a
1
[2(b − a)] = 2.
b−a
!
10. Sketch the graph of f on 1 ≤ x ≤ 3. The integral is the area under the curve, which is a trapezoidal area. So the average
value is
Z 3
1 11 + 19
30
1
(4x + 7) dx = ·
·2=
= 15.
3−1 1
2
2
2
11. The integral represents the area below the graph of f (x) but above the x-axis.
(a) Since each square has area 1, by counting squares and half-squares we find
Z
6
f (x) dx = 8.5.
1
(b) The average value is
1
6−1
Z
6
f (x) dx =
1
8.5
17
=
= 1.7.
5
10
12. Since the average value is given by
Average value =
1
b−a
Z
b
f (x) dx,
a
the units for dx inside the integral are canceled by the units for 1/(b − a) outside the integral, leaving only the units for
f (x). This is as it should be, since the average value of f should be measured in the same units as f (x).
13. The graph of y = ex is above the line y = 1 for 0 ≤ x ≤ 2. See Figure 5.39. Therefore
Area =
Z
0
The integral was evaluated on a calculator.
2
(ex − 1) dx = 4.389.
466
Chapter Five /SOLUTIONS
y
y = ex
y=1
x
2
Figure 5.39
14. The graph of y = 5 ln(2x) is above the line y = 3 for 3 ≤ x ≤ 5. See Figure 5.40. Therefore
5
Z
Area =
(5 ln(2x) − 3) dx = 14.688.
3
The integral was evaluated on a calculator.
y
y = 5 ln(2x)
y=3
x
3
5
Figure 5.40
15. Since x3 ≤ x2 for 0 ≤ x ≤ 1, we have
Z
Area =
1
0
(x2 − x3 ) dx = 0.083.
The integral was evaluated on a calculator.
16. Since x1/2 ≤ x1/3 for 0 ≤ x ≤ 1, we have
Area =
Z
1
0
(x1/3 − x1/2 ) dx = 0.0833.
The integral was evaluated on a calculator.
17. The graph of y = sin x + 2 is above the line y = 0.5 for 6 ≤ x ≤ 10. See Figure 5.41. Therefore
Area =
Z
6
The integral was evaluated on a calculator.
10
sin x + 2 − 0.5 dx = 7.799.
5.4 SOLUTIONS
467
y
y = sin x + 2
y = 0.5
x
6
10
Figure 5.41
18. The graph of y = cos t is above the graph of y = sin t for 0 ≤ t ≤ π/4 and y = cos t is below y = sin t for
π/4 < t < π. See Figure 5.42. Therefore, we find the area in two pieces:
Area =
Z
π/4
(cos t − sin t) dt +
0
Z
π
π/4
(sin t − cos t) dt = 2.828.
The integral was evaluated on a calculator.
y
y = sin x
x
π
π/4
y = cos x
Figure 5.42
19. To find the points of intersection of the two curves, we must solve the equation
e−x = 4(x − x2 ).
This equation cannot be solved exactly, but numerical methods (for example, zooming in on the graph on a calculator)
give x = 0.261 and x = 0.883. See Figure 5.43. Since y = 4(x − x2 ) is above y = e−x on this interval,
Area =
Z
0.883
0.261
The integral was evaluated numerically.
y
1
(4(x − x2 ) − e−x ) dx = 0.172.
y = 4(x − x2 )
y = e−x
x
0.261
0.883
Figure 5.43
468
Chapter Five /SOLUTIONS
20. The curves y = e−x and y = ln x cross where
e−x = ln x.
Numerical methods (for example, zooming in on the graph on a calculator) give x = 1.310. See Figure 5.44. Since
y = e−x is above y = ln x for 1 ≤ x ≤ 1.310 and y = e−x is below y = ln x for 1.310 < x ≤ 2, we break the area
into two parts:
Z 2
Z 1.310
(e−x − ln x) dx +
(ln x − e−x ) dx.
Area =
1
1.310
Evaluating the integrals numerically gives
Area = 0.054 + 0.208 = 0.262.
y
1
y = ln x
y = e−x
x
1 1.310
2
Figure 5.44
Problems
21. (a) For 0 ≤ x ≤ 3, we have
1
3−0
Average value =
Z
3
f (x)dx =
0
1
(6) = 2.
3
(b) If f (x) is even, the graph is symmetric about the x-axis. For example, see Figure 5.45. By symmetry, the area between
x = −3 and x = 3 is twice the area between x = 0 and x = 3, so
Z
3
f (x)dx = 2(6) = 12.
−3
Thus for −3 ≤ x ≤ 3, we have
1
Average value =
3 − (−3)
Z
3
f (x)dx =
−3
1
(12) = 2.
6
The graph confirms that the average value between x = −3 and x = 3 is the same as the average value between
x = 0 and x = 3, which is 2.
x
−3 −1
x
−3
−1
1
1
3
3
Figure 5.45
Figure 5.46
(c) If f (x) is odd, then the graph is symmetric about the origin. For example, see Figure 5.46. By symmetry, the area
above the x-axis cancels out the area below the x-axis, so
Z
3
−3
f (x)dx = 0.
5.4 SOLUTIONS
469
Thus for −3 ≤ x ≤ 3, we have
1
3 − (−3)
Average value =
Z
3
f (x)dx =
−3
1
(0) = 0.
6
The graph confirms that the average value between x = −3 and x = 3 is zero.
22. We know that we can divide the integral up as follows:
Z
3
f (x) dx =
0
1
Z
f (x) dx +
0
3
0
Z
1
f (x) dx =
2
3
f (x) dx.
1
The graph suggests that f is an even function for −1 ≤ x ≤ 1, so
the preceding equation, we have
Z
Z
1
R1
−1
f (x) dx +
−1
f (x) dx = 2
Z
3
R1
0
f (x) dx. Substituting this in to
f (x) dx.
1
23. (a) The integral represents the area of a rectangle with height 1 and base b − a. See Figure 5.47. Thus
(b)
(i)
(ii)
(iii)
R5
R28
1 dx = 3.
R−3
3
1
1 dx = 11.
23 dx = 23
R3
1
1 dx = 46.
y
1
x
a
b
Figure 5.47
24. Using properties of the definite integral, we have:
Z
5
(2f (x) + 3) dx = 17
2
2
Z
5
f (x) dx + 3
2
Z
5
1 dx = 17
2
2
Z
2
5
f (x) dx + 3 · 3 = 17
2
Z
5
f (x) dx = 8
2
Z
5
f (x) dx = 4.
2
25. Since t = 0 in 1975 and t = 35 in 2010, we want:
Z
35
1
225(1.15)t dt
35 − 0 0
1
=
(212,787) = $6080.
35
Average Value =
Rb
a
1 dx = b − a.
470
Chapter Five /SOLUTIONS
26. (a) The integral represents the area above the x-axis and below the line y = x between x = a and x = b. See Figure 5.48.
This area is
A1 + A2 = a(b − a) +
(b)
b2 − a 2
b−a
b+a
1
(b − a)2 = a +
(b − a) =
.
(b − a) =
2
2
2
2
The formula
holds similarly for negative values.
R5
(i) 2 x dx = 21/2.
(ii)
(iii)
R8
R−3
3
1
x dx = 55/2.
5x dx = 5
R3
1
x dx = 20.
✻
y=x
b−a
A2
❄
✛b − a ✲
A1
x
a
b
Figure 5.48
27. We have
30 =
Z
3
f dx =
28. We have
R2
−2
f (x) dx = 0, so
8=
R3
2
Z
R2
−2
f (x) dx +
Z
3
f (x) dx.
2
f (x) dx = 30.
2
−2
Thus
2
−2
−2
Since f is odd,
Z
(f (x) − 3) dx =
Z
f (x) dx − 3
−2
f (x) dx = 8 + 3(2 − (−2)) = 20. Since f is even,
3
2
2
R2
0
Z
2
1 dx.
−2
f (x) dx = (1/2)20 = 10.
29. This integral is 0 because the function x cos(x ) is odd (meaning f (−x) = −f (x)), and so the negative contribution to
the integral from − π4 < x < 0 exactly cancels the positive contribution from 0 < x < π4 . See Figure 5.49.
x3 cos x2
− π4
x
π
4
Figure 5.49
30. On the interval 2 ≤ x ≤ 5,
Average value
=
of f
so
Z
Z
1
5−2
5
f (x) dx = 4,
2
5
f (x) dx = 12.
2
Thus
Z
2
5
(3f (x) + 2) dx = 3
Z
5
f (x) dx + 2
2
Z
2
5
1 dx = 3(12) + 2(5 − 2) = 42.
5.4 SOLUTIONS
31. We know
Z
3
1
Since
R3
1
3x2 dx = 3
R3
1
(x2 − x) dx =
x2 dx and we are given
R3
1
Z
3
1
x2 dx −
R3
1
2xdx = 2
R3
1
xdx and
R3
1
3
x dx.
1
3x2 dx = 26, we have
Z
3
x2 dx =
1
Similarly, since
Z
471
26
.
3
2x dx = 8, we have
Z
3
x dx = 4.
1
Thus,
Z
1
3
(x2 − x) dx =
14
26
−4=
.
3
3
32. (a) Over the interval [−1, 3], we estimate that the total change of the population is about 1.5, by counting boxes between
the curve and the x-axis; we count about 1.5 boxes below the x-axis from x = −1 to x = 1 and about 3 above from
x = 1 to x = 3. So the average rate of change is just the total change divided by the length of the interval, that is
1.5/4 = 0.375 thousand/hour.
(b) We can estimate the total change of the algae population by counting boxes between the curve and the x-axis. Here,
there is about 1 box above the x-axis from x = −3 to x = −2, about 0.75 of a box below the x-axis from x = −2
to x = −1, and a total change of about 1.5 boxes thereafter (as discussed in part (a)). So the total change is about
1 − 0.75 + 1.5 = 1.75 thousands of algae.
R3
33. (a) The integral is the area above the x-axis minus the area below the x-axis. Thus, we can see that −3 f (x) dx is about
−6 + 2 = −4 (the negative of the area from t = −3 to t = 1 plus the area from t = 1 to t = 3.)
(b) Since the integral in part (a) is negative, the average value of f (x) between x = −3 and x = 3 is negative. From the
graph, however, it appears that the average value of f (x) from x = 0 to x = 3 is positive. Hence (ii) is the larger
quantity.
34. (a) At the end of one hour t = 60, and H = 22◦ C.
(b)
Z
60
1
(20 + 980e−0.1t )dt
60 0
1
(10976) = 183◦ C.
=
60
Average temperature =
(c) Average temperature at beginning and end of hour = (1000 + 22)/2 = 511◦ C. The average found in part (b) is
smaller than the average of these two temperatures because the bar cools quickly at first and so spends less time at
high temperatures. Alternatively, the graph of H against t is concave up.
H(◦ C)
1000◦
511◦
H = 20 + 980e−0.1t
22◦
60
t (mins)
35. (a) We have
1
Average population =
40
Evaluating the integral numerically gives
Z
40
112(1.011)t dt.
0
Average population = 140.508 million.
472
Chapter Five /SOLUTIONS
(b) In 2010, t = 0, and P = 112(1.011)0 = 112
In 2050, t = 40, and P = 112(1.011)40 = 173.486.
1
(112 + 173.486) = 142.743 million.
2
(c) If P had been linear, the average value found in part (a) would have been the same as we found in part (b). Since the
population graph is concave up, it is below the secant line. Thus, the actual values of P are less than the corresponding
values on the secant line, so the average found in part (a) is smaller than that found in part (b).
Average =
36. (a) See Figure 5.50. Since the shaded region lies within a rectangle of area 1, the area is less than 1.
(b) Since the area is given by the integral
Area =
Z
1
e−x
2
/2
dx = 0.856.
0
y
1
y = e−x
2
/2
x
0
1
Figure 5.50
√
37. Notice that f (x) = 1 + x3 is increasing for 0 ≤ x ≤ 2, since x3 gets bigger as x increases. This means that f (0) ≤
f (x) ≤ f (2). For this function, f (0) = 1 and f (2) = 3. Thus, the area under f (x) lies between the area under the line
y = 1 and the area under the line y = 3 on the interval 0 ≤ x ≤ 2. See Figure 5.51. That is,
1(2 − 0) ≤
y
Z
0
2
p
1 + x3 dx ≤ 3(2 − 0).
y=3
3
f (x) =
√
1 + x3
y=1
1
x
2
Figure 5.51
38. (a) The integrand is positive, so the integral cannot be negative.
(b) The integrand ≥ 0. If the integral = 0, then the integrand must be identically 0, which is not true.
39. See Figure 5.52.
f (x)
Slope=
f (b)−f (a)
b−a
f (x)
✻
✻
❘
f (b) − f (a)
f (b) − f (a)
❄
a
x
b
Figure 5.52
40. See Figure 5.53.
❄
a
x
b
Figure 5.53
5.4 SOLUTIONS
473
41. See Figure 5.54.
f (x)
f (x)
✻
F (b)−F (a)
b−a
x
a
❄x
a
b
b
Figure 5.54
Figure 5.55
42. See Figure 5.55. Note that we are using the interpretation of the definite integral as the length of the interval times the
average value of the function on that interval, which we developed in Section 5.3.
43. See Figure 5.56. Since
In addition,
R2
0
R1
0
f (x) dx = A1 and
0<
Z
1
0
R2
1
f (x) dx = −A2 and
f (x) dx < −
2
Z
R3
2
f (x) dx <
1
f (x) dx = A3 , we know that
Z
3
f (x) dx.
2
f (x) dx = A1 − A2 , which is negative, but smaller in magnitude than
Z
2
f (x) dx <
2
Z
f (x) dx < 0.
R2
1
f (x) dx. Thus
0
1
The area A3 lies inside a rectangle of height 20 and base 1, so A3 < 20. The area A2 lies inside a rectangle below the
x-axis of height 10 and width 1, so −10 < A2 . Thus:
(viii) < (ii) < (iii) < (vi) < (i) < (v) < ( iv) < (vii).
✛
1
✲
✻
A1
A2
10
❄
−10
f (x)
20
A3
1
✻
10
❄
✛
❄
1
2
3
❄x
✲
Figure 5.56
44. (a)
(i) Since the triangular region under the graphs of f (x) has area 1/2, we have
1
Average(f ) =
2−0
Z
2
1
2−0
Z
2
f (x) dx =
1 1
1
· = .
2 2
4
g(x) dx =
1 1
1
· =
2 2
4
0
(ii) Similarly,
Average(g) =
0
(iii) Since f (x) is nonzero only for 0 ≤ x < 1 and g(x) is nonzero only for 1 < x ≤ 2, the product f (x)g(x) = 0
for all x. Thus
Z 2
Z 2
1
1
f (x)g(x) dx =
0 dx = 0.
Average(f · g) =
2−0 0
2 0
(b) Since the average values of f (x) and g(x) are nonzero, their product is nonzero. Thus the left side of the statement
is nonzero. However, the average of the product f (x)g(x) is zero. Thus, the right side of the statement is zero, so the
statement is not true.
474
Chapter Five /SOLUTIONS
45. (a) Since f (x) = sin x over [0, π] is between 0 and 1, the average of f (x) must itself be between 0 and 1. Furthermore,
since the graph of f (x) is concave down on this interval, the average value must be greater than the average height
of the triangle shown in Figure 5.57, namely, 0.5.
y
1
y = sin x
x
π
Figure 5.57
Z
π
1
sin x dx = 0.64.
π−0 0
46. (a) Splitting the integral in order to make use of the values in the table gives:
(b) Average =
1
√
2π
Z
1
3
e−x
2
/2
1
dx = √
2π
(b) Using the symmetry of ex
2
/2
1
√
2π
Z
3
e−x
2
/2
0
1
dx − √
2π
Z
1
e−x
2
0
/2
dx = 0.4987 − 0.3413 = 0.1574.
, we have
3
Z
e
−x2 /2
−2
1
dx = √
2π
Z
0
e
−x2 /2
−2
2
1
dx + √
2π
Z
2
1
1
= √
e−x /2 dx + √
2π 0
2π
= 0.4772 + 0.4987 = 0.9759.
Z
Z
3
e−x
0
3
e−x
2
2
/2
dx
/2
dx
0
47. There is not enough information to decide. If we had more information—say, for example, we knew that g was periodic
or had even or odd symmetry—we might be able to evaluate this expression.
48. The graph of y = g(−x) is the graph of y = g(x) reflected horizontally across the y-axis. This does not change the total
area. Thus, the two integrals are equal, so
Z
4
g(−x) dx = 12.
−4
49. Since
Z
7
f (x) dx = 25, we have
sZ
7
f (x) dx =
√
25 = 5.
0
0
50. There is not enough information to decide. In terms of area, knowing the area under the graph of f on 0 ≤ x ≤ 7 does not
help us find the area on 0 ≤ x ≤ 3.5. It depends on the shape of the graph and where it lies above and below the x-axis.
51. The graph of g(x) = f (x + 2) is the graph of y = f (x) shifted left by 2 units. Thus, the area under the graph of g on
−2 ≤ x ≤ 5 is the same as the area under the graph of f on 0 ≤ x ≤ 7. Therefore,
Z
5
f (x + 2) dx = 25.
−2
52. We have
Z
0
R7
7
(f (x) + 2) dx =
Z
7
f (x) dx +
7
2 dx = 39.
0
0
|
Z
{z
25
}
| {z }
7·2=14
2 dx is the area of a rectangle of height 2 and width 7.
√
√
√ √
53. (a) Clearly, the points where x = π, 2π, 3π, 4π are where the graph intersects the x-axis because f (x) =
sin(x2 ) = 0 where x is the square root of some multiple of π.
(b) Let f (x) = sin(x2 ), and let A, B, C, and D be the areas of the regions indicated in the figure below. Then we see
that A > B > C > D.
Note that, thinking graphically,
0
5.4 SOLUTIONS
1
A
C
x
B
D
−1
Note that
√
√
Z
√
π
Z
f (x) dx = A,
0
3π
√
π
Z
f (x) dx = A >
0
√
Z
Z
f (x) dx = A − B,
4π
√
Z
and
4π
f (x) dx = A − B + C − D.
0
√
3π
f (x) dx = A − (B − C) = A − B + C >
0
f (x) dx = A − B + C − D >
0
2π
0
f (x) dx = A − B + C,
0
It follows that
Z
Z
√
2π
f (x) dx = (A − B) > 0.
0
And thus the ordering is n = 1, n = 3, n = 4, and n = 2 from largest to smallest. All the numbers are positive.
54. See Figure 5.58.
The slope of
this line is f (a)
✻
F (x)
F (x)
❄
F (b) − F (a)
❄
x
a
x
a
b
Figure 5.58
b
Figure 5.59
55. See Figure 5.59.
56. See 5.60.
F (b)
✻
F (a)
The slope of this line is
✛
F (b) − F (a)
F (b)−F (a)
b−a
❄
a
b
✛
b−a
✲
=
1
b−a
Rb
a
f (x) dx
x
Figure 5.60
57. On the interval a ≤ t ≤ b, we have
Average value
of v(t)
=
1
b−a
Z
b
v(t) dt.
a
Since v(t) = s′ (t), by the Fundamental Theorem of Calculus, we get:
1
b−a
Z
a
b
v(t) dt =
1
(s(b) − s(a)) = Average velocity.
b−a
475
476
Chapter Five /SOLUTIONS
Strengthen Your Understanding
58. It is possible that f (x) has both positive and negative values on the interval [a, b], even if the integral is positive.
R 3 For example, suppose that a = 0 and b = 3, and let f (x) = 2 − x for all x. Then by finding areas, we can see that
f (x)dx = 3/2, which is nonnegative. However, f (x) < 0 for x in the interval (2, 3].
0
59. Using the properties of definite integrals, we have
Z
b
(5 + 3f (x)) dx =
a
b
Z
5 dx +
a
Z
Rb
a
3f (x) dx
a
= 5(b − a) + 3
The value of the integral
b − a.
b
Z
b
f (x) dx.
a
5 dx is 5(b − a) because the integral represents the area of a rectangle with height 5 and width
60. The time needs to be expressed in days in the definite integral, since the population is a function of time t in days. If the
six-month period contains 181 days, the correct integral is
1
181
181
Z
f (t) dt.
0
For a different six-month period, the number 181 may be different.
R1
R1
R1
61. If we let A = 0 f (x) dx, then we have 0 2f (x) dx = 2 0 f (x) dx = 2A. We want to define f (x) so that 2A < A.
In order for this to happen, A must be negative. So we’ll define f (x) to be negative on [0, 1]; for example, f (x) = −1.
R1
R1
R1
Then 0 f (x) dx = −1, and 0 2f (x) dx = 0 −2 dx = −2.
62. Let’s try the function f (x) = 2 − x. The area between the graph of the function f (x) = 2 − x and the x-axis on the
interval [0, 4] consists of two isosceles right triangles with legs of length 2: one above the x-axis and one below. When
R4
we compute the integral 0 (2 − x) dx, the areas of these two congruent triangles cancel out, and the integral is zero.
63. If the car is going v(t) miles per hour at time t hours, then
Average speed =
64. True, since
R2
0
(f (x) + g(x)) dx =
65. False. It is possible that
R2
0
R2
0
66. False. We know that
as
R2
0
0
f (x) dx =
R2
R2
0
0
R2
0
f (x) dx +
2f (x) dx = 2
R2
0
f (x) dx.
R4
R2 2
0
Z
5
v(t) dt.
0
g(x) dx.
R2
0
f (x) dx = 4 and
(f (x) + g(x)) dx =
0
f (x) dx. For example, if f (x) = 3x, then
67. True, since
R2
(f (x) + g(x)) dx = 10 and
if f (x) = 5x − 3 and g(x) = 3, then
R4
f (x) dx +
1
5−0
R2
0
R2
0
g(x) dx = 6, for instance. For example,
5x dx = 10, but
R2
0
f (x) dx, but it is not true that
f (x) dx = 6, but
R4
0
f (x) dx = 4 and
R4
2
R2
0
g(x) dx = 6.
f (x) dx must be the same value
f (x) dx = 24.
68. False. This would be true if h(x) = 5f (x). However, we cannot assume that f (5x) = 5f (x), so for many functions
R2
this statement is false. For example, if f is the constant function f (x) = 3, then h(x) = 3 as well, so 0 f (x) dx =
R2
0
h(x) dx = 6.
69. True. If a = b, then ∆x = 0 for any Riemann sum for f on the interval [a, b], so every Riemann sum has value 0. Thus,
the limit of the Riemann sums is 0.
70. False. For example, let a = −1 and b = 1 and f (x) = x.Then the areas bounded by the graph of f and the x-axis on the
R1
two halves of the interval [−1, 1] cancel with each other and make −1 f (x) dx = 0. See Figure 5.61.
5.4 SOLUTIONS
477
f (x) = x
−1
x
1
Figure 5.61
71. False. Let f (x) = 7 and g(x) = 9 for all x.
R3
R3
R3
R2
Then 1 f (x) dx + 2 g(x) dx = 7 + 9 = 16, but 1 (f (x) + g(x)) dx = 1 16 dx = 32.
72. False. If the graph of f is symmetric about the y-axis, this is true, but otherwise it is usually not true. For example, if
f (x) = x + 1 the area under the graph of f for −1 ≤ x ≤ 0 is less than the area under the graph of f for 0 ≤ x ≤ 1, so
R1
R1
f (x) dx < 2 0 f (x) dx. See Figure 5.62.
−1
f (x) = x + 1
x
−1
1
Figure 5.62
73. True, by Theorem 5.4 on Comparison of Definite Integrals:
1
b−a
Z
b
a
1
f (x) dx ≤
b−a
Z
b
g(x) dx.
a
74. True. We have
Average value of f on [0, 10] =
=
=
1
10 − 0
1
10
1
2
Z
10
Z
f (x) dx
0
5
f (x) dx +
0
10
f (x) dx
5
Z
1
5
Z
5
f (x) dx +
0
1
5
Z
5
10
f (x) dx
The average of the average value of f on [0, 5] and
=
the average value of f on [5, 10].
75. False. If the values of f (x) on the interval [c, d] are larger than the values of f (x) in the rest of the interval [a, b], then the
average value of f on the interval [c, d] is larger than the average value of f on the interval [a, b]. For example, suppose
f (x) =
(
0
x < 1 or x > 2
1
1 ≤ x ≤ 2.
Then the average value of f on the interval [1, 2] is 1, whereas the average value of f on the interval [0, 3] is (1/(3 −
R3
0)) 0 f (x) dx = 1/3.
478
Chapter Five /SOLUTIONS
76. True. We have by the properties of integrals in Theorem 5.3,
Z
9
f (x) dx =
1
Since (1/(4 − 1))
R4
1
4
Z
1
f (x) dx = A and (1/(9 − 4))
Z
9
R9
4
9
Z
f (x) dx +
f (x) dx.
4
f (x) dx = B, we have
f (x) dx = 3A + 5B.
1
Dividing this equation through by 8, we get that the average value of f on the interval [1, 9] is (3/8)A + (5/8)B.
77. True. By the properties of integrals in Theorem 5.3, we have:
Z
b
(f (x) + g(x)) dx =
b
Z
f (x) dx +
a
a
b
Z
g(x) dx.
a
Dividing both sides of this equation through by b − a, we get that the average value of f (x) + g(x) is average value of
f (x) plus the average value of g(x):
1
b−a
b
Z
(f (x) + g(x)) dx =
a
1
b−a
Z
b
b
1
b−a
Z
n
0
3
0≤x≤1
1 < x ≤ 2.
1
2
Z
f (x) dx +
a
g(x) dx.
a
78. False. A counterexample is given by
f (x) =
n
3
0
0≤x≤1
1 II = IV > III
Using the definition of average value and the fact that f is even, we have
Average value
=
of f on II
R2
0
f (x)dx
=
2
=
1
2
R2
R2
−2
−2
f (x)dx
2
f (x)dx
4
= Average value of f on IV.
SOLUTIONS to Review Problems for Chapter Five
485
Since f is decreasing on [0,5], the average value of f on the interval [0, c], where 0 ≤ c ≤ 5, is decreasing as a
function of c. The larger the interval the more low values of f are included. Hence
Average value of f Average value of f Average value of f
>
>
on [0, 1]
on [0, 2]
on [0, 5]
29. (a) A graph of f ′ (x) = sin(x2 ) is shown in Figure 5.70. Since the derivative f ′ (x) is positive between x = 0 and
x = 1, the change in f (x) is positive, so f (1) is larger than f (0). Between x = 2 and x = 2.5, we see that f ′ (x) is
negative, so the change in f (x) is negative; thus, f (2) is greater than f (2.5).
1
x
1
2
3
−1
Figure 5.70: Graph of f ′ (x) = sin(x2 )
(b) The change in f (x) between x = 0 and x = 1 is given by the Fundamental Theorem of Calculus:
f (1) − f (0) =
1
Z
sin(x2 )dx = 0.310.
0
Since f (0) = 2, we have
f (1) = 2 + 0.310 = 2.310.
Similarly, since
f (2) − f (0) =
we have
2
Z
sin(x2 )dx = 0.805,
0
f (2) = 2 + 0.805 = 2.805.
Since
f (3) − f (0) =
we have
Z
3
sin(x2 )dx = 0.774,
0
f (3) = 2 + 0.774 = 2.774.
The results are shown in the table.
x
0
1
2
3
f (x)
2
2.310
2.805
2.774
30. (a) By the product rule
F ′ (t) = 1 · ln t + t ·
(b)
(i) Using a calculator, we find
Z
1
− 1 = ln t.
t
12
ln t dt = 4.793.
10
(ii) The Fundamental Theorem of Calculus tells us that we can get the exact value of the integral by looking at
Z
12
10
ln t dt = F (12) − F (10) = (12 ln 12 − 12) − (10 ln 10 − 10) = 12 ln 12 − 10 ln 10 − 2 = 4.79303.
486
Chapter Five /SOLUTIONS
31. (a) F (0) = 0.
2
(b) F increases because F ′ (x) = e−x is positive.
(c) Using a calculator, we get F (1) ≈ 0.7468, F (2) ≈ 0.8821, F (3) ≈ 0.8862.
32. We have
5
Z
8=
f (x) dx =
R2
33. Since f is even,
−2
2
R
0
f (x) dx = 0, so
R5
−2
5
f (x) dx =
R5
0
Z
18 =
R5
2
f (x) dx −
4 dx = 4(5 − 2) = 12, we have
R5
4
5
Z
f (x) dx +
2
5
4 dx.
2
f (x) dx = 18 − 12 = 6,
5
f (x) dx = 2.
2
f (x) dx = −
f (x) dx =
Z
5
Z
3
f (x) dx = 8/2 = 4 and
5
Z
2
Z
2
f (x) dx = 7 − 3 = 4.
0
(3f (x) + 4) dx = 3
so
R4
2
Z
5
2
35. We have
f (x) dx.
f (x) dx = (1/2)14 = 7. Therefore
2
Thus, since
5
2
5
Z
0
2
34. We have
Z
f (x) dx = 8.
f (x) dx = (1/2)6 = 3 and
Z
f (x) dx +
−2
−2
Since f is odd,
2
Z
4
Z
R4
5
f (x) dx = −1. Thus
f (x) dx +
2
5
Z
f (x) dx = 4 − 1 = 3.
4
36. (a) 0, since the integrand is an odd function and the limits are symmetric around 0.
(b) 0, since the integrand is an odd function and the limits are symmetric around 0.
Rp
37. Let A = 0 f (x) dx. We are told A > 0. We can interpret A as the area under f on this interval. Since the graph of f
repeats periodically every p units, the area under the graph also repeats. This means
Z
p
f (x) dx =
}
2p
Z
5p
Z
7p
{z
0
f (x) dx =
p
Z
p
f (x) dx +
| 0 {z
Z 2p A
f (x) dx =
5p
|
Z
6p
|
2p
}
{z
}
A
A
Z
5p
p
0
7p
f (x) dx
5p
{z
}
area under one
full period
Z
2p
f (x) dx
= 2A
p
}
{z
f (x) dx +
Z
f (x) dx = · · · = A.
|
{z
AZ
|
f (x) dx +
5p
Z
3p
f (x) dx + · · · +
p
so
}
area under one
full period
f (x) dx =
Z
2p
|
area under one
full period
Z
f (x) dx =
p
| 0 {z
We have that
2p
Z
Z
7p
5p
4p
}
f (x) dx = 4A
|
{z
{z
}
A
f (x) dx
6p
|
A
f (x) dx
=
}
= 2A,
2A + 4A
= 3.
2A
SOLUTIONS to Review Problems for Chapter Five
487
38. Since the graph is symmetrical, we know that
0
Z
f (x) dx =
−r
This means that if we let A =
Z
r
Z
1
f (x) dx =
2
0
Z
r
f (x) dx.
−r
0
f (x) dx then
−r
2A
z
Z
r
A
}|
{
f (x) dx +
−r
0
Z
zZ
}|
r
{
f (x) dx
0
2A + A
= 3.
A
=
f (x) dx
−r
|
{z
}
A
39. (a) Since the function is odd, the areas above and below the x-axis cancel. Thus,
Z
0
−3
so
Z
3
2
xe−x dx = −
2
xe−x dx =
Z
0
Z
3
0
2
xe−x dx +
−3
−3
2
xe−x dx,
Z
3
2
xe−x dx = 0.
0
(b) For 0 ≤ x ≤ 3 with n = 3, we have x0 = 0, x1 = 1, x2 = 2, x3 = 3, and ∆x = 1. See Figure 5.71. Thus,
2
2
2
Left sum = f (x0 )∆x + f (x1 )∆x + f (x2 )∆x = 0e−0 · 1 + 1e−1 · 1 + 2e−2 · 1 = 0.4045.
(c) For −3 ≤ x ≤ 0, with n = 3, we have x0 = −3, x1 = −2, x2 = −1, x3 = 0, and ∆x = 1. See Figure 5.71. Thus,
2
2
2
Left sum = f (x0 )∆x + f (x1 )∆x + f (x2 )∆x = −3e−(−3) · 1 − 2e−(−2) · 1 − 1e−(−1) · 1 = −0.4049.
(d) No. The rectangles between −3 and 0 are not the same size as those between 0 and 3. See Figure 5.71. There are
three rectangles with nonzero height on [−3, 0] and only two on [0, 3].
−3
−2
−1
x
1
2
3
Figure 5.71
40. (a) Train A starts earlier than Train B, and stops later. At every moment Train A is going faster than Train B. Both trains
increase their speed at a constant rate through the first half of their trip and slow down during the second half. Both
trains reach their maximum speed at the same time. The area under the velocity graph for Train A is larger than the
area under the velocity graph for Train B, meaning that Train A travels farther—as would be expected, given that its
speed is always higher than B’s.
(b) (i) The maximum velocity is read off the vertical axis. The graph for Train A appears to go about twice as high as
the graph for Train B; see Figure 5.72. So
Maximum velocity of Train A
vA
=
≈ 2.
Maximum velocity of Train B
vB
488
Chapter Five /SOLUTIONS
v (km/hr)
vA
Train A
Train B
vB
✛
✛
tB
tA
t (hr)
✲
✲
Figure 5.72
(ii) The time of travel is the horizontal distance between the start and stop times (the two t-intercepts). The horizontal
distance for Train A appears to be about twice the corresponding distance for Train B; see Figure 5.72. So
tA
Time traveled by Train A
=
≈ 2.
Time traveled by Train B
tB
(iii) The distance traveled by each train is given by the area under its graph. Since the area of triangle is
Height, and since the base and height for Train A is approximately twice that for Train B, we have
Distance traveled by Train A
=
Distance traveled by Train B
1
2
1
2
1
2
· Base ·
· vA · tA
≈ 2 · 2 = 4.
· vB · tB
41. (a) The model indicates that wind capacity was 21,000 in 2000 and 21,000e0.22·10 = 189,525.284 in 2010.
(b) Since the continuous growth rate was 0.22, the wind energy generating capacity was increasing by 22% per year.
(c) We use the formula for average value between t = 0 and t = 10:
Average value =
1
10 − 0
Z
10
1
(766,024.02) = 76,602.402 megawatts.
10
21,000e0.22t dt =
0
42. We know that T is the time at which all the fuel has burned, so
Z
T
r(t) dt is the total amount of fuel burned during the
0
entire time period. We also know that Q is the initial amount of fuel.
Since the fuel is completely burned at time t = T , we conclude that
43. The expression
Z
0.5T
Z
T
r(t) dt = Q.
0
r(t) dt is the amount of fuel burned during the first half of the total time.
0
The expression
Z
T
r(t) dt is the amount of fuel burned during the second half of the total time.
0.5T
Since r ′ (t) < 0, we know the rate fuel burns is going down. Thus, fuel is burned faster at first, then slower later on,
so more fuel is burned during the first half of the total time than during the second half.
We conclude that
44. The expression
Z
T /3
Z
0.5T
r(t) dt >
Z
T
r(t) dt.
0.5T
0
r(t) dt is the amount of fuel burned during the first one-third of the total time. The expression Q/3
0
is one-third of the total fuel burned.
Since r ′ (t) < 0, we know the rate fuel burns is going down. Thus, fuel is burned faster at first, then slower later on,
so more than one-third of the total fuel burned is burned during the first one-third of the total time.
We conclude that
Z
T /3
r(t) dt > Q/3.
0
45. We know that Th is the time at which half the fuel has burned, so
Z
Th
r(t) dt equals half the fuel burned.
0
We know that T is the time at which all the fuel has burned, so 0.5T is halfway through the entire time period. This
means
Z
0
0.5T
r(t) dt is the amount of fuel burned halfway through the total time period.
SOLUTIONS to Review Problems for Chapter Five
489
Since r ′ (t) < 0, we know the rate fuel burns is going down. This means more than half the fuel is burned during the
first half of the total time period than the second half.
We conclude that more than half the fuel is burned during the first half of the time period, so
Z
0.5T
Z
Th
r(t) dt. <
0
r(t) dt
0
46. By the given property,
Z
a
a
a
Z
f (x) dx = −
f (x) dx, so 2
a
Z
a
f (x) dx = 0. Thus
a
47. We know that the average value of v(x) = 4, so
1
6−1
Z
Z
a
f (x) dx = 0.
a
6
v(x) dx = 4, and thus
1
Z
6
v(x) dx = 20.
1
Similarly, we are told that
1
8−6
8
Z
Z
v(x) dx = 5, so
6
Z
1
8
v(x) dx = 10.
6
The average value for 1 ≤ x ≤ 8 is given by
1
Average value =
8−1
8
1
v(x) dx =
7
Z
6
v(x) dx +
1
Z
8
v(x) dx
6
=
30
20 + 10
=
.
7
7
48. (a) Yes.
(b) No, because
R 3 the sum of the left sums has 20 subdivisions. The result is the left-sum approximation with 20 subdivisions to 1 f (x) dx.
49. In Figure 5.73 the area A1 is largest, A2 is next, and A3 is smallest. We have
I=
IV =
Z
b
Za e
b
f (x) dx = A1 ,
II =
Z
c
f (x) dx = A1 − A2 ,
a
f (x) dx = −A2 + A3 ,
V=
Z
III =
a
c
b
Z
e
f (x) dx = A1 − A2 + A3 ,
f (x) dx = −A2 .
The relative sizes of A1 , A2 , and A3 mean that I is positive and largest, III is next largest (since −A2 + A3 is negative,
but less negative than −A2 ), II is next largest, but still positive (since A1 is larger than A2 ). The integrals IV and V are
both negative, but V is more negative. Thus
V < IV < 0 < II < III < I.
f (x)
A1
A3
a
b
A2
c
x
e
Figure 5.73
50. Since f (t) = F ′ (t), we know from the Fundamental Theorem of Calculus that
Z
7
9
f (t) dt = F (9) − F (7) = 1.
|{z}
5
|{z}
4
This tells us that between the hours of 7 am and 9 am, the ice thickness increased by 1 inch.
490
Chapter Five /SOLUTIONS
51. Since f (t) = F ′ (t), we know from the Fundamental Theorem of Calculus that
Z
7
f (t) dt = F (7) − F (4)
4
so F (4) = F (7) −
|{z}
4
This tells us that at 4 am the ice is 2.5 inches thick.
7
Z
f (t) dt = 2.5.
4
{z
|
1.5
}
52. We know that f (t) = F ′ (t), so F ′ (3.5) = f (3.5). We have:
Z
3.5
f ′ (t) dt = 0.75
2
f (3.5) − f (2) = 0.75
Fundamental Theorem of Calculus
f (3.5) = f (2) +0.75 = 1.25,
|{z}
0.5
so F ′ (3.5) = 1.25. This tells us that the ice thickness is growing by 1.25 inches/hour at 3:30 am.
53. (a) For −2 ≤ x ≤ 2, f is symmetrical about the y-axis, so
(b) For any function f ,
(c) Note that
R0
0
f (x) dx =
f (x) dx =
−2
54. (a) We know that
(b)
R2
R5
1
2
R2
f (x) dx =
f (x) dx =
−2
R5
f (x) dx −
(c) Using symmetry again,
R2
0
0
f (x) dx −
R5
f (x) dx, so
−2
f (x) dx−
0
2
R5
R2
1
f
(x)
dx
−
f
(x)
dx.
R2 5 −2
R2
R05
2
R5
−2
0
R2
0
f (x) dx =
1
2
5
2
R0
f (x) dx =
−2
f (x) dx.
f (x) dx =
R5
−2
R5
−2
f (x) dx − 2
f (x) dx −
−2
R5
2
R2
0
f (x) dx and
f (x) dx−
f (x) dx. By symmetry,
f (x) dx =
R
R5
R0
−2
R2
0
R0
−2
R2
−2
f (x) dx =
f (x) dx =
1
2
R2
f (x) dx = 2
R5
−2
−2
R2
0
f (x) dx− 21
f (x) dx, so
R5
2
f (x) dx.
R2
−2
f (x) dx.
f (x) dx =
f (x) dx.
f (x) dx .
55. (a) The mouse changes direction (when its velocity is zero) at about times 17, 23, and 27.
(b) The mouse is moving most rapidly to the right at time 10 and most rapidly
to the left at time 40.
Rt
(c) The mouse is farthest to the right when the integral of the velocity, 0 v(t) dt, is most positive. Since the integral is
the sum of the areas above the axis minus the areas below the axis, the integral is largest when the velocity is zero at
about 17 seconds. The mouse is farthest to the left of center when the integral is most negative at 40 seconds.
(d) The mouse’s speed decreases during seconds 10 to 17, from 20 to 23 seconds, and from 24 seconds to 27 seconds.
(e) The mouse is at the center of the tunnel at any time t for which the integral from 0 to t of the velocity is zero. This is
true at time 0 and again somewhere around 35 seconds.
56. To estimate the distance traveled during the first 15 seconds, we calculate upper and lower bound estimates for this
distance, and then average these estimates. Note that because the speed of the Prius is in miles per hour, we need to
convert seconds to hours (1 sec = 1/3600 hours) to calculate the distance estimates in miles. We then convert miles to
feet (1 mile = 5280 feet).
Total distance traveled
between t = 0 and t = 15
(Lower Estimate)
= 0·
Total distance traveled
between t = 0 and t = 15
(Upper Estimate)
= 20 ·
5
5
5
+ 20 ·
+ 33 ·
= 0.0736 miles or 389 feet
3600
3600
3600
5
5
5
+ 33 ·
+ 45 ·
= 0.1361 miles or 719 feet
3600
3600
3600
Since:
389 + 719
Average of Lower and Upper
= 554 feet,
=
2
Estimates
the 2010 Prius Prototype travels about 554 feet, roughly a tenth of a mile, during the first 15 seconds of movement in
EV-only mode.
57. To estimate the distance between the two cars at t = 5, we calculate upper and lower bound estimates for this distance
and then average these estimates. Note that because the speed of the Prius is in miles per hour, we need to convert seconds
SOLUTIONS to Review Problems for Chapter Five
491
to hours (1 sec = 1/3600 hours) to calculate the distance estimates in miles. We may then convert miles to feet (1 mile
= 5280 feet).
Distance between cars at t = 5
(Upper Estimate)
≈ (Difference in speeds at t = 5) · (Travel time)
= (33 − 20) ·
Distance between cars at t = 5
(Lower Estimate)
5
= 0.018 miles
3600
≈ (Difference in speeds at t = 0) · (Travel time)
= (0 − 0) ·
5
= 0 miles
3600
Since:
0.018 + 0
Average of Lower and Upper Estimates of
= 0.009 miles = 48 feet,
=
2
distance between cars at t = 5
the distance between the two cars is about 48 feet, five seconds after leaving the stoplight.
58. (a) The acceleration is positive for 0 ≤ t < 40 and for a tiny period before t = 60, since the slope is positive over these
intervals. Just to the left of t = 40, it looks like the acceleration is approaching 0. Between t = 40 and a moment just
before t = 60, the acceleration is negative.
(b) The maximum altitude was about 500 feet, when t was a little greater than 40 (here we are estimating the area under
the graph for 0 ≤ t ≤ 42).
(c) The acceleration is greatest when the slope of the velocity is most positive. This happens just before t = 60, where
the magnitude of the velocity is plunging and the direction of the acceleration is positive, or up.
(d) The deceleration is greatest when the slope of the velocity is most negative. This happens just after t = 40.
(e) After the Montgolfier Brothers hit their top climbing speed (at t = 40), they suddenly stopped climbing and started
to fall. This suggests some kind of catastrophe—the flame going out, the balloon ripping, etc. (In actual fact, in their
first flight in 1783, the material covering their balloon, held together by buttons, ripped and the balloon landed in
flames.)
(f) The total change in altitude for the Montgolfiers and their balloon is the definite integral of their velocity, or the total
area under the given graph (counting the part after t = 42 as negative, of course). As mentioned before, the total area
of the graph for 0 ≤ t ≤ 42 is about 500. The area for t > 42 is about 220. So subtracting, we see that the balloon
finished 280 feet or so higher than where it began.
59. (a) About 300 meter3 /sec.
(b) About 250 meter3 /sec.
(c) Looking at the graph, we can see that the 1996 flood reached its maximum just between March and April, for a high
of about 1250 meter3 /sec. Similarly, the 1957 flood reached its maximum in mid-June, for a maximum flow rate of
3500 meter3 /sec.
(d) The 1996 flood lasted about 1/3 of a month, or about 10 days. The 1957 flood lasted about 4 months.
(e) The area under the controlled flood graph is about 2/3 box. Each box represents 500 meter3 /sec for one month. Since
1 month = 30
hours
minutes
seconds
days
· 24
· 60
· 60
month
day
hour
minute
= 2.592 · 106 ≈ 2.6 · 106 seconds,
each box represents
Flow ≈ (500 meter3 /sec) · (2.6 · 106 sec) = 13 · 108 meter3 of water.
So, for the artificial flood,
2
· 13 · 108 = 8.7 · 108 meter3 ≈ 109 meter3 .
3
(f) The 1957 flood released a volume of water represented by about 12 boxes above the 250 meter/sec baseline. Thus,
for the natural flood,
Additional flow ≈
Additional flow ≈ 12 · 13 · 108 = 1.95 · 1010 ≈ 2 · 1010 meter3 .
So, the natural flood was nearly 20 times larger than the controlled flood and lasted much longer.
492
Chapter Five /SOLUTIONS
60. In (a), f ′ (1) is the slope of a tangent line at x = 1, which is negative. As for (c), the rate of change in f (x) is given by
f ′ (x), and the average value of this over 0 ≤ x ≤ a is
1
a−0
Z
a
f ′ (x) dx =
0
f (a) − f (0)
.
a−0
This is the slope of the line through the points (0,
R a1) and (a,
0), which is less
R a negative that the tangent line at x = 1.
Therefore, (a) < (c) < 0. The quantity (b) is 0 f (x) dx /a and (d) is 0 f (x) dx, which is the net area under the
Ra
graph of f (counting the area as negative for f below the x-axis). Since a > 1 and 0 f (x) dx > 0, we have 0 <(b)<(d).
Therefore
(a) < (c) < (b) < (d).
61. This expression gives the tangent-line approximation to f (3.1). Since f ′′ (x) < 0, this tells us that the graph of f is
concave-down, which means that the graph of f lies below its tangent line. Thus, this value overestimates f (3.1).
62. From the Fundamental Theorem of Calculus, we see that
f (3) +
Z
3.1
3
f ′ (t) dt = f (3) + f (3.1) − f (3) = f (3.1),
so this value is exact.
F (3.11) − F (3.1)
gives the slope of the secant line to the graph of F from x = 3.1 to x = 3.11. This
63. The expression
0.01 ′
approximates the value of F (3.1). Since f (x) = F ′ (x), this means the expressions approximates the value of f (3.1).
Notice that f ′ (x) = F ′′ (x); since f ′ (x) > 0, this means F ′′ (x) > 0, which tells us that the graph of F is concaveup. This means the slope of the secant line from x = 3.1 to x = 3.11 is greater (either “less negative” or “more positive”)
than the slope of the tangent line at x = 3.1. Thus, this expression overestimates F ′ (3.1), so it is an overestimate of
f (3.1).
64. We see that
n
X
f ′ (xi ) ∆x is a right-hand Riemann sum. Since f ′′ (x) < 0 for all x, we know that f ′ (x) is a decreasing
i=1
function, which means the right-hand sum is an underestimate of the true value. Thus,
f (3) +
n
X
′
f (xi ) ∆x < f (3) +
i=1
Z
3.1
f ′ (x) dx
3
= f (3) + f (3.1) − f (3)
Fundamental Theorem of Calculus
= f (3.1).
This means the expression is an underestimate of f (3.1).
65. (a) Divide the interval 0 ≤ t ≤ T into small subintervals of length ∆t on which the temperature is approximately
constant. Then if ti is the ith interval, on that interval
∆S ≈ (H − Hmin )∆t = (f (ti ) − Hmin )∆t.
Thus, the total number of degree-days is approximated by
S≈
As n → ∞, we have
S=
n
X
i=1
Z
(f (ti ) − Hmin )∆t.
T
(f (t) − Hmin ) dt.
0
(b) Since Hmin = 15◦ C and S = 125 degree-days, the formula in part (a) gives
125 =
Z
0
T
(f (t) − 15)dt.
SOLUTIONS to Review Problems for Chapter Five
493
We approximate the definite integral using a Riemann sum:
Z
0
T
(f (t) − 15)dt =
n
X
i=1
(f (ti ) − 15)∆t.
In the table, we have ∆t = 1. We want to find how many terms in the sum are required to (approximately) equal the
required number of degree-days, S = 125. From the table, we have
2
X
i=1
(f (ti ) − 15)∆t = (20 − 15)1 + (22 − 15)1 = 12,
so the sum of degree-days over a two-day period is much smaller than the required sum of 125. Likewise, the sum
over a three-day period is only 24:
3
X
i=1
(f (ti ) − 15)∆t = (20 − 15)1 + (22 − 15)1 + (27 − 15)1 = 24.
Summing over a ten-day period and an eleven-day period, we find, respectively, that
10
X
i=1
(f (ti ) − 15)∆t = (20 − 15)1 + (22 − 15)1 + (27 − 15)1 + (28 − 15)1 + (27 − 15)1 + (31 − 15)1
11
X
i=1
+(29 − 15)1 + (30 − 15)1 + (28 − 15)1 + (25 − 15)1 = 117
(f (ti ) − 15)∆t = (20 − 15)1 + (22 − 15)1 + (27 − 15)1 + (28 − 15)1 + (27 − 15)1 + (31 − 15)1
+(29 − 15)1 + (30 − 15)1 + (28 − 15)1 + (25 − 15)1 + (24 − 15)1 = 126.
Thus, the sum reaches 125 degree-days somewhere just short of T = 11 days.
66. (a) The slope of a line is the tangent of the angle the line makes with the horizontal. Thus the slope, f ′ (x), of the tangent
line at x is the tangent of the angle θ.
(b) Using the identity
1
1 + tan2 θ =
cos2 θ
and part (a) we have
1
.
1 + (f ′ (x))2 =
cos2 θ
(c) Using the chain rule, we have
d
1 dθ
(tan θ) =
.
dx
cos2 θ dx
Therefore, using parts (a) and (b), we have
d
dθ
= cos2 θ f ′ (x)
dx
dx
1
f ′′ (x).
=
1 + (f ′ (x))2
(d) By the Fundamental Theorem of Calculus we have
θ(b) − θ(a) =
Z
b
′
θ (x) dx =
Z
a
a
b
f ′′ (x)
dx.
1 + (f ′ (x))2
CAS Challenge Problems
67. (a) We have ∆x = (1 − 0)/n = 1/n and xi = 0 + i · ∆x = i/n. So we get
Right-hand sum =
n
X
i=1
(xi )4 ∆x =
n
X
i 4 1
i=1
n
n
=
n
X
i4
i=1
n5
.
494
Chapter Five /SOLUTIONS
(b) The CAS gives
Right-hand sum =
n
X
i4
i=1
6n4 + 15n3 + 10n2 − 1
.
30n4
=
n5
(The results may look slightly different depending on the CAS you use.)
(c) Using a CAS or by hand, we get
6n4
1
6n4 + 15n3 + 10n2 − 1
= lim
= .
4
n→∞ 30n4
30n
5
lim
n→∞
The numerator is dominated by the highest power term, which is 6n4 , so when n is large, the ratio behaves like
6n4 /30n4 = 1/5 as n → ∞. Thus we see that
Z
1
x4 dx =
0
1
.
5
68. (a) A Riemann sum with n subdivisions of [0, 1] has ∆x = 1/n and xi = i/n. Thus,
Right-hand sum =
n
X
i 5 1
n
i=1
n
=
n
X
i5
i=1
n6
.
(b) A CAS gives
Right-hand sum =
n
X
i5
i=1
(c) Taking the limit by hand or using a CAS gives
lim
n→∞
n6
=
2n4 + 6n3 + 5n2 − 1
.
12n4
2n4 + 6n3 + 5n2 − 1
2n4
1
=
lim
= .
n→∞ 12n4
12n4
6
The numerator is dominated by the highest power term, which is 2n4 , so the ratio behaves like 2n4 /12n4 = 1/6, as
n → ∞. Thus we see that
Z 1
1
x5 dx = .
6
0
69. (a) Since the length of the interval of integration is 2 − 1 = 1, the width of each subdivision is ∆t = 1/n. Thus the
endpoints of the subdivision are
t0 = 1,
t1 = 1 + ∆t = 1 +
ti = 1 + i∆t = 1 +
1
,
n
i
,...,
n
2
,...,
n
n−1
= 1 + (n − 1)∆t = 1 +
.
n
t2 = 1 + 2∆t = 1 +
tn−1
Thus, since the integrand is f (t) = t,
Left-hand sum =
n−1
X
f (ti )∆t =
n−1
X
ti ∆t =
i=0
i=0
n−1
X
1+
i=0
i
n
1
n
=
n−1
X
n+i
i=0
n2
.
(b) The CAS finds the formula for the Riemann sum
n−1
X
n+i
i=0
n2
=
(−1+n) n
2
n2
+ n2
=
3
1
−
.
2
2n
(c) Taking the limit as n → ∞
lim
n→∞
3
1
−
2
2n
= lim
n→∞
3
1
3
3
− lim
= +0= .
2 n→∞ 2n
2
2
(d) The shape under the graph of y = t between t = 1 and t = 2 is a trapezoid of width 1, height 1 on the left and 2 on
the right. So its area is 1 · (1 + 2)/2 = 3/2. This is the same answer we got by computing the definite integral.
SOLUTIONS to Review Problems for Chapter Five
495
70. (a) Since the length of the interval of integration is 2 − 1 = 1, the width of each subdivision is ∆t = 1/n. Thus the
endpoints of the subdivision are
t0 = 1,
1
,
n
t1 = 1 + ∆t = 1 +
i
,...,
n
ti = 1 + i∆t = 1 +
2
,...,
n
n−1
= 1 + (n − 1)∆t = 1 +
.
n
t2 = 1 + 2∆t = 1 +
tn−1
Thus, since the integrand is f (t) = t2 ,
Left-hand sum =
n−1
X
n−1
X
f (ti )∆t =
t2i ∆t =
i=0
i=0
n−1
X
1+
i=0
i
n
2 1
n
=
n−1
X
(n + i)2
n3
i=0
.
(b) Using a CAS to find the sum, we get
n−1
X
(n + i)2
i=0
(−1 + 2 n) (−1 + 7 n)
7
1
3
= +
−
.
6 n2
3
6 n2
2n
=
n3
(c) Taking the limit as n → ∞
lim
n→∞
7
3
+
1
3
−
6 n2
2n
R2
= lim
n→∞
1
3
7
7
7
+ lim
− lim
= +0+0 = .
3 n→∞ 6 n2 n→∞ 2 n
3
3
(d) We have calculated 1 t2 dt using Riemann sums. Since t2 is above the t-axis between t = 1 and t = 2, this integral
is the area; so the area is 7/3.
71. (a) Since the length of the interval of integration is π, the width of each subdivision is ∆x = π/n. Thus the endpoints
of the subdivision are
x0 = 0,
x1 = 0 + ∆x =
xi = 0 + i∆x =
iπ
,
n
π
,
n
...,
2π
,...,
n
nπ
xn = 0 + n∆x =
= π.
n
x2 = 0 + 2∆x =
Thus, since the integrand is f (x) = sin x,
Right-hand sum =
n
X
f (xi )∆x =
n
X
sin(xi )∆x =
i=1
i=1
n
X
sin
i=1
iπ
n
(b) If the CAS can evaluate this sum, we get
n
X
iπ π
sin
n
i=1
n
=
π cot(π/2n)
π cos(π/2n)
=
.
n
n sin(π/2n)
(c) Using the computer algebra system, we find that
lim
n→∞
(d) The computer algebra system gives
Z
π cos(π/2n)
= 2.
n sin(π/2n)
π
sin x dx = 2.
0
72. (a) A CAS gives
Z
b
sin(cx) dx =
a
cos(ac)
cos(bc)
−
.
c
c
(b) If F (x) is an antiderivative of sin(cx), then the Fundamental Theorem of Calculus says that
Z
a
b
sin(cx) dx = F (b) − F (a).
π
.
n
496
Chapter Five /SOLUTIONS
Comparing this with the answer to part (a), we see that
F (b) − F (a) =
cos(bc)
cos(ac)
−
=
c
c
−
cos(cb)
c
−
−
cos(ca)
c
.
This suggests that
cos(cx)
.
c
F (x) = −
Taking the derivative confirms this:
d
dx
−
cos(cx)
c
= sin(cx).
73. (a) Different systems may give different answers. A typical answer is
Z
c
a
x
dx =
1 + bx2
ln
c2 b + 1
a2 b + 1
2b
.
Some CASs may not have the absolute values in the answer; since b > 0, the answer is correct without the absolute
values.
(b) Using the properties of logarithms, we can rewrite the answer to part (a) as
Z
a
c
ln c2 b + 1 − ln a2 b + 1
ln c2 b + 1
ln a2 b + 1
x
dx =
=
−
.
2
1 + bx
2b
2b
2b
If F (x) is an antiderivative of x/(1 + bx2 ), then the Fundamental Theorem of Calculus says that
Z
c
a
x
dx = F (c) − F (a).
1 + bx2
Thus
F (c) − F (a) =
ln a2 b + 1
ln c2 b + 1
−
.
2b
2b
This suggests that
ln 1 + bx2
.
2b
2
= 1 + bx .) Taking the derivative confirms this:
F (x) =
(Since b > 0, we know 1 + bx2
d
dx
ln(1 + b x2 )
2b
=
x
.
1 + b x2
PROJECTS FOR CHAPTER FIVE
1. (a) The work done by the ventricle wall on the blood is the force exerted on the blood times the distance the
wall pushes the blood; that is, the distance the wall moves. In Phases 1 and 3 there is no work done because
the ventricle wall is motionless.
The wall moves in Phases 2 and 4. The work done by the wall is positive if the force from the wall to
the blood is in the direction of the movement, and it is negative if the force and movement are in opposite
directions. In Phase 4, the ventricle is contracting, pushing the blood, so the work is positive. In Phase 2,
the ventricle is expanding, being pushed out by the blood, so the work is negative.
(b) Since pressure is expressed in units of force per unit area (newtons/cm2) and volume is expressed in units
of length cubed (cm3 ), area in the pressure-volume plane has units of
newtons
× (cm)3 = newtons × cm.
(cm)2
The units of area in the pressure-volume plane are force times distance, which are the units of work and
energy.
PROJECTS FOR CHAPTER FIVE
497
(c) (i) Let the area of the LV wall be A. The force exerted by the wall in Phase 4 when the LV has volume V
is the pressure, f (V ), times the area A. When the wall moves inward during Phase 4 a small distance
∆x, the volume, which decreases, changes by approximately ∆V = −A∆x. The work done during
this movement is approximately
Force · Distance = f (V )A · ∆x = −f (V ) ∆V.
Ra
Thus, the total work done in Phase 4 as the volume decreases from b to a is b −f (V ) dV =
Rb
f (V ) dV .
a
(ii) The force exerted by the wall in Phase 2 when the LV has volume V is the pressure, g(V ), times
the area of the LV wall, A. When the wall moves outward during Phase 2 a small distance ∆x, the
volume, which increases, changes by approximately ∆V = A∆x. Since the force from the wall to
the blood is in the direction opposite the direction of motion, the work done during this movement is
approximately
−Force · Distance = −g(V )A · ∆x = −g(V ) ∆V.
Rb
Thus, the total work done in Phase 2 as the volume increases from a to b is a −g(V ) dV =
Rb
− a g(V ) dV .
(d) Since Phases 1 and 3 are vertical lines, the area of the pressure-volume loop is the area between f (V ) and
g(V ) from V = a to V = b. Thus,
Z b
(f (V ) − g(V )) dV.
Area enclosed =
a
Since the LV does no work in Phases 1 and 3, using part (c), we have
Z b
Z b
g(V ) dV
Area enclosed =
f (V ) dV −
a
a
= Work done in Phase 4 + Work done in Phase 2
= Work done by LV in one cardiac cycle.
2. To find distances from the velocity graph, we use the fact that if t is the time measured from noon, and v is the
velocity, then
Z T
Area under velocity
Distance traveled
=
v dt =
by car up to time T
0
graph between 0 and T .
The truck’s motion can be represented on the same graph by the horizontal line v = 50, starting at t = 1. The
distance traveled by the truck is then the rectangular area under this line, and the distance between the two
vehicles is the difference between these areas. Note each small rectangle on the graph corresponds to moving
at 10 mph for a half hour (that is, to a distance of 5 miles).
(a) The distance traveled by the car when the truck starts is represented by the shaded area in Figure 5.74,
which totals about seven rectangles or about 35 miles.
v
80
60
vtruck
40
20
vcar
t
1 2 3 4 5 6 7 8 9 10
Figure 5.74: Shaded area represents distance traveled
by car from noon to 1pm
498
Chapter Five /SOLUTIONS
(b) At 3 pm, the car is traveling with a velocity of about 67 mph, while the truck has a velocity of 50 mph.
Because the car is ahead of the truck at 3 pm and is traveling at a greater velocity, the distance between
the car and the truck is increasing at this time. If dcar and dtruck represent the distance traveled by the car
and the truck respectively, then
Distance apart = dcar − dtruck .
The rate of change of the distance apart is given by its derivative:
(Distance apart)′ = (dcar )′ − (dtruck )′
= vcar − vtruck
At 3 pm, we have (Distance apart)′ = 67 mph − 50 mph = 17 mph. Thus, at 3 pm the car is traveling
with a velocity 17 mph greater than the truck’s velocity, and the distance between them is increasing at 17
miles per hour.
At 2 pm, the car’s velocity is greatest. Because the truck’s velocity is constant, vcar − vtruck is largest
when the car’s velocity is largest. Thus, at 2 pm the distance between the car and the truck is increasing
fastest—that is, the car is pulling away at the greatest rate.
(Note: This only takes into account the time when the truck is moving. When the truck is not moving
(from 12:00 to 1:00), the car pulls away from the truck at an even greater rate.)
(c) The car starts ahead of the truck, and the distance between them increases as long as the velocity of the car
is greater than the velocity of the truck. Later, when the truck’s velocity exceeds the car’s, the truck starts to
gain on the car. In other words, the distance between the car and the truck increases as long as vcar > vtruck ,
and it decreases when vcar < vtruck . Therefore, the maximum distance occurs when vcar = vtruck , that is,
when t ≈ 4.3 hours (at about 4:20 pm). (See Figure 5.75.)
The distance traveled by the car is the area under the vcar graph between t = 0 and t = 4.3; the distance
traveled by the truck is the area under the vtruck line between t = 1 (when it started) and t = 4.3. So the
distance between the car and truck is represented by the shaded area in Figure 5.75. Thus, approximately
Distance between car and truck = 35 miles + 50 miles = 85 miles.
(d) The truck overtakes the car when both have traveled the same distance. This occurs when the area under
the curve (vcar ) up to that time equals the area under the line vtruck up to that time. Since the areas under
vcar and vtruck overlap (see Figure 5.76), they are equal when the lightly shaded area equals the heavily
shaded area (which we know is about 85 miles). This happens when t ≈ 8.3 hours, or about 8:20 pm. At
this time, each has traveled about 365 miles.
Car farthest ahead
at this time
v
80
60
✴
v
Truck overtakes
car at this time
80
vtruck
60
✇
vtruck
40
40
20
20
vcar
t
1 2 3 4 5 6 7 8 9 10
Figure 5.75: Shaded area = distance by which car is
ahead at about 4:20 pm
vcar
t
1 2 3 4 5 6 7 8 9 10
Figure 5.76: Truck overtakes car when dark and light
shaded areas are equal
(e) See Figure 5.77.
PROJECTS FOR CHAPTER FIVE
vel
car
499
truck
✠
✠
time
Figure 5.77
(f) The graphs intersect twice, at about 0.7 hours and 4.3 hours. At each intersection point, the velocity of the
car is equal to the velocity of the truck, so vcar = vtruck . From the time they start until 0.7 hours later, the
truck is traveling at a greater velocity than the car, so the truck is ahead of the car and is pulling farther
away. At 0.7 hours they are traveling at the same velocity, and after 0.7 hours the car is traveling faster
than the truck, so that the car begins to gain on the truck. Thus, at 0.7 hours the truck is farther from the
car than it is immediately before or after 0.7 hours.
Similarly, because the car’s velocity is greater than the truck’s after 0.7 hours, it will catch up with the
truck and eventually pass and pull away from the truck until 4.3 hours, at which point the two are again
traveling at the same velocity. After 4.3 hours the truck travels faster than the car, so that it now gains on
the car. Thus, 4.3 hours represents the point where the car is farthest ahead of the truck.
3. (a) For Operation 1, we have the following:
(i) A plot of current use versus time is given in Figure 5.78.
battery current (amperes)
30
20
10
15 30 45 60 75 90
t, time (minutes)
Figure 5.78: Operation 1
The battery current function is given by the formulas
2πt
10 sin 30 0 ≤ t ≤ 15
D(t) =
0
15 ≤ t ≤ 60
30
60 ≤ t ≤ 90.
(ii) The battery current function gives the rate at which the current is flowing. Thus, the total discharge is
given by the integral of the battery current function:
Total discharge =
Z
0
90
D(t) dt =
Z
0
15
2πt
dt +
10 sin
30
Z
60
0 dt +
15
≈ 95.5 + 900
= 995.5 ampere-minutes = 16.6 ampere-hours.
Z
90
60
30 dt
500
Chapter Five /SOLUTIONS
(iii) The battery can discharge up to 40% of 50 ampere-hours, which is 20 ampere-hours, without damage.
Since Operation 1 can be performed with just 16.6 ampere-hours, it is safe.
(b) For Operation 2, we have the following:
(i) The total battery discharge is given by the area under the battery current curve in Figure 5.79. The
area under the right-most portion of the curve, (when the satellite is shadowed by the earth), is easily
calculated as 30 amps· 30 minutes = 900 ampere-minutes = 15 ampere-hours. For the other part
we estimate by trapezoids, which are the average of left and right rectangles on each subinterval.
Estimated values of the function are in Table 5.3.
Table 5.3
Estimated values of the battery current
Time
0
5
10
15
20
25
30
Current
5
16
18
12
5
12
0
Using ∆t = 5, we see
1
1
1
(5 + 16) · 5 + (16 + 18) · 5 + (18 + 12) · 5
2
2
2
1
1
1
+ (12 + 5) · 5 + (5 + 12) · 5 + (12 + 0) · 5
2
2
2
≈ 330 ampere-minutes = 5.5 ampere-hours.
The total estimated discharge is 20.5 ampere-hours.
Total discharge =
battery current (amperes)
30
20
900 amp-min
10
15 30 45 60 75 90
t, time (minutes)
Figure 5.79: Operation 2
(ii) Since the estimated discharge appears to be an underestimate, Operation 2 probably should not be
performed.
6.1 SOLUTIONS
501
CHAPTER SIX
Solutions for Section 6.1
Exercises
1. (a) If f (x) is positive over an interval, then F (x) is increasing over the interval.
(b) If f (x) is increasing over an interval, then F (x) is concave up over the interval.
2. Since dP/dt is positive for t < 3 and negative for t > 3, we know that P is increasing for t < 3 and decreasing for
t > 3. Between each two integer values, the magnitude of the change is equal to the area between the graph dP/dt and
the t-axis. For example, between t = 0 and t = 1, we see that the change in P is 1. Since P = 0 at t = 0, we must have
P = 1 at t = 1. The other values are found similarly, and are shown in Table 6.1.
Table 6.1
t
0
1
2
3
4
5
P
0
1
2
2.5
2
1
3. Since dP/dt is negative for t < 3 and positive for t > 3, we know that P is decreasing for t < 3 and increasing for
t > 3. Between each two integer values, the magnitude of the change is equal to the area between the graph dP/dt and
the t-axis. For example, between t = 0 and t = 1, we see that the change in P is −1. Since P = 2 at t = 0, we must
have P = 1 at t = 1. The other values are found similarly, and are shown in Table 6.2.
Table 6.2
t
1
2
3
4
5
P
1
0
−1/2
0
1
4. See Figure 6.1.
F (0) = 1
F (0) = 0
1
x
1
Figure 6.1
5. See Figure 6.2.
1
x
1
F (0) = 1
F (0) = 0
Figure 6.2
502
Chapter Six /SOLUTIONS
6. See Figure 6.3.
F (0) = 1
1
F (0) = 0
x
1
Figure 6.3
7. See Figure 6.4.
F (0) = 1
F (0) = 0
1
x
1
Figure 6.4
8. See Figure 6.5.
F (0) = 1
1
x
1
F (0) = 0
Figure 6.5
9. See Figure 6.6.
F (0) = 1
1
F (0) = 0
x
1
Figure 6.6
6.1 SOLUTIONS
503
10. See Figure 6.7.
1
F (0) = 1
x
1
F (0) = 0
Figure 6.7
11. See Figure 6.8
1
F (0) = 1
x
1
F (0) = 0
Figure 6.8
Problems
R5
R5
12. (a) If 2 f (x) dx = 4, and F (5) = 10 then F (2) = 6, since 2 f (x) dx gives the total change in F (x) between x = 2
and
x = 5.
R 100
(b) If 0 f (x) dx = 0, then F (100) = F (0), since the total change in F (x) from x = 0 to x = 100 is 0.
13. By the Fundamental Theorem of Calculus, we know that
f (2) − f (0) =
Using a left-hand sum, we estimate
(18)(2) = 36. Averaging, we have
R2
0
2
f ′ (x)dx.
0
f ′ (x)dx ≈ (10)(2) = 20. Using a right-hand sum, we estimate
Z
0
We know f (0) = 100, so
2
f ′ (x)dx ≈
f (2) = f (0) +
Z
Similarly, we estimate
Z
f ′ (x)dx ≈ 100 + 28 = 128.
4
f ′ (x)dx ≈
2
f (4) = f (2) +
Z
Z
4
(18)(2) + (23)(2)
= 41,
2
4
2
Similarly,
20 + 36
= 28.
2
2
0
so
Z
6
f ′ (x)dx ≈
f ′ (x)dx ≈ 128 + 41 = 169.
(23)(2) + (25)(2)
= 48,
2
R2
0
f ′ (x)dx ≈
504
Chapter Six /SOLUTIONS
so
f (6) = f (4) +
Z
4
The values are shown in the table.
x
f (x)
6
f ′ (x)dx ≈ 169 + 48 = 217.
0
2
4
6
100
128
169
217
R2
14. The change in f (x) between 0 and 2 is equal to 0 f ′ (x) dx. A left-hand estimate for this integral is (17)(2) = 34 and a
right hand estimate is (15)(2) = 30. Our best estimate is the average, 32. The change in f (x) between 0 and 2 is +32.
Since f (0) = 50, we have f (2) = 82. We find the other values similarly. The results are shown in Table 6.3.
Table 6.3
x
f (x)
0
2
4
6
50
82
107
119
15. Between t = 0 and t = 1, the particle moves at 10 km/hr for 1 hour. Since it starts at x = 5, the particle is at x = 15
when t = 1. See Figure 6.9. The graph of distance is a straight line between t = 0 and t = 1 because the velocity is
constant then.
Between t = 1 and t = 2, the particle moves 10 km to the left, ending at x = 5. Between t = 2 and t = 3, it moves
10 km to the right again. See Figure 6.9.
x (km)
15
10
5
1
2
3
4
5
6
t (hr)
Figure 6.9
As an aside, note that the original velocity graph is not entirely realistic as it suggests the particle reverses direction
instantaneously at the end of each hour. In practice this means the reversal of direction occurs over a time interval that is
short in comparison to an hour.
16. (a) We know that
can see that
R3
R3
0
0
f ′ (x)dx = f (3) − f (0) from the Fundamental Theorem of Calculus. From the graph of f ′ we
f ′ (x)dx = 2 − 1 = 1 by subtracting areas between f ′ and the x-axis. Since f (0) = 0, we find that
R7
f (3) = 1. Similar reasoning gives f (7) = 0 f ′ (x)dx = 2 − 1 + 2 − 4 + 1 = 0.
(b) We have f (0) = 0, f (2) = 2, f (3) = 1, f (4) = 3, f (6) = −1, and f (7) = 0. So the graph, beginning at x = 0,
starts at zero, increases to 2 at x = 2, decreases to 1 at x = 3, increases to 3 at x = 4, then passes through a zero as
it decreases to −1 at x = 6, and finally increases to 0 at 7. Thus, there are three zeroes: x = 0, x = 5.5, and x = 7.
y
(c)
3
2
1
x
1
−1
2
3
4
5
6 7
6.1 SOLUTIONS
505
17. Let y ′ (t) = dy/dt. Then y is the antiderivative of y ′ such that y(0) = 0. We know that
y(x) =
x
Z
y ′ (t) dt.
0
Thus, y(x) is the area under the graph of dy/dt from t = 0 to t = x, where regions below the t-axis contribute negatively
to the integral. We see that y(t1 ) = 2, y(t3 ) = 2 − 2 = 0, and y(t5 ) = 2. See Figure 6.10.
Since y ′ is positive on the intervals (0, t1 ) and (t3 , ∞), we know that y is increasing on those intervals. Since y ′ is
negative on the interval (t1 , t3 ), we know that y is decreasing on that interval.
Since y ′ is increasing on the interval (t2 , t4 ), we know that y is concave up on that interval; since y ′ is decreasing on
(0, t2 ), we know that y is concave down there. The point where the concavity changes, t2 , is an inflection point. Finally,
since y ′ is constant and positive on the interval (t4 , ∞), the graph of y is linear with positive slope on this interval. The
value y(t1 ) = 2 is a local maximum and y(t3 ) = 0 is a local minimum.
y
2
t1
t2
t3
t4 t5
t
Figure 6.10
18. Let y ′ (t) = dy/dt. Then y is the antiderivative of y ′ such that y(0) = 0. We know that
y(x) =
Z
x
y ′ (t) dt.
0
Thus y(x) is the area under the graph of dy/dt from t = 0 to t = x, with regions below the t-axis contributing negatively
to the integral. We see that y(t1 ) = −2, y(t3 ) = −2 + 2 = 0, and y(t5 ) = −2. See Figure 6.11.
Since y ′ is positive on the interval (t1 , t3 ), we know that y is increasing on that interval. Since y ′ is negative on the
intervals (0, t1 ) and (t3 , ∞), we know y is decreasing on those intervals.
Since y ′ is increasing on (0, t2 ), we know that y is concave up on that interval. Since y ′ is decreasing on (t2 , t4 ),
we know that y is concave down there. The point where concavity changes, t2 , is an inflection point. In addition, since
y ′ is a negative constant on the interval (t4 , ∞), the graph of y is a line with negative slope on this interval. The value
y(t1 ) = −2 is a local minimum and y(t3 ) = 0 is a local maximum.
t1
t2
−2
Figure 6.11
t3
t4
t5
506
Chapter Six /SOLUTIONS
19. The critical points are at (0, 5), (2, 21), (4, 13), and (5, 15). A graph is given in Figure 6.12.
y
(2, 21)
20
G(t)
(5, 15)
15
(4, 13)
10
5
(0, 5)
1
2
3
4
5
t
Figure 6.12
20. Looking at the graph of g ′ in Figure 6.13, we see that the critical points of g occur when x = 15 and x = 40, since
g ′ (x) = 0 at these values. Inflection points of g occur when x = 10 and x = 20, because g ′ (x) has a local maximum or
minimum at these values. Knowing these four key points, we sketch the graph of g(x) in Figure 6.14.
We start at x = 0, where g(0) = 50. Since g ′ is negative on the interval [0, 10], the value of g(x) is decreasing there.
At x = 10 we have
g(10) = g(0) +
Z
10
g ′ (x) dx
0
= 50 − (area of shaded trapezoid T1 )
10 + 20
= 50 −
· 10 = −100.
2
Similarly,
g(15) = g(10) +
Z
15
g ′ (x) dx
10
= −100 − (area of triangle T2 )
1
= −100 − (5)(20) = −150.
2
Continuing,
g(20) = g(15) +
Z
20
Z
40
15
and
g(40) = g(20) +
20
g ′ (x) dx = −150 +
1
(5)(10) = −125,
2
g ′ (x) dx = −125 +
1
(20)(10) = −25.
2
We now find concavity of g(x) in the intervals [0, 10], [10, 15], [15, 20], [20, 40] by checking whether g ′ (x) increases
or decreases in these same intervals. If g ′ (x) increases, then g(x) is concave up; if g ′ (x) decreases, then g(x) is concave
down. Thus we finally have the graph of g(x) in Figure 6.14.
(20, 10)
g ′ (x)
T2
T1
❄15
−10
x
40
(0, 50)
g(x)
(40, −25)
(10, −100)
(10, −20)
Figure 6.13
x
(20, −125)
(15, −150)
Figure 6.14
6.1 SOLUTIONS
507
21. The rate of change is negative for t < 5 and positive for t > 5, so the concentration of adrenaline decreases until t = 5
and then increases. Since the area under the t-axis is greater than the area over the t-axis, the concentration of adrenaline
goes down more than it goes up. Thus, the concentration at t = 8 is less than the concentration at t = 0. See Figure 6.15.
adrenaline
concentration (µg/ml)
2
4
6
8
t (minutes)
Figure 6.15
22. (a) The total volume emptied must increase with time and cannot decrease. The smooth graph (I) that is always increasing
is therefore the volume emptied from the bladder. The jagged graph (II) that increases then decreases to zero is the
flow rate.
(b) The total change in volume is the integral of the flow rate. Thus, the graph giving total change (I) shows an antiderivative of the rate of change in graph (II).
23. See Figure 6.16. Note that since f (x1 ) = 0 and f ′ (x1 ) < 0, F (x1 ) is a local maximum; since f (x3 ) = 0 and f ′ (x3 ) > 0,
F (x3 ) is a local minimum. Also, since f ′ (x2 ) = 0 and f changes from decreasing to increasing about x = x2 , F has an
inflection point at x = x2 .
F (x)
x1
x2
x3
x
Figure 6.16
24. See Figure 6.17. Note that since f (x1 ) = 0 and f ′ (x1 ) > 0, F (x1 ) is a local minimum; since f (x3 ) = 0 and f ′ (x3 ) < 0,
F (x3 ) is a local maximum. Also, since f ′ (x2 ) = 0 and f changes from decreasing to increasing about x = x2 , F has an
inflection point at x = x2 .
F (x)
x1
x2
Figure 6.17
x3
x
508
Chapter Six /SOLUTIONS
25. See Figure 6.18. Note that since f (x1 ) = 0, F (x1 ) is either a local minimum or a point of inflection; it is impossible to
tell which from the graph. Since f ′ (x3 ) = 0, and f ′ changes sign around x = x3 , F (x3 ) is an inflection point. Also,
since f ′ (x2 ) = 0 and f changes from increasing to decreasing about x = x2 , F has another inflection point at x = x2 .
F (x)
x1
x2
x
x3
Figure 6.18
26. See Figure 6.19. Since f (x1 ) = 0 and f ′ (x1 ) < 0, we see that F (x1 ) is a local maximum. Since f (x3 ) = 0 and
f ′ (x3 ) > 0, we see that F (x3 ) is a local minimum. Since f ′ (x2 ) = 0 and f changes from decreasing to increasing at
x = x2 , we see that F has an inflection point at x = x2 .
x2
x1
x
x3
F (x)
Figure 6.19
27. The graph of f (x) = 2 sin(x2 ) is shown in Figure 6.20. We see that there are roots at x = 1.77 and x = 2.51. These are
the critical points of F (x). Looking at the graph, it appears that of the three areas marked, A1 is the largest, A2 is next,
and A3 is smallest. Thus, as x increases from 0 to 3, the function F (x) increases (by A1 ), decreases (by A2 ), and then
increases again (by A3 ). Therefore, the maximum is attained at the critical point x = 1.77.
What is the value of the function at this maximum? We know that F (1) = 5, so we need to find the change in F
between x = 1 and x = 1.77. We have
Change in F =
Z
1.77
2 sin(x2 ) dx = 1.17.
1
We see that F (1.77) = 5 + 1.17 = 6.17, so the maximum value of F on this interval is 6.17.
2
A1
1
A3
2
A2
−2
Figure 6.20
x
3
6.1 SOLUTIONS
509
28. Both F (x) and G(x) have roots at x = 0 and x = 4. Both have a critical point (which is a local maximum) at x = 2.
However, since the area under g(x) between x = 0 and x = 2 is larger than the area under f (x) between x = 0 and
x = 2, the y-coordinate of G(x) at 2 will be larger than the y-coordinate of F (x) at 2. See below.
G(x)
F (x)
x
1
2
3
4
29. (a) Let r(t) be the leakage rate in liters per second at time t minutes, shown in the graph. Since time is in minutes, it is
helpful to express leakage in units of liters per minute. The leakage rate in liters per minute is 60r(t). The quantity
Rb
leaked during the first b minutes, in liters, is given by 0 60r(t) dt.
We can evaluate the quantity leaked by computing area under the rate graph, by counting grid squares. Each grid
square contributes area
1 grid square = 10
liter
sec
× (10 minutes) = 60 · 10
liters
× (10 minutes) = 6000 liters.
minute
Thus each grid square represents 6000 liters of leakage. So
Total spill over
=
first 10 minutes
Total spill over
20 minutes
1
grid square · 6000 = 3000 liters.
2
= 2 grid squares · 6000 = 12,000 liters.
Continuing, we have
Time t (minutes)
0
10
20
30
40
50
Total spill, over t minutes (liters)
0
3000
12,000
21,000
27,000
30,000
(b) Plotting the values from part (a) and connecting with a smooth curve gives Figure 6.21.
y
30,000
20,000
10,000
x
10
20
30
40
50
Figure 6.21
30. (a) Suppose Q(t) is the amount of water in the reservoir at time t. Then
Q′ (t) =
Rate at which water
in reservoir is changing
=
Inflow
rate
−
Outflow
rate
Thus the amount of water in the reservoir is increasing when the inflow curve is above the outflow, and decreasing
when it is below. This means that Q(t) is a maximum where the curves cross in July 2007 (as shown in Figure 6.22),
and Q(t) is decreasing fastest when the outflow is farthest above the inflow curve, which occurs about October 2007
(see Figure 6.22).
510
Chapter Six /SOLUTIONS
To estimate values of Q(t), we use the Fundamental Theorem which says that the change in the total quantity
of water in the reservoir is given by
Q(t) − Q(Jan 2007) =
Q(t) = Q(Jan 2007) +
or
Z
Z
t
Jan 07
t
Jan 07
(inflow rate − outflow rate) dt
(Inflow rate − Outflow rate) dt.
rate of flow
(millions of gallons/day)
Q(t) is max
Q(t) is min
❄
❄
Outflow
Inflow
April
Jan (07)
July
Oct
Jan(08)
Q(t)
millions of gallons
Q(t) is increasing
most rapidly
❘
Q(t) is decreasing most rapidly
✠
April
Jan (07)
July
Oct
Jan(08)
Figure 6.22
(b) See Figure 6.22. Maximum in July 2007. Minimum in Jan 2008.
(c) See Figure 6.22. Increasing fastest in May 2007. Decreasing fastest in Oct 2007.
(d) In order for the water to be the same as Jan 2007 the total amount of water which has flowed into the reservoir minus
the total amount of water which has flowed out of the reservoir must be 0. Referring to Figure 6.23, we have
Z
July 08
Jan 07
(Inflow − Outflow) dt = −A1 + A2 − A3 + A4 = 0
giving A1 + A3 = A2 + A4
rate of flow
(millions of gallons/day)
A2
A3
A4
Outflow
A1
Jan (‘07)
Inflow
April
July
Oct
Figure 6.23
Jan (‘08)
April
July
6.1 SOLUTIONS
511
Strengthen Your Understanding
31. The statement has f (x) and F (x) reversed. Namely if an antiderivative F (x) is increasing, then F ′ (x) = f (x) ≥ 0.
We can see the statement given is not always true by looking for a counterexample. The function f (x) = 2x is
always increasing, but it has antiderivatives that are less than 0. For example, the antiderivative F (x) with F (0) = −1 is
negative at 0. See Figure 6.24.
A correct statement is: If f (x) > 0 everywhere, then F (x) is increasing everywhere.
f (x) = 2x
F (x)
x
−1
x
Figure 6.24: f (x) = 2x increasing and its antiderivative F (x) is not always positive
32. Consider F (x) = x2 and G(x) = x2 + 1 which are both antiderivatives of f (x) = 2x. Then H(x) = F (x) + G(x) =
2x2 + 1, and H ′ (x) = 4x 6= f (x). Thus, H(x) = F (x) + G(x) is not an antiderivative of f (x).
33. Any function for which the area between the function’s graph above the x-axis is equal to the area between the function’s
graph below the x-axis will work. One such function, f (x) = 1 − x, is shown in Figure 6.25.
y
1
1
2
x
−1
Figure 6.25: f (x) = 1 − x
34. Any positive function will work. One such function is f (x) = 1; see Figure 6.26.
y
1
x
Figure 6.26
35. True. A function can have only one derivative.
36. False. If f (t) is an increasing function, then F ′ is increasing, so F changes at an increasing rate. Thus F is not linear.
512
Chapter Six /SOLUTIONS
Solutions for Section 6.2
Exercises
1. 5x
2.
3.
4.
5.
5 2
t
2
1 3
x
3
1 3
t + 21 t2
3
3
2 2
z
3
6. ln |z|
1
7. −
t
8. sin t
1
d
1
d
=
z −3 = − 2
9.
3
dz z
dz
2z
y5
10.
+ ln |y|
5
z
11. e
12. − cos t
13.
2 3
t
3
4
+ 43 t4 + 54 t5
t3
t2
t
−
−
4
6
2
t2 + 1
1
t2
15.
= t + , which has antiderivative
+ ln |t|
t
t
2
5
2 3
16. x2 − x 2
2
3Z
14.
6t dt = 3t2 + C
17. F (t) =
18. H(x) =
Z
19. F (x) =
20. G(t) =
21. R(t) =
Z
Z
Z
22. F (z) =
Z
23. G(x) =
24. H(x) =
25. P (t) =
26. P (t) =
27. G(x) =
Z
Z
Z
Z
Z
(x3 − x) dx =
x2
x4
−
+C
4
2
(x2 − 4x + 7) dx =
√
t dt =
x3
− 2x2 + 7x + C
3
2 3/2
t
+C
3
(t3 + 5t − 1) dt =
(z + ez ) dz =
t4
5
+ t2 − t + C
4
2
z2
+ ez + C
2
(sin x + cos x) dx = − cos x + sin x + C
(4x3 − 7) dx = x4 − 7x + C
1
√ dt = 2t1/2 + C
t
(2 + sin t) dt = 2t − cos t + C
5
5
dx = − 2 + C
x3
2x
6.2 SOLUTIONS
513
Z
7
dt = 7 tan t + C
cos2 t
29. f (x) = 3, so F (x) = 3x + C. F (0) = 0 implies that 3 · 0 + C = 0, so C = 0. Thus F (x) = 3x is the only possibility.
28. H(t) =
30. f (x) = 2x, so F (x) = x2 + C. F (0) = 0 implies that 02 + C = 0, so C = 0. Thus F (x) = x2 is the only possibility.
−7x2
2
31. f (x) = −7x, so F (x) =
only possibility.
+ C. F (0) = 0 implies that − 27 · 02 + C = 0, so C = 0. Thus F (x) = −7x2 /2 is the
32. f (x) = 2 + 4x + 5x2 , so F (x) = 2x + 2x2 + 53 x3 + C. F (0) = 0 implies that C = 0. Thus F (x) = 2x + 2x2 + 35 x3
is the only possibility.
33. f (x) = 14 x, so F (x) =
possibility.
34. f (x) = x2 , so F (x) =
x2
8
1
8
+ C. F (0) = 0 implies that
· 02 + C = 0, so C = 0. Thus F (x) = x2 /8 is the only
x3
03
x3
+ C. F (0) = 0 implies that
+ C = 0, so C = 0. Thus F (x) =
is the only possibility.
3
3
3
35. f (x) = x1/2 , so F (x) =
only possibility.
2 3/2
x
3
+ C. F (0) = 0 implies that
2
3
· 03/2 + C = 0, so C = 0. Thus F (x) =
2 3/2
x
3
is the
36. f (x) = sin x, so F (x) = − cos x + C. F (0) = 0 implies that − cos 0 + C = 0, so C = 1. Thus F (x) = − cos x + 1
is the only possibility.
37.
38.
39.
40.
41.
5
Z2
Z
Z
Z
x2 + 7x + C
(4t +
1
) dt = 2t2 + ln |t| + C
t
(2 + cos t) dt = 2t + sin t + C
7ex dx = 7ex + C
(3ex + 2 sin x) dx = 3ex − 2 cos x + C
42. We have:
Z
(4ex − 3 sin x) dx =
Z
=4
4ex dx −
Z
Z
ex dx − 3
3 sin x dx
Z
sin x dx
= 4ex − 3(− cos x) + C
= 4ex + 3 cos x + C.
43. We have
Z
√
5x2 + 2 x dx =
Z
=5
5x2 dx +
Z
Z
x2 dx + 2
√
2 x dx
Z
x1/2 dx
2 3/2
1 3
x +2
·x
=5
+C
3
3
5 3 4 3/2
= ·x + ·x
+ C.
3
3
44.
Z
Z
(x + 3)2 dx =
Z
(x2 + 6x + 9) dx =
8
√ dx = 16x1/2 + C
x
2
46. 3 ln |t| + + C
t
47. ex + 5x + C
45.
x3
+ 3x2 + 9x + C
3
514
Chapter Six /SOLUTIONS
48. Expand the integrand and then integrate
Z
3
2
t (t + 1) dt =
Z
(t5 + t3 ) dt =
1 6 1 4
t + t + C.
6
4
2 5/2
x
5
− 2 ln |x| + C
x+1
1
50. Since f (x) =
= 1 + , the indefinite integral is x + ln |x| + C
x
x
49.
51.
Z
3
(x2 + 4x + 3) dx =
0
52.
Z
3
1
53.
Z
1
dt = ln |t|
t
54.
1
π/4
sin x dx = − cos x
55.
2ex dx = 2ex
0
2
2
3ex dx = 3ex
0
56.
Z
57.
Z
0
= (9 + 18 + 9) − 0 = 36
0
√
2
π
+ 1 = 0.293.
= − cos − (− cos 0) = −
4
2
= 3e2 − 3e0 = 3e2 − 3 = 19.167.
(x3 − πx2 ) dx =
x4
πx3
−
4
3
1
1
sin θ dθ = − cos θ
0
0
= 2e − 2 ≈ 3.437.
5
2
3
1
1
0
Z
= ln |3| − ln |1| = ln 3 ≈ 1.0986.
π/4
0
Z
3
x3
+ 2x2 + 3x
3
0
5
=
2
609
− 39π ≈ 29.728.
4
= 1 − cos 1 ≈ 0.460.
1
1 + y2
= + y,
y
y
1 + y2
y2
dy = ln |y| +
y
2
58. Since
Z
2
1
Z
2
3
4
2
= ln 2 +
1
2
4
+ 4 = 16/3 ≈ 5.333.
3
0
0
√
Z π/4
√
π/4
2
2
− (−1 + 0) = 1.
+
60.
(sin t + cos t) dt = (− cos t + sin t)
= −
2
2
0
0
59.
x
+ 2x
3
dx =
x
+ x2
12
3
≈ 2.193.
2
=
Problems
61. The rate at which water is entering the tank (in volume per unit time) is
dV
= 120 − 6t ft3 /min.
dt
Thus, the total quantity of water in the tank at time t = 4, in ft3 , is
V =
Z
0
Since an antiderivative to 120 − 6t is
4
(120 − 6t) dt.
120t − 3t2 ,
we have
V =
Z
0
4
4
2
(120 − 6t) dt = (120t − 3t )
0
= (120 · 4 − 3 · 42 ) − (120 · 0 − 3 · 02 )
= 432 ft3 .
6.2 SOLUTIONS
515
The radius is 5 feet, so if the height is h ft, the volume is V = π52 h = 25πh. Thus, at time t = 4, we have V = 432, so
432 = 25πh
432
= 5.500 ft.
h=
25π
62. (a) The formula v = 6 − 2t implies that v > 0 (the car is moving forward) if 0 ≤ t < 3 and that v < 0 (the car is
moving backward) if t > 3. When t = 3, v = 0, so the car is not moving at the instant t = 3. The car is decelerating
when |v| is decreasing; since v decreases (from 6 to 0) on the interval 0 ≤ t < 3, the car decelerates on that interval.
The car accelerates when |v| is increasing, which occurs on the domain t > 3.
(b) The car moves forward on the interval 0 ≤ t < 3, so it is furthest to the right at t = 3. For all t > 3, the car is
decelerating. There is no upper bound on the car’s distance behind its starting point since it is decelerating for all
t > 3.
(c) Let s(t) be the position of the car at time t. Then
v(t) = s′ (t),
so s(t) is an antiderivative of v(t). Thus,
s(t) =
Z
v(t) dt =
Z
(6 − 2t) dt = 6t − t2 + C.
Since the car’s position is measured from its starting point, we have s(0) = 0, so C = 0. Thus, s(t) = 6t − t2 .
63. (a) Since the rotor is slowing down at a constant rate,
Angular acceleration =
260 − 350
= −60 revs/min2 .
1.5
Units are revolutions per minute per minute, or revs/min2 .
(b) To decrease from 350 to 0 revs/min at a deceleration of 60 revs/min2 ,
Time needed =
350
≈ 5.83 min.
60
(c) We know angular acceleration is the derivative of angular velocity. Since
Angular acceleration = −60 revs/min2 ,
we have
Angular velocity = −60t + C.
Measuring time from the moment when angular velocity is 350 revs/min, we get C = 350. Thus
Angular velocity = −60t + 350.
So, the total number of revolutions made between the time the angular speed is 350 revs/min and stopping is given
by:
Number of revolutions =
=
Z
Z
5.83
(Angular velocity) dt
0
5.83
5.83
0
(−60t + 350)dt = −30t2 + 350t
0
= 1020.83 revolutions.
64. Since C ′ (x) = 4000 + 10x we want to evaluate the indefinite integral
Z
(4000 + 10x) dx = 4000x + 5x2 + K
where K is a constant. Thus C(x) = 5x2 + 4000x + K, and the fixed cost of 1,000,000 riyal means that C(0) =
1,000,000 = K. Therefore, the total cost is
C(x) = 5x2 + 4000x + 1,000,000.
Since C(x) depends on x2 , the square of the depth drilled, costs will increase dramatically when x grows large.
516
65.
Chapter Six /SOLUTIONS
Z
0
3
x2 dx =
x3
3
3
0
= 9 − 0 = 9.
66. Since y = x3 − x = x(x − 1)(x + 1), the graph crosses the axis at the three points shown in Figure 6.27. The two regions
have the same area (by symmetry). Since the graph is below the axis for 0 < x < 1, we have
1
Z
Area = 2 −
x3 − x dx
0
x2
x4
= −2
−
4
2
y
1
0
1
1
−
4
2
= −2
=
1
.
2
y = x3 − x
x
−1
1
Figure 6.27
67. The area we want (the shaded area in Figure 6.28) is symmetric about the y-axis and so is given by
Area = 2
π/3
Z
cos x −
0
=2
Z
π/3
1
2
cos x dx −
0
π/3
= 2 sin x
√
0
Z
3
x
π
π/3
2
dx
9 2
x dx
π2
0
π/3
9 x3
− 2 ·
π
3
0
√
3 π3
3
π
− 2 · 3 = 3− .
= 2·
2
π
3
9
y=
− π3
1
2
2
3
x
π
x
π
3
y = cos x
Figure 6.28
68. Since y < 0 from x = 0 to x = 1 and y > 0 from x = 1 to x = 3, we have
Area = −
Z
1
0
3x2 − 3 dx +
= − x3 − 3x
1
0
Z
3
3x2 − 3 dx
1
+ x3 − 3x
3
1
= −(−2 − 0) + (18 − (−2)) = 2 + 20 = 22.
6.2 SOLUTIONS
517
69. (a) See Figure 6.29. Since f (x) > 0 for 0 < x < 2 and f (x) < 0 for 2 < x < 5, we have
Area =
=
=
Z
Z
2
f (x) dx −
0
2
0
Z
5
f (x) dx
2
Z
(x3 − 7x2 + 10x) dx −
7x3
x4
−
+ 5x2
4
3
2
0
2
−
5
(x3 − 7x2 + 10x) dx
x4
7x3
−
+ 5x2
4
3
56
+ 20 − (0 − 0 + 0) −
=
4−
3
253
=
.
12
h
i
5
2
875
56
−
+ 125 − 4 −
+ 20
4
3
3
h 625
5
2
i
x
Figure 6.29: Graph of f (x) = x3 − 7x2 + 10x
(b) Calculating
R5
0
f (x) dx gives
Z
5
f (x) dx =
0
=
Z
5
0
(x3 − 7x2 + 10x) dx
7x3
x4
−
+ 5x2
4
3
5
0
625
875
−
+ 125 − (0 − 0 + 0)
=
4
3
125
=−
.
12
This integral measures the difference between the area above the x-axis and the area below the x-axis. Since the
definite integral is negative, the graph of f (x) lies more below the x-axis than above it. Since the function crosses
the axis at x = 2,
Z 2
Z 5
Z 5
16
63
−125
f (x) dx +
f (x) dx =
f (x) dx =
−
=
,
3
4
12
0
2
0
whereas
Z 5
Z 2
16
64
253
f (x) dx =
+
=
.
Area =
f (x) dx −
3
4
12
2
0
70. The graph of y = ex − 2 is below the x-axis at x = 0 and above the x-axis at x = 2. The graph crosses the axis where
ex − 2 = 0
x = ln 2.
See Figure 6.30. Thus we find the area by dividing the region at x = ln 2:
Area = −
Z
ln 2
0
(ex − 2) dx +
ln 2
x
Z
2
ln 2
(ex − 2) dx
2
x
= (−e + 2x)
+ (e − 2x)
0
ln 2
= −eln 2 + 2 ln 2 + e0 + e2 − 4 − (eln 2 − 2 ln 2)
= −2eln 2 + 4 ln 2 − 3 + e2
= −2 · 2 + 4 ln 2 − 3 + e2 = e2 + 4 ln 2 − 7.
518
Chapter Six /SOLUTIONS
y
y = ex − 2
x
2
ln 2
Figure 6.30
71. Solving x2 = 2 − x2 shows that the curves intersect at x = ±1. To find the area between the curves for −1 ≤ x ≤ 1 we
integrate the top curve y = 2 − x2 minus the bottom curve y = x2 . Thus
Area between curves =
Z
1
−1
(2 − x2 − x2 ) dx = 2x −
1
2 3
x
3
−1
=2−
2
2
− −2 +
3
3
=
8
.
3
72. The function is a cubic polynomial which crosses the x-axis at x = 1, 2, 3, with f (x) ≥ 0 for 1 ≤ x ≤ 2 and f (x) ≤ 0
for 2 ≤ x ≤ 3. Thus the total area is given by
Total area =
Z
2
f (x) dx +
Z
3
f (x) dx .
2
1
Since f (x) = x3 − 6x2 + 11x − 6, we can evaluate each integral to get
Total area =
11 2
x4
− 2x3 +
x − 6x
4
2
2
+
1
11 2
x4
− 2x3 +
x − 6x
4
2
3
=
2
1
1
1
+ − = .
4
4
2
73. The graph of y = ex − 2 starts below the x-axis at x = 0 and climbs. See Figure 6.31. The graph crosses the axis where
ex = 2
x = ln 2.
Since the integral
Z
we choose c to make the integral 0:
Z
0
c
0
(ex − 2) dx = −A1 + A2 ,
c
c
(ex − 2) dx = (ex − 2x)
= ec − 2c − 1 = 0
0
ec − 2c = 1.
Solving numerically (for example, by zooming in on the graph on a calculator), we get
c=0
and
c = 1.257.
The solution we want is c = 1.257.
y
y = ex − 2
A2
A1
ln 2
Figure 6.31
c
x
6.2 SOLUTIONS
519
74. Since the area under the curve is 6, we have
Z
Thus b
1/2
b
1
1
√ dx = 2x1/2
x
b
= 2b1/2 − 2(1) = 6.
1
= 4 and b = 16.
75. The graph of y = c(1 − x2 ) has x-intercepts of x = ±1. See Figure 6.32. Since it is symmetric about the y-axis, we have
Area =
1
Z
−1
1
4c
= 1,
3
giving
= 2c
We want the area to be 1, so
c(1 − x2 ) dx = 2c
x3
x−
3
Z
0
1
(1 − x2 ) dx
4c
.
3
=
0
c=
3
.
4
y
c
−1
x
1
y = c(1 − x2 )
Figure 6.32
76. The curves intersect at (0, 0) and (π, 0). At any x-coordinate the “height” between the two curves is sin x − x(x − π).
See Figure 6.33.
Height =
y
sin x − x(x − π)
A
✠
x
π
π
2
B
Figure 6.33
Thus the total area is
Z
0
π
[sin x − x(x − π)] dx = =
=
=
π
Z
0
(sin x − x2 + πx) dx
− cos x −
πx2
x3
+
3
2
π3
π3
1−
+
3
2
= 2+
π3
.
6
− (−1)
π
0
520
Chapter Six /SOLUTIONS
Another approach is to notice that the area between the two curves is (area A) + (area B).
Area B = −
=−
Area A =
Z
Z
π
x(x − π) dx since the function is negative on 0 ≤ x ≤ π
0
x3
πx2
−
3
2
π
0
π
π3
π3
π3
−
=
;
2
3
6
=
0
π
sin x dx = − cos x
= 2.
0
π3
.
6
77. See Figure 6.34. The average value of f (x) is given by
Thus the area is 2 +
1
Average =
9−0
Z
9
√
0
2 3/2
x
3
1
x dx =
9
f (x) =
√
9
0
!
=
1
9
2 3/2
1
9
− 0 = 18 = 2.
3
9
x
Average value
2
9
Figure 6.34
78. We have
f x−1 = 4 x−1
so
Z
3
f x
1
−1
= 4x
dx =
Z
3
3
−3
4x3 dx
1
=4
because f (x) = 4x−3
1
x4
4
3
1
= 34 − 14 = 80.
79. By the Fundamental theorem,
Z
1
3
3
f ′ (x) dx = f (3) − f (1) = 4x−3
1
= 4 · 3−3 − 4 · 1−3 = −3.8519.
80. The curves y = x and y = xn cross at x = 0 and x = 1. For 0 < x < 1, the curve y = x is above y = xn . Thus the
area is given by
2
1
Z 1
x
xn+1
1
1
1
−
= −
→ .
An =
(x − xn )d x =
2
n
+
1
2
n
+
1
2
0
0
Since xn → 0 for 0 ≤ x < 1, as n → ∞, the area between the curves approaches the area under the line y = x between
x = 0 and x = 1.
81. (a) Recall that x = eln x . Thus xx = (eln x )x = ex ln x .
(b) Using part (a) and the chain rule
d x ln x
d
d x
(x ) =
(e
) = ex ln x (x ln x) = ex ln x (ln x + 1) = xx (ln x + 1).
dx
dx
dx
(c) By the Fundamental Theorem of Calculus and part (b),
Z
xx (1 + ln x) dx = xx + C.
6.2 SOLUTIONS
521
(d) By the Fundamental Theorem of Calculus,
Z
2
2
xx (1 + ln x)dx = xx
1
Using a calculator, we get
R2
1
= 22 − 11 = 3.
1
xx (1 + ln x) dx = 3.
82. (a) The average value of f (t) = sin t over 0 ≤ t ≤ 2π is given by the formula
1
Average =
2π − 0
=
Z
2π
sin t dt
0
2π
1
(− cos t)
2π
0
1
=
(− cos 2π − (− cos 0)) = 0.
2π
We can check this answer by looking at the graph of sin t in Figure 6.35. The area below the curve and above the
t-axis over the interval 0 ≤ t ≤ π, A1 , is the same as the area above the curve but below the t-axis over the interval
π ≤ t ≤ 2π, A2 . When we take the integral of sin t over the entire interval 0 ≤ t ≤ 2π, we get A1 − A2 = 0.
A1
t
2π
π
A2
Figure 6.35
(b) Since
Z
π
π
0
sin t dt = − cos t
0
= − cos π − (− cos 0) = −(−1) − (−1) = 2,
the average value of sin t on 0 ≤ t ≤ π is given by
Average value =
1
π
Z
π
sin t dt =
0
2
.
π
83. The area beneath the curve in Figure 6.36 is given by
Z
a
y dx =
0
Z
0
a
√
√
4
( a − x)2 dx = a x −
2
√
2
x2
a x3/2
+
3
2
a
0
=
a2
.
6
The area of the square is a so the area above the curve is 5a /6. Thus, the ratio of the areas is 5 to 1.
y
a
0
a
Figure
The √
curve
√ 6.36:
√
x+ y = a
x
522
Chapter Six /SOLUTIONS
Strengthen Your Understanding
84. We cannot integrate the numerator and denominator separately. Using the quotient rule to differentiate we see that
d
dx
x3 + x
+C
x2
=
(3x2 + 1)(x2 ) − (x3 + x)(2x)
,
x4
3x2 + 1
which is not equal to
.
2x
To compute the correct integral, we can split the integrand into two terms:
Z
3x2 + 1
dx =
2x
85. The statement is true for all n 6= −1, since
R
Z
3
1
x+
2
2x
dx =
3 2 1
x + ln |x| + C.
4
2
x−1 dx = ln |x| + C.
86. If two functions F (x) and G(x) differ by a constant, they have the same derivative. Therefore, F (x) = x4 and G(x) =
x4 − 8 both have the same derivative.
87. We are given F (x) = mx + C with m a negative constant. Since F ′ (x) = f (x) = m, any negative constant function
will work, for example f (x) = −1.
√
d
(2(x + 1)3/2 ) = 2 · 23 (x + 1)1/2 = 3 x + 1.
88. True. Check by differentiating dx
89. True. Any antiderivative of 3x2 is obtained by adding a constant to x3 .
90. True. Any antiderivative of 1/x is obtained by adding a constant to ln |x|.
91. False. Differentiating using the product and chain rules gives
d
dx
R
−1 −x2
e
2x
=
2
1 −x2
e
+ e−x .
2x2
R
92. False. It is not true in general
R that xf (x) dx = x f (x)Rdx, so this statement is false for many functions f (x). For
example, if f (x) = 1, then xf (x) dx = x2 /2 + C, but x f (x) dx = x(x + C).
93. True. Adding a constant to an antiderivative gives another antiderivative.
94. True. Since F (x) and G(x) are antiderivatives of the same function on an interval, F (x) − G(x) is a constant function.
Thus F (10) − G(10) = F (5) − G(5) > 0.
95. False. For a counterexample, take f (x) = g(x) = 1. Then F (x) = x and G(x) = x are antiderivatives of f (x) and
g(x), but F (x) · G(x) = x2 is not an antiderivative of f (x) · g(x) = 1.
96. True. The derivative of F (x) − G(x) is (F (x) − G(x))′ = f (x) − f (x) = 0, so F (x) − G(x) is a constant function.
Solutions for Section 6.3
Exercises
1. We differentiate y = xe−x + 2 using the product rule to obtain
dy
= x e−x (−1) + (1)e−x + 0
dx
= −xe−x + e−x
= (1 − x)e−x ,
and so y = xe−x + 2 satisfies the differential equation. We now check that y(0) = 2:
y = xe−x + 2
y(0) = 0e0 + 2 = 2.
6.3 SOLUTIONS
523
2. First, we check that y = sin(2t) satisfies the initial condition y(0) = 0:
sin(2 · 0) = 0.
Next, we substitute y = sin(2t) into each side of the differential equation and check that we get the same result. The
left-hand side gives
dy
= 2 cos(2t).
dt
For 0 ≤ t ≤ π/4, the right-hand side gives
p
p
2
1 − sin2 (2t) = 2 cos(2t).
1 − y2 = 2
Thus for 0 ≤ t ≤ π/4 the initial value problem is satisfied.
3. We need to find a function whose derivative is 2x, so one antiderivative is x2 . The general solution is
y = x2 + C.
To check, we differentiate to get
as required.
d
dy
x2 + C = 2x,
=
dx
dx
4. We need to find a function whose derivative is t2 , so one antiderivative is t3 /3. The general solution is
y=
t3
+ C.
3
To check, we differentiate to get
dy
d
=
dt
dt
as required.
t3
+C
3
= t2 ,
5. We need to find a function whose derivative is x3 + 5x4 , so one antiderivative is x4 /4 + x5 . The general solution is
y=
x4
+ x5 + C.
4
To check, we differentiate to get
dy
d
=
dx
dx
as required.
x4
+ x5 + C
4
= x3 + 5x4 ,
6. We need to find a function whose derivative is et , so one antiderivative is et . The general solution is
y = et + C.
To check, we differentiate to get
as required.
dy
d t
=
e + C = et ,
dt
dt
7. We need to find a function whose derivative is cos x, so one antiderivative is sin x. The general solution is
y = sin x + C.
To check, we differentiate to get
d
dy
=
(sin x + C) = cos x,
dx
dx
as required.
8. We need to find a function whose derivative is 1/x, so one antiderivative is ln x. The general solution is
y = ln x + C.
To check, we differentiate to get
d
1
dy
=
(ln x + C) = ,
dx
dx
x
as required.
524
Chapter Six /SOLUTIONS
9. Integrating gives
Z
dy
dx =
dx
Z
(3x2 ) dx = x3 + C.
If y = 5 when x = 0, then 03 + C = 5 so C = 5. Thus y = x3 + 5.
10. Integrating gives
Z
dy
dx =
dx
Z
(x5 + x6 ) dx =
x6
x7
+
+ C.
6
7
If y = 2 when x = 1, then 16 /6 + 17 /7 + C = 2 so
C =2−
Thus
y=
1
71
1
− =
.
6
7
42
x6
x7
71
+
+
.
6
7
42
11. Integrating gives
Z
dy
dx =
dx
Z
(ex ) dx = ex + C.
If y = 7 when x = 0, then e0 + C = 7 so C = 6. Thus y = ex + 6.
12. Integrating gives
Z
dy
dx =
dx
Z
(sin x) dx = − cos x + C.
If y = 3 when x = 0, then − cos 0 + C = 3 so C = 4. Thus y = − cos x + 4.
Problems
13. The acceleration is a(t) = −32, so the velocity is v(t) = −32t + C. We find C using v(0) = −10, (negative velocity is
downward) so
v(t) = −32t − 10.
Then, the height of the rock above the water is
s(t) = −16t2 − 10t + D.
We find D using s(0) = 100, so
s(t) = −16t2 − 10t + 100.
Now we can find when the rock hits the water by solving the quadratic equation
0 = −16t2 − 10t + 100.
There are two solutions: t = 2.207 seconds and t = −2.832 seconds. We discard the negative solution. So, at t = 2.207
the rock is traveling with a velocity of
v(2.207) = −32(2.207) − 10 = −80.624 ft/sec.
Thus the rock is traveling with a speed of 80.624 ft/sec downward when it hits the water.
14. (a) To find the height of the balloon, we integrate its velocity with respect to time:
h(t) =
=
Z
Z
v(t) dt
(−32t + 40) dt
= −32
t2
+ 40t + C.
2
Since at t = 0, we have h = 30, we can solve for C to get C = 30, giving us a height of
h(t) = −16t2 + 40t + 30.
6.3 SOLUTIONS
525
(b) To find the average velocity between t = 1.5 and t = 3, we find the total displacement and divide by time.
Average velocity =
h(3) − h(1.5)
6 − 54
=
= −32 ft/sec.
3 − 1.5
1.5
The balloon’s average velocity is 32 ft/sec downward.
(c) First, we must find the time when h(t) = 6. Solving the equation −16t2 + 40t + 30 = 6, we get
6 = −16t2 + 40t + 30
0 = −16t2 + 40t + 24
0 = 2t2 − 5t − 3
0 = (2t + 1)(t − 3).
Thus, t = −1/2 or t = 3. Since t = −1/2 makes no physical sense, we use t = 3 to calculate the balloon’s
velocity. At t = 3, we have a velocity of v(3) = −32(3) + 40 = −56 ft/sec. So the balloon’s velocity is 56 ft/sec
downward at the time of impact.
15. Since the acceleration a = dv/dt, where v is the velocity of the car, we have
dv
= −0.6t + 4.
dt
Integrating gives
t2
+ 4t + C.
2
The car starts from rest, so v = 0 when t = 0, and therefore C = 0. If x is the distance from the starting point, v = dx/dt
and
dx
= −0.3t2 + 4t,
dt
so
0.3 3 4 2
x=−
t + t + C = −0.1t3 + 2t2 + C.
3
2
Since x = 0 when t = 0, we have C = 0, so
x = −0.1t3 + 2t2 .
v = −0.6
We want to solve for t when x = 100:
100 = −0.1t3 + 2t2 .
This equation can be rewritten as
0.1t3 − 2t2 + 100 = 0
t3 − 20t2 + 1000 = 0.
The equation can be solved numerically, or by tracing along a graph, or by factoring
The solutions are t = 10 and t =
t = 10 sec.
10±
√
(t − 10)(t2 − 10t − 100) = 0.
500
2
= −6.18, 16.18. Since we are told 0 ≤ t ≤ 12, the solution we want is
16.
√
dy
= k t = kt1/2
dt
2
y = kt3/2 + C.
3
Since y = 0 when t = 0, we have C = 0, so
y=
2 3/2
kt .
3
526
Chapter Six /SOLUTIONS
17. (a) Since
R′ (p) = 25 − 2p
R(p) =
Z
(25 − 2p) dp = 25p − p2 + C.
We assume that the revenue is 0 when the price is 0. Substituting gives
0 = 25 · 0 − 02 + C
C = 0.
Thus
R(p) = 25p − p2 .
(b) The revenue increases with price if R′ (p) > 0, that is 25 − 2p > 0, so p < 12.5 dollars. The revenue decreases with
price if R′ (p) < 0, that is 25 − 2p < 0, so p > 12.5.
18. (a) The marginal cost, M C, is found by differentiating the total cost function, C, with respect to q so M C = C ′ (q).
Thus the differential equation is
C ′ (q) = 3q 2 + 6q + 9.
(b) Solving the differential equation gives
C(q) =
Z
3q 2 + 6q + 9 dq
= q 3 + 3q 2 + 9q + D,
where D is a constant. We can check this by noting
C ′ (q) =
d
q 3 + 3q 2 + 9q + D = 3q 2 + 6q + 9 = M C.
dq
The fixed costs are 400, so C = 400 when q = 0. Thus,
400 = 03 + 3 · 02 + 9 · 0 + D,
so D = 400. The total cost function is
C(q) = q 3 + 3q 2 + 9q + 400.
19. (a) Acceleration = a(t) = −9.8 m/sec2
Velocity = v(t) = −9.8t + 40 m/sec
Height = h(t) = −4.9t2 + 40t + 25 m
.
(b) At the highest point,
so
v(t) = −9.8t + 40 = 0,
40
= 4.082 seconds.
9.8
At that time, h(4.082) = 106.633 m. We see that the tomato reaches a height of 106.633 m, at 4.082 seconds after
it is thrown.
(c) The tomato lands when h(t) = 0, so
−4.9t2 + 40t + 25 = 0.
t=
The solutions are t = −0.583 and t = 8.747 seconds. We see that it lands 8.747 seconds after it is thrown.
6.3 SOLUTIONS
20. (a)
527
v
80 ft/sec
v(t)
A
t
5 sec
(b) The total distance is represented by the shaded region A, the area under the graph of v(t).
(c) The area A, a triangle, is given by
1
1
A = (base)(height) = (5 sec)(80 ft/sec) = 200 ft.
2
2
R5
(d) Using integration and the Fundamental Theorem of Calculus, we have A = 0 v(t) dt or A = s(5) − s(0), where
s(t) is an antiderivative of v(t).
We have that a(t), the acceleration, is constant: a(t) = k for some constant k. Therefore v(t) = kt+C for some
constant C. We have 80 = v(0) = k(0)+C = C, so that v(t) = kt+80. Putting in t = 5, 0 = v(5) = (k)(5)+80,
or k = −80/5 = −16.
Thus v(t) = −16t + 80, and an antiderivative for v(t) is s(t) = −8t2 + 80t + C. Since the total distance
R5
traveled at t = 0 is 0, we have s(0) = 0 which means C = 0. Finally, A = 0 v(t) dt = s(5) − s(0) =
(−8(5)2 + (80)(5)) − (−8(0)2 + (80)(0)) = 200 ft, which agrees with the previous part.
21. (a) The velocity is decreasing at 32 ft/sec2 , the acceleration due to gravity.
(b) The graph is a line because the velocity is decreasing at a constant rate.
(c) The highest point is reached when the velocity is 0, which occurs when
160
= 5 sec.
Time =
32
(d) The object hits the ground at t = 10 seconds, since by symmetry if the object takes 5 seconds to go up, it takes 5
seconds to come back down.
(e) See Figure 6.37.
(f) The maximum height is the distance traveled when going up, which is represented by the area A of the triangle above
the time axis.
Area =
(g) The slope of the line is −32 so
1
(160 ft/sec)(5 sec) = 400 feet.
2
v(t) = −32t + 160.
Antidifferentiating, we get
s(t) = −16t2 + 160t + s0 .
Since the object starts on the ground, s0 = 0, so
s(t) = −16t2 + 160t.
At t = 5, we have
s(t) = −400 + 800 = 400 ft.
velocity (ft/sec)
160
v(t)
A
Highest
point
5
−160
Figure 6.37
Ground
10
t (sec)
528
Chapter Six /SOLUTIONS
2
22. The equation of motion is y = − gt2 +v0 t+y0 = −16t2 +128t+320. Taking the first derivative, we get v = −32t+128.
The second derivative gives us a = −32.
(a) At its highest point, the stone’s velocity is zero:
v = 0 = −32t + 128, so t = 4.
(b) At t = 4, the height is y = −16(4)2 + 128(4) + 320 = 576 ft
(c) When the stone hits the beach,
y = 0 = −16t2 + 128t + 320
0 = −t2 + 8t + 20 = (10 − t)(2 + t).
So t = 10 seconds.
(d) Impact is at t = 10. The velocity, v, at this time is v(10) = −32(10) + 128 = −192 ft/sec. Upon impact, the stone’s
velocity is 192 ft/sec downward.
23. Since the acceleration is constant, a graph of the velocity versus time looks like this:
v (mph)
200 mph
A
30
t (sec)
The distance traveled in 30 seconds, which is how long the runway must be, is equal to the area represented by A.
We have A = 21 (base)(height). First we convert the required velocity into miles per second.
200 miles
1 hour
hour
60 minutes
200 miles
=
3600 second
1
=
miles/second.
18
200 mph =
Therefore A = 12 (30 sec)(200 mph) = 21 (30 sec)
1
18
miles/sec =
5
6
1 minute
60 seconds
miles.
24. The height of an object above the ground which begins at rest and falls for t seconds is
s(t) = −16t2 + K,
where K is the initial height. Here the flower pot falls from 200 ft, so K = 200. To see when the pot hits the ground,
solve −16t2 + 200 = 0. The solution is
r
200
≈ 3.54 seconds.
t=
16
Now, velocity is given by s′ (t) = v(t) = −32t. So, the velocity when the pot hits the ground is
v(3.54) ≈ −113.1 ft/sec,
which is approximately 77 mph downward.
25. The first thing we should do is convert our units. We’ll bring everything into feet and seconds. Thus, the initial speed of
the car is
1 hour
5280 feet
70 miles
≈ 102.7 ft/sec.
hour
3600 sec
1 mile
6.3 SOLUTIONS
529
We assume that the acceleration is constant as the car comes to a stop. A graph of its velocity versus time is given in
Figure 6.38. We know that the area under the curve represents the distance that the car travels before it comes to a stop,
157 feet. But this area is a triangle, so it is easy to find t0 , the time the car comes to rest. We solve
1
(102.7)t0 = 157,
2
which gives
t0 ≈ 3.06 sec.
Since acceleration is the rate of change of velocity, the car’s acceleration is given by the slope of the line in Figure 6.38.
Thus, the acceleration, k, is given by
102.7 − 0
≈ −33.56 ft/sec2 .
k=
0 − 3.06
Notice that k is negative because the car is slowing down.
y
102.7 ft/sec
y = v(t)
t0
t
Figure 6.38: Graph of velocity versus time
26. (a) y =
(b)
Z
(2x + 1) dx, so the solution is y = x2 + x + C.
y
y
C=2
✲
C=2
✲
C=0
✲
C=0
✲
✛
x
x
C = −2
✲
C = −2
✲
(c) At y(1) = 5, we have 12 + 1 + C = 5 and so C = 3. Thus we have the solution y = x2 + x + 3.
27. (a) We are asking for a function whose derivative is sin x + 2. One antiderivative of sin x + 2 is
y = − cos x + 2x.
The general solution is therefore
y = − cos x + 2x + C,
where C is any constant. Figure 6.39 shows several curves in this family.
y
15
C=0
C = −1.99
10
C = 20
C = −5
5
C = −10
x
C = 15
−2π
C = 10
C=5
y(1) = 5
π
−π
−5
−10
−15
Figure 6.39
2π
C = −15
C = −20
530
Chapter Six /SOLUTIONS
(b) We have already seen that the general solution to the differential equation is y = − cos x + 2x + C. The initial
condition allows us to determine the constant C. Substituting y(3) = 5 gives
5 = y(3) = − cos 3 + 2 · 3 + C,
so C is given by
Thus, the (unique) solution is
C = 5 + cos 3 − 6 ≈ −1.99.
y = − cos x + 2x − 1.99.
Figure 6.39 shows this particular solution, marked C = −1.99.
28. (a) a(t) = 1.6, so v(t) = 1.6t + v0 = 1.6t, since the initial velocity is 0.
(b) s(t) = 0.8t2 + s0 , where s0 is the rock’s initial height.
29. (a) s = v0 t − 16t2 , where v0 = initial velocity, and v = s′ = v0 − 32t. At the maximum height, v = 0, so v0 = 32tmax .
2
2
2
5
Plugging into the
distance equation yields 100 = 32tmax − 16tmax = 16tmax , so tmax = 2 seconds, from which we
5
get v0 = 32 2 = 80 ft/sec.
(b) This time g = 5 ft/sec2 , so s = v0 t − 2.5t2 = 80t − 2.5t2 , and v = s′ = 80 − 5t. At the highest point, v = 0, so
= 16 seconds. Plugging into the distance equation yields s = 80(16) − 2.5(16)2 = 640 ft.
tmax = 80
5
30. The velocity as a function of time is given by: v = v0 + at. Since the object starts from rest, v0 = 0, and the velocity
is just the acceleration times time: v = −32t. Integrating this, we get position as a function of time: y = −16t2 + y0 ,
where the last term, y0 , is the initial position at the top of the tower, so y0 = 400 feet. Thus we have a function giving
position as a function of time: y = −16t2 + 400.
To find at what time the object hits the ground, we find t when y = 0. We solve 0 = −16t2 + 400 for t, getting
2
t = 400/16 = 25, so t = 5. Therefore the object hits the ground after 5 seconds. At this time it is moving with a
velocity v = −32(5) = −160 feet/second.
31. In Problem 30 we used the equation 0 = −16t2 + 400 to learn that the object hits the ground after 5 seconds. In a more
general form this is the equation y = − g2 t2 + v0 t + y0 , and we know that v0 = 0, y0 = 400 ft. So the moment the object
hits the ground is given by 0 = − g2 t2 + 400. In Problem 30 we used g = 32 ft/sec2 , but in this case we want to find a g
that results in the object hitting the ground after only 5/2 seconds. We put in 5/2 for t and solve for g:
2(400)
g 5 2
= 128 ft/sec2 .
0 = − ( ) + 400, so g =
2 2
(5/2)2
32. a(t) = −32. Since v(t) is the antiderivative of a(t), v(t) = −32t + v0 . But v0 = 0, so v(t) = −32t. Since s(t) is
the antiderivative of v(t), s(t) = −16t2 + s0 , where s0 is the height of the building. Since the ball hits the ground in 5
seconds, s(5) = 0 = −400 + s0 . Hence s0 = 400 feet, so the window is 400 feet high.
33. Let time t = 0 be the moment when the astronaut jumps up. If acceleration due to gravity is 5 ft/sec2 and initial velocity
is 10 ft/sec, then the velocity of the astronaut is described by
v(t) = 10 − 5t.
Suppose y(t) describes his distance from the surface of the moon. By the Fundamental Theorem,
y(t) − y(0) =
Z
0
t
(10 − 5x) dx
y(t) = 10t −
1 2
5t .
2
since y(0) = 0 (assuming the astronaut jumps off the surface of the moon).
The astronaut reaches the maximum height when his velocity is 0, i.e. when
dy
= v(t) = 10 − 5t = 0.
dt
Solving for t, we get t = 2 sec as the time at which he reaches the maximum height from the surface of the moon. At this
time his height is
1
y(2) = 10(2) − 5(2)2 = 10 ft.
2
6.3 SOLUTIONS
531
When the astronaut is at height y = 0, he either just landed or is about to jump. To find how long it is before he comes
back down, we find when he is at height y = 0. Set y(t) = 0 to get
1 2
5t
2
0 = 20t − 5t2
0 = 10t −
0 = 4t − t2
0 = t(t − 4).
So we have t = 0 sec (when he jumps off) and t = 4 sec (when he lands, which gives the time he spent in the air).
34.
• For [0, t1 ], the acceleration is constant and positive and the velocity is positive so the displacement is positive. Thus,
the work done is positive.
• For [t1 , t2 ], the acceleration, and therefore the force, is zero. Therefore, the work done is zero.
• For [t2 , t3 ], the acceleration is negative and thus the force is negative. The velocity, and thus the displacement, is
positive; therefore the work done is negative.
• For [t3 , t4 ], the acceleration (and thus the force) and the velocity (and thus the displacement) are negative. Thus, the
work done is positive.
• For [t2 , t4 ], the acceleration and thus the force is constant and negative. Velocity both positive and negative; total
displacement is 0. Since force is constant, work is 0.
35. Since
Acceleration =
velocity is the antiderivative of −g, so
Since the initial velocity is v0 , then C = v0 , so
We know that
dv
= −g,
dt
v = −gt + C.
v = −gt + v0 .
ds
= v = −gt + v0 .
dt
Therefore, we can find s by antidifferentiating again, giving:
s=−
gt2
+ v0 t + C.
2
s=−
gt2
+ v0 t + s0 .
2
If the initial position is s0 , then we must have
36. Let the acceleration due to gravity equal −k meters/sec2 , for some positive constant k, and suppose the object falls from
an initial height of s(0) meters. We have a(t) = dv/dt = −k, so that
v(t) = −kt + v0 .
Since the initial velocity is zero, we have
which means v0 = 0. Our formula becomes
v(0) = −k(0) + v0 = 0,
v(t) =
ds
= −kt.
dt
s(t) =
−kt2
+ s0 .
2
This means
Since
s(0) =
−k(0)2
+ s0 ,
2
s(t) =
−kt2
+ s(0).
2
we have s0 = s(0), and our formula becomes
532
Chapter Six /SOLUTIONS
Suppose that the object falls for t seconds. Assuming it has not hit the ground, its height is
s(t) =
−kt2
+ s(0),
2
so that the distance traveled is
s(0) − s(t) =
kt2
meters,
2
which is proportional to t2 .
s
, where t is the time it takes for an object to travel the distance s, starting from rest with uniform
37. (a) t = 1
v
2 max
acceleration a. vmax is the highest velocity the object reaches. Since its initial velocity is 0, the mean of its highest
velocity and initial velocity is 21 vmax .
p
(b) By Problem 36, s = 12 gt2 , where g is the acceleration due to gravity, so it takes 200/32 = 5/2 seconds for the
body to hit the ground. Since v = gt, vmax = 32( 52 ) = 80 ft/sec. Galileo’s statement predicts (100 ft)/(40 ft/sec) =
5/2 seconds, and so Galileo’s result is verified.
(c) If the acceleration is a constant a, then s = 12 at2 , and vmax = at. Thus
s
1
v
2 max
=
1
at2
2
1
at
2
= t.
38. (a) Since the gravitational force is acting downward
−
Hence,
d2 s
GM m
=
m
.
r2
dt2
d2 s
GM
= − 2 = Constant.
dt2
r
If we define g = GM/r 2 , then
d2 s
= −g.
dt2
(b) The fact that the mass cancels out of Newton’s equations of motion reflects Galileo’s experimental observation that
the acceleration due to gravity is independent of the mass of the body.
39. (a) Since s(t) = − 21 gt2 , the distance a body falls in the first second is
s(1) = −
g
1
· g · 12 = − .
2
2
In the second second, the body travels
s(2) − s(1) = −
In the third second, the body travels
s(3) − s(2) = −
and in the fourth second, the body travels
s(4) − s(3) = −
1
1
3g
g · 22 − g · 12 = − (4g − g) = − .
2
2
2
1
5g
1
g · 32 − g · 22 = − (9g − 4g) = − ,
2
2
2
7g
1
1
g · 42 − g · 32 = − (16g − 9g) = − .
2
2
2
(b) Galileo seems to have been correct. His observation follows from the fact that the differences between consecutive
squares are consecutive odd numbers. For, if n is any number, then n2 − (n − 1)2 = 2n − 1, which is the nth odd
number (where 1 is the first).
Strengthen Your Understanding
40. The dropped rock has constant acceleration, dv/dt = −32 ft/sec2 . Assuming that the rock is not thrown, we let initial
velocity be zero, so v = ds/dt = −32t. Thus s = −16t2 + K, where K is the initial height, in feet, from which the rock
was dropped. We compare the two formulas for s:
s1 (t) = −16t2 + 400
6.3 SOLUTIONS
533
and
s2 (t) = −16t2 + 200.
The rock dropped from the 400-foot cliff hits the ground when s1 (t) = 0 or at t = 5 seconds, and the second rock hits
the ground when s2 (t) = 0, approximately t = 3.5 seconds. Thus, the rock dropped from a 400-foot cliff takes less than
twice as long to hit the ground as the rock dropped from a 200-foot cliff.
41. If y = cos(t2 ), then dy/dt = −2t sin(t2 ). Thus, y = cos(t2 ) does not satisfy the differential equation.
42. The differential equation has solutions of the form y = F (x) + C with F ′ (x) = f (x). Two solutions correspond to
different values of C and do not cross at any point.
43. The differential equation dy/dx = 0 has general solution y = C, which is a family of constant solutions.
44. Solutions are of the form
t2
+ 3t + C.
2
Different solutions correspond to different values of C. For example
y=
t2
+ 3t + 1
2
2
t
y=
+ 3t + 2.
2
y=
45. If y = cos (5x), we have dy/dx = −5 sin (5x). The equation dy/dx = −5 sin (5x) is therefore a differential equation
that has y = cos (5x) as a solution.
46. True. If F (x) is an antiderivative of f (x), then F ′ (x) = f (x), so dy/dx = f (x). Therefore, y = F (x) is a solution to
this differential equation.
47. True. If y = F (x) is a solution to the differential equation dy/dx = f (x), then F ′ (x) = f (x), so F (x) is an antiderivative of f (x).
48. True. If acceleration is a(t) = k for some constant k, k 6= 0, then we have
Velocity = v(t) =
Z
a(t) dt =
Z
k dt = kt + C1 ,
for some constant C1 . We integrate again to find position as a function of time:
Position = s(t) =
Z
v(t) dt =
Z
(kt + C1 ) dt =
kt2
+ C1 t + C2 ,
2
for some constant C2 . Since k 6= 0, this is a quadratic polynomial.
49. False. In an initial value problem the value of y is specified at one value of x, but it does not have to be x = 0.
50. False. The solution of the initial value problem dy/dx = 1 with y(0) = −5 is a solution of the differential equation that
is not positive at x = 0.
51. True. If dy/dx = f (x) > 0, then all solutions y(x) have positive derivative and thus are increasing functions.
52. True. Two solutions y = F (x) and y = G(x) of the same differential equation dy/dx = f (x) are both antiderivatives of
f (x) and hence they differ by a constant: F (x) − G(x) = C for all x. Since F (3) 6= G(3) we have C 6= 0.
53. True. If y = f (x) satisfies the differential equation dy/dx = sin x/x, then f ′ (x) = sin x/x. Since (f (x) + 5)′ =
f ′ (x) = sin x/x, the function y = f (x) + 5 is also a solution of the same differential equation.
54. True. All solutions of the differential equation dy/dt = 3t2 are in the family y(t) = t3 + C of antiderivatives of 3t2 . The
initial condition y(1) = π tells us that y(1) = π = 13 + C, so C = π − 1. Thus y(t) = t3 + π − 1 is the only solution
of the initial value problem.
534
Chapter Six /SOLUTIONS
Solutions for Section 6.4
Exercises
1.
Table 6.4
x
0
0.5
1
1.5
2
I(x)
0
0.50
1.09
2.03
3.65
2. Using the Fundamental Theorem, we know that the change in F between x = 0 and x = 0.5 is given by
F (0.5) − F (0) =
Z
0.5
sin t cos t dt ≈ 0.115.
0
Since F (0) = 1.0, we have F (0.5) ≈ 1.115. The other values are found similarly, and are given in Table 6.5.
Table 6.5
b
0
0.5
1
1.5
2
2.5
3
F (b)
1
1.11492
1.35404
1.4975
1.41341
1.17908
1.00996
3. (a) Again using 0.00001 as the lower limit, because the integral is improper, gives Si(4) = 1.76, Si(5) = 1.55.
(b) Si(x) decreases when the integrand is negative, which occurs when π < x < 2π.
4. If f ′ (x) = sin(x2 ), then f (x) is of the form
f (x) = C +
Z
x
Z
x
sin(t2 ) dt.
a
Since f (0) = 7, we take a = 0 and C = 7, giving
f (x) = 7 +
5. If f ′ (x) =
sin(t2 ) dt.
0
sin x
, then f (x) is of the form
x
f (x) = C +
Z
x
sin t
dt.
t
Z
x
sin t
dt.
t
Z
x
Z
x
a
Since f (1) = 5, we take a = 1 and C = 5, giving
f (x) = 5 +
1
6. If f ′ (x) = Si(x), then f (x) is of the form
f (x) = C +
Si(t) dt.
a
Since f (0) = 2, we take a = 0 and C = 2, giving
f (x) = 2 +
Si(t) dt.
0
7. By the Fundamental Theorem, f (x) = F ′ (x). Since f is positive and increasing, F is increasing and concave up. Since
R0
F (0) = 0 f (t)dt = 0, the graph of F must start from the origin. See Figure 6.40.
F (x)
x
Figure 6.40
6.4 SOLUTIONS
535
8. By the Fundamental Theorem, f (x) = F ′ (x). Since f is positive and decreasing, F is increasing and concave down.
R0
Since F (0) = 0 f (t)dt = 0, the graph of F must start from the origin. See Figure 6.41.
F (x)
x
Figure 6.41
9. Since f is always positive, F is always increasing. F has an inflection point where f ′ = 0. Since F (0) =
F goes through the origin. See Figure 6.42.
R0
0
f (t)dt = 0,
F (x)
x
Figure 6.42
10. Since f is always non-negative, F is increasing. F is concave up where f Ris increasing and concave down where f is
0
decreasing; F has inflection points at the critical points of f . Since F (0) = 0 f (t)dt = 0, the graph of F goes through
the origin. See Figure 6.43.
F (x)
x
Figure 6.43
11. cos(x2 ).
12. From
d
dt
we have
Z
d
dt
13. (1 + x)
200
t
Z
t
f (x) dx = f (t)
a
√
√
sin( x) dx = sin( t).
4
.
14. From
d
dx
we have
d
dx
Z
Z
x
f (t) dt = f (x)
a
x
ln(t2 + 1) dt = ln(x2 + 1).
2
15. arctan(x2 ).
16. Considering Si(x2 ) as the composition of Si(u) and u(x) = x2 , we may apply the chain rule to obtain
d(Si(u)) du
d
=
·
dx
du
dx
sin u
· 2x
=
u
2 sin(x2 )
=
.
x
536
Chapter Six /SOLUTIONS
Problems
17. We need to find where F ′′ (x) is positive or negative. First, we compute
2
F ′ (x) = e−x ,
then
2
F ′′ (x) = −2xe−x .
2
Since e−x > 0 for all x, we see that
F ′′ (x) > 0 for x < 0
and
F ′′ (x) < 0 for x > 0.
Thus the graph of F (x) is concave up for x < 0 and concave down for x > 0.
18. The graph of f (x) = xe−x is shown in Figure 6.44. We see that f (x) < 0 when x < 0 and f (x) > 0 when x > 0.
There is some value a with −1 < a < 0, for which the area of f (x) below the x-axis from x = a to x = 0 equals the
area above the x-axis from x = 0 to x = 1. Then for this value of a,
F (a) =
Z
a
f (t) dt =
Z
a
te−t dt = 0.
1
1
Also
1
Z
F (1) =
te−t dt = 0.
1
There are no other values of x for which F (x) = 0, since for x > 1, the area under the graph of f (x) increases, because
f (x) > 0. Similarly, for x < a, we have f (x) < 0 and the area under the graph of f (x) increases as x gets more negative.
f (x)
a
x
−1
2
Figure 6.44
19. By the Fundamental Theorem of Calculus,
F (x) − F (0) =
Z
x
f (t) dt =
0
Since F (0) = 0, we have
F (x) =
Using a calculator, we get
Z
Z
x
sin(t2 ) dt.
0
x
sin(t2 ) dt.
0
F (0) = 0
F (0.5) = 0.041
F (1) = 0.310
F (1.5) = 0.778
F (2) = 0.805
F (2.5) = 0.431.
6.4 SOLUTIONS
537
20. The graph of f (x) = F ′ (x) = sin(x2 ) is in Figure 6.45. The function F (x) is increasing where F ′ (x) = f (x) > 0, that
√
is, for 0 < x < π ≈ 1.772. The function F (x) is decreasing where F ′ (x) = f (x) < 0, that is, for 1.772 < x ≤ 2.5.
1
0.5
√
1
−0.5
π
x
2
f (x) = sin(x2 )
−1
Figure 6.45
21. Using the solution to Problem 20, we see√that F (x) is increasing for x <
F (x) has its maximum value when x = π.
By the Fundamental Theorem of Calculus,
√
F ( π) − F (0) =
√
π
Z
√
√
f (t) dt =
0
Z
π and F (x) is decreasing for x >
√
π. Thus
π
sin(t2 ) dt.
0
Since F (0) = 0, calculating Riemann sums gives
√
√
F ( π) =
Z
π
sin(t2 ) dt = 0.895.
0
22. See Figure 6.46.
F (x)
x2
x1
x3
Figure 6.46
23. We know that F (x) increases for x < 50 because the derivative of F is positive there. See Figure 6.47. Similarly, F (x)
decreases for x > 50. Therefore, the graph of F rises until x = 50, and then it begins to fall. Thus, the maximum value
attained by F is F (50). To evaluate F (50), we use the Fundamental Theorem:
F (50) − F (20) =
Z
50
F ′ (x) dx,
20
which gives
F (50) = F (20) +
Z
50
20
F ′ (x) dx = 150 +
Z
50
F ′ (x) dx.
20
The definite integral equals the area of the shaded region under the graph of F ′ , which is roughly 350. Therefore, the
greatest value attained by F is F (50) ≈ 150 + 350 = 500.
538
Chapter Six /SOLUTIONS
F′
20
10
x
20
40
60
−10
Figure 6.47
R0
24. (a) The definition of g gives g(0) = 0 f (t) dt = 0.
(b) The Fundamental Theorem gives g ′ (1) = f (1) = −2.
(c) The function g is concave upward where g ′′ is positive. Since g ′′ = f ′ , we see that g is concave up where f is
increasing. This occurs on the interval 1 ≤ x ≤ 6.
(d) The function g decreases from x = 0 to x = 3 and increases for 3 < x ≤ 8, and the magnitude of the increase is
more than the magnitude of the decrease. Thus g takes its maximum value at x = 8.
π
π
1
1
d
sin(2t) dt = − cos(2t) = − (1 − 1) = 0.
25. (a) Since (cos(2t)) = −2 sin(2t), we have F (π) =
dt
2
2
0
0
(b) F (π) = (Area above t-axis) − (Area below t-axis) = 0. (The two areas are equal.)
Z
1
sin 2t
π
x
−1
(c) F (x) ≥ 0 everywhere. F (x) = 0 only at integer multiples of π. This can be seen for x ≥ 0 by noting F (x) =
(Area above t-axis) − (Area below t-axis), which is always non-negative and only equals zero when x is an integer
multiple of π. For x > 0
F (−x) =
Z
−x
sin 2t dt
0
=−
=
Z
Z
0
sin 2t dt
−x
x
sin 2t dt = F (x),
0
since the area from −x to 0 is the negative of the area from 0 to x. So we have F (x) ≥ 0 for all x.
1
by the Construction Theorem.
26. (a) F ′ (x) =
ln x
1
′
(b) For x ≥ 2, F (x) > 0, so F (x) is increasing. Since F ′′ (x) = −
< 0 for x ≥ 2, the graph of F (x) is
x(ln x)2
concave down.
(c)
3
F (x)
2
1
x
1
2
3
4
5
6
6.4 SOLUTIONS
539
27. (a) The definition of R gives
Z
R(0) =
0
1 + t2 dt = 0
0
and
R(−x) =
p
Z
−x
p
1 + t2 dt.
0
Changing the variable of integration by letting t = −z gives
Z
−x
p
1 + t2 dt =
0
x
Z
p
1 + (−z)2 (−dz) = −
0
Z
x
0
p
1 + z 2 dz.
Thus R is an odd function.
√
(b) Using the Second Fundamental Theorem gives R′ (x) = 1 + x2 , which is always positive, so R is increasing
everywhere.
(c) Since
x
,
R′′ (x) = √
1 + x2
then R is concave up if x > 0 and concave down if x < 0.
(d) See Figure 6.48.
R(x)
35
x
−10
10
−35
Figure 6.48
(e) We have
R(x)
lim
= lim
x→∞ x2
x→∞
Using l’Hopital’s rule gives
lim
√
x→∞
1 + x2
= lim
x→∞
2x
Thus the limit exists; its value is 1/2.
Rx√
0
1 + t2 dt
x2
.
p
1/x2 + 1
1
= .
2
2
Rx
28. (a) We have F (x) = a f (t) dx = 0 for all x.
(b) We have F ′ (x) = f (x) by the Second Fundamental Theorem. We have F ′ (x) = 0 by part (a). Hence f (x) = 0.
2
29. Since F ′ (x) = e−x and F (0) = 2, we have
F (x) = F (0) +
x
Z
2
e−t dt = 2 +
0
x
Z
2
e−t dt.
0
Substituting x = 1 and evaluating the integral numerically gives
F (1) = 2 +
Z
1
2
e−t dt = 2.747.
0
30. Since G′ (x) = cos(x2 ) and G(0) = −3, we have
G(x) = G(0) +
Z
0
x
2
cos(t ) dt = −3 +
Z
x
cos(t2 ) dt.
0
Substituting x = −1 and evaluating the integral numerically gives
G(−1) = −3 +
Z
0
−1
cos(t2 ) dt = −3.905.
540
Chapter Six /SOLUTIONS
31. We have
Z
w(0.4) =
0.4
q(x) dx
0
≈ 0.1q(0.0) + 0.1q(0.1) + 0.1q(0.2) + 0.1q(0.3)
left-hand sum with ∆t = 0.1
= 0.1(5.3) + 0.1(5.2) + 0.1(4.9) + 0.1(4.5) = 1.99.
Since q ′ (x) < 0, we know a left-hand sum provides an overestimate.
Alternatively, we could use a right-hand sum to provide an underestimate:
Z
w(0.4) =
0.4
q(x) dx
0
≈ 0.1q(0.1) + 0.1q(0.2) + 0.1q(0.3) + 0.1q(0.4)
right-hand sum with ∆t = 0.1
= 0.1(5.2) + 0.1(4.9) + 0.1(4.5) + 0.1(3.9) = 1.85.
32. We have:
v(0.4) =
Z
0.4
q ′ (x) dx,
0
= q(0.4) − q(0)
= 3.9 − 5.3
Fundamental Theorem of Calculus
from the table
= −1.4.
This answer is exact.
33. Since w(t) =
t
Z
q(x) dx, we know from the Construction Theorem that w is an antiderivative of q. This means:
0
w′ (t) = q(t),
′
so w (0.4) = q(0.4) = 3.9
from the table.
This answer is exact.
34. Since v(t) =
Z
t
q ′ (x) dx, we know from the Construction Theorem that v is an antiderivative of q ′ . This means:
0
v ′ (t) = q ′ (t)
′
so v (0.4) = q ′ (0.4)
q(0.5) − q(0.4)
average rate of change of q
≈
0.1
3.1 − 3.9
= −8.
=
0.1
Since q ′ and q ′′ are both negative, we know q is decreasing and its graph is concave-down. This tells us that the secant
line from x = 0.4 to x = 0.5 is steeper (its slope is “more negative”) than the tangent line, making −8 an underestimate.
Alternatively we can write
q(0.4) − q(0.3)
average rate of change of q
0.1
3.9 − 4.5
=
= −6.
0.1
v ′ (0.4) = q ′ (0.4) ≈
Here, the secant line from x = 0.3 to x = 0.4 is less steep (its slope is “less negative”) than the tangent line, making −6
an overestimate.
35. If we let F (x) =
Rx
0
ln(1 + t2 ) dt, using the chain rule gives
d
F (x2 ) = 2xF ′ (x2 ) = 2x ln(1 + (x2 )2 ) = 2x ln(1 + x4 ).
dx
6.4 SOLUTIONS
36. If we let f (t) =
Rt
1
541
cos(x2 ) dx and g(t) = sin t, using the chain rule gives
d
dt
Z
sin t
cos(x2 ) dx = f ′ (g(t)) · g ′ (t) = cos((sin t)2 ) · cos t = cos(sin2 t)(cos t).
1
37. We first write
Z
4
Z
2t
√
sin( x) dx
2t
as
−
Letting
√
sin( x) dx.
4
F (t) = −
and using the chain rule gives
Z
t
√
sin( x) dx,
4
√
√
d
F (2t) = 2F ′ (2t) = 2[− sin( 2t)] = −2 sin( 2t).
dt
38. If we split the integral at x = 0, we have
Z
x2
−x2
2
et dt =
Z
0
2
et dt +
−x2
Z
0
x2
2
et dt = −
If we let
F (x) =
Z
x
Z
−x2
2
et dt +
0
Z
x2
2
et dt.
0
2
et dt,
0
using the chain rule on each part separately gives
2 2
2 2
4
d
[−F (−x2 ) + F (x2 )] = −(−2x)F ′ (−x2 ) + (2x)F ′ (x2 ) = (2x)e(−x ) + (2x)e(x ) = 4xex .
dx
39.
d
d
d
[x erf(x)] = erf(x) (x) + x [erf(x)]
dx
dx
dx
Z x
2
d
2
√
= erf(x) + x
e−t dt
dx
π 0
2
−x2
.
= erf(x) + √ xe
π
40. If we let f (x) = erf(x) and g(x) =
g ′ (x)f ′ (g(x)). Since
√
x, then we are looking for
2
d
√
dx
π
2 −x2
= √ e
π
f ′ (x) =
1
and g ′ (x) = √ , we have
2 x
and so
Z
d
[f (g(x))].
dx
x
2
e−t dt
0
2
f ′ (g(x)) = √ e−x ,
π
√
d
1 2
1
[erf( x)] = √ √ e−x = √ e−x .
dx
2 x π
πx
By the chain rule, this is the same as
542
Chapter Six /SOLUTIONS
41. If we let f (x) =
′
′
Rx
0
2
e−t dt and g(x) = x3 , then we use the chain rule because we are looking for
′
f (g(x)) · g (x). Since f (x) = e
R x3
x
Z
x
Z
d
dx
42. We split the integral
−x
2
d
f (g(x))
dx
=
, we have
3
e
−t2
dt
0
!
= f ′ (x3 ) · 3x2 = e−(x
6
3 )2
· 3x2 = 3x2 e−x .
2
e−t dt into two pieces, say at t = 1 (though it could be at any other point):
x3
e
−t2
Z
dt =
x
We have used the fact that
x3
e
−t2
dt +
1
R1
x
d
dx
e−t dt = −
x3
e
1
e
−t2
Z
dt =
x
2
Z
Z
−t2
x
Rx
1
dt
x3
e
−t2
1
dt −
Z
x
2
e−t dt.
1
2
e−t dt. Differentiating gives
!
Z
d
=
dx
x3
e
−t2
dt
1
!
−
d
dx
Z
x
2
e−t dt
1
For the first integral, we use the chain rule with g(x) = x3 as the inside function, so the final answer is
d
dx
Z
x3
e
−t2
x
dt
!
= e−(x
3 2
)
2
6
2
· 3x2 − e−x = 3x2 e−x − e−x .
Strengthen Your Understanding
Z 5
t2 dt is a constant, therefore its derivative is zero.
43. Note that
0
44. f (x) = x2 has a minimum at x = 0. However, F (x) is non-decreasing for all x since F ′ (x) = x2 ≥ 0 everywhere.
45. The derivative of F (x), which is f (x), has a local minimum at x = 2. The function F (x) has local minimums at x = −1
and x = 3, where f (x) goes from negative to positive.
Rx
46. From the Second Fundamental Theorem of Calculus, we have F (x) = 0 f (t)dt (so that F (0) = 0). In order to assure
that F is a nondecreasing function, we need only pick aRfunction f (t) that is nonnegative for all t. The choices are many,
x
but one possible example is f (t) = t2 , giving F (x) = 0 t2 dt.
Rx
47. From the Second Fundamental Theorem of Calculus, we have G(x) = a g(t)dt. Since G(7) = 0, we let a = 7, and
in order to ensure that G isR concave up, we need a function g(t) that has a positive derivative. One possible example is
x
g(t) = et , giving G(x) = 7 et dt.
48. True. The Construction Theorem for Antiderivatives gives a method for building an antiderivative with a definite integral.
49. True, by the Second Fundamental Theorem of Calculus.
50. True. We see that
F (5) − F (3) =
Z
5
f (t) dt −
0
Z
3
f (t) dt =
0
Z
5
f (t) dt.
3
51. False. If f is positive then F is increasing, but if f is negative then F is decreasing.
52. True. SinceRF and G are both antiderivatives of f , they must differ by a constant. In fact, we can see that the constant C
2
is equal to 0 f (t) dt since
F (x) =
Z
x
f (t) dt =
0
53. True, since
Rx
0
(f (t) + g(t)) dt =
Rx
0
f (t) dt +
Z
x
f (t) dt +
2
Rx
0
g(t) dt.
Z
0
2
f (t) dt = G(x) + C.
SOLUTIONS to Review Problems for Chapter Six
543
Solutions for Chapter 6 Review
Exercises
1. We find the changes in f (x) between any two values of x by counting the area between the curve of f ′ (x) and the x-axis.
Since f ′ (x) is linear throughout, this is quite easy to do. From x = 0 to x = 1, we see that f ′ (x) outlines a triangle of
area 1/2 below the x-axis (the base is 1 and the height is 1). By the Fundamental Theorem,
Z
0
so
1
f ′ (x) dx = f (1) − f (0),
f (0) +
Z
1
f ′ (x) dx = f (1)
0
f (1) = 2 −
3
1
=
2
2
Similarly, between x = 1 and x = 3 we can see that f ′ (x) outlines a rectangle below the x-axis with area −1, so
f (2) = 3/2 − 1 = 1/2. Continuing with this procedure (note that at x = 4, f ′ (x) becomes positive), we get the table
below.
x
0
1
2
3
4
5
6
f (x)
2
3/2
1/2
−1/2
−1
−1/2
1/2
Rb
2. Since F (0) = 0, F (b) = 0 f (t) dt. For each b we determine F (b) graphically as follows:
F (0) = 0
F (1) = F (0) + Area of 1 × 1 rectangle = 0 + 1 = 1
F (2) = F (1) + Area of triangle ( 12 · 1 · 1) = 1 + 0.5 = 1.5
F (3) = F (2) + Negative of area of triangle = 1.5 − 0.5 = 1
F (4) = F (3) + Negative of area of rectangle = 1 − 1 = 0
F (5) = F (4) + Negative of area of rectangle = 0 − 1 = −1
F (6) = F (5) + Negative of area of triangle = −1 − 0.5 = −1.5
The graph of F (t), for 0 ≤ t ≤ 6, is shown in Figure 6.49.
1.5
1
F (t)
1
2
3
4
5
6
t
−1
−1.5
Figure 6.49
3. F is increasing because f is positive; F is concave up because f is increasing. See Figure 6.50.
3
F (x)
2
1
x
1
2
Figure 6.50
3
544
Chapter Six /SOLUTIONS
4. F is increasing because f is positive; F is concave up because f is increasing. See Figure 6.51.
2
F (x)
1
x
1
2
Figure 6.51
5. (a) The value of the integral is negative since the area below the x-axis is greater than the area above the x-axis. We
count boxes: The area below the x-axis includes approximately 11.5 boxes and each box has area (2)(1) = 2, so
5
Z
f (x)dx ≈ −23.
0
The area above the x-axis includes approximately 2 boxes, each of area 2, so
7
Z
5
So we have
Z
7
f (x)dx =
0
Z
f (x)dx ≈ 4.
5
f (x)dx +
0
Z
7
5
(b) By the Fundamental Theorem of Calculus, we have
F (7) − F (0) =
so,
F (7) = F (0) +
Z
0
6.
7.
8.
9.
10.
11.
12.
13.
Z
Z
Z
Z
Z
Z
Z
Z
5x dx =
5 2
x + C.
2
x3 dx =
x4
+C
4
sin θ dθ = − cos θ + C
(x3 − 2) dx =
(t2 +
x4
− 2x + C
4
t3
1
1
) dt =
− +C
t2
3
t
4
4
dt = − + C
t2
t
(x2 + 5x + 8) dx =
x3
5x2
+
+ 8x + C
3
2
√
8
4 w dw = w3/2 + C
3
f (x)dx ≈ −23 + 4 = −19.
Z
7
f (x)dx
0
7
f (x)dx = 25 + (−19) = 6.
SOLUTIONS to Review Problems for Chapter Six
545
14. 2t2 + 7t + C
15. sin θ + C
16.
Z
t3/2 + t−3/2 dt =
2t5/2
− 2t−1/2 + C
5
x2
+ 2x1/2 + C
2
x12
+C
18. πx +
12
19. 3 sin t + 2t3/2 + C
17.
20.
Z
y−
1
y
21. tan x + C
2
dy =
Z
y2 − 2 +
22. 2 ln |x| − π cos x + C
1
y2
dy =
1
y3
− 2y − + C
3
y
1
1
, the indefinite integral is x2 + x + ln |x| + C
x
2
23. Since f (x) = x + 1 +
24. 5ez + C
d x
1 x
2 + C, since
(2 ) = (ln 2) · 2x
25.
ln 2
dx
26. 3 sin x + 7 cos x + C
27. 2ex − 8 sin x + C
28.
Z
−1
−3
2
dr = −r −2
r3
−1
−3
= −1 +
1
= −8/9 ≈ −0.889.
9
29. We have
Z
π/2
π/2
2 cos φdφ = 2
sin φ
−π/2
30. F (x) =
Z
f (x) dx =
31. We have F (x) =
32. F (x) =
33. F (x) =
34. F (x) =
35. F (x) =
Z
Z
Z
Z
√
Z
−π/2
x2 dx =
!
= 2 sin
−π
π
− sin
2
2
= 2 (1 − (−1)) . = 4.
x3
x3
+ C. If F (0) = 4, then F (0) = 0 + C = 4 and thus C = 4. So F (x) =
+ 4.
3
3
x4
x4
+ 2x3 − 4x + C. Since F (0) = 4, we have 4 = 0 + C, so C = 4. So F (x) =
+ 2x3 − 4x + 4.
4
4
x dx =
2 3/2
2
x
+ C. If F (0) = 4, then F (0) = 0 + C = 4 and thus C = 4. So F (x) = x3/2 + 4.
3
3
ex dx = ex + C. If F (0) = 4, then F (0) = 1 + C = 4 and thus C = 3. So F (x) = ex + 3.
sin x dx = − cos x + C. If F (0) = 4, then F (0) = −1 + C = 4 and thus C = 5. So F (x) = − cos x + 5.
cos x dx = sin x + C. If F (0) = 4, then F (0) = 0 + C = 4 and thus C = 4. So F (x) = sin x + 4.
36. Since (xx )′ = xx (1 + ln x), we have
Z
3
3
1
xx (1 + ln x) dx = xx
1
= 33 − 11 = 26.
37. Since y = x + sin x − π, we differentiate to see that dy/dx = 1 + cos x, so y satisfies the differential equation. To show
that it also satisfies the initial condition, we check that y(π) = 0:
y = x + sin x − π
y(π) = π + sin π − π = 0.
38. Differentiating y with respect to x gives
for all values of A.
y ′ = nxn−1
546
Chapter Six /SOLUTIONS
Z
39. y =
Z
40. y =
41. W =
42. r =
(x3 + 5) dx =
Z
Z
43. y =
8x +
1
x
x4
+ 5x + C
4
dx = 4x2 + ln |x| + C
√
8
4 t dt = t3/2 + C
3
3 sin p dp = −3 cos p + C
Z
(6x2 + 4x) dx = 2x3 + 2x2 + C. If y(2) = 10, then 2(2)3 + 2(2)2 + C = 10 and C = 10 − 16 − 8 = −14.
Thus, y = 2x3 + 2x2 − 14.
Z
44. P =
45. s =
Z
10et dt = 10et + C. If P (0) = 25, then 10e0 + C = 25 so C = 15. Thus, P = 10et + 15.
(−32t + 100) dt = −16t2 + 100t + C. If s = 50 when t = 0, then −16(0)2 + 100(0) + C = 50, so C = 50.
Thus s = −16t2 + 100t + 50.
46. Integrating gives
Z
dq
dz =
dz
Z
(2 + sin z) dz = 2z − cos z + C.
If q = 5 when z = 0, then 2(0) − cos(0) + C = 5 so C = 6. Thus q = 2z − cos z + 6.
d
47.
dt
d
48.
dx
Z
π
t
Z
d
cos(z ) dz =
dt
3
1
x
d
ln t dt =
dx
Z
−
Z
−
1
x
t
3
cos(z ) dz
π
ln t dt
= − cos(t3 ).
= − ln x.
Problems
49. We can start by finding four points on the graph of F (x). The first one is given: F (2) = 3. By the Fundamental Theorem
R6
of Calculus, F (6) = F (2) + 2 F ′ (x)dx. The value of this integral is −7 (the area is 7, but the graph lies below the
x-axis), so F (6) = 3 − 7 = −4. Similarly, F (0) = F (2) − 2 = 1, and F (8) = F (6) + 4 = 0. We sketch a graph of
F (x) by connecting these points, as shown in Figure 6.52.
(2, 3)
F (x)
(0, 1)
4
(8, 0)
x
8
(6, −4)
Figure 6.52
50. Between time t = 0 and time t = B, the velocity of the cork is always positive, which means the cork is moving upward.
At time t = B, the velocity is zero, and so the cork has stopped moving altogether. Since shortly thereafter the velocity
of the cork becomes negative, the cork will next begin to move downward. Thus when t = B the cork has risen as far as
it ever will, and is riding on top of the crest of the wave.
From time t = B to time t = D, the velocity of the cork is negative, which means it is falling. When t = D, the
velocity is again zero, and the cork has ceased to fall. Thus when t = D the cork is riding on the bottom of the trough of
the wave.
Since the cork is on the crest at time B and in the trough at time D, it is probably midway between crest and trough
when the time is midway between B and D. Thus at time t = C the cork is moving through the equilibrium position on
its way down. (The equilibrium position is where the cork would be if the water were absolutely calm.) By symmetry,
t = A is the time when the cork is moving through the equilibrium position on the way up.
Since acceleration is the derivative of velocity, points where the acceleration is zero would be critical points of the
velocity function. Since point A (a maximum) and point C (a minimum) are critical points, the acceleration is zero there.
SOLUTIONS to Review Problems for Chapter Six
547
A possible graph of the height of the cork is shown in Figure 6.53. The horizontal axis represents a height equal to
the average depth of the ocean at that point (the equilibrium position of the cork).
height
B
A
time
C
D
Figure 6.53
51. (a) Critical points of F (x) are the zeros of f : x = 1 and x = 3.
(b) F (x) has a local minimum at x = 1 and a local maximum at x = 3.
(c) See Figure 6.54.
x
1
2
3
4
F (x)
Figure 6.54
Notice that the graph could also be above or below the x-axis at x = 3.
52. (a) Critical points of F (x) are x = −1, x = 1 and x = 3.
(b) F (x) has a local minimum at x = −1, a local maximum at x = 1, and a local minimum at x = 3.
(c) See Figure 6.55.
x
−2
2
4
F (x)
Figure 6.55
x3
53.
x1
x2
x4
x5
x
x4
x1
f ′ (x)
(a) f (x) is greatest at x1 .
x2
x3
x5
x
f (x)
548
Chapter Six /SOLUTIONS
(b)
(c)
(d)
(e)
(f)
f (x) is least at x5 .
f ′ (x) is greatest at x3 ..
f ′ (x) is least at x5 .
f ′′ (x) is greatest at x1 .
f ′′ (x) is least at x5 .
54. (a) Starting at x = 3, we are given that f (3) = 0. Moving to the left on the interval 2 < x < 3, we have f ′ (x) = −1,
so f (2) = f (3) − (1)(−1) = 1. On the interval 0 < x < 2, we have f ′ (x) = 1, so
f (0) = f (2) + 1(−2) = −1.
Moving to the right from x = 3, we know that f ′ (x) = 2 on 3 < x < 4. So f (4) = f (3) + 2 = 2. On the interval
4 < x < 6, f ′ (x) = −2 so
f (6) = f (4) + 2(−2) = −2.
′
On the interval 6 < x < 7, we have f (x) = 1, so
f (7) = f (6) + 1 = −2 + 1 = −1.
y
2
x
2
4
6
−2
(b) In part (a) we found that f (0) = −1 and f (7) = −1.
R7
(c) The integral 0 f ′ (x) dx is given by the sum
Z
7
f ′ (x) dx = (1)(2) + (−1)(1) + (2)(1) + (−2)(2) + (1)(1) = 0.
0
Alternatively, knowing f (7) and f (0) and using the Fundamental Theorem of Calculus, we have
Z
7
f ′ (x) dx = f (7) − f (0) = −1 − (−1) = 0.
0
55. We have
Area =
Z
4
x3
3
x2 dx =
1
4
=
1
13
64 − 1
43
−
=
= 21.
3
3
3
56. The graph crosses the x-axis where
7 − 8x + x2 = 0
(x − 7)(x − 1) = 0;
so x = 1 and x = 7. See Figure 6.56. The parabola opens upward and the region is below the x-axis, so
Area = −
Z
7
1
(7 − 8x + x2 ) dx
x3
= − 7x − 4x +
3
2
7
= 36.
1
y = 7 − 8x + x2
1
7
x
Figure 6.56
SOLUTIONS to Review Problems for Chapter Six
549
57. Since y = x3 (1 − x) is positive for 0 ≤ x ≤ 1 and y = 0, when x = 0, 1, the area is given by
Area =
Z
1
3
x (1 − x) dx =
0
1
Z
0
x4
x5
−
4
5
(x3 − x4 ) dx =
1
=
0
1
.
20
58. Since y = 0 only when x = 0 and x = 1, the area lies between these limits and is given by
Area =
Z
0
=
1
x2 (1 − x)2 dx =
2
x5
x3
− x4 +
3
4
5
1
Z
1
x2 (1 − 2x + x2 ) dx =
0
=
0
Z
1
0
(x2 − 2x3 + x4 ) dx
1
.
30
59. The curves cross at the origin and at (1, 1). See Figure 6.57. Since the upper half of x = y 2 is given by y =
curve is above y = x2 , we have
Area =
Z
0
1
√
2
( x − x ) dx =
y
x3
x3/2
−
3/2
3
√
x and this
1
=
1
1
2
− = .
3
3
3
0
y = x2
1
x
1
x = y2
Figure 6.57
60. The graph is shown in Figure 6.58. Since cos θ ≥ sin θ for 0 ≤ θ ≤ π/4, we have
Area =
Z
π/4
0
(cos θ − sin θ) dθ
π/4
= (sin θ + cos θ)
0
√
1
1
= √ + √ − 1 = 2 − 1.
2
2
y = cos θ
❄
y = sin θ
✠
θ
π
4
Figure 6.58
61. The graphs of f (θ) = sin θ and g(θ) = cos θ cross at θ = π/4 and θ = 5π/4. See Figure 6.59. Since sin θ is above
cos θ between these two crossing points and cos θ is above sin θ outside, we have
Area =
Z
0
π/4
(cos θ − sin θ) dθ +
Z
5π/4
π/4
(sin θ − cos θ) dθ +
Z
2π
5π/4
(cos θ − sin θ) dθ
550
Chapter Six /SOLUTIONS
π/4
= (sin θ + cos θ)
0
1
1
√ + √ −1
2
2
√
8
= √ = 4 2.
2
=
5π/4
+ (− cos θ − sin θ)
+
1
− −√
2
−
2π
+ (sin θ + cos θ)
π/4
5π/4
1
−√
2
−
1
1
−√ − √
2
2
+
1−
1
1
−√ − √
2
2
cos θ
5π/4
2π
π
π/4
θ
sin θ
Figure 6.59
62. Since the graph of y = ex is above the graph of y = cos x (see Figure 6.60) we have
Area =
Z
1
(ex − cos x) dx
0
=
Z
0
= ex
1
ex dx −
1
cos x dx
0
1
1
0
Z
− sin x
0
= e1 − e0 − sin 1 + sin 0
= e − 1 − sin 1.
y = ex
y = cos x
1
π
2
x
Figure 6.60
63. The area is given by
A=
Z
1
1
−1
(cosh x − sinh x) dx = (sinh x − cosh x)
−1
= sinh 1 − cosh 1 − (sinh(−1) − cosh(−1))
= 2 sinh 1.
SOLUTIONS to Review Problems for Chapter Six
64. The area under f (x) = 8x between x = 1 and x = b is given by
the integral:
b
Rb
1
(8x)dx. Using the Fundamental Theorem to evaluate
= 4b2 − 4.
Area = 4x2
1
Since the area is 192, we have
4b2 − 4 = 192
4b2 = 196
b2 = 49
b = ±7.
Since b is larger than 1, we have b = 7.
65. The graph of y = x2 − c2 has x-intercepts of x = ±c. See Figure 6.61. The shaded area is given by
Area = −
Z
= −2
= −2
c
(x2 − c2 ) dx
−c
c
Z
0
(x2 − c2 ) dx
x3
− c2 x
3
c
0
= −2
c3
− c3
3
=
4 3
c .
3
3
We want c to satisfy (4c )/3 = 36, so c = 3.
y
y = x2 − c 2
x
c
−c
−c2
Figure 6.61
66. We have
1
Average value =
10 − 0
Z
10
1
(x + 1)dx =
10
2
0
x3
+x
3
10
0
1
=
10
We see in Figure 6.62 that the average value of 103/3 ≈ 34.33 for f (x) looks right.
100
f (x) = x2 + 1
50
34.33
x
5
Figure 6.62
551
10
103
+ 10 − 0
3
=
103
.
3
552
Chapter Six /SOLUTIONS
67. The average value of v(x) on the interval 1 ≤ x ≤ c is
1
c−1
Since
1
c−1
Z
1
c
Z
c
6
6
1
−
dx =
x2
c−1
x
1
c
1
c−1
=
1
−6
6
+6 = .
c
c
6
6
dx = 1, we have = 1, so c = 6.
x2
c
68. Letting y = f (x), we have x = f −1 (y):
√
√
y=5 x
x = 0.2y
x = 0.04y 2
so f −1 (x) = 0.04x2 .
Letting k = 0.04 and n = 2, we have
4
Z
f −1 (x) dx =
1
0.04
· x2+1
2+1
4
4
1
0.04 3
=
·x
3
1
0.04 3
4 − 13
=
3
= 0.84.
69. We have
√ −1
(f (x))−1 = 5 x
1
=
5x0.5
= 0.2x−0.5 .
Letting k = 0.2 and n = −0.5, we have
Z
4
(f (x))−1 dx =
1
0.2
· x−0.5+1
−0.5 + 1
0.2 0.5
=
·x
0.5
4
1
= 0.4 40.5 − 10.5
= 0.4.
70. Letting k = 5 and n = 0.5, we have
Z
4
f (x) dx =
1
5
· x0.5+1
0.5 + 1
4
4
1
5
· x3/2
=
3/2
1
10
=
· 43/2 − 13/2
3
10
70
=
·7=
.
3
3
Thus,
Z
1
4
f (x) dx
−1
=
70
3
−1
=
3
.
70
4
1
SOLUTIONS to Review Problems for Chapter Six
71. We have:
Z
f (2) =
2
2x2 dx
0
=2
Z
2
x2 dx
0
= 2·
1 3
x
3
16
.
3
=
2
0
72. We have:
f (n) =
Z
n
nx2 dx
0
=n
Z
n
x2 dx n is a constant
0
= n·
1 4
n .
3
=
73. If we let f (x) =
74. Since
R3
Rx
2
3
Z
Z
d
dx
et dt = −
d
dx
n
0
sin(t2 ) dt and g(x) = x3 , using the chain rule gives
2
cos x
1 3
x
3
x3
sin(t2 ) dt = f ′ (g(x)) · g ′ (x) = sin((x3 )2 ) · 3x2 = 3x2 sin(x6 ).
2
R cos x
3
d
e dt = −
dx
cos x
t2
Rx
2
et dt, if we let f (x) =
cos x
Z
2
et dt and g(x) = cos x, using the chain rule gives
3
2
2
2
et dt = −f ′ (g(x)) · g ′ (x) = −e(cos x) (− sin x) = sin xecos
3
x
.
75. If we split the integral at x = 0, we have
Z
x
4
e−t dt =
−x
Z
0
x
Z
4
e−t dt +
−x
4
e−t dt = −
0
If we let
F (x) =
x
Z
Z
−x
4
e−t dt +
0
Z
x
4
e−t dt.
0
4
e−t dt,
0
using the chain rule on each part separately gives
4
4
4
d
[−F (−x) + F (x)] = −(−1)F ′ (−x) + (1)F ′ (x) = e−(−x) + e−x = 2e−x .
dx
76. We split the integral at x = 1 (or any other point we choose):
Z
t3
et
Z
p
1 + x2 dx =
1
t3
p
1 + x2 dx +
Z
1
p
1 + x2 dx =
et
Z
t3
1 + x2 dx −
1
Differentiating each part separately and using the chain rule gives
d
dt
Z
t3
et
p
1 + x2 dx =
=
d
dt
t3
p
1 + x2 dx −
1
p
= 3t
Z
1 + (t3 )2 · 3t2 −
p
2
1 + t6 − e
p
t
p
p
d
dt
Z
1
et
p
1
1 + x2 dx
1 + (et )2 · et
1 + e2t .
Z
et
p
1 + x2 dx.
553
554
Chapter Six /SOLUTIONS
77. (a) Inventory at time t is f (t) = Q −
Q
t. See Figure 6.63.
A
Q
f (t)
A
t
Figure 6.63
(b) Over the period 0 ≤ t ≤ A,
1
Average inventory =
A
=
1
A
A
Z
f (t) dt
0
Z
A
0
Q−
Q
t
A
dt
A
Q 2
1
=
Qt −
t
A
2A
=
0
1
QA
QA −
A
2
=
1
Q.
2
Graphically, the average is the y coordinate of the midpoint of the line in the graph above. Thus, the answer should
Q
Q+0
= .
be
2
2
Common sense tells us that since the rate is constant, the average amount should equal the amount at the midpoint
of the interval, which is indeed Q/2.
78. (a) The distance traveled is equal to the area of the region under the graph of v(t) between t = 0 and t = 10, or, the area
of the trapezoid T in Figure 6.64.
Area of T =
(2 + 10 · 0) + (2 + 10 · 10)
v(0) + v(10)
∆t =
10 = 520 ft.
2
2
v
v(t) = 2 + 10t
102 ft/sec
❄
T
2 ft/sec
10 sec
t
Figure 6.64
(b) Since v(t) = 2 + 10t > 0 for all t ≥ 0, the car is always moving in the same direction. If the car’s initial position is
s(0), then its position at time t is simply s(0)+ distance traveled in t seconds. To find this distance we calculate the
area of the trapezoid where the right-hand limit is t rather than 10.
Area of trapezoid =
v(0) + v(t)
(2 + 10 · 0) + (2 + 10t)
4 + 10t
∆t =
t=
t = 5t2 + 2t.
2
2
2
Thus s(t) = s(0) + 5t2 + 2t feet.
(c) The distance traveled by the car between t = 0 and t = 10 seconds is found by substituting into s(t):
s(10) − s(0) = s(0) + 5(10)2 + 2 · 10 − s(0) = 5(10)2 + 2 · 10 = 520 ft.
This is the same answer we found in part (a).
SOLUTIONS to Review Problems for Chapter Six
555
(d) The distance traveled by the car between t = 0 and t = 10 is the area of the region under the graph of v(t) and
between t = 0 and t = 10, that is
Total distance =
10
Z
v(t) dt =
0
Z
10
(2 + 10t) dt.
0
The Fundamental Theorem of Calculus asserts that since s(t) is an antiderivative for v(t),
Z
10
(2 + 10t) dt = s(10) − s(0).
0
The integral on the left is given by the area we found in part (a) and the difference on the right is what we found in
part (c). Thus the Fundamental Theorem guarantees that the distances found in parts (a) and (c) are the same.
79. (a) Since f ′ (t) is positive on the interval 0 < t < 2 and negative on the interval 2 < t < 5, the function f (t) is
increasing on 0 < t < 2 and decreasing on 2 < t < 5. Thus f (t) attains its maximum at t = 2. Since the area
under the t-axis is greater than the area above the t-axis, the function f (t) decreases more than it increases. Thus, the
minimum is at t = 5.
(b) To estimate the value of f at t = 2, we see that the area under f ′ (t) between t = 0 and t = 2 is about 1 box, which
has area 5. Thus,
Z
2
f (2) = f (0) +
0
f ′ (t)dt ≈ 50 + 5 = 55.
The maximum value attained by the function is f (2) ≈ 55.
The area between f ′ (t) and the t-axis between t = 2 and t = 5 is about 3 boxes, each of which has an area of
5. Thus
Z
5
f ′ (t)dt ≈ 55 + (−15) = 40.
f (5) = f (2) +
2
The minimum value attained by the function is f (5) = 40.
(c) Using part (b), we have f (5) − f (0) = 40 − 50 = −10. Alternately, we can use the Fundamental Theorem:
f (5) − f (0) =
Z
5
0
f ′ (t)dt ≈ 5 − 15 = −10.
80. Let v be the velocity and s be the position of the particle at time t. We know that a = dv/dt, so acceleration is the slope
of the velocity graph. Similarly, velocity is the slope of the position graph. Graphs of v and s are shown in Figures 6.65
and 6.66, respectively.
s
v
1
2
3
4
5
6
7
t
1
Figure 6.65: Velocity against time
2
3
4
5
6
7
t
Figure 6.66: Position against time
81. (a) Since 6 sec = 1/10 min,
Angular acceleration =
2500 − 1100
= 14,000 revs/min2 .
1/10
(b) We know angular acceleration is the derivative of angular velocity. Since
Angular acceleration = 14,000,
we have
Angular velocity = 14,000t + C.
Measuring time from the moment at which the angular velocity is 1100 revs/min, we have C = 1100. Thus,
Angular velocity = 14,000t + 1100.
Thus the total number of revolutions performed during the period from t = 0 to t = 1/10 min is given by
Number of
revolutions
=
Z
0
1/10
1/10
(14000t + 1100)dt = 7000t2 + 1100t
= 180 revolutions.
0
556
Chapter Six /SOLUTIONS
82. (a) We find F for each piece, 0 ≤ x ≤ 1 and 1 ≤ x ≤ 2.
For 0 ≤ x ≤ 1, we have f (x) = −x + 1, so F is of the form
Z
(−x + 1) dx = −
x2
+ x + C.
2
Since we want F (1) = 1, we need C = 1/2. See Figure 6.67.
For 1 ≤ x ≤ 2, we have f (x) = x − 1, so F is of the form
Z
(x − 1) dx =
x2
− x + C.
2
Again, since we want F (1) = 1, we have C = 3/2. See Figure 6.67.
(b) Evaluating
2
2
0
2
3
1
− − +0+
= 1.
F (2) − F (0) =
−2+
2
2
2
2
The region under the graph of f consists of two triangles, whose area is
Area =
1
1
+ = 1.
2
2
(c) The Fundamental Theorem of Calculus says
Z
2
0
f (x) dx = F (2) − F (0).
Since the value of the integral is just the area under the curve, we have shown this in part (b).
F (x)
1
x
1
2
Figure 6.67
83. Since the car’s acceleration is constant, a graph of its velocity against time t is linear, as shown below.
v (mph)
80
6
t (seconds)
The acceleration is just the slope of this line:
80 − 0 mph
40
mph
dv
=
=
= 13.33
.
dt
6 sec
3
sec
To convert our units into ft/sec2 ,
40 mph 5280 ft 1 hour
ft
·
·
= 19.55 2
3
sec 1 mile 3600 sec
sec
SOLUTIONS to Review Problems for Chapter Six
557
84. (a) Since the velocity is constantly decreasing, and v(6) = 0, the car stops after 6 seconds.
t (sec)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
v(t) (ft/sec)
30
27.5
25
22.5
20
17.5
15
12.5
10
7.5
5
2.5
0
(b) Over the interval a ≤ t ≤ a + 21 , the left-hand velocity is v(a), and the right-hand velocity is v(a + 21 ). Since we
are considering half-second intervals, ∆t = 12 , and n = 12. The left sum is 97.5 ft., and the right sum is 82.5 ft.
(c) Area A in the figure below represents distance traveled.
A=
1
1
(base)(height) = · 6 · 30 = 90 ft.
2
2
velocity (ft/sec)
30
Deceleration
= 5 ft/sec2
A
6
t (seconds)
(d) The velocity is constantly decreasing at a rate of 5 ft/sec per second, i.e. after each second the velocity has dropped
by 5 units. Therefore v(t) = 30 − 5t.
An antiderivative for v(t) is s(t), where s(t) = 30t − 52 t2 . Thus by the Fundamental Theorem of Calculus,
the distance traveled = s(6) − s(0) = (30(6) − 52 (6)2 ) − (30(0) − 25 (0)2 ) = 90 ft. Since v(t) is decreasing, the
left-hand sum in part (b) overestimates the distance traveled, while the right-hand sum underestimates it.
The area A is equal to the average of the left-hand and right-hand sums: 90 ft = 21 (97.5 ft + 82.5 ft). The
left-hand sum is an overestimate of A; the right-hand sum is an underestimate.
85. (a) Using g = −32 ft/sec2 , we have
t (sec)
0
1
2
3
4
5
v(t) (ft/sec)
80
48
16
−16
−48
−80
(b) The object reaches its highest point when v = 0, which appears to be at t = 2.5 seconds. By symmetry, the object
should hit the ground again at t = 5 seconds.
(c) Left sum = 80(1) + 48(1) + 16( 12 ) = 136 ft , which is an overestimate.
Right sum = 48(1) + 16(1) + (−16)( 12 ) = 56 ft , which is an underestimate.
Note that we used a smaller third rectangle of width 1/2 to end our sum at t = 2.5.
(d) We have v(t) = 80 − 32t, so antidifferentiation yields s(t) = 80t − 16t2 + s0 .
But s0 = 0, so s(t) = 80t − 16t2 .
At t = 2.5, s(t) = 100 ft., so 100 ft. is the highest point.
86. Since A′ (r) = C(r) and C(r) = 2πr, we have
A′ (r) = 2πr.
Thus, we have, for some arbitrary constant K:
A(r) =
Z
2πr dr = 2π
Z
r dr = 2π
Since a circle of radius r = 0 has area = 0, we substitute to find K:
0 = π02 + K
K = 0.
Thus
A(r) = πr 2 .
r2
+ K = πr 2 + K.
2
558
Chapter Six /SOLUTIONS
87. Since V ′ (r) = S(r) and S(r) = 4πr 2 , we have
V ′ (r) = 4πr 2 .
Thus, we have, for some arbitrary constant K:
V (r) =
Z
4πr 2 dr = 4π
Z
r 2 dr = 4π
4
r3
+ K = πr 3 + K.
3
3
Since a sphere of radius r = 0 has volume = 0, we substitute to find K:
4 3
π0 + K
3
K = 0.
0=
Thus
V (r) =
4 3
πr .
3
88. The velocity of the car decreases at a constant rate, so we can write: dv/dt = −a. Integrating this gives v = −at + C.
The constant of integration C is the velocity when t = 0, so C = 60 mph = 88 ft/sec, and v = −at + 88. From this
equation we can see the car comes to rest at time t = 88/a.
Integrating the expression for velocity we get s = − a2 t2 + 88t + C, where C is the initial position, so C = 0. We
can use fact that the car comes to rest at time t = 88/a after traveling 200 feet. Start with
a
s = − t2 + 88t,
2
and substitute t = 88/a and s = 200:
882
a 88 2
88
+ 88
=
2 a
a
2a
882
2
a=
= 19.36 ft/sec
2(200)
200 = −
89. (a) In the beginning, both birth and death rates are small; this is consistent with a very small population. Both rates begin
climbing, the birth rate faster than the death rate, which is consistent with a growing population. The birth rate is then
high, but it begins to decrease as the population increases.
(b)
bacteria/hour
bacteria/hour
B
B−D
D
time (hours)
≈6
10
15
20
≈6
10
15
20
time (hours)
Figure 6.68: Difference between B and D is greatest at t ≈ 6
The bacteria population is growing most quickly when B − D, the rate of change of population, is maximal;
that happens when B is farthest above D, which is at a point where the slopes of both graphs are equal. That point is
t ≈ 6 hours.
(c) Total number born by time t is the area under the B graph from t = 0 up to time t. See Figure 6.69.
Total number alive at time t is the number born minus the number that have died, which is the area under the B
graph minus the area under the D graph, up to time t. See Figure 6.70.
SOLUTIONS to Review Problems for Chapter Six
559
bacteria
bacteria
B
B
D
D
5
10
15
20
N
time (hours)
time (hours)
5
Figure 6.69: Number
born by time t is
Rt
B(x) dx
0
≈ 11 15
20
Figure 6.70:
R t Number alive at time t is
(B(x) − D(x)) dx
0
From Figure 6.70, we see that the population is at a maximum when B = D, that is, after about 11 hours. This
stands to reason, because B − D is the rate of change of population, so population is maximized when B − D = 0,
that is, when B = D.
90. See Figure 6.71.
Suppose t1 is the time to fill the left side to the top of the middle ridge. Since the container gets wider as you go up,
the rate dH/dt decreases with time. Therefore, for 0 ≤ t ≤ t1 , graph is concave down.
At t = t1 , water starts to spill over to right side and so depth of left side does not change. It takes as long for the
right side to fill to the ridge as the left side, namely t1 . Thus the graph is horizontal for t1 ≤ t ≤ 2t1 .
For t ≥ 2t1 , water level is above the central ridge. The graph is climbing because the depth is increasing, but at a
slower rate than for t ≤ t1 because the container is wider. The graph is concave down because width is increasing with
depth. Time t3 represents the time when container is full.
H (height)
t1
t3
2t1
t (time)
Figure 6.71
91. By the Second Fundamental Theorem, we know that N ′ (t) = r(t). Since r(t) > 0, this means N ′ (t) > 0, so N is an
increasing function. Since r ′ (t) < 0, this means N ′′ (t) < 0, so the graph of N is concave down.
92. Since r(x) > 0 and r ′ (x) < 0, we know that the graph of r lies above the axis and is dropping from left to right. From
the definition of N , we see that:
N (20) =
N (10) =
N (20) − N (10) =
N (15) − N (5) =
• The largest area is described by N (20) =
during the first 20 days.
20
Z
r(x) dx = Area from t = 0 to t = 20
Z0 10
Z0 20
Z1015
r(x) dx = Area from t = 0 to t = 10
r(x) dx = Area from t = 10 to t = 20
r(x) dx = Area from t = 5 to t = 15.
5
R 20
0
r(x) dx. This corresponds to the amount of pollutant leeched out
560
Chapter Six /SOLUTIONS
• The remaining three areas all describe 10-day intervals. Since the function is decreasing, areas to the left are larger
than areas to the right. Thus,
Z
Z
Z
15
20
r(x) dx <
10
r(x) dx <
r(x) dx .
5
10
{z
|
}
0
{z
|
N(20)−N(10)
}
|
N(15)−N(5)
{z
}
N(10)
These correspond (from least to greatest) the amount of pollutant leeched out from day 10 to day 20, from day 5 to
day 15, and from day 0 to day 10.
• In conclusion: N (20) − N (10) < N (15) − N (5) < N (10) < N (20).
Z
0
Z
5
93. (a) We know that F (0) =
f (t) dt = 0. so F has a zero at x = 0.
0
We know that
F (5) =
f (t) dt =
0
Z
3
f (t) dt +
0
5
Z
f (t) dt = 0 since
3
R3
f (t) dt = −
0
So F has a zero at x = 5.
We know that f (x) = F ′ (x), so F ′ (3) = f (3) = 0.
This means F has Zat least two zeros, at x = 0, 5, and one critical point, at x = 3.
R5
3
f (t) dt.
1
F (t) dt = 0, so G has a zero at x = 1.
(b) We know that G(1) =
1
We know that F (x) = G′ (x), so G′ (5) = F (5) = 0 and G′ (0) = F (0) = 0.
This means G has at least one zero, at x = 1, and two critical points, at x = 0, 5.
94. (a) The definition of P gives
P (0) =
Z
0
arctan(t2 ) dt = 0
0
and
P (−x) =
Z
−x
arctan(t2 ) dt.
0
Changing the variable of integration by letting t = −z gives
Z
−x
2
arctan(t ) dt =
0
Z
0
x
2
arctan((−z) )(−dz) = −
Z
x
arctan(z 2 ) dz.
0
Thus P is an odd function.
(b) Using the Second Fundamental Theorem gives P ′ (x) = arctan(x2 ), which is greater than 0 for x 6= 0. Thus P is
increasing everywhere.
(c) Since
2x
P ′′ (x) =
,
1 + x4
we have P concave up if x > 0 and concave down if x < 0.
(d) See Figure 6.72.
7
P (x)
x
−5
5
−7
Figure 6.72
SOLUTIONS to Review Problems for Chapter Six
561
CAS Challenge Problems
95. (a) We have ∆x =
(b − a)
b−a
, so, since f (xi ) = xi 3 ,
and xi = a + i(∆x) = a + i
n
n
Riemann sum =
n
X
f (xi )∆x =
n
X
a+i
i=1
i=1
b−a
n
3
b−a
.
n
(b) A CAS gives
n
X
a+
i=1
i(b − a)
n
3
(b − a)
(a − b)(a3 (n − 1)2 + (a2 b + ab2 )(n2 − 1) + b3 (n + 1)3 )
=−
.
n
4n2
Taking the limit as n → ∞ gives
lim
n→∞
n
X
a+i
i=1
(c) The answer to part (b) simplifies to
b − a 3 b − a
n
n
a4
d
b4
−
. Since
4
4
dx
that
Z
b
x3 dx =
a
x4
4
x4
4
=−
b
=
a
(a + b)(a − b)(a2 + b2 )
.
4
= x3 , the Fundamental Theorem of Calculus says
b4
a4
−
.
4
4
96. (a) A CAS gives
Z
e2x dx =
1 2x
e
2
(b) The three integrals in part (a) obey the rule
Z
Z
e3x dx =
1 3x
e
3
eax+b dx =
Z
e3x+5 dx =
1 3x+5
e
.
3
1 ax+b
e
.
a
(c) Checking the formula by calculating the derivative
d
dx
1
a
eax+b
1 d ax+b
e
by the constant multiple rule
a dx
1
d
= eax+b (ax + b) by the chain rule
a
dx
1
= eax+b · a = eax+b .
a
=
97. (a) A CAS gives
Z
1
sin(3x) dx = − cos(3x)
3
Z
1
sin(4x) dx = − cos(4x)
4
Z
1
sin(3x − 2) dx = − cos(3x − 2).
3
(b) The three integrals in part (a) obey the rule
Z
1
sin(ax + b) dx = − cos(ax + b).
a
(c) Checking the formula by calculating the derivative
1
1 d
d
− cos(ax + b) = −
cos(ax + b) by the constant multiple rule
dx
a
a dx
1
d
= − (− sin(ax + b)) (ax + b) by the chain rule
a
dx
1
= − (− sin(ax + b)) · a = sin(ax + b).
a
562
Chapter Six /SOLUTIONS
98. (a) A CAS gives
Z
x−2
dx = x − ln |x − 1|
x−1
Z
x−1
dx = x + ln |x − 2|
x−2
Z
x−3
dx = x − 2 ln |x − 1|
x−1
Although the absolute values are needed in the answer, some CASs may not include them.
(b) The three integrals in part (a) obey the rule
Z
x−a
dx = x + (b − a) ln |x − b|.
x−b
(c) Checking the formula by calculating the derivative
1
d
(x + (b − a) ln |x − b|) = 1 + (b − a)
by the sum and constant multiple rules
dx
x−b
(x − b) + (b − a)
x−a
=
=
x−b
x−b
99. (a) A CAS gives
Z
1
1
dx = (ln |x − 3| − ln |x − 1|)
(x − 1)(x − 3)
2
Z
1
1
dx = (ln |x + 3| − ln |x − 1|).
(x − 1)(x + 3)
4
Z
1
1
dx = (ln |x − 4| − ln |x − 1|)
(x − 1)(x − 4)
3
Although the absolute values are needed in the answer, some CASs may not include them.
(b) The three integrals in part (a) obey the rule
Z
1
1
dx =
(ln |x − b| − ln |x − a|).
(x − a)(x − b)
b−a
(c) Checking the formula by calculating the derivative
d
dx
1
1
1
1
(ln |x − b| − ln |x − a|) =
−
b−a
b−a x−b
x−a
1
(x − a) − (x − b)
=
b−a
(x − a)(x − b)
=
1
b−a
b−a
(x − a)(x − b)
=
1
.
(x − a)(x − b)
PROJECTS FOR CHAPTER SIX
1. (a) If the poorest p% of the population has exactly p% of the goods, then F (x) = x.
(b) Any such F is increasing. For example, the poorest 50% of the population includes the poorest 40%, and
so the poorest 50% must own more than the poorest 40%. Thus F (0.4) ≤ F (0.5), and so, in general, F is
increasing. In addition, it is clear that F (0) = 0 and F (1) = 1.
The graph of F is concave up by the following argument. Consider F (0.05) − F (0.04). This is the
fraction of resources the fifth poorest percent of the population has. Similarly, F (0.20) − F (0.19) is the
fraction of resources that the twentieth poorest percent of the population has. Since the twentieth poorest
percent owns more than the fifth poorest percent, we have
F (0.05) − F (0.04) ≤ F (0.20) − F (0.19).
PROJECTS FOR CHAPTER SIX
563
More generally, we can see that
F (x1 + ∆x) − F (x1 ) ≤ F (x2 + ∆x) − F (x2 )
for any x1 smaller than x2 and for any increment ∆x. Dividing this inequality by ∆x and taking the limit
as ∆x → 0, we get
F ′ (x1 ) ≤ F ′ (x2 ).
So, the derivative of F is an increasing function, i.e. F is concave up.
(c) G is twice the shaded area below in the following figure. If the resource is distributed evenly, then G is
zero. The larger G is, the more unevenly the resource is distributed. The maximum possible value of G is
1.
y=x
F (x)
1
x
2. (a) In Figure 6.73, the area of the shaded region is F (M ). Thus, F (M ) =
Theorem, F ′ (M ) = y(M ).
RM
0
y(t) dt and, by the Fundamental
y (annual yield)
F (M )
t (time in years)
M
Figure 6.73
(b) Figure 6.74 is a graph of F (M ). Note that the graph of y looks like the graph of a quadratic function.
Thus, the graph of F looks like a cubic.
F (total yield)
20000
F (M )
15000
10000
5000
10
20
30
40
Figure 6.74
50
60
M (time in years)
564
Chapter Six /SOLUTIONS
(c) We have
1
1
a(M ) =
F (M ) =
M
M
Z
M
y(t) dt.
0
(d) If the function a(M ) takes on its maximum at some point M , then a′ (M ) = 0. Since
a(M ) =
1
F (M ),
M
differentiating using the quotient rule gives
a′ (M ) =
M F ′ (M ) − F (M )
= 0,
M2
so M F ′ (M ) = F (M ). Since F ′ (M ) = y(M ), the condition for a maximum may be written as
M y(M ) = F (M )
or as
y(M ) = a(M ).
To estimate the value of M which satisfies M y(M ) = F (M ), use the graph of y(t). Notice that
F (M ) is the area under the curve from 0 to M , and that M y(M ) is the area of a rectangle of base M and
height y(M ). Thus, we want the area under the curve to be equal to the area of the rectangle, or A = B
in Figure 6.75. This happens when M ≈ 50 years. In other words, the orchard should be cut down after
about 50 years.
y (annual yield)
Area B
❄
Area A
✲
t (time in years)
50
Figure 6.75
3. (a) (i)
1
sin(t)
t
−3π
−2π
π
−π
t
2π
3π
−1
(ii) Si(x) neither always decreases nor always increases, since its derivative, x−1 sin x, has both positive
and negative values for x > 0. For positive x, Si(x) is the area under the curve sint t and above the
t-axis from t = 0 to t = x, minus the area above the curve and below the t-axis. Looking at the graph
above, one can see that this difference of areas is going to always be positive.
PROJECTS FOR CHAPTER SIX
565
(iii)
It seems that the limit exists: the curve drawn in the slope field,
Z x
sin t
dt,
y = Si(x) =
t
0
seems to approach some limiting height as x → ∞. (In fact, the limiting height is π/2.)
(b)
y
(i)
(ii)
50
15
F (x) =
40
y = xsin x
Z
x
tsin t dt
0
30
y = xsin x
20
10
5
10
15
(iii)
20
x
5
F (x) =
50
Z
10
15
20
x
x
tsin t dt
0
40
30
20
10
5
10
15
x
20
(c) (i) The most obvious feature of the graph of y = sin(x2 ) is its symmetry about the y-axis. This means
the function g(x) = sin(x2 ) is an even function, i.e. for all x, we have g(x) = g(−x). Since
sin(x2 ) is even, its antiderivative F must be odd, that is F (−x) = −F (−x). To see this, set F (t) =
Rt
2
0 sin(x ) dx, then
F (−t) =
Z
0
−t
2
sin(x ) dx = −
Z
0
−t
2
sin(x ) dx = −
Z
t
sin(x2 ) dx = −F (t),
0
since the area from −t to 0 is the same as the area from 0 to t. Thus F (t) = −F (−t) and F is odd.
The second obvious feature of the graph of y = sin(x2 ) is that it oscillates between −1 and 1 with
a “period” which goes to zero as |x| increases. This implies that F ′ (x) alternates between intervals
566
Chapter Six /SOLUTIONS
where it is positive or negative, and increasing or decreasing, with frequency growing arbitrarily
large as |x| increases. Thus F (x) itself similarly alternates between intervals where it is increasing or
decreasing, and concave up or concave down.
Finally, since y = sin(x2 ) = F ′ (x) passes through (0, 0), and F (0) = 0, F is tangent to the
x-axis at the origin.
(ii)
F (x)
Figure 6.76
F never crosses the x-axis in the region x > 0, and lim F (x) exists. One way to see these facts
x→∞
is to note that by the Construction Theorem,
Z x
F (x) = F (x) − F (0) =
F ′ (t) dt.
0
2
So F (x) is just the area between the curve y = sin(t ) and the t-axis for 0 ≤ t ≤ x (with area
above the t-axis counting positively, and area below the t-axis counting negatively). Now looking at
the graph of curve, we see that this area will include alternating pieces above and below the t-axis.
We can also see that the area of these pieces is approaching 0 as we go further out. So we add a piece,
take a piece away, add another piece, take another piece away, and so on.
It turns out that this means that the sums of the pieces converge. To see this, think of walking
from point A to point B. If you walk almost to B, then go a smaller distance toward A, then a yet
smaller distance back toward B, and so on, you will eventually approach some point between A and
B. So we can see that lim F (x) exists. Also, since we always subtract a smaller piece than we just
x→∞
added, and the first piece is added instead of subtracted, we see that we never get a negative sum; thus
F (x) is never negative in the region x > 0, so F (x) never crosses the x-axis there.
7.1 SOLUTIONS
CHAPTER SEVEN
Solutions for Section 7.1
Exercises
1. (a) We substitute w = 1 + x2 , dw = 2x dx.
Z
x=1
x=0
x
1
dx =
1 + x2
2
(b) We substitute w = cos x, dw = − sin x dx.
x= π
4
Z
x=0
Z
sin x
dx = −
cos x
w=2
1
1
dw = ln |w|
w
2
w=1
Z
w=
√
w=1
2/2
√
= − ln |w|
=
1
1
ln 2.
2
1
dw
w
2/2
1
2
= − ln
√
1
2
= ln 2.
2
2
d
d
2. (a) dx
sin(x2 + 1) = 2x cos(x2 + 1);
sin(x3 + 1) = 3x2 cos(x3 + 1)
dx
(ii) 31 sin(x3 + 1) + C
(b) (i) 12 sin(x2 + 1) + C
(c) (i) − 12 cos(x2 + 1) + C
(ii) − 31 cos(x3 + 1) + C
3. We use the substitution w = 3x, dw = 3 dx.
Z
Check:
d 1 3x
( e
dx 3
e3x dx =
1
3
Z
ew dw =
1
1 w
e + C = e3x + C.
3
3
+ C) = 31 e3x (3) = e3x .
4. Make the substitution w = t2 , dw = 2t dt. The general antiderivative is
5. We use the substitution w = −x, dw = − dx.
Z
Check:
d
(−e−x
dx
e−x dx = −
Z
R
2
2
tet dt = (1/2)et + C.
ew dw = −ew + C = −e−x + C.
+ C) = −(−e−x ) = e−x .
6. We use the substitution w = −0.2t, dw = −0.2 dt.
Z
Check:
d
(−125e−0.2t
dt
25e−0.2t dt =
25
−0.2
Z
ew dw = −125ew + C = −125e−0.2t + C.
+ C) = −125e−0.2t (−0.2) = 25e−0.2t .
7. We use the substitution w = 2x, dw = 2 dx.
Z
Check:
d
(− 12
dx
1
sin(2x)dx =
2
cos(2x) + C) =
1
2
Z
1
1
sin(w)dw = − cos(w) + C = − cos(2x) + C.
2
2
sin(2x)(2) = sin(2x).
2
8. We use the substitution w = t , dw = 2t dt.
Z
Check:
t cos(t2 )dt =
1
2
Z
cos(w)dw =
1
d 1
( sin(t2 ) + C) = cos(t2 )(2t) = t cos(t2 ).
dt 2
2
1
1
sin(w) + C = sin(t2 ) + C.
2
2
567
568
Chapter Seven /SOLUTIONS
9. We use the substitution w = 3 − t, dw = − dt.
Check:
d
(cos(3
dt
Z
sin(3 − t)dt = −
Z
sin(w)dw = −(− cos(w)) + C = cos(3 − t) + C.
− t) + C) = − sin(3 − t)(−1) = sin(3 − t).
10. We use the substitution w = −x2 , dw = −2x dx.
Z
Check:
2
d
(− 12 e−x
dx
Z
2
1
1
e−x (−2x dx) = −
2
2
2
1
1
= − ew + C = − e−x + C.
2
2
2
xe−x dx = −
2
Z
ew dw
2
+ C) = (−2x)(− 21 e−x ) = xe−x .
11. Either expand (r + 1)3 or use the substitution w = r + 1. If w = r + 1, then dw = dr and
Z
3
(r + 1) dr =
Z
w3 dw =
1
1 4
w + C = (r + 1)4 + C.
4
4
12. We use the substitution w = y 2 + 5, dw = 2y dy.
Z
1
2
y(y 2 + 5)8 dy =
1
=
2
Z
Z
(y 2 + 5)8 (2y dy)
w8 dw =
=
1 w9
+C
2 9
1 2
(y + 5)9 + C.
18
d 1 2
1
( (y + 5)9 + C) =
[9(y 2 + 5)8 (2y)] = y(y 2 + 5)8 .
dy 18
18
13. We use the substitution w = 1 + 2x3 , dw = 6x2 dx.
Check:
Z
x2 (1 + 2x3 )2 dx =
Z
1
1 w3
1
w2 ( dw) = ( ) + C =
(1 + 2x3 )3 + C.
6
6 3
18
1
1
d
(1 + 2x2 )3 + C =
[3(1 + 2x3 )2 (6x2 )] = x2 (1 + 2x3 )2 .
dx 18
18
14. We use the substitution w = t3 − 3, dw = 3t2 dt.
h
Check:
i
Z
t2 (t3 − 3)10 dt =
=
1
3
Z
(t3 − 3)10 (3t2 dt) =
Z
w10
1 3
1 w11
+C =
(t − 3)11 + C.
3 11
33
1
dw
3
d 1 3
1
[ (t − 3)11 + C] = (t3 − 3)10 (3t2 ) = t2 (t3 − 3)10 .
dt 33
3
15. We use the substitution w = x2 + 3, dw = 2x dx.
Check:
Z
x(x2 + 3)2 dx =
Z
1 w3
1
1
+ C = (x2 + 3)3 + C.
w2 ( dw) =
2
2 3
6
d 1 2
1 2
(x + 3)3 + C =
3(x + 3)2 (2x) = x(x2 + 3)2 .
dx 6
6
16. We use the substitution w = x2 − 4, dw = 2x dx.
h
Check:
i
Z
Check:
d
dx
1 2
(x − 4)9/2 + C
9
=
1
9
Z
Z
1
1
(x2 − 4)7/2 (2xdx) =
w7/2 dw
2
2
1 2 9/2
1
=
w
+ C = (x2 − 4)9/2 + C.
2 9
9
x(x2 − 4)7/2 dx =
9 2
(x − 4)7/2 2x = x(x2 − 4)7/2 .
2
7.1 SOLUTIONS
17. In this case, it seems easier not to substitute.
Z
y 2 (1 + y)2 dy =
=
Check:
d
dy
y5
y4
y3
+
+
+C
5
2
3
Z
y 2 (y 2 + 2y + 1) dy =
y5
y4
y3
+
+
+ C.
5
2
3
Z
(y 4 + 2y 3 + y 2 ) dy
= y 4 + 2y 3 + y 2 = y 2 (y + 1)2 .
18. We use the substitution w = 2t − 7, dw = 2 dt.
Z
(2t − 7)
73
1
dt =
2
Z
w73 dw =
1
1
w74 + C =
(2t − 7)74 + C.
(2)(74)
148
1
74
d
(2t − 7)74 + C =
(2t − 7)73 (2) = (2t − 7)73 .
dt 148
148
19. We use the substitution w = x3 + 1, dw = 3x2 dx, to get
Check:
h
i
Z
x 2 ex
3
+1
dx =
1
3
Z
ew dw =
1 w
1 3
e + C = ex +1 + C.
3
3
3
d 1 x3 +1
1 3
e
+ C = ex +1 · 3x2 = x2 ex +1 .
dx 3
3
20. We use the substitution w = y + 5, dw = dy, to get
Check:
Z
dy
=
y+5
Z
dw
= ln |w| + C = ln |y + 5| + C.
w
1
d
(ln |y + 5| + C) =
.
dy
y+5
21. We use the substitution w = 4 − x, dw = −dx.
Check:
√
√
1
1
dx = −
√ dw = −2 w + C = −2 4 − x + C.
w
4−x
√
1
1
d
1
· −1 = √
.
Check:
(−2 4 − x + C) = −2 · · √
dx
2
4−x
4−x
22. In this case, it seems easier not to substitute.
Z
d
Check:
dx
Z
Z
√
(x2 + 3)2 dx =
Z
(x4 + 6x2 + 9) dx =
x5
+ 2x3 + 9x + C.
5
x5
+ 2x3 + 9x + C = x4 + 6x2 + 9 = (x2 + 3)2 .
5
23. We use the substitution w = cos θ + 5, dw = − sin θ dθ.
Z
sin θ(cos θ + 5)7 dθ = −
Z
1
w7 dw = − w8 + C
8
1
= − (cos θ + 5)8 + C.
8
Check:
1
1
d
− (cos θ + 5)8 + C = − · 8(cos θ + 5)7 · (− sin θ)
dθ
8
8
= sin θ(cos θ + 5)7
h
i
569
570
Chapter Seven /SOLUTIONS
24. We use the substitution w = cos 3t, dw = −3 sin 3t dt.
√
Z
Z
√
1
w dw
cos 3t sin 3t dt = −
3
3
1 2 3
2
= − · w 2 + C = − (cos 3t) 2 + C.
3 3
9
Check:
3
1
2
2 3
d
− (cos 3t) 2 + C = − · (cos 3t) 2 · (− sin 3t) · 3
dt
9
9 2
√
= cos 3t sin 3t.
h
i
25. We use the substitution w = sin θ, dw = cos θ dθ.
d
Check:
dθ
Z
Z
sin6 θ cos θ dθ =
sin7 θ
+ C = sin6 θ cos θ.
7
w6 dw =
sin7 θ
w7
+C =
+ C.
7
7
w3 dw =
sin4 α
w4
+C =
+ C.
4
4
26. We use the substitution w = sin α, dw = cos α dα.
d
Check:
dα
sin4 α
+C
4
Z
=
sin3 α cos α dα =
Z
1
· 4 sin3 α · cos α = sin3 α cos α.
4
27. We use the substitution w = sin 5θ, dw = 5 cos 5θ dθ.
Z
sin6 5θ cos 5θ dθ =
1
5
Z
w6 dw =
1 w7
1
(
)+C =
sin7 5θ + C.
5 7
35
1
d 1
( sin7 5θ + C) =
[7 sin6 5θ](5 cos 5θ) = sin6 5θ cos 5θ.
Check:
dθ 35
35
Note that we could also use Problem 25 to solve this problem, substituting w = 5θ and dw = 5 dθ to get:
Z
1
sin 5θ cos 5θ dθ =
5
6
=
Z
sin6 w cos w dw
1 sin7 w
1
(
)+C =
sin7 5θ + C.
5
7
35
28. We use the substitution w = cos 2x, dw = −2 sin 2x dx.
Z
Z
Z
sin 2x
1
dw
dx = −
cos 2x
2
w
1
1
= − ln |w| + C = − ln | cos 2x| + C.
2
2
tan 2x dx =
Check:
1
1
1
d
· −2 sin 2x
− ln | cos 2x| + C = − ·
dx
2
2 cos 2x
sin 2x
=
= tan 2x.
cos 2x
h
29. We use the substitution w = ln z, dw =
Check:
d
dz
Z
i
1
z
dz.
(ln z)2
dz =
z
Z
w2 dw =
1
(ln z)2
(ln z)3
1
+ C = 3 · (ln z)2 · =
.
3
3
z
z
w3
(ln z)3
+C =
+ C.
3
3
7.1 SOLUTIONS
30. We use the substitution w = et + t, dw = (et + 1) dt.
Z
et + 1
dt =
et + t
d
et + 1
(ln |et + t| + C) = t
.
dt
e +t
31. It seems easier not to substitute.
Z
1
dw = ln |w| + C = ln |et + t| + C.
w
Check:
Z
(t + 1)2
dt =
t2
(t2 + 2t + 1)
dt
t2
Z
Z
=
1+
2
1
+ 2
t
t
(t + 1)2
1
2
1
d
(t + 2 ln |t| − + C) = 1 + + 2 =
.
dt
t
t
t
t2
32. We use the substitution w = y 2 + 4, dw = 2y dy.
dt = t + 2 ln |t| −
1
+ C.
t
Check:
Z
1
y
dy =
y2 + 4
2
Z
dw
1
1
= ln |w| + C = ln(y 2 + 4) + C.
w
2
2
(We can drop
the absolute valueisigns since y 2 + 4 ≥ 0 for all y.)
h
1
1
y
d 1
ln(y 2 + 4) + C = · 2
· 2y = 2
.
Check:
dy 2
2 y +4
y +4
√
√
d
1
arctan w =
, we have
33. Let w = 2x, so dw = 2dx. Then, since
dw
1 + w2
Z
dx
1
= √
1 + 2x2
2
Z
√
dw
1
1
= √ arctan w + C = √ arctan( 2x) + C.
1 + w2
2
2
34. Let w = 2x, then dw = 2 dx so that
Z
√
1
dx
=
2
1 − 4x2
Z
√
dw
1
1
= arcsin w + C = arcsin(2x) + C.
2
2
1 − w2
√
1
x, dw = 2√
dx.
x
Z
Z
√
√
cos x
√
dx =
cos w(2 dw) = 2 sin w + C = 2 sin x + C.
x
√
√
√
1
d
cos x
√
Check:
(2 sin x + C) = 2 cos x
= √ .
dx
2 x
x
√
1
36. We use the substitution w = y, dw = √ dy.
2 y
35. We use the substitution w =
Z
Check:
√
e y
√ dy = 2
y
√
√
√
e y
1
d
(2e y + C) = 2e y · √ = √ .
dy
2 y
y
Z
ew dw = 2ew + C = 2e
√
y
+ C.
37. We use the substitution w = x + ex , dw = (1 + ex ) dx.
Z
Check:
1 + ex
√
dx =
x + ex
Z
√
√
dw
√ = 2 w + C = 2 x + ex + C.
w
1
1
1 + ex
d √
(2 x + ex + C) = 2 · (x + ex )− 2 · (1 + ex ) = √
.
dx
2
x + ex
571
572
Chapter Seven /SOLUTIONS
38. We use the substitution w = 2 + ex , dw = ex dx.
dw
ex
dx =
= ln |w| + C = ln(2 + ex ) + C.
2 + ex
w
(We can drop the absolute value signs since 2 + ex ≥ 0 for all x.)
1
ex
d
[ln(2 + ex ) + C] =
· ex =
.
Check:
dx
2 + ex
2 + ex
39. We use the substitution w = x2 + 2x + 19, dw = 2(x + 1)dx.
Z
Z
Z
(x + 1)dx
1
=
x2 + 2x + 19
2
Z
1
1
dw
= ln |w| + C = ln(x2 + 2x + 19) + C.
w
2
2
(We can drop the absolute value signs, since x2 + 2x + 19 = (x + 1)2 + 18 > 0 for all x.)
1 1
1
1
x+1
Check:
[ ln(x2 + 2x + 19)] =
(2x + 2) = 2
.
dx 2
2 x2 + 2x + 19
x + 2x + 19
40. We use the substitution w = 1 + 3t2 , dw = 6t dt.
Z
Z
t
dt =
1 + 3t2
1
1
1 1
( dw) = ln |w| + C = ln(1 + 3t2 ) + C.
w 6
6
6
(We can drop
1 + 3t2 > 0 for all t).
h the absolute valueisigns1 since
d 1
1
t
Check:
ln(1 + 3t2 ) + C =
(6t) =
.
dt 6
6 1 + 3t2
1 + 3t2
x
−x
x
−x
41. We use the substitution w = e + e , dw = (e − e ) dx.
Z
ex − e−x
dx =
ex + e−x
Z
dw
= ln |w| + C = ln(ex + e−x ) + C.
w
(We can drop the absolute value signs since ex + e−x > 0 for all x).
1
d
[ln(ex + e−x ) + C] = x
(ex − e−x ).
Check:
dx
e + e−x
42. We use the substitution w = sin(x2 ), dw = 2x cos(x2 ) dx.
Z
Check:
x cos(x2 )
1
p
dx =
2
2
sin(x )
Z
1
w− 2 dw =
p
1
1
(2w 2 ) + C = sin(x2 ) + C.
2
x cos(x2 )
1
d p
[cos(x2 )]2x = p
.
( sin(x2 ) + C) = p
2
dx
2 sin(x )
sin(x2 )
43. Since d(cosh 3t)/dt = 3 sinh 3t, we have
Z
sinh 3t dt =
1
cosh 3t + C.
3
44. Since d(sinh x)/dx = cosh x, we have
Z
cosh x dx = sinh x + C.
45. Since d(sinh(2w + 1))/dw = 2 cosh(2w + 1), we have
Z
cosh(2w + 1) dw =
1
sinh(2w + 1) + C.
2
46. Since d(cosh z)/dz = sinh z, the chain rule shows that
d cosh z
(e
) = (sinh z)ecosh z .
dz
Thus,
Z
(sinh z)ecosh z dz = ecosh z + C.
7.1 SOLUTIONS
47. Use the substitution w = cosh x and dw = sinh x dx so
Z
cosh2 x sinh x dx =
Z
w2 dw =
1
1 3
w + C = cosh3 x + C.
3
3
d 1
cosh3 x + C = cosh2 x sinh x.
dx 3
48. We use the substitution w = x2 and dw = 2xdx so
h
Check this answer by taking the derivative:
Z
1
x cosh x dx =
2
2
Z
i
cosh w dw =
1
1
sinh w + C = sinh x2 + C.
2
2
d 1
sinh x2 + C = x cosh x2 .
dx 2
R
49. The general antiderivative is (πt3 + 4t) dt = (π/4)t4 + 2t2 + C.
h
Check this answer by taking the derivative:
i
50. Z
Make the substitution
Z w = 3x, dw = 3 dx. We have
1
1
1
sin w dw = (− cos w) + C = − cos 3x + C.
sin 3x dx =
3
3
3
51. Make
the substitutionZw = x2 , dw = 2x dx. We have
Z
2x cos(x2 ) dx =
cos w dw = sin w + C = sin x2 + C.
52. Make the substitution w = t3 , dw = 3t2 dt. The general antiderivative is
R
12t2 cos(t3 ) dt = 4 sin(t3 ) + C.
53. Z
Make the substitution Zw = 2 − 5x, then dw = −5dx. We have
1
1
1
sin w −
dw = − (− cos w) + C = cos(2 − 5x) + C.
sin(2 − 5x)dx =
5
5
5
54. Make
the substitutionZw = sin x, dw = cos x dx. We have
Z
esin x cos x dx =
ew dw = ew + C = esin x + C.
2
55. Make
the substitution
Z w = x + 1, dw = 2x dx. We have
Z
1
dw
1
1
x
dx =
= ln |w| + C = ln(x2 + 1) + C.
x2 + 1
2
w
2
2
(Notice that since x2 + 1 ≥ 0, |x2 + 1| = x2 + 1.)
56. Make the substitution w = 2x, then dw = 2dx. We have
Z
1
1
dx =
3 cos2 2x
3
=
57.
Z
π
π
cos(x + π) dx = sin(x + π)
Z
1
cos2 w
1
2
dw
1
1
1
dw = tan w + C = tan 2x + C.
cos2 w
6
6
= sin(2π) − sin(π) = 0 − 0 = 0
0
0
1
6
Z
58. We substitute w = πx. Then dw = π dx.
Z
x= 1
2
Z
cos πx dx =
x=0
59.
Z
60.
e− cos θ sin θ dθ = e− cos θ
= e− cos(π/2) − e− cos(0) = 1 −
0
1
w=0
1
1
dw) = (sin w)
π
π
π/2
=
0
1
π
1
e
2
2
61. Let
cos w(
π/2
π/2
0
Z
w=π/2
2xe
√
x2
dx = e
x2
=e
1
22
−e
1
x = w, 21 x− 2 dx = dw,
12
dx
√
x
Z
1
4
= e4 − e = e(e3 − 1)
= 2 dw. If x = 1 then w = 1, and if x = 4 so w = 2. So we have
√
e x
√ dx =
x
Z
1
2
ew · 2 dw = 2ew
2
1
= 2(e2 − e) ≈ 9.34.
573
574
Chapter Seven /SOLUTIONS
62. We substitute w = t + 2, so dw = dt.
Z
t=e−2
1
dt =
t+2
t=−1
63. We substitute w =
√
w=e
w=1
dw
= ln |w|
w
e
1
= ln e − ln 1 = 1.
1 −1/2
x
dx.
2
x. Then dw =
Z
Z
Z w=2
√
cos x
√
dx =
cos w(2 dw)
x
w=1
x=4
x=1
2
= 2(sin w)
1
= 2(sin 2 − sin 1).
64. We substitute w = 1 + x2 . Then dw = 2x dx.
x=2
Z
x
dx =
(1 + x2 )2
x=0
65.
Z
3
(x3 + 5x) dx =
−1
66.
Z
67.
Z
1
−1
3
1
68.
Z
3
1
69.
Z
2
−1
4 3
x
4
+
−1
1
1
dy = tan−1 y
1 + y2
1
dx = ln x
x
=
−1
w=5
w=1
1
w2
2 3
1
1
dw = −
2
2
1
w
5
=
1
2
.
5
= 40.
−1
π
.
2
3
= ln 3.
1
−1
dt
=
(t + 7)2
t+7
√
5x
2
Z
3
1
= −
2
x + 2 dx = (x + 2)3/2
3
1
10
2
1
− −
=
−1
8
=
1
40
2
14
2 3/2
(4)
− (1)3/2 = (7) =
3
3
3
sin x
70. It turns out that
cannot be integrated using elementary methods. However, the function is decreasing on [1,2]. One
x
way to see this is to graph the function on a calculator or computer, as has been done below:
y
1
0.75
0.5
0.25
−0.25
4
1
2
x
3
sin 2
sin 1
−
∆t ≈
2
1
0.61∆t. So with 13 intervals, our error will be less than 0.05. With n = 13, the left sum is about 0.674, and the right sum
is about 0.644. For more accurate sums, with n = 100 the left sum is about 0.6613 and the right sum is about 0.6574. The
actual integral is about 0.6593.
√
71. Let w = y + 1, so y = w2 − 1 and dy = 2w dw. Thus
So since our function is monotonic, the error for our left- and right-hand sums is less than or equal to
Z
y
p
y + 1 dy =
=
Z
(w2 − 1)w2w dw = 2
Z
w4 − w2 dw
2
2
2 5 2 3
w − w + C = (y + 1)5/2 − (y + 1)3/2 + C.
5
3
5
3
7.1 SOLUTIONS
575
72. Let w = (z + 1)1/3 , so z = w3 − 1 and dz = 3w2 dw. Thus
Z
z(z + 1)
1/3
Z
dz =
√
2
(w − 1)w3w dw = 3
Z
w6 − w3 dw
3 7 3 4
3
3
w − w + C = (z + 1)7/3 − (z + 1)4/3 + C.
7
4
7
4
=
73. Let w =
3
t + 1, so t = w2 − 1 and dt = 2w dw. Thus
(w2 − 1)2 + (w2 − 1)
2w dw = 2 w4 − w2 dw
w
2
2
2
2
= w5 − w3 + C = (t + 1)5/2 − (t + 1)3/2 + C.
5
3
5
3
t2 + t
√
dt =
t+1
Z
Z
Z
√
74. Let w = 2 + 2 x, so x = ((w − 2)/2)2 = (w/2 − 1)2 , and dx = 2(w/2 − 1)(1/2) dw = (w/2 − 1) dw. Thus
Z
Z
Z
(w/2 − 1) dw
1
1
=
−
dw
w
2
w
√
√
1
w
− ln |w| + C = (2 + 2 x) − ln |2 + 2 x| + C
=
2
2
√
√
√
√
= 1 + x − ln |2 + 2 x| + C = x − ln |2 + 2 x| + C.
dx
√ =
2+2 x
In the last line, the 1 has been combined with the C.
√
75. Let w = x − 2, so x = w2 + 2 and dx = 2w dw. Thus
Z
x
2
√
x − 2 dx =
Z
2
2
(w + 2) w2w dw = 2
2 7 8 5
w + w +
7
5
2
= (x − 2)7/2 +
7
=
76. Let w =
√
√
(z + 2) 1 − z dz =
=
√
√
8 3
w +C
3
8
8
(x − 2)5/2 + (x − 2)3/2 + C.
5
3
Z
2
(1 − w + 2)w(−2w) dw = 2
Z
w4 − 3w2 dw
2
2 5
w − 2w3 + C = (1 − z)5/2 − 2(1 − z)3/2 + C.
5
5
t + 1, so t = w2 − 1 and dt = 2w dw. Thus
Z
78. Let w =
w6 + 4w4 + 4w2 dw
1 − z, so z = 1 − w2 and dz = −2w dw. Thus
Z
77. Let w =
Z
w2 − 1
2w dw = 2 w2 − 1 dw
w
2
2
= w3 − 2w + C = (t + 1)3/2 − 2(t + 1)1/2 + C.
3
3
t
dt =
√
t+1
Z
Z
2x + 1, so x = 21 (w2 − 1) and dx = w dw. Thus
Z
3 · 21 (w2 − 1) − 2
3 2 7
w dw =
w − dw
w
2
2
1 3 7
1
7
3/2
= w − w + C = (2x + 1)
− (2x + 1)1/2 + C.
2
2
2
2
3x − 2
dx =
√
2x + 1
Z
Z
Problems
79. If we let y = 3x in the first integral, we get dy = 3dx. Also, the limits x = 0 and x = π/3 become y = 0 and
y = 3(π/3) = π. Thus
Z
0
π/3
3 sin2 (3x) dx =
Z
0
π/3
sin2 (3x) 3dx =
Z
π
sin2 (y) dy.
0
576
Chapter Seven /SOLUTIONS
80. If we let t = s2 in the first integral, we get dt = 2s ds, so
1
1
dt = √ dt.
s
t
2 ds =
Also, the limits s = 1 and s = 2 become t = 1 and t = 4. Thus
Z
2
2
2 ln(s + 1) ds =
1
2
Z
Z
2
ln(s + 1)2 ds =
1
4
1
1
ln(t + 1) √ dt.
t
81. If we let w = ez in the first integral, we get dw = ez dz and z = ln w. Also, the limits w = 1 and w = e become z = 0
and z = 1. Thus
Z
Z
1
e
(ln w)3 dw =
z 3 ez dz.
0
1
82. If we let t = π − x in the first integral, we get dt = −dx and x = π − t. Also, the limits x = 0 and x = π become t = π
and t = 0. Thus
Z
Z
Z
0
π
0
83. As x goes from
√
a to
√
x cos(π − x) dx = −
π
(π − t) cos t dt =
π
0
(π − t) cos t dt.
b the values of w = x2 increase from a to b. Since x =
√
Z
√
b
dx =
a
b
Z
a
with
Z
1
√ dw =
2 w
√
√
w we have dx = dw/(2 w). Hence
b
g(w)dw
a
1
g(w) = √ .
2 w
84. As x goes from a to b the values of w = ex increase from ea to eb . We have e−x = 1/w. Since x = ln w we have
dx = dw/w. Hence
Z
b
e
−x
Z
dx =
ea
a
with
eb
Z
1 dw
=
w w
g(w) =
eb
g(w)dw
ea
1
.
w2
85. After the substitution w = ex and dw = ex dx, the first integral becomes
1
2
Z
1
dw,
1 + w2
while after the substitution w = sin x and dw = cos x dx, the second integral becomes
Z
86. R
The substitution w = ln x, dw =
x dx.
1
x
1
dw.
1 + w2
dx transforms the first integral into
87. For the first integral, let w = sin x, dw = cos x dx. Then
Z
e
sin x
For the second integral, let w = arcsin x, dw = √
Z
cos x dx =
1
1−x2
Z
R
w dw, which is just a respelling of the integral
ew dw.
dx. Then
earcsin x
√
dx =
1 − x2
Z
ew dw.
7.1 SOLUTIONS
88. The substitutions w = sin x, dw = cos w and w = x3 + 1, dw = 3x2 dx transform the integrals into
Z
w3 dw
Z
1
3
and
w3 dw.
89. For the first integral, let w = x + 1, dw = dx. Then
For the second integral, let w = 1 +
√
Z
√
x + 1 dx =
1
x, dw = 12 x− 2 dx =
Z
√
1
√
2 x
w dw.
dx. Then,
dx
√
x
= 2 dw, and
Z p
Z
√
√
√
1+ x
dx
dx =
=2
√
1+ x √
w dw.
x
x
Z p
90. Using substitution, we can show
1
1
Z
f ′ (x) sin f (x) dx = − cos f (x)
0
0
= − cos f (1) + cos f (0) = cos 5 − cos 7.
91. Using substitution, we have
Z
3
3
f ′ (x)ef (x) dx = ef (x)
1
1
= ef (3) − ef (1) = e11 − e7 .
92. Using substitution, we have
3
Z
1
f ′ (x)
dx = ln |f (x)|
f (x)
3
1
= ln |f (3)| − ln |f (1)| = ln 11 − ln 7 = ln(11/7).
93. Using substitution, we have
1
1
Z
ex f ′ (ex ) dx = f (ex )
0
0
= f (e1 ) − f (e0 ) = f (e) − f (1) = 10 − 7 = 3.
94. Using substitution, we have
e
Z
1
f ′ (ln x)
dx = f (ln x)
x
e
1
= f (ln e) − f (ln 1) = f (1) − f (0) = 7 − 5 = 2.
95. Using substitution, we can show
Z
1
f ′ (x)(f (x))2 dx =
0
(f (x))3
3
1
=
0
(f (1))3
(f (0))3
73
53
218
−
=
−
=
.
3
3
3
3
3
96. Using substitution, we have
Z
π/2
π/2
0
sin x · f ′ (cos x) dx = −f (cos x)
0
= −f cos
π
2
+ f (cos 0) = −f (0) + f (1) = −5 + 7 = 2.
97. By substitution with w = g(x), we have
Z
g ′ (x)(g(x))4 dx =
Z
w4 dw =
(g(x))5
w5
+C =
+ C.
5
5
577
578
Chapter Seven /SOLUTIONS
98. By substitution with w = g(x), we have
Z
g ′ (x)eg(x) dx =
Z
ew dw = ew + C = eg(x) + C.
99. By substitution with w = g(x), we have
Z
Z
′
g (x) sin g(x) dx =
sin w dw = − cos w + C = − cos g(x) + C.
100. By substitution with w = 1 + g(x), we have
Z
g ′ (x)
p
1 + g(x) dx =
Z
√
Z
w dw =
w3/2
2
+ C = (1 + g(x))3/2 + C.
3/2
3
w1/2 dw =
101. Letting w = 1 − 4x3 , dw = −12x2 dx, we have
Z
x2
p
1 − 4x3 dx =
Z
(1 − 4x3 )1/2 −
so w = 1 − 4x3 , k = −1/12, n = 1/2.
| {z }
w
1
(−12x2 dx) =
12 | {z }
dw
Z
−
1
w1/2 dw,
12
102. Letting w = sin t, dw = cos t dt, we have
Z
cos t
dt =
sin t
Z
Z
−1
(sin
|{z}t ) |cos{zt dt} =
w
so w = sin t, k = 1, n = −1.
dw
1 · w−1 dw,
103. Letting w = x2 − 3, dw = 2x dx, we have
dw
Z
z }| {
2x dx
=
(x2 − 3)2
| {z }
Z
1 · w−2 dw,
w
so w = x2 − 3, k = 1, n = −2.
104. Since w = 5x2 + 7 we have dw = 10x dx so x dx = 0.1 dw. Finding the limits of integration in terms of w gives:
w0 = 5(1)2 + 7 = 12
w1 = 5(5)2 + 7 = 132.
This gives
Z
1
5
√
3x dx
=
5x2 + 7
132
Z
12
Z
3(0.1 dw)
=
w1/2
132
0.3w−1/2 dw,
12
so k = 0.3, n = −1/2, w0 = 12, w1 = 132.
105. Since w = 2x + 3 we have dw = 2x ln 2 dx, so 2x dx = dw/ ln 2. Finding the limits of integration in terms of w gives:
w0 = 20 + 3 = 4
w1 = 25 + 3 = 35.
This gives
Z
0
5
2x dx
=
2x + 3
so k = 1/ ln 2, n = −1, w0 = 4, w1 = 35.
Z
4
35
dw/ ln 2
=
2x + 3
Z
4
35
1 −1
w dw,
ln 2
7.1 SOLUTIONS
579
106. Since w = sin 2x we have dw = 2 cos 2x dx, so cos 2x dx = 0.5 dw. Finding the limits of integration in terms of w
gives:
π
π
= sin
= 0.5
12
6
π
π
= sin
= 1.
w1 = sin 2 ·
4
2
w0 = sin 2 ·
This gives
Z
π/4
sin7 (2x) cos(2x) dx =
so w = −x2 , k = −1/2.
1
0.5w7 dw,
0.5
π/12
so k = 0.5, n = 7, w0 = 0.5, w1 = 1.
107. Letting w = −x2 , dw = −2x dx, we have
Z
Z
xe
−x2
dx =
Z
−
1 w
e dw,
2
Z
1 · ew dw,
108. Letting w = sin φ, dw = cos φ, we have
Z
so w = sin φ, k = 1.
esin φ cos φdφ =
109. We have
Z
√
er dr =
=
Letting w = 0.5r, dw = 0.5 dr, and dr = 2 dw, we have
so w = 0.5r, k = 2.
110. We have
Letting w = z 3 , so dw = 3z 2 dz, we have
so w = z 3 , k = 1/3.
Z
(er )1/2 dr
Z
e0.5r dr.
Z
√
Z
z 2 dz
=
e−z 3
Z
Z
z 2 dz
=
e−z 3
Z
er dr =
Z
2ew dw,
3
ez z 2 dz.
1 w
e dw,
3
111. We have
Z
2t 3t−4
e e
dt =
=
=
so w = 5t and k = 1/(5e4 ).
112. Let w = 2t − 3, dw = 2 dt. Then
Z
Z
Z
Z
e−4 e5t dt
1
e−4 · ew dw
5
1
· ew dw,
5e4
7
e2t−3 dt =
3
so a = 3, b = 11, A = 0.5, w = 2t − 3.
Z
where w = 5t, dw = 5 dt,
t=7
t=3
ew 0.5 dw =
Z
w=11
0.5ew dw,
w=3
580
Chapter Seven /SOLUTIONS
113. Let w = cos(πt), dw = −π sin(πt) dt. Then sin(πt) dt = −(1/π) dw, giving
Z
1
ecos(πt) sin(πt) dt =
0
Z
t=1
ew
t=0
−1
π
so a = 1, b = −1, A = −1/π, w = cos(πt). Note that since
Z
−1
1
−1 w
e dw = −
π
Z
1
−1
dw =
Z
w=−1
Z
1
w=1
−1 w
e dw =
π
−1
−1 w
e dw,
π
1 w
e dw,
π
another possible answer is a = −1, b = 1, A = 1/π, w = cos(πt).
√
114. (a) 2√x + C
(b) 2 x + 1 + C
√
(c) To get this last result, we make the substitution w = x. Normally we would like to substitute dw = 2√1 x dx, but in
this case we cannot since there are no spare √1x terms around. Instead, we note w2 = x, so 2w dw = dx. Then
Z
√
1
dx =
x+1
Z
=2
=2
2w
dw
w+1
Z
(w + 1) − 1
dw
w+1
Z
1−
1
w+1
dw
= 2(w − ln |w + 1|) + C
√
√
= 2 x − 2 ln( x + 1) + C.
√
We also note that we can drop the absolute value signs, since x + 1 ≥ 0 for all x.
115. (a) This integral can be evaluated using integration by substitution. We use w = x2 , dw = 2xdx.
Z
x sin x2 dx =
1
2
Z
1
1
sin(w)dw = − cos(w) + C = − cos(x2 ) + C.
2
2
(b) This integral cannot be evaluated using a simple integration by substitution.
(c) This integral cannot be evaluated using a simple integration by substitution.
(d) This integral can be evaluated using integration by substitution. We use w = 1 + x2 , dw = 2xdx.
Z
1
x
dx =
(1 + x2 )2
2
Z
1
1 −1
−1
dw = (
)+C =
+ C.
w2
2 w
2(1 + x2 )
(e) This integral cannot be evaluated using a simple integration by substitution.
(f) This integral can be evaluated using integration by substitution. We use w = 2 + cos x, dw = − sin xdx.
Z
sin x
dx = −
2 + cos x
Z
1
dw = − ln |w| + C = − ln |2 + cos x| + C.
w
2
116. To find the area under the graph of f (x) = xex , we need to evaluate the definite integral
Z
2
2
xex dx.
0
This is done in Example 9, Section 7.1, using the substitution w = x2 , the result being
Z
2
2
xex dx =
0
1 4
(e − 1).
2
117. Since f (x) = 1/(x + 1) is positive on the interval x = 0 to x = 2, we have
Area =
Z
0
The area is ln 3 ≈ 1.0986.
2
1
dx = ln(x + 1)
x+1
2
0
= ln 3 − ln 1 = ln 3.
7.1 SOLUTIONS
118. The area under f (x) = sinh(x/2) between x = 0 and x = 2 is given by
A=
2
Z
x
sinh
dx = 2 cosh
2
0
2
x
2
0
= 2 cosh 1 − 2.
119. Since (eθ+1 )3 = e3θ+3 = e3θ · e3 , we have
2
Z
Area =
(eθ+1 )3 dθ =
=e
e3θ · e3 dθ
0
0
Z
3
2
Z
2
e
3θ
dθ = e
2
3 1 3θ
0
3
e
=
e3 6
(e − 1).
3
0
120. Since et+1 is larger than et , we have
Area =
Z
2
2
(e
t+1
0
(The integral
R
t
− e ) dt = (e
t+1
t
= e3 − e2 − e + 1.
−e )
0
et+1 dt = et+1 + C can be done by substitution or guess and check.)
121. The curves y = ex and y = 3 cross where
ex = 3
so
x = ln 3,
x
and the graph of y = e is below the line y = 3 for 0 ≤ x ≤ ln 3. (See Figure 7.1.) Thus
Area =
Z
ln 3
ln 3
(3 − ex ) dx = (3x − ex )
0
0
= 3 ln 3 − eln 3 − (0 − 1) = 3 ln 3 − 2.
y
y = ex
y=3
3
x
ln 3
Figure 7.1
122. See Figure 7.2. The period of V = V0 sin(ωt) is 2π/ω, so the area under the first arch is given by
Area =
Z
π/ω
V0 sin(ωt) dt
0
=−
V0
cos(ωt)
ω
π/ω
0
V0
V0
cos(π) +
cos(0)
=−
ω
ω
V0
V0
2V0
= − (−1) +
(1) =
.
ω
ω
ω
581
582
Chapter Seven /SOLUTIONS
V
V0
2π
ω
t
π
ω
V0 sin(ωt)
Figure 7.2
123. If f (x) =
1
, the average value of f on the interval 0 ≤ x ≤ 2 is defined to be
x+1
2
Z
1
2−0
f (x) dx =
0
1
2
Z
2
dx
.
x+1
0
We’ll integrate by substitution. We let w = x + 1 and dw = dx, and we have
x=2
Z
x=0
dx
=
x+1
w=3
Z
w=1
Thus, the average value of f (x) on 0 ≤ x ≤ 2 is
dw
= ln w
w
1
2
3
= ln 3 − ln 1 = ln 3.
1
ln 3 ≈ 0.5493. See Figure 7.3.
f (x) =
1
1+x
0.54931
x
2
Figure 7.3
124. On the interval [0, 2b]
Average value of f =
If we let x = 2u in this integral, we get
1
2b
Z
b
f (2u) 2du =
0
1
b
1
2b
Z
2b
f (x) dx.
0
b
Z
g(u) du = Average value of g on [0, b].
0
125. (a) If w = t/2, then dw = (1/2)dt. When t = 0, w = 0; when t = 4, w = 2. Thus,
Z
4
g(t/2) dt =
Z
2
g(w) 2dw = 2
0
0
Z
2
0
g(w) dw = 2 · 5 = 10.
(b) If w = 2 − t, then dw = −dt. When t = 0, w = 2; when t = 2, w = 0. Thus,
Z
0
2
g(2 − t) dt =
Z
2
0
g(w) (−dw) = +
Z
0
2
g(w) dw = 5.
7.1 SOLUTIONS
583
126. (a) If w = 2t, then dw = 2dt. When t = 0, w = 0; when t = 0.5, w = 1. Thus,
0.5
Z
1
Z
f (2t) dt =
0
0
1
1
f (w) dw =
2
2
Z
1
f (w) dw =
0
3
.
2
(b) If w = 1 − t, then dw = −dt. When t = 0, w = 1; when t = 1, w = 0. Thus,
Z
1
0
0
Z
f (1 − t) dt =
f (w) (−dw) = +
1
Z
1
f (w) dw = 3.
0
(c) If w = 3 − 2t, then dw = −2dt. When t = 1, w = 1; when t = 1.5, w = 0. Thus,
Z
1.5
f (3 − 2t) dt =
1
Z
0
f (w)
1
1
1
− dw = +
2
2
Z
1
f (w) dw =
0
3
.
2
127. (a) The Fundamental Theorem gives
Z
π
−π
cos2 θ sin θ dθ = −
cos3 θ
3
π
−(−1)3
−(−1)3
−
= 0.
3
3
=
−π
2
This agrees with the fact that the function f (θ) = cos θ sin θ is odd and the interval of integration is centered at
x = 0, thus we must get 0 for the definite integral.
(b) The area is given by
Area =
Z
π
cos2 θ sin θ dθ = −
0
128. (a)
Z
4x(x2 + 1) dx =
Z
cos3 θ
3
π
=
0
−(−1)3
−(1)3
2
−
= .
3
3
3
(4x3 + 4x) dx = x4 + 2x2 + C.
(b) If w = x2 + 1, then dw = 2x dx.
Z
4x(x2 + 1) dx =
Z
2w dw = w2 + C = (x2 + 1)2 + C.
(c) The expressions from parts (a) and (b) look different, but they are both correct. Note that (x2 + 1)2 + C = x4 +
2x2 + 1 + C. In other words, the expressions from parts (a) and (b) differ only by a constant, so they are both correct
antiderivatives.
129. (a) Z
We first try the substitution
w = sin θ, dw = cos θ dθ. Then
Z
sin2 θ
w2
+C =
+ C.
w dw =
sin θ cos θ dθ =
2
2
(b) If
w = cos θ, dw = − sin θ dθ, we get
Z
Z we instead try the substitution
cos2 θ
w2
+C = −
+ C.
sin θ cos θ dθ = − w dw = −
2
2
(c) Z
Once we note that sin 2θ
Z = 2 sin θ cos θ, we can also say
1
sin θ cos θ dθ =
sin 2θ dθ.
2
Substituting
w = 2θ, dw = 2 dθ, the above equals
Z
cos 2θ
1
cos w
+C =−
+ C.
sin w dw = −
4
4
4
(d) All these answers are correct. Although they have different forms, they differ from each other only in terms of a
2
2
constant, and thus they are all acceptable antiderivatives. For example, 1 − cos2 θ = sin2 θ, so sin2 θ = − cos2 θ + 21 .
Thus the first two expressions differ only by a constant C.
2
Similarly, cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1, so − cos42θ = − cos2 θ + 14 , and thus the second and third
expressions differ only by a constant. Of course, if the first two expressions and the last two expressions differ only
in the constant C, then the first and last only differ in the constant as well.
√
√
√
130. Letting w = 6x x = 6x3/2 , we have dw = (3/2)6x1/2 dx = 9 xdx, so xdx = (1/9)dw. This means we can write
f (w)
Z
1
9
z
(1/9) dw
}|√ { z√}| {
f 6x x
xdx =
Z
x=9
x=1
f (w) ·
1
dw
9
584
Chapter Seven /SOLUTIONS
Z
=
w=162
w=6
1
· f (w) dw,
9
so k = 1/9, a = 6, b = 162.
−1
131. Letting w = ln x2 + 1 , we have dw = x2 + 1
· 2xdx, so
1
xdx
= dw.
x2 + 1
2
This means we can write
(1/2) dw
f (w)
Z
5
2
f ln x + 1
x dx
x2 + 1
2
=
=
=
5
Z
z
}|
z }| {
{ xdx
f ln x2 + 1
Z2 x=5
f (w) ·
Zx=2
w=ln 26
w=ln 5
x2 + 1
1
dw
2
1
· f (w) dw,
2
so k = 1/2, a = ln 5, b = ln 26.
132. Since tan x = sin x/ cos x, we use the substitution w = cos x. Then we have
y=
Z
tan x + 1 dx =
Z
Z
sin x
dx +
cos x
1 dx = − ln | cos x| + x + C.
Therefore, the general solution of the differential equation is
y = − ln | cos x| + x + C.
The initial condition allows us to determine the constant C. Substituting y(0) = 1 gives
1 = − ln | cos 0| + 0 + C.
Since ln | cos 0| = ln 1 = 0, we have C = 1. The solution is
y = − ln | cos x| + x + 1.
133. We substitute w = 1 − x into Im,n . Then dw = −dx, and x = 1 − w.
When x = 0, w = 1, and when x = 1, w = 0, so
Im,n =
Z
1
n
m
x (1 − x) dx =
0
=−
Z
1
0
n
Z
0
(1 − w)m wn (−dw)
1
m
w (1 − w) dw =
1
Z
wn (1 − w)m dw = In,m .
0
134. (a) Since change of position is the integral of velocity, for t in seconds we have
Change in position =
Z
60
f (t) dt meters.
0
(b) Since 1 minute is 60 seconds, t = 60T . The constant 60 has units sec/min, so 60T has units sec/min × min = sec.
Applying the substitution t = 60T to the integral in part (a), we get
Change in position =
Z
0
1
f (60T ) 60 dT meters.
7.1 SOLUTIONS
585
135. (a) Integrating I, we have
C(t) = 1.3t + C0 .
Substituting t = 0 gives C0 = 311, so
C(t) = 1.3t + 311.
Integrating II, we have
C(t) = 0.5t + 0.03
t2
+ C0 .
2
Substituting t = 0 gives C0 = 311, so
C(t) = 0.5t + 0.015t2 + 311.
Integrating III, we have
0.5 0.02t
e
+ C0
0.02
C(t) =
Substituting t = 0 and C0 = 311, we have
311 = 25e0.02(0) + C0
311 = 25 + C0
C0 = 286.
Thus
C(t) = 25e0.02t + 286.
(b) In 2020, we have t = 70, so
I C(70) = 1.3 · 70 + 311 = 402 ppm.
II C(70) = 0.5 · 70 + 0.015 · 702 + 311 = 419.5 ppm.
III C(70) = 25e0.02(70) + 286 = 387.380 ppm.
136. (a) A time period of ∆t hours with flow rate of f (t) cubic meters per hour has a flow of f (t)∆t cubic meters. Summing
the flows, we get total flow ≈ Σf (t)∆t, so
Total flow =
Z
72
f (t) dt cubic meters.
0
(b) Since 1 day is 24 hours, t = 24T . The constant 24 has units hours per day, so 24T has units hours/day×day = hours.
Applying the substitution t = 24T to the integral in part (a), we get
Total flow =
Z
3
f (24T ) 24 dT cubic meters.
0
137. (a) In 2010, we have P = 6.1e0.012·10 = 6.9 billion people.
In 2020, we have P = 6.1e0.012·20 = 7.8 billion people.
(b) We have
Average population =
1
10 − 0
1
=
10
Z
10
6.1e0.012t dt =
0
1
6.1 0.012t
·
e
10 0.012
10
0
6.1
(e0.12 − e0 ) = 6.5.
0.012
The average population of the world between 2000 and 2010 is predicted to be 6.5 billion people.
138. (a) At time t = 0, the rate of oil leakage = r(0) = 50 thousand liters/minute.
At t = 60, rate = r(60) = 15.06 thousand liters/minute.
(b) To find the amount of oil leaked during the first hour, we integrate the rate from t = 0 to t = 60:
Oil leaked =
Z
0
60
50e−0.02t dt = −
50 −0.02t
e
0.02
60
0
= −2500e−1.2 + 2500e0 = 1747 thousand liters.
586
Chapter Seven /SOLUTIONS
139. (a) E(t) = 1.4e0.07t
(b)
Average Yearly Consumption =
=
Total Consumption for the Century
100 years
1
100
Z
100
1.4e0.07t dt
0
"
1 0.07t
e
= (0.014)
0.07
= (0.014)
h 1
100
0
(e7 − e0 )
#
i
0.07
= 0.2(e − 1) ≈ 219 million megawatt-hours.
7
(c) We are looking for t such that E(t) ≈ 219:
1.4e0.07t ≈ 219
e0.07t = 156.4.
Taking natural logs,
0.07t = ln 156.4
5.05
t≈
≈ 72.18.
0.07
Thus, consumption was closest to the average during 1972.
(d) Between the years 1900 and 2000 the graph of E(t) looks like
E(t)
1.4e7
E(t) = 1.4e0.07t
219
t
1900
1950
(t = 0)
(t = 50)
2000
(t = 100)
From the graph, we can see the t value such that E(t) = 219. It lies to the right of t = 50, and is thus in the
second half of the century.
140. We have Q′ (t) = −I(t) and I(t) = I0 e−t , so
Q′ (t) = −I0 e−t .
We have
Q(t) = I0
Since Q(0) = Q0 , we have
Z
−e−t dt = I0 e−t + C.
Q0 = I0 e−0 + C = I0 + C,
so
Thus,
C = Q0 − I0
Q(t) = I0 e−t + Q0 − I0 .
7.1 SOLUTIONS
141. Since v =
dh
, it follows that h(t) =
dt
Z
v(t) dt and h(0) = h0 . Since
mg
k
v(t) =
we have
Z
h(t) =
The first integral is simply
k
1 − e− m t
v(t) dt =
Z
mg
k
Z
Thus
=
k t
mg
mg − m
−
e
,
k
k
dt −
Z
mg
k
k
e− m t dt.
mg
k
t. Then
t + C. To evaluate the second integral, make the substitution w = − m
k
dw = −
so
587
k
e− m t dt =
Z
h(t) =
=
ew −
Z
m
k
v dt =
k
dt,
m
dw = −
k
m
m w
e + C = − e− m t + C.
k
k
mg
mg
t−
k
k
−
k t
m −m
e
+C
k
k
m2 g
mg
t + 2 e− m t + C.
k
k
Since h(0) = h0 ,
m2 g
mg
· 0 + 2 e0 + C;
k
k
m2 g
C = h0 − 2 .
k
h0 =
Thus
k
m2 g
m2 g
mg
t + 2 e− m t − 2 + h0
k
k
k
k t
mg
m2 g
−m
h(t) =
t− 2 1−e
+ h0 .
k
k
h(t) =
, and it follows that h(t) =
body, the height h is decreasing, so v = − dh
142. Since
dt
Z v is given as the velocity of a falling
√
√
t gk
−t gk
− v(t) dt and h(0) = h0 . Let w = e
+e
. Then
dw =
dw
= (et
so √
gk
√
gk
− e−t
√
gk
−
Z
√
p
gk et
gk
− e−t
gk
dt,
) dt. Therefore,
v(t)dt = −
=−
Z q
q Z
g
k
et
g
k
et
e
t
√
√
gk
√
gk
− e−t
+ e−t
1
gk
−t
+e
q Z
g
1
dw
√
=−
k
w
gk
√
√
gk
√
gk
gk
!
dt
√
et
gk
− e−t
√
r
g
ln |w| + C
gk2
√
√
1
= − ln et gk + e−t gk + C.
k
=−
Since
√
h(0) = −
ln 2
1
ln(e0 + e0 ) + C = −
+ C = h0 ,
k
k
gk
dt
588
Chapter Seven /SOLUTIONS
we have C = h0 +
ln 2
. Thus,
k
√
√
ln 2
1
1
+ h0 = − ln
h(t) = − ln et gk + e−t gk +
k
k
k
et
√
gk
+ e−t
2
√
gk
!
+ h0 .
143. (a) In the first case, we are given that R0 = 1000 widgets/year. So we have R = 1000e0.125t . To determine the total
number sold, we need to integrate this rate over the time period from 0 to 10. Therefore
Total number of widgets sold =
Z
10
1000e0.125t dt =
0
1000 0.125t
e
0.125
10
= 8000(e1.25 − 1) = 19,923 widgets.
0
In the second case,
Total number of widgets sold =
10
10
Z
1,000,000e0.125t dt = 1,000,000e0.125t
0
= 19.9 million widgets.
0
(b) We want to determine T such that
Z
T
1000e0.125t dt =
0
Evaluating both sides, we get
1000 0.0125t
e
0.125
19,923
.
2
T
0
= 8000(e0.125T − 1) = 9,961
8000e0.125T = 17,961
e0.125T = 2.245
0.125T = 0.8088,
T = 6.47 years.
Similarly, in the second case, we want T so that
Z
T
1,000,000e0.125t dt =
0
19,900,000
2
Evaluating both sides, we get
1,000,000 0.125t
e
0.125
T
= 9,950,000
0
8,000,000(e0.125T − 1) = 9,950,000
8,000,000e0.125T = 17,950,000
e0.125T = 2.2438
T = 6.47 years.
So the half way mark is reached at the same time regardless of the initial rate.
(c) Since half the widgets are sold in the last 3 12 years of the decade, if each widget is expected to last at least 3.5 years,
their claim could easily be true.
144. (a) Amount of water entering tank in a short period of time = Rate×Time = r(t)∆t.
(b) Summing the contribution from each of the small intervals ∆t:
Amount of water entering the tank
between t = 0 and t = 5
Taking a limit as ∆t → 0:
≈
n−1
X
r(ti )∆t,
where ∆t = 5/n.
i=0
Amount of water entering the tank
between t = 0 and t = 5
=
Z
0
5
r(t) dt.
7.1 SOLUTIONS
589
(c) If Q(t) is the amount of water in the tank at time t, then Q′ (t) = r(t). We want to calculate Q(5) − Q(0). By the
Fundamental Theorem,
Amount which has
entered tank
= Q(5) − Q(0) =
Z
5
r(t) dt =
0
Z
5
20e0.02t dt =
0
= 1000(e
0.02(5)
20 0.02t
e
0.02
5
0
− 1) ≈ 105.17 gallons.
(d) By the Fundamental Theorem again,
Amount which has
entered tank
= Q(t) − Q(0) =
Z
Q(t) − 3000 =
Z
t
r(t) dt
0
t
20e0.02t dt
0
so
Q(t) = 3000 +
Z
t
20e0.02t dt = 3000 +
0
= 3000 + 1000(e
0.02t
= 1000e0.02t + 2000.
20 0.02t
e
0.02
t
0
− 1)
Strengthen Your Understanding
145. To do guess-and-check or substitution, we need an extra factor of f ′ (x) in the integrand. For example, if f (x) = 2x, the
left side is
Z
Z
4x3
4x2 dx =
(2x)2 dx =
+ C,
3
while the right side is 8x3 /3 + C.
146. If we do guess-and-check and our guess is off by a constant factor, we can fix our guess by dividing our guess by that
constant. This does not work if we are off by a factor which is not constant. In this case, if we check our answer by
differentiating, we get, using the quotient rule:
d
dx
sin(x2 )
2x
=
cos(x2 )(2x)(2x) − 2 sin x2
6= cos x2 .
4x2
147. When we make a substitution in a definite integral, we must either change the limits of integration or return to the original
R 3π/2
variable before evaluating. Since w = 3x, the right side should be (1/3) 0
cos w dw.
148. The inside function is cos θ, so f (θ) should be the derivative of cos θ (or a constant multiple of the derivative). Thus,
f (θ) = − sin θ works.
149. The integrand should have a factor that is a constant multiple of 3x2 − 3, for example
′
150. True. Let w = f (x), so dw = f (x) dx, then
Z
′
f (x) cos(f (x)) dx =
Z
1
d
ln |f (x)| =
· f ′ (x),
dx
f (x)
so, in general
152. True. Let w = 5 − t2 , then dw = −2t dt.
sin(x3 − 3x)(x2 − 1) dx.
cos w dw = sin w + C = sin(f (x)) + C.
151. False. Differentiating gives
Z
R
1
dx 6= ln |f (x)| + C.
f (x)
590
Chapter Seven /SOLUTIONS
Solutions for Section 7.2
Exercises
1. (a) Since we can change x2 into a multiple of x3 by integrating, let v ′ = x2 and u = ex . Using v = x3 /3 and u′ = ex
we get
Z
x2 ex dx =
Z
uv ′ dx = uv −
=
Z
u′ vdx
1
x 3 ex
−
3
3
Z
x3 ex dx.
(b) Since we can change x2 into a multiple of x by differentiating, let u = x2 and v ′ = ex . Using u′ = 2x and v = ex
we have
Z
x2 ex dx =
Z
uv ′ dx = uv −
Z
u′ vdx
= x 2 ex − 2
2. Let u = arctan x, v ′ = 1. Then v = x and u′ =
Z
Z
xex dx.
1
. Integrating by parts, we get:
1 + x2
1 · arctan x dx = x · arctan x −
Z
x·
1
dx.
1 + x2
To compute the second integral use the substitution, z = 1 + x2 .
Z
Thus,
1
x
dx =
1 + x2
2
Z
Z
dz
1
1
= ln |z| + C = ln(1 + x2 ) + C.
z
2
2
arctan x dx = x · arctan x −
1
ln(1 + x2 ) + C.
2
3. Let u = t, v ′ = sin t. Thus, v = − cos t and u′ = 1. With this choice of u and v, integration by parts gives:
Z
Z
t sin t dt = −t cos t −
(− cos t) dt
= −t cos t + sin t + C.
4. Let u = t2 , v ′ = sin t implying v = − cos t and u′ = 2t. Integrating by parts, we get:
Z
t2 sin t dt = −t2 cos t −
Z
2t(− cos t) dt.
Again, applying integration by parts with u = t, v ′ = cos t, we have:
Z
Thus
Z
t cos t dt = t sin t + cos t + C.
t2 sin t dt = −t2 cos t + 2t sin t + 2 cos t + C.
′
5t
5. Let u = t and
u′ =R1 and v = 51 e5t .
R v5t = e ,1so 5t
Then te dt = 5 te − 51 e5t dt = 51 te5t −
1 5t
e
25
+ C.
7.2 SOLUTIONS
6. Let u = t2Rand v ′ = e5t , so u′ = 2t Rand v = 51 e5t .
5t
2
Then t2 e5t dt = 51 t2 e5t −
R 52 5tte dt.1 2 5t 2 1 5t
Using Problem 5, we have t e dt = 5 t e − 5 ( 5 te −
2
2
te5t + 125
e5t + C.
= 15 t2 e5t − 25
7. Let u = p and v ′ = e(−0.1)p , u′ = 1. Thus, v =
by parts gives:
Z
R
1 5t
e )
25
591
+C
e(−0.1)p dp = −10e(−0.1)p . With this choice of u and v, integration
pe(−0.1)p dp = p(−10e(−0.1)p ) −
= −10pe
(−0.1)p
Z
+ 10
Z
(−10e(−0.1)p ) dp
e(−0.1)p dp
= −10pe(−0.1)p − 100e(−0.1)p + C.
8. Let u = z + 1, v ′ = e2z . Thus, v = 12 e2z and u′ = 1. Integrating by parts, we get:
Z
Z
1 2z
1 2z
e −
e dz
2
2
1
1
= (z + 1)e2z − e2z + C
2
4
1
= (2z + 1)e2z + C.
4
(z + 1)e2z dz = (z + 1) ·
9. Integration by parts with u = ln x, v ′ = x gives
Z
x ln x dx =
x2
ln x −
2
Z
1
1
1
x dx = x2 ln x − x2 + C.
2
2
4
Or use the integral table, III-13, with n = 1.
10. Let u = ln x and v ′ = x3 , so u′ =
Z
1
x
and v =
x3 ln x dx =
x4
.
4
x4
ln x −
4
11. Let u = ln 5q, v ′ = q 5 . Then v = 16 q 6 and u′ =
Z
Then
Z
x4
x4
x3
dx =
ln x −
+ C.
4
4
16
1
. Integrating by parts, we get:
q
Z
1
1
1 6
q ln 5q − (5 · ) · q 6 dq
6
5q 6
1
1 6
q + C.
= q 6 ln 5q −
6
36
q 5 ln 5q dq =
12. Let u = θ2Rand v ′ = cos 3θ, so u′ = 2θ andRv = 13 sin 3θ.
Then θ2 cos 3θ dθ = 31 θ2 sin 3θ − 32 θ sin 3θ dθ. The integral on the right hand side is simpler than our original
integral, but to
R evaluate it we need to again use integration by parts.
To find θ sin 3θ dθ, let u = θ and v ′ = sin 3θ, so u′ = 1 and v = − 31 cos 3θ.
This gives
Z
Thus,
1
1
θ sin 3θ dθ = − θ cos 3θ +
3
3
Z
θ2 cos 3θdθ =
Z
1
1
cos 3θ dθ = − θ cos 3θ + sin 3θ + C.
3
9
1 2
2
2
θ sin 3θ + θ cos 3θ −
sin 3θ + C.
3
9
27
592
Chapter Seven /SOLUTIONS
13. Let u = sin θ and v ′ = sin θ, so u′ = cos θ and v = − cos θ. Then
Z
sin2 θ dθ = − sin θ cos θ +
= − sin θ cos θ +
= − sin θ cos θ +
R
Z
Z
Z
cos2 θ dθ
(1 − sin2 θ) dθ
1 dθ −
Z
sin2 θ dθ.
By adding sin2 θ dθ to both sides of the above equation, we find that 2
sin2 θ dθ = − 21 sin θ cos θ + 2θ + C ′ .
R
R
14. Let u = cos(3α + 1) and v ′ = cos(3α + 1), so u′ = −3 sin(3α + 1), and v =
Z
By adding
R
cos2 (3α + 1) dα =
Z
sin2 θ dθ = − sin θ cos θ + θ + C, so
1
3
sin(3α + 1). Then
(cos(3α + 1)) cos(3α + 1) dα
Z
=
1
cos(3α + 1) sin(3α + 1) +
3
=
1
cos(3α + 1) sin(3α + 1) +
3
=
1
cos(3α + 1) sin(3α + 1) + α −
3
sin2 (3α + 1) dα
Z
1 − cos2 (3α + 1) dα
Z
cos2 (3α + 1) dα.
cos2 (3α + 1) dα to both sides of the above equation, we find that
2
which gives
Z
Z
(We use the fact that
Z
1
α
cos(3α + 1) sin(3α + 1) + + C.
6
2
cos2 (3α + 1) dα =
15. Let u = (ln t)2 and v ′ = 1, so u′ =
Z
1
cos(3α + 1) sin(3α + 1) + α + C,
3
cos2 (3α + 1) dα =
2 ln t
and v = t. Then
t
(ln t)2 dt = t(ln t)2 − 2
Z
ln t dt = t(ln t)2 − 2t ln t + 2t + C.
ln x dx = x ln x − x + C, a result which can be derived using integration by parts.)
16. Remember that ln(x2 ) = 2 ln x. Therefore,
Z
Check:
ln(x2 ) dx = 2
Z
ln x dx = 2x ln x − 2x + C.
d
2x
(2x ln x − 2x + C) = 2 ln x +
− 2 = 2 ln x = ln(x2 ).
dx
x
17. Let u = y and v ′ = (y + 3)1/2 , so u′ = 1 and v = 23 (y + 3)3/2 :
Z
y
p
y + 3 dy =
18. Let u = t + 2 and v ′ =
√
Z
2
y(y + 3)3/2 −
3
Z
2
2
4
(y + 3)3/2 dy = y(y + 3)3/2 −
(y + 3)5/2 + C.
3
3
15
2 + 3t, so u′ = 1 and v = 92 (2 + 3t)3/2 . Then
√
2
(t + 2) 2 + 3t dt = (t + 2)(2 + 3t)3/2 −
9
2
= (t + 2)(2 + 3t)3/2 −
9
Z
2
(2 + 3t)3/2 dt
9
4
(2 + 3t)5/2 + C.
135
7.2 SOLUTIONS
593
19. Let u = θ + 1 and v ′ = sin(θ + 1), so u′ = 1 and v = − cos(θ + 1).
Z
(θ + 1) sin(θ + 1) dθ = −(θ + 1) cos(θ + 1) +
Z
cos(θ + 1) dθ
= −(θ + 1) cos(θ + 1) + sin(θ + 1) + C.
20. Let u = z, v ′ = e−z . Thus v = −e−z and u′ = 1. Integration by parts gives:
Z
ze
−z
dz = −ze
−z
−
Z
(−e−z ) dz
= −ze−z − e−z + C
= −(z + 1)e−z + C.
21. Let u = ln x, v ′ = x−2 . Then v = −x−1 and u′ = x−1 . Integrating by parts, we get:
Z
x
−2
ln x dx = −x
−1
ln x −
Z
(−x−1 ) · x−1 dx
= −x−1 ln x − x−1 + C.
1
, so u′ = 1 and v = −2(5 − y)1/2 .
22. Let u = y and v ′ = √5−y
Z
Z
y
4
√
dy = −2y(5 − y)1/2 + 2 (5 − y)1/2 dy = −2y(5 − y)1/2 − (5 − y)3/2 + C.
3
5−y
Z
Z
Z
t+7
t
23.
dt =
dt + 7 (5 − t)−1/2 dt.
√
√
5−t
5−t
1
, so u′ = 1 and v = −2(5−t)1/2 .
To calculate the first integral, we use integration by parts. Let u = t and v ′ = √5−t
Then
Z
Z
4
t
1/2
√
dt = −2t(5 − t)
+ 2 (5 − t)1/2 dt = −2t(5 − t)1/2 − (5 − t)3/2 + C.
3
5−t
We can calculate the second integral directly: 7
Z
Z
(5 − t)−1/2 = −14(5 − t)1/2 + C1 . Thus
4
t+7
√
dt = −2t(5 − t)1/2 − (5 − t)3/2 − 14(5 − t)1/2 + C2 .
3
5−t
24. Let u = (ln x)4 and v ′ = x, so u′ =
Z
4(ln x)3
x
and v =
x2
.
2
Then
2
x (ln x)4
−2
x(ln x) dx =
2
4
Z
x(ln x)3 dx.
x(ln x)3 dx is somewhat less complicated than x(ln x)4 dx. To calculate it, we again try integration by parts, this
time letting u = (ln x)3 (instead of (ln x)4 ) and v ′ = x. We find
R
Z
R
x(ln x)3 dx =
3
x2
(ln x)3 −
2
2
Z
x(ln x)2 dx.
Once again, express the given integral in terms of a less-complicated one. Using integration by parts two more times, we
find that
Z
Z
x2
2
2
(ln x) −
x(ln x) dx
x(ln x) dx =
2
and that
Z
x2
x2
x ln x dx =
ln x −
+ C.
2
4
Putting this all together, we have
Z
x(ln x)4 dx =
x2
3
3
3
(ln x)4 − x2 (ln x)3 + x2 (ln x)2 − x2 ln x + x2 + C.
2
2
2
4
594
Chapter Seven /SOLUTIONS
25. Integrate by parts letting u = (ln r)2 and dv = rdr, then du = (2/r) ln rdr and v = r 2 /2. We get
Z
1 2
r (ln r)2 −
2
r(ln r)2 dr =
Z
r ln r dr.
Then using integration by parts again with u = ln r and dv = rdr, so du = dr/r and v = r 2 /2, we get
Z
1
1 2
1
r ln r dr = r 2 (ln r)2 −
r ln r −
2
2
2
2
26. Let u = arcsin w and v ′ = 1, so u′ = √
Z
1
1−w2
Z
So
7z dz
=7
1 + 49z 2
Z
Z
r dr =
1 2
1
1
r (ln r)2 − r 2 ln r + r 2 + C.
2
2
4
and v = w. Then
arcsin w dw = w arcsin w −
27. Let u = arctan 7z and v ′ = 1, so u′ =
w = 1 + 49z 2 , dw = 98z dz, so
Z
Z
7
1+49z 2
1
98
dw
1
=
w
14
√
p
w
dw = w arcsin w + 1 − w2 + C.
2
1−w
and v = z. Now
Z
R
7z dz
1+49z 2
can be evaluated by the substitution
dw
1
1
=
ln |w| + C =
ln(1 + 49z 2 ) + C
w
14
14
arctan 7z dz = z arctan 7z −
1
ln(1 + 49z 2 ) + C.
14
28. This integral can first be simplified by making the substitution w = x2 , dw = 2x dx. Then
To evaluate
Then
R
Z
x arctan x2 dx =
1
2
Z
arctan w dw.
arctan w dw, we’ll use integration by parts. Let u = arctan w and v ′ = 1, so u′ =
Z
arctan w dw = w arctan w −
Z
1
1+w2
and v = w.
1
w
dw = w arctan w − ln |1 + w2 | + C.
1 + w2
2
Since 1 + w2 is never negative, we can drop the absolute value signs. Thus, we have
Z
1
1
x2 arctan x2 − ln(1 + (x2 )2 ) + C
2
2
1
1
= x2 arctan x2 − ln(1 + x4 ) + C.
2
4
x arctan x2 dx =
2
2
29. Let u = x2 and v ′ = xex , so u′ = 2x and v = 21 ex . Then
Z
2
x3 ex dx =
1 2 x2
x e −
2
Z
2
xex dx =
1 2 x2 1 x2
x e − e + C.
2
2
2
Note
Z that we can also
Z do this problem by substitution and integration by parts. If we let w = x , so dw = 2x dx, then
2
1
x3 ex dx =
wew dw. We could then perform integration by parts on this integral to get the same result.
2
30. To simplify matters, let us try the substitution w = x3 , dw = 3x2 dx. Then
Z
1
x cos x dx =
3
5
3
Z
w cos w dw.
Now we integrate by parts. Let u = w and v ′ = cos w, so u′ = 1 and v = sin w. Then
1
3
Z
Z
1
[w sin w −
sin w dw]
3
1
= [w sin w + cos w] + C
3
1
1
= x3 sin x3 + cos x3 + C
3
3
w cos w dw =
7.2 SOLUTIONS
31. Let u = x, u′ = 1 and v ′ = sinh x, v = cosh x. Integrating by parts, we get
Z
x sinh x dx = x cosh x −
Z
cosh x dx
= x cosh x − sinh x + C.
32. Let u = x − 1, u′ = 1 and v ′ = cosh x, v = sinh x. Integrating by parts, we get
Z
33.
Z
34.
Z
Z
sinh x dx
= (x − 1) sinh x − cosh x + C.
5
5
ln t dt = (t ln t − t)
1
(x − 1) cosh x dx = (x − 1) sinh x −
1
= 5 ln 5 − 4 ≈ 4.047
5
5
x cos x dx = (cos x + x sin x)
3
3
= cos 5 + 5 sin 5 − cos 3 − 3 sin 3 ≈ −3.944.
35. We use integration by parts. Let u = z and v ′ = e−z , so u′ = 1 and v = −e−z . Then
Z
0
10
10
ze−z dz = −ze−z
Z
+
10
e−z dz
0
0
10
= −10e−10 + (−e−z )
0
= −11e−10 + 1
≈ 0.9995.
36.
Z
3
t ln t dt =
1
1 2
1
t ln t − t2
2
4
3
=
1
9
ln 3 − 2 ≈ 2.944.
2
37. We use integration by parts. Let u = arctan y and v ′ = 1, so u′ =
Z
1
1+y 2
1
1
arctan y dy = (arctan y)y
0
0
=
−
and v = y. Thus
Z
π
1
− ln |1 + y 2 |
4
2
1
0
1
y
dy
1 + y2
0
1
π
= − ln 2 ≈ 0.439.
4
2
38.
Z
0
5
5
ln(1 + t) dt = ((1 + t) ln(1 + t) − (1 + t))
0
= 6 ln 6 − 5 ≈ 5.751.
39. We use integration by parts. Let u = arcsin z and v ′ = 1, so u′ = √
Z
0
1
1
arcsin z dz = z arcsin z
0
−
Z
0
1
√
1
and v = z. Then
1 − z2
z
π
dz = −
2
2
1−z
Z
0
1
√
z
dz.
1 − z2
595
596
Chapter Seven /SOLUTIONS
To find
Then
Z
0
1
√
Thus our final
z
dz, we substitute w = 1 − z 2 , so dw = −2z dz.
1 − z2
Z
z=1
z=0
answer is π2
√
1
z
dz = −
2
1 − z2
− 1 ≈ 0.571.
Z
w=0
w
1
−2
w=1
1
dw =
2
Z
1
w=1
1
1
w− 2 dw = w 2
w=0
= 1.
0
40. To simplify the integral, we first make the substitution z = u2 , so dz = 2u du. Then
Z
u=1
1
2
u arcsin u2 du =
u=0
From Problem 39, we know that
R1
0
arcsin z dz =
Z
π
2
arcsin z dz.
z=0
− 1. Thus,
1
u arcsin u2 du =
0
41. (a)
(b)
(c)
(d)
(e)
(f)
(g)
z=1
Z
1 π
( − 1) ≈ 0.285.
2 2
This integral can be evaluated using integration by parts with u = x, v ′ = sin x.
We evaluate this integral using the substitution w = 1 + x3 .
We evaluate this integral using the substitution w = x2 .
We evaluate this integral using the substitution w = x3 .
We evaluate this integral using the substitution w = 3x + 1.
This integral can be evaluated using integration by parts with u = x2 , v ′ = sin x.
This integral can be evaluated using integration by parts with u = ln x, v ′ = 1.
42. A calculator gives
gives
R2
1
ln x dx = 0.386. An antiderivative of ln x is x ln x − 1, so the Fundamental Theorem of Calculus
Z
1
2
2
ln x dx = (x ln x − x)
1
= 2 ln 2 − 1.
Since 2 ln 2 − 1 = 0.386, the value from the Fundamental Theorem agrees with the numerical answer.
Problems
43. Using a log property, we have
Z
ln (5 − 3x)
2
dx =
Z
2 ln(5 − 3x) dx.
So we let
w = 5 − 3x
dw = −3dx
1
dx = − dw,
3
which gives,
w
Z
Thus, w = 5 − 3x, k = −2/3.
ln(5 − 3x)2 dx =
=
Z
Z
− 1 dw
3
z }| { z}|{
2 ln (5 − 3x) dx
−
2
· ln w dw.
3
7.2 SOLUTIONS
44. We have
Z
ln √
Z
1
1
dx =
ln
dx
1/2
(4
−
5x)
4 − 5x
Z
=
ln (4 − 5x)−1/2 dx
Z
=
−
1
· ln(4 − 5x) dx
2
log property.
So we let
w = 4 − 5x
dw = −5dx
1
dx = − dw,
5
which gives
w
−1
dw
5
Z
z }| { z}|{
1
1
dx =
− · ln (4 − 5x) dx
ln √
4 − 5x
Z 2
1
=
ln w dw.
10
Z
Thus w = 4 − 5x, k = 1/10.
45. We have
Z
ln (ln x)3
3 ln (ln x)
dx =
dx
log property
x
x
Z
3 ln (ln x) x−1 dx.
=
Z
So we let
w = ln x
dw = x−1 dx
which gives
w
dw
z }| { z }| {
ln (ln x)3
3 ln (ln x) x−1 dx
dx =
x
Z
=
3 ln w dw.
Z
Z
Thus w = ln x, k = 3.
46. Using integration by parts with u′ = e−t , v = t, so u = −e−t and v ′ = 1, we have
Area =
Z
0
2
2
te
−t
dt = −te
−t
−
0
Z
0
2
−1 · e−t dt
2
= −2e−2 − e−2 + 1 = 1 − 3e−2 .
= (−te−t − e−t )
0
597
598
Chapter Seven /SOLUTIONS
47. Since
R
Z
1 · arctan z dz, we take u = arctan z, v ′ = 1, so u′ = 1/(1 + z 2 ) and v = z. Then
R
arctan z dz =
arctan z dz = z arctan z −
Z
1
z
dz = z arctan z − ln(1 + z 2 ) + C.
1 + z2
2
Thus, we have
2
2
Z
1
arctan z dz = z arctan z − ln(z 2 + 1)
2
0
48. Since
R
R
arcsin y dy =
Thus
y
arcsin y dy = y arcsin y −
1 − y2
Area =
Z
dy =
p
−1/2
1
√ dw = −
2
w
Z
Z
y
p
1 − y2
49. Since ln(x ) = 2 ln x and
R
ln x dx = x ln x − x + C, we have
arcsin y dy = (y arcsin y +
Z
2
Z
2
(ln(x2 ) − ln x) dx =
1
=
dy.
p
1 − y 2 + C.
1
1
Area =
1 − y 2 and v = y. Thus
w−1/2 dw = w1/2 + C =
Z
0
2
p
1 · arcsin y dy, we take u = arcsin y, v = 1, so u = 1/
Substituting w = 1 − y 2 , dw = −2y dy, we have
Z
0
′
Z
1
ln 5.
2
= 2 arctan 2 −
1
Z
p
1−
y2)
= 1 · arcsin 1 − 1 =
π
− 1.
2
0
2
1
2
ln x dx = (x ln x − x)
(2 ln x − ln x) dx
= 2 ln 2 − 2 − (1 ln 1 − 1) = 2 ln 2 − 1.
1
50. Since the graph of f (t) = ln(t2 − 1) is above the graph of g(t) = ln(t − 1) for t > 1, we have
Area =
Z
2
3
2
(ln(t − 1) − ln(t − 1)) dt =
Z
3
ln
2
t2 − 1
t−1
dt =
Z
3
ln(t + 1) dt.
2
We can cancelR the factor of (t − 1) in the last step above because the integral is over 2 ≤ t ≤ 3, where (t − 1) is not zero.
We use ln x dx = x ln x − 1 with the substitution x = t + 1. The limits t = 2, t = 3 become x = 3, x = 4,
respectively. Thus
Area =
Z
3
ln(t + 1) dt =
Z
3
2
4
4
ln x dx = (x ln x − x)
3
= 4 ln 4 − 4 − (3 ln 3 − 3) = 4 ln 4 − 3 ln 3 − 1.
51.
2π
f (x) = x sin x
π
2π
x
−2π
The graph of f (x) = x sin x is shown above. The first positive zero is at x = π, so, using integration by parts,
Area =
Z
0
π
x sin x dx
7.2 SOLUTIONS
π
= −x cos x
= −x cos x
+
0
π
Z
599
π
cos x dx
0
π
+ sin x
0
0
= −π cos π − (−0 cos 0) + sin π − sin 0 = π.
52. From integration by parts in Problem 13, we obtain
Z
1
1
sin2 θ dθ = − sin θ cos θ + θ + C.
2
2
Using the identity given in the book, we have
Z
sin2 θ dθ =
Z
1
1
1 − cos 2θ
dθ = θ − sin 2θ + C.
2
2
4
Although the answers differ in form, they are really the same, since (by one of the standard double angle formulas)
1
1
1
− sin 2θ = − (2 sin θ cos θ) = − sin θ cos θ.
4
4
2
53. Integration by parts: let u = cos θ and v ′ = cos θ, so u′ = − sin θ and v = sin θ.
Z
2
cos θ dθ = sin θ cos θ −
= sin θ cos θ +
Z
(− sin θ)(sin θ) dθ
Z
sin2 θ dθ.
Now use sin2 θ = 1 − cos2 θ.
Z
cos2 θ dθ = sin θ cos θ +
= sin θ cos θ +
Adding
R
(1 − cos2 θ) dθ
Z
dθ −
Z
cos2 θ dθ.
cos2 θ dθ to both sides, we have
2
Z
Z
Use the identity cos2 θ =
cos2 θ dθ = sin θ cos θ + θ + C
cos2 θ dθ =
1
1
sin θ cos θ + θ + C ′ .
2
2
1+cos 2θ
.
2
Z
1
4
Z
cos2 θ dθ =
Z
1 + cos 2θ
1
1
dθ = θ + sin 2θ + C.
2
2
4
The only difference is in the two terms 12 sin θ cos θ and 41 sin 2θ, but since sin 2θ = 2 sin θ cos θ, we have
sin 2θ = 41 (2 sin θ cos θ) = 21 sin θ cos θ, so there is no real difference between the formulas.
54. First, Rlet u = ex and v ′ = sin x, so uR′ = ex and v = − cos x. R
Thus ex sin x dx = −ex cos x + ex cos x dx. To calculate ex cos x dx, we again need to use integration by parts.
Let u = ex and v ′ = cos x, so u′ = ex and v = sin x.
Thus
Z
Z
ex cos x dx = ex sin x −
ex sin x dx.
This gives
Z
ex sin x dx = ex sin x − ex cos x −
Z
ex sin x dx.
600
Chapter Seven /SOLUTIONS
By adding
R
ex sin x dx to both sides, we obtain
2
Thus
Z
Z
ex sin x dx = ex (sin x − cos x) + C.
ex sin x dx =
1 x
e (sin x − cos x) + C.
2
This problem could also be done in other ways; for example, we could have started with u = sin x and v ′ = ex as well.
θ
θ
θ
55. Let u = eθ and v ′ = cos θ, so u′ =
Z e and v = sin θ. Then e cos θ dθ = e sin θ −
1
In Problem 54 we found that
ex sin xdx = ex (sin x − cos x) + C.
2
R
Z
eθ cos θ dθ = eθ sin θ −
h1
Then
Z
R
i
ex sin x dx =
xex sin x dx =
1 x
1
xe (sin x − cos x) −
2
2
=
1 x
1
xe (sin x − cos x) −
2
2
Using Problems 54 and 55, we see that this equals
eθ sin θ dθ.
eθ (sin θ − cos θ) + C
2
1 θ
= e (sin θ + cos θ) + C.
2
56. We integrate by parts. Since in Problem 54 we found that
v ′ = ex sin x, so u′ = 1 and v = 12 ex (sin x − cos x).
R
Z
Z
1 x
e (sin x
2
− cos x), we let u = x and
ex (sin x − cos x) dx
ex sin x dx +
1
2
Z
ex cos x dx.
1
1
1 x
xe (sin x − cos x) − ex (sin x − cos x) + ex (sin x + cos x) + C
2
4
4
1
1
= xex (sin x − cos x) + ex cos x + C.
2
2
57. Again we use Problems 54 and 55. Integrate by parts, letting u = θ and v ′ = eθ cos θ, so u′ = 1 and v =
cos θ). Then
Z
θeθ cos θ dθ =
1
1 θ
θe (sin θ + cos θ) −
2
2
1 θ
θe (sin θ + cos θ) −
2
1
= θeθ (sin θ + cos θ) −
2
1 θ
= θe (sin θ + cos θ) −
2
=
Z
1 θ
e (sin θ
2
+
eθ (sin θ + cos θ) dθ
Z
Z
1
1
eθ sin θ dθ −
eθ cos θ dθ
2
2
1 θ
1
e (sin θ − cos θ) − (sin θ + cos θ) + C
4
4
1 θ
e sin θ + C.
2
58. Using integration by parts on the first integral with u(x) = ln x and v ′ (x) = f ′′ (x), we have u′ (x) = 1/x and
v(x) = f ′ (x), so
Z ′
Z
f (x)
′
′′
dx
f (x) ln x dx = f (x) ln x −
x
Using integration by parts on the second integral with u(x) = f (x) and v ′ (x) = 1/x2 , we have u′ (x) = f ′ (x) and
v(x) = −1/x, so
Z ′
Z
f (x)
f (x)
f (x)
=
−
+
dx.
x2
x
x
Adding the results gives
Z
′′
f (x) ln x dx +
Z
f (x)
dx = f ′ (x) ln x −
x2
Z
f ′ (x)
f (x)
dx −
+
x
x
Z
f ′ (x)
f (x)
dx = f ′ (x) ln x −
+C
x
x
7.2 SOLUTIONS
601
59. Using integration by parts with u(x) = x and v ′ (x) = f ′′ (x) gives u′ (x) = 1 and v(x) = f ′ (x), so
Z
′′
′
xf (x) dx = xf (x) −
Z
f ′ (x) dx = xf ′ (x) − f (x) + C.
60. Using integration by parts, we let u = x, v ′ = f ′ (x). This gives u′ = 1, v = f (x), and we have
Z
′
uv dx = uv −
uv
Z
z }| {
u′ v dx
′
= xf (x) −
Z
Therefore,
0
Z z u}|v {
1 · f (x) dx
since f (x) = F ′ (x).
= xf (x) − F (x) + C
5
5
uv ′ dx = (xf (x) − F (x))
0
= 5f (5) − F (5) − (0 · f (0) − F (0))
= 5 · 27 − 20 − (0 · 2 − 10) = 125.
61. We integrate by parts. Since we know what the answer is supposed to be, it’s easier to choose u and v ′ . Let u = xn and
v ′ = ex , so u′ = nxn−1 and v = ex . Then
Z
xn ex dx = xn ex − n
Z
xn−1 ex dx.
62. We integrate by parts. Let u = xn and v ′ = cos ax, so u′ = nxn−1 and v =
Z
xn cos ax dx =
=
1 n
x sin ax −
a
Z
1 n
n
x sin ax −
a
a
1
a
sin ax. Then
1
(nxn−1 )( sin ax) dx
a
Z
xn−1 sin ax dx.
63. We integrate by parts. Let u = xn and v ′ = sin ax, so u′ = nxn−1 and v = − a1 cos ax. Then
Z
1
xn sin ax dx = − xn cos ax −
a
Z
1
n
= − xn cos ax +
a
a
1
(nxn−1 )(− cos ax) dx
a
Z
xn−1 cos ax dx.
64. We integrate by parts. Since we know what the answer is supposed to be, it’s easier to choose u and v ′ . Let u = cosn−1 x
and v ′ = cos x, so u′ = (n − 1) cosn−2 x(− sin x) and v = sin x. Then
Z
cosn x dx = cosn−1 x sin x + (n − 1)
= cos
n−1
x sin x + (n − 1)
= cosn−1 x sin x − (n − 1)
Thus, by adding (n − 1)
R
n
Z
Z
cosn−2 x sin2 x dx
cosn−2 x(1 − cos2 x) dx
cosn x dx + (n − 1)
cosn x dx to both sides of the equation, we find
Z
so
Z
Z
cosn x dx = cosn−1 x sin x + (n − 1)
Z
n−1
1
cos dx = cosn−1 x sin x +
n
n
n
Z
Z
cosn−2 x dx.
cosn−2 x dx,
cosn−2 x dx.
602
Chapter Seven /SOLUTIONS
65. (a) One way to avoid integrating by parts is to take the derivative of the right hand side instead. Since
the antiderivative of eax sin bx,
R
d ax
[e (A sin bx + B cos bx) + C]
dx
ax
= ae (A sin bx + B cos bx) + eax (Ab cos bx − Bb sin bx)
eax sin bx =
= eax [(aA − bB) sin bx + (aB + bA) cos bx].
Thus aA − bB = 1 and aB + bA = 0. Solving for A and B in terms of a and b, we get
b
a
, B=− 2
.
a 2 + b2
a + b2
A=
Thus
Z
eax sin bx = eax
(b) If we go through the same process, we find
a
b
sin bx − 2
cos bx + C.
a 2 + b2
a + b2
aeax [(aA − bB) sin bx + (aB + bA) cos bx] = eax cos bx.
Thus aA − bB = 0, and aB + bA = 1. In this case, solving for A and B yields
A=
Thus
′
R
b
eax cos bx = eax ( a2 +b
2 sin bx +
a
a2 +b2
b
a
, B= 2
.
a 2 + b2
a + b2
cos bx) + C.
66. Since f (x) = 2x, integration by parts tells us that
Z
10
Z
10
′
f (x)g (x) dx = f (x)g(x)
0
0
−
10
f ′ (x)g(x) dx
0
= f (10)g(10) − f (0)g(0) − 2
We can use left and right Riemann Sums with ∆x = 2 to approximate
R 10
0
Z
10
xg(x) dx.
0
xg(x) dx:
Left sum ≈ 0 · g(0)∆x + 2 · g(2)∆x + 4 · g(4)∆x + 6 · g(6)∆x + 8 · g(8)∆x
= (0(2.3) + 2(3.1) + 4(4.1) + 6(5.5) + 8(5.9)) 2 = 205.6.
Right sum ≈ 2 · g(2)∆x + 4 · g(4)∆x + 6 · g(6)∆x + 8 · g(8)∆x + 10 · g(10)∆x
= (2(3.1) + 4(4.1) + 6(5.5) + 8(5.9) + 10(6.1)) 2 = 327.6.
A good estimate for the integral is the average of the left and right sums, so
Z
0
10
xg(x) dx ≈
205.6 + 327.6
= 266.6.
2
Substituting values for f and g, we have
Z
0
10
f (x)g ′ (x) dx = f (10)g(10) − f (0)g(0) − 2
Z
10
xg(x) dx
0
≈ 102 (6.1) − 02 (2.3) − 2(266.6) = 76.8 ≈ 77.
67. Using integration by parts we have
Z
0
1
1
xf ′′ (x) dx = xf ′ (x)
0
−
Z
1
f ′ (x) dx
0
= 1 · f ′ (1) − 0 · f ′ (0) − [f (1) − f (0)]
= 2 − 0 − 5 + 6 = 3.
eax sin bx dx is
7.2 SOLUTIONS
68. Letting u =
√
603
1
x, u′ = √ , v = f (x), v ′ = f ′ (x), we have:
2 x
Z
√
√
f ′ (x) x dx = f (x) x −
| {z }
| {z }
uv
uv ′
Z
√
1
= f (x) x −
| {z } 2
h(x)
1
f (x) · √ dx
2 x
|
Z
Z
{z
}
vu′
1
f (x) · √ dx
x
|
{z
g ′ (x)
1
g ′ (x) dx
2
1
= h(x) − g(x) + C.
2
= h(x) −
}
69. Letting u = x, v ′ = f ′ (x), we have u′ = 1 and v = f (x). Integration by parts gives:
′
Z z uv
}| {
′
uv
z }| {
xf ′ (x) dx = xf (x) −
This means
Z
7
7
′
xf (x) dx = xf (x)
0
0
−
Z
f (x) dx.
7
f (x) dx
| 0 {z
}
5
= 7 f (7) −0 · f (0) − 5
|{z}
vu
Z z}|{
x = 7 is a zero of f
0
= −5.
70. (a) We have
F (a) =
Z
a
x2 e−x dx
0
a
= −x2 e−x
+
= (−x e
2 −x
= (−x e
2xe−x dx
0
0
2 −x
a
Z
a
− 2xe−x )
− 2xe
−x
+2
0
− 2e
Z
a
e−x dx
0
a
−x
)
0
= −a2 e−a − 2ae−a − 2e−a + 2.
(b) F (a) is increasing because x2 e−x is positive, so as a increases, the area under the curve from 0 to a also increases
and thus the integral increases.
(c) We have F ′ (a) = a2 e−a , so
F ′′ (a) = 2ae−a − a2 e−a = a(2 − a)e−a .
We see that F ′′ (a) > 0 for 0 < a < 2, so F is concave up on this interval.
71. We have
Bioavailability =
Z
3
15te−0.2t dt.
0
We first use integration by parts to evaluate the indefinite integral of this function. Let u = 15t and v ′ = e−0.2t dt, so
u′ = 15dt and v = −5e−0.2t . Then,
Z
15te
−0.2t
dt = (15t)(−5e
−0.2t
)−
= −75te−0.2t + 75
Z
Z
(−5e−0.2t )(15dt)
e−0.2t dt = −75te−0.2 − 375e−0.2t + C.
604
Chapter Seven /SOLUTIONS
Thus,
Z
3
3
0
15te−0.2t dt = (−75te−0.2t − 375e−0.2t )
0
= −329.29 + 375 = 45.71.
The bioavailability of the drug over this time interval is 45.71 (ng/ml)-hours.
72. (a) Increasing V0 increases the maximum value of V , since this maximum is V0 . Increasing ω or φ does not affect the
maximum of V .
(b) Since
dV
= −ωV0 sin(ωt + φ),
dt
the maximum of dV /dt is ωV0 . Thus, the maximum of dV /dt is increased if V0 or ω is increased, and is unaffected
if φ is increased.
(c) The period of V = V0 cos(ωt + φ) is 2π/ω, so
1
Average value =
2π/ω
Z
2π/ω
(V0 cos(ωt + φ))2 dt.
0
Substituting x = ωt + φ, we have dx = ωdt. When t = 0, x = φ, and when t = 2π/ω, x = 2π + φ. Thus,
ω
2π
Average value =
V02
2π
=
Z
2π+φ
V02 (cos x)2
φ
Z
1
dx
ω
2π+φ
(cos x)2 dx.
φ
Using integration by parts and the fact that sin2 x = 1 − cos2 x, we see that
Average value =
V02
2π
h
1
(cos x sin x + x)
2
i2π+φ
φ
V2
= 0 [cos(2π + φ) sin(2π + φ) + (2π + φ) − cos φ sin φ − φ]
4π
V02
V2
=
· 2π = 0 .
4π
2
Thus, increasing V0 increases the average value; increasing ω or φ has no effect.
However, it is not in fact necessary to compute the integral to see that ω does not affect the average value, since all
ω’s dropped out of the average value expression when we made the substitution x = ωt + φ.
dE
= r, so the total energy E used in the first T hours is given by E =
dt
integration by parts. Let u = t, v ′ = e−at . Then u′ = 1, v = − a1 e−at .
73. (a) We know that
E=
Z
Z
T
te−at dt. We use
0
T
te−at dt
0
t
= − e−at
a
T
0
−
Z
T
0
Z
1
− e−at
a
T
dt
1
1
e−at dt
= − T e−aT +
a
a 0
1
1
= − T e−aT + 2 (1 − e−aT ).
a
a
(b)
1
1
1
T
+ 2 1 − lim aT .
lim
T →∞ e
T →∞
a T →∞ eaT
a
Since a > 0, the second limit on the right hand side in the above expression is 0. In the first limit, although both the
numerator and the denominator go to infinity, the denominator eaT goes to infinity more quickly than T does. So in
T
the end the denominator eaT is much greater than the numerator T . Hence lim aT = 0. (You can check this by
T →∞ e
1
T
graphing y = aT on a calculator or computer for some values of a.) Thus lim E = 2 .
T →∞
e
a
lim E = −
7.2 SOLUTIONS
74. Letting u = ln |x| , u′ =
605
1
, v = f (x), v ′ = f ′ (x), we have:
x
Z
′
f (x) ln |x| dx = f (x) ln |x| −
{z
|
}
uv ′
|
{z
}
{z
}
uv
= f (x) ln |x| −
|
h(x)
Z
Z
1
· f (x) dx
x
| {z }
u′ v
f (x)
dx
x}
| {z
g ′ (x)
= h(x) − g(x) + C.
75. (a) We have u = erf(x), v ′ = 1, which means v = x. From the Construction Theorem, we see that erf(x) is an
2
2
2
2
antiderivative of √ e−x , so u′ = √ e−x . Using integration by parts, we have
π
π
′
u′ v
uv
Z z uv
}| {
Z z
}|
{
2
2
erf(x) dx = x erf(x) − x √ e−x dx
π
Z
2
2
√ e−x x dx.
= x erf(x) −
π
z }| {
(b) One possibility is to let w = −x2 , dw = −2xdx, which gives:
Z
2
2
2
√ e−x x dx = √
π
π
2
Z
ew −
1
2
2
1
1
dw = − √ ew + C = − √ e−x + C.
π
π
2
Another possibility is to let w = e−x , dw = −2xe−x dx, which leads to the same result.
(c) Based on our answers to parts (a) and (b), we have:
Z
erf(x) dx = x erf(x) −
Z
2
2
√ e−x x dx
π
2
1
= x erf(x) + √ e−x + C.
π
Checking our answer, we see that:
d
dx
2
1
x erf(x) + √ e−x + C
π
d
d
(x erf(x)) +
dx
dx
2
1
√ e−x
π
2
1
= 1 · erf(x) + x · (erf(x))′ + √ e−x (−2x)
π
| {z }
=
2
√2 e−x
π
2
2
2x
2x
= erf(x) + √ e−x − √ e−x
π
π
= erf(x).
Z
as required
x
1
dt, we see from the Construction Theorem that Li(x) is an antiderivative of 1/ ln(x), which
ln
t
2
′
means that (Li(x)) = 1/ ln(x). We need to perform integration by parts where u = Li(x), v = ln x. We have:
76. Since Li(x) =
u′ =
1
ln x
and
v′ =
1
,
x
which means
Z
Li(x)x−1 dx = uv −
| {z }
uv ′
Z
u′ v dx
= Li(x) ln x −
| {z }
uv
= Li(x) ln x −
Z
Z
integration by parts
1
· ln x dx
ln(x)
|
dx
= Li(x) ln x − x + C.
{z
u′ v
}
606
Chapter Seven /SOLUTIONS
Strengthen Your Understanding
77. To use integration by parts on
R
t ln t dt, use u = ln t, v ′ = t.
78. We can write arctan x = (1)(arctan x). Then using u = arctan x and v ′ = 1, we can evaluate the integral using
integration by parts.
79. Using integration by parts with u = f (x) and v ′ = 1 gives u′ = f ′ (x), v = x, so we get
Z
80. The integral
81.
82.
R
x3 ex dx
R
f (x) dx = xf (x) −
Z
xf ′ (x) dx.
θ2 sin θ dθ requires integration by parts twice. Many other answers are possible.
ex sin x dx, can be evaluated by applying integration by parts twice, both times with u = ex , or by applying integration
by parts once with u = sin x and then a second time with u = cos x.
R
83. True. Let u = t, v ′ = sin(5 − t), so u′ = 1, v = cos(5 − t). Then the integral
guess-and-check or by substituting w = 5 − t.
84. True. If we let u = t2 and v ′ = e3−t and integrate by parts twice, we obtain
Z
85. False. Suppose we have
R
R
1 · cos(5 − t) dt can be done by
t2 e3−t dt = −t2 e3−t + 2(−te3−t − e3−t ) + C.
x2 sin x dx. If we choose u = sin x and v ′ = x2 , the resulting integral is
x3
sin x −
3
Z
x3
cos x dx,
3
which is no simpler than the original integral. Choosing u = x2 and v ′ = sin x make the integral simpler. After integrating
twice, we get −x2 cos x + 2(x sin x + 2 cos x) + C.
Solutions for Section 7.3
Exercises
1 6
1 6
x ln x −
x + C. (Let n = 5 in III-13.)
6
36
1 (−3θ)
e
(−3 cos θ + sin θ) + C.
2.
10
(Let a = −3, b = 1 in II-9.)
1.
3. The integrand, a polynomial, x3 , multiplied by sin 5x, is in the form of III-15. There are only three successive derivatives
of x3 before 0 is reached (namely, 3x2 , 6x, and 6), so there will be four terms. The signs in the terms will be −++−, as
given in III-15, so we get
Z
1
1
1
1
x3 sin 5x dx = − x3 cos 5x +
· 3x2 sin 5x +
· 6x cos 5x −
· 6 sin 5x + C.
5
25
125
625
4. Formula III-13 applies only to functions of the form xn ln x, so we’ll have to multiply out and separate into two integrals.
Z
2
(x + 3) ln x dx =
Z
2
x ln x dx + 3
Z
ln x dx.
Now we can use formula III-13 on each integral separately, to get
Z
(x2 + 3) ln x dx =
x3
x3
ln x −
+ 3(x ln x − x) + C.
3
9
7.3 SOLUTIONS
607
5. Note that you can’t use substitution here: letting w = x3 + 5 does not work, since there is no dw = 3x2 dx in the
integrand. What will work is simply multiplying out the square: (x3 + 5)2 = x6 + 10x3 + 25. Then use I-1:
Z
(x3 + 5)2 dx =
Z
x6 dx + 10
Z
x3 dx + 25
Z
1 dx =
1
1 7
x + 10 · x4 + 25x + C.
7
4
1
6. − cos5 w + C
5
R
R
(Let x = cos w, as suggested in IV-23. Then − sin w dw = dx, and sin w cos4 w dw = − x4 dx.)
3
3
1
7. − sin3 x cos x − sin x cos x + x + C.
4
8
8
(Use IV-17.)
1 3 3 2 3
3 2x
8.
x − x + x−
e + C.
2
4
4
8
3
(Let a = 2, p(x) = x in III-14.)
1 2 2
2
e3x + C.
x − x+
9.
3
9
27
(Let a = 3, p(x) = x2 in III-14.)
1 3
10. ex + C.
3
(Substitute w = x3 , dw = 3x2 dx. It is not necessary to use the table.)
8
8
1 4 4 3 4 2
x − x + x −
x+
e3x + C.
11.
3
9
9
27
81
(Let a = 3, p(x) = x4 in III-14.)
12. Substitute w = 5u, dw = 5 du. Then
Z
Z
1
w5 ln w dw
56
1 6
1 1
w + C)
= 6 ( w6 ln w −
5 6
36
1
1 6
= u6 ln 5u −
u + C.
6
36
u5 ln(5u) du =
Or use ln 5u = ln 5 + ln u.
Z
u5 ln 5u du = ln 5
Z
u5 du +
Z
u5 ln u du
1
1 6
u6
ln 5 + u6 ln u −
u +C
6
6
36
6
u
1 6
=
ln 5u −
u + C.
6
36
=
(using III-13)
y
1
13. √ arctan √ + C.
3
√ 3
(Let a = 3 in V-24).
14. We first factor out the 9 and then use formula V-24:
Z
dx
=
9x2 + 16
Z
dx
x
1
1
+C
= ·
arctan
9(x2 + 16/9)
9 4/3
4/3
1
3x
=
arctan
+ C.
12
4
15. We first factor out the 16 and then use formula V-28 to get
Z
√
dx
=
25 − 16x2
Z
dx
p
16(25/16 −
x2 )
=
1
4
Z
dx
p
(5/4)2 − x2
608
Chapter Seven /SOLUTIONS
1
x
arcsin
+C
4
5/4
4x
1
+ C.
= arcsin
4
5
=
16. The integral suggests formula VI-29, but is not a perfect match because of the coefficient of 9. One way to deal with the
9 is to factor it out, so that, using formula IV-29,
Z
dx
√
=
9x2 + 25
Z
Alternatively, we can write
dx
1
p
=
2
3
9(x + 25/9)
Z
Z
dx
1
p
= ln x +
2
2
3
x + (5/3)
dx
√
=
9x2 + 25
Z
dx
p
(3x)2 + 25
r
x2 +
2
5
3
.
We now use the substitution w = 3x, so that dw = 3dx, and the integral becomes
Z
√
1
dw
3
2
w + 25
=
p
p
1
1
ln w + w2 + 25 + C = ln |3x + 9x2 + 25| + C.
3
3
5
3
sin 3θ sin 5θ +
cos 3θ cos 5θ + C.
16
16
(Let a = 3, b = 5 in II-12.)
3
5
18.
cos 3θ sin 5θ −
sin 3θ cos 5θ + C.
16
16
(Let a = 3, b = 5 in II-10.)
17.
19. Let m = 3 in IV-21.
Z
Z
1
1
1 sin x
1
dx =
+
dx
cos3 x
2 cos2 x
2
cos x
1
1 sin x
sin x + 1
+ ln
=
+ C by IV-22.
2 cos2 x
4
sin x − 1
20. Use long division to reorganize the integral:
Z
t2 + 1
dt =
t2 − 1
Z
1+
2
t2 − 1
To get this second integral, let a = 1, b = −1 in V-26, so
Z
dt =
Z
dt +
Z
2
dt.
(t − 1)(t + 1)
t2 + 1
dt = t + ln |t − 1| − ln |t + 1| + C.
t2 − 1
1 5x
e (5 sin 3x − 3 cos 3x) + C.
34
(Let a = 5, b = 3 in II-8.)
1
(7 cos 2y sin 7y − 2 sin 2y cos 7y) + C.
22.
45
(Let a = 2, b = 7 in II-11.)
21.
23.
Z
1
y 2 sin 2y dy = − y 2 cos 2y +
2
1
= − y 2 cos 2y +
2
(Use a = 2, p(y) = y 2 in III-15.)
1
1
(2y) sin 2y + (2) cos 2y + C
4
8
1
1
y sin 2y + cos 2y + C.
2
4
+ C.
7.3 SOLUTIONS
24. Substitute w = x2 , dw = 2x dx. Then
Z
Z
x3 sin x2 dx =
1
2
Z
w sin w dw. By III-15, we have
1
1
1
1
w sin w dw = − w cos w + sin w + C = − x2 cos x2 + sin x2 + C.
2
2
2
2
25. Substitute w = 7x, dw = 7 dx. Then use IV-21.
Z
1
1
dx =
cos4 7x
7
Z
1
1
dw =
cos4 w
7
1 sin w
2
+
3 cos3 w
3
Z
1
dw
cos2 w
1 sin w
2 sin w
+
+C
21 cos3 w
21 cos w
1 tan w
2
=
+
tan w + C
21 cos2 w
21
2
1 tan 7x
+
tan 7x + C.
=
21 cos2 7x
21
h
=
i
26. Substitute w = 3θ, dw = 3 dθ. Then use IV-19, letting m = 3.
Z
1
1
dθ =
3
sin3 3θ
Z
1
1
1
1 cos w
dw =
+
−
3
2 sin2 w
2
sin3 w
1
1 cos w
+
=−
6 sin2 w
6
=−
Z
1
dw
sin w
1
cos(w) − 1
ln
+C
2
cos(w) + 1
cos(3θ) − 1
1
1 cos 3θ
+ C.
+
ln
6 sin2 3θ
12
cos(3θ) + 1
by IV-20
27. Substitute w = 2θ, dw = 2 dθ. Then use IV-19, letting m = 2.
Z
1
1
dθ =
2
2
sin 2θ
Z
1 cos w
1
1
1
dw = (−
)+C =−
+C =−
+ C.
2
2
sin
w
2
tan
w
2
tan
2θ
sin w
28. Use IV-21 twice to get the exponent down to 1:
Z
Z
1 sin x
3
1
dx =
+
cos5 x
4 cos4 x
4
1 sin x
1
1
dx =
+
cos3 x
2 cos2 x
2
Z
Z
1
dx
cos3 x
1
dx.
cos x
Now use IV-22 to get
Z
Putting this all together gives
Z
29.
30.
Z
(sin x) + 1
1
1
+ C.
dx = ln
cos x
2
(sin x) − 1
(sin x) + 1
1
1 sin x
3 sin x
3
+ C.
dx =
+
+
ln
cos5 x
4 cos4 x
8 cos2 x
16
(sin x) − 1
1
dx =
x2 + 4x + 3
Z
1
1
dx = (ln |x + 1| − ln |x + 3|) + C.
(x + 1)(x + 3)
2
(Let a = −1 and b = −3 in V-26).
You need not use the table.
Z
1
dx =
x2 + 4x + 4
Z
1
1
dx = −
+ C.
(x + 2)2
x+2
609
610
Chapter Seven /SOLUTIONS
31.
Z
1
dz
= − (ln |z| − ln |z − 3|) + C.
z(z − 3)
3
(Let a = 0, b = 3 in V-26.)
32.
Z
dy
=−
4 − y2
Z
dy
1
= − (ln |y − 2| − ln |y + 2|) + C.
(y + 2)(y − 2)
4
(Let a = 2, b = −2 in V-26.)
33. arctan(z + 2) + C.
(Substitute w = z + 2 and use V-24, letting a = 1.)
34.
Z
1
dy =
y 2 + 4y + 5
(Substitute w = y + 2, and let a = 1 in V-24).
Z
1
dy = arctan(y + 2) + C.
1 + (y + 2)2
35. We use the method of IV-23 in the table. Using the Pythagorean Identity, we rewrite the integrand:
sin3 x = sin2 x sin x = 1 − cos2 x sin x = sin x − cos2 x sin x.
Thus, we have
Z
3
sin x dx =
=
Z
Z
sin x − cos2 x sin x dx
sin x dx −
Z
cos2 x sin x dx.
The first of these new integrals can be easily found. The second can be found using the substitution w = cos x so
dw = − sin x dx. The second integral becomes
Z
cos2 x sin x dx = −
Z
w2 dw
1
= − w3 + C
3
1
= − cos3 x + C,
3
so the final answer is
Z
3
sin x dx =
Z
sin x dx −
Z
cos2 x sin x dx
= − cos x + (1/3) cos3 x + C.
36. Using the advice in IV-23, since both m and n are even and since n is negative, we convert everything to cosines, since
cos x is in the denominator.
Z
4
tan x dx =
=
=
By IV-21
Z
Z
Z
sin4 x
dx
cos4 x
Z
1
dx − 2
cos4 x
Z
(1 − cos2 x)2
dx
cos4 x
Z
1
1 sin x
2
dx =
+
cos4 x
3 cos3 x
3
sin x
1
dx =
+ C.
cos2 x
cos x
1
dx +
cos2 x
Z
Z
1
dx,
cos2 x
1 dx.
7.3 SOLUTIONS
Substituting back in, we get
Z
1 sin x
4 sin x
−
+ x + C.
3 cos3 x
3 cos x
tan4 x dx =
37. Since cosh2 x − sinh2 x = 1 we rewrite sinh3 x as sinh x sinh2 x = sinh x(cosh2 x − 1). Then
Z
2
3
sinh x cosh x dx =
=
Z
Z
sinh x(cosh2 x − 1) cosh2 x dx
(cosh4 x − cosh2 x) sinh x dx.
Now use the substitution w = cosh x, dw = sinh x dx to find
Z
(w4 − w2 ) dw =
1 5 1 3
1
1
w − w + C = cosh5 x − cosh3 x + C.
5
3
5
3
38. Since cosh2 x − sinh2 x = 1 we rewrite cosh3 x as cosh x cosh2 x = cosh x(1 + sinh2 x). This gives
Z
Z
sinh2 x cosh3 x dx =
Z
=
sinh2 x(sinh2 x + 1) cosh x dx
(sinh4 x + sinh2 x) cosh x dx.
Now use the substitution w = sinh x, dw = cosh x dx to find
Z
(w4 − w2 ) dw =
1 5 1 3
1
1
w − w + C = sinh5 x + sinh3 x + C.
5
3
5
3
39.
Z
sin3 3θ cos2 3θ dθ =
=
Z
Z
(sin 3θ)(cos2 3θ)(1 − cos2 3θ) dθ
sin 3θ(cos2 3θ − cos4 3θ) dθ.
Using an extension of the tip given in rule IV-23, we let w = cos 3θ, dw = −3 sin 3θ dθ.
Z
sin 3θ(cos2 3θ − cos4 3θ) dθ = −
1
3
Z
(w2 − w4 ) dw
w5
1 w3
−
)+C
=− (
3 3
5
1
1
= − (cos3 3θ) +
(cos5 3θ) + C.
9
15
40. If we make the substitution w = 2z 2 then dw = 4z dz, and the integral becomes:
Z
ze
2z 2
1
cos(2z ) dz =
4
2
Now we can use Formula 9 from the table of integrals to get:
1
4
Z
Z
ew cos w dw
1 1 w
e (cos w + sin w) + C
4 2
1
= ew (cos w + sin w) + C
8
1 2
= e2z (cos 2z 2 + sin 2z 2 ) + C
8
ew cos w dw =
h
i
611
612
Chapter Seven /SOLUTIONS
41. Substitute w = 3α, dw = 3 dα. Then dα =
Z
1
dw. We have
3
π
α= 12
sin 3α dα =
α=0
1
3
Z
w= π
4
sin w dw
w=0
π
4
1
= − cos w
3
√ 0
√
1
1
2
2
−1 =
1−
.
=−
3
2
3
2
42. Let a = 5 and b = 6 in II-12. Then
π
Z
1
(6 sin 5x sin 6x + 5 cos 5x cos 6x)
11
sin 5x cos 6x dx =
−π
π
= 0.
−π
This answer makes sense since sin 5x cos 6x is odd, so its integral over any interval −a ≤ x ≤ a should be 0.
43. Using III-13:
2
Z
Z
(x − 2x3 ) ln x dx =
1
2
1
x ln x dx − 2
Z
1
1
= ( x2 ln x − x2 )
2
4
2
x3 ln x dx
1
2
1
1
− ( x4 ln x − x4 )
2
8
1
2
1
15
3
= 2 ln 2 − − (8 ln 2 −
)
4
8
9
= − 6 ln 2 = −3.034.
8
44. Let a =
√
3 in VI-30 and VI-28:
Z
1
0
Z
1
1 p
3
x
3 − x2 dx = ( x 3 − x2 + arcsin √ )
2
2
3
p
1
0
≈ 1.630.
1
Z
1
1
dx =
dx.
45.
2 + 2x + 1
x
(x
+
1)2
0
0
We substitute w = x + 1, so dw = dx. Note that when x = 1, we have w = 2, and when x = 0, we have w = 1.
Z
x=1
x=0
Z
1
dx =
(x + 1)2
w=2
w=1
1
1
dw = −
w2
w
w=2
w=1
=−
1
1
+1= .
2
2
46. Substitute w = x + 1 and use V-24:
Z
0
1
dx
==
x2 + 2x + 5
Z
1
0
dx
(x + 1)2 + 4
x+1
1
= arctan
2
2
1
=
0
47. Let w = x2 , dw = 2x dx. When x = 0, w = 0, and when x =
Z
1
√
2
0
x dx
√
=
1 − x4
Z
0
1
2
1
2
dw
1
1
1
arctan 1 − arctan ≈ 0.1609.
2
2
2
√1 ,
2
1
√
= arcsin w
2
2
1−w
w = 12 . Then
1
2
=
0
1
1
π
(arcsin − arcsin 0) =
.
2
2
12
7.3 SOLUTIONS
48. Let w = x + 2, giving dw = dx. When x = 0, w = 2, and when x = 1, w = 3. Thus,
1
Z
(x + 2)
dx =
(x + 2)2 + 1
0
Z
3
2
w
dw.
w2 + 1
2
For the last integral, we make the substitution u = w + 1, du = 2w dw. Then, we have
Z
3
2
w
1
dw = ln |w2 + 1|
w2 + 1
2
3
2
1
= (ln |10| − ln |5|)
2
10
1
1
= ln(2)
= ln
2
5
2
49. Using IV-19 with m = 3, followed by IV-20, we find that
Z
π/3
π/4
cos x − 1
dx
1 cos x
1
=−
+ ln
2 sin2 x
4
cos x + 1
sin3 x
π/3
= 0.5398.
π/4
50. Substitute w = x + 3 and then use VI-29:
Z
−1
−3
√
dx
=
x2 + 6x + 10
Z
−1
−3
−1
dx
= ln |x + 3 +
p
(x + 3)2 + 1
√
= ln |2 + 5| ≈ 1.4436.
p
x2 + 6x + 10|
Problems
51. Letting w = 2x + 1, dw = 2 dx, dx = 0.5 dw, we find that k = 0.5, w = 2x + 1, n = 3:
Z
(2x + 1)3 ln(2x + 1) dx =
Z
w3 ln(w)0.5 dw =
Z
0.5w3 ln w dw.
52. Letting w = 2x + 1, dw = 2 dx, dx = 0.5 dw, we see that k = −1/4, w = 2x + 1, n = 3:
Z
(2x + 1)3 ln √
Z
1
1
dx =
w3 ln
0.5 dw
1/2
w
2x + 1
Z
0.5w3 ln w−1/2 dw
=
=
=
Z
Z
0.5w3 −
1
ln w dw
2
log property
1
− w3 ln w dw.
4
53. We can use form (i) by writing this as
Z
where a = −1/6, b = −1/4, c = 5.
dx
=
x
x2
5− −
4
6
Z
dx
,
1 2
1
− x + −
x+5
6
4
.
−3
613
614
Chapter Seven /SOLUTIONS
54. We can use form (ii) by writing this as
Z
dx
5
2x +
7 + 3x
=
Z
dx
5
2x +
7 + 3x
7 + 3x
7 + 3x
·
simplify
Z
3x + 7
dx distribute
5
2x(3x + 7) +
· (3x + 7)
7 + 3x
Z
3x + 7
dx,
=
6x2 + 14x + 5
=
with m = 3, n = 7, a = 6, b = 14, c = 5.
55. We can use form (iii) by writing this as:
Z
Z
dx
dx
=
2 − 5x + 6)3 ((x − 2)(x − 2))2 ((x − 3)(x − 3))2
(x2 − 5x + 6)3 (x2 − 4x + 4)2 (x2 − 6x + 9)2
(x
Z
dx
=
2
3
4
4
Z (x − 5x + 6) (x − 2) (x − 3)
dx
=
4
Z (x2 − 5x + 6)3 ((x − 2)(x − 3))
dx
=
2 − 5x + 6)3 (x2 − 5x + 6)4
(x
Z
dx
=
(x2 − 5x + 6)7
where a = 1, b = −5, c = 6, n = 7.
56. We see that a = e, b = 4, λ = 3:
Z
e6x
dx =
4 + e3x+1
Z
e6x
dx =
4 + e3x e1
Z
e2·3x
dx.
e · e3x + 4
57. Multiplying the numerator and denominator by e−4x , we see that a = 5, b = 4, λ = 2:
Z
e8x
e8x
e−4x
dx =
· −4x dx
4x
6x
4x
6x
4e + 5e
Z 4e + 5e8x−4x e
e
dx
=
4x e−4x + 5e6x e−4x
4e
Z
4x
e
=
dx
4x−4x
6x−4x
Z 4e 2·2x + 5e
e
=
dx,
5e2x + 4
Z
58. (a) Let w = x2 , dw = 2xdx. Then
Z
2
x5 ebx dx =
=
=
Z
Z
Z
2
x4 ebx x dx
w2 ebw
1
dw
2
kw2 ebw dw.
since x4 = w2 and xdx = 21 dw
where k =
1
2
(b) Using the identity from the table, we have
Z
2 bw
kw e
dw = ke
bw
so, since k = 12 , w = x2 ,
Z
2 bw
kw e
1 2
dw = ebx
2
w2
2w
2
− 2 + 3
b
b
b
2x2
2
x4
− 2 + 3
b
b
b
+ C,
+ C.
factor
regroup
regroup
multiply out
simplify,
7.3 SOLUTIONS
59. Using II-10 in the integral table, if m 6= ±n, then
Z
π
sin mθ sin nθ dθ
=
−π
=
1
[m cos mθ sin nθ − n sin mθ cos nθ]
n2 − m2
π
−π
1
[(m cos mπ sin nπ − n sin mπ cos nπ) −
n2 − m2
(m cos(−mπ) sin(−nπ) − n sin(−mπ) cos(−nπ))]
But sin kπ = 0 for all integers k, so each term reduces to 0, making the whole integral reduce to 0.
60. Using formula II-11, if m 6= ±n, then
Z
π
cos mθ cos nθ dθ =
−π
1
(n cos mθ sin nθ − m sin mθ cos nθ)
n2 − m2
We see that in the evaluation, each term will have a sin kπ term, so the expression reduces to 0.
61. (a)
1
1−0
Z
1
V0
sin(120πt)
120π
V0 cos(120πt)dt =
0
1
0
V0
[sin(120π) − sin(0)]
=
120π
V0
=
[0 − 0] = 0.
120π
(b) Let’s find the average of V 2 first.
2
V = Average of V 2 =
=
1
1−0
1
1−0
= V02
Z
Z
1
V 2 dt
0
Z
1
(V0 cos(120πt))2 dt
0
1
cos2 (120πt)dt
0
Now, let 120πt = x, and dt =
dx
. So
120π
V2
V = 0
120π
2
V2
= 0
120π
=
Z
120π
cos2 xdx.
0
1
1
cos x sin x + x
2
2
V02
V2
60π = 0 .
120π
2
120π
II-18
0
p
V0
V2
So, the average of V 2 is 0 and V = average of V 2 = √ .
2
2
√
√
(c) V0 = 2 · V = 110 2 ≈ 156 volts.
62. (a) Since R(T ) is the rate or production, we find the total production by integrating:
Z
0
N
R(t) dt =
Z
N
(A + Be−t sin(2πt)) dt
0
= NA + B
Z
0
N
e−t sin(2πt) dt.
π
.
−π
615
616
Chapter Seven /SOLUTIONS
Let a = −1 and b = 2π in II-8.
B
e−t (− sin(2πt) − 2π cos(2πt))
1 + 4π 2
= NA +
N
.
0
Since N is an integer (so sin 2πN = 0 and cos 2πN = 1),
Z
N
Z
N
R(t) dt = N A + B
0
2π
(1 − e−N ).
1 + 4π 2
−N
2πB
) over the first N years.
Thus the total production is N A + 1+4π
2 (1 − e
(b) The average production over the first N years is
0
R(t) dt
2πB
=A+
N
1 + 4π 2
1 − e−N
N
.
−N
2πB 1−e
→ A, since the second term in the sum goes to 0. This is why A is called the
(c) As N → ∞, A + 1+4π
2
N
average!
(d) When t gets large, the term Be−t sin(2πt) gets very small. Thus, R(t) ≈ A for most t, so it makes sense that the
RN
average of 0 R(t) dt is A as N → ∞.
(e) This model is not reasonable for long periods of time, since an oil well has finite capacity and will eventually “run
dry.” Thus, we cannot expect average production to be close to constant over a long period of time.
Strengthen Your Understanding
63. To find the integral, we factor the denominator
and use V-26 with a =
√
7 − t2 = −(t −
√
7, b = − 7.
√
7)(t +
√
7)
64. If a = 3, then x2 + 4x + 3 = (x + 1)(x + 3), so by Formula V-26, the answer involves ln not arctan. In general, the
antiderivative involves an arctan only if the quadratic has no real roots.
65. To use the table, we first make the substitution w = 2x + 1. Then dw = 2 dx. So the integral becomes
Z
e2x+1 sin(2x + 1) dx =
1
2
Z
ew sin w dw + C.
Now using Formula II-8 we get
Z
e2x+1 sin(2x + 1) dx =
1 2x+1
e
(sin(2x + 1) − cos(2x + 1)) + C.
4
66. Formula II-12 only holds for a 6= b; the integral is defined.
The integral can be solved using substitution. Let sin x = w. Then cos x dx = dw. We have
Z
sin x cos x dx =
Z
w dw =
w2
+ C.
2
Substituting w = sin x back we get
Z
sin x cos x dx =
R
sin2 x
+ C.
2
67. The table only gives formulas for p(x) sin x dx, where p(x) is a polynomial. Since 1/x is not a polynomial, the table
is no help. In fact, it is known that sin x/x does not have an elementary function as an antiderivative.
R √
68. To evaluate 1/ 2x − x2 dx, we must first complete the square for 2x − x2 to get 1 − (x − 1)2 . Then we can substitute
w = x − 1 and use Formula V-28 in the table.
69.
R
1/ sin4 x dx can be evaluated by using Formula IV-19 twice.
7.4 SOLUTIONS
617
70. True. Rewrite sin7 θ = sin θ sin6 θ = sin θ(1 − cos2 θ)3 . Expanding, substituting w = cos θ, dw = − sin θ dθ, and
integrating gives a polynomial in w, which is a polynomial in cos θ.
71. False. Completing the square gives
Z
dx
=
x2 + 4x + 5
Z
dx
= arctan(x + 2) + C.
(x + 2)2 + 1
72. False. Factoring gives
Z
dx
=
x2 + 4x − 5
Z
dx
1
=
(x + 5)(x − 1)
6
Z
1
1
−
x−1
x+5
dx =
1
(ln |x − 1| − ln |x + 5|) + C.
6
73. True. Let w = ln x, dw = x−1 dx. Then
Z
x−1 ((ln x)2 + (ln x)3 ) dx =
Z
(w2 + w3 ) dw =
(ln x)3
(ln x)4
w3
w4
+
+C =
+
+ C.
3
4
3
4
Solutions for Section 7.4
Exercises
1. Since 6x + x2 = x(6 + x), we take
A
B
x+1
=
+
.
6x + x2
x
6+x
So,
x + 1 = A(6 + x) + Bx
x + 1 = (A + B)x + 6A,
giving
A+B = 1
6A = 1.
Thus A = 1/6, and B = 5/6 so
1/6
5/6
x+1
=
+
.
6x + x2
x
6+x
2. Since 25 − x2 = (5 − x)(5 + x), we take
A
B
20
=
+
.
25 − x2
5−x
5+x
So,
20 = A(5 + x) + B(5 − x)
20 = (A − B)x + 5A + 5B,
giving
A−B = 0
5A + 5B = 20.
Thus A = B = 2 and
2
2
20
=
+
.
25 − x2
5−x
5+x
618
Chapter Seven /SOLUTIONS
3. Since w4 − w3 = w3 (w − 1), we have
A
B
C
D
1
=
+
+ 2 + 3.
w4 − w3
w−1
w
w
w
Thus,
1 = Aw3 + B(w − 1)w2 + C(w − 1)w + D(w − 1)
1 = (A + B)w3 + (−B + C)w2 + (−C + D)w + (−D),
giving
A+B = 0
−B + C = 0
−C + D = 0
−D = 1.
Thus A = 1, B = −1, C = −1 and D = −1 so
1
1
1
1
1
=
− − 2 − 3.
w4 − w3
w−1
w
w
w
4. Since y 3 − y 2 + y − 1 = (y − 1)(y 2 + 1), we take
A
By + C
2y
=
+ 2
y3 − y2 + y − 1
y−1
y +1
So,
2y = A(y 2 + 1) + (By + C)(y − 1)
2y = (A + B)y 2 + (C − B)y + A − C,
giving
A+B = 0
−B + C = 2
A − C = 0.
Thus A = C = 1, B = −1 so
1
1−y
2y
=
+ 2
.
y3 − y2 + y − 1
y−1
y +1
5. Since y 3 − 4y = y(y − 2)(y + 2), we take
A
B
C
8
=
+
+
.
y 3 − 4y
y
y−2
y+2
So,
8 = A(y − 2)(y + 2) + By(y + 2) + Cy(y − 2)
8 = (A + B + C)y 2 + (2B − 2C)y − 4A,
giving
A+B+C = 0
2B − 2C = 0
−4A = 8.
Thus A = −2, B = C = 1 so
8
−2
1
1
=
+
+
.
y 3 − 4y
y
y−2
y+2
7.4 SOLUTIONS
619
6. Since s2 + 3s + 2 = (s + 2)(s + 1), we have
2(1 + s)
2(1 + s)
2
=
=
,
s(s2 + 3s + 2)
s(s + 2)(s + 1)
s(s + 2)
so we take
A
B
2
=
+
.
s(s + 2)
s
s+2
Thus,
2 = A(s + 2) + Bs
2 = (A + B)s + 2A,
giving
A+B = 0
2A = 2.
Thus A = 1 and B = −1 and
1
1
2(1 + s)
= −
.
s(s2 + 3s + 2)
s
s+2
7. Since s4 − 1 = (s2 − 1)(s2 + 1) = (s − 1)(s + 1)(s2 + 1), we have
A
B
Cs + D
2
=
+
+ 2
.
s4 − 1
s−1
s+1
s +1
Thus,
2 = A(s + 1)(s2 + 1) + B(s − 1)(s2 + 1) + (Cs + D)(s − 1)(s + 1)
2 = (A + B + C)s3 + (A − B + D)s2 + (A + B − C)s + (A − B − D),
giving
A+B+C = 0
A−B+D = 0
A+B−C = 0
A − B − D = 2.
From the first and third equations we find A + B = 0 and C = 0. From the second and fourth we find A − B = 1 and
D = −1. Thus A = 1/2 and B = −1/2 and
1
1
1
2
=
−
− 2
.
s4 − 1
2(s − 1)
2(s + 1)
s +1
8. Using the result of Problem 1, we have
Z
x+1
dx =
6x + x2
Z
1/6
dx +
x
Z
5/6
1
dx = (ln |x| + 5 ln |6 + x|) + C.
6+x
6
9. Using the result of Problem 2, we have
Z
20
dx =
25 − x2
Z
2
dx +
5−x
Z
2
dx = −2 ln |5 − x| + 2 ln |5 + x| + C.
5+x
10. Using the result of Exercise 3, we have
Z
1
dw =
w4 − w3
Z
1
1
1
1
− − 2 − 3
w−1
w
w
w
dw = ln |w − 1| − ln |w| +
1
1
+
+ C.
w
2w2
620
Chapter Seven /SOLUTIONS
11. Using the result of Problem 4, we have
Z
2y
dy =
y3 − y2 + y − 1
Z
1
dy +
y−1
Z
1−y
1
dy = ln |y − 1| + arctan y − ln y 2 + 1 + C.
y2 + 1
2
12. Using the result of Problem 5, we have
Z
8
dy =
y 3 − 4y
Z
−2
dy +
y
Z
1
dy +
y−2
Z
1
dy = −2 ln |y| + ln |y − 2| + ln |y + 2| + C.
y+2
−
1
s+2
13. Using the result of Exercise 6, we have
Z
2(1 + s)
ds =
s(s2 + 3s + 2)
Z
1
s
ds = ln |s| − ln |s + 2| + C.
14. Using the result of Exercise 7, we have
Z
2
ds =
s4 − 1
Z
1
1
1
−
− 2
2(s − 1)
2(s + 1)
s +1
15. We let
ds =
1
1
ln |s − 1| − ln |s + 1| − arctan s + C.
2
2
A
B
C
3x2 − 8x + 1
=
+
+
x3 − 4x2 + x + 6
x−2
x+1
x−3
giving
3x2 − 8x + 1 = A(x + 1)(x − 3) + B(x − 2)(x − 3) + C(x − 2)(x + 1)
3x2 − 8x + 1 = (A + B + C)x2 − (2A + 5B + C)x − 3A + 6B − 2C
so
A+B+C = 3
−2A − 5B − C = −8
−3A + 6B − 2C = 1.
Thus, A = B = C = 1, so
Z
3x2 − 8x + 1
dx =
x3 − 4x2 + x + 6
Z
dx
+
x−2
Z
dx
+
x+1
Z
dx
= ln |x − 2| + ln |x + 1| + ln |x − 3| + K.
x−3
We use K as the constant of integration, since we already used C in the problem.
16. We let
1
1
A
B
C
= 2
=
+ 2 +
x3 − x2
x (x − 1)
x
x
x−1
giving
1 = Ax(x − 1) + B(x − 1) + Cx2
1 = (A + C)x2 + (B − A)x − B
so
A+C = 0
B−A = 0
−B = 1.
Thus, A = B = −1, C = 1, so
Z
dx
=−
x3 − x2
Z
dx
−
x
Z
dx
+
x2
Z
dx
= − ln |x| + x−1 + ln |x − 1| + K.
x−1
We use K as the constant of integration, since we already used C in the problem.
7.4 SOLUTIONS
17. We let
10x + 2
10x + 2
A
Bx + C
=
=
+ 2
x3 − 5x2 + x − 5
(x − 5)(x2 + 1)
x−5
x +1
giving
10x + 2 = A(x2 + 1) + (Bx + C)(x − 5)
10x + 2 = (A + B)x2 + (C − 5B)x + A − 5C
so
A+B = 0
C − 5B = 10
A − 5C = 2.
Thus, A = 2, B = −2, C = 0, so
Z
10x + 2
dx =
x3 − 5x2 + x − 5
Z
2
dx −
x−5
Z
2x
dx = 2 ln |x − 5| − ln x2 + 1 + K.
x2 + 1
18. Division gives
4x2 + 25x + 11
x4 + 12x3 + 15x2 + 25x + 11
=x+ 3
.
3
2
x + 12x + 11x
x + 12x2 + 11x
Since x3 + 12x2 + 11x = x(x + 1)(x + 11), we write
4x2 + 25x + 11
A
B
C
=
+
+
x3 + 12x2 + 11x
x
x+1
x + 11
giving
4x2 + 25x + 11 = A(x + 1)(x + 11) + Bx(x + 11) + Cx(x + 1)
4x2 + 25x + 11 = (A + B + C)x2 + (12A + 11B + C)x + 11A
so
A+B+C = 4
12A + 11B + C = 25
11A = 11.
Thus, A = B = 1, C = 2 so
Z
x4 + 12x3 + 15x2 + 25x + 11
dx =
x3 + 12x2 + 11x
=
Z
x dx +
Z
dx
+
x
Z
dx
+
x+1
Z
2dx
x + 11
x2
+ ln |x| + ln |x + 1| + 2 ln |x + 11| + K.
2
We use K as the constant of integration, since we already used C in the problem.
19. Division gives
x4 + 3x3 + 2x2 + 1
1
= x2 + 2
.
x2 + 3x + 2
x + 3x + 2
Since x2 + 3x + 2 = (x + 1)(x + 2), we write
1
1
A
B
=
=
+
,
x2 + 3x + 2
(x + 1)(x + 2)
x+1
x+2
giving
1 = A(x + 2) + B(x + 1)
1 = (A + B)x + 2A + B
621
622
Chapter Seven /SOLUTIONS
so
A+B = 0
2A + B = 1.
Thus, A = 1, B = −1 so
Z
x4 + 3x3 + 2x2 + 1
dx =
x2 + 3x + 2
Z
=
x2 dx +
Z
dx
−
x+1
Z
dx
x+2
x3
+ ln |x + 1| − ln |x + 2| + C.
3
20. Since x = (3/2) sin t, we have dx = (3/2) cos t dt. Substituting into the integral gives
Z
√
1
=
9 − 4x2
Z
21. Since x = sin t + 2, we have
1
p
9 − 9 sin2 t
3
cos t dt =
2
Z
1
1
1
2x
dt = t + C = arcsin
2
2
2
3
+ C.
4x − 3 − x2 = 4(sin t + 2) − 3 − (sin t + 2)2 = 1 − sin2 t = cos2 t
and dx = cos t dt, so substitution gives
Z
√
1
=
4x − 3 − x2
Z
√
1
cos t dt =
cos2 t
Z
dt = t + C = arcsin(x − 2) + C.
22. Completing the square gives x2 + 4x + 5 = 1 + (x + 2)2 . Since x + 2 = tan t and dx = (1/ cos2 t)dt, we have
Z
1
dx =
x2 + 4x + 5
Z
1
1
·
dt =
1 + tan2 t cos2 t
Z
dt = t + C = arctan(x + 2) + C.
23. (a) Yes, use x = 3 sin θ.
(b) No; better to substitute w = 9 − x2 , so dw = −2x dx.
24. (a) Substitute w = √
x2 + 10, so dw = 2x dx.
(b) Substitute x = 10 tan θ.
Problems
25. We have
w = (3x + 2)(x − 1)
so
= 3x2 − x − 2
dw = (6x − 1) dx
whichZmeans (12x − 2) dx = 2 Z
dw,
dw
12x − 2
dx = 2
,
giving
(3x + 2)(x − 1)
w
so w = (3x + 2)(x − 1) = 3x2 − x − 2 and k = 2.
26. Using partial fractions, we have
12x − 2
A
B
=
+
(3x + 2)(x − 1)
3x
+
2
x
−1
B
A
+
(3x + 2)(x − 1)
12x − 2 =
3x + 2
x−1
12x − 2 = A(x − 1) + B(3x + 2)
= (A + 3B)x + 2B − A
so A + 3B = 12
multiply by denominator
simplify
regroup
matching coefficients of x
7.4 SOLUTIONS
and 2B − A = −2
matching constants
A + 3B + (2B − A) = 12 + (−2)
giving
adding above two equations
5B = 10
B=2
which means
yielding
Z
A + 3 · 2 = 12
letting B = 2 in A + 3B = 12
A=Z
6
Z
6
2
12x − 2
dx =
+
dx.
(3x + 2)(x − 1)
3x + 2
x−1
Checking our answer, we see that:
2
6
x−1
2
3x + 2
6
+
=
·
+
·
3x + 2
x−1
3x + 2 x − 1
x − 1 3x + 2
6(x − 1) + 2(3x + 2)
=
(3x + 2)(x − 1)
6x − 6 + 6x + 4
=
(3x + 2)(x − 1)
12x − 2
=
,
(3x + 2)(x − 1)
as required.
27. Focusing on the integrand, we have:
2x + 9
2x + 9
=
5
(3x + 5)(4 − 5x)
3 x + 3 (−5) x − 54
2x + 9
=
−15 x + 53 x− 54
2
· x + − 35
− 15
.
=
x + 35 x − 45
So
Z
2x + 9
dx =
(3x − 5)(4 − 5x)
Z
c
d
z}|{
z }| {
2
3
·x + −
15
5 dx.
5
4
x− −
x−
3
5
−
|
We see that a = −5/3, b = 4/5, c = −2/15, d = −3/5.
{z
} | {z }
x−a
x−b
28. We have:
p
12 − 4x2 =
so
Z
√
dx
12 − 4x2
√
3, k = 0.5.
4 (3 − x2 )
q √
2
3 − x2 ,
=2
Z
dx
q √
=
2
=
Thus, a =
p
Z
3
2
0.5 dx
q √
2
3
− x2
− x2
.
623
624
Chapter Seven /SOLUTIONS
29. Taking the hint, we see that
e2x − 4ex + 3 = (ex )2 − 4ex + 3
= w2 − 4w + 3
letting w = ex
= (w − 3)(w − 1). factoring
Notice that dw = ex dx and that the numerator of the integrand can be written 2ex + 1 = 2w + 1. Using the method of
partial fractions, we have:
A
B
2w + 1
=
+
(w − 3)(w − 1)
w−3
w−1
2w + 1 = A(w − 1) + B(w − 3)
multiply through by (w − 3)(w − 1)
= (A + B)w + (−A − 3B). regroup
Identifying the coefficients of w and the constant terms on both sides, we have:
A+B = 2
and
so A = 2 − B
− A − 3B = 1
− (2 − B) − 3B = 1
so
since A = 2 − B
−2 − 2B = 1
3
2
7
and A = 2 − B = .
2
B=−
giving
This means
Z
7/2
−3/2
2w + 1
=
+
, so:
(w − 3)(w − 1)
w−3
w−1
ln 7
0
2ex + 1
ex dx =
2x
e − 4ex + 3
Z
x=ln 7
2w + 1
dw
(w
−
3)(w − 1)
Zx=0
w=7
2w + 1
=
dw
(w
−
3)(w − 1)
Zw=1
7
7/2
−3/2
=
+
dw.
w
−
3
w
−1
1
Thus,
A=
since w = ex
3
7
, B = − , w = ex , dw = ex dx, r = 1, s = 7.
2
2
30. (a) We have
2(x + 3)
x
2
1
3x + 6
=
+
= +
.
x2 + 3x
x(x + 3)
x(x + 3)
x
x+3
Thus
Z
3x + 6
dx =
x2 + 3x
Z
2
1
+
x
x+3
(b) Let a = 0, b = −3, c = 3, and d = 6 in V-27.
Z
3x + 6
dx =
x2 + 3x
=
Z
dx = 2 ln |x| + ln |x + 3| + C.
3x + 6
dx
x(x + 3)
1
(6 ln |x| + 3 ln |x + 3|) + C = 2 ln |x| + ln |x + 3| + C.
3
31. Since x2 + 2x + 2 = (x + 1)2 + 1, we have
Z
1
dx =
x2 + 2x + 2
Substitute x + 1 = tan θ, so x = (tan θ) − 1.
Z
1
dx.
(x + 1)2 + 1
7.4 SOLUTIONS
625
32. Since x2 + 6x + 9 is a perfect square, we write
Z
1
dx =
x2 + 6x + 25
Z
1
dx =
(x2 + 6x + 9) + 16
We use the trigonometric substitution x + 3 = 4 tan θ, so x = 4 tan θ − 3.
Z
1
dx.
(x + 3)2 + 16
33. Since y 2 + 3y + 3 = (y + 3/2)2 + (3 − 9/4) = (y + 3/2)2 + 3/4, we have
Z
dy
=
y 2 + 3y + 3
Substitute y + 3/2 = tan θ, so y = (tan θ) − 3/2.
Z
dy
.
(y + 3/2)2 + 3/4
34. Since x2 + 2x + 2 = (x + 1)2 + 1, we have
Z
Z
x+1
dx =
x2 + 2x + 2
x+1
dx.
(x + 1)2 + 1
Substitute w = (x + 1)2 , so dw = 2(x + 1) dx.
This integral can also be calculated without completing the square, by substituting w = x2 + 2x + 2, so dw =
2(x + 1) dx.
35. Since 2z − z 2 = 1 − (z − 1)2 , we have
Z
4
dz = 4
√
2z − z 2
Z
1
p
1 − (z − 1)2
Substitute z − 1 = sin θ, so z = (sin θ) + 1.
36. Since 2z − z 2 = 1 − (z − 1)2 , we have
Z
z−1
√
dz =
2z − z 2
Z
Substitute w = 1 − (z − 1)2 , so dw = −2(z − 1) dz.
p
z−1
1 − (z − 1)2
dz.
dz.
37. Since t2 + 4t + 7 = (t + 2)2 + 3, we have
Z
(t + 2) sin(t2 + 4t + 7) dt =
Z
(t + 2) sin((t + 2)2 + 3) dt.
Substitute w = (t + 2)2 + 3, so dw = 2(t + 2) dt.
This integral can also be computed without completing the square, by substituting w = t2 + 4t + 7, so dw =
(2t + 4) dt.
38. Since θ2 − 4θ = (θ − 2)2 − 4, we have
Z
(2 − θ) cos(θ2 − 4θ) dθ =
Z
−(θ − 2) cos((θ − 2)2 − 4) dθ.
Substitute w = (θ − 2)2 − 4, so dw = 2(θ − 2) dθ.
This integral can also be computed without completing the square, by substituting w = θ2 −4θ, so dw = (2θ−4) dθ.
39. We write
A
B
1
=
+
,
(x − 5)(x − 3)
x−5
x−3
giving
1 = A(x − 3) + B(x − 5)
1 = (A + B)x − (3A + 5B)
so
A+B = 0
−3A − 5B = 1.
Thus, A = 1/2, B = −1/2, so
Z
1
dx =
(x − 5)(x − 3)
Z
1/2
dx −
x−5
Z
1/2
1
1
dx = ln |x − 5| − ln |x − 3| + C.
x−3
2
2
626
Chapter Seven /SOLUTIONS
40. We write
1
A
B
=
+
,
(x + 2)(x + 3)
x+2
x+3
giving
1 = A(x + 3) + B(x + 2)
1 = (A + B)x + (3A + 2B)
so
A+B = 0
3A + 2B = 1.
Thus, A = 1, B = −1, so
Z
Z
1
dx =
(x + 2)(x + 3)
41. We write
1
dx −
x+2
Z
1
dx = ln |x + 2| − ln |x + 3| + C.
x+3
1
A
B
=
+
,
(x + 7)(x − 2)
x+7
x−2
giving
1 = A(x − 2) + B(x + 7)
1 = (A + B)x + (−2A + 7B)
so
A+B = 0
−2A + 7B = 1.
Thus, A = −1/9, B = 1/9, so
Z
1
dx = −
(x + 7)(x − 2)
Z
1/9
dx +
x+7
Z
1/9
1
1
dx = − ln |x + 7| + ln |x − 2| + C.
x−2
9
9
42. The denominator x2 − 3x + 2 can be factored as (x − 1)(x − 2). Splitting the integrand into partial fractions with
denominators (x − 1) and (x − 2), we have
x
A
B
x
=
=
+
.
x2 − 3x + 2
(x − 1)(x − 2)
x−1
x−2
Multiplying by (x − 1)(x − 2) gives the identity
x = A(x − 2) + B(x − 1)
so
x = (A + B)x − 2A − B.
Since this equation holds for all x, the constant terms on both sides must be equal. Similarly, the coefficient of x on both
sides must be equal. So
−2A − B = 0
A + B = 1.
Solving these equations gives A = −1, B = 2 and the integral becomes
Z
x
dx = −
x2 − 3x + 2
Z
1
dx + 2
x−1
Z
1
dx = − ln |x − 1| + 2 ln |x − 2| + C.
x−2
7.4 SOLUTIONS
627
43. This can be done by formula V–26 in the integral table or by partial fractions
Z
dz
=
z2 + z
Check:
Z
dz
=
z(z + 1)
Z
1
z
−
1
z+1
dz = ln |z| − ln |z + 1| + C.
1
1
1
d
(ln |z| − ln |z + 1| + C) = −
= 2
.
dz
z
z+1
z +z
44. We know x2 + 5x + 4 = (x + 1)(x + 4), so we can use V-26 of the integral table with a = −1 and b = −4 to write
Z
1
dx
= (ln |x + 1| − ln |x + 4|) + C.
x2 + 5x + 4
3
45. We use partial fractions and write
1
A
B
=
+
,
3P − 3P 2
3P
1−P
multiply through by 3P (1 − P ), and then solve for A and B, getting A = 1 and B = 1/3. So
Z
dP
=
3P − 3P 2
=
Z
1
1
+
3P
3(1 − P )
dP =
1
3
Z
dP
1
+
P
3
Z
1
1
P
1
ln |P | − ln |1 − P | + C = ln
+ C.
3
3
3
1−P
46. Using partial fractions, we have:
3x + 1
A
B
3x + 1
=
=
+
.
x2 − 3x + 2
(x − 1)(x − 2)
x−1
x−2
Multiplying by (x − 1) and (x − 2), this becomes
3x + 1 = A(x − 2) + B(x − 1)
= (A + B)x − 2A − B
which produces the system of equations
(
Solving this system yields A = −4 and B = 7. So,
Z
dP
1−P
A+B = 3
−2A − B = 1.
3x + 1
dx =
x2 − 3x + 2
Z
−
Z
4
7
+
dx
x−1
x−2
Z
dx
dx
+7
x−1
x−2
= −4 ln |x − 1| + 7 ln |x − 2| + C.
= −4
47. Since 2y 2 + 3y + 1 = (2y + 1)(y + 1), we write
A
B
y+2
=
+
,
2y 2 + 3y + 1
2y + 1
y+1
giving
y + 2 = A(y + 1) + B(2y + 1)
y + 2 = (A + 2B)y + A + B
so
A + 2B = 1
A + B = 2.
628
Chapter Seven /SOLUTIONS
Thus, A = 3, B = −1, so
Z
y+2
dy =
2y 2 + 3y + 1
Z
3
dy −
2y + 1
Z
3
1
dy = ln |2y + 1| − ln |y + 1| + C.
y+1
2
48. Since x3 + x = x(x2 + 1) cannot be factored further, we write
x+1
A
Bx + C
=
+ 2
.
x3 + x
x
x +1
Multiplying by x(x2 + 1) gives
x + 1 = A(x2 + 1) + (Bx + C)x
x + 1 = (A + B)x2 + Cx + A,
so
A+B = 0
C=1
A = 1.
Thus, A = C = 1, B = −1, and we have
Z
x+1
dx =
x3 + x
Z
1
−x + 1
+ 2
x
x +1
Z
Z
Z
dx
x dx
dx
=
−
+
x
x2 + 1
x2 + 1
1
= ln |x| − ln x2 + 1 + arctan x + K.
2
49. Since x2 + x4 = x2 (1 + x2 ) cannot be factored further, we write
x−2
A
B
Cx + D
=
+ 2 +
.
x2 + x4
x
x
1 + x2
Multiplying by x2 (1 + x2 ) gives
x − 2 = Ax(1 + x2 ) + B(1 + x2 ) + (Cx + D)x2
x − 2 = (A + C)x3 + (B + D)x2 + Ax + B,
so
A+C = 0
B+D = 0
A=1
B = −2.
Thus, A = 1, B = −2, C = −1, D = 2, and we have
Z
Z
2
−x + 2
1
− 2 +
x
x
1 + x2
Z
Z
Z
Z
dx
dx
x dx
dx
−2
−
+2
dx =
x
x2
1 + x2
1 + x2
2
1
= ln |x| + − ln 1 + x2 + 2 arctan x + K.
x
2
We use K as the constant of integration, since we already used C in the problem.
x−2
dx =
x2 + x4
50. Let y = 5 tan θ so dy = (5/cos2 θ) dθ. Since 1 + tan2 θ = 1/ cos2 θ, we have
Z
y2
dy =
25 + y 2
Using 1 + tan2 θ = 1/ cos2 θ again gives
Z
y2
dy = 5
25 + y 2
Z
25 tan2 θ
5
·
dθ = 5
25(1 + tan2 θ) cos2 θ
tan2 θ dθ = 5
In addition, since θ = arctan(y/5), we get
Z
Z
Z
1
−1
cos2 θ
y2
y
dy = y − 5 arctan
25 + y 2
5
Z
tan2 θ dθ.
dθ = 5 tan θ − 5θ + C.
+ C.
7.4 SOLUTIONS
629
√
51. Since (4 − z 2 )3/2 = ( 4 − z 2 )3 , we substitute z = 2 sin θ, so dz = 2 cos θ dθ. We get
Z
Z
Z
2 cos θ dθ
2 cos θ dθ
1
=
=
8 cos3 θ
4
(4 − 4 sin2 θ)3/2
p
√
Since sin θ = z/2, we have cos θ = 1 − (z/2)2 = ( 4 − z 2 )/2, so
Z
dz
=
(4 − z 2 )3/2
Z
dθ
1
= tan θ + C
cos2 θ
4
z/2
dz
1
1 sin θ
1
z
√
= tan θ + C =
+C =
+C = √
+C
4
4 cos θ
4 ( 4 − z 2 )/2
(4 − z 2 )3/2
4 4 − z2
52. We have
A
Bs + C
10
=
+ 2
.
(s + 2)(s2 + 1)
s+2
s +1
Thus,
10 = A(s2 + 1) + (Bs + C)(s + 2)
10 = (A + B)s2 + (2B + C)s + (A + 2C),
giving
A+B = 0
2B + C = 0
A + 2C = 10.
Thus, from the first two equations we have C = −2B = 2A, which, when used in the third, gives 5A = 10, so that
A = 2, B = −2, and C = 4. We now have
2
−2s + 4
2
2s
4
10
=
+ 2
=
− 2
+ 2
,
(s + 2)(s2 + 1)
s+2
s +1
s+2
s +1
s +1
so
Z
10
ds =
(s + 2)(s2 + 1)
Z
2
2s
4
− 2
+ 2
s+2
s +1
s +1
ds = 2 ln |s + 2| − ln s2 + 1 + 4 arctan s + K.
We use K as the constant of integration, since we already used C in the problem.
53. Completing the square, we get
x2 + 4x + 13 = (x + 2)2 + 9.
We use the substitution x + 2 = 3 tan t, then dx = (3/ cos2 t) dt. Since tan2 t + 1 = 1/ cos2 t, the integral becomes
Z
1
dx =
(x + 2)2 + 9
Z
3
dt =
9 tan2 t + 9 cos2 t
1
·
Z
1
1
x+2
dt = arctan
3
3
3
54. Using the substitution w = ex , we get dw = ex dx, so we have
Z
But
ex
dx =
x
(e − 1)(ex + 2)
1
1
=
(w − 1)(w + 2)
3
so
Z
ex
dx =
(ex − 1)(ex + 2)
Z
Z
dw
.
(w − 1)(w + 2)
1
1
−
,
w−1
w+2
1
3
1
1
−
w−1
w+2
dw
1
(ln |w − 1| − ln |w + 2|) + C
3
1
= (ln |ex − 1| − ln |ex + 2|) + C.
3
=
+ C.
630
Chapter Seven /SOLUTIONS
55. Let x = tan θ so dx = (1/cos2 θ)dθ. Since
√
1 + tan2 θ = 1/ cos θ, we have
cos θ
cos θ
1/ cos2 θ
1
√
dθ =
dθ.
dθ =
dx =
2
tan2 θ cos2 θ
sin2 θ
x2 1 + x2
tan θ 1 + tan2 θ
The last integral can be evaluated by guess-and-check or by substituting w = sin θ. The result is
Z
Z
Z
√
Z
x2
√
1
dx =
1 + x2
Z
Z
1
cos θ
dθ = −
+ C.
sin θ
sin2 θ
We√need to know sin θ in terms of x. From the triangle in Figure 7.4, which shows x = tan θ, we see that sin θ =
x/ 1 + x2 . Thus
Z
√
1
1
1 + x2
+C =−
+ C.
√
dx = −
2
2
sin θ
x
x 1+x
√
1 + x2
x
θ
1
Figure 7.4: In this triangle, tan θ = x
56. Let x = 3 sin θ so dx = 3 cos θ dθ, giving
Z
x2
√
dx =
9 − x2
Z
From formula IV-18, we get
9 sin2 θ
p
3 cos θ dθ =
2
9 − 9 sin θ
Z
Z
(9 sin2 θ)(3 cos θ)
dθ = 9
3 cos θ
Z
sin2 θ dθ.
θ
1
sin2 θ dθ = − sin θ cos θ + + C.
2
2
We have√sin θ = x/3 and θ = arcsin(x/3). From the triangle in Figure 7.5, which shows sin θ = x/3, we see that
cos θ = 9 − x2 /3. Therefore
Z
√
x2
dx = 9
9 − x2
Z
9
9
sin2 θ dθ = − sin θ cos θ + θ + C
2
2
√
9 x 9 − x2
9
x
xp
9
x
=− ·
+ arcsin
+C = −
9 − x2 + arcsin
+C
2 3
3
2
3
2
2
3
3
θ
√
x
9 − x2
Figure 7.5: In this triangle,
sin θ = x/3
p
√
57. Let 2x = sin θ. Then dx = (1/2) cos θ dθ and 1 − 4x2 = 1 − sin2 θ = cos θ, so
Z
Z
Z √
1 − 4x2
cos θ
1
cos2 θ
dx
=
cos
θ
dθ
dθ
=
2
2
x2
(sin θ)/4 2
sin2 θ
=2
Z
1 − sin2 θ
dθ = 2
sin2 θ
Z
1
− 1 dθ.
sin2 θ
7.4 SOLUTIONS
631
From formula IV-18, we get
Z
1
cos θ
− θ + C.
sin θ
− 1 dθ = −
sin2 θ
We have√sin θ = 2x and θ = arcsin(2x). From the triangle in Figure 7.6, which shows sin θ = 2x, we see that
cos θ = 1 − 4x2 . Therefore
√
Z √
√
1 − 4x2
cos θ
1 − 4x2
1 − 4x2
dx
=
2
−
θ
−
arcsin(2x)
− 2 arcsin(2x) + C
−
+
C
=
2
−
+
C
=
−
2
x
sin θ
2x
x
1
θ
√
2x
1 − 4x2
Figure 7.6: In this triangle,
sin θ = 2x
p
√
58. Let 3x = 5 sin θ. Then dx = (5/3) cos θ dθ and 25 − 9x2 = 25 − 25 sin2 θ = 5 cos θ, so
Z
Z
Z √
cos2 θ
25 − 9x2
5 cos θ
5
dx =
cos θ dθ = 5
dθ
x
(5/3) sin θ 3
sin θ
=5
1 − sin2 θ
dθ = 5
sin θ
Z
From formula IV-21, we get
Z
1
− sin θ
sin θ
dθ =
Z
1
sin θ
− sin θ
dθ.
1
cos θ − 1
ln
+ cos θ + C.
2
cos θ + 1
We have√
sin θ = 3x/5 and θ = arcsin(3x/5). From the triangle in Figure 7.7, which shows sin θ = 3x/5, we see that
cos θ = 25 − 9x2 /5. Therefore
Z √
√
p
1
cos θ − 1
5
5 − 25 − 9x2
25 − 9x2
+ cos θ + C = ln
+ 25 − 9x2 + C
dx = 5
ln
√
2
x
2
cos θ + 1
2
5 + 25 − 9x
5
θ
√
3x
25 − 9x2
Figure 7.7: In this triangle,
sin θ = 3x/5
59. Let 2x = 3 sin θ. Then dx = (3/2) cos θ dθ and
Z
1
dx =
x 9 − 4x2
√
From formula IV-21, we get
Z
Z
√
9 − 4x2 =
p
9 − 9 sin2 θ = 3 cos θ, so
1
(3 sin θ/2)(3 cos θ)
3
1
cos θ dθ =
2
3
1
1
cos θ − 1
+C
dθ = ln
sin θ
2
cos θ + 1
Z
1
dθ.
sin θ
632
Chapter Seven /SOLUTIONS
√
From the triangle in Figure 7.8, which shows sin θ = 2x/3, we see that cos θ = 9 − 4x2 /3. Therefore
Z
√
√
9 − 4x2 /3 − 1
1
cos θ − 1
1
1
1
9 − 4x2 − 3
√
dx = ln
+ C = ln √
+ C = ln √
+ C.
6
cos θ + 1
6
6
x 9 − 4x2
9 − 4x2 /3 + 1
9 − 4x2 + 3
3
θ
√
2x
9 − 4x2
Figure 7.8: In this triangle,
sin θ = 2x/3
60. Let 4x = tan θ. Then dx = (1/4)/ cos2 θ dθ and
Z
1
√
dx =
x 1 + 16x2
√
1 + 16x2 =
Z
√
cos θ
(1/4) tan θ
1 + tan2 θ = 1/ cos θ, so
1/4
dθ =
cos2 θ
Z
1
dθ.
sin θ
From formula IV-21, we get
Z
cos θ − 1
1
1
+C
dθ = ln
sin θ
2
cos θ + 1
√
From the triangle in Figure 7.9, which shows tan θ = 4x, we see that cos θ = 1/ 1 + 16x2 . Therefore
Z
√
√
1/ 1 + 16x2 − 1
1
cos θ − 1
1
1
1 − 1 + 16x2
1
√
√
√
dx = ln
+ C = ln
+ C = ln
+ C.
2
cos θ + 1
2
2
x 1 + 4x2
1/ 1 + 16x2 + 1
1 + 1 + 16x2
√
1 + 16x2
4x
θ
1
Figure 7.9: In this triangle,
tan θ = 4x
61. Let x = 2 sin θ so dx = 2 cos θdθ. Since
Z
p
4 − 4 sin2 θ = 2 cos θ, we have
1
√
dx =
x2 4 − x2
Z
1
1
2 cos θ dθ =
4
(4 sin2 θ)(2 cos θ)
Z
cos θ
1
dθ = −
+ C.
2
sin θ
sin θ
Z
1
dθ.
sin2 θ
From formula IV-20 with m = 2, we get
We
√ need to know cos θ in terms of x. From the triangle in Figure 7.10, which shows x/2 = sin θ, we see that cos θ =
4 − x2 /2. Thus
Z
√
√
1
1 cos θ
1 4 − x2 /2
1 4 − x2
√
+C =−
+C =−
+ C.
dx = −
4 sin θ
4
x/2
4
x
x2 4 − x2
7.4 SOLUTIONS
2
θ
√
633
x
4 − x2
Figure 7.10: In this triangle,
sin θ = x/2
62. Let 2x = 5 tan θ so dx = (5/2)(1/ cos2 θ)dθ. Since
(25 + 4x2 )3/2 = (25 + 25 tan2 θ)3/2 = (25/ cos2 θ)3/2 = 125/ cos3 θ,
we have
cos3 θ 5 1
1
1
1
dθ =
cos θ dθ =
sin θ + C.
dx =
2
3/2
125
2 cos2 θ
50
50
(25 + 4x )
We need
√ to know sin θ in terms of x. From the triangle in Figure 7.11, which shows 2x/5 = tan θ, we see that sin θ =
2x/ 25 + 4x2 . Thus
Z
Z
Z
Z
1
2x
x
1
1
1
sin θ + C =
√
+C =
√
+C
dx =
50
50 25 + 4x2
25 25 + 4x2
(25 + 4x2 )3/2
√
25 + 4x2
2x
θ
5
Figure 7.11: In this triangle,
tan θ = 2x/5
63. Let x = 4 sin θ so dx = 4 cos θ dθ. Since
(16 − x2 )3/2 = (16 − 16 sin2 θ)3/2 = (16 cos2 θ)3/2 = 64 cos3 θ,
we have
Z
1
dx =
(16 − x2 )3/2
From formula IV-22 with m = 2, we get
Z
Z
1
1
(4 cos θ dθ) =
64 cos3 θ
16
Z
1
dθ.
cos2 θ
sin θ
1
dθ =
+ C = tan θ + C.
cos2 θ
cos θ
We need
√ to know tan θ in terms of x. From the triangle in Figure 7.12, which shows x/4 = sin θ, we see that
tan θ = x/ 16 − x2 . Thus
Z
1
1
1
x
√
dx =
tan θ + C =
+C
16
16 16 − x2
(16 − x2 )3/2
4
θ
√
16 − x2
Figure 7.12: In this triangle,
sin θ = x/4
x
634
Chapter Seven /SOLUTIONS
64. Let 3x = tan θ so dx = (1/3)(1/ cos2 θ) dθ. Since (1 + 9x2 )3/2 = (1 + tan2 θ)3/2 = 1/ cos3 θ, we have
Z
x2
dx =
(1 + 9x2 )3/2
1
1
1 sin2 θ
(cos3 θ)
dθ =
9 cos2 θ
3 cos2 θ
27
Z
Z
sin2 θ
dθ.
cos θ
From formula IV-22 with m = 2, we get
sin2 θ
dθ =
cos θ
Z
Z
1 − cos2 θ
dθ =
cos θ
Z
1
− cos θ
cos θ
dθ =
1
sin θ + 1
− sin θ + C.
ln
2
sin θ − 1
We need
√ to know sin θ in terms of x. From the triangle in Figure 7.13, which shows 3x = tan θ, we see that
sin θ = 3x/ 1 + 9x2 . Thus
Z
√
1
3x
x2
1 1
sin θ + 1
1
3x + 1 + 9x2
√
−√
+ C.
− sin θ + C =
dx =
ln
ln
27 2
sin θ − 1
27 2
(1 + 9x2 )3/2
3x − 1 + 9x2
1 + 9x2
√
1 + 9x2
3x
θ
1
Figure 7.13: In this triangle,
tan θ = 3x
65. Notice that because
3x
(x−1)(x−4)
is negative for 2 ≤ x ≤ 3,
Z
Area = −
2
3
3x
dx.
(x − 1)(x − 4)
Using partial fractions gives
3x
A
B
(A + B)x − B − 4A
=
+
=
.
(x − 1)(x − 4)
x−1
x−4
(x − 1)(x − 4)
Multiplying through by (x − 1)(x − 4) gives
3x = (A + B)x − B − 4A
so A = −1 and B = 4. Thus
−
Z
2
3
3x
dx = −
(x − 1)(x − 4)
Z
3
2
3
4
−1
+
dx = (ln |x − 1| − 4 ln |x − 4|)
x−1
x−4
66. We have
Area =
Z
0
Using partial fractions gives
(x2
1
(x2
3x2 + x
dx.
+ 1)(x + 1)
Ax + B
C
3x2 + x
= 2
+
+ 1)(x + 1)
x +1
x+1
=
(Ax + B)(x + 1) + C(x2 + 1)
(x2 + 1)(x + 1)
=
(A + C)x2 + (A + B)x + B + C
.
(x2 + 1)(x + 1)
Thus
3x2 + x = (A + C)x2 + (A + B)x + B + C,
= 5 ln 2.
2
7.4 SOLUTIONS
635
giving
3 = A + C,
1 = A + B,
0 = B + C,
and
with solution
A = 2, B = −1, C = 1.
Thus
Area =
1
Z
0
=
1
Z
3x2 + x
dx
+ 1)(x + 1)
(x2
2x
x2 + 1
0
1
1
+
x2 + 1
x+1
−
dx
1
2
= ln(x + 1) − arctan x + ln |x + 1|
0
= 2 ln 2 − π/4.
67. We have
Let x = sin θ so dx = cos θ dθ and
Z
1/2
√
0
The integral
68. We have
R
√
1/2
Z
Area =
√
0
1−
x2
=
p
x2
dx.
1 − x2
1 − sin2 θ = cos θ. When x = 0, θ = 0. When x = 1/2, θ = π/6.
x2
dx =
1 − x2
Z
=
π/6
sin2 θ
p
cos θ dθ =
2
1 − sin θ
0
θ
sin θ cos θ
−
2
2
π/6
=
Z
π/6
sin2 θ dθ
0
√
3
π
−
.
12
8
0
sin2 θ dθ is done using parts and the identity cos2 θ + sin2 θ = 1.
√
Z
Area =
√
Let x√ = 2 sin θ so dx = 2 cos θ dθ and
x = 2, θ = π/4.
√
Z
0
2
√
x3
dx =
4 − x2
Z
=8
2
√
0
4−
x2
Z
4 − 4 sin2 θ = 2 cos θ. When x = 0, θ = 0 and when
=
π/4
0
p
x3
dx.
4 − x2
(2 sin θ)3
p
π/4
sin3 θ dθ = 8
0
3
= 8 − cos θ +
69. We have
Area =
Z
x2
+9 =
r
cos θ
3
3
0
Let x = 3 tan θ so dx = (3/ cos2 θ)dθ and
p
2 cos θ dθ
4 − (2 sin θ)2
√
Z
π/4
0
π/4
(sin θ − sin θ cos2 θ) dθ
=8
0
1
dx.
x2 + 9
9 sin2 θ
3
+9=
.
cos2 θ
cos θ
2
5
− √
3
6 2
.
636
Chapter Seven /SOLUTIONS
When x = 0, θ = 0 and when x = 3, θ = π/4. Thus
Z
3
0
Z
π/4
Z
π/4
3
1
1
3
√
√
dx =
dθ =
·
dθ =
2θ
2θ
2
cos
3/
cos
θ
cos
x2 + 9
9
tan
θ
+
9
0
0
π/4
√
√
1
1
sin θ + 1
1
1/ 2 + 1
1+ 2
√
= ln √
.
= ln
= ln
2
sin θ − 1
2
2
1/ 2 − 1
2−1
1
Z
π/4
0
1
dθ
cos θ
0
√
√
This answer can be simplified to ln(1 + 2) by multiplying
the numerator and denominator of the fraction by ( 2 + 1)
R
and using the properties of logarithms. The integral (1/ cos θ)dθ is done using the Table of Integrals.
70. We have
Area =
Let x = 3 tan θ so dx = (3/ cos2 θ)dθ and
x
When x =
√
p
x2
Z
3
√
sin θ
+9=3
cos θ
3
1
√
dx.
x x2 + 9
r
9 sin θ
9 sin2 θ
+9 =
.
cos2 θ
cos2 θ
3, θ = π/6 and when x = 3, θ = π/4. Thus
Z
3
√
3
1
√
dx =
x x2 + 9
π/4
Z
=
π/6
3
1
1
·
dθ =
9 sin θ/ cos2 θ cos2 θ
3
π/4
1 1
cos θ − 1
· ln
3 2
cos θ + 1
=
1
6
π/6
Z
π/4
π/6
1
dθ
sin θ
√
√
1/ 2 − 1
3/2 − 1
ln
√
− ln √
1/ 2 + 1
3/2 + 1
√
√
1− 2
1
3+2
√ + ln √
ln
.
=
6
1+ 2
3−2
√
This answer
√ can be simplified by multiplying the first fraction by (1 − 2) in numerator and denominator and the second
one by ( 3 + 2). This gives
Area =
R
√
√
√
√
1
1
(ln(3 − 2 2) + ln(7 + 4 3)) = ln((3 − 2 2)(7 + 4 3)).
6
6
The integral (1/ sin θ)dθ is done using the Table of Integrals.
71. Using partial fractions, we write
1
A
B
=
+
1 − x2
1+x
1−x
1 = A(1 − x) + B(1 + x) = (B − A)x + A + B.
So, B − A = 0 and A + B = 1, giving A = B = 1/2. Thus
Z
1
dx
=
1 − x2
2
Z
1
1
+
1+x
1−x
dx =
1
(ln |1 + x| − ln |1 − x|) + C.
2
Using the substitution x = sin θ, we get dx = cos θ dθ, we have
Z
dx
=
1 − x2
Z
cos θ
dθ =
1 − sin2 θ
Z
cos θ
dθ =
cos2 θ
Z
1
dθ.
cos θ
The Table of Integrals Formula IV-22 gives
Z
dx
=
1 − x2
Z
(sin θ) + 1
1
1
1
x+1
dθ = ln
+ C = ln
+ C.
cos θ
2
(sin θ) − 1
2
x−1
The properties of logarithms and the fact that |x − 1| = |1 − x| show that the two results are the same:
1
x+1
1
= (ln |1 + x| − ln |1 − x|) .
ln
2
x−1
2
7.4 SOLUTIONS
72. Using partial fractions, we write
A
B
2x
=
+
x2 − 1
x+1
x−1
2x = A(x − 1) + B(x + 1) = (A + B)x − A + B.
So, A + B = 2 and −A + B = 0, giving A = B = 1. Thus
Z
2x
dx =
x2 − 1
Z
1
x+1
+
1
x−1
dx = ln |x + 1| + ln |x − 1| + C.
Using the substitution w = x2 − 1, we get dw = 2x dx, so we have
Z
2x
dx =
x2 − 1
Z
dw
= ln |w| + C = ln x2 − 1 + C.
w
The properties of logarithms show that the two results are the same:
ln |x + 1| + ln |x − 1| = ln |(x + 1)(x − 1)| = ln x2 − 1 .
73. Using partial fractions, we write
3x2 + 1
3x2 + 1
A
Bx + C
=
=
+ 2
3
x +x
x(x2 + 1)
x
x +1
3x2 + 1 = A(x2 + 1) + (Bx + C)x = (A + B)x2 + Cx + A.
So, A + B = 3, C = 0 and A = 1, giving B = 2. Thus
Z
3x2 + 1
dx =
x3 + x
Z
1
x
+
2x
x2 + 1
dx = ln |x| + ln x2 + 1 + K.
Using the substitution w = x3 + x, we get dw = (3x2 + 1)dx, so we have
3x2 + 1
dx =
x3 + x
Z
Z
dw
= ln |w| + K = ln x3 + x + K.
w
The properties of logarithms show that the two results are the same:
ln |x| + ln x2 + 1 + K = ln x(x2 + 1) + K = ln x3 + x + K.
We use K as the constant of integration, since we already used C in the problem.
74. (a) We differentiate:
d
1
−
dθ
tan θ
Thus,
(b) Let y =
√
5 sin θ so dy =
Z
√
=
Z
1
1
·
=
tan2 θ cos2 θ
1
sin2 θ
cos2 θ
·
1
1
=
.
cos2 θ
sin2 θ
1
1
dθ = −
+ C.
tan θ
sin2 θ
5 cos θ dθ giving
Z
Z
√
√
5 cos θ
5 cos θ
dy
1
√
p
p
=
dθ =
dθ
2
2
2
2
2
5
sin θ 5 cos θ
y
5−y
5 sin θ 5 − 5 sin θ
=
1
5
Z
1
1
+ C.
dθ = −
2
5
tan
θ
sin θ
p
p
√
√
√
Since sin θ = y/ 5, we have cos θ = 1 − (y/ 5)2 = 5 − y 2 / 5. Thus,
Z
1
p
+C =−
=−
5 tan θ
y2 5 − y2
dy
p
√
5 − y2/ 5
5 − y2
√
+ C.
+C = −
5y
5(y/ 5)
p
637
638
Chapter Seven /SOLUTIONS
75. (a) If a 6= b, we have
Z
1
dx =
(x − a)(x − b)
(b) If a = b, we have
Z
76. (a) If a 6= b, we have
Z
Z
1
a−b
1
1
−
x−a
x−b
1
dx =
(x − a)(x − a)
x
dx =
(x − a)(x − b)
Z
1
a−b
a
x
dx =
(x − a)2
Z
77. (a) If a > 0, then
−
x−a
(b) If a = b, we have
Z
Z
dx =
1
(ln |x − a| − ln |x − b|) + C.
a−b
1
1
dx = −
+ C.
(x − a)2
x−a
b
x−b
1
a
+
x−a
(x − a)2
x2 − a = (x −
√
1
(a ln |x − a| − b ln |x − b|) + C.
a−b
dx =
dx = ln |x − a| −
a)(x +
√
a
+ C.
x−a
a).
This means that we can use partial fractions:
A
B
1
√ +
√ ,
=
x2 − a
x− a
x+ a
giving
1 = A(x +
√
a) + B(x −
√
√
so A + B = 0 and (A − B) a = 1. Thus, A = −B = 1/(2 a).
So
Z
1
dx =
x2 − a
Z
1
√
2 a
1
1
√ −
√
x− a
x+ a
(b) If a = 0, we have
Z
√
a),
√
√
1
dx = √ (ln x − a − ln x + a ) + C.
2 a
1
1
dx = − + C.
x2
x
(c) If a < 0, then −a > 0 so x2 − a = x2 + (−a) cannot be factored. Thus
Z
1
dx =
x2 − a
Z
1
1
dx = √
arctan
x2 + (−a)
−a
x
√
−a
+ C.
78. (a) We integrate to find
so
Z
b
dx = b
x(1 − x)
t(p) =
Z
a
p
Z
1
x
+
1
1−x
b
dx = b ln
x(1 − x)
(b) We know that t(0.01) = 0 so
dx = b(ln |x| − ln |1 − x|) + C = b ln
p
1−p
0 = b ln
But b > 0 and ln x = 0 means x = 1, so
− b ln
0.01(1 − a)
0.99a
a
1−a
0.01(1 − a)
=1
0.99a
0.01(1 − a) = 0.99a
0.01 − 0.01a = 0.99a
a = 0.01.
.
= b ln
x
+ C,
1−x
p(1 − a)
a(1 − p)
.
7.4 SOLUTIONS
639
(c) We know that t(0.5) = 1 so
1 = b ln
(d) We have
Z
t(0.9) =
0.9
0.01
0.5 · 0.99
0.5 · 0.01
= b ln 99, b =
0.218
1
dx =
ln
x(1 − x)
ln 99
1
= 0.218.
ln 99
0.9(1 − 0.01)
0.01(1 − 0.9)
= 1.478.
79. (a) We want to evaluate the integral
T =
Z
a/2
0
Using partial fractions, we have
k dx
.
(a − x)(b − x)
k
C
D
=
+
(a − x)(b − x)
a−x
b−x
k = C(b − x) + D(a − x)
k = −(C + D)x + Cb + Da
so
0 = −(C + D)
k = Cb + Da,
giving
C = −D =
Thus, the time is given by
T =
Z
a/2
0
k
k dx
=
(a − x)(b − x)
b−a
=
k
.
b−a
Z
a/2
0
1
1
−
a−x
b−x
dx
k
(− ln |a − x| + ln |b − x|)
b−a
a/2
0
b − x a/2
k
ln
=
b−a
a−x 0
k
2b − a
b
=
ln
− ln
b−a
a
a
2b − a
k
=
ln
.
b−a
b
(b) A similar calculation with x0 instead of a/2 leads to the following expression for the time
T =
Z
x0
0
b − x x0
k
k dx
=
ln
(a − x)(b − x)
b−a
a−x 0
k
b − x0
b
=
ln
− ln
.
b−a
a − x0
a
As x0 → a, the value of |a − x0 | → 0, so |b − x0 | /|a − x0 | → ∞. Thus, T → ∞ as x0 → a. In other words, the
time taken tends to infinity.
80. We complete the square in the exponent so that we can make a substitution:
1
m(t) = √
2π
1
= √
2π
1
= √
2π
Z
∞
etx e−x
−∞
∞
Z
Z−∞
∞
−∞
e−(x
2
/2
dx
2 −2tx)/2
e−((x
2
dx
−2tx+t2 )−t2 )/2
dx
640
Chapter Seven /SOLUTIONS
1
= √
2π
2
et
= √
/2
2π
∞
Z
e−(x−t)
2
/2
−∞
Z
∞
e−(x−t)
2
/2
· et
2
/2
dx
dx.
−∞
Substitute w = x − t, then dw = dx and w = ∞ when x = ∞, and w = −∞ when x = −∞. Thus
2
et /2
m(t) = √
2π
m(t) = et
2
/2
Z
2
∞
e−w
2
/2
−∞
et
dw = √
/2
2π
·
√
2π
.
Strengthen Your Understanding
81. The partial fractions required are of the form
A
B
C
1
=
+
+
.
(x − 1)2 (x − 2)
x−1
(x − 1)2
x−2
82. We use x = 2 tan θ.
83. Since
R
1/(1 + x2 ) dx = arctan x + C, so f (x) = 1/(1 + x2 ) works.
84. To evaluate
R
1/(x3 − x) dx, we first factor x3 − x = x(x − 1)(x + 1) and then use partial fractions.
85. If Q(x) does not decompose into linear factors, then we cannot decompose P (x)/Q(x) into partial fractions. So for
example, if P (x) = x and Q(x) = x2 + 1, then the rational function P (x)/Q(x) does not have a partial fraction
decomposition. Alternatively, if Q(x) contains a repeated linear factor, then the partial fraction decomposition may not
be in the given form. For example, if P (x) = x + 1 and Q(x) = x2 + 6x + 9, then the partial fraction decomposition of
P (x)/Q(x) is
1
2
x+1
=
−
.
x2 + 6x + 9
x+3
(x + 3)2
86. An example is the integral
If we make the substitution x =
Z
dx
√
=
9 − 4x2
3
2
sin θ, dx =
Z
3
2
p
3
2
Z
√
dx
.
9 − 4x2
cos θ dθ, then we obtain
cos θ
2
dθ =
9 − 9 sin θ
=
Z
Z
3
2
√
cos θ
9 cos2
θ
dθ =
Z
3
2
cos θ
dθ
3 cos θ
2
1
1
1
dθ = θ + C = · arcsin
x +C
2
2
2
3
87. False. If we use the given substitution the radicand becomes 9 − 9 tan2 θ = 9(1 − tan2 θ). However, 1 − tan2 θ is not
equal to a monomial.
88. True. Since the denominator can be factored into x2 (x + 1), involving the repeated factor x2 , we use partial fractions of
the form
Z
Z
A
B
C
1
dx =
+ 2 +
dx.
x3 + x2
x
x
x+1
89. (e). x = sin θ is appropriate.
90. (c) and (b).
1
1
x2
= −1 +
=1+
.
1 − x2
1 − x2
(1 − x)(1 + x)
641
7.5 SOLUTIONS
Solutions for Section 7.5
Exercises
1. (a) The approximation LEFT(2) uses two rectangles, with the height of each rectangle determined by the left-hand
endpoint. See Figure 7.14. We see that this approximation is an underestimate.
a
x
x
a
b
Figure 7.14
b
Figure 7.15
(b) The approximation RIGHT(2) uses two rectangles, with the height of each rectangle determined by the right-hand
endpoint. See Figure 7.15. We see that this approximation is an overestimate.
(c) The approximation TRAP(2) uses two trapezoids, with the height of each trapezoid given by the secant line connecting the two endpoints. See Figure 7.16. We see that this approximation is an overestimate.
a
x
b
Figure 7.16
(d) The approximation MID(2) uses two rectangles, with the height of each rectangle determined by the height at the
midpoint. Alternately, we can view MID(2) as a trapezoid rule where the height is given by the tangent line at
the midpoint. Both interpretations are shown in Figure 7.17. We see from the tangent line interpretation that this
approximation is an underestimate
a
x
a
b
Figure 7.17
x
b
642
Chapter Seven /SOLUTIONS
2. (a) The approximation LEFT(2) uses two rectangles, with the height of each rectangle determined by the left-hand
endpoint. See Figure 7.18. We see that this approximation is an overestimate.
a
x
x
a
b
Figure 7.18
b
Figure 7.19
(b) The approximation RIGHT(2) uses two rectangles, with the height of each rectangle determined by the right-hand
endpoint. See Figure 7.19. We see that this approximation is an underestimate.
(c) The approximation TRAP(2) uses two trapezoids, with the height of each trapezoid given by the secant line connecting the two endpoints. See Figure 7.20. We see that this approximation is an underestimate.
a
x
b
Figure 7.20
(d) The approximation MID(2) uses two rectangles, with the height of each rectangle determined by the height at the
midpoint. Alternately, we can view MID(2) as a trapezoid rule where the height is given by the tangent line at
the midpoint. Both interpretations are shown in Figure 7.21. We see from the tangent line interpretation that this
approximation is an overestimate.
a
x
a
b
Figure 7.21
x
b
643
7.5 SOLUTIONS
3. (a) The approximation LEFT(2) uses two rectangles, with the height of each rectangle determined by the left-hand
endpoint. See Figure 7.22. We see that this approximation is an underestimate.
a
x
x
a
b
Figure 7.22
b
Figure 7.23
(b) The approximation RIGHT(2) uses two rectangles, with the height of each rectangle determined by the right-hand
endpoint. See Figure 7.23. We see that this approximation is an overestimate.
(c) The approximation TRAP(2) uses two trapezoids, with the height of each trapezoid given by the secant line connecting the two endpoints. See Figure 7.24. We see that this approximation is an underestimate.
a
x
b
Figure 7.24
(d) The approximation MID(2) uses two rectangles, with the height of each rectangle determined by the height at the
midpoint. Alternately, we can view MID(2) as a trapezoid rule where the height is given by the tangent line at
the midpoint. Both interpretations are shown in Figure 7.25. We see from the tangent line interpretation that this
approximation is an overestimate.
a
x
a
b
Figure 7.25
x
b
644
Chapter Seven /SOLUTIONS
4. (a) The approximation LEFT(2) uses two rectangles, with the height of each rectangle determined by the left-hand
endpoint. See Figure 7.26. We see that this approximation is an overestimate.
a
x
x
a
b
Figure 7.26
b
Figure 7.27
(b) The approximation RIGHT(2) uses two rectangles, with the height of each rectangle determined by the right-hand
endpoint. See Figure 7.27. We see that this approximation is an underestimate.
(c) The approximation TRAP(2) uses two trapezoids, with the height of each trapezoid given by the secant line connecting the two endpoints. See Figure 7.28. We see that this approximation is an overestimate.
a
x
b
Figure 7.28
(d) The approximation MID(2) uses two rectangles, with the height of each rectangle determined by the height at the
midpoint. Alternately, we can view MID(2) as a trapezoid rule where the height is given by the tangent line at
the midpoint. Both interpretations are shown in Figure 7.29. We see from the tangent line interpretation that this
approximation is an underestimate.
a
x
a
b
Figure 7.29
x
b
645
7.5 SOLUTIONS
5. (a) The approximation LEFT(2) uses two rectangles, with the height of each rectangle determined by the left-hand
endpoint. See Figure 7.30. We see that this approximation is an underestimate (that is, it is more negative).
a
b
a
x
Figure 7.30
b
x
Figure 7.31
(b) The approximation RIGHT(2) uses two rectangles, with the height of each rectangle determined by the right-hand
endpoint. See Figure 7.31. We see that this approximation is an overestimate (that is, it is less negative).
(c) The approximation TRAP(2) uses two trapezoids, with the height of each trapezoid given by the secant line connecting the two endpoints. See Figure 7.32. We see that this approximation is an overestimate (that is, it is less negative).
a
b
x
Figure 7.32
(d) The approximation MID(2) uses two rectangles, with the height of each rectangle determined by the height at the
midpoint. Alternately, we can view MID(2) as a trapezoid rule where the height is given by the tangent line at
the midpoint. Both interpretations are shown in Figure 7.33. We see from the tangent line interpretation that this
approximation is an underestimate (that is, it is more negative).
a
b
a
x
Figure 7.33
b
x
646
Chapter Seven /SOLUTIONS
6. (a) The approximation LEFT(2) uses two rectangles, with the height of each rectangle determined by the left-hand
endpoint. See Figure 7.34. We see that this approximation is an overestimate (that is, it is less negative).
a
b
a
x
Figure 7.34
b
x
Figure 7.35
(b) The approximation RIGHT(2) uses two rectangles, with the height of each rectangle determined by the right-hand
endpoint. See Figure 7.35. We see that this approximation is an underestimate (that is, it is more negative).
(c) The approximation TRAP(2) uses two trapezoids, with the height of each trapezoid given by the secant line connecting the two endpoints. See Figure 7.36. We see that this approximation is an overestimate (that is, it is less negative).
a
b
x
Figure 7.36
(d) The approximation MID(2) uses two rectangles, with the height of each rectangle determined by the height at the
midpoint. Alternately, we can view MID(2) as a trapezoid rule where the height is given by the tangent line at
the midpoint. Both interpretations are shown in Figure 7.37. We see from the tangent line interpretation that this
approximation is an underestimate (that is, the approximation is more negative).
a
b
a
x
b
x
Figure 7.37
7. (a) Since two rectangles are being used, the width of each rectangle is 3. The height is given by the left-hand endpoint
so we have
LEFT(2) = f (0) · 3 + f (3) · 3 = 02 · 3 + 32 · 3 = 27.
(b) Since two rectangles are being used, the width of each rectangle is 3. The height is given by the right-hand endpoint
so we have
RIGHT(2) = f (3) · 3 + f (6) · 3 = 32 · 3 + 62 · 3 = 135.
(c) We know that TRAP is the average of LEFT and RIGHT and so
TRAP(2) =
27 + 135
= 81.
2
(d) Since two rectangles are being used, the width of each rectangle is 3. The height is given by the height at the midpoint
so we have
MID(2) = f (1.5) · 3 + f (4.5) · 3 = (1.5)2 · 3 + (4.5)2 · 3 = 67.5.
7.5 SOLUTIONS
647
8. (a)
LEFT(2) = 2 · f (0) + 2 · f (2)
= 2·1+2·5
= 12
RIGHT(2) = 2 · f (2) + 2 · f (4)
= 2 · 5 + 2 · 17
= 44
(b)
f (x) = x2 + 1
f (x) = x2 + 1
Area shaded
= LEFT(2)
Area shaded
= RIGHT(2)
x
2
x
4
2
4
LEFT(2) is an underestimate, while RIGHT(2) is an overestimate.
9. (a)
MID(2) = 2 · f (1) + 2 · f (3)
= 2 · 2 + 2 · 10
= 24
LEFT(2) + RIGHT(2)
TRAP(2) =
2
12 + 44
(see Problem 8)
=
2
= 28
(b)
f (x) = x2 + 1
f (x) = x2 + 1
Area shaded
= MID(2)
Area shaded
= TRAP(2)
x
2
x
4
2
4
MID(2) is an underestimate, since f (x) = x2 + 1 is concave up and a tangent line will be below the curve.
TRAP(2) is an overestimate, since a secant line lies above the curve.
10. (a) Since two rectangles are being used, the width of each rectangle is π/2. The height is given by the left-hand endpoint
so we have
π
π
π
π
π
LEFT(2) = f (0) · + f (π/2) · = sin 0 · + sin(π/2) · = .
2
2
2
2
2
(b) Since two rectangles are being used, the width of each rectangle is π/2. The height is given by the right-hand endpoint
so we have
π
π
π
π
π
RIGHT(2) = f (π/2) · + f (π) · = sin(π/2) · + sin(π) · = .
2
2
2
2
2
(c) We know that TRAP is the average of LEFT and RIGHT and so
TRAP(2) =
π
2
+
2
π
2
=
π
.
2
648
Chapter Seven /SOLUTIONS
(d) Since two rectangles are being used, the width of each rectangle is π/2. The height is given by the height at the
midpoint so we have
√
2π
π
π
π
π
.
MID(2) = f (π/4) · + f (3π/4) · = sin(π/4) · + sin(3π/4) · =
2
2
2
2
2
Problems
11. For n = 5, we have ∆t = (2 − 1)/5 = 0.2, so
MID(5) = 0.2f (1.1) + 0.2f (1.3) + 0.2f (1.5) + 0.2f (1.7) + 0.2f (1.9)
= 0.2 (−2.9 − 3.7 − 3.2 − 1.7 + 0.5)
= −2.2.
12. For n = 4, we have ∆x = (2 − 0)/4 = 0.5, so
MID(4) = f (0.25)0.5 + f (0.75)(0.5) + f (1.25)(0.5) + f (1.75)(0.5)
= 0.5(5.8 + 9.3 + 10.8 + 10.3)
= 18.1.
13. We have n = 2 and ∆x = 0.5, so
LEFT(2) = ∆x
(f (2) + f (2.5))
1
1
= 0.5 2 +
2
2.5
2
1
4
= 0.5
+
4
25
41
=
= 0.205
200
RIGHT(2) = ∆x
(f (2.5) + f (3))
1
1
+ 2
= 0.5
2
3
2.5
4
1
= 0.5
+
25
9
61
= 0.1356
=
450
LEFT(2) + RIGHT(2)
TRAP(2) =
2
41
61
+ 450
= 0.1703.
= 200
2
14. We have n = 2 and ∆x = 0.5, so
MID(2) = ∆x
(f (2.25) + f (2.75))
1
1
+
= 0.5
2
2
2.25
2.75
= 0.1649.
15. (a)
(i) Let f (x) =
1
.
1+x2
The left-hand Riemann sum is
2
1
1
f (0) + f
+f
+ ··· + f
8
8
8
1 64
64
64
64
64
64
=
+
+
+
+
+
+
8 64
65
68
73
80
89
≈ 8(0.1020) = 0.8160.
(ii) Let f (x) =
1
.
1+x2
7
8
64
64
+
100
113
The right-hand Riemann sum is
1
2
3
1
f
+f
+f
+ · · · + f (1)
8
8
8
8
1 64
64
64
64
64
64
64
64
=
+
+
+
+
+
+
+
8 65
68
73
80
89
100
113
128
1
= 0.7535.
≈ 0.8160 −
16
7.5 SOLUTIONS
649
(iii) The trapezoid rule gives us that
TRAP(8) =
(b) Since 1 + x2 is increasing for x > 0, so
LEFT(8) + RIGHT(8)
≈ 0.7847.
2
1
is decreasing over the interval. Thus
1 + x2
RIGHT(8) <
Z
0
1
1
dx < LEFT(8)
1 + x2
π
< 0.8160
4
3.014 < π < 3.264.
0.7535 <
16. Let s(t) be the distance traveled at time t and v(t) be the velocity at time t. Then the distance traveled during the interval
0 ≤ t ≤ 6 is
6
s(6) − s(0) = s(t)
=
Z
0
6
s′ (t) dt
(by the Fundamental Theorem)
0
=
Z
6
v(t) dt.
0
We estimate the distance by estimating this integral.
From the table, we find: LEFT(6) = 31, RIGHT(6) = 39, TRAP(6) = 35.
17. Since the function is decreasing, LEFT is an overestimate and RIGHT is an underestimate. Since the graph is concave
down, secant lines lie below the graph so TRAP is an underestimate and tangent lines lie above the graph so MID is an
overestimate. We can see that MID and TRAP are closer to the exact value than LEFT and RIGHT. In order smallest to
largest, we have:
RIGHT(n) < TRAP(n) < Exact value < MID(n) < LEFT(n).
18. For a decreasing function whose graph is concave up, the diagrams below show that RIGHT < MID < TRAP < LEFT.
Thus,
(a) 0.664 = LEFT, 0.633 = TRAP, 0.632 = MID, and 0.601 = RIGHT.
(b) 0.632 < true value < 0.633.
RIGHT = 0.601
MID = 0.632
TRAP = 0.633
LEFT = 0.664
19. f (x) is increasing, so RIGHT gives an overestimate and LEFT gives an underestimate.
20. f (x) is concave down, so MID gives an overestimate and TRAP gives an underestimate.
21. f (x) is decreasing and concave up, so LEFT and TRAP give overestimates and RIGHT and MID give underestimates.
22. f (x) is concave up, so TRAP gives an overestimate and MID gives an underestimate.
650
Chapter Seven /SOLUTIONS
23. (a) Since we are using n = 2 on the interval x = 0 to x = 4, we have ∆x = 2. The midpoint of the first subinterval is 1
and the midpoint of the second subinterval is 3, so we have
√
√
MID(2) = 3 1 · 2 + 3 3 · 2 = 16.392.
(b) We have
Z
4
4
3x1/2 dx = 2x3/2
0
= 2(43/2 ) = 16.
0
(c) Error = Actual − Approximation = 16 − 16.392 = −0.392.
(d) Since n is multiplied by 10 in going from n = 2 to n = 20, we expect the error to be multiplied by approximately
1/102 . We estimate that
1
· (−0.392) = −0.00392.
Error for MID(20) ≈
100
√
(e) We expect the approximation for MID(20) to be an overestimate (since f (x) = 3 x is concave down) and to be
0.00392 away from the exact value of 16. We have
MID(20) ≈ 16.00392.
24. (a) Since f (x) is closer to horizontal (that is, |f ′ | < |g ′ |), LEFT and RIGHT will be more accurate with f (x).
(b) Since g(x) has more curvature, MID and TRAP will be more accurate with f (x).
25. (a) TRAP(4) gives probably the best estimate of the integral. We cannot calculate MID(4).
LEFT(4) = 3 · 100 + 3 · 97 + 3 · 90 + 3 · 78 = 1095
RIGHT(4) = 3 · 97 + 3 · 90 + 3 · 78 + 3 · 55 = 960
1095 + 960
= 1027.5.
TRAP(4) =
2
(b) Because there are no points of inflection, the graph is either concave down or concave up. By plotting points, we see
that it is concave down. So TRAP(4) is an underestimate.
26. (a)
Z
0
2π
2π
sin θ dθ = − cos θ
= 0.
0
(b) See Figure 7.38. MID(1) is 0 since the midpoint of 0 and 2π is π, and sin π = 0. Thus MID(1) = 2π(sin π) = 0.
The midpoints we use for MID(2) are π/2 and 3π/2, and sin(π/2) = − sin(3π/2). Thus MID(2) = π sin(π/2) +
π sin(3π/2) = 0.
2π
θ
Figure 7.38
(c) MID(3) = 0.
In general, MID(n) = 0 for all n, even though your calculator (because of round-off error) might not return
it as such. The reason is that sin(x) = − sin(2π − x). If we use MID(n), we will always take sums where we are
adding pairs of the form sin(x) and sin(2π − x), so the sum will cancel to 0. (If n is odd, we will get a sin π in the
sum which does not pair up with anything — but sin π is already 0.)
27. See Figure 7.39. We observe that the error in the left or right rule depends on how steeply the graph of f rises or falls. A
steep curve makes the triangular regions missed by the left or right rectangles tall and hence large in area. This observation
suggests that the error in the left or right rules depends on the magnitude of the derivative of f .
7.5 SOLUTIONS
Right rule
error
✲
✛
651
Left rule
error
Right rule
error
✲
✛
Left rule
error
Large f ′ : large error
Small f ′ : small error
Figure 7.39: The error in the left and right rules depends on the steepness of the curve
28. From Figure 7.40 it appears that the errors in the trapezoid and midpoint rules depend on how much the curve is bent up
or down. In other words, the concavity, and hence the magnitude of the second derivative, f ′′ , has an effect on the errors
of these two rules.
Trapezoid error
Midpoint error
✲
Trapezoid error
✛
❄
❄
Small |f ′′ |: small error
Midpoint error
Large |f ′′ |: large error
Figure 7.40: The error in the trapezoid and midpoint rules depends on how bent the curve is
√
√
29. (a) The graph of y = 2 − x2 is the upper half of a circle of radius 2 centered at the origin. The integral represents
the area under this curve between the lines x = 0 and x = 1. From Figure 7.41, we see that this area can be split
into 2 parts, A1 and A2 . Notice since OQ = QP = 1, △OQP is isosceles. Thus 6 P OQ = 6 ROP = π4 , and A1
is exactly 18 of the entire circle. Thus the total area is
Area = A1 + A2 =
1 √ 2 1·1
π
1
π( 2) +
= + .
8
2
4
2
R
P = (1, 1)
√
2
1
1
O
Q
Figure 7.41
(b) LEFT(5) ≈ 1.32350, RIGHT(5) ≈ 1.24066, T
TRAP(5) ≈ 1.28208, MID(5) ≈ 1.28705
652
Chapter Seven /SOLUTIONS
Exact value ≈ 1.285398163
Left-hand error ≈ −0.03810,
Trapezoidal error ≈ 0.00332,
Right-hand error ≈ 0.04474,
Midpoint error ≈ −0.001656
Thus right-hand error > trapezoidal error > 0 > midpoint error > left-hand error, and |midpt error| < |trap error| <
|left-error| < |right-error|.
30. We approximate the area of the playing field by using Riemann sums. From the data provided,
LEFT(10) = RIGHT(10) = TRAP(10) = 89,000 square feet.
Thus approximately
89,000 sq. ft.
= 445 lbs. of fertilizer
200 sq. ft./lb.
should be necessary.
31.
✻
...
✲
a = x0
x1
x2
···
xn−1
xn = b
From the diagram, the difference between RIGHT(n) and LEFT(n) is the area of the shaded rectangles.
RIGHT(n) = f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x
LEFT(n) = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x
Notice that the terms in these two sums are the same, except that RIGHT(n) contains f (xn )∆x (= f (b)∆x), and
LEFT(n) contains f (x0 )∆x (= f (a)∆x). Thus
RIGHT(n) = LEFT(n) + f (xn )∆x − f (x0 )∆x
= LEFT(n) + f (b)∆x − f (a)∆x
32.
LEFT(n) + RIGHT(n)
2
LEFT(n) + LEFT(n) + f (b)∆x − f (a)∆x
=
2
1
= LEFT(n) + (f (b) − f (a))∆x
2
TRAP(n) =
7.5 SOLUTIONS
33.
653
✻
...
✲
a = t0 t1
x0
t2
t3
x1
t4
t2n = b
···
x2
xn
Divide the interval [a, b] into n pieces, by x0 , x1 , x2 , . . . , xn , and also into 2n pieces, by t0 , t1 , t2 , . . . , t2n . Then
the x’s coincide with the even t’s, so x0 = t0 , x1 = t2 , x2 = t4 , . . . , xn = t2n and ∆t = 12 ∆x.
LEFT(n) = f (x0 )∆x + f (x1 )∆x + · · · + f (xn−1 )∆x
Since MID(n) is obtained by evaluating f at the midpoints t1 , t3 , t5 , . . . of the x intervals, we get
MID(n) = f (t1 )∆x + f (t3 )∆x + · · · + f (t2n−1 )∆x
Now
LEFT(2n) = f (t0 )∆t + f (t1 )∆t + f (t2 )∆t + · · · + f (t2n−1 )∆t.
Regroup terms, putting all the even t’s first, the odd t’s last:
LEFT(2n) = f (t0 )∆t + f (t2 )∆t + · · · + f (t2n−2 )∆t + f (t1 )∆t + f (t3 )∆t + · · · + f (t2n−1 )∆t
= f (x0 )
So
|
∆x
∆x
∆x
∆x
∆x
∆x
+ f (x1 )
+ · · · + f (xn−1 )
+ f (t1 )
+ f (t3 )
+ · · · + f (t2n−1 )
2
2{z
2} |
2
2 {z
2}
LEFT(n)/2
LEFT(2n) =
MID(n)/2
1
(LEFT(n) + MID(n))
2
1
; f (a) = 1; f (b) = 21 .
34. When n = 10, we have a = 1; b = 2; ∆x = 10
LEFT(10) ≈ 0.71877, RIGHT(10) ≈ 0.66877, TRAP(10) ≈ 0.69377
We have
1 1
RIGHT(10) = LEFT(10) + f (b)∆x − f (a)∆x = 0.71877 + 10
(2) −
1 1 1
LEFT(10) + ∆x
(f
(b)
−
f
(a))
=
0.71877
+
(
−
1)
=
0.69377,
2
10 2 2
so the equations are verified.
1
(1)
10
35. First, we compute:
b−a
(f (b) − f (a))∆x = (f (b) − f (a))
n
= (f (5) − f (2))
= (21 − 13)
=
24
n
3
n
3
n
= 0.66877, and TRAP(10) =
654
Chapter Seven /SOLUTIONS
RIGHT(10) = LEFT(10) + 24 = 3.156 + 2.4 = 5.556.
TRAP(10) = LEFT(10) + 21 (2.4) = 3.156 + 1.2 = 4.356.
LEFT(20) = 12 (LEFT(10) + MID(10)) = 21 (3.156 + 3.242) = 3.199.
RIGHT(20) = LEFT(20) + 2.4 = 3.199 + 1.2 = 4.399.
TRAP(20) = LEFT(20) + 12 (1.2) = 3.199 + 0.6 = 3.799.
36. (a) If f (x) = 1, then
Z
b
a
Also,
h
3
f (x) dx = (b − a).
f (b)
f (a)
+ 2f (m) +
2
2
So the equation holds for f (x) = 1.
If f (x) = x, then
Z
b
f (x) dx =
a
Also,
h
3
f (a)
f (b)
+ 2f (m) +
2
2
b−a
3
=
x2
2
b
=
a
= (b − a).
b2 − a 2
.
2
b−a
3
=
1
1
+2+
2
2
a
a+b
b
+2
+
2
2
2
b−a a
b
+a+b+
3
2
2
b−a 3
3
=
b+ a
3
2
2
(b − a)(b + a)
=
2
b2 − a 2
.
=
2
=
So the equation holds for f (x) = x.
If f (x) = x2 , then
Z
h
3
b
f (x) dx =
a
x3
3
b
=
a
f (b)
f (a)
+ 2f (m) +
2
2
b3 − a 3
. Also,
3
=
b−a
3
=
b−a
3
=
b−a
3
a2
a+b
+2
2
2
2a2 + 2ab + 2b2
2
2
+
b2
2
a2
a2 + 2ab + b2
b2
+
+
2
2
2
b−a 2
a + ab + b2
3
b3 − a 3
=
.
3
=
So the equation holds for f (x) = x2 .
(b) For any quadratic function, f (x) = Ax2 + Bx + C, the “Facts about Sums and Constant Multiples of Integrands”
give us:
Z
b
f (x) dx =
a
Z
b
2
(Ax + Bx + C) dx = A
a
Now we use the results of part (a) to get:
Z
b
f (x) dx = A
a
=
h
3
h
3
a2
b2
+ 2m2 +
2
2
+B
h
3
Z
b
2
x dx + B
a
a
b
+ 2m +
2
2
Z
b
x dx + C
a
+C
Z
b
1 dx.
a
h
3
1
1
+2·1+
2
2
Ab2 + Bb + C
Aa2 + Ba + C
+ 2(Am2 + Bm + C) +
2
2
7.5 SOLUTIONS
=
h
3
f (b)
f (a)
+ 2f (m) +
2
2
655
37. (a) Suppose qi (x) is the quadratic function approximating f (x) on the subinterval [xi , xi+1 ], and mi is the midpoint
of the interval, mi = (xi + xi+1 )/2. Then, using the equation in Problem 36, with a = xi and b = xi+1 and
h = ∆x = xi+1 − xi :
Z
xi+1
xi
f (x)dx ≈
Z
xi+1
qi (x)dx =
xi
∆x
3
qi (xi+1 )
qi (xi )
+ 2qi (mi ) +
2
2
.
(b) Summing over all subintervals gives
Z
a
b
f (x)dx ≈
n−1 Z
X
i=0
xi+1
qi (x)dx =
xi
n−1
X
∆x qi (xi )
i=0
3
2
qi (xi+1 )
+ 2qi (mi ) +
2
.
Splitting the sum into two parts:
=
n−1
n−1
1 X qi (xi ) + qi (xi+1 )
2X
∆x
qi (mi )∆x +
3
3
2
i=0
i=0
2
1
= MID(n) + TRAP(n)
3
3
= SIMP(n).
Strengthen Your Understanding
38. The midpoint rule is exact if the integrand is a linear function.
39. As n → ∞, the error approaches 0, so TRAP(n) → 0 only if the value of the definite integral is 0.
40. It depends on the concavity. If f is concave up, then TRAP(n) ≥ MID(n), but if f is concave down, then TRAP(n) ≤
MID(n).
41. Since the total time for each extra two digits of TRAP goes up by a factor of 10, the total time grows exponentially with
the number of digits. Thus there is much more additional time to go from 8 digits to 10 digits than from 1 digit to 3 digits.
R1
42. Since we want RIGHT(10) to be an underestimate of 0 f (x)dx, we use a function f (x) that is decreasing on [0, 1].
Since we want MID(10) to be an overestimate of the integral, we make f (x) concave down on [0, 1]. The function
f (x) = 1 − x2 has both of these features. Since f (x) = 1 − x2 is both decreasing and concave down on [0, 1], we have
RIGHT(10) <
Z
1
f (x)dx < MID(10).
0
43. In most cases, increasing the number of subintervals in a Trapezoid Rule approximation makes the approximation more
accurate. Therefore, we try a function f (x) for which the Trapezoid Rule overestimates the integral, since it is likely in
R 10
this case that TRAP(40) is a greater overestimate of 0 f (x)dx than TRAP(80). Suppose that f (x) = ex . Then f (x)
is concave up, and TRAP(40) and TRAP(80) are both overestimates. However, TRAP(80) is smaller, since all of the
trapezoids that make up the estimate TRAP(80) fit inside the trapezoids that make up TRAP(40).
44. True. y 2 − 1 is concave up, and the midpoint rule always underestimates for a function that is concave up.
45. False. If the function f (x) is a line, then the trapezoid rule gives the exact answer to
46. False. The subdivision size ∆x = (1/10)(6 − 2) = 4/10.
47. True, since ∆x = (6 − 2)/n = 4/n.
Rb
a
f (x) dx.
48. False. If f is decreasing, then on each subinterval the value of f (x) at the left endpoint is larger than the value at the right
endpoint, which means that LEFT(n) >RIGHT(n) for any n.
49. False. As n approaches infinity, LEFT(n) approaches the value of the integral
R6
2
f (x) dx, which is generally not 0.
656
Chapter Seven /SOLUTIONS
50. True. We have
LEFT(n) − RIGHT(n) = (f (x0 ) + f (x1 ) + · · · + f (xn−1 ))∆x − (f (x1 ) + f (x2 ) + · · · + f (xn ))∆x.
On the right side of the equation, all terms cancel except the first and last, so:
LEFT(n) − RIGHT(n) = (f (x0 ) − f (xn ))∆x = (f (2) − f (6))∆x.
This is also discussed in Section 5.1.
51. True. This follows from the fact that ∆x = (6 − 2)/n = 4/n.
52. False. Since LEFT(n) − RIGHT(n) = (f (2) − f (6))∆x, we have LEFT(n) = RIGHT(n) for any function such that
f (2) = f (6). Such a function, for example f (x) = (x − 4)2 , need not be a constant function.
53. False. Although TRAP(n) is usually a better estimate, it is not always better. If f (2) = f (6), then LEFT(n) = RIGHT(n)
and hence TRAP(n) = LEFT(n) = RIGHT(n), so in this case TRAP(n) is no better.
54. False. This is true if f is an increasing function or if f is a decreasing function, but it isR not true in general. For example,
6
suppose that f (2) = f (6). Then LEFT(n) = RIGHT(n) for all n, which means that if 2 f (x) dx lies between LEFT(n)
and RIGHT(n), then it must equal LEFT(n), which is not always the case.
For example, if f (x) = (x − 4)2 and n = 1, then f (2) = f (6) = 4, so
LEFT(1) = RIGHT(1) = 4 · (6 − 2) = 16.
However
Z
2
6
(x − 4)2 dx =
(x − 4)3
3
6
=
2
23
−
3
−
23
3
=
16
.
3
In this example, since LEFT(n) = RIGHT(n), we have TRAP(n) = LEFT(n). However trapezoids overestimate the
area, since the graph of f is concave up. This is also discussed in Section 7.5.
55. False. Suppose f is the following:
2
x
1
2
3
4
Then LEFT(2) = 0, LEFT(4) = 4, and
Z
b
f (x) dx = 4.
a
56. True. Since f ′ and g ′ are greater than 0, all left rectangles give underestimates. The bigger the derivative, the bigger
the underestimate, so the bigger the error. (Note: if we did not have 0 < f ′ < g ′ , but instead just had f ′ < g ′ , the
statement would not necessarily be true. This is because some left rectangles could be overestimates and some could be
underestimates–so, for example, it could be that the error in approximating g is 0! If 0 < f ′ < g ′ , however, this can’t
happen.
7.6 SOLUTIONS
657
Solutions for Section 7.6
Exercises
1. (a) See Figure 7.42. The area extends out infinitely far along the positive x-axis.
y
y
x
x
1
1
Figure 7.42
Figure 7.43
(b) See Figure 7.43. The area extends up infinitely far along the positive y-axis.
2. We have
Z
∞
e−0.4x dx = lim
0
b→∞
Z
b
e−0.4x dx = lim (−2.5e−0.4x )|b0 = lim (−2.5e−0.4b + 2.5).
b→∞
0
b→∞
As b → ∞, we know e−0.4b → 0 and so we see that the integral converges to 2.5. See Figure 7.44. The area continues
indefinitely out to the right.
e−0.4x
x
Figure 7.44
3. (a) We use a calculator or Rcomputer to evaluate the integrals.
5
When b = 5, we have 0 xe−x dx = 0.9596.
R 10 −x
xe dx = 0.9995.
R020
When b = 20, we have 0 xe−x dx = 0.99999996.
R ∞ −x
When b = 10, we have
(b) It appears from the answers to part (a) that
0
xe
dx = 1.0.
4. (a) See Figure 7.45. The total area under the curve is shaded.
1
x
−4 −3 −2 −1
1
2
3
4
Figure 7.45
R1
2
(b) When a = 1, we use a calculator or computer to see that −1 e−x dx = 1.49365.
Similarly, we have:
When a = 2, the value of the integral is 1.76416.
When a = 3, the value of the integral is 1.77241.
When a = 4, the value of the integral is 1.77245.
When a = 5, the value of the integral is 1.77245.
R∞
2
(c) It appears that the integral −∞ e−x dx converges to approximately 1.77245.
658
Chapter Seven /SOLUTIONS
5. We have
Z
∞
1
1
dx = lim
b→∞
5x + 2
Z
b
1
1
dx = lim
ln (5x + 2)
b→∞ 5
5x + 2
1
b
= lim
b→∞
1
As b ← ∞, we know that ln (5b + 2) → ∞, and so this integral diverges.
1
1
ln (5b + 2) − ln (7) .
5
5
6. We have
∞
Z
1
1
dx = lim
b→∞
(x + 2)2
b
Z
1
−1
1
dx = lim
b→∞ x + 2
(x + 2)2
This integral converges to 1/3.
b
= lim
b→∞
1
−1
−1
−
b+2
3
=0+
1
1
= .
3
3
7. This integral is improper at the lower end, so
1
Z
ln x dx = lim
a→0+
0
Z
1
ln x dx
a
1
= lim (x ln x − x)
a→0+
a
= lim (1 ln 1 − 1) − (a ln a − a))
a→0+
= −1 + lim a(1 − ln a)
a→0+
1 − ln a
a−1
−1/a
= −1 + lim
by l’Hopital
2
a→0+ −1/a
= −1 + lim a
= −1 + lim
a→0+
a→0+
= −1.
If the integral converges, we’d expect it to have a negative value because the logarithm graph is below the x-axis for
0 < x < 1.
√
8. Use the substitution w = x. Since dw = 21 x−1/2 dx, we have dx = 2w dw, so
Z
e−
√
x
dx = 2
Z
we−w dw.
Note that this substitution leaves the limits unchanged. Using integration by parts with u = w and v ′ = e−w , we find that
2
Z
we−w dw = 2 −we−w +
Z
e−w dw
= 2[−we−w − e−w ].
So,
Z
0
∞
we−w dx = lim 2 −we−w − e−w
b→∞
= 2 lim (−be
−b
b→∞
−e
−b
b
0
+ 1)
b+1
+1
eb
i
h
1
= 2 lim − b + 1 by l’Hopital
b→∞
e
= 2.
h
i
= 2 lim −
b→∞
9. We have
Z
0
∞
2
xe−x dx = lim
b→∞
This integral converges to 1/2.
Z
0
b
2
xe−x dx = lim
b→∞
−1 −x2
e
2
b
= lim
0
b→∞
−1 −b2
−1
e
−
2
2
=0+
1
1
= .
2
2
7.6 SOLUTIONS
10. Z
∞
e−2x dx = lim
b→∞
1
Z
b
e−2x dx = lim −
b→∞
1
= lim (−e
−2b
b→∞
/2 + e
−2
e−2x
2
b
1
/2) = 0 + e−2 /2 = e−2 /2,
where the first limit is 0 because limx→∞ e−x = 0.
11. Using integration by parts with u = x and v ′ = e−x , we find that
Z
so
xe−x dx = −xe−x −
∞
Z
0
Z
x
dx = lim
b→∞
ex
−e−x dx = −(1 + x)e−x
Z
b
x
dx
ex
0
b
= lim −1(1 + x)e−x
b→∞
0
= lim 1 − (1 + b)e−b
b→∞
= 1.
12.
Z
1
13.
∞
x
= lim
b→∞
4 + x2
2
Z
b
1
x
1
dx = lim ln |4 + x2 |
b→∞ 2
4 + x2
b
= lim
1
b→∞
1
1
ln |4 + b2 | − ln 5.
2
2
As b → ∞, ln |4 + b | → ∞, so the limit diverges.
Z
0
−∞
ex
dx = lim
b→−∞
1 + ex
Z
b
0
ex
dx
1 + ex
0
= lim ln |1 + ex |
b→−∞
b
= lim [ln |1 + e0 | − ln |1 + eb |]
b→−∞
= ln(1 + 1) − ln(1 + 0) = ln 2.
14. First, we note that 1/(z 2 + 25) is an even function. Therefore,
Z
∞
−∞
Z
dz
=
z 2 + 25
0
−∞
dz
+
z 2 + 25
Z
0
∞
dz
=2
z 2 + 25
Z
∞
0
dz
.
z 2 + 25
We’ll now evaluate this improper integral by using a limit:
Z
∞
0
z2
1
π
dz
1
1 π
= lim
arctan(b/5) − arctan(0) = · =
.
b→∞ 5
+ 25
5
5 2
10
So the original integral is twice that, namely π/5.
√
15. This integral is improper because 1/ x is undefined at x = 0. Then
Z
4
0
The integral converges.
1
√ dx = lim
x
b→0+
Z
b
4
1
√ dx = lim
x
b→0+
√
2 x
4
b
!
= lim
b→0+
√
4 − 2 b = 4.
659
660
Chapter Seven /SOLUTIONS
16.
π/2
Z
π/4
sin x
√
dx =
cos x
lim
b→π/2−
=
lim
b→π/2−
=
lim
b→π/2−
Z
b
sin x
√
dx
cos x
π/4
−
b
Z
(cos x)−1/2 (− sin x) dx
π/4
b
−2(cos x)1/2
lim [−2(cos b)
=
1/2
b→π/2−
√ 21
2
2
=2
π/4
+ 2(cos π/4)1/2 ]
3
= 24.
17. This integral is improper because 1/v is undefined at v = 0. Then
Z
18.
0
1
dv = lim
v
b→0+
Z
1
b
1
1
dv = lim
v
b→0+
ln v
b
!
= − ln b.
As b → 0+ , this goes to infinity and the integral diverges.
1
x4
x4 + 1
dx = lim (
+ ln x)
+
+
x
4
a→0
a→0
a
which diverges as a → 0, since ln a → −∞.
Z
lim
19.
1
∞
Z
1
1
dx = lim
x2 + 1
b→∞
Z
b
1
1
= lim [1/4 − (a4 /4 + ln a)],
a→0+
a
1
dx
x2 + 1
b
= lim arctan(x)
b→∞
1
= lim [arctan(b) − arctan(1)]
b→∞
= π/2 − π/4 = π/4.
20.
Z
1
∞
√
1
x2
+1
dx = lim
b→∞
Z
b
1
√
1
x2
= lim ln |x +
b→∞
= lim ln(b +
b→∞
As b → ∞, this limit does not exist, so the integral diverges.
21. We use V-26 with a = 4 and b = −4:
Z
0
4
−1
du = lim
u2 − 16
b→4−
= lim
b→4−
= lim
b→4−
Z
b
0
Z
0
b
+1
dx
b
p
x2 + 1|
1
p
b2 + 1) − ln(1 +
√
2).
−1
du
u2 − 16
−1
du
(u − 4)(u + 4)
−(ln |u − 4| − ln |u + 4|)
8
b
0
1
= lim − (ln |b − 4| + ln 4 − ln |b + 4| − ln 4) .
8
b→4−
As b → 4− , ln |b − 4| → −∞, so the limit does not exist and the integral diverges.
7.6 SOLUTIONS
22.
Z
∞
y
1
dy = lim
b→∞ 2
y4 + 1
1
Z
b
1
2y
dy
(y 2 )2 + 1
1
= lim arctan(y 2 )
b→∞ 2
b
1
1
= lim [arctan(b2 ) − arctan 1]
b→∞ 2
= (1/2)[π/2 − π/4] = π/8.
23. With the substitution w = ln x, dw =
Z
so
1
dx,
x
dx
=
x ln x
Z
∞
Z
1
dw = ln |w| + C = ln | ln x| + C
w
dx
= lim
b→∞
x ln x
2
Z
b
2
dx
x ln x
b
= lim ln | ln x|
b→∞
2
= lim [ln | ln b| − ln | ln 2|].
b→∞
As b → ∞, the limit goes to ∞ and hence the integral diverges.
24. With the substitution w = ln x, dw =
Z
so
Z
0
1
1
dx,
x
ln x
dx =
x
Z
ln x
dx = lim
x
a→0+
Z
1
1 2
1
w + C = (ln x)2 + C
2
2
w dw =
ln x
1
dx = lim [ln(x)]2
x
a→0+ 2
a
1
a
1
= lim − [ln(a)]2 .
2
a→0+
As a → 0+ , ln a → −∞, so the integral diverges.
25. This is a proper integral; use V-26 in the integral table with a = 4 and b = −4.
Z
20
16
1
dy =
y 2 − 16
Z
20
16
1
dy
(y − 4)(y + 4)
ln |y − 4| − ln |y + 4|
=
8
20
16
ln 16 − ln 24 − (ln 12 − ln 20)
=
8
ln 320 − ln 288
1
=
= ln(10/9) = 0.01317.
8
8
1
1
26. Using the substitution w = −x 2 , −2dw = x− 2 dx,
So
Z
1
2
1
e−x x− 2 dx = −2
Z
0
π
Z
1
2
ew dw = −2e−x + C.
√
1
√ e− x dx = lim
x
b→0+
Z
π
b
√
1
√ e− x dx
x
= lim −2e−
√
b→0+
= 2 − 2e−
√
π
.
π
x
b
661
662
Chapter Seven /SOLUTIONS
27. Letting w = ln x, dw =
1
dx,
x
Z
so
dx
=
x(ln x)2
Z
∞
3
Z
w−2 dw = −w−1 + C = −
dx
= lim
x(ln x)2
b→∞
1
+ C,
ln x
b
Z
dx
x(ln
x)2
3
1
1
+
= lim −
b→∞
ln b
ln 3
1
=
.
ln 3
28.
Z
2
√
0
b
Z
1
dx = lim
b→2−
4 − x2
√
0
1
dx
4 − x2
x
2
= lim arcsin
b→2−
b
0
π
b
= lim arcsin = arcsin 1 = .
−
2
2
b→2
29.
Z
4
30.
Z
∞
dx
= lim
b→∞
(x − 1)2
dx
=
x2 − 1
Z
Z
b
4
b
dx
1
= lim −
b→∞
(x − 1)2
(x − 1)
h
= lim −
b→∞
4
1
1
1
+
= .
b−1
3
3
dx
1
1
= (ln |x − 1| − ln |x + 1|) + C =
(x − 1)(x + 1)
2
2
Z
∞
dx
= lim
b→∞
x2 − 1
4
Z
b
|x − 1|
ln
|x + 1|
|x − 1|
ln
|x + 1|
Z
b
+ C, so
b
4
b−1
1 3
− ln
b+1
2 5
1 5
ln .
2 3
1
ln
= lim
b→∞ 2
1 3
= − ln =
2 5
h
dx
x2 − 1
4
1
= lim
b→∞ 2
i
i
31.
Z
7
∞
√
dy
= lim
b→∞
y−5
dy
y−5
b
p
= lim 2
b→∞
√
7
y−5
7
√
√
= lim (2 b − 5 − 2 2).
b→∞
As b → ∞, this limit goes to ∞, so the integral diverges.
32. The integrand is undefined at y = 3.
Z
0
3
y dy
p
9 − y2
= lim
b→3−
= lim
b→3−
Z
0
b
b
y
p
9 − y2
dy = lim
b→3−
3 − (9 − b2 )1/2 = 3.
2 1/2
−(9 − y )
0
7.6 SOLUTIONS
663
33. The integrand is undefined at θ = 4, so we must split the integral there.
Z
6
4
dθ
= lim
(4 − θ)2
a→4+
6
Z
a
dθ
= lim (4 − θ)−1
(4 − θ)2
a→4+
6
= lim
a→4+
a
1
−2
−
1
.
4−a
Since 1/(4 − a) → −∞ as a → 4 from the right, the integral does not converge. It is not necessary to check the
R 4 dθ
R 6 dθ
R 4 dθ
convergence of 3 (4−θ)
2 . However, we could have started with 3 (4−θ)2 , instead of 4 (4−θ)2 , and arrived at the same
conclusion.
Problems
34. We have
x
Z
f (x) =
et dt
−∞
= lim
a→−∞
x
Z
et dt
a
x
= lim (e − ea )
a→−∞
= ex − lim ea
a→−∞
| {z }
0
= ex .
35. (a) There is no simple antiderivative for this integrand, so we use numerical methods. We find
1
P (1) = √
π
1
Z
2
e−t dt = 0.421.
0
(b) To calculate this improper integral, use numerical methods. If you cannot input infinity into your calculator, increase
the upper limit until the value of the integral settles down. We find
1
P (∞) = √
π
∞
Z
2
e−t dt = 0.500.
0
36. Since the graph is above the x-axis for x ≥ 0, we have
Area =
Z
∞
Z
xe−x dx = lim
b→∞
0
= lim
b→∞
= lim
b→∞
b
−xe−x
−be
−b
+
0
Z
b
xe−x dx
0
b
e−x dx
0
−e
−x
b
0
= lim (−be−b − e−b + e0 ) = 1.
b→∞
37. The curve has an asymptote at t =
π
,
2
and so the area integral is improper there.
Area =
Z
π
2
0
dt
= limπ
cos2 t
b→ 2
Z
b
0
dt
= limπ tan t
cos2 t
b→ 2
b
,
0
which diverges. Therefore the area is infinite.
38. We have:
f (3) =
Z
∞
−t
3
0
(Since lim 3−b = 0.)
b→∞
dt = lim
b→∞
Z
0
b
3−t dt = lim −
b→∞
1 −t
3
ln 3
b
0
= lim −
b→∞
1
1
= 0.910.
3−b − 30 =
ln 3
ln 3
664
Chapter Seven /SOLUTIONS
39. We have:
f (3) =
Z
∞
t−3 dt = lim
b→∞
1
(Since lim b
−2
b→∞
Z
b
b
1
t−3 dt = lim − t−2
b→∞
2
1
= lim −
b→∞
1
= 0.)
1
1 −2
b −1 = .
2
2
40. We have:
f (3) =
Z
∞
Z
3e−3t dt = 3 lim
b→∞
0
41. We have:
f (3) =
b
0
∞
Z
2
2t · 3e−t(3) dt =
0
1
3
e−3t dt = 3 −
b
lim e−3t
b→∞
0
= − lim e−3b − 1 = −(−1) = 1.
∞
Z
6te−9t dt = 6 lim
b→∞
0
Integration by parts with u = t, du = dt, dv = e−9t dt, v = (−1/9)e−9t gives:
Z
te−9t dt = t −
Thus:
Z
1
1 −9t
− te−9t −
e
9
81
1 −9t
e
−
9
f (3) = 6 lim
b→∞
42. (a) We have
∞
Z
0
−
(b) Using integration by parts with u = y and v = (1/α)e
0
ye−y/α
dy = lim
b→∞
α
b
te−9t dt.
0
−y/α
Z
b
0
0
b→∞
1
81
=6 0+
0
2
.
27
b→∞
ye−y/α
dy
α
b
−y/α
+
Z
b
e
−y/α
= lim (−be
−b/α
b→∞
dy
0
0
−be−b/α − αe−y/α
b→∞
=
= lim (1 − e−b/α ) = 1.
b
= lim
, so u′ = 1, v = −e−y/α , we have
−ye
= lim
b
b
e−y/α
dy = lim −e−y/α
b→∞
α
∞
Z
1 −9t
1 −9t
1
e
+ C.
e dt = − te−9t −
9
9
81
′
Z
b→∞
− αe
−b/α
0
!
!
+ α)
Since limb→∞ −be−b/α = limb→∞ e−b/α = 0, we have
Z
ye−y/α
dy = α.
α
(c) Using integration by parts, this time with u = y 2 , v ′ = (1/α)e−y/α , so u′ = 2y, v = −e−y/α , we have
Z
0
∞
y 2 e−y/α
dy = lim
b→∞
α
Z
b
0
y 2 e−y/α
dy
α
b
2 −y/α
= lim
−y e
b→∞
+2
b→∞
Z
Now limb→∞ −b2 e−b/α = 0 and in part (b) we found
Z
∞
0
b
ye
−y/α
0
0
= lim −b2 e−b/α + 2
Z
ye−y/α
dy = α,
α
0
∞
ye−y/α dy
dy
!
7.6 SOLUTIONS
so
Z
∞
ye−y/α dy = α2 .
0
Thus,
∞
Z
y 2 e−y/α
dy = 2
α
0
43. (a) Using a calculator or a computer, the graph is:
Z
∞
ye−y/α dy = 2α2 .
0
r
2000
e
r = 1000te−0.5t
t
2
(b) People are getting sick fastest when the rate of infection is highest, i.e. when r is at its maximum. Since
r ′ = 1000e−0.5t − 1000(0.5)te−0.5t
= 500e−0.5t (2 − t)
this must occur at t = 2.
(c) The total number of sick people =
∞
Z
1000te−0.5t dt.
0
Using integration by parts, with u = t, v ′ = e−0.5t :
−t −0.5t
e
0.5
Total = lim 1000
b→∞
b
−
0
b
0
−1 −0.5t
e
dt
0.5
2 −0.5b
e
0.5
= lim 1000 −2be−0.5b −
b→∞
Z
!
b
0
= lim 1000 −2be−0.5b − 4e−0.5b + 4
b→∞
= 4000 people.
44. The energy required is
E=
Z
∞
1
kq1 q2
1
dr = kq1 q2 lim −
b→∞
r2
r
b
1
= (9 × 109 )(1)(1)(1) = 9 × 109 joules
√
√
45. We let t = (x − a)/ b. This means that dt = dx/ b, and that t = ±∞ when x = ±∞. We have
Z
∞
e−(x−a)
2
/b
dx =
−∞
Z
∞
√
2 √
e−t ( b dt) = b
Z
2 /2
g ′ (x)e−x
2
/2
dx = g(x)e−x
2
/2
−∞
−∞
Since g(x) is bounded, limx→∞ g(x)e−x
Z
2
/2
−
= limx→−∞ g(x)e−x
∞
g ′ (x)e−x
−∞
2
e−t dt =
and v ′ = g ′ (x), so u′ = −xe−x
∞
∞
∞
√
√ √
b π = bπ.
−∞
−∞
46. Applying integration by parts with u = e−x
Z
2
/2
dx =
Z
2
Z
/2
∞
−∞
−xe−x
and v = g(x), we have
2
/2
= 0. Thus
∞
−∞
2 /2
xg(x)e−x
2
/2
dx.
g(x) dx.
665
666
Chapter Seven /SOLUTIONS
47. Make the substitution w = 2x, so dw = 2 dx and x = w/2. When x = 0, w = 0 and when x → ∞, we have w → ∞.
We have
Z
∞
0
x4 e2x
dx =
(e2x − 1)2
Z
∞
0
Z
1
=
32
(w/2)4 ew
dw/2
(ew − 1)2
∞
w 4 ew
dw
− 1)2
(ew
0
1 4π 4
π4
=
=
.
32 15
120
48. (a)
∞
Z
Γ(1) =
e−t dt
0
= lim
b→∞
Z
b
e−t dt
0
b
= lim −e−t
b→∞
0
= lim [1 − e−b ] = 1.
b→∞
Using Problem 11,
Z
Γ(2) =
∞
te−t dt = 1.
0
(b) We integrate by parts. Let u = tn , v ′ = e−t . Then u′ = ntn−1 and v = −e−t , so
Z
So
tn e−t dt = −tn e−t + n
Γ(n + 1) =
Z
Z
tn−1 e−t dt.
∞
tn e−t dt
0
= lim
b→∞
= lim
b→∞
Z
b
tn e−t dt
0
b
− tn e−t
+n
Z
b→∞
= 0+n
Z
∞
t
n−1 −t
e
tn−1 e−t dt
0
0
= lim −bn e−b + lim n
b→∞
b
Z
b
tn−1 e−t dt
0
dt
0
= nΓ(n).
(c) We already have Γ(1) = 1 and Γ(2) = 1. Using Γ(n + 1) = nΓ(n) we can get
Γ(3) = 2Γ(2) = 2
Γ(4) = 3Γ(3) = 3 · 2
Γ(5) = 4Γ(4) = 4 · 3 · 2.
So it appears that Γ(n) is just the first n − 1 numbers multiplied together, so
Γ(n) = (n − 1)!.
Strengthen Your Understanding
49. Let f (x) = g(x) = 1/x. Then both
converges.
R∞
1
f (x) dx and
R∞
1
g(x) dx diverge, but
R∞
1
f (x)g(x) dx =
R∞
1
1/x2 dx, which
7.6 SOLUTIONS
50. If f (x) = 1/x, then
R∞
1
667
f (x) dx diverges, but limx→∞ f (x) = 0.
51. The function f (x) = 1/x is an example. We know that limx→∞ 1/x = 0, but
∞
Z
1
1
dx = lim
b→∞
x
Z
b
1
dx = lim (ln b − ln 1) = ∞.
b→∞
x
1
Thus the improper integral diverges.
52. Since the integral is over a bounded interval, and f (x) is continuous at x = 2 and at x = 5, we need to find a function
f (x) that has a discontinuity in the interval (2, 5). Let’s try f (x) = 1/(x − 3)2 . Because this function is discontinuous at
R5
x = 3, we must split the improper integral 2 1/(x − 3)2 dx into two integrals:
Z
5
2
1
dx =
(x − 3)2
Z
3
1
dx +
(x − 3)2
2
= lim
b→3−
= lim
b→3−
= lim
b→3−
Z
b
2
lim
b→∞
Z
1
1
+
−
(b − 3)
2−3
1
−1
−
(b − 3)
b
f (x) dx =
0
3
1
dx
(x − 3)2
1
dx + lim
(x − 3)2
a→3+
Both of these limits go to ∞. Thus the improper integral
53. True. Since
5
Z
Z
R5
2
Z
5
1
dx
(x − 3)2
a
1
1
−
+
(5 − 3)
(a − 3)
+ lim
a→3+
+ lim
a→3+
1
1
−
(a − 3)
2
f (x)dx diverges.
a
f (x) dx + lim
b→∞
0
Z
b
f (x) dx,
a
the limit on the left side of the equation is finite exactly when the limit on the right side is finite. Thus, if
R∞
converges, then so does a f (x) dx.
R∞
0
f (x) dx
54. diverges.
Rp
R 2p
R 3p
True. Suppose that f has period p. Then 0 f (x) dx, p f (x) dx, 2p f (x) dx,. . . are all equal. If we let k =
Rp
0
f (x) dx, then
∞, the value of
diverges.
R np
R np0
0
f (x) dx = nk, for any positive integer n. Since f (x) is positive, so is k. Thus as n approaches
f (x) dx = nk approaches ∞. That means that limb→∞
55. False. Let f (x) = 1/(x + 1). Then
∞
Z
0
1
dx = lim ln |x + 1|
b→∞
x+1
b
1
dx = lim ln |x + 1|
x+1
b→∞
b
Rb
0
f (x) dx is not finite; that is, the integral
= lim ln(b + 1),
0
b→∞
but limb→∞ ln(b + 1) does not exist.
56. False. Let f (x) = x + 1. Then
Z
∞
0
= lim ln(b + 1),
0
b→∞
but limb→∞ ln(b + 1) does not exist.
57. True. By properties of integrals and limits,
lim
b→∞
Z
0
b
(f (x) + g(x)) dx = lim
b→∞
Z
0
b
f (x) dx + lim
b→∞
Z
b
g(x) dx.
0
Since the two limits on the right side of the equation are finite, the limit on the left side is also finite, that is,
converges.
58. False. For example, let f (x) = x and g(x) = −x. Then f (x) + g(x) = 0, so
R∞
R∞
though 0 f (x) dx and 0 g(x) dx diverge.
R∞
0
R∞
0
(f (x) + g(x)) dx
(f (x) + g(x)) dx converges, even
668
Chapter Seven /SOLUTIONS
59. True. By properties of integrals and limits,
lim
b→∞
Z
b
af (x) dx = a lim
b→∞
0
b
Z
f (x) dx.
0
Thus,
the equation is finite exactly when the limit on the right side of the equation is finite. Thus
R ∞ the limit on the left ofR ∞
(x)
dx
converges
if
f (x) dx converges.
af
0
0
60. True. Make the substitution w = ax. Then dw = a dx, so
Z
b
1
a
f (ax) dx =
0
Z
c
f (w) dw,
0
where c = ab. As b approaches infinity, so does c, since a is constant. Thus the limit of the left side of the equation as b
approaches
infinity is finite exactly when the
R ∞limit of the right side of the equation as c approaches infinity is finite. That
R∞
is, 0 f (ax) dx converges exactly when 0 f (x) dx converges.
61. True. Make the substitution w = a + x, so dw = dx. Then w = a when x = 0, and w = a + b when x = b, so
Z
b
f (a + x) dx =
Z
b+a
f (w) dw =
a
0
Z
c
f (w) dw
a
where c = b + a. As b approaches infinity, so does c, since a is constant. Thus the limit of the left side of the equation as b
infinity is finite exactly when the limit
R cof the right side of the equationRasc c approaches infinity is finite. Since
Rapproaches
∞
converges,
we
know
that
lim
f (w) dw is finite, so limc→∞ a f (w) dw is finite for any positive a.
f
(x)
dx
c→∞
0
0
R∞
Thus, 0 f (a + x) dx converges.
62. False. We have
Z
b
(a + f (x)) dx =
b
a dx +
0
0
R∞
Z
Z
b
f (x) dx.
0
Since 0 f (x) dx converges, the second integral on the right side of the equation has a finite limit as b approaches infinity.
But the first integral on the right side has an infinite
R ∞ limit as b approaches infinity, since a 6= 0. Thus the right side all
together has an infinite limit, which means that 0 (a + f (x)) dx diverges.
Solutions for Section 7.7
Exercises
1. For large x, the integrand behaves like 1/x2 because
x2
x2
1
≈ 4 = 2.
x4 + 1
x
x
Since
Z
∞
Z
∞
1
Since
dx
converges, we expect our integral to converge. More precisely, since x4 + 1 > x4 , we have
x2
x2
x2
1
<
= 2.
x4 + 1
x4
x
1
dx
is convergent, the comparison test tells us that
x2
2. For large x, the integrand behaves like 1/x because
Since
Z
∞
2
Z
1
∞
x2
dx converges also.
+1
x4
x3
1
x3
≈
= .
x4 − 1
x4
x
1
dx does not converge, we expect our integral not to converge. More precisely, since x4 − 1 < x4 , we have
x
x3
1
x3
> 4 = .
−1
x
x
x4
Since
Z
2
∞
1
dx does not converge, the comparison test tells us that
x
Z
2
∞
x3
dx does not converge either.
−1
x4
7.7 SOLUTIONS
669
3. The integrand is continuous for all x ≥ 1, so whether the integral converges or diverges depends only on the behavior of
the function as x → ∞. As x → ∞, polynomials
Z ∞ behave like the highest powered term. Thus, as x → ∞, the integrand
x2
1
1
x2 + 1
behaves like 3 or . Since
dx diverges, we predict that the given integral will diverge.
x3 + 3x + 2
x
x
x
1
4. The integrand is continuous for all x ≥ 1, so whether the integral converges or diverges depends only on the behavior of
the function as x → ∞. As x → ∞, polynomials
behave like the highest powered term. Thus, as x → ∞, the integrand
Z ∞
1
1
1
behaves like 2 . Since
dx converges, we predict that the given integral will converge.
x2 + 5x + 1
x
x2
1
5. The integrand is continuous for all x ≥ 1, so whether the integral converges or diverges depends only on the behavior of
the function as x → ∞. As x → ∞, polynomials
Z ∞ behave like the highest powered term. Thus, as x → ∞, the integrand
1
x
x
1
behaves like 2 or . Since
dx diverges, we predict that the given integral will diverge.
x2 + 2x + 4
x
x
x
1
6. The integrand is continuous for all x ≥ 1, so whether the integral converges or diverges depends only on the behavior of
the function as x → ∞. As x → ∞, polynomials
Z ∞ behave like the highest powered term. Thus, as x → ∞, the integrand
x2
x2 − 6x + 1
behaves like 2 or 1. Since
1 dx diverges, we predict that the given integral will diverge.
x2 + 4
x
1
7. The integrand is continuous for all x ≥ 1, so whether the integral converges or diverges depends only on the behavior of
the function as x → ∞. As x → ∞, polynomials
Z ∞ behave like the highest powered term. Thus, as x → ∞, the integrand
5x
5
5
5x + 2
behaves like 4 or 3 . Since
dx converges, we predict that the given integral will converge.
3
x4 + 8x2 + 4
x
x
x
1
8. For large t, the 2 is negligible in comparison to e5t , so the integrand behaves like e−5t . Thus
1
1
≈ 5t = e−5t .
e5t + 2
e
More precisely, since e5t + 2 > e5t , we have
1
1
< 5t = e−5t .
e5t + 2
e
Z
∞
1
dt converges also.
5t + 2
e
1
9. The integrand is continuous for all x ≥ 1, so whether the integral converges or diverges depends only on the behavior of
the function as x → ∞. As x → ∞, polynomials
Z ∞behave like the highest powered term. Thus, as x → ∞, the integrand
x2 + 4
x2
1
1
behaves like 4 or 2 . Since
dx converges, we predict that the given integral will converge.
2
x4 + 3x2 + 11
x
x
x
1
10. It converges:
Z ∞
Z b
b
dz
dz
1 −2
1
1
1
1
= lim
= lim
− z
=
lim
− 2 =
3
3
2
b→∞
b→∞
b→∞ 50
z
z
2
2
b
5000
50
50
50
Since
11. Since
R∞
1
e−5t dt converges, by the Comparison Theorem
1
1
1
≥
and
1+x
2x
2
Z
∞
0
1
dx diverges, we have that
x
Z
∞
Z
∞
1
dx
diverges.
1+x
Z
∞
dx
dx
1
1
12. If x ≥ 1, we know that 3
≤ 3 , and since
converges, the improper integral
converges.
x +1
x
x3
x3 + 1
1
1
13. The integrand is unbounded as t → 5. We substitute w = t − 5, so dw = dt. When t = 5, w = 0 and when t = 8,
w = 3.
Z 3
Z 8
6
6
√
√ dw.
dt =
w
t
−
5
0
5
Since
Z 3
Z 3
3
√
√
√
6
1
√ dw = lim 6
√ dw = 6 lim 2w1/2 = 12 lim ( 3 − a) = 12 3,
w
w
a→0+
a→0+
a→0+
a
0
a
our integral converges.
14. The integral converges.
Z
1
0
1
x19/20
dx = lim
a→0
Z
a
1
1
x19/20
1
dx = lim 20x1/20
a→0
a
= lim 20 1 − a1/20 = 20.
a→0
670
Chapter Seven /SOLUTIONS
15. This integral diverges. To see this, substitute t + 1 = w, dt = dw. So,
Z
t=5
dt
=
(t + 1)2
t=−1
Z
w=6
dw
,
w2
w=0
which diverges.
16. Since we know the antiderivative of
Since the integrand is even, we write
Z
1
, we can use the Fundamental Theorem of Calculus to evaluate the integral.
1 + u2
∞
−∞
du
=2
1 + u2
Z
b
du
du
= 2 lim
2
b→∞
1
+
u
1
+
u2
0
0
π
= π.
= 2 lim arctan b = 2
b→∞
2
Thus, the integral converges to π.
1
1
< 2 for u ≥ 1, and since
u + u2
u
17. Since
∞
Z
Z
∞
1
du
converges,
u2
Z
∞
1
du
converges.
u + u2
1
1
18. This improper integral diverges. We expect this because, for large θ, √
≈ √ = and
2
θ
θ2 + 1
θ
precisely, for θ ≥ 1
1 1
1
1
1
√
≥ √
= √ √ = √ ·
θ2 + 1
θ2 + θ2
2 θ
2 θ2
and
∞
Z
1
dθ
diverges. (The factor
θ
19. For θ ≥ 2, we have √
verges.
√1
2
1
1
0
1
1
1
≤ √ = 3 , and
3
3
θ +1
θ
θ2
Z
2
∞
dθ
converges (check by integration), so
θ3/2
1
≤ √ , and since
θ3 + θ
θ
1
dθ
√
converges.
θ3 + θ
Z
1
dθ
diverges. More
θ
does not affect the divergence.)
20. This integral is improper at θ = 0. For 0 ≤ θ ≤ 1, we have √
Z
∞
Z
∞
1
1
21. Since
≤ y = e−y and
e−y dy converges, the integral
1 + ey
e
0
22. This integral is convergent because, for φ ≥ 1,
∞
Z
0
Z
Z
∞
2
1
0
√
dθ
conθ3 + 1
1
√ dθ converges,
θ
dy
converges.
1 + ey
2 + cos φ
3
≤ 2,
φ2
φ
and
Z
∞
1
3
dφ = 3
φ2
Z
1
∞
1
dφ converges.
φ2
1
1
23. Since z
< z = e−z for z ≥ 0, and
e + 2z
e
Z
∞
e−z dz converges,
0
2 − sin φ
1
for 0 < φ ≤ π, and since
24. Since 2 ≤
φ
φ2
25. Since
3 + sin α
2
≥ for α ≥ 4, and since
α
α
Z
4
∞
Z
∞
dz
converges.
ez + 2z
Z
π
0
Z
0
π
1
dφ diverges,
φ2
2
dα diverges, then
α
Z
4
0
∞
2 − sin φ
dφ must diverge.
φ2
3 + sin α
dα diverges.
α
Problems
26. (a) The area is infinite. The area under 1/x is infinite and the area under 1/x2 is 1. So the area between the two has to
be infinite also.
(b) Since f (x) is bounded between 0 and 1/x2 , and the area under 1/x2 is finite, f (x) will have finite area by the
comparison test. Similarly, h(x) lies above 1/x, whose area is infinite, so h(x) must have infinite area as well. We
can tell nothing about the area of g(x), because the comparison test tells us nothing about a function larger than a
7.7 SOLUTIONS
671
function with finite area but smaller than one with
R ainfinite area. Finally, k(x) will certainly have infinite area, because
it has a lower bound m, for some m > 0. Thus, 0 k(x) dx ≥ ma, and since the latter does not converge as a → ∞,
neither can the former.
27. The convergence or divergence of an improper integral depends on the long-term behavior of the integrand,
R ∞not on its
short-term behavior. Figure 7.46 suggests that g(x) ≤ f (x) for all values of x beyond x = k. Since k f (x) dx
R∞
converges, we expect k g(x) dx converges also.
R∞
R ∞ However we are interested in a g(x) dx. Breaking the integral into two parts enables us to use the fact that
g(x) dx is finite:
k
∞
Z
g(x) dx =
a
Z
k
g(x) dx +
a
Z
∞
g(x) dx.
k
The first integral is also finite because the interval from a to k is finite. Therefore, we expect
g(x)
R∞
f (x)
✠
x
a
k
Figure 7.46
28. First let’s calculate the indefinite integral
Z
Z
dx
dx
. Let ln x = w, then
= dw. So
x(ln x)p
x
dx
=
x(ln x)p
=
=
Z
dw
wp
ln |w| + C,
1
w1−p + C,
1−p
if p = 1
if p 6= 1
ln | ln x| + C,
1
(ln x)1−p + C,
1−p
if p = 1
if p 6= 1.
Notice that lim ln x = +∞.
x→∞
(a) p = 1:
Z
∞
2
dx
= lim
b→∞
x ln x
ln | ln b| − ln | ln 2|
= +∞.
(b) p < 1:
Z
∞
2
1
dx
=
x(ln x)p
1−p
(c) p > 1:
Z
∞
2
Z
2
∞
lim (ln b)1−p − (ln 2)1−p
b→∞
1
dx
=
x(ln x)p
1−p
= +∞.
lim (ln b)1−p − (ln 2)1−p
b→∞
1
1
lim
− (ln 2)1−p
1 − p b→∞ (ln b)p−1
1
(ln 2)1−p .
=−
1−p
=
Thus,
dx
is convergent for p > 1, divergent for p ≤ 1.
x(ln x)p
a
g(x) dx converges.
672
Chapter Seven /SOLUTIONS
29. The indefinite integral
Z
dx
dx
is computed in Problem 28. Let ln x = w, then
= dw. Notice that lim ln x = 0,
x→1
x(ln x)p
x
and lim ln x = −∞.
x→0+
For this integral notice that ln 1 = 0, so the integrand blows up at x = 1.
(a) p = 1:
Z
1
2
dx
= lim (ln | ln 2| − ln | ln a|)
x ln x a→1+
Since ln a → 0 as a → 1, ln | ln a| → −∞ as b → 1. So the integral is divergent.
(b) p < 1:
Z
2
1
1
dx
=
lim (ln 2)1−p − (ln a)1−p
p
+
x(ln x)
1 − p a→1
1
(ln 2)1−p .
=
1−p
(c) p > 1:
Z
As lim (ln a)1−p = lim
a→1+
Thus,
Z
2
1
a→1+
2
1
1
dx
=
lim (ln 2)1−p − (ln a)1−p
x(ln x)p
1 − p a→1+
1
= +∞, the integral diverges.
(ln a)p−1
dx
is convergent for p < 1, divergent for p ≥ 1.
x(ln x)p
2
30. (a) Since e−x ≤ e−3x for x ≥ 3,
Z
Now
Z
∞
e
−x2
3
∞
e−3x dx = lim
b→∞
3
b
Z
3
∞
e−3x dx
3
1
e−3x dx = lim − e−3x
b→∞
3
b
3
e−3b
e−9
e−9
−
=
.
= lim
b→∞ 3
3
3
Thus
Z
∞
3
(b) By reasoning similar to part (a),
Z
∞
Z
so
2
Z
e−nx dx =
n
∞
n
2
e−x dx ≤
t
e−9
.
3
∞
e−nx dx,
n
∞
Z
t ′
2
e−x dx ≤
e−x dx ≤
n
and
t
dx ≤
Z
1 −n2
e
,
n
1 −n2
e
.
n
31. (a) The tangent line to e has slope (e ) = e . Thus at t = 0, the slope is e0 = 1. The line passes through (0, e0 ) =
(0, 1). Thus the equation of the tangent line is y = 1 + t. Since et is everywhere concave up, its graph is always
above the graph of any of its tangent lines; in particular, et is always above the line y = 1 + t. This is tantamount to
saying
1 + t ≤ et ,
with equality holding only at the point of tangency, t = 0.
7.7 SOLUTIONS
673
1
, then the above inequality becomes
x
(b) If t =
1+
1
1
≤ e1/x , or e1/x − 1 ≥ .
x
x
1
, t is never zero. Therefore, the inequality is strict, and we write
x
1
e1/x − 1 > .
x
1
(c) Since e1/x − 1 > ,
x
1
1
1
= 4.
<
x
x5 (e1/x − 1)
x5 x1
Since t =
Since
∞
Z
1
Z
dx
converges,
x4
∞
1
dx
converges.
x5 (e1/x − 1)
Strengthen Your Understanding
32. We cannot compare the integrals, because the first integrand is sometimes less than the second and sometimes greater than
the second, depending on the sign of sin x.
1
33. We can use the comparison test with √ . Since x > 0, we have
x 2
1
1
< √ ,
0< √
2
x +1
x 2
Z ∞
Z ∞
√
1
1
√ dx is convergent (p =
√
2 > 1), the original integral,
and
dx, converges.
2
2
x
x +1
1
1
34. The integral
Z
∞
f (x) dx might converge or diverge; the comparison test can not be used here.
0
35. Let f (x) =
1
. Then
x2
Z
∞
1
36. We know that the integral
1
dx converges. However
x2
R∞
1
Z
∞
1
1
dx =
f (x)
3/(2x2 ) dx converges because
Z
∞
1
3
dx = lim
b→∞
2x2
Z
b
1
Z
∞
x2 dx diverges.
1
3
3
3
dx = lim − +
b→∞
2x2
2b
2
So we know that if f (x) is positive and f (x) ≤ 3/(2x2 ) for all x ≥ 1, then
f (x) = 3/(2x2 + 1) is a good example.
37. Since −1 ≤ sin x ≤ 1 for all x, we know that, for all x,
R∞
1
=
3
.
2
f (x)dx converges. So the function
7x − 2 ≤ 7x − 2 sin x ≤ 7x + 2.
So since 7x − 2 sin x ≥ 0 for all x ≥ 1, we know that for all x ≥ 1,
3
3
≥
.
7x − 2 sin x
7x + 2
So let
f (x) =
We then have
Z
∞
f (x)dx = lim
b→∞
1
So the integral
R∞
1
b
1
3
3
3
dx = lim
ln |7b + 2| − ln 9 = ∞
b→∞ 7
7x + 2
7
f (x)dx diverges, as desired.
1
1
< x and
38. True. Since x
e +x
e
39. True. Since
Z
1
1
> 2 and
x2 − 3
x
∞
Z
0
Z
0
1
3
.
7x + 2
1
dx converges,
ex
dx
diverges,
x2
Z
1
0
Z
0
∞
dx
converges.
ex + x
dx
diverges.
x2 − 3
674
Chapter Seven /SOLUTIONS
Solutions for Chapter 7 Review
Exercises
1.
1
(t
3
+ 1)3
2. − cos 2θ
5x
3.
ln 5
1
4. et + 5 e5t = et + e5t
5
3 2
w + 7w + C.
2
5. Using the power rule gives
1 2r
e
2
+C
d
cos t = − sin t, we have
7. Since
dt
6.
Z
sin t dt = − cos t + C, where C is a constant.
8. Let 2t = w, then 2dt = dw, so dt = 12 dw, so
Z
where C is a constant.
cos 2t dt =
Z
1
1
1
cos w dw = sin w + C = sin 2t + C,
2
2
2
9. Let 5z = w, then 5dz = dw, which means dz = 15 dw, so
where C is a constant.
Z
e5z dz =
Z
ew ·
1
1
dw =
5
5
Z
ew dw =
1 w
1
e + C = e5z + C,
5
5
10. sin(x + 1) + C
11. Since
Z
sin w dθ = − cos w + C, the substitution w = 2θ, dw = 2 dθ gives
12. Let w = x3 − 1, then dw = 3x2 dx so that
Z
Z
1
(x − 1) x dx =
3
3
4 2
w4 dw =
Z
1
sin 2θ dθ = − cos 2θ + C.
2
1 5
1 3
w +C =
(x − 1)5 + C.
15
15
2 5/2 3 5/3
x
+ x
+C
5
5
14. From the rule for antidifferentiation of exponentials, we get
13. The power rule gives
Z
Z
1
· 3x + C.
ln 3
(ex + 3x ) dx = ex +
Z
1
dz =
e−z dz = −e−z + C
ez
16. Rewrite the integrand as
15.
Z
4
3
− 3
x2
x
dx = 4
Z
x−2 dx − 3
Z
x−3 dx = −4x−1 +
3 −2
x + C.
2
17. Dividing by x2 gives
Z
x3 + x + 1
x2
dx =
Z
x+
1
1
+ 2
x
x
dx =
1
1 2
x + ln |x| − + C.
2
x
SOLUTIONS to Review Problems for Chapter Seven
18. Let w = 1 + ln x, then dw = dx/x so that
(1 + ln x)2
dx =
x
Z
Z
w2 dw =
1 3
1
w + C = (1 + ln x)3 + C.
3
3
19. Substitute w = t2 , so dw = 2t dt.
Z
Check:
2
tet dt =
Z
1
2
d
dt
et 2t dt =
2
1
2
Z
1 t2
e +C
2
= 2t
ew dw =
1 t2
e
2
1 w
1 2
e + C = et + C.
2
2
2
= tet .
20. Integration by parts with u = x, v ′ = cos x gives
Z
x cos x dx = x sin x −
Z
sin x dx + C = x sin x + cos x + C.
Or use III-16 with p(x) = x and a = 1 in the integral table.
21. Integration by parts twice gives
Z
2 2x
x e
x2 e2x
−
dx =
2
Z
x2 2x x 2x 1 2x
e − e + e +C
2
2
4
1 2x
1 2 1
= ( x − x + )e + C.
2
2
4
2xe2x dx =
Or use the integral table, III-14 with p(x) = x2 and a = 1.
22. Using substitution with w = 1 − x and dw = −dx, we get
Z
√
x 1 − x dx = −
Z
√
2
2
2
2
(1 − w) wdw = w5/2 − w3/2 + C = (1 − x)5/2 − (1 − x)3/2 + C.
5
3
5
3
23. Let u = ln y, v ′ = y. Then, v = 12 y 2 and u′ =
Z
1
. Integrating by parts, we get:
y
y ln y dy =
1 2
y ln y −
2
1 2
y ln y −
2
1
= y 2 ln y −
2
=
Z
1 2 1
y · dy
2
y
Z
1
y dy
2
1 2
y + C.
4
24. We integrate by parts, with u = y, v ′ = sin y. We have u′ = 1, v = − cos y, and
Check:
Z
y sin y dy = −y cos y −
Z
(− cos y) dy = −y cos y + sin y + C.
d
(−y cos y + sin y + C) = − cos y + y sin y + cos y = y sin y.
dy
25. We integrate by parts, using u = (ln x)2 and v ′ = 1. Then u′ = 2 lnxx and v = x, so
Z
(ln x)2 dx = x(ln x)2 − 2
But, integrating by parts or using the integral table,
Check:
Z
R
Z
ln x dx.
ln x dx = x ln x − x + C. Therefore,
(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C.
d
2 ln x
1
x(ln x)2 − 2x ln x + 2x + C = (ln x)2 + x
− 2 ln x − 2x + 2 = (ln x)2 .
dx
x
x
675
676
Chapter Seven /SOLUTIONS
26. Using the exponent rules and the chain rule, we have
Z
e0.5−0.3t dt = e0.5
Z
e−0.3t dt = −
e0.5−0.3t
e0.5 −0.3t
e
+C =−
+ C.
0.3
0.3
27. Let sin θ = w, then cos θ dθ = dw, so
Z
2
sin θ cos θ dθ =
Z
w2 dw =
1 3
1
w + C = sin3 θ + C,
3
3
where C is a constant.
28. Substitute w = 4 − x2 , dw = −2x dx:
Z
Check
x
p
4 − x2 dx = −
1
2
Z
√
1
1
w dw = − w3/2 + C = − (4 − x2 )3/2 + C.
3
3
p
d
1
1 3
(4 − x2 )1/2 (−2x) = x 4 − x2 .
− (4 − x2 )3/2 + C = −
dx
3
3 2
h
i
h
i
29. Expanding the numerator and dividing, we have
Z
(u + 1)3
du =
u2
=
Check:
d
du
30. Substitute w =
√
u2
1
+ 3u + 3 ln |u| − + C.
2
u
Z
u+3+
3
1
+ 2
u
u
= u + 3 + 3/u + 1/u2 =
du
(u + 1)3
.
u2
√
y, dw = 1/(2 y) dy. Then
Z
√
cos y
√
dy
=
2
cos w dw = 2 sin w + C = 2 sin y + C.
√
y
Check:
√
√
2 cos y
cos y
√
d
2 sin y + C =
= √ .
√
y
dy
2 y
d
1
(tan z) =
, we have
dz
cos2 z
Check:
32. Denote
(u3 + 3u2 + 3u + 1)
du =
u2
1
u2
+ 3u + 3 ln |u| − + C
2
u
Z
31. Since
Z
Z
1
dz = tan z + C.
cos2 z
(cos z)(cos z) − (sin z)(− sin z)
d sin z
1
d
(tan z + C) =
=
=
.
dz
dz cos z
cos2 z
cos2 z
Z
cos2 θ dθ by A. Let u = cos θ, v ′ = cos θ. Then, v = sin θ and u′ = − sin θ. Integrating by parts, we get:
A = cos θ sin θ −
Z
(− sin θ) sin θ dθ.
Employing the identity sin2 θ = 1 − cos2 θ, the equation above becomes:
A = cos θ sin θ +
Z
dθ −
Z
= cos θ sin θ + θ − A + C.
cos2 θ dθ
SOLUTIONS to Review Problems for Chapter Seven
Solving this equation for A, and using the identity sin 2θ = 2 cos θ sin θ we get:
A=
Z
1
1
sin 2θ + θ + C.
4
2
cos2 θ dθ =
[Note: An alternate solution would have been to use the identity cos2 θ =
1
2
cos 2θ + 12 .]
33. Multiplying out and integrating term by term:
Z
t10 (t − 10) dt =
Z
(t11 − 10t10 ) dt =
Z
t11 dt − 10
Z
t10 dt =
1 12 10 11
t −
t + C.
12
11
34. Substitute w = 2x − 6. Then dw = 2 dx and
Z
35. Let ln x = w, then
1
tan(2x − 6) dx =
2
Z
1
tan w dw =
2
Z
sin w
dw
cos w
1
= − ln | cos w| + C by substitution or by I-7 of the integral table.
2
1
= − ln | cos(2x − 6)| + C.
2
1
x
dx = dw, so
(ln x)2
dx =
x
Z
Z
w2 dw =
1
1 3
w + C = (ln x)3 + C, where C is a constant.
3
3
36. Multiplying out, dividing, and then integrating yields
Z
(t + 2)2
dt =
t3
Z
t2 + 4t + 4
dt =
t3
Z
Z
1
dt +
t
Z
4
dt +
t2
Z
4
4
2
dt = ln |t| − − 2 + C,
t3
t
t
where C is a constant.
37. Integrating term by term:
where C is a constant.
1
x
x2 + 2x +
dx =
1 3
x + x2 + ln |x| + C,
3
38. Dividing and then integrating, we obtain
Z
t+1
dt =
t2
Z
1
dt +
t
39. Let t2 + 1 = w, then 2t dt = dw, tdt =
Z
te
t2 +1
dt =
Z
1
2
Z
1
1
dt = ln |t| − + C, where C is a constant.
t2
t
dw, so
1
1
e · dw =
2
2
w
Z
ew dw =
1 w
1 2
e + C = et +1 + C,
2
2
where C is a constant.
40. Let cos θ = w, then − sin θ dθ = dw, so
Z
where C is a constant.
Z
Z
−1
sin θ
dθ =
dw
cos θ
w
= − ln |w| + C = − ln | cos θ| + C,
tan θ dθ =
677
678
Chapter Seven /SOLUTIONS
41. If u = sin(5θ), du = cos(5θ) · 5 dθ, so
Z
sin(5θ) cos(5θ)dθ =
1
5
=
1
5
Z
sin(5θ) · 5 cos(5θ)dθ =
u2
2
1
5
Z
u du
1
sin2 (5θ) + C
10
+C =
or
Z
Z
1
1
2 sin(5θ) cos(5θ)dθ =
sin(5θ) cos(5θ)dθ =
2
2
−1
=
cos(10θ) + C.
20
Z
sin(10θ)dθ
(using sin(2x) = 2 sin x cos x)
42. Using substitution,
Z
x
dx =
x2 + 1
Z
1
=
2
1/2
dw
w
Z
(x2 + 1 = w, 2x dx = dw, xdx =
1
dw)
2
1
1
1
dw = ln |w| + C = ln |x2 + 1| + C,
w
2
2
where C is a constant.
d
1
43. Since
(arctan z) =
, we have
dz
1 + z2
Z
dz
= arctan z + C, where C is a constant.
1 + z2
44. Let w = 2z, so dw = 2dz. Then, since
Z
1
d
arctan w =
, we have
dw
1 + w2
dz
=
1 + 4z 2
Z
1
dw
2
1 + w2
1
1
arctan w + C = arctan 2z + C.
2
2
=
45. Let w = cos 2θ. Then dw = −2 sin 2θ dθ, hence
Z
cos3 2θ sin 2θ dθ = −
Check:
d
dθ
−
cos4 2θ
8
=−
1
2
Z
w3 dw = −
cos4 2θ
w4
+C =−
+ C.
8
8
(4 cos3 2θ)(− sin 2θ)(2)
= cos3 2θ sin 2θ.
8
46. Let cos 5θ = w, then −5 sin 5θ dθ = dw, sin 5θ dθ = − 15 dw. So
Z
Z
1
1
sin 5θ cos 5θ dθ =
w · (− ) dw = −
5
5
1
= − cos4 5θ + C,
20
3
3
Z
where C is a constant.
47.
Z
sin3 z cos3 z dz =
=
Z
Z
sin z(1 − cos2 z) cos3 z dz
sin z cos3 z dz −
Z
sin z cos5 z dz
w3 dw = −
1 4
w +C
20
SOLUTIONS to Review Problems for Chapter Seven
=
Z
=−
w3 (−dw) −
Z
Z
w3 dw +
Z
w5 (−dw) (let cos z = w, so − sin z dz = dw)
w5 dw
1
1
= − w4 + w6 + C
4
6
1
1
4
= − cos z + cos6 z + C,
4
6
where C is a constant.
48. If u = t − 10, t = u + 10 and dt = 1 du, so substituting we get
Z
Z
(u + 10)u10 du =
(u11 + 10u10 ) du =
1 12 10 11
u +
u +C
12
11
10
1
(t − 10)12 +
(t − 10)11 + C.
12
11
=
49. Let sin θ = w, then cos θ dθ = dw, so
Z
√
cos θ 1 + sin θ dθ =
=
Z
√
1 + w dw
(1 + w)3/2
2
+ C = (1 + sin θ)3/2 + C,
3/2
3
where C is a constant.
50.
Z
xex dx = xex −
Z
ex dx
(let x = u, ex = v ′ , ex = v)
= xex − ex + C,
where C is a constant.
51.
Z
t3 et dt = t3 et −
Z
= t3 et − 3
3 t
3t2 et dt
Z
t2 et dt
2 t
= t e − 3(t e −
Z
= t3 et − 3t2 et + 6
3 t
(let t3 = u, et = v ′ , 3t2 = u′ , et = v)
2 t
Z
(let t2 = u, et = v ′ )
2tet dt)
(let t = u, et = v ′ )
tet dt
t
= t e − 3t e + 6(te −
Z
et dt)
= t3 et − 3t2 et + 6tet − 6et + C,
where C is a constant.
52. Let x2 = w, then 2xdx = dw, x = 1 ⇒ w = 1, x = 3 ⇒ w = 9. Thus,
Z
3
x(x2 + 1)70 dx =
Z
1
1
9
(w + 1)70
1
dw
2
9
1 1
·
(w + 1)71
2 71
1
1
71
71
=
(10 − 2 ).
142
=
679
680
Chapter Seven /SOLUTIONS
53. Let w = 3z + 5 and dw = 3 dz. Then
Z
Z
1
(3z + 5) dz =
3
3
w3 dw =
1 4
1
w +C =
(3z + 5)4 + C.
12
12
54. Rewrite 9 + u2 as 9[1 + (u/3)2 ] and let w = u/3, then dw = du/3 so that
Z
1
du
=
9 + u2
3
Z
dw
1
1
u
= arctan w + C = arctan
1 + w2
3
3
3
+ C.
55. Let u = sin w, then du = cos w dw so that
Z
cos w
dw =
1 + sin2 w
Z
du
= arctan u + C = arctan(sin w) + C.
1 + u2
56. Let w = ln x, then dw = (1/x)dx which gives
Z
1
tan(ln x) dx =
x
Z
Z
tan w dw =
sin w
dw = − ln(| cos w|) + C = − ln(| cos(ln x)|) + C.
cos w
57. Let w = ln x, then dw = (1/x)dx so that
Z
1
sin(ln x) dx =
x
Z
sin w dw = − cos w + C = − cos(ln x) + C.
58. Let u = 16 − w2 , then du = −2w dw so that
Z
√
1
wdw
=−
2
16 − w2
Z
p
√
du
√ = − u + C = − 16 − w2 + C.
u
59. Dividing and then integrating term by term, we get
Z
Z
1
e2y
+ 2y dy =
e2y
e
1
= y − e−2y + C.
2
e2y + 1
dy =
e2y
Z
(1 + e−2y ) dy =
Z
dy + −
1
2
Z
e−2y (−2) dy
60. Let u = 1 − cos w, then du = sin w dw which gives
Z
sin w dw
√
=
1 − cos w
Z
√
√
du
√ = 2 u + C = 2 1 − cos w + C.
u
61. Let w = ln x. Then dw = (1/x)dx which gives
Z
dx
=
x ln x
Z
dw
= ln |w| + C = ln | ln x| + C.
w
62. Let w = 3u + 8, then dw = 3du and
Z
63. Let w =
√
x2 + 1, then dw = √
Z
√
x
x2 + 1
du
=
3u + 8
Z
dw
1
= ln |3u + 8| + C.
3w
3
Z
cos w dw = sin w + C = sin
xdx
so that
x2 + 1
cos
p
x2 + 1 dx =
p
x2 + 1 + C.
SOLUTIONS to Review Problems for Chapter Seven
64. Integrating by parts using u = t2 and dv = √tdt
1+t2
gives du = 2t dt and v =
p
t3
√
dt = t2 1 + t2 −
1 + t2
Z
Z
2t
√
1 + t2 . Now
p
1 + t2 dt
2
(1 + t2 )3/2 + C
3
p
2
= 1 + t2 (t2 − (1 + t2 )) + C
3
p
(t2 − 2)
= 1 + t2
+ C.
3
= t2
p
1 + t2 −
65. Using integration by parts, let r = u and dt = eku du, so dr = du and t = (1/k)eku . Thus
Z
ueku du =
Z
u ku 1
e −
k
k
eku du =
u ku
1
e − 2 eku + C.
k
k
66. Let u = w + 5, then du = dw and noting that w = u − 5 we obtain
Z
Z
4
(w + 5) w dw =
u4 (u − 5) du
Z
=
u5 − 5u4 du
1 6
u − u5 + C
6
1
= (w + 5)6 − (w + 5)5 + C.
6
=
67.
Z
√
e
2x+3
1
dx = √
2
Z
e
√
√
2x+3
2dx. If u =
1
√
2
68.
Z
(ex + x)2 dx =
Z
√
2x + 3, du =
√
2dx, so
1
1 √
eu du = √ eu + C = √ e 2x+3 + C.
2
2
Z
(e2x + 2xex + x2 )dy. Separating into three integrals, we have
Z
Z
1
e dx =
2
2x
2xex dx = 2
Z
Z
e2x 2 dx =
1 2x
e + C1 ,
2
xex dx = 2xex − 2ex + C2
from Formula II-13 of the integral table or integration by parts, and
Z
x2 dx =
x3
+ C3 .
3
Combining the results and writing C = C1 + C2 + C3 , we get
x3
1 2x
e + 2xex − 2ex +
+ C.
2
3
69. Integrate by parts, r = ln u and dt = u2 du, so dr = (1/u) du and t = (1/3)u3 . We have
Z
u2 ln u du =
1
1 3
u ln u −
3
3
Z
u2 du =
1
1 3
u ln u − u3 + C.
3
9
681
682
Chapter Seven /SOLUTIONS
70. The integral table yields
Z
5
6
x
5x + 6
dx = ln |x2 + 4| + arctan + C
x2 + 4
2
2
2
5
x
2
= ln |x + 4| + 3 arctan + C.
2
2
Check:
d
dx
5
6
x
ln |x2 + 4| + arctan + C
2
2
2
5
=
2
1
1
1
(2x) + 3
x2 + 4
1 + (x/2)2 2
6
5x + 6
5x
+ 2
= 2
.
= 2
x +4
x +4
x +4
71. Using Table IV-19, let m = 3, w = 2x, and dw = 2dx. Then
Z
1
1
dx =
2
sin3 (2x)
=
1
2
Z
1
dw
sin3 w
−1 cos w
1
+
(3 − 1) sin2 w
4
Z
1
dw,
sin w
and using Table IV-20, we have
Z
Thus,
Z
1
cos w − 1
1
dw = ln
+ C.
sin w
2
cos w + 1
cos 2x − 1
1
cos 2x
1
dx = −
+ ln
+ C.
8
cos 2x + 1
sin3 (2x)
4 sin2 2x
72. We can factor r 2 − 100 = (r − 10)(r + 10) so we can use Table V-26 (with a = 10 and b = −10) to get
Z
dr
1
=
[ln |r − 10| + ln |r + 10|] + C.
r 2 − 100
20
73. Integration by parts will be used twice here. First let u = y 2 and dv = sin(cy) dy, then du = 2y dy and v =
−(1/c) cos(cy). Thus
Z
Z
2
y2
y cos(cy) dy.
y 2 sin(cy) dy = − cos(cy) +
c
c
Now use integration by parts to evaluate the integral in the right hand expression. Here let u = y and dv = cos(cy)dy
which gives du = dy and v = (1/c) sin(cy). Then we have
Z
y2
2
y sin(cy) dy = − cos(cy) +
c
c
2
=−
y
1
sin(cy) −
c
c
Z
sin(cy) dy
y2
2y
2
cos(cy) + 2 sin(cy) + 3 cos(cy) + C.
c
c
c
74. Integration by parts will be used twice. First let u = e−ct and dv = sin(kt)dt, then du = −ce−ct dt and v =
(−1/k) cos kt. Then
Z
e
−ct
1
c
sin kt dt = − e−ct cos kt −
k
k
1
c
= − e−ct cos kt −
k
k
Z
e−ct cos kt dt
1 −ct
c
e
sin kt +
k
k
c
c2
1
= − e−ct cos kt − 2 e−ct sin kt − 2
k
k
k
Z
Z
e−ct sin kt dt
e−ct sin kt dt
SOLUTIONS to Review Problems for Chapter Seven
Solving for
R
683
e−ct sin kt dt gives
k2 + c2
k2
so
Z
Z
e−ct sin kt dt = −
e−ct sin kt dt = −
e−ct
(k cos kt + c sin kt) ,
k2
e−ct
(k cos kt + c sin kt) + C.
+ c2
k2
75. Using II-9 from the integral table, with a = 5 and b = 3, we have
Z
√
√
dx + ( k)x dx, for the first integral, use Formula I-1 with n = k. For the
√
second integral, use Formula I-3 with a = k. The result is
√
√
Z √
√ x
x( k)+1
( k)x
x
√ + C.
+
(x + ( k) ) dx = √
( k) + 1
ln k
76. Since
Z
77. Factor
(x
√
√
k
√
1
e5x [5 cos(3x) + 3 sin(3x)] + C
25 + 9
1 5x
e [5 cos(3x) + 3 sin(3x)] + C.
=
34
e5x cos(3x) dx =
x
+ ( k) )dx =
Z
√
x
Z
k
3 out of the integrand and use VI-30 of the integral table with u = 2x and du = 2dx to get
Z p
Z
√ p
3 + 12x2 dx =
3 1 + 4x2 dx
√ Z p
3
=
1 + u2 du
2
Z
√ p
1
3
=
u 1 + u2 +
du .
√
4
1 + u2
Then from VI-29, simplify the integral on the right to get
Z p
√ p
p
3
u 1 + u2 + ln |u + 1 + u2 | + C
3 + 12x2 dx =
4
√
p
p
3
=
2x 1 + (2x)2 + ln |2x + 1 + (2x)2 | + C.
4
78. By completing the square, we get
9
3
1
3
x2 − 3x + 2 = (x2 − 3x + (− )2 ) + 2 − = (x − )2 − .
2
4
2
4
Then
Z
1
√
dx =
x2 − 3x + 2
Z
1
p
(x − 32 )2 −
1
4
dx.
Let w = (x − (3/2)), then dw = dx and a2 = 1/4. Then we have
Z
1
√
dx =
2
x − 3x + 2
and from VI-29 of the integral table we have
Z
√
Z
√
1
dw
− a2
w2
p
1
dw = ln w + w2 − a2 + C
2
−a
w2
= ln
= ln
3
x−
2
x−
3
2
+
+
r
x−
p
3
2
2
−
1
+C
4
x2 − 3x + 2 + C.
684
Chapter Seven /SOLUTIONS
79. First divide x2 + 3x + 2 into x3 to obtain
7x + 6
x3
=x−3+ 2
.
x2 + 3x + 2
x + 3x + 2
Since x2 + 3x + 2 = (x + 1)(x + 2), we can use V-27 of the integral table (with c = 7, d = 6, a = −1, and b = −2) to
get
Z
7x + 6
dx = − ln |x + 1| + 8 ln |x + 2| + C.
x2 + 3x + 2
Including the terms x − 3 from the long division and integrating them gives
Z
x3
dx =
2
x + 3x + 2
Z
x−3+
x2
7x + 6
+ 3x + 6
dx =
1 2
x − 3x − ln |x + 1| + 8 ln |x + 2| + C.
2
80. First divide x2 + 1 by x2 − 3x + 2 to obtain
x2
3x − 1
x2 + 1
= 1+ 2
.
− 3x + 2
x − 3x + 2
Factoring x2 − 3x + 2 = (x − 2)(x − 1) we can use V-27 (with c = 3, d = −1, a = 2 and b = 1) to write
Z
3x − 1
dx = 5 ln |x − 2| − 2 ln |x − 1| + C.
x2 − 3x + 2
Remembering to include the extra term of +1 we got when dividing, we get
x2 + 1
dx =
2
x − 3x + 2
Z
Z
1+
x2
3x − 1
− 3x + 2
dx = x + 5 ln |x − 2| − 2 ln |x − 1| + C.
81. We can factor the denominator into ax(x + ab ), so
Z
dx
1
=
ax2 + bx
a
Z
1
x(x + ab )
Now we can use V-26 (with A = 0 and B = − ab to give
1
a
Z
1
1 a
b
= ·
ln |x| − ln x +
a b
a
x(x + ab )
+C =
1
b
ln |x| − ln x +
b
a
+ C.
82. Let w = ax2 + 2bx + c, then dw = (2ax + 2b)dx so that
Z
1
ax + b
dx =
ax2 + 2bx + c
2
Z
1
1
dw
= ln |w| + C = ln |ax2 + 2bx + c| + C.
w
2
2
dx =
83. Multiplying out and integrating term by term,
Z
x
3
+
3
x
2
dx =
Z
x2
9
+2+ 2
9
x
1
9
x3
3
+ 2x + 9
x−1
−1
+C =
x3
9
+ 2x − + C.
27
x
84. If u = 2t + 1, du = 2t (ln 2) dt, so
Z
2t ln 2
1
dt =
2t + 1
ln 2
Z
1
1
1
=
ln |u| + C =
ln |2t + 1| + C.
u
ln 2
ln 2
101−x (−1 dx) = −1
Z
10u du = −1
2t
1
dt =
t
2 +1
ln 2
Z
85. If u = 1 − x, du = −1 dx, so
Z
101−x dx = −1
Z
1
10u
+C =−
101−x + C.
ln 10
ln 10
SOLUTIONS to Review Problems for Chapter Seven
685
86. Multiplying out and integrating term by term gives
Z
(x2 + 5)3 dx =
=
Z
(x6 + 15x4 + 75x2 + 125)dx =
x5
x3
1 7
x + 15
+ 75
+ 125x + C
7
5
3
1 7
x + 3x5 + 25x3 + 125x + C.
7
87. Integrate by parts letting r = v and dt = arcsin v dv then dr = dv and to find t we integrate arcsin v dv by parts letting
x = arcsin v and dy = dv. This gives
t = v arcsin v −
Z
p
(1/
1 − v 2 )v dv = v arcsin v +
Now, back to the original integration by parts, and we have
Z
Adding
R
v arcsin v dv = v 2 arcsin v + v
p
1 − v2 −
Z h
p
v arcsin v +
v arcsin v dv to both sides of the above line we obtain
2
Z
v arcsin v dv = v 2 arcsin v + v
= v 2 arcsin v + v
Dividing by 2 gives
Z
where K = C/2.
v arcsin v dv =
p
1 − v2 −
p
v2
1
−
2
4
1 − v2 −
1 − v2 .
i
p
1 − v 2 dv.
Z p
1 − v 2 dv
1 p
1
v 1 − v 2 − arcsin v + C.
2
2
arcsin v +
1 p
v 1 − v 2 + K,
4
88. By VI-30 in the table of integrals, we have
Z p
4 − x2 dx =
Z
√
x 4 − x2
1
√
+2
dx.
2
4 − x2
The same table informs us in formula VI-28 that
Z
Thus
Z p
4−
89. By long division,
Z
√
x2
1
x
dx = arcsin + C.
2
4 − x2
√
x 4 − x2
x
dx =
+ 2 arcsin + C.
2
2
125
z3
= z 2 + 5z + 25 +
, so
z−5
z−5
z3
dz =
z−5
=
Z
z 2 + 5z + 25 +
z3
5z 2
125
dz =
+
+ 25z + 125
z−5
3
2
5
z3
+ z 2 + 25z + 125 ln |z − 5| + C.
3
2
Z
1
dz
z−5
90. If u = 1 + cos2 w, du = 2(cos w)1 (− sin w) dw, so
Z
Z
1
−2 sin w cos w
1
sin w cos w
dw = −
dw = −
1 + cos2 w
2
1 + cos2 w
2
1
= − ln |1 + cos2 w| + C.
2
Z
1
1
du = − ln |u| + C
u
2
686
91.
Chapter Seven /SOLUTIONS
Z
1
dθ =
tan(3θ)
Z
Z
x
92.
dx =
cos2 x
have
Z
1
sin(3θ)
cos(3θ)
dθ =
cos(3θ)
1
dθ =
sin(3θ)
3
Z
x
Z
Z
cos(3θ)
dθ. If u = sin(3θ), du = cos(3θ) · 3dθ, so
sin(3θ)
3 cos(3θ)
1
dθ =
sin(3θ)
3
Z
1
1
1
du = ln |u| + C = ln | sin(3θ)| + C.
u
3
3
1
1
dx. Using integration by parts with u = x, du = dx and dv =
dx, v = tan x, we
cos2 x
cos2 x
Z
x
1
dx = x tan x −
cos2 x
Z
tan xdx.
Formula I-7 gives the final result of x tan x − (− ln | cos x|) + C = x tan x + ln | cos x| + C.
93. Dividing and integrating term by term gives
Z
94. If u =
x+1
√ dx =
x
√
Z
x
1
√ +√
x
x
Z
(x1/2 + x−1/2 ) dx =
x3/2
3
2
+
x1/2
+C =
1
2
√
2 3/2
x
+ 2 x + C.
3
x + 1, u2 = x + 1 with x = u2 − 1 and dx = 2u du. Substituting, we get
Z
√
x
dx =
x+1
Z
Z p√
x+1
√
=
x
2
Z
Z
(u2 − 1)2u du
=
u
Z
(u2 − 1)2 du = 2
Z
(u2 − 1) du
√
√
2( x + 1)3
2u3
− 2u + C =
− 2 x + 1 + C.
3
3
=
95.
dx =
√
√
1
1
( x + 1)1/2 √ dx; if u = x + 1, du = √ dx, so we have
x
2 x
√
1
( x + 1)1/2 √ dx = 2
2 x
Z
u1/2 du = 2
u3/2
3
2
+C =
4 3/2
4 √
u
+ C = ( x + 1)3/2 + C.
3
3
96. If u = e2y + 1, then du = e2y 2 dy, so
Z
1
e2y
dy =
2y
e +1
2
1
2e2y
dy =
2y
e +1
2
Z
Z
1
1
1
du = ln |u| + C = ln |e2y + 1| + C.
u
2
2
97. If u = z 2 − 5, du = 2z dz, then
Z
z
dz =
(z 2 − 5)3
=
Z
2
(z − 5)
−3
1
z dz =
2
1
+ C.
−4(z 2 − 5)2
Z
2
(z − 5)
−3
1
2z dz =
2
Z
−3
u
1
du =
2
u−2
−2
+C
98. Letting u = z − 5, z = u + 5, dz = du, and substituting, we have
Z
99. If u = 1 + tan x then du =
Z
u+5
u−1
du = (u−2 + 5u−3 )du =
+5
3
u
−1
−5
−1
+
+ C.
=
(z − 5)
2(z − 5)2
z
dz =
(z − 5)3
Z
Z
u−2
−2
+C
1
dx, and so
cos2 x
(1 + tan x)3
dx =
cos2 x
Z
(1 + tan x)3
1
dx =
cos2 x
Z
u3 du =
(1 + tan x)4
u4
+C =
+ C.
4
4
SOLUTIONS to Review Problems for Chapter Seven
100.
Z
2
(2x − 1)ex
dx =
ex
Z
ex
2
−x
687
(2x − 1)dx. If u = x2 − x, du = (2x − 1)dx, so
Z
ex
2
−x
(2x − 1)dx =
Z
eu du
= eu + c
= ex
2
−x
+ C.
101. We use the substitution w = x2 + x, dw = (2x + 1) dx.
Z
2
(2x + 1)ex ex dx =
Z
(2x + 1)ex
= ew + C = ex
2
2
+x
+x
dx =
+ C.
Z
ew dw
2
2
d x2 +x
+ C) = ex +x · (2x + 1) = (2x + 1)ex ex .
(e
Check:
dx
102. Let w = y 2 − 2y + 1, so dw = 2(y − 1) dy. Then
Z p
y 2 − 2y + 1(y − 1) dy =
Z
w1/2
Alternatively, notice the integrand can be written
equivalent answer.
1 2
1
1
dw = · w3/2 + C = (y 2 − 2y + 1)3/2 + C.
2
2 3
3
p
(y − 1)2 (y − 1) = (y − 1)2 . This leads to a different-looking but
1
103. Let w = 2 + 3 cos x, so dw = −3 sin x dx, giving − dw = sin x dx. Then
3
Z
Z
Z
√
√ 1
√
1
sin x
2 + 3 cos x dx =
w −
w dw
dw = −
3
3
1 w 32
= −
3
3
2
3
2
+ C = − (2 + 3 cos x) 2 + C.
9
104. Using Table III-14, with a = −4 we have
Z
1
(x2 − 3x + 2)e−4x dx = − (x2 − 3x + 2)e−4x
4
1
1
− (2x − 3)e−4x −
(2)e−4x + C.
16
64
1 −4x
=
e
(−11 + 20x − 8x2 ) + C.
32
105. Let x = 2θ, then dx = 2dθ. Thus
Z
1
sin (2θ) cos (2θ) dθ =
2
3
2
We let w = sin x and dw = cos x dx. Then
1
2
Z
1
sin x cos x dx =
2
2
3
=
1
2
1
=
2
Z
Z
Z
Z
sin2 x cos3 x dx.
sin2 x cos2 x cos x dx
sin2 x(1 − sin2 x) cos x dx
1
w (1 − w ) dw =
2
2
2
Z
(w2 − w4 ) dw
w5
1
1
1 w3
−
+ C = sin3 x −
sin5 x + C
=
2
3
5
6
10
1
1
sin5 (2θ) + C.
= sin3 (2θ) −
6
10
688
Chapter Seven /SOLUTIONS
106. If u = 2 sin x, then du = 2 cos x dx, so
Z
Z
Z
1
1
cos(2 sin x)2 cos x dx =
cos u du
2
2
1
1
= sin u + C = sin(2 sin x) + C.
2
2
cos(2 sin x) cos x dx =
107. Let w = x + sin x, then dw = (1 + cos x) dx which gives
Z
Z
3
(x + sin x) (1 + cos x) dx =
1 4
1
w + C = (x + sin x)4 + C.
4
4
w3 dw =
108. Using Table III-16,
Z
1
(2x3 + 3x + 4) sin(2x)
2
1
+ (6x2 + 3) cos(2x)
4
3
1
− (12x) sin(2x) − cos(2x) + C.
8
4
3x2
= 2 sin(2x) + x3 sin(2x) +
cos(2x) + C.
2
2x3 + 3x + 4 cos(2x) dx =
109. Use the substitution w = sinh x and dw = cosh xdx so
Z
Z
2
sinh x cosh x dx =
w2 dw =
w3
1
+ C = sinh3 x + C.
3
3
d 1
sinh3 x + C = sinh2 x cosh x.
dx 3
110. We use the substitution w = x2 + 2x and dw = (2x + 2) dx so
h
Check this answer by taking the derivative:
Z
(x + 1) sinh(x2 + 2x) dx =
Check this answer by taking the derivative:
2x).
1
2
i
Z
sinh w dw =
d 1
1
cosh(x2 + 2x) + C = (2x+2) sinh(x2 +2x) = (x+1) sinh(x2 +
dx 2
2
h
i
1
dw, and
2
111. Substitute w = 1 + x2 , dw = 2x dx. Then x dx =
Z
x=1
2 20
x(1 + x )
x=0
1
1
cosh w + C = cosh(x2 + 2x) + C.
2
2
1
dx =
2
w=2
Z
w20 dw =
w=1
w21
42
2
=
1
1
299593
= 49932 .
6
6
112. Substitute w = x2 + 4, dw = 2x dx. Then,
Z
x=1
x=4
x
p
x2
1
+ 4 dx =
2
Z
w=5
1
w 2 dw =
w=20
3
3
1
52 − 8 · 52
=
3
1 32
w
3
5
20
7
35 √
= − · 53/2 = −
5
3
3
113. We substitute w = cos θ + 5, dw = − sin θ dθ. Then
Z
θ=π
θ=0
sin θ dθ(cos θ + 5)7 = −
Z
w=4
w=6
w7 dw =
Z
w=6
w=4
w7 dw =
w8
8
6
= 201,760.
4
SOLUTIONS to Review Problems for Chapter Seven
dw
= x dx. When x = 0, w = 1. When x = 1, w = 6.
10
114. Let w = 1 + 5x2 . We have dw = 10x dx, so
x dx
=
1 + 5x2
689
Z
6
1
1
10
dw
1
=
w
10
Z
6
1
1
dw
=
ln |w|
w
10
6
1
1
ln 6
=
(ln 6 − ln 1) =
10
10
115.
2
Z
1
x2 + 1
dx =
x
Z
2
(x +
1
1
)dx =
x
x2
+ ln |x|
2
2
=
1
3
+ ln 2.
2
116. Using integration by parts, we have
Z
3
3
ln(x ) dx = 3
Z
1
1
3
3
ln x dx = 3(x ln x − x)
1
= 9 ln 3 − 6 ≈ 3.8875.
This matches the approximation given by Simpson’s rule with 10 intervals.
117. In Problem 25, we found that
Z
(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C.
Thus
Z
1
e
e
(ln x)2 dx = [x(ln x)2 − 2x ln x + 2x]
1
= e − 2 ≈ 0.71828.
This matches the approximation given by Simpson’s rule with 10 intervals.
118. Integrating by parts, we take u = e2x , u′ = 2e2x , v ′ = sin 2x, and v = − 12 cos 2x, so
Z
e2x sin 2x dx = −
e2x
cos 2x +
2
Z
e2x cos 2x dx.
Integrating by parts again, with u = e2x , u′ = 2e2x , v ′ = cos 2x, and v =
Z
e2x cos 2x dx =
e2x
sin 2x −
2
Z
1
2
sin 2x, we get
e2x sin 2x dx.
Substituting into the previous equation, we obtain
Z
Solving for
R
e
2x
e2x
e2x
sin 2x dx = −
cos 2x +
sin 2x −
2
2
Z
e2x sin 2x dx.
e2x sin 2x dx gives
Z
e2x sin 2x dx =
1 2x
e (sin 2x − cos 2x) + C.
4
This result can also be obtained using II-8 in the integral table. Thus
Z
π
1
e2x sin 2x = [ e2x (sin 2x − cos 2x)]
4
−π
π
=
−π
1 −2π
(e
− e2π ) ≈ −133.8724.
4
We get −133.37 using Simpson’s rule with 10 intervals. With 100 intervals, we get −133.8724. Thus our answer matches
the approximation of Simpson’s rule.
690
Chapter Seven /SOLUTIONS
119.
Z
10
10
ze−z dz = [−ze−z ]
−
0
0
10
Z
−e−z dz
0
= −10e−10 − [e−z ]
(let z = u, e−z = v ′ , −e−z = v)
10
0
= −10e−10 − e−10 + 1
= −11e−10 + 1.
√
√
3
π
2
π
, and if θ = , then w =
. So we have
120. Let sin θ = w, cos θ dθ = dw. So, if θ = − , then w = −
3
2
4
2
Z
π/4
Z
√
√
2/2
1
w dw = w4
sin θ cos θ dθ =
√
4
−π/3
− 3/2
121. We substitute w =
3
√
3
1
x = x 3 . Then dw =
Z
√
3
8
e
√
3
1
122.
Z
1
0
x
x2
dx =
2/2
3
Z
−
√
3/2
1
=
4
√ 4
2
2
−
√ 4
− 3
2
=−
1
1 − 23
dx = √
dx.
x
3
3
3 x2
x=8
x=8
ew (3 dw) = 3ew
x=1
= 3e
√
3x
x=1
dx
= tan−1 x
x2 + 1
1
0
= tan−1 1 − tan−1 0 =
8
1
= 3(e2 − e).
π
π
−0= .
4
4
123. We put the integral in a convenient form for a substitution by using the fact that sin2 θ = 1−cos2 θ. Thus:
Z
π
4
−π
4
5
.
64
Z
π
4
cos2 θ sin5 θ dθ =
−π
4
cos2 θ(1 − cos2 θ)2 sin θ dθ.
Now, we can make
√ a substitution which helps. We let w = cos θ, so dw = − sin θ dθ.
2
π
π
Note that w =
when θ = − and when θ = . Thus after our substitution, we get
2
4
4
−
Z
w= π
4
w= π
4
w2 (1 − w2 )2 dw.
Since the upper and lower limits of integration are the same, this definite integral must equal 0. Notice that we could have
deduced this fact immediately, since cos2 θ is even and sin5 θ is odd, so cos2 θ sin5 θ is odd.
Thus
Z
0
−π
4
cos2 θ sin5 θ dθ = −
Z
π
4
cos2 θ sin5 θ dθ, and the given integral must evaluate to 0.
0
124. We substitute w = x2 + 4x + 5, so dw = (2x + 4) dx. Notice that when x = −2, w = 1, and when x = 0, w = 5.
Z
x=0
x=−2
2x + 4
dx =
x2 + 4x + 5
Z
w=5
w=1
1
dw = ln |w|
w
w=5
= ln 5.
w=1
1
, let’s imagine that our fraction is the result of adding together two terms, one with a denomi125. Since x21−1 = (x−1)(x+1)
nator of x − 1, the other with a denominator of x + 1:
A
B
1
=
+
.
(x − 1)(x + 1)
x−1
x+1
To find A and B, we multiply by the least common multiple of both sides to clear the fractions. This yields
1 = A(x + 1) + B(x − 1)
= (A + B)x + (A − B).
SOLUTIONS to Review Problems for Chapter Seven
691
Since the two sides are equal for all values of x in the domain, and there is no x term on the left-hand side, A + B = 0.
Similarly, since A and −B are constant terms on the right-hand side, and 1 is the constant term on the left-hand side,
A − B = 1. Therefore, we have the system of equations
A+B = 0
A − B = 1.
Solving this gives us A = 1/2 and B = −1/2, so
1
1
1
=
−
.
x2 − 1
2(x − 1)
2(x + 1)
Now, we find the integral
Z
126. (a) We split
1
x2 −x
1
x(x−1)
=
Z
1
1
dx
−
2(x − 1)
2(x + 1)
1
1
= ln |x − 1| − ln |x + 1| + C.
2
2
1
dx =
x2 − 1
into partial fractions:
1
A
B
=
+
.
x2 − x
x
x−1
Multiplying by x(x − 1) gives
1 = A(x − 1) + Bx = (A + B)x − A,
so −A = 1 and A + B = 0, giving A = −1, B = 1. Therefore,
(b)
Z
1
dx =
x2 − x
Z
Z
1
dx =
x2 − x
Z
1
x−1
−
1
x
dx = ln |x − 1| − ln |x| + C.
1
dx. Using a = 1 and b = 0 in V-26, we get ln |x − 1| − ln |x| + C.
(x − 1)(x)
127. Split the integrand into partial fractions, giving
1
A
B
=
+
x(L − x)
x
L−x
1 = A(L − x) + Bx = (B − A)x + AL.
We have B − A = 0 and AL = 1, so A = B = 1/L. Thus,
Z
1
dx =
x(L − x)
Z
1
L
1
1
+
x
L−x
dx =
1
(ln |x| − ln |L − x|) + C.
L
128. Splitting the integrand into partial fractions with denominators (x − 2) and (x + 2), we have
A
B
1
=
+
.
(x − 2)(x + 2)
x−2
x+2
Multiplying by (x − 2)(x + 2) gives the identity
1 = A(x + 2) + B(x − 2)
so
1 = (A + B)x + 2A − 2B.
Since this equation holds for all x, the constant terms on both sides must be equal. Similarly, the coefficient of x on both
sides must be equal. So
2A − 2B = 1
A + B = 0.
Solving these equations gives A = 1/4, B = −1/4 and the integral becomes
Z
1
1
dx =
(x − 2)(x + 2)
4
Z
1
1
dx −
x−2
4
Z
1
1
dx = (ln |x − 2| − ln |x + 2|) + C.
x+2
4
692
Chapter Seven /SOLUTIONS
129. Let x = 5 sin t. Then dx = 5 cos t dt, so substitution gives
Z
√
1
=
25 − x2
Z
5 cos t
p
25 − 25 sin2 t
dt =
Z
dt = t + C = arcsin
x
5
+ C.
130. Splitting the integrand into partial fractions with denominators x and (x + 5), we have
1
A
B
=
+
.
x(x + 5)
x
x+5
Multiplying by x(x + 5) gives the identity
1 = A(x + 5) + Bx
so
1 = (A + B)x + 5A.
Since this equation holds for all x, the constant terms on both sides must be equal. Similarly, the coefficient of x on both
sides must be equal. So
5A = 1
A + B = 0.
Solving these equations gives A = 1/5, B = −1/5 and the integral becomes
Z
1
1
dx =
x(x + 5)
5
Z
1
1
dx −
x
5
Z
1
1
dx = (ln |x| − ln |x + 5|) + C.
x+5
5
131. We use the trigonometric substitution 3x = sin θ. Then dx =
Z
1
√
dx =
1 − 9x2
Z
=
1
3
1
3
cos θ dθ and substitution gives
1
1
p
· cos θ dθ =
2
3
3
1 − sin θ
1
Z
1 dθ =
Z
cos θ
√
dθ
cos2 θ
1
1
θ + C = arcsin(3x) + C.
3
3
132. Splitting the integrand into partial fractions with denominators x, (x + 2) and (x − 1), we have
A
B
C
2x + 3
=
+
+
.
x(x + 2)(x − 1)
x
x+2
x−1
Multiplying by x(x + 2)(x − 1) gives the identity
2x + 3 = A(x + 2)(x − 1) + Bx(x − 1) + Cx(x + 2)
so
2x + 3 = (A + B + C)x2 + (A − B + 2C)x − 2A.
Since this equation holds for all x, the constant terms on both sides must be equal. Similarly, the coefficient of x on both
sides must be equal. So
−2A = 3
A − B + 2C = 2
A + B + C = 0.
Solving these equations gives A = −3/2, B = −1/6 and C = 5/3. The integral becomes
Z
Z
Z
Z
1
1
1
2x + 3
3
1
5
dx = −
dx −
+
dx
x(x + 2)(x − 1)
2
x
6
x+2
3
x−1
3
1
5
= − ln |x| − ln |x + 2| + ln |x − 1| + K.
2
6
3
We use K as the constant of integration, since we already used C in the problem.
SOLUTIONS to Review Problems for Chapter Seven
693
133. The denominator can be factored to give x(x − 1)(x + 1). Splitting the integrand into partial fractions with denominators
x, x − 1, and x + 1, we have
A
B
C
3x + 1
=
+
+ .
x(x − 1)(x + 1)
x−1
x+1
x
Multiplying by x(x − 1)(x + 1) gives the identity
3x + 1 = Ax(x + 1) + Bx(x − 1) + C(x − 1)(x + 1)
so
3x + 1 = (A + B + C)x2 + (A − B)x − C.
Since this equation holds for all x, the constant terms on both sides must be equal. Similarly, the coefficient of x and x2
on both sides must be equal. So
−C = 1
A−B = 3
A + B + C = 0.
Solving these equations gives A = 2, B = −1 and C = −1. The integral becomes
Z
Z
Z
Z
2
1
1
3x + 1
dx =
dx −
dx −
dx
x(x + 1)(x − 1)
x−1
x+1
x
= 2 ln |x − 1| − ln |x + 1| − ln |x| + K.
We use K as the constant of integration, since we already used C in the problem.
134. Splitting the integrand into partial fractions with denominators (1 + x), (1 + x)2 and x, we have
A
B
C
1 + x2
=
+
+ .
x(1 + x)2
1+x
(1 + x)2
x
Multiplying by x(1 + x)2 gives the identity
1 + x2 = Ax(1 + x) + Bx + C(1 + x)2
so
1 + x2 = (A + C)x2 + (A + B + 2C)x + C.
Since this equation holds for all x, the constant terms on both sides must be equal. Similarly, the coefficient of x and x2
on both sides must be equal. So
C=1
A + B + 2C = 0
A + C = 1.
Solving these equations gives A = 0, B = −2 and C = 1. The integral becomes
Z
1 + x2
dx = −2
(1 + x)2 x
Z
1
dx +
(1 + x)2
Z
2
1
dx =
+ ln |x| + K.
x
1+x
We use K as the constant of integration, since we already used C in the problem.
135. Completing the square, we get
x2 + 2x + 2 = (x + 1)2 + 1.
We use the substitution x + 1 = tan t, so dx = (1/ cos2 t)dt. Since tan2 t + 1 = 1/ cos2 t, the integral becomes
Z
1
dx =
(x + 1)2 + 1
Z
1
1
·
dt =
tan2 t + 1 cos2 t
Z
dt = t + C = arctan(x + 1) + C.
694
Chapter Seven /SOLUTIONS
136. Completing the square in the denominator gives
Z
We make the substitution tan θ = x + 2. Then dx =
Z
Z
dx
=
x2 + 4x + 5
1
cos2 θ
dx
=
(x + 2)2 + 1
=
=
=
dx
.
(x + 2)2 + 1
dθ.
Z
dθ
cos2 θ(tan2 θ + 1)
Z
cos2
Z
dθ = θ + C
dθ
Z
sin2 θ
θ( cos
2θ
+ 1)
dθ
sin2 θ + cos2 θ
But since tan θ = x + 2, θ = arctan(x + 2), and so θ + C = arctan(x + 2) + C.
137. We use the trigonometric substitution bx = a sin θ. Then dx =
Z
1
p
a2 − (bx)2
dx =
Z
1
=
b
1
·
p
Z
a2 − (a sin θ)2
1
cos θ
√
dθ =
2
b
cos θ
a
b
cos θ dθ, and we have
a
cos θ dθ =
b
Z
Z
1
a
p
2
1 − sin θ
But
cos x
dx =
sin3 x + sin x
Z
dw
.
w3 + w
1
1
w
1
=
=
− 2
,
w3 + w
w(w2 + 1)
w
w +1
so
Z
Z
1
w
− 2
dw
w
w +1
1
= ln |w| − ln w2 + 1 + C
2
1
= ln |sin x| − ln sin2 x + 1 + C.
2
cos x
dx =
sin3 x + sin x
139. Using the substitution w = ex , we get dw = ex dx, so we have
Z
But
w2
so
Z
ex
dx =
2x
e −1
Z
1
1
1
=
=
−1
(w − 1)(w + 1)
2
ex
dx =
2x
e −1
Z
1
2
dw
.
w2 − 1
1
1
−
,
w−1
w+1
1
1
−
w−1
w+1
dw
1
(ln |w − 1| − ln |w + 1|) + C
2
1
= (ln |ex − 1| − ln |ex + 1|) + C.
2
=
a
cos θ dθ
b
1
1
bx
1 dθ = θ + C = arcsin
b
b
a
138. Using the substitution w = sin x, we get dw = cos xdx, so we have
Z
·
+ C.
SOLUTIONS to Review Problems for Chapter Seven
140. This is an improper integral because
4
Z
√
16 − x2 = 0 at x = 4. So
dx
= lim
√
b→4−
16 − x2
0
695
Z
b
√
0
dx
16 − x2
b
= lim (arcsin x/4)
b→4−
0
= lim [arcsin(b/4) − arcsin(0)] = π/2 − 0 = π/2.
b→4−
141. This integral is improper because 5/x2 is undefined at x = 0. Then
Z
3
5
dx = lim
x2
b→0+
0
Z
b
3
5
dx = lim
x2
b→0+
3
−5x
−1
b
!
= lim
b→0+
−5
5
.
+
3
b
As b → 0+ , this goes to infinity and the integral diverges.
142. This integral is improper because 1/(x − 2) is undefined at x = 2. Then
Z
0
2
1
dx = lim
x−2
b→2−
Z
b
0
1
dx = lim
x−2
b→2−
b
ln |x − 2|
0
!
= lim (ln |b − 2| − ln 2) .
b→2−
As b → 2− , this goes to negative infinity and the integral diverges.
√
143. This integral is improper because 1/ 3 8 − x is undefined at x = 8. Then
Z
8
0
Z
1
√
dx = lim
3
b→8−
8−x
b
b
(8 − x)
0
−1/3
−1.5(8 − x)
dx = lim
b→8−
2/3
0
!
= lim
b→8−
−1.5(8 − b)2/3 + 1.5(82/3 ) = 6.
The integral converges.
144.
Z
∞
should converge, since
4
145.
Z
∞
dt
converges for n > 1.
tn
t3/2
1
We
Z b
Z ∞calculate its value.
b
dt
2
−3/2
−1/2
√
=
lim
=
lim
t
dt
= 1.
=
lim
−2t
1
−
b→∞
b→∞
b→∞
t3/2
b
4
4
4
dt
Z
dx
= ln | ln x| + C. (Substitute w = ln x, dw = x1 dx).
x ln x
Thus
Z ∞
Z b
dx
dx
= lim
= lim ln | ln x|
b→∞
x
ln
x
x
ln x b→∞
10
10
b
10
= lim ln(ln b) − ln(ln 10).
b→∞
As b → ∞, ln(ln b) → ∞, so this diverges.
146. To find
Then
Z
we−w dw, integrate by parts, with u = w and v ′ = e−w . Then u′ = 1 and v = −e−w .
Z
Thus
Z
we
−w
dw = −we
∞
we
−w
dw = lim
b→∞
0
−w
Z
0
+
Z
e−w dw = −we−w − e−w + C.
b
b
we−w dw = lim (−we−w − e−w )
b→∞
147. The trouble spot is at x = 0, so we write
Z
1
−1
1
dx =
x4
Z
0
−1
1
dx +
x4
Z
0
1
1
dx.
x4
= 1.
0
696
Chapter Seven /SOLUTIONS
However, both these integrals diverge. For example,
Z
1
0
1
dx = lim
x4
a→0+
Since this limit does not exist,
1
Z
0
1
Z
a
1
x−3
dx = lim −
4
+
x
3
a→0
1
= lim
a→0+
a
1
1
−
.
3a3
3
1
dx diverges and so the original integral diverges.
x4
148. Since the value of tan θ is between −1 and 1 on the interval −π/4 ≤ θ ≤ π/4, our integral is not improper and so
converges. Moreover, since tan θ is an odd function, we have
π
4
Z
0
Z
tan θ dθ =
−π
4
tan θ dθ +
−π
4
=−
Z
=−
Z
π
4
Z
tan θ dθ
0
0
tan(−θ) dθ +
−π
4
π
4
Z
tan θ dθ
0
π
4
tan θ dθ +
0
Z
π
4
tan θ dθ = 0.
0
149. It is easy to see that this integral converges:
1
1
< 2,
4 + z2
z
∞
2
1
dz <
4 + z2
2
We can also find its exact value.
Z
∞
Z
and so
1
dz = lim
b→∞
4 + z2
Z
b
2
b→∞
∞
2
1
1
dz = .
z2
2
1
dz
4 + z2
1
z
arctan
2
2
= lim
Z
b
2
!
1
b
1
= lim
arctan − arctan 1
b→∞ 2
2
2
1π
1π
π
=
−
= .
22
24
8
Note that
π
8
< 12 .
150. We find the exact value:
Z
∞
10
1
dz =
z2 − 4
Z
∞
1
dz
(z + 2)(z − 2)
10
= lim
b→∞
Z
b
10
1
dz
(z + 2)(z − 2)
1
= lim (ln |z − 2| − ln |z + 2|)
b→∞ 4
b
10
1
=
lim [(ln |b − 2| − ln |b + 2|) − (ln 8 − ln 12)]
4 b→∞
h
i
b−2
3
1
lim
ln
+ ln
=
4 b→∞
b+2
2
1
ln 3/2
= (ln 1 + ln 3/2) =
.
4
4
151. Substituting w = t + 5, we see that our integral is just
0
0 < p < 1. We find its exact value:
Z
0
15
Z
dw
√ = lim
w
a→0+
Z
15
a
15
dw
√ . This will converge, since
w
1
dw
√ = lim 2w 2
w
a→0+
15
a
√
= 2 15.
Z
b
0
dw
converges for
wp
SOLUTIONS to Review Problems for Chapter Seven
697
152. Since sin φ < φ for φ > 0,
Z
π
2
1
dφ >
sin φ
0
π
2
Z
0
1
dφ,
φ
The integral on the right diverges, so the integral on the left must also. Alternatively, we use IV-20 in the integral table to
get
Z
π
2
0
1
dφ = lim
sin φ
b→0+
Z
π
2
b
1
dφ
sin φ
1
cos φ − 1
= lim
ln
cos φ + 1
b→0+ 2
π
2
b
1
cos b − 1
= − lim ln
.
2 b→0+
cos b + 1
b−1
| → −∞. Thus the integral diverges.
As b → 0+ , cos b − 1 → 0 and cos b + 1 → 2, so ln | cos
cos b+1
153. Let φ = 2θ. Then dφ = 2 dθ, and
Z
π/4
tan 2θ dθ =
0
Z
π/2
1
tan φ dφ =
2
0
=
lim
b→(π/2)−
Z
b
0
Z
π/2
0
1 sin φ
dφ
2 cos φ
1
1 sin φ
dφ = lim − ln |cos φ|
2 cos φ
2
b→(π/2)−
b
.
0
As b → π/2, cos φ → 0, so ln | cos φ| → −∞. Thus the integral diverges.
One could also see this by noting that cos x ≈ π/2 − x and sin x ≈ 1 for x close to π/2: therefore, tan x ≈
1/( π2 − x), the integral of which diverges.
154. The integrand
x
→ 1 as x → ∞, so there’s no way
x+1
Z
∞
1
x
dx can converge.
x+1
155. This function is difficult to integrate, so instead we try to compare it with some other function. Since
that
R∞
0
sin2 θ
θ 2 +1
2
dθ ≥ 0. Also, since sin θ ≤ 1,
∞
Z
0
Thus
156.
Z
0
π
R∞
sin2 θ
θ 2 +1
0
sin2 θ
dθ ≤
θ2 + 1
∞
Z
0
1
dθ = lim arctan θ
b→∞
θ2 + 1
dθ converges, and its value is between 0 and
Z
π
tan2 θdθ =
0
π
2
Z
tan2 θdθ +
0
Z
π
2
b→ 2
3
(sin x) 2 < (sin x)
1
1
so
3 >
(sin
x)
2
(sin x)
3
or (sin x)− 2 > (sin x)−1
Hence,
(sin x)−1 dx = lim ln
a→0
0
R1
0
3
1
1
−
sin x
tan x
(sin x)− 2 dx is infinite.
1
, which is infinite.
a
0
π
.
2
π
,
2
so
tan2 θdθ = limπ [tan θ − θ]b0 + limπ [tan θ − θ]πa
which is undefined.
1
=
π
157. Since 0 ≤ sin x < 1 for 0 ≤ x ≤ 1, we have
Z
b
π
.
2
tan2 θdθ = tan θ − θ + C, by formula IV-23. The integrand blows up at θ =
Thus
sin2 θ
θ 2 +1
a→ 2
≥ 0, we see
698
Chapter Seven /SOLUTIONS
Problems
158. Since (ex )2 = e2x , we have
Area =
1
Z
159. Since (e ) = e
3x
x 2
e2x dx =
0
0
x 3
1
1
Z
(ex )2 dx =
=
1 2
(e − 1).
2
0
and (e ) = e , and the graph of (e ) is above the graph of (ex )2 for x > 0, we have
Area =
2x
1 2x
e
2
Z
x 3
e
x 3
0
x 2
(e ) − (e ) dx =
1
1
= e9 − e6 −
3
2
1
1
−
3
2
160. The curves y = ex and y = 5e−x cross where
3
3
Z
(e
3x
0
2x
− e ) dx =
1
1
1
= e9 − e6 + .
3
2
6
1 3x 1 2x
e − e
3
2
0
ex = 5e−x
e2x = 5
1
x = ln 5.
2
Since the graph of y = 5e−x is above the graph of y = ex for 0 ≤ x ≤
Area =
Z
1
2
ln 5 (see Figure 7.47), we have
(ln 5)/2
(5e−x − ex ) dx
0
(ln 5)/2
= (−5e
−x
x
−e )
0
= −5e−(ln 5)/2 − e(ln 5)/2 − (−5 · 1 − 1)
= −5(eln 5 )−1/2 − (eln 5 )1/2 + 6
= −5(5−1/2 ) − (5)1/2 + 6
√
√
5
= − √ − 5 + 6 = −2 5 + 6.
5
y
5
y = ex
y = 5e−x
x
(ln 5)/2
Figure 7.47
161. As is evident from Figure 7.48 showing the graphs of y = sin x and y = cos x, the crossings occur at x = π4 , 5π
, 9π
, . . .,
4
4
and the regions bounded by any two consecutive crossings have the same area. So picking two consecutive crossings, we
get an area of
Area =
Z
5π
4
π
4
√
= 2 2.
(sin x − cos x) dx
SOLUTIONS to Review Problems for Chapter Seven
(Note that we integrated sin x − cos x here because for
π
4
≤x≤
5π
,
4
699
sin x ≥ cos x.)
5π
4
13π
4
x
9π
4
π
4
Figure 7.48
√
162. In the interval of 0 ≤ x ≤ 2, the equation y = 4 − x2 represents one quadrant of the radius 2 circle centered at the
origin. Interpreting the integral as an area gives a value of 41 π22 = π.
163. Letting w = 3x5 + 2, dw = 15x4 dx, we see that x4 dx = dw/15, so:
Z
3x4
p
3x5 + 2 dx =
=
=
Z
Z
Z
1/2
3 3x5 + 2
3w1/2 ·
· x4 dx
1
· dw
15
1 1/2
w dw,
5
so w = 3x5 + 2, p = 1/2, k = 1/5.
164. Letting w = cos(3θ), dw = −3 sin(3θ) dθ, we see that sin(3θ) dθ = (−1/3) dw, so:
Z
Z
5 sin(3θ) dθ
=
5 (cos(3θ))−3 sin(3θ) dθ
cos3 (3θ)
Z
−dw
5w−3 ·
=
3
Z
5 −3
− w dw,
=
3
so w = cos(3θ), p = −3, k = −5/3.
Alternatively, if we write
5 sin(3θ)
5 sin(3θ)
1
=
·
cos3 (3θ)
cos(3θ) cos2 (3θ)
and let w = tan(3θ), so dw = 3 dθ/ cos2 (3θ), and we have
Z
5 sin(3θ) dθ
=
cos3 (3θ)
Z
5 tan(3θ)
1
dθ =
cos2 (3θ)
Z
5
w dw,
3
so here w = tan(3θ), p = 1, k = 5/3.
165. Using partial fractions we have
A
B
1
=
+
(2x − 3)(3x − 2)
2x − 3
3x − 2
A(3x − 2) + B(2x − 3) = 1
3Ax − 2A + 2Bx − 3B = 1
(3A + 2B)x − 2A − 3B = 1.
We see that:
3A + 2B = 0
700
Chapter Seven /SOLUTIONS
so A = −
and
−2 −
− 2A
− 3B
2
· B − 3B
3
4
· B − 3B
3
5
− B
3
B
2
·B
3
=1
since A = − 32 B
=1
=1
=1
3
= −0.6
5
2
3
A=−
−
3
5
2
= = 0.4.
5
=−
Z
Z
since A = − 32 B
−0.6
0.4
dx
=
+
dx.
(2x − 3)(3x − 2)
2x − 3
3x − 2
166. Let u = 0.5x − 1, du = 0.5 dx. Then dx = 2 du and, solving for x,
Thus,
0.5x − 1 = u
0.5x = u + 1
x = 2u + 2.
This means
x2 + x = (2u + 2)2 + 2u + 2
= 4u2 + 10u + 6
so
Z
2
(x + x) cos(0.5x − 1) dx =
=
where p(u) = 8u2 + 20u + 12.
Z
Z
(4u2 + 10u + 6) cos u(2 du)
(8u2 + 20u + 12) cos u du,
167. Let u = −x2 , du = −2x dx, so that x dx = −0.5du. We have:
u
Z
2
x3 e−x dx =
=
Z
Z
2
x e
|{z}
−u
z}|{
−x2 · x dx
|{z}
−0.5du
0.5ueu du,
so k = 0.5, u = −x2 .
√
168. Letting u = cos x, we have
√
1
du = − x−1/2 sin x dx
2 √
sin x dx
√
=−
.
2 x
So
sin
√
x dx
√
= −2du.
x
Thus,
Z
cos4
√
√
Z
x sin x dx
√
=
x
cos
√ 4
√
x
sin x dx
√
x
SOLUTIONS to Review Problems for Chapter Seven
=
=
where k = −2, n = 4, and u = cos
√
Z
Z
701
√
√ 4 sin x dx
√
·
cos
x
x
−2u4 du,
x.
169. After the substitution w = x2 , the second integral becomes
1
2
Z
√
dw
.
1 − w2
170. After the substitution w = x + 2, the first integral becomes
Z
w−2 dw.
After the substitution w = x2 + 1, the second integral becomes
1
2
Z
w−2 dw.
171. After the substitution w = 1 − x2 , the first integral becomes
−
1
2
Z
w−1 dw.
After the substitution w = ln x, the second integral becomes
Z
w−1 dw.
172. First solution: After the substitution w = x + 1, the first integral becomes
Z
w−1
dw = w −
w
Z
w−1 dw.
With this same substitution, the second integral becomes
Z
w−1 dw.
Second solution: We note that the sum of the integrands is 1, so the sum of the integrals is x. Thus
Z
x
dx = x −
x+1
Z
1
dx.
x+1
173. If we let w = 2x in the first integral, we get dw = 2dx. Also, the limits w = 0 and w = 2 become x = 0 and x = 1.
Thus
Z
Z
2
1
2
e−w dw =
0
2
e−4x 2 dx.
0
174. If we let t = 3u in the first integral, we get dt = 3du, so dt/t = du/u. Also, the limits t = 0 and t = 3 become u = 0
and u = 1. Thus
Z 1
Z 3
sin 3u
sin t
dt =
du.
t
u
0
0
If we rename the variable u as t in the last integral, we have the equality we want.
702
Chapter Seven /SOLUTIONS
175. Since the definition of f is different on 0 ≤ t ≤ 1 than it is on 1 ≤ t ≤ 2, break the definite integral at t = 1.
2
Z
f (t) dt =
0
=
=
Z
Z
1
f (t) dt +
0
2
Z
f (t) dt
1
1
t2 dt +
0
1
t3
3
+
0
Z
2
(2 − t) dt
1
2t −
t2
2
2
1
= 1/3 + 1/2 = 5/6 ≈ 0.833
176. (a)
(i) Multiplying out gives
Z
(x2 + 10x + 25) dx =
x3
+ 5x2 + 25x + C.
3
(ii) Substituting w = x + 5, so dw = dx, gives
Z
(x + 5)2 dx =
Z
w2 dw =
(x + 5)3
w3
+C =
+ C.
3
3
(b) The results of the two calculations are not the same since
(x + 5)3
x3
15x2
75x
125
+C =
+
+
+
+ C.
3
3
3
3
3
However they differ only by a constant, 125/3, as guaranteed by the Fundamental Theorem of Calculus.
177. (a) Since h(z) is even, we know that
R1
R1
R1
0
h(z) dz =
R0
R1
R0
R1
h(z) dz. Since −1 h(z) dz = −1 h(z) dz + 0 h(z) dz, we
−1
R1
see that −1 h(z) dz = 2 0 h(z) dz = 7. Thus 0 h(z) dz = 3.5
(b) If w = z + 3, then dw = dz. When z = −4, w = −1; when z = −2, w = 1. Thus,
Z
−2
5h(z + 3) dt = 5
Z
1
−1
−4
h(w) (dw) = 5 · 7 = 35.
178. The point of intersection of the two curves y = x2 and y = 6 − x is at (2,4). The average height of the shaded area is the
average value of the difference between the functions:
1
(2 − 0)
Z
0
2
((6 − x) − x2 ) dx = (3x −
x2
x3
−
)
4
6
2
=
0
11
.
3
179. The average width of the shaded area in the figure below is the average value of the horizontal distance between the two
functions. If we call this horizontal distance h(y), then the average width is
1
(6 − 0)
6
Z
h(y) dy.
0
y
y = x2
y =6−x
x
We could compute this integral if we wanted to, but we don’t need to. We can simply note that the integral (without
the 16 term) is just the area of the shaded region; similarly, the integral in Problem 178 is also just the area of the shaded
region. So they are the same. Now we know that our average width is just 13 as much as the average height, since we
divide by 6 instead of 2. So the answer is 11
.
9
SOLUTIONS to Review Problems for Chapter Seven
703
180. (a) i. 0
ii. π2
iii. 21
(b) Average value of f (t) < Average value of k(t) < Average value of g(t)
, since they all have periods of 2π (| cos t|
We can look at the three functions in the range − π2 ≤ x ≤ 3π
2
and (cos t)2 also have a period of π, but that does not hurt our calculation). It is clear from the graphs of the three
functions below that the average value for cos t is 0 (since the area above the x-axis is equal to the area below it),
while the average values for the other two are positive (since they are everywhere positive, except where they are 0).
f (t)
f (t)
f (t)
3π
2
−π
2
π
2
t
π
π
2
−π
2
t
π
π
2
−π
2
3π
2
t
π
3π
2
It is also fairly clear from the graphs that the average value of g(t) is greater than the average value of k(t); it is
also possible to see this algebraically, since
(cos t)2 = | cos t|2 ≤ | cos t|
because | cos t| ≤ 1 (and both of these ≤’s are <’s at all the points where the functions are not 0 or 1).
181. This calculation cannot be correct because the integrand is positive everywhere, yet the value given for the integral is
negative.
The calculation is incorrect because the integral is improper but has not been treated as such. The integral is improper
because the integrand 1/x2 is undefined at x = 0. To determine whether the integral converges we split the integral into
two improper integrals:
Z 0
Z 2
Z 2
1
1
1
dx =
dx +
dx.
2
2
2
x
x
x
−2
0
−2
To decide whether the second integral converges, we compute
Z
The limit does not exist, and
182. (a) We have
R2
0
E(x) + F (x) =
2
0
1
dx = lim
x2
a→0+
Z
a
2
1
1
1
dx = lim − +
.
x2
2
a
a→0+
2
(1/x ) dx diverges, so the original the integral
Z
ex
dx +
x
e + e−x
Z
e−x
dx =
x
e + e−x
Z
R2
−2
1/x2 dx diverges.
ex + e−x
dx =
ex + e−x
Z
1 dx = x + C1 .
(b) We have
E(x) − F (x) =
Z
ex
dx −
x
e + e−x
Z
e−x
dx =
x
e + e−x
Z
ex − e−x
dx = ln ex + e−x + C2 ,
ex + e−x
by using the substitution w = ex + e−x in the final integral.
(c) We have
E(x) + F (x) = x + C1
E(x) − F (x) = ln ex + e−x + C2 .
Adding and subtracting we find
1
x
+ ln ex + e−x + C,
2
2
where the arbitrary constant C = (C1 + C2 )/2, and
E(x) =
F (x) =
where C = (C1 − C2 )/2.
1
x
− ln ex + e−x + C,
2
2
704
Chapter Seven /SOLUTIONS
183. Since f (x) is decreasing on [a, b], the left-hand Riemann sums are all overestimates and the right-hand sums are all
underestimates. Because increasing the number of subintervals generally brings an approximation closer to the actual
value, LEFT(10) is closer to the actual value (i.e., smaller, since the left sums are overestimates) than LEFT(5), and
analogously for RIGHT(10) and RIGHT(5). Since the graph of f (x) is concave down, a secant line lies below the curve
and a tangent line lies above the curve. Therefore, TRAP is an underestimate and MID is an overestimate. Putting these
observations together, we have
RIGHT(5) < RIGHT(10) < TRAP(10) < Exact value < MID(10) < LEFT(10) < LEFT(5).
184. (a) f (x) = 1 + e−x is concave up for 0 ≤ x ≤ 0.5, so trapezoids will overestimate
rule will underestimate.
2
(b) f (x) = e−x is concave down for 0 ≤ x ≤ 0.5, so trapezoids will underestimate
Z
0.5
Z
0.5
f (x)dx, and the midpoint
0
f (x)dx and midpoint will
0
overestimate the integral.
(c) Both the trapezoid rule and the midpoint rule will give the exact value of the integral. Note that upper and lower sums
will not, unless the line is horizontal.
185. (a) For the left-hand rule, error is approximately proportional to
for accuracy to p places, then there is a constant k such that
1
2
1
=
2
1
=
2
1
=
2
5 × 10−5 =
5 × 10−9
5 × 10−13
5 × 10−21
1
.
n
If we let np be the number of subdivisions needed
k
n4
k
× 10−8 ≈
n8
k
× 10−12 ≈
n12
k
× 10−20 ≈
n20
× 10−4 ≈
Thus the ratios n4 : n8 : n12 : n20 ≈ 1 : 104 : 108 : 1016 , and assuming the computer time necessary is proportional
to np , the computer times are approximately
4 places:
8 places:
12 places:
20 places:
2 seconds
2 × 104 seconds
2 × 108 seconds
2 × 1016 seconds
≈ 6 hours
≈ 6 years
≈ 600 million years
(b) For the trapezoidal rule, error is approximately proportional to
for accuracy to p places, then there is a constant C such that
1
2
1
=
2
1
=
2
1
=
2
5 × 10−5 =
5 × 10−9
5 × 10−13
5 × 10−21
1
.
n2
If we let Np be the number of subdivisions needed
C
N4 2
C
× 10−8 ≈
N8 2
C
× 10−12 ≈
N12 2
C
× 10−20 ≈
N20 2
× 10−4 ≈
Thus the ratios N4 2 : N8 2 : N12 2 : N20 2 ≈ 1 : 104 : 108 : 1016 , and the ratios N4 : N8 : N12 : N20 ≈ 1 : 102 :
104 : 108 . So the computer times are approximately
4 places:
8 places:
12 places:
20 places:
2 seconds
2 × 102 seconds
2 × 104 seconds
2 × 108 seconds
≈ 3 minutes
≈ 6 hours
≈ 6 years
SOLUTIONS to Review Problems for Chapter Seven
2
2
186. Use integration by parts, with u = x and dv = xe−x . Then v = −(1/2)e−x , and
Z
0
b
2
2
1
x2 e−x dx = − xe−x
2
b
+
0
2
1
1
= − be−b +
2
2
Z
Z
b
0
b
1 −x2
e
dx
2
2
e−x dx.
0
Since the exponential grows faster than any power,
2
lim be−b = lim
b→∞
So
Z
b
b→∞
2
x2 e−x dx = 0 +
0
1
2
Z
∞
b
= 0.
eb2
2
e−x dx =
0
√
√
1
π
π
·
=
.
2 2
4
187. (a) We calculate the integral using partial fractions with denominators P and L − P :
A
B
k
=
+
P (L − P )
P
L−P
k = A(L − P ) + BP
k = (B − A)P + AL.
Thus,
B−A = 0
AL = k,
so A = B = k/L, and the time is given by
T =
Z
L/2
L/4
L/2
Z
k dP
k
=
P (L − P )
L
L/4
1
P
+
1
L−P
k
L
=
ln
− ln
L
2
3L/4
k
=
= ln
L
L/4
dP =
L
L
− ln
2
4
k
ln (3) .
L
k
(ln |P | − ln |L − P |)
L
3L
+ ln
4
L/2
L/4
(b) A similar calculation gives the following expression for the time:
T =
k
(ln |P | − ln |L − P |)
L
P2
=
P1
k
(ln |P2 | − ln |L − P2 | − ln |P1 | + ln |L − P1 |) .
L
If P2 → L, then L − P2 → 0, so ln P2 → ln L, and ln(L − P2 ) → −∞. Thus the time tends to infinity.
188. If I(t) is average per-capita income t years after 2005, then I ′ (t) = r(t).
(a) Since t = 10 in 2015, by the Fundamental Theorem,
I(10) − I(0) =
10
Z
r(t) dt =
0
Z
10
1556.37e0.045t dt =
0
1556.37 0.045t
e
0.045
10
= 19,655.65 dollars.
0
so I(10) = 34,586 + 19,655.65 = 54,241.65.
(b) We have
I(t) − I(0) =
Z
t
r(t) dt =
0
Z
0
t
1556.37e0.045t dt =
1556.37 0.045t
e
0.045
t
0
= 34,586 e0.045t − 1 .
Thus, since I(0) = 34,586,
I(t) = 34,586 + 34,586(e0.045t − 1) = 34,586e0.045t dollars.
705
706
Chapter Seven /SOLUTIONS
189. (a) Since the rate is given by r(t) = 2te−2t ml/sec, by the Fundamental Theorem of Calculus, the total quantity is given
by the definite integral:
Total quantity ≈
∞
Z
2te−2t dt = 2 lim
b→∞
0
Integration by parts with u = t, v ′ = e−2t gives
Z
b
te−2t dt.
0
b
1
− e−2t − e−2t
2
4
b→∞
0
b
1
1 −2b
1
= 2 lim
−
+
e
= 2 · = 0.5 ml.
b→∞ 4
2
4
4
Total quantity ≈ 2 lim
(b) At the end of 5 seconds,
t
Z
Quantity received =
5
0
2te−2t dt ≈ 0.49975 ml.
Since 0.49975/0.5 = 0.9995 = 99.95%, the patient has received 99.95% of the dose in the first 5 seconds.
190. The rate at which petroleum is being used t years after 1990 is given by
r(t) = 1.4 · 1020 (1.02)t joules/year.
Between 1990 and M years later
Total quantity of petroleum used =
M
Z
0
1.4 · 1020 (1.02)t dt = 1.4 · 1020
(1.02)t
ln(1.02)
M
0
1.4 · 1020
=
((1.02)M − 1) joules.
ln(1.02)
Setting the total quantity used equal to 1022 gives
1.4 · 1020
(1.02)M − 1 = 1022
ln(1.02)
100 ln(1.02)
(1.02)M =
+ 1 = 2.41
1.4
ln(2.41)
≈ 45 years.
M =
ln(1.02)
So we will run out of petroleum in 2035.
191. (a) We have
S=
Z
18
0
(30 − 10)dt = 20
18
Z
0
dt = 20(18 − 0) = 360.
The units of S are degree-days, because the integrand f (t) − 10 has units of ◦ C, and dt has units of days.
(b) In Figure 7.49, f (t) and Hmin are represented by the horizontal lines at H = 30 and H = 10, and T is represented
by a vertical line at t = 18. The value of S, given by the definite integral, is represented by the area of the rectangle
bounded by the vertical lines t = 0 (the H-axis) and t = 18, and the horizontal lines H = 10 and H = 30.
(c) The temperature cycles from a high of 30◦ C to a low of 10◦ C once every 6 days. During the 18-day period the
temperature completes 3 complete cycles. The area between this curve and the horizontal line H = Hmin = 10 gives
the value of the definite integral. See Figure 7.50. In order to get the same area as before, (namely S = 360), we see
that T2 must be larger than T = 18. Thus we want T2 to satisfy:
S=
Z
0
T2
(g(t) − 10)dt =
Z
T2
0
10 cos
2πt
6
+ 10
dt = 360.
Notice that, by symmetry, the area on the interval 0 ≤ t ≤ 18 is half the area shown in Figure 7.49. Thus, we expect
that T2 = 2T = 36. We can check this by calculation, using a substitution to evaluate the integral:
S=
Z
0
36
10 cos
2πt
6
+ 10
dt =
10 ·
6
2πt
sin
2π
6
+ 10t
T2
= 360.
0
707
SOLUTIONS to Review Problems for Chapter Seven
Thus, if T2 = 36, the integral evaluates to S = 360, as required. Thus, 36 days are required for development for with
these temperatures.
H, ◦C
30
t = T = 18
H = f (t) = 30
y
20
S = Area = 360
H = Hmin = 10
10
18
H = g(t)
t = T2
30
S = Area = 360
H = Hmin
10
x
t, days
18
36
Figure 7.50
Figure 7.49
192. We want to calculate
Z
1
Cn sin(nπx) · Cm sin(mπx) dx.
0
We use II-11 from the table of integrals with a = nπ, b = mπ. Since n 6= m, we see that
Z
0
1
Ψn (x) · Ψm (x) dx = Cn Cm
=
Z
1
sin(nπx) sin(mπx) dx
0
Cn Cm
m2 π 2 − n2 π 2
1
(nπ cos(nπx) sin(mπx) − mπ sin(nπx) cos(mπx))
Cn Cm
nπ cos(nπ) sin(mπ) − mπ sin(nπ) cos(mπ)
=
(m2 − n2 )π 2
−nπ cos(0) sin(0) + mπ sin(0) cos(0)
=0
since sin(0) = sin(nπ) = sin(mπ) = 0.
CAS Challenge Problems
193. (a) A CAS gives
Z
Z
Z
(ln x)2
ln x
dx =
x
2
(ln x)2
(ln x)3
dx =
x
3
(ln x)4
(ln x)3
dx =
x
4
(b) Looking at the answers to part (a),
(c) Let w = ln x. Then dw = (1/x)dx, and
Z
Z
(ln x)n
dx =
x
(ln x)n
(ln x)n+1
dx =
+ C.
x
n+1
Z
wn dw =
wn+1
(ln x)n+1
+C =
+ C.
n+1
n
0
708
Chapter Seven /SOLUTIONS
194. (a) A CAS gives
Z
Z
ln x dx = −x + x ln x
(ln x)2 dx = 2x − 2x ln x + x(ln x)2
Z
(ln x)3 dx = −6x + 6x ln x − 3x(ln x)2 + x(ln x)3
Z
(ln x)4 dx = 24x − 24x ln x + 12x(ln x)2 − 4x(ln x)3 + x(ln x)4
(b) In each of the cases in part (a),
(ln x)n dx has two parts. The first part isR simply a
R
R the expression for the integral
ln x = −2 ln x dx.
multiple ofRthe expression for (ln x)n−1 dx. For example, (ln x)2 dxRstarts out with 2x−2x
R
3
2
2
4
Similarly,
(ln
x)
dx
starts
out
with
−6x
+
6x
ln
x
−
3(ln
x)
=
−3
(ln
x)
dx,
and
(ln
x)
dx starts out with
R
−4 (ln x)3 dx. The remaining part of each antiderivative is a single term: it’s x(ln x)2 in the case n = 2, it’s
x(ln x)3 for n = 3, and it’s x(ln x)4 for n = 4. The general pattern is
R
Z
n
(ln x) dx = −n
Z
(ln x)n−1 dx + x(ln x)n .
To check this formula, we use integration by parts. Let u = (ln x)n so u′ = n(ln x)n−1 /x and v ′ = 1 so v = x.
Then
Z
n
n
(ln x) dx = x(ln x) −
Z
(ln x)n dx = x(ln x)n − n
This is the result we obtained before.
Alternatively, we can check our result by differentiation:
d
dx
−n
Z
(ln x)
n−1
Z
dx + x(ln x)
n
n
Z
(ln x)n−1
· x dx
x
(ln x)n−1 dx.
= −n(ln x)n−1 +
d
(x(ln x)n )
dx
1
x
= (ln x)n .
= −n(ln x)n−1 + (ln x)n + x · n(ln x)n−1
= −n(ln x)n−1 + (ln x)n + n(ln x)n−1
Therefore,
Z
(ln x)n dx = −n
195. (a) A possible answer from the CAS is
Z
(b) Differentiating
d
dx
sin3 x dx =
−9 cos(x) + cos(3 x)
12
(c) Using the identities, we get
=
Z
(ln x)n−1 dx + x(ln x)n .
−9 cos(x) + cos(3 x)
.
12
3 sin x − sin(3x)
9 sin(x) − 3 sin(3 x)
=
.
12
4
sin(3x) = sin(x + 2x) = sin x cos 2x + cos x sin 2x
= sin x(1 − 2 sin2 x) + cos x(2 sin x cos x)
= sin x − 2 sin3 x + 2 sin x(1 − sin2 x)
= 3 sin x − 4 sin3 x.
Thus,
so
3 sin x − sin(3x) = 3 sin x − (3 sin x − 4 sin3 x) = 4 sin3 x,
3 sin x − sin(3x)
= sin3 x.
4
PROJECTS FOR CHAPTER SEVEN
709
196. (a) A possible answer is
Z
sin x cos x cos(2x) dx = −
cos(4x)
.
16
Different systems may give the answer in a different form.
(b)
cos(4x)
sin(4x)
d
−
=
.
dx
16
4
(c) Using the double angle formula sin 2A = 2 sin A cos A twice, we get
2 sin(2x) cos(2x)
2 · 2 sin x cos x cos(2x)
sin(4x)
=
=
= sin x cos x cos(2x).
4
4
4
197. (a) A possible answer from the CAS is
Z
x4
x
3
dx = x +
− arctan(x).
(1 + x2 )2
2 (1 + x2 )
2
Different systems may give the answer in different form.
(b) Differentiating gives
d
dx
3
x
− arctan(x)
x+
2 (1 + x2 )
2
=1−
1
x2
−
.
2
2
1
+
x2
(1 + x )
(c) Putting the result of part (b) over a common denominator, we get
2
1 + x2 − x2 − (1 + x2 )
1
x2
−
1−
=
2
2
1+x
(1 + x2 )2
(1 + x2 )
=
1 + 2x2 + x4 − x2 − 1 − x2
x4
=
.
2
2
(1 + x )
(1 + x2 )2
PROJECTS FOR CHAPTER SEVEN
1. (a) If et ≥ 1 + t, then
Z
e =1+
x
x
et dt
0
1
(1 + t) dt = 1 + x + x2 .
2
0
We can keep going with this idea. Since et ≥ 1 + t + 21 t2 ,
Z x
et dt
ex = 1 +
Z0 x
1
1
1
≥1+
(1 + t + t2 ) dt = 1 + x + x2 + x3 .
2
2
6
0
n
We notice that each term in our summation is of the form xn! . Furthermore, we see that if we have a sum
2
n
1 + x + x2 + · · · + xn! such that
Z
≥1+
x
ex ≥ 1 + x +
xn
x2
+ ···+
,
2
n!
then
ex = 1 +
Z
x
et dt
Z x
tn
t2
dt
≥1+
1 + t+ + ···+
2
n!
0
x2
x3
xn+1
=1+x+
+
+ ··· +
.
2
6
(n + 1)!
0
710
Chapter Seven /SOLUTIONS
Thus we can continue this process as far as we want, so
n
X xj
1
1
for any n.
ex ≥ 1 + x + x2 + · · · + xn =
2
n!
j!
j=0
(In fact, it turns out that if you let n get larger and larger and keep adding up terms, your values approach
exactly ex .)
Z
Z
x
(b) We note that sin x =
x
cos t dt and cos x = 1 −
sin t dt. Thus, since cos t ≤ 1, we have
0
0
sin x ≤
Z
x
1 dt = x.
0
Now using sin t ≤ t, we have
cos x ≤ 1 −
Z
0
Then we just keep going:
Z
sin x ≤
x
0
Therefore
cos x ≤ 1 −
Z
0
x
x
1
t dt = 1 − x2 .
2
1
1
1 − t2 dt = x − x3 .
2
6
1
1
1
t − t3 dt = 1 − x2 + x4 .
6
2
24
8.1 SOLUTIONS
CHAPTER EIGHT
Solutions for Section 8.1
Exercises
1. (a) The strip stretches between y = 0 and y = 2x, so
Area of region ≈
(b) We have
Z
Area =
X
(2x)∆x.
3
3
(2x) dx = x
2
0
= 9.
0
2. (a) The strip stretches between y = 0 and y = −x2 + 6x, so
Area of region ≈
(b) We have
Area =
6
Z
X
(−x2 + 6x)∆x.
x3
(−x + 6x) dx = −
+ 3x2
3
6
2
0
= 36.
0
3. (a) The strip stretches from x = y/2 to x = 3, so
Area of region ≈
(b) We have
Area =
Z
6
0
y
3−
2
4. (a) The equation y = −x2 + 6x can be solved for x as
X
3−
y
2
∆y.
y2
3y −
4
dy =
6
= 9.
0
x2 − 6x + y = 0
√
p
6 ± 36 − 4y
= 3 ± 9 − y.
x=
2
√
√
The left end of the strip is given by x = 3 − 9 − y, and the right end is given by x = 3 + 9 − y. Thus,
Area of region ≈
(b) We have
X
((3 +
Area =
Z
0
p
9 − y) − (3 −
9
p
2
9 − y dy =
p
9 − y))∆y =
2(9 − y)3/2
−3/2
9
= 36.
0
5. Each strip is a rectangle of length 3 and width ∆x, so
Area of strip = 3∆x,
Area of region =
Z
0
so
5
5
3 dx = 3x
= 15.
0
Check: This area can also be computed using Length × Width = 5 · 3 = 15.
X p
2
9 − y ∆y.
711
712
Chapter Eight /SOLUTIONS
6. Using similar triangles, the height, y, of the strip is given by
y
x
=
3
6
x
.
2
y=
so
Thus,
x
∆x,
2
Area of strip ≈ y∆x =
so
6
Z
Area of region =
0
1
2
Check: This area can also be computed using the formula
6
x
x2
dx =
2
4
= 9.
0
Base · Height =
1
2
7. By similar triangles, if w is the length of the strip at height h, we have
5−h
w
=
3
5
w =3 1−
so
Thus,
Z
5
0
3 1−
h
5
1
2
Check: This area can also be computed using the formula
h
.
5
h
∆h.
5
Area of strip ≈ w∆h = 3 1 −
Area of region =
dh =
3h −
3h2
10
Base · Height =
1
2
8. Suppose the length of the strip shown is w. Then the Pythagorean theorem gives
h2 +
Thus
w 2
2
= 32
3
p
Using VI-30 in the Table of Integrals, we have
Area =
p
h
32 − h2 + 32 arcsin
h
3
=
0
·3·5=
15
.
2
15
.
2
32 − h2 .
32 − h2 ∆h,
32 − h2 dh.
2
−3
5
p
Area of strip ≈ w∆h = 2
Area of region =
p
w=2
so
Z
· 6 · 3 = 9.
3
−3
= 9(arcsin 1 − arcsin(−1)) = 9π.
Check: This area can also be computed using the formula πr 2 = 9π.
2
2
9. The strip has width ∆y, so the variable of integration
p is y. The length of the strip is x. Since x + y = 10 and the region
2
is in the first quadrant, solving for x gives x = 10 − y . Thus
Area of strip ≈ x∆y =
√
The region stretches from y = 0 to y = 10, so
√
Area of region =
Z
p
10 − y 2 dy.
10
p
10 − y 2 dy.
0
Evaluating using VI-30 from the Table of Integrals, we have
√
1
Area =
2
y
p
10 − y 2 + 10 arcsin
y
√
10
0
10
= 5(arcsin 1 − arcsin 0) =
√
Check: This area can also be computed using the formula 14 πr = 14 π( 10)2 = 25 π.
2
5
π.
2
8.1 SOLUTIONS
713
10. The strip has width ∆y, so the variable of integration is y. The length of the strip is 2x for x ≥ 0. For positive x, we have
x = y. Thus,
Area of strip ≈ 2x∆y = 2y∆y.
Since the region extends from y = 0 to y = 4,
Area of region =
Z
4
4
2y dy = y 2
0
Check: The area of the region can be computed by
1
2
= 16.
0
Base · Height =
1
2
· 8 · 4 = 16.
11. The width of the strip is ∆y, so the variable of integration is y. Since the graphs are x = y and x = y 2 , the length of the
strip is y − y 2 , and
Area of strip ≈ (y − y 2 )∆y.
The curves cross at the points (0, 0) and (1, 1), so
Area of region =
Z
1
0
(y − y 2 ) dy =
y3
y2
−
2
3
1
=
0
1
.
6
12. The width of the strip is ∆x, so the variable of integration is x. The line has equation y = 6 − 3x. The length of the strip
is 6 − 3x − (x2 − 4) = 10 − 3x − x2 . (Since x2 − 4 is negative where the graph is below the x-axis, subtracting x2 − 4
there adds the length below the x-axis.) Thus
Area of strip ≈ (10 − 3x − x2 )∆x.
Both graphs cross the x-axis where x = 2, so
Area of region =
Z
0
2
(10 − 3x − x2 ) dx = 10x −
3 2 x3
x −
2
3
2
=
0
34
.
3
13. Each slice is a circular disk with radius r = 2 cm.
Volume of disk = πr 2 ∆x = 4π∆x cm3 .
Summing over all disks, we have
Total volume ≈
Taking a limit as ∆x → 0, we get
Total volume = lim
∆x→0
Evaluating gives
X
X
4π∆x cm3 .
4π∆x =
Z
9
4π dx cm3 .
0
9
= 36π cm3 .
Total volume = 4πx
0
Check: The volume of the cylinder can also be calculated using the formula V = πr 2 h = π22 · 9 = 36π cm3 .
14. Each slice is a circular disk. Since the radius of the cone is 2 cm and the length is 6 cm, the radius is one-third of the
distance from the vertex. Thus, the radius at x is r = x/3 cm. See Figure 8.1.
Volume of slice ≈ πr 2 ∆x =
πx2
∆x cm3 .
9
Summing over all disks, we have
Total volume ≈
Taking a limit as ∆x → 0, we get
Total volume = lim
∆x→0
X x2
π
X x2
π
9
9
∆x cm3 .
∆x =
Z
0
6
π
x2
dx cm3 .
9
714
Chapter Eight /SOLUTIONS
Evaluating, we get
Total volume =
6
π x3
9 3
=
0
π 63
·
= 8π cm3 .
9 3
Check: The volume of the cone can also be calculated using the formula V = 31 πr 2 h = 13 π22 · 6 = 8π cm3 .
✻
2 cm
✻
x/3
✛
0
✲
x
✛
❄
❄x
✲
6 cm
Figure 8.1
15. Each slice is a circular disk. From Figure 8.2, we see that the radius at height y is r = 52 y cm. Thus
Volume of disk ≈ πr 2 ∆y = π
Summing over all disks, we have
Total volume ≈
Taking the limit as ∆y → 0, we get
Total volume = lim
∆y→0
Evaluating gives
2
X 4π
25
X 4π
Total volume =
2
y
5
25
∆y =
y 2 ∆y cm3 .
y 2 ∆y =
Z
0
5
4π y 3
25 3
4
πy 2 ∆y cm3 .
25
5
4π 2
y dy cm3 .
25
20
π cm3 .
3
=
0
Check: The volume of the cone can also be calculated using the formula V = 31 πr 2 h =
✛
2
π 2
2
3
·5=
20
π
3
cm3 .
✲
✻
2
y
✛5✲
5 cm
✻
y
❄
❄
Figure 8.2
16. Each slice is a rectangular slab of length 10 m and width that decreases with height. See Figure 8.3. At height y, the length
x is given by the Pythagorean Theorem
y 2 + x2 = 72 .
Solving gives x =
p
p
72 − y 2 m. Thus the width of the slab is 2x = 2
p
Volume of slab = Length · Width · Height = 10 · 2
72 − y 2 and
p
72 − y 2 · ∆y = 20
72 − y 2 ∆y m3 .
8.1 SOLUTIONS
715
Summing over all slabs, we have
X
Total volume ≈
Taking a limit as ∆y → 0, we get
Total volume = lim
∆y→0
X
p
20
To evaluate, we use the table of integrals or the fact that
Total volume =
Z
Z
p
0
Z
7
p
20
0
7
7
20
72 − y 2 ∆y m3 .
72 − y 2 ∆y =
0
7, so
p
20
72 − y 2 dy m3 .
p
72 − y 2 dy represents the area of a quarter circle of radius
72 − y 2 dy = 20 ·
1 2
π7 = 245π m3 .
4
Check: the volume of a half cylinder can also be calculated using the formula V = 21 πr 2 h = 12 π72 · 10 = 245π m3 .
x
y
7
Figure 8.3
17. Each slice is a circular disk. See Figure 8.4. The radius of the sphere is 5 mm, and the radius r at height y is given by the
Pythagorean Theorem
y 2 + r 2 = 52 .
Solving gives r =
p
52 − y 2 mm. Thus,
Volume of disk ≈ πr 2 ∆y = π(52 − y 2 )∆y mm3 .
Summing over all disks, we have
Total volume ≈
Taking the limit as ∆y → 0, we get
Total volume = lim
∆y→0
Evaluating gives
X
X
2
π(52 − y 2 )∆y mm3 .
2
π(5 − y )∆y =
Total volume = π
y3
25y −
3
Z
5
0
5
=
0
π(52 − y 2 ) dy mm3 .
250
π mm3 .
3
Check: The volume of a hemisphere can be calculated using the formula V = 32 πr 3 = 23 π53 =
r
y
Figure 8.4
5
250
π
3
mm3 .
716
Chapter Eight /SOLUTIONS
18. Each slice is a square; the side length decreases as we go up the pyramid. See Figure 8.5. Since the base of the pyramid is
equal to its vertical height, the slice at distance y from the base, or (2 − y) from the top, has side (2 − y). Thus
Volume of slice ≈ (2 − y)2 ∆y m3 .
Summing over all slices, we get
Total volume ≈
Total volume = lim
∆y→0
Evaluating, we find
Total volume =
Z
0
X
X
(2 − y)2 ∆y m3 .
2
(2 − y) ∆y =
2
(4 − 4y + y 2 ) dy =
Z
0
2
(2 − y)2 dy m3 .
4y − 2y 2 +
y3
3
2
=
0
8 3
m .
3
Check: The volume of the pyramid can also be calculated using the formula V = 13 b2 h = 13 22 · 2 =
✻
8
3
m3 .
✻
(2 − y)
❄
✻
2m
✛ (2 − y) ✲
y
❄
✛
✲
2m
❄
Figure 8.5
Problems
19. Triangle of base and height 1 and 3. See Figure 8.6. (Either 1 or 3 can be the base. A non-right triangle is also possible.)
✻
3
✛
✛
x
❄
✲✛
∆x
1
✲
Figure 8.6
20. Semicircle of radius r = 9. See Figure 8.7.
✛
✲
x ✛
✛ ∆x9 ✲
Figure 8.7
717
8.1 SOLUTIONS
21. Quarter circle of radius r =
√
15. See Figure 8.8.
y
√
✛
15 − h2
h
✲ ✛
h
∆h
Figure 8.8
22. Triangle of base and height 7 and 5. See Figure 8.9. (Either 7 or 5 can be the base. A non-right triangle is also possible.)
✻
5
❄
✛
✛
h
✲
✲✛
∆h
7
✲
Figure 8.9
23. There are at least two possible answers. Since y = x − x2 ≥ 0 when 0 ≤ x ≤ 1, one possibility is that the integral gives
the area between the parabola y = x − x2 and the line y = 0 as shown in Figure 8.10. Alternatively, since x ≥ x2 when
0 ≤ x ≤ 1, the integral gives the area between the curves y = x2 and y = x, as shown in Figure 8.11.
y
y
1
1
x
x
1
Figure 8.10
Figure 8.11
24. (a) The area is shown in Figure 8.12(a) with one vertical slice shown. Each slice is bounded on the top by y = 3x and on
the bottom by y = x2 , so the height of each slice is 3x − x2 . The width of each slice is ∆x, so the area is represented
by
A=
Z
0
3
3 2 1 3 3
x − x
2
3
0
3 2 1 3
= · 3 − 3̇ − (0 − 0)
2
3
= 4.5.
(3x − x2 ) dx =
718
Chapter Eight /SOLUTIONS
y
(a)
y
(b)
9
9
x = y/3
y = 3x
y = x2
x=
√
y
x
x
3
3
Figure 8.12
(b) The area is shown in Figure 8.12(b) with one horizontal slice shown. Each slice is bounded on the right by y = x2 ,
√
which is x = y distance out from the y-axis, and is bounded on the left by y = 3x, which is x = y/3 distance
√
out from the y-axis. The length of each slice is y − y/3. The width of each slice is ∆y, and the area extends from
y = 0 to y = 9 so the area is represented by
A=
Z
9
0
√
y−
9
y
2
1
dy = y 3/2 − y 2
3
3
6
0
2 3/2 1 2
− · 9 − (0 − 0)
= ·9
3
6
= 4.5.
Notice that the two ways of computing the area give the same answer, as we expect.
25. (a) The area is shown in Figure 8.13(a) with one vertical slice shown. Each slice is bounded on the top by y = 12 − x
and on the bottom by y = 2x, so the height of each slice is (12 − x) − 2x. The width of each slice is ∆x and the
area ranges from x = 0 to x = 4, so the area is represented by
A=
Z
4
((12 − x) − 2x) dx =
0
Z
0
4
12 − 3x dx
3 2 4
x
2
0
3 2
= 12 · 4 − · 4 − (0 − 0)
2
= 24.
= 12x −
y
(a)
12
y
(b)
12
y = 12 − x
8
x = 12 − y
8
y = 2x
x = y/2
x
x
4
4
Figure 8.13
(b) The area is shown in Figure 8.13(b) with two horizontal slices shown. Because some of the slices are bounded on the
right by y = 12 − x while others are bounded on the right by y = 2x, we use two separate integrals to calculate
this area using horizontal slices. The slices between y = 0 and y = 8 are bounded on the right by y = 2x, which is
x = y/2 distance from the y-axis, and are bounded on the left by the y-axis, which is x = 0. This part of the area is
given by
Z 8
8
1
1
y
dy = y 2 = · 82 − 0 = 16.
Bottom part of the area =
2
4
4
0
0
The slices between y = 8 and y = 12 are bounded on the right by y = 12 − x, which is x = 12 − y distance from
the y-axis, and are bounded on the left by the y-axis, which is x = 0. This part of the area is given by
Top part of the area =
Z
8
12
(12 − y) dy = 12y −
1 2
y
2
12
8
= 12 · 12 −
1
1
· 122 − 12 · 8 − · 82
2
2
= 8.
8.1 SOLUTIONS
719
The total area is given by
Total area =
Z
8
0
y
dy +
2
12
Z
(12 − y) dy = 16 + 8 = 24.
8
Notice that the two ways of computing the area give the same answer, as we expect.
26. (a) The area is shown in Figure 8.14(a) with two vertical slices shown. Because some of the slices are bounded on the
top by y = x2 while others are bounded on the top by y = 6 − x, we use two separate integrals to calculate this area
using vertical slices. The slices between x = 0 and x = 2 are bounded on the top by y = x2 and are bounded on the
bottom by the x-axis, which is y = 0. This part of the area is given by
Left part of the area =
Z
2
x2 dx =
0
1 3
x
3
2
=
0
1 3
· 2 − 0 = 2.667.
3
The slices between x = 2 and x = 6 are bounded on the top by y = 6 − x and are bounded on the bottom by the
x-axis, which is y = 0. This part of the area is given by
Z
6
Z
6
1 2 6
x
2
2
2
1 2
1 2
= 6 · 6 − · 6 − 6 · 2 − · 2 = 8.
2
2
Right part of the area =
The total area is given by
Total area =
Z
2
2
x dx +
(6 − x) dx = 6x −
(6 − x) dx = 2.667 + 8 = 10.667.
2
0
y
(a)
y
(b)
y = x2
y=
4
√
y
4
x= 6−y
y = 6−x
x
x
2
6
2
6
Figure 8.14
(b) The area is shown in Figure 8.14(b) with one horizontal slice shown. The slices are all bounded on the right by
√
y = 6 − x, which is x = 6 − y distance from the y-axis, and are bounded on the left by y = x2 , which is x = y
distance from the y-axis. The area ranges from y = 0 to y = 4 so the area is given by
A=
Z
4
0
((6 − y) −
√
1 2 2 3/2 4
y − ·y
2
3
0
1 2 2 3/2
= 6·4− ·4 − ·4
−0
2
3
= 10.667.
y) dy = 6y −
Notice that the two ways of computing the area give the same answer, as we expect.
27. (a) The area is shown in Figure 8.15(a) with two vertical slices shown. Because some of the slices are bounded on the
top by y = 2x while others are bounded on the top by y = 6, we need to use two separate integrals to calculate this
area using vertical slices. The slices between x = 0 and x = 3 are bounded on the top by y = 2x and are bounded
on the bottom by the x-axis, which is y = 0. This part of the area is given by
Left part of the area =
Z
0
3
3
2x dx = x2
0
= 32 − 0 = 9.
720
Chapter Eight /SOLUTIONS
The slices between x = 3 and x = 5 are bounded on the top by y = 6 and are bounded on the bottom by the x-axis,
which is y = 0. This part of the area is given by
Right part of the area =
Z
5
5
6 dx = 6x
3
3
= 6 · 5 − 6 · 3 = 12.
The total area is given by
Total area =
3
Z
Z
2x dx +
y
6 dx = 9 + 12 = 21.
3
0
(a)
5
y
(b)
y = 2x
y=6
6
x = y/2
y=6
6
x=5
x=5
x
3
x
5
3
5
Figure 8.15
(b) The area is shown in Figure 8.15(b) with one horizontal slice shown. The slices are all bounded on the right by x = 5
and are bounded on the left by y = 2x, which is x = y/2 distance from the y-axis. The area ranges from y = 0 to
y = 6 so the area is given by
A=
Z
6
0
(5 − (y/2)) dy = 5x −
1 2
y
4
6
0
=5·6−
1 2
· 6 − 0 = 21.
4
Notice that the two ways of computing the area give the same answer, as we expect.
28. Hemisphere with radius 12. See Figure 8.16.
12
h
√
144 − h2
✠
✻
h
❄
12
Figure 8.16
29. Cone with height 12 and radius 12/3 = 4. See Figure 8.17.
4
✛
x
✲
x
3
Figure 8.17
x
12
8.1 SOLUTIONS
721
30. Cone with height 6 and radius 3. See Figure 8.18.
y
6
3 − y/2
✠
✻
y
❄
3
Figure 8.18
31. Hemisphere with radius 2. See Figure 8.19.
y
2
✻
2
(2 − y)
❄
✻
■
y
Figure 8.19
We slice up the sphere in planes perpendicular to the x-axis. Each
√ slice is a circle, with radius y = r 2 − x2 ; that’s
the radius because x2 + y 2 = r 2 when
z = 0. Then the volume is
32.
Radius =
√
p
22 − (2 − y)2
❄
r 2 − x2
V ≈
❘
x
X
π(y 2 ) ∆x =
X
π(r 2 −x2 ) ∆x.
Therefore, as ∆x tends to zero, we get
V =
y
Z
x=r
π(r 2 − x2 ) dx
x=−r
x=r
=2
Z
x=0
π(r 2 − x2 ) dx
πx3
= 2 πr x −
3
2
r
0
4πr 3
=
.
3
33. We slice the cone horizontally into cylindrical disks with radius r and thickness ∆h. See Figure 8.20. The volume of each
disk is πr 2 ∆h. We use the similar triangles in Figure 8.21 to write r as a function of h:
3
r
=
h
12
so r =
1
h.
4
The volume of the disk at height h is π( 14 h)2 ∆h. To find the total volume, we integrate this quantity from h = 0 to
h = 12.
Z 12 2
12
1
π h3
π
h dh =
= 36π = 113.097 m3 .
V =
4
16 3 0
0
722
Chapter Eight /SOLUTIONS
✛
✲
6m
✻
✛
✻
✻
❄∆ h
✻
12 m
✲
3
r
✛✲
12 m
✻
h
h
❄
❄
❄
Figure 8.20
❄
Figure 8.21
This cone is what you get when you rotate the line x = r(h−
y)/h about the y–axis. So slicing perpendicular to the y–axis
yields
34.
Radius =
r(h−y)
h
y
V =
Z
y=h
πx2 dy = π
❘
r2
h2
h
Z
h
0
(h − y)r
h
2
dy
(h2 − 2hy + y 2 ) dy
πr 2
y3
= 2 h2 y − hy 2 +
h
3
x
0
y=0
=π
Z
h
=
0
πr 2 h
.
3
35. (a) A vertical slice has a triangular shape and thickness ∆x. See Figure 8.22.
1
1
Base · Height · ∆x = · 2 · 3∆x = 3∆x cm3 .
2
2
Thus,
∆x→0
X
3∆x =
Z
4
4
= 12 cm3 .
3 dx = 3x
0
0
∆x
✛
✛
✛
✛
✻
✛
2 cm
✛
2 cm
✛
∆x
x
✛
✻
✛
Total volume = lim
✛
Volume of slice = Area of triangle · ∆x =
3 cm
3 cm
❄
❄
Figure 8.22
(b) A horizontal slice has a rectangular shape and thickness ∆h. See Figure 8.23. Using similar triangles, we see that
3−h
w
=
,
2
3
so
w=
2
2
(3 − h) = 2 − h.
3
3
723
8.1 SOLUTIONS
Thus
2
8
h ∆h = 8 − h ∆h.
3
3
Volume of slice ≈ 4w∆h = 4 2 −
So,
Total volume = lim
∆h→0
✛
4 cm
8
h ∆h =
3
3
Z
0
8−
8
h
3
dh =
8h −
4h2
3
3
= 12 cm3 .
0
2
2 cm
✻✻
h
✛
✛
✻
✻
❄
❄
✻
4 cm
❄
❄✛
✛
✻
w
2 cm
3
Figure 8.23
36. We slice the water into horizontal slices, each of which is a rectangle. See Figure 8.24.
Volume of slice ≈ 150w∆h km3 .
To find w in terms of h, we use the similar triangles in Figure 8.25:
h
w
=
3
0.2
w = 15h.
so
So
Volume of slice ≈ 150 · 15h∆h = 2250h∆h km3 .
Summing over all slices and letting ∆h → 0 gives
Total volume = lim
∆h→0
Evaluating the integral gives
X
✛
km
150
✛
3 km
0.2
2250h dh km3 .
0
Total volume = 2250
✛
Z
2250h∆h =
h2
2
0.2
= 45 km3 .
0
✛
✲
3 km
✲
✻
w
0.2 km
w
❄
∆h
✻
✻
h
❄
Figure 8.24
❄
❄
✛
∆h
✛
✛
∆h
3 cm
8−
✛
h
✛
X
❄
Figure 8.25
w
724
Chapter Eight /SOLUTIONS
37. To calculate the volume of material, we slice the dam horizontally. See Figure 8.26. The slices are rectangular, so
Volume of slice ≈ 1400w∆h m3 .
Since w is a linear function of h, and w = 160 when h = 0, and w = 10 when h = 150, this function has slope =
(10 − 160)/150 = −1. Thus
w = 160 − h meters,
so
Volume of slice ≈ 1400(160 − h)∆h m3 .
Summing over all slices and taking the limit as ∆h → 0 gives
Total volume = lim
∆h→0
Evaluating the integral gives
X
1400(160 − h)∆h =
Total volume = 1400
h2
160h −
2
10 m
✛✲
Z
150
0
1400(160 − h) dh m3 .
150
= 1.785 · 107 m3 .
0
✛10 m✲
✻
w
0m
140
160 m
✻ ✻
✛
w
✛
❄
∆h
150 m
h
❄
✛
✲✛
Figure 8.26
160 m
✲
❄
Figure 8.27
Strengthen Your Understanding
38. The horizontal slice at height y goes from x = 0 to a point on the line y = 2x. At this point, x = y/2. The correct
R8
integral for the area is 0 y/2 dy.
39. Slice the sphere x2 +y 2√= 102 into slices of thickness ∆x perpendicular to the x-axis. The typical slice is approximately a
cylinder of radius y = 102 − x2 and height h = ∆x. The volume of the slice is approximately πy 2 h = π(102 −x2 )∆x.
R 10
The volume of the sphere is thus −10 π 102 − x2 dx.
40. It would be hard to use vertical slices to find the area of the shaded region in Figure 8.28 because the vertical slices on the
right portion of the figure are in two pieces.
y
x
Figure 8.28
8.2 SOLUTIONS
725
41. One possible answer is the region between the positive x-axis, the positive y-axis and the line y = 1 − x. For horizontal
slices, the width of a slice at height y is x = 1 − y, and for vertical slices the height of a slice at position x is y = 1 − x.
√
42. True. Since y = ± 9 − x2 represent the top and bottom halves of the sphere, slicing disks perpendicular to the x-axis
gives
Volume of slice ≈ πy 2 ∆x = π(9 − x2 )∆x
Volume =
Z
3
−3
π(9 − x2 ) dx.
43. False. Evaluating does not give the volume of a cone πr 2 h/3:
h
Z
0
π(r − y) dy = π
y2
ry −
2
h
=π
0
h2
rh −
2
Alternatively, you can show by slicing that the integral representing this volume is
Rh
0
.
πr 2 (1 − y/h)2 dy.
44. False. Using the table of integrals (VI-28 and VI-30) or a trigonometric substitution gives
Z
r
0
π
p
r 2 − y 2 dy =
π
2
p
y
r 2 − y 2 + r 2 arcsin
y
r
r
=
0
π2 r2
πr 2
(arcsin 1 − arcsin 0) =
.
2
4
The volume of a hemisphere is 2πr 3 /3.
Rr
Alternatively, you can show by slicing that the integral representing this volume is 0 π(r 2 − y 2 ) dy.
p
r 2 − y 2 . So the volume of
45. True. Horizontal slicing gives rectangular slabs of length l, thickness ∆y, and width w = 2
p
Rr p
one slab is 2l r 2 − y 2 ∆y, and the integral is −r 2l r 2 − y 2 dy.
Solutions for Section 8.2
Exercises
1. (a) The volume of a disk is given by
∆V ≈ π(2x)2 ∆x = 4πx2 ∆x,
so
Volume =
Z
3
4πx2 dx.
0
(b) We have
Volume = 4π
x3
3
3
= 36π.
0
2. (a) The volume of a disk is given by
∆V ≈ π(−x2 + 6x)2 ∆x,
so
Volume =
(b) We have
Volume =
Z
0
Z
6
π(−x2 + 6x)2 dx.
0
6
π(x4 − 12x3 + 36x2 ) dx = π
x5
− 3x4 + 12x3
5
6
=
0
1296π
.
5
726
Chapter Eight /SOLUTIONS
3. (a) The strip stretches from x = y/2 to x = 3. The volume of a disk with a hole in it is
32 −
∆V ≈ π
so
2
y
2
Volume =
Z
π
Volume =
4
6
π
(36 − y 2 ) ∆y,
4
π
(36 − y 2 ) dy.
4
0
(b) We have
∆y =
6
y3
36y −
3
= 36π.
0
4. (a) The equation y = −x2 + 6x can be solved for x as
x2 − 6x + y = 0
√
p
6 ± 36 − 4y
= 3 ± 9 − y.
x=
2
√
√
The left end of the strip is given by x = 3 − 9 − y, and the right end is given by x = 3 + 9 − y. Thus, the
volume of a disk with a hole is
9 − y)2 − (3 −
p
∆V ≈ π((3 +
p
= π((9 + 6
= 12π
so
p
9 − y)2 ) ∆y
p
p
9 − y + 9 − y) − (9 − 6
9 − y ∆y,
Volume = 12π
Z
0
(b) We have
Volume =
9 − y + 9 − y)) ∆y
9
p
9 − y dy.
12π(9 − y)3/2
−3/2
9
= 216π.
0
5. The volume is given by
1
Z
V =
1
Z
2
πy dx =
0
πx4 dx = π
0
x5
5
1
=
0
π
.
5
6. The volume is given by
V =
Z
2
2
πy dx =
1
Z
2
π(x + 1)4 dx =
1
π(x + 1)5
5
2
=
1
211π
.
5
7. The volume is given by
V =
Z
0
2 2
−2
π(4 − x ) dx = π
Z
0
2
−2
4
(16 − 8x + x ) dx = π
8x3
x5
16x −
+
3
5
0
=
−2
8. The volume is given by
V =
Z
1
√
π( x + 1)2 dx = π
Z
1
(x + 1) dx = π
−1
−1
x2
+x
2
1
= 2π.
−1
9. The volume is given by
V =
Z
1
−1
πy 2 dx =
Z
1
−1
π(ex )2 dx =
Z
1
−1
πe2x dx =
π 2x
e
2
1
=
−1
π 2
(e − e−2 ).
2
256π
.
15
727
8.2 SOLUTIONS
10. The volume is given by
V =
π/2
Z
2
πy dx =
0
Integration by parts gives
Z
π/2
π cos2 x dx.
0
π/2
π
(cos x sin x + x)
2
V =
=
0
π2
.
4
11. The volume is given by
Z
V =
1
π
0
1
x+1
2
dx = π
1
Z
0
dx
= −π(x + 1)−1
(x + 1)2
1
= π 1−
0
1
2
=
π
.
2
12. The volume is given by
V =π
Z
1
√
2
( cosh 2x) dx = π
0
1
Z
cosh 2x dx =
0
π
sinh 2x
2
1
=
0
π
sinh 2.
2
13. Since the graph of y = x2 is below the graph of y = x for 0 ≤ x ≤ 1, the volume is given by
V =
Z
0
1
πx2 dx −
Z
1
1
Z
π(x2 )2 dx = π
0
(x2 − x4 ) dx = π
0
x5
x3
−
3
5
1
=
0
2π
.
15
14. Since the graph of y = e3x is above the graph of y = ex for 0 ≤ x ≤ 1, the volume is given by
V =
Z
0
1
3x 2
π(e ) dx −
Z
1
Z
x 2
π(e ) dx =
0
1
π(e
6x
0
2x
− e ) dx = π
e6x
e2x
−
6
2
1
=π
0
e6
e2
1
−
+
6
2
3
.
15. Since f ′ (x) = x, we evaluate the integral numerically or using the table to get
√
Z 2p
ln( 5 + 2) √
2
1 + x dx =
+ 5 = 2.958.
Arc length =
2
0
16. Since f ′ (x) = − sin x, we evaluate the integral numerically to get
Arc length =
Z
2
p
1 + sin2 x dx = 2.508.
0
17. Since f ′ (x) = 1/(x + 1), we evaluate the integral numerically to get
Arc length =
Z
2
r
1+
0
1
x+1
2
dx = 2.302.
18. Note that this function is actually x3/2 in disguise. So
L=
Z
r
2
3 1
1+
x2
2
0
=
4
9
Z
w= 11
2
i2
dx =
Z
x=2
r
1+
x=0
9
x dx
4
1
w 2 dw
w=1
8 32
w
=
27
where we set w = 1 + 94 x, so dx = 49 dw.
h
11
2
1
8
=
27
11
2
32
−1
≈ 3.526,
19. This is a one-quarter of the circumference of a circle of radius 2. That circumference is 2 · 2π = 4π, so the length is
4π
= π.
4
728
Chapter Eight /SOLUTIONS
20. Since f ′ (x) = sinh x, the arc length is given by
L=
Z
2
Z
2
1 + sinh2 x dx =
0
21. The length is
1
p
Z
2
p
cosh2 x dx =
0
p
(x′ (t))2
+
(y ′ (t))2
2
2
cosh x dx = sinh x
0
(z ′ (t))2
+
Z
dt =
Z
= sinh 2.
0
2
1
p
52 + 42 + (−1)2 dt =
√
42.
This is the length of a straight line from the point (8, 5, 2) to (13, 9, 1).
22. We have
D=
Z
1
(−et sin(et ))2 + (et cos(et ))2 dt
0
=
Z
p
1
√
e2t dt =
0
Z
1
et dt
0
= e − 1.
This is the length of the arc of a unit circle from the point (cos 1, sin 1) to (cos e, sin e)—in other words between the
angles θ = 1 and θ = e. The length of this arc is (e − 1).
23. We have
D=
Z
2π
p
(−3 sin 3t)2 + (5 cos 5t)2 dt.
0
We cannot find this integral symbolically, but numerical methods show D ≈ 24.6.
24. Since dx/dt = −3 cos2 t sin t, dy/dt = 3 sin2 t cos t, we have
Arc length =
Z
0
2π
p
9 cos4 t sin2 t + 9 sin4 t cos2 t dt = 3
Z
2π
0
=3
Z
0
p
| cos t sin t|
2π
cos2 t + sin2 t dt
| cos t sin t| dt.
We calculate the integral over the interval 0 ≤ t ≤ π/2, where both cos t and sin t are positive, and multiply by 4:
Arc length = 12
Z
0
π/2
π/2
cos t sin t dt = 6 sin2 t
= 6.
0
Problems
25. The two functions intersect at (0, 0) and (8, 2). We slice the volume with planes perpendicular to the line x = 9. This
divides the solid into thin washers with volume
2
2
Volume of slice = πrout
∆y − πrin
∆y.
The outer radius is the horizontal distance from the line x = 9 to the curve x = y 3 , so rout = 9 − y 3 . Similarly, the inner
radius is the horizontal distance from the line x = 9 to the curve x = 4y, so rin = 9 − 4y. Integrating from y = 0 to
y = 2 we have
Z
2
V =
0
[π(9 − y 3 )2 − π(9 − 4y)2 ] dy.
26. The two functions intersect at (0, 0) and (8, 2). We slice the volume with planes perpendicular to the line y = 3. This
divides the solid into thin washers with
2
2
Volume of slice = πrout
∆x − πrin
∆x.
1
= 3 − 41 x. Similarly, the inner
The outer radius is the vertical distance from the line y = 3 to the curve
√ y = 4 x, so rout √
radius is the vertical distance from the line y = 3 to the curve y = 3 x, so rin = 3 − 3 x. Integrating from x = 0 to
x = 8 we have
Z 8h
√ i
1
V =
π(3 − x)2 − π(3 − 3 x)2 dx.
4
0
8.2 SOLUTIONS
729
27. Note that the lines x = 9 and y = 13 x intersect at (9, 3). We slice the volume with planes that are perpendicular to the
line y = −2. This divides the solid into thin washers with
2
2
Volume of slice = πrout
∆x − πrin
∆x.
Note that the inner radius is the vertical distance from the line y = −2 to the x-axis, so rin = 2. Similarly, the outer
radius is the vertical distance from the line y = −2 to the line y = 31 x, so rout = 2 + 31 x. Integrating from x = 0 to
x = 9 we have
Z 9h
i
1
V =
π(2 + x)2 − π22 dx.
3
0
28. Note that the lines x = 9 and y = 13 x intersect at (9, 3). We slice the volume with planes that are perpendicular to the
line x = −1. This divides the solid into thin washers with
2
2
Volume of slice = πrout
∆y − πrin
∆y.
Note that the inner radius is the horizontal distance from the line x = −1 to the line x = 3y, so rin = 1 + 3y. Similarly,
the outer radius is the horizontal distance from the line x = −1 to the line x = 9, so rout = 1 + 9. Integrating from
y = 0 to y = 3 we have
Z
3
V =
0
[π(1 + 9)2 − π(1 + 3y)2 ] dy.
29. The two functions intersect at (0, 0) and (5, 25). We slice the volume with planes perpendicular to the x-axis. This divides
the solid into thin washers with volume
Volume of slice = π((rout )2 − (rin )2 )∆x.
The outer radius is the vertical distance from the x-axis to the curve y = 5x, so rout = 5x. Similarly, the inner radius is
the vertical distance from the x-axis to the curve y = x2 , so rin = x2 . Integrating from x = 0 to x = 5 we have
V =
Z
5
0
π((5x)2 − (x2 )2 ) dx.
30. The two functions intersect at (0, 0) and (5, 25). We slice the volume with planes perpendicular to the y-axis. This divides
the solid into thin washers with volume
Volume of slice = π((rout )2 − (rin )2 )∆y.
√
√
The outer radius is the horizontal distance from the y-axis to the curve x = y, so rout = y. Similarly, the inner radius
is the horizontal distance from the y-axis to the curve x = y/5, so rin = y/5. Integrating from y = 0 to y = 25 we have
V =
Z
0
25
√
π(( y)2 − (y/5)2 ) dy.
31. The two functions intersect at (0, 0) and (5, 25). We slice the volume with planes perpendicular to the horizontal line
y = −4. This divides the solid into thin washers with volume
Volume of slice = π((rout )2 − (rin )2 )∆x.
The outer radius is the vertical distance from the line y = −4 to the curve y = 5x, so rout = 4 + 5x. Similarly, the inner
radius is the vertical distance from the line y = −4 to the curve y = x2 , so rin = 4 + x2 . Integrating from x = 0 to
x = 5 we have
Z
5
V =
0
π((4 + 5x)2 − (4 + x2 )2 ) dx.
32. The two functions intersect at (0, 0) and (5, 25). We slice the volume with planes perpendicular to the vertical line
x = −3. This divides the solid into thin washers with volume
Volume of slice = π((rout )2 − (rin )2 )∆y.
√
√
The outer radius is the horizontal distance from the line x = −3 to the curve x = y, so rout = 3 + y. Similarly, the
inner radius is the horizontal distance from the line x = −3 to the curve x = y/5, so rin = 3 + y/5. Integrating from
y = 0 to y = 25 we have
Z 25
√
V =
π((3 + y)2 − (3 + y/5)2 ) dy.
0
730
Chapter Eight /SOLUTIONS
33. One arch of the sine curve lies between x = 0 and x = π. Since d(sin x)/dx = cos x, evaluating the integral numerically
gives
Z
π
p
1 + cos2 x dx = 3.820.
Arc length =
0
√
34. The curves cross at (1, 1);
√ see Figure 8.29. The straight side has length 2 = 1.414; as this is the hypotenuse of a right
triangle with sides 1, 1, 2. The curved side has
√
Z 1p
√
ln( 5 + 2)
5
Arc length =
1 + (2x)2 dx =
+
= 1.479.
4
2
0
Thus the perimeter is 1.4142 + 1.479 = 2.893.
y
y = x2
y=x
(1, 1)
x
Figure 8.29
35. (a) Slicing the region perpendicular to the x-axis gives disks of radius y. See Figure 8.30.
Volume of slice ≈ πy 2 ∆x = π(x2 − 1)∆x.
Thus,
Total volume = lim
∆x→0
X
π(x2 − 1)∆x =
= π 9−3−
y
8
−2
3
Z
3
2
π(x2 − 1) dx = π
x3
−x
3
3
2
16π
=
.
3
x2 − y 2 = 1
y
✻
y
❄
1
2
x
3
✲✛
✻
∆x
∆x
Figure 8.30
Rbp
(b) The arc length, L, of the curve y = f (x) is given by L = a 1 + (f ′ (x))2 dx. In this problem y is an implicit
√
function of x. Solving for y gives y = x2 − 1 as the equation of the top half of the hyperbola. Differentiating gives
1
x
dy
= (x2 − 1)−1/2 (2x) = √
.
2
dx
2
x −1
Thus
Arc length =
Z
3
s
1+
2
x
√
x2 − 1
2
dx =
Z
2
3
r
x2
dx =
1+ 2
x −1
Z
2
3
r
2x2 − 1
dx = 1.48.
x2 − 1
8.2 SOLUTIONS
36.
Radius =
√
b
z
a2 −x2
a
y 2 = b2
V =
Z
2
x
a2
1−
.
a
πy 2 dx = π
−a
= 2πb
❘
b2
−a
Z
2
a
0
x
= 2πb2
y
a
Z
x2
1− 2
a
a3
a− 2
3a
dx = 2πb
= 2πb2
2
dx
x3
x− 2
3a
1
a− a
3
a
0
4
πab2 .
3
We slice the region P
perpendicular to the x–axis. The Rieπ(x3 + 1)2 ∆x. So the volume V is
mann sum we get is
the integral
=
37.
x2
a2
1−
Radius = 1 + x3
y
(y = −1)
1
Z
V =
π(x3 + 1)2 dx
−1
❘
x
1
Z
=π
(x6 + 2x3 + 1) dx
−1
=π
x4
x7
+
+x
7
2
= (16/7)π ≈ 7.18.
38.
−1
We slice the region P
perpendicular to theP
y–axis. The Riemann sum we get is
π(1 − x)2 ∆y =
π(1 − y 2 )2 ∆y.
So the volume V is the integral
z
Radius = 1 − x
V =
Z
1
π(1 − y 2 )2 dy
0
y
❘
=π
Z
1
0
(x = 1)
=π
x
(1 − 2y 2 + y 4 ) dy
2y 3
y5
y−
+
3
5
= (8/15)π ≈ 1.68.
39.
1
z
y
1
0
We take slices perpendicular to the x–axis.
Riemann
P The
sum for approximating the volume is
π sin2 x∆x. The
volume is the integral corresponding to that sum, namely
Radius = sin x
V =
Z
π
π sin2 x dx
0
1
1
= π − sin x cos x + x
2
2
❥
x
h
i
π
=
0
π2
≈ 4.935.
2
40. Slice the object into disks horizontally, as in Figure 8.31. A typical disk has thickness ∆y and radius x =
Volume of slice ≈ πx2 ∆y = πy ∆y.
Volume of solid = lim
∆y→0
X
πy ∆y =
Z
1
πy dy = π
0
y2
2
1
=
0
π
.
2
√
y. Thus
731
732
Chapter Eight /SOLUTIONS
y
1
y = x2
✻
❄∆y
x
Figure 8.31
41. Slice the object into rings vertically, as is Figure 8.32. A typical ring has thickness ∆x and outer radius y = 1 and inner
radius y = x2 .
Volume of slice ≈ π12 ∆x − πy 2 ∆x = π(1 − x4 ) ∆x.
Volume of solid = lim
∆x→0
X
4
π(1 − x ) ∆x =
y
Z
1
4
0
π(1 − x ) dx = π
x5
x−
5
1
=
4
π.
5
0
y = x2Outer radius
1
✻
Inner radius
✻
x
1
−1
✲
✛
∆x
Figure 8.32: Cross-section of solid
42. The region is cylindrical with a hole around the axis of rotation, y = −2. Slice it into rings vertically, as in Figure 8.33.
A typical ring has thickness ∆x and outer radius 1 + 2 = 3 and inner radius y + 2 = x2 + 2. Thus
Volume of slice ≈ π32 ∆x − π(x2 + 2)2 ∆x = π(5 − x4 − 4x2 ) ∆x.
Volume of solid =
Z
1
4
0
2
π(5 − x − 4x ) ∆x = π
4
x5
− x3
5x −
5
3
y
y = x2
1
x
1
Axis of rotation: y = −2
−2
❄Inner radius
−4
−5
✲
✛
❄Outer radius
∆x
Figure 8.33: Cross-section of solid
1
=
0
52π
.
15
8.2 SOLUTIONS
733
43. Slicing perpendicularly to the x-axis gives squares whose thickness is ∆x and whose side is 1 − y = 1 − x2 . See
Figure 8.34. Thus
Volume of square slice ≈ (1 − x2 )2 ∆x = (1 − 2x2 + x4 ) ∆x.
Volume of solid =
1
Z
2
x5
(1 − 2x + x ) dx = x − x3 +
3
5
2
0
1
4
=
8
.
15
0
∆x
✛
✲
y
1
y = x2
✻
✛
1−y
Base of square (standing on paper)
❄
x
1
Figure 8.34: Base of solid
44. Slicing perpendicularly to the x-axis gives semicircles whose thickness is ∆x and whose diameter is 1 − y = 1 − x2 . See
Figure 8.35. Thus
1 − x2
2
2
π
π
(1 − 2x2 + x4 ) dx =
4
4
Volume of semicircular slice ≈ π
Volume of solid =
Z
0
1
∆x =
x−
π
(1 − 2x2 + x4 ) ∆x.
4
2 3 x5
x +
3
5
1
=
2π
π 8
·
=
.
4 15
15
0
∆x
✛
✲
y
1
y = x2
✻
✛
1−y
Diameter of semicircle (standing on paper)
❄
x
1
Figure 8.35: Base of solid
45. An equilateral triangle of side s has height
√
3s/2 and
1
Area = · s ·
2
√
√
3s
3 2
=
s .
2
4
Slicing perpendicularly to the y-axis gives equilateral triangles whose thickness is ∆y and whose side is x =
See Figure 8.36. Thus
√
√
3 √ 2
3
( y) ∆y =
y ∆y.
Volume of triangular slice ≈
4
4
√
y.
734
Chapter Eight /SOLUTIONS
Volume of solid =
1
Z
√
√ 2
3
3y
y dy =
4
4 2
0
3
.
8
=
0
y
y = x2
1
✛
∆y
√
1
✲
✛
x
✻
❄
Base of triangle (standing on paper)
x
1
Figure 8.36: Base of solid
46.
This is the volume of revolution gotten from the rotating the
curve y = ex . Take slices perpendicular to the x-axis. They
will be circles with radius ex , so
z
y
r = ex
V =
❘
πe2x
=
2
rin = 3
✲
1
0
Z
=π
=π
Z
Z
[π(3 + y)2 − π32 ] dx
[(3 + ex )2 − 9] dx
1
[e2x + 6ex ] dx = π[
0
z
y
rout = 7
✲
rin = 7 −
0
π(e2 − 1)
=
≈ 10.036.
2
x=1
0
✲
e2x dx
2
2
πrout
dx − πrin
dx = π(3 + y)2 dx − π32 dx.
x=0
1
ex
1
x
(y = −3) So the integral we get from adding all these washers up is
V =
48.
Z
We slice the volume with planes perpendicular to the line y =
−3. This divides the curve into thin washers, as in Example 3 on
page 424 of the text, whose volumes are
y
✲
πy 2 dx = π
x=0
x
47.
rout = ex
x=1
Z
e2x
+ 6ex ]
2
1
0
= π[(e2 /2 + 6e) − (1/2 + 6)] ≈ 42.42.
This problem can be done by slicing the volume into washers
with planes perpendicular to the axis of rotation, y = 7, just like
in Example 3. This time the outside radius of a washer is 7, and
the inside radius is 7 − ex . Therefore, the volume V is
V =
(y = 7)
x
Z
x=1
x=0
[π72 − π(7 − ex )2 ] dx = π
1
= π 14ex − e2x
2
h
≈ 65.54.
i
1
0
Z
0
1
(14ex − e2x ) dx
1
1
= π 14e − e2 − 14 −
2
2
h
i
8.2 SOLUTIONS
735
49. We now slice perpendicular to the x-axis. As stated in the problem, the cross-sections obtained thereby will be squares,
with base length ex . The volume of one square slice is (ex )2 dx. (See Figure 8.37.) Adding the volumes of the slices
yields
Z 1
Z x=1
1
e2x
e2 − 1
y 2 dx =
e2x dx =
=
= 3.195.
Volume =
2 0
2
x=0
0
z
✛
ex
✛
✛
y
✛
ex x
Figure 8.37
50.
We slice perpendicular to the x-axis. As stated in the problem,
the cross-sections obtained thereby will be semicircles, with rax
x 2
dius e2 . The volume of one semicircular slice is 21 π e2
dx.
(Look at the picture.) Adding up the volumes of the slices yields
z
Z
Volume =
π
x=0
x
r=
x=1
✲
ex
2
1
πe2x
=
16
0
y2
π
dx =
2
8
Z
1
e2x dx
0
π(e2 − 1)
=
≈ 1.25.
16
y
51. If we revolve the region 0 ≤ y ≤ g(x) around the x-axis, the volume of the resulting solid is given by
Taking the hint, we can write our integral in this form:
Z
0
π
π(4 − 4 cos2 x) dx =
Z
π
4π 1 − cos2 x dx =
0
Z
π
4π sin2 x dx =
0
Therefore, g(x) = 2 sin x or g(x) = −2 sin x.
Z
0
Z
π
π (g(x))2 dx.
0
π
π (2 sin x)2 dx.
| {z }
(g(x))2
52. At time t = 6, the particle has traveled 3 · 6 = 18 cm. Suppose it is then at the point x = b. Then the arc length from the
origin to this point is 18 cm. Since
d 2 3/2
x
= x1/2 ,
dx 3
we have
Arc length =
Z bp
1 + (x1/2 )2 dx =
0
Z
0
b
√
1 + x dx = 18
736
Chapter Eight /SOLUTIONS
b
2
(1 + x)3/2
3
= 18
0
2
2
(1 + b)3/2 − = 18
3
3
(1 + b)3/2 = 28
b = 282/3 − 1 = 8.221.
When x = 8.221, y = 15.714.
R 120
53. We want to approximate 0 A(h) dh, where h is height, and A(h) represents the cross-sectional area of the trunk at
height h. Since A = πr 2 (circular cross-sections), and c = 2πr, where c is the circumference, we have A = πr 2 =
π[c/(2π)]2 = c2 /(4π). We make a table of A(h) based on this:
Table 8.1
height (feet)
0
20
40
60
80
100
120
Area (square feet)
53.79
38.52
28.73
15.60
2.865
0.716
0.080
We now form left & right sums using the chart:
LEFT(6) = 53.79 · 20 + 38.52 · 20 + 28.73 · 20 + 15.60 · 20 + 2.865 · 20 + 0.716 · 20
= 2804.42.
RIGHT(6) = 38.52 · 20 + 28.73 · 20 + 15.60 · 20 + 2.865 · 20 + 0.716 · 20 + 0.080 · 20
= 1730.22
So
TRAP(6) =
2
54. If y = e−x /2 , then x =
x. See Figure 8.38. So
Thus,
√
RIGHT(6) + LEFT(6)
2804.42 + 1730.22
=
= 2267.32 cubic feet.
2
2
−2 ln y. (Note that since 0 < y ≤ 1, ln y ≤ 0.) A typical slice has thickness ∆y and radius
Volume of slice = πx2 ∆y = −2π ln y ∆y.
Total volume = −2π
Since ln y is not defined at y = 0, this is an improper integral:
Total Volume = −2π
Z
Z
1
ln y dy.
0
1
0
ln y dy = −2π lim
a→0
1
= −2π lim (y ln y − y)
a→0
a
Z
1
ln y dy
a
= −2π lim (−1 − a ln a + a).
a→0
By looking at the graph of x ln x on a calculator, we see that lim a ln a = 0. Thus,
a→0
Total volume = −2π(−1) = 2π.
x
✲
Figure 8.38
∆y
737
8.2 SOLUTIONS
55. (a) We can begin by slicing the pie into horizontal slabs of thickness ∆h located at height h. To find the radius of each
slice, we note that radius increases linearly with height. Since r = 4.5 when h = 3 and r = 3.5 when h = 0, we
should have r = 3.5 + h/3. Then the volume of each slab will be πr 2 ∆h = π(3.5 + h/3)2 ∆h. To find the total
volume of the pie, we integrate this from h = 0 to h = 3:
V =π
Z
3
3.5 +
0
=π
h
3
2
dh
h3
7h2
49h
+
+
27
6
4
3
0
7(32 )
49(3)
33
+
+
=π
27
6
4
≈ 152 in3 .
(b) We use 1.5 in as a rough estimate of the radius of an apple. This gives us a volume of (4/3)π(1.5)3 ≈ 10 in3 . Since
152/10 ≈ 15, we would need about 15 apples to make a pie.
56. (a) The volume can be computed by several methods, not all of them requiring integration. We will slice horizontally,
forming rectangular slabs of length 100 cm, height ∆y, width w and integrate. See Figure 8.39.
y
✻
w
✛
✛
❄
✻
60◦
y
✛
∆y
✛ d✲
❄
∆y
60◦
cm
100
✛
x
5 cm
Figure 8.39
Figure 8.40
From the right triangle, we see
√
y
= tan 60◦ = 3
d
so
y
d= √ .
3
Thus
2y
w = 5 + 2d = 5 + √ .
3
The volume of the slab is
2y
∆V ≈ 100w∆y = 100 5 + √
3
so the total volume is given by
Volume = lim
∆y→0
=
Z
h
0
X
∆V = lim
∆y→0
2y
100 5 + √
3
X
2y
100 5 + √
3
y2
dy = 100 5y + √
3
∆y,
∆y
h
0
h2
= 100 5h + √
3
cm3 .
√
(b) The maximum value of h is h = 5 sin 60◦ = 5 3/2 cm ≈ 4.33 cm.
√
(c) The maximum volume of water that the gutter can hold is given by substituting h = 5 3/2 into the volume:
Maximum volume = 100
√ 2 , !
√
√
√
√
5 3
2500 √
5 3
5·
+
3 =
(2 3 + 3) = 1875 3 ≈ 3247.6 cm3 .
2
2
4
(d) Because the gutter is narrower at the bottom than the top, if it is filled with half the maximum possible volume of
water, the gutter will be filled to a depth of more than half of 4.33 cm.
738
Chapter Eight /SOLUTIONS
(e) We want to solve for the value of h such that
h2
Volume = 100 5h + √
3
57.
√
1
1
· 1875 3 = Vmax
2
2
=
h2
5h + √ = 16.238.
3
Solving gives h = 2.52 and h = −11.18. Since only positive values of h are meaningful, h = 2.52 cm.
y
y = ax2
H
∆y
❄
✻
x
x
p
We divide the interior of the boat into flat slabs of thickness ∆y and width 2x = 2
y
∆y.
a
q
Volume of slab ≈ 2xL∆y = 2L
We are interested in the total volume of the region 0 ≤ y ≤ H, so
Total volume = lim
∆y→0
2L
= √
a
Z
X
2L
H
(1/2)
y
a
y (1/2) dy =
0
If L and H are in meters,
Buoyancy force =
∆y =
Z
H
2L
0
4LH (3/2)
√
.
3 a
y/a. (See above.) We have
(1/2)
y
a
dy
40, 000LH (3/2)
√
newtons.
3 a
58. We can find the volume of the tree by slicing it into a series of thin horizontal cylinders of height dh and circumference
C. The volume of each cylindrical disk will then be
V = πr 2 dh = π
C
2π
2
dh =
C 2 dh
.
4π
Summing all such cylinders, we have the total volume of the tree as
1
Total volume =
4π
Z
120
C 2 dh.
0
We can estimate this volume using a trapezoidal approximation to the integral with ∆h = 20:
1
1
[20(312 + 282 + 212 + 172 + 122 + 82 )] =
(53660).
4π
4π
1
1
[20(282 + 212 + 172 + 122 + 82 + 22 )] =
(34520).
RIGHT estimate =
4π
4π
1
TRAP =
(44090) ≈ 3509 cubic inches.
4π
LEFT estimate =
8.2 SOLUTIONS
739
59. (a) The volume, V , contained in the bowl when the surface has height h is
V =
Z
h
πx2 dy.
0
4
2
However, since y = x , we have x =
√
y so that
V =
Z
0
h
√
2
π y dy = πh3/2 .
3
√
Differentiating gives dV /dh = πh
= π h. We are given that dV /dt = −6 h, where the negative sign reflects
the fact that V is decreasing. Using the chain rule we have
√
1/2
√
6
dh
dh dV
1
dV
1
=
·
=
·
= √ · (−6 h) = − .
dt
dV dt
dV /dh dt
π
π h
Thus, dh/dt = −6/π, a constant.
(b) Since dh/dt = −6/π we know that h = −6t/π + C. However, when t = 0, h = 1, therefore h = 1 − 6t/π. The
bowl is empty when h = 0, that is when t = π/6 units.
60. The problem appears complicated, because we are now working in three dimensions. However, if we take one dimension
at a time, we will see that the solution is not too difficult. For example, let’s just work at a constant depth, say 0. We
apply the trapezoid rule to find the approximate area along the length of the boat. For example, by the trapezoid rule the
= 50. Overall, at depth 0 we
approximate area at depth 0 from the front of the boat to 10 feet toward the back is (2+8)·10
2
have that the area for each length span is as follows:
Table 8.2
length span:
depth
0
0–10
10–20
20–30
30–40
40–50
50–60
50
105
145
165
165
130
We can fill in the whole chart the same way:
Table 8.3
length span:
depth
0–10
10–20
20–30
30–40
40–50
50–60
0
50
105
145
165
165
130
2
25
60
90
105
105
90
4
15
35
50
65
65
50
6
5
15
25
35
35
25
8
0
5
10
10
10
10
Now, to find the volume, we just apply the trapezoid rule to the depths and areas. For example, according to the
trapezoid rule the approximate volume as the depth goes from 0 to 2 and the length goes from 0 to 10 is (50+25)·2
= 75.
2
Again, we fill in a chart:
Table 8.4
length span:
0–10
10–20
20–30
30–40
40–50
50–60
0–2
75
165
235
270
270
220
depth
2–4
40
95
140
170
170
140
span
4–6
20
50
75
100
100
75
6–8
5
20
35
45
45
35
Adding all this up, we find the volume is approximately 2595 cubic feet.
You might wonder what would have happened if we had done our trapezoids along the depth axis first instead of
along the length axis. If you try this, you’d find that you come up with the same answers in the volume chart! For the
trapezoid rule, it does not matter which axis you choose first.
740
Chapter Eight /SOLUTIONS
61. (a) The equation of a circle of radius r around the origin is x2 +y 2 = r 2 . This means that y 2 = r 2 −x2 , so 2y(dy/dx) =
−2x, and dy/dx = −x/y. Since the circle is symmetric about both axes, its arc length is 4 times the arc length in
the first quadrant, namely
s
Z
4
(b) Evaluating this integral yields
4
Z
0
r
s
1+
r
r
1+
0
2
x
−
y
dy
dx
dx = 4
2
dx = 4
= 4r
1+
−
r
r
0
r
2
x
y
x2
dx = 4
1+ 2
r − x2
0
Z
r
0
r
Z
Z
Z
r
0
dx.
r
r2
r2
dx
− x2
1
dx = 4r(arcsin(x/r))
r 2 − x2
r
= 2πr.
0
This is the expected answer.
62. As can be seen in Figure 8.41, the region has three straight sides and one curved one. The lengths of the straight sides are
1, 1, and e. The curved side is given by the equation y = f (x) = ex . We can find its length by the formula
Z
0
1
p
1 + f ′ (x)2 dx =
Z
0
1
p
1 + (ex )2 dx =
Z
0
1
p
1 + e2x dx.
Evaluating the integral numerically gives 2.0035. The total length, therefore, is about 1 + 1 + e + 2.0035 ≈ 6.722.
y
f (x) = ex
1
x
1
Figure 8.41
63. The graph of f is a downward-opening parabola with x-intercepts at x = 0, 4, and f ′ (x) = (−x(x − 4))′ = 4 − 2x.
The arc length of the graph of f from x = 0 to x = 4 is given by
4
Z
Arc length =
q
1 + (f ′ (x))2 dx
Z0 4 q
=
1 + (4 − 2x)2 dx. because f ′ (x) = 4 − 2x
0
This integral can be simplified as
Z
4
0
p
4x2 − 16x + 17 dx.
64. We would like to write the integrand in the form
p
1+
This suggests:
f ′ (x)
2
√
x=
q
1 + (f ′ (x))2 .
= x1/2
f ′ (x) = x1/4
4
f (x) = · x5/4 + C
5
= 0.8x5/4
alternatively f ′ (x) = −x1/4
letting C = 0 for convenience.
8.2 SOLUTIONS
741
Thus,
Z
4
p
1+
1
√
4
Z
x dx =
q
1 + (0.8x5/4 )
1
′ 2
dx,
which is the arc length of the curve y = 0.8x5/4 from x = 1 to x = 4. Other possible solutions are any of the curves
y = ±0.8x5/4 + C.
65. The graph of f is concave down where f ′′ < 0:
2
f (x) = e−x .
f ′ (x) = −2xe−x
2
2
f ′′ (x) = −2e−x + 4x2 e−x
= 2e−x
2
2
2x2 − 1 .
This means that f ′′ (x) < 0 where
2x2 − 1 < 0
1
x2 < .
2
Thus, the graph is concave down for −
p
p
1/2 < x < 1/2. The arc length of this portion of the graph of f is given by
Z √ q
1/2
1 + (f ′ (x))2 dx
√
1/2
Z √1/2 q
2
=
1 + −2xe−x2 dx
√
− 1/2
Z √
Arc length =
−
1/2
=
−
√
p
1 + 4x2 e−2x2 dx.
1/2
66. The graph of f is concave down where f ′′ (x) < 0:
f (x) = x4 − 8x3 + 18x2 + 3x + 7
f ′ (x) = 4x3 − 24x2 + 36x + 3
f ′′ (x) = 12x2 − 48x + 36
= 12(x − 1)(x − 3).
′′
Hence f (x) < 0 for 1 < x < 3. The arc length of this portion of the graph of f is given by
Arc length =
=
Z
3
q
1 + (f ′ (x))2 dx
Z1 3 q
1
1 + (4x3 − 24x2 + 36x + 3)2 dx.
67. The arc length of the catenary between x = −b and x = b is 10 meters. Since
1
d
(cosh x) = sinh x = (ex − e−x ),
dx
2
we have
b
Arc length =
Z
b
=
Z
1+
−b
−b
Thus
r
1 x
(e − e−x )2 dx =
4
1p x
(e + e−x )2 dx =
2
Z
b
−b
Z
b
−b
1p
4 + e2x − 2 + e−2x dx
2
b
1 x
1
(e + e−x ) dx = (ex − e−x )
2
2
= eb − e−b .
−b
eb − e−b = 10.
Solving numerically gives b = 2.312 meters. Thus, the ends of the chain are 2(2.312) = 4.624 meters apart.
742
Chapter Eight /SOLUTIONS
68. Here are many functions which “work.”
dy
dx
• Any linear function y = mx + b “works.” This follows because
Z br
1+
a
• The function y =
x4
8
+
1
4x2
Z r
dx
dy
dx
“works”:
dy 2
1+
dy 2
dx
dx =
=
=
dx =
Z bp
1 + m2 dx = (b − a)
a
= 12 (x3 − 1/x3 ), and
s
Z
Z r
1
4
1
x3
x3 −
4
1+
x3
1
+ 3
x
1+
dy
dx
2
2
2
dx =
dx =
x4
1
− 2 + C.
8
4x
• One more function that “works” is y = ln(cos x); we have
Z r
= m is constant for such functions. So
dx. =
=
=
Z r
1+
Z r
Z
dy
dx
Z
p
1 + m2 .
Z r
x6
1
1
− + 6 dx
4
2
4x
1+
1
1
x3 + 3
2
x
dx
= − sin x/ cos x. Hence
− sin x
cos x
2
dx =
sin2 x + cos2 x
dx =
cos2 x
Z r
1+
Z r
sin2 x
dx
cos2 x
1
dx
cos2 x
1
sin x + 1
1
dx = ln
+ C,
cos x
2
sin x − 1
where the last integral comes from IV-22 of the integral tables.
Rxp
g ′ (t)2 − 1 dt, then, by the Fundamental Theorem of Calculus, f ′ (x) =
69. (a) If f (x) = 0
length of f from 0 to x is
Z
x
p
(f ′ (t))2
1+
0
dt =
Z
x
1+(
0
=
Z
x
Z
0
p
g ′ (x)2 − 1. So the arc
p
g ′ (t)2 − 1)2 dt
1 + g ′ (t)2 − 1 dt
0
=
q
p
x
g ′ (t) dt = g(x) − g(0) = g(x).
(b) If g is the arc length of any function f , then by the Fundamental Theorem of Calculus, g ′ (x) =
So if g ′ (x) < 1, g cannot be the arc length of a function.
(c) We find a function f whose arc length from 0 to x is g(x) = 2x. Using part (a), we see that
f (x) =
Z
0
x
p
(g ′ (t))2 − 1 dt =
Z
x
0
p
22 − 1 dt =
√
p
1 + f ′ (x)2 ≥ 1.
3x.
This is the equation of a line. Does it make sense to you that the arc length of a line segment depends linearly on its
right endpoint?
70. (a) For n = 1, we have
In the first quadrant, the equation is the line
|x| + |y| = 1.
x + y = 1.
By symmetry, the graph in the other quadrants gives the square in Figure 8.42.
For n = 2, the equation is of a circle of radius 1, centered at the origin:
x2 + y 2 = 1.
For n = 4, the equation is
x4 + y 4 = 1.
The graph is similar to a circle, but bulging out more. See Figure 8.42.
8.2 SOLUTIONS
743
(b) For n√= 1, the arc length is the perimeter of the square. Each side is the hypotenuse of a right triangle of sides
1, 1, 2. Thus
√
Arc length = 4 2 = 5.657.
For n = 2, the arc length is the perimeter of the circle of radius 1. Thus
Arc length = 2π · 1 = 2π = 6.283.
For n = 4, we find the arc length using the formula
L=
Z bp
1 + (f ′ (x))2 dx.
a
We find the arc length of the top half of the curve, given by y = (1 − x4 )1/4 , and double it. Since
1
−x3
dy
= (1 − x4 )−3/4 (−4x3 ) =
,
dx
4
(1 − x4 )3/4
Arc length = 2
Z
1
−1
s
1+
−x3
(1 − x4 )3/4
2
dx = 2
Z
1
−1
r
1+
x6
dx.
(1 − x4 )3/2
The integral is improper because the integral is not defined at x = ±1. Using numerical methods, we find
Arc length = 2
Z
1
−1
y
1
r
1+
✛
✛
✛
−1
x6
dx = 7.018.
(1 − x4 )3/2
x4 + y 4 = 1
x2 + y 2 = 1
|x| + |y| = 1
1
x
−1
Figure 8.42
Strengthen Your Understanding
R5
2
71. The volume is
0
π(3x) − π(2x)2 dx, since the slices are disks-with-holes.
72. The arc length is given by
R π/4 √
1 + cos2 x dx.
0
R π/4 p
0
1 + (f ′ (x))2 dx with f (x) = sin x. Thus, the correct formula for the arc length is
√
73. The curve begins at the point (0, 0) and ends at the point (2, 32). The distance between these points is 22 + 322 > 32.
Since the arc length of any curve is greater than or equal to the length of the straight line connecting the points, the arc
length of y = x5 between x = 0 and x = 2 must be greater than 32.
74. The triangular region with vertices (0, 0), (0, 1) and (1, 0) gives the exact same solid shape whether rotated about the xor y-axis.
75. One example is the region bounded by y = 2x and the x-axis for 0 ≤ x ≤ 1. When the region is rotated about the x-axis,
we get a cone of radius 2 and height 1. When it is rotated about the y-axis, we get a cone of radius 1 and height 2.
Since
1
4π
Volume around x-axis = π · 22 · 1 =
3
3
1
2π
2
Volume around y-axis = π · 1 · 2 =
,
3
3
the volume is greater around the x-axis than around the y-axis.
744
Chapter Eight /SOLUTIONS
76. The circle of radius 5 centered at (5, 0) goes through both the points (0, 0) and (10, 0). The upper semicircle and the lower
semicircle are different curves between these two points that have the same arc length, namely half the circumference of
the circle.
√
77. The distance
from (0, 0) to (1, 1) is 2, so any curve other than a straight line between the points has an arc length greater
√
than 2. One possible example is f (x) = x2 .
78. False. The volume also depends on how far away the region is from the axis of revolution. For example, let R be the
rectangle 0 ≤ x ≤ 8, 0 ≤ y ≤ 1 and let S be the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 2. Then rectangle R has area greater
than rectangle S. However, when you revolve R about the x-axis you get a cylinder, lying on its side, of radius 1 and
length 8, which has volume 8π. When you revolve S about the x-axis, you get a cylinder of radius 2 and length 3, which
has volume 12π. Thus the second volume is larger, even though the region revolved has smaller area.
79. False. Suppose that the graph of f starts at the point (0, 100) and then goes down to (1, 0) and from there on goes along
the x-axis. For example, if f (x) = 100(x − 1)2 on the interval [0, 1] and f (x) = 0 on the interval [1, 10], then f is
differentiable on the interval [0, 10]. The arc length of the graph of f on the interval [0, 1] is at least 100, while the arc
length on the interval [1, 10] is 9.
80. p
True. Since f is concave
up, f ′ is an increasing function, so f ′ (x) ≥ f ′ (0) = 3/4 on the interval [0, 4]. Thus
p
1 + (f ′ (x))2 ≥ 1 + 9/16 = 5/4. Then we have:
Arc length =
Z
0
4
p
1 + (f ′ (x))2 dx ≥
Z
0
4
5
dx = 5.
4
81. False. Since f is concave down, this means that f ′ (x) is decreasing, so f ′ (x) ≤ f ′ (0) = 3/4 on the interval [0, 4].
However, it could be that f ′ (x) becomes negative so that (f ′ (x))2 becomes large, making the integral for the arc length
large also. For example, f (x) = (3/4)x − x2 is concave down and f ′ (0) = 3/4, but f (0) = 0 and f (4) = −13, so the
graph of f on the interval [0, 4] has arc length at least 13.
Solutions for Section 8.3
Exercises
√
1. With r = 1 and θ = 2π/3, we find x = r cos√θ = 1 · cos(2π/3) = −1/2 and y = r sin θ = 1 · sin(2π/3) = 3/2.
The rectangular coordinates are (−1/2, 3/2).
√
√
√
√
√
2. √
With r = 3 and θ√= −3π/4,
we find
√
√x = r cos θ = 3 cos(−3π/4) = 3(− 2/2) = − 6/2 and y = r sin θ =
3 sin(−3π/4) = 3(− 2/2) = − √
6/2.
√
The rectangular coordinates are (− 6/2, − 6/2).
√
√
√ √
θ = −π/6, we √
find x = r cos θ = 2 3 cos(−π/6) = 2 3 · 3/2 = 3 and y = r sin θ =
3. With
√ r = 2 3 and √
2 3 sin(−π/6) = 2 3(−1/2) = − 3. √
The rectangular coordinates are (3, − 3).
√
√
4. With r = 2 and θ = 5π/6, we find x = r cos θ = 2 cos(5π/6) = 2(− 3/2) = − 3 and y = r sin θ = 2 sin(5π/6) =
2(1/2) = 1.
√
The rectangular coordinates are (− 3, 1).
p
√
√
5. With x = 1 and y = 1, find r from r = x2 + y 2 = 12 + 12 = 2. Find θ from tan θ = y/x = 1/1 = 1. Thus,
√
θ = tan−1 (1) = π/4. Since (1, 1) is in the first quadrant this is a correct θ. The polar coordinates are ( 2, π/4).
p
p
6. With x = −1 and y = 0, find r = x2 + y 2 = (−1)2 + 02 = 1. Find θ from tan θ = y/x = 0/(−1) = 0. Thus,
θ = tan−1 (0) = 0. Since (−1, 0) is on the x-axis between the second and third quadrant, θ = π. The polar coordinates
are (1, π).
p√
√
√
√
√
√
√ √
7. With√x = 6 and y = − 2, find
r = ( 6)2 + (−
2)2√= 8 = 2 2. Find θ from tan θ = y/x = − 2/ 6 =
√
√
−1
−1/ 3. Thus, θ =
√tan (−1/ 3) = −π/6. Since ( 6, − 2) is in the fourth quadrant, this is the correct θ. The polar
coordinates are (2 2, −π/6).
p √
√
√
√
8. With x = − 3√and y = 1, find r = √ (− 3)2 + 12 = 4 = 2. Find θ from tan θ = y/x = 1/(− 3). Thus,
−1
θ = tan (−1/ 3) = −π/6. Since (− 3, 1) is in the second quadrant, θ = −π/6 + π = 5π/6. The polar coordinates
are (2, 5π/6).
745
8.3 SOLUTIONS
9. (a) Table 8.5 contains values of r = 1 − sin θ, both exact and rounded to one decimal.
Table 8.5
θ
0
r
1 1−
r
1
π/3
π/2
2π/3
π
4π/3
3π/2
5π/3
2π
7π/3
5π/2
8π/3
√
√
√
√
√
√
2
1 + 3/2 1 1 − 3/2
3/2 0 1 − 3/2 1 1 + 3/2
0
1 − 3/2
0.134
0
0.134
1
1.866
2
1.866
1
0.134
0.134
0
(b) See Figure 8.43.
y
y
2
2
1
1
x
−2
−1
1
x
2
−2
−1
1
−1
−1
−2
−2
Figure 8.43
Figure 8.44
2
(c) The circle has equation r = 1/2. The cardioid is r = 1 − sin θ. Solving these two simultaneously gives
1/2 = 1 − sin θ,
or
sin θ = 1/2.
√
Thus, θ = π/6 or 5π/6. This gives the points
(x,
y)
= ((1/2) cos π/6, (1/2) sin π/6) = ( 3/4, 1/4) and (x, y) =
√
((1/2) cos 5π/6, (1/2) sin 5π/6) = (− 3/4, 1/4) as the location of intersection.
(d) The curve r = 1 − sin 2θ, pictured in Figure 8.44, has two regions instead of the one region that r = 1 − sin θ has.
This is because 1 − sin 2θ will be 0 twice for every 2π cycle in θ, as opposed to once for every 2π cycle in θ for
1 − sin θ.
10. There will be n loops. See Figures 8.45-8.48.
Figure 8.45: n = 1
Figure 8.46: n = 2
Figure 8.47: n = 3
Figure 8.48: n = 4
11. The graph will begin to draw over itself for any θ ≥ 2π so the graph will look the same in all three cases. See Figure 8.49.
Figure 8.49
746
Chapter Eight /SOLUTIONS
12. The curve will be a smaller loop inside a larger loop with an intersection point at the origin. Larger n values increase the
size of the loops. See Figures 8.50-8.52.
−1
1
−1
−2
−3
Figure 8.50: n = 2
−1
1
−1
−2
−3
−4
Figure 8.51: n = 3
−1
1
−1
−2
−3
−4
−5
Figure 8.52: n = 4
13. See Figures 8.53 and 8.54. The first curve will be similar to the second curve, except the cardioid (heart) will be rotated
clockwise by 90◦ (π/2 radians). This makes sense because of the identity sin θ = cos(θ − π/2).
Figure 8.53: r = 1 − cos θ
Figure 8.54: r = 1 − sin θ
14. Let 0 ≤ θ ≤ 2π and 3/16 ≤ r ≤ 1/2.
15. A loop starts and ends at the origin, that is, when r = 0. This happens first when θ = π/4 and next when θ = 5π/4. This
can also be seen by using a trace mode on a calculator. Thus restricting θ so that π/4 ≤ θ ≤ 5π/4 will graph the upper
loop only. See Figure 8.55. To show only the other loop use 0 ≤ θ ≤ π/4 and 5π/4 ≤ θ ≤ 2π. See Figure 8.56.
Figure 8.55: π/4 ≤ θ ≤ 5π/4
Figure 8.56: 0 ≤ θ ≤ π/4 and
5π/4 ≤ θ ≤ 2π
16. (a) Let 0 ≤ θ ≤ π/4 and 0 ≤ r ≤ 1.
√
√
(b) Break the√region into two pieces: one with 0 ≤ x ≤ 2/2 and 0 ≤ y ≤ x, the other with 2/2 ≤ x ≤ 1 and
0 ≤ y ≤ 1 − x2 .
√
√
17. The region is given by 8 ≤ r ≤ 18 and π/4 ≤ θ ≤ π/2.
8.3 SOLUTIONS
18. The region is given by 0 ≤ r ≤ 2 and −π/6 ≤ θ ≤ π/6.
747
19. The circular arc has equation r = 1, for 0 ≤ θ ≤ π/2. the vertical line x = 2 has polar equation r cos θ = 2, or
r = 2/ cos θ. So the region is described by 0 ≤ θ ≤ π/2 and 1 ≤ r ≤ 2/ cos θ.
20. Expressing x and y in terms of θ, we have
x = 2 cos θ
and
y = 2 sin θ.
The slope is given by
−2 cos θ
cos θ
dy
=
=−
.
dx
2 sin θ
sin θ
At θ = π/4, we have
dy
dx
=
θ=π/4
√
1/ 2
√ = −1.
1/ 2
21. Expressing x and y in terms of θ, we have
x = eθ cos θ
and
y = eθ sin θ.
The slope is given by
eθ sin θ + eθ cos θ
sin θ + cos θ
dy
= θ
=
.
dx
e cos θ − eθ sin θ
cos θ − sin θ
At θ = π/2, we have sin θ = 1 and cos θ = 0, so
dy
dx
=
θ=π/2
1+0
= −1.
0−1
22. Expressing x and y in terms of θ, we have
x = (1 − cos θ) cos θ
and
y = (1 − cos θ) sin θ.
The slope is given by
−(1 − cos θ) cos θ − sin θ sin θ
dy
=
.
dx
(1 − cos θ) sin θ − sin θ cos θ
At θ = π/2, we have cos θ = 0 and sin θ = 1, so that
dy
dx
θ=π/2
x = eθ cos θ
and
= −1.
23. The curve is given parametrically by
y = eθ sin θ.
Thus, calculating dx/dθ and dy/dθ, gives
Arc length =
=
=
=
Z
Z
Z
π
π/2
π
p
(eθ cos θ − eθ sin θ)2 + (eθ sin θ + eθ cos θ)2 dθ
eθ
π/2
π
p
(cos θ − sin θ)2 + (sin θ + cos θ)2 dθ
√
eθ 2 dθ
π/2
√
2(eπ − eπ/2 ).
748
Chapter Eight /SOLUTIONS
24. The curve is given parametrically by
x = θ2 cos θ
y = θ2 sin θ.
and
Thus, calculating dx/dθ and dy/dθ, gives
Arc length =
Z
2π
(2θ cos θ − θ2 sin θ)2 + (2θ sin θ + θ2 cos θ)2 dθ
0
=
Z
2π
Z
p
4θ2 + θ4 dθ
0
=
p
2π
θ
0
=
p
4 + θ2 dθ
1
1
(4 + 4π 2 )3/2 − 43/2 .
3
3
Problems
25. The formula for area is
1
A=
2
Therefore, since
Z
Z
β
r 2 dθ
α
π/3
Z
π/3
1
1
sin2 (3θ) dθ =
(sin 3θ)2 dθ,
A=
2 0
2 0
we have r = sin 3θ. The integral represents the shaded area inside one petal of the three-petaled rose curve, r = sin 3θ,
in Figure 8.57.
y
1
x
−1
1
−1
Figure 8.57: Graph of r = sin 3θ
26. The spiral is shown in Figure 8.58.
Area =
1
2
Z
2π
θ2 dθ =
0
1 3
θ
6
2π
=
0
y
π/2
−π
x
2π
Figure 8.58: Spiral r = θ
8π 3
.
6
8.3 SOLUTIONS
749
27. The region between the spirals is shaded in Figure 8.59.
Area =
1
2
Z
0
2π
((2θ)2 − θ2 ) dθ =
1
2
2π
Z
3θ2 dθ =
0
1 3
θ
2
2π
= 4π 3 .
0
y
2π
4π
x
Figure 8.59: Region between the
inner spiral, r = θ, and the outer
spiral, r = 2θ
28. The cardioid is shown in Figure 8.60. The following integral can be evaluated using a calculator or by parts or using the
table of integrals.
Area =
1
2
1
=
2
Z
2π
(1 + cos θ)2 dθ =
0
1
2
2π
Z
1 + 2 cos θ + cos2 θ dθ
0
1
1
θ + 2 sin θ + cos θ sin θ + θ
2
2
2π
=
0
1
3π
(2π + 0 + 0 + π) =
.
2
2
y
x
Figure 8.60: Cardioid
r = 1 + cos θ
29. (a) See Figure 8.61. In polar coordinates, the line x = 1 is r cos θ = 1, so its equation is
r=
1
.
cos θ
The circle of radius 2 centered at the origin has equation
r = 2.
(b) The line and circle intersect where
1
=2
cos θ
1
cos θ =
2
π π
θ=− , .
3 3
Thus,
1
Area =
2
Z
π/3
−π/3
2
2 −
1
cos θ
2
dθ.
750
Chapter Eight /SOLUTIONS
(c) Evaluating gives
Z
1
2
Area =
π/3
−π/3
4−
1
cos2 θ
1
(4θ − tan θ)
2
dθ =
π/3
=
−π/3
4π √
− 3.
3
y
θ = π/3
1
2
x
θ = −π/3
x=1
Figure 8.61
Rβ
30. Using the formula Area= (1/2)
α
(f (θ))2 dθ gives
Area =
1
2
Z
2π
a2 dθ =
0
2π
a2
θ
2
= πa2 ,
0
which is the formula for the area of a disk of radius a.
31. Using the formula Arc length =
Rβp
(dx/dθ)2 + (dy/dθ)2 dθ, where x = a cos θ, y = a sin θ and a > 0 gives
α
Arc length =
Z
2π
p
0
(−a sin θ)2 + (a cos θ)2 dθ =
Z
2π
a dθ = aθ|2π
0 = 2πa,
0
which is the formula for the circumference of a circle of radius a.
32. See Figure 8.62. Notice that the curves intersect at (1, 0), where θ = 0, 2π, and at (−1, 0), where θ = π, so
1
Area =
2
Z
π
2π
1
(1 − (1 + sin θ) ) dθ =
2
2
2
Z
2π
(−2 sin θ − sin2 θ) dθ.
π
Using a calculator, integration by parts, or formula IV-17 in the integral table, we have
Area =
1
2
2 cos θ +
1
1
sin θ cos θ − θ
2
2
2π
=
π
1
1
π
2·2+0− π = 2− .
2
2
4
y
r = 1 + sin θ
−1
1
r=1
Figure 8.62
x
8.3 SOLUTIONS
751
33. The two curves intersect where
1
2
1
sin θ =
2
π 5π
θ= ,
.
6 6
1 − sin θ =
See Figure 8.63. We find the area of the right half and multiply that answer by 2 to get the entire area. The integrals can be
computed numerically with a calculator or, as we show, using integration by parts or formula IV-17 in the integral tables.
1
Area of right half =
2
Z
1
2
Z
1
2
Z
=
=
−π/2
π/6
−π/2
π/6
−π/2
2
(1 − sin θ) −
2
1
2
dθ
1 − 2 sin θ + sin2 θ −
1
4
3
− 2 sin θ + sin2 θ
4
dθ
dθ
1
1
3
θ + 2 cos θ − sin θ cos θ + θ
4
2
2
√
7 3
1 5π
+
.
=
2
6
8
=
1
2
π/6
Thus,
Total area =
π/6
−π/2
√
5π
7 3
+
.
6
8
y
r = 1/2
1
2
1
2
π/6
r = 1 − sin θ
x
Figure 8.63
34. Figure 8.64 shows the curves which touch at (2, 0) and the origin. However, the circle lies entirely inside the cardioid, so
we find the area by subtracting the area of the circle from that of the cardioid. To find the areas, we take the integrals.
The cardioid, r = 1 + cos θ, starts at (2, 0) when θ = 0 and traces the top half, reaching the origin when θ = π.
Thus
Z π
1
(1 + cos θ)2 dθ.
Area of cardioid = 2 ·
2 0
The circle starts at (2, 0) when θ = 0 and traces the top half, reaching the origin when θ = π/2. Thus
Area of circle = 2 ·
1
2
Z
0
π/2
(2 cos θ)2 dθ.
752
Chapter Eight /SOLUTIONS
The area, A, we want is therefore
Area = 2 ·
=
Z
1
2
π
0
Z
0
π
(1 + cos θ)2 dθ − 2 ·
1
2
(1 + 2 cos θ + cos2 θ) dθ −
Z
Z
π/2
(2 cos θ)2 dθ
0
π/2
4 cos2 θ dθ
0
π
π/2
4
1
(sin θ cos θ + θ)
− (sin θ cos θ + θ)
2
2
0
0
π
π
3
= π−2· = .
2
2
2
Alternatively, we could compute the area of the cardioid and subtract the area of the circle of radius 1 from it.
The integrals can be computed numerically using a calculator, or, as we show, using integration by parts or formula IV-18 from the integral tables.
=
θ + 2 sin θ +
y
r = 1 + cos θ
r = 2 cos θ
x
Figure 8.64
35. (a) The graph of r = 2 cos θ is a circle of radius 1 centered at (1, 0); the graph of r = 2 sin θ is a circle of radius 1
centered at (0, 1). See Figure 8.65.
(b) The Cartesian coordinates of the points of intersection are at (0, 0) and (1, 1).
The origin corresponds
to θ = π/2 on r = 2 cos θ and to θ = 0 on r = 2 sin θ. The point (1, 1) has polar
√
coordinates r = 2, θ = π/4.
We find the area below the line θ = π/4 and above r = 2 sin θ and double it:
1
Area = 2 ·
2
Z
π/4
2
(2 sin θ) dθ = 4
0
Z
π/4
sin2 θ dθ.
0
Using a calculator, integration by parts or formula IV-17 from the integral tables,
θ
1
Area = 4 − sin θ cos θ +
2
2
y
2
π/4
= −2 ·
0
r = 2 sin θ
θ = π/4
1
r = 2 cos θ
1
Figure 8.65
2
x
π
π
1
+ 2 = − 1.
2
4
2
8.3 SOLUTIONS
753
36. The area is
A=
1
2
Z
a
1
2
r 2 dθ =
0
1
2
Z
a
θ2 dθ = 1
0
θ3
3
a
=1
0
3
a
=1
6
a3 = 6
√
3
a = 6.
37. (a) See Figure 8.66.
(b) The curves intersect when r 2 = 2
4 cos 2θ = 2
1
cos 2θ = .
2
In the first quadrant:
π
π
so θ = .
3
6
Using symmetry, the area in the first quadrant can be multiplied by 4 to find the area of the total bounded region.
2θ =
Area = 4
Z
1
2
π/6
(4 cos 2θ − 2) dθ
0
π/6
4 sin 2θ
− 2θ
2
0
2
π
= 4 sin − π
3
3
√
2
3
− π
=4
2
3
√
2
= 2 3 − π = 1.370.
3
= 2
y
1.5
r=
−2
√
2
2
x
r 2 = 4 cos 2θ
−1.5
Figure 8.66
√
38. The slope of the tangent √
line at θ = π/3 is dy/dx = 3/5. Since x = 3 sin(2θ) cos θ and y = 3 sin(2θ) sin θ, when
θ = π/3, we have x = 3 3/4 and y = 9/4. Thus, the equation of the tangent line is
√
√
9
3
3 3
y− =
x−
4
5
4
√
9
9
3
x−
+
y=
5
20
4
√
3
9
y=
x+ .
5
5
754
Chapter Eight /SOLUTIONS
39. We first find the points with horizontal and vertical tangents in the first quadrant and then use symmetry to obtain the
points in other quadrants.
The slope of the tangent line is
6 cos(2θ) sin θ + 3 sin(2θ) cos θ
dy
=
.
dx
6 cos(2θ) cos θ − 3 sin(2θ) sin θ
The curve has a horizontal tangent where
6 cos(2θ) sin θ + 3 sin(2θ) cos θ = 0.
Solving this equation numerically for 0 < θ < π/2, we have θ = 0.9553; in addition θ = 0 is a solution. Thus, there
are horizontal tangents where x = 1.633 and y = 2.309 and where x = 0, y = 0. Thus, the five points with horizontal
tangents are
(1.633, 2.309);
(−1.633, 2.309);
(−1.633, −2.309);
(1.633, −2.309);
(0, 0).
The curve has vertical tangents where
6 cos(2θ) cos θ − 3 sin(2θ) sin θ = 0.
Solving this equation numerically for 0 < θ < π/2, we have θ = 0.6155; in addition θ = π/2 is a solution. Thus, there
are vertical tangents where x = 2.309, y = 1.633, and where x = 0, y = 0. Thus, there are five points with vertical
tangents:
(2.309, 1.633); (−2.309, 1.633); (−2.309, −1.633); (2.309, −1.633); (0, 0).
40. We can express x and y in terms of θ as a parameter. Since r = θ, we have
x = r cos θ = θ cos θ
y = r sin θ = θ sin θ.
and
Calculating the slope using the parametric formula,
dy/dθ
dy
=
,
dx
dx/dθ
we have
Horizontal tangents occur where dy/dx = 0, so
sin θ + θ cos θ
dy
=
.
dx
cos θ − θ sin θ
sin θ + θ cos θ = 0
θ = − tan θ.
Solving this equation numerically gives
θ = 0, 2.029, 4.913.
Vertical tangents occur where dy/dx is undefined, so
cos θ − θ sin θ = 0
θ=
1
= cot θ.
tan θ
Solving this equation numerically gives
θ = 0.860, 3.426.
8.3 SOLUTIONS
755
41. (a) Expressing x and y parametrically in terms of θ, we have
x = r cos θ =
cos θ
θ
and
y = r sin θ =
sin θ
.
θ
The slope of the tangent line is given by
dy/dθ
dy
=
=
dx
dx/dθ
At θ = π/2, we have
θ cos θ − sin θ
θ2
dy
dx
.
=
θ=π/2
−θ sin θ − cos θ
θ2
=
sin θ − θ cos θ
.
cos θ + θ sin θ
1 − (π/2)0
2
= .
0 + (π/2)1
π
At θ = π/2, we have x = 0, y = 2/π, so the equation of the tangent line is
y=
(b) As θ → 0,
2
2
x+ .
π
π
cos θ
→ ∞ and
θ
Thus, y = 1 is a horizontal asymptote. See Figure 8.67.
x=
y=
sin θ
→ 1.
θ
y
y=1
r = 1/θ
x
Figure 8.67
42. The limaçon is given by r = 1 + 2 cos θ; see Figure 8.68. At θ = 0, the graph is at (3, 0); as θ increases, the graph sweeps
out the top arc (on which the maximum value of y occurs), reaching the origin when
1 + 2 cos θ = 0
1
2
2π
θ=
.
3
cos θ = −
Thus, we want to find the maximum value of y on the interval 0 ≤ θ ≤ 2π/3. Since y = r sin θ, we want to find the
maximum value of
y = (1 + 2 cos θ) sin θ = sin θ + 2 cos θ sin θ.
At a critical point
dy
= cos θ − 2 sin2 θ + 2 cos2 θ = 0
dθ
cos θ − 2(1 − cos2 θ) + 2 cos2 θ = 0
4 cos2 θ + cos θ − 2 = 0
√
−1 ± 33
= 0.593, −0.843.
cos θ =
8
Thus, θ = cos−1 (0.593) = 0.936 and θ = cos−1 (−0.843) = 2.574 are the critical values. Since 2.574 is outside the
interval 0 ≤ θ ≤ 2π/3, there is one critical point θ = 0.963.
At the endpoints of the interval, y = 0. At θ = 0.936, we have y = 1.760, which is the maximum value.
756
Chapter Eight /SOLUTIONS
y
r = 1 + 2 cos θ
x
Figure 8.68: The inner loop has r < 0
43. Since x = θ cos θ and y = θ sin θ, we have
Arc length =
Z
2π
p
sin θ)2
(cos θ − θ
0
+ (sin θ + θ
cos θ)2
dθ =
Z
2π
1 + θ2 dθ = 21.256.
0
44. Since x = cos θ/θ and y = sin θ/θ, we have
2π
Z
Arc length =
π
r
Z 2π r
=
π
−θ sin θ − cos θ
θ2
2
+
θ2 + 1
dθ = 0.712.
θ4
p
θ cos θ − sin θ
θ2
2
dθ
45. Parameterized by θ, the curve r = f (θ) is given by x = f (θ) cos θ and y = f (θ) sin θ. Then
Arc length =
=
=
α
β
r
α
β
p
Z
β
Z
Z
=
β
+
dy
dθ
2
dθ
(f ′ (θ) cos θ − f (θ) sin θ)2 + (f ′ (θ) sin θ + f (θ) cos θ)2 dθ
p
+(f ′ (θ))2 sin2 θ + 2f ′ (θ)f (θ) sin θ cos θ + (f (θ))2 cos2 θ dθ
p
(f ′ (θ))2 (cos2 θ + sin2 θ) + (f (θ))2 (sin2 θ + cos2 θ) dθ
α
β
=
2
(f ′ (θ))2 cos2 θ − 2f ′ (θ)f (θ) cos θ sin θ + (f (θ))2 sin2 θ
α
Z
dx
dθ
Z
p
(f ′ (θ))2 + (f (θ))2 dθ.
α
46. Let f (θ) = θ. Then f ′ (θ) = 1. Using the formula derived in Problem 45, we have
Arclength =
Z
0
2π
p
1 + θ2 dθ =
h
iπ
p
1 p
1
θ 1 + θ2 + ln |θ + 1 + θ2 |
2
2
=
0
47. Let f (θ) = 1 + cos θ. Then f ′ (θ) = − sin θ. Using numerical methods, we have
Arclength =
Z
π/2
p
(− sin θ)2
0
+ (1 +
cos θ)2
dθ =
Z
π/2
√
p
1
1 p
π 1 + π 2 + ln |π + 1 + π 2 |
2
2
2 + 2 cos θ dθ = 2.828.
0
Strengthen Your Understanding
48. The formula θ = tan−1 (y/x) is incorrect if x ≤ 0. If x = 0, then tan−1 (y/x) is not defined and so cannot be
used. If x < 0, then the point (x, y) is in quadrant II or III, but tan−1 (y/x) is an angle in quadrant I or IV, because
−π/2 < tan−1 (y/x) < π/2.
8.4 SOLUTIONS
757
49. The points on the polar curve with π/2 < θ < π are in quadrant IV, because r = sin(2θ) < 0.
50. Since dr/dθ is the rate of change of r with respect to θ, it cannot be interpreted as the slope of the polar curve. The circle
r = 1 has positive slope at θ = 3π/4, yet dr/dθ = 0 everywhere, since the radius does not change as θ increases.
51. The rose with four petals, r = 3 sin 2θ, shown in Figure 8.69 is symmetric about both axes.
y
r = 3 sin 2θ
x
Figure 8.69
52. Changing the polar coordinate θ by adding 2π does not change the point. For example, (r, θ) = (10, π) and (r, θ) =
(10, 3π) correspond to the same point (x, y) = (−10, 0).
53. The circle of radius k centered at the origin has equation r = k. For example, r = 100.
54. One example is the Archimedean spiral r = θ.
55. A polar curve is symmetric about the x-axis if replacing θ by −θ in its formula makes no difference, so any formula
involving only cos θ will work. One such example is the limaçon r = 1 + cos θ.
Solutions for Section 8.4
Exercises
1. Since density is e−x gm/cm,
Mass =
Z
0
10
10
e−x dx = −e−x
= 1 − e−10 gm.
0
2. Strips perpendicular to the x-axis have length 3, area 3∆x, and mass 5 · 3∆x gm. Thus
Mass =
Z
2
Z
3
5 · 3 dx =
0
Z
2
Z
3
15 dx.
0
Strips perpendicular to the y-axis have length 2, area 2∆y, and mass 5 · 2∆y gm. Thus
Mass =
0
5 · 2 dy =
10 dy.
0
3. (a) Suppose we choose an x, 0 ≤ x ≤ 2. If ∆x is a small fraction of a meter, then the density of the rod is approximately
δ(x) anywhere from x to x + ∆x meters from the left end of the rod (see below). The mass of the rod from x to
x + ∆x meters is therefore approximately δ(x)∆x = (2 + 6x)∆x. If we slice the rod into N pieces, then a Riemann
N
X
(2 + 6xi)∆x.
i=1
x
✛
✛
✛
✛
sum is
∆x
2
758
Chapter Eight /SOLUTIONS
(b) The definite integral is
M=
2
Z
δ(x) dx =
0
Z
2
2
(2 + 6x) dx = (2x + 3x2 )
0
= 16 grams.
0
4. We have
Moment =
=
Z
Z
2
xδ(x) dx =
2
Z
x(2 + 6x) dx
0
0
2
2
(6x2 + 2x) dx = (2x3 + x2 )
0
= 20 gram-meters.
0
Now, using this and Problem 3 (b), we have
20 gram-meters
5
Moment
=
= meters (from its left end).
Center of mass =
Mass
16 grams
4
5. (a) Figure 8.70 shows a graph of the density function.
900
300
x
20
Figure 8.70
(b) Suppose we choose an x, 0 ≤ x ≤ 20. We approximate the density of the number of the cars between x and x + ∆x
miles as δ(x) cars per mile. Therefore, the number of cars between x and x + ∆x is approximately δ(x)∆x. If we
slice the 20 mile strip into N slices, we get that the total number of cars is
C≈
N
X
δ(xi )∆x =
N
X
√
600 + 300 sin(4 xi + 0.15) ∆x,
i=1
i=1
where ∆x = 20/N . (This is a right-hand approximation; the corresponding left-hand approximation is
Z
20
X
δ(xi )∆x.)
i=0
(c) As N → ∞, the Riemann sum above approaches the integral
C=
N−1
√
(600 + 300 sin 4 x + 0.15) dx.
0
If we calculate the integral numerically, we find C ≈ 11513. We can also find the integral exactly as follows:
C=
Z
20
√
(600 + 300 sin 4 x + 0.15) dx
0
=
Z
20
600 dx +
0
Z
= 12000 + 300
Let w =
√
2
20
Z
0
20
√
300 sin 4 x + 0.15 dx
√
sin 4 x + 0.15 dx.
0
x + 0.15, so x = w − 0.15 and dx = 2w dw. Then
Z
x=20
x=0
√
sin 4 x + 0.15 dx = 2
Z
w=
w=
√
√
20.15
w sin 4w dw, (using integral table III-15)
0.15
1
1
sin 4w
= 2 − w cos 4w +
4
16
h
≈ −1.624.
√
i
√
20.15
0.15
Using this, we have C ≈ 12000 + 300(−1.624) ≈ 11513, which matches our numerical approximation.
8.4 SOLUTIONS
759
6. (a) Orient the rectangle in the coordinate plane in such a way that the side referred to in the problem—call it S—lies on
the y-axis from y = 0 to y = 5, as shown in Figure 8.71. We may subdivide the rectangle into strips of width ∆x
and length 5. If the left side of a given strip is a distance x away from S (i.e., the y-axis), its density 2 is 1/(1 + x4 ).
If ∆x is small enough, the density of the strip is approximately constant—i.e., the density of the whole strip is about
1/(1 + x4 ). The mass of the strip is just its density times its area, or 5∆x/(1 + x4 ). Thus the mass of the whole
rectangle is approximated by the Riemann sum
X 5∆x
1 + x4
,
y
∆x
5
S
3
x
Figure 8.71
(b) The exact mass of the rectangle is obtained by letting ∆x → 0 in the Riemann sums above, giving us the integral
Z
3
0
5 dx
.
1 + x4
Since it is not easy to find an antiderivative of 5/(1 + x4 ), we evaluate this integral numerically, getting 5.5.
7. The total mass is 7 grams. The center of mass is given by
2(−3) + 5(4)
= 2 cm to right of origin.
7
x=
8. The total mass is 9 gm, and so the center of mass is located at x = 19 (−10 · 5 + 1 · 3 + 2 · 1) = −5.
9. We slice the block horizontally. A slice has area 10·3 = 30 and thickness ∆z. On such a slice, the density is approximately
constant. Thus
Mass of slice ≈ Density · Volume ≈ (2 − z) · 30∆z,
so we have
Mass of block ≈
X
(2 − z)30∆z.
In the limit as ∆z → 0, the sum becomes an integral and the approximation becomes exact. Thus
Mass of block =
Z
0
1
z2
(2 − z)30 dz = 30 2z −
2
1
= 30 2 −
0
1
2
= 45.
Problems
10. According to the table, the number of bats/ha drops from 117 at 3 km to 85 at 4 km. To make an overestimate, we can
assume there are 117 bats/ha everywhere between 3 and 4 km from the cave. We have
Area of ring (km2 ) = Area
of outer
of inner
|
{z circle} − Area
|
{z circle} = 7π.
π·42
π·32
760
Chapter Eight /SOLUTIONS
There are 100 ha per km2 , so the area is 7π(100) = 700π ha. Assuming 117 bats/ha, this gives
Overestimate for
number of bats
= Area of ring (ha) × Number bats per ha = 257,296.
{z
|
}
700π
|
{z
}
117
11. According to the table, the number of bats/ha drops from 117 at 3 km to 85 at 4 km. To make an underestimate, we can
assume there are 85 bats/ha everywhere between 3 and 4 km from the cave. We have
of inner
Area of ring (km2 ) = Area
of outer
|
|
{z circle} − Area
{z circle} = 7π.
π·42
π·32
There are 100 ha per km2 , so the area is 7π(100) = 700π ha. Assuming 85 bats/ha, this gives
Underestimate for
number of bats
= Area of ring (ha) × Number bats per ha = 186,925.
{z
|
|
}
700π
{z
}
85
12. The number of bats in a ring of inner radius r and width ∆r is given by
Number of bats = Area of ring (ha) × Bats per ha
area (km2 )
z }| {
2πr∆r} × f (r)
≈|
100 · {z
|{z}
area (ha)
bats per ha
= 200πf (r)r∆r.
Thus, adding up the contributions from each ring gives:
Total number of bats ≈
Taking the limit of the sum as ∆r → 0, we get
Total number of bats =
X
Z
200πf (r)r∆r.
5
200πf (r)rdr.
0
13. Since the density varies with x, the region must be sliced perpendicular to the x-axis. This has the effect of making the
density approximately constant on each strip. See Figure 8.72. Since a strip is of height y, its area is approximately y∆x.
The density on the strip is δ(x) = 1 + x gm/cm2 . Thus
Mass of strip ≈ Density · Area ≈ (1 + x)y∆x gm.
Because the tops of the strips end on two different lines, one for x ≥ 0 and the other for x < 0, the mass is calculated as
the sum of two integrals. See Figure 8.72. For the left part of the region, y = x + 1, so
Mass of left part = lim
∆x→0
=
Z
0
X
(1 + x)y∆x =
(1 + x)2 dx =
−1
0
Z
(1 + x)(x + 1) dx
−1
0
3
(x + 1)
3
=
1
gm.
3
−1
From Figure 8.72, we see that for the right part of the region, y = −x + 1, so
Mass of right part = lim
∆x→0
=
Z
1
0
X
(1 + x)y∆x =
(1 − x2 ) dx = x −
Z
x3
3
1
(1 + x)(−x + 1) dx
0
1
=
0
2
1
Total mass = + = 1 gm.
3
3
2
gm.
3
8.4 SOLUTIONS
761
y
y =x+1✲
−1
1
y = −x + 1
✠
✲ ✛
✲ ✛
∆x
∆x
x
1
Figure 8.72
14. Since the density δ(y) = 2 + y 2 is constant for fixed y, we take slices parallel to the x-axis of height ∆y and length 2.
Then a slice has area = 2∆y and approximate mass of (2 + y 2 ) · 2∆y. Adding the slices and letting ∆y → 0 gives
Mass =
Z
3
2(2 + y 2 ) dy = 4y +
0
2 3
y
3
3
= 30 gm.
0
15. (a) We must find where the population density is zero. The density is given by the function
10,000(3 − r);
if 10,000(3 − r) = 0, then we must have r = 3. We thus conclude that the radius of Circle City is 3 miles. (Note that
for r > 3, 10,000(3 − r) becomes negative, so at that point, our function no longer gives a meaningful representation
of population density.)
(b) We refer to Example 4 in this section, with f (r) = 10,000(3 − r). The population is approximated by a sum
Population ≈
X
2πr · 10,000(3 − r)∆r.
Since the city radius is 3 miles, r ranges from 0 to 3. Hence as ∆r → 0, the sum is given by the integral
Population =
Z
0
3
2πr · 10,000(3 − r) dr.
This integral evaluates to 9π · 10,000 ≈ 282,743. So we can say that the population of Circle City is approximately
282,743.
16. (a) Partition [0, 10,000] into N subintervalsof width ∆r. The area in the ith subinterval is ≈ 2πri ∆r. So the total mass
PN
50
∆r.
in the slick = M ≈ i=1 2πri 1+r
i
(b) M =
Z
0
10,000
100π
r
r
dr. We may rewrite
as
1+r
1+r
M =
Z
10,000
0
1+r
1+r
−
1
1+r
1
100π(1 −
) dr = 100π
1+r
=1−
1
,
1+r
so that
10,000
r − ln |1 + r|
0
!
6
= 100π(10,000 − ln(10,001)) ≈ 3.14 × 10 kg.
(c) We wish to find an R such that
Z
0
R
100π
r
1
dr =
1+r
2
Z
0
10,000
100π
r
dr ≈ 1.57 × 106 .
1+r
6
So 100π(R − ln |R + 1|) ≈ 1.57 × 10 ; R − ln |R + 1| ≈ 5000. By trial and error, we find R ≈ 5009 meters.
17. (a) We form a Riemann sum by slicing the region into concentric rings of radius r and width ∆r. Then the volume
deposited on one ring will be the height H(r) multiplied by the area of the ring. A ring of width ∆r will have an area
given by
Area = π(r + ∆r)2 − π(r 2 )
= π(r 2 + 2r∆r + (∆r)2 − r 2 )
= π(2r∆r + (∆r)2 ).
762
Chapter Eight /SOLUTIONS
Since ∆r is approaching zero, we can approximate
Area of ring ≈ π(2r∆r + 0) = 2πr∆r.
From this, we have
∆V ≈ H(r) · 2πr∆r.
Thus, summing the contributions from all rings we have
V ≈
Taking the limit as ∆r → 0, we get
V =
X
H(r) · 2πr∆r.
5
Z
2πr 0.115e−2r dr.
0
(b) We use integration by parts:
V = 0.23π
Z
5
re−2r dr
0
= 0.23π
e−2r
re−2r
−
−2
4
5
0
≈ 0.181(millimeters) · (kilometers)2 = 0.181 · 10−3 · 106 meters3 = 181 cubic meters.
(b − a)
; a = x0 < x1 < · · · < xN = b. The mass of the
N
strip on the ith subinterval is approximately mi = δ(xi )[f (xi ) − g(xi )]∆x. If we use a right-hand Riemann sum, the
approximation for the total mass is
18. Partition a ≤ x ≤ b into N subintervals of width ∆x =
N
X
i=1
δ(xi )[f (xi ) − g(xi )]∆x, and the exact mass is M =
Z
a
b
δ(x)[f (x) − g(x)]dx.
19. (a) Use the formula for the volume of a cylinder:
Volume = πr 2 l.
Since it is only a half cylinder
1 2
πr l.
2
(b) Set up the axes as shown in Figure 8.73. The density can be defined as
Volume of shed =
Density = ky.
Now slice the sawdust horizontally into slabs of thickness ∆y as shown in Figure 8.74, and calculate
Volume of slab ≈ 2xl∆y = 2l(
p
r 2 − y 2 )∆y.
Mass of slab = Density · Volume ≈ 2kly
Finally, we compute the total mass of sawdust:
Total mass of sawdust =
Z
r
2kly
0
p
r 2 − y 2 ∆y.
2
r 2 − y 2 dy = − kl(r 2 − y 2 )3/2
3
p
r
=
0
2klr 3
.
3
y
r
∆y
✛
x r
Figure 8.73
✛
l
x
2x
✛
−r
✛
Figure 8.74
8.4 SOLUTIONS
763
20. First we rewrite the chart, listing the density with the corresponding distance from the center of the earth (x km below the
surface is equivalent to 6370 − x km from the center):
3
This gives us spherical shells whose volumes are 43 π(ri3 − ri+1
) for any two consecutive distances from the origin.
We will assume that the density of the earth is increasing with depth. Therefore, the average density of the ith shell is
3
3
) and 34 πDi (ri3 − ri+1
) are upper
between Di and Di+1 , the densities at top and bottom of shell i. So 34 πDi+1 (ri3 − ri+1
and lower bounds for the mass of the shell.
Table 8.6
i
xi
ri = 6370 − xi
Di
0
0
6370
3.3
1
1000
5370
4.5
2
2000
4370
5.1
3
2900
3470
5.6
4
3000
3370
10.1
5
4000
2370
11.4
6
5000
1370
12.6
7
6000
370
13.0
8
6370
0
13.0
To get a rough approximation of the mass of the earth, we don’t need to use all the data. Let’s just use the densities
at x = 0, 2900, 5000 and 6370 km. Calculating an upper bound on the mass,
MU =
4
π[13.0(13703 − 03 ) + 12.6(34703 − 13703 ) + 5.6(63703 − 34703 )] · 1015 ≈ 7.29 × 1027 g.
3
The factor of 1015 may appear unusual. Remember the radius is given in kilometers and the density is given in g/cm3 , so
we must convert kilometers to centimeters: 1 km = 105 cm , so 1 km3 = 1015 cm3 .
The lower bound is
ML =
4
π[12.6(13703 − 03 ) + 5.6(34703 − 13703 ) + 3.3(63703 − 34703 )] · 1015 ≈ 4.05 × 1027 g.
3
Here, our upper bound is just under 2 times our lower bound.
Using all our data, we can find a more accurate estimate. The upper and lower bounds are
7
MU =
4 X
3
Di+1 (ri3 − ri+1
) · 1015 g
π
3
i=0
and
7
ML =
4 X
3
Di (ri3 − ri+1
) · 1015 g.
π
3
i=0
We have
4
π[4.5(63703 − 53703 ) + 5.1(53703 − 43703 ) + 5.6(43703 − 34703 )
3
+ 10.1(34703 − 33703 ) + 11.4(33703 − 23703 ) + 12.6(23703 − 13703 )
MU =
+ 13.0(13703 − 3703 ) + 13.0(3703 − 03 )] · 1015 g
≈ 6.50 × 1027 g
and
4
π[3.3(63703 − 53703 ) + 4.5(53703 − 43703 ) + 5.1(43703 − 34703 )
3
+ 5.6(34703 − 33703 ) + 10.1(33703 − 23703 ) + 11.4(23703 − 13703 )
ML =
+ 12.6(13703 − 3703 ) + 13.0(3703 − 03 )] · 1015 g
≈ 5.46 × 1027 g.
764
Chapter Eight /SOLUTIONS
21. We slice time into small intervals. Since t is given in seconds, we convert the minute to 60 seconds. We consider water
loss over the time interval 0 ≤ t ≤ 60. We also need to convert inches into feet since the velocity is given in ft/sec. Since
1 inch = 1/12 foot, the square hole has area 1/144 square feet. For water flowing through a hole with constant velocity v,
the amount of water which has passed through in some time, ∆t, can be pictured as the rectangular solid in Figure 8.75,
which has volume
Area · Height = Area · Velocity · Time.
1
Area = 144
ft2
❄
Height = g(t)∆t
✲
❄
❄
❄
Figure 8.75: Volume of water passing
through hole
Over a small time interval of length ∆t, starting at time t, water flows with a nearly constant velocity v = g(t)
through a hole 1/144 square feet in area. In ∆t seconds, we know that
Water lost ≈
1
ft2 (g(t) ft/sec)(∆t sec) =
144
Adding the water from all subintervals gives
Total water lost ≈
As ∆t → 0, the sum tends to the definite integral:
Total water lost =
X 1
144
Z
60
0
1
g(t) ∆t ft3 .
144
g(t) ∆t ft3 .
1
g(t) dt ft3 .
144
22. (a) Divide the atmosphere into spherical shells of thickness ∆h. See Figure 8.76. The density on a typical shell, ρ(h), is
approximately constant. The volume of the shell is approximately the surface area of a sphere of radius re + h meters
times ∆h, where re = 6.4 · 106 meters is the radius of the earth,
Volume of Shell ≈ 4π(re + hi )2 ∆h.
A Riemann sum for the total mass is
Mass ≈
X
4π(re + h)2 × 1.28e−0.000124hi ∆h kg.
ith spherical shell
∆h
re + h
Earth
Figure 8.76
8.4 SOLUTIONS
765
(b) This Riemann sum becomes the integral
Mass = 4π
100
Z
(re + h)2 · 1.28e−0.000124h dh
0
= 4π
100
Z
(6.4 · 106 + h)2 · 1.28e−0.000124h dh.
0
Evaluating the integral using numerical methods gives M = 6.5 · 1016 kg.
P
23. We need theP
numerator of x, to be zero, i.e.
xi mi = 0. Since all of the masses are the same, we can factor them out
and write 4
xi = 0. Thus the fourth mass needs to be placed so that all of the positions sum to zero. The first three
positions sum to (−6 + 1 + 3) = −2, so the fourth mass needs to be placed at x = 2.
24. We have
Total mass of the rod =
Z
3
0
In addition,
Z
Moment =
x3
(1 + x ) dx = x +
3
3
2
x(1 + x ) dx =
0
99/4
12
Thus, the center of mass is at the position x̄ =
25. The center of mass is
2
x2
x4
+
2
4
3
=
0
3
= 12 grams.
0
99
gram-meters.
4
= 2.06 meters.
Rπ
x(2 + sin x) dx
.
x̄ = R0 π
0
(2 + sin x) dx
π
The numerator is
Rπ
0
(2x + x sin x) dx = (x2 − x cos x + sin x)
= π 2 + π.
0
π
The denominator is
Rπ
0
(2 + sin x) dx = (2x − cos x)
= 2π + 2. So the center of mass is at
0
π2 + π
π(π + 1)
π
=
= .
2π + 2
2(π + 1)
2
x=
26. (a) We find that
Moment =
Z
1
2
x(1 + kx ) dx =
0
and that
Total mass =
1
Z
(1 + kx2 ) dx =
0
Thus, the center of mass is
x̄ =
(b) Let f (k) =
3
4
2+k
3+k
. Then f ′ (k) =
3
4
1
2
+
1+
1
(3+k)2
k
4
k
3
=
1
kx3
3
2+k
3+k
x2
kx4
+
2
4
3
4
x+
=
0
1
k
+ gram-meters,
2
4
1
=1+
0
k
grams.
3
meters.
, which is always positive, so f is an increasing function of k.
Since f (0) = 0.5, this is the smallest value of f . As k → ∞, f (k) → 3/4 = 0.75. So f (k) is always between 0.5
and 0.75.
27. (a) The density is minimum at x = −1 and increases as x increases, so more of the mass of the rod is in the right half of
the rod. We thus expect the balancing point to be to the right of the origin.
(b) We need to compute
Z
1
−1
x(3 − e−x ) dx =
=
3 2
x + xe−x + e−x
2
3
+ e−1 + e−1 −
2
1
(using integration by parts)
−1
3
2
− e1 + e1 = .
2
e
766
Chapter Eight /SOLUTIONS
We must divide this result by the total mass, which is given by
Z
1
1
−1
(3 − e−x ) dx = (3x + e−x )
We therefore have
x̄ =
−1
= 6−e+
1
.
e
2/e
2
=
≈ 0.2.
6 − e + (1/e)
1 + 6e − e2
28. Since the region is symmetric about the x-axis, ȳ = 0.
To find x̄, we first find the density. The area of the disk is π/2 m2 , so it has √
density 3/(π/2) = 6/π kg/m2 . We find
the mass of the small strip of width ∆x in Figure 8.77. The height of the strip is 1 − x2 , so
Area of the small strip ≈ Ax (x)∆x = 2 ·
When multiplied by the density 6/π, we get
p
1 − x2 ∆x m2 .
12 p
· 1 − x2 ∆x kg.
π
We then sum the product of these masses with x, and take the limit as ∆x → 0 to get
Mass of the strip ≈
Moment =
Z
1
0
12 p
4
x 1 − x2 dx = − (1 − x2 )3/2
π
π
1
=
0
4
meter.
π
Finally, we divide by the total mass 3 kg to get the result x̄ = 4/(3π) meters.
y
∆x
✠
✻
1
√
1 − x2
✲
❄
x
1
−1
Figure 8.77: Area of a small strip
29. (a) Since the density is constant, the mass is the product of the area of the plate and its density.
Area of the plate =
Z
1
x2 dx =
0
1 3
x
3
1
=
0
1
cm2 .
3
Thus the mass of the plate is 2 · 1/3 = 2/3 gm.
(b) See Figure 8.78. Since the region is “fatter” closer to x = 1, x̄ is greater than 1/2.
y
x2
✲ ✛
∆x
Figure 8.78
x
1
8.4 SOLUTIONS
767
(c) To find the center of mass along the x-axis, we slice the region into vertical strips of width ∆x. See Figure 8.78. Then
Area of strip = Ax (x)∆x ≈ x2 ∆x
Then, since the density is 2 gm/cm2 , we have
x=
R1
0
2x3 dx
3 2x4
= ·
2/3
2
4
1
=3
0
1
4
3
cm.
4
=
This is greater than 1/2, as predicted in part (b).
30. (a) Since the density is constant, the mass is the product of the area of the plate and its density.
Area of the plate =
Z
1
√
x dx =
0
2 3/2
x
3
1
=
0
2
cm2 .
3
Thus the mass of the plate is 5 · 2/3 = 10/3 gm.
(b) To find x̄, we slice the region into vertical strips of width ∆x. See Figure 8.79. Then
√
Area of strip = Ax (x)∆x ≈ x∆x cm2 .
Then, since the density is 5 gm/cm2 , we have
x=
R
xδAx (x) dx
=
Mass
R1
0
5x3/2 dx
10/3
3
2x5/2
10
=
1
0
=
3
cm.
5
To find ȳ, we slice the region into horizontal strips of width ∆y
Area of horizontal strip = Ay (y)∆y ≈ (1 − x)∆y = (1 − y 2 )∆y cm2 .
Then, since the density is 5 gm/cm2 , we have
y=
R
yδAy (y) dy
=
Mass
R1
0
5(y − y 3 ) dy
10/3
y
3
=
5
10
√
y2
y4
−
2
4
1
=
0
3 5
3
· = cm.
10 4
8
x
✲ ✛
x
1
∆x
Figure 8.79
31. The triangle is symmetric about the x axis, so ȳ = 0.
To find x̄, we first calculate the density. The area of the triangle is ab/2, so it has density 2m/(ab) where m is the
total mass of the triangle. We need to find the mass of a small strip of width ∆x located at xi (see Figure 8.80).
Area of the small strip ≈ Ax (x)∆x = 2 ·
b(a − x)
∆x.
2a
Multiplying by the density 2m/(ab) gives
Mass of the strip ≈ 2m
(a − x)
∆x.
a2
768
Chapter Eight /SOLUTIONS
We then sum the product of these masses with xi , and take the limit as ∆x → 0 to get
Moment =
a
Z
0
2m
2mx(a − x)
dx = 2
a2
a
x3
ax2
−
2
3
a
=
0
2m
a2
a3
a3
−
2
3
=
ma
.
3
Finally, we divide by the total mass m to get the desired result x̄ = a/3, which is independent of the length of the base b.
y
b
2
b
(a
2a
✻
− x)
❄
x
a
x
■
∆x
− 2b
Figure 8.80
32. Stand the cone with the base horizontal, with center at the origin. Symmetry gives us that x̄ = ȳ = 0. Since the cone is
fatter near its base we expect the center of mass to be nearer to the base.
Slice the cone into disks parallel to the xy-plane.
As we saw in Example 3 on page 416, a disk of thickness ∆z at height z above the base has
Volume of disk = Az (z)∆z ≈ π(5 − z)2 ∆z cm3 .
Thus, since the density is δ,
R5
z · δπ(5 − z)2 dz
zδAz (z) dz
z=
= 0
cm.
Mass
Mass
To evaluate the integral in the numerator, we factor out the constant density δ and π to get
R
Z
5
2
0
z · δπ(5 − z) dz = δπ
Z
0
5
2
z(25 − 10z + z ) dz = δπ
We divide this result by the total mass of the cone, which is
z=
625
δπ
12
1
3δ
π5
3
=
1
π52
3
10z 3
z4
25z 2
−
+
2
3
4
5
=
0
625
δπ.
12
· 5 δ:
5
= 1.25 cm.
4
As predicted, the center of mass is closer to the base of the cone than its top.
33. Since the density is constant, the total mass of the solid is the product of the volume of the solid and its density: δπ(1 −
e−2 )/2. By symmetry, ȳ = 0. To find x̄, we slice the solid into disks of width ∆x, perpendicular to the x-axis. See
Figure 8.81. A disk at x has radius y = e−x , so
Volume of disk = Ax (x)∆x = πy 2 ∆x = πe−2x ∆x.
Since the density is δ, we have
x=
The integral
R
R1
0
x · δπe−2x dx
Total mass
δπ
R1
xe−2x dx
2
=
=
δπ(1 − e−2 )/2
1 − e−2
0
Z
1
xe−2x dx.
0
xe−2x dx can be done by parts: let u = x and v ′ = e−2x . Then u′ = 1 and v = e−2x /(−2). So
Z
xe−2x dx =
xe−2x
−
−2
Z
e−2x
xe−2x
e−2x
dx =
−
.
−2
−2
4
8.4 SOLUTIONS
and then
Z
1
xe−2x dx =
0
The final result is:
xe−2x
e−2x
−
−2
4
1
=
0
e−2
e−2
−
−2
4
1
4
− 0−
=
769
1 − 3e−2
.
4
1 − 3e−2
1 − 3e−2
2
·
=
≈ 0.343.
1 − e−2
4
2 − 2e−2
Notice that x̄ is less that 1/2, as we would expect from the fact that the solid is wider near the origin.
x=
y = e−x
y
Radius = y
✠
x
✲ ✛
∆x
Figure 8.81
34. (a) Position the pyramid so that the center of its base lies at the origin on the xy-plane. Slice the pyramid into square
slabs parallel to its base. We compute the mass of the pyramid by adding the masses of the slabs.
The mass of a slab is its volume multiplied by the density δ. To compute the volume of a slab, we need to get an
expression for the side s of the slab in terms of its height z. Using the similar triangles in Figure 8.82, we see that
(10 − z)
s
=
.
40
10
Thus s = 4(10 − z). Since the area of the square slab’s face is s2 ,
Volume of the slab ≈ Az (z)∆z = s2 ∆z = 16(10 − z)2 ∆z.
Mass of slab = 16δ(10 − z)2 ∆z.
The mass of the pyramid can be found by summing all of the masses of the slabs, and letting the thickness ∆z
approach zero:
Total mass = lim
∆z→0
X
16δ(10 − z)2 ∆z =
Z
10
0
16δ(10 − z)2 dz =
✻
−16δ(10 − z)3
3
✻
10 − z
10 cm
✛
s
✲
❄❄
∆z
✻
z
❄
✛
40 cm
Figure 8.82
✲
❄
10
=
0
16000δ
gm.
3
770
Chapter Eight /SOLUTIONS
(b) From symmetry, we have x̄ = ȳ = 0. Since the pyramid is fatter near its base we expect the center of mass to be
nearer to the base. Since
Volume of slab = Az (z)∆z = 16(10 − z)2 ∆z,
R 10
z · 16δ(10 − z)2 dz
0
.
Total mass
To evaluate the integral in the numerator, we factor out the constant 16δ and expand the integrand to get
z=
16δ
Z
10
2
0
z4
−20z 3
+
50z +
3
4
3
(100z − 20z + z ) dz = 16δ
2
10
=
0
40000δ
.
3
We divide this result by the total mass 16000δ/3 of the pyramid
z=
40000δ/3
40000
=
= 2.5 cm.
16000δ/3
16000
As predicted, the center of mass is closer to the base of the pyramid than its top.
Strengthen Your Understanding
35. Mass density can never be negative, but f (x) < 0 for 0 < x < 5.
36. The center of mass with density δ(x) for 0 ≤ x ≤ 10 is given by
R 10
xδ(x) dx
Center of mass = R0 10
.
δ(x) dx
0
The correct formula for δ(x) = x2 is
R 10
Center of mass = R010
0
x3 dx
x2 dx
.
37. Any density function that is symmetric about the center of the rod will also give a center of mass in the center of the rod.
For example, if a rod has ends at x = −1 and x = 1 and density δ(x) = x2 then its center of mass is at the origin.
38. When the density is constant, for example, if δ(r) = 3 everywhere on the disk, the mass of the disk is the density times
the area of the disk, or 3 · 9π = 27π gm. Since the density is not constant and is less than 3 everywhere except at the
center of the disk, the mass of the disk must be less than 27π gm.
39. Let the rod be the interval 0 ≤ x ≤ 10. Suppose the density is given by
δ(x) =
(
5,
1,
4,
0≤x≤1
1 0 for 0 < t < 10
dt
t
and V ′ (t) < 0 for t > 10, we confirm that this is a maximum. Thus, the best time to sell the wine is in 10 years.
Setting
32. (a) Suppose the oil extracted over the time period [0, M ] is S. (See Figure 8.111.) Since q(t) is the rate of oil extraction,
we have:
Z
Z
Z
M
M
q(t)dt =
S=
0
M
(a − bt)dt =
0
0
(10 − 0.1t) dt.
To calculate the time at which the oil is exhausted, set S = 100 and try different values of M . We find M = 10.6
gives
Z
10.6
(10 − 0.1t) dt = 100,
0
so the oil is exhausted in 10.6 years.
Extraction Curve
✠
q(t)
Area below
the extraction curve
is the total oil extracted
✲
t
0
M
Figure 8.111
(b) Suppose p is the oil price, C is the extraction cost per barrel, and r is the interest rate. We have the present value of
the profit as
Present value of profit =
M
Z
0
=
Z
(p − C)q(t)e−rt dt
10.6
0
(20 − 10)(10 − 0.1t)e−0.1t dt
= 624.9 million dollars.
33. (a) Let’s split the time interval into n parts, each of length ∆t. During the interval from ti to ti+1 , profit is earned at
a rate of approximately (2 − 0.1ti ) thousand dollars per year, or (2000 − 100ti ) dollars per year. Thus during this
period, a total profit of (2000 − 100ti )∆t dollars is earned. Since this profit is earned ti years in the future, its present
value is (2000 − 100ti )∆te−0.1ti dollars. Thus
Total present value ≈
n−1
X
i=0
(2000 − 100ti )e−0.1ti ∆t.
0
∆t
M
✛ ✲
t0
t1
t2
...
ti
ti+1
...
tn
796
Chapter Eight /SOLUTIONS
(b) The Riemann sum corresponds to the integral:
Z
Present value =
M
e−0.10t (2000 − 100t) dt.
0
(c) To find where the present value is maximized, we take the derivative of
P (M ) =
M
Z
e−0.10t (2000 − 100t) dt,
0
with respect to M , and obtain
P ′ (M ) = e−0.10M (2000 − 100M ).
This is 0 when 2000 − 100M = 0, that is, when M = 20 years. The value M = 20 maximizes P (M ), since
P ′ (M ) > 0 for M < 20, and P ′ (M ) < 0 for M > 20. To determine what the maximum is, we evaluate the integral
representation for P (20) by III-14 in the integral table:
P (20) =
=
Z
20
e−0.10t (2000 − 100t) dt
0
(2000 − 100t) −0.10t
e
+ 10000e−0.10t
−0.10
20
0
≈ $11353.35.
34. One good way to approach the problem is in terms of present values. In 1980, the present value of Germany’s loan was
20 billion DM. Now let’s figure out the rate that the Soviet Union would have to give money to Germany to pay off 10%
interest on the loan by using the formula for the present value of a continuous stream. Since the Soviet Union sends gas
at a constant rate, the rate of deposit, P (t), is a constant c. Since they don’t start sending the gas until after 5 years have
passed, the present value of the loan is given by:
Present Value =
Z
∞
P (t)e−rt dt.
5
We want to find c so that
20,000,000,000 =
Z
∞
ce
−rt
dt = c
Z
∞
e−rt dt
5
b
5
= c lim (−10e−0.10t )
b→∞
= ce−0.10(5)
5
≈ 6.065c.
Dividing, we see that c should be about 3.3 billion DM per year. At 0.10 DM per m3 of natural gas, the Soviet Union
must deliver gas at the constant, continuous rate of about 33 billion m3 per year.
35. Measuring money in thousands of dollars, the equation of the line representing the demand curve passes through (50, 980)
and (350, 560). Its slope is (560 − 980)/(350 − 50) = −420/300. See Figure 8.112. So the equation is y − 560 =
(x − 350), i.e. y − 560 = − 57 x + 490. Thus
− 420
300
Consumer surplus =
Z
0
350
7
− x + 1050
5
dx − 350 · 560 = −
7 2
x + 1050x
10
350
0
− 196000
= 85,750.
(Note that 85,750 = 12 · 490 · 350, the area of the triangle in Figure 8.112. We could have used this instead of the integral
to find the consumer surplus.)
Recalling that our unit measure for the price axis is $1000/car, the consumer surplus is $85,750,000.
8.6 SOLUTIONS
797
price
(1000s of dollars/car)
(50, 980)
1050
Demand
(350, 560)
quantity
(number of cars)
Figure 8.112
36. The supply curve, S(q), represents the minimum price p per unit that the suppliers will be willing to supply some quantity
q of the good for. See Figure 8.113. If the suppliers have q ∗ of the good and q ∗ is divided into subintervals of size ∆q,
then if the consumers could offer the suppliers for each ∆q a price increase just sufficient to induce the suppliers to sell
an additional ∆q of the good, the consumers’ total expenditure on q ∗ goods would be
p1 ∆q + p2 ∆q + · · · =
As ∆q → 0 the Riemann sum becomes the integral
Z
q∗
X
S(q) dq. Thus
pi ∆q.
q∗
Z
S(q) dq is the amount the consumers would
0
0
pay if suppliers could be forced to sell at the lowest price they would be willing to accept.
Price
S(q)
P2
P1
∆q
∆q
q∗
∆q
Quantity
Figure 8.113
37.
Z
0
q∗
(p∗ − S(q)) dq =
Z
q∗
0
∗ ∗
p∗ dq −
=p q −
Z
q∗
Z
q∗
S(q) dq
0
S(q) dq.
0
Using Problem 36, this integral is the extra amount consumers pay (i.e., suppliers earn over and above the minimum they
would be willing to accept for supplying the good). It results from charging the equilibrium price.
38. (a) p∗ q ∗ = the total amount paid for q ∗ of the good at equilibrium. See Figure 8.114.
R q∗
(b) 0 D(q) dq = the maximum consumers would be willing to pay if they had to pay the highest price acceptable to
them for each additional unit of the good. See Figure 8.115.
798
Chapter Eight /SOLUTIONS
price
price
Supply : S(q)
Supply : S(q)
p∗
p∗
q∗
Demand : D(q)
quantity
q∗
Figure 8.114
Demand : D(q)
quantity
Figure 8.115
R q∗
S(q) dq = the minimum suppliers would be willing to accept if they were paid the minimum price acceptable to
0
them for each additional unit of the good. See Figure 8.116.
R q∗
(d) 0 D(q) dq − p∗ q ∗ = consumer surplus. See Figure 8.117.
(c)
price
price
Supply : S(q)
Supply : S(q)
p∗
p∗
q∗
Demand : D(q)
quantity
q∗
Figure 8.116
(e) p∗ q ∗ −
(f)
R q∗
0
R q∗
0
Figure 8.117
S(q) dq = producer surplus. See Figure 8.118.
(D(q) − S(q)) dq = producer surplus and consumer surplus. See Figure 8.119.
price
price
Supply : S(q)
Supply : S(q)
p∗
p∗
q∗
Demand : D(q)
quantity
q∗
Figure 8.118
39.
Demand : D(q)
quantity
Figure 8.119
p (price/unit)
p+
p∗
q∗
q (quantity)
Figure 8.120: What effect does the artificially high price, p+ , have?
Demand : D(q)
quantity
8.6 SOLUTIONS
799
(a) A graph of possible demand and supply curves for the milk industry is given in Figure 8.120, with the equilibrium
price and quantity labeled p∗ and q ∗ respectively. Suppose that the price is fixed at the artificially high price labeled
p+ in Figure 8.120. Recall that the consumer surplus is the difference between the amount the consumers did pay
(p+ ) and the amount they would have been willing to pay (given on the demand curve). This is the area shaded in
Figure 8.121(i). Notice that this consumer surplus is clearly less than the consumer surplus at the equilibrium price,
shown in Figure 8.121(ii).
(i)
Artificial price
p (price/unit)
p+
✛
Equilibrium price
p (price/unit)
(ii)
consumer
surplus
p+
✛
consumer
surplus
p∗
p∗
q (quantity)
q+ q∗
q+ q∗
q (quantity)
Figure 8.121: Consumer surplus for the milk industry
(b) At a price of p+ , the quantity sold, q + , is less than it would have been at the equilibrium price. The producer surplus
is the area between p+ and the supply curve at this reduced demand. This area is shaded in Figure 8.122(i). Compare
this producer surplus (at the artificially high price) to the producer surplus in Figure 8.122(ii) (at the equilibrium
price). It appears that in this case, producer surplus is greater at the artificial price than at the equilibrium price.
(Different supply and demand curves might have led to a different answer.)
(i)
Artificial price
p (price/unit)
Equilibrium price
p (price/unit)
(ii)
producer
surplus
p+
p∗
producer
surplus
p+
✠
p∗
q+ q∗
✠
q (quantity)
q+ q∗
q (quantity)
Figure 8.122: Producer surplus for the milk industry
(c) The total gains from trade (Consumer surplus + Producer surplus) at the artificially high price of p+ is the area shaded
in Figure 8.123(i). The total gains from trade at the equilibrium price of p∗ is the area shaded in Figure 8.123(ii). It is
clear that, under artificial price conditions, total gains from trade go down. The total financial effect of the artificially
high price on all producers and consumers combined is a negative one.
(i)
Artificial price
p (price/unit)
(ii)
Equilibrium price
p (price/unit)
p+
p+
p∗
p∗
q+ q∗
q (quantity)
q+ q∗
Figure 8.123: Total gains from trade
q (quantity)
800
Chapter Eight /SOLUTIONS
40.
price
✲
Consumer surplus
p∗
p−
Producer surplus
✲
quantity
q− q∗
(a) The producer surplus is the area on the graph between p− and the supply function. Lowering the price also lowers
the producer surplus.
(b) Note that the consumer surplus—the area between the line p− and the supply curve—increases or decreases depending on the functions describing the supply and demand and on the lowered price. (For example, the consumer surplus
seems to be increased in the graph above, but if the price were brought down to $0 then the consumer surplus would
be zero, and hence clearly less than the consumer surplus at equilibrium.)
(c) The graph above shows that the total gains from the trade are decreased.
Strengthen Your Understanding
41. With no interest, the future value of the stream is equal to the total amount deposited over 10 years, which is $20,000. So
the future value is more than $20,000.
42. The present value of S at a 4% interest rate, compounded annually, is
PV =
S
.
1.04
The present value of S at a 3% interest rate, compounded annually, is
PV =
S
.
1.03
Thus, the present value is smaller with a 4% interest rate.
43. The present value of a continuous stream of payments is found using an integral:
PV =
Z
2
0
P e−rt dt = −
P −rt
e
r
2
=
0
P
(1 − e−2r ).
r
44. Producer surplus is measured in dollars.
45. See Figure 8.124.
y
Supply
Consumer surplus
✲
Producer surplus
✲
Demand
x
Figure 8.124
y
Supply
Consumer surplus
Producer surplus
✲
✲
Demand
x
8.7 SOLUTIONS
801
46. We want 10,000e−r·10 < 5000. Solving
e−10r = 0.5
1
r = − ln(0.5) = 0.0693.
10
We want an interest rate larger than 0.0693, such as 7% or larger.
Larger than 6.93%/ yr
47. A present value of A with an interest rate r, compounded annually, has future value A(1 + r) after one year. Since we
want the future value to be $5000, we have
A(1 + r) = 5000
or
5000
.
1+r
Every choice of r gives a choice of A. For example, a 5% interest rate corresponds to r = 0.05 and gives A =
5000/1.05 = 4761.90. If the annual interest rate is 5%, then $4761.90 now has future value $5000 one year from
now.
A=
48. Suppose there is an annual interest rate of 10%. Then the present value, A, of the $10,000 is:
A=
10, 000
= 3855.
(1 + 0.10)10
Thus, a deposit of roughly $3855 is needed now to have 10,000 dollars in 10 years, assuming a 10% interest rate.
There are two ways to think about the remaining values in the table. We can ask how much the amount needed now,
$3855, will grow to in t years. This means the remaining values in the table arise by calculating:
3855(1 + 0.10)t
for
t = 1, 2, 3, 4.
Alternatively, we realize that a single deposit made t years from now would have 10 − t years to grow to reach our target
investment of $10,000 in 10 years. This means the remaining values in the table arise by calculating:
10, 000
,
(1 + 0.10)10−t
for
t = 1, 2, 3, 4.
If we ignore rounding errors, we see that two approaches yield the same table values:
t (years from now)
0
1
2
3
4
$ (dollars)
3855
4241
4665
5131
5644
Other possible answers to this problem can arise from alternate interest rate choices.
Solutions for Section 8.7
Exercises
1. The two humps of probability in density (a) correspond to two intervals on which its cumulative distribution function is
increasing. Thus (a) and (II) correspond.
A density function increases where its cumulative distribution function is concave up, and it decreases where its
cumulative distribution function is concave down. Density (b) matches the distribution with both concave up and concave
down sections, which is (I). Density (c) matches (III) which has a concave down section but no interval over which it is
concave up.
2.
% of population
per dollar of income
% of population having
at least this income
income
Figure 8.125: Density function
income
Figure 8.126: Cumulative distribution function
802
Chapter Eight /SOLUTIONS
3.
% of population
per dollar of income
% of population having
at least this income
income
income
Figure 8.127: Density function
4.
Figure 8.128: Cumulative distribution function
% of population
per dollar of income
% of population having
at least this income
income
income
Figure 8.129: Density function
Figure 8.130: Cumulative distribution function
5. Since the function takes on the value of 4, it cannot be a cdf (whose maximum value is 1). In addition, the function
decreases for x > c, which means that it is not a cdf. Thus, this function is a pdf. The area under a pdf is 1, so 4c = 1
giving c = 14 . The pdf is p(x) = 4 for 0 ≤ x ≤ 41 , so the cdf is given in Figure 8.131 by
P (x) =
0
for x < 0
4x
1
1
for 0 ≤ x ≤
4
1
for x >
4
P (x)
1
x
1
4
Figure 8.131
6. Since the function is decreasing, it cannot be a cdf (whose values never decrease). Thus, the function is a pdf.
The area under a pdf is 1, so, using the formula for the area of a triangle, we have
1
4c = 1,
2
giving
c=
1
.
2
The pdf is
p(x) =
1
1
− x for
2
8
0 ≤ x ≤ 4,
so the cdf is given in Figure 8.132 by
0
x2
x
P (x) =
−
2
16
1
for
x<0
for
0≤x≤4
for
x > 4.
8.7 SOLUTIONS
803
P (x)
1
x
4
Figure 8.132
7. Since the function levels off at the value of c, the area under the graph is not finite, so it is not 1. Thus, this function cannot
be a pdf.
It is a cdf and c = 1. The cdf is given by
P (x) =
The pdf in Figure 8.133 is given by
p(x) =
0
x
5
1
0
x<0
0≤x≤5
for
x > 5.
for x < 0
1/5
for
for
0
for 0 ≤ x ≤ 5
for x > 5.
P (x)
1/5
x
5
Figure 8.133
8. This function decreases, so it cannot be a cdf. Since the graph must represent a pdf, the area under it is 1. The region
consists of two rectangles, each of base 0.5, and one of height 2c and one of height c, so
Area = 2c(0.5) + c(0.5) = 1
c=
2
1
=
1.5
3
The pdf is therefore
p(x) =
0
for
4/3 for
2/3
0
for
for
x<0
0 ≤ x ≤ 0.5
0.5 < x ≤ 1
x > 1.
The cdf P (x) is the antiderivative of this function with P (0) = 0. See Figure 8.134. The formula for P (x) is
P (x) =
0
4x/3
2/3 + (2/3)(x − 0.5)
1
for
x<0
for
0 ≤ x ≤ 0.5
for
for
0.5 < x ≤ 1
x > 1.
804
Chapter Eight /SOLUTIONS
P (x)
1
2
3
x
0.5
1
Figure 8.134
9. This function increases and levels off to c. The area under the curve is not finite, so it is not 1. Thus, the function must be
a cdf, not a pdf, and 3c = 1, so c = 1/3.
The pdf, p(x) is the derivative, or slope, of the function shown, so, using c = 1/3,
p(x) =
See Figure 8.135.
0
for
x<0
(1/3 − 0)/(2 − 0) = 1/6
for
0≤x≤2
0
for
(1 − 1/3)/(4 − 2) = 1/3 for 2 < x ≤ 4
x > 4.
1
3
p(x)
1
6
x
2
4
Figure 8.135
10. This function does not level off to 1, and it is not always increasing. Thus, the function is a pdf. Since the area under the
curve must be 1, using the formula for the area of a triangle,
1
·c·1=1
2
c = 2.
so
Thus, the pdf is given by
p(x) =
0
4x
2 − 4(x − 0.5) = 4 − 4x
0
for
x<0
for
0 ≤ x ≤ 0.5
for
for
0.5 < x ≤ 1
x > 0.
To find the cdf, we integrate each part of the function separately, making sure that the constants of integration are arranged
so that the cdf
R is continuous.
Since 4xdx = 2x2 + C and P (0) = 0, we have 2(0)2 + C = 0 so C = 0. Thus P (x) = 2x2 on 0 ≤ x ≤ 0.5.
2
At x = 0.5, the cdf has
R value P (0.5) = 2(0.5) 2 = 0.5. Thus, we arrange that the integral of 4 − 4x goes through the
point (0.5, 0.5). Since (4 − 4x) dx = 4x − 2x + C, we have
4(0.5) − 2(0.5)2 + C = 0.5
Thus
P (x) =
0
2x
2
4x − 2x2 − 1
1
giving
C = −1.
for
x<0
for
0 ≤ x ≤ 0.5
for
for
0.5 < x ≤ 1
x > 1.
8.7 SOLUTIONS
805
See Figure 8.136.
P (x)
1
0.5
x
0.5
1
Figure 8.136
11. The statement p(70) = 0.05 means that for some small interval ∆x around 70, the fraction of families with incomes in
that interval around $70,000 is about 0.05∆x.
12. We need to find a nonnegative function p(x) for which the area between y = p(x) and the x axis is 1. One example is
a decreasing linear function p(x) = b + mx with m < 0. We choose b and m so that p(5) = 0 and the triangular area
under the graph of p(x) for 0 ≤ x ≤ 5 is 1. Since b is the height of the triangle, we have
Area =
1
b·5=1
2
or b = 2/5. Then p(5) = 0 gives
2
+ 5m = 0
5
or m = −2/25. Thus
p(x) =
2
2
−
x
5
25
0≤x≤5
when
and
p(x) = 0 otherwise.
Problems
13. For a given energy E, Figure 8.137 shows that the area under the graph to the right of E is larger for graph B than it is for
graph A. Therefore graph B has more molecules at higher kinetic energies, so it is the hotter gas. So graph A corresponds
to 300 kelvins and graph B corresponds to 500 kelvins.
A
B
energy
E
Figure 8.137
14. No. Though the density function has its maximum value at 50, this does not mean that a large fraction of the population
receives scores near 50. The value p(50) can not be interpreted as a probability. Probability corresponds to area under the
graph of a density function. Most of the area in this case is in the broad hump covering the range 0 ≤ x ≤ 40, very little
in the peak around x = 50. Most people score in the range 0 ≤ x ≤ 40.
806
Chapter Eight /SOLUTIONS
15. (a) Let P (x) be the cumulative distribution function of the heights of the unfertilized plants. As do all cumulative
distribution functions, P (x) rises from 0 to 1 as x increases. The greatest number of plants will have heights in the
range where P (x) rises the most. The steepest rise appears to occur at about x = 1 m. Reading from the graph we
see that P (0.9) ≈ 0.2 and P (1.1) ≈ 0.8, so that approximately P (1.1) − P (0.9) = 0.8 − 0.2 = 0.6 = 60% of the
unfertilized plants grow to heights between 0.9 m and 1.1 m. Most of the plants grow to heights in the range 0.9 m to
1.1 m.
(b) Let PA (x) be the cumulative distribution function of the plants that were fertilized with A. Since PA (x) rises the
most in the range 0.7 m ≤ x ≤ 0.9 m, many of the plants fertilized with A will have heights in the range 0.7 m to
0.9 m. Reading from the graph of PA , we find that PA (0.7) ≈ 0.2 and PA (0.9) ≈ 0.8, so PA (0.9) − PA (0.7) ≈
0.8 − 0.2 = 0.6 = 60% of the plants fertilized with A have heights between 0.7 m and 0.9 m. Fertilizer A had the
effect of stunting the growth of the plants.
On the other hand, the cumulative distribution function PB (x) of the heights of the plants fertilized with B rises
the most in the range 1.1 m ≤ x ≤ 1.3 m, so most of these plants have heights in the range 1.1 m to 1.3 m. Fertilizer
B caused the plants to grow about 0.2 m taller than they would have with no fertilizer.
16. (a) F (7) = 0.6 tells us that 60% of the trees in the forest have height 7 meters or less.
(b) F (7) > F (6). There are more trees of height less than 7 meters than trees of height less than 6 meters because every
tree of height ≤ 6 meters also has height ≤ 7 meters.
17. For a small interval ∆x around 68, the fraction of the population of American men with heights in this interval is about
(0.2)∆x. For example, taking ∆x = 0.1, we can say that approximately (0.2)(0.1) = 0.02 = 2% of American men
have heights between 68 and 68.1 inches.
18. We want to find the cumulative distribution function for the age density function. We see that P (10) is equal to 0.15 since
the table shows that 15% of the population is between 0 and 10 years of age. Also,
P (20) =
Fraction of the population
= 0.15 + 0.14 = 0.29
between 0 and 20 years old
and
P (30) = 0.15 + 0.14 + 0.14 = 0.43.
Continuing in this way, we obtain the values for P (t) shown in Table 8.7.
Table 8.7
Cumulative distribution function of ages in the US
t
0
10
20
30
40
50
60
70
80
90
100
P (t)
0
0.15
0.29
0.43
0.60
0.74
0.84
0.92
0.97
0.99
1.00
19. (a) The two functions are shown below. The choice is based on the fact that the cumulative distribution does not decrease.
(b) The cumulative distribution levels off to 1, so the top mark on the vertical scale must be 1.
1
0.8
0.6
0.4
0.2
✛
Cumulative
Density
✠
2
4
6
8
10
The total area under the density function must be 1. Since the area under the density function is about 2.5 boxes,
each box must have area 1/2.5 = 0.4. Since each box has a height of 0.2, the base must be 2.
20. (a) The area under the graph of the height density function p(x) is concentrated in two humps centered at 0.5 m and 1.1 m.
The plants can therefore be separated into two groups, those with heights in the range 0.3 m to 0.7 m, corresponding
to the first hump, and those with heights in the range 0.9 m to 1.3 m, corresponding to the second hump. This grouping
of the grasses according to height is probably close to the species grouping. Since the second hump contains more
area than the first, there are more plants of the tall grass species in the meadow.
8.7 SOLUTIONS
807
(b) As do all cumulative distribution functions, the cumulative distribution function P (x) of grass heights rises from 0
to 1 as x increases. Most of this rise is achieved in two spurts, the first as x goes from 0.3 m to 0.7 m, and the second
as x goes from 0.9 m to 1.3 m. The plants can therefore be separated into two groups, those with heights in the range
0.3 m to 0.7 m, corresponding to the first spurt, and those with heights in the range 0.9 m to 1.3 m, corresponding to
the second spurt. This grouping of the grasses according to height is the same as the grouping we made in part (a),
and is probably close to the species grouping.
(c) The fraction of grasses with height less than 0.7 m equals P (0.7) = 0.25 = 25%. The remaining 75% are the tall
grasses.
21. (a) The percentage of calls lasting from 1 to 2 minutes is given by the integral
Z
2
p(x) dx
Z
2
1
1
0.4e−0.4x dx = e−0.4 − e−0.8 ≈ 22.1%.
(b) A similar calculation (changing the limits of integration) gives the percentage of calls lasting 1 minute or less as
Z
1
p(x) dx =
Z
0
0
1
0.4e−0.4x dx = 1 − e−0.4 ≈ 33.0%.
(c) The percentage of calls lasting 3 minutes or more is given by the improper integral
Z
∞
p(x) dx = lim
b→∞
3
Z
b
Z
h
0.4e−0.4x dx = lim (e−1.2 − e−0.4b ) = e−1.2 ≈ 30.1%.
b→∞
3
(d) The cumulative distribution function is the integral of the probability density; thus,
C(h) =
p(x) dx =
Z
0
0
h
0.4e−0.4x dx = 1 − e−0.4h .
22. (a) The fraction of students passing is given by the area under the curve from 2 to 4 divided by the total area under the
curve. This appears to be about 23 .
(b) The fraction with honor grades corresponds to the area under the curve from 3 to 4 divided by the total area. This is
about 13 .
(c) The peak around 2 probably exists because many students work to get just a passing grade.
(d)
fraction of students
1
GPA
1
2
3
4
23. (a) Most of the earth’s surface is below sea level. Much of the earth’s surface is either around 3 miles below sea level or
exactly at sea level. It appears that essentially all of the surface is between 4 miles below sea level and 2 miles above
sea level. Very little of the surface is around 1 mile below sea level.
(b) The fraction below sea level corresponds to the area under the curve from −4 to 0 divided by the total area under the
curve. This appears to be about 34 .
24. (a) We must have
Z
∞
f (t)dt = 1, for even though it is possible that any given person survives the disease, everyone
0
eventually dies. Therefore,
Z
∞
cte−kt dt = 1.
0
Integrating by parts gives
Z
0
b
c
cte−kt dt = − te−kt
k
b
+
0
Z
0
b
c −kt
e dt
k
b
c
c
= − te−kt − 2 e−kt
k
k
0
c
c
c
= 2 − be−kb − 2 e−kb .
k
k
k
808
Chapter Eight /SOLUTIONS
As b → ∞, we see
Z
∞
c
=1
k2
cte−kt dt =
0
(b) We are told that
Z
5
c = k2 .
so
f (t)dt = 0.4, so using the fact that c = k2 and the antiderivatives from part (a), we have
0
Z
5
2
k te
−kt
dt =
0
k2
k2
− te−kt − 2 e−kt
k
k
5
0
= 1 − 5ke−5k − e−5k = 0.4
so
5ke−5k + e−5k = 0.6.
Since this equation cannot be solved exactly, we use a calculator or computer to find k = 0.275. Since c = k2 , we
have c = (0.275)2 = 0.076.
(c) The cumulative death distribution function, C(t), represents the fraction of the population that have died up to time
t. Thus,
C(t) =
Z
t
0
k2 xe−kx dx = −kxe−kx − e−kx
= 1 − kte−kt − e−kt .
t
0
Strengthen Your Understanding
25. The fact that p(1) = 0.02 tells us that the probability that x falls in a small interval of length ∆x around 1 is 0.02∆x.
26. The cumulative distribution gives the probability that x is less than some value. Thus, P (5) = 0.4 means that the
probability that x is less that 5 is 0.4.
27. Every density function p(t) satisfies the equation
a density function.
R∞
−∞
p(t) dt = 1. Since
R∞
−∞
t2 dt = ∞, the function p(t) = t2 is not
28. Since this function is increasing for x > 0, the area under its graph increases without bound. So the area under the graph
of this function is not equal to 1.
29. As x → ∞ the function P (x) = x2 ex grows without bound, whereas a cumulative distribution function must approach
1.
2
30. Every cumulative distribution is a nondecreasing function, but P (t) = e−t is decreasing for t > 0.
31. A cumulative distribution function is increasing for all values of x; a probability density function is not.
32. We can take
p(x) =
n
1/20,
0,
It is a density function because p(x) ≥ 0 for 0 ≤ x ≤ 20 and
33. We can use
P (t) =
(
0 ≤ x ≤ 20
otherwise.
R∞
−∞
p(x) dx = 1.
0, t < 0
t, 0 ≤ t ≤ 1
1, t > 1.
It is a cumulative distribution function because P (t) is an increasing function that increases from 0 to 1.
34. One possible density function is given by
p(x) =
0
1/5
0
This function is nonzero only between x = 2 and x = 7 and has
x<2
2≤x≤7
7 < x.
R∞
−∞
p(x) dx = 1.
8.8 SOLUTIONS
809
35. The cumulative distribution function increases between x = 3 and x = 7. One possible cumulative distribution function
is
x<3
0
P (x) = 1/4(x − 3)
3≤x≤7
1
x > 7.
36. False. Since p(x) < 0 for x < 0, it cannot be a probability density function.
37. False. It is true that p(x) ≥ 0 for all x, but we also need
check the integral from 0 to ∞. We have
∞
Z
R∞
−∞
2
xe−x dx = lim
b→∞
0
p(x)dx = 1. Since p(x) = 0 for x ≤ 0, we need only
2
1
− e−x
2
b
=
0
1
.
2
Solutions for Section 8.8
Exercises
1.
p(x)
0.24
A3
A1
0.12
0.08
A4
A2
2
0
x (tons of fish)
6
8
Splitting the figure into four pieces, we see that
Area under the curve = A1 + A2 + A3 + A4
1
1
= (0.16)4 + 4(0.08) + (0.12)2 + 2(0.12)
2
2
= 1.
We expect the area to be 1, since
2 ≤ x ≤ 8.
2. Recall that the mean is
Z
∞
p(x) dx = 1 for any probability density function, and p(x) is 0 except when
−∞
Z
∞
−∞
xp(x) dx. In the fishing example, p(x) = 0 except when 2 ≤ x ≤ 8, so the mean is
Z
8
xp(x) dx.
2
Using the equation for p(x) from the graph,
Z
8
xp(x) dx =
Z
6
xp(x) dx +
2
2
=
Z
8
xp(x) dx
6
6
x(0.04x) dx +
2
=
Z
0.04x3
3
Z
8
x(−0.06x + 0.6) dx
6
6
2
+ −0.02x3 + 0.3x2
≈ 5.253 tons.
8
6
810
Chapter Eight /SOLUTIONS
3. (a)
(i)
(ii)
p(x)
p(x)
√1
2π
√1
2π
σ=1
√1
2 2π
√1
3 2π
✛
σ=2
σ=3
✛
✛
µ=4
µ=5
µ=6
x
x
5
4 5 6
(b) Recall that the mean is the “balancing point.” In other words, if the area under the curve was made of cardboard, we’d
expect it to balance at the mean. All of the graphs are symmetric across the line x = µ, so µ is the “balancing point”
and hence the mean.
As the graphs also show, increasing σ flattens out the graph, in effect lessening the concentration of the data near the
mean. Thus, the smaller the σ value, the more data is clustered around the mean.
Problems
4. The mean is the value of the integral
2
Z
x · 0.5(2 − x) dx =
0
The median is the value of T such that
T
Z
0.5(2 − x) dx = 0.5.
0
Integrating gives the equation
x − 0.5
which is a√quadratic with solutions T = 2 ±
T = 2 − 2 = 0.586.
2
.
3
√
x2
2
T
0
=T −
T2
= 0.5
4
2. Since the median must be between 0 and 2, the solution we want is
5. The median is the value of T such that P (T ) = 0.5, so we solve
T−
T2
1
=
4
2
√
√
to get T = 2 ± 2. Since the median is between 0 and 2 we discard the larger solution, so the median is T = 2 − 2.
To find the mean, we first calculate the probability density
Density = p(x) = P ′ (x) = 1 −
x
2
and then evaluate the integral
Mean =
Z
2
xp(x) dx =
0
0
So the mean is 2/3.
Z
2
x−
x2
2
dx = .
2
3
6. (a) Since d(e−ct )/dt = ce−ct , we have
c
Z
0
so
6
e−ct dt = −e−ct
6
0
= 1 − e−6c = 0.1,
1
c = − ln 0.9 ≈ 0.0176.
6
8.8 SOLUTIONS
(b) Similarly, with c = 0.0176, we have
c
Z
6
12
12
e−ct dt = −e−ct
=e
−6c
6
− e−12c = 0.9 − 0.81 = 0.09,
so the probability is 9%.
7. (a) We can find the proportion of students by integrating the density p(x) between x = 1.5 and x = 2:
2
Z
P (2) − P (1.5) =
x3
dx
4
1.5
4 2
x
16
=
1.5
(2)4
(1.5)4
=
−
= 0.684,
16
16
so that the proportion is 0.684 : 1 or 68.4%.
(b) We find the mean by integrating x times the density over the relevant range:
2
Z
Mean =
x
0
2
Z
=
0
x5
20
=
x3
4
x4
dx
4
dx
2
0
25
= 1.6 hours.
=
20
(c) The median will be the time T such that exactly half of the students are finished by time T , or in other words
Z
1
=
2
T
0
1
x4
=
2
16
x3
dx
4
T
0
1
T4
=
2
16
√
4
T = 8 = 1.682 hours.
8. (a) Since
Z
∞
p(x) dx = 1, we have
0
1=
=
Z
∞
ae−0.122x dx
0
a
e−0.122x
−0.122
∞
=
0
a
.
0.122
So a = 0.122.
(b)
P (x) =
Z
x
p(t) dt
0
=
Z
x
0.122e−0.122t dt
0
= −e0.122t
x
0
= 1 − e−0.122x .
811
812
Chapter Eight /SOLUTIONS
(c) Median is the x such that
P (x) = 1 − e−0.122x = 0.5.
So e−0.122x = 0.5. Thus,
x=−
ln 0.5
≈ 5.68 seconds
0.122
and
Mean =
Z
∞
x(0.122)e−0.122x dx = −
0
Z
0
∞
x −0.122e−0.122x dx.
We now use integration by parts. Let u = −x and v ′ = −0.122e−0.122x . Then u′ = −1, and v = e−0.122x .
Therefore,
∞
Mean = −xe
−0.122x
+
Z
∞
e−0.122x dx =
0
0
(d)
1
≈ 8.20 seconds.
0.122
1
P (x)
0.122
p(x)
x
x
9. (a) The cumulative distribution function
P (t) =
Z
0
t
p(x)dx = Area under graph of density function p(x) for 0 ≤ x ≤ t
= Fraction of population who survive t years or less after treatment
= Fraction of population who survive up to t years after treatment.
(b) The probability that a randomly selected person survives for at least t years is the probability that he lives t years or
longer, so
S(t) =
Z
∞
p(x) dx = lim
b→∞
t
b
= lim −e−Ct
b→∞
t
Z
b
Ce−Ct dx
t
= lim −e−Cb − (−e−Ct ) = e−Ct ,
b→∞
or equivalently,
S(t) = 1 −
Z
t
0
p(x) dx = 1 −
Z
t
t
Ce−Ct dx = 1 + e−Ct
0
0
(c) The probability of surviving at least two years is
S(2) = e−C(2) = 0.70
so
ln e−C(2) = ln 0.70
−2C = ln 0.7
1
C = − ln 0.7 ≈ 0.178.
2
= 1 + (e−Ct − 1) = e−Ct .
8.8 SOLUTIONS
813
10. (a) The probability you dropped the glove within a kilometer of home is given by
Z
1
1
0
2e−2x dx = −e−2x
= −e−2 + 1 ≈ 0.865.
0
(b) Since the probability that the glove was dropped within y km =
1 − e−2y = 0.95
Ry
0
p(x)dx = 1 − e−2y , we solve
e−2y = 0.05
ln 0.05
y=
≈ 1.5 km.
−2
11. (a) Since µ = 100 and σ = 15:
1 x−100 2
1
√ e− 2 ( 15 ) .
15 2π
(b) The fraction of the population with IQ scores between 115 and 120 is (integrating numerically)
p(x) =
Z
120
p(x) dx =
120
Z
115
115
(x−100)2
1
√ e− 450 dx
15 2π
Z
120
(x−100)2
1
√
e− 450 dx
15 2π 115
≈ 0.067 = 6.7% of the population.
=
12. (a) The normal distribution of car speeds with µ = 58 and σ = 4 is shown in Figure 8.138.
x
52
60
µ = 58
65
Figure 8.138
The probability that a randomly selected car is going between 60 and 65 is equal to the area under the curve
from x = 60 to x = 65,
1
Probability = √
4 2π
65
Z
e−(x−58)
2
/(2·42 )
60
dx ≈ 0.2685.
We obtain the value 0.2685 using a calculator or computer.
(b) To find the fraction of cars going under 52 km/hr, we evaluate the integral
1
Fraction = √
4 2π
Z
52
e−(x−58)
2
0
Thus, approximately 6.7% of the cars are going less than 52 km/hr.
/32
dx ≈ 0.067.
814
Chapter Eight /SOLUTIONS
13. (a) First, we find the critical points of p(x):
1
d
p(x) = √
dx
σ 2π
=−
2
−2(x − µ) − (x−µ)
e 2σ2
2
2σ
2
(x − µ) − (x−µ)
√ e 2σ2 .
σ 3 2π
This implies x = µ is the only critical point of p(x).
To confirm that p(x) is maximized at x = µ, we rely on the first derivative test. As −
1
√
e
−
(x−µ)2
2σ2
is
σ 3 2π
′
always negative, the sign of p (x) is the opposite of the sign of (x − µ); thus p (x) > 0 when x < µ, and p (x) < 0
when x > µ.
(b) To find the inflection points, we need to find where p′′ (x) changes sign; that will happen only when p′′ (x) = 0. As
′
′
(x−µ)2
d2
1
−
2σ2
√
e
p(x)
=
−
dx2
σ 3 2π
p′′ (x) changes sign when −
when
−
(x − µ)2
+1 ,
σ2
(x − µ)2
+ 1 does, since the sign of the other factor is always negative. This occurs
σ2
−
(x − µ)2
+ 1 = 0,
σ2
−(x − µ)2 = −σ 2 ,
x − µ = ±σ.
′′
Thus, x = µ+σ or x = µ−σ. Since p (x) > 0 for x < µ−σ and x > µ+σ and p′′ (x) < 0 for µ−σ ≤ x ≤ µ+σ,
these are in fact points of inflection.
(c) µ represents the mean of the distribution, while σ is the standard deviation. In other words, σ gives a measure of the
“spread” of the distribution, i.e., how tightly the observations are clustered about the mean. A small σ tells us that
most of the data are close to the mean; a large σ tells us that the data is spread out.
14. The fraction of the population within one standard deviation of the mean is given by
Fraction within σ of mean =
Z
σ
−σ
√
2
2
1
e−x /(2σ ) dx.
2πσ
1
x
Let us substitute w = so that dw = dx, and when x = ±σ, w = ±1. Then we have
σ
σ
Fraction =
Z
σ
−σ
√
2
2
1
e−x /(2σ ) dx =
2πσ
Z
1
−1
√
2
1
e−w /2 · σdw =
2πσ
Z
1
−1
2
1
√ e−w /2 dw.
2π
This integral is independent of σ. Evaluating the integral numerically gives 0.68, showing that about 68% of the population
lies within one standard deviation of the mean.
15. It is not (a) since a probability density must be a non-negative function; not (c) since the total integral of a probability
density must be 1; (b) and (d) are probability density functions, but (d) is not a good model. According to (d), the
probability that the next customer comes after 4 minutes is 0. In real life there should be a positive probability of not
having a customer in the next 4 minutes. So (b) is the best answer.
16. (a) Since P is the cumulative distribution function, the percentage of households that made between $40,000 and
$60,000 is
P (60) − P (40) = 66.8% − 50.1% = 16.7%.
Therefore 16.7% of the households made between $40,000 and $60,000.
The percentage of households making over $100,000 is 100% − 87.1% = 12.9%.
(b) The median income is the income such that half the households make less than this amount. Looking at the table,
we see that the 50% mark occurs between $20,000 and $40,000. Since P (20) = 29.5%, we know 29.5% of the
households made less than $20,000. Assuming local linearity we calculate the slope of the line connecting (20, 29.5)
with (40, 50.1) as (50.1 − 29.5)/20 = 1.03. If x is the additional income it takes to achieve the median, then
1.03x = 50.0 − 29.5, so x = 19.90. Since 20 + 19.90 = 39.90, the median income is approximately $39,900. This
looks sensible since just over 50% of the population earn $40,000.
(c) The percentage of households that made between $40,000 and $75,000 is 76.2 − 50.1 = 26.1. Since this percentage
is less than 1/3, the statement is false.
8.8 SOLUTIONS
815
Rr
17. (a) Let the p(r) be the density function. Then P (r) = 0 p(x) dx, and from the Fundamental Theorem of Calculus,
d
d
P (r) = dr
(1 − (2r 2 + 2r + 1)e−2r ) = −(4r + 2)e−2r + 2(2r 2 + 2r + 1)e−2r , or p(r) = 4r 2 e−2r .
p(r) = dr
′
We have that p (r) = 8r(e−2r ) − 8r 2 e−2r = e−2r · 8r(1 − r), which is zero when r = 0 or r = 1, negative
when r > 1, and positive when r < 1. Thus p(1) = 4e−2 ≈ 0.54 is a relative maximum.
Here are sketches of p(r) and the cumulative position P (r):
1
1
0.5
0.5
P (r)
p(r)
r
1
2
r
3
1
2
3
(b) The median distance is the distance r such that P (r) = 1 − (2r 2 + 2r + 1)e−2r = 0.5, or equivalently, (2r 2 + 2r +
1)e−2r = 0.5.
By experimentation with a calculator, we find that r ≈ 1.33RBohr radii is the medianR distance.
x
∞
The mean distance is equal to the value of the integral 0 rp(r) dr = lim 0 rp(r) dr. We have that
Rx
0
rp(r) dr =
Rx
0
x→∞
4r 3 e−2r dr. Using the integral table, we get
Z
x
4r 3 e−2r dr =
0
h
−
1
1
1
1
(24) e−2x
4r 3 − (12r 2 ) − (24r) −
2
4
8
16
i
3 −2x
3
e
.
= − 2x3 + 3x2 + 3x +
2
2
h
i
x
0
Taking the limit of this expression as x → ∞, we see that all terms involving (powers of x or constants) · e−2x have
limit 0, and thus the mean distance is 1.5 Bohr radii.
The most likely distance is obtained by maximizing p(r) = 4r 2 e−2r ; as we have already seen this corresponds
to r = 1 Bohr unit.
(c) Because it is the most likely distance of the electron from the nucleus.
Strengthen Your Understanding
18. A median satisfies P (T ) = 0.5 where P is the cumulative distribution function.
19. The median is the value which divides the area under the density function graph into halves. The median of this function
cannot be 1 since all of the area is to the left of x = 1. The median is between 0 and 1.
20. We can use the normal distribution
p(x) =
2
2
1
√ e−(x−µ) /(2σ )
σ 2π
with µ = σ = 1/2 to get
2
2
p(x) = √ e−2(x−1/2) .
2π
21. We can take a constant, or uniform distribution
p(x) =
Since
Z
0 for x < 0
1 for
1
x dx =
0
Since
Z
0
1/2
p(x) dx =
0 for
Z
0
1
,
2
0≤x≤1
x > 1.
the mean is
1/2
1 dx =
1
,
2
1
.
2
the median is also
1
.
2
816
Chapter Eight /SOLUTIONS
22. False. Note that p is the density function for the population, not the cumulative density function. Thus p(10) = 1/2 means
that the probability of x lying in a small interval of length ∆x around x = 10 is about (1/2)∆x.
23. True. This follows directly from the definition of the cumulative density function.
24. True. The interval from x = 9.98 to x = 10.04 has length 0.06. Assuming that the value of p(x) is near 1/2 for
R 10.04
9.98 < x < 10.04, the fraction of the population in that interval is 9.98 p(x) dx ≈ (1/2)(0.06) = 0.03.
25. False. Note that p is the density function for the population, not the cumulative density function. Thus p(10) = p(20)
means that x values near 10 are as likely as x values near 20.
26. True. By the definition of the cumulative distribution function, P (20) − P (10) = 0 is the fraction of the population
having x values between 10 and 20.
Solutions for Chapter 8 Review
Exercises
1. Vertical slices are circular. Horizontal slices would be similar to ellipses in cross-section, or at least ovals (a word derived
from ovum, the Latin word for egg).
Figure 8.139
2. The limits of integration are 0 and b, and the rectangle represents the region under the curve f (x) = h between these
limits. Thus,
Area of rectangle =
b
b
Z
h dx = hx
= hb.
0
0
3. The circle x2 + y 2 = r 2 cannot be expressed as a function y = f (x), since for every x with −r < x < r, there are two
corresponding y values on the circle. However, if we consider the top half of the circle√
only, as shown below, we have
x2 + y 2 = r 2 , or y 2 = r 2 − x2 , and taking the positive square root, we have that y = r 2 − x2 is the equation of the
top semicircle.
y=
√
r 2 − x2
✎
−r
r
Then
Area of Circle = 2(Area of semicircle) = 2
Z
r
−r
p
r 2 − x2 dx
SOLUTIONS to Review Problems for Chapter Eight
817
We evaluate this using integral table formula 30.
2
Z
x=r
x=−r
p
r 2 − x2 dx = 2
h p
1
2
x
r 2 − x2 + r 2 arcsin
x
r
2
= r (arcsin 1 − arcsin(−1))
π
π
= πr 2 .
− −
= r2
2
2
i
r
−r
4. Name the slanted line y = f (x). Then the triangle is the region under the line y = f (x) and between the lines y = 0 and
x = b. Thus,
Z
b
Area of triangle =
f (x) dx.
0
Since f (x) is a line of slope h/b which passes through the origin, its equation is f (x) = hx/b. Thus,
Area of triangle =
Z
b
0
hx2
hx
dx =
b
2b
b
=
0
hb2
hb
=
.
2b
2
5. We slice the region vertically. Each rotated slice is approximately a cylinder with radius y = x2 + 1 and thickness ∆x.
See Figure 8.140. The volume of a typical slice is π(x2 + 1)2 ∆x. The volume, V , of the object is the sum of the volumes
of the slices:
X
V ≈
π(x2 + 1)2 ∆x.
As ∆x → 0 we obtain an integral.
V =
Z
0
4
π(x2 + 1)2 dx = π
Z
4
(x4 + 2x2 + 1)dx = π
0
y
2x3
x5
+
+x
5
3
4
=
0
3772π
= 790.006.
15
y = x2 + 1
✲∆x✛
y✻
❄
x
4
Figure 8.140
√
6. We slice the region vertically. Each rotated slice √
is approximately a cylinder with radius y = x and thickness ∆x. See
Figure 8.141. The volume of a typical slice is π( x)2 ∆x. The volume, V , of the object is the sum of the volumes of the
slices:
X √ 2
π( x) ∆x.
V ≈
818
Chapter Eight /SOLUTIONS
As ∆x → 0 we obtain an integral.
V =
Z
2
√
π( x)2 dx = π
1
Z
2
xdx = π
1
x2
2
y
∆x
✲ ✛
y=
√
2
=
1
3π
= 4.712.
2
x
✻
y
❄
1
2
x
Figure 8.141
7. We slice the region vertically. Each rotated slice is approximately a cylinder with radius y = e−2x and thickness ∆x. See
Figure 8.142. The volume of a typical slice is π(e−2x )2 ∆x. The volume, V , of the object is the sum of the volumes of
the slices:
X
V ≈
π(e−2x )2 ∆x.
As ∆x → 0 we obtain an integral.
V =
Z
1
π(e−2x )2 dx = π
0
Z
1
0
e−4x dx = π −
1
(e−4x )
4
y
✲ ∆x✛
✻
y
y = e−2x
❄
1
Figure 8.142
x
1
0
π
= − (e−4 − 1) = 0.771.
4
SOLUTIONS to Review Problems for Chapter Eight
819
8. We slice the region vertically. Each rotated slice is approximately a cylinder with radius y = 4 − x2 and thickness ∆x.
See Figure 8.143. The volume, V , of a typical slice is π(4 − x2 )2 ∆x. The volume of the object is the sum of the volumes
of the slices:
X
V ≈
π(4 − x2 )2 ∆x.
As ∆x → 0 we obtain an integral. Since the region lies between x = −2 and x = 2, we have:
V =
Z
2
−2
π(4 − x2 )2 dx = π
Z
2
−2
(16 − 8x2 + x4 )dx = π
16x −
8x3
x5
+
3
5
2
=
−2
512π
= 107.233.
15
y
∆x
✛
✲
y =4−
x2
✻
y
❄
−2
x
2
Figure 8.143
9. We divide the region into vertical strips of thickness ∆x. As a slice is rotated about the x-axis, it creates a disk of radius
rout from which has been removed a smaller circular disk of inside radius rin . We see in Figure 8.144 that rout = 2x and
rin = x. Thus,
Volume of a slice ≈ π(rout )2 ∆x − π(rin )2 ∆x = π(2x)2 ∆x − π(x)2 ∆x.
To find the total volume, V , we integrate this quantity between x = 0 and x = 3:
V =
Z
0
3
(π(2x)2 − π(x)2 )dx = π
Z
0
3
(4x2 − x2 ) dx = π
y
✻
y = 2x
rout = 2x
y=x
✻
rin = x
❄
❄
3
Figure 8.144
x
Z
0
3
3
3x2 dx = πx3
= 27π = 84.823.
0
820
Chapter Eight /SOLUTIONS
10. The two functions intersect at (0, 0) and (8, 2). We slice the volume with planes perpendicular to the x-axis. This divides
the solid into thin washers with
2
2
Volume of slice = πrout
∆x − πrin
∆x.
The inner radius is the vertical distance from the x-axis to the curve y = 41 x. Similarly, the outer radius is the vertical
√
distances from the x-axis to the curve y = 3 x. Integrating from x = 0 to x = 8 we have
V =
8
Z
0
i
√
1
π( 3 x)2 − π( x)2 dx.
4
h
11. The region is bounded by y = 2, the y-axis and y = x1/3 . The two functions y = 2 and y = x1/3 intersect at (8, 2). We
slice the volume with planes that are perpendicular to the y-axis. This divides the solid into thin cylinders with
Volume ≈ πr 2 ∆y.
The radius is the distance from the y-axis to the curve x = y 3 . Integrating from y = 0 to y = 2 we have
V =
Z
2
π(y 3 )2 dy.
0
12. The region is bounded by y = 2, the y-axis and y = x1/3 . The two functions y = 2 and y = x1/3 intersect at (8, 2). We
slice the volume with planes that are perpendicular to the line y = −2. This divides the solid into thin washers with
2
2
Volume ≈ πrout
∆x − πrin
∆x.
The inner radius is the distance from the line y = −2 to the curve y = x1/3 and the outer radius is the distance from the
line y = −2 to the line y = 2. Integrating from x = 0 to x = 8 we have
V =
Z
8
π(2 − (−2))2 − π(x1/3 − (−2))2 dx.
0
13. The region is bounded by x = 4y, the x-axis and x = 8. The two lines x = 4y and x = 8 intersect at (8, 2). We slice the
volume with planes that are perpendicular to the line x = 10. This divides the solid into thin washers with
2
2
Volume ≈ πrout
dy − πrin
dy.
The inner radius is the distance from the line x = 10 to the line x = 8 and the outer radius is the distance from the line
x = 10 to the line x = 4y. Integrating from y = 0 to y = 2 we have
2
Z
V =
0
1
x,
4
π(10 − 4y)2 − π(2)2 dy.
14. The region is bounded by y =
the x-axis and x = 8. The two lines y = 41 x and x = 8 intersect at (8, 2). We slice
the volume with planes that are perpendicular to the line y = 3. This divides the solid into thin washers with
2
2
Volume ≈ πrout
∆x − πrin
∆x.
The inner radius is the distance from the line y = 3 to the line y = 14 x and the outer radius is the distance from the line
y = 3 to the x-axis. Integrating from x = 0 to x = 8 we have
V =
Z
8
h
π(1 −
0
1 2
x) − π(3)2 dx.
4
i
15. The region is bounded by x = 4x and x = y 3 . The two functions intersect at (0, 0) and (8, 2). We slice the volume with
planes that are perpendicular to the line x = −3. This divides the solid into thin washers with
2
2
Volume = πrout
∆y − πrin
∆y.
The inner radius is the distance from the line x = −3 to x = y 3 and the outer radius is the distance from the line x = −3
to the line x = 4y. Integrating from y = 0 to y = 2 we have
V =
Z
0
2
π(4y + 3)2 − π(y 3 + 3)2 dx.
SOLUTIONS to Review Problems for Chapter Eight
16. Each slice is a circular disk. The radius, r, of the disk increases with h and is given in the problem by r =
√
821
h. Thus
2
Volume of slice ≈ πr ∆h = πh∆h.
Summing over all slices, we have
Total volume ≈
Taking a limit as ∆h → 0, we get
Total volume = lim
∆h→0
Evaluating gives
X
Total volume = π
X
πh∆h.
Z
πh∆h =
12
πh dh.
0
12
h2
2
= 72π.
0
Slice parallel to the base of the cone, or, equivalently, rotate
the line x = (3 − y)/3 about the y–axis. (One can also slice
the other way.) See Figure 8.145. The volume V is given by
17.
y
Radius = 1 − 3
y
V =
Z
y=3
πx2 dy =
y=0
=π
Z
3
0
❘
=π
x
Z
3
π
0
1−
y2
2y
+
3
9
y2
y3
y−
+
3
27
3−y
3
2
dy
dy
3
= π.
0
Figure 8.145
18. (a) We slice the pyramid horizontally. See Figure 8.146. Each slice is a square slab of thickness ∆h, so the volume of a
slice at height h is s2 ∆h, where s is the length of a side. We use the similar triangles in Figure 8.147 to write s as a
function of h:
8
s
=
so s = 0.8(10 − h).
10 − h
10
2
The volume of the slice at height h is (0.8(10 − h)) ∆h. To find the total volume, we integrate this quantity from
h = 0 to h = 10.
10
Z
V =
(0.8(10 − h))2 dh = 0.64
0
(b) As in part (a),
Z
10
0
(h − 10)2 dh =
16
(h − 10)3
75
10
=
0
640
= 213.333 m3 .
3
Volume of a slice at height h ≈ s2 ∆h = (0.8(10 − h))2 ∆h.
The height h ranges from h = 0 to h = 6. We have
V =
Z
0
6
2
(0.8(10 − h)) dh = 0.64
✛
∆h
❄
✻
10 m
Z
(h − 10)2 dh =
16
(h − 10)3
75
6
=
0
4992
= 199.680 m3 .
25
✻
✻
(10 − h)
✛
❄
✻
✛
s
10 m
✲
✛
0
6
h
✛
✛
✛
✛
Figure 8.146
✛
8m
8m
h
❄
✛
8m
Figure 8.147
✲
❄
822
Chapter Eight /SOLUTIONS
19. We slice the tank horizontally. There is an outside radius rout and an inside radius rin and, at height h,
Volume of a slice ≈ π(rout )2 ∆h − π(rin )2 ∆h.
See Figure 8.148. We see that rout = 3 for every slice. We use similar triangles to find rin in terms of the height h:
3
rin
=
h
6
so rin =
1
h.
2
At height h,
Volume of slice ≈ π(3)2 ∆h − π
2
1
h
2
To find the total volume, we integrate this quantity from h = 0 to h = 6.
V =
Z
0
6
2
π(3) − π
1 2
2
h
dh = π
6
Z
0
✛
1
9 − h2 dh = π
4
3m
✲✛
rout
✲
∆h.
h3
9h −
12
6
= 36π = 113.097 m3 .
0
✻
✛rin✲
✻
6m
h
❄
❄
Figure 8.148
20. Since f (x) = sin x, f ′ (x) = cos(x), so
Arc Length =
Z
π
0
p
1 + cos2 x dx.
21. We’ll find the arc length of the top half of the ellipse, and multiply that by 2. In the top half of the ellipse, the equation
(x2 /a2 ) + (y 2 /b2 ) = 1 implies
r
y = +b
1−
x2
.
a2
Differentiating (x2 /a2 ) + (y 2 /b2 ) = 1 implicitly with respect to x gives us
2y dy
2x
+ 2
= 0,
a2
b dx
so
dy
=
dx
−2x
a2
2y
b2
=−
b2 x
.
a2 y
Substituting this into the arc length formula, we get
Arc Length =
Z
a
Z
a
Z
a
1+
−a
=
−a
=
s
−a
v
u
u
t1 +
s
1+
−
b2 x
a2 y
2
dx
b4 x 2
a4 (b2 )(1 −
x2
)
a2
b2 x 2
2
a (a2 − x2 )
!
dx.
dx
SOLUTIONS to Review Problems for Chapter Eight
823
Hence the arc length of the entire ellipse is
2
s
a
Z
−a
1+
b2 x 2
a2 (a2 − x2 )
dx.
22. Since f ′ (x) = cos x, we have
L=
3
Z
p
1 + (f ′ (x))2 dx =
0
Z
3
p
1 + cos2 x dx = 3.621.
0
We see in Figure 8.149 that the length of the curve is slightly longer than the length of the x-axis from x = 0 to x = 3,
so the answer of 3.621 makes sense.
f (x) = sin x
x
3π
Figure 8.149
23. Since f ′ (x) = 10x, we have
L=
Z
0
3
p
1 + (f ′ (x))2 dx =
Z
0
3
Z
p
1 + (10x)2 dx =
0
3
p
1 + 100x2 dx = 45.230.
We see in Figure 8.150 that the length of the curve is definitely longer than 45 and slightly longer than
45.10, so the answer of 45.230 is reasonable.
√
452 + 32 =
f (x) = 5x2
45
x
3
Figure 8.150
√
24. The arc length of 1 − x2 from x = 0 to x = 1 is one quarter of the perimeter of the unit circle. Hence the length is
2π
π
= .
4
2
25. The arc length is given by
Z 2p
1 + e2x dx ≈ 4.785.
L=
1
√
Note that 1 + e2x does not have an obvious elementary antiderivative, so we use an approximation method to find an
approximate value for L.
26. The arc length is given by
L=
Z
2
Z
2
1+
1
=
1
s
1
1
x4 +
−
(16x4 )
2
1
x + x−2 dx =
4
2
dx =
1
x3
−
3
4x
Z
2
1
2
=
1
s
x2
59
.
24
1
+
(4x2 )
2
dx
824
Chapter Eight /SOLUTIONS
27. We have dx/dt = −3 sin t, dy/dt = 2 cos t, so, evaluating the integral numerically, we have
Arc length =
Z
0
The curve is a ellipse.
2π
p
9 cos2 t + 4 sin2 t dt = 15.865.
28. We have dx/dt = −2 sin(2t), dy/dt = 2 cos(2t), so, simplifying the integrand, we have
Arc length =
Z
π
29.
•
Z
2
4 sin (2t) +
0
The curve is a circle of radius 1.
p
4 cos2 (2t) dt
=2
Z
π
dt = 2π.
0
1
f (x) dx gives the area under the graph of f from 0 to 1.
0
• The graph of f is concave up and passes through the points (0, 0) and (1, 1), so it lies below the line y = x.
Z 1
1
f (x) dx < .
• The area under y = x from 0 to 1 is half the area of a square of side 1, or 1/2. Thus,
2
0
30.
31.
• Since f (0) = 0, the Fundamental Theorem gives
Z
0.5
0
f ′ (x) dx = f (0.5) − f (0) = f (0.5).
• The graph of f is concave up and passes through the points (0, 0) and (1, 1), so it lies below the line y = x. This
means f (x) < x for 0 < x < 1.
Z 0.5
1
• Since f (x) < x, we have f (0.5) < 0.5. Hence
f ′ (x)dx < .
2
0
p
• Since f (x) = xp , we know f −1 (x) = x1/p , because (xp )1/p = x1/p = x.
• Since f (0) = 0 and f (1) = 1, we know f −1 (0) = 0 and f −1 (1) = 1. Likewise, since f is increasing, so is f −1 .
• Since p > 1, we know 0 < 1/p < 1, so the graph of f −1 is concave down, and therefore lies above the line y = x
on
Z 0 < x < 1.
1
•
f −1 (x) dx gives the area under f −1 from 0 to 1, so, since the graph lies above y = x on this interval, we have
Z0 1
1
.
2
f −1 (x) dx >
0
32.
•
Z
1
Z
1
π (f (x))2 dx gives the volume of the region formed by rotating the graph of f on 0 ≤ x ≤ 1 about the x-axis.
0
• The graph of f is concave up and contains (0, 0) and (1, 1), so it lies below the line y = x on 0 < x < 1.
• This means the region formed by rotating the graph of f lies within the region formed by rotating the line segment
y = x, which is a cone of base r = 1 and height h = 1. The volume of this cone is (1/3)πr 2 h = π/3.
Z 1
π
• Since this cone contains the region formed by rotating the graph of f , we have
π (f (x))2 dx < .
3
0
33.
•
0
q
1 + (f ′ (x))2 dx gives the arc length of the graph of f from x = 0 to x = 1.
• The graph of f is concave up and contains (0,√
0) and (1, 1),√so it lies below the line y = x on 0 < x < 1. The arc
length of the line between (0, 0), and (1, 1) is 12 + 12 = 2.
Z 1q
√
1 + (f ′ (x))2 dx > 2.
• The line segment between (0, 0) and (1, 1) is shorter than the arc length of f , so
0
Problems
34. (a) The points of intersection are x = 0 to x = 2, so we have
Area =
Z
0
2
(2x − x2 )dx = x2 −
x3
3
2
=
0
4
= 1.333.
3
2
(b) The outside radius is 2x and the inside radius is x , so we have
Volume =
Z
0
2
(π(2x)2 − π(x2 )2 )dx = π
Z
0
2
(4x2 − x4 )dx =
π
(20x3 − 3x5 )
15
2
=
0
64π
= 13.404.
15
SOLUTIONS to Review Problems for Chapter Eight
825
(c) The length of the perimeter is equal to the length of the top plus the length of the bottom. Using the arclength formula,
and the fact that the derivative of 2x is 2 and the derivative of x2 is 2x, we have
L=
Z
0
2
p
1 + 22 dx +
Z
2
p
1 + (2x)2 dx = 4.4721 + 4.6468 = 9.119.
0
√
√
35. There are at least two possible answers. Since √4 − x2 − (− 4 − x2 ) ≥ 0 when 0 ≤ x ≤ 2, one possibility is that the
integral gives the area between
the curve√y = 2 4 − x2 and the line y = 0 as shown in Figure 8.151.
√
4
−
x2 ≥ − 4 − x2 when 0 ≤ x ≤ 2, the integral gives the area between the quarter circles
Alternatively,
since
√
√
2
y = 4 − x and y = − 4 − x2 , as shown in Figure 8.152.
y
y
4
2
y=
√
y = 2 4 − x2
√
4 − x2
x
2
x
−2
2
Figure 8.151
√
y = − 4 − x2
Figure 8.152
36. The two functions intersect at (0, 0) and (5, 25). We slice the volume with planes perpendicular to the horizontal line
y = 30. This divides the solid into thin washers with volume
Volume of slice = π((rout )2 − (rin )2 )∆x.
The outer radius is the vertical distance from the line y = 30 to the curve y = x2 , so rout = 30 − x2 . Similarly, the inner
radius is the vertical distance from the line y = 30 to the curve y = 5x, so rin = 30 − 5x. Integrating from x = 0 to
x = 5 we have
Z
5
V =
0
π((30 − x2 )2 − (30 − 5x)2 ) dx.
37. The two functions intersect at (0, 0) and (5, 25). We slice the volume with planes perpendicular to the vertical line x = 8.
This divides the solid into thin washers with volume
Volume of slice = π((rout )2 − (rin )2 )∆y.
The outer radius is the horizontal distance from the line x = 8 to the curve x = y/5, so rout = 8 − y/5. Similarly, the
√
√
inner radius is the horizontal distance from the line x = 8 to the curve x = y, so rin = 8 − y. Integrating from y = 0
to y = 25 we have
Z 25
√
π((8 − y/5)2 − (8 − y)2 ) dy.
V =
0
826
Chapter Eight /SOLUTIONS
38. (a) See Figure 8.153
r=
x
√
✲
x
x
y
y
Figure 8.153: Rotated
Region
Figure 8.154: Cutaway View
√
(b) Divide [0,1] into N subintervals of width ∆x = N1 . The volume of the ith disc is π( xi )2 ∆x = πxi ∆x. So,
PN
V ≈ i=1 πxi ∆x. See Figure 8.154
(c)
Volume =
1
Z
πx dx =
0
1
π 2
x
2
=
0
π
≈ 1.57.
2
39. (a) See Figure 8.155.
z
rin
z
✲
rout = 1
x
(y
√= 1)
=1− x
✲
✲
rin = y 2
rout = 1
✲
y
x
y
Figure 8.155
Figure 8.156
√
Slice the figure perpendicular to the x–axis. One gets washers of inner radius 1 − x and outer radius 1.
Therefore,
V =
Z
=π
1
π12 − π(1 −
0
Z
1
0
=π
h
√
x)2 dx
√
(1 − [1 − 2 x + x]) dx
4 23
1
x − x2
3
2
i1
=
0
5π
≈ 2.62.
6
(b) See Figure 8.156. Note that x = y 2 . We now integrate over y instead of x, slicing perpendicular to the y–axis. This
gives us washers of inner radius x and outer radius 1. So
V =
Z
y=1
y=0
=
Z
0
(π12 − πx2 ) dy
1
π(1 − y 4 )dy
= πy −
π 5
y
5
1
0
=π−
π
4π
=
≈ 2.51.
5
5
SOLUTIONS to Review Problems for Chapter Eight
827
40. (a) Since y = ax2 is non-negative, we integrate to find the area:
Area =
Z
2
(ax2 )dx = a
0
2
x3
3
=
0
8a
.
3
(b) Each slice of the object is approximately a cylinder with radius ax2 and thickness ∆x. We have
Z
Volume =
2
x5
5
π(ax2 )2 dx = πa2
0
41. (a) Since y = e
−bx
2
=
0
32 2
a π.
5
is non-negative, we integrate to find the area:
Area =
Z
1
(e−bx )dx =
0
(b) Each slice of the object is approximately a cylinder with radius e
Volume =
Z
1
π(e−bx )2 dx = π
Z
=
0
−bx
1
e−2bx dx =
0
0
1
−1 −bx
e
b
1
(1 − e−b ).
b
and thickness ∆x. We have
−π −2bx
e
2b
1
=
0
π
(1 − e−2b ).
2b
42. (a) We divide the region into vertical strips of thickness ∆x. As a slice is rotated about the x-axis, it creates a disk
of radius rout from which has been removed a smaller circular disk of radius rin . We see in Figure 8.157 that
rout = sin x and rin = 0.5x. Thus,
Volume of a slice ≈ π(rout )2 ∆x − π(rin )2 ∆x = π(sin x)2 ∆x − π(0.5x)2 ∆x.
To find the total volume, we integrate this quantity between the points of intersection x = 0 and x = 1.9:
V =
Z
0
1.9
(π(sin x)2 − π(0.5x)2 )dx = π
−
sin x cos x
x3
−
+x
2
12
1.9
= 1.669.
0
y
5
✻
✻
rmin = 5 − sin x
y
rmout = 5 − 0.5x
y = sin x
y = sin x
✻
y = 0.5x
rout = sin x
rin =✻
0.5x
❄
❄
❄
❄
y = 0.5x
x
x
1.9
Figure 8.157
1.9
Figure 8.158
(b) We see in Figure 8.158 that rout = 5 − 0.5x and rin = 5 − sin x. Thus,
Volume of a slice ≈ π(rout )2 ∆x − π(rin )2 ∆x = π(5 − 0.5x)2 ∆x − π(5 − sin x)2 ∆x.
To find the total volume, V , we integrate this quantity between the points of intersection x = 0 and x = 1.9:
V =
Z
1.9
0
(π(5 − 0.5x)2 − π(5 − sin x)2 ) dx =
π
(6(sin x − 20) cos x + x(x2 − 30x − 6))
12
1.9
= 11.550.
0
828
Chapter Eight /SOLUTIONS
43. We divide the region into vertical strips of thickness ∆x. As a slice is rotated about the x-axis, it creates a disk of radius
rout from which has been removed a disk of radius rin . We see in Figure 8.159 that rout = 5 + 2x and rin = 5. Thus,
Volume of a slice ≈ π(rout )2 ∆x − π(rin )2 ∆x = π(5 + 2x)2 ∆x − π(5)2 ∆x.
To find the total volume, V , we integrate this quantity between x = 0 and x = 4:
4
Z
V =
2
2
(π(5 + 2x) − π(5) )dx = π
0
Z
4
0
((5 + 2x)2 − 25) dx = π
y
5
4 3
x + 10x2
3
4
736π
= 770.737.
3
=
0
y = 5 + 2x
rout
✻
✻
r
x
in
4
=5
y = −5
−5
Figure 8.159
44. (a) We divide the region into vertical strips of thickness ∆x. As a slice is rotated about the x-axis, it creates a disk of
radius rout from which has been removed a disk of radius rin . We see in Figure 8.160 that rout = 2 + x2 and rin = 2.
Thus,
Volume of a slice ≈ π(rout )2 ∆x − π(rin )2 ∆x = π(2 + x2 )2 ∆x − π(2)2 ∆x.
To find the total volume, V , we integrate this quantity between x = 0 and x = 3:
V =
Z
3
2 2
0
2
(π(2 + x ) − π(2) )dx = π
Z
3
3
π
((2 + x ) − 4) dx =
(3x5 + 20x3 )
15
2 2
0
=
423π
= 265.778.
5
0
2
(b) We see in Figure 8.161 that rout = 10 and rin = 10 − x . Thus,
Volume of a slice ≈ π(rout )2 ∆x − π(rin )2 ∆x = π(10)2 ∆x − π(10 − x2 )2 ∆x.
To find the total volume, V , we integrate this quantity between x = 0 and x = 3:
V =
Z
0
3
2
2 2
(π(10) − π(10 − x ) )dx = π
y
y = x2
Z
0
3
3
π
(100x3 −3x5 )
(100 − (10 − x ) ) dx =
15
2 2
=
657π
= 412.805.
5
0
y
10
✻
y = 10
y = x2
✻
rout = 10
✻
rout = 2 + x2
✻rin = 2
Figure 8.160
rin = 10 − x2
x
3
y = −2
x
3
Figure 8.161
SOLUTIONS to Review Problems for Chapter Eight
45. Slice the object into disks vertically, as in Figure 8.162. A typical disk has thickness ∆x and radius y =
Volume of disk ≈ πy 2 ∆x = π(1 − x2 ) ∆x.
Volume of solid = lim
∆x→0
X
2
π(1 − x ) ∆x =
1
Z
2
π(1 − x ) dx = π
0
x3
x−
3
√
829
1 − x2 . Thus
1
=
2π
.
3
0
Note: As we expect, this is the volume of a half sphere.
y
1
1
x
✲ ✛
∆x
Figure 8.162
46. Slice the object
pinto rings horizontally, as in Figure 8.163. A typical ring has thickness ∆y, inner radius 2 and outer radius
2 + x = 2 + 1 − y 2 . Thus,
Volume of ring ≈ π(2 +
Volume of solid =
Z
1
p
p
1 − y 2 + 1 − y 2 ) dy
π(4
0
= 4π
Z
1
1
2
= 2πy
p
1 − y 2 dy + π
0
= 4π
y
1
p
1−
y2
p
0
1
Z
+
0
Z
0
1 dy − π
1
1 − y 2 + 1 − y 2 ) ∆y.
1
Z
y 2 dy
0
1
p
1 − y2
1 − y 2 + 2π arcsin y + πy −
= 0 + 2π arcsin 1 + π −
= π2 +
p
1 − y 2 )2 ∆y − π22 ∆y = π(4
πy
3
dy
!!
1
+ πy
0
−
πy 3
3
3 1
0
π
− 0 − 2π arcsin 0 − 0 + 0
3
2π
= 11.964.
3
x2 + y 2 = 1
Axis
y
x = −2
1
✠
❄∆y
✻
−2
✲
Inner radius
Outer radius
Figure 8.163: Cross-section of solid
x
1
✲
1
0
830
Chapter Eight /SOLUTIONS
47. Slice the object
pinto rings horizontally, as in Figure 8.164. A typical ring has thickness ∆y, outer radius 1, and inner radius
1 − x = 1 − 1 − y 2 . Thus,
Volume of ring ≈ π12 ∆y − π(1 −
Volume of solid =
1
Z
= 2π
1
Z
1
= 2π ·
2
= πy
1 − y 2 − (1 − y 2 )) ∆y.
1 − y 2 − 1 + y 2 ) dy
p
0
1 − y 2 )2 ∆y = π(2
p
π(2
0
p
p
1 − y 2 dy − π
y
1
p
1−
Z
1 dy + π
0
2
y2
1
+1
0
p
Z
1
0
Z
1
y 2 dy
0
1
p
1 − y 2 + π arcsin y − πy +
1 − y2
πy
3
dy
!
1
− πy
+
0
πy 3
3
1
0
3 1
0
π
π2
−π+ −0−0+0−0
= 0+
2
3
π2
2π
=
−
= 2.840.
2
3
Outer radius
Inner radius
Axis
❄
y
1
∆y
✛
✛❄
❄
✻
x
1
2
Figure 8.164: Cross-section of solid
48. Slicing perpendicularly to the x-axis gives squares whose thickness is ∆x and whose side is y =
ure 8.165. Thus,
p
Volume of square slice ≈ ( 1 − x2 )2 ∆x = (1 − x2 ) ∆x.
Volume of solid =
Z
0
(1 − x2 ) dx = x −
x3
3
1
=
0
x2 + y 2 = 1
y
1
1
✠
✻
∆y
Base of square (standing on paper)
✠
❄
✲ ✛
x
1
Delta x
Figure 8.165: Base of solid
2
.
3
√
1 − x2 . See Fig-
SOLUTIONS to Review Problems for Chapter Eight
49. Slicing perpendicularly to the y-axis gives semicircles whose thickness is ∆y and whose diameter is x =
Figure 8.166. Thus
p
1 − y2
2
π
Volume of semicircular slice ≈
2
Volume of solid =
Z
1
π
π
(1 − y 2 ) dy =
8
8
0
!2
y
1
π 2
π
· =
.
8 3
24
=
0
y
1
1
Diameter of semicircle (standing on paper)
∆y
1 − y 2 . See
π
(1 − y 2 ) ∆y.
8
∆y =
y3
3
y−
831
p
❄
Leg of isosceles triangle (standing on paper)
✠
✻✛
✲
∆x
✠
❄
✻
✛
∆y
✲
∆x
x
x
1
1
Figure 8.166: Base of Solid
Figure 8.167: Base of solid
50. An isosceles triangle with legs of length s has
Area =
1 2
s .
2
Slicing perpendicularly to the y-axis gives isosceles triangles whose thickness is ∆y and whose leg is x =
Figure 8.167. Thus
1 p
1
Volume of triangular slice ≈ ( 1 − y 2 ) ∆y = (1 − y 2 ) ∆y.
2
2
Volume of solid =
Z
1
1
1
(1 − y 2 ) dy =
2
2
0
y3
y−
3
p
1 − y 2 . See
1
=
1
.
3
0
51. The curve y = x(x − 3)2 has x-intercepts at x = 0, 3 and lies above the x-axis on this interval.
Thus,
Z
0
3
x(x − 3)2 dx gives the area under the graph of f from x = 0 to x = 3.
52. The curve y = x(x − 3)2 has x-intercepts at x = 0, 3 and lies above the x-axis on this interval. Rotating the curve about
the x-axis forms a solid of revolution with
Volume =
Z
3
π (f (x))2 dx =
Z
3
0
0
π x(x − 3)2
2
Z
dx =
3
0
πx2 (x − 3)4 dx.
Thus, this expression represents a volume of revolution about the x-axis between x = 0 and x = 3.
53. Since y = (ex + e−x )/2, y ′ = (ex − e−x )/2. The length of the catenary is
Z
1
−1
p
1+
(y ′ )2
dx =
Z
=
Z
1
−1
1
−1
=
x
r
ex − e−x
1+
2
h
r
h ex + e−x i2
e −e
2
2
−x
1
−1
i2
dx =
dx =
Z
1
−1
= e − e−1 .
Z
1
r
1+
−1
e2x
1
e−2x
− +
dx
4
2
4
ex + e−x
dx
2
832
Chapter Eight /SOLUTIONS
54. (a) Slice the headlight into N disks of height ∆x by cutting perpendicular to the x–axis. The radius of each disk is y;
the height is ∆x. The volume of each disk is πy 2 ∆x. Therefore, the Riemann sum approximating the volume of the
headlight is
N
X
πyi2 ∆x
=
π
Z
0
4
π
4
i=1
i=1
(b)
N
X
9xi
9
9x
dx = π x2
4
8
∆x.
4
= 18π.
0
55. (a) The line y = ax must pass through (l, b). Hence b = al, so a = b/l.
(b) Cut the cone into N slices, slicing perpendicular to the x–axis. Each piece is almost a cylinder. The radius of the ith
bxi
cylinder is r(xi ) =
, so the volume
l
N
2
X
bxi
π
∆x.
V ≈
l
i=1
Therefore, as N → ∞, we get
V =
Z
l
πb2 l−2 x2 dx
0
b2
=π 2
l
x3
3
l
=
0
b2
π 2
l
3
l
3
=
1 2
πb l.
3
56. (a) If you slice the apple perpendicular to the core, you expect that the cross section will be approximately a circle.
f (h)
H
h
1.5
1
If f (h) is the radius of the apple at height h above the bottom, and H is the height of the apple, then
Volume =
Z
H
πf (h)2 dh.
0
Ignoring the stem, H ≈ 3.5. Although we do not have a formula for f (h), we can estimate it at various points.
(Remember, we measure here from the bottom of the apple, which is not quite the bottom of the graph.)
h
0
0.5
1
1.5
2
2.5
3
3.5
f (h)
1
1.5
2
2.1
2.3
2.2
1.8
1.2
Now let g(h) = πf (h)2 , the area of the cross-section at height h. From our approximations above, we get the
following table.
SOLUTIONS to Review Problems for Chapter Eight
h
g(h)
0
0.5
1
1.5
2
2.5
3
3.5
3.14
7.07
12.57
13.85
16.62
13.85
10.18
4.52
833
We can now take left- and right-hand sum approximations. Note that ∆h = 0.5 inches. Thus
LEFT(9) = (3.14 + 7.07 + 12.57 + 13.85 + 16.62 + 13.85 + 10.18)(0.5) = 38.64.
RIGHT(9) = (7.07 + 12.57 + 13.85 + 16.62 + 13.85 + 10.18 + 4.52)(0.5) = 39.33.
57.
Thus the volume of the apple is ≈ 39 cu.in.
(b) The apple weighs 0.03 × 39 ≈ 1.17 pounds, so it costs about 94c/.
z
✛
z
y
rout = 3 + y
y
✛
rin = 3 − y
(y = 3)
(y = 3)
x
x
Figure 8.169: Slice of Torus
Figure 8.168: The Torus
As shown in Figure 8.169, we slice the torus perpendicular to the line y = 3. We obtain washers with width dx, inner
2
2
radius rin = 3 − y, and outer √
radius rout = 3 + y. Therefore, the area of the washer is πrout
− πrin
= π[(3 + y)2 −
(3 − y)2 ] = 12πy. Since y = 1 − x2 , the volume is gotten by summing up the volumes of the washers: we get
Z
1
12π
−1
p
1 − x2 dx = 12π
Z
1
−1
p
1 − x2 dx.
R1 √
But −1 1 − x2 dx is the area of a semicircle of radius 1, which is π2 . So we get 12π ·
could use
Z p
h p
i
1 − x2 dx = x 1 − x2 + arcsin(x) ,
π
2
= 6π 2 ≈ 59.22. (Or, you
by VI-30 and VI-28.)
58. The arc length of the curve y = f (t) from t = 3 to t = 8 is
Z
8
q
R8p
1 + (f ′ (t))2 dt =
3
Thus, we have
f ′ (t)
One possibility is
2
1 + (f ′ (t))2 dt. Thus, we want a function f such that
3
Z
8
3
p
1 + e6t dt.
= e6t .
f ′ (t) = e3t
1
f (t) = e3t + C.
3
1 3t
e + C from t = 3 to t = 8.
3
′
3t
is f (t) = −e , which gives f (t) = − 31 e3t + C.
For any constant C, the original integral is the arc length of the curve y =
Another solution to f ′ (t)
2
= e6t
834
Chapter Eight /SOLUTIONS
59. We take a cross-section of the pipe and cut it up in such a way that the speed of the water is nearly uniform on each slice.
See Figure 8.170.
We use thin rings around the pipe’s center; if a given ring is narrow enough, all points on it will be roughly equidistant
from the center. Since the water speed is a function of the distance from the center, the speed is nearly constant on the
entire ring.
Let r be the distance from the center to the inner boundary of the ring, and let ∆r be the width of the ring, as in
Figure 8.170. By straightening the ring into a thin rectangle, we find that its area is approximately given by the quantity
2πr∆r. The speed across any part of the ring is roughly equal to the speed across the inner boundary, 10(1 − r 2 ) inches
per second. The flow is defined as the speed times the area; thus on any given ring we have
Flow ≈ 10(1 − r 2 ) · 2πr∆r.
The total flow across the pipe cross-section is approximated by a Riemann sum incorporating all of the rings:
Total Flow ≈ 20π
X
(1 − r 2 )r∆r,
where r is in between 0 and 1. Letting ∆r → 0, we obtain the exact solution:
Total Flow = 20π
Z
0
1
(1 − r 2 )r dr = 20π
r2
r4
−
2
4
1
= 5π cubic inches/second.
0
1
r
∆r
Figure 8.170
60. Multiplying r = 2a cos θ by r, converting to Cartesian coordinates, and completing the square gives
r 2 = 2ar cos θ
2
x + y 2 = 2ax
x2 − 2ax + a2 + y 2 = a2
(x − a)2 + y 2 = a2 .
This is the standard form of the equation of a circle with radius a and center (x, y) = (a, 0).
To check the limits on θ note that the circle is in the right half plane, where −π/2 ≤ θ ≤ π/2. Rays from the origin
at all these angles meet the circle because the circle is tangent to the y-axis at the origin.
61. The area is given by
Z
π/2
−π/2
1 2
r dθ =
2
Z
π/2
−π/2
1
(2a cos θ)2 dθ = 2a2
2
Z
π/2
cos2 θ dθ = 2a2
−π/2
1
2
cos θ sin θ +
θ
2
π/2
= πa2 .
−π/2
(We have used formula IV-18 from the integral table. The integral can also be done using a calculator or integration by
parts.)
62. See Figure 8.171. The circles meet where
2a cos θ = a
1
cos θ =
2
π
θ=± .
3
SOLUTIONS to Review Problems for Chapter Eight
The area is obtained by subtraction:
Area =
π/3
Z
−π/3
=
π/3
Z
−π/3
1
1
(2a cos θ)2 − a2
2
2
2a2 cos2 θ −
1 2
a
2
1
θ
cos θ sin θ +
2
2
√
3
π
+
a2 .
=
3
2
=
Since
2a2
dθ
dθ
−
a2
θ
2
π/3
−π/3
√
π/3 + 3/2 a2
= 61%
πa2
the shaded region covers 61% of circle C.
y
π/3
r=a
r = 2a cos θ
x
a
C
−π/3
Figure 8.171
63. (a) Writing C in parametric form gives
x = 2a cos2 θ
and
y = 2a cos θ sin θ,
so the slope is given by
dy
dy/dθ
−2a sin2 θ + 2a cos2 θ
sin2 θ − cos2 θ
=
=
=
.
dx
dx/dθ
−4a cos θ sin θ
2 cos θ sin θ
(b) The maximum y-value occurs where dy/dx = 0, so
sin2 θ − cos2 θ = 0
π
θ=± .
4
The value θ = π/4 gives the maximum y-value; θ = −π/4 gives the minimum y-value.
64. Writing C in parametric form gives
x = 2a cos2 θ
and
y = 2a cos θ sin θ,
so
Arc length =
Z
π/2
(−4a cos θ sin θ)2 + (−2a sin2 θ + 2a cos2 θ)2 dθ
−π/2
= 2a
p
Z
π/2
−π/2
= 2a
Z
π/2
Z
π/2
= 2a
p
p
(sin2 θ + cos2 θ)2 dθ
−π/2
Z
4 cos2 θ sin2 θ + sin4 θ − 2 sin2 θ cos2 θ + cos4 θ dθ
sin4 θ + 2 sin2 θ cos2 θ + cos4 θ dθ
−π/2
= 2a
p
π/2
−π/2
dθ = 2πa.
835
836
Chapter Eight /SOLUTIONS
65. This function has zeros at x = −2 and x = 1. The bounded region lies between these two zeros. Thus,
Volume =
1
Z
−2
π (x − 1)2 (x + 2)
66. The total mass is 12 gm, so the center of mass is located at x =
1
(−5
12
2
dx.
· 3 − 3 · 3 + 2 · 3 + 7 · 3) = 41 .
67. (a) Since the density is constant, the mass is the product of the area of the plate and its density.
Area of the plate =
Z
1
0
√
2 3/2 1 3
( x − x2 ) dx =
x
− x
3
3
1
=
0
1
cm2 .
3
Thus the mass of the plate is 2 · 1/3 = 2/3 gm.
(b) See Figure 8.172. Since the region is “fatter” closer to the origin, x̄ is less than 1/2.
y
1
y=
√
x
y = x2
x
✲ ✛
1
∆x
Figure 8.172
(c) To find x̄, we slice the region into vertical strips of width ∆x. See Figure 8.172.
√
Area of strip = Ax (x)∆x ≈ ( x − x2 )∆x cm2 .
Then we have
x=
R
xδAx (x) dx
=
Mass
R1
0
√
2x( x − x2 ) dx
2/3
3
2
=
Z
1
0
2(x3/2 − x3 ) dx =
2 5/2 1 4
3
·2
x
− x
2
5
4
This is less than 1/2, as predicted in part (b). So x̄ = ȳ = 9/20 cm.
1
=
0
9
cm.
20
68. Let x be the height from ground to the weight. It follows that 0 ≤ x ≤ 20. At height x, to lift the weight ∆x more, the
work needed is 200∆x + 2(20 − x)∆x = (240 − 2x)∆x. So the total work is
W =
Z
0
20
20
(240 − 2x)dx = (240x − x2 )
0
= 240(20) − 202 = 4400 ft-lb.
69. Imagine the pole is divided into n segments of length ∆x. The heights of the segments are given by x1 , x2 , . . . , xi , . . . xn .
20 lb
· ∆x = 2∆x. The work required to raise a segment a vertical distance of xi ft is
A segment of length ∆x weighs
10 ft
Work to raise segment xi ft = Weight · Distance
| {z } = 2xi ∆x.
| {z }
2∆x
The total work is therefore
Total work = lim
n→∞
=
Z
n
X
2xi ∆x
i=1
10
xi
10
2x dx = x2
0
= 100 ft-lbs.
0
To check our answer, notice that the work required to raise the entire 20 lb pole so that it is suspended horizontally 10 ft
above the ground is:
Work = Weight · Distance
| {z } = 200 ft-lbs.
| {z }
20 lbs
10 ft
This is more than 100 lbs, because it should take more work to raise the entire pole 10 ft than to stand it upright.
SOLUTIONS to Review Problems for Chapter Eight
837
70. Let x be the distance from the bucket to the surface of the water. It follows that 0 ≤ x ≤ 40. At x feet, the bucket weighs
(30 − 41 x), where the 41 x term is due to the leak. When the bucket is x feet from the surface of the water, the work done
by raising it ∆x feet is (30 − 14 x) ∆x. So the total work required to raise the bucket to the top is
W =
Z
40
(30 −
0
= 30x −
1
x)dx
4
1 2
x
8
40
0
1
= 30(40) − 402 = 1000 ft-lb.
8
71. Consider lifting a rectangular slab of water h feet from the top up to the top. See Figure 8.173. The area of such a slab is
(10)(20) = 200 square feet; if the thickness is ∆h, then the volume of such a slab is 200 ∆h cubic feet. This much water
weighs 62.4 pounds per ft3 , so the weight of such a slab is (200 ∆h)(62.4) = 12480 ∆h pounds. To lift that much water
h feet requires 12480h ∆h foot-pounds of work. To lift the whole tank, we lift one plate at a time; integrating over the
slabs yields
Z 15
15
12480h2
12480 · 152
=
= 1,404,000 foot-pounds.
12480h dh =
2
2
0
0
10
h
∆h
15
20
Figure 8.173
72. We begin by slicing the oil into slabs at a distance h below the surface with thickness ∆h. We can then calculate the
volume of the slab and the work needed to raise this slab to the surface, a distance of h.
Volume of ∆h disk = πr 2 ∆h = 25π∆h
Weight of ∆h disk = (25π)(50)∆h
Distance to raise = h
Work to raise = (25π)(50)(h)∆h.
Integrating the work over all such slabs, we have
Work =
Z
25
(50)(25π)(h) dh
19
25
= 625πh2
19
= 390,625π − 225,625π
≈ 518,363 ft-lbs.
838
Chapter Eight /SOLUTIONS
A diagram of this tank is shown in Figure 8.174.
✻ ✻
✛
5′
✲
✻
15′
10′
❄
h
✻
❄
❄
6′
∆h
❄
✻
❄
5
h
r
Figure 8.174
Figure 8.175
73. We slice the gasoline horizontally. At a distance h feet below the surface, the horizontal slab is a cylinder with radius r
and thickness ∆h, so
Volume of one slab ≈ πr 2 ∆h.
√
To find the radius r at a depth h from the top as in Figure 8.175, we note that h2 + r 2 = 52 , so r = 25 − h2 . At depth
h
p
Volume of one slice ≈ π( 25 − h2 )2 ∆h = π(25 − h2 )∆h ft3 .
The gasoline at depth h must be lifted a distance of h ft, so
Work to move one slice = ρ · Volume · Distance lifted
≈ ρ(π(25 − h2 )∆h)(h) ft-lb.
The work done, W , to lift all the gasoline is the sum of the work done on the pieces:
W ≈
X
ρ(π(25 − h2 )∆h)h ft-lb..
As ∆h → 0, we obtain a definite integral. Since h varies from h = 0 to h = 5 and ρ = 42, we have:
W =
Z
0
5
ρπ(25h − h3 )dh = 42π
25
h2
h4
−
2
4
5
=
0
13125π
= 20,617 ft-lb.
2
The work to pump all the gasoline out is 20,617 ft-lbs.
74. Let h be height above the bottom of the dam. Then
Water force =
Z
25
0
(62.4)(25 − h)(60) dh
= (62.4)(60)
h2
25h −
2
25
0
= (62.4)(60)(625 − 312.5)
= (62.4)(60)(312.5)
= 1,170,000 lbs.
75. If the weight of the chain were negligible, the work required would be 1000 · 20 = 20,000 ft-lbs. Because of the chain,
the total work is slightly more than 20,000 ft-lbs. When the object is h ft off the ground, the length of chain is 50 − h so
the total weight being lifted is 1000 + 2(50 − h) lb. See Figure 8.176. Thus
Work to lift the weight an addition ∆h higher = Weight · Distance lifted
≈ (1000 + 2(50 − h))∆h ft-lb.
SOLUTIONS to Review Problems for Chapter Eight
839
To find the total work, we integrate this quantity from h = 0 to h = 20:
W =
Z
0
20
(1000 + 2(50 − h))dh =
Z
20
20
(1100 − 2h)dh = (1100h − h2 )
0
= 21,600ft-lbs.
0
✻
50 ft
✻
∆h
20 ft
❄
❄
❄
✻
1000 lbs
h ft
❄
Ground
Figure 8.176
76.
Future Value =
Z
15
3000e0.06(15−t) dt = 3000e0.9
0
= 3000e
Z
15
e−0.06t dt
0
0.9
1
e−0.06t
−0.06
≈ $72,980.16
Present Value =
Z
15
= 3000e0.9
0
15
3000e−0.06t dt = 3000 −
0
≈ $29,671.52.
1
1 0
e−0.9 +
e
−0.06
0.06
1
e−0.06t
0.06
15
0
There’s a quicker way to calculate the present value of the income stream, since the future value of the income stream is
(as we’ve shown) $72,980.16, the present value of the income stream must be the present value of $72,980.16. Thus,
Present Value = $72,980.16(e−.06·15 )
≈ $29,671.52,
which is what we got before.
77. We divide up time between 1971 and 1992 into intervals of length ∆t, and calculate how much of the strontium-90
produced during that time interval is still around.
Strontium-90 decays exponentially, so if a quantity S0 was produced t years ago, and S is the quantity around today,
S = S0 e−kt . Since the half-life is 28 years, 12 = e−k(28) , giving k = − ln(1/2)/28 ≈ 0.025.
We measure t in years from 1971, so that 1992 is t = 21.
1971 (t = 0)
∆t
✲✛
t
1992 (t = 21)
(21 − t)
✲
Since strontium-90 is produced at a rate of 3 kg/year, during the interval ∆t, a quantity 3∆t kg was produced. Since
this was (21 − t) years ago, the quantity remaining now is (3∆t)e−0.025(21−t) . Summing over all such intervals gives
Strontium remaining
in 1992
≈
Z
21
3e−0.025(21−t) dt =
0
3e−0.025(21−t)
0.025
21
= 49 kg.
0
[Note: This is like a future value problem from economics, but with a negative interest rate.]
840
Chapter Eight /SOLUTIONS
78. (a) Slice the mountain horizontally into N cylinders of height ∆h. The sum of the volumes of the cylinders will be
N
X
πr 2 ∆h =
2
N
X
3.5 · 105
√
π
i=1
i=1
h + 600
∆h.
(b)
Volume =
Z
14400
π
400
11
= 1.23 · 10 π
3.5 · 105
√
h + 600
Z
14400
400
2
dh
1
dh
(h + 600)
14400
= 1.23 · 1011 π ln(h + 600)
dh
400
= 1.23 · 1011 π [ln 15000 − ln 1000]
= 1.23 · 1011 π ln(15000/1000)
= 1.23 · 1011 π ln 15 ≈ 1.05 · 1012 cubic feet.
79. Look at the disc-shaped slab of water at height y and of thickness ∆y in Figure 8.177. The rate at which water is flowing
√
out when it is at depth y is k y (Torricelli’s Law, with k constant). Then, if x = g(y), we have
∆t =
Time for water to
drop by this amount
=
π(g(y))2∆y
Volume
=
.
√
Rate
k y
y
(1, 1)
(−1, 1)
x = g(y)
∆y
x
Figure 8.177
If the rate at which the depth of the water is dropping is constant, then dy/dt is constant, so we want
π(g(y))2
= constant,
√
k y
√
√
so g(y) = c 4 y, for some constant c. Since x = 1 when y = 1, we have c = 1 and so x = 4 y, or y = x4 .
80. The statement P (70) = 0.76 means that 76% of the population has ages less than 70.
81. Graph B is more spread out to the right, and so it represents a gas in which more of the molecules are moving at faster
velocities. Thus the average velocity in gas B is larger.
82. Every photon which falls a given distance from the center of the detector has the same probability of being detected. This
suggests that we divide the plate up into concentric rings of thickness ∆r. Consider one such ring having inner radius r
and outer radius r + ∆r. For this ring,
Number of photons hitting ring per unit time ≈ N · Area of ring ≈ N · 2πr∆r.
SOLUTIONS to Review Problems for Chapter Eight
Then,
Number of photons detected on ring per unit time ≈ Number hitting · S(r) ≈ N · 2πr∆r · S(r).
Summing over all rings gives us
Total number of photons detected per unit time ≈
Taking the limit as ∆r → 0 gives
Total number of photons detected per unit time =
X
Z
2πN rS(r)∆r.
R
2πN rS(r)dr.
0
83. The number of houses in a ring of width ∆r a distance ri from the city center is given by:
Number houses = 1000 houses/mi2 × Area of ring
= 1000 · 2πri ∆r = 2000πri ∆r.
The value of the houses in this ring is given by:
Value = Price per house × Number of houses
= p (ri ) · 1000 · 2πri ∆r = 2000πri p (ri ) ∆r.
The total value of the houses within 7 miles of the city center is therefore
Total value = lim
n→∞
=
Z
7
n
X
2000πri p (ri ) ∆r
i=1
2000πrp (r) dr
0
= 2000π
Z
0
= 800,000π
7
2
400e−0.2r rdr
Z
r=7
2
ew (−2.5) dw
r=0
= −2,000,000π
Z
let w = −0.2r 2 , dw = −0.4rdr
r=7
ew dw
r=0
= −2,000,000πe−0.2r
7
2
0
2
= −2,000,000π e−0.2(7) − e−0.2(0)
2
= 6,282,837.
This figure is in $1000s, so the total value of homes is approximately $6.2828 billion.
84. (a) Divide the cross-section of the blood into rings of radius r, width ∆r. See Figure 8.178.
✛
✲
l
✻
R
❄
✛
❄
✻
r
✻
❄
∆r
✲
v=
Figure 8.178
Then
Area of ring ≈ 2πr∆r.
P R2
4ηl
841
842
Chapter Eight /SOLUTIONS
The velocity of the blood is approximately constant throughout the ring, so
Rate blood flows through ring ≈ Velocity · Area
P
(R2 − r 2 ) · 2πr∆r.
=
4ηl
Thus, summing over all rings, we find the total blood flow:
Rate blood flowing through blood vessel ≈
Taking the limit as ∆r → 0, we get
X P
4ηl
Rate blood flowing through blood vessel =
=
πP
2ηl
(b) Since
2 2
4
R r
r
−
2
4
Z
R
=
0
Rate of blood flow =
R
0
(R2 − r 2 )2πr∆r.
πP 2
(R r − r 3 )dr
2ηl
πP R4
.
8ηl
πP R4
,
8ηl
if we take k = πP/(8ηl), then we have
Rate of blood flow = kR4 ,
that is, rate of blood flow is proportional to R4 , in accordance with Poiseuille’s Law.
85. Pick a small interval of time ∆t which takes place at time t. Fuel is consumed at a rate of (25 + 0.1v)−1 gallons per
mile. In the time ∆t, the car moves v ∆t miles, so it consumes v ∆t/(25 + 0.1v) gallons during the instant ∆t. Since
t
, the car consumes
v = 50 t+1
t
50 t+1
∆t
10t ∆t
50t ∆t
v ∆t
=
=
=
t
25 + 0.1v
25(t + 1) + 5t
6t + 5
25 + 0.1 50 t+1
gallons of gas, in terms of the time t at which the instant occurs. To find the total gas consumed, sum up the instants in an
integral:
Z 3
10t
Gas consumed =
dt ≈ 1.25 gallons.
6t + 5
2
86. (a) Slicing horizontally, as shown in Figure 8.179, we see that the volume of one disk-shaped slab is
∆V ≈ πx2 ∆y =
πy
∆y.
a
Thus, the volume of the water is given by
V =
Z
h
0
π y2
π
ydy =
a
a 2
h
=
0
πh2
.
2a
(b) The surface
of the water is a circle of radius x. Since at the surface, y = h, we have h = ax2 . Thus, at the surface,
p
x = (h/a). Therefore the area of the surface of water is given by
A = πx2 =
πh
.
a
(c) If the rate at which water is evaporating is proportional to the surface area, we have
dV
= −kA.
dt
(The negative sign is included because the volume is decreasing.) By the chain rule,
dV
= πh
and A = πh
so
dh
a
a
πh dh
πh
dh
= −k
giving
= −k.
a dt
a
dt
dV
dt
=
dV
dh
·
dh
.
dt
We know
SOLUTIONS to Review Problems for Chapter Eight
843
(d) Integrating gives
h = −kt + h0 .
Solving for t when h = 0 gives
t=
h0
.
k
y
✻
y
2 − (h + b)
❄
✻
y = ax2
h+b
✻
h
❄
✛
x
✲
y✻
∆y
❄
✻
❄
x
x
❄
1
Figure 8.179
Figure 8.180
87. (a) The volume of water in the centrifuge is π(12 )·1 = π cubic meters. The centrifuge has total volume 2π cubic meters,
so the volume of the air in the centrifuge is π cubic meters. Now suppose the equation of the parabola is y = h + bx2 .
We know that the volume of air in the centrifuge is the volume of the top part (a cylinder) plus the volume of the
middle part (shaped like a bowl). See Figure 8.180.
To find the volume of the cylinder of air, we find the maximum water depth. If x = 1, then y = h + b. Therefore
the height of the water at the edge of the bowl, 1 meter away from the center, is h + b. The volume of the cylinder of
air is therefore [2 − (h + b)] · π · (1)2 = [2 − h − b]π.
To find the volume of the bowl of air, we note that the bowl is a volume of rotation with radius x at height y,
where y = h + bx2 . Solving for x2 gives x2 = (y − h)/b. Hence, slicing horizontally as shown in the picture:
Bowl Volume =
Z
h
h+b
πx2 dy =
Z
h
h+b
π
π(y − h)2
y−h
dy =
b
2b
h+b
=
h
bπ
.
2
So the volume of both pieces together is [2 − h − b]π + bπ/2 = (2 − h − b/2)π. But we know the volume of air
should be π, so (2 − h − b/2)π = π, hence h + b/2 = 1 and b = 2 − 2h. Therefore, the equation of the parabolic
cross-section is y = h + (2 − 2h)x2 .
(b) The water spills out the top when h + b = h + (2 − 2h) = 2, or when h = 0. The bottom is exposed when h = 0.
Therefore, the two events happen simultaneously.
88. Any small piece of mass ∆M on either of the two spheres has kinetic energy 12 v 2 ∆M . Since the angular velocity of the
two spheres is the same, the actual velocity of the piece ∆M will depend on how far away it is from the axis of revolution.
The further away a piece is from the axis, the faster it must be moving and the larger its velocity v. This is because if ∆M
is at a distance r from the axis, in one revolution it must trace out a circular path of length 2πr about the axis. Since every
piece in either sphere takes 1 minute to make 1 revolution, pieces farther from the axis must move faster, as they travel a
greater distance.
Thus, since the thin spherical shell has more of its mass concentrated farther from the axis of rotation than does the
solid sphere, the bulk of it is traveling faster than the bulk of the solid sphere. So, it has the higher kinetic energy.
89. Any small piece of mass ∆M on either of the two hoops has kinetic energy 12 v 2 ∆M . Since the angular velocity of the
two hoops is the same, the actual velocity of the piece ∆M will depend on how far away it is from the axis of revolution.
The further away a piece is from the axis, the faster it must be moving and the larger its velocity v. This is because if ∆M
is at a distance r from the axis, in one revolution it must trace out a circular path of length 2πr about the axis. Since every
piece in either hoop takes 1 minute to make 1 revolution, pieces farther from the axis must move faster, as they travel a
greater distance.
The hoop rotating about the cylindrical axis has all of its mass at a distance R from the axis, whereas the other hoop
has a good bit of its mass close (or on) the axis of rotation. So, since the bulk of the hoop rotating about the cylindrical
axis is traveling faster than the bulk of the other hoop, it must have the higher kinetic energy.
844
Chapter Eight /SOLUTIONS
CAS Challenge Problems
90. (a) We need to check that the point with the given coordinates is on the curve, i.e., that
x = a sin2 t,
y=
a sin3 t
cos t
satisfies the equation
r
x3
.
a−x
This can be done by substituting into the computer algebra system and asking it to simplify the difference between
the two sides, or by hand calculation:
y=
Right-hand side =
r
r
(a sin2 t)3
=
a − a sin2 t
r
r
a3 sin6 t
a(1 − sin2 t)
a3 sin6 t
a2 sin6 t
=
=
2
a cos t
cos2 t
3
a sin t
= y = Left-hand side.
=
cos t
We chose the positive square root because both sin t and cos t are nonnegative for 0 ≤ t ≤ π/2. Thus the point always
lies on the curve. In addition, when t = 0, x = 0 and y = 0, so the point starts at x = 0. As t approaches π/2, the
value of x = a sin2 t approaches a and the value of y = a sin3 t/ cos t increases without bound (or approaches ∞),
so the point on the curve approaches the vertical asymptote at x = a. p
(b) We calculate the volume using horizontal slices. See the graph of y = x3 /(a − x) in Figure 8.181.
y
❄∆y
✻
x−a
a
x
Figure 8.181
The slice at y is a disk of thickness ∆y and radius x − a, hence it has volume π(x − a)2 ∆y. So the volume is
given by the improper integral
Z
∞
Volume =
0
π(x − a)2 dy.
(c) We substitute
x = a sin2 t,
and
dy =
d
dt
a sin3 t
cos t
y=
a sin3 t
cos t
dt = a 3 sin2 t +
sin4 t
cos2 t
Since t = 0 where y = 0 and t = π/2 at the asymptote where y → ∞, we get
Volume =
Z
π/2
0
= πa3
π(a sin2 t − a)2 a 3 sin2 t +
Z
0
π/2
sin4 t
cos2 t
dt.
(3 sin2 t cos4 t + sin4 t cos2 t) dt =
dt
π 2 a3
.
8
You can use a CAS to calculate this integral; it can also be done using trigonometric identities.
845
SOLUTIONS to Review Problems for Chapter Eight
91. (a) The expression for arc length in terms of a definite integral gives
Z tp
√
2 t 1 + 4 t2 + arcsinh (2 t)
2
1 + 4x dx =
.
A(t) =
4
0
The integral was evaluated using a computer algebra system; different systems may give the answer in different
forms. Here arcsinh is the inverse function of the hyperbolic sine function.
(b) Figure 8.182 shows that the graphs of A(t) and t2 look very similar. This suggests that A(t) ≈ t2 .
y
y
t2
100
y
A(t)
100
10
t
y = x2
100
10
x
t
10
Figure 8.182
Figure 8.183
(c) The graph in Figure 8.183 is approximately vertical and close to the y axis. Thus, if we measure the arc length up to a
certain y-value, the answer is approximately the same as if we had measured the length straight up the y-axis. Hence
A(t) ≈ y = f (t) = t2 .
So
A(t) ≈ t2 .
92. (a) The expression for arc length in terms of a definite integral gives
A(t) =
Z
t
0
s
1+
1
√
2 x
2
dx =
√√
√
2 t 1 + 4t + arcsinh (2 t)
.
4
The integral was evaluated using a computer algebra system; different systems may give the answer in different
forms. Some may involve ln instead of arcsinh, which is the inverse function of the hyperbolic sine function.
(b) Figure 8.185 shows that the graphs of A(t) and the graph of y = t look very similar. This suggests that A(t) ≈ t.
y
y
100
y
100
A(t)
y=t
10
100
t
Figure 8.184
10
y=
√
x
x
t
100
Figure 8.185
(c) The graph in Figure 8.185 is approximately horizontal and close to the x-axis. Thus, if we measure the arc length up
to a certain x-value, the answer is approximately the same as if we had measured the length straight along the x-axis.
Hence
A(t) ≈ x = t.
So
A(t) ≈ t.
846
Chapter Eight /SOLUTIONS
93. (a) Slice the sphere at right angles to the axis of the cylinder. Consider a slice of thickness ∆x at distance x from
the
√ center of the sphere. The cross-section is an annulus (ring) with internal radius ri = a and outer radius ro =
r 2 − x2 . Thus
Area of annulus = πro 2 − πri 2 = π
p
r 2 − x2
2
− πa2 = π(r 2 − x2 − a2 ).
Volume of slice ≈ π(r 2 − x2 − a2 )∆x.
2
2
2
The lower
√ and upper limits of the integral are where the cylinder meets the sphere, i.e., where x + a = r , or
2
2
x = ± r − a . Thus
Z √
r 2 −a2
Volume of bead =
−
√
r 2 −a2
π(r 2 − x2 − a2 ) dx.
(b) Using a computer algebra system to evaluate the integral, we have
Volume of bead =
3/2
4π 2
r − a2
.
3
PROJECTS FOR CHAPTER EIGHT
1. (a) The hydrostatic pressure ph is a linear function of distance x, so ph = b + mx. At the artery end, x = 0,
and ph = 35. At the vein end, x = L = 0.1 and ph = 15. This means the points (0, 35) and (0.1, 15) are
on the graph of ph , so the slope is given by
m=
15 − 35
mm Hg
= −200
.
0.1 − 0
cm
We know that b = 35, so ph = 35 − 200x mm Hg.
(b) The net pressure, p, is the difference between the hydrostatic pressure and the oncotic pressure. Using
part (a), we have
p = ph − po = (35 − 200x) − 23 = 12 − 200x mm Hg.
(c) Since j = k · p, we have
j units = (k units) × (p units) =
cm
cm
× mm Hg =
.
sec · mm Hg
sec
In addition
Volume/time/area units =
1
cm3
1
cm
Volume units
×
=
×
=
.
Time units
Area units
sec
cm2
sec
Thus, j has units of volume per time per area.
(d) To find the flow rate through a small section of length ∆x of the capillary, we multiply j, the flow rate per
capillary wall area, by the area, A = 2πr ∆x of the section. To find the flow through the wall of the entire
capillary, we integrate. Thus,
Flow rate =
Z
0
L
j · 2πr dx =
Z
L
kp · 2πr dx
0
Using the formula for p from part (b), and the values L = 0.1 cm, r = 0.0004 cm, and k = 10−7 cm/(sec·
mm Hg), evaluating the integral symbolically or numerically, we have
Flow rate =
Z
0
0.1
0.1
2π(10−7 )(0.0004)(12 − 200x) dx = 8 · 10−11 π(12x − 100x2 )
= 5.03 · 10−11
0
cm3
.
sec
PROJECTS FOR CHAPTER EIGHT
847
2. (a) The volume of a cylinder of radius r and height h is V = πr2 h. Since the total volume of the urine sample
is 2000 ml and the diameter is 10 cm (so the radius is 5 cm), we have:
πr2 h = π52 h = 2000
2000
h=
= 25.46 cm.
25π
Thus, the depth of urine in the cylinder is 25.46 cm.
(i) The protein concentration c is a linear function of height y, so c = b + my. At the bottom of the
sample, y = 0, and c = 0.96. At the top, y = 25.46 and c = 0.14. This means the points (0, 0.96)
and (25.46, 0.14) are on the graph of c, so the slope is given by
m=
mg
0.14 − 0.96
= −0.0322
.
25.46 − 0
ml · cm
We know that b = 0.96, so c = 0.96 − 0.0322y mg/ml.
(ii) A cross sectional slice of the cylinder has volume 52 π∆y = 25π∆y. This means the quantity of
protein in a given slice is
∆Q = Volume
| {z } × Concentration
{z
} = 25π∆y (0.96 − 0.0322y) mg.
|
25π∆y
0.96−0.0322y
(iii) To find the total quantity of protein, we integrate:
Z 25.46
25π (0.96 − 0.0322y) dy
Total mass =
y=0
25.46
= 25π 0.96y − 0.0161y 2
= 1099.98 mg.
0
Thus, the patient excreted approximately 1100 mg, or 1.1 gm, of protein during the 24 hour period
in which urine was collected. This is above the threshold of 1 gm per 24 hour period at which active
treatment is recommended.
(b) Let the radius of the collection container be r, the volume of the urine be V, and the concentrations of
protein at the top and bottom of the collected fluid be ct and cb . We compute the quantity of protein by
following the procedure of part (a).
If h is the depth of the sample, we have
V = πr2 h.
Assuming that protein concentration, c, is a linear function of the distance, y, from the bottom of the
container, we have
c = cb + my
where the slope m is given by
m=
ct − cb
.
h
The quantity of protein in a horizontal slice is
∆Q = Volume × Concentration = πr2 ∆y · c.
To find the total quantity of protein in the container, we integrate:
Total mass =
Z
0
h
2
πr c(y) dy =
Z
0
h
πr2 (cb + my) dy
848
Chapter Eight /SOLUTIONS
h
y2
h2
2
= πr cb y + m
= πr cb h + m
2
2
0
2
ct − cb
ct − cb h
2
2
·
= πr h cb +
= πr cb h +
h
2
2
ct + cb
=
V.
2
2
Thus, the quantity of protein in the sample is the product of the average of the top and bottom protein
concentrations by the volume of urine collected.
3. Let us make coordinate axes with the origin at the center of the box. The x and y axes will lie along the central
axes of the cylinders, and the (height) axis will extend vertically to the top of the box. If one slices the cylinders
horizontally, one gets a cross. The cross is what you get if you cut out four corner squares from a square of side
length 2. If h is the height of the cross above (or below) the xy plane, the equation of a cylinder is h2 + y 2 = 1
2
2
2
(or h2 + x2 = 1).
√ Thus the “armpits” of the cross occur
√ where y − 1 = −h = x − 1 for some fixed height
2
2
h—that is, out√ 1 − h units from the center,
or 1 − 1 − h units away from the edge. Each corner square
√
has area (1 − 1 − h2 )2 = 2 − h2 − 2 1 − h2 . The whole big square has area 4. Therefore, the area of the
cross is
p
p
4 − 4(2 − h2 − 2 1 − h2 ) = −4 + 4h2 + 8 1 − h2 .
2
✛
✲
✻
√
1 − h2
❄
✛ √✲
√✛ ✲
1 − 1 − h2
1 − h2
We integrate this from h = −1 to h = 1, and obtain the volume, V :
Z 1
p
V =
−4 + 4h2 + 8 1 − h2 dh
1−
−1
=
"
4h3
1 p
− 4h +
+8·
h 1 − h2 + arcsin h
3
2
#
1
−1
8
16
= −8 + + 4π = 4π −
≈ 7.23.
3
3
This is a reasonable answer, as the volume of the cube is 8, and the volume of one cylinder alone is 2π ≈ 6.28.
4. (a) Let y represent height, and let x represent horizontal distance from the lowest point of the cable. Then the
stretched cable is a parabola of the form y = kx2 passing through the point (1280/2, 143) = (640, 143).
Therefore, 143 = k(640)2 so k ≈ 3.491 × 10−4 . To find the arc length of the parabola, we take twice the
arc length of the part to the right of the lowest point. Since dy/dx = 2kx,
Z 640 p
Z 640 p
1 + (2kx)2 dx = 2
1 + 4k 2 x2 dx.
Arc Length = 2
0
0
The easiest way to find this integral is to substitute the value of k and find the integral’s value numerically, giving
Arc Length ≈ 1321.4 meters.
PROJECTS FOR CHAPTER EIGHT
849
Alternatively, we can make the substitution w = 2kx:
Z 1280k p
2
Arc Length =
1 + w2 dw
2k 0
Z
1 1280k p
1 + w2 dw
=
k 0
!
!
Z 1280k
1280k
p
1
1
1
√
w 1 + w2
dw
=
+
2k
2k
1 + w2
0
0
[Using the integral table, Formula VI-29, or substitute w = tan θ]
!
1280k
p
p
1
1
2
2
1280k 1 + (1280k) +
ln x + 1 + x
=
2k
2k
0
p
p
1
1
1280k 1 + (1280k)2 +
ln 1280k + 1 + (1280k)2
=
2k
2k
≈ 1321.4 meters.
(b) Adding 0.05% to the length of the cable gives a cable length of (1321.4)(1.0005) = 1322.1. We now want
to calculate the new shape of the parabola; that is, we want to find a new k so that the arc length is 1322.1.
Since
Z 640 p
1 + 4k 2 x2 dx
Arc Length = 2
0
we can find k numerically by trial and error. Trying values close to our original value of k, we find
k ≈ 3.52 × 10−4 . To find the sag for this new k, we find the height y = kx2 for which the cable hangs
from the towers. This is
y = k(640)2 ≈ 144.2.
Thus the cable sag is 144.2 meters, over a meter more than on a cold winter day. Notice, though, that
although the length increases by 0.05%, the sag increases by more:144.2/143 ≈ 1.0084, an increase of
0.84%.
√
5. (a) Revolving the semi-circle y = r2 − x2 around the x-axis yields the sphere of radius r. See Figure 8.186.
Differentiating yields:
dy
−1
x
·x =− .
= √
dx
y
r2 − x2
′
Thus, substituting −x/y for f (x), we get
Z r p
Z r s
x2
x2 + y 2 dx
Surface area = 2π
y 1 + 2 dx = 2π
y
−r
−r
Z r
= 2πr
dx = 4πr2 .
−r
y
y=
√
r 2 − x2
✠
r
r
−r
y
x
x
h
Figure 8.186
Figure 8.187
850
Chapter Eight /SOLUTIONS
(b) Revolving the line y = rx/h around the x-axis yields the cone. The base of the cone is a circle with area
πr2 . See Figure 8.187. The area of the rest of the cone is
Surface area = 2π
Z
h
y
0
r h2
= 2π
h 2
r
r2
1 + 2 dx = 2π
h
r
1+
Adding the area of the base, we get
r
r2
1+ 2
h
r
h
Z
h
!
x dx
0
p
r2
= πr r2 + h2
2
h
Total surface area of cone = πr2 + πr
p
r 2 + h2 .
(c) We find the volume of y = 1/x revolved about the x-axis as x runs from 1 to ∞. See Figure 8.188.
y
1
x
y=
❄
x
1
Figure 8.188
Volume =
Z
1
∞
πy 2 dx = π
Z
∞
1
1
dx = π lim
b→∞
x2
Z
1
b
dx
−1
= π lim
2
b→∞ x
x
b
=π
1
Thus, the volume of this solid is finite and equal to π.
(d) Now we show the surface area of this solid is unbounded. We have
s
2
Z ∞ r
Z ∞
1
dy
1
1 + 4 dx
dx = 2π
Surface area = 2π
y 1+
dx
x
x
1
1
r
1
1
1 + 4 , so we bound the integral from below by
We cannot easily compute the antiderivative of
x
x
noticing that
r
1
1 + 4 ≥ 1.
x
Thus we see that
Z b
Z ∞
b
dx
1
dx = 2π lim
= 2π lim ln x .
Surface area ≥ 2π
b→∞ 1 x
b→∞
x
1
1
Since ln x goes to infinity as x goes to infinity, the surface area is unbounded.
Alternatively, we can try calculating
Z b r
1
1
2π
1 + 4 dx
x
x
1
for larger and larger values of b. We would see that the integral seems to diverge.
PROJECTS FOR CHAPTER EIGHT
851
(e) For a solid generated by the revolution of a curve y = f (x) for a ≤ x ≤ b,
Z b
Volume =
πy 2 dx
a
and
Surface area =
Z
a
The volume and the surface area will be equal if
b
2πy
p
1 + (f ′ (x))2 dx.
p
f (x) = 2 1 + (f ′ (x))2 .
We find a function y = f (x) which satisfies this relation:
s
dy
y =2 1+
dx
2
y2
dy
=1+
4
dx
r
dy
y2
=
−1
dx
4
1
dy
p
= dx
2
y2 − 4
Z
Z
dy
1
p
dx
=
2
2
y −4
p
x
ln |y + y 2 − 4| = + C
2
p
y + y 2 − 4 = Aex/2
2
Notice in the third line we have used the fact that dy/dx ≥ 0. Any function, y = f (x), which satisfies this
relationship has the required property.
Rr
R∞
2
6. (a) We want to find a such that 0 p(v) dv = lim a 0 v 2 e−mv /2kT dv = 1. Therefore,
r→∞
Z r
2
1
v 2 e−mv /2kT dv.
= lim
r→∞
a
0
2
To evaluate the integral, use integration by parts with the substitutions u = v and w′ = ve−mv /2kT :
Z r
Z r
2
2
r
e−mv /2kT
e−mv /2kT
−mv 2 /2kT
ve
dv
=
v
−
1
v
{z
}
|{z} −m/kT
|{z} −m/kT dv
|{z} |
0
0
u
u |
{z } 0
{z }
w′
u′ |
w
w
Z
kT r −mr2 /2kT
kT r −mv2 /2kT
=−
e
+
e
dv.
m
m 0
Z ∞
2
1
1
√ e−x /2 dx = , so
From the normal distribution we know that
2
2π
0
√
Z ∞
2
2π
e−x /2 dx =
.
2
0
pm
pm
Therefore in the above integral, make the substitution x =
kT v, so that dx =
kT dv, or dv =
q
kT
m dx. Then
3/2 Z √ m r
Z
kT
2
kT r −mv2 /2kT
kT
e
dv =
e−x /2 dx.
m 0
m
0
852
Chapter Eight /SOLUTIONS
Substituting this into Equation 6a we get
1
= lim
a r→∞
kT r −mr2 /2kT
−
e
+
m
kT
m
3/2 Z √ m r
kT
e
−x2 /2
0
3/2 √
kT
2π
·
.
dx = 0 +
m
2
m 3/2
) . Substituting the values for k, T , and m gives a ≈ 3.4 × 10−8 .
Therefore, a = √22π ( kT
(b) To find the median, we wish to find the speed x such that
Z x
Z x
mv2
1
av 2 e− 2kT dv = ,
p(v) dv =
2
0
0
m 3/2
) . Using a calculator, by trial and error we get x ≈ 441 m/sec.
where a = √22π ( kT
To find the mean, we find
Z
Z
∞
∞
vp(v) dv =
0
mv2
av 3 e− 2kT dv.
0
2
This integral can be done by substitution. Let u = v , so du = 2vdv. Then
Z
Z ∞
mv2
a v=∞ 2 − mv2
v e 2kT 2v dv
av 3 e− 2kT dv =
2 v=0
0
Z
mu
a u=∞ − 2kT
du
ue
=
2 u=0
Z
a r − mu
= lim
ue 2kT du.
r→∞ 2 0
Now, using the integral table, we have
Z
0
∞
av 3 e
2
− mv
2kT
"
#
2
2kT − mu
a
2kT
mu
−
ue 2kT − −
e− 2kT
dv = lim
r→∞ 2
m
m
2
a
2kT
=
−
2
m
≈ 457.7 m/sec.
r
0
The maximum for p(v) will be at a point where p′ (v) = 0.
2
mv2
2mv
′
− mv
2
p (v) = a(2v)e 2kT + av −
e− 2kT
2kT
2
m
3
− mv
.
= ae 2kT 2v − v
kT
r
2kT
′
≈ 405. It’s obvious that p(0) = 0, and that p → 0 as v → ∞.
Thus p (v) = 0 at v = 0 and at v =
m
So v = 405 gives us a maximum: p(405) ≈ 0.002.
a 4k 2 T 2
4 k 1/2 T 1/2
√
(c) The mean, as we found in part (b), is
=
. It is clear, then, that as T increases so
2 m2
2π m1/2
r
2kT
. Thus
does the mean. We found in part (b) that p(v) reached its maximum at v =
m
2 m 3/2 2kT −1
e
The maximum value of p(v) = √
m
2π kT
4 m1/2
= √
.
e 2π kT 1/2
Thus as T increases, the maximum value decreases.
9.1 SOLUTIONS
853
CHAPTER NINE
Solutions for Section 9.1
Exercises
1. The first term is 21 + 1 = 3. The second term is 22 + 1 = 5. The third term is 23 + 1 = 9, the fourth is 24 + 1 = 17, and
the fifth is 25 + 1 = 33. The first five terms are 3, 5, 9, 17, 33.
2. The first term is 1 + (−1)1 = 1 − 1 = 0. The second term is 2 + (−1)2 = 2 + 1 = 3. The third term is 3 − 1 = 2 and
the fourth is 4 + 1 = 5. The first five terms are 0, 3, 2, 5, 4.
3. The first term is 2 · 1/(2 · 1 + 1) = 2/3. The second term is 2 · 2/(2 · 2 + 1) = 4/5. The first five terms are
2/3, 4/5, 6/7, 8/9, 10/11.
4. The first term is (−1)1 (1/2)1 = −1/2. The second term is (−1)2 (1/2)2 = 1/4. The first five terms are
−1/2, 1/4, −1/8, 1/16, −1/32.
5. The first term is (−1)2 (1/2)0 = 1. The second term is (−1)3 (1/2)1 = −1/2. The first five terms are
1, −1/2, 1/4, −1/8, 1/16.
6. The first term is (1 − 1/(1 + 1))(1+1) = (1/2)2 . The second term is (1 − 1/3)3 = (2/3)3 . The first five terms are
(1/2)2 , (2/3)3 , (3/4)4 , (4/5)5 , (5/6)6 .
7. The terms look like powers of 2 so we guess sn = 2n . This makes the first term 21 = 2 rather than 4. We try instead
sn = 2n+1 . If we now check, we get the terms 4, 8, 16, 32, 64, . . ., which is right.
8. We compare with positive powers of 2, which are 2, 4, 8, 16, 32, . . .. Each term is one less, so we take sn = 2n − 1.
9. We observe that if we subtract 1 from each term of the sequence, we get 1, 4, 9, 16, 25, . . ., namely the squares 12 , 22 , 32 , 42 , 52 , . . ..
Thus sn = n2 + 1.
10. First notice that sn = 2n − 1 is a formula for the general term of the sequence
1, 3, 5, 7, 9, . . . .
To obtain the alternating signs in the original sequence, we try multiplying by (−1)n . However, checking (−1)n (2n − 1)
for n = 1, 2, 3, . . . gives
−1, 3, −5, 7, −9, . . . .
To get the correct signs, we multiply by (−1)n+1 and take
sn = (−1)n+1 (2n − 1).
11. The numerator is n. The denominator is then 2n + 1, so sn = n/(2n + 1).
12. The denominators are the even numbers, so we try sn = 1/(2n). To get the signs to alternate, we try multiplying by
(−1)n . That gives
−1/2, 1/4, −1/6, 1/8, −1/10, . . . ,
so we multiply by (−1)n+1 instead. Thus sn = (−1)n+1 /(2n).
Problems
13. Since 2n increases without bound as n increases, the sequence diverges.
854
Chapter Nine /SOLUTIONS
14. Since lim xn = 0 if |x| < 1 and |0.2| < 1, we have lim (0.2)n = 0, so the sequence converges to 0
n→∞
n→∞
n
15. Since lim x = 0 if |x| < 1 and |e
−2
n→∞
so the sequence converges to 3.
| < 1, we have lim (e−2n ) = lim (e−2 )n = 0, so lim (3+e−2n ) = 3+0 = 3,
n→∞
n→∞
n→∞
16. Since lim xn = 0 if |x| < 1 and | − 0.3| < 1, we have lim (−0.3)n = 0, so the sequence converges to 0.
n→∞
n→∞
17. We have:
n
10
n
+
= lim
+ lim 10n.
n→∞ 10
n→∞
n→∞ 10
n
Since n/10 gets arbitrarily large and 10/n approaches 0 as n → ∞, the sequence diverges.
n
n
2
2
2
< 1, we have lim
= lim
= 0, so the sequence converges to 0.
18. Since lim xn = 0 if |x| < 1 and
n
n→∞ 3
n→∞ 3
n→∞
3
19. We have
2n + 1
1
lim
= lim 2 +
= 2,
n→∞
n→∞
n
n
so the sequence converges to 2.
lim
20. We have:
1
= 0.
n
The terms of the sequence alternate in sign, but they approach 0, so the sequence converges to 0.
lim
n→∞
21. Since 1/n approaches zero and ln n becomes arbitrarily large as n → ∞, the sequence diverges.
22. As n increases, the term 2n is much larger in magnitude than (−1)n 5 and the term 4n is much larger in magnitude than
(−1)n 3. Thus dividing the numerator and denominator by n and using the fact that lim 1/n = 0, we have
n→∞
n
lim
n→∞
n
2 + (−1) 5/n
1
2n + (−1) 5
= lim
= .
n→∞ 4 − (−1)n 3/n
4n − (−1)n 3
2
Thus, the sequence converges to 1/2.
23. Since limn→∞ 1/n = 0 and −1 ≤ sin n ≤ 1, the terms approach zero and the sequence converges to 0.
24. Since sn = cos(πn) = 1 if n is even and sn = cos(πn) = −1 if n is odd, the values of sn alternate between 1 and −1,
so the limit does not exist. Thus, the sequence diverges.
25. (a)
(b)
(c)
(d)
matches (IV), since the sequence increases toward 1.
matches (III), since the odd terms increase toward 1 and the even terms decrease toward 1.
matches (II), since the sequence decreases toward 0.
matches (I), since the sequence decreases toward 1.
26. (a)
(b)
(c)
(d)
(e)
matches (II), since limn→∞ (n(n + 1) − 1) = ∞.
matches (III), since limn→∞ (1/(n + 1)) = 0 and 1/(n + 1) is always positive.
matches (I), since limn→∞ (1 − n2 ) = −∞.
matches (IV), since limn→∞ cos(1/n) = cos 0 = 1.
matches (V), since sin n is bounded above and below by ±1, so limn→∞ ((sin n)/n) = 0 and the sign of sin n varies
as n → ∞.
27. (a)
(b)
(c)
(d)
(e)
matches (II), since the sequence increases toward 2.
matches (III), since the even terms decrease toward 2 and odd terms decrease toward −2.
matches (IV), since the even terms decrease toward 2 and odd terms increase toward 2.
matches (I), since the sequence decreases toward 2.
matches (V), since the even terms decrease toward 2 and odd terms increase toward −2.
28. We have s2 = 2s1 + 3 = 2 · 1 + 3 = 5 and s3 = 2s2 + 3 = 2 · 5 + 3 = 13. Continuing, we get
1, 5, 13, 29, 61, 125.
29. We have s2 = s1 + 2 = 3 and s3 = s2 + 3 = 6. Continuing, we get
1, 3, 6, 10, 15, 21.
9.1 SOLUTIONS
30. We have s2 = s1 + 1/2 = 0 + (1/2)1 = 1/2 and s3 = s2 + (1/2)2 = 1/2 + 1/4 = 3/4. Continuing, we get
0,
1 3 7 15 31
, , ,
,
.
2 4 8 16 32
31. We have s3 = s2 + 2s1 = 5 + 2 · 1 = 7 and s4 = s3 + 2s2 = 7 + 2 · 5 = 17. Continuing, we get
1, 5, 7, 17, 31, 65.
32. We have
a2 = a2−1 + 3 · 2 = a1 + 6 = 8 + 6
= 14
a3 = a3−1 + 3 · 3 = a2 + 9 = 14 + 9 = 23
a4 = a4−1 + 3 · 4 = a3 + 12 = 23 + 12 = 35.
33. We have
a2 = a2−1 + 3 · 2 = a1 + 6 = 8 + 6
= 14
a3 = a3−1 + 3 · 3 = a2 + 9 = 14 + 9 = 23
a4 = a4−1 + 3 · 4 = a3 + 12 = 23 + 12 = 35
b2 = b2−1 + a2−1 = b1 + a1 = 5 + 8
= 13
b3 = b3−1 + a3−1 = b2 + a2 = 13 + 14 = 27
b4 = b4−1 + a4−1 = b3 + a3 = 27 + 23 = 50
b5 = b5−1 + a5−1 = b4 + a4 = 50 + 35 = 85.
34. We have:
s1
=0
s2
=0
s3
=1
s4 = s3 + s2 + s1 = 1 + 0 + 0
=1
s5 = s4 + s3 + s2 = 1 + 1 + 0
=2
s6 = s5 + s4 + s3 = 2 + 1 + 1
=4
s7 = s6 + s5 + s4 = 4 + 2 + 1
=7
s8 = s7 + s6 + s5 = 7 + 4 + 2
= 13
s9 = s8 + s7 + s6 = 13 + 7 + 4 = 24
s10 = s9 + s8 + s7 = 24 + 13 + 7 = 44.
35. The first 6 terms of the sequence for the sampling is
(−0.5)2 , (0.0)2 , (0.5)2 , (1.0)2 , (1.5)2 , (2.0)2 ,
= 0.25, 0.00, 0.25, 1.00, 2.25, 4.00.
36. The first 6 terms of the sequence for the sampling is
cos 0.5, cos 1.0, cos 1.5, cos 2.0, cos 2.5, cos 3.0
= 0.878, 0.540, 0.071, −0.416, −0.801, −0.990.
37. The first 6 terms of the sequence for the sampling are
sin 1 sin 2 sin 3 sin 4 sin 5 sin 6
,
,
,
,
,
1
2
3
4
5
6
= 0.841, 0.455, 0.047, −0.189, −0.192, −0.047.
855
856
Chapter Nine /SOLUTIONS
38. The first smoothing gives
0, 6, −6, 6, −6, 6, . . .
The second smoothing gives
3, 0, 2, −2, 2, . . .
The smoothing process diminishes the peaks and valleys of this alternating sequence.
39. The first smoothing gives
0, 0, 6, 6, 6, 0, 0 . . .
The second smoothing gives
0, 2, 4, 6, 4, 2 . . .
The smoothing process spreads out the spike at the fourth term to the neighboring terms.
40. The first smoothing gives
1.5, 2, 3, 4, 5, 6, 7 . . .
The second smoothing gives
1.75, 2.17, 3, 4, 5, 6 . . .
Terms which are already the same as their average with their neighbors are not changed.
41. Each term is 2 more than the previous term, so a recursive definition is sn = sn−1 + 2 for n > 1 and s1 = 1.
42. Each term is 2 more than the previous term, so a recursive definition is sn = sn−1 + 2 for n > 1 and s1 = 2. Notice that
the even positive integers and odd positive integers have the same recursive definition except for the starting term.
43. Each term is twice the previous term minus one, so a recursive definition is sn = 2sn−1 − 1 for n > 1 and s1 = 3. We
also notice that the differences of consecutive terms are powers of 2, so s2 = s1 + 2, s3 = s2 + 22 , and so on. Thus
another recursive definition is sn = sn−1 + 2n−1 for n > 1 and s1 = 3.
44. The differences between consecutive terms are 4, 9, 16, 25, so, for example, s2 = s1 +4 and s3 = s2 +9. Thus, a possible
recursive definition is sn = sn−1 + n2 for n > 1 and s1 = 1.
45. The differences are 2, 3, 4, 5, so, for example, s2 = s1 + 2, s3 = s2 + 3, and s4 = s3 + 4. Thus, a recursive definition is
sn = sn−1 + n for n > 1 and s1 = 1.
46. The numerator and denominator of each term are related to the numerator and denominator of the previous term. The
denominator is the previous numerator and the numerator is the sum of the previous numerator and previous denominator.
For example,
2+3
8
3+5
5
=
and =
.
3
3
5
5
If we simplify, we get
2
8
3
5
= + 1, and = + 1.
3
3
5
5
1
In words, we turn the previous term upside down and add 1. Thus, a recursive definition is sn =
+ 1 for n > 1 and
sn−1
s1 = 1.
47. For n > 1, if sn = 3n − 2, then sn−1 = 3(n − 1) − 2 = 3n − 5, so
sn − sn−1 = (3n − 2) − (3n − 5) = 3,
giving
sn = sn−1 + 3.
In addition, s1 = 3 · 1 − 2 = 1.
48. For n > 1, if sn = n(n + 1)/2, then sn−1 = (n − 1)(n − 1 + 1)/2 = n(n − 1)/2. Since
sn =
n2
n
1 2
(n + n) =
+
2
2
2
and
we have
so
sn − sn−1 =
sn−1 =
n
n
+ = n,
2
2
sn = sn−1 + n.
In addition, s1 = 1(2)/2 = 1.
1 2
n2
n
(n − n) =
− ,
2
2
2
9.1 SOLUTIONS
857
49. For n > 1, if sn = 2n2 − n, then sn−1 = 2(n − 1)2 − (n − 1) = 2n2 − 5n + 3, so
sn − sn−1 = (2n2 − n) − (2n2 − 5n + 3) = 4n − 3,
giving
sn = sn−1 + 4n − 3.
2
In addition, s1 = 2 · 1 − 1 = 1.
50. The sequence seems to converge. By the 25th term it stabilizes to four decimal places at L = 0.7391.
51. The sequence oscillates up and down, but by the 20th term it stabilizes to 4 decimal places at L = 0.5671.
52. The sequence appears to be decreasing toward 0, but at a slower and slower rate. Even after 100 terms, it is hard to guess
what the series will eventually do. It can be shown that it converges to 0.
53. The sequence converges to 1. By the tenth term, it stabilizes to three decimal places at 1.000.
54. (a) Since month 10 is October, V10 is the number of SUVs sold in the US in October 2004.
(b) The difference
P12 Vn − Vn−1 represents the increase in sales between month (n − 1) and month n.Pn
(c) The sum i=1 Vi represents the total sales of SUVs in the year 2004 (twelve months). The sum i=1 Vi represents
the total sales in the n months starting from January 1, 2004.
55. (a) Since you have two parents and four grandparents, s1 = 2 and s2 = 4. In general, sn = 2n .
(b) Solving sn = 6 · 109 gives
2n = 6 · 109
ln(6 · 109 )
= 32.482.
n=
ln 2
Thus, 33 or more generations ago, the number of ancestors is greater than the current population of the world. Since
the population of the world 33 generations ago was much smaller than it is now, there must have been overlap among
our ancestors.
56. In year 1, the payment is
p1 = 10,000 + 0.05(100,000) = 15,000.
The balance in year 2 is 100,000 − 10,000 = 90,000, so
p2 = 10,000 + 0.05(90,000) = 14,500.
The balance in year 3 is 80,000, so
p3 = 10,000 + 0.05(80,000) = 14,000.
Thus,
pn = 10,000 + 0.05(100,000 − (n − 1) · 10,000)
= 15,500 − 500n.
57. (a) The bottom row contains k cans, the next one contains (k − 1) cans, then (k − 2) and so on. Thus, there are k rows.
Since the top row contains 1 can, the second contains 2 cans, etc, we have an = n.
(b) Since the nth row contains n cans, an = n,
Tn = Tn−1 + an
gives
Tn = Tn−1 + n,
for n > 1.
In addition, T1 = 1.
(c) If Tn = 12 n(n + 1), then Tn−1 = 21 (n − 1)n, so
Tn − Tn−1 =
In addition, T1 =
1
2
· 1(2) = 1.
1
n
1
n(n + 1) − n(n − 1) = (n + 1 − (n − 1)) = n.
2
2
2
858
Chapter Nine /SOLUTIONS
58. (a) In the first year, d1 = 20,000(0.12), so the car’s value at the end of the first year is $20,000(0.88). In the second
year, d2 = 20,000(0.88)(0.12), so the car’s value at the end of the second year is $20,000(0.88)2 . Similarly,
d3 = 20,000(0.88)2 (0.12). In general
dn = 20,000(0.88)n−1 (0.12).
(b) The first year r1 = 400; the second year r2 = 400(1.18), the third year r3 = 400(1.18)2 . In general, rn =
400(1.18)n−1 .
(c) We have
Total cost = d1 + d2 + d3 + r1 + r2 + r3
= 20,000(0.12)(1 + 0.88 + (0.88)2 ) + 400(1 + 1.18 + (1.18)2 )
= 7799.52 dollars.
(d) A two-year old car has the same pattern of expenses except that the initial price is $20,000(0.88)2 instead of $20,000
and that the repair costs start at $400(1.18)2 instead of $400. Then
Total cost = 20,000(0.88)2 (0.12)(1 + 0.88 + (0.88)2 ) + 400(1.18)2 (1 + 1.18 + (1.18)2 )
= 6923.05 dollars.
Thus, the two-year-old car costs you less and you should buy it.
59. (a) The first 12 terms are
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144.
(b) The sequence of ratios is
3 5 8 13 21 34 55 89
, , ,
,
,
,
,
....
2 3 5 8 13 21 34 55
To three decimal places, the first ten ratios are
1, 2,
1, 2, 1.500, 1.667, 1.600, 1.625, 1.615, 1.619, 1.618, 1.618.
It appears that the sequence of ratios is converging to r = 1.618. We find (1.618)2 = 2.618 = 1.618 + 1 so
2
2
r seems
√ to satisfy r = r + 1. Alternatively, by the quadratic formula, the positive root of x − x − 1 = 0 is
(1 + 5)/2 = 1.618.
(c) If we multiply both sides of the equation r 2 = r + 1 by Ar n−2 , we obtain
Ar n = Ar n−1 + Ar n−2 .
Thus, if sn = Ar n , then sn−1 = Ar n−1 and sn−2 = Ar n−2 , so the sequence satisfies sn = sn−1 + sn−2 .
60. (a) Since 25/8 = 3 + 1/8, we have f (25/8) = 3 + (1 − 1/8) = 3 + 7/8 = 31/8.
Since 13/9 = 1 + 4/9, we have f (13/9) = 1 + (1 − 4/9) = 1 + 5/9 = 14/9.
Since π = 3 + (π − 3), we have f (π) = 3 + (1 − (π − 3)) = 7 − π.
(b) The terms are given by
s1 = 1
s2 =
s3 =
s4 =
s5 =
s6 =
1
1
1
=
=
f (1)
1 + (1 − 0)
2
1
1
=
=2
f (1/2)
0 + (1 − 1/2)
1
1
1
=
=
f (2)
2 + (1 − 0)
3
1
1
3
=
=
f (1/3)
0 + (1 − 1/3)
2
1
2
1
=
= .
f (3/2)
1 + (1 − 1/2)
3
The first six terms of the sequence are
1
1 3 2
1, , 2, , , .
2
3 2 3
9.1 SOLUTIONS
859
61. We want to define lim sn = L so that sn is as close to L as we please for all sufficiently large n. Thus, the definition
n→∞
says that for any positive ǫ, there is a value N such that
|sn − L| < ǫ
whenever
n ≥ N.
62. We use Theorem 9.1, so we must show that sn is bounded. Since tn converges, it is bounded so there is a number M ,
such that tn ≤ M for all n. Therefore sn ≤ tn ≤ M for all n. Since sn is increasing, s1 ≤ sn for all n. Thus if we let
K = s1 , we have K ≤ sn ≤ M for all n, so sn is bounded. Therefore, sn converges.
Strengthen Your Understanding
63. A decreasing sequence does not have to converge to 0; in fact, it does not have to converge at all (consider the sequence
sn = −n, for example). In this case, the limit of the sequence is
lim
n→∞
3 + 10/n
3+0
3
3n + 10
= lim
=
= .
n→∞ 7 + 3/n
7n + 3
7+0
7
64. Even though each term of the sequence is greater than 2, the terms could be getting progressively closer to 2 and the limit
could equal 2. For example, consider the sequence
sn = 2 +
1
.
n
The limit of this sequence is
lim
n→∞
Thus, the limit of the sequence could be 2.
2+
1
n
= 2 + 0 = 2.
65. Since we want our sequence to be increasing and converging to 0, all of the terms of the sequence must be negative. So
we need to find a sequence whose terms are negative and approaching zero. An example is sn = −1/n.
66. One example is the sequence sn = n. This sequence is increasing and therefore monotone, but it does not converge
because the terms do not get closer and closer to any specific finite value.
67. False. The first 1000 terms could be the same for two different sequences and yet one sequence converges and the other
diverges. For example, sn = 0 for all n is a convergent sequence, but
tn =
is a divergent sequence.
n
0
n
if n ≤ 1000
if n > 1000
68. False. The limit could be zero. For example, sn = 1/n is a convergent sequence of positive terms and lim sn = 0.
n→∞
69. True. If there is no term greater than a million, then the sequence is bounded by 0 < sn < 106 for all n.
70. True. If there is only a finite number of terms greater than a million, then we can choose the largest of them to be an upper
bound M for the sequence. Thus the sequence is bounded by 0 < sn ≤ M for all n.
71. False. The terms sn tend to the limit of the sequence which may not be zero. For example, sn = 1 + 1/n is a convergent
sequence and sn tends to 1 as n increases.
72. False. For example the sequence −2, −1, 0, 1, 2, 3, . . . with sn = n − 3 is monotone increasing and has both positive and
negative terms.
73. True. If a monotone sequence does not converge, then it is unbounded. If moreover the sequence contains only positive
terms then it is bounded below by zero. Thus it is not bounded above, and in particular it is not bounded above by a
million.
74. False. The decreasing sequence −1, −2, −3, . . . has all terms less than a million, but it has no lower bound. Thus it is
unbounded.
75. (b). Sequence (I) is monotone increasing, bounded between 9 and 10.
Sequence (II) is monotone decreasing and bounded between 10 and 11.
Sequence (III) is bounded between −1 and 1 but it is not monotone because it begins 0.54, −0.42, −0.99, −0.65, . . .,
which contains both a jump down, from 0.54 to −0.42 and a jump up, from −0.99 to −0.65. Terms in a monotone
sequence always jump in the same direction.
Sequence (IV) is monotone increasing but is not bounded.
860
Chapter Nine /SOLUTIONS
Solutions for Section 9.2
Exercises
1. Sequence, because the terms are not added together.
2. Series, because the terms are added together.
3. Sequence, because the terms are not added together.
4. Sequence, because the terms are not added together.
5. Series, because the terms are added together.
6. Series, because the terms are added together.
7. Series, because the terms are added together.
8. Yes, a = 5, ratio = −2.
9. No. Ratio between successive terms is not constant:
1/4
1/3
= 0.66 . . ., while
= 0.75.
1/2
1/3
10. Yes, a = 2, ratio = 1/2.
11. Yes, a = 1, ratio = −1/2.
12. No. Ratio between successive terms is not constant:
2x2
3x3
3
= 2x, while 2 = x.
x
2x
2
13. Yes, a = 1, ratio = 2z.
14. No. Ratio between successive terms is not constant:
9z 3
3
6z 2
= 2z, while 2 = z.
3z
6z
2
15. Yes, a = 1, ratio = −x.
16. Yes, a = 1, ratio = −y 2 .
17. Yes, a = y 2 , ratio = y.
z8
−z 4
= −z 2 , while
= −z 4 .
2
z
−z 4
19. The series has 26 terms. The first term is a = 2 and the constant ratio is x = 0.1, so
18. No. Ratio between successive terms is not constant:
Sum =
2(1 − (0.1)26 )
a(1 − x26 )
=
= 2.222.
(1 − x)
0.9
20. The series has 10 terms. The first term is a = 0.2 and the constant ratio is x = 0.1, so
Sum =
0.2(1 − x10 )
0.2(1 − (0.1)10 )
=
= 0.222.
(1 − x)
0.9
21. The series has 9 terms. The first term is a = 0.00002 and the constant ratio is x = 0.1, so
Sum =
0.00002(1 − (0.1)9 )
0.00002(1 − x9 )
=
= 0.0000222.
(1 − x)
0.9
22. The series has 14 terms. The first term is a = 8 and the constant ratio is x = 1/2 = 0.5, so
Sum =
8(1 − x14 )
8(1 − (0.5)14 )
=
= 15.999
(1 − x)
0.5
23. We have
4
4
Sum = 36 + 12 + 4 + + + · · ·
3 92
3
1
1
1
= 36 + 36 · + 36
+ 36
+ ···
3
3
3
36
=
= 54.
1 − 1/3
9.2 SOLUTIONS
861
24. We have
Sum = −810 + 540 − 360 + 240 − 160 + · · ·
2
3
2
2
2
+ (−810) · −
+ (−810) · −
+ ···
= −810 + (−810) · −
3
3
3
−810
=
= −486.
1 − (−2/3)
25. We have
40
20
80
Sum = 80 + √ + 40 + √ + 20 + √ + · · ·
2
2
2 2
3
1
1
1
= 80 + 80 √
+ +80 √
+ ···
+ 80 √
2
2
2
80
=
√ = 273.137.
1 − 1/ 2
26. This is a geometric series with first term 1 and ratio z/2:
1+
z2
z3
z
z
+
+
+··· = 1 + +
2
4
8
2
This series converges for |z/2| < 1, that is for −2 < z < 2.
2
z
2
+
3
z
2
+··· =
1
.
1 − (z/2)
27. This is a geometric series with first term 1 and ratio 3x:
1 + 3x + 9x2 + 27x3 + · · · = 1 + (3x) + (3x)2 + (3x)3 + · · · =
1
.
1 − (3x)
This series converges for |3x| < 1, that is for −1/3 < x < 1/3.
28. This is a geometric series with first term y and ratio −y:
y − y2 + y3 − y4 + · · · =
y
y
=
.
1 − (−y)
1+y
This series converges for | − y| < 1, that is for −1 < y < 1.
29. This is a geometric series with first term 2 and ratio −2z,
2 − 4z + 8z 2 − 16z 3 + · · · =
2
2
=
.
1 − (−2z)
1 + 2z
This series converges for | − 2z| < 1, that is for −1/2 < z < 1/2.
30. We can rewrite the series as 3 + (x + x2 + x3 + · · ·). The terms after the first term define a geometric series with first
term x and ratio x. Therefore, we have
x
.
3 + x + x2 + x3 + · · · = 3 + (x + x2 + x3 + · · ·) = 3 +
1−x
This series converges for |x| < 1, that is for −1 < x < 1.
31. We can rewrite the series as 4 + (y + y 2 /3 + y 3 /9 + · · ·). The terms after the first term define a geometric series with
first term y and ratio y/3. Therefore, we have
y
.
4 + y + y 2 /3 + y 3 /9 + · · · = 4 + (y + y 2 /3 + y 3 /9 + · · ·) = 4 +
1 − (y/3)
This series converges for |y/3| < 1, that is for −3 < y < 3.
Problems
32. Since the amount of ampicillin excreted during the time interval between tablets is 250 mg, we have
Amount of ampicillin excreted = Original quantity − Final quantity
250 = Q − (0.04)Q.
Solving for Q gives, as before,
Q=
250
≈ 260.42.
1 − 0.04
862
Chapter Nine /SOLUTIONS
33. (a)
P1 = 0
P2 = 250(0.04)
P3 = 250(0.04) + 250(0.04)2
P4 = 250(0.04) + 250(0.04)2 + 250(0.04)3
..
.
Pn = 250(0.04) + 250(0.04)2 + 250(0.04)3 + · · · + 250(0.04)n−1
(b) Pn = 250(0.04) 1 + (0.04) + (0.04)2 + (0.04)3 + · · · + (0.04)n−2 = 250
(c)
P = lim Pn
0.04(1 − (0.04)n−1 )
1 − 0.04
n→∞
0.04(1 − (0.04)n−1 )
n→∞
1 − 0.04
(250)(0.04)
= 0.04Q ≈ 10.42
=
0.96
= lim 250
Thus, lim Pn = 10.42 and lim Qn = 260.42. We would expect these limits to differ because one is right
n→∞
n→∞
before taking a tablet, one is right after. We would expect the difference between them to be 250 mg, the amount of
ampicillin in one tablet.
34. (a) The quantity of atenolol in the blood is given by Q(t) = Q0 e−kt , where Q0 = Q(0) and k is a constant. Since the
half-life is 6.3 hours,
1
1
1
= e−6.3k , k = −
ln ≈ 0.11.
2
6.3 2
After 24 hours
Q = Q0 e−k(24) ≈ Q0 e−0.11(24) ≈ Q0 (0.07).
(b)
Thus, the percentage of the atenolol that remains after 24 hours ≈ 7%.
Q0 = 50
Q1 = 50 + 50(0.07)
Q2 = 50 + 50(0.07) + 50(0.07)2
Q3 = 50 + 50(0.07) + 50(0.07)2 + 50(0.07)3
..
.
Qn = 50 + 50(0.07) + 50(0.07)2 + · · · + 50(0.07)n =
50(1 − (0.07)n+1 )
1 − 0.07
(c)
P1 = 50(0.07)
P2 = 50(0.07) + 50(0.07)2
P3 = 50(0.07) + 50(0.07)2 + 50(0.07)3
P4 = 50(0.07) + 50(0.07)2 + 50(0.07)3 + 50(0.07)4
..
.
Pn = 50(0.07) + 50(0.07)2 + 50(0.07)3 + · · · + 50(0.07)n
= 50(0.07) 1 + (0.07) + (0.07)2 + · · · + (0.07)n−1 =
0.07(50)(1 − (0.07)n )
1 − 0.07
9.2 SOLUTIONS
863
35.
q (quantity, mg)
Q0 =
250
Q1
Q2
Q3
Q4
Q5
125
P1
1
P2
P3
2
P4
4
3
P5
t (time,
5
6
days)
36. Let n be the number of days elapsed. As n → ∞, the amount of pollutants right after a dump is given by the sum of an
infinite geometric series:
B = 8 + 8(0.75) + 8(0.75)2 + 8(0.75)3 + · · · =
Thus, the level of pollutants in the bay approaches 32 tons over time.
8
= 32.
1 − 0.75
37. (a) In the first year, consumption is 25 million tons. In the second year this grows to 25(1.01) million tons, and in
subsequent years the consumption grows by a factor of 1.01. During the next n years
Total consumption = 25 + 25(1.01) + 25(1.01)2 + · · · 25(1.01)n−1 million tons.
This is a geometric series with a = 25 and x = 1.01, thus we have
Total consumption = 25
1.01n − 1
.
1.01 − 1
The reserves are exhausted when the total consumption reaches 400 so we find n by solving the equation
25
Simplifying, gives
1.01n − 1
1.01 − 1
= 400.
1.01n − 1
= 16
0.01
so
1.01n − 1 = 0.16
and
1.01n = 1.16.
Taking natural logs of both sides gives
ln 1.01n = ln 1.16
so
n ln 1.01 = ln 1.16
and
n=
ln 1.16
= 14.916.
ln 1.01
The reserves are exhausted after 15 years.
(b) If an r% annual reduction is imposed then
Total consumption = 25 + 25 1 −
r
100
+ 25 1 −
r
100
2
+ · · · million tons.
This is an infinite geometric series with a = 25 and x = 1 − r/100, thus we have
1
1 − (1 − r/100)
2500
=
.
r
Total consumption = 25
For existing reserves never to be exhausted, we need 2500/r < 400, which gives r > 6.25%. Consumption must be
reduced by at least 6.25% annually.
864
Chapter Nine /SOLUTIONS
38.
Total present value, in dollars = 1000 + 1000e−0.01 + 1000e−0.01(2) + 1000e−0.01(3) + · · ·
= 1000 + 1000(e−0.01 ) + 1000(e−0.01 )2 + 1000(e−0.01 )3 + · · ·
This is an infinite geometric series with a = 1000 and x = e(−0.01) , and sum
Total present value, in dollars =
39. (a)
1000
= 100,500.833.
1 − e−0.01
(i) On the night of December 31, 1999:
First deposit will have grown to 2(1.04)7 million dollars.
Second deposit will have grown to 2(1.04)6 million dollars.
···
Most recent deposit (Jan.1, 1999) will have grown to 2(1.04) million dollars.
Thus
Total amount = 2(1.04)7 + 2(1.04)6 + · · · + 2(1.04)
= 2(1.04)(1 + 1.04 + · · · + (1.04)6 )
= 2(1.04)
|
{z
finite geometric series
1 − (1.04)7
1 − 1.04
= 16.43 million dollars.
}
(ii) Notice that if 10 payments were made, there are 9 years between the first and the last. On the day of the last
payment:
First deposit will have grown to 2(1.04)9 million dollars.
Second deposit will have grown to 2(1.04)8 million dollars.
···
Last deposit will be 2 million dollars.
Therefore
Total amount = 2(1.04)9 + 2(1.04)8 + · · · + 2
= 2(1 + 1.04 + (1.04)2 + · · · + (1.04)9 )
=2
|
{z
finite geometric series
1 − (1.04)10
1 − 1.04
}
= 24.01 million dollars.
(b) In part (a) (ii) we found the future value of the contract 9 years in the future. Thus
Present Value =
24.01
= 16.87 million dollars.
(1.04)9
Alternatively, we can calculate the present value of each of the payments separately:
Present Value = 2 +
=2
2
2
2
+
+ ··· +
1.04
(1.04)2
(1.04)9
1 − (1/1.04)10
1 − 1/1.04
= 16.87 million dollars.
Notice that the present value of the contract ($16.87 million) is considerably less than the face value of the contract,
$20 million.
9.2 SOLUTIONS
865
40. The first $200 is invested for 24 months so it has a future value of $200(1 + 0.5/100)24 = $200 · 1.00524 . The second
installment has a future value of $ 200 · 1.00523 , since it is invested for only 23 months. The final installment, paid after
23 months, has a future value of $ 200 · 1.005. So,
Total investment = 200 · 1.00524 + 200 · 1.00523 + . . . + 200 · 1.005
= 200 · 1.005 1.00523 + 1.00522 + . . . + 1
1 − 1.00524
1 − 1.005
= 200 · 1.005 · 25.432
= 200 · 1.005
= $5111.82.
41. To get $20,000 the day he retires he needs to invest a present value P such that P (1 + 5/100)20 = $20,000. Solving for
P gives the present value P = $20,000 · 1.05−20 . To fund the second payment he needs to invest $ 20,000 · 1.05−21 , and
so on. To fund the payment 10 years after his retirement he needs to invest $ 20,000 · 1.05−30 . There are 11 payments in
all, so
Total investment = 20,000 · 1.05−20 + 20,000 · 1.05−21 + . . . + 20,000 · 1.05−30
= 20,000 · 1.05−20 1 + 1.05−1 + . . . + 1.05−10
= 20,000 · 1.05−20
1 − 1.05−11
1 − 1.05−1
= 20,000 · 1.05−20 · 8.7217
= $65,742.60.
42. If the half-life is T hours, then the exponential decay formula Q = Q0 e−kt gives k = ln 2/T . If we start with Q0 = 1
tablet, then the amount of drug present in the body after 5T hours is
Q = e−5kT = e−5 ln 2 = 0.03125,
so 3.125% of a tablet remains. Thus, immediately after taking the first tablet, there is one tablet in the body. Five half-lives
later, this has reduced to 1 · 0.03125 = 0.03125 tablets, and immediately after the second tablet there are 1 + 0.03125
tablets in the body. Continuing this forever leads to
Number of tablets in body = 1 + 0.03125 + (0.03125)2 + · · · + (0.03125)n + · · · .
This is an infinite geometric series, with common ratio x = 0.03125, and sum 1/(1 − x). Thus
Number of tablets in body =
1
= 1.0323.
1 − 0.03125
43. The amount of additional income generated directly by people spending their extra money is $100(0.8) = $80 million.
This additional money in turn is spent, generating another ($100(0.8)) (0.8) = $100(0.8)2 million. This continues
indefinitely, resulting in
Total additional income = 100(0.8) + 100(0.8)2 + 100(0.8)3 + · · · =
100(0.8)
= $400 million
1 − 0.8
44. (a)
50
1.05
50
Present value of second coupon =
, etc.
(1.05)2
Present value of first coupon =
866
Chapter Nine /SOLUTIONS
Total present value =
50
50
1000
50
+
+··· +
+
1.05
(1.05)2
(1.05)10 (1.05)10
|
{z
}
coupons
50
=
1.05
1
1
+··· +
1+
1.05
(1.05)9
1−
50
=
1.05
1
10
1
1.05
1
− 1.05
!
+
| {z }
principal
+
1000
(1.05)10
1000
(1.05)10
= 386.087 + 613.913
= $1000
(b) When the interest rate is 5%, the present value equals the principal.
(c) When the interest rate is more than 5%, the present value is smaller than it is when interest is 5% and must therefore
be less than the principal. Since the bond will sell for around its present value, it will sell for less than the principal;
hence the description trading at discount.
(d) When the interest rate is less than 5%, the present value is more than the principal. Hence the bond will be selling for
more than the principal, and is described as trading at a premium.
45. The total of the spending and respending of the additional income is given by the series: Total additional income =
100(0.9) + 100(0.9)2 + 100(0.9)3 + · · · = 100(0.9)
= $900 million.
1−0.9
Notice the large effect of changing the assumption about the fraction of money spent has: the additional spending more
than doubles.
46. (a) Let hn be the height of the nth bounce after the ball hits the floor for the nth time. Then from Figure 9.1,
Generalizing gives
h0 = height before first bounce = 10 feet,
3
feet,
h1 = height after first bounce = 10
4
2
3
feet.
h2 = height after second bounce = 10
4
hn = 10
n
3
4
.
✻
✻
✻
10
10( 3 )
4
❄
10( 3 )2
4
❄
❄
✻
···
···
hn
❄
Figure 9.1
(b) When the ball hits the floor for the first time, the total distance it has traveled
isjust D1 = 10 feet. (Notice that this
3
is the same as h0 = 10.) Then the ball bounces back to a height of h1 = 10
, comes down and hits the floor for
4
the second time. See Figure 9.1. The total distance it has traveled is
D2 = h0 + 2h1 = 10 + 2 · 10
Then the ball bounces back to a height of h2 = 10
traveled
D3 = h0 + 2h1 + 2h2 = 10 + 2 · 10
2
3
4
3
4
3
4
= 25 feet.
, comes down and hits the floor for the third time. It has
+ 2 · 10
3 2
4
= 25 + 2 · 10
3 2
4
= 36.25 feet.
9.2 SOLUTIONS
867
Similarly,
D4 = h0 + 2h1 + 2h2 + 2h3
3 2
3
3 3
+ 2 · 10
+ 2 · 10
= 10 + 2 · 10
4
4
4
3
3
= 36.25 + 2 · 10
4
≈ 44.69 feet.
(c) When the ball hits the floor for the nth time, its last bounce was of height hn−1 . Thus, by the method used in part
(b), we get
Dn = h0 + 2h1 + 2h2 + 2h3 + · · · + 2hn−1
2
3
n−1
3
3
3
3
+ 2 · 10
+ 2 · 10
+ · · · + 2 · 10
= 10 + 2 · 10
4
4
4
4
|
{z
}
finite geometric series
= 10 + 2 · 10 ·
3
4
1−
= 10 + 15
1
1+
3 n−1
4
− 34
= 10 + 60 1 −
+
!
3 n−1
3
4
4
2
3
4
+ ··· +
n−2
3
4
.
47. (a) The acceleration of gravity is 32 ft/sec2 so acceleration = 32 and velocity v = 32t + C. Since the ball is dropped,
its initial velocity is 0 so v = 32t. Thus the position is s = 16t2 + C. Calling
the initial position s = 0, we have
√
s = 6t. The distance traveled is h so h = 16t. Solving for t we get t = 41 h.
p
√
(b) The first drop from 10 feet takes 14 10 seconds. The first full bounce (to 10 · ( 43 ) feet) takes 41 10 · ( 34 ) seconds
p
to rise, therefore the same time to come down. Thus, the full bounce, up and down, takes 2( 41 ) 10 · ( 43 ) seconds.
√ p 3 2
√ p 3 n
2
The next full bounce takes 2( 14 )10 · ( 43 ) = 2( 14 ) 10
seconds. The nth bounce takes 2( 41 ) 10
4
4
seconds. Therefore the
Total amount of time
1√
10 +
=
4
2√
10
4
|
r
Geometric series with a =
=
1√
4
10 +
1√
2
r
10
3
4
r !2
3
2√
+
10
4
4
2
4
1
1−
p
3/4
3
4
{z
√ p3
10 4 =
!
r !3
2√
+
10
4
seconds.
1
2
√
3
4
p3
10
4
}
and x =
+···
p3
4
Strengthen Your Understanding
48. The formula 4/(1 − 1/4) used is for computing the sum of a geometric series, not the limit of a sequence. The sequence
is given by the formula
1 n−1
sn = 4
, n ≥ 1,
4
so since (1/4) is between 0 and 1, the sequence converges to 0.
49. The formula for the sum of an infinite geometric series does not apply because the common ratio is not between −1 and
1.
50. If the common ratio, x, of a geometric series satisfies |x| ≥ 1 then the series diverges. If x = 3 then the series 3 + 6 +
12 + 24 + · · · diverges.
868
Chapter Nine /SOLUTIONS
51. Two possible examples are the series 1+1+1+1+. . ., which has common ratio 1, and the series 1+(−1)+1+(−1)+. . .,
which has common ratio −1.
52. One way to find an example is to start with a geometric series with four distinct terms, and then rescale the terms so that
their sum is 10.
For example, we might start with the series
1+
1
1
1
+ + ,
2
4
8
whose sum is 15/8. Since we want a sum of 10, we scale these terms up by a factor of
16
10
=
.
15/8
3
This gives the series
8
4
2
16
+ + + = 10.
3
3
3
3
This series is geometric with common ratio 1/2.
53. We know that
∞
X
axn =
n=0
for the geometric series.
a
a
for |x| < 1. Letting x = 1/2 we have
= 2a. Thus a = 5 gives a limit of 10
1−x
1 − 1/2
54. (c). The common ratio in series (I) and (IV) is between −1 and 1. The common ratio in (II) and (III) is greater than one.
Solutions for Section 9.3
Exercises
1. The series is 1 + 2 + 3 + 4 + 5 + · · ·. The sequence of partial sums is
S1 = 1,
S2 = 1 + 2,
S3 = 1 + 2 + 3,
S4 = 1 + 2 + 3 + 4,
S5 = 1 + 2 + 3 + 4 + 5, . . .
which is
1,
3,
6,
10,
15 . . . .
2. The series is −1 + 1/2 − 1/3 + 1/4 − 1/5 + · · ·. The sequence of partial sums is
S1 = −1,
S2 = −1 +
1
,
2
S3 = −1 +
which is
−1,
1
1
− ,
2
3
1
− ,
2
S4 = −1 +
5
− ,
6
−
7
,
12
1
1
1
− + ,
2
3
4
−
S5 = −1 +
1
1
1
1
− + − ,...
2
3
4
5
47
,....
60
3. The series is 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + · · ·. The sequence of partial sums is
S1 =
1
,
2
which is
S2 =
1
1
+ ,
2
6
S3 =
1
1
1
+ +
,
2
6
12
1
,
2
S4 =
1
1
1
1
+ +
+
,
2
6
12
20
2
,
3
3
,
4
4
,
5
5
.
6
S5 =
1
1
1
1
1
+ +
+
+
,...
2
6
12
20
30
9.3 SOLUTIONS
869
4. We use the integral test to determine
Z ∞ whether this series converges or diverges. To do so we determine whether the
1
corresponding improper integral
dx converges or diverges:
(x + 2)2
1
Z
∞
1
1
dx = lim
(x + 2)2
b→∞
= lim
b→∞
Z
Z
Since the integral
∞
Z
b
1
dw
w2
3
b→∞
b→∞
1
dx
(x + 2)2
1
= lim −
= lim
b
1
w
(Substitute w = x + 2)
b
3
1
1
− +
b
3
1
.
3
=
∞
X
1
1
dx converges, we conclude from the integral test that the series
converges.
(x + 2)2
(n + 2)2
1
n=1
5. We use the integral test with f (x) = x/(x2 +
to determine whether this series converges or diverges. We determine
Z 1)
∞
x
whether the corresponding improper integral
dx converges or diverges:
x2 + 1
1
Z
1
∞
x
dx = lim
b→∞
x2 + 1
Since the integral
∞
Z
1
Z
b
1
x
1
dx = lim ln(x2 + 1)
b→∞ 2
x2 + 1
b
= lim
b→∞
1
1
1
ln(b2 + 1) − ln 2 = ∞.
2
2
∞
X n
x
dx
diverges,
we
conclude
from
the
integral
test
that
the
series
diverges.
x2 + 1
n2 + 1
n=1
6. We use the integral test with f (x) = 1/ex to Z
determine whether this series converges or diverges. To do so we determine
∞
1
dx converges or diverges:
whether the corresponding improper integral
x
e
1
∞
Z
1
Since the integral
∞
Z
1
1
dx = lim
b→∞
ex
b
b
Z
1
e−x dx = lim −e−x
b→∞
1
= lim −e−b + e−1 = e−1 .
b→∞
∞
X 1
1
dx converges, we conclude from the integral test that the series
converges. We can also
x
e
en
n=1
observe that this is a geometric series with ratio x = 1/e < 1, and hence it converges.
7. We use the integral test with f (x) = 1/(x(lnZx)2 ) to determine whether this series converges or diverges. We determine
∞
1
whether the corresponding improper integral
dx converges or diverges:
x(ln x)2
2
Z
∞
2
Since the integral
Z
1
dx = lim
b→∞
x(ln x)2
∞
2
8. The improper integral
Z
2
b
−1
1
dx = lim
b→∞ ln x
x(ln x)2
b
= lim
2
b→∞
1
−1
+
ln b
ln 2
=
∞
1
.
ln 2
X
1
1
dx
converges,
we
conclude
from
the
integral
test
that
the
series
converges.
x(ln x)2
n(ln n)2
n=2
Z
1
∞
x
−3
1
dx converges to , since
2
Z
1
b
x−3 dx =
x−2
−2
and
lim
b→∞
b
=
1
b−2
1−2
1
1
−
=
+
−2
−2
−2b2
2
1
1
+
−2b2
2
=
1
.
2
870
Chapter Nine /SOLUTIONS
∞
X
The terms of the series
n−3 form a right hand sum for the improper integral; each term represents the area of a
n=2
rectangle of width 1 fitting completely under the graph of the function x−3 . (See Figure 9.2.) Thus the sequence of partial
sums is bounded above by 1/2. Since the partial sums are increasing (every new term added is positive) the series is
guaranteed to converge to some number less than or equal to 1/2 by Theorem 9.1.
y
y = 1/x3
1/8
1/27
1/64
x
1 2 3 4 5 6 7 8 9
Figure 9.2
9. The improper integral
Z
0
∞
1
π
dx converges to , since
x2 + 1
2
Z
0
b
1
dx = arctan x|b0 = arctan b − arctan 0 = arctan b,
x2 + 1
∞
X 1
π
. The terms of the series
form a right hand sum for the improper integral; each term
b→∞
2
n2 + 1
n=1
1
represents the area of a rectangle of width 1 fitting completely under the graph of the function 2
. (See Figure 9.3.)
x +1
π
Thus the sequence of partial sums is bounded above by . Since the partial sums are increasing (every new term added is
2
positive), the series is guaranteed to converge to some number less than or equal to π/2 by Theorem 9.1.
and lim arctan b =
y
1
y = 1/(x2 + 1)
1/2
1/5
1/10
x
1
2
3
4
5
6
7
8
Figure 9.3
10. The integral test requires that f (x) = x2 , which is not decreasing.
11. The integral test requires that f (x) = (−1)x /x. However (−1)x is not defined for all x.
12. The integral test requires that f (x) = e−x sin x, which is not positive, nor is it decreasing.
9.3 SOLUTIONS
871
Problems
13. Using the integral test, we compare the series with
Z
0
∞
3
dx = lim
b→∞
x+2
Z
b
0
b
3
dx = 3 ln |x + 2| .
x+2
0
Since ln(b + 2) is unbounded as b → ∞, the integral diverges and therefore so does the series.
14. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral:
∞
Z
0
b
Z
4
dx = lim
b→∞
2x + 1
0
b
4
dx = lim 2 ln(2x + 1)
b→∞
2x + 1
= lim 2 ln(2b + 1).
b→∞
0
∞
X
4
diverges.
2n + 1
n=0
R∞ √
15. We use the integral test and calculate the corresponding improper integral, 0 2/ 2 + x dx:
Since the limit does not exist, the integral diverges, so the series
∞
Z
√
0
2
dx = lim
b→∞
2+x
Z
b
2 dx
= lim 4(2 + x)1/2
√
b→∞
2+x
0
Since the limit does not exist, the integral diverges, so the series
∞
X
n=1
√
b
0
= lim 4 (2 + b)1/2 − 21/2 .
b→∞
2
diverges.
2+n
16. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral using the substitution w = x2
Z
∞
0
Z
2x
dx = lim
b→∞
1 + x4
b
0
2x
dx = lim arctan x2
b→∞
1 + x4
∞
Since the limit exists, the integral converges, so the series
X
n=0
b
= lim arctan b2 =
b→∞
0
π
.
2
2n
converges.
1 + n4
17. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral using the substitution w = 1 + x2 :
Z
∞
0
2x
dx = lim
b→∞
(1 + x2 )2
b
Z
0
2x
−1
dx = lim
b→∞ (1 + x2 )
(1 + x2 )2
Since the limit exists, the integral converges, so the series
∞
X
n=0
18. The terms in this series do not tend to 0. We have
lim √
n→∞
b
= lim
b→∞
0
−1
+ 1 = 1.
1 + b2
2n
converges.
(1 + n2 )2
2
2n
= lim p
= 2.
n→∞
4 + n2
4/n2 + 1
Thus the series diverges. It is also possible (although much slower) to use the integral test. We calculate the corresponding
improper integral using the substitution w = x2
Z
∞
0
√
2x
dx = lim
b→∞
4 + x2
Z
0
b
√
p
2x
dx = lim 2 4 + x2
2
b→∞
4+x
Since the limit does not exist, the integral diverges, so the series
∞
X
n=0
√
b
p
= lim (2
0
b→∞
2n
diverges.
4 + n2
4 + b2 − 4).
872
Chapter Nine /SOLUTIONS
R∞
19. We use the integral test and calculate the corresponding improper integral,
Z
∞
1
b
Z
3 dx
= lim
b→∞
(2x − 1)2
1
3 dx
−3/2
= lim
b→∞ (2x − 1)
(2x − 1)2
∞
X
Since the integral converges, the series
n=1
b
= lim
b→∞
1
∞
4 dx
= lim
b→∞
(2x + 1)3
1
Z
b
4 dx
1
= lim −
b→∞
(2x + 1)3
(2x + 1)2
1
∞
X
Since the integral converges, the series
n=1
−3/2
3
+
(2b − 1)
2
=
3
.
2
3
converges.
(2n − 1)2
R∞
20. We use the integral test and calculate the corresponding improper integral,
w = 2x + 1, we have
Z
3/(2x − 1)2 dx:
1
1
b
4/(2x + 1)3 dx. Using the substitution
= lim
b→∞
1
−
1
1
+
(2b + 1)2
9
=
=
3π
,
4
1
.
9
4
converges.
(2n + 1)3
21. Using the integral test, we compare the series with
∞
Z
3
dx = lim
b→∞
x2 + 4
0
Z
b
0
b
3
3
x
dx =
lim arctan
x2 + 4
2 b→∞
2
by integral table V-24. Since the integral converges so does the series.
=
0
3
b
lim arctan
2 b→∞
2
22. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral using the substitution w = 2x:
Z
∞
0
2
dx = lim
1 + 4x2
b→∞
Z
b
2
dx = lim arctan(2x)
1 + 4x2
b→∞
0
Since the limit exists, the integral converges, so the series
∞
X
b
= lim arctan(2b) = π/2.
0
b→∞
2
converges.
1 + 4n2
n=0
23. Writing an = n/(n + 1), we have limn→∞ an = 1 so the series diverges by Property 3 of Theorem 9.2.
24. Writing an = (n + 1)/(2n + 3), we have limn→∞ an = 1/2, so the limit of individual terms is not 0. The series diverges
by Property 3 of Theorem 9.2.
25. Both
∞
X
1 n
2
n=1
∞
X
1 n
n=1
2
26. The series
+
n=1
Therefore
3
4
4
+
1
n
is a convergent geometric series, but
∞
X
3 n
n=1
4
are convergent geometric series. Therefore, by Property 1 of Theorem 9.2, the series
converges.
∞
X
3 n
∞
X
3
3
n=1
2 n
n=1
n
If
∞
X
2 n
and
converged, then
∞
X
3 n
4
n=1
+
1
n
27. The series can be written as
n=1
+
1
n
−
is the divergent harmonic series.
n
∞
X
3 n
n=1
4
=
∞
X
1
n=1
n
would converge by Theorem 9.2.
diverges.
∞
X
n + 2n
∞
X
1
n=1
2n
+
1
n
−
harmonic series, which diverges, then the series
=
n2n
n=1
If this series converges, then
∞
X
1
n=1
∞
X
1
n=1
∞
∞
X
1
2n
=
X n + 2n
n=1
n
2n
∞
X
1
n=1
n
+
1
.
n
would converge by Theorem 9.2. Since this is the
diverges.
9.3 SOLUTIONS
873
28. Let an = (ln n)/n and f (x) = (ln x)/x. We use the integral test and consider the improper integral
Z
∞
c
Since
Z
c
R
ln x
1
dx = (ln x)2
x
2
ln x
dx.
x
R
=
c
1
(ln R)2 − (ln c)2 ,
2
and ln R grows without bound as R → ∞, the integral diverges. Therefore, the integral test tells us that the series,
∞
X
ln n
, also diverges.
n
n=1
29. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral using the substitution w = 1 + ln x:
Z
1
∞
1
dx = lim
b→∞
x(1 + ln x)
Z
1
b
1
dx = lim ln(1 + ln x)
b→∞
x(1 + ln x)
Since the limit does not exist, the integral diverges, so the series
∞
X
n=1
3
∞
x+1
dx = lim
b→∞
x2 + 2x + 2
Z
b
3
= lim ln(1 + ln b).
1
b→∞
1
diverges.
n(1 + ln n)
30. We use the integral test and calculate the corresponding improper integral,
Z
b
R∞
3
(x + 1)/(x2 + 2x + 2) dx:
x+1
1
dx = lim ln |x2 + 2x + 2|
b→∞ 2
x2 + 2x + 2
Since the limit does not exist (it is ∞), the integral diverges, so the series
∞
X
n=3
b
1
(ln(b2 + 2b + 2) − ln 17).
2
= lim
3
b→∞
n+1
diverges.
n2 + 2n + 2
31. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral using the substitution w = 1 + x:
Z
0
∞
1
dx = lim
b→∞
x2 + 2x + 2
Z
0
b
1
1
dx = lim
dx
b→∞ 1 + (x + 1)2
x2 + 2x + 2
b
= lim arctan(x + 1)
b→∞
0
= lim (arctan(b + 1) − arctan 1) = π/2 − arctan 1.
Since the limit exists, the integral converges, so the series
b→∞
∞
X
n=0
1
converges.
n2 + 2n + 2
32. Since the terms in the series are positive and decreasing, we can use the integral test. We calculate the corresponding
improper integral:
Z
∞
x ln x + 4
dx = lim
b→∞
x2
2
Z
b
2
x ln x + 4
dx = lim
b→∞
x2
Z
2
b
ln x
dx +
x
Z
2
b
4
dx .
x2
Using the substitution w = ln x on the first integral, we have
lim
b→∞
Z
2
b
x ln x + 4
dx = lim
b→∞
x2
(ln x)2
4
−
2
x
Since the limit does not exist, the integral diverges, so the series
b
2
∞
X
n ln n + 4
n=2
33. Using ln(2n ) = n ln 2, we see that
X
(ln b)2
(ln 2)2
4
− −
+ 2.
b→∞
2
b
2
= lim
n2
X 1
1
=
.
n
ln(2 )
(ln 2)n
diverges.
The series on the right is the harmonic series multiplied by 1/ ln 2. Since the harmonic series diverges,
diverges.
P∞
n=1
1/ ln(2n )
874
Chapter Nine /SOLUTIONS
34. Using ln(2n ) = n ln 2, we see that
∞
X
∞
X
1
1
=
.
(ln (2n ))2
(ln 2)2 n2
n=1
Since
P
1/n2 converges,
P
n=1
1/((ln (2))2 n2 ) converges by property 1 of Theorem 9.2.
35. (a) With an = ln((n + 1)/n) we have
Sn = a1 + a2 + a3 + · · · + an−1 + an
= ln(2/1) + ln(3/2) + ln(4/3) + · · · + ln(n/(n − 1)) + ln((n + 1)/n)
2 3 4
n
n+1
= ln(n + 1).
· · ···
·
= ln
1 2 3
n−1
n
(b) Since the limit of the partial sums, limn→∞ Sn = limn→∞ ln(n + 1), does not exist, the series diverges.
36. (a) Using r = eln r and n = eln n we have r ln n = e(ln r)(ln n) = nln rP
.
ln n
ln r
− ln r
(b) By
part
(a)
we
have
r
=
n
=
1/n
.
Since
the
p-series
1/np converges if and only if p > P
1, the series
P∞
∞
− ln r
1/n
converges if and only if − ln r > 1, which is equivalent to ln r < −1 or r < 1/e. Thus n=1 r ln n
n=1
converges if 0 < r < 1/e and diverges if r ≥ 1/e.
37. (a) A common denominator is k(k + 1) so
1
k+1
k
k+1−k
1
1
−
=
−
=
=
.
k
k+1
k(k + 1)
k(k + 1)
k(k + 1)
k(k + 1)
(b) Using the result of part (a), the partial sum can be written as
1
1
1
1
1
1
1
1
1
1
S3 =
+
+
= − + − + − =1− .
1·2
2·3
3·4
1
2
2
3
3
4
4
1
1
and Sn = 1−
.
All of the intermediate terms cancel out, leaving only the first and last terms. Thus S10 = 1−
11
n+1
∞
X
1
1
converges to 1.
(c) The limit of Sn as n → ∞ is lim 1 −
= 1 − 0 = 1. Thus the series
n→∞
n+1
k(k + 1)
k=1
38. (a) The partial sum
2·4
1·3
+ ln
2·2
3·3
Using the property ln(A) + ln(B) = ln(AB), we get
S3 = ln
+ ln
3·5
.
4·4
1·3·2·4·3·5
.
2·2·3·3·4·4
5
1 · 5
The intermediate factors cancel out, leaving only ln
, so S3 = ln
.
2·4
8
(b) For the partial sum Sn , similar steps yield
S3 = ln
Sn = ln
1 · 3 · 2 · 4 · 3 · 5 · · · n(n + 2)
2 · 2 · 3 · 3 · 4 · 4 · · · (n + 1) · (n + 1)
As before, most of the factors cancel, leaving Sn = ln
(c) The limit of Sn = ln
converges to ln
1
n+2
2(n + 1)
as n → ∞ is lim ln
n→∞
n+2
2(n + 1)
n+2
2(n + 1)
.
.
= ln
∞
X
1
ln
. Thus the series
2
k=1
k(k + 2)
(k + 1)2
.
2
P
P
39. P
Let Sn be theP
nth partial sum for
an and let Tn be the nth partial sum for
bn . Then the nth partial sums for
P
(an + bn ), (an − bn ), and
kan are Sn + Tn , Sn − Tn , and kSn , respectively. To show that these series converge,
we have to show that the limits of their partial sums exist. By the properties of limits,
lim (Sn + Tn ) = lim Sn + lim Tn
n→∞
n→∞
n→∞
lim (Sn − Tn ) = lim Sn − lim Tn
n→∞
n→∞
n→∞
lim kSn = k lim Sn .
n→∞
n→∞
This proves that the limits of the partial sums exist, so the series converge.
9.3 SOLUTIONS
P
875
P
40. Let Sn be the n-th partial sum for
an and let Tn be the n-th partial sum for
bn . Suppose that SN = TN + k. Since
an = P
bn for n ≥P
N , we have Sn = Tn + k for n ≥ N . Hence if Sn converges to a limit, so does Tn , and vice versa.
an and
bn either both converge or both diverge.
Thus,
P
41. We have an = Sn − Sn−1 . If
an converges, then S = limn→∞ Sn exists. Hence limn→∞ Sn−1 exists and is equal
to S also. Thus
lim an = lim (Sn − Sn−1 ) = lim Sn − lim Sn−1 = S − S = 0.
n→∞
n→∞
n→∞
n→∞
P
P
42. From Property 1 in Theorem
P if an converges, then so does kan .
P
P 9.2, we know that
P Now suppose that an diverges and kan converges for k 6= 0. Thus using Property 1 and replacing an by
kan , we know that the following series converges:
X1
k
(kan ) =
X
an .
Thus, we have arrived at a contradiction, which means our original assumption, that
∞
X
kan converged, must be wrong.
n=1
43. A typical partial sum of the series
∞
X
n=1
(an+1 − an ), say S5 , shows what happens in the general case:
S5 = (a2 − a1 ) + (a3 − a2 ) + (a4 − a3 ) + (a5 − a4 ) + (a6 − a5 ) = a6 − a1
as all of the intermediate terms cancel out. The same thing will happen in the general partial sum: Sn = an+1 − a1 .
Now the series
∞
X
n=1
(an+1 − an ) converges if the sequence of partial sums Sn has a limit as n → ∞. Since we’re as-
suming that the original series
∞
X
an converges, we know that lim an = lim an+1 = 0 by property 3 of Theorem 9.2.
n→∞
n=1
Thus
n→∞
lim Sn = lim (an+1 − a1 ) = 0 − a1 = −a1 .
n→∞
n→∞
Since the sequence of partial sums converges (to −a1 ), the series
P
44. If anP
= 1 for all n, then
P an diverges but
an+1 − an =
1 diverges
and
45. (a) Suppose
Z
∞
P
(an+1 − an ) =
∞
X
n=1
P
(an+1 − an ) converges (also to −a1 ).
0 converges. If an = n for all n, then
P
an diverges,
f (x) dx diverges. In Figure 9.4 we see that for each positive integer k
c
Z
N+k+1
f (x) dx ≤ f (N ) + f (N + 1) + · · · + f (N + k).
N
Since f (n) = an for all n, we have
Z
N+k+1
f (x) dx ≤ aN + aN+1 + · · · + aN+k .
N
Since f (x) is defined for all x ≥ c, if
integral
R N+k+1
N
R∞
c
f (x) dx is divergent, then
f (x) dx diverges, so the partial sums of the series
R∞
∞
X
i=N
N
f (x) dx is divergent. So as k → ∞, the
ai diverge. Thus, the series
∞
X
ai diverges.
i=1
More precisely, suppose the series converged. Then the partial sums would be bounded. (The partial sums would
be less than the sum of the series, since all the terms in the series are positive.) But that would imply that the integral
converged,
by Theorem 9.1 on Convergence of Monotone Bounded Sequences. This contradicts the assumption that
R∞
is divergent.
f
(x)
dx
N
876
Chapter Nine /SOLUTIONS
(b) Suppose
Z
∞
f (x) dx converges. Let N an integer with N ≥ c. Consider the series
c
∞
X
ai . The partial sums of
i=N+1
this series are increasing because all the terms in the series are positive. We show the partial sums are bounded using
the right-hand sum in Figure 9.5. We see that for each positive integer k
f (N + 1) + f (N + 2) + · · · + f (N + k) ≤
Z
N+k
f (x) dx.
N
Since f (n) = an for all n, and c ≤ N , we have
aN+1 + aN+2 + · · · + aN+k ≤
Since f (x) is a positive function,
R∞
c
f (x) dx is convergent,
R N+k
c
R N+k
c
f (x) dx ≤
f (x) dx <
R∞
c
Rb
c
∞
X
N+k
f (x) dx.
c
f (x) dx for all b ≥ N + k. Since f is positive and
f (x) dx, so we have
aN+1 + aN+2 + · · · + aN+k ≤
Thus, the partial sums of the series
Z
Z
∞
f (x) dx
for all k.
c
ai are bounded and increasing, so this series converges by Theorem 9.1.
i=N+1
Now use Theorem 9.2, property 2, to conclude that
∞
X
ai converges.
i=1
f (x)
f (x)
Area = f (N )
✠
Area = f (N + 1)
Area = f (N + 1)
Area = f (N + 2)
✠
✠
✠
c
x
N
c
N +1
Figure 9.4
Area = f (N + 2)
✠
Area = f (N + 3)
✠
x
N N +1
Figure 9.5
46. (a) Show that the sum of each group of fractions is more than 1/2.
(b) Explain why this shows that the harmonic series does not converge.
(a) Notice that
1
1
1
1
+ > +
3
4
4
4
1
1
1
1
+ + +
5
6
7
8
1
1
+
+ ··· +
9
10
2
1
=
4
2
1
1
1
1
4
1
> + + + = =
8
8
8
8
8
2
1
1
1
1
8
1
>
+
+··· +
=
= .
16
16
16
16
16
2
=
In the same way, we can see that the sum of the fractions in each grouping is greater than 1/2.
(b) Since the sum of the first n groups is greater than n/2, it follows that the partial sums of the harmonic series are not
bounded. Thus, the harmonic series diverges.
9.3 SOLUTIONS
47. (a) Since for x > 0,
Z
we have
Z
2
∞
1
dx = lim
b→∞
x ln x
877
1
dx = ln(ln x) + C
x ln x
Z
2
b
1
dx = lim (ln(ln b) − ln(ln 2)) = ∞.
b→∞
x ln x
The series diverges by the integral test.
(b) The terms in each group are decreasing so we can bound each group as follows:
1
1
1
1
1
+
>
+
=
3 ln 3
4 ln 4
4 ln 4
4 ln 4
2 ln 4
and
1
1
1
1
1
1
+
+
+
>4
=
.
5 ln 5
6 ln 6
7 ln 7
8 ln 8
8 ln 8
2 ln 8
Similarly, the group whose final term is 1/(2n ln(2n )) is greater than 1/(2 ln(2n )) = 1/(2(ln 2)n). Thus
N
2
X
n=2
N
X
1
1
>
.
n ln n
2(ln 2)n
n=1
The
P series on the right is the harmonic series multiplied by the constant 1/(2 ln 2). Since the harmonic series diverges,
1/(n ln n) diverges.
R n+1
48. (a) See Figure 9.6. We draw the left-hand sum approximation to 1 (1/x) dx using ∆x = 1. Since the rectangles lie
everywhere above the curve, for every positive integer k, we conclude that
Z
k
k+1
dx
1
< .
x
k
We see that
1
1
an+1 − an =
− ln(n + 2) + ln(n + 1) =
−
n+1
n+1
Z
n+2
n+1
dx
,
x
so we see that an < an+1 for every positive integer n.
(b) For
R n+1every positive integer n, the value of an represents the difference between the left-hand approximation of
(1/x) dx and the integral itself.
1
This difference can be viewed as a sum of areas of n roughly triangular pieces, with the kth piece having a
vertical side running from y = 1/k to y = 1/(k + 1). Each piece has width 1. All the pieces can be moved to the
left and stacked up to fit inside the first rectangle of the left-hand approximation, which runs from x = 1 to x = 2
and from y = 0 to y = 1. Thus the sum of all these stacked areas is less than 1, so an < 1 for all n. See Figure 9.7.
(c) Since an is an increasing sequence bounded above by 1, Theorem 9.1 ensures that limn→∞ an exists.
(d) The sequence converges slowly, but a calculator or computer gives a200 = 0.5747. For comparison, a100 = 0.5723,
a500 = 0.5762. Thus, γ ≈ 0.58. More extensive calculations show that γ ≈ 0.577216.
f (x) = 1/x
f (x) = 1/x
1
1
1
2
...
x
n n+1
Figure 9.6
1
2
···
x
n n+1
Figure 9.7
878
Chapter Nine /SOLUTIONS
49. (a) A calculator or computer gives
20
X
1
1
(b) Since
∞
X
1
1
n2
2
=
1
1
1
+ 2 + · · · + 2 = 1.596.
12
2
20
=
n2
π
, the answer to part (a) gives
6
π2
≈ 1.596
6
√
π ≈ 6 · 1.596 = 3.09
(c) A calculator or computer gives
100
X
1
1
so
n2
1
1
1
+ 2 +··· +
= 1.635,
12
2
1002
=
π2
≈ 1.635
6
√
π ≈ 6 · 1.635 = 3.13.
(d) The error in approximating π 2 /6 by
P20
1
1/n2 is the tail of the series
∞
X
1
21
<
n2
∞
Z
20
∞
1
dx
=−
x2
x
=
20
P∞
21
1
= 0.05.
20
A similar argument leads to a bound for the error in approximating π 2 /6 by
∞
X
1
101
n2
<
Z
∞
100
dx
1
=−
x2
x
∞
=
100
1/n2 . From Figure 9.8, we see that
P100
1
1/n2 as
1
= 0.01.
100
1/x2
20 21 22
...
x
Figure 9.8
50. (a) We have e > 1 + 1 + 1/2 + 1/6 + 1/24 = 65/24 = 2.708.
(b) We have
1
1
1
1
=
≤
= n−1 .
n!
1 · 2 · 3 · 4···n
1 · 2 · 2 · 2···2
2
(c) The inequality in part (b) can be used to replace the given series with a geometric series that we can sum.
e=
∞
X
1
n=0
n!
=1+
∞
X
1
n=1
n!
<1+
∞
X
n=1
1
1
=1+
= 3.
2n−1
1 − 1/2
9.3 SOLUTIONS
51. (a) The right-hand sum for
RN
0
879
xN dx with ∆x = 1 is the sum 15 · 1 + 25 · 1 + 35 · 1 + · · · + N 5 · 1 = SN . This sum is
RN
greater than the integral because the integrand x5 is increasing on the interval 0 < x < N . Since 0 x5 dx = N 6 /6,
we have SN > N 6 /6.
R N+1 N
x dx with ∆x = 1 is the sum 15 · 1 + 25 · 1 + 35 · 1 + · · · + N 5 · 1 = SN .
(b) The left-hand sum for 1
sum is less than the integral because the integrand x5 is increasing on the interval 1 < x < N + 1. Since
RThis
N+1 5
x dx = ((N + 1)6 − 1)/6, we have SN < ((N + 1)6 − 1)/6.
1
(c) By parts (a) and (b) we have
((N + 1)6 − 1)/6
N 6 /6
SN
1
1
=1< 6
<
= (1 + )6 − 6 .
6
N /6
N /6
N 6 /6
N
N
Since both limN→∞ 1 = 1 and limN→∞ ((1 + N1 )6 − N16 ) = 1, we conclude that the limit in the middle also equals
1, limN→∞ SN /(N 6 /6) = 1.
52. (a) Neglecting signs, the table reveals a regular pattern reminiscent of a linear function with slope 4. We see that term n
is given by 4n − 3:
Term
1
2
3
4
n
Value
1
5
9
13
4n − 3
The odd-numbered terms are positive and the even-numbered terms are negative, so
cn =
(
4n − 3
for n odd
−1 · (4n − 3) for n even.
Notice that −(−1) is negative if n is even, and positive if n is odd. Thus we can write cn = −(−1)n (4n − 3).
Since −(−1)n = (−1)1 (−1)n = (−1)n+1 , we can also write this cn = (−1)n+1 (4n − 3).
3
3
3
3
3!!
5!!
7!!
1!!
, b3 =
, b4 =
, b5 =
, . . ..
(b) Using double factorial (!!) notation, we can write: b2 =
2!!
4!!
6!!
8!!
Focusing on the patterns in the table of 1, 3, 5, 7 in the numerator and 2, 4, 6, 8 in the denominator, we see that both
are reminiscent of linear functions with slope 2:
n
This means we can write bn =
Term
2
3
4
5
n
Numerator
1
3
5
7
Denominator
2
4
6
8
2n − 3
(2n − 3)!!
(2n − 2)!!
3
2n − 2
.
n
(c) Putting together our answers for parts (a) and (b), we have an = −(−1) (4n−3)
an = (−1)n+1 (4n − 3)
(2n − 3)!!
(2n − 2)!!
3
and so on, as required.
(2n − 3)!!
(2n − 2)!!
3
, or (equivalently)
. Checking our answer, we have:
3
(2 · 1 − 3)!!
(2 · 1 − 2)!!
3
(2 · 2 − 3)!!
2
a2 = −(−1) (4 · 2 − 3)
(2 · 2 − 2)!!
3
(2
· 3 − 3)!!
3
a3 = −(−1) (4 · 3 − 3)
(2 · 3 − 2)!!
3
(2 · 4 − 3)!!
s4 = −(−1)4 (4 · 4 − 3)
(2 · 4 − 2)!!
a1 = −(−1)1 (4 · 1 − 3)
3
(−1)!!
0!!
2
1!!
= −5
2!!
2
3!!
= 9
4!!
5!! 2
= −13
6!!
=1
1
=
since (−1)!! = 0!! = 1
3
1
since 1!! = 1 and 2!! = 2
2
1×3 3
=
9
2×4
1 × 3 × 5 3
= −13
,
2×4×6
=
−5
Strengthen Your Understanding
53. If the terms of a series do not approach zero, P
the series does not converge. But just because the terms approach zero does
not mean the series converges. For example, (1/n) diverges even though the terms approach zero.
880
Chapter Nine /SOLUTIONS
54. The integral
R∞
1
1/x3 dx converges to 21 , because
Z
∞
1
dx
= lim
x3
b→∞
Z
b
1
1
dx
1
= lim − 2 +
x3
b→∞
2b
2
P∞
=
1
.
2
However, the series n=1 1/n3 = 1 + 1/23 + 1/33 + · · · has a sum which is larger than 1 as its first term is 1 and all
the subsequent terms are positive. Thus, the sum of the series is not 21 .
In general, the sum of a series and the value of an improper integral used to test it for convergence are not the same.
Both converge or both diverge, but if they converge, usually they converge to different values.
P∞
55. The series n=1 1/n is an example. The terms of the series converge to zero, but the series is a p-series with p ≤ 1, and
therefore the series diverges.
∞
X
1
. This series is convergent because it is a p-series with p = 2 > 1.
n2
n=1
P∞
√
If an = 1/n2 , then an = 1/n, and the series n=1 1/n diverges because it is a p-series with p = 1.
56. One example is the series
57. True. Writing out the terms of this series, we have
(1 + (−1)1 ) + (1 + (−1)2 ) + (1 + (−1)3 ) + (1 + (−1)4 ) + · · ·
= (1 − 1) + (1 + 1) + (1 − 1) + (1 + 1) + · · ·
= 0 + 2 + 0 + 2 +···.
58. True. The definition of convergence of a series is that its partial sums are a convergent sequence.
59. P
False. For example, if an = 1/n and bn = −1/n, then |an + bn | = 0, so
|bn | are the harmonic series, which diverge.
60. False. The terms in the series do not go to zero:
2
1
3
4
P
|an + bn | converges. However
P
|an | and
5
2(−1) + 2(−1) + 2(−1) + 2(−1) + 2(−1) + · · · = 2−1 + 21 + 2−1 + 21 + 2−1 + · · ·
= 1/2 + 2 + 1/2 + 2 + 1/2 + · · · .
61. True. If the terms do not tend to zero, the partial sums do not tend to a limit. For example, if the terms are all greater than
0.1, the partial sums will grow without bound.
62. False. Consider the series
∞
X
n=1
63. False. If an = bn = 1/n, then
1/n. This series does not converge, but 1/n → 0 as n → ∞.
X
an and
X
bn do not converge. However, an bn = 1/n2 , so
64. False. If an bn = 1/n2 and an = bn = 1/n, then
65. (d)
X
an bn converges, but
X
an and
X
X
an bn does converge.
bn do not converge.
Solutions for Section 9.4
Exercises
1. Let an = 1/(n − 3), for n ≥ 4. Since n − 3 < n, we have 1/(n − 3) > 1/n, so
an >
The harmonic series
∞
X
1
n=4
n
1
.
n
diverges, so the comparison test tells us that the series
∞
X
n=4
1
also diverges.
n−3
9.4 SOLUTIONS
881
2. Let an = 1/(n2 + 2). Since n2 + 2 > n2 , we have 1/(n2 + 2) < 1/n2 , so
0 < an <
The series
∞
X
1
n2
n=1
converges, so the comparison test tells us that the series
e
1
< 2 , so
n2
n
0 < an <
∞
X
1
n2
n=1
converges, so the comparison test tells us that the series
∞
X
e−n
n=1
n4
P∞
n=1
n3 + 1
+ 2n3 + 2n
1
also converges.
n2 + 2
1
.
n2
4. As n gets large, polynomials behave like the leading term, so for large n,
Since the series
∞
X
n=1
−n
3. Let an = e−n /n2 . Since e−n < 1, for n ≥ 1,we have
The series
1
.
n2
n2
also converges.
n3
1
= .
n4
n
behaves like
1/n diverges, we predict that the given series will diverge.
5. As n gets large, polynomials behave like the leading term, so for large n,
Since the series
P∞
n=1
n+4
n3 + 5n − 3
n
1
= 2.
n3
n
behaves like
1/n2 converges, we predict that the given series will converge.
6. As n gets large, polynomials behave like the leading term, so for large n,
1
n4 + 3n3 + 7
Since the series
P∞
n=1
behaves like
1
.
n4
1/n4 converges, we predict that the given series will converge.
7. As n gets large, polynomials behave like the leading term, so for large n,
√
n−4
n3 + n2 + 8
behaves like
n
1
= 1/2 .
n3/2
n
P∞
1/n1/2 diverges, we predict that the given series will diverge.
n
1
, so
8. Let an = 1/(3n + 1). Since 3n + 1 > 3n , we have 1/(3n + 1) < 1/3n =
3
Since the series
n=1
0 < an <
Thus we can compare the series
∞
X
n=1
n
4
1
3
.
∞
n
X 1
1
with the geometric series
n
3 +1
3
|1/3| < 1, so the comparison test tells us that
4
n
n
4
∞
X
n=1
n=1
1
also converges.
3n + 1
. This geometric series converges since
9. Let an = 1/(n + e ). Since n + e > n , we have
1
1
< 4,
n4 + en
n
so
0 < an <
Since the p-series
∞
X
1
n=1
n4
1
.
n4
converges, the comparison test tells us that the series
∞
X
n=1
1
also converges.
n4 + en
882
Chapter Nine /SOLUTIONS
10. P
Since ln n ≤ n for n ≥ 2, we have 1/ ln n ≥ 1/n, so the series diverges by comparison with the harmonic series,
1/n.
1
1
< 4 , so
11. Let an = n2 /(n4 + 1). Since n4 + 1 > n4 , we have 4
n +1
n
an =
n2
n2
1
< 4 = 2,
+1
n
n
n4
therefore
0 < an <
Since the p-series
∞
X
1
n=1
n2
converges, the comparison test tells us that the series
∞
X
1
n=1
n2
∞
X
n=1
12. We know that | sin n| < 1, so
Since the p-series
1
.
n2
n2
converges also.
+1
n4
n
n
1
n sin2 n
≤ 3
< 3 = 2.
n3 + 1
n +1
n
n
converges, comparison gives that
∞
X
n sin2 n
n=1
n3 + 1
converges.
13. Let an = (2n + 1)/(n2n − 1). Since n2n − 1 < n2n + n = n(2n + 1), we have
2n + 1
1
2n + 1
>
= .
n2n − 1
n(2n + 1)
n
Therefore, we can compare the series
us that
∞
X
2n + 1
n=1
14. Since an =
n2n − 1
∞
X
2n + 1
n2n − 1
n=1
with the divergent harmonic series
∞
X
1
n=1
n
. The comparison test tells
also diverges.
n+1
n
, replacing n by n + 1 gives an+1 = n+1 . Thus
2n
2
|an+1 |
(n + 1)/2n+1
n+1
=
=
,
|an |
n/2n
2n
so
L = lim
n→∞
Since L < 1, the ratio test tells us that
n+1
1 + 1/n
1
|an+1 |
= lim
= lim
= .
n→∞ 2n
n→∞
|an |
2
2
∞
X
n
n=1
2n
converges.
15. Since an = 1/(2n)!, replacing n by n + 1 gives an+1 = 1/(2n + 2)!. Thus
1
(2n + 2)!
(2n)!
(2n)!
|an+1 |
1
=
=
=
,
=
1
|an |
(2n + 2)!
(2n + 2)(2n + 1)(2n)!
(2n + 2)(2n + 1)
(2n)!
so
L = lim
n→∞
Since L = 0, the ratio test tells us that
∞
X
n=1
|an+1 |
1
= lim
= 0.
n→∞ (2n + 2)(2n + 1)
|an |
1
converges.
(2n)!
16. Since an = (n!)2 /(2n)!, replacing n by n + 1 gives an+1 = ((n + 1)!)2 /(2n + 2)!. Thus,
((n + 1)!)2
(2n + 2)!
((n + 1)!)2 (2n)!
|an+1 |
=
=
·
.
2
|an |
(2n + 2)! (n!)2
(n!)
(2n)!
9.4 SOLUTIONS
However, since (n + 1)! = (n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have
(n + 1)2 (n!)2 (2n)!
(n + 1)2
n+1
|an+1 |
=
=
=
,
2
|an |
(2n + 2)(2n + 1)(2n)!(n!)
(2n + 2)(2n + 1)
4n + 2
so
L = lim
n→∞
Since L < 1, the ratio test tells us that
∞
X
(n!)2
n=1
|an+1 |
1
= .
|an |
4
converges.
(2n)!
17. Since an = n!(n + 1)!/(2n)!, replacing n by n + 1 gives an+1 = (n + 1)!(n + 2)!/(2n + 2)!. Thus,
(n + 1)!(n + 2)!
(2n)!
(n + 1)!(n + 2)!
(2n + 2)!
·
.
=
n!(n + 1)!
(2n + 2)!
n!(n + 1)!
(2n)!
|an+1 |
=
|an |
However, since (n + 2)! = (n + 2)(n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have
|an+1 |
(n + 2)(n + 1)
n+2
=
=
,
|an |
(2n + 2)(2n + 1)
2(2n + 1)
so
L = lim
n→∞
Since L < 1, the ratio test tells us that
∞
X
n!(n + 1)!
n=1
(2n)!
|an+1 |
1
= .
|an |
4
converges.
18. Since an = 1/(r n n!), replacing n by n + 1 gives an+1 = 1/(r n+1 (n + 1)!). Thus
1
r n+1 (n + 1)!
|an+1 |
r n n!
1
=
= n+1
=
,
1
|an |
r
(n + 1)!
r(n + 1)
n
r n!
so
L = lim
n→∞
Since L = 0, the ratio test tells us that
∞
X
1
n=1
r n n!
|an+1 |
1
1
=
lim
= 0.
|an |
r n→∞ n + 1
converges for all r > 0.
19. Since an = 1/(nen ), replacing n by n + 1 gives an+1 = 1/(n + 1)en+1 . Thus
1
(n + 1)en+1
|an+1 |
nen
n
1
=
=
=
.
1
|an |
(n + 1)en+1
n+1 e
n
ne
Therefore
L = lim
n→∞
Since L < 1, the ratio test tells us that
∞
X
1
n=1
nen
|an+1 |
1
= < 1.
|an |
e
converges.
20. Since an = 2n /(n3 + 1), replacing n by n + 1 gives an+1 = 2n+1 /((n + 1)3 + 1). Thus
2n+1
(n + 1)3 + 1
2n+1
n3 + 1
n3 + 1
|an+1 |
=
=
·
=2
,
2n
|an |
(n + 1)3 + 1
2n
(n + 1)3 + 1
n3 + 1
883
884
Chapter Nine /SOLUTIONS
so
L = lim
n→∞
Since L > 1 the ratio test tells us that the series
∞
X
n=0
|an+1 |
= 2.
|an |
2n
diverges.
n3 + 1
21. Even though the first term is negative, the terms alternate in sign, so it is an alternating series.
22. Since cos(nπ) = (−1)n , this is an alternating series.
23. Since (−1)n cos(nπ) = (−1)2n = 1, this is not an alternating series.
24. Since an = cos n is not always positive, this is not an alternating series.
√
√
√
√
1
1
< √ ,
25. Let an = 1/ n. Then replacing n by n+1 we have an+1 = 1/ n + 1. Since n + 1 > n, we have √
n
n+1
∞
X
(−1)n
√
converges by the alternating series test.
hence an+1 < an . In addition, limn→∞ an = 0 so
n
n=0
26. Let an = 1/(2n + 1). Then replacing n by n + 1 gives an+1 = 1/(2n + 3). Since 2n + 3 > 2n + 1, we have
0 < an+1 =
1
1
<
= an .
2n + 3
2n + 1
We also have limn→∞ an = 0. Therefore, the alternating series test tells us that the series
∞
X
(−1)n−1
n=1
2n + 1
converges.
27. Let an = 1/(n2 + 2n + 1) = 1/(n + 1)2 . Then replacing n by n + 1 gives an+1 = 1/(n + 2)2 . Since n + 2 > n + 1,
we have
1
1
<
(n + 2)2
(n + 1)2
so
0 < an+1 < an .
We also have limn→∞ an = 0. Therefore, the alternating series test tells us that the series
∞
X
(−1)n−1
n=1
n2 + 2n + 1
28. Let an = 1/en . Then replacing n by n + 1 we have an+1 = 1/en+1 . Since en+1 > en , we have
an+1 < an . In addition, limn→∞ an = 0 so
∞
X
(−1)n
n=1
29.
30.
31.
32.
33.
en
converges.
1
1
< n , hence
en+1
e
converges by the alternating series test. We can also observe
that the series is geometric with ratio x = −1/e can hence converges since |x| < 1.
X (−1)n X −1 n
X 1
X 1 n
X (−1)n
and
are convergent geometric series. Thus
Both
=
=
is abson
n
2
2
2
2
2n
lutely convergent.
X 1
X (−1)n
converges by the alternating series test. However
diverges because it is a multiple of the
The series
2n X
2n
(−1)n
is conditionally convergent.
harmonic series. Thus
2n
Since
n
lim
= 1 6= 0
n→∞ n + 1
X
n
the series
(−1)n
does not converge. It is a divergent series.
n+1
n
X (−1)
X 1
The series
converges by the alternating series test. Moreover, the series
converges by comparison
n4 + 7
n4 + 7
X (−1)n
X 1
. Thus
is absolutely convergent.
with the convergent p-series
n4
n4 + 7
P
1/(n ln n) converges by using the integral test. Since
We first check absolute convergence by deciding whether
Z
2
∞
dx
= lim
b→∞
x ln x
Z
2
b
dx
= lim ln(ln(x))
b→∞
x ln x
b
2
= lim (ln(ln(b)) − ln(ln(2))),
b→∞
9.4 SOLUTIONS
885
1
diverges.
and since this limit does not exist,
n ln n
We now check conditional convergence. The original series is alternating so we check whether an+1 < an . Consider
an = f (n), where f (x) = 1/(x ln x). Since
X
d
dx
1
x ln x
−1
1
1+
ln x
ln x
=
x2
is negative for x > 1, we know that an is decreasing for n ≥ 2. Thus, for n ≥ 2
an+1 =
Since 1/(n ln n) → 0 as n → ∞, we see that
1
1
<
= an .
(n + 1) ln(n + 1)
n ln n
X (−1)n−1
is conditionally convergent.
n ln n
34. Although cos n is sometimes positive and sometimes negative, the series is not alternating because is does not change
sign every term. For example cos 1 > 0, whereas cos 2 and cos 3 are negative. If we use the comparison test on
∞
X
cos n
,
n2
n=1
we find
cos n
1
≤ 2.
n2
n
P∞
The series n=1 1/n2 is a p-series with p = 2, and therefore converges. Thus the original series is absolutely convergent.
35. This is an alternating series that converges by the alternating series test. However
∞
X
(−1)n−1
n=1
√
n
=
∞
X
1
n=1
√ ,
n
which is a p-series with p = 1/2, so this series diverges. Thus the original series is conditionally convergent.
P
36. We first check absolute convergence by deciding whether
arcsin(1/n) converges. Since sin θ ≈ θ for small angles θ,
writing x = sin θ we see that
arcsin x ≈ x.
Since arcsin x ”behaves like“ x for small x, we expect arcsin(1/n) to ”behave like“ 1/n for large n. To confirm this we
calculate
arcsin x
θ
arcsin(1/n)
= lim
= lim
= 1.
lim
1/n
x→0
x
θ→0 sin θ
n→∞
P
P
1/n.
arcsin(1/n) diverges by limit comparison with the harmonic series
We now check conditional convergence. The original series is alternating, so we check whether an+1 < an . Consider
an = f (n), where f (x) = arcsin(1/x). Since
Thus
1
1
d
arcsin = p
dx
x
1 − (1/x)2
−
1
x2
is negative for x > 1, we know that an is decreasing for n > 1. Therefore, for n > 1
an+1 = arcsin(1/(n + 1)) < arcsin(1/n) = an .
Since arcsin(1/n) → 0 as n → ∞, we see that
X
(−1)n−1 arcsin(1/n) is conditionally convergent.
37. We first check absolute convergence by deciding whether
X arctan(1/n)
n2
tween −π/2 and π/2, we have arctan(1/n) < π/2 for all n. We compare
converges. Since arctan x is the angle be-
π/2
arctan(1/n)
< 2 ,
n2
n
and conclude that since (π/2)
absolutely convergent.
P
1/n2 converges,
X arctan(1/n)
n2
converges. Thus
X (−1)n−1 arctan(1/n)
n2
is
886
Chapter Nine /SOLUTIONS
38. We have
(5n + 1)/(3n2 )
an
5n + 1
=
=
,
bn
1/n
3n
so
lim
n→∞
Since
∞
X
1
n=1
n
is a divergent harmonic series, the original series diverges.
39. We have
((1 + n)/(3n))n
an
=
=
bn
(1/3)n
so
lim
n→∞
Since
∞
X
1 n
n=1
5n + 1
5
an
= lim
= = c 6= 0.
n→∞
bn
3n
3
3
n + 1 n
n
1
an
= lim 1 +
n→∞
bn
n
n
= 1+
1
n
n
,
= e = c 6= 0.
is a convergent geometric series, the original series converges.
40. The nth term is an = 1 − cos(1/n) and we are taking bn = 1/n2 . We have
lim
n→∞
1 − cos(1/n)
an
= lim
.
n→∞
bn
1/n2
This limit is of the indeterminate form 0/0 so we evaluate it using l’Hopital’s rule. We have
1 − cos(1/n)
sin(1/n)(−1/n2 )
1 sin(1/n)
1 sin x
1
= lim
= lim
= lim
= .
2
n→∞
1/n
n→∞
−2/n3
n→∞ 2
1/n
x→0 2 x
2
lim
The
P limit comparison test applies with c = 1/2. The p-series
(1 − cos(1/n)) also converges.
P
1/n2 converges because p = 2 > 1. Therefore
41. The nth term an = 1/(n4 − 7) behaves like 1/n4 for large n, so we take bn = 1/n4 . We have
lim
n→∞
an
1/(n4 − 7)
n4
= lim
= lim 4
= 1.
4
bn
n→∞
1/n
n→∞ n − 7
The limit comparison test applies with c = 1. The p-series
P
4
1/(n − 7) also converges.
P
1/n4 converges because p = 4 > 1. Therefore
42. The nth term an = (n + 1)/(n2 + 2) behaves like n/n2 = 1/n for large n, so we take bn = 1/n. We have
lim
n→∞
(n + 1)/(n2 + 2)
n2 + n
an
= lim
= lim 2
= 1.
bn
n→∞
1/n
n→∞ n + 2
The limit comparison test applies with c = 1. Since the harmonic series
also diverges.
P
1/n diverges, the series
P
(n + 1)/(n2 + 2)
43. The nth term an = (n3 − 2n2 + n + 1)/(n4 − 2) behaves like n3 /n4 = 1/n for large n, so we take bn = 1/n. We have
an
(n3 − 2n2 + n + 1)/(n4 − 2)
n4 − 2n3 + n2 + n
= lim
= lim
= 1.
n→∞ bn
n→∞
1/n
n→∞
n4 − 2
lim
The limit comparison test applies with c = 1. The harmonic series
also diverges.
P
1/n diverges. Thus
P
n3 − 2n2 + n + 1 / n4 − 2
44. The nth term an = 2n /(3n − 1) behaves like 2n /3n for large n, so we take bn = 2n /3n . We have
lim
n→∞
2n /(3n − 1)
3n
1
an
= lim
= lim n
= lim
= 1.
n→∞
n→∞ 3 − 1
n→∞ 1 − 3−n
bn
2n /3n
The limit comparison test applies with c = 1. The geometric series
P
n
n
2 /(3 − 1) also converges.
P
2n /3n =
P
(2/3)n converges. Therefore
9.4 SOLUTIONS
45. The nth
887
√
√
√
√
term an = 1/(2 n + n + 2) behaves like 1/(3 n) for large n, so we take bn = 1/(3 n). We have
√
√
√
1/(2 n + n + 2)
3 n
an
= lim
√
= lim √
√
lim
n→∞
n→∞ 2 n +
n→∞ bn
1/(3 n)
n+2
√
3 n
= lim √
p
n→∞
n 2 + 1 + 2/n
= lim
n→∞
3
2+
= 1.
p
1 + 2/n
=
3
√
2+ 1+0
P
√
The limit comparison testP
applies with c√= 1. The series
1/(3 n) diverges because it is a multiple of a p-series with
√
p = 1/2 < 1. Therefore
1/(2 n + n + 2) also diverges.
46. The nth term,
1
1
1
−
=
,
2n − 1
2n
4n2 − 2n
behaves like 1/(4n2 ) for large n, so we take bn = 1/(4n2 ). We have
an =
lim
n→∞
an
1/(4n2 − 2n)
4n2
1
= lim
=
lim
= lim
= 1.
n→∞
n→∞ 4n2 − 2n
n→∞ 1 − 1/(2n)
bn
1/(4n2 )
The limit comparison test applies with c = 1. The series
P 1
1
also converges.
p = 2 > 1. Therefore
− 2n
2n−1
47. The nth term,
P
1/(4n2 ) converges because it is a multiple of a p-series with
n
,
cos n + en
n
n
behaves like n/e for large n, so we take bn = n/e . We have
an =
lim
n→∞
n/ (cos n + en )
en
1
1
an
= lim
= lim
= lim
=
= 1.
n
n→∞
n→∞ cos n + en
n→∞ cos n/en + 1
bn
n/e
0+1
The limit comparison test applies with c = 1. The series
lim
n→∞
which is less than 1. Therefore
48. The nth term,
P
n/en converges by the ratio test because
(n + 1)/en+1
|bn+1 |
n+1 1
1
= lim
= lim
· = ,
n→∞
n→∞
|bn |
n/en
n
e
e
X
n
cos n + en
also converges.
4 sin n + n
,
n2
behaves like 1/n for large n, so we take bn = 1/n. We have
an =
lim
n→∞
(4 sin n + n) /n2
4 sin n + n
4 sin n/n + 1
0+1
an
= lim
= lim
= lim
=
= 1.
n→∞
n→∞
n→∞
bn
1/n
n
1
1
The limit comparison test applies with c = 1. The series
X 4 sin n + n
also diverges.
n2
P
1/n diverges because it is a p-series with p ≤ 1. Therefore
Problems
49. The comparison test requires that an = (−1)n /n2 be positive. It is not.
50. The comparison test requires that an = sin n be positive for all n. It is not.
51. With an = (−1)n , we have |an+1 /an | = 1, and limn→∞ |an+1 /an | = 1, so the test gives no information.
52. With an = sin n, we have |an+1 /an | = | sin(n + 1)/ sin n|, which does not have a limit as n → ∞, so the test does not
apply.
53. The sequence an = n does not satisfy either an+1 < an or limn→∞ an = 0.
888
Chapter Nine /SOLUTIONS
54. The alternating series test requires an = sin n be positive, which it is not. This is not an alternating series.
55. The alternating series test requires an = 2 − 1/n which is positive and satisfies an+1 < an but limn→∞ an = 2. Since
limn→∞ an 6= 0, we cannot use the alternating series test.
56. We cannot use the alternating seriesP
test in this case because the absolute values of the terms are not decreasing; that is,
we do not have a series of the form (−1)n−1 an where an+1 < an for every n. For example, if we look at the fourth
and fifth terms of the series, we notice that 21 < 23 .
In fact if the terms of this series are combined in pairs, we have the harmonic series, which is divergent:
(2 − 1)/1 + (2 − 1)/2 + (2 − 1)/3 + · · · = 1 + 1/2 + 1/3 + · · · .
57. The partial sums are S1 = 1, S2 = −1, S3 = 2, S10 = −5, S11 = 6, S100 = −50, S101 = 51, S1000 = −500,
S1001 = 501, which appear to be oscillating further and further from 0. This series does not converge.
58. The partial sums look like: S1 = 1, S2 = 0.9, S3 = 0.91, S4 = 0.909, S5 = 0.9091, S6 = 0.90909. The series appears
to be converging to 0.909090 . . . or 10/11.
Since an = 10−k is positive and decreasing and lim 10−n = 0, the alternating series test confirms the convergence
n→∞
of the series.
59. The partial sums look like: S1 = 1, S2 = 0, S3 = 0.5, S4 = 0.3333, S5 = 0.375, S10 = 0.3679, S20 = 0.3679, and
higher partial sums agree with these first 4 decimal places. The series appears to be converging to about 0.3679.
Since an = 1/n! is positive and decreasing and limn→∞ 1/n! = 0, the alternating series test confirms the convergence of this series.
60. We use the ratio test with an =
8n+1
8n
. Replacing n by n + 1 gives an+1 =
and
n!
(n + 1)!
|an+1 |
8n+1 /(n + 1)!
8n!
8
=
=
=
.
|an |
8n /n!
(n + 1)!
n+1
Thus
L = lim
n→∞
Since L < 1, the ratio test tells us that
∞
X
8n
n=1
61. We use the ratio test with an =
n!
8
|an+1 |
= lim
= 0.
|an |
n→∞ n + 1
converges.
(n + 1)2n+1
n2n
. Replacing n by n + 1 gives an+1 =
and
3n
3n+1
|an+1 |
((n + 1)2n+1 )/3n+1
2(n + 1)
=
=
.
|an |
n2n /3n
3n
Thus
L = lim
n→∞
Since L < 1, the ratio test tells us that
|an+1 |
2(n + 1)
2(1 + 1/n)
2
= lim
= lim
= .
|an |
n→∞
3n
n→∞
3
3
∞
X
n2n
n=1
62. We use the ratio test and calculate
lim
n→∞
3n
converges.
|an+1 |
(0.1)n+1 /(n + 1)!
0.1
= lim
= lim
= 0.
n→∞
n→∞ n + 1
|an |
(0.1)n /n!
Since the limit is less than 1, the series converges.
63. We use the ratio test and calculate
n!/(n + 1)2
|an+1 |
= lim
= lim
lim
n→∞ (n − 1)!/n2
n→∞
n→∞ |an |
Since the limit does not exist (it is ∞), the series diverges.
n2
n!
·
(n − 1)! (n + 1)2
= lim
n→∞
n2
n·
(n + 1)2
.
889
9.4 SOLUTIONS
64. The first few terms of the series may be written
1 + e−1 + e−2 + e−3 + · · · ;
this is a geometric series with a = 1 and x = e−1 = 1/e. Since |x| < 1, the geometric series converges to
1
e
1
=
=
.
S=
1−x
1 − e−1
e−1
65. The first few terms of the series may be written
e + e2 + e3 + · · · = e + e · e + e · e2 + · · · ;
this is a geometric series with a = e and x = e. Since |x| > 1, this geometric series diverges.
(2n + 2)!
(2n)!
. Replacing n by n + 1 gives an+1 =
and
66. We use the ratio test with an =
(n!)2
((n + 1)!)2
Thus
|an+1 |
((2n + 2)!)/((n + 1)!)2
(2n + 2)!(n!)2
(2n + 2)(2n + 1)(2n)!(n!)2
2(2n + 1)
=
=
=
=
.
2
2
|an |
(2n)!/(n!)
(2n)!((n + 1)!)
(2n)!(n + 1)2 (n!)2
n+1
L = lim
n→∞
Since L > 1, the ratio test tells us that
2(2n + 1)
2(2 + 1/n)
|an+1 |
= lim
= lim
= 4.
n→∞
n→∞ 1 + 1/n
|an |
n+1
∞
X
(2n)!
diverges.
(n!)2
n=1
67. We compare the series with the convergent series
so tan(1/n) < 2 for all n. Thus
P
1/n2 . From the graph of tan x, we see that tan x < 2 for 0 ≤ x ≤ 1,
1
1
1
tan
< 2 2,
2
n
n
n
P
1/n2 converges. Alternatively, we try the integral test. Since the terms in the series
so the series converges, since 2
are positive and decreasing, we can use the integral test. We calculate the corresponding integral using the substitution
w = 1/x:
Z
1
∞
1
1
tan
x2
x
dx = lim
b→∞
Z
1
b
1
1
tan
x2
x
b→∞
∞
Since the limit exists, the integral converges, so the series
X 1
n=1
68. We use the limit comparison test with an =
We have
dx = lim ln cos
n2
1
x
b
= lim
1
b→∞
ln cos
1
b
− ln(cos 1) = − ln(cos 1).
tan (1/n) converges.
n
1
n+1
. Because an behaves like 3 = 2 as n → ∞, we take bn = 1/n2 .
n3 + 6
n
n
lim
n→∞
By the limit comparison test (with c = 1) since
n2 (n + 1)
an
= lim
= 1.
n→∞
bn
n3 + 6
∞
X
1
n=1
69. We use the limit comparison test with an =
bn = 1/n.
We have
lim
n→∞
n2
converges,
∞
X
n+1
n=1
also converges.
n3 + 6
5n + 2
5n
5
. Because an behaves like
=
as n → ∞, we take
2n2 + 3n + 7
2n2
2n
n(5n + 2)
5
an
= lim
= .
bn
n→∞ 2n2 + 3n + 7
2
By the limit comparison test (with c = 5/2) since
∞
X
1
n=1
n
diverges,
∞
X
n=1
5n + 2
also diverges.
2n2 + 3n + 7
p
√
70. Let an = 1/ 3n − 1. Then replacing n by n + 1 gives an+1 = 1/ 3(n + 1) − 1. Since
p
√
3(n + 1) − 1 > 3n − 1,
we have
an+1 < an .
In addition, limn→∞ an = 0 so the alternating series test tells us that the series
∞
X
(−1)n−1
n=1
√
3n − 1
converges.
890
Chapter Nine /SOLUTIONS
71. Since the exponential, 2n , grows faster than the power, n2 , the terms are growing in size. Thus, lim an 6= 0. We conclude
n→∞
that this series diverges.
2
2
2
72. P
Since 0 ≤ | sin n| ≤ 1 for all n, we may be able to compare with 1/n
P . We2 have 0 ≤ | sin
Pn/n | ≤ 21/n for all n. So
2
| sin n/n | converges by comparison with the convergent series (1/n ). Therefore (sin n/n ) also converges,
since absolute convergence implies convergence by Theorem 9.6.
73. We take an =
sin n2
sin n2
1
. Since | sin n2 | ≤ 1 for all n,
≤ 2.
2
n
n2
n
We take bn =
∞
X
sin n2
n=1
n2
∞
∞
∞
n=1
n=1
n=1
X
X 1
X sin n2
1
and
bn =
converges. By the comparison test,
also converges. Therefore,
2
2
n
n
n2
converges by Theorem 9.6.
74. Note that cos(nπ)/n
= (−1)n /n, so this is an alternating series. Therefore, since 1/(n + 1) < 1/n and limn→∞ 1/n =
P
0, we see that (cos(nπ)/n) converges by the alternating series test.
75. As n → ∞, we see that
n
1
n+2
→ 2 = .
n2 − 1
n
n
P
Since (1/n) diverges, we expect our series to have the same behavior.
More precisely, for all n ≥ 2, we have
0≤
so
∞
X
n+2
n=2
n2
−1
n
n+2
1
= 2 ≤ 2
,
n
n
n −1
diverges by comparison with the divergent series
76. Since
X1
n
.
3
3
=
,
ln n2
2 ln n
P
1/ ln n. More precisely, for all n ≥ 2, we have
our series behaves like the series
0≤
so
∞
X
n=2
1
1
3
3
≤
≤
=
,
n
ln n
2 ln n
ln n2
X1
3
diverges by comparison with the divergent series
.
2
ln n
n
p
77. Let an = 1/
n2 (n + 2). Since n2 (n + 2) = n3 + 2n2 > n3 , we have
Since the p-series
0 < an <
∞
X
1
n=1
n3/2
1
.
n3/2
converges, the comparison test tells us that
∞
X
n=1
1
p
n2 (n
+ 2)
also converges.
√
78. Let an = n(n + 1)/ n3 + 2n2 . Since n3 + 2n2 = n2 (n + 2), we have
an =
n+1
n(n + 1)
√
= √
n+2
n n+2
so an grows without bound as n → ∞, therefore the series
∞
X
n(n + 1)
n=1
√
n3 + 2n2
diverges.
9.4 SOLUTIONS
∞
79. Factoring gives
X 2 a
n=1
a ≤ 1.
n
891
∞
= 2a
X 1
n=1
na
. This is a constant times a p-series that converges if a > 1 and diverges if
80. This is a geometric series for all a > 0, with ratio 2/a. Therefore, the series converges when 2/a < 1 and diverges when
∞
X
2 n
2/a ≥ 1. Thus
converges for a > 2 and diverges for a ≤ 2.
a
n=1
81. This is a geometric series with ratio ln a, so it converges when | ln a| < 1 and diverges when | ln a| ≥ 1. We have
| ln a| < 1 if 1/e < a < e and | ln a| ≥ 1 if a ≥ or a ≤ 1/e. Thus, the series
and diverges for a ≥ e and 0 < a ≤ 1/e.
∞
X
(ln a)n converges for 1/e < a < e
n=1
82. For a > 0, the terms of the series are positive and eventually decreasing. We use the integral test and calculate the
corresponding improper integral:
Z b
Z ∞
ln x
ln x
dx
=
lim
dx.
b→∞
xa
xa
1
1
For a 6= 1, use integration by parts with u = ln x and v ′ = x−a :
x−a+1
ln x
dx
=
ln x −
xa
−a + 1
Z
Z
x−a+1
x−a+1
x−a+1
x−a
1
dx =
ln x −
=
ln x −
.
−a + 1
−a + 1
(−a + 1)2
−a + 1
−a + 1
Thus,
lim
b→∞
Z
1
b
b−a+1
1
ln x
ln b −
dx = lim
a
x
b→∞ −a + 1
−a + 1
+
1
1
ln b − 1/(−a + 1)
1
=
lim
+
.
(−a + 1)2
−a + 1 b→∞
ba−1
(−a + 1)2
Use l’Hopital’s Rule to obtain
lim
b→∞
ln b − 1/(−a + 1)
ba−1
= lim
b→∞
1/b
(a − 1)ba−2
= lim
b→∞
1
.
(a − 1)ba−1
This limit exists for a − 1 > 0 and does not exist for a − 1 < 0. Thus the series converges for a > 1 and diverges for
0 < a < 1.
Z ∞
∞
b
X
ln n
ln x
(ln x)2
diverges because
dx = lim
, and this limit diverges. For a ≤ 0,
For a = 1, the series
b→∞
n
x
2
1
1
n=1
∞
X ln n
ln n
does
not
exist,
so
the
series
diverges
by
Property
3
of
Theorem
9.2.
Thus
converges for a > 1 and
lim
n→∞ na
n
n=1
diverges for a ≤ 1.
83. To use the alternating series test, consider an = f (n), where f (x) = arctan(a/x). We need to show that f (x) is
decreasing. Since
1
a
f ′ (x) =
− 2 ,
2
1 + (a/x)
x
we have f ′ (x) < 0 for a > 0, so f (x) is decreasing for all x. Thus an+1 < an for all n, and as lim arctan(a/n) = 0
n→∞
for all a, by the alternating series test,
∞
X
(−1)n arctan(a/n)
n=1
converges.
84. The nth partial sum of the series is given by
Sn = 1 −
(−1)n−1
1
1
+ − ··· +
,
2
3
n
so the absolute value of the first term omitted is 1/(n + 1). By Theorem 9.9, we know that the value, S, of the sum differs
from Sn by less than 1/(n+1). Thus, we want to choose n large enough so that 1/(n+1) ≤ 0.01. Solving this inequality
for n yields n ≥ 99, so we take 99 or more terms in our partial sum.
892
Chapter Nine /SOLUTIONS
85. The nth partial sum of the series is given by
Sn = 1 −
4
2
+ − · · · + (−1)(n−1)
3
9
2 (n−1)
,
3
so the absolute value of the first term omitted is (2/3)n . By Theorem 9.9, we know that the value, S, of the sum differs
from Sn by less than (2/3)n . Thus, we want to choose n large enough so that (2/3)n ≤ 0.01. Solving this inequality
for n yields n ≥ 11.358, so taking 12 or more terms in our partial sum is guaranteed to be within 0.01 of the sum of the
series.
Note: Since this is a geometric series, we know the exact sum to be 1/(1 + 2/3) = 0.6. The partial sum S12 is 0.595,
which is indeed within 0.01 of the sum of the series. Note, however, that S11 = 0.6069, which is also within 0.01 of the
exact sum of the series. Theorem 9.9 gives us a value of n for which Sn is guaranteed to be within a small tolerance of
the sum of an alternating series, but not necessarily the smallest such value.
86. The nth partial sum of the series is given by
Sn =
(−1)n−1
1
1
1
−
+
− ··· +
,
2
24
720
(2n)!
so the absolute value of the first term omitted is 1/(2n + 2)!. By Theorem 9.9, we know that the value, S, of the sum
differs from Sn by less than 1/(2n+2)!. Thus, we want to choose n large enough so that 1/(2n+2)! ≤ 0.01. Substituting
n = 2 into the expression 1/(2n + 2)! yields 1/720 which is less than 0.01. We therefore take 2 or more terms in our
partial sum.
cn converges by the Comparison
87. Since 0 ≤ cn ≤ 2−n for all n, and since P 2−n is a convergent geometric series,
P
series,
an diverges by the Comparison Test.
Test. Similarly, since 2n ≤ an , and since
2n is a divergent geometric
P
P
bn and
dn converge.
We do not have enough information to determine whether or not
P
P
88. (a) The sum
P
a n · bn =
P
1/n5 , which converges, as a p-series with p = 5, or by the integral test:
Z
∞
1
x−4
1
dx
=
lim
b→∞ (−4)
x5
b
1
1
b−4
+ = .
b→∞ (−4)
4
4
= lim
1
P
Since this improper integral converges,
an · bn also converges.
(b) This is an alternating
series
that
satisfies
the
conditions of the alternating series test: the terms are decreasing and
P
√
have limit 0, so (−1)n /P n converges.
(c) We have an bn = 1/n, so
an bn is the harmonic series, which diverges.
89. Since lim an /bn = 0, for large enough n we have |an /bn | < 1/2 and thus 0 ≤ |an | < bn /2 < bn . By the
n→∞
comparison test applied to
thus it converges.
P
|an | and
P
bn , the series
P
|an | converges. The series
P
an converges absolutely and
90. Since lim an /bn = ∞, for large enough n we have an /bn > 1 and thus an > bn . By the comparison test applied to
n→∞
P
an and
P
P
bn , the series
an diverges.
P
91. Each term in
bn is greater than or equal to a1 times a term in the harmonic series:
b1 = a 1 · 1
a1 + a2
1
b2 =
> a1 ·
2
2
1
a1 + a2 + a3
> a1 ·
b3 =
3
3
..
.
1
a1 + a2 + · · · + an
> a1 ·
bn =
n
n
Adding these inequalities gives
Since the
P harmonic series
bn diverges.
series
P
X
bn > a 1
X1
.
n
1/n diverges, a1 times the harmonic series also diverges. Then, by the comparison test, the
9.4 SOLUTIONS
92. Suppose we let cn = (−1)n an . (We have just given the terms of the series
|cn | = |(−1)n an | = |an |.
Thus
P
893
(−1)n an a new name.) Then
P
|cn | converges, and by Theorem 9.6,
X
cn =
X
(−1)n an
converges.
93. (a) Since
|an | = an
if an ≥ 0
|an | = −an if an < 0,
we have
an + |an | = 2|an | if an ≥ 0
an + |an | = 0
Thus, for all n,
(b) If
P
|an | converges, then
P
if an < 0.
0 ≤ an + |an | ≤ 2|an |.
2|an | is convergent, so, by comparison,
X
((an + |an |) − |an |) =
is convergent, as it is the difference of two convergent series.
94. The limit
lim
n→∞
√
n
an = lim
n→∞
P
X
(an + |an |) is convergent. Then
an
2
= 0 < 1,
n
so the series converges.
95. The limit
lim
n→∞
√
n
an = lim
n→∞
5n + 1
= 0 < 1,
3n2
so the series converges.
Strengthen Your Understanding
2n
96. The series
Pis not2alternating since (−1) = 1 for all n > 0, so we cannot use the alternating series test. The series is the
same as
1/n which converges by the p-test with p = 2.
97. We could show the series converges by comparison with the convergent series
lim
n→∞
P
1/n2 . The ratio test gives
1 + 1/n2
(1/((n + 1)2 + 1)
n2 + 1
1+0
= lim
= lim
=
= 1.
n→∞ (n + 1)2 + 1
n→∞ ((n + 1)/n)2 + 1/n2
1/(n2 + 1)
1+0
The ratio test is inconclusive when the limit of the ratios is 1.
98. It is true that the comparison series
1/n2 converges, but in order to make the comparison, we need 1/n3/2 ≤ 1/n2
3/2
for all n; we have instead 1/n
> 1/n2 for all n > 1.
P∞
P
P∞
99. If we want the series n=1 an and n=1 |an | to behave differently, then we need an to be negative for some values of
n. Thus we try an
series.
P∞
Palternating
∞
The series n=1 (−1)n /n works. Because this series is of the form n=1 (−1)n bn where bn > 0 for all n and
bn+1 < bn for all n and also
converges by the alternating series test.
P∞bn → 0, thePseries
∞
However, the series n=1 |an | = n=1 1/n diverges by the p-series test.
100. The following series is alternating:
X
an =
X
(−1)n n,
but since the terms do not tend to 0 as n → ∞, the series does not converge.
101. The geometric series with an = 3n satisfies this condition.
102. False. The sequence −1, 1, −1, 1, . . . given by sn = (−1)n alternates in sign but does not converge.
894
Chapter Nine /SOLUTIONS
2
103. False. It does not tell us anything
Pto know that bn is larger
P than a convergent series. For example, if an = 1/n and
bn = 1, then 0 ≤ an ≤ bn and
an converges, but
bn diverges. Since this statement is not true for all an and bn ,
the statement is false.
104. True. This is one of the statements of the comparison test.
105. True. Consider the series
P
(−bn ) and
P
(−an ). The series
P
(−bn ) converges, since
0 ≤ −an ≤ −bn .
P
bn converges, and
P
P
an converges.
By the comparison test, (−an ) converges, so
P
P
P
106. False. It is true that P
if
an converges. However, knowing that
an converges
|an | converges, then we know that
does not tell us that
|an | converges.
P
P
n−1
For example, if an = (−1)
harmonic series which diverges.
/n, then
an converges by the alternating series test. However,
|an | is the
107. False. For example, if an = 1/n2 , then
lim
n→∞
However,
P
2
1/(n + 1)2
|an+1 |
n2
= lim
= lim
= 1.
2
n→∞
n→∞ (n + 1)2
|an |
1/n
1/n converges.
108. True, since if we write out the terms of the series, using the fact that cos(2πn) = 1 for all n, we have
(−1)0 cos 0 + (−1)1 cos(2π) + (−1)2 cos(4π) + (−1)3 cos(6π) + · · ·
= 1 · 1 − 1 · 1 + 1 · 1 − 1 · 1 + ···
= 1 − 1 + 1 − 1 +···.
This is an alternating series.
109. False. This is an alternating series, but since the terms do not go to zero, it does not converge.
n−1
110. False. For example, if an = (−1)
then
an converges by the alternating series test. But (−1)n an = (−1)n (−1)n−1 /n =
P /n,
n
2n−1
(−1)
/n = −1/n. Thus, (−1) an is the negative of the harmonic series and does not converge.
P
111. This is true. It is a restatement of Theorem 9.9.
112. This statement is false. The statement is true if the series converges by the alternating series test, but not in general.
Consider, for example, the alternating series
S = 10 − 0.01 + 0.8 − 0.7 − 0 + 0 − 0 + · · · .
Since the later terms are all 0, we can find the sum exactly:
S = 10.69.
If we approximated the sum by the first term, S1 = 10, the magnitude of the first term omitted would be 0.01. Thus, if
the statement in this problem were true, we would say that the true value of the sum lay between 10 + 0.01 = 10.01 and
10 − 0.01 = 9.99 which it does not.
113. True. Let cn = (−1)n |an |. Then |cn | = |an | so
P
|cn | converges, and therefore
P
cn =
(−1)n |an | converges.
P
114. True. Since the series is alternating, Theorem 9.9 gives the error bound. Summing the first 100 terms gives S100 , and if
the true sum is S,
1
|S − S100 | < a101 =
< 0.01.
101
115. True. If
P
|an | is convergent, then so is
116. False. The alternating harmonic series
Series test, but the harmonic series
convergent.
117. True. By the comparison
P test, if
diverges, then so does
0.5bn .
P
an .
X (−1)n
n
X (−1)n
P
n
=
is conditionally convergent because it converges by the Alternating
X1
n
is divergent. The alternating harmonic series is not absolutely
an is larger term-by-term than a divergent series, then
118. (b). This series should be compared with
∞
X
1
k=1
k3
.
P
an diverges. If
P
bn
9.5 SOLUTIONS
895
Solutions for Section 9.5
Exercises
1. Yes.
2. No, because it contains negative powers of x.
3. No, each term is a power of a different quantity.
4. Yes. It’s a polynomial, or a series with all coefficients beyond the 7th being zero.
1 · 3 · 5 · · · (2n − 1) n
x for n ≥ 1. Other answers are possible.
5. The general term can be written as
2n · n!
p(p − 1)(p − 2) · · · (p − n + 1) n
6. The general term can be written as
x for n ≥ 1. Other answers are possible.
n!
(−1)k (x − 1)2k
for k ≥ 0. Other answers are possible.
7. The general term can be written as
(2k)!
8. The general term can be written as
are possible.
(−1)k+1 (x − 1)2k+1
(−1)k (x − 1)2k+3
for k ≥ 1 or as
for k ≥ 0. Other answers
(2(k − 1))!
(2k)!
(x − a)n
for n ≥ 1. Other answers are possible.
2n−1 · n!
(k + 1)(x + 5)2k+1
(k + 2)(x + 5)2k+3
10. The general term can be written as
for k ≥ 1 or as
for k ≥ 0. Other answers
(k − 1)!
k!
are possible.
9. The general term can be written as
11. Since Cn = n, replacing n by n + 1 gives Cn+1 = n + 1. Using the ratio test with an = nxn , we have
lim
n→∞
|Cn+1 |
|an+1 |
n+1
= |x| lim
= |x| lim
= |x|.
n→∞ |Cn |
n→∞
|an |
n
Thus the radius of convergence is R = 1.
12. This series may be written as
1 + 5x + 25x2 + · · ·
so Cn = 5n . Using the ratio test, with an = 5n xn , we have
lim
n→∞
|Cn+1 |
5n+1
|an+1 |
= |x| lim
= |x| lim
= 5|x|.
n→∞ |Cn |
n→∞ 5n
|an |
Thus the radius of convergence is R = 1/5.
13. Since Cn = n3 , replacing n by n + 1 gives Cn+1 = (n + 1)3 . Using the ratio test, with an = n3 xn , we have
|Cn+1 |
(n + 1)3
n+1
|an+1 |
= |x|
= |x|
= |x|
|an |
|Cn |
n3
n
We have
lim
n→∞
Thus the radius of convergence is R = 1.
3
.
|an+1 |
= |x|.
|an |
14. Let Cn = 2n + n2 . Then replacing n by n + 1 gives Cn+1 = 2n+1 + (n + 1)2 . Using the ratio test, we have
|Cn+1 |
2n+1 + (n + 1)2
|an+1 |
= |x|
= |x|
= 2|x|
|an |
|Cn |
2n + n2
Since 2n dominates n2 as n → ∞, we have
lim
n→∞
Thus the radius of convergence is R = 12 .
|an+1 |
= 2|x|.
|an |
2n + 21 (n + 1)2
2n + n2
.
896
Chapter Nine /SOLUTIONS
15. Since Cn = (n + 1)/(2n + n), replacing n by n + 1 gives Cn+1 = (n + 2)/(2n+1 + n + 1). Using the ratio test, we
have
|Cn+1 |
(n + 2)/(2n+1 + n + 1)
|an+1 |
n+2
2n + n
n+2
2n + n
= |x|
= |x|
=
|x|
·
=
|x|
·
.
|an |
|Cn |
(n + 1)/(2n + n)
2n+1 + n + 1 n + 1
n + 1 2n+1 + n + 1
Since
lim
n→∞
and
lim
n→∞
2n + n
n+1
2
+n+1
because 2n dominates n as n → ∞, we have
=
lim
n→∞
Thus the radius of convergence is R = 2.
n+2
=1
n+1
1
lim
2 n→∞
2n + n
n
2 + (n + 1)/2
=
1
,
2
|an+1 |
1
= |x|.
|an |
2
16. Since Cn = 2n /n, replacing n by n + 1 gives Cn+1 = 2n+1 /(n + 1). Using the ratio test, we have
so
|Cn+1 |
2n+1 /(n + 1)
2n+1
n
|an+1 |
n
= |x − 1|
= |x − 1|
=
|x
−
1|
·
= 2|x − 1|
,
|an |
|Cn |
2n /n
(n + 1) 2n
n+1
lim
n→∞
Thus the radius of convergence is R = 21 .
17. We use the ratio test:
Therefore, we have
|an+1 |
= 2|x − 1|.
|an |
(x − 3)n+1
an+1
n2n
n
=
·
= |x − 3|
.
n+1
an
(n + 1)2
(x − 3)n
2(n + 1)
|x − 3|
n
an+1
=
.
= |x − 3| lim
n→∞ 2(n + 1
an
2
Thus by the ratio test, the series converges if |x − 3|/2 < 1, that is |x − 3| < 2. The radius of convergence is R = 2.
lim
n→∞
18. We use the ratio test:
(−1)n+1 x2n+2
(2n)!
1
an+1
=
= |x2 |
·
.
an
(2n + 2)!
(−1)n x2n
(2n + 2)(2n + 1)
Therefore we have
an+1
1
= 0.
= x2 lim
n→∞ (2n + 2)(2n + 1)
an
Thus, by the ratio test, the series converges for all x, so the radius of convergence is R = ∞.
lim
n→∞
19. The coefficient of the nth term is Cn = (−1)n+1 /n2 . Now consider the ratio
n2 xn+1
an+1
=
→ |x| as
an
(n + 1)2 xn
n → ∞.
Thus, the radius of convergence is R = 1.
20. Here the coefficient of the nth term is Cn = (2n /n!). Now we have
(2n+1 /(n + 1)!)xn+1
2|x|
an+1
=
=
→ 0 as n → ∞.
an
(2n /n!)xn
n+1
Thus, the radius of convergence is R = ∞, and the series converges for all x.
21. Here Cn = (2n)!/(n!)2 . We have:
(2(n + 1))!/((n + 1)!)2 xn+1
(2(n + 1))!
(n!)2
an+1
=
=
·
|x|
an
(2n)!/(n!)2 xn
(2n)!
((n + 1)!)2
=
Thus, the radius of convergence is R = 1/4.
(2n + 2)(2n + 1)|x|
→ 4|x| as n → ∞.
(n + 1)2
9.5 SOLUTIONS
897
22. Here the coefficient of the nth term is Cn = (2n + 1)/n. Applying the ratio test, we consider:
n
an+1
((2n + 3)/(n + 1))xn+1
2n + 3
·
→ |x| as n → ∞.
=
= |x|
an
((2n + 1)/n)xn
2n + 1 n + 1
Thus, the radius of convergence is R = 1.
23. We write the series as
x−
so
x3
x5
x7
x2n−1
+
−
+ · · · + (−1)n−1
+···,
3
5
7
2n − 1
an = (−1)n−1
Replacing n by n + 1, we have
an+1 = (−1)n+1−1
Thus
so
x2n−1
.
2n − 1
x2(n+1)−1
x2n+1
= (−1)n
.
2(n + 1) − 1
2n + 1
(−1)n x2n+1
|an+1 |
2n − 1
2n − 1 2
=
·
=
x ,
|an |
2n + 1
(−1)n−1 x2n−1
2n + 1
|an+1 |
2n − 1 2
= lim
x = x2 .
n→∞ 2n + 1
|an |
By the ratio test, this series converges if L < 1, that is, if x2 < 1, so R = 1.
L = lim
n→∞
24. As seen in Example 7, we can write the general term of the series as
an = (−1)n−1
so that, replacing n by n + 1, we have
an+1 = (−1)(n+1)−1
Thus,
x2n−1
,
(2n − 1)!
x2(n+1)−1
x2n+1
= (−1)n
.
(2 (n + 1) − 1)!
(2n + 1)!
2n+1
x
(−1)n (2n+1)!
(−1)n x2n+1 (2n − 1)!
(−1) x2
|an+1 |
x2
=
=
=
=
.
n−1
2n−1
|an |
(−1)
x
(2n + 1)!
(2n + 1) 2n
(2n + 1) 2n
x2n−1
(−1)n−1 (2n−1)!
Because
x2
|an+1 |
= lim
= 0,
n→∞ (2n + 1) 2n
n→∞ |an |
we have limn→∞ |an+1 |/|an | = 0 < 1 for all x. Thus, the ratio test guarantees that the power series converges for every
real number x. The radius of convergence is infinite and the interval of convergence is all x.
lim
Problems
25. (a) The general term of the series is xn /n if n is odd and −xn /n if n is even, so Cn = (−1)n−1 /n, and we can use the
ratio test. We have
|(−1)n /(n + 1)|
|an+1 |
n
= |x| lim
= |x| lim
= |x|.
lim
n→∞
n→∞ n + 1
n→∞ |an |
|(−1)n−1 /n|
Therefore the radius of convergence is R = 1. This tells us that the power series converges for |x| < 1 and does not
converge for |x| > 1. Notice that the radius of convergence does not tell us what happens at the endpoints, x = ±1.
(b) The endpoints of the interval of convergence are x = ±1. At x = 1, we have the series
(−1)n−1
1
1
1
+ − + ··· +
+ ···
2
3
4
n
This is an alternating series with an = 1/n, so by the alternating series test, it converges. At x = −1, we have the
series
1
1
1
1
−1 − − − − · · · − − · · ·
2
3
4
n
This is the negative of the harmonic series, so it does not converge. Therefore the right endpoint is included, and the
left endpoint is not included in the interval of convergence, which is −1 < x ≤ 1.
1−
898
Chapter Nine /SOLUTIONS
26. Let Cn = 2n /n. Then replacing n by n + 1 gives Cn+1 = 2n+1 /(n + 1). Using the ratio test, we have
n
|Cn+1 |
2n+1 /(n + 1)
2n+1 n
|an+1 |
= |x|
= |x|
= |x|
·
= 2|x|
.
n
|an |
|Cn |
2 /n
n + 1 2n
n+1
Thus
lim
n→∞
The radius of convergence is R = 1/2.
|an+1 |
= 2|x|.
|an |
For x = 1/2 the series becomes the harmonic series
∞
X
1
n
n=1
For x = −1/2 the series becomes the alternating series
27. We use the ratio test:
which diverges.
∞
X
(−1)n
n
n=1
which converges. See Example 8 on page 517.
|x|
xn+1 3n
an+1
.
= n+1 · n =
an
3
x
3
Since |x|/3 < 1 when |x| < 3, the radius of convergence is 3 and the series converges for −3 < x < 3.
We check the endpoints:
x=3:
∞
X
xn
n=0
∞
x = −3 :
3n
X xn
n=0
3n
∞
X
3n
=
3n
n=0
∞
=
∞
X
3n
n=0
which diverges.
n=0
X (−3)n
=
1n
=
∞
X
(−1)n
which diverges.
n=0
The series diverges at both the endpoints, so the interval of convergence is −3 < x < 3.
28. We use the ratio test:
n
(x − 3)n+1
n
an+1
=
=
·
· |x − 3|.
an
n+1
(x − 3)n
n+1
Since n/(n + 1) → 1 as n → ∞, we have
an+1
= |x − 3|.
an
lim
n→∞
The series converges for |x − 3| < 1. The radius of convergence is 1 and the series converges for 2 < x < 4.
We check the endpoints. For x = 2, we have
∞
X
(x − 3)n
n=2
n
=
∞
X
(2 − 3)n
n=2
n
=
∞
X
(−1)n
n
n=2
.
This is the alternating harmonic series and converges. For x = 4, we have
∞
X
(x − 3)n
n=2
n
=
∞
X
(4 − 3)n
n=2
n
=
∞
X
1
n=2
n
.
This is the harmonic series and diverges. The series converges at x = 2 and diverges at x = 4. Therefore, the interval of
convergence is 2 ≤ x < 4.
29. We use the ratio test:
22n
(n + 1)2 x2(n+1)
an+1
=
· 2 2n =
2(n+1)
an
n x
2
Since (n + 1)/n → 1 as n → ∞, we have
lim
n→∞
2
n + 1 2 x2
n
·
4
.
x2
an+1
=
.
an
4
We have x /4 < 1 when |x| < 2. The radius of convergence is 2 and the series converges for −2 < x < 2.
9.5 SOLUTIONS
899
We check the endpoints. For x = −2, we have
∞
X
n2 x2n
22n
n=1
∞
X
n2 (−2)2n
=
n=1
=
22n
∞
X
n2 ,
n=1
which diverges. Similarly, for x = 2, we have
∞
X
n2 x2n
n=1
22n
=
∞
X
n2 22n
n=1
22n
=
∞
X
n2 ,
n=1
which diverges. The series diverges at both endpoints, so the interval of convergence is −2 < x < 2.
30. We use the ratio test:
(−1)n+1 (x − 5)n+1
an+1
2n n2
=
=
·
n+1
2
n
an
2
(n + 1)
(−1) (x − 5)n
n
n+1
2 |x − 5|
·
2
.
Since n/(n + 1) → 1 as n → ∞, we have
lim
n→∞
|x − 5|
an+1
=
.
an
2
We have |x − 5|/2 < 1 when |x − 5| < 2. The radius of convergence is 2 and the series converges for 3 < x < 7.
We check the endpoints. For x = 3, we have
∞
X
(−1)n (x − 5)n
n=1
2n n2
=
∞
X
(−1)n (3 − 5)n
2n n2
n=1
=
∞
X
(−1)n (−2)n
n=1
2n n2
=
∞
X
1
n=1
n2
.
This is a p-series with p = 2 and it converges. For x = 7, we have
∞
X
(−1)n (x − 5)n
n=1
2n n2
=
∞
X
(−1)n (7 − 5)n
2n n2
n=1
=
∞
X
(−1)n 2n
n=1
2n n2
=
∞
X
(−1)n
n=1
n2
.
X 1
X (−1)n
Since
converges, the alternating series
also converges. The series converges at both its endpoints, so
2
n
n2
the interval of convergence is 3 ≤ x ≤ 7.
31. We use the ratio test to find the radius of convergence;
n!
x2(n+1)+1
x2
an+1
· 2n+1 =
=
an
(n + 1)! x
n+1
Since limn→∞ |x2 |/(n + 1) = 0 for all x, the radius of convergence is R = ∞. There are no endpoints to check. The
interval of convergence is all real numbers −∞ < x < ∞.
32. We use the ratio test to find the radius of convergence
(n + 1)!xn+1
an+1
=
= |x|(n + 1).
an
n!xn
Since limn→∞ |x|(n + 1) = ∞ for all x 6= 0, the radius of convergence is R = 0. There are no endpoints to check. The
series converges only for x = 0, so the interval of convergence is the single point x = 0.
33. We use the ratio test:
Since
r
r
√
(5x)n+1 / n + 1
n
an+1
√
=
= 5|x|
.
an
n+1
(5x)n / n
n
→ 1 as n → ∞, we have
n+1
an+1
= 5|x|.
an
We have 5|x| < 1 when |x| < 1/5. The radius of convergence is 1/5 and the series converges for −1/5 < x < 1/5.
lim
n→∞
900
Chapter Nine /SOLUTIONS
We check the endpoints. For x = −1/5, we have
∞
X
(5x)n
n=1
√
n
∞
X
(−1)n
=
√
n=1
n
.
1
1
1
< √ for all n ≥ 1 and lim √ = 0, by the alternating
n→∞
n
n
n+1
series test, the series converges at the endpoint x = −1/5. For x = 1/5, we have
This is an alternating series. Since we have 0 < √
∞
X
(5x)n
√
n=1
n
=
∞
X
1
n=1
√ .
n
This is a p-series with p = 1/2 and it diverges. Therefore, the interval of convergence is −1/5 ≤ x < 1/5.
34. We use the ratio test:
Since
r
r
√
(5x)2n+2 / n + 1
an+1
n
2 2
√
=5 x
=
.
an
n+1
(5x)2n / n
n
→ 1 as n → ∞, we have
n+1
an+1
= 25x2 .
an
lim
n→∞
We have 25x2 < 1 when |x| < 1/5. The radius of convergence is 1/5 and the series converges for −1/5 < x < 1/5.
We check the endpoints. For x = ±1/5, we have
∞
X
(5x)2n
n=1
√
n
=
∞
X
1
n=1
√ .
n
This is a p-series with p = 1/2 and it diverges. The series diverges at both endpoints x = ±1/5, so the interval of
convergence is −1/5 ≤ x < 1/5.
35. The expression 1/(1 + 2z) has the form a/(1 − x), which is the sum of a geometric series with a = 1, x = −2z. The
power series is
1 + (−2z) + (−2z)2 + (−2z)3 + · · · =
This series converges for | − 2z| < 1, that is, for −1/2 < z < 1/2.
∞
X
(−2z)n .
n=0
36. The expression 2/(1 + y 2 ) has the form a/(1 − x), which is the sum of a geometric series with a = 2, x = −y 2 . The
power series is
2 + (−y 2 ) + (−y 2 )2 + (−y 2 )3 + · · · =
This series converges for | − y 2 | < 1, that is, for −1 < y < 1.
∞
X
2(−y 2 )n .
n=0
37. The expression 3/(1 − z/2) has the form a/(1 − x), which is the sum of a geometric series with a = 3, x = z/2. The
power series is
3 + 3(z/2) + 3(z/2)2 + 3(z/2)3 + · · · =
This series converges for |z/2| < 1, that is, for −2 < z < 2.
∞
X
3(z/2)n .
n=0
38. To compare 8/(4 + y) with a/(1 − x), divide the numerator and denominator by 4. This gives 8/(4 + y) = 2/(1 + y/4),
which is the sum of a geometric series with a = 2, x = −y/4. The power series is
2
3
2 + 2(−y/4) + 2(−y/4) + 2(−y/4) + · · · =
This series converges for |y/4| < 1, that is, for −4 < y < 4.
∞
X
n=0
2(−y/4)n .
9.5 SOLUTIONS
901
39. The coefficient of the nth term of the binomial power series is given by
Cn =
p(p − 1)(p − 2) · · · (p − (n − 1))
.
n!
To apply the ratio test, consider
p(p − 1)(p − 2) · · · (p − (n − 1))(p − n)/(n + 1)!
an+1
= |x|
an
p(p − 1)(p − 2) · · · (p − (n − 1))/n!
p
p−n
n
= |x|
→ |x| as n → ∞.
= |x|
−
n+1
n+1
n+1
Thus, the radius of convergence is R = 1.
40. The kth coefficient in the series
R by the ratio test, so
P
kCk xk is Dk = k·Ck . We are given that the series
|x| lim
k→∞
Thus, applying the ratio test to the new series, we have
lim
k→∞
|Ck+1 |
|x|
=
.
|Ck |
R
P
Ck xk has radius of convergence
|x|
(k + 1)Ck+1
Dk+1 xk+1
= lim
|x| =
.
k→∞
Dk xk
kCk
R
Hence the new series has radius of convergence R.
41. The radius of convergence, R, is between 5 and 7.
42. The series is centered at x = −7. Since the series converges at x = 0, which is a distance of 7 from x = −7, the radius
of convergence, R, is at least 7. Since the series diverges at x = −17, which is a distance of 10 from x = −7, the radius
of convergence is no more than 10. That is, 7 ≤ R ≤ 10.
43. The radius of convergence of the series, R, is at least 4 but no larger than 7.
(a)
(b)
(c)
(d)
False. Since 10 > R the series diverges.
True. Since 3 < R the series converges.
False. Since 1 < R the series converges.
Not possible to determine since the radius of convergence may be more or less than 6.
44. The series is centered at x = 3. Since the series converges at x = 7, which is a distance of 4 from x = 3, we know
R ≥ 4. Since the series diverges at x = 10, which is a distance of 7 from x = 3, we know R ≤ 7. That is, 4 ≤ R ≤ 7.
Since x = 11 is a distance of 8 from x = 3, the series diverges at x = 11.
Since x = 5 is a distance of 2 from x = 3, the series converges there.
Since x = 0 is a distance of 3 from x = 3, the series converges at x = 3.
45. (a) We use the ratio test:
(−1)n+1 x2(n+1)
22n (n!)2
an+1
= 2(n+1)
·
2
an
2
((n + 1)!) (−1)n x2n
For a fixed value of x, we have
=
22n (n!)2
x2n+2
·
22n+2 (n + 1)2 (n!)2
x2n
=
x2
.
4(n + 1)2
x2
→ 0 as
4(n + 1)2
n → ∞.
The series converges for all x, so the domain of J(x) is all real numbers.
(b) Since
x2
+···,
J(x) = 1 −
4
we have J(0) = 1.
902
Chapter Nine /SOLUTIONS
(c) We have
S0 (x) = 1
x2
4
x4
x2
+
S2 (x) = 1 −
4
64
x2
x4
x6
S3 (x) = 1 −
+
−
4
64
2304
x4
x6
x8
x2
+
−
+
.
S4 (x) = 1 −
4
64
2304
147,456
S1 (x) = 1 −
(d) The value of J(1) can be approximated using partial sums. Substituting x = 1 into the partial sum polynomials, we
have
S0 (1) = 1
S1 (1) = 0.75
S2 (1) = 0.765625
S3 (1) = 0.765191
S4 (1) = 0.765198.
We estimate that J(1) ≈ 0.765. Theorem 9.9 can be used to bound the error.
(e) We see from the series that J(x) is an even function, so J(−1) = J(1). Thus, J(−1) ≈ 0.765.
46. (a) We have
f (x) = 1 + x +
so
x2
+ ···,
2
f (0) = 1 + 0 + 0 + · · · = 1.
(b) To find the domain of f , we find the interval of convergence.
|xn+1 /(n + 1)!|
|an+1 |
= lim
= lim
lim
n→∞
n→∞
n→∞ |an |
|xn /n!|
|x|n+1 n!
|x|n (n + 1)!
= |x| lim
n→∞
Thus the series converges for all x, so the domain of f is all real numbers.
(c) Differentiating term-by-term gives
d
f (x) =
dx
′
∞
X
xn
n=0
n!
!
d
=
dx
x3
x4
x2
+
+
+ ···
1+x+
2!
3!
4!
x2
x3
x
+3
+4
+ ···
2!
3!
4!
x2
x3
= 1+x+
+
+ ···.
2!
3!
= 0+1+2
Thus, the series for f and f ′ are the same, so
f (x) = f ′ (x).
(d) We guess f (x) = ex .
47. (a) Since only odd powers are involved in the series for g(x),
g(x) = x −
x5
x7
x3
+
−
+ ···,
3!
5!
7!
we see that g(x) is odd. Substituting x = 0 gives g(0) = 0.
(b) Differentiating term by term gives
x2
x4
x6
+5
−7
+ ···
3!
5!
7!
x4
x6
x2
+
−
+···.
= 1−
2!
4!
6!
g ′ (x) = 1 − 3
1
= 0.
n+1
9.5 SOLUTIONS
3
903
5
x
x
x
+4
−6
+ ···
2!
4!
6!
5
3
x
x
−
+ ···.
= −x +
3!
5!
g ′′ (x) = 0 − 2
So we see g ′′ (x) = −g(x).
(c) We guess g(x) = sin x since then g ′ (x) = cos x and g ′′ (x) = − sin x = g(x). We check g(0) = 0 = sin 0 and
g ′ (0) = 1 = cos 0.
48. (a) We have
(p(x))2 =
2
x2
x4
x6
+
−
+ ···
2!
4!
6!
1−
2
x6
x2 x4
x2
x2
x4
= 1−2·
+ −
−2
−2
·
···
+2
2
2!
4!
6!
2! 4!
1
1
1
1
= 1 − x2 +
+
x4 − x6
+
···
4
12
3 · 5 · 4!
4!
2 6
x4
= 1 − x2 +
−
x ···.
3
45
(q(x))2 =
x−
=x
2
=x
2
= x2
x5
x3
+
− ···
3!
5!
x2
+
1−2
3!
x2
1−
+ x4
3
1−
= x2 −
2
x2
−
3!
= x2
2
1−
!
2
x4
+ 2 ···
5!
1
1
+
(3!)2
5·4·3
2 4
x2
+
x ···
3
45
x4
2 6
+
x ···.
3
45
x4
x2
+
− ···
3!
5!
···
Thus, up to terms in x6 , we have
(p(x))2 + (q(x))2 = 1.
(b) The result of part (a) suggests that p(x) and q(x) could be the sine and cosine. Since p(x) is even and q(x) is odd,
we guess that p(x) = cos x and q(x) = sin x.
Strengthen Your Understanding
49. To find the radius of convergence, we calculate
lim
n→∞
|an+1 |
Cn+1 |x|n+1
Cn+1
= 0.
= lim
= |x| lim
n→∞
n→∞
|an |
Cn |x|n
Cn
Thus, the radius of convergence is ∞, not 0.
50. The series has an interval of convergence centered at x = 0. Since the series diverges at x = 2, the radius of convergence
is 2 or less. This means that the series diverges for all points at a distance of more than 2 from the origin. Thus, the series
cannot converge at x = 3.
51. In order to get a power series that does not converge at x = 0, we need to construct a power series about a point other
than x = 0. We’ll use a = 2, and the series
∞
X
(x − 2)n
.
n
n=0
If we let x = 0 and use the ratio test to determine the convergence or divergence of the series, we get
lim
n→∞
(−2)n+1 /(n + 1)
−2n
= lim
= 2.
n→∞ n + 1
(−2)n /n
Since the limit is greater than 1, the series diverges at x = 0 by the ratio test.
904
Chapter Nine /SOLUTIONS
52. The series
∞
X
n=1
n!(x − 5)n converges at x = 5. The ratio test shows that the series diverges for any other value of x, since
lim
n→∞
for x 6= 5.
53. The series with
P
(n + 1)!(x − 5)n+1
= |x − 5| lim (n + 1) > 1
n→∞
n!(x − 5)n
xn /n2 satisfies this condition. The limit
lim
n→∞
|x|n+1 /(n + 1)2
|an+1 |
n2
= lim
=
|x|
lim
= |x|.
n→∞
n→∞ (n + 1)2
|an |
|x|n /n2
Thus, the radius of convergence
P n is2 1. P
At the endpoints,
|x /n | =
1/n2 , which is a convergent p-series. Thus, the series is absolutely convergent
at the endpoints x = ±1.
54. False. Writing out terms, we have
(x − 1) + (x − 2)2 + (x − 3)3 + · · · .
A power series is a sum of powers of (x − a) for constant a. In this case, the value of a changes from term to term, so it
is not a power series.
55. True. This power series has an interval of convergence about x = 0. If the power series converges for x = 2, the radius
of convergence is 2 or more. Thus, x = 1 is well within the interval of convergence, so the series converges at x = 1.
56. False. This power series has an interval of convergence about x = 0. Knowing the power series converges for x = 1 does
not tell us whether the series converges for x = 2. Since the series converges at x = 1, we know the radius of convergence
is at least 1. However, we do not know whether the interval of convergence extends as far as x = 2, so we cannot say
whether the series converges at x = 2.
X xn
converges for x = 1 (it is a geometric series with ratio of 1/2), but does not converge for
For example,
2n
x = 2 (the terms do not go to 0).
Since this statement is not true for all Cn , the statement is false.
57. True. This power series has an interval of convergence centered on x = 0. If the power series does not converge for x = 1,
then the radius of convergence is less than or equal to 1. Thus, x = 2 lies outside the interval of convergence, so the series
does not converge there.
58. True. The radius of convergence, R, is given by lim |Cn+1 |/|Cn | = 1/R, if this limit exists, and since these series have
n→∞
the same coefficients, Cn , the radii of convergence are the same.
59. P
False. Two series can have the same radius of convergence without having the same coefficients. For example,
nxn both have radius of convergence of 1:
lim
n→∞
1
Cn+1
= lim
=1
n→∞ 1
Cn
60. False. There are power series, such as
61. True. The power series
P
P
n
and
lim
n→∞
P
xn and
Bn+1
n+1
= lim
= 1.
n→∞
Bn
n
xn /n, which converge at one endpoint, −1, but not at the other, 1.
Cn (x − a) converges at x = a.
62. True. Since the power series converges at x = 10, the radius of convergence is at least 10. Thus, x = −9 must be within
the interval of convergence.
63. False. If
Cn xn converges at x = 10, the radius of convergence is at least 10. However, if the radius of convergence
were exactly 10, then x = 10 is the endpoint of the interval of convergence and convergence there does not guarantee
convergence at the other endpoint.
P
64. True. Intervals of convergence can be of any length and centered at any point and can include one endpoint and not the
other.
65. False. The interval of convergence of
P
Cn xn is centered at the origin.
66. True. The interval of convergence is centered on x = a, so a = (−11 + 1)/2 = −5.
67. (d). Since the series is centered at x = 0 and diverges at x = 7, the radius of convergence is R ≤ 7. The series converges
at x = −3, so R ≥ 3. Since | − 9| = 9 > R, the series diverges at x = −9.
SOLUTIONS to Review Problems for Chapter Nine
905
Solutions for Chapter 9 Review
Exercises
3
3
3
1
3
1
1. 3 + + + · · · + 10 = 3 1 + + · · · + 10
2
4
8
2
2
2
2. We have
3 1−
=
1−
1
211
1
2
3 211 − 1
=
210
1
1
1
1
+ − +
− ···
2 4 8
16
3
1
1 2
1
= −2 + (−2) −
+ (−2) −
+ (−2) −
+ ···
2
2
2
4
−2
=− .
=
1 + 1/2
3
Sum = −2 + 1 −
3. This is the sum of a finite geometric series with a = 125, n = 21, x = 0.8:
S21
125 1 − 0.821
= |{z}
125 + 125(0.8) + 125(0.8) + · · · + 125(0.8) =
1 − 0.8
| {z }
| {z } | {z }
20
2
a
ax
ax2
{z
|
axn−1
a(1−xn )
1−x
= 619.235.
}
4. This is a finite geometric series with k − 2 terms in it, so n = k − 2. The initial term is a = (0.5)3 = 0.125 and the
constant ratio is x = 0.5. Using the formula for the sum of a finite geometric series, we get
Sum =
a(1 − xn )
0.125(1 − (0.5)k−2 )
=
= 0.25(1 − (0.5)k−2 ).
1−x
1 − 0.5
5. If b = 1, then the sum is 6. If b 6= 1, we use the formula for the sum of a finite geometric series. We can write the series as
b5 + b5 · b + b5 · b2 + b5 · b3 + b5 · b4 + b5 · b5 .
This is a six-term geometric series (n = 6) with initial term a = b5 and constant ratio x = b :
Sum =
b5 (1 − b6 )
a(1 − xn )
=
.
1−x
1−b
6. Using the formula for the sum of an infinite geometric series,
∞
X
1 n
n=4
3
=
1 4
+
3
1 5
3
+ ··· =
1 4
3
1
1+ +
3
1 2
3
+···
=
( 13 )4
1
=
1
54
1− 3
=
(1/3)4 (1 − (1/3)17 )
317 − 1
=
.
1 − (1/3)
2 · 320
7. Using the formula for the sum of a finite geometric series,
20
X
1 n
n=4
8.
3
=
∞
X
3n + 5
n=0
4n
We have
4 5
=
1
3
∞
X
3 n
n=0
n
∞
X
3
n=0
1
+
3
4
4
=
+
20
1
+· · ·+
3
∞
X
5
n=0
4n
=
4
2
1
3
+ ···
, a sum of two geometric series.
∞
X 5
1
5
20
= 4 and
=
=
, so
1 − 3/4
4n
1 − 1/4
3
n=0
∞
X
3n + 5
n=0
.
1
1+ +
3
1
3
4n
=4+
20
32
=
3
3
16
1
3
906
Chapter Nine /SOLUTIONS
9. We have
S1 = 36.
1
= 48.
3
2
1
1
S3 = 36 + 36
+ 36
= 52.
3
3 2
3
1
1
1
+ 36
S4 = 36 + 36
+ 36
= 53.333.
3
3
3
S2 = 36 + 36
Here we have a = 36 and x = 1/3, so
Sn =
36(1 − (1/3)n )
a(1 − xn )
=
.
1−x
1 − 1/3
As n → ∞, we see that Sn → 36/(2/3) = 54.
10. We have
S1 = 1280.
3
= 320.
4
2
3
3
= 1040.
S3 = 1280 + 1280 −
+ 1280 −
4
4
2
3
3
3
3
S4 = 1280 + 1280 −
+ 1280 −
+ 1280 −
= 500.
4
4
4
S2 = 1280 + 1280 −
Here we have a = 1280 and x = −3/4, so
Sn =
1280(1 − (−3/4)n )
a(1 − xn )
=
.
1−x
1 + 3/4
As n → ∞, we see that Sn → 1280/(7/4) = 731.429.
11. We have
S1 = −810.
2
= −270.
2
3
2
2
= −630.
− 810 −
S3 = −810 − 810 −
3
3
2 2
2 3
2
− 810 −
= −390.
S4 = −810 − 810 −
− 810 −
3
3
3
S2 = −810 − 810 −
Here we have a = −810 and x = −2/3, so
Sn =
a(1 − xn )
−810(1 − (−2/3)n )
=
.
1−x
1 + 2/3
As n → ∞, we see that Sn → −810/(5/3) = −486.
12. We have
S1 = 2.
S2 = 2 + 2(3z).
S3 = 2 + 2(3z) + 2(3z)2 .
S4 = 2 + 2(3z) + 2(3z)2 + 2(3z)3 .
Here we have a = 2 and x = 3z, so
2(1 − (3z)n )
a(1 − xn )
=
.
1−x
1 − 3z
As n → ∞, we see that (3z)n → 0 if |3z| < 1 but diverges if |3z| ≥ 1. Thus,
Sn =
Sn →
2
if |z| < 1/3, but diverges otherwise.
1 − 3z
SOLUTIONS to Review Problems for Chapter Nine
907
13. As n increases, the term 4n is much larger than 3 and 7n is much larger than 5. Thus dividing the numerator and
denominator by n and using the fact that lim 1/n = 0, we have
n→∞
lim
n→∞
(3/n) + 4
4
3 + 4n
= lim
= .
n→∞ (5/n) + 7
5 + 7n
7
Thus, the sequence converges to 4/7.
14. We have:
n+1
= 1.
n→∞
n
The terms of the sequence do not approach 0, and oscillate between values that are getting closer to +1 and −1. Thus the
sequence diverges.
lim
15. The first eight terms of the sequence are:
√
√
√
2
2
2
2
, 1,
, 0, −
, −1, −
, 0.
2
2
2
2
√
The sequence then repeats this pattern, so it diverges.
16. Since the exponential function 2n dominates the power function n3 as n → ∞, the series diverges.
17. We use the integral test with f (x) = 1/x3 to Zdetermine whether this series converges or diverges. To do so we determine
∞
1
dx converges or diverges:
whether the corresponding improper integral
3
x
1
Z
Since the integral
1
1
dx = lim
b→∞
x3
Z
1
b
1
−1
dx = lim
b→∞ 2x2
x3
b
= lim
b→∞
1
1
−1
+
2b2
2
=
1
.
2
∞
∞
Z
∞
X 1
1
dx converges, we conclude from the integral test that the series
converges.
3
x
n3
1
n=1
18. We use the integral test to determine
this series converges or diverges. To do so we determine whether the
Z ∞ whether
3x2 + 2x
dx converges or diverges. The integral can be calculated using the
corresponding improper integral
x3 + x2 + 1
1
3
2
2
substitution w = x + x + 1, dw = (3x + 2x) dx.
Z
1
∞
3x2 + 2x
dx = lim
3
b→∞
x + x2 + 1
Z
1
b
3x2 + 2x
dx
+ x2 + 1
x3
b
= lim ln |x3 + x2 + 1|
b→∞
1
= lim ln |b3 + b2 + 1| − ln 3 = ∞.
b→∞
Z
Since the integral
∞
1
diverges.
∞
X
3x2 + 2x
3n2 + 2n
dx
diverges,
we
conclude
from
the
integral
test
that
the
series
3
2
x +x +1
n3 + n2 + 1
n=1
19. We use the integralZ test to determine whether this series converges or diverges. We determine whether the corresponding
∞
2
xe−x dx converges or diverges:
improper integral
0
Z
∞
2
xe−x dx = lim
b→∞
0
Since the integral
Z
0
∞
2
Z
0
b
2
2
1
xe−x dx = lim − e−x
b→∞
2
b
= lim
0
b→∞
2
1
1
− e−b +
2
2
xe−x dx converges, we conclude from the integral test that the series
∞
X
n=0
=
1
.
2
2
ne−n converges.
908
Chapter Nine /SOLUTIONS
20. We use the integral test to determine
Z ∞ whether this series converges or diverges. To do so we determine whether the
2
corresponding improper integral
dx converges or diverges:
x2 − 1
2
∞
Z
2
2
dx = lim
b→∞
x2 − 1
= lim
b→∞
b
Z
2
Z b
2
b→∞
b→∞
Since the integral
Z
2
∞
x−1
x+1
ln
= lim
b→∞
1
1
−
dx
x−1
x+1
ln |x − 1| − ln |x + 1|
= lim
= lim
2
dx
x2 − 1
b
2
b
2
(Using partial fractions)
!
!
b−1
1
ln
− ln
b+1
3
= ln 1 − ln
1
= ln 3.
3
∞
X 2
2
dx converges, we conclude that the series
converges.
2
x −1
n2 − 1
n=2
21. Since an = 1/(2n n!), replacing n by n + 1 gives an+1 = 1/(2n+1 (n + 1)!). Thus
1
2n+1 (n + 1)!
|an+1 |
2n n!
1
= n+1
=
=
,
1
|an |
2
(n + 1)!
2(n + 1)
n
2 n!
so
L = lim
n→∞
Since L < 1, the ratio test tells us that
∞
X
1
n=1
2n n!
1
|an+1 |
= lim
= 0.
|an |
n→∞ 2n + 2
converges.
22. Since an = (n − 1)!/5n , replacing n by n + 1 gives an+1 = n!/5n+1 . Thus,
so
n!/5n+1
n!
n(n − 1)!
n
|an+1 |
=
=
=
= ,
|an |
(n − 1)!/5n
5(n − 1)!
5(n − 1)!
5
L = lim
n→∞
Since L > 1, the ratio test tells us that
∞
X
(n − 1)!
n=1
5n
|an+1 |
n
= lim
= ∞.
n→∞ 5
|an |
diverges.
23. Since an = (2n)!/(n!(n + 1)!), replacing n by n + 1 gives an+1 = (2n + 2)!/((n + 1)!(n + 2)!). Thus,
(2n + 2)!
(2n + 2)!
n!(n + 1)!
|an+1 |
(n + 1)!(n + 2)!
=
=
·
.
(2n)!
|an |
(n + 1)!(n + 2)!
(2n)!
n!(n + 1)!
However, since (n + 2)! = (n + 2)(n + 1)n! and (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have
so
|an+1 |
(2n + 2)(2n + 1)
2(2n + 1)
=
=
,
|an |
(n + 2)(n + 1)
n+2
L = lim
n→∞
Since L > 1, the ratio test tells us that
∞
X
n=1
|an+1 |
= 4.
|an |
(2n)!
diverges.
n!(n + 1)!
SOLUTIONS to Review Problems for Chapter Nine
909
24. Let an = 1/(n2 + 1). Then replacing n by n + 1 gives an+1 = 1/((n + 1)2 + 1). Since (n + 1)2 + 1 > n2 + 1, we have
1
1
< 2
,
(n + 1)2 + 1
n +1
0<
so
0 < an+1 < an .
We also have limn→∞ an = 0, therefore, the alternating series test tells us that the series
∞
X
(−1)n
n=1
n2 + 1
converges.
p
p
√
25. Let an = 1/ n2 + 1. Then replacing n by n + 1 we have an+1 = 1/ (n + 1)2 + 1. Since (n + 1)2 + 1 >
√
n2 + 1, we have
1
1
,
p
< √
2 +1
2
n
(n + 1) + 1
so
0 < an+1 < an .
In addition, limn→∞ an = 0 so
∞
X
(−1)n
n=0
26. The series
X (−1)n
n1/2
p = 1/2 ≤ 1. Thus
27. Since
√
n2 + 1
converges by the alternating series test.
converges by the alternating series test. However
X (−1)n
n1/2
is conditionally convergent.
lim
n→∞
1+
1
n2
X 1
n1/2
diverges because it is a p-series with
= 1,
1
is divergent.
n2
P
ln n/n converges. Since ln(n) > 1 for all n > 2, we
28. We first check absolute convergence by deciding whether
compare
1
ln n
<
.
n
n
P
P
The harmonic series
1/n diverges, so
ln n/n diverges. Alternatively, we can use the integral test. Since
the nth term an = (−1)n 1 +
Z
1
∞
1
n2
does not tend to zero as n → ∞. Thus, the series
ln x
dx = lim
b→∞
x
Z
b
1
ln x
(ln x)2
dx = lim
b→∞
x
2
X
b
= lim
1
b→∞
(−1)n 1 +
(ln b)2
,
2
X ln n
and since this limit does not exist,
diverges.
n
We now check conditional convergence. The original series is alternating, so we check whether an+1 < an . Consider
an = f (n), where f (x) = ln n/n. Since
d
dx
ln x
x
=
1
(1 − ln x)
x2
is negative for x > e, we know that an is decreasing for n > e. Thus for n ≥ 3,
an+1 =
Since ln n/n → 0 as n → ∞, we see that
29. Since
ln(n)
ln(n + 1)
<
= an .
n+1
n
X (−1)n−1 ln n
n
lim an = lim
n→∞
we know that
X (−1)n−1
arctan n
n→∞
is conditionally convergent.
2
1
= 6= 0
arctan n
π
diverges by Property 3 of Theorem 9.2.
910
Chapter Nine /SOLUTIONS
30. Let an = n2 /(3n2 + 4). Since 3n2 + 4 > 3n2 , we have
n2
1
< , so
3n2 + 4
3
0 < an <
The geometric series
∞
X
1 n
3
n=1
verges.
1 n
3
.
converges, so the comparison test tells us that the series
∞
X
n=1
n2
2
3n + 4
31. Let an = 1/(n sin2 n). Since 0 < sin2 n < 1, for any positive integer n, we have n sin2 n < n, so
an >
The harmonic series
∞
X
1
n=1
32. The nth term an =
√
n
√
an
lim
= lim
n→∞ bn
n→∞
√
also con-
1
1
> , thus
n
n sin2 n
1
.
n
diverges, so the comparison test tells us that the series
n − 1/(n2 + 3) behaves like
n
∞
X
n=1
1
also diverges.
n sin2 n
n/n2 = 1/n3/2 for large n, so we take bn = 1/n3/2 . We have
p
√
n2 1 − 1/n
n − 1/(n2 + 3)
n3/2 n − 1
= lim
= lim 2
= 1.
n→∞
n→∞ n (1 + 3/n2 )
n2 + 3
1/n3/2
The
test applies with c = 1. The p-series
P √limit comparison
n − 1/(n2 + 3) also converges.
P
1/n3/2 converges because p = 3/2 > 1. Therefore
33. The nth term an = (n3 − 2n2 + n + 1)/(n5 − 2) behaves like n3 /n5 = 1/n2 for large n, so we take bn = 1/n2 . We
have
(n3 − 2n2 + n + 1)/(n5 − 2)
an
n5 − 2n4 + n3 + n2
lim
= lim
=
lim
= 1.
n→∞ bn
n→∞
n→∞
1/n2
n5 − 2
The limit comparison test
applies with
c = 1. The p-series
P 3
n − 2n2 + n + 1 / n5 − 2 also converges.
P
1/n2 converges because p = 2 > 1. Therefore the series
34. The nth term is an = sin(1/n2 ). When n is large, 1/n2 is near zero, so sin(1/n2 ) is near 1/n2 . We see that sin(1/n2 )
behaves like 1/n2 for large n, so we take bn = 1/n2 . We have
lim
n→∞
sin(1/n2 )
an
= lim
n→∞
bn
1/n2
sin x
= lim
x→0
x
= 1.
The limit comparison test applies with c = 1. The p-series
1/n2 converges because p = 2 > 1. Therefore
sin(1/n2 )
also converges.
√
√
35. The nth term an = 1/( n3 − 1) behaves like 1/ n3 = 1/n3/2 for large n, so we take bn = 1/n3/2 . We have
√
1/ n3 − 1
1
1
an
n3/2
n3/2
p
= lim p
= √
= 1.
lim
= lim
= lim √
= lim
3/2
3
n→∞
n→∞ bn
n→∞
n→∞
n→∞ n3/2
1/n
1−0
n −1
1 − 1/n3
1 − 1/n3
P
The
√ comparison test applies with c = 1. The p-series
P limit
1/ n3 − 1 also converges.
P
P
1/n3/2 converges because p = 3/2 > 1. Therefore
36. Since f (x) = 1/(x + 1) is continuous, positive and decreasing, we apply the integral test, and we obtain
Z
1
∞
1
dx = lim
x+1
b→∞
Since this improper integral diverges, the series
Z
b
1
∞
X
n=1
1
dx = lim (ln(b + 1) − ln 2) = ∞.
1+x
b→∞
1
also diverges. We can also observe the series is the harmonic
n+1
series, with the first term missing, and hence diverges by Property 2 of Theorem 9.2.
37. This is a p-series with p > 1, so it converges.
SOLUTIONS to Review Problems for Chapter Nine
911
38. We use the integral test to determine
Z ∞ whether this series converges or diverges. To do so we determine whether the
2
√
corresponding improper integral
dx converges or diverges:
x−2
3
Z
∞
3
√
2
dx = lim
b→∞
x−2
= lim
b→∞
b
Z
√
3
Z
b
1
2
√ dw
w
b→∞
n=3
√
2
n−2
1
= ∞.
∞
2
dx diverges, and we conclude from the integral test that the series
x
−2
3
√
diverges. The limit comparison test with bn = 1/ n can also be used.
Since the limit does not exist, the integral
∞
X
(Substitute w = x − 2.)
b
√
= lim 4 w
Z
2
dx
x−2
√
√
39. This is an alternating series. Let an = 1/( n + 1). Then limn→∞ an = 0. Now replace n by n + 1 to give an+1 =
√
√
√
1
1
1/( n + 1 + 1). Since n + 1 + 1 > n + 1, we have √
< √
, so
n
+1
n+1+1
0 < an+1 = √
1
1
= an .
< √
n+1
n+1+1
Therefore, the alternating series test tells us that the series
∞
X
(−1)n−1
n=1
√
n+1
converges.
40. Writing an = n2 /(n2 + 1), we have limn→∞ an = 1 so the series diverges by Property 3 of Theorem 9.2.
41. We use the integral test to determine
this series converges or diverges. To do so we determine whether the
Z ∞ whether
x2
dx converges or diverges:
corresponding improper integral
x3 + 1
1
Z
1
∞
x2
dx = lim
3
x +1
b→∞
Z
1
b
x2
1
dx = lim ln |x3 + 1|
3
x +1
b→∞ 3
Since the limit does not exist, the integral
series
∞
X
n=1
Z
∞
1
b
= lim
1
2
b→∞
1
3
ln(b3 + 1) −
1
ln 2 .
3
x
dx diverges an so we conclude from the integral test that the
x3 + 1
n2
diverges. The limit comparison test with bn = 1/n can also be used.
n3 + 1
42. We use the ratio test. Since an = 3n /(2n)!, replacing n by n + 1 gives an+1 = 3n+1 /(2n + 2)!. Thus
3n+1 /(2n + 2)!
(2n)!
3n+1
an+1
=
=
· n .
n
an
3 /(2n)!
(2n + 2)!
3
Since (2n + 2)! = (2n + 2)(2n + 1)(2n)!, we have
an+1
3
=
,
an
(2n + 2)(2n + 1)
so
L = lim
n→∞
Since L < 1, the ratio test tells us that the series
∞
X
3n
n=1
(2n)!
an+1
= 0.
an
converges.
43. We use the ratio test. Since an = (2n)!/(n!)2 , replacing n by n + 1 gives an+1 = (2n + 2)!/((n + 1)!)2 . Thus
(2n + 2)!
((n + 1)!)2
(2n + 2)!
an+1
n!n!
=
=
·
.
(2n)!
an
(n + 1)!(n + 1)! (2n)!
(n!)2
912
Chapter Nine /SOLUTIONS
Since (2n + 2)! = (2n + 2)(2n + 1)(2n)! and (n + 1)! = (n + 1)n!, we have
(2n + 2)(2n + 1)
an+1
=
,
an
(n + 1)(n + 1)
therefore
an+1
= 4.
an
L = lim
n→∞
As L > 1 the ratio test tells us that the series
∞
X
(2n)!
n=1
44. The series can be written as
∞
X
n2 + 2n
n=1
Since
∞
X
1
n=1
2n
diverges.
(n!)2
=
n2 2n
is a convergent geometric series and
n=1
∞
X
1
n=1
converges by Theorem 9.2.
45. We use the ratio test:
∞
X
1
2n
+
∞
X
1
n=1
n2
.
converges as a p-series with p > 1, we see
n2
∞
X
n2 + 2n
n=1
n2 2n
(2n)!
32n+2
1
an+1
·
.
=
= 32
an
(2n + 2)! 32n
(2n + 2)(2n + 1)
Therefore we have
1
an+1
= 0.
= 32 lim
n→∞ (2n + 2)(2n + 1)
an
Thus by the ratio test, the series converges.
n
(n + 1)
(n + 1)
n+1
1
1
1
46. Let an = 2−n
=
.
Since
<
1
and
=
, we have
(n + 2)
n+2
2n
(n + 2)
2n
2
lim
n→∞
0 < an <
so that we can compare the series
∞
X
2−n
n=1
tells us that
1 n
2
,
∞
n
X 1
(n + 1)
with the convergent geometric series
(n + 2)
2
n=1
∞
X
2−n
n=1
also converges.
. The comparison test
(n + 1)
(n + 2)
47. We have
(2n + 1)!
an+1
2n+1
2
= lim
·
= lim
= 0,
n→∞
n→∞ (2n + 3)!
n→∞ (2n + 3)(2n + 2)
an
2n
so the series converges by the ratio test, since L < 1.
√
48. Since there is an n in the numerator and a n in the denominator, the terms in this series are increasing in magnitude. We
have
n+1
n+1
lim √ (−1)n = lim √ = ∞,
n→∞
n→∞
n
n
√
n
so limn→∞ (−1) (n + 1)/ n does not approach zero. Therefore, the series diverges by Property 3 of Theorem 9.2.
L = lim
49. The series can be written as
∞
X
2 + 3n
n=0
The series
∞
X
1 n
n=0
5
5n
=
∞
X
2
n=0
5n
+
3n
5n
=
∞
X
1 n
2
n=0
5
+
n
3
5
.
is a geometric series which converges because | 15 | < 1. Likewise, the geometric series
converges because | 35 | < 1. Since both series converge, Property 1 of Theorem 9.2 tells us that the series
converges.
∞
X
3 n
n=0
∞
X
2 + 3n
n=0
5n
5
also
SOLUTIONS to Review Problems for Chapter Nine
913
50. The nth term an = ((1 + 5n)/(4n))n behaves like (5/4)n for large n, so we take bn = (5/4)n and use the limit
comparison test. We have
((1 + 5n)/4n)n
1 + 5n n
an
= lim
= lim
,
lim
n
n→∞
n→∞
n→∞ bn
(5/4)
5n
so
1/5
1 5n
an
= lim
1+
= e1/5 .
lim
n→∞
n→∞ bn
5n
The limit comparison
test applies with c = e1/5 . The geometric series
P
Therefore ((1 + 5n)/(4n))n also diverges.
P
(5/4)n diverges because x = 5/4 ≥ 1.
51. Writing an = 1/(2 + sin n), we have limn→∞ an does not exist, so the series diverges by Property 3 of Theorem 9.2.
52. We use the integral test to determine
Z ∞ whether this series converges or diverges. To do so we determine whether the
1
dx converges or diverges:
corresponding improper integral
(2x
−
5)3
3
Z
∞
3
1
1
dx =
lim
(2x − 5)3
2 b→∞
=−
b
Z
1
1
dw
w3
1
1
lim
2 b→∞ 2w2
b
1
1
1
1
= − lim
−
2 b→∞ 2b2
2
Since the integral
Z
3
∞
(Substitute w = 2x − 5)
=
1
.
4
∞
X
1
1
dx
converges,
we
conclude
from
the
integral
test
that
the
series
(2x − 5)3
(2n − 5)3
converges. The limit comparison test, with bn = 1/n3 can also be used.
n=3
53. The nth term an = 1/(n3 − 3) behaves like 1/n3 for large n, so we take bn = 1/n3 . We have
lim
n→∞
1/(n3 − 3)
an
n3
= lim
= lim 3
= 1.
n→∞
n→∞ n − 3
bn
1/n3
The limit comparison test applies with c = 1. The p-series
3 also converges.
54. Note that
∞
X
sin(nπ/2)
n=1
n3
P
=1−
1/n3 converges because p = 3 > 1. Therefore
X
1/(n3 −
1
1
1
+ 3 − 3 + ···
33
5
7
is an alternating series with the absolute values of the terms decreasing to 0. Thus, the series converges by the alternating
series test.
55. Since ln(1 + 1/k) = ln((k + 1)/k) = ln(k + 1) − ln k, the nth partial sum of this series is
Sn =
n
X
ln 1 +
k=1
=
n
X
k=1
1
k
ln(k + 1) −
n
X
ln k
k=1
= (ln 2 + ln 3 + · · · + ln(n + 1)) − (ln 1 + ln 2 + · · · + ln n)
= ln(n + 1) − ln 1
= ln(n + 1).
Thus, the partial sums, Sn , grow without bound as n → ∞, so the series diverges by the definition.
56. The ratio test gives
L = lim
n→∞
so the series converges since L < 1.
(n + 1)/2n+1
an+1
n+1
1
= lim
= lim
= ,
n→∞
n→∞ 2n
an
n/2n
2
914
Chapter Nine /SOLUTIONS
57. Since ln n grows much more slowly than n, we suspect that (ln n)2 < n for large n. This can be confirmed with
L’Hopital’s rule.
2(ln n)/n
2(ln n)
(ln n)2
= lim
= lim
= 0.
lim
n→∞
n→∞
n→∞
n
1
n
Therefore, for large n, we have (ln n)2 /n < 1, and hence for large n,
1
1
<
.
n
(ln n)2
Thus
P∞
n=2
1/(ln n)2 diverges by comparison with the divergent harmonic series
P
1/n.
(2n + 2)!
(2n)!
. Then replacing n by n + 1, we have Cn+1 =
. Thus, with an = (2n)!xn /(n!)2 , we have
58. Let Cn =
(n!)2
((n + 1)!)2
|an+1 |
|Cn+1 |
(2n + 2)!/((n + 1)!)2
(2n + 2)!
(n!)2
= |x|
= |x|
=
|x|
·
.
|an |
|Cn |
(2n)!/(n!)2
(2n)!
((n + 1)!)2
Since (2n + 2)! = (2n + 2)(2n + 1)(2n)! and (n + 1)! = (n + 1)n! we have
(2n + 2)(2n + 1)
|Cn+1 |
=
,
|Cn |
(n + 1)(n + 1)
so
lim
n→∞
|Cn+1 |
(2n + 2)(2n + 1)
|an+1 |
4n + 2
= |x| lim
= |x| lim
= |x| lim
= 4|x|,
n→∞ |Cn |
n→∞ (n + 1)(n + 1)
n→∞ n + 1
|an |
so the radius of convergence of this series is R = 1/4.
59. Let Cn = 1/(n! + 1). Then replacing n by n + 1 gives Cn+1 = 1/((n + 1)! + 1). Using the ratio test, we have
|an+1 |
|Cn+1 |
1/((n + 1)! + 1)
n! + 1
= |x|
= |x|
= |x|
.
|an |
|Cn |
1/(n! + 1)
(n + 1)! + 1
Since n! and (n + 1)! dominate the constant term 1 as n → ∞ and (n + 1)! = (n + 1) · n! we have
lim
n→∞
|an+1 |
= 0.
|an |
Thus the radius of convergence is R = ∞.
60. To find R, we consider the following limit, where the coefficient of the nth term is given by Cn = n2 .
lim
n→∞
(n + 1)2 xn+1
|an+1 |
n2 + 2n + 1
= lim |x|
= lim
2
n
n→∞
n→∞
|an |
n x
n2
= |x| lim
n→∞
1 + (2/n) + (1/n2 )
1
= |x|.
Thus, the radius of convergence is R = 1.
61. Here the coefficient of the nth term is Cn = n/(2n + 1). Now we have
((n + 1)/(2n + 3))xn+1
(n + 1)(2n + 1)
an+1
=
=
|x| → |x| as n → ∞.
an
(n/(2n + 1))xn
n(2n + 3)
Thus, by the ratio test, the radius of convergence is R = 1.
62. We use the ratio test:
3n n2
an+1
xn+1
· n =
= n+1
2
an
3
(n + 1)
x
Since n/(n + 1) → 1 as n → ∞, we have
n 2 |x|
n+1
·
3
.
|x|
an+1
=
.
an
3
We have |x|/3 < 1 when |x| < 3. The radius of convergence is 3 and the series converges for −3 < x < 3.
lim
n→∞
SOLUTIONS to Review Problems for Chapter Nine
915
We check the endpoints. For x = −3, we have
∞
X
xn
n=1
3n n2
=
∞
X
(−3)n
n=1
3n n2
=
∞
X
(−1)n
n2
n=1
.
X (−1)n
X 1
is
a
p-series
with
p
=
2
so
it
converges.
Therefore
the
alternating
series
also converges.
We know
n2
n2
For x = 3, we have
∞
∞
∞
X
X
X
xn
3n
1
=
=
.
3n n2
3n n2
n2
n=1
n=1
n=1
This is a p-series with p = 2 and it converges. The series converges at both its endpoints and the interval of convergence
is −3 ≤ x ≤ 3.
63. We use the ratio test:
an+1
(−1)n+1 (x − 2)n+1
5n
|x − 2|
=
·
=
.
n+1
n
an
5
(−1) (x − 2)n
5
Since |x − 2|/5 < 1 when |x − 2| < 5, the radius of convergence is 5 and the series converges for −3 < x < 7.
We check the endpoints:
x = −3 :
x=7:
∞
X
(−1)n (x − 2)n
5n
n=0
∞
X (−1)n (x − 2)n
5n
n=0
=
∞
X
(−1)n (−3 − 2)n
5n
n=0
∞
=
X (−1)n (7 − 2)n
n=0
5n
=
=
∞
X
1 which diverges.
n=0
∞
X
(−1)n
which diverges.
n=0
The series diverges at both the endpoints, so the interval of convergence is −3 < x < 7.
64. We use the ratio test:
n
(−1)n+1 xn+1
an+1
n
=
=
·
· |x|.
an
n+1
(−1)n xn
n+1
Since n/(n + 1) → 1 as n → ∞, we have
an+1
= |x|.
an
The series converges for |x| < 1. The radius of convergence is 1 and the series converges for −1 < x < 1.
We check the endpoints. For x = −1, we have
lim
n→∞
∞
X
(−1)n xn
n
n=1
=
∞
X
(−1)n (−1)n
n
n=1
=
n=1
This is the harmonic series and diverges. For x = 1, we have
∞
X
(−1)n xn
n=1
n
=
∞
X
(−1)n (1)n
n=1
n
=
∞
X
1
n
.
∞
X
(−1)n
n=1
n
.
This is the alternating harmonic series and converges. The series diverges at x = −1 and converges at x = 1. Therefore,
interval of convergence is −1 < x ≤ 1.
65. We use the ratio test to find the radius of convergence:
n!
xn+1
x
an+1
·
=
=
an
(n + 1)! xn
n+1
Since limn→∞ |x|/(n + 1) = 0 for all x, the radius of convergence is R = ∞. There are no endpoints to check. The
interval of convergence is all real numbers −∞ < x < ∞.
Problems
66. We have
s1 =
(−1)1 (2 · 1 + 1)2
=
22·1−1 + (−1)1+1
(−1)32
21 + 1
=
−9
= −3
3
916
Chapter Nine /SOLUTIONS
(−1)2 (2 · 2 + 1)2
52
25
= 3
=
2·2−1
2+1
2
+ (−1)
2 + (−1)
7
(−1)72
−49
(−1)3 (2 · 3 + 1)2
=
=
s3 = 2·3−1
2
+ (−1)3+1
25 + 1
33
2
4
2
(−1) (2 · 4 + 1)
9
81
s4 = 2·4−1
= 7
=
,
2
+ (−1)4+1
2 + (−1)
127
s2 =
so the first four terms are −3, 25/7, −49/33, 81/127.
67. We see that s1 , s2 , s3 , . . . forms an arithmetic sequence—that is, the values go up by the same amount, 2, each time. We
conclude that
sn = 2n + 3.
68. We see that t1 , t2 , t3 , . . . is a sequence of consecutive odd square numbers. Since t1 = 3, we conclude that
tn = (2n + 1)2 .
69. We have
a2 = a1 + 2 · 2 = 5 + 4 = 9
because a1 = 5
a3 = a2 + 2 · 3 = 9 + 6 = 15
a4 = a3 + 2 · 4 = 15 + 8 = 23.
70. First, we have
a2 = a1 + 2 · 2 = 5 + 4 = 9
because a1 = 5
a3 = a2 + 2 · 3 = 9 + 6 = 15
a4 = a3 + 2 · 4 = 15 + 8 = 23.
Next, we find that
b2 = b1 + a1 = 10 + 5 = 15
because b1 = 10
b3 = b2 + a2 = 15 + 9 = 24
b4 = b3 + a3 = 24 + 15 = 39
b5 = b4 + a4 = 39 + 23 = 62.
71. The series can be written as
∞
X
nr + r n
n=1
nr r n
=
∞
X
1
n=1
rn
+
If 0 < r < 1, both series diverge, but if r > 1 both series converge.
∞
X
n+1
If r = 1 the given series becomes
which diverges.
n
∞
X
1
n=1
nr
.
n=1
By Theorem 9.2 the given series converges if r > 1.
72. The series converges for |x − 2| = 2 and diverges for |x − 2| = 4, thus the radius of convergence of the series, R, is at
least 2 but no larger than 4.
(a)
(b)
(c)
(d)
(e)
False. If x = 7 then |x − 2| = 5, so the series diverges.
False. If x = 1 then |x − 2| = 1, so the series converges.
True. If x = 0.5 then |x − 2| = 1.5, so the series converges.
If x = 5 then |x − 2| = 3 and it is not possible to determine whether or not the series converges at this point.
False. If x = −3 then |x − 2| = 5, so the series diverges.
SOLUTIONS to Review Problems for Chapter Nine
917
73. (a) Using an argument similar to Example 7 in Section 9.5, we take
an = (−1)n
t2n
,
(2n)!
so, replacing n by n + 1,
an+1 = (−1)n+1
Thus,
so
t2(n+1)
t2n+2
= (−1)n+1
.
(2(n + 1))!
(2n + 2)!
|an+1 |
|(−1)n+1 t2n+2 /(2n + 2)!|
t2
=
=
,
n
2n
|an |
|(−1) t /(2n)!|
(2n + 2)(2n + 1)
|an+1 |
t2
= lim
= 0.
n→∞ (2n + 2)(2n + 1)
n→∞ |an |
The radius of convergence is therefore ∞, so the series converges for all t. Therefore, the domain of h is all real
numbers.
(b) Since h involves only even powers,
t4
t6
t2
+
−
+ ···,
h(t) = 1 −
2!
4!
6!
h is an even function.
(c) Differentiating term by term, we have
lim
t
t3
t6
+ 4 − 6 + ···
2!
4!
6!
t5
t3
−
+ ···.
= −t +
3!
5!
h′ (t) = 0 − 2
t2
t4
− 5 + ···
3!
5!
t4
t2
−
+···.
= −1 +
2!
4!
h′′ (t) = −1 + 3
So we see h′′ (t) = −h(t).
74. If a payment M in the future has present value P , then
M = P (1 + r)t ,
where t is the number of periods in the future and r is the interest rate. Thus
P =
M
.
(1 + r)t
The monthly interest rate here is 0.09/12 = 0.0075, so the present value of first payment, made at the end of the first
month, is M/(1.0075). The present value of the second payment is M/(1.0075)2 . Continuing in this way, the sum of the
present value of all of the payments is
M
M
M
+
+ ··· +
.
(1.0075)
(1.0075)2
(1.0075)240
This is a finite geometric series with 240 terms, with sum
M
1.0075
1 − 1.0075−240
1 − 1.0075−1
= 111.145M.
Setting this equal to the loan amount of 200,000 gives a monthly payment of M = $200,000/111.145 = $1799.45.
918
Chapter Nine /SOLUTIONS
75. In the first year, extraction is 12 million tons. In the second year this falls to 12(1 − 0.05) = 12(0.95) million tons, and
in subsequent years the extraction falls by a factor of 0.95. During the next n years
Total extraction = 12 + 12(0.95) + 12(0.95)2 + · · · 12(0.95)n−1 , million tons.
This is a geometric series with a = 12 and x = 0.95, so
Total extraction = 12
Since |x| < 1, when n → ∞
1 − 0.95n
1 − 0.95
.
12
= 240 million tons.
1 − 0.95
World reserves of the mineral must exceed 240 million tons if extraction is to continue indefinitely.
Total extraction →
76. (a) It is easier to work with the value of the car first and then find the yearly losses. The value of the car goes down by
10% a year. Thus, the value at the end of the first years is v1 = 30,000(0.9). The value at the end of the second year
is v2 = 30,000(0.9)2 . The value at the end of n years is vn = 30,000(0.9)n . Thus, the losses in the first four years
are
l1 = 30,000(0.1)
l2 = v1 − v2 = 30,000(0.9) − 30,000(0.9)2 = 30,000(0.9)(0.1)
l3 = v2 − v3 = 30,000(0.9)2 − 30,000(0.9)3 = 30,000(0.9)2 (0.1)
l4 = v3 − v4 = 30,000(0.9)3 − 30,000(0.9)4 = 30,000(0.9)3 (0.1).
Thus,
ln = vn−1 − vn = 30,000(0.9)n−1 (0.1) = 3000(0.9)n−1 .
(b) In the first year, m1 = 500; in the second year, m2 = 500(1.2); in the third year, m3 = 500(1.2)2 . Thus
mn = 500(1.2)n−1 .
(c) We want to find n such that mn ≥ ln , so
500(1.2)n−1 ≥ 3000(0.9)n−1 .
We solve
500(1.2)n−1 = 3000(0.9)n−1
(1.2)n−1
3000
=
(0.9)n−1
500
1.2 n−1
=6
0.9
1.2
(n − 1) ln
= ln 6
0.9
ln 6
n−1 =
ln(1.2/0.9)
n = 6.228 + 1 = 7.228.
So, maintenance first exceeds losses in year 8. In year 7,
In year 8,
l7 = 3000(0.9)6 = $1594,
m7 = 500(1.2)6 = $1493.
l8 = 3000(0.9)7 = $1435,
m8 = 500(1.2)7 = $1792.
SOLUTIONS to Review Problems for Chapter Nine
919
77.
50
1.06
50
, etc.
Present value of second coupon =
(1.06)2
Present value of first coupon =
50
50
50
1000
+
+ ··· +
+
1.06
(1.06)2
(1.06)10 (1.06)10
Total present value =
=
|
{z
}
coupons
50
1.06
1
1
+ ··· +
1.06
(1.06)9
1+
1−
50
=
1.06
1
10
1
1.06
1
− 1.06
!
+
| {z }
principal
+
1000
(1.06)10
1000
(1.06)10
= 368.004 + 558.395
= $926.40
78.
50
1.04
50
, etc.
Present value of second coupon =
(1.04)2
Present value of first coupon =
50
1000
50
50
+
+
+ ··· +
1.04
(1.04)2
(1.04)10 (1.04)10
Total present value =
=
|
{z
}
coupons
50
1.04
1
1
+ ··· +
1.04
(1.04)9
1+
1−
50
=
1.04
1
10
1
1.04
1
− 1.04
!
+
| {z }
principal
+
1000
(1.04)10
1000
(1.04)10
= 405.545 + 675.564
= $1081.11
79. (a) The present value is given by the finite series:
5
5
100
5
+
+ ··· +
+
Total present value =
1.04
(1.04)2
(1.04)100 (1.04)100
=
|
{z
}
coupons
5
1.04
5
=
1.04
1+
1−
1
1
1
+ ··· +
1.04
(1.04)99
100
1
1.04
1
− 1.04
!
+
| {z }
principal
+
100
(1.04)100
100
(1.04)100
= $124.50.
(b) The present value is now given by the infinite series
Total present value =
5
5
5
+
+
+ ···
1.04
(1.04)2
(1.04)3
=
5
1.04
=
5
1.04
1+
1
1
+
+ ···
1.04
(1.04)2
1
1
1 − 1.04
= $125.
Notice how little difference there is between the worth of the bond which pays for 100 years and the one which pays
forever.
920
Chapter Nine /SOLUTIONS
80. The quantity of cephalexin in the body is given by Q(t) = Q0 e−kt , where Q0 = Q(0) and k is a constant. Since the
half-life is 0.9 hours,
1
1
1
= e−0.9k , k = −
ln ≈ 0.8.
2
0.9 2
(a) After 6 hours
Q = Q0 e−k(6) ≈ Q0 e−0.8(6) = Q0 (0.01).
(b)
Thus, the percentage of the cephalexin that remains after 6 hours ≈ 1%.
Q1 = 250
Q2 = 250 + 250(0.01)
Q3 = 250 + 250(0.01) + 250(0.01)2
Q4 = 250 + 250(0.01) + 250(0.01)2 + 250(0.01)3
(c)
250(1 − (0.01)3 )
1 − 0.01
≈ 252.5
250(1 − (0.01)4 )
Q4 =
1 − 0.01
≈ 252.5
Q3 =
Thus, by the time a patient has taken three cephalexin tablets, the quantity of drug in the body has leveled off to 252.5
mg.
(d) Looking at the answers to part (b) shows that
Qn = 250 + 250(0.01) + 250(0.01)2 + · · · + 250(0.01)n−1
250(1 − (0.01)n )
.
=
1 − 0.01
(e) In the long run, n → ∞. So,
Q = lim Qn =
n→∞
250
= 252.5.
1 − 0.01
81. A person should expect to pay the present value of the bond on the day it is bought.
10
1.04
10
Present value of second payment =
, etc.
(1.04)2
Present value of first payment =
Therefore,
Total present value =
This is a geometric series with a =
10
10
10
+
+
+ ···.
1.04
(1.04)2
(1.04)3
10
1
and x =
, so
1.04
1.04
Total present value =
1
10
1.04
1
− 1.04
= £250.
82. (a)
Total amount of money deposited = 100 + 92 + 84.64 + · · ·
= 100 + 100(0.92) + 100(0.92)2 + · · ·
100
= 1250 dollars
=
1 − 0.92
(b) Credit multiplier = 1250/100 = 12.50
The 12.50 is the factor by which the bank has increased its deposits, from $100 to $1250.
SOLUTIONS to Review Problems for Chapter Nine
83. (a)
921
(i) Since the number of bacteria doubles every half hour, the number quadruples every hour. Thus
R1 = B0 · 4
R2 = B0 · 42
..
.
Rn = B0 · 4n .
(ii) Each hour, the number of bacteria is multiplied by a factor a, so
Fn = B0 an .
The bacteria doubles in number in 10 hours, so
F10 = 2B0 .
Thus,
B0 a10 = 2B0
a = 21/10 ,
so
Fn = B0 (21/10 )n = B0 2n/10 .
(iii) The ratio is
B0 4n
Rn
=
=
Fn
B0 2n/10
(b) We want to solve for n making Yn = 1,000,000:
Yn =
21.9
n
4
21/10
n
= 21.9
n
.
= 1,000,000
n=
ln(1,000,000)
= 10.490.
ln(21.9 )
Thus, in about ten and a half hours, there are a million times as many bacteria in the baby formula kept at room
temperature.
84. (a) We have:
2
1
5
s5 =
1
5
=
1
5
=
1 55
·
5 25
=
11
= 0.44.
25
+
2
2
5
+
2
3
5
+
1
4
9
16
25
+
+
+
+
25
25
25
25
25
2
4
5
+
2
5
5
(b) From the pattern, we have:
sn =
=
1
n
2
1
n
2
n
1X i 2
.
n
n
i=1
This can also be written as sn =
+
n
X
1 i 2
i=1
n
n
.
2
n
+
2
3
n
+ ··· +
2
n
n
922
Chapter Nine /SOLUTIONS
(c) We think of sn as a right-hand Riemann sum by writing it as
n
X
i 2
sn =
i=1
|n
{z }
n
·
f (xi )
X
1
=
f (xi ) ∆x,
n
|{z}
i=1
∆x
where f (x) = x2 , ∆x = (1 − 0)/n = 1/n, and where x1 , x2 , . . . , xn is given by 1/n, 2/n, . . . , 1. We see that
x1 , x2 , . . . are evenly-spaced points on the interval from a = 0 to b = 1, separated by gaps of ∆x, as in any standard
right-hand sum. Taking limits, we have:
lim sn = lim
n→∞
n→∞
n
X
f (xi ) ∆x =
Z
1
f (x) dx =
1
x2 dx =
0
0
i=1
Z
1 3
x
3
1
=
0
1
.
3
85. This series converges by the alternating series test, so we can use Theorem 9.9. The nth partial sum of the series is given
by
(−1)n−1
1
1
Sn = 1 − +
−··· +
,
6
120
(2n − 1)!
so the absolute value of the first term omitted is 1/(2n + 1)!. By Theorem 9.9, we know that the true value of the sum
differs from Sn by less than 1/(2n+1)!. Thus, we want to choose n large enough so that 1/(2n+1)! ≤ 0.01. Substituting
n = 2 into the expression 1/(2n + 1)! yields 1/720 which is less than 0.01, so S2 = 1 − (1/6) = 5/6 approximates the
sum to within 0.01 of the actual sum.
86. No. If the series
∞
X
(−1)n−1 an converges then, using Theorem 9.2, part 3, we have lim (−1)n−1 an = 0, which cannot
n→∞
n=1
happen if lim an 6= 0.
n→∞
87. Let bn = (3/2)n . Then 0 ≤ bn ≤ 2n , and
bn diverges since itPis a geometric series with a common ratio greater than
one. Now, let bn = (1/2)n . Then 0 ≤ bn ≤ 2n , but this time
bn converges because it is a geometric series with a
common ratio less than one. Other answers are possible.
P
P
P
P
P
bn ) −
an =
bn would converge by Theorem 9.2. Since we know that
88. If
P (an + bn ) converged, then (an + P
bn does not converge, we conclude that (an + bn ) diverges.
89. We have 0 ≤ an /n ≤ an for all n ≥ 1. Therefore, since
P
P
an converges,
P
an /n converges by the Comparison Test.
an converge, we know that limn→∞ an = 0. Thus limn→∞ (1/an ) does not exist, and it follows that
90. Since
diverges by Property 3 of Theorem 9.2.
P
(1/an )
91. There is not enoughPinformation
case, note that if
Pto determine whether or not nan converges. To 3see thatPthis is the P
an = 1/n2 , then
nan =
(1/n2 ), which
nan =
(1/n), which diverges. However, if an = 1/n then
converges.
92. We have
Pan + (an /2) = (3/2)an , so the series
series
an .
P
P
(an + an /2) converges since it is a constant multiple of the convergent
93. Since
an converges, we know that limn→∞ an = 0. Therefore, we can choose a positive integer N large enough
P so2
that |an | ≤ 1 for all n ≥ N , so we have 0 ≤ a2n P
≤ an for all n ≥ N . Thus, by Property 2 of Theorem 9.2,
an
converges by comparison with the convergent series
an .
94. The series
∞
X
1
n
n=1
diverges by Theorem 9.2 and the fact that
∞
X
1
n=1
The series
n
converges. But
n=1
1
n
=
=
∞
X
2
n=1
n
diverges.
∞
X
1
n=1
∞
X
+
n
−
1
n
∞
X
0=0
n=1
∞
X1
1
− diverges by Theorem 9.2 and the fact that
diverges.
n
n
Thus, if an = 1/n and bn = 1/n, so that
n=1
P
an and
P
bn both diverge, we see that
P
(an + bn ) may diverge.
SOLUTIONS to Review Problems for Chapter Nine
P
P
923
P
If, on the other hand, an = 1/n and bn = −1/n, so that
an and
bn both diverge, we see that (an + bn )
may converge.
P
P
P
an and
bn both diverge, we cannot tell whether
(an + bn ) converges or diverges. Thus the
Therefore, if
statement is true.
95. (a) See Figure 9.9.
C0
C1
C2
C3
Figure 9.9
(b) At the first stage, we remove one segment of length 1/3
At stage 1, length removed =
At the second stage, we remove 2 = 21 segments of length
1
3
1
3
=
1
each.
32
At stage 2, length removed =
At the third stage, we remove 4 = 22 segments of length
1
3
1
9
=
1
.
3
21
.
32
1
each.
33
At stage 3, length removed =
22
.
33
The same reasoning shows that
2n−1
1
=
3n
3
At stage n, length removed =
To find the length of all segments removed by the nth stage, we add
Total length =
1
1
+
3
3
2
3
+
1
3
2 2
3
n−1
+··· +
2
3
1
3
.
2 n−1
3
.
(c) Using part (b), we have an infinite geometric series with a = 1/3 and x = 2/3. Therefore
Total length removed =
∞
X
1 2 i
i=0
3
3
=
1/3
= 1.
1 − 2/3
Notice that even though segments of total length 1 are removed from the initial segment of length 1, there is still an
infinite number of points remaining. The remaining points form a famous fractal, called the Cantor Set.
96. Using a right-hand sum, we have
1
1
1
1
+ + + ··· + <
2
3
4
n
Z
1
n
dx
= ln n.
x
If a computer could add a million terms in one second, then it could add
60
sec
min
hour
days
terms
· 60
· 24
· 365
· 1 million
min
hour
day
year
sec
terms per year. Thus,
1+
1
1
1
+ · · · + < 1 + ln n = 1 + ln(60 · 60 · 24 · 365 · 106 ) ≈ 32.082 < 33.
2
3
n
So the sum after one year is about 32.
924
Chapter Nine /SOLUTIONS
100,000
X 1
97. We want to estimate
using left and right Riemann sum approximations to f (x) = 1/x on the interval 1 ≤
k
k=1
x ≤ 100,000. Figure 9.10 shows a left Riemann sum approximation with 99,999 terms. Since f (x) is decreasing, the left
Riemann sum overestimates the area under the curve. Figure 9.10 shows that the first term in the sum is f (1) · 1 and the
last is f (99,999) · 1, so we have
Z
100,000
1
1
dx < LHS = f (1) · 1 + f (2) · 1 + · · · + f (99,999) · 1.
x
Since f (x) = 1/x, the left Riemann sum is
LHS =
99,999
X 1
1
1
1
· 1 + · 1 +··· +
·1=
,
1
2
99,999
k
k=1
so
Z
100,000
1
99,999
X 1
1
dx <
.
x
k
k=1
Since we want the sum to go k = 100,000 rather than k = 99,999, we add 1/100,000 to both sides:
Z
100,000
1
99,999
100,000
X 1
X 1
1
1
1
dx +
<
+
=
.
x
100,000
k
100,000
k
k=1
k=1
The left Riemann sum has therefore given us an underestimate for our sum. We now use the right Riemann sum in
Figure 9.11 to get an overestimate for our sum.
1
x
1
1
x
x1
x
100,000
x2 · · ·
1
Figure 9.10
x1
x2 · · ·
x
100,000
Figure 9.11
The right Riemann sum again has 99,999 terms, but this time the sum underestimates the area under the curve.
Figure 9.11 shows that the first rectangle has area f (2) · 1 and the last f (100,000) · 1, so we have
RHS = f (2) · 1 + f (3) · 1 + · · · + f (100,000) · 1 <
Z
100,000
1
Since f (x) = 1/x, the right Riemann sum is
RHS =
100,000
X 1
1
1
1
· 1 + · 1 + ··· +
·1=
.
2
3
100,000
k
k=2
So
100,000
X 1
k=2
k
<
Z
100,000
1
1
dx.
x
Since we want the sum to start at k = 1, we add 1 to both sides:
100,000
Z 100,000
100,000
X 1
1
1
= +
<1+
dx.
k
1
k
x
1
X 1
k=1
k=2
1
dx.
x
PROJECTS FOR CHAPTER NINE
925
Putting these under- and overestimates together, we have
Z
100,000
1
Since
Z
100,000
1
Z 100,000
100,000
X 1
1
1
1
dx +
<
<1+
dx.
x
100,000
k
x
1
k=1
1
dx = ln 100,000 − ln 1 = 11.513, we have
x
100,000
11.513 <
X 1
k=1
k
< 12.513.
100,000
Therefore we have
X 1
k=1
k
≈ 12.
98. The argument is false. Property 1 of Theorem 9.2 only applies to convergent series. In addition, by the limits comparison
test with bn = 1/n2 , the series converges.
PROJECTS FOR CHAPTER NINE
1. (a) The undiluted strength is 2 mg/ml, so the concentration of a 10-fold dilution is 0.2 mg/ml and the concentration of a 100-fold dilution is 0.02 mg/ml. Since each of the first 11 steps lasts 15 minutes = 0.25 hour,
the volume infused at each step is given, in ml, by
Volume = Rate × Time = (Rate, in ml/hr) × (0.25 hr).
The dose administered is given, in mg, by
Dose = Concentration × Volume = (Concentration, in mg/ml) × (Rate, in ml/hr) × (0.25 hr).
Thus in the first step, the concentration is 0.02 mg/ml and
Dose administered = 0.02 × 2.5 × 0.25 = 0.0125 mg.
At the second step,
Dose administered = 0.02 × 5.0 × 0.25 = 0.0250 mg,
and so on. See the first 11 rows in Table 9.1.
(b) Using the spreadsheet to find the cumulative dose given in the first 11 steps, we find that 25.5875 mg have
been administered. To reach the target dose of 500 mg,
Dose at 12th step = 500 − 25.5875 = 474.4125 mg.
The time required to deliver the 12th dose is
Time =
474.4125 mg
Dose
=
= 4.633 hours = 278 minutes.
Concentration × Rate
2 × 51.2 mg/hr
The volume infused at the last step is 4.633 · 51.2 = 237.21 ml. See the last row in Table 9.1.
926
Chapter Nine /SOLUTIONS
Table 9.1
Ratio of dose administered
Concentration
Step
Solution
(mg/ml)
1
100-fold dilution
0.02
2
100-fold dilution
3
100-fold dilution
4
Rate
Time
Volume infused
Dose infused
Cumulative
(ml/hr)
(min)
per step (ml)
per step (mg)
dose (mg)
in this step to dose
administered in previous step
2.5
15
0.63
0.0125
0.0125
0.02
5.0
15
1.25
0.0250
0.0375
2
0.02
10.0
15
2.50
0.0500
0.0875
2
100-fold dilution
0.02
20.0
15
5.00
0.1000
0.1875
2
5
10-fold dilution
0.2
4.0
15
1.00
0.2000
0.3875
2
6
10-fold dilution
0.2
8.0
15
2.00
0.4000
0.7875
2
7
10-fold dilution
0.2
16.0
15
4.00
0.8000
1.5875
2
8
10-fold dilution
0.2
32.0
15
8.00
1.6000
3.1875
2
9
undiluted
2
6.4
15
1.60
3.2000
6.3875
2
10
undiluted
2
12.8
15
3.20
6.4000
12.7875
2
11
undiluted
2
25.6
15
6.40
12.8000
25.5875
2
12
undiluted
2
51.2
278
237.21
474.4125
500
(c) The dose starts at 0.0125 mg and increases by a factor of 2 at each step. Hence if Dn is the dose at step n,
we have a geometric series with terms D1 = 0.0125, D2 = 0.0125 · 2, . . ., and Dn = 0.0125 · 2n−1. Thus,
Total dose in 11 steps = 0.0125 + 0.0125 · 2 + · · · + 0.0125 · 210
(211 − 1)
= 0.0125
= 25.5875 mg.
2−1
which agrees with the value computed in the spreadsheet in part (a).
(d) The first 11 fifteen minute doses take 2 hours 45 minutes. Adding the 4.63 hours, or 4 hours and 38
minutes, for the 12th dose, we see that to administer the full target dose of 500 mg requires 7 hours 23
minutes, about 7 and a half hours.
2. (a) To show f is decreasing for x > 1, we look at f ′ (x):
f ′ (x) = n(n + 1)xn−1 − n(n + 1)xn = n(n + 1)xn−1 (1 − x).
Thus, for x > 1, we have f ′ (x) < 0, so f is decreasing. Since f (1) = 1, this means f (x) < 1 for x > 1.
Factoring xn out of f (x), we get
f (x) = (n + 1)xn − nxn+1 = xn (n + 1 − nx) < 1.
(b) We simplify the value of x
x=
(n + 1)/n
(n + 1)2
1 + 1/n
=
=
.
1 + 1/(n + 1)
(n + 2)/(n + 1)
n(n + 2)
Before substituting into xn (n + 1 − nx) < 1, we calculate
(n + 1)2
n(n + 2)
(n + 1)(n + 2) − (n + 1)2
=
n+2
n+1
(n + 1)(n + 2 − (n + 1))
=
.
=
n+2
n+2
n + 1 − nx = n + 1 − n
Thus, substituting into the inequality from part (a), xn (n + 1 − nx) < 1, gives
n+1
< 1.
xn
n+2
PROJECTS FOR CHAPTER NINE
927
(c) We want to show sn < sn+1 . Since sn = (1 + 1/n)n and sn+1 = (1 + 1/(n + 1))n+1 , using the
definition of x, we have
sn
sn+1
(1 + 1/n)n
1
·
(1 + 1/(n + 1))n (1 + 1/(n + 1))
n+1
.
= xn
n+2
=
Thus, by part (b), we have
sn
sn+1
so
= xn
n+1
n+2
< 1,
sn < sn+1 .
Thus, the sequence is increasing.
(d) Substituting x = 1 + 1/2n into the inequality from part (a) gives
n
n
n
1
1
1
1
1
1
n+1−n 1+
1−
= 1+
=
1+
< 1.
1+
2n
2n
2n
2
2
2n
Thus
(e) When we square this inequality, we get
n
1
< 2.
1+
2n
2n
1
1+
< 4,
2n
that is, for all n
s2n < 4.
Thus, the even terms are bounded above by 4. Because we have shown the sequence is increasing, for each
odd term, we have
s2n−1 < 22n < 4,
so the odd terms are also bounded above by 4. Since all terms are bounded below by 0, the sequence is
bounded.
(f) From parts (c) and (e), we know that the sequence is increasing and bounded, and therefore, by Theorem 9.1, it has a limit.
3. (a) (i) p2
(ii) There are two ways to do this. One way is to compute your opponent’s probability of winning two in
a row, which is (1 − p)2 . Then the probability that neither of you win the next points is:
1 − (Probability you win next two + Probability opponent wins next two)
= 1 − (p2 + (1 − p)2 )
= 1 − (p2 + 1 − 2p + p2 )
= 2p2 − 2p
= 2p(1 − p).
The other way to compute this is to observe either you win the first point and lose the second or vice
versa. Both have probability p(1 − p), so the probability you split the points is 2p(1 − p).
(iii)
Probability = (Probability of splitting next two) · (Probability of winning two after that)
= 2p(1 − p)p2
928
Chapter Nine /SOLUTIONS
(iv)
Probability = (Probability of winning next two) + (Probability of splitting next two,
winning two after that)
= p2 + 2p(1 − p)p2
(v) The probability is:
w = (Probability of winning first two)
+ (Probability of splitting first two)·(Probability of winning next two)
+ (Prob. of split. first two)·(Prob. of split. next two)·(Prob. of winning next two)
+ ···
= p2 + 2p(1 − p)p2 + (2p(1 − p))2 p2 + · · · .
This is an infinite geometric series with a first term of p2 and a ratio of 2p(1 − p). Therefore the
probability of winning is
p2
w=
.
1 − 2p(1 − p)
2
(0.5)
= 0.5. This is what we would expect. If you and your opponent
(vi) For p = 0.5, w = 1−2(0.5)(1−(0.5))
are equally likely to score the next point, you and your opponent are equally likely to win the next
game.
(0.6)2
For p = 0.6, w = 1−2(0.6)(0.4)
= 0.69. Here your probability of winning the next point has been
magnified to a probability 0.69 of winning the game. Thus it gives the better player an advantage to
have to win by two points, rather than the “sudden death” of winning by just one point. This makes
sense: when you have to win by two, the stronger player always gets a second chance to overcome the
weaker player’s winning the first point on a “fluke.”
(0.7)2
= 0.84. Again, the stronger player’s probability of winning is
For p = 0.7, w = 1−2(0.7)(0.3)
magnified.
(0.4)2
= 0.31. We already computed that for p = 0.6, w = 0.69. Thus
For p = 0.4, w = 1−2(0.4)(0.6)
the value for w when p = 0.4, should be the same as the probability of your opponent winning for
p = 0.6, namely 1 − 0.69 = 0.31.
(b) (i)
S = (Prob. you score first point)
+(Prob. you lose first point, your opponent loses the next,
you win the next)
+(Prob. you lose a point, opponent loses, you lose,
opponent loses, you win)
+···
= (Prob. you score first point)
+(Prob. you lose)·(Prob. opponent loses)·(Prob. you win)
+(Prob. you lose)·(Prob. opponent loses)·(Prob. you lose)
·(Prob. opponent loses)·(Prob. you win)+ · · ·
2
= p + (1 − p)(1 − q)p + ((1 − p)(1 − q)) p + · · ·
p
=
1 − (1 − p)(1 − q)
(ii) Since S is your probability of winning the next point, we can use the formula computed in part (v) of (a)
PROJECTS FOR CHAPTER NINE
929
for winning two points in a row, thereby winning the game:
w=
• When p = 0.5 and q = 0.5,
S=
S2
.
1 − 2S(1 − S)
0.5
= 0.67.
1 − (0.5)(0.5)
Therefore
(0.67)2
S2
=
= 0.80.
1 − 2S(1 − S)
1 − 2(0.67)(1 − 0.67)
• When p = 0.6 and q = 0.5,
w=
S=
0.6
(0.75)2
= 0.75 and w =
= 0.9.
1 − (0.4)(0.5)
1 − 2(0.75)(1 − 0.75)
4. (a) (i) Amount remaining = Amount taken ·x = 5 · 8x mg.
(ii) Amount remaining = Amount remaining from first dose + Amount of second dose = 5 · 8x + 5 · 7 =
5(8x + 7) mg.
(iii) Amount remaining = Sum of amounts remaining from previous doses = 5 · 8x2 + 5 · 7x + 5 · 6 =
5(8x2 + 7x + 6) mg.
(iv) Amount remaining = 5(8x7 + 7x6 + 6x5 + · · · + 2x + 1) mg.
(v) Amount remaining = 5(8x7 + 7x6 + 6x5 + · · · + 2x + 1)x mg.
(vi) Amount remaining = 5(8x7 + 7x6 + 6x5 + · · · + 2x + 1)xn mg.
(b) Notice that the sum looks like a geometric series, except that each term has been differentiated. If we
differentiate the finite geometric series
S9 = 1 + x + x2 + · · · + x8 =
1 − x9
1−x
we have
d
dS9
= 0 + 1 + 2x + · · · + 8x7 =
dx
dx
1 − x9
1−x
=
−9x8 (1 − x) − (1 − x9 )(−1)
8x9 − 9x8 + 1
=
.
(1 − x)2
(1 − x)2
Thus, we see that
8x9 − 9x8 + 1
.
(1 − x)2
(c) Let n be the number of days since the eighth dose was taken. Then
T = 8x7 + 7x6 + 6x5 + · · · + 2x + 1 =
Amount of prednisone in the body = 5(8x7 + 7x6 + 6x5 + · · · + 2x + 1)xn = 5
8x9 − 9x8 + 1 n
x mg.
(1 − x)2
If the half-life of prednisone is 24 hours, then x = 1/2 = 0.5, so
Amount of prednisone in the body after n days = 5
8(0.5)9 − 9(0.5)8 + 1
(0.5)n = 19.6094(0.5)n mg.
(1 − (0.5))2
We want this to be 3% of one tablet, that is 0.03 × 5 = 0.15 mg, so we need to solve for n:
19.6094(0.5)n = 0.15.
This leads to
(0.5)n =
so
n=
0.15
,
19.6094
ln(0.15/19.6094)
= 7.037 ≈ 7 days.
ln(0.5)
930
Chapter Nine /SOLUTIONS
(d) The amount of prednisone in the body is the sum of the amounts remaining from the previous doses. The
first dose of n tablets was (n − 1) days before, so
Number of tablets remaining from first dose = nxn−1 .
The dose of (n − 1) tablets was (n − 2) days before, so
Number of tablets remaining from second dose = (n − 1)xn−2 .
Adding similar terms, ending with the last dose of 1 table which just occurred, gives the total number of
tablets remaining:
Tn = nxn−1 + (n − 1)xn−2 + · · · + 2x + 1.
To find a closed form for the sum, we differentiate the finite geometric series
Sn+1 = 1 + x + x2 + · · · + xn =
1 − xn+1
,
1−x
giving
dSn+1
d
= 0 + 1 + 2x + · · · + nxn−1 =
dx
dx
1 − xn+1
1−x
=
−(n + 1)xn (1 − x) − (1 − xn+1 )(−1)
.
(1 − x)2
Thus, we see that
Tn = nxn−1 + (n − 1)xn−2 + · · · + 2x + 1 =
nxn+1 − (n + 1)xn + 1
.
(1 − x)2
10.1 SOLUTIONS
CHAPTER TEN
Solutions for Section 10.1
Exercises
1. Let f (x) =
1
= (1 − x)−1 . Then f (0) = 1.
1−x
f ′ (x) = 1!(1 − x)−2
′′
f (x) = 2!(1 − x)
−3
f ′ (0) = 1!,
f ′′ (0) = 2!,
f ′′′ (x) = 3!(1 − x)−4 f ′′′ (0) = 3!,
f (4) (x) = 4!(1 − x)−5 f (4) (0) = 4!,
f (5) (x) = 5!(1 − x)−6 f (5) (0) = 5!,
f (6) (x) = 6!(1 − x)−7 f (6) (0) = 6!,
f (7) (x) = 7!(1 − x)−8 f (7) (0) = 7!.
P3 (x) = 1 + x + x2 + x3 ,
P5 (x) = 1 + x + x2 + x3 + x4 + x5 ,
P7 (x) = 1 + x + x2 + x3 + x4 + x5 + x6 + x7 .
2. Let
1
= (1 + x)−1 . Then f (0) = 1.
1+x
f ′ (x) = −1!(1 + x)−2
′′
f (x) = 2!(1 + x)
−3
f ′ (0) = −1,
f ′′ (0) = 2!,
f ′′′ (x) = −3!(1 + x)−4 f ′′′ (0) = −3!,
f (4) (x) = 4!(1 + x)−5
f (4) (0) = 4!,
f (5) (x) = −5!(1 + x)−6 f (5) (0) = −5!,
f (6) (x) = 6!(1 + x)−7
f
(7)
(x) = −7!(1 + x)
−8
f (8) (x) = 8!(1 + x)−9
f (6) (0) = 6!,
f (7) (0) = −7!,
f (8) (0) = 8!.
P4 (x) = 1 − x + x2 − x3 + x4 ,
P6 (x) = 1 − x + x2 − x3 + x4 − x5 + x6 ,
P8 (x) = 1 − x + x2 − x3 + x4 − x5 + x6 − x7 + x8 .
3. Let f (x) =
√
1 + x = (1 + x)1/2 . Then f (0) = 1, and
f ′ (x) = 21 (1 + x)−1/2
f ′ (0) = 21 ,
f ′′ (x) = − 14 (1 + x)−3/2
f ′′ (0) = − 41 ,
′′′
f (x) =
f (4) (x) =
3
(1 + x)−5/2
8
− 15
(1 + x)−7/2
16
f ′′′ (0) = 83 ,
15
f (4) (0) = − 16
.
931
932
Chapter Ten /SOLUTIONS
Thus,
1
x−
2
1
P3 (x) = 1 + x −
2
1
P4 (x) = 1 + x −
2
P2 (x) = 1 +
4. Let f (x) =
√
3
1 2
x ,
8
1 2
x +
8
1 2
x +
8
1 3
x ,
16
1 3
5 4
x −
x .
16
128
1 − x = (1 − x)1/3 . Then f (0) = 1, and
f ′ (x) = − 31 (1 − x)−2/3
′′
f (x) =
f ′′′ (x) =
f
(4)
(x) =
− 322 (1
− 3103 (1
− 3804 (1
f ′ (0) = − 31 ,
− x)
−5/3
− x)
−11/3
− x)−8/3
f ′′ (0) = − 322 ,
f ′′′ (0) = − 3103 ,
f (4) (0) = − 3804 .
Then,
1 2 2
1
1
1
x−
x = 1 − x − x2 ,
3
2! 32
3
9
1
5 3
1 10
1
3
P3 (x) = P2 (x) −
x = 1 − x − x2 −
x ,
3! 33
3
9
81
1
1
5 3
10 4
1 80 4
x = 1 − x − x2 −
x −
x .
P4 (x) = P3 (x) −
4! 34
3
9
81
243
P2 (x) = 1 −
5. Let f (x) = cos x. Then f (0) = cos(0) = 1, and
f ′ (x) = − sin x
f ′′ (x) = − cos x
′′′
f ′ (0) = 0,
f ′′ (0) = −1,
f (x) = sin x
f ′′′ (0) = 0,
f (4) (x) = cos x
f (4) (0) = 1,
f
(5)
(x) = − sin x f (5) (0) = 0,
f (6) (x) = − cos x f (6) (0) = −1.
Thus,
x2
,
2!
2
x
P4 (x) = 1 −
+
2!
x2
+
P6 (x) = 1 −
2!
P2 (x) = 1 −
x4
,
4!
x4
x6
−
.
4!
6!
6. Let f (x) = ln(1 + x). Then f (0) = ln 1 = 0, and
f ′ (x) = (1 + x)−1
f ′ (0) = 1,
f ′′ (x) = (−1)(1 + x)−2
f ′′ (0) = −1,
′′′
f (x) = 2(1 + x)
−3
f ′′′ (0) = 2,
f (4) (x) = −3!(1 + x)−4
f (4) (0) = −3!,
f (6) (x) = −5!(1 + x)−6
f (6) (0) = −5!,
f (8) (x) = −7!(1 + x)−8
f (8) (0) = −7!,
f
f
f
(5)
(7)
(9)
(x) = 4!(1 + x)
(x) = 6!(1 + x)
(x) = 8!(1 + x)
−5
−7
−9
f (5) (0) = 4!,
f (7) (0) = 6!,
f (9) (0) = 8!.
10.1 SOLUTIONS
So,
x3
x4
x5
x2
+
−
+
,
2
3
4
5
3
4
5
2
x
x
x
x
+
−
+
−
P7 (x) = x −
2
3
4
5
2
3
4
5
x
x
x
x
P9 (x) = x −
+
−
+
−
2
3
4
5
P5 (x) = x −
x6
x7
+
,
6
7
6
7
x
x
x8
x9
+
−
+
.
6
7
8
9
7. Let f (x) = arctan x. Then f (0) = arctan 0 = 0, and
f ′ (x) = 1/(1 + x2 ) = (1 + x2 )−1
′′
2 −2
f (x) = (−1)(1 + x )
f ′ (0) = 1,
f ′′ (0) = 0,
2x
f ′′′ (x) = 2!(1 + x2 )−3 22 x2 + (−1)(1 + x2 )−2 2
f
(4)
2 −4 3 3
(x) = −3!(1 + x )
2 −3 3
2 x + 2!(1 + x )
2 x
2 −3 2
+ 2!(1 + x )
f ′′′ (0) = −2,
f (4) (0) = 0.
2 x
Therefore,
1 3
x .
3
P3 (x) = P4 (x) = x −
8. Let f (x) = tan x. So f (0) = tan 0 = 0, and
f ′ (x) = 1/ cos2 x
f ′ (0) = 1,
′′
f (x) = 2 sin x/ cos3 x
′′′
f ′′ (0) = 0,
2
2
4
f (x) = (2/ cos x) + (6 sin x/ cos x)
f
(4)
(x) = (16 sin x/ cos3 x) + (24 sin3 x/ cos5 x) f (4) (0) = 0.
Thus,
P3 (x) = P4 (x) = x +
9. Let f (x) = √
f ′′′ (0) = 2,
x3
.
3
1
= (1 + x)−1/2 . Then f (0) = 1.
1+x
f ′ (x) = − 21 (1 + x)−3/2
f ′′ (x) =
′′′
f (x) =
f
(4)
(x) =
3
(1 + x)−5/2
22
3·5
− 23 (1 + x)−7/2
3·5·7
(1 + x)−9/2
24
f ′ (0) = − 21 ,
f ′′ (0) =
′′′
f (0) =
f (4) (0) =
3
,
22
− 3·5
,
23
3·5·7
24
Then,
1 3 2
1
3
1
x+
x = 1 − x + x2 ,
2
2! 22
2
8
1 3·5 3
1
3
5 3
P3 (x) = P2 (x) −
x = 1 − x + x2 −
x ,
3! 23
2
8
16
1 3·5·7 4
1
3
5 3
35 4
P4 (x) = P3 (x) +
x = 1 − x + x2 −
x +
x .
4! 24
2
8
16
128
P2 (x) = 1 −
10. Let f (x) = (1 + x)p .
(a) Suppose that p = 0. Then f (x) = 1 and f (k) (x) = 0 for any k ≥ 1. Thus P2 (x) = P3 (x) = P4 (x) = 1.
(b) If p = 1 then f (x) = 1 + x, so
f (0) = 1,
f ′ (x) = 1,
f
Thus P2 (x) = P3 (x) = P4 (x) = 1 + x.
(k)
(x) = 0
k ≥ 2.
933
934
Chapter Ten /SOLUTIONS
(c) In general:
f (x) = (1 + x)p ,
f ′ (x) = p(1 + x)p−1 ,
f ′′ (x) = p(p − 1)(1 + x)p−2 ,
f ′′′ (x) = p(p − 1)(p − 2)(1 + x)p−3 ,
f (4) (x) = p(p − 1)(p − 2)(p − 3)(1 + x)p−4 .
f (0) = 1,
f ′ (0) = p,
f ′′ (0) = p(p − 1),
f ′′′ (0) = p(p − 1)(p − 2),
f (4) (0) = p(p − 1)(p − 2)(p − 3).
p(p − 1) 2
x ,
2
p(p − 1) 2 p(p − 1)(p − 2) 3
x +
x ,
P3 (x) = 1 + px +
2
6
p(p − 1) 2 p(p − 1)(p − 2) 3
P4 (x) = 1 + px +
x +
x
2
6
p(p − 1)(p − 2)(p − 3) 4
+
x .
24
P2 (x) = 1 + px +
√
11. Let f (x) = 1 − x = (1 − x)1/2 . Then f ′ (x) = − 21 (1 − x)−1/2 , f ′′ (x) = − 41 (1 − x)−3/2 , f ′′′ (x) = − 83 (1 − x)−5/2 .
So f (0) = 1, f ′ (0) = − 12 , f ′′ (0) = − 41 , f ′′′ (0) = − 38 , and
1
1 1 2 3 1 3
x−
x −
x
2
4 2!
8 3!
x
x2
x3
= 1− −
−
.
2
8
16
P3 (x) = 1 −
12. Let f (x) = ex . Since f (k) (x) = ex = f (x) for all k ≥ 1, the Taylor polynomial of degree 4 for f (x) = ex about x = 1
is
e1
e1
e1
(x − 1)2 + (x − 1)3 + (x − 1)4
2!
3!
4!
i
h
1
1
1
2
3
(x − 1)4 .
= e 1 + (x − 1) + (x − 1) + (x − 1) +
2
6
24
P4 (x) = e1 + e1 (x − 1) +
1
= (1 + x)−1 . Then f ′ (x) = −(1 + x)−2 , f ′′ (x) = 2(1 + x)−3 , f ′′′ (x) = −6(1 + x)−4 ,
1+x
f (4) (x) = 24(1 + x)−5 . So f (2) = 31 , f ′ (2) = − 312 , f ′′ (2) = 323 , f ′′′ (2) = − 364 , and f (4) (2) = 3245 . Therefore,
13. Let f (x) =
1
2 1
6 1
24 1
1
− (x − 2) + 3 (x − 2)2 − 4 (x − 2)3 + 5 (x − 2)4
3 32
3 2!
3 3!
3 4!
1
(x − 2)2
(x − 2)3
(x − 2)4
x−2
=
+
−
+
1−
.
3
3
32
33
34
P4 (x) =
14. Let f (x) = cos x. f ( π2 ) = 0.
f ′ (x) = − sin x
f ′′ (x) = − cos x
′′′
f ′ ( π2 ) = −1,
f ′′ ( π2 ) = 0,
f (x) = sin x
f ′′′ ( π2 ) = 1,
f (4) (x) = cos x
f (4) ( π2 ) = 0.
10.1 SOLUTIONS
935
So,
1
π
x−
3!
2
π
1
π 3
.
= − x−
+
x−
2
3!
2
P4 (x) = 0 − x −
π
2
+0+
3
15. Let f (x) = sin x.
Then f ′ (x) = cos x, f ′′ (x) = − sin x, and f ′′′ (x) = − cos x, so the Taylor polynomial for sin x of degree three about
x = −π/4 is
π
π
π
+ cos −
x+
P3 (x) = sin −
4
4
4
π
− sin − π4
π 2 − cos − 4
π 3
+
x+
x+
+
2!
4
3!
4
√
1
π
π 2 1
π 3
2
−1 + x +
+
x+
x+
.
=
−
2
4
2
4
6
4
16. Let f (x) = ln(x2 ). Then ln(12 ) = ln 1 = 0.
Then f ′ (x) = 2x−1 , f ′′ (x) = −2x−2 , f ′′′ (x) = 4x−3 , and f (4) (x) = −12x−4 .
The Taylor polynomial of degree 4 for f (x) = ln(x2 ) about x = 1 is
4 · 1−3
−12 · 1−4
−2 · 1−2
(x − 1)2 +
(x − 1)3 +
(x − 1)4
2!
3!
4!
4
12
= 0 + 2(x − 1) − (x − 1)2 + (x − 1)3 −
(x − 1)4
6
24
2
1
= 2(x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 .
3
2
P4 (x) = ln(12 ) + 2 · 1−1 (x − 1) +
Problems
17. The third degree Taylor polynomial of f (x) will have the same terms as the seventh degree polynomial but only up to the
x3 term. So the third degree Taylor polynomial of f (x) is given by
P3 (x) = 1 −
5x2
x
+
+ 8x3 .
3
7
18. Using the fact that
f ′′ (0) 2 f ′′′ (0) 3
x +
x
2!
3!
and identifying coefficients with those given for P3 (x), we obtain the following:
f (x) ≈ P3 (x) = f (0) + f ′ (0)x +
(a) f (0) = constant term which equals 2,
so f (0) = 2.
(b) f ′ (0) = coefficient of x which equals −1,
so f ′ (0) = −1.
f ′′ (0)
2
(c) 2! = coefficient of x which equals −1/3,
so f ′′ (0) = −2/3.
(d)
f ′′′ (0)
3!
= coefficient of x3 which equals 2,
so f ′′′ (0) = 12.
19.
f (x) = 4x2 − 7x + 2
′
f (x) = 8x − 7
′′
f (x) = 8
f (0) = 2
f ′ (0) = −7
f ′′ (0) = 8,
so P2 (x) = 2 + (−7)x + 28 x2 = 4x2 − 7x + 2. We notice that f (x) = P2 (x) in this case.
20. f ′ (x) = 3x2 + 14x − 5, f ′′ (x) = 6x + 14, f ′′′ (x) = 6. Thus, about a = 0,
−5
14 2
6
x+
x + x3
1!
2!
3!
= 1 − 5x + 7x2 + x3
P3 (x) = 1 +
= f (x).
936
Chapter Ten /SOLUTIONS
21. (a) We’ll make the following conjecture:
“If f (x) is a polynomial of degree n, i.e.
f (x) = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 + an xn ,
then Pn (x), the nth degree Taylor polynomial for f (x) about x = 0, is f (x) itself.”
(b) All we need to do is to calculate Pn (x), the nth degree Taylor polynomial for f about x = 0 and see if it is the same
as f (x).
f (0) = a0 ;
f ′ (0) = (a1 + 2a2 x + · · · + nan xn−1 )
x=0
= a1 ;
f ′′ (0) = (2a2 + 3 · 2a3 x + · · · + n(n − 1)an xn−2 )
x=0
= 2!a2 .
If we continue doing this, we’ll see in general
f (k) (0) = k!ak ,
k = 1, 2, 3, · · · , n.
Therefore,
f ′′ (0) 2
f (n) (0) n
f ′ (0)
x+
x + ··· +
x
1!
2!
n!
2
n
= a0 + a1 x + a2 x + · · · + an x
Pn (x) = f (0) +
= f (x).
22. Since the coefficient of (x − 1)5 term of p(x) is given by
C5 =
f (5) (1)
,
5!
we know that f (5) (1) = 5!C5 . Note that
p(x) =
10
X
(x − 1)n
n=0
n!
= 1 + (x − 1) +
(x − 1)2
(x − 1)3
(x − 1)4
(x − 1)5
(x − 1)10
+
+
+
+ ··· +
,
2!
3!
4!
5!
10!
so C5 = 1/5!. Therefore
f (5) (1) =
1
5! = 1.
5!
23. Referring to the table, we have:
f ′′ (0) 2 f ′′′ (0) 3 f (4) (0) 4 f (5) (0) 5
x +
x +
x +
x
2!
3!
4!
5!
−2 2
0
−1 4
4
= −3 + 5x +
x + x3 +
x + x5
2!
3!
4!
5!
1 4
1 5
= −3 + 5x − x2 −
x +
x .
24
30
P5 (x) = f (0) + f ′ (0)x +
24. Referring to the formula for f (n) (0), we have:
f ′ (0) = f (1) (0) = −(−2)1 = 2
f ′′ (0) = f (2) (0) = −(−2)2 = −4
f ′′′ (0) = f (3) (0) = −(−2)3 = 8
f (4) (0) = −(−2)4 = −16
f (5) (0) = −(−2)5 = 32.
10.1 SOLUTIONS
937
Since f (0) = −1, we have
f ′′ (0) 2 f ′′′ (0) 3 f (4) 4 f (5) (0) 5
x +
x +
x +
x
2!
3!
4!
5!
−4 2
8
−16 4 32 5
= −1 + 2x +
x + x3 +
x +
x
2!
3!
4!
5!
2
4 5
4
x .
= −1 + 2x − 2x2 + x3 − x4 +
3
3
15
P5 (x) = f (0) + f ′ (0)x +
25. Since P2 (x) is the second degree Taylor polynomial for f (x) about x = 0, P2 (0) = f (0), which says a = f (0). Since
d
P2 (x)
dx
= f ′ (0),
x=0
b = f ′ (0); and since
d2
P2 (x)
dx2
= f ′′ (0),
x=0
2c = f ′′ (0). In other words, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells us
the concavity of f (x) near x = 0. So c < 0 since f is concave down; b > 0 since f is increasing; a > 0 since f (0) > 0.
26. As we can see from Problem 25, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells
us the concavity of f (x) near x = 0.
So a > 0, b < 0 and c < 0.
27. As we can see from Problem 25, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells
us the concavity of f (x) near x = 0.
So a < 0, b > 0 and c > 0.
28. As we can see from Problem 25, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells
us the concavity of f (x) near x = 0.
So a < 0, b < 0 and c > 0.
29.
3
lim
x→0
x − x3!
x2
sin x
= lim
= lim 1 −
x
x→0
x
x→0
3!
= 1.
30.
2
1 − (1 − x2! +
1 − cos x
=
lim
lim
x→0
x→0
x2
x2
31. For f (h) = eh , P4 (h) = 1 + h +
x4
)
4!
= lim
x→0
1
x2
−
2
4!
h3
h4
h2
+
+
. So
2
3!
4!
(a)
lim
h→0
P4 (h) − 1 − h
eh − 1 − h
= lim
h→0
h2
h2
= lim
h2
2
h→0
+
h3
3!
h2
+
h4
4!
h
h2
1
+
+
2
3!
4!
P4 (h) − 1 − h −
= lim
h→0
h3
h2
2
= lim
h→0
=
1
.
2
(b)
eh − 1 − h −
lim
h→0
h3
h2
2
=
1
.
2
938
Chapter Ten /SOLUTIONS
= lim
h3
3!
h→0
=
4
+ h4!
1
h
= lim
+
3
h→0 3!
h
4!
1
1
= .
3!
6
Using Taylor polynomials of higher degree would not have changed the results since the terms with higher powers of h
all go to zero as h → 0.
32. (a) We use the Taylor polynomial of degree two for f and h about x = 2.
f ′′ (2)
3
(x − 2)2 = (x − 2)2
2!
2
h′′ (2)
7
h(x) ≈ h(2) + h′ (2)(x − 2) +
(x − 2)2 = (x − 2)2
2!
2
f (x) ≈ f (2) + f ′ (2)(x − 2) +
Thus, using the fact that near x = 2 we can approximate a function by Taylor polynomials
lim
x→2
3
(x − 2)2
f (x)
3
= .
= lim 27
x→2 (x − 2)2
h(x)
7
2
(b) We use the Taylor polynomial of degree two for f and g about x = 2.
f ′′ (2)
3
(x − 2)2 = (x − 2)2
2!
2
g ′′ (2)
5
g(x) ≈ g(2) + g ′ (2)(x − 2) +
(x − 2)2 = 22(x − 2) + (x − 2)2 .
2!
2
f (x) ≈ f (2) + f ′ (2)(x − 2) +
Thus,
f (x)
= lim
lim
x→2
x→2 g(x)
− 2)2
22(x − 2) + 5(x − 2)2
3
(x
2
= lim
x→2
3
(x
2
− 2)
22 + 5(x − 2)
=
0
= 0.
22
33. (a) Since the coefficient of the x-term of each f is 1, we know f1′ (0) = f2′ (0) = f3′ (0) = 1. Thus, each of the f s slopes
upward near 0, and are in the second figure.
The coefficient of the x-term in g1 and in g2 is 1, so g1′ (0) = g2′ (0) = 1. For g3 however, g3′ (0) = −1. Thus,
g1 and g2 slope up near 0, but g3 slopes down. The gs are in the first figure.
(b) Since g1 (0) = g2 (0) = g3 (0) = 1, the point A is (0, 1).
Since f1 (0) = f2 (0) = f3 (0) = 2, the point B is (0, 2).
(c) Since g3 slopes down, g3 is I. Since the coefficient of x2 for g1 is 2, we know
g1′′ (0)
=2
2!
so
g1′′ (0) = 4.
By similar reasoning g2′′ (0) = 2. Since g1 and g2 are concave up, and g1 has a larger second derivative, g1 is III and
g2 is II.
Calculating the second derivatives of the f s from the coefficients x2 , we find
f1′′ (0) = 4
f2′′ (0) = −2
f3′′ (0) = 2.
Thus, f1 and f3 are concave up, with f1 having the larger second derivative, so f1 is III and f3 is II. Then f2 is
concave down and is I.
34. Let f (x) be a function that has derivatives up to order n at x = a. Let
Pn (x) = C0 + C1 (x − a) + · · · + Cn (x − a)n
be the polynomial of degree n that approximates f (x) about x = a. We require that Pn (x) and all of its first n derivatives
agree with those of the function f (x) at x = a, i.e., we want
f (a) = Pn (a),
f ′ (a) = Pn′ (a),
f ′′ (a) = Pn′′ (a),
..
.
f (n) (a) = Pn(n) (a).
10.1 SOLUTIONS
939
When we substitute x = a in Pn (x), all the terms except the first drop out, so
f (a) = C0 .
Now differentiate Pn (x):
Pn′ (x) = C1 + 2C2 (x − a) + 3C3 (x − a)2 + · · · + nCn (x − a)n−1 .
Substitute x = a again, which yields
f ′ (a) = Pn′ (a) = C1 .
Differentiate Pn′ (x):
Pn′′ (x) = 2C2 + 3 · 2C3 (x − a) + · · · + n(n − 1)Cn (x − a)n−2
and substitute x = a again:
f ′′ (a) = Pn′′ (a) = 2C2 .
Differentiating and substituting again gives
f ′′′ (a) = Pn′′′ (a) = 3 · 2C3 .
Similarly,
f (k) (a) = Pn (k) (a) = k!Ck .
′′′
f ′′ (a)
So, C0 = f (a), C1 = f ′ (a), C2 = 2! , C3 = f 3!(a) , and so on.
If we adopt the convention that f (0) (a) = f (a) and 0! = 1, then
Ck =
f (k) (a)
, k = 0, 1, 2, · · · , n.
k!
Therefore,
f (x) ≈ Pn (x) = C0 + C1 (x − a) + C2 (x − a)2 · · · + Cn (x − a)n
= f (a) + f ′ (a)(x − a) +
f (n) (a)
f ′′ (a)
(x − a)2 + · · · +
(x − a)n .
2!
n!
35. (a) The first degree Taylor polynomials P (x) and Q(x) for f (x) and g(x) near x = 0 and their product are given by
P (x) = 1 + x
Q(x) = 1 + 2x
P (x)Q(x) = 1 + 3x + 2x2 .
(b) The Taylor polynomial R(x) of degree 2 of h(x) = 1/ (1 − x)(1 − 2x) near x = 0 is
R(x) = 1 + 3x + 7x2 .
(c) The two polynomials R(x) and P (x)Q(x) are not the same. The product of Taylor polynomials of two functions is
usually not a Taylor polynomial of the product.
36. (a) The first degree Taylor polynomials P (x) and Q(x) for f (x) and g(x) near x = 0 and their product are given by
P (x) = f (0) + xf ′ (0)
Q(x) = g(0) + xg ′ (0)
P (x)Q(x) = f (0)g(0) + (f ′ (0)g(0) + f (0)g ′ (0))x + f ′ (0)g ′ (0)x2 .
(b) The Taylor polynomial R(x) of degree 2 of h(x) = f (x)g(x) near x = 0 is
R(x) = f (0)g(0) + (f ′ (0)g(0) + f (0)g ′ (0))x + (f ′′ (0)g(0) + 2f ′ (0)g ′ (0) + f (0)g ′′ (0))x2 /2.
(c) The two polynomials R(x) and P (x)Q(x) are the same if the coefficients of x2 are identical, that is
f ′ (0)g ′ (0) = (f ′′ (0)g(0) + 2f ′ (0)g ′ (0) + f (0)g ′′ (0))/2,
or
f ′′ (0)g(0) + f (0)g ′′ (0) = 0.
940
Chapter Ten /SOLUTIONS
2
37. (a) f (x) = ex .
2
2
2
f ′ (x) = 2xex , f ′′ (x) = 2(1 + 2x2 )ex , f ′′′ (x) = 4(3x + 2x3 )ex ,
2
2
f (4) (x) = 4(3 + 6x2 )ex + 4(3x + 2x3 )2xex .
The Taylor polynomial about x = 0 is
2
0
12 4
0
x + x2 + x3 +
x
1!
2!
3!
4!
1
= 1 + x2 + x4 .
2
P4 (x) = 1 +
(b) f (x) = ex . The Taylor polynomial of degree 2 is
Q2 (x) = 1 +
x
x2
1
+
= 1 + x + x2 .
1!
2!
2
If we substitute x2 for x in the Taylor polynomial for ex of degree 2, we will get P4 (x), the Taylor polynomial for
2
ex of degree 4:
1 2 2
(x )
2
1
= 1 + x2 + x4
2
= P4 (x).
Q2 (x2 ) = 1 + x2 +
(c) Let Q10 (x) = 1 +
x2
x10
x
+
+ ··· +
be the Taylor polynomial of degree 10 for ex about x = 0. Then
1!
2!
10!
P20 (x) = Q10 (x2 )
(x2 )2
(x2 )10
x2
+
+··· +
1!
2!
10!
x2
x4
x20
= 1+
+
+ ··· +
.
1!
2!
10!
= 1+
(d) Let ex ≈ Q5 (x) = 1 +
x
1!
+··· +
x5
.
5!
Then
e−2x ≈ Q5 (−2x)
(−2x)2
(−2x)3
(−2x)4
(−2x)5
−2x
+
+
+
+
1!
2!
3!
4!
5!
4 3 2 4
4 5
2
= 1 − 2x + 2x − x + x −
x .
3
3
15
= 1+
3
38. (a)
t − t3!
t2
sin t
≈
=1−
t
t
6
Z
0
3
(b)
t − t3! +
sin t
≈
t
t
t5
5!
Z
0
=1−
1
1
1
dt = t −
t3
18
t4
t2
+
1−
6
120
dt = t −
t5
t3
+
18
600
Z
0
1
t2
6
sin t
dt ≈
t
1−
0
= 0.94444 · · ·
t2
t4
+
6
120
sin t
dt ≈
t
Z
0
1
1
0
= 0.94611 · · ·
39. (a) The equation sin x = 0.2 has one solution near x = 0 and infinitely many others, one near each multiple of π. See
x3
= 0.2 has three solutions, one near x = 0 and two others. See Figure 10.2.
Figure 10.1. The equation x −
3!
10.1 SOLUTIONS
y
941
y
y = 0.2
x
y = 0.2
x
Figure 10.2: Graph of y = x −
Figure 10.1: Graph of y = sin x and y = 0.2
x3
3!
and y = 0.2
(b) Near x = 0, the cubic Taylor polynomial x − x3 /3! ≈ sin x. Thus, the solutions to the two equations near x = 0 are
approximately equal. The other solutions are not close. The reason is that x − x3 /3! only approximates sin x near
x = 0 but not further away. See Figure 10.3.
y
sin x
y = 0.2
x
x − x3 /3!
Figure 10.3
40. Changing sin θ into θ makes sense if the two values are almost equal. If we measure θ in radians, this is true for values
of θ close to zero. (Recall the first degree Taylor polynomial: sin θ ≈ θ.) In other words, the switch is justified when the
pendulum does not swing very far from the vertical.
41. (a) The graphs of y = cos x and y = 1 − 0.1x cross at x = 0 and for another x-value just to the right of x = 0. (There
are other crossings much further to the right.)
(b) Since
x2
cos x ≈ 1 −
2
the equation becomes
1−
x2
= 1 − 0.1x
2
x2
= 0.1x
2
x = 0, 0.2.
The solution x = 0 is an exact solution to the original equation; x = 0.2 is an approximate solution to the original
equation.
Strengthen Your Understanding
42. The constant term of the Taylor polynomial is f (0), hence substituting 0 into f (x) gives ln(2) which is positive.
43. The coefficient of the x term is given by f ′ (0) and f ′ (0) = 1.
x
44. An example is f (x) = sin x. Another example is f (x) =
. Many other examples are possible.
1 + x2
3
45. An example is p(x) = 1 + 3(x − 1) + (x − 1) . Another example is p(x) = 5 + 3(x − 1) + 2(x − 1)2 + (x − 1)3 .
Many other examples are possible.
46. False. For example, both f (x) = x2 and g(x) = x2 + x3 have P2 (x) = x2 .
942
Chapter Ten /SOLUTIONS
47. False. The approximation sin θ ≈ θ − θ3 /3! holds for θ in radians, not degrees.
48. False. P2 (x) = f (5) + f ′ (5)(x − 5) + (f ′′ (5)/2)(x − 5)2 = e5 + e5 (x − 5) + (e5 /2)(x − 5)2 .
49. False. Since −1 is the coefficient of x2 in P2 (x), we know that f ′′ (0)/2! = −1, so f ′′ (0) < 0, which implies that f is
concave down near x = 0.
50. False. For example the quadratic approximation to cos x for x near 0 is 1 − x2 /2, whereas the linear approximation
is the constant function 1. Although the quadratic approximation is better near 0, for large values of x it takes large
negative values, whereas the linear approximation stays equal to 1. Since cos x oscillates between 1 and −1, the linear
approximation is better than the quadratic for large x (although it is not very good).
51. False. For example, if a = 0 and f (x) = cos x, then P1 (x) = 1, and P1 (x) touches cos x at x = 0, 2π, 4π, . . ..
52. False. Since f (−1) = g(−1) the graphs of f and g intersect at x = −1. Since f ′ (−1) < g ′ (−1), the slope of f is less
than the slope of g at x = −1. Thus f (x) > g(x) for all x sufficiently close to −1 on the left, and f (x) < g(x) for all x
sufficiently close to −1 on the right.
53. True. If
f ′′ (−1)
(x + 1)2
2
g ′′ (−1)
Q2 (x) = Quadratic approximation to g = g(−1) + g ′ (−1)(x + 1) +
(x + 1)2
2
P2 (x) = Quadratic approximation to f = f (−1) + f ′ (−1)(x + 1) +
then P2 (x) − Q2 (x) = (f ′′ (−1) − g ′′ (−1))(x + 1)2 /2 < 0 for all x 6= −1. Thus P2 (x) < Q2 (x) for all x 6= −1. This
implies that for x sufficiently close to −1 (but not equal to −1), we have f (x) < g(x).
Solutions for Section 10.2
Exercises
1. Differentiating (1 + x)3/2 :
f (x) = (1 + x)3/2
f (0) = 1,
f ′ (x) = (3/2)(1 + x)1/2
f ′ (0) = 23 ,
f ′′ (x) = (1/2)(3/2)(1 + x)−1/2 = (3/4)(1 + x)−1/2
′′′
f (x) = (−1/2)(3/4)(1 + x)
−3/2
= (−3/8)(1 + x)
f ′′ (0) = 34 ,
−3/2
f ′′′ (0) = − 38 .
(3/4)x2
(−3/8)x3
3
·x+
+
+ ···
2
2!
3!
2
3
3x
3x
x
= 1+
+
−
+ ···
2
8
16
f (x) = (1 + x)3/2 = 1 +
2. Differentiating
√
4
x + 1:
f (x) =
√
4
x + 1 = (x + 1)1/4
f (0) = 1,
′
f (x) = (1/4)(x + 1)−3/4
f ′ (0) = 41 ,
f ′′ (x) = (−3/4)(1/4)(x + 1)−7/4 = (−3/16)(x + 1)−7/4
′′′
f (x) = (−7/4)(−3/16)(x + 1)
f (x) =
√
4
−11/4
= (21/64)(x + 1)
(−3/16)x2
(21/64)x3
1
·x+
+
+···
4
2!
3!
2
3
x
3x
7x
= 1+ −
+
− ···.
4
32
128
x+1 = 1+
−11/4
3
f ′′ (0) = − 16
,
f ′′′ (0) =
21
.
64
10.2 SOLUTIONS
943
3. Differentiating sin(−x):
f (x) = sin(−x)
f (0) = 0,
′
f ′ (0) = −1,
f (x) = cos(−x)(−1) = − cos(−x)
f ′′ (x) = −(− sin(−x))(−1) = − sin(−x)
f ′′ (0) = 0,
′′′
f (x) = − cos(−x)(−1) = cos(−x)
f
(4)
f ′′′ (0) = 1
f (4) (0) = 0,
(x) = − sin(−x)(−1) = sin(−x)
f (5) (x) = cos(−x)(−1) = − cos(−x)
f (5) (0) = −1,
f (7) (x) = − cos(−x)(−1) = cos(−x)
f (7) (0) = 1.
f
(6)
(x) = −(− sin(−x))(−1) = − sin(−x) f (6) (0) = 0,
1x3
0x4
−1x5
0x6
1x7
0x2
+
+
+
+
+
+ ···
2!
3!
4!
5!
6!
7!
5
7
3
x
x
x
−
+
+ ···.
= −x +
3!
5!
7!
f (x) = sin(−x) = 0 − 1 · x +
Notice that the series for sin(−x) is obtained from the series for sin x by changing the signs. This is expected since
sin(−x) = − sin x.
4. Differentiating ln(1 − x)
f (x) = ln(1 − x)
′
f (x) =
1
(−1)
1−x
f (0) = 0,
= −(1 − x)
−1
f ′ (0) = −1,
f ′′ (x) = −(−(1 − x)−2 )(−1) = −(1 − x)−2
′′′
f (x) = −2(−(1 − x)
−3
)(−1) = −2(1 − x)
f ′′ (0) = −1,
−3
f ′′′ (0) = −2
f (4) (x) = −3(−2(1 − x)−4 )(−1) = −6(1 − x)−4 f (4) (0) = −6.
(−2)x3
(−6)x4
(−1)x2
+
+
+ ···
2!
3!
4!
x2
x3
x4
= −x −
−
−
+ ···.
2
3
4
f (x) = ln(1 − x) = 0 − 1 · x +
5.
f (x) =
1
1−x
′
= (1 − x)−1
f (0) = 1,
f (x) = −(1 − x)−2 (−1) = (1 − x)−2
f ′ (0) = 1,
f ′′ (x) = −2(1 − x)−3 (−1) = 2(1 − x)−3 f ′′ (0) = 2,
f ′′′ (x) = −6(1 − x)−4 (−1) = 6(1 − x)−4 f ′′′ (0) = 6.
f (x) =
2x2
6x3
1
= 1+1·x+
+
+···
1−x
2!
3!
= 1 + x + x2 + x3 + · · ·
6.
f (x) =
′
f (x) =
f ′′ (x) =
′′′
f (x) =
f (x) = √
1
√1
= (1 + x)− 2
1+x
3
− 12 (1 + x)− 2
5
3
(1 + x)− 2
4
7
(1 + x)− 2
− 15
8
f (0) = 1
f ′ (0) = − 21
f ′′ (0) =
3
4
′′′
f (0) = − 15
8
1
( 3 )x2
(− 15
)x3
1
8
x+ 4
+
+···
= 1+ −
2
2!
3!
1+x
x
3x2
5x3
= 1− +
−
+ ···
2
8
16
944
Chapter Ten /SOLUTIONS
7.
f (y) =
√
3
1
1 − y = (1 − y) 3
2
2
f (0) = 1
5
5
f ′ (0) = − 13
f ′ (y) = 13 (1 − y)− 3 (−1) = − 31 (1 − y)− 3
f ′′ (y) = 29 (1 − y)− 3 (−1) = − 92 (1 − y)− 3
′′′
f (y) =
f (y) =
10
(1
27
− y)
−8
3
(−1) =
10
− 27
(1
− y)
10
f (0) = − 27
10
(− 27
(− 92 )y 2
)y 3
1
+
+ ···
y+
3
2!
3!
y
y2
5y 3
= 1− −
−
− ···
3
9
81
1−y = 1+ −
f (x) = sin x
′
f (x) = cos x
f ′′ (x) = − sin x
′′′
f (x) = − cos x
√
2
2
+
sin x =
x−
2
2
√
√
2
2
x−
=
+
2
2
√
f ( π4 ) =
f ′ ( π4 )
f ′′ ( π4 )
f ′′′ ( π4 )
=
=
=
√
2
,
√2
2
,
2√
− 22 ,
√
− 22 .
√ (x − π )2 √ (x − π )3
π
2
2
4
4
−
−···
−
4
2
2!
2
3!
√
2 √
3
π
2
π
2
π
−
x−
x−
−
− ···
4
4
4
12
4
9.
f (θ) = cos θ
f ( π4 ) =
f ′ (θ) = − sin θ
f ′ ( π4 ) =
′′
f (θ) = − cos θ
f ′′′ (θ) = sin θ
√
√
2
2
θ−
−
2
2
√
√
2
2
−
=
θ−
2
2
f ′′ ( π4 ) =
f ′′′ ( π4 ) =
f (t) = cos t
f ( π6 ) =
f ′ (t) = − sin t
f ′ ( π6 ) =
′′
f (t) = − cos t
f ′′′ (t) = sin t
3
1
−
t−
cos t =
2
2
√
1
3
−
=
t−
2
2
√
2
,
2√
− 22 ,
√
− 22 ,
√
2
.
2
√ (θ − π )2 √ (θ − π )3
2
2
π
4
4
−
+
− ···
4
2
2!
2
3!
√
√
π
2
π 2
2
π 3
+
− ···
−
θ−
θ−
4
4
4
12
4
10. Differentiating gives
√
2
9
′′′
p
3
8.
cos θ =
f ′′ (0) =
8
−3
f ′′ ( π6 ) =
f ′′′ ( π6 ) =
√
3
,
2
1
−2,
√
− 23 ,
1
.
2
√ (t − π )2
π 3
3
π
1 (t − 6 )
6
−
+
− ···
6
2
2!
2
3!
√
2
1
3
π
π
π 3
+
−···
−
t−
t−
6
4
6
12
6
11.
f (θ) = sin θ
f ′ (θ) = cos θ
′′
f (θ) = − sin θ
f ′′′ (θ) = − cos θ
√
2
,
√ 2
2
,
√2
2
,
2√
− 22 .
f (− π4 ) = −
f ′ (− π4 ) =
f ′′ (− π4 ) =
f ′′′ (− π4 ) =
10.2 SOLUTIONS
√
2
2
+
sin θ = −
θ+
2
2
√
√
2
2
=−
+
θ+
2
2
√
√
√
π 2
)
4
π 3
)
4
2 (θ +
2 (θ +
−
+ ···
4
2
2!
2
3!
√
√
π
π 2
π 3
2
2
−
+···.
+
θ+
θ+
4
4
4
12
4
π
+
12.
f ( π4 ) = 1,
f (x) = tan x
′
f (x) =
f ′′ (x) =
′′′
f (x) =
1
cos2 x
−2(− sin x)
sin x
= 2cos
3x
cos3 x
−6 sin x(− sin x)
2
+
4
cos x
cos2 x
f ′ ( π4 ) = 2,
f ′′ ( π4 ) = 4,
f ′′′ ( π4 ) = 16.
3
x − π4
(x − π4 )2
π
tan x = 1 + 2 x −
+4
+ 16
+ ···
4
2!
3!
2
3
8
π
π
π
+
+ ···
= 1+2 x−
+2 x−
x−
4
4
3
4
13.
f (x) =
′
f (x) =
′′
f (x) =
f ′′′ (x) =
1
x
− x12
2
x3
− x64
f (1) = 1
f ′ (1) = −1
f ′′ (1) = 2
f ′′′ (1) = −6
2(x − 1)2
6(x − 1)3
1
= 1 − (x − 1) +
−
+ ···
x
2!
3!
= 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · .
14. Again using the derivatives found in Problem 13, we have
f (2) =
1
,
2
1
f ′ (2) = − ,
4
f ′′ (2) =
1
,
4
3
f ′′′ (2) = − .
8
(x − 2)2
3(x − 2)3
1
x−2
1
= −
+
−
+···
x
2
4
4 · 2!
8 · 3!
1
(x − 2)
(x − 2)2
(x − 2)3
= −
+
−
+ ···
2
4
8
16
15. Using the derivatives from Problem 13, we have
f (−1) = −1,
f ′ (−1) = −1,
f ′′ (−1) = −2,
f ′′′ (−1) = −6.
Hence,
2(x + 1)2
6(x + 1)3
1
= −1 − (x + 1) −
−
− ···
x
2!
3!
= −1 − (x + 1) − (x + 1)2 − (x + 1)3 − · · ·
16. The general term can be written as xn for n ≥ 0.
17. The general term can be written as (−1)n xn for n ≥ 0.
18. The general term can be written as −xn /n for n ≥ 1.
945
946
Chapter Ten /SOLUTIONS
19. The general term can be written as (−1)n−1 xn /n for n ≥ 1.
20. The general term can be written as (−1)k x2k+1 /(2k + 1)! for k ≥ 0.
21. The general term can be written as (−1)k x2k+1 /(2k + 1) for k ≥ 0.
22. The general term can be written as x2k /k! for k ≥ 0.
23. The general term can be written as (−1)k x4k+2 /(2k)! for k ≥ 0.
24. The series is
3·2 2 3·2·1 3 3·2·1·0 4
x +
x +
x + ···.
2!
3!
4!
4
The x term and all terms beyond it turn out to be zero, because each coefficient contains a factor of 0. Simplifying gives
(1 + x)3 = 1 + 3x +
(1 + x)3 = 1 + 3x + 3x2 + x3 ,
which is the usual expansion obtained by multiplying out (1 + x)3 .
Problems
25. By looking at Figure 10.4 we can that the Taylor polynomials are reasonable approximations for the function f (x) =
√1
between x = −1 and x = 1. Thus a good guess is that the interval of convergence is −1 < x < 1.
1+x
4
P10 (x)
3
P4 (x)
2
P2 (x)
1
f (x) =
√1
1+x
x
−1
1
2
3
Figure 10.4
26. √
By looking at Figure 10.5, we see that the Taylor polynomials are reasonable approximations for the function f (x) =
1 + x between x = −1 and x = 1. Thus a good guess is that the Taylor series converges to f (x) for −1 < x < 1.
f (x) =
2
√
x+1
P2 (x)
1
P10 (x)
P4 (x)
x
−1
1
2
3
Figure 10.5
27. (a) Figure 10.6 suggests that the Taylor polynomials converge to f (x) =
(b) Since
the ratio test gives
1
on the interval −1 < x < 1.
1−x
1
= 1 + x + x2 + x3 · · · .
1−x
lim
n→∞
|an+1 |
|xn+1 |
= lim
= |x|.
n→∞ |xn |
|an |
10.2 SOLUTIONS
947
Thus, the radius of convergence is R = 1. The series converges if |x| < 1; that is, −1 < x < 1.
f (x) =
1
1−x
−1
✲
P7 (x)
x
1
✲
P5 (x)
P3 (x)✲
Figure 10.6
28. The Taylor series of ex around x = 0 is
∞
ex = 1 + x +
X xk
x2
x3
+
+ ··· =
.
2!
3!
k!
k=0
To find the radius of convergence, we apply the ratio test with ak = xk /k!.
|x|k+1 /(k + 1)!
|x|
|ak+1 |
= lim
= lim
= 0.
k→∞
k→∞ k + 1
k→∞ |ak |
|x|k /k!
lim
Hence the radius of convergence is R = ∞.
29. The Taylor series for ln(1 − x) is
ln(1 − x) = −x −
so
lim
n→∞
x3
xn
x2
−
− ··· −
−···,
2
3
n
1/(n + 1)
|an+1 |
n
= |x| lim
= |x| lim
= |x|.
n→∞
n→∞ n + 1
|an |
1/n
Thus the series converges for |x| < 1, and the radius of convergence is 1. Note: This series can be obtained from the series
for ln(1 + x) by replacing x by −x and has the same radius of convergence as the series for ln(1 + x).
30. (a) We have shown that the series is
1 + px +
p(p − 1) 2 p(p − 1)(p − 2) 3
x +
x + ···
2!
3!
so the general term is
p(p − 1) . . . (p − (n − 1)) n
x .
n!
(b) We use the ratio test
lim
n→∞
|an+1 |
p−n
p(p − 1) . . . (p − (n − 1))(p − n) · n!
= |x| lim
= |x| lim
.
n→∞
n→∞ n + 1
|an |
(n + 1)!p(p − 1) . . . (p − (n − 1))
Since p is fixed, we have
lim
n→∞
p−n
= 1,
n+1
so
R = 1.
948
Chapter Ten /SOLUTIONS
31. We know that the Taylor series for ex around 0 is given by
ex = 1 + x +
x3
x2
+
+ ···.
2!
3!
Using the right hand side of the above equation for ex in the expression
lim
x→0
x2
2!
1+x+
ex − 1
= lim
x→0
x
ex − 1
, we have
x
3
+ x3! + · · · − 1
.
x
Simplifying we get
lim
x→0
x+
x2
2!
3
+ x3! + · · ·
x
x2
= lim 1 +
+
+ · · · = 1.
x→0
x
2!
3!
Hence this limit is equal to 1.
32. The second coefficient of the Taylor expansion is
Similarly, the third coefficient is
g ′′ (0)
= 1,
2
so
g ′′ (0) = 2.
g ′′′ (0)
=0
3!
so
g ′′′ (0) = 0.
so
g (10) (0) =
Finally, the tenth coefficient is
1
g (10) (0)
=
10!
5!
10!
.
5!
33. Let Cn be the coefficient of the nth term in the series. Note that
0 = C1 =
d 2 x2
(x e )
dx
,
x=0
and since
1
= C6 =
2
we have
d6 2 x2
(x e )
dx6
2
d6
(x2 ex )
dx6
x=0
6!
=
x=0
,
6!
= 360.
2
34. (a) From the coefficients of the (x − 1) terms of the f s, we see that
f1′ (1) = 1,
f2′ (1) = −1
f3′ (1) = −2.
From the (x − 1)2 terms of the f s, we see that
f1′′ (1)
= −1,
2!
f2′′ (1)
= 1,
2!
f3′′ (1)
= 1,
2!
so f1′′ (1) = −2, f2′′ (1) = 2, f3′′ (1) = 2.
Thus, f1 slopes up at x = 1 and f2 and f3 slope down; f3 slopes down more steeply than f2 . This means that
the f s are in the first figure, since graphs II and III in the second figure have the same negative slope at point B.
By a similar argument, we find
g1′ (4) = −1,
g2′ (4) = −1,
g3′ (4) = 1, and g1′′ (4) = −2,
g2′′ (4) = 2,
g3′′ (4) = 2.
Thus, two of the gs slope down, one of which is concave up and one is concave down; the third g slopes up and is
concave up. This confirms that the gs are in the second figure.
(b) Since f1 (1) = f2 (1) = f3 (1) = 3, the point A is (1, 3).
Since g1 (4) = g2 (4) = g3 (4) = 5, the point B is (4, 5).
(c) In the first figure, graph I is f1 since it slopes up. Graph II is f2 since it slopes down, but less steeply than graph III,
which is f3 .
In the second figure, graph I is g3 , since it slopes up. Graph II is g2 since it slopes down and is concave up.
Graph III is g1 since it slopes down and is concave down.
10.2 SOLUTIONS
949
35. This is the series for ex with x replaced by 2, so the series converges to e2 .
36. This is the series for sin x with x replaced by 1, so the series converges to sin 1.
37. This is the series for 1/(1 − x) with x replaced by 1/4, so the series converges to 1/(1 − (1/4)) = 4/3.
38. This is the series for cos x with x replaced by 10, so the series converges to cos 10.
39. This is the series for ln(1 + x) with x replaced by 1/2, so the series converges to ln(3/2).
40. The Taylor series for f (x) = 1/(1 + x) is
1
= 1 − x + x2 − x3 + · · · .
1+x
Substituting x = 0.1 gives
1 − 0.1 + (0.1)2 − (0.1)3 + · · · =
1
1
=
.
1 + 0.1
1.1
Alternatively, this is a geometric series with a = 1, x = −0.1.
41. This is the series for ex with x = 3 substituted. Thus
1+3+
27
81
32
33
34
9
+
+
+ ··· = 1 + 3 +
+
+
+ · · · = e3 .
2!
3!
4!
2!
3!
4!
42. This is the series for cos x with x = 1 substituted. Thus
1
1
1
1−
+
−
+ · · · = cos 1.
2!
4!
6!
43. This is the series for ex with −0.1 substituted for x, so
0.001
0.01
−
+ · · · = e−0.1 .
1 − 0.1 +
2!
3!
44. Since 1 + x + x2 + x3 + · · · =
1
1
4
1
, a geometric series, we solve
= 5 giving = 1 − x, so x = .
1−x
1−x
5
5
1 2 1 3
x + x + · · · = ln(1 + x), we solve ln(1 + x) = 0.2, giving 1 + x = e0.2 , so x = e0.2 − 1.
2
3
46. We define eiθ to be
(iθ)3
(iθ)4
(iθ)5
(iθ)6
(iθ)2
+
+
+
+
+···
eiθ = 1 + iθ +
2!
3!
4!
5!
6!
Suppose we consider the expression cos θ + i sin θ, with cos θ and sin θ replaced by their Taylor series:
45. Since x −
cos θ + i sin θ =
1−
θ2
θ4
θ6
+
−
+···
2!
4!
6!
+i θ−
θ3
θ5
+
− ···
3!
5!
Reordering terms, we have
cos θ + i sin θ = 1 + iθ −
iθ3
θ4
iθ5
θ6
θ2
−
+
+
−
− ···
2!
3!
4!
5!
6!
Using the fact that i2 = −1, i3 = −i, i4 = 1, i5 = i, · · ·, we can rewrite the series as
(iθ)3
(iθ)4
(iθ)5
(iθ)6
(iθ)2
+
+
+
+
+ ···
2!
3!
4!
5!
6!
Amazingly enough, this series is the Taylor series for ex with iθ substituted for x. Therefore, we have shown that
cos θ + i sin θ = 1 + iθ +
cos θ + i sin θ = eiθ .
Strengthen Your Understanding
47. The left hand side of the equation is finite, namely −1, whereas the right hand side of the equation is infinite. The statement
is wrong, since
1
= 1 + x + x2 + x3 + · · ·
1−x
only for −1 < x < 1.
950
Chapter Ten /SOLUTIONS
48. First note that this Taylor series is convergent at 3. Substituting x = 4 into the series we get
1 + 1 + ···,
which is divergent, similarly substituting x = 2, we get the alternating geometric series
1 + (−1) + (−1)2 + (−1)3 + · · · ,
which is divergent. Since x = 3 is the center of the interval of convergence of the Taylor series, we see that the radius of
convergence cannot be greater or equal to 1.
49. The function f (x) = cos x is an example. The Taylor series for cos x is
cos x = 1 −
x4
x6
x2
+
−
+ ···
2!
4!
6!
The third-degree term of this series is zero.
50. A possible example is 1 + (x + 1) + (x + 1)2 + . . .. Many other examples are possible.
51. False. The Taylor series for sin x about x = π is calculated by taking derivatives and using the formula
f (a) + f ′ (a)(x − a) +
f ′′ (a)
(x − a)2 + · · · .
2!
The series for sin x about x = π turns out to be
−(x − π) +
(x − π)5
(x − π)3
−
+ ···.
3!
5!
52. True. Since f is even, f (−x) = f (x) for all x. Taking the derivative of both sides of this equation, we get f ′ (−x)(−1) =
f ′ (x), which at x = 0 gives −f ′ (0) = f ′ (0), so f ′ (0) = 0. Taking the derivative again gives f ′′ (−x) = f ′′ (x), i.e., f ′′
is even. Using the same reasoning again, we get that f ′′′ (0) = 0, and, continuing in this way, we get f (n) (0) = 0 for all
odd n. Thus, for all odd n, the coefficient of xn in the Taylor series is f (n) (0)/n! = 0, so all the terms with odd exponent
are zero.
53. True. The coefficient of x7 is −8/7!, so
f (7) (0)
−8
=
7!
7!
giving f (7) (0) = −8.
54. False. For example, the Taylor series
1 + x + x2 + x3 + · · ·
for f (x) = 1/(1 − x) diverges for |x| > 1, but 1/(1 − x) is defined for |x| > 1.
55. True. For large x, the graph of P10 (x) looks like the graph of its highest powered term, x10 /10!. But ex grows faster than
any power, so ex gets further and further away from x10 /10! ≈ P10 (x).
Solutions for Section 10.3
Exercises
1. Substitute y = −x into ey = 1 + y +
y2
2!
+
y3
3!
+ · · ·. We get
(−x)3
(−x)2
+
+ ···
2!
3!
3
2
x
x
−
+ ···.
= 1−x+
2!
3!
e−x = 1 + (−x) +
10.3 SOLUTIONS
2. We’ll use
1
p
1
−1 y 2
1
y+
2
2
2
2!
1 −1 −3 y 3
+···
+
2
2
2
3!
2
3
y
y
y
= 1+ −
+
− ···.
2
8
16
1 + y = (1 + y) 2 = 1 +
Substitute y = −2x.
√
(−2x)
(−2x)2
(−2x)3
−
+
− ···
2
8
16
x2
x3
= 1−x−
−
− ···
2
2
1 − 2x = 1 +
3. Substitute x = θ2 into series for cos x:
(θ2 )4
(θ2 )6
(θ2 )2
+
−
+···
2!
4!
6!
θ8
θ12
θ4
+
−
+ ···.
= 1−
2!
4!
6!
cos (θ2 ) = 1 −
4. Substituting x = −2y into ln(1 + x) = x −
x2
2
+
x3
3
−
x4
4
+ · · · gives
(−2y)3
(−2y)4
(−2y)2
+
−
+ ···
2
3
4
8
= −2y − 2y 2 − y 3 − 4y 4 − · · · .
3
ln(1 − 2y) = (−2y) −
5. Since
d
(arcsin x)
dx
= √
1
1−x2
= 1 + 21 x2 + 38 x4 +
5 6
x
16
arcsin x = c + x +
+ · · ·, integrating gives
3 5
5 7
1 3
x +
x +
x +···.
6
40
112
Since arcsin 0 = 0, c = 0.
6. We substitute 3t into the series for sin x and multiply by t. Since
sin x = x −
x3
x5
x7
+
−
+···,
3!
5!
7!
substituting 3t gives
(3t)5
(3t)7
(3t)3
+
−
+ ···
3!
5!
7!
−9 3 81 5 −243 7
= 3t +
t +
t +
t + ···,
2
40
560
sin(3t) = (3t) −
so
t sin(3t) = 3t2 −
7. Substituting x = −z 2 into
√1
1+x
1
9 4 81 6 243 8
t +
t −
t +···.
2
40
560
= (1 + x)− 2 = 1 − 12 x + 83 x2 −
√
5 3
x
16
+ · · · gives
3(−z 2 )2
5(−z 2 )3
1
(−z 2 )
+
−
+ ···
= 1−
2
2
8
16
1−z
1
3
5 6
= 1 + z2 + z4 +
z +···.
2
8
16
951
952
Chapter Ten /SOLUTIONS
8.
z
−z 2
=z
2 = ze
z
e
1 + (−z 2 ) +
= z − z3 +
(−z 2 )3
(−z 2 )2
+
+···
2!
3!
z7
z5
−
+ ···
2!
3!
9.
φ3 cos(φ2 ) = φ3
1−
= φ3 −
(φ2 )4
(φ2 )6
(φ2 )2
+
−
+···
2!
4!
6!
φ7
φ11
φ15
+
−
+ ···
2!
4!
6!
10. From Example 3, we know the Taylor series for arctan x:
arctan x = x −
x5
x7
x3
+
−
···.
3
5
7
Substituting x = r 2 , we get:
arctan(r 2 ) = r 2 −
r 10
r 14
r6
+
−
+ ···.
3
5
7
11. Multiplying out gives (1 + x)3 = 1 + 3x + 3x2 + x3 . Since this polynomial equals the original function for all x, it must
be the Taylor series. The general term is 0 · xn for n ≥ 4.
12. Substituting t2 into the series for sin x gives
sin(t2 ) = t2 −
= t2 −
(t2 )5
(−1)k (t2 )2k+1
(t2 )3
+
+··· +
+ ···
3!
5!
(2k + 1)!
(−1)k t4k+2
t6
t10
+
+ ··· +
+ ···
3!
5!
(2k + 1)!
Therefore
t sin(t2 ) − t3 =
=−
t3 −
t7
t11
(−1)k t4k+3
+
+··· +
+ ···
3!
5!
(2k + 1)!
(−1)k t4k+3
t7
t11
+
+ ··· +
+ ···
3!
5!
(2k + 1)!
− t3
for k ≥ 1.
13. Using the Binomial theorem:
√
1
1−x
= (1 − x)−1/2
= 1+ −
(−1/2)(−3/2) · · · (− 21 − n + 1)(−x)n
(−1/2)(−3/2)(−x)2
1
(−x) +
+ ··· +
+ · · · for n ≥ 1.
2
2!
n!
Substituting y 2 for x:
1
p
1 − y2
= (1 − y 2 )−1/2
= 1+
(1/2)(3/2) · · · ( 21 + n − 1)y 2n
1 2 3 4
y + y + ··· +
+ · · · for n ≥ 1.
2
8
n!
10.3 SOLUTIONS
14.
1
1
1
x
=
=
1+
2+x
2(1 + x2 )
2
2
1
2
=
1−
x
+
2
x 2
2
−1
x 3
−
2
+ ···
15. Using the binomial expansion for (1 + x)1/2 with x = h/T :
1/2
1/2 √
√
T
h
h 1/2
T +h = T + h
= T 1+
= T 1+
T
T
T
(1/2)(−1/2) 2 (1/2)(−1/2)(−3/2) 3
√
h
h
h
+
+
···
= T 1 + (1/2)
T
2!
T
3!
T
=
√
T
h
T
1
2
1+
−
1
8
2
h
T
+
3
1
16
h
T
··· .
16. Using the binomial expansion for (1 + x)−1 with x = −r/a:
1
1
=
a−r
a−a
=
1
a
=
1
a
=
1
a
1
1
r
=
1+ −
r
a
a
a 1− a
=
r
a
1 + (−1) −
r
1− −
1+
r
a
+
a
2
r
a
+
(−1)(−2)
r
−
2!
a
r 2
+ −
r
a
a
+
r 3
− −
3
r
a
−1
2
+
(−1)(−2)(−3)
r
−
3!
a
3
+ ···
+···
a
+ ··· .
17. Using the binomial expansion for (1 + x)−2 with x = r/a:
1
1
=
(a + r)2
a+a
=
=
1
a2
1
a2
2 =
r
1−2
r
a
r
a
2 =
r
a 1+
a
1 + (−2)
1
+3
+
a
1
a2
2
(−2)(−3)
2!
2
r
a
−4
r
a
3
r
a
r −2
1+
+
a
(−2)(−3)(−4)
3!
3
r
a
+ ···
+ ··· .
18.
√
3
1/3 √
t
t 1/3
3
= P 1+
P
P
t (1/3)(−2/3) t 2 (1/3)(−2/3)(−5/3) t 3
√
3
+
+
···
= P 1 + (1/3)
P
2!
P
3!
P
P +t = P +P
=
√
3
P
1/3
1+
t
P
1
3
t
P
= P 1+
−
1
9
2
t
P
+
5
81
3
t
P
··· .
953
954
Chapter Ten /SOLUTIONS
19.
a
a
=
=
√
2 1
2
2
a +x
a(1 + xa2 ) 2
= 1+ −
1
2
1
1
+
−
3!
2
= 1−
1
2
x2
1+ 2
a
− 12
x2
1
1
+
−
a2
2!
2
3
−
2
2
x
a
+
3
8
5
−
2
4
x
a
−
−
3
2
x2
a2
3
5
16
x2
a2
2
+ ···
6
x
a
+ ···
Problems
20.
p
(1 + t) sin t =
Multiplying and collecting terms yields
t
t2
t3
1+ −
+
− ···
2
8
16
t2
−
2
1
= t + t2 −
2
p
(1 + t) sin t = t +
21.
t
e cos t =
t2
t3
t4
1+t+
+
+
+ ···
2!
3!
4!
Multiplying out and collecting terms gives
= 1+t−
22. Substituting the series for sin θ = θ −
t2
t2
−
2!
2!
+
θ5
5!
+
1+y =1+
√
1
1 + sin θ = 1 +
2
1
+
16
t3
t3
−
3!
2!
θ3
θ5
θ−
+
− ···
3!
5!
θ2
+
8
1 2
θ −
8
3
θ5
θ3
+
− ···
θ−
3!
5!
1
θ−
2
1
= 1+ θ−
2
= 1+
+ ···
t2
t4
t6
1−
+
−
+ ···
2!
4!
6!
+
t4
t4
t4
+
−
4!
4!
(2!)2
+···
1
1 3
1
y − y2 +
y −···
2
8
16
− · · · into
p
gives
t4
t3
−
+···.
3
6
θ3
3!
t3
t5
t−
+
−···
3!
5!
t3
t4
t3
t4
+
−
+
3!
8
16
12
7 3
1 4
t −
t +···.
24
48
et cos t = 1 + t +
θ3
θ3
−
16
2 · 3!
1 3
θ + ···
48
1
−
8
− ···
2
θ3
θ5
θ−
+
−···
3!
5!
+ ···
10.3 SOLUTIONS
955
23. We know that
t2
t3
t4
+
−
+···
2
3
4
So, treating the given function as the sum of a geometric series with initial term 1 and common ratio ln(1 + t), we have
ln(1 + t) = t −
1
= 1+
1 − ln(1 + t)
+ t−
= 1+
t2
t3
t4
t−
+
−
+ ···
2
3
4
t3
t4
t2
+
−
+ ···
2
3
4
t2
t3
+
−···
2
3
t−
= 1+t+
3
t2
t3
+
+···.
2
3
+
t2
t3
t4
t−
+
−
+ ···
2
3
4
2
+ ···
+ t2 − t3 + · · · + t3 − · · · + · · ·
24. (a) Since the Taylor series for ex and e−x are given by
x3
x4
x2
+
+
+ ···
2!
3!
4!
3
4
2
x
x
x
−
+
+ ···,
= 1−x+
2!
3!
4!
ex = 1 + x +
e−x
we have
x2
x3
x4
x4
+0
+2
+ · · · = 2 + x2 +
+ ···.
2!
3!
4!
12
(b) For x near 0, we can approximate ex +e−x by its second degree Taylor polynomial, P2 (x), whose graph is a parabola:
ex + e−x = 2 + 0x + 2
ex + e−x ≈ P2 (x) = 2 + x2 .
25. (a) Since the Taylor series for ex and e−x are given by
x3
x4
x5
x2
+
+
+
+ ···
2!
3!
4!
5!
3
4
5
2
x
x
x
x
−
+
−
+ ···,
= 1−x+
2!
3!
4!
5!
ex = 1 + x +
e−x
we have
x2
x3
x4
x5
x3
x5
+2
+0
+ 2 · · · = 2x +
+
+···.
2!
3!
4!
5!
3
60
(b) For x near 0, we can approximate ex − e−x by its third degree Taylor polynomial, P3 (x):
ex − e−x = 0 + 2x + 0
ex − e−x ≈ P3 (x) = 2x +
x3
.
3
The function P3 (x) is a cubic polynomial whose graph is symmetric about the origin.
26. Recall that the Taylor series for ex around 0 is given by
x3
x2
+
+ ···.
2!
3!
1+x+
2
Hence the Taylor series for ex around 0 is then
1 + x2 +
x6
x4
+
+···.
2!
3!
The first three terms are then
P4 (x) = 1 + x2 +
x4
.
2!
We approximate
Z
0
1
2
ex dx ≈
Z
0
1
1 + x2 +
x4
2!
dx.
956
Chapter Ten /SOLUTIONS
Computing the integral
Z
1
0
27. Notice that
P
x4
1+x +
2!
2
dx = x +
x3
x5
+
3
10
1
=1+
0
1
1
+
≈ 1.433.
3
10
pxp−1 , is the derivative, term-by-term, of a geometric series:
∞
X
p=1
pxp−1 = 1 · x0 + 2 · x1 + 3 · x2 + · · · =
d
(x + x2 + x3 + · · ·).
{z
}
dx |
Geometric series
For |x| < 1, the sum of the geometric series with first term x and common ratio x is
x + x2 + x3 + · · · =
x
.
1−x
Differentiating gives
∞
X
pxp−1 =
p=1
d
dx
x
1−x
=
1(1 − x) − x(−1)
1
=
.
(1 − x)2
(1 − x)2
28. From the series for ln(1 + y),
ln(1 + y) = y −
y2
y3
y4
+
−
+ ···,
2
3
4
we get
ln(1 + y 2 ) = y 2 −
y4
y6
y8
+
−
+ ···
2
3
4
The Taylor series for sin y is
sin y = y −
So
sin y 2 = y 2 −
y5
y7
y3
+
−
+ ···
3!
5!
7!
y6
y 10
y 14
+
−
+ ···
3!
5!
7!
The Taylor series for cos y is
cos y = 1 −
So
y4
y6
y2
+
−
+ ···
2!
4!
6!
y2
y4
y6
−
+
+ ···
2!
4!
6!
Near y = 0, we can drop terms beyond the fourth degree in each expression:
1 − cos y =
ln(1 + y 2 ) ≈ y 2 −
y4
2
sin y 2 ≈ y 2
y2
y4
1 − cos y ≈
−
.
2!
4!
(Note: These functions are all even, so what holds for negative y will hold for positive y.)
Clearly 1 − cos y is smallest, because the y 2 term has a factor of 12 . Thus, for small y,
y2
y4
y4
−
< y2 −
< y2
2!
4!
2
so
1 − cos y < ln(1 + y 2 ) < sin(y 2 ).
10.3 SOLUTIONS
957
29. The Taylor series for 1 + sin θ near θ = 0 is
1 + sin θ = 1 + θ −
θ5
θ3
+
− ···
3!
5!
The Taylor series for eθ is
eθ = 1 + θ +
The Taylor series for
√1
1+θ
θ2
θ3
+
+ ···
2!
3!
is
√
1
θ
3θ2
5θ3
=1− +
−
+ ···
2
8
16
1+θ
Substituting −2θ into this formula yields
√
3
1
5
= 1 + θ + θ2 + θ3 + · · ·
2
2
1 − 2θ
These three series are identical in the constant and linear terms, but they differ in their quadratic terms. For values of
θ near zero, the quadratic terms dominate all of the subsequent terms, so we can use the approximations
1 + sin θ ≈ 1 + θ
1 2
θ
2
1
3
√
≈ 1 + θ + θ2 .
2
1 − 2θ
eθ ≈ 1 + θ +
Clearly√1 + sin θ is smallest, because the θ2 term is zero, and the θ2 terms of the other two series are positive. The
function 1/ 1 − 2θ is largest, because the coefficient of its θ2 term is the greatest. Therefore, for θ near zero,
1 + sin θ ≤ eθ ≤ √
1
.
1 − 2θ
30. We have
P9 (x) =
4
X
x2n+1
n=0
1
=
2n + 1
x3
x5
x7
x9
x
+
+
+
+
.
1
3
5
7
9
So,
1
1
1
1
(2x)3 + (2x)5 + (2x)7 + (2x)9
3
5
7
9
8
32 5 128 7 512 9
= 2x + x3 +
x +
x +
x .
3
5
7
9
P9 (2x) = 2x +
31. The Taylor series about 0 for y =
1
is
1 − x2
y = 1 + x2 + x4 + x6 + · · · .
The series for y = (1 + x)1/4 is, using the binomial expansion,
y =1+
The series for y =
q
1+
3
1
1
−
x+
4
4
4
3
x2
1
−
+
2!
4
4
−
7
4
x3
+ ···.
3!
x
x
= (1 + )1/2 is, again using the binomial expansion,
2
2
y =1+
1 x
1
1
· +
−
2 2
2
2
x2
·
8
+
1
1
−
2
2
3 x3
−
2
·
48
+ ···.
958
Chapter Ten /SOLUTIONS
Similarly for y = √
1
= (1 − x)−(1/2) ,
1−x
y =1+ −
1
1
(−x) + −
2
2
−
3
2
Near 0, let’s truncate these series after their x2 terms:
·
x2
1
+ −
2!
2
1
≈ 1 + x2 ,
1 − x2
1
(1 + x)1/4 ≈ 1 + x −
4
q
x
1
1+ ≈ 1+ x−
2
4
1
1
√
≈ 1+ x+
2
1−x
Thus
Now
II.
1
1−x2
√1
1−x
−
3
2
−
5
2
·
−x3
+ ···.
3!
3 2
x ,
32
1 2
x ,
32
3 2
x .
8
looks like a parabola opening upward near the origin, with y-axis as the axis of symmetry, so (a) = I.
has the largest positive slope ( 21 ), and is concave up (because the coefficient of x2 is positive). So (d) =
1
The last two both have positive slope ( 41 ) and are concave down. Since (1 + x) 4 has the smallest second derivative
(i.e., the most negative coefficient of x2 ), (b) = IV and therefore (c) = III.
32. From the Taylor series for the sine function we know that
sin t ≈ t −
t3
t5
t7
+
− .
3!
5!
7!
Thus, from the definition of Si(x), we have:
Z
2
sin t
dt
Z 2 t
1
t5
t7
t3
+
−
t−
dt
≈
t
3!
5!
7!
Z0 2
t2
t4
t6
=
1−
+
−
dt
3!
5!
7!
0
Si(2) =
0
1 t5
1 t7
1 t3
·
+ ·
− ·
3 3!
5 5!
7 7!
23
25
27
= 2−
+
−
3 · 3!
5 · 5!
7 · 7!
= 1.60526.
= t−
2
0
This is very close to the actual value of Si(2) = 1.60541, found using a computer algebra system.
1
1
1
33. The Taylor series for sin t about t = 0 is sin t = t − t3 + t5 − t7 + · · · . This gives the Taylor series for sin t2
3!
5!
7!
about t = 0:
sin t2 = t2 −
so
1
1
1 2 3
1 2 5
1 2 7
1
t
+
t
−
t
+ · · · = t2 − t6 + t10 − t14 + · · · ,
3!
5!
7!
3!
5!
7!
f (x) =
Z
x
sin t2 dt
Z0 x
1
1
1 6
t + t10 − t14 + · · · dt
3!
5!
7!
0
1 1 7
1 1 11
1 1 15
1
x +
x −
x + ···
= x3 −
3
7 3!
11 5!
15 7!
1
1 7
1 11
1
= x3 −
x +
x −
x15 + · · ·
3
42
1320
75,600
=
t2 −
10.3 SOLUTIONS
959
34. (a) f (t) = tet .
Use the Taylor expansion for et :
t2
t3
+
+ ···
f (t) = t 1 + t +
2!
3!
t3
t4
+
+ ···
2!
3!
= t + t2 +
(b) Z
x
f (t) dt =
0
Z
x
t
te dt =
0
Z
0
=
x
t4
t3
+
+ ···
t+t +
2!
3!
2
t3
t4
t5
t2
+
+
+
+ ···
2
3
4 · 2!
5 · 3!
dt
x
0
x2
x3
x4
x5
=
+
+
+
+ ···
2
3
4 · 2!
5 · 3!
(c) Substitute x = 1 :
Z
1
tet dt =
0
1
1
1
1
+ +
+
+ ···
2
3
4 · 2!
5 · 3!
In the integral above, to integrate by parts, let u = t, dv = et dt, so du = dt, v = et .
Z
1
1
tet dt = tet
0
Hence
0
−
Z
1
et dt = e − (e − 1) = 1
0
1
1
1
1
+ +
+
+ · · · = 1.
2
3
4 · 2!
5 · 3!
35. Since it does not depend on n, we can factor out e−k , giving
∞
X
kn−1
n=1
(n − 1)!
e−k = e−k
∞
X
kn−1
n=1
= e−k
|
(n − 1)!
k
k2
k3
1
+
+
+
+···
0!
1!
2!
3!
{z
This is the series for ek
= e−k · ek
}
= 1.
36. (a) The Taylor approximation to f (x) = cosh x about x = 0 is of the form
cosh x ≈ cosh(0) + f ′ (0)x +
f (n) (0)xn
f ′′ (0)x2
+ ... +
.
2!
n!
We have the following results:
f (x) = cosh x so f (0) = 1,
f ′ (x) = sinh x so f ′ (0) = 0,
f ′′ (x) =
d
(sinh x) = cosh x so f ′′ (0) = 1,
dx
f ′′′ (x) = sinh x so f ′′′ (0) = 0.
The derivatives continue to alternate between cosh x and sinh x, so their values at 0 continue to alternate between 0
and 1. Therefore
x3
x4
x2
+0·
+1·
+ ···,
cosh x ≈ 1 + 0 · x + 1 ·
2!
3!
4!
960
Chapter Ten /SOLUTIONS
so the degree 8 Taylor approximation is given by
cosh x ≈ 1 +
x4
x6
x8
x2
+
+
+
.
2!
4!
6!
8!
(b) We use the polynomial obtained from part (a) to estimate cosh 1,
cosh 1 ≈ 1 +
1
1
1
1
+
+
+
= 1.543080357.
2!
4!
6!
8!
Compared to the actual value of cosh 1 = 1.543080635 . . ., the error is less than 10−6 .
d
(cosh x) = sinh x, we have
(c) Since
dx
sinh x ≈
d
dx
1+
x2
x4
x6
x8
+
+
+
2!
4!
6!
8!
4x3
6x5
8x7
2x
+
+
+
2!
4!
6!
8!
x5
x7
x3
+
+
.
= x+
3!
5!
7!
=
37. Since ex =
∞
X
xn
n=0
n!
and sinh 2x = (e2x − e−2x )/2, the Taylor expansion for sinh 2x is
∞
X
(2x)n
1
sinh 2x =
2
n=0
n!
−
∞
X
(−2x)n
n=0
n!
!
1
=
2
=
∞
X
n=0
(2x)n
(1 − (−1) )
n!
n
∞
X
(2x)2m+1
m=0
(2m + 1)!
!
.
Since cosh 2x = (e2x + e−2x )/2, we have
1
cosh 2x =
2
∞
X
(2x)n
n=0
n!
+
∞
X
(−2x)n
n=0
n!
!
1
=
2
=
∞
X
n=0
(2x)n
(1 + (−1) )
n!
∞
X
(2x)2m
m=0
(2m)!
n
!
.
38. (a) The Taylor series for 1/(1 − x) = 1 + x + x2 + x3 + . . ., so
1
1
=
= 1 + (0.02) + (0.02)2 + (0.02)3 + . . .
0.98
1 − 0.02
= 1.020408 . . .
(b) Since d/dx(1/(1 − x)) = (1/(1 − x))2 , the Taylor series for 1/(1 − x)2 is
d
(1 + x + x2 + x3 + . . .) = 1 + 2x + 3x2 + 4x3 + · · ·
dx
Thus
1
1
=
= 1 + 2(0.01) + 3(0.0001) + 4(0.000001) + · · ·
(0.99)2
(1 − 0.01)2
= 1.0203040506 . . .
10.3 SOLUTIONS
f (x) = (1 + ax)(1 + bx)
39. (a)
−1
2
2
3
= (1 + ax) 1 − bx + (bx) − (bx) + · · ·
= 1 + (a − b)x + (b − ab)x2 + · · ·
961
2
(b) ex = 1 + x + x2 + · · ·
Equating coefficients:
a − b = 1,
1
b − ab = .
2
2
Solving gives a = 12 , b = − 21 .
40. (a) If φ = 0,
left side = b(1 + 1 + 1) = 3b ≈ 0
so the equation is almost satisfied and there could be a solution near φ = 0.
(b) We have
φ5
φ3
+
− ···
3!
5!
φ2
φ4
cos φ = 1 −
+
− ···
2!
4!
sin φ = φ −
So
cos2 φ =
1−
φ4
φ2
+
− ···
2!
4!
Neglecting terms of order φ2 and higher, we get
1−
φ4
φ2
+
− ··· .
2!
4!
sin φ ≈ φ
cos φ ≈ 1
cos2 φ ≈ 1.
So φ + b(1 + 1 + 1) ≈ 0, whence φ ≈ −3b.
mM
.
41. (a) µ =
m+M
(b)
If M >> m, then the denominator m + M ≈ M , so µ ≈
M
=m
µ=m
m+M
m
We can use the binomial expansion since M
< 1.
µ=m 1−
m
+
M
mM
= m.
M
1
M
M
M
m
+
M
M
m
M
2
−
=m
m
M
3
1
m
1+ M
+···
1
1
m
≈ 1836
≈ 0.000545.
M, then M
1836
So a first order approximation to µ would give µ = m(1 − 0.000545). The percentage difference from µ = m is
−0.0545%.
(c) If m ≈
42. (a) Solving for ω0 gives
1 2
=0
ω0 C
1
ω0 L −
=0
ω0 C
ω0 L −
1
ω0 C
1
ω02 =
LC
ω0 L =
ω0 =
r
1
.
LC
(Note: We discarded the negative root, because we need positive frequencies.)
962
Chapter Ten /SOLUTIONS
(b) Set ω = ω0 + ∆ω and get
2
1
ωL −
ωC
1
(ω0 + ∆ω)L −
(ω0 + ∆ω)C
=
We need to find a Taylor expansion for the term
ωL −
1
1
= ω0 L + L∆ω −
ωC
Cω0
=
r
.
1
. Using the binomial expansion we have
C(ω0 + ∆ω)
1
1
=
1+
C(ω0 + ∆ω)
Cω0
1
1−
=
Cω0
Therefore,
2
L
+ L∆ω −
=
C
= 2L∆ω − · · · .
r
1−
1
1
· L + L∆ω −
LC
C
r
∆ω −1
ω0
∆ω
+ ···
ω0
r
∆ω
+ ···
ω0
LC
∆ω LC
+
·
1
C
1
L
+ L∆ω − · · ·
C
So
1 2
≈ 4L2 (∆ω)2 .
ωC
Notice that if ∆ω = 0 in the above expression, we get 0, which is what we expected, since in this case ω = ω0 .
ωL −
43. (a) Factoring the expression for t1 − t2 , we get
2l1
2l2
2l1
2l2
− p
− p
+
c(1 − v 2 /c2 )
c(1 − v 2 /c2 )
c 1 − v 2 /c2
c 1 − v 2 /c2
∆t = t1 − t2 =
2(l1 + l2 )
2(l1 + l2 )
− p
c(1 − v 2 /c2 )
c 1 − v 2 /c2
=
1
1
−p
1 − v 2 /c2
1 − v 2 /c2
2(l1 + l2 )
=
c
!
.
Expanding the two terms within the parentheses in terms of v 2 /c2 gives
v2
1− 2
c
−1
(−1)(−2)
v2
= 1+ 2 +
c
2!
= 1+
v2
1− 2
c
−1/2
−v 2
c2
v2
v4
v6
+ 4 + 6 +···
2
c
c
c
1 v2
= 1+
+
2 c2
= 1+
−1
2
2!
−3
2
2
−v 2
c2
(−1)(−2)(−3)
+
3!
2
3 v4
5 v6
1 v2
+
+
+ ···
2 c2
8 c4
16 c6
+
−1
2
−3
2
3!
−v 2
c2
−5
2
3
+ ···
−v 2
c2
3
+ ···
Thus, we have
∆t =
2(l1 + l2 )
c
=
2(l1 + l2 )
c
∆t ≈
(l1 + l2 )
c
1+
v2
v4
v6
1 v2
3 v4
5 v6
+ 4 + 6 + ··· −1 −
−
−
− ···
2
2
4
c
c
c
2c
8c
16 c6
1 v2
5 v4
11 v 6
+
+
+ ···
2
4
2c
8c
16 c6
5 v4
v2
+
2
c
4 c4
.
10.3 SOLUTIONS
963
(b) For small v. we can neglect all but the first nonzero term, so
∆t ≈
(l1 + l2 ) 2
(l1 + l2 ) v 2
· 2 =
v .
c
c
c3
Thus, ∆t is proportional to v 2 with constant of proportionality (l1 + l2 )/c3 .
44. (a) Since the expression under the square root sign, 1 −
v2
c2
must be positive in order to give a real value of m, we have
v2
>0
c2
v2
<1
c2
v 2 < c2 ,
1−
− c < v < c.
so
In other words, the object can never travel faster that the speed of light.
(b) See Figure 10.7.
m
m0
v
c
−c
Figure 10.7
(c) Notice that m = m0
1−
v2
c2
binomial expansion to get:
−1/2
2
. If we substitute u = − vc2 , we get m = m0 (1 + u)−1/2 and we can use the
m = m0
= m0
(−1/2)(−3/2) 2
1
u + ···
1− u+
2
2!
3 v4
1 v2
+
+ ··· .
1+
2 c2
8 c4
(d) We would expect this series to converge only for values of the original function that exist, namely when |v| < c.
45. (a) To find when V takes on its minimum values, set
−V0
d
dr
2
dV
dr
r0
r
6
= 0. So
−
r0
r
12
=0
−V0 −12r06 r −7 + 12r012 r −13 = 0
12r06 r −7 = 12r012 r −13
r06 = r 6
r = r0 .
6
12r06 V0
r0
1−
, we see that V ′ (r) > 0 for r > r0 and V ′ (r) < 0 for r < r0 . Thus,
r7
r
V = −V0 (2(1)6 − (1)12 ) = −V0 is a minimum.
(Note: We discard the negative root −r0 since the distance r must be positive.)
Rewriting V ′ (r) as
964
Chapter Ten /SOLUTIONS
(b)
V (r) = −V0
r0
2
r
6
−
r0
r
12
V (r0 ) = −V0
V ′ (r0 ) = 0
V ′ (r) = −V0 (−12r06 r −7 + 12r012 r −13 )
′′
V (r) =
−V0 (84r06 r −8
−
V ′′ (r0 ) = 72V0 r0−2
156r012 r −14 )
The Taylor series is thus:
V (r) = −V0 + 72V0 r0−2 · (r − r0 )2 ·
1
+ ···
2
(c) The difference between V and its minimum value −V0 is
V − (−V0 ) = 36V0
(r − r0 )2
+ ···
r02
which is approximately proportional to (r − r0 )2 since terms containing higher powers of (r − r0 ) have relatively
small values for r near r0 .
(d) From part (a) we know that dV /dr = 0 when r = r0 , hence F = 0 when r = r0 . Since, if we discard powers of
(r − r0 ) higher than the second,
(r − r0 )2
V (r) ≈ −V0 1 − 36
r02
giving
F =−
r − r0
r − r0
dV
(−V0 ) = −72V0
.
≈ 72 ·
dr
r02
ro2
So F is approximately proportional to (r − r0 ).
46. Solving for P gives
P =
Expanding P in terms of 1/V gives
n2 a
nRT
− 2.
V − nb
V
nb −1 n2 a
nRT
1−
− 2
V
V
V
nb
n2 a
nRT
1+
+ ··· − 2
P =
V
V
V
nRT
n2 bRT
n2 a
P =
+
− 2
V
V2
V
2
1
1
2
P = nRT
+ n (bRT − a)
+···.
V
V
P =
Strengthen Your Understanding
47. By substituting x = 0, we get 1/2 = 1. So this equation is not correct. The Taylor series for
the Taylor series of
1
and writing
1+x
1
1
=
2+x
2(1 + x/2)
1
x −1
=
1+
2
2
x 2 x 3
1
x
=
−
+ ···
1− +
2
2
2
2
Hence the equation given is missing a factor of 1/2 on the right.
1
can be obtained using
2+x
10.4 SOLUTIONS
965
48. The order of operations has not been respected. In e−x the function ex is composed with the function −x. Therefore we
2
3
2
3
have: e−x = 1 + (−x) + (−x)
+ (−x)
+ · · · = 1 − x + x2! − x3! + · · · .
2!
3!
49. If a function or some of its derivatives are not defined at x = 0 then we cannot write a Taylor series for that function. An
example is ln x. Many other examples are possible.
50. An example is f (x) = |x|. The function |x| is not differentiable at x = 0, hence we cannot write its Taylor series around
0. Many other examples are possible.
51. True. Since the derivative of a sum is the sum of the derivatives, Taylor series add.
52. True. Since the Taylor series for cos x has only even powers, multiplying by x3 gives only odd powers.
53. False. The derivative of f (x)g(x) is not f ′ (x)g ′ (x). If this statement were true, the Taylor series for (cos x)(sin x) would
have all zero terms.
54. True. We have
L1 (x) + L2 (x) = (f1 (0) + f1′ (0)x) + (f2 (0) + f2′ (0)x) = (f1 (0) + f2 (0)) + (f1′ (0) + f2′ (0))x.
The right hand side is the linear approximation to f1 + f2 near x = 0.
55. False. The quadratic approximation to f1 (x)f2 (x) near x = 0 is
f1 (0)f2 (0) + (f1′ (0)f2 (0) + f1 (0)f2′ (0))x +
f1′′ (0)f2 (0) + 2f1′ (0)f2′ (0) + f1 (0)f2′′ (0) 2
x .
2
On the other hand, we have
L1 (x) = f1 (0) + f1′ (0)x,
L2 (x) = f2 (0) + f2′ (0)x,
so
L1 (x)L2 (x) = (f1 (0) + f1′ (0)x)(f2 (0) + f2′ (0)x) = f1 (0)f2 (0) + (f1′ (0)f2 (0) + f2′ (0)f1 (0))x + f1′ (0)f2′ (0)x2 .
The first two terms of the right side agree with the quadratic approximation to f1 (x)f2 (x) near x = 0, but the term of
degree 2 does not.
For example, the linear approximation to ex is 1+x, but the quadratic approximation to (ex )2 = e2x is 1+2x+2x2 ,
not (1 + x)2 = 1 + 2x + x2 .
56. (a) ln(4 − x) = ln(4(1 − x/4)) = ln(4) + ln(1 − x/4), so the Taylor series converges for −1 < x/4 < 1, or
−4 < x < 4. The radius of convergence is 4.
(b) ln(4 + x) = ln(4(1 − (−x/4))) = ln(4) + ln(1 − (−x/4)), so the Taylor series converges for −1 < −x/4 < 1,
or 4 > x > −4. The radius convergence is 4.
(c) ln(1 + 4x2 ) = ln(1 − (−4x2 )), so the Taylor series converges for −1 < −4x2 < 1. This gives x2 < 1/4, or
−1/2 < x < 1/2. The radius of convergence is 1/2.
57. (c). The Taylor series for 3 tan(x/3) = 3(x/3 + (x/3)3 /3 + 21(x/3)5 /120 + · · ·) = x + x2 /27 + 7x5 /3240 + · · ·
Solutions for Section 10.4
Exercises
1. The error bound in approximating e01 using the Taylor polynomial of degree 3 for f (x) = ex about x = 0 is:
|E3 | = |f (0.1) − P3 (0.1)| ≤
M · |0.1 − 0|4
M (0.1)4
=
,
4!
24
where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.1. Now, f (4) (x) = ex . Since ex is increasing for all x, we see that |f (4) (x)| is
maximized for x between 0 and 0.1 when x = 0.1. Thus,
|f (4) | ≤ e0.1 ,
so
|E3 | ≤
e0.1 · (0.1)4
= 0.00000460.
24
966
Chapter Ten /SOLUTIONS
The Taylor polynomial of degree 3 is
P3 (x) = 1 + x +
1 2
1
x + x3 .
2!
3!
The approximation is P3 (0.1), so the actual error is
E3 = e0.1 − P3 (0.1) = 1.10517092 − 1.10516667 = 0.00000425,
which is slightly less than the bound.
2. The error bound in approximating sin(0.2) using the Taylor polynomial of degree 3 for f (x) = sin x about x = 0 is:
|E3 | = |f (0.2) − P3 (0.2)| ≤
M (0.2)4
M · |0.2 − 0|4
=
,
4!
24
where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.2. Now, f (4) (x) = sin x. By looking at the graph of sin x, we see that |f (4) (x)| is
maximized for x between 0 and 0.2 when x = 0.2. Thus,
|f (4) | ≤ sin(0.2),
so
|E3 | ≤
sin(0.2) · (0.2)4
= 0.0000132.
24
The Taylor polynomial of degree 3 is
P3 (x) = x −
1 3
x .
3!
The approximation is P3 (0.1), so the actual error is
E3 = sin(0.2) − P3 (0.2) = 0.19866933 − 0.19866667 = 0.00000266,
which is much less than the bound.
3. The error bound in approximating cos(−0.3) using the Taylor polynomial of degree 3 for f (x) = cos x about x = 0 is:
|E3 | = |f (−0.3) − P3 (−0.3)| ≤
M (−0.3)4
M · | − 0.3 − 0|4
=
,
4!
24
where |f (4) (x)| ≤ M for 0 ≥ x ≥ −0.3. Now, f (4) (x) = cos x, which has its largest value of 1 at x = 0. Thus,
|f (4) | ≤ cos 0 = 1,
so
|E3 | ≤
1 · (−0.3)4
= 0.000338.
24
The Taylor polynomial of degree 3 is
P3 (x) = 1 −
1 2
x .
2!
The approximation is P3 (−0.3), so the actual error is
E3 = cos(−0.3) − P (−0.3) = 0.955336 − 0.955000 = 0.000336,
which is slightly less than the bound.
√
√
0.9 using the Taylor polynomial of degree three for f (x) = 1 + x about x = 0 is:
4. The error bound in approximating
|E3 | = |f (−0.1) − P3 (−0.1)| ≤
M · (−0.1)4
M · | − 0.1 − 0|4
=
,
4!
24
where |f (4) | ≤ M for 0 ≥ x ≥ −0.1. Since
f (4) (x) = −
15
(1 + x)−7/2 ,
16
we see that if x is between 0 and −0.1, the maximum is at x = −0.1. Thus |f (4) x)| ≤ (15/16)(1 − 0.1)−7/2 , so
|E3 | ≤
(−0.1)4
15
(1 − 0.1)−7/2 ·
= 0.00000565.
16
24
10.4 SOLUTIONS
967
The Taylor polynomial of degree 3 is
1
x+
2
1
= 1+ x−
2
P3 (x) = 1 +
1 1 x2
1 1
3 x3
(− )
+ (− )(− )
2 2 2!
2 2
2 3!
1 2
1 3
x +
x .
8
16
The approximation is P3 (−0.1), so the actual error is
√
E3 = 0.9 − P3 (−0.1) = 0.94868330 − 0.94868750 = −0.00000420,
which is slightly less, in absolute value, than the bound.
5. The error bound in approximating ln(1.5) using the Taylor polynomial of degree 3 for f (x) = ln(1 + x) about x = 0 is:
|E4 | = |f (0.5) − P3 (0.5)| ≤
M (0.5)4
M · |0.5 − 0|4
=
,
4!
24
where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.5. Since
f (4) (x) =
−3!
(1 + x)4
and the denominator attains its minimum when x = 0, we have |f (4) (x)| ≤ 3!, so
|E4 | ≤
3! (0.5)4
= 0.0156.
24
The Taylor polynomial of degree 3 is
P3 (x) = 0 + x + (−1)
= x−
x3
x2
+ (−1)(−2)
2!
3!
1 2 1 3
x + x .
2
3
The approximation is P3 (0.5), so the actual error is
E3 = ln(1.5) − P3 (0.5) = 0.4055 − 0.4167 = −0.0112
which is slightly less, in absolute value, than the bound.
√
6. The error bound in approximating 1/ 3 using the Taylor polynomial of degree three for f (x) = (1 + x)−1/2 about
x = 0 is:
M · |2 − 0|4
M · 24
|E3 | = |f (2) − P3 (2)| ≤
=
,
4!
24
(4)
where |f | ≤ M for 0 ≤ x ≤ 2. Since
105
(1 + x)−9/2 ,
f (4) (x) =
16
. Thus,
we see that if x is between 0 and 2, then |f (4) x)| ≤ 105
16
|E3 | ≤
105 24
105
·
=
= 4.375.
16 24
24
This is not a very helpful bound on the error, but that is to be expected as the Taylor series does not converge at x = 2.
(At x = 2, we are outside the interval of convergence.)
The Taylor polynomial of degree 3 is
1
3 x2
1
3
5 x3
1
+ (− )(− )(− )
P3 (x) = 1 + ( )x + (− )(− )
2
2
2 2!
2
2
2 3!
1
3 2
5 3
= 1− x+ x −
x .
2
8
16
The approximation is P3 (2), so the actual error is
1
E3 = √ − P3 (2) = 0.577 − (−1.0) = 1.577,
3
which is much less than the bound, but still very large.
968
Chapter Ten /SOLUTIONS
7. The error bound in approximating tan(1) using the Taylor polynomial of degree three for f (x) = tan x about x = 0 is:
|E3 | = |f (1) − P3 (x)| ≤
M
M · |1 − 0|4
=
4!
24
where |f (4) (x)| ≤ M for 0 ≤ x ≤ 1. Now,
f (4) (x) =
24 sin3 x
16 sin x
+
.
cos3 x
cos5 x
From a graph of f (4) (x), we see that f (4) (x) is increasing for x between 0 and 1. Thus,
|f (4) (x)| ≤ |f (4) (1)| = 396,
so
396
= 16.5.
24
This is not a very helpful error bound! The reason the error bound is so huge is that x = 1 is getting near the vertical
asymptote of the tangent graph, and the fourth derivative is enormous there.
Observing that any term in f ′′ (x) or f ′′′ (x) involving tan x is zero at x = 0, we calculate the Taylor polynomial of
degree 3:
|E3 | ≤
P3 (x) = 0 + x + (0)
= x+
x3
x2
+ (2)
2!
3!
1 3
x .
3
The approximation is P3 (1), so the actual error is
E3 = tan 1 − P3 (1) = 1.557 − 1.333 = 0.224,
which is much less than the bound.
8. The error bound in approximating 0.51/3 using the Taylor polynomial of degree 3 for f (x) = (1 − x)1/3 about x = 0 is:
|E3 | = |f (0.5) − P3 (0.5)| ≤
M (0.5)4
M · |0.5 − 0|4
=
,
4!
24
where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.5. Now,
f (4) (x) = −
80
(1 − x)−11/3 .
81
By looking at the graph of (1 − x)−11/3 , we see that |f (4) (x)| is maximized for x between 0 and 0.5 when x = 0.5.
Thus,
80 1 −11/3
80 11/3
|f (4) | ≤
·2
,
=
81 2
81
so
80 · 211/3 · (0.5)4
= 0.033.
|E3 | ≤
81 · 24
The Taylor polynomial of degree 3 is
1
1 −2 (−x)2
1 −2 −5 (−x)3
(−x) +
+
3
3 3
2!
3 3 3
3!
1 2
5 3
1
x .
= 1− x− x −
3
9
81
P3 (x) = 1 +
The approximation is P3 (0.5), so the actual error is
E3 = (0.5)1/3 − P3 (0.5) = 0.79370 − 0.79784 = −0.00414,
which is much less than the bound.
Problems
9. (a) See the middle row of Table 10.1.
969
10.4 SOLUTIONS
(b) See the bottom row of Table 10.1.
Table 10.1
x
sin x
E1
−0.5
−0.4794
0.0206
−0.4
−0.3894
0.0106
−0.3
−0.2
−0.2955
−0.1987
0.0045
−0.1
0
0
0.0998
0.1987
0.2955
0.3894
0.4794
0.0002
0
−0.0002
−0.0013
−0.0045
−0.0106
−0.0206
−0.0998
0.0013
0.1
0.2
0.3
0.4
0.5
(c) See Figure 10.8.
0.03
E1
x
−0.5
0.5
−0.03
Figure 10.8
The fact that the graph of E1 lies between the horizontal lines at ±0.03 shows that |E1 | < 0.03 for −0.5 ≤
x ≤ 0.5.
√
√
10. Let f (x) = 1 + x. We use a Taylor polynomial
with x = 1 to approximate 2. The error bound for the Taylor
√
approximation of degree three for f (x) = 2 about x = 0 is:
|E3 | = |f (1) − P3 (1)| ≤
where |f (4) (x)| ≤ M for 0 ≤ x ≤ 1.
Now,
f (4) (x) = −
M
M · |1 − 0|4
=
,
4!
24
15
−15
(1 + x)−7/2 =
.
16
16(1 + x)7/2
Since 1 ≤ (1 + x)7/2 for x between 0 and 1, we see that
|f (4) (x)| =
for x between 0 and 1. Thus,
|E3 | ≤
15
7
16(1 + x) 2
≤
15
16
15
< 0.039
16 · 24
11. (a) The third degree Taylor approximation of degree 3 of ex around 0 is
P3 (x) = 1 + x +
x2
x3
+
.
2!
3!
Using the third-degree error bound, if |f (4) (x)| ≤ M for −2 ≤ x ≤ 2, then
|E3 (x)| = |f (x) − P3 (x)| ≤
M 24
M
· |x|4 ≤
.
4!
4!
Since |f (4) (x)| = ex , and ex is increasing on [−2, 2],
f (4) (x) ≤ e2 for all x ∈ [−2, 2].
So we can let M = e2 and we get
|E3 (x)| <
(b) The actual maximum error is |e2 − P3 (2)| = 1.06.
e2 · 24
≈5
4!
970
Chapter Ten /SOLUTIONS
12. For f (x) = cos x, note that f n+1 (x), is ± cos x or ± sin x, no matter what n is. So
f n+1 (x) < 1 for any x.
By the Lagrange error bound with M = 1, we have
En (x) = |f (x) − Pn (x)| ≤
|x|n+1
.
(n + 1)!
The right hand side of the equation above goes to zero as n → ∞.
|x|n+1
Hence we can always find n large enough that
< 10−9 for all x.
(n + 1)!
Note here that 10−9 in the question does not play any role. We could have replaced it by any small number.
13. (a) θ is the first degree approximation of f (θ) = sin θ; it is also the second degree approximation, since the next term in
the Taylor expansion is 0.
P1 (θ) = θ is an overestimate for 0 < θ ≤ 1, and is an underestimate for −1 ≤ θ < 0. (This can be seen easily from
a graph.)
(b) Using the second degree error bound, if |f (3) (θ)| ≤ M for −1 ≤ θ ≤ 1, then
|E2 | ≤
M · |θ|3
M
≤
.
3!
6
For what value of M is |f (3) (θ)| ≤ M for −1 ≤ θ ≤ 1? Well, |f (3) (θ)| = | − cos θ| ≤ 1. So |E2 | ≤
1
6
= 0.17.
3
θ
is the third degree Taylor approximation of f (θ) = sin θ; it is also the fourth degree approximation, since the
14. (a) θ −
3!
next term in the Taylor expansion is 0.
P3 (θ) is an underestimate for 0 < θ ≤ 1, and is an overestimate for −1 ≤ θ < 0. (This can be checked with a
calculator.)
(b) Using the fourth degree error bound, if |f (5) (θ)| ≤ M for −1 ≤ θ ≤ 1, then
|E4 | ≤
M · |θ|5
M
≤
.
5!
120
For what value of M is |f (5) (θ)| ≤ M for −1 ≤ θ ≤ 1? Since f (5) (θ) = cos θ and | cos θ| ≤ 1, we have
|E4 | ≤
1
≤ 0.0084.
120
15. (a) The Taylor polynomial of degree 0 about t = 0 for f (t) = et is simply P0 (x) = 1. Since et ≥ 1 on [0, 0.5], the
approximation is an underestimate.
(b) Using the zero degree error bound, if |f ′ (t)| ≤ M for 0 ≤ t ≤ 0.5, then
′
t
t
|E0 | ≤ M · |t| ≤ M (0.5).
Since |f (t)| = |e | = e is increasing on [0, 0.5],
|f ′ (t)| ≤ e0.5 <
√
4 = 2.
Therefore
|E0 | ≤ (2)(0.5) = 1.
(Note: By looking at a graph of f (t) and its 0th degree approximation, it is easy to see that the greatest error occurs
when t = 0.5, and the error is e0.5 − 1 ≈ 0.65 < 1. So our error bound works.)
16. (a) The second-degree Taylor polynomial for f (t) = et is P2 (t) = 1 + t + t2 /2. Since the full expansion of et =
1 + t + t2 /2 + t3 /6 + t4 /24 + · · · is clearly larger than P2 (t) for t > 0, P2 (t) is an underestimate on [0, 0.5].
(b) Using the second-degree error bound, if |f (3) (t)| ≤ M for 0 ≤ t ≤ 0.5, then
|E2 | ≤
M (0.5)3
M
· |t|3 ≤
.
3!
6
Since |f (3) (t)| = et , and et is increasing on [0, 0.5],
f (3) (t) ≤ e0.5 <
So
|E2 | ≤
√
4 = 2.
(2)(0.5)3
< 0.047.
6
10.4 SOLUTIONS
17. (a)
971
(i) The vertical distance between the graph of y = cos x and y = P10 (x) at x = 6 is no more than 4, so
|Error in P10 (6)| ≤ 4.
Since at x = 6 the cos x and P20 (x) graphs are indistinguishable in this figure, the error must be less than the
smallest division we can see, which is about 0.2 so,
|Error in P20 (6)| ≤ 0.2.
(ii) The maximum error occurs at the ends of the interval, that is, at x = −9, x = 9. At x = 9, the graphs of
y = cos x and y = P20 (x) are no more than 1 apart, so
Maximum error in P20 (x)
≤ 1.
for −9 ≤ x ≤ 9
(b) We are looking for the largest x-interval on which the graphs of y = cos x and y = P10 (x) are indistinguishable.
This is hard to estimate accurately from the figure, though −4 ≤ x ≤ 4 certainly satisfies this condition.
18. The maximum possible error for the nth -degree Taylor polynomial about x = 0 approximating cos x is |En | ≤
M ·|x−0|n+1
,
(n+1)!
where | cos(n+1) x| ≤ M for 0 ≤ x ≤ 1. Now the derivatives of cos x are simply cos x, sin x, − cos x, and − sin x. The
n+1
1
≤ (n+1)!
. The same argument
largest magnitude these ever take is 1, so | cos(n+1) (x)| ≤ 1, and thus |En | ≤ |x|
(n+1)!
works for sin x.
1
19. By the results of Problem 18, if we approximate cos 1 using the nth degree polynomial, the error is at most (n+1)!
.
For the answer to be correct to four decimal places, the error must be less than 0.00005. Thus, the first n such that
1
1
1
< 0.00005 will work. In particular, when n = 7, 8!
= 40370
< 0.00005, so the 7th degree Taylor polynomial
(n+1)!
will give the desired result. For six decimal places, we need
Taylor polynomial is sufficient.
1
(n+1)!
< 0.0000005. Since n = 9 works, the 9th degree
20. The graph of E0 looks like a parabola, and the graph shows
|E0 | < 0.01
|x| ≤ 0.1.
for
(In fact |E0 | < 0.005 on this interval.) Since
x4
x6
x2
+
−
+···,
2!
4!
6!
2
x
x4
x6
E0 = cos x − 1 = −
+
−
+ ···.
2!
4!
6!
cos x = 1 −
So, for small x,
E0 ≈ −
x2
,
2
and therefore the graph of E0 is parabolic. See Figure 10.9.
0.01
−0.1
0.1
x
E0
−0.01
Figure 10.9
21. Since f (x) = ex , the (n + 1)st derivative f (n+1) (x) is also ex , no matter what n is. Now fix a number x and let M = ex ,
then |f (n+1) (t)| ≤ et ≤ ex on the interval 0 ≤ t ≤ x. (This works for x ≥ 0; if x < 0 then we can take M = 1.) The
important observation is that for any x the same number M bounds all the higher derivatives f (n+1) (x).
972
Chapter Ten /SOLUTIONS
By the error bound formula, we now have
|En (x)| = |ex − Pn (x)| ≤
M |x|n+1
(n + 1)!
for every n.
To show that the errors go to zero, we must show that for a fixed x and a fixed number M ,
M
|x|n+1 → 0 as
(n + 1)!
n → ∞.
1
|x|n+1 → 0 as
(n + 1)!
n → ∞.
Since M is fixed, we need only show that
22.
This was shown in the text on page 562. Therefore, the Taylor series 1 + x + x2 /2! + · · · does converge to ex .
x5
x3
+
−···
3!
5!
Write the error in approximating sin x by the Taylor polynomial of degree n = 2k + 1 as En so that
sin x = x −
sin x = x −
x3
x5
x2k+1
+
− · · · (−1)k
+ En .
3!
5!
(2k + 1)!
(Notice that (−1)k = 1 if k is even and (−1)k = −1 if k is odd.) We want to show that if x is fixed, En → 0 as k → ∞.
Since f (x) = sin x, all the derivatives of f (x) are ± sin x or ± cos x, so we have for all n and all x
|f (n+1) (x)| ≤ 1.
Using the bound on the error given in the text on page 562, we see that
|En | ≤
1
|x|2k+2 .
(2k + 2)!
By the argument in the text on page 562, we know that for all x,
|x|n+1
|x|2k+2
=
→ 0 as
(2k + 2)!
(n + 1)!
n = 2k + 1 → ∞.
Thus the Taylor series for sin x does converge to sin x for every x.
23. (a) Since 4 arctan 1 = π, we approximate π by approximating 4 arctan x by Taylor polynomials with x = 1. Let
f (x) = 4 arctan x. We find the Taylor polynomial of f about x = 0.
f (0) = 0
4
so
1 + x2
8x
f ′′ (x) = −
(1 + x2 )2
f ′ (x) =
f ′′′ (x) = −
f ′ (0) = 4
so
f ′′ (0) = 0
8
32x2
+
2
2
(1 + x )
(1 + x2 )3
so
f ′′′ (0) = −8.
Thus, the third degree Taylor polynomial for f is
F3 (x) =
4x
8
4
− x3 = 4x − x3 .
1!
3!
3
In particular,
8
4
= ≈ 2.67.
3
3
Note: If you already have the Taylor series for 1/(1 + x2 ), the Taylor polynomial for arctan x can also be found by
integration.
F3 (1) = 4 −
10.4 SOLUTIONS
973
(b) We now approximate π by looking at g(x) = 2 arcsin x about x = 0 and substituting x = 1.
g(0) = 0
g ′ (x) = √
g ′′ (x) =
g ′′′ (x) =
2
1 − x2
2x
so
3
(1 − x2 ) 2
2
(1 −
g ′ (0) = 2
so
3
x2 ) 2
g ′′ (0) = 0
6x2
+
5
(1 − x2 ) 2
so
g ′′′ (0) = 2.
Thus, the third degree Taylor polynomial for g is
G3 (x) =
2x3
1
2x
+
= 2x + x3 .
1!
3!
3
In particular,
7
G3 (1) = ≈ 2.33.
3
√
2
Note: If you already have the Taylor series for 1/ 1 − x , the Taylor polynomial for arcsin x can also be found by
integration.
(c) To estimate the maximum possible error, |E3 |, in the approximation using the arctangent, we need a bound on the
fourth derivative of f (x) = arctan x on 0 ≤ x ≤ 1. Since
f (4) (x) = −
96x
192x3
+
,
(1 + x2 )4
(1 + x2 )3
now use a graphing calculator to see that the maximum value of |f (4) (x)|, on 0 ≤ x ≤ 1 is about 18.6. Thus,
|E3 | ≤
18.6
≈ 0.78.
4!
(Notice that π ≈ 3.14 is within 0.78 of 2.67.)
(d) To estimate the maximum possible error, |En |, in an approximation using the arcsine, we need a bound on the
derivatives of g(x) = arcsin x on 0 ≤ x ≤ 1. The derivatives of arcsin x contain terms of the form (1 − x2 )−a , for
some positive a. As x gets close to 1, the value of (1 − x2 )−a approaches ∞. Thus, we cannot get a bound on the
derivatives of arcsin x, so the error formula does not give us a bound on |En |.
Strengthen Your Understanding
24. This statement is not correct, since f (x) and its Taylor approximation Pn (x) around a have the same value at x = a. So
f (a) = Pn (a) for all n.
25. We can make the error |f (x) − Pn (x)| as small as we want for any given x by picking a large n. However, we may not
be able to find a Taylor polynomial that produces a small error for all values of x simultaneously.
For example, if f (x) = cos x, then every Taylor polynomial Pn (x) goes to ±∞ as x → ∞ and as x → −∞. This
means that no matter what n we choose, for large values of x, the value of Pn (x) is more than 1 away from f (x).
26. A possible answer is sin x, or ex . Other examples are possible.
27. We’ll try using Taylor polynomials for the function f (x) = 1/x about x = 1. We have
∞
f (x) =
X
1
1
=
=
(−1)n · (x − 1)n
x
1 + (x − 1)
n=0
th
We know that, if Pn (x) is the n -degree Taylor polynomial for f (x) at x = 1, then
|f (x) − Pn (x)| ≤
M
· |x − 1|n+1 ,
(n + 1)!
where M is the maximum value of the absolute value of the (n + 1)st derivative of f (x) on [1, 1.5]. The (n + 1)st
derivative of f (x) is
(−1)n+1 · (n + 1)!
f (n+1) (x) =
,
xn+2
974
Chapter Ten /SOLUTIONS
and the absolute value of this derivative is never more than (n + 1)! on [1, 1.5]. So M = (n + 1)!, and therefore
|f (x) − Pn (x)| ≤ |x − 1|n+1 .
We want to make sure this is less than 0.1 on the interval [1, 1.5], so we want |1.5 − 1|n+1 < 0.1. This is true if n = 3,
so we use the Taylor polynomial
P3 (x) = 1 − (x − 1) + (x − 1)2 − (x − 1)3 .
28. According to the error formula we need max |f (n+1) | ≤ M on the interval between 0 and x. Since we are working with
the second-degree Taylor polynomial we have n = 2. Therefore, we need a function and an interval such that the third
derivative of the function on that interval is less or equal to 4. We can choose, for example, f (x) = ex and c small enough
such that f (3) (x) = ex ≤ 4 on [−c, c]. There are many choices of c, for example c = 1 would work.
29. False. If f is itself a polynomial of degree n then it is equal to its nth Taylor polynomial.
30. True. By Theorem 10.1, |En (x)| < 10|x|n+1 /(n + 1)!. Since limn→∞ |x|n+1 /(n + 1)! = 0, En (x) → 0 as n → ∞,
so the Taylor series converges to f (x) for all x.
True
P∞
31. False. The Taylor series for f near x = 0 always converges at x = 0, since
C0 .
32. True. When x = 1,
∞
X
f (n) (0)
n=0
n!
∞
X
f (n) (0)
n
x =
n!
n=0
n=0
Cn xn at x = 0 is just the constant
.
Since f (n) (0) ≥ n!, the terms of this series are all greater than 1. So the series cannot converge
33. False. For example, if f (n) (0) = n!, then the Taylor series is
∞
X
f (n) (0)
n=0
n!
xn =
∞
X
xn ,
n=0
which converges at x = 1/2.
Solutions for Section 10.5
Exercises
1. No, a Fourier series has terms of the form cos nx, not cosn x.
2. Not a Fourier series because terms are not of the form sin nx.
3. Yes. Terms are of the form sin nx and cos nx.
4. Yes. This is a Fourier series where the cos nx terms all have coefficients of zero.
5.
1
a0 =
2π
a1 =
1
π
Z
1
f (x) dx =
2π
−π
Z
0
1
π
Z
0
Z
π
π
−π
f (x) cos x dx =
−π
−π
"
−1 dx +
Z
π
1
− sin x
=
π
1 dx = 0
0
− cos x dx +
0
Z
π
cos x dx
0
π
+ sin x
−π
0
#
= 0.
10.5 SOLUTIONS
Similarly, a2 and a3 are both 0. See Figure 10.10. R
π
(In fact, notice f (x) cos nx is an odd function, so −π f (x) cos nx = 0.)
1
b1 =
π
π
Z
1
f (x) sin x dx =
π
−π
=
1
b2 =
π
Z
1
b3 =
π
Thus, F1 (x) = F2 (x) =
sin x and F3 (x) =
0
−π
−π
sin x dx
0
π
−π
− sin 2x dx +
0
−π
0
−π
0
π
Z
=
π
F1 (x) = F2 (x) =
sin 3x dx
1
+ (− cos 3x)
3
π
0
=
4
.
3π
−π
x
π
−1
4
π
F3 (x) =
sin x
Figure 10.10
4
π
sin x +
1
f (x) dx =
2π
−π
4
3π
sin 3x
Figure 10.11
6. First,
Z
= 0.
0
−1
1
an =
π
0
sin 3x. See Figure 10.11.
x
π
π
π
1
−π
Z
4
π
sin 2x dx
Z
1
1
a0 =
2π
0
1
+ (− cos 2x)
2
− sin 3x dx +
1 1
cos 3x
π 3
4
3π
π
+ (− cos x)
0
Z
sin x +
Z
0
Z
1
f (x) sin 3x dx =
π
−π
4
π
− sin x dx +
1 1
cos 2x
π 2
π
=
4
π
1
f (x) sin 2x dx =
π
−π
Z
−π
1
cos x
π
π
=
0
Z
975
Z
0
−π
−x dx +
To find the ai ’s, we use the integral table. For n ≥ 1,
π
1
f (x) cos(nx) dx =
π
−π
=
1
π
Z
+
1
=
π
=
0
π
1
x2
−
x dx =
2π
2
0
−π
Z
−
−x cos(nx) dx +
Z
0
−π
x2
+
2
π
x cos(nx) dx
0
x
1
sin(nx) − 2 cos(nx)
n
n
1
x
sin(nx) + 2 cos(nx)
n
n
0
π
0
=
−π
π
0
1
1
1
1
− 2 + 2 cos(−nπ) + 2 cos(nπ) − 2
n
n
n
n
2
(cos nπ − 1)
πn2
π
.
2
4
and a3 = − 9π
. See Figure 10.12. To find the bi ’s, note that f (x) is even, so for n ≥ 1,
Thus, a1 = − π4 , a2 = 0, Z
π
f (x) sin(nx) is odd. Thus,
f (x) sin(nx) = 0, so all the bi ’s are 0. F1 = F2 = π2 − π4 cos x, F3 = π2 − π4 cos x −
4
9π
−π
cos 3x. See Figure 10.13.
976
Chapter Ten /SOLUTIONS
π
π
−π
π
F1 (x) = F2 (x) =
π
2
−
4
π
−π
x
cos x
π
F3 (x) =
π
2
Figure 10.12
−
4
π
cos x −
4
9π
x
cos 3x
Figure 10.13
7. The energy of the function f (x) is
1
E=
π
Z
π
1
(f (x)) dx =
π
−π
2
Z
π
x2 dx =
−π
1 3
x
3π
π
−π
2π 3
2
1 3
(π − (−π 3 )) =
= π 2 = 6.57974.
=
3π
3π
3
From Problem 6, we know all the bi ’s are 0 and a0 =
constant term and first three harmonics is
π
,
2
4
a1 = − π4 , a2 = 0 , a3 = − 9π
. Therefore the energy in the
A20 + A21 + A22 + A23 = 2a20 + a21 + a22 + a23
=2
which means that they contain
π2
4
+
6.57596
= 0.99942 ≈ 99.942% of the total energy.
6.57974
8. First, we find a0 .
a0 =
1
2π
Z
π
x2 dx =
−π
1
2π
To find an , n ≥ 1, we use the integral table (III-15 and III-16).
an =
1
π
Z
16
16
+0+
= 6.57596
π2
81π 2
π
x2 cos nx dx =
−π
x3
3
π
−π
=
π2
3
2x
2
1 x2
sin(nx) + 2 cos(nx) − 3 sin(nx)
π n
n
n
1 2π
2π
=
cos(nπ) + 2 cos(−nπ)
π n2
n
4
= 2 cos(nπ)
n
π
−π
Again, cos(nπ) = (−1)n for all integers n, so an = (−1)n n42 . Note that
bn =
1
π
Z
π
x2 sin nx dx.
−π
x2 is an even function, and sin nx is odd, so x2 sin nx is odd. Thus
th
We deduce that the n
Fourier polynomial for f (where n ≥ 1) is
n
Fn (x) =
−π
x2 sin nx dx = 0, and bn = 0 for all n.
π2 X
4
+
(−1)i 2 cos(ix).
3
i
i=1
In particular, we have the graphs in Figure 10.14.
Rπ
10.5 SOLUTIONS
✛
✛
✐
π2
F3 (x)
f (x) = x2
F2 (x)
✛
π
π
F1 (x)
x
Figure 10.14
9.
1
a0 =
2π
Z
π
1
h(x) dx =
2π
−π
Z
π
x dx =
0
π
4
As in Problem 10, we use the integral table (III-15 and III-16) to find formulas for an and bn .
an =
1
π
Z
π
1
π
h(x) cos(nx) dx =
−π
Z
π
x cos nx dx =
0
1
π
1
=
π
=
1
x
sin(nx) + 2 cos(nx)
n
n
1
1
cos(nπ) − 2
n2
n
1
n2 π
π
0
cos(nπ) − 1 .
Note that since cos(nπ) = (−1)n , an = 0 if n is even and an = − n22π if n is odd.
h(x) cos(nx) dx =
−π
1
π
=
π
Z
1
π
bn =
1
=
π
=−
−
Z
π
π
− cos(nπ)
n
1
cos(nπ)
n
x sin x dx
0
x
1
cos(nx) + 2 sin(nx)
n
n
1
(−1)n+1
n
=
1
π
π
0
if n ≥ 1
We have that the nth Fourier polynomial for h (for n ≥ 1) is
n
Hn (x) =
π X
+
4
i=1
This can also be written as
1
i2 π
cos(iπ) − 1
· cos(ix) +
(−1)i+1 sin(ix)
.
i
[ n2 ]
n
−2
π X (−1)i+1 sin(ix) X
+
cos((2i − 1)x)
Hn (x) = +
4
i
(2i − 1)2 π
i=1
where
n
2
denotes the biggest integer smaller than or equal to
i=1
n
.
2
In particular, we have the graphs in Figure 10.15.
977
978
Chapter Ten /SOLUTIONS
✛
π
✛
h(x)
H3 (x)
✛
✛
H2 (x)
H1 (x)
x
π
−π
Figure 10.15
10. To find the nth Fourier polynomial, we must come up with a general formula for an and bn . First, we find a0 .
1
a0 =
2π
Z
π
π
π
1 x2
x dx =
2π 2
−π
Z
1
g(x) dx =
2π
−π
−π
Now we use the integral table (III-15 and III-16) to find an and bn for n ≥ 1.
1
an =
π
π
Z
1
x cos nx dx =
π
−π
1
π
=
x
1
sin(nx) + 2 cos(nx)
n
n
bn =
1
π
π
x sin nx dx =
−π
1
π
1
=
π
π
−π
1
1
cos(nπ) − 2 cos(−nπ)
n2
n
(Note that since x cos nx is odd, we could have deduced that
Z
=0
Rπ
−
−π
=0
x cos nx = 0.)
x
1
cos(nx) + 2 sin(nx)
n
n
π
π
− cos(nπ) − cos(−nπ)
n
n
2
cos(nπ)
n
Notice that cos(nπ) = (−1)n for all integers n, so bn = (−1)n+1 ( n2 ).
Thus the nth Fourier polynomial for g is
π
−π
=−
Gn (x) =
n
X
(−1)i+1
i=1
In particular, we have the graphs in Figure 10.16.
2
sin(ix).
i
✛
π
✛
✛
✛
−π
π
−π
Figure 10.16
g(x)
G3 (x)
G2 (x)
G1 (x)
x
10.5 SOLUTIONS
979
Problems
11. The period of f (x) is equal to 2. Since a0 , the constant term of the Fourier series of f (x) is the average value of f over
the interval [−1, 1], it is given by the following integral:
1
a0 =
2
Z
1
|x|dx.
−1
By looking at the graph of f (x) between −1 and 1, we see that the area under f is equal to 1. Multiplying this by 1/2 we
have
1
a0 = .
2
We can check this analytically:
1
a0 =
2
Z
1
Z
1
|x|dx =
2
−1
0
(−x)dx +
−1
1
Z
xdx
0
=
1
2
1
1
+
2
2
12. In Problem 10 we found that the Fourier polynomial of g(x) = x of degree n is:
Gn (x) =
n
X
(−1)i+1
i=1
Hence the Fourier series of g(x) is:
g(x) = x =
∞
X
i=1
By substituting x = π/2 and expanding we get
2
sin(ix).
i
2
(−1)i+1 sin(ix).
i
π
2
2
= 2 − + − ···
2
3
5
Factoring 2 out from the right hand side and dividing both sides by 2, we have
∞
X
(−1)k+1
k=1
13. (a)
(i) The graph of y = sin x +
1
3
1
π
= .
2k − 1
4
sin 3x is in Figure 10.17.
y
1
y = sin x +
−π
−2π
2π
π
1
3
sin 3x
x
−1
Figure 10.17
(ii) The graph of y = sin x +
1
3
sin 3x +
1
5
sin 5x is in Figure 10.17.
y
y = sin x +
1
−2π
−π
π
−1
Figure 10.18
2π
1
3
x
sin 3x +
1
5
sin 5x
=
1
.
2
980
Chapter Ten /SOLUTIONS
(b) Following the pattern, we add the term
1
7
sin 7x to get Figure 10.19.
y
−π
−2π
1
3
y = sin x +
1
2π
π
1
5
sin 3x +
sin 5x +
1
7
sin 7x
x
−1
Figure 10.19
(c) The equation is
f (x) =
..
.
1 − 2π
..
.
≤ x < −π
−1
−π ≤ x < 0
.
π ≤ x < 2π
..
.
1
−1
..
0≤x<π
The square wave function is not continuous at x = 0, ± π, ± 2π, . . .. See Figure 10.20.
y
1
−3π −2π
x
π
−π
2π
3π
−1
Figure 10.20
14. (a) The graph of g(x) is in Figure 10.21.
1
−π
g(x)
−π/2
π
π/2
x
Figure 10.21
First find the Fourier coefficients: a0 is the average value of g on [−π, π] so from the graph, it is clear that
a0 =
1
1
(π × 1) = ,
2π
2
or analytically,
1
a0 =
2π
Z
π
1
g(x) dx =
2π
−π
1
1
(π) = ,
=
2π
2
Z
π/2
−π/2
1 dx =
1
x
2π
π/2
=
−π/2
1
2π
π
π
− −
2
2
10.5 SOLUTIONS
ak =
=
bk =
1
π
π
Z
−π
kπ
sin
− sin −
2
2
kπ
π
Z
g(x) sin kx dx =
−π
1
=−
kπ
So,
Z
1
π
Z
cos kx dx =
1
=
kπ
kπ
2 sin
2
1
sin kx
kπ
π/2
−π/2
,
π/2
sin kx dx = −
−π/2
kπ
kπ
cos
− cos −
2
2
π/2
−π/2
kπ
1
1
π
1
π
g(x) cos kx dx =
981
1
cos kx
kπ
π/2
−π/2
1
. = − (0) = 0
kπ
π
2
1
2 sin
= ,
π
2
π
2π
1
2 sin
= 0,
a2 =
2π
2
1
3π
2
a3 =
2 sin
=− ,
3π
2
3π
a1 =
which gives
2
2
1
+ cos x −
cos 3x.
2
π
3π
F3 (x) =
See Figure 10.22.
1
−π
F3 (x)
g(x)
π/2
π/2
x
π
Figure 10.22
(b) There are cosines instead of sines (but the energy spectrum remains the same).
15. We have f (x) = x, 0 ≤ x < 1. Let t = 2πx − π. Notice that as x varies from 0 to 1, t varies from −π to π. Thus
if we rewrite the function in terms of t, we can find the Fourier series in terms of t in the usual way. To do this, let
g(t) = f (x) = x = t+π
on −π ≤ t < π. We now find the fourth degree Fourier polynomial for g.
2π
ao =
1
2π
Z
π
g(t) dt =
−π
1
2π
Z
π
−π
t+π
1
dt =
2π
(2π)2
Notice, a0 is the average value of both f and g. For n ≥ 1,
1
an =
π
=
Z
−π
1
2π 2
= 0.
bn =
=
1
π
π
Z
h
t+π
1
cos(nt)dt =
2π
2π 2
Z
−π
t+π
1
sin(nt) dt =
2π
2π 2
π
=
−π
(t cos(nt) + π cos(nt))dt
−π
Z
i
π
−π
π
(t sin(nt) + π sin(nt)) dt
−π
t
1
π
1
− cos(nt) + 2 sin(nt) − cos(nt)
2π 2
n
n
n
h
π
t
1
π
sin(nt) + 2 cos(nt) + sin(nt)
n
n
n
π
t2
+ πt
2
i
π
−π
4π
2
2
1
(−
cos(πn)) = −
cos(πn) =
(−1)n+1 .
=
2π 2
n
πn
πn
1
2
982
Chapter Ten /SOLUTIONS
We get the integrals for an and bn using the integral table (formulas III-15 and III-16).
Thus, the Fourier polynomial of degree 4 for g is:
G4 (t) =
2
1
2
1
1
+ sin t − sin 2t +
sin 3t −
sin 4t.
2
π
π
3π
2π
Now, since g(t) = f (x), the Fourier polynomial of degree 4 for f can be found by replacing t in terms of x again. Thus,
F4 (x) =
2
1
2
1
1
+ sin(2πx − π) − sin(4πx − 2π) +
sin(6πx − 3π) −
sin(8πx − 4π).
2
π
π
3π
2π
Now, using the fact that sin(x − π) = − sin x and sin(x − 2π) = sin x, etc., we have:
F4 (x) =
1
2
1
2
1
− sin(2πx) − sin(4πx) −
sin(6πx) −
sin(8πx).
2
π
π
3π
2π
See Figure 10.23.
f (x) = x
1
F4 (x)
x
1
Figure 10.23
16. Since the period is 2, we make the substitution t = πx − π. Thus, x = t+π
. We find the Fourier coefficients. Notice
π
that all of the integrals are the same as in Problem 15 except for an extra factor of 2. Thus, a0 = 1, an = 0, and
4
bn = πn
(−1)n+1 , so:
4
2
4
1
sin 3t − sin 4t.
G4 (t) = 1 + sin t − sin 2t +
π
π
3π
π
Again, we substitute back in to get a Fourier polynomial in terms of x:
2
4
sin(πx − π) − sin(2πx − 2π)
π
π
1
4
sin(3πx − 3π) − sin(4πx − 4π)
+
3π
π
4
2
4
1
= 1 − sin(πx) − sin(2πx) −
sin(3πx) − sin(4πx).
π
π
3π
π
F4 (x) = 1 +
See Figure 10.24.
3
2
f (x) = x
1
F4 (x)
x
1
2
−1
Figure 10.24
Notice in this case, the terms in our series are sin(nπx), not sin(2πnx), as in Problem 15. In general, the terms will
x), where b is the period.
be sin(n 2π
b
10.5 SOLUTIONS
983
17. The signal received on earth is in the form of a periodic function h(t), which can be expanded in a Fourier series
h(t) = a0 + a1 cos t + a2 cos 2t + a3 cos 3t + · · ·
+ b1 sin t + b2 sin 2t + b3 sin 3t + · · ·
If the periodic noise consists of only the second and higher harmonics of the Fourier series, then the original signal
contributed the fundamental harmonic plus the constant term, i.e.,
+ a1 cos t + b1 sin t =
a
0
|{z}
|
constant term
{z
}
cos }t
|A {z
.
original signal
fundamental harmonic
In order to find A, we need to find a0 , a1 , and b1 . Looking at the graph of h(t), we see
1
(Area above the x-axis – Area below the x-axis)
2π
h
i
π
π
π
π
π
1
80
− 50
+ 30
+ 30
+ 50
=
2π
2
4
4
4
4
π i
h π
1
1
=
80
− 80
=
· 0, = 0
2π
2
2
2π
a0 = average value of h(t) =
a1 =
1
π
1
=
π
Z
π
h(t) cos t dt
−π
−3π/4
Z
−50 cos t dt +
−π
+
π/4
Z
80 cos t dt +
−π/4
=
"
−π/2
Z
0 cos t dt +
−3π/4
−30 cos t dt +
π/4
1
−50 sin t
π
−3π/4
−π/4
− 30 sin t
−π
+80 sin t
−π/4
− 30 sin t
3π/4
0 cos t dt +
π/2
Z
π
3π/4
−50 cos t dt
π/4
π
− 50 sin t
1
π
1
=
π
Z
√
3π/4
#
√
√
2
2
2
2
1
− 0 − 30 −
− (−1) + 80
− −
−50 −
π
2
2
2
2
√
√
2
2
−30 1 −
− 50 0 −
2
2
√
√
√
√
√
1 √
= [25 2 + 15 2 − 30 + 40 2 + 40 2 − 30 + 15 2 + 25 2]
π
√
1
= [160 2 − 60] = 52.93,
π
=
b1 =
√
Z
−30 cos t dt
−π/2
π/2
π/4
−π/4
−π/2
π/2
Z
Z
π
h(t) sin t dt
−π
Z
+
−3π/4
−π
Z
−50 sin t dt +
π/4
80 sin t dt +
−π/4
"
−3π/4
Z
Z
−π/2
0 sin t dt +
−3π/4
π/2
π/4
−30 sin t dt +
−π/4
Z
−π/4
−π/2
Z
−30 sin t dt
3π/4
π/2
π/4
0 sin t dt +
Z
π
3π/4
π/2
−50 sin t dt
1
50 cos t
=
+ 30 cos t
− 80 cos t
+ 30 cos t
+ 50 cos t
π
−π
−π/2
−π/4
π/4
√
√
√
√
1
2
2
2
2
− (−1) + 30
− 0 − 80
−
=
50 −
π
2
2
2
2
π
3π/4
#
984
Chapter Ten /SOLUTIONS
√
√
2
2
)
+ 50 −1 − (−
2
2
√
√
√
√
1
1
−25 2 + 50 + 15 2 − 0 − 15 2 − 50 + 25 2 = (0) = 0.
=
π
π
+30 0 −
1
π
Also, we could have just noted that b1 =
Substituting in, we get
Rπ
−π
h(t) sin t dt = 0 because h(t) sin t is an odd function.
a0 + a1 cos t + b1 sin t = 0 + 52.93 cos t + 0 = A cos t.
So A = 52.93.
18. The energy spectrum of the flute shows that the first two harmonics have equal energies and contribute the most energy
by far. The higher harmonics contribute relatively little energy. In contrast, the energy spectrum of the bassoon shows the
comparative weakness of the first two harmonics to the third harmonic which is the strongest component.
19. Let f (x) = ak cos kx + bk sin kx. Then the energy of f is given by
1
π
Z
π
(f (x))2 dx =
−π
Z
1
π
π
(ak cos kx + bk sin kx)2 dx
Z−π
π
1
=
π
−π
1 2
ak
=
π
(a2k cos2 kx − 2ak bk cos kx sin kx + b2k sin2 kx) dx
Z
π
−π
cos2 kx dx − 2ak bk
Z
π
cos kx sin kx dx + b2k
−π
Z
π
sin2 kx dx
−π
1 2
ak π − 2ak bk · 0 + b2k π = a2k + b2k .
=
π
20. Since each square in the graph has area ( π4 ) · (0.2),
a0 =
1
2π
Z
π
f (x) dx
−π
π
1
· ( ) · (0.2) [Number of squares under graph above x-axis
2π 4
– Number of squares above graph below x axis]
1
π
≈
· ( ) · (0.2) · [13 + 11 − 14] = 0.25.
2π 4
=
Approximate the Fourier coefficients using Riemann sums.
a1 =
1
π
π
Z
f (x) cos x dx
−π
π
π
1
f (−π) cos(−π) + f −
cos −
+ f (0) cos(0) + f
π
2
2
1
π
= [(0.92)(−1) + (1)(0) + (−1.7)(1) + (0.7)(0)] ·
π
2
= −1.31
≈
h
π
2
cos
π i π
2
·
2
π
2
·
π
2
Similarly for b1 :
1
b1 =
π
Z
π
f (x) sin x dx
−π
1
π
π
f (−π) sin(−π) + f −
sin −
+ f (0) sin(0) + f
π
2
2
π
1
= [(0.92)(0) + (1)(−1) + (−1.7)(0) + (0.7)(1)] ·
π
2
= −0.15.
≈
h
So our first Fourier approximation is
F1 (x) = 0.25 − 1.31 cos x − 0.15 sin x.
π
2
sin
i
10.5 SOLUTIONS
985
See Figure 10.25
Similarly for a2 :
π
Z
1
a2 =
π
f (x) cos 2x dx
−π
1
π
f (−π) cos(−2π) + f −
cos(−π) + f (0) cos(0) + f
π
2
π
1
= [(0.92)(1) + (1)(−1) + (−1.7)(1) + (0.7)(−1)] ·
π
2
= −1.24
h
≈
π
2
cos(−π) ·
π
2
sin(−π) ·
i
π
2
i
π
2
Similarly for b2 :
b2 =
1
π
Z
π
f (x) sin 2x dx
−π
π
1
f (−π) sin(−2π) + f −
sin(−π) + f (0) sin(0) + f
π
2
π
1
= [(0.92)(0) + (1)(0) + (−1.7)(0) + (0.7)(0)] ·
π
2
= 0.
≈
h
So our second Fourier approximation is
F2 (x) = 0.25 − 1.31 cos x − 0.15 sin x − 1.24 cos 2x.
See Figure 10.26.
y
y
2
y = F1 (x)
1
−2π
−π
y = F2 (x)
2
π
x
1
−2π
2π
−1
π
−π
x
2π
−2
−2
Figure 10.25
Figure 10.26
As you can see from comparing our graphs of F1 and F2 to the original, our estimates of the Fourier coefficients are
not very accurate.
There are other methods of estimating the Fourier coefficients such as taking other Riemann sums, using Simpson’s
rule, and using the trapezoid rule. With each method, the greater the number of subdivisions, the more accurate the
estimates of the Fourier coefficients.
The actual function graphed in the problem was
y=
sin( 53 )
2
cos 1
1
− 1.3 cos x −
sin x − cos 2x −
sin 2x
4
π
π
3π
= 0.25 − 1.3 cos x − 0.18 sin x − 0.63 cos 2x − 0.057 sin 2x.
21. The Fourier series for f is
f (x) = a0 +
∞
X
k=1
ak cos kx +
∞
X
bk sin kx.
k=1
Pick any positive integer m. Then multiply through by sin mx, to get
f (x) sin mx = a0 sin mx +
∞
X
k=1
ak cos kx sin mx +
∞
X
k=1
bk sin kx sin mx.
986
Chapter Ten /SOLUTIONS
Now, integrate term-by-term on the interval [−π, π] to get
Z
π
π
Z
f (x) sin mx dx =
−π
a0 sin mx +
−π
= a0
Z
∞
X
ak cos kx sin mx +
k=1
∞
π
Z
π
Z
π
ak
−π
cos kx sin mx dx
−π
k=1
∞
X
bk
+
bk sin kx sin mx
k=1
X
sin mx dx +
∞
X
dx
sin kx sin mx dx .
−π
k=1
!
Rπ
Since m is a positive integer, we know that the first term of the above expression is zero (because −π sin mx dx = 0).
Rπ
Rπ
Since −π cos kx sin mx dx = 0, we know that everything in the first infinite sum is zero. Since −π sin kx sin mx dx =
0 where k 6= m, the second infinite sum reduces down to the case where k = m so
Z
π
f (x) sin mx dx = bm
−π
π
Z
sin mx sin mx dx = bm π.
−π
Divide by π to get
bm
π
Z
1
=
π
f (x) sin mx dx.
−π
22. (a) See Figure 10.27.
1
−3π
−π
−2π
f (x)
− 21
x
π
1
2
2π
3π
Figure 10.27
The energy of the pulse train f is
E=
π
Z
1
π
(f (x))2 dx =
−π
1
π
1/2
Z
12 dx =
−1/2
1
π
1
2
1
− −
2
=
1
.
π
Next, find the Fourier coefficients:
a0 = average value of f on [−π, π] =
1
ak =
π
Z
1
f (x) cos kx dx =
π
−π
1
=
kπ
1
bk =
π
π
Z
k
sin
2
k
− sin −
2
π
k
cos
2
Z
k
− cos −
2
1/2
cos kx dx =
−1/2
1
f (x) sin kx dx =
π
−π
1
=−
kπ
Z
1
1
1
( Area) =
(1) =
,
2π
2π
2π
1
=
kπ
1
sin kx
kπ
k
2 sin
2
1/2
−1/2
sin kx dx = −
1
(0) = 0.
=
kπ
1/2
−1/2
,
1
cos kx
kπ
1/2
−1/2
10.5 SOLUTIONS
987
The energy of f contained in the constant term is
A20 = 2a20 = 2
which is
1
2π
2
1
2π 2
=
1/2π 2
A20
1
=
=
≈ 0.159155 = 15.9155%
E
1/π
2π
The fraction of energy contained in the first harmonic is
A21
E
=
a21
E
=
1
2 sin 2
π
1
π
2
of the total.
≈ 0.292653.
The fraction of energy contained in both the constant term and the first harmonic together is
A2
A20
+ 1 ≈ 0.159155 + 0.292653 = 0.451808%.
E
E
(b) The formula for the energy of the kth harmonic is
A2k = a2k + b2k =
2 sin k2
kπ
2
+ 02 =
4 sin2 k2
.
k2 π 2
By graphing it as a continuous function for k ≥ 1, we see its overall behavior as k gets larger. See Figure 10.28. The
energy spectrum for the first five terms is graphed below as well in Figure 10.29.
A2k
y
0.1
0.08
0.06
0.04
0.02
y=
1
0.1
0.08
0.06
4 sin2 k
2
k2 π 2
5
10
15
20
k
1
2π 2
1
0
Figure 10.28
2
3
4
5
k
Figure 10.29
(c) The constant term and the first five harmonics are needed to capture 90% of the energy off . This was determined by
adding the fractions of energy of f contained in each harmonic until the sum reached at least 90% of the total energy
of f :
A20
A2
A2
A2
A2
A2
+ 1 + 2 + 3 + 4 + 5 ≈ 90.1995%.
E
E
E
E
E
E
(d) F5 (x) =
1
2π
+
1)
2 sin( 2
π
cos x +
sin 1
π
cos 2x +
3)
2 sin( 2
3π
✛
1
−3π
−2π
−π
− 21
cos 3x +
sin 2
2π
cos 4x +
5)
2 sin( 2
5π
F5 (x)
f (x)
1
2
Figure 10.30
π
x
2π
3π
cos 5x. See Figure 10.30.
988
Chapter Ten /SOLUTIONS
23. (a) See Figure 10.31.
1
f (x)
−3π
−π
−2π
− 51
x
π
1
5
2π
3π
Figure 10.31
The energy of the pulse train f is
E=
1
π
Z
π
−π
1/5
Z
1
π
(f (x))2 dx =
12 dx =
−1/5
1
π
1
5
1
− −
5
=
2
.
5π
Next, find the Fourier coefficients:
a0 = average value of f on [−π, π] =
1
ak =
π
Z
1
f (x) cos kx dx =
π
−π
1
=
kπ
1
bk =
π
π
Z
k
sin
5
k
− sin −
5
π
k
cos
5
cos kx dx =
1
=
kπ
Z
k
− cos −
5
The energy of f contained in the constant term is
A20 = 2a20 = 2
which is
1/5
−1/5
1
f (x) sin kx dx =
π
−π
1
=−
kπ
Z
1
1
( Area) =
2π
2π
−1/5
1/5
sin kx dx = −
E
=
a21
E
=
1
5π
1
2 sin 5
π
2
5π
=
1
,
5π
1/5
−1/5
,
1
cos kx
kπ
1/5
−1/5
1
=
(0) = 0.
kπ
2
=
2
25π 2
2/25π 2
1
A20
=
=
≈ 0.063662 = 6.3662%
E
2/5π
5π
The fraction of energy contained in the first harmonic is
A21
5
1
sin kx
kπ
k
2 sin
5
2
2
of the total.
≈ 0.12563.
The fraction of energy contained in both the constant term and the first harmonic together is
A20
A2
+ 1 ≈ 0.06366 + 0.12563 = 0.18929 = 18.929%.
E
E
(b) The formula for the energy of the kth harmonic is
A2k = a2k + b2k =
2 sin k5
kπ
2
+ 02 =
4 sin2 k5
.
k2 π 2
10.5 SOLUTIONS
989
By graphing this formula as a continuous function for k ≥ 1, we see its overall behavior as k gets larger in Figure 10.32. The energy spectrum for the first five terms is shown in Figure 10.33.
A2k
y
0.02
0.02
y=
0.015
4 sin2 k
5
k2 π 2
0.015
0.01
0.01
0.005
0.005
1
10
20
30
40
k
2
52 π
1
0
2
Figure 10.32
3
4
k
5
Figure 10.33
(c) The constant term and the first five harmonics contain
A20
A2
A2
A2
A2
A2
+ 1 + 2 + 3 + 4 + 5 ≈ 61.5255%
E
E
E
E
E
E
of the total energy of f .
(d) The fifth Fourier approximation to f is
2 sin( 1 )
sin( 2 )
2 sin( 3 )
sin( 4 )
1
1
5
+
cos x + π 5 cos 2x + 3π 5 cos 3x + 2π5 cos 4x + 2 sin
cos 5x. See Figure 10.34.
F5 (x) = 5π
π
5π
For comparison, Figure 10.35 shows the thirteenth Fourier approximation to f .
✛
1
f (x)
F13 (x)
1
F5 (x)
−3π
−2π
−π
− 51 15
x
π
2π
3π
−3π
−2π
Figure 10.34
−π
π
− 51 15
Figure 10.35
24. (a) See Figure 10.36.
1
−3π
−π
−2π
−1
f (x)
x
π
1
2π
3π
Figure 10.36
The energy of the pulse train f is
1
E=
π
Z
π
1
(f (x)) dx =
π
−π
2
Z
1
−1
12 =
1
2
(1 − (−1)) = .
π
π
Next, find the Fourier coefficients:
a0 = average value of f on [−π, π] =
1
1
1
( Area) =
(2) = ,
2π
2π
π
x
2π
3π
990
Chapter Ten /SOLUTIONS
ak =
1
π
Z
π
f (x) cos kx dx =
−π
Z
1
Z
1
1
π
cos kx dx =
−1
1
sin kx
kπ
1
−1
1
1
(sin k − sin(−k)) =
(2 sin k),
=
kπ
kπ
1
bk =
π
Z
π
1
f (x) sin kx dx =
π
−π
−1
sin kx dx = −
1
cos kx
kπ
1
−1
1
1
= − (cos k − cos(−k)) =
(0) = 0.
kπ
kπ
The energy of f contained in the constant term is
A20 = 2a20 = 2
which is
2
1
π
=
2
π2
2/π 2
A20
1
=
= ≈ 0.3183 = 31.83%
E
2/π
π
The fraction of energy contained in the first harmonic is
a2
A21
= 1 =
E
E
2 sin 1 2
π
2
π
of the total.
≈ 0.4508 = 45.08%.
The fraction of energy contained in both the constant term and the first harmonic together is
A2
A20
+ 1 ≈ 0.7691 = 76.91%.
E
E
(b) The fraction of energy contained in the second harmonic is
sin 2 2
π
2
π
a2
A22
= 2 =
E
E
≈ 0.1316 = 13.16%
so the fraction of energy contained in the constant term and first two harmonics is
A20
A2
A2
+ 1 + 2 ≈ 0.7691 + 0.1316 = 0.9007 = 90.07%.
E
E
E
Therefore, the constant term and the first two harmonics are needed to capture 90% of the energy of f .
(c)
F3 (x) =
2 sin 1
sin 2
2 sin 3
1
+
cos x +
cos 2x +
cos 3x.
π
π
π
3π
See Figure 10.37.
1
−3π
−2π
−π
−1
✛
F3 (x)
f (x)
1
Figure 10.37
π
x
2π
3π
10.5 SOLUTIONS
991
25. As c gets closer and closer to 0, the energy of the pulse train will also approach 0, since
E=
1
π
Z
π
(f (x))2 dx =
−π
c/2
Z
1
π
1
π
12 dx =
−c/2
c
c
− −
2
2
=
c
.
π
The energy spectrum shows the relative distribution of the energy of f among its harmonics. The fraction of energy carried
by each harmonic gets smaller as c gets closer to 0, as shown by comparing the kth terms of the Fourier series for pulse
trains with c = 2, 1, 0.4. For instance, notice that the fraction or percentage of energy carried by the constant term gets
smaller as c gets smaller; the same is true for the energy carried by the first harmonic.
If each harmonic contributes less energy, then more harmonics are needed to capture a fixed percentage of energy.
For example, if c = 2, only the constant term and the first two harmonics are needed to capture 90% of the total energy of
that pulse train. If c = 1, the constant term and the first five harmonics are needed to get 90% of the energy of that pulse
train. If c = 0.4, the constant term and the first thirteen harmonics are needed to get 90% of the energy of that pulse train.
This means that more harmonics, or more terms in the series, are needed to get an accurate approximation. Compare the
graphs of the fifth and thirteenth Fourier approximations of f in Problem 23.
26. By formula II-11 of the integral table,
Z
π
1
m2 − k 2
cos kx cos mx dx =
−π
m cos(kx) sin(mx) − k sin(kx) cos(mx)
Again, since sin(nπ) = 0 for any integer n, it is easy to see that this expression is simply 0.
1
du.
m
27. We make the substitution u = mx, dx =
Z
π
.
−π
Then
π
cos2 mx dx =
−π
1
m
Z
u=mπ
cos2 u du.
u=−mπ
By Formula IV-18 of the integral table, this equals
1 1
cos u sin u
m 2
mπ
+
−mπ
1 1
m2
Z
mπ
1 du = 0 +
−mπ
mπ
1
u
2m
=
−mπ
1
u
2m
mπ
−mπ
1
=
(2mπ) = π.
2m
28. The easiest way to do this is to use Problem 27.
Z
π
2
sin mx dx =
Z
π
−π
−π
2
(1 − cos mx) dx =
Z
π
−π
dx −
= 2π − π
Z
π
cos2 mx dx
−π
using Problem 27
= π.
29. By formula II-12 of the integral table,
Z
π
sin kx cos mx dx
−π
=
1
m2 − k 2
m sin(kx) sin(mx) + k cos(kx) cos(mx)
1
= 2
m sin(kπ) sin(mπ) + k cos(kπ) cos(mπ)
m − k2
π
−π
−m sin(−kπ) sin(−mπ) − k cos(−kπ) cos(−mπ) .
Since k and m are positive integers, sin(kπ) = sin(mπ) = sin(−kπ) = sin(−mπ) = 0. Also,
R π cos(kπ) = cos(−kπ)
since cos x is even. Thus this expression reduces to 0. [Note: since sin kx cos mx is odd, so −π sin kx cos mx dx must
be 0.]
992
Chapter Ten /SOLUTIONS
30. Using formula II-10 in the integral table,
Z
π
1
k cos(kx) sin(mx) − m sin(kx) cos(mx)
m2 − k 2
sin kx sin mx dx =
−π
π
.
−π
Again, since sin(nπ) = 0 for all integers n, this expression reduces to 0.
31. (a) To show that g(t) is periodic with period 2π, we calculate
g(t + 2π) = f
b(t + 2π)
2π
=f
bt
2π
+b =f
bt
2π
= g(t).
Since g(t + 2π) = g(t) for all t, we know that g(t) is periodic with period 2π. In addition
g
2πx
b
=f
b(2πx/b)
2π
= f (x).
(b) We make the change of variable t = 2πx/b, dt = (2π/b)dx in the usual formulas for the Fourier coefficients of
g(t), as follows:
1
g(t)dt =
2π
t=−π
1
ak =
π
Z
2
b
Z
1
bk =
π
Z
2
b
Z
=
=
π
Z
1
a0 =
2π
Z
b/2
g
x=−b/2
π
1
g(t) cos(kt) dt =
π
t=−π
b/2
f (x) cos
−b/2
π
2πkx
b
f (x) sin
−b/2
2πkx
b
2πx
b
b/2
1
2π
dx =
b
b
Z
b
2
f (x) dx
b
−2
g
2πx
2πkx
cos
b
b
g
2πx
2πkx
sin
b
b
x=−b/2
2π
dx
b
2π
dx
b
dx
1
g(t) sin(kt) dt =
π
t=−π
b/2
Z
Z
b/2
x=−b/2
dx
(c) By part (a), the Fourier series for f (x) can be obtained by substituting t = 2πx/b into the Fourier series for g(t)
which was found in part (b).
Strengthen Your Understanding
32. R
Since sin(kx) is an odd function and cos(mx) is an even function, we know sin(kx) cos(mx) is an odd function. Hence
π
sin(kx) cos(mx) dx is equal to 0.
−π
33. Unlike Taylor series, Fourier series are good global approximations rather than local ones. Thus, a0 is the average of f (x)
on the interval of approximation.
34. Since cos kx is even for each k > 0, we expect f to be even if there are no sine terms. So let f be any even function with
period 2π. Then the coefficients of the sine terms are
bk =
1
π
Z
π
f (x) sin(kx)dx = 0,
for k > 0,
−π
since f (x) sin kx is odd. For example, we could let f (x) = cos x.
35. Since sin kx is an odd function for all k > 0, we expect f to be odd if there are no cosine terms. So let f be any odd
function.with period 2π. Then the coefficients of the cosine terms are
ak =
1
π
Z
π
−π
f (x) cos(kx)dx = 0, for k ≥ 1,
since f (x) cos kx is odd. For example, we could let f (x) = sin x.
SOLUTIONS to Review Problems for Chapter Ten
993
36. True. Since f is even, f (x) sin(mx) is odd for any m, so
bm
1
=
π
Z
π
f (x) sin x(mx) dx = 0.
−π
37. (b). The graph describes an even function, which eliminates (a) and (c). The Fourier series for (d) would have values near
π for x close to 0.
Solutions for Chapter 10 Review
Exercises
e
(x − 1)2
2
1
1
2. ln x ≈ ln 2 + (x − 2) − (x − 2)2
2
8
1. ex ≈ 1 + e(x − 1) +
1
1
π
3. sin x ≈ − √ + √ x +
4
2
2
1
π
+ √ x+
4
2 2
2
4. Differentiating f (x) = tan x, we get f ′ (x) = 1/ cos2 x, f ′′√(x) = 2 sin x/ cos3 x.
√
Since tan(π/4)
= 1, cos(π/4) = sin(π/4) = 1/ 2, we have f (π/4) = 1, f ′ (π/4) = 1/(1/ 2)2 = 2,
√
2(1/ 2)
√
= 4, so
f ′′ (π/4) = (1/
2)3
f ′′ ( π4 )
π 2
x−
2!
4
4
π 2
π 2
π
π
+
x−
+2 x−
=1+2 x−
.
= 1+2 x−
4
2!
4
4
4
tan x ≈ f
π
4
+ f′
π
4
x−
π
4
+
5. f ′ (x) = 3x2 + 14x − 5, f ′′ (x) = 6x + 14, f ′′′ (x) = 6. The Taylor polynomial about x = 1 is
20
6
12
(x − 1) +
(x − 1)2 + (x − 1)3
1!
2!
3!
= 4 + 12(x − 1) + 10(x − 1)2 + (x − 1)3 .
P3 (x) = 4 +
Notice that if you multiply out and collect terms in P3 (x), you will get f (x) back.
1
= (1 − x)−1 . Then f ′ (x) = (1 − x)−2 , f ′′ (x) = 2(1 − x)−3 , f ′′′ (x) = 6(1 − x)−4 , and f (4) (x) =
6. Let f (x) =
1−x
24(1 − x)−5 . The Taylor polynomial of degree 4 about a = 2 is thus
2(1 − 2)−3
(x − 2)2
2!
24(1 − 2)−5
6(1 − 2)−4
(x − 2)3 +
(x − 4)4
+
3!
4!
= −1 + (x − 2) − (x − 2)2 + (x − 2)3 − (x − 2)4 .
P4 (x) = (1 − 2)−1 + (1 − 2)−2 (x − 2) +
√
1 + x = (1 + x)1/2 .
1
1
3
Then f ′ (x) = (1 + x)−1/2 , f ′′ (x) = − (1 + x)−3/2 , and f ′′′ (x) = (1 + x)−5/2 . The Taylor polynomial of degree
2
4
8
three about x = 1 is thus
7. Let f (x) =
P3 (x) = (1 + 1)1/2 +
− 1 (1 + 1)−3/2
1
(1 + 1)−1/2 (x − 1) + 4
(x − 1)2
2
2!
(1 + 1)−5/2
(x − 1)3
3!
√
(x − 1)2
(x − 1)3
x−1
.
= 2 1+
−
+
4
32
128
3
+8
994
Chapter Ten /SOLUTIONS
8. Let f (x) = ln x. Then f ′ (x) = x−1 , f ′′ (x) = −x−2 , f ′′′ (x) = 2x−3 , and f (4) (x) = −3 · 2x−4 .
So,
−2−2
(x − 2)2
2!
2 · 2−3
−3 · 2 · 2−4
+
(x − 2)3 +
(x − 2)4
3!
4!
(x − 2)2
(x − 2)3
(x − 2)4
x−2
−
+
−
.
= ln 2 +
2
8
24
64
P4 (x) = ln 2 + 2−1 (x − 2) +
9. The first four nonzero terms of P7 are given by:
(−1)1+1 31 2·1−1
x
(1 − 1)!
2+1 2
(−1) 3 2·2−1
x
(2 − 1)!
3+1 3
(−1) 3 2·3−1
x
(3 − 1)!
4+1 4
(−1) 3 2·4−1
x
(4 − 1)!
i=1:
i=2:
i=3:
i=4:
Thus, P7 = 3x − 9x3 +
= 3x
= −9x3
27 5
·x
2
27 7
=−
·x .
2
=
27 5 27 7
·x −
·x .
2
2
10. We know that
cos x = 1 −
x2
x4
x6
+
−
+ ···
2!
4!
6!
Therefore, using the hint,
f (x) = 0.5(1 + cos 2x)
(2x)4
(2x)6
(2x)2
+
−
+ ···
= 0.5 + 0.5(1 −
2!
4!
6!
2 2 23 4 25 6
= 1 − x + x − x +···
2!
4!
6!
2 6
1
x + ···
= 1 − x2 + x4 −
3
45
11. We multiply the series for et by t2 . Since
et = 1 + t +
t2
t3
+
+···,
2!
3!
multiplying by t2 gives
t5
t4
+
+ ···
2!
3!
1
1
= t2 + t3 + t4 + t5 + · · · .
2
6
t2 et = t2 + t3 +
12. We substitute 3y into the series for cos x. Since
cos x = 1 −
x2
x4
x6
+
−
+ ···,
2!
4!
6!
substituting x = 3y gives
(3y)4
(3y)6
(3y)2
+
−
+···
2!
4!
6!
9
27 4 81 6
= 1 − y2 +
y −
y + ···.
2
8
80
cos(3y) = 1 −
SOLUTIONS to Review Problems for Chapter Ten
13.
θ2 cos θ2 = θ2
= θ2 −
14. Substituting y = t2 in sin y = y −
(θ2 )4
(θ2 )6
(θ2 )2
+
−
+ ···
2!
4!
6!
1−
θ10
θ14
θ6
+
−
+ ···
2!
4!
6!
y3
y5
y7
+
−
+ · · · gives
3!
5!
7!
sin t2 = t2 −
t10
t14
t6
+
−
+ ···
3!
5!
7!
15.
(−1)(−2) 2 (−1)(−2)(−3) 3
t
= t(1 + t)−1 = t 1 + (−1)t +
t +
t + ···
1+t
2!
3!
= t − t2 + t3 − t4 + · · ·
16. Substituting y = −4z 2 into
1
= 1 − y + y 2 − y 3 + · · · gives
1+y
1
= 1 + 4z 2 + 16z 4 + 64z 6 + · · ·
1 − 4z 2
17.
1
1
√
= p
4−x
2 1−
1
=
1−
2
−
=
1
3!
x
1
=
1−
2
2
x
2
−
1
−
2
1
2
− 12
1
+
2!
3
2
−
x
2
−
5
2
1
−
2
3
x
2
3
−
2
+ ···
1
3 2
5 3
1
+ x+
x +
x + ···
2
8
64
256
2
x
2
√
18. We use the binomial series to expand 1/ 1 − z 2 and multiply by z 2 . Since
√
(−1/2)(−3/2) 2 (−1/2)(−3/2)(−5/2) 3
1
1
= (1 + x)−1/2 = 1 − x +
x +
x + ···
2
2!
3!
1+x
1
3
5 3
= 1 − x + x2 −
x + ···.
2
8
16
Substituting x = −z 2 gives
√
3
5
z2
1
(−z 2 )3 + · · ·
= 1 − (−z 2 ) + (−z 2 )2 −
2
8
16
1 − z2
1
3
5 6
= 1 + z2 + z4 +
z + ···.
2
8
16
Multiplying by z 2 , we have
√
3
5 8
1
z2
z + ···.
= z2 + z4 + z6 +
2
2
8
16
1−z
995
996
Chapter Ten /SOLUTIONS
19.
a
a
=
=
a+b
a(1 + ab )
1+
b
a
−1
=1−
b
+
a
2
b
a
−
3
b
a
+ ···
20. Using the binomial expansion for (1 + x)−3/2 with x = r/a:
1
1
1
1
=
3/2 =
3/2 = a3/2
r
r
(a + r)3/2
a+a a
a 1+ a
=
=
1
a3/2
1
a3/2
1 + (−3/2)
1−
3
2
r
a
r
a
+
r 2
15
8
+
(−3/2)(−5/2)
2!
−
a
35
16
2
r
a
r 3
a
1+
−3/2
r
a
+
(−3/2)(−5/2)(−7/2)
3!
3
r
a
+···
+··· .
21. Using the binomial expansion for (1 + x)3/2 with x = y/B.
2
2 3/2
(B + y )
=
=B
2
B +B
3
= B3
2
y2
B2
1 + (3/2)
1+
3
2
3/2
=
B
1
2
y
B
y 2
B
+
3
8
2
1+
2 3/2
y
B
(3/2)(1/2)
+
2!
y 4
−
1
16
r
R
21
B
=B
2
2
y
B
y 6
B
3
1+
2 3/2
y
B
(3/2)(1/2)(−1/2)
+
3!
··· .
22.
√
R−r =
=
√
√
R 1−
1
r
R 1+
−
2
R
+
=
√
1
3!
1
2
−
1
2
1
+
2!
−
3
2
1
1
r 2
2
−
−
3
r
R
2
−
+ ···
1 r
1 r2
1 r3
R 1−
−
−
− ···
2
2R
8R
16 R3
R
Problems
23. The second degree Taylor polynomial for f (x) around x = 3 is
f (x) ≈ f (3) + f ′ (3)(x − 3) +
= 1 + 5(x − 3) −
f ′′ (3)
(x − 3)2
2!
10
(x − 3)2 = 1 + 5(x − 3) − 5(x − 3)2 .
2!
Substituting x = 3.1, we get
f (3.1) ≈ 1 + 5(3.1 − 3) − 5(3.1 − 3)2 = 1 + 5(0.1) − 5(0.01) = 1.45.
24. Factoring out a 3, we see
3 1+1+
1
1
1
1
+
+
+
+ · · · = 3e1 = 3e.
2!
3!
4!
5!
25. Infinite geometric series with a = 1, x = −1/3, so
Sum =
3
1
= .
1 − (−1/3)
4
3
2
y
B
!
···
SOLUTIONS to Review Problems for Chapter Ten
997
26. This is the series for ex with x = −2 substituted. Thus
1−2+
(−2)2
(−2)3
(−2)4
8
16
4
−
+
+ · · · = 1 + (−2) +
+
+
+ · · · = e−2 .
2!
3!
4!
2!
3!
4!
27. This is the series for sin x with x = 2 substituted. Thus
2−
8
32
128
23
25
27
+
−
+ ··· = 2 −
+
−
+ · · · = sin 2.
3!
5!
7!
3!
5!
7!
28. Factoring out a 0.1, we see
0.1 0.1 −
(0.1)3
(0.1)5
(0.1)7
+
−
+ ···
3!
5!
7!
= 0.1 sin(0.1).
29. (a) Factoring out 7(1.02)3 and using the formula for the sum of a finite geometric series with a = 7(1.02)3 and r =
1/1.02, we see
Sum = 7(1.02)3 + 7(1.02)2 + 7(1.02) + 7 +
= 7(1.02)3
1+
1
1
1
+
+ ··· +
(1.02)
(1.02)2
(1.02)103
1−
1
(1.02)104
= 7(1.02)3
= 7(1.02)3
=
7
7
7
+
+ ··· +
(1.02)
(1.02)3
(1.02)100
1−
1
1.02
104
(1.02)
− 1 1.02
(1.02)104 0.02
7(1.02104 − 1)
.
0.02(1.02)100
(b) Using the Taylor expansion for ex with x = (0.1)2 , we see
7(0.1)6
7(0.1)4
+
+ ···
2!
3!
(0.1)4
(0.1)6
= 7 1 + (0.1)2 +
+
+ ···
2!
3!
Sum = 7 + 7(0.1)2 +
= 7e(0.1)
2
= 7e0.01 .
30. Let Cn be the coefficient of the nth term in the series. C1 = f ′ (0)/1!, so f ′ (0) = 1!C1 = 1 · 1 = 1.
Similarly, f ′′ (0) = 2!C2 = 2! · 12 = 1;
f ′′′ (0) = 3!C3 = 3! · 31 = 2! = 2;
1
f (10) (0) = 10!C10 = 10! · 10
= 10!
= 9! = 362880.
10
31. Write out series expansions about x = 0, and compare the first few terms:
x5
x3
+
+···
3!
5!
2
3
x
x
ln(1 + x) = x −
+
−···
2
3
2
x4
x2
x4
x
+
− ··· =
−
+ ···
1 − cos x = 1 − 1 −
2!
4!
2!
4!
sin x = x −
x2
x3
+
+···
3! Z
Z 2!
dx
arctan x =
= (1 − x2 + x4 − · · ·) dx
1 + x2
ex − 1 = x +
998
Chapter Ten /SOLUTIONS
= x−
x3
x5
+
+···
3
5
(note that the arbitrary constant is 0)
√
1
(1/2)(−1/2) 2
x +···
x 1 − x = x(1 − x)1/2 = x 1 − x +
2
2
x2
x3
+
+···
2
8
So, considering just the first term or two (since we are interested in small x)
√
1 − cos x < x 1 − x < ln(1 + x) < arctan x < sin x < x < ex − 1.
= x−
32. The graph in Figure 10.38 suggests that the Taylor polynomials converge to f (x) =
Taylor expansion is
f (x) =
1
on the interval (−1, 1). The
1+x
1
= 1 − x + x2 − x3 + x4 − · · · ,
1+x
so the ratio test gives
|an+1 |
|(−1)n+1 xn+1 |
= lim
= |x|.
n→∞ |an |
n→∞
|(−1)n xn |
Thus, the series converges if |x| < 1; that is −1 < x < 1.
lim
f (x) =
−1
1
1+x
x
1
✛
✛
P7 (x)
P5 (x)
✛ P3 (x)
Figure 10.38
33. The Taylor series of
1
around x = 0 is
1 − 2x
∞
X
1
= 1 + 2x + (2x)2 + (2x)3 + · · · =
(2x)k .
1 − 2x
k=0
k
To find the radius of convergence, we apply the ratio test with ak = (2x) .
2k+1 |x|k+1
|ak+1 |
= lim
= 2|x|.
k→∞
2k |x|k
k→∞ |ak |
lim
Hence the radius of convergence is R = 1/2.
34. First we use the Taylor series expansion for ln(1 + t),
1 2 1 3 1 4
t + t − t + ···
2
3
4
2
to find the Taylor series expansion of ln(1 + x + x ) by putting t = x + x2 . We get
ln(1 + t) = t −
ln(1 + x + x2 ) = x +
1 2 2 3 1 4
x − x + x +···.
2
3
4
Next we use the Taylor series for sin x to get
sin2 x = (sin x)2 = (x −
1 5
1
1 3
x +
x − · · ·)2 = x2 − x4 + · · · .
6
120
3
Finally,
ln(1 + x + x2 ) − x
=
sin2 x
1 2
x
2
− 32 x3 + 14 x4 + · · ·
1
→ ,
2
x2 − 31 x4 + · · ·
as
x → 0.
SOLUTIONS to Review Problems for Chapter Ten
35. The fourth-degree Taylor polynomial for f at x = 0 is
f ′′ (0) 2 f ′′′ (0) 3 f (4) (0) 4
x +
x +
x
P4 (x) = f (0) + f ′ (0)x +
2
6
24
−3 2 7 3 −15 4
= 0 + 1x +
x + x +
x
2
6
24
3 2 7 3 5 4
= x− x + x − x .
2
6
8
Thus,
0.6
Z
f (x) dx ≈
0
=
=
Z
0.6
P4 (x) dx
0
1 2 3 1 3 7 1 4 5 1 5
x − · x + · x − · x
2
2 3
6 4
8 5
7 4 1 5
1 2 1 3
x − x +
x − x
2
2
24
8
36. (a) Using the Taylor series for ex , we have:
e
0.6
3
0
= 0.10008.
3
2
0.6
0
−x3
−x3
−x3
+
+
= 1 + −x +
2
6
24
x9
x12
x6
3
−
+
+ ···.
= 1−x +
2
6
24
−x3
4
+···
(b) Using the Taylor polynomial of degree 12 from part (a), we have:
′
x6
x9
x12
−
+
+ ···
2
6
24
6x5
9x8
12x11
= −3x2 +
−
+
+ ···
2
6
24
3x8
x11
= −3x2 + 3x5 −
+
+ ···
2
2
f ′ (x) =
1 − x3 +
f ′′ (x) = f ′ (x)
′
′
3x8
x11
+
+···
2
2
11x10
= −6x + 15x4 − 12x7 +
+···.
2
=
−3x2 + 3x5 −
37. We find the Taylor polynomial for cos(x2 ) by substituting into the series for cos x:
4
1
x4
x8
1 2 2
x
+
x2 = 1 −
+
.
2
4!
2
24
x4
x8
+
2
24
cos x2 ≈ 1 −
This means that
Z
1
0
cos x2 dx ≈
Z
0
1
1−
dx =
x−
x5
x9
+
10
216
1
0
=1−
1
1
+
= 0.90463.
10
216
This is a very good estimate; the actual value (found using a computer) is 0.90452 . . ..
38. (a) The series for
sin 2θ
θ
is
sin 2θ
1
=
θ
θ
(2θ)3
(2θ)5
2θ −
+
− ···
3!
5!
sin 2θ
= 2.
θ
(b) Near θ = 0, we make the approximation
so lim
θ→0
sin 2θ
4
≈ 2 − θ2
θ
3
so the parabola is y = 2 − 43 θ2 .
=2−
4θ2
4θ4
+
− ···
3
15
999
1000
Chapter Ten /SOLUTIONS
39. (a) Since (1 − x2 )−1/2 dx = arcsin x, we use the Taylor series for (1 − x2 )−1/2 to find the Taylor series for arcsin x:
R
(1 − x2 )−1/2 = 1 +
so
arcsin x =
Z
5 6
35 8
1 2 3 4
x + x +
x +
x + ···
2
8
16
128
(1 − x2 )−1/2 dx = x +
1 3
3 5
5 7
35 9
x +
x +
x +
x + ···
6
40
112
1152
(b) From Example 3 in Section 10.3, we know
arctan x = x −
so that
1 3 1 5 1 7
x + x − x + ···
3
5
7
x − 31 x3 + 15 x5 − 71 x7 + · · ·
arctan x
→ 1,
=
1 3
3 5
5
35
arcsin x
x + 6 x + 40
x + 112
x7 + 1152
x9 + · · ·
as
x → 0.
40. (a) The Taylor series is given by
1
1
1
· f ′′ (0)x2 +
· f ′′′ (0)x3 +
· f (4) (0)x4 + · · ·
2!
3!
4!
(1 + 1)!
1 (2 + 1)! 2
1 (3 + 1)! 3
1 (4 + 1)! 4
= 1+
x+
x +
x +
x + ···
·
·
·
1
2
3
2
2!
2
3!
2
4!
24 }
| {z }
| {z }
| {z }
| {z
f (x) = f (0) + f ′ (0)x +
f ′ (0)
f ′′ (0)
f ′′′ (0)
f (4) (0)
3! 1
4! 1
5! 1
1
·
· x2 +
·
· x3 +
·
· x4 + · · ·
= 1 + 2! · 1 · x +
2
2! 22
3! 23
4! 24
2
3
4
5
= 1 + 1 · x1 + 2 · x2 + 3 · x3 + 4 · x4 + · · ·
2
2
2
2
k+1 k
1+1 1 2+1 2 3+1 3 4+1 4
·x +
·x +
·x +
· x + ··· +
· x + ···
= 1+
21
22
23
24
2k
We see that the coefficient of xk is (k + 1)!/2k , so
f (x) =
∞
X
k+1
k=0
2k
xk .
Note that the general term works for k = 0, since (0 + 1)!/20 = 1/1 = 1.
(b) Each successive term in this series involves a higher power of 3/2. Since 3/2 > 1, this means each successive term
is larger than the one before.!Therefore the series does not converge.
(c) Given that
Z
0
1
Z
1
0
∞
X
an x
n
dx =
∞ Z
X
n=1
n=1
Z
1
1
n
an x dx , we have:
0
2
3
4
5
· x1 + 2 · x2 + 3 · x3 + 4 · x4 + · · · dx
1
2
2
2
2
Z0 1
Z 1
Z 1
Z 1
Z 1
3
4
5
2
2
3
=
1 dx +
· x dx +
· x dx +
· x dx +
· x4 dx + · · ·
1
2
3
4
2
2
2
2
0
0
0
0
0
f (x) dx =
1+
1
1
1
3 1
4 1
2 1 2
· x
+ 2 · x3 + 3 · x4
1
2
2
2
3
2
4
0
0
0
1
1
1
1
= 1 + 1 + 2 + 3 + 4 + ···.
2
2
2
2
=x
1
+
+
0
5 1 5
· x
24 5
1
0
+ ···
1
= 2.
1 − 1/2
41. (a) See Figure 10.39. The graph of E1 looks like a parabola. Since the graph of E1 is sandwiched between the graph of
y = x2 and the x axis, we have
|E1 | ≤ x2 for |x| ≤ 0.1.
This is the sum of a geometric series, which we know to equal
SOLUTIONS to Review Problems for Chapter Ten
y
0.01
1001
y
y = x2
y = x3
E1
−0.1
E2
x
0.1
−0.1
x
0.1
−0.01
−0.001
Figure 10.39
Figure 10.40
(b) See Figure 10.40. The graph of E2 looks like a cubic, sandwiched between the graph of y = x3 and the x axis, so
|E2 | ≤ x3
for
|x| ≤ 0.1.
(c) Using the Taylor expansion
x3
x2
+
+ ···
2!
3!
ex = 1 + x +
we see that
E1 = ex − (1 + x) =
x3
x4
x2
+
+
+ ···.
2!
3!
4!
Thus for small x, the x2 /2! term dominates, so
E1 ≈
x2
,
2!
and so E1 is approximately a quadratic.
Similarly
E2 = ex − (1 + x +
x2
x3
x4
)=
+
+ ···.
2
3!
4!
Thus for small x, the x3 /3! term dominates, so
E2 ≈
x3
3!
and so E2 is approximately a cubic.
42. (a) We have
1 ′′
f (0)x2 + · · ·
2
1
1
1
1
1
· 0 · x2 + (2!)x3 +
· 0 · x4 + (4!)x5 +
· 0 · x6 + (6!)x7
3!
4!
5!
6!
7!
x7
+
.
7
P7 (x) = f (0) + f ′ (0)x +
1
2
x3
x5
= x+
+
3
5
= 0+1·x+
(b) We infer from the pattern in part (a) that:
x5
x7
x(odd number)
x3
+
+
+ ··· +
+···
3
5
7
same odd number
∞
2k+1
X
x
.
=
2k + 1
f (x) = x +
k=0
43. (a) The Taylor polynomial of degree 2 is
V (x) ≈ V (0) + V ′ (0)x +
V ′′ (0) 2
x .
2
Since x = 0 is a minimum, V ′ (0) = 0 and V ′′ (0) > 0. We can not say anything about the sign or value of V (0).
Thus
V ′′ (0) 2
V (x) ≈ V (0) +
x .
2
1002
Chapter Ten /SOLUTIONS
(b) Differentiating gives an approximation to V ′ (x) at points near the origin
V ′ (x) ≈ V ′′ (0)x.
Thus, the force on the particle is approximated by −V ′′ (0)x.
Force = −V ′ (x) ≈ −V ′′ (0)x.
Since V ′′ (0) > 0, the force is approximately proportional to x with negative proportionality constant, −V ′′ (0). This
means that when x is positive, the force is negative, which means pointing toward the origin. When x is negative, the
force is positive, which means pointing toward the origin. Thus, the force always points toward the origin.
Physical principles tell us that the particle is at equilibrium at the minimum potential. The direction of the force
toward the origin supports this, as the force is tending to restore the particle to the origin.
44. (a) For reference, Figure 10.41 shows the graphs of the two functions.
2
e−x = 1 − x2 +
x6
x4
−
+ ···
2!
3!
1
= 1 − x2 + x4 − x6 + · · ·
1 + x2
Notice that the first two terms are the same in both series.
1
is greater.
1 + x2
(c) Even, because the only terms involved are of even degree.
2
(d) The coefficients for e−x become extremely small for higher powers of x, and we can “counteract” the effect of these
1
powers for large values of x. The series for 1+x
2 has no such coefficients.
(b)
1
y
y=
1
1+x2
y = e−x
2
x
−1
1
Figure 10.41
45. We have:
P4 (x) =
4
X
(−n)n−1
n=1
n!
xn
(−1)1−1 1 (−2)2−1 2 (−3)3−1 3 (−4)4−1 4
x +
x +
x +
x
1!
2!
3!
4!
(−1)0
(−2)1 2 (−3)2 3 (−4)3 4
x+
x +
x +
x
=
1
2
6
24
3 3 8 4
2
= x−x + x − x .
2
3
=
46. We can approximate f (x) using the Taylor polynomial of degree 5:
f ′′ (0) 2 f (3) (0) 3 f (4) (0) 4 f (5) (0) 5
x +
x +
x +
x
P5 (x) = f (0) + f ′ (0)x +
2!
3!
4!
5!
3 4
6 5
−1 2 0 3
x + x −
x +
x
= 2+0·x+
2
6
24
120
1 2 1 4
1 5
= 2− x − x +
x .
2
8
20
Thus,
Z
0
2
f (x) dx ≈
Z
0
2
2−
1 5
1 2 1 4
x − x +
x
2
8
20
1
1 6
1
x
dx = 2x− x3 − x5 +
6
40
120
2
0
1
1 6
1
= 2·2− ·23 − ·25 +
·2 = 2.4.
6
40
120
SOLUTIONS to Review Problems for Chapter Ten
1003
47. We have
f ′ (t) = t−1et
1
1
1
1 + t + · t2 + · t3
≈
t
2
6
1
1
1
= + 1 + · t + · t2 .
t
2
6
Since f is an antiderivative of this function:
f (t) =
Z
Z
1
1
1
+ 1 + · t + · t2 dt
t
2
6
1 1
1 1
= ln t + t + · t2 + · t3 + C
2 2
6 3
1
1 3
= ln t + t + t2 +
t +C
| 4 {z 18 }
f ′ (t) dt ≈
P3 (t)
1 3
1
t .
so P3 (t) = t + t2 +
4
18
Note that the problem does not give enough information for us to find C. The actual definition of f is given in terms
of an improper integral; using this definition, it can be shown that C equals the so-called Euler-Mascheroni constant
λ = 0.57721 . . ..
48. This time we
√ are interested in how a function behaves at large values in its domain. Therefore, we don’t want to expand
V = 2πσ( R2 + a2 − R) about R = 0. We want to find a variable which becomes small as R gets large. Since R > a,
it is helpful to write
!
r
1+
V = R2πσ
a2
−1
R2
.
a 2
We can now expand a series in terms of ( R
) . This may seem strange, but suspend your disbelief. The Taylor series for
q
1+
a2
R2
is
(1/2)(−1/2)
1 a2
+
1+
2 R2
2
So V = R2πσ
order, so
1
1 a2
−
1+
2 R2
8
a2
R2
2
!
a2
R2
2
+ ···
4
a
+ · · · − 1 . For large R, we can drop the − 81 R
4 term and terms of higher
πσa2
.
R
a 2
Notice that what we really did by expanding around ( R ) = 0 was expanding around R = ∞. We then get a series that
converges for large R.
V ≈
GM
Gm
+ (R+r)
2
R2
1
F = GM
+ Gm
r )2
R2
R2 (1+ R
r
Since R < 1, use the binomial
49. (a) F =
(b)
expansion:
1
r
r 2 = 1+
(1 + R
)
R
−2
=1−2
r
Gm
GM
+ 2 1−2
F =
R2
R
R
(c) Discarding higher power terms, we get
r
R
+ (−2)(−3)
2
r
+3
R
r 2
(R
)
+ ···
2!
−··· .
GM
Gm
2Gmr
+ 2 −
R2
R
R3
G(M + m)
2Gmr
=
−
.
R2
R3
F ≈
+m)
is the field strength at a distance R from a single particle
Looking at the expression, we see that the term G(M
R2
2Gmr
of mass M + m. The correction term, − R3 , is negative because the field strength exerted by a particle of mass
(M + m) at a distance R would clearly be larger than the field strength at P in the question.
1004
Chapter Ten /SOLUTIONS
50. (a) For a/h < 1, we have
1
1
1
=
=
h
(a2 + h2 )1/2
h(1 + a2 /h2 )1/2
3 a4
1 a2
+
− ... .
2
2h
8 h4
1−
Thus
2GM mh
F =
a2
2GM mh
=
a2 h
1
1
−
h
h
1 a2
3 a4
1−
+
− ...
2 h2
8 h4
3 a4
1 a2
−
− ...
1−1+
2 h2
8 h4
2GM m 1 a2
=
a2 2 h2
3 a2
...
1−
4 h2
GM m
=
h2
3 a2
− ... .
1−
4 h2
(b) Taking only the first nonzero term gives
GM m
.
h2
F ≈
Notice that this approximation to F is independent of a.
(c) If a/h = 0.02, then a2 /h2 = 0.0004, so
F ≈
3
GM m
GM m
(1 − (0.0004)) =
(1 − 0.0003).
h2
4
h2
Thus, the approximations differ by 0.0003 = 0.03%.
51. (a) If h is much smaller than R, we can say that (R + h) ≈ R, giving the approximation
F =
mgR2
mgR2
≈
= mg.
(R + h)2
R2
(b)
F =
mgR2
mg
=
= mg(1 + h/R)−2
(R + h)2
(1 + h/R)2
= mg
= mg
(−2)
1+
1!
1−
h
R
(−2)(−3)
+
2!
3h2
4h3
2h
+ 2 − 3 + ···
R
R
R
2
h
R
(−2)(−3)(−4)
+
3!
3
h
R
+ ···
(c) The first order correction comes from term −2h/R. The approximation for F is then given by
F ≈ mg 1 −
2h
.
R
If the first order correction alters the estimate for F by 10%, we have
2h
= 0.10
R
so h = 0.05R ≈ 0.05(6400) = 320 km.
The approximation F ≈ mg is good to within 10% — that is, up to about 300 km.
52. Since expanding f (x + h) and g(x + h) in Taylor series gives
f ′′ (x) 2
h + ...,
2!
′′
f (x) 2
h + ...,
g(x + h) = g(x) + g ′ (x)h +
2!
f (x + h) = f (x) + f ′ (x)h +
we substitute to get
f (x + h)g(x + h) − f (x)g(x)
h
SOLUTIONS to Review Problems for Chapter Ten
′
(f (x) + f (x)h +
1 ′′
f (x)h2
2
′
1005
1 ′′
g (x)h2
2
+ ...)(g(x) + g (x)h +
+ ...) − f (x)g(x)
h
f (x)g(x) + (f ′ (x)g(x) + f (x)g ′ (x))h + Terms in h2 and higher powers − f (x)g(x)
=
h
h(f ′ (x)g(x) + f (x)g ′ (x) + Terms in h and higher powers)
=
h
= f ′ (x)g(x) + f (x)g ′ (x) + Terms in h and higher powers.
=
Thus, taking the limit as h → 0, we get
f (x + h)g(x + h) − f (x)g(x)
d
(f (x)g(x)) = lim
h→0
dx
h
= f ′ (x)g(x) + f (x)g ′ (x).
53. Expanding f (y + k) and g(x + h) in Taylor series gives
f ′′ (y) 2
k + ···,
2!
′′
g (x) 2
g(x + h) = g(x) + g ′ (x)h +
h + ···.
2!
f (y + k) = f (y) + f ′ (y)k +
Now let y = g(x) and y + k = g(x + h). Then k = g(x + h) − g(x) so
k = g ′ (x)h +
g ′′ (x) 2
h + ···.
2!
Substituting g(x + h) = y + k and y = g(x) in the series for f (y + k) gives
f (g(x + h)) = f (g(x)) + f ′ (g(x))k +
f ′′ (g(x)) 2
k + ···.
2!
Now, substituting for k, we get
g ′′ (x) 2
f ′′ (g(x)) ′
h + · · ·) +
(g (x)h + . . .)2 + · · ·
2!
2!
= f (g(x)) + (f ′ (g(x))) · g ′ (x)h + Terms in h2 and higher powers.
f (g(x + h)) = f (g(x)) + f ′ (g(x)) · (g ′ (x)h +
So, substituting for f (g(x + h)) and dividing by h, we get
f (g(x + h)) − f (g(x))
= f ′ (g(x)) · g ′ (x) + Terms in h and higher powers,
h
and thus, taking the limit as h → 0,
f (g(x + h)) − f (g(x))
d
f (g(x)) = lim
h→0
dx
h
= f ′ (g(x)) · g ′ (x).
54. (a) Notice g ′ (0) = 0 because g has a critical point at x = 0. So, for n ≥ 2,
g(x) ≈ Pn (x) = g(0) +
g (n) (0) n
g ′′ (0) 2 g ′′′ (0) 3
x +
x + ··· +
x .
2!
3!
n!
(b) The Second Derivative test says that if g ′′ (0) > 0, then 0 is a local minimum and if g ′′ (0) < 0, 0 is a local maximum.
g ′′ (0) 2
x . So, for x near 0,
(c) Let n = 2. Then P2 (x) = g(0) +
2!
g(x) − g(0) ≈
g ′′ (0) 2
x .
2!
If g ′′ (0) > 0, then g(x) − g(0) ≥ 0, as long as x stays near 0. In other words, there exists a small interval around
x = 0 such that for any x in this interval g(x) ≥ g(0). So g(0) is a local minimum.
The case when g ′′ (0) < 0 is treated similarly; then g(0) is a local maximum.
1006
Chapter Ten /SOLUTIONS
55. The situation is more complicated. Let’s first consider the case when g ′′′ (0) 6= 0. To be specific let g ′′′ (0) > 0. Then
g(x) ≈ P3 (x) = g(0) +
g ′′′ (0) 3
x .
3!
g ′′′ (0) 3
g ′′′ (0)
x . (Notice that
> 0 is a constant.) Now, no matter how small an open interval I
3!
3!
g ′′′ (0) 3
x1 < 0
around x = 0 is, there are always some x1 and x2 in I such that x1 < 0 and x2 > 0, which means that
3!
g ′′′ (0) 3
x2 > 0, i.e. g(x1 ) − g(0) < 0 and g(x2 ) − g(0) > 0. Thus, g(0) is neither a local minimum nor a local
and
3!
maximum. (If g ′′′ (0) < 0, the same conclusion still holds. Try it! The reasoning is similar.)
Now let’s consider the case when g ′′′ (0) = 0. If g (4) (0) > 0, then by the fourth degree Taylor polynomial approximation to g at x = 0, we have
So, g(x) − g(0) ≈
g(x) − g(0) ≈
g (4) (0) 4
x >0
4!
for x in a small open interval around x = 0. So g(0) is a local minimum. (If g (4) (0) < 0, then g(0) is a local maximum.)
In general, suppose that g (k) (0) 6= 0, k ≥ 2, and all the derivatives of g with order less than k are 0. In this case
g looks like cxk near x = 0, which determines its behavior there. Then g(0) is neither a local minimum nor a local
maximum if k is odd. For k even, g(0) is a local minimum if g (k) (0) > 0, and g(0) is a local maximum if g (k) (0) < 0.
56. Let us begin by finding the Fourier coefficients for f (x). Since f is odd,
Thus ai = 0 for all i ≥ 0. On the other hand,
bi =
1
π
Z
π
f (x) sin nx dx =
−π
1
π
Z
0
−π
−π
f (x) dx = 0 and
− sin(nx) dx +
1 1
=
cos(nx)
π n
0
−π
Z
π
Rπ
sin(nx) dx
0
1
− cos(nx)
n
π
0
−π
1
cos 0 − cos(−nπ) − cos(nπ) + cos 0
nπ
=
2
1 − cos(nπ) .
nπ
4
nπ
f (x) cos nx dx = 0.
=
Since cos(nπ) = (−1)n , this is 0 if n is even, and
Fn (x) =
Rπ
if n is odd. Thus the nth Fourier polynomial (where n is odd) is
4
4
4
sin x +
sin 3x + · · · +
sin(nx).
π
3π
nπ
Evaluating at x = π/2, we get
π
4
3π
4
5π
4
7π
4
nπ
4
sin +
sin
+
sin
+
sin
+ ··· +
sin
π
2
3π
2
5π
2
7π
2
nπ
2
4
1
1
1
1
2n+1
=
1 − + − + · · · + (−1)
.
π
3
5
7
2n + 1
Fn (π/2) =
But we are assuming Fn (π/2) approaches f (π/2) = 1 as n → ∞, so
1
1
1
1
π
π
π
Fn (π/2) = 1 − + − + · · · + (−1)2n+1
→ ·1= .
4
3
5
7
2n + 1
4
4
57. (a) Expand f (x) into its Fourier series:
f (x) = a0 + a1 cos x + a2 cos 2x + a3 cos 3x + · · · + ak cos kx + · · ·
+ b1 sin x + b2 sin 2x + b3 sin 3x + · · · + bk sin kx + · · ·
Then differentiate term-by-term:
f ′ (x) = −a1 sin x − 2a2 sin 2x − 3a3 sin 3x − · · · − kak sin kx − · · ·
+b1 cos x + 2b2 cos 2x + 3b3 cos 3x + · · · + kbk cos kx + · · ·
SOLUTIONS to Review Problems for Chapter Ten
1007
Regroup terms:
f ′ (x) = +b1 cos x + 2b2 cos 2x + 3b3 cos 3x + · · · + kbk cos kx + · · ·
−a1 sin x − 2a2 sin 2x − 3a3 sin 3x − · · · − kak sin kx − · · ·
which forms a Fourier series for the derivative f ′ (x). The Fourier coefficient of cos kx is kbk and the Fourier coefficient of sin kx is −kak . Note that there is no constant term as you would expect from the formula kak with k = 0.
Note also that if the kth harmonic f is absent, so is that of f ′ .
(b) If the amplitude of the kth harmonic of f is
Ak =
then the amplitude of the kth harmonic of f ′ is
p
(kbk )2 + (−kak )2 =
p
a2k + b2k ,
k ≥ 1,
p
k2 (b2k + a2k ) = k
p
a2k + b2k = kAk .
(c) The energy of the kth harmonic of f ′ is k2 times the energy of the kth harmonic of f .
58. Let rk and sk be the Fourier coefficients of Af + Bg. Then
1
2π
r0 =
Z
π
−π
1
=A
2π
Z
Af (x) + Bg(x) dx
π
1
f (x) dx + B
2π
−π
= Aa0 + Bc0 .
π
Z
g(x) dx
−π
Similarly,
1
rk =
π
π
Z
Af (x) + Bg(x) cos(kx) dx
−π
Z
1
=A
π
π
Z
1
f (x) cos(kx) dx + B
π
−π
= Aak + Bck .
π
g(x) cos(kx) dx
−π
And finally,
1
sk =
π
Z
π
Af (x) + Bg(x) sin(kx) dx
−π
Z
1
=A
π
π
Z
1
f (x) sin(kx) dx + B
π
−π
= Ack + Bdk .
π
g(x) sin(kx) dx
−π
59. Since g(x) = f (x + c), we have that [g(x)]2 = [f (x + c)]2 , so g 2 is f 2 shifted horizontally by c. Since f has period 2π,
so does f 2 and g 2 . If you think of the definite integral as an area, then because of the periodicity, integrals of f 2 over any
interval of length 2π have the same value. So
Energy of f =
Z
π
(f (x))2 dx =
Z
π+c
(f (x))2 dx.
−π+c
−π
Now we know that
1
Energy of g =
π
1
=
π
Z
π
(g(x))2 dx
−π
π
Z
(f (x + c))2 dx.
−π
Using the substitution t = x + c, we see that the two energies are equal.
1008
Chapter Ten /SOLUTIONS
CAS Challenge Problems
60. (a) The Taylor polynomials of degree 10 are
x4
2 x6
x8
2 x10
+
−
+
3
45
315
14175
x4
2 x6
x8
2 x10
2
Q10 (x) = 1 − x +
−
+
−
3
45
315
14175
P10 (x) = x2 −
For sin2 x,
For cos2 x,
(b) The coefficients in P10 (x) are the negatives of the corresponding coefficients of Q10 (x). The constant term of P10 (x)
is 0 and the constant term of Q10 (x) is 1. Thus, P10 (x) and Q10 (x) satisfy
Q10 (x) = 1 − P10 (x).
This makes sense because cos2 x and sin2 x satisfy the identity
cos2 x = 1 − sin2 x.
61. (a) The Taylor polynomials of degree 7 are
For sin x,
For sin x cos x,
x5
x7
x3
+
−
6
120
5040
2 x5
4 x7
2 x3
+
−
Q7 (x) = x −
3
15
315
P7 (x) = x −
(b) The coefficient of x3 in Q7 (x) is −2/3, and the coefficient of x3 in P7 (x) is −1/6, so the ratio is
−2/3
= 4.
−1/6
The corresponding ratios for x5 and x7 are
2/15
= 16
1/120
and
−4/315
= 64.
−1/5040
(c) It appears that the ratio is always a power of 2. For x3 , it is 4 = 22 ; for x5 , it is 16 = 24 ; for x7 , it is 64 = 26 . This
suggests that in general, for the coefficient of xn , it is 2n−1 .
(d) From the identity sin(2x) = 2 sin x cos x, we expect that P7 (2x) = 2Q7 (x). So, if an is the coefficient of xn in
P7 (x), and if bn is the coefficient of xn in Q7 (x), then, since the xn terms P7 (2x) and 2Q7 (x) must be equal, we
have
an (2x)n = 2bn xn .
Dividing both sides by xn and combining the powers of 2, this gives the pattern we observed. For an 6= 0,
bn
= 2n−1 .
an
62. (a) For f (x) = x2 we have f ′ (x) = 2x so the tangent line is
y = f (2) + f ′ (2)(x − 2) = 4 + 4(x − 2)
y = 4x − 4.
For g(x) = x3 − 4x2 + 8x − 7, we have g ′ (x) = 3x2 − 8x + 8, so the tangent line is
y = g(1) + g ′ (1)(x − 1) = −2 + 3(x − 1)
y = 3x − 5.
For h(x) = 2x3 + 4x2 − 3x + 7, we have h′ (x) = 6x2 + 8x − 3. So the tangent line is
y = h(−1) + h′ (−1)(x + 1) = 12 − 5(x + 1)
y = −5x + 7.
SOLUTIONS to Review Problems for Chapter Ten
1009
(b) Division by a CAS or by hand gives
x2
4x − 4
f (x)
=
=1+
2
(x − 2)
(x − 2)2
(x − 2)2
so
r(x) = 4x − 4,
g(x)
x3 − 4x2 + 8x − 7
3x − 5
=
=x−2+
(x − 1)2
(x − 1)2
(x − 1)2
so
h(x)
2x3 + 4x2 − 3x + 7
−5x + 7
=
= 2x +
=
2
(x + 1)
(x + 1)2
(x + 1)2
so
r(x) = 3x − 5,
r(x) = −5x + 7.
(c) In each of these three cases, y = r(x) is the equation of the tangent line. We conjecture that this is true in general.
(d) The Taylor expansion of a function p(x) is
p(x) = p(a) + p′ (a)(x − a) +
p′′′ (a)
p′′ (a)
(x − a)2 +
(x − a)3 + · · ·
2!
3!
Now divide p(x) by (x − a)2 . On the right-hand side, all terms from p′′ (a)(x − a)2 /2! onward contain a power of
(x−a)2 and divide exactly by (x−a)2 to give a polynomial q(x), say. So the remainder is r(x) = p(a)+p′ (a)(x−a),
the tangent line.
63. (a) The Taylor polynomial is
P10 (x) = 1 +
x4
x6
x8
x10
x2
−
+
−
+
12
720
30240
1209600
47900160
(b) All the terms have even degree. A polynomial with only terms of even degree is an even function. This suggests that
f might be an even function.
(c) To show that f is even, we must show that f (−x) = f (x).
f (−x) =
−x
−x
x
x
xex
x
+
=
− = x
−
1
−1
2
2
e −1
2
1 − ex
e−x
xex − 12 x(ex − 1)
ex − 1
x
1
1
xex + 12 x
x(ex − 1) + x
xe − 12 xex + 12 x
=
= 2 x
= 2
x
e −1
e −1
ex − 1
x
x
x
1
= x
+ = f (x)
= x+ x
2
e −1
e −1
2
=
64. (a) The Taylor polynomial is
P11 (x) =
x7
x11
x3
−
+
.
3
42
1320
(b) Evaluating, we get
P11 (1) =
S(1) =
13
17
111
−
+
= 0.310281
3
42
1320
1
Z
sin(t2 ) dt = 0.310268.
0
We need to take about 6 decimal places in the answer as this allows us to see the error. (The values of P11 (1) and
S(1) start to differ in the fifth decimal place.) Thus, the percentage error is (0.310281 − 0.310268)/0.310268 =
0.000013/0.310268 = 0.000042 = 0.0042%. On the other hand,
P11 (2) =
S(2) =
27
211
23
−
+
= 1.17056
3
42
1320
Z
2
sin(t2 ) dt = 0.804776.
0
The percentage error in this case is (1.17056 − 0.804776)/0.804776 = 0.365784/0.804776 = 0.454517, or about
45%.
1010
Chapter Ten /SOLUTIONS
PROJECTS FOR CHAPTER TEN
1. (a) We rewrite the denominator as (1 + (A2 − 2A) cos2 θ)−1/2 and expand using a binomial series. Since we
do not want terms beyond the quadratic, we omit terms with powers higher than A2 .
3
1
1 2
2
2
−1/2
2
2 −2
(1 + (A − 2A) cos θ)
= 1 − (A − 2A) cos θ −
(A2 − 2A)2 cos4 θ + · · ·
2
2!
3
1
= 1 − (A2 − 2A) cos2 θ + (A4 − 4A3 + 4A2 ) cos4 θ + · · ·
2
8
3
1
4
2
2
2
= 1 + A cos θ + A
cos θ − cos θ + · · ·
2
2
Now multiply by (1 − A cos2 θ), again omitting powers higher than A2 :
cos α = (1 − A cos2 θ)(1 + (A2 − 2A) cos2 θ)−1/2
1
3
4
2
2
2
2
cos θ − cos θ + · · ·
= (1 − A cos θ) 1 + A cos θ + A
2
2
1
3
cos4 θ − cos2 θ − cos4 θ + · · ·
= 1 + A2
2
2
1
1
4
2
2
=1+A
cos θ − cos θ + · · ·
2
2
1 2
2
2
= 1 + A cos θ(cos θ − 1) + · · ·
2
1 2
≈ 1 − A cos2 θ sin2 θ.
2
The approximation of part (a) is therefore the approximation of cos α by its quadratic Taylor polynomial.
(b) Since the bulge in the earth is so small, we have α ≈ 0. Therefore to a good accuracy we can approximate
cos α by its Taylor series about 0,
α2
.
cos α ≈ 1 −
2
By part (a) we have
1−
α2
1
≈ 1 − A2 cos2 θ sin2 θ
2
2
α2 ≈ A2 cos2 θ sin2 θ.
Using the identity sin(2θ) = 2 sin θ cos θ we get
α2 ≈ A2 cos2 θ sin2 θ =
1 2 2
A sin (2θ),
4
and therefore, since A > 0, and taking α and θ to be positive, we have
α≈
1
A sin(2θ).
2
(c) The value of α is the exact value and (1/2)A sin(2θ) is the approximation. Thus, we have
Error =
(1/2)A sin(2θ) − α
.
α
Letting A = 0.0034 and expressing all angles in degrees, we have Table 10.2:
PROJECTS FOR CHAPTER TEN
1011
Table 10.2
θ
0◦
20◦
40◦
60◦
80◦
α
0◦
0.06280◦
0.09611◦
0.084425◦
0.033317◦
(1/2)A sin(2θ)
◦
0
◦
0.06261
◦
0.09592
◦
0.084353
0.033314◦
Error
0
0.30%
0.20%
0.085%
0.010%
2. (a) A calculator gives 4 tan−1 (1/5) − tan−1 (1/239) = 0.7853981634, which agrees with π/4 to ten decimal
places. Notice that you cannot verify that Machin’s formula is exactly true numerically (because any calculator has only a finite number of digits.) Showing that the formula is exactly true requires a theoretical
argument.
(b) The Taylor polynomial of degree 5 approximating arctan x is
arctan x ≈ x −
x5
x3
+ .
3
5
Thus,
1
1
− arctan
π = 4 4 arctan
5
239
3
5 !
3
5 !!
1 1 1
1 1
1
1
1
1
1
≈4 4
−
−
+
−
+
5 3 5
5 5
239 3 239
5 239
≈ 3.141621029.
The true value is π = 3.141592653 . . ..
(c) Because the values of x, namely x = 1/5 and x = 1/239, are much smaller than 1, the terms in the series
get smaller much faster.
(d) (i) If A = arctan(120/119) and B = − arctan(1/239), then
tan A =
120
119
and
tan B = −
1
.
239
Substituting
tan(A + B) =
(120/119) + (−1/239)
= 1.
1 − (120/119)(−1/239)
Thus
A + B = arctan 1,
so
arctan
120
119
− arctan
1
239
= arctan 1.
(ii) If A = B = arctan(1/5), then
tan(A + B) =
5
(1/5) + (1/5)
=
.
1 − (1/5)(1/5)
12
Thus
A + B = arctan
so
2 arctan
5
12
,
1
5
= arctan
.
5
12
1012
Chapter Ten /SOLUTIONS
If A = B = 2 arctan(1/5), then tan A = tan B = 5/12, so
tan(A + B) =
120
(5/12) + (5/12)
=
.
1 − (5/12)(5/12)
119
Thus
A + B = arctan
so
120
119
,
120
1
= arctan
.
4 arctan
5
119
(iii) Using the result of part (a) and substituting the results of part (b), we obtain
1
π
1
− arctan
= arctan 1 = .
4 arctan
5
239
4
3. (a) (i) Using a Taylor series expansion, we have
f (x0 − h) = f (x0 ) − f ′ (x0 )h +
f ′′ (x0 ) 2 f ′′′ (x0 ) 3
h −
h + ···.
2
3!
So we have
f ′′ (x0 )
f (x0 ) − f (x0 − h)
− f ′ (x0 ) ≈
h + ···.
h
2
This suggests the following bound for small h:
f (x0 ) − f (x0 − h)
Mh
− f ′ (x0 ) ≤
,
h
2
where |f ′′ (x)| ≤ M for |x − x0 | < |h|.
(ii) We use Taylor series expansions:
f ′′ (x0 ) 2 f ′′′ (x0 ) 3
h +
h + ···
2
3!
f ′′ (x0 ) 2 f ′′′ (x0 ) 3
f (x0 − h) = f (x0 ) − f ′ (x0 )h +
h −
h + ···.
2
3!
f (x0 + h) = f (x0 ) + f ′ (x0 )h +
Subtracting gives
2f ′′′ (x0 ) 3
h + ···
3!
1
= 2f ′ (x0 )h + f ′′′ (x0 )h3 + · · · .
3
f (x0 + h) − f (x0 − h) = 2f ′ (x0 )h +
So
f ′′′ (x0 ) 2
f (x0 + h) − f (x0 − h)
= f ′ (x0 ) +
h + ···.
2h
6
This suggests the following bound for small h:
M h2
f (x0 + h) − f (x0 − h)
− f ′ (x0 ) ≤
,
2h
6
where |f ′′′ (x)| ≤ M for |x − x0 | < |h|.
PROJECTS FOR CHAPTER TEN
1013
(iii) Expanding each term in the numerator is a Taylor series, we have
4
f (x0 + 2h) = f (x0 ) + 2f ′ (x0 )h + 2f ′′ (x0 )h2 + f ′′′ (x0 )h3
3
2
4
+ f (4) (x0 )h4 + f (5) (x0 )h5 + · · ·
3
15
f ′′ (x0 ) 2 f ′′′ (x0 ) 3
h +
h
f (x0 + h) = f (x0 ) + f ′ (x0 )h +
2
3!
f (4) (x0 ) 4 f (5) (x0 ) 5
h +
h + ···,
+
4!
5!
′′
f (x0 ) 2 f ′′′ (x0 ) 3
h −
h
f (x0 − h) = f (x0 ) − f ′ (x0 )h +
2
3!
f (4) (x0 ) 4 f (5) (x0 ) 5
+
h −
h + ···,
4!
5!
4
f (x0 − 2h) = f (x0 ) − 2f ′ (x0 )h + 2f ′′ (x0 )h2 − f ′′′ (x0 )h3
3
2
4
+ f (4) (x0 )h4 − f (5) (x0 )h5 + · · · .
3
15
Combining the expansions in pairs, we have
8
2
8f (x0 + h) − 8f (x0 − h) = 16f ′ (x0 )h + f ′′′ (x0 )h3 + f (5) (x0 )h5 + · · ·
3
15
8
8
f (x0 + 2h) − f (x0 − 2h) = 4f ′ (x0 )h + f ′′′ (x0 )h3 + f (5) (x0 )h5 + · · · .
3
15
Thus,
−f (x0 + 2h) + 8f (x0 + h) − 8f (x0 − h) + f (x0 − 2h) = 12f ′(x0 )h −
6 (5)
f (x0 )h5 + · · ·
15
so
−f (x0 + 2h) + 8f (x0 + h) − 8f (x0 − h) + f (x0 − 2h)
f (5) (x0 ) 4
= f ′ (x0 ) −
h + ···.
12h
30
This suggests the following bound for small h,
M h4
−f (x0 + 2h) + 8f (x0 + h) − 8f (x0 − h) + f (x0 − 2h)
− f ′ (x0 ) ≤
,
12h
30
where |f (5) (x)| ≤ M for |x − x0 | ≤ |h|.
(b) (i)
h
−1
10
−2
10
10−3
−4
10
(f (x0 ) − f (x0 − h))/h
0.951626
0.995017
0.9995
0.99995
Error
4.837 × 10−2
4.983 × 10−3
4.998 × 10−4
5 × 10−5
The errors are roughly proportional to h, agreeing with part (a).
(ii)
h
10−1
(f (x0 + h) − f (x0 − h))/(2h)
1.00167
10−2
1.00001667
10−3
1.0000001667
−4
10
1.000000001667
Error
1.668 × 10−3
1.667 × 10−5
1.667 × 10−7
1.667 × 10−9
The errors are roughly proportional to h2 , agreeing with part (a).
1014
Chapter Ten /SOLUTIONS
(iii)
h
10−1
(−f (x0 + 2h) + 8f (x0 + h) − 8f (x0 − h) + f (x0 − 2h))/(12h)
0.99999667
10−2
0.9999999999667
10−3
0.99999999999999667
−4
10
Error
3.337 × 10−6
3.333 × 10−10
3.333 × 10−14
3.333 × 10−18
0.999999999999999999667
The errors are roughly proportional to h4 , agreeing with part (a). This is the most accurate formula.
(c) (i)
h
10−1
10−2
10−3
−4
10
10−5
−6
10
−7
10
10−8
−9
10
(ii)
h
10−1
10−2
10−3
−4
10
10−5
10−6
10−7
10−8
−9
10
(iii)
h
10−1
10−2
10−3
−4
10
10−5
10−6
10−7
10−8
−9
10
(f (x0 ) − f (x0 − h))/h
Error
1.0001 × 106
1.00 × 1010
1.0101 × 108
1.01 × 1010
1.0001 × 107
1.00 × 1010
9
1.11 × 1010
1.11111 × 10
Undefined
Undefined
10
−1.11111 × 10
−1.11 × 109
−1.001 × 1010
−1.00 × 107
10
−1.01 × 108
−1.0101 × 10
10
−1.00 × 106
−1.0001 × 10
(f (x0 + h) − f (x0 − h))/(2h)
1 × 102
1 × 104
Error
1 × 1010
1 × 1010
1.0001 × 106
1.0001 × 1010
8
1.0101 × 1010
1.0101 × 10
Undefined
Undefined
−1.0101 × 1010
−1.01 × 108
−1.0001 × 1010
−1.00 × 106
−1.000001 × 1010
−1.00 × 104
10
−1.00 × 102
−1.00000001 × 10
(−f (x0 + 2h) + 8f (x0 + h) − 8f (x0 − h) + f (x0 − 2h))/(12h)
1.25 × 102
1.25 × 104
1.00 × 1010
1.25013 × 106
1.00 × 1010
8
1.01 × 1010
1.26326 × 10
Undefined
Undefined
−9.99579 × 109
4.21 × 106
−9.9999995998 × 1010
−9.99999999996 × 1010
Error
1.00 × 1010
4.00 × 102
10
−9.999999999999996 × 10
4.00 × 10−2
4.00 × 10−6
For relatively large values of h, these approximation formulas fail miserably. The main reason is that
f (x) = 1/x changes very quickly at x0 = 10−5 . In fact, f (x) → ±∞ as x → 0. So we must use very
small values for h when estimating a limit (involving f and x0 = 10−5 ) as h → 0. Here, h > 10−5 is too
big, since the values of x0 − h cross over the discontinuity at x = 0. For smaller values of h, that make
sure we stay on the good side of the abyss, these formulas work quite well. Already by h = 10−6 , formula
(c) is the best approximation.
11.1 SOLUTIONS
CHAPTER ELEVEN
Solutions for Section 11.1
Exercises
1. Since y = x3 , we know that y ′ = 3x2 . Substituting y = x3 and y ′ = 3x2 into the differential equation we get
0 = xy ′ − 3y
= x(3x2 ) − 3(x3 )
= 3x3 − 3x3
= 0.
Since this equation is true for all x, we see that y = x3 is in fact a solution.
2. (a) We substitute y = x4 and its derivative dy/dx = 4x3 into the differential equation:
dy
= 4y
dx
3
x(4x ) = 4(x4 )?
x
4x4 = 4x4 .
Yes:
The function y = x4 satisfies the differential equation so it is a solution.
(b) We substitute y = x4 + 3 and its derivative dy/dx = 4x3 into the differential equation:
dy
= 4y
dx
3
x(4x ) = 4(x4 + 3)?
x
4x4 6= 4x4 + 12.
The function y = x4 + 3 does not satisfy the differential equation so it is not a solution.
(c) We substitute y = x3 and its derivative dy/dx = 3x2 into the differential equation:
dy
= 4y
dx
x(3x2 ) = 4(x3 )?
x
3x3 6= 4x3 .
The function y = x3 does not satisfy the differential equation so it is not a solution.
(d) We substitute y = 7x4 and its derivative dy/dx = 28x3 into the differential equation:
dy
= 4y
dx
x(28x3 ) = 4(7x4 )?
x
Yes:
28x4 = 28x4 .
The function y = 7x4 satisfies the differential equation so it is a solution.
3. (a) We substitute y = 4x3 and its derivative dy/dx = 12x2 into the differential equation:
dy
= 6x2
dx
(4x3 ) · (12x2 ) = 6x2 ?
y
48x5 6= 6x2 .
The function y = 4x3 does not satisfy the differential equation so it is not a solution.
1015
1016
Chapter Eleven /SOLUTIONS
(b) We substitute y = 2x3/2 and its derivative dy/dx = 3x1/2 into the differential equation:
dy
= 6x2
dx
(2x3/2 ) · (3x1/2 ) = 6x2 ?
y
6x2 = 6x2 .
Yes:
The function y = 2x3/2 satisfies the differential equation so it is a solution.
(c) We substitute y = 6x3/2 and its derivative dy/dx = 9x1/2 into the differential equation:
dy
= 6x2
dx
(6x3/2 ) · (9x1/2 ) = 6x2 ?
y
54x2 6= 6x2 .
The function y = 6x3/2 does not satisfy the differential equation so it is not a solution.
4. Differentiating y(x) = Aeλx gives
y ′ (x) = Aλeλx = λ(Aeλx ) = λy.
Therefore, y(x) is a solution of y ′ = λy for any value of A.
5. If y = sin 2t, then
Thus
d2 y
dt2
dy
dt
= 2 cos 2t, and
d2 y
dt2
= −4 sin 2t.
+ 4y = −4 sin 2t + 4 sin 2t = 0.
6. If P = P0 et , then
dP
d
= (P0 et ) = P0 et = P.
dt
dt
7. Differentiating x2 + y 2 = r 2 implicitly, with r a constant, gives
2x + 2y
dy
= 0.
dx
Solving for dy/dx, we get
dy
2x
x
=−
=− .
dx
2y
y
8. (a) To determine whether Q is increasing or decreasing, we check to see whether dQ/dt is positive or negative. Substituting Q = 8 and t = 2 into the differential equation, we have:
t
dQ
=
− 0.5
dt
Q
dQ
2
= − 0.5
dt
8
dQ
= −0.25 < 0.
dt
Since dQ/dt is negative, we see that Q is decreasing at t = 2.
(b) Since dQ/dt = −0.25, the rate of change of Q at t = 2 is −0.25 per unit of t. If the rate of change stays approximately constant over the interval, then Q changes by approximately −0.25 in going from t = 2 to t = 3. We
have:
Value of Q at 3 ≈ Value of Q at 2 + Change in Q
= 8 + (−0.25)
= 7.75.
9. We know that at time t = 0 the value of y is 8. Since we are told that dy/dt = 0.5y, we know that at time t = 0 the
derivative of y is .5(8) = 4. Thus, as t goes from 0 to 1, y will increase by 4, so at t = 1, y = 8 + 4 = 12.
Likewise, at t = 1, we get dy/dt = 0.5(12) = 6 so that at t = 2, we obtain y = 12 + 6 = 18.
At t = 2, we have dy/dt = 0.5(18) = 9 so that at t = 3, we obtain y = 18 + 9 = 27.
At t = 3, we have dy/dt = 0.5(27) = 13.5 so that at t = 4, we obtain y = 27 + 13.5 = 40.5.
Thus, we get the values in the following table
11.1 SOLUTIONS
t
0
1
2
3
4
y
8
12
18
27
40.5
1017
10. At the end of 5 days, dy
= 100 − 67.2 = 32.8% per week. Thus during the next working day, which is the first day of
dt
the second week, the amount learned is about 32.8( 15 ) = 6.6%.
At the end of 6 working days,
y ≈ 67.2% + 6.6% = 73.8%.
Continuing in this manner gives the data in Table 11.1.
Table 11.1
Time (days)
learned (approximate)
Time (days)
learned (approximate)
6
7
8
9
10
11
12
73.8
79.0
83.2
86.6
89.3
91.4
93.1
13
14
15
16
17
18
19
94.5
95.6
96.5
97.2
97.7
98.2
98.6
11. Since x(0) = 5, we have Ce3·0 = 5; that is, C = 5. So the particular solution is x(t) = 5e3t .
12. Since P = 5 when t = 3, we have 5 = C/3; therefore, C = 15. So the particular solution is P = 15/t.
√
13. Because y = 3 √
when t = 1, we know that 3 = 2 · 1 + C. Therefore, 2 + C = 9, and thus C = 7. So the particular
solution is y = 2t + 7.
14. Because Q = 4 when t = 2, we have 4 = 1/(2C + C); therefore, 3C = 1/4. So C = 1/12. Thus, the particular solution
is Q = 12/(t + 1).
Problems
15. In order to prove that y = A + Cekt is a solution to the differential equation
dy
= k(y − A),
dt
we must show that the derivative of y with respect to t is in fact equal to k(y − A):
y = A + Cekt
dy
= 0 + (Cekt )(k)
dt
= kCekt
= k(Cekt + A − A)
= k (Cekt + A) − A
= k(y − A).
16. If y = cos ωt, then
d2 y
= −ω 2 cos ωt.
dt2
dy
= −ω sin ωt,
dt
Thus, if
d2 y
dt2
+ 9y = 0, then
−ω 2 cos ωt + 9 cos ωt = 0
(9 − ω 2 ) cos ωt = 0.
Thus 9 − ω 2 = 0, or ω 2 = 9, so ω = ±3.
17. Differentiating and using the fact that
d
(cosh t) = sinh t
dt
and
d
(sinh t) = cosh t,
dt
1018
Chapter Eleven /SOLUTIONS
we see that
dx
= ωC1 sinh ωt + ωC2 cosh ωt
dt
d2 x
= ω 2 C1 cosh ωt + ω 2 C2 sinh ωt
dt2
= ω 2 (C1 cosh ωt + C2 sinh ωt) .
Therefore, we see that
d2 x
= ω 2 x.
dt2
18. If Q = Cekt , then
dQ
= Ckekt = k(Cekt ) = kQ.
dt
= −0.03Q, so we know that kQ = −0.03Q. Thus we either have Q = 0 (in which case C = 0
We are given that dQ
dt
and k is anything) or k = −0.03. Notice that if k = −0.03, then C can be any number.
19. Since y = x2 + k, we know that y ′ = 2x. Substituting y = x2 + k and y ′ = 2x into the differential equation, we get
10 = 2y − xy ′
= 2(x2 + k) − x(2x)
= 2x2 + 2k − 2x2
= 2k.
Thus, k = 5 is the only solution.
20. If y satisfies the differential equation, then we must have
d (5 + 3ekx )
= 10 − 2(5 + 3ekx )
dx
3kekx = 10 − 10 − 6ekx
3kekx = −6ekx
k = −2.
So, if k = −2 the formula for y solves the differential equation.
21. (a) If y = Cxn is a solution to the given differential equation, then we must have
d (Cxn )
− 3(Cxn ) = 0
dx
x(Cnxn−1 ) − 3(Cxn ) = 0
x
Cnxn − 3Cxn = 0
C(n − 3)xn = 0.
Thus, if C = 0, we get y = 0 is a solution, for every n. If C 6= 0, then n = 3, and so y = Cx3 is a solution.
(b) Because y = 40 for x = 2, we cannot have C = 0. Thus, by part (a), we get n = 3. The solution to the differential
equation is
y = Cx3 .
To determine C if y = 40 when x = 2, we substitute these values into the equation.
40 = C · 23
40 = C · 8
C = 5.
So, now both C and n are fixed at specific values.
11.1 SOLUTIONS
22. (a) Differentiating y(x) = A + Bex
2
/2
gives y ′ (x) = Bxex
y ′ − xy − x = Bxex
2
/2
2
/2
so that
− x A + Bex
2
2
/2
− x = −x(A + 1).
For y(x) = A + Bex /2 to be a solution we need −x(A + 1) = 0 so A = −1.
2
(b) Given the solution y(x) = −1 + Bex /2 , we have y(0) = −1 + B. Since y(0) = 1, we have B = 2.
23. (a) If
y=
then
ex + e−x
,
2
dy
ex − e−x
=
,
dx
2
and
ex + e−x
d2 y
=
.
2
dx
2
If k = 1, then
k
r
1+
dy 2
dx
=
=
=
=
(b) y =
eAx +e−Ax
,
2A
r
1+
r
ex − e−x 2
2
1
e−2x
+ +
=
4
2
4
e2x
r
1
e−2x
e2x
− +
4
2
4
ex + e−x
2
2
(since ex + e−x > 0)
d2 y
.
dx2
so
dy
eAx − e−Ax
=
dx
2
1+
dy
dx
2
= 1+
=
This means
k
r
1+
and
eAx − e−Ax
2
d2 y
=A
dx2
d2 y
dx2
=k
q
1+
dy 2
dx
dy 2
,
dx
2
=1+
eAx + e−Ax
2
.
1 2Ax
e
+ e−2Ax − 2
4
1 Ax
2
1 2Ax
e
+ e−2Ax + 2 =
e + e−Ax .
4
4
=k
r
1 Ax
k
(e + e−Ax )2 = · eAx + e−Ax
4
2
eAx + e−Ax
2
=k
24.
1+
ex + e−x
ex + e−x
=
2
2
Therefore we have
Since we want
=
r
(since eAx + e−Ax > 0).
we must have A = k.
(I) y = ex ,
y ′ = ex ,
3
′
y ′′ = ex
2
y ′′ = 6x
(II) y = x ,
y = 3x ,
(III) y = e−x ,
y ′ = −e−x ,
(IV) y = x
−2
,
′
y = −2x
−3
y ′′ = e−x
, y ′′ = 6x−4
and so:
(a) (I),(III) because y ′′ = y in each case.
(b) (IV) because x2 y ′′ + 2xy ′ − 2y = x2 (6x−4 ) + 2x(−2x−3 ) − 2x−2 = 6x−2 − 4x−2 − 2x−2 = 0.
(c) (II),(IV) because x2 y ′′ = 6y in each case.
1019
1020
Chapter Eleven /SOLUTIONS
25.
(I) y = 2 sin x, dy/dx = 2 cos x,
(II) y = sin 2x,
(III) y = e2x ,
(IV) y = e
−2x
dy/dx = 2 cos 2x,
dy/dx = 2e2x ,
,
dy/dx = −2e
−2x
d2 y/dx2 = −2 sin x
d2 y/dx2 = −4 sin 2x
d2 y/dx2 = 4e2x
,
d2 y/dx2 = 4e−2x
and so:
(a)
(b)
(c)
(d)
26. (a)
(b)
(c)
(d)
(IV)
(III)
(III), (IV)
(II)
(IV) because dy/dx = k = kx/x = y/x.
(III) because dy/dx = kekx = ky.
(I) because dy/dx = kxekx + ekx = ky + xekx /x = ky + y/x.
(II) because dy/dx = kxk−1 = kxk /x = ky/x.
27. No, it is not the general solution since it does not contain an arbitrary constant. It is a particular solution since y ′ =
3e3x = 3y, but it is not the general solution.
28. (a) We substitute y = A + Be−2t and its derivative dy/dt = −2Be−2t into the differential equation:
dy
= 100 − 2y
dt
−2t
−2Be
= 100 − 2(A + Be−2t )
−2Be−2t = 100 − 2A − 2Be−2t
0 = 100 − 2A
A = 50.
If A = 50, the function satisfies the differential equation. There are no conditions on B. The function y = 50+Be−2t
is the general solution to the differential equation.
(b) We substitute y = 85 and t = 0 into the general solution:
y = 50 + Be−2t
85 = 50 + Be−2(0)
85 = 50 + B
B = 35.
We see that B = 35. The particular solution satisfying the differential equation and the initial condition is y =
50 + 35e−2t .
Strengthen Your Understanding
29. Although Q = 6e4t satisfies the differential equation dQ/dt = 4Q, it is not the general solution. The general solution is
the family of all possible solutions to the differential equation and contains an arbitrary constant. Q = 6e4t is only one
particular solution, not a general solution.
30. The differential equation dx/dt = 1/x represents a function x(t) whose rate of change with respect to t is 1/x. Since
x > 0 when t = 0, 1/x is also positive, and thus, x is increasing when t is near 0.
31. An example is dy/dx = x/y with the condition that y = 100 when x = 0. Many other examples are possible.
32. A second-order differential equation is one involving the second derivative, so an example is d2 y/dx2 = 3y or y ′′ = xy 2
or y ′′ + 2y ′ + 4 = 0. Many other examples are possible.
33. An example is dy/dx = 2x with solutions y = x2 and y = x2 + 5. Other examples are possible.
34. An example is dy/dx = sin x. We find the solution by integrating and see that the solution is the function y = − cos x +
C. One solution is the trigonometric function y = − cos x. Other examples are possible.
35. An example is dy/dx = 1/x. We find the solution by integrating and see that the solution is the function y = ln |x| + C.
One solution is the logarithmic function y = ln |x|. Other examples are possible.
11.2 SOLUTIONS
1021
36. An example is dy/dx = ex . In fact, if f (x) is any increasing positive function, then the solutions of dy/dx = f (x) are
increasing since f (x) > 0 and concave up since d2 y/dx2 = f ′ (x) > 0.
37. We want to have dy/dx = 0 when y − x2 = 0, so let dy/dx = y − x2 .
38. False. The general solution contains an arbitrary constant, but the constant may not be added to the particular solution.
For example, the differential equation dy/dt = y has y = et as a particular solution, but the general solution is
y = Cet , which cannot be rewritten in the form y = et + C.
39. False. The function y = t2 is a solution to y ′′ = 2.
40. True. Since f ′ (x) = g(x), we have f ′′ (x) = g ′ (x). Since g(x) is increasing, g ′ (x) > 0 for all x, so f ′′ (x) > 0 for all x.
Thus the graph of f is concave up for all x.
41. False. We just need an example of a function f (x) which is decreasing for x > 0, but whose derivative f ′ (x) = g(x) is
increasing for x > 0. An example is f (x) = 1/x. Clearly f (x) is decreasing for x > 0 but its derivative f ′ (x) = −1/x2
is clearly increasing for x > 0.
42. True. Since g(x) is increasing, g(x) ≥ g(0) = 1 for all x ≥ 0. Since f ′ (x) = g(x), this means that f ′ (x) > 0 for all
x ≥ 0. Therefore f (x) is increasing for all x ≥ 0.
43. False. If g(x) > 0 for all x, then f (x) would have to be increasing for all x so f (x + p) = f (x) would be impossible.
For example, let g(x) = 2 + cos x. Then a possibility for f is f (x) = 2x + sin x. Then g(x) is periodic, but f (x) is not.
44. False. Let g(x) = 0 for all x and let f (x) = 17. Then f ′ (x) = g(x) and limx→∞ g(x) = 0, but limx→∞ f (x) = 17.
45. True. Since limx→∞ g(x) = ∞, there must be some value x = a such that g(x) > 1 for all x > a. Then f ′ (x) > 1 for
all x > a. Thus, for some constant C, we have f (x) > x + C for all x > a, which implies that limx→∞ f (x) = ∞.
More precisely, let C = f (a) − a and let h(x) = f (x) − x − C. Then h(a) = 0 and h′ (x) = f ′ (x) − 1 > 0 for all
x > a. Thus h is increasing so h(x) > 0 for all x > a, which means that f (x) > x + C for all x > a.
46. False. Let f (x) = x3 and g(x) = 3x2 . Then y = f (x) satisfies dy/dx = g(x) and g(x) is even while f (x) is odd.
47. False. The example f (x) = x3 and g(x) = 3x2 shows that you might expect f (x) to be odd. However, the additive
constant C can mess things up. For example, still let g(x) = 3x2 , but let f (x) = x3 + 1 instead. Then g(x) is still even,
but f (x) is not odd (for example, f (−1) = 0 but −f (1) = −2).
Solutions for Section 11.2
Exercises
1. (a) The slope at any point is given by the derivative, so we find the slope by substituting the x- and y-coordinates into
the differential equation dy/dx = x2 − y 2 to find dy/dx.
The slope at (1, 0) is
The slope at (0, 1) is
The slope at (1, 1) is
The slope at (2, 1) is
The slope at (1, 2) is
The slope at (2, 2) is
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
= 12 − 02 = 1.
= 02 − 12 = −1.
= 12 − 12 = 0.
= 22 − 12 = 3.
= 12 − 22 = −3.
= 22 − 22 = 0.
1022
Chapter Eleven /SOLUTIONS
(b) See Figure 11.1.
y
2
1
x
1
2
Figure 11.1
2. We substitute x- and y-coordinates into the differential equation to determine the slopes. At the point (1, 1), we have
dy/dx = 1/1 = 1 so the slope is 1. At the point (1, 0), we have dy/dx = 1/0, which is undefined. An undefined slope
corresponds to a vertical line segment. Continuing in this way, we create Figure 11.2.
y
1
x
−1
1
−1
Figure 11.2
3. We substitute x- and y-coordinates into the differential equation to determine the slopes. At the point (−1, 2), we have
dy/dx = 22 = 4 so the slope is 4. At the point (0, 2), we have dy/dx = 22 = 4, so the slope is again 4. Continuing in
this way, we create Figure 11.3.
y
2
1
x
−1
1
Figure 11.3
4. Figure (I) shows a line segment at (4, 0) with positive slope. The only possible differential equation is (b), since at (4, 0)
we have y ′ = cos 0 = 1. Note that (a) is not possible as y ′ (4, 0) = e−16 = 0.0000001, a much smaller positive slope
than that shown.
11.2 SOLUTIONS
1023
Figure (II) shows a line segment at (0, 4) with zero slope. The possible differential equations are (d), since at (0, 4)
we have y ′ = 4(4 − 4) = 0, and (f), since at (0, 4) we have y ′ = 0(3 − 0) = 0.
Figure (III) shows a line segment at (4, 0) with negative slope of large magnitude. The only possible differential
equation is (f), since at (4, 0) we have y ′ = 4(3−4) = −4. Note that (c) is not possible as y ′ (4, 0) = cos(4−0) = −0.65,
a negative slope of smaller magnitude than that shown.
Figure (IV) shows a line segment at (4, 0) with a negative slope of small magnitude. The only possible differential
equation is (c), since at (4, 0) we have y ′ = cos(4−0) = −0.65. Note that (f) is not possible as y ′ (4, 0) = 4(3−4) = −4,
a negative slope of larger magnitude than that shown.
Figure (V) shows a line segment at (0, 4) with positive slope. Possible differential equations are (a), since at (0, 4)
2
we have y ′ = e0 = 1, and (c), since at (0, 4) we have y ′ = cos(4 − 4) = 1.
Figure (VI) shows a line segment at (0, 4) with a negative slope of large magnitude. The only possible differential
equation is (e), since at (0, 4) we have y ′ = 4(3 − 4) = −4. Note that (b) is not possible as y ′ (0, 4) = cos 4 = −0.65, a
negative slope of smaller magnitude than that shown.
5. There are many possible answers. One possibility is shown in Figures 11.4 and 11.5.
y
y
x
x
Figure 11.4
Figure 11.5
6. See Figure 11.6. Other choices of solution curves are, of course, possible.
y
y
x
x
Figure 11.6
1024
Chapter Eleven /SOLUTIONS
7. The first graph has the equation y ′ = x2 − y 2 . We can see this by looking along the line y = x. On the first slope field,
it seems that y ′ = 0 along this line, as it should if y ′ = x2 − y 2 . This is not the case for the second graph. Another way
to see this is to look along the y-axis on both graphs. The slope lines change with y on the first graph, but are constant on
the second. But at x = 0, the slope should be −y 2 , which varies with y, so the first graph is the one that fits.
At (0, 1), y ′ = −1, and at (1, 0), y ′ = 1, so we are looking for points on the axes where the line segment is sloped
◦
at 45 . See Figure 11.7.
y
x=1
(0, 1)
x
(1, 0)
Figure 11.7
8. (a) See Figure 11.8.
y
1
x
0
−1
Figure 11.8
(b) The solution is y(x) = 1.
(c) Since y ′ = 0 and x(y − 1) = 0, this is a solution.
9. (a) See Figure 11.9.
y
4
i
x
−4
4
ii
−4
iii
Figure 11.9
(b) The solution through (−1, 0) appears to be linear, so its equation is y = −x − 1.
(c) If y = −x − 1, then y ′ = −1 and x + y = x + (−x − 1) = −1, so this checks as a solution.
11.2 SOLUTIONS
1025
Problems
10. We draw line segments with positive slope to the left of the y-axis, line segments with negative slope to the right of the
y-axis, and horizontal line segments on the y-axis. A slope field satisfying these conditions is shown in Figure 11.10.
Other answers are possible.
y
x
Figure 11.10
11. We draw line segments with positive slope when P is between 2 and 5, line segments with negative slope in the sections
below P = 2 and above P = 5, and horizontal line segments on the horizontal lines P = 2 and P = 5. A slope field
satisfying these conditions is shown in Figure 11.11. Other answers are possible.
P
5
2
t
Figure 11.11
12. (a) See Figure 11.12.
Figure 11.12
(b) From the graph, the solution through (1, 0) appears linear with the equation approximately y = x − 1.
In fact, if y = x − 1, then x − y = x − (x − 1) = 1 = y ′ , so y = x − 1 is the solution through (1, 0).
1026
Chapter Eleven /SOLUTIONS
13. (a) See Figure 11.13.
P
(−2, 12)
(−2, 10)
(1, 4)
(4, 1)
(−5, 1)
t
(0, 0)
Figure 11.13
(b) If 0 < P < 10, the solution is increasing; if P > 10, it is decreasing. If P (0) = 5, then P tends to 10.
14. (a) and (b) See Figure 11.14
(vi) y
(i)
(iii)
x
(iv)
(ii)
(v)
Figure 11.14
(c) Figure 11.14 shows that a solution will be increasing if its y-values fall in the range −1 < y < 2. This makes sense
since if we examine the equation y ′ = 0.5(1 + y)(2 − y), we will find that y ′ > 0 if −1 < y < 2. Notice that if the
y-value ever gets to 2, then y ′ = 0 and the function becomes constant, following the line y = 2. (The same is true if
ever y = −1.)
From the graph, the solution is decreasing if y > 2 or y < −1. Again, this also follows from the equation, since
in either case y ′ < 0.
The curve has a horizontal tangent if y ′ = 0, which only happens if y = 2 or y = −1. This also can be seen on
the graph in Figure 11.14.
x+y
15. Notice that y ′ =
is zero when x = −y and is undefined when x = y. A solution curve will be horizontal
x−y
(slope= 0) when passing through a point with x = −y, and will be vertical (slope undefined) when passing through a
point with x = y. The only slope field for which this is true is slope field (b).
11.2 SOLUTIONS
1027
16. (a) See Figure 11.15.
y
6
(ii)
x
−6
6
(i)
−6
Figure 11.15
(b) We can see that the slope lines are horizontal when y is an integer multiple of π. We conclude from Figure 11.15 that
the solution is y = nπ in this case.
To check this, we note that if y = nπ, then (sin x)(sin y) = (sin x)(sin nπ) = 0 = y ′ . Thus y = nπ is a
solution to y ′ = (sin x)(sin y), and it passes through (0, nπ).
17. (a) Since y ′ = −y, the slope is negative above the x-axis (when y is positive) and positive below the x-axis (when y is
negative). The only slope field for which this is true is II.
(b) Since y ′ = y, the slope is positive for positive y and negative for negative y. This is true of both I and III. As y gets
larger, the slope should get larger, so the correct slope field is I.
(c) Since y ′ = x, the slope is positive for positive x and negative for negative x. This corresponds to slope field V.
1
(d) Since y ′ = , the slope is positive for positive y and negative for negative y. As y approaches 0, the slope becomes
y
larger in magnitude, which corresponds to solution curves close to vertical. The correct slope field is III.
(e) Since y ′ = y 2 , the slope is always positive, so this must correspond to slope field IV.
18. The slope fields in (I) and (II) appear periodic. (I) has zero slope at x = 0, so (I) matches y ′ = sin x, whereas (II) matches
2
y ′ = cos x. The slope in (V) tends to zero as x → ±∞, so this must match y ′ = e−x . Of the remaining slope fields,
′
−x
only (III) shows negative slopes, matching y = xe . The slope in (IV) is zero at x = 0, so it matches y ′ = x2 e−x . This
leaves field (VI) to match y ′ = e−x .
19. Judging from the figure, we see that:
• The slope depends on y, not on x.
• The slope is positive for y < 0 and y > 10.
• The slope is zero at y = 0 and y = 10.
This corresponds to equation (e): y ′ = 0.05y(y − 10).
20. Judging from the figure, we see that:
• The slope depends on y, not on x.
• The slope is positive for 0 < y < 10.
• The slope is zero at y = 0 and y = 10.
This corresponds to equation (a): y ′ = 0.05y(10 − y).
21. Judging from the figure, we see that:
• The slope depends on x, not on y.
• The slope is positive for 0 < x < 10.
• The slope is zero at x = 0 and x = 10.
This corresponds to equation (b): y ′ = 0.05x(10 − x).
22. Judging from the figure, we see that:
• The slope depends on x, not on y.
• The slope is positive for 0 < x < 5.
• The slope is zero at x = 0 and x = 5.
This corresponds to equation (d): y ′ = 0.05x(5 − x).
1028
Chapter Eleven /SOLUTIONS
2
dy
23. (a) We know that
= ex > 0 for all x. As x → ∞, the slopes tend toward infinity. Similarly, as x → −∞, the
dx
slopes tend to infinity. Thus the appropriate slope field is (IV).
dy
> 0 for all x. As x → ±∞, the slopes tend toward 0. Only slope fields I and III meet these
(b) We know that
dx
2
dy
conditions. We know that when x = 0,
= e−2x = e0 = 1.
dx
2
dy
= e−2x = e−2 ≈ 0.14. Thus the appropriate slope field is (I).
We know that when x = 1,
dx
dy
> 0 for all x. As x → ±∞, the slopes tend toward 0. Only slope fields I and III meet these
(c) We know that
dx
2
2
dy
dy
conditions. We know that when x = 0,
= e−x /2 = e0 = 1. We know that when x = 1,
= e−x /2 =
dx
dx
e−1/2 ≈ 0.61. Thus the appropriate slope field is (III).
dy
(d) The slope field is both positive and negative. In fact, when x = 0,
= e−0.5x cos x = e0 cos 0 = 1.
dx
dy
dy
Also, when x = 1,
= e−0.5x cos x = e−1/2 cos 1 ≈ 0.33, and when x = 2,
= e−0.5x cos x =
dx
dx
e−1 cos 2 ≈ −0.153. Thus the appropriate slope field is (V).
1
dy
=
≈ 0.44. Thus the appropriate
(e) The slope field is positive for all values of x. When x = 0,
dx
(1 + 0.5 cos x)2
slope field is (II).
(f) The slope field is negative for all values of x. Thus the appropriate slope field is (VI).
24. When a = 1 and b = 2, the Gompertz equation is y ′ = −y ln(y/2) = y ln(2/y) = y(ln 2 − ln y). This differential
equation is similar to the differential equation y ′ = y(2 − y) in certain ways. For example, in both equations y ′ is positive
for 0 < y < 2 and negative for y > 2. Also, for y-values close to 2, the quantities (ln 2 − ln y) and (2 − y) are both
close to 0, so y(ln 2 − ln y) and y(2 − y) are approximately equal to zero. Thus around y = 2 the slope fields look almost
the same. This happens again around y = 0, since around y = 0 both y(2 − y) and y(ln 2 − ln y) go to 0. (Note that
lim (y ln y) = 0.) For y values close to 1 the slope fields look similar since the local linearization of ln y near y = 1 is
y→0+
y − 1; hence, near y = 1, y(ln 2 − ln y) ≈ y(ln 2 − (y − 1)) ≈ y(1.69 − y) ≈ y(2 − y). Finally, for y > 2, ln y grows
much slower than y, so the slope field for y ′ = y(ln 2 − ln y) is less steep, negatively, than for y ′ = y(2 − y).
Strengthen Your Understanding
25. Since the slope is zero at the point (1, 1), the slope field has a horizontal line segment there. On the other hand, the solution
y = x goes through the point (1, 1) with a slope of 1. See Figure 11.16. This is impossible since the slope field is tangent
to the solution curves.
y
1
x
1
Figure 11.16
26. Note that y ′ = y has a slope field with all positive slopes when y > 0. The given slope field, however, has negative slopes
in the second quadrant.
27. If the slopes are all positive, then dy/dx is always positive. Some examples are dy/dx = x2 + 1 or dy/dx = y 2 + 1 or
dy/dx = x2 + y 2 + 1 or dy/dx = ex .
28. The sign of the derivative depends on the y-coordinate, since the y-coordinate determines whether a point is above or
below the x-axis. Since we want a positive derivative when y is positive and a negative derivative when y is negative, one
example is dy/dx = y.
11.2 SOLUTIONS
1029
29. If a derivative dy/dx depends only on x, then the derivative is constant for any fixed value of x. In other words, the slope
is the same for all points on any vertical line. One possibility is the slope field for dy/dx = x in Figure 11.17.
y
x
Figure 11.17
30. If a derivative dy/dx depends only on y, then the derivative is constant for any fixed value of y. In other words, the slope
is the same for all points on any horizontal line. One possibility is the slope field for dy/dx = y in Figure 11.18.
y
x
Figure 11.18
31. False. If y(0) ≤ 0, then limx→∞ y = −∞.
32. True. No matter what initial value you pick, the solution curve has the x-axis as an asymptote.
33. False. There appear to be two equilibrium values dividing the plane into regions with different limiting behavior.
34. True. We have dy/dx > 0 at every point because x2 + y 2 + 1 > 0, and a positive derivative indicates increasing function.
35. False. We have
d(x2 + y 2 + 1)
d2 y
=
2
dx
dx
dy
= 2x + 2y
dx
= 2x + 2y(x2 + y 2 + 1)
= 2x + 2y + 2x2 y + 2y 3 .
At the point (x, y) = (−1, 0) we have d2 y/dx2 = −2 < 0. A negative second derivative indicates function concave
down. The solution curve of the differential equation that passes through the point (−1, 0) is concave down at (−1, 0).
36. True. The slope of the graph of f is dy/dx = 2x − y. Thus when x = a and y = b, the slope is 2a − b.
37. True. Saying y = f (x) is a solution for the differential equation dy/dx = 2x − y means that if we substitute f (x) for y,
the equation is satisfied. That is, f ′ (x) = 2x − f (x).
38. False. Since f ′ (x) = 2x − f (x), we would have 1 = 2x − 5 so x = 3 is the only possibility.
1030
Chapter Eleven /SOLUTIONS
39. True. Differentiate dy/dx = 2x − y, to get:
d
dy
d2 y
=
(2x − y) = 2 −
= 2 − (2x − y).
dx2
dx
dx
40. False. Since f ′ (1) = 2(1) − 5 = −3, the point (1, 5) could not be a critical point of f .
41. True. Since dy/dx = 2x − y, the slope of the graph of f is negative at any point satisfying 2x < y, that is any point
lying above the line y = 2x. The slope of the graph of f is positive at any point satisfying 2x > y, that is any point lying
below the line y = 2x.
42. True. When we differentiate dy/dx = 2x − y, we get:
dy
d2 y
=2−
= 2 − (2x − y).
dx2
dx
Thus at any inflection point of y = f (x), we have d2 y/dx2 = 2 − (2x − y) = 0. That is, any inflection point of f must
satisfy y = 2x − 2.
43. False. Suppose that g(x) = f (x) + C, where C 6= 0. In order to be a solution of dy/dx = 2x − y we would need
g ′ (x) = 2x − g(x). Instead we have:
g ′ (x) = f ′ (x) = 2x − f (x) = 2x − (g(x) − C) = 2x − g(x) + C.
Since C 6= 0, this means g(x) is not a solution of dy/dx = 2x − y.
44. True. We will use the hint. Let w = g(x) − f (x). Then:
dw
= g ′ (x) − f ′ (x) = (2x − g(x)) − (2x − f (x)) = f (x) − g(x) = −w.
dx
Thus dw/dx = −w. This equation is the equation for exponential decay and has the general solution w = Ce−x . Thus,
lim (g(x) − f (x)) = lim Ce−x = 0.
x→∞
x→∞
Solutions for Section 11.3
Exercises
1. Using Euler’s method, we have:
At (x, y) = (0, 4): y ′ = (0 − 2)(4 − 3)
so at x = 0.1:
At (x, y) = (0.1, 3.8):
so at x = 0.2:
At (x, y) = (0.2, 3.648):
= −2
y = 4.0 − 2(0.1)
= 3.8
y = 3.8 − 1.52(0.1)
= 3.648
′
y = (0.1 − 2)(3.8 − 3)
= −1.52
′
y = (0.2 − 2)(3.648 − 3) = −1.1664.
Thus, the completed table is
x
y
0.0
4.0
0.1
3.8
0.2
3.648
y′
−2
−1.52
−1.1664
because ∆x = 0.1
because ∆x = 0.1
11.3 SOLUTIONS
1031
2. Using Euler’s method, we have:
At (x, y) = (1, −3): y ′ = 4(1)(−3)
so at x = 1.01:
= −12
y = −3 − 12(0.01)
= −3.12
At (x, y) = (1.01, −3.12):
y ′ = 4(1.01)(−3.12)
At (x, y) = (1.02, −3.246):
y ′ = 4(1.02)(−3.246)
so at x = 1.02:
because ∆x = 0.01
= −12.6048
y = −3.12 − 12.6048(0.01) = −3.246
because ∆x = 0.01
= −13.244.
Thus, the completed table is
x
y
y′
1.00
−3
−12
1.01
1.02
−3.12
−3.246
−12.6048
−13.244
3. We know P = 1500 at time t = 0. This means
dP
= 0.00008(1500)(1900 − 1500) = 48.
dt
From time t = 0 to t = 1 we have ∆t = 1, so the new value of P is given by
Value of P
at t = 1
= 1500 + 48(1) = 1548.
We repeat this process to find the values of P at t = 2 and t = 3. See the table.
t
P
dP/ dt
0
1500
1
1500 + 48(1) = 1548
0.00008(1500)(1900 − 1500) = 48
0.00008(1548)(1900 − 1548) = 43.59168
2
1548 + 43.59168(1) = 1591.59168
3
1591.59168 + 39.26881 = 1630.86049
0.00008(1591.59168)(1900 − 1591.59168) = 39.26881
(no calculation necessary)
Rounding gives values of P = 1548, P = 1591.5917, P = 1630.860 at t = 1, 2, 3.
4. (a) Since dy/dx is always 3 and ∆x = 0.2, at every step we have
∆y =
dy
· ∆x = 3(0.2) = 0.6.
dx
The results are in Table 11.2. We see that Euler’s method gives an approximate value of y = 5 at x = 1.
Table 11.2
x
0
0.2
0.4
0.6
0.8
1.0
y
2
2.6
3.2
3.8
4.4
5
(b) The general solution to dy/dx = 3 is y = 3x + C. We use the initial condition to see that C = 2 so the particular
solution is y = 3x + 2.
(c) The exact value of the solution at x = 1 is y = 3(1) + 2 = 5. Since the approximate value from Euler’s method is
also 5, the error is 0.
(d) Euler’s method approximates a solution using line segments. Since in this case the exact solution is itself a line,
Euler’s method gives exact values.
1032
Chapter Eleven /SOLUTIONS
5. (a) At point P0 = (0, 10), we have
∆y = slope at P0 · ∆x = (10 − 0)(0.2) = 2.
Therefore, point P1 is (0.2, 12). At point P1 , we have
∆y = slope at P1 · ∆x = (12 − 0.2)(0.2) = 2.36.
So point P2 is (0.4, 14.36). Continuing in this way, we obtain the approximate solution shown in Table 11.3.
Table 11.3
x
0
0.2
0.4
0.6
0.8
1.0
y
10
12
14.36
17.15
20.46
24.39
(b) Since the slopes are getting larger, we expect the solution to be concave up.
(c) Since the solution is concave up, we expect our approximations to be underestimates. It can be shown that the true
value of y at x = 1 is 26.46.
6. (a) The results from Euler’s method with ∆x = 0.1 are in Table 11.4.
(b) We have
x4
+ C,
y(x) =
4
so that y(0) = 0 gives C = 0, and the required solution is therefore
y(x) =
x4
.
4
This is shown in the 3rd column of Table 11.4.
(c) The computed solution underestimates the real solution since the solution is concave up and is approximated in every
interval by the tangent which is beneath the curve. See Figure 11.19.
Table 11.4
y
Computed Solution
xn
Approx. y(xn )
y(xn )
0
0
0
0.1
0
0.000025
0.2
0.0001
0.0004
0.3
0.0009
0.002025
0.4
0.0036
0.0064
0.5
0.01
0.015625
0.6
0.0225
0.0324
0.7
0.0441
0.060025
0.8
0.0784
0.1024
0.9
0.1296
0.164025
1.0
0.2025
0.25
7. (a) In Table 11.5, we see that y(0.4) ≈ 1.5282.
(b) In Table 11.6, we see that y(0.4) = −1.4. (This answer is exact.)
x
Figure 11.19
11.3 SOLUTIONS
1033
Table 11.6 Euler’s method for
y ′ = x + y with y(−1) = 0
Table 11.5 Euler’s method for
y ′ = x + y with y(0) = 1
x
y
0
1
∆y =(slope)∆x
0.1
1.1
0.12 = (1.2)(0.1)
0.2
1.22
0.142 = (1.42)(0.1)
0.3
1.362
0.1662 = (1.662)(0.1)
0.4
1.5282
0.1 = (1)(0.1)
x
y
∆y =(slope)∆x
−1
0
−0.1 = (−1)(0.1)
−0.8
−0.2
−0.9
−0.1
−0.7
..
.
−0.3
..
.
0
..
.
−1
..
.
0.4
−1.4
−0.1 = (−1)(0.1)
−0.1 = (−1)(0.1)
Notice that y
decreases by 0.1
for every step
8. (a) See Figure 11.20.
(b) See Table 11.7. At x = 1, Euler’s method gives y ≈ 0.16.
(c) Our answer to (a) appears to be an underestimate. This is as we would expect, since the solution curve is concave up.
y
✛
0.25
True
answer
Table 11.7
0.16
✛
Approximation
x
1
Figure 11.20
x
y
∆y = (slope)∆x
0
0
0
0.2
0
0.0016
0.4
0.0016
0.0128
0.6
0.0144
0.0432
0.8
0.0576
0.1024
1
0.1600
9. (a) See Figure 11.21.
(b) y(0) = 1,
y(0.1) ≈ y(0) + 0.1y(0) = 1 + 0.1(1) = 1.1
y(0.2) ≈ y(0.1) + 0.1y(0.1) = 1.1 + 0.1(1.1) = 1.21
y(0.3) ≈ y(0.2) + 0.1y(0.2) = 1.21 + 0.1(1.21) = 1.331
y(0.4) ≈ 1.4641
y(0.5) ≈ 1.61051
y(0.6) ≈ 1.77156
y(0.7) ≈ 1.94872
y(0.8) ≈ 2.14359
y(0.9) ≈ 2.35795
y(1.0) ≈ 2.59374
(c) See Figure 11.21. A smooth curve drawn through the solution points seems to match the slope field.
(d) For y = ex , we have y ′ = ex = y and y(0) = e0 = 1. See Table 11.8.
1034
Chapter Eleven /SOLUTIONS
Table 11.8
Computed Solution
Figure 11.21
xn
Approx. y(xn )
y(xn )
0
1
1
0.1
1.1
1.10517
0.2
1.21
1.22140
0.3
1.331
1.34986
0.4
1.4641
1.49182
0.5
1.61051
1.64872
0.6
1.77156
1.82212
0.7
1.94872
2.01375
0.8
2.14359
2.22554
0.9
2.35795
2.45960
1.0
2.59374
2.71828
Problems
10. (a)
(i)
Table 11.9 Euler’s method for
y ′ = (sin x)(sin y), starting at (0, 2)
(ii)
x
y
∆y =(slope)∆x
0
2
0.1
2
0.009 = (sin 0.1)(sin 2)(0.1)
0 = (sin 0)(sin 2)(0.1)
0.2
2.009
0.018 = (sin 0.2)(sin 2.009)(0.1)
0.3
2.027
Table 11.10 Euler’s method for
y ′ = (sin x)(sin y), starting at (0, π)
x
y
∆y =(slope)∆x
0
π
0 = (sin 0)(sin π)(0.1)
0.1
π
0 = (sin 0.1)(sin π)(0.1)
0.2
π
0 = (sin 0.2)(sin π)(0.1)
0.3
π
(b) The slope field shows that the slope of the solution curve through (0, π) is always 0. Thus the solution curve is the
horizontal line with equation y = π.
11. (a)
Table 11.11
x
y
0
1.000
0.2
0.800
0.4
0.680
0.6
0.620
0.8
0.615
1
0.667
∆y = (slope)∆x
−0.200
−0.120
−0.060
−0.005
0.052
11.3 SOLUTIONS
1035
(b)
Table 11.12
x
y
∆y = (slope)∆x
−0.100
0
1.000
0.1
0.900
0.2
0.820
0.3
0.757
0.4
0.709
0.5
0.674
0.6
0.654
0.7
0.647
0.8
0.654
0.021
0.9
0.675
0.035
1
0.710
−0.080
−0.063
−0.048
−0.034
−0.020
−0.007
0.007
12. By looking at the slope fields, or by computing the second derivative
d2 y
dy
= 2x − 2y
= 2x − 2x2 y + 2y 3 ,
dx2
dx
we see that the solution curve is concave up, so Euler’s method gives an underestimate.
13. Since the error is proportional to one over the number of subintervals, the error using 10 intervals should be roughly half
the error obtained using 5 intervals. Since both the estimates are underestimates, if we let A be the actual value we have:
1
(A − 0.667) = A − 0.710
2
A − 0.667 = 2A − 1.420
A = 0.753
Therefore, 0.753 should be a better approximation.
14. (a)
Table 11.13
t
y
slope =
1
t
∆y = (slope)∆t =
1
0
1
0.1
1.1
0.1
0.909
0.091
1.2
0.191
0.833
0.083
1.3
0.274
0.769
0.077
1.4
0.351
0.714
0.071
1.5
0.422
0.667
0.067
1.6
0.489
0.625
0.063
1.7
0.552
0.588
0.059
1.8
0.610
0.556
0.056
1.9
0.666
0.526
0.053
2
0.719
1
(0.1)
t
(b) If dy
= 1t , then y = ln |t| + C.
dt
Starting at (1, 0) means y = 0 when t = 1, so C = 0 and y = ln |t|.
After ten steps, t = 2, so y = ln 2 ≈ 0.693.
(c) Approximate y = 0.719, Exact y = 0.693.
Thus the approximate answer is too big. This is because the solution curve is concave down, and so the tangent lines
are above the curve. Figure 11.22 shows the slope field of y ′ = 1/t with the solution curve y = ln t plotted on top of
it.
1036
Chapter Eleven /SOLUTIONS
y
t
1
Figure 11.22
15. (a) ∆x = 0.5
Table 11.14 Euler’s method for
y ′ = 2x, with y(0) = 1
x
y
∆y =(slope)∆x
0
1
0.5
1
0 = (2 · 0)(0.5)
1
1.5
0.5 = (2 · 0.5)(0.5)
∆x = 0.25
Table 11.15 Euler’s method for
y ′ = 2x, with y(0) = 1
x
y
0
1
0.25
1
0.50
1.125
0.75
1.375
1
1.75
∆y =(slope)∆x
0 = (2 · 0)(0.25)
0.125 = (2 · 0.25)(0.25)
0.25 = (2 · 0.5)(0.25)
0.375 = (2 · 0.75)(0.25)
(b) General solution is y = x2 + C, and y(0) = 1 gives C = 1. Thus, the solution is y = x2 + 1. So the true value of y
when x = 1 is y = 12 + 1 = 2.
(c) When ∆x = 0.5, error = 0.5.
When ∆x = 0.25, error = 0.25.
Thus, decreasing ∆x by a factor of 2 has decreased the error by a factor of 2, as expected.
16. For ∆x = 0.2, we get the following results.
y(1.2) ≈ y(1) + 0.2 sin(1 · y(1)) = 1.168294
y(1.4) ≈ y(1.2) + 0.2 sin(1.2 · y(1.2)) = 1.365450
y(1.6) ≈ y(1.4) + 0.2 sin(1.4 · y(1.4)) = 1.553945
y(1.8) ≈ y(1.6) + 0.2 sin(1.6 · y(1.6)) = 1.675822
y(2.0) ≈ y(1.8) + 0.2 sin(1.8 · y(1.8)) = 1.700779
Repeating this with ∆x = 0.1 and 0.05 gives the results in Table 11.16 below
11.3 SOLUTIONS
1037
Table 11.16
Computed Solution
x-value
∆x = 0.2
∆x = 0.1
∆x = 0.05
1.0
1
1
1
1.084147
1.086501
1.177079
1.181232
1.275829
1.280619
1.375444
1.379135
1.469214
1.469885
1.549838
1.546065
1.611296
1.602716
1.650458
1.637809
1.667451
1.652112
1.664795
1.648231
1.1
1.2
1.168294
1.3
1.4
1.365450
1.5
1.6
1.553945
1.7
1.8
1.675822
1.9
2.0
1.700779
The computed approximations for y(2) using step sizes ∆x = 0.2, 0.1, 0.05 are 1.700779, 1.664795, and 1.648231,
respectively. Plotting these points we see that they lie approximately on a straight line.
y(2)
2
1.5
1
0.5
0.1
0.2
0.3
∆x
Figure 11.23
In the limit, as ∆x tends to zero, the results produced by Euler’s method should converge to the exact value of y(2).
This limiting value is the vertical intercept of the line drawn in Figure 11.23. This gives y(2) ≈ 1.632.
= 0.05, so ∆B =
17. (a) Using one step, ∆B
∆t
year.
(b) With two steps, ∆t = 0.5 and we have
∆B
∆t
∆t = 50. Therefore we get an approximation of B ≈ 1050 after one
Table 11.17
t
B
∆B = (0.05B)∆t
0
1000
25
0.5
1025
25.63
1.0
1050.63
(c) Keeping track to the nearest hundredth with ∆t = 0.25, we have
Table 11.18
t
B
∆B = (0.05B)∆t
0
1000
12.5
0.25
1012.5
12.66
0.5
1025.16
12.81
0.75
1037.97
12.97
1
1050.94
1038
Chapter Eleven /SOLUTIONS
(d) In part (a), we get our approximation by making a single increment, ∆B, where ∆B is just 0.05B. If we think in
terms of interest, ∆B is just like getting one end of the year interest payment. Since ∆B is 0.05 times the balance
B, it is like getting 5% interest at the end of the year.
(e) Part (b) is equivalent to computing the final amount in an account that begins with $1000 and earns 5% interest
compounded twice annually. Each step is like computing the interest after 6 months. When t = 0.5, for example, the
interest is ∆B = (0.05B) · 21 , and we add this to $1000 to get the new balance.
Similarly, part (c) is equivalent to the final amount in an account that has an initial balance of $1000 and earns
5% interest compounded quarterly.
18. Assume that x > 0 and that we use n steps in Euler’s method. Label the x-coordinates we use in the process
x0 , x1 , . . . , xn , where x0 = 0 and xn = x. Then using Euler’s method to find y(x), we get
Table 11.19
x
y
∆y = (slope)∆x
P0
0 = x0
0
f (x0 )∆x
P1
x1
f (x0 )∆x
f (x1 )∆x
P2
.
..
x2
.
..
f (x0 )∆x + f (x1 )∆x
.
..
f (x2 )∆x
.
..
Pn
x = xn
n−1
X
f (xi )∆x
i=0
Thus the result from Euler’s method is
mates
Rx
0
n−1
X
f (xi )∆x. We recognize this as the left-hand Riemann sum that approxi-
i=0
f (t) dt.
Strengthen Your Understanding
dy
19. For differential equations of the form
= k, where k is constant, Euler’s method traces the exact solution for any initial
dx
condition.
20. If x(0) > 0, then the statement is true. However, if x(0) < 0, then the solution curve is decreasing. Since we are using the
value of x at the beginning of a subinterval to estimate the rate of change of x on an interval, we obtain an overestimate
for x(1).
21. The approximate values lie on a line if the slope ∆y/∆x is constant, which occurs if dy/dx is constant. One example is
dy/dx = 5. The approximate values found using Euler’s method lie on a line for any initial condition and value of ∆x.
Other examples are possible.
22. One step of Euler’s method gives an underestimate when the solution is concave up. Thus, we look for a differential
equation whose solution curves are concave up everywhere. An example is given by the equation dy/dx = y with
y(0) = 1 which has solution y = ex .
23. False. Euler’s method approximates y-values of points on the solution curve.
24. True. Both lead to
y(1) ≈ f (0) · 0.2 + f (0.2) · 0.2 + f (0.4) · 0.2 + f (0.6) · 0.2 + f (0.8) · 0.2.
Solutions for Section 11.4
Exercises
1. (a)
(d)
(g)
(j)
Yes
No
No
Yes
(b)
(e)
(h)
(k)
No
Yes
Yes
Yes
(c) Yes
(f) Yes
(i) No
(l) No
11.4 SOLUTIONS
2. Separating variables gives
Z
so
1
dP = −
P
Z
2dt,
ln |P | = −2t + C.
Therefore
P = ±e−2t+C = Ae−2t .
The initial value P (0) = 1 gives 1 = A, so
P = e−2t .
3. Separating variables gives
Z
so
dP
=
P
Z
0.02 dt,
ln |P | = 0.02t + C.
Thus
|P | = e0.02t+C
and
P = Ae0.02t , where A = ±eC .
We are given P (0) = 20. Therefore, P (0) = Ae(0.02)·0 = A = 20. So the solution is
P = 20e0.02t .
4. Separating variables and integrating both sides gives
Z
or
1
1
dL =
L
2
ln |L| =
Z
dp
1
p + C.
2
This can be written
L = ±e(1/2)p+C = Aep/2 .
The initial condition L(0) = 100 gives 100 = A, so
L = 100ep/2 .
5. Separating variables gives
so
Z
dQ
=
Q
ln |Q| =
So
and
Z
dt
,
5
1
t + C.
5
1
1
|Q| = e 5 t+C = e 5 t eC
1
Q = Ae 5 t , where A = ±eC .
1
From the initial conditions we know that Q(0) = 50, so Q(0) = Ae( 5 )·0 = A = 50. Thus
1
Q = 50e 5 t .
1039
1040
Chapter Eleven /SOLUTIONS
6. Separating variables gives
Z
so that
P dP =
Z
dt
P2
= t+C
2
√
P = ± 2t + D
or
(where D = 2C).
The initial condition P (0) = 1 implies we must take the positive root and that 1 = D, so
√
P = 2t + 1.
7. Separating variables gives
Z
dm
=
m
Z
3 dt
ln |m| = 3t + C
m = ±eC e3t = Ae3t .
Since m = 5 when t = 1, we have 5 = Ae3 , so A = 5/e3 . Thus
m=
5 3t
e = 5e3t−3 .
e3
8. Separating variables gives
so
Thus,
Z
dI
=
I
Z
0.2 dx,
ln |I| = 0.2x + C.
I = Ae0.2x , where A = ±eC .
According to the given boundary condition, I(−1) = 6. Therefore, I(−1) = Ae0.2(−1) = Ae−0.2 = 6, so A = 6e0.2 .
Thus
I = 6e0.2 e0.2x = 6e0.2(x+1) .
9. Separating variables gives
Z
Z
dz
5 dt
=
z
ln |z| = 5t + C.
Solving for z, we have
z = Ae5t , where A = ±eC .
Using the fact that z(1) = 5, we have z(1) = Ae5 = 5, so A = 5/e5 . Therefore,
z=
5 5t
e = 5e5t−5 .
e5
10. Separating variables gives
Hence
Z
1
dm =
m
Z
ds.
ln |m| = s + C
11.4 SOLUTIONS
which gives
m = ±es+C = Aes .
The initial condition m(1) = 2 gives 2 = Ae1 or A = 2/e, so
m=
2 s
e = 2es−1 .
e
11. Separating variables gives
Z
or
Z
1
du =
u2
−
1
dt
2
1
1
= t + C.
u
2
The initial condition gives C = −1 and so
u=
1
.
1 − (1/2)t
12. Separating variables and integrating gives
Z
which gives
1
dz =
z
z = ±e(1/2)y
ydy
1 2
y +C
2
ln |z| =
or
Z
2
+C
= Aey
2
/2
.
The initial condition y = 0, z = 1 gives A = 1. Therefore
z = ey
2
/2
.
13. Separating variables gives
Z
dy
=−
y
Z
1
dx
3
1
ln |y| = − x + C.
3
Solving for y, we have
1
y = Ae− 3 x , where A = ±eC .
Since y(0) = A = 10, we have
1
y = 10e− 3 x .
14. Separating variables gives
Z
Z
dy
=
0.5dt
y − 200
ln |y − 200| = 0.5t + C
y = 200 + Ae0.5t ,
where A = ±eC .
The initial condition, y(0) = 50, gives
50 = 200 + A,
so
A = −150.
Thus,
y = 200 − 150e0.5t .
1041
1042
Chapter Eleven /SOLUTIONS
15. Separating variables gives
Z
so
dP
=
P +4
Z
dt,
ln |P + 4| = t + C
P + 4 = Aet
P = Aet − 4.
Since P = 100 when t = 0, we have P (0) = Ae0 − 4 = 100, and A = 104. Therefore
P = 104et − 4.
16. Factoring out a 2 on the right makes the integration easier:
dy
= 2y − 4 = 2(y − 2)
dx
Z
Z
dy
2 dx,
=
y−2
giving
ln |y − 2| = 2x + C.
Thus,
|y − 2| = e2x+C ,
so
y − 2 = Ae2x , where A = ±eC .
The curve passes through (2, 5), which means 3 = Ae4 , so A = 3/e4 . Thus,
y =2+
3 2x
e = 2 + 3e2x−4 .
e4
17. Factoring and separating variables gives
dQ
= 0.3(Q − 400)
dt
Z
Z
dQ
0.3 dt
=
Q − 400
ln |Q − 400| = 0.3t + C
Q = 400 + Ae0.3t ,
where A = ±eC .
The initial condition, Q(0) = 50, gives
50 = 400 + A
so
A = −350.
Thus
Q = 400 − 350e0.3t .
18. Factoring out the 0.1 gives
so
dm
= 0.1m + 200 = 0.1(m + 2000)
dt
Z
Z
dm
=
0.1 dt,
m + 2000
ln |m + 2000| = 0.1t + C,
11.4 SOLUTIONS
and
m = Ae0.1t − 2000, where A = ±eC .
Using the initial condition, m(0) = Ae(0.1)·0 − 2000 = 1000, gives A = 3000. Thus
m = 3000e0.1t − 2000.
19. Rearrange and write
Z
or
Z
1
dR =
1−R
dy
− ln |1 − R| = y + C
which can be written as
1 − R = ±e−C−y = Ae−y
or
R = 1 − Ae−y .
The initial condition R(1) = 0.1 gives 0.1 = 1 − Ae−1 and so
A = 0.9e.
Therefore
R = 1 − 0.9e1−y .
20. Rewriting gives
dB
+ 2B = 50
dt
and
dB
= −2B + 50 = −2(B − 25),
dt
so
Z
dB
=−
B − 25
Z
2 dt
ln |B − 25| = −2t + C.
Thus, we have
B − 25 = Ae−2t , where A = ±eC .
Using the initial condition, B(1) = 100, we have 75 = Ae−2 , so A = 75e2 . Thus
B = 25 + 75e2 e−2t = 25 + 75e2−2t .
21. Write
and so
Z
1
dy =
y
Z
1
dt
3+t
ln |y| = ln |3 + t| + C
or
ln |y| = ln D|3 + t|
where ln D = C. Therefore
y = D(3 + t).
The initial condition y(0) = 1 gives D =
1
,
3
so
y=
1
(3 + t).
3
1043
1044
Chapter Eleven /SOLUTIONS
22. Separating variables gives
dz
= tez
dt
−z
Z e dz = tdt
Z
−z
e dz =
t dt,
so
t2
+ C.
2
Since the solution passes through the origin, z = 0 when t = 0, we must have
−e−z =
−e−0 =
0
+ C, so C = −1.
2
Thus
−e−z =
or
t2
− 1,
2
t2
z = − ln 1 −
2
.
23. Separating variables gives
5y
dy
=
dx
Z
Zx
dy
5
=
dx
y
x
ln |y| = 5 ln |x| + C.
Thus
5
|y| = e5 ln |x| eC = eC eln |x| = eC |x|5 ,
giving
y = Ax5 ,
Since y = 3 when x = 1, so A = 3. Thus
where
A = ±eC .
y = 3x5 .
24. Separating variables gives
dy
= y 2 (1 + t)
dt
Z
Z
dy
=
(1 + t) dt,
y2
so
−
1
t2
=t+
+ C,
y
2
giving
y=−
1
.
t + t2 /2 + C
Since y = 2 when t = 1, we have
2=−
1
,
1 + 1/2 + C
Thus
y=−
so
2C + 3 = −1,
and
1
2
=− 2
.
t2 /2 + t − 2
t + 2t − 4
C = −2.
11.4 SOLUTIONS
25. Separating variables gives
dz
= z + zt2 = z(1 + t2 )
Z dt Z
dz
= (1 + t2 )dt,
z
so
ln |z| = t +
t3
+ C,
3
giving
z = Aet+t
3
/3
.
We have z = 5 when t = 0, so A = 5 and
z = 5et+t
3
/3
.
26. Separating variables gives
dw
= θw2 sin θ2
dθ
Z
Z
dw
=
θ sin θ2 dθ,
w2
so
1
1
= − cos θ2 + C.
w
2
According to the initial conditions, w(0) = 1, so −1 = − 21 + C and C = − 21 . Thus,
−
−
1
1
1
= − cos θ2 −
w
2
2
cos θ2 + 1
1
=
w
2
2
.
w=
cos θ2 + 1
27. Separating variables and integrating gives
Z
1
dw = −
w2
Z
tan ψdψ.
To integrate the right side, write tan ψ = sin ψ/ cos ψ and use the substitution w = cos ψ giving
−
1
= ln | cos ψ| + C
w
so
w=
Using the initial condition w(0) = 2 we have
2=
−1
.
ln | cos ψ| + C
−1
−1
−1
=
=
ln | cos 0| + C
ln 1 + C
0+C
so
1
C=− .
2
Thus the solution is
w=
−1
.
ln | cos ψ| − 1/2
1045
1046
Chapter Eleven /SOLUTIONS
28. Separating variables gives
du
= u2
dx
Z
Z
Z
du
dx
1
1
=
=
−
dx,
u2
x(x + 1)
x
1+x
x(x + 1)
so
1
= ln |x| − ln |x + 1| + C.
u
We have u(1) = 1, so − 11 = ln |1| − ln |1 + 1| + C. So C = ln 2 − 1. Solving for u yields
−
−
1
2|x|
= ln |x| − ln |x + 1| + ln 2 − 1 = ln
− 1,
u
|x + 1|
so
u=
−1
.
2x
|−1
ln | x+1
Problems
29. (a) We separate variables and integrate:
4x
dy
= 2
dx
y
y 2 dy = 4x dx
Z
2
y dy =
Z
4x dx
y3
= 2x2 + C
3
y=
p
3
6x2 + B.
Here, we use B as the arbitrary constant, replacing 3C when we multiply through by 3.
(b) When we substitute y = 1 when x = 0, we have:
1=
1=
p
3
√
3
6(02 ) + B
B
3
B = 1 = 1.
√
The particular solution satisfying y(0) = 1 is y = 3 6x2 + 1.
When we substitute y = 2 when x = 0, we have:
2=
2=
p
3
√
3
6(02 ) + B
B
3
B = 2 = 8.
√
The particular solution satisfying y(0) = 2 is y = 3 6x2 + 8.
When we substitute y = 3 when x = 0, we have:
3=
3=
p
3
√
3
3
6(02 ) + B
B
B = 3 = 27.
√
The particular solution satisfying y(0) = 3 is y = 3 6x2 + 27.
The three solutions are shown in Figure 11.24.
11.4 SOLUTIONS
y
4
x
−4
4
Figure 11.24
30. (a) We separate variables and integrate:
dP
= 0.2(P − 50)
dt
1
dP = 0.2 dt
P − 50
Z
Z
1
dP =
0.2 dt
P − 50
ln |P − 50| = 0.2t + C
|P − 50| = e0.2t+C
P − 50 = Be0.2t
P = 50 + Be0.2t .
Here, we use B as the arbitrary constant,replacing ±eC .
(b) When we substitute P = 40 when t = 0, we have:
40 = 50 + Be0.2(0)
40 = 50 + B
B = −10.
The particular solution satisfying P (0) = 40 is P = 50 − 10e0.2t .
When we substitute P = 50 when t = 0, we have:
50 = 50 + Be0.2(0)
50 = 50 + B
B = 0.
The particular solution satisfying P (0) = 50 is the constant solution P = 50.
When we substitute P = 60 when t = 0, we have:
60 = 50 + Be0.2(0)
60 = 50 + B
B = 10.
The particular solution satisfying P (0) = 60 is P = 50 + 10e0.2t .
The three solutions are shown in Figure 11.25.
P
70
30
4
Figure 11.25
t
1047
1048
Chapter Eleven /SOLUTIONS
31. (a) Separating variables and integrating gives
Z
1
dy =
100 − y
Z
dt
so that
− ln |100 − y| = t + C
or
y(t) = 100 − Ae−t .
(b) See Figure 11.26.
y
110
25
t
Figure 11.26
(c) The initial condition y(0) = 25 gives A = 75, so the solution is
y(t) = 100 − 75e−t .
The initial condition y(0) = 110 gives A = −10 so the solution is
y(t) = 100 + 10e−t .
(d) The increasing function, y(t) = 100 − 75e−t .
32. By separating variables, we obtain r dr = k dt. Integrating yields
r2
= kt + C,
2
where C is a constant. If t = 0 is the time when the spill begins, then r = 0 when t = 0; therefore, we must have C = 0.
Therefore, we have
√
r = 2kt.
√
Using the fact that r = 400 when t = 16, we obtain 400 = 2k · 16; this yields k = 5000. Since the units of dr/dt are
feet/hour and r is measured in feet, the units of k must be feet2 /hour.
33. (a) See Figure 11.27.
y
(0, 1)
(0, 0)
(0, −1)
Figure 11.27
x
11.4 SOLUTIONS
1049
(b) It appears that the solution curves in the upper half plane are asymptotic to y = 0 as x tends to −∞, and that they
are unbounded above as x increases.
It appears that the solution curves in the lower half plane are asymptotic to y = 0 as x tends to +∞, and that
they are unbounded below as x decreases.
The solution curve through the origin is the x-axis, so it is asymptotic to y = 0 on both ends.
(c) We separate variables. From dy/dx = y 2 we get, for y 6= 0,
Z
Z
1
dy =
dx
y2
1
− = x+C
y
−1
y=
.
x+C
(d) For a given number C, the y-value is not defined for x = −C, so the formula gives two solution curves. The first is
−1
x+C
y=
for x < −C.
This curve is in the upper half plane, has a vertical asymptote at x = −C and satisfies limx→−∞ y = 0. The second
curve is
−1
y=
for x > −C,
x+C
which is in the lower half plane, has a vertical asymptote at x = −C, and satisfies limx→+∞ y = 0.
34. Separating variables gives
Z
Integrating gives
dR
=
R
Z
k dt.
ln |R| = kt + C,
so
|R| = ekt+C = ekt eC
R = Aekt ,
where A = ±eC
or
A = 0.
35. Separating variables gives
Q
dQ
=
dt
Z
Zk
dQ
1
=
dt
Q
k
Integrating gives
t
+C
k
Q = Aet/k ,
ln |Q| =
where A = ±eC
or
A = 0.
36. Separating variables gives
Integrating yields
so
Z
dP
=
P −a
Z
dt.
ln |P − a| = t + C,
|P − a| = et+C = et eC
P = a + Aet ,
where A = ±eC
or A = 0.
1050
Chapter Eleven /SOLUTIONS
37. Separating variables gives
Z
Integrating yields
dQ
=
b−Q
Z
dt.
− ln |b − Q| = t + C,
so
|b − Q| = e−(t+C) = e−t e−C
Q = b − Ae−t ,
where A = ±e−C
or A = 0.
38. Separating variables gives
Z
Integrating yields
dP
=
P −a
Z
k dt.
ln |P − a| = kt + C,
so
P = a + Aekt
where A = ±eC
or A = 0.
39. Factoring and separating variables gives
b
dR
= a R+
dt
a
Z
Z
dR
=
a dt
R + b/a
b
= at + C
ln R +
a
b
R = − + Aeat ,
a
where A can be any constant.
40. Separating variables and integrating gives
Z
This gives
1
dP =
aP + b
1
ln |aP + b| = t + C
a
ln |aP + b| = at + aC
aP + b = ±eat+aC = Aeat ,
or
P =
Z
dt.
where A = ±eaC
1
(Aeat − b).
a
41. Separating variables and integrating gives
or
Z
Z
1
dy =
y2
−
1
1
= k t + t3 + C.
y
3
Hence,
y=
k(1 + t2 )dt
−1
.
k(t + 31 t3 ) + C
or
A = 0,
11.4 SOLUTIONS
42. Separating variables and integrating gives
Z
or
1
dR =
R2 + 1
Z
adx
arctan R = ax + C
so that
R = tan(ax + C).
43. Separating variables and integrating gives
Z
or
1
dL =
L−b
ln |L − b| = k
Solving for L gives
1
L = b + Aek( 2 x
2
+ax)
,
Z
1
2
k(x + a)dx
x2 + ax + C.
where A can be any constant.
44. Separating variables gives
dy
= y(2 − y),
dt
so
Z
so
−
Integrating yields
1
2
dy
=−
y(y − 2)
Z
1
1
−
y
y−2
Z
dt,
dy = −
ln
Exponentiating both sides yields
1−
|y − 2|
= −2t + 2C.
|y|
2
= e−2t+2C
y
2
= 1 − Ae−2t ,
y
2
y=
.
1 − Ae−2t
But
y(0) =
so A = −1, and
y=
where A = ±e2C
2
= 1,
1−A
2
.
1 + e−2t
45. Separating variables gives
dx
x ln x
=
,
dt
t
and thus
dt.
1
(ln |y − 2| − ln |y|) = −t + C,
2
so
so
Z
Z
dx
=
x ln x
Z
dt
,
t
ln | ln x| = ln t + C,
1051
1052
Chapter Eleven /SOLUTIONS
so
| ln x| = eC eln t = eC t.
Therefore
ln x = At,
where A = ±eC
or
A = 0,
so
x = eAt .
46. Separating variables gives
dx
= (1 + 2 ln t) tan x
dt
1 + 2 ln t
dx
=
dt
tan x
Z
Z t
cos x
1
2 ln t
dt.
dx =
+
sin x
t
t
t
Integrating gives
ln | sin x| = ln t + (ln t)2 + C
| sin x| = eln t+(ln t)
2
= t(eln t )ln t eC = t(tln t )eC .
+C
So
sin x = At(ln t)+1 ,
where A = ±eC
Therefore
or
x = arcsin(At(ln t)+1 ).
47. Since
y
dy
,
= −y ln
dt
2
we have
dy
= − dt,
y ln(y/2)
so that
Substituting w = ln(y/2), dw =
1
dy gives:
y
Z
dy
=
y ln(y/2)
Z
Z
Z
so
dw
=−
w
(− dt).
dt
ln |w| = −t + C
y
= −t + C.
ln ln
2
Since y(0) = 1, we have C = ln | ln 12 | = ln(| − ln 2|) = ln(ln 2). Thus
ln ln
or
ln
Since eln(ln 2) = ln 2, this simplifies to
y
2
= −t + ln(ln 2),
y
= e−t+ln(ln 2)
y
2
= (ln 2)e−t ,
2
ln
so
ln
y
2
= ±(ln 2)e−t .
A = 0.
11.4 SOLUTIONS
1053
Since y(0) = 1, and ln(1/2) = − ln 2, we take the − sign, giving
ln
Thus,
y
2
= −(ln 2)e−t .
−t
y = 2e−(ln 2)e
−t
y = 2(e− ln 2 )e
−t
y = 2(2−e
(Note that ln(y/2) = (ln 2)e
−t
−t
= 2(2−1 )e
).
does not satisfy y(0) = 1.)
48. (a) See Figure 11.28.
y
x
Figure 11.28
(b) It appears that the solution curves are unbounded below as x decreases and unbounded above as x increases.
(c) In the region −1 ≤ y ≤ 1, we have dy/dx = 1, a constant slope, so the solution curves in this region are straight
lines of slope 1: the equations are y = x + C. This formula holds for −C − 1 ≤ x ≤ −C + 1.
In the regions where |y| ≥ 1, we solve the differential equation by separation of variables,
Z
Z
1
dy =
dx
y2
1
− = x+C
y
−1
.
y=
x+C
Thus, the formula for the general solution is y = −1/(x + C) for both y ≤ −1 and y ≥ 1. The corresponding
x-values are determined as follows. We have y ≤ −1 if
−1
≤ −1
x+C
1
≥1
x+C
0 α, we have dy/dt < 0, so the function is decreasing.
For y < α, we have dy/dt > 0, so the function is increasing. See Figure 11.33.
y
α
t
Figure 11.33
6. When the derivative dw/dt = 0, the function is constant. This occurs when w = 3 or w = 7. So, w = 3 and w = 7
are solutions. Their graphs are horizontal lines. When 3 < w < 7, the derivative is negative. Therefore, solutions, w,
are decreasing in this region. For w < 3 or w > 7, the derivative is positive. Thus, solutions, w, are increasing in these
regions. See Figure 11.34.
w
7
6
3
1
Figure 11.34
t
1058
Chapter Eleven /SOLUTIONS
R
R
dH
dH
7. (a) Separating variables, we have H−200
= −k dt, so H−200
= −k dt, whence ln |H − 200| = −kt + C, and
−kt
C
H − 200 = Ae , where A = ±e . The initial condition is that the yam is 20◦ C at the time t = 0. Thus
20 − 200 = A, so A = −180. Thus H = 200 − 180e−kt .
−80
, giving
(b) Using part (a), we have 120 = 200 − 180e−k(30) . Solving for k, we have e−30k = −180
k=
ln 49
≈ 0.027.
−30
Note that this k is correct if t is given in minutes. (If t is given in hours, k =
4
ln 9
−1
2
≈ 1.62.)
8. (a) We know that the equilibrium solution is the solution satisfying the differential equation whose derivative is everywhere 0. Thus we have
dy
=0
dt
0.5y − 250 = 0
y = 500.
(b) We use separation of variables. Since
dy
= 0.5y − 250,
dt
we have
Z
Z
1
dy =
dt
0.5y − 250
2 ln |0.5y − 250| = t + C
0.5y − 250 = e(t+C)/2
y = Aet/2 + 500,
where A = 2eC/2 .
(c) Using initial value y(0) = 500, we have y = 500, the equilibrium solution. Using initial value y(0) = 400, we have
A = −100 and so y = 500 − 100et/2 . Using initial value y(0) = 600, we have A = 100 and so y = 500 + 100et/2 .
These three solutions are shown below.
y
y = 500 + 100et/2
y = 500
500
y = 500 − 100et/2
1
t
(d) We see above that the equilibrium solution y = 500 is unstable.
9. (a) To find the equilibrium solutions, we must set
dy/dx = 0.5y(y − 4)(2 + y) = 0
which gives three solutions: y = 0, y = 4, and y = −2.
(b) From Figure 11.35, we see that y = 0 is stable and y = 4 and y = −2 are both unstable.
11.5 SOLUTIONS
1059
y
x
Figure 11.35
10. (a) A very hot cup of coffee cools faster than one near room temperature. The differential equation given says that the
rate at which the coffee cools is proportional to the difference between the temperature of the surrounding air and the
temperature of the coffee. Since dH/dt < 0 (the coffee is cooling) and H − 20 > 0 (the coffee is warmer than room
temperature), k must be positive.
(b) Separating variables gives
Z
Z
1
dH =
−kdt
H − 20
and so
ln |H − 20| = −kt + C
and
H(t) = 20 + Ae−kt .
If the coffee is initially boiling (100◦ C), then A = 80 and so
H(t) = 20 + 80e−kt .
When t = 2, the coffee is at 90◦ C and so 90 = 20 + 80e−2k so that k =
Let the time when the coffee reaches 60◦ C be Hd , so that
1
2
ln 87 .
60 = 20 + 80e−kHd
e−kHd =
Therefore,
Hd =
1
.
2
2 ln 2
1
ln 2 =
≈ 10minutes.
k
ln 87
11. Since it takes 6 years to reduce the pollution to 10%, another 6 years would reduce the pollution to 10% of 10%, which is
equivalent to 1% of the original. Therefore it takes 12 years for 99% of the pollution to be removed. (Note that the value
of Q0 does not affect this.) Thus the second time is double the first because the fraction remaining, 0.01, in the second
instance is the square of the fraction remaining, 0.1, in the first instance.
12. Michigan:
so
r
158
dQ
=− Q=−
Q ≈ −0.032Q
dt
V
4.9 × 103
Q = Q0 e−0.032t .
We want to find t such that
0.1Q0 = Q0 e−0.032t
so
t=
− ln(0.1)
≈ 72 years.
0.032
1060
Chapter Eleven /SOLUTIONS
Ontario:
so
r
−209
dQ
=− Q=
Q = −0.131Q
dt
V
1.6 × 103
Q = Q0 e−0.131t .
We want to find t such that
0.1Q0 = Q0 e−0.131t
so
− ln(0.1)
≈ 18 years.
0.131
Lake Michigan will take longer because it is larger (4900 km3 compared to 1600 km3 ) and water is flowing through
it at a slower rate (158 km3 /year compared to 209 km3 /year).
t=
13. Lake Superior will take the longest, because the lake is largest (V is largest) and water is moving through it most slowly
(r is smallest). Lake Erie looks as though it will take the least time because V is smallest and r is close to the largest. For
Erie, k = r/V = 175/460 ≈ 0.38. The lake with the largest value of r is Ontario, where k = r/V = 209/1600 ≈ 0.13.
Since e−kt decreases faster for larger k, Lake Erie will take the shortest time for any fixed fraction of the pollution to be
removed.
For Lake Superior,
r
65.2
dQ
=− Q=−
Q ≈ −0.0053Q
dt
V
12,200
so
Q = Q0 e−0.0053t .
When 80% of the pollution has been removed, 20% remains so Q = 0.2Q0 . Substituting gives us
0.2Q0 = Q0 e−0.0053t
so
ln(0.2)
≈ 301 years.
0.0053
65.2
, rather than 0.0053. Using 0.0053 gives 304 years.)
(Note: The 301 is obtained by using the exact value of Vr = 12,200
For Lake Erie, as in the text
dQ
r
175
=− Q=−
Q ≈ −0.38Q
dt
V
460
so
Q = Q0 e−0.38t .
t=−
When 80% of the pollution has been removed
0.2Q0 = Q0 e−0.38t
ln(0.2)
≈ 4 years.
t=−
0.38
So the ratio is
301
Time for Lake Superior
≈
≈ 75.
Time for Lake Erie
4
In other words it will take about 75 times as long to clean Lake Superior as Lake Erie.
Problems
14. (a) We define P to be the population of India, in billions of people, in year t, where t represents the number of years
since 2010.
(b) We have dP/dt = 0.0135P with initial condition P (0) = 1.15.
(c) The general solution is P = Ce0.0135t and the particular solution satisfying the initial condition is P = 1.15e0.0135t .
15. (a) We define N to be the amount of nicotine in the body, in mg, at time t, where t represents the number of hours since
smoking the cigarette.
(b) We have dN/dt = −0.347N with initial condition N (0) = 0.4. Notice that the constant −0.347 is negative since
the quantity of nicotine is decreasing.
(c) The general solution is N = Ce−0.347t and the particular solution satisfying the initial condition is N = 0.4e−0.347t .
11.5 SOLUTIONS
1061
16. (a) We define S to be world solar PV market installations, in megawatts, in year t, where t represents the number of
years since 2007.
(b) We have dS/dt = 0.48S with initial condition S(0) = 2826.
(c) The general solution is S = Ce0.48t and the particular solution satisfying the initial condition is S = 2826e0.48t .
17. (a) We define G to be the size, in acres, of Grinnell Glacier in year t, where t represents the number of years since 2007.
(b) We have dG/dt = −0.043G with initial condition G(0) = 142. Notice that the constant −0.043 is negative because
the size is decreasing.
(c) The general solution is G = Ce−0.043t and the particular solution satisfying the initial condition is G = 142e−0.043t .
18. (a) The differential equation is
1 dB
= 0.067
B dt
or
dB
= 0.067B.
dt
(b) The differential equation is
1 dP
dP
= 0.033 or
= 0.033P.
P dt
dt
(c) For initial prices B0 and P0 , solving the differential equations in part (a) and part (b) gives
B = B0 e0.067t ,
and
P = P0 e0.033t .
(d) Doubling time for textbook price is given by
2B0 = B0 e0.067t
ln 2
= 10.345 years.
t=
0.067
(e) Doubling time for inflation is given by
2P0 = P0 e0.033t
ln 2
= 21.004 years.
t=
0.033
(f) The doubling times are in the ratio
(ln 2)/0.067
10.345
0.033
Inflation growth rate
Doubling time:Textbook
=
=
=
=
.
Doubling time:Inflation
21.004
(ln 2)/0.033
0.067
Textbook growth rate
The ratio of the doubling times is the reciprocal of the ratio of the growth rates.
19. We find the temperature of the orange juice as a function of time. Newton’s Law of Heating says that the rate of change of
the temperature is proportional to the temperature difference. If S is the temperature of the juice, this gives us the equation
dS
= −k(S − 65)
dt
for some constant k.
Notice that the temperature of the juice is increasing, so the quantity dS/dt is positive. In addition, S = 40 initially,
making the quantity (S − 65) negative.
Separating variables gives:
Z
Z
dS
= − k dt
S − 65
ln |S − 65| = −kt + C
So
S − 65 = Ae−kt , where A = ±eC .
S = 65 + Ae−kt .
Since at t = 0, S = 40, we have 40 = 65 + C, so C = −25. Thus, S = 65 − 25e−kt for some positive constant k. See
Figure 11.36 for the graph.
1062
Chapter Eleven /SOLUTIONS
S (◦ F)
65◦
40◦
t
Figure 11.36: Graph of
S = 65 − 25e−kt for k > 0
20. According to Newton’s Law of Cooling, the temperature, T , of the roast as a function of time, t, satisfies
T ′ (t) = k(350 − T )
T (0) = 40.
Solving this differential equation, we get that T = 350 − 310e−kt for some k > 0. To find k, we note that at t = 1 we
have T = 90, so
90 = 350 − 310e−k(1)
260
= e−k
310
260
k = − ln
310
≈ 0.17589.
Thus, T = 350 − 310e−0.17589t . Solving for t when T = 140, we have
140 = 350 − 310e−0.17589t
210
= e−0.17589t
310
ln(210/310)
t=
−0.17589
t ≈ 2.21 hours.
21. (a)
Q0
Q
1
Q
2 0
Q = Q0 e−0.0187t
37
t
dQ
= −kQ
dt
(c) Since 25% = 1/4, it takes two half-lives = 74 hours for the drug level to be reduced to 25%. Alternatively, Q =
Q0 e−kt and 12 = e−k(37) , we have
ln(1/2)
≈ 0.0187.
k=−
37
Therefore Q = Q0 e−0.0187t . We know that when the drug level is 25% of the original level that Q = 0.25Q0 .
Setting these equal, we get
0.25 = e−0.0187t .
(b)
giving
t=−
ln(0.25)
≈ 74 hours ≈ 3 days.
0.0187
11.5 SOLUTIONS
1063
22. (a) Since the amount leaving the blood is proportional to the quantity in the blood,
dQ
= −kQ for some positive constant k.
dt
Thus Q = Q0 e−kt , where Q0 is the initial quantity in the bloodstream. Only 20% is left in the blood after 3 hours.
0.20
Thus 0.20 = e−3k , so k = ln−3
≈ 0.5365. Therefore Q = Q0 e−0.5365t .
(b) Since 20% is left after 3 hours, after 6 hours only 20% of that 20% will be left. Thus after 6 hours only 4% will be
left, so if the patient is given 100 mg, only 4 mg will be left 6 hours later.
23. (a) Suppose Y (t) is the quantity of oil in the well at time t. We know that the oil in the well decreases at a rate proportional
to Y (t), so
dY
= −kY.
dt
Integrating, and using the fact that initially Y = Y0 = 106 , we have
Y = Y0 e−kt = 106 e−kt .
In six years, Y = 500, 000 = 5 · 105 , so
5 · 105 = 106 e−k·6
so
0.5 = e−6k
ln 0.5
= 0.1155.
k=−
6
When Y = 600, 000 = 6 · 105 ,
Rate at which oil decreasing =
dY
dt
= kY = 0.1155(6 · 105 ) = 69,300 barrels/year.
(b) We solve the equation
5 · 104 = 106 e−0.1155t
0.05 = e−0.1155t
ln 0.05
t=
= 25.9 years.
−0.1155
24. (a) If C ′ = −kC, and then C = C0 e−kt . Since the half-life is 5730 years, 21 C0 = C0 e−5730k . Solving for k, we have
≈ 0.000121.
−5730k = ln(1/2) so k = − ln(1/2)
5730
(b) From the given information, we have 0.91 = e−kt , where t is the age of the shroud. Solving for t, we have t =
− ln 0.91
≈ 779.4 years.
k
25. The rate of disintegration is proportional to the quantity of carbon-14 present. Let Q be the quantity of carbon-14 present
at time t, with t = 0 in 1977. Then
Q = Q0 e−kt ,
where Q0 is the quantity of carbon-14 present in 1977 when t = 0. Then we know that
Q0
= Q0 e−k(5730)
2
so that
k=−
ln(1/2)
= 0.000121.
5730
Thus
Q = Q0 e−0.000121t .
The quantity present at any time is proportional to the rate of disintegration at that time so
Q0 = c8.2
and
Q = c13.5
1064
Chapter Eleven /SOLUTIONS
where c is a constant of proportionality. Thus substituting for Q and Q0 in
Q = Q0 e−0.000121t
gives
c13.5 = c8.2e−0.000121t
so
ln(13.5/8.2)
≈ −4120.
0.000121
Thus Stonehenge was built about 4120 years before 1977, in about 2150 B.C.
dT
26. (a)
= −k(T − A), where A = 68◦ F is the temperature of the room, and t is time since 9 am.
dt
(b)
t=−
Z
Z
dT
= − kdt
T −A
ln |T − A| = −kt + C
T = A + Be−kt .
Using A = 68, and T (0) = 90.3, we get B = 22.3. Thus
T = 68 + 22.3e−kt .
At t = 1, we have
89.0 = 68 + 22.3e−k
21 = 22.3e−k
21
≈ 0.06.
k = − ln
22.3
Thus T = 68 + 22.3e−0.06t .
We want to know when T was equal to 98.6◦ F, the temperature of a live body, so
98.6 = 68 + 22.3e−0.06t
30.6
= −0.06t
ln
22.3
30.6
1
t= −
ln
0.06
22.3
t ≈ −5.27.
The victim was killed approximately 5 14 hours prior to 9 am, at 3:45 am.
27. (a) The differential equation is
dT
= −k(T − A),
dt
where A = 10◦ F is the outside temperature.
(b) Integrating both sides yields
Z
dT
=−
T −A
Then ln |T − A| = −kt + C, so T = A + Be−kt . Thus
Z
k dt.
T = 10 + 58e−kt .
Since 10:00 pm corresponds to t = 9,
57 = 10 + 58e−9k
47
= e−9k
58
47
ln
= −9k
58
1 47
k = − ln
≈ 0.0234.
9 58
11.5 SOLUTIONS
1065
At 7:00 the next morning (t = 18) we have
T ≈ 10 + 58e18(−0.0234)
= 10 + 58(0.66)
≈ 48◦ F,
so the pipes won’t freeze.
(c) We assumed that the temperature outside the house stayed constant at 10◦ F. This is probably incorrect because the
temperature was most likely warmer during the day (between 1 pm and 10 pm) and colder after (between 10 pm and
7 am). Thus, when the temperature in the house dropped from 68◦ F to 57◦ F between 1 pm and 10 pm, the outside
temperature was probably higher than 10◦ F, which changes our calculation of the value of the constant k. The house
temperature will most certainly be lower than 48◦ F at 7 am, but not by much—not enough to freeze.
= kD.
28. (a) Since speed is the derivative of distance, Galileo’s mistaken conjecture was dD
dt
(b) We know that if Galileo’s conjecture were true, then D(t) = D0 ekt , where D0 would be the initial distance fallen.
But if we drop an object, it starts out not having traveled any distance, so D0 = 0. This would lead to D(t) = 0 for
all t.
29. (a) Letting k be the constant of proportionality, by Newton’s Law of Cooling, we have
dH
= k(68 − H).
dt
(b) To find the equilibrium solution, we solve dH/dt = k(68 − H) = 0. We find that H = 68 is an equilibrium
solution. This makes sense because the temperature of an object at 68◦ F in a 68◦ F room will not change. To determine
whether the equilibrium is stable, we consider a solution with an initial condition near H = 68. If H = 69, then
dH/dt = k(68 − 69) = −k, which is negative; therefore, H will decrease and move towards the equilibrium.
Similarly, a solution with an initial condition less than H = 68 will increase towards the equilibrium. Therefore, the
equilibrium is stable.
(c) We solve this equation by separating variables:
Z
Z
dH
=
k dt
68 − H
− ln |68 − H| = kt + C
68 − H = ±eC−kt
H = 68 − Ae−kt .
(d) We are told that H = 40 when t = 0; this tells us that
40 = 68 − Ae−k(0)
40 = 68 − A
A = 28.
Knowing A, we can solve for k using the fact that H = 48 when t = 1:
48 = 68 − 28e−k(1)
20
= e−k
28
20
= 0.33647.
k = − ln
28
So the formula is H(t) = 68 − 28e−0.33647t . We calculate H when t = 3, by
H(3) = 68 − 28e−0.33647(3) = 57.8◦ F.
30. (a) Since we are told that the rate at which the quantity of the drug decreases is proportional to the amount of the drug
left in the body, we know the differential equation modeling this situation is
dQ
= kQ.
dt
Since we are told that the quantity of the drug is decreasing, we know that k < 0.
1066
Chapter Eleven /SOLUTIONS
(b) To find the equilibrium, we solve dQ/dt = 0 and get Q = 0. Thus, Q = 0 is an equilibrium. This makes sense
because patients with no hydrocodone bitartrate in their body will remain that way unless they take the drug. We
expect the equilibrium to be stable because patients who have taken the drug will gradually lose it, moving towards
the equilibrium. We can confirm this by checking solutions with initial conditions near Q = 0.
(c) We know that the general solution to the differential equation
dQ
= kQ
dt
is
Q = Cekt .
(d) We are told that the half life of the drug is 3.8 hours. This means that at t = 3.8, the amount of the drug in the body
is half the amount that was in the body at t = 0, or, in other words,
0.5Q(0) = Q(3.8).
Solving this equation gives
0.5Q(0) = Q(3.8)
0.5Cek(0) = Cek(3.8)
0.5C = Cek(3.8)
0.5 = ek(3.8)
ln(0.5) = k(3.8)
ln(0.5)
=k
3.8
k ≈ −0.182.
(e) From part (c) we know that the formula for Q is
Q = Ce−0.182t .
We are told that initially there are 10 mg of the drug in the body. Thus, at t = 0, we get
10 = Ce−0.182(0)
so
C = 10.
Thus, our equation becomes
Q(t) = 10e−0.182t .
Substituting t = 12, we get
Q(t) = 10e−0.182t
Q(12) = 10e−0.182(12)
= 10e−2.184
Q(12) ≈ 1.126 mg.
31. (a) The differential equation is
r
dB
=
· B.
dt
100
The constant of proportionality is
r
.
100
(b) To find the equilibrium, we solve
r
dB
=
·B =0
dt
100
to get B = 0.
If we have a solution with initial condition B > 0, then dB/dt > 0, and the solution increases and moves away
from the equilibrium. Thus, the equilibrium is unstable. In terms of a bank account, this means that an empty bank
account remains empty unless somebody deposits money into it, but a bank account with money in it accumulates
more money.
11.5 SOLUTIONS
1067
(c) Solving, we have
r dt
dB
=
B
Z
Z100
dB
r
=
dt
B
100
r
ln |B| =
t+C
100
B = ±e(r/100)t+C = Ae(r/100)t ,
A = ±eC .
A is the initial amount in the account, since A is the amount at time t = 0.
(d) See Figure 11.37.
B = 1000e0.15t
20,000
B = 1000e0.10t
10,000
B = 1000e0.04t
1000
15
30
t
Figure 11.37
Strengthen Your Understanding
32. When y = 2, we have dy/dx = 8 − 8x 6= 0. An equilibrium solution must have derivative equal to zero for all values of
x.
33. An equilibrium is a constant solution, so y = x2 cannot be an equilibrium solution.
34. Newton’s Law of Heating and Cooling says that the rate of change of the temperature of an object is proportional to the
difference between the temperature of the object and the temperature of the surrounding air. Once the roast is in the oven,
the temperature of the surrounding air is 350◦ F, so the correct differential equation is dH/dt = k(H − 350).
35. The differential equation is of the form dy/dt = ky, where k is negative. One possible answer is dy/dt = −2y.
36. Some possible answers are dQ/dt = Q − 500 or dQ/dx = 2(Q − 500). Other answers are possible. We must have Q
as the dependent variable and the derivative equal to zero when Q = 500.
37. Since the graph has an equilibrium solution at P = 25, the solution with initial condition P0 = 25 must be a horizontal
line. Since the equilibrium solution is unstable, it is likely that the solutions with P0 = 20 and P0 = 30 bend away from
the horizontal line. See Figure 11.38. Other answers are possible.
P
25
t
Figure 11.38
1068
Chapter Eleven /SOLUTIONS
Solutions for Section 11.6
Exercises
1. (a) (III) An island can only sustain the population up to a certain size. The population will grow until it reaches this
limiting value.
(b) (V) The ingot will get hot and then cool off, so the temperature will increase and then decrease.
(c) (I) The speed of the car is constant, and then decreases linearly when the breaks are applied uniformly.
(d) (II) Carbon-14 decays exponentially.
(e) (IV) Tree pollen is seasonal, and therefore cyclical.
2. The balance is increasing at a rate of 0.05 times the current balance and is decreasing at a rate of 12,000 dollars per year.
The differential equation is
dB
= 0.05B − 12,000.
dt
3. The balance is increasing at a rate of 0.037 times the current balance and is also increasing at a rate of 5000 dollars per
year. The differential equation is
dB
= 0.037B + 5000.
dt
4. The balance is decreasing at a rate of 0.08 times the current balance and is increasing at a rate of 2000 dollars per year.
The differential equation is
dB
= −0.08B + 2000.
dt
5. The balance is decreasing at a rate of 0.065 times the current balance and is also decreasing at a rate of 50,000 dollars per
year. The differential equation is
dB
= −0.065B − 50,000.
dt
6. (a) If B = f (t), where t is in years,
dB
= Rate of money earned from interest + Rate of money deposited
dt
dB
= 0.10B + 1000.
dt
(b) We use separation of variables to solve the differential equation
dB
= 0.1B + 1000.
dt
Z
1
dB =
0.1B + 1000
Z
dt
1
ln |0.1B + 1000| = t + C1
0.1
0.1B + 1000 = C2 e0.1t
B = Ce0.1t − 10,000
For t = 0, B = 0, hence C = 10,000. Therefore, B = 10,000e0.1t − 10,000.
7. (a) By Newton’s Law of Cooling, we have
dH
= k(H − 50)
dt
for some k. Furthermore, we know the juice’s original temperature H(0) = 90.
11.6 SOLUTIONS
1069
(b) Separating variables, we get
Z
We then integrate:
dH
=
(H − 50)
Z
k dt.
ln |H − 50| = kt + C
H − 50 = ekt · A
H = 50 + Aekt .
Thus, H(0) = 90 gives A = 40, and H(5) = 80 gives
50 + 40e5k = 80
30
e5k =
40
5k = ln(0.75)
1
k = ln(0.75) ≈ −0.05754.
5
Therefore
H(t) = 50 + 40e−0.05754t .
(c) We now solve for t at which H(t) = 60:
60 = 50 + 40e−0.05754t
1
= e−0.05754t
4
ln(0.25) = −0.05754t
t = 24 minutes.
8. Since mg is constant and a = dv/dt, differentiating ma = mg − kv gives
m
dv
da
= −k
= −ma.
dt
dt
Thus, the differential equation is
k
da
= − a.
dt
m
Solving for a gives
a = a0 e−kt/m .
At t = 0, we have a = g, the acceleration due to gravity. Thus, a0 = g, so
a = ge−kt/m .
Problems
9. (a) There are two factors that are affecting B: the money leaving the account, which is at a constant rate of −2000 per
year, and the interest accumulating in it, which accrues at a rate of (0.08)B. Since
Rate of change of balance = Rate in − Rate out,
the differential equation for B is
dB
= 0.08B − 2000.
dt
(b) To find the equilibrium solution, we solve dB/dt = 0 to get 0.08B−2000 = 0, or B = 25000. To determine whether
the equilibrium is stable, we consider what happens to a solution with initial condition near the equilibrium. If the
initial condition is B = 26000, then initially dB/dt = 0.08(26000) − 2000 > 0. Thus, the interest accumulated by
the account exceeds the amount being spent, and the balance increases. If a solution has initial condition B = 24000,
then dB/dt = 0.08(24000) − 2000 < 0. Thus, more money is being spent than the interest can replenish, so the
balance decreases. The equilibrium is unstable.
1070
Chapter Eleven /SOLUTIONS
(c) We solve the differential equation by separating variables and then integrating:
Z
Z
dB
=
dt
0.08B − 2000
12.5 ln |0.08B − 2000| = t + C
t
+C
ln |0.08B − 2000| =
12.5
0.08B − 2000 = ±e0.08t+C
B = 25,000 + Ae0.08t .
(d)
(i) If the initial deposit is 20,000, then we have B = 20,000 when t = 0, which leads to A = −5000. Knowing
A, we can find B(5) as:
B(5) = 25,000 − 5000e0.08(5) = $17,540.88.
(ii) Now B = 30,000 when t = 0 leads to A = 5000, giving B(5) = $32,459.12.
10. Let D(t) be the quantity of dead leaves, in grams per square centimeter. Then
factor out −0.75 and then separate variables.
dD
dt
= 3 − 0.75D, where t is in years. We
dD
= −0.75(D − 4)
dt
Z
Z
dD
−0.75 dt
=
D−4
ln |D − 4| = −0.75t + C
|D − 4| = e−0.75t+C = e−0.75t eC
D = 4 + Ae−0.75t , where A = ±eC .
If initially the ground is clear, the solution looks like the following graph:
D
4
t
The equilibrium level is 4 grams per square centimeter, regardless of the initial condition.
11. (a) Letting P represent the quantity of pollutant in the lake, in metric tons, in year t, we have
dP
= −0.16P + 8.
dt
Substituting P0 = 45 and P0 = 55, we get
dP
dt
P =45
= −0.16(45) + 8 = 0.8
and
dP
dt
P =55
= −0.16(55) + 8 = −0.8.
Since the derivative is positive for P0 = 45, the quantity of pollutant is increasing then. For P0 = 55, the derivative
is negative, so the quantity of pollutant is decreasing then.
(b) Writing the differential equation as
dP
= −0.16(P − 50),
dt
we use separation of variables to see that the general solution is
P = 50 + Ce−0.16t .
In all cases, the quantity of pollutant levels off at 50 metric tons after a long time.
11.6 SOLUTIONS
1071
12. Caffeine is leaving the body at a rate of 17% per hour and is entering the body at a rate of 130 mg per hour, so the
differential equation is
dA
= −0.17A + 130.
dt
Writing the differential equation as
dA
= −0.17(A − 764.706),
dt
we use separation of variables to see that the general solution is A = 764.706 + Ce−0.17t . Since the initial condition is
A0 = 0, we have C = −764.706 so the particular solution is
A = 764.706 − 764.706e−0.17t .
We substitute t = 10 to find the amount of caffeine at 5 pm:
A = 764.706 − 764.706e−0.17(10) = 625.007.
At 5 pm, the person has about 625 mg of caffeine in the body.
13. (a) If I is intensity and l is the distance traveled through the water, then for some k > 0,
dI
= −kI.
dl
(The proportionality constant is negative because intensity decreases with distance). Thus I = Ae−kl . Since I = A
when l = 0, A represents the initial intensity of the light.
(b) If 50% of the light is absorbed in 10 feet, then 0.50A = Ae−10k , so e−10k = 21 , giving
k=
− ln 12
ln 2
=
.
10
10
In 20 feet, the percentage of light left is
e−
so
3
4
ln 2 ·20
10
= e−2 ln 2 = (eln 2 )−2 = 2−2 =
1
,
4
or 75% of the light has been absorbed. Similarly, after 25 feet,
e−
ln 2 ·25
10
5
5
= e−2.5 ln 2 = (eln 2 )− 2 = 2− 2 ≈ 0.177.
Approximately 17.7% of the light is left, so 82.3% of the light has been absorbed.
14. (a) The differential equation for the population has the form
Rate of change = Birth rate − Death rate
The birth rate is 140 m/yr = 0.14 bn/yr. Since we assume the death rate is increasing linearly from 57 m/year to
80 m/year over 30 years, we have
Rate of change of death rate =
80 − 57
= 0.767(million/year)/year.
30
Then at time t, the death rate is given by
Death rate = 57 + 0.767t million/year = 0.057 + 0.000767t billion/year.
The differential equation is
dP
= 0.14 − (0.057 + 0.000767t)
dt
dP
= 0.083 − 0.000767t.
dt
(b) The differential equation can be solved by direct integration:
P = 0.083t − 0.000767
t2
+ P0 .
2
Since P0 = 6.9 billion is the population at t = 0, we have
P = 0.083t − 0.000767
t2
+ 6.9.
2
1072
Chapter Eleven /SOLUTIONS
(c) In 2050, we have t = 40, so
P = 0.083(40) − 0.000767
15. (a) If P = pressure and h = height,
dP
dh
402
2
+ 6.9 = 9.61 bn.
−5
= −3.7 × 10−5 P, so P = P0 e−3.7×10
at sea level (when h = 0) is 29.92, so P = 29.92e
−5
−3.7×10
h
h
. Now P0 = 29.92, since pressure
. At the top of Mt. Whitney, the pressure is
−5
(14500)
≈ 17.50 inches of mercury.
−5
(29000)
≈ 10.23 inches of mercury.
P = 29.92e−3.7×10
At the top of Mt. Everest, the pressure is
P = 29.92e−3.7×10
(b) The pressure is 15 inches of mercury when
−5
15 = 29.92e−3.7×10
Solving for h gives h =
−1
3.7×10−5
h
15
ln( 29.92
) ≈ 18,661.5 feet.
16. (a) Since the rate of change of the weight is given by
1
dW
=
(Intake − Amount to maintain weight)
dt
3500
we have
dW
1
=
(I − 20W ).
dt
3500
(b) To find the equilibrium, we solve dW/dt = 0, or
1
(I − 20W ) = 0.
3500
Solving for W , we get W = I/20.
This means that if an athletic adult male weighing I/20 pounds has a constant caloric intake of I calories per
day, his weight remains constant. We expect the equilibrium to be stable because an athletic adult male slightly over
the equilibrium weight loses weight because his caloric intake is too low to maintain the higher weight. Similarly, an
adult male slightly under the equilibrium weight gains weight because his caloric intake is higher than required to
maintain his weight.
(c) Factoring the right side of the differential equation
1
1
I
dW
=
(I − 20W ) = −
W−
,
dt
3500
175
20
we separate variables and integrate:
Thus we have
Z
dW
=−
W − I/20
ln W −
so that
W−
Z
1
dt.
175
I
1
=−
t+C
20
175
I
= Ae−(1/175)t
20
or in other words
I
+ Ae−(1/175)t .
20
Let us call the person’s initial weight W0 at t = 0. Then W0 = I/20 + Ce0 , so C = W0 − I/20. Thus
W =
W =
I
I
+ W0 −
e−(1/175)t .
20
20
11.6 SOLUTIONS
1073
(d) Using part (c), we have W = 150 + 10e−(1/175)t . This means that W → 150 as t → ∞. See Figure 11.39.
W
160
150
t
100 days
Figure 11.39
17. (a) We know that the rate at which morphine leaves the body is proportional to the amount of morphine in the body at
that particular instant. If we let Q be the amount of morphine in the body, we get that
Rate of morphine leaving the body = kQ,
where k is the rate of proportionality. The solution is Q = Q0 ekt (neglecting the continuously incoming morphine).
Since the half-life is 2 hours, we have
1
Q0 = Q0 ek·2 ,
2
so
ln(1/2)
k=
= −0.347.
2
(b) Since
Rate of change of quantity = Rate in − Rate out,
we have
dQ
= −0.347Q + 2.5.
dt
(c) Equilibrium occurs when dQ/dt = 0, that is, when 0.347Q = 2.5 or Q = 7.2 mg.
√
dy
18. Let the depth of the water at time t be y. Then
= −k y, where k is a positive constant. Separating variables,
dt
so
Z
dy
√ =−
y
Z
k dt,
√
2 y = −kt + C .
√
When t = 0, y = 36; 2√36 = −k · 0 + C, so C = 12.
When t = 1, y = 35; 2 35 = −k + 12, so k ≈ 0.17.
√
Thus, 2 y ≈ −0.17t + 12. We are looking for t such that y = 0; this happens when t ≈
days.
12
0.17
≈ 71 hours, or about 3
19. We are given that the rate of change of pressure with respect to volume, dP/dV is proportional to P/V , so that
P
dP
=k .
dV
V
Using separation of variables and integrating gives
Z
dP
=k
P
Z
dV
.
V
Evaluating these integral gives
ln P = k ln V + c
or equivalently,
P = AV k .
1074
Chapter Eleven /SOLUTIONS
20. (a) If A is surface area, we know that for some constant K
dV
= −KA.
dt
If r is the radius of the sphere, V = 4πr 3 /3 and A = 4πr 2 . Solving for r in terms of V gives r = (3V /4π)1/3 , so
dV
= −K(4πr 2 ) = −K4π
dt
where k is another constant, k = K(4π)1/3 32/3 .
(b) Separating variables gives
Z
1/3
Since V = V0 when t = 0, we have 3V0
3V
4π
2/3
dV
=−
V 2/3
Z
so
k dt
3V 1/3 = −kt + C.
= C, so
1/3
3V 1/3 = −kt + 3V0
Solving for V gives
k
1/3
V = − t + V0
3
This function is graphed in Figure 11.40.
3
.
.
V
V0
1/3
3V0
Figure 11.40
(c) The snowball disappears when V = 0, that is when
k
1/3
− t + V0 = 0
3
giving
1/3
t=
3V0
k
.
21. Let V (t) be the volume of water in the tank at time t, then
√
dV
=k V
dt
This is a separable equation which has the solution
V (t) = (
kt
+ C)2
2
Since V (0) = 200 this gives 200 = C 2 so
V (t) = (
kt √
+ 200)2 .
2
t
/k
dV
= −kV 2/3
dt
11.6 SOLUTIONS
However, V (1) = 180 therefore
180 = (
so that k = 2
√
180 −
√
1075
k √
+ 200)2 ,
2
200 = −1.45146. Therefore,
V (t) = (−0.726t +
√
200)2 .
The tank will be half-empty when V (t) = 100, so we solve
100 = (−0.726t +
√
200)2
to obtain t = 5.7 days. The tank will be half empty in 5.7 days.
The volume after 4 days is V (4) which is approximately 126.32 liters.
dy
= −k(y − a), where k > 0 and a are constants.
22. (a) We have
dt
(b) Separating variables and integrating we have
Z
so
dy
=
y−a
Z
−k dt,
ln |y − a| = ln(y − a) = −kt + C.
Thus,
y − a = Ae−kt
where A = eC .
Initially nothing has been forgotten, so y(0) = 1. Therefore, 1 − a = Ae0 = A, so y − a = (1 − a)e−kt or
y = (1 − a)e−kt + a.
(c) As t → ∞, e−kt → 0, so y → a.
Thus, a represents the fraction of material which is remembered in the long run. The constant k tells us about the rate
at which material is forgotten.
23. (a) We have
dp
= −k(p − p∗ ),
dt
where k is constant. Notice that k > 0, since if p > p∗ then dp/dt should be negative, and if p < p∗ then dp/dt
should be positive.
(b) Separating variables, we have
Z
Z
dp
−k dt.
=
p − p∗
Solving, we find p = p∗ + (p0 − p∗ )e−kt , where p0 is the initial price.
(c) See Figure 11.41.
p
p0 > p∗
p∗
p0 < p∗
t
Figure 11.41
(d) As t → ∞, p → p∗ . We see this in the solution in part (b), since as t → ∞, e−kt → 0. In other words, as t → ∞, p
approaches the equilibrium price p∗ .
1076
Chapter Eleven /SOLUTIONS
24. We have
The rate that people
first go see the movie
{z
|
dN/ dt
=k×
}
like to see it but haven’t yet
!
{z
}
The number of people who would
|
L−N
dN
= k(L − N )
Z
dt
dN
= − k dt
N −L
ln |N − L| = −kt + C
Z
separation of variables
|N − L| = Be−kt
N − L = Ae
B = eC
−kt
A = −B
N = L + Ae−kt .
Presumably no one sees the movie before it is released, so N = 0 on day t = 0, and we have
0 = L + Ae−k·0
A = −L
so N = L − Le−kt .
25. (a)
dQ
r
= r − αQ = −α(Q − )
dt
α
Z
Z
dQ
dt
= = −α
Q − r/α
r
= −αt + C
α
r
Q − = Ae−αt
α
ln Q −
When t = 0, Q = 0, so A = − αr and
Q=
r
(1 − e−αt )
α
So,
Q∞ = lim Q =
t→∞
r
.
α
Q
r
α
Q=
r
(1
α
− e−αt )
t
(b) Doubling r doubles Q∞ . Since Q∞ = r/α, the time to reach 21 Q∞ is obtained by solving
r
r
= (1 − e−αt )
2α
α
1
= 1 − e−αt
2
1
e−αt =
2
ln(1/2)
ln 2
t=−
=
.
α
α
So altering r does not alter the time it takes to reach 12 Q∞ . See Figure 11.42.
11.6 SOLUTIONS
Q
2r
α
2r
(1
α
Q=
− e−αt )
r
α
Q=
r
(1
α
− e−αt )
t
ln 2
α
Figure 11.42
(c) Q∞ is halved by doubling α, and so is the time, t = lnα2 , to reach 12 Q∞ .
Quantity in room
.
26. (a) Concentration of carbon monoxide =
Volume
If Q(t) represents the quantity of carbon monoxide in the room at time t, c(t) = Q(t)/60.
Rate quantity of
carbon monoxide in room = rate in − rate out
changes
Now
Rate in = 5%(0.002m3 /min) = 0.05(0.002) = 0.0001m3 /min.
Since smoky air is leaving at 0.002m3 /min, containing a concentration c(t) = Q(t)/60 of carbon monoxide
Rate out = 0.002
Q(t)
60
Thus
dQ
0.002
= 0.0001 −
Q
dt
60
Since c = Q/60, we can substitute Q = 60c, giving
d(60c)
0.002
= 0.0001 −
(60c)
dt
60
0.0001
0.002
dc
=
−
c
dt
60
60
(b) Factoring the right side of the differential equation and separating gives
0.0001
dc
=−
(c − 0.05) ≈ 3 × 10−5 (c − 0.05)
dt
Z
Z 3
dc
= − 3 × 10−5 dt
c − 0.05
ln |c − 0.05| = −3 × 10−5 t + K
−5
c − 0.05 = Ae−3×10
t
whereA = ±eK .
Since c = 0 when t = 0, we have A = −0.05, so
−5
c = 0.05 − 0.05e−3×10
−5
t
(c) As t → ∞, e−3×10 t → 0 so c → 0.05.
Thus in the long run, the concentration of carbon monoxide tends to 5%, the concentration of the incoming air.
1077
1078
Chapter Eleven /SOLUTIONS
−5
27. We have c = 0.05 − 0.05e−3×10
t
. We want to solve for t when c = 0.0002:
−5
0.0002 = 0.05 − 0.05e−3×10
−5
e
−0.0498 = −0.05e−3×10
t
t
−3×10−5 t
= 0.996
− ln(0.996)
t=
= 133.601 min.
3 × 10−5
28. (a) Now
dS
= (Rate at which salt enters the pool) − (Rate at which salt leaves the pool),
dt
and, for example,
Rate at which salt
enters the pool
!
Concentration of
=
salt solution
!
×
Flow rate of
salt solution
!
(grams/minute) = (grams/liter) × (liters/minute)
so
Rate at which salt enters the pool =
(10 grams/liter) × (60 liters/minute) = (600 grams/minute)
The rate at which salt leaves the pool depends on the concentration of salt in the pool. At time t, the concentration is
S(t)
, where S(t) is measured in grams.
2×106 liters
Thus
Rate at which salt leaves the pool =
3S(t) grams
S(t) grams
60 liters
×
=
.
2 × 106 liters
minute
105 minutes
Thus
(b)
3S
dS
= 600 −
.
dt
100,000
dS
3
= − 100,000
(S − 20,000,000)
R
3
dS
= − 100,000
dt
S−20,000,000
3
t+C
ln |S − 20,000,000| = − 100,000
Rdt
3
S = 20,000,000 − Ae− 100,000 t
3
Since S = 0 at t = 0, A = 20,000,000. Thus S(t) = 20,000,000 − 20,000,000e− 100,000 t .
−
3
t
(c) As t → ∞, e 100,000 → 0, so S(t) → 20,000,000 grams. The concentration approaches 10 grams/liter. Note that
this makes sense; we’d expect the concentration of salt in the pool to become closer and closer to the concentration
of salt being poured into the pool as t → ∞.
29. We are given that
BC = 2OC.
If the point A has coordinates (x, y) then OC = x and AC = y. The slope of the tangent line, y ′ , is given by
y′ =
y
AC
=
,
BC
BC
so
BC =
Substitution into BC = 2OC gives
so
y
.
y′
y
= 2x,
y′
y′
1
=
.
y
2x
11.6 SOLUTIONS
1079
Separating variables to integrate this differential equation gives
Z
dy
=
y
Z
dx
2x
p
1
ln |x| + C = ln |x| + ln A
2
p
|y| = A |x|
√
y = ±(A x).
√
Thus, in the first quadrant, the curve has equation y = A x.
ln |y| =
30. (a) Newton’s Law of Motion says that
Force = (mass) × (acceleration).
Since acceleration, dv/dt, is measured upward and the force due to gravity acts downward,
−
so
(b) Since v =
dv
mgR2
=m
(R + h)2
dt
dv
gR2
=−
.
dt
(R + h)2
dh
,
dt
the chain rule gives
dv dh
dv
dv
=
·
=
· v.
dt
dh dt
dh
Substituting into the differential equation in part (a) gives
v
gR2
dv
=−
.
dh
(R + h)2
(c) Separating variables gives
Z
Since v = v0 when h = 0,
v dv = −
Z
gR2
dh
(R + h)2
gR2
v2
=
+C
2
(R + h)
v0 2
gR2
=
+C
2
(R + 0)
gives
C=
v0 2
− gR,
2
so the solution is
gR2
v0 2
v2
=
+
− gR
2
(R + h)
2
v 2 = v0 2 +
2gR2
− 2gR
(R + h)
(d) The escape velocity v0 ensures that v 2 ≥ 0 for all h ≥ 0. Since the positive quantity
ensure that v 2 ≥ 0 for all h, we must have
When v0 2 = 2gR so v0 =
√
2gR2
→ 0 as h → ∞, to
(R + h)
v0 2 ≥ 2gR.
2gR, we say that v0 is the escape velocity.
31. (a) For this situation,
Rate money added
to account
!
=
Rate money added
via interest
Translating this into an equation yields
dB
= 0.05B + 1200.
dt
!
+
Rate money
deposited
!
.
1080
Chapter Eleven /SOLUTIONS
(b) Solving this equation via separation of variables gives
dB
= 0.05B + 1200
dt
= 0.05(B + 24000).
So
Z
and
dB
=
B + 24000
Z
0.05 dt,
ln |B + 24000| = 0.05t + C.
Solving for B gives
|B + 24000| = e0.05t+C = eC e0.05t ,
so
B = Ae0.05t − 24000, (where A = ec ).
We may find A using the initial condition B0 = f (0) = 0
A − 24000 = 0 or
A = 24000.
The solution is
B = 24,000e0.05t − 24,000.
(c) After 5 years, the balance is
B = f (5) = 24,000(e0.05(5) − 1) = 6816.61 dollars.
32. (a) The balance in the account at the beginning of the month is given by the following sum
balance in
account
=
previous month’s
balance
+
interest on
previous month’s balance
+
monthly deposit
of $100
Denote month i’s balance by Bi . Assuming the interest is compounded continuously, we have
previous month’s
balance
!
+
interest on previous
month’s balance
Since the interest rate is 10% = 0.1 per year, interest is
0.1
12
0.1
Explicitly, we have for the five years (60 months) the equations:
B0 = 0
0.1
B1 = B0 e 12 + 100
0.1
B2 = B1 e 12 + 100
0.1
B3 = B2 e 12 + 100
.. ..
. .
0.1
B60 = B59 e 12 + 100
In other words,
B1 = 100
0.1
B2 = 100e 12 + 100
0.1
B3 = (100e 12 + 100)e 12 + 100
= 100e
(0.1)2
12
0.1
= Bi−1 e0.1/12 .
per month. So at month i, the balance is
Bi = Bi−1 e 12 + 100
0.1
!
+ 100e 12 + 100
11.7 SOLUTIONS
B4 = 100e
.. ..
. .
B60 = 100e
B60 =
59
X
(0.1)3
12
(0.1)59
12
100e
+ 100e
(0.1)2
12
+ 100e
+ 100e
(0.1)58
12
(0.1)
12
1081
+ 100
+ · · · + 100e
(0.1)1
12
+ 100
(0.1)k
12
k=0
(b) The sum B60 =
Z
59
X
100e
(0.1)k
12
can be written as B60 =
k=0
5
59
X
1200e
(0.1)k
12
(
k=0
1
) which is the left Riemann sum for
12
1
1200e0.1t dt, with ∆t =
and N = 60. Evaluating the sum on a calculator gives B60 = 7752.26.
12
0
(c) The situation described by this problem is almost the same as that in Problem 31, except that here the money is being
deposited once a month rather than continuously; however the nominal yearly rates are the same. Thus we would
expect the balance after 5 years to be approximately the same in each case. This means that the answer to part (b)
of this problem should be approximately the same as the answer to part (c) to Problem 31. Since the deposits in this
problem start at the end of the first month, as opposed to right away, we would expect the balance after 5 years to be
slightly smaller than in Problem 31, as is the case.
Alternatively, we can use the Fundamental Theorem of Calculus to show that the integral can be computed
exactly
Z
5
0
1200e0.1t dt = 12000(e(0.1)5 − 1) = 7784.66
R5
Thus 0 1200e0.1t dt represents the exact solution to Problem 31. Since 1200e0.1t is an increasing function, the left
hand sum we calculated in part (b) of this problem underestimates the integral. Thus the answer to part (b) of this
problem should be less than the answer to part (c) of Problem 31.
Strengthen Your Understanding
33. When we substitute B = 5000 into the differential equation, we see that dB/dt = 0.08(5000) − 250 = 150 > 0. Since
the rate of change is positive, the balance is increasing.
34. The units do not match up. If time t is measured in days, we need to convert 25 mg per hour to the equivalent rate of 600
mg per day. If time t is measured in hours, we need to convert the rate 15% per day into the equivalent rate of 0.625% per
hour.
35. We use Q for the quantity of drug in the body at time t. The rate in is constant (such as 50) and the rate out is proportional
(such as 0.08Q). One possible answer is dQ/dt = 50 − 0.08Q. Other answers are possible.
36. We use Q for
the size of the quantity at time t. The rate due to its growth on its own is proportional to the cube root
√
3
(such
as
0.2
Q)
and the rate due to the external contribution is constant (such as 100). One possible answer is dQ/dt =
√
0.2 3 Q + 100.
37. We use Q for the size of the quantity at time t. Since the rate of growth dQ/dt goes down as Q goes up, one possible
model is that the rate of growth is inversely proportional to Q. We have dQ/dt = k/Q with k positive. One example is
0.5
dQ
=
.
dt
Q
Other answers are possible.
Solutions for Section 11.7
Exercises
1. (a) P =
dP
dt
1
1+e−t
= (1 + e−t )−1
= −(1 + e−t )−2 (−e−t ) =
1
1+e−t
−t
Then P (1 − P ) =
(b) As t tends to ∞, e
1−
e−t
.
(1+e−t )2
1
1+e−t
goes to 0. Thus
1
1+e−t
1
lim
−t =
t→∞ 1+e
=
1.
e−t
1+e−t
=
e−t
(1+e−t )2
=
dP
dt
.
1082
Chapter Eleven /SOLUTIONS
2. We know that there is an equilibrium solution at the carrying capacity P = 100 and that P increases toward 100 if
0 < P < 100 and P decreases toward 100 if P > 100. Furthermore, we know that there is an inflection point at a height
of P = 100/2 = 50 and that the graph is concave up for 0 < P < 50 and is concave down for 50 < P < 100. See
Figure 11.43.
P
(c)
100
(b)
50
(a)
t
Figure 11.43
3. We know that there is an equilibrium solution at the carrying capacity Q = 1/0.0004 = 2500 and that Q increases toward
2500 if 0 < Q < 2500 and Q decreases toward 2500 if Q > 2500. Furthermore, we know that there is an inflection
point at a height of Q = 2500/2 = 1250 and that the graph is concave up for 0 < Q < 1250 and is concave down for
1250 < Q < 2500. See Figure 11.44.
y
(c)
2500
(b)
1250
(a)
x
Figure 11.44
4. Since dP/dt is a quadratic function of P with negative leading coefficient, the graph of dP/dt against P is a parabola
opening down. Since dP/dt = 0 when P = 0 and when P = 250, the horizontal intercepts of the parabola are at 0 and
250. See Figure 11.45. Since we don’t know the value of k, we can’t put a scale on the vertical axis.
dP/dt
250
P
Figure 11.45
5. Since dA/dt is a quadratic function of A with negative leading coefficient, the graph of dA/dt against A is a parabola
opening down. Since dA/dt = 0 when A = 0 and when A = 1/0.0002 = 5000, the horizontal intercepts of the parabola
are at 0 and 5000. See Figure 11.46. Since we don’t know the value of k, we can’t put a scale on the vertical axis.
dA/dt
5000
Figure 11.46
A
1083
11.7 SOLUTIONS
6. We see in Figure 11.69 that dP/dt = 0 when P = 0 and when P = 45. Thus, there are equilibrium solutions at P = 0
and at P = 45. If 0 < P < 45, the derivative dP/dt is positive so P is an increasing function of t. If P < 0 or P > 45,
the derivative dP/dt is negative so P is a decreasing function of t. See Figure 11.47.
P
45
t
Figure 11.47
7. We see in Figure 11.70 that there appear to be equilibrium solutions at Q = 0 and Q = 800. Thus, the graph of dQ/dt
against Q will have horizontal intercepts at Q = 0 and Q = 800. Since Q is growing logistically, the graph of dQ/dt
against Q is a parabola opening down. See Figure 11.48.
dQ/dt
Q
800
Figure 11.48
8. (a) See Figure 11.49.
(b) The value P = 1 is a stable equilibrium. (See part (d) below for a more detailed discussion.)
(c) Looking at the solution curves, we see that P is increasing for 0 < P < 1 and decreasing for P > 1. The values of
P = 0, P = 1 are equilibria. In the long run, P tends to 1, unless you start with P = 0. The solution curves with
initial populations of less than P = 1/2 have inflection points at P = 1/2. (This will be demonstrated algebraically
in part (d) below.) At the inflection point, the population is growing fastest.
(d) See Figure 11.50.
Since dP/dt = 3P − 3P 2 = 3P (1 − P ), the graph of dP/dt against P is a parabola, opening downward
with P intercepts at 0 and 1. The quantity dP/dt is positive for 0 < P < 1, negative for P > 1 (and P < 0). The
quantity dP/dt is 0 at P = 0 and P = 1, and maximum at P = 1/2. The fact that dP/dt = 0 at P = 0 and P = 1
tells us that these are equilibria. Further, since dP/dt > 0 for 0 < P < 1, solution curves starting here increase
toward P = 1.
If the population starts at a value P < 1/2, it increases at an increasing rate up to P = 1/2. After this, P
continues to increase, but at a decreasing rate. The fact that dP/dt has a maximum at P = 1/2 tells us that there is a
point of inflection when P = 1/2. Similarly, since dP/dt < 0 for P > 1, solution curves starting with P > 1 will
decrease to P = 1. Thus, P = 1 is a stable equilibrium.
P
dP
dt
1
0.75
1
1
Figure 11.49
2
t
Figure 11.50
P
1084
Chapter Eleven /SOLUTIONS
9. (a) Equilibrium values are values where dP/dt = 0, so the equilibrium values are at P = 0 and P = 2000.
(b) At P = 500, we see that dP/dt is positive so P is increasing.
10. (a) Equilibrium values are values where dP/dt = 0, so the equilibrium values are at P = 0 and P = 400.
(b) At P = 500, we see that dP/dt is negative so P is decreasing.
11. (a) We see that k = 0.035 which tells us that the quantity P grows by about 3.5% per unit time when P is very small
relative to L. We also see that L = 6000 which tells us the upper limit on the value of P if P is initially below 6000.
(b) The largest rate of change occurs when P = L/2 = 3000.
12. (a) We factor out 0.1P on the right hand side to obtain dP/dt = 0.1P (1 − 0.0008P ). We see that k = 0.1 which
tells us that the quantity P grows by about 10% per unit time when P is very small relative to L. We also see that
L = 1/0.0008 = 1250 which tells us the upper limit on the value of P if P is initially below 1250.
(b) The largest rate of change occurs when P = L/2 = 625.
13. We see from the differential equation that k = 0.05 and L = 2800, so the general solution is
P =
2800
.
1 + Ae−0.05t
14. We see from the differential equation that k = 0.012 and L = 5700, so the general solution is
P =
5700
.
1 + Ae−0.012t
15. We see from the differential equation that k = 0.68 and L = 1/0.00025 = 4000, so the general solution is
P =
4000
.
1 + Ae−0.68t
16. We factor out 0.2P to obtain dP/dt = 0.2P (1 − 0.004P ). We see that k = 0.2 and L = 1/0.004 = 250, so the general
solution is
250
.
P =
1 + Ae−0.2t
17. We rewrite
P
,
2
so k = 10 and L = 2. Since P0 = L/4, we have A = (L − P0 )/P0 = 3. Thus
10P − 5P 2 = 10P 1 −
P =
2
.
1 + 3e−10t
The time to peak dP/dt is
t=
1
ln A = ln(3)/10.
k
18. We rewrite
0.02P − 0.0025P 2 = 0.02P
1−
P
8
so k = 0.02 and L = 8. Since P0 = 1, we have A = (8 − 1)/1 = 7. Thus
P =
8
.
1 + 7e−0.02t
The time of peak dP/dt is
t=
1
1
ln A =
ln 7 = 50 ln 7.
k
0.02
,
11.7 SOLUTIONS
1085
19. We can immediately read off k = 0.3, L = 100. Since P0 = 75, we have A = (L − P0 )/P0 = (100 − 75)/75 = 1/3.
Thus
100
.
P =
1 + (1/3)e−0.3t
The time to peak dP/dt is
1
t = ln A = − ln(3)/0.3.
k
Note that the time to peak dP/dt is negative since P0 > L/2.
20. Rewriting the differential equation:
1 dP
= 0.12 − 0.02P,
P dt
so k = 0.12 and −k/L = −0.02. Thus L = 0.12/0.02 = 6. Since A = (6 − 2)/2 = 2, we have
P =
6
.
1 + 2e−0.12t
The time to peak dP/dt is
ln 2
1
ln A =
.
k
0.12
t=
21. We see from the differential equation that k = 0.8 and L = 8500, so the general solution is
8500
.
1 + Ae−0.8t
We substitute t = 0 and P = 500 to solve for the constant A:
8500
500 =
1 + Ae0
1+A
1
=
500
8500
17 = 1 + A
P =
A = 16.
The solution to this initial value problem is
P =
8500
.
1 + 16e−0.8t
22. We see from the differential equation that k = 0.04 and L = 1/0.0001 = 10,000, so the general solution is
10000
.
1 + Ae−0.04t
We substitute t = 0 and P = 200 to solve for the constant A:
10000
200 =
1 + Ae0
1+A
1
=
200
10000
50 = 1 + A
P =
A = 49.
The solution to this initial value problem is
P =
10,000
.
1 + 49e−0.04t
Problems
23. (a) Substituting the value t = 0 we get
N (0) =
400
400
=
= 1.
1 + 399(1)
1 + 399e−0.4(0)
The fact that N (0) = 1 tells us that at the moment the rumor begins spreading, there is only one person who knows
the content of the rumor.
1086
Chapter Eleven /SOLUTIONS
(b) Substituting t = 2 we get
N (2) =
400
400
=
= 2.
1 + 399(0.449)
1 + 399e−0.4(2)
Substituting in t = 10 we get
400
400
= 48.
=
1 + 7.308
1 + 399e−0.4(10)
(c) The graph of N (t) is shown in Figure 11.51.
N (10) =
400
N (t)
200
5
10
15
20
t
Figure 11.51
(d) We are asked to find the time t at which 200 people will have heard the rumor. We can use the formula in the text
1
1
ln 399 = 14.972.
t = ln A =
k
0.4
Alternatively we can solve the equation
400
.
200 =
1 + 399e−0.4t
We get
400
1 + 399e−0.4t =
=2
200
−0.4t
399e
=1
1
−0.4t
e
=
399
1
ln e−0.4t = ln
399
−0.4t = ln(1/399)
ln(1/399)
t=
= 14.972.
−0.4
Thus, after about 15 hours half the people have heard the rumor.
We are asked to solve for the time at which 399 people (that is, virtually everyone) will have heard the rumor.
We solve
400
399 = N (t) =
1 + 399e−0.4t
400
1 + 399e−0.4t =
399
400
−0.4t
−1
399e
=
399
400/399 − 1
400
1
e−0.4t =
−1 =
399 399
399
400/399 − 1
−0.4t
ln e
= ln
399
400/399 − 1
−0.4t = ln
399
ln((400/399 − 1)/399)
t=
= 29.945.
−0.4
Thus, after approximately 30 hours 399 people (virtually everyone) will have heard the rumor.
(e) The rumor is spreading fastest at L/2 = 400/2 = 200 or when 200 people have already heard the rumor, so after
about 15 hours.
11.7 SOLUTIONS
1087
24. (a) At t = 0, which corresponds to 1935, we have
P =
1
= 0.252
1 + 2.968e−0.0275(0)
showing that about 25% of the land was in use in 1935.
(b) This model predicts that as t gets very large, P will approach 1. That is, the model predicts that in the long run, all
the land will be used for farming.
(c) The time when half the land is in use is the time at which dP/dt is greatest. This occurs when
t=
1
1
ln A =
ln(2.968) = 39.6 years.
k
0.0275
Since t = 0 corresponds to 1935, t = 39.6 corresponds to 1935 + 39.6 = 1974.6. According to this model, the
Tojolobal were using half their land in 1974.
(d) The inflection point occurs when P = L/2 or at one-half the carrying capacity. In this case, P = 21 in 1974, as
shown in part (c).
25. (a) The equilibrium population occurs when dP /dt is zero. Solving
dP
= 1 − 0.0004P = 0
dt
gives P = 2500 fish as the equilibrium population.
(b) The solution of the differential equation is
P (t) =
2500
1 + Ae−0.25t
subject to P (−10) = 1000 if t = 0 represents the present time. So we have
1000 =
2500
(1 + Ae2.5 )
from which A = 0.123127 and
2500
≈ 2230.
(1 + 0.123127)
Therefore, the current population is approximately 2230 fish.
(c) The effect of losing 10% of the fish each year gives the revised differential equation
P (0) =
dP
= (0.25 − 0.0001P )P − 0.1P
dt
or
dP
= (0.15 − 0.0001P )P.
dt
The revised equilibrium population occurs where dP/dt = 0, or about 1500 fish.
26. (a) The maximum rate of change occurs at approximately t = 50 and the rate of change at that point is approximately
(270 − 195)/(50 − 40) = 7.5.
(b) The maximum rate of change occurs at the point where P = 270 so we estimate that the carrying capacity is
approximately 270 · 2 = 540.
27. (a) The limiting value of f (t) is about 36, so the total number of infected computers is about 36,000.
(b) The curve has an inflection point at about t ≈ 16 hours, and then n = f (16) ≈ 18.
(c) The virus was spreading fastest at about when t = 16, that is at 4 pm on July 19, 2001. At that time, about 18,000
computers were infected.
(d) At the inflection point, the number of computers infected is about half the number infected in the long run.
28. Let r be the relative growth rate, that is r = (1/N )dN/ dt. Since r is a linear function of N , its graph contains the points
(N, r) = (5, 15%) and (N, r) = (10, 14.5%), so its slope is
m=
0.145 − 0.15
−0.005
∆r
=
=
= −0.001.
∆N
10 − 5
5
Using the point (5, 0.15), we have
r = 0.15 − 0.001(N − 5)
r = 0.155 − 0.001N.
1088
Chapter Eleven /SOLUTIONS
This gives us the differential equation
1 dN
= 0.155 − 0.001N
N dt
dN
= N (0.155 − 0.001N ) .
dt
We see that dN/ dt = 0 where N = 0 or where
0.155 − 0.001N = 0
0.001N = 0.155
N = 155.
Our model predicts the spread of pigweed will halt when 155 million acres are afflicted.
29. (a) Using the approximation
P (t + 1) − P (t)
dP
≈
,
dt
1
we see that dP/dt and annual production are approximately equal.
For 1993 we have dP/dt = 22.0 billion barrels per year.
For 2008 we have dP/dt = 26.9 billion barrels per year.
(b) For 1993 we have
22.0
1 dP
=
= 0.0304/year = 3.04% per year.
P dt
724
For 2008 we have
26.9
1 dP
=
= 0.0245/year = 2.45% per year.
P dt
1100
(c) We are finding the equation of the line that contains the two points (724, 0.0304) and (1100, 0.0245). The slope
between these points is
0.0245 − 0.0304
Slope =
= −0.0000157.
1100 − 724
Thus, the fitted line has an equation of the form
1 dP
= k − 0.0000157P.
P dt
Using the point P = 724, (1/P )dP/dt = 0.0304, we solve for the vertical intercept k:
0.0304 = k − 0.0000157(724)
k = 0.0304 + 0.0000157(724) = 0.0418.
Thus, the equation of the line is
1 dP
= −0.0000157P + 0.0418.
P dt
(d) The total quantity, L. of world oil reserves in 1859 is the value of P making dP/dt = 0, so it is the horizontal
intercept of the line we computed in (c). We solve
0 = −0.0000157P + 0.0418
for P to obtain P = 2662. This model estimates the total world oil reserves in 1859 to be L = 2662 billion barrels.
(e) Using t = 0 for 1993, we have the differential equation
dP
= 0.0418P
dt
The solution is
1−
P
.
2662
2662
,
1 + Ae−0.0418t
where A = (2662 − 724)/724 = 2.677. So, in billions of barrels, we have
P =
P =
2662
.
1 + 2.677e−0.0418t
11.7 SOLUTIONS
1089
30. (a) Using the formula on page 632 of the text for the time which gives the maximum value of dP/dt we have
1
1
ln A =
ln 2.677 = 23.6 years.
k
0.0418
So peak worldwide oil production is projected to occur about the year 1993 + 24 = 2017.
(b) Using the formula on page 632 of the text for the maximum value of (1/P )dP/dt with k = 0.0418 and L = 3500
we have
1 L − P0
1
3500 − 724
t = ln
=
ln
= 32.2 years.
k
P0
0.0418
724
So peak worldwide oil production is projected to occur about the year 1993 + 32 = 2025.
t=
31. (a) The differential equation is
dP
= 0.0418P
dt
The predicted values of dP/dt for 1998 and 2003 are
dP
dt
dP
dt
1998
2003
= 0.0418 · 841 1 −
= 0.0418 · 964 1 −
1−
841
2662
964
2662
P
.
2662
= 24.0 billion barrels per year
= 25.7 billion barrels per year.
Comparison with the actual annual production values, 24.4 bn barrel in 1998 and 25.3 bn barrel in 2003, we see the
model fits the data well. In each case, the prediction is about 0.4/25 ≈ 0.016 ≈ 2% from the the actual values. See
Figure 11.52.
dP/dt (bn barrel/yr)
30
1998
1993
20
2008
2003
10
500
1000
1500
P (bn barrel)
Figure 11.52
(b) The point (0, 724) lies exactly on the curve, since the data for 1993 was used to find the solution to the differential
equation. The predicted values are
2662
= 839.2 billion barrels
1 + 2.677e−0.0418·5
2662
= 963.6 billion barrels
In 2003, P =
1 + 2.677e−0.0418·10
2662
= 1095.5 billion barrels.
In 2008, P =
1 + 2.677e−0.0418·15
In 1998, P =
These values very close to the actual values of 841, 940, 1100 billion barrels—less than 1/2% difference. See Figure 11.53.
P (bn barrel)
2662
2000
1000
2003
1998
1993
1993
2008
2008
2043
Figure 11.53
t (year)
1090
Chapter Eleven /SOLUTIONS
32. (a) Since t = 0 is in 1993, for 2010, we substitute t = 17 into the logistic function
P =
2662
1 + 2.677e−0.0418t
so
P =
2662
= 1149.722 bn barrels.
1 + 2.677e−0.0418·17
(b) To estimate the quantity of oil produced during 2010, we can use a difference quotient or a derivative. We can find
P = 1177.098 billion barrels when t = 18 (in 2011) and use the approximation
dP
dt
2010
≈
P (t + 1) − P (t)
1
=
t=17
P (18) − P (17)
1177.098 − 1149.722
=
= 27.376 bn barrels.
1
1
The derivative gives a similar value:
dP
dt
= 27.302 bn barrels.
t=17
(c) Since the original reserves were estimated to be L = 2662 bn barrels, the oil projected to remain is 2662 −
1149.722 = 1512.278 billion barrels.
(d) The difference between the projection and the actual value is 27.376 − 26.9 = 0.476 billion barrels, or about half a
billion in 2010.
More precisely, the estimate is 0.476/26.9 = 0.018 ≈ 2% too high—quite close.
33. (a) Since t = 0 is in 1993, for 2020, we substitute t = 27 into the logistic function
P =
2662
1 + 2.677e−0.0418t
so
P =
2662
= 1426.603 bn barrels.
1 + 2.677e−0.0418·27
(b) When 300 billion barrels of oil remain, then 2662 − 300 = 2362 billion barrels of oil have been produced. We are
solving P (t) = 2362 for t:
2662
= 2362
1 + 2.677e−0.0418t
−0.0418t
2362 + 2362(2.677e
) = 2662
2362(2.677e−0.0418t ) = 300
e−0.0418t = 0.0474
−0.0418t = ln 0.0474
t = 72.9 years.
So, in about 1993 + 73 = 2066, only 300 billion barrels of oil are projected to remain in the ground.
34. (a) We know that a logistic curve can be modeled by the function
P =
L
1 + Ae−kt
where A = (L − P0 )/(P0 ) and P is the number of people infected by the virus at a particular time t. We know that
L is the limiting value, or the maximal number of people infected with the virus, so in our case
L = 5000.
We are also told that initially there are only ten people infected with the virus so that we get
P0 = 10.
Thus we have
L − P0
P0
5000 − 10
=
10
= 499.
A=
We are also told that in the early stages of the virus, infection grows exponentially with k = 1.78. Thus we get that
the logistic function for people infected is
5000
P =
.
1 + 499e−1.78t
1091
11.7 SOLUTIONS
(b) See Figure 11.54.
y
5000
2500
t
3.5
Figure 11.54
(c) Looking at the graph we see that the the point at which the rate changes from increasing to decreasing, the inflection
point, occurs at roughly t = 3.5 giving a value of P = 2500. Thus after roughly 2500 people have been infected, the
rate of infection starts dropping. See above.
35. (a) Let I be the number of informed people at time t, and I0 the number who know initially. Then this model predicts
that dI/dt = k(M − I) for some positive constant k. Solving this, we find the solution is
I = M − (M − I0 )e−kt .
We sketch the solution with I0 = 0. Notice that dI/dt is largest when I is smallest, so the information spreads fastest
in the beginning, at t = 0. In addition, Figure 11.55 shows that I → M as t → ∞, meaning that everyone gets the
information eventually.
(b) In this case, the model suggests that dI/dt = kI(M − I) for some positive constant k. This is a logistic model with
carrying capacity M . We sketch the solutions for three different values of I0 in Figure 11.56.
I
I
M
M
I0 = 0.75M
0.5M
t
Figure 11.55
I0 = 0.05M
I0 = 0
t
Figure 11.56
(i) If I0 = 0 then I = 0 for all t. In other words, if nobody knows something, it does not spread by word of mouth!
(ii) If I0 = 0.05M, then dI/dt is increasing up to I = M/2. Thus, the information is spreading fastest at I = M/2.
(iii) If I0 = 0.75M, then dI/dt is always decreasing for I > M/2, so dI/dt is largest when t = 0.
36. (a) Figure 11.57 shows that the yeast population seems to stabilize at about 13, so we take this to be the limiting value,
L.
P
dP
= kP (1 − ) for k, we get:
(b) Solving
dt
L
dP/dt
k=
.
P · (1 − P/L)
We now find dP /dt from the first two data points.
P (10) − P (0)
∆P
8.87 − 0.37
dP
≈
=
=
= 0.85.
dt
∆t
10 − 0
10
Putting in our values for dP/dt, L, and P (10), we get:
k≈
dP/dt
0.85
=
= 0.30.
P (10) · (1 − P (10)/L)
(8.87)(1 − 8.87/13)
1092
Chapter Eleven /SOLUTIONS
(c) For k = 0.3 and L = 13,
A=
L − P0
13 − 0.37
=
= 34.1.
P0
0.37
Putting this into the equation for P we get:
P =
13
13
=
,
1 + Ae−kt
1 + 34.1e−0.3t
which is plotted in Figure 11.58.
population
population
14
12
10
8
6
4
2
14
12
10
8
6
4
2
time (hours)
time (hours)
5 10 15 20 25 30 35 40 45 50
5 10 15 20 25 30 35 40 45 50
Figure 11.58: P = 13/ 1 + 34.1e−0.3t
Figure 11.57
37. (a) The population seems to level off around 5.8, which leads us to believe that the population is growing logistically. If
it were growing exponentially, we would expect the rate of increase to continue increasing with time.
(b) To find k, we solve the logistic equation, dP /dt = kP (1 − P/L), for k:
k=
dP/dt
.
P · (1 − P/L)
We now need to estimate dP/dt and L from the data. The population seems to level off at 5.8, so we take this as as
the carrying capacity, L. We use the first two data points to find dP/dt:
P (13) − P (0)
dP
∆P
1.7 − 1.00
≈
=
=
= 0.054
dt
∆t
13 − 0
13
Putting in our values for dP/dt, L, and P (13), we get:
k≈
0.054
dP/dt
=
= 0.045
P (13) · (1 − P (13)/L)
(1.7)(1 − 1.7/5.8)
(c) The solution curve for dP /dt = kP (1 − P/L) with k = 0.045, and L = 5.8 has equation
P =
Since P0 = 1 then A =
5.8−1
1
5.8
(1 + Ae−kt )
where
A=
L − P0
.
P0
= 4.8, we have
P =
5.8
1 + 4.8e−0.045t
The data and the curve are sketched in Figure 11.59.
population
6
5
4
3
2
1
time (hours)
10 20 30 40 50 60 70 80 90 100110120
Figure 11.59: P = 5.8/ 1 + 4.8e−0.045t
11.7 SOLUTIONS
1093
38. Let I be the number of infected people. Then, the number of healthy people in the population is M − I. The rate of
infection is
0.01
(M − I)I.
Infection rate =
M
and the rate of recovery is
Recovery rate = 0.009I.
Therefore,
0.01
dI
=
(M − I)I − 0.009I
dt
M
or
I
dI
= 0.001I(1 − 10 ).
dt
M
This is a logistic differential equation, and so the solution will look like the following graph:
I
M
10
t
The limiting value for I is
1
M,
10
so 1/10 of the population is infected in the long run.
dp
dt
39. (a)
= kp(B − p), where k > 0.
(b) To find when dp
is largest, we notice that dp
= kp(B − p), as a function of p, is a parabola opening downward with
dt
dt
the maximum at p = B2 , i.e. when 21 the tin has turned to powder. This is the time when the tin is crumbling fastest.
t
B
B
2
t
= 0, so we would expect p to remain 0 forever. However, since many organ pipes get tin
(c) If p = 0 initially, then dp
dt
pest, we must reconcile the model with reality. There are two possible ideas which solve this problem. First, we could
assume that p is never 0. In other words, we assume that all tin pipes, no matter how new, must contain some small
amount of tin pest. Assuming this means that all organ pipes must deteriorate due to tin pest eventually. Another
explanation is that the powder forms at a slow rate even if there was none present to begin with. Since not all organ
pipes suffer, it is possible that the conversion is catalyzed by some other impurities not present in all pipes.
40. (a) At equilibrium dP/dt = 0, so
dP
P
= kP 1 −
dt
L
Thus if P 6= 0,
− cP = 0,
kL − kP − cL = 0
k−c
L.
P =
k
At equilibrium, the annual harvest is
H = cP =
c(k − c)
L.
k
1094
Chapter Eleven /SOLUTIONS
(b) Since k and L are constant and the annual harvest is H = c(k − c)L/k, at the maximum where dH/dc = 0, we have
(k − 2c)
dH
=
L=0
dc
k
k
c= .
2
The maximum value of H is then
(k − k/2)
k
k
L = L.
2
k
4
(c) The equilibrium population is represented by the P -value (horizontal coordinate) at the point of intersection of the
line and parabola. The annual harvest at the equilibrium, cP , is represented by the vertical coordinate at the point of
intersection.
As c increases toward k, the slope of the line dP/dt = cP gets steeper and the intersection point between the
line and parabola moves closer to the origin. For c > k/2, the equilibrium population, P = (k − c)L/k, and the
annual harvest, H = c(k − c)L/k, get smaller and smaller. At c = k, the population becomes extinct.
H=
41. The population dies out if H is large enough that dP/dt < 0 for all P . The largest value for dP/dt = kP (1 − P/L)
occurs when P = L/2; then
L/2
P
L
kL
dP
=k
1−
=
= kP 1 −
.
dt
L
2
L
4
Thus if H > kL/4, we have dP/dt < 0 for all P and the population dies out if the quota is met.
Strengthen Your Understanding
42. Since
dP
P
= 0.08P − 0.0032P 2 = 0.08P 1 −
,
dt
25
we see that when we set dP/dt equal to zero, there are two solutions: one at P = 25 and one at P = 0. There are two
equilibrium solutions, with the second one at P = 0.
43. The maximum rate of change occurs at Q = 25 not at t = 25.
44. The curve has the shape we expect of a logistic curve. However, since the graph is leveling off at a carrying capacity of
100, we expect the inflection point to be at a height of 50, but in the graph the inflection point is above 60.
45. Any quantity that might increase exponentially at first and then eventually level off at a maximum value is a reasonable
answer: for example, the growth of a population in a confined space or sales of a new product with a saturation level.
46. If the maximum rate of change of a logistic function occurs at P = 75, then the carrying capacity is 2 · 75 = 150. An
answer is in the form
P
dP
= kP 1 −
,
dt
150
where k is any positive constant.
47. Since Q is growing logistically, the graph of dQ/dt against Q is a parabola opening down which passes through the
origin. Since Q has an equilibrium value at Q = 500, the other horizontal intercept is at Q = 500. See Figure 11.60.
dQ/dt
500
Figure 11.60
Q
11.8 SOLUTIONS
1095
48. Since P is growing logistically, the graph of dP/dt against P is a parabola opening down which passes through the
origin. Since P increases when 0 < P < 20, we know that dP/dt is positive when 0 < P < 20. Since P decreases
when P < 0 or P > 20, we know that dP/dt is negative for those values of P . See Figure 11.61.
dP/dt
20
P
Figure 11.61
49. False. This is a logistic equation with equilibrium values P = 0 and P = 2. Solution curves do not cross the line P = 2
and do not go from (0, 1) to (1, 3).
50. True. This is a logistic differential equation. Any solution with P (0) > 0 tends toward the carrying capacity, L, as t → ∞.
Solutions for Section 11.8
Exercises
1. Here x and y both increase at about the same rate.
2. Initially x = 0, so we start with only y. Then y decreases while x increases. Then x continues to increase while y starts
to increase as well. Finally y continues to increase while x decreases.
3. x decreases quickly while y increases more slowly.
4. The closed trajectory represents populations which oscillate repeatedly.
5. We set each derivative equal to zero and solve:
dx
=0
dt
−3x + xy = 0
x(−3 + y) = 0
x = 0 or y = 3.
Also,
dy
=0
dt
5y − xy = 0
y(5 − x) = 0
y = 0 or x = 5.
Since both derivatives must be zero at an equilibrium point, the equilibrium points are ordered pairs for which x = 0 or
y = 3 and y = 0 or x = 5. The equilibrium points are (0, 0) and (5, 3).
6. We set each derivative equal to zero and solve:
dx
=0
dt
−2x + 4xy = 0
x(−2 + 4y) = 0
x = 0 or y = 0.5.
1096
Chapter Eleven /SOLUTIONS
Also,
dy
=0
dt
−8y + 2xy = 0
y(−8 + 2x) = 0
y = 0 or x = 4.
Since both derivatives must be zero at an equilibrium point, the equilibrium points are ordered pairs for which x = 0 or
y = 0.5 and y = 0 or x = 4. The equilibrium points are (0, 0) and (4, 0.5).
7. We set each derivative equal to zero and solve:
dx
=0
dt
15x − 5xy = 0
x(15 − 5y) = 0
x = 0 or y = 3.
Also,
dy
=0
dt
10y + 2xy = 0
y(10 + 2x) = 0
y = 0 or x = −5.
Since both derivatives must be zero at an equilibrium point, the equilibrium points are ordered pairs for which x = 0 or
y = 3 and y = 0 or x = −5. The equilibrium points are (0, 0) and (−5, 3). (While a negative value might not make
sense if x represents the size of a population, it is reasonable in other scenarios in which systems of differential equations
are relevant such as if x represents the net worth of a company.)
8. We set each derivative equal to zero and solve:
dx
=0
dt
2
x − xy = 0
x(x − y) = 0
x = 0 or y = x.
Also,
dy
=0
dt
2
15y − 3y = 0
y(15 − 3y) = 0
y = 0 or y = 5.
Since both derivatives must be zero at an equilibrium point, the equilibrium points are ordered pairs for which x = 0 or
y = x and y = 0 or y = 5. The equilibrium points are (0, 0) and (0, 5) and (5, 5).
9. (a) At the point x = 3 and y = 2, we have
dx
= 5(3) − 3(3)(2) = −3 < 0
dt
dy
= −8(2) + (3)(2) = −10 < 0,
dt
so both x and y are decreasing.
(b) At the point x = 5 and y = 1, we have
dx
= 5(5) − 3(5)(1) = 10 > 0
dt
dy
= −8(1) + (5)(1) = −3 < 0,
dt
so x is increasing and y is decreasing.
11.8 SOLUTIONS
1097
10. (a) At the point P = 2 and Q = 3, we have
dP
= 2(2) − 10 = −6 < 0
dt
dQ
= 3 − 0.2(2)(3) = 1.8 > 0,
dt
so P is decreasing and Q is increasing.
(b) At the point P = 6 and Q = 5, we have
dP
= 2(6) − 10 = 2 > 0
dt
dQ
= 5 − 0.2(6)(5) = −1 < 0,
dt
so P is increasing and Q is decreasing.
Problems
11. (a) The values are given by
Time t (days)
8
12
16
20
Susceptibles
1950
1850
1550
1000
750
550
350
20
80
240
460
500
460
320
Infecteds
22
24
28
32
36
40
44
250
200
200
200
180
100
40
20
(b) The peak of the epidemic is on day 22 when about 500 people are infected.
(c) About 2000 people are susceptible at the onset of the epidemic and only 200 are still susceptible at its end. About
1800 catch the disease and about 200 are spared.
12. (a) On day 20 about 750 are infected.
(b) On day 20 about 2600 are susceptible. Since there are 4000 people in the population, there are 4000 − 2600 = 1400
who have already had the disease.
(c) At day 60 the epidemic is over. There are about 400 people who are still susceptible, because they never got the
disease. The rest, 4000 − 400 = 3600 caught the disease sometime during the epidemic.
13. (a) The human population shrinks as humans are turned into zombies, so parameter a is negative. The interaction term
for zombies is positive, since an interaction between a human and a zombie increases the zombie population, so c is
positive. The zombie population shrinks by a certain percentage each time interval because some zombies will starve
to death, so b is negative.
(b) The terms aHZ and cHZ both indicate the rate of human to zombie conversions. Since each loss of one human is
directly a gain of one zombie, aHZ = −cHZ. Therefore, the parameter a is exactly the negative of the parameter c.
Thus, a = −c.
14. (a) This is an example of a predator-prey relationship, so this system goes with IV, with x representing the fox population
and y representing the hare population.
(b) This system models the relationship in II, with x representing the tree population and y representing the owl population.
(c) This system models the relationship in I with x and y being either bees or flowers.
(d) We write a system of differential equations for III with elk and buffalo. In each case, the species would do fine on its
own but is hurt by the other. The terms have signs as follows, with proportionality constants included as desired.
dx
= x − xy
dt
dy
= y − xy
dt
Notice that x and y can each represent either the elk or the buffalo population.
15. Since
dS
= −aSI,
dt
dI
= aSI − bI,
dt
dR
= bI
dt
1098
Chapter Eleven /SOLUTIONS
we have
dI
dR
dS
+
+
= −aSI + aSI − bI + bI = 0.
dt
dt
dt
d
(S + I + R) = 0, so S + I + R = constant.
Thus dt
16. This is an example of a predator-prey relationship. Normally, we would expect the worm population, in the absence of
predators, to increase without bound. As the number of worms w increases, so would the rate of increase dw/dt; in other
words, the relation dw/dt = w might be a reasonable model for the worm population in the absence of predators.
However, since there are predators (robins), dw/dt won’t be that big. We must lessen dw/dt. It makes sense that
the more interaction there is between robins and worms, the more slowly the worms are able to increase their numbers.
Hence we lessen dw/dt by the amount wr to get dw/dt = w − wr. The term −wr reflects the fact that more interactions
between the species means slower reproduction for the worms.
Similarly, we would expect the robin population to decrease in the absence of worms. We’d expect the population
decrease at a rate related to the current population, making dr/dt = −r a reasonable model for the robin population in
absence of worms. The negative term reflects the fact that the greater the population of robins, the more quickly they are
dying off. The wr term in dr/dt = −r + wr reflects the fact that the more interactions between robins and worms, the
greater the tendency for the robins to increase in population.
17. If there are no worms, then w = 0, and dr
= −r giving r = r0 e−t , where r0 is the initial robin population. If there are
dt
dw
no robins, then r = 0, and dt = w giving w = w0 et , where w0 is the initial worm population.
18. There is symmetry across the line r = w. Indeed, since
dr
dw
dr
dw
=
r(w−1)
,
w(1−r)
if we switch w and r we get
dw
dr
=
w(r−1)
,
r(1−w)
so
r(1−w)
.
w(r−1)
=
Since switching w and r changes nothing, the slope field must be symmetric across the line r = w. The
slope field shows that the solution curves are either spirals or closed curves. Since there is symmetry about the line r = w,
the solutions must in fact be closed curves.
19. If w = 2 and r = 2, then dw
= −2 and dr
= 2, so initially the number of worms decreases and the number of robins
dt
dt
increases. In the long run, however, the populations will oscillate; they will even go back to w = 2 and r = 2. See
Figure 11.62.
r (robins in thousands)
3
(2500 robins)
2
1
1
2
3
w (worms in millions)
Figure 11.62
20. Sketching the trajectory through the point (2, 2) on the slope field given shows that the maximum robin population is
about 2500, and the minimum robin population is about 500. When the robin population is at its maximum, the worm
population is about 1,000,000.
21.
Worms
population
❄
1.5
Robins
✠
1
P0
P2
P0
P2
Figure 11.63
P0
P2
P0
t
11.8 SOLUTIONS
1099
22. It will work somewhat; the maximum number the robins reach will increase. However, the minimum number the robins
reach will decrease as well. (See graph of slope field.) In the long term, the robin-worm populations will again fall into
a cycle. Notice, however, if the extra robins are added during the part of the cycle where there are the fewest robins, the
new cycle will have smaller variation. See Figure 11.64.
Note that if too many robins are added, the minimum number may get so small the model may fail, since a small
number of robins are more susceptible to disaster.
r (robins in thousands)
3
✛
✛
2
New trajectory
Old trajectory
1
1
2
3
w (worms in millions)
Figure 11.64
23. The numbers of robins begins to increase while the number of worms remains approximately constant. See Figure 11.65.
The numbers of robins and worms oscillate periodically between 0.2 and 3, with the robin population lagging behind
the worm population.
r
3
2
(3, 1)
1
w
1
2
3
Figure 11.65
24. Estimating from the phase plane, we have
0.18 < r < 3
so the robin population lies between 180 and 3000. Similarly
0.2 < w < 3,
so the worm population lies between 200,000 and 3,000,000.
When the robin population is at its minimum r ≈ 0.2, then w ≈ 0.87, so that there are approximately 870,000
worms.
Robins
3
P
✠
✠
Worms
1
t
Figure 11.66
1100
Chapter Eleven /SOLUTIONS
25. (a) Symbiosis, because both populations decrease while alone but are helped by the presence of the other.
y
(b)
4
3
2
1
x
1
2
3
4
Both populations tend to infinity or both tend to zero.
26. (a) Competition, because both populations grow logistically when alone, but are harmed by the presence of the other.
(b) See Figure 11.67. In the long run, x → 2, y → 0. In other words, y becomes extinct.
y
4
3
2
1
x
1
2
3
4
Figure 11.67
27. (a) Predator-prey, because x decreases while alone, but is helped by y, whereas y increases logistically when alone, and
is harmed by x. Thus x is predator, y is prey.
(b) See Figure 11.68. Provided neither initial population is zero, both populations tend to about 1. If x is initially zero,
but y is not, then y → ∞. If y is initially zero, but x is not, then x → 0.
y
4
3
2
1
x
1
2
Figure 11.68
3
4
11.8 SOLUTIONS
28. (a) Thinking of y as a function of x and x as a function of t, then by the chain rule:
1101
dy
dy dx
=
, so:
dt
dx dt
dy/dt
−0.01x
x
dy
=
=
=
dx
dx/dt
−0.05y
5y
y (thousand Japanese troops)
30
20
10
10
20
30
40
50
60
x (thousand US troops)
(b) The figure above shows the slope field for this differential equation and the trajectory starting at x0 = 54, y0 = 21.5.
The trajectory goes to the x-axis, where y = 0, meaning that the Japanese troops were all killed or wounded before
the US troops were, and thus predicts the US victory (which did occur). Since the trajectory meets the x-axis at
x ≈ 25, the differential equation predicts that about 25,000 US troops would survive the battle.
(c) The fact that the US got reinforcements, while the Japanese did not, does not alter the predicted outcome (a US
victory). The US reinforcements have the effect of changing the trajectory, altering the number of troops surviving
the battle. See the graph below.
y (thousand Japanese troops)
30
20
10
10
20
30
40
50
60
x (thousand US troops)
29. (a) Thinking of y as a function of x and x as a function of t, then by the chain rule:
dy
dy dx
=
, so:
dt
dx dt
dy/dt
−bx
bx
dy
=
=
=
dx
dx/dt
−ay
ay
(b) Separating variables,
Z
ay dy =
Z
bx dx
x2
y2
=b
+k
2
2
2
2
ay − bx = C where C = 2k
a
30. (a) Lanchester’s square law for the battle of Iwo Jima is
0.05y 2 − 0.01x2 = C.
If we measure x and y in thousands, x0 = 54 and y0 = 21.5, so 0.05(21.5)2 −0.01(54)2 = C giving C = −6.0475.
Thus the equation of the trajectory is
0.05y 2 − 0.01x2 = −6.0475
giving
x2 − 5y 2 = 604.75.
(b) Assuming that the battle did not end until all the Japanese were dead or wounded, that is, y = 0, then the number
of US soldiers remaining is given by x2 − 5(0)2 = 604.75. This gives x = 24.59, or about 25,000 troops. This is
approximately what happened.
1102
Chapter Eleven /SOLUTIONS
31. (a) Since the guerrillas are hard to find, the rate at which they are put out of action is proportional to the number of chance
encounters between a guerrilla and a conventional soldier, which is in turn proportional to the number of guerrillas
and to the number of conventional soldiers. Thus the rate at which guerrillas are put out of action is proportional to
the product of the strengths of the two armies.
(b)
dx
= −xy
dt
dy
= −x.
dt
(c) Thinking of y as a function of x and x a function of of t, then by the chain rule:
dy
dy dx
=
so:
dt
dx dt
dy/dt
−x
1
dy
=
=
= .
dx
dx/dt
−xy
y
Separating variables:
Z
y dy =
Z
dx
y2
= x + C.
2
The value of C is determined by the initial strengths of the two armies. Note that C could be written on the opposite
side of the equation, giving it the opposite sign.
y2
= x + C, we see that if
(d) The sign of C determines which side wins the battle. Looking at the general solution
2
√
C > 0 the y-intercept is at 2C, so y wins the battle by virtue of the fact that it still has troops when x = 0. If
C < 0 then the curve intersects the axes at x = −C, so x wins the battle because it has troops when y = 0. If C = 0,
then the solution goes to the point (0, 0), which represents the case of mutual annihilation.
(e) We assume that an army wins if the opposing force goes to 0 first. Figure 11.69 shows that in our formulation, the
conventional force wins if C > 0 and the guerrillas win if C < 0. Neither side wins if C = 0 (all soldiers on both
sides are killed in this case).
y (conventional)
4
y2
=x
2
(i.e. C = 0)
C>0
3
conventional wins
2
1
C<0
guerrilla wins
1
2
3
4
x (guerrilla)
Figure 11.69
32. (a) Taking the constants of proportionality to be a and b, with a > 0 and b > 0, the equations are
dx
= −axy
dt
dy
= −bxy.
dt
(b) Thinking of x and y as functions of t, then by the chain rule:
dy/dt
−bxy
b
dy
=
=
= .
dx
dx/dt
−axy
a
11.8 SOLUTIONS
1103
Solving the differential equation gives
b
x + C,
a
where C depends on the initial sizes of the two armies. Note that C could be written on the opposite side of the
equation, giving it the opposite sign.
b
(c) The sign of C determines which side wins the battle. Looking at the general solution y = x + C, we see that if
a
C > 0 the y-intercept is at C, so y wins the battle by virtue of the fact that it still has troops when x = 0. If C < 0
a
then the curve intersects the axes at x = − C, so x wins the battle because it has troops when y = 0. If C = 0,
b
then the solution goes to the point (0, 0), which represents the case of mutual annihilation.
(d) We assume that an army wins if the opposing force goes to 0 first. Figure 11.70 shows that in our formulation, y wins
if C > 0 and x wins if C < 0.
y=
y (guerrilla)
y = bx/a
(i.e. C = 0)
4
3
C>0
y wins
2
C<0
x wins
1
1
2
3
4
x (guerrilla)
Figure 11.70
33. (a) At an equilibrium point both w and r are constant, so
dr
dw
= 0 and
= 0.
dt
dt
Therefore, we need to solve
Rearranging gives
aw − cwr = 0 and − br + kwr = 0.
w(a − cr) = 0 and − r(b − kw) = 0
so the only equilibrium points are w = 0, r = 0 and
w=
b
a
and r = .
k
c
(b) If the insecticide causes a decline in the number of worms then the model becomes
dr
dw
= aw − cwr − pw and
= −br + kwr
dt
dt
where p is a positive constant. Solving as before,
w(a − p − cr) = 0 and − r(b − kw) = 0,
so the equilibrium points are w = 0, r = 0 and
w=
a−p
b
and r =
.
k
c
So, the equilibrium worm population is unchanged but the equilibrium robin population falls.
Strengthen Your Understanding
34. It is possible for dx/dt and dy/dt to be positive simultaneously. For example, if x = y = 1, then dx/dt = 2.6 and
dy/dt = 3.5, meaning that both x and y are increasing.
1104
Chapter Eleven /SOLUTIONS
35. Using the chain rule, we know that
dy
dy/dt
=
.
dx
dx/dt
Thus, if dx/dt < 0 and dy/dt > 0, we must have dy/dx < 0.
36. Since X is indifferent to Y and thrives on its own, we have dx/dt = kx with k positive. Since Y needs X to survive, we
have dy/dt = −k1 y + k2 xy, with k1 and k2 positive. One possible example is
dx
= 0.5x
dt
dy
= −0.1y + 0.3xy.
dt
37. We have dx/dt = k1 x − k2 xy with k1 and k2 positive. Similarly for dy/dt. One possible example is
dx
= 0.5x − 0.2xy
dt
dy
= 0.1y − 0.3xy.
dt
38. The parameter a is larger if a disease is more contagious, so we pick a very contagious disease for D1 and a disease that
is not very contagious for D2 . The flu is spread easily through such things as sneezing, whereas a disease such as HIV
requires something like an exchange of bodily fluids, so we might pick a flu for D1 and HIV for D2 .
39. True. Specifying x(0) and y(0) corresponds to picking a starting point in the plane and thereby picking the unique solution
curve through that point.
40. False. Competitive exclusion, in which one population drives out another, is modeled by a system of differential equations.
Solutions for Section 11.9
Exercises
1. Equilibrium points occur where both derivatives dx/dt and dy/dt are zero. Thus, equilibrium are located at the intersection of a nullcline with vertical line segments and a nullcline with horizontal line segments. We see in Figure 11.71 there
is only one equilibrium point located at the point (4, 10).
y
15
10
5
x
2
4
6
Figure 11.71
2. (a) Since the point (4, 7) is in Region I, dx/dt < 0 and dy/dt > 0.
(b) Since the point (4, 10) is an equilibrium point, dx/dt = 0 and dy/dt = 0.
(c) Since the point (6, 15) is in Region II, dx/dt < 0 and dy/dt < 0.
11.9 SOLUTIONS
1105
3. Since (2, 5) is in Region IV, dx/dt > 0 and dy/dt > 0. Thus, the trajectory goes up and to the right toward the
equilibrium (4, 10). See Figure 11.72.
y
15
10
5
x
2
4
6
Figure 11.72
4. Since dx/dt > 0 and dy/dt < 0 in Region III, a trajectory that starts at (2, 10) goes down and to the right toward the
equilibrium (4, 10) located at the intersection of the nullclines. See Figure 11.73.
y
15
10
5
x
2
4
6
Figure 11.73
5. We can see in Figure 11.89 that any trajectory eventually tends toward the equilibrium point (4, 10).
6. Equilibrium points occur where both derivatives dx/dt and dy/dt are zero. Thus, equilibrium are located at the intersection of a nullcline with vertical line segments and a nullcline with horizontal line segments. We see in Figure 11.71 there
are three equilibrium points located at (0, 0), (0, 6), and approximately (13, 0).
y
6
4
2
x
5
10
Figure 11.74
15
1106
Chapter Eleven /SOLUTIONS
7. (a) The point (5, 2) has dx/dt > 0 and dy/dt > 0.
(b) The point (10, 2) has dx/dt < 0. Since the point (10, 2) is on a nullcline horizontal line segments, dy/dt = 0.
(c) The point (10, 1) has dy/dt > 0. Since the point (10, 1) is on a nullcline with vertical line segments, dx/dt = 0.
8. At the point (2, 5), dx/dt < 0 and dy/dt > 0. The trajectory goes up and to the left, tending toward the equilibrium
point (0, 6). See Figure 11.75.
y
6
4
2
x
5
10
15
Figure 11.75
9. At the point (10, 4), we have dx/dt < 0 and dy/dt < 0. The trajectory goes down and to the left, crossing the top
nullcline with dy/dt = 0. The trajectory then goes up and to the left since dy/dt has changed from negative to positive.
The trajectory eventually tends toward the equilibrium point (0, 6). See Figure 11.76.
y
6
4
2
x
5
10
15
Figure 11.76
10. Any trajectory eventually tends toward the equilibrium point (0, 6).
11. On the phase plane, it appears that slopes are vertical along the lines y = 3 and x = 0 (the y-axis) and that the slopes are
horizontal along the lines x = 5 and y = 0 (the x-axis). See Figure 11.77.
y
3
x
5
Figure 11.77
11.9 SOLUTIONS
1107
12. (a) To find the equilibrium points we set
20x − 10xy = 0
25y − 5xy = 0.
So, x = 0, y = 0 is an equilibrium point. Another one is given by
10y = 20
5x = 25.
Therefore, x = 5, y = 2 is the other equilibrium point.
(b) At x = 2, y = 4,
dx
= 20x − 10xy = 40 − 80 = −40
dt
dy
= 25y − 5xy = 100 − 40 = 60.
dt
Since these are not both zero, this point is not an equilibrium point.
Problems
13. We first find the nullclines. Again, we assume x, y ≥ 0.
Vertical nullclines occur where dx/dt = 0, which happens when dx
= x(2 − x − y) = 0,
dt
i.e. when x = 0 or x + y = 2.
= y(1−x−y) = 0, i.e. when y = 0 or x+y = 1.
Horizontal nullclines occur where dy/dt = 0, which happens when dy
dt
These nullclines are shown in Figure 11.78.
Equilibrium points (also shown in Figure 11.78) occur where both dy/dt and dx/dt are 0, i.e. at the intersections of
vertical and horizontal nullclines. There are three such points for these equations: (0, 0), (0, 1), and (2, 0).
y
2
y
2
x+y = 2
dx/dt = 0
✠
(III)
x+y = 1
dy/dt = 0
1
(II)
1
✠
(I)
x
1
2
Figure 11.78: Nullclines and equilibrium points
(dots)
x
1
2
Figure 11.79: General directions of trajectories
and equilibrium points (dots)
Looking at sectors in Figure 11.79, we see that no matter in what sector the initial point lies, the trajectory will head
toward the equilibrium point (2, 0).
= 0, which happens when x = 0 or y = 31 (2 − x).
14. We first find the nullclines. Vertical nullclines occur where dx
dt
dy
Horizontal nullclines occur where dt = y(1 − 2x) = 0, which happens when y = 0 or x = 21 . These nullclines are
shown in Figure 11.80.
Equilibrium points (also shown in Figure 11.80) occur at the intersections of vertical and horizontal nullclines. There
are three such points for this system of equations; (0, 0), ( 12 , 12 ) and (2, 0).
The nullclines divide the positive quadrant into four regions as shown in Figure 11.80. Trajectory directions for these
regions are shown in Figure 11.81.
1108
Chapter Eleven /SOLUTIONS
y
y
2/3
II
I
III
IV
2/3
x
✻■
✠
✻
✒
✲
❘
✲
2
1/2
x
2
1/2
Figure 11.81: General directions of
trajectories and equilibrium points
(dots)
Figure 11.80: Nullclines and
equilibrium points (dots)
= x(2 − x − 2y) = 0, which happens when x = 0 or
15. We first find nullclines. Vertical nullclines occur where dx
dt
=
y(1
−
2x − y) = 0, which happens when y = 0 or y = 1 − 2x.
y = 12 (2 − x). Horizontal nullclines occur where dy
dt
These nullclines are shown in Figure 11.82.
Equilibrium points (also shown in the figure below) occur at the intersections of vertical and horizontal nullclines.
There are three such points for this system; (0, 0), (0, 1), and (2, 0).
The nullclines divide the positive quadrant into three regions as shown in the figure below. Trajectory directions for
these regions are shown in Figure 11.83.
y
1
1
III
I
✠
✻
✻
✒
✲
II
0.5
❘
2
Figure 11.82: Nullclines and equilibrium
points (dots)
✲
x
0.5
2
Figure 11.83: General directions of
trajectories and equilibrium points (dots)
= x(1 − y − x3 ) = 0, which happens when x = 0 or
16. We first find the nullclines. Vertical nullclines occur where dx
dt
dy
y
x
y = 1 − 3 . Horizontal nullclines occur where dt = y(1 − 2 − x) = 0, which happens when y = 0 or y = 2(1 − x).
These nullclines are shown in Figure 11.84.
Equilibrium points (also shown in Figure 11.84) occur at the intersections of vertical and horizontal nullclines. There
are four such points for this system: (0, 0), (0, 2), (3, 0), and ( 53 , 45 ).
The nullclines divide the positive quadrant into four regions as shown in Figure 11.84. Trajectory directions for these
regions are shown in Figure 11.85.
y
y
❄
2
1
2
I
II
III
IV
1
1
x
3
Figure 11.84: Nullclines and
equilibrium points (dots)
✻
❘
✠
✻
✒
✲
❘
1
✲
✲
x
3
Figure 11.85: General directions
of trajectories and equilibrium
points (dots)
11.9 SOLUTIONS
1109
17. We first find the nullclines. Again, we assume x, y ≥ 0.
dx
= x(1 − x − y3 ) = 0 when x = 0 or x + y/3 = 1.
dt
dy
= y(1 − y − x2 ) = 0 when y = 0 or y + x/2 = 1.
dt
These nullclines are shown in Figure 11.86. There are four equilibrium points for these equations. Three of them are the
points, (0, 0), (0, 1), and (1, 0). The fourth is the intersection of the two lines x + y/3 = 1 and y + x/2 = 1. This point
is ( 45 , 35 ).
y
y
3
3
✛
x + y/3 = 1
dx/dt = 0
(II)
(I)
y + x/2 = 1
dy/dt = 0
1
( 45 , 35 )
1
( 45 , 35 )
✠
(IV)
(III)
x
1
2
1
Figure 11.86: Nullclines and equilibrium points
(dots)
x
2
Figure 11.87: General directions of trajectories
and equilibrium points (dots)
Looking at sectors in Figure 11.87, we see that no matter in what sector the initial point lies, the trajectory will head
toward the equilibrium point ( 45 , 53 ). Only if the initial point lies on the x- or y-axis, will the trajectory head toward the
equilibrium points at (1, 0), (0, 1), or (0, 0). In fact, the trajectory will go to (0, 0) only if it starts there, in which case
x(t) = y(t) = 0 for all t. From direction of the trajectories in Figure 11.87, it appears that if the initial point is in sectors
(I) or (III), then it will remain in that sector as it heads toward the equilibrium.
= x(1 − x2 − y) = 0 when x = 0 or y + x2 = 1.
18. We assume that x, y ≥ 0 and then find the nullclines. dx
dt
dy
y
y
= y(1 − 3 − x) = 0 when y = 0 or x + 3 = 1.
dt
We find the equilibrium points. They are (2, 0), (0, 3), (0, 0), and ( 54 , 35 ). The nullclines and equilibrium points are shown
in Figure 11.88.
y
y
3
3
✛
x + y/3 = 1
dy/dt = 0
(II)
(I)
y = 1 − x/2
dx/dt = 0
1
( 45 , 35 )
1
( 45 , 35 )
✠
x
1
2
Figure 11.88: Nullclines and equilibrium points
(dots)
(IV)
(III)
1
x
2
Figure 11.89: General directions of trajectories
and equilibrium points (dots)
Figure 11.89 shows that if the initial point is in sector (I), the trajectory heads toward the equilibrium point (0, 3).
Similarly, if the trajectory begins in sector (III), then it heads toward the equilibrium (2, 0) over time. If the trajectory
begins in sector (II) or (IV), it can go to any of the three equilibrium points (2, 0), (0, 3), or ( 45 , 53 ).
1110
Chapter Eleven /SOLUTIONS
19. (a) dS/dt = 0 where S = 0 or I = 0 (both axes).
dI/dt = 0.0026I(S − 192), so dI/dt = 0 where I = 0 or S = 192.
Thus every point on the S axis is an equilibrium point (corresponding to no one being sick).
dI
dS
< 0 and
> 0.
(b) In region I, where S > 192,
dt
dt
dS
dI
In region II, where S < 192,
< 0 and
< 0. See Figure 11.90.
dt
dt
I
I
II
I
✛
✠❄
■
✛✻
S
192
S
192
Figure 11.90
Figure 11.91
(c) If the trajectory starts with S0 > 192, then I increases to a maximum when S = 192. If S0 < 192, then I always
decreases. See Figure 11.90. Regardless of the initial conditions, the trajectory always goes to a point on the S-axis
(where I = 0). The S-intercept represents the number of students who never get the disease. See Figure 11.91.
20. The nullclines are where dw
= 0 or dr
= 0.
dt
dt
dw
=
0
when
w
−
wr
=
0,
so
w(1
−
r)
= 0 giving w = 0 or r = 1.
dt
dr
=
0
when
−r
+
rw
=
0,
so
r(w
−
1)
= 0 giving r = 0 or w = 1.
dt
r
r
✛
1
w=1
dr/dt = 0
(I)
(IV)
(II)
(III)
1
■
r=1
dw/dt = 0
w
1
w
1
Figure 11.92: Nullclines and equilibrium points (dots)
Figure 11.93
The equilibrium points are where the nullclines intersect: (0, 0) and (1, 1). The nullclines split the first quadrant into
four sectors. See Figure 11.92. We can get a feel for how the populations interact by seeing the direction of the trajectories
in each sector. See Figure 11.93. If the populations reach an equilibrium point, they will stay there. If the worm population
dies out, the robin population will also die out, too. However, if the robin population dies out, the worm population will
continue to grow.
Otherwise, it seems that the populations cycle around the equilibrium (1, 1). The trajectory moves from sector to
sector: trajectories in sector (I) move to sector (II); trajectories in sector (II) move to sector (III); trajectories in sector
(III) move to sector (IV); trajectories in sector (IV) move back to sector (I). The robins keep the worm population down
by feeding on them, but the robins need the worms (as food) to sustain the population. These conflicting needs keep the
populations moving in a cycle around the equilibrium.
21. (a) If B were not present, then we’d have A′ = 2A, so company A’s net worth would grow exponentially. Similarly,
if A were not present, B would grow exponentially. The two companies restrain each other’s growth, probably by
competing for the market.
(b) To find equilibrium points, find the solutions of the pair of equations
A′ = 2A − AB = 0
B ′ = B − AB = 0
11.9 SOLUTIONS
1111
The first equation has solutions A = 0 or B = 2. The second has solutions B = 0 or A = 1. Thus the equilibrium
points are (0,0) and (1,2).
(c) In the long run, one of the companies will go out of business. Two of the trajectories in the figure below go toward the
A axis; they represent A surviving and B going out of business. The trajectories going toward the B axis represent
A going out of business. Notice both the equilibrium points are unstable.
B
4
3
2
1
1
2
3
4
A
22. (a) See Figure 11.94.
dx
10.5
= 0 when x =
= 23.3
dt
0.45
dy
= 0 when 8.2x − 0.8y − 142 = 0
dt
There is an equilibrium point where the trajectories cross at x = 23.3, y = 61.7
dx
dy
In region I,
> 0,
< 0.
dt
dt
dy
dx
< 0,
< 0.
In region II,
dt
dt
dx
dy
In region III,
< 0,
> 0.
dt
dt
dx
dy
In region IV,
> 0,
> 0.
dt
dt
(b) See Figure 11.95.
Region II
y (US)
dx
dt
=0
✛
✠ ❄
✲
✛
Region I
y (US)
dy
dt
=0
Region III
✲
❄
❘
■
✛ ✻
61.7
61.7
Region IV
✛
23.3
✻
✒
✲
x (Soviet)
Figure 11.94: Nullclines and equilibrium point (dot) for
US-Soviet arms race
23.3
x (Soviet)
Figure 11.95: Trajectories for US-Soviet arms race.
(c) All the trajectories tend toward the equilibrium point x = 23.3, y = 61.7. Thus the model predicts that in the longrun the arms race will level off with the Soviet Union spending 23.3 billion dollars a year on arms and the US 61.7
billion dollars.
(d) As the model predicts, yearly arms expenditure did tend toward 23 billion for the Soviet Union and 62 billion for the
US.
1112
Chapter Eleven /SOLUTIONS
23. (a) The nullclines are P = 0 or P1 +3P2 = 13 (where dP1 /dt = 0) and P = 0 or P2 +0.4P1 = 6 (where dP2 /dt = 0).
(b) The phase plane in Figure 11.96 shows that P2 will eventually exclude P1 regardless of where the experiment starts
so long as there were some P2 originally. Consequently, the data points would have followed a trajectory that starts
at the origin, crosses the first nullcline and goes left and upward between the two nullclines to the point P1 = 0,
P2 = 6.
P2
6
✠
4
✻
✒
✲
2
✛✻
■
P2 + 0.4P1 = 6
dP2 /dt = 0
✛
✠ ❄
✲
✠
5
10
P1 + 3P2 = 13
dP1 /dt = 0
15
P1
Figure 11.96: Nullclines and equilibrium points (dots) for
Gause’s yeast data (hollow dots)
Strengthen Your Understanding
24. The trajectory has no arrow to indicate direction.
25. Equilibrium occurs when both dx/dt = 0 and dy/dt = 0. Both nullclines have horizontal line segments indicating
points that dy/dt = 0. There are no nullclines that indicate where dx/dt = 0. Thus, the point (6, 6) has dy/dt = 0 but
dx/dt 6= 0 and is not an equilibrium point.
26. An equilibrium point is a point where a nullcline with vertical trajectories intersects a nullcline with horizontal trajectories.
There are many possible ways to draw this graph. One example is shown in Figure 11.97.
y
1
x
1
Figure 11.97
27. Nullclines show where trajectories are vertical or horizontal. We see that the trajectory shown is vertical at approximately
the points (0.4, 1) and (2.1, 1) so it is reasonable to draw a nullcline showing vertical trajectories on the line y = 1.
The trajectory shown is horizontal at approximately the points (1, 2.1) and (1, 0.4) so it is reasonable to draw a nullcline
showing horizontal trajectories on the line x = 1. See Figure 11.98. Other answers are possible.
SOLUTIONS to Review Problems for Chapter Eleven
1113
y
3
2
1
x
1
2
3
Figure 11.98
28. We follow the directions, and use the fact that if the trajectory crosses the nullcline x = 2, it is horizontal there and then
bends down. See Figure 11.99. Other answers are possible.
y
3
2
1
x
1
2
3
4
Figure 11.99
Solutions for Chapter 11 Review
Exercises
1. For (I): y = xex , y ′ = ex + xex , and y ′′ = 2ex + xex .
For (II): y = xe−x , y ′ = e−x − xe−x , and y ′′ = −2e−x + xe−x .
Thus (I) satisfies equation (d).
(II) satisfies equation (c).
2. (a) Since y ′ = 1 + y 2 , the slope is everywhere positive. This is true for slope field (III).
(b) Since y = x, the slopes are positive to the right of the y-axis and negative to the left of the y-axis. Solution curves
are parabolas, as in slope field (VI).
(c) Since y ′ = sin x, the slopes oscillate between −1 and 1. This corresponds to slope field (V).
(d) Since y ′ = y, the slopes are positive above the x-axis and negative below the x-axis. This is true for slope field (I).
(e) Since y ′ = x − y, the slopes are zero when x = y. This corresponds to slope field (IV).
(f) Since y ′ = 4 − y, the slopes are positive when y < 4 and negative when y > 4. This corresponds to slope field (II).
3. (a) = (I), (b) = (IV), (c) = (III). Graph (II) represents an egg originally at 0◦ C which is moved to the kitchen table (20◦
C) two minutes after the egg in part (a) is moved.
4. This equation is separable, so we integrate, giving
so
Z
dP =
P (t) =
Z
t dt
t2
+ C.
2
1114
Chapter Eleven /SOLUTIONS
5. This equation is separable, so we integrate, giving
Z
so
1
dy =
0.2y − 8
Z
dx
1
ln |0.2y − 8| = x + C.
0.2
Thus
y(x) = 40 + Ae0.2x .
6. This equation is separable, so we integrate, giving
Z
so
1
dP =
10 − 2P
Z
dt
1
ln |10 − 2P | = t + C.
−2
Thus
P = 5 + Ae−2t .
7. This equation is separable, so we integrate, giving
Z
so
1
dH =
10 + 0.5H
Z
dt
1
ln |10 + 0.5H| = t + C.
0.5
Thus
H = Ae0.5t − 20.
8. This equation is separable, so we integrate, using the table of integrals or partial fractions, to get
Z
Z
so
1
dR = 2
R − 3R2
1
dR +
R
Z
Z
dt
3
dR = 2
1 − 3R
Z
dt
ln |R| − ln |1 − 3R| = 2t + C
R
= 2t + C
ln
1 − 3R
R
= Ae2t
1 − 3R
Ae2t
R=
.
1 + 3Ae2t
9. This equation is separable, so we integrate, using the table of integrals or partial fractions, to get:
250
100
so
Z
Z
250
dP =
100P − P 2
1
dP +
P
Z
Z
1
dP
100 − P
dt
=
Z
2.5(ln |P | − ln |100 − P |) = t + C
P
2.5 ln
=t+C
100 − P
dt
SOLUTIONS to Review Problems for Chapter Eleven
1115
P
= Ae0.4t
100 − P
100Ae0.4t
P =
1 + Ae0.4t
10.
dy
dx
so C = − 12 . Thus,
11.
dP
dt
dy
= −xy 2 , so ydy2
dx
2
− y1 = − x2 − 12 giving
+ xy 2 = 0 means
= 0.03P + 400 so
dP
P + 40000
3
R
=
2
−x dx giving − y1 = − x2 + C. Since y(1) = 1 we have −1 = − 21 + C
R
=
R
R
0.03dt.
2
.
x2 +1
y=
| = 0.03t + C giving P = Ae0.03t − 40000
. Since P (0) = 0, A =
ln |P + 40000
3
3
dy
12. 1 + y 2 − dx
= 0 gives
y = tan x.
dy
dx
= y 2 + 1, so
R
dy
1+y 2
dy
dy
13. 2 sin x − y 2 dx
= 0 giving 2 sin x = y 2 dx
.
14.
15.
R
−2 = 9 + C, so C = −11. Thus, −2 cos x =
dk
dt
=
R
y
3
therefore P =
40000
(e0.03t − 1).
3
dx and arctan y = x + C. Since y(0) = 0 we have C = 0, giving
y 2 dy so −2 cos x =
√
− 11 giving y = 3 33 − 6 cos x.
2 sin x dx =
3
40000
,
3
R
y3
3
+ C. Since y(0) = 3 we have
= (1 + ln t)k gives dk
= (1 + ln t)dt so ln |k| = t ln t + C. k(1) = 1, so 0 = 0 + C, or C = 0. Thus,
k
ln |k| = t ln t and |k| = et ln t = tt , giving k = ±tt .
But recall k(1) = 1, so k = tt is the solution.
dy
dx
R
=
y(3−x)
1 y−4)
x( 2
1 y−4)
(2
dy = (3−x)
dx so
y
x
5
have 2 − 4 ln 5 = ln |1| − 1 +
gives
Since y(1) = 5, we
R
R
R
R
( 12 − y4 )dy =
C so C =
7
2
R
( x3 − 1)dx. Thus 21 y − 4 ln |y| = 3 ln |x| − x + C.
− 4 ln 5. Thus,
7
1
y − 4 ln |y| = 3 ln |x| − x + − 4 ln 5.
2
2
We cannot solve for y in terms of x, so we leave the equation in this form.
16.
dy
dx
=
0.2y(18+0.1x)
x(100+0.5y)
giving
R
(100+0.5y)
0.2y
dy =
Z
R
18+0.1x
x
5
500
+
y
2
dx, so
dy =
Z
1
18
+
x
10
dx.
1
Therefore, 500 ln |y| + 52 y = 18 ln |x| + 10
x + C. Since the curve passes through (10,10), 500 ln 10 + 25 = 18 ln 10 +
1 + C, so C = 482 ln 10 + 24. Thus, the solution is
500 ln |y| +
1
5
y = 18 ln |x| +
x + 482 ln 10 + 24.
2
10
We cannot solve for y in terms of x, so we leave the answer in this form.
17. This equation is separable and so we write it as
dz
1
= 1.
z(z − 1) dt
We integrate with respect to t, giving
Z
so that
Z
1
dz =
z(z − 1)
Z
Z
Z
dt
1
1
dz −
dz =
dt
z−1
z
ln |z − 1| − ln |z| = t + C
z −1
ln |
| = t + C,
z
z−1
= et+C = ket .
z
Solving for z gives
z(t) =
1
.
1 − ket
1116
Chapter Eleven /SOLUTIONS
The initial condition z(0) = 10 gives
1
= 10
1−k
or k = 0.9. The solution is therefore
z(t) =
1
.
1 − 0.9et
18. Using the solution of the logistic equation given on page 631 in Section 11.7, and using y(0) = 1, we get y =
10
1+9e−10t
.
100−x
= y(100−x)
gives ( 20−y
) dy =
dx. Thus, 20 ln |y| − y = 100 ln |x| − x + C. The curve passes through
x(20−y)
y
x
(1, 20), so 20 ln 20 − 20 = −1 + C giving C = 20 ln 20 − 19. Therefore, 20 ln |y| − y = 100 ln |x| − x + 20 ln 20 − 19.
We cannot solve for y in terms of x, so we leave the equation in this form.
p
p
R
R√
3
df
20. dx
x dx, so 2 f (x) = 32 x 2 + C. Since f (1) = 1, we have 2 = 32 + C so C = 34 .
= xf (x) gives √df =
19.
dy
dx
R
R
f (x)
21.
p
3
3
Thus, 2 f (x) = 23 x 2 + 43 , so f (x) = ( 31 x 2 + 23 )2 .
(Note: this is only defined for x ≥ 0.)
dy
dx
y
= ex−y giving ey dy = ex dx so ey = ex + C. Since y(0) = 1, we have e1 = e0 + C so C = e − 1. Thus,
e = ex + e − 1, so y = ln(ex + e − 1).
[Note: ex + e − 1 > 0 always.]
R
R
dy
dx
= ex+y = ex ey implies e−y dy = ex dx implies −e−y = ex + C. Since y = 0 when x = 1, we have
−1 = e + C, giving C = −1 − e. Therefore −e−y = ex − 1 − e and y = − ln(1 + e − ex ).
R
R
√
= 1 − z 2 sin θ implies √ dz 2 = ecos θ sin θ dθ implies arcsin z = −ecos θ + C. According to the
23. e− cos θ dz
dθ
22.
initial conditions: z(0) =
sin(−ecos θ + π6 + e).
R
1
,
2
R
so arcsin
1−z
1
=
2
−ecos 0 + C, therefore
π
6
= −e + C, and C =
π
6
+ e. Thus z =
y dy
dt
dt
1
24. (1+t2 )y dy
= 1−y implies that 1−y
= 1+t
−1 + 1−y
dy = 1+t
2 implies that
2 . Therefore −y−ln |1−y| =
dt
π
arctan t + C. y(1) = 0, so 0 = arctan 1 + C, and C = − 4 , so −y − ln |1 − y| = arctan t − π4 . We cannot solve for
y in terms of t.
R
R
R
25. We have
R
dy
= 2y sin3 t,
dt
so
Z
Using Integral Table Formula 17, gives
−
2−y dy =
Z
sin3 t dt.
1 −y
1
2
2 = − sin2 t cos t − cos t + C.
ln 2
3
3
According to the initial conditions: y(0) = 0 so
−
Thus,
−
2
1
= − + C,
ln 2
3
and
C=
2
1
−
.
3
ln 2
1 −y
1
2
2
1
2 = − sin2 t cos t − cos t + −
.
ln 2
3
3
3
ln 2
Solving for y gives:
2 ln 2
2 ln 2
ln 2
sin2 t cos t +
cos t −
+ 1.
3
3
3
It can be shown that the right side is always > 0, so we can take natural logs.
2−y =
y ln 2 = − ln
so
y=
− ln
ln 2
3
ln 2
3
sin2 t cos t +
2 ln 2
2 ln 2
cos t −
+1 ,
3
3
2
sin2 t cos t + 2 ln
cos t −
3
ln 2
2 ln 2
3
+1
.
SOLUTIONS to Review Problems for Chapter Eleven
1117
Problems
26. (a) To find equilibrium values, we solve
dP
=0
dt
0.025P − 0.00005P 2 = 0
0.025P (1 − 0.002P ) = 0
P = 0 and
P = 500.
The equilibrium values are P = 0 and P = 500.
(b) Between 0 and 500, P is increasing. Above 500, P is decreasing. We have:
(i) For P0 = 100, the population will increase to a limiting value of 500.
(ii) For P0 = 400, the population will increase to a limiting value of 500.
(iii) For P0 = 500, the population is at equilibrium and will stay at 500.
(iv) For P0 = 800, the population will decrease to a limiting value of 500.
27. (a) We know that the equilibrium solutions are the functions satisfying the differential equation whose derivative everywhere is 0. Thus we have
dy
=0
dt
0.2(y − 3)(y + 2) = 0
(y − 3)(y + 2) = 0.
(b)
The solutions are y = 3 and y = −2.
y
t
Figure 11.100
Looking at Figure 11.100, we see that the line y = 3 is an unstable solution, while the line y = −2 is a stable
solution.
28. Since f (x) is a solution to y ′ = xy − y, we know that
f ′ (x) = xf (x) − f (x).
Letting y = 2f (x), we want to find out if the left-hand side of this differential equation, y ′ , equals the right-hand side,
xy − y. First we consider the left-hand side:
y ′ = (2f (x))′
= 2f ′ (x)
= 2 (xf (x) − f (x)) since f ′ (x) = xf (x) − f (x)
= 2xf (x) − 2f (x).
1118
Chapter Eleven /SOLUTIONS
Turning to the right-hand side, we see that:
xy − y = x (2f (x)) − 2f (x)
because y = 2f (x)
= 2xf (x) − 2f (x),
We see that the left-hand side and the right-hand side both equal 2xf (x) − 2f (x), so y = 2f (x) is also a solution to the
equation.
29. Since y = f (x) is a solution to y ′ = xy − y, we know that
f ′ (x) = xf (x) − f (x).
Letting y = 2 + f (x), we want to find out if the left-hand side of this differential equation, y ′ , equals the right-hand side,
xy − y. First we consider the left-hand side:
y ′ = (f (x) + 2)′
= f ′ (x)
= xf (x) − f (x) since f ′ (x) = xf (x) − f (x).
Turning to the right-hand side, we see that:
xy − y = x (2 + f (x)) − (2 + f (x))
because y = 2 + f (x)
= xf (x) − f (x) + 2x − 2.
We see that the left-hand side, xf (x)−f (x), is not the same as the right-hand side, xf (x)−f (x)+2x−2, so y = 2+f (x)
is not a solution to the equation.
1
1
1
1
∆x =
= .
30. (a) (i) 1 step: ∆y =
(cos x)(cos y)
(cos 0)(cos 0) 2
2
Thus, using 1 step, we get ( 21 , 21 ) as our approximation.
(ii) 2 steps: ∆x = 14 .
x
y
∆y =
0
0
0.25
0.25
0.25
0.266
0.5
0.516
1
∆x
(cos x)(cos y)
Thus, using 2 steps, we get (0.5,0.516) as our approximation.
(iii) 4 steps: ∆x = 81
x
y
∆y =
0
0
0.125
0.125
0.125
0.127
0.25
0.252
0.133
0.375
0.385
0.145
0.5
0.530
1
∆x
(cos x)(cos y)
Thus, using 4 steps, we get (0.5,0.530) as our approximation.
(b) We have
1
dy
=
,
dx
(cos x)(cos y)
so
Z
cos y dy =
Z
dx
.
cos x
SOLUTIONS to Review Problems for Chapter Eleven
1119
The integral table gives
sin y =
(sin x) + 1
1
ln
+ C.
2
(sin x) − 1
Our curve passes through (0,0), so, 0 = 0 + C, and C = 0. Therefore
y = arcsin
(sin x) + 1
1
ln
2
(sin x) − 1
.
1
,
When x = 12 , y ≈ 0.549. Our answers in parts (a)-(c) are all underestimates. In each case, the error is about n+1
1
where n is the number of steps. We expect the error to be approximately proportional to n , so this seems reasonable.
31. (a) ∆x = 15 = 0.2.
At x = 0:
y0 = 1, y ′ = 4; so ∆y = 4(0.2) = 0.8. Thus, y1 = 1 + 0.8 = 1.8.
At x = 0.2:
y1 = 1.8, y ′ = 3.2; so ∆y = 3.2(0.2) = 0.64. Thus, y2 = 1.8 + 0.64 = 2.44.
At x = 0.4:
y2 = 2.44, y ′ = 2.56; so ∆y = 2.56(0.2) = 0.512. Thus, y3 = 2.44 + 0.512 = 2.952.
At x = 0.6:
y3 = 2.952, y ′ = 2.048; so ∆y = 2.048(0.2) = 0.4096. Thus, y4 = 3.3616.
At x = 0.8:
y4 = 3.3616, y ′ = 1.6384; so ∆y = 1.6384(0.2) = 0.32768. Thus, y5 = 3.68928. So y(1) ≈ 3.689.
y
(b)
5
x
5
Since solution curves are concave down for 0 ≤ y ≤ 5, and y(0) = 1 < 5, the estimate from Euler’s method
will be an overestimate.
(c) Z
Solving by separation:
Z
dy
=
dx, so − ln |5 − y| = x + C.
5−y
Then 5 − y = Ae−x where A = ±e−C . Since y(0) = 1, we have 5 − 1 = Ae0 , so A = 4.
Therefore, y = 5 − 4e−x , and y(1) = 5 − 4e−1 ≈ 3.528.
(Note: as predicted, the estimate in (a) is too large.)
(d) Doubling the value of n will probably halve the error and, therefore, give a value half way between 3.528 and 3.689,
which is approximately 3.61.
32. A continuous growth rate of 0.38% means that
1 dP
= 0.38% = 0.0038.
P dt
Separating variables and integrating gives
Z
dP
=
P
Z
0.0038 dt
P = P0 e0.0038t = (7.5 · 106 )e0.0038t .
1120
Chapter Eleven /SOLUTIONS
33. (a) Assuming that the world’s population grows exponentially, satisfying dP/dt = cP, and that the land in use for crops
is proportional to the population, we expect A to satisfy dA/dt = kA.
(b) We have A(t) = A0 ekt = 4.55 · 109 ekt , where t is the number of years after 1966. Since 30 years later the amount
of land in use is 4.93 billion hectares, we have
4.93 · 109 = (4.55 · 109 )ek(30) ,
so
e30k =
4.93
.
4.55
Solving for k gives
k=
ln (4.93/4.55)
= 0.00267.
30
Thus,
We want to find t such that
Taking logarithms gives
A = (4.55 · 109 )e0.00267t .
6 · 109 = A(t) = (4.55 · 109 )e0.00267t .
ln (6/4.55)
= 103.608 years.
0.00267
This model predicts land will have run out 104 years after 1966, that is by the year 2070.
t=
34. (a) Since the growth rate of the tumor is proportional to its size, we should have
dS
= kS.
dt
(b) We can solve this differential equation by separating variables and then integrating:
Z
dS
=
S
Z
k dt
ln |S| = kt + B
S = Cekt .
(c) This information is enough to allow us to solve for C:
5 = Ce0t
C = 5.
(d) Knowing that C = 5, this second piece of information allows us to solve for k:
8 = 5e3k
1
8
k = ln
≈ 0.1567.
3
5
So the tumor’s size is given by
S = 5e0.1567t .
35. (a) The rate of growth of the money in the account is proportional to the amount of money in the account. Thus
dM
= rM.
dt
(b) Solving, we have dM/M = r dt.
Z
Z
dM
=
r dt
M
ln |M | = rt + C
M = ert+C = Aert ,
When t = 0 (in 2010), M = 2000, so A = 2000 and M = 2000ert .
A = eC .
SOLUTIONS to Review Problems for Chapter Eleven
1121
(c) See Figure 11.101.
M
40000
M = 2000e0.10t
10000
M = 2000e0.05t
2000
t
t = 30
2040
t=0
2010
Figure 11.101
36. We know that
Rate at which quantity of
= −k(current quantity).
carbon-14 is changing
If Q is the quantity of carbon-14 at time t (in years),
Rate at which quantity is changing =
dQ
= −kQ.
dt
This differential equation has solution
Q = Q0 e−kt
where Q0 is the initial quantity. Since at the end of one year 9999 parts are left out of 10,000, we know that
9999 = 10,000e−k(1) .
Solving for k gives
Thus Q = Q0 e
−0.0001t
. See Figure 11.102.
k = ln 0.9999 ≈ 0.0001.
Q
Q0
Q = Q0 e−0.0001t
t
Figure 11.102: Exponential decay
37. Using (Rate balance changes) = (Rate interest is added)− (Rate payments are made), when the interest rate is i, we have
dB
= iB − 100.
dt
Solving this equation, we find:
100
dB
=i B−
dt
i
Z
Z
dB
=
i dt
B − 100
i
100
= it + C
i
100
= Aeit , where A = ±eC .
B−
i
ln B −
1122
Chapter Eleven /SOLUTIONS
At time t = 0 we start with a balance of $1000. Thus
= Ae0 , so A = 1000 − 100
.
1000 − 100
i
i
Thus B = 100
+ (1000 − 100
)eit .
i
i
When i = 0.05, B = 2000 − 1000e0.05t .
When i = 0.1, B = 1000.
When i = 0.15, B = 666.67 + 333.33e0.15t .
We now look at the graph in Figure 11.103 when i = 0.05, i = 0.1, and i = 0.15.
i = 0.15
B
i = 0.1
1000
i = 0.05
years
t ≈ 13.86
Figure 11.103
38. (a) c′ (t) = a(k − c(t)) where a > 0 is a constant.
(b)
Z
Z
dc
=
a dt
k−c
− ln |k − c| = at + C, C is a constant of integration
k − c = Ae−at
If c = c0 when t = 0, then k − c0 = A, so
k − c = (k − c0 )e−at
c = k + (c0 − k)e−at
(c) If c0 = 0, then c = k − ke−at = k(1 − e−at ).
concentration
k
c(t) = k(1 − e−at )
t (time)
39. (a) Use the fact that
Rate balance
changing
=
Rate interest
accrued
−
Rate payments
made
Thus
.
dB
= 0.05B − 12,000.
dt
(b) We solve the equation by separation of variables. First, however, we factor out a 0.05 on the right hand side of the
equation to make the work easier.
dB
= 0.05(B − 240000)
dt
dB
= 0.05 dt
B − 240000
Z
Z
dB
0.05 dt,
=
B − 240000
SOLUTIONS to Review Problems for Chapter Eleven
1123
so
ln |B − 240000| = 0.05t + C
|B − 240000| = e0.05t+C = e0.05t eC ,
so B − 240000 = Ae0.05t , where A = ±eC .
If the initial balance is B0 , then B0 − 240000 = Ae0 = A, thus B − 240000 = (B0 − 240000)e0.05t , so
B = (B0 − 240000)e0.05t + 240000.
(c) To find the initial balance such that the account has a 0 balance after 20 years, we solve
0 = (B0 − 240,000)e(0.05)20 + 240,000 = (B0 − 240,000)e1 + 240,000,
B0 = 240,000 −
240,000
≈ $151,708.93.
e
40. (a) Since the rate of change is proportional to the amount present, dy/dt = ky for some constant k.
(b) Solving the differential equation, we have y = Aekt , where A is the initial amount. Since 100 grams become 54.9
grams in one hour, 54.9 = 100ek , so k = ln(54.9/100) ≈ −0.5997.
Thus, after 10 hours, there remains 100e(−0.5997)10 ≈ 0.2486 grams.
41. Let C(t) be the current flowing in the circuit at time t, then
dC
= −αC
dt
where α > 0 is the constant of proportionality between the rate at which the current decays and the current itself.
The general solution of this differential equation is C(t) = Ae−αt but since C(0) = 30, we have that A = 30, and
so we get the particular solution C(t) = 30e−αt .
When t = 0.01, the current has decayed to 11 amps so that 11 = 30e−α0.01 which gives α = −100 ln(11/30) =
100.33 so that,
C(t) = 30e−100.33t .
42. Since the rate at which the volume, V , is decreasing is proportional to the surface area, A, we have
dV
= −kA,
dt
4
where the negative sign reflects the fact that V is decreasing. Suppose the radius of the sphere is r. Then V = πr 3 and,
3
dV
dr
using the chain rule,
= 4πr 2 . The surface area of a sphere is given by A = 4πr 2 . Thus
dt
dt
4πr 2
dr
= −k4πr 2
dt
so
dr
= −k.
dt
Since the radius decreases from 1 cm to 0.5 cm in 1 month, we have k = 0.5 cm/month. Thus
dr
= −0.5
dt
so
Since r = 1 when t = 0, we have r0 = 1, so
r = −0.5t + r0 .
r = −0.5t + 1.
We want to find t when r = 0.2, so
and
0.2 = −0.5t + 1
t=
0.8
= 1.6 months.
0.5
1124
Chapter Eleven /SOLUTIONS
43. By rewriting the equation, we see that it is logistic:
(100 − P )
1 dP
=
.
P dt
1000
Before looking at its solution, we explain why there must always be at least 100 individuals. Since the population begins
at 200, the quantity dP/dt is initially negative, so the population initially decreases. It continues to do so while P > 100.
If the population ever reached 100, then dP/dt would be 0. This would mean the population stopped changing—so if the
population ever decreased to 100, that’s where it would stay. The fact that dP/dt is always negative for P > 100 also
shows that the population is always under 200, as shown in Figure 11.104.
P
200
100
t
Figure 11.104
The solution, as given by the formula derived in the chapter, is
P =
100
1 − 0.5e−0.1t
44. (a) The logistic model is a reasonable one because in the 1990s, very few households had DVD players. Since DVD
disc media is more convenient than VHS tape media, as DVD players dropped in price, more people bought them.
However, the rate of DVD-player ownership had to start slowing down at some point as it is impossible for more than
100% of households to have DVD players.
(b) To find the point of inflection, we find the year at which the increase in DVD-player ownership alters from increasing
to decreasing. The following table from 1998 to 2006.
Year
1998-99
1999-2000
2000-01
2001-02
2002-03
2003-04
2004-05
2005-06
Change in %
4
8
8
14
15
20
5
6
Looking at the table, we see that the change in percent alters from increasing to decreasing in about 2003. At this
time 50% of households own DVD players. The inflection point is approximately (2003, 50). Since at the inflection
point on a logistic curve, the vertical coordinate is L/2, we have
L/2 = 50
L = 100.
Thus the limiting value is about 100%. Ownership may be expected to rise until the next new media format replaces
it.
(c) Since the general form of a logistic function is
P =
L
1 + Ae−kt
where L is the limiting value, in this case we have L = 86.395, predicting a limiting value of 86.395%.
45. (a) Quantity of A present at time t equals (a − x).
Quantity of B present at time t equals (b − x).
So
Rate of formation of C = k(Quantity of A)(Quantity of B)
gives
dx
= k(a − x)(b − x)
dt
SOLUTIONS to Review Problems for Chapter Eleven
1125
(b) Separating gives
Z
dx
=
(a − x)(b − x)
Z
k dt.
Rewriting the denominator as (a − x)(b − x) = (x − a)(x − b) enables us to use Formula 26 in the Table of Integrals
provided a 6= b. For some constant K, this gives
1
(ln |x − a| − ln |x − b|) = kt + K.
a−b
Thus
ln
x−a
= (a − b)kt + K(a − b)
x−b
x−a
= eK(a−b) e(a−b)kt
x−b
x−a
= M e(a−b)kt where M = ±eK(a−b) .
x−b
Since x = 0 when t = 0, we have M =
a
.
b
Thus
x−a
a
= e(a−b)kt .
x−b
b
Solving for x, we have
x(b − ae
bx − ba = ae(a−b)kt (x − b)
(a−b)kt
) = ab − abe(a−b)kt
x=
ab(ebkt − eakt )
ab(1 − e(a−b)kt )
=
.
(a−b)kt
bebkt − aeakt
b − ae
46. Quantity of A left at time t = Quantity of B left at time t equals (a − x).
Thus
Rate of formation of C = k(Quantity of A)(Quantity of B)
gives
dx
= k(a − x)(a − x) = k(a − x)2 .
dt
Separating gives
Integrating gives, for some constant K,
Z
dx
=
(x − a)2
Z
k dt
−(x − a)−1 = kt + K.
When t = 0, x = 0 so K = a−1 . Solving for x:
−(x − a)−1 = kt + a−1
1
x−a = −
kt + a−1
a2 kt
a
=
x = a−
akt + 1
akt + 1
47. (a) If alone, the x population grows exponentially, since if y = 0 we have dx/dt = 0.01x. If alone, the y population
decreases to 0 exponentially, since if x = 0 we have dy/dt = −0.2y.
(b) This is a predator-prey relationship: interaction between populations x and y decreases the x population and increases
the y population. The interaction of lions and gazelles might be modeled by these equations.
48. (a) If alone, the x and y populations each grow exponentially, because the equations become dx/dt = 0.01x and
dy/dt = 0.2y.
(b) For each population, the presence of the other decreases their growth rate. The two populations are therefore competitors—
they may be eating each other’s food, for instance. The interaction of gazelles and zebras might be modeled by these
equations.
1126
Chapter Eleven /SOLUTIONS
49. (a) The x population is unaffected by the y population—it grows exponentially no matter what the y population is, even
if y = 0. If alone, the y population decreases to zero exponentially, because its equation becomes dy/dt = −0.1y.
(b) Here, interaction between the two populations helps the y population but does not effect the x population. This is not
a predator-prey relationship; instead, this is a one-way relationship, where the y population is helped by the existence
of x’s. These equations could, for instance, model the interaction of rhinoceroses (x) and dung beetles (y).
50. (a) See Figure 11.105.
(b) The two equilibrium values are P = 0 and P = 100. Given any positive initial condition, the shrimp population will
level off at 100 tons of shrimp in the bay.
(c) The new differential equation is
dP
= 0.8P (1 − 0.01P ) − 10.
dt
(d) Notice that subtracting 10 just moves the graph of dP/dt against P down 10 units. See Figure 11.106.
(e) We see in Figure 11.106 that the equilibrium values are at approximately P = 14.6 and P = 85.4.
(f) We see in Figure 11.106 that if P = 12, then dP/dt is negative. The shrimp population will decrease from P = 12.
If P = 25 or P = 75, we see in Figure 11.106 that dP/dt is positive, so the shrimp populations will increase from
either of these populations.
dP/dt
dP/dt
20
20
10
10
−20
−10
P
20 40 60 80 100
−20
20
40
60
80
100
P
−10
Figure 11.105
Figure 11.106
51. (a) We have
dy
dt
dx
dt
=
−3y − xy
y(x + 3)
dy
=
=
.
dx
−2x − xy
x(y + 2)
y+2
y
Thus,
so
Z
So,
dy =
2
(1 + ) dy =
y
Z
x+3
dx
x
(1 +
3
) dx.
x
y + 2 ln |y| = x + 3 ln |x| + C.
Since x and y are non-negative,
y + 2 ln y = x + 3 ln x + C.
This is as far as we can go with this equation – we cannot solve for y in terms of x, for example. We can, however,
put it in the form
ey+2 ln y = ex+3 ln x+C , or y 2 ey = Ax3 ex .
(b) An equilibrium state satisfies
dx
= −2x − xy = 0 and
dt
dy
= −3y − xy = 0.
dt
Solving the first equation, we have
−x(y + 2) = 0,
so x = 0 or
The second equation has solutions
y = 0 or
Since x, y ≥ 0, the only equilibrium point is (0, 0).
x = −3.
y = −2.
SOLUTIONS to Review Problems for Chapter Eleven
(c) We can use either of our forms for the solution. Looking at
y 2 ey = Ax3 ex ,
we see that if x and y are very small positive numbers, then
ex ≈ ey ≈ 1.
Thus,
y 2 ≈ Ax3 ,
or
y2
≈ A, a constant.
x3
Looking at
y + 2 ln y = x + 3 ln x + C,
we note that if x and y are small, then they are negligible compared to ln y and ln x. Thus,
2 ln y ≈ 3 ln x + C,
giving
ln y 2 − ln x3 ≈ C,
so
ln
and therefore
y2
≈C
x3
y2
≈ eC , a constant.
x3
(d) If
x(0) = 4
and y(0) = 8,
then
8 + 2 ln 8 = 4 + 3 ln 4 + C.
Note that
2 ln 8 = 3 ln 4 = ln 64,
giving
4 = C.
So the phase trajectory is
y + 2 ln y = x + 3 ln x + 4.
2 y
4 3 x
3 x+4
(Or equivalently, y e = e x e = x e
(e) If the concentrations are equal, then
.)
y + 2 ln y = y + 3 ln y + 4,
giving
−4
− ln y = 4
or
y = e−4 .
Thus, they are equal when y = x = e ≈ 0.0183.
(f) Using part (c), we have that if x is small,
y2
≈ e4 .
x3
Since x = e−10 is certainly small,
y2
e−30
≈ e4 ,
and
y ≈ e−13 .
52. (a) Using the chain-rule we get
dy/dt
dy
=
dx
dx/dt
giving
so
−ax
x
dy
=
= .
dx
−ay
y
Z
y dy =
Z
x dx,
1127
1128
Chapter Eleven /SOLUTIONS
giving
y 2 − x2 = C.
Figure 11.107 shows the graph of the slope field of this equation with the solution satisfying y = 46 when x = 40
sketched in.
When t = 0, y = 46, x = 40, so C = 462 − 402 = 516. Therefore, the solution is y 2 − x2 = 516.
(b) The battle is over when x = 0. The number of French/Spanish ships remaining is the y-intercept, thus, y 2 −02 = 516
giving y ≈ 22.7 French/Spanish ships remaining.
(c) For both sub-battles, the solution trajectory will have the form y 2 − x2 = C. Each sub-battle will have a different
value of C.
For the 32 versus 23 sub-battle, C = 232 − 322 = −495, so the trajectory is:
y 2 − x2 = −495
or
x2 − y 2 = 495.
√
This has no y-intercept, but an x-intercept of x = 495 ≈ 22.2, meaning that the model predicts the British won
the sub-battle with about 22.2 ships remaining.
For the 8 versus 23 sub-battle,
C = 232 − 82 = 465 so the trajectory is y 2 − x2 = 465. This has no x-intercept,
√
but a y-intercept of y = 465 ≈ 21.6, meaning that the model predicts a French/Spanish victory with about 21.6
ships remaining.
(d) If the remaining ships from these two sub-battles then fight each other, the British have a slight advantage (22.2 versus
21.6). Thus the British could be expected to win the overall battle, although they started with a weaker fleet. This is
in fact what happened.
The trajectory for this last battle has C = 495 − 460 = 30, so the equation is
This has an x-intercept of x =
√
x2 − y 2 = 30.
30 ≈ 5.5, so the model predicts a British victory with about 5 21 ships remaining.
(French/Spanish)
(40, 46)
Direct confrontation
✲
40
2nd sub-battle
30
22.7
21.6
❄
(22.2, 21.6)
(32, 23)
✛
(8, 23)
✛
1st sub-battle
Final battle
10
(British)
5.5 10
20 22.2 30
40
Figure 11.107
53. (a) The insects grow exponentially with no birds around (the equation becomes dx/dt = 3x); the birds die out exponentially with no insects to feed on dy/dt = −10y). The interaction increases the birds’ growth rate (the +0.001xy
term is positive), but decreases the insects’ growth rate (the −0.02xy term is negative). This is as we would expect:
having the insects around helps the birds; having birds around hurts the insects.
(b) Equilibrium solutions occur where both derivatives are zero:
(3 − 0.02y)x = 0
−(10 − 0.001x)y = 0.
We see that the solutions are (0, 0) and (10, 000, 150)
SOLUTIONS to Review Problems for Chapter Eleven
1129
(c) The chain rule gives an equation for dy/dx:
dy/dt
y(−10 + 0.001x)
dy
=
=
.
dx
dx/dt
x(3 − 0.02y)
Separation of variables gives
Z
−10 + 0.001x
dx =
x
which yields 3 ln y − 0.02y = −10 ln x + 0.001x + C.
Using the initial point A = (10, 000, 160), we have
Z
3 − 0.02y
dy,
y
3 ln 160 − 0.02(160) = −10 ln 10,000 + 0.001(10,000) + C.
Thus, C ≈ 94.13 and the solution is 3 ln y − 0.02y = −10 ln x + 0.001x + 94.1
(d) We can check that the equation is satisfied by points B, C, D by substituting the coordinates into the equation
3 ln y − 0.02y = −10 ln x + 0.001x + 94.1.
(e) The phase plane and the trajectory is in Figure 11.108.
(f) Consider point A. We have
dy
= 0,
dt
and
dx
= 3(10, 000) − 0.02(10, 000)(160) = −2000 < 0.
dt
Thus x is decreasing at point A. Hence the rotation is counterclockwise in the phase plane, and the order of traversal
is A −→ B −→ C −→ D.
(g) The graphs of x and y versus t are in Figure 11.109.
y
160
y
A
x (in hundreds)
100
B
✐
D
equilibrium point (10000, 150)
C
t
x
Figure 11.108
Figure 11.109
(h) At points A and C, we have dy/dx = 0, and at B and D, we have dx/dy = 0, so these points are extrema: y is
maximized at A, minimized at C; x is maximized at D, minimized at B.
54. (a) In the equation for dx/dt, the term involving x, namely −0.2x, is negative meaning that as x increases, dx/dt
decreases. This corresponds to the statement that the more a country spends on armaments, the less it wants to
increase spending.
On the other hand, since +0.15y is positive, as y increases, dx/dt increases, corresponding to the fact that the
more a country’s opponent arms, the more the country will arm itself.
The constant term, 20, is positive means that if both countries are unarmed initially, (so x = y = 0), then dx/dt
is positive and so the country will start to arm. In other words, disarmament is not an equilibrium situation in this
model.
(b) The nullclines are shown in Figure 11.110. When dx/dt = 0, the trajectories are vertical (on the line −0.2x +
0.15y + 20 = 0); when dy/dt = 0 the trajectories are horizontal (on 0.1x − 0.2y + 40 = 0). There is only one
equilibrium point, x = y = 400.
1130
Chapter Eleven /SOLUTIONS
(c) In region I, try x = 400, y = 0, giving
dx
= −0.2(400) + 0.15(0) + 20 < 0
dt
dy
= 0.1(400) − 0.2(0) + 4 − 0 > 0
dt
In region II, try x = 500, y = 500, giving
dx
= −0.2(500) + 0.15(500) + 20 < 0
dt
dy
= 0.1(500) − 0.2(500) + 40 < 0
dt
In region III, try x = 0, y = 400, giving
dx
= −0.2(0) + 0.15(400) + 20 > 0
dt
dy
= 0.1(0) − 0.2(400) + 40 < 0
dt
In region IV, try x = 0, y = 0, giving
dx
= −0.2(0) + 0.15(0) + 20 > 0
dt
dy
= 0.1(0) − 0.2(0) + 40 > 0
dt
See Figure 11.110.
(d) The one equilibrium point is stable.
y
(billion $)
600
−0.2x + 0.15y + 20 = 0o
dx/dt = 0
Region III
✲
❄
❘
Region II
✛
✠ ❄
❘
✛
400
n
0.1x − 0.2y + 40 = 0
dy/dt = 0
Region I
200
Region IV
■
✛ ✻
✻
✒
✲
100
400
600
x (billion $)
Figure 11.110: Nullclines and equilibrium point(dot) for arms race
(e) If both sides disarm, then both sides spend $0. Thus initially x = y = 0, and dx/dt = 20 and dy/dt = 40. Since
both dx/dt and dy/dt are positive, both sides start arming. Figure 11.110 shows that they will both arm until each is
spending about $400 billion.
(f) If the country spending $y billion is unarmed, then y = 0 and the corresponding point on the phase plane is on the
x-axis. Any trajectory starting on the x-axis tends toward the equilibrium point x = y = 400. Similarly, a trajectory
starting on the y-axis represents the other country being unarmed; such a trajectory also tends to the same equilibrium
point.
Thus, if either side disarms unilaterally, that is, if we start out with one of the countries spending nothing, then
over time, they will still both end up spending roughly $400 billion.
(g) This model predicts that, in the long run, both countries will spend near to $400 billion, no matter where they start.
SOLUTIONS to Review Problems for Chapter Eleven
1131
55. If 0 < P < L, then P/L < 1 and P/(2L) < 1:
dP
P
= − |{z}
k
1−
dt
L
+
| {z } |
+
1−
P
2L
{z
< 0.
}
+
Thus, if initially there are fewer than L animals, dP/ dt < 0 and the population will decrease.
56. If L < P < 2L, then P/L > 1 and P/(2L) < 1:
P
dP
1−
= − |{z}
k
dt
L
+
| {z } |
−
1−
P
2L
{z
> 0.
}
+
Thus, if initially there are between L and 2L animals, dP/ dt > 0 and the population will increase. Since P = 2L is an
equilibrium solution, the population will increase towards P = 2L.
57. If P > 2L, then P/L > 1 and P/(2L) > 1:
P
dP
1−
= − |{z}
k
dt
L
+
| {z } |
−
1−
P
2L
{z
< 0.
}
−
Thus, if initially there are more than 2L animals, dP/ dt < 0 and the population will decrease. Since P = 2L is an
equilibrium solution, the population will decrease towards P = 2L.
CAS Challenge Problems
58. (a) We find the equilibrium solutions by setting dP/dt = 0, that is, P (P − 1)(2 − P ) = 0, which gives three solutions,
P = 0, P = 1, and P = 2.
(b) To get your computer algebra system to check that P1 and P2 are solutions, substitute one of them into the equation
and form an expression consisting of the difference between the right and left hand sides, then ask the CAS to simplify
that expression. Do the same for the other function. In order to avoid too much typing, define P1 and P2 as functions
in your system.
(c) Substituting t = 0 gives
1
P1 (0) = 1 − √ = 1/2
4
1
P2 (0) = 1 + √ = 3/2.
4
We can find the limits using a computer algebra system. Alternatively, setting u = et , we can use the limit laws to
calculate
u
et
= lim √
= lim
lim √
u→∞
u→∞
t→∞
3 + e2t
3 + u2
=
Therefore, we have
r
lim
u→∞
v
u
=u
t
u2
=
3 + u2
r
1
=
3
lim 2 + 1
u→∞ u
r
lim
u→∞
r
u2
3 + u2
3
u2
1
+1
1
= 1.
0+1
lim P1 (t) = 1 − 1 = 0
t→∞
lim P2 (t) = 1 + 1 = 2.
t→∞
To predict these limits without having a formula for P , looking at the original differential equation. We see if 0 <
P < 1, then P (P − 1)(2 − P ) < 0, so P ′ < 0. Thus, if 0 < P (0) < 1, then P ′ (0) < 0, so P is initially decreasing,
and tends toward the equilibrium solution P = 0. On the other hand, if 1 < P < 2, then P (P − 1)(2 − P ) > 0, so
P ′ > 0. So, if 1 < P (0) < 2, then P ′ (0) > 0, so P is initially increasing and tends toward the equilibrium solution
P = 2.
1132
Chapter Eleven /SOLUTIONS
59. (a) Using the integral equation with n + 1 replaced by n, we have
yn (a) = b +
Z
a
(yn−1 (t)2 + t2 ) dt = b + 0 = b.
a
(b) We have a = 1 and b = 0, so the integral equation tells us that
yn+1 (s) =
Z
s
(yn (t)2 + t2 ) dt.
1
With n = 0, since y0 (s) = 0, the CAS gives
y1 (s) =
Z
s
0 + t2 dt = −
1
Then
y2 (s) =
and
y3 (s) =
Z
Z
s
1
(y1 (t)2 + t2 ) dt = −
1
s3
+ .
3
3
s
s3
s4
s7
17
+ +
−
+
,
42
9
3
18
63
s
(y2 (t)2 + t2 ) dt
1
157847
289 s
17 s2
82 s3
17 s4
s5
s6
s7
11 s8
+
−
+
−
+
−
+
−
+
374220
1764
378
243
252
42
486
63
1764
2 s11
s12
s15
5 s9
+
−
+
.
6804
2079
6804
59535
=−
(c) The solution y, and the approximations y1 , y2 , y3 are graphed in Figure 11.111. The approximations appear to be
accurate on the range 0.5 ≤ s ≤ 1.5.
y3 (s)
y
y(s)
3
✠
y2 (s)
y1 (s)
2
1
x
1
−0.5
2
Figure 11.111
60. (a) See Figure 11.112.
y
− π2
π
2
Figure 11.112
x
PROJECTS FOR CHAPTER ELEVEN
1133
(b) Different CASs give different answers, for example they might say y = sin x, or they might say
y = sin x,
−
π
π
≤x≤ .
2
2
(c) Both the sample CAS answers in part (b) are wrong. The first one, y = sin x, is wrong because sin x starts decreasing
at x = π/2, where the slope field clearly shows that y should be increasing at all times. The second answer is better,
but it does not give the solution outside the range −π/2 ≤ x ≤ π/2. The correct answer is the one sketched in
Figure 11.112, which has formula
y=
−1
sin x
1
x ≤ − π2
− π2 ≤ x ≤
x≤
π
2
π
2
≤ x.
PROJECTS FOR CHAPTER ELEVEN
1. Assuming dTr /dt is proportional to Tr , we have:
dTr
= −kTr
dt
where k is a positive constant of proportionality.
The solution to this differential equation is the exponential decay function,
Tr = Ae−kt .
The solution was derived by separating variables:
Z
Z
1
dTr = −k dt
Tr
ln |Tr | = −kt + C
|Tr | = eC e−kt
Tr = Ae−kt .
where |A| = eC .
We find the values of A and k from the data in Table 11.12.We know that:
so
Ae−k·4 = 37
since Tr = 37 at t = 4
Ae−k·19.5 = 13
since Tr = 13 at t = 19.5
Ae−k·4
37
dividing
=
Ae−k·19.5
13
37
e15.5k =
13
1
37
k=
ln
= 0.06748.
15.5
13
Having found k, we can now find A:
since Tr = 37 at t = 4
Ae−0.06748(4) = 37
A = 48.5 ng/ml.
Thus, the concentration of tryptase t hours after surgery is modeled by the exponential decay function
Tr = 48.5e−0.06748t ng/ml.
The greatest concentration is 48.5 ng/ml, which occurs at t = 0 when the patient leaves surgery. We can
diagnose anaphylaxis since this peak level is above 45 ng/ml.
1134
Chapter Eleven /SOLUTIONS
2. (a) (i) Integrating we have
dP
= 30.2
dt
P = 30.2t + C.
Since P (0) = 95, we have C = 95, so
P = 30.2t + 95.
Substituting t = 87 and rounding to the nearest person gives
P = 30.2 · 87 + 95 = 2722.
The linear model predicts that 2722 people would have contracted SARS by June 12, 2003.
(ii) Separating variables, we have
1 dP
= 0.12
P
Z
Z dt
dP
= 0.12 dt
P
ln |P | = 0.12t + C
P = Ae0.12t .
Since P (0) = 95, we have A = 95, so
P = 95e0.12t .
Substituting t = 87 and rounding to the nearest person gives
P = 95e0.12·87 = 3, 249, 062.
The exponential model predicts that 3.249 million people would have contracted SARS by June 12,
2003.
(iii) Writing the differential equation in the form
dP
= 0.19P − 0.0002P 2
dt
0.0002
dP
= 0.19P 1 −
P
dt
0.19
dP
P
= 0.19P 1 −
,
dt
950
we use the analytic solution derived on page 630 of the text to obtain
P =
950
,
1 + Ae−0.19t
with A =
950 − 95
=9
95
so
950
.
1 + 9e−0.19t
Substituting t = 87 and rounding to the nearest person gives
P =
P =
950
= 950.
1 + 9e−0.19·87
The logistic model predicts that 950 people will have contracted SARS by June 12, 2003.
PROJECTS FOR CHAPTER ELEVEN
1135
(b) (i) The three methods give very different predictions. The linear and logistic are about 3000 and 1000,
respectively, while the exponential model is 3 million, nearly half the population of Hong Kong.
(ii) The number of new cases per day is approximated by the derivative, dP /dt. The linear model predicts
a constant number of new cases each day; the exponential model predicts an increasing number of new
cases each day; the logistic model predicts that the number of new cases per day will first increase
and then decrease.
(iii) The general trend in the figure shows that the number of new cases per day first climbed and then fell,
suggesting that the logistic model fits best. The high values are largely Mondays, and represent two
days of data recorded as one, since no new cases were reported on Sundays.
(c) (i) The formula
950
P =
1 + 9e−0.19t
has limiting value P = 950 as t → ∞. Thus, this formula predicts that the maximum number of cases
expected is 950.
(ii) The graph allows us to estimate (very roughly) when the daily increase was largest, namely about
April 10. Since the maximum rate of change of P (and the maximum daily increase in P ) occurs at
L/2, where L is the maximum value of P , we expect the maximum value of P to be about 2 · 998 ≈
2000.
(d) See Figures 11.113–11.115. The dots represent the actual data.
P
P
P = 30.2t + 95
2000
P = 95e0.12t
6
3 · 10
2 · 106
1 · 106
1000
t (days since
100 March 17)
50
Figure 11.113: Linear predictions and actual data
50
100
t (days since
March 17)
Figure 11.114: Exponential predictions and actual
data
P
2000
P =
1000
50
100
950
1 + 9e−0.19t
t (days since
March 17)
Figure 11.115: Logistic predictions and actual data
3. (a) Since I0 is the number of infecteds on day t = 0, March 17, we have I0 = 95. Since S0 is the initial
number of susceptibles, which is the whole population of Hong Kong, S0 ≈ 6.8 million.
(b) For a = 1.25 · 10−8 and b = 0.06, the system of equations is
dS
= −1.25 · 10−8 SI
dt
dI
= 1.25 · 10−8 SI − 0.06I.
dt
So, by the chain rule,
dI/dt
1.25 · 10−8 SI − 0.06I
4.8 · 106
dI
=
=
= −1 +
.
−8
dS
dS/dt
−1.25 · 10 SI
S
1136
Chapter Eleven /SOLUTIONS
The slope field and trajectory are in Figure 11.116.
I (infecteds)
0.4 · 106
0.3 · 106
0.2 · 106
(6.8 · 106 , 95)
0.1 · 106
✠
6
S (susceptibles)
5 · 10
Figure 11.116
(c) The maximum value of I is about 300,000; this gives us the maximum number of infecteds at any one
time—the total number of people infected during the course of the disease is much greater than this. The
trajectory meets the S-axis at about 3.3 million; this tells us that when the disease dies out, there are still
3.3 million susceptibles who have never had the disease. Therefore 6.8 − 3.3 = 3.5 million people are
predicted to have had the disease.
The threshold value of S occurs where dI/dt = 0 and I 6= 0, so, for a = 1.25 · 10−8 and b = 0.06,
dI
= 1.25 · 10−8 SI − 0.06I = 0,
dt
giving
0.06
= 4.8 · 106 people.
1.25 · 10−8
The threshold value tells us that if the initial susceptible population, S0 is more than 4.8 million, there will
be an epidemic. If S0 is less than 4.8 million, there will not be an epidemic. Since the population of Hong
Kong is over 4.8 million, an epidemic is predicted.
(d) The value of b represents the rate at which infecteds are removed from circulation. Quarantine increases
the rate people are removed and thus increases b.
(e) For a = 1.25 · 10−8 and b = 0.24, the system of differential equations is
Threshold value = S =
dS
= −1.25 · 10−8 SI
dt
dI
= 1.25 · 10−8 SI − 0.24I.
dt
So, by the chain rule,
dI/dt
1.25 · 10−8 SI − 0.24I
19.2 · 106
dI
=
=
=
−1
+
.
dS
dS/dt
−1.25 · 10−8 SI
S
The slope field is in Figure 11.117. The solution trajectory does not show as the disease dies out right
away.
I (infecteds)
0.4 · 106
0.3 · 106
0.2 · 106
0.1 · 106
5 · 106
Figure 11.117
S (susceptibles)
PROJECTS FOR CHAPTER ELEVEN
1137
(f) The threshold value of S occurs where dI/dt = 0 and I 6= 0, so, for b = 0.24 and the same value of a,
dI
= 1.25 · 10−8 SI − 0.24I = 0,
dt
giving
0.24
= 19.2 · 106 people.
1.25 · 10−8
The threshold value tells us that if S0 is less than 19.2 million, there will be no epidemic. The population
of Hong Kong is 6.8 million, so S0 is below this value. Thus no epidemic is predicted.
Policies, such as quarantine, which raise the value of b can be effective in preventing an epidemic. In
this case, the value of b increased sufficiently that the population of Hong Kong fell below the threshold
value, and a potential epidemic was averted. However, we do not have evidence that the quarantine policy
was responsible for the increase in b.
(g) Policy I: Closing off the city changes the initial values of S0 and I0 but not the values of a and b. If not
one infected person enters the city, then I0 = 0 and the solution trajectory is an equilibrium point on the
S-axis. However, in practice it is almost impossible to cut off a city completely, so usually I0 > 0. Also,
by the time a policy to close off a city is put into effect, there may already be infected people inside the
city, so again I0 > 0. Thus, whether or not there is an epidemic depends on whether S0 is greater than the
threshold value, not on the value of I0 (provided I0 > 0).
For example, in the case of Hong Kong with the March values of a and b, changing the value of I0 to 1
leaves the solution trajectory much as before; see Figure 11.118. The main difference is that the epidemic
occurs slightly later. So a policy of isolating a city only works if it keeps the disease out of the city of the
city entirely. Thus, Policy I does not help the city except in the exceptional case that every infected person
is kept out.
Threshold value = S =
I (infecteds)
0.4 · 106
0.3 · 106
0.2 · 106
(6.8 · 106 , 1)
0.1 · 106
✠
6
S (susceptibles)
5 · 10
Figure 11.118
Policy II: From the analysis of the Hong Kong data, we see that a quarantine policy can help prevent
an epidemic if the value of b is increased enough to bring S0 below the threshold value. Thus, Policy II
can be very effective.
4. (a)
p(x) = the number of people with incomes ≥ x.
p(x + ∆x) = the number of people with incomes ≥ x + ∆x.
So the number of people with incomes between x and x + ∆x is
p(x) − p(x + ∆x) = −∆p.
Since all the people with incomes between x and x + ∆x have incomes of about x (if ∆x is small),
the total amount of money earned by people in this income bracket is approximately x(−∆p) = −x∆p.
1138
Chapter Eleven /SOLUTIONS
(b) Pareto’s law claims that the average income of all the people with incomes ≥ x is kx. Since there are p(x)
people with income ≥ x, the total amount of money earned by people in this group is kxp(x).
The total amount of money earned by people with incomes ≥ (x + ∆x) is therefore k(x + ∆x)p(x +
∆x). Then the total amount of money earned by people with incomes between x and x + ∆x is
kxp(x) − k(x + ∆x)p(x + ∆x).
Since ∆p = p(x + ∆x) − p(x), we can substitute p(x + ∆x) = p(x) + ∆p. Thus the total amount of
money earned by people with incomes between x and x + ∆x is
kxp(x) − k(x + ∆x)(p(x) + ∆p).
Multiplying out, we have
kxp(x) − kxp(x) − k(∆x)p(x) − kx∆p − k∆x∆p
Simplifying and dropping the second order term ∆x∆p gives the total amount of money earned by people
with incomes between x and x + ∆x as
−kp∆x − kx∆p.
(c) Setting the answers to parts (a) and (b) equal gives
−x∆p = −kp∆x − kx∆p.
Dividing by ∆x, and letting ∆x → 0 so that
x
∆p
∆x
→ p′ , we have
∆p
∆p
= kp + kx
∆x
∆x
xp′ = kp + kxp′
so
(1 − k)xp′ = kp.
(d) We solve this equation by separating variables
Z
Z
k
dx
dp
=
p
(1 − k) x
k
ln x + C (no absolute values needed since p, x > 0)
ln p =
(1 − k)
ln p = ln xk/(1−k) + ln A (writing C = ln A)
ln p = ln[Axk/(1−k) ] (using ln(AB) = ln A + ln B)
p = Axk/(1−k)
(e) We take A = 1. For k = 10, p = x−10/9 ≈ x−1 . For k = 1.1, p = x−11 . The functions are graphed in
Figure 11.119. Notice that the larger the value of k, the less negative the value of k/(1 − k) (remember
k > 1), and the slower p(x) → 0 as x → ∞.
p(x)
✛
x−10/9 (k = 10)
x−11 (k = 1.1)
1
✠
1
Figure 11.119
2
x
PROJECTS FOR CHAPTER ELEVEN
5. (a) Writing F = b
3
a2 −ar
r3
1139
= 0 shows F = 0 when r = a, so r = a gives the equilibrium position.
(b) Expanding 1/r about r = a gives
1
1
1
=
= 3
r3
(a + r − a)3
a
=
1
a3
=
1
a3
−3
r−a
1+
a
!
2
r−a
(−3)(−4) r − a
1−3
+
− ···
a
2!
a
3(r − a) 6(r − a)2
+
− ··· .
1−
a
a2
Similarly, expanding 1/r2 about r = a gives
−2
r−a
1
1
1
=
= 2 1+
r2
(a + r − a)2
a
a
!
2
1
r−a
(−2)(−3) r − a
= 2 1−2
− ···
+
a
a
2!
a
!
2
1
r−a
r−a
= 2 1−2
+3
− ··· .
a
a
a
Thus, combining gives
3(r − a) 6(r − a)2
1
2(r − a) 3(r − a)2
1
1−
+
−
·
·
·
−
1
−
+
−
·
·
·
F =b
a
a
a2
a
a
a2
(r − a) 3(r − a)2
b
−
+
− ···
=
a
a
a2
b
3(r − a)2
= 2 −(r − a) +
− ··· .
a
a
(c) Setting x = r − a gives
F ≈
b
a2
3x2
−x +
.
a
(d) For small x, we discard the quadratic term in part (c), giving
F ≈
−b
x.
a2
The acceleration is d2 x/dt2 . Thus, using Newton’s Second Law:
Force = Mass · Acceleration
we get
d2 x
−bx
=
m
.
a2
dt2
So
d2 x
b
+ 2 x = 0.
dt2
a m
This differential equation
represents an oscillation of the form x = C1 cos ωt + C2 sin ωt, where ω 2 =
p
2
2
b/(a m) so ω = b/(a m). Thus, we have
r
2π
m
Period =
= 2πa
.
ω
b
12.1 SOLUTIONS
1141
CHAPTER TWELVE
Solutions for Section 12.1
Exercises
√
√
√
1. The point P√is 12 + 22 + 12 = 6 = 2.45 units from the origin, and Q is 22 + 02 + 02 = 2 units from the origin.
Since 2 < 6, the point Q is closer.
p
2. The distance formula: d =
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 gives us the distance between any pair of points
(x1 , y1 , z1 ) and (x2 , y2 , z2 ). Thus, we find
√
Distance from P1 to P2 = 2 2
√
Distance from P2 to P3 = 6
√
Distance from P1 to P3 = 10
So P2 and P3 are closest to each other.
3. The distance of a point P = (x, y, z) from the yz-plane is |x|, from the xz-plane is |y|, and from the xy-plane is |z|.
So, B is closest to the yz-plane, since it has the smallest x-coordinate in absolute value. B lies on the xz-plane, since its
y-coordinate is 0. B is farthest from the xy-plane, since it has the largest z-coordinate in absolute value.
4. Your final position is (1, −1, 1). This places you in front of the yz-plane, to the left of the xz-plane, and above the
xy-plane.
5. An example is the line y = z in the yz-plane. See Figure 12.1.
z
y
x
Figure 12.1
6. The midpoint is found by averaging coordinates:
Midpoint =
−1 + 5 3 + 6 9 − 3
,
,
2
2
2
= (2, 4.5, 3).
7. The graph is a horizontal plane at height 4 above the xy-plane. See Figure 12.2.
1142
Chapter Twelve /SOLUTIONS
z
x
y
Figure 12.2
8. The graph is a plane parallel to the yz-plane, and passing through the point (−3, 0, 0). See Figure 12.3.
z
x = −3
−3
y
x
31
Figure 12.3
9. The graph is a plane parallel to the xz-plane, and passing through the point (0, 1, 0). See Figure 12.4.
z
y=1
y
1
x
Figure 12.4
10. The graph is all points with y = 4 and z = 2, i.e., a line parallel to the x-axis and passing through the points
(0, 4, 2); (2, 4, 2); (4, 4, 2) etc. See Figure 12.5.
z
(0, 4, 2)
(2, 4, 2)
2
4
y
x
Figure 12.5
(4, 4, 2)
12.1 SOLUTIONS
1143
11. The radius is 7 − (−1) = 8, so the highest point is at (2, 3, 15).
12. The equation is x2 + y 2 + z 2 = 25
13. The sphere has equation (x − 1)2 + y 2 + z 2 = 4.
14. The plane has equation y = 3.
15. (a) 80-90◦ F
(b) 60-72◦ F
(c) 60-100◦ F
16.
predicted high temperature
100
90
80
70
distance from
Topeka
Topeka
✛
south
north
✲
Figure 12.6
17.
80
✠
80
✠
60
60
North
Boise
100
Boise
100
South
West
Figure 12.7
East
Figure 12.8
18. Beef consumption by households making $20,000/year is given by Row 1 of Table 12.1 on page 667 of the text.
Table 12.1
p
3.00
3.50
4.00
4.50
f (20, p)
2.65
2.59
2.51
2.43
For households making $20,000/year, beef consumption decreases as price goes up.
Beef consumption by households making $100, 000/year is given by Row 5 of Table 12.1.
Table 12.2
p
3.00
3.50
4.00
4.50
f (100, p)
5.79
5.77
5.60
5.53
For households making $100,000/year, beef consumption also decreases as price goes up.
Beef consumption by households when the price of beef is $3.00/lb is given by Column 1 of Table 12.1.
1144
Chapter Twelve /SOLUTIONS
Table 12.3
I
20
40
60
80
100
f (I, 3.00)
2.65
4.14
5.11
5.35
5.79
When the price of beef is $3.00/lb, beef consumption increases as income increases.
Beef consumption by households when the price of beef is $4.00/lb is given by Column 3 of Table 12.1.
Table 12.4
I
20
40
60
80
100
f (I, 4.00)
2.51
3.94
4.97
5.19
5.60
When the price of beef is $4.00/lb, beef consumption increases as income increases.
19. Table 12.5 gives the amount M spent on beef per household per week. Thus, the amount the household spent on beef in a
year is 52M . Since the household’s annual income is I thousand dollars, the proportion of income spent on beef is
P =
52M
M
= 0.052 .
1000I
I
Thus, we need to take each entry in Table 12.5, divide it by the income at the left, and multiply by 0.052. Table 12.6 shows
the results.
Table 12.5 Money spent on beef
($/household/week)
Table 12.6
Proportion of annual income spent on beef
Price of Beef ($)
Price of Beef ($)
3.00
Income
($1,000)
3.50
4.00
4.50
3.00
3.50
4.00
4.50
20
0.021
0.024
0.026
0.028
40
0.016
0.018
0.020
0.023
60
0.013
0.015
0.017
0.019
20
7.95
9.07
10.04
10.94
40
12.42
14.18
15.76
17.46
60
15.33
17.50
19.88
21.78
80
0.010
0.012
0.013
0.015
80
16.05
18.52
20.76
22.82
100
0.009
0.011
0.012
0.013
100
17.37
20.20
22.40
24.89
Income
($1,000)
20. If the price of beef is held constant, beef consumption for households with various incomes can be read from a fixed
column in Table 12.1 on page 667 of the text. For example, the column corresponding to p = 3.00 gives the function
h(I) = f (I, 3.00); it tells you how much beef a household with income I will buy at $3.00/lb. Looking at the column
from the top down, you can see that it is an increasing function of I. This is true in every column. This says that at any
fixed price for beef, consumption goes up as household income goes up—which makes sense. Thus, f is an increasing
function of I for each value of p.
Problems
21. (a) According to Table 12.2 of the problem, it feels like −19◦ F.
(b) A wind of 20 mph, according to Table 12.2.
(c) About 17.5 mph. Since at a temperature of 25◦ F, when the wind increases from 15 mph to 20 mph, the temperature
adjusted for wind chill decreases from 13◦ F to 11◦ F, we can say that a 5 mph increase in wind speed causes a 2◦ F
decrease in the temperature adjusted for wind chill. Thus, each 2.5 mph increase in wind speed brings about a 1◦ F
drop in the temperature adjusted for wind chill. If the wind speed at 25◦ F increases from 15 mph to 17.5 mph, then
the temperature you feel will be 13 − 1 = 12◦ F.
(d) Table 12.2 shows that with wind speed 20 mph the temperature will feel like 0◦ F when the air temperature is somewhere between 15◦ F and 20◦ F. When the air temperature drops 5◦ F from 20◦ F to 15◦ F, the temperature adjusted for
wind-chill drops 6◦ F from 4◦ F to −2◦ F. We can say that for every 1◦ F decrease in air temperature there is about a
6/5 = 1.2◦ F drop in the temperature you feel. To drop the temperature you feel from 4◦ F to 0◦ F will take an air
temperature drop of about 4/1.2 = 3.3◦ F from 20◦ F. With a wind of 20 mph, approximately 20 − 3.3 = 16.7◦ F
would feel like 0◦ F.
12.1 SOLUTIONS
22.
Table 12.7
20◦ F
Temperature adjusted for wind chill at
Wind speed (mph)
5
10
15
20
25
Adjusted temperature (◦ F)
13
9
6
4
3
Temperature adjusted for wind chill at 0◦ F
Table 12.8
Wind speed (mph)
Adjusted temperature (◦ F)
23.
Table 12.9
5
10
15
20
25
−11
−16
−19
−22
−24
Temperature adjusted for wind chill at 5 mph
Temperature (◦ F)
35
30
25
20
15
10
5
0
Adjusted temperature (◦ F)
31
25
19
13
7
1
−5
−11
Table 12.10
1145
Temperature adjusted for wind chill at 20 mph
Temperature (◦ F)
35
30
25
20
15
10
5
0
Adjusted temperature (◦ F)
24
17
11
4
−2
−9
−15
−22
24. (a) The total cost in dollars of renting a car is 40 times the number of days plus 0.15 times the number of miles driven,
so
C = f (d, m) = 40d + 0.15m.
(b) We have
f (5, 300) = 40(5) + 0.15(300) = $245.
Renting a car for 5 days and driving it 300 miles costs $245.
25. The gravitational force on a 100 kg object which is 7, 000, 000 meters from the center of the earth (or about 600 km above
the earth’s surface) is about 820 newtons.
26. (a) The acceleration due to gravity decreases as h increases, because the gravitational force gets weaker the farther away
you are from the planet. (In fact, g is inversely proportional to the square of the distance from the center of the planet.)
(b) The acceleration due to gravity increases as m increases. The more massive the planet, the larger the gravitational
force. (In fact, g is proportional to m.)
27. By drawing the top four corners, we find that the length of the edge of the cube is 5. See Figure 12.9. We also notice that
the edges of the cube are parallel to the coordinate axis. So the x-coordinate of the the center equals
−1 +
5
= 1.5.
2
−2 +
5
= 0.5.
2
The y-coordinate of the center equals
The z-coordinate of the center equals
2−
5
= −0.5.
2
1146
Chapter Twelve /SOLUTIONS
(1.5, 0.5, 2)
(−1, −2, 2)
✠
✻
(4, −2, 2)
(−1, 3, 2)
(4, 3, 2)
2.5
❄✛
(1.5, 0.5, −0.5)
Figure 12.9
28. The equation for the points whose distance from the x-axis is 2 is given by
a cylinder of radius 2 along the x-axis. See Figure 12.10.
p
y 2 + z 2 = 2, i.e. y 2 + z 2 = 4. It specifies
z
y
x
Figure 12.10
p
29. The distance of any point with coordinates (x, y, z) from the x-axis is y 2 + z 2 . The distance of the point from the
xy-plane is |x|. Since the condition states that these distances are equal, the equation for the condition is
p
y 2 + z 2 = |x|
y 2 + z 2 = x2 .
i.e.
This is the equation of a cone whose tip is at the origin and which opens along the x-axis with a slope of 1 as shown in
Figure 12.11.
z
x
y
Figure 12.11
30. The coordinates
of points on the y-axis are (0, y, 0). The distance from any such point (0, y, 0) to the point (a, b, c) is
p
d = a2 + (b − y)2 + c2 . Therefore, the closest point will have y = b in order to minimize d. The resulting distance is
√
then: d = a2 + c2 .
12.1 SOLUTIONS
1147
31. (a) The sphere has center at (2, 3, 3) and radius 4. The planes parallel to the xy-plane just touching the sphere are 4
above and 4 below the center. Thus, the planes z = 7 and z = −1 are both parallel to the xy-plane and touch the
sphere at the points (2, 3, 7) and (2, 3, −1).
(b) The planes x = 6 and x = −2 just touch the sphere at (6, 3, 3) and at (−2, 3, 3) respectively and are parallel to the
yz-plane.
(c) The planes y = 7 and y = −1 just touch the sphere at (2, 7, 3) and at (2, −1, 3) respectively and are parallel to the
xz-plane.
32. The edges of the cube have length 4. Thus, the center of the sphere is the center of the cube which is the point (4, 7, 1)
and the radius is r = 2. Thus an equation of this sphere is
(x − 4)2 + (y − 7)2 + (z − 1)2 = 4.
33. (a) The vertex at the opposite end of a diagonal across the base is (12, 7, 2). The other two points are (5, 7, 2) and
(12, 1, 2).
(b) The vertex at the opposite end of a diagonal across the top is (5, 1, 4). The other two points are (5, 7, 4) and (12, 1, 4).
34. Using the distance formula, we find that
Distance from P1 to P =
Distance from P2 to P =
Distance from P3 to P =
Distance from P4 to P =
√
√
√
√
206
152
170
113
So P4 = (−4, 2, 7) is closest to P = (6, 0, 4).
35. (a) To find the intersection of the sphere with the yz-plane, substitute x = 0 into the equation of the sphere:
(−1)2 + (y + 3)2 + (z − 2)2 = 4,
therefore
This equation represents a circle of radius
On the xz-plane y = 0:
therefore
√
(y + 3)2 + (z − 2)2 = 3
3.
(x − 1)2 + 32 + (z − 2)2 = 4,
(x − 1)2 + (z − 2)2 = −5
The negative sign on the right side of this equation shows that the sphere does not intersect the xz-plane, since the
left side of the equation is always non-negative.
On the xy-plane, z = 0:
(x − 1)2 + (y + 3)2 + (−2)2 = 4,
therefore
(x − 1)2 + (y + 3)2 = 0.
This equation has the unique solution x = 1, y = −3, so the xy-plane intersects the sphere in the single point
(1, −3, 0).
(b) Since the sphere does not intersect the xz-plane, it cannot intersect the x or z axes. On the y-axis, we have x = z = 0.
Substituting this into the equation for the sphere we get
(−1)2 + (y + 3)2 + (−2)2 = 4,
therefore
(y + 3)2 = −1.
This equation has no solutions because the right hand side is negative, and the left-hand side is always non-negative.
Thus the sphere does not intersect any of the coordinate axes.
1148
Chapter Twelve /SOLUTIONS
13
= −5.5, 7.5. See
2
5
Figure 12.12. The height corresponds to the z-axis, therefore the z-coordinates of the corners must be −2± = 0.5, −4.5.
2
The width corresponds to the x-axis, therefore the x-coordinates of the corners must be 1 ± 3 = 4, −2. The coordinates
of those eight corners are therefore
36. The length corresponds to the y-axis, therefore the y-coordinates of the corners must be 1 ±
(4, 7.5, 0.5), (−2, 7.5, 0.5), (−2, −5.5, 0.5), (4, −5.5, 0.5),
(4, 7.5, −4.5), (−2, 7.5, −4.5), (−2, −5.5, −4.5), (4, −5.5, −4.5).
z
(−2, −5.5, 0.5)
(4, −5.5, 0.5)
(−2, 7.5, 0.5)
y
x
(4, 7.5, 0.5)
(−2, −5.5, −4.5)
(4, −5.5, −4.5)
(−2, 7.5, −4.5)
(4, 7.5, −4.5)
Figure 12.12
37. The length of the side of the triangle is 2, so its height is
√
3. The coordinates of the highest point are (8, 0,
√
3).
38. (a) We find the midpoint by averaging
Midpoint =
1 + 5 5 + 13 7 + 19
,
,
2
2
2
= (3, 9, 13).
(b) We use a weighted average, with the coordinates of point A weighted three times more heavily than point B:
Point =
3 · 1 + 5 3 · 5 + 13 3 · 7 + 19
,
,
4
4
4
= (2, 7, 10).
= (4, 11, 16).
(c) We find the point in a similar way to part (b), but weighting B more heavily
Point =
1 + 3 · 5 5 + 3 · 13 7 + 3 · 19
,
,
4
4
4
Strengthen Your Understanding
39. The graph of the equation y = 1 is a plane perpendicular to the y-axis, not a line. The x-axis is parallel to the plane.
40. The xy-plane has equation z = 0,
The equation xy = 0 means either x = 0 (the equation of the yz-plane) or y = 0 (the equation of the xz-plane).
Points on the xy-plane all have z = 0; this is its equation.
41. The closest point on the x-axis to (2, 3, 4) is (2, 0, 0). The distance from (2, 3, 4) to this point is
p
√
d = (2 − 2)2 + (3 − 0)2 + (4 − 0)2 = 25 = 5.
42. One possible function that is increasing in x and decreasing in y is given by the formula f (x, y) = x −y. For a fixed value
of x, the value of x − y decreases as y increases, and for a fixed value of y, the value of x − y increases as x increases.
There are many other possible answers.
43. If we pick a point with z = −5, its distance from the plane z = −5 is zero. The distance of a point from the xz-plane is
the magnitude of the y-coordinate. So the point (−2, −1, −5) is a distance of 1 from the xz-plane and a distance of zero
from the plane z = −5. There are many other possible points.
12.2 SOLUTIONS
1149
44. True. Since each choice of x and y determines a unique value for f (x, y), choosing x = 10 yields a unique value of
f (10, y) for any choice of y.
45. True. Since each choice of h > 0 and s > 0 determines a unique value for the volume V , we can say V is a function of h
and s. In fact, this function has a formula: V (h, s) = h · s2 .
46. False. If, for example, d = 2 meters and H = 57◦ C, there could be many times t at which the water temperature is 57 ◦ C
at 2 meters depth.
47. False. A function may have different inputs that yield equal outputs.
48. True. Since each of f (x) and g(y) has at most one output for each input, so does their product.
49. True. All points in the z = 2 plane have z-coordinate 2, hence are below any point of the form (a, b, 3).
50. False. The plane z = 2 is parallel to the xy-plane.
√
51. True. Both are distance 2 from the origin.
52. False. The point (2, −1, 3) does not satisfy the equation. It is at the center of the sphere, and does not lie on the graph.
53. True. The origin is the closest point in the yz-plane to the point (3, 0, 0), and its distance to (3, 0, 0) is 3.
54. False. There is an entire circle (of radius 4) of points in the yz-plane that are distance 5 from (3, 0, 0).
55. False. The value of b can be ±4.
56. True. Otherwise f would have more than one value for a given pair (x, y), which cannot happen if f is a function.
57. False. For example, the y-axis intersects the graph of f (x, y) = 1 − x2 − y 2 twice, at y = ±1.
Solutions for Section 12.2
Exercises
1. (a) The value of z decreases as x increases. See Figure 12.13.
(b) The value of z increases as y increases. See Figure 12.14.
z
z
y
x
Figure 12.13
2. (a)
(b)
(c)
(d)
(e)
Figure 12.14
is (IV), since z = 2 + x2 + y 2 is a paraboloid opening upward with a positive z-intercept.
is (II), since z = 2 − x2 − y 2 is a paraboloid opening downward.
is (I), since z = 2(x2 + y 2 ) is a paraboloid opening upward and going through the origin.
is (V), since z = 2 + 2x − y is a slanted plane.
is (III), since z = 2 is a horizontal plane.
3. (a) The value of z only depends on the distance from the point (x, y) to the origin. Therefore the graph has a circular
symmetry around the z-axis. There are two such graphs among those depicted in the figure in the text: I and V. The
1
one corresponding to z = x2 +y
2 is I since the function blows up as (x, y) gets close to (0, 0).
(b) For similar reasons as in part (a), the graph is circularly symmetric about the z-axis, hence the corresponding one
must be V .
(c) The graph has to be a plane, hence IV.
(d) The function is independent of x, hence the corresponding graph can only be II. Notice that the cross-sections of this
graph parallel to the yz-plane are parabolas, which is a confirmation of the result.
(e) The graph of this function is depicted in III. The picture shows the cross-sections parallel to the zx-plane, which have
the shape of the cubic curves z = x3 − constant.
1150
Chapter Twelve /SOLUTIONS
4. The graph is a horizontal plane 3 units above the xy-plane. See Figure 12.15.
z
1
y
x
Figure 12.15
5. The graph is a sphere of radius 3, centered at the origin. See Figure 12.16.
z
3
3
3
x
y
Figure 12.16
6. The graph is a bowl opening up, with vertex at the point (0, 0, 4). See Figure 12.17.
z
4
y
x
Figure 12.17
7. Since z = 5−(x2 +y 2 ), the graph is an upside-down bowl moved up 5 units and with vertex at (0, 0, 5). See Figure 12.18.
z
5
y
x
Figure 12.18
12.2 SOLUTIONS
1151
8. In the yz-plane, the graph is a parabola opening up. Since there are no restrictions on x, we extend this parabola along the
x-axis. The graph is a parabolic cylinder opening up, extended along the x-axis. See Figure 12.19.
z
y
x
Figure 12.19
9. The graph is a plane with x-intercept 6, and y-intercept 3, and z-intercept 4. See Figure 12.20.
z
4
3 y
x 6
Figure 12.20
10. In the xy-plane, the graph is a circle of radius 2. Since there are no restrictions on z, we extend this circle along the z-axis.
The graph is a circular cylinder extended in the z-direction. See Figure 12.21.
z
y
x
Figure 12.21
11. In the xz-plane, the graph is a circle of radius 2. Since there are no restrictions on y, we extend this circle along the y-axis.
The graph is a circular cylinder extended in the y-direction. See Figure 12.22.
z
2
2
x
y
Figure 12.22
1152
Chapter Twelve /SOLUTIONS
12. All the points on the cylinder are at a distance
√
7 from the y-axis. Since this distance is given by
p
√
x2 + z 2 = 7
√
x2 + z 2 , we have
x2 + z 2 = 7.
√
13. A sphere of radius 3 centered at the origin has equation x2 + y 2 + z 2 = 32 = 9, so shifting the center to (0. 7, 0) gives
x2 + (y −
√
7)2 + z 2 = 9.
14. A paraboloid with vertex at the origin but opening along the positive x-axis is
x = y2 + z2 .
A parabola opening toward the negative x-axis is
x = −y 2 − z 2
so moving the vertex to (1, 3, 5) gives
x = 1 − (y − 3)2 − (z − 5)2 .
Problems
15. (a) Cross-sections with x fixed at x = b are in Figure 12.23.
10
x=1
3
x=0
✛
y=0
1
−1
✠
y
−2
1
2
y=1
2
x = −1
5
−2
−1
−5
−1
2
✛
−2
−10
x
1
y = −1
−3
Figure 12.23: Cross-section
f (a, y) = y 3 + ay, with a = −1, 0, 1
Figure 12.24: Cross-section
f (x, b) = b3 + bx, with b = −1, 0, 1
(b) Cross-section with y fixed at y = 6 are in Figure 12.24.
16. We have f (3, 2) = 2e−2(5−3) = 0.037. We see that 2 hours after the injection of 3 mg of this drug, the concentration of
the drug in the blood is 0.037 mg per liter.
17. (a) Holding x fixed at 4 means that we are considering an injection of 4 mg of the drug; letting t vary means we are
watching the effect of this dose as time passes. Thus the function f (4, t) describes the concentration of the drug in
the blood resulting from a 4 mg injection as a function of time. Figure 12.25 shows the graph of f (4, t) = te−t .
Notice that the concentration in the blood from this dose is at a maximum at 1 hour after injection, and that the
concentration in the blood eventually approaches zero.
C (mg per liter)
0.3
C = f (4, t)
0.2
0.1
1
2
3 4 5
t (hours)
Figure 12.25: The function f (4, t)
shows the concentration in the blood
resulting from a 4 mg injection
C (mg per liter)
0.5
0.4
0.3
0.2
0.1
1
C = f (x, 1)
2
3
4
5
x (mg)
Figure 12.26: The function f (x, 1)
shows the concentration in the blood 1
hour after the injection
12.2 SOLUTIONS
1153
(b) Holding t fixed at 1 means that we are focusing on the blood 1 hour after the injection; letting x vary means we
are considering the effect of different doses at that instant. Thus, the function f (x, 1) gives the concentration of
the drug in the blood 1 hour after injection as a function of the amount injected. Figure 12.26 shows the graph of
f (x, 1) = e−(5−x) = ex−5 . Notice that f (x, 1) is an increasing function of x. This makes sense: If we administer
more of the drug, the concentration in the bloodstream is higher.
18. The one-variable function f (a, t) represents the effect of an injection of a mg at time t. Figure 12.27 shows the graphs of
the four functions f (1, t) = te−4t , f (2, t) = te−3t , f (3, t) = te−2t , and f (4, t) = te−t corresponding to injections of
1, 2, 3, and 4 mg of the drug. The general shape of the graph is the same in every case: The concentration in the blood is
zero at the time of injection t = 0, then increases to a maximum value, and then decreases toward zero again. We see that
if a larger dose of the drug is administered, the peak of the graph is later and higher. This makes sense, since a larger dose
will take longer to diffuse fully into the bloodstream and will produce a higher concentration when it does.
C (mg per liter)
0.3
a=4
0.2
a=3
✛
0.1
a=2
✛
a=1
1
2
3
4
5
t (hours)
Figure 12.27: Concentration C = f (a, t) of the drug resulting from an a mg injection
19. (a) is (IV), (b) is (IX), (c) is (VII), (d) is (I), (e) is (VIII), (f) is (II), (g) is (VI), (h) is (III), (i) is (V).
20. (a) This is a bowl; z increases as the distance from the origin increases, from a minimum of 0 at x = y = 0.
(b) Neither. This is an upside-down bowl. This function decreases from 1, at x = y = 0, to arbitrarily large negative
values as x and y increase due to the negative squared terms of x and y. It looks like the bowl in part (a) except
flipped over and raised up slightly.
(c) This is a plate. Solving the equation for z gives z = 1 − x − y which describes a plane whose x and y slopes are −1.
It is perfectly flat, but not horizontal.
(d) Within its domain, this function is a bowl. It is√undefined at points at which x2 + y 2 > 5, but within those limits it
describes the bottom half of a sphere of radius 5 centered at the origin.
(e) This function is a plate. It is perfectly flat and horizontal.
21. (a)
z
(i)
z
(ii)
25
25
✛x=1
✛y=1
16
16
■
■
x=0
y=0
4
4
y
−4
−2
0
2
4
Figure 12.30: Cross-sections of
z = x2 + y 2
x
−4
−2
0
2
4
Figure 12.31: Cross-sections of
z = x2 + y 2
1154
Chapter Twelve /SOLUTIONS
(b)
z
(i)
2
1
z
(ii)
4
y
5
−4
2
1
−4
x=0
y=0
✠
✠
−16
−25
x
4 5
−16
✒
−25
x=1
Figure 12.34: Cross-sections of
z = 1 − x2 − y 2
✒
y=1
Figure 12.35: Cross-sections of
z = 1 − x2 − y 2
(c)
x=0
(i)
y=0
(ii)
z
✠
z
✠
4
✒
4
✒
x=1
y=1
y
−4
x
4
−4
−4
4
−4
Figure 12.38: Cross-sections of
x+y+z =1
Figure 12.39: Cross-sections of x + y + z = 1
(d)
(i)
z
(ii)
z
2
2
−2.5
−2.5
x
2.5
y
2.5
✒
−2
✒
■
x=0
Figure 12.42:
pCross-sections of
z = − 5 − x2 − y 2
−2
■
y=0
x=1
Figure 12.43:
pCross-sections of
z = − 5 − x2 − y 2
y=1
12.2 SOLUTIONS
1155
(e)
(i)
z
x=0
❘
3
(ii)
x=1
z
y=0
✠
❘
3
y=1
✠
y
−4
x
4
−4
−3
4
−3
Figure 12.46: Cross-section of z = 3
Figure 12.47: Cross-section of z = 3
22. (a) If we have iron stomachs and can consume cola and pizza endlessly without ill effects, then we expect our happiness
to increase without bound as we get more cola and pizza. Graph (IV) shows this since it increases along both the
pizza and cola axes throughout.
(b) If we get sick upon eating too many pizzas or drinking too much cola, then we expect our happiness to decrease once
either or both of those quantities grows past some optimum value. This is depicted in graph (I) which increases along
both axes until a peak is reached, and then decreases along both axes.
(c) If we do get sick after too much cola, but are always able to eat more pizza, then we expect our happiness to decrease
after we drink some optimum amount of cola, but continue to increase as we get more pizza. This is shown by graph
(III) which increases continuously along the pizza axis but, after reaching a maximum, begins to decrease along the
cola axis.
23. One possible equation: z = x2 + y 2 + 5. See Figure 12.48.
z
5
x
y
Figure 12.48
24. One possible equation: x + y + z = 1. See Figure 12.49.
z
x
y
Figure 12.49
1156
Chapter Twelve /SOLUTIONS
25. One possible equation: z = (x − y)2 . See Figure 12.50.
z
x
y
Figure 12.50
26. One possible equation: z = −
p
x2 + y 2 . See Figure 12.51.
z
y
x
Figure 12.51
27. When h is fixed, say h = 1, then
V = f (r, 1) = πr 2 1 = πr 2
Similarly,
2
4
f (r, ) = πr 2
3
9
When r is fixed, say r = 1, then
and
1
π
f (r, ) = r 2
3
9
f (1, h) = π(1)2 h = πh
Similarly,
f (2, h) = 4π
h=
Volume
2
3
f (r,
25
f (3, h) = 9πh.
and
r=3
Volume
2
)
3
f (3, h)
50
h=1
20
1
3
f (r, 31 )
h=
f (r, 1)
r=2
40
f (2, h)
15
30
10
20
f (1, h)
r=1
5
0
10
r
1
2
3
Figure 12.52
4
5
0
1
2
3
Figure 12.53
4
5
h
12.2 SOLUTIONS
1157
28. (a) The plane y = 0 intersects the graph in the curve z = 4x2 + 1, which is a parabola opening upward.
(b) The plane x = 0 intersects the graph in z = −y 2 + 1, which is a parabola opening downward because of the negative
coefficient of y 2 .
(c) The plane z = 1 intersects the graph in 4x2 − y 2 = 0. Since this factors as (2x − y)(2x + y) = 0, it is the equation
for the two lines y = 2x and y = −2x.
29.
pizza fixed at 4
(a)
happiness
cola fixed at 4
(b)
happiness
❄
1
❄
1
0.5
0.5
✻
✻
pizza fixed at 1
(or pizza fixed at 7)
0
cola fixed at 1
(or cola fixed at 7)
0
cola
4
8
pizza
4
8
Figure 12.55: Cross-sections of graph I
Figure 12.54: Cross-sections of graph I
pizza fixed at 2
(a)
happiness
cola fixed at 2
(b)
❄
happiness
1
❄
3
pizza fixed at 4
(or pizza fixed at 0)
2
❄
0.5
1
✻
cola fixed at 1
0
0
cola
2
4
happiness
4
pizza fixed at 2
(b)
happiness
2
❄
cola
fixed at 2
❘
2
1
■
✻
cola fixed at 1
(or cola fixed at 3)
pizza fixed at 1
0
4
Figure 12.57: Cross-sections of graph II
Figure 12.56: Cross-sections of graph II
(a)
pizza
2
cola
2
4
Figure 12.58: Cross-sections of graph III
0
pizza
1
2
Figure 12.59: Cross-sections of graph III
1158
Chapter Twelve /SOLUTIONS
(a)
happiness
4
happiness
(b)
pizza
fixed at 2
4
❘
cola
fixed at 2
❘
2
2
■
■
pizza fixed at 1
0
cola fixed at 1
0
cola
1
2
Figure 12.60: Cross-sections of graph IV
pizza
1
2
Figure 12.61: Cross-sections of graph IV
30. (a) Figures 12.62-12.65 show the wave profile at time t = −1, 0, 1, 2.
z
z
t = −1
t=0
x
x
Figure 12.62
Figure 12.63
z
z
t=2
t=1
x
x
Figure 12.64
Figure 12.65
(b) Increasing x
(c) The graph in Figure 12.66 represents a wave traveling in the opposite direction.
z
t
x
Figure 12.66
31. (a) Cross-sections with t fixed are in Figure 12.67. The equations are
f (x, 0) = cos 0 sin x = sin x,
1
f (x, π/4) = cos(π/4) sin x = √ sin x.
2
1159
12.2 SOLUTIONS
Cross-sections with t fixed are in Figure 12.68. The equations are
1
f (π/4, t) = cos t sin(π/4) = √ cos t,
2
f (π/2, t) = cos t sin(π/2) = cos t.
t=0
1
❄
t = π/4
x = π/2
1
✠
❄
π
x
x = π/4
π
−1
❄
2π
t
−1
Figure 12.67: Cross-sections with t
fixed
Figure 12.68: Cross-sections with x
fixed
(b) If x = 0, π, then f (0, t) = cos t · sin 0 = 0 = f (π, t). The ends of the string are at x = 0, π, which do not move so
the displacement is 0 there for all t.
(c) The cross-sections with t fixed are snapshots of the string at different instants in time. Graphs of these cross-sections
are the curves obtained when the the graph of f in Figure 12.69 is sliced perpendicular to the t axis. Every plane
perpendicular to the t-axis intersects the surface in one arch of a sine curve. The amplitude of the arch changes with
t as a cosine curve.
The cross-sections with x fixed show how a single point on the string moves as time goes by. Graphs of these
cross-sections are obtained by slicing the graph perpendicular to the x axis. Notice in Figure 12.69 that the crosssections with x = 0 and x = π are flat lines since the endpoints of the string don’t move. The cross-section with
x = π/2 is a cosine curve with amplitude 1, because the midpoint of the string oscillates back and forth. Crosssections with x fixed between 0 and π/2 and between π/2 and π are cosine curves with amplitude between 0 and 1,
representing the fact that these points on the string oscillate back and forth with the same period as x = π/2, but a
smaller amplitude.
f (x, t)
1
π/2
x
π
3π/2
2π
t
π
Figure 12.69: Graph of vibrating string function
f (x, t) = cos t sin x
Strengthen Your Understanding
32. The graph of a function f (x, y) is a parabolic surface in 3-space, not a circle.
33. If we hold x fixed, then z = f (x, y) = x2 is also fixed, so the cross-section is a line parallel to the y-axis.
34. We know that z = x2 + y 2 + 2 is positive everywhere and that the surface intersects the plane z = 2 only at (0, 0, 2). So
let f (x, y) = x2 + y 2 + 2.
35. The function f (x, y) = x2 − 1 intersects the xz-plane (and any plane parallel to the xz-plane) in the parabola z = x2 − 1.
It also intersect the yz-plane in the line z = −1. So f is a possible example. The function g(x, y) = x2 + y intersects the
xz-plane in the parabola z = x2 and the yz-plane in the line y = z. So g is another possible example. There are many
others.
36. The function f (x, y) = 1 − x2 − y 2 intersects the xy-plane in the circle x2 + y 2 = 1. So f is a possible example. There
are many others.
1160
Chapter Twelve /SOLUTIONS
37. False. Fixing w = k gives the one-variable function g(v) = ev /k, which is an increasing exponential function if k > 0,
but is decreasing if k < 0.
38. True. For example, consider the weekly beef consumption C of a household as a function of total income I and the cost of
beef per pound p. It is possible that consumption increases as income increases (for fixed p) and consumption decreases
as the price of beef increases (for fixed I).
39. True. For example, consider f (x, y) = ex · (6 − y). Then g(x) = f (x, 5) = ex , which is an increasing function of x. On
the other hand, h(x) = f (x, 10) = −4ex , which is a decreasing function of x.
40. False. The point (0, 0, 0) does not satisfy the equation.
41. True. The x-axis is where y = z = 0.
42. False. If x = 10, substituting gives 102 + y 2 + z 2 = 10, so y 2 + z 2 = −90. Since y 2 + z 2 cannot be negative, a point
with x = 10 cannot satisfy the equation.
43. True. The cross-section with y = 1 is the line z = x + 1.
44. True. The cross-sections with x = c are all of the form z = 1 − y 2 .
45. True. The cross-sections with y = c are of the form z = 1 − c2 , which are horizontal lines.
46. True. For any a and b, we have f (a, b) 6= g(a, b). The graph of g is same as the graph of f , except it is shifted 2 units
vertically.
47. True. The intersection,
where f (x, y) = g(x, y), is given by x2 + y 2 = 1 − x2 − y 2 , or x2 + y 2 = 1/2. This is a circle
√
of radius 1/ 2 parallel to the xy-plane at height z = 1/2.
48. False. For example, f (x, y) = x2 (or any cylinder along the y-axis) is not a plane but has lines for x = c cross-sections.
49. False. Wherever f (x, y) = 0 the graphs of f (x, y) and −f (x, y) will intersect.
50. True. The graph is the bowl-shaped g(x, y) = x2 + y 2 turned upside-down and shifted upward by 10 units.
51. (c), a plane. While x is fixed at 2, y and z can vary freely.
Solutions for Section 12.3
Exercises
1. We’ll set z = 4 at the peak. See Figure 12.70.
y
x
1
2
3
4
Figure 12.70
12.3 SOLUTIONS
2. See Figure 12.71.
y
1
6
1
x
Figure 12.71
3. We will take z = 4 to be the flat area. See Figure 12.72.
y
z=3
z=2
z=1
x
Figure 12.72
4. See Figure 12.73.
y
−3
−2
−1
3 2
1
1
2 3
x
−1
−2
−3
Figure 12.73
1161
1162
Chapter Twelve /SOLUTIONS
5. The contour where f (x, y) = x + y = c, or y = −x + c, is the graph of the straight line with slope −1 as shown in
Figure 12.74. Note that we have plotted the contours for c = −3, −2, −1, 0, 1, 2, 3. The contours are evenly spaced.
y
2
c
=
3
c
1
=
2
c
=
1
−2
x
1
−1
2
c
c
=
=
−
0
2
c
−1
=
c
3
=
−
−
1
−2
Figure 12.74
6. The contour where f (x, y) = 3x + 3y = c or y = −x + c/3 is the graph of the straight line of slope −1 as shown in
Figure 12.75. Note that we have plotted the contours for c = −9, −6, −3, 0, 3, 6, 9. The contours are evenly spaced.
y
2
c
=
9
c
1
=
6
c
=
3
−2
−1
x
1
2
=
6
0
−
=
=
c
c
c
−1
c
9
=
−
−
3
−2
Figure 12.75
√
7. The contour where f (x, y) = x2 + y 2 = c, where c ≥ 0, is the graph of the circle centered at (0, 0), with radius c
as shown in Figure 12.76. Note that we have plotted the contours for c = 0, 1, 2, 3, 4. The contours become more closely
packed as we move further from the origin.
y
2
c
=
c
4
=
c
3
=
2
c
=
1
1
c=0
−2
−1
x
1
−1
−2
Figure 12.76
2
12.3 SOLUTIONS
1163
2
2
8. The
√ contour where f (x, y) = −x − y + 1 = c, where c ≤ 1, is the graph of the circle centered at (0, 0), with radius
1 − c as shown in Figure 12.77. Note that we have plotted the contours for c = −3, −2, −1, 0, 1. The contours become
more closely packed as we move further from the origin.
y
2
3
− 2
= −
c = −1
c = 0
c
c
1
=
c=1
−2
x
−1
1
2
−1
−2
Figure 12.77
9. The contour where f (x, y) = xy = c, is the graph of the hyperbola y = c/x if c 6= 0 and the coordinate axes if c = 0,
as shown in Figure 12.78. Note that we have plotted contours for c = −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5. The contours
become more closely packed as we move further from the origin.
−
=
2
−
1
2
−
c
3
=
=
c
=
c
4
c
=
=
c
c
3
−
4
y
=
1
=
c
c
=
c
0
c
−
1
x
3
−
4
−
c
=
3
=
c
c
=
=
2
c
=
=
c
1
−
2
=
=
c
c
4
Figure 12.78
10. The contour where f (x, y) = y − x2 = c is the graph of the parabola y = x2 + c, with vertex (0, c) and symmetric about
the y-axis, shown in Figure 12.79. Note that we have plotted the contours for c = −2, −1, 0, 1. The contours become
more closely packed as we move farther from the y-axis.
y
c
=
−1
1
−
=
−
2
c
−2
1
2
x
=
−1
c
−2
0
1
c
=
1
2
Figure 12.79
1164
Chapter Twelve /SOLUTIONS
11. The contour where f (x, y) = x2 + 2y 2 = c, where c ≥ 0, is the graph of the ellipse with focuses (− 2c , 0), ( 2c , 0)
and axes lying on x- and y-axes as shown in Figure 12.80. Note that we have plotted the contours for c = 0, 1, 2, 3, 4.
The contours become more closely packed as we move further from the origin.
p
p
y
2
c=
1
c=
1
c=0
−2
4
x
−1
1
2
−1
−2
Figure 12.80
√
p
√
12. The contour where f (x, y) = x2 + 2y 2 = c, where c ≥ 0, is the graph of the ellipse with focuses (− c 2 2 , 0), ( c 2 2 , 0)
and axes lying on x- and y-axes as shown in Figure 12.81. Note that we have plotted the contours for c = 0, 1, 2, 3, 4. See
Figure 12.81.
y
4
c
=
3
c
=
4
c
=
2
x
2
−4
−2
2
4
−2
−4
Figure 12.81
p
13. The contour where f (x, y) = cos( x2 + y 2 ) = c, where −1 ≤ c ≤ 1, is a set of circles centered at (0, 0), with radius
cos−1 c + 2kπ with k = 0, 1, 2, .. and − cos−1 c + 2kπ, with k = 1, 2, 3, ... as shown in Figure 12.82. Note that we have
plotted contours for c = 0, 0.2, 0.4, 0.6, 0.8, 1.
5π
2
y
c=1
✠
✛
=
0
✛
c
x
− 5π
2
5π
2
− 5π
2
Figure 12.82
12.3 SOLUTIONS
1165
14. Since f (5, 10) = 3 · 52 · 10 + 7 · 5 + 20 = 805, an equation for the contour is
3x2 y + 7x + 20 = 805.
15. (a) Level curves are in Figure 12.83.
y
z = −2
z=2
z=1
x
z = −1
z=1
z
z = 2y (x = 2)
z = y (x = 1)
z = −1
z = −2
z=2
y
z = −y (x = −1)
z = −2y (x = −2)
Figure 12.83
Figure 12.84
(b) Cross-sections with x constant are in Figure 12.84
(c) Setting y = x gives the curve z = x2 in Figure 12.85
z
Curve z = x2
Line x = y
Figure 12.85
16. (a)
(b)
(c)
(d)
(e)
(III)
(I)
(V)
(II)
(IV)
17. The values in Table 12.5 are not constant along rows or columns and therefore cannot be the lines shown in (I) or (IV).
Also observe that as you move away from the origin, whose contour value is 0, the z-values on the contours increase.
Thus, this table corresponds to diagram (II).
The values in Table 12.6 are also not constant along rows or columns. Since the contour values are decreasing as you
move away from the origin, this table corresponds to diagram (III).
Table 12.7 shows that for each fixed value of x, we have constant contour value, suggesting a straight vertical line at
each x-value, as in diagram (IV).
Table 12.8 also shows lines, however these are horizontal since for each fixed value of y we have constant contour
values. Thus, this table matches diagram (I).
18. Superimposing the surface z = 1/2 on the graph of f (x, y) gives Figure 12.86. The contour f (x, y) = 1/2 is the
intersection of the two surfaces; that is, the collection of closed curves as shown in Figure 12.87
1166
Chapter Twelve /SOLUTIONS
y
2π
π
z
x
π
−π
−2π
x
2π
−π
y
−2π
Figure 12.86
Figure 12.87
Problems
19. We expect total sales to decrease as the price increases and to increase as advertising expenditures increase. Moving
parallel to the x-axis, the Q-values on the contours decrease, whereas moving parallel to the y-axis, the Q-values increase.
Thus, x is the price and y is advertising expenditures
20. To find a value, evaluate f (x, y) = 100ex − 50y 2 at any point (x, y) on the contour. Check by evaluating the function at
a couple of points on each contour. Starting from the left and estimating points on the contour, we have
First contour: f (0, 0) = 100, f (0.2, 0.65) = 101
Second contour: f (0.4, 0) = 149, f (0.6, 0.8) = 150
Third contour: f (0.7, 0) = 201, f (0.8, 0.65) = 201
Fourth contour: f (0.92, 0) = 251, f (1, 0.65) = 251.
Since the true values of f are equally spaced multiples of 10, it seems that they must be 100, 150, 200, and 250. See
Figure 12.88.
y
1
0.8
150
100
0.6
0.4
200
0.2
250
x
0.2
0.4
0.6
0.8
1
Figure 12.88
21. (a) We have
f (x, y) = x2 − y 2 − 2x + 4y − 3 = (x − 1)2 − (y − 2)2 .
Thus, the graph of f has the same saddle shape as that of z = x2 − y 2 but centered at x = 1, y = 2. The function
increases in the x-direction and decreases in the y-direction, so f corresponds to III.
(b) We have
g(x, y) = x2 + y 2 − 2x − 4y + 15 = (x − 1)2 + (y − 2)2 + 10.
Thus, the graph of g is a paraboloid opening upward, with vertex at (1, 2, 10). So h corresponds to VI.
(c) We have
h(x, y) = −x2 − y 2 + 2x + 4y − 8 = −(x − 1)2 − (y − 2)2 − 3.
Thus, the graph of h is a paraboloid opening downward, with vertex at (1, 2, −3). So h corresponds to I.
12.3 SOLUTIONS
1167
(d) We have
j(x, y) = −x2 + y 2 + 2x − 4y + 3 = −(x − 1)2 + (y − 2)2 .
Thus, the graph of j has the same saddle shape as that of z = −x2 + y 2 but centered at x = 1, y = 2. The function
decreases in the x-direction
and increases in the y-direction, so j corresponds to IV.
p
(e) Since k(x, y) = (x − 1)2 + (y − 2)2 , the graph of k is a cone opening upward with vertex at (1, 2, 0). Thus, the
graph of k corresponds
p to II.
(f) Since l(x, y) = − (x − 1)2 + (y − 2)2 , the graph of l is a cone opening downward with vertex at (1, 2, 0). Thus,
the graph of l corresponds to V.
22. (a) Find the point where the horizontal line for 15 mph meets the contour for −20◦ F wind chill. The actual temperature
is about 0◦ F.
(b) The horizontal line for 10 mph meets the vertical line for 0◦ F about 1/5 of the way from the contour for −20◦ F to
the contour for 0◦ F wind chill. We estimate the wind chill to be about −16◦ F.
(c) We look for the point on the vertical line for −20◦ F where the wind chill is −50◦ F, the danger point for humans.
This is a point on the line that is about half way between the contours for −60◦ F and −40◦ F. The point can not be
determined exactly, but we estimate that it occurs where the wind speed is about 23 mph.
(d) A temperature drop of 20◦ F corresponds to moving left from one vertical grid line to the next on the horizontal
line for 15 mph. This horizontal movement appears to correspond to about 1 1/4 the horizontal distance between
contours crossing the line. Since contours are spaced at 20◦ F wind chill, we estimate that the wind chill drops about
25◦ F when the air temperature goes down 20◦ F during a 15 mph wind.
23. To sketch the curve, first put dots on the point where an f contour crosses a g contour of the same value. Then connect
the dots with a smooth curve. See Figure 12.89.
y
16
18
16
14
12
10
10
14
8
12
10
6
8
4
6
4
2
2
0
x
10
Figure 12.89: Black: f (x, y). Blue; g(x, y)
24. Many different answers are possible. Answers are in degrees Celsius.
(a) Minnesota in winter. See Figure 12.90.
10
15
0
10
−10
5
Figure 12.90
Figure 12.91
(b) San Francisco in winter. See Figure 12.91.
(c) Houston in summer. See Figure 12.92.
30
22
40
24
55
26
Figure 12.92
Figure 12.93
1168
Chapter Twelve /SOLUTIONS
(d) Oregon in summer. See Figure 12.93.
25. The point x = 10, t = 5 is between the contours H = 70 and H = 75, a little closer to the former. Therefore, we
estimate H(10, 5) ≈ 72, i.e., it is about 72◦ F. Five minutes later we are at the point x = 10, t = 10, which is just above
the contour H = 75, so we estimate that it has warmed up to 76◦ F by then.
26. The line t = 5 crosses the contour H = 80 at about x = 4; this means that H(4, 5) ≈ 80, and so the point (4, 80) is
on the graph of the one-variable function y = H(x, 5). Each time the line crosses a contour, we can plot another point
on the graph of H(x, 5), and thus get a sketch of the graph. See Figure 12.94. Each data point obtained from the contour
map has been indicated by a dot on the graph. The graph of H(x, 20) was obtained in a similar way.
H
85
80
t = 20
75
70
t=5
65
0
x
5
10
15
20
25
30
Figure 12.94: Graph of H(x, 5) and H(x, 20): heat as a function of distance from the
heater at t = 5 and t = 20 minutes
These two graphs describe the temperature at different positions as a function of x for t = 5 and t = 20.
Notice that the graph of H(x, 5) descends more steeply than the graph of H(x, 20); this is because the contours are
quite close together along the line t = 5, whereas they are more spread out along the line t = 20. In practical terms the
shape of the graph of H(x, 5) tells us that the temperature drops quickly as you move away from the heater, which makes
sense, since the heater was turned on just five minutes ago. On the other hand, the graph of H(x, 20) descends more
slowly, which makes sense, because the heater has been on for 20 minutes and the heat has had time to diffuse throughout
the room.
27. (a) The contour lines are much closer together on path A, so path A is steeper.
(b) If you are on path A and turn around to look at the countryside, you find hills to your left and right, obscuring the
view. But the ground falls away on either side of path B, so you are likely to get a much better view of the countryside
from path B.
(c) There is more likely to be a stream alongside path A, because water follows the direction of steepest descent.
28. (a) The point representing 13% and $6000 on the graph lies between the 120 and 140 contours. We estimate the monthly
payment to be about $137.
(b) Since the interest rate has dropped, we will be able to borrow more money and still make a monthly payment of $137.
To find out how much we can afford to borrow, we find where the interest rate of 11% intersects the $137 contour
and read off the loan amount to which these values correspond. Since the $137 contour is not shown, we estimate its
position from the $120 and $140 contours. We find that we can borrow an amount of money that is more than $6000
but less than $6500. So we can borrow about $250 more without increasing the monthly payment.
(c) The entries in the table will be the amount of loan at which each interest rate intersects the 137 contour. Using the
$137 contour from (b) we make table 12.11.
Table 12.11
Amount borrowed at a monthly payment of $137.
Interest Rate (%)
0
1
2
3
4
5
6
7
Loan Amount ($)
8200
8000
7800
7600
7400
7200
7000
6800
Interest rate (%)
8
9
10
11
12
13
14
15
Loan Amount ($)
6650
6500
6350
6250
6100
6000
5900
5800
(a) The point representing 8% and $6000 on the graph lies between the 120 and 140 contours. We estimate the monthly
payment to be about $122.
12.3 SOLUTIONS
1169
(b) Since the interest rate has dropped, we will be able to borrow more money and still make a monthly payment of $122.
To find out how much we can afford to borrow, we find where the interest rate of 6% intersects the $122 contour
and read off the loan amount to which these values correspond. Since the $122 contour is not shown, we estimate its
position from the $120 and $140 contours. We find that we can borrow an amount of money that is more than $6000
but less than $6500. So we can borrow about $350 more without increasing the monthly payment.
(c) The entries in the table will be the amount of loan at which each interest rate intersects the 122 contour. Using the
$122 contour from (b) we make table 12.12.
Table 12.12
Amount borrowed at a monthly payment of $122.
Interest Rate (%)
0
1
2
3
4
5
6
7
Loan Amount ($)
7400
7200
7000
6800
6650
6500
6350
6200
Interest rate (%)
8
9
10
11
12
13
14
15
Loan Amount ($)
6000
5850
5700
5600
5500
5400
5300
5200
29. The vertical spacing between the contours just north and just south of the trail increases as you move eastward along the
trail. A possible contour diagram is in Figure 12.95.
10
20
Elevation in meters
0
101
Trail
1000
0
98
990
Figure 12.95
30. (a)
(b)
(c)
(d)
I
IV
II
III
See Figure 12.96.
1170
Chapter Twelve /SOLUTIONS
y
(a)
y
(b)
x
x
y
(c)
y
(d)
x
x
Figure 12.96
31. Figure 12.97 shows an east-west cross-section along the line N = 50 kilometers.
Figure 12.98 shows an east-west cross-section along the line N = 100 kilometers.
Density of the
fox population P
2.5
2
Density of the
fox population P
1.5
1
1.5
0.5
1
0.5
East
60
120
180
East
30
60
Figure 12.97
90
120
150
180
Figure 12.98
Figure 12.99 shows a north-south cross-section along the line E = 60 kilometers.
Figure 12.100 shows a north-south cross-section along the line E = 120 kilometers.
Density of the
fox population P
Density of the
fox population P
2.5
2
1
1.5
1
0.5
0.5
North
35
Figure 12.99
70
North
50
100
Figure 12.100
150
12.3 SOLUTIONS
1171
32. (a) The profit is given by the following:
π = Revenue from q1 + Revenue from q2 − Cost.
Measuring π in thousands, we obtain:
π = 3q1 + 12q2 − 4.
(b) A contour diagram of π follows. Note that the units of π are in thousands.
q2
3
2
1
π=
0
π=
10
2
π=
30
π=
20
4
6
q1
8
33. For any Cobb-Douglass function F (K, L) = bLα K β , if we increase the inputs by a factor of m, from (K, L) to
(mK, mL) we get:
F (mK, mL) = b(mL)α (mK)β
= mα+β bLα K β
= mα+β F (K, L)
Thus we see that increasing inputs by a factor of m increases outputs by a factor of mα+β .
If α + β < 1, then increasing each input by a factor of m will result in an increase in output of less than a factor of
m. This applies to statements (a) and (E). In statement (a), α + β = 0.25 + 0.25 = 0.5, so increasing inputs by a factor
of m = 4, as in statement (E), increases output by a factor of 40.5 = 2. We can match statements (a) and (E) to graph
(II) by noting that when (K, L) = (1, 1), we have F = 1 and when we double the inputs (m = 2) to (K, L) = (2, 2),
F increases by less than a factor of 2. This is called decreasing returns to scale.
If α + β = 1, then increasing K and L by a factor of m will result in an increase in F by the same factor m. This
applies to statements (b) and (D). In statement (b), α + β = 0.5 + 0.5 = 1, and in statement (D), an increase in inputs
by a factor of 3 results in an increase in F by the same factor. We match these statements to graph (I) where we see that
increasing (K, L) from (1, 1) to (3, 3) results in an increase in F from F = 1 to F = 3. This is called constant returns
to scale.
If α + β > 1, then we have increasing returns to scale, i.e. an increase in K and L by a factor of m results in an
increase in F by more than a factor of m. This is the case for equation (c), where α + β = 0.75 + 0.75 = 1.5. Statement
(G) also applies an increase in inputs by a factor of m = 2 results in an increase in output by more than 2, in this case
by a factor of almost 3. We can match statements (c) and (G) to graph (III), where we see that increasing (K, L) from
(1, 1) to (2, 2) results in a change in F by more than a factor of 2 (but less than a factor of 3). This is called increasing
returns to scale.
This information is summarized in Table 12.13.
Table 12.13
Function
Graph
F (L, K) = L0.25 K 0.25
(II)
Statement
(E)
F (L, K) = L0.5 K 0.5
(I)
(D)
F (L, K) = L0.75 K 0.75
(III)
(G)
1172
Chapter Twelve /SOLUTIONS
34. Suppose P0 is the production given by L0 and K0 , so that
β
P0 = f (L0 , K0 ) = cLα
0 K0 .
We want to know what happens to production if L0 is increased to 2L0 and K0 is increased to 2K0 :
P = f (2L0 , 2K0 )
= c(2L0 )α (2K0 )β
β
β
= c2α Lα
0 2 K0
β
= 2α+β cLα
0 K0
= 2α+β P0 .
Thus, doubling L and K has the effect of multiplying P by 2α+β . Notice that if α + β > 1, then 2α+β > 2, if α + β = 1,
then 2α+β = 2, and if α + β < 1, then 2α+β < 2. Thus, α + β > 1 gives increasing returns to scale, α + β = 1 gives
constant returns to scale, and α + β < 1 gives decreasing returns to scale.
35. (a) If f (x, y) = x0.2 y 0.8 = c, then solving for y gives
y 0.8 =
y=
c
x0.2
c
x0.2
1/0.8
=
A
.
x1/4
Here A is another constant, A = c1/0.8 .
Similarly, if g(x, y) = x0.8 y 0.2 = k, then solving for y gives
y 0.2 =
k
x0.8
k 1/0.2
B
= 4.
x0.8
x
Since the y-values in Figure (I) decay more quickly than those in Figure (II), we see that Figure (I) is g(x, y) and
Figure (II) is f (x, y).
(b) Since the y-values in Figure (III) decrease slower than in Figure (I) and faster than in Figure (II), we have 0.2 < α <
0.8.
y=
36. Using the rules of logarithms on f and g gives
f (x, y) = ln(x0.7 y 0.3 )
g(x, y) = ln(x0.3 y 0.7 ).
Thus the level curves of f are of the form
ln(x0.7 y 0.3 ) = c
so
x0.7 y 0.3 = ec = A
or
y=
A1/0.3
.
x7/3
so
x0.3 y 0.7 = ec = A
or
y=
A1/0.7
.
x3/7
The level curves of g are of the form
ln(x0.3 y 0.7 ) = c
The level curves of h and j are ellipses. For any constant c, the level curve
h(x, y) = 0.3x2 + 0.7y 2 = c
p
p
cuts the x-axis at x =
c/0.3 and the y-axis at y =
c/0.7. Thus the x-intercept is larger than the y-intercept. A
similar argument tells us that the x-intercept of j(x, y) = 0.7x2 + 0.3y 2 = c is smaller than its y-intercept.
Thus Graph (I) is h(x, y); Graph (II) is j(x, y); Graph (III) is f (x, y); Graph (IV) is g(x, y).
1173
12.3 SOLUTIONS
37. (a) Multiply the values on each contour of the original contour diagram by 3. See Figure 12.101.
y
y
6
−8
3
−9
x
0
−3
−11
−6
−12
Figure 12.101: 3f (x, y)
x
−10
Figure 12.102: f (x, y) − 10
(b) Subtract 10 from the values on each contour. See Figure 12.102.
(c) Shift the diagram 2 units to the right and 2 units up. See Figure 12.103.
y
y
0
2
1
−1
x
−2
−1
−4
−2
Figure 12.103: f (x − 2, y − 2)
x
0
−3
Figure 12.104: f (−x, y)
(d) Reflect the diagram about the y-axis. See Figure 12.104.
38. (a) See Figure 12.105.
y
y
0
1
0
1
2
−1
2
3
3
4
4
5
5
Figure 12.105
x
−2
−3
−4
−5
1
2
3
4
5
x
Figure 12.106
(b) See Figure 12.106.
39. Since f (x, y) = x2 − y 2 = (x − y)(x + y) = 0 gives x − y = 0 or x + y = 0, the contours f (x, y) = 0 are the lines
y = x or y = −x. In the regions between them, f (x, y) > 0 or f (x, y) < 0 as shown in Figure 12.107. The surface
z = f (x, y) is above the xy-plane where f > 0 (that is on the shaded regions containing the x-axis) and is below the
1174
Chapter Twelve /SOLUTIONS
xy-plane where f < 0. This means that a person could sit on the surface facing along the positive or negative x-axis, and
with his/her legs hanging down the sides below the y-axis. Thus, the graph of the function is saddle-shaped at the origin.
y
f =0
f=0
f<0
f>0
f>0
x
f<0
y=x
y = −x
Figure 12.107
40. We need three lines with g(x, y) = 0, so that the xy-plane is divided into six regions. For example
g(x, y) = y(x − y)(x + y)
has the contour map in Figure 12.108. (Many other answers to this question are possible.)
y
y = −x
y=x
g<0
g>0
g>0
x
g=0
g<0
g<0
g>0
g=0
g=0
Figure 12.108
41. To read off the cross-sections of f with t fixed, we choose a t value and move horizontally across the diagram looking at
the values on the contours. For t = 0, as we move from the left at x = 0 to the right at x = π, we cross contours of 0.25,
0.50, 0.75 and reach a maximum at x = π/2, and then decrease back to 0. That is because if time is fixed at t = 0, then
f (x, 0) is the displacement of the string at that time: no displacement at x = 0 and x = π and greatest displacement at
x = π/2. For cross-sections with t fixed at larger values, as we move along a horizontal line, we cross fewer contours
and reach a smaller maximum value: the string is becoming less curved. At time t = π/2, the string is straight so we see
a value of 0 all the way across the diagram, namely a contour with value 0. For t = π, the string has vibrated to the other
side and the displacements are negative as we read across the diagram reaching a minimum at x = π/2.
The cross-sections of f with x fixed are read vertically. At x = 0 and x = π, we see vertical contours of value 0
because the end points of the string have 0 displacement no matter what time it is. The cross-section for x = π/2 is found
by moving vertically up the diagram at x = π/2. As we expect, the contour values are largest at t = 0, zero at t = π/2,
and a minimum at t = π.
Notice that the spacing of the contours is also important. For example, for the t = 0 cross-section, contours are most
closely spaced at the end points at x = 0 and x = π and most spread out at x = π/2. That is because the shape of the
string at time t = 0 is a sine curve, which is steepest at the end points and relatively flat in the middle. Thus, the contour
diagram shows the steepest terrain at the end points and flattest terrain in the middle.
42. (a) Since P is proportional to d2 and to v 3 , a formula for P is P (d, v) = kd2 v 3 , where k is the constant of proportionality.
12.3 SOLUTIONS
1175
(b) Let d be the diameter of the original windmill, and let v1 be the wind speed at which the windmill produces 100 kW.
Then
100
kd2 v13 = 100, and thus k = 2 3 .
d v1
The second windmill has diameter 2d and we want to find a speed v2 such that k(2d)2 v23 = 100. We solve for v2 :
d2 v13
100
v3
100
=
=
= 1
3
2
2
2
2
k(2d)
4d
4
100/(d v1 ) · 4d
v1
v2 = √
3
4
v23 =
√
So v2 needs to be 1/ 3 4 of v1 , or about 63% of v1 .
p
(c) Contours of P are curves of the form kd2 v 3 = c, or v = 3 c/(kd2 ). Thus, a contour diagram for P looks like the
diagram in Figure 12.109.
y
40
10
,00
0,0
00
100
,000
1000
x
50
Figure 12.109
Strengthen Your Understanding
43. A contour diagram for z = f (x, y) is a collection of curves in the xy-plane. The contour diagram is like a 2-dimensional
map of the graph of f (x, y), which is a surface in 3-space.
44. The contours of both functions are concentric circles centered at (x, y) = (0, 0). However, for the equally spaced zvalues, such as z = 1, 2, 3, 4 . . ., the contour diagram of f consists of equally spaced concentric circles, whereas the
contour diagram of g consists of circles that get closer and closer together as z increases in value.
45. The function f (x, y) = x2 has contours that are two
√ parallel lines for positive values of z. In particular for z = 10, the
contour of f consists of two parallel lines: x = ± 10. The functions g(x, y) = |y| and h(x, y) = x3 − 9x + 5 also
work. The z = 10 contour of h consists of three parallel lines. There are many others possibilities.
46. If z = f (x, y) = y − x2 , then all the contours have the form y − x2 = c, so y = x2 + c, which are parabolas for every
value of c.
47. Could not be true. If the origin is on the level curve z = 1, then z = f (0, 0) = 1 6= −1. So (0, 0) cannot be on both
z = 1 and z = −1.
48. Might be true. One may consider the function
z = f (x, y) = (x2 + y 2 − 2)(x2 + y 2 − 3) + 1
49. Might be true. The function z = x2 − y 2 + 1 has this property. The level curve z = 1 is the lines y = x and y = −x.
50. Not true. There are no level curves for z > 1 or z ≤ 0.
51. True. For every point (x, y), compute the value z = e−(x
to that value goes through the point (x, y).
2 +y 2 )
at that point. The level curve obtained by getting z equal
52. True. If there were such an intersection point, that point would have two different temperatures simultaneously.
53. True. Different regions that are isolated from each other can have the same temperature.
1176
Chapter Twelve /SOLUTIONS
54. True. If f = c then the contours are of the form c = y 2 + (x − 2)2 , which are circles centered at (2, 0) if c > 0. But if
c = 0 the contour is the single point (2, 0).
55. False. The graph could be a hemisphere, a bowl-shape, or any surface formed by rotating a curve about a vertical line.
56. False. Contours get closer together in a direction if the function is increasing or decreasing at an increasing rate in that
direction.
57. False. As a counterexample, consider any function with one variable missing, e.g. f (x, y) = x2 . The graph of this is not
a plane (it is a parabolic cylinder) but has contours which are lines of the form x = c.
58. False. The fact that the f = 10 and g = 10 contours are identical only says that one horizontal slice through each graph
is the same, but does not imply that the entire graphs are the same. A counterexample is given by f (x, y) = x2 + y 2 and
g(x, y) = 20 − x2 − y 2 .
59. True. The graph of g is the same as the graph of f translated down by 5 units, so the horizontal slice of f at height 5 is
the same as the horizontal slice of g at height 0.
Solutions for Section 12.4
Exercises
1.
Table 12.14
x\y
0.0
1.0
0.0
−1.0
1.0
2.0
2.
3.0
5.0
Table 12.15
x\y
2.0
3.0
−1.0
0.0
1.0
4.0
6.0
8.0
1.0
3.0
5.0
3. A table of values is linear if the rows are all linear and have the same slope and the columns are all linear and have the
same slope. The table does not represent a linear function since none of the rows or columns is linear.
4. A table of values is linear if the rows are all linear and have the same slope and the columns are all linear and have the
same slope. We see that the table might represent a linear function since the slope in each row is 3 and the slope in each
column is −4.
5. A table of values is linear if the rows are all linear and have the same slope and the columns are all linear and have the
same slope. The table might represent a linear function since the slope in each row is 5 and the slope in each column is 2.
6. A table of values is linear if the rows are all linear and have the same slope and the columns are all linear and have the
same slope. The table does not represent a linear function since different rows have different slopes.
7. A contour diagram is linear if the contours are parallel straight lines, equally spaced for equally spaced values of z. This
contour diagram could represent a linear function.
8. A contour diagram is linear if the contours are parallel straight lines, equally spaced for equally spaced values of z. We
see that the contour diagram in the problem does not represent a linear function.
9. Since
0
= c+m·0+n·0
−1 = c + m · 0 + n · 2
c + 2n = −1
−4 = c + m · (−3) + n · 0 c − 3m = −4
we get:
c = 0, m =
Thus, z =
c=0
1
4
x − y.
3
2
4
1
,n = − .
3
2
12.4 SOLUTIONS
1177
10. Let the equation of the plane be
z = c + mx + ny
Since we know the points: (4, 0, 0), (0, 3, 0), and (0, 0, 2) are all on the plane, we know that they satisfy the same equation.
We can use these values of (x, y, z) to find c, m, and n. Putting these points into the equation we get:
0 = c + m · 4 + n · 0 so c = −4m
0 = c + m · 0 + n · 3 so c = −3n
2 = c + m · 0 + n · 0 so c = 2
Because we have a value for c, we can solve for m and n to get
2
1
c = 2, m = − , n = − .
2
3
So the linear function is
z =2−
1
2
x − y.
2
3
11. Figure 12.110 shows the two lines the plane must contain.
Both lines are parallel to the x-axis; thus our plane must have x-slope zero. On the other hand, the line in the xy-plane
is 2 units down and one unit to the right of the line in the xz-plane; hence the y-slope of our plane must be −2. Thus the
equation is
z = 0x − 2y + c = −2y + c,
for some constant c. Since the plane contains the point (0, 0, 2), the value of c must be 2. So the equation is
z = −2y + 2.
z
y
x
Figure 12.110
12. When y = 0, c + mx = 3x + 4, so c = 4, m = 3. Thus, when x = 0, we have 4 + ny = y + 4, so n = 1. Thus,
z = 4 + 3x + y.
13. (a) Since z is a linear function of x and y with slope 2 in the x-direction, and slope 3 in the y-direction, we have:
z = 2x + 3y + c
We can write an equation for changes in z in terms of changes in x and y:
∆z = (2(x + ∆x) + 3(y + ∆y) + c) − (2x + 3y + c)
= 2∆x + 3∆y
Since ∆x = 0.5 and ∆y = −0.2, we have
∆z = 2(0.5) + 3(−0.2) = 0.4
So a 0.5 change in x and a −0.2 change in y produces a 0.4 change in z.
1178
Chapter Twelve /SOLUTIONS
(b) As we know that z = 2 when x = 5 and y = 7, the value of z when x = 4.9 and y = 7.2 will be
z = 2 + ∆z = 2 + 2∆x + 3∆y
where ∆z is the change in z when x changes from 4.9 to 5 and y changes from 7.2 to 7. We have ∆x = 4.9 − 5 =
−0.1 and ∆y = 7.2 − 7 = 0.2. Therefore, when x = 4.9 and y = 7.2, we have
z = 2 + 2 · (−0.1) + 5 · 0.2 = 2.4
14. (a) Substituting in the values for the slopes, we see that the formula for the plane is z = c + 5x − 3y for some value of
c. Substituting the point (4, 3, −2) gives c = −13. The formula for the plane is
z = −13 + 5x − 3y.
(b) When z = 0, we have
0 = −13 + 5x − 3y
3y = 5x − 13
13
5
.
y = x−
3
3
The contour for z = 0 is a line with slope 5/3 and y-intercept 13/3. Similarly we find other contours. See Figure 12.111.
y
5
−48
−36
−24
x
−5
5
−12
0
12
−5
24
Figure 12.111
Problems
15. The revenue function, R, is linear and so we may write it as:
R = (p1 )c + (p2 )d
where p1 is the price of CDs and p2 is the price of DVDs, in dollars. From the diagram, we can pick two points, such as
c = 100, d = 100 on the contour R = 2000, and c = 50, d = 300 on the contour R = 4000. These points give the
following system of linear equations:
2000 = 100p1 + 100p2
4000 = 50p1 + 300p2 .
Solving gives p1 = 8 dollars and p2 = 12 dollars.
16. (a) Yes.
(b) The coefficient of m is 15 dollars per month. It represents the monthly charge to use this service. The coefficient of t
is 0.05 dollars per minute. Each minute the customer is on-line costs 5 cents.
(c) The intercept represents the base charge. It costs $35 just to get hooked up to this service.
(d) We have f (3, 800) = 120. A customer who uses this service for three months and is on-line for a total of 800 minutes
is charged $120.
12.4 SOLUTIONS
1179
17. (a) Expenditure, E, is given by the equation:
E = (price of raw material 1)m1 + (price of raw material 2)m2 + C
where C denotes all the other expenses (assumed to be constant). Since the prices of the raw materials are constant,
but m1 and m2 are variables, we have a linear function.
(b) Revenue, R, is given by the equation:
R = (p1 )q1 + (p2 )q2 .
Since p1 and p2 are constant, while q1 and q2 are variables, we again have a linear function.
(c) Revenue is again given by the equation,
R = (p1 )q1 + (p2 )q2 .
Since p2 and q2 are now constant, the term (p2 )q2 is also constant. However, since p1 and q1 are variables, the (p1 )q1
term means that the function is not linear.
18. The data in Table 12.10 is apparently linear with a slope in the w direction of about 0.9 calories burned for every extra
20 lbs of weight, and a slope in the s direction of about 1.6 calories burned for every extra mile per hour of speed. Since
B = 4.2 when w = 120 and s = 8, a formula for B is
B = 4.2 + 0.9(w − 120) + 1.6(s − 8).
The formula does not make sense for low weights or speeds. For example, it says that a person weighing 120 pounds
going 5 mph burns a negative number of calories per minute, as would a person (child) weighing 60 lbs and going 7 mph.
19. The time in minutes to go 10 miles at a speed of s mph is (10/s)(60) = 600/s. Thus the 120 lb person going 10 mph
uses (7.4)(600/10) = 444 calories, and the 180 lb person going 8 mph uses (7.0)(600/8) = 525 calories. The 120 lb
person burns 444/120 = 3.7 calories per pound for the trip, while the 180 lb person burns 525/180 = 2.9 calories per
pound for the trip.
20. A trip of 10 miles at s mph takes 10/s hours = 600/s minutes. Since the number of calories burned per minute is B, the
total number of calories burned on the trip is B · 600/s. Thus
P =
600(4.2 + 0.9(w − 120) + 1.6(s − 8))
B(600/s)
=
w
sw
21. The function, g, has a slope of 3 in the x direction and a slope of 1 in the y direction, so g(x, y) = c + 3x + y. Since
g(0, 0) = 0, the formula is g(x, y) = 3x + y.
22. The function h decreases as y increases: each increase of y by 2 takes you down one contour and hence changes the
function by 2, so the slope in the y direction is −1. The slope in the x direction is 2, so the formula is h(x, y) = c+2x−y.
From the diagram we see that h(0, 0) = 4, so c = 4. Therefore, the formula for this linear function is h(x, y) = 4+2x−y.
23. For each column in the table, we find that as x increases by 1, f (x, y) increases by 2, so the x slope is 2. For each row in
the table, we find that as y increases by 1, f (x, y) decreases by 0.5, so the y slope is −0.5. So the function has the form
f (x, y) = 2x − 0.5y + c. Also note that f (0, 0) = 1, so c = 1. Therefore, the function is f (x, y) = 2x − 0.5y + 1.
24. For each column in the table, we find that as x increases by 100, f (x, y) decreases by 1, so the x slope is −0.01. For
each row in the table, we find that as y increases by 10, f (x, y) increases by 3, so the y slope is 0.3. So the function
has the form f (x, y) = −0.01x + 0.3y + c. Also note that f (100, 10) = 3, so c = 1. Therefore, the function is
f (x, y) = −0.01x + 0.3y + 1
25. See Figure 12.112.
z
2
−2
y
x
Figure 12.112
1
1180
Chapter Twelve /SOLUTIONS
26. See Figure 12.113.
z
2
1
x
y
2
Figure 12.113
27. See Figure 12.114.
z
4
−4
x
2 y
Figure 12.114
28. See Figure 12.115.
z
6
y
2
3
x
Figure 12.115
29. (a) The contours of f have equation
k = c + mx + ny,
where k is a constant.
Solving for y gives:
k−c
m
x+
n
n
Since c, m, n and k are constants, this is the equation of a line. The coefficient of x is the slope and is equal to −m/n.
y=−
12.4 SOLUTIONS
1181
(b) Substituting x + n for x and y − m for y into f (x, y) gives
f (x + n, y − m) = c + m(x + n) + n(y − m)
Multiplying out and simplifying gives
f (x + n, y − m) = c + mx + mn + ny − nm
f (x + n, y − m) = c + mx + ny = f (x, y)
(c) Part (b) tells us that if we move n units in the x direction and −m units in the y direction, the value of the function
f (x, y) remains constant. Since contours are lines where the function has a constant value, this implies that we remain
on the same contour. This agrees with part (a) which tells us that the slope of any contour line will be −m/n. Since
the slope is ∆y/∆x, it follows that changing y by −m and x by n will keep us on the same contour.
30. (a) We see always the same change in z, namely ∆z = 7, for each step through the table in this diagonal direction. For
example, in the third step of the diagonal starting at 3 we get 24 − 17 = 7, and in the second step of the diagonal
starting at 6 we get 20 − 13 = 7.
(b) We see always the same change in z, namely ∆z = −5, for each step in this direction. For example, in the second
step starting from 19 we get 9 − 14 = −5, and in the first step starting at 22 we get 17 − 22 = −5.
(c) For a linear function, z = mx + ny + c, we have:
z1 − z2 = (mx1 + ny1 + c) − (mx2 + ny2 + c) = m(x1 − x2 ) + n(y1 − y2 ).
Writing ∆z = z1 − z2 , and ∆x = x1 − x2 , and ∆y = y1 − y2 , we have
∆z = m∆x + n∆y.
For the particular linear function in this problem, we have
∆z =
3
4
∆x + ∆y.
5
2
In part (a), as we move down the diagonal, we are taking steps with the same ∆x = 5 and same ∆y = 2. Therefore
we will get the same change in z for each step,
∆z =
3
4
(5) + (2) = 7.
5
2
In part (b), for each step we have ∆x = −10 and ∆y = 2, so for each step
∆z =
3
4
(−10) + (2) = −5.
5
2
31. (a) We have ∆z = 7. Thus
Slope = √
(b) We have ∆z = −5. Thus
7
7
= √ .
52 + 22
29
−5
−5
.
= √
Slope = p
104
(−10)2 + 22
32. Graph (I) has contour lines that slope upward from left to right, so it corresponds to h, j, k, or m. Since the values on the
contour lines are increasing with x and decreasing with y, Graph (I) corresponds to h or j. Since (0, 0, 12) is a point on
the contours of h but not of j for −2 ≤ x, y ≤ 2, the values on the contour lines show that Graph (I) corresponds to h.
Graph (II) has contour lines that slope downward from left to right, so it corresponds to f , g, n, or p. Since the values
on the contour lines are decreasing with x and with y, Graph (II) corresponds to n or p. Since (0, 0, 14) is a point on the
contours of n but not of p for −2 ≤ x, y ≤ 2, the values on the contour lines show that Graph (II) corresponds to n.
Graph (III) has contour lines that slope downward from left to right, so it corresponds to f , g, n, or p. Since the
values on the contour lines are increasing with x and with y, Graph (III) corresponds to f or g. Since (0, 0, 10) is a point
on the contours of f but not of g for −2 ≤ x, y ≤ 2, the values on the contour lines show that Graph (III) corresponds to
f.
Graph (IV) has contour lines that slope upward from left to right, so it corresponds to h, j, k, or m. Since the values
on the contour lines are increasing with y and decreasing with x, Graph (IV) corresponds to k or m. Since (0, 0, 60) is
a point on the contours of m but not of k for −2 ≤ x, y ≤ 2, the values on the contour lines show that Graph (IV)
corresponds to m.
1182
Chapter Twelve /SOLUTIONS
Strengthen Your Understanding
33. The function f (x, y) = ex+y has contours that are parallel lines x + y = c, but it is not linear. This example generalizes
to g(x + y) for any function g(t). The family of functions h(x, y) = r(x) also works, for any function r. There are other
examples.
34. The function f (x, y) = xy has linear cross-sections for both x and y fixed, but it is not linear. Any function of the form
g(x, y) = (mx + b)(ny + c) also satisfies this condition.
35. A possible example is in Table 12.16, where the rows have slopes 1, 2, 3, respectively, and the columns have slopes 1, 2,
3. Notice that the function is not linear since the slopes in each row (and in each column) are different.
Table 12.16
x\y
1
2
3
1
1
2
3
2
2
4
6
3
3
6
9
36. If the linear function is z = mx + ny + c, then the contour for z = 0 is:
mx + ny + c = 0.
We want this line to have slope 2, so we rewrite it in slope-intercept form:
y=−
c
m
x− .
n
n
Thus, we want −m/n = 2, for example m = −2, n = 1. So z = −2x + y is one example. There are others.
37. False. At every point (x, y) the z coordinate on the first plane is 2 units lower than the second so these planes are parallel
and do not intersect.
38. False. The first row is linear with slope 1/0.1 = 10. The second row is linear with slope 1.07/0.1 = 10.7. Since the
slope of the first row is not the same as the slope of the second row, the function is not linear.
39. False. The contours are of the form c = 3x + 2y which are lines with slope −3/2.
40. True. The contours of a linear function f (x, y) = c + mx + ny look like k = c + mx + ny which are parallel lines with
slope −m/n.
41. True. f (0, 0) = 0, f (0, 1) = 4 give a y slope of 4, but f (0, 0) = 0, f (0, 3) = 5 give a y slope of 5/3. Since linearity
means the y slope must be the same between any two points, this function cannot be linear.
42. True. A linear function has constant slopes in the x and y directions, so its graph is a plane.
43. True. Since the graph of a linear function is a plane, any vertical slice parallel to the yz-plane will yield a line.
44. False. Any function of the form f (x, y) = c is linear (with zero slope in both the x and y directions) and has a graph
which is parallel to the xy-plane.
45. True. Functions can have only one value for a given input, so their graphs can intersect a vertical line at most once. A
vertical plane would not satisfy this property, so cannot be the graph of a function.
46. False. There is at least one point where f (a, b) = 0, for example (a, b) = (1, 1). There are an infinite number of other
points lying on the straight-line contour f (a, b) = 0.
47. False. All of the columns have to have the same slope, as do the rows, but the row slopes can differ from the column
slopes.
48. False. Simply knowing where the plane intersects the xy-plane does not determine the plane uniquely. There are an infinite
number of linear functions whose graph intersects the xy-plane in this line. Two examples: f (x, y) = −1 − 2x + y and
g(x, y) = −2 − 4x + 2y.
12.5 SOLUTIONS
1183
Solutions for Section 12.5
Exercises
√
1. (a) Observe that setting f (x, y, z) = c gives a cylinder about the x-axis, with radius c. These surfaces are in graph (I).
(b) By the same reasoning the level curves for h(x, y, z) are cylinders about the y-axis, so they are represented in graph
(II).
2. Points on one of the nested spheres in II have constant distance from the origin, so these spheres are level surfaces
f (x, y, z) = x2 + y 2 + z 2 = c. Points on one of the nested cylinders in I have constant distance from the y-axis, so these
cylinders are level surfaces g(x, y, z) = x2 + z 2 = k.
3. If we solve for z, we get z = 13 (5 − x − 2y), so the level surface is the graph of f (x, y) = 31 (5 − x − 2y).
4. We are looking for all points (x, y, z) whose distance from the origin is 2, that is, (x − 0)2 + (y − 0)2 + (z − 0)2 = 4,
or x2 + y 2 + z 2 = 4, which is a level surface of f (x, y, z) = x2 + y 2 + z 2 .
5. If we solve for z, we get z = (1 − x2 − y)2 , so the level surface is the graph of f (x, y) = (1 − x2 − y)2 .
6. We are looking for all points (x, y, z) whose distance from (a, b, c) is a constant k, that is, (x−a)2 +(y −b)2 +(z −c)2 =
k2 , which is a level surface of f (x, y, z) = (x − a)2 + (y − b)2 + (z − c)2 .
7. Only the elliptical paraboloid, the hyperbolic paraboloid and the plane. These are the only surfaces in the catalog that
satisfy the “vertical line test,” that is, they have at most one z-value for each x and y.
8. An elliptic paraboloid.
9. A hyperboloid of two sheets.
10. A plane.
11. An ellipsoid.
12. Yes,
z = f (x, y) = x2 + 3y 2 .
13. Yes,
z = f (x, y) =
3
2
x + y − 2.
5
5
14. No, because some z values correspond to two points on the surface.
15. No, because z =
Problems
p
x2 + 3y 2 and z = −
p
x2 + 3y 2 , so some z-values correspond to two points on the surface.
16. The plane is represented by
z = f (x, y) = 2x −
y
−3
2
and
g(x, y, z) = 4x − y − 2z = 6.
Other answers are possible
17. The top half of the sphere is represented by
z = f (x, y) =
and
p
10 − x2 − y 2
g(x, y, z) = x2 + y 2 + z 2 = 10,
Other answers are possible.
z ≥ 0.
18. The bottom half of the ellipsoid is represented by
z = f (x, y) = −
p
2(1 − x2 − y 2 )
g(x, y, z) = x2 + y 2 +
Other answers are possible
z2
= 1,
2
z ≤ 0.
1184
Chapter Twelve /SOLUTIONS
19. (a) The isothermal surfaces of f are parallel planes. Each plane is described by the equation
2x − 3y + z = c + 20,
for each value of the constant c.
(b) We have:
fz (0, 0, 0) = 1.
This means that if we start at the point (0, 0, 0) and move slightly upwards in the direction of the positive z-axis, our
temperature is increasing by one degree Fahrenheit for each additional unit we move.
(c) To increase our temperature the fastest we should move away from the isothermal plane passing through (0, 0, 0)
in a direction that allows us to reach the warmer isothermal planes as fast as possible. This means that we should
follow a normal vector to the isothermal plane passing through (0, 0, 0) that has a positive ~k component (temperature
increases with c and c + 20 gives the z-intercept of each isothermal plane). We have:
Isothermal plane through (0, 0, 0) : 2x − 3y + z = 20,
Normal vector to isothermal plane through (0, 0, 0) : ~
n = 2~i − 3~j + ~k .
So, we must move away from the origin in the direction of the vector 2~i − 3~j + ~k .
(d) Isothermal surfaces of f are of the form
z = c + 20 − 2x + 3y,
so, setting c = −3, we see that f (x, y) = −2x + 3y + 17 is an isothermal surface of f . On this surface the
temperature is −3 degrees Fahrenheit.
20. (a) We expect B to be an increasing function of all three variables.
(b) A deposit of $1250 at a 1% annual interest rate leads to a balance of $1276 after 25 months.
21. We expect P to be an increasing function of A and r. (If you borrow more, your payments go up; if the interest rates go
up, your payments go up.) However, P is a decreasing function of t. (If you spread out your payments over more years,
you pay less each month.)
22. The graph of g(x, y) = x + 2y is the set of all points (x, y, z) satisfying z = x + 2y, or x + 2y − z = 0. This is a level
surface, but we want the surface equal to the constant value 1, not 0, so we can add 1 to both sides to get x+2y−z+1 = 1.
Thus, f (x, y, z) = x + 2y − z + 1 has level surface f = 1 identical to the graph of g(x, y) = x + 2y.
p
p
23. If we solve x2 +y 2 /4+z 2 /9 = 1 for z we get z = ±3
p
and g(x, y) = −3 1 − x2 − y 2 /4.
1 − x2 − y 2 /4. Thus we can take f (x, y) = 3
1 − x2 − y 2 /4
24. The equation of any plane parallel to the plane z = 2x+3y−5 has x-slope 2 and y-slope 3, so has equation z = 2x+3y−c
for any constant c, or 2x + 3y − z = c. Thus we could take g(x, y, z) = 2x + 3y − z. Other answers are possible.
25. (a) The graph of f (x, y) is obtained by plotting points
root function is
p (x, y, z), where z = f (x, y). Since the square
1 − x2 − y 2 and squaring both sides leads to x2 + y 2 + z 2 = 1,
never negative, we have z ≥ 0. Setting z =
which is the equation for a sphere of radius 1. The graph of the function includes only those points where z ≥ 0, that
is, the upper hemisphere of radius 1, centered
at the origin.
p
(b) If we take g(x, y, z) = f (x, y) − z = 1 − x2 − y 2 − z, then the level surface g(x, y, z) = 0 is the surface S.
26. (a) The graph of f (x, y) is obtained by plotting p
points (x, y, z), where z = f (x, y). Since the square root function is
never negative, we have z ≥ 0. Setting z = 1 − y 2 and squaring both sides leads to y 2 + z 2 = 1, which is the
equation for a circular cylinder of radius 1 lying along the x-axis (since x is missing from the equation). The graph
of the function includes only those points
p where z ≥ 0, that is, the upper half of the cylinder.
(b) If we take g(x, y, z) = f (x, y) − z = 1 − y 2 − z, then the level surface g(x, y, z) = 0 is the surface S.
p
p
27. Starting with the equation z = x2 + y 2 , we flip the cone and shift it up one, yielding z = 1 − x2 + y 2 . This is a
cone with√vertex at (0, 0, 1) that intersects the xy-plane in a circle of radius 1. Interchanging the variables, we see that
equation whose graph includes the desired cone C. Finally, we express this equation as a level
y = 1 − x2 + z 2 is an √
surface g(x, y, z) = 1 − x2 + z 2 − y = 0.
28. In the xz-plane, the equation x2 /4 + z 2 = 1 is an ellipse, with widest points at x = ±2 on the x-axis and crossing the
z-axis at z = ±1. Since the equation has no y term, the level surface is a cylinder of elliptical cross-section, centered
along the y-axis.
29. Setting y to a constant c yields the equation x2 + z 2 = 1 − c2 /4, which, for −2 ≤ c ≤ 2 gives circular cross-sections.
Fixing x = c yields the equation y 2 /4 + z 2 = 1 − c2 , which for −1 ≤ c ≤ 1 yields elliptical cross-sections. A similar
result is true for cross-sections with constant z. Thus the level surface appears to be a unit sphere, centered at the origin,
that has been stretched by a factor of two in the y-direction (this shape is called an ellipsoid).
12.5 SOLUTIONS
1185
30. The level surfaces are graphs of the equations x + y + z = c for different values of the constant c. These are all parallel
planes.
31. The level surfaces are the graphs of sin(x + y + z) = k for constant k (with −1 ≤ k ≤ 1). This means x + y + z =
sin−1 (k) + 2πn, or π − sin−1 (k) + 2nπ for all integers n. Therefore for each value of k, with −1 ≤ k ≤ 1, we get an
infinite family of parallel planes. So the level surfaces are families of parallel planes.
32. Let’s consider the function y = 2 + sin z drawn in the yz-plane in Figure 12.116.
z
2
y
x
y = 2 + sin z
Figure 12.116
Now rotate this graph around the z-axis. Then, a point (x, y, z) is on the surface if and only if x2 +y 2 = (2+sin z)2 .
Thus, the surface generated is a surface of rotation with the profile shown in Figure 12.116.
Similarly, the surface with equation x2 + y 2 = (f (z))2 is the surface obtained rotating the graph of y = f (z) around
the z-axis.
33. f (x, y, z) = x2 − y 2 + z 2 has 3 types of level surfaces depending on the values of c in the equation x2 − y 2 + z 2 = c.
We write this as x2 + z 2 = y 2 + c and think of what happens as we take a cross-section of the surface, perpendicular to
the y-axis by holding y fixed.
(i) For c > 0, the level surface is a hyperboloid of 1 sheet.
(ii) For c < 0, the level surface is a hyperboloid of 2 sheets.
(iii) For c = 0, the level surface is a cone.
2
2
2
34. The level surfaces are the graphs of g(x, y, z) = e−(x +y +z ) = k for constant values of k such that 0 < k ≤ 1. So
x2 + y 2 + z 2 = − ln k, which is the graph of a sphere since − ln k ≥ 0.
35. The level surfaces are all planes described as follows:
When h(x, y, z) = 1, the plane is given by
ez−y = 1,
so
z − y = ln 1 = 0.
so
z − y = ln e = 1.
so
z − y = ln e2 = 2.
When h(x, y, z) = e, the plane is given by
ez−y = e,
When h(x, y, z) = e2 , the plane is given by
ez−y = e2 ,
See Figure 12.117.
1186
Chapter Twelve /SOLUTIONS
z
✛
✛
h = e2
h=e
y
✛
h=1
x
Figure 12.117
36. For values of f < 4, the level surfaces are spheres, with larger f giving smaller radii. See Figure 12.118.
z
f =0
❘
f =1
✲
x
✒
y
f=2
Figure 12.118
37. For values of g < 1, the level surfaces are cylinders centered on the z-axis, with larger g values giving smaller radii. See
Figure 12.119.
z
g=0
✠
g = −1
✙
✒
g = −2
y
x
Figure 12.119
Strengthen Your Understanding
38. The graph of f (x, y, z) is all points (x, y, z, w) in 4-space such that w = f (x, y, z). This graph cannot be drawn in
3-space; it would need 4 dimensions.
39. Since z is missing in the formula for f (x, y, z), the level surface f (x, y, z) = x2 − y 2 = c is a hyperbolic cylinder. All
its cross-sections perpendicular to the z-axis are the same hyperbola x2 − y 2 = c.
12.6 SOLUTIONS
1187
40. Since z is missing in the formula for f (x, y, z), the level surface f (x, y, z) = x2 + y 2 = c is a cylinder running along
the z-axis.
41. A linear function of three variables has level surfaces that are equally spaced planes. Choosing a linear function that does
not depend on x gives level surfaces perpendicular to the yz-plane. The function f (x, y, z) = y + z, for example, works.
Its level surfaces are the planes: c = y + z, or z = −y + c.
42. A cylinder centered on the y-axis has equation x2 + z 2 = c, so we take f (x, y, z) = x2 + z 2 . There are other possible
answers.
√
43. Let f (x, y, z) = (x + y + z)2 . Then f (x, y, z) = c ≥ 0 means x + y + z = c, which for different c are parallel planes.
44. One family of paraboloids is given by equations of the form z = x2 + y 2 − c, where c is a constant. Rearranging this
equation, we obtain x2 + y 2 − z = c. Therefore, the level sets of the function f (x, y, z) = x2 + y 2 − z are paraboloids.
45. True. Both are the set of all points (x, y, z) in 3-space satisfying z = x2 + y 2 .
p
46. False. The graph of f (x, y) = 1 − x2 − y 2 is the upper unit hemisphere, while the graph of g = 1 is x2 +y 2 +z 2 = 1,
which is the entire unit sphere (both spheres with center at the origin).
47. True. The graph of f (x, y) is the set of all points (x, y, z) satisfying z = f (x, y). If we define the three-variable function
g by g(x, y, z) = f (x, y) − z, then the level surface g = 0 is exactly the same as the graph of f (x, y).
48. False. For example, the function g(x, y, z) = x2 + y 2 + z 2 has level surface g = 1 which is a sphere of radius 1, centered
at the origin. This surface cannot be the graph of any function f (x, y), since a vertical line intersects it in more than one
place.
49. True. The level surfaces are of the form x + 2y + z = k, or z = k − x − 2y. These are the graphs of the linear functions
f (x, y) = k − x − 2y, each of which has x-slope of −1 and y-slope equal to −2. Thus they form parallel planes.
50. False. The level surfaces are of the form x2 + y + z 2 = k, or y = k − x2 − z 2 . These are paraboloids centered on the
y-axis, not cylinders.
51. False. The level surface g = 0 of the function g(x, y, z) = x2 + y 2 + z 2 consists of only the origin.
52. True. The level surfaces g = k are of the form ax + by + cz + d = k, or
z=
1
(−ax − by + (k − d)).
c
Thus z is a linear function of x and y, whose graph is a plane.
53. False. For example, the function g(x, y, z) = sin(x + y + z) has level surfaces of the form x + y + z = k, where
k = arcsin(c) + nπ, for n = 0, ±1, ±2, . . .. These surfaces are planes (for −1 ≤ c ≤ 1).
54. True. If there is a point (a, b, c) lying on both g(x, y, z) = k1 and g(x, y, z) = k2 , then we must have g(a, b, c) = k1
and g(a, b, c) = k2 . Since g is a function, it can only have a single value at a point, so k1 = k2 .
Solutions for Section 12.6
Exercises
1. No, 1/(x2 + y 2 ) is not defined at the origin, so is not continuous at all points in the square −1 ≤ x ≤ 1, −1 ≤ y ≤ 1.
2. The function 1/(x2 + y 2 ) is continuous on the square 1 ≤ x ≤ 2, 1 ≤ y ≤ 2. The functions x2 and y 2 are continuous
everywhere, and so is their sum. The constant function 1 is continuous, and thus so is the ratio 1/(x2 + y 2 ), as long as
x2 + y 2 6= 0. Since the only place x2 + y 2 = 0 is at the origin, and the origin is not included in the square, the function
is continuous in the square.
3. The function y/(x2 + 2) is continuous on the disk x2 + y 2 ≤ 1. The functions x2 + 2 and y are continuous everywhere,
and so is their ratio, as long as the denominator is not 0. But x2 + 2 is always at least 2, so the function is continuous on
the disk (actually at all points in the plane).
4. The function esin x /cos y is continuous on the rectangle − π2 ≤ x ≤ π2 , 0 ≤ y ≤ π4 . The functions sin x and ex
are continuous everywhere, and so is their composition esin x . Then the ratio esin x /cos y is continuous as long as the
denominator is not 0. But cos y is not 0 in the interval 0 ≤ y ≤ π4 , so the function is continuous on the given rectangle.
5. The function tan(θ) is undefined when θ = π/2 ≈ 1.57. Since there are points in the square −2 ≤ x ≤ 2, −2 ≤ y ≤ 2
with x · y = π/2 (e.g. x = 1, y = π/2) the function tan(xy) is not defined inside the square, hence not continuous.
1188
Chapter Twelve /SOLUTIONS
√
2
2
6. The function 2x − y is undefined
√ when 2x − y < 0. Since there are points in the disk x + y ≤ 4 with 2x − y < 0
(e.g. x = 0, y = 1) the function 2x − y is not defined at all points inside the disk and hence is not continuous.
7. Since the composition of continuous functions is continuous, the function f is continuous at (0, 0) and we have
lim
(x,y)→(0,0)
f (x, y) =
lim
(x,y)→(0,0)
e−x−y = e−0−0 = 1
8. Since the composition of continuous functions is continuous, the function f is continuous at (0, 0). We have:
lim
f (x, y) =
(x,y)→(0,0)
lim
(x,y)→(0,0)
(x2 + y 2 ) = 0 + 0 = 0.
9. Since f does not depend on y we have:
lim
(x,y)→(0,0)
f (x, y) = lim
x→0
x
0
=
= 0.
x2 + 1
0+1
10. Since the composition of continuous functions is continuous, the function f is continuous at (0, 0). We have:
lim
(x,y)→(0,0)
f (x, y) =
lim
(x,y)→(0,0)
0+0
x+y
=
= 0.
sin y + 2
0+2
11. We want to compute
lim
(x,y)→(0,0)
f (x, y) =
sin(x2 + y 2 )
.
x2 + y 2
(x,y)→(0,0)
lim
p
As r = x2 + y 2 is the distance from (x, y) to (0, 0) we have that (x, y) → (0, 0) is equivalent to r → 0. Hence the
limit becomes:
sin r 2
lim
f (x, y) = lim
= 1.
r→0
r2
(x,y)→(0,0)
Problems
12. We want to show that f does not have a limit as (x, y) approaches (0, 0). So let us suppose that (x, y) tends to (0, 0)
along the line y = mx, where the slope m 6= 1. Then
f (x, y) = f (x, mx) =
(1 + m)x
x + mx
1+m
=
=
.
x − mx
(1 − m)x
1−m
Therefore
lim f (x, mx) =
x→0
and so for m = 2 we get
1+m
1−m
lim
f (x, y) =
3
1+2
=
= −3
1−2
−1
lim
f (x, y) =
1+3
4
=
= −2.
1−3
−2
(x,y)→(0,0)
y=2x
and for m = 3
(x,y)→(0,0)
y=3x
Thus no matter how close they are to the origin, there will be points (x, y) where the value f (x, y) is close to −3 and
points (x, y) where f (x, y) is close to −2. So the limit:
lim
(x,y)→(0,0)
f (x, y) does not exist.
12.6 SOLUTIONS
1189
13. We want to show that f does not have a limit as (x, y) approaches (0, 0). Let us suppose that (x, y) tends to (0, 0) along
the line y = mx. Then
1 − m2
x 2 − m2 x 2
=
.
f (x, y) = f (x, mx) = 2
x + m2 x 2
1 + m2
Therefore
1 − m2
lim f (x, mx) =
x→0
1 + m2
and so for m = 1 we get
0
1−1
= =0
f (x, y) =
lim
(x,y)→(0,0)
1
+
1
2
y=x
and for m = 0
lim
(x,y)→(0,0)
y=0
f (x, y) =
1−0
= 1.
1+0
Thus no matter how close they are to the origin, there will be points (x, y) such that f (x, y) is close to 0 and points (x, y)
where f (x, y) is close to 1. So the limit:
lim
(x,y)→(0,0)
f (x, y) does not exist.
14. Points along the positive x-axis are of the form (x, 0); at these points the function looks like 2x/2x = 1 everywhere
(except at the origin, where it is undefined). On the other hand, along the y-axis, the function looks like −y 2 /y 2 = −1.
Since approaching the origin along two different paths yields numbers that are not the same, the limit does not exist.
15. We want to show that f does not have a limit as (x, y) approaches (0, 0). For this let us consider x > 0, y > 0, which
gives
xy
= 1.
lim
f (x, y) =
lim
(x,y)→(0,0) |xy|
(x,y)→(0,0)
x>0,y>0
x>0,y>0
On the other hand, if x > 0, y < 0, we get
lim
(x,y)→(0,0)
x>0,y<0
f (x, y) =
lim
(x,y)→(0,0)
x>0,y<0
xy
xy
=
lim
= −1.
|xy| (x,y)→(0,0) −xy
x>0,y<0
Thus no matter how close to the origin they are, there will be points (x, y) such that f (x, y) is close to 1 and points (x, y)
such that f (x, y) is close to −1. So the limit
lim
(x,y)→(0,0)
f (x, y) does not exist.
16. Let us suppose that (x, y) tends to (0, 0) along the curve y = kx2 , where k 6= −1. We get
f (x, y) = f (x, kx2 ) =
1
x2
=
.
x2 + kx2
1+k
Therefore:
lim f (x, kx2 ) =
x→0
1
1+k
and so for k = 0 we get
lim
f (x, y) = 1
lim
f (x, y) =
(x,y)→(0,0)
y=0
and for k = 1
(x,y)→(0,0)
y=x2
1
.
2
Thus no matter how close they are to the origin, there will be points (x, y) where the value f (x, y) is close to 1 and points
(x, y) where f (x, y) is close to 21 . So the limit:
lim
(x,y)→(0,0)
does not exist.
f (x, y)
1190
Chapter Twelve /SOLUTIONS
17. For x > 0, we have
f (x, y) = y.
Thus, the surface representing f for x > 0 is the plane z = y.
For x < 0, we have
f (x, y) = −y.
Thus, the surface representing f for x < 0 is the plane z = −y.
For x = 0, we have
f (x, y) = 0.
Thus, the surface representing f is two half-planes and the y-axis.
(a) The function is continuous at every point on the x-axis.
(b) The function is not continuous at any point on the y-axis, except at the origin, because f (x, y) = 0 on the y-axis and
not nearby unless y = 0.
(c) The function is continuous at the origin.
(a) Yes
(b) No
(c) Yes
18. The function, f is continuous at all points (x, y) with x 6= 3. We analyze the continuity of f at the point (3, a). We have:
lim
f (x, y) = lim (c + y) = c + a
lim
f (x, y) =
(x,y)→(3,a),x<3
(x,y)→(3,a),x>3
y→a
lim
(5 − x) = 2.
x>3,x→3
We want to see if we can find one value of c such that c + a = 2 for all a. This would mean that c = 2 − a, but then c
would be dependent on a. Therefore, we cannot make the function continuous everywhere.
19. The function f is continuous at all points (x, y) with x 6= 3. So let’s analyze the continuity of f at the point (3, a). We
have
lim
f (x, y) = lim (c + y) = c + a
lim
f (x, y) = lim (5 − y) = 5 − a.
(x,y)→(3,a)
x<3
(x,y)→(3,a)
x>3
y→a
y→a
So we need to see if we can find one value for c such that c + a = 5 − a for all a. This would require that c = 5 − 2a,
but then c would depend on a, which is exactly what we don’t want. Therefore, we cannot make the function continuous
everywhere.
20. It is not continuous at (0, 0). The function f (x, y) = x2 + y 2 gets closer and closer to 0 as (x, y) gets closer to the
origin; but the value of f (0, 0) is not 0, it is 2. Since the value of the function is not equal to the limit, the function is not
continuous at the origin.
21. The function f (x, y) = x2 + y 2 + 1 gets closer and closer to 1 as (x, y) gets closer to the origin. To make f continuous
at the origin, we need to have f (0, 0) = 1. Thus c = 1 will make the function continuous at the origin.
22. (a) The graphs are shown in Figure 12.120.
y
z
x
x
y
Figure 12.120
(b) Yes, it seems that if x and y are both close to 0, the values of the function are both close to 0 = f (0, 0).
12.6 SOLUTIONS
1191
23. (a) We have f (x, 0) = 0 for all x and f (0, y) = 0 for all y, so these are both continuous (constant) functions of one
variable.
(b) The contour diagram suggests that the contours of f are lines through the origin. Providing it is not vertical, the
equation of such a line is
y = mx.
To confirm that such lines are contours of f , we must show that f is constant along these lines. Substituting into the
function, we get
f (x, y) = f (x, mx) =
x(mx)
mx2
m
=
=
= constant.
x2 + (mx)2
x 2 + m2 x 2
1 + m2
Since f (x, y) is constant along the line y = mx, such lines are contained in contours of f .
(c) We consider the limit of f (x, y) as (x, y) → (0, 0) along the line y = mx. We can see that
lim f (x, mx) =
x→0
m
.
1 + m2
Therefore, if m = 1 we have
1
2
lim
f (x, y) =
lim
f (x, y) = 0.
(x,y)→(0,0)
y=x
whereas if m = 0 we have
(x,y)→(0,0)
y=0
Thus, no matter how close we are to the origin, we can find points (x, y) where the value f (x, y) is 1/2 and points
(x, y) where the value f (x, y) is 0. So the limit lim(x,y)→(0,0) f (x, y) does not exist. Thus, f is not continuous at
(0, 0), even though the one-variable functions f (x, 0) and f (0, y) are continuous at (0, 0). See Figures 23 and 23
Strengthen Your Understanding
24. For continuity, one also needs the value of the limit to be the same as f (a, b).
25. For the quotient f /g, one also needs g(a, b) 6= 0.
26. Let
f (x, y) =
27. Let
1
1
+
.
x2 + y 2
(x − 1)2 + (y − 2)2
x2 + 2y 2
.
x2 + y 2
Approaching along the x-axis means setting y = 0, so then
f (x, y) =
f (x, y) =
x2
=1
x2
for all x 6= 0.
Thus, the limit approaching (0, 0) along the x-axis is 1.
Approaching along the y-axis means setting x = 0, so then
f (x, y) =
2y 2
= 2,
y2
Thus, the limit approaching (0, 0) along the y-axis is 2.
28. One possible answer is f (x, y) =
(
0 if x < 2
1 if x ≥ 2.
29. One possible answer if f (x, y) = 1/((x − 2)2 + y 2 ).
30. One possible answer is f (x, y) = 1/(x2 + y 2 − 1).
for all y 6= 0.
1192
Chapter Twelve /SOLUTIONS
Solutions for Chapter 12 Review
Exercises
1. The distance of a point P = (x, y, z) from the yz-plane is |x|, from the xz-plane is |y|, and from the xy-plane is |z|.
So A is closest to the yz-plane, since it has the smallest x-coordinate in absolute value. B lies on the xz-plane, since its
y-coordinate is 0. C is farthest from the xy-plane, since it has the largest z-coordinate in absolute value.
2. Your final position is (1, −1, −3). Therefore, you are in front of the yz-plane, to the left of the xz-plane, and below the
xy-plane.
3. An example is the line z = −x in the xz-plane. See Figure 12.121.
z
y
x
Figure 12.121
4. Given (x, y) we can solve uniquely for z, namely z = 5 − 3x + 2y. Thus, z is a function of x and y:
z = f (x, y) = 5 − 3x + 2y.
5. The equation x2 + y 2 + z 2 = 100 does not determine z uniquely from x and y. For example, the points (0, 0, 10) and
(0, 0, −10) both satisfy the equation. Therefore z is not a function of (x, y).
6. Given (x, y) we can solve uniquely for z, namely z = 2 +
z = f (x, y) = 2 +
y
3x2
x
+ −
+ y 2 . Thus, z is a function of x and y:
5
5
5
y
3x2
x
+ −
+ y2.
5
5
5
7. Planes perpendicular to the positive y-axis should yield the graphs of upright parabolas f (x, y), which widen as y decreases (giving f (x, 2) and f (x, 1)). When y = 0, the parabola flattens out, creating a horizontal line for f (x, 0). The
graphs then turn downward, creating the parabolas f (x, −1) and f (x, −2) which become narrower as y decreases. So the
graph (IV) bests fits this information.
8. (a) is (IV). The level curves of f and g are lines, with slope of f = −1 and slope of g = 1. See Figure 12.122.
(b) is (II). The level curves of f and g are lines, with slope of f = −2/3 and slope of g = 2/3. See Figure 12.123.
(c) is (I). The level curves of f are parabolas opening upward; the level curves of g are the shape of ln x, but upside
down and for both positive and negative x-values. See Figure 12.124.
(d) is (III). The level curves of f are hyperbolas centered on the x- or y-axes; the level curves of g are rectangular
hyperbolas in quadrants (I) and (III) or quadrants (II) and (IV). See Figure 12.125.
SOLUTIONS to Review Problems for Chapter Twelve
y
y
g = c2
g = c2
f = c1
x
f = c1
x
Figure 12.122
y
1193
Figure 12.123
y
✛
f = c1
✛
g = c2
✛
x
x
✛
Figure 12.124
g = c2
f = c1
Figure 12.125
9. (a) is (I), because there is a minimum at the origin and the surface slopes steadily upward.
(b) is (IV), because there is a maximum at the origin and the surface slopes increasingly steeply downward as we
move away from the origin.
(c) is (II), because there is a maximum at the origin and the surface slopes steadily downward.
(d) is (III), because there is a minimum at the origin and the surface slopes increasingly fast upward as we move
away from the origin.
10. Contours are lines of the form 3x − 5y + 1 = c as shown in Figure 12.126. Note that for the regions of x and y given, the
c values range from −12 < c < 12 and are evenly spaced.
y
2
1
2
−1
−8
−4
0
x
4
8
−1
−2
−2
12
−1
1
2
Figure 12.126
11. Since setting z = c, with −1 ≤ c ≤ 1 gives y = sin−1 c + 2nπ or y = π − sin−1 c + 2nπ =constant, where n is any
integer, contours are horizontal lines as shown in Figure 12.127.
1194
Chapter Twelve /SOLUTIONS
y
2
0.95
0.95
1
0.8
0.2
0.6
0.4
x
−0.2
−0.4
−0.6
−1
−0.8
−0.95
−2 −0.95
−2
−1
1
2
Figure 12.127
12. Contours are ellipses of the form 2x2 + y 2 = c as shown in Figure 12.128. Note that for the ranges of x and y given, the
range of c value is 1 ≤ c < 9 and are closer together farther from the origin.
y
2
1
1
x
7
5
3
−1
−2
−2
−1
1
2
Figure 12.128
13. The contours are ellipses of the form 2x2 + y 2 = − ln c as shown in Figure 12.129. For the ranges of x and y given, the
c values range from just above 0 to 1.
y
2
1
0
0 .8
0. 0.2 .5
0
0.
01 7
x
−1
−2
−2
−1
1
2
Figure 12.129
14. These conditions describe a line parallel to the z-axis which passes through the xy-plane at (2, 1, 0).
15. The equation is (x − 1)2 + (y − 2)2 + (z − 3)2 = 25
SOLUTIONS to Review Problems for Chapter Twelve
1195
16. The equation will be of the form mx + ny + ez = d, but you can divide through by d to get an equation of the form
ax + by + cz = 1 (d can not be zero, as the origin is not in the plane). Now plug in the points: From (0, 0, 2), we get
a(0) + b(0) + c(2) = 1. From this we get c = 12 . Similarly we get a = 51 , and b = 13 . So the equation that fits these
points is
y
z
x
+ + = 1.
5
3
2
The equation of this plane can also be obtained by calculating the normal as the cross product of two vectors lying in the
plane.
17. We complete the square
x2 + 4x + y 2 − 6y + z 2 + 12z = 0
2
x + 4x + 4 + y 2 − 6y + 9 + z 2 + 12z + 36 = 4 + 9 + 36
(x + 2)2 + (y − 3)2 + (z + 6)2 = 49
The center is (−2, 3, −6) and the radius is 7.
18. A contour diagram is linear if the contours are parallel straight lines, equally spaced for equally spaced values of z. This
contour diagram does not represent a linear function.
19. A contour diagram is linear if the contours are parallel straight lines, equally spaced for equally spaced values of z. This
contour diagram could represent a linear function.
20. (a) Since the function is linear, the increment between successive entries in the same column is constant. From the third
column we see that the increment is 2 − 8 = −6. Subtract 6 to go from any entry in the table to the entry below it,
and add 6 to get the entry above it. See Table 12.17.
Table 12.17
y
2.5
x
3.0
3.50
8
−1
6
7
1
0
1
2
3
−6
−5
−4
(b) From the third column of the table we calculate
Slope in x-direction = m =
2−8
= −3.
1 − (−1)
From the first row of the table we calculate
Slope in y-direction = n =
8−6
= 2.
3.5 − 2.5
The equation of the linear function is
f (x, y) = z0 + m(x − x0 ) + n(y − y0 )
= f (−1, 2.5) − 3(x − (−1)) + 2(y − 2.5) = −2 − 3x + 2y.
21. The level surfaces appear to be circular cylinders centered on the z-axis. Since they don’t change with z, there is no z in
the formula, and we can use the formula for a circle in the xy-plane, x2 + y 2 = r 2 . Thus the level surfaces are of the
form f (x, y, z) = x2 + y 2 = c for c > 0.
22. The paraboloid is z = x2 + y 2 + 5, so it is represented by
z = f (x, y) = x2 + y 2 + 5
and
Other answers are possible.
g(x, y, z) = x2 + y 2 + 5 − z = 0.
1196
Chapter Twelve /SOLUTIONS
23. Plane is (x/2) + (y/3) + (z/4) = 1, so it is represented by
z = f (x, y) = 4 − 2x −
and
g(x, y, z) =
4
y
3
x
y
z
+ + = 1.
2
3
4
Other answers are possible.
24. The upper half of the sphere is represented by
z = f (x, y) =
and
p
1 − x2 − y 2
g(x, y, z) = x2 + y 2 + z 2 = 1.
Other answers are possible.
25. The sphere is (x − 3)2 + y 2 + z 2 = 4, so the lower half is represented by
z = f (x, y) = −
and
Other answers are possible.
p
4 − (x − 3)2 − y 2
g(x, y, z) = (x − 3)2 + y 2 + z 2 = 4.
26. The level surfaces have equation cos(x + y + z) = c. For each value of c between −1 and 1, the level surface is an infinite
family of planes parallel to x + y + z = arccos(c). For example, the level surface cos(x + y + z) = 0 is the family of
planes
π
x + y + z = ± 2nπ, n = 0, 1, 2, . . . .
2
27. A cylindrical surface.
28. A cone.
29. (a) The contours of g are parallel straight lines, and equally spaced function values correspond to equally spaced contours.
These are the characteristics of the contour diagram of a linear function.
(b) The zero contour goes through the origin, so g(0, 0) = 0 is one value of the function.
The slope m in the x-direction, obtained from the function values at (0, 0) and (50, 0), is
m=
10000 − 0
g(50, 0) − g(0, 0)
=
= 200.
50 − 0
50 − 0
The slope n in the y-direction, obtained from the function values at (0, 0) and (0, 50), is
n=
g(0, 50) − g(0, 0)
5000 − 0
=
= 100.
50 − 0
50 − 0
We have the formula
g(x, y) = z0 + m(x − x0 ) + n(y − y0 )
= g(0, 0) + m(x − 0) + n(y − 0) = 200x + 100y.
Problems
30. The cross-sections perpendicular to the t-axis are sine curves of the form g(x, b) = (cos b) sin 2x; these have period π.
The cross-sections perpendicular to the x-axis are cosine curves of the form g(a, t) = (sin 2a) cos t; these have period
2π.
SOLUTIONS to Review Problems for Chapter Twelve
1197
g(x, t)
1
π
2
π
2
x
π
π
2π
t
Figure 12.130: Graph g(x, t) = cos t sin 2x
1
✛
0.5
✛
0
π
4
1
t = 0, 2π
t=
π
2
0.5
π 5π
,
3 3
x
π
3π
4
✛
x=
✛
π
4
x=
π 5π
,
12 12
0
−0.5
−0.5
−1
−1
Figure 12.131: Cross-section
g(x, b) = (cos b) sin 2x,
with b = 0, π/3, 5π/3, 2π
π
2
π
3π
2
2π
t
Figure 12.132: Cross-section
g(a, t) = (sin 2a) cos t
with a = π/12, π/4, 5π/12
31. If
P0 = f (L0 , K0 ) = 1.01L0.75
K00.25
0
then replacing L0 and K0 by 2L0 and 2K0 gives
f (2L0 , 2K0 ) = 1.01(2L0 )0.75 (2K0 )0.25
= 20.75 20.25 · 1.01L0.75
K00.25
0
= 2f (L0 , K0 )
= 2P0 .
So, doubling labor and capital doubles production.
32. (a) The level curve f = 1 is given by
p
x2 + y 2 + x = 1
Since
p
p
x2 + y 2 = 1 − x.
x2 + y 2 ≥ 0, we must have x ≤ 1. Squaring gives
x2 + y 2 = (1 − x)2 = 1 − 2x + x2
So the level curve is given by
1
1
x = − y2 +
2
2
with x ≤ 1. Looking at the equation for the level curve, x always satisfies x ≤ 1 since x ≤ 21 . This means the
level curve f = 1 is the parabola x = − 12 y 2 + 12 . See Figure 12.133.
Similarly, the level curve f = 2 has equation, valid for x ≤ 2,
p
x2 + y 2 = 2 − x
x2 + y 2 = 4 − 4x + x2
1
x = − y2 + 1
4
1198
Chapter Twelve /SOLUTIONS
The level curve f = 3 has equation, valid for x ≤ 3,
p
x2 + y 2 = 3 − x
x2 + y 2 = 9 − 6x + x2
3
1
x = − y2 + .
6
2
Both f = 2 and f = 3 are valid for all x and y satisfying the respective equations, so the level curves are parabolas.
See Figure 12.133.
(b) The level curve f = c has equation, valid for x ≤ c,
p
x2 + y 2 = c − x
x2 + y 2 = c2 − 2cx + x2
c
1
x = − y2 + .
2c
2
If c > 0, then any x satisfying this equation satisfies x ≤ 2c , so we have x < c. Thus, the level curve exists for c > 0.
If c < 0, then any x satisfying the level curve equation also satisfies x ≥ 2c , so x > c (since c is negative). Thus, the
level curves do not exist for c < 0. If c = 0, we get the level curve y = 0 with x ≤ 0. Summarizing, we have that
level curves exist only for c ≥ 0.
y
f =c
x
Figure 12.133
33. (a) You can see the sequence of values 1, 2, 3, 4, 5, 6, . . . as you follow diagonal paths in the table upward to the right,
changing to the next lower diagonal after reaching the top x = 1 row. The pattern continues in the same way, giving
Table 12.18
Table 12.18
y
1
x
1
1
2
2
3
4
4
7
5
11
6
16
2
3
ր
5
ր
ր
ր
ր
8
12
17
23
3
ր
ր
ր
ր
ր
6
9
13
18
24
31
4
ր
ր
ր
ր
ր
10
14
19
25
32
40
5
ր
ր
ր
ր
ր
15
20
26
33
41
50
6
ր
ր
ր
ր
ր
21
27
34
42
51
61
(b) It appears that the value of f increases by 1 whenever x is decreased by 1 and y is increased by 1. To check this,
compute
f (x − 1, y + 1) = (1/2)((x − 1) + (y + 1) − 2)((x − 1) + (y + 1) − 1) + (y + 1)
= (1/2)(x + y − 2)(x + y − 1) + y + 1
= f (x, y) + 1
SOLUTIONS to Review Problems for Chapter Twelve
1199
It appears that the value of f increases by 1 when moving from a point (1, y) to the point (y + 1, 1). To check
this, compute
f (y + 1, 1) = (1/2)((y + 1) + 1 − 2)((y + 1) + 1 − 1) + 1
1
1
= y2 + y + 1
2
2
= (1/2)(1 + y − 2)(1 + y − 1) + y + 1
= f (1, y) + 1
34. Let us suppose that (x, y) approaches (0, 0) along the line y = x. Then
f (x, y) = f (x, x) =
x
x3
= 2
.
x4 + x2
x +1
Therefore
lim
(x,y)→(0,0)
y=x
f (x, y) = lim
x→0
x
= 0.
x2 + 1
On the other hand, if (x, y) approaches (0, 0) along the parabola y = x2 we have
f (x, y) = f (x, x2 ) =
x4
1
=
2x4
2
and
lim
(x,y)→(0,0)
y=x2
f (x, y) = lim f (x, x2 ) =
x→0
1
.
2
Thus no matter how close they are to the origin, there will be points (x, y) such that f (x, y) is close to 0 and points (x, y)
such that f (x, y) is close to 12 . So the limit
lim
f (x, y)
(x,y)→(0,0)
does not exist.
35. Points along the positive x-axis are of the form (x, 0); at these points the function looks like x/2x = 1/2 everywhere
(except at the origin, where it is undefined). On the other hand, along the y-axis, the function looks like y 2 /y = y, which
approaches 0 as we get closer to the origin. Since approaching the origin along two different paths yields numbers that
are not the same, the limit does not exist.
36. We will study the continuity of f at (a, 0). Now f (a, 0) = 1 − a. In addition:
lim
f (x, y) = lim (1 − x) = 1 − a
lim
f (x, y) = lim −2 = −2.
(x,y)→(a,0)
y>0
(x,y)→(a,0)
y<0
x→a
x→a
If a = 3, then
lim
(x,y)→(3,0)
y>0
f (x, y) = 1 − 3 = −2 =
lim
(x,y)→(3,0)
y<0
f (x, y)
and so lim(x,y)→(3,0) f (x, y) = −2 = f (3, 0). Therefore f is continuous at (3, 0).
On the other hand, if a 6= 3, then
lim
(x,y)→(a,0)
y>0
f (x, y) = 1 − a 6= −2 =
lim
(x,y)→(a,0)
y<0
f (x, y)
so lim(x,y)→(a,0) f (x, y) does not exist. Thus f is not continuous at (a, 0) if a 6= 3.
Thus, f is not continuous along the line y = 0. (In fact the only point on this line where f is continuous is the point
(3, 0).)
1200
Chapter Twelve /SOLUTIONS
37. (a) A student with SATs of 1050 and a GPA of 3.0 has a z-value given by
z = 0.003 · 1050 + 0.8 · 3.0 − 4 = 1.55
Since 1.55 < 2.3, this student will not be admitted.
(b) A student with SATs of 1600 and GPA of y has a z-value given by
z = 0.003 · 1600 + 0.8y − 4 = 0.8 + 0.8y = 0.8(y + 1)
Since 0.8(y +1) may be greater than or less than 2.3, not all of the students with SAT scores of 1600 will be admitted.
(c) A student with GPA of 4.3 and SATs of x has a z-value given by
z = 0.003x + 0.8 · 4.3 − 4 = 0.003x − 0.56
Since 0.003x − 0.56 may be greater than or less than 2.3, not all of the students with a high school GPA of 4.3 will
be admitted.
(d) See Figure 12.134.
y
4.3
4
4
✛
3
Admission area
2
3
1
2
0
−1
1
−2
0
x
400
1000
1600
Figure 12.134
(e) If ∆x = 100, then ∆z = 0.003∆x = 0.003 · 100 = 0.3.
If ∆y = 0.5, then ∆z = 0.8 · 0.5 = 0.4.
An extra 0.5 of high school GPA increases a student’s z-value by more than an extra 100 points on the SAT. Thus,
the increase in GPA is more important.
38. (a) The plane y = 1 intersects the graph in the parabola z = (x2 + 1) sin(1) + x = x2 sin(1) + x + sin(1). Since sin(1)
is a constant, z = x2 sin(1) + x + sin(1) is a quadratic function whose graph is a parabola.
Any plane of the form y = a will do as long as a is not a multiple of π.
(b) The plane y = π intersects the graph in the straight line z = π 2 x. (Since sin π = 0, the equation becomes linear,
z = π 2 x if y = π.)
(c) The plane x = 0 intersects the graph in the curve z = sin y.
39. (a) To find the level curves, we let T be a constant.
T = 100 − x2 − y 2
x2 + y 2 = 100 − T,
√
which is an equation for a circle of radius 100 − T centered at the origin. At T = 100◦ , we have
√ a circle of radius
◦
0 (a point). At T = 75 √
, we have a circle of radius 5. At T = 50◦ , we have a circle of radius 5 2. At T = 25◦ , we
have a circle of radius 5 3. At T = 0◦ , we have a circle of radius 10.
SOLUTIONS to Review Problems for Chapter Twelve
1201
y
✛
10
T =0
✛
✛
T = 100
❥
✒
T = 50
T = 75
x
10
✒
✒
✛
Motion of bug
✒
T = 25
Figure 12.135
(b) No matter where we put the bug, it should go straight toward the origin—the hottest point on the xy-plane. Its
direction of motion is perpendicular to the tangent lines of the level curves, as can be seen in Figure 12.135.
40. Let the equation of the plane be z = ax + by + c. When z = 0, the line on the xy-plane is ax + by + c = 0. Since we
know that the plane intersects the xy-plane along the line y = 2x + 2 we have b 6= 0 and
−
a
=2
b
c
=2
b
−
Since (1, 2, 2) lies on the plane, we can use the equation z = ax + by + c to get
2 = a + 2b + c
Solving the equations gives
a = 2,
b = −1,
c = 2.
Hence z = 2x − y + 2 and the linear function is f (x, y) = 2x − y + 2.
p
41. (a) Since z = c, where −1 ≤ c ≤ 1 is a constant, gives x2 + y 2 = ± cos−1 (c) + 2kπ, where k is any integer
such that ± cos−1 (c) + 2kπ is non-negative, or x2 + y 2 = r 2 , where r = ± cos−1 (c) + 2kπ, which represents
a family of circles of radius r centered at (0, 0), the level curves of the function are families of circles, as shown in
Figure 12.136.
.5
−
−
−
.5
0
.5
0
.5
−
.5
y
.5
1
0
1
x
Figure 12.136
(b) The plane containing the x- and z-axes is the plane y = 0. Thus the cross-section is z = cos
cos x, as shown in Figure 12.137.
√
x2 + 02 = cos(|x|) =
1202
Chapter Twelve /SOLUTIONS
(c) Denote the line y = x in the xy-plane as r-axis and put units on it such that the units on the√r-axis coincide with the
units on the x-axis and y-axis, namely, r 2 = x2 + y 2 . Thus, the cross-section is z = cos r 2 = cos(|r|) = cos r,
as shown in Figure 12.138.
z
z
1
1
π
−π
x
Figure 12.137
r
π
−π
Figure 12.138
42. The function y = f (x, 0) = cos 0 sin x = sin x gives the displacement of each point of the string when time is held fixed
at t = 0. The function f (x, 1) = cos 1 sin x = 0.54 sin x gives the displacement of each point of the string at time t = 1.
Graphing f (x, 0) and f (x, 1) gives in each case an arch of the sine curve, the first with amplitude 1 and the second with
amplitude 0.54. For each different fixed value of t, we get a different snapshot of the string, each one a sine curve with
amplitude given by the value of cos t. The result looks like the sequence of snapshots shown in Figure 12.139.
f (x, 0) = sin x
f (x, 1) = 0.54 sin x
y
1
✠
✠
0.54
π
−0.54
x
−1
Figure 12.139
43. The function f (0, t) = cos t sin 0 = 0 gives the displacement of the left end of the string as time varies. Since that point
remains stationary, the displacement is zero. The function f (1, t) = cos t sin 1 = 0.84 cos t gives the displacement of the
point at x = 1 as time varies. Since cos t oscillates back and forth between 1 and −1, this point moves back and forth with
maximum displacement of 0.84 in either direction. Notice the maximum displacements are greatest at x = π/2 where
sin x = 1.
44. (a) For t = 0, we have y = f (x, 0) = sin x, 0 ≤ x ≤ π, as in Figure 12.140.
y
1
√
y
2
2
x
π
π/2
π/2
Figure 12.140
π
x
Figure 12.141
√
For t = π/4, we have y = f (x, π/4) = 22 sin x, 0 ≤ x ≤ π, as in Figure 12.141.
For t = π/2, we have y = f (x, π/2) = 0, as in Figure 12.142.
y
y
π
Figure 12.142
x
−
π/2
√
2
2
Figure 12.143
π
x
SOLUTIONS to Review Problems for Chapter Twelve
1203
√
For t = 3π/4, we have y = f (x, 3π/4) = −2 2 sin x, 0 ≤ x ≤ π, as in Figure 12.143.
For t = π, we have y = f (x, π) = − sin x, 0 ≤ x ≤ π, as in Figure 12.144.
y
π/2
π
x
−1
Figure 12.144
(b) The graphs show an arch of a sine wave which is above the x-axis, concave down at t = 0, is straight along the x-axis
at t = π/2, and below the x-axis, concave up at t = π, like a guitar string vibrating up and down.
45. (a) For g(x, t) = cos 2t sin x, our snapshots for fixed values of t are still one arch of the sine curve. The amplitudes,
which are governed by the cos 2t factor, now change twice as fast as before. That is, the string is vibrating twice as
fast.
(b) For y = h(x, t) = cos t sin 2x, the vibration of the string is more complicated. If we hold t fixed at any value, the
snapshot now shows one full period, i.e. one crest and one trough, of the sine curve. The magnitude of the sine curve
is time dependent, given by cos t. Now the center of the string, x = π/2, remains stationary just like the end points.
This is a vibrating string with the center held fixed, as shown in Figure 12.145.
y
✠
t=π
t=
3π
4
✠
π
2
x
■π
■ t= π
2
■
π
t=
t=0
4
Figure 12.145: Another vibrating string: y = h(x, t) = cos t sin 2x
CAS Challenge Problems
46. (a) Let C = (x, y, 0). Since distance AC = 2 we have x2 + y√2 = 22 , and since distance
BC = 2 we have (x − 2)2 +
√
2
2
y = 2 . Solving these two equations, we have C = (1, 3, 0) or C = (1, − 3, 0). We will pick the first choice
(the second choice gives different answers in the next part).
(b) Let D = (x, y, z). Distance DA = 2 implies that x2 + y 2 + z 2√= 4. Distance DB = 2 implies that (x − 2)2 +
2
2
2
4. Solving these three equations, we
y 2 + z 2 = 4 . Distance
√ DC = 2√implies
√ that (x − 1) + (y√− 3) + z√ =√
have: x = 1,
√ y =√1/ √3, z = 2 2/ 3 or x = 1, y = 1/ 3, z = −2 2/ 3. Picking the first choice we have
D = (1, 1/ 3, 2 2/ 3).
(c) The figure is a tetrahedron, that is, a polyhedron with four faces, each of which is an equilateral triangle: ABC,
ABD, ACD, BCD.
47. (a)
f (x, f (x, y)) = 3 + x + 2(3 + x + 2y) = (3 + 2 · 3) + (1 + 2)x + 22 y = 9 + 3x + 4y
f (x, f (x, f (x, y))) = 3 + x + 2(3 + x + 2(3 + x + 2y))
= (3 + 2 · 3 + 22 · 3) + (1 + 2 + 22 )x + 23 y = 21 + 7x + 8y
(b) From part (a) we guess that the general pattern for k nested f s is
(3 + 2 · 3 + 22 · 3 + · · · + 2k−1 · 3) + (1 + 2 + 22 + · · · + 2k−1 )x + 2k y
Thus
f (x, f (x, f (x, f (x, f (x, f (x, y)))))) =
(3 + 2 · 3 + 22 · 3 + · · · + 25 · 3) + (1 + 2 + 22 + · · · + 25 )x + 26 y = 189 + 63x + 64y.
1204
Chapter Twelve /SOLUTIONS
48. (a) Since f (1, 1, 1) = 16, f (1, 1, 2) = 21, an increase of 1 in z increases the value of f by 5. Thus we estimate
f (1, 1, 3) ≈ 21 + 5 = 26. Similarly, since f (1, 0, 1) = 20, f (1, 1, 1) = 16, an increase of 1 in y decreases the value
of f by 4. So we estimate f (1, 2, 1) ≈ 16 − 4 = 12.
(b) When x and y are fixed at 1, f is a linear function of z, thus the linear approximation will give a precise answer for
f (1, 1, 3). However, when x and z are fixed at 1, f is the sum of an exponential function of y and a linear function,
thus the linear approximation will not be accurate for f (1, 2, 1).
(c) Since f (x, y, z) = ax2 +byz +czx3 +d2x−y and f (1, 0, 1) = 20, f (1, 1, 1) = 16, f (1, 1, 2) = 21, f (0, 0, 1) = 6,
we have
a + c + 2d = 20
a + b + c + d = 16
a + 2b + 2c + d = 21
d=6
Solving for a, b, c, d, we get f (x, y, z) = 5x2 + 2yz + 3zx3 + 6 · 2x−y .
(d) f (1, 1, 3) = 26, which matches the estimate in part (a). f (1, 2, 1) = 15, which does not agree with the estimate in
part (a).
PROJECTS FOR CHAPTER TWELVE
1. (a) The Leq is greatest near the runways. The largest contour marked is 72 dB; Heathrow’s two main runways
are located within this contour and run east-west. The noise level on the runways exceeds 72 dB. For
comparison, the noise level 50 feet from the edge of a freeway in mid-morning is about 76 dB.
(b) Since the prevailing wind is from the west, the planes take off towards the west and land coming in
from the east, which explains why the contours are aligned east-west. Many of the planes taking off want
eventually to go east, so they turn off to the southwest to start a U-turn back to the east. A limited number,
particularly those heading for transatlantic flights, turn right; fewer still head due west on a straight-out
departure. When planes are approaching or departing, they have to use considerable power at low altitude,
and hence are significantly noisier; this noise is concentrated at the end of the runways.
(c) The noise level falls off rapidly to the north and south of the runways. This is reflected in the fact that the
contours are very close together along the length of both runways.
(d) Suppose the decibel measure of the sound, B1 , on one contour is given by
L1
B1 = 10 log10
L0
and the decibel measure of the sound on the next higher contour, B2 , is
L2
.
B2 = 10 log10
L0
Since contours are labeled at 3 dB intervals,
L2
L1
L2 /L0
L2
3 = B2 − B1 = 10 log10
− log10
= 10 log10
= 10 log10
.
L0
L0
L1 /L0
L1
Solving for L2 /L1 gives
L2
= 103/10 ≈ 2.
L1
Thus, moving from one contour to the next at 3 dB higher corresponds to approximately doubling the
sound intensity.
(e) We have shown that an increase of 3 dB corresponds approximately to doubling the sound intensity, so a
decrease of 3 dB corresponds approximately to halving the sound intensity. We are told that the new jets
will make 50% less noise, so sound intensity will be halved. Thus, we will subtract 3 dB from each contour
value; for example, the present 57 dB contour will be labeled as 54 dB.
PROJECTS FOR CHAPTER TWELVE
1205
2. (a) About 15 feet along the wall, because that’s where there are regions of cold air (55◦ F and 65◦ F).
(b) Roughly between 10 am and 12 noon, and between 4 pm and 6 pm.
(c) Roughly between midnight and 2 am, between 10 am and 1 pm, and between 4 pm and 9 pm, since that is
when the temperature near the heater is greater than 80◦ F.
(d)
◦
◦
F
85
6 am
75
75
65
65
55
85
11 am
55
10
◦
F
85
20
30
ft
10
◦
F
75
75
65
65
55
30
ft
F
85
3 pm (15 hours)
20
5 pm (17 hours)
55
10
20
30
ft
10
20
30
ft
Figure 12.146
(e)
◦
◦
F
◦
F
85
85
85
75
75
75
65
65
65
55
55
hr
8
16 24
Figure 12.147: Temp. vs.
Time at heater
F
55
hr
8
16 24
Figure 12.148: Temp. vs.
Time at window
hr
8
16 24
Figure 12.149: Temp. vs.
Time midway between
heater and window
(f) The temperature at the window is colder at 5 pm than at 11 am because the outside temperature is colder
at 5 pm than at 11 am.
(g) The thermostat is set to roughly 70◦ F. We know this because the temperature in the room stays close to
70◦ F until we get close (a couple of feet) to the window.
(h) We are told that the thermostat is about 2 feet from the window. Thus, the thermostat is either about 13 feet
or about 17 feet from the wall. If the thermostat is set to 70◦ F, every time the temperature at the thermostat
goes over or under 70◦ F, the heater turns off or on. Look at the point at which the vertical lines at 13 feet
or about 17 feet cross the 70◦ F contours. We need to decide which of these crossings correspond best with
the times that the heater turns on and off. (These times can be seen along the wall.) Notice that the 17
foot line does not cross the 70◦ F contour after 16 hours (4 pm). Thus, if the thermostat were 17 feet from
the wall, the heater would not turn off after 4 pm. However, the heater does turn off at about 21 hours (9
pm). Since this is the time that the 13 foot line crosses the 70◦ F contour, we estimate that the thermostat
is about 13 feet away from the wall.
1206
Chapter Twelve /SOLUTIONS
3. (a) Let x = distance (microns) from center of waveguide, t = time (nanoseconds) as shown in the problem,
and I = intensity of light as marked on the given level curves.
I
1.2
0.9
0.6
0.3
−10
−5
0
I
−5
−10
−5
Time
0
I
−5
x
10
5
x
10
5
−10
t=8
0
I
−10
Time
0
I
−5
x
10
5
t=6
1.2
0.9
0.6
0.3
x
10
5
0
−5
Time
Time
t=2
1.2
0.9
0.6
0.3
t=4
1.2
0.9
0.6
0.3
−10
1.2
0.9
0.6
0.3
t=0
1.2
0.9
0.6
0.3
−10
I
Time
x
10
5
Time
t = 10
0
x
10
5
Figure 12.150
(b) Two waves would start out at opposite ends of the screen. The wave on the left would be slightly taller
and narrower than the wave on the right. The waves would move toward one another, the wave on the right
moving a little faster. They would meet to the left of the center and appear to merge, becoming taller. They
would then proceed in the directions they were initially going, ultimately leaving the screen on the side
opposite to where they began.
(c) Let x = distance (microns), t = time (nanoseconds), and I = intensity.
I
I
1.2
I
1.2
Location
x = −5
0.9
1.2
Location
x=0
0.9
0.6
0.6
0.6
0.3
0.3
0.3
2
4
6
8
10 12
t
2
4
Location
x=5
0.9
6
8
10 12
t
2
4
6
8
10 12
t
Figure 12.151
(d) Two pulses of light are traveling down a wave-guide toward one another. They meet in the center and,
as they pass through one another, appear brighter. They then continue along in the wave-guide in the
directions they were going.
13.1 SOLUTIONS
CHAPTER THIRTEEN
Solutions for Section 13.1
Exercises
1. The vectors are ~a = ~i + 3~j , ~b = 3~i + 2~j , ~v = −2~i − 2~j , and w
~ = −~i + 2~j .
2. ~a = 2~i + ~j ,
~b = 2~i ,
~c = −2~i ,
d~ = −2~i + 2~j ,
~e = −2~i − ~j
3. The vector we want is the displacement from Q to P , which is given by
~ = (1 − 4)~i + (2 − 6)~j = −3~i − 4~j
QP
4. The vector we want is the displacement from P to Q, which is given by
P~Q = (4 − 1)~i + (6 − 2)~j = 3~i + 4~j
5. ~a = ~b = ~c = 3~k , d~ = 2~i + 3~k , ~e = ~j ,
6. ~
u = ~i + ~j + 2~k and ~v = −~i + 2~k .
7. 4~i + 2~j − 3~i + ~j = ~i + 3~j
f~ = −2~i
8. ~i + 2~j − 6~i − 3~j = −5~i − ~j
9. −4~i + 8~j − 0.5~i + 0.5~k = −4.5~i + 8~j + 0.5~k
10. (0.9~i − 1.8~j − 0.02~k ) − (0.6~i − 0.05~k ) = 0.3~i − 1.8~j + 0.03~k
11. 3~i − 4~j + 2~k − 6~i − 8~j + ~k = −3~i − 12~j + 3~k
12. 4~i − 3~j + 7~k − 10~i − 2~j + 4~k = −6~i − 5~j + 11~k
13. 0.6~i + 0.2~j − ~k + 0.3~i + 0.3~k = 0.9~i + 0.2~j − 0.7~k
14. ~i − 12 ~j + 23 ~k + 3~i − 21 ~j + 32 ~k = 4~i − ~j + 3~k
p
√
15. k~v k = 12 + (−1)2 + 22 = 6.
16. The length is given by
k~
zk=
17. k~v k =
18. k~v k =
19. k~v k =
p
12 + (−1)2 + 32 =
p
7.22
p
+
(−1.5)2
+
√
11.
2.12
=
1.22 + (−3.6)2 + 4.12 =
p
√
√
(1)2 + (−3)2 + (−1)2 =
√
1+9+1=
√
11.
58.5 ≈ 7.6.
31.21 ≈ 5.6.
20. 4~
z = 4(~i − 3~j − ~k ) = 4~i − 12~j − 4~k .
21.
5~a + 2~b = 5(2~j + ~k ) + 2(−3~i + 5~j + 4~k )
= (5(2)~j + 5(1)~k ) + (2(−3)~i + 2(5)~j + 2(4)~k )
= (10~j + 5~k ) + (−6~i + 10~j + 8~k ) = (0 − 6)~i + (10 + 10)~j + (5 + 8)~k
= −6~i + 20~j + 13~k .
1207
1208
Chapter Thirteen /SOLUTIONS
22. ~a + ~z = (2~j + ~k ) + (~i − 3~j − ~k ) = (0 + 1)~i + (2 − 3)~j + (1 − 1)~k = ~i − ~j
23. 2~c + ~
x = 2(~i + 6~j ) + (−2~i + 9~j ) = (2~i + 12~j ) + (−2~i + 9~j ) = (2 − 2)~i + (12 + 9)~j = 21~j .
24.
2~a + 7~b − 5~
z = 2(2~j + ~k ) + 7(−3~i + 5~j + 4~k ) − 5(~i − 3~j − ~k )
= (4~j + 2~k ) + (−21~i + 35~j + 28~k ) − (5~i − 15~j − 5~k )
= (−21 − 5)~i + (4 + 35 + 15)~j + (2 + 28 + 5)~k = −26~i + 54~j + 35~k .
25.
k~
y −~
x k = k(4~i − 7~j ) − (−2~i + 9~j )k = k(4 − (−2))~i + (−7 − 9)~j k = k6~i − 16~j k
p
√
√
√
= 62 + (−16)2 = 36 + 256 = 292 = 2 73.
26. (a) See Figure
√
√ 13.1.
(b) k~v k = 52 + 72 = 74 = 8.602.
(c) We see in Figure 13.2 that tan θ = 57 and so θ = 54.46◦ .
y
y
5
x
x
θ
5
7
~v
−7
Figure 13.1
27. Since
Figure 13.2
p
(0.06)2 + (0.08)2 = 0.1, the vector is
1
(0.06~i − 0.08~k ) = 0.6~i − 0.8~k .
0.1
~ ~ ~
28. To find a vector in the opposite direction
can take the scalar multiple (−1)~v = −~i + ~j − ~k . The
p to ~v = i − j + k , we
√
magnitude of the vector (−1)~v is (−1)2 + 12 + (−1)2 = 3. So
√
(−~i + ~j − ~k )/ 3
is the unit vector in the opposite direction to ~i − ~j + ~k .
p
√
29. Since ||~v ||2 = 22 + (−1)2 + (− 11)2 = 16, we have ||~v || = 4. Thus, the vector we want is
√
√
~v
1
11 ~
1
1
−
= − (2~i − ~j − 11~k ) = − ~i + ~j +
k.
||~v ||
4
2
4
4
30. The length of the vector ~i −~j +2~k is 12 + (−1)2 + 22 =
it by √26 . So the answer is √26~i − √26 ~j + √46 ~k .
p
√
6. We can scale the vector down to length 2 by multiplying
Problems
31. If two vectors are parallel, they are scalar multiples of one another. Thus
Solving for a gives
6
a2
=
.
5a
−3
a2 = −2 · 5a
so
a = 0, −10.
1209
13.1 SOLUTIONS
32. (a) The displacement from P to Q is given by
−
−
→
P Q = (4~i + 6~j ) − (~i + 2~j ) = 3~i + 4~j .
Since
p
−
−
→
kP Qk = 32 + 42 = 5,
−
−
→
a unit vector ~
u in the direction of P Q is given by
~
u =
1−
1
3
4
−
→
P Q = (3~i + 4~j ) = ~i + ~j .
5
5
5
5
(b) A vector of length 10 pointing in the same direction is given by
4
3
10~
u = 10( ~i + ~j ) = 6~i + 8~j .
5
5
√
33. The vector ~v = −~i + ~j points northwest. Since k~v k = 2, the unit vector pointing northwest is ~
u = − √12~i + √12 ~j .
√
√
√
√
34. (a) The components are v1 = 2 cos π/4 = 2, v2 = 2 sin π/4 = 2. See Figure 13.3. Thus ~v√ = 2~i + 2~j .
(b) Since the vector lies in the xz-plane, its y-component is 0. Its x-component is 1 cos( π6 )~i = 23~i and its z-component
√
is 1 sin( π6 )~j = 21 ~j . See Figure 13.4. So the vector is 23~i + 12 ~k .
y
z
π
6
45◦ =
π
4
= 30◦
x
x
Figure 13.3
y
Figure 13.4
35. The coordinates of the points are:
A = (0, 0),
B = (2, 2),
C = (7, 0),
D = (3, 4),
E = (4, 2).
−→
−−→
(a) We find AB = 2~i + 2~j , CD = −4~i + 4~j . Therefore,
~
u = (2.5)(2~i + 2~j ) + (−0.8)(−4~i + 4~j ) = 5~i + 5~j + 3.2~i − 3.2~j = 8.2~i + 1.8~j ,
~v = (2.5)(−2~i − 2~j ) − (−0.8)(−4~i + 4~j ) = −5~i − 5~j − 3.2~i + 3.2~j = −8.2~i − 1.8~j .
−→
−→
(b) We see that ~v = −~
u . We know that −AB is equivalent to BA. In other words,
−→
−−→
~v = −(2.5)AB + (0.8)CD.
By factoring out a −1, we get
−→
−−→
~v = −((2.5)AB + (−0.8)CD) = −~
u.
−→
−
→
36. We find EA to be −4~i − 2~j . A unit vector on the EA direction is
−4i − 2j
~n = √
42 + 22
1
−2~
= √ i − √ ~j .
5
5
So a vector of length 2 in this direction is
−4
2
p
~ = 2~n = √ ~i − √ ~j .
5
5
√
Thus, p
~ = − 4 5 5~i −
√
2 5~
j
5
.
1210
Chapter Thirteen /SOLUTIONS
37. (a)
(b)
(c)
(d)
True, by the property of commutativity.
It does not make sense, we cannot add vectors and scalars.
True, by the property of commutativity.
This is not always true. For example, let ~a = ~i + 2~j , ~b = −2~i + ~j . Then
k~a + ~b k = k~i + 2~j − 2~i + ~j k = k − ~i + 3~j k =
p
√
k~a k = 12 + 22 = 5
p
√
k~b k = (−2)2 + 12 = 5.
38.
So, k~a k + k~b k =
√
5+
√
p
(−1)2 + 32 =
√
10
√
√
√
5 = 2 5. But 10 6= 2 5 and k~a + ~b k 6= k~a k + k~b k.
~
x
y
~
~
z
w
~
w
~
−~
v
w
~ = ~v + (−~
u)
w
~
✲
~v
~
v
−~
u
~
u
Figure 13.5
Figure 13.6
Break the hexagon up into 6 equilateral triangles, as shown in Figure 13.5.
Then ~
u − ~v + w
~ = ~0 , so w
~ = ~v − ~
u
Similarly, ~
x = −~
u,~
y = −~v , ~
z = −w
~ =~
u − ~v .
39. (a) We need 6~i + 8~j + 3~k = λ(2~i + (t2 + 23 t + 1)~j + t~k ) for some λ. This gives
6 = 2λ
8 = (t2 +
2
t + 1)λ
3
3 = tλ
From the first equation, we have λ = 3. Substituting λ = 3 into the third equation gives t = 1. Check the second
equation, it says 8 = 8, if t = 1 and λ = 3. So for t = 1, the two vectors are parallel to each other.
(b) Similar to part (a), we need to solve
2 = tλ
−4 = λ
1 = λ(t − 1)
From the first two equations we have λ = −4 and t = − 12 . Substituting this into the third equation gives 1 = 6.
Thus this system of equations has no solution, so the pair of vectors is not parallel to each other for any value of t.
(c) 2t~i + t~j + t~k = 3t (6~i + 3~j + 3~k ). For any t, the two vectors are parallel to each other.
√
40. Since the component of ~v in the ~i -direction is 3, we have ~v = 3~i + b~j for some b. Since k~v k = 5, we have 32 + b2 =
5, so b = 4 or b = −4. There are two vectors satisfying the properties given: ~v = 3~i + 4~j and ~v = 3~i − 4~j .
41. Let ~v = x~i + y~j .
We want k~v k = 1 and k~v + ~i k = 1, i.e.,
p
x2 + y 2 = 1
and
p
(x + 1)2 + y 2 = 1.
13.1 SOLUTIONS
1211
Setting these equations equal and solving for x gives:
p
x2 + y 2 =
2
2
p
(x + 1)2 + y 2
x + y = (x + 1)2 + y 2
(after squaring both sides)
x2 + y 2 = x2 + 2x + 1 + y 2
0 = 2x + 1
1
x=− .
2
Since we know x = − 21 , we can use the fact that k~v k = 1 to solve for y:
p
k~v k = 1
x2
+ y2 = 1
2
x + y2 = 1
y 2 = 1 − x2
y=±
=±
p
1 − x2
r
1−
√
3
.
=±
2
Thus the vectors we are looking for are:
√
1
3
~v = − ~i +
2
2
2
1
2
√
1
3~
and ~v = − ~i −
j.
2
2
−→ −
−→
42. We must check that all the points are the same distance apart, i.e., the magnitude of the displacement vectors OA, OB,
−
−
→ −→ −
−→
−→
OC, BA, CB and CA is the same. Here goes:
p
−→
kOAk = k(2~i + 0~j + 0~k ) − (0~i + 0~j + 0~k )k = 22 + 02 + 02 = 2
q
√
√
−−→
kOBk = k(1~i + 3~j + 0~k ) − (0~i + 0~j + 0~k )k = 12 + ( 3)2 + 02 = 2
p
p
√
−
−
→
kOCk = k(1~i + 1/ 3~j + 2 2/3~k ) − (0~i + 0~j + 0~k )k = 1 + 1/3 + 4(2/3) = 2
√
√
−→
kBAk = k(2~i + 0~j + 0~k ) − (1~i + 3~j + 0~k )k = 1 + 3 + 0 = 2
p
√
√
−−→
kCBk = k(1~i + 3~j + 0~k ) − (1~i + 1/ 3~j + 2 2/3~k )k
q
p
√
√
= 02 + ( 3 − 1/ 3)2 + 4(2/3) = 3 − 2 + 1/3 + 8/3 = 2
p
p
√
−→
kCAk = k(2~i + 0~j + 0~k ) − (1~i + 1/ 3~j + 2 2/3~k )k = 1 + 1/3 + 4(2/3) = 2.
43. In Figure 13.7 let O be the origin, points A, B, and C be the vertices of the triangle, point D be the midpoint of BC, and
Q be the point in the line segment DA that is 31 |DA| away from D.
y
C
A
Q
D
B
x
O
Figure 13.7
1212
Chapter Thirteen /SOLUTIONS
From Figure 13.7 we see that
−
−→ −−→ −−→ −−→ 1 −−→
OQ = OD + DQ = OD + DA
3
−−→ 1 −→ −−→
= OD + (OA − OD)
3
−−→ 1 −→ 1 −−→
= OD + OA − OD
3
3
1 −→ 2 −−→
= OA + OD.
3
3
−−→
Because the diagonals of a parallelogram meet at their midpoint, and 2OD is a diagonal of the parallelogram formed by
−−→
−
−
→
OB and OC, we have:
1 −−→ −
−→
−−→
OD = (OB + OC),
2
so we can write:
1 −→ 2
−
−
→
OQ = OA +
3
3
1 −
1 −→ −−→ −
−
→
−→ −
−
→
(OB + OC) = (OA + OB + OC).
2
3
−→ −−→
Thus a vector from the origin to a point 13 of the way along median AD from D, the midpoint, is given by 31 (OA + OB +
−
−
→
OC).
In a similar manner we can show that the vector from the origin to the point 13 of the way along any median from the
−→ −
−→ −
−
→
midpoint of the side it bisects is also 13 (OA + OB + OC). See Figure 13.8 and 13.9.
y
y
C
A
C
A
Q
Q
B
B
x
x
Figure 13.8
Thus the medians of a triangle intersect at a point
Figure 13.9
1
3
of the way along each median from the side that each bisects.
Strengthen Your Understanding
44. The number k~
u + ~v k could be less than 1. For example, for ~
u = ~i and ~v = −0.5~i , ~
u + ~v = 0.5~i and k~
u + ~v k =
0.5 < 1.
45. If c < 0, then c~
u has the opposite direction of ~
u . Also, if c = 0, then c~v = ~0 , which has no direction at all.
46. If the angle between the vectors ~
u and ~v is more than 90◦ , then k~v − ~
u k is the length of the longer diagonal of the
parallelogram.
47. Since ~
u +w
~ =~
u , we see that w
~ =~
u −~
u = ~0 . Since w
~ is the zero vector, this means ~v + w
~ = ~v + ~0 = ~v .
48. Since ~v lies on a plane parallel to the yz-plane it must have a zero ~i -component. So ~v = a~j + b~k . If we choose a = 1,
then ~v = ~j + b~k . Now we set:
p
√
k~v k = 1 + b2 = 2, which gives b = ± 3.
√
So, ~v = ~j + 3 ~k is a possible answer.
−
−
→
−→
49. Take any equilateral triangle P QR whose sides have length one. Let ~
u = P Q and ~v =
√ P R, which are unit vectors.
−→
~
~
Then ~v − ~
u = QR, which is also a unit vector. The vectors ~
u = i and ~v = (1/2)i + ( 3/2)~j work, for example.
50. Since w
~ is the difference between ~
u and ~v we have w
~ = ~v − ~
u , or 2~i + 3~j = ~v − ~
u . Solving for ~v we get
~v = 2~i + 3~j + ~
u . We can choose any vector we wish for ~
u and compute the corresponding vector ~v . For example, if
we let u = ~i then we get ~v = 3~i + 3~j .
13.2 SOLUTIONS
1213
51. False. There are exactly two unit vectors: one in the same direction as ~v and the other in the opposite direction. Explicitly,
1
the unit vectors parallel to ~v are ±
~v .
k~v k
p
√
52. False. The length of this vector is 1/3 + 1/3 + 4/3 = 2, not 1.
53. True. Multiplying by a scalar greater than one stretches the length of the vector by the scalar.
54. False. If ~v and w
~ are not parallel, the three vectors ~v , w
~ and ~v + w
~ can be thought of as three sides of a triangle. (If
the tail of w
~ is placed at the head of ~v , then ~v + w
~ is a vector from the tail of ~v to the head of w
~ .) The length of one
~
side of a triangle is less than
√the sum of the lengths of the other two sides. Alternatively, a counterexample is ~v = i and
w
~ = ~j . Then k~i + ~j k = 2 but k~i k + k~j k = 2.
55. False. If ~v and w
~ are not parallel, the three vectors ~v , w
~ and ~v − w
~ can be thought of as three sides of a triangle. (If
the tails of ~v and w
~ are placed together, then ~v − w
~ is a vector from the head of w
~ to the head of ~v .) The length of one
side of a triangle is less than
the
sum
of
the
lengths
of
the
other
two
sides.
Alternatively,
a counterexample is ~v = ~i and
√
w
~ = ~j . Then k~i − ~j k = 2 but k~i k − k~j k = 0.
56. False. Two vectors are parallel if and only if one is a nonzero scalar multiple of the other. If c(~i − 2~j + ~k ) = 2~i − ~j + ~k ,
then c = 2 so that the ~i components are equal, but multiplication by 2 does not make the ~j or ~k components equal. Thus,
there is no scalar multiple of ~i − 2~j + ~k that is equal to 2~i − ~j + ~k .
57. False. As a counterexample, take 3~i and −~i . Then the sum is 2~i which has magnitude 2 (smaller than k3~i k = 3).
58. False. Since magnitudes are nonnegative this cannot be true when c < 0. The correct statement is kc~v k = |c|k~v k.
59. False. To find the displacement vector from (1, 1, 1) to (1, 2, 3) we subtract ~i + ~j + ~k from ~i + 2~j + 3~k to get
(1 − 1)~i + (2 − 1)~j + (3 − 1)~k = ~j + 2~k .
60. False. The displacement vector from (a, b) to (c, d) has the same magnitude but opposite direction as the displacement
vector from (c, d) to (a, b).
Solutions for Section 13.2
Exercises
1. Scalar
2. Scalar
3. Temperature is measured by a single number, and so is a scalar.
4. The magnetic field is a vector because it has both a magnitude (the strength of the field) and a direction (the direction of
the compass).
~ = (P1 , P2 , · · · , P50 ) where Pi is the population of the ith state, shows that P
~ can be thought of as a vector
5. Writing P
with 50 components.
√
√
~j = 7.07~i − 7.07~j . Notice that the
2)~i − (5 2)√
6. In components, we have ~v = 10 cos(45◦ )~i − 10 sin(45◦ )~j = (5 √
coefficient in the ~j -direction must be negative. The components are 5 2~i and −5 2 ~j .
7. In components, we have ~v = −40 cos(20◦ )~i − 40 sin(20◦ )~j = −37.59~i − 13.68~j . Notice that both coefficients are
negative. The components are −37.59~i and −13.68~j .
8. (a) If the car is going east, it is going solely in the positive x direction, so its velocity vector is 50~i .
(b) If the car is going south, it is going solely in the negative y direction, so its velocity vector is −50~j .
(c) If the car is going southeast, the angle between the x-axis and the velocity vector is −45◦ . Therefore
velocity vector = 50 cos(−45◦ )~i + 50 sin(−45◦ )~j
√
√
= 25 2~i − 25 2~j .
(d) If the car is going northwest, the velocity vector is at a 45◦ angle to the y-axis, which is 135◦ from the x-axis.
Therefore:
√
√
velocity vector = 50(cos 135◦ )~i + 50(sin 135◦ )~j = −25 2~i + 25 2~j .
1214
Chapter Thirteen /SOLUTIONS
9. We need to calculate the length of each vector.
p
k21~i + 35~j k = 212 + 352 =
√
k40~i k = 402 = 40.
√
1666 ≈ 40.8,
So the first car is faster.
10. See Figure 13.10. Since
tan θ =
we have
θ = arctan
18
15
18
,
15
= 50.194◦ .
y
18~j
~
F
θ
15~i
x
Figure 13.10
Problems
11. (a) We start by finding the velocity, w
~ , of the boat relative to the riverbed. It is given by
w
~ = ~v + ~c = 8~i + (0.6~i + 0.8~j ) = 8.6~i + 0.8~j .
The speed of the boat relative to the riverbed is the magnitude of w
~:
Speed = kw
~k=
p
8.62 + 0.82 ≈ 8.64 km/hr.
(b) Since the velocity, ~v = 8~i , is parallel to the x-axis, we want to find the angle between w
~ and the x-axis. From
Figure 13.11, this angle is
0.8
θ = arctan
≈ 0.093 radians.
8.6
θ
w
~
❘
0.8
8.6
Figure 13.11: Angle between ~v and w
~
In practical terms, this angle tells us that if you set your boat parallel to the x-axis at 8 km/hr, the current will
take you about 0.093 radians ≈ 5◦ off course.
12. Now ~c = −2(0.6~i + 0.8~j ) = −1.2~i − 1.6~j . The velocity vector for the boat relative to the riverbed is
w
~ = ~v + ~c = (8 − 1.2)~i − 1.6~j = 6.8~i − 1.6~j
so the speed is
p
6.82 + (−1.6)2 = 6.986 km/hr.
13.2 SOLUTIONS
1215
13. (a) The velocity vector for the boat is ~b = 25~i and the velocity vector for the current is
~c = −10 cos(45◦ )~i − 10 sin(45◦ )~j = −7.07~i − 7.07~j .
The actual velocity of the boat is
(b) k~b + ~c k = 19.27 km/hr.
~b + ~c = 17.93~i − 7.07~j .
(c) We see in Figure 13.12 that tan θ =
7.07
, so θ = 21.52◦ south of east.
17.93
✛
✲
17.93
✻
θ
~c
~b
7.07
~b + ~c
❄
Figure 13.12
14.
✛
North
Police car
40 km/hr
y
P
C
Police car
km/hr
✻ 30truck
~
r
O
x East
Truck
Figure 13.13
Since both vehicles reach the crossroad in exactly one hour, at the present the truck is at O in Figure 13.13; the police
car is at P and the crossroads is at C. If ~r is the vector representing the line of sight of the truck with respect to the police
car.
~r = −40~i − 30~j
15. Suppose ~
u represents the velocity of the plane relative to the air and w
~ represents the velocity of the wind. We can add
these two vectors by adding their components. Suppose north is in the y-direction and east is the x-direction. The vector
representing the airplane’s velocity makes an angle of 45◦ with north; the components of ~
u are
u = 700 sin 45◦~i + 700 cos 45◦~j ≈ 495~i + 495~j .
~
Since the wind is blowing from the west, w
~ = 60~i . By adding these we get a resultant vector ~v = 555~i + 495~j . The
direction relative to the north is the angle θ shown in Figure 13.14 given by
θ = tan−1
555
x
= tan−1
y
495
≈ 48.3◦
The magnitude of the velocity is
k~v k =
p
4952 + 5552 =
= 744 km/hr.
p
553,050
1216
Chapter Thirteen /SOLUTIONS
N
45◦
✻
~
v
u
~
y
θ
0
w
~
✛
✲
x
❄
E
Figure 13.14: Note that θ is the angle between north and the vector ~v
~ 3 so that F
~ 1+F
~ 2+F
~ 3 = 0. Since
16. We want the total force on the object to be zero. We must choose the third force F
~ 2 = 11~i − 4~j , we need F
~ 3 = −11~i + 4~j .
F~ 1 + F
17. The velocity vector for the wind is w
~ = 60 cos(45◦ )~i − 60 sin(45◦ )~j = 42.43~i − 42.43~j . If the airplane is to head
due east, then the component in the ~j -direction of p
~ +w
~ must be zero (where p
~ represents the velocity vector for the
airspeed of the plane.) Thus, we have p
~ = A~i + 42.43~j , for some value of A. Since the airplane is flying at an airspeed
of 500 km/hr, we have
k~
p k = 500
p
A2 + 42.432 = 500
A = 498.20.
We have
p
~ = 498.20~i + 42.43~j .
This is the direction the plane should head in order to go due east. We use tan θ =
should head 4.87◦ north of east. Since
42.43
, so θ = 4.87◦ . The plane
498.20
p
~ +w
~ = 540.63~i ,
the airplane’s speed relative to the ground is k~
p +w
~ k = 540.63 km/hr.
18. Let the x-axis point east and the y-axis point north. Since the wind is blowing from the northeast at a speed of 50 km/hr,
the velocity of the wind is
w
~ = −50 cos 45◦~i − 50 sin 45◦~j ≈ −35.4~i − 35.4~j .
Let ~a be the velocity of the airplane, relative to the air, and let φ be the angle from the x-axis to ~a ; since k~a k =
600 km/hr, we have ~a = 600 cos φ~i + 600 sin φ~j . (See Figure 13.15.)
y
φ
❄
w
~
~a
x
~
v
Figure 13.15
Now the resultant velocity, ~v , is given by
~v = ~a + w
~ = (600 cos φ~i + 600 sin φ~j ) + (−35.4~i − 35.4~j )
= (600 cos φ − 35.4)~i + (600 sin φ − 35.4)~j .
Since the airplane is to fly due east, i.e., in the x direction, then the y-component of the velocity must be 0, so we must
have
600 sin φ − 35.4 = 0
35.4
.
sin φ =
600
Thus φ = arcsin(35.4/600) ≈ 3.4◦ .
13.2 SOLUTIONS
1217
~,W
~ , and E
~ to represent the current, wind, and engine
19. Let the x-axis point east and the y-axis point north. We use C
vectors, respectively. We resolve the current and wind velocity vectors into components. Since the current points 25◦
north of east with a speed of 12, we have
~ = 12 cos(25◦ )~i + 12 sin (25◦ )~j = 10.876~i + 5.071~j .
C
~ lies in the first quadrant, both coefficients are positive.
Since C
The wind points 80◦ south of east with a speed of 7 km/hr, so we have
~ = 7 cos(80◦ )~i − 7 sin (80◦ )~j = 1.216~i − 6.894~j .
W
~ lies in the fourth quadrant, the coefficient of ~i is positive and the coefficient of ~j is negative.
Since W
The combined velocity on the boat is due east at a speed of 40 km/hr, so we want
~ +W
~ +E
~ = 40~i .
C
~:
We solve for E
~ = 40~i − (C
~ +W
~ )
E
= 40~i − ((10.876~i + 5.071~j ) + (1.216~i − 6.894~j ))
= 40~i − (12.092~i − 1.823~j )
= 27.908~i + 1.823~j .
~k =
The engine should push the boat with a speed of kE
arctan(1.823/27.908) = 3.74◦ north of east.
√
27.9082 + 1.8232 = 27.97 km/hr, and in direction
20. (a) Let x-axis be the East direction and y-axis be the North direction. From Figure 13.16,
θ = sin−1 (4/5) = 53.1◦ .
That is, he should steer at 53.1◦ east of south.
N
θ
✻
5 km/hr
y
✻
*
*
Steering direction
✲
*
4 km/hr
x
Figure 13.16
(b)
θ
5 km/hr
N
10 km/hr
~
R
y
✻ ✻
✲
~
R
π
4
4 km/hr
Figure 13.17
x
1218
Chapter Thirteen /SOLUTIONS
~ be the resultant of the wind and river velocities, that is
Let R
~ = −4~i + (10 cos( π )~i + 10 cos( π )~j )
R
4
4
√
√
~
~
= (−4 + 5 2)i + 5 2j .
~ to
From Figure 13.17, we see that to get the the x-component of his rowing velocity and the x-component of R
cancel each other, we must have
√
5 sin θ = −4 + 5 2
√
−4 + 5 2
−1
= 37.9◦ .
θ = sin
5
However for this value of θ, the y-component of the velocity is
√
5 2 − 5 cos(37.9◦ ) = 3.1.
Since the y-component is positive, the man will not move across the river in a southward direction.
~ be the resultant force, and let F
~ 1 and F
~ 2 be the forces exerted by the larger and smaller tugs. See Figure 13.18.
21. Let R
~ 1 and F
~ 2 must cancel each other in order to ensure that the
Then kF~1 k = 54 kF~2 k. The y components of the vectors F
ship travels due east, hence
kF~1 k sin 30◦ = kF~2 k sin θ,
so
giving sin θ = 58 , and hence θ = sin−1
5
8
5 ~
kF2 k sin 30◦ = kF~2 k sin θ,
4
= 38.7◦ .
N
y
✻✻
~
F
✲
1
~
R
30◦
θ
~
F
x
2
Figure 13.18
22. The known force is
F~1 = 15 cos 20◦~i + 15 sin 20◦~j = 14.1~i + 5.13~j pounds.
Let the unknown force be F~2 = a~i + b~j .
Because F~1 and F~2 together pull due east, we have F~1 + F~2 = k~i for a positive constant k. Thus
(15 cos 20◦ + a)~i + (15 sin 20◦ + b)~j = k~i + 0~j .
Hence
b = −15 sin 20◦ = −5.1 pounds.
√
Because F~2 has magnitude 20 pounds, we have a2 + b2 = 20. Thus
a=
p
202 − b2 = 19.3 pounds
where the positive square root is required because the sum F~1 + F~2 points east and not west. Thus
F~2 = 19.3~i − 5.1~j pounds.
13.2 SOLUTIONS
1219
~ 1 = 100 cos(30◦ )~i + 100 sin(30◦ )~j = 86.60~i + 50~j and the force
23. The force exerted on the object from the first rope F
◦ ~
~
exerted from the second rope is F 2 = 70 cos(80 )i − 70 sin(80◦ )~j = 12.16~i − 68.94~j . The sum of these two forces
~ 1 +F
~ 2 = 98.76~i − 18.94~j . See Figure 13.19. In order for the object to move vertically, the total force on the object
is F
must be in the form F~ = 0~i + 0~j + b~k for some b. Thus the force vector for the crane is
~
F
c
= −98.76~i + 18.94~j + b~k
~ c k = 3000. Thus,
for some b. To find b, we use the fact that kF
~ c k = 3000
kF
p
(98.76)2 + (18.94)2 + b2 = 3000
b = ±2998.31
We use the positive value of b since we want the object to go up rather than down. The force exerted by the crane is
~
F
c
= −98.76~i + 18.94~j + 2998.31~k .
The total force acting on the object is 2998.31~k , or 2998.31 newtons straight up.
N
~
F
c
~
F
1
= 100 newtons
30◦
80◦
W
S
~
F
2
E
= 70 newtons
Figure 13.19: Horizontal forces on object
24. (a) The displacement vector of the moon relative to the earth is
~r = 384~i .
The displacement vector of the spaceship relative to the earth is
r~E = 280~i + 90~j .
The displacement vector of the spaceship relative to the moon is
r~L = r~E − ~r = −104~i + 90~j .
See Figure 13.20.
√
√
2
2
(b) Distance of spaceship from Earth = ||r~E || = 280
p + 90 = 86500√= 294.109 thousand km.
Distance of spaceship from the moon = ||r~L || = (−104)2 + 902 = 18916 = 137.535 thousand km.
−→
(c) See Figure 13.20. The gravitational force of the earth, FE , is parallel to r~E but of length 461 and in the opposite
direction:
461
−→
FE = − √
(280~i + 90~j ) = −438.885~i − 141.070~j .
186500
−
→
The gravitational force of the moon, FL , is parallel to r~L but of length 26 and in the opposite direction:
26
−
→
FL = − √
(−104~i + 90~j ) = 19.660~i − 17.041~j .
18916
1220
Chapter Thirteen /SOLUTIONS
The resulting force, F~ is
−→ −
→
F~ = FE + FL = 419.225~i − 158.084~j .
Spacecraft
~r E
=
~
90j
~i +
0
28
~
rL
Earth
Moon
~
r = 384~i
Figure 13.20
25. The speed of the particle before impact is v, so the speed after impact is 0.8v. If we consider the barrier as being along
the x-axis (see Figure 13.21), then the ~i -component is 0.8v cos 60◦ = 0.8v(0.5) = 0.4v.
Similarly, the ~j -component is 0.8v sin 60◦ = 0.8v(0.8660) ≈ 0.7v. Thus
~v after = 0.4v~i + 0.7v~j .
y
~
v before
v sin 60◦
0.8v sin 60◦
~v after
✛
60◦
v cos 60◦
✲✛
60◦
✲
x
0.8v cos 60◦
Figure 13.21
26. The total scores are out of 300 and are given by the total score vector ~v + 2w
~:
~v + 2w
~ = (73, 80, 91, 65, 84) + 2(82, 79, 88, 70, 92)
= (73, 80, 91, 65, 84) + (164, 158, 176, 140, 184)
= (237, 238, 267, 205, 268).
To get the scores as a percentage, we divide by 3, giving
1
(237, 238, 267, 205, 268) ≈ (79.00, 79.33, 89.00, 68.33, 89.33).
3
27. Since there are 16 ounces in a pound, we multiply the vector by 1/16 to get 0.01875~i + 0.0125~j + 0.03125~k in dollars
per ounce.
28. (a) Since the radius of the circle is 1 meter, the circumference is 2π meters. Thus, the object is moving at 2π meters/minute, or π/30 meters/second ≈ 0.11 meters/second.
(b) 30 seconds after passing the point (0, 1), the object is at the point (−1, 0). (Since it completes 1 revolution each
minute, it will move π radians in 30 seconds.) This is true regardless of whether the point is moving clockwise
or counterclockwise. However, since the velocity vector, ~v , is tangential to the curve in the direction of motion, it
will have an opposite sign if the motion is in the opposite direction. So, moving clockwise ~v = 2π~j , and moving
counterclockwise ~v = −2π~j , if the speed is measured in meters/minute.
13.2 SOLUTIONS
1221
29. The speed is a scalar which equals 30 times the circumference of the circle per minute. So it is a constant. The velocity is a
vector. Since the direction of the motion changes all the time, the velocity is not constant. This implies that the acceleration
is nonzero.
30.
B
~v
C
w
~
w
~
A
~v
O
Figure 13.22
−→
−→
The vector ~v + w
~ is equivalent to putting the vectors OA and AB end-to-end as shown in Figure 13.22; the vector
−
−
→
−
−→
w
~ + ~v is equivalent to putting the vectors OC and CB end-to-end. Since they form a parallelogram, ~v + w
~ and w
~ + ~v
−−→
are both equal to the vector OB, we have ~v + w
~ =w
~ + ~v .
31.
~v
β~
v
α~
v
β~
v
Figure 13.23
The vectors ~v , α~v and β~v are all parallel. Figure 13.23 shows them with α, β > 0, so all the vectors are in the same
direction. Notice that α~v is a vector α times as long as ~v and β~v is β times as long as ~v . Therefore α~v + β~v is a vector
(α + β) times as long as ~v , and in the same direction. Thus,
α~v + β~v = (α + β)~v .
32.
scaling
by α
❄
~
v +w
~
α(~
v +w
~)
αw
~
w
~
~
v
α~
v
Figure 13.24
The effect of scaling the left-hand picture in Figure 13.24 is to stretch each vector by a factor of α (shown with
α > 1). Since, after scaling up, the three vectors α~v , αw
~ , and α(~v + w
~ ) form a similar triangle, we know that α(~v + w
~)
is the sum of the other two: that is
α(~v + w
~ ) = α~v + αw
~.
33. Assume α, β > 0. The vector β~v is in the same direction and β times as long as ~v . The vector α(β~v ) is in the same
direction and α times as long as β~v , and so is αβ times as long as ~v and in the same direction as ~v . Thus,
α(β~v ) = (αβ)~v .
34. Since the zero vector has zero length, adding it to ~v has no effect.
1222
Chapter Thirteen /SOLUTIONS
35. According to the definition of scalar multiplication, 1 · ~v has the same direction and magnitude as ~v , so it is the same as
~v .
36. By Figure 13.25, the vectors ~v + (−1)w
~ and ~v − w
~ are equal.
~
v −w
~
w
~
~v
(−1)w
~
~
v + (−1)
w
~
(−1)w
~
Figure 13.25
37.
C
w
~
B
~v
O
~
u
A
Figure 13.26
−−→
−−→
−
−→
The vector ~
u + ~v is represented by OB. The vector (~
u + ~v ) + w
~ is represented by OB followed by BC, which
−
−
→
−→
−→
−→
−
−
→
is therefore OC. Now ~v + w
~ is represented by AC. So ~
u + (~v + w
~ ) is OA followed by AC, which is OC. Since we
−
−→
get the vector OC by both methods, we know
(~
u + ~v ) + w
~ =~
u + (~v + w
~)
38. (a) Target A is at the point (30, 0, 3); Target B is at the point (20, 15, 0); Target C is the point (12, 30, 8). You fire
−→
−
−
→
from the point P = (0, 0, 5). The vectors to each of these targets are P A = 30~i − 2~k , P B = 20~i + 15~j − 5~k ,
−
−
→
P C = 12~i + 30~j + 3~k .
−→
−
−→
−
−→
(b) You fire from the point Q = (0, −1, 3), so QA = 30~i + ~j , QB = 20~i + 16~j − 3~k , QC = 12~i + 31~j + 5~k .
Strengthen Your Understanding
√
39. The vectors ~v = 2~i + 3~j + ~k and w
~ = 3~i + 2~j + ~k have both magnitude 14 and the same ~k -component, but
different ~i -components and ~j -components, so they are not the same vector.
40. If the ~j -component of ~v is larger than or equal to 2 (or smaller than or equal to −2), then the magnitude of ~v is greater
than the magnitude of w
~ . For example, ~v = 0.5~i + 2~j .
~
~
~
41. Writing F = ai + bj , we have:
~ = F~ + G
~ = (a + 1)~i + (b + 1)~j .
R
~ . So F~ = ~i − 2~j is a
Choosing a = 1 and b = −2, results in a positive ~i -component and a negative ~j -component for R
possible vector. There are many possible others.
42. If ~
u and ~v are not parallel, then ~
u , ~v and ~
u + ~v form a triangle. In any triangle, the length of any side is less than the
sum of the lengths of the other two sides: k~
u + ~v k < k~
u k + k~v k. Instead, choose ~v = c~
u , for any c > 0. For example,
let ~
u = ~i and ~v = 2~i . Then
k~i + 2~i k = k 3~i k = 3 = 1 + 2 = k~i k + k2~i k.
13.3 SOLUTIONS
1223
43. Yes. Velocity describes how fast something moves and its direction of travel.
44. No. Speed describes only how fast something moves. It does not specify a direction.
45. Yes. Force describes how hard something is pushed or pulled and the direction of the push or pull.
46. No. Area measures the size of the region occupied by a two-dimension figure or surface, and gives no information about
direction.
47. Yes. Acceleration measures the rate of change of velocity and the direction of the change.
48. No. Volume measures the size of the region occupied by a three-dimension object and gives no information about direction.
Solutions for Section 13.3
Exercises
1. ~a · ~
y = (2~j + ~k ) · (4~i − 7~j ) = −14.
2. ~c · ~
y = (~i + 6~j ) · (4~i − 7~j ) = (1)(4) + (6)(−7) = 4 − 42 = −38.
3. ~a · ~b = (2~j + ~k ) · (−3~i + 5~j + 4~k ) = (0)(−3) + (2)(5) + (1)(4) = 0 + 10 + 4 = 14.
4. ~a · ~
z = (2~j + ~k ) · (~i − 3~j − ~k ) = (0)(1) + (2)(−3) + (1)(−1) = 0 − 6 − 1 = −7.
5. ~c · ~a + ~a · ~
y = (~i + 6~j ) · (2~j + ~k ) + (2~j + ~k ) · (4~i − 7~j ) = 12 − 14 = −2.
6. ~c + ~
y = (~i + 6~j ) + (4~i − 7~j ) = 5~i − ~j , so
~a · (~c + ~
y ) = (2~j + ~k ) · (5~i − ~j ) = −2.
7. Since ~a · ~b is a scalar and ~a is a vector, the answer to this equation is a vector parallel to ~a . We have
~a · ~b = (2~j + ~k ) · (−3~i + 5~j + 4~k ) = 0(−3) + 2(5) + 1(4) = 14.
Thus,
(~a · ~b ) · ~a = 14~a = 14(2~j + ~k ) = 28~j + 14~k
8. Since ~a · ~
y and ~c · ~z are both scalars, the answer to this equation is the product of two numbers and therefore a number.
We have
~a · ~
y = (2~j + ~k ) · (4~i − 7~j ) = 0(4) + 2(−7) + 1(0) = −14
Thus,
~c · ~z = (~i + 6~j ) · (~i − 3~j − ~k ) = 1(1) + 6(−3) + 0(−1) = −17
(~a · ~
y )(~c · ~
z ) = 238
9. Since ~c · ~c is a scalar and (~c · ~c )~a is a vector, the answer to this equation is another scalar. We could calculate ~c · ~c , then
(~c · ~c )~a , and then take the dot product ((~c · ~c )~a ) · ~a . Alternatively, we can use the fact that
((~c · ~c )~a ) · ~a = (~c · ~c )(~a · ~a ).
Since
~c · ~c = (~i + 6~j ) · (~i + 6~j ) = 12 + 62 = 37
we have,
~a · ~a = (2~j + ~k ) · (2~j + ~k ) = 22 + 12 = 5,
(~c · ~c )(~a · ~a ) = 37(5) = 185
10. A normal vector can be obtained from the coefficients: ~n = 2~i + ~j − ~k .
1224
Chapter Thirteen /SOLUTIONS
11. Rewriting the equation as
2x − 2z = 3x + 3y
or
x + 3y + 2z = 0
tells us that a normal vector is
~n = ~i + 3~j + 2~k .
12. A normal vector can be obtained from the coefficients of x, y, z in the equation of the plane and is: ~n = 1.5~i +3.2~j + ~k .
13. Writing the equation in the form
3x + 4y − z = 7
shows that a normal vector is
~n = 3~i + 4~j − ~k
14. Rewriting the equation as
πx − π = (1 − π)y − (1 − π)z + π
gives
πx + (π − 1)y + (1 − π)z = 2π
so a normal vector is
~n = π~i + (π − 1)~j + (1 − π)~k .
15. The plane is
3x − y + 4z = 3 · 1 − 1 · 5 + 4 · 2
3x − y + 4z = 6.
16. The plane is
5x + 4y − z = 5 · 2 + 4(−1) − 1 · 3
5x + 4y − z = 3.
17. The equation of the plane is
which can be written as x − y + z = 3.
(x − 1) − (y − 3) + (z − 5) = 0,
18. Since the plane is normal to the vector 5~i + ~j − 2~k and passes through the point (0, 1, −1), an equation for the plane is
5x + y − 2z = 5 · 0 + 1 · 1 + (−2) · (−1) = 3
5x + y − 2z = 3.
19. Two planes are parallel if their normal vectors are parallel. Since the plane 2x + 4y − 3z = 1 has normal vector
~
n = 2~i + 4~j − 3~k , the plane we are looking for has the same normal vector and passes through the point (1, 0, −1).
Thus the plane we want has equation:
2x + 4y − 3z = 2 · 1 + 4 · 0 + (−3) · (−1) = 5
20. Two planes are parallel if their normal vectors are parallel. Since the plane 3x + y + z = 4 has normal vector ~n =
3~i + ~j + ~k , the plane we are looking for has the same normal vector and passes through the point (−2, 3, 2). Thus, it
has the equation
3x + y + z = 3 · (−2) + 3 + 2 = −1.
13.3 SOLUTIONS
21. The normal vector to the plane is ~n = 2~i − 3~j + 5~k , so, for some d, the equation of the plane is
2x − 3y + 5z = d.
Substituting the point (4, 5, −2) we see that d = −17. Thus the equation of the plane is
2x − 3y + 5z = −17.
22.
cos θ =
(~i + ~j + ~k ) · (~i − ~j − ~k )
k~i + ~j + ~k kk~i − ~j − ~k k
= √
23.
(1)(1) + (1)(−1) + (1)(−1)
p
11 + 12 + 12 12 + (−1)2 + (−1)2
1
=− .
3
So, θ = arccos(−1/3) ≈ 1.911 radians, or about 109.471◦ .
cos θ =
(~i + ~k ) · (~j − ~k )
k~i + ~k kk~j − ~k k
= √
(1)(0) + (0)(1) + (1)(−1)
11 + 02 + 12
1
=− .
2
So, θ = arccos(−1/2) = 2π/3 radians, or 120◦ .
p
02 + (1)2 + (−1)2
24.
cos θ =
(~i + ~j − ~k ) · (2~i + 3 ~j + ~k )
k~i + ~j − ~k kk2~i + 3 ~j + ~k k
(1)(2) + (1)(3) + (−1)(1)
= p
√
12 + 12 + (−1)2 22 + 32 + 12
4
= √ .
42
25.
√
So, θ = arccos(4/ 42) ≈ 0.906 radians, or ≈ 51.887◦ .
(~i + ~j ) · (~i + 2 ~j − ~k )
k~j + ~j kk~i + 2 ~j − ~k k
(1)(1) + (1)(2) + (0)(−1)
= √
p
2
1 + 12 + 02 12 + 22 + (−1)2
√
3
3
3
= √ √ = √ =
.
2
2 6
2 3
cos θ =
√
So, θ = arccos( 3/2) = π/6 radians, or 30◦ .
26.
cos θ =
(~i ) · (2~i + 3 ~j − ~k )
k~i kk2~i + 3 ~j − ~k k
= √
(1)(2) + (0)(3) + (0)(−1)
12 + 02 + 02
2
= √ .
14
√
So, θ = arccos(2/ 14) = 1.007 radians, or 57.688◦ .
p
22 + 32 + (−1)2
1225
1226
Chapter Thirteen /SOLUTIONS
Problems
27. (a) Dividing ~v by its magnitude produces a unit vector ~
u in the same direction as ~v :
~
u =
3
1
2
1
(2~i + 3~j ) = √ ~i + √ ~j .
~v = √
k~v k
22 + 32
13
13
(b) Any vector w
~ = a~i + b~j such that ~v · w
~ = 2a + 3b = 0 is perpendicular to ~v . For example, w
~ = 3~i − 2~j has
~
~
this property, as do all scalar multiples of 3i − 2j .
28. (a) The plane can be written as 5x − 2y − z + 7 = 0, so the vector 5~i − 2~j − ~k is normal to the plane. The vector
λ~i + ~j + 0.5~k is parallel to 5~i − 2~j − ~k if one is a scalar multiple of the other. This occurs if the coefficients are
in proportion:
λ
1
0.5
=
=
.
5
−2
−1
Solving gives λ = −2.5.
(b) Substituting x = a + 1, y = a, z = a − 1 into the equation of the plane gives
a − 1 = 5(a + 1) − 2a + 7
a − 1 = 5a + 5 − 2a + 7
−13 = 2a
a = −6.5.
. So the only such point is ( 21
, 0, 0).
On the x-axis, y = z = 0, so 5x = 21, giving x = 21
5
5
Other points are (0, −21, 0), and (0, 0, 3). There are many other possible answers.
~
n = 5~i − ~j + 7~k . It is the normal vector.
The vector between two points in the plane is parallel to the plane. Using the points from part (b), the vector 3~k −
(−21~j ) = 21~j + 3~k is parallel to the plane.
30. (a) The vector ~n = 3~i − ~j − ~k is perpendicular to the plane since the plane can be written in the form (3~i − ~j − ~k ) ·
(x~i + y~j + z~k ) = −2.
(b) Find two points in the plane by putting x and y-values into the equation and calculating the corresponding z-values.
If x = 1 and y = 0, then z = 2 + 3(1) − 0 = 5, so the point (1, 0, 5) is on the plane. So is the point (0, 1, 1).
Therefore the vector (~i + 5~k ) − (~j + ~k ) = ~i − ~j + 4~k is parallel to the plane. (To check, we take the dot product
with ~
n = 3~i − ~j − ~k and see if we get zero: (3~i − ~j − ~k ) · (~i − ~j + 4~k ) = 3 + 1 − 4 = 0. Therefore ~i − ~j + 4~k
is parallel to the plane.)
29. (a)
(b)
(c)
(d)
31. (a) Writing the plane in the form 2x + 3y − z = 0 shows that a normal vector is
~n = 2~i + 3~j − ~k .
Any multiple of this vector is also a correct answer.
(b) Any vector perpendicular to ~n is parallel to the plane, so one possible answer is
~v = 3~i − 2~j .
Many other answers are possible.
32. (a)
(b)
(c)
(d)
goes with (I).
goes with (III), (IV).
goes with (II), (III).
goes with (II).
33. (a) Perpendicular vectors have a dot product of 0. Since ~a · ~c = 1(−2) − 3(−1) − 1 · 1 = 0, and ~b · d~ = 1(−1) +
1(−1) + 2 · 1 = 0, the pairs we want are ~a , ~c and ~b , d~ .
(b) Parallel vectors are multiples of one another, so there are no parallel vectors in this set.
(c) Since ~v · w
~ = ||~v ||||w
~ || cos θ, the dot product of the vectors we want is positive. We have
~a · ~b = 1 · 1 − 3 · 1 − 1 · 2 = −4
~a · d~ = 1(−1) − 3(−1) − 1 · 1 = 1
~b · ~c = 1(−2) + 3(−1) + 2 · 1 = −1
~c · d~ = −2(−1) − 1(−1) + 1 · 1 = 4,
13.3 SOLUTIONS
1227
and we already know ~a · ~c = ~b · d~ = 0. Thus, the pairs of vectors with an angle of less than π/2 between them are
~a , d~ and ~c , d~ .
(d) Vectors with an angle of more than π/2 between them have a negative dot product, so pairs are ~a , ~b and ~b , ~c .
34. Vectors ~v 1 , ~v 4 , and ~v 8 are all parallel to each other. Vectors ~v 3 , ~v 5 , and ~v 7 are all parallel to each other, and are all
perpendicular to the vectors in the previous sentence. Vectors ~v 2 and ~v 9 are perpendicular.
35. (a) Any multiple of ~v will work, for example, 8~i + 6~j .
(b) Any vector w
~ such that ~v · w
~ = 0 will work, such as −3~i + 4~j .
36. In general, ~
u and ~v are perpendicular when ~
u · ~v = 0.
In this case, ~
u · ~v = (t~i − ~j + ~k ) · (t~i + t~j − 2~k ) = t2 − t − 2.
This is zero when t2 − t − 2 = 0, i.e. when (t − 2)(t + 1) = 0, so t = 2 or −1.
In general, ~
u and ~v are parallel if and only if ~v = α~
u for some real number α.
Thus we need αt~i − α~j + α~k = t~i + t~j − 2~k , so we need αt = t, and −α = t, and α = −2. But if α = −2, we
can’t have αt = t unless t = 0, and if t = 0, we can’t have −α = t, so there are no values of t for which ~
u and ~v are
parallel.
37. (a) Increasing ||~v || increases ~v · w
~ because ~v · w
~ = ||~v || ||w
~ || cos θ, and cos θ is positive.
(b) Increasing θ decreases ~v · w
~ because cos θ is a decreasing function.
38. Let
~a = ~a parallel + ~a perp
where ~a parallel
d~ :
is parallel to d~ , and ~a perp is perpendicular to d~ . Then ~a parallel is the projection of ~a in the direction of
~a parallel =
d~
~a ·
kd~ k
d~
kd~ k
(2~i − 4~j + ~k )
= (3~i + 2~j − 6~k ) · √
22 + 42 + 12
8
= − (2~i − 4~j + ~k )
21
8
= − d~
21
(2~i − 4~j + ~k )
√
22 + 42 + 12
Since we now know ~a and ~a parallel , we can solve for ~a perp :
~a perp = ~a − ~a parallel
8
(2~i − 4~j + ~k )
= (3~i + 2~j − 6~k ) − −
21
10
118 ~
79
k.
= ~i + ~j −
21
21
21
Thus we can now write ~a as the sum of two vectors, one parallel to d~ , the other perpendicular to d~ :
~a = −
8 ~
d +
21
79~
10
118 ~
i + ~j −
k
21
21
21
−→
−→
39. We first find displacement vectors AB = (4 − 2)~i + (2 − 2)~j + (1 − 2)~k = 2~i − ~k and AC = (2 − 2)~i + (3 − 2)~j +
(1 − 2)~k = ~j − ~k . Then
−→ −→
AB · AC
cos(6 BAC) = −→ −→
kABkkACk
1
= √ √
5 2
= 0.3162.
Thus angle BAC is 71.57◦ (or 1.25 radians.)
1228
Chapter Thirteen /SOLUTIONS
40. If P = (5, 0, 0), Q = (0, −3, 0), R = (0, 0, 2), then vectors along the three sides of the triangle are
−
−→
QP = (5 − 0)~i + (0 − (−3))~j = 5~i + 3~j
−→
RP = (5 − 0)~i + (0 − 2)~k = 5~i − 2~k
−→
QR = (0 − (−3))~j + (2 − 0)~k = 3~j + 2~k
Thus, the lengths of the sides of the triangle are
√
√
−
−→
kQP k = k5~i + 3~j k = 25 + 9 = 34
√
√
−→
kRP k = k5~i − 2~k k = 25 + 4 = 29
√
√
−→
kQRk = k3~j + 2~k k = 9 + 4 = 13.
The angle between the vectors ~v and w
~ is given by
~v · w
~
cos θ =
k~v kkw
~k
so
θ = arccos
~v · w
~
k~v kkw
~k
.
Thus,
−−→ −→
QP · RP
−
−
→ −→
kQP kkRP k
(5~i + 3~j ) · (5~i − 2~k )
√ √
= arccos
34 29
Angle at P = arccos
= arccos
√
25
√
34 29
= 37.235◦ .
−−→ −→
QP · QR
−
−
→ −→
kQP kkQRk
(5~i + 3~j ) · (3~j + 2~k )
√ √
= arccos
34 13
Angle at Q = arccos
= arccos
9
√ √
34 13
= 64.654◦ .
Now we use the fact that the angles must add up to 180◦ . Thus
Angle at R = 180◦ − (37.235◦ + 64.654◦ ) = 78.111◦ .
41. (a) The points A, B and C are shown in Figure 13.27.
z
A
C
x
B
Figure 13.27
y
13.3 SOLUTIONS
1229
First, we calculate the vectors which form the sides of this triangle:
−→
AB = (4~i + 2~j + ~k ) − (2~i + 2~j + 2~k ) = 2~i − ~k
−
−→
BC = (2~i + 3~j + ~k ) − (4~i + 2~j + ~k ) = −2~i + ~j
−→
AC = (2~i + 3~j + ~k ) − (2~i + 2~j + 2~k ) = ~j − ~k
Now p
we calculate the lengths of each of the sides of the triangles:
√
−→
kABk = 22 + (−1)2 = 5
p
√
−−→
kBCk = (−2)2 + 12 = 5
p
√
−→
kACk = 12 + (−1)2 = 2
√
Thus the length of the shortest side of S is 2.
−→ −→
2·0+0·1+(−1)·(−1)
·AC
√ √
≈ 0.32
(b) cos 6 BAC = −AB
→ −→ =
5· 2
kAB k·kAC k
√
42. (a) We first find the unit vector in direction ~v . Since ||~v || = 32 + 42 = 5, the unit vector in direction of ~v is
u = 0.6~i + 0.8~j . Then
~
~
~ u )~
F
u
parallel = (F · ~
= (4 · 0.6 + 1 · 0.8)~
u
= 3.2~
u
= 1.92~i + 2.56~j .
(b) We have
~ perp = F
~ −F
~
~ ~
~
~
~
~
F
parallel = (4i + j ) − (1.92i + 2.56j ) = 2.08i − 1.56j .
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = 4 · 3 + 1 · 4 = 16.
W =F
√
43. (a) We first find the unit vector in direction ~v . Since ||~v || = 32 + 42 = 5, the unit vector in direction of ~v is
u = 0.6~i + 0.8~j . Then
~
~ ·~
F~ parallel = (F
u )~
u
= (0.2 · 0.6 − 0.5 · 0.8)~
u
= −0.28~
u
= −0.168~i − 0.224~j .
(b) We have
~ −F
~
~
~
~
~
~
~
F~ perp = F
parallel = (0.2i − 0.5j ) − (−0.168i − 0.224j ) = 0.368i − 0.276j .
(c) Since work is the dot product of the force and displacement vectors, we have
W = F~ · ~v = 0.2 · 3 − 0.5 · 4 = −1.4.
√
44. (a) We first find the unit vector in direction ~v . Since ||~v || = 32 + 42 = 5, the unit vector in direction of ~v is
u = 0.6~i + 0.8~j . Then
~
~
~ u )~
F
u
parallel = (F · ~
= (9 · 0.6 + 12 · 0.8)~
u
= 15~
u
= 9~i + 12~j .
~ . This makes sense (and could have been predicted)
Notice that the component of F~ in direction ~v is equal to F
~ is parallel to ~v .
since F
(b) We have
~ perp = F
~ −F
~
~
F
parallel = 0 .
(c) Since work is the dot product of the force and displacement vectors, we have
W = F~ · ~v = 9 · 3 + 12 · 4 = 75.
1230
Chapter Thirteen /SOLUTIONS
45. (a) We first find the unit vector in direction ~v . Since ||~v || =
u = 0.6~i + 0.8~j . Then
~
√
32 + 42 = 5, the unit vector in direction of ~v is
~ ·~
F~ parallel = (F
u )~
u
= (−0.4 · 0.6 + 0.3 · 0.8)~
u
~
= 0.
~ in direction ~v is equal to ~0 . This makes sense (and could have been predicted) since
Notice that the component of F
F~ is perpendicular to ~v .
(b) We have
~ − F~
~
F~ perp = F
parallel = F .
(c) Since work is the dot product of the force and displacement vectors, we have
W = F~ · ~v = −0.4 · 3 + 0.3 · 4 = 0.
Notice that since the force is perpendicular to the displacement, the work done is zero.
√
46. (a) We first find the unit vector in direction ~v . Since ||~v || = 32 + 42 = 5, the unit vector in direction of ~v is
u = 0.6~i + 0.8~j . Then
~
~ ·~
F~ parallel = (F
u )~
u
= (−3 · 0.6 − 5 · 0.8)~
u
= −5.8~
u
= −3.48~i − 4.64~j .
(b) We have
~ perp = F
~ −F
~
~
~
~
~
~
~
F
parallel = (−3i − 5j ) − (−3.48i − 4.64j ) = 0.48i − 0.36j .
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = −3 · 3 − 5 · 4 = −29.
W =F
47. (a) We first find the unit vector in direction ~v . Since ||~v || =
u = 0.6~i + 0.8~j . Then
~
√
32 + 42 = 5, the unit vector in direction of ~v is
~ ·~
F~ parallel = (F
u )~
u
= (−6 · 0.6 − 8 · 0.8)~
u
= −10~
u
= −6~i − 8~j
~.
=F
~ is in the opposite direction from ~v so it makes sense (and could have been predicted) that F~
Notice that F
parallel =
F~ .
(b) We have
~ perp = F
~ −F
~
~
F
parallel = 0 .
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = −6 · 3 − 8 · 4 = −50.
W =F
48. (a) We first √
find the unit vector in direction ~v . Since ||~v || =
u = ~v / 13. Then
~
√
22 + 32 =
√
~
~ u )~
F
u
parallel = (F · ~
√
u
= (−60/ 13)~
−120~
−180 ~
=
i +
j
13
13
= −9.231~i − 13.846~j .
13, the unit vector in direction of ~v is
13.3 SOLUTIONS
1231
(b) We have
~ − F~
~
~
~
~
~
F~ perp = F
parallel = (−20j ) − (−9.231i − 13.846j ) = 9.231i − 6.154j ).
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = −60.
W =F
49. (a) We first find the unit vector in direction ~v . Since ||~v || =
√
u = ~v / 26. Then
~
p
52 + (−1)2 =
√
26, the unit vector in direction of ~v is
~
~ u )~
F
u
parallel = (F · ~
√
= (20/ 26)~
u
−20 ~
100~
i +
j
=
26
26
= 3.846~i − 0.769~j .
(b) We have
~ − F~
~
~
~
~
~
F~ perp = F
parallel = (−20j ) − (3.846i − 0.769j ) = −3.846i − 19.231j .
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = 20.
W =F
~
~
50. (a) Notice that ~v is parallel to F~ (although in the opposite direction.) Therefore, F
parallel = F .
(b) We have
~ perp = F
~ −F
~
~
F
parallel = 0 .
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = −60.
W =F
~ . Therefore, F
~
~
51. (a) Notice that ~v is perpendicular to F
parallel = 0 .
(b) We have
~ − F~
~
F~ perp = F
parallel = F .
(c) Since work is the dot product of the force and displacement vectors, we have
~ · ~v = 0.
W =F
The work is zero since the direction of the force is perpendicular to the direction of the displacement.
52. Let the room be put in the coordinate system as shown in Figure 13.28.
B = (0, 80, 25)
A = (0, 0, 25)
✛
θ
25
D = (200, 80, 0)
✛
80
✛
200
✛
✛
✛
0 = (0, 0, 0)
C = (200, 0, 0)
Figure 13.28
1232
Chapter Thirteen /SOLUTIONS
Then the vectors of the two strings are given by:
−−→
AD = (200~i + 80~j + 0~k ) − (0~i + 0~j + 25~k ) = 200~i + 80~j − 25~k
−
−→
BC = (200~i + 0~j + 0~k ) − (0~i + 80~j + 25~k ) = 200~i − 80~j − 25~k .
−
−→
−
−→
Let the angle between AD and BC be θ. Then we have
−−→ −
−→
AD · BC
cos θ = −−→ −−→
kADk kBCk
200(200) + (80)(−80) + (−25)(−25)
p
= p
2
200 + 802 + (−25)2 (200)2 + (−80)2 + (−25)2
34225
=
47025
= 0.727804
53. We need to find the speed of the wind in the direction of the track. Looking at Figure 13.29, we see that we want the
component of w
~ in the direction of ~v . We calculate
(5~i + ~j ) · (2~i + 6~j )
w
~ · ~v
=
k~v k
k2~i + 6~j k
16
= √ ≈ 2.53
40
<5
kw
~ parallel k = kw
~ k cos θ =
Therefore, the race results will not be disqualified.
y
Track
~
v = 2~i + 6~j
Wind
w
~ = 5~i + ~j
θ
x
0
Figure 13.29
54. We find the component of the wind w
~ in the direction of the airplane. A direction vector for the airplane is ~v = ~i − ~j .
The component of w
~ 1 in the direction of ~v is given by
−4 · 1 + (−1) · (−1)
w
~ 1 · ~v
=
√
= −2.12.
k~v k
2
Similarly, we find the component for each wind vector in the direction of the airplane. We see that, in the direction of the
airplane, the component of w
~ 1 is −2.12, of w
~ 2 is 2.12, of w
~ 3 is −6.36, of w
~ 4 is 5.66, and of w
~ 5 is 4.95. The vector
w
~ 4 increases the plane’s speed the most and the vector w
~ 3 slows the plane down the most.
√
55. (a) The speed of the current is k~c k = 5 = 2.24 m/sec.
(b) The speed of the current in the direction of the canoe’s motion is the component of ~c in the direction of ~v . This is
given by:
Speed of current in direction of canoe’s motion =
(1)(5) + (2)(3)
~c · ~v
√
=
k~v k
52 + 32
11
= √
34
= 1.89 m/sec.
Notice that the speed of the current in the direction of the canoe is less than the speed of the current in the direction
in which the current is moving.
13.3 SOLUTIONS
1233
56. Let ~
u = 3~i + 4~j and ~v = 5~i − 12~j . We seek a vector w
~ = x~i + y~j such that the cosine of the angle between ~
u and
w
~ equals the cosine of the angle between ~v and w
~ . Thus
~v · w
~
~
u ·w
~
=
k~
u kkw
~k
k~v kkw
~k
or
3x + 4y
p
5
x2 + y 2
=
5x − 12y
p
13
x2 + y 2
.
Simplifying, we have x = −8y. The vector we want is of the form w
~ = −8y~i + y~j , but should we take y > 0 or y < 0?
The smaller of the two angles formed by ~
u and ~v is between 0◦ and 1800 , and so w
~ must make an acute angle with ~u
and ~v . If y > 0 then ~
u ·w
~ = −20y < 0 indicating an obtuse angle and if y < 0 then ~
u ·w
~ = −20y > 0 indicating an
acute angle. We have w
~ = −8y~i + y~j with y < 0. Thus w
~ can be any positive multiple of the vector 8~i − ~j .
57. The planes are parallel, with normal vectors ~n = 2~i − 5~j + ~k . Pick any point on 2x − 5y + z = 10, say (5, 0, 0), and
any point on z = 5y − 2x, say (0, 0, 0). The vector between them is d~ = 5~i , so we want to find the magnitude of the
component of d~ in the direction of ~n . A unit vector in the direction of ~
n is
1
~v = p
2~i − 5~j + ~k ,
2
2
2
2 + (−5) + 1
so
5~i · (2~i − 5~j + ~k )
10
√
= √ .
Distance = |d~ · ~v | =
30
30
58. We have
p
~ ·~
q = (1.00)(43) + (3.50)(57) + (4.00)(12) + (2.75)(78) + (5.00)(20) + (3.00)(35)
= 710 dollars.
The vendor took in $710 in from sales. The quantity ~
p ·~
q represents the total revenue earned.
59. The vector ~a represents the averages of the exams, written as decimals. The vector w
~ represents the weightings.
w
~ · ~a = 0.1 · 0.75 + 0.15 · 0.91 + 0.25 · 0.84 + 0.5 · 0.87 = 0.8565 = 85.65%
The dot product, 86.65%, represents the class average of the four exams in the course.
60. If ~
x and ~
y are two consumption vectors corresponding to points satisfying the same budget constraint, then
p ·~
~
x =k=p
~ ·~
y.
Therefore we have
p
~ · (~
x −~
y)=p
~ ·~
x −p
~ ·~
y = 0.
Thus p
~ and ~
x −~
y are perpendicular; that is, the difference between two consumption vectors on the same budget
constraint is perpendicular to the price vector.
61. Property 2 says that multiplying one of the vectors by a scalar simply multiplies the dot product by the same scalar. If
λ > 0, then when one vector is multiplied by λ, the angle between the vectors does not change, but the length of one
vector, and hence the dot product, is multiplied by λ. The result remains true when λ < 0. For a justification in the case
when λ < 0, see Problem 67 on page 1235.
62. We want to show that (~b · ~c )~a − (~a · ~c )~b and ~c are perpendicular. We do this by taking their dot product:
((~b · ~c )~a − (~a · ~c )~b ) · ~c = (~b · ~c )(~a · ~c ) − (~a · ~c )(~b · ~c ) = 0.
Since the dot product is 0, the vectors (~b · ~c )~a − (~a · ~c )~b and ~c are perpendicular.
~ = w1~i + w2~j + w3~k .
63. Suppose ~v = v1~i + v2~j + v3~k and w
• Property 1:
We calculate both ~v · w
~ and w
~ · ~v using the algebraic definition of the dot product:
~v · w
~ = v1 w1 + v2 w2 + v3 w3
w
~ · ~v = w1 v1 + w2 v2 + w3 v3
But since ordinary multiplication of scalars is commutative, v1 w1 = w1 v1 and so on. Therefore
~v · w
~ =w
~ · ~v .
1234
Chapter Thirteen /SOLUTIONS
• Property 2:
First we observe that
λw
~ = λ(w1~i + w2~j + w3~k ) = (λw1 )~i + (λw2 )~j + (λw3 )~k
λ~v = λ(v1~i + v2~j + v3~k ) = (λv1 )~i + (λv2 )~j + (λv3 )~k .
Now we calculate the three quantities ~v · (λw
~ ) and λ(~v · w
~ ) and (λ~v ) · w
~
~v · (λw
~ ) = v1 (λw1 ) + v2 (λw2 ) + v3 (λw3 )
λ(~v · w
~ ) = λ(v1 w1 + v2 w2 + v3 w3 )
(λ~v ) · w
~ = (λv1 )w1 + (λv2 )w2 + (λv3 )w3
Since ordinary multiplication is associative and commutative, we know that
v1 (λw1 ) = λv1 w1 = (λv1 )w1 and so on. Thus, we have ~v · (λw
~ ) = (λ~v ) · w
~.
In addition, the distributive property of ordinary multiplication tells us that
λ(v1 w1 + v2 w2 + v3 w3 ) = λv1 w1 + λv2 w2 + λv3 w3
Thus, we know that all three quantities are equal
~v · (λw
~ ) = λ(~v · w
~ ) = (λ~v ) · w
~
• Property 3:
First we observe that
~v + w
~ = (v1 + w1 )~i + (v2 + w2 )~j + (v3 + w3 )~k .
Next we calculate the quantities ((~v + w
~ )·~
u ) and (~v · ~
u +w
~ ·~
u ).
(~v + w
~ )·~
u = (v1 + w1 )u1 + (v2 + w2 )u2 + (v3 + w3 )u3
~v · ~
u +w
~ ·~
u = (v1 u1 + v2 u2 + v3 u3 ) + (w1 u1 + w2 u2 + w3 u3 ).
The distributive law of ordinary multiplication shows that (v1 + w1 )u1 = v1 u1 + w1 u1 , and so on. Thus, the dot
product is distributive also:
(~v + w
~ )·~
u = ~v · ~
u +w
~ ·~
u
64. Since ~
u ·w
~ = ~v · w
~ , (~
u − ~v ) · w
~ = 0. This equality holds for any w
~ , so we can take w
~ =~
u − ~v . This gives
k~
u − ~v k2 = (~
u − ~v ) · (~
u − ~v ) = 0,
that is,
This implies ~
u − ~v = 0, that is, ~
u = ~v .
k~
u − ~v k = 0.
65. Since the dot product of a vector with itself is the square of the vector’s magnitude, we can check that two vectors have
the same magnitude by computing dot products. We have
~v
~
u
−
k~
u k2
k~v k2
·
~v
~
u
−
k~
u k2
k~v k2
=
=
~u · ~
u
~
u · ~v + ~v · ~
u
~v · ~v
−
+
k~
u k4
k~
u k2 k~v k2
k~v k4
2~
u · ~v
1
1
−
+
.
k~
u k2
k~
u k2 k~v k2
k~v k2
and
~
u
~v
−
k~
u kk~v k
k~
u kk~v k
·
~
u
~v
−
k~
u kk~v k
k~
u kk~v k
=
=
~u · ~
u
~
u · ~v + ~v · ~
u
~v · ~v
−
+
k~
u k2 k~v k2
k~
u k2 k~v k2
k~
u k2 k~v k2
2~
u · ~v
1
1
−
+
k~v k2
k~
u k2 k~v k2
k~
u k2
The two dot products are equal, which shows that the vectors ~
u /k~
u k2 − ~v /k~v k2 and ~
u /(k~
u kk~v k) − ~v /(k~
u kk~v k)
have the same magnitude.
13.3 SOLUTIONS
1235
66. If ~
u = ~0 , then both sides of the equation are zero. If ~
u 6= ~0 , write ~v parallel , w
~ parallel , and (~v + w
~ )parallel for the
components of ~v , w
~ , and ~v + w
~ in the direction of ~
u . Then Figure 13.34 shows that
~v parallel + w
~ parallel = (~v + w
~ )parallel .
So
u
~v · ~
k~
u k2
Thus, since ~
u 6= ~0 , we deduce that
~
u +
w
~ · ~u
k~
u k2
~
u =
(~v + w
~ )·~
u
k~
u k2
~
u.
w
~ ·~
u
(~v + w
~ )·~
u
~v · ~
u
+
−
= 0,
k~
u k2
k~
u k2
k~
u k2
so
~v · ~
u +w
~ ·~
u = (~v + w
~ )·~
u.
67. Suppose θ is the angle between ~
u and ~v .
(a) By the definition of scalar multiplication, we know that −~v is in the opposite direction of ~v , so the angle between ~u
and −~v is π − θ. (See Figure 13.30.) Hence,
~
u · (−~v ) = k~
u kk − ~v k cos(π − θ)
= k~
u kk~v k(− cos θ)
= −(~
u · ~v )
λ~
v
u
~
π−θ
π−θ
λ<0
θ
θ
−~
v
π−θ
λ~
u
~
v
Figure 13.30
~
u
~v
Figure 13.31
(b) If λ < 0, the angle between ~
u and λ~v is π − θ, and so is the angle between λ~
u and ~v . (See Figure 13.31.) So we
have,
u · (λ~v ) = k~
~
u kkλ~v k cos(π − θ)
= |λ|k~
u kk~v k(− cos θ)
= −λk~
u kk~v k(− cos θ) since |λ| = −λ
= λk~
u kk~v k cos θ
= λ(~
u · ~v )
By a similar argument, we have
(λ~
u ) · ~v = kλ~
u kk~v k cos(π − θ)
= −λk~
u kk~v k(− cos θ)
= λ(~
u · ~v )
68. Let ~
u and ~v be the displacement vectors from C to the other two vertices. Then
c2 = k~
u − ~v k2
= (~
u − ~v ) · (~
u − ~v )
=~
u ·~
u − ~v · ~
u −~
u · ~v + ~v · ~v
= k~
u k2 − 2kukkvk cos C + k~v k2
= a2 − 2ab cos C + b2
1236
Chapter Thirteen /SOLUTIONS
69. We substitute ~
u = u1~i + u2~j + u3~k and by the result of Problem 66, we expand as follows:
(~
u · ~v )geom = (u1~i + u2~j + u3~k ) · ~v
= (u1~i ) · ~v + (u2~j ) · ~v + (u3~k ) · ~v
where all the dot products are defined geometrically By the result of Problem 67 we can write
(~
u · ~v )geom = u1 (~i · ~v )geom + u2 (~j · ~v )geom + u3 (~k · ~v )geom .
Now substitute ~v = v1~i + v2~j + v3~k and expand, again using Problem 66 and the geometric definition of the dot
product:
(~
u · ~v )geom = u1 ~i · (v1~i + v2~j + v3~k )
geom
+u2 ~j · (v1~i + v2~j + v3~k )
+u3
~k · (v1~i + v2~j + v3~k )
geom
geom
= u1 v1 (~i · ~i )geom + u1 v2 (~i · ~j )geom + u1 v3 (~i · ~k )geom
+u2 v1 (~i · ~i )geom + u2 v2 (~i · ~j )geom + u2 v3 (~i · ~k )geom
+u3 v1 (~i · ~i )geom + u3 v2 (~i · ~j )geom + u3 v3 (~i · ~k )geom
The geometric definition of the dot product shows that
~i · ~i = k~i k k~i k cos 0 = 1
~i · ~j = k~i k k~j k cos π = 0.
2
Similarly ~j · ~j = ~k · ~k = 1 and ~i · ~k = ~j · ~k = 0. Thus, the expression for (~
u · ~v )geom becomes
(~
u · ~v )geom = u1 v1 (1) + u1 v2 (0) + u1 v3 (0)
+u2 v1 (0) + u2 v2 (1) + u2 v3 (0)
+u3 v1 (0) + u3 v2 (0) + u3 v3 (1)
= u1 v1 + u2 v2 + u3 v3 .
70. (a) Since q(t) = (~v + tw
~ ) · (~v + tw
~ ) = k~v + tw
~ k2 and since the length of any vector is nonnegative, we must have
q(t) = k~v + tw
~ k2 ≥ 0
for all real t.
(b) Using the distributive law
q(t) = (~v + tw
~ ) · (~v + tw
~ ) = ~v · ~v + tw
~ · ~v + ~v · tw
~ + t2 w
~ ·w
~
= k~v k2 + 2(~v · w
~ )t + kw
~ k2 t2 .
If w
~ 6= 0, then kw
~ k 6= 0 and q(t) is quadratic in t.
(c) Since q(t) ≥ 0, the quadratic has one repeated root or no roots, so the discriminant must be less than or equal to zero.
Thus,
(2~v · w
~ )2 − 4k~v k2 kw
~ k2 ≤ 0.
Taking square roots, we have
|~v · w
~ | ≤ k~v kkw
~ k.
If w
~ = 0, then q(t) is no longer a quadratic. However, in that case,
|~v · w
~ | = 0 = k~v kkw
~k
so the inequality still holds.
Strengthen Your Understanding
71. The expression (~
u · ~v ) · w
~ does not make sense. You cannot take the dot product of the scalar ~
u · ~v and the vector w
~.
13.4 SOLUTIONS
1237
72. The formula ~v parallel = (~
u · ~v )~
u works only when ~
u is a unit vector. We can see that ~v parallel 6= 3(~i + ~j ) because
~v perp = ~v − ~v parallel = (2~i + ~j ) − (3~i + 3~j ) = −~i − 2~j ,
which is not perpendicular to ~
u = ~i + ~j .
73. Given an equation for a plane, to find the normal vector, we put x, y, z all on the same side of the equation. This gives
2x + 3y − z = 0. Then a normal vector is 2~i + 3~j − ~k .
74. The displacement vector from (1, 1) to (a, b) is (a − 1)~i + (b − 1)~j , so we want
((a − 1)~i + (b − 1)~j ) · (~i + 2~j ) = (a − 1) + 2(b − 1) = a + 2b − 3 = 0.
For example, if b = 3, then a = 3 − 2(3) = −3. In general, any (a, b) with a = 3 − 2b is a possible example.
75. A plane perpendicular to ~i + 2~j + 3~k is x + 2y + 3z = 0. Solving for z, we get z = (−1/3)x + (−2/3)y, so we can
let f (x, y) = (−1/3)x + (−2/3)y.
76. False. The dot product is a scalar.
77. True. Components of a normal vector can be read directly from coefficients of x, y and z in the equation for a plane.
78. True. The cosine of the angle between the vectors is negative when the angle is between π/2 and π.
79. False. The equation z = x + y has normal ~i + ~j − ~k , which is not parallel to ~i + ~j + ~k . An equation satisfying the
given conditions is x + y + z = 6.
80. True. The vector from (0, 1, 0) to (1, 1, 0) is ~i , while the vector from (0, 1, 0) to (0, 1, 1) is ~k , and ~i · ~k = 0.
81. True. ~v · ~v = k~v k2 , which cannot be negative.
82. False. If the vectors are nonzero and perpendicular, the dot product will be zero (e.g. ~i · ~j = 0).
83. False. If ~v and w
~ are different vectors, but both are perpendicular to ~
u , then both ~
u · ~v and ~
u ·w
~ are zero, yet ~v 6= w
~.
For example, take ~
u = ~i , ~v = ~j and w
~ = ~k .
84. True. Using the distributive property, and the fact that ~
u · (−~v ) = −~
u · ~v , we have
(~
u + ~v ) · (~
u − ~v ) = ~
u ·~
u +~
u · (−~v ) + ~v · ~
u − ~v · ~v = k~
u k2 − k~v k2 .
85. True. This vector is ~v
Perp
u . To check, calculate the dot product
, the component of ~v perpendicular to the unit vector ~
~
u · (~v − (~v · ~
u )~
u)=~
u · ~v − (~v · ~
u )(~
u ·~
u) =~
u · ~v − ~v · ~
u = 0,
since ~
u ·~
u = k~
u k2 = 1.
Solutions for Section 13.4
Exercises
1. ~v × w
~ = ~k × ~j = −~i (remember ~i , ~j , ~k are unit vectors along the axes, and you must use the right hand rule.)
2. ~v = −~i , and w
~ = ~j + ~k
~i ~j ~k
~v × w
~ = −1 0 0 = ~j − ~k
0 11
3. ~v = ~i + ~k , and w
~ = ~i + ~j
~i ~j ~k
~v × w
~ = 1 0 1 = −~i + ~j + ~k
110
1238
Chapter Thirteen /SOLUTIONS
4. ~v = ~i + ~j + ~k , and w
~ = ~i + ~j − ~k
~i ~j ~k
= −2~i + 2~j
~v × w
~ = 11 1
1 1 −1
5. ~v = 2~i − 3~j + ~k , and w
~ = ~i + 2~j − ~k
~k
~i ~j
= ~i + 3~j + 7~k
~v × w
~ = 2 −3 1
1 2 −1
6. ~v = 2~i − ~j − ~k , and w
~ = −6~i + 3~j + 3~k
~i
~v × w
~ =
~j
~k
2 −1 −1 = 0~i + 0~j + 0~k = ~0 .
−6 3
3
7. Since ~v = −3~i + 5~j + 4~k and w
~ = ~i − 3~j − ~k ,
~i ~j
~v × w
~ = −3 5
~k
4
1 −3 −1
= (5(−1) − 4(−3))~i − ((−3)(−1) − 4(1))~j + ((−3)(−3) − 5(1))~k
= (−5 + 12)~i − (3 − 4)~j + (9 − 5)~k
= 7~i + ~j + 4~k .
z
8.
~k
~j
~i
2~i
y
~i + ~j
x
Figure 13.32
By the definition of cross product, 2~i × (~i + ~j ) is in the direction of ~k . The magnitude of it equals to the area of
the parallelogram which is
√
√
√
π
2
π
k2~i k · k~i + ~j k sin = 2 2 sin = 2 2 ·
= 2.
4
4
2
So 2~i × (~i + ~j ) = 2~k . See Figure 13.32.
13.4 SOLUTIONS
z
9.
~k
~i − ~j
~j
y
~i + ~j
~i
x
Figure 13.33
By definition, (~i + ~j ) × (~i − ~j ) is in the direction of −~k . The magnitude is
√ √
π
k~i + ~j k · k~i − ~j k sin = 2 · 2 · 1 = 2.
2
10.
So (~i + ~j ) × (~i − ~j ) = −2~k . See Figure 13.33.
[(~i + ~j ) × ~i ] × ~j = (~i × ~i + ~j × ~i ) × ~j
= (~0 − ~k ) × ~j
= −~k × ~j
= ~j × ~k = ~i .
11.
(~i + ~j ) × (~i × ~j ) = (~i + ~j ) × ~k
= (~i × ~k ) + (~j × ~k )
= −~j + ~i = ~i − ~j .
12.
~i
~
~a × b = 3
~j
1 −1
1 −4
=
~k
2
3 −1 ~
3 1 ~
1 −1 ~
i −
j +
k
1 2
1 −4
−4 2
= −2~i − 7~j − 13~k .
Since
and
~a · (~a × ~b ) = 3(−2) + (−7) − (−13) = 0
~b · (~a × ~b ) = 1(−2) − 4(−7) + 2(−13) = 0,
~a × ~b is perpendicular to both ~a and ~b .
13. We find that ~v × w
~ = −6~i + 7~j + 8~k and w
~ × ~v = 6~i − 7~j − 8~k . Notice that
~v × w
~ = −(w
~ × ~v ).
1239
1240
Chapter Thirteen /SOLUTIONS
14. We can form the displacement vectors ~a = −~i + ~j + 0~k from (1, 0, 0) to (0, 1, 0) and ~b = −~i + 0~j + ~k from (1, 0, 0)
to (0, 0, 1). A normal vector to the plane is ~a × ~b = ~i + ~j + ~k . Using the point (1, 0, 0), the plane can be written as
(x − 1) + y + z = 0 or x + y + z = 1.
15. The displacement vector from (3, 4, 2) to (−2, 1, 0) is:
~a = −5~i − 3~j − 2~k .
The displacement vector from (3, 4, 2) to (0, 2, 1) is:
~b = −3~i − 2~j − ~k .
Therefore the vector normal to the plane is:
~n = ~a × ~b = −~i + ~j + ~k .
Using the first point, the equation of the plane can be written as:
−(x − 3) + (y − 4) + (z − 2) = 0.
The equation of the plane is thus:
−x + y + z = 3.
16. We first calculate the cross product:
~i ~j ~k
~b × ~c = 5 4 3 = (4 − 3)~i − (5 − 3)~j + (5 − 4)~k = ~i − ~j + ~k .
111
Then
Volume = |(~b × ~c ) · ~a | = |(~i − ~j + ~k ) · (3~i + 4~k + 5~) | = |3 − 4 + 5| = 4.
17. We first calculate the cross product:
~i ~j ~k
~b × ~c = 1 −1 1
= (1 − 1)~i − (−1 − 1)~j + (1 + 1)~k = 0~i + 2~j + 2~k .
1 1 −1
Then
Volume = |(~b × ~c ) · ~a | = |(0~i + 2~j + 2~k ) · (−~i + ~k + ~k )| = |0 + 2 + 2| = 4.
18. We first calculate the cross product:
~i ~j ~k
~b × ~c = 0 2 9 = (6 − 0)~i − (0 − 0)~j + (0 + 0)~k = 6~i .
003
Then
Volume = |(~b × ~c ) · ~a | = |(6~i ) · (~i + 8~j + 7~k )| = |6 + 0 + 0| = 6.
19. We first calculate the cross product:
~i ~j ~k
~b × ~c = 1 0 1 = (0 − 1)~i − (1 − 0)~j + (1 − 0)~k = −~i − ~j + ~k .
011
Then
Volume = |(~b × ~c ) · ~a | = |(−~i − ~j + ~k ) · (~i + ~k + 2~k )| = | − 1 − 1 + 2| = 0.
The volume is zero because ~a = ~b +~c , so the three vectors lie in a plane; they do not make a 3-dimensional parallelepiped
Problems
20. Normal vectors of the planes are ~
n 1 = 2~i − 3~j + 5~k and ~n 2 = 4~i + ~j − 3~k respectively. The line of intersection is
perpendicular to both normals. (Picture the pages in a partially open book.) We can use ~
n 1 × ~n 2 = 4~i + 26~j + 14~k .
13.4 SOLUTIONS
1241
21. We use the same normal ~n = 4~i + 26~j + 14~k and the point (0, 0, 0) to get 4(x − 0) + 26(y − 0) + 14(z − 0) = 0, or
4x + 26y + 14z = 0.
22. We use the same normal ~n = 4~i + 26~j + 14~k and the point (4, 5, 6) to get 4(x − 4) + 26(y − 5) + 14(z − 6) = 0, or
4x + 26y + 14z = 230.
23. The origin (0, 0, 0) is on the plane. A vector normal to the plane is given by the cross product of the vectors from the
origin to the given points: (~i + 3~j ) × (2~i + 4~j + ~k ) = 3~i − ~j − 2~k . Thus, an equation for the plane is 3x − y − 2z = 0.
24. The normal vectors to the two planes are ~n 1 = 4~i − 3~j + 2~k and ~n 2 = ~i + 5~j − ~k . A vector parallel to the line of
intersection of the two planes is perpendicular to both these normal vectors, so
Vector parallel to line = ~n 1 × ~n 2 = −7~i + 6~j + 23~k .
25. The normal vectors to the planes are n~1 = 2~i − 3~j + 5~k and n~2 = 4~i + ~j − 3~k . The line of intersection is
perpendicular to both normal vectors (picture the pages in a partially open book). Hence the vector we need is n~1 × n~2 =
4~i + 26~j + 14~k .
26. The vector parallel to the line of intersection is 4~i + 26~j + 14~k and this is normal to the desired plane. Therefore,
4x + 26y + 14z = 0 is the equation of the plane.
27. We use the same normal vector ~n = 4~i + 26~j + 14~k and the point (4, 5, 6) to get 4(x − 4) + 26(y − 5) + 14(z − 6) = 0.
28. Normal vectors to the planes are
~n 1 = ~i − ~j + ~k
and
~
n 2 = 2~i + ~j − 2~k .
The vector ~n 1 × ~n 2 is perpendicular to both planes and is normal to the plane we want:
~i ~j
~k
~n 1 × ~
n 2 = 1 −1 1
= ~i + 4~j + 3~k .
2 1 −2
The plane through the origin with normal ~n 1 × ~n 2 is
x + 4y + 3z = 0.
29. (a) Since
and
−
−
→
P Q = (3~i + 5~j + 7~k ) − (~i + 2~j + 3~k ) = 2~i + 3~j + 4~k ,
−→
P R = (2~i + 5~j + 3~k ) − (~i + 2~j + 3~k ) = ~i + 3~j ,
~i ~j ~k
−
−
→ −→
P Q × P R = 2 3 4 = −12~i + 4~j + 3~k ,
130
which is a vector perpendicular to the plane containing P , Q and R. Since
p
−
−→ −→
kP Q × P Rk = (−12)2 + 42 + 32 = 13,
the unit vectors which are perpendicular to a plane containing P , Q, and R are
4
3
12~
i + ~j + ~k ,
13
13
13
or the unit vector pointing to the opposite direction,
−
12~
4
3
i − ~j − ~k .
13
13
13
(b) The angle between P Q and P R is θ for which
−
−
→ −→
2·1+3·3+4·0
11
PQ · PR
√
= √
,
cos θ = −
−
→
−→ = √ 2
2 + 32 + 42 · 12 + 32 + 02
290
kP Qk · kP Rk
so
11
θ = cos−1 ( √
) ≈ 49.76◦ .
290
1242
Chapter Thirteen /SOLUTIONS
−
−
→ −→
(c) The area of triangle P QR = 21 kP Q × P Rk = 13
.
2
(d) Let d be the distance from R to the line through P and Q (see Figure 13.34), then
13
1
−
−→
d · kP Qk = the area of △ P QR =
.
2
2
Therefore,
13
13
13
d= −
= √ .
−
→ = √ 2
2 + 32 + 42
29
kP Qk
Q
P
d
R
Figure 13.34
−→
30. (a) We first find two displacement vectors: AB = (3 − (−1))~i + (2 − 3)~j + (4 − 0)~k = 4~i − ~j + 4~k and
−→
AC = 2~i − 4~j + 5~k . The normal vector, ~n , to the plane is perpendicular to these two vectors, so we have
−→ −→
~n = AB × AC = 11~i − 12~j − 14~k .
Using the normal vector, we see that the equation of the plane is 11x−12y−14z = d for some number d. Substituting
one of the points gives d = −47. Therefore, an equation for the plane is
11x − 12y − 14z = −47.
(b) The area of the triangle is given by
Area =
1√
1 −→ −→
461 = 10.74.
kAB × ACk =
2
2
31. Since ~v × w
~ is perpendicular to both ~v and w
~ , we can conclude that ~v × w
~ is parallel to the z-axis.
32. (a) Since ~v · w
~ = ||~v || ||w
~ || cos θ and ||~v × w
~ || = ||~v || ||w
~ || sin θ,
tan θ =
sin θ
||~v × w
~ ||
3
=
= = 0.6.
cos θ
~v · w
~
5
(b) Then θ = tan−1 (0.6) = 0.540.
33. Since
and
so
so
k~v × w
~ k = k~v k · kw
~ k sin θ,
~v · w
~ = k~v k · kw
~ k cos θ,
k~v k · kw
~ k sin θ
k~v × w
~k
=
= tan θ,
~v · w
~
k~v k · kw
~ k cos θ
tan θ =
k2~i − 3~j + 5~k k
=
3
√
38
= 2.055.
3
13.4 SOLUTIONS
1243
34. (a) Since ~v · w
~ = ||~v || ||w
~ || cos θ and ||~v × w
~ || = ||~v || ||w
~ || sin θ, we find
||~v × w
~ || = ||12~i − 3~j + 4~k || =
Then
tan θ =
(b) Then θ = tan−1 (1.625) = 1.019.
p
122 + (−3)2 + 42 = 13.
sin θ
||~v × w
~ ||
13
=
=
= 1.625.
cos θ
~v · w
~
8
35. (a) Increases the force.
(b) The Magnus force is perpendicular to the direction the ball is travelling (which is into the page) and the axis of spin.
Thus F~M points down and to the left. The ball curves to the left and drops faster than a ball that is not spinning.
36. (a) The vector ~k × ~r is in the xy-plane because it is perpendicular to ~k . It is also perpendicular to ~r , so it points in
either the same or opposite direction as ~v . The right hand rule shows that it is in the same direction. Finally, ~k × ~r
has magnitude k~k × ~r k = k~k kk~r k sin (90◦ ) = k~r k = k~v k. Thus ~k × ~r and ~v have the same directions and
magnitudes, and so they are equal.
(b) Since ~k × ~r = ~k × (x~i + y~j ) = −y~i + x~j is the position vector of P , we have P = (−y, x).
37. (a) We have
−−−→
P1 P2 = 2~i + 4~j + 2~k ,
and
−−−→
P3 P4 = 2~i + 4~j + 2~k .
so these two displacement vectors are equal. Also,
−−−→
−−−→
P1 P3 = 3~i
and
P2 P4 = 3~i ,
so these two vectors are also equal. These points form a parallelogram.
−−−→
−−−→
(b) Vectors along adjacent sides of the parallelogram are P1 P2 and P1 P3 . Since
~i ~j ~k
−−−→ −−−→
P1 P2 × P1 P3 = 2 4 2 = 6~j − 12~k ,
300
we have
√
−−−→ −−−→
Area of parallelogram = kP1 P2 × P1 P3 k = k6~j − 12~k k = 180.
−−−→
−−−→
−−−→
38. The three vectors ~a = P1 P2 = 2~i + 4~j + 2~k and ~b = P1 P3 = 3~i and ~c = P1 P5 = ~i + 4~k determine the
parallelepiped. To find the volume of the parallelepiped, we first compute the cross product ~b × ~c :
~i ~j ~k
~b × ~c = 3 0 0 = −12~j .
104
We have
Volume of parallelepiped = (~b × ~c ) · ~a = (−12~j ) · ~a = |−48| = 48.
39. First let
~a = a1~i + a2~j + a3~k ~b = b1~i + b2~j + b3~k ~c = c1~i + c2~j + c3~k
so ~b + ~c = (b1 + c1 )~i + (b2 + c2 )~j + (b3 + c3 )~k . Now, using the general formula for cross products, we have:
~a × (~b + ~c )
= [a2 (b3 + c3 ) − a3 (b2 + c2 )]~i + [a3 (b1 + c1 ) − a1 (b3 + c3 )]~j + [a1 (b2 + c2 ) − a2 (b1 + c1 )]~k
= (a2 b3 + a2 c3 − a3 b2 − a3 c2 )~i + (a3 b1 + a3 c1 − a1 b3 − a1 c3 )~j
+(a1 b2 + a1 c2 − a2 b1 − a2 c1 )~k
= (a2 b3 − a3 b2 )~i + (a2 c3 − a3 c2 )~i + (a3 b1 − a1 b3 )~j + (a3 c1 − a1 c3 )~j
+(a1 b2 − a2 b1 )~k + (a1 c2 − a2 c1 )~k
= (a2 b3 − a3 b2 )~i + (a3 b1 − a1 b3 )~j + (a1 b2 − a2 b1 )~k + (a2 c3 − a3 c2 )~i + (a3 c1 − a1 c3 )~j
+(a1 c2 − a2 c1 )~k
= (~a × ~b ) + (~a × ~c )
1244
Chapter Thirteen /SOLUTIONS
Thus, ~a × (~b + ~c ) = ~a × ~b + ~a × ~c .
40. If λ = 0, then all three cross products are ~0 , since the cross product of the zero vector with any other vector is always 0.
If λ > 0, then λ~v and ~v are in the same direction and w
~ and λw
~ are in the same direction. Therefore the unit
normal vector ~n is the same in all three cases. In addition, the angles between λ~v and w
~ , and between ~v and w
~ , and
between ~v and λw
~ are all θ. Thus,
(λ~v ) × w
~ = kλ~v kkw
~ k sin θ~n
= λk~v kkw
~ k sin θ~n
= λ(~v × w
~)
= k~v kkλw
~ k sin θ~n
= ~v × (λw
~)
If λ < 0, then λ~v and ~v are in opposite directions, as are w
~ and λw
~ in opposite directions. Therefore if ~
n is the
normal vector in the definition of ~v × w
~ , then the right-hand rule gives −~n for (λ~v ) × w
~ and ~v × (λw
~ ). In addition,
if the angle between ~v and w
~ is θ, then the angle between λ~v and w
~ and between ~v and λw
~ is (π − θ). Since if λ < 0,
we have |λ| = −λ, so
(λ~v ) × w
~ = kλ~v kkw
~ k sin(π − θ)(−~n )
= |λ| k~v kkw
~ k sin(π − θ)(−~n )
= −λk~v kkw
~ k sin θ(−~n )
= λk~v kkw
~ k sin θ~n
= λ(~v × w
~ ).
Similarly,
~v × (λw
~ ) = k~v kkλw
~ k sin(π − θ)(−~n )
= −λk~v kkw
~ k sin θ(−~n )
= λ(~v × w
~ ).
41. The quantities ~a · (~b × ~c ) and (~a × ~b ) · ~c both represent the volume of the same parallelepiped, namely that defined
by the three vectors ~a , ~b , and ~c , and therefore must be equal. Thus, the two triple products ~a · (~b × ~c ) and (~a × ~b ) · ~c
must be equal except perhaps for their sign. In fact, both are positive if ~a , ~b , ~c are right-handed and negative if ~a , ~b , ~c
are left-handed. This can be shown by drawing a picture:
~b
~c
right-handed
left-handed
~b
~a
~a
Figure 13.35
42. If θ is the angle between ~a and ~b , then
k~a × ~b k2 = (k~a kk~b k sin θ)2
= k~a k2 k~b k2 sin2 θ
= k~a k2 k~b k2 (1 − cos2 θ)
= k~a k2 k~b k2 − k~a k2 k~b k2 cos2 θ
= k~a k2 k~b k2 − (~a · ~b )2 .
~c
13.4 SOLUTIONS
43. Solve for ~c to get ~c = −(~a + ~b ). So
1245
~b × ~c = ~b × (−(~a + ~b ))
= −(~b × (~a + ~b ))
= −(~b × ~a + ~b × ~b )
= −(~b × ~a + ~0 )
= −(~b × ~a )
= ~a × ~b .
Also,
~c × ~a = −(~a + ~b ) × ~a
= −((~a + ~b ) × ~a )
Therefore, ~a × ~b = ~b × ~c = ~c × ~a .
= −(~a × ~a + ~b × ~a )
= −(~0 + ~b × ~a )
= ~a × ~b .
k~b k sin θ
~b
~c
θ
~b
~a
Figure 13.36
Geometrically, the magnitude of the cross product of two vectors is equal to the area of the parallelogram formed by
the vectors. If ~a + ~b + ~c = 0, then we can think of the vectors ~a , ~b , and ~c as forming a triangle. (See Figure 13.36.) So
by showing that
~a × ~b = ~b × ~c = ~c × ~a ,
we are showing that the areas of the parallelograms formed by any two sides of the same triangle are equal.
44. The cross product is given by
~i ~j ~k
~b × ~c = b b b = b2 b3 ~i − b1 b3 ~j + b1 b2 ~k
1 2 3
c2 c3
c1 c3
c1 c2
c1 c2 c3
so
a1 a2 a3
b2 b3
b1 b3
b1 b2
− a2
+ a3
= b1 b2 b3 .
~a · (~b × ~c ) = a1
c2 c3
c1 c3
c1 c2
c1 c2 c3
~ in components and expand using the distributive property of the cross product.
45. Write ~v and w
~v × w
~ = (v1~i + v2~j + v3~k ) × (w1~i + w2~j + w3~k )
= v1 w1~i × ~i + v1 w2~i × ~j + v1 w3~i × ~k
+v2 w1~j × ~i + v2 w2~j × ~j + v2 w3~j × ~k
+v3 w1~k × ~i + v3 w2~k × ~j + v3 w3~k × ~k
Now we use the fact that ~i × ~i = ~0 ,~i × ~j = ~k ,~i × ~k = −~j , ~j × ~i = −~k , ~j × ~j = ~0 , ~j × ~k = ~i , ~k × ~i =
~j , ~k × ~j = −~i , ~k × ~k = ~0 . Thus we have
~v × w
~ = ~0 + v1 w2~k + v1 w3 (−~j ) + v2 w1 (−~k ) + ~0 + v2 w3~i + v3 w1~j + v3 w2 (−~i ) + ~0
= (v2 w3 − v3 w2 )~i + (v3 w1 − v1 w3 )~j + (v1 w2 − v2 w1 )~k .
1246
Chapter Thirteen /SOLUTIONS
46. Any vector ~v that is perpendicular to both ~a and ~b will have the property that its dot product with ~a and ~b is 0, that is
~a · ~v = a1 x + a2 y + a3 z = 0,
~b · ~v = b1 x + b2 y + b3 z = 0.
Multiply the first equation by b1 and the second by a1 and subtract to get
(b1 a2 − a1 b2 )y + (b1 a3 − a1 b3 )z = 0
or y =
−(b1 a3 − a1 b3 )z
(b1 a2 − a1 b2 )
(for b1 a2 6= a1 b2 )
Multiply the second equation by a2 and the first by b2 and subtract to get
(b2 a1 − a2 b1 )x + (b2 a3 − a2 b3 )z = 0
or
x=
−(b2 a3 − a2 b3 )z
.
(b2 a1 − a2 b1 )
So
(b1 a3 − a1 b3 )z ~
−(b2 a3 − a2 b3 )z~
i −
j + z~k .
(b2 a1 − a2 b1 )
(b1 a2 − a1 b2 )
Pick z = b2 a1 − b1 a2 and multiply out, and we see that the algebraic method of finding a cross product yields the same
result as our standard method.
47. (a) Since ~c is perpendicular to ~a × ~b , and since ~a × ~b is normal to the plane containing ~a and ~b , it follows that ~c must
be in the plane containing ~a and ~b .
(b) Using the expression given in the problem for ~c , we get
~v =
~a · ~c = ~a · (~a × (~b × ~a ))
= (~a × ~a ) · (~b × ~a )
= ~0 · (~b × ~a ) = 0.
and
~b · ~c = ~b · (~a × (~b × ~a ))
= (~b × ~a ) · (~b × ~a )
= k~b × ~a k2
= k~a k2 k~b k2 − (~a · ~b )2 .
(c) Since ~c lies in the plane containing ~a and ~b , it is of the form ~c = x~a + y~b for some scalars x and y. Thus, using
the fact that ~a · ~c = 0 from part (b), we have
~a · ~c = ~a · (x~a + y~b ) = xk~a k2 + y(~a · ~b ) = 0.
Similarly, using the fact that ~b · ~c = k~a k2 k~b k2 − (~a · ~b )2 from part (b), we have
~b · ~c = ~b · (x~a + y~b ) = x(~a · ~b ) + yk~b k2 = k~a k2 k~b k2 − (~a · ~b )2 .
Solving these two linear equations in x and y, we find x = −~a · ~b and y = k~a k2 .
48. Problem 41 tells us that (~
u × ~v ) · w
~ =~
u · (~v × w
~ ). Using this result on the triple product of (~a + ~b ) × ~c with any
~
vector d together with the fact that the dot product distributes over addition gives us:
[(~a + ~b ) × ~c ] · d~ = (~a + ~b ) · (~c × d~ )
= ~a · (~c × d~ ) + ~b · (~c × d~ )
= (~a × ~c ) · d~ + (~b × ~c ) · d~
= [(~a × ~c ) + (~b × ~c )] · d~ .
(dot product is distributive)
(using Problem 41 again)
(dot product is distributive)
So, since [(~a + ~b ) × ~c ] · d~ = [(~a × ~c ) + (~b × ~c )] · d~ , then
[(~a + ~b ) × ~c ] · d~ − [(~a × ~c ) + (~b × ~c )] · d~ = 0,
13.4 SOLUTIONS
1247
Since the dot product is distributive, we have
[((~a + ~b ) × ~c ) − (~a × ~c ) − (~b × ~c )] · d~ = 0.
Since this equation is true for all vectors d~ , by letting
d~ = ((~a + ~b ) × ~c ) − (~a × ~c ) − (~b × ~c ),
we get
k(~a + ~b ) × ~c − ~a × ~c − ~b × ~c k2 = 0
and hence
(~a + ~b ) × ~c − (~a × ~c ) − (~b × ~c ) = ~0 .
Thus
(~a + ~b ) × ~c = (~a × ~c ) + (~b × ~c ).
49. The area vector for face OAB =
The area vector for face OBC =
The area vector for face OAC =
The area vector for face ABC =
1~
b × ~a .
2
1
~a × ~c .
2
1~
b × ~c .
2
1 ~
(b − ~a )
2
× (~c − ~a ).
1
1
1
1~
b × ~a + ~c × ~b + ~a × ~c + (~b − ~a ) × (~c − ~a ) =
2
2
2
2
1~
1
1
1
b × ~a + ~c × ~b + ~a × ~c + (~b × ~c − ~b × ~a − ~a × ~c − ~a × ~a ) = 0.
2
2
2
2
50. First let two adjoining sides of the rectangle be our vectors ~a and ~b . See Figure 13.37. So we have
~a = ~j
and ~b = 2~i
~ = ~a × ~b . So, by the
Since it faces downward (that is, in the negative z direction), according to the right hand rule A
formula for cross products
~i ~j ~k
~
A = 0 1 0 = −2~k .
200
z
y
~a
~b
~
A
Figure 13.37
x
1248
Chapter Thirteen /SOLUTIONS
51. The area vector for a circle has a magnitude πr 2 and direction normal to the plane of the circle. See Figure 13.38. Thus,
since our circle is facing in the positive x direction, and r = 2,
~ = π(2)2~i = 4π~i .
A
z
y
x
Figure 13.38
52. First choose vectors ~a and ~b along the sides of the triangle. Let’s choose
−→
−→
~a = AB = 2~i − ~j − ~k , ~b = AC = ~i − ~j .
These vectors then form two sides of a parallelogram whose area is k~a ×~b k. Our triangle forms half of this parallelogram,
so the area of triangle ABC = 12 k~a × ~b k. Since there are two possible orientations, the area of the triangle ABC is
represented by one of the following two vectors:
1
1
1
1
1
± (~a × ~b ) = ± (2~i − ~j − ~k ) × (~i − ~j ) = ± − ~i − ~j − ~k
2
2
2
2
2
We want the upward orientation, so we pick the negative sign, giving
~ = 1~i + 1 ~j + 1 ~k .
A
2
2
2
53. (a) Since
we have
~
u × ~v = (u2 v3 − u3 v2 )~i + (u3 v1 − u1 v3 )~j + (u1 v2 − u2 v1 )~k ,
Area of S = k~
u × ~v k = (u2 v3 − u3 v2 )2 + (u3 v1 − u1 v3 )2 + (u1 v2 − u2 v1 )2
1/2
.
~ and V
~ ,
u and ~v onto the xy-plane. These are the vectors U
(b) The two edges of R are given by the projections of ~
~ = u1~i + u2~j and V
~ = v1~i + v2~j , Thus
obtained by omitting the ~k -components of ~
u and ~v : we have U
~ ×V
~ k = k(u1 v2 − u2 v1 )~k k = |u1 v2 − u2 v1 | .
Area of R = kU
(c) The vector m~i + n~j − ~k is normal to the plane z = mx + ny + c. Since the vectors ~
u and ~v are in the plane
~ ), the vector ~
(they’re the sides of S
u × ~v is also normal to the plane. Thus, these two vectors are scalar multiples of
one another. Suppose
~
u × ~v = λ(m~i + n~j − ~k )
Since the ~k component of ~
u × ~v is (u1 v2 − u2 v1 )~k , comparing the ~k -components tells us that
λ = −(u1 v2 − u2 v1 ).
Thus,
−(u1 v2 − u2 v1 )(m~i + n~j − ~k ) = ~
u × ~v = (u2 v3 − u3 v2 )~i + (u3 v1 − u1 v3 )~j + (u1 v2 − u2 v1 )~k ,
so
u2 v3 − u3 v2
u2 v1 − u1 v2
u3 v1 − u1 v3
n=
.
u2 v1 − u1 v2
m=
SOLUTIONS to Review Problems for Chapter Thirteen
1249
(d) We have
(1 + m2 + n2 ) · (Area of R)2 =
1+
u v − u v 2
2 3
3 2
u2 v1 − u1 v2
+
u v − u v 2
3 1
1 3
u2 v1 − u1 v2
(u1 v2 − u2 v1 )2
= (u2 v3 − u3 v2 )2 + (u3 v1 − u1 v3 )2 + (u1 v2 − u2 v1 )2
= (Area of S)2 .
Strengthen Your Understanding
54. If ~n is unit vector perpendicular to ~
u and ~v , then so is −~n . There are exactly two possible unit vectors perpendicular to
two given nonparallel vectors. The right-hand rule gives one of the two.
55. ~
u × ~v = ~0 when ~
u and ~v are parallel, not perpendicular.
56. Since ~
u × ~v is perpendicular to the plane containing ~
u and ~v , we want ~k to be perpendicular to the plane containing ~u
and ~v . That is, we want ~
u and ~v to both lie in the xy-plane. Thus choose ~
u to be any nonzero vector in the xy-plane not
parallel to ~v . For example, let ~
u = 2~i + ~j .
57. We could let ~v = a~i + b~j + c~k , compute k~
u × ~v k in terms of a, b, c, set the length equal to 10, and solve to find values
of a, b, c. It is much easier to think geometrically. Let ~v be any vector perpendicular to ~
u , forming a rectangle of area 10.
Since k~
u k = 5, that means we can choose v to be any vector perpendicular to ~
u of length 2. So first find a unit vector
perpendicular to ~
u , say 4~i − 3~j , unitize it to get (4/5)~i − (3/5)~j and then scale it by 2. We get (8/5)~i − (6/5)~j .
58. True. The cross product yields a vector.
59. False. ~
u × ~v has direction perpendicular to both ~
u and ~v .
60. False. This is only true when ~
u and ~v are perpendicular. In general, k~
u × ~v k = k~
u kk~v k sin θ, where θ is the angle
between ~
u and ~v . The value of k~
u × ~v k is the area of the parallelogram with sides ~
u and ~v .
61. True. The left-hand side evaluates to ~k · ~k = 1, while the right-hand side evaluates to ~i · ~i = 1.
62. False. If ~
u and w
~ are two different vectors both of which are parallel to ~v , then ~v × ~
u = ~v × w
~ = ~0 , but ~
u 6= w
~ .A
counterexample is ~v = ~i , ~
u = 2~i and w
~ = 3~i .
63. True. Since (~v × w
~ ) is perpendicular to ~v , the dot product with ~v is zero.
64. True. The cross product is a vector in 3-space, while the dot product is a scalar, so they cannot be equal.
p
65. True. The cross product (~i + ~j ) × (~j + 2~k ) = 2~i − 2~j + ~k , which has magnitude 22 + (−2)2 + 12 = 3. Since the
triangle has area of 1/2 the parallelogram with the given vectors as sides, the triangle has area 3/2.
66. True. Any vector w
~ that is parallel to ~v will give ~v × w
~ = ~0 .
67. False. It is not true in general, but there are special cases when ~v × w
~ =w
~ × ~v . For example, when ~v is parallel to w
~,
or when one of the vectors is ~0 . In either case the cross products ~v × w
~ and w
~ × ~v are both the zero vector.
Solutions for Chapter 13 Review
Exercises
1. Scalar. ~
u · ~v = (2~i − 3~j − 4~k ) · (~k − ~j ) = 2 · 0 − 3(−1) − 4 · 1 = −1.
2. Vector. We calculate
~i ~j
~k
~
u × ~v = 2 −3 −4 = −7~i − 14~j + 7~k .
3 −1 1
3. 4 − 9 + 4 = −1
4. ~i · (−~i ) = −1
5. ~a = −2~j , ~b = 3~i , ~c = ~i + ~j , d~ = 2~j , ~e = ~i − 2~j , f~ = −3~i − ~j .
1250
Chapter Thirteen /SOLUTIONS
6. Resolving ~v into components gives ~v = 8 cos(40◦ )~i − 8 sin(40◦ )~j = 6.13~i − 5.14~j . Notice that the component in
the ~j direction must be negative.
7. 5~c = 5~i + 30~j
8. ~c + ~
x +~
y = ~i + 6~j − 2~i + 9~j + 4~i − 7~j = 3~i + 8~j .
√
9. ||~
x − ~c || = || − 2~i + 9~j − (~i + 6~j )|| = || − 3~i + 3~j || = 3 2.
10. ~v + 2w
~ = 2~i + 3~j − ~k + 2(~i − ~j + 2~k ) = 4~i + ~j + 3~k .
11. 3~v − w
~ − ~v = 2~v − w
~ = 2(2~i + 3~j − ~k ) − (~i − ~j + 2~k ) = 3~i + 7~j − 4~k .
√
√
12. ||~v + w
~ || = ||3~i + 2~j + ~k || = 32 + 22 + 12 = 14.
13. ~v · w
~ = (2~i + 3~j − ~k ) · (~i − ~j + 2~k ) = 2 · 1 + 3 · (−1) + (−1) · 2 = −3.
~i ~j
~k
14. ~v × w
~ = 2 3 −1 = (6 − 1)~i − (4 + 1)~j + (−2 − 3)~k = 5~i − 5~j − 5~k .
1 −1 2
15. For any vector ~v , we have ~v × ~v = ~0 .
16. Since ~v · w
~ = 2 · 1 + 3(−1) + (−1)2 = −3, we have (~v · w
~ )~v = −6~i − 9~j + 3~k .
17. Since ~v × w
~ is perpendicular to w
~ , we have (~v × w
~ )·w
~ = 0.
~
18. We have ~v × w
~ = 5~i − 5~j − 5k , so
~i ~j ~k
(~v × w
~ )×w
~ = 5 −5 −5 = (−10 − 5)~i − (10 + 5)~j + (−5 + 5)~k = −15~i − 15~j .
1 −1 2
19. The cross product of two parallel vectors is ~0 , so the cross product of any vector with itself is ~0 .
20. A normal vector can be obtained from the coefficients of x, y, z in the equation of the plane and is: ~n = 2~i + ~j − ~k .
21. The equation can be rewritten as
z − 5x + 10 = 15 − 3y
−5x + 3y + z = 5
so ~
n = −5~i + 3~j + ~k .
22. If the planes are parallel, they have a common normal vector ~n . Rewrite the equation of the plane as 4x − 3y − z = −8
so that ~n = 4~i − 3~j − ~k and the desired plane is 4(x − 0) − 3(y − 0) − (z − 0) = 0 or 4x − 3y − z = 0.
23. (a) We have ~v · w
~ = 3 · 4 + 2 · (−3) + (−2) · 1 = 4.
(b) We have ~v × w
~ = −4~i − 11~j − 17~k .
(c) A vector of length 5 parallel to ~v is
5
5
~v = √ (3~i + 2~j − 2~k ) = 3.64~i + 2.43~j − 2.43~k .
k~v k
17
(d) The angle between vectors ~v and w
~ is found using
cos θ =
~v · w
~
4
= √ √ = 0.190,
k~v kkw
~k
17 26
so θ = 79.0◦ .
(e) The component of vector ~v in the direction of vector w
~ is
4
~v · w
~
= √ = 0.784.
kw
~k
26
(f) The answer is any vector ~a such that ~a · ~v = 0. One possible answer is 2~i − 2~j + ~k .
(g) A vector perpendicular to both is the cross product:
~v × w
~ = −4~i − 11~j − 17~k .
SOLUTIONS to Review Problems for Chapter Thirteen
24. Since ||2~i + 3~j − ~k || =
p
22 + 32 + (−1)2 =
√
1251
14, vectors of length 10 are
10
± √ (2~i + 3~j − ~k ).
14
25. We take the cross product of ~i + ~j and ~i − ~j − ~k and then make a unit vector parallel to the cross product.
~i
~j
~k
(~i + ~j ) × (~i − ~j − ~k ) = 1
1
0
= −~i + ~j − 2~k .
1 −1 −1
Since || − ~i + ~j − 2~k || =
p
(−1)2 + 12 + (−2)2 = 6, unit vectors are
±
−~i + ~j − 2~k
√
.
6
26. We want a unit vector of the form a~i + b~j such that
(a~i + b~j ) · (3~i − 2~j ) = 3a − 2b = 0.
Let’s take a = 2√and b = 3. Then the vector 2~i + 3~j is perpendicular to 3~i − 2~j , but 2~i + 3~j is not a unit vector. Since
||2~i + 3~j || = 13, unit vectors are
2~i + 3~j
.
± √
13
27. ~
n = 4~i + 6~k (the coefficients of x, y, z are the same as the coefficients of ~i , ~j , and ~k .)
28. First, we rewrite the equation of the plane as
x − y − z = 1.
In this form, the coefficients of x, y, and z are the coefficients of ~i , ~j , and ~k in a vector that is perpendicular to the plane.
So any scalar multiple of ~i − ~j − ~k is perpendicular to the plane.
29. The vector w
~ we want is shown in Figure 13.39, where the given vector is ~v = 4~i + 3~j . The vectors ~v and w
~ are the
same length and the two angles marked α are equal, so the two right triangles shown are congruent. Thus
a = −3
b = 4.
and
Therefore
w
~ = −3~i + 4~j .
y
(a, b)
(4, 3)
α
w
~
4
~
v
3
α
3
x
4
Figure 13.39
30. The cross product of two vectors is perpendicular to both of them, so a possible answer is
~i ~j ~k
~v × w
~ = 3 −1 1 = ~i − 2~j − 5~k .
1 −2 1
1252
Chapter Thirteen /SOLUTIONS
31. To determine if two vectors are parallel, we need to see if one vector is a scalar multiple of the other one. Since ~
u = −2w
~,
q and no other pairs have this property, only ~
u and w
~ , and ~v and ~
q are parallel.
and ~v = 14 ~
~ = 2d~ , the two vectors are parallel in the same direction, so
32. Since F
~
~
~
~
F
parallel = F and F perp = 0 .
The work done is
~ · d~ = 2 + 8 = 10.
W =F
√
√
~ k = 20, times the distance traveled, kd~ k = 5, since the
Notice that this is the same as the magnitude of the force, kF
force is the same direction as the displacement.
~ = −2d~ , the two vectors are parallel in opposite directions, so
33. Since F
~
~
~
~
F
parallel = F and F perp = 0 .
The work done is
~ · d~ = −10.
W =F
Note that work done is negative since the force is in the opposite direction to the displacement.
~ · d~ = 0, the two vectors are perpendicular, so
34. Since F
~
~
~
~
F
parallel = 0 and F perp = F .
The work done is
~ · d~ = 0.
W =F
No work is done since the force is perpendicular to the displacement.
35. The unit vector in the direction of d~ is ~
u = (3/5)~i − (4/5)~j . Thus
10
6
8
~
~ u ~
u =− ~
u = − ~i + ~j ,
F
parallel = F · ~
5
5
5
12 ~
16~
~
~
~
i +
j.
F perp = F − F parallel =
5
5
~ perp · ~
Notice that F
u = 0, as we expect. The work done is
~ · d~ = 6 − 16 = −10.
W =F
~
~
The work is negative since F
parallel is in the opposite direction of the displacement vector d .
√
36. The unit vector in the direction of d~ is ~
u = (1/ 2)(~i + ~j ). Thus
2
~
~
u = ~i + ~j ,
F
u = √ ~
parallel = F · ~u ~
2
~ −F
~
~ ~
F~ perp = F
parallel = i − j .
~ perp · ~
Notice that F
u = 0, as we expect. The work done is
W = F~ · d~ = 2 − 0 = 2.
37. The unit vector in the direction of d~ = 3~j is ~
u = ~j . Thus, the parallel component of F~ is just its j component, and the
perpendicular component is its i component:
~
~
~
~
F
parallel = 2j and F perp = 5i .
The work done is
~ · d~ = 6.
W =F
SOLUTIONS to Review Problems for Chapter Thirteen
1253
38. The area of the triangle is half the area of the parallelogram created by these two vectors, and the area of the parallelogram
is the magnitude of the cross product. We first calculate the cross product:
~i ~j ~k
~
~a × b = 1 2 −1 = 0~i − 5~j − 10~k .
4 −2 1
The area of the triangle is given by:
Area of triangle =
1
1
1√
1
Area of parallelogram = k~a × ~b k = k − 5~j − 10~k k =
125 = 5.590.
2
2
2
2
Problems
True, since vectors ~c and f~ point in the same direction and have the same length.
False, since vectors ~a and d~ point in opposite directions. We have ~a = −d~ .
False, since −~b points in the opposite direction to ~b , the vectors −~b and ~a are perpendicular.
True. The vector f~ can be ”moved” to point directly up the z-axis.
True. We move in the positive x-direction following vector ~a and then in the positive y-direction following vector
−~b . The resulting sum is the vector ~e .
(f) False, vector d~ is the negative of the vector ~g − ~c . It is true that d~ = ~c − ~g .
~
~
~
~
40. Let the velocity vector of the
pplane is traveling
pairplane be V = xi +y j +z k in km/hr. We know that x = −y because the
~ k = x2 + y 2 + z 2 = 200 km/hr and z = 300 m/min = 18 km/hr. We have x2 + y 2 + z 2 =
northwest. Also, kV
√
x2 + x2 + 182 = 200, so x = −140.8, y = 140.8, z = 18. (The value of x is negative and y is positive because the
plane is heading northwest.) Thus,
~v = −140.8~i + 140.8~j + 18~k .
39. (a)
(b)
(c)
(d)
(e)
41. The velocity vector of the plane with respect to the air has the form
(See Figure 13.40.) Therefore
80~k .
The wind vector is
√
~v = a~i + 80~k where k~v k = 480.
√
a2 + 802 = 480 so a = 4802 − 802 ≈ 473.3 km/hr. We conclude that ~v ≈ 473.3~i +
w
~ = 100(cos 45◦ )~i + 100(sin 45◦ )~j
≈ 70.7~i + 70.7~j
The velocity vector of the plane with respect to the ground is then
~v + w
~ = (473.3~i + 80~k ) + (70.7~i + 70.7~j )
= 544~i + 70.7~j + 80~k
From Figure 13.41, we see that the velocity relative to the ground is
544~i + 70.7~j .
The ground speed is therefore
√
5442 + 70.72 ≈ 548.6 km/hr.
~
v
50~k
a~i
Figure 13.40: Side view
Velocity relative
to ground
w
~
a~i
Figure 13.41: Top view
1254
Chapter Thirteen /SOLUTIONS
>
>
42. (a) See Figure 13.42. Notice that the velocity vectors are tangent to the curve, they point in the direction of motion, and
they are longer when the rocket is moving faster.
> >
>
>
>
Figure 13.42
(b) If the rocket has a parachute, it comes down more slowly. The velocity vectors on the downward part of the graph are
shorter for this rocket.
43. See Figure 13.43.
P
R
Q
Figure 13.43
44. At the point P , the velocity of the car is changing the quickest; not in magnitude, but in direction only. The acceleration
vector is therefore the longest at this point. The direction of the vector is directed in toward the center of the track because
the difference in velocity vectors at nearby points is a vector pointing toward the center.
√
√ √
√
√
45. Since 3~i + 3~j = 3( 3~i + ~j ), we know that 3~i + 3~j and 3~i + ~j are scalar multiples of one another, and
therefore parallel.
√
√
√
√
√
√
+ ~j ) · (~i √
− 3~j ) = 3 − 3 = 0, √
we know that √3~i + ~j and ~i − 3~j are perpendicular.
Since ( 3~i √
Since 3~i + 3~j and 3~i + ~j are parallel, 3~i + 3~j and ~i − 3~j are perpendicular, too.
46. Let the x-axis point east and the y-axis point north. We resolve the forces into components. Since the first force points
50◦ south of east with a force of 25 newtons, we have
F~1 = 25 cos(50◦ )~i − 25 sin 50◦~j = 16.070~i − 19.151~j .
Since F~1 lies in the fourth quadrant, the coefficient of ~i is positive and the coefficient of ~j is negative.
The second force points 70◦ north of west with a force of 60 newtons, so we have
F~2 = −60 cos(70◦ )~i + 60 sin 70◦~j = −20.521~i + 56.382~j .
Since F~2 lies in the second quadrant, the coefficient of ~i is negative and the coefficient of ~j is positive.
The third force must make the total force equal to zero, so we have
F~1 + F~2 + F~3 = ~0
F~3 = −(F~1 + F~2 )
= −((16.070~i − 19.151~j ) + (−20.521~i + 56.382~j ))
= −(−4.451~i + 37.231~j )
The magnitude of this force is kF~3 k =
83.20◦ south of east.
= 4.451~i − 37.231~j .
√
4.4512 + 37.2312 = 37.50 newtons. The direction is arctan(37.231/4.451) =
47. If the vectors are perpendicular, we need
~v · w
~ = (2a~i − a~j + 16~k ) · (5~i + a~j − ~k ) = 10a − a2 − 16 = 0.
Solving 10a − a2 − 16 = −(a − 2)(a − 8) = 0 gives a = 2, 8.
SOLUTIONS to Review Problems for Chapter Thirteen
1255
48. Since a normal vector of the plane is ~n = −~i + 2~j + ~k , an equation for the plane is
−x + 2y + z = −1 + 2 · 0 + 2 = 1
−x + 2y + z = 1.
49. Since the plane is normal to the vector 2~i − 3~j + 7~k and passes through the point (1, −1, 2), an equation for the plane is
2x − 3y + 7z = 2 · 1 − 3 · (−1) + 7 · 2 = 19
2x − 3y + 7z = 19.
−→
−→
−→
50. See Figure 13.44. One way to find the angle at A is to find the angle between vectors AB and AC. Since AB = −1~i −7~j
−→
~
~
and AC = −5i − 3j , we have
−→ −→
AB · AC
cos(6 BAC) = −→ −→
kABkkACk
(−1)(−5) + (−7)(−3)
√ √
=
50 34
= 0.6306.
Thus the angle at vertex A is 50.91◦ . Similarly, we see that the angle at vertex B is 53.13◦ and (since the angles of a
triangle add up to 180◦ ) the angle at vertex C is 75.96◦ .
y
6
4
C
A
2
x
−6 −4 −2
−2
B
−4
2
4
6
−6
Figure 13.44
−
−
→
−→
51. Let ~r 1 be the displacement vector P Q and let ~r 2 be the displacement vector P R. Then
~r 1 = (1 + 2)~i + (3 − 2)~j + (−1 − 0)~k = 3~i + ~j − ~k ,
~r 2 = (−4 + 2)~i + (2 − 2)~j + (1 − 0)~k = −2~i + ~k ,
~i ~j ~k
~r 1 × ~r 2 =
3 1 −1 = ~i − (3 − 2)~j + 2~k = ~i − ~j + 2~k .
−2 0 1
√
√
The area of the triangle = 12 k~r 1 × ~r 2 k = 21 12 + 12 + 22 = 26 .
−→
52. (a) The displacement vector AB lies in the plane and is given by
−→
AB = (0 − 2)~i + (1 − 1)~j + (3 − 0)~k = −2~i + 3~k .
−→
Similarly, the displacement vector AC also lies in the plane,
−→
AC = (1 − 2)~i + (0 − 1)~j + (1 − 0)~k = −~i − ~j + ~k .
−→ −→
−→
−→
(b) The vector ~n = AB × AC is perpendicular to both AB and AC and is therefore perpendicular to the plane.
−→ −→
AB × AC = (−2~i + 3~k ) × (−~i − ~j + ~k ) = 3~i − ~j + 2~k .
1256
Chapter Thirteen /SOLUTIONS
(c) The normal vector to the plane is ~n = 3~i − ~j + 2~k , so the equation is of the form
3x − y + 2z = d.
Substituting, for example, x = 1, y = 0, z = 1 gives d = 5:
3x − y + 2z = 5.
−
−
→
−→
53. (a) If we let P Q in Figure 13.45 be the vector from point P to point Q and P R be the vector from P to R, then
−
−
→
P Q = −~i + 2~k
−→
P R = 2~i − ~k ,
−
−
→
−→
then the area of the parallelogram determined by P Q and P R is:
Area of
−
−→ −→
parallelogram = kP Q × P Rk =
~i ~j ~k
= k3~j k = 3.
−1 0 2
2 0 −1
Thus, the area of the triangle P QR is
Area of
triangle
!
1
=
2
Area of
parallelogram
!
=
3
= 1.5.
2
z
Q = (−1, 1, 2)
x
P = (0, 1, 0)
y
(2, 1, −1) = R
Figure 13.45
−
−→ −→
(b) Since ~n = P Q × P R is perpendicular to the plane P QR, and from above, we have ~n = 3~j , the equation of the
plane has the form 3y = C. At the point (0, 1, 0) we get 3 = C, therefore 3y = 3, i.e., y = 1.
54. Find an arbitrary point on the plane 2x + 4y − z = −1, say A = (0, 0, 1). The normal ~n to the plane at B is ~n =
−→
2~i + 4~j − ~k and P A = −2~i + ~j − 2~k . See Figure 13.46.
P
θ
~
n
B
Figure 13.46
A
SOLUTIONS to Review Problems for Chapter Thirteen
So the distance d from the point P to the plane is
−
−→
−→
d = kP Bk = kP Ak cos θ
−→
PA · ~
n
−→
−→
since P A · ~n = kP Akk~n k cos θ)
=
k~n k
(−2~i + ~j − 2~k ) · (2~i + 4~j − ~k )
p
=
22 + 42 + (−1)2
2
= √ .
21
55. The displacement from (1, 1, 1) to (1, 4, 5) is
r~1 = (1 − 1)~i + (4 − 1)~j + (5 − 1)~k = 3~j + 4~k .
The displacement from (−3, −2, 0) to (1, 4, 5) is
r~2 = (1 + 3)~i + (4 + 2)~j + (5 − 0)~k = 4~i + 6~j + 5~k .
A normal vector is
~i ~j ~k
~n = r~1 × r~2 = 0 3 4 = (15 − 24)~i − (−16)~j + (−12)~k = −9~i + 16~j − 12~k .
465
The equation of the plane is
−9x + 16y − 12z = −9 · 1 + 16 · 1 − 12 · 1 = −5
9x − 16y + 12z = 5.
We pick a point A on the plane, A = ( 95 , 0, 0) and let P = (0, 0, 0). (See Figure 13.47.) Then P~A = (5/9)~i .
P
θ
~
n
B
A
Figure 13.47
So the distance d from the point P to the plane is
−
−→
−→
d = kP Bk = kP Ak cos θ
−→
PA · ~
n
−→
−→
=
since P A · ~n = kP Akk~n k cos θ)
k~n k
( 5~i ) · (−9~i + 16~j − 12~k )
= 9 √
92 + 162 + 122
5
= √
= 0.23.
481
1257
1258
Chapter Thirteen /SOLUTIONS
56. (a) 500 km/hr in the west direction, so ~v = −500~i .
(b) While traveling at constant altitude, the plane travels 250 km westward. Thus the coordinates of the point where the
plane begins to descend are (550, 60, 4) − (250, 0, 0) = (300, 60, 4).
(c) The vector from the plane to the airport at the time it begins its descent is (200~i + 10~j ) − (300~i + 60~j + 4~k ) =
~
~
~
~i − 50~j − 4~k . Since
−100
p i − 50j − 4k . Velocity is a vector of length 200 km/hr in the direction of −100
100 ~
50 ~
4 ~
(−100)2 + (−50)2 + (−4)2 ≈ 111.9, a unit vector in the direction of descent is − 111.9
i − 111.9
j − 111.9
k.
Thus
50 ~
4 ~
100 ~
i −
j −
k = −178.7~i − 89.4~j − 7.2~k .
Velocity vector = 200 −
111.9
111.9
111.9
57. Let ~v = vx~i + vy~j + vz ~k be the vector. We will use the properties given in the problem to find vx , vy , and vz . If ~v has
magnitude 10, then k~v k = 10.
If ~v makes an angle of 45◦ with the x-axis, then its x-component, vx , is given by:
√
2
= 7.0710.
vx = ~v · ~i = k~v k cos 45◦ = 10
2
Similarly, if ~v makes a 75◦ angle with the y-axis, then its y-component, vy , is given by:
vy = ~v · ~j = k~v k cos 75◦ = 10(0.25882) = 2.5882.
We now have two components of ~v :
~v = 7.0710~i + 2.5882~j + vz ~k .
√
We only need to find vz . To do this we use the fact that ~v · ~v = k~v k = 10.
~v · ~v = 100
vx2 + vy2 + vz2 = 100
vz2 = 100 − vx2 − vy2
vz2 = ±
p
100 − vx2 − vy2
vz = ±6.580
Since the problem tells us that the ~k -component is positive, vz = +6.580. Thus
~v = 7.0710~i + 2.5882~j + 6.580~k .
−
−
→
−
−
→
−
−
→
58. (a) Suppose ~v = OP as in Figure 13.48. The ~i component of OP is the projection of OP on the x-axis:
−→
OT = v cos α~i .
−
−
→
−
−→
Similarly, the ~j and ~k components of OP are the projections of OP on the y-axis and the z-axis respectively. So:
−→
OS = v cos β~j
−
−→
OQ = v cos γ~k
−→ −→ −
−→
Since ~v = OT + OS + OQ, we have
~v = v cos α~i + v cos β~j + v cos γ~k .
(b) Since
v 2 = ~v · ~v = (v cos α~i + v cos β~j + v cos γ~k ) ·
(v cos α~i + v cos β~j + v cos γ~k )
= v 2 (cos2 α + cos2 β + cos2 γ)
so
cos2 α + cos2 β + cos2 γ = 1.
1259
SOLUTIONS to Review Problems for Chapter Thirteen
z
Q
γ
β
α
S
T
y
x
Figure 13.48
~
59. Let the x-axis point east and the y-axis point north. Denote the forces exerted by Charlie, Sam and Alice by F
F~ A (see Figure 13.49).
~
C, F S
and
y
◦
62◦ 43
F~C
F~S
φ
x
F~A
Figure 13.49
~
Since kF
Ck
= 175 newtons and the angle θ from the x-axis to F~
~
F
Similarly,
C
is 90◦ + 62◦ = 152◦ , we have
C
= 175 cos 152◦~i + 175 sin 152◦~j ≈ −154.52~i + 82.16~j .
F~
S
= 200 cos 47◦~i + 200 sin 47◦~j ≈ 136.4~i + 146.27~j .
~
Now Alice is to counterbalance Sam and Charlie, so the resultant force of the three forces F
that is,
~ C + F~ S + F
~ A = 0.
F
Thus, we have
~
F
A
~
= −F
C
− F~
S
≈ −(−154.52~i + 82.16~j ) − (136.4~i + 146.27~j )
= 18.12~i − 228.43~j
~ A k = 18.122 + (−228.43)2 ≈ 229.15 newtons.
and, kF
~ A , then
If φ is the angle from the x-axis to F
p
φ = arctan
−228.43
≈ −85.5◦ .
18.12
~
C, F S
~
and F
A
must be 0,
1260
Chapter Thirteen /SOLUTIONS
CAS Challenge Problems
60. (~a × ~b ) · ~c = 0 , (~a × ~b ) × (~a × ~c ) = ~0
Since ~c is the sum of a scalar multiple of ~a and a scalar multiple of ~b , it lies in the plane containing ~a and ~b . On
the other hand, ~a × ~b is perpendicular to this plane, so ~a × ~b is perpendicular to ~c . Therefore, (~a × ~b ) · ~c = 0. Also,
~a × ~c is also perpendicular to the plane, thus parallel to ~a × ~b , and thus (~a × ~b ) × (~a × ~c ) = ~0 .
61. The first parallelepiped has volume
|(~a × ~b ) · ~c | = |ywr − vzr + zus − xws + xvt − yut|.
The second has volume |(~a × ~b ) · (2~a − ~b + ~c )|, which also simplifies to |ywr − vzr + zus − xws + xvt − yut|.
Both parallelepipeds have base with edges ~a and ~b . The third edge of the first one is ~c and the third edge of the second
one is ~c + 2~a − ~b . Thus the top face of the second parallelepiped is obtained by shifting the top face of the first by
2~a − ~b . Since this is parallel to the base, the second parallelepiped has the same altitude as the first. Since the volume of
a parallelepiped is product of the area of its base with its height, the two parallelepipeds have the same volume.
62. (a) From the geometric definition of the dot product, we have
cos θ =
|~a · ~b |
10
= √ √ .
14 9
k~a kk~b k
Using sin2 θ = 1 − cos2 θ, we get
x + 2y + 3z = 0
2x + y + 2z = 0
x2 + y 2 + z 2 = k~a k2 k~b k2 (1 − cos2 θ) = (14)(9) 1 −
100
(14)(9)
Solving these equations we get x = −1, y = −4, z = 3 or x = 1, y = 4, and z = −3. Thus ~c = −~i − 4~j + 3~k
or ~c = ~i + 4~j − 3~k .
(b) ~a × ~b = ~i + 4~j − 3~k . This is the same as one of the answers in part (a). The conditions in part (a) ensured that
~c is perpendicular to ~a and ~b and that it has magnitude k~a kk~b k| sin θ|. The cross product is the solution that, in
addition, satisfies the right-hand rule.
63. (a) We have
−→
kABk
−→
kACk
−−→
kADk
−−→
kBCk
−−→
kBDk
−−→
kCDk
= k2~i k = 2
√
√
= k~i + 3~j k = 1 + 3 = 2
p
p
√
√
= k~i + (1/ 3)~j + 2 2/3~k k = 1 + (1/3) + (8/3) = 4 = 2
√
√
= k − ~i + 3~j k = 1 + 3 = 2
p
p
√
√
= k − ~i + (1/ 3)~j + 2 2/3~k k = 1 + (1/3) + (8/3) = 4 = 2
p
p
√
√
√
= k(1/ 3 − 3)~j + 2 2/3~k )k = (1/3 − 2 + 3) + 8/3 = 4 = 2
Thus all the points are 2 units apart.
(b) By solving the equations
x2 + y 2 + z 2 = (x − 2)2 + y 2 + z 2
√
x2 + y 2 + z 2 = (x − 1)2 + (y − 3)2 + z 2
p
√
x2 + y 2 + z 2 = (x − 1)2 + (y − 1/ 3)2 + (z − 2 2/3)2
√ √
we get P = (1, 1/ 3, 6/6).
(c) The cosine of the angle AP B is 1/3 and the angle is 109.471◦ .
−
−
→ −→
64. (a) P Q× P R is perpendicular to the plane containing P , Q, R, and therefore parallel to the normal vector a~i +b~j +c~k .
(b)
−
−→ −→
P Q × P R = (tv − sw − ty + wy + sz − vz)~i +
(−tu + rw + tx − wx − rz + uz)~j + (su − rv − sx + vx + ry − uy)~k
PROJECTS FOR CHAPTER THIRTEEN
1261
(c) After substituting z = (d − ax − by)/c, w = (d − au − bv)/c, t = (d − ar − bs)/c into the result of part (a), and
simplifying the expression, we obtain:
a(s(u − x) + vx − uy + r(−v + y))~
−
−
→ −→
i+
PQ × PR =
c
b(s(u − x) + vx − uy + r(−v + y))~
j + (s(u − x) + vx − uy + r(−v + y))~k
c
(s(u − x) + vx − uy + r(−v + y)) ~
(ai + b~j + c~k ).
=
c
−
−
→ −→
Thus P Q × P R is a scalar multiple of a~i + b~j + c~k , and hence parallel to it.
PROJECTS FOR CHAPTER THIRTEEN
1. (a) Let r = k~a k and s = k~b k, and let α, β, be the angles between ~a , ~b , and the x-axis as shown in the figure.
Suppose θ is the angle between ~a and ~b . We drew the figure with α < β and thus β − α = θ. If α > β,
then α − β = θ. In both cases we know that
Area of parallelogram = k~a kk~b k sin θ.
Using the formula
sin(β − α) = sin β cos α − cos β sin α,
and the fact that a1 = r cos α, a2 = r sin α, b1 = s cos β, and b2 = s sin β, we get
a1 b2 − a2 b1 = (r cos α)(s sin β) − (r sin α)(s cos β)
= rs(cos α sin β − sin α cos β)
= rs sin(β − α)
(from sin(β − α) = sin β cos α − cos β sin α)
~
= k~a kkb k sin(β − α)
If β > α, we have β − α = θ, so
a1 b2 − a2 b1 = k~a kk~b k sin θ = Area of parallelogram.
If β < α, we have α − β = θ, so
|a1 b2 − a2 b1 | = k~a kk~b k| sin(β − α)| = k~a kk~b k sin θ = Area of parallelogram.
(b) The sign of a1 b2 − a2 b1 is the same as the sign of β − α, so the sign of a1 b2 − a2 b1 tells us whether the
rotation from ~a to ~b is counterclockwise (then a1 b2 − a2 b1 is positive) or clockwise (then a1 b2 − a2 b1 is
negative).
(c) Part (a) tells us that
Area of the parallelogram = |a1 b2 − a2 b1 |.
The algebraic definition of the cross product is
~a × ~b = (a1 b2 − a2 b1 )~k .
The geometric definition has magnitude given by k~a × ~b k = Area of parallelogram. So the magnitude
of the algebraic definition agrees with the magnitude of the geometric definition. To check agreement of
the direction of ~a × ~b for the two definitions, we notice that (a1 b2 − a2 b1 )~k is perpendicular to ~a and
~b since ~a and ~b are in the ~i , ~j -plane. Also, part (b) says (a1 b2 − a2 b1 )~k will point up (down) when the
rotation from ~a to ~b is counterclockwise (clockwise). So the direction of the algebraic definition obeys
the right-hand rule.
1262
Chapter Thirteen /SOLUTIONS
2. (a) Since the vectors ~a 1 , . . . , ~a 4 show the square roots of the relative frequencies of the alleles, and the
relative frequencies in a population add to 1, we have
q
√
√
√
√
2
2
2
2
0.10 + 0.09 + 0.12 + 0.69 = 1
k~a 2 k =
q
√
√
√
√
2
2
2
2
k~a 3 k =
0.21 + 0.07 + 0.06 + 0.66 = 1
q
√
√
√
√
2
2
2
2
k~a 4 k =
0.22 + 0.00 + 0.21 + 0.57 = 1
√
√
√
√
√
√
√
√
0.10 · 0.21 + 0.09 · 0.07 + 0.12 · 0.06 + 0.69 · 0.66 = 0.9840
√
√
√
√
√
√
√
√
~a 3 · ~a 4 = 0.21 · 0.22 + 0.07 · 0.00 + 0.06 · 0.21 + 0.66 · 0.57 = 0.9405.
~a 2 · ~a 3 =
The distance between the English and the Bantus is given by θ where
cos θ =
0.9840
~a 2 · ~a 3
=
= 0.9840
k~a 2 kk~a 3 k
1·1
so θ = 10.3o .
The distance between the English and the Koreans is given by φ where
cos φ =
~a 3 · ~a 4
0.9405
=
= 0.9405
k~a 3 kk~a 4 k
1·1
so φ ≈ 19.9o . Hence the English are genetically closer to the Bantus than to the Koreans.
(b) Let f~ 1 be the 4-vector showing the relative frequencies of the alleles in the Eskimo population. Let f~ 2 ,
f~ 3 , f~ 4 be the corresponding vectors for the Bantu, English, and Korean populations, respectively. Let f~ 5
be the 4-vector for the relative frequencies for the half Eskimo, half Bantu population, and let ~a 5 be the
4-vector for the square roots of the relative frequencies . So
1
1
f~ 5 = f~ 1 + f~ 2 = (0.195, 0.045, 0.075, 0.685)
2√
2√
√
√
~a 5 = ( 0.195, 0.045, 0.075, 0.685).
Then
q
√
√
√
√
2
2
2
2
k~a 5 k =
0.195 + 0.045 + 0.075 + 0.685 = 1
√
√
√
√
√
√
√
√
~a 3 · ~a 5 = 0.21 · 0.195 + 0.07 · 0.045 + 0.06 · 0.075 + 0.66 · 0.685 = 0.9980.
So the distance between the English population and the half Eskimo, half Bantu population is
0.9980
~a 3 · ~a 5
= arccos
k~a 3 kk~a 5 k
1·1
= arccos 0.9980 = 3.6o .
θ = arccos
Since 3.6 < 10.3, the English are closer to the Bantu/Eskimo mix than to the Bantu alone.
(c) Suppose that x is the fraction of the population that is Eskimo, where 0 ≤ x ≤ 1. Then (1 − x) is the
fraction that is Bantu. (For example, x = 0.5, in part (b).) Let f~ 6 be the 4-vector of relative frequencies
and ~a 6 the 4-vector of square roots of relative frequencies for a population that is x Eskimo and (1 − x)
Bantu. We have
f~ 6 = xf~ 1 + (1 − x)f~ 2 = f~ 2 + x(f~ 1 − f~ 2 )
= (0.10 + 0.19x, 0.09 − 0.09x, 0.12 − 0.09x, 0.69 − 0.01x).
Then, as before
k~a 6 k = 1
PROJECTS FOR CHAPTER THIRTEEN
1263
and
~a 3 · ~a 6 =
√
√
√
√
0.21 · 0.10 + 0.19x + 0.07 · 0.09 − 0.09x
√
√
√
√
+ 0.06 · 0.12 − 0.09x + 0.66 · 0.69 − 0.01x.
Since cos θ is a decreasing function of θ for 0 ≤ θ ≤ π, to minimize the angle θ = arccos
we must maximize
f (x) =
~a 3 · ~a 6
,
k~a 3 kk~a 6 k
~a 3 · ~a 6
= ~a 3 · ~a 6 .
k~a 3 kk~a 6 k
Using a calculator or computer, we find that the maximum of this function for 0 ≤ x ≤ 1 is
f (0.4788) = 0.9980.
So the minimum distance of θ = arccos(0.9980) = 3.6o from the English occurs at a mix of about 47.88%
Eskimo and 52.12% Bantu.
3. (a) Let the forces F~ 1 from bar AB on joint A and F~ 2 from AE on A be given by
F~
F~
1
2
= a~i
= f (cos 65.38◦ ~i + sin 65.38◦ ~j ).
The sum of F~ 1 , F~ 2 , and the upward supporting force at A must be the zero vector. Hence
F~ 1 + F~ 2 + 12500~j = ~0
(a + f cos 65.38◦ )~i + (f sin 65.38◦ + 12500)~j = ~0
a + f cos 65.38◦ = 0
f sin 65.38◦ + 12500 = 0.
Solving the last two equations for f and a gives
−12500
= −13750 lb
sin 65.38◦
a = −f cos 65.38◦ = 5730 lb.
f=
There is a 5730 lb force from AB acting to the right on joint A. Since the bar is pulling the joint, AB
is under 5730 lb tension.
There is a 13750 lb force from AE acting downward on joint A. Since the bar is pushing the joint,
AE is under 13750 lb compression.
~ 2 from CD on C be given by
~ 1 from bar BC on joint C and G
(b) Let the forces G
~ 1 = b~i
G
~ 2 = g(cos 114.62◦ ~i + sin 114.62◦ ~j ).
G
~ 2 , and the upward supporting force at C must be the zero vector. Hence
~ 1, G
The sum of G
~ 1+G
~ 2 + 17500~j = ~0
G
(b + g cos 114.62◦)~i + (g sin 114.62◦ + 17500)~j = ~0
b + g cos 114.62◦ = 0
g sin 114.62◦ + 17500 = 0.
Solving the last two equations for g and b gives
−17500
= −19250 lb
sin 114.62◦
b = −g cos 114.62◦ = −8020 lb.
g=
1264
Chapter Thirteen /SOLUTIONS
There is an 8020 lb force from BC acting to the left on joint C. Since the bar is pulling the joint, BC
is under 8020 lb tension.
There is a 19250 lb force from CD acting downward on joint C. Since the bar is pushing the joint,
CD is under 19250 lb compression.
~ 2 from BD on D be given by
~ 1 from bar DE on joint D and H
(c) Let the forces H
~ 1 = c~i
H
~ 2 = h(cos 65.38◦ ~i + sin 65.38◦ ~j ).
H
~ 3 from CD on D is the opposite of force G
~ 2 of CD on C computed in part (c). The sum of
The force H
~
~
~
the forces H 1 , H 2 , H 3 and the downward force from the weight at D must be the zero vector. Hence
~ 2+H
~ 2 − 20000~j = ~0
~ 1+H
H
(c + h cos 65.38 − g cos 114.62 )~i + (h sin 63.38 − g sin 114.62◦ − 20000)~j = ~0
◦
◦
◦
c + h cos 65.38◦ − g cos 114.62◦ = 0
h sin 63.38◦ − g sin 114.62◦ − 20000 = 0.
Since we found g in part (b) we can solve the last two equations for h and c. We have
g sin 114.62◦ + 20000
= 2750 lb
sin 65.38◦
◦
c = g cos 114.62 − h cos 65.38 = 6880 lb.
h=
There is an 6880 lb force from DE acting to the right on joint D. Since the bar is pushing the joint,
DE is under 6880 lb compression.
There is a 2750 lb force from BD acting upward on joint D. Since the bar is pushing the joint, BD is
under 2750 lb compression.
(d) Let the force P~ from bar BE on joint E be given by
P~ = p(cos 114.62◦ ~i + sin 114.62◦ ~j ).
~ 1 from DE computed in
The other forces acting on E are −F~ 2 from AE computed in part (a), −H
part (c), and the downward force from the weight at E. The sum of these four forces must be the zero
vector. Hence
P~ − F~
~ 1 − 10000~j = ~0
−H
(p cos 114.62 − f cos 65.38 − c)~i + (p sin 114.62 − f sin 65.38◦ − 10000)~j = ~0
p cos 114.62◦ − f cos 65.38◦ − c = 0
◦
◦
2
◦
p sin 114.62◦ − f sin 65.38◦ − 10000 = 0.
Since we found f in part (a) and c in part (c) we can solve either of the last two equations for p. Using the
last equation, we have
f sin 65.38◦ + 10000
= −2750 lb
p=
sin 114.62◦
There is an 2750 lb force from BE acting downward right on joint E. Since the bar is pulling the joint,
BE is under 2750 lb tension.
14.1 SOLUTIONS
CHAPTER FOURTEEN
Solutions for Section 14.1
Exercises
1. Using difference quotients to approximate the partial derivatives
∆z
0−2
2
=
=−
∆x
6−1
5
2 − (−1)
3
∆z
=
= .
fy (3, 2) ≈
∆y
5−0
5
fx (3, 2) ≈
2. If h is small, then
fx (3, 2) ≈
f (3 + h, 2) − f (3, 2)
.
h
With h = 0.01, we find
fx (3, 2) ≈
f (3.01, 2) − f (3, 2)
=
0.01
3.012
(2+1)
−
32
(2+1)
0.01
= 2.00333.
With h = 0.0001, we get
f (3.0001, 2) − f (3, 2)
fx (3, 2) ≈
=
0.0001
3.00012
(2+1)
32
(2+1)
−
0.0001
= 2.0000333.
Since the difference quotient seems to be approaching 2 as h gets smaller, we conclude
fx (3, 2) ≈ 2.
To estimate fy (3, 2), we use
fy (3, 2) ≈
f (3, 2 + h) − f (3, 2)
.
h
With h = 0.01, we get
f (3, 2.01) − f (3, 2)
fy (3, 2) ≈
=
0.01
32
(2.01+1)
−
32
(2+1)
0.01
= −0.99668.
With h = 0.0001, we get
fy (3, 2) ≈
f (3, 2.0001) − f (3, 2)
=
0.0001
32
(2.0001+1)
−
32
(2+1)
0.0001
Thus, it seems that the difference quotient is approaching −1, so we estimate
fy (3, 2) ≈ −1.
3. Using first ∆x = 0.1 and ∆y = 0.1, we have the estimates:
f (1.1, 3) − f (1, 3)
0.1
0.0470 − 0.0519
=
= −0.0493,
0.1
fx (1, 3) ≈
= −0.9999667.
1265
1266
Chapter Fourteen /SOLUTIONS
and
f (1, 3.1) − f (1, 3)
0.1
0.0153 − 0.0519
=
= −0.3660.
0.1
fy (1, 3) ≈
Now, using ∆x = 0.01 and ∆y = 0.01, we have the estimates:
f (1.01, 3) − f (1, 3)
0.01
0.0514 − 0.0519
=
= −0.0501,
0.01
fx (1, 3) ≈
and
f (1, 3.01) − f (1, 3)
0.01
0.0483 − 0.0519
= −0.3629.
=
0.01
fy (1, 3) ≈
4. (a)
(b)
(c)
(d)
Dollars/Year.
Negative. You expect to pay less for an older car.
Dollars/Dollar
Positive. You expect to pay more for a car that was more expensive new.
5. ∂P/∂t: The unit is dollars per month. This is the rate at which payments change as the number of months it takes to pay
off the loan changes. The sign is negative because payments decrease as the pay-off time increases.
∂P/∂r: The unit is dollars per percentage point. This is the rate at which payments change as the interest rate
changes. The sign is positive because payments increase as the interest rate increases.
6. (a) The units of ∂c/∂x are units of concentration/distance. (For example, (gm/cm3 )/cm.) The practical interpretation of
∂c/∂x is the rate of change of concentration with distance as you move down the blood vessel at a fixed time. We
expect ∂c/∂x < 0 because the further away you get from the point of injection, the less of the drug you would expect
to find (at a fixed time).
(b) The units of ∂c/∂t are units of concentration/time. (For example, (gm/cm3 )/sec.) The practical interpretation of
∂c/∂t is the rate of change of concentration with time, as you look at a particular point in the blood vessel. We would
expect the concentration to first increase (as the drug reaches the point) and then decrease as the drug dies away.
Thus, we expect ∂c/∂t > 0 for small t and ∂c/∂t < 0 for large t.
7. (a) If you borrow $8000 at an interest rate of 1% per month and pay it off in 24 months, your monthly payments are
$376.59.
(b) The increase in your monthly payments for borrowing an extra dollar under the same terms as in (a) is about 4.7
cents.
(c) If you borrow the same amount of money for the same time period as in (a), but if the interest rate increases by 1%,
the increase in your monthly payments is about $44.83.
8. (a) We expect fp to be negative because if the price of the product increases, the sales usually decrease.
(b) If the price of the product is $8 per unit and if $12000 has been spent on advertising, sales increase by approximately
150 units if an additional $1000 is spent on advertising.
9. (a) Negative. As the price of beef goes up, we expect people to buy less beef and so the quantity of beef sold goes down.
If b increases, we expect Q to decrease.
(b) Positive. As the price of chicken goes up, we expect people to buy more beef. As c increases, we expect Q to increase.
(c) We estimate that
−213
∆Q
≈
kg/dollar.
∆b
1
If the price of beef rose by one dollar, the store would sell approximately 213 fewer kilograms of beef.
10. Moving right from P in the direction of increasing x increases f , so fx (P ) > 0.
Moving up from P in the direction of increasing y increases f , so fy (P ) > 0.
11. Moving right from Q in the direction of increasing x increases f , so fx (Q) > 0.
Moving up from Q in the direction of increasing y decreases f , so fy (Q) < 0.
14.1 SOLUTIONS
1267
12. Moving right from R in the direction of increasing x decreases f , so fx (R) < 0.
Moving up from R in the direction of increasing y decreases f , so fy (R) < 0.
13. Moving right from S in the direction of increasing x decreases f , so fx (S) < 0.
Moving up from S in the direction of increasing y increases f , so fy (S) > 0.
14. For fw (10, 25) we get
fw (10, 25) ≈
f (10 + h, 25) − f (10, 25)
.
h
Choosing h = 5 and reading values from Table 12.2 on page 672 of the text, we get
fw (10, 25) ≈
f (15, 25) − f (10, 25)
13 − 15
=
= −0.4◦ F/mph
5
5
This means that when the wind speed is 10 mph and the true temperature is 25◦ F, as the wind speed increases from
10 mph by 1 mph we feel an approximately 0.4◦ F drop in temperature. This rate is negative because the temperature you
feel drops as the wind speed increases.
15. Using a difference quotient with h = 5, we get
fT (5, 20) ≈
f (5, 20 + 5) − f (5, 20)
19 − 13
=
= 1.2◦ F/◦ F.
5
5
This means that when the wind speed is 5 mph and the true temperature is 20◦ F, the apparent temperature increases by
approximately 1.2◦ F for every increase of 1◦ F in the true temperature. This rate is positive because the true temperature
you feel increases as true temperature increases.
16. Since the average rate of change of the temperature adjusted for wind-chill is about −0.8 (drops by 0.8◦ F), with every 1
mph increase in wind speed from 5 mph to 10 mph, when the true temperature stays constant at 20◦ F, we know that
fw (5, 20) ≈ −0.8.
Problems
17. The values of z increase as we move in the direction of increasing x-values, so fx is positive. The values of z decrease as
we move in the direction of increasing y-values, so fy is negative. We see in the contour diagram that f (2, 1) = 10. We
estimate the partial derivatives:
14 − 10
∆z
=
= 2,
fx (2, 1) ≈
∆x
4−2
6 − 10
∆z
=
= −4.
fy (2, 1) ≈
∆y
2−1
18. From the contour diagram, approximate values of f at nearby x-values are f (3, 5) = 10, f (6.3, 5) = 8, f (0.4, 5) = 12.
Difference quotient approximations are
fx (3, 5) ≈
f (6.3) − f (3)
= −0.61
6.3 − 3
fx (3, 5) ≈
f (0.4) − f (3)
= −0.77.
0.4 − 3
Another reasonable approximation is obtained by averaging the two difference quotients:
fx (3, 5) ≈ Average =
−0.61 − 0.77
= −0.7.
2
1268
Chapter Fourteen /SOLUTIONS
19. The partial derivative, ∂Q/∂b is the rate of change of the quantity of beef purchased with respect to the price of beef,
when the price of chicken stays constant. If the price of beef increases and the price of chicken stays the same, we expect
consumers to buy less beef and more chicken. Thus when b increases, we expect Q to decrease, so ∂Q/∂b < 0.
On the other hand, ∂Q/∂c is the rate of change of the quantity of beef purchased with respect to the price of chicken,
when the price of beef stays constant. An increase in the price of chicken is likely to cause consumers to buy less chicken
and more beef. Thus when c increases, we expect Q to increase, so ∂Q/∂c > 0.
∂q1
< 0. Similarly, an
20. (a) An increase in the price of a new car will decrease the number of cars bought annually. Thus
∂x
∂q2
< 0.
increase in the price of gasoline will decrease the amount of gas sold, implying
∂y
(b) Since the demands for a car and gas complement each other, an increase in the price of gasoline will decrease the
∂q1
∂q2
total number of cars bought. Thus
< 0. Similarly, we may expect
< 0.
∂y
∂x
21. We have
90 − 93
∆P
=
= −1.5 percent/month.
ft (18, 6) ≈
∆t
20 − 18
Eighteen months after rats are exposed to a formaldehyde concentration of 6 ppm, the percent of rats surviving is decreasing at a rate of about 1.5 per month. In other words, during the eighteenth month, an additional 1.5% of the rats
die.
We have
∆P
82 − 93
fc (18, 6) ≈
=
= −1.22 percent/ppm.
∆c
15 − 6
If the original concentration increases by 1 ppm, the percent surviving after 18 months decreases by about 1.22.
22. The fact that fx (P ) > 0 tells us that the values of the function on the contours increase as we move to the right in
Figure 14.3 past the point P . Thus, the values of the function on the contours
(a) Decrease as we move upward past P . Thus fy (P ) < 0.
(b) Decrease as we move upward past Q (since Q and P are on the same contour line.) Thus fy (Q) < 0.
(c) Decrease as we move to the right past Q. Thus fx (Q) < 0
23. (a) The partial derivative gx (x, y) is zero at points where a contour has a horizontal tangent line. See Figure 14.1.
(b) The partial derivative gy (x, y) is zero at points where a contour has a vertical tangent line. See Figure 14.2.
y
y
10
10
60
60 50
50
50
50
40
30
40
60
60
30
20
20
10
10
x
x
10
Figure 14.1
10
Figure 14.2
(i) Near A, the value of z increases as x increases, so fx (A) > 0.
(ii) Near A, the value of z decreases as y increases, so fy (A) < 0.
(b) fx (P ) changes from positive to negative as P moves from A to B along a straight line, because after P crosses the
y-axis, z decreases as x increases near P .
fy (P ) does not change sign as P moves from A to B along a straight line; it is negative along AB.
24. (a)
25. (a) For points near the point (0, 5, 3), moving in the positive x direction, the surface is sloping down and the function is
decreasing. Thus, fx (0, 5) < 0.
(b) Moving in the positive y direction near this point the surface slopes up as the function increases, so fy (0, 5) > 0.
14.1 SOLUTIONS
1269
26. Locating the points (3, 2, f (3, 2)) and (1, 2, f (1, 2)) on the graph in Figure 14.3 we see that fx (1, 2) and fx (3, 2) are
both negative with fx (1, 2) < fx (3, 2). Similarly, fy (3, 2) and fy (1, 2) are both positive with fy (3, 2) < fy (1, 2).
Therefore,
fx (1, 2) < fx (3, 2) < 0 < fy (3, 2) < fy (1, 2).
z
✲
(1, 2, f (1, 2))
(3, 2, f (3, 2))
x
✲
y
Figure 14.3
27. (a) Estimate ∂P/∂r and ∂P/∂L by using difference quotients and reading values of P from the graph:
P (16, 4000) − P (8, 4000)
∂P
(8, 4000) ≈
∂r
16 − 8
100 − 80
=
= 2.5,
8
and
P (8, 5000) − P (8, 4000)
∂P
≈
∂L
5000 − 4000
100 − 80
=
= 0.02.
1000
Pr (8, 4000) ≈ 2.5 means that at an interest rate of 8% and a loan amount of $4000 the monthly payment increases
by approximately $2.50 for every one percent increase of the interest rate. PL (8, 4000) ≈ 0.02 means the monthly
payment increases by approximately $0.02 for every $1 increase in the loan amount at an 8% rate and a loan amount
of $4000.
(b) Using difference quotients and reading from the graph
P (14, 6000) − P (8, 6000)
∂P
(8, 6000) ≈
∂r
14 − 8
140 − 120
=
= 3.33,
6
and
P (8, 7000) − P (8, 6000)
∂P
(8, 6000) ≈
∂L
7000 − 6000
140 − 120
=
= 0.02.
1000
Again, we see that the monthly payment increases with increases in interest rate and loan amount. The interest rate
is r = 8% as in part (a), but here the loan amount is L = $6000. Since PL (8, 4000) ≈ PL (8, 6000), the increase
in monthly payment per unit increase in loan amount remains the same as in part a). However, in this case, the effect
of the interest rate is different: here the monthly payment increases by approximately $3.33 for every one percent
increase of interest rate at r = 8% and loan amount of $6000.
(c)
P (19, 7000) − P (13, 7000)
∂P
(13, 7000) ≈
∂r
19 − 13
180 − 160
= 3.33,
=
6
1270
Chapter Fourteen /SOLUTIONS
and
P (13, 8000) − P (13, 7000)
∂P
(13, 7000) ≈
∂L
8000 − 7000
180 − 160
=
= 0.02.
1000
The figures show that the rates of change of the monthly payment with respect to the interest rate and loan amount
are roughly the same for (r, L) = (8, 6000) and (r, L) = (13, 7000).
28. The sign of ∂f /∂P1 tells you whether f (the number of people who ride the bus) increases or decreases when P1 is
increased. Since P1 is the price of taking the bus, as it increases, f should decrease. This is because fewer people will be
∂f
tells you
willing to pay the higher price, and more people will choose to ride the train. On the other hand, the sign of
∂P2
the change in f as P2 increases. Since P2 is the cost of riding the train, as it increases, f should increase. This is because
fewer people will be willing to pay the higher fares for the train, and more people will choose to ride the bus.
∂f
∂f
< 0 and
> 0.
Therefore,
∂P1
∂P2
29. (i) Statement (i) indicates that, as v increases at a constant temperature, W will decrease. Therefore, fv (T, v) < 0, and
so statement (i) matches formula (c).
(ii) Statement (ii) indicates that, as T increases at a constant riding speed, W also increases. Therefore, fT (T, v) > 0,
and so statement (ii) matches formula (a).
We now see that formula (c) does not match either given statement. In words, the statement “f (0, v) ≤ 0” is saying
that, if the air temperature is held constant at 0◦ F then, no matter what speed you are biking at, you will feel at least as
cold as 0◦ F.
30. (a) Estimating T (x, t) from the figure in the text at x = 15, t = 20 gives
∂T
∂x
∂T
∂t
(15,20)
(15,20)
≈
T (23, 20) − T (15, 20)
20 − 23
3
=
= − ◦ C per m,
23 − 15
8
8
≈
T (15, 25) − T (15, 20)
25 − 23
2
=
= ◦ C per min.
25 − 20
5
5
At 15 m from heater at time t = 20 min, the room temperature decreases by approximately 3/8◦ C per meter and
increases by approximately 2/5◦ C per minute.
(b) We have the estimates,
∂T
∂x
∂T
∂t
(5,12)
(5,12)
≈
T (7, 12) − T (5, 12)
25 − 27
=
= −1 ◦ C per m,
7−5
2
≈
T (5, 40) − T (5, 12)
30 − 27
3 ◦
=
=
C per min.
40 − 12
28
28
At x = 5, t = 12 the temperature decreases by approximately 1◦ C per meter and increases by approximately 3/28◦ C
per minute.
31. The quantity HT (10, 0.1) is approximated by a difference quotient. The first partial derivative with respect to T is approximated by
H(10 + ∆T, 0.1) − H(10, 0.1)
for small ∆T .
HT (10, 0.1) ≈
∆T
We are free to choose ∆T . If we take H(10, 0.1) = 110 and H(20, 0.1) = 100, we get the approximation
HT (10, 0.1) ≈
H(20, 0.1) − H(10, 0.1)
95 − 120
=
= −2.5.
10
10
(Note that you may get a different answer if you read different values from the graph.) The geometric meaning of
the partial derivative HT (10, 0.01) that we just approximated is the slope of the curve shown in the figure in the text
corresponding to w = 0.1 at the point where T = 10. In practical terms, we have found that for fog at 10◦ C containing
0.1 g water per m3 of fog, a 1◦ C increase in temperature will reduce the heat requirement for dissipating the fog by about
1 calories per cubic meter of fog.
14.1 SOLUTIONS
1271
32. Reading values of H from the graph gives Table 14.1. In order to compute HT (T, w) at T = 30, it is useful to have
values of H(T, w) for T = 40◦ C. The column corresponding to w = 0.4 is not used in this problem.
Table 14.1 Estimated values of
H(T, w) (in calories/meter3 )
Table 14.2 Estimated values of
HT (T, w) (in calories/meter3 /◦ C)
w (gm/m3 )
T (◦ C)
0.1
0.2
0.3
0.4
10
110
240
330
450
20
100
180
260
350
30
70
150
220
300
40
65
140
200
270
w (gm/m3 )
10
T (◦ C)
20
30
0.1
0.2
0.3
−1.0
−6.0
−7.0
−0.5
−1.0
−2.0
−3.0
−3.0
−4.0
The estimates for HT (T, w) in Table 14.2 are now computed using the formula
HT (T, w) ≈
H(T + 10, w) − H(T, w)
.
10
33. Values of H from the graph are given in Table 14.3. In order to compute Hw (T, w) for w = 0.3, it is useful to have the
column corresponding to w = 0.4. The row corresponding to T = 40 is not used in this problem. The partial derivative
Hw (T, w) can be approximated by
Hw (10, 0.1) ≈
H(10, 0.1 + h) − H(10, 0.1)
h
for small h.
We choose h = 0.1 because we can read off a value for H(10, 0.2) from the graph. If we take H(10, 0.2) = 240, we get
the approximation
H(10, 0.2) − H(10, 0.1)
240 − 110
=
= 1300.
Hw (10, 0.1) ≈
0.1
0.1
In practical terms, we have found that for fog at 10◦ C containing 0.1 g/m3 of water, an increase in the water content of
the fog will increase the heat requirement for dissipating the fog at the rate given by Hw (10, 0.1). Specifically, a 1 g/m3
increase in the water content will increase the heat required to dissipate the fog by about 1300 calories per cubic meter of
fog.
Wetter fog is harder to dissipate. Other values of Hw (T, w) in Table 14.4 are computed using the formula
Hw (T, w) ≈
H(T, w + 0.1) − H(T, w)
,
0.1
where we have used Table 14.3 to evaluate H.
Table 14.3 Estimated values of
H(T, w) (in calories/meter3 )
Table 14.4
w (gm/m3 )
T
(◦ C)
0.1
0.2
0.3
0.4
10
110
240
330
450
20
100
180
260
350
30
70
150
220
300
40
65
140
200
270
Table of values of Hw (T, w) (in cal/gm)
w (gm/m3 )
T (◦ C)
0.1
0.2
0.3
10
1300
900
1200
20
800
800
900
30
800
700
800
∂p
= fc (c, s) = rate of change in blood pressure as cardiac output increases while systemic vascular resistance
∂c
remains constant.
(b) Suppose that p = kcs. Note that c (cardiac output), a volume, s (SVR), a resistance, and p, a pressure, must all be
positive. Thus k must be positive, and our level curves should be confined to the first quadrant. Several level curves
are shown in Figure 14.4. Each level curve represents a different blood pressure level. Each point on a given curve is
a combination of cardiac output and SVR that results in the blood pressure associated with that curve.
34. (a)
1272
Chapter Fourteen /SOLUTIONS
s (SVR)
s (SVR)
A
After
Nitroglycerine,
SVR drops
✠
3
p = 9k
2
p = 1k
1
❄
p = 4k
✲
B
✻
c (vol)
1
2
c (vol)
3
After Dopamine,
cardiac output increases
Figure 14.4
Figure 14.5
(c) Point B in Figure 14.5 shows that if the two doses are correct, the changes in pressure will cancel. The patient’s
cardiac output will have increased and his SVR will have decreased, but his blood pressure won’t have changed.
(d) At point F in Figure 14.6, the patient’s blood pressure is normalized, but his/her cardiac output has dropped and his
SVR is up.
s (SVR)
F
resident
gives drug
increasing
SVR
✲
✻
during heart attack
cardiac output drops
E
✛
c2
✠
D
c1
c (vol)
Figure 14.6
Note: c1 and c2 are the cardiac outputs before and after the heart attack, respectively.
35. (a) Since fx > 0, the values on the contours increase as you move to the right. Since fy > 0, the values on the contours
increase as you move upward. See Figure 14.7.
y
0
x
−
1
0
1
x
−
1
2
−
2
3
−
3
4
−
4
5
−
5
y
1
2
−
2
3
−
3
4
−
4
5
−
5
Figure 14.7: fx > 0 and fy > 0
Figure 14.8: fx > 0 and fy < 0
(b) Since fx > 0, the values on the contours increase as you move to the right. Since fy < 0, the values on the contours
decrease as you move upward. See Figure 14.8.
1273
14.1 SOLUTIONS
(c) Since fx < 0, the values on the contours decrease as you move to the right. Since fy > 0, the values on the contours
increase as you move upward. See Figure 14.9.
y
5
y
5
4
−
4
3
−
3
2
−
−
0
2
1
−
1
x
−
4
5
−
3
4
−
2
3
−
1
2
0
−
1
x
5
Figure 14.9: fx < 0 and fy > 0
Figure 14.10: fx < 0 and fy < 0
(d) Since fx < 0, the values on the contours decrease as you move to the right. Since fy < 0, the values on the contours
decrease as you move upward. See Figure 14.10.
Strengthen Your Understanding
36. The units of ∂f /∂x and ∂f /∂y are the same if x and y have the same units. They are different if x and y have different
units.
37. If a formula for f (x, y) does not contain y explicitly, then all difference quotients are zero:
f (x, y + h) − f (x, y)
= 0 for h 6= 0.
h
Hence the partial derivative with respect to y is defined and fy (x, y) = 0.
38. Since fx < 0, the slope of f in the x-direction is negative. Since fy > 0, the slope of f in the y-direction is positive. A
possible table of a linear function satisfying these conditions is:
y
x
0
5
10
0
50
52
54
10
40
42
44
20
30
32
34
39. One type of function that would satisfy the given condition would be one with fx = 4 and fy = −1. A function with
these partial derivatives is f (x, y) = 4x − y.
40. True. This is the instantaneous rate of change of f in the x-direction at the point (10, 20).
41. False. The graph of f is a paraboloid or bowl-shape, with lowest point at the origin and opening in the positive z-direction.
At (1, 1) the function is increasing in the y-direction, so fy (1, 1) > 0.
42. True. Slicing the graph of f in the xz-plane yields a semicircle, and at the point x = 0 this semicircle has a horizontal
tangent line. Thus fx (0, 0) = 0. A similar argument shows that fy (0, 0) = 0.
43. False. The units of ∂P /∂V are the units of P divided by the units of V , which gives (grams/cm3 )/cm3 , or grams/cm6 .
44. True. The property fx (a, b) > 0 means that f increases in the positive x-direction near (a, b), so f must decrease in the
negative x-direction near (a, b).
45. True. When we fix s = k, then g(r, k) = r 2 + k, whose graph is a parabola. Since this parabola is concave up, we know
its slope increases as r increases.
1274
Chapter Fourteen /SOLUTIONS
46. True. Using difference quotients to approximate gu (1, 1), we find that
(2.01)1.01 − 21
g(1 + 0.01, 1) − g(1, 1)
=
≈ 2.40816
0.01
0.01
g(1 + 0.001, 1) − g(1, 1)
(2.001)1.001 − 21
=
≈ 2.38847
0.001
0.001
(2.0001)1.0001 − 21
g(1 + 0.0001, 1) − g(1, 1)
=
≈ 2.38651.
0.0001
0.0001
47. False. Increasing m, the number of miles on the engine, for fixed d generally lowers the price of the car, so ∂P /∂m < 0.
A higher original cost, d, with fixed m usually gives an increased sales price, so ∂P /∂d > 0.
48. True. A function with constant fx (x, y) and fy (x, y) has constant x-slope and constant y-slope, and therefore has a graph
which is a plane.
49. False. Having zero for the x and y partial derivatives at a single point means only that cross-sections through the graph
have horizontal tangents at that point. For example, f (x, y) = x2 + y 2 has fx (0, 0) = fy (0, 0) = 0, but f is not constant.
Solutions for Section 14.2
Exercises
1. (a) Make a difference quotient using the two points (3, 2) and (3, 2.01) that have the same x-coordinate 3 but whose
y-coordinates differ by 0.01. We have
fy (3, 2) ≈
f (3, 2.01) − f (3, 2)
28.0701 − 28
=
= 7.01.
2.01 − 2
0.01
(b) Differentiating gives fy (x, y) = x + 2y, so fy (3, 2) = 3 + 2 · 2 = 7.
2. fx (x, y) = 10xy 3 + 8y 2 − 6x and fy (x, y) = 15x2 y 2 + 16xy.
3. We have
fx (x, y) = 3x2 + 6xy
and
so
fx (1, 2) = 15
and
fy (x, y) = 3x2 − 4y,
fy (1, 2) = −5.
∂
(3x5 y 7 − 32x4 y 3 + 5xy) = 21x5 y 6 − 96x4 y 2 + 5x
∂y
∂z
∂ 2
5.
=
(x + x − y)7 = 7(x2 + x − y)6 (2x + 1) = (14x + 7)(x2 + x − y)6 .
∂x
∂x
∂z
∂ 2
=
(x + x − y)7 = −7(x2 + x − y)6 .
∂y
∂y
6. Thinking of A, α, β as constants,
4.
fx = (α + β)Aα xα+β−1 y 1−α−β
fy = (1 − α − β)Aα xα+β y −α−β .
7. The differentiation is easier if we rewrite f using the rules of logarithms:
f (x, y) = 0.6 ln x + 0.4 ln y.
Then
0.6
x
0.4
.
fy =
y
fx =
14.2 SOLUTIONS
4
4
2
7
1
1
15x abc
1
15abcx
15a bcx − 1
(−2) 3 +
=− 3 +
=
2ay
x
y
ax y
y
ax3 y
4
9. zx = 2xy + 10x y
8. zx =
10. Vr = 23 πrh
2πr
∂
2πr
11.
=− 2
∂T
T
T
1 −1/2
∂ √
a
12.
(a x) = a · x
= √
∂x
2
2 x
√
√
√
√
√
√
xy
1
xy
∂
13.
(xe xy ) = e xy + xe xy · (xy)−1/2 y = e xy (1 + √ ) = e xy (1 +
)
∂x
2
2 xy
2
14. We have
∂ sin(x+ct)
e
= esin(x+ct) · cos(x + ct) · c.
∂t
15. Fm = g
2v
16. av =
r
1
∂A
= (a + b)
17.
∂h
2
1
1
∂
mv 2 = v 2
18.
∂m 2
2
1
2B
∂
19.
B2 =
∂B u0
u0
2π
∂ 2πr
=
20.
∂r
v
v
2mv
21. Fv =
r
∂
(v0 + at) = 1 + 0 = 1
22.
∂v0
Gm1
∂F
=
23.
∂m2
r2
24. With a constant, we have
∂
∂x
1 −x2 /a2
e
a
=
1 −x2 /a2
2x
e
− 2
a
a
=−
2x −x2 /a2
e
.
a3
25. With x constant, we use the product and chain rules
∂
∂a
1 −x2 /a2
e
a
2
2
2
2
1
1
= − 2 e−x /a + e−x /a
a
a
2x2
a3
2
=−
e−x /a
a4
2
a2 − 2x2 .
∂ xy
∂exy
∂f
=
[e (ln y)] =
ln y = yexy (ln y).
∂x
∂x
∂x
∂
1
1
27.
(v0 t + at2 ) = v0 + · 2at = v0 + at
∂t
2
2
28. Since we take the partial derivative with respect to θ, we think of π and φ as constant. Thus,
26.
1
∂
(cos (πθφ) + ln(θ2 + φ)) = cos(πθφ) · πφ + 2
· 2θ.
∂θ
θ +φ
G
2πr 3/2
1
πr 3/2
√
√
= 2πr 3/2 (− )(GM )−3/2 (G) = −πr 3/2 ·
=− √
2
GM
GM GM
M GM
a
a
30. fa = e sin(a + b) + e cos(a + b)
29.
∂
∂M
31. zx = cos(5x3 y − 3xy 2 ) · (15x2 y − 3y 2 ) = (15x2 y − 3y 2 ) cos(5x3 y − 3xy 2 )
∂
∂
∂
ln(yexy ) = (yexy )−1
(yexy ) = (yexy )−1 · y (exy ) = (yexy )−1 · y · y · exy = y
32. gx (x, y) =
∂x
∂x
∂x
1275
1276
Chapter Fourteen /SOLUTIONS
3
∂ √
∂F
∂
3K
=
=
3 LK = 3
(LK)1/2 = √
33.
∂L
∂L
∂L
2
2 LK
8
∂V
4
∂V
= πrh and
= πr 2 .
34.
∂r
3
∂h
3
35. uE = 12 ǫ0 · 2E + 0 = ǫ0 E
36.
∂
∂x
√
2
2
1
e−(x−µ) /(2σ )
2πσ
= √
2
2
1
e−(x−µ) /(2σ ) ·
2πσ
∂Q
= cγ(a1 K b1 + a2 Lb2 )γ−1 · a1 b1 K b1 −1
∂K
38. zx = 7x6 + yxy−1 , and zy = 2y ln 2 + xy ln x
r
K
.
L
−2(x − µ)
2σ 2
−(x − µ) −(x−µ)2 /(2σ 2 )
= √
e
.
2πσ 3
37.
39. We regard x as constant and differentiate with respect to y using the product rule:
∂z
= 2ex+2y sin y + ex+2y cos y
∂y
Substituting x = 1, y = 0.5 gives
∂z
∂y
= 2e2 sin(0.5) + e2 cos(0.5) = 13.6.
(1,0.5)
40.
1
∂f
= ln (y cos x) + x
(−y sin x) = ln(y cos x) − x tan x
∂x
y cos x
∂f
∂x
(π/3,1)
= ln (cos π/3) − π/3 tan π/3 ≈ −2.51
Problems
41. (a) From contour diagram,
f (2.3, 1) − f (2, 1)
2.3 − 2
6−5
= 3.3,
=
0.3
f (2, 1.4) − f (2, 1)
fy (2, 1) ≈
1.4 − 1
6−5
= 2.5.
=
0.4
(b) A table of values for f is given in Table 14.5.
fx (2, 1) ≈
Table 14.5
y
x
0.9
1.0
1.1
1.9
4.42
4.61
4.82
2.0
4.81
5.00
5.21
2.1
5.22
5.41
5.62
From Table 14.5 we estimate fx (2, 1) and fy (2, 1) using difference quotients:
5.41 − 5.00
= 4.1
2.1 − 2
5.21 − 5.00
= 2.1.
fy (2, 1) ≈
1.1 − 1
fx (2, 1) ≈
We obtain better estimates by finer data in the table.
(c) fx (x, y) = 2x, fy (x, y) = 2y. So the true values are fx (2, 1) = 4, fy (2, 1) = 2.
1277
14.2 SOLUTIONS
42. (a) The difference quotient for evaluating fw (2, 2) is
2.01
f (2 + 0.01, 2) − f (2, 2)
e(2.01) ln 2 − e2 ln 2
eln(2
=
=
h
0.01
2(2.01) − 22
≈ 2.78
=
0.01
)
2
− eln(2
0.01
fw (2, 2) ≈
)
The difference quotient for evaluating fz (2, 2) is
f (2, 2 + 0.01) − f (2, 2)
h
e2 ln(2.01) − e2 ln 2
(2.01)2 − 22
=
=
= 4.01
0.01
0.01
fz (2, 2) ≈
(b) Using the derivative formulas we get
∂f
= ln z · ew ln z = z w · ln z
∂w
∂f
w
fz =
= ew ln z ·
= w · z w−1
∂z
z
fw =
so
fw (2, 2) = 22 · ln 2 ≈ 2.773
fz (2, 2) = 2 · 22−1 = 4.
43. (a) Since ∂z/∂x = 6x − ay if the surface is sloping up at (1, 2) we have 6 − 2a > 0, so a < 3.
(b) Since ∂z/∂y = 8y − ax, at (1, 2) we have ∂z/∂y = 16 − a > 0 if a < 3. Thus the surface is sloping upward in the
y-direction at (1, 2).
44. (a) To calculate ∂B/∂t, we hold P constant and differentiate B with respect to t:
∂
∂B
=
(P ert ) = P rert .
∂t
∂t
In financial terms, ∂B/∂t represents the change in the amount of money in the bank as one unit of time passes by.
(b) To calculate ∂B/∂P , we hold t constant and differentiate B with respect to P :
∂
∂B
=
(P ert ) = ert .
∂P
∂P
In financial terms, ∂B/∂P represents the change in the amount of money in the bank at time t as you increase the
amount of money that was initially deposited by one unit.
G
∂g
2Gm
∂g
= 2 and
=− 3
45. (a)
∂m
r
∂r
r
∂g
G
(b) For constant r, the graph of g against m is a straight line through the origin with slope
= 2 . Thus g increases
∂m
r
as m increases, while r is constant. See Figure 14.11.
g
g
m
Figure 14.11
r
Figure 14.12
1278
Chapter Fourteen /SOLUTIONS
For constant m, the graph of g against r has the shape shown in Figure 14.12, (the same shape as the graph of
∂g
2Gm
y = x12 ). The slope of the graph in Figure 14.12 is
= − 3 . So as r increases, g decreases, since the slope is
∂r
r
negative.
46. Substituting w = 65 and h = 160, we have
f (65, 160) = 0.01(650.25 )(1600.75 ) = 1.277 m2 .
This tells us that a person who weighs 65 kg and is 160 cm tall has a surface area of about 1.277 m2 . Since
fw (w, h) = 0.01(0.25w−0.75 )h0.75 m2 /kg,
we have fw (65, 160) = 0.005 m2 /kg. Thus, an increase of 1 kg in weight increases surface area by about 0.005 m2 .
Since
fh (w, h) = 0.01w0.25 (0.75h−0.25 ) m2 /cm,
we have fh (65, 160) = 0.006 m2 /cm. Thus, an increase of 1 cm in height increases surface area by about 0.006 m2 .
∂E
= c2
47. (a)
∂m
1
!
− 1 . We expect this to be positive because energy increases with mass.
1 − v 2 /c2
∂E
1
2v
mv
(b)
= mc2 · −
(1 − v 2 /c2 )−3/2 − 2 =
. We expect this to be positive because energy
∂v
2
c
(1 − v 2 /c2 )3/2
increases with velocity.
p
48. The function in h(x, t) tells us the depth of the water in cm at position x meters and time t seconds. Thus, hx (2, 5) is in
cm per meter and ht (2, 5) is in cm per second. So
hx (x, t) = −0.5 sin(0.5x − t)
hx (2, 5) = −0.5 sin(0.5(2) − 5) = −0.378 cm/meter.
ht (x, t) = sin(0.5x − t)
ht (2, 5) = sin(0.5(2) − 5) = 0.757 cm/second.
At position x = 2 and time t = 5, the value of hx (2, 5) is the rate of change of depth of the water with distance—the
slope of the surface of the wave. The value of ht (2, 5) is the vertical velocity of the surface of the water of the x = 2 at
time t = 5.
49. (a) Substituting t = 0 and t = 1 into the formula for H gives:
H(x, 0) = 100 sin(πx)
H(x, 1) = 100e−0.1 sin(πx) = 90.5 sin(πx).
The graphs of H(x, 0) and H(x, 1) are shown below.
temp (◦ C)
100
✛
H(x, 0) =
100 sin(πx)
■
H(x, 1) =
100e−0.1 sin(πx)
= 90.5 sin(πx)
0.5
1
x (meters)
(b) To calculate Hx (x, t), we hold t constant and differentiate with respect to x:
∂
∂
H(x, t) =
(100e−0.1t sin(πx)) = 100πe−0.1t cos(πx)
∂x
∂x
Hx (0.2, t) = 100πe−0.1t cos(0.2π) = 254.2e−0.1t ◦ C/meter
Hx (x, t) =
Hx (0.8, t) = 100πe−0.1t cos(0.8π) = −254.2e−0.1t ◦ C/meter.
14.2 SOLUTIONS
1279
The practical interpretation of these partial derivatives is the rate of change in temperature at x = 0.2 and x = 0.8 as
we increase the distance from the end x = 0. Notice that e−0.1t is positive for all t. Given the formula for H(x, t),
we see that the closer the position to the center of the rod, the hotter the temperature. The partial derivative Hx (0.2, t)
has a positive sign because, at x = 0.2 as we increase x, we get closer to the center of the rod which is hottest. The
partial derivative Hx (0.8, t) has a negative sign because, at x = 0.8 as we increase x, we get further away from the
center of the rod which is hottest.
(c) To calculate Ht (x, t), we hold x constant and differentiate with respect to t:
∂
∂H
=
(100e−0.1t sin(πx)) = −10e−0.1t sin(πx) ◦ C/second.
∂t
∂t
For all t, and for 0 < x < 1 (that is, for all t and all x inside the rod), the partial derivative Ht (x, t) is negative. In
terms of heat, Ht (x, t) represents the rate at which the temperature of the rod is changing as time passes at position
x and time t. Thus, the temperature inside the rod is always decreasing.
50. We compute the partial derivatives:
∂Q
∂Q
= bαK α−1 L1−α so K
= bαK α L1−α
∂K
∂K
∂Q
∂Q
= b(1 − α)K α L−α so L
= b(1 − α)K α L1−α
∂L
∂L
Adding these two results, we have:
K
∂Q
∂Q
+L
= b(α + 1 − α)K α L1−α = Q.
∂K
∂L
51. Since fx (x, y) = 4x3 y 2 − 3y 4 , we could have
f (x, y) = x4 y 2 − 3xy 4 .
In that case,
fy (x, y) =
∂ 4 2
(x y − 3xy 4 ) = 2x4 y − 12xy 3
∂y
as expected. More generally, we could have f (x, y) = x4 y 2 − 3xy 4 + C, where C is any constant.
Strengthen Your Understanding
52. The variable with respect to which the partial derivative is taken needs to be specified. The function f (x, y) = x2 y 3 has
two partial derivatives, fx = 2xy 3 and fy = 3y 2 x2 .
53. The partial derivative fx (0, 0) is defined as a limit of difference quotients:
fx (0, 0) = lim
h→0
f (h, 0) − f (0, 0)
.
h
Evaluating a difference quotient for a specific value of h, such as h = 0.01, does not tell us the value or the sign of the
partial derivative.
For example, if f (x, y) = x2 + y 2 , then f (0.01, 0) − f (0, 0) = 0.012 > 0, but fx (x, y) = 2x, so fx (0, 0) = 0.
54. A linear function with the required partial derivatives is g(x, y) = 2x + 3y. Adding to g a nonlinear function with both
partial derivatives equal to zero at the origin solves the problem. For example, f (x, y) = 2x + 3y + x2 has fx (0, 0) = 2
and fy (0, 0) = 3.
55. If f (x, y) = xy and g(x, y) = xy + y 2 , then
fx = gx = y
fy = x
gy = x + 2y 6= fy .
56. If f (x, y) is defined by a formula that does not contain x, then fx = 0 everywhere. For example, if f (x, y) = y 2 + 1,
then fx = 0 and f is not constant.
1280
Chapter Fourteen /SOLUTIONS
57. True. Any function of the form f (x, y) = xy + c, where c is a constant satisfies fx = y and fy = x.
58. True. Substituting ∂z/∂u = cos v and ∂z/∂v = −u sin v into the equation yields
cos v
sin v ∂z
∂z
−
= cos2 v + sin2 v = 1.
∂u
u ∂v
59. True. Since g is a function of x only, it can be treated like a constant when taking the y partial derivative.
60. False. Differentiating with respect to s gives
ks = rses + res .
At the point (−1, 2) we have ks (−1, 2) = −3e2 , which is negative, so k is decreasing in the s-direction.
61. False. In order to have fx (x, y) = y 2 , we would need f to have the form f (x, y) = xy 2 + h(y), where h(y) is a function
of y only. Then the y-partial derivative of f would be fy (x, y) = 2xy + h′ (y), where h′ (y) is the derivative of h(y). No
matter what function h(y) is, it will contain no x’s, so it is impossible for fy to equal x2 .
62. False. The function f could be any function of x only. For example, f (x, y) = x2 has fy (x, y) = 0.
63. False. Treating y constant and taking the derivative with respect to x yields fx (x, y) = yeg(x) g ′ (x), using the chain rule.
64. False. For example, consider f (x, y) = xy. This function is symmetric, since f (y, x) = yx = xy = f (x, y), but
fx (x, y) = y and fy (x, y) = x are not equal.
65. Calculate the partial derivatives and check. The answer is (d), and
∂ 0.4 0.6
(x y ) = x0.4x−0.6 y 0.6 = 0.4x0.4 y 0.6
∂x
∂
y (x0.4 y 0.6 ) = yx0.4 0.6y −0.4 = 0.6x0.4 y 0.6 .
∂y
x
Thus
xfx + yfy = 0.4x0.4 y 0.6 + 0.6x0.4 y 0.6 = x0.4 y 0.6 = f.
Solutions for Section 14.3
Exercises
1. The partial derivatives are
zx = ex/y
and zy =
−x x/y
e
+ ex/y ,
y
so
The tangent plane to z = yex/y
zx (1, 1) = e and zy (1, 1) = −e1 + e1 = 0.
at (x, y) = (1, 1) has equation
z = z(1, 1) + zx (1, 1)(x − 1) + zy (1, 1)(y − 1)
= e + e(x − 1) + 0(y − 1)
= ex.
2. Since
3π
3π
3π
=
= 0.
cos 2 ·
4
4
3π
3π4
zy = x cos xy, we have zy 2,
= 2 cos 2 ·
= 0.
4
4
zx = y cos xy, we have zx 2,
Since
z 2,
the tangent plane is
z = z 2,
z = −1.
3π
4
3π
4
+ zx 2,
= sin 2 ·
3π
4
3π
4
= −1,
(x − 2) + zy 2,
3π
4
y−
3π
4
14.3 SOLUTIONS
3. Differentiating gives
∂z
2x
= 2
,
∂x
x +1
∂z
= 2y,
∂y
∂z
∂x
so
so
∂z
∂y
=0
(0,3,9)
= 6.
(0,3,9)
Thus, the tangent plane is
z = 9 + 0(x − 0) + 6(y − 3)
z = 6y − 9.
4. We have
z = ey + x + x2 + 6.
The partial derivatives are
∂z
∂x
∂z
∂y
= (2x + 1)
(x,y)=(1,0)
=3
(x,y)=(1,0)
= ey
= 1.
(x,y)=(1,0)
(x,y)=(1,0)
So the equation of the tangent plane is
z = 9 + 3(x − 1) + y = 6 + 3x + y.
5. The partial derivatives are
zx = x
and zy = 4y,
so
z(2, 1) = 4,
zx (2, 1) = 2 and zy (2, 1) = 4.
The tangent plane to z = 12 (x2 + 4y 2 ) at (x, y) = (2, 1) has equation
z = z(2, 1) + zx (2, 1)(x − 2) + zy (2, 1)(y − 1)
= 4 + 2(x − 2) + 4(y − 1)
= −4 + 2x + 4y.
6. The surface is given by
z = f (x, y) = x2 + y 2 − 1,
where f (1, 3) = 9 and
fx = 2x
and fy = 2y
and
fx (1, 3) = 2
and fy (1, 3) = 6.
Thus, the tangent plane is
z = f (1, 3) + fx (1, 3)(x − 1) + fy (1, 3)(y − 3)
z = 9 + 2(x − 1) + 6(y − 3)
z = 2x + 6y − 11.
7. The surface is given by
and f (2, 3) = 40 − 22 · 32 = 4. We have
z = f (x, y) = 40 − x2 y 2 ,
fx (x, y) = −2xy 2 ,
so
fx (2, 3) = −36.
fy (x, y) = −2x2 y,
so
fy (2, 3) = −24.
and
1281
1282
Chapter Fourteen /SOLUTIONS
Thus, the tangent plane is
z = f (2, 3) + fx (2, 3)(x − 2) + fy (2, 3)(y − 3)
z = 4 − 36(x − 2) − 24(y − 3)
z = −36x − 24y + 148.
8. The surface is given by
z = f (x, y) = 6 − x2 y − ln(xy),
where f (4, 0.25) = 2 and
fx = −2xy −
1
1
· y = −2xy −
xy
x
fy = −x2 −
1
1
· x = −x2 − .
xy
y
and
Thus
fx (4, 0.25) = −2 · 4(0.25) −
and
fy (4, 0.25) = −42 −
1
= −2.25,
4
1
= −20,
0.25
so the tangent plane is
z = f (4, 0.25) + fx (4, 0.25)(x − 4) + fy (4, 0.25)(y − 0.25)
z = 2 − 2.25(x − 4) + 20(y − 0.25)
z = −2.25x − 20y + 16.
9. df = y cos(xy) dx + x cos(xy) dy
10. Since gu = 2u + v and gv = u, we have
dg = (2u + v) du + u dv
11. Since zx = −e−x cos(y) and zy = −e−x sin(y), we have
dz = −e−x cos(y)dx − e−x sin(y)dy.
12. dh = e−3t cos(x + 5t) dx + (−3e−3t sin(x + 5t) + 5e−3t cos(x + 5t)) dt
13. We have dg = gx dx + gt dt. Finding the partial derivatives, we have gx = 2x sin(2t) so gx (2, π4 ) = 4 sin(π/2) = 4,
and gt = 2x2 cos(2t) so gt (2, π4 ) = 8 cos( π2 ) = 0. Thus dg = 4 dx.
14. We have df = fx dx+fy dy. Finding the partial derivatives, we have fx = e−y so fx (1, 0) = e−0 = 1, and fy = −xe−y
so fy (1, 0) = −1e−0 = −1. Thus, df = dx − dy.
15. We have dP = PL dL + PK dK.
Now
PK = (1.01)(0.75)K −0.25 L0.25
and
PK (100, 1) ≈ 2.395,
PL = (1.01)(0.25)K 0.75 L−0.75
Thus
PL (100, 1) ≈ 0.008
dP ≈ 2.395 dK + 0.008 dL.
14.3 SOLUTIONS
1283
16. We have dF = Fm dm + Fr dr.
Fm =
Fr =
G
G
G
, Fm (100, 10) =
=
= 0.01G
r2
(10)2
100
−2G100
−G
−2Gm
, Fr (100, 10) =
=
= −0.2G.
r3
(10)3
5
Thus,
dF = 0.01G dm − 0.2G dr.
Problems
17. (a) The units are dollars/square foot.
(b) The price of land 300 feet from the beach and of area near 1000 square feet is greater for larger plots by about $3 per
square foot.
(c) The units are dollars/foot.
(d) The price of a 1000 square foot plot about 300 feet from the beach is less for plots farther from the beach by about
$2 per extra foot from the beach.
(e) Compared to the 998 ft2 plot at 295 ft from the beach, the other plot costs about 7 × 3 = $21 more for the extra
7 square feet but about 10 × 2 = $20 less for the extra 10 feet you have to walk to the beach. The net difference is
about a dollar, and the smaller plot nearer the beach is cheaper.
18. (a) Since the equation of a tangent plane should be linear, this answer is wrong.
(b) The student did not substitute the values x = 2, y = 3 into the formulas for the partial derivatives used in the formula
of a tangent plane.
(c) Let f (x, y) = z = x3 − y 2 . Since fx (x, y) = 3x2 and fy (x, y) = −2y, substituting x = 2, y = 3 gives
fx (2, 3) = 12 and fy (2, 3) = −6. Then the equation of the tangent plane is
z = 12(x − 2) − 6(y − 3) − 1,
or
z = 12x − 6y − 7.
19. (a) The two tables of values are Table 14.6 and 14.7.
Table 14.6
Table 14.7
y
x
y
1.9
2.0
2.1
0.9
0.3847
0.3697
0.3510
1.0
0.3481
0.3345
0.3176
1.1
0.3150
0.3027
0.2873
x
1.99
2.00
2.01
0.99
0.3394
0.3379
0.3363
1.00
0.3360
0.3345
0.3330
1.01
0.3327
0.3312
0.3297
Both tables are nearly linear. To check this, observe that the increments in each row (column) are equal, or nearly
so. Table 14.7 is more linear due to finer data.
(b) Table 14.7 shows f (1, 2) ≈ 0.3345. Also
f (1.01, 2) − f (1, 2)
1.01 − 1
0.3312 − 0.3345
=
= −0.3300,
0.01
f (1, 2.01) − f (1, 2)
fy (1, 2) ≈
2.01 − 2
0.3330 − 0.3345
=
= −0.1500.
0.01
fx (1, 2) ≈
Using the estimates for the partial derivatives that we just made from the Table 14.7, we get that the local linearization
of f around (1, 2) is
f (x, y) ≈ f (1, 2) + fx (1, 2)(x − 1) + fy (1, 2)(y − 2)
= 0.3345 − 0.33(x − 1) − 0.15(y − 2).
1284
Chapter Fourteen /SOLUTIONS
Now we use fx = −e−x sin(y) and fy = e−x cos(y), giving
fx (1, 2) = −0.3345,
fy (1, 2) = −0.1531.
These values of the partial derivatives tell us that the local linearization of f around (1, 2) is
f (x, y) ≈ 0.3345 − 0.3345(x − 1) − 0.1531(y − 2).
Notice that the two linearizations agree up to two decimal places.
20. We have
∂f
∂x
fx (3, 1) =
= 2xy|(3,1) = 6,
(3,1)
and
∂f
∂y
fy (3, 1) =
(3,1)
= x2 |(3,1) = 9.
Also f (3, 1) = 9. So the local linearization is,
z = 9 + 6(x − 3) + 9(y − 1).
21. The tangent plane approximation gives
∆h ≈ hx ∆x + hy ∆y,
or
h(x, y) ≈ h(a, b) + hx ∆x + hy ∆y.
With (a, b) = (600, 100) and (x, y) = (605, 98), we see that ∆x = 5 and and ∆x = −2. Thus
h(605, 98) ≈ 300 + 12(5) − 8(−2) = 376.
Using the information given, this is our best estimate for h(605, 98). However, it may not be a good estimate if the
derivatives are changing rapidly near the point (600, 100).
22. We need the partial derivatives, fx (1, 0) and fy (1, 0). We have
fx (x, y) = 2xexy + x2 yexy ,
3 xy
fy (x, y) = x e ,
so fx (1, 0) = 2
so fy (1, 0) = 1.
(a) Since f (1, 0) = 1, the tangent plane is
z = f (1, 0) + 2(x − 1) + 1(y − 0) = 1 + 2(x − 1) + y.
(b) The linear approximation can be obtained from the equation of the tangent plane:
f (x, y) ≈ 1 + 2(x − 1) + y.
(c) At (1, 0), the differential is
df = fx dx + fy dy = 2dx + dy.
23. Since fx (x, y) = √
fx (1, 2) =
x
2
and fy (x, y) = √3y2
x2 +y 3
√ 1
= 31 and
12 +23
fy (1, 2) =
2 x +y 3
2
√3·2
2
12 +23
,
= 2.
Thus the differential at the point (1, 2) is
1
dx + 2dy.
3
Using the differential at the point (1, 2), we can estimate f (1.04, 1.98). Since
df = df (1, 2) = fx (1, 2)dx + fy (1, 2)dy =
△f ≈ fx (1, 2)△x + fy (1, 2)△y
where △f = f (1.04, 1.98) − f (1, 2) and △x = 1.04 − 1 and △y = 1.98 − 2, we have
f (1.04, 1.98) ≈ f (1, 2) + fx (1, 2)(1.04 − 1) + fy (1, 2)(1.98 − 2)
p
0.04
− 2(0.02) ≈ 2.973.
= 12 + 23 +
3
14.3 SOLUTIONS
1285
24. (a) Calculating partial derivatives, we obtain gu (u, v) = 2u + v and gv (u, v) = u. Therefore, dg = gu (u, v) du +
gv (u, v) dv = (2u + v) du + u dv, and so our final answer is dg = (2u + v) du + u dv.
(b) Let (u, v) = (1, 2). In moving from (1, 2) to (1.2, 2.1), we see that ∆u = du = 0.2 and ∆v = dv = 0.1.
Therefore, we have
dg = (2(1) + 2)(0.2) + (1)(0.1)
= 0.8 + 0.1
= 0.9,
which is our final answer.
25. Local linearization gives us the approximation
T (x, y) ≈ T (2, 1) + Tx (2, 1)(x − 2) + Ty (2, 1)(y − 1)
T (x, y) ≈ 135 + 16(x − 2) − 15(y − 1).
Thus,
T (2.04, 0.97) ≈ 135 + 16(2.04 − 2) − 15(0.97 − 1) = 136.09◦ C.
26. We first recall that the formula for a volume of a right circular cylinder is V = πr 2 h, where r is the radius of the cylinder
and h is the height of the cylinder. In our case, we are being asked to calculate dV, the approximate change in V, when r
changes from 50 to 51 cm and h changes from 100 to 101 cm. We have
dV = Vr (r, h) dr + Vh (r, h) dh
= 2πrh dr + πr 2 dh
= 2π(50)(100) · 1 + π(50)2 · 1
= 10000π + 2500π,
so we conclude that the volume changes by approximately 12500π or 39270 cm3 .
27. Making use of the values of Pr and PL from the solution to Problem 27 on page 1270, we have the local linearizations:
For (r, L) = (8, 4000),
P (r, L) ≈ 80 + 2.5(r − 8) + 0.02(L − 4000),
For (r, L) = (8, 6000),
For (r, L) = (13, 7000),
P (r, L) ≈ 120 + 3.33(r − 8) + 0.02(L − 6000),
P (r, L) ≈ 160 + 3.33(r − 13) + 0.02(L − 7000).
28. A linear approximation near (480, 20) is given by
f (T, P ) ≈ f (480, 20) + fT (480, 20)(T − 480) + fP (480, 20)(P − 20).
Directly from the table on page 774, we have
f (480, 20) − f (500, 20)
27.85 − 28.46
=
= 0.0305
−20
−20
f (480, 20) − f (480, 22)
27.85 − 25.31
=
= −1.27
fp (480, 20) ≈
−2
−2
fT (480, 20) ≈
This yields the linear approximation near (480, 20)
f (T, P ) ≈ 27.85 + 0.0305(T − 480) − 1.27(P − 20)
= 0.0305T − 1.27P + 38.61.
1286
Chapter Fourteen /SOLUTIONS
29. (a) The linear approximation gives
f (520, 24) ≈ 24.20,
f (500, 22) ≈ 25.52,
f (480, 24) ≈ 23.18,
f (500, 26) ≈ 21.86.
The approximations for f (520, 24) and f (500, 26) agree exactly with the values in the table; the other two do not.
The reason for this is that the partial derivatives were estimated using difference quotients with these values.
(b) We could get a more balanced estimate by using a difference quotient that uses the values on both sides. Thus, we
could estimate the partial derivatives as follows:
f (520, 24) − f (480, 24)
40
(24.20 − 23.19)
= 0.02525,
=
40
fT (500, 24) ≈
and
f (500, 26) − f (500, 22)
4
(21.86 − 25.86)
= −1.
=
4
fp (500, 24) ≈
This yields the linear approximation
V = f (T, p) ≈ 23.69 + 0.02525(T − 500) − (p − 24) ft3 .
This approximation yields values
f (520, 24) ≈ 24.195,
f (500, 22) ≈ 25.69,
f (480, 24) ≈ 23.185,
f (500, 26) ≈ 21.69.
Although none of these predictions are accurate, the error in the predictions that were wrong before has been reduced.
This new linearization is a better all-round approximation for values near (500, 24).
30. Letting ∆T denote the change in temperature between these two points, we have
∆T ≈ fx (3, 5)∆x + ft (3, 5)∆t
=
◦
C
−2
m
(−0.5 m) +
◦
C
1.2
min
(1 min)
= 1◦ C + 1.2◦ C
= 2.2◦ C.
We therefore see that the temperature would be about 2.2◦ C warmer at our destination.
31. (a) Solving for P gives
P = f (T, V ) =
(b) Since
n2 a
nRT
− 2.
V − nb
V
nR
∂P
= fT (T, V ) =
∂T
V − nb
∂P
−nRT
2n2 a
= fV (T, V ) =
+
,
2
∂V
(V − nb)
V3
we have
∆P ≈ fT (T0 , V0 )∆T + fV (T0 , V0 )∆V
nR
∆T +
∆P ≈
V0 − nb
nRT0
2n2 a
−
+
2
(V0 − nb)
V03
∆V.
14.3 SOLUTIONS
1287
32. (a) When V = 25 and P = 1, we have T = 304.9. The differential dT is
dT =
∂T
1
1
∂T
dV +
dP = (−16.574 2 + 1.06 3 + 12.187P ) dV + (−0.3879 + 12.187V ) dP.
∂V
∂P
V
V
When V = 25 and P = 1 this is
dT = 12.16 dV + 304.29 dP.
(b) If ∆P = 0.1 and ∆T = 0, then
0 ≈ (12.16)∆V + (304.29)(0.1),
so
30.429
≈ −2.5.
12.16
Thus the volume would have to decrease by about 2.5 dm3 , or about 10%.
∆V ≈ −
33. (a) For a mass m of liquid, we have ρ = m/V , so
dρ =
−m
m
−m
dV = 2 βV dT = −β dT = −βρ dT.
V2
V
V
(b) From part (a), we have ∆ρ ≈ −βρ∆T , so
1 ∆ρ
·
.
ρ ∆T
Thus, in the limit as ∆T and ∆ρ become very small, we have
β≈−
1 dρ
1
β=−
=−
ρ dT
ρ
Slope of tangent line
in Figure 14.13
!
.
We use Figure 14.13 to estimate ρ, ∆ρ, and ∆T . We use these values to approximate β.
From Figure 14.13 we see that ρ ≈ 997 when T = 20. In addition, we see that ρ ≈ 1000 when T = 0. Between
these points, the temperature change is ∆T = 20 − 0 = 20, and the density change is ∆ρ = 997 − 1000 = −3.
Thus, ∆ρ ≈ −βρ∆T
1 ∆ρ
1
(−3)
β≈− ·
=−
·
≈ 0.00015.
ρ ∆T
997
20
At T = 80 we have ρ ≈ 973 and at T = 60, we have ρ ≈ 983. Thus ∆T = 80 − 60 = 20, and ∆ρ =
973 − 983 = −10, so when T = 80 we have,
β≈−
(−10)
1
1 ∆ρ
·
=−
·
≈ 0.0005.
ρ ∆T
973
20
As you can see from Figure 14.13, using ∆T = 20 may not give a very good approximation. To get a better
approximation, use a smaller value of ∆T .
ρ (kg/m3 )
1000
(20, 997)
990
(60, 983)
980
(80, 973)
970
960
T (◦ C)
0
20
40
60
Figure 14.13
80
100
1288
Chapter Fourteen /SOLUTIONS
34. The error in η is approximated by dη, where
dη =
We need to find
∂η
∂η
dr +
dp.
∂r
∂p
π p4r 3
∂η
=
∂r
8 v
and
∂η
π r4
=
.
∂p
8 v
For r = 0.005 and p = 105 we get
∂η
(0.005, 105 ) = 0.39270,
∂p
∂η
(0.005, 105 ) = 3.14159 · 107 ,
∂r
so that
dη =
∂η
∂η
dr +
dp.
∂r
∂p
is largest when we take all positive values to give
dη = 3.14159 · 107 · 0.00025 + 0.39270 · 1000 = 8246.68.
This seems quite large but η(0.005, 105 ) = 39269.9 so the maximum error represents about 20% of any value
computed by the given formula. Notice also the relative error in r is ±5%, which means the relative error in r 4 is ±20%.
35. At temperature t0 , the length is l0 and the period is
T0 = 2π
r
l0
.
g
Now suppose the temperature changes by ∆t, causing a change in length, ∆l. Since l = l0 (1 + α(t − t0 )), we have
∆l ≈
dl
∆t = l0 α∆t.
dt
The change in period, ∆T , caused by this change in length is given by
∆T ≈
dT
2π 1
π
∆l = √ · l−1/2 ∆l = √ ∆l.
dl
g 2
gl
At t = t0 , we have l = l0 , so substituting for ∆l and using the fact that T0 = 2π
r
π
π
∆T ≈ √ ∆l = √ l0 α∆t = πα
gl0
gl0
p
l0 /g, we have
l0
T0
∆t = α ∆t.
g
2
Now we have to figure out how many seconds a day the clock gains or loses as a result of this change in period. We
assume the units of l and g are chosen to given T in seconds. When the period is T , the pendulum executes N oscillations
per day, where
24 · 60 · 60
86400
Number of seconds in a day
=
=
.
N=
Period in seconds
T
T
Thus, when the period changes from T0 to T0 + ∆T , we have
∆N ≈
86400
dN
∆T = −
∆T.
dT
T02
Substituting for ∆T gives
86400 αT0
43200α
·
∆t = −
∆t.
2
T0
T02
The clock records T0 seconds as having passed for each oscillation executed. Therefore the number of seconds gained or
lost per day when N changes by ∆N is given by
∆N ≈ −
Number of seconds gained/lost per day = T0 ∆N = −43200∆t.
The number of seconds is negative and the clock is slow if ∆t > 0; the number of seconds is positive and the clock is fast
if ∆t < 0. The answer −43200∆t is independent of l0 .
14.3 SOLUTIONS
1289
36. (a) The local linearization of f (x, y) at (a, b) is
g(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b).
The contour of g through (a, b) is g(a, b) = c, where
c = f (a, b) + fx (a, b)(a − a) + fy (a, b)(b − b) = f (a, b).
Thus, the contour of g through (a, b) is
f (a, b) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
0 = fx (a, b)(x − a) + fy (a, b)(y − b).
Since the local linearization g(x, y) gives the best linear approximation to f at (a, b), the contour of g through (a, b)
must be the tangent line to the contour of f through (a, b). Thus, the tangent line is
fx (a, b)(x − a) + fy (a, b)(y − b) = 0.
(b) Putting the equation in part (a) in slope-intercept form we have
y=−
fx (a, b)
fx (a, b)
x+a
+ b.
fy (a, b)
fy (a, b)
The slope is the coefficient of x, which is −fx (a, b)/fy (a, b).
(c) We have fx (3, 4) = 10 and fy (3, 4) = 3. The tangent line has the equation 10(x − 3) + 3(y − 4) = 0.
Strengthen Your Understanding
37. A correct equation for the tangent plane at the point (3, 4) is
z = f (3, 4) + fx (3, 4)(x − 3) + fy (3, 4)(y − 4).
38. The equations of the tangent planes are
z = f (0, 0) + fx (0, 0)x + fy (0, 0)y
z = g(0, 0) + gx (0, 0)x + gy (0, 0)y.
If the function values at the origin are not equal, f (0, 0) 6= g(0, 0), then the two planes are parallel but not identical. For
example, if f (x, y) = x2 + y 2 and g(x, y) = x2 + y 2 + 2, then the tangent planes to the graphs of these functions at the
origin are z = 0 and z = 2, respectively.
39. The given equation is not linear, so it is not the equation of a plane.
Let f (x, y) = x2 y. The tangent plane to the surface z = f (x, y) at the point (1, 2) has equation
z = f (1, 2) + fx (1, 2)(x − 1) + fy (1, 2)(y − 2).
The derivatives fx (x, y) = 2xy and fy (x, y) = x2 must be evaluated at the point (1, 2). The correct tangent plane
equation is
z = 2 + 4(x − 1) + (y − 2).
40. For any function f (x, y) and any constant C, the two functions f (x, y) and g(x, y) = f (x, y) + C have the same
differential. For example, f (x, y) = xy 2 and g(x, y) = xy 2 + 10 have the same differential:
df = y 2 dx + 2xy dy
and
dg = y 2 dx + 2xy dy.
41. The plane z = 3 is a horizontal plane parallel to the xy-plane. This plane is tangent to the sphere of radius 3 centered at
the origin. Other examples are possible, the graph of the paraboloid z = 3 − x2 − y 2 also has tangent plane z = 3 at
(0, 0, 3).
1290
Chapter Fourteen /SOLUTIONS
2
2
42. True. The partials are fx = 2xyex and fy = ex , so fx (0, 1) = 0 and fy (0, 1) = 1. Then the tangent plane has equation
z = f (0, 1) + 0(x − 0) + 1(y − 1) = 1 + y − 1 = y.
43. True. The change in f is approximately df = 2 · 2 · (−0.1) + sin(2) · 0.0002. Since −1 ≤ sin x ≤ 1, the term
sin(2) · 0.0002 is small in comparison to the first term, which is −0.4.
In fact, df = 2 · 2(−0.1) + sin(2) · 0.0002 ≈ −0.3998.
44. False. The graph of f is a paraboloid, opening upward. The tangent plane to this surface at any point lies completely under
the surface (except at the point of tangency). So the local linearization underestimates the value of f at nearby points.
45. False. As a counterexample, consider f (x, y) = x2 + y 2 and g(x, y) = 2x + y 2 at the point (1, 1). Then both functions
have differential 2 dx + 2 dy at the point (1, 1).
46. False. As a counterexample, consider f (x, y) = x2 + y 2 and g(x, y) = 2x + y 2 − 1 at the point (1, 1). Then both
functions have tangent plane z = 2 + 2(x − 1) + 2(y − 1) at the point (1, 1).
47. True. If f is a constant function, there is no change in f between any two points. Alternatively, fx = fy = 0, so
df = 0 dx + 0 dy = 0.
48. True. If f is linear, then f (x, y) = mx + ny + c for some m, n and c. So fx = m and fy = n giving df = m dx + n dy,
which is linear in the variables dx and dy.
49. False. The property of local linearity is that most functions behave approximately linearly close to a point. While the
contours may appear parallel and equally spaced, unless the function is already linear they won’t be exactly parallel or
evenly spaced, no matter how closely we zoom.
Solutions for Section 14.4
Exercises
1. Since the partial derivatives are
∂f
15 4
15 4
=
x −0=
x
∂x
2
2
∂f
24 5
24
= 0−
y = − y5
∂y
7
7
we have
grad f =
∂f ~
∂f ~
i +
j =
∂x
∂y
15 4 ~
x i −
2
24 5 ~
y j.
7
2. Since the partial derivatives are
∂Q
= 50
∂K
and
∂Q
= 100
∂L
we have
∇f = 50~i + 100~j .
3. Since the partial derivatives are
∂f
= 2m + 0 = 2m
∂m
∂f
= 0 + 2n = 2n
∂n
we have
grad f =
∂f ~
∂f ~
i +
j = 2m~i + 2n~j .
∂m
∂n
14.4 SOLUTIONS
4. Since the partial derivatives are
zx = ey
and
zy = xey
we have
∇z = ey~i + xey~j .
5. Since the partial derivatives are
1
5α
∂f
= (5α2 + β)−1/2 (10α + 0) = p
∂α
2
5α2 + β
1
1
∂f
= (5α2 + β)−1/2 (0 + 1) = p
,
∂β
2
2 5α2 + β
we have
5α
∂f ~
∂f ~
i +
j =
grad f =
∂α
∂β
p
5α2 + β
6. Since the partial derivatives are
!
1
~i +
p
5α2 + β
2
and fh = πr 2 ,
fr = 2πrh
we have
∇f = 2πrh~i + πr 2~j .
7. Since the partial derivatives are
zx = ey
zy = xey + ey + yey ,
and
we have
∇z = ey~i + ey (1 + x + y)~j .
8. Since the partial derivatives are
fK = 0.3K −0.7 L0.7
we have
∇f = 0.3
L
K
and fL = 0.7K 0.3 L−0.3 ,
0.7
0.3
~i + 0.7 K
L
~j .
9. Since the partial derivatives are
fr = sin θ
fθ = r cos θ,
and
we have
∇f = sin θ~i + r cos θ~j .
10. Since the partial derivatives are
2x
x2 + y 2
fx =
we have
∇f =
and
fy =
2y
,
x2 + y 2
2
(x~i + y~j ).
x2 + y 2
11. Since the partial derivatives are
zx =
1
y
cos
we have
∇z =
x
,
y
zy =
−x
y2
cos
1
x
x
x
cos ( )~i − 2 cos ( )~j .
y
y
y
y
x
y
!
~j .
1291
1292
Chapter Fourteen /SOLUTIONS
12. Since the partial derivatives are
zx =
we have
1
1+
x 2
y
1
y
=
y
,
y 2 + x2
and
zy =
1
1+
−
x 2
y
x
y2
=−
x
,
y 2 + x2
x ~
y ~
i − 2
j.
y 2 + x2
y + x2
∇z =
13. Since the partial derivatives are
∂f
(2α − 3β)(2 + 0) − (2 − 0)(2α + 3β)
=
∂α
(2α − 3β)2
4α − 6β − (4α + 6β)
=
(2α − 3β)2
12β
=−
(2α − 3β)2
∂f
(2α − 3β)(0 + 3) − (0 − 3)(2α + 3β)
=
∂β
(2α − 3β)2
(6α − 9β) + (6α + 9β)
=
(2α − 3β)2
12α
=
(2α − 3β)2
we have
grad f =
∂f ~
∂f ~
i +
j =
∂α
∂β
−
12β
(2α − 3β)2
~i +
12α
(2α − 3β)2
14. Since the partial derivatives are
zx =
ey (x + y) − xey
yey
=
2
(x + y)
(x + y)2
zy =
xey (x + y) − xey
ey (x2 + xy − x)
=
(x + y)2
(x + y)2
we have
∇z =
ey (x2 + xy − x)~
yey ~
i +
j
2
(x + y)
(x + y)2
15. Since the partial derivatives are
∂f
= 2xy + 7y 3
∂x
∂f
= x2 + 21xy 2
∂y
we have
=
grad f
(1,2)
∂f ~
∂f ~
i +
j
∂x
∂y
(1,2)
= (2xy + 7y 3 )~i + (x2 + 21xy 2 )~j
3
(1,2)
= 2(1)(2) + 7(2) ~i + (1)2 + 21(1)(2)2 ~j
= (4 + 56)~i + (1 + 84)~j
= 60~i + 85~j
~j .
14.4 SOLUTIONS
16. Since the partial derivatives are
∂f
= 10m + 0 = 10m
∂m
∂f
= 0 + 12n3 = 12n3
∂n
we have
=
grad f
(5,2)
∂f ~
∂f ~
i +
j
∂m
∂n
(5,2)
= 10m~i + 12n3~j
(5,2)
= 10(5)~i + 12(2)3~j
= 50~i + 96~j
17. Since the partial derivatives are
fr = 2π(h + r)
and fh = 2πr,
we have
∇f (2, 3) = 10π~i + 4π~j .
18. The gradient of esin y at x = 0, y = π is given by
= 0~i + (cos yesin y )~j
grad(esin y )
(0,π)
(0,π)
= (−1)e0~j = −~j .
19. Since the partial derivatives are
∂f
= cos (x2 ) · (2x) + 0 = 2x cos (x2 )
∂x
∂f
= 0 − sin y = − sin y
∂y
we have
grad f
√
=
π
( 2 ,0)
∂f ~
∂f ~
i +
j
∂x
∂y
√
π
( 2 ,0)
2x cos(x2 ) ~i + (− sin y) ~j
=
√
1
π( √ ) ~i + 0
2
q
π ~
i
=
2
=
√
π
=
( 2 ,0)
20. Since the partial derivatives are
2x + y
x2 + xy
x
fy = 2
x + xy
fx =
at the point (4, 1), we have
9
= 0.45
20
4
fy (4, 1) =
= 0.2.
20
fx (4, 1) =
√
2(
π
π
) cos( ) ~i + (− sin 0)~j
2
4
1293
1294
Chapter Fourteen /SOLUTIONS
Then
gradf (4, 1) = 0.45~i + 0.2~j .
21. Since the partial derivatives are
fx =
−2x
(x2 + y 2 )2
we have
fy =
and
−2y
,
(x2 + y 2 )2
1 ~
(2i − 6~j ).
100
∇f =
22. Since the partial derivatives are
1
∂f
= (tan x + y)−1/2
∂x
2
and
1
1
√
+0 =
,
cos2 x
2 cos2 x tan x + y
1
1
∂f
,
= (tan x + y)−1/2 (0 + 1) = √
∂y
2
2 tan x + y
then
∂f ~
∂f ~
i +
j =
grad f =
∂x
∂y
Hence we have
2(cos(0))2
(0,1)
=
1
~i +
√
2 cos2 x tan x + y
1
=
grad f
1
√
2
p
tan(0) + 1
2(1) 0 + 1
1
1~
= i + ~j
2
2
~i +
!
1
√
2 tan x + y
1
~i +
1
√
2 0+1
p
2
~j
tan(0) + 1)
!
~j .
~j
23. Since fx = y and fy = x + 3y 2 , at (1, 2) we have grad f = 2~i + 13~j . Thus
f~u (1, 2) = grad f ·
4
3~
i − ~j
5
5
=
2 · 3 + 13(−4)
46
=− .
5
5
=
3 · 3 + (−4) · (−4)
= 5.
5
24. Since fx = 3 and fy = −4, we have grad f = 3~i − 4~j . Thus
f~u (1, 2) = grad f ·
3~
4
i − ~j
5
5
25. Since fx = 2x and fy = −2y, at (1, 2) we have grad f = 2~i − 4~j . Thus
f~u (1, 2) = grad f ·
3
~i − 4 ~j
5
5
=
2 · 3 − 4(−4)
22
=
.
5
5
26. Since fx = 2 cos(2x − y) and fy = − cos(2x − y), at (1, 2) we have grad f = 2 cos(0)~i − cos(0)~j = 2~i − ~j . Thus
f~u (1, 2) = grad f ·
4
3~
i − ~j
5
5
=
2 · 3 − 1(−4)
10
=
= 2.
5
5
14.4 SOLUTIONS
1295
27. Since k~v k = 5, we see that ~v is not a unit vector. The unit vector ~
u in the direction of ~v is
~
u =
4
3
~v
= ~i − ~j .
k~v k
5
5
The partial derivatives are fx (x, y) = 2xy and fy (x, y) = x2 . So
f~u (2, 6) = fx (2, 6) ·
4
5
3
+ fy (2, 6) · −
5
= 24
4
5
3
+4 −
5
=
84
.
5
28. Since grad f = fx~i + fy~j and df = fx dx + fy dy we have
df = ydx + xdy.
29. Since grad f = fx~i + fy~j and df = fx dx + fy dy we have
df = (2x + 3ey )dx + 3xey dy.
30. Since df = fx dx + fy dy and grad f = fx~i + fy~j we have
grad f = 2x~i + 10y~j .
31. Since df = fx dx + fy dy and grad f = fx~i + fy~j we have
grad f = (x + 1)yex~i + xex~j .
32. Since the values of z decrease as we move in direction ~i from the point (−2, 2), the directional derivative is negative.
33. Since the values of z decrease as we move in direction ~j from the point (0, −2), the directional derivative is negative.
34. Since the values of z decrease as we move in direction ~i +2~j from the point (0, −2), the directional derivative is negative.
35. Since the values of z increase as we move in direction ~i −2~j from the point (0, −2), the directional derivative is positive.
36. Since the values of z stay approximately the same (since the direction is tangent to the contour) as we move in direction
~i + ~j from the point (−1, 1), the directional derivative is approximately zero.
37. Since the values of z increase as we move in direction −~i +~j from the point (−1, 1), the directional derivative is positive.
38. The approximate direction of the gradient vector at point (−2, 0) is −~i , since the gradient vector is perpendicular to the
contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
39. The approximate direction of the gradient vector at point (0, −2) is −~j , since the gradient vector is perpendicular to
the contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
40. The approximate direction of the gradient vector at point (2, 0) is ~i , since the gradient vector is perpendicular to the
contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
41. The approximate direction of the gradient vector at point (0, 2) is ~j , since the gradient vector is perpendicular to the
contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
42. The approximate direction of the gradient vector at point (−2, 2) is −~i + ~j , since the gradient vector is perpendicular to
the contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
43. The approximate direction of the gradient vector at point (−2, −2) is −~i − ~j , since the gradient vector is perpendicular
to the contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and
any positive multiple of the vector given is also correct.
1296
Chapter Fourteen /SOLUTIONS
44. The approximate direction of the gradient vector at point (2, 2) is ~i + ~j , since the gradient vector is perpendicular to
the contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
45. The approximate direction of the gradient vector at point (2, −2) is ~i − ~j , since the gradient vector is perpendicular to
the contour and points in the direction of increasing z-values. Answers may vary since answers are approximate and any
positive multiple of the vector given is also correct.
Problems
46. The distance from P to Q is
p
(3.03 − 3)2 + (3.96 − 4)2 = 0.05, so the directional derivative is approximately
f~u (P ) ≈
f (Q) − f (P )
20 − 15
=
= 100.
0.05
0.05
47. (a) The unit vector ~
u in the same direction as ~v is
~
u =
1
3
−1
1
~v = p
~v = √ ~i + √ ~j = −0.316228~i + 0.948683~j .
2
2
k~v k
10
10
(−1) + 3
The vector w
~ of length 0.1 in the direction of ~v is
w
~ = 0.1~
u = −0.0316228~i + 0.0948683~j .
The displacement vector from P to Q is w
~ . Hence
Q = (4 − 0.0316228, 5 + 0.0948683) = (3.96838, 5.09487).
(b) Since the distance from P to Q is 0.1, the directional derivative of f at P in the direction of Q is approximately
f~u ≈
f (Q) − f (P )
3.01052 − 3
=
= 0.1052.
0.1
0.1
(c) We have
1
~i + √ 1
~j
grad f (x, y) = √
2 x+y
2 x+y
1
1
grad f (4, 5) = ~i + ~j .
6
6
The directional derivative at P = (4, 5) in the direction of ~
u is
f~u (4, 5) = grad f (4, 5) · ~
u =
1 −1
1 3
1
√ + √ = √ = 0.1054.
6 10
6 10
3 10
48. (a) The partial derivatives are given by
fx = ex (tan y) + 4xy,
Thus
fx (0,
π
)=1
4
fy = ex (sec2 y) + 2x2 .
and
fy (0,
π
) = 2,
4
and so
π
) = ~i + 2~j .
4
The unit vector u~1 in the direction of ~i − ~j is √12 (~i − ~j ). Then the directional derivative of f at (0, π4 ) in the
direction of ~i − ~j is
grad f (0,
fu~1 (0,
π
π
) = grad f (0, ) · u~1
4
4
1
1
~
~
= (i + 2j ) · ( √ ~i − √ ~j )
2
2
√
√
2
1
= √ − 2=−
.
2
2
14.4 SOLUTIONS
(b) The unit vector u~2 in the direction of ~i +
√
3~j is u~2 = 21 (~i +
√
1297
3~j ). From part (a),
π
) = ~i + 2~j .
4
√
Then the directional derivative of f at (0, π4 ) in the direction of ~i + 3~j is
grad f (0,
fu~2 (0,
√
π
π
1
3~
) = grad f (0, ) · u~2 = (~i + 2~j ) · ( ~i +
j)
4
4
2
2
1 √
= + 3.
2
49. We want the directional derivative in the direction of ~
u at (1, 2), so we want to calculate f~u (1, 2). Since fx = 2x and
fy = 2y, at the point (1, 2), we have grad f (1, 2) = 2~i + 4~j . Since ~
u is a unit vector, we obtain
f~u (1, 2) = grad f (1, 2) · ~
u
= (2~i + 4~j ) · (0.6~i + 0.8~j )
= 2(0.6) + 4(0.8)
= 4.4.
50. (a) √
First we will find a unit vector in the same direction as the vector ~v = 3~i − 2~j . Since this vector has magnitude
13, a unit vector is
3
2
1
~v = √ ~i − √ ~j .
~
u1 =
k~v k
13
13
The partial derivatives are
(1 + x2 ) − (x + y) · 2x
1 − x2 − 2xy
=
,
2
2
(1 + x )
(1 + x2 )2
1
,
and fy (x, y) =
1 + x2
fx (x, y) =
then, at the point P , we have
1 − 12 − 2 · 1 · (−2)
= 1,
(1 + 12 )2
1
1
= .
fy (P ) = fy (1, −2) = fy (1, −2) =
1 + 12
2
fx (P ) = fx (1, −2) =
Thus
f~u 1 (P ) = grad f (P ) · ~
u1
1
3
2
= (~i + ~j ) · ( √ ~i − √ ~j )
2
13
13
1
2
3
= √ −√ = √ .
13
13
13
(b) The unit vector in the same direction as the vector ~v = −~i + 4~j is
1
1
~v = p
(−~i + 4~j )
k~v k
(−1)2 + 42
4
1
= − √ ~i + √ ~j .
17
17
~
u2 =
Since we have calculated from part (a) that fx (P ) = 1 and fy (P ) = 1/2,
u2
f~u 2 (P ) = grad f (P ) · ~
4
1
1
= (~i + ~j ) · (− √ ~i + √ ~j )
2
17
17
1
2
1
= −√ + √ = √ .
17
17
17
1298
Chapter Fourteen /SOLUTIONS
(c) The direction of greatest increase is grad f at P . By part (a) we have found that
fx (P ) = 1 and fy (P ) =
1
.
2
Therefore the direction of greatest increase is
1
grad f (P ) = ~i + ~j .
2
51. We have
∇f = (2xy 3 )~i + (3x2 y 2 )~j .
(a) A vector in the direction of maximum rate of change is ∇f (−1, 2) = −16~i + 12~j .
(b) A vector in the direction of minimum rate of change is −∇f (−1, 2) = 16~i − 12~j .
(c) Any vector perpendicular to ∇f gives a direction in which the rate of change is zero. One possible answer is 12~i +
16~j .
√
√
52. gradf (π/4, 1) = (ı + )/ 2. The displacement is ı + , so the direction is ı + j)/ 2 and the directional derivative is 1.
53. (a) The average rate of change of the function is the “rise over the run”, or the change in the z-values divided by the
horizontal distance. The vector from the point (3, 1) to the point (1, 2) is ~v = −2~i + ~j . We have
Average rate of change =
f (1, 2) − f (3, 1)
1.6931 − 9
√
=
= −3.268.
5
k − 2~i + ~j k
(b) The instantaneous rate of change is given by the directional derivative in the direction ~v = −2~i + ~j . We first find
the gradient vector:
1
grad f = (2x)~i + ~j ,
y
so grad f (3, 1) = 6~i + ~j . In the direction of ~v = −2~i + ~j ,
Directional derivative = grad f ·
−12 + 1
~v
=
√
= −4.919.
k~v k
5
54. We see that
grad f = (3y)~i + (3x + 2y)~j ,
so at the point (2, 3), we have
grad f = 9~i + 12~j .
(9)(3) + (12)(−1)
15
~v
√
=
= √ ≈ 4.74.
(a) The directional derivative is ∇f ·
k~v k
10
10
(b) The direction of maximum rate of change √
is ∇f (2, 3) = 9~i + 12~j .
(c) The maximum rate of change is k∇f k = 225 = 15.
55. (a) The directional derivative should be a number, not a vector.
(b) Since the partial derivatives are
fx (x, y) = 2xey ,
fy (x, y) = x2 ey ,
fx (1, 0) = 2,
fy (1, 0) = 1.
we have
grad f (1, 0) = 2~i + ~j .
+ 35 ~j . Thus, the correct answer is
The unit vector in the direction of ~v is u
~ =
4~
i
5
f~u (1, 0) = grad f (1, 0) · ~
u =2·
4
3
8
3
11
+1· = + =
.
5
5
5
5
5
14.4 SOLUTIONS
1299
56. f~i (4, 1) means the rate of change of f in the x direction at (4, 1). Thus,
f~i (4, 1) ≈
f (5, 1) − f (4, 1)
.
1
The point (5, 1) is about 2/3 of the way from the contour for f = 3 to the contour for f = 4, so we estimate f (5, 1) = 3.7.
Thus
3.7 − 2
= 1.7.
f~i (4, 1) ≈
1
We can get a better estimate if we average this result with the difference quotient obtained by going the other way, that is
f~i (4, 1) ≈
2−1
f (4, 1) − f (3, 1)
=
= 1.
1
1
Averaging the two estimates gives f~i (4, 1) ≈ 1.35
57. f~j (4, 1) means the rate of change of f in the y direction at (4, 1). Thus
f~j (4, 1) ≈
f (4, 2) − f (4, 1)
.
1
The point (4, 2) is about 1/3 of the way from the contour for f = 2 to the contour for f = 3, so we estimate f (4.2) = 1.3.
Thus
1.3 − 2
f~j (4, 1) ≈
= −0.7.
1
We can get a better estimate by averaging this with the difference quotient obtained by going the other way, that is,
f~j (4, 1) ≈
2−4
f (4, 1) − f (4, 0)
=
= −2.
1
1
Averaging the two estimates we get f~j (4, 1) ≈ −1.35.
√
58. Since ~
u = (~i √− ~j )/ 2, we head away from the point (4, 1) toward the point (5, 0). Since the points (4, 1) and (5, 0)
are a distance 2 apart,
f (5, 0) − f (4, 1)
√
f~u (4, 1) ≈
.
2
The point (5, 0) is about half way between the contour for f = 5 and the contour for f = 6, so we estimate f (5, 0) = 5.5.
Thus
5.5 − 2
= 2.5.
f~u (4, 1) ≈ √
2
We can get a better estimate by averaging this with the difference quotient obtained by going the other way, that is,
f~u (4, 1) ≈
f (4, 1) − f (3, 2)
2 − 0.5
√
= √
= 1.1
2
2
Averaging these two results gives an estimate f~u (4, 1) ≈ 1.8.
√
~i + ~j )/ 2, we head away from the point (4, 1) toward the point (3, 2). Since the points (4, 1) and (3, 2)
59. Since ~
u = (−√
are a distance 2 apart,
f (3, 2) − f (4, 1)
√
f~u (4, 1) ≈
.
2
The point (3, 2) is about half way between the point where f = and the contour for f = 1, so we estimate f (3, 2) = 0.5.
Thus
0.5 − 2
= −1.1
f~u (4, 1) ≈ √
2
We can get a better estimate by averaging this with the difference quotient obtained by going the other way, that is,
f~u (4, 1) ≈
f (4, 1) − f (5, 0)
2 − 5.5
= √
= −2.5.
√
2
2
Averaging these two results gives an estimate f~u (4, 1) ≈ −1.8.
1300
Chapter Fourteen /SOLUTIONS
√
√
60. Since ~
u = (−2~i + ~j )/ 5, we head away from the point (4, 1) in the direction (−2~i + ~j )/ 5, that is, toward the point
√
(2, 2). From the graph, we see that f (4, 1) = 2 and f (2, 2) = 0. Since the points (4, 1) and (2, 2) are a distance 5
apart, we have
f (2, 2) − f (4, 1)
0−2
= √ = −0.9
√
f~u (4, 1) ≈
5
5
We can get a better estimate by averaging this with the difference quotient obtained by going the other way, that is,
f (4, 1) − f (6, 0)
2−8
√
= √ = −2.7.
5
5
Averaging these two results gives an estimate f~u (4, 1) ≈ −1.8.
f~u (4, 1) ≈
61. First, we check that 22 + 32 = 13. Then let f (x, y) = x2 + y 2 so that the given curve is the contour f (x, y) = 13. Since
fx = 2x and fy = 2y, we have grad f (2, 3) = 4~i + 6~j . Since gradients are perpendicular to contours, a vector normal
to the curve at (2, 3) is ~n = 4~i + 6~j . Using the normal vector to a line the same way we use the normal vector to a plane,
we get that an equation of the tangent line is 4(x − 2) + 6(y − 3) = 0.
62. First, we check that (2)(3) = 6. Then let f (x, y) = xy so that the given curve is the contour f (x, y) = 6. Since fx = y
and fy = x, we have grad f (2, 3) = 3~i + 2~j . Since gradients are perpendicular to contours, a vector normal to the curve
at (2, 3) is ~
n = 3~i + 2~j . Using the normal vector to a line the same way we use the normal vector to a plane, we get that
the equation of the tangent line is 3(x − 2) + 2(y − 3) = 0.
63. First, we check that 3 = 22 − 1. Then let f (x, y) = y − x2 + 1 so that the given curve is the contour f (x, y) = 0.
Since fx = −2x and fy = 1, we have grad f (2, 3) = −4~i + ~j . Since gradients are perpendicular to contours, a vector
normal to the curve at (2, 3) is ~
n = −4~i + 1~j . Using the normal vector to a line the same way we use the normal vector
to a plane, we get that an equation of the tangent line is −4(x − 2) + (y − 3) = 0. Notice, if we had instead found the
slope of the tangent line using dy/dx = 2x, we get (y − 3) = 4(x − 2), which agrees with the equation we got using the
gradient.
64. First, we check that (3 − 2)2 + 2 = (2)(3) − 3. Then let f (x, y) = (y − x)2 − xy + 5 so that the given curve is the
contour f (x, y) = 0. Since fx = −2(y − x) − y and fy = 2(y − x) − x, we have grad f (2, 3) = −5~i + 0~j . Since
gradients are perpendicular to contours, a vector normal to the curve at (2, 3) is ~
n = −5~i ; in other words, the tangent
line is a vertical line. Thus the equation of the tangent line is x = 2.
65. (a) In the ~i − ~j direction the function is decreasing, so the value of g~u (2, 5) is negative.
(b) In the ~i + ~j direction the function is decreasing, so the value of g~u (2, 5) negative as well.
66. At the point (1.2, 0), the value of the function is 4.2. Nearby, the largest value is 8.9 at the point (1.4, −1). Since the
gradient vector points in the direction of maximum increase, it points into the fourth quadrant.
67. (a) Negative. ∇f is perpendicular to the level curve at the point P , so its x-component which is ∇f · ~i is negative.
(b) Positive. The y-component of ∇f is in the same direction as ~j at P and hence the dot product will be positive.
(c) Positive. The partial derivative with respect to x at Q is positive because the value of f is increasing in the positive x
direction at Q. (Note that Q lies between the level curves with values 3 and 4 and that the one with value 4 is further
in the positive x direction from Q.)
(d) Negative. Again, Q lies between the level curves with values 3 and 4 and the one with value 3 is further from Q in
the positive y direction, so the partial derivative with respect to y at Q is negative.
68. k∇f k at P is larger because the level curves are closer there.
69. At points (x, y) where the gradients are defined and are not the zero vector, the level curves of f and g intersect at right
angles if and only grad f · grad g = 0.
We have grad f · grad g = (~i + ~j ) · (~i − ~j ) = 0. The level curves of f and g are straight lines that cross at right
angles. See Figure 14.14.
y
g = c2
f = c1
x
Figure 14.14
14.4 SOLUTIONS
1301
70. At points (x, y) where the gradients are defined and are not the zero vector, the level curves of f and g intersect at right
angles if and only grad f · grad g = 0.
We have grad f · grad g = (2~i + 3~j ) · (2~i − 3~j ) = −5 6= 0. The level curves of f and g are straight lines that do
not cross at right angles. See Figure 14.15.
y
g = c2
f = c1
x
Figure 14.15
71. At points (x, y) where the gradients are defined and are not the zero vector, the level curves of f and g intersect at right
angles if and only grad f · grad g = 0.
We have grad f · grad g = (2x~i − ~j ) · ((1/x)~i + 2~j ) = 0 at all points where both functions are defined. The level
curves of f are parabolas that intersect the level curves of g in right angles. See Figure 14.16.
y
f = c1
g = c2
x
Figure 14.16
72. At points (x, y) where the gradients are defined and are not the zero vector, the level curves of f and g intersect at right
angles if and only grad f · grad g = 0.
We have grad f · grad g = (2x~i − 2y~j ) · (y~i + x~j ) = 0. The level curves of f and g are hyperbolas that cross at
right angles except at the point (0, 0). See Figure 14.17.
y
g = c2
x
■
f = c1
Figure 14.17
1302
Chapter Fourteen /SOLUTIONS
73. (a) The graph of z = y 2 is in Figure 14.18.
√
(b) If z = c, then y 2 = c, so the level curves are y = ± c. See Figure 14.19.
(c) The level curves in part (b) show that the direction of the greatest increase is in the y direction if the point is in the
upper half xy-plane (where y > 0). Since the point (2, 3, 9) is in the upper half xy-plane, we climb fastest in the
direction ~j , parallel to the y-axis.
y
4
z = 16
z=9
z
z=4
z=1
z=0
y
x
x
z=1
−2
Figure 14.18
z=4
Figure 14.19
74. (a) The fastest descent is in the direction of −∇f , so
−∇f (1, 3) = (4x~i + 2y~j )
= 4~i + 6~j .
(1,3)
Any positive multiple of this vector points in the same direction.
(b) If you start to √
move in this direction,
the slope of the path is the rate of change in your height with distance, which is
√
−||∇f || = − 42 + 62 = − 52.
75. (a) We have f (x, y) = C for points (x, y) at a distance C from P , points on the circle of radius C centered at P . The
level curves of f are circles centered at P
(b) The gradient of f at the point (x, y) points in the direction that you should move that point to increase its distance
from P most rapidly, away from P on the line from P through (x, y). The vector grad f points directly away from
P at every point.
(c) The magnitude k grad f (x, y)k is the rate change of f as you go in the direction of grad g(x, y), which is directly
away from P . Every unit farther away from P increases f by 1 because f is the distance from P , so k grad f (x, y)k =
1.
76. The vector from (2, 1) to (1, 3) is v~1 = (1 − 2)~i + (3 − 1)~j = −~i + 2~j . A unit vector in this direction is u~1 =
− √15~i + √25 ~j .
A vector from (2, 1) to (5, 5) is v~2 = (5 − 2)~i + (5 − 1)~j = 3~i + 4~j . A unit vector in this direction is u~2 = 53~i + 54 ~j .
The directional derivative along u~1 is
zu~1 (1, 2) = ∇z ·
So
2
1
− √ ~i + √ ~j
5
5
1 ∂z
2 ∂z
= −√
+√
.
5 ∂x
5 ∂y
1 ∂z
2 ∂z
2
−√
+√
= −√ ,
5 ∂x
5 ∂y
5
that is,
(1)
−
∂z
∂z
+2
= −2.
∂x
∂y
The directional derivative along u~2 is
zu~2 (1, 2) = ∇z ·
3~
4
i + ~j
5
5
=
3 ∂z
4 ∂z
+
,
5 ∂x
5 ∂y
14.4 SOLUTIONS
so
1303
3 ∂z
4 ∂z
+
= 1,
5 ∂x
5 ∂y
that is,
(2)
∂z
∂z
+4
= 5.
∂x
∂y
Now we solve the system of equations (1) and (2). Multiplying equation (1) by 3 gives
(3)
∂
∂z
+6
= −6.
−3
∂x
∂y
Adding (2) and (3) we get:
∂z
10
= −1.
∂y
So
∂z
= −0.1
∂y
and from equation (1)
∂z
∂z
1
=2
+ 2 = 2(− ) + 2 = 1.8.
∂x
∂y
10
3
77. Directional derivative = ∇f · ~
u , where ~
u = unit vector. If we move from (4, 5) to (5, 6), we move in the direction ~i + ~j
1 ~
1~
so ~
u = √ i + √ j . So,
2
2
1
1
∇f · ~
u = fx √
+ fy √
= 2.
2
2
2
1
u = √ ~i + √ ~j . So
Similarly, if we move from (4, 5) to (6, 6), the direction is 2~i + ~j so ~
5
5
∇f · ~
u = fx
Solving the system of equations for fx and fy
2
√
5
+ fy
1
√
5
= 3.
√
fx + fy = 2 2
√
2fx + fy = 3 5
gives
Thus at (4, 5),
√
√
fx = 3 5 − 2 2
√
√
fy = 4 2 − 3 5.
√
√
√
√
∇f = (3 5 − 2 2)~i + (4 2 − 3 5)~j .
78. (a) We have
x
gx = p
,
x2 + 3y + 3
3/2
,
gy = p
x2 + 3y + 3
so gx (1, 4) =
1
4
so gy (1, 4) =
3
8
1
3
grad g(1, 4) = ~i + ~j .
4
8
(b) Using the value g(1, 4) = 4 and the partial derivatives in part (a) we have
g(x, y) =
p
x2 + 3y + 3 ≈ 4 +
1
3
(x − 1) + (y − 4).
4
8
(c) Using the linearization in part (b), we have
g(1.01, 3.98) ≈ 4 +
3
1
(0.01) + (−0.02) = 3.995.
4
8
1304
Chapter Fourteen /SOLUTIONS
79. fx = 2x, fy = −2y, so grad f (3, −1) = 6~i + 2~j . For the direction θ = π/4, the direction is ~
u =
√
1 ~
1 ~
8
~
~
√
√
√
f~u (3, −1) = (6i + 2j ) · ( i +
j)=
= 4 2.
2
2
√1 ~
i
2
+
√1 ~
j
2
, so
2
The directional derivative is largest in the direction of the gradient vector grad f (3, −1) = 6~i + 2~j .
80. The temperature (Fahrenheit) as a function of position y in miles and time t in hours is given by
H = f (y, t) = 30 − 0.05y − 5t.
At time t the moose is at position y = 20t + C where C is an unknown constant that depends on where the moose was at
t = 0. At time t the moose perceives temperature H = g(t) = f (20t + C, t). The rate of change, g ′ (t), of temperature
with time, can be evaluated with the chain rule:
g ′ (t) =
∂f
∂f dy
·
+
= −0.05(20) − 5 = −6 ◦ /hour.
∂y dt
∂t
81. Assume that the x-axis points
√ east and the y-axis points north. We are given that k∇f k = 5 and that ∇f is in the direction
~i + ~j . Since k~i + ~j k = 2 and ∇f is a multiple of ~i + ~j , we have
5
∇f = √ (~i + ~j ).
2
The rate of change toward the north is the directional derivative in direction ~j , which is
5
5
∇f · ~j = √ (~i + ~j ) · ~j = √ .
2
2
82. Let’s put a coordinate plane on the area you are hiking, with your trail along the x-axis and the second trail branching off
at the origin as in Figure 14.20. You are moving in the positive x direction. Let h(x, y) be the elevation at the point (x, y)
on the mountain.
y
Branch trail
30◦
x
Your trail
Figure 14.20: Two trails
Since the trail along the x-axis ascends at a 20◦ angle, we have hx (0, 0) = tan 20◦ . Since the trail is the steepest
path, grad h must point along your trail in the positive x direction. Thus
grad h = hx~i + 0~j = tan 20◦~i .
We must compute the rate of change of elevation in the direction of the branch trail. The unit vector in this direction
is ~
u = cos 30◦~i + sin 30◦~j , and thus the directional derivative is
h~u = (grad h) · ~
u = (tan 20◦~i ) · (cos 30◦~i + sin 30◦~j ) = (tan 20◦ )(cos 30◦ ) = 0.3152.
The angle of ascent of the branch trail is thus tan−1 (0.3152) = 17.5◦ .
83. (a) P corresponds to greatest rate of increase of f and Q corresponds to greatest rate of decrease of f . See Figure 14.21.
(b) The points are marked in Figure 14.22.
(c) Amplitude is k grad f k. The equation is
f~u = k grad f k cos θ.
14.4 SOLUTIONS
Greatest rate of increase
of f in direction of P
f~u
✙
R (Zero f~u )
P (Max f~u )
❥
u
~
P
R
Q
1305
S
P
θ
grad f
θ
■
Q (Min f~u )
Greatest rate of decrease
of f in direction of Q
Figure 14.21
S (Zero f~u )
Figure 14.22
84. (a) Since f (x, y) = 5y − x2 − y 2 , we have ∇f (x, y) = −2x~i + (5 − 2y) ~j , so ∇f (1, 1) = √
−2~i + 3 ~j , which is
the direction of steepest ascent. Therefore, the initial rate of steepest ascent is k∇f (1, 1)k = 13 meters ascended
for each horizontal meter covered.
~ ~j . A unit vector that points in the
(b) In order to go straight northwest, we want to travel along the vector
√ ~v = −i +
√
same direction as ~v is therefore given by ~
u = ~v /k~v k = −(1/ 2)~i + (1/ 2) ~j , so
f~u (1, 1) = ∇f (1, 1) · ~
u
√
√
= (−2)(−1/ 2) + 3(1/ 2) = 3.54
meters ascended for each horizontal meter traveled.
(c) Here, we are being asked to determine vectors ~
u such that f~u (1, 1) = 0. Since f~u (1, 1) = 0 if and only if
(−2~i + 3~j ) · (u1~i + u2~j ) = 0 or −2u1 + 3u2 = 0, we see by inspection that ~
u = 3~i + 2~j and ~
u = −3~i − 2~j
give the two possible directions.
85. (a) To estimate the change in f , we use the gradient vector to estimate the change in f in moving from P to Q. Because
the contours are approximately parallel, moving from P to Q takes you to the same contour as moving from P to R.
(See Figure 14.23.) If θ is the angle between ~
u and grad f (a, b), then
Change in f
between P and Q
= Change in f
=
Rate of change
in direction P R
!
Distance traveled
between P and R
!
≈ k grad f k(h cos θ).
grad f
~
u
Q
✻
h θ
✛
h
sθ
co
P = (a, b)
R
✛
❄
Figure 14.23
(b) Since ~
u is a unit vector, we use the definition of f~u (a, b) to estimate
k grad f (a, b)kh cos θ
Change in f
≈
h
h
= k grad f (a, b)k cos θ = k grad f kk~
u k cos θ = grad f (a, b) · ~
u.
f~u (a, b) ≈
1306
Chapter Fourteen /SOLUTIONS
This approximation gets better as we choose h smaller and smaller, and in the limit we get the formula:
f~u (a, b) = grad f (a, b) · ~
u.
86. (a) The ellipse is a level curve of the function f (x, y) = x2 /2 + y 2 . The vector grad f = x~i + 2y~j is perpendicular
to the contours of f at the point (x, y). Thus, the vector w
~ = a~i + 2b~j is perpendicular to the line L tangent to the
ellipse at the point (a, b).
(b) Let ~
u be the vector from P = (−1, 0) to (a, b), so ~
u = (a + 1)~i + b~j . The distance from P to the line is
p=
a(a + 1) + 2b2
|a(a + 1) + 2(1 − a2 /2)|
|2 + a|
|~
u ·w
~|
√
= p
=
=
= √
2
2
kw
~k
a + 4b
4 − a2
a2 + 4(1 − a2 /2)
r
2+a
2−a
where the last equality follows from the substitution b2 = 1 − a2 /2 and the fact that |a| < 2.
(c) Let ~v be the vector from Q = (1, 0) to (a, b), so ~v = (a − 1)~i + b~j . The distance from Q to L is
(d) We have
a(a − 1) + 2b2
|a(a − 1) + 2b2 |
|2 − a|
|~v · w
~|
√
=
= p
=
= √
q=
2
2
kw
~k
a + 4b
4 − a2
a2 + 4(1 − a2 /2)
pq =
r
2+a
2−a
r
r
2−a
.
2+a
2−a
= 1.
2+a
87. (a) Let ~v = −fy (a, b)~i + fx (a, b)~j . Then ~v 6= 0 because grad f 6= 0. Since ~v · grad f (a, b) = (−fy (a, b)~i +
fx (a, b)~j ) · (fx (a, b)~i + fy (a, b)~j ) = 0 we see that the vector ~v is perpendicular to grad f (a, b). The contour C is
perpendicular to grad f (a, b) at the point (a, b) by the geometric property of the gradient vector. Since the vector ~v
and the contour C at (a, b) are both perpendicular to the same vector they are parallel to each other, which is another
way of saying that ~v is tangent to C at (a, b).
(b) A line parallel to a vector r~i + s~j with r 6= 0 has slope s/r. By part (a) the line tangent to C at (a, b) is parallel to
~v = r~i + s~j where r = −fy (a, b) and s = fx (a, b). If the tangent line is not vertical, then fy (a, b) 6= 0. Thus the
tangent line has slope s/r = −fx (a, b)/fy (a, b).
88. The direction of most rapid increase of the sum f + g is given by the vector grad(f (x, y) + g(x, y)) = grad f (x, y) +
grad g(x, y).
(a) A unit vector in the direction of w
~ is the vector ~
u =w
~ /kw
~ k. The rates of change of f and g in the direction of w
~
are the directional derivatives
f~u (x, y) = grad f (x, y) · ~
u = grad f (x, y) · w
~ /kw
~k
g~u (x, y) = grad g(x, y) · ~
u = grad g(x, y) · w
~ /kw
~ k.
The two directional derivatives are equal because
grad f (x, y) · w
~ = grad f (x, y) · (grad f (x, y) + grad g(x, y))
= k grad f (x, y)k2 + grad f (x, y) · grad g(x, y)
= k grad g(x, y)k2 + grad f (x, y) · grad g(x, y)
= grad g(x, y) · (grad f (x, y) + grad g(x, y))
= grad g(x, y) · w
~
(b) From the calculation in the solution to part (a), we have grad f (x, y)· w
~ = grad g(x, y)· w
~ . Since k grad f (x, y)k =
k grad g(x, y)k we have
grad g(x, y) · w
~
grad f (x, y) · w
~
=
k grad f (x, y)kkw
~k
k grad g(x, y)kkw
~k
which shows that the angle between w
~ and grad f (x, y) is the same as the angle between w
~ and grad g(x, y). Since
grad f (x, y) is perpendicular to the contour of f through P and grad g(x, y) is perpendicular to the contour of g
through P , this shows the w
~ makes equal angles with the two contours. Thus w
~ bisects the angle between the two
contours.
Strengthen Your Understanding
89. Directional derivatives are scalars, not vectors.
14.4 SOLUTIONS
1307
90. Gradients are vectors, not scalars.
91. The closer together the contours, the longer the gradient vector.
92. We have
f~u (0, 0) = ~
u · gradf (0, 0) = ~
u · 2~i + 3~j .
Many unit vectors make this dot product negative, for example ~
u = −i or ~
u = −j.
93. A possible answer is in Figure 14.24, where the gradient at P is shorter than the gradient at Q because the contours are
closer at Q than at P .
y
3
6
Q
2
4
2
8
P
1
x
6
4
8
−2
2
−1
−3
−3 −2 −1
1
2
3
Figure 14.24
94. (a) The gradient vector ∇f (P ) is perpendicular to the contour of f that goes through the point P . It points in the
direction of maximal positive rate of change of f at the point P .
(b) The magnitude k∇f (P )k is the directional derivative of f at the point P in the direction of the gradient vector
∇f (P ) itself. Thus the magnitude equals the maximum directional rate of change of f at the point P .
(c) Given a unit vector ~
u , the dot product ∇f (P ) · ~
u equals the directional derivative f~u (P ) of f at the point P in the
direction of ~
u.
95. False. For example, suppose f is the linear function f (x, y) = x + y. Then fx = fy = 1 at all points. Consider the
contour f = 1, which is the line x + y = 1 of slope −1. There is no point on this contour where the slope is fy /fx = 1.
96. False. The gradient vector of a function of two variables, f (x, y), is a vector in 2-space given by fx (a, b)~i + fy (a, b)~j .
97. False. Left side is a vector, right side is a scalar.
98. False. The gradient is perpendicular to the contour of f at (a, b).
99. True. If ~
u is a unit vector, then the directional derivative is given by the formula f~u (a, b) = grad f (a, b) · ~
u.
100. True. The components of the gradient vector are the x and y partial derivatives, fx (a, b) and fy (a, b). These are the
directional derivatives of f in the ~i and ~j directions, respectively.
101. False. The directional derivative is a scalar, not a vector.
102. False. The gradient vector at (3, 4) has no relation to the direction of the vector 3~i +4~j . For example, if f (x, y) = x+2y,
then grad f = ~i + 2~j , which is not perpendicular to 3~i + 4~j . The gradient vector grad f (3, 4) is perpendicular to the
contour of f passing through the point (3, 4).
103. True. The gradient points in the direction of maximal increase of f , and the opposite direction gives the direction of
maximum decrease for f .
104. False. The gradient points in the direction of greatest local maximum increase. This means that f increases in the ~i
direction only near the point (1, 2). We cannot conclude that f keeps increasing in that direction as far away from (1, 2)
as (10, 2).
105. True. The length of the gradient gives the maximal rate of increase. We have grad g(3, 0) = 6~i and grad h(3, 0) = 6~j ,
so k grad g(3, 0)k = k grad h(3, 0)k = 6.
106. True. The length of the gradient
gives the maximal directional derivative in any direction. The gradient vector is grad f (0, 0) =
√
~i + ~j , which has length 2.
107. True. It is the rate of change of f in the direction of ~
u at the point (x0 , y0 ).
1308
Chapter Fourteen /SOLUTIONS
108. False. f~u (a, b) = k∇f (a, b)k cos θ, where θ is the angle between grad f and ~
u.
109. Must be true, because at any point grad f is perpendicular to level curves through that point.
110. True. Take the direction perpendicular to grad f at that point. If grad f = 0, any direction will do.
111. Is never true. If k grad f k = 0, then grad f = 0, so grad f · ~
u = 0 for any unit vector ~
u . Thus the directional derivative
must be zero.
Solutions for Section 14.5
Exercises
1. Since fx = 2x, fy = 0 and fz = 0, we have
grad f = 2x~i .
2. We have fx = 2x, fy = 3y 2 , and fz = −4z 3 . Thus
grad f = 2x~i + 3y 2~j − 4z 3~k .
3. Since f (x, y, z) = ex+y+z = ex ey ez , we have fx = ex ey ez , fy = ex ey ez and fz = ex ey ez , so
grad f = ex ey ez (~i + ~j + ~k ).
4. Since fx = − sin(x + y), fy = − sin(x + y) + cos(y + z), fz = cos(y + z), we have
grad f = − sin(x + y)~i + (cos(y + z) − sin(x + y))~j + cos(y + z)~k .
5. Since fx = −2xyz 2 /(1 + x2 )2 , fy = z 2 /(1 + x2 ), and fz = 2yz/(1 + x2 ),
grad f =
z2 ~
2yz ~
−2xyz 2 ~
i +
j +
k.
2
2
(1 + x )
1 + x2
1 + x2
6. We have fx = −2x/(x2 + y 2 + z 2 )2 , fy = −2y/(x2 + y 2 + z 2 )2 , and fz = −2z/(x2 + y 2 + z 2 )2 . Thus
−2
(x~i + y~j + z~k ).
(x2 + y 2 + z 2 )2
grad f =
p
7. We have fx = x/
p
x2 + y 2 + z 2 , fy = y/
p
x2 + y 2 + z 2 , and fz = z/
x2 + y 2 + z 2 . Thus
1
grad f = p
(x~i + y~j + z~k ).
x2 + y 2 + z 2
8. We have fx = ey sin z, fy = xey sin z, and fz = xey cos z. Thus
grad f = ey sin z~i + xey sin z~j + xey cos z~k .
9. Since fx = y, fy = x, and fz = ez cos (ez ),
grad f = y~i + x~j + ez cos (ez ) ~k .
10. Since fx1 = 2x1 x32 x43 , fx2 = 3x21 x22 x43 , fx3 = 4x21 x32 x33 , we have
grad f = (2x1 x32 x43 )~i + (3x21 x22 x43 )~j + (4x21 x32 x33 )~k .
14.5 SOLUTIONS
1309
2
11. Since fp = ep , fq = 1/q, fr = 2rer , we have
2
1
grad f = ep~i + ~j + 2rer ~k .
q
12. We have
2
grad(ez + y ln(x2 + 5)) = y
2
2x ~
i + ln(x2 + 5)~j + 2zez ~k .
x2 + 5
13. We have fx = 0, fy = 2yz and fz = y 2 . Thus grad f = 2yz~j + y 2~k and grad f (1, 0, 1) = ~0 .
14. We have fx = 2, fy = 3 and fz = 4 so grad f = 2~i + 3~j + 4~k at all points.
15. We have
grad(x2 + y 2 − z 4 )
(3,2,1)
= 2x~i + 2y~j − 4z 3~k
(3,2,1)
= 6~i + 4~j − 4~k .
16. We have fx = yz, fy = xz, and fz = yz. Thus fx (1, 2, 3) = 6, fy (1, 2, 3) = 3, and fz (1, 2, 3) = 2, so
grad f = 6~i + 3~j + 2~k .
17. We have
= y cos(xy)~i + (x cos(xy) + z cos(yz))~j + y cos(yz)~k
grad(sin(xy) + sin(yz))
(1,π,−1)
(1,π,−1)
= −π~i − π~k .
18. We have
grad(x ln(yz))
(2,1,e)
x
x
= ln(yz)~i + ~j + ~k
y
z
(2,1,e)
2
= ~i + 2~j + ~k .
e
19. We have grad f = y~i + x~j + 2z~k , so grad f (1, 2, 3) = 2~i + ~j + 6~k . A unit vector in the direction we want is
√
u = (1/ 3)(~i + ~j + ~k ). Therefore, the directional derivative is
grad f (1, 2, 3) · ~
u =
9
2·1+1·1+6·1
√
= √ .
3
3
20. We have grad f = y~i + x~j + 2z~k , so grad f (1, 1, 1) = ~i + ~j + 2~k . A unit vector in the direction we want is
√
u = (1/ 14)(~i + 2~j + 3~k ). Therefore, the directional derivative is
grad f (1, 1, 1) · ~
u =
9
1·1+1·2+2·3
√
= √ .
14
14
21. We have grad f = y~i + x~j + 2z~k , so grad f (1, 1, 0) = ~i + ~j . A unit vector in the direction we want is u =
√
(1/ 2)(−~i + ~k ). Therefore, the directional derivative is
grad f (1, 1, 0) · ~
u =
−1
1(−1) + 1 · 0 + 0 · 1
√
= √ .
2
2
22. We have grad f = y~i + x~j + 2z~k , so grad f (0, 1, 1) = ~i + 2~k . A unit vector in the direction we want is u =
√
(1/ 2)(−~i + ~k ). Therefore, the directional derivative is
grad f (0, 1, 1) · ~
u =
1(−1) + 0 · 0 + 2 · 1
1
√
= √ .
2
2
1310
Chapter Fourteen /SOLUTIONS
23. We have grad f = y~i + x~j + 2z~k , so grad f (2, 3, 4) = 3~i + 2~j + 8~k . Let ~
u be a unit vector making an angle of
3π/4 with grad f (2, 3, 4). Then, the directional derivative is
3π
grad f (2, 3, 4) · u
~ = k grad f (2, 3, 4)kk~
u k cos
4
=
√
77(1)
−1
√
2
=−
r
77
.
2
24. We have grad f = y~i + x~j + 2z~k , so grad f (2, 3, 4) = 3~i + 2~j + 8~k . The maximum rate of change of f at (2, 3, 4) is
in the direction of grad
√ f (2, 3, 4) and the directional derivative in that direction is the maximum rate of change, namely
k grad f (2, 3, 4)k = 77.
25. First, we check that (−1)2 − (1)2 + 22 = 4. Then let f (x, y, z) = x2 − y 2 + z 2 so that the given surface is the level
surface f (x, y, z) = 4. Since fx = 2x, fy = −2y, and fz = 2z, we have grad f (−1, 1, 2) = −2~i − 2~j + 4~k . Since
gradients are perpendicular to level surfaces, a vector normal to the surface at (−1, 1, 2) is ~n = −2~i − 2~j + 4~k . Thus
an equation for the tangent plane is
−2(x + 1) − 2(y − 1) + 4(z − 2) = 0.
26. First, we check that 2 = (−1)2 + (1)2 . Then let f (x, y, z) = z − x2 − y 2 so that the given surface is the level surface
f (x, y, z) = 0. Since fx = −2x, fy = −2y, and fz = 1, we have grad f (−1, 1, 2) = 2~i − 2~j + ~k . Since gradients
are perpendicular to level surfaces, a vector normal to the surface at (−1, 1, 2) is ~n = 2~i − 2~j + ~k . Thus an equation
for the tangent plane is
2(x + 1) − 2(y − 1) + (z − 2) = 0.
Note that you could also view the surface as the graph of the function z = g(x, y) = x2 + y 2 and get the equation of the
tangent plane using the local linearization of g.
27. First, we check that 1 = 22 − 3. Then we let f (x, y, z) = y 2 − z 2 + 3, so that the given surface is the level surface
f (x, y, z) = 0. Since fx = 0, fy = 2y, and fz = −2z, we have grad f (−1, 1, 2) = 2~j − 4~k . Since gradients are
perpendicular to level surfaces, a vector normal to the surface at (−1, 1, 2) is ~
n = 2~j − 4~k . Thus an equation for the
tangent plane is
2(y − 1) − 4(z − 2) = 0.
28. First, we check that (−1)2 − (−1)(1)(2) = 3. Then let f (x, y, z) = x2 − xyz so that the given surface is the level
surface f (x, y, z) = 3. Since fx = 2x − yz, fy = −xz, and fz = −xy, we have grad f (−1, 1, 2) = −4~i + 2~j + ~k .
Since gradients are perpendicular to level surfaces, a vector normal to the surface at (−1, 1, 2) is ~n = −4~i + 2~j + ~k .
Thus an equation for the tangent plane is
−4(x + 1) + 2(y − 1) + (z − 2) = 0.
29. First, we check that cos(−1 + 1) = e−2+2 . Then we let f (x, y, z) = cos(x + y) − exz+2 , so that the given surface
is the level surface f (x, y, z) = 0. Since fx = − sin(x + y) − zexz+2 , fy = − sin(x + y), and fz = −xexz+2 , we
have grad f (−1, 1, 2) = −2~i + ~k . Since gradients are perpendicular to level surfaces, a vector normal to the surface at
(−1, 1, 2) is ~
n = −2~i + ~k . Thus an equation for the tangent plane is
−2(x + 1) + (z − 2) = 0.
30. First, we check that 1 = 4/(2(−1) + 3(2)). We could let f (x, y, z) = y − 4/(2x + 3y), but instead let’s try f (x, y, z) =
y(2x + 3z) so that the given surface is the level surface f (x, y, z) = 4. Since fx = 2y, fy = 2x + 3z, and fz = 3y,
we have grad f (−1, 1, 2) = 2~i + 4~j + 3~k . Since gradients are perpendicular to level surfaces, a vector normal to the
surface at (−1, 1, 2) is ~n = 2~i + 4~j + 3~k . Thus an equation for the tangent plane is
2(x + 1) + 4(y − 1) + 3(z − 2) = 0.
14.5 SOLUTIONS
31. (a) The unit vector ~
u 1 in the direction of ~v 1 = ~i − ~k is ~
u1 =
1 ~
√
i
2
fx (x, y, z) = 6xy 2 ,
−
√1 ~
k
2
1311
. We have
and fx (−1, 0, 4) = 0
2
fy (x, y, z) = 6x y + 2z, and fy (−1, 0, 4) = 8
fz (x, y, z) = 2y,
and fz (−1, 0, 4) = 0.
So,
f~u 1 (−1, 0, 4) = fx (−1, 0, 4)
=0
1
√
2
= 0.
u 2 = − √119~i +
(b) The unit vector ~
derivatives from part (a),
√3 ~
j
19
1
√
2
1
+ 8(0) + 0 − √
2
√3 ~
k
19
+
1
+ fy (−1, 0, 4)(0) + fz (−1, 0, 4) − √
2
is in the direction of ~v 2 = −~i + 3~j + 3~k . Using the partial
1
f~u 2 (−1, 0, 4) = 0 − √
19
+8
3
√
19
+0
3
√
19
24
= √ .
19
32. The unit vector ~
u in the direction of ~v = 2~i + ~j − 2~k is
~
u =
~v
=
k~v k
The partial derivatives are
2 ~
i +
3
1 ~
j −
3
2 ~
k.
3
fx (x, y, z) = 2x + 3y,
fy (x, y, z) = 3x,
fz (x, y, z) = 2.
Thus, we have
f~u (2, 0, −1) = fx (2, 0, −1)
=4
2
3
+6
2
3
1
3
+ fy (2, 0, −1)
+2 −
2
3
=
1
3
+ fz (2, 0, −1) −
10
3
2
3
33. (a) We have fx = 2x − yz, fy = 2y − xz and fz = −xy so
grad f = (2x − yz)~i + (2y − xz)~j − xy~k .
(b) At the point (2, 3, 1) we have
grad f (2, 3, 1) = ~i + 4~j + 6~k .
Thus an equation of the tangent plane to the level surface at the point (2, 3, 1) is
(x − 2) + 4(y − 3) + 6(z − 1) = 0
or
x + 4y + 6z = 20.
34. Since
we have
−x
zx = p
17 − x2 − y 2
z(3, 2) =
√
17 − 9 − 4 = 2,
and
−y
zy = p
,
17 − x2 − y 2
zx (3, 2) = −
3
2
zy (3, 2) =
−2
= −1.
2
1312
Chapter Fourteen /SOLUTIONS
The tangent plane to z =
p
17 − x2 − y 2 at (x, y) = (3, 2) is
z = z(3, 2) + zx (3, 2)(x − 3) + zy (3, 2)(y − 2) = 2 + (
17
3
−3
)(x − 3) + (−1)(y − 2) =
− x − y,
2
2
2
or
2z + 3x + 2y = 17.
35. Since
zx = −8/yx2
we have
and
zy = −8/xy 2 ,
zx (1, 2) = −8/(2)(1)2 = −4,
z(1, 2) = 8/(1)(2) = 4,
The tangent plane to z = 8/xy at (x, y) = (1, 2) is
zy (1, 2) = −8/(1)(2)2 = −2.
z = z(1, 2) + zx (1, 2)(x − 1) + zy (1, 2)(y − 2) = 4 − 4(x − 1) − 2(y − 2) = 12 − 4x − 2y.
36. The surface is given by F (x, y, z) = 0 where F (x, y, z) = x − y 3 z 7 . The normal direction is
grad F =
∂F ~
∂F ~
∂F ~
i +
j +
k = ~i − 3y 2 z 7~j − 7y 3 z 6~k .
∂x
∂y
∂z
Thus, at (1, −1, −1) a normal vector is ~i + 3~j + 7~k . The tangent plane has the equation
1(x − 1) + 3(y − (−1)) + 7(z − (−1)) = 0
x + 3y + 7z = −9.
37. At the point P = (1, 2, 3) we have grad f = 2 · 3~i + 1 · 3~j + 1 · 2~k = 6~i + 3~j + 2~k . Hence the tangent plane to the
level surface f (x, y, z) = 0 at the point P is given by the equation
6(x − 1) + 3(y − 2) + 2(z − 1) = 0.
38. At the point P = (10, −10, 30) we have grad f = 2 · 10~i + 302~j + 2(−10)30~k = 20~i + 900~j − 600~k . Hence the
tangent plane to the level surface f (x, y, z) = 0 at the point P is given by the equation
20(x − 10) + 900(y + 10) − 600(z − 30) = 0.
39. A normal to the surface is 2x~i + 2y~j + 2z~k ; a normal to the tangent plane at (2, 3, 2) is
~
n = 4~i + 6~j + 4~k .
The tangent plane can be written as
or
4(x − 2) + 6(y − 3) + 4(z − 2) = 0
4x + 6y + 4z = 4 · 2 + 6 · 3 + 4 · 2 = 34
2x + 3y + 2z = 17.
14.5 SOLUTIONS
1313
40. Let f (x, y, z) = x2 + y 2 so that the surface is the level surface f (x, y, z) = 1. Since
grad f = 2x~i + 2y~j
we have
grad f (1, 0, 1) = 2~i .
Thus an equation of the tangent plane at the point (1, 0, 1) is
2(x − 1) + 0(y − 0) + 0(z − 1) = 0
or
2(x − 1) = 0.
The tangent plane is given by the equation
x = 1.
41. Since z = 2x + y + 3 is a plane it is its own tangent plane.
42. Let f (x, y, z) = 3x2 − 4xy + z 2 so that the surface is the level surface f (x, y, z) = 0. Since
grad f = (6x − 4y)~i − 4x~j + 2z~k ,
we have
grad f (a, a, a) = 2a~i − 4a~j + 2a~k .
Thus an equation of the tangent plane at (a, a, a) is
2a(x − a) − 4a(y − a) + 2a(z − a) = 0
or
2ax − 4ay + 2az = 0.
Since a 6= 0, we can write the plane more simply as
2x − 4y + 2z = 0.
Notice that it is the same plane for every a.
43. The point on the surface z = 9/(x + 4y) where x = 1 and y = 2 has third coordinate z = 9/(1 + 8) = 1. We want the
tangent plane to the surface at the point (1, 2, 1).
Let f (x, y, z) = z(x + 4y) so that the given surface is the level surface f (x, y, z) = 9. Since
grad f = z~i + 4z~j + (x + 4y)~k
we have
grad f (1, 2, 1) = ~i + 4~j + 10~k .
Thus an equation of the tangent plane at (1, 2, 1) is
(x − 1) + 4(y − 2) + 9(z − 1) = 0
so
x + 4y + 9z = 18.
Problems
44. The point (−1, −1, 3) lies above the point (−1, −1). The vector grad g(−1, −1) points horizontally in the direction in
which g increases most rapidly and lies directly under the path of steepest ascent. (See Figures 14.25 and 14.26.)
1314
Chapter Fourteen /SOLUTIONS
y
1 2
3
z
4
3
2 1
x
✲
x
y
Figure 14.25: Contour diagram for
z = g(x, y) = 4 − x2 showing direction of
grad g(−1, −1)
Figure 14.26: Graph of g(x, y) = 4 − x2
showing path of steepest ascent from the
point (−1, −1, 3)
45. The gradient of (a) is 2x~i + 2y~j + 2z~k , which points radially outward from the origin, so (a) goes with (III).
The gradient of (c) is parallel to the gradient of (a) but pointing inward, so (c) goes with (IV).
The gradient of (b) is 2x~i + 2y~j , which points radially outward from the z-axis, so (b) goes with (I).
The gradient of (d) is parallel to the gradient of (b) but pointing inward, so (d) goes with (II).
46. (a) The surface is the level surface F (x, y, z) = 7, where F (x, y, z) = x2 + y 2 − xyz. Thus the normal vector to the
tangent plane is grad F = (2x − yz)~i + (2y − xz)~j + (−xy)~k . Evaluated at (2, 3, 1), we get the normal to the
plane
~n = ~i + 4~j − 6~k .
Thus the equation of the plane is
(x − 2) + 4(y − 3) − 6(z − 1) = 0.
(b) Solving x2 + y 2 − xyz = 7 for z, we get
x2 + y 2 − 7
.
z=
xy
Thus, we have
x
y
7
x2 + y 2 − 7
= + −
.
f (x, y) =
xy
y
x
xy
We have
1
y
7
fx (x, y) = − 2 + 2
y
x
x y
x
1
7
fy (x, y) = − 2 + +
.
y
x
xy 2
Thus fx (2, 3) = 1/3 − 3/4 + 7/12 = 1/6 and fy (2, 3) = −2/9 + 1/2 + 7/18 = 2/3. Thus the equation of the
tangent plane is
z = 1 + (1/6)(x − 2) + (2/3)(y − 3).
This is the same as the answer to part (a) when that equation is solved for z.
47. At the point (1, 2, 1), we have
grad f (1, 2, 1) = 2 · 1~i + 2 · 2~j + 2 · 1~k = 2~i + 4~j + 2~k .
The normal to the plane pointing away from the origin is
~n = ~i + 2~j + 3~k .
Thus we find the rate of change of f in the direction of the unit vector
~n
1
~
n
= √
= √ (~i + 2~j + 3~k ).
2
2
2
||~n ||
1 +2 +3
14
The rate of change is given by the directional derivative
~
u =
(~i + 2~j + 3~k )
22
= √ .
√
f~u = grad f · ~
u = 2~i + 4~j + 4~k ·
14
14
14.5 SOLUTIONS
1315
48. (a) To get a normal vector to the surface z = cos x sin y at the point (0, π/2, 1), we first represent the surface S by the
equation F (x, y, z) = z − cos x sin y = 0. Then we calculate the gradient of F which is normal to S
grad F = sin x sin y~i − cos x cos y~j + ~k .
At the point (0, π/2, 1),
grad F (0, π/2, 1) = ~k .
(b) The plane with normal ~k and through the point (0, π/2, 1) is
z = 1.
So z = 1 is the equation of the tangent plane at the point (0, π/2, 1)
49. (a) Since f (x, y, z) = sin(x2 + y 2 + z 2 ), the level surfaces of f are spheres centered at the origin.
(b) Since fx = 2x sin(x2 + y 2 + z 2 ) and fy = 2y sin(x2 + y 2 + z 2 ) and fz = 2z sin(x2 + y 2 + z 2 ), we have
grad f = 2x sin(x2 + y 2 + z 2 )~i + 2y sin(x2 + y 2 + z 2 )~j + 2z sin(x2 + y 2 + z 2 )~k .
(c) We can write the formula for grad f as
grad f = 2~r sin(x2 + y 2 + z 2 ),
so grad f is parallel to ~r at the point (x, y, z). Thus, the angle between grad f and ~r is 0◦ or 180◦ . The angle is 0◦
if sin(x2 + y 2 + z 2 ) is positive at that point and 180◦ if sin(x2 + y 2 + z 2 ) is negative at that point. (We are assuming
that sin(x2 + y 2 + z 2 ) 6= 0.)
50. (a) To do this you need to imagine the surfaces with the normal vector ~n at P .
For (I),
For (II),
~n is (−, −, +)
~
n is (+, +, +)
or (+, +, −)
or (−, −, −)
so (E)
so (F)
(b) Since the equation of the tangent plane is of the form
For (III),
~n is (+, −, +)
or (−, +, −)
so (G)
For (IV),
~n is (−, +, +)
or (+, −, −)
so (H)
n1 x + n2 y + n3 z = k,
the coefficients of x, y, z in the plane must have the same sign as the components of the normal. Hence (I)-(E)-(L);
(II)-(F)-(J); (III)-(G)-(M); (IV)-(H)-(K).
51. (a) A normal vector at (x, y, z) is given by
∂F ~
∂F ~
∂F ~
i +
j +
k
∂x
∂y
∂z
1
2y
= 2x~i − 2 ~j + 3 ~k
z
z
∇F (x, y, z) =
The direction of maximum increase of F is ∇F . At (0, 0, 1) this is −~j , which is a unit vector. So u~1 = −~j .
~
~ ~
At (1, 1, 1) this is 2~i − ~j + 2~k , so the unit vector u~2 = √ 22i −j +22k 2 = 32~i − 13 ~j + 32 ~k .
2 +(−1) +2
2·0 ~
k =
13
(b) A normal vector at (0, 0, 1) is ∇f (0, 0, 1) = (2 · 0)~i − 112 ~j +
The equation of the tangent plane at the point (0, 0, 1) is
−~j
((x − 0)~i + (y − 0)~j + (z − 1)~k ) · (−~j ) = 0,
i.e.,
−y = 0,
so y = 0.
A normal vector at (1, 1, 1) is ∇f (1, 1, 1) = (2 · 1)~i − ( 112 )~j + ( 2·1
)~k = 2~i − ~j + 2~k .
13
The equation of the tangent plane at (1, 1, 1) is
((x − 1)~i + (y − 1)~j + (z − 1)~k ) · (2~i − ~j + 2~k ) = 0,
i.e.,
so
2x − 2 − y + 1 + 2z − 2 = 0,
2x − y + 2z = 3.
1316
Chapter Fourteen /SOLUTIONS
(c) The vector ∇F is parallel to the xy-plane when it is perpendicular to ~k , i.e. when
2y
1
(2x~i − 2 ~j + 3 ~k ) · ~k = 0,
z
z
that is,
2y
= 0 so y = 0.
z3
Points that meet these requirements are those (x, y, z) such that
x2 −
y
= 0 and
z2
y = 0,
i.e., points such that x = y = 0 and z 6= 0, that is, the z-axis minus the origin. (Observe that we must exclude the
points where z = 0 because the surface is not defined there: the expression x2 − (y/z 2 ) is undefined when z = 0.)
52. (a) The vector grad f (x, y) is perpendicular to the level curve of f through (x, y):
grad f (x, y) = (ex − 1) cos y~i − (ex − x) sin y~j
Thus, at the point (2, 3),
grad f (2, 3) = (e2 − 1) cos 3~i − (e2 − 2) sin 3~j
The vector grad f points in the direction of greatest increase in f , so the vector we want is
− grad f (2, 3) = −(e2 − 1) cos 3~i + (e2 − 2) sin 3~j
= 6.33~i + 0.76~j
(b) To find a vector normal to the surface, we write the surface in the form
F (x, y, z) = (ex − x) cos y − z = 0.
Then
So, at P ,
grad F = (ex − 1) cos y~i − (ex − x) sin y~j − ~k .
grad F = (e2 − 1) cos 3~i − (e2 − 2) sin y~j − ~k
= −6.33~i − 0.76~j − ~k .
The vector ~v is perpendicular to grad F , so
~v · grad F = (5~i + 4~j + a~k ) · (−6.33~i − 0.76~j − ~k ) = 0.
This gives
so
−5(6.33) − 4(0.76) − a = 0
a = −34.69
53. (a) A normal to the surface is given by the gradient of the function f (x, y, z) = x2 + y 2 + 3z 2 ,
grad f = 2x~i + 2y~j + 6z~k .
At the point (0.6, 0.8, 1), a normal to the surface and to the tangent plane is
~n = 1.2~i + 1.6~i + 6~k .
Since the plane goes through the point (0.6, 0.8, 1), its equation is
1.2(x − 0.6) + 1.6(y − 0.8) + 6(z − 1) = 0
1.2x + 1.6y + 6z = 8.
14.5 SOLUTIONS
1317
(b) We want to know if there are x, y, z values such that a normal to the surface is parallel to the normal to the plane.
That is, is 2x~i + 2y~j + 6z~k parallel to 8~i + 6~j + 30~k ? Yes, if 2x = 8t, 2y = 6t, 6z = 30t for some value
of t. That is, if
x = 4t, y = 3t, z = 5t.
Substituting these equations into the equation for the surface and solving for t, we get
(4t)2 + (3t)2 + 3 (5t)2 = 4
100t2 = 4
t=±
r
4
= ±0.2.
100
So there are two points on the surface at which the tangent plane is parallel to the plane 8x + 6y + 30z = 1. They
are
±(0.8, 0.6, 1).
54. (a) The height is given by f (4, 3) = 2(42 ) − (32 ) = 23. Your house is 23 units above the xy-plane.
(b) We want a directional derivative so we start by computing the gradient. Since the partial derivatives are fx = 4x
and fy = −2y, we have fx (4, 3) = 16 and fy (4, 3) = −6. The gradient vector is grad f = 16~i − 6~j . Letting
~v = −4~i −3~j , we have k~v k = 5, so a unit vector in the same direction is ~
u = (−4/5)~i −(3/5)~j = −0.8~i −0.6~j .
The directional derivative in the direction of ~
u at the point (4, 3) is
f~u (4, 3) = grad f (4, 3) · ~
u
~
~
= (16i − 6j ) · (−0.8~i − 0.6~j )
= −9.2.
The ground slopes down in that direction quite steeply, going down at a rate of 9.2 units for every one horizontal unit
moved.
(c) The water runs off in the direction of minimum slope, which is the direction of the negative of the gradient vector at
that point. We already saw that grad f (4, 3) = 16~i − 6~j , so the water runs off in the direction of −16~i + 6~j .
(d) We can write the surface z = 2x2 − y 2 as F (x, y, z) = 2x2 − y 2 − z = 0. Then Fx = 4x and Fy = −2y and
Fz = −1. At the point (4, 3, 23), we have Fx (4, 3, 23) = 16 and Fy (4, 3, 23) = −6 and Fz (4, 3, 23) = −1. The
equation of the tangent plane is
16(x − 4) − 6(y − 3) − 1(z − 23) = 0
16x − 6y − z = 23
55. (a) See Figure 14.27.
y
3
1
✻
2
2
1
✒
0
■
■
−
1
−1
−2
π
2
π
x
3π
2
2π
Figure 14.27
(b) The bug is walking parallel to the y-axis. Looking to the right or left, the bug sees higher contours — thus it is in a
valley.
(c) See Figure 14.27.
1318
Chapter Fourteen /SOLUTIONS
56. (a) The plane z = 5 is horizontal. The surface z = 1 + x2 + y 2 is bowl-shaped, with its lowest point at (0, 0, 1). At this
point its tangent plane is horizontal and therefore parallel to the plane z = 5.
(b) The tangent plane to the surface z = f (x, y) at the point where (x, y) = (a, b) has equation
z = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b).
Thus, for z = 1 + x2 + y 2 , the tangent plane is
z = (1 + a2 + b2 ) + 2a(x − a) + 2b(y − b)
= (1 − a2 − b2 ) + 2ax + 2by.
This is parallel to z = 5 + 6x − 10y when 6 = 2a, and −10 = 2b so a = 3, b = −5. Then z = f (3, −5) =
1 + 32 + (−5)2 = 35, so the point on the surface whose tangent plane is parallel to z = 5 + 6x − 10y is (3, −5, 35).
57. (a) In the direction of grad F :
= (2x + 2xz 2 )~i + (4y 3 )~j + (2x2 z)~k
grad F
(−1,1,1)
(−1,1,1)
(b) The rate of change in the direction of grad F with respect to distance = k∇F k =
rate of change with respect to time. If we move at 4 units/sec:
Rate of change of
= −4~i + 4~j + 2~k .
√
16 + 16 + 4 = 6. Now we want
Conc
Dist
Conc
= Rate of change of
× Rate of change of
Time
Dist
Time
= 6 × 4 = 24mg/cm3 /sec.
58. The tangent plane is given by
so
3(x − 2) + 10(y − 1) − 5(z − 7) = 0
3x + 10y − 5z + 19 = 0.
59. Since f~u = ∇f · ~
u , we see that f~u 1 and fu~ 4 are both positive because the angles between u
~ 1 and ∇f and between ~u 4
and ∇f are both between 0 and π/2. In addition, f~u 1 is larger because the angle between ~u 1 and ∇f is smaller than the
angle between ~
u 4 and f~ . Similarly, f~u 2 and fu~ 3 are both negative, and f~u 3 is more negative. Thus,
fu~ 3 < f~u 2 < 0 < f~u 4 < f~u 1
60. (a) We have ∇G = (2x − 5y)~i + (−5x + 2yz)~j + (y 2 )~k , so ∇G(1, 2, 3) = −8~i + 7~j + 4~k . The rate of change is
given by the directional derivative in the direction ~v :
(2~i + ~j − 4~k )
~v
√
= (−8~i + 7~j + 4~k ) ·
k~v k
21
−25
−16 + 7 − 16
√
= √ ≈ −5.455.
=
21
21
Rate of change in density = ∇G ·
(b) The direction of maximum rate of change is ∇G(1,p
2, 3) = −8~i + 7~j + 4~k .
√
(c) The maximum rate of change is k∇G(1, 2, 3)k = (−8)2 + 72 + 42 = 129 ≈ 11.36.
61. (a) The function T (x, y, z) = constant where x2 + y 2 + z 2 = constant. These surfaces are spheres centered at the origin.
(b) Calculating the partial derivative with respect to x gives
2
2
2
∂T
= −2xe−(x +y +z ) .
∂x
Similar calculations for the other variables shows that
2
2
2
grad T = (−2x~i − 2y~j − 2z~k )e−(x +y +z ) .
14.5 SOLUTIONS
(c) At the point (1, 0, 0)
1319
grad T (1, 0, 0) = −2e−1~i .
Moving from the point (1, 0, 0) to (2, 1, 0), you move in the direction
(2 − 1)~i + (1 − 0)~j = ~i + ~j .
A unit vector in this direction is
~i + ~j
~
u = √ .
2
The directional derivative of T (x, y, z) in this direction at the point (1, 0, 0) is
√
~i + ~j
= − 2e−1 .
T~u (1, 0, 0) = −2e−1~i · √
2
Since you are moving at a speed of 3 units per second,
√
√
Rate of change of temperature = − 2e−1 · 3 = −3 2e−1 degrees/second.
62. We have
√
√
2
2
2
2
2
2
∂P
1
x
,
= 5e−0.1 x +y +z (−0.1)(2x) (x2 + y 2 + z 2 )−1/2 = −0.5e−0.1 x +y +z p
2
∂x
2
x + y2 + z2
and similarly
√
2
2
2
∂P
y
= −0.5e−0.1 x +y +z p
,
∂y
x2 + y 2 + z 2
and
Thus the gradient of P at (0, 0, 1) is
grad P =
∂P
∂x
(x,y,z)=(0,0,1)
~i + ∂P
∂y
√
2
2
2
∂P
z
= −0.5e−0.1 x +y +z p
.
∂z
x2 + y 2 + z 2
~j + ∂P
∂z
(x,y,z)=(0,0,1)
Let ~
u be a unit vector in the direction of ~v , so
u =
~
Then
(x,y,z)=(0,0,1)
~k = −0.5e−0.1~k .
~v
.
k~v k
Rate of change of pressure in atm/sec = (Directional derivative in direction ~
u in atm/mi)(Speed of spacecraft in mi/sec)
~v
k~v k = grad P · ~v
= P~u k~v k = (grad P · ~
u )k~v k = grad P ·
k~v k
= −0.5e−0.1~k · (~i − 2.5~k ) = 0.5 · 2.5e−0.1 = 1.131 atm/sec.
63. (a) is (V) since ~r + ~a is a vector not a scalar.
(b) is (IV) since grad(~r · ~a ) = grad(a1 x + a2 y + a3 z) = ~a .
(c) is (V) since ~r × ~a is a vector not a scalar.
64. We must calculate the gradient of ϕ.
grad ϕ(x, y, z) = grad
GmM
k~r k
1
= GM m grad p
2
x + y2 + z2
Now
∂
∂x
1
p
x2
+
y2
+
z2
!
=
−x
∂
1
.
(x2 + y 2 + z 2 )−1/2 = − (x2 + y 2 + z 2 )−3/2 2x = 2
∂x
2
(x + y 2 + z 2 )3/2
1320
Chapter Fourteen /SOLUTIONS
The partial derivatives with respect to y and z are similar, so
x~i + y~j + z~k
1
.
=− 2
grad p
2
2
2
(x + y 2 + z 2 )3/2
x +y +z
Thus,
x~i + y~j + z~k
(x2 + y 2 + z 2 )3/2
~r
= −GM m
k~r k3
~
=F
grad ϕ = −GM m
65. The tangent plane to z =
p
2x2 + 2y 2 − 25 at (x, y) = (4, 3) is
z = z(4, 3) + zx (4, 3)(x − 4) + zy (4, 3)(y − 3)
p
2(4)
2(3)
= 2(4)2 + 2(3)2 − 25 + p
(x − 4) + p
(y − 3)
2
2
2
2(4) + 2(3) − 25
2(4) + 2(3)2 − 25
8
6
= 5 + (x − 4) + (y − 3).
5
5
The tangent plane to z = 15 (x2 + y 2 ) at (x, y) = (4, 3) is
z = z(4, 3) + zx (4, 3)(x − 4) + zy (4, 3)(y − 3)
1
2
2
= (42 + 32 ) + (4)(x − 4) + (3)(y − 3)
5
5
5
6
8
= 5 + (x − 4) + (y − 3).
5
5
Thus the two surfaces are tangential at the point (4, 3, 5).
66. The points of intersection are (x, y) = (a, b) such that
1
1 2
(a + b2 − 1) = (1 − a2 − b2 )
2
2
a2 + b2 − 1 = 1 − (a2 + b2 )
2(a2 + b2 ) = 2
a2 + b2 = 1,
which are points on the unit circle.
The tangent plane to z = 12 (x2 + y 2 − 1) at (x, y) = (a, b) such that a2 + b2 = 1 is
z = z(a, b) + zx (a, b)(x − a) + zy (a, b)(y − b)
= 0 + a(x − a) + b(y − b)
= ax + by − (a2 + b2 )
= ax + by − 1, or ax + by − z = 1.
Similarly, the tangent to z =
− x2 − y 2 ) at (x, y) = (a, b) such that a2 + b2 = 1 is z = −ax − by + 1 or
ax + by + z = 1. The normal vector to the plane ax + by − z = 1 is n~1 = a~i + b~j − ~k and the normal vector to the
plane ax + by + z = 1 is n~2 = a~i + b~j + ~k .
Since n~1 · n~2 = a2 + b2 − 1 = 1 − 1 = 0, n~1 and n~2 are perpendicular, hence the two surfaces are orthogonal at
all points of intersection.
1
(1
2
67.
grad(~
µ · ~r ) = grad(µ1 x + µ2 y + µ3 z)
= µ1~i + µ2~j + µ3~k = ~
µ.
14.5 SOLUTIONS
1321
68.
grad(k~r ka ) = grad((x2 + y 2 + z 2 )a/2 )
a
a
= (x2 + y 2 + z 2 )(a/2)−1 (2x)~i + (x2 + y 2 + z 2 )(a/2)−1 (2y)~j
2
2
a
+ (x2 + y 2 + z 2 )(a/2)−1 (2z)~k
2
= a(x2 + y 2 + z 2 )(a−2)/2 (x~i + y~j + z~k )
= ak~r ka−2 ~r .
69. If write ~r = x~i + y~j + z~k , then we know
grad f (x, y, z) = g(x, y, z)(x~i + y~j + z~k ) = g(x, y, z)~r
so grad f is everywhere radially outward, and therefore perpendicular to a sphere centered at the origin. If f were not
constant on such a sphere, then grad f would have a component tangent to the sphere. Thus, f must be constant on any
sphere centered at the origin.
Strengthen Your Understanding
70. The gradient vector grad f (x, y) points in the direction perpendicular to the level curves f (x, y) = C in the xy-plane.
71. The correct equation of the tangent plane is
fx (0, 0, 0)x + fy (0, 0, 0)y + fz (0, 0, 0)z = 0.
72. The surface z = f (x, y) can be rewritten f (x, y) − z = 0 with normal at (0, 0) given by
~n = fx (0, 0)~i + fy (0, 0)~j − ~k .
To have normal vector ~n = ~i − 2~j − ~k , we take fx (0, 0) = 1 and fy (0, 0) = −2. One example is f (x, y) = x − 2y.
73. Any function f (x, y, z) = 2x + 3y + 4z + C where C is a constant has grad f = 2~i + 3~j + 4~k . For example, we can
take
f (x, y, z) = 2x + 3y + 4z + 100.
74. We have
grad f = 2~i − 3~j .
We want vectors ~
u and ~v which are perpendicular to grad f . Two possibilities are ~k and 3~i + 2~j . Creating unit vectors
gives
1
3~i + 2~j .
~
u = ~k
and
~v = √
32 + 22
Then grad f · ~
u = grad f · ~v = 0, so f~u = f~v = 0.
75. False. The equation z = 2 + 2x(x − 1) + 3y 2 (y − 1) is not linear. The correct equation is z = 2 + 2(x − 1) + 3(y − 1),
which is obtained by evaluating the partial derivatives at the point (1, 1).
76. True. For example, the function f (x, y) = x2 + y 2 has a horizontal tangent plane at (0, 0).
77. This is never true, because, if θ is the angle between grad f and the z-axis at any point, then f~k = k grad f k cos θ ≤
k grad f k.
√
78. Can be true, For example for f (x, y, z) = 4x − 3z, we have grad f = 4~i − 3~k then k grad f k = 42 + 32 = 5 and
f~k = fz = −3.
79. (a) The units of || grad f || are ◦ C per meter. It represents the rate of change of temperature with distance as you move in
the direction of grad f .
(b) The units of grad f · ~v are ◦ C per second. It represents the rate of change of temperature with time as you move with
velocity ~v .
(c) The units of || grad f || · ||~v || are ◦ C per second. It represents the rate of change of temperature with time if you move
in the direction of grad f with speed ||~v ||.
1322
Chapter Fourteen /SOLUTIONS
Solutions for Section 14.6
Exercises
1. Using the chain rule we see:
dz
∂z dx
∂z dy
=
+
dt
∂x dt
∂y dt
= −y 2 e−t + 2xy cos t
= −(sin t)2 e−t + 2e−t sin t cos t
= sin(t)e−t (2 cos t − sin t)
We can also solve the problem using one variable methods:
z = e−t (sin t)2
d
dz
= (e−t (sin t)2 )
dt
dt
d(sin t)2
de−t
(sin t)2 + e−t
=
dt
dt
= −e−t (sin t)2 + 2e−t sin t cos t
= e−t sin t(2 cos t − sin t)
2. Using the chain rule we see:
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= 2t(sin y + y cos x) +
1
(x cos y + sin x)
t
= 2t sin(ln t) + 2t ln(t) cos(t2 ) + t cos(ln t) +
sin t2
t
This problem can also be solved using one variable methods. Attempting to solve the problem that way will demonstrate
the advantage of using the chain rule.
3. Substituting into the chain rule gives
∂z dx
∂z dy
dz
=
+
= cos
dt
∂x dt
∂y dt
= cos
x
y
2y + 2xt
y2
x
y
= 2 cos
1
y
(2) + cos
x
y
2t
1 + t2
2
1 − t (1 − t2 )2
4. This is a case where substituting is easier:
z = ln(t−2 + t)
1 − 2t−3
dz
= −2
dt
t +t
t3 − 2
=
t + t4
If you use the chain rule the solution is:
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
y
−2x
+√
= 2 2
t (x + y 2 )
t(x2 + y 2 )
−x
y2
(−2t)
14.6 SOLUTIONS
=
1323
1
−2
+
t3 ((1/t)2 + t)
(1/t)2 + t
−2
t2
+
t + t4
1 + t3
t3 − 2
=
t + t4
=
5. Substituting into the chain rule gives
∂z dx
∂z dy
dz
=
+
= ey (2) + xey (−2t)
dt
∂x dt
∂y dt
2
= 2ey (1 − xt) = 2e1−t (1 − 2t2 ).
6. Substituting into the chain rule gives
∂z dx
∂z dy
dz
=
+
= ey (2) + (xey + ey + yey )(−2t)
dt
∂x dt
∂y dt
2
= 2ey (1 − xt − t − yt) = 2e1−t (1 − 2t2 − 2t + t3 ).
7. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z
∂z ∂x
∂z ∂y
=
+
=
∂u
∂x ∂u
∂y ∂u
cos
x
y
1
y
1
+
u
cos
x
y
−x
y2
·0
1
ln u
cos
.
vu
v
∂z ∂x
∂z ∂y
x
1
−x
ln u
∂z
x
ln u
=
+
= cos
· 0 + cos
· 1 = − 2 cos
.
2
∂v
∂x ∂v
∂y ∂v
y
y
y
y
v
v
=
8. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z ∂x
∂z ∂y
∂z
=
+
∂u
∂x ∂u
∂y ∂u
1
1
· 2(u3 + v 3 )(3u2 )
= · 2(u2 + v 2 )(2u) +
x
y
4u(u2 + v 2 )
6u2 (u3 + v 3 )
=
+
(u2 + v 2 )2
(u3 + v 3 )2
=
u2
4u
6u2
+ 3
.
2
+v
u + v3
Similarly, we have
∂z ∂x
∂z ∂y
∂z
=
+
∂v
∂x ∂v
∂y ∂v
1
1
· 2(u2 + v 2 )(2v) +
· 2(u3 + v 3 )(3v 2 )
x
y
4v(u2 + v 2 )
6v 2 (u3 + v 3 )
=
+
2
2
2
(u + v )
(u3 + v 3 )2
=
=
4v
6v 2
+
.
u2 + v 2
u3 + v 3
1324
Chapter Fourteen /SOLUTIONS
9. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
ev
1
∂z ∂x
∂z ∂y
∂z
+ xey · 0 =
=
+
= ey
.
∂u
∂x ∂u
∂y ∂u
u
u
∂z
∂z ∂x
∂z ∂y
=
+
= ey (0) + xey · 1 = ev ln u.
∂v
∂x ∂v
∂y ∂v
10. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z ∂x
∂z ∂y
1
ev
∂z
=
+
= ey
.
+ ey (1 + x + y) · 0 =
∂u
∂x ∂u
∂y ∂u
u
u
∂z
∂z ∂x
∂z ∂y
=
+
= ey (0) + ey (1 + x + y) · 1 = (1 + ln u + v)ev .
∂v
∂x ∂v
∂y ∂v
11. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z ∂x
∂z ∂y
∂z
=
+
= ey (2u) + xey (2u)
∂u
∂x ∂u
∂y ∂u
2
2
= 2uey (1 + x) = 2ue(u −v ) (1 + u2 + v 2 ).
∂z ∂x
∂z ∂y
∂z
=
+
= ey (2v) + xey (−2v)
∂v
∂x ∂v
∂y ∂v
= 2vey (1 − x) = 2ve(u
2
−v 2 )
(1 − u2 − v 2 ).
12. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z ∂x
∂z ∂y
∂z
=
+
= ey (2u) + (xey + ey + yey )(2u)
∂u
∂x ∂u
∂y ∂u
= 2uey (1 + x + 1 + y) = 2uey (x + y + 2)
2
2
2
2
= 2ue(u −v ) (u2 + v 2 + u2 − v 2 + 2) = 2ue(u −v ) (2u2 + 2)
∂z ∂x
∂z ∂y
∂z
=
+
= ey (2v) + (xey + ey + yey )(−2v)
∂v
∂x ∂v
∂y ∂v
= 2vey (1 − x − 1 − y) = −2vey (x + y)
= −2ve(u
2
−v 2 )
(u2 + v 2 + u2 − v 2 ) = −4u2 ve(u
2
−v 2 )
.
13. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z
∂z ∂x
∂z ∂y
=
+
∂u
∂x ∂u
∂y ∂u
∂z ∂x
∂z ∂y
∂z
=
+
∂v
∂x ∂v
∂y ∂v
First to find ∂z/∂u
∂z
= (e−y − ye−x ) sin v + (−xe−y + e−x )(−v sin u)
∂u
= (e−v cos u − v(cos u)e−u sin v ) sin v − (−u(sin v)e−v cos u + e−u sin v )v sin u
Now we find ∂z/∂v using the same method.
∂z
= (e−y − ye−x )u cos v + (−xe−y + e−x ) cos u
∂v
= (e−v cos u − v(cos u)e−u sin v )u cos v + (−u(sin v)e−v cos u + e−u sin v ) cos u
14.6 SOLUTIONS
1325
14. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z ∂x
∂z ∂y
∂z
=
+
∂u
∂x ∂u
∂y ∂u
∂z
∂z ∂x
∂z ∂y
=
+
∂v
∂x ∂v
∂y ∂v
This problem is most easily solved by substitution:
z = cos(u2 ((cos v)2 + (sin v)2 ))
= cos u2
∂z
= −2u sin u2
∂u
∂z
=0
∂v
This problem can also be solved using the chain rule but it is more difficult.
15. Since z is a function of two variables x and y which are functions of two variables u and v, the two chain rule identities
which apply are:
∂z ∂x
∂z ∂y
1
1
1
−x
∂z
=
+
=
)(2u)
( )(2u) +
(
∂v
∂x ∂u
∂y ∂u
1 + ( xy )2 y
1 + ( xy )2 y 2
y−x
−2uv 2
)
=
y 2 + x2
u4 + v 4
1
−x
∂z
∂z ∂x
∂z ∂y
1
1
( )(2v) +
(
=
+
=
)(−2v)
∂v
∂x ∂v
∂y ∂v
1 + ( xy )2 y
1 + ( xy )2 y 2
= 2u(
= 2v(
2vu2
y+x
)
=
.
y 2 + x2
u4 + v 4
Problems
16. By the chain rule
dz
∂z dx
∂z dy
=
+
dt
∂x dt
∂y dt
= (3x2 y 2 )(3t2 ) + (2x3 y)(2t)
= (3t6 t4 )3t2 + (2t9 t2 )2t
= 9t12 + 4t12
= 13t12
Directly, we have z = t9 · t4 = t13 , so dz/dt = 13t12 .
17. Using the chain rule we have
∂w ∂x
∂w ∂y
∂w ∂z
∂w
=
+
+
∂ρ
∂x ∂ρ
∂y ∂ρ
∂z ∂ρ
= (2x) sin φ cos θ + (2y) sin φ sin θ − (2z) cos φ
= 2ρ sin2 φ cos2 θ + 2ρ sin2 φ sin2 θ − 2ρ cos2 φ,
= 2ρ sin2 φ(cos2 θ + sin2 θ) − 2ρ cos2 φ
= 2ρ sin2 φ − 2ρ cos2 φ
= −2ρ cos 2φ.
∂w ∂x
∂w ∂y
∂w ∂z
∂w
=
+
+
∂θ
∂x ∂θ
∂y ∂θ
∂z ∂θ
= (2x)(−ρ sin φ sin θ) + (2y)ρ sin φ cos θ
= −2ρ2 sin2 φ cos θ sin θ + 2ρ2 sin2 φ sin θ cos θ
=0
1326
Chapter Fourteen /SOLUTIONS
18. (a) ∂f /∂t
(b) (∂f /∂x)(dx/dt)
(c) (∂f /∂y)(dy/dt)
19. When t = 1,
x = g(1) = 3
and
y = h(1) = 10,
so z = f (3, 10) = 7. Thus
∂f ′
∂f ′
∂z
=
g (t) +
h (t)
∂t
∂x
∂y
gives
∂z
∂t
t=1
= fx (3, 10) · g ′ (1) + fy (3, 10) · h′ (1)
= 100 · 4 + 0.1 · 11
= 400 + 1.1 = 401.1.
20. The voltage at any time t is given by V = IR where R is the resistance for the whole circuit. (In this case R =
R1 R2 /(R1 + R2 ).) So the rate at which the voltage is changing is
dI
dR
dV
=
R+I
dt
dt
dt
∂R dR2
dI
∂R dR1
R+I
+
=
dt
∂R dt
∂R2 dt
1 2
dI
R2
dR1
R12
dR2
=
R+I
+
dt
(R1 + R2 )2 dt
(R1 + R2 )2 dt
25
15
9
= 0.01
+2
(0.5) +
(−0.1)
8
64
64
= 0.3812.
So the voltage is increasing by 0.3812 volts/sec.
21. Let p(x, t) be the air pressure in pascals at x km east of the island at time t hours after the ship passes the island. We want
to compute ∂p/∂t.
Let S(t) be the air pressure on the ship at time t, so that S(t) = p(10t, t). By the chain rule we have
dS
∂p dx
∂p dt
∂p dx
∂p
=
+
=
+
dt
∂x dt
∂t dt
∂x dt
∂t
Since dS/dt = (−50 pascal)/(2 hour) = −25 pascal/hour, and ∂p/∂x = −2 pascal/km, and dx/dt = 10 km/hour,
solving for ∂p/∂t, we have
∂p
= −25 pascal/hour − (−2 pascal/km)(10 km/hour) = −5 pascal/hour
∂t
22. Let the square cross section have side length x(T ) and the bar have length L(T ) at temperature T ◦ C; then
Volume of the bar = V = x2 L.
Using the chain rule we have:
∂V dx
∂V dL
dV
=
+
dT
∂x dT
∂L dT
dx
dL
+ x2
= 2xL
dT
dT
= (2xL + x2 ) · 13 · 10−6 ,
since dx/dT = dL/dT = 13 · 10−6 . When L = 3 and x = 0.05, the rate of increase of volume is
dV
= (2 · (0.05) · 3 + 0.052 ) · 13 × 10−6 = 3.93 · 10−6 m3 /◦ C.
dT
14.6 SOLUTIONS
23.
1327
T (◦ C)
R (in)
15
✻
27
14
2
25
13
❄
✛
30
2020
✲
✻
3
✛
40
✲❄
2020
2040
23
t (years)
2040
Figure 14.28: Global warming predictions:
Rainfall as a function of time
t (years)
Figure 14.29: Global warming predictions:
Temperature as a function of time
We know that, as long as the temperature and rainfall stay close to their current values of R = 15 inches and
T = 30◦ C, a change, ∆R, in rainfall and a change, ∆T , in temperature produces a change, ∆C, in corn production given
by
∆C ≈ 3.3∆R − 5∆T.
Now both R and T are functions of time t (in years), and we want to find the effect of a small change in time, ∆t, on R
and T . Figure 14.28 shows that the slope of the graph for R versus t is about −2/30 ≈ −0.07 in/year when t = 2020.
Similarly, Figure 14.29 shows the slope of the graph of T versus t is about 3/40 ≈ 0.08◦ C/year when t = 2020. Thus,
around the year 2020,
∆R ≈ −0.07∆t and ∆T ≈ 0.08∆t.
Substituting these into the equation for ∆C, we get
∆C ≈ (3.3)(−0.07)∆t − (5)(0.08)∆t ≈ −0.6∆t.
Since at present C = 100, corn production will decline by about 0.6 % between the years 2020 and 2021. Now ∆C ≈
−0.6∆t tells us that when t = 2020,
∆C
≈ −0.6,
∆t
and therefore, that
dC
≈ −0.6.
dt
24. (a) The level surfaces of f are concentric spheres centered at the origin. This is because if f (x, y, z) is fixed, then g(ρ)
is fixed, which means ρ must be fixed. Fixed ρ gives a sphere.
(b) By the chain rule, we find
dg ∂ρ
1
x
∂f
=
= g ′ (ρ) (x2 + y 2 + z 2 )−1/2 · 2x = g ′ (ρ) p
.
∂x
dρ ∂x
2
x2 + y 2 + z 2
Thus
~
~
~
~ = grad f = dg · xi + y j + z k ,
F
dρ
ρ
~
~
~
~
so F (x, y, z) is parallel to xi + y j + z k . A unit vector in the direction of F~ at (1, 2, 2) is
~i + 2~j + 2~k
2
2
1
~
u = √
= ~i + ~j + ~k .
2
2
2
3
3
3
1 +2 +2
(c) We have
Since ρ =
1/3. Thus
√
~ || =
||F
dg
·
dρ
x~i + y~j + z~k
ρ
=
dg
.
dρ
12 + 22 + 22 = 3 at (1, 2, 3), we estimate g ′ (3) from the graph, obtaining g ′ (ρ(1, 2, 2)) = g ′ (3) ≈
1
.
3
(d) From the answers to parts (b) and (c), we know that F~ (1, 2, 2) is a vector of length about 1/3 in the direction of
~i + 2~j + 2~k . Thus, we estimate
~ (1, 2, 2) ≈ 1 (~i + 2~j + 2~k ).
F
9
~ (1, 2, 2)|| ≈
||F
1328
Chapter Fourteen /SOLUTIONS
~ (3, 0, 0) ≈ 1~i , because ||F
~ (3, 0, 0)|| = g ′ (ρ(3, 0, 0)) = g ′ (3) ≈ 1/3, and the direction of
(e) We estimate F
3
F~ (x, y, z) is parallel to x~i + y~j + z~k = 3~i .
~ (P )|| = ||F
~ (Q)|| because ||F
~ (P )|| = |g ′ (ρ(P ))| = |g ′ (ρ(Q))| = ||F
~ (Q)||.
(f) (i) We have ||F
~ (~r ) points in the same direction as ~r , where x~i + y~j + z~k .
(ii) At each point, F
25. By the chain rule,
∂z
= gu ut + gv vt + gw wt .
∂t
Thus, there are three terms.
26.
w
∂w
∂y
∂w
∂x
x
∂w
∂z
y
∂x ∂y
∂v ∂u
z
∂y ∂z
∂v ∂u
∂z
∂v
∂x
∂u
u
v
Figure 14.30
The tree diagram in Figure 14.30 tells us that
∂w
∂w ∂x
∂w ∂y
∂w ∂z
=
+
+
,
∂u
∂x ∂u
∂y ∂u
∂z ∂u
∂w ∂x
∂w ∂y
∂w ∂z
∂w
=
+
+
.
∂v
∂x ∂v
∂y ∂v
∂z ∂v
27.
w
∂w
∂y
∂w
∂x
x
∂w
∂z
y
dy
dt
dx
dt
z
dz
dt
t
Figure 14.31
From the tree diagram in Figure 14.31, we get
∂w dx
∂w dy
∂w dz
dw
=
+
+
.
dt
∂x dt
∂y dt
∂z dt
14.6 SOLUTIONS
1329
28. All are done using the chain rule.
(a) We have u = x, v = 3. Thus du/dx = 1 and dv/dx = 0 so
f ′ (x) = Fu (x, 3)(1) + Fv (x, 3)(0) = Fu (x, 3).
(b) We have u = 3, v = x. Thus du/dx = 0 and dv/dx = 1 so
f ′ (x) = Fu (3, x)(0) + Fv (3, x)(1) = Fv (3, x).
(c) We have u = x, v = x. Thus du/dx = dv/dx = 1 so
f ′ (x) = Fu (x, x)(1) + Fv (x, x)(1) = Fu (x, x) + Fv (x, x).
(d) We have u = 5x, v = x2 . Thus du/dx = 5 and dv/dx = 2x so
f ′ (x) = Fu (5x, x2 )(5) + Fv (5x, x2 )(2x).
29. Using the chain rule,
zu (u, v) = fx (x, y) · xu (u, v) + fy (x, y) · yu (u, v).
Since x(1, 2) = 5 and y(1, 2) = 3, substituting gives
zu (1, 2) = fx (5, 3) · xu (1, 2) + fy (5, 3) · yu (1, 2) = b · e + d · p.
30. Using the chain rule,
zv (u, v) = fx (x, y) · xv (u, v) + fy (x, y) · yv (u, v).
Since x(1, 2) = 5 and y(1, 2) = 3, substituting gives
zv (1, 2) = fx (5, 3) · xv (1, 2) + fy (5, 3) · yv (1, 2) = b · k + d · q.
31. Implicit differentiation with respect to x of the equation f (x, y) = f (a, b) gives fx (x, y)dx/dx + fy (x, y)dy/dx = 0.
Thus fy (x, y)dy/dx = −fx (x, y). If fy (a, b) 6= 0 we can solve for the slope of the level curve at the point (a, b):
dy/dx = −fx (a, b)/fy (a, b).
32. We have z = h(x, y) where h(x, y) = f (x)g(y), x = t, and y = t. The chain rule gives dz/dt = (∂h/∂x)dx/dt +
(∂h/∂y)dy/dt = ∂h/∂x +∂h/∂y. Since ∂h/∂x = f ′ (x)g(y) and ∂h/∂y = f (x)g ′ (y) we have dz/dt = f ′ (x)g(y)+
f (x)g ′ (y) = f ′ (t)g(t) + f (t)g ′ (t).
33. Let g(t) = f (tx, ty). We use the chain rule, with u = tx and v = ty as our variables. Then we have
∂f (u, v) dv
∂f (u, v) du
+
∂u
dt
∂v
dt
= fu (u, v) x + fv (u, v) y.
g ′ (t) =
At t = 1, we have u = x and v = y. So
g ′ (1) = x fx (x, y) + y fy (x, y).
On the other hand, since f (x, y) is homogeneous of degree p we also have g(t) = tp f (x, y). Thus we have
g ′ (t) = p tp−1 f (x, y)
and
g ′ (1) = p f (x, y).
Thus,
x fx (x, y) + y fy (x, y) = p f (x, y).
1330
Chapter Fourteen /SOLUTIONS
34. Use chain rule for the equation 0 = F (x, y, f (x, y)). Differentiating both sides with respect to x, remembering z =
f (x, y) and regarding y as a constant gives:
0=
∂F dz
∂F dx
+
.
∂x dx
∂z dx
Since dx/dx = 1, we get
−
so
∂F ∂z
∂F
=
,
∂x
∂z ∂x
∂F/∂x
∂z
=−
.
∂x
∂F/∂z
Similarly, differentiating both sides of the equation 0 = F (x, y, f (x, y)) with respect to y gives:
0=
∂F dy
∂F dz
+
.
∂y dy
∂z dy
Since dy/dy = 1, we get
−
so
∂F ∂z
∂F
=
,
∂y
∂z ∂y
∂F/∂y
∂z
=−
.
∂y
∂F/∂z
35. Using the chain rule,
zu (u, v) = fx (x, y) · xu (u, v) + fy (x, y) · yu (u, v).
Since x(4, 5) = 2 and y(4, 5) = 3, substituting gives
zu (4, 5) = fx (2, 3) · xu (4, 5) + fy (2, 3) · yu (4, 5) = b · e + d · p.
36. Using the chain rule,
zv (u, v) = fx (x, y) · xv (u, v) + fy (x, y) · yv (u, v).
Since x(4, 5) = 2 and y(4, 5) = 3, substituting gives
zv (4, 5) = fx (2, 3) · xv (4, 5) + fy (2, 3) · yv (4, 5) = b · k + d · q.
37. (a) We will use the chain rule identities,
∂z
∂z ∂x
∂z ∂y
=
+
∂r
∂x ∂r
∂y ∂r
∂z
∂z ∂x
∂z ∂y
=
+
.
∂θ
∂x ∂θ
∂y ∂θ
and
These equations are to be in terms of ∂z/∂x and ∂z/∂y, so we may calculate the other terms, switching from
Cartesian to polar coordinates. Recall polar coordinates :
x = r cos θ,
y = r sin θ
Thus we have
∂x
∂r
∂y
∂r
∂x
∂θ
∂y
∂θ
∂(r cos θ)
∂r
∂(r sin θ)
=
∂r
∂(r cos θ)
=
∂θ
∂(r sin θ)
=
∂θ
=
= cos θ
= sin θ
= −r sin θ
= r cos θ
Now, substituting into the equations for ∂z/∂r and ∂z/∂θ, we get
(1)
(2)
We will call these equations (1) and (2).
∂z
∂z
∂z
= cos θ
+ sin θ
∂r
∂x
∂y
∂z
∂z
∂z
= −r sin θ
+ r cos θ .
∂θ
∂x
∂y
14.6 SOLUTIONS
(b) Now we solve for ∂z/∂x and ∂z/∂y. From (2) we get:
∂z
=
∂x
(3)
∂z
∂z
− r cos θ
∂θ
∂y
Now substitute (3) into (1):
∂z
= cos θ
∂r
=−
∂z
∂z
− r cos θ
∂θ
∂y
−1
,
r sin θ
−1
r sin θ
+ sin θ
∂z
∂y
cos2 θ ∂z
∂z
cos θ ∂z
+
+ sin θ
r sin θ ∂θ
sin θ ∂y
∂y
Now solve for ∂z/∂y:
∂z
∂y
sin2 θ
cos2 θ
+
sin θ
sin θ
=
cos θ ∂z
∂z
+
∂r
r sin θ ∂θ
∂z 1
∂z
cos θ ∂z
(
)=
+
∂y sin θ
∂r
r sin θ ∂θ
∂z
∂z
cos θ ∂z
= sin θ
+
.
∂y
∂r
r ∂θ
Now, substitute ∂z/∂y into equation (3) and solve for ∂z/∂x.
∂z
=
∂x
∂z
∂z
− r cos θ
∂θ
∂y
−1
r sin θ
cos θ
∂z
cos θ ∂z
−1 ∂z
+
sin θ
+
r sin θ ∂θ
sin θ
∂r
r ∂θ
cos2 θ − 1 ∂z
∂z
+
= cos θ
∂r
r sin θ ∂θ
sin2 θ ∂z
∂z
−
= cos θ
∂r
r sin θ ∂θ
∂z
sin θ ∂z
= cos θ
−
.
∂r
r ∂θ
=
(c) Now we use the chain rule to get ∂z/∂x and ∂z/∂y.
(4)
∂z ∂r
∂z ∂θ
∂z
=
+
,
∂y
∂r ∂y
∂θ ∂y
∂z
∂z ∂r
∂z ∂θ
=
+
∂x
∂r ∂x
∂θ ∂x
We will call this equation (4).
p
As before, we will calculate some of these partials using r = x2 + y 2 and θ = arctan(y/x)
∂
∂r
=
∂y
p
x2 + y 2
y
= p
= sin θ
∂y
x2 + y 2
∂ arctan(y/x)
1
x
cos θ
∂θ
=
=
(x−1 ) = 2
=
∂y
∂y
1 + ( xy )2
(x + y 2 )
r
x
∂r
= p
= cos θ
∂x
x2 + y 2
1
y
sin θ
∂θ
y
=
=−
(− 2 ) = − 2
∂x
1 + ( xy )2
x
(x + y 2 )
r
Now, substituting these into (4), we get:
∂z
∂z
cos θ ∂z
= sin θ
+
∂y
∂r
r ∂θ
∂z
sin θ ∂z
∂z
= cos θ
−
∂x
∂r
r ∂θ
Note that these equations match with those found in part (b).
1331
1332
Chapter Fourteen /SOLUTIONS
38. Using x = r cos θ and y = r sin θ we compute ∂z/∂r and ∂z/∂θ in terms of ∂z/∂x and ∂z/∂y:
∂z ∂x
∂z
∂z
=
+
∂r
∂x ∂r
∂y
∂z ∂x
∂z
∂z
=
+
∂θ
∂x ∂θ
∂y
∂y
∂z
∂z
=
cos θ +
sin θ
∂r
∂x
∂y
∂y
∂z
∂z
=
(−r sin θ) +
r cos θ
∂θ
∂x
∂y
So we have
∂z 2
∂r
In addition we have,
=
∂z 2
∂x
∂z ∂z
cos θ sin θ +
cos θ + 2
∂x ∂y
2
∂z
∂y
2
∂z
∂z
1 ∂z
=
(− sin θ) +
cos θ
r ∂θ
∂x
∂y
thus,
1
r2
Adding we get
39. Since
Then
∂U
∂P
40. To calculate
sin2 θ
∂z
∂θ
2
∂z
∂x
2
2
1
+ 2
r
=
∂z
∂r
∂z ∂z
sin θ − 2
sin θ cos θ +
∂x ∂y
2
∂z
∂θ
2
=
∂z
∂x
2
+
∂z
∂y
∂z
∂y
2
cos2 θ
2
involves the variables P and V , we are viewing U as a function of these two variables, so U = U3 (P, V ).
V
∂U
∂P
∂U
∂P
=
V
∂U3 (P, V )
.
∂P
, we think of U as a function of P and T , as in U1 (T, P ). Thus
T
∂U
∂P
=
dP +
∂U
∂V
T
∂U1
.
∂P
41. From the example, we know that for this gas
dU =
In addition, we have
∂U
∂P
V
dV = 7dP + 8dV.
P
P dV + V dP = 3dV + 4dP = 2dT.
We substitute for dP = (2dT − 3dV )/4 into the expression for dU , giving
2dT − 3dV
4
7
11
dU = dT +
dV.
2
4
dU = 7
+ 8dV
Comparing with the formula for dU obtained from the function U2 , where U = U2 (T, V ):
dU =
we have
∂U
∂T
=
V
∂U
∂T
dT +
V
7
2
∂U
∂V
and
dV,
T
∂U
∂V
=
T
11
.
4
42. The partial derivative on the left side of the equation is obtained by thinking of T as a function of V and P . The partial
derivative on the right side is obtained by thinking of V as a function of T and P .
Thinking of T as a function of V and P , we have
dT =
∂T
∂V
P
dV +
∂T
∂P
V
dP.
14.6 SOLUTIONS
Thinking of V as a function of T and P . we also have
dV =
Solving for dT in terms of dV and dP gives
∂V
∂T
1
dT =
∂V
∂T
P
dT +
P
∂V
∂P
dP.
T
∂V
dV −
∂P
dP .
P
Comparing coefficients of dV in the two expressions for dT gives
∂T
∂V
.
=1
P
∂V
∂T
.
P
43. From Example 7, we know that
∂U
∂T
Interchanging the roles of T and V , we get
Using the result of Problem 42, namely
∂T
P
∂U
∂V
∂U
=
∂V
P
∂T
∂V
∂U
∂V
+
=
P
T
∂V
∂T
∂U
∂V
V
∂V
∂P
∂U
∂P
we get
dU =
Rearranging terms gives
dU =
∂U
∂P
∂U
∂P
dP +
∂U
∂T
∂V
∂T
∂V
∂T
∂U
∂V
dP +
+
T
dP +
dP +
T
V
+
V
∂U
∂V
V
∂U
∂V
·
P
P
∂V
.
P
, gives
(b) Substituting for dV in the following expression for dU ,
dU =
.
P
P
44. (a) Thinking of V as a function of P and T gives
dV =
∂T
∂U ∂T
∂T
T
.
=1
P
∂V
V
∂U
=
∂U ∂V
+
∂V
∂P
∂V
∂P
V .
P
dT.
P
dV,
P
T
dP +
T
∂U
∂V
(c) The formula for dU obtained by thinking of U as a function of P and T is
dU =
∂U
∂T
dT +
P
∂U
∂P
(d) Comparing coefficients of dP and dT in the two formulas gives
∂U
=
∂U
·
∂V
∂V
∂T
P
·
dP.
T
∂T P
∂V P
∂T P
∂U
∂U
∂V
∂U
=
+
·
.
∂P T
∂P V
∂V P
∂P T
P
dT .
∂V
∂T
dT.
P
1333
1334
Chapter Fourteen /SOLUTIONS
45. As the introduction to this problem indicates, we can differentiate with respect to x inside the integral:
′
f (x) =
Z
b
Fu (x, y) dy.
0
46. By the Fundamental Theorem of Calculus, we substitute y = x in the integrand:
f ′ (x) = F (b, x).
47. (a) By Problem 45, treating w as the constant b:
Gu (u, w) =
Z
w
Fu (u, y) dy.
0
By Problem 46, treating u as the constant b:
Gw (u, w) = F (u, w).
(b) To differentiate f (x) = G(x, x), we apply the chain rule to G(u, w) with u = x, w = x. Since ∂u/∂x = ∂w/∂x =
1, we get
f ′ (x) = Gu (x, x)(1) + Gw (x, x)(1).
By part (a),
′
f (x) =
Z
x
Fx (x, y) dy + F (x, x).
0
Strengthen Your Understanding
48. Writing z = f (x, y), with x = g(t) and y = h(t), the chain rule gives
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= fx (g(t), h(t))g ′ (t) + fy (g(t), h(t))h′ (t).
49. According to the chain rule, we evaluate Cx , Rx and Tx at points (x, y) and we evaluate CR and TR at points (R, T ).
Since we know R(0, 2) = 5 and T (0, 2) = 1, the chain rule gives:
Cx (0, 2) = CR (5, 1)Rx (0, 2) + CT (5, 1)Tx (0, 2).
50. The partial derivatives fx and fy should be evaluated at (2, 3):
∂z
∂t
= fx (2, 3)g ′ (0) + fy (2, 3)h′ (0).
t=0
51. We have
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
dx
dy
= 2xy
+ x2
dt
dt
′
= 2g(t)h(t)g (t) + g(t)2 h′ (t).
If h(0) = 0, h′ (0) = 1, and g(0) = 3, then (dz/dt)|t=0 = 9. For example, we can take h(t) = t and g(t) = 3 + t.
14.7 SOLUTIONS
1335
52. We have
dz
∂z dx
∂z dy
=
+
dt
∂x dt
∂y dt
= fx (x, y)2e2t + fy (x, y) cos t
dz
|t=0 = fx (1, 0) · 2 + fy (1, 0) · 1 = 2fx (1, 0) + fy (1, 0).
dt
If fx (1, 0) = 4 and fy (1, 0) = 2, then dz/dt|t=0 = 10. For example, we can take f (x, y) = 4x + 2y.
53. A possible answer is z = x + y, x = et and y = t2 .
54. A possible answer is w = uv, u = 2s2 + t and v = est .
55. A possible answer is z = x + y, x = t and y = t.
56. By the chain rule,
dz
= gu ux x′ + gu uy y ′ + gu ut + gv vx x′ + gv vy y ′ + gv vt .
dt
Thus, the answer is (c).
Solutions for Section 14.7
Exercises
1. Calculating the partial derivatives:
∂2f
= 2.
∂x2
∂f
= 2(x + y),
∂x
Therefore, we get
∂2f
= 2,
∂y 2
∂f
= 2(x + y),
∂y
∂2f
= 2,
∂y∂x
∂2f
= 2.
∂x∂y
2. Calculating the partial derivatives:
∂f
= 3(x + y)2 ,
∂x
∂f
= 3(x + y)2 ,
∂y
∂2f
= 6(x + y).
∂x2
∂2f
= 6(x + y).
∂y 2
Consequently, we get
∂2f
= 6(x + y),
∂x∂y
∂2f
= 6(x + y).
∂y∂x
3. We have fx = 6xy + 5y 3 and fy = 3x2 + 15xy 2 , so fxx = 6y, fxy = 6x + 15y 2 , fyx = 6x + 15y 2 , and fyy = 30xy.
4. We have fx = 2ye2xy and fy = 2xe2xy , so fxx = 4y 2 e2xy , fxy = 4xye2xy + 2e2xy , fyx = 4xye2xy + 2e2xy , and
fyy = 4x2 e2xy .
5. Since f = (x + y)ey , the partial derivatives are
fx = ey ,
fy = ey (x + 1 + y)
fyx = ey = fxy
fxx = 0,
y
fyy = xe + ey + ey + yey = ey (x + 2 + y).
6. Since f (x, y) = xey , the partial derivatives are
fx = ey ,
fxx = 0,
fy = xey
fxy = ey = fyx ,
fyy = xey .
1336
Chapter Fourteen /SOLUTIONS
7. Since f (x, y) = sin(x/y), the first partial derivatives are:
x 1
fx = (cos ( )) ,
y y
x −x
fy = (cos ( ))( 2 ).
y
y
Thus, the second partial derivatives are
1
x
fxx = −(sin ( ))( 2 )
y
y
x −1
x −x 1
fxy = −(sin ( ))( 2 )( ) + (cos ( ))( 2 ) = fyx
y
y
y
y
y
x −x 2
x 2x
fyy = −(sin ( ))( 2 ) + (cos ( ))( 3 ).
y
y
y
y
8. Since f (x, y) =
p
x2 + y 2 , we have
y
fy = p
x2 + y 2
x
,
fx = p
x2 + y 2
fxx =
fxy
p
x2 + y 2 − x( √
x2
x
)
x2 +y 2
=
y2
(x2
y2
+ y 2 )3/2
+
x(2y)
−xy
1
= 2
= fyx
=−
2 (x2 + y 2 )3/2
(x + y 2 )3/2
fyy =
p
x2 + y 2 − y √
y
x2 +y 2
(x2 + y 2 )
=
x2
.
(x2 + y 2 )3/2
9. We have
fx = 15x2 y 2 − 7y 3 + 18x
so
fxx = 30xy 2 + 18
and
fxy = 30x2 y − 21y 2
fy = 10x3 y − 21xy 2 ,
and
fyx = 30x2 y − 21y 2
and
and
fyy = 10x3 − 42xy.
10. Since f (x, y) = sin(x2 + y 2 ), we have
fx = (cos (x2 + y 2 ))2x
2
2
, fy = (cos (x2 + y 2 ))2y
2
fxx = −(sin (x + y ))4x + 2 cos (x2 + y 2 )
fxy = −(sin (x2 + y 2 ))4xy = fyx
fyy = −(sin (x2 + y 2 ))4y 2 + 2 cos (x2 + y 2 ).
11. We have fx = 6 cos 2x cos 5y and fy = −15 sin 2x sin 5y, so fxx = −12 sin 2x cos 5y, fxy = −30 cos 2x sin 5y,
fyx = −30 cos 2x sin 5y, and fyy = −75 sin 2x cos 5y.
12. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = (y − 1)(x + 1)2
fx (x, y) = 2(y − 1)(x + 1)
fy (x, y) = (x + 1)2
fxx (x, y) = 2(y − 1)
fyy (x, y) = 0
fxy (x, y) = 2(x + 1)
14.7 SOLUTIONS
1337
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = −1 − 2x + y − x2 + 2xy
Notice this is the same as what you get if you expand (y − 1)(x + 1)2 and then keep only the terms of degree 2 or less.
13. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = (x − y + 1)2
fx (x, y) = 2(x − y + 1)
fy (x, y) = −2(x − y + 1)
fxx (x, y) = 2
fyy (x, y) = 2
fxy (x, y) = −2
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = 1 + 2x − 2y + x2 − 2xy + y 2
Notice this is the same as what you get if you expand (x − y + 1)2 .
14. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = e−2x
2
−y 2
fx (x, y) = −4xe−2x
fy (x, y) = −2ye−2x
fxx (x, y) = −4e−2x
fyy (x, y) = −2e−2x
2
−y 2
2 −y 2
2
−y 2
+ 16x2 e−2x
2
−y 2
+ 4y 2 e−2x
fxy (x, y) = 8xye−2x
2
2
2
−y 2
−y 2
−y 2
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = 1 − 2x2 − y 2
15. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = ex cos y
fx (x, y) = ex cos y
fy (x, y) = −ex sin y
fxx (x, y) = ex cos y
fyy (x, y) = −ex cos y
fxy (x, y) = −ex sin y
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
1 2 1 2
x − y
2
2
Notice this is the same as what you get if you multiply the quadratic approximations for ex and cos y, that is (1 + x +
x2 /2)(1 − y 2 /2), and then keep only the terms of degree 2 or less.
Q(x, y) = 1 + x +
1338
Chapter Fourteen /SOLUTIONS
16. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = (1 + 2x − y)−1
fx (x, y) = −2(1 + 2x − y)−2
fy (x, y) = (1 + 2x − y)−2
fxx (x, y) = 8(1 + 2x − y)−3
fyy (x, y) = 2(1 + 2x − y)−3
fxy (x, y) = −4(1 + 2x − y)−3
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = 1 − 2x + y + 4x2 − 4xy + y 2
Notice this is the same as what you get if you substitute u = y − 2x in the quadratic approximation Q(u) = 1 + u + u2
for 1/(1 − u).
17. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = cos(x + 3y)
fx (x, y) = − sin(x + 3y)
fy (x, y) = −3 sin(x + 3y)
fxx (x, y) = − cos(x + 3y)
fyy (x, y) = −9 cos(x + 3y)
fxy (x, y) = −3 cos(x + 3y)
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = 1 −
1 2
9
x − 3xy − y 2
2
2
Notice this is the same as what you get if you substitute x + 3y for u in the single variable quadratic approximation
Q(u) = 1 − u2 /2 for cos u.
18. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2
2
2
So first we find all the relevant derivatives
f (x, y) = sin 2x + cos y
fx (x, y) = 2 cos 2x
fy (x, y) = − sin y
fxx (x, y) = −4 sin 2x
fyy (x, y) = − cos y
fxy (x, y) = 0
We substitute into the formula to get for our answer:
Q(x, y) = 1 + 2x −
1 2
y
2
14.7 SOLUTIONS
19. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
So first we find all the relevant derivatives:
f (x, y) = ln(1 + x2 − y)
2x
fx (x, y) =
1 + x2 − y
−1
fy (x, y) =
1 + x2 − y
2(1 + x2 − y) − 4x2
fxx (x, y) =
(1 + x2 − y)2
−1
fyy (x, y) =
(1 + x2 − y)2
2x
fxy (x, y) =
(1 + x2 − y)2
Substituting into the formula we get as our answer:
Q(x, y) = −y + x2 −
y2
2
20. Since f (x, y) = ln(1 + x − 2y), the first and second derivatives are
fx =
fy =
fxx =
fxy =
fyy =
1
1 + x − 2y
−2
1 + x − 2y
−1
(1 + x − 2y)2
2
(1 + x − 2y)2
−4
,
(1 + x − 2y)2
so we find that
f (0, 0) = 0
fx (0, 0) = 1
fy (0, 0) = −2
fxx (0, 0) = −1
fxy (0, 0) = 2
fyy (0, 0) = −4.
The best quadratic approximation for f (x, y) for (x, y) near (0, 0) is
f (x, y) ≈ x − 2y −
21. Since f (x, y) =
√
1 2
x + 2xy − 2y 2 .
2
1 + 2x − y, the first and second derivatives are
1
1 + 2x − y
−1
fy = √
2 1 + 2x − y
−1
fxx =
(1 + 2x − y)3/2
fx = √
1339
1340
Chapter Fourteen /SOLUTIONS
1
2(1 + 2x − y)3/2
−1
=
,
4(1 + x − 2y)3/2
fxy =
fyy
so we find that
f (0, 0) = 1
fx (0, 0) = 1
fy (0, 0) = −1/2
fxx (0, 0) = −1
fxy (0, 0) = 1/2
fyy (0, 0) = −1/4.
The best quadratic approximation for f (x, y) for (x, y) near (0, 0) is
f (x, y) ≈ 1 + x −
1
1
1
1
y − x2 + xy − y 2 .
2
2
2
8
22. (a) fx (P ) < 0 because f decreases as you go to the right.
(b) fy (P ) = 0 because f does not change as you go up.
(c) fxx (P ) > 0 because fx increases as you go to the right (fx changes from a large negative number to a small negative
number).
(d) fyy (P ) = 0 because fy does not change as you go up.
(e) fxy (P ) = 0 because fx does not change as you go up.
23. (a) fx (P ) < 0 because f decreases as you go to the right.
(b) fy (P ) = 0 because f does not change as you go up.
(c) fxx (P ) < 0 because fx decreases as you go to the right (fx changes from a small negative number to a large negative
number).
(d) fyy (P ) = 0 because fy does not change as you go up.
(e) fxy (P ) = 0 because fx does not change as you go up.
24. (a) fx (P ) > 0 because f increases as you go to the right.
(b) fy (P ) = 0 because f does not change as you go up.
(c) fxx (P ) < 0 because fx decreases as you go to the right. (Since the level curves are further apart as you go to the
right, the rate of change of f decreases. Thus, fx changes from a large positive to a small positive number.)
(d) fyy (P ) = 0 because fy does not change as you go up.
(e) fxy (P ) = 0 because fx does not change as you go up.
25. (a) fx (P ) > 0 because f increases as you go to the right.
(b) fy (P ) = 0 because f does not change as you go up.
(c) fxx (P ) > 0 because fx increases as you go to the right. (The rate of change of f is larger when you go to the right
since the level curves are closer together. Thus fx changes from a small positive to a large positive number.)
(d) fyy (P ) = 0 because fy does not change as you go up.
(e) fxy (P ) = 0 because fx does not change as you go up.
fx (P ) = 0 because f does not change as you go to the right.
fy (P ) > 0 because f increases as you go up.
fxx (P ) = 0 because fx does not change as you go to the right.
fyy (P ) < 0 because fy decreases as you go up (Since the level curves are further apart as you move up, the rate of
change of f is slower, that is, fy decreases as you move up.)
(e) fxy (P ) = 0 because fx does not change as you go up.
26. (a)
(b)
(c)
(d)
fx (P ) = 0 because f does not change as you go to the right.
fy (P ) < 0 because f decreases as you go up.
fxx (P ) = 0 because fx does not change as you go to the right.
fyy (P ) < 0 because fy decreases as you go up. (fy changes from a negative number with smaller magnitude to a
negative number with larger magnitude.)
(e) fxy (P ) = 0 because fx does not change as you go up.
27. (a)
(b)
(c)
(d)
14.7 SOLUTIONS
1341
28. (a) fx (P ) < 0 because f decreases as you go to the right.
(b) fy (P ) < 0 because f decreases as you go up.
(c) fxx (P ) = 0 because fx does not change as you go to the right. (Notice that the level curves are equidistant and
parallel, so the partial derivatives of f do not change if you move horizontally or vertically.)
(d) fyy (P ) = 0 because fy does not change as you go up.
(e) fxy (P ) = 0 because fx does not change as you go up.
29. (a) fx (P ) > 0 because f increases as you go to the right.
(b) fy (P ) > 0 because f increases as you go up.
(c) fxx (P ) = 0 because fx does not change as you go to the right. (Notice that the level curves are equidistant and
parallel, so the partial derivatives of f do not change if you move horizontally or vertically.)
(d) fyy (P ) = 0 because fy does not change as you go up.
(e) fxy (P ) = 0 because fx does not change as you go up.
30. (a) fx (P ) < 0 because f decreases as you go to the right.
(b) fy (P ) > 0 because f increases as you go up.
(c) fxx (P ) > 0 because fx increases as you move right (fx changes from negative numbers with larger magnitude to
negative numbers with smaller magnitude).
(d) fyy (P ) > 0 because the level curves are closer together as you move up, so fy increases as you go up.
(e) fxy (P ) < 0 because the rate of change of f with respect to x is a negative number at P and a negative number with
larger magnitude higher up. Therefore fx decreases as the point moves up.
31. (a) fx (P ) > 0 because f increases as you go to the right.
(b) fy (P ) < 0 because f decreases as you go up.
(c) fxx (P ) < 0 because the level curves are further apart as you go to the right, so the rate of increase of f is slower as
you move to the right. Therefore, fx decreases as you go to the right.
(d) fyy (P ) < 0 because fy decreases as you move up (fy changes from a negative number with smaller magnitude to a
negative number with larger magnitude).
(e) fxy (P ) > 0 because the rate of change of f with respect to x at P is lower than at points above P . Therefore fx
increases as you move up.
Problems
32. We have f (1, 0) = 1 and the relevant derivatives are:
fx =
fy =
fxx =
fxy =
fyy =
1
1
(x + 2y)−1/2 so fx (1, 0) =
2
2
(x + 2y)−1/2 so fy (1, 0) = 1
1
1
− (x + 2y)−3/2 so fxx (1, 0) = −
4
4
1
1
− (x + 2y)−3/2 so fxy (1, 0) = −
2
2
−(x + 2y)−3/2 so fyy (1, 0) = −1 .
Thus the linear approximation, L(x, y) to f (x, y) at (1, 0), is given by:
f (x, y) ≈ L(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0)
1
= 1 + (x − 1) + y .
2
The quadratic approximation, Q(x, y) to f (x, y) near (1, 0), is given by:
f (x, y) ≈ Q(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0) +
1
fxx (1, 0)(x − 1)2
2
1
fyy (1, 0)(y − 0)2
2
1
1
1
1
= 1 + (x − 1) + y − (x − 1)2 − (x − 1)y − y 2 .
2
8
2
2
The values of the approximations are
+ fxy (1, 0)(x − 1)(y − 0) +
L(0.9, 0.2) = 1 − 0.05 + 0.2 = 1.15
Q(0.9, 0.2) = 1 − 0.05 + 0.2 − 0.00125 + 0.01 − 0.02 = 1.13875
1342
Chapter Fourteen /SOLUTIONS
and the exact value is
√
f (0.9, 0.2) = 1.3 ≈ 1.14018.
Observe that the quadratic approximation is closer to the exact value.
33. We have f (1, 0) = 0 and the relevant derivatives are:
fx = 2xy
fy = x2
so fx (1, 0) = 0
so fy (1, 0) = 1
fxx = 2y
so
fxx (1, 0) = 0
fxy = 2x so
fxy (1, 0) = 2
fyy = 0 so fyy (1, 0) = 0 .
Thus the linear approximation, L(x, y) to f (x, y) at (1, 0), is given by:
f (x, y) ≈ L(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0)
=y.
The quadratic approximation, Q(x, y) to f (x, y) near (1, 0), is given by:
f (x, y) ≈ Q(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0) +
+ fxy (1, 0)(x − 1)(y − 0) +
1
fxx (1, 0)(x − 1)2
2
1
fyy (1, 0)(y − 0)2
2
= y + 2(x − 1)y .
The values of the approximations are
L(0.9, 0.2) = 0.2
Q(0.9, 0.2) = 0.2 + 2(−0.1)(0.2) = 0.16
and the exact value is
f (0.9, 0.2) = (0.81)(0.2) = 0.162.
Observe that the quadratic approximation is closer to the exact value.
34. We have f (1, 0) = 1 and the relevant derivatives are:
fx = e−y
so
fy = −xe−y
fx (1, 0) = 1
so fy (1, 0) = −1
fxx = 0 so fxx (1, 0) = 0
fxy = −e−y
fyy = xe
−y
so fxy (1, 0) = −1
so fyy (1, 0) = 1 .
Thus the linear approximation, L(x, y) to f (x, y) at (1, 0), is given by:
f (x, y) ≈ L(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0)
= 1 + (x − 1) − y .
The quadratic approximation, Q(x, y) to f (x, y) near (1, 0), is given by:
f (x, y) ≈ Q(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0) +
1
fxx (1, 0)(x − 1)2
2
1
fyy (1, 0)(y − 0)2
2
1
= 1 + (x − 1) − y − (x − 1)y + y 2 .
2
The values of the approximations are
+ fxy (1, 0)(x − 1)(y − 0) +
L(0.9, 0.2) = 1 − 0.1 − 0.2 = 0.7
Q(0.9, 0.2) = 1 − 0.1 − 0.2 + 0.02 + 0.02 = 0.74
and the exact value is
f (0.9, 0.2) = (0.9)e−0.2 ≈ 0.737
Observe that the quadratic approximation is closer to the exact value.
14.7 SOLUTIONS
35. Differentiating, we get
Fx = ex sin y + ey cos x Fy = ex cos y + ey sin x
Fxx = ex sin y − ey sin x
Fyy = −ex sin y + ey sin x = −Fxx
Thus, Fxx + Fyy = 0.
36. We have f (1, 0) = 0 and the relevant derivatives are:
fx = cos(x − 1) cos y
fy = − sin(x − 1) sin y
so
fx (1, 0) = 1
so fy (1, 0) = 0
fxx = − sin(x − 1) cos y
so
fxx (1, 0) = 0
so
fxy (1, 0) = 0
fyy = − sin(x − 1) cos y
so
fyy (1, 0) = 0 .
fxy = − cos(x − 1) sin y
Thus the linear approximation, L(x, y) to f (x, y) at (1, 0), is given by:
f (x, y) ≈ L(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0)
= x − 1.
The quadratic approximation, Q(x, y) to f (x, y) near (1, 0), is given by:
f (x, y) ≈ Q(x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (1, 0)(y − 0) +
+ fxy (1, 0)(x − 1)(y − 0) +
1
fxx (1, 0)(x − 1)2
2
1
fyy (1, 0)(y − 0)2
2
= x − 1.
Thus the linear and quadratic approximations are the same. The values of the approximations are
L(0.9, 0.2) = Q(0.9, 0.2) = −0.1,
and the exact value is
f (0.9, 0.2) = sin(−0.1) cos(0.2) ≈ −0.098
37. Differentiating, we get
Fx = −e−x sin y, Fy = e−x cos y, Fxx = e−x sin y, Fyy = −e−x sin y = −Fxx .
Thus, Fxx + Fyy = 0.
38. Differentiating, we get
1
y
−y
(− ) = 2
1 + ( xy ) x2
x + y2
1
x
1
( )= 2
Fy =
1 + ( xy )2 x
x + y2
−1
2xy
Fxx = (−y) 2
(2x) = 2
(x + y 2 )2
(x + y 2 )2
−x
−2xy
Fyy = 2
(2y) = 2
= −Fxx
(x + y 2 )2
(x + y 2 )2
Fx =
Thus, Fxx + Fyy = 0.
39. First let us take the partial derivatives:
ut = aeat sin (bx)
ux = beat cos (bx)
uxx = −b2 eat sin (bx)
Substituting into the equation, we have
aeat sin (bx) = ut = uxx = −b2 eat sin (bx)
So, a = −b2 .
1343
1344
Chapter Fourteen /SOLUTIONS
40. (a) Taking partial derivatives of u, we get
2
2
3
π
x2
1
ut = − (πt)− 2 e−x /(4t) + 2 e−x /(4t) √
4
4t
2 πt
=−
=
uxx =
=
4(πt)
e−x
3
2
1
√
2
2
/(4t)
e−x
/(4t)
+
x2
√
8t2
πt
e−x
x2
√
2
/(4t)
2
e−x /(4t)
4t πt
8t2 πt
1
−2x −x2 /(4t)
√
e
4t
2 πt
2
x
− √ e−x /(4t)
4t πt
2
1
x
−2x −x2 /(4t)
− √ e−x /(4t) − √
e
4t
4t πt
4t πt
2
2
x2
1
√ e−x /(4t)
− √ e−x /(4t) +
2
4t πt
8t πt
=−
ux =
π
+
So we have
ut = uxx ,
showing that u satisfies the heat equation.
(b) See Figure 14.32. Note that as time progresses the heat at the origin decreases and flows out toward the ends of the
rod, until at t = 10 the temperature appears to be leveling out toward being constant throughout the rod.
u(x, 0.01)
u(x, 10)
u(x, 0.1)
✙ u(x, 1)
❄
x
Figure 14.32
41. The graph of f is concave up as we move parallel to the x-axis from the point (0, 0), so fxx (0, 0) is positive. The graph
of f is concave down as we move parallel to the y-axis from the point (0, 0), so fyy (0, 0) is negative.
42. Since zy = g(x), zyy = 0, because g is a function of x only.
43. (a) zyx = zxy = 4y
∂
∂
(zxy ) =
(4y) = 0
(b) zxyx =
∂x
∂x
∂
(4y) = 4
(c) zxyy = zyxy =
∂y
44. (a) Moving parallel to the x-axis means that the z-labels on the contours increase, so z is an increasing function of x.
Moving parallel to the y-axis, the z-labels decrease, so z is a decreasing function of y.
(b) Since z is an increasing function of x, we have fx > 0. Similarly, fy < 0.
(c) Since the contours get closer together as we move parallel to the x-axis, we have fxx > 0. This means that z is
increasing faster and faster as x increases. Similar reasoning shows that fyy < 0.
(d) The vector grad f is perpendicular to the level curves and points in the direction of increasing f values. See Figure 14.33.
(e) The vector grad f is longer at P because the contours are closer together at P than at Q.
14.7 SOLUTIONS
1345
y
P
6
❯
1
2
3
4
5
4
3
5
6
7
8
9 10
2
Q
1
s
x
1
2
3
4
5
6
Figure 14.33
45. The table of values is linear, because all the rows are linear and have the same slope, 1, and all the columns are linear and
have the same slope, 2. A linear function has no quadratic terms, so the coefficients of x2 , xy, and y 2 are all zero. We
have d = e = f = 0.
46. The rows of the table are not linear, so there is at least one nonzero quadratic term in the polynomial P (x, y). The
coefficients of the quadratic terms are closely related to the second order partial derivatives.
Since the x-slope increases as we move along a row in the direction of increasing x, we have ∂ 2 P/∂x2 > 0. Hence
d = (∂ 2 P/∂x2 )/2 > 0.
Since the y-slope is constant as we move across the table in the direction of increasing x, we have ∂/∂x(∂P/∂y) =
0. Hence e = ∂ 2 P/∂x∂y = 0.
Since the y-slope is constant as we move down a column in the direction of increasing y, we have ∂ 2 P/∂y 2 = 0.
Hence f = (∂ 2 P/∂y 2 )/2 = 0.
47. The rows of the table are linear but they do not all have the same slope, so the polynomial P (x, y) is not linear. Alternatively, since the columns of the table are not linear, P (x, y) is not linear. There is at least one nonzero quadratic term in
P . The coefficients of the quadratic terms are closely related to the second order partial derivatives.
Since the x-slope is constant as we move across a row in the direction of increasing x, we have ∂ 2 P/∂x2 = 0.
Hence d = (∂ 2 P/∂x2 )/2 = 0.
Since the x-slope increases as we move down the table in the direction of increasing y, we have ∂/∂y(∂P/∂x) > 0.
Hence e = ∂ 2 P/∂y∂x > 0.
Since the y-slope decreases as we move down a column in the direction of increasing y, we have ∂ 2 P/∂y 2 < 0.
Hence f = (∂ 2 P/∂y 2 )/2 < 0.
48. Neither the rows nor the columns of the table are linear, so the polynomial P (x, y) is not linear. There is at least one
nonzero quadratic term in P . The coefficients of the quadratic terms are closely related to the second order partial derivatives.
Since the x-slope increases as we move across a row in the direction of increasing x, we have ∂ 2 P/∂x2 > 0. Hence
d = (∂ 2 P/∂x2 )/2 > 0.
Since the x-slope decreases as we move down the table in the direction of increasing y, we have ∂/∂y(∂P/∂x) < 0.
Hence e = ∂ 2 P/∂y∂x < 0.
Since the y-slope increases as we move down a column in the direction of increasing y, we have ∂ 2 P/∂y 2 > 0.
Hence f = (∂ 2 P/∂y 2 )/2 > 0.
49. (a) The vertical spacing between the contours just north and just south of the trail increases as you move eastward along
the trail.
1346
Chapter Fourteen /SOLUTIONS
10
20
Elevation in meters
0
101
Trail
1000
0
98
990
Figure 14.34
(b) Let h = f (x, y), where h is elevation in meters, and x and y are distances in meters east and north of the start of the
trail. Hence the trail begins at (0, 0) and lies along the line y = 0. We also know:
• The trail is level. Hence ∂h/∂x = 0 at all points of the trail.
• There is a mountain to the left. Hence ∂h/∂y > 0 at all points of the trail.
• The slope up to the left is getting more gentle as you hike east. Thus ∂h/∂y is a decreasing function of x. Hence
∂/∂x(∂h/∂y) = (∂ 2 h)/(∂x∂y) < 0.
(c) The decision to delay turning off the trail was based on the second mixed partial, (∂ 2 h)/(∂x∂y).
50. (a) Increasing L causes production Y to increase, so ∂Y /∂L > 0.
(b) The increase in production resulting from hiring an additional worker is approximately ∂Y /∂L. This number is larger
for larger values of K, so ∂Y /∂L is an increasing function of K. Its derivative with respect to K is positive. Hence
∂
∂K
∂Y
∂L
=
∂2Y
> 0.
∂K∂L
51. (a) Person A. Weight gain of 1 pound results in an approximate surface area increase of ∂S/∂w. We know that ∂S/∂w
is an increasing function of height h, because
∂
∂h
∂S
∂w
∂S
∂w
> 0.
< 0.
Thus ∂S/∂w is larger for the taller person, A.
(b) Person B. Weight gain of 1 pound results in an approximate surface area increase of ∂S/∂w. We know that ∂S/∂w
is a decreasing function of weight w, because
∂
∂w
Thus ∂S/∂w is greater for the lighter person, B.
52. (a) Since P and Q lie on the same level curve, we have a = k.
(b) We have b = fx and c = fy . Since the gradient of f at P (respectively Q) points toward M or away from M , from
the figure, we see fx (P ) and fy (P ) have opposite signs, while fx (Q) and fy (Q) have the same signs. Thus Q is the
point (x1 , y1 ), so P is (x2 , y2 ).
(c) Since b = fx (Q) > 0 and c = fy (Q) > 0, the value of f must increase as we go away from M . Thus, M must be a
minimum (the surface is a valley).
(d) Since M is a minimum, m = fx (P ) < 0 and n = fy (P ) > 0.
53. (a) Calculate the partial derivatives:
f (x, y) = sin x sin y
f (0, 0) = 0
f ( π2 , π2 ) = 1
fx (x, y) = cos x sin y
fx (0, 0) = 0 fx ( π2 , π2 ) = 0
fy (x, y) = sin x cos y
fy (0, 0) = 0 fy ( π2 , π2 ) = 0
fxx (x, y) = − sin x sin y fxx (0, 0) = 0 fxx ( π2 , π2 ) = −1
fxy (x, y) = cos x cos y
fxy (0, 0) = 1 fxy ( π2 , π2 ) = 0
fyy (x, y) = − sin x sin y fyy (0, 0) = 0 fyy ( π2 , π2 ) = −1
14.7 SOLUTIONS
1347
Thus, the Taylor polynomial about (0, 0) is
f (x, y) ≈ Q1 (x, y) = xy.
The Taylor polynomial about ( π2 , π2 ) is
f (x, y) ≈ Q2 (x, y) = 1 −
π
1
x−
2
2
y
(b)
π
1
y−
2
2
−
2
.
y
π
2
x
0
2
x
π
2
0
Figure 14.35: f (x, y) ≈ xy: Quadratic
approximation about (0, 0)
Figure 14.36: f (x, y) ≈
1 − 12 (x − π2 )2 − 21 (y − π2 )2 : Quadratic
approximation about ( π2 , π2 )
54. (a) The definition of fx is:
fx (a, b) = lim
h→0
f (a + h, b) − f (a, b)
.
h
(b) We define fxy = (fx )y as follows:
fxy (a, b) = (fx )y (a, b) = lim
k→0
fx (a, b + k) − fx (a, b)
.
k
(c) Substituting the expression for fx into the definition fxy :
fx (a, b + k) − fx (a, b)
k
f
(a
+ h, b + k) − f (a, b + k)
f (a + h, b) − f (a, b)
1
lim
= lim
− lim
h→0
k→0 k
h→0
h
h
fxy (a, b) = lim
k→0
= lim lim
k→0 h→0
f (a + h, b + k) − f (a, b + k) − f (a + h, b) + f (a, b)
.
hk
(d) Similarly,
f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b)
.
hk
(e) The numerators in the two expressions in part (c) and (d) are the same (just swap the middle terms), so the only
difference between them is the order in which the limits are taken. To be sure fxy and fyx are equal, we have to
assume we can swap the order of the limits. Swapping limits can be a tricky business, but it can be done in this case
if fxy and fyx are continuous.
fyx (a, b) = lim lim
h→0 k→0
55. (a)
(i) Dollars/Year. Negative. The value of the car decreases with age.
(ii) Dollars/Dollar. Positive. For two cars of the same age, the one that had the highest value when new costs more
now.
(b) The experts say that ∂P/∂C is a decreasing function of age. This means that ∂/∂A(∂P/∂C) = ∂ 2 P/(∂A∂C) < 0.
(c) The term eCA, because ∂ 2 P/(∂A∂C) = e.
1348
Chapter Fourteen /SOLUTIONS
56. (a) (II) At first ∂T /∂S > 0, but after increasing S, we have ∂T /∂S < 0. Hence ∂T /∂S is a decreasing function of S.
This means that ∂ 2 T /∂S 2 < 0.
(b) (I) At first ∂T /∂V > 0, but after increasing V , we have ∂T /∂V < 0. Hence ∂T /∂V is a decreasing function of V .
This means that ∂ 2 T /∂V 2 < 0.
(c) (III) At first ∂T /∂S < 0, but after increasing V , we have ∂T /∂S > 0. Hence ∂T /∂S is an increasing function of
V . This means that
∂T
∂2T
∂
> 0.
=
∂V ∂S
∂V ∂S
57. Let us first calculate the values of all the partial derivatives at (0, 0) that we need:
(x + 2y + 1)1/2 ,
f (x, y) =
1
2
fx (x, y) =
fxy (x, y) =
f (0, 0) =
,
− 14
− 12
(x + 2y + 1)
−3/2
,
(x + 2y + 1)−3/2 ,
fyy (x, y) = − (x + 2y + 1)
−3/2
,
1,
fx (0, 0) = 1/2,
(x + 2y + 1)−1/2 ,
fy (x, y) =
fxx (x, y) =
(x + 2y + 1)
−1/2
fy (0, 0) =
1,
fxx (0, 0) = −1/4,
fxy (0, 0) = −1/2,
fyy (0, 0) = −1.
(a) The local linearization L(x, y) of f at (0, 0) is given by
f (x, y) ≈ L(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y = 1 +
1
x + y.
2
(b) The second-order Taylor polynomial, Q(x, y), for f at (0, 0) is given by
f (x, y) ≈ Q(x, y)
= f (0, 0) + fx (0, 0)x + fy (0, 0)y +
= 1+
fxx (0, 0) 2
fyy (0, 0) 2
x + fxy (0, 0)xy +
y
2
2
1
1
1
1
x + y − x2 − xy − y 2 .
2
8
2
2
Notice that the local linearization of f is the same as the linear part of the Taylor polynomial of degree 2 for f . The
extra terms in the Taylor polynomial of degree 2 can be thought of as “correction terms” to the linear approximation.
(c) Table 14.8 records the values of f (x, y), L(x, y), and Q(x, y). Observe that the quadratic approximations Q(x, y)
are closer to the true values f (x, y) than are the linear approximations L(x, y). Of course both approximations are
exact at (0, 0).
Table 14.8 Linear and quadratic approximations to
f near (0, 0)
Point
Linear
Quadratic
True
(x, y)
L(x, y)
Q(x, y)
f (x, y)
(0, 0)
1
1
1
(0.1, 0.1)
1.15
1.13875
1.140175
(−0.1, 0.1)
1.05
1.04875
1.048809
(0.1, −0.1)
0.95
0.94875
0.948683
0.85
0.83875
0.836660
(−0.1, −0.1)
58. The contour diagrams in Figures 14.37–14.42 use the fact that
f (x, y) =
p
x + 2y + 1,
1
L(x, y) = 1 + x + y,
2
1
1
1
1
Q(x, y) = 1 + x + y − x2 − xy − y 2 .
2
8
2
2
1349
14.7 SOLUTIONS
y
y
y
0.6
0.6
0.6
1.7
1.4
1.4
1.4
1.1
−0.6
1.1
x
0.6
0.8
−0.6
0.8
−0.6
−0.6
−0.6
Figure 14.38: L(x, y)
y
Figure 14.39: Q(x, y)
y
2
3.8
3.5
3.2
2.9
2.6
2.3
2.0
1.7
1.4
1.1
0.8
2.0
1.7
1.4
1.1
x
0.8
y
2
2.6
2.3
2
−2
2
0.8
1.1
1.4
1.4
1.1
0.8
x
2
−2
−2
−2
Figure 14.40: f (x, y)
x
0.6
0.8
−0.6
Figure 14.37: f (x, y)
−2
1.1
x
0.6
x
2
−2
Figure 14.41: L(x, y)
Figure 14.42: Q(x, y)
The contours for f (x, y) and L(x, y) are straight lines; those for L(x, y) are equally spaced because L(x, y) is a
linear function. The contours for f (x, y) are straight lines because if we set
f (x, y) =
then
p
x + 2y + 1 = constant
x + 2y + 1 = constant.
However, the contours of f (x, y) are not equally spaced because f (x, y) is not linear.
In the “close up” diagram [−0.6, 0.6]×[−0.6, 0.6], the contours of Q(x, y) look like lines (though they are not). The
contour diagram of Q(x, y) is more similar to the contour diagram of f (x, y) than is L(x, y). This is because Q(x, y) is
a better approximation to f (x, y) than is L(x, y).
In the [−2, 2] × [−2, 2] diagram, the values on the level curves of L(x, y) and Q(x, y) show that neither of them is
a good approximation to f (x, y) away from the origin.
59. Letting G(t) = f (t, 0) so that G′ (t) = fx (t, 0), the Fundamental Theorem tells us that
Thus
Z a
fx (t, 0)dt = f (a, 0) − f (0, 0)
Ra
G′ (t)dt = G(a) − G(0).
Rb
H ′ (t)dt = H(b) − H(0).
t=0
t=0
′
Letting H(t) = f (a, t) so that H (t) = fy (a, t), the Fundamental Theorem tells us that
Thus
Z
b
t=0
fy (a, t)dt = f (a, b) − f (a, 0)
t=0
Thus
f (0, 0) +
Z
a
fx (t, 0)dt +
t=0
Z
b
fy (a, t)dt
t=0
= f (0, 0) + (f (a, 0) − f (0, 0)) + (f (a, b) − f (a, 0))
= f (a, b).
1350
Chapter Fourteen /SOLUTIONS
60. We have
Z
|f (a, b)| = f (0, 0) +
≤
Z
a
Z
a
t=0
fx (t, 0)dt +
Z
fx (t, 0)dt +
t=0
b
fy (a, t)dt
Z
Z
b
fy (a, t)dt
t=0
|fx (t, 0)| dt +
Adt +
Z
t=0
t=0
a
≤ |f (0, 0)| +
≤ 0+
a
b
Z
b
t=0
|fy (a, t)| dt
Bdt
t=0
t=0
= A |a| + B |b|
Strengthen Your Understanding
61. The function f (x, y) = x3 + y 4 is not the zero function. Nevertheless, the values of f , fx , fy , fxx , fxy , and fyy at (0, 0)
are all zero. Therefore, the quadratic approximation of f near (0, 0) is the zero function.
62. If fx = xy and fy = y 2 , then
∂
xy = x
∂y
∂ 2
y = 0.
=
∂x
fxy =
fyx
Since fxy and fyx are continuous, we expect fxy = fyx . Thus there is no function f with the given partial derivatives.
63. One example is f (x, y) = x2 + y 2 . More generally, functions of the form f (x, y) = g(x) + h(y) with g ′′ 6= 0 and
h′′ 6= 0 all have fxx 6= 0, fyy 6= 0, and fxy = 0.
64. Since the quadratic approximation of a function near (0, 0) depends only on the values of the function and its first and
second order partial derivatives at (0, 0), changing a function by adding another function for which all these values are
zero does not change the quadratic approximation. For example, adding x3 to a function f (x, y) does not change the
quadratic approximation near (0, 0).
For example, f (x, y) = 2x + y 2 and g(x, y) = 2x + y 2 + x3 both have quadratic approximation Q(x, y) = 2x + y 2
near (0, 0).
65. If f and g have exactly the same contour diagrams inside a circle about the origin, then any partial derivative of f of
any order has the same value at the point (0, 0) as the corresponding partial derivative of g. Since the Taylor polynomials
of degree 2 for f and g near (0, 0) are constructed from the values of the partial derivatives at (0, 0), the two Taylor
polynomials are identical. It makes no difference what the contour diagrams look like outside the circle. See Figure 14.43
for one possible solution.
y
y
f (x, y)
x
g(x, y)
x
Figure 14.43
14.8 SOLUTIONS
1351
Solutions for Section 14.8
Exercises
1. Not differentiable at (0, 0).
2. Not differentiable at (−1, 0).
3. Not differentiable at all points on the x or y axes.
4. Not differentiable where x = −2 or where y = 3; that is not differentiable at all points on the lines x = −2 and y = 3.
5. Differentiable at all points.
6. Not differentiable where x = 0; that is, not differentiable on y-axis.
7. Differentiable everywhere, since |x − 3|2 = (x − 3)2 .
8. Differentiable at all points, since cos |y| = cos y.
9. Not differentiable at (1, 2).
10. Differentiable at all points.
Problems
11. (a) The contour diagram for f (x, y) =
x
y
+
y
x
is shown in Figure 14.44.
y
2
3
2
4
−4
−3
−3
−4
−2
3 4
−4
−3
3
2
2
x
−3
−4
4
3
4
−2
Figure 14.44
(b) If x 6= 0 and y 6= 0 then f is differentiable at (x, y). Now we need to look at points of the form (x, 0), where x 6= 0
and (0, y), where y 6= 0. The function f is not differentiable at these points as it is not continuous.
(c) For x 6= 0 and y 6= 0,
1
y
fx (x, y) = − 2 .
y
x
So fx exists for x 6= 0, y 6= 0, and it is continuous.
For all points (x0 , 0) on the x-axis we have:
fx (x0 , 0) = lim
x→x0
f (x, 0) − f (x0 , 0)
= 0.
x − x0
Thus, fx exists but is not continuous at these points.
For points (0, y0 ) on the y-axis we have:
lim
x→0
f (x, y0 ) − f (0, y0 )
1
y0
= lim ( + 2 ).
x→0 y0
x
x
This limit does not exist, so the partial derivative fx (0, y0 ) does not exist.
1352
Chapter Fourteen /SOLUTIONS
Similarly, for x 6= 0 and y 6= 0,
x
1
+ .
y2
x
For points (0, y0 ) on the y-axis we have fy (0, y0 ) = 0, while fy (x0 , 0) does not exist for x0 6= 0.
Both fx (x, y) and fy (x, y) are continuous at (x, y) only for x 6= 0 and y 6= 0.
(d) We claim f is not continuous at (0, 0). Let x = t and y = t, where t → 0, t 6= 0. Then
fy (x, y) = −
f (x, y) = f (t, t) = 2,
t 6= 0.
for
So,
lim f (t, t) = 2 6= f (0, 0) = 0,
t→0
and therefore
lim
(x,y)→(0,0)
f (x, y) 6= f (0, 0).
Thus, f is not differentiable at (0, 0) since f is not continuous at (0, 0).
(e) From part (c) we have fx (0, 0) = 0 and fy (0, 0) = 0. The functions fx and fy are not continuous at (0, 0).
12. (a) The contour diagram for f (x, y) = 2xy/(x2 + y 2 )2 is shown in Figure 14.45.
y
2
−.25
−.5
−1
−2
.25
1
.5
2
.5
1
x
−1
−.5
−.25
.25
−2
Figure 14.45
(b) The function f is differentiable at all points (x, y) 6= (0, 0), as it is a rational fraction with denominator (x2 +y 2 )2 =
0 only when (x, y) = (0, 0).
(c) The partial derivatives of f at points (x, y) 6= (0, 0) are given by
fx (x, y) =
2y(y 2 − 3x2 )
,
(x2 + y 2 )3
fy (x, y) =
2x(x2 − 3y 2 )
.
(x2 + y 2 )3
Both fx and fy are continuous for (x, y) 6= (0, 0).
(d) The function f is not continuous at (0, 0). To see this, let x = y = t for t 6= 0. Then,
1
2t2
= 2,
4t4
2t
and so limt→0 f (t, t) does not exist. Hence, f is not differentiable at (0, 0).
(e) At (0, 0), the partial derivatives of f are given by
f (x, y) = f (t, t) =
fx (0, 0) = lim
x→0
0−0
f (x, 0) − f (0, 0)
= lim
= 0,
x→0
x
x
f (0, y) − f (0, 0)
0−0
= lim
= 0.
y
y→0
y
We claim that lim(x,y)→(0,0) fx (x, y) does not exist. To see this, let x = y = t for t 6= 0. Then,
fy (0, 0) = lim
y→0
fx (x, y) = fx (t, t) =
2t(t2 − 3t2 )
−4t3
1
=
=− 3
2
3
(2t )
8t6
2t
and so the limit
−1
2t3
does not exist. Hence, fx is not continuous at (0, 0). Similarly, fy is not continuous at (0, 0).
lim fx (t, t) = lim
t→0
t→0
14.8 SOLUTIONS
1353
13. (a) The contour diagram for f (x, y) = x2 y/(x4 + y 2 ) is shown in Figure 14.46.
y
2
.2
.4
.2
.4
.2
.2
.4
.4
2
−.
.4
4
2
−.
x
−.
−.2
−. 4
2
−
−.2
−.4
−2
−2
Figure 14.46
(b) The function f is differentiable at all (x, y) 6= (0, 0) as it is a rational fraction with denominator which is zero only
when (x, y) = (0, 0).
(c) The partial derivatives of f are given by
2xy(y 2 − x4 )
,
(x4 + y 2 )2
fx (x, y) =
fy (x, y) =
x2 (x4 − y 2 )
,
(x4 + y 2 )2
(x, y) 6= (0, 0),
for
for
(x, y) 6= (0, 0).
Both fx and fy are continuous at (x, y) 6= (0, 0).
(d) We use the definition of differentiability. If f were differentiable at (0, 0), then the linear approximation of f at (0, 0)
would be L(x, y) = mx + ny, where m = fx (0, 0) and n = fy (0, 0). We have
fx (0, 0) = lim
f (x, 0) − f (0, 0)
= 0,
x
fy (0, 0) = lim
f (0, y) − f (0, 0)
= 0.
y
x→0
y→0
So, we need to compute the limit:
lim
(x,y)→(0,0)
f (x, y) − L(x, y)
p
x2 + y 2
=
lim
(x,y)→(0,0)
x2 y
(x4 + y 2 )
This limit is not zero since if we choose x = y = t, t > 0, we have
x2 y
(x4
y2)
p
x2
y2
= √
p
x2 + y 2
1
t
= √
,
2
2 · |t| · (t + 1)
2(t2 + 1)
.
+
+
√
which converges to 1/ 2 6= 0, as t → 0, t > 0. Hence, f is not differentiable at (0, 0).
(e) The partial derivative fx is not continuous at (0, 0) since if we choose x = y = t 6= 0, we have
fx (x, y) = fx (t, t) =
2(1 − t2 )
2t2 (t2 − t4 )
= 2
,
4
2
2
(t + t )
(t + 1)2
and so,
lim fx (t, t) = 2 6= fx (0, 0) = 0.
t→0
Similarly fy is not continuous at (0, 0).
1354
Chapter Fourteen /SOLUTIONS
p
x2 + y 2 is shown in Figure 14.47.
14. (a) The contour diagram for f (x, y) = xy/
y
2
−1
−2
1
−.5
−.25
.5
.25
2
−.25
−.5
.25
.5
x
−1
1
−2
Figure 14.47
(b) By the chain rule, f is differentiable at all points (x, y) where x2 + y 2 6= 0, and so at all points (x, y) 6= (0, 0).
(c) The partial derivatives of f are given by
fx (x, y) =
y3
,
(x2 + y 2 )3/2
for
(x, y) 6= (0, 0),
x3
,
+ y 2 )3/2
for
(x, y) 6= (0, 0).
and
fy (x, y) =
(x2
Both fx and fy are continuous at (x, y) 6= (0, 0).
(d) If f were differentiable at (0, 0), the chain rule would imply that the function
g(t) =
would be differentiable at t = 0. But
n
f (t, t), t 6= 0
0,
t=0
1
1 t2
t2
= √ · |t|,
= √ ·
g(t) = √
2
|t|
2
2
2t
which is not differentiable at t = 0. Hence, f is not differentiable at (0, 0).
(e) The partial derivatives of f at (0, 0) are given by
fx (0, 0) = lim
x→0
f (x, 0) − f (0, 0)
= lim
x→0
x
f (0, y) − f (0, 0)
fy (0, 0) = lim
= lim
y→0
y→0
y
The limit
lim
(x,y)→(0,0)
√ x·0
x2 +02
−0
x
√ 0·y
02 +y 2
= lim
0−0
= 0,
x
= lim
0−0
= 0.
y
x→0
−0
y
y→0
fx (x, y) does not exist since if we choose x = y = t, t 6= 0, then
t3
t3
fx (x, y) = fx (t, t) =
= √
=
2
3/2
(2t )
2 2 · |t|3
1
√
,
2 2
1
− 2√
,
2
Thus, fx is not continuous at (0, 0). Similarly, fy is not continuous at (0, 0).
t > 0,
t < 0.
14.8 SOLUTIONS
1355
y
15. (a)
2
−.8
.8
.6
.4
−.6
−.4
−.2
−2
.2
2
−.2
x
.2
−.4
−.6
.4
.6
−.8
.8
−2
Figure 14.48
(b) f is differentiable at all (x, y) 6= (0, 0) as it is a rational function with nonvanishing denominator.
(c)
f (x, 0) − f (0, 0)
=0
fx (0, 0) = lim
x→0
x
f (0, y) − f (0, 0)
fy (0, 0) = lim
=0
y→0
y
(d) Let us use the definition. If f were differentiable, the linear approximation of f would be L(x, y) = mx + ny, where
m = fx (0, 0) = 0 and n = fy (0, 0) = 0. So let’s compute
lim
(x,y)→(0,0)
f (x, y) − L(x, y)
p
x2 + y 2
=
xy 2
.
(x,y)→(0,0) (x2 + y 2 )3/2
lim
This limit is not zero as, for x = y = t → 0, t > 0,
t→0
(x2
xy 2
t3 t > 0 1
= √
−→ √ .
2
3/2
+y )
2 2|t|3
2 2
Hence f is not differentiable at (0, 0).
(e)
g(t) = f (x(t), y(t)) =
ab2
ab2 t3
= 2
t
2
2
+ b )t
a + b2
(a2
So
ab2
.
+ b2
(f) fx (0, 0) · x′ (0) + fy (0, 0) · y ′ (0) = 0, as fx (0, 0) = fy (0, 0) = 0. Suppose the chain rule holds, then
g ′ (0) =
a2
g ′ (t) = fx (x(t), y(t)) · x′ (t) + fx (x(t), y(t))y ′ (t).
2
But g ′ (0) = a2ab+b2 from part (e) and g ′ (0) 6= 0 since a 6= 0 and b 6= 0. Hence the chain rule does not hold. This
happens because f was not differentiable at (0, 0).
(g) If ~
u = a~i + b~j , then a2 + b2 = 1 as ~
u is a unit vector. Thus,
f~u (0, 0) = lim
t→0
f (at, bt)
g(t)
a2 b
= lim
= g ′ (0) = 2
= a2 b.
t→0 t
t
a + b2
1356
Chapter Fourteen /SOLUTIONS
16. (a) The contour diagram of f (x, y) =
p
|xy| is shown in Figure 14.49.
z
y
2
.5
.5
1
1
1.5
−2
1.5
2
1.5
x
y
1.5
1
1
.5
.5
x
−2
Figure 14.49
Figure 14.50
p
(b) The graph of f (x, y) = |xy| is shown in Figure 14.50.
(c) f is clearly differentiable at (x, y) where x 6= 0 and y 6= 0. So we need to look at points (x0 , 0), x0 6= 0 and (0, y0 ),
y0 6= 0. At (x0 , 0):
f (x, 0) − f (x0 , 0)
=0
fx (x0 , 0) = lim
x→x0
x − x0
f (x0 , y) − f (x0 , 0)
= lim
fy (x0 , 0) = lim
y→0
y→0
y
(d)
p
|x0 y|
y
which does not exist. So f is not differentiable at the points (x0 , 0), x0 6= 0. Similarly, f is not differentiable at the
points (0, y0 ), y0 6= 0.
f (x, 0) − f (0, 0)
=0
x
f (0, y) − f (0, 0)
fy (0, 0) = lim
=0
y→0
y
fx (0, 0) = lim
x→0
√
(e) Let ~
u = (~i + ~j )/ 2:
f~u (0, 0) = lim
t→0+
f ( √t2 ,
√t )
2
t
− f (0, 0)
= lim
t→0+
q
t
t2
2
1
= √ .
2
We know that ∇f (0, 0) = ~0 because both partial derivatives are 0. But if f were differentiable,
f~u (0, 0) =
√
∇f (0, 0) · ~
u = fx (0, 0) · √12 + fy (0, 0) · √12 = 0. But since, in fact, f~u (0, 0) = 1/ 2, we conclude that f
is not differentiable.
17. (a) The contour diagram of f (x, y) = xy 2 /(x2 + y 4 ) is shown in Figure 14.51.
14.8 SOLUTIONS
y
.2
2
4
−
−.
.4
.2
−. 4
−. 2
−2
.4
.2
2
−. 2
−. 4
.2
.2
.4
−
4
.2
−.
x
.4
−2
Figure 14.51
(b) Let ~
u = a~i + b~j be the unit vector. Then
f~u (0, 0) = lim
t→0
ab2 t3
ab2
b2
f (at, bt)
= lim
= 2 =
t→0 t(a2 t2 + b4 t4 )
t
a
a
if a 6= 0
and
f~u (0, 0) = 0
if a = 0.
2
(c) f is not continuous at (0, 0). To see this let x = t , y = t, t → 0, t 6= 0. Then
f (x, y) =
t4
t4
1 t→0 1
= −→ 6= f (0, 0) = 0.
+ t4
2
2
So f is not differentiable at (0, 0) either.
18. (a) If f were differentiable at (0, 0), then
1
1
f~u (0, 0) = grad f (0, 0) · ~
u = fx (0, 0) · √ + fy (0, 0) · √ = 0
2
2
which contradict the information that f~u (0, 0) = 3.
(b) Let
2
(
y2
x
3
+
, x 6= 0 and y 6= 0,
f (x, y) = √2
y
x
0,
x = 0 or y = 0.
Then fx (0, 0) = 0, fy (0, 0) = 0:
2 √
√
f ( √t2 , √t2 ) − 0
3
t2
2
2
t
= lim √
·
+
·
= 3.
f~u (0, 0) = lim
t→0
t→0
t
2
t
2
t
2t
19. (a) Differentiating gives
fx (x, y) =
x4 y + 4x2 y 3 − y 5
(3x2 y 3 − y 3 )(x2 + y 2 ) − 2x(x3 y − xy 3 )
=
(x2 + y 2 )2
(x2 + y 2 )2
similarly,
x5 − 4x3 y 2 − xy 4
for (x, y) 6= (0, 0).
(x2 + y 2 )2
(b) We find the partial derivatives at the origin by using the limit definition:
fy (x, y) =
fx (0, 0) = lim
f (x, 0) − f (0, 0)
=0
x
fy (0, 0) = lim
f (0, y) − f (0, 0)
=0
y
x→0
y→0
1357
1358
Chapter Fourteen /SOLUTIONS
(c) Let’s compute:
lim
(x,y)→(0,0)
fx (x, y) =
x4 y + 4x2 y 3 − y 5
.
(x2 + y 2 )2
(x,y)→(0,0)
lim
Let’s switch to polar coordinates: x = r cos θ, y = r sin θ. Then (x, y) → (0, 0) is equivalent to r → 0. Therefore:
lim
(x,y)→(0,0)
fx (x, y) = lim
r→0
r 5 (cos4 θ sin θ + 4 cos4 sin3 θ − sin5 θ)
= 0 = fx (0, 0).
r4
Similarly, lim(x,y)→(0,0) fy (x, y) = 0 = fy (0, 0). So fx and fy are continuous.
(d) From part (c) f is differentiable at (0, 0).
20. The fact that f is differentiable says that its graph is well approximated by a plane near (a, b). Since a plane is a graph of
continuous function, it is reasonable to expect f to be continuous too. To prove this, we have to show that
lim
(x,y)→(a,b)
f (x, y) = f (a, b) or equivalently
lim f (a + h, b + k) = f (a, b).
h→0
k→0
Suppose f is differentiable at (a, b). Then there is a linear function
L(x, y) = f (a, b) + m(x − a) + n(y − b)
such that
lim
h→0
k→0
f (a + h, b + k) − L(a + h, b + k)
√
= 0.
h2 + k2
If this limit is 0, then we also have
lim f (a + h, b + k) − L(a + h, b + k) = 0.
h→0
k→0
Substituting for L(a + h, b + k), this gives
lim (f (a + h, b + k) − f (a, b) − mh − nk) = 0.
h→0
k→0
So,
lim f (a + h, b + k) − f (a, b) = 0.
h→0
k→0
Therefore f is continuous at (a, b).
Strengthen Your Understanding
21. The converse of this statement is true. If a function is differentiable at the origin, then it is continuous
at the origin.
p
However, a function can be continuous and not differentiable at a point; for example, f (x, y) = x2 + y 2 is continuous
but not differentiable at (0, 0).
22. A counterexample is provided by
f (x, y) =
(
xy
, (x, y) 6= (0, 0),
x2 + y 2
0,
(x, y) = (0, 0).
We have fx (0, 0) = fy (0, 0), but f (x, y) is not continuous or differentiable at (0, 0).
23. The function f (x, y) =
p
x2 + y 2 whose graph is a cone with vertex at the origin is not differentiable at the origin.
24. Taking f (x, y) = |x − 1| gives a function that is not differentiable on the line x = 1.
25. (a)
(b)
(c)
(d)
Differentiable; point is at top of hemisphere.
Not differentiable; hemisphere vertical at this point.
Not differentiable; top point of cone with ellipse-shaped cross-section.
Differentiable; point on side of cone.
SOLUTIONS to Review Problems for Chapter Fourteen
1359
Solutions for Chapter 14 Review
Exercises
1. Vector. Taking the gradient and substituting x = 1, y = 2 gives
3 −y/2
grad(x e
)
=
(1,2)
x3 −y/2~
e
j
i −
2
2 −y/2~
3x e
(1,2)
= 3e−1~i −
1 −1~
e j.
2
2. Scalar. The gradient of f (x, y) at x = 1, y = 1 is given by
= 2xy 3~i + 3x2 y 2~j
grad(x2 y 3 )
(1,1)
Thus the directional derivative is
= 2~i + 3~j .
(1,1)
~i + ~j
2+3
5
= √ = √
f~u = (2~i + 3~j ) · √
2
2
2
3. Vector. grad((cos x)ey + z) = −(sin x)ey~i + (cos x)ey~j + ~k .
4. Scalar. Differentiating with respect to x using the chain rule gives
2
∂f
= 2xy 2 ex y
∂x
so, differentiating again using the product rule, we have
2
2
∂2f
= 2y 2 ex y + 4x2 y 3 ex y .
2
∂x
5. Taking the derivative of f with respect to x and treating y as a constant we get fx = 2xy + 3x2 − 7y 6 . To find fy we
take the derivative of f with respect to y and treat x as a constant. Thus, fy = x2 − 42xy 5 .
∂w
= 12,800πgh − 960πgh2 .
6. First we distribute to get w = 6400πgh2 −320πgh3 . Taking the partial with respect to h we get
∂h
∂T
2π 1
π
7.
= √ · l−1/2 = √ .
∂l
g 2
lg
8. To find the derivative of B with respect to t, note that the variable is only in the exponent. Thus,
∂B
= P (1 + r)t ln(1 + r).
∂t
To find the derivative of B with respect to r, note that this time the exponent is a constant, so we can use the power rule
to obtain
∂B
= tP (1 + r)t−1 .
∂r
9. For both partial derivatives we use the quotient rule. Thus,
fx =
2xy(x2 + y 2 ) − x2 y2x
2x3 y + 2xy 3 − 2x3 y
2xy 3
=
=
,
(x2 + y 2 )2
(x2 + y 2 )2
(x2 + y 2 )2
and
fy =
x2 (x2 + y 2 ) − x2 y2y
x4 + x2 y 2 − 2x2 y 2
x4 − x2 y 2
=
= 2
.
2
2
2
2
2
2
(x + y )
(x + y )
(x + y 2 )2
1360
Chapter Fourteen /SOLUTIONS
10. To find the derivative of F with respect to r we first rewrite F = Gµmy(r 2 + y 2 )−3/2 and then use the chain rule. Thus,
3
∂F
= Gµmy −
(r 2 + y 2 )−5/2 · 2r.
∂r
2
To find the derivative of F with respect to y we use the quotient rule to obtain
Gµm(r 2 + y 2 )3/2 − Gµmy( 23 )(r 2 + y 2 )1/2 · 2y
∂F
=
.
∂y
(r 2 + y 2 )3
1
∂ p/q
∂f
= ep/q .
e
=
∂p
∂p
q h
i
−1
−1
∂ p/q
∂f
∂
p
=
e
epq
=
= (−pq −2 )epq = − 2 ep/q .
∂q
∂q
∂q
q
12. zx = − sin x, zx (2, 3) = − sin 2 ≈ −0.9
11.
13. Since we take the derivative with respect to N , we use the power rule to obtain fN = cαN α−1 V β .
14. This function is symmetric with respect to x and y. Therefore the answers look very similar. Using the chain rule we have
2(x − a)
(x − a)
fx = p
= p
2
2
2 (x − a) + (y − b)
(x − a)2 + (y − b)2
and
(y − b)
2(y − b)
= p
.
fy = p
2 (x − a)2 + (y − b)2
(x − a)2 + (y − b)2
15. Using the chain rule
√
∂
1
x
1
−1/2
tan ωx =
·x= √
√ · (ωx)
√ .
∂ω
2
cos2
ωx
2 ωx cos2
ωx
∂y
= cos(ct − 5x)c = c cos(ct − 5x)
∂t
17. Using the quotient rule
16.
zy =
=
=
(15xy − 8)(21x2 y 6 − 2y) − 15x(3x2 y 7 − y 2 )
(15xy − 8)2
315x3 y 7 − 168x2 y 6 − 30xy 2 + 16y − 45x3 y 7 + 15xy 2
(15xy − 8)2
270x3 y 7 − 168x2 y 6 − 15xy 2 + 16y
(15xy − 8)2
18. Using the quotient rule,
(2yβ + 5)exβ−3 x − 2yexβ−3
[(2yβ + 5)x − 2y]exβ−3
(5x − 2y + 2xyβ)exβ−3
∂α
=
=
=
2
2
∂β
(2yβ + 5)
(2yβ + 5)
(2yβ + 5)2
19. Using the chain rule,
∂ p
1
( 2πxyw − 13x7 y 3 v) = (2πxyw − 13x7 y 3 v)−1/2 (2πxy − 0)
∂w
2
πxy
.
= p
2πxyw − 13x7 y 3 v
SOLUTIONS to Review Problems for Chapter Fourteen
1361
20. Using the product and chain rules,
∂
∂λ
x2 yλ − 3λ5
√
λ2 − 3λ + 5
= (x2 y − 15λ4 )
=
=
=
p
λ2 − 3λ + 5 −
(2λ − 3)(x2 yλ − 3λ5 )
√
2 λ2 − 3λ + 5
(x2 y − 15λ4 ) · 2(λ2 − 3λ + 5) − (2λ − 3)(x2 yλ − 3λ5 )
√
2(λ2 − 3λ + 5) λ2 − 3λ + 5
1
λ2 − 3λ + 5
x2 y[2(λ2 − 3λ + 5) − (2λ − 3)λ] − 15λ4 · 2(λ2 − 3λ + 5) + (2λ − 3) · 3λ5
√
2(λ2 − 3λ + 5) λ2 − 3λ + 5
x2 y(−3λ + 10) − 3λ4 (8λ2 − 27λ + 50)
√
2(λ2 − 3λ + 5) λ2 − 3λ + 5
21. Using the chain rule and the quotient rule,
∂
∂w
x2 yw − xy 3 w7
w−1
7
2
x2 yw − xy 3 w7
w−1
7
=−
2
w−1
x2 yw − xy 3 w7
=−
=
7
2
w−1
x2 yw − xy 3 w7
−7/2
−9/2
−9/2
−9/2
·
(w − 1)(x2 y − 7xy 3 w6 ) − (x2 yw − xy 3 w7 )(1)
(w − 1)2
(w − 1)(x2 y − 7xy 3 w6 ) − (x2 yw − xy 3 w7 )
(w − 1)2
x2 y + 6xy 3 w7 − 7xy 3 w6
(w − 1)2
∂
(ex cos(xy) + ay 2 ) = ex cos(xy) + ex ·
22. To find the partial derivative with respect to x we treat y and a as constants: ∂x
(− sin(xy)) · y.
∂
To find the partial derivative with respect to y we treat x and a as constants: ∂y
(ex cos(xy)+ay 2) = ex ·(− sin(xy))·
x + 2ay.
∂
(ex cos(xy) + ay 2 ) = y 2 .
To find the partial derivative with respect to a we treat x and y as constants: ∂a
∂f0
1
1
1
C
1
√
√
23.
=
(− )(LC)−3/2 C = −
=−
∂L
2π
2
4π LC LC
4πL LC
24. The first order partial derivative fx is
fx = −x(x2 + y 2 )−3/2
Thus the second order partials are
fxx = −(x2 + y 2 )−3/2 + 3x2 (x2 + y 2 )−5/2 = (2x2 − y 2 )(x2 + y 2 )−5/2
and
fxy = 3xy(x2 + y 2 )−5/2 .
25. The first order partial derivatives are
ux = ex sin y,
uy = ex cos y.
Thus the second order partials are
uxx = ex sin y,
uyy = −ex sin y.
26. The first order partial derivative Vr is
Vr = 2πrh.
Thus the second order partials are
Vrr = 2πh,
Vrh = 2πr.
1362
Chapter Fourteen /SOLUTIONS
27. The first order partials are
fx = cos(x − 2y),
fy = −2 cos(x − 2y).
The second order partials fxx and fyx are
fxx = − sin(x − 2y),
fyx = 2 sin(x − 2y).
fxxy = 2 cos(x − 2y),
fyxx = 2 cos(x − 2y).
The third order partials are then
Note that these are equal as we would expect, since the order does not matter for higher order partial derivatives of smooth
functions.
28. We have
∂2
∂ 2 ax−bt
(e
) + 2 (eax−bt ) = (a2 + b2 )eax−bt .
2
∂x
∂t
29. Since fx = 2x, fy = 2y + 3y 2 and fz = 0, we have
grad f = 2x~i + (2y + 3y 2 )~j .
30. Since the partial derivatives are
fx = 3x2 − yz,
we have
fy = −xz,
fz = 3z 2 − xy,
∇f = (3x2 − yz)~i − xz~j + (3z 2 − xy)~k .
31. Since f (x, y, z) =
1
, we have
xyz
fx = −
1
,
x2 yz
fy = −
we have
grad f = −
1
xyz
1
xy 2 z
fz = −
1~
1
1
i + ~j + ~k
x
y
z
1
,
xyz 2
.
32. Since the partial derivatives are
fx = −y cos(y 2 − xy) and
fy = (2y − x) cos(y 2 − xy),
we have
∇f = cos(y 2 − xy)(−y~i + (2y − x)~j ).
33. Since the partial derivatives are
zx = (2x) cos (x2 + y 2 ),
and zy = (2y) cos (x2 + y 2 ),
∇z = 2x cos (x2 + y 2 )~i + 2y cos (x2 + y 2 )~j .
34. Since f (x, y, z) = xey + ln x + ln z, we have fx = ey + 1/x, fy = xey , fz = 1/z, so
grad f = ey +
1
1 ~
i + xey~j + ~k .
x
z
35. Since fx = 2x cos(x2 + y 2 + z 2 ), fy = 2y cos(x2 + y 2 + z 2 ) and fz = 2z cos(x2 + y 2 + z 2 ), we have
grad f = 2x cos(x2 + y 2 + z 2 )~i + 2y cos(x2 + y 2 + z 2 )~j + 2z cos(x2 + y 2 + z 2 )~k .
SOLUTIONS to Review Problems for Chapter Fourteen
1363
36. Since the partial derivatives are
fρ = sin φ cos θ,
we have
fφ = ρ cos φ cos θ,
fθ = −ρ sin φ sin θ,
∇f = sin φ cos θ~i + ρ cos φ cos θ~j − ρ sin φ sin θ~k .
37. Since the partial derivatives are
(t2 − 2t + 4)
1
∂f
√
= (t2 − 2t + 4)(− )s−3/2 = −
∂s
2
2s s
∂f
1
= √ (2t − 2)
∂t
s
we have
∂f ~
∂f ~
i +
j =
grad f =
∂s
∂t
38. Since the partial derivatives are
we have
(t2 − 2t + 4) ~
√
i +
−
2s s
x
fx = p
x2 + y 2
and
1
√ (2t − 2) ~j .
s
y
fy = p
,
x2 + y 2
1
(x~i + y~j ).
∇f = p
2
x + y2
39. Since the partial derivatives are
∂f
= cos(xy) · (y) − sin(xy) · (y) = y[cos(xy) − sin(xy)]
∂x
∂f
= cos(xy) · (x) − sin(xy) · (x) = x[cos(xy) − sin(xy)]
∂y
we have
grad f =
∂f ~
∂f ~
i +
j
∂x
∂y
= y[cos(xy) − sin(xy)]~i + x[cos(xy) − sin(xy)]~j .
40. We have fx = 2x, fy = 0 and fz = 0. Thus grad f = 2x~i and grad f (0, 0, 0) = ~0 .
41. We have fx = 2xz, fy = 0 and fz = x2 . Thus grad f = 2xz~i + x2~k and grad f (1, 1, 1) = 2~i + ~k .
42. We √
have grad f = 3x2~i − 3y 2~j , so grad f (2, −1) = 12~i − 3~j . A unit vector in the direction we want is ~u =
(1/ 2)(~i − ~j ). Therefore, the directional derivative is
grad f (−2, 1) · ~
u =
12 · 1 − 3(−1)
15
√
= √ .
2
2
43. We have grad f = ey~i +xey~j , so grad f (3, 0) = ~i +3~j . A unit vector in the direction we want is ~
u = (1/5)(4~i −3~j ).
Therefore, the directional derivative is
grad f (3, 0) · ~
u =
1 · 4 + 3(−3)
= −1.
5
44. We have grad f = 2x~i + 2y~j − 2z~k , so grad f (2, 3, 4) = 4~i + 6~j − 8~k . A unit vector in the direction we want is
u = (1/3)(2~i − 2~j + ~k ). Therefore, the directional derivative is
~
grad f (2, 3, 4) · ~
u =
4 · 2 + 6(−2) − 8 · 1
= −4.
3
1364
Chapter Fourteen /SOLUTIONS
45. The unit vector ~
u in the direction of ~v = ~i − ~k is ~
u =
√1 ~
i
2
fx (x, y, z) = 6xy 2 ,
−
√1 ~
k
2
. We have
and fx (−1, 0, 4) = 0
2
fy (x, y, z) = 6x y + 2z, and fy (−1, 0, 4) = 8
fz (x, y, z) = 2y,
and fz (−1, 0, 4) = 0.
So,
f~u (−1, 0, 4) = fx (−1, 0, 4)
=0
1
√
2
= 0.
u = − √119~i +
46. The unit vector ~
√3 ~
j
19
+
1
√
2
1
+ fy (−1, 0, 4)(0) + fz (−1, 0, 4) − √
2
1
+ 8(0) + 0 − √
2
√3 ~
k
19
is in the direction of ~v = −~i + 3~j + 3~k . We have
fx (x, y, z) = 6xy 2 ,
and fx (−1, 0, 4) = 0
2
fy (x, y, z) = 6x y + 2z, and fy (−1, 0, 4) = 8
fz (x, y, z) = 2y,
and fz (−1, 0, 4) = 0.
So,
1
f~u (−1, 0, 4) = 0 − √
19
+8
3
√
19
+0
3
√
19
24
= √ .
19
47. We have grad f = (ex+z cos y)~i − (ex+z sin y)~j + (ex+z cos y)~k , so grad f (1, 0, −1) = ~i + ~k . A unit vector in the
√
direction we want is ~
u = (1/ 3)(~i + ~j + ~k ). Therefore, the directional derivative is
grad f (1, 0, −1) · ~
u =
1·1+1·1
2
√
= √ .
3
3
48. The given curve is the contour for f (x, y) = x2 −y 2 passing through the point (2, 1). Thus a normal vector is grad f (2, 1).
We have grad f = 2x~i − 2y~j , so a normal vector is grad f (2, 1) = 4~i − 2~j .
49. The given surface is the level surface for f (x, y, z) = xy + xz + yz passing through the point (1, 2, 3). Thus a normal
vector is grad f (1, 2, 3). We have grad f = (y + z)~i + (x + z)~j + (x + y)~k , so a normal vector is grad f (1, 2, 3) =
5~i + 4~j + 3~k .
50. The given surface is the level surface for f (x, y, z) = z 2 − 2xyz − x2 − y 2 passing through the point (1, 2, −1). Thus a
normal vector is grad f (1, 2, −1). We have
grad f = (−2x − 2yz)~i + (−2y − 2xz)~j + (2z − 2xy)~k ,
so a normal vector is grad f (1, 2, −1) = 2~i − 2~j − 6~k .
51. Let f (x, y, z) = z 2 − 4x2 − 3y 2 so that the surface (a hyperboloid) is the level surface f (x, y, x) = 9. Since
grad f = −8x~i − 6y~j + 2z~k
we have
grad f (1, 1, 4) = −8~i − 6~j + 8~k .
Thus an equation of the tangent plane at (1, 1, 4) is
−8(x − 1) − 6(y + 6) + 8(z − 4) = 0
so
−4x − 3y + 4z = 9.
SOLUTIONS to Review Problems for Chapter Fourteen
52. Let f (x, y, z) = x3 − 2y 2 + z so that the given surface is the level surface f (x, y, z) = 0. Since
grad f = 3x2~i − 4y~j + ~k
we have
grad f (1, 0, −1) = 3~i + ~k .
Thus an equation of the tangent plane at (1, 0, −1) is
3(x − 1) + 0(y − 0) + 1(z − (−1)) = 0
or
3x + z = 2.
53. Let f (x, y, z) = z − 1/(xy). We have
Thus
grad f =
1
1 ~
i + 2 ~j + ~k .
x2 y
xy
grad f (1, 1, 1) = ~i + ~j + ~k .
The tangent plane to the level surface at (1, 1, 1) is
(x − 1) + (y − 1) + (z − 1) = 0
or
x + y + z = 3.
54. The first and second derivatives are
fx = 2xy 2 − 5y 3
fy = 2x2 y − 15xy 2
fxx = 2y 2
fxy = 4xy − 15y 2
fyx = 4xy − 15y 2
fyy = 2x2 − 30xy.
55. Using the chain rule we see:
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= sin y (cos t) + x(cos y) (− sin t)
= sin(cos t) cos t − sin t cos(cos t) sin t
= sin(cos t) cos t − sin2 t cos(cos t).
We can also solve the problem using one variable methods:
z = sin t sin(cos t)
d
dz
= (sin t sin(cos t))
dt
dt
d(sin(cos t))
d sin t
=
sin(cos t) + sin t
dt
dt
= cos t sin(cos t) + sin t cos(cos t)(− sin t)
= cos t sin(cos t) − sin2 t cos(cos t).
1365
1366
Chapter Fourteen /SOLUTIONS
56. This is a case where substituting is easier:
z = sin((2t)2 + (t2 )2 )
= sin(4t2 + t4 )
dz
= (8t + 4t3 ) cos(4t2 + t4 ).
dt
If you use the chain rule the solution is:
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= 2x cos(x2 + y 2 )(2) + 2y cos(x2 + y 2 )(2t)
= 2(2t) cos(4t2 + t4 )(2) + 2(t2 ) cos(4t2 + t4 )(2t)
= (2 · 2t · 2 + 2 · t2 · 2t) cos(4t2 + t4 )
= (8t + 4t3 ) cos(4t2 + t4 ).
57. Substituting into the chain rule gives
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= 2(x2 + y) · 2x · 2 + 2(x2 + y) · 2t
= (x2 + y)(8x + 4t)
= ((2t)2 + t2 )(8 · 2t + 4t)
= (4t2 + t2 )(20t)
= 5t2 (20t)
= 100t3 .
58. Substituting into the chain rule gives
∂z dx
∂z dy
dz
=
+
dt
∂x dt
∂y dt
= ((x + y)ex + ex ) · (2t) + ex · (−2t)
2
2
2
= (t2 + (1 − t2 ))et · (2t) + et · (2t) + et · (−2t)
2
= 2tet (t2 + 1 − t2 + 2t − 2t)
2
= 2tet .
This is a case where substituting is much easier:
z = (t2 + (1 − t2 ))et
= et
2
2
2
dz
= 2tet .
dt
59. Using the chain rule we see:
dz
∂z dx
∂z dy
=
+
dt
∂x dt
∂y dt
1
1
= · 3t2 + · 2t
x
y
1
1
· 2t
= 3 · 3t2 + 2
t
t +1
3
2t
= + 2
.
t
t +1
SOLUTIONS to Review Problems for Chapter Fourteen
1367
This problem can also be solved using one variable methods.
z = ln(t2 + 1) + ln t3
= ln(t2 + 1) + 3 ln t
2t
3
dz
= 2
+ .
dt
t +1
t
60. Substituting into the chain rule gives
∂z dp
∂z dq
dz
=
+
dt
∂p dt
∂q dt
= q cos(pq) · cos t + p cos(pq) · (− sin(t2 ) · 2t)
= cos(pq) cos t2 cos t − 2t sin t sin(t2 )
= cos(sin t cos t2 ) cos t2 cos t − 2t sin t sin(t2 ) .
61. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = (x + 1)3 (y + 2)
fx (x, y) = 3(x + 1)2 (y + 2)
fy (x, y) = (x + 1)3
fxx (x, y) = 6(x + 1)(y + 2)
fyy (x, y) = 0
fxy (x, y) = 3(x + 1)2
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = 2 + 6x + y + 6x2 + 3xy
Notice this is the same as what you get if you expand (x + 1)3 (y + 2) and keep only the terms of degree 2 or less.
62. The quadratic Taylor expansion about (0, 0) is given by
f (x, y) ≈ Q(x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y +
1
1
fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2 .
2
2
First we find all the relevant derivatives
f (x, y) = cos x cos 3y
fx (x, y) = − sin x cos 3y
fy (x, y) = −3 cos x sin 3y
fxx (x, y) = − cos x cos 3y
fyy (x, y) = −9 cos x cos 3y
fxy (x, y) = 3 sin x sin 3y
Now we evaluate each of these derivatives at (0, 0) and substitute into the formula to get as our final answer:
Q(x, y) = 1 −
1 2 9 2
x − y
2
2
Notice this is the same as what you get if you multiply together the quadratic approximations for cos x and cos 3y and
then keep only the terms of degree 2 or less.
1368
Chapter Fourteen /SOLUTIONS
63. The quadratic Taylor expansion about (3, 5) is given by
f (x, y) ≈ Q(x, y) = f (3, 5) + fx (3, 5)(x − 3) + fy (3, 5)(y − 5) +
fxy (3, 5)(x − 3)(y − 5) +
1
fxx (3, 5)(x − 3)2 +
2
1
fyy (3, 5)(y − 5)2 .
2
First we find all the relevant derivatives
f (x, y) = (2x − y)1/2
fx (x, y) = (2x − y)−1/2
1
fy (x, y) = − (2x − y)−1/2
2
fxx (x, y) = −(2x − y)−3/2
1
fyy (x, y) = − (2x − y)−3/2
4
1
fxy (x, y) = (2x − y)−3/2
2
Now we evaluate each of these derivatives at (3, 5) and substitute into the formula to get as our final answer:
Q(x, y) = 1 + (x − 3) −
1
1
1
1
(y − 5) − (x − 3)2 + (x − 3)(y − 5) − (y − 5)2 .
2
2
2
8
Problems
64. The gradient of (a) is 2x~i + 2y~j + 2z~k , which points radially outward from the origin and increases in magnitude, so
(a) goes with (IV).
The gradient of (b) is
x~i + y~j + z~k
p
,
x2 + y 2 + z 2
which points radially outward and has magnitude 1. Thus (b) goes with (V).
The gradient of (c) is 3~i + 4~j , which is constant and parallel to the xy-plane, so (c) goes with (I).
The gradient of (d) is 3~i + 4~k , which is constant and parallel to xz-plane, so (d) goes with (II).
65. (a) Let f (x, y, z) = 2x2 − 2xy 2 + az so that the given surface is the level surface f (x, y, z) = a. Since
grad f = (4x − 2y 2 )~i − 4xy~j + a~k
we have
grad f (1, 1, 1) = 2~i − 4~j + a~k .
Thus an equation of the tangent plane at (1, 1, 1) is
2(x − 1) − 4(y − 1) + a(z − 1) = 0
or
2x − 4y + az = a − 2.
(b) Substituting x = 0, y = 0, z = 0 into the equation for the tangent plane in part (a) we have 0 = a − 2. The tangent
plane passes through the origin if a = 2.
66. (a) This means you must pay a mortgage payment of $1090.08/month if you have borrowed a total of $92,000 at an
interest rate of 14%, on a 30-year mortgage.
(b) This means that the rate of change of the monthly payment with respect to the interest rate is $72.82; i.e., your
monthly payment will go up by approximately $72.82 for one percentage point increase in the interest rate for the
$92,000 borrowed under a 30-year mortgage.
(c) It should be positive, because the monthly payments will increase if the total amount borrowed is increased.
(d) It should be negative, because as you increase the number of years in which to pay the mortgage, you should have to
pay less each month.
SOLUTIONS to Review Problems for Chapter Fourteen
1369
67. (a) At Q, R, we have fx < 0 because f decreases as we move in the x-direction.
(b) At Q, P , we have fy > 0 because f increases as we move in the y-direction.
(c) At all four points, P, Q, R, S, we have fxx > 0, because fx is increasing as we move in the x-direction. (At P, S,
we see that fx is positive and getting larger; at Q, R, we see that fx is negative and getting less negative.)
(d) At all four points, P, Q, R, S, we have fyy > 0, so there are none with fyy < 0. The reasoning is similar to part (c).
68. Estimating from the contour diagram, using positive increments for ∆x and ∆y, we have, for point A,
∂n
∂x
∂n
∂y
(A)
(A)
foxes/km2
km
≈
1/2
1
1.5 − 1
=
=
≈ 0.06
67 − 59
8
16
≈
1/2
0.5 − 1
1
=−
=−
≈ −0.06
60 − 51
9
18
foxes/km2
.
km
So, from point A the fox population density increases as we move eastward. The population density decreases as we move
north from A.
At point B,
∂n
∂x
∂n
∂y
(B)
(B)
≈
1/4
0.75 − 1
1
=−
=−
≈ −0.01
135 − 115
20
80
foxes/km2
km
≈
1/2
0.5 − 1
1
=−
=−
≈ −0.05
120 − 110
10
20
foxes/km2
.
km
So, fox population density decreases as we move both east and north of B. However, notice that the partial derivative
∂n/∂x at B is smaller in magnitude than the others. Indeed if we had taken a negative ∆x we would have obtained an
estimate of the opposite sign. This suggests that better estimates forB are
∂n
∂x
∂n
∂y
(B)
(B)
≈0
foxes/km2
km
≈ −0.05
foxes/km2
.
km
At point C,
∂n
∂x
∂n
∂y
(C)
(C)
≈
2 − 1.5
1/2
1
=
=
≈ 0.02
135 − 115
20
40
≈
1/2
2 − 1.5
1
=
=
≈ 0.02
80 − 55
25
50
foxes/km2
km
foxes/km2
.
km
So, the fox population density increases as we move east and north of C. Again, if these estimates were made using
negative values for ∆x and ∆y we would have had estimates of the opposite sign. Thus, better estimates are
∂n
∂x
∂n
∂y
(C)
(C)
≈0
foxes/km2
km
≈0
foxes/km2
.
km
69. The derivative ∂c/∂x = b is the rate of change of the cost of producing one unit of the product with respect to the amount
of labor used (in man hours) when the amount of raw material used stays the same. Thus ∂c/∂x = b represents the hourly
wage.
70. (a) The difference quotient for approximating fu (u, v) is given by
fu (u, v) ≈
f (u + h, v) − f (u, v)
.
h
1370
Chapter Fourteen /SOLUTIONS
Putting (u, v) = (1, 3) and h = 0.001, the difference quotient is
1.001(1.0012 + 32 )3/2 − 1(12 + 32 )3/2
0.001
31.6639 − 31.6228
0.0411
≈
≈
≈ 41.1
0.001
0.001
fu (1, 3) ≈
(b) Using the derivative formulas
fu =
∂f
= (u2 + v 2 )3/2 + 3u2 (u2 + v 2 )1/2 = (u2 + v 2 )1/2 (4u2 + v 2 )
∂u
so
fu (1, 3) = (12 + 32 )1/2 · (4 · 12 + 32 ) ≈ 41.11
We see that the approximation in part (a) was reasonable.
71. Substituting for G, M , m, and r, we have
F =
(6.67 · 10−11 )(6 · 1024 )(70)
GM m
=
= 684 newtons.
2
r
(6.4 · 106 )2
The gravitational force on this person is about 684 newtons.
Differentiating gives
GM
∂F
= 2 = 9.77 newtons/kg
∂m
r
and
∂F
−2GM m
=
= −0.000214 newtons/meter.
∂r
r3
These partial derivatives tell us that the gravitational force increases by about 9.77 newtons for an increase of 1 kg in
the mass, and the gravitational force decreases by about 0.000214 newtons if the distance from the center of the earth
increases by 1 meter.
72. (a) The area of a circle of radius r is given by
A = πr 2
and the perimeter is
L = 2πr.
Thus we get
r=
and
A=π
Thus we get
L 2
2π
π=
L
2π
=
L2 π
L2
=
.
2
4π
4π
L2
.
4A
(b) We will treat π as a function of L and A.
dπ =
∂π
2L
L2
∂π
dL +
dA =
dL −
dA.
∂L
∂A
4A
4A2
If L is in error by a factor λ, then ∆L = λL, and if A is in error by a factor µ, then ∆A = µA. Therefore,
2L
L2
∆L −
∆A
4A
4A2
2
2L
L
=
λL −
µA
4A
4A2
2
2
µL
L2
2λL
−
= (2λ − µ)
= (2λ − µ)π,
=
4A
4A
4A
∆π ≈
so π is in error by a factor of 2λ − µ.
SOLUTIONS to Review Problems for Chapter Fourteen
1371
73. (a) The linear approximation
∆F ≈ Fx ∆x + Fy ∆y
gives, for x = 400 and y = 50,
∆F ≈ 40x−1/3 y 1/3 ∆x + 20x2/3 y −2/3 ∆y = 20∆x + 80∆y.
The company increases its skilled labor by 5 hours, so ∆y = 5. Since output is to remain constant, we have ∆F = 0.
Making these substitutions in the linear approximation we get
20∆x + 80 · 5 = 0,
so ∆x = −20.
With 5 additional hours per day of skilled labor and 20 fewer hours per day of unskilled labor, the company can
keep its output at the current level.
(b) When x = 400 and y = 50, output is given by
F (400, 50) = 60 · 4002/3 · 501/3 = 12,000
When y is increased to 55, if output remains constant, we have
12,000 = 60 · x2/3 · 551/3 .
Solving for x gives
3/2
12000
= 381.385 hours per day.
60 · 551/3
Thus, the number of hours of unskilled labor is reduced by 400 − 381.385 = 18.615 hours per day.
x=
74. (a) If the volume is held constant, ∆V = 0, so ∆U ≈ 27.32∆T . Thus the energy increases if the temperature increases.
(b) If the temperature is held constant, then ∆T = 0, so ∆U ≈ 840∆V . Thus the energy increases if the volume
increases (yes, it sounds bizarre, but remember the temperature is being held constant).
(c) First, we convert 100 cm3 to 0.0001 m3 . Now, using the differential approximation,
∆U ≈ 840 ∆V + 27.32 ∆T
= (840)(−0.0001) + (27.32)(2)
= −0.084 + 54.64 ≈ 55 joules.
75. Since grad f = fx~i + fy~j , we see that fx is given by the ~i -component of grad f and fy is given by the ~j -component
of grad f . Also, fxx is the rate of change of fx in the x-direction and fyy is the rate of change of fy in the y-direction.
(a) At P, S, we have fx > 0.
(b) At R, S, we have fy < 0.
(c) At all four points, P, Q, R, S, we have fxx > 0 because fx increases as we move in the x-direction. (At P, S, we
see fx is positive and gets larger; at Q, R, we see fx is negative and getting less negative.)
(d) No points. At P, Q, the value of fy is positive and increasing as y increases. At R, S, the value of fy is negative and
increasing (getting less negative) as y increases.
76. The directional derivative is approximately the change in z (as we move in direction ~v ) divided by the horizontal change
1
2−1
= ≈ 0.3.
in position. The directional derivative is f~i ≈
4−1
3
77. The directional derivative is approximately the change in z (as we move in direction ~v ) divided by the horizontal change
1
2−1
= ≈ 0.3.
in position. The directional derivative is f~j ≈
4−1
3
78. The directional derivative is approximately the change in z (as we move in direction ~v ) divided by the horizontal change
2−1
1
in position. In the direction of ~v , the directional derivative ≈
= √ ≈ 0.7.
~
~
2
ki + j k
79. The directional derivative is approximately the change in z (as we move in direction ~v ) divided by the horizontal change
1.2
3.2 − 2
= √ ≈ 0.8.
in position. In the direction of ~v , the directional derivative ≈
2
k~i + ~j k
80. The directional derivative is approximately the change in z (as we move in direction ~v ) divided by the horizontal
change in position. In the direction of ~v we go from point (3, 3) to point (1, 4). We have the directional derivative
2−3
−1
≈
= √ ≈ −0.4.
5
k − 2~i + ~j k
1372
Chapter Fourteen /SOLUTIONS
81. The directional derivative is approximately the change in z (as we move in direction ~v ) divided by the horizontal change
in position. In the direction of ~v we go from point (4, 1) to point (2, 2), and the value of z remains unchanged. We have
the directional derivative ≈ 0.
82. The gradient vectors are perpendicular to the level curves. To determine the length of the gradient vector, we estimate √
the
rate of change of the function from the contour diagram. At (1, 1), the value of f changes from 1 to 2 in a distance of 2
(as it moves from (1, 1) to (2, 2)), so the length of grad f is √12 ≈ 0.7. At (1, 4), the value of f changes slightly faster
(the lines are closer together), so grad f is slightly longer here. In fact, the value of f changes from 2 to 3 as we move
from (1, 4) to (2.2, 4.2) a distance of about 1.2. So the new length is 1/1.2 = 0.8.
y
5
✿
4
4
3
3
2
2
✒
1
1
x
0
1
2
3
4
5
Figure 14.52: Gradient Vectors
83. (a) The gradient vector at the point x = 1, y = 2 is
∇z = ∇(x2 y) = 2xy~i + x2~j = 4~i + ~j .
The unit vector making an angle of 5π/4 with the x-axis is
~
u = cos
√
√
5π
2
2~
5π~
i + sin ~j = − ~i −
j.
4
4
2
2
The directional derivative in this direction is
f~u (1, 2) = ∇z · ~
u = (4~i + ~j ) ·
√
√
√
2 ~ ~
2
5 2
(−i − j ) = −
(4 + 1) = −
.
2
2
2
(b) The directional derivative is a maximum in the direction of the gradient vector ∇z = 4~i + ~j .
84. (a) Incorrect. k grad Hk is not the rate of change of H. In fact, there’s no such thing as the rate of change of H, although
directional derivatives can give its rate of change in a particular direction. For example, this expression would give
the wrong answer if the ant was crawling along a contour of H, since then the rate of change of the temperature it
experiences is zero even though k grad Hk and ~v might not be zero.
(b) Correct. If ~
u = ~v /k~v k, then
~v
k~v k = (grad H · ~
u )k~v k = H~u k~v k
k~v k
= (Rate of change of H in direction ~
u in deg/cm)(Speed of ant in cm/sec)
grad H · ~v = grad H ·
= Rate of change of H in deg/sec.
(c) Incorrect, this is the directional derivative, which gives the rate of change with respect to distance, not time.
85. (a) We choose a coordinate system with the origin at the buoy, the x-axis pointing east and the y-axis pointing north.
Then the boat is at the point with position vector ~i − 2~j . So a vector pointing in the direction
√ of the buoy from the
boat’s position is ~0 − (~i − 2~j ) = −~i + 2~j , and a unit vector in this direction is ~
u = (1/ 5)(−~i + 2~j ). We have
grad h = −30 · 2x~i − 20 · 2y~j = −60x~i − 40y~j ,
SOLUTIONS to Review Problems for Chapter Fourteen
1373
so at the boat’s position, x = 1 and y = −2, we have
grad h = −60 · 1~i − 40(−2)~j = −60~i + 80~j .
So the directional derivative of the depth in the direction of the buoy is
60 + 160
1
√
= 98.387 ft/mile.
grad h · ~
u = (−60~i + 80~j ) · √ (−~i + 2~j ) =
5
5
So the depth is increasing at a rate of 98.387 ft/mile.
(b) The boat is moving at 3 mph and from part (a) we know that the depth is changing at 98.387 ft/mi. So
Rate of change of depth with respect to time = (Speed)(Rate of change of depth with respect to distance)
ft
miles
98.387
= 295.161 ft/hour.
=3
hour
mile
86. (a) The vector grad f = 2~i − 5~j is perpendicular to the level curve at the point (1, 3). Now the vector 5~i + 2~j is
perpendicular to the vector 2~i − 5~j . Thus, the vector 5~i + 2~j is tangent to curve. (There are many other vectors with
this property, such as −5~i − 2~j , 10~i + 4~j , etc.) The slope of the tangent line is therefore 2/5. Since the tangent
line goes through the point (1, 3), its equation is
y−3 =
or
y=
2
(x − 1)
5
13
2
x+
.
5
5
(b) The surface z = f (x, y) can be written in the form
F (x, y, z) = f (x, y) − z = 0.
The normal the this surface is
Thus, at the point (1, 3, 7), the normal is
grad F = fx~i + fy~j − ~k .
grad F (1, 3, 7) = 2~i − 5~j − ~k .
Thus, the equation of the tangent plane is
2x − 5y − z = 2(1) − 5(3) − 7 = −20
2x − 5y − z + 20 = 0.
87. (a) We want the partial derivative with respect to x at the point (2, 1, 5), so
∂H
∂x
(2,1,5)
= −2xe−(x
2
+2y 2 +3z 2 )
(2,1,5)
= −4e−81 ◦ C/meter.
(b) By the chain rule,
∂H ∂x
∂H
=
·
= −4e−81 · 10 = −40e−81 ◦ C/sec.
∂t
∂x ∂t
(c) The magnitude of the gradient gives the maximum rate of change, so
grad H
(2,1,5)
= −e−(x
2
+2y 2 +3z 2 )
Thus
|| grad H|| = e−81
(2x~i + 4y~j + 6z~k )
(2,1,5)
p
42 + 42 + 302 =
√
= −e−81 (4~i + 4~j + 30~k ).
932e−81 ◦ C/meter.
1374
Chapter Fourteen /SOLUTIONS
88. The point (4, 1, 3) lies on the surface. The surface is the level surface of the function
F (x, y, z) = f (x, y) − z = 0.
The normal to the surface at the point (4, 1, 3) is
grad F (4, 1, 3) = fx (4, 1)~i + fy (4, 1)~j − ~k = 2~i − ~j − ~k .
Thus the equation of the tangent plane is
2x − y − z = 2(4) − 1 − 3 = 4
2x − y − z = 4
89. The temperature increases fastest in the direction of the gradient vector, namely grad T = −2x~i − 2y~j . Figure 14.53
shows that the gradient vector points radially inward.
grad T = −2x~i − 2y~j
T = 100
64 6
3
✛
Level curves of T
are circles
0
Figure 14.53
90. (a) If ~j points north and ~i points east, then the direction the car is driving is ~j − ~i . A unit vector in this direction is
1
~
u = √ (~j − ~i ).
2
The gradient of the height function is
grad h =
∂h ~
∂h ~
i +
j = 50~i + 100~j .
∂E
∂N
So the directional derivative is
100 − 50
1
√
= 35.355 ft/mi.
h~u = grad g · ~h = (50~i + 100~j ) · √ (~j − ~i ) =
2
2
(b) The car is traveling at v mi/hr, so
Rate of change of with respect to time = v
miles
ft
35.355
= 35.355v ft/hour.
hour
mile
91. At points (x, y) where the gradients are defined and are not the zero vector, the level curves of f and g intersect at right
angles if and only grad f · grad g = 0.
We have
x
grad f · grad g =
=
p
x2 + y 2
y
~j
+ 1 ~i + p
x2 + y 2
x2
−1
x2 + y 2
!
+
!
·
x
p
x2 + y 2
y2
= 2
− 1 = 0.
x2 + y 2
x + y2
x2 + y 2
!
y
~j
− 1 ~i + p
x2 + y 2
The level curves of f and g are parabolas that cross at right angles except at the point (0, 0). See Figure 14.54.
!
SOLUTIONS to Review Problems for Chapter Fourteen
1375
y
f = c1
x
g = c2
Figure 14.54
92. Since V = m/r, where r is the distance from (x, y, z) to (x0 , y0 , z0 ), we have r =
so
m
,
V = p
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
−m (x − x0 )
∂V
=
3/2 ,
2
∂x
(x − x0 ) + (y − y0 )2 + (z − z0 )2
p
(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ,
Similarly,
3m (x − x0 )2
m
∂2V
=
−
3/2
2
2 5/2
2
2
∂x2
(x − x0 ) + (y − y0 ) + (z − z0 )
(x − x0 ) + (y − y0 )2 + (z − z0 )2
and,
m
∂2V
3m (y − y0 )2
=
−
3/2
2
2
2
2 5/2
2
∂y
(x − x0 ) + (y − y0 ) + (z − z0 )
(x − x0 ) + (y − y0 )2 + (z − z0 )2
So
3m (z − z0 )2
∂2V
m
=
−
3/2 ,
2
2
2 5/2
2
∂z 2
(x − x0 ) + (y − y0 ) + (z − z0 )
(x − x0 ) + (y − y0 )2 + (z − z0 )2
3m (x − x0 )2 + (y − y0 )2 + (z − z0 )2
∂2V
∂2V
m
∂2V
+
+
=
−3
3/2
2
2
2 5/2
2
∂x2
∂y 2
∂z 2
(x − x0 ) + (y − y0 ) + (z − z0 )
(x − x0 ) + (y − y0 )2 + (z − z0 )2
=0
93. Write V = f (u) where u = x + ct, then using the chain rule
df ∂u
∂V
=
·
= f ′ (u)(1).
∂x
du ∂x
Similarly,
df ∂u
∂V
=
·
= f ′ (u)(c) = cf ′ (u).
∂t
du ∂t
Thus
∂V
∂V
= cf ′ (u) = c
.
∂t
∂x
94. (a) The chain rule gives
∂z
∂z ∂u
∂z ∂v
=
+
= (2u − ev )1 + (−uev )2.
∂x
∂u ∂x
∂v ∂x
At (x, y) = (1, 2), we have u = 1 + 2 · 2 = 5 and v = 2 · 1 − 2 = 0, so
∂z
∂y
(x,y)=(1,2)
= (2 · 5 − e0 )1 − 5e0 · 2 = −1.
1376
Chapter Fourteen /SOLUTIONS
(b) The chain rule gives
∂z ∂u
∂z ∂v
∂z
=
+
= (2u − ev )2 + (−uev )(−1).
∂y
∂u ∂y
∂v ∂y
At (x, y) = (1, 2), we have (u, v) = (5, 0), so
∂z
∂y
(x,y)=(1,2)
= (2 · 5 − e0 )2 + 5e0 = 23.
95. All are done using the chain rule.
(a) We have u = x, v = y, w = 3. Thus ∂u/∂x = 1, ∂v/∂x = 0, ∂w/∂x = 0, so
Gx (x, y) = Fu (x, y, 3)(1) + Fv (x, y, 3)(0) + Fw (x, y, 3)(0) = Fu (x, y, 3).
(b) We have u = 3, v = y, w = x. Thus ∂u/∂x = 0, ∂v/∂x = 0, ∂w/∂x = 1, so
Gx (x, y) = Fu (3, y, x)(0) + Fv (3, y, x)(0) + Fw (3, y, x)(1) = Fw (3, y, x).
(c) We have u = x, v = y, w = x. Thus ∂u/∂x = 1, ∂v/∂x = 0, ∂w/∂x = 1, so
Gx (x, y) = Fu (x, y, x)(1) + Fv (x, y, x)(0) + Fw (x, y, x)(1) = Fu (x, y, x) + Fw (x, y, x).
(d) We have u = x, v = y, w = xy. Thus ∂u/∂x = 1, ∂v/∂x = 0, ∂w/∂x = y, so
Gx (x, y) = Fu (x, y, xy)(1) + Fv (x, y, xy)(0) + Fw (x, y, xy)(y)
= Fu (x, y, xy) + yFw (x, y, xy).
96. Average productivity increases as x1 increases if
∂
(average
∂x1
productivity) > 0. Now
∂
∂
P
(average productivity) =
∂x1
∂x1 x1
1
1 ∂P
∂
+P
=
x1 ∂x1
∂x1 x1
P
1 ∂P
P
1 ∂P
− 2 =
−
=
x1 ∂x1
x1
x1 ∂x1
x1
So
∂
(average productivity) > 0 means that
∂x1
P
∂P
−
∂x1
x1
> 0, i.e.,
∂P
P
>
.
∂x1
x1
97. The differential is
∂P
∂P
dL +
dK = 10L−0.75 K 0.75 dL + 30L0.25 K −0.25 dK.
∂L
∂K
When L = 2 and K = 16, this is
dP ≈ 47.6 dL + 17.8 dK.
dP =
98. The error, dT , in the period T is given by
dT =
∂T
∂T
dl +
dg,
∂l
∂g
where
∂T
=
∂l
q
and
∂T
=−
∂g
l
g
π
l
q
l
g
g
π
,
SOLUTIONS to Review Problems for Chapter Fourteen
1377
so that
Tl (2, 9.8) = 0.7096,
We also have that
dl = −0.01,
Tg (2, 9.8) = −0.1448.
dg = 0.01.
The maximum discrepancy in the period is then given by
dT = 0.7096(−0.01) − 0.1448(0.01) = −0.008544.
99. Looking at the contour diagram, we see that the contours are almost straight lines that are reasonably evenly spaced, so
treating the function as linear is not a bad approximation. The monthly payment m is a function of two variables, P , the
dollars you borrow, and r, the interest rate. We are going to use the formula
m(P, r) ≈ m(P0 , r0 ) + mp (P0 , r0 )(P − P0 ) + mr (P0 , r0 )(r − r0 )
to approximate the payment m(P, r). The amount of $6000 and the interest rate 11% would probably be reasonable
choices for our P0 and r0 .
Directly from the Figure 12.8 on page 691, we have
m(6000, 0.11) = $130
Next we approximate mp (6000, 0.11) by a difference quotient.
mp (6000, 0.11) ≈
m(6000 + h, 0.11) − m(6000, 0.11)
h
Here we choose h = 500, and we get m(6500, 0.11) = 140 from the figure, so
mp (6000, 0.11) ≈
140 − 130
= 0.02
500
Next we approximate mr (6000, 0.11) by taking h = 0.03 and m(6000, 0.11 + 0.03) = m(6000, 0.14) = 140 from the
figure,
m(6000, 0.11 + h) − m(6000, 0.11)
140 − 130
=
≈ 333.33
mr (6000, 0.11) ≈
h
0.03
Thus we have:
m(P, r) ≈ 130 + 0.02(P − 6000) + 333.33(r − 0.11)
= −26.67 + 0.02P + 333.33r
in dollars
The constants in the answer tell us several useful things. The slope of 0.02 along the P axis tells us how much our monthly
payment will increase if we decide to borrow more money; we can expect to pay about 2 cents on each extra dollar every
month. The slope in the r direction tells us how much our payment will change if the interest rate changes. We can expect
to pay an extra $333.33 each month for each point the interest rate goes up. The constant c is a gauge of the non-linearity
of this function. We know that if we borrow no money at zero percent interest, we should expect to not have a monthly
payment (m = 0). The constant we calculated is negative which implies that the bank will pay us $26.67 each month if
we do not borrow any money! So, we should bear in mind that the function we have calculated is only useful close to the
point at which we made the approximation, in this case P0 = 6000 and r0 = 11%.
100. (a) The local linearization of f near (2, 1) is 7 − 3(x − 2) + 4(y − 1), so the equation of the tangent plane is:
z = 7 − 3(x − 2) + 4(y − 1).
(b) We use local linearity. The contour is f (x, y) = 7, so to find the tangent line to the contour, we set the local
linearization for f equal to 7:
7 − 3(x − 2) + 4(y − 1) = 7
Thus, an equation for the tangent line is
−3(x − 2) + 4(y − 1) = 0.
1378
Chapter Fourteen /SOLUTIONS
101. (a) Rewrite the equation of the tangent plane:
−3(x − 2) + 4(y − 1) − (z − 7) = 0.
Then we see a normal vector to the plane is −3~i + 4~j − ~k .
(b) Since the gradient is perpendicular to contours, a normal vector to the tangent line is −3~i + 4~j . Or, setting the local
linearization for f equal to 7, we get that the equation for the tangent line is −3(x − 2) + 4(y − 1) = 0.Thus a
normal vector to the tangent line is −3~i + 4~j .
102. The directional derivative, or slope, of f at (2, 1) in the direction perpendicular to these contours is the length of the
gradient. At the point (2, 1), we have grad f = −3~i + 4~j , so k grad f k = 5. Thus if d is the distance between the
contours, we have
7.3 − 7
= Slope in direction perpendicular to contours = 5.
d
Thus d = 0.3/5 = 0.06.
103. Using local linearity with the slope in the x direction of −3 and slope in the y direction of 4, we get the values in
Table 14.9.
Table 14.9
y
x
0.9
1.0
1.1
1.8
7.2
7.6
8.0
2.0
6.6
7.0
7.4
2.2
6.0
6.4
6.8
104. See Figure 14.55.
y
2
13
12
11
10
9
8
1
7
6
5
4
3
2
1
1
x
3
Figure 14.55
105. The bug is heading in the direction of − grad f . In that direction, the rate of cooling in degrees per centimeter is given by
the magnitude of the gradient, which is 5. Since the bug is moving 3 centimeters per minute, in degrees per minute, we
have:
cm
degrees ◦ C
degrees ◦ C
Rate of cooling = 5
·3
= 15
.
cm
min
min
106. By the chain rule,
fr (2, 1) = fx (2, 1) cos θ + fy (2, 1) sin θ
fθ (2, 1) = fx (2, 1)(−r sin θ) + fy (2, 1)(r cos θ).
√
√
√
√
Since x + y = r , we have r = 22 + 12 = 5, and cos θ = x/r = 2/ 5, and sin θ = y/r = 1/ 5. Thus
2
2
2
1
2
2
fr (2, 1) = (−3) √ + (4) √ = − √
5
5
5
fθ (2, 1) = (−3)(−1) + (4)(2) = 11.
√
√
For ~
u = (1/ 5)(2~i + ~j ), we have f~u (2, 1) = −2/ 5. Since ~
u is pointing radially out from the origin to the point
(2, 1), we expect that f~u (2, 1) = fr (2, 1).
SOLUTIONS to Review Problems for Chapter Fourteen
1379
107. To increase f as much as possible, we should head in the direction of the gradient from the point (2, 1). The rate of
increase of f in the direction of the gradient is the magnitude of the gradient. Since at the point (2, 1), we know grad f =
−3~i + 4~j , the magnitude is 5. The furthest we can go from (2, 1), inside the circle, is 0.1 units, so the most we can
increase f is (0.1)(5) = 0.5. Thus
Largest value of function ≈ f (2, 1) + 0.5 = 7.5.
This value is achieved at the point obtained from (2, 1) by a displacement of 0.1 units in the direction of grad f , that is,
a displacement by the vector
(0.1)
grad f
= (0.1)((−3/5)~i + (4/5)~j ) = −0.06~i + 0.08~j .
k grad f k
Thus, the largest value of f is achieved at the point (2 − 0.06, 1 + 0.08) = (1.94, 1.08).
108. (a) Fix y = 3. When x changes from 2.00 to 2.01, f (x, 3) decreases from 7.56 to 7.42. So
∂f
∂x
(2,3)
≈
∆f
∆x
=
(2,3)
7.42 − 7.56
−0.14
=
= −14.
2.01 − 2.00
0.01
Fix x = 2, when y changes from 3.00 to 3.02, f (2, y) increases from 7.56 to 7.61. So
∂f
∂y
(2,3)
≈
∆f
∆y
=
(2,3)
0.05
7.61 − 7.56
=
= 2.5.
3.02 − 3.00
0.02
(b) Since the unit vector ~
u of the direction ~i + 3~j is
~
u =
~i + 3~j
1
3
= √ ~i + √ ~j ,
~
10
10
ki + 3~j k
f~u (2, 3) = grad f (2, 3) · ~
u ≈
= (−14~i + 2.5~j ) ·
(c) Maximum rate equals k grad f k ≈
mately equal to −14~i + 2.5~j .
(d) The equation of the level curve is
p
∆f
∆x
~i + ∆f
∆y
(2,3)
1
3
√ ~i + √ ~j
10
10
~j
(2,3)
!
· ~u
6.5
= − √ ≈ −2.055.
10
(−14)2 + (2.5)2 ≈ 14.221 in the direction of the gradient which is approxi-
f (x, y) = f (2, 3) = 7.56.
(e) The vector must be perpendicular to grad f , so ~v = 2.5~i + 14~j is a possible answer. (There are many others.).
(f) The differential at the point (2, 3) is
df = −14 dx + 2.5 dy.
If dx = 0.03, dy = 0.04, we get
df = −14(0.03) + 2.5(0.04) = −0.32.
The df approximates the change in f when (x, y) changes from (2, 3) to (2.03, 3.04).
109. Let us first collect the computations that we will need.
f (x, y) = cos (x + 2y) sin (x − y),
fx (x, y) = cos (x + 2y) cos (x − y) − sin (x − y) sin (x + 2y)
= cos (x + 2y + x − y) = cos (2x + y),
fy (x, y) = − cos (x + 2y) cos (x − y) − 2 sin (x − y) sin (x + 2y)
= − cos (x + 2y − (x − y)) − sin (x + 2y) sin (x − y)
1
= − cos (3y) + [cos (2x + y) − cos (3y)]
2
1380
Chapter Fourteen /SOLUTIONS
3
1
cos (2x + y) − cos (3y),
2
2
fxx (x, y) = −2 sin (2x + y),
=
fxy (x, y) = − sin (2x + y),
1
9
fyy (x, y) = − sin (2x + y) + sin (3y).
2
2
Then
f (0, 0) = 0,
fx (0, 0) = 1,
fy (0, 0) = −1,
fxx (0, 0) = 0,
fxy (0, 0) = 0,
fyy (0, 0) = 0.
Hence the quadratic Taylor polynomial P (x, y) of f (x, y) at (0, 0) is
P (x, y) = f (0, 0) + fx (0, 0)x + fy (0, 0)y
1
1
+ fxx (0, 0)x2 + fxy (0, 0)xy + fyy (0, 0)y 2
2
2
= x − y.
110. (a) The first-order Taylor polynomial of a function f about a point (a, b) is equal to
f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b).
Computing the partial derivatives, we get:
2
+(y−3)2
2
+(y−3)2
fx = 2(x − 1)e(x−1)
fy = 2(y − 3)e(x−1)
fx (0, 0) = 2(−1)e(−1)
= −2e10
fy (0, 0) = 2(−3)e(−1)
2
+(−3)2
2 +(−3)2
= −6e10
Thus,
f (x, y) ≈ e10 − 2e10 x − 6e10 y
(b) The second-order Taylor polynomial of a function f about the point (1, 3) is given by
f (1, 3) + fx (1, 3)(x − 1) + fy (1, 3)(y − 3)
1
1
+ fxx (1, 3)(x − 1)2 + fxy (1, 3)(x − 1)(y − 3) + fyy (1, 3)(y − 3)2 .
2
2
Computing the partial derivatives, we get:
2
+(y−3)2
2
+(y−3)2
fx = 2(x − 1)e(x−1)
fy = 2(y − 3)e(x−1)
fxx = (4(x − 1)2 + 2)e(x−1)
2
fxy = 4(x − 1)(y − 3)e(x−1)
fyy = (4(y − 3)2 + 2)e(x−1)
Substituting in the point (1, 3) to these partial derivatives, we get:
fx (1, 3) = 0
+(y−3)2
2
2
+(y−3)2
+(y−3)2
SOLUTIONS to Review Problems for Chapter Fourteen
1381
fy (1, 3) = 0
fxy (1, 3) = 0
2
fxx (1, 3) = (4(0)2 + 2)e0
2
fyy (1, 3) = (4(0) + 2)e
Thus,
+02
02 +02
=2
=2
2
2
(x − 2)2 + 0(x − 1)(y − 3) + (y − 3)2
2
2
f (x, y) ≈ 1 + (x − 1)2 + (y − 3)2 .
f (x, y) ≈ e0 + 0(x − 1) + 0(y − 3) +
(c) A vector perpendicular to the level curve is grad f . At the point (0, 0), we have
grad f = fx (0, 0)~i + fy (0, 0)~j
Computing partial derivatives, we have
fx = 2(x − 1)e(x−1)
fy = 2(y − 3)e(x−1)
fx (0, 0) = 2(−1)e(−1)
= −2e10
fy (0, 0) = 2(−3)e(−1)
2 +(y−3)2
2
+(y−3)2
2
+(−3)2
2
+(−3)2
= −6e10
Therefore, a perpendicular vector is grad f = −2e10~i − 6e10~j . Any multiple of grad f , say −2~i − 6~j , will
do.
(d) Since the surface can be represented by the level surface
F (x, y, z) = f (x, y) − z = 0,
a vector perpendicular to the surface at (0, 0) is given by
grad F = fx (0, 0)~i + fy (0, 0)~j − ~k = −2e10~i − 6e10~j − ~k
111. (a) ft ( π2 , π2 ) is negative since we move from zero contour of f to negative contours as t slightly increases and x = π2 .
ft ( π2 , π) is positive since f increases with time t from t = π and x = π2 .
When ft ( π2 , b) is positive, the point on the string at x = π2 is moving upward at t = b. It moves downward for
π
ft ( 2 , b) negative at t = b.
(b) ft ( π2 , t) is positive for π < t < 2π. For these t, f moves to larger and larger contours at x = π2 .
(c) For any fixed values of t between 0 and π2 , f increases with x for x between 0 and 3π
. Also for any fixed t between
2
3π
5π
3π
and
,
f
increases
with
x
for
x
between
0
and
.
Hence,
f
(x,
t)
is
positive
for
0 < x < 3π/2 and t either
x
2
2
2
satisfying 0 < t < π/2 or 3π/2 < t < 5π/2.
CAS Challenge Problems
112. (a) Using a CAS to calculate the partial derivatives, we find
f (0, 0) =
fxx (0, 0) =
1
6
−1
54
fx (0, 0) =
1
9
fxy (0, 0) =
fy (0, 0) = 1
2
3
fyy (0, 0) = 3
Thus the quadratic approximation is
Q(x, y) =
x
x2
2xy
3y 2
1
+ +y−
+
+
.
6
9
108
3
2
(b) The CAS gives
1
x
x2
ex
= + −
+ ···
2x
5+e
6
9
108
(1 + sin 3y)2 = 1 + 6y + 9y 2 + · · ·
1382
Chapter Fourteen /SOLUTIONS
Multiplying the two together, we get
1
1
x
x2
x
x2
2xy
3y 2
+ −
+ · · · (1 + 6y + 9y 2 + · · ·) = + + y −
+
+
+ ···
6
9
108
6
9
108
3
2
which is the quadratic approximation we obtained in part (a).
(c) We have
g ′′ (0) 2
x + ···
2
′′
h (0) 2
y + ···
h(y) = h(0) + h′ (0)y +
2
g(x) = g(0) + g ′ (0)x +
Multiplying these together, we get
g(0)h(0) + g ′ (0)h(0)x + g(0)h′ (0)y +
g ′′ (0)
h′′ (0) 2
h(0)x2 + g ′ (0)h′ (0)xy + g(0)
y + ···
2
2
On the other hand,
f (0, 0) = g(0)h(0),
′′
fxx (0, 0) = g (0)h(0),
fx (0, 0) = g ′ (0)h(0),
′
′
fxy (0, 0) = g (0)h (0),
fy (0, 0) = g(0)h′ (0),
fyy (0, 0) = g(0)h′′ (0)
Thus the quadratic approximation to f is the same as the product of the approximations to g and h.
113. (a) We have f (1, 2) = A0 +A1 +2A2 +A3 +2A4 +4A5 , fx (1, 2) = A1 +2A3 +2A4 , and fy (1, 2) = A2 +A4 +4A5 ,
so the linear approximation is
L(x, y) = A0 + A1 + 2A2 + A3 + 2A4 + 4A5 + (A1 + 2A3 + 2A4 )(x − 1) + (A2 + A4 + 4A5 )(y − 2).
Also, m(t) = 1 + B1 t and n(t) = 2 + C1 t.
(b) Using a CAS to compute the derivatives, we find that they are both the same:
d
d
f (x(t), y(t))|t=0 = l(m(t), n(t))|t=0 = A1 B1 + 2A3 B1 + 2A4 B1 + A2 C1 + A4 C1 + 4A5 C1 .
dt
dt
This can be explained using the chain rule and the fact that the derivative of a function at a point is the same as the
derivative of its linear approximation there:
∂
d
f (x(t), y(t))|t=0 =
f (x(0), y(0))x′ (0) +
dt
∂x
∂
l(x(0), y(0))m′ (0) +
=
∂x
∂
f (x(0), y(0))y ′ (0)
∂y
∂
d
l(x(0), y(0))n′ (0) = l(m(t), n(t))|t=0
∂y
dt
114. (a) We have
f (1, 2) = A0 + A1 + 2A2 + A3 + 2A4 + 4A5 ,
fx (1, 2) = A1 + 2A3 + 2A4 ,
fy (1, 2) = A2 + A4 + 4A5 ,
fxx (1, 2) = 2A3 ,
fxy (1, 2) = A4 ,
fyy (1, 2) = 2A5 .
Thus Q(x, y) = (A0 + A1 + 2A2 + A3 + 2A4 + 4A 5) + (A1 + 2A3 + 2A4 )(x − 1) + (A2 + A4 + 4A5 )(y −
2) + A3 (x − 1)2 + A4 (x − 1)(y − 2) + A5 (y − 2)2 . Expanding this expression in powers of x and y, we find
Q(x, y) = A0 + A1 x + A2 y + A3 x2 + A4 xy + A5 y 2 = f (x, y)
.
(b) The quadratic expansion for f about (1, 2) is equal to f itself. This is because f is already a quadratic function, and
is true about any point (a, b).
(c) The linear approximation of f (x, y) at (1, 2) is:
A0 + A1 + 2A2 + A3 + 2A4 + 4A5 + (A1 + 2A3 + 2A4 )(x − 1) + (A2 + A4 + 4A5 )(y − 2).
When we expand this we get
A0 − A3 + A4 − 4A5 + (A1 + 2A3 + 2A4 )x + (A2 + A4 + 4A5 )y.
This is not the same as f (x, y), and it is not even the same as the linear part of f (x, y), namely A0 + A1 x + A2 y.
PROJECTS FOR CHAPTER FOURTEEN
1383
115. The computer algebra system gives:
d
∂w ∂x du
∂w ∂y du
∂w ∂x dv
∂w ∂y dv
∂w dz
dw
= f (x(u(t), v(t)), y(u(t), v(t)), z(t)) =
+
+
+
+
dt
dt
∂x ∂u dt
∂y ∂u dt
∂x ∂v dt
∂y ∂v dt
∂z dt
This agrees with the result obtained using the chain rule and the tree diagram in Figure 14.56. The diagram shows four
paths from w to t, each corresponding to one of the terms in the expression for dw/dt.
w
∂w
∂y
∂w
∂x
∂w
∂z
y
x
∂x
∂v
∂y
∂u
∂x
∂u
u
∂y
∂v
v
du
dt
z
dz
dt
dv
dt
t
Figure 14.56
PROJECTS FOR CHAPTER FOURTEEN
1. (a) Calculating the necessary partial derivatives:
2
2
2
1
e−(x +y +z )/4Kt
3/2
(4πKt)
2
2
2
∂T
1
31
−(x2 +y 2 +z 2 )/4Kt x + y + z
=
e
−
∂t
4Kt2
2t
(4πKt)3/2
2
2
2
2
∂2T
1
x
1
=
e−(x +y +z )/4Kt
−
∂x2
4K 2 t2
2Kt
(4πKt)3/2
T (x, y, z, t) =
Thus,
K(Txx + Tyy + Tzz ) =
2
2
2
3
1
−(x2 +y 2 +z 2 )/4Kt x + y + z
e
−
= Tt
4Kt2
2t
(4πKt)3/2
(b) For t = constant, the level surfaces are given by
x2 + y 2 + z 2 = c2 ,
c = constant
which is an equation of a sphere centered at origin.
(c) With t = constant, grad T = Tx~i + Ty~j + Tz~k . Since
2x
x
∂T
=−
T =−
T
∂x
4Kt
2Kt
1384
Chapter Fourteen /SOLUTIONS
and similarly for Ty and Tz . Thus we have
2
2
2
1
1
e−(x +y +z )/4Kt
grad T (x, y, z) = −(x~i + y~j + z~k )
2Kt (4πKt)3/2
2
1
1
e−r /4Kt , where ~r = x~i + y~j + z~k
= (−~r )
3/2
2Kt (4πKt)
The heat flows toward lower temperatures, that is in the direction of − grad T . Since grad T is in the
direction of −~r , heat is flowing outward and with exponentially decreasing magnitude.
2. (a) First, note that m, n, and (1 − q) are all positive.
Suppose we increase m, the number of students. Then there are more birthdays to match in a fixed
size year, and so q increases. Thus we expect ∂q/∂m to be positive, which is the case. On the other hand,
if we increase n, the number of days in a year, then there are more slots for birthdays in a year, and so less
chance of matching. This means q decreases, and so we expect ∂q/∂n to be negative, which is the case.
(b) Since we do not have an exact formula for q, we find approximate values for q using q(23, 365) = 0.5073
and local linearization. The local linearization of q at (23, 365) is
∂q
∂q
∆m +
∆n
∂m
∂n
23
232
=
(1 − 0.5073)(m − 23) −
(1 − 0.5073)(n − 365).
365
(2)(3652 )
∆q ≈
When m = 21 and n = 365 we have
∆q ≈
232
23
(1 − 0.5073)(21 − 23) −
(1 − 0.5073)(365 − 365) = −0.0621.
365
(2)(3652 )
Thus
q(21, 365) = q(23, 365) + ∆q
≈ 0.5073 − 0.0621
= 0.4452.
(c) When m = 24 and n = 358, we have
∆q ≈
23
232
(1 − 0.5073)(24 − 23) −
(1 − 0.5073)(358 − 365) = 0.0379.
365
(2)(3652 )
Thus
q(24, 358) = q(23, 365) + ∆q
≈ 0.5073 + 0.0379
= 0.5452.
(d) We want to compare the values of q when m = 25 and n = 365 and when m = 23 and n = 365 − 31 =
334. When m = 25 and n = 365 we have
∆q ≈
232
23
(1 − 0.5073)(25 − 23) −
(1 − 0.5073)(365 − 365) = 0.0621.
365
(2)(3652 )
On the other hand, when m = 23 and n = 334 we have
∆q ≈
232
23
(1 − 0.5073)(23 − 23) −
(1 − 0.5073)(334 − 365) = 0.0312.
365
(2)(3652 )
So adding 2 more students gives you a greater increase in probability than no birthdays in December.
PROJECTS FOR CHAPTER FOURTEEN
1385
(e) The number of possible choices of birthdays for all the students is nm since each of the m students could
choose (have a birthday on) any of the n days. We count the number of these choices for which there are no
matching birthdays as follows. The first student can choose any of n days for his/her birthday. The second
student can choose any birthday that does not match the first student’s, that is, any of n − 1 days. The third
student can choose any birthday that does not match the first two students’, that is, any of n − 2 days; and
so on for all m students. So the number of choices of birthdays for all the students for which there is no
matching is n(n − 1)(n − 2) · · · (n − (m − 1)). Thus, the fraction of possible choices for which there is
no matching is
n(n − 1)(n − 2) · · · (n − (m − 1))
.
nm
Then q is one minus the chance of no matching birthdays:
q(m, n) = 1 −
n(n − 1)(n − 2) · · · (n − (m − 1))
.
nm
15.1 SOLUTIONS
1387
CHAPTER FIFTEEN
Solutions for Section 15.1
Exercises
1. The contour diagrams in (I) and (V) show that the value of the function decreases as we move in any direction from the
critical point. Therefore, the function has a local maximum at P . The contour diagrams in (II) and in (VI) show that the
value of the function increases as we move in any direction from the critical point, and therefore P is a local minimum.
The contour lines in (III) and (IV) are hyperbolas, and therefore, the critical point, P , is a saddle point.
2. The point A is not a critical point and the contour lines look like parallel lines. The point B is a critical point and is a local
maximum; the point C is a saddle point.
3. (a) None.
(b) Points E and G.
(c) Points D and F .
4. At the origin g(0, 0) = 0. Since y 3 ≥ 0 for y > 0 and y 3 < 0 for y < 0, the function g takes on both positive and negative
values near the origin, which must therefore be a saddle point. The second derivative test does not tell you anything since
D = 0.
5. At the origin f (0, 0) = 0. Since x6 ≥ 0 and y 6 ≥ 0, the point (0, 0) is a local (and global) minimum. The second
derivative test does not tell you anything since D = 0.
6. At the origin, the second derivative test gives
D = kxx kyy − (kxy )2 = (− sin x sin y)(− sin x sin y) − (cos x cos y)2
= sin2 0 sin2 0 − cos2 0 cos2 0
x=0,y=0
= −1 < 0.
Thus k(0, 0) is a saddle point.
7. At the origin h(0, 0) = 1. Since cos x and cos y are never above 1, the origin must be a local (and global) maximum. The
second derivative test
D = hxx hyy − (hxy )2 = (− cos x cos y)(− cos x cos y) − (sin x sin y)2
= cos2 x cos2 y − sin2 x sin2 y
=1>0
x=0,y=0
x=0,y=0
and hxx (0, 0) < 0, so (0, 0) is a local maximum.
8. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = 2x − 2y = 0,
fy = −2x + 6y − 8 = 0.
We see from the first equation that x = y. Substituting this into the second equation shows that y = 2. The only critical
point is (2, 2).
We have
D = (fxx )(fyy ) − (fxy )2 = (2)(6) − (−2)2 = 8.
Since D > 0 and fxx = 2 > 0, the function f has a local minimum at the point (2, 2).
9. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = 6 − 2x + y = 0,
fy = x − 2y = 0.
1388
Chapter Fifteen /SOLUTIONS
We see from the second equation that x = 2y. Substituting this into the first equation shows that y = 2. The only critical
point is (4, 2).
We have
D = (fxx )(fyy ) − (fxy )2 = (−2)(−2) − 12 = 3.
Since D > 0 and fxx = −2 < 0, the function f has a local maximum at (4, 2).
10. The partial derivatives give
fx = 2x + 4 = 0 and fy = −2y + 2 = 0.
Thus we have x = −2 and y = 1, so (−2, 1) is a critical point. We use the discriminant:
D = (fxx )(fyy ) − (fxy )2 = 2(−2) − 0 = −4 < 0.
Since D < 0, we have a saddle point.
11. Setting fx = 0 and fy = 0 to find the critical point, we have
fx = −6x − 4 + 2y = 0
2y − 6x = 4
and
and
fy = 2x − 10y + 48 = 0, or
10y − 2x = 48.
Solving these equations simultaneously gives x = 1 and y = 5.
Since fxx = −6, fyy = −10 and fxy = 2 for all (x, y), at (1, 5) the discriminant
D = (−6)(−10) − (2)2 = 56 > 0, and fxx < 0.
Thus f (x, y) has a local maximum value at (1, 5).
12. The partial derivatives give
fx = −2x + 6 = 0 and fy = 4y − 8 = 0.
Thus we have x = 3 and y = 2, so (3, 2) is a critical point. We use the discriminant:
D = (fxx )(fyy ) − (fxy )2 = (−2)2 = −4 < 0.
Since D < 0, we have a saddle point.
13. At a critical point, fx = 0 and fy = 0.
fx = 2xy − 2y = 2y(x − 1) = 0,
fy = x2 + 4y − 2x = 0.
In order for fx to be zero, we need y = 0 or x = 1. If y = 0, then fy = x2 − 2x = x(x − 2) = 0 so x = 0 or x = 2.
Therefore (0, 0) and (2, 0) are both critical points. If x = 1, then fy = 4y − 1 = 0 so y = 0.25. The three critical points
are (0, 0), (2, 0), and (1, 0.25).
We use the discriminant to classify the three points. We have
2
D(x, y) = fxx fyy − fxy
= (2y)(4) − (2x − 2)2
= 8y − (2x − 2)2 .
Checking each critical point, we see that
D(0, 0) = −4 < 0, so (0, 0) is a saddle point.
D(2, 0) = −4 < 0, so (2, 0) is also a saddle point.
D(1, 0.25) = 2 > 0 and fxx (1, 0.25) = 0.5 > 0, so (1, 0.25) is a local minimum.
14. At a critical point
fx (x, y) = 6x2 − 6xy + 12x = 0
fy (x, y) = −3x2 − 12y = 0
From the second equation, we conclude that −3(x2 + 4y) = 0, so y = − 14 x2 . Substituting for y in the first equation
gives
1
6x2 − 6x − x2 + 12x = 0
4
15.1 SOLUTIONS
1389
or
x
1 3
x + 2x = (4x + x2 + 8) = 0.
4
4
Thus x = 0 or x2 + 4x + 8 = 0. The quadratic has no real solutions, so the only one critical point is (0, 0).
At (0, 0), we have
D(0, 0) = fxx fyy − (fxy )2 = (12)(−12) − 02 = −144 < 0,
x2 +
so (0, 0) is a saddle point.
15. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = 3x2 − 3 = 0
fy = 3y 2 − 3 = 0
shows that x = ±1 and y = ±1. There are four critical points: (1, 1), (−1, 1), (1, −1) and (−1, −1).
We have
D = (fxx )(fyy ) − (fxy )2 = (6x)(6y) − (0)2 = 36xy.
At the points (1, −1) and (−1, 1), we have D = −36 < 0, so f has saddle points at (1, −1) and (−1, 1). At (1, 1),
we have D = 36 > 0 and fxx = 6 > 0, so f has a local minimum at (1, 1). At (−1, −1), we have D = 36 > 0 and
fxx = −6 < 0, so f has a local maximum at (−1, −1).
16. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = 3x2 − 6x = 0
fy = 3y 2 − 3 = 0
shows that x = 0 or x = 2 and y = −1 or y = 1. There are four critical points: (0, −1), (0, 1), (2, −1), and (2, 1).
We have
D = (fxx )(fyy ) − (fxy )2 = (6x − 6)(6y) − (0)2 = (6x − 6)(6y).
At the point (0, −1), we have D > 0 and fxx < 0, so f has a local maximum.
At the point (0, 1), we have D < 0, so f has a saddle point.
At the point (2, −1), we have D < 0, so f has a saddle point.
At the point (2, 1), we have D > 0 and fxx > 0, so f has a local minimum.
17. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = 3x2 − 3 = 0
fy = 3y 2 − 12y = 0
shows that x = −1 or x = 1 and y = 0 or y = 4. There are four critical points: (−1, 0), (1, 0), (−1, 4), and (1, 4).
We have
D = (fxx )(fyy ) − (fxy )2 = (6x)(6y − 12) − (0)2 = (6x)(6y − 12).
At critical point (−1, 0), we have D > 0 and fxx < 0, so f has a local maximum at (−1, 0).
At critical point (1, 0), we have D < 0, so f has a saddle point at (1, 0).
At critical point (−1, 4), we have D < 0, so f has a saddle point at (−1, 4).
At critical point (1, 4), we have D > 0 and fxx > 0, so f has a local minimum at (1, 4).
18. Find the critical point(s) by setting
fx = (xy + 1) + (x + y) · y = y 2 + 2xy + 1 = 0,
fy = (xy + 1) + (x + y) · x = x2 + 2xy + 1 = 0,
then we get x2 = y 2 , and so x = y or x = −y.
If x = y, then x2 + 2x2 + 1 = 0, that is, 3x2 = −1, and there is no real solution. If x = −y, then x2 − 2x2 + 1 = 0,
which gives x2 = 1. Solving it we get x = 1 or x = −1, then y = −1 or y = 1, respectively. Hence, (1, −1) and (−1, 1)
are critical points.
Since
fxx (x, y) = 2y,
fxy (x, y) = 2y + 2x and
fyy (x, y) = 2x,
1390
Chapter Fifteen /SOLUTIONS
the discriminant is
2
D(x, y) = fxx fyy − fxy
= 2y · 2x − (2y + 2x)2
= −4(x2 + xy + y 2 ).
thus
D(1, −1) = −4(12 + 1 · (−1) + (−1)2 ) = −4 < 0,
D(−1, 1) = −4((−1)2 + (−1) · 1 + 12 ) = −4 < 0.
Therefore (1, −1) and (−1, 1) are saddle points.
19. At a critical point, fx = 0, fy = 0.
fx = 8y − (x + y)3 = 0, we know 8y = (x + y)3 .
fy = 8x − (x + y)3 = 0, we know 8x = (x + y)3 .
Therefore we must have x = y. Since (x + y)3 = (2y)3 = 8y 3 , this tells us that 8y − 8y 3 = 0. Solving gives y = 0, ±1.
Thus the critical points are (0, 0), (1, 1), (−1, −1).
fyy = fxx = −3(x + y)2 , and fxy = 8 − 3(x + y)2 .
The discriminant is
2
D(x, y) = fxx fyy − fxy
= 9(x + y)4 − 64 − 48(x + y)2 + 9(x + y)4
= −64 + 48(x + y)2 .
D(0, 0) = −64 < 0, so (0, 0) is a saddle point.
D(1, 1) = −64 + 192 > 0 and fxx (1, 1) = −12 < 0, so (1, 1) is a local maximum.
D(−1, −1) = −64 + 192 > 0 and fxx (−1, −1) = −12 < 0, so (−1, −1) is a local maximum.
20. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = e2x
fy = e
2
+y 2
2x2 +y 2
(4x) = 0,
(2y) = 0,
shows that the only critical point is (0, 0).
We have
D = (fxx )(fyy ) − (fxy )2 = e2x
2
+y 2
(4 + (4x)2 ) · e2x
2
+y 2
(2 + (2y)2 ) − (e2x
2
+y 2
(4x · 2y))2 .
At (0, 0), we have D = 4 · 2 − 02 > 0 and fxx = 4 > 0, so the function has a local minimum at the point (0, 0).
Problems
21. At the critical point x = 1, y = 0,
fx = 2x + A = 0,
fy = 2y = 0.
so 2 + A = 0 or A = −2
Thus, f (x, y) = x2 − 2x + y 2 + B has a critical point at (1, 0). Since fxx = 2 and fyy = 2 and fxy = 0 at (1, 0),
D = fxx fyy − (fxy )2 = 2 · 2 − 02 = 4,
so the second derivative test shows the critical point at (1, 0) is a local minimum. The value of the minimum is
f (1, 0) = 12 − 2 · 1 + 02 + B = 20,
so B = 21.
15.1 SOLUTIONS
1391
22. At a local minimum,
fx = 2x + y + a = 0
fy = x + 2y + b = 0.
Since these equations must be satisfied by x = 2, y = 5, we have
a = −2.2 − 5 = −9
b = −2 − 2 · 5 = −12.
The point (2, 5) gives a local minimum since
D = fxx fyy − (fxy )2 = 2 · 2 − 12 > 0 and fxx = 2 > 0.
To ensure that f (2, 5) = 11, we substitute
f (2, 5) = 22 + 2 · 5 + 52 − 9 · 2 − 12 · 5 + c = 11
c = 50.
23. (a) This function has only one critical point, a local maximum, where (x − a)2 + (y − b)2 = 0; that is, at (a, b). Taking
partial derivatives gives the same result:
fx (x, y) = e−(x−a)
fy (x, y) = e
2
2
−(y−b)2
−(x−a)2 −(y−b)2
(−2(x − a))
(−2(y − b)).
2
Since e−(x−a) −(y−b) is never zero, fx and fy are only 0 at x = a, y = b.
(b) We have a = −1, b = 5.
2
2
(c) The point (−1, 5) is a local maximum because e−(x−a) −(y−b) is largest where −(x − a)2 − (y − b)2 is zero.
24. We have fx = 2kx − 4y and fy = 2y − 4x, so fxx = 2k, fxy = −4, and fyy = 2. The discriminant is
D = (fxx )(fyy ) − (fxy )2 = (2k)(2) − (−4)2 = 4k − 16.
Since D = 4k − 16, we see that D < 0 when k < 4. The function has a saddle point at the point (0, 0) when k < 4.
When k > 4, we have D > 0 and fxx > 0, so the function has a local minimum at the point (0, 0). When k = 4,
the discriminant is zero, and we get no information about this critical point. By looking at the values of the function in
Table 15.1, it appears that f has a local minimum at the point (0, 0) when k = 4.
Table 15.1
x
y
−0.1
−0.1
0.01
0
0.1
0.01
0.09
0
0.04
0
0.04
0.1
0.09
0.01
0.01
(a) The function f (x, y) has a saddle point at (0, 0) if k < 4.
(b) There are no values of k for which this function has a local maximum at the point (0, 0).
(c) The function f (x, y) has a local minimum at (0, 0) if k ≥ 4.
25. (a)
(b)
(c)
(d)
P is a local maximum.
Q is a saddle point.
R is a local minimum.
S is none of these.
1392
Chapter Fifteen /SOLUTIONS
26. Figure 15.1 shows the gradient vectors around P and Q pointing perpendicular to the contours and in the direction of
increasing values of the function.
y
6
5
4
3
−
1
2
3
1
0
0
−4
−
−2
1
S
❨
✒
■ ✐
✒■
✛
Q
✮
❘✠
❘
✠ ✙
−1
−5
−6
−6
x
−
0
4
1
2
1
−
P
❘
✲ ✠
✛
✶ 6✐
2
3
−
−
4
−5
−6
0
■ R✒✿
③
✙
☛ ❲❘
3
5
6
Figure 15.1
27. Figure 15.2 shows the direction of ∇f at the points where k∇f k is largest, since at those points the contours are closest
together.
y
5
3
−6
1
−
3
−
0
−6
−5
−6 x
−
4
2
−
−
4
1
5
3
6
2
1
P
−
3
Q
0
−5
−1
0
R
✠
1
0
−2
−4
1
S
4
2
6
Figure 15.2
2
28. First, we identify the critical points. The partials are fx (x, y) = 3x2 and fy (x, y) = −2ye−y . These will vanish
simultaneously when x = 0 and y = 0, so our only critical point is (0, 0). The discriminant is
2
2
2
2
D = fxx (x, y)fyy (x, y) − fxy
(x, y) = (6x)(4y 2 e−y − 2e−y ) − 0 = 12xe−y (2y 2 − 1).
Unfortunately, the discriminant is zero at the origin so the second derivative test can tell us nothing about our critical point.
We can, however, see that we are at a saddle point by looking at the behavior of f (x, y) along the line y = 0. Here we have
f (x, 0) = x3 + 1, so for positive x, we have f (x, 0) > 1 = f (0, 0) and for negative x, we have f (x, 0) < 1 = f (0, 0).
So f (x, y) has neither a maximum nor a minimum at (0, 0).
15.1 SOLUTIONS
29. At a critical point,
fx = cos x sin y = 0
so
cos x = 0 or sin y = 0;
fy = sin x cos y = 0
Case 1: Assume cos x = 0. This gives
so
sin x = 0 or cos y = 0.
and
π π 3π
3π
,− , ,
,···
2
2 2 2
(This can be written more compactly as: x = kπ + π/2, for k = 0, ±1, ±2, · · ·.)
If cos x = 0, then sin x = ±1 6= 0. Thus in order to have fy = 0 we need cos y = 0, giving
x = ··· −
π π 3π
3π
,− , ,
,···
2
2 2 2
(More compactly, y = lπ + π/2, for l = 0, ±1, ±2, · · ·)
Case 2: Assume sin y = 0. This gives
y = · · · − 2π, −π, 0, π, 2π, · · ·
(More compactly, y = lπ, for l = 0, ±1, ±2, · · ·)
If sin y = 0, then cos y = ±1 6= 0, so to get fy = 0 we need sin x = 0, giving
y = ··· −
x = · · · , −2π, −π, 0, π, 2π, · · ·
(More compactly, x = kπ for k = 0, ±0, ±1, ±2, · · ·)
Hence we get all the critical points of f (x, y). Those from Case 1 are as follows:
π
π π
π 3π
π
, − ,
, − ,
···
··· − ,−
2
2
2 2
2 2
π π π π π 3π
,
,
···
···
,−
,
,
2 2 2 2 2 2
π
3π π
3π 3π
3π
···
,−
,
,
,
,
···
2
2
2 2
2 2
Those from Case 2 are as follows:
· · · (−π, −π), (−π, 0), (−π, π), (−π, 2π) · · ·
· · · (0, −π), (0, 0), (0, π), (0, 2π) · · ·
· · · (π, −π), (π, 0), (π, π), (π, 2π) · · ·
More compactly these points can be written as,
(kπ, lπ), for k = 0, ±1, ±2, · · · , l = 0, ±1, ±2, · · ·
π
π
and (kπ + , lπ + ), for k = 0, ±1, ±2, · · · , l = 0, ±1, ±2, · · ·
2
2
To classify the critical points, we find the discriminant. We have
fxx = − sin x sin y,
fyy = − sin x sin y,
and
fxy = cos x cos y.
Thus the discriminant is
2
D(x, y) = fxx fyy − fxy
= (− sin x sin y)(− sin x sin y) − (cos x cos y)2
= sin2 x sin2 y − cos2 x cos2 y
= sin2 y − cos2 x. (Use: sin2 x = 1 − cos2 x and factor.)
At points of the form (kπ, lπ) where k = 0, ±1, ±2, · · · ; l = 0, ±1, ±2, · · ·, we have
D(x, y) = −1 < 0 so (kπ, lπ) are saddle points.
At points of the form (kπ + π2 , lπ + π2 ) where k = 0, ±1, ±2, · · · ; l = 0, ±1, ±2, · · ·
D(kπ + π2 , lπ + π2 ) = 1 > 0, so we have two cases:
If k and l are both even or k and l are both odd, then
fxx = − sin x sin y = −1 < 0, so (kπ + π2 , lπ + π2 ) are local maximum points.
If k is even but l is odd or k is odd but l is even, then
fxx = 1 > 0 so (kπ + π2 , lπ + π2 ) are local minimum points.
1393
1394
Chapter Fifteen /SOLUTIONS
30. To find critical points, set partial derivatives equal to zero:
fx = sin x = 0 when x = 0, ±π, ±2π, · · ·
fy = y = 0 when
y = 0.
The critical points are
· · · (−2π, 0), (−π, 0), (0, 0), (π, 0), (2π, 0), (3π, 0) · · ·
To classify, calculate D = fxx fyy − (fxy )2 = cos x.
At the points (0, 0), (±2π, 0), (±4π, 0), (±6π, 0), · · ·
D = (1) > 0 and
fxx > 0 (Sincefxx (0, 2kπ) = cos(2kπ) = 1).
Therefore (0, 0), (±2π, 0), (±4π, 0), (±6π, 0), · · · are local minima.
At the points (±π, 0), (±3π, 0), (±5π, 0), (±7π, 0), · · ·, we have cos(2k + 1)π = −1, so
D = (−1) < 0.
Therefore (±π, 0), (±3π, 0), (±5π, 0), (±7π, 0), · · · are saddle points.
31. To find the critical points, we must solve the equations
∂f
= ex (1 − cos y) = 0
∂x
∂f
= ex (sin y) = 0.
∂y
The first equation has solution
The second equation has solution
Since x can be anything, the lines
y = 0, ±2π, ±4π, . . . .
y = 0, ±π, ±2π, ±3π, . . . .
y = 0, ±2π, ±4π, . . .
are lines of critical points.
We calculate
D = (fxx )(fyy ) − (fxy )2 = ex (1 − cos y)ex cos y − (ex sin y)2
= e2x (cos y − cos2 y − sin2 y)
= e2x (cos y − 1)
At any critical point on one of the lines y = 0, y = ±2π, y = ±4π, . . .,
D = e2x (1 − 1) = 0.
Thus, D tells us nothing. However, all along these critical lines, the value of the function, f , is zero. Since the function f
is never negative, the critical points are all both local and global minima.
32. (a) (1, 3) is a critical point. Since fxx > 0 and the discriminant
2
D = fxx fyy − fxy
= fxx fyy − 02 = fxx fyy > 0,
the point (1, 3) is a minimum.
(b) See Figure 15.3.
15.1 SOLUTIONS
y
1395
1
4
32
16
64
12
0
0
3
✠
x
1
Figure 15.3
2
2
33. (a) (a, b) is a critical point. Since the discriminant D = fxx fyy − fxy
= −fxy
< 0, (a, b) is a saddle point.
(b) See Figure 15.4.
y
−7
−5
−3
97
−1
5
0
0
3
1
1
3
5
1
0
1
7
9
0 −1
x
−3
−5
−7
Figure 15.4
34. Begin by constructing little pieces of the contour diagram around each of the points (−1, 0), (3, 3), and (3, −3) where
some information about f is given. The general shape will be as in Figure 15.5, and the directions of increasing contour
values are indicated for each part. Then complete the diagram in any way. One possible solution is given in Figure 15.6.
0
0
1
1
2
2
3
4
Figure 15.5: Part of contour
diagram with arrows showing
direction of increasing
function values
Figure 15.6: Contour diagram
of f (x, y)
1396
Chapter Fifteen /SOLUTIONS
35. Since (2, 4) is a local minimum, the contours around (2, 4) are closed curves with increasing values as we go away from
the point (2, 4). We assume that the function values continue to increase as we move parallel to the y-axis to the point
(2, 1). Since (2, 1) is a saddle point, we draw the contours so that the values go down as we move up or down from this
point, and up as we move to the left or right. One possible contour diagram is shown in Figure 15.7.
y
6
5
4
−4
−3
−2
−1
3
2
1
1 2 34
0
−1
−1
−1
1
x
2
3
4
5
6
Figure 15.7
36. (a) We set partial derivatives to zero:
fx = 2ax − 2y − 4 = 0
fy = 2by − 2x − 6 = 0.
So we have ax − y = 2 and −x + by = 3, leading to
x=
2b + 3
ab − 1
and
y=
3a + 2
.
ab − 1
(b) The discriminant is given by
D = (fxx ) (fyy ) − (fxy )2 = (2a)(2b) − 22 = 4(ab − 1)
If a = b = 2, then D = 4(2 · 2 − 1) = 12. Since fxx = 4 > 0, the critical point is a local minimum.
(c) We use the fact that D = 4(ab − 1) and fxx = 2a.
If a > 0 and ab > 1, then D > 0 and fxx > 0 so we have a local minimum.
If a < 0 and ab > 1, then D > 0 and fxx > 0 so we have a local maximum.
If ab < 1, then D < 0 so we have a saddle point.
37. (a) Setting the partial derivatives equal to 0, we have
fx (x, y) = 2x(x2 + y) + 2x(x2 − y) = 4x3 = 0
fy (x, y) = −(x2 + y) + (x2 − y) = −2y = 0.
Thus, (0, 0) is the only critical point.
(b) Calculating D gives
D = (fxx )(fyy ) − (fxy )2 = (12x2 )(−2) − 02 = −24x2 .
At x = 0, y = 0, we have
D(0, 0) = 0.
Thus the second derivative test tells us nothing about the nature of the critical point.
(c) Since f (0, 0) = 0, we sketch contours with values near 0. The contour f = 0 is given by
(x2 − y)(x2 + y) = 0,
that is, the two parabolas
y = x2
and y = −x2
15.1 SOLUTIONS
1397
We also sketch the contours f = 1 and f = −1. See Figure 15.8.
Since there are values of the function which are both positive (above f (0, 0)) and negative (below (f (0, 0)),
near the critical point (0, 0), the origin is neither a local maximum nor a local minimum; it is a saddle point.
y
f
=
−
=
=
=
1
0
f
f
f
1
1
(0, 0)
f
=
−
1
f
=
0
x
Figure 15.8
38. The first order partial derivatives are
fx (x, y) = 2kx − 2y
and
fy (x, y) = 2ky − 2x.
And the second order partial derivatives are
fxx (x, y) = 2k
fxy (x, y) = −2
fyy (x, y) = 2k
Since fx (0, 0) = fy (0, 0) = 0, the point (0, 0) is a critical point. The discriminant is
D = (2k)(2k) − 4 = 4(k2 − 1).
For k = ±2, the discriminant is positive, D = 12. When k = 2, fxx (0, 0) = 4 which is positive so we have a local
minimum at the origin. When k = −2, fxx (0, 0) = −4 so we have a local maximum at the origin. In the case k = 0,
D = −4 so the origin is a saddle point.
Lastly, when k = ±1 the discriminant is zero, so the second derivative test can tell us nothing. Luckily, we can factor
f (x, y) when k = ±1. When k = 1,
f (x, y) = x2 − 2xy + y 2 = (x − y)2 .
This is always greater than or equal to zero. So f (0, 0) = 0 is a minimum and the surface is a trough-shaped parabolic
cylinder with its base along the line x = y.
When k = −1,
f (x, y) = −x2 − 2xy − y 2 = −(x + y)2 .
This is always less than or equal to zero. So f (0, 0) = 0 is a maximum. The surface is a parabolic cylinder, with its top
ridge along the line x = −y.
1398
Chapter Fifteen /SOLUTIONS
y
k = −2
y
k = −1
y
k=0
−16
−12
−8
16
12
−30
−20
8
−10
−5
−4
−16
−12
−8
4
−4
1
−1
−1
−1
x
x
x
−1
1
−1
−5
−10
8
−8
−12
−16
−20
−30
y
k=1
4
−4
k=2
12
16
y
16
12
30
20
8
10
4
5
1
1
x
x
1
5
10
20
30
Figure 15.9
39. The partial derivatives are
fx (x, y) = 3x2 − 3y 2
and
fy (x, y) = −6xy.
Now fx (x, y) will vanish if x = ±y and fy (x, y) will vanish if either x = 0 or y = 0. Since the partial derivatives
are defined everywhere, the only critical points are where fx (x, y) and fy (x, y) vanish simultaneously. (0, 0) is the only
critical point.
To find the contour for f (x, y) = 0, we solve the equation x3 − 3xy 2 = 0. This can be factored into
√
√
f (x, y) = x(x − 3 y)(x + 3 y) = 0
√
√
whose roots are x = 0, x = 3 y and x = − 3 y. Each of these roots describes a line through the origin; the three of
them divide the plane into six regions. Crossing any one of these lines will change the sign of only one of the three factors
of f (x, y), which will change the sign of f (x, y).
15.1 SOLUTIONS
1399
y
f>0
f<0
✛ y = x/√3
2
−2
11
−1
0
−2
−1
f <0
0
0
0
0
11
2
f>0
1
2
f>0
−1
x
✛ y = −x/√3
−2
f<0
Figure 15.10
40. Contours near a local extremum are approximate ellipses and contours near a saddle point are approximate hyperbolas.
The functions in the contour plots (a) and (b) have a local extremum at the origin; those in (c) and (d) have a saddle point
at the origin.
All four functions have value zero at the critical point at the origin. Graph (I) corresponds to a local maximum,
because the values near the origin are all below zero. Graph (III) corresponds to a local minimum, because the values near
the origin are all above zero. Graphs (II) and (IV) correspond to saddle points because there are both positive and negative
values near the critical point at the origin.
Contour plot (a) shows that on circles centered at the origin, z takes values farther from zero at θ = π/2 and
θ = 3π/2 than at θ = 0 and θ = π. Thus (a) matches (III), and the critical point at the origin is a local minimum.
The function represented by Contour plot (b) takes values farther from zero at θ = 0 and θ = π than at θ = π/2 and
θ = 3π/2, so corresponds to (I) and has a local maximum at the origin.
The function in Contour plot (c) has value zero on the asymptotes of the hyperbolas, which are at θ = π/4, 3π/4,
5π/4 and 7π/4. This matches Graph (II), which represents a saddle point.
The function in Contour plot (d) has value zero on the asymptotes of the hyperbolas, which are at θ = 0, π/2, π and
3π/2. This matches Graph (IV),which represents a saddle point.
Strengthen Your Understanding
41. If fx = fy = 0 at (1, 3), then (1, 3) is a critical point of f . However, (1, 3) may be a saddle point and not a local
maximum or minimum.
42. If D = 0 at (a, b), the point (a, b) can be a maximum, a minimum, or a saddle point. We are only sure that (a, b) is a
saddle point if D < 0.
43. If both cross-sections are concave up, then fxx (a, b) > 0 and fyy (a, b) > 0, but we could still have D < 0 when
fxy (a, b) is large enough, giving a saddle point.
For example, let f (x, y) = x2 + y 2 + 7xy. Then (0, 0) is a critical point. We have fxx (0, 0) = fyy (0, 0) = 2 > 0,
but fxy (0, 0) = 7, so D = (2)(2) − 72 < 0.
44. Let f (x, y) = ex Then fx = ex , which is never 0.
45. The function f (x, y) = 4 − (x − 2)2 − (y + 3)2 has a local maximum at (2, −3, 4).
46. True. By definition, a critical point is either where the gradient of f is zero or does not exist.
47. False. The point P0 could be a saddle point of f .
48. False. The point P0 could be a saddle point of f .
49. True. If P0 were not a critical point of f , then grad f (P0 ) would point in the direction of maximum increase of f , which
contradicts the fact that P0 is a local maximum or minimum.
50. True. The graph of this function is a cone that opens upward with its vertex at the origin.
1400
Chapter Fifteen /SOLUTIONS
51. False. The graph of this function is a saddle shape, with a saddle point at the origin. The function increases in the ~i
direction and decreases in the ~j direction.
52. True. Adding 5 to the function shifts the graph 5 units vertically, which leaves the (x, y) coordinates of the local extrema
intact.
53. True. Multiplying by −1 turns the graph of f upside down, so local maxima become local minima and vice-versa.
54. False. For example, the linear function f (x, y) = x + y has no local extrema at all.
55. False. The statement is only true for points sufficiently close to P0 .
56. False. Local maxima are only high points for f when compared to nearby values; the global maximum is the largest of
any values of f over its entire domain.
Solutions for Section 15.2
Exercises
1. Mississippi lies entirely within a region designated as 80s so we expect both the maximum and minimum daily high
temperatures within the state to be in the 80s. The southwestern-most corner of the state is close to a region designated as
90s, so we would expect the temperature here to be in the high 80s, say 87-88. The northern-most portion of the state is
located near the center of the 80s region. We might expect the high temperature there to be between 83-87.
Alabama also lies completely within a region designated as 80s so both the high and low daily high temperatures
within the state are in the 80s. The southeastern tip of the state is close to a 90s region so we would expect the temperature
here to be about 88-89 degrees. The northern-most part of the state is near the center of the 80s region so the temperature
there is 83-87 degrees.
Pennsylvania is also in the 80s region, but it is touched by the boundary line between the 80s and a 70s region. Thus
we expect the low daily high temperature to occur there and be about 80 degrees. The state is also touched by a boundary
line of a 90s region so the high will occur there and be 89-90 degrees.
New York is split by a boundary between an 80s and a 70s region, so the northern portion of the state is likely to be
about 74-76 while the southern portion is likely to be in the low 80s, maybe 81-84 or so.
California contains many different zones. The northern coastal areas will probably have the daily high as low as
65-68, although without another contour on that side, it is difficult to judge how quickly the temperature is dropping off
to the west. The tip of Southern California is in a 100s region, so there we expect the daily high to be 100-101.
Arizona will have a low daily high around 85-87 in the northwest corner and a high in the 100s, perhaps 102-107 in
its southern regions.
Massachusetts will probably have a high daily high around 81-84 and a low daily high of 70.
2. The maximum value, which is about 11, occurs at (5.1, 4.9). The minimum value, which is about −1, occurs at (1, 3.9).
3. The maximum value, which is slightly above 30, say 30.5, occurs approximately at the origin. The minimum value, which
is about 20.5, occurs at (2.5, 5).
4. The maxima occur at about (π/2, 0) and (π/2, 2π). The minimum occurs at (π/2, π). The maximum value is about 1,
the minimum value is about −1.
5. To maximize z = x2 + y 2 , it suffices to maximize x2 and y 2 . We can maximize both of these at the same time by
taking the point (1, 1), where z = 2. It occurs on the boundary of the square. (Note: We also have maxima at the points
(−1, −1), (−1, 1) and (1, −1) which are on the boundary of the square.)
To minimize z = x2 + y 2 , we choose the point (0, 0), where z = 0. It does not occur on the boundary of the square.
6. To maximize z = −x2 − y 2 it suffices to minimize x2 and y 2 . Thus, the maximum is at (0, 0), where z = 0. It does not
occur on the boundary of the square.
To minimize z = −x2 − y 2 , it suffices to maximize x2 and y 2 . Do this by taking the point (1, 1), (−1, −1), (−1, 1),
or (1, −1) where z = −2. These occur on the boundary of the square.
7. To maximize this function, it suffices to maximize x2 and minimize y 2 . We can do this by choosing the point (1, 0), or
(−1, 0) where z = 1. These occur on the boundary of the square.
To minimize z = x2 − y 2 , it suffices to maximize y 2 and minimize x2 . We can do this by taking the point (0, 1), or
(0, −1) where z = −1. These occur on the boundary of the square.
8. The function f has no global maximum or global minimum.
9. The function g has a global minimum (it is 0) but no global maximum.
15.2 SOLUTIONS
1401
10. The function h has no global maximum or minimum.
11. Since f (x, y) ≤ 0 for all x, y and sincef (0, 0) = 0, the function has a global maximum (it is 0) and no global minimum.
12. Suppose x is fixed. Then for large values of y the sign of f is determined by the highest power of y, namely y 3 . Thus,
f (x, y) → ∞ as
f (x, y) → −∞ as
y→∞
y → −∞.
So f does not have a global maximum or minimum.
Problems
13. (a) The critical points of f are the point(s) at which the partial derivatives, fx and fy , are zero. We have
fx = 4x − 3y + 1
fy = −3x + 16y − 1.
Solving the linear system fx = 0, fy = 0, we find (x, y) = (−13/55, 1/55). To classify this point, we have to find
2
the sign of fxx · fyy − fxy
there. We calculate
2
fxx · fyy − fxy
= 4 · 16 − 9 = 55 > 0.
Thus, (−13/55, 1/55) is a local minimum.
(b) We complete the square in the following way:
2x2 − 3xy + 8y 2 + x − y = 2x2 − x(3y − 1) + 8y 2 − y
1
1
= 2(x − (3y − 1))2 − (3y − 1)2 + 8y 2 − y
4
8
1
1
= 2(x − (3y − 1))2 + (55y 2 − 2y − 1)
4
8
7
1
55
1 2
−
= 2(x − (3y − 1))2 +
y−
.
4
8
55
55
Therefore the function has a global minimum located at the point (x, y) where both the two squares vanish. The
coordinates of that point satisfy:
x−
1
(3y − 1) = 0
4
y−
and
1
= 0.
55
The two conditions again give the point (x, y) = (−13/55, 1/55). The contour diagram for f is shown in Figure 15.11. The fact that f can be written in this way as the sum of two squares shows that the point (x, y) =
(−13/55, 1/55) is a global minimum.
0.5
,
( −13
55
1
)
55
0
0.5
1
2
3
-0.5
Figure 15.11: The contour diagram for
f (x, y) = 2x2 − 3xy + 8y 2 + x − y
0.5
1402
Chapter Fifteen /SOLUTIONS
14. We calculate the partial derivatives and set them to zero.
∂ (range)
= −10t − 6h + 400 = 0
∂t
∂ (range)
= −6t − 6h + 300 = 0.
∂h
10t + 6h = 400
6t + 6h = 300
Solving, we obtain
4t = 100
so
t = 25
Solving for h, we obtain 6h = 150, yielding h = 25. Since the range is quadratic in h and t, the second derivative test
tells us this is a local and global maximum. So the optimal conditions are h = 25% humidity and t = 25◦ C.
15. Let the sides be x, y, z cm. Then the volume is given by V = xyz = 32.
The surface area S is given by
S = 2xy + 2xz + 2yz.
Substituting z = 32/(xy) gives
S = 2xy +
64
64
+
.
y
x
At a critical point,
∂S
= 2y −
∂x
∂S
= 2x −
∂y
64
=0
x2
64
= 0,
y2
The symmetry of the equations (or by dividing the equations) tells us that x = y and
2x −
64
=0
x2
x3 = 32
x = 321/3 = 3.17 cm.
Thus the only critical point is x = y = (32)1/3 cm and z = 32/ (32)1/3 · (32)1/3 = (32)1/3 cm. At the critical point
Sxx Syy − (Sxy )2 =
(128)2
128 128
· 3 − 22 = 3 3 − 4.
3
x
y
x y
Since D > 0 and Sxx > 0 at this critical point, the critical point x = y = z = (32)1/3 is a local minimum. Since
S → ∞ as x, y → ∞, the local minimum is a global minimum.
16. If the coordinates of the corner on the plane are (x, y, z), the volume of the box is V = xyz. Since z = 1 − 3x − 2y on
the plane, the volume is given by
V = xy(1 − 3x − 2y) = xy − 3x2 y − 2xy 2 .
The domain is the triangular region 0 ≤ x ≤ 13 , 0 ≤ y ≤ (1 − 3x)/2. At a critical point,
∂V
= y − 6xy − 2y 2 = y(1 − 6x − 2y) = 0
∂x
∂V
= x − 3x2 − 4xy = x(1 − 3x − 4y) = 0,
∂y
One solution is x = y = 0. Another is x = 0, y = 21 ; another is y = 0, x = 13 . Another is the solution of
1 − 6x − 2y = 0
1 − 3x − 4y = 0,
15.2 SOLUTIONS
1403
namely x = 19 , y = 61 .
If either x = 0 or y = 0, then V = 0, so these solutions do not give the maximum volume. Since
D = Vxx Vyy − (Vxy )2 = (−6y)(−4x) − (1 − 6x − 4y)2
4
1
1
1
1 2
1
1
= − = > 0,
−4 ·
− 1−6· −4·
9 6
6
9
9
6
9
9
3
and Vxx ( 19 , 61 ) = −1 < 0, the point x = 91 , y = 16 , is a local maximum at which V = (1/9)(1/6) − 3(1/9)2 (1/6) −
2(1/9)(1/6)2 = 1/162.
Since all points on the boundary of the domain give V = 0, the local maximum is a global maximum.
D
1 1
,
= −6 ·
17.
h
w
l
Figure 15.12
Let w, h and l be width, height and length of the suitcase in cm. Then its volume V = lwh, and w + h + l ≤ 135.
To maximize the volume V , choose w + h + l = 135, and thus l = 135 − w − h,
V = wh(135 − w − h)
= 135wh − w2 h − wh2
Differentiating gives
Vw = 135h − 2wh − h2 ,
Vh = 135w − w2 − 2wh.
Find the critical points by solving Vw = 0 and Vh = 0:
Vw = 0 gives
Vh = 0 gives
135h − h2 = 2wh,
135w − w2 = 2wh.
As hw 6= 0, we cancel h (and w respectively) in the above equations and get
135 − h = 2w
135 − w = 2h
Subtracting gives
w − h = 2(w − h)
hence w = h. Therefore, substituting into the equation Vw = 0
135h − h2 = 2h2
and therefore
3h2 = 135h.
Since h 6= 0, we have
135
= 45.
3
So w = h = 45 cm. Thus, l = 135 − w − h = 45 cm. To check that this critical point is a maximum, we find
h=
Vww = −2h,
Vhh = −2w,
Vwh = 135 − 2w − 2h,
so
2
D = Vww Vhh − Vwh
= 4hw − (135 − 2w − 2h)2 .
At w = h = 45, we have Vww = −2(45) < 0 and D = 4(45)2 − (135 − 90 − 90)2 = 6075 > 0, hence V is maximum
at w = h = l = 45.
Therefore, the suitcase with maximum volume is a cube with dimensions width = height = length = 45 cm.
1404
Chapter Fifteen /SOLUTIONS
18. The box is shown in Figure 15.13. Cost of four sides = (2hl + 2wh)(1)c/. Cost of two bottoms = (2wl)(2)c/. Thus the
total cost C (in cents) of the box is
C = 2(hl + wh) + 4wl.
But volume wlh = 512, so l = 512/(wh), thus
C=
2048
1024
+ 2wh +
.
w
h
To minimize C, find the critical points of C by solving
2048
= 0,
h2
1024
Cw = 2h −
= 0.
w2
Ch = 2w −
We get
2wh2 = 2048
2hw2 = 1024.
Since w, h 6= 0, we can divide the first equation by the second giving
2048
2wh2
=
,
2hw2
1024
so
h
= 2,
w
thus
h = 2w.
Substituting this in Ch = 0, we obtain h3 = 2048, so h = 12.7 cm. Thus w = h/2 = 6.35 cm, and l = 512/(wh) =
6.35 cm. Now we check that these dimensions minimize the cost C. We find that
2
D = Chh Cww − Chw
=(
4096 2048
)( 3 ) − 22 ,
h3
w
and at h = 12.7, w = 6.35, Chh > 0 and D = 16 − 4 > 0, thus C has a local minimum at h = 12.7 and w = 6.35.
Since C increases without bound as w, h → 0 or ∞, this local minimum must be a global minimum.
Therefore, the dimensions of the box that minimize the cost are w = 6.35 cm, l = 6.35 cm and h = 12.7 cm.
h
w
l
Figure 15.13
19. The square of the distance from the point (x, y, z) to the origin is
S = x2 + y 2 + z 2 .
If the point is on the plane, z = 1 − 3x − 2y, we have
S = x2 + y 2 + (1 − 3x − 2y)2 .
15.2 SOLUTIONS
1405
At the critical point
∂S
= 2x + 2(1 − 3x − 2y)(−3) = 2(10x + 6y − 3) = 0
∂x
∂S
= 2y + 2(1 − 3x − 2y)(−2) = 2(6x + 5y − 2) = 0.
∂y
Simplifying gives
10x + 6y = 3
6x + 5y = 2,
with solution x = 3/14, y = 1/7. At this point z = 1/14. We have
D = Sxx Syy − (Sxy )2 = (20)(10) − 122 = 56,
so D > 0 and Sxx > 0. Thus, the point x = 3/14, y = 1/7 is a local minimum. Since S → ∞ as x, y → ±∞, the local
minimum is a global minimum. Thus, x = 3/14, y = 1/7, z = 1/14 is the closest point to the origin on the plane.
20. We minimize the square of the distance from the point (x, y, z) to the origin:
S = x2 + y 2 + z 2 .
Since z 2 = 9 − xy − 3x, we have
S = x2 + y 2 + 9 − xy − 3x.
At a critical point
∂S
= 2x − y − 3 = 0
∂x
∂S
= 2y − x = 0,
∂y
so x = 2y, and
2(2y) − y − 3 = 0
2
giving y = 1, so x = 2 and z = 9 − 2 · 1 − 3 · 2 = 1, so z = ±1. We have
D = Sxx Syy − (Sxy )2 = 2 · 2 − (−1)2 = 4 − 1 > 0,
so, since D > 0 and Sxx > 0, the critical points are local minima. Since S → ∞ as x, y → ±∞, the local minima are
global minima.
√
If x = 2, y = 1, z = ±1, we have S = 22 + 12 + 12 = 6, so the shortest distance to the origin is 6.
21. (a) We draw the level curves (parallel straight lines) of f (x, y) = ax + by + c. We can see that the level lines with the
maximum and minimum f -values which intersect with the disk are the level lines that are tangent to the boundary of
the disk. Therefore, the maximum and minimum occur at the boundary of the disk. See Figure 15.14.
f = max
f = min
▼
f increases
Figure 15.14
1406
Chapter Fifteen /SOLUTIONS
f = max
f = max
f = min
f increases❪
f increases
✻
Figure 15.15
f = min
Figure 15.16
(b) Similar to part (a), we see the level lines with the largest and smallest f -values which intersect with the rectangle must
pass the corner of the rectangle. So the maximum and minimum occur at the corners of rectangle. See Figure 15.15.
When the level curves are parallel to a pair of the sides, then the points on the sides are all maximum or minimum, as
shown below in Figure 15.16.
(c) The graph of f is a plane. The part of the graph lying above a disk R is either a flat disk, in which case every point is
a maximum, or is a tilted ellipse, in which case you can see that the maximum will be on the edge. Similarly, the part
lying above a rectangle is either a rectangle or a tilted parallelogram, in which case the maximum will be at a corner.
22. (a) The revenue R = p1 q1 + p2 q2 . Profit = P = R − C = p1 q1 + p2 q2 − 2q12 − 2q22 − 10.
p1
∂P
= p1 − 4q1 = 0 gives q1 =
∂q1
4
∂P
p2
= p2 − 4q2 = 0 gives q2 =
∂q2
4
Since
∂2P
2
∂q1
∂2 P
2
∂q1
= −4,
∂2 P
2
∂q2
= −4 and
∂2P
∂q1 ∂q2
= 0, at (p1 /4, p2 /4) we have that the discriminant, D = (−4)(−4) > 0
< 0, thus P has a local maximum value at (q1 , q2 ) = (p1 /4, p2 /4). Since P is quadratic in q1 and q2 , this
and
is a global maximum. So the maximum profit is
P =
p21
p2
p2
p2
p2
p2
+ 2 − 2 1 − 2 2 − 10 = 1 + 2 − 10.
4
4
16
16
8
8
(b) The rate of change of the maximum profit as p1 increases is
∂
2p1
p1
(max P ) =
=
.
∂p1
8
4
23. The total revenue is
and as q = q1 + q2 , this gives
R = pq = (60 − 0.04q)q = 60q − 0.04q 2 ,
R = 60q1 + 60q2 − 0.04q12 − 0.08q1 q2 − 0.04q22 .
Therefore, the profit is
P (q1 , q2 ) = R − C1 − C2
= −13.7 + 60q1 + 60q2 − 0.07q12 − 0.08q22 − 0.08q1 q2 .
15.2 SOLUTIONS
1407
At a local maximum point, we have grad P = ~0 :
∂P
= 60 − 0.14q1 − 0.08q2 = 0,
∂q1
∂P
= 60 − 0.16q2 − 0.08q1 = 0.
∂q2
Solving these equations, we find that
q1 = 300
and q2 = 225.
To see whether or not we have found a local maximum, we compute the second-order partial derivatives:
∂2P
= −0.14,
∂q12
Therefore,
D=
∂2P
= −0.16,
∂q22
∂2P
= −0.08.
∂q1 ∂q2
∂2P
∂2P ∂2P
−
= (−0.14)(−0.16) − (−0.08)2 = 0.016,
2
2
∂q1 ∂q2
∂q1 ∂q2
and so we have found a local maximum point. The graph of P (q1 , q2 ) has the shape of an upside down paraboloid since
P is quadratic in q1 and q2 , hence (300, 225) is a global maximum point.
24. (a) This tells us that an increase in the price of either product causes a decrease in the quantity demanded of both
products. An example of products with this relationship is tennis rackets and tennis balls. An increase in the price of
either product is likely to lead to a decrease in the quantity demanded of both products as they are used together. In
economics, it is rare for the quantity demanded of a product to increase if its price increases, so for q1 , the coefficient
of p1 is negative as expected. The coefficient of p2 in the expression could be either negative or positive. In this case,
it is negative showing that the two products are complementary in use. If it were positive, however, it would indicate
that the two products are competitive in use, for example Coke and Pepsi.
(b) The revenue from the first product would be q1 p1 = 150p1 − 2p21 − p1 p2 , and the revenue from the second product
would be q2 p2 = 200p2 − p1 p2 − 3p22 . The total sales revenue of both products, R, would be
R(p1 , p2 ) = 150p1 + 200p2 − 2p1 p2 − 2p21 − 3p22 .
Note that R is a function of p1 and p2 . To find the critical points of R, set ∇R = 0, i.e.,
∂R
∂R
=
= 0.
∂p1
∂p2
This gives
∂R
= 150 − 2p2 − 4p1 = 0
∂p1
and
∂R
= 200 − 2p1 − 6p2 = 0
∂p2
Solving simultaneously, we have p1 = 25 and p2 = 25. Therefore the point (25, 25) is a critical point for R. Further,
∂2R
∂2R
∂2R
=
−4,
=
−6,
= −2,
∂p21
∂p22
∂p1 ∂p2
so the discriminant at this critical point is
D = (−4)(−6) − (−2)2 = 20.
Since D > 0 and ∂ 2 R/∂p21 < 0, this critical point is a local maximum. Since R is quadratic in p1 and p2 , this is a
global maximum. Therefore the maximum possible revenue is
R = 150(25) + 200(25) − 2(25)(25) − 2(25)2 − 3(25)2
= (6)(25)2 + 8(25)2 − 7(25)2
= 4375.
This is obtained when p1 = p2 = 25. Note that at these prices, q1 = 75 units, and q2 = 100 units.
1408
Chapter Fifteen /SOLUTIONS
25. Let P (K, L) be the profit obtained using K units of capital and L units of labor. The cost of production is given by
C(K, L) = kK + ℓL,
and the revenue function is given by
R(K, L) = pQ = pAK a Lb .
Hence, the profit is
P = R − C = pAK a Lb − (kK + ℓL).
In order to find local maxima of P , we calculate the partial derivatives and see where they are zero. We have:
∂P
= apAK a−1 Lb − k,
∂K
∂P
= bpAK a Lb−1 − ℓ.
∂L
The critical points of the function P (K, L) are solutions (K, L) of the simultaneous equations:
k
= pAK a−1 Lb ,
a
ℓ
= pAK a Lb−1 .
b
Multiplying the first equation by K and the second by L, we get
ℓL
kK
=
,
a
b
and so
ℓa
L.
kb
Substituting for K in the equation k/a = pAK a−1 Lb , we get:
K=
ℓa
k
= pA
a
kb
We must therefore have
L1−a−b = pA
Hence, if a + b 6= 1,
L = pA
and
a−1
La−1 Lb .
a a−1
ℓ
b
a
k
.
a a ℓ (a−1) 1/(1−a−b)
k
b
,
a (a−1) 1/(1−a−b)
ℓa
ℓa
a
L=
K=
pA
kb
kb
k
ℓ
b
.
To see if this is really a local maximum, we apply the second derivative test. We have:
∂2P
= a(a − 1)pAK a−2 Lb ,
∂K 2
∂2P
= b(b − 1)pAK a Lb−2 ,
∂L2
∂2P
= abpAK a−1 Lb−1 .
∂K∂L
Hence,
D=
∂2P ∂2P
−
∂K 2 ∂L2
∂2P
∂K∂L
2
= ab(a − 1)(b − 1)p2 A2 K 2a−2 L2b−2 − a2 b2 p2 A2 K 2a−2 L2b−2
= ab((a − 1)(b − 1) − ab)p2 A2 K 2a−2 L2b−2
= ab(1 − a − b)p2 A2 K 2a−2 L2b−2 .
15.2 SOLUTIONS
1409
Now a, b, p, A, K, and L are positive numbers. So, the sign of this last expression is determined by the sign of 1 − a − b.
(a) We assumed that a + b < 1, so D > 0, and as 0 < a < 1, then ∂ 2 P/∂K 2 < 0 and so we have a unique local
maximum. To verify that the local maximum is a global maximum, we focus on the cost. Let C = kK + ℓL. Since
K ≥ 0 and L ≥ 0, K ≤ C/k and L ≤ C/ℓ. Therefore the profit satisfies:
P = pAK a Lb − (kK + ℓL)
≤ pA
C a C b
k
ℓ
−C
= mC a+b − C
where m = pA(1/k)a (1/ℓ)b . Since a + b < 1, the profit is negative for large costs C, say C ≥ C0 (C0 = m1−a−b
will do). Therefore, in the KL-plane for K ≥ 0 and L ≥ 0, the profit is less than or equal to zero everywhere on or
above the line kK + ℓL = Co . Thus the global maximum must occur inside the triangle bounded by this line and the
K and L axes. Since P ≤ 0 on the K and L axes as well, the global maximum must be in the interior of the triangle
at the unique local maximum we found.
In the case a + b < 1, we have decreasing returns to scale. That is, if the amount of capital and labor used is
multiplied by a constant λ > 0, we get less than λ times the production.
(b) Now suppose a + b ≥ 1. If we multiply K and L by λ for some λ > 0, then
Q(λK, λL) = A(λK)a (λL)b = λa+b Q(K, L).
We also see that
C(λK, λL) = λC(K, L).
So if a + b = 1, we have
P (λK, λL) = λP (K, L).
Thus, if λ = 2, so we are doubling the inputs K and L, then the profit P is doubled and hence there can be no
maximum profit.
If a + b > 1, we have increasing returns to scale and there can again be no maximum profit: doubling the inputs
will more than double the profit. In this case, the profit increases without bound as K, L go toward infinity.
26. We have
fx = 2x(y + 1)3 = 0 only when x = 0 or y = −1
fy = 3x2 (y + 1)2 + 2y = 0 never when y = −1 and only for y = 0 when x = 0
We conclude that fx = 0 and fy = 0 only when x = 0, y = 0, so f has only one critical point, namely (0, 0).
The second derivative test at (0, 0) gives
D = fxx fyy − (fxy )2 = 2(y + 1)3 (6x2 (y + 1) + 2) − (6x(y + 1)2 )2
= 2(1)(2) − 0 > 0 when x = 0, y = 0
Since fxx > 0 at (0, 0), this means f has a local minimum at (0, 0).
[Alternatively, if we expand (y + 1)3 , then we can view f (x, y) as x2 + y 2 + (terms of degree 3 or greater in x and
y), which means that f behaves likes x2 + y 2 near (0, 0).]
Although (0, 0) is a local minimum, it cannot be a global minimum since for fixed x, say x = 1, the function f (x, y)
is a cubic polynomial in y and cubics take on arbitrarily large positive and negative values.
In the single-variable case, suppose a function f defined on the real line is differentiable and its derivative is continuous. Then if f has only one critical point, say x = 0, and if that critical point is a local minimum, it must also be a
global minimum. This is because f ′ cannot change sign without f ′ = 0 so we must have f ′ < 0 for x < 0 and f ′ > 0
for x > 0. Thus f is decreasing for all x < 0 and increasing for all x > 0, so x = 0 is the global minimum for f .
27. The variables are a and b, so we set
∂S
= 2(a + b) + 8(4a + b − 2) + 18(9a + b − 4) = 0
∂a
∂S
= 2(a + b) + 2(4a + b − 2) + 2(9a + b − 4) = 0,
∂b
so, collecting terms and dividing by 4 and 2 respectively,
49a + 7b − 22 = 0
14a + 3b − 6 = 0.
1410
Chapter Fifteen /SOLUTIONS
Solving gives a = 24/49, b = −2/7.
Since there is only one critical point and S is unbounded as a, b → ∞, this critical point is the global minimum.
Therefore, the best fitting parabola is
24 2 2
x − .
y=
49
7
28. Let the line be in the form y = b + mx. Then, when x equals 0, 1, and 2, y equals b, b + m, and b + 2m respectively.
The sum of the squares of the vertical distances, which is what we want to minimize, is
f (m, b) = (4 − b)2 + (3 − (b + m))2 + (1 − (b + 2m))2
To find critical points, set each partial derivative equal to zero.
fm = 0 + 2(3 − (b + m))(−1) + 2(1 − (b + 2m))(−2)
= 6b + 10m − 10
fb = 2(4 − b)(−1) + 2(3 − (b + m))(−1) + 2(1 − (b + 2m))(−1)
= 6b + 6m − 16
Setting both partial derivatives equal to zero and dividing by 2, we get a system of equations:
3b + 5m = 5
3b + 3m = 8
with solutions m = − 23 and b =
25
.
6
Thus, the line is y =
25
6
− 32 x.
29. (a) We have f (2, 1) = 120.
(i) If x > 20 then f (x, y) > 10x > 200 > f (2, 1).
(ii) If y > 20 then f (x, y) > 20y > 400 > f (2, 1).
(iii) If x < 0.01 and y ≤ 20 then f (x, y) > 80/(xy) > 80/((0.01)(20)) = 400 > f (2, 1).
(iv) If y < 0.01 and x ≤ 20 then f (x, y) > 80/(xy) > 80/((20)(0.01)) = 400 > f (2, 1).
(b) The continuous function f must achieve a minimum at some point (x0 , y0 ) in the closed and bounded region R′ :
0.01 ≤ x ≤ 20, 0.01 ≤ y ≤ 20. Since (2, 1) is in R′ , we must have f (x0 , y0 ) ≤ f (2, 1). By part (a), f (x0 , y0 )
is less than all values of f in the part of R that is outside R′ , so f (x0 , y0 ) is a minimum for f on all of R. Since
(x0 , y0 ) is not on the boundary of R, it must be a critical point of f .
(c) The only critical point of f in R is the point (2, 1), so by part (b) f has a global minimum there.
30. (a) The function f is continuous in the region R, but R is not closed and bounded so a special analysis is required.
Notice that f (x, y) tends to ∞ as (x, y) tends farther and farther from the origin or tends toward any point on
the x or y axis. This suggests that a minimum for f , if it exists, can not be too far from the origin or too close to the
axes. For example, if x > 10 then f (x, y) > 4x > 40, and if y > 10 then f (x, y) > 5y > 50. If 0 < x < 0.1 then
f (x, y) > 2/x > 20, and if 0 < y < 0.1 then f (x, y) > 3/y > 30.
Since f (1, 1) = 14, a global minimum for f if it exists must be in the smaller region R′ : 0.1 ≤ x ≤ 10,
0.1 ≤ y ≤ 10. The region R′ is closed and bounded and so f does have a minimum value at some point in R′ , and
since that value is at most 14, it is also a global minimum for all of R.
(b) Since the region R has no boundary, the minimum value must occur at a critical point of f . At a critical point we
have
3
2
fy = − 2 + 5 = 0.
fx = − 2 + 4 = 0
x
y
p
p
The only critical point is ( 1/2, 3/5) ≈ (0.7071, 0.7746), at which f achieves the minimum value
p
p
√
√
f ( 1/2, 3/5) = 4 2 + 2 15 ≈ 13.403.
31. (a) There is one variable, p, in this problem; P0 and PF are constants. At the minimum energy,
dE
2p
2P 2
= 2 − 3F = 0
dp
P0
p
p4 = P02 PF2
√
p = P0 PF .
This value of p gives a minimum for E because it is the only critical point and the value of E grows toward ∞ as
p → ∞.
15.2 SOLUTIONS
1411
(b) There are now two variables, p1 and p2 , so at the minimum energy,
2p1
2p2
∂E
= 2 − 32 = 0
∂p1
P0
p1
∂E
2p2
2P 2
= 2 − 3F = 0.
∂p2
p1
p2
Solving these equations simultaneously:
p1
p2
= 23 ,
2
P0
p1
so p41 = P02 p22
and taking square roots gives
p21 = P0 p2 .
P2
p2
= F3 ,
2
p1
p2
so p42 = PF2 p21
and taking square roots gives
p22 = PF p1 .
Substituting p22 = PF p1 into p41 = P02 p22 gives
p41 = P02 PF p1
p31 = P02 PF
p1 =
so
Substituting p21 = P0 p2 into p42 = PF2 p21 gives
p42 = PF2 P0 p2
so
p
3
P02 PF .
p
3
P0 PF2 .
p2 =
The critical value of E is again a minimum because the critical point is unique and the value of E tends to ∞ as
p1 , p2 → ∞.
32. (a) Look at the formula q = Kp−E aθ . Since price, p, has a negative exponent, when the price is increased, less is sold.
Since advertising, a, has a positive exponent, when the amount of advertising is increased, more is sold.
(b) Since q = Kp−E aθ , we have
∂q
∂
EKp−E aθ
Eq
= Kaθ (p−E ) = Kaθ (−Ep−E−1 ) = −
=−
∂p
∂p
p
p
and
∂q
∂ θ
θKp−E aθ
θq
= Kp−E
(a ) = Kp−E θaθ−1 =
=
.
∂a
∂a
a
a
(c) Profit is revenue minus cost. Revenue is pq. Cost is the sum of the cost cq of producing the items and the cost pa a of
advertising. Hence
π = Profit = Revenue − Cost = pq − cq − pa a.
(d) At maximum profit, both partial derivatives are zero:
∂π
=0
∂p
and
∂π
= 0.
∂a
(e) Using π = pq − cq − pa a, we have
∂q
∂q
∂q
∂π
= q+p
−c
= q + (p − c)
∂p
∂p
∂p
∂p
∂π
E
= q + (p − c) − q
∂p
p
=q
1 − (p − c)
E
p
In addition
∂q
∂q
θ
∂π
=p
−c
− pa = q(p − c) − pa .
∂a
∂a
∂a
a
(f) At maximum profit, ∂π/∂p = 0, so solving for p gives
E
=0
p
(p − c)E
=1
p
1
p−c
=
p
E
1 − (p − c)
.
1412
Chapter Fifteen /SOLUTIONS
In addition, at maximum profit, ∂π/∂a = 0, so we have
θ
= pa
a
a
p−c
=
.
pa
qθ
q(p − c)
(g) By parts (e) and (f), at maximum profit we have
p−c
pa
1 qθ
·
=
·
p
p−c
E a
and hence
θ
pa a
= .
pq
E
The numerator, pa a, is the amount the company spends on advertising. The denominator, pq, is the company’s
revenue. The monopoly spends a fixed fraction, θ/E of its revenue for advertising, no matter how the price of
advertising might change.
Strengthen Your Understanding
33. If the region is closed and bounded and the function is continuous, then it must have a global maximum, even if it has no
critical points.
34. The Extreme Value Theorem does not say what happens when R is unbounded. For example, suppose the region R is the
whole xy-plane. Some functions may have a global minimum, such as f (x, y) = x2 + y 2 , and others may not, such as
f (x, y) = x + y.
35. The local maximum is not necessarily the global maximum. For example, let f (x, y) be a function with the contour
diagram in Figure 15.17.
y
0
1
1
3
2
2
3
x
1
2
1
3
−2
0
−1
−3
−3 −2 −1
1
2
3
Figure 15.17
This function has a local maximum value of 1 at the origin. However, 1 is definitely not the global maximum value
since the function attains higher values at other input values.
36. Let f (x, y) = x + y, which tends to +∞ as x → ∞ and tends to −∞ as x → −∞.
37. Let f (x, y) = x2 + y 2 and let R be the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Then the maximum value of f occurs at (1, 1).
38. True. For unconstrained optimization, global extrema occur at one (or more) of the local extrema.
39. False. For example, the linear function f (x, y) = x + y has neither a global minimum or global maximum on all of
2-space.
40. True. The region is the unit disk without its boundary (the unit circle), and the distance between any two points in this
region is less than 2—it does not stretch off to infinity in any direction.
41. False. The region is the unit disk without its boundary (the unit circle), so it is not closed (in fact, it is open).
42. True. The global minimum is 0, which occurs at the origin. This is clear since the function f (x, y) = x2 + y 2 is greater
than or equal to zero everywhere, and is only zero at the origin.
15.3 SOLUTIONS
1413
43. False. On the given region the function f is always less than one. By picking points closer and closer to the circle
x2 + y 2 = 1 we can make f larger and larger (although never larger than one). There is no point in the open disk that
gives f its largest value.
44. False. While f can only have (at most) one largest value, it may attain this value at more than one point. For example, the
function f (x, y) = sin(x + y) has a global maximum of 1 at both (π/2, 0) and (0, π/2).
45. True. The region is both closed and bounded, guaranteeing both a global maximum and minimum.
46. True. The global minimum could occur on the boundary of the region.
Solutions for Section 15.3
Exercises
1. Our objective function is f (x, y) = x + y and our equation of constraint is g(x, y) = x2 + y 2 = 1. To optimize f (x, y)
with Lagrange multipliers, we solve ∇f (x, y) = λ∇g(x, y) subject to g(x, y) = 1. The gradients of f and g are
∇f (x, y) = ~i + ~j ,
∇g(x, y) = 2x~i + 2y~j .
So the equation ∇f = λ∇g becomes
~i + ~j = λ(2x~i + 2y~j )
Solving for λ gives
1
1
=
,
2x
2y
which tells us that x = y. Going back to our equation of constraint, we use the substitution x = y to solve for y:
λ=
g(y, y) = y 2 + y 2 = 1
2y 2 = 1
1
y2 =
2
y=±
√
√
√
√
r
√
1
2
=±
.
2
2
Since x = y, our critical points are ( 22 , 22 ) and (− 22 , − 22 ). Since the constraint is closed and bounded, maximum
and minimum
values
of f subject to the constraint exist. √
Evaluating
f at the critical points we find that the maximum
√
√
√
√
√
value is f ( 22 , 22 ) = 2 and the minimum value is f (− 22 , − 22 ) = − 2.
2. The Lagrange conditions give:
1 = λ2x,
3 = λ2y.
Thus 2λ = 1/x = 3/y, so y = 3x. Substituting this into the constraint, we get x2 + (3x)2 = 10, so x = ±1. Since
y = 3x, the points satisfying the Lagrange conditions are (1, 3) and (−1, −3). Since f (1, 3) = 12 and f (−1, −3) = −8,
the maximum value is 12 at (1, 3) and the minimum value is −8 at (−1, −3).
3. The Lagrange conditions give:
2(x − 1) = λ2x,
2(y + 2) = λ2y.
We can’t have x = 0, since then the first equation becomes −2 = 0. Similarly, y 6= 0. Thus we can divide by x and y.
Solving both equations for λ and setting the expressions equal, we get
y+2
x−1
=
.
x
y
Thus, we have y(x − 1) = x(y + 2), so y = −2x. Substituting this into the constraint, we get x2 + (−2x)2 = 5, so
x = ±1. Since y = −2x, the points satisfying the Lagrange conditions are (1, −2) and (−1, 2). Since f (1, −2) = 0 and
f (−1, 2) = 20, the maximum value is 20 at (−1, 2) and the minimum value is 0 at (1, −2).
1414
Chapter Fifteen /SOLUTIONS
4. The Lagrange conditions give:
3x2 = λ6x,
1 = λ2y.
If x = 0 in the first equation, then from the constraint, y = ±2. Thus (0, ±2) are two points satisfying the Lagrange
conditions. If x 6= 0, we can solve for λ from both equations and setting the expressions equal, we get
1
1
x=
,
2
2y
so y = 1/x. Substituting this into the constraint, we get 3x2 + (1/x)2 = 4. Therefore 3x4 + 1 = 4x2 , so
3x4 − 4x2 + 1 = (x2 − 1)(3x2 − 1) = 0.
√
Therefore x = ±1, x = ±1/ 3. Since y = 1/x, we get
(1, 1),
(−1, −1),
√ √
(1/ 3, 3),
√
√
(−1/ 3, − 3),
as points satisfying the Lagrange conditions. The corresponding values of f (x, y) are 2, −2, 1.92, −1.92. There are also
the points (0, ±2) we found for the case x = 0, and the values for f there are ±2. Thus the maximum value is 2 at (1, 1)
and (0, 2) and the minimum value is −2 at (−1, −1) and (0, −2).
5. Our objective function is f (x, y) = 3x − 2y and our equation of constraint is g(x, y) = x2 + 2y 2 = 44. Their gradients
are
∇f (x, y) = 3~i − 2~j ,
∇g(x, y) = 2x~i + 4y~j .
So the equation ∇f = λ∇g becomes 3~i − 2~j = λ(2x~i + 4y~j ). Solving for λ gives us
λ=
3
−2
=
,
2x
4y
which we can use to find x in terms of y:
−2
3
=
2x
4y
−4x = 12y
x = −3y.
Using this relation in our equation of constraint, we can solve for y:
x2 + 2y 2 = 44
(−3y)2 + 2y 2 = 44
9y 2 + 2y 2 = 44
11y 2 = 44
y2 = 4
y = ±2.
Thus, the critical points are (−6, 2) and (6, −2). Since the constraint is closed and bounded, maximum and minimum
values of f subject to the constraint exist. Evaluating f at the critical points, we find that the maximum is f (6, −2) =
18 + 4 = 22 and the minimum value is f (−6, 2) = −18 − 4 = −22.
6. The objective function is f (x, y) = 2xy and the constraint equation is g(x, y) = 5x + 4y = 100, so grad f =
(2y)~i + (2x)~j and grad g = 5~i + 4~j . Setting grad f = λ grad g gives
2y = 5λ,
2x = 4λ.
From the first equation we have λ = 2y/5, and from the second equation we have λ = x/2. Setting these equal gives
y = 1.25x.
Substituting this into the constraint equation 5x + 4y = 100 gives x = 10 and y = 12.5. A maximum or minimum value
for f subject to the constraint can occur only at (10, 12.5).
15.3 SOLUTIONS
1415
We have f (10, 12.5) = 250. From Figure 15.18, we see that the point (10, 12.5) gives a maximum.
y
30
400
250
20
(10, 12.5)
100
10
5x + 4y = 100
x
10
20
Figure 15.18
7. Let f (x1 , x2 ) = x1 2 + x2 2 and g(x1 , x2 ) = x1 + x2 . Then grad f = λ grad g gives
2x1 = λ
2x2 = λ,
so x1 + x2 = 1 gives
λ
λ
+ =1
2
2
λ = 1.
or
Thus
1
.
2
Since f (x1 , x2 ) becomes arbitrarily large as x1 , x2 → ∞, there is no global maximum. The global minimum is given by
x1 = x2 =
f
1 1
,
2 2
2
1
2
=
+
2
1
2
=
1
.
2
8. Our objective function is f (x, y) = x2 + y and our equation of constraint is g(x, y) = x2 − y 2 = 1. Their gradients are
∇f (x, y) = 2x~i + ~j ,
∇g(x, y) = 2x~i − 2y~j .
Thus ∇f = λ∇g gives
2x = λ2x
1 = −λ2y
But x cannot be zero, since the constraint equation, −y 2 = 1, would then have no real solution for y. So the equation
∇f = λ∇g becomes
1
2x
=
2x
−2y
1
1=
−2y
−2y = 1
1
y=− .
2
λ=
Substituting this into our equation of constraint we find
1
1
g(x, − ) = x2 − −
2
2
2
=1
x2 =
5
4
x=±
√
5
.
2
1416
Chapter Fifteen /SOLUTIONS
√
√
√
√
So the critical points are ( 25 , − 12 ) and (− 25 , − 12 ). Evaluating f at these points we find f ( 25 , − 21 ) = f (− 25 , − 21 ) =
5
− 21 = 43 . This is the minimum value for f (x, y) constrained to g(x, y) = 1. To see this, note that for x2 = y 2 + 1,
4
f (x, y) = y 2 + 1 + y = (y + 1/2)2 + 3/4 ≥ 3/4. Alternatively, see Figure 15.19. To see that f has no maximum on
g(x, y) = 1, note that f → ∞ as x → ∞ and y → ∞ on the part of the graph of g(x, y) = 1 in quadrant I.
y
2
2.25
x2 − y 2 = 1
1.5
0.75
−2
(− 5/2, −1/2)
x
2
√
( 5/2, 1/2)
0
√
−2
Figure 15.19: Graph of x2 − y 2 = 1
9. The objective function is f (x, y, z) = x + 3y + 5z and the equation of constraint is g(x, y, z) = x2 + y 2 + z 2 = 1.
Their gradients are
∇f (x, y, z) = ~i + 3~j + 5~k ,
∇g(x, y, z) = 2x~i + 2y~j + 2z~k .
So the equation ∇f = λ∇g becomes ~i + 3~j + 5~k = λ(2x~i + 2y~j + 2z~k ). Solving for λ we find
λ=
3
5
1
=
=
.
2x
2y
2z
Which provides us with the equations
2y = 6x
10x = 2z.
Solving the first equation for y gives us y = 3x. Solving the second equation for z gives us z = 5x. Substituting these
into the equation of constraint, we can find x:
x2 + (3x)2 + (5x)2 = 1
x2 + 9x2 + 25x2 = 1
35x2 = 1
1
x2 =
35
x=±
√
√
r
√
1
35
=±
.
35
35
√
Since y = 3x and z = 5x, the critical points are at ±( 3535 , 3 3535 , 735 ). Since the constraint is closed and bounded, maximum√and minimum
values of f subject to the constraint exist. Evaluating√f at the√critical√points, we find the maximum
√
√
√
√
√
is f ( 3535 , 3 3535 , 735 ) = 35 35
= 35, and the minimum value is f (− 3535 , −3 3535 , − 735 ) = − 35.
35
10. Our objective function is f (x, y, z) = x2 − y 2 − 2z and our equation of constraint is g(x, y, z) = x2 + y 2 − z = 0.
To optimize f (x, y, z) with Lagrange multipliers, we solve ∇f (x, y, z) = λ∇g(x, y, z) subject to g(x, y, z) = 0. The
gradients of f and g are
∇f (x, y, z) = 2x~i − 2y~j − 2~k ,
∇g(x, y, z) = 2x~i + 2y~j − ~k .
15.3 SOLUTIONS
1417
We get
2x = 2λx
−2y = 2λy
−2 = −λ
x2 + y 2 = z.
The third equation gives λ = 2 and from the first x = 0, from the second y = 0 and from the fourth z = 0. So the only
solution is (0, 0, 0), and f (0, 0, 0) = 0.
To see what kind of extreme point is (0, 0, 0), let (a, b, c) be a point which satisfies the constraint, i.e. a2 + b2 = c.
Then f (a, b, c) = a2 − b2 − 2c = −a2 − 3b2 ≤ 0. The conclusion is that 0 is the maximum value of f and that there is
no minimum.
11. The Lagrange conditions give:
yz = λ2x,
xz = λ2y,
xy = λ8z.
We note that if x = 0, then the objective function f (x, y, z) = xyz has value 0 and this cannot be the maximum or
minimum value since xyz can take on both positive and negative values. Similarly, we can assume that y 6= 0 and z 6= 0.
Solving for λ and setting expressions equal, we get:
xz
xy
yz
=
=
.
x
y
4z
Thus y 2 z = x2 z so y 2 = x2 , and 4xz 2 = xy 2 , so y 2 = 4z 2 . Therefore
x2 + y 2 + 4z 2 = y 2 + y 2 + y 2 = 12,
so y = ±2. Since x = ±y and z = ±y/2, there are eight points satisfying the Lagrange conditions, each of the
form (±2, ±2, ±1). Thus the maximum value of the objective function f (x, y, z) = xyz is 4 at (2, 2, 1),(2, −2, −1),
(−2, 2, −1),(−2, −2, 1), and the minimum value is −4 at (−2, −2, −1),(2, 2, −1),(2, −2, 1),(−2, 2, 1).
12. The region x2 + y 2 ≤ 4 is the shaded disk of radius 2 centered at the origin (including the circle x2 + y 2 = 4) shown in
Figure 15.20.
y
4
16
8
1
4
−4
4 x
−4
Figure 15.20
We will first find the local maxima and minima in the interior of the disk. So we need to find the extrema of
f (x, y) = x2 + 2y 2
in the region
x2 + y 2 < 4.
For this we compute the critical points:
fx = 2x = 0
fy = 4y = 0
So the critical point is (0, 0). As fxx (0, 0) = 2, fyy (0, 0) = 4 and fxy (0, 0) = 0 we have
D = fxx (0, 0) · fyy (0, 0) − (fxy (0, 0))2 = 8 > 0 and fxx (0, 0) = 2 > 0.
Therefore (0, 0) is a minimum point and f (0, 0) = 0.
1418
Chapter Fifteen /SOLUTIONS
Now let’s find the local extrema of f on the boundary of the disk, hence this time we have to solve a constraint
problem. We want the extrema of f (x, y) = x2 + 2y 2 subject to g(x, y) = x2 + y 2 − 4 = 0. We use Lagrange
multipliers:
grad f = λ grad g and x2 + y 2 = 4,
which give
2x = 2λx
4y = 2λy
2
x + y 2 = 4.
From the first equation we have x = 0 or λ = 1. If x = 0, from the last equation y 2 = 4 and therefore (0, 2) and
(0, −2) are solutions.
If x 6= 0 then λ = 1 and from the second equation y = 0. Substituting this into the third equation we get x2 = 4 so
(2, 0) and (−2, 0) are the other two solutions.
The region x2 + y 2 ≤ 4 is closed and bounded, so maximum and minimum values of f in the the region exist.
Therefore, as f (0, 2) = f (0, −2) = 8 and f (2, 0) = f (−2, 0) = 4, (0, 2) and (0, −2) are global maxima and (0, 0) is
the global minimum on the whole region. The maximum value of f is 8 and the minimum value of f is 0.
√
13. The region x2 + y 2 ≤ 2 is the shaded disk of radius 2 centered at the origin (including the circle x2 + y 2 = 2) as shown
in Figure 15.21.
√
y
2
5
3
1
√
− 2
√ x
2
−1
−3
−5
√
− 2
Figure 15.21
We first find the local maxima and minima of f in the interior of our disk. So we need to find the extrema of
f (x, y) = x + 3y,
in the region
x2 + y 2 < 2.
As
fx = 1
fy = 3
f does not have critical points. Now let’s find the local extrema of f on the boundary of the disk. We want to find the
extrema of f (x, y) = x + 3y subject to the constraint g(x, y) = x2 + y 2 − 2 = 0. We use Lagrange multipliers
grad f = λ grad g
and
x2 + y 2 = 2,
which give
1 = 2λx
3 = 2λy
2
x + y 2 = 2.
As λ cannot be zero, we solve for x and y in the first two equations and get x =
equation gives
8λ2 = 10
1
2λ
and y =
3
.
2λ
Plugging into the third
15.3 SOLUTIONS
so λ =
√
± 25
√3 )
5
and we get the solutions ( √15 ,
1419
and (− √15 , − √35 ). Evaluating f at these points gives
√
1
3
f ( √ , √ ) = 2 5 and
5
5
√
3
1
f (− √ , − √ ) = −2 5
5
5
The region x2 + y 2 ≤ 2 is closed and bounded, so maximum and minimum values of f in the region exist. Therefore
( √15 , √35 ) is a global maximum of f and (− √15 , − √35 ) is a global minimum of f on the whole region x2 + y 2 ≤ 2.
14. The domain x2 +2y 2 ≤ 1 is the shaded interior of the ellipse x2 +2y 2 = 1 including the boundary, shown in Figure 15.22.
− √1
y
2
0.5
0.3
−0.5
−0.3
−0.1
0.1
x
−1
1
− √1
2
Figure 15.22
First we want to find the local maxima and minima of f in the interior of the ellipse. So we need to find the extrema
of
f (x, y) = xy,
in the region
x2 + 2y 2 < 1.
For this we compute the critical points:
fx = y = 0 and fy = x = 0.
So there is one critical point, (0, 0). As fxx (0, 0) = 0, fyy (0, 0) = 0 and fxy (0, 0) = 1 we have
D = fxx (0, 0) · fyy (0, 0) − (fxy (0, 0))2 = −1 < 0
so (0, 0) is a saddle and f does not have local extrema in the interior of the ellipse.
Now let’s find the local extrema of f on the boundary, hence this time we’ll have a constraint problem. We want the
extrema of f (x, y) = xy subject to g(x, y) = x2 + 2y 2 − 1 = 0. We use Lagrange multipliers:
grad f = λ grad g
and x2 + 2y 2 = 1
which give
y = 2λx
x = 4λy
x2 + 2y 2 = 1
From the first two equations we get
xy = 8λ2 xy.
2
So x = 0 or y = 0 or 8λ = 1.
√
√
√
If x = 0, from the last equation 2y 2 = 1 so y = ± 22 and we get the solutions (0, 22 ) and (0, − 22 ).
If y = 0, from the last equation we get x2 = 1 and so the solutions are (1, 0) and (−1, 0).
1
1
If x 6= 0 and y 6= 0 then 8λ2 = 1, hence λ = ± 2√
. For λ = 2√
2
2
x=
√
2y
and plugging into the third equation gives 4y 2 = 1 so we get the solutions (
1
For λ = − 2√
we get
2
√
x = − 2y
√
2 1
, 2)
2
and (−
√
2
, − 21 ).
2
1420
Chapter Fifteen /SOLUTIONS
and plugging into the third equation gives 4y 2 = 1, and the solutions (
√
√
√
√
solutions: (1, 0), (−1, 0), ( 22 , 12 ), (− 22 , − 12 ), ( 22 , − 21 ), (− 22 , 12 ).
Evaluating f at these points gives:
√
2
, − 21 )
2
and (−
√
2 1
, 2 ).
2
So finally we have the
√
2
2
) = f (0, −
) = f (1, 0) = f (−1, 0) = 0
f (0,
2
2
√
√
√
2 1
2
2
1
f(
, ) = f (−
,− ) =
2 2
2
2
4
√
√
√
1
2
2 1
2
, − ) = f (−
, )=−
.
f(
2
2
2 2
4
√
The region x2 + 2y 2 ≤ 1 is√closed and bounded, so the maximum
and minimum values of f in the region exist. Hence
√
the maximum value of f is 42 and the minimum value of f is − 42 .
15. The region x + y ≥ 1 is the shaded half plane (including the line x + y = 1) shown in Figure 15.23.
y
3
2
3
1
5
1
x
1
2
3
−1
−3
−5
Figure 15.23
Let’s look for the critical points of f in the interior of the region. As
fx = 3x2
fy = 1
there are no critical points inside the shaded region. Now let’s find the extrema of f on the boundary of our region. We
want the extrema of f (x, y) = x3 + y subject to the constraint g(x, y) = x + y − 1 = 0. We use Lagrange multipliers
grad f = λ grad g
and x + y = 1,
which give
3x2 = λ
1=λ
x + y = 1.
From the first two equations we get 3x2 = 1, so the solutions are
1
1
(√ ,1 − √ )
3
3
1
1
and (− √ , 1 + √ ).
3
3
Evaluating f at these points we get
1
f( √ , 1 −
3
1
f (− √ , 1 +
3
1
√ ) = 1−
3
1
√ ) = 1+
3
2
√
3 3
2
√ .
3 3
15.3 SOLUTIONS
1421
From the contour diagram in Figure 15.23, we see that ( √13 , 1 − √13 ) is a local minimum and (− √13 , 1 + √13 ) is a local
maximum of f on x + y = 1. Are they global extrema as well?
If we take x very big and y = 1 − x then f (x, y) = x3 + y = x3 − x + 1 which can be made as big as we want (if
we choose x big enough). So there will be no global maximum.
Similarly, taking x negative with big absolute value and y = 1 − x, f (x, y) = x3 + y = x3 − x + 1 can be made as
small as we want (if we choose x small enough). So there is no global minimum. This can also be seen from Figure 15.23.
16. First, we look for critical points for f :
2(x + 3) = 0,
2(y − 3) = 0.
Thus the only critical point for f is (−3, 3), but this point does not satisfy the constraint x2 + y 2 ≤ 2, so we do not use
it. The Lagrange conditions are
2(x + 3) = λ2x, 2(y − 3) = λ2y.
If x = 0, the first equation becomes 3 = 0, so x 6= 0. Similarly, y 6= 0. Solving for λ and setting expressions equal, we
have
x+3
y−3
=
.
x
y
Thus, (x + 3)y = (y − 3)x, so y = −x. Substituting this into the constraint equation, we get 2x2 = 2, so x = ±1. Since
y = −x, the points satisfying the Lagrange conditions are (1, −1) and (−1, 1). Since f (1, −1) = 32 and f (−1, 1) = 8,
the maximum value is 32 at (1, −1) and the minimum value is 8 at (−1, 1).
17. We first find the critical points of f :
fx = 2xy = 0,
fy = x2 + 6y − 1 = 0.
From the first equation, we get either x = 0 or y = 0. If x = 0, from the second equation we get 6y − 1 = 0 so y = 1/6.
If instead y = 0, then from the second equation x = ±1. We conclude that the critical points are (0, 1/6), (1, 0), and
(−1, 0). All three critical points satisfy the constraint x2 + y 2 ≤ 10.
The Lagrange conditions, grad f = λ grad g, are:
2xy = λ2x,
x2 + 6y − 1 = λ2y
From the first equation, when x 6= 0, we divide by x to get λ = y. Substituting into the second equation, we get
x2 + 6y − 1 = 2y 2 .
Then using x2 = 10 − y 2 from the constraint, we have
10 − y 2 + 6y − 1 = 2y 2 ,
so 3y 2 − 6y − 9 = 0. Factoring, we get 3(y − 3)(y + 1) = 0. From the constraint, we get x = ±1 when y = 3
and x = ±3 for y√= −1. If instead x = 0, so that we cannot divide by x in the first Lagrange equation, then from the
constraint, y = ± 10. Summarizing, the following points are either critical points or satisfy the Lagrange conditions:
√
(1, 0), (−1, 0), (0, 1/6), (±1, 3), (±3, −1), (0, ± 10).
These are the candidates for global maximum or minimum points. The corresponding values for f (x, y) = x2 y + 3y 2 − y
are:
√
0, 0, −1/12, 27, −5, 30 ∓ 10.
√
√
The largest value is 30 + 10 at the point (0, − 10) and the smallest value is −5 at (±3, −1).
18. (a) Minimum. The minimum value of f on the constraint is f (P ) = 20.
(b) Neither. We have f (Q) = 30, which is neither a minimum nor a maximum of f on the constraint because f (P ) =
20 < f (Q) < f (R) = 40. The fact that a contour of f and the constraint curve are tangent at Q is not enough to
conclude that f (Q) is either a maximum or minimum of f subject to the constraint.
(c) Neither. We have f (R) = 40, which is neither a minimum nor a maximum of f on the constraint because f (P ) =
20 < f (R) < f (S) = 50.
(d) Maximum. The maximum value of f on the constraint is f (S) = 50.
1422
Chapter Fifteen /SOLUTIONS
Problems
19. The function f (x, y) = x + y − (x − y)2 attains a maximum value at a critical point inside the triangle or somewhere on
its boundary.
At a critical point we have
fx = 1 − 2(x − y) = 0,
fy = 1 + 2(x − y) = 0,
so 2(x − y) = 1
so 2(x − y) = −1.
The equations have no solution, so f has no critical points.
We next find the maximum of f on each edge of the triangle separately. Each edge is a constraint and the maximum
on an edge can be found by the method of Lagrange multipliers. It is also very easy to use the equation of an edge to
change the 2-variable constrained maximum problem into a 1-variable maximum problem, which we do for the two edges
lying on the x and y-axes.
On the boundary segment x = 0, 0 ≤ y ≤ 1, we have f (0, y) = y − y 2 which attains a maximum at the point
(0, 1/2).
On the boundary segment 0 ≤ x ≤ 1, y = 0, we have f (x, 0) = x − x2 which attains a maximum at the point
(1/2, 0).
On the boundary segment x + y = 1, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, we use the method of Lagrange multipliers with
constraint g(x, y) = x + y = 1. The equations
fx = λgx
fy = λgy
g=1
give
1 − 2(x − y) = λ
1 + 2(x − y) = λ
x+y = 1
with solution (x, y) = (1/2, 1/2). On the edge we have f (x, y) = x + y − (x − y)2 = 1 − (x − y)2 so we see that
(1/2, 1/2) gives a constrained maximum. Since
f 0,
1
2
=
1
4
f
1
2
,0 =
1
4
f
1 1
,
2 2
=1
we learn that f attains a maximum value of 1 at the point (x, y) = (1/2, 1/2) on the boundary of the triangle.
20. (a) The contour for z = 1 is the line 1 = 2x + y, or y = −2x + 1. The contour for z = 3 is the line 3 = 2x + y, or
y = −2x + 3. The contours are all lines with slope −2. See Figure 15.24.
y
y
5
5
4
4
3
3
2
−7
−5
−3
−5
1
−3
x
−1
−5 −4 −3 −2 −1
−1
2
−7
1
1
23
−2
−3
−4
−5
Figure 15.24
3
4
5
5
7
1
x
−1
−5 −4 −3 −2 −1
−1
1
1
23
−2
3
4
5
5
7
−3
−4
−5
Figure 15.25
√
(b) The graph of x2 + y 2 = 5 is a circle of radius 5 = 2.236 centered at the origin. See Figure 15.25.
(c) The circle representing the constraint equation in Figure 15.25 appears to be tangent to the contour close to z = 5 at
the point (2, 1), and this is the contour with the highest z-value that the circle intersects. The circle is tangent to the
contour z = −5 approximately at the point (−2, −1), and this is the contour with the lowest z-value that the circle
intersects. Therefore, subject to the constraint x2 + y 2 = 5, the function f has a maximum value of about 5 at the
point (2, 1) and a minimum value of about −5 at the point (−2, −1).
Since the radius vector, 2~i + ~j , at the point (2, 1) is perpendicular to the line 2x + y = 5, the maximum is
exactly 5 and occurs at (2, 1). Similarly, the minimum is exactly −5 and occurs at (−2, −1).
15.3 SOLUTIONS
1423
(d) The objective function is f (x, y) = 2x + y and the constraint equation is g(x, y) = x2 + y 2 = 5, and so grad f =
2~i + ~j and grad g = (2x)~i + (2y)~j . Setting grad f = λ grad g gives
2 = λ(2x),
1 = λ(2y).
On the constraint, x 6= 0 and y 6= 0. Thus, from the first equation, we have λ = 1/x, and from the second equation
we have λ = 1/(2y). Setting these equal gives
x = 2y.
Substituting this into the constraint equation x2 + y 2 = 5 gives (2y)2 + y 2 = 5 so y = −1 and y = 1. Since x = 2y,
the maximum or minimum values occur at (2, 1) or (−2, −1). Since f (2, 1) = 5 and f (−2, −1) = −5, the function
f (x, y) = 2x + y subject to the constraint x2 + y 2 = 5 has a maximum value of 5 at the point (2, 1) and a minimum
value of −5 at the point (−2, −1). This confirms algebraically what we observed graphically in part (c).
21. Let g(x, y) = 2x + 3y, so the line is g(x, y) = 6. At the maximum on the line grad f = λ grad g, so
fx = αxα−1 y 1−α = λ · 2
fy = (1 − α)xα y −α = λ · 3.
Dividing to eliminate λ we have
2
α
x−1 y 1 = .
1−α
3
Since (1.5, 1) is a critical point, we have
α
2
(1.5)−1 11 =
1−α
3
3α = 2(1.5)(1 − α) = 3(1 − α)
3α = 3 − 3α
α = 0.5.
22. We know that a maximum or minimum value of f subject to the constraint equation g(x, y) = c occurs where grad f
is parallel to grad g, or at the endpoints of the constraint. The vectors grad f and grad g are parallel where the graph of
g(x, y) = c is tangent to the contours of f , which occurs at approximately x = 6 and y = 6. At the point (6, 6), we
have f = 400. The graph of g(x, y) = c crosses the contours f = 300, f = 200, f = 100 but does not cross any
contours with f -values greater than 400. We see that the maximum of f subject to the constraint is 400 at the point (6, 6).
It appears that f takes on its minimum value (less than 100) at one of the endpoints, which are approximately (10.5, 0)
and (0, 13.5).
23. (a) The curves are shown in Figure 15.26.
s
1500
III
II
I
1000
500
(50, 500)
20
40
s = 1000 − 10l
60
80
100
Figure 15.26
(b) The income equals $10/hour times the number of hours of work:
s = 10(100 − l) = 1000 − 10l.
l
1424
Chapter Fifteen /SOLUTIONS
(c) The graph of this constraint is the straight line in Figure 15.26.
(d) For any given salary, curve III allows for the most leisure time, curve I the least. Similarly, for any amount of leisure
time, curve III also has the greatest salary, and curve I the least. Thus, any point on curve III is preferable to any point
on curve II, which is preferable to any point on curve I. We prefer to be on the outermost curve that our constraint
allows. We want to choose the point on s = 1000 − 10l which is on the most preferable curve. Since all the curves
are concave up, this occurs at the point where s = 1000 − 10l is tangent to curve II. So we choose l = 50, s = 500,
and work 50 hours a week.
24. (a) The gradient vectors, ∇f , point inward around a local maximum. See the two points marked A in Figure 15.27.
(b) Some of the gradient vectors around a saddle are pointing inward toward the point; some are pointing outward away
from the point. See the point marked B in Figure 15.27.
(c) The critical points on g = 1 are at points where ∇f is perpendicular to the curve g = 1. There are four of them, all
marked with a dot in Figure 15.27. Imagine the level surfaces of f sketched in everywhere perpendicular to ∇f ; the
maximum value of f is at the point marked C in Figure 15.27
(d) Again imagine level curves of f . The minimum value of f is at the point marked D.
C
B
A
A
D
Figure 15.27
(e) At C, the maximum on g = 1, the vector ∇g points outward (because it points toward g = 2), while ∇f points
inward. The Lagrange multiplier, λ, is defined so that ∇f = λ∇g, so λ must be negative.
25. (a) The point P gives a minimum; the maximum is at one of the end points of the line segment (either the x- or the
y-intercept). The value of λ is negative, since f decreases in the direction in which g increases.
(b) The point P gives a maximum; the minimum is at the x- or y-intercept. The value of λ is positive, since f and g
increase in the same direction.
26. Since λ is the additional quantity of f that is obtained by relaxing the constraint by 1 unit, λ is larger if the level curves
of f are close together near the optimal point. The answer is I < II < III.
27. The maximum and minimum values change by approximately λ∆c. The Lagrange conditions give:
3 = λ2x,
−2 = λ4y.
Solving for λ and setting the expressions equal, we get x = −3y. Substituting into the constraint, we get y = ±2, so the
points satisfying the Lagrange conditions are (−6, 2) and (6, −2). The corresponding values of f (x, y) = 3x − 2y are
−22 and 22. From the first equation, we have λ = 3/(2x). Thus the minimum value changes by 3/(−12)∆c = −∆c/4
and the maximum changes by 3/(12)∆c = ∆c/4.
28. The maximum and minimum values change by approximately λ∆c. The Lagrange conditions give:
y = λ8x,
2
x = λ2y.
Solving for λ and setting the expressions equal, we get 4x = y 2 . Substituting into the constraint, we get x = ±1. Since
y = ±2x, the points satisfying the Lagrange conditions are (1, 2), (−1, 2), (1, −2), (−1, −2). Since f (x, y) = xy, we
get a maximum value of 2 at (1, 2), (−1, −2) and a minimum value of −2 at (1, −2), (−1, 2). Since λ = y/(8x), the
maximum value changes by (2/8)∆c = ∆c/4 and the minimum changes by −(2/8)∆c/) = −∆c/4.
15.3 SOLUTIONS
1425
29. (a) The company wishes to maximize P (x, y) given the constraint C(x, y) = 50, 000. The objective function is P (x, y)
and the constraint equation is C(x, y) = 50, 000. The Lagrange multiplier λ is approximately equal to the change in
P (x, y) given a one unit increase in the budget constraint. In other words, if we increase the budget by $1, we can
produce about λ more units of the good.
(b) The company wishes to minimize C(x, y) given the constraint equation P (x, y) = 2000. The objective function is
C(x, y) and the constraint equation is P (x, y) = 2000. The Lagrange multiplier λ is approximately equal to the
change in C(x, y) given a one unit increase in the production constraint. In other words, it costs about λ dollars to
produce one more unit of the good.
30.
r
✲
✻
h
❄
Figure 15.28
Let V be the volume and S be the surface area of the container. Then
V = πr 2 h and
S = 2πrh + 2πr 2
where h is the height and r is the radius as shown in Figure 15.28. We have V = 100 cm3 as our constraint. Since
and
∇S = (2πh + 4πr)~i + 2πr~j = π((2h + 4r)~i + 2r~j )
∇V = 2πrh~i + πr 2~j = π(2rh~i + r 2~j ),
at the optimum
∇S = λ∇V, we have
~
π((2h + 4r)i + 2r~j ) = πλ(2rh~i + r 2~j ),
2h + 4r = 2λrh
that is
and 2r = λr 2 ,
hence
λ=
2
.
r
We assume r 6= 0 or else we have a very awkward cylinder. Then, plug λ = 2/r into the first equation to obtain:
2h + 4r = 2
2
rh
r
2h + 4r = 4h
h = 2r.
Finally, solve for r and h using the constraint:
V = πr 2 h = 100
πr 2 (2r) = 100
50
r3 =
π
r=
Solving for h, we obtain h = 2r = 2
r
3
50
.
π
r
3
50
.
π
1426
Chapter Fifteen /SOLUTIONS
31. (a) We want to minimize C subject to g = x + y = 39. Solving ∇C = λ∇g gives
10x + 2y = λ
2x + 6y = λ
so y = 2x. Solving with x + y = 39 gives x = 13, y = 26, λ = 182. Therefore C = $4349.
(b) Since λ = 182, increasing production by 1 will cause costs to increase by approximately $182. (because λ =
k∇ Ck
= rate of change of C with g). Similarly, decreasing production by 1 will save approximately $182.
k∇ gk
32. Using Lagrange
multipliers,
let G = 2000
− 5x −10y = 0 be
the constraint.
2yx2 ~
xy 2 ~
yx2 ~
2xy 2 ~
i
+
2
+
j
=
1
+
i
+
2
+
j.
∇P = 1 +
2 · 108
2 · 108
108
108
∇G = −5~i − 10~j .
Now, ∇P = λ∇G, so
1+
xy 2
= −5λ and
108
Thus
2+
yx2
= −10λ.
108
2+
2xy 2
yx2
=2+ 8.
8
10
10
Solving, we get 2y = x or x = 0 or y = 0.
y
200
Constraint
G=0
x
400
Figure 15.29
From G = 0 we have: when x = 0, y = 200, when y = 0, x = 400, and when x = 2y, x = 200, y = 100. So
(0,200), (400,0) and (200,100) are the critical points and they include the end points.
Substitute into P : P (0, 200) = 400, P (400, 0) = 400, P (200, 100) = 402 so the organization should buy 200 sacks of
rice and 100 sacks of beans.
33. (a) Let c be the cost of producing the product. Then c = 10W + 20K = 3000. At optimum production,
∇q = λ∇c.
∇q =
1
1
9
W−4 K 4
2
~i +
3
3
3
W 4 K− 4
2
~j , and ∇c = 10~i + 20~j . Equating we get
1
1
9
W−4 K 4
2
= λ10,
and
3
3
3
W 4 K− 4
2
= λ20.
Dividing yields K = 16 W , so substituting into c gives
10W + 20
1
W
6
=
40
W = 3000.
3
Thus W = 225 and K = 37.5. Substituting both answers to find λ gives
λ=
1
1
9
(225)− 4 (37.5) 4
2
10
3
= 0.2875.
1
We also find the optimum quantity produced, q = 6(225) 4 (37.5) 4 = 862.57.
15.3 SOLUTIONS
1427
(b) At the optimum values found above, marginal productivity of labor is given by
∂q
∂W
=
(225,37.5)
9 − 41 14
W K
2
= 2.875,
(225,37.5)
and marginal productivity of capital is given by
∂q
∂K
=
(225,37.5)
3 43 − 34
W K
2
= 5.750.
(225,37.5)
The ratio of marginal productivity of labor to that of capital is
∂q
∂W
∂q
∂K
=
1
10
cost of a unit of L
=
=
.
2
20
cost of a unit of K
(c) When the budget is increased by one dollar, we substitute the relation K1 = 61 W1 into 10W1 + 20K1 = 3001 which
W1 = 3001. Solving yields W1 = 225.075 and K1 = 37.513, so q1 = 862.86 =
gives 10W1 + 20( 16 W1 ) = 40
3
q + 0.29. Thus production has increased by 0.29 ≈ λ, the Lagrange multiplier.
34. (a) The problem is to maximize
V = 1000D0.6 N 0.3
subject to the budget constraint in dollars
40000D + 10000N ≤ 600000
or (in thousand dollars)
40D + 10N ≤ 600
(b) Let B = 40D + 10N = 600 (thousand dollars) be the budget constraint. At the optimum
so
∇V = λ∇B,
∂V
∂B
=λ
= 40λ
∂D
∂D
∂B
∂V
=λ
= 10λ.
∂N
∂N
Thus
∂V /∂D
= 4.
∂V /∂N
Therefore, at the optimum point, the rate of increase in the number of visits with respect to an increase in the number
of doctors is four times the corresponding rate for nurses. This factor of four is the same as the ratio of the salaries.
(c) Differentiating and setting ∇V = λ∇B yields
600D−0.4 N 0.3 = 40λ
300D0.6 N −0.7 = 10λ
Thus, we get
600D−0.4 N 0.3
300D0.6 N −0.7
=λ=
40
10
So
N = 2D.
To solve for D and N , substitute in the budget constraint:
600 − 40D − 10N = 0
So D = 10 and N = 20.
600 − 40D − 10 · (2D) = 0
600(10−0.4 )(200.3 )
≈ 14.67
40
Thus the clinic should hire 10 doctors and 20 nurses. With that staff, the clinic can provide
λ=
V = 1000(100.6 )(200.3 ) ≈ 9,779 visits per year.
1428
Chapter Fifteen /SOLUTIONS
(d) From part c), the Lagrange multiplier is λ = 14.67. At the optimum, the Lagrange multiplier tells us that about 14.67
extra visits can be generated through an increase of $1,000 in the budget. (If we had written out the constraint in
dollars instead of thousands of dollars, the Lagrange multiplier would tell us the number of extra visits per dollar.)
(e) The marginal cost, MC, is the cost of an additional visit. Thus, at the optimum point, we need the reciprocal of the
Lagrange multiplier:
1
1
≈ 0.068 (thousand dollars),
MC = ≈
λ
14.67
that is, at the optimum point, an extra visit costs the clinic 0.068 thousand dollars, or $68.
This production function exhibits declining returns to scale (e.g. doubling both inputs less than doubles output,
because the two exponents add up to less than one). This means that for large V , increasing V will require increasing
D and N by more than when V is small. Thus the cost of an additional visit is greater for large V than for small. In
other words, the marginal cost will rise with the number of visits.
35. (a) The solution to Problem 33 gives λ = 0.29. We recalculate λ with a budget of $4000.
The condition that grad q = λ grad(budget) in Problem 33 gives
9 −1/4 1/4
W
K
= λ(10)
2
and
3 3/4 −3/4
W K
= λ(20),
2
so K = 16 W . Substituting into the budget constraint after replacing the budget of $3000 by $4000 gives
1
40
10W + 20( W ) =
W = 4000.
6
3
Thus, W = 300 and K = 50 and q = 1150.098.
Multiplying the first equation by W and the second by K and adding gives
3
9
W ( W −1/4 K 1/4 ) + K( W 3/4 K −3/4 ) = W (10λ) + K(20λ).
2
2
So
9
2
+
3
W 3/4 K 1/4 = λ(10W + 20K)
2
6W 3/4 K 1/4 = λ(4000)
Thus,
λ=
1150.098
6W 3/4 K 1/4
=
= 0.29
4000
4000
Thus, the value of λ remains unchanged.
(b) The solution to Problem 34 shows that λ = 14.67. We solve the problem again with a budget of $700,000.
The condition that grad V = λ grad B in Problem 34 gives
600D−0.4 N 0.3 = 40λ
300D0.6 N −0.7 = 10λ
Thus, N = 2D. Substituting in the budget constraint after replacing the budget of 600 by 700 (the budget in measured
in thousands of dollars) gives
40D + 10(2D) = 700
so D = 11.667 and N = 23.337 and V = 11234.705. As in part a), we multiply the first equation by D and the
second by N and add:
D(600D−0.4 N 0.3 ) + N (300D0.6 N −0.7 ) = D(40λ) + N (10λ),
so
(600 + 300)D0.6 N 0.3 = λ(400 + 10N )
900D0.6 N 0.3 = λ(700)
Since V = 1000D0.6 N 0.3 = 11234.705, we have
λ=
9 V
900D0.6 N 0.3
= (
) = 14.44.
700
7 1000
Thus, the value of λ has changed with the budget.
15.3 SOLUTIONS
1429
(c) We are interested in the marginal increase of production with budget (that is, the value of λ) and whether it is affected
by the budget.
Suppose $B is the budget. In part (a) we found
λ=
6W 3/4 K 1/4
B
and in part (b) we found
900D0.6 N 0.3
.
B
In part (a), both W and K are proportional to B. Thus, W = c1 B and K = c2 B, so
λ=
6(c1 B)3/4 (c2 B)1/4
B
3/4 1/4
6c1 C2 B 3/4 B 1/4
=
B
3/4 1/4
= 6c1 c2 .
λ=
So we see λ is independent of B.
In part (b), both D and N are proportional to B, so D = c3 B and N = c4 B. Thus,
900(c3 B)0.6 (c4 B)0.3
B
0.3 0.6 0.3
900c0.6
C
B
3
4 B
=
B
0.3 1
= 900c0.6
.
3 c4
B 0.1
λ=
So we see λ is not independent of B.
The crucial difference is that the exponents in Problem 33 add to 1, that is 3/4+1/4 = 1, whereas the exponents
in Problem 34 do not add to 1, since 0.6 + 0.3 = 0.9.
Thus, the condition that must be satisfied by the Cobb-Douglas production function
Q = cK a Lb
to ensure that the value of λ is not affected by production is that
a + b = 1.
This is called constant returns to scale.
36. (a) For a given budget, maximum production is achieved at the point on the budget line that is tangent to a production
contour. Approximate points are shown in Figure 15.30
(b) Estimating production quantities from the contours gives the following values, or close to them:
Budget, B (in dollars)
Max production, M (in pairs of skis per week)
2000
4000
6000
8000
10000
17
33
50
67
83
(c) We estimate the derivative dM/dB at B = $6000 with a difference quotient. Taking values from the table for
B = 6000 and B = 8000 we have
λ=
dM
67 − 50
≈
= 0.0085 pairs of skis per week per dollar.
dB
8000 − 6000
For example, increasing the budget from $6000 to $7000 would increase maximum production by approximately
0.0085(1000) = 8.5 pairs of skis per week.
1430
Chapter Fifteen /SOLUTIONS
L, labor
10,
000
20 B =
800
0
15 60
00
P = 90
80
70
60
50
40
30
20
10 40
00
5
200
0
5
10
15
K , capital
20
Figure 15.30
37. Since patient 1 has a visit every x1 weeks, this patient has 1/x1 visits per week. Similarly, patient 2 has 1/x2 visits per
week. Thus, the constraint is
1
1
+
=m
g(x1 , x2 ) =
x1
x2
To minimize
v1
x1
v2
x2
f (x1 , x2 ) =
·
+
·
v1 + v2 2
v1 + v2 2
subject to g(x1 , x2 ), we solve the equations
grad f = λ grad g
g(x1 , x2 ) = m.
This gives us the equations
1
∂f
v1
1
=
· =λ − 2
∂x1
v1 + v2 2
x1
1
∂f
v2
1
=
· =λ − 2
∂x2
v1 + v2 2
x2
1
1
+
= m.
x1
x2
=λ
∂g
∂x1
=λ
∂g
∂x2
Dividing the first equation by the second gives
v1
x2
= 22 .
v2
x1
As v1 , v2 , x1 , x2 , m are strictly positive we have
x2
=
x1
Substituting for x2 in the constraint occasion gives
1
+
x1
solving for x1 gives
1
x1
1+(
v2
v1
v1 12
)
v2
v1
v2
21
·
12
.
1
=m
x1
=m
1
x1 =
1
(v1 ) 2 + (v2 ) 2
1
m · (v1 ) 2
and similarly
1
1
x2 =
(v1 ) 2 + (v2 ) 2
1
m · (v2 ) 2
.
,
15.3 SOLUTIONS
38. We want to optimize
f (x1 , x2 ) =
1431
v1
x1
v2
x2
·
+
·
v1 + v2 2
v1 + v2 2
subject to
1
1
+
= m.
x1
x2
At the optimum point, x1 , x2 , and the Lagrange multiplier λ must satisfy the equations
g(x1 , x2 ) =
1
λ
v1
· =− 2
v1 + v2 2
x1
v2
1
λ
· =− 2
v1 + v2 2
x2
1
1
+
= m.
x1
x2
Solving the first and second equations for 1/x1 and 1/x2 , respectively, gives
1
v1
1
=−
·
x21
2λ (v1 + v2 )
1
v2
1
=−
·
x22
2λ (v1 + v2 )
substituting into the constraint gives (note that λ < 0):
−
1
v1
·
2λ (v1 + v2 )
1/2
+
1/2
1/2 (v )1/2 + (v )1/2
1
2
−1
v2
·
2λ (v1 + v2 )
−
1 v1 + v2 + 2(v1 v2 )1/2
·
= m2 .
2λ
v1 + v2
So
and thus
λ=−
1
2m2
1+
= −
1
2λ
2(v1 v2 )1/2
v1 + v2
·
(v1 + v2 )1/2
= m.
.
The units of λ are weeks2 (since the units of m are 1/weeks). The Lagrange multiplier measures df /dm, which
represents the rate of change in the expected delay in tumor detection as the available number of visits per week increases.
The negative sign represents the fact that as the number of visits per week increases, the delay decreases.
39. (a) The objective function is the complementary energy,
function is
f12
f2
+ 2 , and the constraint is f1 +f2 = mg. The Lagrangian
2k1 2k2
f2
f12
+ 2 − λ(f1 + f2 − mg).
2k1
2k2
We look for solutions to the system of equations we get from grad L = ~0 :
L(f1 , f2 , λ) =
∂L
f1
=
−λ = 0
∂f1
k1
∂L
f2
=
−λ = 0
∂f2
k2
∂L
= −(f1 + f2 − mg) = 0.
∂λ
Combining
∂L
∂L
f1
f2
∂L
−
=
−
= 0 with
= 0 gives the two equation system
∂f1
∂f2
k1
k2
∂λ
f2
f1
−
=0
k1
k2
f1 + f2 = mg.
1432
Chapter Fifteen /SOLUTIONS
Substituting f2 = mg − f1 into the first equation leads to
k1
mg
k1 + k2
k2
f2 =
mg.
k1 + k2
f1 =
(b) Hooke’s Law states that for a spring
Force of spring = Spring constant · Distance stretched or compressed from equilibrium.
Since f1 = k1 · λ and f2 = k2 · λ, the Lagrange multiplier λ equals the distance the mass stretches the top spring
and compresses the lower spring.
40. (a) Let f (x1 , x2 , x3 ) =
gives
P3
i=1
xi 2 = x1 2 + x2 2 + x3 2 and g(x1 , x2 , x3 ) =
2x1 = λ
and
2x2 = λ
so
x1 = x2 = x3 =
P3
i=1
xi = 1. Then grad f = λ grad g
2x3 = λ.
and
λ
.
2
Then x1 + x2 + x3 = 1 gives
3
λ
=1
2
so
λ=
2
3
so x1 = x2 = x3 =
1
.
3
These values of x1 , x2 , x3 give the minimum (rather than maximum) because the value of f increases without bound
as x2 , x2 , x3 → ∞.
Pn
Pn
(b) A similar argument shows that i=1 xi has its minimum value subject to i=1 xi = 1 when
x1 = x2 = · · · = xn =
1
.
n
41. The maximum of f (x, y) = ax2 + bxy + cy 2 subject to the constraint g(x, y) = 1 where g(x, y) = x2 + y 2 occurs
where grad f = λ grad g. Since grad f = (2ax + by)~i + (bx + 2cy)~j and grad g = 2x~i + 2y~j we have
2ax + by = 2xλ
bx + 2cy = 2yλ
x2 + y 2 = 1
Adding x times the first equation to y times the second gives x(2ax + by) + y(bx + 2cy) = (2x2 + 2y 2 )λ. Dividing by
2 and using the constraint equation gives f (x, y) = ax2 + bxy + cy 2 = (x2 + y 2 )λ = λ. This equation holds for all
solutions (x, y, λ) of the three equations, including the solution that corresponds to the maximum value of f subject to
the constraint. Thus the maximum value is f (x, y) = λ.
42. Let (x, y, z) be a point on the paraboloid. The square of the distance from (x, y, z) to the point (1, 2, 10) is given by
f (x, y, z) = (x − 1)2 + (y − 2)2 + (z − 10)2 ,
and so we wish to minimize f (x, y, z) subject to the constraint
g(x, y, z) = x2 + y 2 − z = 0.
We look for solutions to the equations grad f = λ grad g and g = 0:
2(x − 1) = 2λx,
2(y − 2) = 2λy,
2(z − 10) = −λ,
x2 + y 2 − z = 0.
1433
15.3 SOLUTIONS
If λ = 0, the first three equations would imply that (x, y, z) = (1, 2, 10), which does not satisfy the fourth equation and
so λ 6= 0. The first equation then implies that x 6= 0 and the second equation implies that y 6= 0, so we can eliminate λ
from the first three equations to get:
y−2
x−1
=
x
y
and
y−2
= −2(z − 10).
y
These give
y = 2x
and z =
2−y
+ 10.
2y
Substituting for x and z in the equation z = x2 + y 2 , we obtain
y2
2−y
+ 10 =
+ y2,
2y
4
which simplifies to give
5y 3 − 38y − 4 = 0.
Let h(y) = 5y 3 − 38y − 4. We find that h(−3) < 0 and h(−1) > 0, and so the cubic h(y) has a root between −3
and −1. Similarly, since h(−1) > 0 and h(0) < 0, then h(y) has one a between −1 and 0. Finally, as h(0) < 0 and
h(3) > 0, we see that h(y) has a root between 0 and 3. Let’s find the root lying between 0 and 3. Using a calculator, we
find that this root is approximately given by
y1 ≈ 2.808,
and so, using y = 2x, the corresponding point (x, y, z) on the paraboloid z = x2 + y 2 is given by
(x1 , y1 , z1 ) ≈ (1.404, 2.808, 9.856).
The remaining roots of h(y) are given by y2 ≈ −0.1055 and y3 ≈ −2.7026. The corresponding points on the paraboloid
z = x2 + y 2 which satisfy y = 2x are then
(x3 , y3 , z3 ) ≈ (−1.3513, −2.7026, 9.1301),
(x2 , y2 , z2 ) ≈ (−0.0528, −0.1055, 0.0139),
and so we easily see that the point (1.404, 2.808, 9.856) will be closest to (1, 2, 10). Therefore, the minimum distance is
d=
p
(1.404 − 1)2 + (2.808 − 2)2 + (9.856 − 10)2 ≈ 0.9148.
43. (a) The objective function f (x, y) = px + qy gives the cost to buy x units of input 1 at unit price p and y units of input
2 at unit price q.
The constraint g(x, y) = u tells us that we are only considering the cost of inputs x and y that can be used to
produce quantity u of the product.
Thus the number C(p, q, u) gives the minimum cost to the company of producing quantity u if the inputs it
needs have unit prices p and q.
(b) The Lagrangian function is
L(x, y, λ) = px + qy − λ(xy − u).
We look for solutions to the system of equations we get from grad L = ~0 :
∂L
= p − λy = 0
∂x
∂L
= q − λx = 0
∂y
∂L
= −(xy − u) = 0.
∂λ
We see that λ = p/y = q/x so y = px/q. Substituting for y in the constraint xy = u leads to x =
p
p
y = pu/q and λ = pq/u. The minimum cost is thus
C(p, q, u) = p
r
qu
+q
p
r
√
pu
= 2 pqu.
q
p
qu/p,
1434
Chapter Fifteen /SOLUTIONS
44. (a) The objective function U (x, y) gives the utility to the consumer of x units of item 1 and y units of item 2.
Since px + qy gives the cost to buy x units of item 1 at unit price p and y units of item 2 at unit price q, the
constraint px + qy = I tells us that we are only considering the utility of inputs x and y that can be purchased with
budget I.
Thus the number V (p, q, I) gives the maximum utility the consumer can get with a budget of I if the two items
have unit prices p and q.
The indirect utility function tells how much utility the consumer can buy, depending on his budget and the prices
of the two items.
(b) The value of the Lagrange multiplier λ is the rate of change of the maximum utility V the consumer can get with
his budget as the budget increases. This means that for small changes ∆I in the budget, smart buying will result in a
change ∆V ≈ λ∆I in the utility to the consumer of his purchases.
(c) The Lagrangian function is
L(x, y, λ) = xy − λ(px + qy − I).
We look for solutions to the system of equations we get from grad L = ~0 :
∂L
= y − λp = 0
∂x
∂L
= x − λq = 0
∂y
∂L
= −(px + qy − I) = 0.
∂λ
We see that λ = y/p = x/q so y = px/q. Substituting for y in the constraint px + qy = I leads to x = I/(2p),
y = I/(2q) and λ = I/(2pq). The maximum utility is thus
V (p, q, I) = U (
I I
I
I
I2
, )=
·
=
.
2p 2q
2p 2q
4pq
The marginal utility of money is
λ(p, q, I) = λ =
I
.
2pq
45. (a) The critical points of h(x, y) occur where
hx (x, y) = 2x − 2λ = 0
hy (x, y) = 2y − 4λ = 0.
The only critical point is (x, y) = (λ, 2λ) and it gives a minimum value for h(x, y). That minimum value is m(λ) =
h(λ, 2λ) = λ2 + (2λ)2 − λ(2λ + 4(2λ) − 15) = −5λ2 + 15λ.
(b) The maximum value of m(λ) = −5λ2 + 15λ occurs at a critical point, where m′ (λ) = −10λ + 15 = 0. At this
point, λ = 1.5 and m(λ) = −5 · 1.52 + 15 · 1.5 = 11.25.
(c) We want to minimize f (x, y) = x2 + y 2 subject to the constraint g(x, y) = 15, where g(x, y) = 2x + 4y. The
Lagrangian function is L(x, y, λ) = x2 + y 2 − λ(2x + 4y − 15) so we solve the system of equations
∂L
= 2x − 2λ = 0
∂x
∂L
= 2y − 4λ = 0
∂y
∂L
= −(2x + 4y − 15) = 0.
∂λ
The first two equations give x = λ and y = 2λ. Substitution into the third equation gives 2λ + 4(2λ) − 15 = 0
or λ = 1.5. Thus x = 1.5 and y = 3. The minimum value of f (x, y) subject to the constraint is f (1.5, 3) =
1.53 + 32 = 11.25.
(d) The two question have the same answer.
46. (a) By the method of Lagrange multipliers, the point (2, 1) is a candidate when the gradient for f at (2, 1) is a multiple of
the gradient of the constraint function at (2, 1). The constraint function is g(x, y) = x2 +y 2 , so grad g = 2x~i +2y~j .
We have grad g(2, 1) = 4~i + 2~j . This is not a multiple of grad f (2, 1) = −3~i + 4~j , so (2, 1) is not a candidate.
15.3 SOLUTIONS
1435
(b) The constraint function is g(x, y) = (x−5)2 +(y+3)2 , so grad g = 2(x−5)~i +2(y+3)~j . We have grad g(2, 1) =
−6~i + 8~j . This is a multiple of grad f (2, 1) = −3~i + 4~j , so (2, 1) is a candidate.
The contours near (2, 1) are parallel straight lines with increasing f -values as we move in the direction of
−3~i + 4~j (approximately toward the northwest). The center of the constraint circle is at (5, −3), approximately
southeast of the point (2, 1). Thus the point (2, 1) is a candidate for a maximum.
In general, if the constraint is a circle and grad f points outside the circle, then the point is a candidate for a
maximum.
(c) The constraint function is g(x, y) = (x + 1)2 + (y − 5)2 . Thus grad g(2, 1) = 6~i − 8~j , which is again a multiple
of grad f , so (2, 1) is a candidate.
This time the center of the constraint circle, (−1, 5) is approximately northwest of (2, 1), the same general
direction in which grad f is pointing. This means the point (2, 1) is a candidate for a minimum.
In general, if the constraint is a circle and grad f points inside the circle, then the point is a candidate for a
minimum.
47. (a) If the prices are p1 and p2 and the budget is b, the quantities consumed are constrained by
p1 x1 + p2 x2 = b.
We want to maximize
u(x1 , x2 ) = a ln x1 + (1 − a) ln x2
subject to the constraint
p1 x1 + p2 x2 = b.
Using Lagrange multipliers, we solve
a
∂u
=
= λp1
∂x1
x1
∂u
1−a
=
= λp2 ,
∂x2
x2
giving x1 = a/(λp1 ) and x2 = (1 − a)/(λp2 ). Substituting into the constraint, we get
(1 − a)
a
+
=b
λ
λ
so
λ=
Thus
x1 =
ab
p1
1
.
b
x2 =
(1 − a)b
p2
so the maximum satisfaction is given by
S = u(x1 , x2 ) = u
ab (1 − a)b
,
p1
p2
= a ln
ab
p1
+ (1 − a) ln
(1 − a)b
p2
= a ln a + a ln b − a ln p1 + (1 − a) ln(1 − a) + (1 − a) ln b − (1 − a) ln p2
= a ln a + (1 − a) ln(1 − a) + ln b − a ln p1 − (1 − a) ln p2 .
(b) We want to calculate the value of b needed to achieve u(x1 , x2 ) = c. Thus, we solve for b in the equation
c = a ln a + (1 − a) ln(1 − a) + ln b − a ln p1 − (1 − a) ln p2 .
Since
we have
ln b = c − a ln a − (1 − a) ln(1 − a) + a ln p1 + (1 − a) ln p2 ,
b=
ec pa1 p21−a
ec · ea ln p1 · e(1−a) ln p2
=
.
ea ln a · e(1−a) ln(1−a)
aa (1 − a)(1−a)
1436
Chapter Fifteen /SOLUTIONS
48. (a) The constraints x = c are the vertical lines on the contour diagram of f . Maxima occur where a vertical line is
tangent to a contour. Four contours with vertical tangents are shown and we can imagine others. The curve we want
is shown in Figure 15.31.
(b) The constraints x = c are the vertical lines on the graph of cross-sections of f . Since f is on the vertical axis, the
maximum value of f on the constraint occurs on the cross section that crosses the constraint at the highest point. Not
all cross-sections are shown, so we imagine the others to find the highest point. The curve we want goes across the
top of all the cross-sections, as shown in Figure 15.32.
(c) The curve in part (b) shows the maximum value of f as a function of the constraining value x = c. Since the Lagrange
multiplier λ is the rate of change of the maximum value of f with respect to c, the value of λ is the slope of that
envelope curve.
y
4
3
17
z
y=4
20
2
2
y=3
14
15
10
11
1
8
y=5
y=2
y=1
5
5
x
1
2
3
x
0
4
1
Figure 15.31
2
3
4
Figure 15.32
Strengthen Your Understanding
49. Let g(x, y) = x + y. The critical points occur where
∂f
=y=λ
∂x
∂f
= x = λ.
∂y
and
Thus x = y. Since the critical point must lie on the line x + y = 2, we have x = y = 1.
The value of f decreases as we move away from (1, 1), so (1, 1) is the maximum on the constraint. The maximum
value of f is f (1, 1) = 1.
50. The maximum value of f can occur at the endpoints. For example, the level curves of f in Figure 15.33 show that f has
a maximum at (0, 3) even though the level curves of f are nowhere tangent to the constraint.
−34
−10
−18
6
−2
3
−26
y
x
6
Figure 15.33
15.3 SOLUTIONS
1437
51. It is helpful to think of a contour diagram for f . For example, if the contours are all parallel with the highest contour
tangent to x2 + y 2 = 5 at (3, 4), that would work. Since the tangent at (3, 4) has normal 3~i + 4~j , this suggests the linear
function f (x, y) = 3x + 4y
52. Let f be any function having a global minimum at (0, 0), for example f (x, y) = x2 + y 2 .
53. Let f (x, y) = x2 + y 2 . Then
grad f = 2x~i + 2y~j .
So at (1, 1),
grad f = 2~i + 2~j .
If g(x, y) = x + y,
grad f = 2 grad g.
The point (1, 1) is a critical point. It is a minimum on the line x + y = 2 since the value of f increases without bound as
we move away from (1, 1).
54. Let f (x, y) = 10 − x2 − y 2 . Then
grad f = −2x~i − 2y~j .
If g(x, y) = x + y, then
grad g = ~i + ~j .
Then grad f = λ grad g gives
−2x = λ
−2y = λ.
So, x = y and since x + y = 4, we have x = y = 2.
Thus f (x, y) = 10 − x2 − y 2 has a critical point at (2, 2). This point gives a maximum. There is no minimum value
as f decreases as we go away from (2, 2).
55. The constraint is the line segment in the first quadrant joining the points (0, 3) and (6, 0). A possible contour diagram is
shown in Figure 15.34. The maximum value of f is 6 at the point (0, 3), an endpoint of the constraint.
−34
−10
−18
6
−2
3
−26
y
x
6
Figure 15.34
56. The maximum is f = 4 and occurs at (0, 4). The minimum is f = 2 and occurs at about (4, 2).
57. The maximum is f = 4 and occurs at (0, 4). The minimum is f = 0 and occurs at the origin.
58. True. The point (a, b) must lie on the constraint g(x, y) = c, so g(a, b) = c.
59. False. The point (a, b) is not necessarily a critical point of f , since it is a constrained extremum.
60. True. The constraint is the same as x = y, so along the constraint f = 2x, which grows without bound as x → ∞.
61. False. The condition grad f = λgrad g yields the two equations 1 = λ2x and 2 = λ2y. Substituting x = 2 in the first
equation gives λ = 1/4, while setting y = −1 in the second gives λ = −1, so the point (2, −1) is not a local extremum
of f constrained to x + 2y = 0.
1438
Chapter Fifteen /SOLUTIONS
62. False. Since grad f and grad g point in opposite directions, they are parallel. Therefore (a, b) could be a local maximum
or local minimum of f constrained to g = c. However the information given is not enough to determine that it is a
minimum. If the contours of g near (a, b) increase in the opposite direction as the contours of f , then at a point with
grad f (a, b) = λgrad g(a, b) we have λ ≤ 0, but this can be a local maximum or minimum.
For example, f (x, y) = 4 − x2 − y 2 has a local maximum at (1, 1) on the constraint g(x, y) = x + y = 2. Yet at
this point, grad f = −2~i − 2~j and grad g = ~i + ~j , so grad f and grad g point in opposite directions.
63. False. A maximum for f subject to a constraint need not be a critical point of f
64. False. The condition for the Lagrange multiplier λ is grad f (a, b) = λ grad g(a, b).
65. False. Just as a critical point need not be a maximum or minimum for unconstrained optimization, a point satisfying the
Lagrange condition need not be a maximum or minimum for a constrained optimization.
66. True. Since f (a, b) = M , we must satisfy the Lagrange conditions that fx (a, b) = λg(a, b) and fy (a, b) = λgy (a, b),
for some λ. Thus fx (a, b)/fy (a, b) = gx (a, b)/gy (a, b).
67. True. Since f (a, b) = m, the point (a, b) must satisfy the Lagrange condition that fx (a, b) = λg(a, b), for some λ. In
particular, if gx (a, b) = 0, then fx (a, b) = 0.
68. False. Whether increasing c will increase M depends on the sign of λ at a point (a, b) where f (a, b) = M
69. True. The value of λ at a maximum point gives the proportional change in M for a change in c.
70. False. The value of λ at a minimum point gives the proportional change in m for a change in c. If λ > 0 and the change
in c is positive, the change in m will also be positive.
Solutions for Chapter 15 Review
Exercises
1. At a critical point,
fx = 2x + 2y − 4 = 0
fy = 2x − 2y − 8 = 0
Solving these equations gives, the critical point x = 3, y = −1. To classify the critical point, we find
D = fxx fyy − (fxy )2 = 2(−2) − 22 = −8.
Since D < 0, we have a saddle point at (3, −1).
2. The critical points of f are obtained by solving fx = fy = 0, that is
fx (x, y) = 2y 2 − 2x = 0
and
fy (x, y) = 4xy − 4y = 0,
so
2(y 2 − x) = 0 and 4y(x − 1) = 0
The second equation gives either y = 0 or x = 1. If y = 0 then x = 0 by the first equation, so (0, 0) is a critical point. If
x = 1 then y 2 = 1 from which y = 1 or y = −1, so two further critical points are (1, −1), and (1, 1).
Since
D = fxx fyy − (fxy )2 = (−2)(4x − 4) − (4y)2 = 8 − 8x − 16y 2 ,
we have
D(0, 0) = 8 > 0, D(1, 1) = D(1, −1) = −16 < 0,
and fxx = −2 < 0. Thus, (0, 0) is a local maximum; (1, 1) and (1, −1) are saddle points.
3. To find the critical points, we solve fx = 0 and fy = 0 for x and y. Solving
fx = 3x2 − 6x = 0
fy = 2y + 10 = 0
shows that x = 0 or x = 2 and y = −5. There are two critical points: (0, −5) and (2, −5).
We have
D = (fxx )(fyy ) − (fxy )2 = (6x − 6)(2) − (0)2 = 12x − 12.
When x = 0, we have D = −12 < 0, so f has a saddle point at (0, −5). When x = 2, we have D = 12 > 0 and
fxx = 6 > 0, so f has a local minimum at (2, −5).
SOLUTIONS to Review Problems for Chapter Fifteen
1439
4. At a critical point
fx (x, y) = 2xy − 2y = 0
fy (x, y) = x2 + 4y − 2x = 0.
From the first equation, 2y(x − 1) = 0, so either y = 0 or x = 1. If y = 0, then x2 − 2x = 0, so x = 0 or x = 2. Thus
(0, 0) and (2, 0) are critical points. If x = 1, then 12 + 4y − 2 = 0, so y = 1/4. Thus (1, 1/4) is a critical point. Now
D = fxx fyy − (fxy )2 = 2y · 4 − (2x − 2)2 = 8y − 4(x − 1)2 ,
so
D(0, 0) = −4, D(2, 0) = −4, D(1, 14 ) = 2
so (0, 0) and (2, 0) are saddle points. Since fyy = 4 > 0, we see that (1, 1/4) is a local minimum.
5. Critical points occur where fx = fy = 0:
fx (x, y) =
−80
+ 10 + 10y.
x2 y
fy (x, y) =
−80
+ 10x + 20.
xy 2
Substituting x = 2, y = 1 gives
−80
+ 10 + 10.1 = 0
22 · 1
−80
fy (2, 1) =
+ 10.2 + 20 = 0.
2 · 12
fx (2, 1) =
So (2, 1) is a critical point.
To determine if this critical point is a minimum we use the second derivative test.
160
, fxx (2, 1) = 20,
x3 y
160
=
, fyy (2, 1) = 80,
xy 3
80
= 2 2 + 10, fxy (2, 1) = 30.
x y
fxx =
fyy
fxy
So D = 20 · 80 − 302 = 700 > 0 and fxx (2, 1) > 0, therefore the point (2, 1) is a local minimum.
6. The partial derivatives are
fx = cos x + cos (x + y).
fy = cos y + cos (x + y).
Setting fx = 0 and fy = 0 gives
cos x = cos y
For 0 < x < π and 0 < y < π, cos x = cos y only if x = y. Then, setting fx = fy = 0:
cos x + cos 2x = 0,
cos x + 2 cos2 x − 1 = 0,
(2 cos x − 1)(cos x + 1) = 0.
So cos x = 1/2 or cos x = −1, that is x = π/3 or x = π. For the given domain 0 < x < π, 0 < y < π, we only
consider the solution when x = π/3 then y = x = π/3. Therefore, the critical point is ( π3 , π3 ).
Since
√
fxx (x, y) = − sin x − sin (x + y) fxx ( π3 , π3 ) = − sin π3 − sin 2π
=− 3
3
fxy (x, y) = − sin (x + y)
the discriminant is
fxy ( π3 , π3 ) = − sin
fyy (x, y) = − sin y − sin (x + y) fyy ( π3 , π3 ) = − sin
2
D(x, y) = fxx fyy − fxy
√
√
√
= (− 3)(− 3) − (− 23 )2 =
√
Since fxx ( π3 , π3 ) = − 3 < 0, ( π3 , π3 ) is a local maximum.
9
4
2π
3
π
−
3
sin
> 0.
2π
3
=−
√
3
2
√
=− 3
1440
Chapter Fifteen /SOLUTIONS
7. We find critical points:
fx (x, y) = 12 − 6x = 0
fy (x, y) = 6 − 2y = 0
so (2, 3) is the only critical point. At this point
D = fxx fyy − (fxy )2 = (−6)(−2) = 12 > 0,
and fxx < 0, so (2, 3) is a local maximum. Since this is a quadratic, the local maximum is a global maximum.
Alternatively, we complete the square, giving
f (x, y) = 10 − 3(x2 − 4x) − (y 2 − 6y) = 31 − 3(x − 2)2 − (y − 3)2 .
This expression for f shows that its maximum value (which is 31) occurs where x = 2, y = 3.
8. The partial derivatives are fx = 2x − 3y, fy = 3y 2 − 3x. For critical points, solve fx = 0 and fy = 0 simultaneously.
From 2x − 3y = 0 we get x = 23 y. Substituting it into 3y 2 − 3x = 0, we have that
3
9
9
3y 2 − 3( y) = 3y 2 − y = y(3y − ) = 0.
2
2
2
So y = 0 or 3y − 92 = 0, that is, y = 0 or y = 3/2. Therefore the critical points are (0, 0) and ( 49 , 32 ).
The contour diagram for f in Figure 15.35 (drawn by a computer), shows that (0, 0) is a saddle point and that ( 49 , 23 ) is a
local minimum.
y
110
90
4
70
3
50
30
2
0
( 94 , 23 )
10
1
x
−4
−3
0
−2
−2
−30
−70
1
2
−1
−10
−90
−1
−50
3
4
10
0
30
−3
−4
Figure 15.35: Contour map of f (x, y) = x2 + y 3 − 3xy
We can also see that (0, 0) is a saddle point and ( 49 , 32 ) is a local minimum analytically. Since fxx = 2, fyy =
6y, fxy = −3, the discriminant is
2
D(x, y) = fxx fyy − fxy
= 12y − (−3)2 = 12y − 9.
D(0, 0) = −9 < 0, so (0, 0) is a saddle point.
D( 94 , 32 ) = 9 > 0 and fxx = 2 > 0, we know that ( 49 , 32 ) is a local minimum. The point ( 49 , 32 ) is not a global minimum
since f ( 94 , 32 ) = −1.6875, whereas f (0, −2) = −8.
1441
SOLUTIONS to Review Problems for Chapter Fifteen
9. Note that the x-axis and the y-axis are not in the domain of f . Since x 6= 0 and y 6= 0, by setting fx = 0 and fy = 0 we
get
1
= 0 when x = ±1
x2
4
fy = 1 − 2 = 0 when y = ±2
y
fx = 1 −
So the critical points are (1, 2), (−1, 2), (1, −2), (−1, −2). Since fxx = 2/x3 and fyy = 8/y 3 and fxy = 0, the
discriminant is
8
16
2
2
− 02 =
.
D(x, y) = fxx fyy − fxy
=
x3
y3
(xy)3
Since D < 0 at the points (−1, 2) and (1, −2), these points are saddle points. Since D > 0 at (1, 2) and (−1, −2) and
fxx (1, 2) > 0 and fxx (−1, −2) < 0, the point (1, 2) is a local minimum and the point (−1, −2) is a local maximum.
No global maximum or minimum, since f (x, y) increases without bound if x and y increase in the first quadrant; f (x, y)
decreases without bound if x and y decrease in the third quadrant.
10. The partial derivatives are
1
, fy = x + 2y.
x
For critical points, solve fx = 0 and fy = 0 simultaneously. From fy = x + 2y = 0 we get that x = −2y. Substituting
into fx = 0, we have
1
1
1
1
= (y 2 − ) = 0
y+ =y−
x
2y
y
2
fx = y +
Since
1
y
6= 0, y 2 −
1
2
= 0, therefore
√
1
2
y = ±√ = ±
,
2
2
√
√
√
√
√ √
and x = ∓ 2. So the critical points are − 2, 22 and
2, − 22 . But x must be greater than 0, so − 2,
√
2
2
not in the domain.
√
√
2, − 22 is a saddle point of f (x, y).
The contour diagram for f in Figure 15.36 (drawn by computer), shows that
y
4
10
0
−10
2
1
√
√
√
+ ln 2
f ( 2, − 22 ) = − 21
2
−1
30
20
3
❄
✲
−4
5
6
7
✻
−2
−3
x
4
−10
−20
0
Figure 15.36: Contour map of f (x, y) = xy + ln x + y 2 − 10
We can also see that
Since fxx =
− x12 , fyy
√
2, −
√
2
2
is a saddle point analytically.
= 2, fxy = 1, the discriminant is:
2
D(x, y) = fxx fyy − fxy
2
= − 2 − 1.
x
D
√
2, −
√
2
2
= −2 < 0, so
√
√
2, −
2
2
is a saddle point.
8
is
1442
Chapter Fifteen /SOLUTIONS
11. The objective function is f (x, y) = 3x − 4y and the constraint equation is g(x, y) = x2 + y 2 = 5, so grad f = 3~i − 4~j
and grad g = (2x)~i + (2y)~j . Setting grad f = λ grad g gives
3 = λ(2x),
−4 = λ(2y).
From the first equation we have λ = 3/(2x), and from the second equation we have λ = −2/y. Setting these equal gives
3
x = − y.
4
√
√
Substituting this into the constraint equation x2 + y 2 = 5 gives y 2 = 16/5, so y = 4/ 5 and y = −4/ 5. Since
x = − 34 y, there are two points where a maximum or a minimum might occur:
√
√
√
√
(−3/ 5, 4/ 5) and (3/ 5, −4/ 5).
Since the
is closed
maximum
and
√ constraint
√
√ and bounded,
√
√
√ minimum values of f subject to the constraint
√ exist.
√ Since
f (−3/ 5, 4/ 5) = −5√ 5 and f√(3/ 5, −4/ 5) = 5 5, we see that f has a minimum value at (−3/ 5, 4/ 5) and
a maximum value at (3/ 5, −4/ 5).
12. The objective function is f (x, y) = x2 + y 2 and the equation of constraint is g(x, y) = x4 + y 4 = 2. Their gradients are
∇f (x, y) = 2x~i + 2y~j ,
∇g(x, y) = 4x3~i + 4y 3~j .
So the equation ∇f = λ∇g becomes 2x~i + 2y~j = λ(4x3~i + 4y 3~j ). This tells us that
2x = 4λx3 ,
2y = 4λy 3 .
Now if x = 0, the first equation is true for any value of λ. In particular, we can choose λ which satisfies the second
equation. Similarly, y = 0 is solution.
Assuming both x 6= 0 and y 6= 0, we can divide to solve for λ and find
λ=
2y
2x
=
4x3
4y 3
1
1
=
2x2
2y 2
y 2 = x2
y = ±x.
Going back to our equation of constraint, we find
g(0, y) = 04 + y 4 = 2,
g(x, 0) = x4 + 04 = 2,
√
4
so y = ± 2
√
4
so x = ± 2
g(x, ±x) = x4 + (±x)4 = 2, so x = ±1.
√
√
Thus, the critical points are (0, ± 4 2), (± 4 2, 0), (1, ±1) and (−1, ±1). Since the constraint is closed and bounded,
maximum and minimum values of f subject to the constraint exist. Evaluating f at the critical points, we find
f (1, 1) = f (1, −1) = f (−1, 1) = f (−1, −1) = 2,
√
√
√
√
√
4
4
4
4
2) = f (0, − 2) = f ( 2, 0) = f (− 2, 0) = 2.
√
Thus, the minimum value of f (x, y) on g(x, y) = 2 is 2 and the maximum value is 2.
f (0,
13. The objective function is f (x, y) = x2 + y 2 and the constraint equation is g(x, y) = 4x − 2y = 15, so grad f =
(2x)~i + (2y)~j and grad g = 4~i − 2~j . Setting grad f = λ grad g gives
2x = 4λ,
2y = −2λ.
SOLUTIONS to Review Problems for Chapter Fifteen
1443
From the first equation we have λ = x/2, and from the second equation we have λ = −y. Setting these equal gives
y = −0.5x.
Substituting this into the constraint equation 4x − 2y = 15 gives x = 3. The only critical point is (3, −1.5).
We have f (3, −1.5) = (3)2 + (1.5)2 = 11.25. One way to determine if this point gives a maximum or minimum
value or neither for the given constraint is to examine the contour diagram of f with the constraint sketched in, Figure 15.37. It appears that moving away from the point P = (3, −1.5) in either direction along the constraint increases the
value of f , so (3, −1.5) is a point of minimum value.
y
Constraint: 4x − 2y = 15
1
5
10 15
20
x
1
2
3
−1
4
P = (3, −1.5)
−2
−3
Figure 15.37
14. The objective function is f (x, y) = x2 − xy + y 2 and the equation of constraint is g(x, y) = x2 − y 2 = 1. The gradients
of f and g are
∇f (x, y) = (2x − y)~i + (−x + 2y)~j ,
∇g(x, y) = 2x~i − 2y~j .
Therefore the equation ∇f (x, y) = λ∇g(x, y) gives
2x − y = 2λx
−x + 2y = −2λy
x2 − y 2 = 1.
Let us suppose that λ = 0. Then 2x = y and 2y = x give x = y = 0. But (0, 0) is not a solution of the third equation,
so we conclude that λ 6= 0. Now let’s multiply the first two equations
−2λy(2x − y) = 2λx(−x + 2y).
As λ 6= 0, we can cancel it in the equation above and after doing the algebra we get
√
√
which gives x = (2
√ + 3)y or x = (2 − 3)y.
If x = (2 + 3)y, the third equation gives
x2 − 4xy + y 2 = 0
(2 +
√
3)2 y 2 − y 2 = 1
so y ≈ ±0.278 and
√ x ≈ ±1.038. These give the critical points (1.038, 0.278), (−1.038, −0.278).
If x = (2 − 3)y, from the third equation we get
√
(2 − 3)2 y 2 − y 2 = 1.
√
But (2 − 3)2 − 1 ≈ −0.928 < 0 so the equation has no solution. Evaluating f gives
f (1.038, 0.278) = f (−1.038, −0.278) ≈ 0.866
1444
Chapter Fifteen /SOLUTIONS
Since y → ∞ on the constraint, rewriting f as
f (x, y) = x −
y
2
2
+
3 2
y
4
shows that f has no maximum on the constraint. The minimum value of f is 0.866. See Figure 15.38.
y
x2 − y 2 = 1
−2
0.2
0.866
(−1.038, −0.278)
(1.038, 0.278)
x
2
2
Figure 15.38
15. The objective function is f (x, y) = x2 + 2y 2 and the constraint equation is g(x, y) = 3x + 5y = 200, so grad f =
(2x)~i + (4y)~j and grad g = 3~i + 5~j . Setting grad f = λ grad g gives
2x = 3λ,
4y = 5λ.
From the first equation, we have λ = 2x/3, and from the second equation we have λ = 4y/5. Setting these equal gives
x = 1.2y.
Substituting this into the constraint equation 3x + 5y = 200 gives y = 23.256. Since x = 1.2y, we have x = 27.907. A
maximum or minimum value of f can occur only at (27.907, 23.256).
We have f (27.907, 23.256) = 1860.484. From Figure 15.39, we see that the point (27.907, 23.256) is a minimum
value of f subject to the given constraint.
y
50
4000
40
3000
30
20
1860
(27.9, 23.6)
1000
3x + 5y = 200
10
x
10
20
30
40
Figure 15.39
50
SOLUTIONS to Review Problems for Chapter Fifteen
1445
16. Our objective function is f (x, y) = xy and our equation of constraint is g(x, y) = 4x2 + y 2 = 8. Their gradients are
∇f (x, y) = y~i + x~j ,
∇g(x, y) = 8x~i + 2y~j .
So the equation ∇f = λ∇g becomes y~i + x~j = λ(8x~i + 2y~j ). This gives
8xλ = y
and
2yλ = x.
Multiplying, we get
8x2 λ = 2y 2 λ.
If λ = 0, then x = y = 0, which does not satisfy the constraint equation. So λ 6= 0 and we get
2y 2 = 8x2
y 2 = 4x2
y = ±2x.
To find x, we substitute for y in our equation of constraint.
4x2 + y 2 = 8
4x2 + 4x2 = 8
x2 = 1
x = ±1
So our critical points are (1, 2), (1, −2), (−1, 2) and (−1, −2). Since the constraint is closed and bounded, maximum
and minimum values of f subject to the constraint exist. Evaluating f (x, y) at the critical points, we have
f (1, 2) = f (−1, −2) = 2
f (1, −2) = f (1, −2) = −2.
Thus, the maximum value of f on g(x, y) = 8 is 2, and the minimum value is −2.
17. We will use the Lagrange multipliers with:
Objective function: f (x, y) = −3x2 − 2y 2 + 20xy
Constraint: g(x, y) = x + y − 100
We first find
∇f = (−6x + 20y)~i + (−4y + 20x)~j
∇g = ~i + ~j .
To optimize f , we must solve the equations
∇f = λ∇g
~
(−6x + 20y)i + (−4y + 20x)~j = λ(~i + ~j ) = λ~i + λ~j
We have a vector equation, so we equate the coordinates:
−6x + 20y = λ
So
20x − 4y = λ.
− 6x + 20y = 20x − 4y
24y = 26x
13
x
y=
12
Substituting into the constraint equation x + y = 100, we obtain:
13
x+
x = 100
12
25
x = 100
12
x = 48.
Consequently, y = 52, and f (48, 52) = 37,600. The point (48, 52) leads to the extreme value of f (x, y), given that
x + y = 100. Note that f has no minimum on the line x + y = 100 since f (x, 100 − x) = −3x2 − 2(100 − x)2 +
20x(100 − x) = −25x2 + 2400x − 20000 which goes to −∞ as x goes to ±∞. Therefore, the point (48, 52) gives the
maximum value for f on the line x + y = 100.
1446
Chapter Fifteen /SOLUTIONS
18. Our objective function is f (x, y, z) = x2 − 2y + 2z 2 and our equation of constraint is g(x, y, z) = x2 + y 2 + z 2 − 1 = 0.
To optimize f (x, y, z) with Lagrange multipliers, we solve ∇f (x, y, z) = λ∇g(x, y, z) subject to g(x, y, z) = 0. The
gradients of f and g are
∇f (x, y, z) = 2x~i − 2~j + 4z~k ,
∇g(x, y) = 2x~i + 2y~j + 2z~k .
We get,
x = λx
−1 = λy
2z = λz
x2 + y 2 + z 2 = 1.
From the first equation we get x = 0 or λ = 1.
If x = 0 we have
−1 = λy
2z = λz
2
y + z 2 = 1.
From the second equation z = 0 or λ = 2. So if z = 0, we have y = ±1 and √
we get the solutions
(0, 1, 0),(0, −1, 0). If
√
z 6= 0 then λ = 2 and y = − 21 . So z 2 = 43 which gives the solutions (0, − 21 , 23 ), (0, − 21 , − 23 ).
If x 6= 0, then λ = 1, so y = −1, which implies, from the equation x2 + y 2 + z 2 = 1, that x = 0, which contradicts
the assumption.
Since the constraint is closed and bounded, maximum and minimum values of f subject to the
constraint exist. √
There√
fore, evaluating f at the critical points, we get f (0, 1, 0) = −2, f (0, −1, 0) = 2 and f (0, − 12 , 23 ) = f (0, − 21 , − 23 ) =
5
. So the maximum value of f is 25 and the minimum is −2.
2
19. Our objective function is f (x, y, z) = 2x + y + 4z and our equation of constraint is g(x, y, z) = x2 + y + z 2 = 16.
Their gradients are
∇f (x, y, z) = 2~i + 1~j + 4~k ,
∇g(x, y, z) = 2x~i + 1~j + 2z~k .
So the equation ∇f = λ∇g becomes 2~i + 1~j + 4~k = λ(2x~i + 1~j + 2z~k ). Solving for λ we find
1
4
2
= =
2x
1
2z
2
1
λ= =1= .
x
z
λ=
Which tells us that x = 1 and z = 2. Going back to our equation of constraint, we can solve for y.
g(1, y, 2) = 16
2
1 + y + 22 = 16
y = 11.
So our one critical point is at (1, 11, 2). The value of f at this point is f (1, 11, 2) = 2 + 11 + 8 = 21. This is the
maximum value of f (x, y, z) on g(x, y, z) = 16. To see this, note that for y = 16 − x2 − z 2 ,
f (x, y, z) = 2x + 16 − x2 − z 2 + 4z = 21 − (x − 1)2 − (z − 2)2 ≤ 21.
√
√
As y → −∞, the point (− 16 − y, y, 0) is on the constraint and f (− 16 − y, y, 0) → −∞, so there is no minimum
value for f (x, y, z) on g(x, y, z) = 16.
SOLUTIONS to Review Problems for Chapter Fifteen
1447
20. We first find the critical points in the disk
∇z = (8x − y)~i + (8y − x)~j
Setting ∇z = 0 gives 8x − y = 0 and 8y − x = 0. The only solution is x = y = 0. So (0, 0) is the only critical point in
the disk.
Next we find the extremal values on the boundary using Lagrange multipliers. We have objective function z =
4x2 − xy + 4y 2 and constraint G = x2 + y 2 − 2 = 0.
∇z = (8x − y)~i + (8y − x)~j
∇G = 2x~i + 2y~j
∇z = λ∇G gives
8x − y = 2λx
8y − x = 2λy
If λ = 0 we get
8x − y = 0
8y − x = 0
with only solutions x = y = 0, which does not satisfy the constraint: x2 + y 2 − 2 = 0. Therefore λ 6= 0 and we get:
2λy(8x − y) = 2λx(8y − x)
and
2
y(8x − y) = x(8y − x).
2
So x = y , x = ±y.
Substitute into G = 0, we get 2x2 − 2 = 0 so x = ±1. The extremal points on the boundary are therefore
(1, 1), (1, −1), (−1, 1), (−1, −1). The region x2 + y 2 ≤ 2 is closed and bounded, so minimum values of f in the region
exist. We check the values of z at these points :
z(1, 1) = 7,
z(−1, −1) = 7,
z(1, −1) = 9,
z(−1, 1) = 9,
z(0, 0) = 0
Thus (−1, 1) and (1, −1) give the maxima over the closed disk and (0, 0) gives the minimum.
21. The region x2 ≥ y is the shaded region in Figure 15.40 which includes the parabola y = x2 .
y
70
−70
−50
−30
30
−10
50
10
x
Figure 15.40
We first want to find the local maxima and minima of f in the interior of our region. So we need to find the extrema
of
f (x, y) = x2 − y 2 ,
in the region
For this we compute the critical points:
fx = 2x = 0
fy = −2y = 0.
x2 > y.
1448
Chapter Fifteen /SOLUTIONS
As (0, 0) does not belong to the region x2 > y, we have no critical points. Now let’s find the local extrema of f on
the boundary of our region, hence this time we have to solve a constraint problem. We want to find the extrema of
f (x, y) = x2 − y 2 subject to g(x, y) = x2 − y = 0. We use Lagrange multipliers:
grad f = λ grad g
and x2 = y.
This gives
2x = 2λx
2y = λ
x2 = y.
From the first equation we get x = 0 or λ = 1.
If x = 0, from the third equation we get y = 0, so one solution is (0, 0). If x 6= 0, then λ = 1 and from the second
equation we get y = 21 . This gives x2 = 21 so the solutions ( √12 , 12 ) and (− √12 , 21 ).
So f (0, 0) = 0 and f ( √12 , 12 ) = f (− √12 , 12 ) = 14 . From Figure 15.40 showing the level curves of f and the region
x2 ≥ y, we see that (0, 0) is a local minimum of f on x2 = y, but not a global minimum and that ( √12 , 12 ) and (− √12 , 12 )
are global maxima of f on x2 = y but not global maxima of f on the whole region x2 ≥ y.
So there are no global extrema of f in the region x2 ≥ y.
22. The region x2 + y 2 ≤ 1 is the shaded disk of radius 1 centered at the origin (including the circle x2 + y 2 = 1) shown in
Figure 15.41.
Let’s first compute the critical points of f in the interior of the disk. We have
fx = 3x2 = 0
fy = −2y = 0,
whose solution is x = y = 0. So the only one critical point is (0, 0). As fxx (0, 0) = 0, fyy (0, 0) = −2 and fxy (0, 0) =
0,
D = fxx (0, 0) · fyy (0, 0) − (fxy (0, 0))2 = 0
which does not tell us anything about the nature of the critical point (0, 0).
But, if we choose x,y very small in absolute value and such that x3 > y 2 , then f (x, y) > 0. If we choose x,y very
small in absolute value and such that x3 < y 2 , then f (x, y) < 0. As f (0, 0) = 0, we conclude that (0, 0) is a saddle
point.
We can get the same conclusion looking at the level curves of f around (0, 0), as shown in Figure 15.42.
So, f does not have extrema in the interior of the disk.
Now, let’s find the local extrema of f on the circle x2 + y 2 = 1. So we want the extrema of f (x, y) = x3 − y 2
subject to the constraint g(x, y) = x2 + y 2 − 1 = 0. Using Lagrange multipliers we get
grad f = λ grad g
and
x2 + y 2 = 1,
which gives
3x2 = 2λx
2
−2y = 2λy
x + y 2 = 1.
From the second equation y = 0 or λ = −1.
If y = 0, from the third equation we get x2 = 1, which gives the solutions (1, 0), (−1, 0).
If y 6= 0 then λ = −1 and from the first equation we get 3x2 = −2x, hence x = 0 or x = − 32 . If x = 0, from the
third equation we get y 2 = 1, so the solutions (0, 1),(0, −1). If x = − 23 , from the third equation we get y 2 = 95 , so the
√
√
solutions (− 32 , 35 ), (− 32 , − 35 ).
Evaluating f at these points we get
f (1, 0) = 1,
and
f
f (−1, 0) = f (0, 1) = f (0, −1) = −1
√
√
5
2
5
23
2
=f − ,
=− .
− ,−
3
3
3 3
27
SOLUTIONS to Review Problems for Chapter Fifteen
1449
The region x2 + y 2 ≤ 1 is closed and bounded, so maximum and minimum values of f in the region exist. Therefore
the maximum value of f is 1 and the minimum value is −1.
y
y
1
1
1
−0.7
−0.3
−0.1
x
0
0.3 0.7
−1
x
1
−1
Figure 15.42: Level curves of f
Figure 15.41
23. If x = 10 then f (x, y) = 100 − y 2 is a parabola opening downward, so it has a maximum but no minimum.
24. If y = 10 then f (x, y) = x2 − 100 is a parabola opening upward, so it has a minimum but no maximum.
√
√
25. If x2 + y 2 = 10, then f (x, y) = x2 − y 2 = 2x2 √
− 10 and x has values in the interval − 10
√≤ x ≤ 10. Hence f (x, y)
has a maximum on the constraint at (x, y) = (± 10, 0) and a minimum at (x, y) = (0, ± 10).
26. If xy = 10, then f (x, y) = x2 − y 2 = x2 − 100/x2 and x can take any nonzero value. Since
lim
x→∞
we see f has no maximum on the constraint. Since
x2 −
lim x2 −
x→0
we see f has no minimum on the constraint.
100
x2
100
x2
= ∞,
= −∞,
Problems
27. The function f (x, y) = 0.3 ln x + 0.7 ln y has domain x > 0, y > 0. (It is not defined for x ≤ 0 or y ≤ 0.)
At a critical point grad f = λ grad g, so
0.3
= 2λ
x
0.7
= 3λ.
y
Dividing gives
0.3 y
2
·
=
x 0.7
3
y=
2 0.7
14
·
x=
x.
3 0.3
9
Substituting into 2x + 3y = 6 gives
2x + 3
14
x =6
9
20
x=6
3
so x = 0.9.
Thus, y = 14(0.9)/9 = 1.4.
This is the only critical point on the constraint, and value of f decreases toward −∞ as x or y → 0. Thus (0.9, 1.4)
gives the maximum:
f (0.9, 1.4) = 0.3 ln(0.9) + 0.7 ln(1.4) = 0.204.
1450
Chapter Fifteen /SOLUTIONS
28. (a) The distance is
p
D = (x − 3)2 + (y − 4)2 .
(b) We want to minimize D subject to the constraint x2 + y 2 = 1. The same values of x and y which minimize D also
minimize D2 = (x − 3)2 + (y − 4)2 , so we work with D2 . Thus, at a critical point
grad(D2 ) = λ grad(x2 + y 2 )
2(x − 3) = 2λx
2(y − 4) = 2λy
so we see λ 6= 1 and
x=
giving
3
1−λ
and y =
y=
4
,
1−λ
4
x·
3
Substituting into x2 + y 2 = 1, we have
x2 +
4 2
x =1
3
25 2
x =1
9
x=±
Since y = 4x/3, the critical points are
that is
3 4
,
5 5
D(0.6, 0.8) =
D(−0.6, −0.8) =
and
(0.6, 0.8)
The distances between (3, 4) and these points are
r
and
9
3
=± .
25
5
3
4
− ,−
,
5
5
(−0.6, −0.8),
p
(3 − 0.6)2 + (4 − 0.8)2 = 4
p
(3 + 0.6)2 + (4 + 0.8)2 = 6.
Thus, the minimum distance is 4 and occurs at (0.6, 0.8).
(c) The maximum distance is 6 and occurs at (−0.6, −0.8).
29. The maximum and minimum values change by approximately λ∆c. The Lagrange conditions give:
2x = λ4x3 ,
2y = λ4y 3 .
If x = 0, then y = ±21/4 from the constraint. If y = 0, then x = ±21/4 . If x 6= 0 and y 6= 0, we can solve for λ and set
the expressions equal to get x2 = y 2 , so y = ±x.
Thus, there are eight points satisfying the Lagrange conditions: four of the form (0, ±21/4 ) or (±21/4 , 0), and four
of the form (±1, ±1). Since f (x,√
y) = x2 + y 2 , we get a maximum value of 2 at the four points of the form (±1, ±1)
2/4
and a minimum value of 2
= 2 at the other four points. For the maximum value, we use λ = 1/(2x2 ) = 1/2, so
2
the change is approximately
∆c/2. For the minimum value at (0, ±21/4 ), we
√
√ use λ = 1/(2y ), so there the change is
approximately ∆c/(2 2). Similarly, the change at ±21/4 , 0) is also ∆c/(2 2).
30. Let the line be in the form y = b + mx. When x equals −1, 0 and 1, then y equals b − m, b, and b + m, respectively. The
sum of the squares of the vertical distances, which is what we want to minimize, is
f (m, b) = (2 − (b − m))2 + (−1 − b)2 + (1 − (b + m))2 .
To find the critical points, we compute the partial derivatives with respect to m and b,
fm = 2(2 − b + m) + 0 + 2(1 − b − m)(−1)
= 4 − 2b + 2m − 2 + 2b + 2m
= 2 + 4m,
fb = 2(2 − b + m)(−1) + 2(−1 − b)(−1) + 2(1 − b − m)(−1)
= −4 + 2b − 2m + 2 + 2b − 2 + 2b + 2m
= −4 + 6b.
1451
SOLUTIONS to Review Problems for Chapter Fifteen
Setting both partial derivatives equal to zero, we get a system of equations:
2 + 4m = 0,
−4 + 6b = 0.
The solution is m = −1/2 and b = 2/3. One can check that it is a minimum. Hence, the regression line is y =
2
3
− 21 x.
2
31. Since fxx < 0 and D = fxx fyy − fxy
> 0, the point (1, 3) is a maximum. See Figure 15.43.
y
3
−
− 4
1
32
−
16
64
−
−
−
12
0
0
✠
x
1
Figure 15.43
32. At a local maximum value of f ,
∂f
= −2x − B = 0.
∂x
We are told that this is satisfied by x = −2. So −2(−2) − B = 0 and B = 4. In addition,
∂f
= −2y − C = 0
∂y
and we know this holds for y = 1, so −2(1) − C = 0, giving C = −2. We are also told that the value of f is 15 at the
point (−2, 1), so
15 = f (−2, 1) = A − ((−2)2 + 4(−2) + 12 − 2(1)) = A − (−5), so A = 10.
Now we check that these values of A, B, and C give f (x, y) a local maximum at the point (−2, 1). Since
fxx (−2, 1) = −2,
fyy (−2, 1) = −2
and
fxy (−2, 1) = 0,
we have that fxx (−2, 1)fyy (−2, 1) −
maximum value 15 at (−2, 1).
33. (a)
2
fxy
(−2, 1)
= (−2)(−2) − 0 > 0 and fxx (−2, 1) < 0. Thus, f has a local
(i) Suppose N = kAp . Then the rule of thumb tells us that if A is multiplied by 10, the value of N doubles. Thus
2N = k(10A)p = k10p Ap .
Thus, dividing by N = kAp , we have
2 = 10p
so taking logs to base 10 we have
p = log 2 = 0.3010.
(where log 2 means log 10 2). Thus,
N = kA0.3010 .
1452
Chapter Fifteen /SOLUTIONS
(ii) Taking natural logs gives
ln N = ln(kAp )
ln N = ln k + p ln A
ln N ≈ ln k + 0.301 ln A
Thus, ln N is a linear function of ln A.
(b) Table 15.2 contains the natural logarithms of the data:
Table 15.2
ln N and ln A
ln A
ln N
Redonda
1.1
1.6
Saba
3.0
2.2
Montserrat
2.3
2.7
Puerto Rico
9.1
4.3
Island
9.3
4.2
Hispaniola
11.2
4.8
Cuba
11.6
4.8
Jamaica
Using a least squares fit we find the line:
ln N = 1.20 + 0.32 ln A
This yields the power function:
N = e1.20 A0.32 = 3.32A0.32
Since 0.32 is pretty close to log 2 ≈ 0.301, the answer does agree with the biological rule.
34. We want to minimize
C = f (q1 , q2 ) = 2q12 + q1 q2 + q22 + 500
subject to the constraint q1 + q2 = 200 or g(q1 , q2 ) = q1 + q2 − 200 = 0.
Since ∇f = (4q1 + q2 )~i + (2q2 + q1 )~j and ∇g = ~i + ~j , ∇f = λ∇g gives
4q1 + q2 = λ
2q2 + q1 = λ.
Solving we get
4q1 + q2 = 2q2 + q1
so
3q1 = q2 .
We want
q1 + q2 = 200
q1 + 3q1 = 4q1 = 200.
Therefore
q1 = 50 units,
q2 = 150 units.
35. (a) Let g(x, y) = x + y. We are minimizing f (x, y) = x2 + 2y 2 subject to the constraint g(x, y) = c.
The method of Lagrange multipliers is to solve the equations
fx = λgx
fy = λgy
g = c,
which are
2x = λ
4y = λ
x + y = c.
SOLUTIONS to Review Problems for Chapter Fifteen
1453
We have
c
4c
2c
y=
λ=
,
3
3
3
so there is a critical point at (2c/3, c/3). Since moving away from the origin increases values of f in Figure 15.44,
we see that f has a minimum on the constraint. The minimum value is
x=
m(c) = f
2c c
,
3 3
=
2c2
.
3
y
Constraint
x+y = c
✠( 2c
, c)
3 3
x
Figure 15.44
(b) Calculations in part (a) showed that λ = 4c/3.
(c) The multiplier λ is the rate of change of m(c) as c increases and the constraint moves: that is, λ = m′ (c).
36. We are comparing the maxima of f (x, y) subject to the two different constraints g(x, y) = 240 and g(x, y) = 242. As c
changes in the constraint g(x, y) = c, we have
Change in maximum of f ≈ λ × Change in c = 20 · 2 = 40.
Since the maximum with c = 240 is 6300, we have
Maximum of f constrained by g(x, y) = 242 ≈ 6300 + 40 = 6340.
37. Constraint is G = P1 x + P2 y − K = 0.
Since ∇Q = λ∇G, we have
caxa−1 y b = λP1
and
cbxa y b−1 = λP2 .
a−1 b
λP1
ay
P1
bP1
cax
y
=
, or simplifying,
=
. Hence, y =
x.
cbxa y b−1
λP2
bx
P
aP
2
2
bP1
a+b
x = P1
Substitute into the constraint to obtain P1 x + P2
x = K, giving
aP2
a
Dividing the two equations yields
x=
aK
(a + b)P1
and
y=
bK
.
(a + b)P2
We now check that this is indeed the maximization point. Since x, y ≥ 0, possible maximization points are (0,
K
),
P2
aK
bK
K
, 0), and (
,
). Since Q = 0 for the first two points and Q is positive for the last point, it follows
P1
(a + b)P1 (a + b)P2
aK
bK
that (
,
) gives the maximal value.
(a + b)P1 (a + b)P2
(
1454
Chapter Fifteen /SOLUTIONS
38. (a) To be producing the maximum quantity Q under the cost constraint given, the firm should be using K and L values
given by
∂Q
= 0.6aK −0.4 L0.4 = 20λ
∂K
∂Q
= 0.4aK 0.6 L−0.6 = 10λ
∂L
20K + 10L = 150.
L
20λ
4
0.6aK −0.4 L0.4
= 1.5
=
= 2, so L = K. Substituting in 20K + 10L = 150, we obtain
0.4aK
K
10λ
3
4 0.6
L−0.6
9
1
20K + 10
K = 150. Then K = and L = 6, so capital should be reduced by unit, and labor should be
3
2
2
increased by 1 unit.
a4.50.6 60.4
New production
=
≈ 1.01, so tell the board of directors, “Reducing the quantity of capital by 1/2 unit
(b)
Old production
a50.6 50.4
and increasing the quantity of labor by 1 unit will increase production by 1% while holding costs to $150.”
Hence
39. (a) Points A, B, C, D, E; that is, where a level curve of f and the constraint curve are parallel.
(b) Point F since the value of f is greatest at this point.
(c) Point D has the greatest f value of the points A, B, C, D, E.
g=c
❘
A
E
F
13
D
12
B
11
10
C
9
40. We want to minimize the function h(x, y) subject to the constraint that
g(x, y) = x2 + y 2 = 1,0002 = 1,000,000.
Using the method of Lagrange multipliers, we obtain the following system of equations:
10x + 4y
= 2λx,
10,000
4x + 4y
= 2λy,
hy = −
10,000
x2 + y 2 = 1,000,000.
hx = −
Multiplying the first equation by y and the second by x we get
−x(4x + 4y)
−y(10x + 4y)
=
.
10,000
10,000
Hence:
2y 2 + 3xy − 2x2 = (2y − x)(y + 2x) = 0,
and so the climber either moves along the line x = 2y or y = −2x.
SOLUTIONS to Review Problems for Chapter Fifteen
1455
We must now choose one of these lines and the direction along that line which will lead to the point of minimum
height on the circle. To do this we find the points of intersection of these lines with the circle x2 + y 2 = 1,000,000,
compute the corresponding heights, and then select the minimum point.
If x = 2y, the third equation gives
5y 2 = 1,0002 ,
√
so that y = ±1,000/ 5 ≈ ±447.21 and x = ±894.43. The corresponding height is h(±894.43, ±447.21) = 2400 m.
If y = −2x, we find that x = ±447.21 and y = ∓894.43. The corresponding height is h(±447.21, ∓894.43) =
2900 m. Therefore, she should travel along the line x = 2y, in either of the two possible directions.
41. The objective function is
f (x, y, z) =
and the constraint is
p
(x − a)2 + (y − b)2 + (z − c)2 ,
g(x, y, z) = Ax + By + Cz + D = 0.
Partial derivatives of f and g are
fx =
fy =
fz =
1
2
· 2 · (x − a)
x−a
=
,
f (x, y, z)
f (x, y, z)
1
2
· 2 · (y − b)
y−b
=
,
f (x, y, z)
f (x, y, z)
1
2
· 2 · (z − c)
z−c
=
,
f (x, y, z)
f (x, y, z)
gx = A, gy = B, and gz = C.
Using Lagrange multipliers, we need to solve the equations
grad f = λ grad g
where grad f = fx~i + fy~j + fz ~k and grad g = gx~i + gy~j + gz ~k . This gives a system of equations:
x−a
= λA
f (x, y, z)
y−b
= λB
f (x, y, z)
z−c
= λC
f (x, y, z)
Ax + By + Cz + D = 0.
Now
x−a
A
=
y−b
B
=
z−c
C
= λf (x, y, z) gives
A
(y − b) + a,
B
C
z = (y − b) + c,
B
x=
Substitute into the constraint,
A
Hence
A
(y − b) + a + By + C
B
A2
C2
+B+
B
B
y=
C
(y − b) + c + D = 0,
B
A2
C2
b − Aa +
b − Cc − D.
B
B
(A2 + C 2 )b − B(Aa + Cc + D)
,
A2 + B 2 + C 2
−B(Aa + Bb + Cc + D)
y−b =
A2 + B 2 + C 2
y=
1456
Chapter Fifteen /SOLUTIONS
A
(y − b)
B
−A(Aa + Bb + Cc + D)
=
A2 + B 2 + C 2
C
z − c = (y − b)
B
−C(Aa + Bb + Cc + D)
=
A2 + B 2 + C 2
x−a=
Thus the minimum f (x, y, z) is
p
(x − a)2 + (y − b)2 + (z − c)2
f (x, y, z) =
=
+
−A(Aa + Bb + Cc + D)
A2 + B 2 + C 2
2
−C(Aa + Bb + Cc + D)
A2 + B 2 + C 2
+
2
−B(Aa + Bb + Cc + D)
A2 + B 2 + C 2
2
1/2
|Aa + Bb + Cc + D|
√
.
A2 + B 2 + C 2
The geometric meaning is finding the shortest distance from a point (a, b, c) to the plane Ax + By + Cz + D = 0.
=
42. We first express the revenue R in terms of the prices p1 and p2 :
R(p1 , p2 ) = p1 q1 + p2 q2
= p1 (517 − 3.5p1 + 0.8p2 ) + p2 (770 − 4.4p2 + 1.4p1 )
= 517p1 − 3.5p21 + 770p2 − 4.4p22 + 2.2p1 p2 .
At a local maximum we have grad R = 0, and so:
∂R
= 517 − 7p1 + 2.2p2 = 0,
∂p1
∂R
= 770 − 8.8p2 + 2.2p1 = 0.
∂p2
Solving these equations, we find that
p1 = 110 and p2 = 115.
To see whether or not we have a found a local maximum, we compute the second-order partial derivatives:
∂2R
= −7,
∂p21
∂2R
= −8.8,
∂p22
∂2R
= 2.2.
∂p1 ∂p2
Therefore,
∂2R ∂2R
∂2R
−
= (−7)(−8.8) − (2.2)2 = 56.76,
2
2
∂p1 ∂p2
∂p1 ∂p2
and so we have found a local maximum point. The graph of P (p1 , p2 ) has the shape of an upside down paraboloid. Since
P is quadratic in q1 and q2 , (110, 115) is a global maximum point.
D=
43. We want to minimize cost C = 100L + 200K subject to Q = 900L1/2 K 2/3 = 36000. Using Lagrange multipliers, we
get
∇Q = 450L−1/2 K 2/3 ~i + 600L1/2 K −1/3 ~j .
∇C = 100~i + 200~j
∇C = λ∇Q gives
100 = λ450L−1/2 K 2/3
and 200 = λ600L1/2 K −1/3 .
Since λ 6= 0 this gives
450L−1/2 K 2/3 = 300L1/2 K −1/3 .
Solving, we get L = (3/2)K. Substituting into Q = 36,000 gives
900
h
Solving yields K = 40 ·
2 1/2
3
i6/7
and L = 30 which gives C = $7, 000.
3 1/2
2
K
≈ 19.85, so L ≈
K 2/3 = 36,000.
3
(19.85)
2
= 29.78. We can thus calculate cost using K = 20
SOLUTIONS to Review Problems for Chapter Fifteen
1457
44. We wish to minimize the objective function
C(x, y, z) = 20x + 10y + 5z
subject to the budget constraint
Q(x, y, z) = 20x1/2 y 1/4 z 2/5 = 1, 200.
Therefore, we solve the equations grad C = λ grad Q and Q = 1, 200:
20 = 10λx−1/2 y 1/4 z 2/5
or
λ = 2x1/2 y −1/4 z −2/5 ,
,
or
λ = 2x−1/2 y 3/4 z −2/5 ,
5 = 8λx1/2 y 1/4 z −3/5 ,
or
λ = 0.625x−1/2 y −1/4 z 3/5 ,
10 = 5λx
20x
1/2 1/4 2/5
y
z
1/2 −3/4 2/5
y
z
= 1, 200.
The first and second equations imply that
x = y,
while the second and third equations imply that
3.2y = z.
Substituting for x and z in the constraint equation gives
20y 1/2 y 1/4 (3.2y)2/5 = 1200
y ≈ 23.47,
and so
x ≈ 23.47
and
z ≈ 75.1.
45. Cost of production, C, is given by
C = p1 W + p2 K = b. At the optimal point, ∇q = λ∇C.
Since ∇q = c(1 − a)W −a K a ~i + caW 1−a K a−1 ~j and ∇C = p1~i + p2~j , we get
c(1 − a)W −a K a = λp1 and caW 1−a K a−1 = λp2 .
∂q
= c(1 − a)W −a K a and marginal productivity of capital is given by
Now, marginal productivity of labor is given by ∂W
∂q
1−a a−1
= caW
K
, so their ratio is given by
∂K
∂q
∂W
∂q
∂K
=
c(1 − a)W −a K a
λp1
p1
=
=
caW 1−a K a−1
λp2
p2
which is the ratio of the cost of one unit of labor to the cost of one unit of capital.
46. (a) The objective function is the energy loss, i21 R1 + i22 R2 , and the constraint is i1 + i2 = I, where I is a constant. The
Lagrangian function is
L(i1 , i2 , λ) = i21 R1 + i22 R2 − λ(i1 + i2 − I).
We look for solutions to the system of equations we get from grad L = ~0 :
∂L
= 2i1 R1 − λ = 0
∂i1
∂L
= 2i2 R2 − λ = 0
∂i2
∂L
= −(i1 + i2 − I) = 0.
∂λ
Combining
∂L
∂L
∂L
−
= 2(i1 R1 − i2 R2 ) = 0 with
= 0 gives the two equation system
∂i1
∂i2
∂λ
i1 R1 − i2 R2 = 0
i1 + i2 = I.
1458
Chapter Fifteen /SOLUTIONS
Substituting i2 = I − i1 into the first equation leads to
R2
I
R1 + R2
R1
i2 =
I.
R1 + R2
i1 =
(b) Ohm’s Law states that across a resistor
Voltage = Current · Resistance.
Since λ/2 = i1 · R1 = i2 · R2 , the Lagrange multiplier λ equals twice the voltage across the resistors.
47. Let the sides of the base be x and y cm. Let the height be z cm. Then the volume is given by xyz = 32 and the surface
area, S, is given by
S = xy + 2xz + 2yz.
Substituting z = 32/(xy) gives
S = xy +
64
64
+
.
y
x
At a critical point
∂S
=y−
∂x
∂S
= x−
∂y
64
=0
x2
64
= 0.
y2
The symmetry of the equations tells us that x = y and
x−
64
=0
x2
3
x = 64
x = 4 cm.
Thus the only critical point is x = y = 4 cm and z = 32/(4 · 4) = 2 cm. At the critical point
D = Sxx Syy − (Sxy )2 =
(128)2
128 128
· 3 − 12 = 3 3 − 1.
x3
y
x y
Since D > 0 and Sxx > 0 at this critical point, the critical point x = y = 4, z = 2 is a local minimum. Since S → ∞ as
x, y → ∞, the local minimum is a global minimum.
48. The point P is the solution to the constraint optimization problem of maximizing the square of the distance function.
D = x2 + y 2 + x2
subject to the constraint
g(x, y, z) = f (x, y) − z = 0.
(We take the square of the distance between the point (x, y, z) and the origin, which is
Distance =
p
(x − 0)2 + (y − 0)2 + (z − 0)2 =
p
x2 + y 2 + z 2 ,
because it makes the calculations easier.) Therefore, at point P , we have ∇D = λ∇g, so ∇D is parallel to ∇g.
We know that ∇g is perpendicular to the surface g(x, y, z) = 0; that is, perpendicular to the surface z = f (x, y).
Also
∇D = 2x~i + 2y~j + 2z~k .
At point P , whose position vector is p
~ = a~i + b~j + c~k , we have
∇D = 2(a~i + b~j + c~k ) = 2~
p.
Thus, p
~ is parallel to ∇D and therefore p
~ is also perpendicular to the surface.
SOLUTIONS to Review Problems for Chapter Fifteen
1459
49. You should try to anticipate your opponent’s choice. After you choose a value λ, your opponent will use calculus to
find the point (x, y) that maximizes the function f (x, y) = 10 − x2 − y 2 − 2x − λ(2x + 2y). At that point, we have
fx = −2x − 2 − 2λ = 0 and fy = −2y − 2λ = 0, so your opponent will choose x = −1 − λ and y = −λ. This gives a
value L(−1 − λ, −λ, λ) = 10 − (−1 − λ)2 − (−λ)2 − 2(−1 − λ) − λ(2(−1 − λ) + 2(−λ)) = 11 + 2λ + 2λ2 which you
want to make as small as possible. You should choose λ to minimize the function h(λ) = 11 + 2λ + 2λ2 . You choose λ
so that h′ (λ) = 2 + 4λ = 0, or λ = −1/2. Your opponent then chooses (x, y) = (−1 − λ, −λ) = (−1/2, 1/2), giving
a final score of L(−1/2, 1/2, −1/2) = 10.5. No choice of λ that you can make can force the value of L below 10.5. But
your choice of λ = −1/2 makes it impossible for your opponent to force the value of L above 10.5.
50. The wetted perimeter of the trapezoid is given by the sum of the lengths of the three walls, so
p=w+
2d
sin θ
We want to minimize p subject to the constraint that the area is fixed at 50 m2 . A trapezoid of height h and with parallel
sides of lengths b1 and b2 has
(b1 + b2 )
A = Area = h
.
2
In this case, d corresponds to h and b1 corresponds to w. The b2 term corresponds to the width of the exposed surface of
the canal. We find that b2 = w + (2d)/(tan θ). Substituting into our original equation for the area along with the fact that
the area is fixed at 50 m2 , we arrive at the formula:
Area =
d
2d
w+w+
2
tan θ
=d w+
d
tan θ
= 50
We now solve the constraint equation for one of the variables; we will choose w to give
w=
d
50
−
.
d
tan θ
Substituting into the expression for p gives
p=w+
50
d
2d
2d
=
−
+
.
sin θ
d
tan θ
sin θ
We now take partial derivatives:
∂p
50
1
2
=− 2 −
+
∂d
d
tan θ
sin θ
d
1
2d
∂p
· cos θ
=
·
−
∂θ
tan2 θ cos2 θ
sin2 θ
From ∂p/∂θ = 0, we get
d · cos2 θ
1
2d
·
· cos θ.
=
2
2θ
cos
sin θ
sin2 θ
Since sin θ =
6 0 and cos θ =
6 0, canceling gives
1 = 2 cos θ
so
cos θ =
1
.
2
π
, we get
2
Substituting into the equation ∂p/∂d = 0 and solving for d gives:
Since
0<θ<
θ=
π
.
3
1
2
−50
−√ +√
=0
d2
3
3/2
which leads to
d=
Then
r
50
√ ≈ 5.37m.
3
d
50
5.37
50
−
≈
− √ ≈ 6.21 m.
d
tan θ
5.37
3
When θ = π/3, w ≈ 6.21 m and d ≈ 5.37 m, we have p ≈ 18.61 m.
Since there is only one critical point, and since p increases without limit as d or θ shrink to zero, the critical point
must give the global minimum for p.
w=
1460
Chapter Fifteen /SOLUTIONS
CAS Challenge Problems
51. (a) The partial derivatives of f are
√
√
a + x + −1 + a + x y
∂f
= √
2
√
√
∂x
2 a+x a+x+y 1+ a+x+y
√
∂f
1 − 2a − 2x + a + x −y
= √
2
√
∂y
2 a+x+y 1+ a+x+y
Solving ∂f /∂x = 0, ∂f /∂y = 0, we get x = 1/4 − a, y = 1. The discriminant at this point is D = −16/625.
Thus, by the second derivative test, the point is a saddle point.
(b) The y coordinate of the critical points stays the same and the x coordinate is a √
units to the left of its position when
x+y
√ by substituting x + a for
a = 0. The type is always a saddle point. This is because f is obtained from
1+y+ x
x, so that the graph is shifted a units in the negative x-direction but its shape remains the same.
52. (a) We have grad f = 2x~i + ~j and grad g = (2x + 2y)~i + (2x + 2y)~j . So the equations to be solved in the method
of Lagrange multipliers are
2x = λ(2x + 2y)
1 = λ(2x + 2y)
2
2
x + 2xy + y − 9 = 0
Solving these with a CAS, we get two solutions:
x = 1/2, y = −7/2, λ = −1/6,
or
x = 1/2, y = 5/2, λ = 1/6
Student A reasons that since f (1/2, −7/2) = −13/4 and f (1/2, 5/2) = 11/4 , the (global) maximum and
minimum values are 11/4 and 13/4, respectively. Student B graphs the constraint curve g = 0 and a contour diagram
of f . The constraint curve turns out to be two straight lines, since the constraint x2 + 2xy + y 2 − 9 = 0, which can
be rewritten as (x + y)2 = 9, or x + y = ±3. The value of f goes to infinity on each of these straight lines. On the
line y = −x + 3, f (x, y) = x2 + y = x2 − x + 3, and on the line y = −x − 3, f (x, y) = x2 + y = x2 − x − 3.
Thus Student B is correct. The points Student A found are actually local maximum and local minimum values, not
global. Since the constraint is not bounded, there is no guarantee that there is a local maximum or minimum. See
Figure 15.45.
y
4
2
x
−4
−2
2
4
g
−2
−4
g
Figure 15.45: Contours of f and two straight
lines giving constraint g = 0
53. (a) We have grad f = 3~i + 2~j and grad g = (4x − 4y)~i + (−4x + 10y)~j , so the Lagrange multiplier equations are
3 = λ(4x − 4y)
2 = λ(−4x + 10y)
2
2x − 4xy + 5y 2 = 20
PROJECTS FOR CHAPTER FIFTEEN
1461
Solving these with a CAS we get λ = −0.4005, x = −3.9532, y = −2.0806 and λ = 0.4005, x = 3.9532, y =
2.0806. We have f (−3.9532, −2.0806) = −11.0208, and f (3, 9532, 2.0806) = 21.0208. The constraint equation
is 2x2 − 4xy + 5y 2 = 20, or, completing the square, 2(x − y)2 + 3y 2 = 20. This has the shape of a skewed ellipse,
so the constraint curve is bounded, and therefore the local maximum is a global maximum. Thus the maximum value
is 21.0208.
(b) The maximum value on g = 20.5 is ≈ 21.0208 + 0.5(0.4005) = 21.2211. The maximum value on g = 20.2 is
≈ 21.0208 + 0.2(0.4005) = 21.1008.
(c) We use the same commands in the CAS from part (a), with 20 replaced by 20.5 and 20.2, and get the maximum
values 21.2198 for g = 20.5 and 21.1007 for g = 20.2. These agree with the approximations we found in part (b) to
2 decimal places.
PROJECTS FOR CHAPTER FIFTEEN
1. (a) The price of p/q units of B at unit price q is p/q · q = p, the same as the price of one unit of A. On a fixed
budget, p/q units of B can substitute for one unit of A. Thus the ERS is p/q.
(b) The TRS measures the rate at which y increases with respect to x as a point (x, y) slides in the direction
of decreasing x along a fixed contour f (x, y) = Q of the production function. Thus TRS = −dy/dx, the
negative of the slope of the contour. On the contour the differential df = fx dx + fy dy is zero because f
has constant value there. Thus fx dx + fy dy = 0 which gives TRS = −dy/dx = fx /fy .
(c) We want to maximize the production function f (x, y) subject to a budget constraint px + qy = C, where
C is the fixed budget. At a maximum we have grad f = λ grad g or
fx = λgx
fy = λgy .
Dividing the first equation by the second gives fx /fy = gx /gy or TRS = ERS.
(d) We are asked to minimize the cost function g(x, y) subject to a production constraint f (x, y) = Q, where
Q is the fixed quantity to be produced. At a minimum we have grad g = λ grad f or
gx = λfx
gy = λfy .
Dividing the first equation by the second gives gx /gy = fx /fy or ERS = TRS.
2.
y = b + mx
(xi , b + mxi )
(x1 , y1 )
(x1 , b + mx1 )
(xi , yi )
Figure 15.46
(a) Points which are directly above or below each other share the same x coordinate, therefore, the point on
the least squares line which is directly above or below the point in question will have x coordinate xi and
from the formula for the least squares line, it will have ypcoordinate b + mxi . (See Figure 14.1.)
(b) The general distance formula in two dimensions is d = (x2 − x1 )2 + (y2 − y1 )2 , so d2 = (x2 − x1 )2 +
(y2 − y1 )2 . Since the x coordinates are identical for the two points in question, the first term in the square
root is zero. This yields d2 = (yi − (b + mxi ))2 .
(c) In both cases we use the chain rule and our knowledge of summations to show the relationship.
!
n
n
X
∂
∂f
∂ X
2
=
(yi − (b + mxi ))
=
(yi − (b + mxi ))2
∂b
∂b i=1
∂b
i=1
1462
Chapter Fifteen /SOLUTIONS
=
n
X
∂
(yi − (b + mxi ))
∂b
2(yi − (b + mxi )) ·
i=1
=
n
X
2(yi − (b + mxi )) · (−1)
i=1
= −2
n
X
(yi − (b + mxi ))
i=1
n
X
∂
∂f
=
∂m
∂m
=
n
X
(yi − (b + mxi ))2
i=1
2(yi − (b + mxi )) ·
i=1
=
n
X
!
=
n
X
∂
(yi − (b + mxi ))2
∂m
i=1
∂
(yi − (b + mxi ))
∂m
2(yi − (b + mxi )) · (−xi )
i=1
= −2
n
X
(yi − (b + mxi )) · xi
i=1
(d) We can separate
∂f
into three sums as shown:
∂b
n
n
n
X
X
X
∂f
= −2
yi − b
1−m
xi
∂b
i=1
i=1
i=1
Similarly we can separate
!
∂f
after multiplying through by xi :
∂m
n
n
n
X
X
X
∂f
= −2
yi xi − b
xi − m
xi 2
∂m
i=1
i=1
i=1
Setting
!
∂f
∂f
and
equal to zero we have:
∂b
∂m
bn + m
n
X
xi =
i=1
b
n
X
xi + m
n
X
n
X
x2i =
i=1
i=1
yi
i=1
n
X
xi yi
i=1
(e) To P
solve this pair of linear equations, we multiply the first equation by
by ni=1 xi , and subtract; we get
bn
n
X
i=1
So,
b=
n
X
Similarly,
m=
n
i=1
x2i , multiply the second one
n
n
n
n
n
X
X
X
X
X
x2i − b(
xi )2 =
yi
x2i −
xi yi
xi ,
x2i
i=1
Pn
i=1
n
X
yi −
i=1
n
X
i=1
xi yi −
n
X
i=1
xi
i=1
n
X
i=1
n
X
i=1
xi
n
X
i=1
i=1
i=1
! n
X
xi yi / n
x2i −
i=1
yi
!
/ n
n
X
i=1
x2i −
i=1
n
X
i=1
n
X
i=1
!2
xi
xi
!2
PROJECTS FOR CHAPTER FIFTEEN
1463
(f) Applying the formulas to the given data, we have b = − 31 , m = 1 which gives y = −(1/3) + x, in
agreement with the example.
3. (a) If p = e−x where x → ∞ then p → 0 with p > 0 and
lim
(p ln p) = lim (−xe−x ) = 0,
p→0
x→∞
p>0
since the exponential decreases faster than any power of x. Alternatively, use l’Hopital’s rule:
lim
(p ln p) = lim
p→0
p→0
p>0
p>0
1/p
ln p
= lim
=0
p→0
1/p
−1/p2
p>0
(b) We apply the method of Lagrange multipliers to find the critical points of S(p1 , · · · , p30 ). The constraint
function is g(p1 , · · · , p30 ) = p1 + · · · + p30 . We have
!
30
X
∂S
∂
ln pi
1
=
−
=−
(ln pj + 1),
pi
∂pj
∂pj
ln 2
ln 2
i=1
therefore
30
grad S = −
1 X
(ln pj + 1)k~j
ln 2 j=1
where k~1 , · · · , k~30 are the unit vectors corresponding 30 independent directions of the pj -axes. Also,
grad g =
30
X
k~j
j=1
so the condition grad S = λ grad g becomes
−
1
(ln pj + 1) = λ,
ln 2
for i = 1, · · · , 30.
Thus,
ln pj = −λ ln 2 − 1
and, in particular, all the pj s must be equal. Since the pj s have to satisfy the constraint g(p1 , · · · , p30 ) = 1,
1
1
1
1
we see that pj = 30
and that the point ( 30
, 30
, · · · , 30
) is the only critical point of S. We have
1
1 (− ln 30)
ln 30
1 1
, ,···,
= −30 ·
=
.
S
30 30
30
30 ln 2
ln 2
We will not prove that this is indeed the maximum value of S (this requires a higher-dimensional analogue
of the second derivative test). Since in part (c) we show that the minimum value of S is 0, the critical point
we have found here is not a global minimum; the maximum of S has to be attained somewhere and it is
reasonable to believe that it is attained at the unique critical point. The maximum entropy corresponds to
maximum uncertainty in the outcome of the competition.
(c) We already know that S ≥ 0. However, S can be zero: For example, if p1 = 1 and p2 = · · · = p30 = 0,
we have S(1, 0, · · · , 0) = 0. Therefore the minimum value of S is 0. Now we determine all the values of
pi s for which S(p1 , · · · , p30 ) = 0. The condition
S=−
30
X
pi ln pi
i=1
2
=0
together with the restrictions − ln pi ≥ 0 shows that, for S to vanish, each individual term in the above
sum has
vanish. This means pi ln pi = 0 for all i = 1, · · · , 30, that is, pi = 0 or pi = 1 for i = 1, · · · , 30.
Pto
30
Since i=1 pi = 1, only one of the pi s is 1 whereas the other 29 are 0. This corresponds to the case where
one of the teams is certain to win, that is, there is no uncertainty. The result can be interpreted by saying
that zero entropy implies zero uncertainty.
16.1 SOLUTIONS
1465
CHAPTER SIXTEEN
Solutions for Section 16.1
Exercises
1. Using ∆x = 3 and ∆y = 0.5:
Lower estimate = (4 + 5 + 3 + 4)∆x ∆y = 16 · 3 · 0.5 = 24,
Upper estimate = (7 + 10 + 5 + 7)∆x ∆y = 29 · 3 · 0.5 = 43.5.
2. Mark the values of the function on the plane, as shown in Figure 16.1, so that you can guess respectively at the smallest
and largest values the function takes on each small rectangle.
X
Lower sum =
f (xi , yi )∆x∆y
= 4∆x∆y + 6∆x∆y + 3∆x∆y + 4∆x∆y
= 17∆x∆y
= 17(0.1)(0.2) = 0.34.
Upper sum =
X
f (xi , yi )∆x∆y
= 7∆x∆y + 10∆x∆y + 6∆x∆y + 8∆x∆y
= 31∆x∆y
= 31(0.1)(0.2) = 0.62.
y
↓ 3
2.4
2.2
4
5
4
6
8
✻
∆y = 0.2
❄
2.0
5
7
1.0
1.1
✛
∆x = 0.1
✲
10
1.2 ← x
Figure 16.1
3. There are nine squares. Using the largest value of g in each square for the overestimate and the smallest value for the
underestimate, we have
Overestimate ≈ (2 + 2.8 + 3.5 + 2.8 + 4 + 4.9 + 3.4 + 4.9 + 6)∆x∆y = 34.3 · 2 · 2 = 137.2.
Underestimate ≈ (0 + 0.8 + 0.9 + 0.8 + 2 + 2.8 + 0.9 + 2.8 + 4)∆x∆y = 15 · 2 · 2 = 60.
1466
Chapter Sixteen /SOLUTIONS
4. In the subrectangle in the top left of the figure given, it appears that f (x, y) has a maximum value of about 9. In the
subrectangle in the top middle, f (x, y) has a maximum value of 10. Continuing in this way, and multiplying by ∆x and
∆y, we have
Overestimate = (9 + 10 + 12 + 7 + 8 + 10 + 5 + 7 + 8) · 10 · 5 = 3800.
Similarly, we find
Thus, we expect that
Underestimate = (7 + 7 + 8 + 4 + 5 + 7 + 1 + 3 + 6) · 10 · 5 = 2400.
2400 ≤
Z
R
f (x, y)dA ≤ 3800.
5. Let R be Rthe rectangle, and let f (x, y) be the population density at the point (x, y) in the rectangle. The population is
given by R f (x, y) dA. We approximate the integral with a 6 term Riemann sum using six 1 km × 1 km squares in the
region. To make a Riemann sum, choose a population density value at one point of each of the squares. For accuracy, we
choose the midpoint of each square, but other choices are possible.
The densities at the midpoints can only be estimated from the contour diagram, and different people may make
different judgments. The value at the point (2.5, 1.5) is particularly difficult, because the diagram tells us only that the
density is between 0 and 200. The table gives one reasonable set of values.
x
y
0.5
1.5
2.5
1.5
500
450
100
0.5
650
200
400
Each square has area 1 km2 . We have
Z
R
f (x, y) dA ≈ 500 · 1 + 450 · 1 + 100 · 1 + 650 · 1 + 200 · 1 + 400 · 1 = 2300 people.
The population is approximately 2300 people.
Problems
6. The function being integrated is f (x, y) = 1, which is positive everywhere. Thus, its integral over any region is positive.
7. The function being integrated is f (x, y) = 5x. Since x > 0 in R, f is positive in R and thus the integral is positive.
8. The function being integrated is f (x, y) = 5x, which is an odd function in x. Since B is symmetric with respect to x, the
contributions to the integral cancel out, as f (x, y) = −f (−x, y). Thus, the integral is zero.
9. The function being integrated, f (x, y) = y 3 + y 5 , is an odd function in y while D is symmetric with respect to y. Then,
by symmetry, the positive and negative contributions of f will cancel out and thus its integral is zero.
10. In a region such as B in which y < 0, the quantity y 3 + y 5 is less than zero. Thus, its integral is negative.
11. The function being integrated, f (x, y) = y − y 3 , is an odd function in y while D is symmetric with respect to y. By
symmetry, the integral is zero.
12. The function being integrated, f (x, y) = y − y 3 is always negative in the region B since in that region −1 < y < 0 and
|y 3 | < |y|. Thus, the integral is negative.
13. The total area of the square R is (1.5)(1.5) = 2.25. See Figure 16.2. On a disk of radius ≈ 0.5 the function has a value of 3
or more, giving a total contribution to the integral of at least (3)·(π ·0.52 ) ≈ 2.3. On less than half of the rest of the square
the function has a value between −2 and 0, giving a contribution to the integral of between (1/2·2.25)(−2) = −2.25
R and
0. Since the positive contribution to the integral is therefore greater in magnitude than the negative contribution, R f dA
is positive.
16.1 SOLUTIONS
1467
y
2.0
1.5
1.0
0.5
−1
2
0
3
−0.5
−1.0
−1.0 −0.5
4
x
0
1
2
0
0.5
1.0
1.5
2.0
Figure 16.2
14. We use four subrectangles to find an overestimate and underestimate of the integral:
Overestimate = (15 + 9 + 9 + 5)(4)(3) = 456,
Underestimate = (5 + 2 + 3 + 1)(4)(3) = 132.
A better estimate of the integral is the average of the two:
Z
R
f (x, y)dA ≈
456 + 132
= 294.
2
The units of the integral are milligrams, and the integral represents the total number of mg of mosquito larvae in this 8
meter by 6 meter section of swamp.
15. Let’s break up the room into 25 sections, each of which is 1 meter by 1 meter and has area ∆A = 1.
We shall begin our sum as an upper estimate starting with the lower left corner of the room and continue across the
bottom and moving upward using the highest temperature, Ti , in each case. So the upper Riemann sum becomes
P25
i=1
Ti ∆A = T1 ∆A + T2 ∆A + T3 ∆A + · · · + T25 ∆A
= ∆A(T1 + T2 + T3 + · · · + T25 )
= (1)( 31 +29 + 28 + 27 + 27+
29 +28 + 27 + 27 + 26+
27 +27 + 26 + 26 + 26+
26 +26 + 25 + 25 + 25+
25 +24 + 24 + 24 + 24)
= (1)(659) = 659.
In the same way, the lower Riemann sum is formed by taking the lowest temperature, ti , in each case:
P25
t ∆A
i=1 i
= t1 ∆A + t2 ∆A + t3 ∆A + · · · + t25 ∆A
= ∆A(t1 + t2 + t3 + · · · + t25 )
= (1)( 27 +27 + 26 + 26 + 25+
26 +26 + 25 + 25 + 25+
25 +24 + 24 + 24 + 24+
24 +23 + 23 + 23 + 23+
23 +21 + 20 + 21 + 22)
= (1)(602) = 602.
So, averaging the upper and lower sums we get: 630.5.
1468
Chapter Sixteen /SOLUTIONS
To compute the average temperature, we divide by the area of the room, giving
630.5
≈ 25.2◦ C.
(5)(5)
Average temperature =
Alternatively we can use the temperature at the central point of each section ∆A. Then the sum becomes
P25
i=1
P25
Ti′ ∆A = ∆A
i=1
Ti′
= (1)( 29 +28 + 27 + 26.5 + 26+
27 +27 + 26 + 26 + 25.5+
26 +25.5 + 25 + 25 + 25+
25 +24 + 24 + 24 + 24+
24 +23 + 22 + 22.5 + 23)
= (1)(630) = 630.
Then we get
Average temperature =
P25
Ti′ ∆A
630
=
≈ 25.2◦ C.
Area
(5)(5)
i=1
16. We divide the base region into four subrectangles as shown in Figure 16.3. The height of the object at each point (x, y)
is given by f (x, y) = x + y, we label each corner of the subrectangles with the value of the function at that point. (See
Figure 16.3.) Since Volume = Height × Length × Width, and ∆x = 2 and ∆y = 3, we have
Overestimate = (8 + 10 + 5 + 7)(2)(3) = 360,
and
Underestimate = (3 + 5 + 0 + 2)(2)(3) = 60.
We average these to obtain
Volume ≈
360 + 60
= 210.
2
y
6
3
f =4
f =7
f = 10
f =2
f =5
f =8
f =0
f =3
2
f =6
x
4
Figure 16.3
Strengthen Your Understanding
17. This is true only if f is nonnegative everywhere. Integrals can be negative, but volumes never can be negative.
18. R
The sign of an integral depends on the values of f on the region R. If f is positive everywhere on the region R, then
f (x, y) dA is positive.
R
19. We want the value of f at the lower left-hand corner of each subrectangle to be the largest value in the subrectangle. For
example, let f (x, y) be a linear function with negative slopes, say f (x, y) = 5 − x − y. Let R be the square with vertices
(±1, ±1).
16.2 SOLUTIONS
1469
20. A function whose values are negative everywhere has a negative average value on any region. For example, f (x, y) = −1
has average value −1 on the square.
21. False. For example, if f (x, y) < 0 for all (x, y) in the region R, then
22. True. The double integral is the limit of the sum
X
X
f (x, y)∆A =
R
f dA is negative.
R
k∆A = k
X
∆A
over rectangles that lie inside the region R. As the area ∆A → 0, this sum approaches k · Area(R).
23. False. The function f (x, y) = exy is largest at the (1, 1) corner of R, so for any (x, y) in R we have exy ≤ e1·1 = e.
Then
Z
exy dA = lim
∆A→0
R
So
R
R
exy dA ≤ e ≈ 2.7.
24. False. For example, if f = 1, then
X
R
R
exy ∆A ≤ lim
∆A→0
X
e∆A = e lim
∆A→0
1 dA = Area(R) = 6 and
P
R
S
X
∆A = e · Area(R) = e.
1 dA = Area(S) = 6.
25. True. The double integral is the limit of the sum ∆A→0 ρ(x, y)∆A. Each of the terms ρ(x, y)∆A is an approximation
of the total population inside a small rectangle of area ∆A. Thus the limit of the sum of all of these numbers as ∆A → ∞
gives the total population of the region R.
26. False. If the graph of f has equal volumes above and below the xy-plane over the region R, the double integral is zero
without having f (x, y) = 0 everywhere.
27. True. Writing the definition of the integral of g, we have
Z
g dA = lim
∆A→0
R
X
g(x, y)∆A = lim
∆A→0
X
kf (x, y)∆A = k lim
∆A→0
X
f (x, y)∆A = k
Z
28. False. As a counterexample, let R be a rectangle with area 2 and take f (x, y) = g(x, y) = 1. Then
R
R
R
1 dA = Area(R) = 2, but R f dA = R g dA = Area(R) · Area(R) = 4.
R
f dA.
R
R
R
f · g dA =
29. False. There is no reason to expect this to be true, since the behavior of f on one half of R can be completely unrelated
to the behavior of f on the other half. As a counterexample, suppose that f is defined so that f (x,
R y) = 0 for points
(x, y) lying in S, and f (x, y) = 1 for points (x, y) lying in the part of R that is not in S. Then S f dA = 0, since
R
f = 0 on all of S. To evaluate R f dA, note that f = 1 on the square S1 which is 0 ≤ x ≤ 1, 1 ≤ y ≤ 2. Then
R
R
f dA = S f dA = Area(S1 ) = 1, since f = 0 on S.
R
1
30. True. Since all points in the region R satisfy x < y, it is true that at every point in R, f (x, y) = x + x < x + y = g(x, y).
Since all of the values of f in R are less than those of g, the average of the values of f is less than the average of the
values of g.
Solutions for Section 16.2
Exercises
1. See Figure 16.4.
y
π
Z
0
π
Z
x
y sin x dy dx
0
π
Figure 16.4
x
1470
Chapter Sixteen /SOLUTIONS
2. See Figure 16.5.
3. See Figure 16.6.
4. See Figure 16.7.
Z
y
1
1
y
xy dx dy
y2
0
Z 2Z
y
Z
0
2
y2
y 2 x dx dy
0
Z 1Z
y
1
0
cos πx
y dy dx
x−2
x
1
1
−1
x
x
1
2
Figure 16.5
−2
4
Figure 16.6
Figure 16.7
5. We evaluate the inside integral first:
4
Z
(4x + 3y) dx = (2x2 + 3yx)
4
= 32 + 12y.
0
0
Therefore, we have
Z
3
Z
0
4
(4x + 3y) dxdy =
3
Z
(32 + 12y) dy = (32y + 6y 2 )
= 150.
0
0
0
3
6. We evaluate the inside integral first:
3
Z
(x2 + y 2 ) dy =
0
Therefore, we have
Z
0
2
Z
3
(x2 + y 2 ) dydx =
0
x2 y +
Z
y3
3
y=3
= 3x2 + 9.
y=0
2
(3x2 + 9) dx = (x3 + 9x)
2
= 26.
0
0
7. We evaluate the inside integral first:
Z
2
(6xy) dy = (3xy 2 )
3
Z
0
2
Z
= 12x.
0
0
Therefore, we have
2
(6xy) dydx =
0
Z
3
(12x) dx = (6x2 )
3
= 54.
0
0
8. We evaluate the inside integral first:
Z
2
(x2 y) dy =
0
x2 y 2
2
y=2
= 2x2 .
y=0
Therefore, we have
Z
0
1
Z
0
2
(x2 y) dydx =
Z
0
1
(2x2 ) dx =
2x3
3
1
=
0
2
.
3
16.2 SOLUTIONS
1471
9. Calculating the inner integral first, we have
Z
1
0
Z
1
ye
xy
dx dy =
0
1
1
Z
e
xy
0
0
!
1
Z
dy =
0
1
1
Z
ey − e0 dy =
0
(ey −1) dy = (ey −y)
0
= e1 −1−(e0 −0) = e−2.
10. Calculating the inner integral first, we have
2
Z
0
Z
y
y dx dy =
Z
y
2
0
0
Z
yx dy =
2
y 2 dy =
0
0
y3
3
2
=
0
8
.
3
11. Calculating the inner integral first, we have
Z
3
0
y
Z
Z
sin x dx dy =
3
0
0
y
− cos x
0
dy =
3
3
Z
(− cos y + 1) dy = (− sin y + y)
0
0
= − sin 3 + 3.
12. Calculating from the inside and using integration by parts, we have
Z
π/2
0
13.
Z
3
1
Z
Z
sin x
x dy dx =
e
x+y
dxdy =
0
xy
0
0
4
sin x
π/2
Z
4
3
Z
x y
dx =
e e
1
Z
π/2
π/2
x sin x dx = (−x cos x + sin x)
0
0
1
0
dx =
Z
3
ex (e4 − 1) dx = (e4 − 1)(e2 − 1)e. See Figure 16.8.
y
y=4
x
1
3
Figure 16.8
14.
Z
0
2
Z
0
x
2
ex dydx =
Z
x
2
2
ex y dx =
0
0
Z
2
2
xex dx =
0
1 x2
e
2
2
=
0
1 4
(e − 1). See Figure 16.9.
2
y
y=x
x
2
Figure 16.9
15.
Z
1
5
Z
= 1.
0
2x
sin x dy dx =
x
Z
5
sin x · y
1
=
Z
1
2x
x
dx
5
sin x · x dx
= (sin x − x cos x)
5
1
= (sin 5 − 5 cos 5) − (sin 1 − cos 1) ≈ −2.68.
1472
Chapter Sixteen /SOLUTIONS
See Figure 16.10.
y
y = 2x
y=x
x
1
5
Figure 16.10
16.
Z
4
Z
y
√
1
x2 y 3 dx dy =
y
Z
4
1
3
Z
y3
1
=
=
1
3
1
=
3
y
x3
3
√
dy
y
4
1
9
(y 6 − y 2 ) dy
y7
y 11/2
−
7
11/2
4
1
47
411/2 × 2
−
7
11
−
See Figure 16.11.
y
x=
√
y
1
2
−
7
11
≈ 656.082
y=x
y=4
4
1
x
1
2
4
Figure 16.11
17.
Z
1
4
Z
1
2
f dy dx
or
Z
1
2
Z
4
f dx dy
1
18. This region lies between x = 0 and x = 4 and between the lines y = 3x and y = 12, and so the iterated integral is
Z
4
0
Z
12
f (x, y) dydx.
3x
Alternatively, we could have set up the integral as follows:
Z
12
0
Z
0
y/3
f (x, y) dxdy.
16.2 SOLUTIONS
19. The line connecting (−1, 1) and (3, −2) is
1473
3x + 4y = 1
or
1 − 3x
4
y=
So the integral becomes
3
Z
−1
Z
(1−3x)/4
Z
f dy dx or
1
−2
−2
Z
(1−4y)/3
f dx dy
−1
20. The line on the left (through points (0, 0) and (3, 6)) is the line y = 2x; the line on the right (through points (3, 6) and
(5, 0)) is the line y = −3x + 15. See Figure 16.12. One way to set up this iterated integral is:
Z
0
6
(15−y)/3
Z
f (x, y) dxdy.
y/2
The other option for setting up this integral requires two separate integrals, as follows:
Z
3
0
Z
2x
Z
f (x, y) dydx +
5
3
0
y
Z
−3x+15
f (x, y) dydx.
0
y = 2x
y = −3x + 15
R
x
Figure 16.12
21. Two of the sides of the triangle have equations x =
3
Z
1
y−5
y−1
and x =
. So the integral is
2
−2
−1
(y−5)
2
Z
f dx dy
1 (y−1)
2
22. The line connecting (1, 0) and (4, 1) is
1
(x − 1)
3
y=
So the integral is
Z
4
1
Z
2
f dy dx
(x−1)/3
23.
Z
√
x + y dA =
R
2
Z
Z
0
=
Z
=
2
3
√
x + y dx dy
0
2
3
2
(x + y) 2
3
0
=
1
Z
0
2
3
1
dy
0
3
((1 + y) 2 − y 2 ) dy
5
5
2 2
· [(1 + y) 2 − y 2 ]
3 5
2
0
5
5
4
((3 2 − 2 2 ) − (1 − 0))
=
15
√
4 √
(9 3 − 4 2 − 1) = 2.38176
=
15
1474
Chapter Sixteen /SOLUTIONS
24. In the other order, the integral is
Z
1
0
2
Z
√
x + y dy dx.
0
First we keep x fixed and calculate the inside integral with respect to y:
2
Z
√
x + y dy =
0
0
1
y=0
2
(x + 2)3/2 − x3/2 .
=
3
Then the outside integral becomes
Z
y=2
2
(x + y)3/2
3
h
i
2
2
2 2
(x + 2)5/2 − x5/2
(x + 2)3/2 − x3/2 dx =
3
3 5
5
1
0
2 2 5/2
= ·
3
− 1 − 25/2 = 2.38176
3 5
Note that the answer is the same as the one we got in Exercise 23.
25.
Z
Z
2
(5x + 1) sin 3y dA =
1
−1
1
R
Z
=
Z
π/3
(5x2 + 1) sin 3y dy dx
0
2
(5x + 1)
−1
2
3
=
Z
1
π/3
1
− cos 3y
3
0
dx
(5x2 + 1) dx
−1
2 5 3
( x + x)
3 3
=
1
−1
2 10
32
= (
+ 2) =
3 3
9
26. The region of integration, R, is shown in Figure 16.13.
y
1
y =1−x
x
1
Figure 16.13
Integrating first over y, as shown in the diagram, we obtain
Z
R
xy dA =
Z
1
0
Z
1−x
0
Now integrating with respect to x gives
Z
R
dx =
1
x2 −
xy dy
xy dA =
4
Z
0
1
xy 2
2
1−x
dx =
0
0
1 3 1 4
x + x
3
8
Z
1
=
0
1
.
24
1
1
x(1 − x)2 dx
2
1475
16.2 SOLUTIONS
27. It would be easier to integrate first in the x direction from x = y − 1 to x = −y + 1, because integrating first in the y
direction would involve two separate integrals.
Z
2
(2x + 3y) dA =
1
Z
0
R
=
=
Z
1
1
0
=
(4x2 + 12xy + 9y 2 ) dx dy
y−1
0
=
−y+1
Z
0
Z
(2x + 3y)2 dx dy
y−1
1
Z
−y+1
Z
4 3
x + 6x2 y + 9xy 2
3
−y+1
dy
y−1
8
[ (−y + 1)3 + 9y 2 (−2y + 2)] dy
3
−
2
9
(−y + 1)4 − y 4 + 6y 3
3
2
9
13
2
= − (−1) − + 6 =
3
2
6
1
0
See Figure 16.14.
y
1
y =x+1
y = −x + 1
x
−1
1
Figure 16.14
Problems
28. The diagonal line has equation y = 2x. Integrating with respect to y first gives
Z
1
0
Z
−1
0
xy dy dx =
0
Z
1
x
0
y2
2
2x
dx =
Z
1
2x3 dx =
0
0
p
x4
2
1
=
0
1
.
2
1 − y 2 . Integrating with respect to x first, we have
√
Z 1
Z 1 2
1−y 2
1−y 2
x y
y2
y4
1
xy dx dy =
(1 − y 2 )y dy =
−
dy =
2 0
2 −1
4
8
−1
29. The right half of the circle is x =
Z Z √
1
2x
1
= 0.
−1
y
30.
(1, 1)
R
x
2
The edges of the triangle are the lines y = x on the left and y = 2 − x on the right. We integrate with respect to x
first, giving
Z
0
1
Z
y
2−y
xy dx dy =
Z
0
1
x2 y
2
2−y
dy =
y
1
2
Z
1
((2−y)2 y−y 3 ) dy =
0
1
2
Z
0
1
(4y−4y 2 +y 3 −y 3 )dy = y 2 −
2y 3
3
1
=
0
1
.
3
1476
Chapter Sixteen /SOLUTIONS
y
31.
1
R
x
2
p
The circle has equation (x − 1)2 + y 2 = 1; thus the top half has equation y = 1 − (x − 1)2 . Integrating with
respect to y first gives
Z 2 2 √1−(x−1)2
Z 2
Z 2 Z √1−(x−1)2
xy
x
xy dy dx =
dx =
(1 − (x − 1)2 )dx
2
2
0
0
0
0
0
Z
=
0
2
x3
2
x2 −
dx =
x3
3
−
x4
8
2
=
0
2
.
3
32. (a) We divide the base region into four subrectangles as shown in Figure 16.15. The height of the object at each point
(x, y) is given by f (x, y) = xy, we label each corner of the subrectangles with the value of the function at that point.
(See Figure 16.15.) Since Volume = Height × Length × Width, and ∆x = 2 and ∆y = 3, we have
Overestimate = (12 + 24 + 6 + 12)(2)(3) = 324,
and
Underestimate = (0 + 6 + 0 + 0)(2)(3) = 36.
We average these to obtain
Volume ≈
324 + 36
= 180.
2
y
6
3
f =0
f = 12
f = 24
f =0
f =6
f = 12
f =0
f =0
f =0
x
2
4
Figure 16.15
(b) We have f (x, y) = xy, so
Volume =
Z
0
4
Z
0
6
xy dydx =
Z
4
0
xy 2
2
y=6
dx =
y=0
Z
0
4
18x dx = 9x2
4
= 144.
0
The volume of this object is 144. Notice that 144 is between the over- and underestimates, 324 and 36, found in
part (a).
16.2 SOLUTIONS
1477
33. As given, the region of integration is as shown in Figure 16.16.
y
1
x=y
x=1
x
1
Figure 16.16
Reversing the limits gives
1
Z
0
Z
0
x
2
ex dydx =
1
Z
x
yex
0
2
0
1
ex
=
2
2
dx =
1
Z
2
xex dx
0
e−1
=
.
2
0
34. The function sin (x2 ) has no elementary antiderivative, so we try integrating with respect to y first. The region of integration is shown in Figure 16.17. Changing the order of integration, we get
Z
0
1
Z
1
sin (x2 ) dx dy =
1
Z
0
y
=
Z
x
dx
0
1
0
=−
sin (x2 ) dy dx
sin (x2 ) · y
0
=
x
0
1
Z
Z
sin (x2 ) · x dx
cos (x2 )
2
1
0
cos 1
1
1
=−
+ = (1 − cos 1) = 0.23.
2
2
2
y
y=x
x
1
Figure 16.17
1478
Chapter Sixteen /SOLUTIONS
35. As given, the region of integration is as shown in Figure 16.18.
y
1
x=
√
y
x=1
x
1
Figure 16.18
Reversing the limits gives
Z
0
1
x2
Z
p
2+
0
x3
dydx =
Z
x2
1
(y
0
=
Z
1
p
2+
x2
0
) dx
0
p
2 + x3 dx
3
2
(2 + x3 ) 2
9
=
x3
1
=
0
√
2 √
(3 3 − 2 2).
9
36. As given, the region of integration is as shown in Figure 16.19.
y
3
x = y2
x=9
x
9
Figure 16.19
Reversing the limits gives
9
Z
0
√
x
Z
Z
9
1
=
2
Z
2
y sin (x ) dydx =
0
y 2 sin (x2 )
2
0
=−
√
x
0
!
dx
9
x sin (x2 ) dx
0
cos (x2 )
4
9
0
cos (81)
1
= 0.056.
= −
4
4
37. The region of the integration is shown in Figure 16.20. To make the integration easier, we want to change the order of the
integration and get
Z
0
1
Z
e
ey
x
dx dy =
ln x
=
Z eZ
1
Z
0
e
1
=
Z
1
ln x
x
dy dx
ln x
x
·y
ln x
e
x dx =
ln x
dx
0
x2
2
e
=
1
1 2
(e − 1).
2
1479
16.2 SOLUTIONS
y
(e, 1)
e
1
x
Figure 16.20
38. The region is bounded by x = 1, x = 4, y = 2, and y = 2x. Thus
Volume =
4
Z
1
Z
2x
(6x2 y) dydx.
2
To evaluate this integral, we evaluate the inside integral first:
Z
2x
(6x2 y) dy = (3x2 y 2 )
2x
2
2
= 3x2 (2x)2 − 3x2 (22 ) = 12x4 − 12x2 .
Therefore, we have
Z
4
1
Z
2x
(6x2 y) dydx =
Z
4
1
2
(12x4 − 12x2 ) dx =
The volume of this object is 2203.2.
12 5
x − 4x3
5
4
= 2203.2.
1
39. (a) The volume is given by
V =
Z 1Z
1
−1 −1
x2 + y 2 dy dx =
Z
1
x2 y +
−1
y3
3
1
dx = 2
Z
1
−1
−1
x2 +
1
dx = 2
3
x
x3
+
3
3
1
=
−1
8
.
3
(b) The region above the surface, together with the region whose volume we found in part (a), make up a box with a
square base of side 2 and height 2. The volume of the box is 2 · 2 · 2 = 8, the volume under the surface is
V =8−
8
16
=
.
3
3
40. The region is bounded by x = 1 − y, x = y − 1, y = 1, and y = 3. See Figure 16.21. To evaluate this integral we
integrate with respect to x first, giving
y
(-2,3)
3
x=1−y
y=3
(2,3)
x=y−1
1
x
−2
2
−1
Figure 16.21
1480
Chapter Sixteen /SOLUTIONS
Z Z
2
(2x + y) dA =
3
Z
1
R
=
Z
(2x2 + y) dx dy
1−y
3
y−1
2x3
+ yx
3
1
=2
y−1
Z
3
Z
3
Z
dy = 2
2(y − 1)3
+ y(y − 1) dy
3
1
1−y
2y 3
2
− y 2 + y − dy
3
3
1
44
=
3
41. (a) See Figure 16.22.
y
(1/2, 1/2)
y=x
x+y = 1
R
x
Figure 16.22
(b) If we integrate with respect to x first, we have
Z
Z
f (x, y) dA =
R
1/2
0
Z
1−y
f (x, y) dx dy.
y
If we integrate with respect to y first, the integral must be split into two parts, so
Z
f (x, y) dA =
Z
1/2
Z
1/2
1
2
Z
0
R
Z
x
Z
1−y
f (x, y) dy dx +
0
1
Z
1/2
Z
1−x
f (x, y) dy dx.
0
(c) If f (x, y) = x,
Z
x dA =
0
R
=
=
x dx dy =
y
1/2
Z
1/2
0
0
1/2
1
(y − y 2 )
2
=
0
Z
1
2
(1 − y)2 − y 2 dy =
1−y
x2
2
dy
y
1/2
1 − 2y dy
0
1
.
8
Alternatively,
Z
x dA =
1/2
Z
0
R
=
Z
Z
x
x dy dx +
0
1/2
=
Z
x
dx +
x2 dx +
0
Z
Z
+
0
1−x
x dy dx
0
xy
dx
0
1
1/2
1/2
Z
1−x
1
1/2
0
1/2
x3
=
3
1
1/2
xy
0
Z
x(1 − x) dx
x2
x3
−
2
3
1
1/2
1
1
1
1
1
1
=
+ − − +
= .
24
2
3
8
24
8
16.2 SOLUTIONS
1481
42. (a) The line x = y/2 is the line y = 2x, and y = x and y = 2x intersect at x = 0. Thus, R is the shaded region in
Figure 16.23. One expression for the integral is
Z
f dA =
3
Z
0
R
2x
Z
2
x2 ex dy dx.
x
Another expression is obtained by reversing the order of integration. When we do this, it is necessary to split R into
two regions on a line parallel to the x-axis along the point of intersection of y = x and x = 3; this line is y = 3.
Then we obtain
Z Z
Z
Z Z
3
y
6
2
3
x2 ex dx dy +
f dA =
0
R
2
x2 ex dx dy.
y/2
3
y/2
(b) We evaluate the first integral. Integrating with respect to y first:
Z
3
0
Z
2x
2 x2
x e
Z
dy dx =
x
2x
3
2 x2
dx =
(x e y)
0
Z
3
2
x3 ex dx.
0
x
2
2
We use integration by parts with u = x2 , v ′ = xex . Then u′ = 2x and v = 21 ex , so
Z
3
2
x3 ex dx =
0
=
3
1 2 x2
x e
2
0
−
Z
3
2
xex dx =
0
1
(9)e9 − 0 −
2
1 9 1
e −
2
2
1 2 x2
x e
2
3
0
−
1
= + 4e9 .
2
y
7
(3, 6)
6
5
4
3
y = 2x
(3, 3)
2
y=x
1
x
1
2
3
Figure 16.23
43. To find the average value, we evaluate the integral
Z
0
3
Z
6
(x2 + 4y) dydx,
0
and then divide by the area of the base region.
To evaluate this integral, we evaluate the inside integral first:
Z
6
(x2 + 4y) dy = (x2 y + 2y 2 )
0
y=6
y=0
1 x2
e
2
= 6x2 + 72.
3
0
1482
Chapter Sixteen /SOLUTIONS
Therefore, we have
Z
0
3
Z
6
3
Z
(x2 + 4y)dydx =
0
(6x2 + 72)dx = (2x3 + 72x)
3
= 270.
0
0
The value of the integral is 270. The area of the base region is 3 · 6 = 18. To find the average value of the function, we
divide the value of the integral by the area of the base region:
1
Area
Average value =
3
Z
6
Z
0
1
· 270 = 15.
18
(x2 + 4y) dydx =
0
The average value is 15. This is reasonable, since the smallest value of f (x, y) on this region is 0, and the largest value is
32 + 4 · 6 = 33.
44. To find the average value, we first find the value of the integral
Z
4
0
3
Z
(xy 2 ) dydx.
0
We evaluate the inside integral first:
Z
3
2
(xy ) dy =
0
Therefore, we have
Z
0
4
3
Z
(xy 2 ) dydx =
Z
xy 3
3
y=3
= 9x.
y=0
4
(9x) dx =
0
0
9x2
2
4
= 72.
0
The value of the integral is 72. To find the average value, we divide the value of the integral by the area of the region:
1
Area
Average value =
Z
4
0
Z
3
(xy 2 ) dydx =
0
72
= 6.
3·4
The average value of f (x, y) on this rectangle is 6. This is reasonable since the smallest value of xy 2 on this region is 0
and the largest value is 4 · 32 = 36.
45. The intersection of the graph of f (x, y) = 25 − x2 − y 2 and xy-plane is a circle x2 + y 2 = 25. The given solid is shown
in Figure 16.24.
z
f (x, y) = 25 − x2 − y 2
y
x
Figure 16.24
Thus the volume of the solid is
V =
Z
f (x, y) dA
R
=
Z
5
−5
Z √25−y2
−
√
25−y 2
(25 − x2 − y 2 ) dx dy.
16.2 SOLUTIONS
1483
46. The intersection of the graph of f (x, y) = 25 − x2 − y 2 and the plane z = 16 is a circle, x2 + y 2 = 32 . The given solid
is shown in Figure 16.25.
f (x, y) = 25 − x2 − y 2
z
✠
z = 16
y
x
Figure 16.25
Thus, the volume of the solid is
Z
V =
R
Z √9−y2
3
Z
=
(f (x, y) − 16) dA
−3
−
√
9−y 2
(9 − x2 − y 2 ) dx dy.
47. The solid is shown in Figure 16.26, and the base of the integral is the triangle as shown in Figure 16.27.
z
y
✛
y=0
4
y−x =4
backside
✲
2x + y + z = 4
✠
y−x=4
−4
2x + y = 4
y
4
2
x
x
−4
Figure 16.26
2
Figure 16.27
Thus, the volume of the solid is
V =
=
Z
ZR
z dA
(4 − 2x − y) dA
R
4
=
Z
0
Z
(4−y)/2
y−4
(4 − 2x − y) dx dy.
1484
Chapter Sixteen /SOLUTIONS
48.
Volume =
Z
2
0
2
Z
Z
2
y=x
xy dy dx =
0
0
=
1 2
xy
2
2
dx
0
2
Z
2x dx
0
2
= x2
0
=4
49. The region of integration is shown in Figure 16.28. Thus
Volume =
Z
0
1
Z
x
(x2 + y 2 ) dy dx =
Z
1
x2 y +
0
0
y3
3
dx =
Z
1
0
y=0
x4
4 3
x dx =
3
3
y
1
x=y
x=1
x
1
Figure 16.28
50. The region of integration is shown in Figure 16.29. Thus,
Volume =
9
Z
=
x
3/2
(x + y) dy dx =
0
0
Z
√
Z
9
0
x
Z
0
x
+
2
dx =
9
y2
xy +
2
2 5/2 x2
x
+
5
4
√
y= x
dx
y=0
9
=
0
y
3
x = y2
x=9
x
9
Figure 16.29
2349
= 117.45.
20
1
=
0
1
.
3
16.2 SOLUTIONS
1485
51. The plane 2x + y + z = 4 cuts the xy-plane in the line 2x + y = 4, so the region of integration is the triangle shown in
Figure 16.30. We want to find the volume under the graph of z = 4 − 2x − y. Thus,
Volume =
Z
2
0
=
Z
=
0
−2x+4
(4 − 2x − y) dy dx =
0
2
0
Z
Z
2
Z
0
2
4y − 2xy −
(−2x + 4)2
4(−2x + 4) − 2x(−2x + 4) −
2
(2x2 − 8x + 8) dx =
2 3
x − 4x2 + 8x
3
2
=
0
y2
2
−2x+4
dx
0
dx
16
.
3
y
4
x = −(y − 4)/2 or y = −2x + 4
x
2
Figure 16.30
52. Let R be the triangle with vertices (1, 0), (2, 2) and (0, 1). Note that (3x + 2y + 1) − (x + y) = 2x + y + 1 > 0 for
x, y > 0, so z = 3x + 2y + 1 is above z = x + y on R. We want to find
Z
Volume =
R
((3x + 2y + 1) − (x + y)) dA =
Z
(2x + y + 1) dA.
R
We need to express this in terms of double integrals.
y
(2, 2)
2
y = 1 + 0.5x
R2
1
R1
y = 2x − 2
y =1−x
x
O
1
Figure 16.31
2
1486
Chapter Sixteen /SOLUTIONS
To do this, divide R into two regions with the line x = 1 to make regions R1 for x ≤ 1 and R2 for x ≥ 1. See
Figure 16.31. We want to find
Z
(2x + y + 1) dA =
Z
(2x + y + 1) dA +
R1
R
Z
(2x + y + 1) dA.
R2
Note that the line connecting (0, 1) and (1, 0) is y = 1 − x, and the line connecting (0, 1) and (2, 2) is y = 1 + 0.5x. So
Z
(2x + y + 1) dA =
R1
Z
1
Z
2
0
The line between (1, 0) and (2, 2) is y = 2x − 2, so
Z
(2x + y + 1) dA =
1
R2
Z
1+0.5x
Z
1+0.5x
(2x + y + 1) dy dx.
1−x
(2x + y + 1) dy dx.
2x−2
We can now compute the double integral for R1 :
Z
1
0
1+0.5x
Z
(2x + y + 1) dy dx =
1
Z
2xy +
0
1−x
=
1
Z
0
=
=
y2
+y
2
21 2
x + 3x
8
7 3 3 2
x + x
8
2
19
,
8
1+0.5x
dx
1−x
dx
1
dx
0
and the double integral for R2 :
Z
1
2
Z
1+0.5x
(2x + y + 1) dy dx =
1+0.5x
2
Z
(2xy + y 2 /2 + y)
1
2x−2
=
Z
dx
2x−2
1
0
= −
39
−
8
x2 + 9x +
3
2
13 3 9 2 3
x + x + x
8
2
2
29
=
.
8
dx
2
1
19
29
48
+
=
= 6.
8
8
8
53. We want to calculate the volume of the tetrahedron shown in Figure 16.32.
So, Volume =
z
1/c
1/b
1/a
x
y
=1
ax + by
Figure 16.32
We first find the region in the xy-plane where the graph of ax + by + cz = 1 is above the xy-plane. When z = 0
we have ax + by = 1. So the region over which we want to integrate is bounded by x = 0, y = 0 and ax + by = 1.
Integrating with respect to y first, we have
Volume =
Z
1/a
0
Z
0
(1−ax)/b
z dy dx =
Z
0
1/a
Z
0
(1−ax)/b
1 − by − ax
dy dx
c
16.2 SOLUTIONS
Z
=
0
Z
=
1/a
1/a
y
by 2
axy
−
−
c
2c
c
1487
y=(1−ax)/b
dx
y=0
1
(1 − 2ax + a2 x2 ) dx
2bc
0
1
.
6abc
=
54. The region R is shaded in Figure 16.33. The integral is
Z Z √
Z
a2 −y 2
a
xy dA =
xy dx dy
R
a−y
0
=
=
Z
a
a2 −y 2
1 2
x y
2
0
1
2
√
1
2
dy =
a−y
Z
a
0
(2ay 2 − 2y 3 ) dy =
y
a
Z
a
((a2 − y 2 )y − (a − y)2 y)y dy
0
ay 3
y4
−
3
4
a
=
a4
.
12
0
x2 + y 2 = a2
✠
✛
x+y = a
x
a
Figure 16.33
55. (a) The contour f (x, y) = 1 lies in the xy-plane and has equation
2e−(x−1)
2
−y 2
= 1,
so
−(x − 1)2 − y 2 = ln(1/2)
(x − 1)2 + y 2 = ln 2 = 0.69.
This is the equation of a circle centered at (1, 0) in the xy-plane.
Other contours are of the form
2e−(x−1)
2
−y 2
2
=c
2
−(x − 1) − y = ln(c/2).
Thus, all the contours are circles centered at the point (1, 0).
2
(b) The cross-section
has equation z = f (1, y) = e−y . If x = 1, the base region in the xy-plane extends from
√
√
y = − 3 to y = 3. See Figure 16.34, which shows the circular region below W in the xy-plane. So
√
Area =
Z
−
3
√
2
e−y dy.
3
1488
Chapter Sixteen /SOLUTIONS
(c) Slicing parallel to the y-axis, we get
Z
Volume =
2
−2
y=
√
Z √4−x2
−
√
e−(x−1)
y
4 − x2
(1,
2
−2
√
y = − 4 − x2
2
−y 2
dy dx.
4−x2
√
3)
x
2
−2
√
(1, − 3)
Figure 16.34: Region beneath W in the
xy-plane
56. The region bounded by the x-axis and the graph of y = x − x2 is shown in Figure 16.35. The area of this region is
A=
1
Z
0
(x − x2 )dx = (
x2
x3
−
)
2
3
1
0
1
1
1
= − = .
2
3
6
y
1
4
x
0
1
1
2
Figure 16.35
So the average distance to the x-axis for points in the region is
R
R
Average distance =
Z
y dA =
R
=
Therefore the average distance is
1/60
1/6
Z
Z
1
Z
0
0
1
0
x−x2
y dy
!
x4
x2
− x3 +
2
2
y dA
area(R)
dx
dx =
1
1
1
1
− +
=
.
6
4
10
60
= 1/10.
57. (a) One solution would be to arrange that the minimum values of f on the square occur at the corners, so that the corner
values give an underestimate of the average. See Figure 16.36.
1489
16.2 SOLUTIONS
y
y
94
96
1
0.3
0.2
1
98
0.1
100
0
x
x
1
1
Figure 16.36
Figure 16.37
(b) One solution would be to arrange that the maximum values of f on the square occur at the corners, so that the corner
values give an overestimate of the average. See Figure 16.37.
58. (a) We have
Z
1
f dA
Average value of f =
Area of Rectangle Rectangle
=
=
1
6
Z
1
6
Z
2
x=0
0
2
Z
3
(ax + by) dydx =
y=0
3ax +
1
= (6a + 9b)
6
3
= a + b.
2
9
b
2
dx =
1
6
Z
1
6
2
axy + b
0
3 2 9
ax + bx
2
2
2
y2
2
y=3
dx
y=0
0
The average value will be 20 if and only if a + (3/2)b = 20.
This equation can also be expressed as 2a + 3b = 40, which shows that f (x, y) = ax + by has average value
of 20 on the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3 if and only if f (2, 3) = 40.
(b) Since 2a + 3b = 40, we must have b = (40/3) − (2/3)a. Any function f (x, y) = ax + ((40/3) − (2/3)a)y where
a is any real number is a correct solution. For example, a = 1 leads to the function f (x, y) = x + (38/3)y, and
a = −3 leads to the function f (x, y) = −3x + (46/3)y, both of which have average value 20 on the given rectangle.
See Figure 16.38 and 16.39.
y
y
x
Figure 16.38: f (x, y) = x +
x
38
y
3
Figure 16.39: f (x, y) = −3x +
46
y
3
59. (a) We have
Average value of f =
1
Area of Square
1
=
4
Z
0
2
Z
0
2
Z
f dA
Square
1
(ax + bxy + cy ) dydx =
4
2
2
Z
0
2
y3
y2
ax y + bx + c
2
3
2
y=2
dx
y=0
1490
Chapter Sixteen /SOLUTIONS
=
1
4
Z
2
2ax2 + 2bx +
0
8
c
3
1 16
16
=
a + 4b +
c
4 3
3
4
4
= a+b+ c
3
3
dx =
1
4
8
2 3
ax + bx2 + cx
3
3
2
0
The average value will be 20 if and only if (4/3)a + b + (4/3)c = 20.
(b) Since (4/3)a + b + (4/3)c = 20, we must have b = 20 − (4/3)a − (4/3)c. Any function f (x, y) = ax2 + (20 −
(4/3)a − (4/3)c)xy + cy 2 where a and c are any real numbers is a correct solution. For example, a = 1, c = 3 leads
to the function f (x, y) = x2 + (44/3)xy + 3y 2 , and a = −3, c = 0 leads to the function f (x, y) = −3x2 + 24xy,
both of which have average value 20 on the given square. See Figures 16.40 and 16.41.
y
y
x
Figure 16.40: f (x, y) = x
2
x
+ 44
xy+3y 2
3
2
Figure 16.41: f (x, y) = −3x + 24xy
60. Assume the length of the two legs of the right triangle are a and b, respectively. See Figure 16.42. The line through (a, 0)
and (0, b) is given by yb + xa = 1. So the area of this triangle is
A=
1
ab.
2
y
b
a
x
Figure 16.42
Thus the average distance from the points in the triangle to the y-axis (one of the legs) is
Average distance =
=
1
A
2
ab
a
Z
0
Z
Z
b x+b
−a
x dy dx
0
a
0
b
− x2 + bx dx
a
b
b
2
− x3 + x2
=
ab
3a
2
=
2
ab
2
a b
6
=
a
.
3
a
0
16.2 SOLUTIONS
1491
Similarly, the average distance from the points in the triangle to the x-axis (the other leg) is
Z bZ
1
Average distance =
A
=
=
0
0
2
ab
y dx dy
0
Z b
2
ab
−a
y+a
b
ab
6
a
− y 2 + ay
b
2
=
b
.
3
dy
61. The force, ∆F , acting on ∆A, a small piece of area, is given by
∆F ≈ p∆A,
where p is the pressure at that point. Thus, if R is the rectangle, the total force is given by
F =
Z
p dA.
R
We choose coordinates with the origin at one corner of the plate. See Figure 16.43.
y
(a, b)
b
a
x
Figure 16.43
Suppose p is proportional to the square of the distance from the corner represented by the origin. Then we have
p = k(x2 + y 2 ),
Thus, we want to compute
Z
for some positive constant k.
k(x2 + y 2 )dA. Rewriting as an iterated integral, we have
R
F =
Z
2
2
k(x + y ) dA =
Z bZ
0
R
=k
2
2
k(x + y ) dxdy = k
0
Z b
0
=
a
Z b
0
a3
+ ay 2
3
dy = k
x3
+ xy 2
3
y3
a3 y
+a
3
3
b
0
!
a
0
dy
k 3
(a b + ab3 ).
3
62. The outer circle is a semicircle of radius 4. This is shown in Figure 16.44, with center at D. Thus, CE = 2 and DC = 2,
while AD = 4. Notice that angle ADO is a right angle.
1492
Chapter Sixteen /SOLUTIONS
A
r
4
r−2
O
2
D
2
C
E
B
Figure 16.44
Suppose the large circle has center O and radius r. Then OA = r and OD = OC − DC = r − 2. Applying
Pythagoras’ Theorem to triangle OAD gives
r 2 = 42 + (r − 2)2
r 2 = 16 + r 2 − 4r + 4
r = 5.
If we put the origin at O, the equation of the large circle is x2 + y 2 = 25. In the same coordinates, the equation of the
small circle, which has center at D = (3, 0), is (x − 3)2 + y 2 = 16. The right hand side of the two circles are given by
x=
p
25 − y 2
and
x=3+
p
16 − y 2 .
Since the y-coordinate of A is 4 and the y-coordinate of B is −4, we have
Z Z √
4
3+
Area =
√
−4
=
Z
16−y 2
1 dx dy
25−y 2
4
(3 +
−4
= 13.95.
p
16 − y 2 −
p
25 − y 2 ) dy
Strengthen Your Understanding
63. The region of integration for the first integral is the triangle bounded by y = 0, y = x and x = 1. The region for the
second integral is the triangle bounded by x = 0, x = y and y = 1. These are not the same triangle.
64. The integral on the right does not make sense. The outside limits of integration contain a variable, but the outside limits
should always be constants.
65. Suppose the cylinder has height 2 and the base has boundary x2 + y 2 = 1. Then, integrating with respect to y first gives
Z Z √
Z
1−x2
1
Volume =
2 dA =
Base
−1
−
√
2 dx dy.
1−x2
66. The region of integration is bounded by the x-axis and the lines y = x and x = 1. Integrating with respect to y first, we
have
Z Z
1
x
f (x, y) dy dx = 4.
0
0
We start by trying f (x, y) = x. Since
Z
0
1
Z
0
x
x dy dx =
Z
0
y=x
1
xy
dx =
y=0
we take f (x, y) = 12x. Many other answers are possible.
Z
0
1
x2 dx =
x3
3
1
=
0
1
,
3
16.3 SOLUTIONS
1493
67. Since the height of the prism is not specified, we can choose any height we wish, for example 1. If the base triangle has
area 6 and the height of the prism is 1 then the volume of the prism is 6. Any right triangle with legs of length 6 and 2 will
work. For example, the triangle with vertices (0, 0), (0, 2) and (6, 0) has area 6. Therefore,
2
Z
Z
0
6−3y
1 dx dy
0
represents the volume of a triangular prism whose base has area 6.
68. False. Since the inside integral is performed with respect to x and the outside integral with respect to y, the region of
integration is the rectangle 5 ≤ x ≤ 12, 0 ≤ y ≤ 1.
69. False. The iterated integral
R2R1
0
f dxdy is over a rectangle. The correct limits are
0
R 1 R 2−x
0
f dydx.
x
70. True. For any point in the region of integration we have 1 ≤ x ≤ 2, and so y is between the positive numbers 1 and 8.
71. False. The sign of
R2R2
1
RbRd
a
c
f dydx depends on the behavior of the function f on the region of integration. For example,
(−x)dydx = − 23 .
1
72. True. Since f does not depend on x, the inside integral (which is with respect to x) evaluates to
(f − 0) = f . Thus
Z bZ
a
1
f dx dy =
0
b
Z
R1
0
x=1
f dx = xf
=
x=0
f dy.
a
73. False. The given limits describe only the upper half disk where y ≥ 0. The correct limits are
R a R √a2 −x2
−a
−
√
a2 −x2
f dydx.
74. True. In the inner integral with respect to y, the function g(x) can be treated as a constant, so
Z bZ
a
The result of the integral
Rd
b
g(x)
Z
d
Z
h(y) dy
c
75. False. As a counterexample, consider
0
b
g(x)
a
c
a
2
g(x)h(y) dxdy =
Z
d
Z
h(y) dy
c
dx.
h(y) dy is a constant, so may be factored out of the integral with respect to x. Thus we have
c
Z
Z
d
2
R2R2
0
0
0
Z
2
0
Z
2
x dx +
h(y) dy
c
Z
x2
+ yx
2
2
x2
2
y dy =
0
0
d
Z
·
b
g(x) dx .
a
x=2
+
x=0
x=2
dy =
x=0
y2
2
Z
2
(2 + 2y) dy = 8
0
y=2
= 2 + 2 = 4.
y=0
Solutions for Section 16.3
Exercises
1.
Z
(x + y) dxdy. We have
(x + y) dxdy =
and
dx =
Z
f dV =
Z
=
Z
2
0
W
0
Z
1
−1
2
Z
1
−1
Z
2
3
(x2 + 5y 2 − z) dz dy dx
(x2 z + 5y 2 z −
1 2
z )
2
3
dy dx
2
1494
Chapter Sixteen /SOLUTIONS
=
2
Z
0
=
−1
2
Z
1
Z
(x2 + 5y 2 −
5 3 5
y − y)
3
2
(x2 y +
0
=
2
Z
5
) dy dx
2
1
dx
−1
10
− 5) dx
3
(2x2 +
0
2
5
2
= ( x3 − x)
3
3
0
16
10
=
−
=2
3
3
2.
Z
1
Z
f dV =
0
W
0
(2ax + 2by + 2c) dy dx
0
1
Z
=
(ax + by + cz) dz dy dx
0
1
Z
2
Z
0
1
Z
=
1
Z
(2ax + b + 2c) dx
0
= a + b + 2c
3.
Z
f dV =
Z
π
0
W
=
Z
0
π
=
Z
Z
π
sin x cos(y + z) dz dy dx
0
π
Z
0
=
π
Z
π
sin x sin(y + z)
0
π
π
Z
Z0 π Z0 π
0
= −2
= −2
Z
sin x[sin(y + π) − sin y] dy dx
sin x(−2 sin y) dy dx
0
π
π
sin x(− cos y)
0
Z
dy dx
0
dx
0
π
2 sin x dx
0
π
= −4(− cos x)
0
= (−4)(2) = −8
4.
Z
f dV =
Z
a
0
W
=
Z
0
a
0
=
Z
0
Z bZ
Z
e−x−y−z dz dy dx
0
Z bZ
0
a
c
c
e−x e−y e−z dz dy dx
0
c
b
e
0
−x −y
e
(−e−z )
dy dx
0
16.3 SOLUTIONS
=
Z
0
aZ
b
e−x e−y (−e−c + 1) dy dx
0
= (1 − e
−c
)
Z
b
a
e−x (−e−y )
0
dx
0
= (1 − e−b )(1 − e−c )
Z
a
e−x dx
0
= (1 − e−a )(1 − e−b )(1 − e−c )
5. The region is the half cylinder in Figure 16.45.
z
1
x
1
y
1
Figure 16.45
6. The region is the half cylinder in Figure 16.46.
z
1
1
1
x
y
Figure 16.46
7. The region is the cylinder in Figure 16.47.
z
1
1
1
x
Figure 16.47
y
1495
1496
Chapter Sixteen /SOLUTIONS
8. The region is the half cylinder in Figure 16.48.
z
1
x 1
y
1
Figure 16.48
9. The region is the hemisphere in Figure 16.49.
z
1
x
1
1
y
Figure 16.49
10. The region is the quarter sphere in Figure 16.50.
z
1
y
1
x
1
Figure 16.50
11. The region is the quarter sphere in Figure 16.51.
z
1
x
1
1
Figure 16.51
y
16.3 SOLUTIONS
1497
12. The region is the hemisphere in Figure 16.52.
z
1
y
1
x 1
Figure 16.52
13. The region is the quarter sphere in Figure 16.53.
z
1
x
1
1
y
Figure 16.53
Problems
14. Positive. Since ez is positive on T , its integral is positive.
15. Positive. Since ez is positive on B, its integral is positive.
16. Zero. Since sin z is positive on T , and negative (and of equal absolute value) on B the integral is zero because the integrals
over the two halves of the sphere cancel.
17. Positive. Since sin z is positive on T , its integral is positive.
18. Zero. Since sin z is positive on the upper half of R and negative ( and of equal absolute value) on the lower half of R, the
integral of sin z is zero because the integrals over the two halves cancel.
19. Zero. The value of y is positive on the half of the cone above the second and third quadrants and negative (of equal
absolute value) on the half of the cone above the third and fourth quadrants. The integral of y over the entire solid cone is
zero because the integrals over the four quadrants cancel.
20. Zero. The value of x is positive above the first and fourth quadrants in the xy-plane, and negative (and of equal absolute
value) above the second and third quadrants. The integral of x over the entire solid cone is zero because the integrals over
the two halves of the cone cancel.
21. Positive. Since z is positive on W , its integral is positive.
22. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first and
third quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above
the second and fourth quadrants (where x and y have opposite signs). These add up to zero in the integral of xy over all
of W .
Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first
with respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such an
interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.
23. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integral
is over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in an
iterated integral is zero, then the triple integral is zero.
1498
Chapter Sixteen /SOLUTIONS
24. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is its
integral.
25. Positive. The function
p
x2 + y 2 is positive, so its integral over the solid W is positive.
26. Positive. The function e
−xyz
is a positive function everywhere so its integral over W is positive.
27. Positive. If (x, y, z) is any point inside the solid W then
its integral over the solid W is positive.
p
x2 + y 2 < z. Thus the integrand z −
28. Figure 16.54 shows a slice through the region for a fixed x. The required volume, V , is given by
V =
2
Z
1
=
Z
1
Z
0
2
Z
3y
Z
dz dy dx =
1
y
1
y2
Z
dx =
1
2
2
Z
3y
1
z
dy dx =
0
Z
1
y
2
Z
p
x2 + y 2 > 0, and so
1
2y dy dx
0
dx = 1.
1
0
z
3
2
z = 3y
1
z=y
y
0
1
Figure 16.54
29. The required volume, V , is given by
V =
5
Z
0
=
0
=
Z
=
x2
dz dy dx
0
3
x2 dy dx
y=3
x2 y
0
Z
Z
0
5
Z
3
0
5
Z
Z
dx
y=0
5
3x2 dx
0
= 125.
30. We have z = 2 − x − y, and the region under the surface in the xy-plane is bounded by the axes and the line x + y = 2
or y = 2 − x. Thus the volume is given by
V =
Z
0
2
Z
0
2−x
Z
0
2−x−y
dz dy dx =
Z
0
2
Z
0
2−x
(2 − x − y) dy dx
16.3 SOLUTIONS
=
Z
2
=
Z
=
dx
0
2
2(2 − x) − x(2 − x) −
0
Z
y
2
2y − xy −
0
2
0
1499
2 2−x
x2
− 2x + 2
2
(2 − x)2
dx
2
x3
− x2 + 2x
6
dx =
2
=
0
4
3
31. (a) Since the trough is symmetric about the xz-plane, we find the mass of sludge in one half and double it. Thus the total
mass is given by
Z Z Z
10
1
1
e−3x dz dy dx.
Mass of sludge = 2
0
0
y
Other orders of integration are possible.
(b) Evaluating gives
Mass = 2
Z
10
0
=2
Z
=2
1
0
10
0
Z
Z
Z
0
1
e−3x dz dy dx = 2
1
Z
10
0
y
e−3x
e−3x
dx = 2 −
2
6
Z
1
1
ze−3x
0
e−3x − ye−3x dy dx = 2
0
10
Z
10
=
0
Z
10
0
dy dx
y
ye−3x −
y 2 −3x
e
2
1 − e−30
.
3
32. The required volume, V , is given by
V =
10
Z
0
=
0
10
Z
0
=
=
Z
10
dzdydx
x+y
(10 − (x + y)) dydx
h
10y − xy −
0
Z
10−x
Z
0
10
Z
10−x
Z
10
1 2
y
2
1
(10 − x)2 dx
2
0
iy=10−x
dx
y=0
500
=
3
33. Since z = x + y is below z = 1 + 2x + 2y for x, y ≥ 0, we have
V =
Z
1
Z
0
0
2
Z
1+2x+2y
1 dz dy dx.
x+y
The order of integration of x and y can be reversed.
34. For x2 + y 2 ≤ 1, the paraboloid z = x2 + y 2 is below the sphere x2 + y 2 + z 2 = 4, so
Z √
Z Z √
1−x2
1
V =
−1
−
√
4−x2 −y 2
1 dz dy dx.
1−x2
x2 +y 2
The order of integration of x and y can be reversed.
35. The two surfaces are planes given by
z = 6 − 2x − 2y
z = 6 − 3x − 4y.
1
dx
0
1500
Chapter Sixteen /SOLUTIONS
For x, y ≥ 0, the plane z = 6 − 2x − 2y is above z = 6 − 3x − 4y. The region in the xy-plane is shown in Figure 16.55.
Thus
Z Z
Z
1
1−x
6−2x−2y
V =
1 dz dy dx.
0
0
6−3x−4y
The order of integration of x and y can be reversed.
y
1
y =1−x
x
1
Figure 16.55
36. The top half of the sphere is given by
z=
p
9 − x2 − y 2 .
The region in the xy-plane is shown in Figure 16.56. If we integrate with respect to y first, we have to break the region in
two pieces. Thus it is easier to integrate with respect to x first, giving
Z Z
Z √
2
9−x2 −y 2
(y+2)/2
1 dz dx dy.
V =
0
y
0
y
2
y=x
y = 2x − 2
x
1
2
Figure 16.56
37. The sphere x2 + y 2 + z 2 = 9 intersects the plane z = 2 in the circle
x2 + y 2 + 22 = 9
x2 + y 2 = 5.
The upper half of the sphere is given by z =
p
√
V =
Z
5
√
− 5
9 − x2 − y 2 . Thus, using the limits from Figure 16.57 gives
Z √
Z √
5−x2
−
√
The order of integration of x and y can be reversed.
9−x2 −y 2
1 dz dy dx.
5−x2
2
16.3 SOLUTIONS
√ y
5
y=
√
1501
5 − x2
√x
5
√
− 5
√
y = − 5 − x2
√
− 5
Figure 16.57
38. The top half of the sphere is given by
z=
p
4 − x2 − y 2 .
The region in the xy-plane is shown in Figure 16.58. If we integrate with respect to x first, we have to break the region
into two pieces. Thus, it is easier to integrate with respect to y first, giving
Z √
Z Z √
4−x2
1
4−x2 −y 2
1 dz dy dx.
V =
0
0
0
y
2
x2 + y 2 = 4
x
1
2
Figure 16.58
39. A slice through W for a fixed value of y is a semi-circle the boundary of which is z 2 = r 2 − x2 , for z ≥ 0, so the inner
integral is
√
r 2 −x2
Z
f (x, y, z) dz.
0
Lining up these stacks parallel to x-axis gives a slice from x = −r to x = r giving
Z Z √
r 2 −x2
r
f (x, y, z) dz dx.
−r
0
Finally, there is a slice for each y between 0 and 1, so the integral we want is
Z Z Z √
1
r 2 −x2
r
f (x, y, z) dz dx dy.
0
−r
0
40. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = 4 − z 2 , for y ≥ 0, so the inner
integral is
√
Z
0
4−z 2
f (x, y, z) dy.
1502
Chapter Sixteen /SOLUTIONS
Lining up these stacks parallel to z-axis gives a slice from z = −2 to z = 2 giving
Z Z √
4−z 2
2
f (x, y, z) dy dz.
−2
0
Finally, there is a slice for each x between 0 and 1, so the integral we want is
Z Z Z √
1
4−z 2
2
f (x, y, z) dy dz dx.
0
−2
0
41. A slice through W for a fixed value of x is a semi-circle the boundary of which is y 2 = r 2 − x2 − z 2 , for y ≥ 0, so the
inner integral is
√
r 2 −x2 −z 2
Z
f (x, y, z) dy.
0
√
√
Lining up these stacks parallel to z-axis gives a slice from z = − r 2 − x2 to z = r 2 − x2 giving
Z √
Z √
r 2 −x2
−
√
r 2 −x2 −z 2
f (x, y, z) dy dz.
r 2 −x2
0
Finally, there is a slice for each x between −r and r, so the integral we want is
Z √
Z Z √
r 2 −x2
r
−r
−
√
r 2 −x2 −z 2
f (x, y, z) dy dz dx.
r 2 −x2
0
42. A slice through W for a fixed value of x is a semi-circle the boundary of which is z 2 = r 2 − x2 − y 2 , for z ≥ 0, so the
inner integral is
√
r 2 −x2 −y 2
Z
f (x, y, z) dz.
0
√
√
Lining up these stacks parallel to y-axis gives a slice from y = − r 2 − x2 to y = r 2 − x2 giving
Z √
Z √
r 2 −x2
−
√
r 2 −x2 −y 2
f (x, y, z) dz dy.
r 2 −x2
0
Finally, there is a slice for each x between 0 and r, so the integral we want is
Z Z √
Z √
0
r 2 −x2 −y 2
r 2 −x2
r
−
√
f (x, y, z) dz dy dx.
r 2 −x2
0
43. Figure 16.59 shows a slice through the region for a fixed value of y.
z
1
z=x
z = x2
x
0
1
Figure 16.59
16.3 SOLUTIONS
1503
We break the region into small cubes of volume ∆V = ∆x∆y∆z. A stack of cubes vertically above the point (x, z)
in the xz-plane gives the strip shown in Figure 16.59 and so the inner integral is
Z
x
Z
x
dz
x2
The plane and the surface meet when x = x2 , giving x(1 − x) = 0, so x = 0 or x = 1. Lining up the stacks parallel to
the z-axis gives a slice from x = 0 to x = 1. Thus, the limits on the middle integral are
1
Z
0
dz dx.
x2
Finally, there is a slice for each y between 0 and 3, so the integral we want is
3
Z
0
The required volume, V , is given by
V =
Z
1
3
Z
0
=
Z
0
=
Z
Z
1
dz dx dy
x − x2 dx dy
0
1
x3
x2
−
2
3
3
1
dy
6
Z
3
0
x
x2
0
0
=
1
3Z
3
Z
dz dx dy.
x2
0
Z
x
Z
dy
0
1
dy
6 0
1
= ·3
6
1
= .
2
=
44. The required volume, V , is given by
V =
Z
5
0
=
Z
0
=
Z
=
0
=
Z
0
5Z
5−x
x+y
dz dy dx
0
(x + y) dy dx
0
5
xy +
0
Z
5−x
Z
5
1 2
y
2
y=5−x
dx
y=0
x(5 − x) +
1
(5 − x)2
2
125
.
3
dx
45. The pyramid is shown in Figure 16.60. The planes y = 0, and y − x = 4, and 2x + y + z = 4 intersect the plane z = −6
in the lines y = 0, y − x = 4, 2x + y = 10 on the z = −6 plane as shown in Figure 16.61.
These three lines intersect at the points (−4, 0, −6), (5, 0, −6), and (2, 6, −6). Let R be the triangle in the planes
z = −6 with the above three points as vertices. Then, the volume of the solid is
V =
Z
6
0
Z
(10−y)/2
y−4
Z
4−2x−y
−6
dz dx dy
1504
Chapter Sixteen /SOLUTIONS
=
=
=
Z
Z
Z
6
Z
0
(10−y)/2
(10 − 2x − y) dx dy = 162
y−4
6
(10−y)/2
(10x − x2 − xy)
0
6
(
0
dy
y−4
9y 2
− 27y + 81) dy
4
= 162
z
y
✛
✲
y=0
(2, 6, −6)
y − x = 4 backside
✛
z = −6 plane
2x + y + z = 4
−4
y
4
2
y−x=4
x
(−4, 0, −6)
✛
2x + y − 6 = 4
(2, 6, −6)
z = −6 bottom
(5, 0, −6)
x
(5, 0, −6)
(−4, 0, −6)
Figure 16.60
Figure 16.61
46. Since x + y + z = 1 can be written as
z = 1 − x − y,
the plane z = 1 + x + y is above the plane z = 1 − x − y for x ≥ 0, y ≥ 0. The region of integration in the xy-plane is
the triangle shown in Figure 16.62. Thus
Volume =
Z
1
0
=
Z
1
Z
1
0
1−x
0
0
=
Z
Z
0
1−x
Z
1+x+y
1 dz dy dx =
1
Z
0
1−x−y
Z
1+x+y
1−x
z
1
dy dx
1−x−y
((1 + x + y) − (1 − x − y)) dy dx =
2
(2x(1 − x) + (1 − x) ) dx =
Z
Z
1
0
1
2
0
(1 − x ) dx =
1−x
Z
(2x + 2y) dy dx =
0
0
x3
x−
3
y
1
y =1−x
x
1
Figure 16.62
Z
1
=
0
2
.
3
1−x
1
2xy + y 2
dx
0
1505
16.3 SOLUTIONS
47. The plane x + y + z = 1 cuts the xy-plane in the line x + y = 1. For x + y < 1, the plane x + y + z = 1 is above the
xy-plane. For 1 < x + y ≤ 2, the plane x + y + z = 1 is below the xy-plane. Therefore, z = 1 − x − y is positive for
x + y < 1 and negative for 1 < x + y < 2. Thus
Volume =
Z
x+y≤1
Z
(1 − x − y) dA −
1≤x+y≤2
(1 − x − y) dA
From Figure 16.63, we see that the region 1 ≤ x + y ≤ 2 must be split into two, for example as shown. Integrating with
respect to x first, we have
Z
1
Volume =
1
=
Z
Z
1
0
0
0
=
1−y
Z
0
(1 − x − y) dx dy −
2
x
− xy
x−
2
1−y
dy −
0
1
Z
0
1
Z
0
1−y
(2 − (1 − y) − 2y) dy −
2
2−y
Z
1−y
(1 − x − y) dx dy +
2
x
− xy
x−
2
Z
1
0
2−y
dy −
1−y
Z
2
1
Z
1
2
Z
0
1
(1 − y)2
dy +
2
(1 − y)3
=−
6
1
Z
1
0
1
dy −
2
1
y
+
2
0
1
−
2
0
Z
2
0
y3
− y2
3
x2
− xy
x−
2
2−y
dy
0
Z
2
1
2
=
1
1
1
+ + = 1.
6
2
3
1
2
x=2−y
1
x=1−y
x
1
2
Figure 16.63
48. (a) The solid is shown in Figure 16.64. The density, δ, is given by δ = 4z, so
Mass =
Z
4z dV.
W
The top of the solid has equation
z = 2 − x − y.
The base in the xy-plane is shown in Figure 16.65, so the iterated integral must be split into two pieces.
Z
W
4z dV =
Z
0
1/2
Z
0
x
Z
0
2−x−y
4z dz dy dx +
Z
1
1/2
Z
0
1−x
Z
0
dy
2−y
(2 − (2 − y) − 2y) dy
2
y
Mass =
2−y
1−y
(2 − (2 − y) − 2y) −
(2 − (1 − y) − 2y)
2
2
y(y − 2)
dy
2
1
(1 − x − y) dx dy
−
=
2−y
Z
2−x−y
4z dz dy dx.
1506
Chapter Sixteen /SOLUTIONS
(b) Evaluating gives
Mass =
1/2
Z
0
2−x−y
x
Z
2z
2
dy dx +
0
1/2
=2
Z
= 2
Z
0
x
(2 − x − y)2 dy dx + 2
0
1/2
−
0
1/2
0
Z
(2 − x − y)
3
3
x
dx + 2
Z
2
=
3
3
(2 − x) − (2 − x − x)
0
4
Z
Z
2z 2
(2 − x)
8(1 − x)
−
+
4
4
4
2 33
88
2 55
+
=
.
= ·
3 64
3 64
96
3
1/2
0
dy dx
0
0
1
1/2
Z
2−x−y
1−x
1
1/2
0
1/2
2
=
3
1
Z
1−x
Z
0
−
(2 − x − y)3
3
2
dx +
3
2
+
3
(2 − x − y)2 dy dx
Z
1−x
dx
0
1
1/2
(2 − x)3 − 13 dx
(2 − x)4
−
−x
4
1
1/2
z
x+y+z = 2
y
x+y =1
y
(1/2, 1/2)
y=x
R
x+y = 1
x
x
Figure 16.64: The solid
Figure 16.65: Base of solid
49. The region looks like an upside-down trough, with cross section in the xz-plane shown in Figure 16.66, extending a
distance of 3 cm in the y-direction.
Since the region and the density function are symmetric about the z-axis, we find the mass of the right side and
double it. In grams
Mass = 2
Z
3
Z
3
Z
3
0
=2
0
1
Z
1
Z
1
0
0
=2
Z
0
0
0
1−x
Z
(10 − z) dz dx dy
z2
10z −
2
10(1 − x) −
1−x
dx dy
0
(1 − x)2
2
dx dy
16.3 SOLUTIONS
=
3
Z
1
Z
0
(19 − 18x − x2 ) dx dy
0
= 3·
x3
19x − 9x −
3
2
1
= 3 19 − 9 −
0
1
3
= 29 gm.
z
1
z = 1+x
z = 1−x
−1
1
x
Figure 16.66
50. The region of integration is shown in Figure 16.67, and the mass of the given solid is given by
z
6
+ y2 + z6 = 1
or z = −2x − 3y + 6
x
3
2
✛
y
x
3
y
2
+
=1
x+2
or y = − 2
3
3
x
Figure 16.67
mass =
Z
δ dV
R
=
Z
3
0
=
Z
0
3
0
=
Z
Z
=
Z
3
3
0
−2x−3y+6
x+2
−2
3
Z
Z
3
dydx
0
(x + y)(−2x − 3y + 6) dydx
−2
x+2
3
0
3
0
Z
(x + y) dz dy dx
(x + y)z
0
0
=
−2x−3y+6
0
2 x+2
−3
Z
Z
0
0
=
−2
x+2
3
Z
(−2x2 − 3y 2 − 5xy + 6x + 6y) dydx
5
−2x y − y − xy 2 + 6xy + 3y 2
2
2
3
14 3 8 2
x − x + 2x + 4
27
3
dx
2 x+2
−3
dx
0
1507
1508
Chapter Sixteen /SOLUTIONS
=
7 4 8 3
x − x + x2 + 4x
54
9
3
0
7
8
21
15
=
· 34 − · 33 + 32 + 12 =
−3=
.
54
9
2
2
51. The pyramid is shown in Figure 16.68. The planes y = 0, and y − x = 4, and 2x + y + z = 4 intersect the plane z = −6
in the lines y = 0, y − x = 4, 2x + y = 10 as shown in Figure 16.69.
z
y
✛
y=0
✲
(2, 6, −6)
y − x = 4 backside
✛
z = −6 plane
2x + y + z = 4
−4
y
4
2
y−x=4
x
(−4, 0, −6)
✛
2x + y − 6 = 4
(2, 6, −6)
z = −6 bottom
(5, 0, −6)
x
(5, 0, −6)
(−4, 0, −6)
Figure 16.68
Figure 16.69
These three lines (the edges of the pyramid) intersect the plane z = −6 at the points (−4, 0, −6), (5, 0, −6), and
(2, 6, −6). Let R be the triangle in the plane z = −6 with these three points as vertices. Then, the mass of the solid is
Mass =
Z
6
0
=
Z
Z
6
=
6
Z
(10−y)/2
Z
(10−y)/2
y−4
6
Z
4−2x−y
δ(x, y, z) dz dx dy
−6
Z
4−2x−y
6
(
0
y dz dx dy
−6
y(10 − 2x − y) dx dy
y(10x − x2 − xy)
0
=
Z
y−4
0
Z
(10−y)/2
y−4
0
=
Z
x=(10−y)/2
dy
x=y−4
9y 3
− 27y 2 + 81y) dy
4
= 243.
52. (a) Since δ(x, y, z) = z is measured in grams per cm3 , and since volume is measured in cm3 , the integral represents the
mass of the pyramid, in grams.
(b) The pyramid is bounded below by the plane z = 0 and is bounded above by the four planes z = 6 − x, z = 6 − y,
z = x, and z = y. Therefore, since there are four different surfaces bounding the pyramid on the top, we would need
four separate triple integrals.
(c) We integrate in the y-direction first. The pyramid is bounded on the left and right sides by the planes y − z = 0 and
y + z = 6, so our y limits of integration are y = z and y = 6 − z. After projecting into the xz-plane, we obtain the
region bounded by the lines z = x, z = 6 − x, and z = 0, as shown in Figure 16.70:
16.3 SOLUTIONS
1509
z
3
x
3
6
Figure 16.70
Therefore, if we integrate in the x direction next, our limits are x = z and x = 6 − z, so we have
Z
zdV =
Z
3
0
E
=
Z
=
Z
6−z
z
3
0
0
Z
Z
z
3
Z
6−z
z dy dx dz =
z
6−z
3
Z
0
z(6 − 2z)dxdz =
Z
3
Z
6−z
6−z
zy
z
z(6 − 2z)x
0
dx dz
z
6−z
dz =
0
z
3
(36z − 24z 2 + 4z 3 ) dz = (18z 2 − 8z 3 + z 4 )
Z
3
z(6 − 2z)2 dz
= 27,
0
and we conclude that the mass of the pyramid is 27 grams.
53. (a) The vectors ~
u = ~i − ~j and ~v = ~i − ~k lie in the required plane so p
~ =~
u × ~v = ~i + ~j + ~k is perpendicular
to this plane. Let (x, y, z) be a point in the plane, then (x − 1)~i + y~j + z~k is perpendicular to ~
p , so ((x − 1)~i +
y~j + z~k ) · (~i + ~j + ~j ) = 0 and so
(x − 1) + y + z = 0.
Therefore, the equation of the required plane is x + y + z = 1.
(b) The required volume, V , is given by
V =
Z
1
0
=
Z
1
Z
0
1
Z
1
=
0
dz dy dx
(1 − x − y) dy dx
1 2
y
2
1−x
dx
0
1 − x − x(1 − x) −
0
Z
1−x−y
y − xy −
0
=
1−x
Z
Z
0
0
0
=
1−x
Z
1
1
(1 − x)2 dx
2
1
(1 − x)2 dx
2
1
= .
6
54. We will integrate in the z-direction first. Observe that the bottom of E is given by the plane z = 0, and the top of E is
given by the plane z = 4 − 2x − 4y. After projecting E into the xy-plane, we obtain the region bounded by the lines
3x − 2y = 0, 2x + 4y = 4, and y = 0 (see Figure 16.71).
1510
Chapter Sixteen /SOLUTIONS
y
3
4
x
2
1
2
Figure 16.71
We can see from the picture that, if we integrate in the x-direction next, our limits of integration will be x = (2/3)y
and x = 2 − 2y. Therefore, our final answer is
Z
3/4
0
Z
2−2y
2y
3
Z
4−2x−4y
f (x, y, z) dz dx dy.
0
55. We will integrate in the x-direction first. Observe that the “back” side of E is given by the plane x = 0, and the “front”
side of E is given by the plane x = 2 − 2y − (1/2)z. After projecting E into the yz-plane, we obtain the region bounded
by the lines y = 0, z = 0, z = 2, and 4y + z = 4 (see Figure 16.72).
y
2
x
1
1
2
Figure 16.72
We can see from the picture that, if we integrate in the y-direction next, our limits of integration will be y = 0 and
x = 1 − (1/4)z. Therefore, our final answer is
Z
2
0
Z
0
1− z
4
Z
2−2y− z
2
f (x, y, z) dx dy dz.
0
56. We will integrate in the y-direction first. Observe that the “left” side of E is given by the plane 3x − 2y = 0, and the
“right” side of E is given by the plane 2x+4y +z = 4. Solving these equations for y gives us a lower limit of y = (3/2)x
and an upper limit of y = 1 − (1/2)x − (1/4)z. After projecting E into the xz-plane, we obtain the region shown in
Figure 16.73.
y
4
x
1
2
Figure 16.73
16.3 SOLUTIONS
1511
We note that the x and the z intercepts in the picture can be determined by observing that the slanted boundary of
the projected region is given by the line of intersection of the planes y = (3/2)x and 2x + 4y + z = 4. Substituting
y = (3/2)x into 2x + 4y + z = 4 yields the following:
3x
+z =4
2
2x + 6x + z = 4
2x + 4
8x + z = 4
Therefore, the projected region is bounded by the lines x = 0, z = 0, and 8x + z = 4 in the xz-plane, and so our final
answer is
Z
Z 1Z
x
z
4−8x
2
1− 2 − 4
f (x, y, z) dy dz dx.
3x
2
0
0
57. Orient the region as shown in Figure 16.74 and use Cartesian coordinates with origin at the center of the sphere. The
equation of the sphere is x2 + y 2 + z 2 = 25, and we want the volume between the planes z = 3 and z = 5. The plane
z = 3 cuts the sphere in the circle x2 + y 2 + 32 = 25, or x2 + y 2 = 16.
Z Z √
Z √
16−x2
4
Volume =
−4
−
√
25−x2 −y 2
dz dy dx.
16−x2
3
z
Sphere is
x2 + y 2 + z 2 = 25
5
✠
✻
2
❄
✻
■Circle is
x2 + y 2 = 16
3
❄
y
■
x
x2 + y 2 = 16
Figure 16.74
58. (a) The equation of the surface of the whole cylinder along the y-axis is x2 + z 2 = 1. The part we want is
z=
See Figure 16.75.
p
1 − x2
0 ≤ y ≤ 10.
z
x
1
10
y
Figure 16.75
(b) The integral is
Z
D
f (x, y, z) dV =
Z
0
10
Z
1
−1
Z √1−x2
0
f (x, y, z) dz dx dy.
1512
Chapter Sixteen /SOLUTIONS
59. The intersection of two cylinders x2 + z 2 = 1 and y 2 + z 2 = 1 is shown in Figure 16.76. This region is bounded by four
surfaces:
p
p
p
p
z = − 1 − x2 , z = 1 − x2 , y = − 1 − z 2 , and y = 1 − z 2
So the volume of the given solid is
V =
Z
1
−1
Z √1−x2 Z √1−z2
−
√
1−x2
−
√
dy dz dx
1−z 2
z
y
x
Figure 16.76
60. Integrating in the z-direction first, our lower limit is the plane z = 0, and the upper limit is the surface z = 6y 2 . After
projecting E into the xy-plane, we obtain the region in the first quadrant that lies inside the ellipse x2 + 3y 2 = 12 (see
Figure 16.77).
y
2
√
2 3
x
Figure 16.77
Therefore, our x-limits of integration are the line x = 0 and the ellipse x =
Z 2
Z Z √
12−3y 2
2
6y
p
12 − 3y 2 , so our answer is
f (x, y, z) dz dx dy.
0
0
0
61. Integrating
p in the x-direction first, our lower limit of integration is the plane x = 0, and the upper limit is the cylinder
12 − 3y 2 . After projecting E into the yz-plane, we obtain the two-dimensional region bounded between the
x =
curves z = 0, y = 2, and z = 6y 2 . Substituting y = 2 into z = 6y 2 gives z = 24, which is the z coordinate of the point
of intersection of the two curves (see Figure 16.78).
16.3 SOLUTIONS
1513
y
24
x
2
Figure 16.78
Therefore, our final answer is
Z
2
0
Z
6y 2
0
Z √12−3y2
f (x, y, z) dx dz dy.
0
p
62. Integrating in the y-direction first, one bounding surface isz = 6y 2 , which, after solving for y, becomes y = z/6, our
p
lower limit of integration. Similarly, the other bounding surface isx2 + 3y 2 = 12, which becomes y = (12 − x2 )/3,
our upper limit. After projecting E into the xz-plane, we obtain a region bounded by two line segments and a curve (see
Figure 16.79).
y
24
√ x
2 3
Figure 16.79
We can find the equation of the curve by finding the intersection of the cylinders y =
projecting into the xz-plane:
p
z/6 and x2 + 3y 2 = 12 and
12 = x2 + 3y 2
12 = x2 + 3
z
6
24 = 2x2 + z
z = 24 − 2x2
Therefore, our z-limits of integration are the line z = 0 and the parabola z = 24 − 2x2 , and so our final answer is
Z
0
√
12
Z
0
24−2x2
Z
p
√z
12−x2
3
f (x, y, z) dy dz dx.
6
63. From the problem, we know that (x, y, z) is in the cube which is bounded by the three coordinate planes, x = 0, y = 0,
z = 0 and the planes x = 2, y = 2, z = 2. We can regard the value x2 + y 2 + z 2 as the density of the cube. The average
1514
Chapter Sixteen /SOLUTIONS
value of x2 + y 2 + z 2 is given by
R
average value =
V
(x2 + y 2 + z 2 ) dV
volume(V )
R2R2R2
0
=
0
0
(x2 + y 2 + z 2 ) dx dy dz
8
R 2 R 2 x3
0
=
3
0
R2R2
0
=
0
8
y
3
+ 32 y 3 + 2z 2 y
0
R2
16
3
0
=
64
3
=
16
3
8
32
z
3
=
+
+ 34 z 3
+
8
8
32
3
2
0
dydz
8
+ 2y 2 + 2z 2 dydz
8
8
3
R2
=
+ (y 2 + z 2 )x
2
0
dz
8
+ 4z 2 dz
2
0
= 4.
64. As an aid in finding limits of integration, we begin by finding the coordinates of the four points highlighted on the
diagram. Clearly, one of them is the origin (0, 0, 0). The point at the top of E is formed by the intersection of the plane
x + 2y + z = 4 with the z-axis, so we substitute x = 0 and y = 0 into the equation of the plane to obtain z = 4, yielding
a point with coordinates (0, 0, 4). Similarly, the intersection of the plane x + 2y + z = 4 with the y-axis produces a
point with coordinates (0, 2, 0). Finally, the last point highlighted on the diagram occurs at the intersection of the surfaces
x + 2y + z = 4, x = 2y 2 , and z = 0. Substituting z = 0 and x = 2y 2 into the equation x + 2y + z = 4 yields
2y 2 + 2y = 4
y2 + y − 2 = 0
(y + 2)(y − 1) = 0
Since the desired point is in the first octant, its y-coordinate must be y = 1, which yields x = 2y 2 = 2, and we see that
the coordinates of the point are given by (2, 1, 0).
(a) Integrating in the z-direction first, our region is bounded by the plane z = 0 on the bottom and the plane x+2y+z = 4
on the top. Solving for z therefore yields a lower limit of z = 0 and an upper limit of z = 4−x −2y. After projecting
E into the xy-plane, we obtain the picture in Figure 16.80.
The curve at the top of this projected region is formed by the intersection of the plane x + 2y + z = 4 and z = 0,
yielding an equation of x + 2y = 4, or y = (4 − x)/2. The curve at the bottom of the projected region is simply the
portion ofpthe cylinder x = 2y 2 that lies in the xy-plane. Therefore, our y-limits of integration are y = (4 − x)/2
and y = x/2, yielding a final answer of
Z
0
2
Z
4−x
2
√x
2
Z
4−x−2y
f (x, y, z) dz dy dx.
0
2
(b) Integrating in the y-direction first, our region is bounded by the parabolic cylinder
p x = 2y on one side and the plane
x + 2y + z = 4 on the other side. Solving these equations for y yields y = x/2 and y = (4 − x − z)/2 as our
y-limits of integration. After projecting E into the xz-plane, we obtain the picture in Figure 16.81.
We can find the equation of the curve bounding the top of the projected region by finding the intersection of the
surfaces x = 2y 2 and x + 2y + z = 4 and projecting into the xz-plane:
x + 2y + z = 4
q
x
+z = 4
x+2
2
√
x + 2x + z = 4
z = 4−x−
√
2x
1515
16.3 SOLUTIONS
Therefore, our z-limits of integration are the line z = 0 and the curve z = 4 − x −
Z
0
2Z
4−x−
√
2x, and so our final answer is
4−x−z
2
2x Z
f (x, y, z) dy dz dx.
√x
0
√
2
y
z
2
4
1
x
1
x
2
2
Figure 16.80
Figure 16.81
65. The mass m is given by
m=
=
=
=
Z
Z
Z
Z
1 dV =
1
Z
0
W
1
1
Z
0
1
Z
x+y+1
Z
0
1 dz dy dx
0
(x + y + 1) dy dx
0
1
1
xy + y 2 /2 + y)
dx
0
0
1
(x + 3/2) dx = 2 gm.
0
Then the x-coordinate of the center of mass is given by
Z
1
x̄ =
2
=
=
=
W
1
Z
1
2
0
1
2
Z
1
Z
1
0
1
Z
0
x dz dy dx
0
x(x + y + 1) dy dx
x2 y + xy 2 /2 + xy)
0
1
0
dx
1
Z
1
2
x+y+1
Z
0
1
Z
1
x dV =
2
(x2 + 3/2x) dx = 13/24 cm.
0
An essentially identical calculation (since the region is symmetric in x and y) gives ȳ = 13/24 cm.
Finally, we compute z̄:
z̄ =
=
=
=
1
2
1
2
1
2
Z
z dV =
W
1
Z
0
Z
1
12
Z
1
1
2
Z
0
1
Z
1
0
Z
x+y+1
z dz dy dx
0
(x + y + 1)2 /2 dy dx
0
1
(x + y + 1)3 /6
0
Z
0
So (x̄, ȳ, z̄) = (13/24, 13/24, 25/24).
1
0
dx
1
((x + 2)3 − (x + 1)3 ) dx = 25/24 cm.
1516
Chapter Sixteen /SOLUTIONS
66. The mass m is given by
Z
m=
1 dV =
W
1
Z
=
0
1
3
=
1
3
=
1
=
3
1
3
=
1
Z
0
(1−x)/2
Z
(1−x)/2
1
dx
0
1−x
1−x
−x
−
2
2
0
1
0
1 dz dy dx
1 − x − 2y
dy dx
3
(y − xy − y 2 )
Z
(1−x−2y)/3
Z
0
1
0
Z
(1−x)/2
0
0
Z
Z
1 − x 2
2
(1 − x)2
(1 − x)2
−
2
4
(1 − x)
12
1
3
dx
dx
= 1/36 gm.
0
Then the coordinates of the center of mass are given by
x̄ = 36
Z
x dV = 36
Z
y dV = 36
W
Z
1
0
and
ȳ = 36
and
z̄ = 36
z dV = 36
Z
0
W
(1−x)/2
Z
(1−x)/2
0
1
0
W
Z
Z
Z
1
(1−x−2y)/3
Z
(1−x−2y)/3
x dz dy dx = 1/4 cm.
0
0
Z
Z
y dz dy dx = 1/8 cm.
0
(1−x)/2
(1−x−2y)/3
Z
0
z dz dy dx = 1/12 cm.
0
67. The volume V of the solid is 1 · 2 · 3 = 6. We need to compute
Z
m
6
W
0
0
Z
3
x2 + y 2 dz dy dx
0
2
3(x2 + y 2 ) dy dx
(x2 y + y 3 /3)
0
Z
1
m
8abc
Z
m
2
=
Z
0
1
Z
m
2
=
2
Z
0
1
Z
m
6
=
1
Z
m
6
x2 + y 2 dV =
2
0
dx
(2x2 + 8/3) dx = 5m/3
0
68. The volume of the solid is 8abc, so we need to evaluate
m
8abc
Z
(y 2 + z 2 ) dV =
W
m
=
8abc
m
=
4bc
=
m
2c
−c
c
Z
Z
Z
c
−c
c
Z
b
−b
Z
b
Z
a
(y 2 + z 2 ) dx dy dz
−a
2a(y 2 + z 2 ) dy dz
−b
(y 3 /3 + yz 2 )
−c
c
(b2 /3 + z 2 ) dz
−c
= m(b2 + c2 )/3
b
−b
dz
16.3 SOLUTIONS
1517
69. By the definition, we have that
m
V
a+b =
m
=
V
m
V
=
Z
(y 2 + z 2 ) dV +
ZW
ZW
Strengthen Your Understanding
R
W
2
Z
(x2 + z 2 ) dV
W
2
(x + y + 2z ) dV
(x2 + y 2 ) dV +
W
m
= c+
V
Since z 2 is always positive, the integral
2
m
V
Z
m
V
Z
(2z 2 ) dV
W
2
(2z ) dV
W
(2z 2 ) dV will be positive, thus a + b > c.
70. This would be true if the function f were even in z; that is, if f (x, y, −z) = f (x, y, z), so that
R the integral over the lower
Rhalf of the sphere and the upper half of the sphere were equal. But if f (x, y, z) = z, then S f (x, y, z) dV = 0 while
f (x, y, z) dV is positive.
U
71. The limits for the innermost integrals should be the same since both integrate first with respect to z.
72. The volume of R is π · 22 · 3 = 12π, so we choose f to be the constant function f (x, y, z) = 7/(12π).
73. Since f is not constant, the integral over different parts of the spherical region must cancel. For example, if we take
f (x, y, z) = z, the integral over the top and bottom halves of the region cancel. Many other answers are possible.
74. False. The integral gives the total mass of the material contained in W.
75. True. The region lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 and below the plane z = x.
76. False. The given limits only cover the part of the unit ball in the first octant where x ≥ 0, y ≥ 0, and z ≥ 0. To cover the
entire unit ball the limits are
√
√
Z
1
−1
Z
1−x2
−
√
1−x2
Z
−
1−x2 −y 2
√
f dzdydx.
1−x2 −y 2
77. True. Both sets of limits describe the solid region lying above the triangle x + y ≤ 1, x ≥ 0, y ≥ 0, z = 0 and below the
plane x + y + z = 1.
78. True. Both sets of limits describe the solid region lying above the rectangle −1 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0 and below
the parabolic cylinder z = 1 − x2 .
RbRdRk
79. False. The iterated integral is of the form like a c e f dz dy dx only if the rectangular region has faces parallel to the
coordinate axes. More general rectangular regions, such as a cube with one corner at the origin and the opposite corner at
(0, 0, 1) will need to be written as the sum of iterated integrals where the limits are not constant.
80. False. As a counterexample, consider f (x, y, z) = 12 − x. Then f is positive on half the cube and negative on the other
R1R1R1
half. Symmetry can be used to show that 0 0 0 ( 21 − x)dz dy dx = 0.
81. True. Since
R
W
f dV = lim
and since f > g, we have
X
f (xi , yj , zk )∆V, where (xi , yj , zk ) is a point inside the ijk-th sub-box of volume ∆V,
i,j,k
lim
X
f (xi , yj , zk )∆V > lim
X
i,j,k
i,j,k
g(xi , yj , zk )∆V =
Z
g dV.
W
82. False. As a counterexample, let W1 be the solid cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1, and let W2 be the solid cube
1
1
1
1
−
1 ) = 1 and volume(W2 ) = 8 . Now if f (x, y, z) = −1, then
R 2 ≤ x ≤ 0, − 2 ≤ y ≤ 0, − 2 ≤ z ≤R 0. Then volume(W
1
f dV = 1 · −1 which is less than W f dV = 8 · −1.
W
1
2
83. True. If W is the solid region lying under theRgraph of f and above the region RRin the xy-plane, we can compute the
volume of W either using the double integral R f dA, or using the triple integral W 1 dV.
1518
Chapter Sixteen /SOLUTIONS
Solutions for Section 16.4
Exercises
Z π/2 Z
1/2
f rdr dθ
1.
0
2.
Z
2π
0
3.
Z
0
√
Z
Z
f rdr dθ
0
3π/4
π/4
4.
2
3π/2
π/2
Z
2
f rdr dθ
0
Z
2
f rdr dθ
1
5. Since this is a rectangular region, we use Cartesian coordinates. The rectangle is described by the inequalities 1 ≤ x ≤ 5
and 2 ≤ y ≤ 4, so the integral is
Z Z
5
4
f (x, y) dy dx.
1
2
6. A circle is best described in polar coordinates. The radius is 5, so r goes from 0 to 5. To include the entire circle, we need
θ to go from 0 to 2π. The integral is
Z Z
2π
5
f (r cos θ, r sin θ) r dr dθ.
0
0
7. This is a portion of a circle so it is best described in polar coordinates. The region is a piece of a ring in which r goes from
2 to 4. Since we include only the portion of the ring below the x-axis, we need θ to go from π to 2π. The integral is
Z
π
2π
Z
4
f (r cos θ, r sin θ) r dr dθ.
2
8. Since this is a triangular region we can use Cartesian coordinates. The bottom boundary of the triangle is the line y = x+1
and the top boundary is the line y = 5 − x. The x limits are 0 to 2. The integral is
Z
2
0
Z
5−x
f (x, y) dy dx.
x+1
9. See Figure 16.82.
y
r=4
θ = π/2
✠
x
θ = −π/2
Figure 16.82
16.4 SOLUTIONS
10. See Figure 16.83.
y
r=1
1
❘
x
−1
Figure 16.83
11. See Figure 16.84.
r=2
y
✠
r=1
✙
1
2
x
Figure 16.84
12. See Figure 16.85.
y
θ = π/3
r=1
θ = π/6
x
Figure 16.85
1519
1520
Chapter Sixteen /SOLUTIONS
13. See Figure 16.86.
y
θ = π/4
x
r = 1/ cos θ
or r cos θ = 1
or x = 1
Figure 16.86
14. See Figure 16.87.
y
θ = 3π/4
x
✒
✒
r=4
θ = 3π/2
r=3
Figure 16.87
15. See Figure 16.88.
y
θ = π/4
r = 2/ sin θ
or r sin θ = 2
or y = 2
x
Figure 16.88
Problems
16. The presence of the term x2 +y 2 suggests that we should convert the integral into polar coordinates. Since
the integral becomes
Z p
x2 + y 2 dxdy =
R
Z
0
2π
Z
2
3
r 2 drdθ =
Z
0
2π
r3
3
3
dθ =
2
Z
2π
0
19
38π
dθ =
.
3
3
p
x2 + y 2 = r,
16.4 SOLUTIONS
17. By using polar coordinates, we get
Z
2
2
sin(x + y )dA =
Z
2π
0
R
=
Z
Z
2
sin(r 2 )r dr dθ
0
2π
0
1
− cos(r 2 )
2
Z
2
dθ
0
2π
1
(cos 4 − cos 0) dθ
2 0
1
= − (cos 4 − 1) · 2π = π(1 − cos 4)
2
=−
18. The region is pictured in Figure 16.89.
y
2
1
1
2
x
Figure 16.89
Using polar coordinates, we get
Z
R
2
2
(x − y )dA =
Z
0
π/2
Z
2
2
1
2
2
r (cos θ − sin θ)rdr dθ =
=
15
4
15
=
4
=
Z
π/2
0
(cos2 θ − sin2 θ) ·
1 4
r
4
2
dθ
1
π/2
(cos2 θ − sin2 θ) dθ
0
Z
Z
π/2
cos 2θ dθ
0
15 1
· sin 2θ
4 2
π/2
= 0.
0
√
√
19. By the given limits 0 ≤ x ≤ −1, and − 1 − x2 ≤ y ≤ 1 − x2 , the region of integration is in Figure 16.90.
y
x
−1
Figure 16.90
1521
1522
Chapter Sixteen /SOLUTIONS
In polar coordinates, we have
Z
3π/2
π/2
Z
1
r cos θ r dr dθ =
0
Z
3π/2
1
3
Z
cos θ
π/2
=
3π/2
1 3
r
3
cos θ
1
dθ
0
dθ
π/2
1
sin θ
3
=
3π/2
=
π/2
1
2
(−1 − 1) = −
3
3
20. From the given limits, the region of integration is in Figure 16.91.
√
y
6
y=x
√x
6
y = −x
√
− 6
Figure 16.91
In polar coordinates, −π/4 ≤ θ ≤ π/4. Also,
√
Z
0
6
Z
x
dy dx =
−x
=
Z
Z
√
6 = x = r cos θ. Hence, 0 ≤ r ≤
√
π/4
Z
−π/4
√
6/ cos θ. The integral becomes
6/cos θ
r dr dθ
0
π/4
√
r2
2
−π/4
6/cos θ
0
!
dθ =
Z
π/4
−π/4
6
dθ
2 cos2 θ
π/4
= 3 tan θ
−π/4
= 3 · (1 − (−1)) = 6.
Notice that we can check this answer because the integral gives the area of the shaded triangular region which is
√
(2 6) = 6.
21. From the given limits, the region of integration is in Figure 16.92.
y
2
x=y
√
2
π/4
√
Figure 16.92
x
2
2
1
2
·
√
6·
16.4 SOLUTIONS
1523
So, in polar coordinates, we have,
π/4
Z
0
2
Z
Z
(r 2 cos θ sin θ)r dr dθ =
0
π/4
cos θ sin θ
0
=4
Z
π/4
sin(2θ)
2
0
2
1
4
r4
dθ
0
dθ
π/4
= − cos(2θ)
= 0 − (−1) = 1.
0
y
22. (a)
y = x/3
1
x
3
Figure 16.93
(b)
Z
0
1
Z
3y
f (x, y) dx dy.
0
(c) For polar coordinates, on the line y = x/3, tan θ = y/x = 1/3, so θ = tan−1 (1/3). On the y-axis, θ = π/2. The
quantity r goes from 0 to the line y = 1, or r sin θ = 1, giving r = 1/ sin θ and f (x, y) = f (r cos θ, r sin θ). Thus
the integral is
Z
π/2
tan−1 (1/3)
23. (a)
Z
1/ sin θ
f (r cos θ, r sin θ)r dr dθ.
0
√
(i) The two planes contain the z-axis and form the sides of the orange wedge. Since x ≥ 0, the plane y = x/ 3
forms an angle of π/6 radians with the y = 0 plane, hence 0 ≤ θ ≤ π/6. We also have 0 ≤ ρ ≤ 5 and
0 ≤ φ ≤ π. Hence:
Volume =
π/6
Z
0
Evaluating this integral, we get:
Volume =
Z
π/6
0
Z
π
sin φ
0
π
Z
0
Z
5
ρ2 sin φ dρ dφ dθ.
0
53
−0
3
Z
π/6
125
(cos π − cos 0) dθ
3 0
125π
250 π
=
.
=
3 6
9
dφ dθ = −
(ii) In cylindrical coordinates, the sphere is given by r 2 + z 2 = 25.
√ If we integrate first
√ with respect to z, then z varies between the top and bottom halves of the sphere, or
− 25 − r 2 ≤ z ≤ 25 − r 2 . Then, the shadow of the wedge over the xy-plane gives as a circular sector
where 0 ≤ r ≤ 5, and 0 ≤ θ ≤ π/6. Hence:
Z
Z Z √
π/6
25−r 2
5
Volume =
0
0
−
√
r dz dr dθ.
25−r 2
Evaluating this integral, we get:
Volume =
Z
0
π/6
Z
0
5
p
r
25 −
r2
+
p
25 −
r2
dr dθ =
Z
π/6
Z
0
=
Z
0
π/6
250
3
p
2r
−
0
=
5
Z
0
25 − r 2
(25 − r 2 )
3/2
π/6
dθ =
3/2
dr dθ
5
dθ
0
250 π
125π
=
.
3 6
9
1524
Chapter Sixteen /SOLUTIONS
If we integrate first with respect to r and consider only the half of the wedge above the xy-plane then, for each
value of z,√r varies between zero and the radius of the horizontal cross section of the sphere at height z. Hence:
0 ≤ r ≤ 25 − z 2 . We also have 0 ≤ z ≤ 5 and 0 ≤ θ ≤ π/6. This gives:
Z
Z Z √
π/6
25−z 2
5
r dr dz dθ,
Volume half wedge =
0
0
0
for the top portion of the wedge, or for the full wedge
Volume = 2
Z
π/6
Z
0
5
0
Z √25−z2
r dr dz dθ.
0
Evaluating the last integral, we get:
Volume = 2
Z
π/6
0
Z
5
0
25 − z 2
−0
2
dz dθ =
Z
π/6
0
=
250
3
z3
− 25z
3
π/6
Z
dθ =
0
5
dθ
0
250 π
125π
=
.
3 6
9
Spherical coordinates provide the most efficient integration method for calculating the volume of the wedge.
This makes sense because the wedge is cut out by spherical fundamental surfaces, and hence, in these coordinates, all integration endpoints are constant.
(b) Since the interior angle of the wedge is π/6, we need a total of 12 wedges to recover a full sphere of radius 5. Hence:
Volume of wedge =
1
1
Volume of sphere of radius 5 =
12
12
4 3
125π
π5 =
.
3
9
√
24. Since r = 2/ cos θ we have x = r cos θ = 2. Since θ ranges from 0 to π/6, y ranges from 0 to y = x/ 3. Converting
to Cartesian coordinates we have
Z
2
0
Z
√
x/ 3
dy dx =
0
√
x/ 3
2
Z
dx =
y
0
2
0
0
2
Z
2
x
x2
√ dx = √
3
2 3
2
2
0
2
= √ .
3
2
25. (a) The region (shaded) is between the circles x + y = 1 and x + y = 4; see Figure 16.94. The first integral is to
the left of the dashed line x = 1; the second integral is to the right of the dashed line.
y
2
x2 + y 2 = 4
x2 + y 2 = 1
❘
1
x
1
2
Figure 16.94
(b) Converting to polar coordinates, we find the quantity in part (a) is given by
Z Z √
Z
Z
Z Z √
4−x2
1
0
√
4−x2
2
1−x2
π/2
x dy dx =
x dy dx +
1
0
0
=
r3
3
2
r cos θ r dr dθ
1
2
1
· sin θ
π/2
=
0
7
7
·1= .
3
3
16.4 SOLUTIONS
1525
26. The graph of f (x, y) = 25−x2 −y 2 is an upside down bowl, and the region whose volume we want is contained between
the bowl (above) and the xy-plane (below). We must first find the region in the xy-plane where f (x, y) is positive. To do
that, we set f (x, y) ≥ 0 and get x2 + y 2 ≤ 25. The disk x2 + y 2 ≤ 25 is the region R over which we integrate.
Volume =
Z
2
R
Z
2
(25 − x − y ) dA =
Z
=
2π
0
Z
625
4
625π
=
2
=
2π
Z
0
25 2 1 4
r − r
2
4
2π
5
0
(25 − r 2 ) rdr dθ
5
dθ
0
dθ
0
27. First, let’s find where the two surfaces intersect.
p
p
8 − x2 − y 2 =
2
x2 + y 2
2
2
8 − x − y = x + y2
x2 + y 2 = 4
So z = 2 at the intersection. See Figure 16.95.
z
✛
✛
2
✒
y
✛
x2 + y 2 = 4
2
R
x
Figure 16.95
The volume of the ice cream cone has two parts. The first part (which is the volume of the cone)
Z is the volume of the
solid bounded by the plane z = 2 and the cone z =
p
x2 + y 2 . Hence, this volume is given by
R
(2 −
where R is the disk of radius 2 centered at the origin, in the xy-plane. Using polar coordinates, we have:
Z
R
2−
p
x2 + y 2
dA =
2π
Z
Z
0
=
(2 − r) · r dr dθ
0
Z
2π
4
3
Z
"
r3
r −
3
0
=
2
2
2π
0
= 8π/3
dθ
2#
0
dθ
p
x2 + y 2 ) dA,
1526
Chapter Sixteen /SOLUTIONS
TheZsecond part is the volume of the region above the plane z = 2 but inside the sphere x2 + y 2 + z 2 = 8, which is given
(
by
R
p
8 − x2 − y 2 − 2) dA where R is the same disk as before. Now
8 − x2 − y 2 − 2) dA =
(
R
2π
Z
Z p
0
0
=
Z
2
p
8 − r 2 − 2)rdr dθ
r
0
2π
Z
(
0
2π
Z
=
2
Z
p
8 − r 2 dr dθ −
1
− (8 − r 2 )3/2
3
0
Z
2
0
!
2π
0
2π
Z
Z
2
2r dr dθ
0
dθ −
1
(43/2 − 83/2 ) dθ −
3 0
√
1
= − · 2π(8 − 16 2) − 8π
3
√
2π
(16 2 − 8) − 8π
=
3
√
8π(4 2 − 5)
=
3
√
Thus, the total volume is the sum of the two volumes, which is 32π( 2 − 1)/3.
=−
Z
Z
2
2π
r2
0
dθ
0
2π
4 dθ
0
28. (a) The volume, V , is given by
V =
Z
e−(x
2
+y 2 )
dA.
x2 +y 2 ≤a2
Converting to polar coordinates gives
V =
2π
Z
0
Z
2π
a
2
e−r r dr dθ = θ
0
(b) As a → ∞, the value of e
0
−a2
2
1
− e−r
2
a
= 2π
0
2
1
1
− e−a
2
2
2
= π(1 − e−a ).
→ 0, so the volume tends to π.
29. (a) Using polar coordinates, we have
Mass =
Z
2π
0
(b) Integrating with respect to θ first
Mass = 2π
Z
0
3
Z
3
0
1
r dr dθ.
r2 + 1
3
1
r
dr = 2π ln |r 2 + 1|
r2 + 1
2
= π(ln 10 − ln 1) = π ln 10.
0
30. (a)
Total Population =
Z
3π/2
Z
π/2
4
δ(r, θ) rdr dθ.
1
(b) We know that δ(r, θ) decreases as r increases, so that eliminates (iii). We also know that δ(r, θ) decreases as the
x-coordinate decreases, but x = r cos θ. With a fixed r, x is proportional to cos θ. So as the x-coordinate decreases,
cos θ decreases and (i) δ(r, θ) = (4 − r)(2 + cos θ) best describes this situation.
(c)
Z
3π/2
π/2
Z
1
4
(4 − r)(2 + cos θ) rdr dθ =
3π/2
Z
π/2
=9
Z
(2 + cos θ)(2r 2 −
3π/2
(2 + cos θ) dθ
π/2
= 9 2θ + sin θ
3π/2
π/2
= 18(π − 1) ≈ 39
Thus, the population is around 39,000.
1 3
r )
3
4
dθ
1
16.4 SOLUTIONS
1527
31. The density function is given by
ρ(r) = 10 − 2r
where r is the distance from the center of the disk. So the mass of the disk in grams is
Z
ρ(r) dA =
2π
Z
0
R
=
0
=
Z
(10 − 2r)rdr dθ
0
2π
Z
5
Z
2π
0
5r 2 −
2 3
r
3
5
dθ
0
250π
125
dθ =
(grams)
3
3
32. The charge density is δ = k/r, where k is a constant.
Total charge =
Z
δ dA =
Z
0
Disk
R
Z
2π
0
k
r dθ dr = k
r
Z
R
0
2π
Z
dθ dr = k
0
Z
R
2π dr = 2kπR.
0
Thus the total charge is proportional to R with constant of proportionality 2kπ.
33. (a) The curve r = 1/(2 cos θ) or r cos θ = 1/2 is the line x = 1/2. The curve r = 1 is the circle of radius 1 centered at
the origin. See Figure 16.96.
y
θ = π/3
✠
1
1
2
x
■
x = 1/2 θ = −π/3
Figure 16.96
(b) The line intersects the circle where 2 cos θ = 1, so θ = ±π/3. From Figure 16.96 we see that
Area =
Z
π/3
−π/3
Z
1
r dr dθ.
1/(2 cos θ)
Evaluating gives
Area =
Z
π/3
−π/3
r2
2
1
1/(2 cos θ)
tan θ
1
θ−
=
2
4
π/3
−π/3
!
dθ =
1
=
2
1
2
Z
π/3
−π/3
1−
1
4 cos2 θ
dθ
√
√
π
4π − 3 3
3
2 −2
=
.
3
4
12
34. (a) The circle r = 2 cos θ has radius 1 and is centered at (1, 0); the circle r = 1 has radius 1 and is centered at the origin.
See Figure 16.97.
1528
Chapter Sixteen /SOLUTIONS
θ = π/3
y
❄
r=1
r = 2 cos θ
1
x
2
✻
θ = −π/3
Figure 16.97
(b) The circles intersect where 2 cos θ = 1, so θ = ±π/3. From Figure 16.97 we see that
Z
Area =
π/3
−π/3
2 cos θ
Z
r dr dθ.
1
Evaluating gives
Area =
Z
π/3
−π/3
=
r2
2
2 cos θ
1
!
dθ =
1
(2 cos θ sin θ + 2θ − θ)
2
1
2
Z
π/3
4 cos2 θ − 1 dθ
−π/3
π/3
=
−π/3
1
2
4·
√
√
1
3
π
3
π
·
+2
=
+ .
2 2
3
2
3
35. (a) We must first decide where to put the origin. We locate the origin at the center of one disk and locate the center of the
second disk at the point (1, 0). See Figure 16.98. (Other choices of origin are possible.)
y
√
x2 + y 2 = 1
✠
y
(x − 1)2 + y 2 = 1
✠
r=1
✠
r = 2 cos θ
✠
3/2
1
2
1
π/3
x
x
√
− 3/2
Figure 16.98
Figure 16.99
By symmetry, the points of intersection of the circles are half-way between the centers, at x = 1/2. The y-values
at these points are given by
r
√
2
p
1
3
2
=±
.
y =± 1−x =± 1−
2
2
We integrate in the x-direction first, so that it is not necessary to set up two integrals. The right-side of the circle
x2 + y 2 = 1 is given by
p
x = 1 − y2.
The left side of the circle (x − 1)2 + y 2 = 1 is given by
x=1−
Thus the area of overlap is given by
√
Area =
Z
−
3/2
√
3/2
p
1 − y2.
Z √1−y2
1−
√
1−y 2
dxdy.
16.4 SOLUTIONS
1529
(b) In polar coordinates, the circle centered at the origin has equation r = 1. See Figure 16.99. The other circle, (x −
1)2 + y 2 = 1, can be written as
x2 − 2x + 1 + y 2 = 1
x2 + y 2 = 2x,
so its equation in polar coordinates is
r 2 = 2r cos θ,
and, since r 6= 0,
r = 2 cos θ.
√
√
At the top point of intersection of the two circles, x = 1/2, y = 3/2, so tan θ = 3, giving θ = π/3.
Figure 16.99 shows that if we integrate with respect to r first, we have to write the integral as the sum of two
integrals. Thus, we integrate with respect to θ first. To do this, we rewrite
r = 2 cos θ
θ = arccos
as
This gives the top half of the circle; the bottom half is given by
θ = − arccos
Thus the area is given by
Area =
1
Z
Z
r
2
r
.
2
.
arccos(r/2)
r dθdr.
− arccos(r/2)
0
36. The required region is shaded in Figure 16.100. The limits of integration will correspond to the two points where the two
curves intersect. These points are given by solving the equation 2+3 cos θ = 2. Simplifying this equation gives cos θ = 0
so the required solutions are θ = π/2 and θ = −π/2 as shown in Figure 16.100.
y
1
x
−2
2
4
−1
Figure 16.100
The required area is given by
Z
1 dA =
Z
π
2
−π
2
R
=
=
1
2
Z
1
2
Z
Z
2+3 cos θ
r dr dθ
2
π
2
−π
2
π
2
−π
2
(2 + 3 cos θ)2 − 22 dθ
12 cos θ + 9 cos2 θ dθ
1530
Chapter Sixteen /SOLUTIONS
=
=
3
2
Z
3
2
Z
π
2
(4 cos θ + 3 cos2 θ) dθ
−π
2
π
2
4 cos θ +
−π
2
3
(1 + cos 2θ) dθ
2
3
3
3
4 sin θ + θ + sin 2θ
2
2
4
=
= 12 +
9π
.
4
π
2
−π
2
Strengthen Your Understanding
37. The part of the boundary of R corresponding to x = 1 in terms of polar coordinates is not r = 1. Rather, it is r cos θ = 1,
which gives r = 1/ cos θ. The angle between y = 0 and y = x is π/4, so we have
Z
x dA =
R
π/4
Z
Z
0
1/ cos θ
(r cos θ)r dr dθ.
0
38. When converting to polar coordinates, we need an extra factor of r, because dA = r dr dθ. Thus, we should have:
Z
Z
(x2 + y 2 ) dA =
R
2π
0
Z
0
2
r 2 · r dr dθ =
Z
0
2π
Z
2
r 3 dr dθ
0
39. Any region that is a sector
of a circle centered at the origin suggests polar coordinates. An example is the quarter disk
√
0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x2 .
40. Any integrand that is a function of
41. (a), (c), (e)
p
p
x2 + y 2 suggests polar coordinates. For example, let f (x, y) = 1/
x2 + y 2 .
42. The region lies in the first quadrant and is bounded by four lines. The equations r = 1/sin θ and r = 4/sin θ are the
horizontal lines y = r sin θ = 1 and y = r sin θ = 4. The equation θ = π/4 gives the line y = x, and the equation
θ = π/2 gives the y-axis.
Solutions for Section 16.5
Exercises
1. (a) is (IV); (b) is (II); (c) is (VII); (d) is (VI); (e) is (III); (f) is (V).
2. The plane has equation θ = π/4.
3. The top half of the sphere has equation z =
4. The cone has equation z = r.
p
1 − x2 − y 2 =
√
1 − r2 .
5. The cone has equation φ = π/4.
6. The plane has equation ρ cos φ = 10 or ρ = 10/ cos φ.
7. The plane has equation ρ cos φ = 4 or ρ = 4/ cos φ.
8.
Z
f dV =
Z
=
Z
W
3
−1
3
−1
Z
2π
0
Z
0
Z
1
(sin (r 2 )) rdr dθ dz
0
2π
1
(− cos r 2 )
2
1
dθ dz
0
16.5 SOLUTIONS
=−
1
2
= −π
Z
Z
3
−1
3
−1
2π
Z
(cos 1 − cos 0) dθ dz
0
(cos 1 − 1) dz = −4π(cos 1 − 1) = 4π(1 − cos 1)
9.
Z
f dV =
Z
=
Z
1
−1
W
1
−1
1
Z
=
3π/4
Z
π/4
(r 2 + z 2 ) rdr dθ dz
0
3π/4
Z
4
Z
(64 + 8z 2 ) dθ dz
π/4
π
(64 + 8z 2 ) dz
2
−1
8
200
π=
π
3
3
= 64π +
10.
Z
f dV =
2π
Z
0
W
=
Z
2π
7
3
7
=
3
7
=
6
=
=
7
6
0
2π
Z
2
ρ2 sin2 φ dρ dφ dθ
π
4
sin2 φ dφ dθ
0
2π
π/4
Z
0
0
2π
Z
Z
1
Z
0
Z
(sin φ)ρ2 sin φ dρ dφ dθ
1
π/4
Z
2
Z
0
0
=
π/4
Z
1
sin 2φ)
2
(φ −
0
Z
2π
Z
5
(
0
1 − cos 2φ
dφ dθ
2
π/4
dθ
0
π
1
− ) dθ
4
2
7
π
1
7π(π − 2)
· 2π( − ) =
6
4
2
12
11. We have:
Z
f dV =
W
=
=
Z
Z
0
0
5
0
0
Z
Z
π
π/2
2π
Z
0
5
= 2π
2π
Z
Z
1 2
· ρ sin φ dφ dθ dρ
ρ
π
ρ sin φ dφ dθ dρ
π/2
2π
ρ dθ dρ
0
5
Z
ρ dρ = 25π
0
Note that the integral is improper, but it can be shown that the result is correct.
12. Using Cartesian coordinates, we get:
Z
3
0
Z
0
1
Z
0
5
f dz dy dx
1531
1532
Chapter Sixteen /SOLUTIONS
13. Using cylindrical coordinates, we get:
1
Z
2π
Z
0
0
4
Z
f · rdr dθ dz
0
14. Using cylindrical coordinates, we get:
Z
4
π/2
Z
0
0
2
Z
0
f · rdr dθ dz
15. Using spherical coordinates, we get:
Z
π
0
π
Z
0
3
Z
f · ρ2 sin φ dρ dφ dθ
2
16. Using spherical coordinates, we get:
Z
2π
0
π/6
Z
0
Z
3
f · ρ2 sin φ dρ dφ dθ
0
17. We use Cartesian coordinates, oriented as shown in Figure 16.101. The slanted top has equation z = mx, where m is the
slope in the x-direction, so m = 1/5. Then if f is an arbitrary function, the triple integral is
Z
5
0
2
Z
0
x/5
Z
f dz dy dx.
0
Other answers are possible.
z
(5, 0, 1)
1
y
2
❄
(5, 2, 1)
(5, 2, 0)
5
x
Figure 16.101
18. We choose cylindrical coordinates oriented as in Figure 16.102. The cone has equation z = r. Since we have a half cone
scooped out of a half cylinder, θ varies between 0 and π. Thus, if f is an arbitrary function, the integral is
Z
0
π
Z
2
0
Z
r
f r dzdrdθ.
0
Other answers are possible.
z
z=r
2
π/4
x
2
Figure 16.102
16.5 SOLUTIONS
1533
Problems
19. In cylindrical coordinates, the sphere has equation r 2 + z 2 = K 2 . Thus
Z Z Z √
2π
K 2 −r 2
K
V =
0
−
0
√
r dz dr dθ.
K 2 −r 2
20. In spherical coordinates, the sphere has equation ρ = K. Thus
V =
π
Z
0
K
Z
0
2π
Z
ρ2 sin φ dθ dρ dφ.
0
21. We use cylindrical coordinates. The cone has radius r = 2 when z = 4, so its equation is z = 2r. Thus, the integral is
Z
2π
2
Z
0
4
Z
0
f (r, θ, z)r dz dr dθ.
2r
22. We use spherical coordinates. The cone has radius 2 when z = 4, so ρ sin φ = 2 when ρ cos φ = 4. Thus tan φ = 1/2,
so φ = arctan(1/2). The top of the cone, z = 4, is given by ρ cos φ = 4. Thus, the integral is
Z
0
2π
Z
arctan(1/2)
0
4/ cos φ
Z
g(ρ, φ, θ)ρ2 sin θ dρ dφ dθ.
0
23. In rectangular coordinates, a cone has equation z = k
√
22 = 2, we have k = 2. Thus, the integral is
Z 2 Z √4−x2 Z
−2
−
√
4−x2
p
x2 + y 2 for some constant k. Since z = 4 when
x2 + y 2 =
4
√
2
h(x, y, z) dz dy dx.
x2 +y 2
24. (a) In Cartesian coordinates, the bottom half of the sphere x2 + y 2 + z 2 = 1 is given by z = −
Z Z √
Z
Z
1−x2
1
0
dV =
0
W
p
0
−
√
p
1 − x2 − y 2 . Thus
dz dy dx.
1−x2 −y 2
√
(b) In cylindrical coordinates, the sphere is r 2 + z 2 = 1 and the bottom half is given by z = − 1 − r 2 . Thus
Z
dV =
W
Z
π/2
0
Z
1
0
Z
0
Z
1
−
√
r dz dr dθ.
1−r 2
(c) In spherical coordinates, the sphere is ρ = 1. Thus,
Z
dV =
Z
π/2
0
W
Z
π
π/2
ρ2 sin φ dρ dφ dθ.
0
25. (a) Since the cone has a right angle at its vertex, it has equation
z=
p
x2 + y 2 .
The sphere has equation x2 + y 2 + z 2 = 1, so the top half is given by
z=
The cone and the sphere intersect in the circle
p
x2 + y 2 =
1 − x2 − y 2 .
1
,
2
1
z= √ .
2
1534
Chapter Sixteen /SOLUTIONS
See Figure 16.103. Thus
Z
dV =
W
Z
1/
√
−1/
2
√
2
Z √(1/2)−x2 Z √1−x2 −y2
−
√
√
(1/2)−x2
dz dy dx.
x2 +y 2
z
1
✻
■Circle is
√1
2
x2 + y 2 =
π
4
❄
1
2
y
■
x
x2 + y 2 =
1
2
Figure 16.103
(b) In cylindrical coordinates, the cone has equation z = r and the sphere has equation z =
Z Z √ Z √
Z
2π
1/
√
1 − r 2 . Thus
1−r 2
2
dV =
r dz dr dθ.
0
W
0
r
(c) In spherical coordinates, the cone has equation φ = π/4 and the sphere is ρ = 1. Thus
Z
dV =
2π
Z
0
W
π/4
Z
0
Z
1
ρ2 sin φ dρ dφ dθ.
0
26. (a) Since the cone has a right angle at its vertex, it has equation
p
z = x2 + y 2 .
√
Figure 16.104 shows the plane with equation z = 1/ 2. The plane and the cone intersect in the circle x2 +y 2 = 1/2.
Thus,
√
√
√
Z
dV =
Z
1/
2
−1/
W
√
2
Z
(1/2)−x2
−
√
(1/2)−x2
Z
1/
√
2
dz dy dx.
x2 +y 2
z
√1
2
√1
2
π
4
1
y
x
Figure 16.104
(b) In cylindrical coordinates the cone has equation z = r, so
Z
dV =
Z
0
W
2π
Z
1/
0
√
2
Z
r
1/
√
2
r dz dr dθ.
√
√
(c) In spherical coordinates, the cone has equation φ = π/4 and the plane z = 1/ 2 has equation ρ cos φ = 1/ 2.
Thus
Z
Z 2π Z π/4 Z 1/(√2 cos φ)
dV =
ρ2 sin φ dρ dφ dθ.
W
0
0
0
16.5 SOLUTIONS
1535
2
2
2
2
27. (a) In cylindrical coordinates,
the cone
√
√ is z = r and the sphere is r + z = 4. The surfaces intersect where z + z =
2
2z = 4. So z = 2 and r = 2.
Z Z √ Z √
4−r 2
2
2π
r dzdrdθ.
Volume =
0
0
r
(b) In spherical coordinates, the cone is φ = π/4 and the sphere is ρ = 2.
2π
Z
Volume =
0
Z
2π
Z
π
Z
0
0
0
Z
2
ρ2 sin φ dρdφdθ.
0
2
ρ2 sin φ dρdφdθ.
0
0
1
Z
Z 2π Z 2 Z √4−r2
(b)
r
dzdrdθ
−
√
28. (a)
π/4
Z
−
4−r 2
2π
0
Z
Z √1−r2
1
0
−
√
r dzdrdθ.
1−r 2
29. We want the volume of the region above the cone φ = π/3 and below the sphere ρ = 3:
V =
Z
2π
Z
0
π/3
0
3
Z
ρ2 sin φ dρ dφ dθ.
0
The order of integration can be altered and other coordinates can be used.
30. We want the volume of the region between the sphere ρ = 3 and the cone z = r. The sphere can also be written
x2 + y 2 + z 2 = 32 or r 2 + z 2 = 9. The cone can also be written as √
φ = π/4.
√
The sphere cuts the cone z = r in the circle 2r 2 = 9, or r = 3/ 2, lying in the plane z = 3/ 2.
In cylindrical coordinates,
√
V =
Z
2π
Z
2π
0
In spherical coordinates
V =
√
Z
3/
Z
π/4
2
r dz dr dθ.
r
0
0
9−r 2
Z
0
Z
3
ρ2 sin φ dρ dφ dθ.
0
The order of integration can be altered and other coordinates can be used.
31. In cylindrical coordinates, the region is the half cylinder given by 5 ≤ z ≤ 10,
π
Z
V =
√
Z
√
0
3
2
Z
√
2≤r≤
√
3, 0 ≤ θ ≤ π. Thus
10
r dz dr dθ.
5
The order of integration can be altered and other coordinates can be used.
32. The cone can be√
written z = r, and the first quadrant of the xy-plane is given by 0 ≤ θ ≤ π/2. The region x2 + y 2 ≤ 7
is given by r ≤ 7. Thus
√
V =
Z
π/2
7
Z
0
0
Z
r
r dz dr dθ.
0
The order of integration can be altered and other coordinates can be used.
33. We use cylindrical coordinates since the sphere x2 + y 2 + z 2 = 10, or r 2 + z 2 = 10, and the plane z = 1 can both be
simply expressed. The plane cuts the sphere in the circle r 2 + 12 = 10, or r = 3. Thus
Z Z Z √
2π
10−r 2
3
r dz dr dθ,
V =
0
or
V =
Z
2π
0
1
0
√
Z
1
10
Z √10−z2
0
Order of integration can be altered and other coordinates can be used.
r dr dz dθ.
1536
Chapter Sixteen /SOLUTIONS
34. In spherical coordinates, the cone z = r is given by φ = π/4 and the sphere x2 + y 2 + z 2 = 8 is given by ρ =
Since φ is measured from the positive z-axis, the region we are interested in has π/4 ≤ φ ≤ π/2. Thus
Z
V =
2π
0
Z
π/2
π/4
√
Z
√
8.
8
ρ2 sin φ dρ dφ dθ.
0
The order of integration can be altered and other coordinates can be used.
√
√
35. (a) In cylindrical coordinates, the cone has equation z = 3r. When z = 1, we have r = 1/ 3, so
Volume =
2π
Z
1/
Z
√
3
0
0
Z
1
√
3r
r dz dr dθ.
(b) Evaluating gives
Volume = 2π
Z
√
1 3
1
z
√
3r
0
√
1 3
Z
r dr = 2π
(1 −
0
= 2π
√
3r)r dr
√
3 3
r2
−
r
2
3
1/
√
3
= 2π
0
1
1
−
6
9
=
π
.
9
36. For x2 + y 2 ≤ 1, the cone is below the plane z = 10 + x. In cylindrical coordinates, the plane is z = 10 + r cos θ, and
the cone is z = r. Thus
Volume =
Z
2π
Z
2π
Z
2π
Z
2π
Z
2π
0
=
0
=
1
Z
1
r dz dr dθ
r
10+r cos θ
rz
dr dθ
r
(10r + r 2 cos θ − r 2 ) dr dθ
0
0
=
Z
10+r cos θ
Z
0
0
=
1
0
0
=
Z
r3
r3
cos θ −
5r +
3
3
2
5+
1
1
cos θ −
3
3
2π
14
1
θ + sin θ
3
3
=
1
dθ
0
dθ
28π
.
3
0
37. The cone is centered along the positive x-axis and intersects the sphere in the circle
(y 2 + z 2 ) + y 2 + z 2 = 4
y 2 + z 2 = 2.
We use spherical coordinates with φ measured from thep
x-axis and θ measured in the yz-plane. (Alternatively, the volume
we want is equal to the volume between the cone z = x2 + y 2 and the sphere x2 + y 2 + z 2 = 4.) The cone is given
by φ = π/4. The sphere has equation ρ = 2. Thus
Z
2π
=
Z
=
Z
Volume =
Z
π/4
2π
Z
π/4
2π
Z
π/4
0
0
0
0
2
ρ2 sin φ dρ dφ dθ
0
0
0
Z
2
ρ3
sin φ dφ dθ
3
0
8
sin φ dφ dθ
3
1537
16.5 SOLUTIONS
=
π/4
2π
Z
8
− cos φ
3
0
16π
=
3
1
1− √
2
dθ =
0
2π
Z
8
3
0
1
1− √
2
dθ
.
2
2
38. Using cylindrical coordinates,
√ the equation of the sphere is r + z = 4. The top of the sphere has equation z =
When z = 1 we have r = 3. Figure 16.105 shows the limits of integration on the integral.
Z Z √ Z √
4 − r2 .
4−r 2
3
2π
√
r dz dr dθ
Volume =
Z
= 2π
√
√
4−r 2
dr = 2π
rz
0
= 2π
1
0
√
3
0
1
(4 − r 2 )3/2
r2
−
−3
2
Z
√
3
3
(r
0
p
4 − r 2 − r) dr
5
π.
3
=
0
z
2
1✻
1
❄
√
3
π/6
π/3
r
Figure 16.105
39. Use cylindrical coordinates: when r 2 = x2 + y 2 = 1, then x2 + y 2 + z 2 = 1 + z 2 = 2 so z = ±1. The region W is
shown in Figure 16.106.
Z Z Z √
Z
1
2−z 2
2π
(x2 + y 2 ) dV =
−1
W
=
Z
1
−1
2π
=
4
=
π
2
0
Z
Z
1
2π
r4
4
0
1
√
r 2 · r drdθdz
2−z 2
1
3 − 4z 2 + z
−1
3z −
4 3 z5
z +
3
5
z
1
dθdz =
4
4
1
−1
Z
2π
0
dz
1
Z
=
−1
28π
.
15
Cylinder
x2 + y 2 = 1
✠
1
Sphere
x2 + y 2 + z 2 = 2
✠
y
x
−1
Figure 16.106
(2 − z 2 )2 − 1 dθdz
1538
Chapter Sixteen /SOLUTIONS
40. The region whose volume we want is shown in Figure 16.107:
z
✻
5
2
❄
π
6
θ=
θ=
π
3
y
x
Figure 16.107
Using cylindrical coordinates, the volume is given by the integral:
Z
V =
Z
=
2
0
π/6
2
0
Z
Z
π/3
2
0
Z
2
Z
Z
5
r dr dθ dz
0
π/6
25
2
=
π/3
Z
5
r2
2
dθ dz
0
π/3
dθ dz
π/6
25
π
π
=
−
dz
2 0
3
6
25π
25 π
· ·2=
.
=
2 6
6
41. Orient the cone as shown in Figure 16.108 and use cylindrical coordinates with the origin at the vertex of the cone. Since
the√angle at the vertex of the cone is a right angle, the angles AOB and COB are both π/4. Thus, OB = 5 cos π/4 =
5/ 2. The curved surface of the cone has equation z = r, so
Volume =
Z
2π
Z
2π
0
=
√
Z
5/
Z
5/
2
0
0
2π
=θ
0
r
√
2
5/
√
r dz dr dθ
√
z=5/ 2
dr dθ =
1
1
−
2
3
Z
0
z=r
5 r2
r3
√
−
3
2 2
53
= 2π · √
2 2
2
rz
0
Z
5/
√
2
= 2π
0
2π
Z
5/
0
√
2
r
5
√ −r
2
5 52
53
√ · 2 −
√
2 2
2·3· 2
53 π
= √ = 46.28 cm3 .
6 2
dr dθ
16.5 SOLUTIONS
1539
z
C
B
A
π/4 π/4
5
❫ ✢
√ r
5/ 2
O
Figure 16.108
42. (a) The angle φ takes on values in the range 0 ≤ φ ≤ π. Thus, sin φ is nonnegative everywhere in W1 , and so its integral
is positive.
(b) The function φ is symmetric across the xy plane, such that for any point (x, y, z) in W1 , with z 6= 0, the point
(x, y, −z) has a cos φ value with the same magnitude but opposite sign of the cos φ value for (x, y, z). Furthermore,
if z = 0, then (x, y, z) has a cos φ value of 0. Thus, with cos φ positive on the top half of the sphere and negative on
the bottom half, the integral will cancel out and be equal to zero.
43. (a) The integral is negative. In W2 , we have 0 < z < 1. Thus, z 2 − z is negative throughout W2 and thus its integral is
negative.
(b) On the top half of the sphere, z is nonnegative, but x can be both positive and negative. Thus, since W2 is symmetric
with respect to the yz plane, the contribution of a point (x, y, z) will be canceled out by its reflection (−x, y, z).
Thus, the integral is zero.
44. We must first decide on coordinates. We pick cylindrical coordinates with the z-axis along the axis of the cylinders. The
insulation stretches from z = 0 to z = l. See Figure 16.109. The volume is given by the integral
Volume =
Z
2π
0
Z lZ
0
a+h
r drdzdθ.
a
Evaluating the integral gives
Volume =
Z
0
2π
l
Z
0
r2
2
l
a+h
dzdθ = 2πz
0
a
(a + h)2
a2
−
2
2
= πl((a + h)2 − a2 ).
To check our answer, notice that the volume is the difference between the volume of two cylinders of radius a and a + h.
These cylinders have volumes πl(a + h)2 and πla2 , respectively.
z
✻
l
❄ ✛
h
✲✛
a
✲
y
a+h
a
x
Figure 16.109
45. The plane (x/p) + (y/q) + (z/r) = 1 cuts the axes at the points (p, 0, 0); (0, q, 0); (0, 0, r). Since p, q, r are positive,
the region between this plane and the coordinate planes is a pyramid in the first octant. Solving for z gives
z=r
1−
y
x
−
p
q
=r−
ry
rx
−
.
p
q
The volume, V , is given by the double integral
V =
Z
R
r−
ry
rx
−
p
q
dA,
1540
Chapter Sixteen /SOLUTIONS
where R is the region shown in Figure 16.110. Thus
V =
p
Z
0
=
q−qx/p
Z
p
Z
p
r
0
=
p
Z
rxy
ry 2
ry −
−
p
2q
0
=
qx
q−
p
rq −
0
ry
rx
−
p
q
r−
0
Z
dydx
y=q−qx/p
y=0
r
qx
− x q−
p
p
2rqx
rqx2
rq 2
+ 2 −
p
p
2q
1−
!
dx
r
−
2q
qx
q−
p
2x
x2
+ 2
p
p
rqx2
rqx3
rqx
rqx2
rqx3
+ 2 −
+
− 2
p
p 3
2
p2
2p 3
pqr
pqr
pqr
pqr
pqr
= pqr − pqr +
−
+
−
=
.
3
2
2
6
6
=
rqx −
2 !
dx
dx
p
0
x
y
+ =1
p
q
y = q − qx
p
y
q
✠
R
x
p
Figure 16.110
46. We must first decide on coordinates. We imagine the vertex of the cone downward, at the origin, with the flat base in the
plane z = h, as in Figure 16.111. Then, using cylindrical coordinates as in Figure 16.112, we see that the curved surface
of the cone has equation z = hr/a. Thus the volume is given by
Volume =
Z
2π
0
Z
0
a
Z
h
r dzdrdθ.
hr/a
Evaluating gives
Volume =
Z
2π
0
= 2π
Z
0
z=h
a
rz
drdθ =
0
z=hr/a
hr 3
hr 2
−
2
3a
Z
a
= 2πh
0
2π
Z
0
a
a2
a2
−
2
3
hr 2
a
πha2
.
3
=
z
z
✛
hr −
drdθ
z = hr/a
h
a
✲
✻
h
❄
y
a
x
Figure 16.111
Figure 16.112
r
16.5 SOLUTIONS
1541
47. We must first decide on coordinates. We pick spherical coordinates with the common center of the two spheres as the
origin. We imagine the half-melon with the flat side horizontal and the positive z-axis going through the curved surface.
See Figure 16.113. The volume is given by the integral
Volume =
2π
Z
0
Z
π/2
Z
0
b
ρ2 sin φ dρdφdθ.
a
Evaluating gives
Volume =
Z
2π
0
Z
0
π/2
ρ3
sin φ
3
p=b
π/2
dφdθ = 2π(− cos φ)
0
p=a
a3
b3
−
3
3
=
2π 3
(b − a3 ).
3
To check our answer, notice that the volume is the difference between the volumes of two half spheres of radius a and b.
These half spheres have volumes 2πb3 /3 and 2πa3 /3, respectively.
z
a
b
y
x
Figure 16.113
48. (a) We use the axes shown in Figure 16.114. Then the sphere is given by r 2 + z 2 = 25, so
Z Z Z √
2π
25−r 2
5
Volume =
0
−
1
√
r dzdrdθ.
25−r 2
(b) Evaluating gives
Volume = 2π
Z
1
z=
5
√
rz
z=−
25−r 2
dr = 2π
√
25−r 2
Z
5
2r
1
p
25 − r 2 dr
5
2
(25 − r 2 )3/2
3
1
√
4π
(24)3/2 = 64 6π = 492.5 mm3 .
=
3
= 2π
−
r 2 + z 2 = 25
x2 + y 2 + z 2 = 25
y
✠
5
1
Figure 16.114
5
x
1542
Chapter Sixteen /SOLUTIONS
49. (a) To find the mass, we integrate the density over the region, W . Converting to cylindrical coordinates, the surface of
the pile is z = 2 − r 2 , so we have
Mass =
Z
W
(2 − z) dV =
Z
2π
0
√
Z
0
2
Z
2−r 2
(2 − z)r dz dr dθ.
0
(b) Evaluating gives
Mass =
Z
2π
0
= 2π
Z
Z
√
2
0
√
2
= 2π
0
z2
2z −
2
2(2 − r 2 ) −
0
Z
√
2
2−r 2
r dr dθ
0
(2 − r 2 )2
2
r dr
√
r4
r6
2r 3
− r2 +
−
4r −
3
2
12
r5
dr = 2π
4 − 2r − 2r + 2r −
2
2
3
0
2
!
=
√
4
16 2
−
π.
3
3
50. The density function can be rewritten as δ(ρ, φ, θ) = ρ. So the mass is
Z
δ(P ) dV =
2π
Z
0
W
=
0
2π
Z
π/4
Z
0
=
81
4
=
81
4
0
π/4
Z
3
Z
ρ · ρ2 sin φ dρ dφ dθ
81
sin φ dφ dθ
4
Z
√
2
(−
+ 1) dθ
2
0
√
√
2
81
· 2π · (−
+ 1) =
π(− 2 + 2)
2
4
0
2π
51. We use spherical coordinates because we are integrating over a sphere and the density has spherical symmetry. D = 2ρ.
M=
Z
2π
0
Z
π
0
Z
3
(2ρ)ρ2 sin φ dρ dφ dθ.
0
52. Using cylindrical coordinates, the density is given by δ = kr 2 gm/cm3 , where k is a constant. Since δ = 2 when r = 2,
we have
2 = k22 so k = 0.5.
The equation of the sphere is x2 + y 2 + z 2 = 32 , and in cylindrical coordinates,
r 2 + z 2 = 9.
Thus r =
√
9 − z 2 on the sphere, so
2π
Z
3
Mass =
Z
2π Z
3
=
Z
Z
2π
3
0
0
=
−3
0
4
r
0.5
4
−3
0
1
=
8
Z √9−z2
Z
2π
=
8
Z
√
(0.5r 2 )r dr dz dθ
9−z 2
dz dθ
0
1
(9 − z 2 )2 dz dθ
8
−3
2π
3
0
Z
−3
(81 − 18z 2 + z 4 ) dz dθ
18z 3
z5
81z −
+
3
5
3
=
−3
324π
gm.
5
16.5 SOLUTIONS
1543
53. We use spherical coordinates. The density, δ, of the sphere at a distance ρ from the center is
δ = kρ2
for k a positive constant.
Thus, for a sphere of radius 1,
Mass =
Z
2π
0
π
Z
0
Z
= 2π
1
Z
kρ2 · ρ2 sin θ dρ dφ dθ
0
π
1
5
π
ρ
sin φ
5
k
0
dφ =
2πk
(− cos φ)
5
0
=
4πk
.
5
0
For a sphere of radius 2, a similar calculation gives
2π
Z
Mass =
0
π
Z
2
Z
0
kρ2 · ρ2 sin θ dρ dφ dθ
0
25
= 2πk (− cos φ)
5
π
=
128πk
.
5
0
Therefore
Ratio of masses =
4πk/5
1
=
.
128πk/5
32
54. (a) We use spherical coordinates. Since δ = 9 where ρ = 6 and δ = 11 where ρ = 7, the density increases at a rate
of 2 gm/cm3 for each cm increase in radius. Thus, since density is a linear function of radius, the slope of the linear
function is 2. Its equation is
δ − 11 = 2(ρ − 7) so δ = 2ρ − 3.
(b) Thus,
Mass =
Z
2π
0
(c) Evaluating the integral, we have
Mass = 2π
π
− cos φ
0
Z
π
0
Z
7
7
!
6
3ρ3
2ρ4
−
4
3
6
(2ρ − 3)ρ2 sin φ dρ dφ dθ.
= 2π · 2(425.5) = 1702π gm = 5346.991 gm.
55. (a) First we must choose a coordinate system, since none is given. We pick the xy-plane to be the fixed plane and the
z-axis to be the line perpendicular to the plane. Then the distance from a point to the plane is |z|, so the density at a
point is given by
Density = ρ = k|z|.
Using cylindrical coordinates for the integral, we find
Mass =
Z
2π
0
Z
a
0
Z √a2 −r2
−
√
k|z|r dzdrdθ.
a2 −r 2
(b) By symmetry, we can evaluate this integral over the top half of the sphere, where |z| = z. Then
√
Z 2π Z a 2 z= a2 −r2
Z 2π Z a Z √a2 −r2
z
kzr dzdrdθ = 2k
r
drdθ
Mass = 2
2 z=0
0
0
0
0
0
=k
Z
2π
0
= 2πk
Z
0
a
2
2
r(a − r ) drdθ = k2π
a4
a4
−
2
4
πka4
.
=
2
r2 2 r4
a −
2
4
a
0
1544
Chapter Sixteen /SOLUTIONS
56. The distance from a point (x, y, z) to the origin is given by
R p
p
x2 + y 2 + z 2 . Thus we want to evaluate
x2 + y 2 + z 2 dV
Vol(R)
R
where R is the region bounded by the hemisphere z =
We will use spherical coordinates.
p
8 − x2 − y 2 and the cone z =
p
x2 + y 2 . See Figure 16.115.
z
✛
✛
2
y
✛
x2 + y 2 = 4
x
Figure 16.115
In spherical coordinates, the quantity ρ goes from 0 to
(because the angle of the cone is π/4). Thus we have
x2 + y 2 + z 2 dV =
Z
=
Z
Z p
R
2π
0
0
=
2π
Z
Z
√
Z
Z
ρ(ρ2 sin φ) dρdφdθ
0
π/4
√
ρ4
sin φ ·
4
8
dφdθ
0
π/4
16 sin φ dφdθ
π/4
16(− cos φ)
0
=
8
0
2π
Z
π/4
0
0
=
Z
8, and θ goes from 0 to 2π, and φ goes from 0 to π/4
0
2π
Z
√
dθ
0
2π
16 1 −
0
√
2
dθ
2
√
2
π
2
√
From Problem 27 of Section 16.4 we know that Vol(R) = 32π( 2 − 1)/3, therefore
= 32 1 −
Average distance =
R p
x2 + y 2 + z 2 dV
Vol(R)
R
√
32 1 − 22 π
3
√
=
= √ .
[32( 2 − 1)π/3]
2
16.5 SOLUTIONS
1545
57. The total volume of the cone is 13 πr 2 h = 31 π · 12 · 1 = 31 π, so the total mass is 31 π (since the density is always 1). The
center of mass z-coordinate is given by
Z
3
z dV
z̄ =
π C
Using cylindrical coordinates to evaluate this integral gives
z̄ =
=
0
0
R
C
zr dr dz dθ
0
1
Z
z
Z
z3
dz dθ
2
0
2π
Z
1
Z
0
2π
Z
3
π
3
=
π
58. (a) The mass m of the cone is given by
2π
Z
3
π
1
3
dθ =
8
4
0
δ dV . In cylindrical coordinates this is
Z
m=
0
2π
5
z̄ =
π
Z
0
0
2π
z
Z
z 2 r dr dz dθ
0
1
Z
=
0
1
Z
Z
Z
=
2π
z4
dz dθ
2
π
1
dθ =
10
5
0
(b) The center of mass z-coordinate is given by
C
z · z 2 dV
Using cylindrical coordinates to evaluate this integral gives
5
π
z̄ =
5
π
=
5
π
=
2π
Z
0
0
2π
Z
Z
0
Z
1
Z
z
z 3 r dr dz dθ
0
1
0
2π
Z
z5
dz dθ
2
1
5
dθ =
12
6
0
Comparing this answer with the center of mass in Problem 57, where the density was constant, it makes sense
that the center of mass would be higher in this problem, since more mass is concentrated near the top of the cone.
59. We first need to find the mass of the solid, using cylindrical coordinates:
m=
2π
Z
0
=
0
=
1
0
2π
Z
Z
Z √z/a
0
r dr dz dθ
0
0
2π
Z
1
Z
z
dz dθ
2a
1
π
dθ =
4a
2a
It makes sense that the mass would vary inversely with a, since increasing a makes the paraboloid skinnier. Now for
the z-coordinate of the center of mass, again using cylindrical coordinates:
2a
z̄ =
π
Z
0
2π
Z
1
0
Z √z/a
0
zr dr dz dθ
1546
Chapter Sixteen /SOLUTIONS
Z
0
2π
Z
2a
π
=
2π
Z
2a
π
=
1
z2
dz dθ
2a
0
1
2
dθ =
6a
3
0
60. The volume of the hemisphere is 23 πa3 so its mass is 32 πa3 b. To find the location of the center of mass, we place the
base of the hemisphere on the xy-plane with the origin at its center, so we can describe it in spherical coordinates by
0 ≤ ρ ≤ a, 0 ≤ φ ≤ π2 and 0 ≤ θ ≤ 2π. Then the x-coordinate of the center of mass is, integrating using spherical
coordinates:
Z
3
x̄ =
2πa3 b
since the first integral
R 2π
0
a
0
π
2
Z
2π
Z
0
ρ sin(φ) cos(θ) · ρ2 sin(φ) dθ dφ dρ = 0
0
cos(θ) dθ is zero. A similar computation shows that ȳ = 0. Now for the z-coordinate:
a
Z
3
z̄ =
2πa3 b
0
3
=
· 2π
2πa3 b
=
=
a
Z
3
a3 b
Z
Z
0
Z
ρ cos(φ) · ρ2 sin(φ) dθ dφ dρ
a
Z
π
2
ρ3 cos(φ) sin(φ) dφ dρ
0
π
2
sin2 (φ)
2
ρ3
a
Z
2π
0
0
0
3
2a3 b
π
2
ρ3 dρ =
0
dρ
0
3a
8b
So the x and y-coordinates are located at the center of the base, while the z-coordinate is located
the base.
3a
8b
above the center of
61. The sum of the three moments of inertia I for the ball B will be
3
3I =
4πa3
3
=
4πa3
Z
Z
3
(y + z ) dV +
3
4πa
B
Z
2
3
(x + z ) dV +
3
4πa
B
2
2
2
2
2
Z
(x2 + y 2 ) dV
B
2
(2x + 2y + 2z ) dV,
B
which, in spherical coordinates is
3
2πa3
Z
(x2 + y 2 + z 2 ) dV =
B
3
= 3
a
=
Thus 3I =
6 2
a ,
5
so I =
3
2πa3
6
a3
Z
Z
0
a
Z0 a
0
a
Z
π
0
Z
Z
2π
0
π
ρ2 · ρ2 sin(φ) dθ dφ dρ
4
ρ sin(φ) dφ dρ
0
ρ4 dρ =
6 2
a .
5
2 2
a .
5
62. First we need to find the volume of the cone. In spherical coordinates we find:
V =
Z
a
0
Z
π
3
0
Z
0
2π
ρ2 sin(φ) dθ dφ dρ =
πa3
3
Now, to find the moment of inertia about the z-axis we need to compute the integral
this in spherical coordinates as
3
πa3
Z
W
3
x + y dV =
πa3
2
2
Z
0
a
Z
π
3
0
Z
0
3
πa3
R
W
x2 + y 2 dV . We can do
2π
(ρ2 sin2 (φ) cos2 (θ) + ρ2 sin2 (φ) sin2 (θ)) · ρ2 sin(φ) dθ dφ dρ
16.5 SOLUTIONS
3
πa3
=
Z
6
= 3
a
Z
0
a
0
6 5
= 3
a 24
aZ
Z
π
3
0
ρ4 sin3 (φ) dθ dφ dρ
0
π
3
ρ sin3 (φ) dφ dρ
4
0
a
Z
1547
2π
Z
a2
.
4
ρ4 dρ =
0
63. Using spherical coordinates,
Z bZ πZ
1
Stored energy =
2
a
q2
=
8πǫ
Z
0
b
2π
q2
ǫE ρ sin φ dθ dφ dρ =
32π 2 ǫ
2
0
2
1
q2
dρ
=
ρ2
8πǫ
a
1
1
−
.
a
b
Z bZ πZ
a
0
2π
0
1
sin φ dθ dφ dρ
ρ2
64. Use cylindrical coordinates, with the z-axis being the axis of the cable. Consider a piece of cable of length 1. Then
Stored energy =
=
Z bZ 1Z
1
2
a
q2
4πǫ
0
b
Z
a
2π
ǫE 2 r dθ dz dr =
0
q2
8π 2 ǫ
Z bZ 1Z
a
0
2π
0
1
dθ dz dr
r
1
q2
q2
b
dr =
(ln b − ln a) =
ln .
r
4πǫ
4πǫ a
So the stored energy is proportional to ln(b/a) with constant of proportionality q 2 /4πǫ.
65. The surface z = 4 − x2 − y 2 cuts the xy-plane in the circle x2 + y 2 = 4. Thus
2
2
Z
Z Z √
4−x2
2
Mass =
−2
−
√
4−x −y
e−x−y dz dy dx gm.
4−x2
0
66. The cylinder has radius 2. Using cylindrical coordinates to find the mass and integrating with respect to r first, we have
Mass =
Z
0
2π
Z
0
3
Z
2
(1 + r)r dr dz dθ =
Z
0
0
2π
Z
0
3
r2
r3
+
2
3
2
dz dθ = 2π · 3 ·
0
67. In cylindrical coordinates, the density, δ is given by δ = kr for some positive constant k.
For the smaller cylinder, x2 + y 2 ≤ 1, 0 ≤ z ≤ 2, whose radius is 1,
Mass =
Z
2π
0
2
Z
2
0
Z
1
0
kr 3
kr · r dr dz dθ = 2π · 2 ·
3
1
=
4πk
.
3
0
2
For the larger cylinder, x + y ≤ 9, 0 ≤ z ≤ 2, whose radius is 3,
Mass =
Z
0
Thus, the ratio of the masses is
2π
Z
2
0
1
4πk/3
=
.
36πk
27
Z
3
0
kr 3
kr · r dr dz dθ = 2π · 2 ·
3
3
= 36πk.
0
4
8
+
2
3
= 28π gm.
1548
Chapter Sixteen /SOLUTIONS
68. Integrating with respect to z first, we have
Z Z Z √
1
2π
W =
0
9−r 2 )−1
(
√
0
r dz dθdr +
1−r 2
1
or integrating with respect to r first, we have
Z Z Z √
1
0
9−(z+1)2
2π
√
0
√
2 2
Z
Z
r dr dθ dz +
1−z 2
2
1
Z
2π
0
Z
2π
0
Z (√9−r2 )−1
r dz dθ dr
0
Z √9−(z+1)2
r dr dθ dz.
0
69. Assume the base of the cylinder sits on the xy-plane with center at the origin. Because the cylinder is symmetric about the
z-axis, the force in the horizontal x or y direction is 0. Thus we need only compute the vertical z component of the force.
We are going to use cylindrical coordinates; since the force is G · mass/(distance)2 , a piece of the cylinder of volume
dV located at (r, θ, z) exerts on the unit mass a force with magnitude G(δ dV )/(r 2 + z 2 ). See Figure 16.116.
Vertical component
=
of force
G(δ dV )
Gδ dV
z
Gδz dV
· cos φ = 2
·√
= 2
.
r2 + z 2
r + z2
(r + z 2 )3/2
r2 + z 2
Adding up all the contributions of all the dV ’s, we obtain
Vertical force =
H
Z
0
=
H
Z
0
=
H
Z
0
H
Z
Z
Z
2π
0
Z
R
0
2π
0
1
(Gδz) − √
2
r + z2
2π
0
(Gδz) ·
2πGδ
= 2πGδ(z −
p
=
Z
Gδzr
drdθdz
(r 2 + z 2 )3/2
0
−√
1− √
= 2πGδ(H −
√
R
dθdz
0
1
1
+
z
R2 + z 2
z
R2 + z 2
H
dθdz
dz
R2 + z 2 )
0
p
R2
+
H2
+ R) = 2πGδ(H + R −
p
R2 + H 2 )
r2 + z 2
z
φ
Figure 16.116
70. The charge density is δ = kz, where k is a constant. In cylindrical coordinates,
Total charge =
Z
δ dV =
Cylinder
= kπ
Z
0
h
Z hZ
0
0
R
Z
R2 z dz = k(πR2 )
2π
kzr dθ dr dz = k
0
h2
kπ 2 2
=
R h .
2
2
Z hZ
Thus, the total charge is proportional to R2 h2 with constant of proportionality kπ/2.
0
0
R
2πzr dr dz
16.6 SOLUTIONS
1549
71. The charge density is δ = k/ρ. Integrating in spherical coordinates,
2π
Z
Total charge =
0
Z πZ
0
R
k 2
ρ sin φ dρ dφ dθ = k
ρ
0
R2
= 4πk
= 2πkR2 .
2
Z
2π
0
Z
π
0
R2
sin φ dφ dθ
2
Thus, the total charge is proportional to R2 with constant of proportionality 2πk.
72. In the system used in this book the volume element is dV = ρ2 sin φ dρ dφ dθ. In the system shown in the problem, φ
and θ have been interchanged and ρ changed to r. So the volume element is dV = r 2 sin θ dr dθ dφ.
Strengthen Your Understanding
Z π Z 2π Z 1
ρ2 sin φ dρ dθ dφ
73. (c)
0
0
0
74. The integral is missing part of the volume element in spherical coordinates. The integral
Z
0
gives the volume inside the sphere of radius 1.
2π
Z
0
π
Z
1
ρ2 sin φ dρ dφ dθ
0
75. We cannot switch the order of integration without rewriting the limits of integration if doing so produces limits that are
not constant on the outer integral.
76. In spherical coordinates, the upper half of a sphere of radius r is given by 0 ≤ ρ ≤ r, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π/2.
Therefore, for a hemisphere of radius 5, we have
Volume =
Z
2π
0
Z
π/2
0
5
Z
ρ2 sin φ dρ dφ dθ.
0
77. Let W be the unit ball x2 + y 2 + z 2 ≤ 1. It is not easy to integrate
Z p
x2 + y 2 + z 2 dx dy dz.
W
In spherical coordinates, this integral becomes easy to integrate:
Z
2π
0
Z
π
0
Z
1
ρρ2 sin φ dρ dφ dθ =
0
Z
2π
0
Z
0
π
Z
1
ρ3 sin φ dρ dφ dθ.
0
Solutions for Section 16.6
Exercises
1. Yes, p is a joint density function. The values of p(x, y) are nonnegative, since p(x, y) = 1/2 for all points inside R and
p(x, y) = 0 for all other points. The volume under the graph of p over the region R is (1/2)(5 − 4)(0 − (−2)) = 1.
2. No, p is not a joint density function. Since p(x, y) = 0 outside the region R, the volume under the graph of p is the same
as the volume under the graph of p over the region R, which is 2 not 1.
3. No, p is a not joint density function, because p(x, y) < 0 for some points (x, y) in the region R. For example,
p(−0.7, 0.1) = −0.6.
4. Yes, p is a joint density function. Since x ≤ y everywhere in the region R, we have p(x, y) = 6(y − x) ≥ 0 for all x
and y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volume
under the graph of p over the region R is 1:
Z
R
p(x, y) dA =
Z
1
0
Z
0
y
6(y − x) dx dy =
Z
0
1
6 yx −
x2
2
y
dy =
0
Z
0
1
1
3y 2 dx = y 3
= 1.
0
1550
Chapter Sixteen /SOLUTIONS
5. Yes, p is a joint density function. In the region R we have 1 ≥ x2 + y 2 , so p(x, y) = (2/π)(1 − x2 − y 2 ) ≥ 0 for all x
and y in R, and p(x, y) = 0 for all other (x, y). To check that p is a joint density function, we check that the total volume
under the graph of p over the region R is 1. Using polar coordinates, we get:
Z
p(x, y)dA =
R
Z
2
π
2π
0
1
Z
2
π
(1 − r 2 )r dr dθ =
0
Z
2π
0
1
r2
r4
−
2
4
dθ =
0
2
π
2π
Z
1
dθ = 1.
4
0
6. Yes, p is a joint density function. Since e−x−y is always positive, p(x, y) = xye−x−y ≥ 0 for all x and y in R, and
hence for all x and y. To check that p is a joint density function, we check that the total volume under the graph of p over
the region R is 1. Since e−x−y = e−x e−y , we have
Z
xye
−x−y
dA =
∞
Z
0
R
∞
Z
xye
−x−y
Z
dx dy =
∞
ye
−y
0
0
∞
Z
xe
−x
dx
0
dy.
Using integration by parts:
Z
xe−x dx = lim (−xe−x − e−x )
b→∞
0
Thus
Z
b
∞
xye−x−y dA =
Z
∞
ye−y
0
R
Z
0
= (0 − 0) − (0 − 1) = 1.
∞
xe−x dx
0
dy =
Z
∞
ye−y dy = 1.
0
7. We have p(x, y) = 0 for all points (x, y) satisfying x ≥ 3, since all such points lie outside the region R. Therefore the
fraction of the population satisfying x ≥ 3 is 0.
8. The fraction is 0, since
R1
1
R∞ R1
xy dx = 0, so
−∞
p(x, y) dx dy =
1
R1R1
0
1
xy dx dy=0.
9. Since x + y ≤ 3 for all points (x, y) in the region R, the fraction of the population satisfying x + y ≤ 3 is 1.
10. Since p(x, y) = 0 for any (x, y) with x < 0 and also p(x, y) = 0 for any (x, y) with y > 1 or y < 0, the fraction of the
population is given by the double integral:
Z
1
0
1
Z
xy dx dy =
Z
1
0
0
1
x2 y
2
dy =
1
Z
0
0
1
y2
y
dy =
2
4
=
0
1
.
4
11. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p over
the region inside the rectangle R and to the right of the line x = y. Therefore the fraction of the population is given by
the double integral:
Z
0
1
Z
2
xy dx dy =
Z
1
0
y
2
x2 y
2
dy =
Z
1
y3
2
2y −
0
y
dy =
y2 −
y4
8
1
=
0
7
.
8
12. Since p(x, y) = 0 for all (x, y) outside the rectangle R, the population is given by the volume under the graph of p over
the region inside the rectangle R and below the line x + y = 1. This is the same as the region bounded by the x-axis, the
y-axis, and the line x + y = 1. Therefore the fraction of the population is given by the double integral:
Z
0
1
Z
1−y
xy dx dy =
0
Z
1
0
x2 y
2
1−y
dy =
Z
1
0
0
(1 − y)2 y
dy =
2
y3
y4
y2
−
+
4
3
8
13. The fraction of the population is given by the double integral:
Z
0
1/2
Z
1
xy dx dy =
0
Z
0
1/2
x2 y
2
1
dy =
0
Z
0
1/2
y2
y
dy =
2
4
1/2
=
0
1
.
16
1
=
0
1
.
24
16.6 SOLUTIONS
1551
14. We are looking for points inside the circle x2 + y 2 = 1 and inside the rectangle R. In the first quadrant, all of the circle
and its interior lies inside the rectangle R. Thus the fraction of the population we want is given by the volume under the
graph of p over the region inside the circle x2 + y 2 = 1 in the first quadrant. We evaluate this double integral using polar
coordinates:
Z
0
π/2
Z
1
(r cos θ)(r sin θ) r dr dθ =
Z
π/2
0
0
r4
cos θ sin θ
4
1
1
dθ =
4
0
Z
π/2
cos θ sin θ dθ.
0
Making the substitution w = sin θ, we get:
Z
π/2
cos θ sin θ dθ =
Z
1
w dw =
0
0
1
.
2
Thus the fraction is (1/4)(1/2) = 1/8.
Problems
15. (a)
1
Z
0
Z
1
1/3
2
(x + 2y) dx dy =
3
=
1
Z
0
Z
2 1 2
( x + 2xy)
3 2
1
1/3
dy
1
1
2
2 1
( + 2y) − (
+ y) dy
3 2
18
3
Z
1
0
h
i
2
4
4
=
( + y) dy
3 0 9
3
2 2 1
2 4
y+ y
=
0
3 9
3
20
2 10
=
.
=
3 9
27
(b) It is easier to calculate the probability that x < (1/3) + y does not happen, that is, the probability that x ≥ (1/3) + y,
and subtract it from 1. The probability that x ≥ (1/3) + y is
Z
1
1/3
Z
0
x−(1/3)
2
(x + 2y) dy dx =
3
Z
=
2
3
1
1/3
2
=
3
Z
2
(xy + y 2 )
3
1
1/3
Z
(x(x −
x−(1/3)
0
1
1
) + (x − )2 ) dx
3
3
1
1/3
dx
(2x2 − x +
1
) dx
9
1
1 1
2 2
= ( x3 − x2 + x) 1/3
3 3
2
9
i
h
1
1
2
1
1
2 2
−
+
)
=
( − + )−(
3 3
2
9
81
18
27
= 44/243.
Thus, the probability that x < (1/3) + y is 1 − (44/243) = 199/243.
1552
Chapter Sixteen /SOLUTIONS
1
3
x<
+y
1
3
Figure 16.117
16. (a) We know that
R∞ R∞
−∞
−∞
f (x, y)dydx = 1 for a joint density function. So,
1=
∞
Z
−∞
Z
∞
1
Z
f (x, y)dydx =
−∞
0
1
Z
kxydydx
x
=
1
k
8
y=
√
hence k = 8.
√
(b) The region where x < y < x is sketched in Figure 16.118
y
y=x
1
x
x
0
1
Figure 16.118
So the probability that (x, y) satisfies x < y <
Z
0
1Z
√
x is given by:
√
x
8xy dy dx =
x
1
Z
√
4x(y 2 )
0
=
Z
dx
1
4x(x − x2 )dx
0
=4
x
x
1 3 1 4
x − x
3
4
1
1
=4
−
3
4
1
=
3
1
0
This tells us that in choosing points from the√region defined by 0 ≤ x ≤ y ≤ 1, that 1/3 of the time we would
pick a point from the region defined by x < y < x. These regions are shown in Figure 16.118.
17. (a) For a density function,
1=
Z
∞
−∞
Z
∞
f (x, y) dy dx =
−∞
Z
0
2
Z
0
1
kx2 dy dx
16.6 SOLUTIONS
=
Z
1553
2
kx2 dx
0
kx3
3
=
2
=
0
8k
.
3
So k = 3/8.
(b)
Z
1
Z
0
(c)
2−y
0
Z
0
Z
1
3 2
x dx dy =
8
Z
3 2
x dx dy =
8
1/2 Z
1
0
0
−1
1
(2 − y)3 dy =
(2 − y)4
8
32
1/2
0
1 3
x
8
1
dy =
0
1/2
Z
0
1
0
=
15
32
1
1
dy =
.
8
16
18. (a) The area of S is (2)(4) = 8. Because the density function p(x, y) is constant on S and the total volume under a
density function above the xy-plane is 1, p(x, y) = 1/8 for (x, y) in S, and p(x, y) = 0 for (x, y) outside S.
(b) The probability that (x, y) is in T is
Z
f (x, y) dy dx =
T
1
8
Z
19. Since
XX
x
y
area(T )
α
= .
8
8
dy dx =
T
Z
f (x, y) ∆x ∆y ≈
f (x, y) dx dy
R
and since x never exceeds 1, and we can assume that no one lives to be over 100, so y does not exceed 100, we have
Fraction of
policies
=
Z
f (x, y) dx dy =
100
Z
65
R
Z
1
f (x, y) dx dy,
0.8
where R is the rectangle: 0.8 ≤ x ≤ 1, 65 ≤ y ≤ 100.
20. (a) Since the exponential function is always positive and λ is positive, p(t) ≥ 0 for all t, and
b
∞
Z
p(t)dt = lim −e−λt
b→∞
0
= lim −e−bt + 1 = 1.
b→∞
0
(b) The density function for the probability that the first substance decays at time t and the second decays at time s is
p(t, s) = λe−λt µe−µs = λµe−λt−µs ,
for s ≥ 0 and t ≥ 0, and is zero otherwise.
(c) We want the probability that the decay time t of the first substance is less than or equal to the decay time s of the
second, so we want to integrate the density function over the region 0 ≤ t ≤ s. Thus, we compute
Z
∞
0
∞
Z
λµe−λt e−µs ds dt =
Z
∞
λe−λt (−e−µs )
0
t
=
=
Z
∞
t
dt
∞
λe−λt e−µt dt
Z0 ∞
λe(−λ+µ)t dt
0
=
−λ −(λ+µ)t
e
λ+µ
∞
0
=
λ
.
λ+µ
So for example, if λ = 1 and µ = 4, then the probability that the first substance decays first is 1/5 .
21. (a)
Z
π
6
θ=0
(b)
Z
π+ π
6
12
θ= π
6
Z
Z
4
4
1
r= cos
θ
p(r, θ)r dr dθ
1
r= cos
θ
p(r, θ)r dr dθ +
Z
2π
6
π
θ= π
+ 12
6
Z
4
1
r= sin
θ
p(r, θ)r dr dθ
1554
Chapter Sixteen /SOLUTIONS
22. (a) If t ≤ 0, then F (t) = 0 because the average of two positive numbers can not be negative. If 1 < t then F (t) = 1
because Rthe average of two numbers each at most 1 is certain to be less than or equal to 1. For any t, we have
F (t) = R p(x, y)dA where R is the region of the plane defined by (x + y)/2 ≤ t. Since p(x, y) = 0 outside the
unit square, we need integrate only over the part of R that lies inside the square, and since p(x, y) = 1 inside the
square, the integral equals the area of that part of the square. Thus, we can calculate the area using area formulas.
For 0 ≤ t ≤ 1, we draw the line (x + y)/2 = t, which has x- and y-intercepts of 2t. Figure 16.119 shows that for
0 < t ≤ 1/2,
1
F (t) = Area of triangle = · 2t · 2t = 2t2 .
2
In Figure 16.120, when x = 1, we have y = 2t − 1. Thus, the vertical side of the unshaded triangle is 1 − (2t − 1) =
2 − 2t. The horizontal side is the same length, so for 1/2 < t ≤ 1,
F (t) = Area of Square − Area of triangle = 12 −
The final result is:
0
2t2
F (t) =
1 − 2(1 − t)2
1
if t ≤ 0
if 0 < t ≤ 1/2
.
if 1/2 < t ≤ 1
if 1 < t
y
2t
y
1
x+y
2
1
(2 − 2t)2 − 1 − 2(1 − t)2 .
2
1
=t
2t
2t
✛ 2 − 2t ✲
✛✲
✻
2t − 1
2 − 2t
2t − 1✻
❄
x
1
1
Figure 16.119: For 0 < t ≤
Figure 16.120: For
1
2
❄
x+y
2
=t
x
2t
1
0, and so its integral
x2 + y 2 is positive, so its integral over the solid W is positive.
33. Negative. If (x, y, z) is any point inside the cone then z < 2. Hence the function z − 2 is negative on W and so is its
integral.
34. Zero. y is positive on the half of the half-cone above the first quadrant in the xy-plane and negative (of equal absolute
value) on the half of the half-cone above the fourth quadrant. The integral of y over W is zero because the integrals over
each half add up to zero.
35. Zero. You can see this in several ways. One way is to observe that xy is positive on part of the cone above the first
quadrant (where x and y are of the same sign) and negative (of equal absolute value) on the part of the cone above the
fourth quadrant (where x and y have opposite signs). These add up to zero in the integral of xy over all of W .
Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first
with respect to y. For fixed x and z, the y-integral is over an interval symmetric about 0. The integral of y over such an
interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.
36. Zero. Write the triple integral as an iterated integral, say integrating first with respect to y. For fixed x and z, the y-integral
is over an interval symmetric about 0. The integral of y over such an interval is zero. If any of the inner integrals in an
iterated integral is zero, then the triple integral is zero.
37. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive.
38. (a) The top of the tetrahedron is z = x − y + 2, and its triangular base is in the second quadrant of the xy-plane bounded
by the x- and y-axes, and the line 0 = x − y + 2, or y = 2 + x. Thus the volume is given by
V =
Z
0
−2
2+x
Z
0
Z
x−y+2
dz dy dx.
0
Other orders of integration are possible.
(b) Evaluating gives
V =
Z
0
−2
0
=
Z
−2
=
Z
0
−2
2+x
Z
0
(x − y + 2) dy dx =
x(2 + x) −
Z
0
−2
xy −
y2
+ 2y
2
2+x
dx
0
1
(2 + x)2 + 2(2 + x) dx
2
x2
+ 2x + 2
2
dx =
x3
+ x2 + 2x
6
0
=
−2
4
.
3
39. (a) Since z is positive above the xy-plane and negative below the xy-plane, the contributions to
cancel; both R
these integrals are zero. Hence (i) and (iii) are zero.
The integral T z is positive.
(b) The equation of the sphere is x2 + y 2 + z 2 = 1, so the top half is z =
to z first, we have
Z √
Z
Z Z √
1−x2
1
zdV =
T
−1
−
√
p
1−x2 −y 2
0
B
dV and
R
R
zdV
1 − x2 − y 2 . Thus, integrating with respect
z dz dy dx
1−x2
R
SOLUTIONS to Review Problems for Chapter Sixteen
Z
1
1
=
2
Z
1
=
2
Z
1
2
Z
=
Z √1−x2
=
√
1−x2
Z √
−
−1
1
−1
z2
2
√
1−x2 −y 2
dy dx
0
1−x2
−
√
1−x2
(1 − x2 − y 2 ) dy dx
1
y3
y−x y−
3
−1
√
2
1
p
2
−1
= 1.571
1565
−
1−x2
√
dx
1−x2
1 − x2 − 2x2
p
1 − x2 +
2
(1 − x2 )3/2
3
dx
40. R is one eighth of a sphere of radius 1, below the xy-plane and under the first quadrant. See Figure 16.139.
y
z
x
Figure 16.139
41. (a) The region of integration is the region between the cone z = r, the xy-plane and the cylinder r = 3. In spherical
coordinates, r = 3 becomes ρ sin φ = 3, so ρ = 3/ sin φ. The cone is φ = π/4 and the xy-plane is φ = π/2. See
Figure 16.140. Thus, the integral becomes
Z
2π
0
Z
π/2
π/4
Z
3/ sin φ
ρ2 sin φ dρdφdθ.
0
z
z=r
r=3
π
4
x
3
Figure 16.140: Region of integration is
between the cone and the xy-plane
(b) The original integral is easier to evaluate, so
Z
0
2π
Z
0
3
Z
0
r
r dzdrdθ =
Z
2π
0
Z
0
3
z=r
drdθ =
zr
z=0
Z
0
2π
Z
0
3
r 2 drdθ = 2π ·
r3
3
3
= 18π.
0
1566
Chapter Sixteen /SOLUTIONS
42. The region stands on a rectangular base in the xy-plane, with vertical sides and a slanting top, the plane z = 1 + x. See
Figure 16.141. The integral is
Z Z Z
2
1
1+x
f (x, y, z) dz dy dx.
0
0
0
The order of integration can be altered.
z = 1+x
❄
z
1
y
2
x
Figure 16.141
43. The region is a solid ring between the planes z = 2 and z = 3, with inner radius r =
Figure 16.142. In cylindrical coordinates, the integral is
Z
2π
0
3
Z
Z
5 and outer radius r =
√
6. See
√
6
√
2
√
r dr dz dθ.
5
The order of integration can be changed.
z
z=3
z=2
Figure 16.142
44. The region is a hollow half-sphere, with inner radius
coordinates, the integral is
Z πZ πZ
0
0
√
3 and outer radius
2
√
ρ2 sin φ dρ dφ dθ.
3
√
4 = 2. See Figure 16.143. In spherical
SOLUTIONS to Review Problems for Chapter Sixteen
1567
The order of integration can be altered.
z
y
√
3
2
❄✠
x
Figure 16.143
45. The region stands on a circular base of radius 1 in the xy-plane and has cylindrical sides. The top is part of a sphere
of
√
radius 3. The cylinder meets the sphere where x2 + y 2 = 1 and x2 + y 2 + z 2 = 9, so 1 + z 2 = 9, z = 8. See
Figure 16.144. In cylindrical coordinates, the integral is
Z Z Z √
2π
9−r 2
1
r dz dr dθ.
0
z
0
0
Sphere: x2 + y 2 + z 2 = 9
✠
Cylinder: x2 + y 2 = 1
✠
x
Figure 16.144
46. Positive, since e−x is always positive.
47. Negative, since y 3 is negative on B, where y < 0.
48. Positive, since (x + y 2 ) is positive on R, where x > 0.
49. Can’t tell, since y 3 is both positive and negative for x < 0.
50. Can’t tell, since x < 0 and y 2 ≥ 0 on L, where x < 0.
51. Zero. The solid sphere is symmetric and z is positive on the top half and negative (of equal absolute value) on the bottom
half. The integral of z over the entire solid is zero because the integrals over each half add up to zero.
52. Zero. x is positive on the hemisphere x2 +y 2 +z 2 ≤ 1, x > 0 and negative (of equal absolute value) on x2 +y 2 +z 2 ≤ 1,
x < 0. The integral of x over the entire solid is zero because the integrals over each half add up to zero.
53. Zero. You can see this in several ways. One way is to observe that xy is positive on the part of the sphere above and below
the first and third quadrants (where x and y are of the same sign) and negative (of equal absolute value) on the part of the
sphere above and below the second and fourth quadrants (where x and y have opposite signs). These add up to zero in the
integral of xy over all of W .
Another way to see that the integral is zero is to write the triple integral as an iterated integral, say integrating first
with respect to x. For fixed y and z, the x-integral is over an interval symmetric about 0. The integral of x over such an
interval is zero. If any of the inner integrals in an iterated integral is zero, then the triple integral is zero.
1568
Chapter Sixteen /SOLUTIONS
54. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. Then sin( π2 xy) is integrated
over an integral symmetric about the origin, and this integral is zero because sin( π2 xy) is an odd function. Since the
innermost integral is zero so is the triple integral.
55. Zero. Write the triple integral as an iterated integral, say integrating first with respect to x. For fixed y and z, the x-integral
is over an interval symmetric about 0. The integral of x over such an interval is zero. If any of the inner integrals in an
iterated integral is zero, then the triple integral is zero.
56. Positive. The function e−xyz is a positive function everywhere so its integral over W is positive.
57. Negative. Since z 2 − 1 ≤ 0 in the sphere, its integral is negative.
58. Positive.
p
x2 + y 2 + z 2 is positive on W , so its integral is positive.
59. The integral is over the region 0 ≤ x2 + y 2 ≤ 3, 1 ≤ z ≤ 4 − x2 − y 2 . Using cylindrical coordinates, we get
2π
Z
0
√
Z
3
0
4−r 2
Z
1
rdz dr dθ =
z2
1
2π
Z
0
=
Z
=
Z
0
2π
0
Z
r
(− )
z
√
3
(−
0
2π
0
2π
Z
√
3
Z
4−r 2
dr dθ
1
r
r
+ )dr dθ
4 − r2
1
1
1
ln(4 − r 2 ) + r 2
2
2
√3
dθ
0
3
1
1
( ln 1 + − ln 4 − 0)dθ
2
2
2
0
3
= 2π( − ln 2) = π(3 − 2 ln 2)
2
=
60. The integral is over the region x, y ≥ 0, x2 + y 2 ≤ 1, 0 ≤ z ≤
Z
0
π/2
Z
0
1
Z
r
(z + r) rdz dr dθ =
Z
p
x2 + y 2 . Using cylindrical coordinates, we get
π/2
0
0
=
Z
=
π/2
Z
0
π/2
0
=
1
Z
0
0
Z
Z
r
(rz + r 2 ) dz dr dθ
0
1
1
( r 3 + r 3 ) dr dθ
2
3 4
r
8
1
dθ
0
3π
3 π
· =
8 2
16
61. The region is a hemisphere 0 ≤ x2 +y 2 +z 2 ≤ 32 , z ≥ 0, so spherical coordinates are appropriate. Recall the conversion
formula x = ρ sin φ cos θ. Then the integral in spherical coordinates becomes
Z
2π
0
=
Z
2π
Z
2π
0
243
=
5
243
5
Z
Z
π/2
π/2
0
0
ρ4 sin3 φ cos2 θ dρ dφ dθ
243
sin3 φ cos2 θ dφ dθ
5
Z
π/2
cos2 θ · sin φ(1 − cos2 φ) dφ dθ
0
2π
2π
1
cos θ − cos φ + cos3 φ
3
2
0
Z
3
Z
0
0
2π
Z
(ρ sin φ cos θ)2 ρ2 sin φ dρ dφ dθ
0
0
243
=
5
=
Z
3
Z
0
0
=
π/2
Z
cos2 θ[−(−1) +
1
(−1)] dθ
3
π2
0
dθ
SOLUTIONS to Review Problems for Chapter Sixteen
=
=
243 2
·
5
3
Z
2π
0
1569
1 + cos 2θ
dθ
2
1
81
(θ + sin 2θ)
5
2
2π
81
162π
(2π + 0) =
5
5
=
0
62. W is a cylindrical shell, so cylindrical coordinates should be used. See Figure 16.145.
z
✛1✲✛1✲
✻
4
❄
y
x
Figure 16.145
Z
W
z
dV =
(x2 + y 2 )3/2
=
4
Z
0
0
=
0
=
0
=
Z
Z
Z
0
2
z
rdr dθ dz
r3
Z
2
z
dr dθ dz
r2
1
2π
2π
0
4
Z
1
2π
0
4
Z
Z
0
4
Z
2π
0
4
Z
Z
z
(− )
r
2
dθ dz
1
z
dθ dz
2
1
z
· 2π dz = π · z 2
2
2
4
= 8π
0
63. (a) The region (shaded) is one eighth of the circle x2 + y 2 = 8; see Figure 16.146. The first integral is above the dashed
line y = 2; the second integral is below the dashed line.
y
y=x
√
8
2
x2 + y 2 = 8
√ x
8
2
Figure 16.146
(b) Converting to polar coordinates, we find the quantity in part (a) is given by
Z √ Z √
Z Z
8−y 2
8
e−x
2
0
2
−y 2
2
y
e−x
dx dy +
0
0
2
−y 2
dx dy
1570
Chapter Sixteen /SOLUTIONS
π/2
Z
=
π/4
Z
2
e−r r dr dθ
0
π
π
−
2
4
=
√
8
1
− e−r
2
2
√
8
=
0
π
4
1
2
e0 −
π
1 −8
=
1 − e−8 .
e
2
8
64. The depth of the lake is given in meters and the diameter in kilometers. We should work with a single unit of length. In
this solution we work with kilometers, but meters would work just as well.
The shape of the lake suggests integration in polar coordinates, with r km measured from the center of the island.
Thus t = r − 1 is the distance in kilometers from the island when r varies between 1 and 5. The depth of the lake r km
from the center of the island is
Depth =
100(r − 1)(4 − (r − 1))
1
3r
r2
=− +
−
km.
1000
2
5
10
Volume of the lake =
2π
Z
0
Z
5
1
−
1
3r
r2
+
−
2
5
10
r dr dθ =
32π
= 20.1 km3 .
5
65. (a) The equation of the curved surface of this half cylinder along the x-axis is (y − 1)2 + z 2 = 1. The part we want is
z=
(b) The integral
Z
p
1 − (y − 1)2
f (x, y, z) dV =
D
Z
10
0
0 ≤ y ≤ 2 0 ≤ x ≤ 10.
2
Z
Z √1−(y−1)2
f (x, y, z) dz dy dx.
0
0
66. Figure 16.147 shows a slice through the region for a fixed x. Breaking the volume into small cubes each of volume
∆V = ∆x ∆y ∆z and stacking the cubes above (x, y, 0), starting at z = 0 and going up to z = x + y, tells us the inner
integral is
Z
x+y
dz.
0
Lining up the stacks parallel to the z axis gives a slice, for each fixed value of x, from y = 0 to x + y = 5, thus the middle
integral is
Z
Z
5−x
x+y
dz dy.
0
0
Finally, adding up the contributions for x = 0 to x = 5 gives the volume, V , as
V =
Z
5
0
=
Z
0
5
Z
5−x
0
(xy +
Z
x+y
dz dy dx =
Z
5
0
0
1 2
y )
2
y=5−x
dx =
y=0
Z
0
Z
5−x
(x + y) dy dx
0
5
x(5 − x) +
z
z = x+y
y
Figure 16.147
1
(5 − x)2
2
dx =
125
.
3
SOLUTIONS to Review Problems for Chapter Sixteen
1571
67. (a) The region is the half cylinder in Figure 16.148.
z
1
1
x
y
1
Figure 16.148
(b) Use cylindrical coordinates with x replacing z and y in place of x and z in place of y. Then
Z
1
−1
Z
1
−1
Z √1−z2
1
Z
f (x, y, z) dy dz dx =
0
−1
Z
π/2
Z
−π/2
1
1
0
r 3 · r dr dθ dx = x
π/2
θ
−1
−π/2
r5
5
1
=
0
2π
.
5
68. (a) The density increases at a rate of (25 − 1)/12 = 2 gm/cm2 for each cm of radius. Thus, at radius r,
Density = 1 + 2r gm/cm2 .
Thus
Mass =
Z
2π
0
(b) Evaluating gives
Mass =
Z
2π
0
Z
r2
2
+ r3
2
3
12
(1 + 2r) r dr dθ gm.
0
12
0
dθ = 2π · 1224 = 2448π gm.
69. Orient the region as shown in Figure 16.149 and use cylindrical coordinates with the origin at the center of the sphere.
The equation of the sphere is x2 + y 2 + z 2 = 25, or r 2 + z 2 = 25. If z = 3, then r 2 + 32 = 25, so r 2 = 16 and r = 4.
√
√
Volume =
Z
2π
Z
2π
0
=
0
= 2π
Z
4
Z
4
0
0
25−r 2
Z
r dzdrdθ =
Z
2π
0
3
Z
(r
25 − r 2 − 3r) drdθ = θ
125
27
−
3
3
drdθ
3r 2
(25 − r 2 )3/2
−
3
2
52π
− 24 =
= 54.45 cm3 .
3
z
5
2
Sphere
r 2 + z 2 = 25
✻
✠
❄
✻
5
3
❄
✛
4
✲
Figure 16.149
r
25−r 2
z=3
−
0
rz
0
2π
p
z=
4
4
0
1572
Chapter Sixteen /SOLUTIONS
70. The region of integration is shown in Figure 16.150, and the mass of the given solid is given by
z
12
z = 12 − 4x − 3y
4
y
4x + 3y = 12
✛
or y = 1
(12 − 4x)
3
3
x
Figure 16.150
Mass =
Z
δ dV
R
Z
3
Z
3
Z
3
=
3
=
Z
=
0
0
=
1 (12−4x)
3
Z
Z
1 (12−4x)
3
12−4x−3y
x2 dz dy dx
0
z=12−4x−3y
2
x z
0
0
0
Z
dydx
z=0
1 (12−4x)
3
0
0
Z
x2 (12 − 4x − 3y) dydx
3
x2 (12y − 4xy − y 2 )
2
y= 1
(12−4x)
3
dx
0
3
8 5
x
= 8x − 4x +
15
3
4
108
.
=
5
0
71. Orient the region as shown in Figure 16.151 and use spherical coordinates with the origin at the center of the sphere. The
equation of the sphere is x2 + y 2 + z 2 = 25, or ρ = 5. The plane z = 3 is the plane ρ cos φ = 3, so ρ = 3/ cos φ. In
Figure 16.151, angle AOB is given by
cos φ =
3
,
5
so
φ = arccos(3/5).
The volume is given by
V =
Z
2π
arccos(3/5)
0
0
= 2π
Z
Z
arccos(3/5)
0
Z
2π
5
2
ρ sin φ dρdφdθ = θ
3/ cos φ
9
125
−
3
cos3 φ
0
sin φ dφ
Z
0
arccos(3/5)
ρ3
sin φ
3
p=5
dφ
p=3/ cos φ
SOLUTIONS to Review Problems for Chapter Sixteen
= 2π
Z
arccos(3/5)
0
= 2π
125
sin φ dφ −
3
125
cos φ
−
3
arccos(3/5)
0
arccos(3/5)
Z
0
1
φ
−9
2 cos2
1
9
125 3
−1 −
−1
= 2π −
3
5
2 (3/5)2
2
9 16
125
−
−
= 2π −
3
5
2 9
50
52π
= 2π
−8 =
≈ 54.45 cm3 .
3
3
9
sin φ dφ
cos3 φ
z
3
0
!
Sphere
ρ=5
✻
✠
❄A
✻
✛
2
arccos(3/5)
1573
B
5
φ
❄✛
x
O
Figure 16.151
72. Let the lower left part of the forest be at (0, 0). Then the other corners have coordinates as shown. The population density
function is then given by
ρ(x, y) = 10 − 2y
The equations of the two diagonal lines are x = −2y/5 and x = 6 + 2y/5. So the total rabbit population in the forest is
Z
5
Z
2y
6+ 5
2y
−5
0
(10 − 2y) dx dy =
Z
=
Z
5
(10 − 2y)(6 +
0
0
5
(60 − 4y −
= (60y − 2y 2 −
4
y) dy
5
8 2
y ) dy
5
8 3
y )
15
5
0
8
· 125
= 300 − 50 −
15
2750
550
=
=
≈ 183
15
3
73. We use spherical coordinates. Since the density, δ, is equal to the distance from the point to the origin, we have
δ = ρ gm/cm3 .
Therefore the mass of the hemisphere is given by
Mass =
Z
2π
0
=4
Z
π/2
0
Z
0
2π
Z
2
0
Z
0
ρ · ρ2 sin φ dρdφdθ =
π/2
sin φ dφdθ = 4
Z
0
2π
Z
2π
0
Z
π/2
0
ρ4
4
2
sin φ dφdθ
0
π/2
(− cos φ)
0
dθ = 4 · 2π · 1 = 8π gm.
74. Since the hole resembles a cylinder, we will use cylindrical coordinates. Let the center of the sphere be at the origin, and
let the center of the hole be the z-axis (see Figure 16.152).
1574
Chapter Sixteen /SOLUTIONS
√
R
z
a2 − R2
✠
❘
a
y
x
Figure 16.152
√
√
Then we will integrate from z = − a2 − R2 to z = a2 − R2 , and each cross-section will be an annulus. So the
volume is
Z √a2 −R2 Z 2π
Z √a2 −R2 Z 2π Z √a2 −z2
1 2
r dr dθ dz =
(a − z 2 − R2 ) dθ dz
√
√
2
− a2 −R2 0
R
− a2 −R2 0
Z √
a2 −R2
=π
−
√
a2 −R2
2
(a2 − z 2 − R2 ) dz
2
p
= π (a − R )(2
=
3
4π 2
(a − R2 ) 2
3
a2
−
R2 )
3
1
− (2(a2 − R2 ) 2 )
3
75. We must first decide on coordinates. We pick Cartesian coordinates with the smaller sphere centered at the origin, the
larger one centered at (0, 0, −1). A vertical cross-section of the region in the xz-plane is shown in Figure 16.153. The
smaller sphere has equation x2 + y 2 + z 2 = 1. The larger sphere has equation x2 + y 2 + (z + 1)2 = 2.
z
x2 + y 2 + z 2 = 1
✠
x2 + y 2 + (z + 1)2 = 2
✠x
−1
Figure 16.153
Let R represent the region in the xy-plane which lies directly underneath (or above) the region whose volume we
want. The curve bounding this region is a circle, and we find its equation by solving the system:
x2 + y 2 + z 2 = 1
x2 + y 2 + (z + 1)2 = 2
SOLUTIONS to Review Problems for Chapter Sixteen
1575
Subtracting the equations gives
(z + 1)2 − z 2 = 1
2z + 1 = 1
z = 0.
2
2
2
2
Since z = 0, the two surfaces intersect in the xy-plane in the
pcircle x + y = 1. Thus R is x + y ≤ 1.
2
2
The top half of the small sphere is represented by z = 1 − x − y ; the top half of the large sphere is represented
p
by z = −1 + 2 − x2 − y 2 . Thus the volume is given by
Z √
Z Z √
1−x2
1
Volume =
−1
−
√
1−x2 −y 2
1−x2
−1+
√
dz dy dx.
2−x2 −y 2
Starting to evaluate the integral, we get
Volume =
1
Z
−1
Z √1−x2 p
−
√
1 − x2 − y 2 + 1 −
(
1−x2
We simplify the integral by converting to polar coordinates
Volume =
2π
Z
0
=
1
Z
1 − r2 + 1 −
0
2π
Z
−
0
= 2π
p
p
2 − x2 − y 2 ) dydx.
p
2 − r 2 r drdθ
(1 − r 2 )3/2
r2
(2 − r 2 )3/2
+
+
3
2
3
1
1
+ −
2
3
23/2
1
− +
3
3
= 2π
1
dθ
0
√
2 2
7
−
= 1.41.
6
3
76. Suppose the brick is set up as shown in Figure 16.154.
z
✛
5
3
✛
✻
✛
✛
1
y
❄
x
Figure 16.154
The brick has m/v = density = 1. The moment of inertia about the z-axis is
Iz =
Z
=
Z
5/2
−5/2
5/2
−5/2
=
Z
5/2
−5/2
Z
3/2
−3/2
Z
3/2
Z
1/2
1(x2 + y 2 ) dz dy dx
−1/2
(x2 + y 2 ) dy dx
−3/2
(3x2 +
9
) dx
4
125
45
85
=
+
=
4
4
2
1576
Chapter Sixteen /SOLUTIONS
The moment of inertia about the y-axis is
Iy =
5/2
Z
−5/2
=
5/2
Z
−5/2
=
5/2
Z
3/2
Z
−3/2
3/2
Z
Z
1/2
1(x2 + z 2 ) dz dy dx
−1/2
(x2 +
−3/2
(3x2 +
−5/2
1
) dy dx
12
1
) dx
4
5
65
125
+ =
=
4
4
2
The moment of inertia about the x-axis is
Ix =
5/2
Z
−5/2
=
Z
=
Z
5/2
−5/2
5/2
3/2
Z
−3/2
3/2
Z
Z
1/2
1(y 2 + z 2 ) dz dy dx
−1/2
(y 2 +
−3/2
(
−5/2
1
) dy dx
12
9
1
+ ) dx
4
4
25
10
=
4
2
= 5·
77. Let the ball be centered at the origin. Since a ball looks the same from all directions, we can choose the axis of rotation;
in this case, let it be the z-axis. It is best to use spherical coordinates, so then
x2 + y 2 = (ρ sin φ cos θ)2 + (ρ sin φ sin θ)2
= ρ2 sin2 φ
Then m/v = Density = 1, so the moment of inertia is
Iz =
Z
R
0
=
Z
=
=
=
0
Z
Z
π
1(ρ2 sin2 φ)ρ2 sin φ dφ dθ dρ
0
2π
Z
0
2π
π
ρ4 (sin φ)(1 − cos2 φ) dφ dθ dρ
4
ρ (− cos φ +
0
R
0
Z
Z
0
R
0
Z
2π
0
R
0
Z
Z
Z
0
R
2π
1
cos3 φ)
3
π
dθ dρ
0
4 4
ρ dθ dρ
3
8
8π 4
ρ dρ =
πR5
3
15
78. Set up the cylinder with the base centered at the origin on the xy plane, facing up. (See Figure 16.155.) Newton’s Law of
Gravitation states that the force exerted between two particles is
F =G
m1 m2
ρ2
where G is the gravitational constant, m1 and m2 are the masses, and ρ is the distance between the particles. We take a
small volume element, so m1 = m, and m2 = δdV . In cylindrical coordinates,
if m is at (0,0,0) and δdV is at (r, θ, z),
√
(see Figure 16.155), then the distance from m to δdV is given by ρ = r 2 + z 2 for r1 ≤ r ≤ r2 and 0 ≤ z ≤ h.
SOLUTIONS to Review Problems for Chapter Sixteen
1577
z
✛
✲✛
r2
✲
r1
✛
z✻
❄
φ
r
m
δdV
Figure 16.155
Due to the symmetry of the cylinder the sum of all the horizontal forces is zero; the net force on m is vertical. The force
acting on the particle as a result of the small piece dV makes an angle φ with the vertical and therefore has vertical
component
Vertical force on
GmδdV
GmδdV
Gmδz
z
particle from small = √
· cos φ = 2
=
·√
3 dV.
2
2
2
2
2
2
2
r
+
z
( r +z )
r +z
(r + z 2 ) 2
piece of cylinder
Thus, since dV = rdzdrdθ,
Total force =
Z
2π
r2
Z
0
r1
= 2πGmδ
h
Z
0
r2
Z
Z
r1
r2
= 2πGmδ
Gmδz
r dzdrdθ
(r 2 + z 2 )3/2
h
0
Z
r1
zr
drdz
(r 2 + z 2 )3/2
1−
r
1
(r 2 + h2 ) 2
r2
1
= 2πGmδ(r − (r 2 + h2 ) 2 )
= 2πGmδ(r2 − r1 −
R
p
r22
dr
r1
+ h2 +
p
r12 + h2 ).
79. (a) The constant k is determined by the condition that R k(x + y)dA = 1 where the region R is the quarter disk with
radius 10
x2 + y 2 ≤ 100
x≥0
y ≥ 0.
Using polar coordinates gives the integral
Z
k(x + y)dA =
Z
π/2
0
R
=k
Z
π/2
Z
π/2
Z
Z
10
r 2 (cos θ + sin θ)dr dθ
0
0
=
k(r cos θ + r sin θ)r dr dθ
0
0
=k
10
1000
(cos θ + sin θ)dθ
3
1000k
(sin θ − cos θ)
3
π2
=
0
1000k
2000k
2=
.
3
3
Since 2000k/3 = 1, Rwe have k = 3/2000.
(b) Evaluate the integral S f dA where S is the region 0 ≤ r ≤ 7, 0 ≤ θ ≤ π/2. We have
Z
3
f dA =
2000
S
3
=
Z
0
π/2
Z
0
7
3
r (cos θ + sin θ)dr dθ =
2000
2
7
(sin θ − cos θ)
2000
π/2
3
=
0
Z
π/2
0
343
7
2=
.
2000
1000
The probability that the point is closer than 7 units from the origin is 343/1000.
73
(cos θ + sin θ)dθ
3
1578
Chapter Sixteen /SOLUTIONS
CAS Challenge Problems
80. The region is the triangle to the right of the y-axis, below the line y = 1, and above the line y = x. Thus the integral can
R1Ry 2
R1R1 2
be written as 0 x ey dydx or as 0 0 ey dxdy. The second of these integrals can be evaluated easily by hand:
Z
1
0
Z
y
2
ey dxdy =
0
1
Z
0
=
1 y2
e
2
x=y
2
ey x
x=0
1
=
0
dy =
1
Z
2
yey dy
0
1
(e − 1)
2
The other integral cannot be done by hand with the methods you have learned, but some computer algebra systems will
compute it and give the same answer.
81. In Cartesian coordinates the integral is
Z p
3
x2 + y 2 dA =
1
Z
1
−1
D
In polar coordinates it is
Z
Z p
3
Z
x2 + y 2 dA =
D
Z
=
2π
0
Z √1−x2 p
3
−
√
3
√
x2 + y 2 dydx.
1−x2
r 2 rdrdθ =
6π
3
dθ =
8
7
0
2π
0
0
2π
Z
Z
1
r 5/3 drdθ
0
The Cartesian coordinate version requires the use of a computer algebra system. Some CASs may be able to handle it
and may give the answer in terms of functions called hypergeometric functions. To compare the answers are the same you
may need to ask the CAS to give a numerical value for the answer. It’s possible your CAS will not be able to handle the
integral at all.
82.
Z
1
Z
0
R 0 R 1 x+y
x+y
dydx = 1/2 and −1 0 (x−y)
3 dxdy = −1/2. This does not contradict the theorem because the
3
(x − y)
0
−1
function is not continuous everywhere inside the region of integration. In fact, it is not even defined at the origin.
83.
Average value for F =
=
1
Area
1
4h2
= a+
h
Z
−h
(a + bx4 + cy 4 + dx2 y 2 + ex3 y 3 ) dx dy
−h
4ah2 +
4bh6
4ch6
4dh6
+
+
5
5
9
1
(9b + 9c + 5d) h4
45
The limit is
lim (a +
h→0
h
Z
1
(9b + 9c + 5d) h4 ) = a.
45
Notice that F (0, 0) = a.
Average value for G =
1
4h2
Z
−h
1
=
4h2
= c+
The limit is
lim
h→0
c+
h
Z
h
(a sin(kx) + b cos(ky) + c) dx dy
−h
4 ch2 k + bh sin(hk)
k
!
b sin(hk)
.
hk
b sin(hk)
hk
= c + b lim
h→0
sin(hk)
= b + c.
hk
PROJECTS FOR CHAPTER SIXTEEN
1579
Notice that G(0, 0) = b + c.
Finally,
Average value for H =
(a + b) −1 + e2h
−2 − 2h − h2 + e2h 2 − 2h + h2
4e2h h2
.
You may need to simplify the answer given by your CAS to get this form. The limit of this as h → 0 (calculated with a
CAS) is 0. This is equal to H(0, 0).
In each case the limit of the average values over smaller and smaller squares centered at the origin is equal to the value
of the function at the origin. We conjecture that this is true in general for a continuous function. This makes sense because
when the square is small, the function is approximately constant on the square with value equal to its value at the origin.
Therefore the integral is approximately the area times the value of the function, so the average value is approximately the
value of the function. This approximation gets better and better as h → 0.
PROJECTS FOR CHAPTER SIXTEEN
1. (a) We are integrating over the whole plane, so converting to polar coordinates gives
Z
∞
−∞
Z
∞
e−(x
2
+y 2 )
dxdy =
2π
Z
0
−∞
Z
∞
2
e−r rdrdθ =
0
0
(b) Rewriting the integrand as a product gives
Z
Z ∞Z ∞
2
2
e−(x +y ) dxdy =
−∞
−∞
2π
Z
∞
−∞
Z
2
1
− e−r
2
∞
2
∞
dθ =
Z
0
0
2π
1
dθ = π.
2
2
e−x e−y dxdy.
−∞
2
Now e−y is a constant as far as the integral with respect to x is concerned, so
Z ∞
Z ∞Z ∞
Z ∞
−x2 −y 2
−y 2
−x2
e
e dxdy =
e
e
dx dy.
−∞
−∞
−∞
−∞
We assume that the integral with respect to x converges, and so is a constant as far as the integral with
respect to y is concerned. Thus, we have
Z ∞
Z ∞
Z ∞
Z ∞
−y 2
−x2
−x2
−y 2
dx dy =
dx
e
e
e
e dy .
−∞
−∞
But
R∞
−∞
2
e−x dx and
Z
∞
−∞
Z
∞
R∞
−∞
2
−∞
e−(x
−∞
2
e−y dy are the same number, so we can write
+y 2 )
dxdy =
−∞
Z
∞
−∞
Z
2
e−x dx
∞
2
e−y dy
−∞
=
Z
∞
−∞
2
2
e−x dx .
(c) Using the results of parts (a) and (b), we have
Z
∞
e
−∞
−x2
2 Z
dx =
∞
−∞
Z
∞
e−(x
2
+y 2 )
dxdy = π.
−∞
Taking square roots and observing that the integral we are looking for is positive, we have
Z ∞
√
2
e−x dx = π.
−∞
1580
Chapter Sixteen /SOLUTIONS
2. (a) We want to find the average value of |x − y| over the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1:
Average distance between gates =
Z
1
0
Let’s fix x, with 0 ≤ x ≤ 1. Then |x − y| =
Z
1
|x − y| dy =
Z
x
Z
(x − y) dy +
0
0
y2
xy −
2
1
= x2 − x + .
2
y−x
x−y
Z
1
|x − y| dy dx.
0
for y ≥ x
. Therefore
for y ≤ x
1
(y − x) dy
x
x
+
=
0
y2
− xy
2
1
= x2 −
x
x2
1
x2
+ −x−
+ x2
2
2
2
So,
Z
1
Z
1
|x − y| dy dx
Z 1
Z 1 Z 1
1
(x2 − x + ) dx
=
|x − y| dy dx =
2
0
0
0
Average distance between gates =
0
=
0
x2
1
x3
−
+ x
3
2
2
1
=
0
1
.
3
(b) There are (n + 1)2 possible pairs (i, j) of gates, i = 0, . . . , n, j = 0, . . . , n, so the sum given represents
the average distances apart of all such gates. The Riemann sum with ∆x = ∆y = 1/n, if we choose the
least x and y-values in each subdivision is
n−1
X
X n−1
i=0 j=0
i
j 1
−
,
n n n2
which for large n is just about the same as the other sum. For n = 5 the sum is about 0.389; for n = 10
the sum is about 0.364.
17.1 SOLUTIONS
1581
CHAPTER SEVENTEEN
Solutions for Section 17.1
Exercises
1. Since we are moving on the y-axis, x = 0, and y goes from −2 to 1. Thus a possible parameterization is
x = 0,
y = t,
−2 ≤ t ≤ 1.
2. We want a quarter-circle of radius 2 starting at (2, 0) and ending at (0, 2). The equations x = 2 cos t, y = 2 sin t describe
counterclockwise motion in a circle of radius 2 centered at the origin, passing (2, 0) when t = 0 and (0, 2) when t = π/2.
So a possible parameterization is
x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π/2.
3. As we move down the straight line from (0, 3) to (1, 0), x increases linearly from 0 to 1 and y decreases linearly from 3
to 0. Thus a possible parameterization is
x = t,
y = 3 − 3t,
0 ≤ t ≤ 1.
4. We want the bottom half of a semicircle of radius 1 centered at (0, 1). The equations x = cos t, y = 1 + sin t describe
clockwise motion in this circle, passing (−1, 1) when t = π and (1, 1) when t = 2π. So a possible parameterization is
x = cos t,
y = 1 + sin t,
π ≤ t ≤ 2π.
5. We want the straight line segment from (1, 1) to (3, 2). The position vector of (1, 1) is ~i + ~j and the displacement vector
from (1, 1) to (3, 2) is 2~i + ~j , so the line has equation
~r = ~i + ~j + t(2~i + ~j ),
or
x = 1 + 2t, y = 1 + t.
This passes (1, 1) when t = 0 and (3, 2) when t = 1, so a possible parameterization is
x = 1 + 2t,
y = 1 + t,
0 ≤ t ≤ 1.
6. The curve is a segment of a parabola y = ax2 starting at (0, 0) and ending up at (2, 2). Thus the parabola has equation
y = 21 x2 . Since x goes from 0 to 2, a possible parameterization is
x = t,
y=
1 2
t ,
2
0 ≤ t ≤ 2.
7. One possible parameterization is
x = t,
y = 1,
z = −t.
x = 3 + t,
y = 2t,
z = −4 − t.
8. One possible parameterization is
9. One possible parameterization is
x = 1,
y = 0,
z = t.
10. One possible parameterization is
x = 5,
y = −1 + 5t,
z = 1 + 2t.
11. One possible parameterization is
x = 1 + 3t,
y = 2 − 3t,
z = 3 + t.
1582
Chapter Seventeen /SOLUTIONS
12. One possible parameterization is
x = −3 + 2t,
y = 4 + 2t,
z = −2 − 3t.
13. The displacement vector from the first point to the second is ~v = (−1 − (−3))~i + (−3 − (−2))~j + (−1 − 1)~k =
2~i − ~j − 2~k . The line through point (−3, −2, 1) and with direction vector ~v = 2~i − ~j − 2~k is given by parametric
equations
x = −3 + 2t,
y = −2 − t,
z = 1 − 2t.
Other parameterizations of the same line are also possible.
14. The displacement vector from the first point to the second is ~v = 4~i − 5~j − 3~k . The line through point (1, 5, 2) and
with direction vector ~v = 4~i − 5~j − 3~k is given by parametric equations
x = 1 + 4t,
y = 5 − 5t,
z = 2 − 3t.
Other parameterizations of the same line are also possible.
15. The vector connecting the two points is 3~i − ~j + ~k . So a possible parameterization is
x = 2 + 3t,
y = 3 − t,
z = −1 + t.
16. The line passes through (3, −2, 2) and (0, 2, 0). The displacement vector from the first of these points to the second is
~v = (0 − 3)~i + (2 − (−2))~j + (0 − 2)~k = −3~i + 4~j − 2~k . The line through point (3, −2, 2) and with direction
vector ~v = −3~i + 4~j − 2~k is given by parametric equations
x = 3 − 3t,
y = −2 + 4t,
z = 2 − 2t.
Other parameterizations of the same line are also possible.
17. The line passes through (3, 0, 0) and (0, 0, −5). The displacement vector from the first of these points to the second is
~v = −3~i − 5~k . The line through point (3, 0, 0) and with direction vector ~v = −3~i − 5~k is given by parametric
equations
x = 3 − 3t,
y = 0,
z = −5t.
Other parameterizations of the same line are also possible.
18. A vector along the line through these points is ~v = 2~i + 2~j − ~k . Since the line goes through the point (2, 1, 3), a
parametric equation for the line segment is
x = 2 + 2t, y = 1 + 2t, z = 3 − t with 0 ≤ t ≤ 1.
19. Since the radius is 3, the center is (0, 0, 5), and the circle lies in the z = 5 plane, a parametric equation is
x = 3 cos t, y = 3 sin t, z = 5, for 0 ≤ t < 2π.
20. The vector ~n = 2~i − 3~j − ~k is normal to the plane, so the line is in the direction of ~n and through the point (1, 1, 6).
A possible equation is
x = 1 + 2t, y = 1 − 3t, z = 6 − t.
21. The xy-plane is where z = 0, and to make the particle go in the clockwise direction we start at (2, 0, 0) and head in the
negative y-direction. Thus one possible answer is
x = 2 cos t,
y = −2 sin t,
z = 0.
17.1 SOLUTIONS
1583
22. Since the radius is 2, the circle must be of the form x = 2 cos t, y = 2 sin t, z = 1. But this parameterization traces out the
circle clockwise when viewed from below. Therefore, the parameterization we want is x = 2 cos t, y = −2 sin t, z = 1.
23. The xz-plane is y = 0, so one possible answer is
x = 2 cos t,
y = 0,
z = 2 sin t.
24. The circle lies in the plane z = 2, so one possible answer is
x = 3 cos t,
y = 3 sin t,
z = 2.
25. The yz-plane is x = 0, so the circle of radius 3 in the yz-plane centered at the origin would have equations
x = 0,
y = 3 cos t,
z = 3 sin t.
To move the center to (0, 0, 2) we add 2 to the equation for z, so one possible answer is
x = 0,
y = 3 cos t,
z = 2 + 3 sin t.
26. The circle of radius 5 in the yz-plane centered at the origin has equations
x = 0,
y = 5 cos t,
z = 5 sin t.
To move the center to (−1, 0, −2), we add −1 to the equation for x and −2 to the equation for z, so one possible answer
is
x = −1, y = 5 cos t, z = −2 + 5 sin t.
27. The xy-plane is z = 0, so a possible answer is
x = t2 ,
y = t,
z = 0.
x = t,
y = t3 ,
z = 0.
28. The xy-plane is z = 0, so a possible answer is
29. The xz-plane is y = 0, so a possible answer is
x = −3t2 ,
30. The plane z = 2 cuts the cone z =
p
y = 0,
z = t.
x2 + y 2 in the circle
2=
p
x2 + y 2 .
This circle is centered on the z-axis, lies in the plane z = 2, and has radius 2. A parametric equation is
x = 2 cos t, y = 2 sin t, z = 2, for 0 ≤ t < 2π.
31. Since the curve is parallel to the xy-plane, z is constant, and since it passes through (0, 4, 4), we have z = 4. One possible
answer is
x = t, y = 4 − 5t4 , z = 4.
1584
Chapter Seventeen /SOLUTIONS
32. Since its diameters are parallel to the y and z-axes and its center is in the yz-plane, the ellipse must lie in the yz-plane,
x = 0. The ellipse with the same diameters centered at the origin would have its y-coordinate range between −5/2 and
5/2 and its z-coordinate range between −1 and 1. Thus this ellipse has equation
x = 0,
y=
5
cos t,
2
z = sin t.
To move the center to (0, 1, −2), we add 1 to the equation for y and −2 to the equation for z, so one possible answer for
our ellipse is
5
x = 0, y = 1 + cos t, z = −2 + sin t.
2
33. Since its diameters lie along the x and y-axes and its center is the origin, the ellipse must lie in the xy-plane, hence at
z = 0. The x-coordinate ranges between −3 and 3 and the y-coordinate between −2 and 2. One possible answer is
x = 3 cos t,
y = 2 sin t,
z = 0.
34. Since its diameters are parallel to the x and z-axes, the ellipse must be parallel to the xz-plane. The ellipse with the same
diameters, but centered at the origin, would have its x-coordinate range between −3/2 and 3/2 and its z-coordinate range
between −1 and 1. Thus this ellipse has equation
x=
3
cos t,
2
y = 0,
z = sin t.
Since our ellipse has center (0, 1, −2), it must be in the plane y = 1. To move the center to (0, 1, −2), we add 1 to the
equation for y and −2 to the equation for z, so one possible answer for our ellipse is
x=
3
cos t,
2
y = 1,
z = −2 + sin t.
35. The displacement vector between the points is ~
u = 3~i + 5~k , so a possible parameterization of the line is
x = −1 + 3t,
y = 2,
z = −3 + 5t.
36. The vector from P0 to P1 is ~v = (5 + 1)~i + (2 + 3)~j = 6~i + 5~j . Since P0 = −~i − 3~j , the line is
~r (t) = −~i − 3~j + t(6~i + 5~j )
for 0 ≤ t ≤ 1.
In coordinate form, the equations are x = −1 + 6t, y = −3 + 5t, 0 ≤ t ≤ 1
37. The vector from P0 to P1 is ~v = (4 − 1)~i + (1 + 3)~j + (−3 − 2)~k = 3~i + 4~j − 5~k . Since P0 has position vector
~i − 3~j + 2~k , the line is
~r (t) = ~i − 3~j + 2~k + t(3~i + 4~j − 5~k )
for 0 ≤ t ≤ 1.
In coordinate form the equations are x = 1 + 3t, y = −3 + 4t, z = 2 − 5t.
38. Since the semicircle is in the yz-plane we have x = 0. A circle of radius 5 in the yz-plane, centered at the origin and
parameterized in the clockwise direction (from the positive z-axis toward the positive y-axis), goes from (0, 0, 5) to
(0, 0, −5). It has equations y = 5 cos t and z = −5 sin t. The semicircle where y ≥ 0 is the obtained by restricting t to
−π/2 ≤ t ≤ π/2. Thus a possible answer is
x = 0,
y = 5 cos t,
z = −5 sin t,
−π/2 ≤ t ≤ π/2.
39. Since the semicircle is in the xy-plane we have z = 0. A circle of radius 1 in the xy-plane, centered at the origin
and parameterized in the counterclockwise direction, goes from (1, 0, 0) to (−1, 0, 0). It has equations x = cos t and
y = sin t. The semicircle where y ≥ 0 is the obtained by restricting t to 0 ≤ t ≤ π. Thus a possible answer is
x = cos t,
y = sin t,
z = 0,
0 ≤ t ≤ π.
17.1 SOLUTIONS
1585
√
40. The graph is parameterized by x = t, y = t. To obtain the segment, we restrict t to 1 ≤ t ≤ 16. Thus one possible
answer is
√
1 ≤ t ≤ 16.
x = t, y = t,
41. The line segment P Q has length 10, so it must be a diameter of the circle. The center of the circle is therefore the midpoint
of P Q, which is the point (5, 0). The upper arc of the circle between P and Q can be parameterized as follows:
~r (t) = 5~i + 5(− cos t~i + sin t~j ),
0 ≤ t ≤ π.
The lower arc can be parameterized as follows:
~r (t) = 5~i + 5(cos t~i + sin t~j ),
π ≤ t ≤ 2π.
42. The equation for z is z = 3. The x-coordinate goes from 4 to 0 and the y-coordinate from 0 to −3, so possible equations
for x and y are x = 4 cos t and y = −3 sin t, with t from 0 to π/2. Thus one possible answer is
x = 4 cos t,
y = −3 sin t,
z=3
0 ≤ t ≤ π/2.
Problems
−
−
→
43. We find the parameterization in terms of the displacement vector OP = 2~i + 5~j from the origin to the point P and the
−
−
→
displacement vector P Q = 10~i + 4~j from P to Q.
−
−
→
−
−→
~r (t) = OP + tP Q or, expressed in coordinates, ~r (t) = (2 + 10t)~i + (5 + 4t)~j . To see that this is correct, note
−
−
→
−
−
→ −
−
→
that the equation parameterizes a line because it is linear, that t = 0 corresponds to OP + 0P Q = OP , the vector from
−
−
→
−
−
→ −
−→
the origin to P , and that t = 1 corresponds to OP + 1P Q = OQ, the vector from the origin to Q.
−
−
→
44. We find the parameterization in terms of the displacement vector OP = 2~i + 5~j from the origin to the point P and the
−
−
→
~
~
displacement vector P Q = 10i + 4j from P to Q.
−
−→
−
−
→
~r (t) = OP + (t/5)P Q or ~r (t) = (2 + (t/5)10)~i + (5 + (t/5)4)~j
−
−
→
45. We find the parameterization in terms of the displacement vector OP = 2~i + 5~j from the origin to the point P and the
−
−
→
~
~
displacement vector P Q = 10i + 4j from P to Q.
−
−
→
~r (t) = OP +
t − 20 −
−
→
PQ = 2 +
10
t − 20
10 ~i + 5 +
10
t − 20
4 ~j .
10
−
−
→
46. We find the parameterization in terms of the displacement vector OP = 2~i + 5~j from the origin to the point P and the
−
−
→
displacement vector P Q = 10~i + 4~j from P to Q.
−
−→
−
−
→
~r (t) = OP + (t − 10)P Q or ~r (t) = (2 + (t − 10)10)~i + (5 + (t − 10)4)~j
−
−
→
47. We find the parameterization in terms of the displacement vector OP = 2~i + 5~j from the origin to the point P and the
−
−
→
displacement vector P Q = 10~i + 4~j from P to Q.
−
−→
−
−
→
~r (t) = OP − tP Q or ~r (t) = (2 − 10t)~i + (5 − 4t)~j
48. Substituting t = −1 into the parametric equations tells us that the plane passes through the point
(x, y, z) = (8, −12, −6).
A vector parallel to the line is ~v = −3~i + 5~j + 6~k . This vector is normal to the plane, so an equation for the plane is
−3(x − 8) + 5(y + 12) + 6(z + 6) = 0,
or
−3x + 5y + 6z = −120.
49. The line is in the direction of the vector ~v = 7~i + 3~j − 2~k and the vector ~
n = 2~i − 3~j + 5~k is normal to the given
plane. If the vectors ~v and ~n are perpendicular then the line and the plane are parallel. Since
~v · ~n = (7)(2) + (3)(−3) + (−2)(5) = −5,
the line and the plane are not parallel.
1586
Chapter Seventeen /SOLUTIONS
50. These equations parameterize a line. Since (3 + t) + (2t) + 3(1 − t) = 6, we have x + y + 3z = 6. Similarly,
x − y − z = (3 + t) − 2t − (1 − t) = 2. That is, the curve lies entirely in the plane x + y + 3z = 6 and in the plane
x − y − z = 2. Since the normals to the two planes, n~1 = ~i + ~j + 3~k and n~2 = ~i − ~j − ~k are not parallel, the line is
the intersection of two nonparallel planes, which is a straight line in 3-dimensional space.
51. (a) A vector on the line will lie in both planes and will therefore be orthogonal to both normal vectors. To produce a
vector orthogonal to two given vectors, you can take their cross product.
(b) The vector (~i + 2~j − 3~k ) × (3~i − ~j + ~k ) = −~i − 10~j − 7~k is parallel to the line.
(c) We need a point on the line and a vector parallel to the line. We found a vector in part (b). To find a point, we
set z = 0 and solve for x and y in the equations for the planes. We have x + 2y = 7 and 3x − y = 0 from
which x = 1 and y = 3. Hence, the point (1, 3, 0) is on the line. Finally, a parametric equation for the line is
~r = (1 − t)~i + (3 − 10t)~j − 7t~k . Other answers are possible.
52. The vector 2~i − ~j + ~k is in the direction of the line and therefore parallel to the plane. The line (and thus the plane)
contains the point (1, 3, 4). The displacement vector between (1, 3, 4) and (2, 3, 4) is the vector ~i , and this vector is also
parallel to the plane. A normal, ~n , to the plane is the cross product
~i ~j ~k
~
~
n = (2~i − ~j + k ) × ~i = 2 −1 1 = ~j + ~k .
1 0 0
The equation of the plane has the form y + z = d. Substituting either of the points gives
y + z = 7.
53. (a) The line is parallel to a normal vector, ~n , to the plane
~n = 2~i − 3~j − ~k .
Since the line goes through the point (1, 3, 7), its equation is
~r = (~i + 3~j + 7~k ) + t(2~i − 3~j − ~k ).
(b) Rewriting the equation of the line as
x = 1 + 2t,
y = 3 − 3t,
y = 7 − t,
we substitute into the equation of the plane 2x − 3y = z to get
2(1 + 2t) − 3(3 − 3t) = 7 − t
4t + 9t + t = 7 − 2 + 9
t = 1.
Thus, the point of intersection is x = 3, y = 0, z = 6.
(c) The distance between (1, 3, 7) and the plane is measured along the line perpendicular to the plane. Thus, it is the
distance from (1, 3, 7) to (3, 0, 6):
p
√
Distance = (1 − 3)2 + (3 − 0)2 + (7 − 6)2 = 14.
54. (a) The line segment starting at P0 and ending at P1 is parametrized by
−−→
−−−→
~r (t) = OP0 + tP0 P1 , 0 ≤ t ≤ 1.
We write this in coordinates: Let P0 = (x0 , y0 , z0 ) and P1 = (x1 , y1 , z1 ). Then a vector between the points is
−−−→
P0 P1 = (x1 − x0 )~i + (y1 − y0 )~j + (z1 − z0 )~k , so
~r (t) = (x0~i + y0~j + z0~k ) + t((x1 − x0 )~i + (y1 − y0 )~j + (z1 − z0 )~k ),
or
x(t) = x0 + t(x1 − x0 ) = (1 − t)x0 + tx1
y(t) = y0 + t(y1 − y0 ) = (1 − t)y0 + ty1
z(t) = z0 + t(z1 − z0 ) = (1 − t)z0 + tz1
17.1 SOLUTIONS
1587
Thus we have
−−→
−−→
~r (t) = (1 − t)OP0 + tOP1 .
−−→
−−→
(b) The parametric equation ~r (t) = tOP0 + (1 − t)OP1 , 0 ≤ t ≤ 1 is the line segment from P1 to P0 , the same line
segment as in part (a), but traversed in the opposite direction.
55. (a) Normal vectors to the two planes are
~
n 1 = 2~i − ~j − 3~k
and
~n 2 = ~i + ~j + ~k .
The vector ~n 1 × ~n 2 is perpendicular to both planes and parallel to the line of intersection:
~i ~j ~k
~n 1 × ~n 2 = 2 −1 −3 = 2~i − 5~j + 3~k .
1 1
1
(b) To check that the point (1, −1, 1) lies on the planes, substitute into each equation.
2x − y − 3z = 2 · 1 − (−1) − 3 · 1 = 0
x + y + z = 1 − 1 + 1 = 1.
Thus, the point lies on both planes.
(c) Parametric equations of the line are
x = 1 + 2t,
y = −1 − 5t,
z = 1 + 3t.
56. Since the point of intersection is on the plane 2x − 3y + 5z = −7 and on the line
x = 5 + 7t,
y = 4 + 3t,
z = −3 − 2t
for some t, substituting the equations of the line into the plane equation gives
2(5 + 7t) − 3(4 + 3t) + 5(−3 − 2t) = −7.
Solving for t gives t = −2 and so the point of intersection is x = 5 + 7(−2) = −9, y = 4 + 3(−2) = −2,
z = −3 − 2(−2) = 1.
57. The coefficients of t in the parameterizations show that line ~r 1 is parallel to the vector −3~i + 2~j + ~k and line ~r 2 is
parallel to −6~i + 4~j + 3~k . Since these vectors are not parallel, the lines are not parallel, so the lines are different.
58. The coefficients of t in the parameterizations show that line ~r 1 is parallel to the vector −3~i + ~j + 2~k and line ~r 2 is
parallel to 6~i − 2~j − 4~k . Since these vectors are parallel, the lines are parallel. To see if they are the same line, check
whether they have a common point. If so, every point is common. Pick any point on ~r 1 , say where t = 0, which shows that
the point (5, 1, 0) is on line ~r 1 . To determine whether this point is on line ~r 2 , search for a solution of the simultaneous
equations
2 + 6t = 5 2 − 2t = 1 2 − 4t = 0.
The solution of the first equation is t = 1/2, which also solves the other two equations, which shows that the point
(5, 1, 0) is on line ~r 2 , corresponding to t = 1/2.
Since the two lines are parallel and go through a common point, they are the same line.
59. The coefficients of t in the parameterizations show that line ~r 1 is parallel to the vector −3~i + ~j + 2~k and line ~r 2 is
parallel to 6~i − 2~j − 4~k . Since these vectors are parallel, the lines are parallel. To see if they are the same line, check
whether they have a common point. If so, every point is common., Pick any point on ~r 1 , say where t = 0, which shows
that the point (5, 1, 0) is on line ~r 1 . To determine whether this point is on line ~r 2 , search for a solution of the simultaneous
equations
2 + 6t = 5 2 − 2t = 1 3 − 4t = 0.
The solution of the first equation is t = 1/2, which is not a solution of the third equation, so there is no common solution.
The two lines are parallel but they are different, because one line contains the point (5, 1, 0) and the other does not.
60. The lines intersect if
c+t = s
1+t = 1−s
5 + t = 3 + s.
Solving the last two equations gives t = −1 and s = 1. Substituting into the first equation gives c = 2.
1588
Chapter Seventeen /SOLUTIONS
61. (a) The line can be written as
x = 2 + 3t,
y = 5 + t,
z = 2t.
We substitute into x + y + z = 1 and solve
(2 + 3t) + (5 + t) + 2t = 1
6t + 7 = 1
t = −1.
Thus, the point is (x, y, z) = (2 + 3(−1), 5 − 1, 2(−1)) = (−1, 4, −2).
(b) The vector ~v = 3~i + ~j + 2~k is parallel to the line; the normal vector ~n = ~i + ~j + ~k is perpendicular to the plane.
Thus
~k
~i ~j
~v × ~n = 3 1
2 = −~i − ~j + 2~k
1
1
1
is perpendicular to the line and lies in the plane. Other answers are possible.
(c) The line passes through (−1, 4, −2) and is parallel to −~i − ~j + 2~k . Its equation can be written
~r = −~i + 4~j − 2~k + t(−~i − ~j + 2~k ).
62. Add the two equations to get 3x = 8, or x = 83 . Then we have
−y + z =
1
.
3
y = t,
z=
So a possible parameterization is
x=
8
,
3
1
+ t.
3
63. Add the two equations to get 2x +3z = 5, or x = − 32 z + 25 . Subtract the two equations to get 2y −z = 1, or y = 21 z + 21 .
So a possible parameterization is
3
5
1
1
x = − t + , y = t + , z = t.
2
2
2
2
64. Let f (x, y, z) = x2 + y 2 − z. Then the surface z = x2 + y 2 is a level surface of f at the value 0. The gradient of f is
perpendicular to the level surface.
grad f = 2x~i + 2y~j − ~k = 2~i + 4~j − ~k .
So a possible parameterization is
x = 1 + 2t,
y = 2 + 4t,
z = 5 − t.
65. Parametric equations for a line in 3-space are in the form
x = x0 + at
y = y0 + bt
z = z0 + ct
where (x0 , y0 , z0 ) is a point on the line and the direction vector is ~v = a~i +b~j +c~k . We are given the point as (−4, 2, 3).
The vector ~j + ~k is parallel to the yz-plane and at an angle of 45◦ to both positive y- and z-axes. Thus, the direction
vector for this line is ~v = ~j + ~k . Parametric equations for this line are
x = −4,
y = 2 + t,
z = 3 + t.
17.1 SOLUTIONS
1589
66. The question is equivalent to asking if the line through (−3, −4, 2) and (4, 5, 0) enters the sphere x2 + y 2 + z 2 = 1. A
parameterization for this line is given by
x = −3 + 7t,
y = −4 + 9t,
2
2
z = 2 − 2t.
2
We want to see whether the line intersects the sphere x + y + z = 1. Substituting we have
(−3 + 7t)2 + (−4 + 9t)2 + (2 − 2t)2 = 1
29 − 122t + 134t2 = 0
2
Since (122) − 4(29)134 < 0, this equation has no real solutions. Thus, the line does not enter the sphere and the point
is visible.
67. (a) Both paths are straight lines, the first passes through the point (−1, 4, −1) in the direction of the vector ~i − ~j + 2~k
and the second passes through (−7, −6, −1) in the direction of the vector 2~i + 2~j + ~k . The two paths are not
parallel.
(b) Is there a time t when the two particles are at the same place at the same time? If so, then their coordinates will be
the same, so equating coordinates we get
−1 + t = −7 + 2t
4 − t = −6 + 2t
−1 + 2t = −1 + t.
Since the first equation is solved by t = 6, the second by t = 10/3, and the third by t = 0, no value of t solves all
three equations. The two particles never arrive at the same place at the same time, and so they do not collide.
(c) Are there any times t1 and t2 such that the position of the first particle at time t1 is the same as the position of the
second particle at time t2 ? If so then
−1 + t1 = −7 + 2t2
4 − t1 = −6 + 2t2
−1 + 2t1 = −1 + t2 .
We solve the first two equations and get t1 = 2 and t2 = 4. This is a solution for the third equation as well, so the
three equations are satisfied by t1 = 2 and t2 = 4. At time t = 2 the first particle is at the point (1, 2, 3), and at time
t = 4 the second is at the same point. The paths cross at the point (1, 2, 3), and the first particle gets there first.
68. (a) The particle moves clockwise around a circle with center (a, a) and radius b, starting at (a, a + b). The motion has
period 2π/k.
(b) (i) Increasing b increases the radius.
(ii) Increasing a moves the center away from the origin along the line y = x.
(iii) Increasing k makes the particle move faster and reduces the period.
(iv) If a = b, the circle touches both the x- and y-axes at the points (a, 0) and (0, a), respectively.
69. (a)
(b)
(c)
(d)
The curve is a loop because temperature and salinity go through the same changes every year.
At t = 8, mid-August.
At t = 4, mid-April.
From the graph, we have (approximately) T (5) = 15.9, T (6) = 17.8, and T (7) = 20.4. Using a difference quotient,
we have
T (7) − T (5)
dT
≈
= 2.25 ◦ C/month.
dt t=6
2
The seawater temperature in mid-June is increasing at a rate of about 2◦ C per month.
70. (a) The particle is moving in the direction of the vector ~v = (−2 − 0)~i + (2 − 1)~j + (−2 − 0)~k = −2~i + ~j − 2~k .
Since at the point (0, 1, 0)
grad c
(0,1,0)
= −2x~i − 2y~j − 2z~k e−(x
2 +y 2 +z 2 )
(0,1,0)
= −2e−1~j ,
the directional derivative, which gives the rate of change we want, is given by
−2~i + ~j − 2~k
2
c~v = (−2e−1~j ) · p
= − e−1 microgr/m3 per meter.
2
2
2
3
(−2) + 1 + (−2)
1590
Chapter Seventeen /SOLUTIONS
(b) Substituting the parametric equation of the curve into the concentration, we have the concentration as a function of t:
c(t) = e
−
(1−t2 )2 +t2 +(1−t2 )2
Differentiating gives
= e−(2(1−t
c′ (t) = − 4 1 − t2 (−2t) + 2t e−(2(1−t
Setting c′ (t) = 0 and observing that e−(2(1−t
2 2
) +t
2
) +t2 )
2 2
) +t2 )
2 2
.
.
) 6= 0, we have
− 4 1 − t2 (−2t) + 2t = 0
t −4 1 − t2 + 1 = 0
2
t 4t − 3 = 0
t = 0, ±
√
3
.
2
Since c(t) → 0 as t → ∞, one of the critical points gives the global maximum. Substituting gives
=e
√c(0)
3
c ±
2
=e
−2
= 0.135
−(2(1−3/4)2 +3/4)
= e−7/8 = 0.417.
√
Thus, the maxima occur at t = ± 3/2 seconds.
71. The three shadows appear as a circle, a cosine wave and a sine wave, respectively.
z
y
x = cos t
y = sin t
z
10
10
1
y = sin t
z=t
x = cos t
z=t
5
5
x
−1
1
x
−1
−1
0
1
y
−1
0
1
Figure 17.1
72. (a)
(b)
(c)
(d)
(e)
Equations II represent the line y = x.
Equations IV represent the line x + y = a. Since a > 0, this line is not through the origin.
Equations V give the hyperbola x2 − y 2 = a2 .
Equations I represent the circle x2 + y 2 = a2 , traversed clockwise starting at (0, a).
Equations III represent the circle x2 + y 2 = a2 , traversed counterclockwise starting at (a, 0).
73. (a) Parametric equations are
x = 2 + at,
y = 1 + bt, z = 3 + ct.
(b) The line goes through the origin if the position vector 2~i + ~j + 3~k is parallel to the vector a~i + b~j + c~k . This
occurs if a, b, c are in the ratio 2 : 1 : 3; that is if
a
b
c
= = .
2
1
3
74. (a) The vector −2~i + 7~j + 4~k is parallel to the line. A normal to the plane is a~i + b~j + c~k . We want the normal to
the plane to be parallel to the line, so we take a = −2, b = 7, c = 4. Any value of d will do, for example d = 0.
(b) The same values of a, b, c as in part (a) work, though now we need to choose d so that the point (5, 3, 0) lies on the
plane. So a = −2, b = 7, c = 4 and
d = −2(5) + 7(3) + 4(0) = 11.
1591
17.1 SOLUTIONS
(c) The normal a~i + b~j + c~k must be perpendicular to the vector −2~i + 7~j + 4~k , so
−2a + 7b + 4c = 0
We can choose any values of a, b, c which satisfy this equation, so a = 7, b = 2, c = 0 work. To ensure that the point
(5, 3, 0), which lies on the line, also lies on the plane, substitute the coordinates of the point into the plane, giving
d = 7x + 2y + 0z = 7(5) + 2(3) = 41.
75. The helices wind around a cylinder of radius α, which explains the significance of α. As t increases from 0 to 2π, the
helix winds once around the cylinder, climbing upward a distance of 2πβ. Thus β controls how stretched out the helix is
in the vertical direction. See Figure 17.2 and Figure 17.3.
z
z
z
y
x
y
x
y
x
Figure 17.2: Three values of α with the same β
z
z
z
x
y
x
y
x
y
Figure 17.3: Three values of β with the same α
76. The displacement from the point (1, 2, 3) to the point (3, 5, 7) is 3~i + 5~j + 7~k − (~i + 2~j + 3~k ) = 2~i + 3~j + 4~k . So
the equation of the line is
x~i + y~j + z~k = 1~i + 2~j + 3~k + t(2~i + 3~j + 4~k )
or
x~i + y~j + z~k = (1 + 2t)~i + (2 + 3t)~j + (3 + 4t)~k .
The square of the distance from a point (x, y, z) on the line to the origin, denoted by D(t) is
D(t) = (x − 0)2 + (y − 0)2 + (z − 0)2
= (1 + 2t)2 + (2 + 3t)2 + (3 + 4t)2
= 1 + 4t + 4t2 + 4 + 12t + 9t2 + 9 + 24t + 16t2
1592
Chapter Seventeen /SOLUTIONS
= 14 + 40t + 29t2
14
40
t+
= 29 t2 +
29 2 29 2
14
20
20
−
+
.
= 29
t+
29
29
29
Since D(t) is minimum when t = −20/29 and
2
20
D(−20/29) = 29 −
the shortest distance is
p
29
14
+
29
=
6
,
29
6/29.
77. The line ~r = ~a + t~b is parallel to the vector ~b and through the point with position vector ~a .
(a) is (vii). The equation ~b · ~r = 0 is a plane perpendicular to ~b and satisfied by (0, 0, 0).
(b) is (ii). For any constant k, the equation ~b · ~r = k is a plane perpendicular to ~b . If k = ||~a || 6= 0, the plane does not
contain the origin.
(c) is (iv). The equation (~a × ~b ) · (~r − ~a ) = 0 is the equation of a plane which is satisfied by ~r = ~a , so the point with
position vector ~a lies on the plane. Since ~a × ~b is perpendicular to ~b , the plane is parallel to the line, and therefore
it contains the line.
78. (a) Parametric equations are
x = 1 + 2t,
y = 5 + 3t,
z = 2 − t.
(b) We want to minimize D, the square of the distance of a point to the origin, where
D = (x − 0)2 + (y − 0)2 + (z − 0)2 = (1 + 2t)2 + (5 + 3t)2 + (2 − t)2 .
Differentiating to find the critical points gives
dD
= 2(1 + 2t)2 + 2(5 + 3t)3 + 2(2 − t)(−1) = 0
dt
2 + 4t + 15 + 9t − 2 + t = 0
−15
.
t=
14
Thus
−8
−15
=
14
7
25
−15
=
y = 5+3
14
14
−15
43
z = 2−
=
.
14
14
x = 1+2
Since the distance of the point on the line from the origin increases without bound as the magnitude of x, y, z increase,
the only critical point of D must be a global minimum. Therefore, the point (−8/7, 25/14, 43/14) is the point on
the line closest to the origin.
79. Since the origin is beneath Denver and 1650 meters = 1.65 km, Denver’s coordinates, in kilometers, are (0, 0, 1.65). From
Figure 17.4, we see the x and y coordinates of Bismark are given by
x = 850 cos 60◦ = 425 km
and y = 850 sin 60◦ = 736 km.
Since 550 meters = 0.55 km, the coordinates of Bismark in kilometers are (425, 736, 0.55).
−−→
−−→
The velocity vector, ~v , of the plane is parallel to the vector DB joining Denver to Bismark, where DB = 425~i +
~
~
~
~
736~j + (0.55 − 1.65)
√k = 425i + 736j − 1.1k .
−−→
Since ||DB|| = 4252 + 7362 + 1.12 ≈ 850 km and the plane is moving at 650 km/hr, the velocity vector is given
by
650
(425~i + 736~j − 1.1~k ) = 325~i + 563~j − 0.84~k .
~v =
850
Since the plane is 8000 m = 8 km above Denver, it passes through the point (0, 0, 9.65). Therefore the parametric
equation is
~r = 9.65~k + t(325~i + 563~j − 0.84~k ).
17.1 SOLUTIONS
y
Bismark
850 km
Denver
1593
y
60◦
x
x
Figure 17.4
80. (a) We look along the line that passes through P = (1, −2, −1) and is parallel to ~v = ~i + 2~j + ~k . The question is
which plane, the blue or the yellow, this line first meets.
Parametric equations for the line are
x = 1 + t,
y = −2 + 2t,
z = −1 + t.
We substitute these into the equations of the respective planes and solve for t in each case:
(1 + t) + 3(−2 + 2t) − 2(−1 + t) = 6 2(1 + t) + (−2 + 2t) + (−1 + t) = 3
5t − 3 = 6
t=
5t − 1 = 3
9
5
t=
4
5
From this we see that the line first intersects the yellow plane 2x + y + z = 3, when t = 4/5. So you see the yellow
plane. (Note that we did not need to find the points of intersection of the line with the planes.)
(b) A vector from P to a point on the green line gives a direction looking directly at the line. If we get a parametric
equation for the green line then we can write down a vector from P to any variable point on the line.
To get a parametric equation we need a vector parallel to the green line and a point that lies on the green line.
We take the cross product of the normal of the blue plane, n~b = ~i + 3~j − 2~k , and the normal of the yellow plane,
n~y = 2~i + ~j + ~k . This gives a vector 5~i − 5~j − 5~k , so we take ~
u = ~i − ~j − ~k as a vector parallel to the green
line.
We also need one point on the line. For that, we can choose a value of z, and find the corresponding values of x
and y on both the blue and yellow planes. Taking z = 0, say, gives the equations x + 3y = 6 and 2x + y = 3, which
have x = 3/5 and y = 9/5 as solutions. So a point on the green line is Q = (3/5, 9/5, 0). Therefore a parametric
equation for the green line is
9
3
x = + t, y = − t, z = −t.
5
5
A vector from P = (1, −2, −1) to a variable point on the line is then, for −∞ < t < ∞,
2
19
9
3
+ t ~i + −2 −
− t ~j + (−1 − (−t))~k = − + t ~i + −
+ t ~j + (−1 + t)~k .
5
5
5
5
Thus, if we look in the direction of ~v , for any value of t, we look at the line.
(c) Consider the plane that contains the point P and the green line; let’s call it the green plane. The green plane divides
3-space into two half-spaces. From P , if we look in a direction pointing into one of the half-spaces we see the yellow
plane (as in part (a)) and if we look in a direction pointing into the other half-space we see the blue plane. We have
to figure out which half-space is which.
We need a normal vector to the green plane. We know the point P = (1, −2, −1) on the plane and the equation
of the green line. We find that a normal vector to the green plane is ~
n = 2~i + ~j + ~k .
From part (a) we know that the vector ~v = ~i + 2~j + ~k points from P into the half-space where we see the
yellow plane. The dot product of ~n and ~v is
~v = 1 −
~n · ~v = 2 · 1 + 1 · 2 + 1 · 1 = 5 > 0.
This means that any vector pointing into this half-space has a positive dot product with ~n . Thus the condition on a
general vector w
~ = a~i + b~j + c~k to point into this half-space is
2a + b + c > 0;
Similarly, w
~ points into the half-space where we see the blue plane if
2a + b + c < 0.
1594
Chapter Seventeen /SOLUTIONS
81. (a) If ~n · ~v = 0, then ~n and ~v are perpendicular. Since P1 is perpendicular to ~n and L is parallel to ~v , we see that P1
and L are parallel. In fact, L may lie in the plane.
(b) Since ~
n × ~v is perpendicular to ~n and to ~v , the vector ~n × ~v is parallel to P1 and perpendicular to L. Thus, P2 ,
which is perpendicular to ~
n × ~v , is
(i) Perpendicular to P1 .
(ii) Parallel to L.
82. (a)
(i)
(ii)
(iii)
(iv)
(b) (i)
(ii)
(iii)
(iv)
(v)
is the original graph reflected in both the x- and y-axes, so (C).
is the original graph reflected in the y-axis, so (A).
is the original graph shifted right by 1, so (D).
is the original graph shifted right by 1 and up by 1, so (G).
Not possible; all points would lie on a circle
Not possible; spiral would be equally spaced.
Possible; spirals increase in diameter as t increases.
Not possible; spiral would be equally spaced.
Not possible; all points would lie on a circle
Strengthen Your Understanding
83. The two parameterizations are different, but the curves they describe are the same. A shift of the curve in space by two
units in the ~i -direction can be parameterized by ~r 2 (t) = ~r (t) + 2~i .
84. The curve is a helix centered on the z-axis. All its points are at distance |R| from the z-axis. The distance from the origin
to the point on the curve with position vector ~r (t) is given by
Distance =
p
~r (t) · ~r (t) =
p
R 2 + t2 .
85. The curves x = cos t, y = sin t, z = 0 and x = 0, y = cos t, z = sin t are both unit circles centered at the origin. The
first is in the xy-plane, and the second is in the yz-plane.
86. Examples of two different lines through the point (1, 2, 3) are given by
~r (t) = ~i + 2~j + 3~k + t ~i + 2~j
and
~r (t) = ~i + 2~j + 3~k + t ~i − ~k .
87. The line
x = t, y = 2t, z = 3 + 4t
can also be parameterized by
x = t3 , y = 2t3 , z = 3 + 4t3 ,
and the functions x = t3 , y = 2t3 , and z = 3 + 4t3 are not linear functions of t.
88. False. The y coordinate is zero when t = 0, but when t = 0 we have x = 2 so the curve never passes through (0, 0).
89. True. Every point (x, y) on this curve satisfies y = (t2 )2 = x2 .
90. False. For example, the graph of x = cos t, y = sin t for 0 ≤ t ≤ 2π is a circle. A circle is not the graph of a function,
since for some values of x there are two values of y.
91. True. Every y-coordinate is one less than every x-coordinate, so the equation of the line is y = x − 1.
92. False. When t = 0, we have (x, y) = (0, −1). When t = π/2, we have (x, y) = (−1, 0). Thus the circle is being traced
out clockwise.
93. True. The functions et and ln t are inverses, so ln et = t. Thus if x = et , y = t, we have y = t = ln et = ln x.
94. True. Taking two values for t, say t = 0 and t = 1 give the points (1, 0) and (0, 2), which lie on a line with equation
y = −2x+2. The second parameterization describes the same set of points, since y = −4s+2 = −2(2s)+2 = −2x+2.
95. True. Adding the equations z = x + y and z = 1 − x − y gives 2z = 1 or z = 12 . Thus the line of intersection is parallel
to the xy-plane at height z = 12 . Letting x be the parameter t and z = 21 in the first plane’s equation gives 21 = t + y or
y = 12 − t. The same result is obtained by setting x = t and z = 21 in the second plane’s equation.
17.2 SOLUTIONS
1595
96. True. To find an intersection point, we look for values of s and t that make the coordinates in the first line the same as the
coordinates in the second. Setting x = t and x = 2s equal, we see that t = 2s. Setting y = 2 + t equal to y = 1 − s, we
see that t = −1 − s. Solving both t = 2s and t = −1 − s yields t = − 32 , s = − 31 . These values of s and t will give equal
x and y coordinates on both lines. We need to check if the z coordinates are equal also. In the first line, setting t = − 32
gives z = 73 . In the second line, setting s = − 31 gives z = − 31 . As these are not the same, the lines do not intersect.
97. False. All points on this line lie in the plane x = 1, so the line is parallel to the yz-plane.
98. True. The ~j component of ~r is always one more than twice the ~i component, so the line is y = 2x + 1.
99. False. The line ~r 1 (t) is in the direction of the vector ~i − 2~j , while the line ~r 2 (t) is in the direction of the vector ~2 i − ~j .
Since these vectors are not parallel (they are not scalar multiples of one another) the lines are not parallel.
Solutions for Section 17.2
Exercises
1. The velocity vector ~v is given by:
d
d
d
(2 + 3t)~i + (4 + t)~j + (1 − t)~k = 3~i + ~j − ~k .
dt
dt
dt
The acceleration vector ~a is given by:
~v =
~a =
d(3)~
d(1)~
d(1) ~
d~v
=
i +
j −
k = ~0
dt
dt
dt
dt
2. The velocity vector ~v is given by:
d
d
d
(2 + 3t2 )~i + (4 + t2 )~j + (1 − t2 )~k = 6t~i + 2t~j − 2t~k .
dt
dt
dt
The acceleration vector ~a is given by:
~v =
~a =
d(2t)~
d(2t) ~
d(6t)~
i +
j −
k = 6~i + 2~j − 2~k .
dt
dt
dt
3. The velocity vector ~v is given by:
d
d
d ~
ti + t2~j + t3~k = ~i + 2t~j + 3t2~k .
dt
dt
dt
The acceleration vector ~a is given by:
~v =
~a =
d(1) ~
d(2t)~
d(3t2 ) ~
d~v
=
ti +
j +
k = 2~j + 6t~k .
dt
dt
dt
d
4. The velocity vector ~v is given by:
d(t)~
i +
dt
The acceleration vector ~a is given by:
~v =
~a =
d 3
(t − t) ~j = ~i + (3t2 − 1)~j .
dt
d(1)~
d~v
=
i +
dt
dt
d
(3t2 − 1) ~j = 6t~j .
dt
5. The velocity vector ~v is given by:
d
d
(3 cos t)~i + (4 sin t)~j = −3 sin t~i + 4 cos t~j .
dt
dt
The acceleration vector ~a is given by:
~v =
~a =
d
d
d~v
= (−3 sin t)~i + (4 cos t)~j = −3 cos t~i − 4 sin t~j .
dt
dt
dt
6. Since ~r (t) = 3 cos(t2 )~i + 3 sin(t2 )~j + t2~k , we have
~v (t) = −6t sin (t2 )~i + 6t cos (t2 )~j + 2t~k ,
~a (t) = (−6 sin (t2 ) − 12t2 cos (t2 ))~i + (6 cos (t2 ) − 12t2 sin (t2 ))~j + 2~k .
1596
Chapter Seventeen /SOLUTIONS
7. The velocity vector ~v is given by:
d
d
d ~
(t)i + (t2 )~j + (t3 )~k
dt
dt
dt
= ~i + 2t~j + 3t2~k .
~v =
The speed is given by:
2
k~v k =
4
p
1 + 4t2 + 9t4 .
Now k~v k is never zero since 1 + 4t + 9t ≥ 1 for all t. Thus, the particle never stops.
8. The velocity vector ~v is given by:
~v =
d
d
(cos 3t)~i + (sin 5t)~j = −3 sin 3t~i + 5 cos 5t~j .
dt
dt
The speed is given by
p
9 sin2 (3t) + 25 cos2 (5t).
k~v k =
Thus, k~v k = 0 when sin(3t) = cos(5t) = 0 but there are no values of t for which this is true, so the particle never stops.
9. To find ~v (t) we first find dx/dt = 6t and dy/dt = 3t2 . Therefore, the velocity vector is ~v = 6t~i + 3t2~j . The speed of
the particle is given by the magnitude of the vector,
k~v k =
r
dx
dt
2
+
dy
dt
2
=
p
(6t)2 + (3t2 )2 = 3|t| ·
p
4 + t2 .
The particle stops when ~v = ~0 , so when 6t = 3t2 = 0. Therefore, the particle stops when t = 0.
10. The velocity vector ~v is given by:
d
d
d
((t − 1)2 )~i + (2)~j + (2t3 − 3t2 )~k
dt
dt
dt
= 2(t − 1)~i + (6t2 − 6t)~k .
~v =
The speed is given by:
k~v k =
p
p
(2(t − 1))2 + (6t2 − 6t)2 = 2|t − 1|
1 + 9t2 .
The particle stops when ~v = ~0 , so when 2(t − 1) = (6t2 − 6t) = 0. Since these are all satisfied only by t = 1, this is
the only time that the particle stops.
11. To find ~v (t) we first find dx/dt = 6t cos(t2 ) and dy/dt = −6t sin(t2 ). Therefore, the velocity is ~v = 6t cos(t2 )~i −
6t sin(t2 )~j . The speed of the particle is given by
k~v k =
=
p
(6t cos(t2 ))2 + (−6t sin(t2 ))2
p
36t2 (cos(t2 ))2 + 36t2 (sin(t2 ))2
p
= 6|t|
cos2 (t2 ) + sin2 (t2 )
= 6|t|.
The particle comes to a complete stop when speed is 0, that is, if 6|t| = 0, and so when t = 0 .
12. The velocity vector ~v is given by:
~v =
d
d
d
(3 sin2 t)~i + (cos t − 1)~j + (t2 )~k = 6 sin t cos t~i − sin t~j + 2t~k .
dt
dt
dt
The speed is given by:
p
k~v k = 36 sin2 t cos2 t + sin2 t + 4t2 .
The particle comes to a stop when ~v = ~0 , so when when 6 sin t cos t = − sin t = 2t = 0, and so the particle stops when
t = 0.
13. We have
Length =
Z
1
2
p
(x′ (t))2
+
(y ′ (t))2
+
(z ′ (t))2
dt =
Z
1
This is the length of a straight line from the point (8, 5, 2) to (13, 9, 1).
2
p
52 + 42 + (−1)2 dt =
√
42.
17.2 SOLUTIONS
1597
14. We have
Length =
2π
Z
p
(−3 sin 3t)2 + (5 cos 5t)2 dt.
0
We cannot find this integral symbolically, but numerical methods show Length ≈ 24.6.
15. We have
Length =
Z
1
(−et sin(et ))2 + (et cos(et ))2 dt
0
=
Z
p
1
√
e2t dt =
Z
1
et dt
0
0
= e − 1.
This is the length of the arc of a unit circle from the point (cos 1, sin 1) to (cos e, sin e)—in other words between the
angles θ = 1 and θ = e. The length of this arc is (e − 1).
16. The velocity vector is
1
~v = ~r ′ (t) = 2~i + ~j + 2t~k ,
t
so
Length of curve =
Z
2
||~v || dt =
1
=
Z
1
2
r
Z
2
1
2t2 )2
(1 +
t2
r
1
4 + 2 + 4t2 dt =
t
dt =
Z
1
2
2
Z
1 + 2t
dt =
t
2
1
Z
1
r
2
4t2 + 1 + 4t4
dt
t2
1
+ 2t
t
dt = 3 + ln 2.
Note that when we took the square root, we used the fact that (1 + 2t2 )/t is positive for 1 ≤ t ≤ 2.
17. The velocity vector ~v is
~v =
dy
dz ~
dx~
i + ~j +
k = 3(2π)(− sin(2πt))~i + 3(2π) cos(2πt)~j + 0~k
dt
dt
dt
= −6π sin(2πt)~i + 6π cos(2πt)~j .
The acceleration vector ~a is
~a =
d2 y
d2 z
d2 x~
i + 2 ~j + 2 ~k = −6π(2π) cos(2πt)~i + 6π(2π)(− sin(2πt))~j
2
dt
dt
dt
= −12π 2 cos(2πt)~i − 12π 2 sin(2πt)~j .
To check that ~v and ~a are perpendicular, we check that the dot product is zero:
~v · ~a = (−6π sin(2πt)~i + 6π cos(2πt)~j ) · (−12π 2 cos(2πt)~i − 12π 2 sin(2πt)~j )
= 72π 3 sin(2πt) cos(2πt) − 72π 3 cos(2πt) sin(2πt) = 0
The speed is
k~v k = k − 6π sin(2πt)~i + 6π cos(2πt)~j k = 6π
and so is constant. The magnitude of the acceleration is
p
k~a k = k − 12π 2 cos(2πt)~i − 12π 2 sin(2πt)~j k = 12π 2
which is also constant.
sin2 (2πt) + cos2 (2πt) = 6π,
p
cos2 (2πt) + sin2 (2πt) = 12π 2 ,
18. The velocity vector ~v is
~v =
dy
dz
dx~
i + ~j + ~k = 0~i + 2(3) cos(3t)~j + 2(3)(− sin(3t))~k
dt
dt
dt
= 6 cos(3t)~j − 6 sin(3t)~k .
1598
Chapter Seventeen /SOLUTIONS
The acceleration vector ~a is
~a =
d2 y ~
d2 z ~
d2 x~
i
+
j
+
k = 6(3)(− sin(3t))~j − 6(3) cos(3t)~k
dt2
dt2
dt2
= −18 sin(3t)~j − 18 cos(3t)~k .
To check that ~v and ~a are perpendicular, we check that the dot product is zero:
~v · ~a = (6 cos(3t)~j − 6 sin(3t)~k ) · (−18 sin(3t)~j − 18 cos(3t)~k )
= −108 cos(3t) sin(3t) + 108 sin(3t) cos(3t) = 0.
The speed is
p
k~v k = k6 cos(3t)~j − 6 sin(3t)~k k = 6
and so is constant. The magnitude of the acceleration is
sin2 (3t) + cos2 (3t) = 6,
p
k~a k = k − 18 sin(3t)~j − 18 cos(3t)~k k = 18
which is also constant.
sin2 (3t) + cos2 (3t) = 18,
19. In vector form the parameterization is
~r = 2~i + 3~j + 5~k + t2 (~i − 2~j − ~k ).
Thus the motion is along the straight line through (2, 3, 5) in the direction of ~i − 2~j − ~k . The velocity vector ~v is
~v =
The acceleration vector ~a is
~a =
The speed is
dy
dz ~
dx~
i + ~j +
k = 2t(~i − 2~j − ~k )
dt
dt
dt
d2 y
d2 z
d2 x~
i + 2 ~j + 2 ~k = 2(~i − 2~j − ~k ).
2
dt
dt
dt
√
k~v k = 2|t|k~i − 2~j − ~k k = 2 6|t|.
The acceleration vector is constant and points in the direction of ~i − 2~j − ~k . When t < 0 the absolute value |t| is
decreasing, hence the speed is decreasing. Also, when t < 0 the velocity vector 2t(~i − 2~j − ~k ) points in the direction
opposite to ~i − 2~j − ~k . When t > 0 the absolute value |t| is increasing and hence the speed is increasing. Also, when
t > 0 the velocity vector points in the same direction as ~i − 2~j − ~k .
20. In vector form the parameterization is
~r = ~i + −5~j − 2~k + (2t3 + 3t)(−~i + 2~j + 3~k ).
Thus the motion is along the straight line through (1, −5, −2) in the direction of −~i + 2~j + 3~k . The velocity vector ~v is
~v =
dy
dz
dx~
i + ~j + ~k = (6t2 + 3)(−~i + 2~j + 3~k )
dt
dt
dt
The acceleration vector ~a is
~a =
The speed is
d2 y ~
d2 z ~
d2 x~
i
+
j
+
k = 12t(−~i + 2~j + 3~k ).
dt2
dt2
dt2
√
√
k~v k = |6t2 + 3|k − ~i + 2~j + 3~k k = 3 14|2t2 + 1| = 3 14(2t2 + 1).
The graph of the speed is a parabola opening upward with vertex at t = 0. Thus the speed is decreasing when t < 0
and increasing when t > 0. The velocity vector always points in the same direction −~i + 2~j + 3~k , since 6t2 + 3 is
always positive. The acceleration vector points in the opposite direction to −~i + 2~j + 3~k when t < 0 and in the same
direction when t > 0. Thus the acceleration vector points in the opposite direction to the speed when t < 0 and in the
same direction when t > 0.
1599
17.2 SOLUTIONS
21. At t = 2, the position and velocity vectors are
~r (2) = (2 − 1)2~i + 2~j + (2 · 23 − 3 · 22 )~k = ~i + 2~j + 4~k ,
~v (2) = 2 · (2 − 1)~i + (6 · 22 − 6 · 2)~k = 2~i + 12~k .
So we want the line going through the point (1, 2, 4) at the time t = 2, in the direction 2~i + 12~k :
x = 1 + 2(t − 2),
y = 2 z = 4 + 12(t − 2).
Problems
22. A parameterization is
or equivalently
~r (t) = 5~i + 4~j − 2~k + (t − 4)(2~i − 3~j + ~k )
x = 5 + 2(t − 4),
y = 4 − 3(t − 4),
z = −2 + (t − 4).
23. The velocity vector for this motion is
~v = (2t − 6)~i + (3t2 − 3)~j .
The motion is vertical when the component in the ~i direction is 0 and motion in ~j direction is not 0. Motion in ~i direction
is 0 when
2t − 6 = 0,
t = 3.
At that time, motion in ~j direction is not 0. The motion is horizontal when the component in the ~j direction is 0 and
motion in ~i direction is not 0. Motion in ~j direction is 0 when
3t2 − 3 = 0,
t = 1, −1.
At these times, motion in ~i direction is not 0. To determine the end behavior, recall that a polynomial is approximated by
its highest powered term for large values (positive or negative) of the independent variable. Thus, as t → ±∞, we have
x ≈ t2 and y ≈ t3 . The end behavior, and the x and y coordinates when the motion is vertical or horizontal, are shown in
Table 17.1. The graph is shown in Figure 17.5.
y
190
140
Table 17.1
t
90
x
y
−∞
+∞
−∞
1
−5
−2
+∞
+∞
−1
3
+∞
7
−9
40
2
18
−10
−10
x
10
Figure 17.5
20
1600
Chapter Seventeen /SOLUTIONS
24. The velocity vector for this motion is
~v =
dy
dx~
i + ~j = (3t2 − 12)~i + (2t + 10)~j .
dt
dt
The motion is vertical when the component in the ~i direction is 0 and motion in ~j direction is not 0. Motion in ~i direction
is 0 when
3t2 − 12 = 0,
t = 2, −2.
At these times, motion in ~j direction is not 0. The motion is horizontal when the component in the ~j direction is 0 and
motion in ~i direction is not 0. Motion in ~j direction is 0 when
2t + 10 = 0,
t = −5.
At this time, the motion in ~i direction is not 0. To determine the end behavior, recall that a polynomial is approximated by
its highest powered term for large values (positive or negative) of the independent variable. Thus, as t → ±∞, we have
x ≈ t3 and y ≈ t2 . The end behavior, and the x and y coordinates when the motion is vertical or horizontal, are shown in
Table 17.2. The graph is shown in Figure 17.6.
y
80
Table 17.2
t
−∞
−5
−2
2
+∞
40
x
y
−∞
+∞
16
−16
−65
−25
−16
24
+∞
−800
−400
x
−40
+∞
Figure 17.6
25. Plotting the positions on the xy plane and noting their times gives the graph shown in Figure 17.7.
y
t=3
t = 2.5, 3.5
10
t=4
8
t=2
6
t = 1.5
4
t=0
t=1
2
0
t = 0.5
x
2
4
6
Figure 17.7
8
17.2 SOLUTIONS
1601
(a) We approximate dx/dt by ∆x/∆t calculated between t = 1.5 and t = 2.5:
∆x
3−7
−4
dx
≈
=
=
= −4.
dt
∆t
2.5 − 1.5
1
Similarly,
dy
∆y
10 − 5
5
≈
=
= = 5.
dt
∆t
2.5 − 1.5
1
So,
~v (2) ≈ −4~i + 5~j
and Speed = k~v k =
√
41.
(b) The particle is moving vertically at about time t = 1.5. Note that the particle is momentarily stopped at about t = 3;
however it is not moving parallel to the y-axis at this instant.
(c) The particle stops at about time t = 3 and reverses course.
−
−
→
26. (a) The vector P Q between the points is given by
−
−→
P Q = 2~i + 5~j + 3~k .
√
√
−
−
→
Since ||P Q|| = 22 + 52 + 32 = 38, the velocity vector of the motion is
5
~v = √ (2~i + 5~j + 3~k ).
38
(b) The motion is along a line starting at the point (3, 2, −5) and with the velocity vector from part (a). The equation of
the line is
5
~r = 3~i + 2~j − 5~k + t~v = 3~i + 2~j − 5~k + √ (2~i + 5~j + 3~k )t,
38
so
10
25
15
x = 3 + √ t,
y = 2 + √ t,
z = −5 + √ t.
38
38
38
27. (a) The particle starts at (2, −1, 5) so ~r 0 = 2~i − ~j + 5~k . In 5 seconds, the particle moves through a displacement given
−
−
→
by P Q = 3~i + 4~j − 6~k . Its velocity, ~v , is given by
4
6
3
~v = ~i + ~j − ~k = 0.6~i + 0.8~j − 1.2~k .
5
5
5
Thus, the equation of the motion is
~r = 2~i − ~j + 5~k + t(0.6~i + 0.8~j − 1.2~k )
or
x = 2 + 0.6t,
y = −1 + 0.8t,
z = 5 − 1.2t
where 0 ≤ t ≤ 5.
(b) The velocity vector in part (a), ~v = 0.6~i + 0.8~j − 1.2~k , means that the particle is moving with
Speed = ||~v || =
p
(0.6)2 + (0.8)2 + (1.2)2 = 1.562.
To make the speed equal 5, take a new velocity vector given by
~v =
5
(0.6~i + 0.8~j − 1.2~k ) = 1.92~i + 2.56~j − 3.84~k .
1.562
Thus, the equation of the motion is
~r = 2~i − ~j + 5~k + t(1.92~i + 2.56~j − 3.84~k )
or
x = 2 + 1.92t,
The particle reaches Q = (5, 3, −1) when
y = −1 + 2.56 t,
z = 5 − 3.84t.
2 + 1.92t = 5
or t = 1.56 seconds. The parametric equations describe the motion from P to Q when 0 ≤ t ≤ 1.56.
1602
Chapter Seventeen /SOLUTIONS
28. (a) We substitute x = 1 + t, y = 5 + 2t, z = −7 + t into x + y + z = 1 and solve for t:
(1 + t) + (5 + 2t) + (t − 7) = 1
4t − 1 = 1
t = 0.5 sec.
When t = 0.5, the particle is at the point (x, y, z) = (1 + (0.5), 5 + 2(0.5), (0.5) − 7) = (1.5, 6, −6.5).
(b) The particle’s velocity is
~v = ~i + 2~j + ~k ,
so
Speed = ||12 + 22 + 12 || =
√
6 meters/sec.
29. (a) At t = 0, we have ~r (0) = 0~i + 0~j + 6.4~k , so the stone’s initial position is (0, 0, 6.4). Thus the rooftop is 6.4
meters above the ground.
(b) The stone hits the ground when the height above the ground is 0; that is, when its z coordinate is 0:
6.4 − 4.9t2 = 0
t=±
r
6.4
= ±1.14.
4.9
Since t must be positive, the stone hits the ground about 1.14 seconds after it is thrown.
(c) The velocity of the stone at time t is given by
~v (t) = ~r ′ (t) = 10~i − 5~j − 9.8t~k ,
so when the stone hits the ground at t = 1.14 seconds,
~v (1.14) = 10~i − 5~j − 9.8(1.14)~k = 10~i − 5~j − 11.172~k .
√
The stone’s speed is given by ||~v (1.14)|| = 102 + 52 + 11.1722 = 15.81 meters/sec.
(d) The stone hits the ground at the point with position vector
~r (1.14) = 10(1.14)~i − 5(1.14)~j + (6.4 − 4.9(1.14)2 )~k ,
which is the point (11.4, −5.7, 0).
(e) The acceleration of the stone at time t is given by
~a (t) = ~v ′ (t) = −9.8~k .
Thus, the acceleration is constant; the stone hits the ground at an acceleration of −9.8 meters/sec2 ; that is 9.8 meters/sec2
downward.
30. (a) Since z = 90 feet when t = 0, the tower is 90 feet high.
(b) The child reaches the bottom when z = 0, so t = 90/5 = 18 minutes.
(c) Her velocity is given by
d~r
= −(10 sin t)~i + (10 cos t)~j − 5~k ,
~v =
dt
so
p
p
√
Speed = ||~v || = (−10 sin t)2 + (10 cos t)2 + (−5)2 = 102 + 52 = 125 ft/min.
(d) Her acceleration is given by
~a =
d~v
= −(10 cos t)~i − (10 sin t)~j ft/min2 .
dt
31. (a) The height at time t is given by z = 100 − (t − 5)2 , so the maximum height occurs at t = 5 secs, when r =
10~i + 15~j + 100~k , so the point is (10, 15, 100).
(b) The velocity of the particle at time t is given by
~v = 2~i + 3~j − 2(t − 5)~k ,
so the speed is
||~v || =
p
22 + 32 + 22 (t − 5)2 =
p
13 + 4(t − 5)2 cm/sec.
Thus, the maximum speed occurs when t = 0 secs and when t = 10 secs and is given by ||~v || =
√
113 = 10.630 cm/sec.
√
(c) The minimum speed occurs when t = 5 and is given by ||~v || = 13 = 3.606 cm/sec.
p
13 + 4(52 ) =
17.2 SOLUTIONS
1603
32. We want dw/dt at t = 0. The chain rule gives
dw
∂w dx
∂w dy
∂w dz
=
+
+
.
dt
∂x dt
∂y dt
∂z dt
Since ~r ′ (0) = 2~i + 3~j + 6~k and
~r ′ (t) = x′ (t)~i + y ′ (t)~j + z ′ (t)~k ,
we have
x′ (0) = 2,
y ′ (0) = 3,
z ′ (0) = 6.
Since grad f = 4~i − 3~j + ~k at the point (7, 2, 5) which the particle reaches at time t = 0, we have
∂w
∂x
∂w
∂y
= 4,
(7,2,5)
Thus
dw
dt
t=0
33. (a) The ball hits the ground when y = 0, so
−25 ±
= 1.
(7,2,5)
= 4·2−3·3+1·6 = 5
2 + 25t − 4.9t2 = 0.
The quadratic formula gives
t=
(7,2,5)
∂w
∂z
= −3,
p
252 − 4 · 2(−4.9)
= −0.079 or 5.181 sec.
−2(4.9)
We need the positive answer, t = 5.181 sec.
(b) At the time the ball hits the ground, x = 20(5.181) = 103.616 meters. Thus, the ball hits the ground after 5.181
seconds at a point 103.616 meters horizontally from where it was thrown.
(c) h = 2 meters.
(d) g = 9.8 meters/sec2 .
(e) Since v cos θ = 20 and v sin θ = 25, we have
tan θ =
v sin θ
25
=
= 1.25,
v cos θ
20
so
θ = arctan(1.25) = 0.896.
Then
v=
20
= 32.016 meters/sec.
cos 0.896
34. (a) To eliminate t, substitute t = x/20 into the equation for z. This gives
z=5
x
20
− 0.5
x 2
20
=
x2
x
−
.
4
800
(b) To decide when the particle is at ground level, set the equation for z equal to 0 and solve for t:
5t − 0.5t2 = 0
−0.5t(t − 10) = 0,
so t = 0 and t = 10 seconds.
(c) The particle’s velocity is ~v = x′ (t)~i + z ′ (t)~k , so
~v = 20~i + (5 − t)~k .
(d) The particle’s speed is
p
||~v || =
p
202 + (5 − t)2 =
p
400 + (5 − t)2 m/s.
(e) No. The quantity 400 + (5 − t)2 is never 0.
(f) The particle is at its highest point halfway between the times when it is at ground level, or when t = 5. Alternatively,
the highest point occurs when z ′ = 0, that is
z ′ (t) = 5 − t = 0
so
t = 5 seconds.
1604
Chapter Seventeen /SOLUTIONS
35. (a) The top of the tower is at the point (0, 0, 20), so we want ~r (0) = 20~k . This is (I) and (IV). Only (IV) is going
downward.
√
2
Projectile (IV) hits the ground when z = 0, which occurs when 20 − t√
= 0, so t = 20 = 4.5. (We take the
positive root since the projectile is launched when t = 0.) At this time, ~r ( 20) = 8.9~j , so the projectile hits the
ground at the point (0, 8.9, 0), which is 8.9 meters from the base of the tower in the direction of the tree.
(b) To hit the top of the tree, the projectile must go through the point (0, 20, 20). This is (II).
√
The projectile reaches the top of the tree when 2t2 = 20, so (taking the positive root) t = 10 = 3.2 sec. The
projectile is launched from ~r (0) = ~0 , the base of the tower.
(c) Projectiles launched from somewhere on the tower have x(0) = y(0) = 0 and 0 ≤ z(0) ≤ 20. Only (III) and (V)
have nonzero x(0) and y(0).
To hit the tree, there must be a time for which the projectile is at a point (0, 20, z) for some 0 ≤ z ≤ 20.
Since (III) has x(t) = 20 for all t, it does not hit the tree. So (V) is the answer.
For (V), we have 2t = 20, when t = 10 sec. Then ~r (10) = 20~j + 10~k , so the projectile hits the tree at
(0, 20, 10), which is half way up.
36. (a) For any positive constant k, the parameterization
x = −5 sin(kt) y = 5 cos(kt)
moves counterclockwise on a circle of radius 5 starting at the point (0, 5). We choose k to make the period 8 seconds.
If k · 8 = 2π, then k = π/4 and the parameterization is
x = −5 sin
πt
4
y = 5 cos
(b) Since it takes 8 seconds for the particle to go around the circle
Speed =
πt
.
4
2π(5)
Circumference of circle
5π
=
=
cm/sec.
8
8
4
37. Since the acceleration due to gravity is −9.8 m/sec2 , we have ~r ′′ (t) = −9.8~k . Integrating gives
~r ′ (t) = C1~i + C2~j + (−9.8t + C3 )~k ,
~r (t) = (C1 t + C4 )~i + (C2 t + C5 )~j + (−4.9t2 + C3 t + C6 )~k .
The initial condition, ~r (0) = ~0 , implies that C4 = C5 = C6 = 0, thus
~r (t) = C1 t~i + C2 t~j + (−4.9t2 + C3 t)~k .
To find the position vector, we need to find the values of C1 , C2 , and C3 . This we do using the coordinates of the
highest point. When the rocket reaches its peak, the vertical component of the velocity is zero, so −9.8t + C3 = 0. Thus,
at the highest point, t = C3 /9.8. At that time
~r (t) = 1000~i + 3000~j + 10000~k ,
so, for the same value of t:
C1 t = 1000,
C2 t = 3000,
−4.9t2 + C3 t = 10, 000,
Substituting t = C3 /9.8 into the third equation gives
−4.9
C3
9.8
2
+
C32
= 10,000
9.8
2
C3 = 2(9.8)10,000
C3 = 442.7
Then C1 =
1000
C3 /9.8
= 22.1 and C2 =
3000
C3 /9.8
= 66.4. Thus,
~r (t) = 22.1t~i + 66.4t~j + (442.7t − 4.9t2 )~k .
17.2 SOLUTIONS
1605
38. (a) The parametric equation describing Emily’s motion is
x = 10 cos
Her velocity vector is
2π
20
t = 10 cos
~v =
π
10
t ,
y = 10 sin
2π
20
t = 10 sin
π
10
t
z = constant.
dx~
dy
dz ~
π ~
π ~
i + ~j +
k = −π sin
t i + π cos
t j.
dt
dt
dt
10
10
Her speed is given by:
k~v k =
r
−π sin
=π
r
sin2
π 2
10
t
π
+ π cos
t + cos2
10
√
= π 1 = π m/sec,
π
10
π 2
10
t
+ 02
t
which is independent of time (as we expected). This is certainly the long way to solve this problem though, since we
could have simply divided the circumference of the circle (20π) by the time taken for a single rotation (20 seconds)
to arrive at the same answer.
(b) When Emily drops the ball, it initially has Emily’s velocity vector, but it immediately begins accelerating in the zdirection due to the force of gravity. The motion of the ball will then be tangential to the merry-go-round, curving
down to the ground. In order to find the tangential component of the ball’s motion, we must know Emily’s velocity
at the moment she dropped the ball. Then we can integrate the velocity and obtain the position of the ball. Assuming
Emily drops the ball at time t = 0, her position and velocity vector are
~r (0) = 10~i + 3~k and ~v (0) = π~j .
Thus, the ball has velocity only in the y-direction when it is dropped. In the z-direction, we have
Acceleration =
d2 z
= −9.8 m/sec2 .
dt2
Since the initial velocity 0 and initial height 3, we have
z = 3 − 4.9t2 .
The ball touches the ground when z = 0, that is, when t = 0.78 sec. In that time, the ball also travels π(0.78) = 2.45
meters in the y-direction. So, the final position is (10, 2.45, 0). The distance between this point and P = (10, 0, 0)
is 2.45 meters.
(c) The distance of the ball from Emily when it hits the ground is found by finding Emily’s position at t = 0.78 sec and
using the distance formula. Emily’s position when the ball hits the ground is (10 cos(0.078π), 10 sin(0.078π), 3) =
(9.70, 2.43, 3). The distance between this point and the point where the ball struck the ground is:
d≈
p
(10 − 9.70)2 + (2.45 − 2.43)2 + (0 − 3)2 = 3.01 meters.
Note that the merry-go-round does not rotate very much in the 0.78 sec needed for the ball to reach the ground, so
our answer makes sense.
39. Since the particle moves in a circle of radius a we have ||~r (t)|| = a so
~r (t) · ~r (t) = a2 .
Differentiating with respect to t gives
d
d
~r (t) · ~r (t) + ~r (t) · ~r (t) = 0
dt
dt
so
~r ′ (t) · ~r (t) = 0.
Thus, the position vector, ~r (t), and the velocity vector, ~v = ~r ′ (t), are perpendicular at all times t.
1606
Chapter Seventeen /SOLUTIONS
40. (a) The center of the wheel moves horizontally, so its y-coordinate will never change; it will equal 1 at all times. In one
second, the wheel rotates 1 radian, which corresponds to 1 meter on the rim of a wheel of radius 1 meter, and so the
rolling wheel advances at a rate of 1 meter/sec. Thus the x-coordinate of the center, which equals 0 at t = 0, will
equal t at time t. At time t the center will be at the point (x, y) = (t, 1).
(b) By time t the spot on the rim will have rotated t radians clockwise, putting it at angle −t as in Figure 17.8. The
coordinates of the spot with respect to the center of the wheel are (cos(−t), sin(−t)). Adding these to the coordinates
(t, 1) of the center gives the location of the spot as (x, y) = (t + cos t, 1 − sin t). See Figure 17.9.
y
y
(t, 1)
−t
x
x
Figure 17.8
Figure 17.9
41. (a) No. The height of the particle is given by 2t; the vertical velocity is the derivative d(2t)/dt = 2. Because this is a
positive constant, the vertical component of the velocity vector is upward at a constant speed of 2.
(b) When 2t = 10, so t = 5.
(c) The velocity vector is given by
dx~
dy
dz ~
d~r
=
i + ~j +
k
dt
dt
dt
dt
= −(sin t)~i + (cos t)~j + 2~k .
~v (t) =
From (b), the particle is at 10 units above the ground when t = 5, so at t = 5,
~v (5) = 0.959~i + 0.284~j + 2~k .
Therefore, ~v (5) = − sin(5)~i + cos(5)~j + 2~k .
(d) At this point, t = 5, the particle is located at
~r (5) = (cos(5), sin(5), 10) = (0.284, −0.959, 10).
The tangent vector to the helix at this point is given by the velocity vector found in part (c), that is, ~v (5) = 0.959~i +
0.284~j + 2~k . So, the equation of the tangent line is
~r (t) = 0.284~i − 0.959~j + 10~k + (t − 5)(0.959~i + 0.284~j + 2~k ).
42. We have velocity vector ~v (t) = −α sin t~i + α cos t~j + β~k . For the speed we compute
Speed = (α2 cos2 t + α2 sin2 t + β 2 )1/2 =
which does not depend on t.
p
α2 + β 2
43. (a) Let the ant begin the trip at time t = 0, and let’s place the origin of our coordinate system at the center of the disk.
We align the axes so that at time t = 0 the radius along which the ant crawls falls on the positive x-axis. At time t
seconds, the ant is at a distance of r = t cm from the origin and at angle θ = 2πt radians from the positive x-axis.
The Cartesian coordinates of this point are (x, y) = (r cos θ, r sin θ) = (t cos(2πt), t sin(2πt)). We can write the
parametric equations of the ant’s motion in vector form as
~r (t) = t cos(2πt)~i + t sin(2πt)~j , 0 ≤ t ≤ 100.
17.2 SOLUTIONS
1607
(b) The velocity vector of the ant is the derivative
~v (t) = ~r ′ (t) = (cos(2πt) − 2πt sin(2πt))~i + (sin(2πt) + 2πt cos(2πt))~j .
The speed is the magnitude of the velocity vector
k~v k = ((cos(2πt) − 2πt sin(2πt))2 + (sin(2πt) + 2πt cos(2πt))2 )1/2
= (1 + 4π 2 t2 )1/2 cm/sec.
Observe that the speed of the ant is increasing. Even though the ant is crawling at constant rate on the disk, the turning
of the disk moves the ant faster and faster as it gets closer to the edge.
(c) The acceleration vector is
~a = ~v ′ (t) = (−4π sin(2πt) − 4π 2 t cos(2πt))~i + (4π cos(2πt) − 4π 2 t sin(2πt))~j .
The magnitude of the acceleration is
k~a k = ((−4π sin(2πt) − 4π 2 t cos(2πt))2 + (4π cos(2πt) − 4π 2 t sin(2πt))2 )1/2
= 4π(1 + π 2 t2 )1/2 cm/sec2 .
44. (a) Since x = R cos(ωt) and y = R sin(ωt), and x2 + y 2 = R2 cos2 (ωt) + R2 sin2 (ωt) = R2 , we have motion around
a circle of radius R centered at the origin. The particle moves counterclockwise, completing one revolution in time
2π/ω. Thus, the period = 2π/ω.
(b) The velocity vector is
d~r
= −ωR sin(ωt)~i + ωR cos(ωt)~j .
~v =
dt
We expect the velocity, ~v , to be tangent to the circle. To verify that this, we compute
~v · ~r = (−ωR sin(ωt)~i + ωR cos(ωt)~j ) · (R cos(ωt)~i + R sin(ωt)~j )
= −ωR2 sin(ωt) cos(ωt) + ωR2 cos(ωt) sin(ωt) = 0.
This shows that the velocity vector is perpendicular to the radius from the center of the circle to the particle, which
moves counterclockwise.
The speed is k~v k = ωR, which is constant. Notice that this makes sense, because in time 2π/ω, the particle
travels a distance of 2πR, giving a speed of 2πR/(2π/ω) = ωR.
(c) The acceleration vector is
~a =
d~v
= −ω 2 R cos(ωt)~i − ω 2 R sin(ωt)~j = −ω 2~r .
dt
The acceleration vector points in the direction opposite to the position vector ~r , and thus points toward the center of
the circle. It has constant magnitude k~a k = ω 2 R = kvk2 /R.
45. (a) Let x represent horizontal displacement (in cm) from some starting point and y the distance (in cm) above the ground.
Since
25 · 105
25 km/hr =
= 694.444 cm/sec,
602
if t is in seconds, the motion of the center of the pedal is given by
x~i + y~j = 694.444t~i + 30~j .
The circular motion of your foot relative to the center is described by
h~i + k~j = 20 cos(2πt)~i + 20 sin(2πt)~j ,
so the motion of the light on your foot relative to the ground is described by
x~i + y~j = (694.444t + 20 cos(2πt))~i + (30 + 20 sin(2πt))~j .
1608
Chapter Seventeen /SOLUTIONS
(b) See Figure 17.10.
y
40
20
x
600
1200
Figure 17.10
(c) Suppose your pedal is rotating with angular velocity ω radians/sec, so that the motion is described by
x~i + y~j = (694.444t + 20 cos ωt)~i + (30 + 20 sin ωt)~j .
The light moves backward if dx/dt is negative. Since
dx
= 694.444 − 20ω sin ωt,
dt
the minimum value of dx/dt occurs when ωt = π/2, and then
dx
= 694.444 − 20ω < 0
dt
giving
ω ≥ 34.722 radians/sec.
Since there are 2π radians in a complete revolution, an angular velocity of 34.722 radians/sec means 34.722/2π ≈
5.526 revolutions/sec.
46. At time t object B is at the point with position vector ~r B (t) = ~r A (2t), which is exactly where object A is at time 2t.
Thus B visits the same points as A, but does so at different times; A gets there later. While B covers the same path as A,
it moves twice as fast. To see this, note for example that between t = 1 and t = 3, object B moves along the path from
~r B (1) = ~r A (2) to ~r B (3) = ~r A (6) which is traversed by object A during the time interval from t = 2 to t = 6. It takes
A twice as long to cover the same ground.
In the case where ~r A (t) = t~i + t2~j , both objects move on the parabola y = x2 . Both A and B are at the origin at
time t = 0, but B arrives at the point (2, 4) at time t = 1, whereas A does does not get there until t = 2.
47. In uniform circular motion the velocity vector is tangent to the circle of motion and the acceleration vector is directed
toward the center of the circle. At all times the velocity ~v and acceleration ~a are perpendicular. Since ~v · ~a = (2~i + ~j ) ·
(~i + ~j ) = 3 6= 0, ~v and ~a are not perpendicular, and so the object can not be in uniform circular motion.
48. The acceleration vector points from the object to the center of the orbit, and the velocity vector points from the object
tangent to the circle in the direction of motion. From Figure 17.11 we see that the movement is counterclockwise.
Center of orbit
~a
~v
Figure 17.11
17.2 SOLUTIONS
1609
49. (a) Using the product rule for differentiation we get
d~r
d~r
d~r
d
(~r · ~r ) = ~r ·
+
· ~r = 2~r ·
.
dt
dt
dt
dt
(b) Since ~a is a constant, d~a /dt = 0 so the product rule gives
d~r
d
(~a × ~r ) = ~a ×
.
dt
dt
(c) The product rule gives
d~r
d
d~r
d 3
(r ~r ) = r 3
+ (r 3 )~r = r 3
+ 3r 2 ~r .
dt
dt
dt
dt
√
50. (a) Since grad f (1, 7, 2) = ~i − ( 6)~j + ~k is normal to the tangent plane, and since the plane goes through the point
(1, 7, 2), an equation is
√
√
√
x − 6y + z = 1 · 1 − 6 · 7 + 1 · 2 = 3 − 7 6
√
√
x − 6y + z = 3 − 7 6.
√
(b) A vector normal to the level surface is ~n = ~i − ( 6)~j + ~k . The curve C pass through the point (1, 7, 2) at t = 0,
so we find the tangent to the curve at this point:
~r ′ (t) = 2(t + 1)~i − 7 sin t~j + 2et~k ,
′
~r ′ (0) = 2~i − 7 · 0~j + 2e0~k = 2~i + 2~k .
The angle, θ, between ~n and ~r (0) is given by
√
(~i − ( 6)~j + ~k ) · (2~i + 2~k )
~n · ~r ′ (0)
1·2+1·2
4
1
= p
= √ √ = .
=
cos θ =
√
√
√
~
~
~
~
~
||~n || · ||~r ′ (0)||
2
2
2
2
2
2
8 8
||i − ( 6)j + k || · ||2i + 2k ||
1 + ( 6) + 1 · 2 + 2
Thus, θ = arccos(1/2) = π/3.
(c) By the chain rule, the rate at which the concentration, c, is changing is
∂f dx
∂f dy
∂f dz
dc
=
+
+
= grad f · ~r ′ (t).
dt
∂x dt
∂y dt
∂z dt
Thus, at t = 0,
√
dc
= grad f · ~r ′ (0) = (~i − ( 6)~j + ~k ) · (2~i + 2~k ) = 4 ppm per second.
dt
51. Since ~v = s cos θ ~i + s sin θ ~j , the unit vector in the direction of ~v is
T~ = cos θ ~i + sin θ ~j
and
~ = ~k × T~ = − sin θ ~i + cos θ ~j .
N
Using the chain rule to differentiate ~v with respect to t, we have
~a = (s′ cos θ − s(sin θ)θ′ )~i + (s′ sin θ + s(cos θ)θ′ )~j
= s′ (cos θ ~i + sin θ ~j ) + sθ′ (− sin θ ~i + cos θ ~j )
~ .
= s′ T~ + sθ′ N
~ be the force towards O. Then F~ is parallel to ~r (t) but in the opposite direction, so F
~ = −k~r (t) for some constant
52. Let F
~ = m~a , where ~a (t) = ~r ′′ (t) is the acceleration of the particle. Now consider
k. However, by Newton’s second law, F
d
d
d
(~r (t) × ~v (t)) =
(~r (t)) × ~v (t) + ~r (t) ×
(~v (t))
dt
dt
dt
= ~v (t) × ~v (t) + ~r (t) × ~a (t) = ~0 .
since ~r (t) and ~a (t) are parallel and ~v (t) × ~v (t) = ~0 . Thus, ~r (t) × ~v (t) = ~c , a constant.
Since ~r (t) × ~v (t) = ~c , the particle moves so that ~r (t) is perpendicular to ~c . Thus, ~r (t) is always in the plane
perpendicular to ~c that contains the fixed point O.
1610
Chapter Seventeen /SOLUTIONS
53. (a) If ∆t = ti+1 − ti is small enough so that Ci is approximately a straight line, then we can make the linear approximations
x(ti+1 ) ≈ x(ti ) + x′ (ti )∆t,
y(ti+1 ) ≈ y(ti ) + y ′ (ti )∆t,
z(ti+1 ) ≈ z(ti ) + z ′ (ti )∆t,
and so
Length of Ci ≈
≈
=
p
(x(ti+1 ) − x(ti ))2 + (y(ti+1 ) − y(ti ))2 + (z(ti+1 ) − z(ti ))2
p
x′ (ti )2 (∆t)2 + y ′ (ti )2 (∆t)2 + z ′ (ti )2 (∆t)2
p
x′ (ti )2 + y ′ (ti )2 + z ′ (ti )2 ∆t.
(b) From point (a) we obtain the approximation
Length of C =
≈
X
length of Ci
Xp
x′ (ti )2 + y ′ (ti )2 + z ′ (ti )2 ∆t.
The approximation gets better and better as ∆t approaches zero, and in the limit the sum becomes a definite integral:
Length of C = lim
∆t→0
=
Xp
x′ (ti )2 + y ′ (ti )2 + z ′ (ti )2 ∆t
Z bp
x′ (t)2 + y ′ (t)2 + z ′ (t)2 dt.
a
Strengthen Your Understanding
54. Velocity and acceleration are orthogonal for uniform circular motion, which is motion at constant speed. If the speed is
not constant, then velocity and acceleration are not orthogonal. For example, if
~r (t) = (cos t2 )~i + (sin t2 )~j
then
~v (t) · ~a (t) = 4t 6= 0.
55. Acceleration is a vector, not a scalar.
56. The length of the parameter interval is not the same as the length of the curve parameterized. The length of the curve is
given by
Length =
Z
B
A
57. The particle with position
k~v (t)k dt.
~r (t) = t + 2t2 ~i + 2t~j + 3t2~k
has velocity ~v (0) = ~i + 2~j and acceleration ~a (0) = 4~i + 6~k .
58. The curve
~r (t) = cos t~i + sin t~j + t~k ,
with velocity
a≤t≤b
~v (t) = − sin t~i + cos t~j + ~k
has
Length =
Z bp
a
~v (t) · ~v (t) dt =
Z
a
b
√
√
2 dt = (b − a) 2.
√
√
If b − a = 10/ 2, the length is 10. For example, the interval 0 ≤ t ≤ 10/ 2 corresponds to a piece of the helix of
length 10.
17.3 SOLUTIONS
1611
59. False. The velocity vector is ~v (t) = ~r ′ (t) = 2t~i − ~j . Then ~v (−1) = −2~i − ~j and ~v (1) = 2~i − ~j , which are not
equal.
√
√
60. True. The velocity vector is ~v (t) = ~r ′ (t) = 2t~i − ~j , so the speed is s(t) = 4t2 + 1. Then s(−1) = s(1) = 5.
61. False. While this is true for motion in a circle with constant speed, it is not true in general. For a counterexample, consider
motion along a parabola ~r (t) = t~i + t2~j . Then ~v (t) = ~i + 2t~j and ~a (t) = 2~j . Taking the dot product gives
~v (t) · ~a (t) = 4t, which is not zero for all t. Thus the velocity and acceleration vectors are not always perpendicular.
62. False. If a particle is moving along a line with nonconstant speed, then the acceleration and velocity vectors are parallel.
For a counterexample, consider motion along the line ~r (t) = t2~i + t2~j . Then ~v (t) = 2t~i + 2t~j and ~a (t) = 2~i + 2~j ,
so ~v (t) = t~a (t). Thus in this case the velocity vector and acceleration vectors are parallel at all points.
63. False. As a counterexample, consider the curve ~r (t) = t2~i + t2~j for 0 ≤ t ≤ 1. In this case, when t is replaced by −t,
the parameterization is the same, and is not reversed.
64. True. The length of the curve C is given by
Rb
a
||~v (t)|| dt =
Rb
a
1 dt = b − a.
65. False.
The velocity√of the particle is given by ~v (t) = ~r (t) = 3~i + 2~j + ~k , so speed is constant: ||~v (t)|| =
√
32 + 22 + 12 = 14. So the particle never stops.
66. False. As ap
counterexample, consider motion along the helix ~r (t) = cos t~i + sin t~j + t~k . In this case the speed is
√
||~v (t)|| = (− sin t)2 + (cos t)2 + 12 = 2. Thus the particle has constant speed, but is traveling along a helix, not a
line.
′
67. True. We find the tangent vector to a parametrically defined curve by differentiating ~r (t).
68. False. Suppose ~r (t) = (cos t)~i + (sin t)~j , so ~r (t) traces out the unit circle. Then ~r (t) lies along the radius of the circle
and
~r ′ (t) = (− sin t)~i + (cos t)~j
and ~r ′ (t) is tangent to the circle. Thus ~r (t) and ~r ′ (t) are perpendicular, so their cross product is not zero.
69. False. Suppose ~r (t) = t~i + t~j . Then ~r ′ (t) = ~i + ~j and
~r ′ (t) · ~r (t) = (t~i + t~j ) · (~i + ~j ) = 2t.
So ~r ′ (t) · ~r (t) 6= 0 for t 6= 0.
70. False. This result is true for uniform circular motion, but is not true in general.
Solutions for Section 17.3
Exercises
~ = x~i
1. V
~ = −y~i
2. V
~ = x~i + y~j = ~r
3. V
~ = −y~i + x~j
4. V
~ = −x~i − y~j = −~r
5. V
~ = ~r : vectors are of unit length and point outward.
6. V
k~r k
7. (a) Parallel to y-axis.
(b) Length increasing in x-direction.
(c) Length not dependent on y.
8. (a) Not parallel to either axis.
(b) Length does not change as x increases.
(c) The length increases as y increases.
9. (a) Parallel to x-axis
(b) Length increases as x increases
(c) Length decreases as y increases.
1612
Chapter Seventeen /SOLUTIONS
10. (a) Since grad(x4 + e3y ) = 4x3~i + 3e3y~j , the vector field is parallel to neither axis.
(b) The length increases as x increases.
(c) The length increases as y increases.
11. See Figure 17.12.
y
x
~ (x, y) = 2~i + 3~j
Figure 17.12: F
12. See Figure 17.13.
y
x
~ (x, y) = y~i
Figure 17.13: F
13. See Figure 17.14.
y
x
Figure 17.14: F~ (x, y) = −y~j
17.3 SOLUTIONS
14. See Figure 17.15.
y
x
~ (~r ) = 2~r
Figure 17.15: F
15. See Figure 17.16.
y
x
~ (~r ) =
Figure 17.16: F
~r
k~r k
16. See Figure 17.17.
~ (~r ) = −~r /k~r k3
Figure 17.17: F
1613
1614
Chapter Seventeen /SOLUTIONS
17. The vector field points in a clockwise direction around the origin. Since ky~i − x~j k =
as you go away from the origin. See Figure 17.18.
p
y 2 + x2 , the vectors get longer
y
x
Figure 17.18
18. See Figure 17.19.
y
x
Figure 17.19: F~ (x, y) = 2x~i + x~j
19.
y
x
~ (x, y) = (x+y)~i +(x−y)~j
Figure 17.20: F
20. (a) III. The vector field ~r points outward and increases in length farther from the origin.
p
(b) II. The vector field −y~i + x~j is tangent to a circle centered at the origin. Since || − y~i + x~j || = x2 + y 2 , the
vector field given has length 1 everywhere.
(c) IV. The vector field −~r points toward the origin and increases in length farther from the origin.
(d) VI. The vector field y~i − x~j is tangent to a circle centered at the origin and points clockwise.
17.3 SOLUTIONS
1615
21. Notice that for a repulsive force, the vectors point outward, away from the particle at the origin, for an attractive force, the
vectors point toward the particle. So we can match up the vector field with the description as follows:
(a)
(b)
(c)
(d)
IV
III
I
II
Problems
~ (x, y) = 3~i − 4~j .
22. F~ (x, y) = a~i + b~j for any real numbers a and b is a constant vector field. For example, F
~ has
23. If F~ (x, y) = f (x, y)~v , where f (x, y) is any positive nonconstant function and ~v is any nonzero vector, then F
~ k = f k~v k. It has constant direction because all its vectors are in the same direction as ~v . For
nonconstant magnitude kF
~ (x, y) = (1 + x2 )(3~i + 2~j ).
example F
24. If
1
(f (x, y)~i + g(x, y)~j ),
F~ (x, y) = p
f (x, y)2 + g(x, y)2
~ k = 1. For many choices of f and g the vector field F~ is nonconstant because its
then f~ has constant magnitude kF
direction is nonconstant. For example,
~ (x, y) = √ 1
(~i − x~j ).
F
1 + x2
~ (x, y) = x~i + y~j has nonconstant magnitude kF
~k =
x2 + y 2 . Its
25. Many answers are possible. For example, F
~ (1, 0) = ~i and F~ (0, 1) = ~j .
direction is also nonconstant, since F
~ (x, y) = f (x, y)((1 + y 2 )~i − (x + y)~j ) where f (x, y) is any function, then F~ · G
~ = 0, which shows that F
~ is
26. If F
~
perpendicular to G .
~ (x, y) = (y + cos x)((1 + y 2 )~i − (x + y)~j ).
For example F
p
27. Vector fields (B) and (C) both appear to be constant, and therefore correspond to the equally spaced level curves in (I) and
(II). Since the gradient points toward increasing values of the function, (B) corresponds to (II) and (C) corresponds to (I).
Vector field (A) points away from the center, so it corresponds to (IV), which has a minimum in the center.
Vector field (D) points toward the center, so it corresponds to (III) which has a maximum at the center.
28. The sketches show that the vector fields point in different directions on the y-axis, so we examine the formulas for the
vector fields on the y-axis. On the y-axis, where x = 0, we have:
~ (0, y) = y~j , a vector pointing up if y > 0 and down if y < 0, as in I
F
~ (0, y) = −y~i , a vector pointing left if y > 0 and right if y < 0, as in II
G
~ (0, y) = −y~j , a vector pointing down if y > 0 and up if y < 0, as in III
H
~ is II, and H
~ is III.
So F~ is I, G
29. The sketches show that the vector fields can be distinguished by the directions they point on the coordinate axes, so we
examine the formulas for the vector fields on the axes.
~ (0, y) = −y~i + y~j = y(−~i + ~j ), a vector pointing up to the left if y > 0 and down to the right if
(a) F~ (0, y) + G
y < 0, as in II.
~ (0, y) = ~0 , the zero vector as in III.
(b) F~ (0, y) + H
~ (0, y) + H
~ (0, y) = −y~i − y~j = −y(~i + ~j ), a vector pointing down to the left if y > 0 and up to the right if
(c) G
~ (x, 0) + H
~ (x, 0) = x~i + x~j = x(~i + ~j ), a vector pointing up to the right if x > 0 and
y < 0, as in I and IV. G
~ +H
~ is IV.
down to the left if x < 0, as in II and IV, so G
~ (x, 0) + G
~ (x, 0) = −x~i + x~j = x(−~i + ~j ), a vector pointing up to the left if x > 0 and down to the right if
(d) −F
x < 0, as in I.
1616
Chapter Seventeen /SOLUTIONS
~ (x, y) = x~i . See Figure 17.21.
30. One possible solution is F
y
3
2
1
x
−3
−2
−1
1
2
3
−1
−2
−3
Figure 17.21
~i − y~j
~ (x, y) = −x
, then all vectors will be of unit length and will point toward the origin. See Figure 17.22.
31. If we let F
p
x2 + y 2
y
2
1
x
−3
−2
−1
1
2
3
−1
−2
−3
Figure 17.22
~ (x, y) = A~i +B~j such that F
~ ·~r = Ax +By = 0.
32. The position vector at each point is ~r = x~i +y~j . We want to find F
One possible answer is let A = y and B = −x. So F~ (x, y) = y~i − x~j . Since the vectors are of unit length, we get
y~i − x~j
.
F~ (x, y) = p
x2 + y 2
y
3
2
1
x
−3 −2 −1
−1
−2
−3
1
2
3
1617
17.3 SOLUTIONS
~ is a multiple
33. (a) The line l is parallel to the vector ~v = ~i − 2~j − 3~k . The vector field F~ is parallel to the line when F
of ~v . Taking the multiple to be 1 and solving for x, y, z we find a point at which this occurs:
x=1
x + y = −2
x − y + z = −3
gives x = 1, y = −3, z = −7, so a point is (1, −3, −7). Other answers are possible.
~ · ~v = 0, that is
(b) The line and vector field are perpendicular if F
(x~i + (x + y)~j + (x − y + z)~k ) · (~i − 2~j − 3~k ) = 0
x − 2x − 2y − 3x + 3y − 3z = 0
−4x + y − 3z = 0.
One point which satisfies this equation is (0, 0, 0). There are many others.
(c) The equation for this set of points is −4x + y − 3z = 0. This is a plane through the origin.
~ = 0F
~ +G
~ = −y~i + x~j is shown in Figure 17.23.
34. (a) The vector field L
~ = aF
~ +G
~ = (ax − y)~i + (ay + x)~j where a > 0 is shown in Figure 17.24.
(b) The vector field L
~
~
~
(c) The vector field L = aF + G = (ax − y)~i + (ay + x)~j where a < 0 is shown in Figure 17.25.
y
y
x
Figure 17.23
y
x
Figure 17.24
x
Figure 17.25
~ = F~ + 0G
~ = x~i + y~j is shown in Figure 17.26.
35. (a) The vector field L
~
~
~
(b) The vector field L = F + bG = (x − by)~i + (y + bx)~j where b > 0 is shown in Figure 17.27.
~ = F~ + bG
~ = (x − by)~i + (y + bx)~j where b < 0 is shown in Figure 17.28.
(c) The vector field L
y
y
x
Figure 17.26
y
x
Figure 17.27
x
Figure 17.28
36. (a) Since the velocity of the water is the sum of the velocities of the individual fields, then the total field should be
~v = ~v stream + ~v fountain .
1618
Chapter Seventeen /SOLUTIONS
It is reasonable to represent ~v stream by the vector field ~v stream = A~i , since A~i is a constant vector field flowing in
the i-direction (provided A > 0). It is reasonable to represent ~v fountain by
~v fountain = K~r r /r 2 = K(x2 + y 2 )−1 (x~i + y~j ),
since this is a vector field flowing radially outward (provided K > 0), with decreasing velocity as r gets larger.
We would expect the velocity to decrease as the water from the fountain spreads out. Adding the two vector fields
together, we get
~v = A~i + K(x2 + y 2 )−1 (x~i + y~j ), A > 0, K > 0.
(b) The constants A and K signify the strength of the individual components of the field. A is the strength of the flow of
the stream alone (in fact it is the speed of the stream), and K is the strength of the fountain acting alone.
y
y
(c)
x
Figure 17.29: A = 1, K = 1
x
Figure 17.30: A = 2, K = 1
y
x
Figure 17.31: A = 0.2, K = 2
37. (a) The gradient is perpendicular to the level curves. See Figures 17.33 and 17.32. A function always increases in the
direction of its gradient; this is why the values on the level curves of f and g increase as we approach the origin.
y
y
3
2
1
0
0
4
5
5
4
3
2
1
x
Figure 17.32: Level curves z = f (x, y)
x
Figure 17.33: Level curves z = g(x, y)
17.3 SOLUTIONS
1619
(b) f climbs faster at outside, slower at center; g climbs slower at outside, faster at center:
z
f (x, y)
g(x, y)
x
Figure 17.34
This can be understood if we notice that the magnitude of the gradient of f decreases as one approaches the
origin whereas the magnitude of the gradient of g increases (at least for a while - what happens very close to the
origin depends on the behavior of grad g in the region. One possibility for g is shown in Figure 17.34; the graph of g
could also have a sharp peak at 0 or even blow up.)
~ by its magnitude always produces the unit vector in the same direction as F~ .
38. (a) Dividing a vector F
(b) Since
p
~ k = k(1/F )(−v~i + u~j )k = (1/F ) v 2 + u2 = (1/F )F = 1,
kN
~ is a unit vector. We check that N
~ is perpendicular to F~ using the dot product of N
~ and F~ :
then N
~ · F~ = (1/F )(−v~i + u~j ) · (u~i + v~j ) = 0.
N
~ does N
~ point? The vector ~k is pointing out of the diagram. Since the cross product ~k × F~ is
Which side of F
~ , then N
~ lies in the xy-plane and points at a right angle to the direction of F
~ . By the
perpendicular to both ~k and F
~ points to the left as shown in the figure.
right-hand rule, N
Strengthen Your Understanding
~ (x, y, z) = F
~ (2x, 2y, 2z), move the vectors in a plot of F
~ (x, y, z) halfway to
39. To obtain a plot of the vector field G
~ (1, 2, 3), move the vector
the origin, without changing their directions or magnitudes. For example, to plot the vector G
F~ (2, 4, 6) to the point (1, 2, 3).
~ (x, y, z) =
Doubling the lengths of the arrows in the vector field F~ (x, y, z) gives a plot of the vector field H
~ (x, y, z).
2F
~ (x, y, z) = x2 − yz is not a vector field because the
40. The values of a vector field are vectors, not scalars. The formula F
2
expression x − yz is not a vector.
41. The vector field
~ (x, y, z) = (x2 + 1) ~i + ~j + ~k
F
is nonconstant and is parallel to ~i + ~j + ~k at every point.
42. Any vector field that is not the zero vector at any point and that has nonconstant direction can be divided by its magnitude
to produce a nonconstant vector field of unit vectors. For example, we have
1
~ (x, y, z) = p
F
(x2 + 1)~i + z~j + y~k .
(x2 + 1)2 + z 2 + y 2
1620
Chapter Seventeen /SOLUTIONS
Solutions for Section 17.4
Exercises
1. Since x′ (t) = 0 and y ′ (t) = 2, we have x = x0 and y = 2t + y0 . Thus, the solution curves are x = constant.
y
y
12
12
x
−12
12
x
−12
−12
12
−12
Figure 17.35: The field ~v = 2~j
Figure 17.36: The flow x = constant
2. Since x′ (t) = 3 and y ′ (t) = 0, we have x = 3t + x0 and y = y0 . Thus, the solution curves are y = constant.
y
y
18
18
x
−18
18
−18
Figure 17.37: The field ~v = 3~i
x
−18
18
−18
Figure 17.38: The flow y =constant
3. Since x′ (t) = 3 and y ′ (t) = −2, we have x = 3t + x0 and y = −2t + y0 . Thus the flow lines are straight lines parallel
dy
= − 23 . Thus, y = − 32 x + c, where c is a constant.
to the vector 3~i − 2~j . Alternatively, we have dx
17.4 SOLUTIONS
y
y
9
9
x
x
−9
9
−9
9
−9
−9
Figure 17.39: The field ~v = 3~i − 2~j
Figure 17.40: The flow y = − 32 x + c
4. As
~v (t) =
dx~
dy
i + ~j ,
dt
dt
the system of differential equations is
(
Since
dx
dt
dy
dt
= x
= 0.
d
d
(x(t)) = (aet ) = x
dt
dt
and
d
d
(y(t)) = (b) = 0,
dt
dt
the given flow satisfies the system. The solution curves are the horizontal lines y = b. See Figures 17.41 and 17.42.
y
18
y
x
−18
x
18
−18
Figure 17.41: ~v (t) = x~i
Figure 17.42: The flow x(t) = aet , y(t) = b
5. As
~v (t) =
dx~
dy
i + ~j ,
dt
dt
the system of differential equations is
(
dx
dt
dy
dt
= 0
= x.
1621
1622
Chapter Seventeen /SOLUTIONS
Since
d(x(t))
d
= (a) = 0
dt
dt
and
d(y(t))
d
= (at + b) = a = x,
dt
dt
the given flow satisfies the system. The solution curves are the vertical lines x = a. See Figures 17.43 and 17.44.
y
12
y
x
−12
x
12
−12
Figure 17.43: ~v (t) = x~j
Figure 17.44: The flow x(t) = a, y(t) = at + b
y
y
6.
x
x
Figure 17.46: The flow x = aet ,
y = bet .
Figure 17.45: ~v (t) = x~i + y~j
As
~v (t) =
dy
dx~
i + ~j ,
dt
dt
the system of differential equations is
(
Since
and
dx
dt
dy
dt
= x
= y.
dx(t)
d
= [aet ] = aet = x(t)
dt
dt
dy(t)
d
= [bet ] = bet = y(t),
dt
dt
the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained are
y = ab x.
17.4 SOLUTIONS
7. As
~v (t) =
1623
dy
dx~
i + ~j ,
dt
dt
the system of differential equations is
(
Since
dx
dt
dy
dt
= x
= −y.
dx(t)
d
= [aet ] = aet = x(t)
dt
dt
and
dy(t)
d
= [be−t ] = −be−t = y(t),
dt
dt
the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained are
xy = ab. See Figures 17.47 and 17.48.
y
y
x
x
Figure 17.47: ~v (t) = x~i − y~j
Figure 17.48: The flow x = aet , y = be−t
8. As
~v (t) =
dy
dx~
i + ~j ,
dt
dt
the system of differential equations is
(
Since
and
dx
dt
dy
dt
= y
= −x.
d
dx(t)
= [a sin t] = a cos t = y(t)
dt
dt
dy(t)
d
= [a cos t] = −a sin t = −x(t),
dt
dt
the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained are
x2 + y 2 = a2 . See Figures 17.49 and 17.50.
1624
Chapter Seventeen /SOLUTIONS
y
y
x
x
Figure 17.50: The flow x = a sin t,
y = a cos t
Figure 17.49: ~v (t) = y~i − x~j
9. As
~v (t) =
dy
dx~
i + ~j ,
dt
dt
the system of differential equations is
(
Since
dx
dt
dy
dt
= y
= x.
dx(t)
d
= [a(et + e−t )] = a(et − e−t ) = y(t)
dt
dt
and
dy(t)
d
= [a(et − e−t )] = a(et + e−t ) = x(t),
dt
dt
the given flow satisfies the system. By eliminating the parameter t in x(t) and y(t), the solution curves obtained are
x2 − y 2 = 4a2 . See Figures 17.51 and 17.52.
y
y
x
x
Figure 17.52: The flow
x(t) = a(et + e−t ), y(t) = a(et − e−t )
Figure 17.51: ~v (t) = y~i + x~j
10. The vector field is given by ~v = y 2~i + 2x2~j , that is, the flow line (x(t), (y(t)) satisfies
x′ (t) = y 2
y ′ (t) = 2x2
We’ll use Euler’s method with ∆t = 0.1 to find the parameterized curve (x(t), y(t)) through (1, 2). So
xn+1 = xn + 0.1yn2
yn+1 = yn + (0.1)2x2n .
17.4 SOLUTIONS
1625
Initially, that is when t = 0, we have (x0 , y0 ) = (1, 2). Then
x1 = x0 + 0.1y02 = 1 + 0.1 · 22 = 1.4
y1 = y0 + 0.1 · 2x20 = 2 + 0.1 · 2 · 12 = 2.2.
Thus, we see that after one step, x1 = 1.4 and y1 = 2.2. Further values are given in the Table 17.3.
Table 17.3
x
1.4
1.884
2.556
3.646
5.770
y
2.2
2.592
3.302
4.609
7.268
Problems
11. This corresponds to area A in Figure 17.53.
km
600
✲
A
D
500
✠
✛
400
B
300
C
✲
200
100
0
0
100
200
300
400
500
600
km
Figure 17.53
12. This corresponds to area B in Figure 17.53 in Problem 11.
13. This corresponds to area C in Figure 17.53 in Problem 11.
14. This corresponds to area D in Figure 17.53 in Problem 11.
~ (x, y) has the same direction as F
~ (x, y) but G
~ (x, y) is twice as long.
15. (a) At every point (x, y) in the plane, the vector G
~ (x, y) = −y~i + x~j see Figures 17.54 and 17.55.
For the case where F
1626
Chapter Seventeen /SOLUTIONS
y
y
x
x
Figure 17.55: G(x, y) = 2(−y~i + x~j )
~ = −y~i + x~j
Figure 17.54: F
~ and G
~ have the same direction. Therefore the flow lines of the two
(b) At every point in the plane the two vector fields F
vector fields have the same slopes at every point. By the uniqueness of solutions of differential equations with initial
conditions, the flow lines of the two vector fields through any given point must be the same. This means that if two
~ and the other into the flow of G
~ , they will move on exactly
objects are placed at the same point, one into the flow of F
the same paths. However, they will move at different speeds. The two flows will have different parameterizations.
~ is twice as
For the case where F~ (t) = −y~i + x~j both flows are circular about the origin, but the flow of G
~.
fast as the flow of F
16. The directions of the flow lines are as shown.
(a)
(b)
(c)
(d)
(e)
(f)
III
I
II
V
VI
IV
(I)
1
y
(III)
y
(II)
1
x
−2
−1
1
−2
−1
1
−2
2
−1
1
y
−1
1
2
−1
y
(V)
y
x
x
2
−1
(IV)
1
y
(VI)
5
5
x
−2
−1
1
2
x
−5
−1
x
5
5
−5
−5
−5
d~r
(t) = F~ (x(t), y(t)) = u(x(t),
17. The object’s motion is described by a function ~r (t) = x(t)~i + y(t)~j , where
dt
y(t))~i + v(x(t), y(t))~j . Using the chain rule to differentiate, we have
d2~r
dt2
du~
dv
=
i + ~j
dt
dt
dx
dy
dx
dy
= (ux
+ uy )~i + (vx
+ vy )~j
dt
dt
dt
dt
= (ux u + uy v)~i + (vx u + vy v)~j
~a (t) =
17.4 SOLUTIONS
1627
18. (a) Perpendicularity is indicated by zero dot product. We have ~v · grad H = (−Hy~i + Hx~j ) · (Hx~i + Hy~j ) = 0.
(b) If ~r (t) = x(t)~i + y(t)~j is a flow line we have, using the chain rule,
dx
dy
d
H(x(t), y(t)) = Hx
+ Hy
= Hx (−Hy ) + Hy (Hx ) = 0.
dt
dt
dt
Thus H(x(t), y(t)) is constant which shows that a flow line stays on a single level curve of H.
For a different solution, use geometric reasoning. The vector field ~v is tangent to the level curves of H because,
by part (a), ~v and the level curves are both perpendicular to the same vector field grad H. Thus the level curves of
H and the flowlines of ~v run in the same direction.
19. Let ~r (t) = x(t)~i + y(t)~j be a flow line of ~v . If f (x, y) has the same value at all points (x(t), y(t)) then the flow line
lies on a level curve of f . We can check whether
g(t) = f (x(t), y(t)) = x(t)y(t)
is constant by computing the derivative g (t). Since ~v = x~i − y~j , we have dx/dt = x and dy/dt = −y. Thus,
′
g ′ (t) = x
dx
dy
+y
= −xy + yx = 0
dt
dt
and g(t) is constant. This means that the flow line lies on a level curve of f . The flow lines are parameterized hyperbolas
with equation xy = c.
20. Let ~r (t) = x(t)~i + y(t)~j be a flow line of ~v . If f (x, y) has the same value at all points (x(t), y(t)) then the flow line
lies on a level curve of f . We can check whether
g(t) = f (x(t), y(t)) = x(t)2 − y(t)2
is constant by computing the derivative g ′ (t). Since ~v = y~i + x~j , we have dx/dt = y and dy/dt = x. Thus,
g ′ (t) = 2x
dy
dx
− 2y
= 2xy − 2yx = 0
dt
dt
and g is constant. This means that the flow line lies on a level curve of f . The flow lines are parameterized hyperbolas
with equation x2 − y 2 = c.
21. Let ~r (t) = x(t)~i + y(t)~j be a flow line of ~v . If f (x, y) has the same value at all points (x(t), y(t)) then the flow line
lies on a level curve of f . We can check whether
g(t) = f (x(t), y(t)) = bx(t)2 − ay(t)2
is constant by computing the derivative g ′ (t). Since ~v = ay~i + bx~j , we have dx/dt = ay and dy/dt = bx. Thus,
g ′ (t) = 2bx
dy
dx
− 2ay
= 2abxy − 2abyx = 0
dt
dt
and g is constant. This means that the flow line lies on a level curve of f . The flow lines are parameterized conic sections
with equation bx2 − ay 2 = c, hyperbolas if a and b are both positive or both negative, and ellipses if a and b have opposite
sign.
22. (a) Each vector in the vector field ~v is horizontal, tangent to a circle whose center is on the z-axis, and pointing counterclockwisepwhen viewed from above. Thus, ~v is parallel to −y~i + x~j . The point (x, y, z) is moving on a circle of
radius r = x2 + y 2 and has
πr
2πr
=
.
Speed =
24
12
Since the vector
pat the point (x, y, z) has magnitude πr/12 and is parallel to the unit vector
(−y~i + x~j )/ x2 + y 2 , we have
πr
~v =
12
−y~i + x~j
p
x2
+
y2
!
=
π
(−y~i + x~j ) meters/hr.
12
(b) A point moves in a horizontal circle, centered on the z-axis, and oriented counter-clockwise when viewed from above.
These circles are the flow lines.
1628
Chapter Seventeen /SOLUTIONS
23. (a) The flow line ~r (t) = x(t)~i + y(t)~j is the path with ~r ′ (t) = ~v = (ax − y)~i + (x + ay)~j . Thus
x′ = ax − y
y ′ = x + ay.
We show h(t) is constant on a flow line by showing that h′ (t) = 0. Using the chain rule, we have
h′ (t) = e−2at (2xx′ + 2yy ′ ) − 2ae−2at (x2 + y 2 )
= e−2at (2x(ax − y) + 2y(x + ay)) − 2ae−2at (x2 + y 2 ) = 0.
The function h(t) = e−at (x2 + y 2 ) is constant because its derivative is zero.
(b) We have h(0) = ea·0 (x2 + y 2 ) = 1 for points on the unit circle x2 + y 2 = 1 at t = 0 . Hence, along the same flow
line, h(t) = e−2at (x2 + y 2 ) = 1 for all t. Thus at time t, the particle’s coordinates satisfy
x2 + y 2 = e2at ,
which is the equation of the circle of radius eat centered at the origin.
This result matches what can be seen in the vector field plot in Figure 17.56. If a < 0, the flow lines move
toward the origin. If a > 0, the flow lines move away from the origin, and if a = 0 they circle the origin.
y
y
x
y
x
x
a<0
a>0
a=0
Figure 17.56
Strengthen Your Understanding
~ = −y~i + x~j has linear components and circular flow lines.
24. The vector field F
~ (x, y, z) =
25. Although the vectors in the vector field all point in the same direction, their lengths could vary. For example F
(x2 + 1)~i has flow lines that are all lines parallel to the x-axis pointing in the positive direction, but it is not constant.
~ (x, y, z), we have
26. Since ~r (t) = t~i + t2~j + t3~k is a flow line of the vector field F
~ (t, t2 , t3 ) = ~v (t) = ~i + 2t~j + 3t2~k
F
where ~v (t) is the velocity of ~r (t). Let
~ (x, y, z) = F1 (x, y, z)~i + F2 (x, y, z)~j + F3 (x, y, z)~k .
F
We must determine three functions, F1 , F2 , and F3 such that
F1 (t, t2 , t3 ) = 1 F2 (t, t2 , t3 ) = 2t
F3 (t, t2 , t3 ) = 3t2 .
One solution is given by
F1 (x, y, z) = 1 F2 (x, y, z) = 2x F3 (x, y, z) = 3y.
Thus, ~r (t) is a flow line of the vector field
F~ (x, y, z) = ~i + 2x~j + 3y~k .
~ (x, y, z) = x~i + y~j + z~k points away from the origin at every point other than the origin, so its flow
27. The vector field F
lines are rays pointing from the origin.
28. True, since the vectors x~j are parallel to the y-axis.
SOLUTIONS to Review Problems for Chapter Seventeen
1629
29. False. The flow lines are circles centered at the origin.
30. False. The flow lines are lines parallel to the x-axis.
31. False. Each flow line stays in the quadrant in which it originates.
32. True. Any flow line which stays in the first quadrant has x, y → ∞.
33. False. The flow lines for F~ are perpendicular to the contours for f , since the flow lines follow ∇f , which is perpendicular
to the contours of f .
~ (~r ) is tangent
34. True. Since a vector tangent to a circle centered at the origin is perpendicular to the radius vector ~r and F
~ (~r ) must be perpendicular to ~r . Thus F
~ (~r ) · ~r = 0 for all ~r .
to its flow lines, F
~ (x, y) to be parallel to ~v for all x and y. That
35. False. If the flow lines are all straight lines parallel to ~v , we need F
~ (x, y) must be equal to ~v ; it only needs to be a scalar multiple ~v . For example, the vector field
does not mean that F
F~ (x, y) = 6~i + 10~j has all its flow lines parallel to ~v . Another example is F~ (x, y) = 3ex−y~i + 5ex−y~j = ex−y ~v ,
where the scalar multiplied times ~v varies as x and y vary.
36. True. If (x, y) were a point where the y-coordinate along a flow line reached a relative maximum, then the tangent vector
to the flow line, namely F~ (x, y), there would have to be horizontal (or ~0 ), that is its ~j component would have to be 0.
~ is always 2.
But the ~j component of F
~ (x, 0) = ex~i , pointing to the right,
37. False. At all points on the x-axis, y = 0, so the vector field is a horizontal vector, F
~
~
since it is a positive multiple of i . Thus the x-axis itself is a flow line for F . Since there can be only one flow line through
any point, no flow line can cross the x-axis.
Solutions for Chapter 17 Review
Exercises
1. The line has equation
~r = 2~i − ~j + 3~k + t(5~i + 4~j − ~k ),
or, equivalently
x = 2 + 5t
y = −1 + 4t
z = 3 − t.
2. The displacement vector from the point (1, 2, 3) to the point (3, 5, 7) is:
3~i + 5~j + 7~k − (~i + 2~j + 3~k ) = 2~i + 3~j + 4~k .
So the equations are
x = 1 + 2t,
y = 2 + 3t,
z = 3 + 4t.
3. x = t, y = 5.
4. The parameterization x~i + y~j = 2 cos t~i + 2 sin t~j has the right radius but starts at the point (2, 0). To start at (0, 2),
we need x~i + y~j = 2 cos(t + π2 )~i + 2 sin(t + π2 )~j = 2 sin t~i + 2 cos t~j .
5. The parameterization x~i + y~j = (4 + 4 cos t)~i + (4 + 4 sin t~) j gives the correct circle, but starts at (8, 4). To start on
the x-axis we need
π
π
x~i + y~j = (4 + 4 cos(t − ))~i + (4 + 4 sin(t − ))~j = (4 + 4 sin t)~i + (4 − 4 cos t)~j .
2
2
6. The parametric equation of a circle is
When t = 0, x = 1, y = 0, and when t =
equation is correct.
π
,
2
x = cos t, y = sin t.
x = 0, y = 1. This shows a counterclockwise movement, so our original
1630
Chapter Seventeen /SOLUTIONS
7. The vector (~i + 2~j + 5~k ) − (2~i − ~j + 4~k ) = −~i + 3~j + ~k is parallel to the line, so a possible parameterization is
x = 2 − t,
y = −1 + 3t,
z = 4 + t.
8. A line perpendicular to the xz-plane will have x = constant, z = constant, y = anything: This is given by x = 1, y =
t, z = 2.
9. Since the vector ~n = grad(2x − 3y + 5z) = 2~i − 3~j + 5~k is perpendicular to the plane, this vector is parallel to the
line. Thus the equation of the line is
x = 1 + 2t,
y = 1 − 3t,
z = 1 + 5t.
10. The xy-plane is where z = 0, so one possible answer is
x = 3 cos t,
y = 3 sin t,
z = 0.
This goes in the counterclockwise direction because it starts at (3, 0, 0) and heads in the positive y-direction.
11. Since the circle has radius 3, the equation must be of the form x = 3 cos t, y = 5, z = 3 sin t. But since the circle is being
viewed from farther out on the y-axis, the circle we have now would be seen going clockwise. To correct this, we add a
negative to the third component, giving us the equation x = 3 cos t, y = 5, z = −3 sin t.
12. We can find this equation in two ways. First we could find two points on the line of intersection and then proceed as in
Example 7 on page 921. To find two points just substitute two different values for z and solve for x and y for each value
of z. Alternatively, assuming the line is not horizontal (which it turns out not to be), we could take z to be the parameter
t, so z = t. To find x and y as functions of t we solve the two equations for x and y in terms of t. We have
t = 4 + 2x + 5y
t = 3 + x + 3y.
Eliminating x we get
−t = −2 − y
and
y = −2 + t.
Substituting −2 + t for y in the second equation and solving for x, we get
x = 3 − 2t.
Our equations are therefore
or
x = 3 − 2t, y = −2 + t, z = t.
~r = x~i + y~j + z~k = 3~i − 2~j + t(−2~i + ~j + ~k ).
13. See Figure 17.57. The parameterization is
~r = 10 cos
2πt ~
2πt ~
i − 10 sin
j + 7~k .
30
30
z
(0, 0, 7)
*
y
x
Figure 17.57
1631
SOLUTIONS to Review Problems for Chapter Seventeen
14. The velocity vector ~v is given by:
~v =
d
d
(3 cos t)~i + (4 sin t)~j = −3 sin t~i + 4 cos t~j .
dt
dt
15. The velocity vector ~v is given by:
~v =
d
d ~
ti + (t3 − t)~j = ~i + (3t2 − 1)~j .
dt
dt
16. The velocity vector ~v is given by:
~v =
d
d
d
(2 + 3t)~i + (4 + t)~j + (1 − t)~k = 3~i + ~j − ~k .
dt
dt
dt
17. The velocity vector ~v is given by:
~v =
d
d
d
(2 + 3t2 )~i + (4 + t2 )~j + (1 − t2 )~k = 6t~i + 2t~j − 2t~k .
dt
dt
dt
18. The velocity vector ~v is given by:
~v =
d
d
d ~
ti + t2~j + t3~k = ~i + 2t~j + 3t2~k .
dt
dt
dt
19. Vector. Differentiating using the chain rule gives
√
1
1
· 2 ~i − 3 sin 2t + 1 · √
· 2 ~j +
2t + 1 · √
2 2t + 1
2 2t + 1
√
√
3 cos 2t + 1~
3 sin 2t + 1 ~
1
~k .
√
=
i − √
j +√
2t + 1
2t + 1
2t + 1
Velocity =
3 cos
√
1
√
· 2 ~k
2 2t + 1
20. Scalar. The velocity vector is
~v = 2t~i + et~j ,
and
Speed = ||~v || =
21. Vector. Differentiating using the chain rule gives
Velocity =
p
(2t)2 + (et )2 =
− cos t
~i +
√
2 3 + sin t
p
4t + e2t .
− sin t
√
2 3 + cos t
~j .
22. Vector. Differentiating using the product and chain rule gives
Velocity = (et + tet )~i + 2e2t~j
Acceleration = (2et + tet )~i + 4e2t~j .
23. No. The first is parallel to the vector 2~i − ~j + 3~k and the second is parallel to ~i + 2~j + 2~k .
24. Yes. They are both parallel to the vector 2~i − ~j + 3~k .
25. The direction vectors of the lines, −~i + 4~j − 2~k and 2~i − 8~j + 4~k , are multiplies of each other (the second is −2 times
the first). Thus the lines are parallel. To see if they are the same line, we take the point corresponding to t = 0 on the first
line, which has position vector 3~i + 3~j − ~k , and see if it is on the second line. So we solve
(1 + 2t)~i + (11 − 8t)~j + (4t − 5)~k = 3~i + 3~j − ~k .
This has solution t = 1, so the two lines have a point in common and must be the same line, parameterized in two different
ways.
1632
Chapter Seventeen /SOLUTIONS
26. (a) We get the part of the line with x < 0 and y < 0 and z < 10.
(b) We get the part of the line between the points (0, 0, 10) and (1, 2, 13).
27. See Figure 17.58.
y
x
~ (x, y) = −y~i + x~j
Figure 17.58: F
~
28. At each point, all these
√ vector fields point in the same direction (rotating clockwise around the origin). Since kF k =
2 +x2
y
1
~
~
= √ 1
, the vectors in the field shrink as you go away from the origin. See Figure 17.59.
2
2 ky i −xj k =
2
2
x +y
x +y
x2 +y 2
y
x
Figure 17.59
29. The vector field points in a clockwise direction around the origin. Since
k
y
p
x2 + y 2
!
~i −
the length of the vectors is constant everywhere.
x
p
x2 + y 2
!
p
x2 + y 2
~j k = p
= 1,
x2 + y 2
y
x
Figure 17.60
SOLUTIONS to Review Problems for Chapter Seventeen
1633
Problems
30. Substitute x = 2t + 1, y = 3t − 2, z = −t + 3 into the equation of the sphere:
(x − 1)2 + (y − 1)2 + (z − 2)2 = 2
(2t)2 + (3t − 1)2 + (−t + 1)2 = 2
4t2 + 9t2 − 6t + 1 + t2 − 2t + 1 = 2
14t2 − 8t = 0
Thus, t = 0, giving x = 1, y = −2, z = 3, and when t = 4/7, x = 15/7, y = −2/7, z = 17/7.
31. (a) To find where the particle is at time equal to 0, we simply substitute 0 in for all t in the equation. Therefore, the
particle is at the point with position vector
~r (0) = [2 + 5(0)]~i + (3 + 0)~j + 2(0)~k
= 2~i + 3~j + 0~k .
Thus, the particle is at the point (2, 3, 0).
(b) To find the time at which the particle is at the point (12, 5, 4), we solve for t for each component, and the t should be
the same, if the curve goes through this point. For the x-component, we get
2 + 5t = 12
t = 2.
For the y-component, we get
3+t = 5
t = 2.
And for the z-component, we get
2t = 4
t = 2.
Therefore, at t = 2 the particle reaches (12, 5, 4).
(c) The particle never reaches (12, 4, 4), because the equation
~r = (2 + 5t)~i + (3 + t)~j + 2t~k = 12~i + 4~j + 4~k
has no solution. Thus, the point does not lie on the line.
(I) has radius 1 and traces out a complete circle, so I = C4 .
(II) has radius 2 and traces out the top half of a circle, so II = C1
(III) has radius 1 and traces out a quarter circle, so III = C2 .
(IV) has radius 2 and traces out the bottom half of a circle, so IV = C6 .
(b) C3 has radius 1/2 and traces out a half circle below the x-axis, so
32. (a)
~r = 0.5 cos t~i − 0.5 sin t~j .
C5 has radius 2 and traces out a quarter circle below the x-axis starting at the point (−2, 0). Thus we have
~r = −2 cos(t/2)~i − 2 sin(t/2)~j .
33. (a) A vector field associates a vector to every point in a region of the space. In other words, a vector field is a vector-valued
function of position given by ~v = f~ (~r ) = f~ (x, y, z)
(b) (i) Yes, ~r + ~a = (x + a1 )~i + (y + a2 )~j + (z + a3 )~k is a vector-valued function of position.
(ii) No, ~r · ~a is a scalar.
(iii) Yes.
(iv) x2 + y 2 + z 2 is a scalar.
1634
Chapter Seventeen /SOLUTIONS
34. Vector fields (A) and (D) both point radially outward, so they correspond to (I) and (II). Since (A) has vectors that are of
constant length, it corresponds to (II), where the level curves are equally spaced. (D) corresponds to (I).
Vector field (B) corresponds to (III) since the vectors in (B) point away from the origin on the x-axis, and the function
in (III) increases in this direction. To confirm, the vectors in (B) point toward the origin on the y-axis, and the function
decreases away from the origin on the y-axis.
In vector field (C), vectors point away toward the origin on the x-axis and away from the origin on the y-axis. This
corresponds to (IV), in which the function decreases away from the origin on the x-axis and increases on the y-axis.
~ is tangent to (IV), F
~ is tangent to (I), G
~ is tangent to (II), and
35. Sketches of the vector fields in Figure 17.61 show that E
~
H is tangent to (III).
y
y
y
x
~ = x~i + y~j
E
x
y
x
~ = x~i − y~j
F
~ = y~i − x~j
G
x
~ = y~i + x~j
H
Figure 17.61
36. At time t the particle is s = t − 7 seconds from P , so the displacement vector from the point P to the particle is d~ = s~v .
To find the position vector of the particle at time t, we add this to the position vector ~r 0 = 5~i + 4~j + 3~k for the point
P . Thus a vector equation for the motion is:
~r = ~r 0 + s~v
= (5~i + 4~j + 3~k ) + (t − 7)(3~i + ~j + 2~k ),
or equivalently,
x = 5 + 3(t − 7),
y = 4 + 1(t − 7),
z = 3 + 2(t − 7).
Notice that these equations are linear. They describe motion on a straight line through the point (5, 4, 3) that is parallel to
the velocity vector ~v = 3~i + ~j + 2~k .
37. The displacement vector from (1, 1, 1) to (2, −1, 3) is d~ = (2~i − ~j + 3~k ) − (~i + ~j + ~k ) = ~i − 2~j + 2~k meters. The
velocity vector has the same direction as d~ and is given by
~v =
d~
= 0.2~i − 0.4~j + 0.4~k meters/sec.
5
Since ~v is constant, the acceleration ~a = ~0 .
38. Parametric equations for a line in 2-space are
x = x0 + at
y = y0 + bt
where (x0 , y0 ) is a point on the line and ~v = a~i + b~j is the direction of motion. Notice that the slope of the line is equal
to ∆y/∆x = b/a, so in this case we have
b
= Slope = −2,
a
b = −2a.
In addition, the speed is 3, so we have
p
k~v k = 3
a 2 + b2 = 3
a2 + b2 = 9.
SOLUTIONS to Review Problems for Chapter Seventeen
1635
Substituting b = −2a gives
a2 + (−2a)2 = 9
5a2 = 9
3
3
a = √ ,−√ .
5
5
√
√
If we use a = 3/ 5, then b = −2a = −6/ 5. The point (x0 , y0 ) can be any point on the line: we use (0, 5). The
parametric equations are
3
6
x = √ t, y = 5 − √ t.
5
5
√
√
Alternatively, we can use a = −3/ 5 giving b = 6/ 5. An alternative answer, which represents the particle moving in
the opposite direction is
6
3
x = − √ t, y = 5 + √ t.
5
5
39. (a) The quantity || grad f || represents the maximum rate of change of temperature with distance at each point. Its units
◦
are C per cm.
p
(b) The speed of the particle is (g ′ (t))2 + (k′ (t))2 . Its units are cm/sec.
(c) The rate of change of the particle’s temperature with time is given by the chain rule
∂f dx
∂f dy
dH
=
+
= fx · g ′ (t) + fy · k′ (t).
dt
∂x dt
∂y dt
◦
Its units are C/sec.
40. We should have x = x0 − 2t. Since no initial position at time t = 0 is given, we can use any point on the line y = 3x + 7
as (x0 , y0 ). We choose the y-intercept (0, 7). Then x = 0 − 2t and y = 7 + bt. Since the slope of the line is 3 and the
x-coordinate decreases by 2 units for each unit of time, we know that the y-coordinate decreases by 6 units for each unit
of time. Therefore, b = −6. Our equations are x = −2t, y = 7 − 6t.
41. (a) In order for the particle to stop, its velocity ~v = (dx/dt)~i + (dy/dt)~j must be zero, so we solve for t such that
dx/dt = 0 and dy/dt = 0, that is
dx
= 3t2 − 3 = 3(t − 1)(t + 1) = 0,
dt
dy
= 2t − 2 = 2(t − 1) = 0.
dt
The value t = 1 is the only solution. Therefore, the particle stops when t = 1 at the point (t3 − 3t, t2 − 2t)|t=1 =
(−2, −1).
(b) In order for the particle to be traveling straight up or down, the x-component of the velocity vector must be 0. Thus,
we solve dx/dt = 3t2 − 3 = 0 and obtain t = ±1. However, at t = 1 the particle has no vertical motion, as we saw
in part (a). Thus, the particle is moving straight up or down only when t = −1. Since the velocity at time t = −1 is
~v (−1) =
dx
dt
~i + dy
dt
t=−1
~j = −4~j ,
t=−1
the motion is straight down. The position at that time is (t3 − 3t, t2 − 2t)|t=−1 = (2, 3).
(c) For horizontal motion we need dy/dt = 0. That happens when dy/dt = 2t − 2 = 0, and so t = 1. But from part (a)
we also have dx/dt = 0 also at t = 1, so the particle is not moving at all when t = 1. Thus, there is no time when
the motion is horizontal.
42. (a) The velocity vector is given by differentiating ~r (t) component by component to give
~v =
d~r
= (−4 sin 4t~i + 4 cos 4t~j + 3~k ).
dt
Similarly, differentiating the velocity vector gives the acceleration vector to be
~a =
d2~r
= (−16 cos 4t~i − 16 sin 4t~j ).
dt2
1636
Chapter Seventeen /SOLUTIONS
(b) The speed of the particle is
k~v k =
p
√
√
~v .~v = 42 (sin2 4t + cos2 4t) + 32 = 25 = 5.
(c) Since k~v k = 5 the particle’s speed is constant.
(d) The angle between the position and acceleration vectors, θ, is given
cos(θ) =
~r .~a
k~r kk~a k
~r (0) = cos(0)~i + sin(0)~j + 0~k = ~i
~a (0) = (−16 cos(0)~i − 16 sin(0)~j = −16~i
~r .~a = −16,
so that
k~r k| = 1,
cos(θ) = −1,
k~a k = 16
θ = 180◦ .
43. (a) With the center at (0, 0, 8) and a point of the circle at (0, 5, 8), we know that the radius is 5. When t = 0, we have
x = 0 and y = 5. Since the stone is rotating horizontally, z = 0 for all t. The period is 2π. Thus, the parameterization
is:
x(t) = 5 sin t
y(t) = 5 cos t
z(t) = 8
This parameterization has the correct period (if t is in seconds) and satisfies the initial conditions.
(b) From our parameterization with t in seconds, we can see that the stone reaches (5, 0, 8) at time π/2. Thus at t = π/2,
~v = xt (π/2)~i + yt (π/2)~j
= 5 cos(π/2)~i − 5 sin(π/2)~j
= −5~j .
The acceleration of an object is the second derivative of its position. Thus, at t = π/2,
~a = xtt (π/2)~i + ytt (π/2)~j
= −5 sin(π/2)~i − 5 cos(π/2)~j
= −5~i
(c) At the moment in which the stone has left the circle, the only acceleration that acts on the stone is that of gravity.
From that, assuming a gravity vector field oriented in the −z direction, we get the differential equations
ztt (t) = −g
xtt (t) = ytt (t) = 0.
If we now measure t from the instant the string breaks, then the initial conditions are the velocity and position of the
stone at t = 0. Since the velocity at the moment of release is ~v = −5~j , we have
xt (0) = 0,
yt (0) = −5, zt (0) = 0.
The initial position at t = 0 is:
x(0) = 5,
y(0) = 0, z(0) = 8.
44. (a) Differentiating we have
x′ (t) = 5,
x′′ (t) = 0,
Thus at time t = 0,
y ′ (t) = 3,
z ′ (t) = −2t + 2,
y ′′ (t) = 0,
z ′′ (t) = −2.
Position = (x(0), y(0), z(0)) = (0, 0, 15)
Velocity = x′ (0)~i + y ′ (0)~j + z ′ (0)~k = 5~i + 3~j + 2~k
Acceleration = x′′ (0)~i + y ′′ (0)~j + z ′′ (0)~k = −2~k .
SOLUTIONS to Review Problems for Chapter Seventeen
1637
(b) The particle hits the ground when z(t) = 0, so
15 − t2 + 2t = 0
−(t − 5)(t + 3) = 0
t = −3, 5.
Since t ≥ 0, the particle hits the ground when t = 5. At that time
Position = (x(5), y(5), z(5)) = (25, 15, 0)
Velocity = x′ (5)~i + y ′ (5)~j + z ′ (5)~k = 5~i + 3~j − 8~k
so
Speed = kVelocityk =
p
52 + 32 + (−8)2 =
√
98.
2x(x2 + y 2 ) − 2x(x2 − y 2 )
4xy 2
=
.
2
2 2
(x2 + y 2 )2
(x + y )
2
2
2
2
2
−2y(x + y ) − 2y(x − y )
−4yx
= 2
.
fy =
(x2 + y 2 )2
(x + y 2 )2
∇f (1, 1) = ~i − ~j , i.e., south-east.
(b) We need a vector ~
u such that ∇f (1, 1) · ~
u = 0, i.e., such that (~i − ~j ) · ~
u = 0. The vector ~
u = ~i + ~j clearly works;
so does ~
u = −~i − ~j . Dividing by the length to get a unit vector, we have ~
u = √12~i + √12 ~j or ~
u = − √12~i − √12 ~j .
(c) f is a function of x and y, which in turn are functions of t. Thus, the chain rule can be used to show how f changed
with t.
∂f dx
∂f dy
4xy 2
4x2 y
df
=
·
+
·
= 2
· 2e2t − 2
· (6t2 + 6).
dt
∂x dt
∂y dt
(x + y 2 )2
(x + y 2 )2
df
4
4
At t = 0, x = 1, y = 1; so,
= · 2 − · 6 = −4.
dt
4
4
46. (a) Separate the ant’s path into three parts: from (0, 0) to (1, 0) along the x-axis; from (1, 0) to (0, 1) via the circle; and
= π2 , and 1 respectively.
from (0, 1) to (0, 0) along the y-axis. (See Figure 17.62.) The lengths of the paths are 1, 2π
4
Thus, the time it takes for the ant to travel the three paths are (using the formula t = vd ) 12 , 31 , and 12 seconds.
45. (a) fx =
y
(0, 1)
x
(1, 0)
Figure 17.62
From t = 0 to t = 21 , the ant is heading toward (1, 0) so its coordinate is (2t, 0). From t = 21 to t = 12 + 31 = 65 ,
the ant is veering to the left and heading toward (0, 1). At t = 12 , it is at (1, 0) and at t = 65 , it is at (0, 1). Thus its
(t − 21 )], sin[ 3π
(t − 12 )]). Finally, from t = 65 to t = 65 + 12 = 34 , the ant is headed home. Its
position is (cos[ 3π
2
2
4
coordinates are (0, −2(t − 3 )).
In summary, the function expressing the ant’s coordinates is
(2t, 0)
1
)), sin( 3π
(t
2
2
1
))
2
(t − 21 )), cos( 3π
(t
sin( 3π
2
2
(−2(t − 43 ), 0)
1
))
2
(t
cos( 3π
2
when 01 ≤ t ≤
1
2
5
6
4
.
3
when 01 ≤ t ≤
1
2
5
6
4
.
3
−
−
when 2 < t ≤
(0, −2(t − 43 ))
when 65 ≤ t ≤
(b) To do the reverse path, observe that we can reverse the ant’s path by interchanging the x and y coordinates (flipping
it with respect to the line y = x), so the function is
(x(t), y(t)) =
(x(t), y(t)) =
(0, 2t)
−
when
when
2
5
6
0).
d(V 2 )
2 tan A(15 tan A − 4) − 15(tan2 A + 1) 3600
=
·
=0
dA
(15 tan A − 4)2
cos2 A
3600 15 tan2 A − 8 tan A − 15
=0
cos2 A
(15 tan A − 4)2
15 tan2 A − 8 tan A − 15 = 0
√
8 + 964
tan A =
30
≈ 1.30
A ≈ 52◦ .
1647
PROJECTS FOR CHAPTER SEVENTEEN
2. (a) The product rule gives
~
d
d~r
d~v
dL
= (~r × ~v ) =
× ~v + ~r ×
dt
dt
dt
dt
= ~v × ~v + ~r × ~a .
L
But the cross product of any vector with itself is ~0 . So ~v × ~v = ~0 . Hence ddt
= ~r × ~a .
(b) The area swept out by the planet is approximately a triangle, with sides ~r , ~r + ∆~r , and ∆~r . Since
k∆~r × ~r k is the area of the parallelogram formed by ∆~r and ~r , and since the triangle is half the size of
the parallelogram, we have ∆A ≈ 12 k∆~r × ~r k.
(c) Dividing by ∆t gives
∆A
1 ∆~r
≈
× ~r .
∆t
2 ∆t
~
~ = ~r × ~v , we get
Taking the limit as ∆t → 0 and recalling that L
1 ~
dA
= kL
k.
dt
2
(d) Since ~a is directed from the earth to the sun, and ~r from the sun to the earth, we see that ~r and ~a are
~ /dt = ~0 .
parallel. So ~r × ~a = ~0 , as the cross product of parallel vectors is ~0 . By part (a), this means dL
~ must be a constant.
So L
~ k is a constant, part (c) implies that
(e) Since kL
area swept out between
=
t = t0 and t = t1
Z
t1
t0
1 ~
dA
dt = kL
k
dt
2
Z
t1
t0
dt =
1 ~
kL k(t1 − t0 ).
2
So the area swept out over a time interval t1 − t0 only depends on t1 − t0 , not t0 and t1 individually.
(f) Let’s compare the triangles of are swept out by the planet when it is closest to and furthest from the sun,
for a given size time interval. Since the ~r and ~r + ∆~r sides are shorter when the planet is closest to the
sun, the central angle and the third side must be larger then. So ∆~r , and hence ~v = ∆~r /∆t, are larger
when the planet is closest to the sun, compared to when the planet is furthest from the sun (for a fixed ∆t).
3.
(a)
y
(b)
y
x
Figure 17.69
x
Figure 17.70
1648
Chapter Seventeen /SOLUTIONS
(d)
(c)
y
y
x
x
Figure 17.71
Figure 17.72
18.1 SOLUTIONS
1649
CHAPTER EIGHTEEN
Solutions for Section 18.1
Exercises
1. Negative because the vector field points in the opposite direction to the path.
2. Positive, because the vectors are longer on the portion of the path that goes in the same direction as the vector field.
3. Zero, because, by symmetry, the positive integral along the left half of the path cancels the negative integral along the
right half.
4. Positive, because the vector field points in the same direction as the path.
5. Zero, because the positive contributions on the upper half of the path cancel the negative contributions on the lower half
of the path.
6. Negative, because the vector field points in the opposite direction to the path.
~ is perpendicular to the curve at every point along it,
7. Since F
Z
~ · d~r = 0.
F
C
~ = 2~j and is parallel to the curve. Thus,
8. At every point along the curve, F
Z
C
~ · d~r = 2 · Length of curve = 2 · 5 = 10.
F
~ is perpendicular to the line, the line integral is 0.
9. Since F
10. Only the ~i -component contributes to the integral, so
Z
C
~ · d~r = 6 · Length of path = 6 · (7 − 3) = 24.
F
~ is a constant vector field and the curve is a line,
11. Since F
Z
R
C
~ · ∆~r , where ∆~r = 7~j . Therefore,
F~ · d~r = F
~ · d~r = (3~i + 4~j ) · 7~j = 28
F
C
12. At every point, the vector field is parallel to segments ∆~r = ∆x~i of the curve. Thus,
Z
C
~ · d~r =
F
Z
6
2
x~i · dx~i =
Z
6
xdx =
2
x2
2
6
= 16.
2
~ does not contribute to the line integral. Since ∆~r = ∆x~i , we have
13. The ~j -component of F
Z
C
~ · d~r =
F
Z
2
6
(x~i + y~j ) · dx~i =
Z
6
xdx =
2
x2
2
6
= 16.
2
~ is parallel to ∆~r . Suppose r is the distance from the point (x, y) to the origin, so k~r k = r.
14. At every point on the path, F
√
√
√
~ · ∆~r = kF
~ kk∆~r k = r∆r. At the start of the path, r = 22 + 22 = 2 2 and at the end r = 6 2. Thus,
Then F
Z
C
~ · d~r =
F
Z
√
6 2
√
2 2
r2
rdr =
2
√
6 2
√
2 2
= 32.
1650
Chapter Eighteen /SOLUTIONS
15. The path is along the y-axis, so only the ~j -component contributes to the line integral. Since C is oriented in the −~j
direction, we have
Z
(x~i + 6~j − ~k ) · d~r = −6 · Length of path = −6 · 8 = −48.
C
16. Only the ~i -component of the vector field contributes to the integral. This component, 5~i , points in the opposite direction
to the orientation of the path, which has length 8. Thus,
Z
C
(5~i + 7~j ) · d~r = −5 · Length of path = −5 · 8 = −40.
17. Only the ~i -component contributes to the integral, so
Z
C
~ · d~r =
F
Z
3
2~
x i · ~i dx =
2
Z
2
3
x3
x dx =
3
3
2
=
19
.
3
2
18. Only the ~j component contributes to the integral. On the y-axis, x = 0, so
Z
C
F~ · d~r =
Z
5
y3
y j · ~j dy =
3
5
2~
3
=
98
.
3
3
19. Since the curve is along the y-axis, only the ~j component of the vector field contributes to the integral:
Z
C
(2~j + 3~k ) · d~r =
Z
C
2~j · d~r = 2 · Length of C = 2 · 10 = 20.
20. The path is parallel to the z-axis, so the vector field is perpendicular to the path at every point. Thus, the line integral is 0.
21. Only the ~i -component contributes to the line integral, so d~r = ~i dx. On C we have y = 0, so
Z
C
((2y + 7)~i + 3x~j ) · d~r =
Z
(5,0,0)
(1,0,0)
(7~i + 3x~j ) · ~i dx =
Z
1
5
7 dx = 7 · 4 = 28.
22. The vector field x~i + y~j + z~k points radially outward and is everywhere perpendicular to the unit circle. Thus, the line
integral is 0.
Problems
23. The line integral along C1 is positive; the line integrals along C2 and C3 appear to be zero.
24. The line integral along C1 appears to be zero, the line integral along C2 is positive, and the line integral along C3 is
negative.
25. The line integral along C1 is negative, the line integral along C2 is negative, and the line integral along C3 appears to be
zero.
26. The line integral along C1 appears to be 0, the line integral along C2 is negative, and the line integral along C3 is positive.
27. Since itR appears that C1 is everywhere perpendicular to the vector field, all of the dot products in the line integral are zero,
~ with ∆~
ri are all positive, so their sum is positive and
hence C F~ · d~r ≈ 0. Along the path C2 the dot products of F
R1
R
~ · d~r . For C3 the vectors ∆~
~ , so the dot
we have
F
ri are in the opposite direction to the vectors of F
F~ · d~r <
C1
C2
products F~ · ∆~
ri are all negative; so,
R
C3
Z
C3
F~ · d~r < 0. Thus, we have
~ · d~r <
F
Z
C1
~ · d~r <
F
Z
C2
~ · d~r
F
18.1 SOLUTIONS
28. The integral
separately:
R
C
1651
~ · d~r is a sum of the line integrals of F
~ over each of its three straight segments, which we can compute
F
Z
PQ
Z
QR
Z
RS
Z
C
~ = (4~i + 2~j ) · ~i = 4
~ · d~r = P~Q · F
F
~ · d~r = QR
~ ·F
~ = (−~i + 2~j ) · (2~i − ~j ) = −4
F
~ ·F
~ = (−2~i − 2~j ) · (3~i + ~j ) = −8
~ · d~r = RS
F
~ · d~r = 4 − 4 − 8 = −8.
F
29. (a) See Table 18.1.
Table 18.1
(x, y)
~ (x, y)
F
(0, −1)
−~i
~
−i + ~j
(1, −1)
(2, −1)
(3, −1)
(4, −1)
(4, 0)
−~i + 4~j
−~i + 9~j
−~i + 16~j
✍
3
(4, 2)
(4, 3)
3~i + 16~j
✗
2
✗
1
16~j
~i + 16~j
2~i + 16~j
(4, 1)
y
−1
1
2
3
■
❑
❖
❖
✻
x
Figure 18.1
(b) See Figure 18.1.
(c) From the point (0, −1) to the point (4, −1), the x-component of the force field is always −1, i.e., it is pushing the
object backward with a constant force of 1. Thus, the work done on that part of the path is −1 · 4 = −4, because
only the horizontal component of the force field contributes to work.
From the point (4, −1) to the point (4, 3), the y-component of the force field is always 16, so it is pushing the
object forward with force of 16. Thus, the work done on that part of the path is 16 · 4 = 64, because only the vertical
component of the force field contributes to work.
So the total work done is −4 + 64 = 60.
30. The force has no horizontal component. Therefore the (positive) work done in the first half of C1 will be exactly canceled
by the (negative) work done in the second half, so the total work over the path C1 is zero. The same holds true for C2 ,
~ is constant and because the horizontal part of C2
again by virtue of the vertical symmetry of the path and the fact that F
contributes zero work. For C3 , the total work will be greater than zero, since the diagonal part of C3 is in the same general
~ and the horizontal part of C3 contributes zero work.
direction as F
31. The dot product of F~ and 10~i is positive if a > 0. There are no restrictions on b and c.
~ does not contribute to the line integral. Since the line integral of y~i around C is negative, for the
32. The ~k component of F
~ to be positive, we need a < 0. No restriction on c is needed.
line integral of F
~ is in the same direction as C if b > 0, so we want b < 0. No restriction is needed on c.
33. The vector field F
34. For any value of a, the vector field ay~i − ax~j is perpendicular to the vector ~i + ~j + ~k which is in the direction of C.
~ is in the direction of C if the coefficient of ~k is positive, that is, if
Thus a can take any value. The ~k component of F
c > 1.
1652
Chapter Eighteen /SOLUTIONS
35. The line C is parallel to the z-axis, so a~i + b~j does not contribute to the line integral. Thus, there are no restrictions on
a and b. The dot product of F~ and −~k is negative if c > 3.
36. (a) See Figure 18.2.
(i)
y
(ii)
y
x
(iii)
x
y
(iv)
x
y
x
Figure 18.2
(b) For (i) and (iii) a closed curve can be drawn; not for the others.
37. The vector field is F (~r ) = ~r . See Figure 18.3. The vector field is perpendicular to the circular arcs at every point, so
Z
Also, since it is radially symmetric,
C1
Z
~ · d~r =
F
C2
So,
Z
=
C
Z
~ · d~r = −
F
Z
+
C1
Z
+
C2
~ · d~r = 0.
F
C3
Z
C4
Z
C3
F~ · d~r .
+
Z
= 0.
C4
y
2
C4
1
C1
x
1
2C 3
3
−1
−2
C2
−3
Figure 18.3
4
18.1 SOLUTIONS
1653
~ (x, y) · ~r (x, y) = 0
38. This vector field is illustrated in Figure 18.4. It is perpendicular to C2 and C4 at every point, since F
and C2 and C4 are radial line segments, then
Z
C2
~ · d~r =
F
Z
C4
~ · d~r = 0.
F
Since C3 is longer than C1 , and the vector field is larger in magnitude along C3 , the line integral along C3 has greater
absolute value than that along C1 . The line integral along C3 is positive and the line integral along C1 is negative, so
Z
C
F~ · d~r =
Z
C3
~ · d~r +
F
Z
C1
F~ · d~r > 0.
See Figure 18.4.
y
2
C4
1
C1
x
1
2
3
C3
−1
C2
−2
Figure 18.4
~ (x, y) · ~r (x, y) = 0
39. This vector field is illustrated in Figure 18.5. It is perpendicular to C2 and C4 at every point, since F
and C2 and C4 are radial line segments, then
Z
C2
~ · d~r =
F
Z
C4
~ · d~r = 0.
F
Since C3 is longer than C1 , and the vector field is larger in magnitude along C3 , the line integral along C3 has greater
absolute value than that along C1 . The line integral along C1 is positive and the line integral along C3 is negative, so
Z
C
F~ · d~r =
Z
C3
~ · d~r +
F
Z
C1
F~ · d~r < 0.
See Figure 18.5.
y
2
C4
1
C1
x
1
2
−1
C2
−2
Figure 18.5
C3
3
1654
Chapter Eighteen /SOLUTIONS
40. Since it does not depend on y, this vector field is constant along vertical lines, x = constant. Now let us consider two
points P and Q on C1 which lie on the same vertical line. Because C1 is symmetric with respect to the x-axis, the tangent
vectors at P and Q will be symmetric with respect to the vertical axis so their sum is a vertical vector. But F~ has only
~ · (∆~r (P ) + ∆~r (Q)) = 0. As F
~ is constant along vertical lines (so F
~ (P ) = F
~ (Q)),
horizontal component and thus F
we obtain
~ (Q) · ∆~r (Q) = 0.
F~ (P ) · ∆~r (P ) + F
Summing these products and making k∆~r k → 0 gives us
Z
C1
F~ · d~r = 0.
~ · d~r = 0.
The same thing happens on C3 , so C F
3
Now let P be on C2 and Q on C4 lying on the same vertical line. The respective tangent vectors are symmetric with
respect to the vertical axis hence they add up to a vertical vector and a similar argument as before gives
R
~ (P ) · ∆~r (P ) + F
~ (Q) · ∆~r (Q) = 0
F
and
Z
C2
and so
~ · d~r +
F
Z
C
See Figure 18.6.
Z
C4
~ · d~r = 0
F
~ · d~r = 0.
F
y
2
C4
1
C1 2
1
x
3
C3
−1
C2
−2
Figure 18.6
~ (x, y) is perpendicular to the position vector ~r (x, y) = x~i + y~j because
41. First of all, F
F~ (x, y) · ~r (x, y) =
xy
−xy
+ 2
= 0.
x2 + y 2
x + y2
Also the magnitude of F~ is inversely proportional to the distance from the origin because
~ (x, y)k =
kF
p
x2 + y 2
1
=
.
x2 + y 2
k~r (x, y)k
So F~ is perpendicular to C2 and C4 and therefore
Z
C2
~ · d~r =
F
Z
C4
~ · d~r = 0.
F
18.1 SOLUTIONS
1655
Suppose R is the radius of C3 . On C3 , the vector field F~ has the same direction as the tangent vector which is approximated by ∆~r , so we have
~ · ∆~r = kF
~ k · k∆~r k = 1 k∆~r k.
F
R
When all these products are summed and the limit is taken as k∆~r k → 0, we get
Z
~ · d~r = 1
F
R
C3
=
Z
C3
kd~r k
1
(length of C3 ) = measure of the arc C3 in radians.
R
~ is in the opposite direction to the tangent vector
Similarly, suppose r is the radius of C1 . On C1 , the vector field F
which is approximated by ∆~r . Hence we have
Z
1
F~ · d~r = −
r
C1
Z
C1
kd~r k
1
= −( (length of C1 )) = −(measure of C1 in radians).
r
Since C1 and C3 have the same measure in radians, we have
Z
C
~ · d~r =
F
Z
C1
=−
~ · d~r +
F
Z
C2
~ · d~r +
F
π
π
+ 0 + (+ ) + 0 = 0.
2
2
Z
C3
F~ · d~r +
Z
C4
~ · d~r
F
See Figure 18.7.
y
2
C4
1
C1
1
x
2
3
C3
−1
−2
C2
Figure 18.7
42. Figure 18.8 shows the wind velocity vectors on each side of the square, where the speed is v meter/sec on the south side
and (v − 12) meter/sec on the north side. The circulation is the sum of the line integrals along the four sides of the square.
The line integrals along the eastern and western edges are both zero, since the wind velocity is perpendicular to these
edges. The integral to the right along the south side equals (1000 km)(−v meter/sec) = −v × 106 meter2 /sec, and the
integral to the left along the north side equals (1000 km)((v − 12) meter/sec) = (v − 12) × 106 meter2 /sec.
Total circulation = −v × 106 + (v − 12) × 106 = −1.2 × 107 meter2 /sec.
1656
Chapter Eighteen /SOLUTIONS
(v − 12) meter/sec
✛
N
✻
✛
1000 km
✛
1000 km
✛
v meter/sec
Figure 18.8: Wind velocity across a
square
43. Yes, the line integral over C1 is the negative of the line integral over C2 . One way to see this is to observe that the vector
field x~i + y~j is symmetric in the y-axis and that C1 and C2 are reflections in the y axis (except for orientation). See
Figure 18.9. Since the orientation of C2 is the reverse of the orientation of a mirror image of C1 , the two line integrals are
opposite in sign.
y
(0, 2)
C2
C1
x
(−1, 0)
(1, 0)
Figure 18.9
~ || ≤ 7, the line integral cannot be larger than 7 times the length of the curve. Thus
44. Since ||F
Z
C
~ · d~r ≤ 7 · Circumference of circle = 7 · 2π = 14π.
F
The line integral is equal to 14π if F~ is everywhere of magnitude 7, tangent to the curve, and pointing in the direction in
which the curve is traversed.
The smallest possible value occurs if the vector field is everywhere of magnitude 7, tangent to the curve and pointing
opposite to the direction in which the curve is transversed. Thus
Z
C
F~ · d~r ≥ −14π.
45. The line integral is defined by chopping the curve C into little pieces, Ci , and forming the sum
X
Ci
F~ · ∆~r .
When the pieces are small, ∆~r is approximately tangent to Ci , and its magnitude is approximately equal to the length of
~ and ∆~r are almost parallel, the dot product is approximately equal to the
the little piece of curve Ci . This means that F
product of their magnitudes, i.e.,
~ · ∆~r ≈ m · (Length of Ci ).
F
18.1 SOLUTIONS
1657
When we sum all the dot products, we get
X
Ci
~ · ∆~r ≈
F
X
m · (Length of Ci )
Ci
X
= m·
(Length of Ci )
Ci
= m · (Length of C)
~ · d~r = 0 for every closed curve C. Pick any two fixed points P1 , P2 and curves C1 , C2 each going from
46. Suppose C F
P1 to P2 . See Figure 18.10. Define −C2 to be the same curve as C2 except in the opposite direction. Therefore, the
curve formedRby traversing C1 , followed by C2 in the opposite direction, written as C1 − C2 , is a closed curve, so by our
~ · d~r = 0. However, we can write
assumption, C −C F
R
1
2
Z
~ · d~r =
F
C1 −C2
Z
~ · d~r −
F
C1
since C2 and −C2 are the same except for direction. Therefore,
Z
C1
so
F~ · d~r −
Z
C1
Z
C2
~ · d~r
F
F~ · d~r = 0,
C2
F~ · d~r =
Z
Z
C2
~ · d~r .
F
Since C1 and C2 are any two curves with the endpoints
P1 , P2 , this gives the desired result – namely, that
R
R fixing endpoints
~ · d~r . In other words, the value of the integral
~ · d~r does not
and direction uniquely determines the value of C F
F
C
depend on the path taken. We say the line integral is path-independent.
P2
C1
C2
P1
Figure 18.10
47. Pick any closed curve C. Choose two distinct points P1 , P2 on C. Let C1 , C2 be the two curves from P1 to P2 along C.
See Figure 18.11. Let −C2 be the same as C2 , except in the opposite direction. Thus, C1 − C2 = C. Therefore,
Z
C
~ · d~r =
F
Z
~ · d~r =
F
C1 −C2
Z
~ · d~r −
F
C1
Z
C2
~ · d~r
F
since C2 and −C2 differ only in direction.
But C1R and C2 have the same endpoints (P1 and P2 ) and same direction (P1
R
~ · d~r =
~ · d~r . Therefore,
to P2 ), so by assumption we have C F
F
C
1
2
Z
C
F~ · d~r =
Z
C1
~ · d~r −
F
Z
C2
P2
C1
C2
P1
Figure 18.11
F~ · d~r = 0.
1658
Chapter Eighteen /SOLUTIONS
~ , we expect the answer to be negative.
48. Let r = k~r k. Since ∆~r points outward, in the opposite direction to F
Z
C
~ · d~r =
F
Z
GM m~r
−
· d~r =
r3
C
GM m
=
r
10000
Z
= GM m
8000
= −2.5 · 10−5 GM m.
10000
8000
−
GM m
dr
r2
1
1
−
10000
8000
49. Let r = k~r k. Since ∆~r points outward, in the opposite direction to F~ , we expect a negative answer. We take the upper
limit to be r = ∞, so the integral is improper.
Z
C
~ · d~r =
F
Z
GM m~r
−
· d~r =
r3
C
= lim
b→∞
Z
b
8000
−
Z
∞
8000
−
GM m
dr
r2
GM m
GM m
dr = lim
b→∞
r2
r
b
= lim GM m
8000
b→∞
GM m
=−
8000
1
1
−
b
8000
~ , so the force applied in moving the particle against the field is
50. The force of the field on the particle at each point is E
~
−E , so
Z
~ · d~r
φ(P ) = −
E
C
where C is a path from P0 to P .
~
51. Any point P which is a units from
R the origin can be reached from P0 by a path C lying on the sphere of radius a. Since E
~ ·d~r = 0, so φ(P ) = 0. On the other hand, if P does not lie on the sphere of radius a,
is perpendicular to the sphere, C E
it can be reached by a path consisting of two pieces, C1 and
R C2 , one lying on the sphereR of radius a and one going straight
~ ·d~r = 0 as before, but
~ ·d~r 6= 0, since E
~ is parallel
along a line radiating from the origin (see Figure 18.12). C E
E
C2
1
R
R
~ · d~r =
~ · d~r .
to C2 and always points out. Thus, if C is the path consisting of C1 followed by C2 , we have
E
E
C
C2
~ · d~r is always positive or always negative along the path C which joins P0 to P . Hence the set of points with
Thus C E
potential zero is the sphere of radius a.
R
P
z
C2
C1
P0
y
x
Figure 18.12
52. In Problem 51 we saw that the surface where the potential is zero is a sphere of radius a. Let S be any sphere centered at
the origin, and let P1 be a point on S, and C1 a path from P0 to P1 . If P is any point on S, then P can be reached from
18.1 SOLUTIONS
1659
P
R 0 by a path, C, consisting of C1 followed by C2 , where C2 is a path from P1 to P lying entirely on the sphere, S. Then
~ · d~r = 0, since E
~ is perpendicular to the sphere. So
E
C
2
φ(P ) = −
Z
C
~ · d~r = −
E
Z
~ · d~r −
E
C1
Z
C2
~ · d~r = −
E
Z
C1
~ · d~r = φ(P1 ).
E
Thus, φ is constant on S. The equipotential surfaces are spheres centered at the origin.
53. (a) Suppose P is b units from the origin. Then P can be reached by a path, C, consisting of two pieces, C1 and C2 , one
lying on the sphere of radius a and one going straight along a line radiating from the origin (see Figure 18.13). We
~ · ∆~r = 0 on C1 , and, writing r = k~r k, we have E
~ · ∆~r = ||E
~ ||∆r on C2 , so
have E
φ(P ) = −
Z
~ · d~r = −
E
C
= 0−
Z
b
C1
~ · d~r −
E
~ || dr = 0 −
||E
a
Q 1
=
4πǫ r
Z
b
a
Z
b
a
Z
C2
~ · d~r
E
Q 1
dr
4πǫ r 2
Q 1
Q 1
−
.
=
4πǫ b
4πǫ a
Let P be the point with position vector ~r . Then
φ(~r ) = −
Q 1
Q 1
+
.
4πǫ a
4πǫ ||~r ||
P
z
C2
C1
P0
y
x
Figure 18.13
(b) If we let a → ∞ in the formula for φ, the first term goes to zero and we get the simpler expression
φ(~r ) =
Strengthen Your Understanding
R
54. This is only true if
C
~ · d~r > 0. However, if
F
R
C
55. The value of a line integral is a scalar, not a vector.
Q 1
.
4πǫ ||~r ||
~ · d~r < 0, then
F
R
−C
F~ · d~r > 0.
~
~ ~
~ ~
56. The curve C is parallel at every
R point to the vector i + j . The vector field F = i − j is perpendicular to C, because
~
~
~
~
F · i + j = 0. We have C F · d~r = 0 because the vector field and the curve are orthogonal at every point.
57. Choose for C1 a curve going in the direction of the vectors in the vector field, and choose for C2 a curve going in the
opposite direction of the vectors in the vector field. See Figure 18.14. A second option for C2 consists of using C1 oriented
in the downward direction.
1660
Chapter Eighteen /SOLUTIONS
y
C2
C1
x
Figure 18.14
~ · ∆~r is a scalar quantity,
58. False. Because F
R
C
~ · d~r is also a scalar quantity.
F
59. True. You can trace Rout C2 using the same
R subdivisions, but each ∆~r will have the opposite sign as before and will be
~ · d~r = −2
F~ · d~r = −6.
traced out twice, so C F
C
1
2
~ = x~i + y~j = ~r has radial direction, pointing everywhere perpendicular to the path of
60. True. The vector field F
integration, so the line integral is zero.
61. True. The line integral is the limit of a sum of dot products, hence is a scalar.
~ along the
62. False. The relative sizes of the line integrals along C1 and C2 depend on the behavior of the vector field F
curves. As a counterexample, take the vector field F~ = ~i , and C1 to be the line from the origin to (0, 2), while C2 is
the
R line from the origin to (1, 0). Then the length of C1 isR 2, which is greater than the length of C2 , which is 1. However
~ · d~r > 0 (since F
~ points along C2 ).
~ · d~r = 0 (since F~ is perpendicular to C1 ) while
F
F
C
C
2
1
~ = ~j and C be the
63. False. For example, the vector field F~ could be perpendicular to C everywhere. For instance, let F
curve t~i , for 0 ≤ t ≤ 1. Alternatively, F~ might point along part of C and in the opposite direction on another part of
~ = x~i and C be the curve t~i , for
C and so that the sum cancels out, yielding a zero line integral. For instance, let F
−1 ≤ t ≤ 1.
~ (~r i ) · ∆~r i in this line integral are positive, since the vector field (the constant ~i ) points
64. True. All of the dot products F
in the same direction as ∆~r i .
65. False. All of the dot products F~ (~r i ) · ∆~r i in this line integral are zero, since the vector field (the constant ~i ) points
perpendicular to ∆~r i .
66. False. The relation between these two line integrals depends on the behavior of the vector field along each of the curves,
so there is no reason to expect one to be the negative of the other. As an example, if F~ (x, y) = y~i , then, by symmetry,
both line integrals are equal to the same negative number.
67. False. The vector field swirls counterclockwise about the origin, and the path is oriented clockwise, so the line integral is
negative.
Solutions for Section 18.2
Exercises
1. If we use the parameterization x = sin t, y = cos t for 0 ≤ t ≤ π, we have x′ = cos t, y ′ = − sin t, so
Z
C
~ · d~r =
F
Z
0
π
((cos t)~i + (sin t)~j ) · ((cos t)~i − (sin t)~j ) dt =
Z
0
π
(cos2 t − sin2 t) dt.
Other answers are possible
2. If we use the parameterization x = t, y = 1 + t, z = t for 0 ≤ t ≤ 2, we have x′ = 1, y ′ = 1, z ′ = 1, so
Z
C
F~ · d~r =
Z
0
2
(t~i + t2~k ) · (~i + ~j + ~k )dt =
Z
2
(t + t2 ) dt.
0
18.2 SOLUTIONS
1661
3. If we use the parameterization x = cos t, y = sin t, z = 10 for 0 ≤ t ≤ 2π, we have x′ = − sin t, y ′ = cos t, z ′ = 0, so
Z
F~ · d~r =
C
=
2π
Z
(cos(cos t)~i + cos(sin t)~j + (cos 10)~k ) · (− sin t)~i + (cos t)~j ) dt
0
2π
Z
(− sin t cos(cos t) + cos t cos(sin t)) dt.
0
4. Only the y-component of the vector field contributes to the line integral. On the curve, d~r = ~j dy, so
Z
C
(3~i + (y + 5)~j ) · d~r =
Z
3
(y + 5) dy =
0
y2
+ 5y
2
3
=
0
39
.
2
5. Only the ~i -component contributes to the line integral, so d~r = ~i dx and
Z
C
(2x~i + 3y~j ) · d~r =
Z
(5,0,0)
(1,0,0)
(2x~i + 3y~j ) · ~i dx =
Z
5
5
2x dx = x2
= 24.
1
1
6. We will find the line integral from (0, 0) to (3, 1) and then take the negative. The line segment is parameterized by
x = 3t
y = t,
for 0 ≤ t ≤ 1.
Then ~r ′ (t) = 3~i + ~j , so
Z
C
(2y 2~i + x~j ) · d~r = −
Z
1
2t2~i + 3t~j
0
· (3~i + ~j ) dt = −
Z
0
1
(6t2 + 3t) dt = − 2t3 +
3 2
t
2
1
0
7
=− .
2
7. The semicircle has radius 1 and is centered at (2, 0). It can be parameterized by
x = 2 + cos t y = sin t,
for 0 ≤ t ≤ π.
Then ~r (t) = − sin t~i + cos t~j , so
′
Z
C
(x~i + y~j ) · d~r =
=
π
Z
(2 + cos t)~i + sin t~j
0
Z
π
· (− sin t~i + cos t~j ) dt
π
(−2 sin t − cos t sin t + sin t cos t) dt = 2 cos t
0
0
= −4.
~ = (x2 + y)~i + y 3~j , the line integral along the third segment, which is parallel to the z-axis, is zero. On the first
8. Since F
segment, which is parallel to the y-axis, only the ~j -component contributes. On the second segment, which is parallel to
the x-axis, only the ~i -component contributes. On the first segment x = 4 and y varies from 0 to 3; on the second segment
y = 3 and x varies from 4 to 0. Thus, we have
Z
C
~ · d~r =
F
=
Z
3
2
((4 + y)~i + y 3~j ) · ~j dy +
0
Z
3
3
y dy +
0
Z
4
0
Z
4
y4
(x + 3) dx =
4
2
0
((x2 + 3)~i + 33~j ) · ~i dx
3
0
−
x3
+ 3x
3
4
=
0
81
64
157
−
− 12 = −
.
4
3
12
~ contributes to the line integral. Since C goes a distance of 3 in the −~i direction, we have
9. Only the ~i component of F
Z
C
F~ · d~r = (2~i ) · (−3~i ) = −6.
10. Parameterizing C by x(t) = t, y(t) = t for 1 ≤ t ≤ 5, we have ~r ′ (t) = ~i + ~j ,
Z
~ · d~r =
F
Z
1
5
(3~j − ~i ) · (~i + ~j ) dt =
Z
1
5
2 dt = 8.
1662
Chapter Eighteen /SOLUTIONS
11. The curve C is parameterized by (x, y) = (t, t) for 0 ≤ t ≤ 3. Thus,
Z
C
~ · d~r =
F
3
Z
(t~i + t~j ) · (~i + ~j )dt =
0
Z
3
3
2tdt = t2
0
= 9.
0
12. Parameterize the curve: ~r (t) = sin t~i + cos t~j , 0 ≤ t ≤ π. Then
Z
C
π
Z
~ · d~r =
F
cos t~i − sin t~j
0
π
Z
=
· cos t~i − sin t~j
(cos t)2 + (− sin t)2 dt =
0
Z
π
dt
1 dt = π.
0
13. The line can be parameterized by (1 + 2t, 2 + 2t), for 0 ≤ t ≤ 1, so the integral looks like
Z
~ · d~r =
F
C
=
1
Z
F~ (1 + 2t, 2 + 2t) · (2~i + 2~j ) dt
0
1
Z
[(1 + 2t)2~i + (2 + 2t)2~j ] · (2~i + 2~j ) dt
0
=
1
Z
2(1 + 4t + 4t2 ) + 2(4 + 8t + 4t2 ) dt
0
=
1
Z
(10 + 24t + 16t2 ) dt
0
= (10t + 12t2 + 16t3 /3)
1
0
= 10 + 12 + 16/3 − (0 + 0 + 0) = 82/3
14. Use x(t) = t, y(t) = t2 , so x′ (t) = 1, y ′ (t) = 2t, with 0 ≤ t ≤ 2. Then
Z
~ · d~r =
F
=
2
Z
(−t2 sin t~i + cos t~j ) · (~i + 2t~j ) dt
0
Z
2
(−t2 sin t + 2t cos t) dt = t2 cos t
0
2
0
= 4 cos 2.
15. Parameterizing C by x(t) = 3t, y(t) = 2t for 0 ≤ t ≤ 1, we have ~r ′ (t) = 3~i + 2~j , so
Z
C
~ · d~r =
F
=
Z
1
((2t)3~i + (3t)2~j ) · (3~i + 2~j ) dt
0
Z
1
1
(24t3 + 18t2 ) dt = 6t4 + 6t3
0
16. The curve C is parameterized by
~r = cos t~i + sin t~j ,
so,
= 12.
0
for 0 ≤ t ≤ 2π,
~r ′ (t) = − sin t~i + cos t~j .
Thus,
Z
C
~ · d~r =
F
=
Z
2π
(2 sin t~i − sin (sin t)~j ) · (− sin t~i + cos t~j )dt
0
Z
0
2π
(−2 sin2 t − sin (sin t) cos t)dt
= sin t cos t − t + cos (sin t)
= −2π.
2π
0
18.2 SOLUTIONS
17. The parameterization is given, so
Z
C
F~ · d~r =
=
=
=
Z
Z
Z
Z
4
~ (2t, t3 ) · (2~i + 3t2~j ) dt
F
2
4
[(ln(t3 )~i + ln(2t)~j ] · (2~i + 3t2~j ) dt
2
4
(2 ln(t3 ) + 3t2 ln(2t)) dt
2
4
(6 ln(t) + 3t2 ln(2t)) dt
since ln(t3 ) = 3 ln(t).
2
This integral can be computed numerically, or using integration by parts or the integral table, giving
Z
C
Z
~ · d~r =
F
4
(6 ln(t) + 3t2 ln(2t)) dt
2
= (6(t ln(t) − t) + t3 ln(2t) − t3 /3)
4
2
44
136
− (28 ln 2 −
)
3
3
= 212 ln 2 − 92/3 ≈ 116.28.
= 240 ln 2 −
The expression containing ln 2 was obtained using the properties of the natural log.
18. Parameterizing C by x(t) = t, y(t) = t, z(t) = t for 0 ≤ t ≤ 2, we have ~r ′ (t) = ~i + ~j + ~k , so
Z
C
(x~i + 6~j − ~k ) · d~r =
=
Z
Z
2
(t~i + 6~j − ~k ) · (~i + ~j + ~k ) dt
0
2
0
(t + 6 − 1) dt =
t2
+ 5t
2
2
= 12.
0
19. The triangle C consists of the three paths shown in Figure 18.15.
(3, 2)
C3
(0, 0)
C2
(3, 0)
C1
Figure 18.15
Write C = C1 + C2 + C3 where C1 , C2 , and C3 are parameterized by
C1 : (t, 0) for 0 ≤ t ≤ 3;
Then
Z
C
where
Z
C1
C2 : (3, t) for 0 ≤ t ≤ 2;
F~ · d~r =
~ · d~r =
F
Z
0
Z
C1
~ · d~r +
F
3
~ (t, 0) · ~i dt =
F
Z
0
Z
C2
C3 : (3 − 3t, 2 − 2t) for 0 ≤ t ≤ 1.
F~ · d~r +
Z
C3
~ · d~r
F
3
(2t + 4)dt = (t2 + 4t)
3
0
= 21
1663
1664
Chapter Eighteen /SOLUTIONS
Z
Z
C2
~ · d~r =
F
C3
~ · d~r =
F
=
Z
2
~ (3, t) · ~j dt =
F
0
Z
1
0
= 50
2
(5t + 3)dt = (5t2 /2 + 3t)
0
2
0
= 16
~ (3 − 3t, 2 − 2t) · (−3~i − 2~j )dt
F
0
Z
Z
1
((−4t + 8)~i + (−19t + 13)~j ) · (−3~i − 2~j )dt
1
Z
(t − 1)dt = −25.
0
So
Z
C
~ d~r = 21 + 16 − 25 = 12.
F
20. Since ~r = x(t)~i + y(t)~j + z(t)~k = t~i + t2~j + t3~k , for 1 ≤ t ≤ 2,
we have ~r ′ (t) = x′ (t)~i + y ′ (t)~j + z ′ (t)~k = ~i + 2t~j + 3t2~k . Then
Z
C
2
Z
~ · d~r =
F
(t~i + 2t3 t2~j + t~k ) · (~i + 2t~j + 3t2~k ) dt
1
Z
=
2
(t + 4t6 + 3t3 ) dt
1
t2
4t7
3t4
=
+
+
2
7
4
21. We parameterize C by
2
=
1
2389
≈ 85.32
28
~r = 2t~i + 3t~j + 4t~k ,
Then ~r (t) = 2~i + 3~j + 4~k and so
′
Z
C
~ · d~r =
F
=
1
Z
(2t)3~i + (3t)2~j + (4t)~k · (2~i + 3~j + 4~k )dt
0
Z
for 0 ≤ t ≤ 1.
1
(16t3 + 27t2 + 16t)dt
0
1
= 4t4 + 9t3 + 8t2
= 21.
0
22. Since C is given by ~r = cos t~i + sin t~j + t~k , we have ~r ′ (t) = − sin t~i + cos t~j + ~k . Thus,
Z
C
F~ · d~r =
=
Z
Z
4π
(− sin t~i + cos t~j + 5~k ) · (− sin t~i + cos t~j + ~k )dt
0
4π
2
2
(sin t + cos t + 5)dt =
Z
4π
6dt = 24π.
0
0
23. The first step is to parameterize C by
(x(t), y(t), z(t)) = (0, 2 cos t, −2 sin t),
Thus, we have
0 ≤ t ≤ 2π.
~r ′ (t) = x′ (t)~i + y ′ (t)~j + z ′ (t)~k = −2 sin t~j − 2 cos t~k .
So we have
Z
C
F~ · d~r =
Z
2π
0
(e2 cos t~i + ~k ) · ((−2 sin t)~j + (−2 cos t)~k )dt
18.2 SOLUTIONS
=
Z
2π
−2 cos tdt
0
2π
= −2 sin t
0
=0
24.
25.
R
RC
C
3xdx − y sin xdy
y 2 dx + z 2 dy + (x2 − 5)dz
26. F~ = (x + 2y)~i + x2 y~j
27. F~ = e−3y~i − yz(sin x)~j + (y + z)~k
28. From x = t2 and y = t3 we get dx = 2tdt and dy = 3t2 dt. Hence
Z
ydx + xdy =
Z
5
t3 (2t)dt + t2 (3t2 )dt =
1
C
Z
5
1
5t4 dt = 55 − 1 = 3124.
29. From
x = cos t,
y = sin t,
z = 3t
we get
dx = − sin t dt
Hence
Z
dx + ydy + zdz =
dy = cos t dt,
Z
2π
− sin t dt + sin t cos t dt + 3t(3dt)
0
C
dz = 3dt.
= cos t +
9
1
2
sin2 t + t2 |2π
0 = 18π .
2
2
30. Parameterize C:
x = 1 + 4t,
y = 3 + 6t,
0≤t≤1
so that dx = 4dt and dy = 6dt. Hence
Z
3ydx + 4xdy =
Z
1
3(3 + 6t)4dt + 4(1 + 4t)6dt =
0
C
Z
1
60 + 168tdt = 144.
0
31. Parameterize C:
x = 0,
y = 3 cos t,
z = 3 sin t,
so that
dx = 0dt,
dy = −3 sin t dt,
Hence
Z
C
xdx + zdy − ydz =
Z
0
0 ≤ t ≤ 2π
dz = 3 cos t dt.
2π
0dt + 3 sin t(−3 sin t)dt − 3 cos t(3 cos t)dt =
Problems
32. (a) Since ~r (t) = t~i + t2~j , we have ~r ′ (t) = ~i + 2t~j . Thus,
Z
C
~ · d~r =
F
=
Z
1
F~ (t, t2 ) · (~i + 2t~j ) dt
0
Z
0
1
[(3t − t2 )~i + t~j ] · (~i + 2t~j ) dt
Z
0
2π
−9dt = −18π.
1665
1666
Chapter Eighteen /SOLUTIONS
Z
=
1
(3t + t2 ) dt
0
= (
3t2
t3
+ )
2
3
1
=
0
3
1
11
+ − (0 + 0) =
2
3
6
2~
(b) Since ~r (t) = t i + t~j , we have ~r ′ (t) = 2t~i + ~j . Thus,
Z
C
1
Z
~ · d~r =
F
~ (t2 , t) · (2t~i + ~j ) dt
F
0
1
Z
=
[(3t2 − t)~i + t2~j ] · (2t~i + ~j ) dt
0
Z
=
1
0
= (
(6t3 − t2 ) dt
t3
3t4
− )
2
3
1
0
3
1
7
= − − (0 − 0) =
2
3
6
33. (a) Figure 18.16 shows the curves.
y
1
C2
C1
x
−1
Figure 18.16
(b) On C1 , only the ~j -component of F~ contributes to the integral. There d~r = ~j dy, so
Z
~ · d~r =
F
C1
1
1
y2
y~j · ~j dy =
y dy =
2
−1
−1
Z
Z
1
= 0.
−1
On C2 , we have ~r ′ (t) = − sin t~i + cos t~j , so
Z
C2
~ · d~r =
F
Z
=
Z
3π/2
π/2
((cos t + 3 sin t)~i + sin t~j ) · (− sin t~i + cos t~j ) dt
3π/2
π/2
− cos t sin t − 3 sin2 t + cos t sin t dt =
3π/2
sin t cos t
−
= −3
2
2
t
=−
Z
3π/2
π/2
−3 sin2 t dt
3π
.
2
π/2
34. First, check that each of these gives a parameterization of L: each has both coordinates equal (as do all points on L) and
~ = (3x − y)~i + x~j using
each begins at (0, 0) and ends at (1, 1). Now we calculate the line integral of the vector field F
each parameterization.
(a) Using B(t) gives
Z
L
~ · d~r =
F
Z
0
1/2
((6t − 2t)~i + 2t~j ) · (2~i + 2~j ) dt =
Z
1/2
1/2
12t dt = 6t2
0
=
0
3
.
2
18.2 SOLUTIONS
1667
(b) Now we use C(t):
Z
L
F~ · d~r =
=
=
Z
Z
2
1
2
1
2
3
3(t2 − 1)
(t2 − 1) ~
t2 − 1 ~
−
i +
j
3
3
3
2t 2
2
(t − 1) dt =
3
3
t4
t2
−
4
2
2
=
1
Z
1
2
·
2t~
2t
i + ~j
3
3
dt
(t3 − t) dt
3
.
2
~ is
35. (a) The unit circle centered
at the origin has equation x2 + y 2 = 1. At any point in the plane, the magnitude of F
p
2
2
~
~
given by kF k = (−y) + x . Along the unit circle, kF k = 1.
(b) Suppose ~r = x~i + y~j is a radius vector to a point (x, y) on the unit circle centered at the origin. See Figure 18.17.
Then
~r · F~ = (x~i + y~j ) · (−y~i + x~j ) = −xy + xy = 0.
So the vector field is perpendicular to any corresponding radius vector, that is, the vector field is tangent to the circle
at every point.
y
~ = −y~i + x~j
F
~
r
x
x2 + y 2 = 1
Figure 18.17
(c) We can parameterize C by (cos t, sin t), for 0 ≤ t ≤ 2π. Then
Z
C
~ · d~r =
F
=
Z
2π
~ (cos t, sin t) · (− sin t~i + cos t~j ) dt
F
0
Z
2π
(− sin t~i + cos t~j ) · (− sin t~i + cos t~j ) dt
0
=
Z
2π
(sin2 t + cos2 t) dt
0
=
Z
2π
1 dt
0
= 2π
Thus,
Z
C
~ · d~r = 2π = Circumference of the unit circle.
F
36. We parameterize the helical staircase by observing that
x = 5 cos t,
y = 5 sin t,
z=t
1668
Chapter Eighteen /SOLUTIONS
has the correct radius, but climbs 2π in one revolution. To make it climb 4 meters in one revolution, we write:
x = 5 cos t,
y = 5 sin t,
z=
4t
2t
= .
2π
π
Thus,
2~
k.
π
~ = −70g~k , and we want to go around 2 turns of the staircase, so we take 0 ≤ t ≤
The gravitational force is given by F
4π. Thus,
~r ′ (t) = −5 sin t~i + 5 cos t~j +
Work done by gravity =
=
Z
Z
~ · d~r =
F
4π
−
0
4π
Z
0
2
−70g~k · (−5 sin t~i + 5 cos t~j + ~k )dt
π
140g
140g
dt = −
t
π
π
4π
= −560g.
0
Notice that the result can also be obtained by multiplying the force by the vertical distance:
Gravitational force · Vertical distance moved = (−70g)8 = −560g.
Now
Work done by person = −Work done by gravity = 560g.
37. The original integral is along the line from (1, 0, 0) to (2, 2, 3).
(a) This integral uses the same parameterization, but starting at (2, 2, 3) and ending at (1, 0, 0). Thus, the value of the
integral is −5.
(b) This integral is along the same line segment from (1, 0, 0) to (2, 2, 3) but using the parameterization x = t2 + 1,
y = 2t3 , z = 3t2 . Thus, the value of the integral is the same, 5.
(c) Using the parameterization in part(b), this integral traverses the line segment twice, from (2, 2, 3) to (1, 0, 0) and
then back to (2, 2, 3). Since the segment is traversed once in each direction, the value of the integral is 0.
38. The integral corresponding to A(t) = (t, t) is
1
Z
3t dt.
0
The integral corresponding to B(t) = (2t, 2t) is
1/2
Z
12t dt.
0
The substitution s = 2t has ds = 2 dt and s = 0 when t = 0 and s = 1 when t = 1/2. Thus, substituting t =
integral corresponding to B(t) gives
Z
1/2
12t dt =
Z
1
0
0
s 1
12( )( ds) =
2 2
Z
s
into the
2
1
3s ds.
0
The integral on the right-hand side is now the same as the integral corresponding to A(t). Therefore we have
Z
1/2
12t dt =
0
Z
1
3s ds =
0
Z
1
3t dt.
0
Alternatively, a similar calculation shows that the substitution t = 2w converts the integral corresponding to A(t)
into the integral corresponding to B(t).
39. The integral corresponding to A(t) = (t, t) is
Z
1
3t dt.
0
The integral corresponding to C(t) = ( t
2 −1
3
,
t2 −1
)
3
is
2
3
Z
1
2
(t3 − t) dt.
18.2 SOLUTIONS
1669
2
t −1
2
has ds = t dt. Also s = 0 when t = 1 and s = 1 when t = 2. Thus, substituting into the
3
3
integral corresponding to C(t) gives
The substitution s =
2
3
Z
2
(t3 − t) dt =
1
2
Z
1
1
Z
2
(t2 − 1) t dt =
3
3s ds.
0
The integral on the right-hand side is the same as the integral corresponding to A(t). Therefore we have
2
3
Alternatively, the substitution t =
to C(t).
Z
2
3
(t − t) dt =
1
1
Z
3s ds =
0
Z
1
3t dt.
0
w2 − 1
converts the integral corresponding to A(t) into the integral corresponding
3
40. The integral corresponding to A(t) = (t, t) is
1
Z
3t dt.
0
The integral corresponding to D(t) = (et − 1, et − 1) is
3
Z
ln 2
(e2t − et ) dt.
0
The substitution s = et − 1 has ds = et dt. Also s = 0 when t = 0 and s = 1 when t = ln 2. Thus, substituting into the
integral corresponding to D(t) and using the fact that e2t = et · et gives
3
Z
ln 2
(e2t − et ) dt = 3
0
Z
ln 2
(et − 1)et dt =
0
Z
1
3s ds.
0
The integral on the right-hand side is the same as the integral corresponding to A(t). Therefore we have
3
Z
ln 2
(e2t − et ) dt =
0
Z
1
3s ds =
0
Z
1
3t dt.
0
Alternatively, the substitution t = ew − 1 converts the integral corresponding to A(t) into the integral corresponding to
B(t).
R
41. (a) The line integral C (xy~i + x~j ) · d~r is positive. This follows from the fact that all of the vectors of xy~i + x~j
at points along C point approximately in the same direction as C (meaning the angles between the vectors and the
direction of C are less than π/2).
(b) Using the parameterization x(t) = t, y(t) = 3t, with x′ (t) = 1, y ′ (t) = 3, we have
Z
C
F~ · d~r =
=
Z
4
~ (t, 3t) · (~i + 3~j ) dt
F
0
Z
4
(3t2~i + t~j ) · (~i + 3~j ) dt
0
=
Z
4
(3t2 + 3t) dt
0
= t3 +
= 88.
3 2
t
2
4
0
(c) Figure 18.18 shows the oriented path C ′ , with the “turn around” points P and Q. The particle first travels from the
origin to the point P (call this path C1 ), then backs up from P to Q (call this path C2 ), then goes from Q to the point
(4, 12) in the original direction (call this path C3 ). See Figure 18.19. Thus, C ′ = C1 + C2 + C3 . Along the parts of
C1 and C2 that overlap, the line integrals cancel, so we are left with the line integral over the part of C1 that does not
overlap with C2 , followed by the line integral over C3 . Thus, the line integral over C ′ is the same as the line integral
over the direct route from the point (0, 0) to the point (4, 12).
1670
Chapter Eighteen /SOLUTIONS
y
12
11
10
9
8
7
6
5
4
3
2
1
0
(4, 12)
C2
P
C3
C1
Q
x
1
2
3
4
Figure 18.18
Figure 18.19
(d) The parameterization
1 3
(t − 6t2 + 11t), (t3 − 6t2 + 11t)
3
has (x(0), y(0)) = (0, 0) and (x(4), y(4)) = (4, 12). The form of the parameterization we were given shows that
the second coordinate is always three times the first. Thus all points on the parameterized curve lie on the line y = 3x.
We have to do a bit more work to guarantee that all points on the curve lie on the line between the point (0, 0)
and the point (4, 12); it is possible that they might shoot off to, say, (100, 300) before returning to (4, 12). Let’s
investigate the maximum and minimum values of f (t) = t3 − 6t2 + 11t on the interval 0 ≤ t ≤ 4. We can do this on
a graphing calculator or computer, or use single-variable calculus. We already know the values of f at the endpoints,
namely 0 and 12. We’ll look for local extrema:
(x(t), y(t)) =
0 = f ′ (t) = 3t2 − 12t + 11
which has roots at t = 2 ± √13 . These are the values of t where the particle changes direction: t = 2 − √13
corresponds to point P and t = 2 + √13 corresponds to point Q of C ′ . At these values of t we have f (2 − √13 ) ≈ 6.4,
and f (2 + √13 ) ≈ 5.6. The fact that these values are between 0 and 12 shows that f takes on its maximum and
minimum values at the endpoints of the interval and not in between.
(e) Using the parameterization given in part (d), we have
1
~r ′ (t) = x′ (t)~i + y ′ (t)~j = (3t2 − 12t + 11)~i + (3t2 − 12t + 11)~j .
3
Thus,
Z
C′
=
~ · d~r
F
Z
4
1
1
F~ ( (t3 − 6t2 + 11t), t3 − 6t2 + 11t) · ( (3t2 − 12t + 11)~i + (3t2 − 12t + 11)~j ) dt
3
3
4
1
1
1
( (t3 − 6t2 + 11t)2~i + (t3 − 6t2 + 11t)~j ) · ( (3t2 − 12t + 11)~i + (3t2 − 12t + 11)~j ) dt
3
3
3
0
=
Z
0
=
4
Z
1 3
1
(t − 6t2 + 11t)(3t2 − 12t + 11) ((t3 − 6t2 + 11t)~i + ~j ) · ( ~i + ~j )
3
3
0
=
Z
4
1 3
(t − 6t2 + 11t)(3t2 − 12t + 11)
3
0
=
1
9
Z
0
4
n
n1
3
o
(t3 − 6t2 + 11t) + 1
o
dt
(t3 − 6t2 + 11t)(3t2 − 12t + 11)(t3 − 6t2 + 11t + 3) dt
Numerical integration yields an answer of 88, which agrees with the answer found in part b).
dt
18.2 SOLUTIONS
1671
Strengthen Your Understanding
~ was done correctly, but instead of computing d~r =
42. The substitution of the parameterization into the vector field F
~r ′ (t)dt, the second integral used ~r (t) dt. Since ~r (t) = cos t~i + sin t~j , we have ~r ′ (t)dt = − sin t~i + cos t~j , so the
correct parameterized line integral is
Z
C
F~ · d~r =
Z
0
π/2
(cos t~i − sin t~j ) · (− sin t~i + cos t~j ) dt.
R
43. The sign and value of C 3 dx + 4 dy depend on the oriented curve C. For example, if moving along C in the direction
of its orientation increases x and y, then the integral is positive, and if it decreases x and y, then the integral is negative.
If the curve is in the direction of the vector 3~i + 4~j , then the integral is positive, but if the curve is in the opposite
direction, then the integral is negative.
44. The path is a semicircle of radius 3 centered at the origin, so a vector field that points away from the origin
will be
R
~ (x, y) = x~i +y~j , then
~ ·d~r =
perpendicular to the path at all points, and the integral will be zero. For example, if F
F
C
0.
45. The vector field points parallel to the x-axis at every point and is constant along any horizontal line, since the magnitude
only depends on y. The parameterized path y = π/2, x = t, 0 ≤ t ≤ 3, has length 3, and at every
R point on the path the
~ · d~r = 3 · 1 = 3.
vector field points in the direction of the path and has magnitude k sin y~i k = sin(π/2) = 1. So C F
46. False. The relation between these two line integrals depends on the behavior of the vector field along each of the curves, so
there is no reason to expect one to be larger than the other. If, for example, the line integral along C2 is negative, then the
line integral along both taken together (C1 + C2 ) will be less than the line integral over C1 by itself. A specific
example
R
~ = ~i , with C1 the line from (0, 0) to (1, 0), and C2 the line from (1, 0) to (0, 1). Then
~ · d~r =
is given by F
F
C1
R1
R
R
~ · d~r = 1/2 + 1 ~i · (−~i + ~j ) dt = 1/2 − 1 = −1/2.
~i · ~i dt = 1/2, and
F
0
C1 +C2
0
47. True. The dot product of the integrand 4~i with ~r (t) = ~i + 2t~j is 4, so the integral has value
′
R2
0
4 dt = 8.
48. False. The curves C1 and C2 are different. The curve C1 starts at the point (1, 0) and travels around the unit circle
counterclockwise to (−1, 0). The curve C2 starts at the point (1, 0) and travels around the unit circle clockwise to (−1, 0).
49. True. The curves C1 and C2 both parameterize the upper unit semicircle with the same orientation (but at different speeds).
Since the line integral is independent of parameterization, the integrals over C1 and C2 are the same.
50. False. As a counterexample, consider the unit circle C, centered at the origin, oriented counterclockwise and the vector
~ = −y~i + x~j . The vector field is always tangent to the circle, and in the same direction as C, so the line integral
field F
is positive.
~ = x~i . Then if we parameterize C1 by ~r (t) = t~i , with 0 ≤ t ≤ 1,
51. False. As a counterexample, consider the vector field F
we get
Z 1
Z 1
Z
1
1
t2
= .
x~i · d~r =
t~i · ~i dt =
t dt =
2
2
0
0
C1
0
A similar computation for C2 gives a line integral with value 2.
52. True. If we parameterize C by ~r (t) = a cos t~i + a sin t~j , with 0 ≤ t ≤ 2π, then
Z
C
(2x~i + y~j ) · d~r =
Z
0
2π
(2a cos t~i + a sin t~j ) · (−a sin t~i + a cos t~j ) dt =
=
Z
2π
0
−a2 cos t sin t dt
a2 cos2 t
2
2π
= 0.
0
53. False. If we parameterize C by ~r (t) = a cos t~i + a sin t~j , with 0 ≤ t ≤ 2π, then
Z
C
(2y~i +x~j )·d~r =
Z
2π
(2a sin t~i +a cos t~j )·(−a sin t~i +a cos t~j ) dt =
0
Z
0
2π
−2a2 sin2 t + a2 cos2 t dt = −πa2 .
54. True. The curves C1 and C2 are the same (they follow the graph of y = x2 between (0, 0) and (2, 4)), except that their
orientations are opposite.
~
55. (a). The two parameterizations give the same path, but one goes
R in the opposite directionR to the other. Since F points
~ · d~r is negative.
~ · d~r is positive, so
F
away from the origin, and C1 is oriented away from the origin, C F
C
1
2
1672
Chapter Eighteen /SOLUTIONS
56. (a). The two parameterizations move along the line y = x in the direction away from the origin, but they have different
endpoints. The path C1 goes from (0, 0) to (1, 1) and the path C2 goes from (0, 0) to (sin 1, sin 1) ≈ (0.84, 0.84). Since
F~ points away from the origin, and both pathsRare oriented soRthat the direction of travel is oriented away from the origin,
~ · d~r .
the integral along the longer path is larger, so C F~ · d~r > C F
1
2
Solutions for Section 18.3
Exercises
~ is a gradient field, with F
~ = grad f where f (x, y) = x2 + y 4 , we use the Fundamental Theorem of Line
1. Since F
Integrals. The starting point of the path C is (2, 0) and the end is (0, 2). Thus,
Z
C
~ · d~r = f (0, 2) − f (2, 0) = 16 − 4 = 12.
F
~
2. Since, if f (x, y, z) = sin(xy) + ez , we have gradf
√ =√ F , we use the Fundamental Theorem of Line Integrals. The
starting point of the path is (0, 0, 0) and the end is ( 2, 5, 2) so
Z
C
√ √
√
F~ · d~r = f ( 2, 5, 2) − f (0, 0, 0) = sin 10 + e2 − 1.
Notice that since F~ is a gradient field, the intermediate points on the path do not affect the answer.
3. Negative, not path-independent.
4. Zero, path-independent (a constant vector field is path-independent).
5. Negative, not path-independent.
6. Zero, appears path-independent
7. The field appears to be path-independent. If the path between any two points is a line segment, the line inegral will have
a value V . Any detour up or down appears to be perpendicular to the field, thus having no effect. Any detour left or right
will be compensated for because the vector field appears to take on the same values for all points along any vertical line.
Alternatively, the field could be imagined to be the gradient field of a function like z = x2 .
8. The field appears to be path-dependent. If we choose two points lying on a line that passes through the center, it appears
that the line integral on a straight path between them has value zero, but a line integral on a circular path clockwise has
positive value.
9. The field appears to be path-independent, because the vector field appears constant and could thus be the gradient field of
a linear function.
10. The field appears to be path-independent. If the path between any two points is a line segment, the line inegral will have
a value V . Any detour left or right appears to be perpendicular to the field, thus having no effect. Any detour up or down
will be compensated for because the vector field appears to take on the same values for all points along any horizontal
line. Alternatively, the field could be imagined to be the gradient field of a function like z = y 2 .
11. The field appears to be path-independent. If the path between any two points is a line segment, the line inegral will have
a value V . Any detour up-left or down-right appears to be perpendicular to the field, thus having no effect. Any detour
up-right or down-left will be compensated for because the vector field appears to take on the same values for all points
along any line going from top-left to bottom-right on a 45◦ angle to the image of the field.
12. The field appears to be path-dependent. If we choose two points lying on a line that passes through the center, it appears
that the line integral on a straight path between them has value zero, but a line integral on a circular path counterclockwise
has positive value.
13. We know that
∂f
∂f
= 2xy and
= x2 ,
∂x
∂y
so, integrating with respect to x, thinking of y as a constant gives
f (x, y) = x2 y + C(y).
18.3 SOLUTIONS
1673
Differentiating with respect to y gives
∂f
= x2 + C ′ (y),
∂y
so we take C(y) = k for some constant K 2 . Thus
f (x, y) = x2 y + K.
14. We know that
∂f
∂f
= 2xy and
= x2 + 8y 3
∂x
∂y
Now think of y as a constant in the equation for ∂f /∂x and integrate, giving
f (x, y) = x2 y + C(y).
Since the constant of integration may depend on y, it is written C(y). Differentiating this expression for f (x, y) with
respect to y and using the fact that ∂f /∂y = x2 + 8y 3 , we get
∂f
= x2 + C ′ (y) = x2 + 8y 3 .
∂y
Therefore
C ′ (y) = 8y 3
for some constant K. Thus,
so
C(y) = 2y 4 + K.
f (x, y) = x2 y + 2y 4 + K.
15. Integrating
∂f
= yzexyz + z 2 cos(xz 2 )
∂x
with respect to x and thinking of y and z as constant gives
f (x, y, z) = exyz + sin(xz 2 ) + C(y, z).
Differentiating with respect to y and using the fact that ∂f /∂y = xzexyz gives
∂C
∂f
= xzexyz +
= xzexyz .
∂y
∂y
Thus, ∂C/∂y = 0. This means C does not depend on y and can be written C(z), giving:
f (x, y, z) = exyz + sin(xz 2 ) + C(z).
Differentiating with respect to z, we get
∂f
= xyexyz + 2zx cos(xz 2 ) + C ′ (z).
∂z
The expression for grad f tells us that
∂f
= xyexyz + 2xz cos(xz 2 ).
∂z
Thus, we have C ′ (z) = 0 so C = constant, giving
f (x, y, z) = exyz + sin(xz 2 ) + C.
~ = grad f is a gradient vector field, the Fundamental Theorem of Line Integrals give us
16. Since F
Z
C
~ · d~r = f (end) − f (start) = x2 + 2y 3 + 3z 4
F
(0,0,5)
(4,0,0)
= 3 · 54 − 42 = 1859.
17. Since F~ = 3x2~i + 4y 3~j = grad(x3 + y 4 ), we take f (x, y) = x3 + y 4 . Then by the Fundamental Theorem of Line
Integrals,
Z
~ · d~r = f (−1, 0) − f (1, 0) = (−1)3 − 13 = −2.
F
C
1674
Chapter Eighteen /SOLUTIONS
~ = (x + 2)~i + (2y + 3)~j = grad
18. Since F
Z
C
F~ · d~r =
x2
2
+ 2x + y 2 + 3y , the Fundamental Theorem of Line Integrals gives
x2
+ 2x + y 2 + 3y
2
(3,1)
=
(1,0)
9
2
+6+1+3 −
1
2
+ 2 = 12.
~ = 2 sin(2x + y)~i + sin(2x + y)~j = grad(− cos(2x + y)), we take
19. Since F
f (x, y) = − cos(2x + y).
Then, using the Fundamental Theorem of Line Integrals,
Z
C
F~ · d~r = f (0, 5π) − f (π, 0) = − cos(5π) − (− cos(2π)) = −(−1) − (−1) = 2.
Notice that only the endpoints of the curve affect the answer.
~ = 2x~i − 4y~j + (2z − 3)~k = grad(x2 − 2y 2 + z 2 − 3z), the Fundamental Theorem of Line Integrals gives
20. Since F
Z
C
(2,3,−1)
F~ · d~r = (x2 − 2y 2 + z 2 − 3z)
~ = x2/3~i + e7y~j = grad
21. Since F
3 5/3
x
5
+ 71 e7y , we see F~ is a gradient vector field. Therefore,
Z
C
~ = x2/3~i + e7y~j = grad
22. Since F
Z
C
3 5/3
x
5
= (4 − 2 · 32 + (−1)2 + 3) − (12 − 2 · 12 + 12 − 3) = −7.
(1,1,1)
(x2/3~i + e7y~j ) · d~r = 0.
+ 71 e7y , we have
(x2/3~i + e7y~j ) · d~r =
3 5/3 1 7y
x
+ e
5
7
(0,1)
(1,0)
3
1
3
1
= · 05/3 + e7·1 − · 15/3 − e7·0
5
7
5
7
1
3
= (e7 − 1) − .
7
5
~ = grad(exy + sin z), we take f (x, y, z) = exy + sin z and use the Fundamental Theorem of Line Integrals
23. Since F
Z
C
~ · d~r = f (3, 1, 9π) − f (0, 0, π) = e3 + sin (9π) − e0 − sin π
F
= e3 − 1.
~ = y sin(xy)~i + x sin(xy)~j = grad(− cos(xy)), the Fundamental Theorem of Line Integrals gives
24. Since F
Z
C
(3,18)
~ · d~r = − cos(xy)
F
(1,2)
= − cos(54) + cos(2) = cos(2) − cos(54).
~ = 2xy 2 zex2 y 2 z~i +2x2 yzex2 y 2 z~j +x2 y 2 ex2 y 2 z ~k = grad(ex2 y 2 z ) and the curve C is closed, the Fundamental
25. Since F
R
~ · d~r = 0, since
Theorem of Line Integrals tells us that C F
Z
C
(1,0,1)
~ · d~r = ex
F
2 2
y z
(1,0,1)
= e0 − e0 = 0.
Problems
26. Since ~v is a gradient field, the end point should be (5, 4), the point with largest value of f = x2 + y 2 , and the starting
point should be (0, 0), the point with smallest f .
18.3 SOLUTIONS
1675
27. Since F~ is a gradient field, to maximize C ~v · d~r we need to choose the path with the end point that makes 2x2 + 3y 2 as
large as possible and with the starting point that makes 2x2 + 3y 2 as small as possible. Path P Q satisfies these conditions.
28. The vector field is a gradient field since
R
cos(xy)esin(xy) (y~i + x~j ) + ~k = grad(esin(xy) + z),
so we use the Fundamental Theorem of Line Integrals:
Z
C
(0.5,π,7)
cos(xy)esin(xy) (y~i + x~j ) + ~k · d~r = esin(xy) + z
=e
sin(π/2)
(π,2,5)
+ 7 − esin 2π − 5
= e+7−1−5
= e + 1.
~ points radially outward, and so is everywhere perpendicular to A; thus,
~ · d~r = 0.
29. The vector field F
F
A
~
~ · ∆~r are positive. By
Along the first half of B, the terms F · ∆~r are negative; along the second half the terms F
symmetry the positive and negative contributions cancel out, giving a Riemann sum and a line integral of 0.
~ along the x-axis have the same magnitude
The line integral is also 0 along C, by cancellation. Here the values of F
as those along the y-axis. On the first half of C the path is traversed in the opposite direction to F~ ; on the second half of
~ . So the two halves cancel.
C the path is traversed in the same direction as F
R
30. We parameterize A by x = t, y = t, where 0 ≤ t ≤ 1. Then
Z
A
~ · d~r =
F
=
Z
1
(t~i + t~j ) · (~i + ~j ) dt
0
Z
1
1
2t dt = t2
= 1.
0
0
The path B has the parameterization x = t, y = t2 , where 0 ≤ t ≤ 1. Then we have
Z
B
~ · d~r =
F
=
Z
1
(t~i + t2~j ) · (~i + 2t~j ) dt
0
Z
1
Z
1
t2
2t4
+
2
4
(t + 2t3 ) dt =
0
1
= 1.
0
We have to break the path C into two separate parameterizations: x = t, y = 0, where 0 ≤ t ≤ 1 and x = 1, y = t,
where 0 ≤ t ≤ 1. Then
Z
C
~ · d~r =
F
=
(t~i · ~i ) dt +
0
Z
0
1
t dt +
Z
1
Z
0
1
(~i + t~j ) · ~j dt
t dt =
0
1
1
+ = 1.
2
2
31. Yes. If f (x, y) = 12 x2 , then grad f = x~i .
32. Yes. If f (x, y) = 31 x3 − xy 2 , then grad f = (x2 − y 2 )~i − 2xy~j .
33. Yes. Let
f (~r ) = −
1
= −(x2 + y 2 + z 2 )−1/2
r
Then
∂f
= x(x2 + y 2 + z 2 )−3/2
∂x
∂f
= y(x2 + y 2 + z 2 )−3/2
∂y
∂f
= z(x2 + y 2 + z 2 )−3/2
∂z
So grad f = (x2 + y 2 + z 2 )−3/2 (x~i + y~j + z~k ) = ~r /r 3
1676
Chapter Eighteen /SOLUTIONS
~ . Then we would have
34. No. Suppose there were a function f such that grad f = F
−z
∂f
= √
.
∂x
x2 + z 2
Hence we would have
∂
−z
∂2f
=
(√
) = 0.
∂y∂x
∂y
x2 + z 2
~ , we have that
In addition, since grad f = F
y
∂f
.
= √
2
∂y
x + z2
Thus we also know that
∂2f
∂
=
∂x∂y
∂x
y
p
x2 + y 2
Notice that
Since we expect
!
= −xy(x2 + z 2 )−3/2 .
∂2f
∂2f
6=
.
∂y∂x
∂x∂y
∂2 f
∂y∂x
=
∂2f
,
∂x∂y
we have got a contradiction. The only way out of this contradiction is to conclude there
~
is no function f with grad f = F . Thus F~ is not a gradient vector field.
~ (x, y, z) = grad(ex2 +yz ), we use the Fundamental Theorem of line integrals
35. Since F
Z
C
~ · d~r =
F
Z
grad ex
C
2
+yz
· d~r = ex
2
+yz
(3,0,0)
(0,0,0)
= e9 − e0 = e9 − 1.
36. (a) To find the change in f by computing a line integral, we first choose a path C between the points; the simplest is a
line. We parameterize the line by (x(t), y(t)) = (t, πt/2), with 0 ≤ t ≤ 1. Then (x′ (t), y ′ (t)) = (1, π/2), so the
Fundamental Theorem of Line Integrals tells us that
f (1,
π
) − f (0, 0) =
2
=
Z
grad f · d~r
C
1
Z
0
=
1
Z
1
Z
2
2
1
Z
0
2
· i +
2
2tet sin
0
=
2tet sin
0
=
πt
~
grad f t,
= et sin
πt
2
1
dt
2
πt ~
πt ~
i + et cos
j
2
2
πt
2
2
πt
d
et sin
dt
2
π~
j
2
2
+
πet
πt
cos
2
2
π
· ~i + ~j
2
dt
dt
dt
= e = 2.718.
0
This integral can also be approximated numerically.
(b) The other way to find the change in f between these two points is to first find f . To do this, observe that
2
2
2
∂
∂
ex sin y ~i +
2xex sin y~i + ex cos y~j =
∂x
∂y
2
So one possibility for f is f (x, y) = ex sin y. Thus,
2
= ex sin y
(0,0)
= e1 sin
(0,0)
2
2
ex sin y ~j = grad ex sin y .
(1,π/2)
(1,π/2)
Change in f
π
2
− e0 sin 0 = e.
The exact answer confirms our calculations in part (a) which show that the answer is e.
18.3 SOLUTIONS
1677
37. The unit circle cuts the negative x-axis at (−1, 0, 0), and it cuts the negative y-axis at (0, −1, 0). There is a quarter circle
between these points if the circle is traversed counterclockwise.
(a) Since 2πx~i + y 2~j = grad(πx2 + y 3 /3), we use the Fundamental Theorem of Line Integrals:
Z
C
(2πx~i + y 2~j ) · d~r =
πx2 +
y3
3
(0,−1,0)
=
(−1,0,0)
(−1)3
3
π(02 ) +
−
π(−1)2 +
03
3
=−
1
− π.
3
(b) Since F~ is not a gradient field, we parameterize c. If x = cos t and y = sin t, then π ≤ t ≤ 3π/2 parametrizes C.
Thus
Z
C
3π/2
Z
(−2y~i + x~j ) · d~r =
(−2 sin t~i + cos t~j ) · (− sin t~i + cos t~j ) dt
π
Z
=
3π/2
(2 sin2 t + cos2 t) dt =
π
~ = grad
38. Since F
x2 + y
2
3π/2
(1 + sin2 t) dt
π
= t+
Z
t
sin t cos t
−
2
2
2
3π/2
=
π
3π
.
4
, the line integral can be calculated using the Fundamental Theorem of Line Integrals:
(3/
2
2
~ · d~r = x + y
F
2
c
Z
√
2,3/
√
2)
=
(0,0)
9
.
2
39. This vector field is not a gradient field, so we evaluate the
directly. Let C1 be the path along the x-axis from
√
√ line integral
(0, 0) to (3, 0) and let C2 be the path from (3, 0) to (3/ 2, 3/ 2) along x2 + y 2 = 9. Then
Z
C
Z
~ · d~r =
H
C1
~ · d~r +
H
Z
~ · d~r .
H
C2
~ is therefore perpendicular to the path. Thus,
On C1 , the vector field has only a ~j component (since y = 0), and H
Z
C1
~ · d~r = 0.
H
On C2 , the vector field is tangent to the path. The path is one eighth of a circle of radius 3 and so has length 2π(3/8) =
3π/4.
Z
~ k · Length of path = 3 · 3π = 9π .
~ · d~r = kH
H
4
4
C2
Thus,
Z
~ · d~r = 9π .
H
4
C
~ = grad(y ln(x + 1)), we evaluate the line integral using the Fundamental Theorem of Line Integrals:
40. Since F
Z
C
(3/
~ · d~r = y ln(x + 1)
F
√
2,3/
√
2)
(0,0)
3
= √ ln
2
3
√ +1
2
3
− 0 ln 1 = √ ln
2
3
√ +1 .
2
~ = grad(exy + sin(x + y)), the line integral can be calculated using the Fundamental Theorem of Line Integrals:
41. Since G
Z
c
(3/
F~ · d~r = exy + sin(x + y)
√
2,3/
√
2)
=e
(0.0)
9/2
+ sin
6
√
2
√
− e0 = e9/2 + sin(3 2) − 1.
1678
Chapter Eighteen /SOLUTIONS
~ = yz 2 exyz 2~i + xz 2 exyz 2~j + 2xyzexyz 2~k = grad(exyz 2 ), we can use the Fundamental Theorem of Line
42. Since F
Integrals.
start of the path, where t = 0, is (1, 0, 0). The end of the path is (cos(1.25π), sin(1.25π), 1.25π) =
√ The √
(−1/ 2, −1/ 2, 1.25π). Thus
Z
(−1/
~ · d~r = e
F
C
√
2,−1/
√
2,1.25π)
xyz 2
= e(1.25π)
2
/2
− 1.
(1,0,0)
~ = grad(x2 + y 2 + z 2 ), by the Fundamental Theorem of Line Integrals we have
43. Since F
Z
C
~ · d~r =
F
(1,5,9)
Z
grad(x2 + y 2 + z 2 ) · d~r = (x2 + y 2 + z 2 )
C
= 12 + 52 + 92 = 107.
(0,0,0)
~ =F
~ + y~i , we have
Since G
Z
~ · d~r =
G
C
To find
Z
C
Z
F~ · d~r +
C
Z
C
y~i · d~r = 107 +
Z
y~i · d~r .
C
y~i · d~r , we parameterize the line by ~r = t~i + 5t~j + 9t~k , for 0 ≤ t ≤ 1. We have ~r ′ (t) = ~i + 5~j + 9~k , so
Z
C
Thus,
y~i · d~r =
Z
Z
1
0
~ · d~r =
G
C
Z
5t~i · (~i + 5~j + 9~k ) dt =
Z
C
~ · d~r +
F
Z
1
5t2
2
5t dt =
0
1
=
0
5
.
2
5
y~i · d~r = 107 + = 109.5.
2
C
~ = grad(ye ), we can use the Fundamental Theorem which says that
44. (a) Since F
x
Z
(3,7)
C
F~ · d~r = yex
= 7e3 − 2e1 .
(1,2)
It does not matter how the curve goes because the Fundamental Theorem gives the line integral in terms of the values
of the function f (x, y) = yex at the end points only.
(b) The line is given by ~r = ~i + 2~j + t(2~i + 5~j ) = (1 + 2t)~i + (2 + 5t)~j . Thus, r ′ (t) = 2~i + 5~j , so
Z
C
~ · d~r =
F
=
Z
1
(2 + 5t)e
1+2t~
i +e
1+2t~
j
0
9e1+2t
2
1
+ 10
0
Z
1
te1+2t dt.
· (2~i + 5~j ) dt =
Z
1
(4 + 10t + 5)e1+2t dt
0
0
Using integration by parts with u = t, u′ = 1, v ′ = e1+2t , v = e1+2t /2 for the second integral, we get
Z
~ · d~r = 9 (e3 − e1 ) + 10
F
2
C
te1+2t
2
1
0
−
1+2t 1
=
5e
9 3
(e − e1 ) + 5e3 −
2
2
= 7e3 − 2e.
Z
0
=
0
1
e1+2t
dt
2
!
5
9 3
(e − e1 ) + 5e3 − (e3 − e1 )
2
2
18.3 SOLUTIONS
1679
45. Although this curve is complicated, the vector field is a gradient field since
x
y ~
x
y ~
x
y
F~ = sin
sin
i − cos
cos
j = grad −2 cos
sin
2
2
2
2
2
2
.
Thus, only the endpoints of the curve, P and Q, are needed. Since P = (−3π/2, 3π/2) and Q = (−3π/2, −3π/2) and
F~ = grad(−2 cos(x/2) sin(y/2)), we have
Q=(−3π/2,−3π/2)
Z
~ · d~r = −2 cos x sin y
F
2
2
C
P =(−3π/2,3π/2)
3π
3π
3π
3π
sin −
+ 2 cos −
sin
= −2 cos −
4
4
4
4
3π 3π
3π 3π
= 2 cos
sin
+ 2 cos
sin
4
4
4
4
1
1
1
1
= −2 · √ · √ − 2 · √ · √ = −2.
2
2
2
2
46. (a) By the Fundamental Theorem of Line Integrals
Z
(3,4)
(0,2)
grad f · d~r = f (3, 4) − f (0, 2) = 66 − 57 = 9.
(b) By the Fundamental Theorem of Line Integrals, since C is a closed path,
47. (a) Three possible paths are shown in Figure 18.20.
R
C
grad f · d~r = 0.
y
C2
C1
Q
x
P
C3
Figure 18.20
~ · dr
~ = 0 along C1 .
Since F~ is perpendicular to the horizontal axis everywhere, F
~ , the first leg of C2 will have a positive line integral. The second
Since C2 starts out in the direction of F
horizontal part of C2 will have a 0 line integral, and the third leg that ends at Q will have a positive line integral. Thus
the line integral along C2 is positive.
A similar argument shows that the line integral along C3 < 0.
(b) No, F~ is not a gradient field, since the line integrals along these three paths joining P and Q do not have the same
value.
48. (a) See Figure 18.21.
y
9
7
5
Q
3
1
x
P
Figure 18.21
1680
Chapter Eighteen /SOLUTIONS
(b) Vector at P is shorter than vector at Q.
(c) By the Fundamental Theorem of Line Integrals
Z
C
grad f · d~r = f (Q) − f (P ) = 9 − 3 = 6.
~ is a gradient vector field, we use the Fundamental Theorem of line integrals, giving
49. Since F
Z
C
Z
~ · d~r =
F
grad f · d~r = f (end) − f (start).
C
(a) The line integral C F~ · d~r is 0, since the curve begins and ends on the same contour, so f (end) = f (start).
2
(b) Since C1 crosses more contours than C4 , and since both curves are oriented in the direction of increasing f ,
R
0<
Z
F~ · d~r <
Z
~ · d~r .
F
F~ · d~r < 0 =
Z
~ · d~r .
F
C4
C1
Since C3 goes from higher to lower values of f ,
Z
C3
Thus, we have
Z
C3
Z
~ · d~r <
F
C2
~ · d~r <
F
C2
Z
C4
~ · d~r <
F
Z
C1
F~ · d~r .
(c) Since C3 and C4 have endpoints on the same contours, but with start and finish reversed,
Z
C3
The line integral
R
C3
~ · d~r = −
F
Z
C4
~ · d~r .
F
F~ · d~r is negative because Q3 is on a contour of lower value than P3 .
50. (a) The integral is positive, because the portion of the path that goes with the vector field is longer than the portion of the
path that goes against it, and in addition the vectors are larger in magnitude along the former and smaller in magnitude
along the latter.
~ = grad f for some function f , then the integral around every closed path would be zero. But
(b) If it were true that F
~ cannot be a gradient vector field.
in part (a) we saw that the integral around one closed path was not zero, so F
(c) The region shown is in the first quadrant. In that quadrant, the vectors of F~1 point away from the origin, so F~1 does
not fit. The vectors of both F~2 and F~3 point up and to the left, so they are both possibilities; of these, F~2 fits best
because its vectors get larger in magnitude as you move away from the origin, which fits the diagram. The vectors in
F~ 3 shrink as you move away from the origin.
51. Since the vector field is path independent, the line integral around the closed curve (0, 0) to (1, 0) to (1, 1) to (0, 1) to
(0, 0) is zero. Thus
Z
(0,0)
(0,1)
F~ · d~r = −
Z
(1,0)
(0,0)
~ · d~r +
F
(1,1)
Z
(1,0)
Z
~ · d~r +
F
(0,1)
(1,1)
~ · d~r
F
= −(5.1 + 3.2 − 4.7) = −3.6.
52. (a) To maximize the line integral, we choose C to be parallel to grad f = 3~i + 4~j . Thus C has parametric equation
~r = (2~i + ~j ) + t~v where ~v = 3~i + 4~j , so
x = 2 + 3t
y = 1 + 4t.
If the other end of C is at (x1 , y1 ), since the length of C is 10, we have
p
(x1 − 2)2 + (y1 − 1)2 = 10
p
(3t)2 + (4t)2 = 10
t
p
32 + 42 = 10
5t = 10
t = 2.
18.3 SOLUTIONS
1681
Thus t = 2 at (x1 , y1 ), so
x1 = 2 + 2 · 3 = 8
Thus C ends at the point (8, 9).
(b) By the Fundamental Theorem of Line Integrals,
Z
C
and
y1 = 1 + 2 · 4 = 9.
grad f · d~r = f (8, 9) − f (2, 1) = (3 · 8 + 4 · 9) − (3 · 2 + 4 · 1) = 50.
Alternately, since grad f and C are parallel,
Z
C
grad f · d~r = k grad f k · Length of C = 5 · 10 = 50.
53. (a) Since ~r · ~a = a1 x + a2 y + a3 z, we have
grad(~r · ~a ) = a1~i + a2~j + a3~k = ~a .
(b) By the Fundamental Theorem of Line Integrals, if r~0 = x0~i + y0~j + z0~k , we have
Z
(x0 ,y0 ,z0 )
C
grad(~r · ~a ) · d~r = ~r · ~a
(0,0,0)
= ~r 0 · ~a .
(c) Since ~r 0 · ~a = ||~r 0 ||||~a || cos θ = 10||~a || cos θ, where the angle between ~r 0 and ~a , the maximum value of ~r 0 · ~a
occurs if ~r 0 is parallel to ~a . Then θ = 0 and
Z
C
grad(~r · ~a ) · d~r = 10||~a || cos 0 = 10||~a ||.
54. (a) Work done by the force is the line integral, so
Work done against force = −
Z
C
~ · d~r = −
F
Z
C
(−mg ~k ) · d~r .
Since ~r = (cos t)~i + (sin t)~j + t~k , we have ~r ′ = −(sin t)~i + (cos t)~j + ~k ,
Work done against force =
Z
2π
mg ~k · (− sin t~i + cos t~j + ~k )dt
0
=
Z
2π
mg dt = 2π mg.
0
(b) We know from physical principles that the force is conservative. (Because the work done depends only on the vertical
distance moved, not on the path taken.) Alternatively, we see that
~ = −mg ~k = grad(−mgz),
F
so F~ is a gradient field and therefore path independent, or conservative.
55. (a) We parameterize the path by (cos t, sin t) for π/2 ≤ t ≤ π. Since t = π/2 gives the end point, (0, 1) and t = π
gives the starting point (−1, 0), we have
Z
C
~ =
~ · dr
F
Z
~ (cos t, sin t) · (− sin t~i + cos t~j ) dt
F
π
=−
=−
=
π/2
Z
Z
Z
π
π/2
π
π/2
(sin t~i − cos t~j ) · (− sin t~i + cos t~j ) dt
(− sin2 t − cos2 t) dt
π
π
π/2
1 dt = t
= π/2.
π/2
The work done by the force is +π/2. The work is positive since the force is always in the direction of the path (in
fact it is always tangent to C since F~ · ~r = 0).
(b) If we redo our computations using the entire unit circle, the only change will be the limits of integration: they’ll
change to 0 to 2π. This yields an answer of 2π (or −2π, depending on orientation). Since the work around a closed
path is not zero, the force is not path-independent.
1682
Chapter Eighteen /SOLUTIONS
~ (x, y) is parallel to grad h(x, y), it is perpendicular to the level curves of h. Since the oriented
56. (a) Since F~ (x, y) − G
~ (x, y) is perpendicular to C at every point of C. Hence
path C is on a level curve of h, F~ (x, y) − G
Z
C
Therefore
~ (x, y)) · d~r = 0.
~ (x, y) − G
(F
Z
C
~ · d~r =
F
Z
C
~ · d~r .
G
(b) By the Fundamental Theorem of Calculus for Line Integrals we have
Z
C
~ · d~r =
G
Z
C
grad φ · d~r = φ(Q) − φ(P ).
Using part (a) we have
Z
C
~ · d~r = φ(Q) − φ(P ).
F
57. (a) We have
grad h = ~j
grad φ = y~i + x~j
~
F − grad φ = x~j = x grad h.
~ − grad φ is a multiple of grad h.
Thus, F
(b) By part (a) the vector fields F~ and grad φ have the same components perpendicular to grad h, which is to say the
same components in the direction of the level curve C of h. Thus, the line integrals of F~ and grad φ along C are
equal. Using the Fundamental Theorem of Calculus for Line Integrals, we have
Z
C
~ · d~r =
F
Z
C
grad φ · d~r = φ(Q) − φ(P ) = 80 − 30 = 50.
58. (a) We have
grad h = ~i
grad φ = 2y~i + 2x~j
~
F − grad φ = −y~i = −y grad h.
~ − grad φ is a multiple of grad h.
Thus, F
(b) By part (a) the vector fields F~ and grad φ have the same components perpendicular to grad h, which is to say the
same components in the direction of the level curve C of h. Thus, the line integrals of F~ and grad φ along C are
equal. Using the Fundamental Theorem of Calculus for Line Integrals, we have
Z
C
~ · d~r =
F
Z
C
grad φ · d~r = φ(Q) − φ(P ) = 60 − 30 = 30.
59. (a) We have
grad h = −2x~i + ~j
grad φ = (2x2 + y)~i + x~j
~ − grad φ = −2x2~i + x~j = x grad h.
F
~ − grad φ is a multiple of grad h.
Thus, F
(b) By part (a) the vector fields F~ and grad φ have the same components perpendicular to grad h, which is to say the
same components in the direction of the level curve C of h. Thus, the line integrals of F~ and grad φ along C are
equal. Using the Fundamental Theorem of Calculus for Line Integrals, we have
Z
C
~ · d~r =
F
Z
C
grad φ · d~r = φ(Q) − φ(P ) = 384 − 0 = 384.
18.3 SOLUTIONS
1683
60. (a) We have
grad h = ~i + ~j
grad φ = (x + 3y)~i + (3x + 2y)~j
~
F − grad φ = (−x − 2y)~i + (−x − 2y)~j = −(x + 2y) grad h.
~ − grad φ is a multiple of grad h.
Thus, F
(b) By part (a) the vector fields F~ and grad φ have the same components perpendicular to grad h, which is to say the
same components in the direction of the level curve C of h. Thus, the line integrals of F~ and grad φ along C are
equal. Using the Fundamental Theorem of Calculus for Line Integrals, we have
Z
C
~ · d~r =
F
Z
C
grad φ · d~r = φ(Q) − φ(P ) = 1800 − 1850 = −50.
61. (a) By the chain rule
dh
∂f dx
∂f dy
=
+
= fx x′ (t) + fy y ′ (t),
dt
∂x dt
∂y dt
which is the result we want.
(b) Using the parameterization of C that we were given,
Z
C
grad f · d~r =
=
Z
b
(fx (x(t), y(t))~i + fy (x(t), y(t))~j ) · (x′ (t)~i + y ′ (t)~j )dt
a
Z
b
(fx (x(t), y(t))x′ (t) + fy (x(t), y(t))y ′ (t))dt.
a
Using the result of part (a), this gives us
Z
C
grad f · d~r =
Z
b
h′ (t)dt
a
= h(b) − h(a) = f (Q) − f (P ).
62. (a) The level surfaces are horizontal planes given by gz = c, so z = c/g. The potential energy increases with the height
above the earth. This means that more energy is stored as “potential to fall” as height increases.
(b) The gradient of φ points upward (in the direction of increasing potential energy), so ∇φ = g~k . The gravitational
~ = −g~k . The negative sign represents the fact that the
force acts toward the earth in the direction of −~k . So, F
gravitational force acts in the direction of the decreasing potential energy.
63. (a) We have
ϕ(~r ) =
p1 x + p2 y + p3 z
.
(x2 + y 2 + z 2 )3/2
Taking partial derivatives gives
p1 (x2 + y 2 + z 2 )3/2 − (3/2)(p1 x + p2 y + p3 z)(2x)(x2 + y 3 + z 2 )1/2
(x2 + y 2 + z 3 )3
(~
p · ~r )x
p1
−3
=
||~r ||3
||~r ||5
= −D1 (~r ).
ϕx (~r ) =
~ .
Similar computations give ϕy = −D2 and ϕz = −D3 , so grad ϕ = −D
~
(b) The field D is necessarily path-independent since it is a gradient vector field.
Strengthen Your Understanding
~ · d~r is a scalar and F
~ (Q) − F
~ (P ) is a vector. The correct statement of
64. The statement cannot be correct, because C F
R
~
~ · d~r = f (Q) − f (P ).
the Fundamental Theorem of Calculus for Line Integrals is as follows: if F = grad f , then C F
~
~
Evaluate the potential function, f , for F , not F itself, at the endpoints of C.
R
1684
Chapter Eighteen /SOLUTIONS
~ is path-independent; if it is not, then f is not well-defined.
65. This is only true if F
R
~ · d~r can’t be evaluated using the Fundamental Theorem of
~ is not a gradient vector field, then
F
66. If a vector field F
C
Calculus for Line Integrals. However, other methods such as parameterizing C might work to evaluate the integral.
67. The integral is required to have the same value Rfor all oriented paths from (0, 0) to (1, 2). This is achieved by choosing
F~ = grad f to be a gradient vector field. Since C grad f · d~r = f (1, 2) − f (0, 0), we seek a function f (x, y) such that
f (1, 2) − f (0, 0) = 100. For example, we can take f (x, y) = 50xy. With
F~ = grad f = 50y~i + 50x~j
we have
Z
C
F~ · d~r = 100.
68. Every gradient field is path-independent. For example,
~ = grad xy = y~i + x~j
F
is path-independent.
~ (x, y) =
69. A path-independent vector field must have zero circulation around all closed paths. Consider a vector field like F
|x|~j , shown in Figure 18.22.
A rectangular path that is symmetric about the y-axis will have zero circulation: on the horizontal sides, the field is
perpendicular, so the line integral is zero. The line integrals on the vertical sides are equal in magnitude and opposite in
sign, so they cancel out, giving a line integral of zero. However, this field is not path-independent, because it is possible to
find two paths with the same endpoints but different values of the line integral of F~ . For example, consider the two points
(0, 0) and (0, 1). The path C1 in Figure 18.23 along the y axis gives zero for the line integral, because the field is 0 along
the y axis, whereas a path like C2 will have a nonzero line integral. Thus the line integral depends on the path between
~ is not path-independent.
the points, so F
y
y
(0, 1)
C1
x
Figure 18.22
(0, 0)
C2
x
Figure 18.23
~ would be a vector. In the special case
70. This is false, because the line integral yields a scalar whereas the total change in F
~ happens to be the gradient of a scalar function f , the line integral does give the total change of the scalar function
when F
f along the path—but not of the vector function F~ .
~ but that both are gradient fields. For example,
71. You can easily come up with counterexamples: suppose that F~ 6= G
~ = ~j . Then, if C is a closed curve, the line integral around C of both F
~ and G
~ will equal to zero. But this
F~ = ~i and G
~.
does not mean that F~ = G
72. The total change of f along C depends only on the endpoints of C. If f has the same values at each endpoint (or if C is
closed, so that the endpoints coincide) then the total change will be zero. This in no way restricts the shape of the curve
C. For example, take f (x, y) = x2 + y 2 and C to be the straight line from the point (1, 0) to the point (0, 1). Then
f (1, 0) = f (0, 1) = 1 so the change in f along C is zero, but C is not a contour of f .
18.4 SOLUTIONS
1685
73. False. The line integral of a gradient vector field around this circle would be 0, but the converse is not necessarily true.
~ is necessarily a gradient field.
That is, the fact that the line integral around this one circle is zero does not mean F
2
2
1
74. True. Since x~i + y~j = grad (x + y ) , the Fundamental Theorem of Line Integrals gives
2
Z
1
(x~i + y~j ) · d~r = (x2 + y 2 )
2
C
(a,b)
1 2
(a + b2 ).
2
(0,0)
75. False. The line integrals of many vector fields (so called path independent or conservative fields) are zero around closed
curves, but this is not true of all fields. For example, a vector field that is flowing in the same direction as the curve C all
~ = −y~i + x~j , where C is the unit circle
along the curve has a positive line integral. A specific example is given by F
centered at the origin, oriented counterclockwise.
76. True. By the Fundamental Theorem for Line Integrals, if C is a path from P to Q, then
so the value of the line integral
R
C
Z
C
grad f · d~r = f (Q) − f (P ),
grad f · d~r depends only on the endpoints and not the path.
~ (x, y) = ~i is path77. False. The statement is true if C1 and C2 have the same initial and final points. For example, F
~ over a path from (0, 0) to (1, 0) is
independent (since it is the gradient of f (x, y) = x), but the line integral of F
f (1, 0) − f (0, 0) = 1, but the line integral over a path from (0, 0) to (0, 1) is f (0, 1) − f (0, 0) = 0.
~ is a gradient field, the line integral gives the total change in the potential function f , where F~ =
78. False. However, if F
grad f .
79. True. The construction at the end of Section 18.3 shows how to make a potential function from a path-independent vector
field.
80. True. Since a gradient field is path-independent, and C1 and C2 have the same initial and final points, the two line integrals
are equal.
~
81. True.RSince the curve
R can be thought of as having the same initial and final points, if f is a potential function for F we
~ · d~r =
grad
f
·
d~
r
=
f
(P
)
−
f
(P
)
=
0.
have C F
C
82. False. If there were a potential function f , then fx = y 2 , so f (x, y) = xy 2 + g(y), where g(y) is a function of y only.
Then fy = 2xy + g ′ (y), which depends on x no matter what g ′ (y) is. Thus fy cannot be equal to a constant k, and so
~ = grad f .
there is no potential function f such that F
~ = y~i . By symmetry, the line integral of F
~ over any circle centered at the origin is zero.
83. False. For example, take F
But the curve consisting of the upper semicircle connecting (−a, 0) to (a, 0) has a positive line integral, while the line
connecting these points along the x-axis has a zero line integral, so the field cannot be path-independent.
Solutions for Section 18.4
Exercises
∂Fi
∂F2
∂F1
∂F2
= 1 and
= −1, so
6=
and this cannot be a gradient vector field.
∂y
∂x
∂y
∂x
~ = 2xy~i + x2~j = grad(x2 y).
2. Yes, since F
~ (x, y) = y~i + y~j is the whole xy-plane. In order to see if F
~ is a gradient let us apply
3. The domain of the vector field F
the curl test:
∂F1
=1
∂y
and
∂F2
=0
∂x
~ is not the gradient of any function.
So F
1. We have
~ = 2xy~i + 2xy~j ,
4. No, since if F
∂F2
= 2y
∂x
and
∂F1
= 2x,
∂y
1686
Chapter Eighteen /SOLUTIONS
so
∂F2
∂F1
−
6= 0.
∂x
∂y
~ (x, y) = (x2 + y 2 )~i + 2xy~j is the whole xy-plane. Let us apply the curl test:
5. The domain of the vector field F
∂F2
∂F1
= 2y =
∂y
∂x
so F~ is the gradient of some function f . In order to compute f we first integrate
∂f
= x2 + y 2
∂x
with respect to x, thinking of y as a constant.
We get
x3
+ xy 2 + C(y)
3
Differentiating with respect to y and using the fact that ∂f /∂y = 2xy gives
f (x, y) =
∂f
= 2xy + C ′ (y) = 2xy
∂y
Thus C ′ (y) = 0 so C is a constant and
f (x, y) =
x3
+ xy 2 + C.
3
~ = (2xy 3 + y)~i + (3x2 y 2 + x)~j is the whole xy-plane. We apply the curl test:
6. The domain of the vector field F
∂F1
∂F2
= 6xy 2 + 1 =
∂y
∂x
so F~ is the gradient of a function f . In order to compute f we first integrate
∂f
= 2xy 3 + y
∂x
with respect to x thinking of y as a constant. We get
f (x, y) = x2 y 3 + xy + C(y)
Differentiating with respect to y and using the fact that ∂f /∂y = 3x2 y 2 + x gives
∂f
= 3x2 y 2 + x + C ′ (y) = 3x2 y 2 + x
∂y
Thus C ′ (y) = 0 so C is constant and
f (x, y) = x2 y 3 + xy + C.
~k
~j
~i
+ +
is the set of points (x, y, z) in the three space such that x 6= 0, y 6= 0
x
y
z
and z 6= 0. This is what is left in the three space after removing the coordinate planes. This domain has the property that
every closed curve is the boundary of a surface entirely contained in it, hence we can apply the curl test.
~ =
7. The domain of the vector field F
~i
~ =
curl F
~j ~k
∂
∂
∂
∂x ∂y ∂z
1
1
1
x
y
z
~ = ~0 and thus F
~ is the gradient of a function f . In order to compute f we first integrate
So curl F
∂f
1
=
∂x
x
18.4 SOLUTIONS
1687
with respect to x, thinking of y and z as constants. We get
f (x, y, z) = ln |x| + C(y, z)
Differentiating with respect to y and using the fact that ∂f /∂y = 1/y gives
∂C
1
∂f
=
=
∂y
∂y
y
We integrate this relation with respect to y thinking of z as a constant. We get
f (x, y, z) = ln |xy| + K(z)
Differentiating with respect to z and using the fact that ∂f /∂z = 1/z gives
1
∂f
= K ′ (z) =
∂z
z
Now we integrate with respect to z and get
f (x, y, z) = ln A|xyz|
where A is a positive constant.
~k
~i
~j
+ +
is the set of points in the three space, (x, y, z) such that x 6= 0 and
x
y
xy
y 6= 0. This is the set of points in the three space left after removing the planes x = 0 and y = 0. This domain has the
property that every closed curve is the boundary of a surface entirely contained in it, hence we can apply the curl test.
~ =
8. The domain of the vector field F
~i ~j ~k
curl F~ =
∂
∂
∂
∂x ∂y ∂z
1
1
1
x
y xy
∂ 1
∂ 1
( )−
( )
∂y xy
∂z y
1
1
= − 2~i + 2 ~j 6= 0
xy
x y
= ~i
− ~j
∂ 1
∂ 1
( )−
( )
∂x xy
∂z x
+ ~k
∂ 1
∂ 1
( )−
( )
∂x y
∂y x
~ is not the gradient of any function.
Therefore F
9. The domain of the vector field F~ = 2x cos(x2 + z 2 )~i + sin(x2 + z 2 )~j + 2z cos(x2 + z 2 )~k is the whole three space
so we can apply the curl test.
~ =
curl F
~i
~j
~k
∂
∂x
2
∂
∂y
2
∂
∂z
2
2x cos(x + y 2 ) sin(x + y 2 ) 2z cos(x + y 2 )
= −4yz sin(x2 + y 2 )~i + 4xz sin(x2 + y 2 )~j + (2x cos(x2 + y 2 ) + 4xy sin(x2 + y 2 ))~k 6= 0
~ 6= ~0 , F
~ is not the gradient of any function.
As curl F
10. We have
(x2 + y 2 )1 − y(2y)
x2 − y 2
∂F1
=
= 2
2
2
2
∂y
(x + y )
(x + y 2 )2
(x2 + y 2 )1 − x(2x)
∂F2
y 2 − x2
x2 − y 2
=−
=
−
=
.
∂x
(x2 + y 2 )2
(x2 + y 2 )2
(x2 + y 2 )2
∂F1
∂F2
=
. However, the domain of the vector field contains a “hole” at the origin, so the curl test does not apply.
∂y
∂x
This is not a gradient field. See Example 7 on page 992 of the text.
Thus
11. We have F1 = y 2 and F2 = x. By Green’s Theorem
Z
C
y 2~i + x~j
· d~r =
Z
R
∂F1
∂F2
−
∂x
∂y
dx dy =
Z
0
3
Z
0
2
(1 − 2y) dx dy = −12.
1688
Chapter Eighteen /SOLUTIONS
12. By Green’s Theorem, with R representing the interior of the circle,
Z
C
F~ · d~r =
Z
R
∂
∂
(−x) −
(y)
∂x
∂y
dA = −2
dA =
Z
dA
R
= −2 · Area of circle = −2π(12 ) = −2π.
13. By Green’s Theorem, with R representing the interior of the square,
Z
C
~ · d~r =
F
=
Z
∂
∂
(xy) −
(0)
∂x
∂y
R
Z
1
0
Z
1
y dy dx =
1
Z
0
0
y2
2
Z
y dA
R
1
dx =
0
1
.
2
14. By Green’s Theorem, with R representing the interior of the triangle,
Z
C
Z
∂
∂
(2x + 3y 2 ) −
(2x2 + 3y) dA =
∂x
∂y
R
1
= − Area of triangle = − · 4 · 3 = −6.
2
~ · ~r =
F
Z
R
(2 − 3) dA = −
Z
R
(y − 3) dA.
Z
dA
R
15. By Green’s Theorem, with R representing the interior of the circle,
Z
~ · d~r =
F
C
Z
R
∂
∂
(xy) −
(3y)
∂x
∂y
dA =
The integral of y over the interior of the circle is 0, by symmetry, because positive contributions of y from the top half of
the circle cancel those from the bottom half. Thus
Z
y dA = 0.
R
So
Z
F~ · d~r =
Z
((3x + 5y)~i + (2x + 7y)~j ) · d~r =
C
16. Green’s theorem gives
C
Z
R
(y − 3) dA =
Z
R
−3 dA = −3 · Area of circle = −3 · π(1)2 = −3π.
=
Z Z
Z ZR
R
−
∂
∂
(3x + 5y) +
(2x + 5y) dA
∂y
∂x
−3dA = −3 · Area of R = −3πm2 .
17. (a) The vector field points in the opposite direction to the orientation of the curve, hence the circulation is negative. See
Figure 18.24.
y
1
x
−1
1
−1
Figure 18.24
18.4 SOLUTIONS
1689
~ = y~i , we have ∂F1 /∂y = 1 and ∂F2 /∂x = 0 and ∂F1 /∂y = 1. Thus, using Green’s Theorem if R is
(b) Since F
the region enclosed by the closed curve C (the unit circle centered at the origin and traversed counterclockwise), we
have
Z
Z
Z
∂F1
∂F2
−
dx dy =
(−1) dx dy = −Area of R = −π.
F~ · d~r =
∂x
∂y
R
R
C
Problems
18. The perimeter of the rectangle is a closed curve, C, so we can use Green’s Theorem. See Figure 18.25. The curve is
traversed in the correct direction to apply Green’s Theorem directly. Let R be the interior of the rectangle,
Z
C
~ · d~r =
F
Z
R
∂(x + y)
∂(sin x)
−
∂x
∂y
Z
dxdy =
R
1 dxdy = Area of rectangle = 4 · 5 = 20.
y
(−1, 5)
(3, 5)
✻
5
R
❄
✛
4
(−1, 0)
x
✲ (3, 0)
Figure 18.25
19. By Green’s Theorem, with R as the interior of the square, we have
Z
C
(sin(x2 ) cos y)~i + (sin(y 2 ) + ex )~j · d~r =
Z
R
1
=
Z
1
=
Z
Z
1
Z
0
∂
∂
(sin(y 2 ) + ex ) −
(sin(x2 ) + cos y)
∂x
∂y
1
0
1
(ex + x sin y)
dy
0
0
(e + sin y) − 1) dy
1
= e − cos 1 − 1 + cos 0 = e − cos 1.
= (ey − cos y − y)
0
20. The curve is closed, so we can use Green’s Theorem. If R represents the interior of the region
Z
C
~ · d~r =
F
=
Z
R
Z
R
∂F1
∂F2
−
∂x
∂y
(1 − (−1)) dA =
dA =
Z
R
Z
R
∂(x − y)
∂(x)
−
∂x
∂y
2 dA = 2 · Area of sector.
Since R is 1/8 of a circle, R has area π(32 )/8. Thus,
Z
dA
(ex + sin y) dx dy
0
=
~ · d~r = 2 · 9π = 9π .
F
8
4
C
dA
1690
Chapter Eighteen /SOLUTIONS
21. The curve is closed, so we can use Green’s Theorem. If R represents the interior of the region
Z
C
~ · d~r =
F
=
Z
ZR
R
∂F2
∂F1
−
∂x
∂y
dA =
Z
R
∂(sin y)
∂(x + y)
−
∂x
∂y
dA
(−1) dA = (−1) · Area of sector.
Since R is 1/8 of a circle, R has area π(32 )/8. Thus,
Z
9π
9π
=− .
F~ · d~r = (−1) ·
8
8
C
~ = grad(x2 ey ), the Fundamental Theorem of Line Integrals gives
22. (a) Since F
Z
(2,4)
C
~ · d~r = x2 ey
F
= 4e4 .
(0,0)
(b) Since
∂G1
∂
∂
∂G2
−
=
(x + y) −
(x − y) = 2,
∂x
∂y
∂x
∂y
~ is not a gradient field. Parameterizing C by x(t) = t, y(t) = 2t for 0 ≤ t ≤ 2, we have
we know that G
~r ′ (t) = ~i + 2~j , so
Z
C
~ · d~r =
G
=
2
Z
((t − 2t)~i + (t + 2t)~j ) · (~i + 2~j ) dt
0
2
Z
0
((t − 2t) + (2t + 4t)) dt =
Z
2
5t dt =
0
5 2
t
2
2
= 10.
0
23. (a) By symmetry between quadrants II and IV, this integral is zero. To confirm, we can parameterize. This line is given
by
~r = (−~i + ~j ) + t(2~i − 2~j ) = (−1 + 2t)~i + (1 − 2t)~j for 0 ≤ t ≤ 1.
′
~
Then ~r (t) = 2i − 2~j , so
Z
C
~ · d~r =
F
Z
1
0
(1 − 2t)~i + (−1 + 2t)~j
(b) The curve C is closed, so we use Green’s Theorem:
Z
C
~ · d~r =
F
Z
R
· (2~i − 2~j ) dt =
∂
∂
(x) −
(y)
∂x
∂y
dA =
Z
Z
1
0
(4 − 8t) dt = 0.
0 dA = 0.
R
(c) Since C = C2 − C1 , we have
Z
C
Z
~ · d~r =
F
C2
Z
0=
C2
So
Z
C2
~ is constant on C2 :
(d) The magnitude of G
~ || =
||G
p
F~ · d~r −
Z
C1
~ · d~r
F
F~ · d~r − 0.
~ · d~r = 0.
F
√
y 2 + x2 = 3 2.
p
(3y)2 + (−3x)2 = 3
Thus, since G is everywhere tangent to C2 and points in the opposite direction to C2 , we have
Z
C2
√
√
~ · d~r = −||G
~ || · Length of C2 = −(3 2)π 2 = −6π.
G
18.4 SOLUTIONS
1691
(e) The curve C is closed, so we use Green’s Theorem:
Z
C
Z
Z
∂
∂
(−3x) −
(3y) dA =
(−6) dA
∂x
∂y
R
R
√
π( 2)2
= −6π.
= −6 · Area of R = −6
2
~ · d~r =
G
(f) Since C = C2 − C1 , we have
Z
C
~ · d~r =
G
Z
~ · d~r −
G
C2
−6π = −6π −
Hence,
Z
C1
Z
C1
Z
C1
~ · d~r
G
~ · d~r .
G
~ · d~r = 0.
G
(g) The curve C is closed, so we use Green’s Theorem:
Z
C
~ +G
~ ) · d~r =
(F
Z
C
Z
Z
∂
∂
(−2x) −
(4y) dA =
−6 dA
∂x
∂y
R
R
√ 2
π( 2)
= −6π.
= −6 · Area of R = −6 ·
2
(4y~i − 2x~j ) · d~r =
Alternately,
Z
C
~ +G
~ ) · d~r =
(F
Z
C
Z
~ · d~r +
F
C
~ · d~r = 0 − 6π = −6π.
G
24. (a) The curve, C, is closed and oriented in the correct direction for Green’s Theorem. See Figure 18.26. Writing R for
the interior of the circle, we have
Z
C
(x − y)~i + (y 2 + x)~j · d~r =
2
=
Z
ZR
R
∂(y 2 + x)
∂(x2 − y)
−
∂x
∂y
(1 − (−1)) dxdy = 2
Z
dxdy
dxdy
R
2
= 2 · Area of circle = 2(π · 3 ) = 18π.
(b) The circle given has radius R and center (a, b). The argument in part (a) works for any circle of radius R, oriented
counterclockwise. So the line integral has the value 2πR2 .
y
(x − 5)2 + (y − 4)2 = 9
✠
4
R
C
x
5
Figure 18.26
1692
Chapter Eighteen /SOLUTIONS
25. (a) The particle moving along C1 starts at the origin and moves simultaneously radially outward and counterclockwise in
the xz-plane, describing a spiral. The spiral completes two full revolutions around the origin, starting at (0, 0, 0) and
ending at the point (2, 0, 0). The particle moving along C2 starts at the origin and moves simultaneously outward,
counterclockwise, and forward along the positive y-axis, describing a 3-dimensional spiral coil growing in diameter
as it wraps around the y-axis. The coil completes two full revolutions around the y-axis, starting at (0, 0, 0) and
ending at the point (2, 2, 0).
(b) We suspect F~ is a gradient field and look for a potential function f . We have:
f (x, y, z) =
Z
fx dx =
Z
yz dx = yzx + g(y, z).
We now have two expressions for fy , which we set equal to each other:
z(x + 1) = fy = zx + gy .
This means
zx + z = zx + gy
Thus,
g(y, z) =
and also
Z
gy dy =
Z
, or
gy = z.
z dy = zy + h(z),
f (x, y, z) = yzx + zy + h(z).
Lastly, we set both expressions for fz equal to each other
xy + y + 1 = fz = yx + y + h′ (z).
This means h(z) = z which gives us
f (x, y, z) = xyz + zy + z.
~ = ∇f and we can calculate the line integral using the Fundamental Theorem of Calculus for Line Integrals:
Thus, F
Z
C2
~ · dr = f (2, 2, 0) − f (0, 0, 0) = 0 − 0 = 0.
F
(c) The beginning points and the endpoints of C1 and C2 , respectively, have identical z-values. Since this common z~ whose potential function has a factor of z, then use the Fundamental
value is zero, we may look for a gradient field G
Theorem of Calculus for Line Integrals to calculate both integrals. Note that
~ =F
~ = ∇(xyz + zy + z)
G
~
G = ∇(ze−xyz )
and
both work, for example. In each case we have:
Z
C1
There are many other possible answers.
(d) Reasoning as in (c) above, we choose
~ · dr =
G
Z
C2
~ · dr = 0.
G
~1 = F
~ = ∇(xyz + zy + z)
H
~
H2 = ∇(ze−xyz ),
and
for example. These fields work because, using the Fundamental Theorem of Calculus for Line Integrals, we get
Z
C1
The fields
~1 · d~r =
H
Z
C1
~2 · d~r = 0.
H
~1 = ∇(yx2 ) and
H
~2 = ∇(y(x + z))
H
also work, since y = 0 at the beginning and endpoints of C1 and the Fundamental Theorem of Calculus for Line
Integrals yields:
Z
Z
~2 · d~r = 0.
~1 · d~r =
H
H
C1
C1
There are many other possible answers, and not all of them yield a zero value for the integrals.
18.4 SOLUTIONS
1693
~ = x~j , we have ∂F2 /∂x = 1 and ∂F1 /∂y = 0. Thus, using Green’s Theorem if R is the region enclosed by the
26. Since F
closed curve C, we have
Z
~ · d~r =
F
C
Z
R
∂F1
∂F2
−
∂x
∂y
Z
dx dy =
1 dx dy = Area of R
R
~ = x~j = a cos t~j and ~r ′ (t) = −a sin t~i + b cos t~j , we have
27. Using F
A=
Z
C
= ab
~ · d~r =
F
Z
2π
2π
Z
(a cos t)(b cos t) dt
0
cos2 t dt
0
Z
2π
1 + cos 2t
dt
2
0
2π
ab
sin 2t 0 = πab
= πab +
4
= ab
The ellipse is shown in Figure 18.27.
y
b
a
Figure 18.27:
x
x2
y2
+ 2 =1
2
a
b
~ = x~j = a cos3 t and ~r ′ (t) = −3a cos2 t sin t~i + 3a sin2 t cos t~j , we have
28. Using F
A=
Z
C
= 3a
~ · d~r =
F
2
2π
Z
4
=
3a2
16
=
3a2
16
2π
(a cos3 t)(3a sin2 t cos t) dt
0
2
cos t sin t dt = 3a
Z
2π
0
2π
2
2
cos t(sin t cos t) dt = 3a
2
Z
2π
0
2π
(1 + cos 2t − cos 4t − cos 2t cos 4t) dt
0
Z
Z
(1 + cos 2t)(1 − cos 4t) dt
0
Z
2
0
0
3a2
=
16
Z
2π
(1 + cos 2t − cos 4t −
1
1
cos 6t − cos 2t) dt
2
2
1
1
1
1
3a2
(t − sin 2t − sin 4t +
sin 6t + sin 2t)
=
16
2
4
12
4
2π
=
0
3πa2
8
For the last integral we use the trigonometric formula cos 2t cos 4t = 12 (cos 6t + cos 2t).
The hypocycloid is shown in Figure 18.28.
cos2 t
sin2 2t
dt
4
1694
Chapter Eighteen /SOLUTIONS
y
a
a
x
Figure 18.28: x2/3 + y 2/3 = a2/3
~ = x~j =
29. Using F
3t(2 − t3 ) ~
3t2 ~
1 − 2t3 ~
j and ~r ′ (t) =
i +
j , we have
3
3
2
1+t
(1 + t )
(1 + t3 )2
A=
Z
C
~ · d~r =
F
Z
∞
0
=9
Z
3t(2 − t3 )
3t
·
dt
1 + t3 (1 + t3 )2
∞
0
t2 (2 − t3 )
dt
(1 + t3 )3
We make the change of variables u = 1 + t3 so du = 3t2 dt and 2 − t3 = 3 − u. So
A=3
Z
∞
1
3−u
du.
u3
This is an improper integral, so it can be computed as follows
A=3
Z
∞
3−u
du = lim 3
b→∞
u3
1
= lim
b→∞
"
1
u−2
9 −
2
h 91
= lim −
b
Z b
1
1
+3
u
1
−1 +3
1
2 b2
b
3
9
= − (0 − 1) + 3(0 − 1) = .
2
2
The Folium of Descartes is shown in Figure 18.29.
b→∞
b
1
3
1
− 2
u3
u
#
i
−1
y
2
x
2
Figure 18.29: x3 + y 3 = 3xy
du
18.4 SOLUTIONS
1695
30. Suppose C encloses a region R. Then, using Green’s Theorem, we have
Z
C
~ · d~r =
F
=
=
=
Z
ZR
ZR
ZR
R
∂F2
∂F1
−
∂x
∂y
dA
∂
∂
(4x(1 − y 2 ) + x sin(xy)) −
(−y 3 + y sin(xy)) dA
∂x
∂y
4(1 − y 2 ) + sin(xy) + xy cos(xy) + 3y 2 − sin(xy) − xy cos(xy) dA
(4 − y 2 ) dA
This integral over R is largest if C encloses the maximum possible region where 4 − y 2 > 0, that is, where −2 ≤ y ≤ 2.
Therefore C should be the curve with two sides along the lines y = −2 and y = 2, as well as two arcs of the circle
x2 + y 2 = 25. See Figure 18.30.
5
y
x2 + y 2 = 25
2
x
−2
−5
Figure 18.30
31. Since the level curves must be perpendicular to the gradient vectors, if there were a contour diagram fitting this gradient
field, it would have to look like Figure 18.31. However, this diagram could not be the contour diagram because the origin
is on all contours. This means that f (0, 0) would have to take on more than one value, which is impossible. At a point P
other than the origin, we have the same problem. The values on the contours increase as you go counterclockwise around,
since the gradient vector points in the direction of greatest increase of a function. But, starting at P , and going all the way
around the origin, you would eventually get back to P again, and with a larger value of f , which is impossible.
An additional problem arises from the fact that the vectors in the original vector field are longer as you go away from
the origin. This means that if there were a potential function f then || grad f || would increase as you went away from the
origin. This would mean that the level curves of f would get closer together as you go outward which does not happen in
the contour diagram in Figure 18.31.
y
x
Figure 18.31
1696
Chapter Eighteen /SOLUTIONS
32. The drawing of the contour diagrams in Figure 18.32 fitting this gradient field would look like Figure 18.32. The values
on the contours would increase both as y increases (for positive x) and as y decreases (for negative x), following the rule
that the gradient vector points in the direction of greatest increase of a function. Therefore, it is impossible for this to be a
contour diagram.
y
✻
Increasing
f -values
o
Increasing
x
f -values
❄
Figure 18.32
~ ,G
~ ,H
~ are all gradient vector fields, since
33. (a) We see that F
grad(xy) = F~
~
grad(arctan(x/y)) = G
for all x, y
except where y = 0
~
grad (x2 + y 2 )1/2 = H
except at (0, 0).
~ for x 6= 0.
Other answer are possible. For example grad(− arctan(y/x)) = G
(b) Parameterizing the unit circle, C, by x = cos t, y = sin t, 0 ≤ t ≤ π, we have ~r ′ (t) = − sin t~i + cos t~j , so
Z
C
~ · d~r =
F
Z
2π
((sin t)~i + (cos t)~j ) · ((− sin t)~i + (cos t)~j ) dt =
0
Z
2π
cos(2t) dt = 0.
0
~ is tangent to the circle, pointing in the opposite direction to the parameterization, and of length 1
The vector field G
everywhere. Thus
Z
C
~ · d~r = −1 · Length of circle = −2π.
G
~ points radially outward, so it is perpendicular to the circle everywhere. Thus
The vector field H
Z
C
~ · d~r = 0.
H
~ and H
~ because their domains do not
(c) Green’s Theorem does not apply to the computation of the line integrals for G
~ = y~i + x~j .
include the origin, which is in the interior, R, of the circle. Green’s Theorem does apply to F
Z
C
~ · d~r =
F
Z
R
∂F2
∂F1
−
∂x
∂y
dx dy =
Z
0 dx dy = 0.
34. (a)
I Green’s Theorem can be used. The curve is closed and the vector field is smooth throughout the interior of the
region enclosed
II Green’s Theorem cannot be used. The vector field is not defined at the origin which is inside the curve.
III Green’s Theorem cannot be used. The curve is not closed.
(b) For the integral in [I], let R be the region enclosed by C. See Figure 18.33. Green’s Theorem gives
Z
(x2 + y 2 ) dx + (x2 + y 2 ) dy =
Z
R
C
=
Z
=
Z
=
Z
∂F1
∂F2
−
∂x
∂y
(2x − 2y) dA =
R
1
x
(2xy − y 2 )
0
0
3
Z
1
0
Z
Z
4
(x − 2x + x ) dx =
∂ 2
∂ 2
(x + y 2 ) −
(x + y 2 )
∂x
∂y
x
x2
1
0
x2
Z
R
dx =
1
2
dA =
(2x − 2y) dy dx
(2x2 − x2 − (2x3 − x4 )) dx
2
x5
x3
− x4 +
3
4
5
1
=
0
1
.
30
dA
18.4 SOLUTIONS
1697
y
1
y=x
y = x2
x
1
Figure 18.33
35. Green’s theorem says that for a closed curve C oriented counterclockwise, bounding region R,
Z
C
Z
F~ · d~r =
R
∂F1
∂F2
−
∂x
∂y
dA =
Z
R
(x2 + y 2 − 1) dA.
If R is a region contained strictly inside the unit circle, then x2 + y 2 < 1 for any point (x, y) in R, so x2 + y 2 − 1 < 0,
which gives
Z
Z
R
(x2 + y 2 − 1) dA < 0, which implies that
C
~ · d~r < 0.
F
Now, let C be the curve C1 − C2 . Since
Z
C
~ · d~r =
F
Z
C1 −C2
~ · d~r =
F
Z
C1
~ · d~r −
F
Z
C2
~ · d~r = L1 − L2 < 0,
F
we have L1 < L2 . Similarly, if we let C = C2 − C3 , then
Z
C
F~ · d~r =
Z
C2
~ · d~r −
F
Z
C3
F~ · d~r = L2 − L3 < 0,
which gives L2 < L3 . Thus
L1 < L2 < L3 .
36. (a) Writing R1 for the interior of the circle, Green’s Theorem gives
Z
C1
~ · d~r =
F
Z
R1
∂F2
∂F1
−
∂x
∂y
dA =
Z
S
3 dA = 3 · Area of disk = 3 · π12 = 3π.
(b) Writing R2 as the interior of the rectangle, Green’s Theorem gives
Z
C2
~ · d~r =
F
Z
R2
3 dA = 3 · Area of rectangle = 3 · 3 · 2 = 18.
(c) In parts (a) and (b), we see that the line integral is three times the area enclosed in the curve. Since C3 encloses a disk
of radius 7 and area π · 72 = 153.9, and C4 encloses a disk of radius 8 and area π · 82 = 201.1, and C5 encloses a
square of side 14 and area 142 = 196, we have
Z
C3
F~ · d~r <
Z
C5
F~ · d~r <
Z
C4
~ · d~r .
F
1698
Chapter Eighteen /SOLUTIONS
37. (a) We use Green’s Theorem. Let R be the region enclosed by the circle C. Then
Z
C
~ · d~r =
F
=
Z
R
Z
R
∂F2
∂F1
−
∂x
∂y
Z
dA =
R
(12 − (3x2 + 3y 2 ) dA =
Z
R
∂
∂ y2
(e + 12x) −
(3x2 y + y 3 + ex )
∂x
∂y
dA
(12 − 3(x2 + y 2 )) dA.
Converting to polar coordinates, we have
Z
C
F~ · d~r =
Z
0
2π
1
Z
0
(12 − 3r 2 )r dr dθ = 2π 6r 2 −
2
3 4
r
4
1
0
= 2π 6 −
3
4
=
21π
.
2
2
(b) The integrand of the integral over the disk R is 12 − 3(x + y ). Since the integrand is positive for x2 + y 2 < 4 and
negative for x2 + y 2 > 4, the integrand is positive inside the circle of radius 2 and negative outside that circle. Thus,
the integral over R increases with a until a = 2 and then decreases. The maximum value of the line integral occurs
when a = 2.
38. (a) Taking partial derivatives using the product rule, we have
∂
∂F2
=
∂x
∂x
x
x2 + y 2
=
∂
∂F1
=
∂y
∂y
−y
x2 + y 2
=
x2
x · 2x
y 2 − x2
1
− 2
= 2
.
2
2
2
+y
(x + y )
(x + y 2 )2
Similarly,
Thus,
y · 2y
y 2 − x2
−1
+
=
.
x2 + y 2
(x2 + y 2 )2
(x2 + y 2 )2
~ = ∂F2 − ∂F1 = 0.
curl of F
∂x
∂y
(b) Let C be the clock-wise oriented closed path consisting of four pieces
• C1
• BD, the straight line path from B to D
• −C2
• CA, the straight line path from C to A.
By Green’s Theorem and part (a),
Z
C
Hence
Z
C1
~ · d~r +
F
Z
BD
F~ · d~r = 0.
~ · d~r −
F
Z
C2
~ · d~r +
F
Z
CA
~ · d~r = 0.
F
~ is perpendicular to each of the radial paths BD and CA. To see
The key observation is that the vector field F
this, consider the radial vector field ~r = x~i + y~j which is tangent to the paths BD and CA. Since
~ =0
~r · F
the vector fields RF~ and ~r are orthogonal.
R
~ · d~r = 0 and it follows that
~ · d~r =
F
Therefore, BD F
CA
Z
C1
~ · d~r =
F
Z
C2
~ · d~r .
F
~ is tangent to the circle and ||F
~ || = 1.
(c) Compute the integral over C2 , which is on the unit circle. On the unit circle, F
Thus,
Z
~ || · Length of curve = 1 · θ = θ.
~ · d~r = ||F
F
C
18.4 SOLUTIONS
1699
39. (a) We can show
~ı
~ =
curl E
~k
~
∂
∂
∂
∂x
∂y
∂z
qy
qz
qx
||~
r ||3 ||~
r ||3 ||~
r ||3
= ~0 .
~:
Let’s check, for instance, the ~ı component of curl E
qz
qy
∂
(−3/2)2qyz − (−3/2)2qzy
∂
−
=
= 0.
∂y (x2 + y 2 + z 2 )3/2
∂z (x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )5/2
~ is a gradient vector field, as curl E
~ = 0 and E
~ is defined everywhere in 3-space except at the
The vector field E
origin. This domain satisfies the criteria for the curl test in 3-space. Every closed curve in 3-space which does not
pass through 0 bounds a surface not containing the origin.
~ , since
(b) The function ϕ(~r ) = q/||~r || is a potential for E
∂
∂ϕ
= q (x2 + y 2 + z 2 )−1/2 = −qx(x2 + y 2 + z 2 )−3/2 = −E1
∂x
∂x
~ = − grad ϕ.
and similarly for ∂ϕ/∂y and ∂ϕ/∂z; hence E
Strengthen Your Understanding
~ . It
~ is path-independent, we must know that
~ · d~r = 0 for every closed path in the domain of F
40. To conclude that F
F
C
is not sufficient to check the equality for a single specific closed path.
R
41. Since C is not a closed curve, Green’s Theorem does not apply. Calculating the line integral requires more information
~.
about F
~ = xy~i + Q(x, y)~j is a gradient field, then by the curl test we see that
42. If F
∂Q
∂
(xy) =
.
∂y
∂x
Therefore, ∂Q/∂x = x. If Q(x, y) = x2 /2, then
x2
F~ = xy~i + ~j = grad
2
x2 y
2
.
~ is zero everywhere except at the origin, the vector field F~ is path independent in any region
43. Because the scalar curl of F
that does not contain or encircle the origin. Let C1 be the counterclockwise path on the unit circle from (1, 0) to (0, 1)
and let C2 be the clockwise path on the unit circle from (1, 0) to (0, 1). See Figure 18.34.
The paths C1 and C2 are not both contained in a single region of the plane that does not contain or encircle the origin,
~ over C1 and C2 are not equal. Since F
~ is tangent to the unit circle, points
so it is possible that the line integrals of F
~ || = 1, we have
counterclockwise, and ||F
Z
π
F~ · d~r = 1 · Length C1 =
2
C1
Z
~ · d~r = −1 · Length C2 = − 3π .
F
2
C2
and
(0, 1)
y
C1
x
(1, 0)
C2
Figure 18.34
1700
Chapter Eighteen /SOLUTIONS
~ = x~i + x2~j is not a gradient field because its scalar curl is not zero:
44. The vector field F
∂F1
∂ 2
∂
∂F2
−
=
x −
x = 2x 6= 0.
∂x
∂y
∂x
∂y
∂F2
∂F1
−
is 0 − 0 = 0, so the field is path-independent.
∂x
∂y
R
~ = grad f means that F~ is a potential field, hence F~ is path-independent. Thus
46. False. The fact that F
F~ · d~r = 0
C
since C is closed.
~ and G
~ are both path-independent, we know F
~ = grad f and G
~ = grad g for some scalar functions f
47. True. Since F
~
~
~
~
and g. Then grad(f + g) = grad f + grad g = F + G , so F + G is a gradient field, hence path-independent.
~ = x~j and G
~ = y~i . Then both of these are path-dependent (they each have
48. False. As a counterexample, consider F
~ +G
~ = y~i + x~j is zero everywhere, so F~ + G
~ is path-independent.
nonzero curl), but the curl of F
45. True. The value of
49. True. This vector field has components F1 = x, F2 = y, and F3 = z. Using the 3-space curl test gives zero for all of the
components of curlF~ , so the field is path-independent.
∂F1
∂F2
−
is 0 − 0 = 0, so the field is path-independent.
50. True. The value of
∂x
∂y
~ is path-independent, we know F
~ = grad f for some scalar function f . Then grad(kf ) = k grad f = kF~ ,
51. True. Since F
~
so kF is a gradient field, hence path-independent.
~ = 2x~i , which is path-independent, since it is the gradient of
52. False. As a counterexample, consider the vector field F
~ = 2xy~i . The curl of this vector field is
f (x, y) = x2 . Multiplying F~ by the function h(x, y) = y gives the field y F
~
−2x 6= 0, so y F is path-dependent.
Solutions for Chapter 18 Review
Exercises
1. On the top half of the circle, the angle between the vector field and the curve is less than 90◦ , so the line integral is positive.
On the bottom half of the circle, the angle is more than 90◦ , so the line integral is negative. However the magnitude of the
vector field is larger onR the top half of the curve, so the positive contribution to the line integral is larger than the negative.
~ · d~r is positive.
Thus the line integral C F
2. The angle between the vector field and the curve is more than 90◦ at all points on C, so the line integral is negative.
3. (a) The line integral around A is zero, because the curve is perpendicular to the field everywhere.
(b) The line integral along C1 or C3 is zero because the curves are everywhere perpendicular to the vector field. Along
C2 , the line integral is negative, since F~ points along the opposite direction to the curve. Along C4 , the line integral
~ points in the same direction as the curve.
is positive, since F
(c) The line integral around C is zero because C1 and C3 are perpendicular to the field and the contributions from C2
and C4 cancel out.
4. (a) The line integral around A is negative, because the vectors of the field are all pointing in the opposite direction to the
direction of the path.
(b) Along C1 , the line integral is positive, since F~ points in the same direction as the curve. Along C2 or C4 , the line
~ is perpendicular to the curve everywhere. Along C3 , the line integral is negative, since F
~
integral is zero, since F
points in the opposite direction to the curve.
(c) The line integral around C is negative because C3 is longer than C1 and the magnitude of the field is bigger along
C3 than C1 .
5. Scalar. The displacement along the line from (5, 2) to (1, 8) is given by −4~i + 6~j , so
Z
C
(3~i + 4~j ) · d~r = (3~i + 4~j ) · (−4~i + 6~j ) = 12.
SOLUTIONS to Review Problems for Chapter Eighteen
1701
6. Scalar. Since the path is along the y-axis, the ~i component does not contribute to the line integral. On the path, d~r = ~j dy,
so
Z 6
Z
Z 6
6
y2
(x~i + y~j ) · ~j dy =
= 16.
(x~i + x~j ) · d~r =
y dy =
2 2
2
C
2
~ = 6~i − 7~j , consider the function f
7. Since F
f (x, y) = 6x − 7y.
~
~
Then we see that grad f = 6i − 7j , so we use the Fundamental Theorem of Calculus for Line Integrals:
Z
C
~ · d~r =
F
Z
C
grad f · d~r
= f (4, 4) − f (2, −6) = (−4) − (54) = −58.
x2
y2
~ . Thus,
+ , then grad f = x~i + y~j = F
2
2
8. We know that if f (x, y) =
Z
C
~ is a gradient field, F
~ = grad
9. Since F
Z
C
~ · d~r = 0.
F
y2
x2
+
, we have
2
2
~ · d~r =
F
x2
y2
+
2
2
(0,10)
=
(0,0)
100
− 0 = 50.
2
10. We can parameterize the curve C by (t, t2 + 1), for 0 ≤ t ≤ 1. Then
Z
C
~ · d~r =
F
=
1
Z
~ (t, t2 + 1) · (~i + 2t~j )dt =
F
0
1
Z
Z
(−1 + 2t(t4 + 2t2 + t + 1))dt =
~ = grad
11. Since F
4t4
2t3
2t6
+
+
+ t2
6
4
3
x2
y2
z2
+
+
2
2
2
C
Z
1
(−1 + 2t5 + 4t3 + 2t2 + 2t)dt
0
−t +
Z
((−1)~i + (t4 + 2t2 + t + 1)~j ) · (~i + 2t~j )dt
0
0
=
1
1
=2
0
, the Fundamental Theorem of Line Integrals gives
F~ · d~r =
x2
y2
z2
+
+
2
2
2
(0,0,7)
=
(2,3,0)
72
−
2
22
32
+
2
2
= 18.
12. The path can be broken into three line segments: C1 , from (1, 0) to (−1, 0), and C2 , from (−1, 0) to (0, 1), and C3 , from
(0, 1) to (1, 0). (See Figure 18.35.)
y
(0, 1)
C2
C3
x
(−1, 0)
C1
Figure 18.35
(1, 0)
1702
Chapter Eighteen /SOLUTIONS
Along C1 we have y = 0 so the vector field xy~i + (x − y)~j is perpendicular to C1 ; Thus, the line integral along
C1 is 0.
C2 can be parameterized by (−1 + t, t), for 0 ≤ t ≤ 1 so the integral is
Z
C2
F~ · d~r =
=
=
Z
Z
Z
1
~ (−1 + t, t) · (~i + ~j ) dt
F
0
1
[t(−1 + t)~i + (−1)~j ] · (~i + ~j ) dt
0
1
(−t + t2 − 1) dt
0
1
= (−t2 /2 + t3 /3 − t)
0
= −1/2 + 1/3 − 1 − (0 + 0 + 0) = −7/6
C3 can be parameterized by (t, 1 − t), for 0 ≤ t ≤ 1 so the integral is
Z
C3
~ · d~r =
F
=
1
Z
~ (t, 1 − t) · (~i − ~j ) dt
F
0
1
Z
(t(1 − t)~i + (2t − 1)~j ) · (~i − ~j ) dt
0
=
1
Z
(−t2 − t + 1) dt
0
1
= (−t3 /3 − t2 /2 + t)
0
= −1/3 − 1/2 + 1 − (0 + 0 + 0) = 1/6
So the total line integral is
Z
C
~ · d~r =
F
Z
F~ · d~r +
C1
Z
C2
~ · d~r +
F
Z
~ · d~r = 0 + (− 7 ) + 1 = −1
F
6
6
C3
13. Using x as the parameter we have dy = 2xdx. Thus
Z
3x2 dx + 4ydy =
Z
5
3x2 dx + 4x2 (2xdx) =
1
C
Z
5
5
x2 + 8x3 dx = x3 + 2x4
1
= 1372.
1
14. Using x as the parameter we have dy = cos x dx. Thus
Z
ydx + xdy =
Z
π/2
Z
sin x dx + x(cos x dx) =
0
C
π/2
sin x + x cos x dx
0
π/2
= − cos x + cos x + x sin x
=
π
.
2
0
15. The domain is all 3-space. Since F1 = y,
curl y~i =
∂F2
∂F3
−
∂y
∂z
~i +
∂F
so F~ is not path-independent.
1
∂z
−
∂F3 ~
j +
∂x
∂F1
∂F2
−
∂x
∂y
~k = −~k 6= ~0 ,
16. The domain is all 3-space. Since F2 = y,
curl y~j =
so F~ is path-independent.
∂F3
∂F2
−
∂y
∂z
~i +
∂F1
∂F3 ~
j +
−
∂z
∂x
∂F2
∂F1
−
∂x
∂y
~k = ~0 ,
SOLUTIONS to Review Problems for Chapter Eighteen
1703
17. The domain is all 3-space. Since F3 = z,
curl z~k =
so F~ is path-independent.
∂F3
∂F2
−
∂y
∂z
~i +
∂F
∂F3 ~
−
j +
∂z
∂x
1
∂F2
∂F1
−
∂x
∂y
~k = ~0 ,
18. Since F2 = F3 = z,
curl (z~j + z~k ) =
so F~ is not path-independent.
∂F3
∂F2
−
∂y
∂z
~i +
∂F
1
∂z
∂F3 ~
j +
∂x
−
∂F2
∂F1
−
∂x
∂y
~k = −~i 6= ~0 ,
19. The domain is all 3-space. Since F1 = y, F2 = x,
curl y~i + x~j =
so F~ is path-independent
∂F3
∂F2
−
∂y
∂z
~i +
∂F1
∂F3 ~
j +
−
∂z
∂x
∂F2
∂F1
−
∂x
∂y
~k = ~0 ,
20. The domain is all 3-space. Since F1 = x + y,
curl (x + y)~i =
so F~ is not path-independent.
∂F2
∂F3
−
∂y
∂z
~i +
∂F1
∂F3 ~
−
j +
∂z
∂x
∂F2
∂F1
−
∂x
∂y
~k = −~k 6= ~0 ,
21. The domain is all 3-space. Since F1 = yz, F2 = zx, F3 = xy,
curl (yz~i + zx~j + xy~k ) =
∂F2
∂F3
−
∂y
∂z
~i +
∂F3 ~
∂F1
−
j +
∂z
∂x
= (x − x)~i + (y − y)~j + (z − z)~k = ~0 ,
so F~ is path-independent.
∂F1
∂F2
−
∂x
∂y
~k
22. Since the line is parallel to the y-axis, only the ~j -component contributes to the line integral. On C, we have x = 2, so
F~ = 10~i + 6~j and d~r = ~j dy. Thus,
Z
~ · d~r =
F
C
Z
8
3
6~j · ~j dy = 6 · 5 = 30.
23. Since the line is parallel to the x-axis, only the ~i -component contributes to the line integral. On C, we have d~r = ~i dx,
so
Z
C
F~ · d~r =
Z
2
12
12
5
5x~i · ~i dx = x2
2
= 350.
2
24. Parameterizing C by x(t) = t, y(t) = t2 , with 1 ≤ t ≤ 2, we have ~r ′ (t) = ~i + 2t~j . Thus,
Z
C
Z
~ · d~r =
F
2
1
(5t~i + 3t~j ) · (~i + 2t~j ) dt =
Z
2
(5t + 6t2 ) dt =
1
5t2
+ 2t3
2
2
=
43
.
2
1
25. Parameterizing C by x(t) = 3 cos t, y = 3 sin t, with 0 ≤ t ≤ π, we have ~r ′ (t) = −(3 sin t)~i + (3 cos t)~j . Thus,
Z
C
~ · d~r =
F
Z
=9
π
(15 cos t~i + 9 cos t~j ) · (−3 sin t~i + 3 cos t~j ) dt
0
Z
π
(−5 cos t sin t + 3 cos2 t) dt
0
=9
π
5
3
cos2 t + (cos t sin t + t)
2
2
27π
=
.
2
The integral
R
cos2 t was calculated using Formula IV-18.
0
1704
Chapter Eighteen /SOLUTIONS
26. We use Green’s Theorem:
Z
C
F~ · d~r =
Z
C
(5x~i + 3x~j ) · d~r =
=
Z
ZR
R
∂(3x)
∂(5x)
−
∂x
∂y
dx dy
1
63
3 · 3) =
.
2
2
3 dx dy = 3 · Area of region = 3(3 · 2 +
27. We can calculate this line integral either by calculating a separate line integral for each side, or by adding a line segment,
C1 , from (1, 4) to (1, 1) to form the closed curve C +C1 . Since we now have a closed curve, we can use Green’s Theorem:
Z
C+C1
F~ · d~r =
Z
C+C1
(5x~i + 3x~j ) · d~r =
=
Z
∂
∂
(3x) −
(5y)
∂x
∂y
R
Z
R
dx dy
3 dx dy = 3 · Area of region = 3 2 · 3 +
Since d~r = −~j dy on C1 , we have
Z
C1
Since
Z
C+C1
we have
~ · d~r =
F
~ · d~r =
F
Z
C
Z
1
3 · 4 = 36.
2
1
3 · 1~j · (−~j dy) = −3 · 3 = −9.
4
~ · d~r +
F
Z
C
Z
C1
~ · d~r =
F
Z
C
~ · d~r − 9 = 36
F
F~ · d~r = 45.
~ = grad(5x + 4y), we have
28. Since F
Z
F~ · d~r = 0
Z
F~ · d~r = (5x + 4y)
C1
C2
~ = grad
29. Since F
5 2
x
2
Z
C1
Z
C2
(7,4)
= (5 · 7 + 4 · 4) − (5 · 3 + 4 · 4) = 20.
(3,4)
+ 2y 2 , we have
~ · d~r = 0
F
~ · d~r =
F
5
2
x2 + 2y 2
(7,4)
=
(3,4)
5
2
· 72 + 2 · 42 −
5
2
· 32 + 2 · 42
= 100.
~ is not a gradient vector field, and
30. Since F
∂
∂
(4x) −
(5y) = 4 − 5 = −1,
∂x
∂y
we find the line integral around C1 by Green’s Theorem. The path C1 is oriented clockwise, so with R1 as the disk inside
C1 , we have
Z
Z
x =−
−1 dx dy = Area of disk = 9π.
F~ · d~
C1
R1
For C2 , we parameterize the curve
x = 5 − 2 cos t,
y = 4 + 2 sin t,
0 ≤ t ≤ π.
SOLUTIONS to Review Problems for Chapter Eighteen
1705
Then ~r ′ (t) = (2 sin t)~i + (2 cos t)~j , so
Z
C2
~ · d~r =
F
=
Z
C2
π
Z
(5(4 + 2 sin t)~i + 4(5 − 2 cos t)~j ) · (2 sin t~i + 2 cos t~j ) dt
0
=
Z
(5y~i + 4x~j ) · d~r
π
(40 cos t − 16 cos2 t + 40 sin t + 20 sin2 t) dt
0
= 88.283.
The integral has been computed numerically. The exact answer, obtained using the Table of Integrals, is 80 + 2π.
Problems
31. If a vector field is a gradient vector field, it has zero circulation around every closed curve. Vector fields (i) and (iii) do
not have this property. Therefore, (ii) and (iv) could represent gradient vector fields.
32. (a) The curves C1 and C3 give line integrals which we expect to be zero because at every point, the curve looks perpendicular to the vector field.
(b) The curve C4 gives a negative line integral because the path is traversed in the direction opposite to the vector field.
(c) The line integrals along C2 , C5 , C6 and C7 are all positive. The vector field is path-independent; it is the gradient
of a function f whose contours appear to be equally spaced circles centered at the origin; the value of f increases
going outward. By the Fundamental Theorem of Line Integrals, the value of a line integral is the difference between
the values of f at the two endpoints. The difference between the radii of the circles containing the endpoints of C2
and the difference between the radii of the circles containing the endpoints of C6 look about the same, so the line
integrals along C2 and C6 are approximately equal. Since C6 and C7 have the same endpoints, their line integrals are
also equal. The difference between the radii of the circles containing the endpoints of C2 is less than the difference
between the radii of the circles containing the endpoints of C5 , so the line integral along C2 is smaller than the line
integral along C5 . Thus
C2 = C6 = C7 < C5 .
33. The original integral is around the unit circle, oriented counterclockwise.
(a) This integral uses the same parameterization, but goes twice around the circle. The value of the integral is 24.
(b) This integral uses the same parameterization, but with the limits reversed and with −d~r instead of d~r . Thus, the
value of the integral does not change; it is 12.
(c) This integral uses a different parameterization of the circle x = sin t, y = cos t, which goes once around the circle
clockwise. The value of the integral is −12.
34. (a) The path C is a line segment, tangent to T~ = ~i + ~j at every point. Because the path C is on the line y = x we have
F~ (x, y) = 2~i + 2~j = 2T~ on C. Thus F~ is tangent to C at every point and points in the direction of the orientation
~ is 0.
of C. The angle between C and F
√
~ k = k2~i + 2~j k = 2 2.
(b) On C we have kF
√
~ is everywhere tangent to C in the direction of the orientation
(c) The path C has length 5 2.√Since the vector field F
and of constant magnitude 2 2 we have
Z
C
√
√
~ k · Length of C = 2 2 · 5 2 = 20.
F~ · d~r = kF
35. (a) For path (i), we have x(t) = t, y(t) = t2 , so x′ (t) = 1, y ′ (t) = 2t. Thus,
Z
C
~ · d~r =
F
=
Z
1
F~ (t, t2 ) · (~i + 2t~j ) dt
0
Z
1
[(t + t2 )~i + t~j ] · (~i + 2t~j ) dt
0
=
Z
0
1
(t + 3t2 ) dt
1706
Chapter Eighteen /SOLUTIONS
= (
t2
+ t3 )
2
1
0
1
3
= + 1 − (0 + 0) = .
2
2
For path (ii), we have x(t) = t2 , y(t) = t, so x′ (t) = 2t, y ′ (t) = 1. Thus,
Z
C
F~ · d~r =
=
1
Z
~ (t2 , t) · (2t~i + ~j ) dt
F
0
1
Z
[(t2 + t)~i + t2~j ] · (2t~i + ~j ) dt
0
=
Z
1
(2t3 + 3t2 ) dt
0
= (
t4
+ t3 )
2
1
0
3
1
= + 1 − (0 + 0) = .
2
2
For path (iii), we have x(t) = t, y(t) = tn , so x′ (t) = 1, y ′ (t) = ntn−1 . Thus,
Z
C
F~ · d~r =
=
=
=
=
Z
Z
Z
Z
1
~ (t, tn ) · (~i + ntn−1~j ) dt
F
0
1
[(t + tn )~i + t~j ] · (~i + ntn−1~j ) dt
0
1
(t + tn + ntn ) dt
0
1
(t + (n + 1)tn ) dt
0
t2
+ tn+1
2
1
0
1
3
= + 1 − (0 + 0) = .
2
2
~ = grad f . Each path goes from (0, 0) to (1, 1). Thus in each case
(b) If f (x, y) = xy + x2 /2, we have F
Z
3
F~ · d~r = f (1, 1) − f (0, 0) = .
2
C
36. The path C is the displacement ~v = (4 − 5)~i + (7 − 5)~j = −~i + 2~j . Since the vector field is constant, the line integral
is
Z
(2~i + 13~j ) · d~r = (2~i + 13~j ) · (−~i + 2~j ) = −2 + 26 = 24
C
37. This is a gradient vector field,
3
4x~i + 3y~j = grad 2x2 + y 2 .
2
Thus,
Z
C
(4x~i + 3y~j ) · d~r = (2x2 +
3 2
y )
2
(4,7)
(5,5)
= 2 · 42 +
3
3 2
· 7 − 2 · 52 − · 52 = 18.
2
2
38. This is not a gradient field, so we parameterize the line segment. The vector from (5, 5) to (4, 7) is ~v = −~i + 2~j , so the
parameterization is
x = 5 − t, y = 5 + 2t, 0 ≤ t ≤ 1,
SOLUTIONS to Review Problems for Chapter Eighteen
1707
and d~r = (−~i + 2~j ) dt. The integral is
Z
C
Z
=
Z
=
Z
=
((4x + 5y)~i + (2x + 3y)~j ) · d~r
1
(4(5 − t) + 5(5 + 2t))~i + (2(5 − t) + 3(5 + 2t)~j ) · (−~i + 2~j ) dt
0
1
−(20 − 4t + 25 + 10t) + 2(10 − 2t + 15 + 6t) dt
0
1
(5 + 2t)dt = 5t + t2
0
1
0
= 6.
39. Since the curve is closed, we use Green’s Theorem:
Z
C
((4x + 5y)~i + (2x + 3y)~j ) · d~r =
Each of the four points of the star has area
Thus,
Z
C
Z
R
1
2
∂
∂
(2x + 3y) −
(4x + 5y) dA =
∂x
∂y
Z
R
−3 dA = −3 · Area of star.
· 2 · 2 = 2; the square center has area 22 = 4, so area of star is 4 · 2 + 4 = 12.
((4x + 5y)~i + (2x + 3y)~j ) · d~r = −3 · 12 = −36.
40. (a) By Green’s Theorem, if R is the interior of C,
Z
C
~ · d~r =
F
Z
R
∂
∂
(5x) −
(2y) dA =
∂x
∂y
Z
R
3dA = 3 · Area of region = 3 2 ·
1
10(9 − 1) = 240.
2
~ = 2~i + 5x~j . Only the ~i component contributes to the line integral, so
(b) On C1 , we have y = 1, so F
Z
C1
~ · d~r = 2 · Length of C1 = 2 · 20 = 40.
F
(c) Since C = C1 + C2 , we have
Z
C2
F~ · d~r =
Z
C
~ · d~r −
F
Z
C1
F~ · d~r = 240 − 40 = 200.
~ = grad(x2 ey ) is a gradient vector field and C is a closed curve,
~ · d~r = 0.
41. (a) Since F
F
C
(b) Since
∂G2
∂G1
∂
∂
−
=
(x + y) −
(x − y) = 2,
∂x
∂y
∂x
∂y
by Green’s Theorem,
R
Z
C
~ · d~r =
G
Z
R
∂G2
∂G1
−
∂x
∂y
dA = 2 · Area of triangle =
2·3·8
= 24.
2
~ = grad x3 /3 + x3 y 4 , the Fundamental Theorem of Line Integrals gives
42. (a) Since F
3
~ · d~r = x + x3 y 4
F
3
C1
Z
(−2,0)
=−
(2,0)
8
+0−
3
8
16
+0 = − .
3
3
(b) Since a gradient field is path-independent, and the endpoints of C1 and C2 are the same:
Z
C2
F~ · d~r =
Z
16
F~ · d~r = − .
3
C1
1708
Chapter Eighteen /SOLUTIONS
~ is not a gradient vector field, so we parameterize C1 . Using x(t) = t, y(t) = 0, from t = 2 to
(c) The vector field G
t = −2 gives
Z
Z −2
Z −2
−2
64
t5
~ · d~r =
=− .
G
(t4 + 0)~i + (0~j ) · ~i dt =
t4 dt =
5
5
C1
2
2
2
(d) Parameterizing C2 by x(t) = 2 cos t, y(t) = 2 sin t for 0 ≤ t ≤ π gives
Z
C2
~ · d~r =
G
Z
π
(2 cos t)4 + (2 cos t)3 (2 sin t)2 ~i + (2 cos t)2 (2 sin t)3~j
0
= 32
Z
π
(− cos4 t sin t − 2 cos3 t sin3 t + 2 cos3 t sin3 t) dt
0
= −32
π
Z
cos4 t sin t dt = 32
0
cos5 t
5
π
0
=−
· (−2 sin t~i + 2 cos t~j ) dt
64
.
5
43. (a) The vector field is everywhere perpendicular to the radial line from the origin to (2, 3), so the line integral is 0.
(b) Since the path is parallel to the x-axis, only the ~i component of the vector field contributes to the line integral. The ~i
component is −3~i on this line, and the displacement along this line is −2~i , so
Line integral = (−3~i ) · (−2~i ) = 6.
~ || = (−y 2 ) + x2 =
(c) The circle of radius 5 has equation x2 + y 2 = 25. On this curve, ||F
F~ is everywhere tangent to the circle, and the path is 3/4 of the circle. Thus
p
√
25 = 5. In addition,
~ || · Length of curve = 5 · 3 · 2π(5) = 75 π.
Line integral = ||F
4
2
(d) Use Green’s Theorem. Writing C for the curve around the boundary of the triangle, we have
∂F1
∂F2
−
= 1 − (−1) = 2,
∂x
∂y
so
Z
C
F~ · d~r =
Z
Triangle
2 dA = 2 · Area of triangle = 2 · 7 = 14.
~ = (6x + y 2 )~i + 2xy~j = grad(3x2 + xy 2 ), the vector field F
~ is path independent, so
44. (a) Since F
Z
C1
~ · d~r = 0.
F
(b) Since C1 is closed, we use Green’s Theorem, so
Z
C1
~ · d~r =
G
Z
=2
Interior of C1
Z
C1
∂
∂
(x + y) −
(x − y)
∂x
∂y
dA = 2 · Area inside C1 = 2 ·
dA
1
· 2 · 2 = 4.
2
~ = grad(3x2 + xy 2 ), using the Fundamental Theorem of Line Integrals gives
(c) Since F
Z
C2
~ · d~r = (3x2 + xy 2 )
F
(0,−2)
(2,0)
= 0 − 3 · 22 = −12.
(d) Parameterizing the circle by
x = 2 cos t
y = 2 sin t
0≤t≤
gives
x′ = −2 sin t
y ′ = 2 cos t,
3π
,
2
SOLUTIONS to Review Problems for Chapter Eighteen
1709
so the integral is
Z
C2
~ · d~r =
G
=
3π/2
Z
(2 cos t − 2 sin t)~i + (2 cos t + 2 sin t)~j
0
Z
3π/2
· (−2 sin t~i + 2 cos t~j ) dt
4(− cos t sin t + sin2 t + cos2 t + sin t cos t) dt
0
=4
Z
3π/2
0
~ = x~i + y~j = grad
45. (a) Since F
Theorem of Line Integrals,
3π
= 6π.
2
dt = 4 ·
x2 + y 2
2
~ is a gradient vector field. Thus, by the Fundamental
, we know that F
x2 + y 2
F~ · d~r =
2
OA
Z
(3,0)
=
(0,0)
9
.
2
(b) We know that F~ is path independent. If C is the closed curve consisting of the line in part (a) followed by the
two-part curve in part (b), then
Z
F~ · d~r = 0.
C
Thus, if ABO is the two-part curve of part (b) and OA is the line in part (a),
Z
ABO
~ · d~r = −
F
Z
~ · d~r = − 9 .
F
2
OA
46. (a) C1 is a line along the vertical axis; C2 is a half circle from the positive y to the negative y-axis. See Figure 18.36.
y
1
C2
C1
x
−1
Figure 18.36
(b) Either use Green’s Theorem or calculate directly. Using Green’s Theorem, with R as the region inside C, we get
Z
C1 +C2
~ · d~r =
F
=
Z
R
Z
R
∂
∂
(y) −
(x + 3y)
∂x
∂y
dA
−3 dA = −3(Area of region) = −3
π · 12
3π
=− .
2
2
~ (x, y) = y~j and the path C is the line from (0, −1) to (0, 1).
47. See Figure 18.37. The example chosen is the vector field F
Since the vectors are symmetric about the x-axis, the dot products F~ · ∆~r cancel out along C to give 0 for the line
integral. Many other answers are possible.
1710
Chapter Eighteen /SOLUTIONS
y
(0, 1)
C
x
(0, −1)
Figure 18.37
48. (a) See Figure 18.38. Notice that C = C1 + C2 + C3 is a closed curve.
(b) See Figure 18.39.
~ in the direction of C1 is −~i ,
(c) (i) Since the component of F
Z
C1
F~ · d~r = −Length of C1 = −1.
~ || =
(ii) Since F~ is parallel to C2 and in the same direction, and ||F
Z
C2
~ · d~r =
F
√
√
2 · Length of C2 =
2,
√
2·
√
2 = 2.
~ in the direction of C3 is ~j , and a vector in the direction of C3 is −~j ,
(iii) Since the component of F
Z
C3
F~ · d~r = −Length of C3 = −1.
(iv) Since F~ is constant, it is a gradient field and C is closed,
Z
y
C2
C1
Figure 18.38
49. (a) Since
~ · d~r = 0.
F
y
(0, 1)
C3
C
(1, 0)
x
x
Figure 18.39
∂ 5
∂
(y +x)− (x3 −y) = 1+1 = 2, any closed curve oriented counterclockwise will do. See Figure 18.40.
∂x
∂y
1711
SOLUTIONS to Review Problems for Chapter Eighteen
y
y
x
x
Figure 18.40
Figure 18.41
∂ 3
∂ 5
(y − xy) −
(x ) = −y, any closed curve in the lower half-plane oriented counterclockwise or any
∂x
∂y
closed curve in the upper half-plane oriented clockwise will do. See Figure 18.41. Other answers are possible.
(b) Since
50. A contour of f is a set on which f does not change, so the total change of f from P to Q, f (P ) − f (Q), is zero. If C is
a part of a contour of f , we know that grad f is perpendicular to C. This means that the line integral of grad f along C,
which also computes the total change in f between its endpoints, must be zero, since the dot products in its definition are
all zero.
51. (a) The vector field ∇f is perpendicular to the level curves, in direction of increasing f . The length of ∇f is the rate of
change of f in that direction. See Figure 18.42
y
23.3
23
1
P
22.7
Q
x
1
2
Figure 18.42
(b) Longer.
(c) Using the Fundamental Theorem of Calculus for Line Integrals, we have
Z
C
52. (a)
∇f · d~r = f (Q) − f (P ) = 22.7 − 23 = −0.3.
(i) The curve C is the line given by ~r = x~i + y~j , which we can parameterize by x = t, y = −t + 1 for 0 ≤ t ≤ 1.
Then ~r ′ (t) = ~i − ~j so
Z
C
~v · d~r =
Z
0
1
((1 − t)~i + 2t~j ) · (~i − ~j )dt =
1
Z
0
1
(1 − 3t)dt = − .
2
(ii) The curve C is the circle given by ~r = x~i + y~j where x = sin t, y = cos t for 0 ≤ t ≤
cos t~i − sin t~j and
Z
C
~v · d~r =
Z
0
π/2
(cos t~i + 2 sin t~j ) · (cos t~i − sin t~j )dt =
Z
0
π/2
π
.
2
Thus ~r ′ (t) =
π
(cos2 t − 2 sin2 t)dt = − .
4
(b) Since the value of the integral along two paths gives different results, ~v is not path independent.
1712
Chapter Eighteen /SOLUTIONS
53. (a) The path C is a line segment, tangent to T~ = ~i + ~j at every point. Because the path C is on the line y = x we have
~ = 0, which shows that C and F
~ are perpendicular at every point of C. The
F~ (x, y) = 2~i − 2~j on C. Hence T~ · F
angle
between
them
in
π/2.
R
~ and C are perpendicular at every point of C.
~ · d~r = 0 because F
(b) C F
54. By Green’s Theorem, if Ra is the interior of Ca
Z
Ca
The quantity
sign where
F~ · d~r =
Z
Ra
∂F2
∂F1
−
∂x
∂y
dA.
∂F1
∂F2
−
is positive for points (x, y) near the origin and negative farther away. This quantity changes
∂x
∂y
∂F1
∂F2
−
= 3(x2 + y 2 ) − (x2 + y 2 )3/2 = 0
∂x
∂y
3(x2 + y 2 ) = (x2 + y 2 )3/2
(x2 + y 2 )1/2 = 3.
∂F1
∂F2
−
is positive within C3 , the circle of radius 3, and negative outside. The maximum value of the line
∂x
∂y
integral occurs when a = 3. Converting to polars,
Thus
Z
C3
~ · d~r =
F
=
Z
3(x2 + y 2 ) − (x2 + y 2 )3/2 dA
R3
Z
2π
0
= 2π
Z
0
3
(3r 2 − r 3 )r dr dθ
r5
3r 4
−
4
5
3
=
0
35 π
.
10
~ in the xy-plane. If C is a circle
55. We’ll assume that the rod is positioned along the z-axis, and look at the magnetic field B
of radius r in the plane, centered at the origin, then we are told that the magnetic field is tangent to the circle and has
~ k. We divide the curve C into little pieces Ci and then we sum B
~ · ∆~r computed on each piece
constant magnitude kB
~
Ci . But ∆~r points nearly in the same direction as B , that is, tangent to C, and has magnitude nearly equal to the length
~ k× length of Ci . When all of these dot products are summed and the limit
of Ci . So the dot product is nearly equal to kB
is taken as k∆~r k → 0, we get
Z
C
~ k × length of C = kB
~ k × 2πr
~ · d~r = kB
B
Now Ampère’s Law also tells us that
Z
C
~ · d~r = kI
B
~ k gives kI = 2πrkB
~ k, so
Setting these expressions for the line integral equal to each other and solving for kB
~ k=
kB
56. (a) An example of a central field is in Figure 18.43.
kI
.
2πr
1713
SOLUTIONS to Review Problems for Chapter Eighteen
y
y
x
x
Figure 18.43
Figure 18.44
(b) The vectors of F~ are radial and the contours of f must be perpendicular to the vectors. Therefore, every contour
must be a circle centered at the origin. Sketching some contours results in a diagram like that in Figure 18.44.
(c) No, not every gradient field is a central field, because there are gradient fields which are not perpendicular to circles.
~ = ~j
An example is the gradient of f (x, y) = y, where grad f = ~j , so the gradient is parallel to the y axis. Thus, F
is an example of a gradient field which is not a central field.
~ is tangent to the circle. Thus the
(d) When a particle moves around a circle centered at O, no work is done, because F
~ is central, the work done on any
only work done in moving from P to Q is in moving between the circles. Since F
~ is parallel to this path and
radial line between C1 and C3 , for example, depends on only the radii of C1 and C3 (F
its magnitude is a function of the distance to the center of the circle only). For that reason, on a path which goes from
C1 to C2 and then from C2 to C3 , the same amount of work will be done as on a path direct from C1 to C3 .
(e) Pick any two points P and Q. Any path between them can be well-approximated by a path which is partly radial and
partly around a circle centered at O. By the answer to part d), the work along any such path depends only on the radii
~ must
of the circles on which P and Q sit, not on the path. Thus, the work done is independent of the path. Hence, F
be path-independent and therefore a gradient field.
57. (a) Since −y~i + x~j is a counterclockwise rotation, both ω and K must be positive. In order to find the values of ω and
K, we
pmust look at the velocity field where we know the magnitude. At a radius of 100 m from the center, we know
that x2 + y 2 = 100, and that k~v k = 3 · 105 . Thus, using ~v = ω(−y~i + x~j ) we get
k~v k = ω
so
p
(−y)2 + x2 = 100ω = 3 · 105 meters/hr,
ω = 3000 rad/hr.
2
2 −1
Using ~v = K(x + y )
(−y~i + x~j ) gives
k~v k = |K|(x2 + y 2 )−1
(b)
so K = 3 · 107 meters2 ·rad/hr.
Figure 18.45
p
(−y)2 + x2 =
K100
= 3 · 105 meters/hr,
1002
Figure 18.46
1714
Chapter Eighteen /SOLUTIONS
The vector field in Figure 18.45 shows the velocity vectors inside the tornado, (i.e. r < 100 meters). The vector
field in Figure 18.46 shows the velocity vectors as seen from a great distance (i.e. r >> 100 meters) with the tornado
at the origin.
(c) Let C be the circle of radius r around the origin. If r < 100 meters, the velocity vectors at distance r from the origin
have magnitude ωr. Since they are tangent, and point counterclockwise, the circulation is
Z
C
~v · d~r = k~v k · Length of C = 2ωπr 2 .
If r ≥ 100 meters, the vectors at distance r from the origin have magnitude K/r and are again tangent to the circle.
The circulation here is
Z
K
~v · d~r = kvk · Length of C = ( )2πr = 2Kπ.
r
C
58. The free vortex appears to starts at about r = 200 meters (that’s where the graph changes its behavior) and the tangential
5
velocity at this point is about 200 km/hr
p = 2 · 10 meters/hr.
Since ~v = ω(−y~i + x~j ) for x2 + y 2 ≤ 200, at r = 200 we have
k~v k = ω
so
p
(−y)2 + x2 = ω(200) = 2 · 105 meters/hr,
ω = 103 rad/hr.
2
2 −1
Since ~v = K(x + y )
(−y~i + x~j ) for
p
x2 + y 2 ≥ 200, at r = 200 we have
k~v k = K(2002 )−1 (200) =
K
= 2 · 105 meters/hr
200
so
K = 4 · 107 m2 ·rad/hr.
CAS Challenge Problems
59. (a) We parameterize Ca by ~r (t) = a cos t~i + a sin t~j . Then, using a CAS, we find
Z
Ca
F~ (~r (t)) · ~r ′ (t) dt =
Z
2π
0
a3 cos t3
a cos t 2a cos t −
+ a3 cos tsin t2
3
−a sin t −(a sin t) +
π
= − (−6a2 + a4 )
2
2a3 sin t3
3
dt
√
The derivative of the expression on the right with respect to a is −(2π)(−3a
+ a3 ), which is zero at a = 0, ± 3.
√
Checking at a = 0 and as a → ∞, we find the maximum is at a = 3.
(b) We have
∂F2
∂F1
−
= (2 − x2 + y 2 ) − (−1 + 2y 2 ) = 3 − x2 − y 2 .
∂x
∂y
So, by Green’s theorem,
Z Z
Z
Ca
F~ · d~r =
Da
(3 − x2 − y 2 ) dA,
where Da is the disk of √
radius a centered at the origin. The integrand is positive for x2 + y 2 < 3, so it√is positive
inside the disk of radius 3 and negative outside it. Thus the integral has its maximum value when a = 3.
60. We parameterize the line from (0, 0) to (x, y) by ~r (t) = t(x~i + y~j ). Using a CAS to compute the integral, we get
(a)
f (x, y) =
Z
0
1
~ (~r (t)) · ~r ′ (t) dt =
F
Z
0
1
2axyt dt = axy + Constant
PROJECTS FOR CHAPTER EIGHTEEN
1715
(b)
f (x, y) =
1
Z
F~ (~r (t)) · ~r ′ (t)dt
0
=
Z
1
(abebt
2
xy
txy + (c + abebt
2
xy
tx)y) dt = aebxy + cy + Constant
0
61. We have
Z
C1
3
Z
~ · d~r =
F
~ (~r (t)) · ~r ′ (t)dt
F
0
Z
=
3
(2 2at + bt2 + 2t 2ct + dt2 ) dt = 18a + 18b + 36c + (81d/2)
0
and
Z
C2
~ · d~r =
F
=
Z
3
F~ (~r (t)) · ~r ′ (t)dt
0
Z
0
3
−2 2a (3 − t) + b(3 − t)2 − 2 2c (3 − t) + d(3 − t)2 (3 − t) dt
= −18a − 18b − 36c − (81d/2)
The second integral is the negative of the first. This is because C2 is the same curve as C1 but traveling in the opposite
direction.
PROJECTS FOR CHAPTER EIGHTEEN
1. (a) Since k~v (t)k2 = ~v (t) · ~v (t) and since ~v (t) = ~r ′ (t) = x′ (t)~i + y ′ (t)~j + z ′ (t)~k , we have
1 d
1 d
k~v (t)k2 =
(~v (t) · ~v (t))
2 dt
2 dt
1 d ′ 2
(x (t) + y ′ (t)2 + z ′ (t)2 )
=
2 dt
1
= (2x′ (t)x′′ (t) + 2y ′ (t)y ′′ (t) + 2z ′ (t)z ′′ (t))
2
= x′ (t)x′′ (t) + y ′ (t)y ′′ (t) + z ′ (t)z ′′ (t)
= (x′ (t)~i + y ′ (t)~j + z ′ (t)~k ) · (x′′ (t)~i + y ′′ (t)~j + z ′′ (t)~k )
= ~v (t) · ~a (t) (Since ~a (t) = x′′ (t)~i + y′′ (t)~j + z′′ (t)~k .)
= ~a (t) · ~v (t).
(b) (i) We use F~ = m~a and the parameterization of C given by r(t) for t0 ≤ t ≤ t1 . In addition, we need
d
k~v (t)k2 = ~a · ~v :
the fact that 12 dt
Z t1
Z
Z
~
m~a · d~r =
m~a · ~r ′ dt
F · d~r =
C
C
t0
t1
=
=
Z
t0
Z t1
t0
m (~a · ~v ) dt
m
2
d
k~v (t)k2 dt
dt
t
1
m
= k~v (t)k2
2
t0
m
m
2
= k~v (t1 )k − k~v (t0 )k2
2
2
= Kinetic energy at Q − Kinetic energy at P.
1716
Chapter Eighteen /SOLUTIONS
(ii) Since F~ = −∇f we use the Fundamental Theorem of Line Integrals:
Z
Z
~
F · d~r =
−∇f · d~r = −(f (~r (t1 )) − f (~r (t0 )))
C
C
= −(Potential energy at Q − Potential energy at P )
= Potential energy at P − Potential energy at Q.
(iii) In parts (a) and (b) we derived two expressions for the work done by F~ as the particle moves from P
to Q. These two expressions must be equal, so
Kinetic energy at Q − Kinetic energy at P = Potential energy at P − Potential energy at Q.
Rewriting this equation we have,
(Potential energy + Kinetic energy) at P = (Potential energy + Kinetic energy) at Q.
This shows that the total energy is the same at P as at Q. Since P and Q are arbitrary points in space,
the total energy of a particle moving in a force vector field F~ = −∇f is a constant.
2. (a) We have
m
~ = (x − a)~i + y~j .
Since m
~ has magnitude L we have
(x − a)2 + y 2 = L2
and so
x−a=
where we have used the fact that a < x. Thus
p
L2 − y 2
~k × m
~ = −y~i + (x − a)~j .
Using once more the fact that m
~ has magnitude L, we see that the unit vector F~ in the direction of ~k × m
~
is
−y~
1p 2
1
L − y 2~j .
~ )=
i +
F~ = (~k × m
L
L
L
(b) We have
∂F1
∂F2
curl F~ =
−
∂x
∂y
p
2
1 ∂(−y)
1
1 ∂ L − y2
−
= .
=
L
∂x
L ∂y
L
(c) By Green’s Theorem and part (b), we have
Z
Z
F~ · d~r =
curl F~ dA
C
ZR
1
1
=
dA = · (Area of R).
L
R L
Since F~ is by definition in part (a) a vector field of unit vectors in the direction of the wheel vector along
C, we have
Z
(Area of R) = L
F~ · d~r = L · (Total roll of planimeter wheel).
C
PROJECTS FOR CHAPTER EIGHTEEN
1717
3. (a) We take the surface to be a disk of radius r, parallel to the xy-plane and centered on the z-axis. The
~ , is tangent to the circle and has
boundary of the disk is the circle C. We know that the magnetic field, B
~
constant magnitude, kB k, along each circle. Thus, for any such circle,
Z
~ k · Length of C = 2πrkB
~ k.
~ · d~r = kB
B
C
If r ≥ r0 (where r0 is the radius of the wire) then the current through the surface is I. Therefore
Z
~
~ · d~r = kI,
2πrkB k =
B
C
so
kI
.
2πr
If r < r0 , then the current flowing through the surface is not I, but the amount of current passing through
a disk of radius r. Such a disk has an area which is (πr2 )/(πr02 ) of the cross-sectional area of the wire. So
the current to use in Ampère’s Law is (πr2 )/(πr02 ) I. Thus,
Z
2
~k=
~ · d~r = k πr I.
2πrkB
B
πr02
C
~k=
kB
~ k gives
Solving for kB
~k=
kB
kIr
.
2πr02
(b) We again use Ampère’s Law on a disk of radius r, lying perpendicular to, and centered on, the z-axis (See
Figure 18.47). If the boundary, C, of this disk lies inside the torus, then the wire goes through the disk N
times, and so the net current through the disk is N I. Thus,
Z
~ k=
~ · d~r = kN I,
2πrkB
B
C
which gives
kN I
.
2πr
On the other hand, if the boundary, C, lies outside the torus, then the net current through the disk is 0. (The
wire goes into the disk N times and out of the disk N times, and so the currents cancel.) Hence we have
Z
~ · d~r = 0.
~
B
2πrkB k =
~k=
kB
C
~ k = 0.
So kB
Wires forming
solenoid
z
(0, β, 0)
❘
✠
y
❘
α
✒ x
(β, 0, 0)
Figure 18.47
■
Circle C of radius r;
boundary of horizontal
disk
1718
Chapter Eighteen /SOLUTIONS
4. (a) (i) For each plane F~ = −c||~v ||~v and since ~r ′ (t) = ~v (t), the work done is given by
Z π
Z
Z π
~
F · d~r =
(−c||~v (t)||~v (t)) · ~v (t) dt = −c
||~v (t)||3 dt.
0
C
0
The velocity of the first airplane is
~v 1 (t) = (−2 sin t)~ı + (2 cos t)~ + 3~k .
Since
||~v 1 (t)|| =
p
√
(−2 cos t)2 + (2 sin t)2 + 32 = 13,
the work done by the force F~ 1 is
Z
Z
W1 =
F~ 1 · d~r 1 = −c
For the second plane
and the work done by the force F~
F~
C
r 2 = −c
2 · d~
Z
Z
π
133/2 dt = −133/2 πc.
0
4
~v 2 (t) = − ~i + 3~k ,
π
r
16
+ 9,
||~v 2 (t)|| =
π2
so
Z
||~v 1 (t)||3 dt = −c
0
C
W2 =
π
2
is
π
||~v 2 (t)||3 dt = −c
0
Z
0
π
16
+9
π2
3/2
dt = −
16
+9
π2
3/2
πc.
(ii) At time t = 0, both airplanes are at the point (2, 0, 0). At t = π, both airplanes are at the point
(−2, 0, 3π); therefore, their trajectories have the same endpoints. The fact that W1 6= W2 shows that
the drag force is not conservative, since the work done by a conservative force is the same on all paths
with the same endpoints.
(b) (i) The first plane’s path is given by
~r 1 (t) = 2 cos t~i + 2 sin t~j + 3t~k ,
so
||~r 1 (t)|| =
and
p
p
(2 cos t)2 + (2 sin t)2 + (3t)2 = 4 + 9t2 ,
~v 1 (t) = ~r ′1 (t) = (−2 sin t)~i + (2 cos t)~j + 3~k .
Thus, the gravitational force acting on the first airplane at time t is
F~ 1 (t) = −
GM m
((2 cos t)~ı + (2 sin t)~ + 3t~k ),
(4 + 9t2 )3/2
so the total work done by the gravitational force on the first plane is
Z π
Z π
GM m
−
(2 cos t~i + 2 sin t~j + 3~k ) · (−2 sin t~i + 2 cos t~j + 3t~k ) dt
W1 =
F~ 1 (t) · ~v 1 (t) dt =
(4 + 9t2 )3/2
0
0
Z π
GM m
=
−
· 9t dt.
(4 + 9t2 )3/2
0
Using the substitution x = 4 + 9t2 , we get
π
2 −1/2
W1 = GM m(4 + 9t )
= GM m
0
√
1
1
−√
4 + 9π 2
2
.
PROJECTS FOR CHAPTER EIGHTEEN
1719
Similarly, for the second plane, we have
4 ~
~r 2 (t) = 2 − t i + 3t~k ,
π
so
s
s
2
2
4
4
2 − t + (3t)2 =
2 − t + 9t2 ,
||~r 2 (t)|| =
π
π
and
4
~v 2 (t) = ~r ′2 (t) = − ~i + 3~k .
π
Thus, the gravitational force on the second plane is
4 ~
GM m
~k ,
2
−
t
i
+
3t
F~ 2 (t) = −
π
((2 − 4t/π)2 + 9t2 )3/2
so the work done by the gravitational force on the second plane is
Z π
F~ 2 (t) · ~v 2 (t) dt
W2 =
0
Z π
4 ~
4~
GM m
~
~
2 − t i + 3tk · − i + 3k dt
=
−
π
π
((2 − 4t/π)2 + 9t2 )3/2
0
Z π
GM m
16
8
=
−
− + 2 t + 9t dt.
2 + 9t2 )3/2
π
π
((2
−
4t/π)
0
Using the substitution x = (2 − 4t/π)2 + 9t2 , we get
π
2
2 −1/2
W2 = GM m((2 − (4t/π)) + 9t )
= GM m
0
1
1
√
−√
4 + 9π 2
2
.
(ii) The fact that W1 = W2 suggests that F~ might be conservative, since a conservative force does the
same work along two paths with the same endpoints. That this is indeed the case can be seen from the
fact that F~ is a gradient vector field with potential function
ϕ(~r ) = −
(so that F~ = − grad ϕ).
GM m
,
||~r ||
19.1 SOLUTIONS
1721
CHAPTER NINETEEN
Solutions for Section 19.1
Exercises
~ is
1. The surface is a right triangle in the yz-plane with sides 2 and 3, so its area is (1/2) · 2 · 3 = 3. The area vector A
~ is a multiple of ~i with magnitude 3. There are two such vectors, 3~i and −3~i .
perpendicular to the plane yz-plane, so A
~ points in the negative x direction, we have A
~ = −3~i .
Since A
2
~ has magnitude 25π and is perpendicular to the xy-plane. Since it points
2. The disc has area πR = 25π. The area vector A
~ = 25π~k .
upward, the area vector is A
~ is perpendicular to the plane y = 10, so A
~ is a multiple of
3. The surface is a 5 × 3 rectangle of area 15. The area vector A
~ points away from the xz-plane,
~j with magnitude 15. There are two such vectors, 15~j and −15~j . Since y > 0 and A
~ = 15~j .
we have A
~ is perpendicular to the plane y = −10, so A
~ is a multiple
4. The surface is a 5 × 3 rectangle of area 15. The area vector A
~ points away from the xz-plane,
of ~j with magnitude 15. There are two such vectors, 15~j and −15~j . Since y < 0 and A
~ = −15~j .
we have A
5. We need a flat surface with area 150 that is perpendicular to the vector ~j . There are many examples. For instance, we can
take any 3 × 50 rectangle in the xz-plane oriented in the positive y direction.
~ = ~j dA, and on the disk y = 6, so
6. Scalar. Only the ~j -component of the vector field contributes to the flux and dA
Z
S
~ = 6 · Area of disk = 6 · π32 = 54π.
(x~i + 6~j ) · dA
~ = ~k dA, so
7. (a) Only the ~k -component of the vector field contributes to the flux and dA
Z
S
~ = 5 · Area of square = 5 · 32 = 45.
(4~i + 5~k ) · dA
(b) This flux is the negative of the flux in part (a) because the surface is oriented in the opposite direction, so its value is
−45.
~ = −~k dA, so
Calculating directly, here dA
Z
S
~ = −5 · Area of square = −5 · 32 = −45.
(4~i + 5~k ) · dA
~ = −~i dA, so
8. (a) Only the ~i -component of the vector field contributes to the flux and dA
Z
S
~ = −2 · Area of disk = −2 · π42 = −32π.
(2~i + 3~j ) · dA
(b) This flux is the negative of the flux in part (a) because the surface is oriented in the opposite direction, so its value is
32π.
~ = ~i dA, so
Calculating directly, here dA
Z
S
~ = 2 · Area of disk = 2 · π42 = 32π.
(2~i + 3~j ) · dA
~ points in direction of positive x-axis, the same direction as the normal vector.
9. (a) The flux is positive, since F
~ points toward negative x-axis, which is opposite the orientation of
(b) The flux is negative, since below the xy-plane F
the surface.
(c) The flux is zero. Since F~ has only an x-component, there is no flow across the surface.
(d) The flux is zero. Since F~ has only an x-component, there is no flow across the surface.
(e) The flux is zero. Since F~ has only an x-component, there is no flow across the surface.
1722
Chapter Nineteen /SOLUTIONS
10. The vector field F~ = F1~i + F2~j + F3~k = −z~i + x~k is a field parallel to the xz-plane that suggests swirling around
the origin from the positive x-axis to the positive z-axis.
~ · ~n = (−z~i + x~k ) · ~i = −z, z is positive here.
(a) The flux going through this surface is negative, because F
~ · ~n = −z, z is negative here.
(b) The flux going through this surface is positive, because F
(c) The flux through this surface is negative, because F~ · ~n = (−z~i + x~k ) · (−~k ) = −x, x is positive.
(d) The flux through this surface is negative, because F~ · ~n = −x, x is positive.
(e) The flux through this surface is zero, because it is in the xz-plane, which is parallel to the vector field.
11. The vector field ~r is a field that always points away from the origin.
(a)
(b)
(c)
(d)
The flux through this surface is zero, because the plane is parallel to the field.
The flux through this surface is zero also, for the same reason.
The flux through this surface is zero also, for the same reason.
The flux through this surface is negative, because the field in that quadrant is going up and away from the origin, and
since the orientation is downward, the flux is negative.
(e) The flux through this surface is zero also.
~ = 4~k , and the flux is ~v · A
~ = (~i + 2~j −
12. The area of the rectangular region is 4, and the orientation vector is ~k . So, A
~
~
3k ) · 4k = −12.
~ = 8~i and the flux is
13. The rectangular region is parallel to the yz-plane and has area 8. The orientation vector is ~i , so A
~ = (~i + 2~j − 3~k ) · 8~i = 8.
~v · A
14. The rectangle lies in the plane z + 2y = 4. So a normal vector is 2~j + ~k and a unit normal vector is √15 (2~j + ~k ). Since
√
this points in the positive z-direction it is indeed an orientation for the rectangle. Since the area of this rectangle is 4 5
~ = 8~j + 4~k ,
we have A
~
~v · A = (~i + 2~j − 3~k ) · (8~j + 4~k ) = 16 − 12 = 4.
15. The rectangle lies in the plane 3x + 2z = 6. So 3~i + 2~k is a normal vector and √113 (3~i + 2~k ) is a unit normal vector.
Since this points in both √
the positive x-direction and the positive z-direction, it is an orientation for this surface. Since the
~ = 6~i + 4~k and ~v · A
~ = (~i + 2~j − 3~k ) · (6~i + 4~k ) = 6 − 12 = −6.
area of the rectangle is 2 13, we have A
~ = ~k dA, so only the ~k component of ~v contributes to the flux:
16. On the surface, dA
Flux =
Z
S
~ =
~v · dA
Z
S
(~i − ~j + 3~k ) · ~k dA = 3 · Area of disk = 3 · π22 = 12π.
~ = ~i dA, so only the ~i component of ~v contributes to the flux:
17. On the surface, dA
Flux =
Z
S
~ =
~v · dA
Z
S
(~i − ~j + 3~k ) · ~i dA = Area of triangle = 4.
~ = ~i dA, so only the ~i component of ~v contributes to the flux:
18. On the surface, dA
Flux =
Z
S
~ =
~v · dA
Z
S
(~i − ~j + 3~k ) · ~i dA = Area of square = 4.
√
19. The triangle lies in the plane x + y + z = 1, with normal ~i + ~j + ~k . A unit normal is ~n = (~i + ~j + ~k )/ 3, so
~i + ~j + ~k
3
(~i − ~j + 3~k ) ·
√
dA = √ Area of triangle .
3
3
S
S
p
√
√
The base of the triangle in the xy-plane has length 2; the height is 3/2, so the area is 3/2. Thus
√
3
3
3
= .
Flux = √ ·
2
2
3
Flux =
Z
~ =
~v · dA
Z
20. Since the surface is closed and the vector field is constant, the flux is zero.
21. Since the surface is in the plane x = 2, only the ~i -component contributes to the flux. The area vector of the surface is
π12~i = π~i . Thus,
Z
~ = 2~i · π~i = 2π.
(2~i + 3~j + 5~k ) · dA
S
19.1 SOLUTIONS
1723
~ ~ ~
22. Since the surface is in the plane
√ x + y + x = 1, whose normal vector is i 2+ j + k , a unit
√ normal in the direction
√ of the
~
~
~
orientation is (i + j + k )/ 3. Thus, the area vector of the surface is π1 (~i + ~j + ~k )/ 3 = π(~i + ~j + ~k )/ 3. The
flux is given by
Z
S
~ = (2~i + 3~j + 5~k ) · π
(2~i + 3~j + 5~k ) · dA
~i + ~j + ~k
π(2 + 3 + 5)
10π
√
√
=
= √ .
3
3
3
23. Since the surface is closed, the flux of a constant vector field out of it is 0.
24. The only contribution to the flux is from the face x = 1, since the vector field is zero or parallel to the other faces. On this
~ = ~i . This face has area 6, so its area vector A
~ = 6~i . Thus
face, G
~ = 6.
Flux = ~i · A
25. The only contribution to the flux is from the face z = 3, since the vector field is zero or parallel to the other faces. On this
~ = 3x~k . The vector field is everywhere perpendicular to the face z = 3 but varies in magnitude from point to
face, H
~ = ~k dx dy. Thus
point. On this surface, dA
Flux =
2
Z
0
Z
1
3x~k · ~k dx dy =
0
Z
0
2
Z
1
3x dx dy =
0
3x2
2
1
0
2
·y
= 3.
0
~ = −~i dA, so
26. Only the x-component of the vector field contributes to the flux. On the surface, F~ = 3~i + 4~j and dA
Z
S
~ = −3 · Area of disk = −3 · π52 = −75π.
(3~i + 4~j ) · dA
27. Since ~r is perpendicular to S and ||~r || = 3 on S, we have
Z
S
~ = 3 · Area of surface = 3 · 4π32 = 108π.
~r · dA
~ = ~i dA and x = 3π/2, so
28. Only the ~i component contributes to the flux. On S, we have dA
Z
S
~ = sin(3π/2) · Area of disk = −1 · π(π 2 ) = −π 3 .
(sin x~i + (y 2 + z 2 )~j + y 2~k ) · dA
~k is perpendicular to S and in the same direction as the orientation, and since ||5~i + 5~j + 5~k || =
29. √
Since 5~i + 5~j + 5√
2
2
2
5 + 5 + 5 = 75 on S, we have
Z
S
~ =
(5~i + 5~j + 5~k ) · dA
√
75 · Area of circle =
√
√
75 · π32 = 9π 75.
~ lies in the plane of the square, the flux is 0.
30. Since the vector field F
~ = ~i dA and only the ~i component of F
~ contributes to the flux:
31. Since the disk is in the yz-plane, dA
~ = 2 · Area of disk = 2π.
Flux = F~ · A
~ contributes to the flux. This component is 5~j on the
32. The square is in the plane y = 5, so only the ~j component of F
~
~
square and the area vector A is parallel to j , so
~ ·A
~ = 5 · Area of square = 5(1.6)2 = 12.8.
Flux = F
33. The plane is z = −2. Since the vector field is −2~k on the plane, it is parallel to the normal and in the same direction, and
of length 2 there. Thus,
Z
√
~ = 2 · Area of square = 2 · ( 14)2 = 28.
z~k · dA
Flux =
S
1724
Chapter Nineteen /SOLUTIONS
34. See Figure 19.1. The vector field is a vortex going around the z-axis, and the square is centered on the x-axis, so the flux
going across one half of the square is balanced by the flux coming back across the other half. Thus, the net flux is zero, so
Z
S
~ · dA
~ = 0.
F
z
y
x
~
n
Figure 19.1
35. Only the ~i -component contributes to the flux, so
Z
S
~ · dA
~ = 7 · Area of disk = 7 · π22 = 28π.
F
36. In the plane y = 3, we have F~ = x~i + 6~j + 3z~k . Only the ~j -component contributes to the flux, so
Z
S
~ = 6 · Area of square = 6 · 22 = 24.
F~ · dA
37. Since the vector field is constant, the flux is zero.
~ || = 10 and points inward everywhere (opposite to the orientation of
38. On the sphere of radius 2, the vector field has ||F
the surface). So
Z
Flux =
S
~ || · Area of sphere = −10 · 4π22 = −160 π.
~ = −||F
F~ · dA
~ = ~k dA, and z = 4, so,
39. We have dA
Z
S
~ · dA
~ =
F
Z
S
(x~i + y~j + (42 + 3)~k ) · ~k dA =
Z
19 dA
S
= 19(Area of rectangle) = 19(6) = 114.
√
40. The normal to the plane is ~n = ~i + ~j ; a unit vector in this direction and in the direction of orientation is (~i + ~j )/ 2.
Thus,
~n
6+7
130
Flux = (6~i + 7~j ) ·
· Area of triangle = √ · 10 = √ .
||~n ||
2
2
~ = ~j dx dz on the square, S, we have
41. The only contribution to the flux is from the ~j -component, and since dA
Flux =
Z
S
~ =
(6~i + x2~j − ~k ) · dA
Z
2
−2
Z
2
−2
x2~j · ~j dx dz =
Z
2
−2
x3
3
2
dz =
−2
64
16
·4=
.
3
3
~ = ~i dA, and x = 4, so,
42. We have dA
Z
S
~ · dA
~ =
F
Z
S
((4 + 3)~i + (y + 5)~j + (z + 7)~k ) · ~i dA =
= 7 · Area of rectangle = 7 · 6 = 42.
Z
7 dA
S
~ || = 21 and points outward everywhere. So
43. On the sphere of radius 3, the vector field has ||F
Flux =
Z
S
~ || · Area of sphere = 21 · 4π32 = 756 π.
~ · dA
~ = ||F
F
19.1 SOLUTIONS
1725
~ and the area vector on the surface of the sphere are parallel, but in opposite directions. Since F
~ is
44. The vector field F
~
pointing inward and ||F || = 6 on the surface
Flux =
Z
S
~ · dA
~ =
F
Z
S
~ || cos π dA = −6 · Area of sphere = −6 · 4π22 = −96π.
||F
~ = ~i dA, so
45. We have dA
Z
S
~ · dA
~ =
F
=
Z
(2z~i + x~j + x~k ) · ~i dA =
S
2
Z
Z
0
3
Z
2z dA
S
2z dzdy = 18.
0
~ is the area vector of the square
46. Since the vector field is constant, if A
~.
Flux = F~ · A
√
An upward normal to the plane is ~i + ~j + ~k , so a unit vector in this direction is (~i + ~j + ~k ) 3. The area vector has
√
~ = 4(~i + ~j + ~k )/ 3. Thus
magnitude 4, so A
√
4(~i + ~j + ~k )
4(1 + 2)
√
√
Flux = (~i + 2~j ) ·
=
= 4 3.
3
3
~ = ~k dxdy and
47. Since the disk is in the xy-plane and oriented upward, dA
Z
Disk
~ · dA
~ =
F
Z
Disk
Z
(x2 + y 2 )~k · ~k dxdy =
(x2 + y 2 ) dxdy.
Disk
Using polar coordinates
Z
~ =
F~ · dA
Disk
Z
2π
0
Z
3
0
3
r4
4
r 2 · r drdθ = 2π
=
0
81π
.
2
~ = ~k dxdy, so
48. Since the disk is horizontal and oriented upward, dA
Z
Disk
~ =
F~ · dA
Z
Disk
Z
cos(x + y )~k · ~k dxdy =
2
2
cos(x2 + y 2 ) dxdy.
Disk
Using polar coordinates, since the disk has radius 3, we have
Z
Disk
Z
~ =
F~ · dA
cos(x2 + y 2 )dxdy =
Disk
Z
2π
0
=
Z
3
cos(r 2 )rdrd θ
0
2π
0
= 2π
Z
3
1
1
sin(r 2 ) dθ = 2π · sin(r 2 )
2
2
0
3
0
1
1
sin(32 ) − sin(02 ) = π sin 9.
2
2
~ = ~i dydz, so we have
49. Since the disk is oriented in the positive x-direction, dA
Flux =
Z
Disk
~ · dA
~ =
F
Z
e
Disk
Z
y 2 +z 2~
i · ~i dydz =
ey
2
+z 2
dydz.
Disk
To calculate this integral, we use polar coordinates with y = r cos θ and z = r sin θ. Then r 2 = y 2 + z 2 and
Z
Disk
~ · dA
~ =
F
Z
0
2π
Z
0
2
2
2
er · rdrdθ = 2π ·
er
2
2
0
= π(e4 − 1).
1726
Chapter Nineteen /SOLUTIONS
~ is parallel to the xy plane, there is no flux across the surface, so
50. See Figure 19.2. Since F
Z
S
~ = 0.
~ · dA
F
~
n
z
x
y
Figure 19.2
~ is the vector pointing in the direction normal to the surface
51. See Figure 19.3. The area vector of a small area element ∆A
~ = ~k ∆A. Thus,
with magnitude ∆A. The unit vector normal to the surface is ~k , so ∆A
Z
S
~ =
F~ · dA
lim
~ k→0
k∆A
X
~ =
~r · ∆A
lim
k∆Ak→0
X
~ =
~r · ~k ∆A
Z
S
~r · ~k dA.
Now ~r · ~k = 2 for all the points on S because all such points have z-coordinate equal to 2. Thus, we have
Z
S
~ · dA
~ =
F
Z
S
~r · ~k dA =
Z
S
2 dA = 2 · Area of S = 8π.
z
~
n
~
r
y
x
Figure 19.3
52. See Figure 19.4. Since the vector field is parallel to the x-axis, only the two sides perpendicular to the x-axis contribute to
the flux integral. On the side where x = 0, the vector field is 2~i , and hence the flux through that side is −(2)(32 ) = −18
(negative because the flow is inward and the normal vector is pointing out). The flow out the other side is at x = 3, so
F~ = −~i , so the flux out that side is (−~i ) · (32~i ) = −9. So the net flux is −18 − 9 = −27. So
Z
S
~ · dA
~ = −27.
F
z
x
y
Figure 19.4
19.1 SOLUTIONS
1727
Problems
53. (a)
(i) The flux
R
~ is positive. The vectors x~i all point out of the box.
x~i · dA
RB
~ is zero. The vector field is parallel to the x-axis. For each y-value, the flux entering the box
(ii) The flux B y~i · dA
at one endR cancels the flux leaving at the other end.
~ is zero. The flux entering the sphere where x < 0 cancels the flux leaving where x > 0.
(iii) The flux S |x|~i · dA
R
R
R
~ = y~i · dA
~ − x~i · dA
~ = Zero − Positive , this flux is negative.
(iv) Since S (y − x)~i · dA
R
~
~
(b) From the Divergence Theorem, B xi · dA is greater. Since div(x~i ) = 1,
Z
S
Z
B
~ =
x~i · dA
~ =
x~i · dA
Z
1 · dV = Volume of sphere =
4
π.
3
1 · dV = Volume of box = 8 >
4
π.
3
Inside sphere
Z
Inside box
~ 1 = 4~ı and A
~ 2 = −4~ı
54. (a) Consider two opposite faces of the cube, S1 and S2 . The corresponding area vectors are A
~ is constant, we find the flux by taking the dot product, giving
(since the side of the cube has length 2). Since E
~ ·A
~ 1 = (a~i + b~j + c~k ) · 4~i = 4a.
Flux through S1 = E
~ ·A
~ 2 = (a~i + b~j + c~k ) · (−4~i ) = −4a.
Flux through S2 = E
Thus the fluxes through S1 and S2 cancel. Arguing similarly, we conclude
R that, for any pair of opposite faces, the
~ through these faces is zero. Hence, by addition,
~ · dA
~ = 0.
sum of the fluxes of E
E
S
(b) The basic idea is the same as in part (a), except that we now need to use Riemann sums. First divide S into two
~ . For a tiny patch S1 in the hemisphere
hemispheres H1 and H2 by the equator C located in a plane perpendicular to E
H1 , consider the patch S2 in the opposite hemisphere which is symmetric to S1 with respect to the center O of the
~ 1 and ∆A
~ 2 satisfy ∆A
~ 2 = −∆A
~ 1 , so if we consider S1 and S2 to be approximately
sphere. The area vectors ∆A
~ · ∆A
~ 1 = −E
~ · ∆A
~ 2 . By decomposing H1 and H2 into small patches as above and using Riemann
flat, then E
sums, we get
Z
Z
Z
H1
~ · dA
~ =−
E
H2
~ · dA
~,
E
so
S
~ · dA
~ = 0.
E
~ through any surface with a center of symmetry is
(c) The reasoning in part (b) can be used to prove that the flux of E
zero. For instance, in the case of the cylinder, cut it in half with a plane z = 1 and denote the two halves by H1 and
~ · ∆A1 = −E
~ · ∆A
~ 2 . Thus, we get
H2 . Just as before, take patches in H1 and H2 with ∆A1 = −∆A2 , so that E
which shows that
Z
H1
~ · dA
~ =−
E
Z
S
Z
H2
~ · dA
~,
E
~ · dA
~ = 0.
E
55. Notice that the speed is 3 cm/sec at the center of the pipe and 0 cm/sec at the sides. Suppose ~i is the unit vector parallel
to the direction of flow. Then, at a distance r from the center of the pipe, the velocity is given by
3
~v = 3 − r 2 ~i cm/sec.
4
Divide the circular cross-section into concentric rings of width ∆r, so that the velocity is approximately constant on
~ are parallel (see Figure 19.5), we have
each one. The area of a typical ring is ∆A ≈ 2πr∆r. Then since ~v and ∆A
~ = k~v kk∆A
~ k ≈ 3 − 3 r 2 cm · (2πr∆r) cm2 .
Flux through ring ≈ ~v ∆A
4
sec
✻
4 cm
✛
❄
✻
r
✻
❄
∆r
~
v
~
∆A
❄
Figure 19.5: Flux through pipe when
velocity varies with distance from the center
1728
Chapter Nineteen /SOLUTIONS
Thus, the flux through the circular cross-section of the pipe is given by
Flux =
lim
~ k→0
k∆A
= lim
X
∆r→0
r=2
=
Z
r=0
X
~
~v · ∆A
3−
3 2
r 2πr∆r
4
3
3 − r 2 2πr dr = 6π
4
Z
0
2
r3
r−
4
dr = 6π cm3 /sec.
56. (a) The net electric flux through this surface is zero, because the surface is placed so that it is always parallel with the
electric field, and there is no flow through the surface.
(b) The net flux is zero, because the flow in through one half of the cylinder is canceled by the flow out through the other
half.
~ and H
~ do not have flux through S because they are parallel to the cylinder that S lies on, and
57. The ~j components of G
~,G
~ , and H
~ have the same ~i and ~k components, so they have the
are therefore tangent to S. The three vector fields, F
same flux through S.
58. Since this vector field points radially out from the origin, it is everywhere parallel to the vector representing the surface
~ . Thus since kF
~ (~r )k = 1/R2 on the surface, S,
area, dA
~ (~r ) · dA
~ = 1 dA,
F
R2
so
Z
~ = 1 · Surface area of sphere = 1 (4πR2 ) = 4π.
~ (~r ) · dA
F
R2
R2
S
~ . Thus since
59. Since this vector field points radially out from the origin, it is everywhere parallel to the area vector, ∆A
~ (~r )k = 1/R on the surface, S,
kF
~ (~r ) · ∆A
~ = 1 ∆A
F
R
so
Z
X
1
1
~ = 1 lim
∆A =
· Surface area of sphere = (4πR2 ) = 4πR.
F~ (~r ) · dA
R ∆A→0
R
R
S
60. (a) The vector field is perpendicular to the surface of a sphere centered at the origin. Thus the magnitude of the flux
~ || decreases as p
depends on the magnitude of the vector field on the surface. Since for fixed ~r , the value of ||F
increases, the maximum flux occurs when p = 0.
~ || = 2 on the surface. Thus
(b) For a sphere of radius 2 with p = 0, we have ||F
Flux =
Z
S
~ = 2 · Area of surface = 2 · 4π22 = 32π.
F~ · dA
~ is ~v · A
~ . Therefore, the flux through each face of the cube is equal to (−~i + 2~j +
61. (a) For a flat surface, flux through A
~k ) · (A
~ of the face).
First we shall find the flux through the two faces parallel to the xy-plane, beginning with the one with negative
~ = −4~k . The
z. The unit vector normal to this face and pointing outward is −~k . The area of the face equals 4, so A
flux through the face with negative z equals
(−~i + 2~j + ~k ) · (−4~k ) = 0 + 0 − 4 = −4
~ = 4~k . The flux through
For the face with positive z, the unit normal vector that points outward is ~k . Therefore A
this face is given by
(−~i + 2~j + ~k ) · 4~k = 0 + 0 + 4 = 4
Next, we will find the flux through the two faces parallel to the xz-plane, beginning with the one with negative
~ = −4~j . The flux then equals
y. A unit vector normal to this face pointing outward is −~j . Therefore A
(−~i + 2~j + ~k ) · (−4~j ) = 0 − 8 + 0 = −8
~ = 4~j . The flux then equals
For the face with positive y, the unit normal vector pointing outward is ~j . Therefore A
(−~i + 2~j + ~k ) · (4~j ) = 0 + 8 + 0 = 8
19.1 SOLUTIONS
1729
Next, we will find the flux through the two faces parallel to the yz plane, beginning with the one with negative
~ = −4~i . The flux then equals
x. A unit vector normal to this plane pointing outward is −~i . Therefore A
(−~i + 2~j + ~k ) · (−4~i ) = 4 + 0 + 0 = 4
~ = 4~i . The flux then equals
For the face with positive x, the unit normal vector pointing outward is ~i . Therefore A
(−~i + 2~j + ~k ) · (4~i ) = −4 + 0 + 0 = −4
Adding up all of these fluxes to get the flux out of the entire cube, we get
Total flux = −4 + 4 − 8 + 8 + 4 − 4 = 0
(b) For any constant vector field ~v = a~i + b~j + c~k , we can calculate the flux out of the cube by the same method.
First we shall find the flux out of the two faces parallel to the xy plane, beginning with the one with negative z.
A unit vector normal to this plane, that points negative (because of the orientation of the face) is −~k . The area of the
~ = −4~k . The flux through A
~ then equals
face equals 4, therefore A
(a~i + b~j + c~k ) · (−4~k ) = 0 + 0 − 4c = −4c
~ = 4~k . The flux then equals
For the face with positive z, the unit normal vector pointing outward is ~k . Therefore A
(a~i + b~j + c~k ) · (4~k ) = 0 + 0 + 4c = 4c.
Next, we will find the flux through the two faces parallel to the xz plane, beginning with the one with negative
~ = −4~j . The flux then equals
y. A unit vector normal to this plane pointing outward is −~j . Therefore A
(a~i + b~j + c~k ) · (−4~j ) = 0 − 4b + 0 = −4b
~ == 4~j . The flux then equals
For the face with positive y, the unit normal vector pointing outward is ~j . Therefore A
(a~i + b~j + c~k ) · (4~j ) = 0 + 4b + 0 = 4b
Next, we will find the flux through the two faces parallel to the yz plane, beginning with the one with negative
~ = −4~i . The flux then equals
x. A unit vector normal to this plane pointing outward is −~i . Therefore A
(a~i + b~j + c~k ) · (−4~i ) = −4a + 0 + 0 = −4a
~ = 4~i . The flux then equals
For the face in the positive x, the unit normal vector pointing outward is ~i . Therefore A
(a~i + b~j + c~k ) · (4~i ) = 4a + 0 + 0 = 4a
Adding up all of these fluxes to get the flux out of the entire cube, we get
Total flux = −4c + 4c − 4b + 4b + 4a − 4a = 0
(c) The answers in parts (a) and (b) make sense because the vector field is constant, and so it does not change as it comes
in the one side of the cube, and exits the other side. Therefore the two fluxes cancel each other out, making the total
flux zero.
~ be the area vector of any face of the tetrahedron in Figure 19.6. The flux through the face equals ~v · A
~ because
62. (a) Let A
~,
~
~
the vector field is constant. Therefore, the flux through each face of the tetrahedron is equal to (−i + 2j + ~k ) · A
~ is the area of that face.
where A
First we shall find the flux out of the triangle in the xy plane. A unit vector normal to that plane, that points
~ =
negative (because of the orientation of the face), is equal to −~k . The area of the face equals 0.5, therefore A
~ then equals
−0.5~k . The flux through A
(−~i + 2~j + ~k ) · (−0.5~k ) = 0 + 0 − 0.5 = −0.5.
Next, we will find the flux out of the triangle in the xz plane. A unit vector normal to that plane, that points
~ = −0.5~j . The flux through A
~ then equals
negative, is equal to −~j . The area of the face equals 0.5, therefore A
(−~i + 2~j + ~k ) · (−0.5~j ) = 0 − 1 + 0 = −1.
1730
Chapter Nineteen /SOLUTIONS
Next, we will find the flux out of the triangle in the yz plane. A unit vector normal to that plane, that points
~ = −0.5~i . The flux through A
~ then equals
negative, is equal to −~i . The area of the face equals 0.5, therefore A
(−~i + 2~j + ~k ) · (−0.5~i ) = 0.5 + 0 + 0 = 0.5.
Last, we will find the flux out of the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1). A unit vector normal to
√
that plane, that points positive, is equal to √13 (~i + ~j + ~k ). The area of the face equals 3/2, since it is an equilateral
√
triangle with side 2. Therefore:
√
~ = √1 (~i + ~j + ~k )( 3/2) = 0.5(~i + ~j + ~k ).
A
3
~ then equals
The flux through A
(−~i + 2~j + ~k ) · (0.5~i + 0.5~j + 0.5~k ) = −0.5 + 1 + 0.5 = 1.
The total flux out of the tetrahedron is −0.5 − 1 + 0.5 + 1 = 0. Therefore the flux equals zero.
z
y
x
Figure 19.6
(b) For any constant vector field ~v = a~i + b~j + c~k , we can find the flux out of the tetrahedron.
First we shall find the flux out of the triangle in the xy plane. A unit vector normal to that plane, that points
~ =
negative (because of the orientation of the face), is equal to −~k . The area of the face equals 0.5, therefore A
~
~
−0.5k . The flux through A then equals
(a~i + b~j + c~k ) · (−0.5~k ) = 0 + 0 − 0.5 = −0.5c.
Next, we will find the flux out of the triangle in the xz plane. A unit vector normal to that plane, that points
~ = −0.5~j . The flux through A
~ then equals
negative, is equal to −~j . The area of the face equals 0.5, therefore: A
(a~i + b~j + c~k ) · (−0.5~j ) = 0 − 0.5b + 0 = −0.5b.
Next, we will find the flux out of the triangle in the yz plane. A unit vector normal to that plane, that points
~ = −0.5~i . The flux through A
~ then equals
negative, is equal to (−~i ). The area of the face equals 0.5, therefore A
(a~i + b~j + c~k ) · (−0.5~i ) = −0.5a + 0 + 0 = −0.5a.
Last, we will find the flux out of the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1). A unit vector normal to
√
that plane, that points positive, is equal to √13 (~i + ~j + ~k ). The area of the face equals 3/2, therefore:
√
~ = √1 (~i + ~j + ~k )( 3/2) = 0.5(~i + ~j + ~k ).
A
3
~ then equals
The flux through A
(a~i + b~j + c~k ) · (0.5~i + 0.5~j + 0.5~k ) = 0.5a + 0.5b + 0.5c.
The total flux out of the tetrahedron is −0.5c − 0.5b − 0.5a + 0.5a + 0.5b + 0.5c = 0. Therefore, the flux is
equal to zero.
(c) The answers in (a) and (b) make sense because the vector field is constant, so it does not change as it enters through
one side of the tetrahedron, and exits the other side. Therefore the two cancel each other out,causing the flux to be
equal to zero.
19.1 SOLUTIONS
1731
63. Since Pressure = Force/Area, we have
~ at point (x, y, z) ≈ P (x, y, z)k∆A
~ k.
Force on a small patch with area ∆A
~ (if S is oriented with the inward
This force is directed inward and normal to the surface, so the force is P (x, y, z)∆A
normal). For buoyancy, take the upward component of this force, so
~ · ~k .
Buoyancy force = P (x, y, z)∆A
Then:
Total buoyancy =
=
lim
~ k→0
k∆A
Z
S
=
Z
S
X
s
~ · ~k
P (x, y, z)∆A
~
P (x, y, z)~k · dA
~
F~ · dA
64. (a) From Newton’s law of cooling, we know that the temperature gradient will be proportional to the heat flow. If the
~ = k grad T . Since grad T points in the direction of
constant of proportionality is k then we have the equation F
increasing T , but heat flows toward lower temperatures, the constant k must be negative.
(b) This form of Newton’s law of cooling is saying that heat will be flowing in the direction in which temperature is decreasing most rapidly, in other words, in the direction exactly opposite to grad T . This agrees with our intuition which
tells us that a difference in temperature causes heat to flow from the higher temperature to the lower temperature, and
the rate at which it flows depends on the temperature gradient.
(c) The rate of heat loss from W is given by the flux of the heat flow vector field through the surface of the body. Thus,
Rate of heat
loss from W
=
~
Flux of F
out of S
=
Z
S
~ · dA
~ =k
F
Z
S
~
(grad T ) · dA
~ . Note that E
~ points radially outward from the z-axis.
65. (a) Figure 19.7 shows the electric field E
y
2
x
−2
2
−2
Figure 19.7: The electric field in the xy-plane due to a line of positive charge uniformly
~
~
~ (x, y, 0) = 2λ xi + y j
distributed along the z-axis: E
x2 + y 2
~ points in the same direction as the outward normal ~n , and
(b) On the cylinder x2 + y 2 = R2 , the electric field E
~k=
kE
2λ ~
2λ
kxi + y~j k =
.
R2
R
1732
Chapter Nineteen /SOLUTIONS
So
Z
S
~ · dA
~ =
E
Z
S
Z
~ · ~n dA =
E
2λ
=
R
Z
~ k dA =
kE
S
Z
S
2λ
dA
R
2λ
2λ
· Area of S =
· 2πRh = 4πλh,
R
R
dA =
S
which is positive, as we expected.
~ is sketched in Figure 19.8 for I > 0.
66. (a) The vector field B
y
4
2
x
−4
−2
2
4
−2
−4
Figure 19.8
(b) The disk S can be parameterized as z = h (viewed as a constant function of x and y), for x, y in the region
{x2 + y 2 ≤ a2 }. Hence
Z
S
~ · dA
~ =
B
Z
I −y~ı + x~ ~
· 2
· k dx dy = 0,
2π
x + y2
{x2 +y 2 ≤a2 }
~ is tangent to the surface
since ~ı · ~k = 0 and ~ · ~k = 0. The answer is as we would expect, since the vector field B
S, hence there is no flux through S. R
~ through S2 is given by
~ · dA
~ . On S2 we have
(c) The flux of B
B
S
2
~ (x, y, z) = I · −y~ı = − I ~ı ,
B
2π y 2
2πy
and
~ = ~n dA = −~ı dy dz
dA
Hence,
Z
S2
~ · dA
~ =
B
Z
h
0
I
=
2π
I
=
2π
Z
b
a
I (−~ı )
·
· (−~ı ) dy dz
2π
y
Z hZ
0
Z
a
1
dy dz
y
h
[ln |y|]ba dz
0
Z
b
h
I
=
(ln |b| − ln |a|) dz
2π 0
b
I
h ln
.
=
2π
a
~ is everywhere parallel to the orientation of S2 . For 0 <
This time we get a non-zero flux since the direction of B
a < b the flux is positive since |b/a| > 1 and increases as the area S2 increases. This is as Figure 19.8 would lead
us to expect. For a < b < 0, the flux is negative since |b/a| < 1. If a < 0 < b, the flux can be either positive or
negative.
19.1 SOLUTIONS
1733
67. (a) If ~r = x~ı + y~ + z~k is the position vector of a point on the sphere, then
~ (~r ) = 3zp ~r − p ~k .
D
a5
a3
The second term is a constant vector field. Hence, by symmetry,
Z
S
−
p~
k
a3
~ = 0.
· dA
(See the solution to Problem 54 on page 1727). Let us also apply a symmetry argument to
Z
(
S
3zp
~.
~r ) · dA
a5
~ through the upper hemisphere H1 equals minus the flux of D
~ through the lower
We will show that the flux of D
~
hemisphere H2 . The flux of D through H1 and H2 will be computed as limits of Riemann sums.
Consider a small patch P1 in H1 and call its reflection about the xy-plane P2 . The contribution of P1 to the flux
~ through S is
of D
3z1 p
~ 1 = 3z1 p ~r · Area(P1 ) ~r = 3z1 p · Area(P1 ),
~r · dA
a5
a5
k~r k
a4
whereas the contribution from P2 is
3z2 p
~ 1 = 3z2 p Area(P2 ).
~r · dA
a5
a4
But Area(P1 ) = Area(P2 ) and z2 = −z1 , so the contributions from P1 and P2 cancel each other. Dividing H1 and
H2 into symmetric patches as above, and taking the limit as the areas of the patches become smaller and smaller, one
gets
Z
Z
3zp
3zp
~
~.
~
r
~
r
·
d
A
=
−
· dA
a5
a5
H2
H1
i.e.
Z
3zp
~r
a5
~ = 0.
· dA
Z
p~
k
a3
~ = 0,
· dA
S
Since we also know that
S
we can conclude that
−
Z
S
(b) By Gauss’s Law,
Z
S
~ · dA
~ = 0.
D
~ · dA
~ = 4π(q − q) = 0,
E
~ through S.
which is the same as the flux of D
Strengthen Your Understanding
68. The value of a flux integral is a scalar, not a vector.
R
~ · dA
~ is determined by the interaction between the vector field F
~ and the oriented surface
69. The sign of a flux integral S F
S. It can not be determined from either one alone. For example, if S is oriented in the +~k direction, then the flux through
~ = ~k is positive and the flux through S of F
~ = −k is negative.
S of F
R
~
~
70. The integral S F · dA is zero if the vector field F~ is tangent to the surface S at every point of S. The displacement
vector F~ = −~i + ~j between
R the two points (1, 0, 0) and (0, 1, 0) in S is tangent to S. Hence, with the constant vector
~ = −~i + ~j we have
~ = 0.
field F
F~ · dA
S
~ = z~k , and let S be the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 1, oriented upwards. The vector field is normal to
71. Let F
the surface, in the direction of the orientation, and has constant magnitude 1 on the surface, even though it has varying
magnitude elsewhere.
Z
S
~ = kF
~ k · Area of S = 1 · 1 = 1.
~ dA
F
1734
Chapter Nineteen /SOLUTIONS
72. True. By definition, the flux integral is the limit of a sum of dot products, hence is a scalar.
~ is perpendicular to the flat surface.
73. False. A
74. False. The flux integral measures the net flow through the surface. There could be as much flow into the sphere as out,
~ = ~i has zero flux integral over the entire
which would give a flux integral of zero. As an example, the constant field F
sphere, yet is not the zero vector field.
75. True. The flow of this field is in the same direction as the orientation of the surface everywhere on the surface, so the flux
is positive.
76. True. The flow of this field is in the same direction as the orientation of the surface everywhere on the surface, so the flux
is positive.
77. True. Since the vector field is constant, the negative flux into the bottom of the cube is equal in magnitude to the positive
flux out of the top, so these cancel in the sum defining the flux integral. The other four faces of the cube each have zero
flux from the field, since the field is parallel to each of them.
~ in the sum defining the flux integral with their
78. True. Reversing the orientation on S replaces all of the area vectors ∆A
negatives, so that the flux integral over −S is the negative of the flux integral over S.
79. False. There is no reason to expect a relationship between the flux integrals over S1 and S2 simply based on their relative
areas. The value of the flux integral over a surface depends both on the shape of the surface and the behavior of the vector
field at points on the surface. For example, let S1 be the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 0, oriented upward, and let
S2 be the rectangle 0 ≤ y ≤ 1, 0 ≤ z ≤ R2, x = 0 with positive orientation in the ~i direction.
R The area of S1 = 1 and the
~ is parallel to S1 ) and
~ = 2. These values
~ · dA
~ = 0 (since F
F~ · dA
area of S2 = 2. Then if F~ = ~i we have S F
S2
1
R
R
~ =
~ · dA
~.
do not satisfy 2
F~ · dA
F
S1
S2
~ · ∆A
~ = (2G
~ ) · ∆A
~ = 2(G
~ · ∆A
~ ). So each
80. True. In the sum defining the flux integral for F~ , we have terms like F
~ is twice the corresponding term in the sum approximating the flux of G
~,
term in the sum approximating the flux of F
~ . Thus the flux of F
~ is twice the flux of G
~.
making the sum for F~ twice that of the sum for G
~ could be large in magnitude on
81. False. The flux integral measures the net flow through the surface S. The vector field G
~ ||), but be parallel to the surface S, and so contribute nothing to the flux. Put another way, a “small”
S (larger than ||F
vector field, flowing directly across S, can have greater flux than a much “larger” field flowing parallel to S.
~
~
~
~
For example, take S to be
R the square 0 ≤ x R≤ 1, 0 ≤ y ≤ 1, z = 0, oriented upward. ThenRif F = k andR G = 5i ,
~
~
~
~
~
~
~
~
~
the flux integrals have values S F ·dA = 1 and S G ·dA = 0 (since G is parallel to S). Thus S F ·dA > S G ·dA ,
~ || = 1 < 5 = ||G
~ ||.
but ||F
82. For (a), we want the vector field with the largest ~i component so F~ 1 .
~ 1.
For (b), we want the vector field with the largest ~i component so F
~ 4.
For (c), we want the vector field with the most negative ~k component, so F
~
For (d), we want the vector field with the most negative k component, so F~ 4 .
~ 3.
For (e), we want the vector field with the largest ~j component so F
Solutions for Section 19.2
Exercises
1. We have
2. We have
3. We have
4. We have
~ = −fx~i − fy~j + ~k dx dy = −3~i + 5~j + ~k dx dy.
dA
~ = −fx~i − fy~j + ~k dx dy = −8~i − 7~j + ~k dx dy.
dA
~ = −fx~i − fy~j + ~k dx dy = −4x~i + 6y~j + ~k dx dy.
dA
~ = −fx~i − fy~j + ~k dx dy = −y~i + (−x − 2y)~j + ~k dx dy.
dA
19.2 SOLUTIONS
5. We have
We have
1735
~ = −fx~i − fy~j + ~k dx dy = −2~i + 3~j + ~k dx dy.
dA
~ = (−20 + 60 + 30)dx dy = 70 dx dy.
F~ · dA
~ through the surface S is given by
The flux of F
Z
S
6. We have
We have
3
Z
~ · dA
~ =
F
−2
Z
5
70 dy dx.
0
~ = −fx~i − fy~j + ~k dx dy = 4~i − 10~j + ~k dx dy.
dA
~ · dA
~ = (4z − 10x + y) dx dy.
F
~ through the surface S is given by
The flux of F
Z
S
7. We have
We have
~ = −fx~i − fy~j + ~k
dA
8
Z
~ · dA
~ =
F
0
4
Z
(4z − 10x + y) dx dy.
0
dx dy = sin x~i − 2 cos 2y~j + ~k
dx dy.
~ · dA
~ = (yz sin x − 2xy cos 2y + xy)dx dy.
F
The region R is a triangle bounded by the x-axis, the y-axis, and the line y = 5 − x. Thus, R is described by the
~ through the surface S is given by
inequalities 0 ≤ y ≤ 5 − x, 0 ≤ x ≤ 5. The flux of F
Z
S
8. We have
We have
~ · dA
~ =
F
Z
0
5
Z
5−x
(yz sin x − 2xy cos 2y + xy) dy dx.
0
~ = −fx~i − fy~j + ~k
dA
dx dy = −e3y~i − 3xe3y~j + ~k
dx dy.
~ = −3xe3y cos(x + 2y) dx dy.
F~ · dA
2
2
2
The region
√ R is the part of the disk x + y ≤ 5 where x ≥ 0 and y ≥ 0. Thus, R is described by the inequalities
~ through the surface S is given by
0 ≤ y ≤ 52 − x2 , 0 ≤ x ≤ 5. The flux of F
Z
Z Z √
25−x2
5
S
9. We have
Hence
~ · dA
~ =
F
0
−3xe3y cos(x + 2y) dy dx.
0
~ = −fx~i − fy~j + ~k dx dy = −4~i + 2~j + ~k dx dy.
dA
~ · dA
~ = (−12 − 4 + 6)dx dy = −10 dx dy.
F
~
The flux of F through the surface S is given by
Z
5
~ = −fx~i − fy~j + ~k
dA
Z
S
10. We have
Since z = xy we have
~ · dA
~ =
F
0
Z
10
0
−10 dy dx = −500.
dx dy = −y~i − x~j + ~k
~ · dA
~ = (−y + 2x + xy) dx dy.
F
~ through the surface S is given by
The flux of F
Z
S
~ · dA
~ =
F
=
Z
10
0
Z
0
Z
10
(−y + 2x + xy) dx dy
0
10
100 + 40y dy = 3000.
dx dy.
1736
Chapter Nineteen /SOLUTIONS
11. We have
2
~ = −fx~i − fy~j + ~k
dA
Since z = x + 2y have
dx dy = −2x~i − 2~j + ~k
~ · dA
~ = (−2x cos y − 2x2 − 4y + 1) dx dy.
F
dx dy.
~ through the surface S is given by
The flux of F
Z
S
~ =
F~ · dA
=
12. We have
Since z = x + y + 2 have
Z
Z
1
0
Z
1
(−2x cos y − 2x2 − 4y + 1)dx dy
0
1
0
5
1
− 4y − cos y dy = − − sin 1 = −2.508.
3
3
~ = −fx~i − fy~j + ~k
dA
dx dy = −~i − ~j + ~k
~ · dA
~ = (y + 2) dx dy.
F
dx dy.
The (x, y) values for points on S are in the triangle with vertices (−1, 0), (1, 0), (0, 1). The edges of the triangle are
given by the x-axis, y = 1 + x and y = 1 − x. For this region, it makes sense to integrate on x first, so we describe the
~ through the surface S is
region by the inequalities y − 1 ≤ x ≤ 1 − y, 0 ≤ y ≤ 1. The flux of F
Z
S
~ · dA
~ =
F
=
1
Z
0
Z
Z
1−y
(y + 2) dx dy
y−1
1
4 − 2y − 2y 2 dy =
0
7
.
3
13. Since the cylinder radius is 10, we have
~ = 10 cos θ~i + sin θ~j
dA
Hence
dz dθ.
~ = 10 (cos θ + 2 sin θ) dz dθ.
F~ · dA
~ through the surface S is given by
The θz-region corresponding to S is given by 0 ≤ θ ≤ π/2, 0 ≤ z ≤ 5. The flux of F
Z
S
~ · dA
~ =
F
Z
π/2
0
Z
5
10 (cos θ + 2 sin θ) dz dθ.
0
14. Since the cylinder radius is 10, we have
~ = 10 cos θ~i + sin θ~j
dA
Using x = 10 cos θ and y = 10 sin θ we have
dz dθ.
~ = 10 cos θ~i + 20 sin θ~j + 3z~k .
F
Hence
~ · dA
~ = 10 10 cos2 θ + 20 sin2 θ dz dθ.
F
The θz-region corresponding to S is given by 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 5. The flux of F~ through the surface S is given by
Z
S
~ · dA
~ =
F
Z
0
2π
Z
0
5
10 10 cos2 θ + 20 sin2 θ dz dθ.
19.2 SOLUTIONS
1737
15. Since the cylinder radius is 6, we have
~ = 6 cos θ~i + sin θ~j
dA
Using x = 6 cos θ and y = 6 sin θ we have
dz dθ.
F~ = z 2~i + e6 cos θ~j + ~k .
Hence
~ = 6z 2 cos θ + 6 sin θe6 cos θ dz dθ.
F~ · dA
The surface S is the part of the cylinder x2 + y 2 = 62 that is inside the sphere x2 + y 2 + z 2 = 102 . The cylinder and the
sphere intersect at points where z 2 = 102 − 62 = 64, where z = 8 or z = −8. Therefore, the θz-region corresponding
~ through the surface S is given by
to S is given by 0 ≤ θ ≤ 2π, −8 ≤ z ≤ 8. The flux of F
Z
S
~ · dA
~ =
F
2π
Z
0
8
Z
6z 2 cos θ + 6 sin θe6 cos θ dz dθ.
−8
16. Since the cylinder radius is 2, we have
~ = 2 cos θ~i + sin θ~j
dA
Using x = 2 cos θ and y = 2 sin θ we have
dz dθ.
~ = 8z cos2 θ sin θ~j + z 3~k .
F
Hence
~ · dA
~ = 16z cos2 θ sin2 θ dz dθ.
F
2
The surface S is the part of the cylinder x + y = 22 that is between the xy-plane z = 0 and the paraboloid z = x2 + y 2 .
The cylinder and the paraboloid intersect at points where z = 22 . Therefore, the θz-region corresponding to S is given
~ through the surface S is given by
by 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 4. The flux of F
2
Z
S
~ · dA
~ =
F
2π
Z
0
Z
4
16z cos2 θ sin2 θ dz dθ.
0
17. Since the cylinder radius is 5, we have
~ = 5 cos θ~i + sin θ~j
dA
Hence
dz dθ.
~ · dA
~ = 5z sin θ dz dθ.
F
The θz-region corresponding to S, the part of the cylinder with y ≥ 0 and 0 ≤ z ≤ 20, is given by 0 ≤ θ ≤ π,
0 ≤ z ≤ 20. The flux of F~ through the surface S is given by
Z
S
~ · dA
~ =
F
Z
π
0
Z
20
5z sin θ dz dθ =
Z
π
1000 sin θ dθ = 2000.
0
0
18. Since the cylinder radius is 10, we have
~ = 10 cos θ~i + sin θ~j
dA
dz dθ.
Using x = 10 cos θ and y = 10 sin θ we have
~ = 10 sin θ~i + 10z cos θ~k .
F
Hence
~ = 100 cos θ sin θ dz dθ.
F~ · dA
The surface S is the part of the cylinder with x ≥ 0, y ≥ 0 and 0 ≤ z ≤ 3, so the θz-region corresponding to S is given
by 0 ≤ θ ≤ π/2, 0 ≤ z ≤ 3. The flux of F~ through the surface S is given by
Z
S
~ · dA
~ =
F
=
Z
π/2
0
Z
Z
3
100 cos θ sin θ dz dθ
0
π/2
300 cos θ sin θ dθ
0
π/2
= 150 sin2 θ|θ=0 = 150.
1738
Chapter Nineteen /SOLUTIONS
19. Since the cylinder radius is 2, we have
~ = 2 cos θ~i + sin θ~j
dA
Using x = 2 cos θ and y = 2 sin θ we have
dz dθ.
~ = 4z cos θ sin θ~j + 2ez cos θ~k .
F
Hence
~ · dA
~ = 8z cos θ sin2 θ dz dθ.
F
The surface S is the part of the cylinder with 0 ≤ y ≤ x and 0 ≤ z ≤ 10, so the θz-region corresponding to S is given
~ through the surface S is given by
by 0 ≤ θ ≤ π/4, 0 ≤ z ≤ 10. The flux of F
Z
S
Z
~ · dA
~ =
F
π/4
0
Z
=
Z
10
8z cos θ sin2 θ dz dθ
0
π/4
400 cos θ sin2 θ dθ
0
√
100 2
400
π/4
sin3 θ|θ=0 =
.
=
3
3
20. Since the cylinder radius is 1, we have
~ = cos θ~i + sin θ~j
dA
Using x = cos θ and y = sin θ we have
dz dθ.
~ = cos θ sin θ~i + 2z~j .
F
Hence
~ · dA
~ = cos2 θ sin θ + 2z sin θ dz dθ.
F
The surface S is the part of the cylinder with x ≥ 0, 0 ≤ y ≤ 1/2 and 0 ≤ z ≤ 2, so the θz-region corresponding to S is
~ through the surface S is given by
given by 0 ≤ θ ≤ π/6, 0 ≤ z ≤ 2. The flux of F
Z
S
~ · dA
~ =
F
=
Z
π/6
0
Z
Z
2
cos2 θ sin θ + 2z sin θ dz dθ
0
π/6
2 cos2 θ sin θ + 4 sin θ dθ
0
14
9√
2
π/6
−
3 = 0.770.
= − cos3 θ − 4 cos θ|θ=0 =
3
3
4
21. Since the sphere radius is 10, we have
~ = 102 sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Hence
~ · dA
~ = 100 (sin φ cos θ + 2 sin φ sin θ + 3 cos φ) sin φ dφ dθ.
F
~
The θφ-region corresponding to S, the upper hemisphere z ≥ 0, is given by 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2. The flux of F
through the surface S is given by
Z
S
~ · dA
~ =
F
Z
0
2π
Z
π/2
100 (sin φ cos θ + 2 sin φ sin θ + 3 cos φ) sin φ dφ dθ.
0
19.2 SOLUTIONS
1739
22. Since the sphere radius is 5, we have
~ = 52 sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Using x = 5 sin φ cos θ, y = 5 sin φ sin θ and z = 5 cos φ we have
F~ = 5 sin φ cos θ~i + 10 sin φ sin θ~j + 15 cos φ~k .
Hence
~ · dA
~ = 52 5 sin2 φ cos2 θ + 10 sin2 φ sin2 θ + 15 cos2 φ sin φ dφ dθ.
F
~ through the
The θφ-region corresponding to S, the entire sphere, is given by 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. The flux of F
surface S is given by
Z
S
~ · dA
~ =
F
2π
Z
0
Z
π
25 5 sin2 φ cos2 θ + 10 sin2 φ sin2 θ + 15 cos2 φ sin φ dφ dθ.
0
23. Since the sphere radius is 2, we have
~ = 22 sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Using z = 2 cos φ we have
~ = 4 cos2 φ~i .
F
Hence
~ · dA
~ = 16 cos2 φ sin2 φ cos θ dφ dθ.
F
The θφ-region corresponding to S, the part of the sphere with x ≥ 0, is given by −π/2 ≤ θ ≤ π/2, 0 ≤ φ ≤ π. The
flux of F~ through the surface S is given by
Z
S
~ · dA
~ =
F
π/2
Z
−π/2
Z
π
16 cos2 φ sin2 φ cos θ dφ dθ.
0
24. Since the sphere radius is 3, we have
~ = 32 sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Using x = 3 sin φ cos θ we have
~ = e3 sin φ cos θ ~k .
F
Hence
~ · dA
~ = 9 cos φ sin φe3 sin φ cos θ dφ dθ.
F
The θφ-region corresponding to S, the part of the sphere with y ≥ 0 and z ≤ 0, is given by 0 ≤ θ ≤ π, π/2 ≤ φ ≤ π.
~ through the surface S is given by
The flux of F
Z
S
~ · dA
~ =
F
Z
π
0
π
Z
9 cos φ sin φe3 sin φ cos θ dφ dθ.
π/2
25. Since the sphere radius is 20, we have
~ = 202 sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Using z = 20 cos φ we have
F~ = 20 cos φ~i .
Hence
~ · dA
~ = 8000 sin2 φ cos φ cos θ dφ dθ.
F
The θφ-region corresponding to S, the part of the sphere where x ≥ 0, y ≥ 0 and z ≥ 0 , is given by 0 ≤ θ ≤ π/2,
~ through the surface S is given by
0 ≤ φ ≤ π/2. The flux of F
Z
S
~ · dA
~ =
F
=
Z
π/2
0
Z
0
Z
π/2
8000 sin2 φ cos φ cos θ dφ dθ
0
π/2
8000
8000
cos θ dθ =
.
3
3
1740
Chapter Nineteen /SOLUTIONS
26. Since the sphere radius is 4, we have
~ = 42 sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Using x = 4 sin φ cos θ, y = 4 sin φ sin θ and z = 4 cos φ we have
~ = 4 sin φ sin θ~i − 4 sin φ cos θ~j + 4 cos φ~k .
F
Hence
~ = 64 cos2 φ sin φ dφ dθ.
F~ · dA
The θφ-region corresponding to the entire sphere is given by 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π. The flux of F~ through the surface
S is given by
Z
S
~ · dA
~ =
F
=
2π
Z
0
−
0
=
Z
π
64 cos2 φ sin φ dφ dθ
0
2π
Z
Z
2π
0
64
cos3 φ|πφ=0 dθ
3
256
128
dθ =
π.
3
3
27. Since the sphere radius is 1, we have
~ = sin φ cos θ~i + sin φ sin θ~j + cos φ~k sin φ dφ dθ.
dA
Using x = sin φ cos θ and y = sin φ sin θ we have
~ = sin φ cos θ~i + sin φ sin θ~j .
F
Hence
~ · dA
~ = sin3 φ cos2 θ + sin3 φ sin2 θ dφ dθ = sin3 φ dφ dθ.
F
~
The θφ-region corresponding to region above the cone φ = π/4 is given by 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4. The flux of F
through the surface S is given by
Z
S
~ =
F~ · dA
Z
π/4
0
= 2π
Z
0
Z
2π
sin3 φ dθ dφ = 2π
0
π/4
Z
π/4
sin3 φ dφ
0
sin φ(1 − cos2 φ) dφ
1
π/4
= 2π(− cos φ + cos3 φ)|φ=0
3
√
8−5 2
π = 0.486.
=
6
28. We find the equation for the plane S in the form z = f (x, y). The rectangle lies in the plane z + 2y = 4 we have
z = f (x, y) = 4 − 2y.
Thus, we have
~ = (−fx~i − fy~j + ~k ) dx dy = (0~i + 2~j + ~k ) dx dy = (2~j + ~k ) dx dy.
dA
The flux integral is therefore
Z
S
~ · dA
~ =
F
Z 2Z
0
0
2
(4 − 2y)~k · (2~j + ~k ) dx dy =
Z 2Z
0
0
2
4 − 2y dx dy = 8.
19.2 SOLUTIONS
1741
29. We find the equation for the plane S in the form z = f (x, y). The rectangle lies in the plane 3x + 2z = 6 so we have
z = f (x, y) = 3 −
3
x.
2
Thus, we have
The flux integral is therefore
Z
S
~ · dA
~ =
F
Z 2Z
0
2
0
3
3 − x ~k ·
2
3~
i + 0~j + ~k
2
~ = (−fx~i − fy~j + ~k ) dx dy =
dA
3~ ~
i +k
2
dx dy =
dx dy =
Z 2Z
0
3~ ~
i +k
2
2
3−
0
dx dy.
3
x dx dy = 6.
2
Problems
30. We find the equation for the plane S in the form z = f (x, y). We have
z = f (x, y) = 1 − x − y.
Thus, we have
~ = (−fx~i − fy~j + ~k ) dy dx = (−(−1)~i − (−1)~j + ~k ) dy dx = (~i + ~j + ~k ) dy dx.
dA
The surface S intersects the xy-plane in the line x + y = 1, so the region R in the xy-plane below S is bounded by the
x-axis, y-axis, and the line y = 1 − x. The flux integral is, therefore,
Z
S
Z 1Z
~ =
F~ · dA
0
0
(1 − x − y)~k · (~i + ~j + ~k ) dy dx
0
Z 1Z
=
1−x
1−x
(1 − x − y) dy dx =
0
Z
0
1
y2
y − xy −
2
1−x
dx =
Z
1
0
0
1
1
1
− x + x2 dx = .
2
2
6
~ = (−~i − ~j + ~k ) dx dy. As S is oriented upward, we have
31. Using z = f (x, y) = x + y, we have dA
Z
S
~ · dA
~ =
F
=
Z
3
0
Z
0
Z
2
((x − y)~i + (x + y)~j + 3x~k ) · (−~i − ~j + ~k ) dxdy
0
3
Z
0
2
(−x + y − x − y + 3x) dxdy =
Z
0
3
Z
2
x dxdy = 6.
0
32. Writing the surface S as z = f (x, y) = −y + 1, we have
~ = (−fx~i − fy~j + ~k )dxdy.
dA
Thus,
Z
S
~ =
F~ · dA
=
Z
Z
=
Z
=
Z
~ (x, y, f (x, y)) · (−fx~i − fy~j + ~k ) dxdy
F
R
1
0
Z
1
(2x~j + y~k ) · (~j + ~k ) dxdy
0
1
0
Z
1
(2x + y) dxdy =
0
1
(1 + y) dy = (y +
0
2
Z
y
)
2
1
1
(x2 + xy) dy
0
1
0
=
0
3
.
2
1742
Chapter Nineteen /SOLUTIONS
33. Writing the surface S as z = f (x, y) = y 2 + 5, we have
~ = (−fx~i + −fy~j + ~k )dxdy.
dA
Thus,
Z
S
Z
~ =
F~ · dA
Z
=
Z
=
R
~ (x, y, f (x, y)) · (−fx~i − fy~j + ~k ) dxdy
F
(−y~j + (y 2 + 5)~k ) · (−2y~j + ~k ) dxdy
R
1
0
Z
1
2
(3y + 5) dxdy =
−2
Z
1
(9y 2 + 15) dy
0
1
= (3y 3 + 15y)
= 18.
0
~ = (−zx~i − zy~j + ~k )dx dy = (~j + ~k )dx dy. The flux is
34. One the surface S we have z = −y + 1 and dA
Z
S
~ · dA
~ =
F
=
Z
1
0
Z
Z
0
((ln(x2 )~i + ex~j + cos(1 − (−y + 1))~k ) · (~j + ~k ) dx dy
0
1
0
=
1
Z
Z
1
(ex + cos y) dx dy =
1
1
(ex + x cos y)
0
0
1
Z
dy
0
1
(e + cos y − 1) dy = (ye + sin y − y)
0
= e + sin 1 − 1.
~ does not contribute to the flux. Since the ~i and ~j components
35. On the curved sides of the cylinder, the ~k component of F
are constant, these components contribute 0 to the flux on the entire cylinder. Therefore the only nonzero contribution to
~ = ~k dA, and from the ~k component through
the flux results from the ~k component through the top, where z = 2 and dA
~ = −~k dA:
the bottom, where z = −2 and dA
Flux =
Z
Top
=
Z
Top
=4
Z
~ · dA
~ +
F
Z
Bottom
2~k · ~k dA +
Top
~ · dA
~
F
Z
Bottom
(−2~k ) · (−~k dA)
dA = 4 · Area of top = 4 · π(32 ) = 36π.
36. On the curved side of the cylinder, only the components x~i + z~k contribute to the flux. Since x~i + z~k is perpendicular
to the curved surface and ||x~i + z~k || = 2 there (because the cylinder has radius 2), we have
Flux through sides = 2 · Area of curved surface = 2 · 2π · 2 · 6 = 48π.
~ = ~j dA; on the other end, y = −3 and
On the flat ends, only y~j contributes to the flux. On one end, y = 3 and dA
~ = −~j dA. Thus
dA
Flux through ends = Flux through top + Flux through bottom
= 3~j · ~j π(22 ) + (−3~j ) · (−~j π(22 )) = 24π.
So,
Total flux = 48π + 24π = 72π.
37. Writing the surface S as z = f (x, y) = −2x − 4y + 1, we have
~ = (−fx~i − fy~j + ~k )dxdy.
dA
19.2 SOLUTIONS
With R as shown in Figure 19.9, we have
Z
S
~ · dA
~ =
F
=
=
=
=
Z
Z
Z
Z
Z
R
~ (x, y, f (x, y)) · (−fx~i − fy~j + ~k ) dxdy
F
R
(3x~i + y~j + (−2x − 4y + 1)~k ) · (2~i + 4~j + ~k ) dxdy
(4x + 1) dxdy =
R
1
1
Z
Z
−2x+2
(4x + 1) dydx
0
0
(4x + 1)(−2x + 2) dx
0
1
(−8x2 + 6x + 2) dx = (−
0
8x3
+ 3x2 + 2x)
3
1
=
0
7
.
3
y
2
y = −2x + 2
R
✠
x
1
Figure 19.9
38. Writing the surface S as z = f (x, y) = 25 − x2 − y 2 , we have
~ = (−fx~i − fy~j + ~k )dxdy.
dA
Thus,
Z
S
~ =
F~ · dA
=
=
=
Z
Z
Z
Z
R
~ (x, y, f (x, y)) · (−fx~i − fy~j + ~k ) dxdy
F
R
(x~i + y~j ) · (2x~i + 2y~j + ~k ) dxdy
2(x2 + y 2 ) dxdy =
0
R
2π
0
2π
Z
5
r4
2
dθ =
0
Z
5
2r 2 r drdθ
0
625
(2π) = 625π.
2
39. Writing the surface S as z = f (x, y) = 25 − x2 − y 2 , we have
~ = (−fx~i − fy~j + ~k )dxdy = (2x~i + 2y~j + ~k )dxdy.
dA
Thus
Z
S
~ =
F~ · dA
=
Z
R
Z
cos(x2 + y 2 )~k · (2x~i + 2y~j + ~k ) dxdy
2
2
cos(x + y ) dxdy =
Z
0
2π
2π
0
R
=
Z
sin r 2
2
5
dθ = π sin 25.
0
Z
5
0
cos r 2 · r drdθ
1743
1744
Chapter Nineteen /SOLUTIONS
~ = (~i + ~j + ~k ) dx dy, so the downward one is
40. Using z = 1 − x − y, the upward pointing area element is dA
~
~
~
~
dA = (−i − j − k ) dx dy. Since S is oriented downward, we have
Z
S
Z
~ · dA
~ =
F
~
(x~i + y~j + z~k ) · dA
S
3
Z
=
0
Z
=
2
Z
(x~i + y~j + (1 − x − y)~k ) · (−~i − ~j − ~k ) dxdy
0
3
0
2
Z
0
(−x − y − 1 + x + y) dxdy = −6.
~ = (−2x~i − 2y~j + ~k ) dx dy. Since S is
41. Using z = x2 + y 2 , we find that the upward pointing area element is dA
~
~
~
~
oriented downward, we have dA = (2xi + 2y j − k ) dx dy, so
Z
S
~ =
F~ · dA
=
=
Z
S
~
(x~i + y~j + z~k ) · dA
Z
(x~i + y~j + (x2 + y 2 )~k ) · (2x~i + 2y~j − ~k ) dx dy
ZDisk
(2x2 + 2y 2 − x2 − y 2 ) dx dy =
Disk
=
2π
Z
0
42. Here z =
p
9 − x2 − y 2 , so
S
~ =
F~ · dA
=
Z p
9−
x
S
Z
3
−3
r4
r r dr dθ = 2π ·
4
0
x2
Z √9−x2
−
√
−
1
2
x
zx = − p
9 − x2 − y 2
The flux integral is given by
Z
1
Z
+ y~k
y 2~i
·
(x2 + y) dydx
0
Z
(x2 + y 2 ) dx dy
Disk
π
= .
2
y
zy = − p
.
9 − x2 − y 2
x
y
p
9 − x2 − y 2
~i + p
~j + ~k
9 − x2 − y 2
!
dxdy
9−x2
Changing to polar coordinates gives
Z
S
~ · dA
~ =
F
=
2π
Z
Z
0
Z
3
(r 2 cos2 θ + r sin θ) rdrdθ
0
2π
81
4
0
cos2 θ +
27
sin θ
3
dθ =
81
π.
4
43. We have 0 ≤ z ≤ 6 so 0 ≤ x2 + y 2 ≤ 36. Let R be the disk of radius 6 in the xy-plane centered at the origin. Because
of the cone’s point, the flux integral is improper; however, it does converge. We have
Z
S
~ · dA
~ =
F
=
Z
R
Z
R
(−x
p
x2 + y 2~i − y
x
·
=
~ (x, y, f (x, y)) · (−fx~i − fy~j + ~k ) dxdy
F
Z
R
p
x2 + y 2~j + (x2 + y 2 )~k )
y
~i − p
~j + ~k
−p
x2 + y 2
x2 + y 2
2(x2 + y 2 ) dxdy
!
dxdy
19.2 SOLUTIONS
6
Z
=2
0
= 4π
1745
2π
Z
r 3 dθdr
0
Z
6
r 3 dr = 1296π.
0
44. Since y = f (x, z) = x2 + z 2 , we have
~ = (−fx~i + ~j − fz ~k ) dxdz = (−2x~i + ~j − 2z~k ) dxdz.
dA
~ , we have
Thus, substituting y = x2 + z 2 into F
Z
S
~ · dA
~ =
F
=
Z
x2 +z 2 ≤1
Z
x2 +z 2 ≤1
=
Z
1
Z
1
−1
=
−1
((x2 + z 2 )~i + ~j − xz~k ) · (−2x~i + ~j − 2z~k ) dxdz
(−2x3 − 2xz 2 + 1 + 2xz 2 ) dxdz
Z √1−z2
(1 − 2x3 ) dxdz
√
1−z 2
Z 1 Z √1−z2
Z √1−z2
dxdz −
2x3 dxdz
√
√
−
1−z 2
−
= Area of disk −
−1
Z
1
−1
x
4
2
√
−
−
1−z 2
1−z 2
√
1−z 2
dz = π − 0 = π
45. The plane through the points (1, 0, 0), (0, 1, 0), and (0, 0, 1) is given by x + y + z = 1, so S is the part of the graph of
z = f (x, y) = 1 − x − y above the region R in the xy-plane where x ≥ 0, y ≥ 0, and x + y ≤ 1. Thus
Z
~ · dA
~ =
F
S
=
=
Z
R
Z
ZR
F~ (x, y, f (x, y)) · (−fx~i − fy~j + ~k ) dx dy
(x2~i + y 2~j + (1 − x − y)2~k ) · (~i + ~j + ~k ) dx dy
(x2 + y 2 + (1 − x − y)2 ) dx dy
R
1
=
Z
0
=
Z
1
Z
1−x
(1 + 2x2 + 2y 2 − 2x − 2y + 2xy) dy dx
0
2
(1 − x)3 − 2x(1 − x)
3
1
−(1 − x)2 + x(1 − x)2 ] dx = .
4
[(1 − x) + 2x2 (1 − x) +
0
46. Since the radius of the cylinder is 1, using cylindrical coordinates we have
~ = (cos θ~i + sin θ~j )dθdz.
dA
Thus,
Z
S
~ =
F~ · dA
=
Z
6
0
Z
0
Z
2π
(cos θ~i + sin θ~j ) · (cos θ~i + sin θ~j ) dθ dz
0
6
Z
0
2π
1 dθ dz = 12π.
1746
Chapter Nineteen /SOLUTIONS
~ = (cos θ~i + sin θ~j )dθdz. Thus,
47. Since the radius of the cylinder is 1, using cylindrical coordinates we have dA
Z
S
6
Z
~ · dA
~ =
F
0
(z cos θ~i + z sin θ~j + z 3~k ) · (cos θ~i + sin θ~j ) dθ dz
0
6
Z
=
2π
Z
2π
Z
0
z dθ dz = 2π
0
z2
2
6
= 36π.
0
~ through S is given by
48. The flux of F
Z
S
Z
~ · dA
~ =
F
2π
0
Z
=
π/2
Z
(2 cos φ~k ) · (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )22 sin φ dφ dθ
0
2π
π/2
Z
8 sin φ cos2 φdφdθ = 16π(
φ=0
θ=0
− cos3 φ
)
3
π/2
=
φ=0
16π
.
3
√
√
49. A parameterization for the surface S is given by z = 1 − x2 − y 2 over R for − 1 − x2 ≤ y ≤ 1 − x2 , −1 ≤ x ≤
1. Thus
x
y
zx = − p
and zy = − p
,
1 − x2 − y 2
1 − x2 − y 2
so
p
Z
S
~ · dA
~ =
F
Z
=
Z
R
(y~i − x~j + z~k ) · (−zx~i − zy~j + ~k ) dx dy
1
−1
=
Z
Z √1−x2
−
2π
0
√
1
Z
yx − xy
p
1 − x2 − y 2
1−x2
p
+ z dx dy =
h
1 − r 2 r dr dθ = 2π −
0
Z
1
−1
Z √1−x2 p
−
√
1 2
· (1 − r 2 )3/2
2 3
1−x2
i1
0
=
1 − x2 − y 2 dy dx
2π
.
3
50. Since the radius of the sphere is 5, using spherical coordinates we have
~ = (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )25 sin φ dθ dφ.
dA
Thus,
Z
S
~ =
F~ · dA
Z
π
2
0
= 625
Z
2π
(25 cos2 φ~k ) · (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )25 sin φ dθ dφ
0
Z
π
2
0
2π
Z
cos3 φ sin φ dθdφ
0
(cos φ)4
= −1250π
4
π
2
=
0
625
π.
2
51. Since the radius of the sphere is a, using spherical coordinates we have
~ = (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )a2 sin φ dφ dθ.
dA
Thus,
Z
S
~ =
F~ · dA
Z
2π
0
π
Z
(a sin φ cos θ~i + a sin φ sin θ~j + a cos φ~k ) ·
0
(sin φ cos θ~i + sin φ sin θ~j + cos φ~k )a2 sin φ dφ dθ
=a
3
Z
2π
0
= 2πa3
Z
π
sin φ dφ dθ
0
Z
π
sin φ dφ = (2πa3 )(2) = 4πa3 .
0
19.2 SOLUTIONS
1747
~ does not contribute to the flux as it is perpendicular to the surface. The vector field x~i + y~j is
52. The ~k -component of F
everywhere perpendicular to S and has constant magnitude ||x2 + y 2 || = 1 on the surface S. Thus
Z
S
~ =
F~ · dA
Z
~ = 1 · Area of S = 1 π = π .
(x~i + y~j ) · dA
2
2
S
Alternatively, the flux can be computed
√ by integrating with respect to x and z, treating y as a function of x and z. A
parameterization of S is given by y = 1 − x2 , 0 ≤ x ≤ 1,0 ≤ y ≤ 1, 0 ≤ z ≤ 1. Thus,
Z
S
~ · dA
~ =
F
=
1
Z
0
1
Z
(x~i +
Z
1
(x~i +
0
1
Z
0
=
1
0
0
=
Z
Z
1
0
1
Z
0
Z
1
√
√
0
p
1 − x2~j + z~k ) · (−yx~i + ~j − yz ~k ) dx dz
p
1 − x2~j + z~k ) ·
p
x2
+ 1 − x2
2
1−x
√
x
~i + ~j + 0~k
1 − x2
dx dz
dx dz
1
dx dz
1 − x2
1
= 1 · arcsin x
π
.
2
=
0
53. We integrate with respect the y and z, thinking of x as a function of y and z. Since x = sin y sin z, we have xy =
cos y sin z and xz = sin y cos z. The surface is oriented in the direction of increasing x, so
Z
S
~ =
F~ · A
=
Z
π/2
0
Z
π/2
Z
π/2
Z
π/2
Z
π/2
0
(sin y sin z~i + ~j + ~k ) · (~i − cos y sin z~j − sin y cos z~k ) dy dz
Z
π/2
(sin y sin z − cos y sin z − sin y cos z) dy dz
0
π/2
− cos y sin z − sin y sin z + cos y cos z
0
=
π/2
Z
0
0
=
~ · (~i − xy~j − xz ~k ) dy dz
F
0
0
=
π/2
Z
dz
0
π/2
(sin z − sin z − cos z) dz = − sin z
= −1.
0
54. We integrate with respect the x and z, treating y as a function of x and z. Since y = x2 + z 2 , we have yx = 2x and
yz = 2z. The region of integration in the xz-plane is given by x2 + z 2 = 1, x ≥ 0, z ≥ 0. The orientation is toward the
xz-plane, so we have
Z
Z Z √
1−z 2
1
S
~ · dA
~ =
F
=
0
Z
1
0
=
Z
1
0
0
~ · (yx~i − ~j + yz ~k ) dx dz
F
0
((x + z)~i + ~j + z~k ) · (2x~i − ~j + 2z~k ) dx dz
0
(2x2 + 2xz − 1 + 2z 2 ) dx dz.
Z √1−z2
Z √1−z2
Using polar coordinates with x = r cos θ, z = r sin θ,
Z
S
~ · dA
~ =
F
Z
0
π/2
Z
0
1
(2r 2 + 2r 2 cos θ sin θ − 1)r dr dθ
1748
Chapter Nineteen /SOLUTIONS
=
Z
π/2
Z
π/2
0
=
1
r4
r4
r2
+
cos θ sin θ −
2
2
2
0
1
1 (sin θ)
cos θ sin θ dθ =
2
2
2
0
2
!
dθ
π/2
=
1 2
1
1 = .
4
4
0
~ = ~k dx dy, so
55. On the disk, z = 0 and dA
Z
S
~ =
F~ · dA
=
Z
x2 +y 2 ≤1
Z
(xzeyz~i + x~j + (5 + x2 + y 2 )~k ) · ~k dx dy
(5 + x2 + y 2 ) dx dy =
x2 +y 2 ≤1
= 2π
0
r4
5r 2
+
2
4
56. The plane is x − z = 0 over region 0 ≤ x ≤
√
Z
1
=
0
2π
Z
1
(5 + r 2 )r dr dθ
0
11π
.
2
2, 0 ≤ y ≤ 2. See Figure 19.10.
z
2
2
y
√
2
x
Figure 19.10
Flux =
Z
2
0
=
Z
0
√
Z
0
2
Z
0
2
(exy + 3z + 5)~i + (exy + 5z + 3)~j + (3z + exy )~k · (~i − ~k ) dx dy
√
2
√
√
(exy + 3z + 5 − 3z − exy ) dx dy = 5(2)( 2) = 10 2
√
√
~ ||, we have
Alternatively, since a unit normal to the surface is ~n / 2 = (~i − ~j )/ 2, writing dA = ||dA
~ − ~k
5
~ · i√
dA =
√ dA
H
2
2
S
√
5
5
= √ (Area of slanted square) = √ 4 = 10 2.
2
2
Flux =
Z
~ · dA
~ =
H
Z
Z
57. (a) The charge is contained in a sphere of radius a centered at the origin, and uniformly distributed through the region
enclosed by the sphere.
~ = E(ρ)~e ρ . Let S be a sphere of
(b) Since ~e ρ is the unit vector outward normal to the sphere of radius ρ, we have E
fixed radius ρ, centered at the origin. Then
Z
W
δ dV =
(
4
πρ3 δ0
3
4
πa3 δ0
3
ρ≤a
ρ > a.
19.2 SOLUTIONS
1749
~ = ~e ρ dA, we have
On the other hand, since on the sphere dA
Z
S
~ · dA
~ =
E
Z
S
E(ρ)~e ρ · ~e ρ dA = E(ρ)
Z
dA = E(ρ)4πρ2 .
S
Therefore, by Gauss’s Law,
2
E(ρ)4πρ =
(
k 43 πρ3 δ0
ρ≤a
k 43 πa3 δ0
ρ > a.
~ = E(ρ)~e ρ , simplifying gives
Since E
~ =
E
k δ0 ρ~e ρ
3 3
k δ0 a ~e ρ
3r 3
ρ≤a
ρ > a.
58. (a) The charge is contained in a cylinder of radius a centered at the origin, and uniformly distributed through the region
enclosed by the cylinder.
~ = E(r)~e r . Let S be a cylinder
(b) Since ~e r is the unit outward pointing normal to the cylinder of radius r, we have E
of fixed radius r, height 1, centered along the z-axis. If r ≤ a,
Z
δ dV = πr 2 δ0 ,
Z
δ dV = πa2 δ0 .
W
and if r > a
W
~ = ~e r dA on S, we can write
On the other hand, since dA
Z
S
~ · dA
~ =
E
Z
S
E(r)(~e r · ~e r )dA = E(r)
Z
dA = E(r)2πr.
S
(The flux across the top and bottom of the cylinder is zero.) So, by Gauss’s Law
E(r)2πr =
2
kπr δ0
~ = E(r)~e r , simplifying gives
Since E
~ =
E
kπa2 δ0
1 kδ0 r~e r
2
a
1
kδ0
2
2
r
~e r
if r ≤ a
if r > a.
if r ≤ a
if r > a.
Strengthen Your Understanding
~ = cos θ~i + sin θ~j R dz dθ applies only to surfaces that lie on a cylinder centered on the z-axis of
59. The formula dA
constant radius R. The distance from the cone to the z-axis is not constant, but depends on the point on the cone from
which you measure the distance.
~ , so
60. The orientation vector ~n is a unit vector in the direction of dA
and
1
−fx~i − fy~j + ~k
~
n = p
2
2
fx + fy + 1
dA =
p
fx2 + fy2 + 1 dx dy.
1750
Chapter Nineteen /SOLUTIONS
61. For a surface z = f (x, y) oriented upwards, we have
~ = −fx~i − fy~j + ~k
dA
If f (x, y) = −x − y, then fx = −1, fy = −1, and
dx dy.
~ = (~i + ~j + ~k ) dx dy.
dA
~ = x~i + y~j in polar coordinates is
62. The vector field F
~ (r, θ, z) = r cos θ~i + r sin θ~j .
F
For a surface S on the cylinder r = 10, oriented outwards, we have
Z
S
~ · dA
~ =
F
=
=
Z
ZT
ZT
T
F~ (10, θ, z) · cos θ~i + sin θ~j 10 dz dθ
10 cos θ~i + 10 sin θ~j
100 dz dθ = 100 ·
Z
· cos θ~i + sin θ~j 10 dz dθ
dz dθ,
T
R
where T is the θz-region corresponding to S. If, for example, 0 ≤ θ ≤ 2 and 0 ≤ z ≤ 3, then T dz dθ is 6. Hence
~ · dA
~ = 600 for the surface S parameterized by r = 10, 0 ≤ θ ≤ 2, 0 ≤ z ≤ 3, oriented outwards.
F
S
An intuitive geometric solution is also possible. On the cylinder x2 + y 2 = 102 of radius 10, the vector field
~ k = 10 and is normal to the cylinder, pointing in the orientation direction. Hence, for a
F~ = x~i + y~j has magnitude kF
surface S on the cylinder oriented away from the z-axis, we have
R
Z
S
~ · dA
~ = 10 · Area of S.
F
The region S on the cylinder r = 10R with 0 ≤ θ ≤ 2, 0 ≤ z ≤ 3 is a curved rectangle with base 10∆θ = 10 · 2 = 20
~ · dA
~ = 600.
and height 3, so has area 60. Hence S F
~
~
~
~
63. True. The area vector for the graph of f (x, y), parametrized in the usual
p way, is given by A = fx i + fy j + k . The
2
2
~
surface area is then the double integral of the magnitude of A , namely fx + fy + 1 over the given rectangle.
~ (x, y) and B
~ (x, y) both point upward. But they could point in different
64. False. Both surfaces are oriented upward, so A
directions, since the graph of z = −f (x, y) is the graph of z = f (x, y) turned upside down.
65. False. The total flux can be 0 without the vector field always being perpendicular to the surface. For example, if F (x, y, z) =
~k , then the flux is zero over the sphere, but F is not perpendicular to the sphere except at the north and south poles.
66. (a) The surface can be considered as made up of two rectangles, the original one and another “next door” along the
x-axis. Because the vector field is independent of x, the flux through both rectangles are the same. Thus, the flux has
doubled.
(b) Because the vector field does not depend on x, the flux is unchanged.
(c) The sign is reversed, but the magnitude of the flux remains the same.
(d) Tripling each side of the rectangle multiplies its area by 9. However, the surface now extends further up the z-axis,
where the vector field is not given. If the vector field is larger further up the z-direction (as suggested by the diagram),
then the flux has multiplied by a factor of more than 9.
Solutions for Section 19.3
Exercises
1. Scalar. div (x2 + y)~i + (xyez )~j − (ln(x2 + y 2 ))~k
= 2x + xez
2
2
2. Scalar. div((2 sin(xy) + tan z)~i + (tan y)~j + (ex +y )~k ) = 2y cos(xy) + 1/ cos2 y.
3. The first vector field appears to be diverging more at the origin, since both fields are zero at the origin and the vectors near
the origin are larger in field (I) than they are in field (II).
19.3 SOLUTIONS
1751
4. (a) Positive. The inflow from the lower left is less than the outflow from the upper right. The net outflow is positive.
(b) Zero. The inflow on the right side is equal to outflow on the left.
(c) Negative. The inflow from above is greater than the outflow below. The net outflow is negative.
~ = ∂ (−y) + ∂ (x) = 0
5. div F
∂x
∂y
∂
∂
~ =
6. div F
(−x) +
(y) = −1 + 1 = 0
∂x
∂y
~ = ∂ (−x + y) + ∂ (y + z) + ∂ (−z + x) = −1 + 1 − 1 = −1
7. div F
∂x
∂y
∂z
2
2
∂
∂
~
(x − y ) + (2xy) = 2x + 2x = 4x
8. div F =
∂x
∂y
9. We have
div(3x2~i − sin(xz)(~i + ~k )) = 6x − z cos(xz) − x cos(xz).
10. We have
2x
∂F1
∂
ln x2 + 1 = 2
=
,
∂x
∂x
x +1
∂F2
∂
=
(cos y) = − sin y,
∂y
∂y
∂
∂F3
=
(xyez ) = xyez .
∂z
∂z
So,
~
div F
∂F1
∂F2
∂F3
+
+
∂x
∂y
∂z
2x
− sin y + xyez .
= 2
x +1
=
11. Using the formula for ~a × ~r in Cartesian coordinates, we get
~ =
div F
∂
∂
∂
(a2 z − a3 y) +
(a3 x − a1 z) +
(a1 y − a2 x) = 0
∂x
∂y
∂z
12. Taking partial derivatives, we get
~ = ∂
div F
∂x
−y
(x2 + y 2 )
+
∂
∂y
x
(x2 + y 2 )
=
2yx
2xy
− 2
= 0.
(x2 + y 2 )2
(x + y 2 )2
13. In coordinates, we have
(x − x0 )
(y − y0 )
~ (x, y, z) = p
~i + p
~j
F
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
So if (x, y, z) 6= (x0 , y0 , z0 ), then
(z − z0 )
~k .
+p
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
(x − x0 )2
p
−
2
((x − x0 ) + (y − y0 )2 + (z − z0 )2 )3/2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
1
div F~ =
+
+
!
(y − y0 )2
p
−
((x − x0 )2 + (y − y0 )2 + (z − z0 )2 )3/2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
1
(z − z0 )2
p
−
2
((x − x0 ) + (y − y0 )2 + (z − z0 )2 )3/2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
1
!
!
1752
Chapter Nineteen /SOLUTIONS
=
+
+
(x − x0 )2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
−
2
2
2
3/2
2
((x − x0 ) + (y − y0 ) + (z − z0 ) )
((x − x0 ) + (y − y0 )2 + (z − z0 )2 )3/2
(y − y0 )2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
−
2
2
2
3/2
2
((x − x0 ) + (y − y0 ) + (z − z0 ) )
((x − x0 ) + (y − y0 )2 + (z − z0 )2 )3/2
(x − x0 )2 + (y − y0 )2 + (z − z0 )2
(z − z0 )2
−
2
2
2
3/2
2
((x − x0 ) + (y − y0 ) + (z − z0 ) )
((x − x0 ) + (y − y0 )2 + (z − z0 )2 )3/2
3((x − x0 )2 + (y − y0 )2 + (z − z0 )2 ) − ((x − x0 )2 + (y − y0 )2 + (z − z0 )2 )
((x − x0 )2 + (y − y0 )2 + (z − z0 )2 )3/2
2
2
.
= p
=
2
2
2
k~
r
−
~r 0 k
(x − x0 ) + (y − y0 ) + (z − z0 )
=
Problems
14. Two vector fields that have positive divergence everywhere are in Figures 19.11 and 19.12.
y
y
x
Figure 19.11
x
Figure 19.12
15. Two vector fields that have negative divergence everywhere are in Figures 19.13 and 19.14.
y
y
x
Figure 19.13
x
Figure 19.14
16. Two vector fields that have zero divergence everywhere are in Figures 19.15 and 19.16.
19.3 SOLUTIONS
y
1753
y
x
x
Figure 19.15
Figure 19.16
~ at a point is approximately equal to the flux density out of a small region around the point, at (2, 3, −1) we
17. Since div G
have
~ ≈ Flux out of small region = −0.00004π = −0.030.
div G
Volume of region
4π(0.1)3 /3
Note that since the original flux was given into the region, we take the negative to get the flux out of the region.
18. Since div F (1, 2, 3) is the flux density out of a small region surrounding the point (1, 2, 3), we have
~ (1, 2, 3) ≈ Flux out of small region around (1, 2, 3)
div F
Volume of region.
So
Flux out of region ≈ (div F~ (1, 2, 3)) · Volume of region
4
= 5 · π(0.01)3
3
0.00002π
.
=
3
19. (a)
(i) By the definition of divergence, if S is a sphere centered at (2, 0, 0), we have
~ · dA
~
F
S
.
div F~ = lim
Vol→0 Volume of S
R
Thus, if the sphere is small
Z
S
~ · Volume of S.
~ · dA
~ ≈ div F
F
~ = x2 + y 2 − z, if S is the sphere given
Since div F
Z
~ · dA
~ ≈ (22 + 02 − 0) · 4 π(0.1)3 = 0.016 π.
F
3
3
S
(ii) The relationship between flux, divergence, and volume holds when S is cube. Thus
Z
S
~ · Volume of S,
~ · dA
~ ≈ div F
F
gives
Z
S
~ · dA
~ ≈ (02 + 02 − 10) · (0.2)3 = −0.08.
F
(b) The point (2, 0, 0) is a source because the flux out of a small region around the point is positive; (0, 0, 10) is a sink
because the flux out of a small region around the point is negative.
1754
Chapter Nineteen /SOLUTIONS
~ at a point is approximately equal to the flux density out of a small region around the point, at (4, 5, 2),
20. (a) Since div F
we have
0.0125
Flux out of small region
=
= 2.984.
div F~ ≈
Volume of region
(4/3)π(0.1)3
(b) The flux through the sphere is approximated by
Z
Flux through sphere =
S
~ · dA
~ ≈ div F
~ · Volume of sphere = 2.984 · 4 π(0.2)3 = 0.100.
F
3
We could also estimate the flux by noticing that it is eight times the original flux, that is, 8(0.0125) = 0.100.
21. (a) On S1 , x = a and normal is in negative x-direction, so
~ · ∆A
~ = (2a~i − 3y~j + 5z~k ) · (−∆A~i ) = −2a∆A.
F
Thus
Z
~ · dA
~ =
F
S1
Z
S1
−2adA = −2a(Area of S1 ) = −2aw2 .
On S2 , x = a + w and normal is in positive x-direction, so
~ · ∆A
~ = (2(a + w)~i − 3y~j + 5z~k ) · (∆A~i ) = 2(a + w)∆A.
F
Thus
Z
S2
~ · dA
~ =
F
Z
2(a + w)dA = 2(a + w)(Area of S2 ) = 2(a + w)w2
S2
Calculating the flux through the other sides similarly, we get
=
Total
flux
Z
S1
~ · dA
~ +
F
Z
S2
~ +
F~ · dA
Z
S3
~ · dA
~ +
F
Z
S4
~ · dA
~ +
F
Z
S5
~ · dA
~ +
F
= −2aw2 + 2(a + w)w2 + 3bw2 − 3(b + w)w2 − 5cw2 + 5(c + w)w2
Z
S6
~
F~ · dA
= (2w − 3w + 5w)w2 = 4w3 .
(b) To find div F~ at the point (a, b, c), let the box shrink to the point by letting w → 0. Then
~ = lim
div F
w→0
= lim
w→0
Flux through box
Volume of box
4w3
= 4.
w3
(c)
~ = ∂ (2x) + ∂ (−3y) + ∂ (5z) = 2 − 3 + 5 = 4.
div F
∂x
∂y
∂z
22. Using flux: On S1 , x = a and normal is in negative x-direction, so
~ · ∆A
~ = ((3a + 2)~i + 4a~j + (5a + 1)~k ) · (−∆A~i ) = −(3a + 2)∆A
F
Thus
Z
S1
~ =
F~ · dA
Z
S1
−(3a + 2)dA = −(3a + 2)(Area of S1 ) = −(3a + 2)w2 .
On S2 , x = a + w and normal is in the positive x-direction, so
~ = [(3(a + w) + 2)~i + 4(a + w)~j + (5(a + w) + 1) ~k ] · (∆A~i ) = (3a + 3w + 2)∆A.
F~ · ∆A
Thus
Z
S2
~ =
F~ · dA
Z
(3a + 3w + 2)dA = (3a + 3w + 2)(Area of S2 ) = (3a + 3w + 2)w2 .
S2
19.3 SOLUTIONS
S3
~ =
F~ · dA
1755
~ · dA
~ =
F
4xdA. Since these two are integrated over the same
S4
R
R
R
~ cancels out
~
~ =
~ =
−(5x + 1)dA
region in the xz-plane, the two integrals cancel. Similarly, S F · dA
F~ · dA
S5
S6
5
R
~ . Therefore,
(5x + 1)dA
Next, we have
R
R
S3
−4xdA and
R
R
S4
S6
Total
Z flux
=
~ · dA
~ +
F
S1
Z
S2
~ · dA
~ +
F
Z
S3
2
~ · dA
~ +
F
2
= −(3a + 2)w + (3a + 3w + 2)w +
+
Z
S5
−(5x + 1)dA +
Z
Z
S3
Z
S4
~ +
F~ · dA
−4xdA +
Z
Z
S5
~ · dA
~ +
F
Z
S6
~ · dA
~
F
4xdA
S4
(5x + 1)dA = 3w3 .
S6
To find div F~ at the point (a, b, c), let the box shrink to the point by letting w → 0. Then
Flux through box
w→0
Volume of box
3
3w
= 3.
= lim
w→0
w3
~ = lim
div F
Using partial derivatives:
div F~ =
∂
∂
∂
(3x + 2) +
(4x) +
(5x + 1) = 3
∂x
∂y
∂z
23. Figure 19.17 shows a two dimensional cross-section of the vector field ~v = −2~r . The vector field points radially inward,
so if we take S to be a sphere of radius R centered at the origin, oriented outward, we have
~ = −2R k∆A
~ k,
~v · ∆A
~ on the sphere. Therefore,
for a small area vector ∆A
Z
S
~ =
~v · dA
Z
S
~ k = −2R(Surface area of sphere) = −2R(4πR2 ) = −8πR3 .
−2R kdA
Thus, we find that
div ~v (0, 0, 0) = lim
vol→0
~
~v · dA
Volume of sphere
R
S
!
= lim
R→0
−8πR3
4
πR3
3
= −6.
Notice that the divergence is negative. This is what you would expect, since the vector field represents an inward flow at
the origin.
Since ~v = −2~r = −2x~i − 2y~j − 2z~k , the coordinate definition give
div ~v =
∂
∂
∂
(−2x) +
(−2y) +
(−2z) = −2 − 2 − 2 = −6.
∂x
∂y
∂z
y
−R
R
x
Figure 19.17: The vector field ~v = −2~r
1756
Chapter Nineteen /SOLUTIONS
24. (a) Since a is a constant,
∂
div grad f =
∂x
∂f
∂x
∂
+
∂y
∂f
∂y
=
∂
∂
(ay + 2axy) +
(ax + ax2 + 3y 2 ) = 2ay + 6y.
∂x
∂y
(b) Since div grad f = (2a + 6)y, we have div grad f = 0 for all x, y if a = −3.
25. (a) Since
~ = ∂ (9a2 x + 10ay 2 ) + ∂ (10z 3 − 6ay) + ∂ (−3z − 10x2 − 10y 2 ) = 9a2 − 6a − 3,
div F
∂x
∂y
∂z
~ = 0 when
we have div F
9a2 − 6a − 3 = 0
3(3a + 1)(a − 1) = 0
1
a=−
or a = 1.
3
(b) Taking the derivative with respect to a to locate the minimum gives
d
(9a2 − 6a − 3) = 18a − 6 = 0
da
1
.
3
~ is an upward-opening parabola.
This value of a gives a minimum because the expression for div F
a=
26. Take S to be a small sphere of radius R centered at the origin. Then on S we have
~k=
kF
~k
kR
R
1
= 3 = 2.
R3
R
R
~ points radially outward, F
~ is parallel to the outward normal on the surface, so
In addition, since F
~ = kF
~ k∆A = 1 ∆A.
F~ · ∆A
R2
Thus,
Z
S
~ · dA
~ =
F
=
Z
S
1
1
dA = 2
R2
R
Z
dA
S
1
1
· (Area of sphere) = 2 (4πR2 ) = 4π.
R2
R
The divergence is therefore given by
~ = lim
div F
Vol→0
Flux out of sphere
= lim
Volume inside sphere
R→0
4π
4
3
πR
3
= lim
R→0
3
R3
.
Since this limit is infinite, or undefined, we say that divergence of this vector field is not defined at the origin.
~ = ∂ (−y) + ∂ (x) + ∂ (x + y) = 0, so this could be a magnetic field.
27. (a) div B
∂x
∂y
∂z
∂
∂
∂
~
(−z) +
(y) +
(x) = 0 + 1 + 0 = 1, so this could not be a magnetic field.
(b) div B =
∂x
∂y
∂z
~ = ∂ (x2 − y 2 − x) + ∂ (y − 2xy) + ∂ (0) = 2x − 1 + 1 − 2x + 0 = 0, so this could be a magnetic field.
(c) div B
∂x
∂y
∂z
28. (a) We have
~ = ∂z = 1.
div F
∂z
(b) Above the xy plane, the vector field consists of vectors pointing vertically upward, getting longer as you go up.
Below the xy plane, it consists of vectors pointing vertically downward, getting longer as you go down. You can
clearly see the divergence on the xy plane, since vectors on either side of it point in opposite directions, but it is not
so clear elsewhere. However, the fact that the vectors are getting longer as you go up means that the flux through a
cube situated above the xy plane will be non-zero, since the flux out of its top face will be greater than the flux into
the bottom face.
19.3 SOLUTIONS
1757
29. (a) In Cartesian coordinates,
~ (x, y, z) =
F
x
y
z
~k .
~i +
~j +
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
So if (x, y, z) 6= (0, 0, 0), then
div F~ (x, y, z) =
+
=
+
+
1
3x2
− 2
2
2
2
3/2
(x + y + z )
(x + y 2 + z 2 )5/2
3z 2
1
− 2
2
2
2
3/2
(x + y + z )
(x + y 2 + z 2 )5/2
x2 + y 2 + z 2
3y 2
− 2
2
2
2
5/2
(x + y + z )
(x + y 2 + z 2 )5/2
3x2
x2 + y 2 + z 2
− 2
2
2
2
5/2
(x + y + z )
(x + y 2 + z 2 )5/2
3z 2
x2 + y 2 + z 2
− 2
2
2
2
5/2
(x + y + z )
(x + y 2 + z 2 )5/2
3(x2 + y 2 + z 2 ) − 3(x2 + y 2 + z 2 )
(x2 + y 2 + z 2 )5/2
= 0.
+
1
3y 2
− 2
2
2
2
3/2
(x + y + z )
(x + y 2 + z 2 )5/2
=
z
(b)
x
Figure 19.18: The vector field F~ (~r ) =
shown in the xz-plane
~r
k~r k3
The vector field is radial (all the arrows point out), so you might think that it is has non-zero divergence. (See
Figure 19.18.) However the fact that the divergence is 0 at every point shows that flux density out of any small
volume around a point must be 0. This is possible because the arrows also get shorter as you go out.
30. The charges that produce this electric field are concentrated along two vertical lines, one near x = −1 and the another
one near x = 1. This is seen by the change in direction of the field at those lines. Near x = −1 the field is being repulsed
by the line (seen by the field going away from the line), and the charge is therefore positive. Near x = 1 the field is being
attracted to the line (seen by the field going toward the line), and the charge is therefore negative.
31. (a) Translating the vector field into rectangular coordinates gives, if (x, y, z) 6= (0, 0, 0)
~ (x, y, z) =
E
ky
kz
kx
~k .
~i +
~j +
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
We now take the divergence of this to get
~ =k
div E
−3
3
x2 + y 2 + z 2
+ 2
2
(x + y 2 + z 2 )5/2
(x + y 2 + z 2 )3/2
= 0.
(b) Let S be the surface of a sphere centered at the origin. We have seen that for this field, the flux
R
~ · dA
~ . Then
for all such spheres, regardless of their radii. So let the constant c stand for E
~ (0, 0, 0) = lim
div E
vol→0
~ · dA
~
E
c
= lim
.
Volume inside S
vol→0 Volume
R
R
~ · dA
~ is the same
E
1758
Chapter Nineteen /SOLUTIONS
(c) For a point charge, the charge density is not defined. The charge density is 0 everywhere else.
32. (a) The velocity vector for the traffic flow would look like:
(b) When 0 ≤ x < 2000, the velocity is decreasing linearly from 55 to 15, so its formula is (55 − x/50)~i mph. Then,
when 2000 ≤ x < 7000, the speed is constant, so ~v (x) = 15~i mph. Next, when 7000 ≤ x < 8000, the velocity
is increasing linearly from 15 to 55, so ~v (x) = (15 + (x − 7000)/25)~i mph. Finally, when x ≥ 8000, the speed is
constant, so ~v (x) = 55~i mph.
(c) div ~v = dv(x)/dx.
At x = 1000, v(x) = 55 − x/50, so div ~v = −1/50.
At x = 5000, v(x) = 15, so div ~v = 0.
At x = 7500, v(x) = 15 + (x − 7000)/25, so div ~v = 1/25.
At x = 10, 000, v(x) = 55, so div ~v = 0.
In each case the units of div ~v are miles/hour
.
feet
33. (a) Usually, the distance between cars is more at higher speeds and less at lower speeds. The cars are traveling the fastest
at x = 0, so at that point, the traffic should be the least dense. Thus,
ρ(0) < ρ(1000) < ρ(5000)
(b) Since ρ is in cars/mile, ~v is in miles/hour~v is in km/hour ρ~v is in cars/hour. The vector quantity ρ~v gives the number
of cars passing through a fixed point in a time interval.
(c) Pick any two points on the highway, x = a and x = b (a < b). We expect ρ~v to be the same at both places. This
is because if more cars pass through a than b, that would mean cars are disappearing (or at least stopping, which we
know is not the case since the velocity field is not 0) between a and b. On the other hand, if more cars pass through b
than a, that would mean cars are being created between a and b. So we expect ρ~v to be the same at a and b. Since a
and b were chosen arbitrarily, we can say that ρ~v is constant at all x. This means div(ρ~v ) = 0.
(d) At x = 0, ~v (0) = 55~i and ρ(0) = 75. We have ρ~v (0) = 4125~i = constant. So ρ(x) = kρ~v k/k~v (x)k = 4125/v.
4125
x if 0 ≤ x < 2000
55 − 50
4125
= 275 if 2000 ≤ x < 7000
ρ(x) =
15
4125
if 7000 ≤ x < 8000
ρ(x) =
15 + x−7000
25
4125
ρ(x) =
= 75 if x ≥ 8000
55
(e) We have ρ(0) = 75, ρ(1000) = 118, ρ(5000) = 275, where ρ is given in cars/mile. At x = 0, there are 75 cars in
a 1-mile stretch of highway. Since there are two lanes, there are about 38 cars in a mile in one lane. A mile is 5280
feet, so that says on average, one car occupies 139 feet. So at x = 0, the distance between two cars is 139 feet.
Similarly, we find that at x = 1000, the distance is 89 feet, and at x = 5000, the distance is 38 feet.
ρ(x) =
34. (a) We calculate
~i ~j ~k
~r × ~c = x y z
= (c3 y − c2 z)~i − (c3 x − c1 z)~j + (c2 x − c1 y)~k .
c1 c2 c3
Thus
div(~r × ~c ) = 0.
(b) By the Divergence Theorem
Z
S
~ = 0.
(~r × ~c ) · dA
19.3 SOLUTIONS
1759
35. Let ~a = a1~i + a2~j + a3~k with a1 , a2 , and a3 constant. Then f~a = f (x, y, z)(a1~i + a2~j + a3~k ) = f (x, y, z)a1~i +
f (x, y, z)a2~j + f (x, y, z)a3~k = f a1~i + f a2~j + f a3~k . So
∂(f a2 )
∂(f a3 )
∂(f a1 )
+
+
∂x
∂y
∂z
∂f
∂f
∂f
= a1
+ a2
+ a3
since a1 , a2 , a3 are constants
∂x
∂y
∂z
∂f
∂f ~
∂f ~
= ( ~i +
j +
k ) · (a1~i + a2~j + a3~k )
∂x
∂y
∂z
= (grad f ) · ~a .
div(f~a ) =
~ = F1~i + F2~j + F3~k . Then
36. Let F
~ ) = div(gF1~i + gF2~j + gF3~k )
div(g F
∂
∂
∂
=
(gF1 ) +
(gF2 ) +
(gF3 )
∂x
∂y
∂z
∂F1
∂g
∂F2
∂g
∂F3
∂g
F1 + g
+
F2 + g
+
F3 + g
=
∂x
∂x
∂y
∂y
∂z
∂z
∂g
∂g
∂g
=
F1 +
F2 +
F3 + g
∂x
∂y
∂z
~.
= (grad g) · F~ + g div F
∂F1
∂F2
∂F3
+
+
∂x
∂y
∂z
37. Now grad f = fx~i + fy~j + fz ~k and grad g is similar. Thus
~i ~j ~k
grad f × grad g = fx fy fz = (fy gz − fz gy )~i − (fx gz − fz gx )~j + (fx gy − fy gx )~k .
gx gy gz
Therefore
div(grad f × grad g) =
∂
∂
∂
(fy gz − fz gy ) +
(fz gx − fx gz ) +
(fx gy − fy gx ).
∂x
∂y
∂z
Expanding using the product rule gives
div(grad f × grad g) = fyx gz + fy gzx − fzx gy − fz gyx + fzy gx + fz gxy
−fxy gz − fx gzy + fxz gy + fx gyz − fyz gx − fy gxz .
Now consider pairs of terms such as fyx gz − fxy gz . Since fyx = fxy provided the second derivatives are continuous,
these two terms cancel out. All the other terms cancel in pairs, showing that
div(grad f × grad g) = 0.
~ ) = (grad g) · F
~ + g div F~ , we have
38. Using div(g F
1
1
div(~a × ~r ) + grad(
) · ~a × ~r
k~r kp
k~r kp
−p
1
0+
~r · (~a × ~r )
=
k~r kp
k~r kp+2
= 0 since ~r and ~a × ~r are perpendicular.
~ =
div F
~ ) = (grad g) · F
~ + g div F~ , we have
39. Using div(g F
~ = grad
div B
1
xa
· ~r +
1
div ~r = −ax−(a+1)~i · ~r + x−a (3) = (3 − a)x−a .
xa
~ ) = (grad g) · F
~ + g div F~ , we have
40. Using div(g F
~ = grad(~b · ~r ) · (~a × ~r ) + ~b · ~r div(~a × ~r ) = ~b · (~a × ~r ) + ~v · ~r 0 = ~b · (~a × ~r ).
div G
1760
Chapter Nineteen /SOLUTIONS
~ . Since T~ is in the direction of F
~ we have
41. First compute the unit vectors T~ and N
T~ =
1
1 ~
F = (u~i + v~j ).
~
F
kF k
~ is the unit vector in the direction of ~k × F
~ we have
Since N
~k × F
~ = ~k × (u~i + v~j )
= −v~i + u~j
1
(−v~i + u~j )
k − v~i + u~j k
1
= (−v~i + u~j ).
F
~ =
N
The chain rule for partial differentiation of the formulas u = F cos θ and v = F sin θ gives
ux = (cos θ)Fx − F (sin θ)θx
vy = (sin θ)Fy + F (cos θ)θy .
We have
~ = ux + vy
div F
= ((cos θ)Fx − F (sin θ)θx ) + ((sin θ)Fy + F (cos θ)θy )
1
= (−vθx + uθy ) + (uFx + vFy )
F
1
= (θx~i + θy~j ) · (−v~i + u~j ) + (Fx~i + Fy~j ) · (u~i + v~j )
F
~ + grad F · T~ .
= F grad θ · N
~ is θ ~ = grad θ · N
~ and the directional derivative of F in
Since the directional derivative of θ in the direction of N
N
the direction of T~ is FT~ = grad F · T~ we have
div F~ = F θN~ + FT~ .
42. We have now our temperature a function depending on t, x, y, z, hence T = T (t, x, y, z). For a fixed moment, say t0 , T
is a function of only x, y, z. For this moment, t = t0 , we have:
Rate of heat loss
from volume V = k
Z
S
where grad T =
∂T ~
i
∂x
+
∂T ~
j
∂y
+
∂T ~
k
∂z
~.
(grad T ) · dA
. Now the rate of change, with respect to time, in the average temperature
t=t0
in the region, at t = t0 , is proportional to the average rate at which heat is being lost per unit volume at t = t0 , so
∂Tavg
∂t
t=t0
= −c
Rate heat lost
Volume V
=
−ck
t=t0
R
S
~
(grad T ) · dA
Volume V
Taking the limit as V shrinks around the point, the average temperature through the region becomes the temperature at
that point. Thus using the definition of the divergence (with respect to x, y, z), we have
∂T
∂t
t=t0
= −ck lim
V →0
R
S
~
(grad T ) · dA
Volume V
!
= (−ck div grad T )t=t0
As this holds at every moment t0 , one has:
∂T
= B · div grad T,
∂t
where B = −ck is a function of time only, and the gradient and divergence are taken with respect to the variables x, y, z.
19.3 SOLUTIONS
1761
43. (a) At any point ~r = x~i + y~j , the direction of the vector field ~v is pointing away from the origin, which means it is of
the form ~v = f~r for some positive function f , whose value can vary depending on ~r . The magnitude of ~v depends
2
only on the distance r, thus f must be a function
depending only on r, which is equivalent to depending only on r
since r ≥ 0. So ~v = f (r 2 )~r = f (x2 + y 2 ) (x~i + y~j ).
(b) At (x, y) 6= (0, 0) the divergence of ~v is
div ~v =
∂(K(x2 + y 2 )−1 x)
∂(K(x2 + y 2 )−1 y)
Ky 2 − Kx2
Kx2 − Ky 2
+
=
+
= 0.
2
2
2
∂x
∂y
(x + y )
(x2 + y 2 )2
Therefore, ~v is a point source at the origin.
(c) The magnitude of ~v is
K
k~v k = K(x2 + y 2 )−1 |x~i + y~j | = K(x2 + y 2 )−1 (x2 + y 2 )1/2 = K(x2 + y 2 )−1/2 = .
r
(d) The vector field looks like that in Figure 19.19:
y
x
Figure 19.19
(e) We need to show that grad φ = ~v .
∂ K
∂ K
( log(x2 + y 2 ))~i +
( log(x2 + y 2 ))~j
∂x 2
∂y 2
Kx ~
Ky ~
= 2
i + 2
j
x + y2
x + y2
= K(x2 + y 2 )−1 (x~i + y~j )
grad φ =
= ~v
44. (a) At any point ~r = x~i + y~j , the direction of the vector field ~v is pointing toward the origin, which means it is of
the form ~v = f~r for some negative function f whose value can vary depending on ~r . The magnitude of ~v depends
2
only on the distance r, thus f must be a function
depending only on r, which is equivalent to depending only on r
2
2
2
since r ≥ 0. So ~v = f (r )~r = f (x + y ) (x~i + y~j ).
(b) At (x, y) 6= (0, 0) the divergence of ~v is
div ~v =
∂(K(x2 + y 2 )−1 x)
∂(K(x2 + y 2 )−1 y)
Ky 2 − Kx2
Kx2 − Ky 2
+
=
+
= 0.
2
2
2
∂x
∂y
(x + y )
(x2 + y 2 )2
Therefore, ~v is a point sink at the origin.
(c) The magnitude of ~v is
k~v k = |K|(x2 + y 2 )−1 |x~i + y~j | = |K|(x2 + y 2 )−1 (x2 + y 2 )1/2 = |K|(x2 + y 2 )−1/2 =
(remember, K < 0)
|K|
.
r
1762
Chapter Nineteen /SOLUTIONS
(d) The vector field looks like the following:
Figure 19.20
(e) We need to show that grad φ = ~v .
∂ K
∂ K
( log(x2 + y 2 ))~i +
( log(x2 + y 2 ))~j
∂x 2
∂y 2
Kx ~
Ky ~
= 2
i + 2
j
x + y2
x + y2
= K(x2 + y 2 )−1 (x~i + y~j )
grad φ =
= ~v
Strengthen Your Understanding
45. Divergence of a vector field is a scalar not a vector. We have div(2x~i ) = 2.
46. The divergence of a vector field is a scalar function, not a vector field. The correct divergence is
~ = 2x + 2 − 2z.
div F
47. Only vector fields have a divergence. A scalar function such as f (x, y, z) = x2 + yz does not have a divergence.
~ (x, y, z) = 2x~i + 3y~j + 4z~k . then
48. If F
~ (x, y, z) = ∂ (2x) + ∂ (3y) + ∂ (4z) = 2 + 3 + 4 = 9.
div F
∂x
∂y
∂z
~ (x, y, z) = (2x + x2 )~i − 2xy~j , giving
More complicated examples work too, such as if F
~ (x, y, z) = ∂ 2x + x2 + ∂ (−2xy) + ∂ 0 = (2 + 2x) − 2x + 0 = 2.
div F
∂x
∂y
∂z
49. If
then
~ (x, y, z) = y 2~i + xz~j + x~k
F
∂
∂
∂ 2
y +
(xz) +
x = 0.
div F~ (x, y, z) =
∂x
∂y
∂z
~ (x, y) = 2x~i is not divergence free since div F~ (x, y) = 2 6= 0.
50. The vector field F
51. True.
~ +G
~ ) = ∂(F1 + G1 ) + ∂(F2 + G2 ) + ∂(F3 + G3 )
div(F
∂x
∂y
∂z
∂F1
∂F2
∂F3
∂G1
∂G2
∂G3
=
+
+
+
+
+
∂x
∂y
∂z
∂x
∂y
∂z
~
~
= div F + div G .
19.4 SOLUTIONS
1763
~ ·G
~ ) is given by
52. False. Let’s compare the x–components of each side of the equation. The x–component of grad(F
∂(F1 G1 + F2 G2 + F3 G3 )
∂x
∂F1
∂G1
∂F2
∂G2
∂F3
∂G3
=
G1 + F1
+
G2 + F2
+
G3 + F3
.
∂x
∂x
∂x
∂x
∂x
∂x
~ ·G
~ ))1 =
(grad(F
~ ) + (div F
~ )·G
~ is
However, the x–component of F~ · (div G
~ (div G
~ ) + (div F~ )G
~ )1 = F1 (div G
~ ) + (div F
~ )G1
(F
= F1
∂G1
∂G2
∂G3
+
+
∂x
∂y
∂z
+
∂F1
∂F2
∂F3
+
+
∂x
∂y
∂z
G1 .
These two x–components are different and therefore
~ ·G
~ ) 6= F~ (div G
~ ) + (div F
~ )G
~.
grad(F
~ is a scalar whose value depends on the point at which it is calculated.
53. True. div F
54. False. The divergence is a scalar function that gives flux density at a point.
55. True. The net flow through a small volume is zero since the flow in is canceled by the equal flow out. Alternatively,
~ = ∂a + ∂b + ∂c = 0.
divF
∂x
∂y
∂z
~ = x~k . Then divF
~ = ∂x = 0.
56. False. As a counterexample, consider F
∂z
~ = ~i and f (x, y, z) = x. Then div(f F
~ ) = div x~i = 1, and f divF
~ = x ·0 = 0.
57. False. As a counterexample, consider F
~ = 2x~i +2y~j +2z~k . Then F~ = grad(x2 +y 2 +z 2 ), and div F~ = 2+2+2 6= 0.
58. False. As a counterexample, consider F
~ = x2~i Then div F
~ = 2x, and grad 2x = 2~i 6= ~0 .
59. False. As a counterexample, consider F
~ can be written F
~ (x, y, z) = x~i + y~j + z~k , the divergence of F
~ is 3.
60. False. Since F
~
61. True. Here is a way of constructing a vector field F . The idea is to think of f as a function of x (with y and z constant)
and take the antiderivative. We define
Z x
f (t, y, z) dt.
g(x, y, z) =
0
By the Fundamental Theorem of one-variable calculus, we know
~ = f.
g(x, y, z)~i has divF
∂g
~ =
= f. So, if f is given, the vector field F
∂x
~ = ~j both have divergence zero, but are not the same vector fields.
62. False. As a counterexample, note that F~ = ~i and G
63. False. The left-hand side of the equation, div(grad f ), is a scalar function and the right hand side, grad(div F~ ), is a
vector. There cannot be an equality between a‘ scalar and a vector.
64. (a), (b), and (e) all depend on the point (x, y, z), so they are vector fields. Since div ~r = 3 and div ~i = 0, the vectors in
(c) and (d) are constant vector fields.
Solutions for Section 19.4
Exercises
~ = dA~i .
1. First directly: On the faces x = 0, y = 0, z = 0, the flux is zero. On the face x = 2, a unit normal is ~i and dA
So
Z
Z
~ =
~r · dA
(2~i + y~j + z~k ) · (dA~i )
Sx=2
(since on that face, x = 2)
=
Z
Sx=2
Sx=2
2dA = 2 · (Area of face) = 2 · 4 = 8.
1764
Chapter Nineteen /SOLUTIONS
In exactly the same way, you get
Z
Sy=2
so
~ =
~r · dA
Z
S
Now using divergence:
2
Z
Flux =
Z
0
2
0
Z
0
Sz=2
~ = 8,
~r · dA
−
→
~r · dA = 3 · 8 = 24.
∂x
∂y
∂z
+
+
= 3,
∂x
∂y
∂z
div F~ =
so
Z
2
3 dx dy dz = 3 · (Volume of Cube) = 3 · 8 = 24
2. First directly, since the vector field is totally in the ~j direction, there is no flux through the ends. On the side of the
cylinder, a normal vector at (x, y, z) is x~i + y~j . This is in fact a unit normal, since x2 + y 2 = 1 (the cylinder has radius
1). Also, using x = cos θ, y = sin θ, in this case, the element of area dA equals 1dθdz. So
Flux =
=
Z
~ · dA
~ =
F
2
Z
0
Z
2π
Z
2
0
Z
2π
(y~j ) · (x~i + y~j ) dθ dz
0
Z
y 2 dθ dz =
2
0
0
2π
Z
Z
sin2 θ dθ dz =
0
2
πdz = 2π.
0
Now we calculate the flux using the divergence theorem. The divergence of the field is given by the sum of the
∂y
= 1. Since the divergence is constant, we can
∂y
simply calculate the volume of the cylinder and multiply by the divergence
respective partials of the components, so the divergence is simply
Flux = 1πr 2 h = 2π
3. Finding flux directly:
1) On bottom face, z = 0 so F~ = x2~i + 2y 2~j is parallel to face so flux is zero.
2) On front face, y = 0 so F~ = x2~i + 3z 2~k is parallel to face so flux is zero.
~ = x2~i + 2~j + 3z 2~k and A
~ = ~j so flux is 2.
3) On back face, y = 1 so F
2~
2~
~
~
~
4) On top face, z = 1 so F = x i + 2y j + 3k and A = ~k so flux is 3.
~ = ~i + 2y 2~j + 3z 2~k and A
~ = −~i so flux is −1.
5) On side x = 1, F
~ = 4~i + 2y 2~j + 3z 2~k and A
~ = ~i so flux is 4.
6) On side x = 2, F
Total flux is thus 8.
By the Divergence Theorem:
~ = 2x + 4y + 6z
div F
So, if W is the interior of the box, we have
Z
S
~ · dA
~ =
F
Z
(2x + 4y + 6z)dV = 2
2
Z
1
=2
Z
2
1
=
Z
2
2
Z
h
1
0
3z 2
xz + 2yz +
2
xy + y 2 +
3y
2
i1
1
0
Z
0
(2x + 5)dx = (x2 + 5x)
(x + 2y + 3z)dz dy dx
dy dx = 2
Z
2
1
Z
2
1
2
1
0
0
dx = 2
1
1
Z
1
W
=2
Z
x+1+
Z
1
0
3
x + 2y +
2
!
3
dx
2
=8
1
4. Since div F~ = 1 + 1 + 1 = 3, the Divergence Theorem gives
Z
S
~ · dA
~ =
F
Z
W
3 dV = 3
Z
W
dV = 3 · Volume of the cylinder = 3π.
dy dx
19.4 SOLUTIONS
1765
5. The location of the pyramid has not been completely specified. For instance, where is it centered on the xy plane? How
is base oriented with respect to the axes? Thus, we cannot compute the flux by direct integration with the information we
~.
have. However, we can calculate it using the divergence theorem. First we calculate the divergence of F
~ =
div F
∂(−z)
∂0
∂x
+
+
=0+0+0 =0
∂x
∂y
∂z
Thus for any closed surface the flux will be zero, so the flux through our pyramid, regardless of its location or orientation,
is zero.
6. Since the surface is closed, the flux of a constant vector field out of it is 0.
~ = 1, if W is the interior of the box, the Divergence Theorem gives
7. Since div G
Flux =
Z
W
1 dV = 1 · Volume of box = 1 · 2 · 3 · 4 = 24.
~ = y, if W is the interior of the box, the Divergence Theorem gives
8. Since div H
Z
Flux =
y dV =
4
Z
0
W
Z
0
3
y2
2
3
y2
2
0
3
2
Z
y dx dy dz = 4 · 2 ·
0
= 36.
0
9. Since div J~ = 2xy, if W is the interior of the box, the Divergence Theorem gives
Flux =
Z
2xy dV =
4
Z
0
W
Z
0
3
Z
2
2
2xy dx dy dz = x2
0
4
z
0
= 72.
0
~ = 0, the flux through the closed surface of the box is 0.
10. Since div N
11. We have
div((3x + 4y)~i + (4y + 5z)~j + (5z + 3x)~k ) = 3 + 4 + 5 = 12.
Let W be the interior of the cube. Then by the divergence theorem,
Z
S
~ =
((3x + 4y)~i + (4y + 5z)~j + (5z + 3x)~k ) · dA
Z
W
12 dV = 12 · Volume of cube = 12 · (2 · 3 · 4) = 288.
12. By the Divergence Theorem,
Flux =
Z
S
~ · dA
~ =
M
=
Z
~ dV =
div M
0
4
Z
0
3
(xy + 5) dV =
W
2
W
Z
Z
Z
0
x2 y
+ 5x dy dz =
2
0
Z
4
0
Z
3
4
Z
0
3
Z
2
(xy + 5) dx dy dz
0
(2y + 10) dy dz =
0
Problems
13. Since
div F~ = 0 + 1 + 0 = 1,
we have
Z
S
~ · dA
~ = (div F~ ) · Vol = 4 π33 = 36π.
F
3
14. Since div F~ = y + z + x, the flux is given by
Z
Sphere
~ · dA
~ =
F
Z
Sphere
(x + y + z)dV.
Z
0
4
3
(y 2 + 10y) dz = 156.
0
1766
Chapter Nineteen /SOLUTIONS
We calculate the first term of the integral
Z
x dV =
Sphere
Z
1
Z
1
x dx dy dz
x2
2
dy dz
√
1−z 2
√
Z 1−z2
−1
=
Z √1−z2 Z √1−y2 −z2
−
√
−1 − 1−z 2
√
Z Z
1−z 2
1
=
−1
=
−
√
1−z 2
−
√
1−z 2
Z √1−z2
1
Z
−
−1
√
(
1−y 2 −z 2
√
−
1−y 2 −z 2
√
1−y 2 −z 2
p
(−
1 − y 2 − z 2 )2
−
2
p
1 − y 2 − z 2 )2
2
!
dy dz
0 dydz = 0.
The other terms in the integral are zero by a similar calculation.
The same result can be obtained by a symmetry argument, which is much shorter: Since x, y, and z each take equal
positive and negative values on half the sphere, the integral of each term is 0. Thus, the flux is zero:
Z
Sphere
~ =
F~ · dA
Z
(x + y + z)dV = 0.
Sphere
15. Since div F~ = 3x2 + 3y 2 + 3z 2 , the Divergence Theorem gives
Z
~ · dA
~ =
F
S
Z
~ dV =
div F
Z
3(x2 + y 2 + z 2 ) dV.
W
W
Since W is the interior of a cylinder of radius 2 centered on the z-axis, we use cylindrical coordinates, giving
Z
S
~ · dA
~ =
F
Z
3(x2 + y 2 + z 2 ) dV = 3
0
W
=3
Z
2π
Z
2π
0
= 3 · 2π
Z
2
r3 z +
0
5
rz 3
3
125 r 2
5r 4
+
·
4
3
2
Z
2
0
dr dθ = 3
0
2
5
Z
(r 2 + z 2 )r dz dr dθ
0
Z
2π
0
Z
2
(5r 3 +
0
125
r) dr dθ
3
= 620π.
0
16. Since div F~ = 3x2 + 3y 2 + 3z 2 , the Divergence Theorem gives
Flux =
Z
S
~ · dA
~ =
F
Z
(3x2 + 3y 2 + 3z 2 ) dV.
W
In spherical coordinates, the region W lies between the spheres ρ = 2 and ρ = 3 and inside the cone φ = π/4. Since
3x2 + 3y 2 + 3z 2 = 3ρ2 , we have
Flux =
Z
S
~ =
F~ · dA
3
Z
2π
0
Z
3 5
ρ (− cos φ)
5
2
= 2π ·
π/4
0
π/4
Z
2
=
0
3
3ρ2 · ρ2 sin φ dρ dφ dθ
633(2 −
5
√
2)
π = 232.98.
17. We calculate div F~ and use the Divergence Theorem:
~ =
div F
∂
∂
∂
(x + 3eyz ) +
(ln(x2 z 2 + 1) + y) +
(z) = 3.
∂x
∂y
∂z
Thus, with S representing the surface of the cylinder and W the region inside, we have
Z
S
~ · dA
~ =
F
Z
W
3 dV = 3 · Volume of cylinder = 3 · π22 4 = 48π.
19.4 SOLUTIONS
1767
2 2
~ = div(ey z ~i +(tan(0.001x2 z 2 )+y 2 )~j +(ln(1+x2 y 2 )+z 2 )~k ) = 2y +2z, by the Divergence Theorem,
18. Since div F
if S is the surface of the box W ,
Flux =
Z
S
Z
~ · dA
~ =
F
div F~ dV =
W
3
Z
=
0
Z
=
Z
0
Z
4
Z
4
Z
0
5
(2y + 2z) dx dy dz
0
5
(2xy + 2xz) dy dz =
0
3
3
0
0
4
2
(5y + 10yz) dz =
0
3
Z
Z
4
(10y + 10z) dy dz
0
(80 + 40z)dz
0
0
3
= (80z + 20z 2 )
3
Z
= 420.
0
19. We use the Divergence Theorem, with
∂
∂
∂ 2
(x ) +
(z) +
(y) = 2x.
div F~ =
∂x
∂y
∂z
Let W be the interior of the cone. Then
Z
S
~ · dA
~ =
F
Z
~ dV =
div F
Z
2x dV.
W
W
The region W is shown in Figure 19.21. Evaluating the integral over W as an iterated integral gives
Z
Z
Z Z √
Z Z √
1−z 2
1
2x dV =
W
−1
=
Z
1
−1
=
Z
1
−1
−
√
1−z 2
Z √1−z2
−
√
1−z 2
√
2x dx dy dz =
y 2 +z 2
2
−1
2
1 − (y + z ) dy dz =
2(1 − z 2 )1/2 −
Z
1
1−z 2
1
1
−
1
−1
√
1−z
x2 √
2
dy dz
y 2 +z 2
1
y − y3 − z2y
3
2
(1 − z 2 )3/2 − 2z 2 (1 − z 2 )1/2
3
√
−
1−z 2
√
dz
1−z 2
dz = 1.571.
The last integral was computed numerically. It can also be done by trigonometric substitutions; the exact value is π/2.
z
x
1
Figure 19.21: The cone x =
with 0 ≤ x ≤ 1
p
y2 + z2 ,
1768
Chapter Nineteen /SOLUTIONS
20. By the Divergence Theorem, if W is the cylinder and S is its surface:
Z
S
~ · dA
~ =
F
Z
Z
~ dV =
div F
W
W
10 dV = 10 · Volume of cylinder = 10πa3 .
21. Apply the Divergence Theorem to the solid cone, whose interior we call W . The surface of W consists of S and D. Thus
Z
~ · dA
~ +
F
S
Z
D
~ · dA
~ =
F
Z
~ dV.
div F
W
But div F~ = 0 everywhere, since F~ is constant. Thus
Z
D
~ =−
F~ · dA
Z
S
~ · dA
~ = −3.22.
F
22. Since div ~r = div(x~i + y~j + z~k ) = 3, applying the Divergence Theorem to the vector field F~ = ~r gives
Z
S
Thus
1
3
R
S
~ = V.
~r · dA
~ =
~r · dA
Z
3dV = 3
V
Z
dV = 3V.
V
~ dV =
~ · dA
~ =
0dV = 0 for a closed surface S, where W is the region
23. By the Divergence Theorem, S F
divF
W
W
enclosed by S.
~ || = ||~r ||/||~r ||3 , we have ||F
~ || = 1 on the unit sphere. Thus
24. (a) Since ||F
R
R
R
Flux =
Z
Sphere
~ · dA
~ = 1 · Surface area of sphere = 4π.
F
(b) Since
x
y
z
~r
~k ,
~i +
~j +
= 2
k~r k3
(x + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
we compute the partial derivative of each component
∂
∂x
∂
∂y
∂
∂z
x
(x2 + y 2 + z 2 )3/2
(x2
(x2
+
+
y
+ z 2 )3/2
y2
z
+ z 2 )3/2
y2
=
(x2
=
(x2
=
(x2
1
3x2
− 2
2
3/2
2
+z )
(x + y + z 2 )(3/2)+1
+
y2
+
y2
+
y2
1
3y 2
− 2
2
3/2
2
+z )
(x + y + z 2 )(3/2)+1
1
3z 2
− 2
.
2
3/2
2
+z )
(x + y + z 2 )(3/2)+1
So
div
~r
k~r k3
=
(x2
+
3(x2 + y 2 + z 2 )
3
− 2
2
3/2
+z )
(x + y 2 + z 2 )(3/2)+1
y2
2
=
3(x + y 2 + z 2 ) − 3(x2 + y 2 + z 2 )
0
= 2
= 0.
(x2 + y 2 + z 2 )5/2
(x + y 2 + z 2 )5/2
(c) Consider the region W between the box, B, and the sphere, S, with the box oriented outward and the sphere oriented
inward. Then the Divergence Theorem says
Z
~ dV =
div F
Z
B
W
~ +
F~ · dA
Since the sphere is now oriented inward,
Z
S
~ = −4π.
F~ · dA
Z
S
~.
F~ · dA
19.4 SOLUTIONS
~ = 0, we have
Since div F
0=
Z
B
Z
div
~r
||~r ||2
~ · dA
~ − 4π
F
~ · dA
~ = 4π
F
B
25. (a) Since ~r = x~i + y~j , for ~r 6= ~0 , we have
1769
x~i + y~j
x2 + y 2
x
y
∂
∂
= div
=
+
∂x x2 + y 2
∂y x2 + y 2
x · 2x
1
y · 2y
1
− 2
+ 2
− 2
= 2
x + y2
(x + y 2 )2
x + y2
(x + y 2 )2
x2 + y 2 − 2x2 + x2 + y 2 − 2y 2
(x2 + y 2 )2
= 0.
=
(b) Since the cylinder contains points where ~r = ~0 (on the z-axis) and div(~r /||~r ||2 ) is undefined there, we cannot use
the Divergence Theorem.
(c) Since the cylinder is closed, it is oriented outward. Calculating the flux directly, the flux through the ends of the
cylinder is 0 since the vector field has no ~k -component. (The flux is zero even though the vector field is undefined on
the z-axis.) On the curved sides of the cylinder, the vector field is perpendicular to the surface and of length 1. Thus
Flux =
Z
S
~r
~ = 1 · Area of curved side = 1 · 2π1 · 2 = 4π.
· dA
||~r ||2
(d) The cylinder, S, in part (c) lies inside this one, S1 . In the space, W, between the cylinders, the divergence is 0. The
surface of W consists of S oriented inward and S1 oriented outward. The Divergence Theorem can be applied to the
region W, so
Z
S1 −S
Z
~r
~ =
· dA
||~r ||2
Z
S1
Z
S1
div
W
~r
~ −
· dA
||~r ||2
~r
~ =
· dA
||~r ||2
Z
S
Z
S
~r
||~r ||2
dV = 0
~r
~ =0
· dA
||~r ||2
~r
~ = 4π.
· dA
||~r ||2
Thus, the flux through this cylinder is the same as the flux through the cylinder in part (c).
26. We use the Divergence Theorem to compare the integrals. We have
~
div F
~
div F
~
div F
1
= y 2 + 3z 2 + 3x2 + 2z 2 + 3y 2 = 3x2 + 4y 2 + 5z 2
2
= y 2 + z 2 + x2
3
= z 2 + x2 + z 2 + y 2 + y 2 + z 2 = x2 + 2y 2 + 3z 2 .
For all x, y, z,
~
div F
2
≤ div F~
3
~ 1,
≤ div F
with equality only at the origin. Since the flux integrals are all through the same surface S, we have
Z
~
F
~ <
· dA
2
S
Z
~
F
3
S
~ <
· dA
Z
~
F
S
1
~.
· dA
27. (a) By the Divergence Theorem, if S is the surface of the rectangular solid W ,
Flux =
Z
S
~ · dA
~ =
F
Z
~ dV =
div F
Z cZ bZ
0
W
=
0
Z cZ
0
0
0
a
2(6 − x) dx dy dz
b
12x − x2
a
0
dy dz = cb(12a − a2 ).
1770
Chapter Nineteen /SOLUTIONS
(b) To maximize the flux cb(12a − a2 ), we take b and c to be as large as possible and choose a to maximize (12a − a2 )
on the interval 0 ≤ a ≤ 10. Thus we take a = 6, b = c = 10. Then
Flux = cb(12a − a2 )
= 3600.
a=6,b=10,c=10
~ is radial, it is everywhere parallel to the area vector, ∆A
~ . Also, ||F
~ || = 1 on the surface of the sphere
28. (a) Since F
x2 + y 2 + z 2 = 1, so
Flux through the sphere =
Z
S
=
~ =
F~ · dA
lim
~ ||→0
||∆A
X
lim
~ ||→0
||∆A
X
~
F~ · ∆A
~ || ||∆A
~ || =
||F
lim
~ ||→0
||∆A
= Surface area of sphere = 4π · 12 = 4π.
X
~ ||
||∆A
(b) In Cartesian coordinates,
~ (x, y, z) =
F
y
z
x
~k .
~i +
~j +
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2
So,
~ (x, y, z) =
div F
+
+
=
+
+
1
3x2
−
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )5/2
3y 2
1
−
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )5/2
3z 2
1
−
(x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )5/2
x2 + y 2 + z 2
3y 2
−
(x2 + y 2 + z 2 )5/2
(x2 + y 2 + z 2 )5/2
3x2
x2 + y 2 + z 2
−
(x2 + y 2 + z 2 )5/2
(x2 + y 2 + z 2 )5/2
x2 + y 2 + z 2
3z 2
−
(x2 + y 2 + z 2 )5/2
(x2 + y 2 + z 2 )5/2
3(x2 + y 2 + z 2 ) − 3(x2 + y 2 + z 2 )
(x2 + y 2 + z 2 )5/2
= 0.
=
~ is not
(c) We cannot apply the Divergence Theorem to the whole region within the box, because the vector field F
defined at the origin. However, we can apply the Divergence Theorem to the region, W , between the sphere and the
~ = 0 there, the theorem tells us that
box. Since div F
Z
~ +
F~ · dA
Box
(outward)
Z
~ · dA
~ =
F
Sphere
(inward)
Z
~ dV = 0.
div F
W
Therefore, the flux through the box and the sphere are equal if both are oriented outward:
Z
Box
(outward)
~ · dA
~ =−
F
Z
Sphere
(inward)
~ · dA
~ =
F
Z
Sphere
(outward)
~ · dA
~ = 4π.
F
19.4 SOLUTIONS
29. We have:
1771
x~i + y~j + z~k
(x2 + y 2 + z 2 )3/2
~ (x, y, z) =
F
~ = 0, we can replace the
Calculating the flux of F~ through the ellipsoid directly would be difficult. However, since div F
~ = 0. Let T be the surface of a sphere centered at the origin
ellipsoid by a sphere. Except at the origin, we have div F
inside the ellipsoid S, and let W be the region between S and T . Suppose both S and T are oriented away from the origin.
By the Divergence Theorem, we have
Flux out of W =
Z
W
and therefore
~ · dV = 0,
div F
Z
Flux out of W = (Flux out − Flux in) =
S
Thus, we have
Z
S
~ · dA
~ =
F
Z
~ · dA
~ −
F
Z
T
~ · dA
~ = 0.
F
~ on sphere) · (Surface area)
~ · dA
~ = (Magnitude of F
F
T
=
1
radius2
· (4π · radius2 ) = 4π.
30. Use the Divergence Theorem. Since
~ (x, y, z) =
F
1
(x~i + y~j + z~k ),
(x2 + y 2 + z 2 )3/2
~ = 0, except at the origin.
we have div F
Let T be the surface of a sphere inside the cylinder S, and let W be the region between S and T . By the Divergence
Theorem,
Z
Z
~ dV = 0.
~ · dA
~ =
F
div F
Flux out of W =
S+T
W
Since S is oriented outward and T is oriented inward,
Z
Net flux out of W = Flux out − Flux in =
S
so
Z
S
~ · dA
~ =
F
Z
T
~ · dA
~ −
F
Z
T
~ = 0.
F~ · dA
~ on sphere · Surface area
~ = Magnitude of F
F~ · dA
=
1
Radius2
· (4π Radius2 ) = 4π.
31. (a) Let W1 be the ball inside S1 . By the Divergence Theorem,
Z
S1
~ =
F~ · dA
Z
(x2 + y 2 + z 2 + 3) dV.
W1
Using spherical coordinates, we have
Z
S1
~ · dA
~ =
F
Z
(x2 + y 2 + z 2 + 3) dV =
2π
0
W1
= 2π ·
Z
ρ5
+ ρ3
5
π
1
(− cos φ)
0
0
Z
π
0
= 2π ·
Z
1
(ρ2 + 3)ρ2 sin φ dρ dφ dθ
0
6
24
·2=
π.
5
5
~ through
(b) If we were to calculate all three integrals using the Divergence Theorem, we would be integrating div F
the interior of each of these regions. Since S2 lies entirely inside S3 , and S3 lies entirely inside S4 , and since
~ = x2 + y 2 + z 2 + 3 is positive everywhere,
div F
Z
S2
~ <
F~ · dA
Z
S3
~ · dA
~ <
F
Z
S4
~ · dA
~.
F
1772
Chapter Nineteen /SOLUTIONS
~ = 1 · 2 · 12 = 2.
32. (a) At the point (1, 2, 1), we have div F
(b) Since the box is small, we use the approximation
div F~ = Flux density ≈
Flux out of box
.
Volume of box
Thus
Flux out of box ≈ (div F~ ) · (Volume of box) = 2(0.2)3 = 0.016.
(c) To calculate the flux exactly, we use the Divergence Theorem,
Flux out of box =
Z
~ dV =
div F
Z
xyz 2 dV.
Box
Box
Since the box has side 0.2, it is given by 0.9 < x < 1.1, 1.9 < y < 2.1, 0.9 < z < 1.1, so
Flux =
Z
1.1
0.9
=
Z
2.1
1.9
Z
1.1
xyz 2 dzdydx =
0.9
0.9
(1.1)3 − (0.9)3
3
1.1
Z
Z
1.1
0.9
2.1
xy 2
2
dx =
1.9
Z
2.1
xy
1.9
z3
3
1.1
dydx
0.9
(1.1)3 − (0.9)3 (2.1)2 − (1.9)2 x2
·
·
3
2
2
1.1
0.9
(1.1)3 − (0.9)3 (2.1)2 − (1.9)2 (1.1)2 − (0.9)2
=
·
·
= 0.016053 . . . .
3
2
2
~ .
Notice that you can calculate the flux without knowing the vector field, F
33. Any closed surface, S, oriented inward, will work. Then,
Z
S (inward)
~ · dA
~ =−
F
Z
S (outward)
~,
F~ · dA
so, by the Divergence Theorem, with W representing the region inside S,
Z
S (inward)
~ · dA
~ =−
F
Z
W
div F~ dV = −
Z
(x2 + y 2 + 3)dV.
W
The integral on the right is positive because the integrand is positive everywhere. Therefore the flux through S oriented
inward is negative.
~ at that point. So div F
~ = 30 watts/km3 .
34. (a) The rate at which heat is generated at any point in the earth is div F
~
~ = α~r
~
~
(b) Differentiating gives div(α(xi + y j + z k )) = α(1 + 1 + 1) = 3α so α = 30/3 = 10 watts/km3 . Thus, F
~
has constant divergence. Note that F = α~r has flow lines going radially outward, and symmetric about the origin.
(c) The vector grad T gives the direction of greatest increase in temperature. Thus, −grad T gives the direction of
~ = −k grad T says that heat will flow in the direction of greatest
greatest decrease in temperature. The equation F
decrease in temperature (i.e. from hot regions to cold), and at a rate proportional to the temperature gradient.
~ is given by the answer to part (b). Then, using part (c), we have
(d) We assume that F
~ = 10(x~i + y~j + z~k ) = −30,000 grad T,
F
so
grad T = −
Integrating we get
T =
10
(x~i + y~j + z~k ).
30,000
−10
(x2 + y 2 + z 2 ) + C.
2(30,000)
At the surface of the earth, x2 + y 2 + z 2 = 64002 , and T = 20◦ C, so
T =
−1
(64002 ) + C = 20.
6000
Thus,
C = 20 +
64002
= 6847.
6000
At the center of the earth, x2 + y 2 + z 2 = 0, so
T = 6847◦ C.
19.4 SOLUTIONS
1773
35. (a) Using the expression given for the force, we have
~ · ~i = −
Force in ~i direction = F
Z
S
~
δgz~i · dA
Z
= −δg
S
~.
z~i · dA
Now apply the Divergence Theorem to this integral. (Notice that in order to do this, you need to orient S outward,
hence the minus sign disappears.)
Z
∂z
~
~
dV = 0.
F · i = δg
∂x
V
Similarly:
~ · ~j = −δg
Force in ~j direction = F
= δg
Z
S
Z
~
z~j · dA
∂z
dV = 0
∂y
V
(b)
~ · ~k = −δg
Force in ~k direction = F
= δg
Z
Z
S
V
~
z~k · dA
∂z
dV = δg
∂z
Z
dV = δgV.
V
~ gives
36. (a) Taking partial derivatives of E
∂E1
∂
=
[qx(x2 + y 2 + z 2 )−3/2 ] = q[(x2 + y 2 + z 2 )−3/2 + x(−3/2)(2x)(x2 + y 2 + z 2 )−5/2 ]
∂x
∂x
= q(y 2 + z 2 − 2x2 )(x2 + y 2 + z 2 )−5/2 .
Similarly,
∂E2
= q(x2 + z 2 − 2y 2 )(x2 + y 2 + z 2 )−5/2
∂x
∂E3
= q(x2 + y 2 − 2z 2 )(x2 + y 2 + z 2 )−5/2 .
∂x
~ = 0.
Summing, we obtain div E
~ and the area vector ∆A
~ are parallel,
(b) Since on the surface of the sphere, the vector field E
~ · ∆A
~ = kE
~ kk∆A
~ k.
E
Now, on the surface of a sphere of radius a,
~k=
kE
Thus,
Z
Sa
~ · dA
~ =
E
qk~r k
q
= 2.
k~r k3
a
Z
q
~ k = q · Surface area of sphere = q · 4πa2 = 4πq.
kdA
a2
a2
a2
0=
Z
~ is not defined at the origin (which lies inside
(c) It is not possible to apply the Divergence Theorem in part (b) since E
the region of space bounded by Sa ), and the Divergence Theorem requires that the vector field be defined everywhere
inside S.
(d) Let R be the solid region lying between a small sphere Sa , centered at the origin, and the surface S. Applying the
Divergence Theorem and the result of part (a), we get:
~ dV =
div E
R
Z
Sa
~ · dA
~ +
E
Z
S
~ · dA
~,
E
1774
Chapter Nineteen /SOLUTIONS
where S is oriented with the outward normal vector, and Sa with the inward normal vector (since this is “outward”
with respect to the region R). Since
Z
Sa , inward
~ · dA
~ =−
E
Z
Sa , outward
~ · dA
~,
E
the result of part (b) yields
Z
S
~ · dA
~ = 4πq.
E
~ is defined everywhere
[Note: It is legitimate to apply the Divergence Theorem to the region R since the vector field E
in R.]
~ = 0 by taking partial derivatives. For instance,
37. Check that div E
∂
∂E1
=
[q(x − x0 )[(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]−3/2 ]
∂x
∂x
= q[(y − y0 )2 + (z − z0 )2 − 2(x − x0 )2 ][(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]−5/2
and similarly,
∂E2
= q[(x − x0 )2 + (z − z0 )2 − 2(y − y0 )2 ][(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]−5/2
∂y
∂E3
= q[(x − x0 )2 + (y − y0 )2 − 2(z − z0 )2 ][(x − x0 )2 + (y − y0 )2 + (z − z0 )2 ]−5/2 .
∂z
Therefore,
∂E2
∂E3
∂E1
+
+
= 0.
∂x
∂y
∂z
~ is defined everywhere but at the point with position vector ~r 0 . If this point lies outside the surface S,
The vector field E
the Divergence Theorem can be applied to the region R enclosed by S, yielding:
Z
S
~ · dA
~ =
E
Z
~ dV = 0.
div E
R
If the charge q is located inside S, consider a small sphere Sa centered at q and contained in R. The Divergence Theorem
for the region R′ between the two spheres yields:
Z
S
~ · dA
~ +
E
Z
Sa
~ · dA
~ =
E
Z
~ dV = 0.
div E
R′
In this formula,
the Divergence Theorem requires S to be given the outward orientation, and Sa the inward orientation. To
R
~ · dA
~ , we use the fact that on the surface of the sphere, E
~ and ∆A
~ are parallel and in opposite directions,
compute S E
a
so
~ · ∆A
~ = −kE
~ kk∆A
~k
E
since on the surface of a sphere of radius a,
~k=q
kE
Then,
Z
Sa
~ · dA
~ =
E
Z
−
k~r − ~r 0 k
q
= 2.
k~r − ~r 0 k3
a
q
~ k = −q · Surface area of sphere = − q · 4πa2 = −4πq.
kdA
a2
a2
a2
Z
S
Z
Sa
~ · dA
~ = −4πq.
E
~ · dA
~ −
E
Z
Sa
~ · dA
~ = 4πq.
E
Strengthen Your Understanding
38. The surface S is not the boundary of a solid region, so the Divergence Theorem does not apply.
19.4 SOLUTIONS
1775
39. The correct statement of the Divergence Theorem is:
Z
S
Z
~ · dA
~ =
F
~ dV.
div F
W
40. Since div F~ = 0 everywhere, the Divergence Theorem shows that
Z
~ · dA
~ =
F
S
Z
~ dV = 0
div F
W
for every surface S that bounds a solid region W . For example, we can take S to be a sphere with any center and radius.
41. If S is a sphere of radius 1 centered at the origin and W is the region inside it, then
Flux out of S =
Z
~ dA =
F
Z
~ dV = 3.
div F
W
S
~ to be a vector field with constant divergence, div F
~ = k, and calculate the constant k so as to get a flux of 3.
We pick F
~ = kx~i . Then div F
~ = k everywhere and
Let F
Flux out of S =
Z
W
k dV = k · (Volume of the sphere) =
So we have
Flux out of S =
4
πk.
3
9
4
πk = 3, which gives k =
.
3
4π
~ = (9x/4π)~i is one possible answer.
Thus, F
~ has a net outflow per unit volume everywhere.
42. False. Since the divergence is positive, F
~
43. True. The divergence of the field xi + (3y)~j + (y − 5x)~k is equal to 1 + 3 + 0 = 4 at all points.
44. False. The divergence of any constant field is zero at all points.
~ = 4x~i , which is parallel to the xy-plane and hence has zero flux through the circle.
45. False. The vector field could be F
Note that the Divergence Theorem cannot be used to calculate the flux in this case since a circle is not a closed surface
enclosing a volume.
~ is constant, the flux of F
~ through the cylinder is
46. True. The Divergence Theorem applies in this case and, since divF
equal to 4 times the volume of the cylinder, or 4(3π).
R
~ = x2~i ,
~ , which is a function, would always be constant ( F
~ · dA
~ is a constant). Take F
47. False. If this were true, div F
S
~ = 2x, which is not constant.
so div F
~ is a scalar, and we cannot take the flux integral of a scalar. On the
48. False. Neither side of this equation makes sense: div F
other side, F~ is a vector field, and we cannot take the triple integral of a vector field.
R
R
~ · dA
~ =−
div F~ dV , where W is the solid interior of S and the negative
49. True. By the Divergence theorem, S F
W
R
~ = 0, we have
~ · dA
~ = 0.
sign is due to the inward orientation of S. Since div F
F
S
~ · dA
~ =
50. True. By the Divergence theorem, S F
div F~ dV , where W is the solid interior of S. Since div F~ = 1,
W
R
R
~
~
we have S F · dA = W 1 dV which is equal to the volume enclosed by S.
R
R
~ · dA
~ , where S is the outward oriented boundary of W .
51. True. The Divergence theorem says that W div F~ dV = S F
In this case, the boundary of W consists of the surfaces S1 and S2 . To give this boundary surface a consistent outward
orientation, we
on S1 thatR points toward the origin, and a normal on S2 that points away from the
R
R use a normal vector
~ +
~ · dA
~ , with S2 oriented outward and S1 oriented inward. Reversing
F
origin. Thus W div F~ dV = S F~ · dA
S1
2
R
R
R
~ dV =
~ −
~ · dA
~.
the orientation on S1 so that both spheres are oriented outward yields
div F
F~ · dA
F
R
R
W
S2
S1
52. False. The boundary of the cube W consists of six squares, so the Divergence theorem requires adding the flux integrals
over the four remaining sides.
53. True. The boundary of the cube W consists of six squares, but four of them are parallel to the xz or yz-planes and so
contribute zero flux for this particular vector field. The only two surfaces of the boundary with nonzero flux are S1 and
S2 , which are parallel to the xy-plane.
1776
Chapter Nineteen /SOLUTIONS
~ dV = 0, where W is the solid ball with
div F
~ can be positive
boundary S. But this does not necessarily mean that div F~ = 0 at all points in W . The divergence of F
~ = x2~i .
at some points in W and negative at other points in W , yielding a triple integral of zero. For example, let F
The flux of this vector field through the sphere is 0. (The flux out for x > 0 cancels the flux in for x < 0.) However,
~ = 2x, which is not 0.
div F
54. False. The Divergence theorem tells us in this case that
R
S
~ · dA
~ =
F
R
W
55. True. Let D be the disk that forms the bottom of the cylinder, x2 + y 2 ≤ 1, z = 0, oriented
R downward. Then
R the surface
~ ·dA
~ =
consisting of Sh and D is closed and oriented outward, so the Divergence Theorem says S +D F
div F~ dV ,
W
h
R
~ = 0, we have
~ = 0. Writing the flux integral
where W is the solid interior of the cylinder. Since div F
F~ · dA
Sh +D
~ = 0, so
~ · dA
~ =−
~ · dA
~ +
F~ · dA
F
as the sum of integrals over Sh and D gives S F
Sh
h
R
R D
~
~
~
~
integral D F · dA does not depend on the height h, so S F · dA is independent of h.
R
R
h
R
R
56. Since div F~ = 5 + 7 + 9 = 21, by the Divergence Theorem, if Wi is the region inside Si , we have
Qi =
Z
Wi
21 dV = 21 · Vol Wi .
Thus, we arrange Qi by the volume of Wi , so
Q4 < Q3 < Q2 < Q1 .
Solutions for Chapter 19 Review
Exercises
~ = −~j dA, so
1. Scalar. Only the ~j -component of the vector field contributes to the flux and dA
Z
S
~ = −4 · Area of disk = −4 · π52 = −100π.
(3~i + 4~j ) · dA
2. Scalar. Since
div
y~i − x~j
x2 + y 2
=
−y · 2x
x · 2y
+ 2
= 0.
(x2 + y 2 )2
(x + y 2 )2
~ = ~i dA, so
3. (a) Only the ~i -component of the vector field contributes to the flux and dA
Z
~ = 1 · Area of square = 16.
(~i + 2~j + ~k ) · dA
Z
~ = 1 · Area of square = 16.
(~i + 2~j + ~k ) · dA
S
~ = ~i dA, so
(b) Only the ~i -component of the vector field contributes to the flux and dA
S
~ = −~k dA, so
(c) Only the ~k -component of the vector field contributes to the flux and dA
Z
~ = −1 · Area of square = −16.
(~i + 2~j + ~k ) · dA
Z
~ = −1 · Area of square = −16.
(~i + 2~j + ~k ) · dA
S
~ = −~k dA, so
(d) Only the ~k -component of the vector field contributes to the flux and dA
S
~ = ~j dA, so
(e) Only the ~j -component of the vector field contributes to the flux and dA
Z
S
~ = 2 · Area of square = 32.
(~i + 2~j + ~k ) · dA
D
~ · dA
~ . The flux
F
SOLUTIONS to Review Problems for Chapter Nineteen
1777
~ is parallel to the x-axis, so (x~j + y~k ) · dA
~ = 0.
4. Zero, since dA
~ = −~i dy dz and x = 7, so (x~i + y~k ) · dA
~ = −7 dy dz. Thus, the flux integral is negative.
5. Negative. On S, we have dA
~ = −~i dA, we have
6. Zero. Since dA
Z
Z
S
~ =−
(y~i + x~k ) · dA
y dA = 0.
S
The integral is 0 because the disk is centered on the x-axis, so the contributions from the parts of the disk where y is
~ = −~i dx dy on S, we have
positive and where y is negative cancel. In more detail, since dA
Z
S
Z
~ =
(y~i + x~k ) · dA
S
−y dx dy = 0.
~ = −~i dA and (x − 10) = −3 on S,
7. Positive. Since dA
Z
S
~ =
((x − 10)~i + (x + 10)~j ) · dA
Z
(−3~i + (x + 10)~j ) · (−~i dA) =
Z
S
3 dA = 3 · Area of S.
8. The disk has area 25π, so its area vector is 25π~j . Thus
Flux = (2~i + 3~j ) · 25π~j = 75π.
~ is a constant vector field, the flux through a closed surface is zero. (The flux that enters one side, exits the other
9. Since F
side.)
~ = 5~j on the square,
10. The square has area 16, so its area vector is 16~j . Since F
Flux = 5~j · 16~j = 80.
~ = −5~i on the square,
11. The square has area 9, so its area vector is 9~i . Since F
Flux = −5~i · 9~i = −45.
~ = −~j dxdz and
12. Since the square, S, is in the plane y = 0 and oriented in the negative y-direction, dA
Z
S
~ · dA
~ =
F
Z
S
(0 + 3)~j · (−~j dxdz) = −3
Z
dxdz = −3 · Area of square = −3(22 ) = −12.
S
~ = ~k dxdy and
13. Since the square, S, is oriented upward, dA
Flux =
Z
S
~ · dA
~ =
F
Z
S
x~k · ~k dxdy =
Z
0
3
Z
3
x dxdy =
0
Z
3
0
x2
2
3
dy =
0
9
2
Z
3
dy =
0
27
.
2
14. Since the vector field is constant, Flux = 0. The flux through opposite faces of the cube cancel.
15. The only contribution to the flux is from the ~k -component, and since the square, S, is oriented upward, we have
Flux =
Z
S
~ =
(6~i + x2~j − ~k ) · dA
Z
S
~ = −Area of square = −4.
−~k · dA
16. All the vectors in the vector field point horizontally (because their z-component is zero), and the surface is horizontal, so
there is no flow through the surface and the flux is zero.
~ = ~k dA, so
17. We have dA
Z
S
~ · dA
~ =
F
=
Z
(z~i + y~j + 2x~k ) · ~k dA =
S
3
Z
0
Z
2
2x dxdy = 12.
0
Z
S
2x dA
1778
Chapter Nineteen /SOLUTIONS
~ = ~i dA, so
18. We have dA
Z
S
Z
~ =
F~ · dA
((2 + cos z)~i + y~j + 2x~k ) · ~i dA =
S
4
Z
=
0
3
Z
Z
(2 + cos z) dA
S
(2 + cos z) dydz = 3(8 + sin 4)
0
~ = −~j dA, so,
19. On the surface S, y is constant, y = −1, and dA
Z
S
Z
~ · dA
~ =
F
S
=−
(x2~i + (x + e−1 )~j − ~k ) · (−~j ) dA = −
Z
4
0
Z
2
(x + e
−1
0
) dx dz = −4(2 + 2e
−1
Z
(x + e−1 ) dA
S
) = −8(1 + e−1 ).
~ are parallel to the surface S, so they contribute nothing to the flux integral.
20. Observe that the ~j and ~k components of F
~
~
On the surface S, theRi component R
of F equals 5~i , because x = 0 on S. Since 5~i is normal to S and in the direction of
~ · dA
~ =
~ = k5~i k(Area of S) = 20.
the orientation of S, S F
5~i · dA
S
21. There is no flux through the base or top of the cylinder because the vector field is parallel to these faces. For the curved
~ . The vector field is pointing radially outward from the z-axis and so is
surface, consider a small patchp
with area ∆A
2
2
~ = kF
~ kk∆A
~k=
~
~
parallel to ∆A . Since kF k = x + y = 2 on the curved surface of the cylinder, we have F~ · ∆A
2∆A. Replacing ∆A with dA, we get
Z
S
Z
~ =
F~ · dA
Curved
surface
2 dA = 2(Area of curved surface) = 2(2π · 2 · 3) = 24π.
~ = −y~i + x~j + z~k is tangent to the curved surface of the cylinder. (The area vector is parallel to the
22. The vector field F
vector pointing radially outward from the z-axis, namely x~i + y~j and (−y~i + x~j + z~k ) · (x~i + y~j ) = 0.) Thus the
~ = dA~k , so
only contributions to the flux integral are from the top and the bottom. On the top, z = 1 and dA
~ · dA
~ = (−y~i + x~j + ~k ) · dA~k = dA.
F
Thus
Z
~ · dA
~ =
F
Top
Z
dA = Area of top = π(1)2 = π.
Top
~ = (−dA ~k ), so
Similarly, on the base, z = −1 and dA
~ = (−y~i + x~j − ~k ) · (−dA~k ) = dA.
F~ · dA
Z
Base
Therefore,
~ · dA
~ =
F
Z
dA = Area of base = π.
Base
Total flux through cylinder = Flux through top + Flux through base = 2π.
23. First we have
y
zy = p
.
2
x + y2
x
zx = p
2
x + y2
Although z is not a smooth function of x and y at (0, 0), the improper integral that we get converges:
Z
S
~ · dA
~ =
F
Z
=
Z
(x2~i + y 2~j +
S
S
x3 + y 3
x
~i − p y
~j + ~k ) dA
x2 + y 2~k ) · (− p
2
2
x +y
x2 + y 2
p
−p
+
x2 + y 2
p
x2
+
y2
!
dA
SOLUTIONS to Review Problems for Chapter Nineteen
1779
Changing to polar coordinates we have
Z
S
π/2
~ · dA
~ =
F
Z
π/2
=
Z
Z
π/2
0
0
=
π/2
Z
(−r 2 cos3 θ − r 2 sin3 θ + r)r drdθ
0
1
r4
− (cos3 θ + sin3 θ) + r 3
4
3
0
=
1
Z
0
1
1
− (cos3 θ + sin3 θ) +
4
3
1
r=1
r=0
!
dθ
dθ
1
− (cos θ − cos θ sin2 θ + sin θ − sin θ cos2 θ) +
4
3
π/2
1
1
1
θ
= − (sin θ − sin3 θ − cos θ + cos3 θ) +
4
3
3
3
dθ
0
1
π
= − .
6
3
~ and ~j is positive if b > 0. There are no conditions on a and c.
24. The vector normal to S is ~j ; the dot product of F
~ do not contribute to the flux; only the ~k component
~
~
25. By the symmetry of the sphere, the i and j components of F
contributes. The vector normal to S has a negative ~k component, so we need c > 0. There are no conditions on a and b.
26. The sphere is oriented outward. Provided a > 0, the vector field points outward, giving positive flux.
√
~ = (~i +~j + ~k )/ 3 dA.
27. The normal to the plane is ~i +~j + ~k . Converting this to a unit vector, we see that on the plane dA
The flux integrates the dot product
~ ~j + ~k
a+b+c
~ · dA
~ = (a~i + b~j + c~k ) · i +√
F
√
dA =
dA.
3
3
Thus we need a + b + c > 0.
28. (a) By direct calculation, the flux of F~ through the plane x = 0 is 0 because the ~i -component of the vector field is 0
~ through S, the square 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 in the plane x = 1 is
there. The flux of F
Flux =
Z
S
~ · dA
~ =
F
Z
0
1
Z
1
(1 · y~i + yz~j + z · 1~k ) · ~i dy dz =
0
Z
1
0
Z
1
y dy dz =
0
Z
1
0
y2
2
1
dz =
0
1
.
2
Similarly, the faces y = 0 and y = 1 contribute a flux of 0 and 1/2, respectively, as do the faces z = 0 and z = 1.
Therefore
1
1
3
1
Total flux = 0 + + 0 + + 0 + = .
2
2
2
2
(b) Since div F~ = y + z + x, the flux is given by
Z
Box
~ · dA
~ =
F
1
1
Z
(x + y + z)dV.
Box
We calculate the integral
Z
(x + y + z) dV =
Box
Z
1
0
=
Z
0
1
0
=
Z
0
Z
Z
1
(x + y + z) dx dy dz =
1
+ y + z dy dz =
2
1
1
+ +z
2
2
Z
0
0
1
0
Z
dz =
Z
0
1
Z
0
1
1
Z
1
0
x2
+ yx + zx
2
y2
y
+
+ zy
2
2
(1 + z) dz =
z+
z2
2
1
dz
0
1
=
0
3
.
2
1
dy dz
0
1780
Chapter Nineteen /SOLUTIONS
~ = x3~i + 2y~j + 3~k through each of the six faces of the
29. (a) We will compute separately the flux of the vector field F
cube.
~ has flux through
The face SI where x = 1, which has normal vector ~i . Only the ~i component x3~i = ~i of F
SI .
Z
SI
Z
~ =
F~ · dA
SI
~ = k~i k(area of SI ) = 4.
~i · dA
~ has flux
The face SII where x = −1, which has normal vector −~i . Only the ~i component x3~i = −~i of F
through SII .
Z
~ · dA
~ =
F
Z
~ =
F~ · dA
SII
Z
SII
~ = k − ~i k(area of SII ) = 4.
−~i · dA
~ has flux
The face SIII where y = 1, which has normal vector ~j . Only the ~j component 2y~j = 2~j of F
through SIII .
SIII
Z
~ = k2~j k(area of SIII ) = 8.
2~j · dA
SIII
~ has flux
The face SIV where y = −1, which has normal vector −~j . Only the ~j component 2y~j = −2~j of F
through SIV .
Z
Z
~ · dA
~ =
F
SIV
~ = k − 2~j k(area of SIV ) = 8.
−2~j · dA
SIV
~ has flux through SV .
The face SV where z = 1, which has normal vector ~k . Only the ~k component 3~k of F
Z
SV
~ · dA
~ =
F
Z
SV
~ = k3~k k(area of SV ) = 12.
3~k · dA
~ has flux through
The face SV I where z = −1, which has normal vector −~k . Only the ~k component 3~k of F
SV I .
Z
SV I
Z
~ =
F~ · dA
~ = −k3~k k(area of SV I ) = −12.
3~k · dA
SV I
(Total flux through S) = 4 + 4 + 8 + 8 + 12 − 12 = 24.
~ = 3x2 + 2,
(b) Since S is a closed surface the Divergence Theorem applies. Since divF
Z
S
~ · dA
~ =
F
Z
1
x=−1
Z
1
y=−1
Z
1
(3x2 + 2)dzdydx = 24.
z=−1
30. Since div F~ = −1 + 2 + 2 = 3, we use the Divergence Theorem:
Flux =
Z
S
~ · dA
~ =
F
Z
W
~ dV = 3 · Volume of Sphere = 3 ·
div F
4 3
π2 = 32π.
3
~ = 2 + 3 + 4 = 9, the Divergence Theorem gives
31. Since divF
Z
S
~ =
F~ · dA
Z
Interior
of sphere
div F~ dV = 9 · Volume of sphere = 9 ·
~ = 2x + 2y, the Divergence Theorem gives
32. Since divF
Z
S
~ · dA
~ =
F
Z
=2
(2x + 2y) dV
Interior
of cube
Z
0
3
Z
3
0
Z
0
3
(x + y) dx dy dz
4 3
π5 = 1500π.
3
SOLUTIONS to Review Problems for Chapter Nineteen
Z
=2
0
3
Z
=2
3Z
0
3
Z
=2
0
Z
0
3
3
0
x2
+ xy
2
9
+ 3y
2
dy dz
9
3
y + y2
2
2
1781
3
dy dz
0
3
dz = 2
3
Z
27 dz = 162.
0
0
~ = 1 + 2 − 1 = 2, the Divergence Theorem gives
33. Since divF
Z
S
~ · dA
~ =
F
Z
~ dV = 2 · Volume of sphere = 2 · 4 π13 = 8π .
div F
3
3
Interior
of sphere
~ = 3x2 + 3y 2 , using cylindrical coordinates to calculate the triple integral gives
34. Since divF
Z
S
~ · dA
~ =
F
Z
2
Z
2
(3x + 3y ) dV = 3
Interior
of cylinder
2π
0
Z
5
0
Z
0
2
r 2 · r dr dz dθ = 3 · 2π · 5
r4
4
2
= 120π.
0
35. Since div F~ = 1 + 1 + 1 = 3, the flux through the closed cylinder, S1 , with interior W , is
Z
S1
~ · dA
~ =
F
Z
W
3 dV = 3 · Volume of cylinder = 3π.
With the base of the cylinder oriented downward and the top of the cylinder oriented upward,
Z
S
~ · dA
~ =
F
Z
S1
~ −
F~ · dA
Z
Base
~ −
F~ · dA
Z
Top
~.
F~ · dA
~ is parallel to the base, the flux through the base is 0. The flux through the top is contributed entirely by the ~k
Since F
component. Since z = 1, we have
Flux through top =
Z
Top
~ · dA
~ =
F
Thus
Z
S
Z
Top
~ =
(x~i + y~j + ~k ) · dA
Z
Top
~k · dA
~ = π.
~ · dA
~ = 3π − π = 2π.
F
Problems
z
36.
z
(0, 0, 2)
✛
✛
1
✛
(0, 0, 3)
1
✛
S1
S2
y
−x
x
Figure 19.22
−y
Figure 19.23
1782
Chapter Nineteen /SOLUTIONS
z
(1, 0, 1)
2
x
✛
2
√
2
✛y
√
2
✛
S3
✛
✛
√
√
✛
y
✛
✛
z
x
Figure 19.24
Figure 19.25
~ = (−~i − ~j + ~k ) · (~k ) = 1
Flux through S1 = F~ · A
~ = (−~i − ~j + ~k ) · (~k ) = 1
Flux through S2 = F~ · A
~
~ = (−~i − ~j + ~k ) · (−2~j ) = 2
Flux through S3 = F · A
√
√
~
~
~
For S4 , a normal is −i + k and the area is 2, so A = − 2~i + 2~k
√
√
√
~ ·A
~ = (−~i − ~j + ~k ) · (− 2~i + 2~k ) = 2 2.
Flux through S4 = F
So,
Flux through S1 = Flux through S2 < Flux through S3 < Flux through S4 .
37. (a) We have grad f = (y + yzexyz )~i + (x + xzexyz )~j + xyexyz~k .
(b) By the Fundamental Theorem of Line Integrals, we have
Z
(2,3,4)
C
grad f · d~r = (xy + exyz )
(1,1,1)
= (2 · 3 + e2·3·4 − (1 · 1 + e1·1·1 ) = 5 + e24 − e1 .
~ = ~k dx dy and z = 0, so
(c) Only the k-component of grad f contributes to the flux integral. On the xy-plane, dA
√
2
Z 2 Z √4−x2
Z 2 2 4−x2 Z 2
Z
x4
xy
x
~
=
xye0 dy dx =
(4 − x2 ) dx = x2 −
= 2.
grad f · dA =
2 0
2
8
0
0
0
0
S
0
~ = 4~i . Since the vector
38. The square of side 2 in the plane x = 5, oriented in the positive x-direction, has area vector A
field is constant
Flux = (a~i + b~j + c~k ) · 4~i = 4a = 24.
Thus, a = 6 and we cannot say anything about the values of b and c.
39. (a) In the first and third integrals, only the ~i component contributes to the flux integral. The orientation is in the positive
~i directions in both cases, the disks are the same size, and the (x2 + 4) is larger on the surface S3 , so the flux through
S3 is larger than the flux through S1 . In the case of S2 , only the ~j -component contributes. Since the disks in S2 and
S3 are the same size, and S2 oriented in the positive y-direction, the fact that (x2 + 4) is larger on S3 than y is on S2
tells us that the flux through S3 is largest.
~ = ~i dA, so
(b) On S3 , the vector field is ((−3)2 + 4)~i + y~j = 13~i + y~j and dA
Z
S3
~ =
((x2 + 4)~i + y~j ) · dA
Z
S3
(13~i + y~j ) · ~i dA =
Z
S3
13 dA = 13 · Area of disk = 13π.
40. (a) At the north pole, the area vector of the plate is upward (away from the center of the earth), and so is in the opposite
direction to the magnetic field. Thus the magnetic flux is negative.
(b) At the south pole, the area vector of the plate is again away from the center of the earth (because that is upward in the
southern hemisphere), and so is in the same direction as the magnetic field. Thus, the magnetic flux is positive.
(c) At the equator the magnetic field is parallel to the plate, so the flux is zero.
SOLUTIONS to Review Problems for Chapter Nineteen
1783
41. (a) By the definition of divergence as flux density, we have
Flux out of small region ≈ Divergence · Volume.
~ = 22 + 12 − 12 = 4, so
(i) At (2, 1, 1), we have div F
0.016
4
π = 0.0168.
Flux ≈ 4 · π(0.1)3 =
3
3
~ = 02 + 02 − 12 = −1, so
(ii) At (0, 0, 1), we have div F
Flux ≈ −1 ·
4
π(0.1)3 = −0.004.
3
(b) The fact that the flux in part (i) is positive tells us that the vector field is pointing, on average, outward (rather than
inward) on the sphere around (2, 1, 1). The fact that the flux in part (ii) is negative tells us that, on average, the vector
field is pointing inward (rather than outward) on the sphere around the point (0, 0, 1).
42. (a) The cube is in Figure 19.26. The vector field is parallel to the x-axis and zero on the yz-plane. Thus the only
~ is constant on S2 , the flux through
contribution to the flux is from S2 . On S2 , x = c, the normal is outward. Since F
face S2 is
Z
S2
~ · dA
~ =F
~ ·A
~ S2
F
= c~i · c2~i
= c3 .
Thus, total flux through box = c3 .
(b) Using the geometric definition of divergence
Flux through box
c→0
Volume of box
3
c
= lim
c→0
c3
=1
~ = lim
div F
(c) Using partial derivatives,
∂
∂
∂
(x) +
(0) +
(0) = 1 + 0 + 0 = 1.
div F~ =
∂x
∂y
∂z
S1 (back)
❄
z
✛ S4 (back)
S6
y
S3
S2
■
x
Figure 19.26
S5 (bottom)
1784
Chapter Nineteen /SOLUTIONS
43. See Figure 19.26.
R
~ · dA
~ =
(a) Since 2~i + 3~k is a constant field, its contribution to the flux is zero (flux in cancels flux out). Therefore F
R
R
R
~
~
~
~
~
~
~
(y j ) · dA = S y j · dA + S y j · dA since only S3 and S4 are perpendicular to y j . On S3 , y = 0 so
3
4
R
R
~ = c(Area of S4 ) =
~ = 0. On S4 , y = c and normal is in the positive y-direction, so
y~j · dA
y~j · dA
S
S
4
3
c · c2 = c3 . Thus, total flux = c3 .
(b) Using the geometric definition of divergence
Flux through box
c→0
Volume of box
3
c
= 1.
= lim
c→0
c3
~ = lim
div F
(c)
∂
∂
∂
(2) +
(y) +
(3) = 0 + 1 + 0 = 1.
∂x
∂y
∂z
44. See Figure 19.26.
(a) This vector field points radially outward from the z-axis. Thus, the vector field is parallel to the surface on S1 , S3 , S5
and S6 , so the only contributions to the flux integral are from S2 and S4 .
On S2 , x = c and normal is in the positive x-direction, so the flux is
Z
S2
~ · dA
~ =
F
Z
S2
(c~i + y~j ) · (dA~i ) =
Z
Z
S4
(x~i + c~j ) · (dA~j ) =
Z
c dA = c(Area of S2 ) = c3 .
S2
Similarly, the flux through S4 is
Z
S4
~ · dA
~ =
F
c dA = c(Area of S4 ) = c3 .
S4
Thus, the total flux through the box = 2c3 .
(b) Using the geometric definition of divergence, we have
Flux through surface of box
c→0
Volume of box
3
2c
= 2.
= lim
c→0
c3
~ = lim
div F
(c)
∂
∂
∂
(x) +
(y) +
(0) = 1 + 1 + 0 = 2.
∂x
∂y
∂z
45. We close the cylinder, S, by adding the circular disk, S1 , at the top, z = 3. The surface S + S1 is oriented outward, so S1
is oriented upward. Applying the Divergence Theorem to the closed surface S + S1 enclosing the region W , we have
Z
S+S1
Z
S
Since
Z
~ =
F~ · dA
~ · dA
~ +
F
Z
S1
div F~ dV
W
~ =
F~ · dA
Z
~ dV.
div F
div F~ = div(z 2~i + x2~j + 5~k ) = 0,
we have
Z
S
~ =−
F~ · dA
Z
S1
~.
F~ · dA
~ contributes to the flux through S1 , and dA
~ = ~k dx dy on S1 , so
Only the ~k -component of F
Z
S
~ · dA
~ =−
F
Z
S1
(z 2~i + x2~j + 5~k ) · ~k dx dy = −
Z
S1
√
5 dx dy = −5 · Area of S1 = −5π( 2)2 = −10π.
SOLUTIONS to Review Problems for Chapter Nineteen
1785
46. We close the cylinder, S, by adding the circular disk, S1 , at the top, z = 3. The surface S + S1 is oriented outward, so S1
is oriented upward. Applying the Divergence Theorem to the closed surface S + S1 enclosing the region W , we have
Z
S+S1
Z
S
Since
Z
~ =
F~ · dA
Z
~ · dA
~ +
F
S1
div F~ dV
W
~ =
F~ · dA
Z
~ dV.
div F
~ = div(y 2~i + z 2~j + (x2 + y 2 )~k ) = 0,
div F
we have
Z
S
Z
~ =−
F~ · dA
S1
~.
F~ · dA
~ contributes to the flux through S1 , and dA
~ = ~k dx dy on S1 , so
Only the ~k -component of F
Z
S
~ · dA
~ =−
F
Z
S1
(y 2~i + z 2~j + (x2 + y 2 )~k ) · ~k dx dy = −
Converting to polar coordinates, since the cylinder has radius
Z
S
2π
Z
~ =−
F~ · dA
0
Z
√
2
√
Z
S1
2, we have
√
r4
r · r dr dθ = −2π
4
2
2
0
(x2 + y 2 ) dx dy.
= −2π.
0
47. We close the cylinder, S, by adding the circular disk, S1 , at the top, z = 3. The surface S + S1 is oriented outward, so S1
is oriented upward. Applying the Divergence Theorem to the closed surface S + S1 enclosing the region W , we have
Z
S+S1
Z
S
Since
Z
~ =
F~ · dA
~ · dA
~ +
F
Z
S1
div F~ dV
W
~ =
F~ · dA
Z
~ dV.
div F
~ = div(z~i + x~j + y~k ) = 0,
div F
we have
Z
S
Z
~ =−
F~ · dA
S1
~.
F~ · dA
~ contributes to the flux through S1 , and dA
~ = ~k dx dy on S1 , so
Only the ~k -component of F
Z
S
~ · dA
~ =−
F
Z
S1
(z~i + x~j + y~k ) · ~k dx dy = −
Z
y dx dy.
S1
Since y is an odd function, by symmetry, its integral over S1 is zero. Thus,
Z
S
~ · dA
~ =−
F
Z
y dx dy = 0.
S
48. We close the cylinder, S, by adding the circular disk, S1 , at the top, z = 3. The surface S + S1 is oriented outward, so S1
is oriented upward. Applying the Divergence Theorem to the closed surface S + S1 enclosing the region W , we have
Z
S+S1
Z
S
Since
~ =
F~ · dA
~ · dA
~ +
F
Z
S1
Z
div F~ dV
W
~ =
F~ · dA
Z
~ dV.
div F
~ = div(y 2~i + x2~j + 7z~k ) = 7,
div F
1786
Chapter Nineteen /SOLUTIONS
we have
Z
S
~ =
F~ · dA
Z
W
Z
7 dV −
S1
~ · dA
~ = 7 · Volume of cylinder −
F
√
= 7 · π( 2)2 6 −
Z
S1
Z
S1
~ · dA
~
F
~ · dA
~ = 84π −
F
Z
S1
~ · dA
~.
F
~ contributes to the flux through S1 , and dA
~ = ~k dx dy and z = 3 on S1 , so
Only the ~k -component of F
Z
S
~ = 84π −
F~ · dA
Z
(y i + x j + 7 · 3~k ) · ~k dx dy = 84π −
2~
S1
2~
Z
√
= 84π − 21 · Area of S1 = 84π − 21 · π( 2)2 = 42π.
21 dx dy
S1
49. We close the cylinder, S, by adding the circular disk, S1 , at the top, z = 3. The surface S + S1 is oriented outward, so S1
is oriented upward. Applying the Divergence Theorem to the closed surface S + S1 enclosing the region W , we have
Z
~ =
F~ · dA
S+S1
Z
~ · dA
~ +
F
S
Since
Z
Z
div F~ dV
W
~ =
F~ · dA
S1
Z
~ dV.
div F
~ = div(x3~i + y 3~j + ~k ) = 3x2 + 3y 2 ,
div F
we have
Z
S
~ · dA
~ =
F
Z
2
W
2
(3x + 3y ) dV −
Z
S1
~ · dA
~.
F
~ = ~k dx dy,
To find the integral over W , we use cylindrical coordinates. For the integral over S1 , we use the fact that dA
~ contributes to the flux.
so only the k-component of F
Z
S
~ · dA
~ =
F
Z
2π
0
Z
3
−3
2π
=θ
3
0
2
3r 2 · r dr dz dθ −
√
3 4
r
4
z
0
√
Z
−
0
−3
2
Z
Z
S1
(x3~i + y 3~j + ~k ) · ~k dx dy
dx dy
S1
3 √ 4
( 2) − Area of S1
4
√
= 36π − π( 2)2 = 34π.
= 2π · 6 ·
~ = 0, the flux through the cone equals the flux upward through the disk r ≤ 4
50. By the Divergence Theorem, since div F
~ = π42~k . Since the flux is negative, F
~ = −c(~i + ~k ) with c > 0.
in the plane z = 4. The area vector of the disk is A
Thus
~ ·A
~ = −c(~i + ~k ) · π42~k = −16πc = −7.
Flux = F
so
7
~ = − 7 (~i + ~k ).
and F
c=
16π
16π
~
~
~
51. (a) Since ~r = xi + y j , for ~r 6= 0 , we have
div
~r
||~r ||2
x~i + y~j
x2 + y 2
∂
∂
x
y
= div
=
+
∂x x2 + y 2
∂y x2 + y 2
1
x · 2x
1
y · 2y
= 2
− 2
+ 2
− 2
x + y2
(x + y 2 )2
x + y2
(x + y 2 )2
x2 + y 2 − 2x2 + x2 + y 2 − 2y 2
(x2 + y 2 )2
= 0.
=
SOLUTIONS to Review Problems for Chapter Nineteen
1787
(b) For ~r = 0, which is the z-axis.
(c) Since the cylinder does not include the z-axis, we can use the Divergence Theorem. If S is the cylinder and W is its
interior, we have
Z
Z
Z
~r
~r
~
dV =
· dA =
div
0 dV = 0.
||~r ||2
||~r ||2
W
W
S
(d) Since the vector field ~r /||~r ||2 has no component perpendicular to the flux there, it has no flux through the ends. Thus
the flux through the sides of this cylinder is also 0.
52. (a) The divergence of F~ must be zero.
(b) We calculate the divergence:
div(a(ex + y − x)~i + 12y(1 − ex )~j ) = aex − a + 12(1 − ex ) = (a − 12)ex + 12 − a.
Thus, the divergence is 0 if a = 12.
53. (a) First we calculate
~i ~j ~k
~r × ~a = x y z
= (ya3 − za2 )~i + (za1 − xa3 )~j + (xa2 − ya1 )~k .
a1 a2 a3
Then we calculate
div(~r × ~a ) = div (ya3 − za2 )~i + (za1 − xa2 )~j + (xa2 − ya1 )~k
(b) The cube is closed and oriented outward. By the Divergence Theorem,
Z
Box
~ =
(~r × ~a ) · dA
Z
Interior of box
= 0.
div(~r × ~a )dV = 0.
54. We use the Divergence Theorem, where W is the region inside S:
Z
S
~ =
F~ · dA
Z
div F dV.
W
Since W = W1 + W2 , where W1 is the region inside the cube and W2 is the region outside the cube and inside the sphere
S, we find the integrals over W1 and W2 separately:
Z
W1
Z
W2
Thus,
Z
S
div F dV = 3 · Volume W1 = 3 · 43 = 192.
div F dV = 5 · Volume W2 = 5 ·
~ · dA
~ =
F
Z
div F dV = 192 +
W
4
π(10)3 − 43
3
=
20,000
π − 320.
3
20,000
20,000
π − 320 =
π − 128.
3
3
55. (a) The disk S2 is in the plane x = 2. It is oriented away from the origin because S is closed and oriented outward. See
Figure 19.27.
1788
Chapter Nineteen /SOLUTIONS
z
S2
S1
x
2
p
Figure 19.27: The cone x =
0≤x≤2
(b)
y 2 + z 2 for
~ = ~i dA, so only the ~i component of F~ contributes to the flux through S2 . On S2 , we have
(i) On S2 , we have dA
x = 2 so
Z
Z
~ · dA
~ =
F
3 · 2~i · ~i dA = 6 · Area of S2 = 6(π22 ) = 24π.
S2
S2
(ii) Since it is difficult to calculate the flux through S1 directly, we use the Divergence Theorem, with
∂
∂
∂
(3x) +
(4y) +
(5z) = 3 + 4 + 5 = 12.
∂x
∂y
∂z
~ =
div F
Let W be the interior of the cone. Then by the Divergence Theorem,
Z
S
so
Z
~ · dA
~ =
F
S
Thus
Z
S1
Z
~ · dA
~ =
F
Z
W
S1
~ · dA
~ +
F
Z
S2
Z
~ · dA
~ =
F
12 dV = 12 · Volume of cone = 12 ·
~ · dA
~ =
F
Z
W
~ dV −
div F
Z
S2
~ dV,
div F
W
1 2
π2 · 2 = 32π.
3
~ = 32π − 24π = 8π.
F~ · dA
56. (a) On the disk, only the ~i -component, 5~i , contributes to the integral, so
Z
S1
~ =
~r · dA
Z
S1
√
5~i · ~i dA = 5 · Area of disk = 5π( 7)2 = 35π.
(b) Using the Divergence Theorem, with W the region within the cylinder and div ~r = 3,
Z
Z
√
~ =
~r · dA
3 dV = 3 · Volume of cylinder = 3 · π( 7)2 · 5 = 105π.
S2
W
(c) The closed cylinder S2 has sides consisting of S1 and S3 , and also a disk, S4 , in the yz-plane y 2 + z 2 ≤
Since the ~i component is 0 on S4 , the flux through S4 is 0. Thus,
Z
S2
~ =
~r · dA
Z
S1
~ +
~r · dA
105π = 35π +
so
Z
S3
Z
S3
Z
S3
~ +
~r · dA
Z
~ + 0,
~r · dA
~ = 105π − 35π = 70π.
~r · dA
S4
~.
~r · dA
√
7, x = 0.
SOLUTIONS to Review Problems for Chapter Nineteen
57. We cannot find
Z
S
1789
~ · dA
~ directly as we do not know f1 and f2 . We close the surface by adding D, the disk x2 + y 2 ≤ 9
F
in the xy-plane, oriented upward. If W is the solid region enclosed, the Divergence Theorem tells us
Z
S+D
Z
~ · dA
~ =
F
~ dV =
div F
Z
5 dV = 5 · Volume of W = 5 ·
W
W
Thus,
Z
S
~ +
F~ · dA
Z
D
1 4 3
· π3 = 90π.
2 3
~ = 90π.
F~ · dA
~ contributes to the flux through D, where dA
~ = ~k dA, so
Only the ~k component of F
Z
~ =
F~ · dA
Z
~ · dA
~ = 90π −
F
D
Thus,
S
~ =
58. (a) Since F
Z
D
~k · ~k dA = Area of D = π32 = 9π.
Z
D
~ · dA
~ = 90π − 9π = 81π.
F
x~i + y~j + z~k
,
(x2 + y 2 + z 2 )3/2
∂
∂F1
=
x(x2 + y 2 + z 2 )−3/2
∂x
∂x
3
= 1(x2 + y 2 + z 2 )−3/2 − x(x2 + y 2 + z 2 )−5/2 · 2x
2
x2 + y 2 + z 2 − 3x2
=
.
(x2 + y 2 + z 2 )5/2
Similar calculations for ∂F2 /∂y and ∂F3 /∂z show that
~ =
div F
3(x2 + y 2 + z 2 ) − 3x2 − 3y 2 − 3z 2
∂F2
∂F3
∂F1
+
+
=
= 0.
∂x
∂y
∂z
(x2 + y 2 + z 2 )5/2
~ is undefined at the origin, but it is zero everywhere else.
Thus div F
~ is everywhere perpendicular to the surface of the sphere, and on the sphere we have
(b) The vector field F
~ || =
||F
Thus
Z
S
1
10
= 2.
103
10
~ · dA
~ = 1 · Area of sphere = 1 · 4π102 = 4π.
F
102
102
(c) Since div F~ = 0 throughout B1 , by the Divergence Theorem
Z
B1
~ · dA
~ =
F
Z
~ dV = 0.
div F
Interior of B1
~ is not defined at the origin, we cannot use the Divergence Theorem in the same way as we did in part (c).
(d) Since div F
However, we can apply the Divergence Theorem to W , the space between B2 and S, with B2 oriented inward:
Z
S
Thus
Z
~ +
F~ · dA
B2 (outward)
Z
B2 (inward)
~ · dA
~ =−
F
Z
~ =
F~ · dA
B2 (inward)
Z
div F~ dV = 0,
W
~ · dA
~ =
F
Z
S
~ · dA
~ = 4π.
F
(e) If the origin is outside the surface, the flux through the surface is 0, by the Divergence Theorem. If the origin is inside
the surface, then we apply the Divergence Theorem to the region between the surface and a sphere. The sphere is
taken large enough to contain the surface entirely within it, or small enough to lie entirely within the surface. We find
that the flux is 4π. The only case we do not consider is when the origin actually lies on the surface.
1790
Chapter Nineteen /SOLUTIONS
~ in terms of (x, y, z) as
59. We can rewrite F
−Gmx
~k
~j + p −Gmz
~i + p −Gmy
F~ = p
(x2 + y 2 + z 2 )3
(x2 + y 2 + z 2 )3
(x2 + y 2 + z 2 )3
Now we find the divergence of F~ in the usual manner
−Gmy
−Gmz
∂
∂
~ = ∂ p −Gmx
div F
+
p
+
p
∂x (x2 + y 2 + z 2 )3
∂y (x2 + y 2 + z 2 )3
∂z (x2 + y 2 + z 2 )3
= −3Gm
3Gm
x2 + y 2 + z 2
+ 2
+ y 2 + z 2 )5/2
(x + y 2 + z 2 )3/2
(x2
= 0.
~ = 0 for all points except the origin. We consider a region enclosed by two concentric spheres. Since the
Thus, div F
divergence of the field is zero at all points except the origin, the volume enclosed contains only points with zero divergence.
Consequently, the flux through the surface of the enclosed volume must be zero. Since the field is always inward pointing,
this is equivalent to saying that the flux into the outer sphere must equal the flux out of the inner sphere, and so we see
that for any two spheres, the flux must be equal, which shows that the flux is independent of the radius of the spheres.
60. Since the divergence is zero at all points not containing the charge, the flux must be zero through any closed surface
containing no charge. We imagine a surface composed of two concentric cylinders and their end-caps, where the axis of
both cylinders is the z-axis. Then, since no charge is contained in the region enclosed, the flux through the surface must
be zero. Now, we know that the field points away from the axis, which means it is parallel to the end-caps. Consequently,
there must be no flux through the end-caps. This implies that the flux through the inner cylinder must equal the flux out
of the outer cylinder. Since the strength of the field only depends upon the distance from the z axis, the flux through each
cylinder is a constant. This implies that the following equation must hold
Flux through each cylinder = Ea 2πra L = Eb 2πrb L
where Ea and Eb are the strengths of the field at ra and rb respectively, and L is the length of the cylinders. Dividing
through, we can arrive at the following relationship:
Ea /Eb = rb /ra
If we take Eb to be a constant at a fixed value of rb , then the equation can be simplified to
Ea = k/ra
where k = Eb rb . Thus we see that the strength of the field is proportional to 1/r.
61. (a) If we examine the equation for ~v , we see that when r = 0, that is, at the center of the pipe, ~v (0) becomes u~i . So u is
the speed at the center of the pipe; it is also the maximum speed since u(1 − r 2 /a2 ) reaches its maximum at r = 0.
(b) The flow rate at the wall of the pipe (where r = a) is
~v (a) = u(1 − a2 /a2 )~i = ~0 .
(c) To find the flux through a circular cross-sectional area, we use polar coordinates in the plane perpendicular to the
~ becomes r dr dθ~i . So the flux is given by
velocity. In these coordinates, an infinitesimal area, dA
Flux =
Z
S
~ =
~v · dA
= 2πu
Z
0
62. (a)
a
Z
S
3
u(1 − r 2 /a2 )~i · r dr dθ~i =
r
(r − 2 ) dr = 2πu
a
a2
a2
−
2
4
=
Z
2π
0
Z
0
a
u(1 − r 2 /a2 )r dr dθ
πua2
.
2
R
(i) The integral W ρ dV represents the total charge in the volume W .
R
~ represents the total current flowing out of the surface S.
(ii) The integral S J~ · dA
(b) The total current flowing out of the surface S is the rate at which the total charge inside the surface S (i.e., in the
volume W ) is decreasing. In other words,
Rate current flowing out of S = −
so
Z
~ =−∂
J~ · dA
∂t
S
Z
W
∂
(charge in W ),
∂t
ρ dV
.
SOLUTIONS to Review Problems for Chapter Nineteen
1791
63. (a) Since ~v = grad φ we have
~v =
y 2 − x2
(x2 + y 2 )2
1+
−2xy ~
j
(x2 + y 2 )2
~i +
(b) Differentiating the components of ~v , we have
div ~v =
∂
∂x
1+
y 2 − x2
(x2 + y 2 )2
∂
∂y
+
−2xy
(x2 + y 2 )2
=
2x(x2 − 3y 2 )
2x(3y 2 − x2 )
+
=0
2
2
3
(x + y )
(x2 + y 2 )3
(c) The vector ~v is tangent to the circle x2 +y 2 = 1, if and only if the dot product of the field on the circle with any radius
vector of that circle is zero. Let (x, y) be a point on the circle. We want to show: ~v · ~r = ~v (x, y) · (x~i + y~j ) = 0.
We have:
~v (x, y) · (x~i + y~j ) =
y 2 − x2
1+ 2
(x + y 2 )2
= x+x
=
~i +
−2xy ~
j
(x2 + y 2 )2
2xy 2
y 2 − x2
− 2
2
2
2
(x + y )
(x + y 2 )2
· (x~i + y~j )
x(x2 + y 2 − 1)
,
x2 + y 2
but we know that for any point on the circle, x2 + y 2 = 1, thus we have ~v · ~r = 0. Therefore, the velocity field is
tangent to the circle. Consequently, there is no flow through the circle and any water on the outside of the circle must
flow around it.
(d) See Figure 19.28.
Figure 19.28
p
64. (a) The distance from any point to the origin is given by x2 + y 2 + z 2 , so the denominator is simply r 3 . The field can
~i + ky3 ~j + kz3 ~k ). Its magnitude is thus:
then be rewritten in components as ( kx
r3
r
r
k~v k =
r
K2
x2 + y 2 + z 2
=
r6
r
K2
K
= 2
r4
r
which is only a function of the distance from the origin. It is clear that the vector field points away from the origin for
all points (x, y, z), because it is the radius vector x~i + y~j + z~k , multiplied by the positive scalar K/r 3 . Suppose
(x, y, z) 6= (0, 0, 0), then
div ~v (x, y, z) =
∂
∂x
=K
Kx
r3
+
∂
∂y
Ky
r3
+
∂
∂z
Kz
r3
−2x2 + y 2 + z 2
x2 − 2y 2 + z 2
x2 + y 2 − 2z 2
+
+
r5
r5
r5
Hence, indeed ~v is a point source at the origin.
= 0.
1792
Chapter Nineteen /SOLUTIONS
(b) The dependence of ~v on r, the distance from the origin, is shown in part (a).
(c) The flux through a sphere centered at the origin is calculated as:
Flux =
Z
S
~
~v · dA
Since the vector field’s magnitude is a function only of the distance from the origin, it will be constant over the
~ will be
surface of a sphere centered at the origin. Furthermore, since it is pointed away from the origin, ~v · dA
~
simply k~v k · kdA k. Thus
~
Total flux out of a sphere of radius r = k~v r k · kA
sphere k =
K4 2
4
πr = πK.
r2 3
3
So the flux does not even depend upon r since the rate at which the area of the sphere is increasing is exactly equal
to the rate at which the magnitude of the field is decreasing.
(d) This is best handled by observing (as in part (a)) that the divergence of the vector field at any point besides the origin
is zero. Since the divergence anywhere but the origin is zero, the net flux through any closed surface not enclosing
the origin must also be zero.
CAS Challenge Problems
65. (a) When x > 0, the vector x~i points in the positive x-direction, and when x < 0 it points in the negative x-direction.
Thus it always points from the inside of the ellipsoid to the outside, so we expect the flux integral to be positive. The
upper half of the ellipsoid is the graph of z = f (x, y) = √12 (1 − x2 − y 2 ), so the flux integral is
Z
S
~ · dA
~ =
F
Z
=
Z
1/2
−1/2
1/2
−1/2
=
1/2
Z
x~i · (−fx~i − fy~j + ~k ) dxdy
−1/2
1/2
Z
(−xfx ) dxdy =
−1/2
Z
1/2
−1/2
Z
1/2
−1/2
x2
p
1 − x2 − y 2
dxdy
√
− 2 + 11 arcsin( √13 ) + 10 arctan( √12 ) − 8 arctan( √52 )
12
= 0.0958.
Different CASs may give the answer in different forms. Note that we could have predicted the integral was positive
without evaluating it, since the integrand is positive everywhere in the region of integration.
(b) For x > −1, the quantity x + 1 is positive, so the vector field (x + 1)~i always points in the direction of the positive
x-axis. It is pointing into the ellipsoid when x < 0 and out of it when x > 0. However, its magnitude is smaller when
−1/2 < x < 0 than it is when 0 < x < 1/2, so the net flux out of the ellipsoid should be positive. The flux integral
is
Z
S
~ =
F~ · dA
Z
=
Z
1/2
−1/2
1/2
−1/2
√
=
1/2
Z
−1/2
(x + 1)~i · (−fx~i − fy~j + ~k ) dxdy
1/2
Z
−1/2
−(x + 1)fx dxdy =
Z
1/2
−1/2
Z
1/2
−1/2
x(1 + x)
p
1 − x2 − y 2
2 − 11 arcsin( √13 ) − 10 arctan( √12 ) + 8 arctan( √52 )
12
dxdy
= 0.0958
The answer is the same as in part (a).p
This makes sense because the difference between the integrals in parts (a) and
R 1/2 R 1/2
(b) is the integral of −1/2 −1/2 (x/ 1 − x2 − y 2 ) dxdy, which is zero because the integrand is odd with respect
to x.
(c) This integral should be positive for the same reason as in part (a). The vector field y~j points in the positive y-direction
when y > 0 and in the negative y-direction when y < 0, thus it always points out of the ellipsoid. Evaluating the
integral we get
Z
S
~ · dA
~ =
F
Z
=
Z
1/2
−1/2
1/2
−1/2
Z
1/2
−1/2
Z
y~j · (−fx~i − fy~j + ~k ) dxdy
1/2
−1/2
(−yfy ) dxdy =
Z
1/2
−1/2
Z
1/2
−1/2
y2
p
1 − x2 − y 2
dxdy
PROJECTS FOR CHAPTER NINETEEN
√
=
2 − 2 arcsin( √13 ) − 19 arctan( √12 ) + 8 arctan( √52 )
12
1793
= 0.0958.
The symbolic answer appears different but has the same numerical value as in parts (a) and (b). In fact the answer is
the same because the integral here is the same as in part (a) except that the roles of x and y have been exchanged.
Different CASs may give different symbolic forms.
66. (a) The surface has a shape of a flower or trumpet opening in the direction of the positive y-axis. See Figure 19.29.
The outer rim is a circle of radius 4, so the surface lies above z = −4. Thus z + 4 > 0 on the surface, so the
vector field (z + 4)~k points in the positive z direction everywhere on the surface. Thus it crosses the surface in the
opposite direction as the orientation when it is below the xy-plane, and in the same direction when it is above the
xy-plane. Also, it has smaller magnitude when −2 ≤ z ≤ 2 than it does when 0 ≤ z ≤ 2, so we expect the negative
contribution to the flux integral to be smaller than the positive contribution, so the flux integral should be positive.
z
−x
y
Figure 19.29
(b) The area vector element is
∂~r
∂~r
×
= s2 cos t~i − (2s3 cos(t)2 + 2s3 sin2 t)~j + s2 sin(t)~k .
∂s
∂t
This points in the direction of the negative y-axis, as required for computing the flux integral. The flux integral is
Z
S
~ =
F~ · dA
=
Z
2π
0
Z
0
Z
2
0
2π
Z
dsdt
(s2 sin t + 2)s2 sin t dsdt =
32π
5
~ (~r (s, t)) ·
F
2
0
∂~r
∂~r
×
∂s
∂t
PROJECTS FOR CHAPTER NINETEEN
1. (a) Taking partial derivatives using the product rule, we have
∂
x
∂F1
=
∂x
∂x (x2 + y 2 + z 2 )3/2
1
3
x · 2x
= 2
−
2
2
3/2
2
2 (x + y 2 + z 2 )5/2
(x + y + z )
y 2 + z 2 − 2x2
.
= 2
(x + y 2 + z 2 )5/2
Similarly,
∂F2
∂
=
∂y
∂y
y
2
2
(x + y + z 2 )3/2
=
x2 + z 2 − 2y 2
(x2 + y 2 + z 2 )5/2
1794
Chapter Nineteen /SOLUTIONS
and
∂F3
∂
=
∂z
∂z
Thus,
z
2
2
(x + y + z 2 )3/2
=
x2 + y 2 − 2z 2
.
(x2 + y 2 + z 2 )5/2
∂F2
∂F3
∂F1
+
+
= 0.
divergence of F~ =
∂x
∂y
∂z
(b) Let S be the closed surface, oriented outward, consisting of three pieces
• S1
• −S2
• S3 , the surface connecting the boundaries of S1 and S2 , consisting of line segments on lines through
the origin.
By the Divergence Theorem and part (a),
Z
~ = 0.
F~ · dA
S
Hence
Z
~ −
F~ · dA
S1
Z
~ +
F~ · dA
S2
Z
~ = 0.
F~ · dA
S3
The key observation is that the vector field F~ is tangent to the lateral surface S3 . Therefore,
~ = 0 and it follows that
dA
Z
~ =
F~ · dA
S1
Z
~.
F~ · dA
R
S3
F~ ·
S2
(c) Compute the integral over S2 , which is on the unit sphere. Notice that F~ is perpendicular to the sphere
and that on the sphere, ||F~ || = 1. Thus
Z
Z
~ =
~ = ||F~ || · Area of S2 = Area ofS2 .
F~ · dA
F~ · dA
S1
S2
2. (a) Since ~e ρ is a unit vector pointing radially away from the origin, ~e ρ = ~r /k~r k = ~r /ρ. Thus, we have
f (ρ)
f (ρ) ~
f (ρ) ~
f (ρ) ~
F~ =
~r =
xi +
yj +
zk .
ρ
ρ
ρ
ρ
p
Let g(ρ) = f (ρ)/ρ. Since ρ = k~r k = x2 + y 2 + z 2 , and ∂ρ/∂x = 21 (x2 + y 2 + z 2 )−1/2 · 2x =
p
x/ x2 + y 2 + z 2 , using the chain rule, we have
d
∂ρ
x
∂
g(ρ) =
g(ρ) ·
= p
g ′ (ρ)
2
2
2
∂x
dρ
∂x
x +y +z
∂
y
g(ρ) = p
g ′ (ρ)
∂y
x2 + y 2 + z 2
∂
z
g(ρ) = p
g ′ (ρ).
2
∂z
x + y2 + z 2
So
∂
∂
∂
(g(ρ)x) +
(g(ρ)y) +
(g(ρ)z)
∂x
∂y
∂z
∂
∂
∂
= x g(ρ) + g(ρ) + y g(ρ) + g(ρ) + z g(ρ) + g(ρ)
∂x
∂y
∂z
div F~ =
PROJECTS FOR CHAPTER NINETEEN
1795
x2
y2
z2
= p
g ′ (ρ) + p
g ′ (ρ) + p
g ′ (ρ) + 3g(ρ)
x2 + y 2 + z 2
x2 + y 2 + z 2
x2 + y 2 + z 2
ρ2
x2 + y 2 + z 2 ′
g (ρ) + 3g(ρ) = g ′ (ρ) + 3g(ρ) = ρg ′ (ρ) + 3g(ρ)
= p
ρ
x2 + y 2 + z 2
′
3f (ρ)
f (ρ) f (ρ)
f (ρ)
+
=ρ
− 2
= f ′ (ρ) + 2
.
ρ
ρ
ρ
ρ
On the other hand,
1
f (ρ)
1 d 2
(ρ f (ρ)) = 2 (2ρf (ρ) + ρ2 f ′ (ρ)) = 2
+ f ′ (ρ).
2
ρ dρ
ρ
ρ
Thus,
1
div F~ = 2 (ρ2 f (p)).
ρ
(b) Applying the formula from part (a), we have
1 d
div F~ = 2 (ρ2 f (ρ)) = 0.
ρ dρ
Since ρ 6= 0, multiplying by ρ2 gives us (d/dρ)(ρ2 f (ρ)) = 0, hence ρ2 f (ρ) is a constant, say k. So
f (ρ) = k/ρ2 , and hence
k
F~ = 2 ~e ρ .
ρ
(c) Suppose the sphere, S, has radius R. Since F~ has constant magnitude
on SRand points perpendicularly
to
R
R
~ = f (R) dA, so the flux of F~ out of S is F~ · dA
~ = f (R) dA = f (R) dA =
S, we have F~ · dA
S
S
S
f (R) · Area of S = 4πR2 f (R). On the other hand, using spherical coordinates, part (a) tells us that:
div F~ =
so if W is the region enclosed by S then
Z
div F~ dV =
W
Z
0
1 d
ρ2 f (ρ) ,
2
ρ dρ
R Z π Z 2π
= 4π
Z
0
R
0
0
d
dρ
1 d 2
(ρ f (ρ)) ρ2 sin φ dθ dφ dρ
ρ2 dρ
R
ρ2 f (ρ) dρ = 4πρ2 f (ρ) = 4πR2 f (R).
0
R
R
~ , confirms the Divergence Theorem.
The fact that W div F~ dV equals the flux integral, S F~ · dA
~ is spherically symmetric, we can write
(d) Since we are assuming E
~ = E(ρ)~e ρ .
E
By part (a) and Gauss’s Law,
~ = 1 d (ρ2 E(ρ)) =
div E
ρ2 dρ
(
δ0
0
Suppose ρ ≤ a. Then, since ρ 6= 0,
d 2
(ρ E(ρ)) = δ0 ρ2 .
dρ
ρ≤a
ρ>a
1796
Chapter Nineteen /SOLUTIONS
Therefore
ρ2 E(ρ) =
and so
Z
δ 0 ρ3
+ C,
3
δ0 ρ2 dρ =
E(ρ) =
C
δ0 ρ
+ 2.
3
ρ
~ , and hence E, is assumed to be continuous, we must have C = 0, so E(ρ) = δ0 ρ/3.
Since E
Now suppose that ρ > a. Then
d 2
(ρ E(ρ)) = 0,
dρ
therefore
ρ2 E(ρ) = k,
and so
E(ρ) =
k
.
ρ2
Since E is assumed to be continuous,
lim E(ρ) = lim E(ρ).
ρ→a+
ρ→a−
Now,
k
a2
ρ→a
δ0 a
lim E(ρ) =
3
ρ→a−
lim+ E(ρ) =
so k = δ0 a3 /3. Thus
δ ρ
0
E(ρ) = δ3a3
0
3
ρ≤a
ρ>a
3. (a) (i) Since the direction of the electric field is perpendicular to the surface of any cylinder with the wire
as an axis, it is parallel to the surfaces of the two washers. Consequently, there is no flux through the
washers.
(ii) Gauss’s Law tells us that the total flux through the surface must be zero, since no charge is contained
within it. (Note that the region within the surface S lies between the cylinders.) Since the flux through
the washers is zero, the flux into the inner surface must equal the flux out of the outer surface in order
for the net flux through the surface to be zero.
(iii) Since the surface area of a cylinder is given by A = 2πRL where R is the radius of the cylinder and
L is its length, and we know that Ea Aa = Eb Ab (since the fluxes are equal), we have
Eb (2πbL) = Ea (2πaL)
Eb
2πaL
=
Ea
2πbL
a
Eb
= .
Ea
b
(iv) The equations in part (iii) imply that
aEa = bEb .
PROJECTS FOR CHAPTER NINETEEN
1797
Since a, b are arbitrary radii we can say:
rEr = Constant
Er = Constant
1
,
r
for any radius r. This statement tells us that the strength of the electric field at r is proportional to
1/r.
(b) Since the electric field points perpendicular to the sheet, it is parallel to all sides of the box, except for the
two sides parallel to the sheet. Additionally, since there is no charge contained in the box, Gauss’s Law
tells us the net flux through the surface of the box must be zero. This implies that the flux into the near
face must equal the flux out of the far face. Since the faces have the same area, the field must have equal
strengths at the two faces in order for their fluxes to be equal. Since we did not use the values of a or b, we
see that for all points in space on the same side of the sheet, the field has the same magnitude.
4. (a) (i) In cylindrical coordinates, the position vector of a point (R, θ, z) on the cylinder is given by
~r = R cos θ~i + R sin θ~j + z~k .
So k~r k =
√
R2 + z 2 . For an area element on the cylinder we have
~ = (cos θ~i + sin θ~j )R dz dθ,
dA
so the flux integral is:
Z
Z
~ · dA
~ =
E
2π
0
S
=q
Z
q
−H
2π
0
~r
H
Z
Z
H
−H
k~r k
3
· (cos θ~i + sin θ~j )R dz dθ
R cos2 θ + R sin2 θ
R dz dθ = 2πq
(R2 + z 2 )3/2
Z
H
−H
R2 dz
.
(R2 + z 2 )3/2
To compute this one variable integral, we write:
Z H
Z H
Z H
(R2 + z 2 )dz
z 2 dz
R2 dz
=
−
2
2 3/2
2
2 3/2
2
2 3/2
−H (R + z )
−H (R + z )
−H (R + z )
Z
H
z 2 dz
=
(R2 + z 2 )3/2
Z
We calculate the integral
−H
Z
H
−H
Note that the integral
(ii)
z 2 dz
=
(R2 + z 2 )3/2
H
−H
(R2
dz
+ z 2 )1/2
2H
= √
.
R2 + H 2
Z
H
−H
(R2
H
zdz
using integration by parts:
2 + z 2 )3/2
(R
−H
!
Z H
H
z
dz
−
− 2
2
2 1/2
(R + z 2 )1/2 −H
−H (R + z )
Z
z
dz
has canceled. Therefore we have
+ z 2 )1/2
Z
~ · dA
~ = 4πq √ H
E
.
R2 + H 2
S
• Let R be fixed. We have
~ · dA
~ = lim 4πq √ H
E
= 0.
H→0
H 2 + R2
S
Z
~ · dA
~ = lim 4πq √ H
lim
E
= 4πq.
H→∞ S
H→∞
H 2 + R2
lim
H→0
Z
1798
Chapter Nineteen /SOLUTIONS
• Now let H be fixed. We have
~ · dA
~ = lim 4πq √ H
= 4πq.
E
R→0 S
R→0
H 2 + R2
Z
~ · dA
~ = lim 4πq √ H
lim
E
= 0.
R→∞
R→∞ S
H 2 + R2
Z
lim
Each of these results is as we would expect from Gauss’s Law.
(b) Let S denote the side of the cylinder. We have
Z
Z
Z
Z
~ · dA
~ =
~ · dA
~ +
~ · dA
~ +
E
E
E
T
Top
S
~ · dA
~.
E
Bottom
The result of part (a) shows that
~ · dA
~ = 4πq √ H
.
E
R2 + H 2
S
Z
Let’s compute
top, z = H so
R
Top
Therefore k~r k =
~ = ~k r drdθ. On the
~ · dA
~ . The normal at any point on the top is ~n = ~k and so dA
E
~r = r cos θ~i + r sin θ~j + H ~k .
√
Z
r2 + H 2 and we have
Top
~ · dA
~ =
E
2π
Z
Z
0
=q
Z
q
0
2π
0
= qH
R
Z
0
Z
R
~r
· ~k r dr dθ
k~r k3
(r cos θ~i + r sin θ~j + H ~k ) · ~k
r dr dθ
(r2 + H 2 )3/2
0
2π Z R
0
1
r dr dθ
= −2πqH √
2
2
3/2
2
(r + H )
r + H2
R
0
H
= −2πq √
+ 2πq.
2
R + H2
Similarly, or using a symmetry argument, we find that the flux through the bottom is given by
Z
~ · dA
~ = −2πq √ H
+ 2πq.
E
R2 + H 2
Bottom
Thus, the total flux is given by
Z
H
H
~ · dA
~ = 4πq √ H
E
− 2πq √
+ 2πq − 2πq √
+ 2πq
2
2
2
2
2
R +H
R +H
R + H2
S
= 4πq.
Job: chap20-sols Sheet: 1 Page: 1799 (August 24, 2012 11 : 19) [ex-1]
20.1 SOLUTIONS
CHAPTER TWENTY
Solutions for Section 20.1
Exercises
1. Vector. We have
curl(z~i − x~j + y~k ) =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= ~i + ~j − ~k .
z −x y
2. Vector. We have
~i
∂
curl(−2z~i − z~j + xy~k ) =
∂x
−2z
~k
∂
= (x + 1)~i − (y + 2)~j .
∂z
xy
~j
∂
∂y
−z
3. We have
curl(3x~i − 5z~j + y~k ) =
~i
~j
∂
∂x
∂
∂y
~k
∂
∂z
= 6~i .
3x −5z y
4. Using the definition in Cartesian coordinates, we have
curl F~ =
~i
~j
∂
∂x
∂
∂y
x2 − y 2 2xy
=
~k
∂
∂z
0
∂
∂
∂
∂ 2
(0) −
(2xy) ~i + − (0) +
(x − y 2 ) ~j +
∂y
∂z
∂x
∂z
= 4y~k .
∂
∂ 2
(2xy) −
(x − y 2 ) ~k
∂x
∂y
5. Using the definition in Cartesian coordinates,
~ =
curl F
~i
~j
∂
∂x
∂
∂y
(−x + y)
=
(y + z)
~k
∂
∂z
(−z + x)
∂
∂
∂
∂
(−z + x) −
(y + z) ~i + − (−z + x) +
(−x + y) ~j
∂y
∂z
∂x
∂z
+
∂
∂
(y + z) −
(−x + y) ~k
∂x
∂y
= −~i − ~j − ~k .
6. Using the definition in Cartesian coordinates,
~ =
curl F
~k
~i
~j
∂
∂x
∂
∂y
∂
∂z
2yz
3xz
7xy
= (7x − 3x)~i − (7y − 2y)~j + (3z − 2z)~k
= 4x~i − 5y~j + z~k .
1799
Job: chap20-sols Sheet: 2 Page: 1800 (August 24, 2012 11 : 19) [ex-1]
1800
Chapter Twenty /SOLUTIONS
7. Using the definition in Cartesian coordinates,
curl F~ =
~i
~j
∂
∂x
2
∂
∂y
3
x
=
~k
∂
∂z
4
y
z
∂ 4
∂ 3 ~
∂
∂ 2 ~
(z ) −
(y ) i + − (z 4 ) +
(x ) j +
∂y
∂z
∂x
∂z
= ~0 .
∂ 3
∂ 2 ~
(y ) −
(x ) k
∂x
∂y
8. Using the definition in Cartesian coordinates,
~ =
curl F
~i
~j
~k
∂
∂x
x
∂
∂y
∂
∂z
z2
e
=
cos y e
2
∂ z2
∂
∂
∂ x ~
(e ) −
(cos y) ~i + − (ez ) +
(e ) j +
∂y
∂z
∂x
∂z
= ~0 .
∂
∂ x ~
(cos y) −
(e ) k
∂x
∂y
9. Using the definition in Cartesian coordinates
~ =
curl F
~k
~i
~j
∂
∂x
∂
∂y
y 2 + xzy
x + yz
∂
∂z
3 2
zx y + x7 y 6
= (2x3 yz + 6x7 y 5 − xy)~i + (−3x2 y 2 z − 7x6 y 6 + y)~j + (yz − z)~k
10. This vector field points radically outward and has unit length everywhere (except the origin). Thus, we would expect its
curl to be ~0 . Computing the curl directly we get
curl
~r
k~r k
~i
∂
∂x
=
x
(x2 +y 2 +z 2 )1/2
The ~i -component is given by =
−
∂
∂y
∂
∂z
y
(x2 +y 2 +z 2 )1/2
z
(x2 +y 2 +z 2 )1/2
1
2yz
·
−
2 (x2 + y 2 + z 2 )3/2
=0
~k
~j
−
2yz
1
·
2 (x2 + y 2 + z 2 )1/2
~i
Similarly, the ~j and ~k components are also both 0.
11. The circulation of the vector field around the boundary of any square centered at the origin with sides parallel to the axes
is positive because the line integrals on the top and bottom sides are positive and the line integrals on the left and right
sides are zero. Therefore, we suspect a nonzero curl.
12. The circulation around any square centered on the origin with sides parallel to the axes is zero because the line integrals
on all four sides are zero, so we suspect a zero curl.
13. The circulation around any rectangle centered on the origin with sides parallel to the axes is zero because the line integrals
on the top and bottom sides add to zero and the line integrals on the left and right sides are zero. We suspect that the vector
field has zero curl.
14. The vector field is swirling counterclockwise, so the circulation around any circle centered at the origin is positive. Thus
we suspect a nonzero curl.
Job: chap20-sols Sheet: 3 Page: 1801 (August 24, 2012 11 : 19) [ex-1]
20.1 SOLUTIONS
1801
Problems
15. (a) The vector field always points perpendicularly to both ~k and ~r , in the direction determined by the right hand rule,
and its magnitude is twice the magnitude of ~r . Thus
~ (~r ) = 2~k × ~r = −2y~i + 2x~j .
F
(b) Using the definition in Cartesian coordinates
curl ~v =
~i
~j
∂
∂x
∂
∂y
~k
∂
∂z
−2y 2x 0
∂
∂
∂
∂
∂
∂
(2x))~i + ( (−2y) −
(0))~j + ( (2x) −
(−2y))~k = 4~k .
= ( (0) −
∂y
∂z
∂z
∂x
∂x
∂y
This makes sense, because we computed the circulation density of this vector field in the z-direction and found it was
4, and we would expect the z-direction to give the maximum circulation density from the symmetry of the vector
field.
16. The part of this vector field in the xy-plane looks like Figure 19.29 on page 1025, and shows no rotational tendency. Thus
we expect the curl to be ~0 . In fact it is, because the circulation around every closed curve C is zero, since
~ = x~i + y~j + z~k = grad(x2 /2 + y 2 /2 + z 2 /2),
F
~ is a gradient field. Thus the circulation density is zero in any direction, and hence curl F~ (P ) = ~0 for every point
so F
P . Using the formula, we see that
~ =
curl F~ = ∇ × F
=
~k
~i
~j
∂
∂x
∂
∂y
∂
∂z
x
y
z
∂z
∂y ~
i +
−
∂y
∂z
∂x
∂z ~
j +
−
∂z
∂x
∂y
∂x ~
−
k = 0~i + 0~j + 0~k = ~0 .
∂x
∂y
17. The curl is defined in such a way that if ~n is a unit vector and C is a small circle in the plane perpendicular to ~n and with
orientation induced by ~
n , then
so
~ ) · ~n = Circulation density
(curl G
R
~ · d~r
G
C
≈
Area inside C
Circulation =
Z
C
~ ) · ~n · Area inside C.
~ · d~r ≈ (curl G
G
(a) Let C be the circle in the xy-plane, and let ~n = ~k . Then
Circulation ≈ (2~i − 3~j + 5~k ) · ~k · π(0.01)2
= 0.0005π.
(b) By a similar argument to part (a), with ~
n = ~i , we find the circulation around the circle in the yz-plane:
Circulation ≈ (2~i − 3~j + 5~k ) · ~i · π(0.01)2
= 0.0002π.
(c) Similarly for circulations around the circle in the xz-plane,
Circulation ≈ −0.0003π.
Job: chap20-sols Sheet: 4 Page: 1802 (August 24, 2012 11 : 19) [ex-1]
1802
Chapter Twenty /SOLUTIONS
18. The vector curl F~ has its component in the x-direction given by
(curl F~ )x ≈
=
Circulation around small circle around x-axis
Area inside circle
0.5π
Circulation around C2
=
= 50.
Area inside C2
π(0.1)2
Similar reasoning leads to
Circulation around C3
3π
=
= 300,
Area inside C3
π(0.1)2
~ )z ≈ Circulation around C1 = 0.02π = 2.
(curl F
Area inside C1
π(0.1)2
(curl F~ )y ≈
Thus,
~ ≈ 50~i + 300~j + 2~k .
curl F
~ depends only on x, the second component depends only on y, and
19. The conjecture is that when the first component of F
the third component depends only on z, that is, if
F~ = F1 (x)~i + F2 (y)~j + F3 (z)~k
then
~ = ~0
curl F
~ = F1 (x)~i + F2 (y)~j + F3 (z)~k , then
The reason for this is that if F
curl F~ =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
F1 (x)
=
F2 (y)
F3 (z)
∂
∂
∂
∂
F3 (z) −
F2 (y) ~i + − F3 (z) +
F1 (x) ~j +
∂y
∂z
∂x
∂z
= ~0 .
∂
∂
F2 (y) −
F1 (x) ~k
∂x
∂y
20. (a) We have
~ =
curl G
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= (f − c)~i + (bez − e cos x)~j + (2dx − 3ay 2 )~k .
ay 3 + bez cz + dx2 e sin x + f y
~ is parallel to the yz-plane, then it has no ~i component. Thus, f − c = 0 or f = c.
(b) If curl G
~ is parallel to the z-axis, then there are no ~i and ~j components. Thus,
(c) If curl G
f −c=0
and
bez − e cos x = 0.
Since the second equation holds for all z and x, we have b = e = 0 as well as f = c.
21. (a) A thin twig at the origin along the x-axis would only feel the velocity along that axis, and thus go counterclockwise.
(b) Clockwise.
(c) Using the Cartesian coordinate definition, we get
~i ~j ~k
curl F~ =
∂
∂
∂
∂x ∂y ∂z
= 0~i + 0~j + (1 − 1)~k = ~0 .
y x 0
This is as expected, since a paddle-wheel (instead of a twig) placed in the field would not rotate at all.
Job: chap20-sols Sheet: 5 Page: 1803 (August 24, 2012 11 : 19) [ex-1]
1803
20.1 SOLUTIONS
22. We have
so
~ (t, x, y, z) = (cos t~j + sin t~k ) × ~r = (z cos t − y sin t)~i + x sin t~j − x cos t~k ,
F
(curl F~ )(t, x, y, z) = 2 cos t~j + 2 sin t~k .
~ )(0, x, y, z) = 2~j is horizontal, pointing in the y direction.
(a) At t = 0 the vector (curl F
~ )(π/2, x, y, z) = 2~k is vertical, pointing in the z direction.
(b) At t = π/2 the vector (curl F
~ )(t, x, y, z) = 2 cos t~j + 2 sin t~k is parallel to the yz-plane, making an angle of
(c) At 0 < t < π/2 the vector (curl F
t radians with the horizontal plane. Thus as t goes from 0 to π/2, the curl goes steadily from horizontal to vertical.
23. Investigate the velocity vector field of the atmosphere near the fire. If the curl of this vector field is non-zero, there is
circulatory motion. Consequently, if the magnitude of the curl of this vector field is large near the fire, a fire storm has
probably developed.
24. (a) Figure 20.1 shows a cross-section of the vector field in xy-plane with ω = 1, so ~v = −y~i + x~j .
Figure 20.2 shows a cross-section of vector field in xy-plane with ω = −1, so ~v = y~i − x~j .
y
y
x
x
Figure 20.1: ~v = −y~i + x~j
Figure 20.2: ~v = y~i − x~j
p
x2 + y 2 . The velocity of the vortex at any point
(b) The distance from the center of the vortex is given by r =
is −ωy~i + ωx~j , and thep
speed of the vortex at any point is the magnitude of the velocity, or s = k~v k =
p
(−ωy)2 + (ωx)2 = |ω| x2 + y 2 = |ω| r.
(c) The divergence of the velocity field is given by:
div ~v =
∂(ωx)
∂(−ωy)
+
=0
∂x
∂y
The curl of the field is:
curl ~v = curl(−ωy~i + ωx~j ) =
∂
∂
(ωx) −
(−ωy) ~k = 2ω~k
∂x
∂y
(d) We know that ~v has constant magnitude |ω|R everywhere on the circle and is everywhere tangential to the circle.
In addition, if ω > 0, the vector field rotates counterclockwise; if ω < 0, the vector field rotates clockwise. Thus if
ω > 0, ~v and ∆~r are parallel and in the same direction, so
Z
C
~ = |~v | · (Length of C) = ωR · 2πR = 2πωR2
~v · dr
If ω < 0, then |ω| = −ω and ~v and ∆~r are in opposite directions, so
Z
C
~ = − |~v | · (Length of C) = − |ω| R · (2πR) = 2πωR2 .
~v · dr
Job: chap20-sols Sheet: 6 Page: 1804 (August 24, 2012 11 : 19) [ex-1]
1804
Chapter Twenty /SOLUTIONS
25. To show that the force field is irrotational we must show that its curl is zero. Let us do this in Cartesian coordinates:
F~ = f (r)~r = f (
~ is
The third component of curl F
p
x2 + y 2 + z 2 )(x~i + y~j + z~k )
p
p
∂
∂
(f ( x2 + y 2 + z 2 )y) −
(f ( x2 + y 2 + z 2 )x)
∂x
∂y
p
p
2xy
2xy
′
− f ′ ( x2 + y 2 + z 2 ) · p
= f ( x2 + y 2 + z 2 ) · p
2 x2 + y 2 + z 2
2 x2 + y 2 + z 2
= 0.
~ are 0 too.
A similar computation shows that the other components of curl F
~
~
~
~
26. Let C = ai + bj + ck . Then
~ +C
~)=
curl(F
∂
∂
(F3 + c) −
(F2 + b) ~i +
∂y
∂z
+
=
∂F3
∂F2
−
∂y
∂z
~.
= curl F
~i +
∂
∂z
(F1 + a) −
∂
(F3 + c) ~j
∂x
∂
∂
(F2 + b) −
(F1 + a) ~k
∂x
∂y
∂F1
∂F3 ~
j +
−
∂z
∂x
∂F2
∂F1
−
∂x
∂y
~k
27.
~ =(
curl F
∂F2 ~
∂F1
∂F3 ~
∂F2
∂F1 ~
∂F3
−
)i + (
−
)j + (
−
k)
∂y
∂z
∂z
∂x
∂x
∂y
∂F2
∂ ∂F1
∂F3
∂ ∂F2
∂F1
∂ ∂F3
(
−
)+
(
−
)+
(
−
)
div curl F~ =
∂x ∂y
∂z
∂y ∂z
∂x
∂z ∂x
∂y
∂ 2 F3
∂ 2 F2
∂ 2 F1
∂ 2 F3
∂ 2 F2
∂ 2 F1
=
−
+
−
+
−
∂x∂y
∂x∂z
∂y∂z
∂y∂x
∂z∂x
∂z∂y
~ has continuous second partial derivatives,
Since, if F
∂ 2 F3
∂ 2 F3
=
,
∂x∂y
∂y∂x
∂ 2 F2
∂ 2 F2
=
,
∂x∂z
∂z∂x
and
∂ 2 F1
∂ 2 F1
=
∂y∂z
∂z∂y
~ = 0.
everything cancels out and we get div curl F
28. The Fundamental Theorem of Calculus for Line Integrals states that if C is a path from P to Q, then
Z
C
grad f · d~r = f (Q) − f (P ).
Since C is a closed path we have
Z
c
(a) For any unit vector ~
n
circ~n grad f =
grad f · d~r = f (P ) − f (P ) = 0
lim
Area→0
R
grad f · d~r
Area of C
=
0
lim
Area
Area→0
=0
where the limit is taken over curves C in a plane perpendicular to ~n , and oriented by the right hand rule. Thus the
circulation density of grad f is zero in every direction, and hence curl grad f = ~0 .
Job: chap20-sols Sheet: 7 Page: 1805 (August 24, 2012 11 : 19) [ex-1]
20.1 SOLUTIONS
1805
(b) Using the Cartesian coordinate definition
curl grad f = curl(
=
∂f ~
∂f ~
∂f ~
i +
j +
k)
∂x
∂y
∂z
∂2f ~
∂2f
−
i +
∂y∂z
∂z∂y
= 0~i + 0~j + 0~k = ~0 .
∂2f
∂2f
−
∂z∂x
∂x∂z
~j +
∂2f
∂2f
−
∂x∂y
∂y∂x
~k
29.
~)
curl(φF
=
=
∂
∂
(φF3 ) −
(φF2 ) ~i +
∂y
∂z
φ
∂z
(φF1 ) −
∂
(φF3 ) ~j +
∂x
φ
∂φ
∂F1
∂φ
∂F2
+
F2 − φ
−
F1 ~k
∂x
∂x
∂y
∂y
∂F3
∂F2
−
∂y
∂z
+
= φ curl F~ +
~i +
∂F1
∂F3 ~
j +
−
∂z
∂x
∂φ
∂φ
F3 −
F2 ~i +
∂y
∂z
∂φ~
∂φ ~
∂φ ~
i +
j +
k
∂x
∂y
∂z
= φ curl F~ + (grad φ) × F~ .
∂F2
∂F1
−
∂x
∂y
∂φ
∂φ
F1 −
F3 ~j +
∂z
∂x
∂
∂
(φF2 ) −
(φF1 ) ~k
∂x
∂y
∂φ
∂F2
∂φ
∂φ
∂F3
∂φ
∂F3
∂F1
+
F3 − φ
−
F2 ~i + φ
+
F1 − φ
−
F3 ~j
∂y
∂y
∂z
∂z
∂z
∂z
∂x
∂x
+
=φ
∂
~k
∂φ
∂φ
F2 −
F1 ~k
∂x
∂y
× (F1~i + F2~j + F3~k )
~ = grad f × grad g + f curl grad g = grad f × grad g, since curl grad g = 0. Since the cross
30. By Problem 29, curl F
~ is a scalar times grad g,
product of two vectors is perpendicular to both vectors, curl F~ is perpendicular to grad g. But F
~
~
so curl F is perpendicular to F .
~ = F1~i + F2~j + F3~k , ~
u = u1~i + u2~j + u3~k , ~v = v1~i + v2~j + v3~k , then
31. If F
~ · ~v ) · ~
~ ·~
grad(F
u − grad(F
u ) · ~v
= grad(F1 v1 + F2 v2 + F3 v3 ) · (u1~i + u2~j + u3~k ) − grad(F1 u1 + F2 u2 + F3 u3 ) · (v1~i + v2~j + v3~k )
∂F2
∂F3
∂F1
∂F2
∂F3
∂F1
v1 u1 +
v2 u1 +
v3 u1 +
v1 u2 +
v2 u2 +
v3 u2 +
=
∂x
∂x
∂x
∂y
∂y
∂y
∂F1
∂F2
∂F3
v1 u3 +
v2 u3 +
v3 u3 −
∂z
∂z
∂z
∂F2
∂F3
∂F1
∂F2
∂F3
∂F1
u1 v1 +
u2 v1 +
u3 v1 +
u1 v2 +
u2 v2 +
u3 v2 +
∂x
∂x
∂x
∂y
∂y
∂y
∂F1
∂F2
∂F3
u1 v3 +
u2 v3 +
u3 v3
∂z
∂z
∂z
∂F3
∂F1
∂F2
∂F2
∂F3
∂F1
=
−
(u2 v3 − u3 v2 ) +
−
(u3 v1 − u1 v3 ) +
−
(u1 v2 − u2 v1 )
∂y
∂z
∂z
∂x
∂x
∂y
= (curl F~ ) · ~
u × ~v .
~ is in the xy-plane, curl F
~ is parallel to ~k (because F3 = 0 and F1 , F2 have no z-dependence). Imagine
32. (a) Since F
~
computing the circulation of F counterclockwise around a small rectangle R at the point P with sides of length h
~ and sides of length t perpendicular to F
~ as shown in Figure 20.3. Since F
~ is perpendicular to C2 and
parallel to F
~ is approximately constant on C1 and C3 , its value
C4 , the line integral over these two sides is zero. Assuming that F
Job: chap20-sols Sheet: 8 Page: 1806 (August 24, 2012 11 : 19) [ex-1]
1806
Chapter Twenty /SOLUTIONS
~ is parallel to C1 and C3 , the line integral over
on these sides is F (Q)T~ and −F (P )T~ , respectively. Thus, since F
C1 is approximately F (Q)h and the line integral over C3 is approximately −F (P )h. Finally
F (Q)h − F (P )h
F (Q) − F (P )
Circulation around R
≈
=
Area of R
ht
t
−
−
→
≈ Directional derivative of F in the direction of P Q.
~ (P ) ≈
curl F
~
F
C3
t
✸
C2
h
P
h
t
C4
C1
Q
Figure 20.3: Path R used to find curl F~ at P
~ = F (x, y)T~ = F (x, y)a~i + F (x, y)b~j , with a, b constant, we have
(b) Since F
~ = (bFx − aFy )~k .
curl F
Also T~ × ~k = (a~i + b~j ) × ~k = b~i − a~j , so
bFx − aFy = (gradF ) · (b~i − a~j ) = gradF · ((a~i + b~j ) × ~k ) = FT~
×~
k
,
where FT~ ×~k is the directional derivative of F in the direction of the unit vector T~ × ~k , which is perpendicular to
F~ . The right-hand rule applied to T~ × ~k shows that T~ × ~k is obtained by a clockwise rotation of T~ through 90◦ .
33. (a) Using r = (x2 +y 2 )1/2 , we calculate rx = (1/2)(x2 +y 2 )−1/2 2x. Notice that rx = x/r and, by a similar argument,
ry = y/r. We have
A
curl r · (−y~i + x~j ) =
=
~i
~j
~k
∂
∂x
A
∂
∂y
A
∂
∂z
−r y
r x
0
A
∂(0)
∂(r x) ~
−
i +
∂y
∂z
= 0~i + 0~j + c~k = c~k
∂(−r A y)
∂(0)
−
∂z
∂x
~j +
∂(r A x)
∂(−r A y)
−
∂x
∂y
~k
where
c=
∂
∂
(xr A ) −
(−yr A ) = r A + Axr A−1 rx + r A + Ayr A−1 ry
∂x
∂y
= 2r A + Ar A−1 (xrx + yry )
= 2r A + Ar A−1
= 2r A + Ar A−1
A
= 2r + Ar
A−1
= (2 + A)r A .
x2
y2
+
r
r
r2
r
x2 + y 2
r
(b) The curl is in the direction of ~k for A = −1, is the zero vector for A = −2, and is in the direction of −~k for
A = −3.
Job: chap20-sols Sheet: 9 Page: 1807 (August 24, 2012 11 : 19) [ex-1]
20.1 SOLUTIONS
1807
The counterclockwise flow in the figure suggests the curl is in the +~k direction in all three cases. However, it
is not easy to see all the effects in the picture. The counterclockwise flow indicates that the circulation on any path
around ~k is positive, but the curl is the circulation density, not the circulation, in the ~k -direction. To understand the
curl, we must see how the circulation around the ~k -direction changes when the circles get smaller and smaller. We
need to take into account the changing length of the vector field as well as the changing direction. This is too subtle
an effect to be measured accurately by eye. The three vector fields differ in the rate at which the magnitude of the
vector field changes as you move perpendicular to the flow (the shear) and this accounts for the differing directions
of their curls. When A = −1 the magnitude of the vector field is constant, when A = −2 or −3 the magnitude
decreases as you move farther from the origin, and the decrease is most rapid for A = −3.
(c) The curl has a component only in the ~k direction. Think of the this component as the limit of the circulation density
around a small circle in the xy-plane, as the circle shrinks to zero. Thus the sign of the component of the curl tells us
that this circulation around the circle centered at (1, 1, 1) is positive for A = −1, zero for A = −2, and negative for
A = −3. (This assumes that the circle is small enough that it does not go around the origin, where the vector fields
are not defined.)
Since the vector fields and their curls are not defined at (0, 0, 0), the curl calculated in part (b) does not tell us
anything about the circulation around a circle centered at (0, 0, 0).
Strengthen Your Understanding
34. The curl of a vector field is another vector field, not a scalar function.
~ = y~i is parallel to the x-axis at every point, but curl F~ = −~k 6= ~0 .
35. The vector field F
~ consists of parallel straight lines, the vector field does exhibit shear, which can produce curl.
Though the flow of F
2~
2~
2~
~
36. If F = x i + y j + z k then
~ =
curl F
∂ 2 ~
∂ 2
z −
y i +
∂y
∂z
∂ 2
∂ 2 ~
x −
z j +
∂z
∂x
∂ 2
∂ 2 ~
y −
x k = ~0 .
∂x
∂y
~ = F1~i + F2~j + F3~k , we have
37. For F
~ =
curl F
∂F2
∂F3
−
∂y
∂z
~i +
∂F1
∂F3 ~
−
j +
∂z
∂x
∂F2
∂F1
−
∂x
∂y
~k .
~ = z~i and curl F
~ = ~j .
If F1 = z and F2 = F3 = 0, then F
38. True. The circulation density is obtained by dividing the circulation around a circle C (a scalar) by the area enclosed by
C (also a scalar), in the limit as the area tends to zero.
39. True.
curl grad f =
=
|
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
∂f
∂x
∂f
∂y
∂f
∂z
∂2f
∂2f
−
∂y∂z
∂z∂y
{z
0
}
~i +
|
∂2f
∂2f
−
∂z∂x
∂x∂z
{z
0
}
~j +
|
∂2f
∂2f
−
∂x∂y
∂y∂x
{z
0
~k
}
~ = 0 and curlF
~ = ~0 .
40. False. As a counterexample, any constant vector field F~ = a~i + b~j + c~k has divF
~ = F1~i + F2~j + F3~k and G
~ = G1~i + G2~j + G3~k , we have F~ + G
~ = (F1 + G1 )~i + (F2 + G2 )~j +
41. True. Writing F
~
~
~
~
(F3 + G3 )k . Then the i component of curl(F + G ) is
∂(F2 + G2 )
∂(F3 + G3 )
∂F3
∂F2
∂G3
∂G2
−
=
−
+
−
∂y
∂z
∂y
∂z
∂y
∂z
~ . The ~j and ~k components work out in a similar
which is the ~i component of curlF~ plus the ~i component of curlG
manner.
Job: chap20-sols Sheet: 10 Page: 1808 (August 24, 2012 11 : 19) [chap20-sols]
1808
Chapter Twenty /SOLUTIONS
~ ·G
~ ) is a scalar, so we cannot compute
42. False. The left-hand side of the equation does not make sense. The quantity (F
the curl of it.
~ = z~i and G
~ = x~j . Then F
~ ×G
~ = xz~k , and curl(F
~ ×G
~ ) = −z~j . However,
43. False. For example, take F
~
~
~
~
~
(curlF ) × (curlG ) = j × k = i .
44. True. We calculate the x–components for each side of the equation:
∂(f G2 )
∂(f G3 )
−
∂y
∂z
∂f
∂G3
∂f
∂G2
=
G3 + f
−
G2 − f
∂y
∂y
∂z
∂z
∂f
∂f
∂G3
∂G2
= ( G3 −
G2 ) + f (
−
)
∂y
∂z
∂y
∂z
~ )1 + (f (curl G
~ ))1 .
= ((grad f ) × G
~ ))1 =
(curl(f G
Computations for the other two components are similar, so
~ ) = (grad f ) × G
~ + f · (curl G
~ ).
curl(f G
~ = z~i + x~j . Then curlF
~ = ~j + ~k , which is not perpendicular to F
~ , since (z~i + x~j ) · (~j +
45. False. For example, take F
~k ) = x 6= 0.
46. True. F~ is rotating around the y axis, so by the right hand rule curl F~ has a positive y component. Therefore taking the
dot product of curl F~ and ~j will give a positive number.
47. (a)
curl(y~i − x~j + z~k ) =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= −2~k
y −x z
(b)
~i ~j ~k
curl(y~i + z~j + x~k ) =
∂
∂
∂
∂x ∂y ∂z
y
= −~i − ~j − ~k
z x
(c)
~i
curl(−z~i + y~j + x~k ) =
~j ~k
∂
∂
∂
∂x ∂y ∂z
= −2~j
−z y x
(d)
~i
curl(x~i + z~j − y~k ) =
~j
∂
∂
∂x ∂y
~k
∂
∂z
= −2~i
x z −y
(e)
~i
curl(z~i + x~j + y~k ) =
~j ~k
∂
∂
∂
∂x ∂y ∂z
z
x y
So (a), (c), (d) are parallel to the z-, y-, and x-axes, respectively.
= ~i + ~j + ~k
Job: chap20-sols Sheet: 11 Page: 1809 (August 24, 2012 11 : 19) [ex-2]
20.2 SOLUTIONS
1809
Solutions for Section 20.2
Exercises
1. To calculate
terized by
R
C
~ · d~r directly, we compute the integral along the paths C1 and C2 in Figure 20.4. Now C1 is parameF
x(t) = 3 − t,
y(t) = 0 z(t) = 0
and C2 is parameterized by
x(t) = −3 cos t,
y(t) = 0,
z(t) = 3 sin t
so r ′ (t) = −~i ,
for 0 ≤ t ≤ 6,
so ~r ′ (t) = 3 sin t~i + 3 cos t~k .
for 0 ≤ t ≤ π,
Thus
Z
C
~ · d~r =
F
=
Z
F~ · d~r +
C1
Z
6
0
=−
Z
~ · d~r
F
C2
(3 − t)(~i + ~j ) · (−~i ) dt +
Z
6
(3 − t) dt + 9
0
t2
= −3t +
2
6
+9
0
π
Z
π
Z
0
((−3 cos t + 3 sin t)~i − 3 cos t~j ) · (3 sin t~i + 3 cos t~k ) dt
(− sin t cos t + sin2 t) dt
0
sin2 t
1
t
−
− sin t cos t +
2
2
2
π
9π
.
2
=
0
Formula IV-17 was
R used to calculate the last integral.
To calculate C F~ · d~r using Stokes’ Theorem, we find
~ =
curl F
~i
~j ~k
∂
∂x
∂
∂
∂y ∂z
x+z x
= ~i + ~j + ~k .
y
~ = ~j dx dz. Thus,
For Stokes’ Theorem, the semicircular region S in Figure 20.4 is oriented into the page, so dA
Z
C
~ · d~r =
F
Z
S
~ · dA
~ =
curl F
Z
S
= Area of semicircle =
(~i + ~j + ~k ) · ~j dx dz =
1 2
9
π3 = π.
2
2
Z
dx dz
S
z
C2
S
C1
(−3, 0, 0)
x
(3, 0, 0)
Figure 20.4
2. Since C is the curve x2 + y 2 = 4, oriented counterclockwise, we calculate
x(t) = 2 cos t,
y(t) = 2 sin t,
z(t) = 0,
0 ≤ t ≤ 2π,
R
C
~ · d~r directly using the parameterization
F
so ~r ′ (t) = −2 sin t~i + 2 cos t~j .
Thus,
Z
C
~ · d~r =
F
Z
2π
0
= −4
(2 sin t~i − 2 cos t~j ) · (−2 sin t~i + 2 cos t~j ) dt
Z
0
2π
(sin2 t + cos2 t) dt = −4 · 2π = −8π.
Job: chap20-sols Sheet: 12 Page: 1810 (August 24, 2012 11 : 19) [ex-2]
1810
Chapter Twenty /SOLUTIONS
Since S is given by z = 4 − x2 − y 2 , we have
∂z
= −2x
∂x
∂z
= −2y,
∂y
and
~ = (2x~i + 2y~j + ~k ) dx dy.
dA
so
Also
curl F~ =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= −2~k ,
y −x 0
so, writing R for the disk below S in the xy-plane, Stokes’ Theorem gives
Z
C
~ · d~r =
F
Z
~ =
curl F~ · dA
S
= −2
Z
Z
R
−2~k · (2x~i + 2y~j + ~k ) dx dy
dx dy = −2 · Area of disk = −2 · π22 = −8π.
3. To calculate C F~ · d~r directly, we compute the integral along each of the sides C1 , C2 , C3 in Figure 20.5. Now C1 is
parameterized by
x(t) = t, y(t) = 0, z(t) = 0 for 0 ≤ t ≤ 5, so r ′ (t) = ~i .
Similarly, C2 is parameterized by
R
x(t) = 5,
y(t) = t,
z(t) = 0
for 0 ≤ t ≤ 5,
so ~r ′ (t) = ~j .
Also, C3 is parameterized by
x(t) = 5 − t,
y(t) = 5 − t,
z(t) = 0
for 0 ≤ t ≤ 5,
so ~r ′ (t) = −~i − ~j .
Thus
Z
C
F~ · d~r =
=
=
To calculate
R
C
Z
Z
Z
C1
~ · d~r +
F
5
Z
C2
F~ · d~r +
t(~i + ~j ) · ~i dt +
0
5
t dt +
0
Z
0
5
Z
5
0
5 − t dt +
Z
C3
~ · d~r
F
(5 − t)(~i + ~j ) · ~j dt +
Z
5
0 dt =
0
5
0
5
((5 − t) − (5 − t))(~i + ~j ) · (−~i − ~j ) dt
5 dt = 25.
0
~ · d~r using Stokes’ Theorem, we find
F
~ =
curl F
Z
Z
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= −~i + ~j + (1 − (−1))~k = −~i + ~j + 2~k .
x−y+z x−y+z 0
~ = ~k dx dy. Thus
For Stokes’ Theorem, the triangular region S in Figure 20.5 is oriented upward, so dA
Z
C
~ · d~r =
F
=
Z
S
Z
S
~ · dA
~ =
curl F
Z
S
(−~i + ~j + 2~k ) · ~k dx dy
2 dx dy = 2 · Area of triangle = 2 ·
y
(5, 5, 0)
C3
(0, 0, 0)
S
C1
Figure 20.5
C2
x
(5, 0, 0)
1
· 5 · 5 = 25.
2
Job: chap20-sols Sheet: 13 Page: 1811 (August 24, 2012 11 : 19) [ex-2]
20.2 SOLUTIONS
1811
4. A sketch of the surface S and curve C which is the union of four curves C1 , C2 , C3 , and C4 , and the region R is shown
in Figure 20.6.
y
R
x
z
C3
1
C2
C4
y
x
2
C1
Figure 20.6
~ , oriented upward.
To compute the flux integral, we find dA
~ = (2x~i + ~k )dxdy
dA
~ = −y~i − z~j − x~k .
curl F
and
Thus,
Z
S
~ · dA
~ =
curl F
=
Z
(−y~i − z~j − x~k ) · (2x~i + ~k )dxdy
S
1
Z
0
The line integral
Z
S
Z
2
−2
(−2xy − x)dydx = −2.
F~ · d~r is the sum of four integrals along C1 , C2 , C3 , and C4 .
On C1 : x = 1, z = 0, dx = 0, dz = 0, so
Z
S
~ · d~r =
F
Z
2
0 dy = 0.
−2
On C2 : y = 2, z = 1 − x2 , dy = 0, dz = −2xdx, so
Z
C2
Z
F~ · d~r =
2
C2
2
2xdx + 2(1 − x )0 + x(1 − x )(−2x)dx =
Z
0
(2x − 2x2 + 2x4 )dx = −
1
11
.
15
On C3 : x = 0, z = 1, dx = 0, dz = 0, so
Z
C3
~ · d~r =
F
Z
0 + ydy + 0 =
Z
−2
ydy = 0.
2
C3
On C4 : y = −2, z = 1 − x2 , dy = 0, dz = −2xdx, so
Z
C4
Hence
Thus,
~ · d~r =
F
Z
C4
2
2
−2xdx − 2(1 − x )0 + x(1 − x )(−2x)dx =
Z
Z
0
1
(−2x − 2x2 + 2x4 )dx = −
~ · d~r = 0 − 11 + 0 − 19 = −2.
F
15
15
C
Z
C
~ · d~r =
F
Z
S
~ · dA
~.
curl F
19
.
15
Job: chap20-sols Sheet: 14 Page: 1812 (August 24, 2012 11 : 19) [ex-2]
1812
Chapter Twenty /SOLUTIONS
5. The boundary of S is C, the circle x2 + y 2 = 1, z = 0, oriented counterclockwise and parameterized in polar coordinates
by
~r (θ) = cos θ~i + sin θ~j , 0 ≤ θ ≤ 2π,
so,
~r ′ (θ) = − sin θ~i + cos θ~j .
Hence
Z
C
F~ · d~r =
=
2π
Z
(sin θ~i + 0~j + cos θ~k ) · (− sin θ~i + cos θ~j + 0~k )dθ
0
Z
2π
− sin2 θdθ = −π.
0
Now consider the integral
given by
2
R
S
~ . Here curl F
~ = −~i − ~j − ~k and the area vector dA
~ , oriented upward, is
curl F~ · dA
~ = 2x~i + 2y~j + ~k dxdy.
dA
2
If R is the disk x + y ≤ 1, then we have
Z
S
Z
~ · dA
~ =
curl F
R
(−~i − ~j − ~k ) · (2x~i + 2y~j + ~k )dxdy.
Converting to polar coordinates gives:
Z
~ =
curl F~ · dA
S
=
Z
2π
0
Z
Z
2π
Z
1
0
2π
0
= −π.
Thus, we confirm that
(−~i − ~j − ~k ) · (2r cos θ~i + 2r sin θ~j + ~k )rdrdθ
0
0
=
1
Z
Z
(−2r cos θ − 2r sin θ − 1)rdrdθ
1
2
(− cos θ − sin θ) −
dθ
3
2
C
Z
~ · d~r =
F
S
~ · dA
~.
curl F
6. Since
~i ~j ~k
~ =
curl F
∂
∂
∂
∂x ∂y ∂z
2
2
2
x y
the line integral
7. Since
R
C
= ~0
z
~ · d~r = 0 around any closed curve, including this unit circle.
F
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= −~i − ~j − ~k ,
y−x z−y x−z
√
~ = π( 5)2~k = 5π~k , Stokes’ Theorem gives
and the area vector of the disk x + y ≤ 5 is A
2
Z
C
~ · d~r =
F
Z
Disk
2
~ = (−~i − ~j − ~k ) · Area vector = (−~i − ~j − ~k ) · 5π~k = −5π.
(−~i − ~j − ~k ) · dA
Job: chap20-sols Sheet: 15 Page: 1813 (August 24, 2012 11 : 19) [ex-2]
20.2 SOLUTIONS
1813
8. If C is the circle x2 + y 2 = 9 oriented counterclockwise when viewed from above, Stokes’ Theorem gives
Z
S
Z
~ · dA
~ =
curl F
C
~ · d~r .
F
~ , namely −y~i + x~j , contribute to the line integral. Since −y~i + x~j
Only the ~i and ~j components of F
~
~
circle, and −y i + xj is tangent to the circle and points in the direction of the orientation of C we have
Z
S
~ · dA
~ =
curl F
Z
F~ · d~r = −y~i + x~j
C
= 3 on the
· Length of curve = 3 · 2π3 = 18π.
Alternatively, the line integral can be evaluated by parameterizing the curve using x = 3 cos t, y = 3 sin t for 0 ≤ t ≤
2π.
9. If C is the rectangular path around the rectangle, traversed counterclockwise when viewed from above, Stokes’ Theorem
gives
Z
Z
S
~ · dA
~ =
curl F
C
~ · d~r .
F
~ does not contribute to the line integral, and the ~j component contributes to the line integral only
The ~k component of F
along the segments of the curve parallel to the y-axis. Thus, if we break the line integral into four parts
Z
S
Z
~ =
curl F~ · dA
(3,0)
(0,0)
F~ · d~r +
Z
(3,2)
(3,0)
~ · d~r +
F
Z
(0,2)
Z
(0,0)
(3,2)
F~ · d~r +
Z
(0,0)
(0,2)
~ · d~r ,
F
~ by its ~j component in the other two
we see that the first and third integrals are zero, and we can replace F
Z
S
~ =
curl F~ · dA
Z
(3,2)
(3,0)
(x + 7)~j · d~r +
(0,2)
(x + 7)~j · d~r .
Now x = 3 in the first integral and x = 0 in the second integral and the variable of integration is y in both, so
Z
S
~ =
curl F~ · dA
Z
2
10 dy +
0
Z
2
0
7 dy = 20 − 14 = 6.
10. (a) A counterclockwise parameterization of the circle is
x(t) = cos t,
y(t) = sin t
0 ≤ t ≤ 2π.
Since x′ (t) = − sin t, and y ′ (t) = cos t, we have
Z
C
~ · d~r =
F
=
2π
Z
(sin t~i − cos t~j ) · (− sin t~i + cos t~j )dt
0
Z
2π
0
(b) curl F~ =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
(− sin2 t − cos2 t)dt = −2π.
= 0~i + 0~j − 2~k = −2~k .
y −x 0
(c) Using Stokes’ Theorem, where R is the region inside the circle oriented upward
Z
C
~ · d~r =
F
Z
R
~ =
curl F~ · dA
Z
R
~ = −2 · Area of circle = −2π.
−2~k · dA
(d) Stokes’ Theorem, which says that if C is a closed curve which is the boundary of a surface R, and F~ is a smooth
vector field, then
Z
Z
~ · d~r =
~ · dA
~.
F
curl F
C
R
Here, the orientations of C and of R are related by the right-hand rule.
Job: chap20-sols Sheet: 16 Page: 1814 (August 24, 2012 11 : 19) [ex-2]
1814
Chapter Twenty /SOLUTIONS
11. (a) We have
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= −~i − ~j .
cos x ey x − y − z
(b) If S is the disk on the plane within the circle C, Stokes’ Theorem gives
Z
C
~ · d~r =
F
Z
~.
curl F~ · dA
S
√
For Stokes’ Theorem, the disk is oriented upward. Since the unit normal to the plane is (~i + ~j + ~k )/ 3 and
the disk has radius 3, the area vector of the disk is
√
~ ~j + ~k
~ = i +√
π(32 ) = 3 3π(~i + ~j + ~k ).
A
3
Thus, using curl F~ = −~i − ~j , we have
Z
C
F~ · d~r = −~i − ~j
√
√
· 3 3π(~i + ~j + ~k ) = −6 3π.
12. No, because the curve C over which the integral is taken is not a closed curve, and so it is not the boundary of a surface.
Problems
~ around C is the flux of curl F
~ through the disk S in the yz-plane enclosed by
13. By Stokes’ theorem, the circulation of F
C. By the right hand rule, a positive normal vector to the disk points in the direction of the negative x-axis, −~i . Thus
~ = curl F · (−~i )dA = −dA, so the flux through S is negative. So the circulation is negative.
curl F~ · dA
14. (a) We use Stokes’ theorem, with S the interior of the circle in the xy-plane, oriented upward. Then
Z
C
~ · d~r =
F
Z
S
~ · dA
~ =
curl F
Z
S
~.
((x2 + z 2 )~j + 5~k ) · dA
~ = ~k dA, we have
Since dA
Z
C
~ · d~r =
F
Z
S
5 dA = 5 · Area of S = 5 · π32 = 45π.
(b) Using Stokes’ theorem again, this time with S as the interior of the circle in the xz-plane, we have
Z
C
~ · d~r =
F
Z
S
~ · dA
~ =
curl F
Z
S
~.
((x2 + z 2 )~j + 5~k ) · dA
~ = ~j dx dz, only the ~j component of curl F
~ contributes to the flux. Thus, converting to polars gives
Since dA
Z
C
~ · d~r =
F
Z
(x2 + z 2 ) dx dz =
S
Z
0
2π
Z
3
r 2 r dr dθ = 2π
0
r4
4
3
=
0
81π
.
2
15. (a) We have
~i ~j ~k
curl(y~i + z~j + x~k ) =
∂
∂
∂
∂x ∂y ∂z
= −~i − ~j − ~k .
y z x
(b) Let S be the triangular interior of the curve C, oriented upward. Then, by Stokes’ Theorem,
Z
C
(y~i + z~j + x~k ) · d~r =
Z
S
~ =
curl(y~i + z~j + x~k ) · dA
Z
S
~.
−(~i + ~j + ~k ) · dA
~ = ~k dA, so
On the triangle dA
Z
S
~ =
−(~i + ~j + ~k ) · dA
Z
1
−(~i + ~j + ~k ) · ~k dA = − Area of triangle = − · 4 · 3 = −6.
2
S
Job: chap20-sols Sheet: 17 Page: 1815 (August 24, 2012 11 : 19) [ex-2]
20.2 SOLUTIONS
1815
16. (a) We have
~j ~k
~i
~ =
curl F
= ~i (−1) − ~j (1) + ~k (−1) = −~i − ~j − ~k .
∂
∂
∂
∂x ∂y ∂z
y
z x
(i) Using Stokes’ Theorem, with S representing the disk inside the circle, oriented upward, we have
(b)
Z
C
~ · d~r =
F
Z
S
~ · dA
~ =
curl F
Z
(−~i − ~j − ~k ) · ~k dA = − Area of disk = −4π.
S
(ii) This is a right triangle in the plane x = 2; it has height 5 and base length 3. Using Stokes’ Theorem, with S
representing the triangle, oriented toward the origin (in the direction −~i ), we have
Z
C
~ · d~r =
F
Z
S
~ =
curl F~ · dA
Z
S
(−~i − ~j − ~k ) · (−~i dA) =
Z
dA = Area of triangle =
S
1
15
·3·5 =
.
2
2
17. (a) We have
~i ~j ~k
curl(z~i + x~j + y~k ) =
= ~i + ~j + ~k .
∂
∂
∂
∂x ∂y ∂z
z
x y
(b) Let S be the square interior of the curve C, oriented toward the origin. Then, by Stokes’ Theorem,
Z
C
(z~i + x~j + y~k ) · d~r =
Z
S
~ =
curl(z~i + x~j + y~k ) · dA
Z
S
~.
(~i + ~j + ~k ) · dA
√
Since the plane has normal vector ~n = ~i + ~j + ~k , a unit normal in this direction is (~i + ~j + ~k )/ 3. But S is
√
~ = −(~i + ~j + ~k )/ 3 dA. Thus
oriented toward the origin, so dA
Z
S
~ =
(~i + ~j + ~k ) · dA
√
−(~i + ~j + ~k )
3
3
√
(~i + ~j + ~k ) ·
dA = − √ · Area of square = − √ · 22 = −4 3.
3
3
3
S
Z
~ is constant, curl F
~ = ~0 , so if S is the disk in the plane enclosed by the circle, Stokes’ Theorem gives
18. Since F
Z
~ · d~r =
F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
C
Z
S
~ = 0.
curl F~ · dA
19. Since
~ =
curl F
−y x
z
~ · d~r =
F
Z
=
∂
∂
(x) −
(−y) ~k = 2~k ,
∂x
∂y
writing S for the disk in the plane enclosed by the circle, Stokes’ Theorem gives
Z
C
S
~ · dA
~ =
curl F
Z
S
~.
2~k · dA
~ = ~n dA, where ~
Now dA
n is the unit vector perpendicular to the plane, so
1
~n = √ (~i + ~j + ~k ).
3
Thus
Z
C
F~ · d~r =
Z
1
2
2~k · √ (~i + ~j + ~k ) dA = √
3
3
S
Z
2
2
8π
dA = √ · Area of disk = √ · π22 = √ .
3
3
3
S
Job: chap20-sols Sheet: 18 Page: 1816 (August 24, 2012 11 : 19) [ex-2]
1816
Chapter Twenty /SOLUTIONS
20. Since
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
∂
∂
=
(y − x)~i −
(y − x)~j +
∂y
∂x
y −x y − x
∂
∂
(−x) −
(y) ~k = ~i + ~j − 2~k ,
∂x
∂y
writing S for the disk in the plane enclosed by the circle, Stokes’ Theorem gives
Z
C
~ · d~r =
F
Z
S
~ · dA
~ =
curl F
Z
S
~.
(~i + ~j − 2~k ) · dA
~ = ~n dA, where ~
Now dA
n is the unit vector perpendicular to the plane, so
1
~n = √ (~i + ~j + ~k ).
3
Thus
Z
~ · d~r =
F
C
~i + ~j + ~k
√
dA =
(~i + ~j − 2~k ) ·
3
S
Z
Z
0
√ dA = 0.
3
S
21. Since
~i
~ =
curl F
∂
∂x
2y + e
x
~j
~k
∂
∂y
∂
∂z
(sin y) − x 2y − x + cos z
2
= 2~i − (−1)~j + (−1 − 2)~k = 2~i + ~j − 3~k ,
writing S for the disk in the plane enclosed by the circle, Stokes’ Theorem gives
Z
C
~ · d~r =
F
Z
S
~ =
curl F~ · dA
Z
S
~.
(2~i + ~j − 3~k ) · dA
~ = ~n dA, where ~
Now dA
n is the unit vector perpendicular to the plane, so
1
~n = √ (~i + ~j + ~k ).
3
Thus
Z
C
F~ · d~r =
~i + ~j + ~k
√
(2~i + ~j − 3~k ) ·
dA =
3
S
Z
Z
0 dA = 0.
S
22. Since
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
=
0 −z y
∂
∂
(y) −
(−z) ~i − 0~j + 0~k = 2~i ,
∂y
∂z
writing S for the disk in the plane enclosed by the circle, Stokes’ Theorem gives
Z
C
~ · d~r =
F
Z
S
~ =
curl F~ · dA
Z
S
~.
2~i · dA
~ = ~n dA, where ~
Now dA
n is the unit vector perpendicular to the plane, so
1
~n = √ (~i + ~j + ~k ).
3
Thus
Z
C
F~ · d~r =
~i + ~j + ~k
dA =
√
2~i ·
3
S
Z
Z
S
2
2
8π
2
√ dA = √ · Area of disk = √ π22 = √ .
3
3
3
3
Job: chap20-sols Sheet: 19 Page: 1817 (August 24, 2012 11 : 19) [ex-2]
20.2 SOLUTIONS
1817
23. Since
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= 2~i + 2~j + 2~k ,
(z − y) (x − z) (y − x)
writing S for the disk in the plane enclosed by the circle, Stokes’ Theorem gives
Z
C
F~ · d~r =
Z
~ =
curl F~ · dA
S
Z
S
~.
(2~i + 2~j + 2~k ) · dA
~ = ~n dA, where ~
Now dA
n is the unit vector perpendicular to the plane, so
1
~n = √ (~i + ~j + ~k ).
3
Thus
Z
C
~ · d~r =
F
Z
~i + ~j + ~k
dA =
√
(2~i + 2~j + 2~k ) ·
3
S
2
2
Z
S
√
√
√
6
√ dA = 2 3 · Area of disk = 2 3π22 = 8 3π.
3
2
24. (a) All 3-space, because (1 + ax + by + cz ) is always positive.
(b) We have
2ax~i + 2by~j + 2cz~k
.
1 + ax2 + by 2 + cz 2
(c) We expect curl(grad f ) = 0 because all gradient vector fields satisfy the curl test. Direct calculation gives the same
result.
~ = grad f = grad(ln(1 + ax2 + by 2 + cz 2 )) with a = 1, b = 2, c = 3. So, if P and Q are the start
(d) Vector field F
and end points of C:
grad f = grad(ln(1 + ax2 + by 2 + cz 2 )) =
Z
C
F~ · d~r =
Z
C
Q
grad f · d~r = f
.
P
Since P is the point on C where t = 0, we have P = (cos 0, sin 0, 0) = (1, 0, 0). Since Q is the point on C with
t = 13π/2, we have Q = (cos(13π/2), sin(13π/2), 13π/2) = (0, 1, 13π/2). Thus,
Z
C
(0,1,13π/2)
~ · d~r = ln(1 + x2 + 2y 2 + 3z 2 )
F
(1,0,0)
= ln(3 + 507π 2 /4) − ln(2)
25. (a) The equation of the rim, C, is x2 + y 2 = 9, z = 2. This is a circle of radius 3 centered on the z-axis, and lying in the
plane z = 2.
(b) Use Stokes’ Theorem, with C oriented clockwise when viewed from above:
Z
S
~ =
curl(−y~i + x~j + z~k ) · dA
Z
C
(−y~i + x~j + z~k ) · d~r .
Since C is horizontal, the ~k component does not contribute to the integral. The remaining vector field, −y~i + x~j , is
tangent to C, of constant magnitude || − y~i + x~j || = 3 on C, and points in the opposite direction to the orientation.
Thus
Z
S
~ =
curl(−y~i + x~j + z~k ) · dA
Z
(−y~i + x~j ) · d~r = −3 · Length of curve = −3 · 2π3 = −18π.
26. The curl F~ (x, y, z) of this vector field is equal to −2~j . Notice, as a check, that this field rotates in a direction opposite
to the direction of C. Therefore we expect a negative line integral. The surface S is the disk parallel to the xz plane with
radius 2. The curl points in the opposite direction to the normal vector to the surface with this orientation, so by Stokes’
Theorem,
Circulation around S =
Z
S
~ =
~ · dA
curl F
Z
S
~ =
(−2~j ) · dA
= −2(Area of circle) = −2π22 = −8π.
Z
S
−2 dA
Job: chap20-sols Sheet: 20 Page: 1818 (August 24, 2012 11 : 19) [ex-2]
1818
Chapter Twenty /SOLUTIONS
27. We calculate
~i
~ = ∂
curl G
∂x
xy
~j
∂
∂y
z
~k
∂
= (3 − 1)~i − (0 − 0)~j + (0 − x)~k = 2~i − x~k .
∂z
3y
Let S be the surface of the square, oriented in the positive x-direction. Then, by Stokes’ Theorem
Z
C
~ · d~r =
G
Z
S
~ ) · dA
~ =
(curl G
Z
S
~.
(2~i − x~k ) · dA
~ = ~i dydz, so
On the square we have, dA
Z
C
~ · d~r =
G
Z
S
(2~i − x~k ) · ~i dydz = 2 · Area of square = 2 · 36 = 72.
~ is the curl of a vector field , we can use Stokes’ Theorem to replace the spherical surface, S, by the disk, D, of
28. Since F
radius 2 centered at the origin and oriented upward. Calculating F~ gives
~ =
F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= (2y sin(x2 ))~i − (2xy 2 cos(x2 ) + 2z sin(z 2 ))~j + ~k .
x3 + cos(z 2 ) x + sin(y 2 ) y 2 sin(x2 )
~ contributes to the flux, so
On the horizontal disk D, only the ~k component of F
Flux =
Z
S
~ =
F~ · dA
Z
D
~ · dA
~ = 1 · Area of disk = 4π.
F
~ = ~k for F~ =
29. Use Stokes’ theorem, applied to the surface R, oriented upward. Since curlF
R
R
1
~
~
~
~
~
(−y i + xj ) · d~r = R k · dA = kk k(area of R) = area of R.
2 C
1
(−y~i
2
+ x~j ), we have
30. The region is shown in Figure 20.7, so C2 − C1 is the boundary of the region, traversed with the region on the left. Thus,
~ = ~k dx dy and
Stokes’ Theorem applies with the region oriented upward, so dA
Z
C2
F~ · d~r −
Z
C1
~ · d~r =
F
Z
R
~ · dA
~ =
curl F
Z
R
3~k · ~k dx dy = 3 · Area of ring = 3(π52 − π22 ) = 63π.
y
C2 : x2 + y 2 = 25
x
✒
C1 : x2 + y 2 = 4
,
Figure 20.7
31. Let S be the curved surface of the cylinder. The boundary of S consists of the curves C1 and C2 , so Stokes’ Theorem
gives
Z
Z
Z
~ · d~r .
~
~
~
F · d~r +
F
curl F · dA =
S
C1
C2
~ contribute to this flux. The surface S
We calculate the flux of curl F~ through S. Only the ~i and ~j components of curl F
~ perpendicular to S has constant magnitude on S
has equation x2 + y 2 = 4, so the component of curl F
Job: chap20-sols Sheet: 21 Page: 1819 (August 24, 2012 11 : 19) [ex-2]
20.2 SOLUTIONS
~ || = ||3x~i + 3y~j || =
||F
1819
p
(3x)2 + (3y)2 = 3 · 2 = 6.
~ has no ~k -component on S, and it is parallel to 3x~i + 3y~j on S. Evaluating the flux integral
The vector dA
Z
S
~ · dA
~ = ||3x~i + 3y~j || · Area of S = 6 · (2π2 · 5) = 120π.
curl F
Thus,
Z
C1
F~ · d~r +
Z
C2
~ · d~r = 120π.
F
~ = x3~i + sin(y 3 )~j + ez 3 ~k . Then
32. (a) Let F
~ =
curl F
~i
~j
~k
∂
∂x
3
∂
∂y
∂
∂z
z3
x
= ~0 .
sin(y 3 ) e
~ · d~r = 0 by Stokes’ Theorem.
(b) Since C is a closed curve, C F
(c) The line integral is 0 for any closed curve C in 3-space.
~ is everywhere parallel to the xy-plane, and
33. (a) F~ has only ~i and ~j components, and they do not depend on z. Thus F
takes the same values for every value of z.
(b) We have
~k
~i
~j
∂
∂
∂
∂F2
∂F1 ~
~
k.
curl F =
=
−
∂x
∂y
∂z
∂x
∂y
F1 (x, y) F2 (x, y) 0
R
~ in S, we
(c) Since C is in the xy-plane, oriented counterclockwise when viewed from above, for an area element dA
~ = ~k dx dy. Thus Stokes’ Theorem says
have dA
Z
C
~ · d~r =
F
Z
S
~ =
curl F~ · dA
Z
S
∂F1
∂F2
−
∂x
∂y
~k · ~k dx dy =
Z
S
∂F1
∂F2
−
∂x
∂y
dx dy.
(d) Green’s Theorem.
34. (a) Notice that the denominators (z 2 + 1)2 and (z 2 + 1) are always positive and so affect the magnitude (but not the
direction) of the motion of each of the terms.
The (−y~i + x~j ) term represents rotation around the z-axis (counterclockwise when viewed from above). The
−z(x~i + y~j ) term represents radial motion (toward the z-axis when z > 0 and away when z < 0). The ~k term
is downward motion. So F~ is a flow rotating inward and downward around the z-axis (for z > 0), like an actual
bathtub drain.
(b) Let D be the disk representing the drain, oriented downward. Then the rate at which the water is leaving the bathtub
is the flux of water flowing out of the drain:
Z
D
~ · dA
~ =
F
Z
D
F~ · (−dA~k ) =
Z
D
1
dA.
z2 + 1
Because D is in the xy-plane, z = 0, so
Flux out of D =
Z
dA = π cm3 /sec.
D
(c) We have, in units/sec,
~ =−
div F
z
2z
z
− 2
+ 2
= 0.
(z 2 + 1)2
(z + 1)2
(z + 1)2
Job: chap20-sols Sheet: 22 Page: 1820 (August 24, 2012 11 : 19) [ex-2]
1820
Chapter Twenty /SOLUTIONS
(d) Let W be the closed region bounded by the hemisphere S of radius 1 lying below the xy-plane and the disk, D, in
the xy-plane representing the drain. Both S and D are oriented downward, so by the Divergence Theorem, we have:
0=
=
=
Z
Z
Z
div F~ dV = Flux out of W − Flux into W
W
Z
S
~ · dA
~ −
F
S
~ · dA
~ − π.
F
Thus,
Z
S
~ · dA
~
F
D
~ = π cm3 /sec.
F~ · dA
(e) Since C is oriented clockwise, we parameterize the circle by ~r (t) = (sin t)~i + (cos t)~j . In addition, on the drain,
z = 0. Thus,
Z
C
~ · d~r =
G
Z
1 ~
(y i − x~j − (x2 + y 2 )~k ) · d~r
2
C
1
=
2
1
2
=
Z
2π
(cos t~i − sin t~j − ~k ) · (cos t~i − sin t~j ) dt
0
Z
2π
(sin2 t + cos2 t)dt = π.
0
So,
Z
C
~ · d~r = π cm3 /sec.
G
(f) Computing the partial derivatives, we find that
~.
~ = − y + xz ~i − yz − x ~j − 1 ~k = F
curl G
(z 2 + 1)2
(z 2 + 1)2
z2 + 1
(g) By Stokes’ Theorem, we have:
Z
C
~ · d~r =
G
Z
D
~ · dA
~ =
curl G
Z
D
~.
F~ · dA
~ = F~ , Stoke’s Theorem tells us that the answers to parts (d) and (e) should be equal.
Thus, since curl G
Strengthen Your Understanding
35. The curve C is not the boundary of a surface, so Stokes’ Theorem does not apply.
36. The orientations of the surface S and its boundary curve C are not compatible. For Stokes’ Theorem to apply with the
downward orientation of the surface, the boundary C must be oriented clockwise in the xy-plane. With the orientations
of S and C as given in the problem statement, we have
Z
C
~ · d~r = −
F
Z
S
~ · dA
~.
curl F
~ = 0 everywhere, Stokes’ Theorem shows that
37. Since curl F
Z
C
F~ · d~r =
Z
S
~ · dA
~ =0
curl F
for every closed curve C that is the boundary of a surface S. For example, we can take C to be any circle with either
orientation.
38. The surface S could be the flat disk x2 + y 2 ≤ 1, z = 0 oriented upward, since using the right hand rule would orient the
boundary circle counterclockwise.
Job: chap20-sols Sheet: 23 Page: 1821 (August 24, 2012 11 : 19) [chap20-sols]
20.2 SOLUTIONS
1821
~ through the flat disc S in the xy-plane
39. True. By Stokes’ Theorem, the circulation of F~ around C is the flux of curl F
~ = ±~k dA, where the sign depends on the orientation of the circle.
enclosed by the circle. An area element for S is dA
~ is perpendicular to the z-axis, curl F
~ · dA
~ = ±(curl F
~ · ~k )dA = 0, so the flux of curl F
~ through S is
Since curl F
~
zero, hence the circulation of F around C is zero.
40. True. On a small patch of S that includes the boundary circle, the positive normal is outward. Letting the thumb of the
right hand point in this direction makes the fingers curl in the counterclockwise direction.
41. False. Using the right-hand rule gives C1 oriented clockwise and C2 oriented counterclockwise when viewed from the
positive y-axis.
R
~ ·d~r =
42. False. The curl needs to be in the flux integral, not the line integral, for a correct statement of Stokes’ theorem: C F
R
~
~
curl F · dA .
S
43. True. By Stokes’ theorem, both flux integrals are equal to the line integral
oriented counterclockwise when viewed from the positive z-axis.
R
C
F~ · d~r , where C is the circle x2 + y 2 = 1,
~ · d~r , where C is the boundary of the closed
44. True. By Stokes’ theorem, the flux integral is equal to the line integral C F
sphere. Since the sphere has no boundary curve, the line integral is zero. Alternatively, the closed sphere can be divided
into two hemispheres S1 (the top half with upward orientation) and S2 (the bottom half with downward orientation.) Then
S1 and S2 both have the circle C (x2 + y 2 = 1, z = 0) as their common boundary, except that for S1 , C is oriented
counterclockwise when viewed
using Stokes’
R as the boundary Rof S2 , C is oriented
R clockwise. Then,
R
R from above, and
~ =
~ · dA
~ +
~ · dA
~ =
~ · d~r −
~ · d~r = 0.
curl
F
F
curl
F
F
theorem on S1 and S2 gives S curlF~ · dA
S1
S2
C
C
~ = 0.
The same result can be obtained using the Divergence theorem and the fact that div curl F
R
~ · d~r =
45. True. Let D be the interior disk of C, oriented by the right hand rule. By Stokes’ theorem, C F
R
R
R
R
~ · dA
~ . Since curlF
~ = curlG
~ , we have
~ · d~r =
~ · d~r .
~ · d~r =
curl
G
F
G
and C G
D
C
C
R
R
46. True. Let S be the rectangular region inside C, oriented by the right hand rule. By Stokes’ theorem,
R
~ · dA
~ = 0.
curlF
S
D
R
~,
curlF~ · dA
C
~ · d~r =
F
~ · d~r = 0 implies, by Stokes’ theorem, that
~ · dA
~ = 0. However, curl F~ need
47. False. The condition that C F
curl F
S
~ = y~i , and let S be the upper unit hemisphere x2 + y 2 + z 2 = 1, z ≥ 0
not be ~0 for this to occur. For example, let F
2
orientedRupward. Then C is the circle x + y 2 = 1, z = 0 oriented counterclockwise when viewed from above. The line
integral C y~i · d~r = 0, but curl F~ = −~k .
R
R
48. Since
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= (6 − (−1))~i − (−7 − 2)~j + (1 − 3)~k = 7~i + 9~j − 2~k ,
2z + 3y x − z 6y − 7x
~,
by Stokes’ Theorem, if S is the flat interior of the circle, with area vector A
Z
C
~ · d~r =
F
Z
S
~ · dA
~ =
curl F
Z
S
~ = (7~i + 9~j − 2~k ) · A
~.
(7~i + 9~j − 2~k ) · dA
~ is parallel to curl F~ = 7~i + 9~j − 2~k , so curl F~ is the normal to the plane of the circle. Since
This is a maximum if A
the plane containing the circle is through the origin, its equation is
7x + 9y − 2z = 0.
~ , notice that curl F~ points downward (because the z-component is negative).
To give the orientation of S shown by curl F
Therefore, the circle must be oriented counterclockwise when seen from the negative z-axis, or clockwise from the positive
z-axis.
Job: chap20-sols Sheet: 24 Page: 1822 (August 24, 2012 11 : 19) [ex-3]
1822
Chapter Twenty /SOLUTIONS
Solutions for Section 20.3
Exercises
1. We have
~j ~k
~i
~ =
curl F
= (1 − 1)~i + (0 − 0)~j + (0 − 0)~k = ~0 .
∂
∂
∂
∂x ∂y ∂z
2x z
y
~ = ~0 and F
~ is defined everywhere, we know by the curl test that F
~ is a gradient field. In fact, F
~ = gradf ,
Since curl F
where f (x, y, z) = x2 + yz, so f is a potential function for F~ .
2. We have
~j ~k
~i
~ =
curl F
= (0 − 1)~i + (0 − 1)~j + (0 − 1)~k 6= ~0 .
∂
∂
∂
∂x ∂y ∂z
y
z x
Since curl F~ 6= ~0 , by the curl test F~ is not a gradient field.
3. We have
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= (1 − 1)~i + (2 − 2)~j + (1 − 1)~k = ~0 .
y + 2z x + z 2x + y
~ is a gradient field. In fact,
Since curl F~ = ~0 and F~ is defined everywhere, we know by the curl test that F
~.
F~ = gradf , where f (x, y, z) = xy + yz + 2xz, so f is a potential function for F
4. We have
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= (−1 − (−1))~i + (−2 − 2)~j + (1 − 1)~k 6= ~0 .
y − 2z x − z 2x − y
~ 6= ~0 , by the curl test F
~ is not a gradient field.
Since curl F
~
~
~ is not a gradient field.
~
5. Since curl G = 2k 6= 0 , the vector field G
6. We have
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
yz xz + z 2 yx + 2yz
= ((x + 2z) − (x + 2z))~i − (y − y)~j + (z − z)~k = ~0 .
~ = ~0 and F
~ is defined everywhere, we know by the curl test that F
~ is a gradient field. In fact, F
~ = gradf ,
Since curl F
2
~
where f (x, y, z) = xyz + yz , so f is a potential function for F .
7. We have
div F~ =
∂x
∂y
∂z
+
+
= 0 + 0 + 0 = 0.
∂x
∂y
∂z
~ is a curl field.
Since div F~ = 0, by the divergence test F
8. We have
∂z
∂y
∂x
div F~ =
+
+
= 0 + 1 + 0 = 1.
∂x
∂y
∂z
~ =
Since div F
6 0, by the divergence test F~ is not a curl field.
9. We have
~ = ∂(2x) + ∂(−y) + ∂(−z) = 2 − 1 − 1 = 0.
div F
∂x
∂y
∂z
~ = 0, by the divergence test F~ is a curl field.
Since div F
Job: chap20-sols Sheet: 25 Page: 1823 (August 24, 2012 11 : 19) [ex-3]
20.3 SOLUTIONS
1823
10. We have
~ = ∂(x + y) + ∂(y + z) + ∂(x + z) = 1 + 1 + 1 = 3.
div F
∂x
∂y
∂z
~ =
Since div F
6 0, by the divergence test F~ is not a curl field.
11. We have
∂(−xy)
∂(2yz)
∂(yz − z 2 )
+
+
= −y + 2z + (y − 2z) = 0.
∂x
∂y
∂z
~ = 0, by the divergence test F~ is a curl field.
Since div F
div F~ =
12. We have
∂(xy)
∂(xy)
∂(xy)
+
+
= y + x + 0.
div F~ =
∂x
∂y
∂z
~ =
Since div F
6 0, by the divergence test F~ is not a curl field.
13. The region is all points above the xy-plane. Any curve in that region can be contracted in that region to a point, so the curl
test can be used. Also, any surface in that region bounds a solid in that region, so the divergence test can be used.
14. A small circle in the xz-plane centered at the origin surrounds the y-axis and cannot be contracted in the region to a point,
so the curl test cannot be applied. A closed surface in the region bounds a solid in the region so the divergence test can be
applied.
15. Any closed curve in the region can be contracted in the region to a point; even a small circle around the z-axis can be
contracted by pulling it around the end of the positive z-axis. Thus, the curl test can be applied. A closed surface in the
region bounds a solid in the region so the divergence test can be applied
16. Any closed curve be contracted in the region to a point. Even a small circle around the missing segment of the x-axis
can be contracted by pulling it around either end of the segment. Thus, the curl test can be applied. A sphere of radius 2
centered at the origin contains inside it the missing segment, so it does not bound a solid in the region. Thus, the divergence
test cannot be applied.
Problems
17. (a) We calculate the curl of each of these vector fields.
~ = curl(−by~i ) =
curl A
~k
~i
~j
∂
∂x
∂
∂y
∂
∂z
−by 0
0
=−
∂
(−by)~k = b~k .
∂y
(b)
~ = curl(bx~j ) =
curl A
(c)
~k
~i
~j
∂
∂x
∂
∂y
∂
∂z
0
bx
0
=
∂
(bx)~k = b~k .
∂x
1
~ = − 1 yb~i + 1 xb~j
− ~r × B
2
2
2
~ = curl
curl A
1~
B × ~r
2
~i
~j
∂
∂x
=
∂
∂y
~k
∂
∂z
−(1/2)by (1/2)bx 0
=
∂
∂x
1
∂
bx −
2
∂y
1
by
2
−
~k = b~k .
Job: chap20-sols Sheet: 26 Page: 1824 (August 24, 2012 11 : 19) [ex-3]
1824
Chapter Twenty /SOLUTIONS
18. Let ~v = a~i + b~j + c~k and try
~ = ~v × ~r = (a~i + b~j + c~k ) × (x~i + y~j + z~k ) = (bz − cy)~i + (cx − az)~j + (ay − bx)~k .
F
Then
curl F~ =
Taking a = 1, b =
z)~j + (y + 32 x)~k .
− 32 ,
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
bz − cy
cx − az
= 2a~i + 2b~j + 2c~k .
ay − bx
~ = 2~i − 3~j + 4~k , so the desired vector field is F~ = (− 3 z − 2y)~i + (2x −
c = 2 gives curl F
2
~.
19. In Example 3 on page 1051 we showed that curl(~b × ~r ) = 2~b . Thus (1/2)~b × ~r is a vector potential for B
20. Note that
∂
∂
(2y) +
(4x) = 0
div(2y~i + 4x~j ) =
∂x
∂y
and
~i ~j ~k
curl(3x~i + 9y~j ) =
= ~0 ,
∂
∂
∂
∂x ∂y ∂z
3x 9y 0
so (3x~i + 9y~j ) + (2y~i + 4x~j ) is the required decomposition.
~ = 2x + 2y + 2z 6= 0, there is not a vector potential for G
~.
21. Since div G
~ = 2 + 3 − 5 = 0, a vector potential does exist. One such is the vector field H
~ = (−xy + 5yz)~i + (2xy +
22. Since div F
xz 2 )~k , but there are many others.
23. (a) Yes. To show this, we use a version of the product rule for curl (Problem 29 on page 1805):
~ ) = φ curl F
~ + (grad φ) × F
~,
curl(φF
~ is a vector field. So
where φ is a scalar function and F
curl q
~r
k~r k3
= curl
q
~r
k~r k3
= ~0 + q grad
=
1
k~r k3
q
curl ~r + grad
k~r k3
q
k~r k3
× ~r
× ~r
Since the level surfaces of 1/k~r k3 are spheres centered at the origin, grad(1/k~r k3 ) is parallel to ~r , so grad(1/k~r k3 )×
~ = ~0 .
~r = ~0 . Thus, curl E
~ is 3-space minus (0, 0, 0). Any closed curve in this region is the boundary of a surface
(b) Yes. The domain of E
contained entirely in the region. (If the first surface you pick happens to contain (0, 0, 0), change its shape slightly to
avoid it.)
~ satisfies both conditions of the curl test, it must be a gradient field. In fact,
(c) Yes. Since E
~ = grad −q 1
E
k~r k
.
~ =B
~.
24. We must show curl A
~ =
curl A
∂
∂y
=
−I
c
2I
=
c
~.
=B
−I
∂
ln(x2 + y 2 ) ~i −
c
∂x
−y~i + x~j
x2 + y 2
2y
x2 + y 2
~i + I
c
−I
ln(x2 + y 2 ) ~j
c
2x
(x2 + y 2 )
~j
Job: chap20-sols Sheet: 27 Page: 1825 (August 24, 2012 11 : 19) [ex-3]
20.3 SOLUTIONS
1825
25. (a) Yes. This is the case p = 2 of Example 5 on page 1052.
~ is 3-space minus the z-axis. A closed curve C which surrounds the z-axis cannot be contracted
(b) No. The domain of B
to a point without hitting the z-axis, so it cannot remain at all times within the domain.
(c) No. In Example 2 on page 1058 we found that if C is a circle around the origin,
Z
~ · d~r = 4πI .
B
c
C
~ has non-zero circulation around C, and hence cannot be a gradient field.
Thus B
26. (a) Using the product rule from Problem 29 on page 1805, we find
~ = curl
curl E
Now curl ~r = ~0 and grad
1
k~
r kp
~r
k~r kp
1
=
curl ~r + grad
k~r kp
1
k~r kp
× ~r .
~ = ~0 .
is parallel to ~r , so both terms are zero. Thus curl E
~ is 3-space minus the origin if p > 0, and it is all of 3-space if p ≤ 0.
(b) The domain of E
~
(c) Both domains have the property that any closed curve can be contracted to a point without hitting the origin, so E
~ has constant magnitude r 1−p on the sphere of radius r centered at the origin,
satisfies the curl test for all p. Since E
and is parallel to the outward normal at every point of the sphere, the sphere must be a level surface of the potential
~ k = r 1−p , a good guess is
function φ, that is, φ is a function of r alone. Further, since kE
φ(r) =
that is,
Z
r 1−p dr,
r 2−p
2−p
if p 6= 2
ln r
if p = 2.
~ by checking that grad φ = E
~.
You can check that this is indeed a potential function for E
~
~
~
~
27. We apply Stokes’ Theorem to the vector field F (x, y, z) = u(x, y)i + v(x, y)j + 0k . Since ∂u/∂z = ∂v/∂z = 0, it
is straightforward to compute that curlF~ = (∂v/∂x − ∂u/∂y)~k . Therefore, by Stokes’ Theorem,
φ(r) =
Z
C
(u~i + v~j ) · d~r =
Z
R
~
(∂v/∂x − ∂u/∂y)~k · dA
~ = dxdy~k . The result follows.
Since the surface R is in the xy-plane, oriented upward, dA
~
~
28. (a) Although div B = 0, the vector field B does not satisfy the divergence test because its domain is 3-space minus the
origin, which does not have the required property that every closed surface is the boundary of a solid region which
is entirely contained within the domain. For example, the solid region inside a sphere centered at the origin contains
~.
the origin, hence is not in the domain of B
(b) Using the product rule from Problem 29 on page 1805, we find
~ =
curl A
By Example 3 on page 1051,
1
k~r k3
curl(µ × ~r ) + grad
1
k~r k3
× (~
µ × ~r ).
curl(~
µ × ~r ) = 2~
µ
and by Problem 68 on page 796
grad
So
~ =2
curl A
From Problem 47 on page 750, we have
1
k~r k3
= −3
1
~r .
k~r k5
1
k~r k5
~
µ
−3
k~r k3
~r × (~
µ × ~r ).
~r × (~
µ × ~r ) = k~r k2 µ
~ − (~
µ · ~r )~r .
So
3(~
µ · ~r )~r
1
µ
~
~
~ =2 µ
(k~r k2 µ
~ − (~
µ · ~r )~r ) = −
curl A
−3
+
.
k~r k3
k~r k5
k~r k3
k~r k5
(c) No. The divergence test says a vector field must be a curl field if it satisfies the conditions of the test; it does not say
the vector field cannot be a curl field if the vector field fails to satisfy the test.
Job: chap20-sols Sheet: 28 Page: 1826 (August 24, 2012 11 : 19) [chap20-sols]
1826
Chapter Twenty /SOLUTIONS
~ + grad ψ) = curl A
~ + curl grad ψ = curl A
~ =B
~.
29. (a) Since curl grad ψ = 0 for any function ψ, curl(A
(b) We have
~ + grad ψ) = div A
~ + div grad ψ = div A
~ + ∇2 ψ.
div(A
Thus ψ should be chosen to satisfy the partial differential equation
~.
∇2 ψ = − div A
Strengthen Your Understanding
~ = 0 for any smooth vector field F
~ , and div x~i = 1 6= 0, there can be no vector field F
~ such that
30. Since div curl F
~
~
curl F = xi .
~ is meaningless, because div F
~ is a scalar function, but curl is only defined for vector fields.
31. The expression curl div F
~
~
~
~
~
32. If F = xi + y j , then div F 6= 0, so F is not the curl of another vector field.
33. If f (x, y, z) = x2 , then grad f = 2x~i and div grad f = 2 6= 0.
34. True. For example, take F~ = y~k .
~ = F1~i + F2~j + F3~k . Then curlF
~ = x~i gives
35. False. To see why, write F
∂F2
∂F1
∂F3
∂F2
∂F1
∂F3
−
= x;
−
= 0;
−
= 0.
∂y
∂z
∂z
∂x
∂x
∂y
Now take the partial
∂
∂x
of the first equation,
∂
∂y
of the second and
∂
∂z
of the third. This gives
∂ 2 F3
∂ 2 F2
∂ 2 F1
∂ 2 F3
∂ 2 F2
∂ 2 F1
−
= 1;
−
= 0;
−
= 0.
∂x∂y
∂x∂z
∂y∂z
∂y∂x
∂z∂x
∂z∂y
Assuming the continuity of the second order partials, the equality of mixed partials in the second and third equations
2
∂ 2 F2
F3
~ = x~i .
= ∂x∂z
, which contradicts the first equation. Thus there cannot be a vector field F~ with curlF
shows that ∂∂x∂y
~ · dA
~ = 0 implies, by Stokes’ theorem, that
~ · d~r = 0. However, F
~ need not be a
F
curlF
C
2
2
~
~
gradient field for this to occur. For example, let F = xk , and let S be the upper unit hemisphere x + y + z 2 = 1, z ≥ 0
oriented Rupward. Then C is the circle x2 + y 2 = 1, z = 0 oriented counterclockwise when viewed from above. The line
~ is everywhere perpendicular to C. The curl of F~ is the constant field −~j ,
integral C x~k · d~r = 0, since the field F
R
~
~ = 0, since the constant field −~j flows in, and then out of the
so F is not a gradient field. Yet we have S −~j · dA
hemisphere S.
R
R
~ · dA
~ =
~ · d~r , where C is one or more closed curves that form the boundary
37. True. By Stokes’ theorem, S curl F
F
C
~ is a gradient field, its line integral over any closed curve is zero.
of S. Since F
36. False. The condition that
R
R
S
∂f
∂
∂z ∂y
(b) Is not zero. Let F~ = x~j + x~k , then
38. (a) Is zero, since, for example,
−
∂
∂y
∂f
∂z
= 0.
~i ~j ~k
curl F~ =
∂
∂
∂
∂x ∂y ∂z
= −~j + ~k ,
0 x x
~ 6= ~0 .
so F~ × curl F
∂F2
∂F3
∂
∂F1
+
+
(c) Has components like
, not zero.
∂x
∂x
∂y
∂z
∂F3
∂F2
∂ ∂F1
∂F3
∂
∂F2
∂F1
∂
−
+
−
+
−
= 0.
(d) Is zero, since
∂x
∂y
∂z
∂y ∂z
∂x
∂z
∂x
∂y
∂2f
∂2f
∂2f
(e) Is
+
+
, not zero.
∂x2
∂y 2
∂z 2
Job: chap20-sols Sheet: 29 Page: 1827 (August 24, 2012 11 : 19) [ex-misc]
SOLUTIONS to Review Problems for Chapter Twenty
1827
Solutions for Chapter 20 Review
Exercises
1. We have
curl((x + y)~i − (y + z)~j + (x + z)~k ) =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= ~i − ~j − ~k
x + y −(y + z) (x + z)
2. We want the ~j component of curl ~n . Since
curl ~
n =
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
,
2x + 3y 4y + 5z 6z + 7x
the ~j component is
(curl ~n ) · ~j = −
∂
∂
(6z + 7y) −
(2x + 3y) = −7.
∂x
∂z
3. Fields with zero curl: (c), (d), (f ) because these don’t appear to be swirling.
4. Fields (a), (c), (e) because they do not appear to be exploding or collapsing.
5. C2 , C3 , C4 , C6 , since line integrals around C1 and C5 are clearly nonzero. You can see directly that
R
~ · d~r are zero, because C2 and C6 are perpendicular to their fields at every point.
F
C6
R
C2
~ · d~r and
F
6. Not defined, since we can’t take the gradient of a vector function.
~ (~r ) × G
~ (~r ) is a vector field; the integral represents the flux—a scalar—through the surface
7. Defined; scalar because F
S.
8. Defined; scalar because grad f is a vector field, so (grad f ) × ~r is a vector field, so we can calculate its divergence,
giving a scalar.
~ is a vector and the cross product of two vectors is a vector.
9. Defined; vector since curl F
10. The circulation around any square with sides parallel to the axes and centered at the point is zero because the line integrals
on the top and bottom sides add to zero and the line integrals on the left and right sides add to zero. We suspect that the
vector field has zero curl.
11. The circulation around the boundary of any square with sides parallel to the axes and enclosing the point is nonzero
because the vectors at the bottom are larger than those at the top, so we suspect a nonzero curl at this point.
12. We have
div F~ =
∂
∂ 4
∂
x2 +
y3 +
z = 2x + 3y 2 + 4z 3 ,
∂x
∂y
∂z
~i ~j ~k
curl F~ =
∂
∂
∂
∂x ∂y ∂z
2
3
4
= ~0 .
x y z
So F~ is not solenoidal, but F~ is irrotational.
13. We have
div F~ =
∂
∂
∂
(xy) +
(yz) +
(zx) = y + z + x,
∂x
∂y
∂z
~i ~j ~k
~ =
curl F
∂
∂
∂
∂x ∂y ∂z
xy yz zx
So F~ is not solenoidal and not irrotational.
= −y~i − z~j − x~k .
Job: chap20-sols Sheet: 30 Page: 1828 (August 24, 2012 11 : 19) [ex-misc]
1828
Chapter Twenty /SOLUTIONS
14. We have
~ = ∂ (cos x) + ∂ (ey ) + ∂ (x + y + z) = − sin x + ey + 1
div F
∂x
∂y
∂z
~k
~i
~j
∂
∂
∂
= ~i − ~j .
curl F~ =
∂x
∂y
cos x
∂z
ey x + y + z
So F~ is not solenoidal and not irrotational.
15. We have
~i
~ =
curl F
∂
∂x
y+z
~ = ∂ (ey+z ) + ∂ (sin(x + z)) + ∂ (x2 + y 2 ) = 0
div F
∂x
∂y
∂z
~k
~j
∂
∂
= (2y − cos(x + z))~i − 2x − ey+z ~j + cos(x + z) − ey+z ~k .
∂y
∂z
sin(x + z) x2 + y 2
~ is not irrotational.
So F~ is solenoidal, but F
e
16. (a) Direct method:
curl F~ =
~i
~j
∂
∂x
∂
∂y
0
~k
∂
∂z
= −2x~i − (−y)~j + z~k = −2x~i + y~j + z~k .
xz −xy
~ has no ~i -component, so the ~i -component of curl F
~ does not contribute to the flux. Thus
On the surface, dA
Z
S
~ · dA
~ =
curl F
Since y~j + z~k is perpendicular to S and ||y~j + z~k || =
Z
S
~ · dA
~ =
curl F
√
Z
S
~.
(y~j + z~k ) · dA
p
y2 + z2 =
5 · Area of S =
√
√
5 on S, we have
√
5 · 2π 5 · 3 = 30π.
(b) Using Stokes’ theorem, we replace the flux integral by two line integrals around the circular boundaries, C1 and C2 ,
of S. See Figure 20.8.
Z
Z
Z
S
curl F~ · ds =
C1
F~ · dr +
C2
~ · d~r .
F
~ = ~0 , and therefore
On C1 , the left boundary, x = 0, so F
F~ · d~r = 0. On C2 , the right boundary, x = 3, so
C1
p
p
√
~ || = (3z)2 + (−3y)2 = 9(z 2 + y 2 ) = 3 5. and F
~ is tangent to
F~ = 3z~j − 3y~k . This vector field has ||F
the boundary C2 and pointing in the same direction as C2 . Thus
Z
√
√
~ || · Length of C2 = 3 5 · 2π 5 = 30π.
~ · d~r = ||F
F
R
C2
z
C2
C1
x
Figure 20.8
Job: chap20-sols Sheet: 31 Page: 1829 (August 24, 2012 11 : 19) [ex-misc]
SOLUTIONS to Review Problems for Chapter Twenty
17. (a) Let us parameterize the curve C by ~r (t) = 3 cos t~i + 3 sin t~j ,
Then d~r = (−3 sin t~i + 3 cos t~j )dt and so
Z
C
1829
0 ≤ t ≤ 2π.
Z
((yz 2 − y)~i + (xz 2 + x)~j + 2xyz~k ) · d~r =
(−3 sin t~i + 3 cos t~j ) · d~r
C
2π
Z
=
9dt = 18π.
0
(b) Since C is a closed curve, Stokes’ Theorem applies. We choose the surface S to be the disk in the xy-plane bounded
by C, and it must be oriented upward. Since curlF~ = 2~k ,
Z
C
Z
~ · d~r =
F
S
~ = k2~k k(area of S) = 2(π32 ) = 18π.
2~k · dA
18. (a) Let C be the boundary of the disc D, given by y 2 + z 2 ≤ 1 and x = 0, oriented counterclockwise when viewed from
the positive x-axis. Using Stokes’ Theorem, we have
Z
2
curl ex ~i + (x + y)~k
D
~ =
· dA
Z
2
ex ~i + (x + y)~k
C
· d~r .
Parameterize the curve C by ~r (t) = cos t~j + sin t~k for 0 ≤ t ≤ 2π. Then, ~r ′ (t) = − sin t~j + cos t~k . Thus,
Z
x2~
e i + (x + y)~k
C
· d~r =
=
2π
Z
e0~i + (0 + cos t)~k · (− sin t~j + cos t~k )dt
0
Z
2π
cos2 t dt =
0
(b) We have
2
curl ex ~i + (x + y)~k
~ =~
and dA
n dA = ~i dA. So,
Z
D
x2~
curl e i + (x + y)~k
19. First C is parameterized by
~ =
· dA
Z
D
1
(t + sin t cos t)
2
2π
= π.
0
= ~i − ~j ,
(~i − ~j ) · ~i dA =
Z
dA = Area of disc D = π.
D
~r (θ) = 2 cos θ~i + 2 sin θ~j + ~k .
Note that C bounds the disk S given by x + y 2 ≤ 4, z = 1. Then
2
~r ′ (θ) = −2 sin θ~i + 2 cos θ~j .
Now,
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= 3~i + ~j + 5~k ,
z − 2y 3x − 4y z + 3y
~
~
and dA = k dA. Using Stokes’ Theorem we get
Z
C
~ · d~r =
F
=
Z
S
Z
S
~
curl F~ · dA
(3~i + ~j + 5~k ) · ~k dA =
Z
5dA
S
= 5(Area of circle) = 5(4π) = 20π.
20. First note that curl = 2~k .
~ where S is the disk of radius 10 in the xy-plane centered at the
2~k · dA
R
~ = −k2~k k(area of S) = −200π.
origin, oriented downward. Since this orientation is opposite to 2~k , S 2~k · dA
R
R
~ · d~r =
~ where S is the disk of radius 10 in the yz-plane centered at the
(b) By Stokes’ Theorem, C F
2~k · dA
S
origin, oriented in the negative x direction. Since the vector field 2~k is parallel to the surface S, its flux through the
surface is zero.
(a) By Stokes’ Theorem,
R
C
~ · d~r =
F
R
S
Job: chap20-sols Sheet: 32 Page: 1830 (August 24, 2012 11 : 19) [ex-misc]
1830
Chapter Twenty /SOLUTIONS
~ = ~r /k~r k3 consists of vectors pointing radially outward. There is no swirl, so curl F
~ = ~0 . From Stokes’
21. The graph of F
Theorem,
Z
Z
Z
~ · dA
~ =
~ =0
~ · d~r =
~0 · dA
curl F
F
S
C
S
22. The circulation is the line integral C F~ · d~r which can be evaluated directly by parameterizing the circle, C. Or, since
C is the boundary of a flat disk S, we can use Stokes’ Theorem:
R
Z
C
~ · d~r =
F
Z
S
~
curl F~ · dA
where S is the disk x2 + y 2 ≤ 1, z = 2 and is oriented upward (using the right hand rule). Then curl F~ = −y~i − x~j + ~k
and the unit normal to S is ~k . So
Z
~ · dA
~ =
curl F
S
=
Z
S
Z
(−y~i − x~j + ~k ) · ~k dxdy
1 dxdy
S
= Area of S = π
Problems
23. (a) is (I) since div(~r + ~a ) = 3.
(b) is (I) since
div(~r × ~a ) = div((a3 y − a2 z)~i + (a1 z − a3 x)~j + (a2 x − a1 y)~k ) = 0.
(c) is (V) since ~r · ~a is a scalar.
(d) is (III) since curl(~r + ~a ) = ~0 .
(e) is (IV) since
curl(~r × ~a ) = curl((a3 y − a2 z)~i + (a1 z − a3 x)~j + (a2 x − a1 y)~k ) = −2a1~i − 2a2~j − 2a3~k = −2~a .
(f) is (V) since ~r · ~a is a scalar.
24. (a)
(b)
(c)
(d)
(e)
grad(~r · ~a ) = grad(a1 x + a2 y + a3 z) = ~a .
Not defined. Can’t take the divergence of a scalar.
Not defined. Can’t take the curl of a scalar.
Not defined. Can’t take the gradient of a vector.
Now
~k
~i ~j
~r × ~a = x y
z = (a3 y − a2 z)~i − (a3 x − a1 z)~j + (a2 x − a1 y)~k .
a1 a2
a3
Thus,
div(~r × ~a ) = 0.
(f) Using the expression for ~r × ~a in part (e), we have
curl(~r × ~a ) =
~i
~j
∂
∂x
∂
∂y
~k
∂
∂z
= −2a1~i − 2a2~j − 2a3~k = −2~a .
a3 y − a2 z a1 z − a3 x a2 x − a1 y
~ contributes to the flux integral. On S we have y = 5 and dA
~ = −~j dA (since S is
25. (a) Only the y-component of F
oriented toward the origin), so
Z
S
~ · dA
~ =
F
Z
S
53~j · (−~j dA) = −125 · Area of S = −125(π32 ) = −1125π.
Job: chap20-sols Sheet: 33 Page: 1831 (August 24, 2012 11 : 19) [ex-misc]
1831
SOLUTIONS to Review Problems for Chapter Twenty
(b) Not defined; we cannot integrate a vector field over a solid region in space.
(c) Since
~k
~i ~j
∂
∂
∂
~
curl F = ∂x ∂y ∂z = ~0 ,
x3 y 3
z3
~ · dA
~ = 0.
we have S curl F
(d) Not defined; we cannot calculate the gradient of a vector field.
(e) Since
~ = div(x3~i + y 3~j + z 3~k ) = 3x2 + 3y 2 + 3z 2 ,
div F
R
we have
Z
div F~ dV = 3
div F~ dV = 3
Z
2π
0
W
π
Z
0
Z
(x2 + y 2 + z 2 )dV.
W
W
Converting to spherical coordinates gives
Z
Z
2
2 2
ρ ρ sin φ dρ dφ dθ = 3 · 2π
0
π
− cos φ
0
ρ5
5
2
0
!
=
384π
.
5
~ = grad((x + y + z )/4), we use the Fundamental Theorem of Line Integrals to get
(f) Since F
4
4
4
4
4
4
~ · d~r = x + y + z
F
4
C
Z
(2,3,4)
=
(0,0,0)
353
24 + 34 + 44
=
.
4
4
Alternatively, we parameterize the line by x = 2t, y = 3t, z = 4t for 0 ≤ t ≤ 1. Then on the line F~ =
8t3~i + 27t3~j + 64t3~k and r ′ (t) = 2~i + 3~j + 4~k , giving
Z
C
~ · d~r =
F
Z
1
0
(8t i + 27t j + 64t k ) · (2~i + 3~j + 4~k ) dt =
3~
3~
3~
Z
1
(16t3 + 81t3 + 256t3 ) dt = 353
0
t4
4
1
=
0
353
.
4
~ is a vector field and we cannot integrate a vector field over a solid region in space.
(g) Not defined; curl F
~ = x3~i + y 3~j + z 3~k so
(h) We have F
Z
W
~ · (~i + ~j + ~k )dV =
F
Z
0
3
Z
0
2
Z
0
1
(x3 + y 3 + z 3 ) dx dy dz = 2 · 3
x4
4
1
0
+1·3
y4
4
2
0
+1·2
z4
4
3
0
14
24
34
= 2·3 +1·3 +1·2
= 54.
4
4
4
26. (a)
(b)
(c)
(d)
(e)
(f)
(g)
Defined; equal to (e) and (f) by Stokes’ Theorem.
Not defined.
Not defined.
Defined; not equal to others in list.
Defined; equal to (a) and (f) by Stokes’ Theorem.
Defined; equal to (a) and (e) by Stokes’ Theorem.
Defined; not equal to others in list.
~ = 6~i + 5~j − 8~k , so
27. (a) At P , we have curl F
~ · (~i + ~j + ~k ) = 6 + 5 − 8 = 3.
curl F
(b) The definition of the curl tells us that if ~n is the unit vector normal to the plane containing the circle and pointing
away from the origin, then
R
F~ · d~r
C
~
.
curl F · ~n ≈
Area enclosed by C
√
Since ~n = (~i + ~j + ~k )/ 3, we have
Z
C
~ · d~r ≈ (curl F
~ · ~n )Area of circle = √3 π(0.01)2 = 0.0003π
√
= 0.000544.
F
3
3
Job: chap20-sols Sheet: 34 Page: 1832 (August 24, 2012 11 : 19) [ex-misc]
1832
Chapter Twenty /SOLUTIONS
28. .
~ has its component in the x-direction given by
The vector curl G
~ )x ≈ Circulation around small square around x-axis
(curl G
Area inside square
6
Circulation around S2
=
= 600.
=
Area inside S2
(0.1)2
Similar reasoning leads to
~ )y ≈ Circulation around S3 = −5 = −500.
(curl G
Area inside S3
(0.1)2
Circulation around S1
−0.02
=
= −2.
Area inside S1
(0.1)2
~ )z ≈
(curl G
Thus,
~ ≈ 600~i − 500~j − 2~k .
curl G
~ < 0, and div G
~ < 0; div G
~ is larger in magnitude (more negative) if the scales are the same.
It appears that div F
~ both appear to be zero at the origin (and elsewhere).
curl F~ and curl G
Yes, the cylinder with axis along the z-axis will have negative flux through it (ends parallel to xy-plane).
Same as part(c).
No, you cannot draw a closed curve around the origin such that F~ has a non-zero circulation around it because curl
is zero. By Stokes’ theorem, circulation equals the integral of the curl over the surface bounded by the curve.
(f) Same as part(e)
√
~ contributes to the integral. Since the disk is
of F
30. (a) Since the disk is in the plane z = 5, only the ~k component √
~ = −~k dA on the disk; in addition, z = 5 on the disk. Thus,
oriented downward, dA
29. (a)
(b)
(c)
(d)
(e)
Z
S
~ =
F~ · dA
Z
S
√
√
√
√
√
( 5~k ) · (−~k dA) = − 5 · Area of disk = − 5π( 10)2 = −10 5π.
~
~
~
(b) The line
√ is parallel to the x-axis, so only the i component of F contributes to the integral. Since d~r = i dx and
y = 3 on the line,
Z
Z
√
√
√
~ · d~r =
( 3~i ) · (~i dx) = 3 · Length of line = 2 3.
F
C
C
(c) Using Stokes’ Theorem, if C is the boundary of S with induced orientation,
Z
S
~ =
curl F~ · dA
Z
C
~ · d~r .
F
The boundary of the rectangle C has four parts. The orientation is counterclockwise when viewed from above. The
integral along the x-axis is 0 because the ~i component is 0 there. Similarly, the integral along the y-axis is 0 because
the ~j component is 0 there.
√The integral along the top of the rectangle uses the reverse orientation to the answer to part (b), so its value is
the right side depends only on the ~j component at x = 2, so the value is of this integral is
−2 3. The integral along √
π · 2 · Length of line = 2 3π. Thus,
Z
S
~ · dA
~ =
curl F
Z
C
√
√
√
~ · d~r = 0 + 0 − 2 3 + 2 3π = 2 3(π − 1).
F
Alternatively, direct calculations gives
~i ~j ~k
curl F =
∂
∂
∂
∂x ∂y ∂z
= (π − 1)~k .
y πx z
Thus,
Z
S
~ =
curl F~ · dA
Z
S
√
(π − 1)~k · ~k dA = (π − 1) · Area of S = 2 3(π − 1).
Job: chap20-sols Sheet: 35 Page: 1833 (August 24, 2012 11 : 19) [ex-misc]
SOLUTIONS to Review Problems for Chapter Twenty
1833
31. We calculate
curl x i + y j + (x + y + z)~k
2~
2~
=
~i ~j
~k
∂
∂
∂x ∂y
2
2
∂
∂z
= ~i − ~j ,
x y x+y+z
2
and use Stokes’ Theorem with D as the surface (x − 1) + (y − 2)2 ≤ 4 in the xy-plane, oriented in the positive
z-direction:
Z
Z
2~
2~
~
~.
(~i − ~j ) · dA
x i + y j + (x + y + z)k · d~r =
D
C
~ = ~k dA, we have
Since dA
Z
D
~ =
(~i − ~j ) · dA
Z
0 dA = 0.
D
32. We use Stokes’ Theorem. We have
~j ~k
~i
3~
3~
z~
∂
∂
∂
∂x ∂y ∂z
3
3
z
curl(−y i + x j + e k ) =
= (3x2 + 3y 2 )~k .
−y x e
So if D is the disk x2 + y 2 ≤ 3, z = 4, oriented upward, we have
Z
C
(−y 3~i + x3~j + ez~k ) · d~r =
Z
D
~ =3
(3x2 + 3y 2 )~k · dA
Z
(x2 + y 2 ) dA.
D
Converting to polar coordinates, we have
Z
3~
C
z~
3~
(−y i + x j + e k ) · d~r = 3
Z
2π
0
√
Z
3
r4
r · r dr dθ = 3 · 2π
4
√
3
2
0
=
0
27π
.
2
33. We calculate
curl sin x i + cos y j + (x + y)~k
2~
2~
=
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
sin x2 cos y 2 x + y
= ~i − ~j ,
and use Stokes’ Theorem with D as the surface (y − 1)2 + (z − 2)2 ≤ 4 in the yz-plane, oriented in the positive
x-direction:
Z
Z
~.
(~i − ~j ) · dA
sin x2~i + cos y 2~j + (x + y)~k · d~r =
D
C
~ = ~i dA, we have
Since dA
Z
D
Z
~ =
(~i − ~j ) · dA
1 dA = Area D = π22 = 4π.
D
34. We use Stokes’ Theorem. If C is the circle y 2 + z 2 = 3, oriented counterclockwise when viewed from the positive
x-direction, then Stokes’ Theorem gives
Z
S
~ · dA
~ =
curl F
Z
C
F~ · d~r =
Z
C
(z + y)~i − (z + x)~j + (y + x)~k · d~r .
Since C is in the yz-plane, the ~i component does not contribute to the line integral. On the yz-plane, x = 0 so
Z
S
~ · dA
~ =
curl F
Z
C
F~ · d~r =
Z
C
(−z~j + y~k ) · d~r .
Job: chap20-sols Sheet: 36 Page: 1834 (August 24, 2012 11 : 19) [ex-misc]
1834
Chapter Twenty /SOLUTIONS
−z~j + y~k
On the circle y 2 + z 2 = 3, we have
counterclockwise direction, so
Z
S
~ · dA
~ =
curl F
Z
35. Since
C
√
=
(−z~j + y~k ) · d~r =
3 and −z~j + y~k is tangent to the curve pointing in the
√
3 · Length of curve =
√
√
3 · 2π 3 = 6π.
div(3x~i + 4y~j + xy~k ) = 3 + 4 + 0 = 7,
we calculate the flux using the Divergence Theorem:
Flux =
Z
S
~ =
(3x~i + 4y~j + xy~k ) · dA
Z
7 dV = 7 · Volume of box = 7 · 3 · 5 · 2 = 210.
W
36. We use the Divergence Theorem.
~ = 3 − 1 + 2 = 4 and a closed surface is oriented outward, the Divergence Theorem gives
Since divF
Z
S
~ · dA
~ =
F
Z
~ dV = 4 · Volume of sphere = 4 · 4 π13 = 16π .
div F
3
3
Interior
of sphere
~ = 3x2 +3y 2 +3z 2 and a closed surface is oriented outward, the Divergence
37. We use the Divergence Theorem. Since divF
Theorem gives
Z
S
~ · dA
~ =
F
Z
=
Z
=3
~ dV
div F
Interior
of sphere
3x2 + 3y 2 + 3z 2 dV
Interior
of sphere
Z
2π
Z
π
0
0
1
Z
ρ2 · ρ2 sin φ dρ dφ dθ
0
π
= 3 · 2π(− cos φ)
ρ5
5
0
1
=
0
12π
.
5
38. We use Stokes’ Theorem. Since
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= −2~i − 3~j − ~k ,
x + y y + 2z z + 3x
if S is the interior of the square, then
Z
C
Z
~ · d~r =
F
S
~ · dA
~ =
curl F
Since the area vector of S is 49~j , we have
Z
C
~ · d~r =
F
Z
S
Z
S
~
(−2~i − 3~j − ~k ) · dA
~ = −3~j · 49~j = −147.
(−2~i − 3~j − ~k ) · dA
39. We use Stokes’ Theorem. Since
~ =
curl F
~i
~j
~k
∂
∂x
3
∂
∂y
∂
∂z
= (3x2 + 3y 2 )~k ,
x − y + z x3 + y + z x + y + z 3
if S is the disk x2 + y 2 ≤ 10 oriented upward,
Z
C
F~ · d~r =
Z
S
~ =3
(3x + 3y )~k · dA
2
2
Z
0
2π
√
Z
0
√
10
r4
r · r dr dθ = 3 · 2π ·
4
2
10
= 150π.
0
Job: chap20-sols Sheet: 37 Page: 1835 (August 24, 2012 11 : 19) [ex-misc]
SOLUTIONS to Review Problems for Chapter Twenty
1835
~ = 3y 2 + 3z 2 , we have
40. We use the Divergence Theorem. Since divF
Z
Z
~ =
F~ · dA
S
(3y 2 + 3z 2 ) dV.
Interior
of cylinder
Converting to cylindrical coordinates with y = r cos θ, z = r sin θ, we have y 2 + z 2 = r 2 , so
Z
S
~ · dA
~ =3
F
1
Z
−1
2π
Z
0
Z
4
r 2 · r dr dθ dx = 3 · 2 · 2π
0
r4
4
4
= 768π.
0
41. We use the fact that if S1 and S2 are surfaces which share a common boundary C, and if S1 and S2 determine the same
orientation of C, then
Z
Z
~ · dA
~ =
~ · dA
~.
curl F
curl F
S1
S2
We replace the surface S by S0 , the base of the cube, oriented upward because S0 is easier to integrate over. (Both
surfaces have the unit square in the xy-plane as the boundary.) Since S0 is horizontal, we need only find the ~k component
of curl F~ :
x
y
~ = ∂(ye ) − ∂(−xe ) = (yex + xey )~k .
z-component of curl F
∂x
∂y
Thus, we have
Z
S
~ · dA
~ =
curl F
Z
=
Z
(yex + xey ) dA =
S0
Z
0
1
1
Z
1
(yex + xey ) dxdy =
0
0
1
yex +
0
y2
ey
ey
− y) dy = (e − 1)
+
2
2
2
(ey +
Z
1
0
x2 y
e
2
1
dy
0
= e − 1.
42. Let S be the disk inside the unit circle, oriented upward. By Stokes’ Theorem,
Z
C
F~ · d~r =
Z
S
~ = 4 · Area of disk = 4 · π12 = 4π.
4~k · dA
43. (a) Let D be the face of the box which was removed from the yz-plane. We use Stokes’ Theorem with D as the surface.
~ =
curl F
~i
~j
~k
∂
∂x
∂
∂y
∂
∂z
= 2~i − 2~k ,
x + y −z y
so
Z
C
~ · d~r =
F
Z
D
~ =
curl F~ · dA
Z
D
~ = 2(Area of D) = 2 · 32 = 18.
(2~i − 2~k ) · dA
~ = 1. Since the Divergence Theorem requires a closed
(b) To use the Divergence Theorem, we calculate that div F
surface, we close S by adding D, oriented in the direction of the negative x-axis. Then
Z
S+D
~ =
F~ · dA
Z
S
~ · dA
~ +
F
Z
D
~ · dA
~ =
F
Z
1 dv = Volume of box = 27.
Interior
~ = −~i dy dz, so
Now on D, where x = 0, we have F~ = y~i − z~j + y~k . In addition, on D, we have dA
Z
D
~ =
F~ · dA
Z
=−
Thus,
Z
S
(y~i − z~j + y~k ) · (−i dy dz) = −
Z
0
3
y2
2
3
−9
dz =
2
0
~ = 27 −
F~ · dA
Z
Z
0
3
dz = −
Z
0
Z
3
y dy dz
0
27
.
2
~ · dA
~ = 27 − − 27
F
2
D
3
=
81
.
2
Job: chap20-sols Sheet: 38 Page: 1836 (August 24, 2012 11 : 19) [ex-misc]
1836
Chapter Twenty /SOLUTIONS
44. (a) Can be computed. If W is the interior of the sphere, by the Divergence Theorem, we have
Z
S
~ =
F~ · dA
Z
W
128π
4
.
div F~ dV = 4 · Volume of sphere = 4 · π · 23 =
3
3
(b) Cannot be computed.
~ ) = 0. If W is the inside of the sphere, then by the Divergence
(c) Can be computed. Use the fact that div(curl F
Theorem,
Z
Z
Z
~ =
div(curl F~ )dV =
0 dV = 0
curl F~ · dA
W
S
W
~k
~i
~j
∂
∂
∂
~
45. (a) curl F~ =
= (x − 2xy)~i − (y − y 2 )~j + (2yz − 2yz)~k 6= ~0 . Since curl F~ 6= 0, we know F
∂x
∂y
∂z
y 2 z 2xyz xy
is not conservative.
~ is not conservative.
(b) A line integral round the closed path shown in Figure 20.9 is not zero, so F
Note: To show that a vector field is not conservative, we only need to find one path whose line integral is nonzero.
To show that a vector field is conservative, we must show the line integral is zero on all paths.
y
x
Figure 20.9
46. (a) We have
~k
∂
x
∂
y
~k .
~i + 0~j + ∂
=
0
+
∂z
∂x x2 + y 2
∂y x2 + y 2
0
~i
~j
∂
∂
~ =
curl F
∂x
∂y
−y
x
x2 + y 2 x2 + y 2
Since
∂
∂x
and similarly
∂
∂y
x
x2 + y 2
y
x2 + y 2
=
=
x2
x(2x)
1
x2 + y 2 − 2x2
y 2 − x2
− 2
=
= 2
2
2
2
2
2
2
+y
(x + y )
(x + y )
(x + y 2 )2
x2 − y 2
, we have, provided x2 + y 2 6= 0,
(x2 + y 2 )2
~ = ~0 .
curl F
~ is all points in 3-space except the z-axis.
The domain of curl F
~ is tangent to the circle and ||F
~ || = 1. Thus
(b) On C1 , the unit circle x2 + y 2 = 1 in the xy-plane, the vector field F
Circulation =
Z
C1
~ || · Perimeter of circle = 2π.
~ · d~r = ||F
F
Note that Stokes’ Theorem cannot be used to calculate this circulation since the z-axis pierces any surface which has
this circle as boundary.
Job: chap20-sols Sheet: 39 Page: 1837 (August 24, 2012 11 : 19) [ex-misc]
SOLUTIONS to Review Problems for Chapter Twenty
1837
~ = ~0 everywhere
(c) Consider the disk (x − 3)2 + y 2 ≤ 1 in Rthe plane z = 4. This disk has C2 as boundary and curl F
~ · d~r = 0.
on this disk. Thus, by Stokes’ Theorem C F
2
(d) The square S has an interior region which is pierced by the z-axis, so we cannot use Stokes’ Theorem. We consider
the region, D, between the circle C1 and the square S. See Figure 20.10. Stokes’ Theorem applies to the region D,
provided C1 is oriented clockwise. Then we have
Z
C1 (clockwise)
Thus,
Z
S
~ · d~r = −
F
~ · d~r +
F
Z
C1 (clockwise)
Z
S
~ · d~r =
F
Z
D
~ = 0.
curl F~ · dA
Z
~ · d~r =
F
C1 (counterclockwise)
~ · d~r = 2π.
F
(e) If a simple closed curve goes around the z-axis, then it contains a circle C of the form x2 + y 2 = a2 . The circulation
around C is 2π or −2π, depending on its orientation. A calculation similar to that in part (d) then shows that the
circulation around the curve is 2π or −2π, again depending on its orientation. If the closed curve does not go around
the z-axis, then curl F~ = ~0 everywhere on its interior and the circulation is zero.
y
(−2, 2, 0)
(2, 2, 0)
S
C1
x
D
(−2, −2, 0)
(2, −2, 0)
Figure 20.10
~ is tangent to the circle, because
47. (a) Notice that F
(−y~i + x~j )
· (x~i + y~j ) = 0,
F~ · ~r =
x2 + y 2
~ has constant magnitude on the circle x2 + y 2 = 9, since
and F
~ || =
||F
−y~i + x~j
x2 + y 2
=
p
(−y)2 + x2
1
= .
x2 + y 2
3
Thus, the line integral can be calculated as follows:
Z
C1
~ · d~r = ||F
~ || · Circumference of circle = 1 · 2π3 = 2π.
F
3
(b) We have
~i
curl F~ =
=
=
~j
~k
∂
∂
∂
∂x
∂y
∂z
−y
x
0
x2 +y 2 x2 +y 2
=
∂
∂x
x
x2 + y 2
∂
−
∂y
−y
x2 + y 2
2x2
1
2y 2
1
− 2
+ 2
− 2
2
2
2
2
2
x +y
(x + y )
x +y
(x + y 2 )2
x2 + y 2 − 2x2 + x2 + y 2 − 2y 2 ~
k = ~0 .
(x2 + y 2 )2
~ = ~0 except where x = y = 0, on the z-axis, where it is not defined.
Thus, curl F
~k
~k
Job: chap20-sols Sheet: 40 Page: 1838 (August 24, 2012 11 : 19) [ex-misc]
1838
Chapter Twenty /SOLUTIONS
R
(c) Stokes’ Theorem cannot be used to calculate C because the z-axis cuts any surface whose boundary is C1 .
1
(d) Since the disk D2 inside C2 does not intersect the z-axis, Stokes’ Theorem can be used:
Z
C2
~ · d~r =
F
Z
~ · dA
~ =
curl F
D2
~ is not a gradient field since
(e) The vector field F
R
C1
Z
D2
~ = 0.
~0 · dA
F~ · d~r 6= 0.
48. Note that planes of the form mx + ny = d are vertical, so their normals have no ~k component.
(a) Since
~ =
curl F
∂F1
∂F2
−
∂x
∂y
~k =
∂(−y)
∂(x)
−
∂x
∂y
~k = 2~k ,
by Stokes’ Theorem, the circulation around curves in the xy-plane and the plane x + y + z = 0 can be nonzero. The
circulation is 0 for (II), (III), (V).
~ = grad(xy), the circulation around any closed curve is 0. So the circulation is 0 around (I), (II), (III), (IV),
(b) Since G
(V).
(c) Since
~ = − ∂H2~i = − ∂(z)~i = −~i ,
curl H
∂z
∂z
the circulation can be nonzero around any curve in a plane whose normal has a nonzero ~i component. The circulation
is 0 for (I), (III).
~ contributes to the line integral, so
49. (a) Only the ~k -component of F
Z
C
~ · d~r =
F
10
Z
z~k · ~k dz =
0
Z
10
z dz =
0
z2
2
10
= 50.
0
~ contributes to the flux integral; this component is 10~k . Thus,
(b) Only the ~k component of F
Z
S
√
~ · dS
~ = 10 · Area of S = 10π( 3)2 = 30π.
F
~ = 1. Thus, if W is the region inside the box,
(c) We use the Divergence Theorem, with div F
Z
S
~ · dA
~ =
F
Z
W
1 dV = 1 · Volume of box = 23 = 8.
(d) Only the ~i and ~j components contribute to the line integral. The horizontal component of the vector field, y~i − x~j ,
is tangent to the circle, and points in the clockwise direction. On C, the length of the horizontal component is
||y~i − x~j || =
Thus,
Z
C
p
y 2 + x2 = 3.
~ · d~r = −3 · Circumference of circle = −3 · 2π3 = −18π.
F
CAS Challenge Problems
50. (a) R
curl F~ (1, 2, 1) = −6~i − 8~j + ~k .
~ · d~r = − 3 (4a2 + a4 )π.
(b) C F
2
a
lim
a→0
(c)
R
Ca
F~ · d~r
πa2
= −6.
~ (1, 2, 1) according to the circulation density formula.
~
This
of curlF
R is the i component
~ · d~r = −8a2 π.
F
Da
R
F~ · d~r
Ca
lim
= −8.
a→0
πa2
~ (1, 2, 1).
This is the ~j component of curlF
Job: chap20-sols Sheet: 41 Page: 1839 (August 24, 2012 11 : 19) [ex-project]
PROJECTS FOR CHAPTER TWENTY
(d)
R
Ea
~ · d~r = − 1 (−4a2 + 3a4 )π.
F
4
lim
a→0
R
Ca
F~ · d~r
πa2
1839
= 1.
~ (1, 2, 1).
This is the ~k component of curlF
(e) The curves Ca are circles of radius a around the point (1, 2, 1) facing in the positive x-direction, so the limit as a → 0
of the circulation around them is the ~i -component of the curl, according to the geometric definition. Similarly, the
curves Da are circles facing in the y-direction and the curves Ea are circles facing in the z-direction, so they give the
~j and ~k components of the curl.
~ = 2ax+bz +2cy +p+q so by the Divergence Theorem
51. (a) R
Let W be the region
enclosed by the sphere. We haveR div F
R
R
R
~ ·dA
~ =
dV
.
Now
x
dV
= W y dV = W z dV = 0, because W is symmetric
F
(2ax+bz+2cy+p+q)
W
S
W
R
about the origin and x, y, and z are odd functions. So W (2ax+bz +2cy +p+q) dV =
(b) Using spherical coordinates, we calculate the flux integral directly as
Z
2π
0
Z
π
R
B
(p+q)dV =
4(p+q)πR3
.
3
((bR2 cos(θ) cos(φ) sin(φ) + aR2 cos(θ)2 sin(φ)2 )~i
0
+(pR sin(θ) sin(φ) + cR2 sin(θ)2 sin(φ)2 )~j + (qR cos(φ)
4(p + q)πR3
+rR3 cos(θ)3 sin(φ)3 )~k ) · (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )R2 sin φ dφdθ =
.
3
Rather than entering this integral directly into your CAS, it is better to define the vector field and parameterization
separately and enter the formula for flux integral through a sphere.
~ (1, 1, 1) = 5.
52. (a) div F
(b) On a small sphere centered at (1, 1, 1) the divergence is approximately constant and equal to its value at (1, 1, 1),
namely 5. Geometrically
the divergence is the flux density, so the total flux is approximately the divergence multiplied
R
~ ≈ 5( volume of S) = 20 π(0.1)3 ≈ 0.02094...
by the volume, so S F~ · dA
3
~ = (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )a2 sin φ dφdθ. Using a CAS to
(c) The area element on the sphere Ris dA
~ = (20πa3 /3) + (4πa5 /5), which is .02097 when a = .1. This is close
calculate the flux integral we get S F~ · dA
to the approximation we found in part (b). The limit
lim
a→0
R
Sa
~ · dA
~
F
Volume inside Sa
= lim
a→0
(20πa3 /3) + (4πa5 /5)
3a2
= lim (5 +
) = 5.
3
(4/3)πa
a→0
5
This agrees with the value of the divergence we found in part (a). This makes sense, because the limit is just the
geometric definition of divergence.
PROJECTS FOR CHAPTER TWENTY
~ = ~0 at
1. (a) Since S does not touch the wire, it is in a region of space where there is no current. Hence curl B
every point of S. Therefore,
Z
~ · dA
~ = 0.
curl B
S
~ through S is 0.
The flux of curl B
(b) The boundary of S is a curve, C, in two pieces, C1 and C2 . Given the upward orientation of S, the curve
C1 is oriented clockwise and C2 is oriented counterclockwise when viewed from above. (See Figure 20.26
in the text.)
The vector field has constant magnitude on each circle and is parallel to the circle. Therefore,
Z
~ (P2 )k · Length of C2 = 2πR2 kB
~ (P2 )k.
~ · d~r = kB
B
C2
Because C1 is oriented in the opposite direction
Z
~ (P1 )k.
~ (P1 )k · Length of C1 = −2πR1 kB
~ · d~r = −kB
B
C1
Job: chap20-sols Sheet: 42 Page: 1840 (August 24, 2012 11 : 19) [ex-project]
1840
Chapter Twenty /SOLUTIONS
(c) From Stokes’ Theorem,
Z
~ · dA
~ =
curl B
S
Since
Z
~ · d~r .
B
C
Z
~ · dA
~ =0
curl B
S
and
Z
~ · d~r =
B
~ · d~r +
B
C1
C
we have
Z
Z
Hence,
and thus
~ · d~r
B
C2
~ · d~r = −
B
C2
Z
Z
~ · d~r .
B
C1
~ (P1 )k)
~ (P2 )k = −(−2πR1 kB
2πR2 kB
~ (P2 )k = (R1 kB
~ (P1 )k) 1 .
kB
R2
~ (P1 )k be constant and thinking of R2 = r as a variable, we have
Letting k = R1 kB
~ (P2 )k = k 1 .
kB
r
~ (P2 )k is proportional to the reciprocal of the distance r from P2 to the
This relationship shows that kB
wire.
(d) If the distance from P to the wire is r, then the distance from Q to the wire is 2r. Using the proportionality
in part (c), we have
~ (P )k = k
kB
r
and
~ (Q)k = k = 1 kB
~ (P )k.
kB
2r
2
~ in half.
Doubling the distance from the wire cuts the magnitude of B
(e) Suppose the distance from P to the wire is r, so that
~ (P )k =
kB
k
.
r
~ (Q)k = 0.8kB
~ (P )k, then
If the distance from Q to the wire is R and if kB
~ (Q)k =
kB
Solving gives
k
k
= 0.8 .
R
r
r
= 1.25r,
0.8
so the distance from the wire must be increased by 25%.
R=
Job: chap20-sols Sheet: 43 Page: 1841 (August 24, 2012 11 : 19) [chap20-sols]
PROJECTS FOR CHAPTER TWENTY
1841
~ . Since T~ is in the direction of F~ we have
2. First compute the unit vectors T~ and N
T~ =
1
1 ~
F = (u~i + v~j ).
~
F
kF k
~ is the unit vector in the direction of ~k × F~ we have
Since N
~k × F~ = ~k × (u~i + v~j )
= −v~i + u~j
1
~ =
(−v~i + u~j )
N
~
k − v i + u~j k
=
1
(−v~i + u~j ).
F
The chain rule for partial differentiation of the formulas u = F cos θ and v = F sin θ gives
uy = (cos θ)Fy − F (sin θ)θy
vx = (sin θ)Fx + F (cos θ)θx .
Since, for a 2-dimensional vector field, curlF~ = (vx − uy )~k we have
c = vx − uy
= ((sin θ)Fx + F (cos θ)θx ) − ((cos θ)Fy − F (sin θ)θy )
1
= (uθx + vθy ) + (vFx − uFy )
F
1
= (θx~i + θy~j ) · (u~i + v~j ) − (Fx~i + Fy~j ) · (−v~i + u~j )
F
~ .
= F grad θ · T~ − grad F · N
Since the directional derivative of θ in the direction of T~ is θT~ = grad θ · T~ and the directional derivative
~ we have
~ is F ~ = grad F · N
of F in the direction of N
N
c = F θT~ − FN~ .
21.1 SOLUTIONS
1843
CHAPTER TWENTY-ONE
Solutions for Section 21.1
Exercises
1. There is just one parameter, s, so the parameterization describes a curve.
2. There are two parameters, s and t, so the parameterization describes a surface.
3. There are two parameters, s and t, so the parameterization describes a surface.
4. There is just one parameter, s, so the parameterization describes a curve.
5. A horizontal disk of radius 5 in the plane z = 7.
6. A cylinder of radius 5 centered around the z-axis and stretching around from z = 0 to z = 7.
7. A helix (curve) of radius 5 which makes one turn about the z-axis, starting at the point (5, 0, 0) and ending at the point
(5, 0, 10π).
p
8. Since z = r = x2 + y 2 , we have a cone around the z-axis. Since 0 ≤ r ≤ 5, we have 0 ≤ z ≤ 5, so the cone has
height and maximum radius of 5.
9. The top half of the sphere (z ≥ 0).
10. The half of the sphere with y ≤ 0.
11. A vertical segment lying between two longitudinal lines (θ =
π
4
π
) and stretching between the poles.
3
π
= 4 and φ = π3 ) in the northern hemisphere.
and θ =
12. Half the horizontal ring around the sphere between two latitude lines (φ
Problems
13. Two vectors in the plane containing P = (0, 0, 0), Q = (1, 2, 3), and R = (2, 1, 0) are the displacement vectors
~v 1 = P~Q = ~i + 2~j + 3~k
~v 2 = P~R = 2~i + ~j .
Letting ~r 0 = 0~i + 0~j + 0~k = ~0 we have the parameterization
~r (s, t) = ~r 0 + s~v 1 + t~v 2
= (s + 2t)~i + (2s + t)~j + 3s~k .
14. Two vectors in the plane containing P = (1, 2, 3), Q = (2, 5, 8), and R = (5, 2, 0) are the displacement vectors
~v 1 = P~Q = ~i + 3~j + 5~k
~v 2 = P~R = 4~i − 3~k .
Letting ~r 0 = ~i + 2~j + 3~k we have the parameterization
~r (s, t) = ~r 0 + s~v 1 + t~v 2
= (1 + s + 4t)~i + (2 + 3s)~j + (3 + 5s − 3t)~k .
15. To parameterize the plane we need two nonparallel vectors ~v 1 and ~v 2 that are parallel to the plane. Such vectors are
perpendicular to the normal vector to the plane, ~n = ~i + ~j + ~k . We can choose any vectors ~v 1 and ~v 2 such that
~v 1 · ~n = ~v 2 · ~n = 0.
One choice is
~v 1 = ~i − j
~v 2 = ~i − ~k .
Letting ~r 0 = 3~i + 5~j + 7~k we have the parameterization
~r (s, t) = ~r 0 + s~v 1 + t~v 2
= (3 + s + t)~i + (5 − s)~j + (7 − t)~k .
1844
Chapter Twenty-One /SOLUTIONS
16. To parameterize the plane we need two nonparallel vectors ~v 1 and ~v 2 that are parallel to the plane. Such vectors are
perpendicular to the normal vector to the plane, ~n = ~i + 2~j + 3~k . We can choose any vectors ~v 1 and ~v 2 such that
~v 1 · ~n = ~v 2 · ~n = 0.
One choice is
~v 1 = 2~i − j
~v 2 = 3~i − ~k .
Letting ~r 0 = 5~i + ~j + 4~k we have the parameterization
~r (s, t) = ~r 0 + s~v 1 + t~v 2
= (5 + 2s + 3t)~i + (1 − s)~j + (4 − t)~k .
17. (a) We want to find s and t so that
2+s = 4
3+s+t = 8
4t = 12
Since s = 2 and t = 3 satisfy these equations, the point (4, 8, 12) lies on this plane.
(b) Are there values of s and t corresponding to the point (1, 2, 3)? If so, then
1 = 2+s
2 = 3+s+t
3 = 4t
From the first equation we must have s = −1 and from the third we must have t = 3/4. But these values of s and t
do not satisfy the second equation. Therefore, no value of s and t corresponds to the point (1, 2, 3), and so (1, 2, 3)
is not on the plane.
18. If the planes are parallel, then their normal vectors will also be parallel. The equation of the first plane can be written
~r = 2~i + 4~j + ~k + s(~i + ~j + 2~k ) + t(~i − ~j ).
A normal vector to the first plane is ~n 1 = (~i + ~j + 2~k ) × (~i − ~j ) = 2~i + 2~j − 2~k . The second plane can be written
~r = 2~i + s(~i + ~k ) + t(2~i + ~j − ~k ).
A normal vector to the second plane is ~n 2 = (~i + ~k ) × (2~i + ~j − ~k ) = −~i + 3~j + ~k . Since ~n 1 and ~n 2 are not
parallel, neither are the two planes.
19. The surface is the plane z = 1. The family of parameter curves with s constant and t varying consists of lines in the plane
parallel to the y-axis. The family with t constant and s varying consists of lines in the plane parallel to the x-axis.
20. The surface is the cylinder of radius 1 centered on the x-axis with equation y 2 + z 2 = 1. The family of parameter curves
with s constant and t varying consists of circles on the cylinder, the cross-sections of the cylinder parallel to the yz-plane.
The family with t constant and s varying consists of lines on the cylinder parallel to the x-axis.
21. The surface is the graph of the equation z = x2 + y 2 , a parabola. The family of parameter curves with s constant and t
varying consists of cross-sections of the graph with x fixed. If s = s0 , the cross-section has equation z = s20 + y 2 in the
plane x = s0 , so it is a parabola. The family of parameter curves with t constant and s varying consists of cross-sections
of the graph with y fixed. If t = t0 , the cross-section has equation z = x2 + t20 in the plane y = t0 , so it is a parabola.
22. The parametric equations are the spherical coordinate parameterization of the unit sphere centered at the origin, with
s = θ and t = φ. The family of parameter curves with s = θ constant and t = φ varying are meridians of constant
longitude, semicircles from the north pole to the south pole. The family of parameter curves with t = φ constant and
s = θ varying consists of latitude circles.
23. Since you walk 5 blocks east and 1 block west, you walk 5 blocks in the direction of ~v 1 , and 1 block in the opposite
direction. Thus,
s = 5 − 1 = 4,
Similarly,
t = 4 − 2 = 2.
21.1 SOLUTIONS
1845
Hence
x~i + y~j + z~k = (x0~i + y0~j + z0~k ) + 4v~1 + 2v~2
= (x0~i + y0~j + z0~k ) + 4(2~i − 3~j + 2~k ) + 2(~i + 4~j + 5~k )
= (x0 + 10)~i + (y0 − 4)~j + (z0 + 18)~k .
Thus the coordinates are:
x = x0 + 10,
y = y0 − 4,
z = z0 + 18.
24. (a) Nearer to the equator.
(b) Farther from the north pole.
(c) Farther from Greenwich.
25. A horizontal circle in the northern hemisphere at a latitude of 45◦ north of the equator.
26. A vertical half-circle, going from the north to south poles.
27. Set up the coordinates as in Figure 21.1. The surface is the revolution surface obtained by revolving the curve shown in
Figure 21.2 about the z axis. From the measurements given, we obtain the equation of the curve in Figure 21.2:
x = cos
π
z + 3,
3
0 ≤ z ≤ 48
(a) Rotating this around the z-axis, and taking z = t as the parameter, we get the parametric equations
π
t + 3) cos θ
3
π
y = (cos
t + 3) sin θ
3
z=t
0 ≤ θ ≤ 2π, 0 ≤ t ≤ 48
x = (cos
(b) We know that the points in the curve consists of cross-sections of circles parallel to the xy plane and of radius
cos((π/3)z + 3). Thus,
2
π
z+3
Area of cross-section = π cos
3
Integrating over z, we get
Volume = π
=π
Z
Z
48
0
48
0
cos
2
π
z+3
3
cos2
dz
π
π
z + 6 cos z + 9 dz
3
3
= 456π in3 .
z
z
✛
4′′
✲
✻
✛
′′
2✲
6′′
❄
y
x
Figure 21.1
3
Figure 21.2
x
1846
Chapter Twenty-One /SOLUTIONS
28. The sphere (x − a)2 + (y − b)2 + (z − c)2 = d2 has center at the point (a, b, c) and radius d. We use spherical coordinates
θ and φ as the two parameters. The parameterization of the sphere with center at the origin and radius d is
x = d sin φ cos θ,
y = d sin φ sin θ,
z = d cos φ.
Since the given sphere has center at the point (a, b, c) we add the displacement vector a~i + b~j + c~k to the radial vector
corresponding to a parameterization of the sphere with center at the origin and radius d to give
x = a + d sin φ cos θ,
0 ≤ φ ≤ π,
y = b + d sin φ sin θ,
0 ≤ θ ≤ 2π,
z = c + d cos φ.
To check that this is a parameterization for the given sphere, we substitute for x, y, z:
(x − a)2 + (y − b)2 + (z − c)2
= d2 sin2 φ cos2 θ + d2 sin2 φ sin2 θ + d2 cos2 φ
= d2 sin2 φ + d2 cos2 φ = d2 .
29. Let (θ, π/2) be the original coordinates. If θ < π, then the new coordinates will be (θ + π, π/4). If θ ≥ π, then the new
coordinates will be (θ − π, π/4).
30. If we set z = u, x2 + y 2 = u2 is the equation of a circle with radius |u|. Hence a parameterization of the cone is:
x = u cos v,
y = u sin v,
z = u.
0 ≤ v ≤ 2π,
31. Since the parameterization in Example 6 on page 1079 was r = (1 − hz )a and since the cone is given by z = r, we have
z = (1 − ar )h. The parameterization we want is
x = r cos θ,
0 ≤ r ≤ a,
y = r sin θ,
0 ≤ θ ≤ 2π,
r
h.
z = 1−
a
32. The plane in which the circle lies is parameterized by
~r (p, q) = x0~i + y0~j + z0~k + p~
u + q~v .
Because ~
u and ~v are perpendicular unit vectors, the parameters p and q establish a rectangular coordinate system on this
plane exactly analogous to the usual xy-coordinate system, with (p, q) = (0, 0) corresponding to the point (x0 , y0 , z0 ).
Thus the circle we want to describe, which is the circle of radius a centered at (p, q) = (0, 0), can be parameterized by
p = a cos t,
q = a sin t.
Substituting into the equation of the plane gives the desired parameterization of the circle in 3-space,
~r (t) = x0~i + y0~j + z0~k + a cos t~
u + a sin t~v ,
where 0 ≤ t ≤ 2π.
33. (a) Add second and third equations to get y + z = 1 + 2s. Thus, y + z = 1 + x or −x + y + z = 1, which is the
equation of a plane. Now, s = x/2, and t = (y − z + 1)/2, so the conditions 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 are equivalent
to 0 ≤ x ≤ 2, 0 ≤ y − z + 1 ≤ 2 or 0 ≤ x ≤ 2, −1 ≤ y − z ≤ 1.
21.1 SOLUTIONS
1847
(b) The surface is shown in Figure 21.3.
z
y
x
Figure 21.3: The surface x = 2s,
y = s + t, z = 1 + s − t, for
0 ≤ s ≤ 1, 0 ≤ t ≤ 1
34. (a) z 2 = 1 − s2 − t2 = 1 − x2 − y 2 . So x2 + y 2 + z 2 = 1 which is the equation of a sphere. The conditions s2 + t2 ≤ 1,
s, t ≥ 0 are equivalent to x2 + y 2 ≤ 1 and x, y ≥ 0. But if x2 + y 2 + z 2 = 1, then x2 + y 2 ≤ 1 is satisfied
automatically, so our surface is defined by:
x2 + y 2 + z 2 = 1,
(b) The surface x = s, y = t, z =
√
x, y, z ≥ 0.
1 − s2 − t2 for s2 + t2 ≤ 1, s, t ≥ 0 is shown in Figure 21.4.
z
x
y
Figure 21.4
35. (a) From the first two equations we get:
s=
x+y
,
2
t=
x−y
.
2
Hence the equation of our surface is:
z=
x+y
2
2
+
x−y
2
2
=
x2
y2
+
,
2
2
which is the equation of a paraboloid.
The conditions: 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 are equivalent to: 0 ≤ x + y ≤ 2, 0 ≤ x − y ≤ 2. So our surface is
defined by:
y2
x2
+
,
0≤ x+y ≤2 0 ≤x−y ≤2
z=
2
2
(b) The surface is shown in Figure 21.5.
1848
Chapter Twenty-One /SOLUTIONS
z
x
y
Figure 21.5: The surface x = s + t,
y = s − t, z = s2 + t2 for 0 ≤ s ≤ 1,
0≤t≤1
Strengthen Your Understanding
36. A counter example is provided by an unusual parameterization of the xy-plane:
~r (s, t) = s~i + (s + t) ~j .
The parameter curves with t constant are lines parallel to ~v 1 = ~i + ~j , and the parameter curves with s constant are lines
parallel to ~v 2 = ~j . Since ~v 1 and ~v 2 are not orthogonal, neither are the parameter curves.
37. A parameter curve for constant φ on the sphere is parameterized by
r~1 (θ) = R sin φ cos θ~i + R sin φ sin θ~j + R cos φ~k .
It is a circle with radius R sin φ.
38. The point on the unit sphere where θ = π/4 and φ = π/4 has position vector
√
2~
1
1
~r 0 = sin φ cos θ~i + sin φ sin θ~j + cos φ~k = ~i + ~j +
k.
2
2
2
~ ~
Vectors
√ perpendicular to the radius vector ~r 0 are parallel to the tangent plane. Two such vectors are ~v 1 = i − j
and ~v 2 = 2~i − ~k . A parameterization of the tangent plane can be given by
~r (s, t) = ~r 0 + s~v 1 + t~v 2 .
39. Since
x = s + 1, y = t + 2, z = s + t
we have x + y = z + 3. Therefore, an equation for the plane is given by
f (x, y, z) = x + y − z − 3 = 0.
40. To give a parameterized curve on the sphere, we assign a point on the sphere to every parameter value t, by giving values
of θ and φ. For example, letting θ = t and φ = t2 we have the curve
~r 1 (t) = sin t2 cos t~i + sin t2 sin t~j + cos t2~k
which is not a parameter curve because neither θ nor φ is constant.
41. True. The plane passes through the point (1, −2, 3) and contains the vectors ~i and ~j .
42. False. There is only one parameter, s. The equations parameterize a line.
21.2 SOLUTIONS
1849
43. True. The position vector of a point on the lower hemisphere is the negative of the position vector of the opposite point
on the upper hemisphere. As ~r ranges over all points in the upper hemisphere, −~r ranges over all points in the lower
hemisphere.
√
44. False. For example, if ~r (s, t) = s~i + t~j + 1 − s2 − t2~k with (s, t) inside the unit disk s2 + t2 ≤ 1 then ~r (−s, −t)
parameterizes the same upper hemisphere.
45. True. Adding a constant vector shifts the plane by a corresponding displacement, keeping it parallel to the original plane.
46. True. If the surface is parameterized by ~r (s, t) and the point has parameters (s0 , t0 ) then the parameter curves ~r (s0 , t)
and ~r (s, t0 ) pass through (s0 , t0 ).
47. False. For example, the lines of longitude on a sphere correspond to different values of the parameter θ, but all pass
through the north and south poles.
48. (I) Part of a plane, since z = x + y.
(II) Part of a cylinder of radius 1 with center on the z-axis, since x2 + y 2 = 1.
(III) Part of a sphere of radius 1 centered at the origin, since x2 + y 2 + z 2 = 1. With s = φ and t = θ this is the
usual parameterization of the sphere using spherical coordinates.
p
(IV) Part of a cone with vertex at the origin and central axis on the z-axis, since z = x2 + y 2 .
The match-up is: I-b, II-a, III-c, IV-d
Solutions for Section 21.2
Exercises
1. We have
∂(x, y)
xs xt
52
=
=
= −1.
∂(s, t)
ys yt
31
Therefore,
∂(x, y)
= 1.
∂(s, t)
2. We have
∂(x, y)
xs xt
2s −2t
=
=
= 4s2 + 4t2 .
∂(s, t)
ys yt
2t 2s
Therefore,
∂(x, y)
= 4s2 + 4t2 .
∂(s, t)
Notice we can drop the absolute value signs because in this case the Jacobian is nonnegative for all s and t.
3. We have
∂(x, y)
xs xt
es cos t −es sin t
=
=
= (es cos t)2 + (es sin t)2 .
∂(s, t)
es sin t es cos t
ys yt
Therefore,
∂(x, y)
= e2s (cos2 t + sin2 t) = e2s .
∂(s, t)
Notice we can drop the absolute value signs because the Jacobian in this case is positive for all s and t.
4. We have
∂(x, y)
xs xt
3s2 − 3t2 −6st
=
=
= 9(s2 − t2 )2 + 36s2 t2 .
∂(s, t)
6st
3s2 − 3t2
ys yt
Therefore, multiplying out and simplifying
∂(x, y)
= 9 s4 − 2s2 t2 + t4 + 4s2 t2 = 9(s2 + t2 )2 .
∂(s, t)
Notice we can drop the absolute value sign since the Jacobian in this case is nonnegative for all s and t.
1850
Chapter Twenty-One /SOLUTIONS
5. The square T is defined by the inequalities
0 ≤ s = ax ≤ 1
0 ≤ t = by ≤ 1
that correspond to the inequalities
0 ≤ x ≤ 1/a = 10
that define R. Thus a = 1/10 and b = 1.
0 ≤ y ≤ 1/b = 1
6. The square T is defined by the inequalities
0 ≤ s = ax ≤ 1
0 ≤ t = by ≤ 1
that correspond to the inequalities
0 ≤ x ≤ 1/a = 1
0 ≤ y ≤ 1/b = 1/4
that define R. Thus a = 1 and b = 4.
7. The square T is defined by the inequalities
0 ≤ s = ax ≤ 1
0 ≤ t = by ≤ 1
that correspond to the inequalities
0 ≤ x ≤ 1/a = 50
0 ≤ y ≤ 1/b = 10
that define R. Thus a = 1/50 and b = 1/10.
8. Inverting the change of coordinates gives x = s − at, y = t.
The four edges of R are
1
1
y = 0, y = 3, y = x, y = (x − 10).
4
4
The change of coordinates transforms the edges to
t = 0, t = 3, t =
1
1
1
10
1
s − at, t = s − at −
.
4
4
4
4
4
These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). This
happens when the t terms drop out, or a = −4. With a = −4 the change of coordinates gives
Z Z
T
∂(x, y)
ds dt
∂(s, t)
over the rectangle
T : 0 ≤ t ≤ 3, 0 ≤ s ≤ 10.
9. Inverting the change of coordinates gives x = s − at, y = t.
The four edges of R are
1
1
y = 0, y = 5, y = − x, y = − (x − 10).
3
3
The change of coordinates transforms the edges to
1
1
1
10
1
.
t = 0, t = 5, t = − s + at, t = − s + at +
3
3
3
3
3
These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). This
happens when the t terms drop out, or a = 3. With a = 3 the change of coordinates gives
Z Z
T
∂(x, y)
ds dt
∂(s, t)
over the rectangle
T : 0 ≤ t ≤ 5, 0 ≤ s ≤ 10.
21.2 SOLUTIONS
Problems
10. Given T = {(s, t) | 0 ≤ s ≤ 3, 0 ≤ t ≤ 2} and
(
x = 2s − 3t
y = s − 2t
The shaded area in Figure 21.6 is the corresponding region R in the xy-plane.
y
(6, 3)
1
x
2
y=
y=
(0, 0)
2
x
3
y=
−1
x
R2
R1
y=
2
x
3
(0, −1)
−1
1
x
2
(−6, −4)
Figure 21.6
Since
∂x
∂s
∂y
∂s
∂(x, y)
=
∂(s, t)
∂x
∂t
∂y
∂t
2 −3
=
= −1,
1 −2
∂(x, y)
= 1.
∂(s, t)
Thus we get
Z
T
Since
Z
dx dy =
Z
∂(x, y)
ds dt =
∂(s, t)
dx dy +
R1
R
=
Z
0
−6
Z
dx dy =
Z
R2
1
6
x+1
thus
Z
3
ds
0
Z
dx +
6
dt = 6.
dx
Z
1
2x
3
dy +
1 x−1
2
− x+1
6
0
dx dy =
2
Z
0
0
−6
R
11. Given
Z
Z
T
Z
6
0
dx = 3 + 3 = 6,
∂(x, y)
ds dt.
∂(s, t)
x = ρ sin φ cos θ
y = ρ sin φ sin θ
∂(x, y, z)
=
∂(ρ, φ, θ)
∂x
∂ρ
∂y
∂ρ
∂z
∂ρ
∂x
∂φ
∂y
∂φ
∂z
∂φ
∂x
∂θ
∂y
∂θ
∂z
∂θ
z = ρ cos φ,
sin φ cos θ ρ cos φ cos θ −ρ sin φ sin θ
= sin φ sin θ ρ cos φ sin θ ρ sin φ cos θ
cos φ
−ρ sin φ
dx
0
Z
1x
2
dy
2 x−1
3
1851
1852
Chapter Twenty-One /SOLUTIONS
ρ cos φ cos θ −ρ sin φ sin θ
= cos φ
ρ cos φ sin θ
ρ sin φ cos θ
2
2
2
sin φ cos θ −ρ sin φ sin θ
+ ρ sin φ
sin φ sin θ
ρ sin φ cos θ
2
= cos φ(ρ cos θ cos φ sin φ + ρ sin θ cos φ sin φ)
+ρ sin φ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)
= ρ2 cos2 φ sin φ + ρ2 sin3 φ
= ρ2 sin φ.
12. Given
(
we have
(
x = 3s − 4t
y = 5s + 2t,
s=
t=
Since
∂(x, y)
=
∂(s, t)
∂s
∂x
∂t
∂x
∂(s, t)
=
∂(x, y)
∂s
∂y
∂t
∂y
=
So
1
(x + 2y)
13
1
(3y − 5x).
26
∂x
∂s
∂y
∂s
1
2
13 13
5 3
− 26
26
∂x
∂t
∂y
∂t
=
=
3 −4
5 2
= 26,
3 1
26
13
+
5 2
26
13
=
1
.
26
∂(x, y) ∂(s, t)
1
·
= 26 ·
= 1.
∂(s, t) ∂(x, y)
26
13. Given
(
we have
∂(x, y)
=
∂(s, t)
x = 2s + t
y = s − t,
∂x
∂s
∂y
∂s
∂x
∂t
∂y
∂t
=
hence
2 1
1 −1
= −3,
∂(x, y)
= 3.
∂(s, t)
We get
Z
(x + y) dA =
Z
T
R
3s
∂(x, y)
dsdt =
∂(s, t)
where T is the region in the st-plane corresponding to R.
Z
(3s)(3) ds dt = 9
T
Z
s ds dt,
T
t
y
(0, 3)
(1, 3)
(2, 1)
T
x
(0, 0)
R
(5, −2)
(3, −3)
Figure 21.7
(0, 0)
(1, 0)
Figure 21.8
s
21.2 SOLUTIONS
1853
Now, we need to find T .
As
(
x = 2s + t
(
or
y = s−t
1
(x + y)
3
1
(x
− 2y),
3
s=
t=
so from the above transformation and Figure 21.7, T is the shaded area in Figure 21.8. Therefore
Z
(x + y) dA = 9
Z
1
s ds
0
R
3
Z
0
1
dt = (27)( ) = 13.5.
2
dx dy where R is the region x2 + 2xy + 2y 2 ≤ 1. We must change coordinates in both
14. The area of the ellipse is
R
the area element dA = dx dy and the region R.
Inverting the coordinate change gives x = s − t, y = t. Thus
RR
∂(x, y)
=
∂(s, t)
∂x
∂s
∂y
∂s
∂x
∂t
∂y
∂t
1 −1
=
0
= 1.
1
Therefore
∂(x, y)
ds dt = ds dt.
∂(s, t)
dx dy =
The region of integration is
x2 + 2xy + 2y 2 = (s − t)2 + 2(s − t)t + 2t2 = s2 + t2 ≤ 1.
Let T be the unit disc s2 + t2 ≤ 1. We have
Z Z
dx dy =
Z Z
ds dt = Area of T = π.
T
R
15. We must change coordinates in the area element √
dA = dx dy, the integrand x and the region R.
Inverting the coordinate change gives x = s − t, y = s where we use the positive square root because the region
R is in the first quadrant where x ≥ 0. Thus
∂x
∂s
∂y
∂s
∂(x, y)
=
∂(s, t)
∂x
∂t
∂y
∂t
√
1/(2 s − t)
=
√
−1/(2 s − t)
1
0
1
.
= √
2 s−t
Therefore
∂(x, y)
1
ds dt = √
ds dt.
∂(s, t)
2 s−t
dx dy =
√
The integrand is x = s − t.
The region of integration can be transformed by examination of its boundaries. The left and right boundaries of R
are given by y − x2 = t = 0 and y − x2 = t = −9. The bottom and top boundaries of R are given by y = s = 0 and
y = s = 16.
Let T be the rectangle 0 ≤ s ≤ 16, −9 ≤ t ≤ 0 of area (9)(16) = 144. We have
Z Z
x dx dy =
√
T
R
16. Let
Z Z
(
1
1
s−t √
ds dt = (Area of T ) = 72.
2
2 s−t
s = x−y
(
that is
t = x + y,
x=
y=
we get
1
∂(x, y)
2
=
∂(s, t)
− 12
Hence
I=
Z
R
cos
x−y
x+y
dx dy =
Z
T
cos
1
2
1
2
=
s ∂(x, y)
t
∂(s, t)
1
(s + t)
2
1
(t − s),
2
1
.
2
ds dt =
1
2
Z
T
cos
s
t
ds dt,
1854
Chapter Twenty-One /SOLUTIONS
where R is the triangle bounded by x + y = 1, x = 0, y = 0 and T is its image which is the triangle bounded by t = 1,
s = −t, s = t.
Then
1
2
I=
1
=
2
Z
Z
1
0
Z
t
cos
−t
1
s
t
ds dt =
t · 2 sin 1 dt = sin 1
0
Z
1
2
1
Z
0
1
t[sin(1) − sin(−1)] dt
t dt =
0
sin 1
= 0.42.
2
17. (a) The probability that (x + y)/2 is less than or equal to t is the integral of the joint density function p(x, y) over the
infinite region (half plane) where (x + y)/2 ≤ t. Thus
F (t) =
1
2πσ 2
Z
∞
−∞
Z
2t−x
e−(x
2
+y 2 )/(2σ 2 )
dy dx.
−∞
(b) With u = (x + y)/2, v = (x − y)/2, we have x = u + v, y = u − v. Thus
∂(x, y)
1 1
= −2
=
1 −1
∂(u, v)
so
dx dy =
∂(x, y)
du dv = 2du dv.
∂(u, v)
Also x2 + y 2 = 2(u2 + v 2 ). After writing the limits of integration in the uv-coordinates, we have
1
F (t) =
2πσ 2
Z
t
−∞
Continuing using the fact that
Z
∞
e
−∞
R∞
−∞
F (t) =
−(u2 +v 2 )/σ 2
e−x
2
2πσ 2
2
/a2
Z
2
2dv du =
2πσ 2
Z
t
e
−u2 /σ 2
−∞
Z
∞
e−v
2 /σ 2
dv du.
−∞
√
dx = a π with a replaced by σ, we obtain
t
e−u
2
/σ 2
−∞
√
1
(σ π)du = √
πσ
Z
t
e−u
2
/σ 2
du.
−∞
(c) The probability density function of z is the derivative of its cumulative distribution function F (t). By the Fundamental
Theorem of Calculus,
2
2
1
p(t) = F ′ (t) = √ e−t /σ .
πσ
(d) Since the density function for z can be written in the form
√ 2
2
1
√ e−t /(2(σ/ 2) ) ,
2π(σ/ 2)
√
the distribution of z is normal, with mean 0 and standard deviation σ/ 2.
Notice that the standard deviation of the average z = (x + y)/2 is less than the standard deviations of the
individual numbers x and y. The average of two random numbers is more likely to be near the mean than are either
of the two numbers individually.
p(t) = √
18. Let’s denote the (x, y) coordinates of the points in the lagoon by L. Since x and y are measured in kilometers and d is
measured in meters, and 1 km = 1000 m, the volume of a small piece of the lagoon is given by
∆V ≈ d(x, y)(1000∆x)(1000∆y)m3 .
Thus, the total volume of the lagoon is given by
V = 10002
Z
d(x, y) dxdy.
L
Changing coordinates using u = x/2 and v = y − f (x) converts the depth function to:
d(x(u, v), y(u, v)) = 40 − 160v 2 − 160u2 = 160(
1
− u2 − v 2 ) meters.
4
21.2 SOLUTIONS
1855
Thus, the points in the lagoon have (u, v) coordinates in the disk, D, given by u2 + v 2 ≤ 1/4.
The Jacobian of the transformation is:
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
2
=
0
= 2.
′
2f (2u) 1
Thus, the integral in u, v coordinates is
2
V = 1000
Z
6
d(x, y) dxdy = 10
L
Z
1
160( − u2 − v 2 )2 dudv = 320 · 106
4
D
Z
(
D
1
− u2 − v 2 ) dudv.
4
Converting to polar coordinates, we have
V = 320 · 106
Z
2π
0
Z
1/2
(
0
1
1 r2
r4
− r 2 )r drdθ = 320 · 106 2π(
− )
4
4 2
4
1/2
= 107 π m3 .
0
Strengthen Your Understanding
19. The region R does not correspond to the region T . The region R corresponds separately to both
T1 : 0 ≤ s ≤ 1, −2 ≤ t ≤ 0
T2 : 0 ≤ s ≤ 1, 0 ≤ t ≤ 2.
and to
In a change of coordinates for the integral over R, we use only one of these two regions. Both the following integrals are
correct:
Z
Z
∂(x, y)
f (x, y) dx dy =
ds dt
f (s, t2 )
∂(s, t)
T1
R
and
Z
f (x, y) dx dy =
Z
f (s, t2 )
T2
R
∂(x, y)
ds dt.
∂(s, t)
20. The Jacobian for the change of coordinates is given by
∂(x, y)
0 3t2
=
= −3t2 .
∂(s, t)
1 0
The change of coordinates formula requires the absolute value of the Jacobian. The correct formula is
Z
(x + 2y) dx dy =
Z
t3 + 2s
T
R
3t2 ds dt
21. The change of coordinates x = 2s, y = 3t transforms the region 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 in the st-plane into the rectangle
0 ≤ x ≤ 2, 0 ≤ y ≤ 3 in the xy-plane.
22. Let x = 2s, y = t. Geometrically, this change of coordinates stretches every region in the horizontal direction by a factor
of 2, while leaving the vertical distances the same, which doubles the area.
To see this analytically, let T be a region in the st-plane, corresponding to region R in the xy-plane. We have
∂(x, y)
=
∂(s, t)
Hence
Area of R =
Z
R
dx dy =
Z
T
∂x
∂s
∂y
∂s
∂x
∂t
∂y
∂t
=
20
= 2.
01
∂(x, y)
ds dt =
∂(s, t)
Z
T
2 ds dt = 2 · Area of T .
23. False. The change of variable leaves the value of the integral unchanged. Thus, calculating the value of the s, t-integral
gives you the value of the x, y-integral.
24. False. The Jacobian can be negative; we use the absolute value of the Jacobian in the integral.
1856
Chapter Twenty-One /SOLUTIONS
Solutions for Section 21.3
Exercises
1. Since
~r (s, t) = (s + t)~i + (s − t)~j + st ~k ,
we have
∂~r
= ~i + ~j + t~k
∂s
∂~r
= ~i − ~j + s~k ,
∂t
and
so
~i
∂~r
∂~r
~
dA =
×
ds dt = 1
∂s
∂t
1
~j ~k
1 t ds dt = ((s + t)~i − (s − t)~j − 2~k ) ds dt.
−1 s
~ has the opposite sign:
For the opposite orientation, dA
~ = −((s + t)~i − (s − t)~j − 2~k ) ds dt.
dA
2. Since
~r (s, t) = sin t~i + cos t ~j + (s + t) ~k ,
we have
∂~r
= 0~i + 0 ~j + ~k
∂s
and
so
~ =
dA
∂~r
∂~r
×
ds dt =
∂s
∂t
∂~r
= cos t~i − sin t ~j + ~k ,
∂t
~i
~j
~k
0
0
1 ds dt = (sin t~i + cos t ~j ) ds dt.
cos t − sin t 1
~ has the opposite sign:
For the opposite orientation, dA
~ = −(sin t~i + cos t ~j ) ds dt.
dA
3. Since
we have
~r (s, t) = es ~i + cos t ~j + sin t ~k ,
∂~r
= es ~i + 0 ~j + 0 ~k
∂s
and
so
~ =
dA
∂~r
= 0~i − sin t ~j + cos t ~k ,
∂t
∂~r
∂~r
×
ds dt
∂s
∂t
~k
~i
~j
= es
0
0
ds dt
0 − sin t cos t
= (−es cos t ~j − es sin t ~k ) ds dt
= −es (cos t ~j + sin t ~k ) ds dt.
~ has the opposite sign:
For the opposite orientation, dA
~ = es (cos t ~j + sin t ~k ) ds dt.
dA
21.3 SOLUTIONS
4. Since
1857
~r (u, v) = (u + v) ~j + (u − v) ~k ,
we have
∂~r
= ~j + ~k
∂u
so
∂~r
= ~j − ~k ,
∂v
and
~i ~j ~k
∂~r
∂~r
×
du dv = 0 1 1 du dv = −2~i du dv.
∂u
∂v
0 1 −1
~ =
dA
~ has the opposite sign:
For the opposite orientation, dA
~ = 2~i du dv.
dA
5. Since S is given by
~r (s, t) = (s + t)~i + (s − t)~j + (s2 + t2 )~k ,
we have
∂~r
= ~i + ~j + 2s~k
∂s
∂~r
= ~i − ~j + 2t~k ,
∂t
and
and
~i
∂~r
∂~r
×
= 1
∂s
∂t
1
~j ~k
1 2s = (2s + 2t)~i + (2s − 2t)~j − 2~k .
−1 2t
Since the ~k component of this vector is negative, it points down, and so has the opposite orientation to the one specified.
Thus, we use
~ = − ∂~r × ∂~r ds dt,
dA
∂s
∂t
and so we have
Z
S
~ · dA
~ =−
F
=2
1
Z
0
1
Z
Z
0
=2
(
0
6. Since S is parameterized by
(s2 + t2 )~k · (2s + 2t)~i + (2s − 2t)~j − 2~k
0
1
(s2 + t2 ) ds dt = 2
0
1
Z
1
Z
1
Z
0
1
t3
1
+ t2 ) dt = 2( t + )
3
3
3
s3
+ st2
3
1
= 2(
0
s=1
ds dt
dt
s=0
1
1
4
+ )= .
3
3
3
~r (s, t) = 2s~i + (s + t)~j + (1 + s − t)~k ,
we have
∂~r
= 2~i + ~j + ~k
∂s
so
~i ~j
∂~r
∂~r
×
= 21
∂s
∂t
01
∂~r
= ~j − ~k ,
∂t
and
~k
= −2~i + 2~j + 2~k ,
1
−1
which points upward, in the direction opposite to the orientation given. Thus,
Z
S
~ · dA
~ =−
F
=−
=
Z
Z
1
0
Z
0
0
1
Z
1
(2s~i + (s + t)~j )(−2~i + 2~j + 2~k ) ds dt
0
1
Z
1
(−4s + 2s + 2t) dsdt =
0
0
s=1
2
s − 2st
Z
s=0
!
dt =
Z
0
1
1
Z
0
1
(2s − 2t) ds dt
(1 − 2t) dt = t − t2
1
= 0.
0
1858
Chapter Twenty-One /SOLUTIONS
7. The cross-product ∂~r /∂s × ∂~r /∂t is given by
~i
∂~r
∂~r
×
=
∂s
∂t
e
~k
~j
s
0
= 18 sin(3t)~i − 3es sin(3t)~k .
6
0 −3 sin(3t) 0
s
Since the z-component, −3e sin(3t), of ∂~r /∂s × ∂~r /∂t is always negative for 0 ≤ s ≤ 4 and 0 < t < π/6, the vector
∂~r /∂s × ∂~r /∂t points downward and so in the direction of the given orientation of S.
Thus,
~ = ∂~r × ∂~r
ds dt
dA
∂s
∂t
and
Z
S
4
Z
~ =
F~ · dA
0
=
= −6
we have
π/6
Z
0
8. Since S is parameterized by
(es~i ) · (18 sin(3t)~i − 3es sin(3t)~j ) dt ds
0
4
Z
π/6
Z
Z
18es sin(3t) dt ds = −18
0
4
es (0 − 1) ds = 6
0
Z
0
4
Z
4
0
es cos(3t)
3
π/6
ds
0
es ds = 6(e4 − 1).
~r (s, t) = 3 sin s~i + 3 cos s~j + (t + 1)~k ,
∂~r
= 3 cos s~i − 3 sin s~j
∂s
∂~r
= ~k .
∂t
and
So
~k
~i
~j
∂~r
∂~r
×
= 3 cos s −3 sin s 0 = −3 sin s~i − 3 cos s~j ,
∂s
∂t
0
0
1
which points toward the z-axis and thus opposite to the orientation we were given. Hence, we use
~ =−
dA
∂~r
∂~r
×
ds dt,
∂s
∂t
and so we have
Z
S
~ =−
F~ · dA
=9
Z
Z
1
0
1
0
=9
Z
0
Z
Z
π
(3 cos s~i + 3 sin s~j ) · (−3 sin s~i − 3 cos s~j ) ds dt
0
π
2 sin s cos s ds dt = 9
0
0
1
Z
−
cos 2s
2
s=π
s=0
1
Z
π
sin 2s ds dt
0
dt = 0.
9. Using cylindrical coordinates, we see that the surface S is parameterized by
~r (r, θ) = r cos θ~i + r sin θ~j + r~k .
We have
∂~r
∂~r
×
=
∂r
∂θ
~i
cos θ
~j
~k
sin θ 1 = −r cos θ~i − r sin θ~j + r~k .
−r sin θ r cos θ 0
~ =
Since the vector ∂~r /∂r × ∂~r /∂θ points upward, in the direction opposite to the specified orientation, we use dA
− (∂~r /∂r × ∂~r /∂θ) dr dθ. Hence
21.3 SOLUTIONS
Z
S
~ · dA
~ =−
F
=−
2π
Z
0
Z
Z
0
=−
R7
7
=−
R7
7
R7
=−
7
= −(
R
(r 5 cos2 θ sin2 θ~k ) · (−r cos θ~i − r sin θ~j + r~k ) dr dθ
0
2π
Z
1859
Z
R
r 6 cos2 θ sin2 θ dr dθ
0
2π
sin2 θ cos2 θ dθ
0
2π
Z
sin2 θ(1 − sin2 θ) dθ
0
Z
2π
(sin2 θ − sin4 θ) dθ
0
R7 π
−π 7
)( ) =
R .
7 4
28
The cone is not differentiable at the point (0, 0). However the flux integral, which is improper, converges.
10. Place the cylinder S of radius a and length L so that its central axis is the z axis between z = 0 and z = L. A
parameterization of S is
x = a cos θ,
y = a sin θ,
z = t,
for 0 ≤ θ ≤ 2π,
0 ≤ t ≤ L.
We compute
∂~r
∂~r
×
= (−a sin θ~i + a cos θ~j ) × ~k = a cos θ~i + a sin θ~j
∂θ
∂t
∂~r
∂~r
×
=a
∂θ
∂t
Surface area =
Z
dA =
Z
R
S
∂~r
∂~r
×
∂θ
∂t
dA =
Z
2π
Z
θ=0
L
adtdθ = 2πaL.
t=0
11. A parameterization of S is
x = s,
y = t,
z = 3s + 2t,
for 0 ≤ s ≤ 10,
0 ≤ t ≤ 20.
We compute
∂~r
∂~r
×
= (~i + 3~k ) × (~j + 2~k ) = −3~i − 2~j + ~k
∂s
∂t
√
∂~r
∂~r
= 14
×
∂θ
∂t
Surface area =
Z
dA =
S
Z
R
∂~r
∂~r
×
∂θ
∂t
dA =
Z
20
t=0
Z
10
√
√
14dsdt = 200 14.
s=0
Problems
12. The surface S is parameterized by
~r (x, z) = x~i + (x2 + z 2 )~j + z~k .
The surface S, together with its given orientation ~n , is graphed in Figure 21.9. Using the right-hand rule we see that the
vector ~r x × ~r z points in the direction of ~n . Thus,
~ =
dA
∂~r
∂~r
×
∂x
∂z
~i ~j ~k
dx dz = 1 2x 0 dx dz = (2x~i − ~j + 2z~k ) dx dz.
0 2z 1
1860
Chapter Twenty-One /SOLUTIONS
Thus
Z
S
~ =
F~ · dA
=
Z
Z
S
((x + z)~i + ~j + z~k ) · (2x~i − ~j + 2z~k ) dx dz
S
(2x2 + 2xz − 1 + 2z 2 )dx dz.
Changing to polar coordinates, x = r cos θ, z = r sin θ, where 1/2 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, we obtain
Z
S
~ =
F~ · dA
Z
=
Z
2π
0
Z
2π
2π
r4
r2
r4
+
sin θ cos θ −
2
2
2
(2r 2 + 2r 2 sin θ cos θ − 1)r dr dθ
r=1
dθ
r=1/2
15
3
dθ
sin θ cos θ +
32
32
0
=
1
1/2
0
=
Z
15
3
(sin θ)2 +
θ
64
32
2π
=
0
3π
.
16
z
r~z
~
n
r~x
y
x
S
Figure 21.9
13. The plane is parameterized by
~r = x~i + y~j + z~k = x~i + y~j + (2 − 2x − y)~k ,
where (x, y) is in the disk R lying inside the circle x2 + y 2 = 2x. By completing the square, this circle can be rewritten
as (x − 1)2 + y 2 = 1 and so the disk has area π.
r
r
× ∂~
k dxdy, where
We have dA = k ∂~
∂x
∂y
~i
∂~r
∂~r
×
=
∂x
∂y
~k
~j
1
0
0
1
= 2~i + ~j + ~k
−2
−1
and so
∂~r
∂~r
×
∂x
∂y
=
√
6.
Thus, the surface area of the ellipse S is given by
Surface area =
Z
1 dA =
S
=
=
√
√
Z
√
6 dxdy
R
6 × (Area of disk x2 + y 2 = 2x)
6π.
21.3 SOLUTIONS
1861
z
S
y
x
Figure 21.10
14. The surface S is parameterized by
~r = x~i + f (x) cos θ~j + f (x) sin θ~k ,
a ≤ x ≤ b, 0 ≤ θ ≤ 2π.
The area element on A is
dA =
∂~r
∂~r
×
∂x
∂θ
dx dθ
= k(~i + f ′ (x) cos θ~j + f ′ (x) sin θ~k ) × (−f (x) sin θ~j + f (x) cos θ~k )k dx dθ
= kf (x)f ′ (x)~i − f (x) cos θ~j − f (x) sin θ~k k dx dθ
= f (x)
= f (x)
p
f ′ (x)2 + cos2 θ + sin2 θ dx dθ
p
1 + f ′ (x)2 dx dθ.
So
Surface area =
Z
S
dA =
Z
2π
0
Z
b
f (x)
a
p
1+
f ′ (x)2
dx dθ = 2π
b
Z
f (x)
a
p
1 + f ′ (x)2 dx.
15. Let x be the distance d1 . Since w is the total width of the channel, we have d2 = w − x. The flux through a rectangle
with dimensions A = w × h, is given by
Z
Flux =
A
~.
~v · dA
~ = (h dx)~j . Thus
For a thin section of the channel of width dx, we have dA
Flux =
=
Z
w
Z0 w
0
~v · (h dx)~j =
Z
w
0
kx(w − x)~j · (h · dx)~j =
kh(wx − x2 ) dx = kh
1 3 1 3
w − w
2
3
=
Z
w
0
khx(w − x) dx
1
khw3 meter3 /sec.
6
16. (a) Building on the parameterization x = cos u, y = sin u, z = 0 of the circular base of the cone, we get
x = (1 − v) cos u + av
y = (1 − v) sin u + bv
z = cv
0 ≤ u ≤ 2π,
0 ≤ v ≤ 1.
Note that v = 0 corresponds to the base of the cone and v = 1 is its vertex.
(b) Writing ~r = x~i + y~j + z~k we have
∂~r
= −(1 − v) sin u~i + (1 − v) cos u~j
∂u
∂~r
= (a − cos u)~i + (b − sin u)~j + c~k .
∂v
1862
Chapter Twenty-One /SOLUTIONS
Thus
~k
~i
~j
∂~r
∂~r
×
= −(1 − v) sin u (1 − v) cos u 0
∂u
∂v
a − cos u
b − sin u
c
= c(1 − v) cos u~i + c(1 − v) sin u~j + (1 − v)(1 − a cos u − b sin u)~k
so
∂~r
∂~r
×
∂u
∂v
Thus
Surface area =
Z
2π
0
Z
1
= (1 − v)
∂~r
∂~r
×
∂u
∂v
0
p
c2 + (1 − a cos u − b sin u)2 .
du dv =
1
2
Z
0
2π
p
c2 + (1 − a cos u − b sin u)2 du.
This is an elliptic integral that can
notpbe evaluated in terms of elementary functions.
R 2π
(c) We have Surface Area = (1/2) 0
1 + (1 − 2 cos u)2 du = 5.805.
17. If S is the part of the graph of z = f (x, y) lying over a region R in the xy-plane, then S is parameterized by
~r (x, y) = x~i + y~j + f (x, y)~k ,
So
(x, y) in R.
∂~r
∂~r
×
= (~i + fx~k ) × (~j + fy ~k ) = −fx~i − fy~j + ~k .
∂x
∂y
Since the ~k component is positive, this points upward, so if S is oriented upward
~ = (−fx~i − fy~j + ~k ) dx dy
dA
and therefore we have the expression for the flux integral obtained on page 1018:
Z
S
~ · dA
~ =
F
Z
R
F~ (x, y, f (x, y)) · (−fx~i − fy ~k + ~k ) dx dy.
18. If S is the part of the cylinder of radius R corresponding to the region T in θz-space, then S is parameterized in cylindrical
coordinates by
~r (θ, z) = R cos θ~i + R sin θ~j + z~k ,
(θ, z) in T .
So
∂~r
∂~r
×
= (−R sin θ~i + R cos θ~j ) × ~k = R cos θ~i + R sin θ~j .
∂θ
∂z
This points outward, so
~ = (R cos θ~i + R sin θ~j ) dθ dz = (cos θ~i + sin θ~j )R dθ dz
dA
and therefore we obtain the expression for the flux integral in cylindrical coordinates on page 1019:
Z
S
~ · dA
~ =
F
Z
T
~ (R cos θ, R sin θ, z) · (cos θ~i + sin θ~j )R dθ dz.
F
19. If S is the part of the sphere of radius R corresponding to the region T in θφ-space, then S is parameterized in spherical
coordinates by
~r (θ, φ) = R sin φ cos θ~i + R sin φ sin θ~j + R cos φ~k ,
(θ, φ) in T .
So
~k
~i
~j
∂~r
∂~r
×
= −R sin φ sin θ R sin φ cos θ
0
∂θ
∂φ
R cos φ cos θ R cos φ sin θ −R sin φ
= −R2 sin2 φ cos θ~i − R2 sin2 φ sin θ~j − R2 sin φ cos φ~k
= −R2 sin φ(sin φ cos θ~i + sin φ sin θ~j + cos φ~k ).
21.3 SOLUTIONS
1863
This points inward, so the outward area element is
~ = (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )R2 sin φ dθ dφ,
dA
and therefore we obtain the expression for the flux integral in spherical coordinates on page 1020:
Z
S
=
Z
T
~
F~ · dA
F~ (R sin φ cos θ, R sin φ sin θ, R cos φ) · (sin φ cos θ~i + sin φ sin θ~j + cos φ~k )R2 sin φ dθ dφ.
20. In terms of the st-parameterization,
~ =
dA
∂~r
∂~r
×
ds dt.
∂s
∂t
By the chain rule, we have
∂~r ∂u
∂~r
∂~r
=
+
∂s
∂u ∂s
∂v
∂~r ∂u
∂~r
∂~r
=
+
∂t
∂u ∂t
∂v
∂v
∂s
∂v
.
∂t
So taking the cross product, we get
∂~r
∂~r
×
=
∂s
∂t
=
∂~r ∂v
∂~r ∂u
+
∂u ∂s
∂v ∂s
∂u ∂v
−
∂s ∂t
∂u ∂v
∂t ∂s
×
∂~r
∂u
×
∂~r ∂v
∂~r ∂u
+
∂u ∂t
∂v ∂t
∂~r
.
∂v
Now suppose we are going to change coordinates in a double integral from uv-coordinates to st-coordinates. The
Jacobian is
∂u ∂v
∂u ∂v
∂(u, v)
∂u ∂v
= ∂s ∂s =
−
.
∂u ∂v
∂(s, t)
∂s ∂t
∂t ∂s
∂t
∂t
Since the Jacobian is assumed to be positive, converting from a uv-integral to an st-integral gives:
Z
T
~ · ∂~r × ∂~r dudv =
F
∂u
∂v
=
Z
~ · ∂~r × ∂~r ∂(u, v) dsdt
F
∂u
∂v ∂(s, t)
R
Z
~ · ∂~r × ∂~r
F
∂u
∂v
R
However, we know that this gives us
Z
T
~ · ∂~r × ∂~r du dv =
F
∂u
∂v
Z
~ · ∂~r × ∂~r
F
∂u
∂v
R
∂u ∂v
∂s ∂t
∂u ∂v
∂u ∂v
−
∂s ∂t
∂t ∂s
−
∂u ∂v
∂t ∂s
dsdt =
Z
dsdt.
~ · ∂~r × ∂~r dsdt.
F
∂s
∂t
R
Thus, the flux integral in uv-coordinates equals the flux integral in st-coordinates.
Strengthen Your Understanding
R1R1
21. The integral 0 0 f (s, t) ds dt gives the volume of the region between the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 in the xy-plane
and the surface z = f (x, y).
R3R2
22. The area of the region R given by 0 ≤ s ≤ 2, 0 ≤ t ≤ 3 in st-space is 0 0 ds dt = 6, but the area of the parameterized
surface S in xyz-space depends on its parameterization.
The surface S is parameterized by ~r = f (s, t)~i + g(s, t)~j + h(s, t)~k . The surface area of S is given by
Area =
Z
S
dA =
Z
0
3
Z
0
2
∂~r
∂~r
×
∂s
∂t
ds dt.
1864
Chapter Twenty-One /SOLUTIONS
R1R1
23. The region 0 ≤ s ≤ 1, 0 ≤ t ≤ 1 in the parameter space must correspond to surface area 0 0 2 ds dt = 2 on the
xy-plane being parameterized. This suggests a parameterization that is a scaling of the usual xy coordinates.
Let x = 2s, y = t, z = 0 so that the parameterization is
~r (s, t) = 2s~i + t~j .
We have
∂~r
∂~r
×
= 2~i × ~j = 2~k
∂s
∂t
∂~r
∂~r
×
ds dt = 2ds dt.
dA =
∂s
∂t
24. Since ~r = (s − t)~i + t2~j + (s + t)~k we have
∂~r
∂~r
×
= −2t~i − 2~j + 2t~k .
∂s
∂t
~ (x, y, z) = −~j , we have
Therefore, if F
so
Z
S
~ =
F~ · dA
∂~r
∂~r
×
∂s
∂t
=2>0
F~ (~r (s, t)) ·
∂~r
∂~r
×
∂s
∂t
F~ ·
1
Z
0
Z
0
1
ds dt = 2 > 0.
~ is perpendicular to the area vector dA
~.
25. True. At every point on S the vector field F
2
2
26. True. The vector 2~i − 4~j + 6~k is the gradient vector for f (x, y, z) = x − y + z 2 at (1, 2, 3) so it is perpendicular to
the surface. Thus it is parallel to the area vector.
~ and 3~i + 4~j + 5~k are perpendicular to the plane at every point, so they are multiples of each
27. False. It is true that both dA
other. However, the ratio between them might not be a constant. For example, x = s3 , y = t3 , z = (1/5)(7 − 3s3 − 4t3 )
is a parameterization of the plane, but
~ = (3s2~i − (9/5)s2~k ) × (3t2~j − (12/5)t2~k )dsdt
dA
= ((27/5)s2 t2~i + (36/5)s2 t2~j + 9s2 t2~k )dsdt
= (9/5)s2 t2 (3~i + 4~j + 5~k )dsdt.
p
28. Since the half sphere is x = − 1 − y 2 − z 2 , we parameterize in the form x = x(y, z). Thus, the answer is either (e)
p
p
or (f). We have ∂~r /∂y = (y/ 1 − y 2 − z 2 )~i + ~j and ∂~r /∂z = (z/ 1 − y 2 − z 2 )~i + ~k , and ∂~r /∂y × ∂~r /∂z =
p
p
p
~i − (y/ 1 − y 2 − z 2 )~j − (z/ 1 − y 2 − z 2 )~k = −~r / 1 − y 2 − z 2 . Our surface is oriented away from the origin,
p
so we want (∂~r /∂z) × (∂~r /∂y) = ~r / 1 − y 2 − z 2 , so the answer is (f).
Solutions for Chapter 21 Review
Exercises
1. Since x2 + y 2 = 4z 2 , we have z =
around the z-axis.
1
2
p
x2 + y 2 . Thus we have a cone of height 7 and maximum radius 14, centered
2. This is a parabolic cylinder y = x2 , between x = −5 and x = 5, with its axis along the z-axis, stretching from z = 0 to
z = 7.
3. The disc S is defined by the inequality
s 2 + t 2 = a 2 x 2 + b2 y 2 ≤ 1
that corresponds to the inequality x + y ≤ 152 or equivalently
2
2
1
1 2
x + 2 y2 ≤ 1
152
15
that defines R. Thus a2 = 1/152 and b2 = 1/152 . We have a = 1/15, b = 1/15.
SOLUTIONS to Review Problems for Chapter Twenty-One
1865
4. The disc S is defined by the inequality
s 2 + t 2 = a 2 x 2 + b2 y 2 ≤ 1
that corresponds to the inequality x /4 + y /9 ≤ 1 that defines R. Thus a2 = 1/4 and b2 = 1/9. We have a = 1/2,
b = 1/3.
2
2
5. Inverting the change of coordinates gives x = s − at, y = t.
The four edges of R are
y = 15, y = 35, y = 2(x − 10) + 15, y = 2(x − 30) + 15.
The change of coordinates transforms the edges to
t = 15, t = 35, t = 2s − 2at − 5, t = 2s − 2at − 45.
These are equations for the edges of a rectangle in the st-plane if the last two are of the form: s = (Constant). This
happens when the t terms drop out, or a = −1/2. With a = −1/2 the change of coordinates gives
Z Z
over the rectangle
∂(x, y)
ds dt
∂(s, t)
T
T : 15 ≤ t ≤ 35,
45
5
≤s≤
.
2
2
6. The cross product ∂~r /∂s × ∂~r /∂t is given by
~i
∂~r
∂~r
×
=
∂s
∂t
~k
~j
2s
2
0
0
2t
5
= 10~i − 10s~j + 4st~k .
Since the z-component, 4st, of the vector ∂~r /∂s × ∂~r /∂t is positive for 0 < s ≤ 1, 1 ≤ t ≤ 3, we see that ∂~r /∂s ×
∂~r /∂t points upward, in the direction of the orientation of S we were given. Thus, we use
~ =
dA
and so we have
Z
S
~ · dA
~ =
F
=
1
Z
0
0
=
Z
Z
0
ds dt,
(5t~i + s2~j ) · (10~i − 10s~j + 4st~k ) dt ds
3
3
1
1
∂~r
∂~r
×
∂s
∂t
3
Z
1
1
Z
(50t − 10s ) dt ds =
Z
0
t=3
1
(25t2 − 10s3 t)
ds
t=1
1
(200 − 20s3 ) ds = (200s − 5s4 )
0
= 200 − 5 = 195.
7. Since
~r (a, θ) = a cos θ~i + a sin θ~j + sin a2~k ,
we have
∂~r
∂~r
×
=
∂a
∂θ
~i
~k
~j
2
2
2
2
sin θ 2a cos a2 = −2a cos θ cos a ~i − 2a sin θ cos a ~j + a ~k .
cos θ
−a sin θ a cos θ
0
The z-component, a, of the vector ∂~r /∂a × ∂~r /∂θ is positive for 1 ≤ a ≤ 3, 0 ≤ θ ≤ π, so ∂~r /∂a × ∂~r /∂θ points
~ = (∂~r /∂a × ∂~r /∂θ) da dθ, giving
upward, in the direction of the orientation of S we were given. Thus, we use dA
Z
S
~ · dA
~ =
F
=
Z
Z
3
1
π
Z
0
(−
0
3
1
Z
π
2
2
)~i + (
) ~j
a cos θ
a sin θ
∂~r
·
∂a
(4a cos a2 − 4a cos a2 ) dθ da = 0.
×
∂~r
dθ da
∂θ
1866
Chapter Twenty-One /SOLUTIONS
8. Two vectors in the plane containing P = (5, 5, 5), Q = (10, −10, 10), and R = (0, 20, 40) are the displacement vectors
~v 1 = P~Q = 5~i − 15~j + 5~k
~v 2 = P~R = −5~i + 15~j + 35~k .
Letting ~r 0 = 5~i + 5~j + 5~k we have the parameterization
~r (s, t) = ~r 0 + s~v 1 + t~v 2
= (5 + 5s − 5t)~i + (5 − 15s + 15t)~j + (5 + 5s + 35t)~k .
9. We use spherical coordinates φ and θ as the two parameters. The parameterization of the sphere center at the origin and
radius 5 is:
x = 5 sin φ cos θ, y = 5 sin φ sin θ, z = 5 cos φ.
We have to shift the center of the sphere from the origin to the point (2, −1, 3). This gives
x = 2 + 5 sin φ cos θ,
y = −1 + 5 sin φ sin θ,
z = 3 + 5 cos φ.
Problems
10. (a) The cone of height h, maximum radius a, vertex at the origin and opening upward is shown in Figure 21.11.
z
h
a
y
x
Figure 21.11
By similar triangles, we have
r
a
= ,
z
h
so
z=
hr
.
a
Therefore, one parameterization is
x = r cos θ,
y = r sin θ,
hr
.
z=
a
0 ≤ r ≤ a,
0 ≤ θ < 2π,
(b) Since r = az/h, we can write the parameterization in part (a) as
az
cos θ,
h
az
sin θ,
y=
h
z = z.
x=
0 ≤ z ≤ h,
0 ≤ θ < 2π,
SOLUTIONS to Review Problems for Chapter Twenty-One
1867
11. (a) The surface is the cylinder x2 + y 2 = 1 of radius 1 centered on the z-axis.
(b) The parameter curves with constant s and varying t are helices that wind clockwise around the cylinder as they
advance up the cylinder with increasing t. See Figure 21.12.
The parameter curves with constant t and varying s are helices that wind counterclockwise up the cylinder. See
Figure 21.13.
z
x
z
y
y
x
Figure 21.12: Constant s, varying t
Figure 21.13: Constant t, varying s
12. The plane through (1, 3, 4) and orthogonal to ~n = 2~i + ~j − ~k is given by 2(x − 1) + (y − 3) − (z − 4) = 0, that is,
2x + y − z − 1 = 0.
Thus, thinking of the plane as z = 2x + y − 1, one possible parameterization is
x = u,
y = v,
z = 2u + v − 1.
13. The parameterization for a sphere of radius a using spherical coordinates is
x = a sin φ cos θ,
y = a sin φ sin θ,
z = a cos φ.
Thinkof the ellipsoid as a sphere whose radius is different along each axis and you get the parameterization:
x = a sin φ cos θ,
y = b sin φ sin θ,
0 ≤ φ ≤ π,
0 ≤ θ ≤ 2π,
z = c cos φ.
To check this parameterization, substitute into the equation for the ellipsoid:
y2
z2
a2 sin2 φ cos2 θ
b2 sin2 φ sin2 θ
c2 cos2 φ
x2
+ 2 + 2 =
+
+
2
2
2
a
b
c
a
b
c2
2
2
2
2
= sin φ(cos θ + sin θ) + cos φ = 1.
√
14. The vase obtained by rotating the curve z = 10 x − 1, 1 ≤ x ≤ 2, around the z-axis is shown in Figure 21.14.
√
z = 10 x − 1
10 z
y
−2
2
Figure 21.14
x
1868
Chapter Twenty-One /SOLUTIONS
At height z, the cross-section is a horizontal circle of radius a. Thus, a point on this horizontal circle is given by
~r = a cos θ~i + a sin θ~j + z~k .
However, the radius a varies, so we need to express it in terms of√
the other parameters θ and z. If you look at the xz-plane,
the radius of this circle is given by x, so solving for x in z = 10 x − 1 gives
a=x=
Thus, a parameterization is
~r =
so
x=
z
10
where 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 10.
z
10
2
2
z
10
+ 1 cos θ~i +
y=
+ 1 cos θ,
2
+ 1.
z
10
2
z
10
+ 1 sin θ~j + z~k
2
+ 1 sin θ,
z = z,
15. (a) As x2 + y 2 = 9 and s ∈ [0, π] is equivalent to x ≥ 0, and t ∈ [0, 1] is equivalent to z ∈ [1, 2]. So, x2 + y 2 = 9 is
the equation of a cylinder, and our surface is defined by:
x2 + y 2 = 9,
x ≥ 0,
1 ≤ z ≤ 2.
(b) The surface x = 3 sin s, y = 3 cos s, z = t + 1 for 0 ≤ s ≤ π, 0 ≤ t ≤ 1 is shown in Figure 21.15.
z
x
y
Figure 21.15
y
16.
2
R
x
4
Figure 21.16
Given T = {(s, t) | 0 ≤ s ≤ 2, s ≤ t ≤ 2} and
(
x = s2
y = t,
SOLUTIONS to Review Problems for Chapter Twenty-One
√
R = {(x, y) | 0 ≤ x ≤ 4,
∂x
∂t
∂y
∂t
∂x
∂s
∂y
∂s
∂(x, y)
=
∂(s, t)
∂(x, y)
= 2s
∂(s, t)
Z
So,
T
∂(x, y)
ds dt = 2
∂(s, t)
Z
2
Z
s ds
0
dt = 2
s
dx dy =
R
Z
dx
0
Z
2
Z
s(2 − s) ds = 2 s2 −
2
Z
√
dy =
x
2 3/2
x
3
h
= 2x −
Thus
0 ≤ s ≤ 2.
0
4
dx dy =
R
Z
= 2s,
0 1
since
2
Z
x ≤ y ≤ 2}.
2s 0
=
i4
0
1869
Z
4
(2 −
0
=8−
√
s3
3
2
0
=
8
.
3
x) dx
16
8
= .
3
3
∂(x, y)
ds dt.
∂(s, t)
T
17. We must change coordinates in the area element dA = dx dy, the integrand sin(x + y) and the region R.
Inverting the coordinate change gives x = (s + t)/2, y = (t − s)/2. Thus
∂x
∂s
∂y
∂s
∂(x, y)
=
∂(s, t)
∂x
∂t
∂y
∂t
=
1/2 1/2
−1/2 1/2
=
1
.
2
Therefore
dx dy =
∂(x, y)
1
ds dt = ds dt.
∂(s, t)
2
The integrand is sin(x + y) = sin t.
The region of integration is
x2 + y 2 =
2
2
Let T be the disc s + t ≤ 2 of radius
Z Z
s + t 2
2
√
+
2. We have
t − s 2
sin(x + y) dx dy =
R
=
2
Z Z
s2 + t2
≤ 1.
2
1
(sin t) ds dt = 0.
2
T
The final integral is zero by symmetry, the integral over the part of the disc where t < 0 canceling the integral over the
part where t > 0.
18. Given
(
s = xy
t = xy 2 ,
we have
∂(s, t)
=
∂(x, y)
Since
∂s
∂x
∂t
∂x
=
y
x
y 2 2xy
= xy 2 = t.
∂(s, t) ∂(x, y)
·
= 1,
∂(x, y) ∂(s, t)
∂(x, y)
=t
∂(s, t)
So
∂s
∂y
∂t
∂y
Z
R
xy 2 dA =
Z
T
t
so
∂(x, y)
ds dt =
∂(s, t)
∂(x, y)
1
=
∂(s, t)
t
Z
T
1
t( ) ds dt =
t
Z
T
ds dt,
1870
Chapter Twenty-One /SOLUTIONS
where T is the region bounded by s = 1, s = 4, t = 1, t = 4.
Then
Z
Z
4
2
xy dA =
1
R
ds
Z
4
dt = 9.
1
19. The elliptic cylindrical surface is parameterized by
~r = x~i + y~j + z~k = a cos θ~i + b sin θ~j + z~k
where 0 ≤ θ ≤ 2π, −c ≤ z ≤ c.
We have
~i
~j ~k
∂~r
∂~r
×
= −a sin θ b cos θ 0 = b cos θ~i + a sin θ~j .
∂θ
∂z
0
0 1
~ = (b cos θ~i + a sin θ~j ) dθdz, giving
This vector points away from the z-axis, so we use dA
Z
S
~ · dA
~ =
F
=
c
Z
−c
c
Z
−c
Z
Z
2π
0
a
b
( (a cos θ)~i + (b sin θ~j )) · (b cos θ~i + a sin θ~j ) dθ dz
a
b
2π
(b2 cos2 θ + a2 sin2 θ) dθdz
0
2
= 2πc(a + b2 ).
20. The surface of S is parameterized by
where
~r (θ, φ) = x~i + y~j + z~k ,
x = a + d sin φ cos θ,
y = b + d sin φ sin θ,
for
0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.
z = c + d cos φ,
The vector ∂~r /∂φ × ∂~r /∂θ points outward by the right-hand rule, so
~ =
dA
∂~r
∂~r
×
∂φ
∂θ
dφdθ.
Thus,
~ · dA
~ =F
~ ·
F
∂~r
∂~r
×
∂φ
∂θ
dφdθ
x2 y 2 z 2
∂y
∂φ
∂y
∂θ
∂x
∂φ
∂x
∂θ
=
∂z
∂φ
∂z
∂θ
dφdθ
(a + d sin φ cos θ)2 (b + d sin φ sin θ)2 (c + d cos φ)2
=
d cos φ cos θ
d cos φ sin θ
−d sin φ sin θ
d sin φ cos θ
−d sin φ
dφdθ.
0
Hence,
Z
S
~ · dA
~ = d2
F
Z
0
2π
Z
π
( a2 sin2 φ cos θ + 2ad sin3 φ cos2 θ + d2 sin4 φ cos3 θ
0
+ b2 sin2 φ sin θ + 2bd sin3 φ sin2 θ + d2 sin4 φ sin3 θ
+ c2 sin φ cos φ + 2cd sin φ cos2 φ + d2 sin φ cos3 φ) dφdθ.
PROJECTS FOR CHAPTER TWENTY-ONE
Since
Z
2π
cos θ dθ =
0
and
Z
2π
sin θ dθ =
0
sin φ cos φ dφ =
Z
S
~
F~ · A
d2
Z
2π
= 2πd
3
2π
sin3 θ dθ = 0,
0
π
sin φ cos3 φ dφ = 0,
π
Z
0
Z
Z
0
0
we have
cos3 θ dθ =
0
π
Z
2π
Z
1871
(2ad sin3 φ cos2 θ + 2bd sin3 φ sin2 θ + 2cd sin φ cos2 φ) dφdθ
0
π
Z
(a sin3 φ + b sin3 φ + 2c sin φ cos2 φ) dφ
0
= 4πd3
Z
π/2
(a sin3 φ + b sin3 φ + 2c sin φ cos2 φ) dφ
0
=
8 3
πd (a + b + c).
3
PROJECTS FOR CHAPTER TWENTY-ONE
1. The sphere x2 + y 2 + z 2 = 1 is shown in Figure 21.17.
(x, y, z)
(0, 0, 1)
(p, q)
z
❘
❘
y
x
(0, 0, −1)
Figure 21.17
(a)
(b)
(c)
(d)
(e)
The origin corresponds to the south pole.
The circle x2 + y 2 = 1 corresponds to the equator.
We get all the points of the sphere by this parameterization except the north pole itself.
x2 + y 2 > 1 corresponds to the upper hemisphere.
x2 + y 2 < 1 corresponds to the lower hemisphere.
2. We use the angles θ and φ shown in the figure in the problem. The angle θ is in the xy-plane measured
counterclockwise from the positive x-axis; the angle φ is measured perpendicular to the xy-plane.
(a) The circle of radius b in the xy-plane is parameterized by
~r = b cos θ~i + b sin θ~j
0 ≤ θ ≤ 2π.
(b) One vector is always ~k . The other vector is in the same direction as ~r in part (a) but has length 1. Therefore,
we take the other vector to be m
~ = cos θ~i + sin θ~j . Thus, relative to its center, the small circle of radius
a can be parameterized by
~s = a cos φm
~ + a sin φ~k = a cos φ(cos θ~i + sin θ~j ) + a sin φ~k
0 ≤ φ ≤ 2π.
(c) The parameterization of the torus with parameters 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ 2π, is given by
x~i + y~j + z~k = ~r + ~s
= (b cos θ + a cos φ cos θ)~i + (b sin θ + a cos φ sin θ)~j + a sin φ~k .
APPENDIX A SOLUTIONS
1873
APPENDIX
Solutions for Section A
1. The graph is
40
20
-3
-2
-1
1
2
3
-20
-40
(a) The range appears to be y ≤ 30.
(b) The function has two zeros.
2. (a) The root is between 0.3 and 0.4, at about 0.35.
(b) The root is between 1.5 and 1.6, at about 1.55.
(c) The root is between −1.8 and −1.9, at about −1.85.
3. The root occurs at about −1.05
4. The root is between −1.7 and −1.8, at about −1.75.
5. The largest root is at about 2.5.
6. There is one root at x = −1 and another at about x = 1.35.
7. There is one real root at about x = −1.1.
8. The root occurs at about 0.9, since the function changes sign between 0.8 and 1.
9. Using a graphing calculator, we see that when x is around 0.45, the graphs intersect.
10. The root occurs between 0.6 and 0.7, at about 0.65.
11. The root occurs between 1.2 and 1.4, at about 1.3.
12. Zoom in on graph: t = ±0.824. [Note: t must be in radians; one must zoom in two or three times.]
13. (a) Only one real zero, at about x = −1.15.
(b) Three real zeros: at x = 1, and at about x = 1.41 and x = −1.41.
14. First, notice that f (3) ≈ 0.5 > 0 and that f (4) ≈ −0.25 < 0.
1st iteration: f (3.5) > 0, so a zero is between 3.5 and 4.
2nd iteration: f (3.75) < 0, so a zero is between 3.5 and 3.75.
3rd iteration: f (3.625) < 0, so a zero is between 3.5 and 3.625.
4th iteration: f (3.588) < 0, so a zero is between 3.5 and 3.588.
5th iteration: f (3.545) > 0, so a zero is between 3.545 and 3.588.
6th iteration: f (3.578) < 0, so a zero is between 3.567 and 3.578.
7th iteration: f (3.572) > 0, so a zero is between 3.572 and 3.578.
8th iteration: f (3.575) > 0, so a zero is between 3.575 and 3.578.
Thus we know that, rounded to two places, the value of the zero must be 3.58. We know that this is the largest zero
of f (x) since f (x) approaches −1 for larger values of x.
1874
APPENDIX /SOLUTIONS
15. (a) Let F (x) = sin x − 2−x . Then F (x) = 0 will have a root where f (x) and g(x) cross. The first positive value of x
for which the functions intersect is x ≈ 0.7.
(b) The functions intersect for x ≈ 0.4.
16. The graph is
8
6
4
2
-4 -3 -2 -1-2
-4
-6
-8
1 2 3 4
We find one zero at about 0.6. It looks like there might be another one at about −1.2, but zoom in close. . . closer. . . closer,
and you’ll see that though the graphs are very close together, they do not touch, and so there is no zero near −1.2. Thus
the zero at about 0.6 is the only one. (How do you know there are no other zeros off the screen?)
17. (a) Since f is continuous, there must be one zero between θ = 1.4 and θ = 1.6, and another between θ = 1.6 and
θ = 1.8. These are the only clear cases. We might also want to investigate the interval 0.6 ≤ θ ≤ 0.8 since f (θ)
takes on values close to zero on at least part of this interval. Now, θ = 0.7 is in this interval, and f (0.7) = −0.01 < 0,
so f changes sign twice between θ = 0.6 and θ = 0.8 and hence has two zeros on this interval (assuming f is not
really wiggly here, which it’s not). There are a total of 4 zeros.
(b) As an example, we find the zero of f between θ = 0.6 and θ = 0.7. f (0.65) is positive; f (0.66) is negative. So
this zero is contained in [0.65, 0.66]. The other zeros are contained in the intervals [0.72, 0.73], [1.43, 1.44], and
[1.7, 1.71].
(c) You’ve found all the zeros. A picture will confirm this; see Figure A.1.
f (θ) = sin 3θ cos 4θ + 0.8
1.6
0.2
0.4
0.6
0.8
1
1.2
1.4
1.8
θ
Figure A.1
18. (a) There appear to be two solutions: one on the interval from 1.13 to 1.14 and one on the interval from 1.08 to 1.09. From
3
1.13 to 1.14, πx 3 increases from 0.0465 to 0.0478 while (sin 3x)(cos 4x) decreases from 0.0470 to 0.0417, so they
3
must cross in between. Similarly, going from 1.08 to 1.09, πx3 increases from 0.0406 to 0.0418 while (sin 3x)(cos 4x)
increases from 0.0376 to 0.0442. Thus the difference between the two changes sign over that interval, so their difference must be zero somewhere in between.
(b) Reasonable estimates are x = 1.085 and x = 1.131.
19. (a) The first ten results are:
n
0
1
2
3
4
5
6
7
8
1
3.14159
5.05050
5.50129
5.56393
5.57186
5.57285
5.57297
5.57299
5.57299
APPENDIX A SOLUTIONS
1875
(b) The solution is x ≈ 5.573. We started with an initial guess of 1, and kept repeating the given procedure until our
values converged to a limit at around 5.573. For each number on the table, the procedure was in essence asking the
question “Does this number equal 4 times the arctangent of itself?” and then correcting the number by repeating the
question for 4 times the arctangent of the number.
(c) P0 represents our initial guess of x = 1 (on the line y = x). P1 is 4 times the arctangent of 1. If we now use take
this value for P1 and slide it horizontally back to the line y = x, we can now use this as a new guess, and call it
P2 . P3 , of course, represents 4 times the arctangent of P2 , and so on. Another way to make sense of this diagram is
to consider the function F (x) = 4 arctan x − x. On the diagram, this difference is represented by the vertical lines
connecting P0 and P1 , P2 and P3 and so on. Notice how these lines (and hence the difference between arctan x and
x) get smaller as we approach the intersection point, where F (x) = 0.
(d) For an initial guess of x = 10, the procedure gives a decreasing sequence which converges (more quickly) to the same
value of about 5.573. Graphically, our initial guess of P0 will lie to the right of the intersection on the line y = x.
The iteration procedure gives us a sequence of P1 ,P2 , . . . that zigzags to the left, toward the intersection point. For
an initial guess of x = −10, the procedure gives an increasing sequence converging to the other intersection point
of these two curves at x ≈ −5.573. Graphically, we get a sequence which is a reflection through the origin of the
sequence we got for an initial guess of x = 10. This is so because both y = x and y = arctan x are odd functions.
20. Starting with x = 0, and repeatedly taking the cosine, we get the numbers below. Continuing until the first three decimal
places remain fixed under iteration, we have this list and diagram:
y
1.2
x
cos x
0
0.735069
1
0.7401473
0.5403023
0.7356047
0.8575532
0.7414251
0.6542898
0.7375069
0.7934804
0.7401473
0.7013688
0.7383692
0.7639597
0.7395672
0.7221024
0.7387603
0.7504178
0.7393039
0.7314043
0.7389378
0.7442374
etc.
1
0.8
0.6
0.4
0.2
x
0.2
0.4
0.6
21.
4
−1
x
2
−5
Bounded and −5 ≤ f (x) ≤ 4.
4
f (x) = 4x − x2
0.8
1
1.2
1876
APPENDIX /SOLUTIONS
22.
8
h(θ) = 5 + 3 sin θ
5
2
π
−π
−2π
θ
2π
Bounded and 2 ≤ h(θ) ≤ 8.
23.
f (t) =
sin t
t2
t
Not bounded because f (t) goes to infinity as t goes to 0.
Solutions for Section B
1. 2eiπ/2
2. 5eiπ
√
3. 2eiπ/4
4. 5ei4.069
5. 0eiθ , for any θ.
6. e3πi/2
√
7. 10eiθ , where θ = arctan(−3) + π = 1.893 is an angle in the second quadrant.
) ≈ −1.176 is an angle in the fourth quadrant.
8. 13eiθ , where θ = arctan(− 12
5
9. −3 − 4i
10. −11 + 29i
11. −5 + 12i
12. 1 + 3i
13.
1
4
−
9i
8
14. 3 − 6i
15. cos
18.
+ i sin
2π
3
= − 21 + i
√
√
3
2
π
= 23 + 2i is one solution.
6
3π
5 (cos 2 + i sin 3π
) = −125i
2
√
√
4
4
π
10 cos 8 + i 10 sin π8 is one solution.
16. cos
17.
2π
3
π
6
+ i sin
3
19. One value of
√
20. One value of
√
i is
√
−i is
π
π
1
π
ei 2 = (ei 2 ) 2 = ei 4 = cos
p
ei
3π
2
= (ei
3π
2
1
) 2 = ei
3π
4
π
4
+ i sin
= cos
3π
4
π
4
=
√
2
2
+ i sin
3π
4
√
+i
=
2
2
√
− 22
√
+i
2
2
APPENDIX B SOLUTIONS
21. One value of
√
3
22.
√
23.
24.
25.
26.
27.
28.
√
3
π
iπ
2
1
3
√
iπ
6
1877
ei 2 = (e ) = e = cos π6 + i sin π6 = 23 + 2i
√
√
√ π
√
√ π
√
π 1
One value of 7i is 7ei 2 = (7ei 2 ) 2 = 7ei 4 = 7 cos π4 + i 7 sin π4 = 214 + i 214
√ π
1
π
(1 + i)100 = ( 2ei 4 )100 = (2 2 )100 (ei 4 )100 = 250 · ei·25π = 250 cos 25π + i250 sin 25π = −250
√
√ π
√
√
√ √
√
π
1
π 2
One value of (1 + i)2/3 is ( 2ei 4 )2/3 = (2 2 ei 4 ) 3 = 3 2ei 6 = 3 2 cos π6 + i 3 2 sin π6 = 3 2 · 23 + i 3 2 · 21
√
√
√
One value of (−4 + 4i)2/3 is [ 32e(i3π/4) ](2/3) = ( 32)2/3 e(iπ/2) = 25/3 cos π2 + i25/3 sin π2 = 2i 3 4
√
√
√
√
π
π
π
π
+ i 2 sin 12
≈ 1.366 + 0.366i
One value of ( 3 + i)1/2 is (2ei 6 )1/2 = 2ei 12 = 2 cos 12
√
π
π
−1/2
i 6 −1/2
π
π
1 i(− 12 )
1
√
√
√
) ≈ 0.683 − 0.183i
One value of ( 3 + i)
is (2e )
= 2e
= 2 cos(− 12 ) + i 12 sin(− 12
√
√
√
√
√
√
2
= 3 2 ei 2θ =
Since 5 + 2i = 3eiθ , where θ = arctan √25 ≈ 0.730, one value of ( 5 + 2i) 2 is (3eiθ )
√
√
√
√
√
√
3 2 cos 2θ + i3 2 sin 2θ ≈ 3 2 (0.513) + i3 2 (0.859) ≈ 2.426 + 4.062i
i is
29. We have
1
1 i
= · = −i,
i
i i
1
= 2 = −1,
i
1 i
1
· = i,
= 3 =
i
−i i
1
= 4 = 1.
i
i−1 =
i−2
i−3
i−4
The pattern is
n
i =
−36
−i
−1
Since 36 is a multiple of 4, we know i
= 1.
Since 41 = 4 · 10 + 1, we know i−41 = −i.
i
1
n = −1, −5, −9, · · ·
n = −2, −6, −10, · · ·
n = −3, −7, −11, · · ·
n = −4, −8, −12, · · · .
30. Substituting A1 = 2 − A2 into the second equation gives
(1 − i)(2 − A2 ) + (1 + i)A2 = 0
so
2iA2 = −2(1 − i)
−(1 − i)
−i(1 − i)
A2 =
=
= i(1 − i) = 1 + i
i
i2
Therefore A1 = 2 − (1 + i) = 1 − i.
31. Substituting A2 = 2 − A1 into the second equation gives
(i − 1)A1 + (1 + i)(2 − A1 ) = 0
iA1 − A1 − A1 − iA1 + 2 + 2i = 0
−2A1 = −2 − 2i
A1 = 1 + i
Substituting, we have
A2 = 2 − A1 = 2 − (1 + i) = 1 − i.
32. (a) To divide complex numbers, multiply top and bottom by the conjugate of 1 + 2i, that is, 1 − 2i:
3 − 4i 1 − 2i
3 − 4i − 6i + 8i2
−5 − 10i
3 − 4i
=
·
=
=
= −1 − 2i.
1 + 2i
1 + 2i 1 − 2i
12 + 22
5
Thus, a = −1 and b = −2.
(b) Multiplying (1 + 2i)(a + bi) should give 3 − 4i, as the following calculation shows:
(1 + 2i)(a + bi) = (1 + 2i)(−1 − 2i) = −1 − 2i − 2i − 4i2 = 3 − 4i.
1878
APPENDIX /SOLUTIONS
33. To confirm that z =
a + bi
, we calculate the product
c + di
ac + bd
bc − ad
= 2
i (c + di)
c2 + d2
c + d2
2
ac + bcd − bcd + ad2 + (bc2 − acd + acd + bd2 )i
=
c2 + d2
a(c2 + d2 ) + b(c2 + d2 )i
=
= a + bi.
c2 + d2
z(c + di) =
34. (a)
√
√
√
√
√
√
z1 z2 = (−3 − i 3)(−1 + i 3) = 3 + ( 3)2 + i( 3 − 3 3) = 6 − i2 3.
√
√
√
√
√
√
√
3 − ( 3)2 + i( 3 + 3 3)
i·4 3
z1
−3 − i 3 −1 − i 3
√ ·
√ =
√
=
=
= i 3.
2
2
z2
4
−1 + i 3 −1 − i 3√
(−1) + ( 3)
(b) We find (r1 , θ1 ) corresponding to z1 = −3 − i 3:
√
√
√
(−3)2 + ( 3)2 = 12 = 2 3;
√
√
− 3
7π
3
tan θ1 =
=
, so θ1 =
.
−3
3
6
(See Figure B.2.) Thus,
√
√
7π
−3 − i 3 = r1 eiθ1 = 2 3 ei 6 .
√
We find (r2 , θ2 ) corresponding to z2 = −1 + i 3:
r1 =
q
r2 =
√
(−1)2 + ( 3)2 = 2;
q
√
√
3
2π
= − 3, so θ2 =
.
−1
3
√
2π
−1 + i 3 = r2 eiθ2 = 2ei 3 .
tan θ2 =
(See Figure B.3.) Thus,
z2
θ1 = 7π/6
−3
◆
r1
√
√
r2
✮
3
θ2 = 2π/3
−1
− 3
z1
Figure B.2
We now calculate z1 z2 and
Figure B.3
z1
.
z2
√ 7π 2π
√
√ 11π
7π
2π
2ei 3 = 4 3ei( 6 + 3 ) = 4 3ei 6
z1 z2 = 2 3ei 6
√
i
√
√
√ h
3
11π
11π
1
+ i sin
=4 3
−i
= 6 − i2 3.
= 4 3 cos
6
6
2
2
√ i 7π
√
√ π
7π
2π
2 3e 6
z1
=
= 3ei( 6 − 3 ) = 3ei 2
i 2π
z2
2e 3
√
√
π
π
= i 3.
= 3 cos + i sin
2
2
These agrees with the values found in (a).
APPENDIX B SOLUTIONS
1879
35. First we calculate
z1 z2 = (a1 + b1 i)(a2 + b2 i) = a1 a2 − b1 b2 + i(a1 b2 + a2 b1 ).
Thus, z1 z2 = a1 a2 − b1 b2 − i(a1 b2 + a2 b1 ).
Since z̄1 = a1 − b1 i and z̄2 = a2 − b2 i,
z̄1 z̄2 = (a1 − b1 i)(a2 − b2 i) = a1 a2 − b1 b2 − i(a1 b2 + a2 b1 ).
Thus, z1 z2 = z̄1 z̄2 .
36. If the roots are complex numbers, we must have (2b)2 − 4c < 0 so b2 − c < 0. Then the roots are
x=
−2b ±
p
p
(2b)2 − 4c
= −b ± b2 − c
2
p
= −b ± −1(c − b2 )
p
= −b ± i
√
Thus, p = −b and q = c − b2 .
√
37. True, since a is real for all a ≥ 0.
c − b2 .
38. True, since (x − iy)(x + iy) = x2 + y 2 is real.
39. False, since (1 + i)2 = 2i is not real.
40. False. Let f (x) = x. Then f (i) = i but f (ī) = ī = −i.
41. True. We can write any nonzero complex number z as reiβ , where r and β are real numbers with r > 0. Since r > 0, we
can write r = ec for some real number c. Therefore, z = reiβ = ec eiβ = ec+iβ = ew where w = c + iβ is a complex
number.
42. False, since (1 + 2i)2 = −3 + 4i.
43.
1 = e0 = ei(θ−θ) = eiθ ei(−θ)
= (cos θ + i sin θ)(cos(−θ) + i sin(−θ))
= (cos θ + i sin θ)(cos θ − i sin θ)
= cos2 θ + sin2 θ
44. Using Euler’s formula, we have:
ei(2θ) = cos 2θ + i sin 2θ
On the other hand,
2
ei(2θ) = (eiθ ) = (cos θ + i sin θ)2 = (cos2 θ − sin2 θ) + i(2 cos θ sin θ)
Equating imaginary parts, we find
sin 2θ = 2 sin θ cos θ.
45. Using Euler’s formula, we have:
ei(2θ) = cos 2θ + i sin 2θ
On the other hand,
2
ei(2θ) = (eiθ ) = (cos θ + i sin θ)2 = (cos2 θ − sin2 θ) + i(2 cos θ sin θ)
Equating real parts, we find
cos 2θ = cos2 θ − sin2 θ.
1880
APPENDIX /SOLUTIONS
46. Differentiating Euler’s formula gives
d iθ
(e ) = ieiθ = i(cos θ + i sin θ) = − sin θ + i cos θ
dθ
d iθ
d
d
d
Since in addition
(e ) =
(cos θ + i sin θ) =
(cos θ) + i (sin θ), by equating imaginary parts, we conclude
dθ
dθ
dθ
dθ
d
sin θ = cos θ.
that
dθ
47. Differentiating Euler’s formula twice gives
d2
d2
d2
d2 iθ
(e
)
=
(cos
θ
+
i
sin
θ)
=
(cos
θ)
+
i
(sin θ).
dθ2
dθ2
dθ2
dθ2
But
d2 iθ
(e ) = i2 eiθ = −eiθ = − cos θ − i sin θ.
dθ2
Equating real parts, we find
d2
(cos θ) = − cos θ.
dθ2
48. Replacing θ by −x in the formula for sin θ:
sin(−x) =
1 ix
1 −ix
e
e − e−ix = − sin x.
− eix = −
2i
2i
49. Replacing θ by (x + y) in the formula for sin θ:
1 i(x+y)
1 ix iy
e
− e−i(x+y) =
e e − e−ix e−iy
2i
2i
1
((cos x + i sin x) (cos y + i sin y) − (cos (−x) + i sin (−x)) (cos (−y) + i sin (−y)))
=
2i
1
((cos x + i sin x) (cos y + i sin y) − (cos x − i sin x) (cos y − i sin y))
=
2i
= sin x cos y + cos x sin y.
sin(x + y) =
50. Since x1 , y1 , x2 , y2 are each functions of the variable t, differentiating the sum gives
(z1 + z2 )′ = (x1 + iy1 + x2 + iy2 )′ = (x1 + x2 + i (y1 + y2 ))′
= (x1 + x2 )′ + i (y1 + y2 )′
= x′1 + x′2 + i y1′ + y2′
= (x1 + iy1 )′ + (x2 + iy2 )′
= z1′ + z2′ .
Differentiating the product gives
(z1 z2 )′ = ((x1 + iy1 ) (x2 + iy2 ))′ = (x1 x2 − y1 y2 + i (y1 x2 + x1 y2 ))′
= (x1 x2 − y1 y2 )′ + i (y1 x2 + x1 y2 )′
= x′1 x2 + x1 x′2 − y1′ y2 − y1 y2′ + i y1′ x2 + y1 x′2 + x′1 y2 + x1 y2′
= [x′1 x2 − y1′ y2 + i(x′1 y2 + y1′ x2 )] + [x1 x′2 − y1 y2′ + i(y1 x′2 + x1 y2′ )]
= x′1 + iy1′ (x2 + iy2 ) + (x1 + iy1 ) x′2 + iy2′
= z1′ z2 + z1 z2′ .
APPENDIX C SOLUTIONS
1881
Solutions for Section C
1. (a) f ′ (x) = 3x2 + 6x + 3 = 3(x + 1)2 . Thus f ′ (x) > 0 everywhere except at x = −1, so it is increasing everywhere
except perhaps at x = −1. The function is in fact increasing at x = −1 since f (x) > f (−1) for x > −1, and
f (x) < f (−1) for x < −1.
(b) The original equation can have at most one root, since it can only pass through the x-axis once if it never decreases.
It must have one root, since f (0) = −6 and f (1) = 1.
(c) The root is in the interval [0, 1], since f (0) < 0 < f (1).
(d) Let x0 = 1.
x0 = 1
x1 = 1 −
f (1)
1
11
=1−
=
≈ 0.917
f ′ (1)
12
12
11
12
11
12
f
11
−
x2 =
12
f′
x3 = 0.913 −
≈ 0.913
f (0.913)
≈ 0.913.
f ′ (0.913)
Since the digits repeat, they should be accurate. Thus x ≈ 0.913.
√
√
2. Let f (x) = x3 − 50. Then f ( 3 50) = 0, so we can use Newton’s method to solve f (x) = 0 to obtain x = 3 50.
Since f ′ (x) = 3x2 , f ′ is always positive, and f is therefore increasing. Consequently, f has only one zero. Since
33 = 27 < 50 < 64 = 43 , let x0 = 3.5. Then
x0 = 3.5
x1 = 3.5 −
f (3.5)
≈ 3.694
f ′ (3.5)
Continuing, we find
x2 ≈ 3.684
x3 ≈ 3.684.
Since the digits repeat, x3 should be correct, as can be confirmed by calculator.
√
√
3. Let f (x) = x4 − 100. Then f ( 4 100) = 0, so we can use Newton’s method to solve f (x) = 0 to obtain x = 4 100.
′
3
4
4
f (x) = 4x . Since 3 = 81 < 100 < 256 = 4 , try 3.1 as an initial guess.
x0 = 3.1
f (3.1)
≈ 3.164
f ′ (3.1)
f (3.164)
x2 = 3.164 − ′
≈ 3.162
f (3.164)
f (3.162)
x3 = 3.162 − ′
≈ 3.162
f (3.162)
x1 = 3.1 −
Thus
√
4
100 ≈ 3.162.
1
4. Let f (x) = x3 − 10
. Then f (10−1/3 ) = 0, so we can use Newton’s method to solve f (x) = 0 to obtain x = 10−1/3 .
p1
p1
p
′
2
f (x) = 3x . Since 3 27
< 3 10
< 3 18 , try x0 = 21 . Then x1 = 0.5 − ff′(0.5)
≈ 0.467. Continuing, we find
(0.5)
x2 ≈ 0.464.x3 ≈ 0.464. Since x2 ≈ x3 , 10−1/3 ≈ 0.464.
5. Let f (x) = sin x − 1 + x; we want to find all zeros of f , because f (x) = 0 implies sin x = 1 − x.
Graphing sin x and 1 − x in Figure C.4, we see that f (x) has one solution at x ≈ 21 .
1882
APPENDIX /SOLUTIONS
y = sin x
1
π/2
y = 1−x
−1
Figure C.4
Letting x0 = 0.5, and using Newton’s method, we have f ′ (x) = cos x + 1, so that
x1 = 0.5 −
sin(0.5) − 1 + 0.5
≈ 0.511,
cos(0.5) + 1
sin(0.511) − 1 + 0.511
≈ 0.511.
cos(0.511) + 1
Thus sin x = 1 − x has one solution at x ≈ 0.511.
x2 = 0.511 −
6. Let f (x) = cos x−x. We want to find all zeros of f , because f (x) = 0 implies that cos x = x. Since f ′ (x) = − sin x−1,
f ′ is always negative (as − sin x never exceeds 1). This means f is always decreasing and consequently has at most 1
root. We now use Newton’s method. Since cos 0 > 0 and cos π2 < π2 , cos x = x for 0 < x < π2 . Thus, try x0 = π6 .
cos π6 − π6
π
≈ 0.7519,
−
6
− sin π6 − 1
x2 ≈ 0.7391,
x1 =
x3 ≈ 0.7390.
x2 ≈ x3 ≈ 0.739. Thus x ≈ 0.739 is the solution.
7. Let f (x) = e−x − ln x. Then f ′ (x) = −e−x − x1 . We want to find all zeros of f , because f (x) = 0 implies
that e−x = ln x. Since e−x is always decreasing and ln x is always increasing, there must be only 1 solution. Since
e−1 > ln 1 = 0, and e−e < ln e = 1, then e−x = ln x for some x, 1 < x < e. Try x0 = 1. We now use Newton’s
method.
e−1 − 0
≈ 1.2689,
−e−1 − 1
x2 ≈ 1.309,
x1 = 1 −
x3 ≈ 1.310.
Thus x ≈ 1.310 is the solution.
8. Let f (x) = ex cos x − 1. Then f ′ (x) = −ex sin x + ex cos x. Now we use Newton’s method, guessing x0 = 1 initially.
x1 = 1 −
f (1)
≈ 1.5725
f ′ (1)
Continuing: x2 ≈ 1.364, x3 ≈ 1.299, x4 ≈ 1.293, x5 ≈ 1.293. Thus x ≈ 1.293 is a solution. Looking at a graph of
f (x) suffices to convince us that there is only one solution.
9. Let f (x) = ln x − x1 , so f ′ (x) = x1 + x12 .
Now use Newton’s method with an initial guess of x0 = 2.
ln 2 − 21
≈ 1.7425,
1
+ 41
2
x2 ≈ 1.763,
x1 = 2 −
x3 ≈ 1.763.
Thus x ≈ 1.763 is a solution. Since f ′ (x) > 0 for positive x, f is increasing: it must be the only solution.
10. (a) One zero in the interval 0.6 < x < 0.7.
(b) Three zeros in the intervals −1.55 < x < −1.45, x = 0, 1.45 < x < 1.55.
(c) Two zeros in the intervals 0.1 < x < 0.2, 3.5 < x < 3.6.
APPENDIX D SOLUTIONS
1883
11. f ′ (x) = 3x2 + 1. Since f ′ is always positive, f is everywhere increasing. Thus f has only one zero. Since f (0) < 0 <
f (1), 0 < x0 < 1. Pick x0 = 0.68.
x0 = 0.68,
x1 = 0.6823278,
x2 ≈ 0.6823278.
Thus x ≈ 0.682328 (rounded up) is a root. Since x1 ≈ x2 , the digits should be correct.
12. Let f (x) = x2 − a, so f ′ (x) = 2x.
x2 −a
n
Then by Newton’s method, xn+1 = xn − 2x
n
For a = 2 :
√
x0 = 1, x1 = 1.5, x2 ≈ 1.416, x3 ≈ 1.414215, x4 ≈ 1.414213 so 2 ≈ 1.4142.
For a = 10 :
√
x0 = 5, x1 = 3.5, x2 ≈ 3.17857, x3 ≈ 3.162319, x4 ≈ 3.162277 so 10 ≈ 3.1623.
For a = 1000 :
x0 = 500, x1 = 251,
√ x2 ≈ 127.49203, x3 ≈ 67.6678, x4 ≈ 41.2229, x5 ≈ 32.7406, x6 ≈ 31.6418, x7 ≈ 31.62278,
x8 ≈ 31.62277 so 1000 ≈ 31.6228.
For a = π :
√
x0 = π2 , x1 ≈ 1.7853, x2 ≈ 1.7725 x3 ≈ 1.77245, x4 ≈ 1.77245 so π ≈ 1.77245.
Solutions for Section D
Exercises
√
1. The magnitude is k3~i k = 32 + 02 = 3.
The angle of 3~i is 0 because the vector lies along the positive x-axis.
√
√
2. The magnitude is k2~i + ~j k = 22 + 12 = 5.
Since 2~i + ~j is in the first quadrant, the angle is arctan(1/2) = 0.464 radians, or about 26.6 degrees.
q
√
√
√ 2
√
~
~
3. The magnitude is k − 2 i + 2 j k = (− 2)2 + 2 = 2.
√
√
The direction of the vector − 2~i + 2 ~j is given by the angle θ = 3π/4 as the vector bisects the second quadrant.
4. ~v + w
~ = (1 − 2)~i + (2 + 3)~j = (−1)~i + 5~j = −~i + 5~j .
5. 2~v + w
~ = (2 − 2)~i + (4 + 3)~j = 7~j .
6. w
~ + (−2)~
u = −2~i + (3 − 4)~j = (−2)~i + (−1)~j = −2~i − ~j .
7. Calculating magnitudes of each vector yields:
k3~i + 4~j k =
p
k − 5~i k =
p
k~i + ~j k =
32 + 42 = 5
√
12 + 12 = 2
p
(−5)2 + 02 = 5
p
k5~j k = 02 + 52 = 5
q
√ 2 √
√
~
2 = 2
k 2j k =
p
√
k2~i + 2~j k = 22 + 22 = 8
k − 6~j k =
p
~
~
Thus 3~i + 4~j , −5
√ i and 5j have the same magnitude.
Also ~i + ~j and 2~j have the same magnitude.
02 + (−6)2 = 6.
1884
APPENDIX /SOLUTIONS
8. Two vectors are in the same direction if one is a positive scalar multiple of the other.
Since
2(~i + ~j ) = 2~i + 2~j
~
~
~
~
the vectors i + j and 2i + 2j are in the same direction.
Also
5 √
5~j = √ ( 2~j )
2
√
so 5~j and 2~j are in the same direction.
9. Two vectors have opposite direction if one is a negative scalar multiple of the other. Since
−5
5~j =
(−6~j )
6
the vectors 5~j and −6~j have opposite direction. Similarly, −6~j and
√
2~j have opposite direction.
10. We work out the left hand side kk~r k and show that it is the same as the right hand side |k|k~r k. We have
p
p
√ p
kk~r k = kka~i + kb~j k = (ka)2 + (kb)2 = k2 (a2 + b2 ) = k2 a2 + b2 = |k|k~r k.
11. (a) The magnitude of −3~i + 4~j is (−3)2 + 42 = 5, so we want to scale down the magnitude by a factor of 5. Scalar
multiplying by 1/5 does not change the vector’s direction, and gives the unit vector (−3/5)~i + (4/5)~j .
(b) Scalar multiplying by −1 reverses a vector’s direction without changing its magnitude. So
p
(−1)
4
−3~
i + ~j
5
5
3
4
= ~i − ~j
5
5
is a unit vector in the direction opposite to the original.
~
12. A vector making an angle of 90◦ with
√ the positive x-axis lies along the positive y-axis, so it is of the form ~v = aj with
a positive. Its magnitude is k~v k = 02 + a2 = a, so a = 5. The vector is 5~j .
13. Scalar multiplication by 2 doubles the magnitude of a vector without changing its direction. Thus, the vector is 2(4~i −
3~j ) = 8~i − 6~j .
14. Scalar multiplication by −1 reverses the direction of a vector without changing its magnitude. Thus, the vector is
(−1)(4~i − 3~j ) = −4~i + 3~j .
15. The vector is (4 − 3)~i + (4 − 2)~j = ~i + 2~j .
16. In components, the vector from (6, 6) to (−6, −6) is ((−6) − 6)~i + ((−6) − 6)~j = −12~i − 12~j , which is not equal
to 6~i − 6~j .
17. In components, the vector from (7, 7) to (9, 11) is (9 − 7)~i + (11 − 9)~j = 2~i + 2~j .
In components, the vector from (8, 10) to (10, 12) is (10 − 8)~i + (12 − 10)~j = 2~i + 2~j .
The two vectors are equal.
√
18. In components, the vector of length 2 making an angle of π/4 with the positive x-axis is ~i + ~j , which is not the same
as −~i + ~j .
19. In components, the vector from (1, 12) to (6, 10) is (6 − 1)~i + (10 − 12)~j = 5~i − 2~j . The two vectors are equal.
20. The velocity is ~v (t) = 1~i √+ 2t~j . When√t = 1 the velocity vector is ~v = ~i + 2~j .
The speed is k~v k = 12 + 22 = 5.
The acceleration is ~a (t) = 2~j .
~
~ ~
21. The velocity is ~v (t) = et~i√+ (1/(1 + t))
√ j . When t = 0, the velocity vector is ~v = i + j .
The speed is k~v k = 12 + 12 = 2.
The acceleration is ~a (t) = et~i − 1/((1 + t)2 )~j . When t = 0, the acceleration vector is ~a = ~i − ~j .
22. The velocity is ~v (t) = −5 sin t~i + 5 cos t~j . When t = π/2, the velocity vector is ~v = −5~i .
p
The speed is k~v k = (−5)2 + 02 = 5.
The acceleration is ~a (t) = −5 cos(t)~i − 5 sin(t)~j . For t = π/2, the acceleration vector is ~a = −5~j .
23. The position vector is
√
√
~r (π/4) = cos(π/4)~i + sin(π/4)~j = (1/ 2)~i + (1/ 2)~j .
The velocity vector is
~v (t) =
d
d
cos t~i +
sin t ~j = − sin t~i + cos t ~j ,
dt
dt
APPENDIX D SOLUTIONS
so
The speed is
1885
√
√
~v (π/4) = − sin(π/4)~i + cos(π/4)~j = (−1/ 2)~i + (1/ 2)~j .
||~v || =
√
√
(−1/ 2)2 + (1/ 2)2 = 1.
q
We recognize ~r (t) as the parameterization of a unit circle, centered at the origin. Figure D.5 shows the curve together
with the position and velocity vectors when t = π/4. We see that the velocity vector is tangent to the circle.
y
1
1
~
v = − √ ~i + √ ~j
2
2
✛
1
1
~
r = √ ~i + √ ~j
2
2
x
Figure D.5: Position and velocity vectors
for motion along a circle
Source Exif Data:
File Type : PDF File Type Extension : pdf MIME Type : application/pdf PDF Version : 1.7 Linearized : No Author : Modify Date : 2014:09:09 13:32:13-04:00 Subject : Create Date : 2014:09:09 09:58:09-04:00 PXC Viewer Info : PDF-XChange Viewer;2.5.196.0;Jun 30 2011;20:40:17;D:20140909133213-04'00' Page Count : 1886 Has XFA : No XMP Toolkit : XMP Core 4.1.1 Metadata Date : 2014:09:09 13:31:14-04:00 Creator Tool : -|- this layout: pidus -|- Format : application/pdf Description : Title : Creator : Document ID : uuid:cf904ebf-6b75-44f2-92e9-c6cb9afabfc0 Instance ID : uuid:90234b9f-0e95-42d2-88eb-3631238f50e1 Producer : -|- this layout: pidus -|- Keywords : Page Mode : UseOutlines Page Layout : SinglePageEXIF Metadata provided by EXIF.tools