Howard Anton Student Solutions Manual Set For Calculus Early Transcendentals Single Variable 8th E
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CONTENTS
Introduction
.............................................................. 1
Chapter 1.
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
Chapter 2.
Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Chapter 3.
The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
Chapter 4.
Logarithmic and Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . 99
Chapter 5.
Analysis of Functions and Their Graphs . . . . . . . . . . . . . . . . . . . . . . . . 139
Chapter 6.
Applications of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
Chapter 7.
Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
Chapter 8.
Applications of the Definite Integral
in Geometry, Science, and Engineering . . . . . . . . . . . . . . . . . . . . . . . . . 256
Chapter 9.
Principles of Integral Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
Chapter 10.
Mathematical Modeling with Differential Equations . . . . . . . . . . . . . . 343
Chapter 11.
Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
Chapter 12.
Analytic Geometry in Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .408
Chapter 13.
Three-Dimensional Space; Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448
Chapter 14.
Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 490
Chapter 15.
Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524
Chapter 16.
Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
Chapter 17.
Topics in Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 608
Appendix A. Real Numbers, Intervals, and Inequalities . . . . . . . . . . . . . . . . . . . . . . . 640
Appendix B. Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647
Appendix C. Coordinate Planes and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 650
Appendix D. Distance, Circles, and Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . 658
Appendix E. Trigonometry Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668
Appendix F.
Solving Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674
CALCULUS:
A New Horizon from Ancient Roots
EXERCISE SET FOR INTRODUCTION
1.
(a)
x = 0.123123123 . . .; 1000x = 123.123123123 . . . = 123 + x; 999x = 123; x =
(b)
x = 12.7777 . . .; 10x = 127.7777 . . ., so 9x = 10x − x = 115; x =
(c)
x = 38.07818181 . . .; 100x = 3807.818181 . . .; 99x = 3769.74;
376974
41886
20943
3769.74
=
=
=
x=
99
9900
1100
550
537
4296
=
0.4296000 . . . = 0.4296 =
10000
1250
(d)
41
123
=
999
333
115
9
repeats
2.
3.
z }| {
22
= 3. 142857 . . .
(a) π is irrational, and thus has a nonrepeating decimal expansion, whereas
7
22
(b)
>π
7
Ã
Ã
√ !
√ !
333
63 17 + 15 5
22
63 17 + 15 5
223
355
√
√
<
<
<
(b)
(a)
<
71
106
25 7 + 15 5
113
7
25 7 + 15 5
Ã
√ !
63 17 + 15 5
333
√
(d)
(c)
106
25 7 + 15 5
¶2 µ
¶2
16
256 2
8
D =
r =
r . The area of a circle of radius r is
9
9
81
πr2 so 256/81 was the approximation used for π.
µ
4.
5.
6.
7.
(a)
If r is the radius, then D = 2r so
(b)
256/81 ≈ 3.16049, 22/7 ≈ 3.14268, and π ≈ 3.14159 so 256/81 is worse than 22/7.
The first series, taken to ten terms, adds to 3.0418; the second, as printed, adds to 3.1416.
1
1
1
1
1
1
1
= 0.111111 . . . =
+
+
+
+
+
+ ...
9
10 100 1000 10000 100000 1000000
1
8
5
1
8
5
2
= 0.185185 . . . =
+
+
+
+
+
+ ...
(b)
27
10 100 1000 10000 100000 1000000
3
1
1
1
1
1
14
= 0.311111 . . . =
+
+
+
+
+
+ ...
(c)
45
10 100 1000 10000 100000 1000000
(a)
6
3
6
3
6
3
7
= 0.636363 . . . =
+
+
+
+
+
+ ...
11
10 100 1000 10000 100000 1000000
2
4
2
4
2
4
8
= 0.242424 . . . =
+
+
+
+
+
+ ...
(b)
33
10 100 1000 10000 100000 1000000
4
1
6
6
6
6
5
= 0.416666 . . . =
+
+
+
+
+
+ ...
(c)
12
10 100 1000 10000 100000 1000000
(a)
8.
(a) 1, 2, 1.75, 1.7321
(b) 1, 3, 2.33, 2.238, 2.2361
9.
(a) 1, 4, 2.875, 2.6549, 2.6458
(b) 1, 25.5, 13.7, 8.69, 7.22, 7.0726, 7.0711
10.
(a)
Let x1 = 12 (a + b), x2 = 12 (a + x1 ), x3 = 12 (a + x2 ), etc. Then b > x1 > x2 > · · · > xn−1 > xn > a
so all the xi ’s are distinct, there are infinitely many of them and they all lie between a and b.
(b) x = 0.99999 . . ., 10x = 9.99999 . . ., 9x = 9, x = 1
(c)
(1.999999 . . .)/2 = 0.999999 . . . = 1; yes it is consistent, as all three are equal.
(d) 10x = 9 + x, so x = 9/9 = 1. They are equal.
1
CHAPTER 1
Functions
EXERCISE SET 1.1
1.
(a)
(c)
around 1943
no; you need the year’s population
(b) 1960; 4200
(d) war; marketing techniques
(e) news of health risk; social pressure, antismoking campaigns, increased taxation
2.
(a) 1989; $35,600
(b) 1983; $32,000
(c) the first two years; the curve is steeper (downhill)
3.
(a) −2.9, −2.0, 2.35, 2.9
(d) −1.75 ≤ x ≤ 2.15
4.
(a) x = −1, 4
(d) x = 0, 3, 5
5.
(a) x = 2, 4
(b) none
(c) x ≤ 2; 4 ≤ x
(d) ymin = −1; no maximum value
6.
(a) x = 9
(b) none
(c) x ≥ 25
(d) ymin = 1; no maximum value
7.
(a)
(b) none
(c) y = 0
(e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2
(b) none
(c) y = −1
(e) ymax = 9 at x = 6; ymin = −2 at x = 0
Breaks could be caused by war, pestilence, flood, earthquakes, for example.
(b) C decreases for eight hours, takes a jump upwards, and then repeats.
8.
(a)
Yes, if the thermometer is not near a window or door or other source of sudden temperature
change.
(b) The number is always an integer, so the changes are in movements (jumps) of at least one unit.
9.
(a)
If the side adjacent to the building has length x then L = x + 2y. Since A = xy = 1000,
L = x + 2000/x.
(b)
x > 0 and x must be smaller than the width of the building, which was not given.
(c)
(d) Lmin ≈ 89.44
120
20
80
80
10.
(a) V = lwh = (6 − 2x)(6 − 2x)x
(b) From the figure it is clear that 0 < x < 3.
(c)
(d) Vmax ≈ 16
20
0
3
0
2
3
Chapter 1
11.
500
. Then
πr2
500
C = (0.02)(2)πr2 + (0.01)2πrh = 0.04πr2 + 0.02πr 2
πr
10
= 0.04πr2 + ; Cmin ≈ 4.39 at r ≈ 3.4, h ≈ 13.8.
r
(a) V = 500 = πr2 h so h =
7
1.5
6
4
10
(b) C = (0.02)(2)(2r)2 + (0.01)2πrh = 0.16r2 + . Since
r
0.04π < 0.16, the top and bottom now get more weight.
Since they cost more, we diminish their sizes in the solution,
and the cans become taller.
7
1.5
5.5
4
12.
(c)
r ≈ 3.1, h ≈ 16.0, C ≈ 4.76
(a)
The length of a track with straightaways of length L and semicircles of radius r is
P = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π = 1222.65 ft.
Since this is less than 1320 ft (a quarter-mile), a solution is possible.
(b) P = 2L + 2πr = 1320 and 2r = 2x + 160, so
L = 12 (1320 − 2πr) = 12 (1320 − 2π(80 + x)) = 660 − 80π − πx.
450
0
100
0
(c)
The shortest straightaway is L = 360, so x = 15.49 ft.
(d) The longest straightaway occurs when x = 0, so L = 660 − 80π = 408.67 ft.
EXERCISE SET 1.2
1.
2.
3.
f (0) = 3(0)2 − 2 = −2; f (2) = 3(2)2 − 2 = 10; f (−2) = 3(−2)2 − 2 = 10; f (3) = 3(3)2 − 2 = 25;
√
√
f ( 2) = 3( 2)2 − 2 = 4; f (3t) = 3(3t)2 − 2 = 27t2 − 2
√
√
(b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f (−2) = 2(−2) = −4; f (3) = 2(3) = 6; f ( 2) = 2 2;
f (3t) = 1/3t for t > 1 and f (3t) = 6t for t ≤ 1.
(a)
−1 + 1
π+1
−1.1 + 1
−0.1
1
3+1
= 2; g(−1) =
= 0; g(π) =
; g(−1.1) =
=
=
;
3−1
−1 − 1
π−1
−1.1 − 1
−2.1
21
t2
t2 − 1 + 1
= 2
g(t2 − 1) = 2
t −1−1
t −2
√
√
(b) g(3) = 3 + 1 = 2; g(−1) = 3; g(π) = π + 1; g(−1.1) = 3; g(t2 − 1) = 3 if t2 < 2 and
√
g(t2 − 1) = t2 − 1 + 1 = |t| if t2 ≥ 2.
(a)
g(3) =
(a)
x 6= 3
(b)
√
√
x ≤ − 3 or x ≥ 3
Exercise Set 1.2
(c)
4
x2 − 2x + 5 = 0 has no real solutions so x2 − 2x + 5 is always positive or always negative. If
x = 0, then x2 − 2x + 5 = 5 > 0; domain: (−∞, +∞).
(d) x 6= 0
4.
(a)
x 6= −
sin x 6= 1, so x 6= (2n + 12 )π,
n = 0, ±1, ±2, . . .
(e)
7
5
1
(b) x − 3x2 must be nonnegative; y = x − 3x2 is a parabola that crosses the x-axis at x = 0, and
3
1
opens downward, thus 0 ≤ x ≤
3
2
x −4
> 0, so x2 − 4 > 0 and x − 4 > 0, thus x > 4; or x2 − 4 < 0 and x − 4 < 0, thus
(c)
x−4
(d)
−2 < x < 2
x 6= −1
(e)
cos x ≤ 1 < 2, 2 − cos x > 0, all x
5.
(a) x ≤ 3
6.
(a) x ≥
7.
(a) yes
(c) no (vertical line test fails)
8.
The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse), so that L = 20 sin(θ/2).
9.
The cosine of θ is (L − h)/L (side adjacent over hypotenuse), so h = L(1 − cos θ).
10.
(b) −2 ≤ x ≤ 2
2
3
(b) −
T
3
3
≤x≤
2
2
(c) x ≥ 0
(d) all x
(e) all x
(c) x ≥ 0
(d) x 6= 0
(e) x ≥ 0
(b) yes
(d) no (vertical line test fails)
11.
h
12.
w
t
t
t
5
13.
(a)
10
15
If x < 0, then |x| = −x so f (x) = −x + 3x + 1 = 2x + 1. If x ≥ 0, then |x| = x so
f (x) = x + 3x + 1 = 4x + 1;
½
2x + 1, x < 0
f (x) =
4x + 1, x ≥ 0
(b) If x < 0, then |x| = −x and |x − 1| = 1 − x so g(x) = −x + 1 − x = 1 − 2x. If 0 ≤ x < 1, then
|x| = x and |x − 1| = 1 − x so g(x) = x + 1 − x = 1. If x ≥ 1, then |x| = x and |x − 1| = x − 1
so g(x) = x + x − 1 = 2x − 1;
x<0
1 − 2x,
1,
0≤x<1
g(x) =
2x − 1,
x≥1
14.
(a)
If x < 5/2, then |2x−5| = 5−2x so f (x) = 3+(5−2x) = 8−2x. If x ≥ 5/2, then |2x−5| = 2x−5
so f (x) = 3 + (2x − 5) = 2x − 2;
½
8 − 2x, x < 5/2
f (x) =
2x − 2, x ≥ 5/2
5
Chapter 1
(b)
15.
If x < −1, then |x − 2| = 2 − x and |x + 1| = −x − 1 so g(x) = 3(2 − x) − (−x − 1) = 7 − 2x. If
−1 ≤ x < 2, then |x − 2| = 2 − x and |x + 1| = x + 1 so g(x) = 3(2 − x) − (x + 1) = 5 − 4x. If
x ≥ 2, then |x − 2| = x − 2 and |x + 1| = x + 1 so g(x) = 3(x − 2) − (x + 1) = 2x − 7;
x < −1
7 − 2x,
5 − 4x, −1 ≤ x < 2
g(x) =
2x − 7,
x≥2
(a) V = (8 − 2x)(15 − 2x)x
(b) −∞ < x < +∞, −∞ < V < +∞
(c) 0 < x < 4
(d) minimum value at x = 0 or at x = 4; maximum value somewhere in between (can be approximated by zooming with graphing calculator)
16.
(a) x = 3000 tan θ
(b) θ 6= nπ + π/2 for n an integer, −∞ < n < ∞
(c) 0 ≤ θ < π/2, 0 ≤ x < +∞
(d) 3000 ft
6000
6
0
0
17.
(i) x = 1, −2 causes division by zero
(ii) g(x) = x + 1, all x
18.
(i) x = 0 causes division by zero
(ii) g(x) =
19.
(a) 25◦ F
20.
If v = 48 then −60 = WCI = 1.6T − 55; thus T = (−60 + 55)/1.6 ≈ −3◦ F.
21.
√
If v = 8 then −10 = WCI = 91.4 + (91.4 − T )(0.0203(8) − 0.304 8 − 0.474); thus
√
T = 91.4 + (10 + 91.4)/(0.0203(8) − 0.304 8 − 0.474) and T = 5◦ F
√
x + 1 for x ≥ 0
(b) 2◦ F
(c) −15◦ F
22.
The WCI is given by three formulae, but the first
work with the data. Hence
√ and third don’t √
−15 = WCI = 91.4 + (91.4 − 20)(0.0203v − 0.304 v − 0.474); set x = v so that v = x2 and obtain
0.0203x2 − 0.304x − 0.474 + (15 + 91.4)/(91.4 − 20) = 0. Use the quadratic formula to find the two
roots. Square them to get v and discard the spurious solution, leaving v ≈ 25.
23.
Let t denote time in minutes after 9:23 AM. Then D(t) = 1000 − 20t ft.
EXERCISE SET 1.3
1.
(e) seems best, though only (a) is bad.
y
2.
(e) seems best, though only (a) is bad and (b) is
not good.
y
0.5
0.5
x
-1
1
-1
1
-0.5
-0.5
x
Exercise Set 1.3
3.
6
(b) and (c) are good; (a) is very bad.
4.
(b) and (c) are good; (a) is very bad.
y
y
15
-11
14
-12
13
-13
12
-14
x
-2
5.
-1
0
1
-2
2
[−3, 3] × [0, 5]
6.
-1
0
1
2
x
[−4, 2] × [0, 3]
y
y
2
4
1
2
-3
7.
-2
-3
-1
1
2
-2
-1
1
x
x
3
(a) window too narrow, too short
(c) good window, good spacing
(b) window wide enough, but too short
(d) window too narrow, too short
y
-5
-100
5
10
15
20
x
-200
-300
-400
-500
(e) window too narrow, too short
8.
(a) window too narrow
(c) good window, good tick spacing
(b) window too short
(d) window too narrow, too short
y
50
-16 -12
-8
-4
4
x
-50
-100
-150
-200
-250
(e) shows one local minimum only, window too narrow, too short
7
Chapter 1
9.
[−5, 14] × [−60, 40]
10.
[6, 12] × [−100, 100]
y
y
100
40
20
50
-5
5
x
10
8
-20
10
x
12
-50
-40
-100
-60
11.
[−0.1, 0.1] × [−3, 3]
12.
[−1000, 1000] × [−13, 13]
y
y
3
10
2
5
1
x
-0.1
x
1000
-1000
0.1
-1
-5
-2
-10
-3
13.
[−250, 1050] × [−1500000, 600000]
14.
[−3, 20] × [−3500, 3000]
y
-1000
y
1000
x
1000
-500000
5
10
x
15
-1000
-2000
15.
[−2, 2] × [−20, 20]
16.
[1.6, 2] × [0, 2]
y
y
20
2
10
1.5
x
-2
-1
1
1
2
0.5
-10
x
-20
17.
1.6
depends on graphing utility
18.
1.7
1.8
y
6
6
4
4
2
-2
2
2
-2
2
depends on graphing utility
y
-4
1.9
4
x
-4
-2
2
-2
-4
-4
-6
-6
4
x
Exercise Set 1.3
19.
8
(a) f (x) =
√
√
(b) f (x) = − 16 − x2
16 − x2
y
y
4
3
2
1
-4
-4
-2
2
4
x
4
y
(d)
4
4
3
2
-4
2
-1
-2
-3
-4
x
y
(c)
-2
-2
2
4
2
x
1
-2
1
-4
2
3
4
(e) No; the vertical line test fails.
20.
p
(a) y = ±3 1 − x2 /4
√
(b) y = ± x2 + 1
y
4
2
-4
-2
2
4
x
-2
-4
21.
y
(a)
y
(b)
(c)
y
1
1
x
-1
x
x
-1
(d)
1
1
-1
2
y
(e)
y
1
(f )
y
1
1
π
2π
x
−π
1
π
—1
x
x
-1
1
-1
y
22.
1
x
-1
1
-1
23.
The portions of the graph of y = f (x) which lie below the x-axis are reflected over the x-axis to give the
graph of y = |f (x)|.
9
Chapter 1
24.
Erase the portion of the graph of y = f (x) which lies in the left-half plane and replace it with the reflection
over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and you obtain the graph
of y = f (|x|).
25.
(a) for example, let a = 1.1
(b)
y
y
3
2
1
µ
1
26.
They are identical.
27.
2
3
x
y
y
15
10
x
5
x
-1
28.
1
2
3
This graph is very complex. We show three views, small (near the origin), medium and large:
(a)
(b)
y
(c)
y
y
10
x
2
-1
x
-50
-1
1000
x
40
29.
(a)
(b)
y
y
1
1.5
0.5
1
x
-3
-2
-1
1
2
3
-0.5
0.5
-1
-3
(c)
1
-1
-2
2
3
x
y
(d)
y
1.5
1.5
1
0.5
-1
1
1
2
3
x
0.5
x
-2
-1
1
Exercise Set 1.3
10
30.
y
1
x
2
31.
(a)
stretches or shrinks the graph in the
y-direction; flips it if c changes sign
(b)
As c increases, the parabola moves down
and to the left. If c increases, up and right.
y
y
c=2
4
c=1
2
-1
c=2
6
c=1
4
x
1
8
2
c = -1.5
-2
c = -1.5
-2
-1
1
x
2
(c) The graph rises or falls in the y-direction with changes in c.
y
c = +2
8
c = .5
6
c = -1
4
2
-2
32.
-1
1
2
x
y
(a)
2
x
1
2
—2
(b)
x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of the curve
come together. If a moves past b then a and b switch roles.
y
y
y
3
3
2
2
2
1
1
3
x
-1
-2
-3
1
2
1
1
3
-1
a = 0.5
b = 1.5
-2
-3
2
3
x
1
-1
a=1
b = 1.5
-2
-3
a = 1.5
b = 1.6
2
3
x
11
33.
Chapter 1
The curve oscillates between the lines
y = x and y = −x with increasing rapidity as |x| increases.
34.
The curve oscillates between the lines
y = +1 and y = −1, infinitely many times
in any neighborhood of x = 0.
y
y
30
20
10
x
-30 -20 -10
-10
10
20
x
30
—2
4
-20
—1 —1
-30
35.
Plot f (x) on [−10, 10]; then on [−1, 0], [−0.7, −0.6], [−0.65, −0.64], [−0.646, −0.645]; for the other
root use [4, 5], [4.6, 4.7], [4.64, 4.65], [4.645, 4.646]; roots −0.6455, 4.6455.
36.
Plot f (x) on [−10, 10]; then on [−4, −3], [−3.7, −3.6], [−3.61, −3.60], [−3.606, −3.605]; for the other
root use [3, 4], [3.6, 3.7], [3.60, 3.61], [3.605, 3.606]; roots 3.6055, −3.6055.
EXERCISE SET 1.4
1.
(a)
(b)
y
-1
y
2
1
1
0
1
x
2
-1
(c)
1
2
(d)
y
y
1
-1
2.
2
1
x
2
-4
y
(a)
-2
2
1
x
-2
2
y
3
y
(d)
1
1
x
2
-1
x
2
y
(b)
x
-2
(c)
x
3
3
x
-1
1
Exercise Set 1.4
3.
12
(a)
y
(b)
1
y
1
-2
-1
1
2
x
-0.5
0.5
1
1.5
1
2
3
x
-1
-1
(c)
(d)
y
y
1
1
-1
1
2
3
x
-1
x
-1
-1
4.
5.
Translate right 2 units, and up one unit.
y
y
1
-2
x
2
10
-2
6.
Translate left 1 unit, reflect over x-axis, and
translate up 2 units.
7.
y
2
4
6
x
Translate left 1 unit, stretch vertically by
a factor of 2, reflect over x-axis, translate
down 3 units.
y
-8 -6 -4 -2
2
4
6
x
-20
x
-40
1
-60
-80
–2
-100
8.
Translate right 3 units, compress vertically
by a factor of 12 , and translate up 2 units.
9.
y = (x + 3)2 − 9; translate left 3 units and
down 9 units.
y
y
15
10
5
-8
2
-6
-4
-2
2
-5
4
x
x
13
10.
Chapter 1
y = (x + 3)2 − 19; translate left 3 units
and down 19 units.
11.
y = −(x − 1)2 + 2; translate right 1 unit,
reflect over x-axis, translate up 2 units.
y
y
–4
x
2
x
–5
-2
-1
1
2
3
4
-2
-4
-6
12.
y = 12 [(x − 1)2 + 2]; translate left 1 unit
and up 2 units, compress vertically by
a factor of 12 .
13.
Translate left 1 unit, reflect over x-axis,
translate up 3 units.
y
y
2
1
2
1
2
4
6
8 10 12
x
x
1
14.
Translate right 4 units and up 1 unit.
15.
Compress vertically by a factor of 12 ,
translate up 1 unit.
y
y
2
4
1
4
x
10
1
16.
Stretch vertically by a factor of
reflect over x-axis.
√
3 and
17.
2
3
x
Translate right 3 units.
y
10
y
2
x
-1
2
-10
4
6
x
Exercise Set 1.4
18.
14
Translate right 1 unit and reflect
over x-axis.
19.
Translate left 1 unit, reflect over x-axis,
translate up 2 units.
y
y
12
10
8
6
4
2
2
x
-2
2
-4
-3
-2
-1
-4
20.
y = 1 − 1/x; reflect over x-axis, translate
up 1 unit.
21.
1
-2
-4
-6
-8
x
2
Translate left 2 units and down 2 units.
y
y
5
-4
-2
x
2
x
-2
-5
22.
Translate right 3 units, reflect over x-axis,
translate up 1 unit.
23.
Stretch vertically by a factor of 2, translate
right 1 unit and up 1 unit.
y
y
1
4
x
5
-1
2
x
2
24.
y = |x − 2|; translate right 2 units.
25.
Stretch vertically by a factor of 2, reflect
over x-axis, translate up 2 units.
y
y
2
4
1
3
2
x
2
4
1
-2
2
-1
x
15
26.
Chapter 1
Translate right 2 units and down 3 units.
27.
Translate left 1 unit and up 2 units.
y
y
x
2
3
2
–2
1
x
-3
28.
-2
-1
1
Translate right 2 units, reflect over x-axis.
y
1
x
4
–1
½
29.
y
(a)
(b) y =
2
0 if x ≤ 0
2x if 0 < x
x
-1
30.
1
y
x
2
–5
31.
x2 + 2x + 1, all x; 2x − x2 − 1, all x; 2x3 + 2x, all x; 2x/(x2 + 1), all x
32.
3x − 2 + |x|, all x; 3x − 2 − |x|, all x; 3x|x| − 2|x|, all x; (3x − 2)/|x|, all x 6= 0
33.
√
√
3 x − 1, x ≥ 1; x − 1, x ≥ 1; 2x − 2, x ≥ 1; 2, x > 1
34.
(2x2 + 1)/x(x2 + 1), all x 6= 0; −1/x(x2 + 1), all x 6= 0; 1/(x2 + 1), all x 6= 0; x2 /(x2 + 1), all x 6= 0
35.
(a) 3
(b) 9
(c) 2
(d) 2
36.
(a) π − 1
(b) 0
(c) −π 2 + 3π − 1
(d) 1
Exercise Set 1.4
37.
38.
16
(a) t4 + 1
(b) t2 + 4t + 5
(c) x2 + 4x + 5
(e) x2 + 2xh + h2 + 1
(f ) x2 + 1
(g) x + 1
(a)
(e)
√
√
4
p√
5s + 2
(b)
x
(f ) 0
√
(c) 3 5x
√
(g) 1/ 4 x
x+2
1
+1
x2
(h) 9x2 + 1
(d)
√
(d) 1/ x
(h) |x − 1|
39.
2x2 −2x+1, all x; 4x2 +2x, all x
41.
1 − x, x ≤ 1;
43.
1
1
1
1
, x 6= , 1; −
− , x 6= 0, 1
1 − 2x
2
2x 2
44.
45.
x−6 + 1
46.
47.
(a) g(x) =
48.
(a) g(x) = x + 1, h(x) = x2
(b) g(x) = 1/x, h(x) = x − 3
49.
(a) g(x) = x2 , h(x) = sin x
(b) g(x) = 3/x, h(x) = 5 + cos x
50.
(a) g(x) = 3 sin x, h(x) = x2
(b) g(x) = 3x2 + 4x, h(x) = sin x
51.
(a) f (x) = x3 , g(x) = 1 + sin x, h(x) = x2
(b) f (x) =
52.
(a) f (x) = 1/x, g(x) = 1 − x, h(x) = x2
(b) f (x) = |x|, g(x) = 5 + x, h(x) = 2x
53.
√
√
1 − x2 , |x| ≤ 1
40. 2 − x6 , all x; −x6 + 6x4 − 12x2 + 8, all x
42.
x2
x2 + 3 − 3, |x| ≥
x
1
, x 6= 0; + x, x 6= 0
+1
x
x
x+1
54.
1
-1
1
2
3
x
-1
-2
-3
-4
55.
Note that f (g(−x)) = f (−g(x)) = f (g(x)), so f (g(x)) is even.
y
f (g(x))
1
–3
1
–1
–1
–3
x
√
{−2, −1, 0, 1, 2, 3}
2
-3 -2
√ √
6; x, x ≥ 3
(b) g(x) = |x|, h(x) = x2 − 3x + 5
x, h(x) = x + 2
y
p√
x, g(x) = 1 − x, h(x) =
√
3
x
17
56.
Chapter 1
Note that g(f (−x)) = g(f (x)), so g(f (x)) is even.
y
3
g( f (x))
1
x
–3
1
–1
3
–1
–2
57.
f (g(x)) = 0 when g(x) = ±2, so x = ±1.4; g(f (x)) = 0 when f (x) = 0, so x = ±2.
58.
f (g(x)) = 0 at x = −1 and g(f (x)) = 0 at x = −1
59.
6xh + 3h2
3(x + h)2 − 5 − (3x2 − 5)
=
= 6x + 3h
h
h
60.
2xh + h2 + 6h
(x + h)2 + 6(x + h) − (x2 + 6x)
=
= 2x + h + 6
h
h
61.
1/(x + h) − 1/x
x − (x + h)
−1
=
=
h
xh(x + h)
x(x + h)
62.
1/(x + h)2 − 1/x2
x2 − (x + h)2
2x + h
= 2
=− 2
h
x h(x + h)2
x (x + h)2
63.
(a) the origin
64.
(a)
(b) the x-axis
(c) the y-axis
(b)
y
(c)
y
x
65.
(a)
66.
(a)
x
f (x)
−3
1
−2
−5
(d) none
y
x
x
−1
−1
0
0
1
−1
2
−5
3
1
(b)
y
x
f (x)
(b)
−3
1
−2
5
−1
−1
0
0
1
1
2
−5
y
x
x
67.
(a) even
(b) odd
68.
neither; odd; even
69.
(a)
(c)
f (−x) = (−x)2 = x2 = f (x), even
f (−x) = | − x| = |x| = f (x), even
(e)
f (−x) =
(f )
f (−x) = 2 = f (x), even
(c) odd
(d) neither
(b) f (−x) = (−x)3 = −x3 = −f (x), odd
(d) f (−x) = −x + 1, neither
x3 + x
(−x)3 − (−x)
=
−
= −f (x), odd
1 + (−x)2
1 + x2
3
-1
Exercise Set 1.4
70.
71.
18
(a)
x-axis, because x = 5(−y)2 + 9 gives x = 5y 2 + 9
(b)
x-axis, y-axis, and origin, because x2 − 2(−y)2 = 3, (−x)2 − 2y 2 = 3, and
(−x)2 − 2(−y)2 = 3 all give x2 − 2y 2 = 3
(c)
origin, because (−x)(−y) = 5 gives xy = 5
(a)
y-axis, because (−x)4 = 2y 3 + y gives x4 = 2y 3 + y
(b)
origin, because (−y) =
(c)
x-axis, y-axis, and origin because (−y)2 = |x| − 5, y 2 = | − x| − 5, and (−y)2 = | − x| − 5 all give
y 2 = |x| − 5
72.
(−x)
x
gives y =
2
3 + (−x)
3 + x2
73.
3
-4
4
2
-3
3
–3
74.
(a)
-2
Whether we replace x with −x, y with −y, or both, we obtain the same equation, so by Theorem
1.4.3 the graph is symmetric about the x-axis, the y-axis and the origin.
(b) y = (1 − x2/3 )3/2
(c)
For quadrant II, the same; for III and IV use y = −(1 − x2/3 )3/2 . (For graphing it may be helpful
to use the tricks that precede Exercise 29 in Section 1.3.)
75.
76.
y
y
5
2
4
3
2
1
-1
77.
1
2
3
4
x
x
2
(a)
(b)
y
y
2
1
O
C
c
o
x
O
C
c
o
x
19
78.
Chapter 1
(a)
(b)
y
y
1
1
x
-1
-1
1
1
2 3
-1
(c)
x
(d)
y
y
1
3
−π
π/2
x
-1
1
x
-1
1
79.
Yes, e.g. f (x) = xk and g(x) = xn where k and n are integers.
80.
If x ≥ 0 then |x| = x and f (x) = g(x). If x < 0 then f (x) = |x|p/q if p is even and f (x) = −|x|p/q if p
is odd; in both cases f (x) agrees with g(x).
EXERCISE SET 1.5
1.
2.
3 3 − (8/3)
1 0 − (8/3)
2
3−0
=− ,
=− ,
=
0−2
2 0−6
18
2−6
3
(b) Yes; the first and third slopes above are negative reciprocals of each other.
(a)
−1 − 3
3−3
−1 − 3
−1 − (−1)
= 0,
= 2,
= 0,
=2
−3 − 5
5−7
7 − (−1)
−3 − (−1)
(b) Yes; there are two pairs of equal slopes, so two pairs of parallel lines.
(a)
3. III < II < IV < I
4.
III < IV < I < II
5.
(a)
−5 − (−1)
1 − (−1)
1 − (−5)
= 2,
= 2,
= 2. Since the slopes connecting all pairs of points are
1 − (−2)
−2 − 0
1−0
equal, they lie on a line.
(b)
4−2
2−5
4−5
1
= −1,
= 3,
= . Since the slopes connecting the pairs of points are not
−2 − 0
0−1
−2 − 1
3
equal, the points do not lie on a line.
6.
The slope, m = −2, is obtained from
(a) If x = 9 then y = 1.
7.
The slope, m = 3, is equal to
(a) If x = 5 then y = 14.
y−5
, and thus y − 5 = −2(x − 7).
x−7
(b) If y = 12 then x = 7/2.
y−2
, and thus y − 2 = 3(x − 1).
x−1
(b) If y = −2 then x = −1/3.
Exercise Set 1.5
20
y−0
1
y−5
= or y = x/2. Also
= 2 or y = 2x − 9. Solve simultaneously
x−0
2
x−7
to obtain x = 6, y = 3.
8.
(a)
Compute the slopes:
9.
(a)
5−0
2−0
and the second is
. Since they are negatives of each other we get
1−x
4−x
2(4 − x) = −5(1 − x) or 7x = 13, x = 13/7.
The first slope is
10.
(a) 27◦
(b) 135◦
(c) 63◦
(d) 91◦
11.
(a) 153◦
(b) 45◦
(c) 117◦
(d) 89◦
12.
√
(a) m = tan φ = − 3/3, so φ = 150◦
13.
(a) m = tan φ =
14.
y = 0 and x = 0 respectively
(b) m = tan φ = 4, so φ = 76◦
√
3, so φ = 60◦
(b) m = tan φ = −2, so φ = 117◦
15.
y = −2x + 4
6
-1
1
0
16.
y = 5x − 3
17.
2
Parallel means the lines have equal slopes, so
y = 4x + 7.
12
–1
2
–8
-1
1
0
18.
The slope of both lines is −3/2, so y − 2 = (−3/2)(x − (−1)),
or 3x + 2y = 1
4
-2
1
–4
19.
1
The negative reciprocal of 5 is −1/5, so y = − x + 6.
5
12
-9
9
0
21
20.
Chapter 1
The slope of x − 4y = 7 is 1/4 whose negative reciprocal
is −4, so y − (−4) = −4(x − 3) or y + 4x = 8.
9
0
18
–3
21.
4 − (−7)
y − (−7)
=
, or y = 11x − 18
x−1
2−1
7
0
4
-9
22.
6−1
y−1
=
, or y = −5x − 9
x − (−2)
−3 − (−2)
15
–4
0
–5
23.
24.
(a) m1 = m2 = 4, parallel
(c) m1 = m2 = 5/3, parallel
(b)
m1 = 2 = −1/m2 , perpendicular
(d)
If A 6= 0 and B 6= 0 then m1 = −A/B = −1/m2 , perpendicular; if A = 0 or B = 0 (not both)
then one line is horizontal, the other vertical, so perpendicular.
(e)
neither
(a)
m1 = m2 = −5, parallel
m1 = −4/5 = −1/m2 , perpendicular
(c)
(b) m1 = 2 = −1/m2 , perpendicular
(d) If B 6= 0, m1 = m2 = −A/B; if B = 0 both
are vertical, so parallel
(e) neither
25.
(a) m = (0 − (−3))/(2 − 0)) = 3/2 so y = 3x/2 − 3
(b) m = (−3 − 0)/(4 − 0) = −3/4 so y = −3x/4
26.
(a) m = (0 − 2)/(2 − 0)) = −1 so y = −x + 2
(b) m = (2 − 0)/(3 − 0) = 2/3 so y = 2x/3
27.
5 − (−4)
= 9/10 ft/s.
10 − 0
(a)
The velocity is the slope, which is
(b)
x = −4
(c)
The line has slope 9/10 and passes through (0, −4), so has equation x = 9t/10 − 4; at t = 2,
x = −2.2.
(d)
t = 80/9
Exercise Set 1.5
22
5−1
= 2 m/s
4−2
(b) x − 1 = 2(t − 2) or x = 2t − 3
28.
(a) v =
29.
(a) The acceleration is the slope of the velocity, so a =
4
4
13
(b) v − 3 = − (t − 1), or v = − t +
3
3
3
30.
31.
(c) x = −3
4
3 − (−1)
= − ft/s2 .
1−4
3
13
(c) v =
ft/s
3
5
−1
0−5
=−
=
ft/s2 .
10 − 0
10
2
(d) t = 4 s
(a)
The acceleration is the slope of the velocity, so a =
(b)
v = 5 ft/s
(a)
It moves (to the left) 6 units with velocity v = −3 cm/s, then remains motionless for 5 s, then
moves 3 units to the left with velocity v = −1 cm/s.
(b)
vave =
(c)
Since the motion is in one direction only, the speed is the negative of the velocity, so
9
save =
cm/s.
10
(c)
v = 4 ft/s
9
0−9
=−
cm/s
10 − 0
10
32.
It moves right with constant velocity v = 5 km/h; then accelerates; then moves with constant, though
increased, velocity again; then slows down.
33.
(a)
If x1 denotes the final position and x0 the initial position, then v = (x1 − x0 )/(t1 − t0 ) = 0 mi/h,
since x1 = x0 .
(b) If the distance traveled in one direction is d, then the outward journey took t = d/40 h. Thus
2d
80t
total dist
=
=
= 48 mi/h.
save =
total time
t + (2/3)t
t + (2/3)t
(c)
34.
35.
t + (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip
(a) down, since v < 0
(a)
(b)
v
100
80
60
40
20
20 40 60 80 100 120
36.
(c) It’s constant at 32 ft/s2 .
(b) v = 0 at t = 2
x
t
t
if
0 ≤ t ≤ 10
10t
100
if 10 ≤ t ≤ 100
v=
600 − 5t if 100 ≤ t ≤ 120
23
37.
Chapter 1
(a)
y = 20 − 15 = 5 when x = 45, so 5 = 45k,
k = 1/9, y = x/9
(b)
y
0.6
0.4
0.2
2
38.
4
6
x
(c)
l = 15 + y = 15 + 100(1/9) = 26.11 in.
(d) If ymax = 15 then solve 15 = kx = x/9 for
x = 135 lb.
(a)
Since y = 0.2 = (1)k, k = 1/5 and y = x/5
(b)
y
1
2
(c)
y = 3k = 3/5 so 0.6 ft.
(d)
4
6
x
ymax = (1/2)3 = 1.5 so solve 1.5 = x/5 for
x = 7.5 tons
39.
Each increment of 1 in the value of x yields the increment of 1.2 for y, so the relationship is linear. If
y = mx + b then m = 1.2; from x = 0, y = 2, follows b = 2, so y = 1.2x + 2
40.
Each increment of 1 in the value of x yields the increment of −2.1 for y, so the relationship is linear.
If y = mx + b then m = −2.1; from x = 0, y = 10.5 follows b = 10.5, so y = −2.1x + 10.5
41.
(a) With TF as independent variable, we have
(b) 5/9
(c) Set TF = TC =
(d) 37◦ C
TC − 100
5
0 − 100
=
, so TC = (TF − 32).
TF − 212
32 − 212
9
5
(TF − 32) and solve for TF : TF = TC = −40◦ (F or C).
9
(a)
One degree Celsius is one degree Kelvin, so the slope is the ratio 1/1 = 1. Thus TC = TK −273.15.
(b)
TC = 0 − 273.15 = −273.15◦ C
43.
(a)
5.9 − 1
p−1
=
, or p = 0.098h + 1
h−0
50 − 0
44.
(a)
42.
(b)
45.
(a)
(b)
46.
(b)
when p = 2, or h = 1/0.098 ≈ 10.20 m
133.9 − 123.4
R − 123.4
=
, so R = 0.42T + 115.
T − 20
45 − 20
T = 32.38◦ C
0.75 − 0.80
r − 0.80
=
, so r = −0.0125t + 0.8
t−0
4−0
64 days
(a)
Let the position at rest be y0 . Then y0 + y = y0 + kx; with x = 11 we get y0 + kx = y0 + 11k = 40,
and with x = 24 we get y0 + kx = y0 + 24k = 60. Solve to get k = 20/13 and y0 = 300/13.
(b)
300/13 + (20/13)W = 30, so W = (390 − 300)/20 = 9/2 g.
Exercise Set 1.6
47.
(a)
24
For x trips we have C1 = 2x and
C2 = 25 + x/4
(b)
2x = 25 + x/4, or x = 100/7, so the commuter pass becomes worthwhile at x = 15.
C
60
40
20
x
5 10 15 20 25 30
48.
If the student drives x miles, then the total costs would be CA = 4000 + (1.25/20)x and
CB = 5500 + (1.25/30)x. Set 4000 + 5x/80 = 5500 + 5x/120 and solve for x = 72, 000 mi.
49.
(a)
H ≈ 20000/110 ≈ 181
(b) One light year is 9.408 × 1012 km and t =
1
1
9.408 × 1018 km
d
=
=
=
v
H
20km/s/Mly
20km/s
= 4.704 × 1017 s = 1.492 × 1010 years.
(c)
The Universe would be even older.
EXERCISE SET 1.6
1.
(a)
(c)
y = 3x + b
(b)
y
y = 3x + 6
10
y = 3x + 2
5
-2
y = 3x + 6
y = 3x - 4
-1
1
2
x
-5
-10
2.
3.
1
Since the slopes are negative reciprocals, y = − x + b.
3
(a)
(c)
y = mx + 2
(b)
m = tan φ = tan 135◦ = −1, so y = −x + 2
(b)
y = m(x − 1)
(d)
2x + 4y = C
y
m = 1.5
5
m = –1
4
m=1
3
2
1
-2
-1
1
2
x
-1
4.
(a)
y = mx
(c) y = −2 + m(x − 1)
25
5.
Chapter 1
The slope is −1.
(a)
(b)
The y-intercept is y = −1.
y
y
4
5
4
3
2
1
-2
(c)
-1
2
-2
1
-1
-2
-3
-1
1
-2
x
2
x
2
-4
-6
They pass through the point (−4, 2).
(d)
The x-intercept is x = 1.
y
y
3
6
2
4
1
x
2
0.5
-4
2
1.5
-1
x
-6
1
-2
-2
-2
-3
6.
(a)
horizontal lines
(b)
The y-intercept is y = −1/2.
y
y
2
x
x
2
-1
(c)
The x-intercept is x = −1/2.
(d)
They pass through (−1, 1).
y
y
(-1,1)
1
1
x
1
x
-2
1
7.
Let the line be tangent to the circle at the point (x0 , y0 ) where x20 + y02 = 9. The slope of the tangent
line is the negative reciprocal of y0 /x0 (why?), so m = −x0 /y0 and y = −(x0 /y0 )x + b. Substituting
p
9 − x0 x
.
the point (x0 , y0 ) as well as y0 = ± 9 − x20 we get y = ± p
9 − x20
8.
Solve the simultaneous equations to get the point (−2, 1/3) of intersection. Then y =
1
+ m(x + 2).
3
Exercise Set 1.6
9.
26
The x-intercept is x = 10 so that with depreciation at 10% per year the final value is always zero, and
hence y = m(x − 10). The y-intercept is the original value.
y
x
2
4
8
6
10
10.
A line through (6, −1) has the form y +1 = m(x−6). The intercepts are x = 6+1/m and y = −6m−1.
Set −(6+1/m)(6m+1) = 3, or 36m2 +15m+1 = (12m+1)(3m+1) = 0 with roots m = −1/12, −1/3;
thus y + 1 = −(1/3)(x − 6) and y + 1 = −(1/12)(x − 6).
11.
(a) VI
12.
In all cases k must be positive, or negative values would appear in the chart. Only kx−3 decreases, so
that must be f (x). Next, kx2 grows faster than kx3/2 , so that would be g(x), which grows faster than
h(x) (to see this, consider ratios of successive values of the functions). Finally, experimentation (a
spreadsheet is handy) for values of k yields (approximately) f (x) = 10x−3 , g(x) = x2 /2, h(x) = 2x1.5 .
13.
(a)
(b) IV
(c) III
(d) V
y
-2
y
-1
1
2
x
30
-10
20
-20
10
-30
x
-2
-1
1
2
-10
-40
-20
-50
-30
-60
y
(b)
y
10
4
8
2
6
x
-4
-2
2
4
4
-2
2
-4
x
-4
-2
2
4
y
(c)
y
2
1.5
1
0.5
1
x
-1
1
-1
-2
2
3
x
1
2
3
(e) I
(f ) II
27
14.
Chapter 1
y
(a)
y
80
40
40
-2
x
x
2
-2
2
-40
y
(b)
y
4
x
-2
2
-2
x
-2
2
-4
-4
y
(c)
y
4
x
-1
x
-3
15.
-2
-1 -2
(a)
(b)
y
10
10
6
5
4
x
-2
y
(c)
y
-1
1
5
2
2
x
-2
-1
x
-5
-2
-1
1
1
2
-5
2
-2
-10
-10
16.
(a)
(b)
y
(c)
y
y
3
2
2
x
2
x
2
x
2
17.
y
(a)
80
60
40
20
6
x
4
2
x
-3
-2
-1
y
(b)
8
1
-1
-20
-40
-60
-80
1
2
3
4
5
Exercise Set 1.6
28
(c)
(d)
y
-5 -4 -3 -2 -1
1 2 3
x
y
40
-10
20
-20
x
1
-30
2
3
4
5
-20
-40
-40
-50
18.
(a)
(b)
y
y
1
1
x
1
x
-2
y
(c)
1
y
(d)
4
4
x
-2
-6
2
x
-2
-4
19.
y
(a)
(b)
1.5
y
1
1
0.5
0.5
-3
-2
-1
1
x
x
1
-0.5
-1.5
30
30
20
20
10
x
x
1
-10
2
-2
3
-1
-10
-20
-20
-30
-30
y
(a)
6
5
y
(d)
10
20.
4
-1
y
-1
3
-0.5
-1
(c)
2
1
2
2
3
y
(b)
30
20
10
10
x
-10
1
3
4
x
-1
1
-10
-20
-30
29
Chapter 1
y
(c)
y
(d)
106
15
10
x
-1
1
5
x
-106
-4
-2
2
y
21.
15
10
5
x
-4
22.
-2
2
y
(a)
1.5
1
1
0.5
0.5
-2
23.
y
(b)
1.5
2
x
x
-2
2
√
t = 0.445 d
2.3
0
25
0
24.
(a) t = 0.373r1.5
25.
(a)
(c)
(b) 238,000 km
N·m
V (L)
0.25
P (N/m2 ) 80 × 103
(c) 1.89 days
(b) 20 N·m
0.5
40 × 103
1.0
20 × 103
1.5
13.3 × 103
2.0
10 × 103
26.
If the side of the square base is x and the height of the container is y then V = x2 y = 100; minimize
A = 2x2 + 4xy = 2x2 + 400/x. A graphing utility with a zoom feature suggests that the solution is a
1
cube of side 100 3 cm.
27.
(a)
F = k/x2 so 0.0005 = k/(0.3)2 and
k = 0.000045 N·m2 .
(b)
0.000005 N
Exercise Set 1.6
(c)
30
F
(d)
When they approach one another, the force
becomes infinite; when they get far apart it
tends to zero.
(b)
W = C/50002 = (3.2 × 1010 )/(25 × 106 ) =
1280 lb.
(d)
No, but W is very small when x is large.
10-5
5
28.
(a)
10
d
2000 = C/(4000)2 , so C = 3.2 × 1010 lb·mi2
(c)
W
20000
15000
10000
5000
x
2000 4000 6000 8000 10000
29.
(a) II; y = 1, x = −1, 2
(c) IV; y = 2
30.
The denominator has roots x = ±1, so x2 − 1 is the denominator. To determine k use the point (0, −1)
to get k = 1, y = 1/(x2 − 1).
31.
Order the six trigonometric functions as sin, cos, tan, cot, sec,
(a) pos, pos, pos, pos, pos, pos
(b) neg,
(c) pos, neg, neg, neg, neg, pos
(d) neg,
(e) neg, neg, pos, pos, neg, neg
(f ) neg,
32.
(a)
(c)
(e)
neg, zero, undef, zero, undef, neg
zero, neg, zero, undef, neg, undef
neg, neg, pos, pos, neg, neg
(b)
(d)
(f )
33.
(a)
(c)
sin(π − x) = sin x; 0.588
sin(2π + x) = sin x; 0.588
p
sin 2x = ±2 sin x 1 − sin2 x; use the +
sign for x small and positive; 0.951
(b) cos(−x) = cos x; 0.924
(d) cos(π − x) = − cos x; −0.924
(a)
sin(3π + x) = − sin x; −0.588
(b)
(c)
sin(8π + x) = sin x ; 0.588
(e)
cos(3π + 3x) = −4 cos3 x + 3 cos x; −0.384
(a)
−a
(c)
(e)
−c
−b
(e)
34.
35.
√
(g) ±2b 1 − b2
(i) 1/b
(k) 1/c
(b) I; y = 0, x = −2, 3
(d) III; y = 0, x = −2
(f )
csc:
zero, undef, zero, undef, neg
pos, neg, neg, pos, neg
pos, neg, neg, pos, neg
pos, neg, neg, neg, neg, pos
pos, zero, undef, zero, undef, pos
neg, neg, pos, pos, neg, neg
cos2 x = 1 − sin2 x; 0.654
cos(−x − 2π) = cos x; 0.924
p
(d) sin(x/2) = ± (1 − cos x)/2; use the negative sign for x small and negative; −0.195
1 − cos2 x
; 0.172
(f ) tan2 x =
cos2 x
(b)
b
(h)
(j)
(l)
2b2 − 1
−1/a
(1 − b)/2
√
(d) ± 1 − a2
(f ) −a
31
Chapter 1
36.
(a)
The distance is 36/360 = 1/10th of a great
circle, so it is (1/10)2πr = 2, 513.27 mi.
37.
If the arc length is 1, then solve the ratio
38.
The distance travelled is equal to the length of that portion of the circumference of the wheel which
touches the road, and that is the fraction 225/360 of a circumference, so a distance of (225/360)(2π)3 =
11.78 ft
39.
The second quarter revolves twice (720◦ ) about its own center.
40.
Add r to itself until you exceed 2πr; since 6r < 2πr < 7r, you can cut off 6 pieces of pie, but there’s
not enough for a full seventh piece. We conclude that there is no exact solution of the equation
‘One pie = 2πr’.
41.
(a) y = 3 sin(x/2)
(b) y = 4 cos 2x
(c) y = −5 sin 4x
42.
(a) y = 1 + cos πx
(b) y = 1 + 2 sin x
(c) y = −5 cos 4x
43.
(a) y = sin(x + π/2)
(b) y = 3 + 3 sin(2x/9)
(c) y = 1 + 2 sin(2(x − π/4))
44.
√
V = 120 2 sin(120πt)
45.
(a) 3, π/2, 0
(b) 36/360 = 1/10
2πr
x
=
to get x ≈ 80, 936 km.
1
29.5
(b) 2, 2, 0
y
(c) 1, 4π, 0
y
2
3
3
1
x
2
x
π/2
2
4
1
-2
-2
2π
46.
(a) 4, π, 0
y
π/4
4
2
x
3π/4
5π/4
7π/4
π/3
-2
2π/3
π
x
x
3π/2
-0.2
9π/2 15π/2 21π/2
-2
-4
-4
47.
x
y
0.4
2
6π
(c) 4, 6π, −6π
(b) 1/2, 2π/3, π/3
y
4π
(a) A sin(ωt + θ) = A sin(ωt) cos θ + A cos(ωt) sin θ = A1 sin(ωt) + A2 cos(ωt)
p
(b) A1 = A cos θ, A2 = A sin θ, so A = A21 + A22 and θ = tan−1 (A2 /A1 ).
√
1
(c) A = 5 13/2, θ = tan−1 √ ;
10
2 3
√
µ
¶
1
5 13
sin 2πt + tan−1 √
x=
2
2 3
^
-10
6
Exercise Set 1.7
48.
32
three; x = 0, x = ±1.8955
3
-3
3
–3
EXERCISE SET 1.7
1.
x + 1 = t = y − 1, y = x + 2
(a)
(c)
t
0 1 2 3 4 5
x −1 0 1 2 3 4
y
1 2 3 4 5 6
(c)
t 0 0.2500 0.50
0.7500
1
x 1 0.7071 0.00 −0.7071 −1
y 0 0.7071 1.00
0.7071
0
y
t=5
6
t=4
4
t=3
t=2
2
t=1
t=0
x
2
2.
4
x2 + y 2 = 1
(a)
y
1
t=.75
t=.5
t=.25
t=0
x
t=1
-1
3.
1
t = (x + 4)/3; y = 2x + 10
4.
t = x + 3; y = 3x + 2, −3 ≤ x ≤ 0
y
y
12
(0, 2)
-8
5.
6
x
x
(-3, -7)
cos t = x/2, sin t = y/5; x2 /4 + y 2 /25 = 1
6.
t = x2 ; y = 2x2 + 4, x ≥ 0
y
y
5
8
4
x
x
-5
5
-5
1
33
7.
Chapter 1
cos t = (x − 3)/2, sin t = (y − 2)/4;
(x − 3)2 /4 + (y − 2)2 /16 = 1
8.
sec2 t − tan2 t = 1; x2 − y 2 = 1,
x ≤ −1 and y ≥ 0
y
y
6
x
-1
x
7
-2
9.
cos 2t = 1 − 2 sin2 t; x = 1 − 2y 2 , −1 ≤ y ≤ 1
10.
t = (x − 3)/4; y = (x − 3)2 − 9
y
y
1
x
x
1
-1
-1
(3, -9)
11.
x/2 + y/3 = 1, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3
12.
y = x − 1, x ≥ 1, y ≥ 0
y
y
3
1
x
2
13.
1
x
x = 5 cos t, y = −5 sin t, 0 ≤ t ≤ 2π
14.
x = cos t, y = sin t, π ≤ t ≤ 3π/2
5
1
-7.5
7.5
-1
-5
15.
1
–1
x = 2, y = t
16.
x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ 2π
2
3
-2
0
-1
2
3
–3
Exercise Set 1.7
17.
34
x = t2 , y = t, −1 ≤ t ≤ 1
18.
x = 1 + 4 cos t, y = −3 + 4 sin t, 0 ≤ t ≤ 2π
2
1
-7
0
8
1
-8
-1
19.
(a)
IV, because x always increases whereas y oscillates.
(b) II, because (x/2)2 + (y/3)2 = 1, an ellipse.
20.
21.
(c)
V, because x2 + y 2 = t2 increases in magnitude while x and y keep changing sign.
(d)
VI; examine the cases t < −1 and t > −1 and you see the curve lies in the first, second and
fourth quadrants only.
(e)
III because y > 0.
(f )
I; since x and y are bounded, the answer must be I or II; but as t runs, say, from 0 to π, x goes
directly from 2 to −2, but y goes from 0 to 1 to 0 to −1 and back to 0, which describes I but
not II.
(a)
from left to right
(b)
counterclockwise
(c)
counterclockwise
(d)
As t travels from −∞ to −1, the curve goes from (near) the origin in the third quadrant and
travels up and left. As t travels from −1 to +∞ the curve comes from way down in the second
quadrant, hits the origin at t = 0, and then makes the loop clockwise and finally approaches the
origin again as t → +∞.
(e)
from left to right
(f )
Starting, say, at (1, 0), the curve goes up into the first quadrant, loops back through the origin
and into the third quadrant, and then continues the figure-eight.
(a)
(b)
14
-35
8
0
(c)
22.
√
x = 0 when t = 0, 2 3.
(a)
t 0 1 2 3
4
5
x 0 5.5 8 4.5 −8 −32.5
y 1 1.5 3 5.5
9
13.5
(d)
5
-2
14
0
√
for 0 < t < 2 2
(e)
at t = 2
(b)
y is always ≥ 1 since cos t ≤ 1
(c)
greater than 5, since cos t ≥ −1
35
23.
Chapter 1
(a)
o
(b)
3
0
20
-1
1
O
-5
24.
(a)
-2.3
6
(b)
1.7
2.3
-10
10
-1.7
^
(c)
x − x0
y − y0
=
x1 − x0
y 1 − y0
x = 1 + t, y = −2 + 6t
26.
(a)
x = −3 − 2t, y = −4 + 5t
27.
(a)
|R − P |2 = (x − x0 )2 + (y − y0 )2 = t2 [(x1 − x0 )2 + (y1 − y0 )2 ] and |Q − P |2 = (x1 − x0 )2 + (y1 − y0 )2 ,
so r = |R − P | = |Q − P |t = qt.
(b)
t = 1/2
25.
28.
29.
30.
(a)
x = 2 + t, y = −1 + 2t
(a) (5/2, 0)
(b) (9/4, −1/2)
(b)
Set t = 0 to get (x0 , y0 ); t = 1 for (x1 , y1 ).
(d) x = 2 − t, y = 4 − 6t
(b)
(c)
x = at, y = b(1 − t)
t = 3/4
(c) (11/4, 1/2)
The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1 coincide.
y − y0
y1 − y0
t − t0
to obtain
=
.
t1 − t0
x − x0
x1 − x0
(c) x = 3 − 2(t − 1), y = −1 + 5(t − 1)
(b) from (x0 , y0 ) to (x1 , y1 )
(a) Eliminate
5
0
5
-2
31.
(a)
y−d
x−b
=
a
c
y
(b)
3
2
1
1
2
3
x
Exercise Set 1.7
32.
36
(a) If a = 0 the line segment is vertical; if c = 0 it is horizontal.
(b) The curve degenerates to the point (b, d).
y
33.
2
1.5
1
0.5
0.5
34.
35.
1
x
x = 1/2 − 4t,
x = −1/2,
x = −1/2 + 4(t − 1/2),
y = 1/2
y = 1/2 − 4(t − 1/4)
y = −1/2
for 0 ≤ t ≤ 1/4
for 1/4 ≤ t ≤ 1/2
for 1/2 ≤ t ≤ 3/4
x = 1/2,
y = −1/2 + 4(t − 3/4) for 3/4 ≤ t ≤ 1
(b) x = −1 + 4 cos t, y = 2 + 3 sin t
(a) x = 4 cos t, y = 3 sin t
3
(c)
5
-4
4
-5
3
-3
36.
-1
(a) t = x/(v0 cos α), so y = x tan α − gx2 /(2v02 cos2 α).
y
(b)
12000
10000
8000
6000
4000
2000
40000
80000
x
37.
√
√
(a) From Exercise 36, x = 400 2t, y = 400 2t − 4.9t2 .
38.
(a)
(b)
15
-25
(b) 16,326.53 m
15
-25
25
25
–15
(c)
–15
15
15
-25
25
–15
a = 3, b = 2
(c) 65,306.12 m
-25
15
25
–15
a = 2, b = 3
-25
25
–15
a = 2, b = 7
37
39.
Chapter 1
Assume that a 6= 0 and b 6= 0; eliminate the parameter to get (x − h)2 /a2 + (y − k)2 /b2 = 1. If |a| = |b|
the curve is a circle with center (h, k) and radius |a|; if |a| =
6 |b| the curve is an ellipse with center
(h, k) and major axis parallel to the x-axis when |a| > |b|, or major axis parallel to the y-axis when
|a| < |b|.
(a) ellipses with a fixed center and varying axes of symmetry
(b) (assume a 6= 0 and b 6= 0) ellipses with varying center and fixed axes of symmetry
(c) circles of radius 1 with centers on the line y = x − 1
40.
Refer to the diagram to get bθ = aφ, θ = aφ/b but θ − α = φ + π/2
so α = θ − φ − π/2 = (a/b − 1)φ − π/2
x = (a − b) cos φ − b sin α
µ
= (a − b) cos φ + b cos
y = (a − b) sin φ − b cos µ
α
= (a − b) sin φ − b sin
41.
a−b
b
a−b
b
y
a−b
¶
θ
φ
φ,
α
bθ
aφ
x
¶
φ.
y
(a)
a
-a
a
x
-a
(b)
Use b = a/4 in the equations of Exercise 40 to get
1
3
1
3
x = a cos φ + a cos 3φ, y = a sin φ − a sin 3φ;
4
4
4
4
but trigonometric identities yield cos 3φ = 4 cos3 φ − 3 cos φ, sin 3φ = 3 sin φ − 4 sin3 φ,
so x = a cos3 φ, y = a sin3 φ.
(c)
x2/3 + y 2/3 = a2/3 (cos2 φ + sin2 φ) = a2/3
y
42.
y
y
1
50
50
0
-3
-2
-1
x
x
-2
-1
-1
1
-50
-50
-1
a = −1
a = −2
a=0
y
y
50
50
x
1
0
2
1
2
-50
-50
a=1
a=2
3
x
x
Supplementary Exercises 1
38
CHAPTER 1 SUPPLEMENTARY EXERCISES
1.
1940-45; the greatest five-year slope
2.
(a)
(c)
f (−1) = 3.3, g(3) = 2
x < −2, x > 3
(b)
(d)
(e)
the domain is −4 ≤ x ≤ 4.1, the range is
−3 ≤ y ≤ 5
(f )
3.
4.
T
x = −3, 3
the domain is −5 ≤ x ≤ 5 and the range is
−5 ≤ y ≤ 4
f (x) = 0 at x = −3, 5; g(x) = 0 at
x = −3, 2
x
70
60
50
40
0
2
4
6
t
t
5
8
13
5.
If the side has length x and height h, then V = 8 = x2 h, so h = 8/x2 . Then the cost C = 5x2 +
2(4)(xh) = 5x2 + 64/x.
6.
Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paint
to be used is proportional to the area covered. If P is the amount of paint to be used, P = kπr2 . The
constant k depends on physical factors, such as the thickness of the paint, absorption of the wood,
etc.
7.
y
5
-5
-1
5
x
8.
Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. Then the plastic has
4
4
4
volume equal to the difference of the volumes, i.e. V = π(r + h)3 − πr3 = πh[3r2 + 3rh + h2 ] in3 .
3
3
3
9.
(a)
The base has sides (10 − 2x)/2 and 6 − 2x, and the height is x, so V = (6 − 2x)(5 − x)x ft3 .
(b) From the picture we see that x < 5 and 2x < 6, so 0 < x < 3.
(c)
3.57 ft ×3.79 ft ×1.21 ft
10.
{x 6= 0} and ∅ (the empty set)
11.
impossible; we would have to solve 2(3x − 2) − 5 = 3(2x − 5) − 2, or −9 = −17
12.
(a)
13.
1/(2 − x2 )
15.
(3 − x)/x
(b) no; f (g(x)) can be defined at x = 1,
whereas g, and therefore f ◦ g, requires
x 6= 1
14.
g(x) = x2 + 2x
x
−4 −3 −2 −1
0
1
2
3
4
f (x)
0 −1
2
1
3 −2 −3
4 −4
g(x)
3
2
1 −3 −1 −4
4 −2
0
(f ◦ g)(x)
4 −3 −2 −1
1
0 −4
2
3
(g ◦ f )(x) −1 −3
4 −4 −2
1
2
0
3
39
16.
Chapter 1
(a)
y = |x − 1|, y = |(−x) − 1| = |x + 1|,
y = 2|x + 1|, y = 2|x + 1| − 3,
y = −2|x + 1| + 3
y
(b)
3
1
-3
-1
2
x
-1
17.
18.
19.
20.
(a) even × odd = odd
(c) even + odd is neither
(b) a square is even
(d) odd × odd = even
π 3π π 5π 7π 11π
(a) y = cos x − 2 sin x cos x = (1 − 2 sin x) cos x, so x = ± , ± , ,
,− ,−
2
2 6 6
6
6
π
3π
π √
5π √
7π √
11π √
(b) (± , 0), (± , 0), ( , 3/2), ( , − 3/2), (− , − 3/2), (−
, 3/2)
2
2
6
6
6
6
(a)
If x denotes the distance from A to the base of the tower, and y the distance from B to the
base, then x2 + d2 = y 2 . Moreover h = x tan α = y tan β, so d2 = y 2 − x2 = h2 (cot2 β − cot2 α),
d2
d2 sin2 α sin2 β
. The trigonometric identity
h2 =
=
2
2
2
cot β − cot α
sin α cos2 β − cos2 α sin2 β
d sin α sin β
sin(α + β) sin(α − β) = sin2 α cos2 β − cos2 α sin2 β yields h = p
.
sin(α + β) sin(α − β)
(b)
295.72 ft.
(a)
3π
2π
(t − 101) =
, or t = 374.75,
when
365
2
which is the same date as t = 9.75, so during the night of January 10th-11th
(b)
y
60
40
20
t
100
200
300
-20
(c)
from t = 0 to t = 70.58 and from t = 313.92 to t = 365 (the same date as t = 0) , for a total of
about 122 days
21.
C is the highest nearby point on the graph; zoom to find that the coordinates of C are (2.0944, 1.9132).
Similarly, D is the lowest nearby point, and its coordinates are (4.1888, 1.2284). Since f (x) = 12 x−sin x
is an odd function, the coordinates of B are (−2.0944, −1.9132) and those of A are (−4.1888, −1.2284).
22.
Let y = A + B sin(at + b). Since the maximum and minimum values of y are 35 and 5, A + B = 35
and A − B = 5, so A = 20, B = 15. The period is 12 hours, so 12a = 2π and a = π/6. The
maximum occurs at t = 2, so 1 = sin(2a + b) = sin(π/3 + b), π/3 + b = π/2, b = π/2 − π/3 = π/6 and
y = 20 + 15 sin(πt/6 + π/6).
23.
(a)
The circle of radius 1 centered at (a, a2 ); therefore, the family of all circles of radius 1 with
centers on the parabola y = x2 .
(b)
All parabolas which open up, have latus rectum equal to 1 and vertex on the line y = x/2.
24.
(a) x = f (1 − t), y = g(1 − t)
y
25.
2
1
-2
-1
1
-1
-2
2
x
Supplementary Exercises 1
26.
40
Let y = ax2 + bx + c. Then 4a + 2b + c = 0, 64a + 8b + c = 18, 64a − 8b + c = 18, from which b = 0
3 2 6
and 60a = 18, or finally y =
x − .
10
5
y
27.
2
1
x
-1
1
2
-1
-2
28.
29.
(a)
R = R0 is the R-intercept, R0 k is the slope,
and T = −1/k is the T -intercept
(c) 1.1 = R0 (1 + 20/273), or R0 = 1.025
p
√
d = (x − 1)2 + ( x − 2)2 ;
d = 9.1 at x = 1.358094
(b)
−1/k = −273, or k = 1/273
(d)
T = 126.55◦ C
y
2
1
1
30.
p
d = (x − 1)2 + 1/x2 ;
d = 0.82 at x = 1.380278
x
2
y
2
1.8
1.6
1.4
1.2
1
0.8
0.5
1
1.5
2
2.5
3
31.
w = 63.9V , w = 63.9πh2 (5/2 − h/3); h = 0.48 ft when w = 108 lb
32.
(a)
W
(b)
w = 63.9πh2 (5/2 − h/3); at h = 5/2,
w = 2091.12 lb
(b)
N = 80 when t = 9.35 yrs
(c)
220 sheep
4000
3000
2000
1000
1
33.
2
3
4
h
5
N
(a)
200
150
100
50
t
10
20
30
40
50
41
34.
Chapter 1
(a)
T
(b)
T = 17◦ F, 27◦ F, 32◦ F
(b)
T = 3◦ F, −11◦ F, −18◦ F, −22◦ F
(c)
v = 35, 19, 12, 7 mi/h
10
v
20
35.
(a)
WCI
20
v
10
20
30
40
50
-20
36.
The domain is the set of all x, the range is −0.1746 ≤ y ≤ 0.1227.
37.
The domain is the set −0.7245 ≤ x ≤ 1.2207, the range is −1.0551 ≤ y ≤ 1.4902.
38.
(a)
39.
(a)
The potato is done in the interval
27.65 < t < 32.71.
v
(b)
91.54 min.
(b)
As t → ∞, (0.273)t → 0, and thus
v → 24.61 ft/s.
(d)
No; but it comes very close (arbitrarily
close).
25
20
15
10
5
t
1
(c)
2
3
4
5
For large t the velocity approaches c.
(e) 3.013 s
CHAPTER 1 HORIZON MODULE
0.25, 6.25 × 10−2 , 3.91 × 10−3 , 1.53 × 10−5 , 2.32 × 10−10 , 5.42 × 10−20 , 2.94 × 10−39 , 8.64 × 10−78 ,
7.46 × 10−155 , 5.56 × 10−309 ;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
4, 16, 256, 65536, 4.29 × 109 , 1.84 × 1019 , 3.40 × 1038 , 1.16 × 1077 , 1.34 × 10154 , 1.80 × 10308
1.
(a)
2.
2, 2.25, 2.2361111, 2.23606798, 2.23606798, . . .
3.
(a)
1 1 1 1 1 1
, , , , ,
2 4 8 16 32 64
4.
(a)
yn+1 = 1.05yn
(b)
y0 =$1000, y1 =$1050, y2 =$1102.50, y3 =$1157.62, y4 =$1215.51, y5 =$1276.28
(c)
yn+1 = 1.05yn for n ≥ 1
(b)
(d)
yn =
1
2n
yn = (1.05)n 1000; y15 =$2078.93
Horizon Module 1
5.
(a)
42
x1/2 , x1/4 , x1/8 , x1/16 , x1/32
(b)
They tend to the horizontal line y = 1, with
a hole at x = 0.
1.8
0
3
0
6.
(a)
(b)
7.
1
2
2
3
3
5
5
8
8
13
13
21
21
34
34
55
55
89
89
144
1, 2, 3, 5, 8, 13, 21, 34, 55, 89 ;
each new numerator is the sum of the previous two numerators.
(c)
144 233 377 610 987 1597 2584 4181 6765 10946
,
,
,
,
,
,
,
,
,
233 377 610 987 1597 2584 4181 6765 10946 17711
(d)
F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥ 2.
(e)
the positive solution
(a)
y1 = cr, y2 = cy1 = cr2 , y3 = cr3 , y4 = cr4
(b)
yn = crn
(c)
If r = 1 then yn = c for all n; if r < 1 then yn tends to zero; if r > 1, then yn gets ever larger
(tends to +∞).
8.
The first point on the curve is (c, kc(1 − c)), so y1 = kc(1 − c) and hence y1 is the first iterate. The
point on the line to the right of this point has equal coordinates (y1 , y1 ), and so the point above it on
the curve has coordinates (y1 , ky1 (1 − y1 )); thus y2 = ky1 (1 − y1 ), and y2 is the second iterate, etc.
9.
(a)
0.261, 0.559, 0.715, 0.591, 0.701
(b) It appears to approach a point somewhere near 0.65.
CHAPTER 2
Limits and Continuity
EXERCISE SET 2.1
1.
(a) −1
(d) 1
(b) 3
(e) −1
(c) does not exist
(f ) 3
2.
(a) 2
(d) 2
(b) 0
(e) 0
(c)
(f )
3.
(a)
1
(b) 1
(c)
1
(d) 1
(e)
−∞
(f )
+∞
4.
(a)
3
(b) 3
(c)
3
(d)
3
(e)
+∞
(f )
+∞
5.
(a)
0
(b)
0
(c)
0
(d) 3
(e)
+∞
(f )
+∞
6.
(a)
2
(b)
2
(c)
2
(d)
(e)
−∞
(f )
+∞
7.
(a) −∞
(d) undef
8.
(a)
+∞
(b)
+∞
(c)
+∞
(d)
9.
(a)
−∞
(b)
−∞
(c)
−∞
(d) 1
3
(b) +∞
(e) 2
does not exist
2
(c) does not exist
(f ) 0
undef
(e)
0
(f )
−1
(e)
1
(f )
2
10.
(a) 1
(d) −2
(b) −∞
(e) +∞
(c) does not exist
(f ) +∞
11.
(a) 0
(d) 0
(b) 0
(e) does not exist
(c) 0
(f ) does not exist
12.
(a) 3
(d) 3
(b) 3
(e) does not exist
(c) 3
(f ) 0
13.
for all x0 6= −4
15.
(a)
At x = 3 the one-sided limits fail to exist.
(b)
At x = −2 the two-sided limit exists but is not equal to F (−2).
(c)
At x = 3 the limit fails to exist.
(a)
At x = 2 the two-sided limit fails to exist.
(b)
At x = 3 the two-sided limit exists but is not equal to F (3).
(c)
At x = 0 the two-sided limit fails to exist.
16.
14.
43
for all x0 6= −6, 3
Exercise Set 2.1
17.
(a)
44
2
1.5
1.1
1.01
1.001
0
0.5
0.9
0.99
0.999
0.1429 0.2105 0.3021 0.3300 0.3330 1.0000 0.5714 0.3690 0.3367 0.3337
1
The limit is 1/3.
0
2
0
(b)
2
1.5
1.1
1.01 1.001 1.0001
0.4286 1.0526 6.344 66.33 666.3 6666.3
50
The limit is +∞.
1
2
0
(c)
0
0.5
0.9
0.99
0.999
0.9999
−1 −1.7143 −7.0111 −67.001 −667.0 −6667.0
0
0
1
The limit is −∞.
-50
18.
(a)
−0.25 −0.1 −0.001 −0.0001 0.0001 0.001
0.1
0.25
0.5359 0.5132 0.5001
0.5000 0.5000 0.4999 0.4881 0.4721
0.6
The limit is 1/2.
-0.25
0.25
0
45
Chapter 2
(b)
0.25
0.1
0.001 0.0001
8.4721 20.488 2000.5 20001
100
The limit is +∞.
0
0.25
0
(c)
−0.25
−0.1
−0.001 −0.0001
−7.4641 −19.487 −1999.5 −20000
0
-0.25
0
The limit is −∞.
-100
19.
(a)
−0.25 −0.1 −0.001 −0.0001 0.0001 0.001
0.1
0.25
2.7266 2.9552 3.0000
3.0000 3.0000 3.0000 2.9552 2.7266
3
The limit is 3.
-0.25
0.25
2
(b)
0 −0.5
−0.9 −0.99 −0.999
−1.5
−1.1
−1.01 −1.001
1 1.7552 6.2161 54.87
541.1 −0.1415 −4.536 −53.19 −539.5
60
The limit does not exist.
-1.5
0
-60
Exercise Set 2.1
20.
(a)
46
0
−0.5
−0.9 −0.99 −0.999 −1.5
−1.1 −1.01 −1.001
1.5574 1.0926 1.0033 1.0000 1.0000 1.0926 1.0033 1.0000 1.0000
1.5
The limit is 1.
-1.5
(b)
1
0
−0.25 −0.1 −0.001 −0.0001 0.0001 0.001
0.1
0.25
1.9794 2.4132 2.5000
2.5000 2.5000 2.5000 2.4132 1.9794
2.5
The limit is 5/2.
-0.25
0.25
2
21.
(a)
−100,000,000 −100,000 −1000 −100
−10
10
100
1000
2.0000
2.0001
2.0050 2.0521 2.8333 1.6429 1.9519 1.9950
100,000 100,000,000
2.0000
2.0000
40
asymptote y = 2 as x → ±∞
-14
6
-40
(b)
−100,000,000 −100,000 −1000
−100
−10
10
100
1000
20.0855
20.0864 20.1763 21.0294 35.4013 13.7858 19.2186 19.9955
100,000 100,000,000
20.0846
20.0855
70
asymptote y = 20.086.
-160
160
0
47
Chapter 2
−100,000,000 −100,000 −1000 −100
−10
10 100 1000 100,000 100,000,000
−100,000,001 −100,000 −1001 −101.0 −11.2 9.2 99.0 999.0 99,999 99,999,999
(c)
50
no horizontal asymptote
-20
20
–50
22.
−100,000,000 −100,000 −1000 −100
−10
10
100
1000 100,000 100,000,000
0.2000
0.2000
0.2000 0.2000 0.1976 0.1976 0.2000 0.2000 0.2000
0.2000
(a)
0.2
-10
10
asymptote y = 1/5 as x → ±∞
-1.2
−100,000,000 −100,000 −1000 −100
−10
10
100
0.0000
0.0000
0.0000 0.0000 0.0016 1668.0 2.09 × 1018
(b)
1000
1.77 × 10301
100,000 100,000,000
?
?
10
asymptote y = 0 as x → −∞, none as
x → +∞
-6
(c)
0
6
−100,000,000 −100,000 −1000
−100
−10
10
100
0.0000
0.0000
0.0008 −0.0051 −0.0544 −0.0544 −0.0051
1000 100,000 100,000,000
0.0008 0.0000
0.0000
1.1
asymptote y = 0 as x → ±∞
-30
30
-0.3
23.
(a)
lim
x→0+
sin x
x
(b)
lim
x→0+
x−1
x+1
(c)
lim (1 + 2x)1/x
x→0−
Exercise Set 2.2
24.
(a)
25.
(a)
lim+
x→0
48
cos x
x
(b)
lim+
x→0
1
x+1
(c)
(b)
y
¶x
µ
2
lim− 1 +
x→0
x
yes; for example f (x) = (sin x)/x
x
26.
29.
(a)
no
(b)
yes; tan x and sec x at x = nπ + π/2, and cot x and csc x at x = nπ, n = 0, ±1, ±2, . . .
(a)
The plot over the interval [−a, a] becomes subject to catastrophic subtraction if a is small enough
(the size depending on the machine).
(c)
It does not.
EXERCISE SET 2.2
1.
(a)
−6
(b)
13
(c)
−8
(d) 16
(e)
2
(f )
−1/2
(g) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
2.
(h)
The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
(a)
0
(b)
The limit doesn’t exist because lim f doesn’t exist and lim g does.
(c)
0
(d) 3
(e)
0
(f )
The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.
p
(g) The limit doesn’t exist because f (x) is not defined for 0 ≤ x < 2.
(h)
1
3.
(a)
7
(b)
−3
(c)
π
(d)
−6
(e)
36
(f )
−∞
4.
(a)
1
(b)
−1
(c)
1
(d)
−1
(e)
1
(f )
−1
5.
0
6.
3/4
10.
12
11.
15.
0
20.
p
3
3/2
−3
9.
0
13. 3/2
14.
4/3
0
18.
19.
√
− 5
24.
√
−1/ 6
7.
8
−4/5
12.
16.
0
17.
21.
√
1/ 6
22.
√
√
8.
5
23.
5/3
√
3
4
27. does not exist
28.
−∞
29. −∞
+∞
32.
does not exist
33.
does not exist
34.
−∞
36.
−∞
37.
−∞
38. does not exist
39.
−1/7
41.
6
42.
+∞
43.
44.
4
25. +∞
26.
30. +∞
31.
35. +∞
40. +∞
3
+∞
49
Chapter 2
46.
+∞
47.
−∞
45.
+∞
48. +∞
49.
(a)
2
(b)
2
(c)
2
50.
(a)
−2
(b) 0
(c)
does not exist
51.
(a)
3
y
(b)
4
1
52.
(a)
53.
(a)
−6
56.
57.
58.
59.
60.
61.
(b) F (x) = x − 3
Theorem 2.2.2(a) doesn’t apply; moreover one cannot add/subtract infinities.
µ
¶
µ
¶
1
x−1
1
− 2 = lim+
= −∞
(b) lim+
x→0
x→0
x x
x2
µ
54.
x
lim
x→0−
lim
x→0
x
1
1
+
x x2
¡√
¶
= lim−
x→0
x+1
= +∞
x2
55.
lim
x→0
x
¡√
1
x
¢=
4
x+4+2
x2
¢ =0
x+4+2
√
p
x2 + 3 + x
3
2
√
lim ( x + 3 − x)
= lim √
=0
2
2
x→+∞
x→+∞
x +3+x
x +3+x
√
p
x2 − 3x + x
−3x
2
lim ( x − 3x − x) √
= lim √
= −3/2
2
2
x→+∞
x→+∞
x − 3x + x
x − 3x + x
´ √x2 + ax + x
³p
ax
x2 + ax − x √
= lim √
= a/2
lim
x→+∞
x2 + ax + x x→+∞ x2 + ax + x
´ √x2 + ax + √x2 + bx
³p
p
a−b
(a − b)x
2
2
√
√
x + ax − x + bx √
= lim √
=
lim
2
2
2
2
x→+∞
x→+∞
2
x + ax + x + bx
x + ax + x + bx
lim p(x) = (−1)n ∞ and lim p(x) = +∞
x→+∞
x→−∞
62.
If m > n the limits are both zero. If m = n the limits are both 1. If n > m the limits are (−1)n+m ∞
and +∞, respectively.
63.
If m > n the limits are both zero. If m = n the limits are both equal to am , the leading coefficient of
p. If n > m the limits are ±∞ where the sign depends on the sign of am and whether n is even or
odd.
64.
(a)
p(x) = q(x) = x
(b)
p(x) = x, q(x) = x2
(c)
p(x) = x2 , q(x) = x
(d)
p(x) = x + 3, q(x) = x
65.
The left and/or right limits could be plus or minus infinity; or the limit could exist, or equal any
preassigned real number. For example, let q(x) = x − x0 and let p(x) = a(x − x0 )n where n takes on
the values 0, 1, 2.
Exercise Set 2.3
66.
50
If m > n the limit is zero. If m = n the limit is cm /dm . If n > m the limit is ±∞, where the sign
depends on the signs of cn and dm .
EXERCISE SET 2.3
1.
2.
3.
4.
(a)
|f (x) − f (0)| = |x + 2 − 2| = |x| < 0.1 if and only if |x| < 0.1
(b)
|f (x) − f (3)| = |(4x − 5) − 7| = 4|x − 3| < 0.1 if and only if |x − 3| < (0.1)/4 = 0.0025
(c)
|f (x) − f (4)| = |x2 − 16| < ² if |x − 4| < δ. We get f (x) = 16 + ² = 16.001 at x = 4.000124998,
which corresponds to δ = 0.000124998; and f (x) = 16 − ² = 15.999 at x = 3.999874998, for
which δ = 0.000125002. Use the smaller δ: thus |f (x) − 16| < ² provided |x − 4| < 0.000125 (to
six decimals).
(a)
|f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.1 if and only if |x| < 0.05
(b)
|f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.01 if and only if |x| < 0.005
(c)
|f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.0012 if and only if |x| < 0.0006
(a)
x1 = (1.95)2 = 3.8025, x2 = (2.05)2 = 4.2025
(b)
δ = min ( |4 − 3.8025|, |4 − 4.2025| ) = 0.1975
(a)
x1 = 1/(1.1) = 0.909090 . . . , x2 = 1/(0.9) = 1.111111 . . .
(b)
δ = min( |1 − 0.909090|, |1 − 1.111111| ) = 0.0909090 . . .
5.
|2x − 8| = 2|x − 4| < 0.1 if |x − 4| < 0.05, δ = 0.05
6.
|x/2 + 1| = (1/2)|x − (−2)| < 0.1 if |x + 2| < 0.2, δ = 0.2
7.
|7x + 5 − (−2)| = 7|x − (−1)| < 0.01 if |x + 1| <
8.
|5x − 2 − 13| = 5|x − 3| < 0.01 if |x − 3| <
1
1
,δ=
700
700
1
1
,δ=
500
500
¯ 2
¯ ¯ 2
¯
¯x − 4
¯ ¯ x − 4 − 4x + 8 ¯
¯ = |x − 2| < 0.05 if |x − 2| < 0.05, δ = 0.05
9. ¯¯
− 4¯¯ = ¯¯
¯
x−2
x−2
10.
¯ 2
¯ ¯ 2
¯
¯x − 1
¯ ¯ x − 1 + 2x + 2 ¯
¯
¯
¯
¯ = |x + 1| < 0.05 if |x + 1| < 0.05, δ = 0.05
¯ x + 1 − (−2)¯ = ¯
¯
x+1
11.
¯
¯
if δ < 1 then ¯x2 − 16¯ = |x − 4||x + 4| < 9|x − 4| < 0.001 if |x − 4| <
1
1
,δ=
9000
9000
¯√
¯
¯ x + 3¯
√
1
|x − 9|
|x − 9|
¯= √
< |x − 9| < 0.001 if |x − 9| < 0.004, δ = 0.004
if δ < 1 then | x − 3| ¯¯ √
<√
¯
4
x+3
| x + 3|
8+3
¯
¯
¯ 1 1 ¯ |x − 5|
|x − 5|
≤
< 0.05 if |x − 5| < 1, δ = 1
13. if δ ≤ 1 then ¯¯ − ¯¯ =
x 5
5|x|
20
12.
14.
|x − 0| = |x| < 0.05 if |x| < 0.05, δ = 0.05
16.
|(4x − 5) − 7| = |4x − 12| = 4|x − 3| < ² if |x − 3| < 14 ², δ = 14 ²
17.
|2x − 7 − (−3)| = 2|x − 2| < ² if |x − 2| < 12 ², δ = 12 ²
15. |3x − 15| = 3|x − 5| < ² if |x − 5| < 13 ², δ = 13 ²
51
Chapter 2
18.
|2 − 3x − 5| = 3|x + 1| < ² if |x + 1| < 13 ², δ = 13 ²
19.
¯ 2
¯
¯x + x
¯
¯
¯
¯ x − 1¯ = |x| < ² if |x| < ², δ = ²
21.
if δ < 1 then |2x2 − 2| = 2|x − 1||x + 1| < 6|x − 1| < ² if |x − 1| < 16 ², δ = min(1, 16 ²)
22.
if δ < 1 then |x2 − 5 − 4| = |x − 3||x + 3| < 7|x − 3| < ² if |x − 3| < 17 ², δ = min(1, 17 ²)
23.
¯
¯ 3|x − 1 |
¯1
¯
1
3 < 18|x − 1 | < ² if |x − 1 | < 1 ², δ = min( 1 , 1 ²)
if δ < then ¯¯ − 3¯¯ =
6 18
6
x
|x|
3
3
18
24.
20.
¯ 2
¯
¯x − 9
¯
¯
¯
¯ x + 3 − (−6)¯ = |x + 3| < ² if |x + 3| < ², δ = ²
5
3
1
1
1
If δ < and |x − (−2)| < δ then − < x < − , x + 1 < − , |x + 1| > ; then
2
2
2
2
2
¯
¯
¯ |x + 2|
¯ 1
1
1 1
¯=
¯
−
(−1)
¯ |x + 1| < 2|x + 2| < ² if |x + 2| < 2 ², δ = min( 2 , 2 ²)
¯x + 1
25.
¯
¯ ¯
¯
√
¯√
√
x + 2 ¯¯ ¯¯ x − 4 ¯¯ 1
¯
= √
< |x − 4| < ² if |x − 4| < 2², δ = 2²
| x − 2| = ¯( x − 2) √
x + 2¯ ¯ x + 2¯ 2
26.
¯
¯√
¯ x + 3 + 3¯
√
¯ = √ |x − 6| ≤ 1 |x − 6| < ² if |x − 6| < 3², δ = 3²
¯
| x + 3 − 3| ¯ √
3
x + 3 + 3¯
x+3+3
27.
|f (x) − 3| = |x + 2 − 3| = |x − 1| < ² if 0 < |x − 1| < ², δ = ²
28.
If δ < 1 then |(x2 + 3x − 1) − 9| = |(x − 2)(x + 5)| < 8|x − 2| < ² if |x − 2| <
29.
30.
√
√
1
< 0.1 if x > 10, N = 10
2
x
1
x
− 1| = |
| < 0.01 if x + 1 > 100, N = 99
(b) |f (x) − L| = |
x+1
x+1
¯ ¯
¯1¯
1
(c) |f (x) − L| = ¯¯ 3 ¯¯ <
if |x| > 10, x < −10, N = −10
x
1000
¯
¯ ¯
¯
¯ ¯ 1 ¯
¯ x
¯
¯
¯
¯ < 0.01 if |x + 1| > 100, −x − 1 > 100, x < −101, N = −101
(d) |f (x) − L| = ¯
− 1¯ = ¯
x+1
x + 1¯
(a)
(a)
(c)
|f (x) − L| =
¯ ¯
¯1¯
¯ ¯ < 0.1, x > 101/3 , N = 101/3
¯ x3 ¯
¯ ¯
¯1¯
¯ ¯ < 0.001, x > 10, N = 10
¯ x3 ¯
r
31.
1
², δ = min (1, 18 ²)
8
(a)
(b)
x21
= 1 − ², x1 = −
1 + x21
r
1−²
N=
²
x1 = −1/²3 ; x2 = 1/²3
32.
(a)
33.
1
< 0.01 if |x| > 10, N = 10
x2
(b)
¯ ¯
¯1¯
¯ ¯ < 0.01, x > 1001/3 , N = 1001/3
¯ x3 ¯
1−²
x22
;
= 1 − ², x2 =
²
1 + x22
r
1−²
²
r
1−²
(c) N = −
²
(b)
N = 1/²3
(c) N = −1/²3
Exercise Set 2.3
34.
35.
52
1
< 0.005 if |x + 2| > 200, x > 198, N = 198
x+2
¯
¯ ¯
¯
¯ x
¯ ¯ 1 ¯
¯
¯=¯
¯
−
1
¯x + 1
¯ ¯ x + 1 ¯ < 0.001 if |x + 1| > 1000, x > 999, N = 999
36.
¯
¯ ¯
¯
¯ 4x − 1
¯ ¯ 11 ¯
¯
¯=¯
¯
−
2
¯ 2x + 5
¯ ¯ 2x + 5 ¯ < 0.1 if |2x + 5| > 110, 2x > 105, N = 52.5
37.
¯
¯
¯ 1
¯
¯
¯ < 0.005 if |x + 2| > 200, −x − 2 > 200, x < −202, N = −202
−
0
¯x + 2
¯
38.
¯ ¯
¯1¯
¯ ¯ < 0.01 if |x| > 10, −x > 10, x < −10, N = −10
¯ x2 ¯
39.
¯ ¯
¯
¯
¯ ¯ 11 ¯
¯ 4x − 1
¯
¯
¯
¯
¯ 2x + 5 − 2¯ = ¯ 2x + 5 ¯ < 0.1 if |2x + 5| > 110, −2x − 5 > 110, 2x < −115, x < −57.5, N = −57.5
40.
¯ ¯
¯
¯
¯ ¯ 1 ¯
¯ x
¯
¯
¯
¯
¯ x + 1 − 1¯ = ¯ x + 1 ¯ < 0.001 if |x + 1| > 1000, −x − 1 > 1000, x < −1001, N = −1001
41.
¯ ¯
¯1¯
¯ ¯ < ² if |x| > √1 , N = √1
¯ x2 ¯
²
²
43.
¯
¯
¯ 1 ¯
1
1
1
1
¯
¯
¯ x + 2 ¯ < ² if |x + 2| > ² , −x − 2 < ² , x > −2 − ² , N = −2 − ²
42.
¯ ¯
¯1¯
¯ ¯ < ² if |x| > 1 , −x > 1 , x < − 1 , N = − 1
¯x¯
²
²
²
²
¯
¯
¯ 1 ¯
¯
¯ < ² if |x + 2| > 1 , x + 2 > 1 , x > 1 − 2, N = 1 − 2
44. ¯
x + 2¯
²
²
²
²
45.
¯
¯ ¯
¯
¯ x
¯ ¯ 1 ¯
1
1
1
¯
¯
¯
¯
¯ x + 1 − 1¯ = ¯ x + 1 ¯ < ² if |x + 1| > ² , x > ² − 1, N = ² − 1
46.
¯
¯ ¯
¯
¯ x
¯ ¯ 1 ¯
1
1
1
1
¯
¯
¯
¯
¯ x + 1 − 1¯ = ¯ x + 1 ¯ < ² if |x + 1| > ² , −x − 1 > ² , x < −1 − ² , N = −1 − ²
47.
¯
¯ ¯
¯
¯ 4x − 1
¯ ¯ 11 ¯
11
11
11
11 5
5 11
¯
¯
¯
¯
¯ 2x + 5 − 2¯ = ¯ 2x + 5 ¯ < ² if |2x+5| > ² , −2x−5 > ² , 2x < − ² −5, x < − 2² − 2 , N = − 2 − 2²
48.
¯
¯ ¯
¯
¯ 4x − 1
¯ ¯ 11 ¯
11
11
11 5
11 5
¯
¯
¯
¯
¯ 2x + 5 − 2¯ = ¯ 2x + 5 ¯ < ² if |2x + 5| > ² , 2x > ² − 5, x > 2² − 2 , N = 2² − 2
49.
(a)
(c)
50.
51.
1
1
> 100 if |x| <
2
x
10
−1
1
< −1000 if |x − 3| < √
(x − 3)2
10 10
(b)
(d)
(a)
1
1
> 10 if and only if |x − 1| < √
(x − 1)2
10
(c)
1
1
√
> 100000 if and only if |x − 1| <
2
(x − 1)
100 10
if M > 0 then
(b)
1
1
> 1000 if |x − 1| <
|x − 1|
1000
1
1
1
, |x| <
− 4 < −10000 if x4 <
x
10000
10
1
> 1000 if and only if
(x − 1)2
1
|x − 1| < √
10 10
1
1
1
1
, 0 < |x − 3| < √ , δ = √
> M , 0 < (x − 3)2 <
2
(x − 3)
M
M
M
53
Chapter 2
52.
if M < 0 then
1
−1
1
1
,δ=√
< M , 0 < (x − 3)2 < − , 0 < |x − 3| < √
(x − 3)2
M
−M
−M
53.
if M > 0 then
1
1
1
> M , 0 < |x| <
,δ=
|x|
M
M
54.
if M > 0 then
1
1
1
> M , 0 < |x − 1| <
,δ=
|x − 1|
M
M
55.
if M < 0 then −
56.
if M > 0 then
57.
if x > 2 then |x + 1 − 3| = |x − 2| = x − 2 < ² if 2 < x < 2 + ², δ = ²
58.
if x < 1 then |3x + 2 − 5| = |3x − 3| = 3|x − 1| = 3(1 − x) < ² if 1 − x < 13 ², 1 − 13 ² < x < 1, δ = 13 ²
59.
if x > 4 then
60.
if x < 0 then
61.
if x > 2 then |f (x) − 2| = |x − 2| = x − 2 < ² if 2 < x < 2 + ², δ = ²
62.
if x < 2 then |f (x) − 6| = |3x − 6| = 3|x − 2| = 3(2 − x) < ² if 2 − x < 13 ², 2 − 13 ² < x < 2, δ = 13 ²
63.
64.
65.
66.
1
1
1
1
< M , 0 < x4 < − , |x| <
,δ=
4
1/4
x
M
(−M )
(−M )1/4
1
1
1
1
, x < 1/4 , δ = 1/4
> M , 0 < x4 <
4
x
M
M
M
√
√
x − 4 < ² if x − 4 < ²2 , 4 < x < 4 + ²2 , δ = ²2
−x < ² if −x < ²2 , −²2 < x < 0, δ = ²2
1
1
1
1
< M, x − 1 < − , 1 < x < 1 −
,δ=−
1−x
M
M
M
1
1
1
1
(b) if M > 0 and x < 1 then
> M, 1 − x <
, 1−
< x < 1, δ =
1−x
M
M
M
(a)
if M < 0 and x > 1 then
1
1
1
1
> M, x <
,0 0 and x > 0 then
(a)
Given any M > 0 there corresponds N > 0 such that if x > N then f (x) > M , x + 1 > M ,
x > M − 1, N = M − 1.
(b)
Given any M < 0 there corresponds N < 0 such that if x < N then f (x) < M , x + 1 < M ,
x < M − 1, N = M − 1.
(a)
Given
corresponds N > 0 such that if x > N then f (x) > M , x2 − 3 > M ,
√
√ any M > 0 there
x > M + 3, N = M + 3.
(b) Given any M < 0 there corresponds N < 0 such that if x < N then f (x) < M , x3 + 5 < M ,
x < (M − 5)1/3 , N = (M − 5)1/3 .
67.
2
if δ ≤ 2 then |x − 3| <
¡ 2,1 −2
¢ < x − 3 < 2, 1 < x < 5, and |x − 9| = |x + 3||x − 3| < 8|x − 3| < ² if
1
|x − 3| < 8 ², δ = min 2, 8 ²
68.
(a)
We don’t care about the value of f at x = a, because the limit is only concerned with values of
x near a. The condition that f be defined for all x (except possibly x = a) is necessary, because
if some points were excluded then the limit may not exist; for example, let f (x) = x if 1/x is
not an integer and f (1/n) = 6. Then lim f (x) does not exist but it would if the points 1/n were
x→0
(b)
excluded.
√
when x < 0 then x is not defined
(c)
yes; if δ ≤ 0.01 then x > 0, so
√
x is defined
Exercise Set 2.4
69.
54
|(x3 −4x+5)− 2| < 0.05, −0.05 < (x3 − 4x+ 5)− 2 < 0.05, 1.95 < x3 − 4x+ 5 < 2.05; x3 − 4x+ 5 = 1.95
at x = 1.0616, x3 − 4x + 5 = 2.05 at x = 0.9558; δ = min (1.0616 − 1, 1 − 0.9558) = 0.0442
2.2
0.9
1.9
70.
√
1.1
5x + 1 = 3.5 at x = 2.25,
√
5x + 1 = 4.5 at x = 3.85, so δ = min(3 − 2.25, 3.85 − 3) = 0.75
5
2
4
0
EXERCISE SET 2.4
1.
(a)
no, x = 2
(b)
no, x = 2
(e)
yes
(f )
yes
(a)
no, x = 2
(b)
no, x = 2
(e)
no, x = 2
(f )
yes
(a)
no, x = 1, 3
(b)
yes
(e)
no, x = 3
(f )
yes
(a)
no, x = 3
(b)
yes
(e)
no, x = 3
(f )
yes
5.
(a)
3
(b)
3
6.
−2/5
7.
(a)
2.
3.
4.
(b)
y
(c) no, x = 2
(d) yes
(c)
(d) yes
no, x = 2
(c) no, x = 1
(d)
(c) yes
(d) yes
y
1
3
x
1
3
x
yes
55
Chapter 2
(c)
y
(d)
y
1
x
1
2
-1
(
8.
f (x) = 1/x, g(x) =
9.
(a)
0
if x = 0
1
sin
x
if x 6= 0
(b)
C
3
x
One second could cost you one dollar.
$4
t
1
10.
2
(a)
no; disasters (war, flood, famine, pestilence, for example) can cause discontinuities
(b)
continuous
(c)
not usually continuous; see Exercise 9
(d)
continuous
11.
none
14.
f is not defined at x = ±1
15. f is not defined at x = ±4
√
−7 ± 57
f is not defined at x =
2
17. f is not defined at x = ±3
f is not defined at x = 0, −4
19. none
16.
18.
12. none
13.
none
20. f is not defined at x = 0, −3
16
is continuous on 4 < x;
x
lim− f (x) = lim+ f (x) = f (4) = 11 so f is continuous at x = 4
21. none; f (x) = 2x + 3 is continuous on x < 4 and f (x) = 7 +
x→4
22.
23.
x→4
lim f (x) does not exist so f is discontinuous at x = 1
x→1
(a)
f is continuous for x < 1, and for x > 1; lim− f (x) = 5, lim+ f (x) = k, so if k = 5 then f is
x→1
x→1
continuous for all x
(b) f is continuous for x < 2, and for x > 2; lim− f (x) = 4k, lim+ f (x) = 4 + k, so if 4k = 4 + k,
x→2
x→2
k = 4/3 then f is continuous for all x
24.
no, f is not defined for x ≤ 2
(a)
no, f is not defined at x = 2
(b)
(c)
yes
(d) no, f is not defined for x ≤ 2
Exercise Set 2.4
25.
(a)
56
(b)
y
x
c
26.
(a)
f (c) = lim f (x)
(b)
lim f (x) = 2
y
x
c
x→c
lim g(x) = 1
x→1
x→1
y
y
1
1
x
-1
1
27.
x
(c)
Define f (1) = 2 and redefine g(1) = 1.
(a)
x = 0, lim− f (x) = −1 6= +1 = lim+ f (x) so the discontinuity is not removable
(b)
x = −3; define f (−3) = −3 = lim f (x), then the discontinuity is removable
(c)
f is undefined at x = ±2; at x = 2, lim f (x) = 1, so define f (2) = 1 and f becomes continuous
x→0
x→0
x→−3
x→2
there; at x = −2, lim does not exist, so the discontinuity is not removable
x→−2
28.
f is not defined at x = 2; lim f (x) = lim
(b)
continuous there
lim− f (x) = 1 6= 4 = lim+ f (x), so f has a nonremovable discontinuity at x = 2
(c)
29.
1
1
x+2
= , so define f (2) = and f becomes
x2 + 2x + 4
3
3
(a)
(a)
x→2
x→2
x→2
x→2
lim f (x) = 8 6= f (1), so f has a removable discontinuity at x = 1
x→1
discontinuity at x = 1/2, not removable; at
x = −3, removable
y
5
5
-5
x
(b)
2x2 + 5x − 3 = (2x − 1)(x + 3)
57
30.
Chapter 2
(a)
there appears to be one discontinuity near
x = −1.52
(b)
one discontinuity at x = −1.52
4
-3
3
–4
31.
For x > 0, f (x) = x3/5 = (x3 )1/5 is the composition (Theorem 2.4.6) of the two continuous functions
g(x) = x3 and h(x) = x1/5 and is thus continuous. For x < 0, f (x) = f (−x) which is the composition
of the continuous functions f (x) (for positive x) and the continuous function y = −x. Hence f (−x) is
continuous for all x > 0. At x = 0, f (0) = lim f (x) = 0.
x→0
32.
4
2
4
2
x
√ + 7x + 1 ≥ 1 > 0, thus f (x) is the composition of the polynomial x + 7x + 1, the square root
x, and the function 1/x and is therefore continuous by Theorem 2.4.6.
33.
(a)
Let f (x) = k for x 6= c and f (c) = 0; g(x) = l for x 6= c and g(c) = 0. If k = −l then f + g is
continuous; otherwise it’s not.
(b)
f (x) = k for x 6= c, f (c) = 1; g(x) = l 6= 0 for x 6= c, g(c) = 1. If kl = 1, then f g is continuous;
otherwise it is not.
34.
A rational function is the quotient f (x)/g(x) of two polynomials f (x) and g(x). By Theorem 2.4.2 f
and g are continuous everywhere; by Theorem 2.4.3 f /g is continuous except when g(x) = 0.
35.
Since f and g are continuous at x = c we know that lim f (x) = f (c) and lim g(x) = g(c). In the
x→c
x→c
following we use Theorem 2.2.2.
(a) f (c) + g(c) = lim f (x) + lim g(x) = lim(f (x) + g(x)) so f + g is continuous at x = c.
x→c
x→c
x→c
(b)
same as (a) except the + sign becomes a − sign
(c)
lim f (x)
f
f (c)
f (x)
= x→c
= lim
so
is continuous at x = c
x→c
g(c)
lim g(x)
g(x)
g
x→c
36.
h(x) = f (x) − g(x) satisfies h(a) > 0, h(b) < 0. Use the Intermediate Value Theorem or Theorem
2.4.9.
37.
Of course such a function must be discontinuous. Let f (x) = 1 on 0 ≤ x < 1, and f (x) = −1 on
1 ≤ x ≤ 2.
38.
A square whose diagonal has length r has area f (r) = r2 /2. Note that
f (r) = r2 /2 < πr2 /2 < 2r2 = f (2r). By the Intermediate Value Theorem there must be a value
c between r and 2r such that f (c) = πr2 /2, i.e. a square of diagonal c whose area is πr2 /2.
39.
The cone has volume πr2 h/3. The function V (r) = πr2 h (for variable r and fixed h) gives the volume
of a right circular cylinder of height h and radius r, and satisfies V (0) < πr2 h/3 < V (r). By the
Intermediate Value Theorem there is a value c between 0 and r such that V (c) = πr2 h/3, so the
cylinder of radius c (and height h) has volume equal to that of the cone.
40. If f (x) = x3 − 4x + 1 then f (0) = 1, f (1) = −2. Use Theorem 2.4.9.
41.
If f (x) = x3 + x2 − 2x then f (−1) = 2, f (1) = 0. Use the Intermediate Value Theorem.
Exercise Set 2.4
42.
58
Since lim p(x) = −∞ and lim p(x) = +∞ (or vice versa, if the leading coefficient of p is negative),
x→−∞
x→+∞
it follows that for M = −1 there corresponds N1 < 0, and for M = 1 there is N2 > 0, such that
p(x) < −1 for x < N1 and p(x) > 1 for x > N2 . Choose x1 < N1 and x2 > N2 and use Theorem 2.4.9
on the interval [x1 , x2 ] to find a solution of p(x) = 0.
43. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f (−1.3) < 0 and
f (−1.2) > 0, the midpoint x = −1.25 of [−1.3, −1.2] is the required approximation of the root.
For the positive root use the interval [0, 1]; since f (0.7) < 0 and f (0.8) > 0, the midpoint x = 0.75 of
[0.7, 0.8] is the required approximation.
44.
x = −1.25 and x = 0.75.
10
1
0.7
-2
0.8
-1
-5
-1
45.
For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f (−1.7) < 0 and
f (−1.6) > 0, use the interval [−1.7, −1.6]. Since f (−1.61) < 0 and f (−1.60) > 0 the midpoint
x = −1.605 of [−1.61, −1.60] is the required approximation of the root. For the positive root use
the interval [1, 2]; since f (1.3) > 0 and f (1.4) < 0, use the interval [1.3, 1.4]. Since f (1.37) > 0 and
f (1.38) < 0, the midpoint x = 1.375 of [1.37, 1.38] is the required approximation.
46.
x = −1.605 and x = 1.375.
1
-1.7
1
-1.6
1.3
-2
47.
48.
1.4
-0.5
x = 2.24.
b
a
+
. Since lim+ f (x) = +∞ and lim− f (x) = −∞ there exist x1 > 1 and x2 < 3
x→1
x→3
x−1 x−3
(with x2 > x1 ) such that f (x) > 1 for 1 < x < x1 and f (x) < −1 for x2 < x < 3. Choose x3 in (1, x1 )
and x4 in (x2 , 3) and apply Theorem 2.4.9 on [x3 , x4 ].
Set f (x) =
49.
The uncoated sphere has volume 4π(x−1)3 /3 and the coated sphere has volume 4πx3 /3. If the volume
of the uncoated sphere and of the coating itself are the same, then the coated sphere has twice the
volume of the uncoated sphere. Thus 2(4π(x − 1)3 /3) = 4πx3 /3, or x3 − 6x2 + 6x − 2 = 0, with the
solution x = 4.847 cm.
50.
Let g(t) denote the altitude of the monk at time t measured in hours from noon of day one, and let f (t)
denote the altitude of the monk at time t measured in hours from noon of day two. Then g(0) < f (0)
and g(12) > f (12). Use Exercise 36.
51. We must show lim f (x) = f (c). Let ² > 0; then there exists δ > 0 such that if |x − c| < δ then
x→c
|f (x) − f (c)| < ². But this certainly satisfies Definition 2.3.3.
59
Chapter 2
EXERCISE SET 2.5
none
4.
x = nπ + π/2, n = 0, ±1, ±2, . . .
6.
none
9.
2nπ + π/6, 2nπ + 5π/6, n = 0, ±1, ±2, . . .
11.
sin x, x3 + 7x + 1
√
x, 3 + x, sin x, 2x
(d)
(a)
13.
cos
5. x = nπ, n = 0, ±1, ±2, . . .
8. x = nπ + π/2, n = 0, ±1, ±2, . . .
7. none
(a)
12.
3. x = nπ, n = 0, ±1, ±2, . . .
2. x = π
1.
10.
(b) |x|, sin x
(c)
x3 , cos x, x + 1
(e)
(f )
x5 − 2x3 + 1, cos x
sin x, sin x
Use Theorem 2.4.6.
µ
1
x→+∞ x
lim
none
(b)
g(x) = cos x, g(x) =
¶
µ
= cos 0 = 1
µ
πx
15. sin lim
x→+∞ 2 − 3x
¶
14.
√
³ π´
3
= sin −
=−
3
2
sin
2
x→+∞ x
sin x
= −1
x
19.
− lim−
21.
√
sin x
1
lim+ x lim+
=0
x→0
5 x→0
x
23.
tan 7x
7
sin 7x 3x
7
7
tan 7x
=
so lim
=
(1)(1) =
x→0
sin 3x
3 cos 7x 7x sin 3x
sin 3x
3(1)
3
µ
¶
lim
= sin 0 = 0
sin h
1
1
lim
=
2 h→0 h
2
µ
¶
1
sin θ
18.
lim
lim
= +∞
θ→0+ θ
θ→0+ θ
µ
¶2
1
1
sin x
20.
lim
=
3 x→0 x
3
sin 3θ
=3
θ→0 3θ
24.
22.
¶
sin θ
=0
lim sin θ lim
θ→0
θ→0 θ
sin 6x
6 sin 6x 8x
6
3
sin 6x
=
, so lim
= =
x→0 sin 8x
sin 8x
8 6x sin 8x
8
4
µ
25.
¶
lim cos h lim
h
=1
h→0 sin h
h→0
26.
sin h 1 + cos h
sin h(1 + cos h)
1 + cos h
sin h
=
=
=
; no limit
1 − cos h
1 − cos h 1 + cos h
1 − cos2 h
sin h
27.
θ2 (1 + cos θ)
θ2
1 + cos θ
=
=
1 − cos θ 1 + cos θ
1 − cos2 θ
28.
cos( 12 π − x) = sin( 12 π) sin x = sin x, so lim
29.
31.
µ
θ
sin θ
x→0
0
1
, g(x) = x2 + 1
+1
16.
17. 3 lim
x→0
x2
¶2
θ2
= (1)2 2 = 2
θ→0 1 − cos θ
(1 + cos θ) so lim
x
¡1
¢ =1
cos 2 π − x
30.
(1 − cos 5h)(1 + cos 5h)(1 + cos 7h)
1 − cos 5h
=
cos 7h − 1
(cos 7h − 1)(1 + cos 5h)(1 + cos 7h)
1 − cos 5h
25
=−
lim
h→0 cos 7h − 1
49
t2
=
1 − cos2 t
25
= −
49
µ
µ
t
sin t
sin 5h
5h
¶2
, so lim
t→0
¶2 µ
7h
sin 7h
t2
=1
1 − cos2 t
¶2
1 + cos 7h
so
1 + cos 5h
Exercise Set 2.5
32.
34.
36.
37.
60
µ ¶
1
= lim sin t;
t→+∞
x→0
x
limit does not exist
lim+ sin
lim x − 3 lim
x→0
x→0
sin x
= −3
x
k = f (0) = lim
x→0
38.
No; sin x/|x| has unequal one-sided limits.
39.
(a)
(c)
lim
35.
2 + lim
sin t
=1
t
(b)
sin(π − t) = sin t, so lim
x→π
lim
t→0−
1 − cos t
= 0 (Theorem 2.5.3)
t
³π
´
π
cos(π/x)
(π − 2t) sin t
π − 2t
sin t
− t = sin t, so lim
= lim
= lim
lim
=
x→2
t→0
t→0
t→0
2
x−2
4t
4
t
4
cos
41.
t = x − 1; sin(πx) = sin(πt + π) = − sin πt; and lim
42.
t = x − π/4; tan x − 1 =
x→1
µ
45.
sin x
=3
x
π−x
t
= lim
=1
t→0 sin t
sin x
40.
43.
x→0
1
sin kx
= k, lim+ f (x) = 2k 2 , so k = 2k 2 , k =
x→0
kx cos kx
2
x→0
t→0+
µ ¶
1
= lim cos t;
t→+∞
x→0
x
limit does not exist
lim+ cos
sin 3x
sin 3x
= 3 lim
= 3, so k = 3
x→0
x
3x
lim f (x) = k lim
x→0−
33.
−x ≤ x cos
50π
x
sin(πx)
sin πt
= − lim
= −π
t→0
x−1
t
tan x − 1
2 sin t
2 sin t
; lim
= lim
=2
t→0 t(cos t − sin t)
cos t − sin t x→π/4 x − π/4
¶
µ
≤x
44. −x ≤ x sin
2
lim f (x) = 1 by the Squeezing Theorem
x→0
46.
2
50π
√
3
x
¶
≤ x2
lim f (x) = 0 by the Squeezing Theorem
x→+∞
y
1
y = cos x
-1
x
0
1
y = f(x)
4
y = 1 – x2
x
-1
-1
47.
48.
Let g(x) = −
1
1
sin x
and h(x) = ; thus lim
= 0 by the Squeezing Theorem.
x→+∞ x
x
x
y
y
x
x
61
49.
50.
Chapter 2
(a)
πx
sin x = sin t where x is measured in degrees, t is measured in radians and t =
. Thus
180
sin x
sin t
π
lim
= lim
=
.
x→0 x
t→0 (180t/π)
180
πx
. Thus
cos x = cos t where x is measured in degrees, t in radians, and t =
180
1 − cos x
1 − cos t
= lim
= 0.
lim
x→0
t→0 (180t/π)
x
sin 10◦ = 0.17365
(b) sin 10◦ = sin
π
π
≈
= 0.17453
18
18
51.
(a)
52.
(a)
cos θ = cos 2α = 1 − 2 sin2 (θ/2)
≈ 1 − 2(θ/2)2 = 1 − 12 θ2
1 ³ π ´2
≈ 0.98477
(c) cos 10◦ = 1 −
2 18
(b)
cos 10◦ = 0.98481
53.
(a)
0.08749
(b)
tan 5◦ ≈
54.
(a)
h = 52.55 ft
(b)
πα
is a good
Since α is small, tan α◦ ≈
180
approximation.
(c)
h ≈ 52.36 ft
(a)
Let f (x) = x − cos x; f (0) = −1, f (π/2) = π/2. By the IVT there must be a solution of
f (x) = 0.
55.
(b)
π
= 0.08727
36
(c) 0.739
y
1.5
y=x
1
0.5
y = cos x
56.
(a)
x
π/2
0
f (x) = x + sin x − 1; f (0) = −1, f (π/6) = π/6 − 1/2 > 0. By the IVT there must be a solution
of f (x) = 0.
(b)
(c) x = 0.511.
y
y = 1– sin x
y=x
0.5
0
57.
58.
π/6
x
(a)
There is symmetry about the equatorial plane.
(b)
Let g(φ) be the given function. Then g(38) < 9.8 and g(39) > 9.8, so by the Intermediate Value
Theorem there is a value c between 38 and 39 for which g(c) = 9.8 exactly.
(a)
does not exist
(b)
the limit is zero
Supplementary Exercises 2
(c)
62
For part (a) consider the fact that given any δ > 0 there are infinitely many rational numbers x
satisfying |x| < δ and there are infinitely many irrational numbers satisfying the same condition.
Thus if the limit were to exist, it could not be zero because of the rational numbers, and it could
not be 1 because of the irrational numbers, and it could not be anything else because of all the
numbers. Hence the limit cannot exist. For part (b) use the Squeezing Theorem with +x and
−x as the ‘squeezers’.
CHAPTER 2 SUPPLEMENTARY EXERCISES
1.
2.
(a)
1
(b)
no limit
(c)
no limit
(f )
0
(g)
0
(h) 2
(a)
f (x) = 2x/(x − 1)
(d) 1
(i)
(e)
3
1/2
y
(b)
10
x
10
4.
f (x) = −1 for a ≤ x <
5.
(a)
a+b
a+b
and f (x) = 1 for
≤ x ≤ b.
2
2
0.222 . . . , 0.24390, 0.24938, 0.24994, 0.24999, 0.25000; for x 6= 2, f (x) =
the limit is 4.
(b)
1.15782, 4.22793, 4.00213, 4.00002, 4.00000, 4.00000; to prove, use
6.
(a)
y=0
7.
(a)
x
f(x)
(b)
1
1.000
(b)
0.1
0.443
0.01
0.409
0.001
0.406
sin 4x
4 sin 4x
tan 4x
=
=
.
x
x cos 4x
cos 4x 4x
none
0.0001
0.406
1
, so the limit is 1/4;
x+2
(c)
0.00001
0.405
y=2
0.000001
0.405
y
0.5
-1
1
x
·
8.
(a)
0.4 amperes
(d) 0.0187
9.
y
(a)
(b)
[0.3947, 0.4054]
(e)
It becomes infinite.
(b)
1
0.4
0.2
0.8
x
(c)
3
3
,
7.5 + δ 7.5 − δ
¸
Let g(x) = x − f (x). Then g(1) ≥ 0 and
g(0) ≤ 0; by the Intermediate Value Theorem there is a solution c in [0, 1] of g(c) = 0.
63
Chapter 2
µ
10.
(a)
(b)
(c)
(d)
µ
¶
µ
¶
1 − cos θ
1 − cos2 θ
= tan lim
= tan lim
= tan 0 = 0
θ→0
θ→0
θ→0 θ(1 + cos θ)
θ
√
√
√
t−1
t−1
t−1 t+1
(t − 1)( t + 1) √
√
√
√
= t + 1; lim √
=
=
= lim( t + 1) = 2
t→1
t−1
t−1
t−1 t+1
t − 1 t→1
lim tan
1 − cos θ
θ
¶
(2 − 1/x)5
(2x − 1)5
=
→ 25 /3 = 32/3 as x → +∞
3
+ 2x − 7)(x − 9x)
(3 + 2/x − 7/x2 )(1 − 9/x2 )
¶
¶
µ
µ
sin(θ + π)
− sin θ
= lim cos
sin(θ + π) = sin θ cos π − cos θ sin π = − sin θ, so lim cos
θ→0
θ→0
2θ
2θ
¶
µ
¡ 1¢
− sin θ
= cos − 2
= cos lim
θ→0
2θ
(3x2
11.
If, on the contrary, f (x0 ) < 0 for some x0 in [0, 1], then by the Intermediate Value Theorem we would
have a solution of f (x) = 0 in [0, x0 ], contrary to the hypothesis.
12.
For x < 2 f is a polynomial and is continuous; for x > 2 f is a polynomial and is continuous. At
x = 2, f (2) = −13 6= 13 = lim+ f (x) so f is not continuous there.
x→2
13.
f (−6) = 185, f (0) = −1, f (2) = 65; apply Theorem 2.4.9 twice, once on [−6, 0] and once on [0, 2]
14.
3.317
15.
Let ² = f (x0 )/2 > 0; then there corresponds δ > 0 such that if |x − x0 | < δ then |f (x) − f (x0 )| < ²,
−² < f (x) − f (x0 ) < ², f (x) > f (x0 ) − ² = f (x0 )/2 > 0 for x0 − δ < x < x0 + δ.
16.
y
1
4
x
1.449 (x must be ≥ −1)
(b) x = 0, ±1.896
17.
(a)
18.
Since lim sin(1/x) does not exist, no conclusions can be drawn.
19.
(a)
5, no limit,
no limit
20.
(a)
−1/5, +∞, −1/10, −1/10, no limit, 0, 0
21.
a/b
24.
2
x→0
√
√
10,
√
10, no limit, +∞,
(b)
5, 10, 0, 0, 10, −∞, +∞
(b)
−1, +1, −1, −1, no limit, −1, +1
22. 1
25.
0
23.
26. k 2
does not exist
27.
3−k
The numerator satisfies: |2x + x sin 3x| ≤ |2x| + |x| = 3|x|. Since the denominator grows like x2 , the
limit is 0.
√
√
x2
1
x2 + 4 − 2 x2 + 4 + 2
√
√
30. (a)
=
=√
, so
2
2
2
2
x2
x + 4 + 2 x ( x + 4 + 2)
x +4+2
√
1
x2 + 4 − 2
1
= lim √
=
lim
x→0
x→0
x2
4
x2 + 4 + 2
28.
Supplementary Exercises 2
x
f (x)
(b)
1
0.236
64
0.1
0.2498
0.01
0.2500
0.001
0.2500
0.0001
0.25000
0.00001
0.00000
The division may entail division by zero (e.g. on an HP 42S), or the numerator may be inaccurate
(catastrophic subtraction, e.g.).
(c)
in the 3d picture, catastrophic subtraction
31.
x
f (x)
0.1
2.59
0.01
2.70
0.001
2.717
0.0001
2.718
0.00001
2.7183
0.000001
2.71828
32.
x
f (x)
3.1
5.74
3.01
5.56
3.001
5.547
3.0001
5.545
3.00001
5.5452
3.000001
5.54518
33.
x
f (x)
1.1
0.49
1.01
0.54
1.001
0.540
1.0001
0.5403
1.00001
0.54030
1.000001
0.54030
34.
x
f (x)
0.1
99.0
0.01
9048.8
35.
x
f (x)
100
0.48809
0.001
368063.3
0.0001
4562.7
104
0.49876
1000
0.49611
0.00001
3.9 × 10−34
105
0.49961
106
0.49988
0.000001
0
107
0.49996
36.
For large values of x (not much more than 100) the computer can’t handle 5x or 3x , yet the limit is 5.
37.
δ = 0.07747
39.
(a)
38. $2,001.60, $2,009.66, $2,013.62, $2013.75
x3 − x − 1 = 0, x3 = x + 1, x =
√
3
x + 1.
(b)
y
2
-1
1
x
-1
y
(c)
x1
40.
1, 1.26, 1.31, 1.322, 1.324, 1.3246, 1.3247
(b)
0, −1, −2, −9, −730
x
x2 x3
y
(a)
(d)
20
10
x
-1
41.
x=
√
5
1
x1
2
x2
3
x + 2; 1.267168
42. x = cos x; 0.739085 (after 33 iterations!).
CHAPTER 3
The Derivative
EXERCISE SET 3.1
1.
(a)
(4)2 /2 − (3)2 /2
7
f (4) − f (3)
=
=
4−3
1
2
f (x1 ) − f (3)
x2 /2 − 9/2
= lim
= lim 1
x1 →3
x1 →3
x1 − 3
x1 − 3
msec =
(b) mtan
x21 − 9
(x1 + 3)(x1 − 3)
x1 + 3
= lim
= lim
=3
x1 →3 2(x1 − 3)
x1 →3
x1 →3
2(x1 − 3)
2
= lim
(c)
mtan = lim
x1 →x0
f (x1 ) − f (x0 )
x1 − x0
(d)
y
10
x2 /2 − x20 /2
= lim 1
x1 →x0
x1 − x0
Tangent
x21 − x20
x1 →x0 2(x1 − x0 )
x1 + x0
= x0
= lim
x1 →x0
2
= lim
2.
Secant
5
x
23 − 13
f (2) − f (1)
=
=7
2−1
1
f (x1 ) − f (1)
x3 − 1
(x1 − 1)(x21 + x1 + 1)
= lim 1
= lim
= lim
x1 →1
x1 →1 x1 − 1
x1 →1
x1 − 1
x1 − 1
(a)
msec =
(b)
mtan
= lim (x21 + x1 + 1) = 3
x1 →1
(c)
mtan = lim
x1 →x0
f (x1 ) − f (x0 )
x1 − x0
y
(d)
9
x3 − x30
= lim 1
x1 →x0 x1 − x0
Tangent
= lim (x21 + x1 x0 + x20 )
x1 →x0
= 3x20
3.
(a)
Secant
5
x
1/3 − 1/2
1
f (3) − f (2)
=
=−
3−2
1
6
f (x1 ) − f (2)
1/x1 − 1/2
= lim
= lim
x1 →2
x1 →2
x1 − 2
x1 − 2
msec =
(b) mtan
2 − x1
−1
1
= lim
=−
x1 →2 2x1 (x1 − 2)
x1 →2 2x1
4
= lim
(c) mtan = lim
x1 →x0
f (x1 ) − f (x0 )
x1 − x0
(d)
1/x1 − 1/x0
x1 →x0
x1 − x0
x0 − x1
= lim
x1 →x0 x0 x1 (x1 − x0 )
y
4
= lim
= lim
x1 →x0
Secant
1
−1
=− 2
x0 x1
x0
1
65
x
Tangent
Exercise Set 3.1
4.
(a)
66
f (2) − f (1)
1/4 − 1
3
=
=−
2−1
1
4
f (x1 ) − f (1)
1/x21 − 1
= lim
= lim
x1 →1
x1 →1 x1 − 1
x1 − 1
msec =
(b) mtan
= lim
1 − x21
−(x1 + 1)
= −2
= lim
− 1) x1 →1
x21
x1 →1 x21 (x1
(c)
mtan = lim
x1 →x0
= lim
x1 →x0
f (x1 ) − f (x0 )
x1 − x0
(d)
y
1/x21 − 1/x20
x1 − x0
x20 − x21
2
x1 →x0 x x2 (x1 − x0 )
0 1
1
= lim
2
2
−(x1 + x0 )
=− 3
= lim
2
2
x1 →x0
x0 x1
x0
5.
(a)
mtan = lim
x1 →x0
Tangent
x
Secant
f (x1 ) − f (x0 )
(x21 + 1) − (x20 + 1)
= lim
x1 →x0
x1 − x0
x1 − x0
x21 − x20
= lim (x1 + x0 ) = 2x0
x1 →x0 x1 − x0
x1 →x0
= 2(2) = 4
= lim
6.
(b)
mtan
(a)
mtan = lim
x1 →x0
f (x1 ) − f (x0 )
(x21 + 3x1 + 2) − (x20 + 3x0 + 2)
= lim
x1 →x0
x1 − x0
x1 − x0
(x21 − x20 ) + 3(x1 − x0 )
= lim (x1 + x0 + 3) = 2x0 + 3
x1 →x0
x1 →x0
x1 − x0
= 2(2) + 3 = 7
= lim
7.
(b)
mtan
(a)
mtan = lim
x1 →x0
f (x1 ) − f (x0 )
= lim
x1 →x0
x1 − x0
= lim √
x1 →x0
8.
9.
√
√
x1 − x0
x1 − x0
1
1
√ = √
x1 + x0
2 x0
(b)
1
1
mtan = √ =
2
2 1
(a)
mtan = lim
(b)
mtan = −
(a)
mtan = (50 − 10)/(15 − 5)
= 40/10
= 4 m/s
√
√
1/ x1 − 1/ x0
f (x1 ) − f (x0 )
= lim
x1 →x0
x1 →x0
x1 − x0
x1 − x0
√
√
x0 − x1
1
−1
= lim √ √
= − 3/2
= lim √ √
√
√
x1 →x0
x0 x1 (x1 − x0 ) x1 →x0 x0 x1 ( x1 + x0 )
2x0
1
1
=−
16
2(4)3/2
(b)
Velocity
(m/s)
4
10
20
t (s)
67
10.
11.
Chapter 3
(a)
(10 − 10)/(3 − 0) = 0 cm/s
(b)
t = 0, t = 2, and t = 4.2 (horizontal tangent line)
(c)
maximum:
(d)
(3 − 18)/(4 − 2) = −7.5 cm/s (slope of estimated tangent line to curve at t = 3)
t = 1 (slope > 0) minimum:
t = 3 (slope < 0)
From the figure:
s
t0 t1
(a)
t
t2
The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 is
greater than that at t2 .
(b) The initial velocity is 0 because the slope of a horizontal line is 0.
(c)
The particle is speeding up because the slope increases as t increases from t0 to t1 .
(d) The particle is slowing down because the slope decreases as t increases from t1 to t2 .
s
12.
t0
t
t1
13.
It is a straight line with slope equal to the velocity.
14.
(a)
decreasing (slope of tangent line decreases with increasing time)
(b)
increasing (slope of tangent line increases with increasing time)
(c)
increasing (slope of tangent line increases with increasing time)
(d)
decreasing (slope of tangent line decreases with increasing time)
(a)
72◦ F at about 4:30 P.M.
(c)
decreasing most rapidly at about 9 P.M.; rate of change of temperature is about −7◦ F/h (slope
of estimated tangent line to curve at 9 P.M.)
15.
(b)
about (67 − 43)/6 = 4◦ F/h
16. For V = 10 the slope of the tangent line is about −0.25 atm/L, for V = 25 the slope is
about −0.04 atm/L.
17.
(a)
during the first year after birth
(b)
about 6 cm/year (slope of estimated tangent line at age 5)
(c)
the growth rate is greatest at about age 14; about 10 cm/year
(d)
Growth rate
(cm/year)
40
30
20
10
t (yrs)
5
10
15
20
Exercise Set 3.2
18.
68
(a)
The rock will hit the ground when
16t2 = 576, t2 = 36, t = 6 s (only t ≥ 0
is meaningful)
(c)
vave =
16(3)2 − 16(0)2
= 48 ft/s
3−0
16(6)2 − 16(0)2
= 96 ft/s
6−0
(b)
vave =
(d)
vinst = lim
16t21 − 16(6)2
16(t21 − 36)
= lim
t1 →6
t1 →6
t1 − 6
t1 − 6
= lim 16(t1 + 6) = 192 ft/s
t1 →6
19.
20.
21.
(a)
5(40)3 = 320,000 ft
(b) vave = 320,000/40 = 8,000 ft/s
(c)
5t3 = 135 when the rocket has gone 135 ft,
so t3 = 27, t = 3 s;
vave = 135/3 = 45 ft/s.
(d)
5t31 − 5(40)3
5(t31 − 403 )
= lim
t1 →40
t1 →40
t1 − 40
t1 − 40
vinst = lim
= lim 5(t21 + 40t1 + 1600) = 24, 000 ft/s
t1 →40
[3(3)2 + 3] − [3(1)2 + 1]
= 13 mi/h
3−1
(a)
vave =
(b)
vinst = lim
(a)
vave =
(b)
vinst = lim
(3t21 + t1 ) − 4
(3t1 + 4)(t1 − 1)
= lim
= lim (3t1 + 4) = 7 mi/h
t1 →1
t1 →1
t1 →1
t1 − 1
t1 − 1
6(4)4 − 6(2)4
= 720 ft/min
4−2
6t41 − 6(2)4
6(t41 − 16)
= lim
t1 →2
t1 →2
t1 − 2
t1 − 2
6(t21 + 4)(t21 − 4)
= lim 6(t21 + 4)(t1 + 2) = 192 ft/min
t1 →2
t1 →2
t1 − 2
= lim
EXERCISE SET 3.2
1.
f 0 (1) = 2, f 0 (3) = 0, f 0 (5) = −2, f 0 (6) = −1/2
2.
f 0 (4) < f 0 (0) < f 0 (2) < 0 < f 0 (−3)
3. (b)
4.
f 0 (2) = m = 3
f 0 (−1) = m =
5.
(c)
the same, f 0 (2) = 3
4−3
=1
0 − (−1)
6.
y
y
1
0
x
x
7.
y − (−1) = 5(x − 3), y = 5x − 16
9.
f (x + h) − f (x)
3(x2 + 2xh + h2 ) − 3x2
= lim
= lim 3(2x + h) = 6x; f (3) = 3(3)2 = 27,
h→0
h→0
h→0
h
h
f 0 (3) = 18 so y − 27 = 18(x − 3), y = 18x − 27
f 0 (x) = lim
8. y − 3 = −4(x + 2), y = −4x − 5
69
10.
Chapter 3
f (x + h) − f (x)
(x + h)2 − (x + h) − (x2 − x)
= lim
= lim (2x + h − 1) = 2x − 1;
h→0
h→0
h→0
h
h
f (2) = 22 − 2 = 2, f 0 (2) = 3 so y − 2 = 3(x − 2), y = 3x − 4
f 0 (x) = lim
f (x + h) − f (x)
(x + h)3 − x3
= lim
= 3x2 ; f (0) = 03 = 0,
h→0
h→0
h
h
f 0 (0) = 0 so y − 0 = (0)(x − 0), y = 0
11. f 0 (x) = lim
12.
13.
14.
15.
f (x + h) − f (x)
2(x + h)3 + 1 − (2x3 + 1)
= lim
= lim (6x2 + 6xh + 2h2 ) = 6x2 ;
h→0
h→0
h→0
h
h
f (−1) = 2(−1)3 + 1 = −1, f 0 (−1) = 6 so y + 1 = 6(x + 1), y = 6x + 5
f 0 (x) = lim
√
√
f (x + h) − f (x)
x+1+h− x+1
= lim
f (x) = lim
h→0
h→0
h
h
√
√
√
√
1
x+1+h− x+1 x+1+h+ x+1
1
√
√
√
= lim
= lim √
= √
;
h→0
h
2 x+1
x + 1 + h + x + 1 h→0 x + 1 + h + x + 1
√
1
1
5
1
f (8) = 8 + 1 = 3, f 0 (8) = so y − 3 = (x − 8), y = x +
6
6
6
3
0
f (x + h) − f (x)
(x + h)4 − x4
= lim
= lim (4x3 + 6x2 h + 4xh2 + h3 ) = 4x3 ;
h→0
h→0
h→0
h
h
f (−2) = (−2)4 = 16, f 0 (−2) = −32 so y − 16 = −32(x + 2), y = −32x − 48
f 0 (x) = lim
x − (x + ∆x)
1
1
−
x(x + ∆x)
f 0 (x) = lim x + ∆x x = lim
∆x→0
∆x→0
∆x
∆x
= lim
∆x→0
16.
−∆x
1
1
= lim −
=− 2
x∆x(x + ∆x) ∆x→0 x(x + ∆x)
x
1
1
x2 − (x + ∆x)2
−
(x + ∆x)2
x2
x2 (x + ∆x)2
= lim
f 0 (x) = lim
∆x→0
∆x→0
∆x
∆x
2
x2 − x2 − 2x∆x − ∆x2
−2x∆x − ∆x2
−2x − ∆x
= lim 2
= lim 2
=− 3
2
2
∆x→0
∆x→0 x ∆x(x + ∆x)2
∆x→0 x (x + ∆x)2
x ∆x(x + ∆x)
x
= lim
17.
[a(x + ∆x)2 + b] − [ax2 + b]
ax2 + 2ax∆x + a(∆x)2 + b − ax2 − b
= lim
∆x→0
∆x→0
∆x
∆x
f 0 (x) = lim
2ax∆x + a(∆x)2
= lim (2ax + a∆x) = 2ax
∆x→0
∆x→0
∆x
= lim
18.
1
1
(x + 1) − (x + ∆x + 1)
−
(x + ∆x) + 1 x + 1
(x + 1)(x + ∆x + 1)
f 0 (x) = lim
= lim
∆x→0
∆x→0
∆x
∆x
= lim
x + 1 − x − ∆x − 1
−∆x
= lim
∆x(x + 1)(x + ∆x + 1) ∆x→0 ∆x(x + 1)(x + ∆x + 1)
= lim
−1
1
=−
(x + 1)(x + ∆x + 1)
(x + 1)2
∆x→0
∆x→0
19.
f 0 (x) = lim
∆x→0
= lim
∆x→0
√
1
1
√
√
−√
x − x + ∆x
x
x + ∆x
= lim
√ √
∆x→0 ∆x x x + ∆x
∆x
1
x − (x + ∆x)
−1
√
√
= lim √ √
= − 3/2
√ √
√
√
2x
∆x x x + ∆x( x + x + ∆x) ∆x→0 x x + ∆x( x + x + ∆x)
Exercise Set 3.2
20.
70
(x + ∆x)1/3 − x1/3
, but a3 − b3 = (a − b)(a2 + ab + b2 ) so with a = (x + ∆x)1/3
∆x→0
∆x
and b = x1/3
(x + ∆x) − x
f 0 (x) = lim
∆x→0 ∆x[(x + ∆x)2/3 + (x + ∆x)1/3 x1/3 + x2/3 ]
f 0 (x) = lim
= lim
∆x→0
21.
1
1
= 2/3
(x + ∆x)2/3 + (x + ∆x)1/3 x1/3 + x2/3
3x
f (t + h) − f (t)
[4(t + h)2 + (t + h)] − [4t2 + t]
= lim
h→0
h→0
h
h
f 0 (t) = lim
4t2 + 8th + 4h2 + t + h − 4t2 − t
h→0
h
= lim
8th + 4h2 + h
= lim (8t + 4h + 1) = 8t + 1
h→0
h→0
h
= lim
22.
4
4
4
π(r + h)3 − πr3
π(r3 + 3r2 h + 3rh2 + h3 − r3 )
dV
3
3
3
= lim
= lim
h→0
h→0
dr
h
h
4
= lim π(3r2 + 3rh + h2 ) = 4πr2
h→0 3
23.
(a)
D
(b)
F
(c)
B
(d) C
(e)
A
(f )
E
24.
Any function of the form f (x) = x + k has slope 1, and thus the derivative must be equal to 1
everywhere.
y
3
1
-2
x
-2
25.
(a)
(b)
y
(c)
y
x
y
x
1
2
x
-1
26.
(a)
(b)
y
x
27.
(a)
f (x) = x2 and a = 3
(c)
y
x
(b) f (x) =
√
y
x
x and a = 1
71
28.
29.
Chapter 3
(a)
(b) f (x) = x7 and a = 1
f (x) = cos x and a = π
[4(x + h)2 + 1] − [4x2 + 1]
4x2 + 8xh + 4h2 + 1 − 4x2 − 1
dy
= lim
= lim
= lim (8x + 4h) = 8x
h→0
h→0
dx h→0
h
h
¯
dy ¯¯
= 8(1) = 8
dx ¯
x=1
µ
30.
dy
= lim
dx h→0
¶ µ
¶
5
5
5x − 5(x + h)
5
5
+1 −
+1
−
x+h
x
x(x + h)
= lim x + h x = lim
h→0
h→0
h
h
h
5x − 5x − 5h
−5
5
= lim
=− 2
= limh→0
h→0 x(x + h)
hx(x + h)
x
¯
5
5
dy ¯¯
=−
=−
dx ¯x=−2
(−2)2
4
31. y = −2x + 1
32.
1.5
5
-2
2
0
2.5
0
-3
33. (b)
h
(f (1 + h) − f (1))/h
0.5
0.1
0.01
0.001 0.0001 0.00001
1.6569 1.4355 1.3911 1.3868 1.3863 1.3863
34. (b)
h
(f (1 + h) − f (1))/h
0.5
0.1
0.01
0.001
0.0001 0.00001
0.50489 0.67060 0.70356 0.70675 0.70707 0.70710
35.
(a)
dollars/ft
(b)
(c)
As you go deeper the price per foot may increase dramatically, so f 0 (x) is roughly the price per
additional foot.
If each additional foot costs extra money (this is to be expected) then f 0 (x) remains positive.
(d)
From the approximation 1000 = f 0 (300) ≈
so the extra foot will cost around $1000.
36.
f (301) − f (300)
we see that f (301) ≈ f (300) + 1000,
301 − 300
(a)
gallons/dollar
(b)
The increase in the amount of paint that would be sold for one extra dollar.
(c)
It should be negative since an increase in the price of paint would decrease the amount of paint
sold.
f (11) − f (10)
we see that f (11) ≈ f (10) − 100, so an increase of one
From −100 = f 0 (10) ≈
11 − 10
dollar would decrease the amount of paint sold by around 100 gallons.
(d)
37.
(a)
F ≈ 200 lb, dF/dθ ≈ 60 lb/rad
(b) µ = (dF/dθ)/F ≈ 60/200 = 0.3
38.
(a)
dN/dt ≈ 34 million/year; in 1950 the world population was increasing at the rate of about 34
million per year.
(b)
dN/dt
34
≈
≈ 0.014 = 1.4 %/year
N
2490
Exercise Set 3.2
39.
41.
72
(a)
T ≈ 120◦ F, dT /dt ≈ −4.5◦ F/min
(b)
k = (dT /dt)/(T − T0 ) ≈ (−4.5)/(120 − 75) = −0.1
lim f (x) = lim
√
3
x = 0 = f (0), so f is continuous at x = 0.
√
3
f (0 + h) − f (0)
h−0
1
= lim
= lim 2/3 = +∞, so
lim
h→0
h→0
h→0 h
h
h
f 0 (0) does not exist.
x→0
y
x→0
2
-2
42.
lim f (x) = lim(x − 2)2/3 = 0 = f (2) so f is continuous at x = 2.
x→2
x
2
y
x→2
f (2 + h) − f (2)
h −0
1
= lim
= lim 1/3 which does not exist
h→0
h→0 h
h
h
so f 0 (2) does not exist.
2/3
5
lim
h→0
x
2
43.
lim f (x) = lim+ f (x) = f (1), so f is continuous at x = 1.
x→1−
x→1
y
f (1 + h) − f (1)
[(1 + h) + 1] − 2
= lim−
= lim− (2 + h) = 2;
h→0
h→0
h→0
h
h
f (1 + h) − f (1)
2(1 + h) − 2
= lim+
= lim+ 2 = 2, so f 0 (1) = 2.
lim
h→0+
h→0
h→0
h
h
2
5
lim−
-3
44.
lim f (x) = lim+ f (x) = f (1) so f is continuous at x = 1.
x→1−
x
y
x→1
f (1 + h) − f (1)
[(1 + h)2 + 2] − 3
= lim−
= lim− (2 + h) = 2;
h→0
h→0
h→0
h
h
f (1 + h) − f (1)
[(1 + h) + 2] − 3
= lim+
= lim+ 1 = 1, so f 0 (1)
lim
h→0+
h→0
h→0
h
h
does not exist.
5
lim−
-3
45.
3
3
x
f is continuous at x = 1 because it is differentiable there, thus lim f (1 + h) = f (1) and so f (1) = 0
h→0
f (1 + h)
f (1 + h) − f (1)
f (1 + h)
because lim
exists; f 0 (1) = lim
= lim
= 5.
h→0
h→0
h→0
h
h
h
f (x + h) − f (x)
, but (with
h
f (h) + 5xh
y = h) f (x + h) = f (x) + f (h) + 5xh so f (x + h) − f (x) = f (h) + 5xh and f 0 (x) = lim
=
h→0
h
f (h)
lim (
+ 5x) = 3 + 5x.
h→0
h
46.
Let x = y = 0 to get f (0) = f (0) + f (0) + 0 so f (0) = 0. f 0 (x) = lim
47.
f 0 (x) = limh→0
h→0
f (x + h) − f (x)
f (x)f (h) − f (x)
f (x)[f (h) − 1]
f (h) − f (0)
= lim
= lim
= f (x) lim
h→0
h→0
h→0
h
h
h
h
= f (x)f 0 (0) = f (x)
73
Chapter 3
EXERCISE SET 3.3
1.
28x6
2.
5.
0
6.
2
9. 3ax + 2bx + c
13.
−3x−4 − 7x−8
15.
f 0 (x) = (3x2 + 6)
10.
−36x11
3.
24x7 + 2
4.
2x3
√
1
7. − (7x6 + 2)
3
8.
2
x
5
1
a
2
µ
1
2x +
b
¶
√
11. 24x−9 + 1/ x
14.
5
12. −42x−7 − √
2 x
1
1
√ − 2
2 x x
µ
¶ µ
¶
¶
µ
1
1 d
d
1
2x −
+ 2x −
(3x2 + 6) = (3x2 + 6)(2) + 2x −
(6x)
dx
4
4 dx
4
= 18x2 − 32 x + 12
d
d
(7 + x5 ) + (7 + x5 ) (2 − x − 3x3 ) = (2 − x − 3x3 )(5x4 ) + (7 + x5 )(−1 − 9x2 )
dx
dx
= −24x7 − 6x5 + 10x4 − 63x2 − 7
16. f 0 (x) = (2 − x − 3x3 )
d
d
(2x−3 + x−4 ) + (2x−3 + x−4 ) (x3 + 7x2 − 8)
dx
dx
= (x3 + 7x2 − 8)(−6x−4 − 4x−5 ) + (2x−3 + x−4 )(3x2 + 14x) = −15x−2 − 14x−3 + 48x−4 + 32x−5
17.
f 0 (x) = (x3 + 7x2 − 8)
18.
f 0 (x) = (x−1 + x−2 )
19.
12x(3x2 + 1)
21.
d
d
(5x − 3) (1) − (1) (5x − 3)
5
dy
dx
dx
=
=−
; y 0 (1) = −5/4
dx
(5x − 3)2
(5x − 3)2
22.
23.
24.
25.
26.
d
d
(3x3 + 27) + (3x3 + 27) (x−1 + x−2 )
dx
dx
= (x−1 + x−2 )(9x2 ) + (3x3 + 27)(−x−2 − 2x−3 ) = 3 + 6x − 27x−2 − 54x−3
20. f (x) = x10 + 4x6 + 4x2 , f 0 (x) = 10x9 + 24x5 + 8x
√
d √
d
( x + 2) (3) − 3 ( x + 2)
√ √
dy
dx√
dx
=
= −3/(2 x( x + 2)2 ; y 0 (1) = −3/18 = −1/6
2
dx
( x + 2)
d
d
(2t + 1) (3t) − (3t) (2t + 1)
dx
(2t + 1)(3) − (3t)(2)
3
dt
dt
=
=
=
dt
(2t + 1)2
(2t + 1)2
(2t + 1)2
d
d
(3t) (t2 + 1) − (t2 + 1) (3t)
dx
(3t)(2t) − (t2 + 1)(3)
t2 − 1
dt
dt
=
=
=
2
2
dt
(3t)
9t
3t2
d
d
¯
(x + 3) (2x − 1) − (2x − 1) (x + 3)
dy
(x + 3)(2) − (2x − 1)(1)
7
dy ¯¯
7
dx
dx
=
=
=
;
=
¯
2
2
2
dx
(x + 3)
(x + 3)
(x + 3) dx x=1
16
d
d
(x2 − 5) (4x + 1) − (4x + 1) (x2 − 5)
(x2 − 5)(4) − (4x + 1)(2x)
4x2 + 2x + 20
dy
dx
dx
=
=
=
−
;
dx ¯
(x2 − 5)2
(x2 − 5)2
(x2 − 5)2
13
dy ¯¯
=
¯
dx x=1
8
Exercise Set 3.3
27.
28.
µ
¶
¢ ¡
¢ d 3x + 2
d ¡ −5
x + 1 + x−5 + 1
dx
dx
x
¶
¶
¸ µ
¶
·
µ
µ
¢ x(3) − (3x + 2)(1)
¢
¢ ¡ −5
¢ ¡ −5
3x + 2 ¡
2
3x + 2 ¡
−6
−6
−5x
−5x
=
+ x +1
+ x +1 − 2 ;
=
x
x2
x
x
¯
dy ¯¯
= 5(−5) + 2(−2) = −29
dx ¯x=1
dy
=
dx
µ
74
3x + 2
x
¶
µ
¶ µ
¶
x−1
x−1 d
dy
7
2 d
= (2x − x )
+
(2x7 − x2 )
dx
dx x + 1
x + 1 dx
¶
·
¸ µ
(x + 1)(1) − (x − 1)(1)
x−1
7
2
(14x6 − 2x)
+
= (2x − x )
(x +µ1)2 ¶
x+1
2
x−1
(14x6 − 2x);
+
= (2x7 − x2 ) ·
2
(x + 1)
x+1
¯
1
2
dy ¯¯
= (2 − 1) + 0(14 − 2) =
dx ¯
4
2
x=1
29. 32t
30.
2π
31.
3πr2
33.
(a)
35.
(a)
g 0 (x) =
√ 0
1
1
xf (x) + √ f (x), g 0 (4) = (2)(−5) + (3) = −37/4
4
2 x
(b)
g 0 (x) =
xf 0 (x) − f (x) 0
(4)(−5) − 3
= −23/16
, g (4) =
x2
16
(a)
g 0 (x) = 6x − 5f 0 (x), g 0 (3) = 6(3) − 5(4) = −2
36.
(b) g 0 (x) =
37.
−2α−2 + 1
¯
dV ¯¯
= 4π(5)2 = 100π
dr ¯r=5
dV
= 4πr2
dr
(b)
32.
2f (x) − (2x + 1)f 0 (x) 0
2(−2) − 7(4)
, g (3) =
= −8
2
f (x)
(−2)2
(a)
F 0 (x) = 5f 0 (x) + 2g 0 (x), F 0 (2) = 5(4) + 2(−5) = 10
(b)
F 0 (x) = f 0 (x) − 3g 0 (x), F 0 (2) = 4 − 3(−5) = 19
(c)
F 0 (x) = f (x)g 0 (x) + g(x)f 0 (x), F 0 (2) = (−1)(−5) + (1)(4) = 9
(d)
F 0 (x) = [g(x)f 0 (x) − f (x)g 0 (x)]/g 2 (x), F 0 (2) = [(1)(4) − (−1)(−5)]/(1)2 = −1
¯
2
dy ¯¯
2
(1 + x)(−1) − (1 − x)(1)
1
dy
=−
,
= − and y = − for x = 2 so an equation of
=
38.
2 dx ¯
dx
(1 + x)2 µ
(1
+
x)
9
3
x=2
¶
2
2
1
1
= − (x − 2), or y = − x + .
the tangent line is y − −
3
9
9
9
39.
y − 2 = 5(x + 3), y = 5x + 17
40.
·
¸
d
1
1
λ0 + 6λ5
d λλ0 + λ6
(λλ0 + λ6 ) =
=
(λ0 + 6λ5 ) =
dλ 2 − λ0
2 − λ0 dλ
2 − λ0
2 − λ0
41.
(a)
dy/dx = 21x2 − 10x + 1, d2 y/dx2 = 42x − 10
(b)
dy/dx = 24x − 2, d2 y/dx2 = 24
(c)
dy/dx = −1/x2 , d2 y/dx2 = 2/x3
(d)
y = 35x5 − 16x3 − 3x, dy/dx = 175x4 − 48x2 − 3, d2 y/dx2 = 700x3 − 96x
75
42.
43.
44.
45.
Chapter 3
(a)
y 0 = 28x6 − 15x2 + 2, y 00 = 168x5 − 30x
(b)
y 0 = 3, y 00 = 0
(c)
y0 =
(d)
y = 2x4 + 3x3 − 10x − 15, y 0 = 8x3 + 9x2 − 10, y 00 = 24x2 + 18x
(a)
y 0 = −5x−6 + 5x4 , y 00 = 30x−7 + 20x3 , y 000 = −210x−8 + 60x2
(b)
y = x−1 , y 0 = −x−2 , y 00 = 2x−3 , y 000 = −6x−4
(c)
y 0 = 3ax2 + b, y 00 = 6ax, y 000 = 6a
(a)
dy/dx = 10x − 4, d2 y/dx2 = 10, d3 y/dx3 = 0
(b)
dy/dx = −6x−3 − 4x−2 + 1, d2 y/dx2 = 18x−4 + 8x−3 , d3 y/dx3 = −72x−5 − 24x−4
(c)
dy/dx = 4ax3 + 2bx, d2 y/dx2 = 12ax2 + 2b, d3 y/dx3 = 24ax
f 0 (x) = 6x, f 00 (x) = 6, f 000 (x) = 0, f 000 (2) = 0
¯
dy
d2 y
d2 y ¯¯
4
3
(b)
= 30x − 8x,
= 120x − 8,
= 112
dx
dx2
dx2 ¯x=1
(a)
(c)
46.
(a)
(b)
47.
2
4
, y 00 = − 3
5x2
5x
d2 £ −3 ¤
d3 £ −3 ¤
d4 £ −3 ¤
d £ −3 ¤
−5
−6
x
,
,
= −3x−4 ,
x
=
12x
x
=
−60x
x
= 360x−7 ,
dx
dx2
dx3
dx4
¯
d4 £ −3 ¤¯¯
= 360
x ¯
dx4
x=1
y 0 = 16x3 + 6x2 , y 00 = 48x2 + 12x, y 000 = 96x + 12, y 000 (0) = 12
3
4
dy
d2 y
−6 d y
−7 d y
= −24x−5 ,
=
120x
,
=
−720x
,
= 5040x−8 ,
y = 6x−4 ,
dx
dx2
dx3
dx4
¯
d4 y ¯¯
= 5040
dx4 ¯x=1
y 0 = 3x2 + 3, y 00 = 6x, and y 000 = 6 so
y 000 + xy 00 − 2y 0 = 6 + x(6x) − 2(3x2 + 3) = 6 + 6x2 − 6x2 − 6 = 0
48.
y = x−1 , y 0 = −x−2 , y 00 = 2x−3 so
x3 y 00 + x2 y 0 − xy = x3 (2x−3 ) + x2 (−x−2 ) − x(x−1 ) = 2 − 1 − 1 = 0
49.
51.
F 0 (x) = xf 0 (x) + f (x), F 00 (x) = xf 00 (x) + f 0 (x) + f 0 (x) = xf 00 (x) + 2f 0 (x)
dy
= 0, but
The graph has a horizontal tangent at points where
dx
dy
= x2 − 3x + 2 = (x − 1)(x − 2) = 0 if x = 1, 2. The
dx
corresponding values of y are 5/6 and 2/3 so the tangent line is
horizontal at (1, 5/6) and (2, 2/3).
1.5
0
3
0
Exercise Set 3.3
52.
76
dy
dy
9 − x2
;
= 2
= 0 when x2 = 9 so x = ±3. The points are
dx
(x + 9)2 dx
(3, 1/6) and (−3, −1/6).
0.2
-5.5
5.5
-0.2
53.
f 0 (1) ≈
54.
f 0 (1) ≈
0.999699 − (−1)
f (1.01) − f (1)
=
= 0.0301, and by differentiation, f 0 (1) = 3(1)2 − 3 = 0
0.01
0.01
1.01504 − 1
f (1.01) − f (1)
=
= 1.504, and by differentiation,
0.01
0.01
µ
¶¯
√
x
¯
x+ √
= 1.5
f 0 (1) =
¯
2 x x=1
55. f 0 (1) = 0
57.
56. f 0 (1) = 1
The y-intercept is −2 so the point (0, −2) is on the graph; −2 = a(0)2 + b(0) + c, c = −2. The
x-intercept is 1 so the point (1,0) is on the graph; 0 = a + b − 2. The slope is dy/dx = 2ax + b; at
x = 0 the slope is b so b = −1, thus a = 3. The function is y = 3x2 − x − 2.
dy
58. Let P (x0 , y0 ) be the point where y = x2 + k is tangent to y = 2x. The slope of the curve is
= 2x
dx
and the slope of the line is 2 thus at P , 2x0 = 2 so x0 = 1. But P is on the line, so y0 = 2x0 = 2.
Because P is also on the curve we get y0 = x20 + k so k = y0 − x20 = 2 − (1)2 = 1.
59.
The points (−1, 1) and (2, 4) are on the secant line so its slope is (4 − 1)/(2 + 1) = 1. The slope of
the tangent line to y = x2 is y 0 = 2x so 2x = 1, x = 1/2.
60. The √
points (1, 1) and
√ (4, 2) are on
√ the secant√line so its slope is 1/3. The slope of the tangent line to
y = x is y 0 = 1/(2 x) so 1/(2 x) = 1/3, 2 x = 3, x = 9/4.
61.
y 0 = −2x, so at any point (x0 , y0 ) on y = 1 − x2 the tangent line is y − y0 = −2x0 (x − x0 ), or
y = −2x0 x + x20 + 1. The point (2, 0) is√to be on the line, so 0 = −4x0 + x20 + 1, x20 − 4x0 + 1 = 0. Use
√
4 ± 16 − 4
= 2 ± 3.
the quadratic formula to get x0 =
2
62.
Let P1 (x1 , ax21 ) and P2 (x2 , ax22 ) be the points of tangency. y 0 = 2ax so the tangent lines at P1 and P2
are y − ax21 = 2ax1 (x − x1 ) and y − ax22 = 2ax2 (x − x2 ). Solve for x to get x = 12 (x1 + x2 ) which is the
x-coordinate of a point on the vertical line halfway between P1 and P2 .
63.
y 0 = 3ax2 + b; the tangent line at x = x0 is y − y0 = (3ax20 + b)(x − x0 ) where y0 = ax30 + bx0 . Solve
with y = ax3 + bx to get
(ax3 + bx) − (ax30 + bx0 ) = (3ax20 + b)(x − x0 )
ax3 + bx − ax30 − bx0 = 3ax20 x − 3ax30 + bx − bx0
x3 − 3x20 x + 2x30 = 0
(x − x0 )(x2 + xx0 − 2x20 ) = 0
(x − x0 )2 (x + 2x0 ) = 0, so x = −2x0 .
64.
Let (x0 , y0 ) be the point of tangency. Refer to the solution to Exercise 65 to see that the endpoints of
the line segment are at (2x0 , 0) and (0, 2y0 ), so (x0 , y0 ) is the midpoint of the segment.
77
65.
Chapter 3
1
1
x
2
; the tangent line at x = x0 is y − y0 = − 2 (x − x0 ), or y = − 2 + . The tangent line
x2
x0
x0
x0
1
crosses the x-axis at 2x0 , the y-axis at 2/x0 , so that the area of the triangle is (2/x0 )(2x0 ) = 2.
2
y0 = −
tangent where f 0 (x) = 0. Use the quadratic formula on
66. f 0 (x) = 3ax2 + 2bx + c; there is a horizontal
√
3ax2 + 2bx + c = 0 to get x = (−b ± b2 − 3ac)/(3a) which gives two real solutions, one real solution,
or none if
(b) b2 − 3ac = 0
(c) b2 − 3ac < 0
(a) b2 − 3ac > 0
67. F = GmM r−2 ,
dF
2GmM
= −2GmM r−3 = −
dr
r3
68.
dR/dT = 0.04124 − 3.558 × 10−5 T which decreases as T increases from 0 to 700. When T = 0,
dR/dT = 0.04124 Ω/◦ C; when T = 700, dR/dT = 0.01633 Ω/◦ C. The resistance is most sensitive to
temperature changes at T = 0◦ C, least sensitive at T = 700◦ C.
69.
f 0 (x) = 1 + 1/x2 > 0 for all x
6
-6
6
-6
70.
x2 − 4
;
(x2 + 4)2
f 0 (x) > 0 when x2 < 4, i.e. on −2 < x < 2
1.5
f 0 (x) = −5
-5
5
-1.5
71.
(f · g · h)0 = [(f · g) · h]0 = (f · g)h0 + h(f · g)0 = (f · g)h0 + h[f g 0 + f 0 g] = f gh0 + f g 0 h + f 0 gh
72.
(f1 f2 · · · fn )0 = (f10 f2 · · · fn ) + (f1 f20 · · · fn ) + · · · + (f1 f2 · · · fn0 )
73.
(a)
2(1 + x−1 )(x−3 + 7) + (2x + 1)(−x−2 )(x−3 + 7) + (2x + 1)(1 + x−1 )(−3x−4 )
(b) (x7 + 2x − 3)3 = (x7 + 2x − 3)(x7 + 2x − 3)(x7 + 2x − 3) so
d 7
(x + 2x − 3)3 = (7x6 + 2)(x7 + 2x − 3)(x7 + 2x − 3)
dx
+(x7 + 2x − 3)(7x6 + 2)(x7 + 2x − 3)
+(x7 + 2x − 3)(x7 + 2x − 3)(7x6 + 2)
= 3(7x6 + 2)(x7 + 2x − 3)2
74.
(a)
−5x−6 (x2 + 2x)(4 − 3x)(2x9 + 1) + x−5 (2x + 2)(4 − 3x)(2x9 + 1)
+x−5 (x2 + 2x)(−3)(2x9 + 1) + x−5 (x2 + 2x)(4 − 3x)(18x8 )
Exercise Set 3.3
(b)
75.
78
(x2 + 1)50 = (x2 + 1)(x2 + 1) · · · (x2 + 1), where (x2 + 1) occurs 50 times so
d 2
(x + 1)50 = [(2x)(x2 + 1) · · · (x2 + 1)] + [(x2 + 1)(2x) · · · (x2 + 1)]
dx
+ · · · + [(x2 + 1)(x2 + 1) · · · (2x)]
= 2x(x2 + 1)49 + 2x(x2 + 1)49 + · · · + 2x(x2 + 1)49
= 100x(x2 + 1)49 because 2x(x2 + 1)49 occurs 50 times.
f is continuous at 1 because lim− f (x) = lim+ f (x) = f (1), also lim− f 0 (x) = lim− 2x = 2 and
x→1
x→1
1
1
lim+ f (x) = lim+ √ = so f is not differentiable at 1.
x→1
x→1 2 x
2
x→1
x→1
0
76. f is continuous at 1/2 because lim − f (x) = lim + f (x) = f (1/2), also
x→1/2
0
x→1/2
0
lim f (x) = lim − 3x = 3/4 and lim + f (x) = lim + 3x/2 = 3/4 so f 0 (1/2) = 3/4.
x→1/2−
2
x→1/2
x→1/2
x→1/2
77. If f is differentiable at x = 1, then f is continuous there;
lim+ f (x) = lim− f (x) = f (1) = 3, a + b = 3; lim+ f 0 (x) = a and
x→1
x→1
x→1
lim− f 0 (x) = 6 so a = 6 and b = 3 − 6 = −3.
x→1
78.
(a)
lim f 0 (x) = lim− 2x = 0 and lim+ f 0 (x) = lim+ 2x = 0; f 0 (0) does not exist because f is not
x→0−
x→0
x→0
x→0
continuous at x = 0.
(b) lim− f 0 (x) = lim+ f 0 (x) = 0 and f is continuous at x = 0, so f 0 (0) = 0;
x→0
x→0
x→0
x→0
lim− f 00 (x) = lim− (2) = 2 and lim+ f 00 (x) = lim+ 6x = 0, so f 00 (0) does not exist.
79.
(a)
x→0
x→0
f (x) = 3x − 2 if x ≥ 2/3, f (x) = −3x + 2 if x < 2/3 so f is differentiable everywhere except
perhaps at 2/3. f is continuous at 2/3, also lim − f 0 (x) = lim − (−3) = −3 and lim + f 0 (x) =
x→2/3
x→2/3
x→2/3
lim + (3) = 3 so f is not differentiable at x = 2/3.
x→2/3
(b) f (x) = x2 − 4 if |x| ≥ 2, f (x) = −x2 + 4 if |x| < 2 so f is differentiable everywhere except
perhaps at ±2. f is continuous at −2 and 2, also lim− f 0 (x) = lim− (−2x) = −4 and lim+ f 0 (x) =
x→2
x→2
x→2
lim+ (2x) = 4 so f is not differentiable at x = 2. Similarly, f is not differentiable at x = −2.
x→2
f 0 (x) = −(1)x−2 , f 00 (x) = (2 · 1)x−3 , f 000 (x) = −(3 · 2 · 1)x−4
n(n − 1)(n − 2) · · · 1
f (n) (x) = (−1)n
xn+1
0
−3
00
(b) f (x) = −2x , f (x) = (3 · 2)x−4 , f 000 (x) = −(4 · 3 · 2)x−5
(n + 1)(n)(n − 1) · · · 2
f (n) (x) = (−1)n
xn+2
·
¸
·
¸
·
¸
d
d
d2
d
d
d
d
d2
81. (a)
[cf
(x)]
=
[f (x)]
[cf
(x)]
=
c
[f
(x)]
=
c
[f
(x)]
=
c
dx2
dx dx
dx dx
dx dx
dx2
·
¸
·
¸
d
d
d2
d
d2
d2
d
d
[f
(x)
+
g(x)]
=
[f
(x)]
+
[g(x)]
=
[f
(x)
+
g(x)]
=
[f
(x)]
+
[g(x)]
dx2
dx dx
dx dx
dx
dx2
dx2
80.
(a)
(b)
yes, by repeated application of the procedure illustrated in part (a)
82.
(f · g)0 = f g 0 + gf 0 , (f · g)00 = f g 00 + g 0 f 0 + gf 00 + f 0 g 0 = f 00 g + 2f 0 g 0 + f g 00
83.
(a)
f 0 (x) = nxn−1 , f 00 (x) = n(n − 1)xn−2 , f 000 (x) = n(n − 1)(n − 2)xn−3 , . . .,
f (n) (x) = n(n − 1)(n − 2) · · · 1
(b)
from part (a), f (k) (x) = k(k − 1)(k − 2) · · · 1 so f (k+1) (x) = 0 thus f (n) (x) = 0 if n > k
79
Chapter 3
(c)
from parts (a) and (b), f (n) (x) = an n(n − 1)(n − 2) · · · 1
84.
f 0 (2 + h) − f 0 (2)
= f 00 (2); f 0 (x) = 8x7 − 2, f 00 (x) = 56x6 , so f 00 (2) = 56(26 ) = 3584.
h→0
h
85.
(a)
lim
If a function is differentiable at a point then it is continuous at that point, thus f 0 is continuous
on (a, b) and consequently so is f .
(b) f and all its derivatives up to f (n−1) (x) are continuous on (a, b)
EXERCISE SET 3.4
1.
f 0 (x) = −2 sin x − 3 cos x
2.
f 0 (x) = sin x(− sin x) + cos x(cos x) = cos2 x − sin2 x = cos 2x
3.
f 0 (x) =
4.
f 0 (x) = x2 (− sin x) + (cos x)(2x) = −x2 sin x + 2x cos x
x(cos x) − sin x(1)
x cos x − sin x
=
x2
x2
5. f 0 (x) = x3 (cos x) + (sin x)(3x2 ) − 5(− sin x) = x3 cos x + (3x2 + 5) sin x
7.
cos x
x(− csc2 x) − (cot x)(1)
x csc2 x + cot x
cot x
(because
= cot x), f 0 (x) =
=
−
x
sin x
x2
x2
√
f 0 (x) = sec x tan x − 2 sec2 x
8.
f 0 (x) = (x2 + 1) sec x tan x + (sec x)(2x) = (x2 + 1) sec x tan x + 2x sec x
9.
f 0 (x) = sec x(sec2 x) + (tan x)(sec x tan x) = sec3 x + sec x tan2 x
6. f (x) =
10.
f 0 (x) =
=
11.
(1 + tan x)(sec x tan x) − (sec x)(sec2 x)
sec x tan x + sec x tan2 x − sec3 x
=
(1 + tan x)2
(1 + tan x)2
sec x(tan x + tan2 x − sec2 x)
sec x(tan x − 1)
=
2
(1 + tan x)
(1 + tan x)2
f 0 (x) = (csc x)(− csc2 x) + (cot x)(− csc x cot x) = − csc3 x − csc x cot2 x
12. f 0 (x) = 1 + 4 csc x cot x − 2 csc2 x
13.
f 0 (x) =
(1 + csc x)(− csc2 x) − cot x(0 − csc x cot x)
csc x(− csc x − csc2 x + cot2 x)
=
but
2
(1 + csc x)
(1 + csc x)2
1 + cot2 x = csc2 x (identity) thus cot2 x − csc2 x = −1 so
csc x(− csc x − 1)
csc x
=−
f 0 (x) =
(1 + csc x)2
1 + csc x
csc x(1 + sec2 x)
tan x(− csc x cot x) − csc x(sec2 x)
=−
2
tan x
tan2 x
14.
f 0 (x) =
15.
f (x) = sin2 x + cos2 x = 1 (identity) so f 0 (x) = 0
16.
f (x) =
1
= tan x, so f 0 (x) = sec2 x
cot x
Exercise Set 3.4
17.
18.
80
tan x
(because sin x sec x = (sin x)(1/ cos x) = tan x),
1 + x tan x
(1 + x tan x)(sec2 x) − tan x[x(sec2 x) + (tan x)(1)]
f 0 (x) =
(1 + x tan x)2
sec2 x − tan2 x
1
=
=
(because sec2 x − tan2 x = 1)
(1 + x tan x)2
(1 + x tan x)2
f (x) =
f (x) =
(x2 + 1) cot x
(because cos x csc x = (cos x)(1/ sin x) = cot x),
3 − cot x
f 0 (x) =
(3 − cot x)[2x cot x − (x2 + 1) csc2 x] − (x2 + 1) cot x csc2 x
(3 − cot x)2
=
6x cot x − 2x cot2 x − 3(x2 + 1) csc2 x
(3 − cot x)2
19.
dy/dx = −x sin x + cos x, d2 y/dx2 = −x cos x − sin x − sin x = −x cos x − 2 sin x
20.
dy/dx = − csc x cot x, d2 y/dx2 = −[(csc x)(− csc2 x) + (cot x)(− csc x cot x)] = csc3 x + csc x cot2 x
21. dy/dx = x(cos x) + (sin x)(1) − 3(− sin x) = x cos x + 4 sin x,
d2 y/dx2 = x(− sin x) + (cos x)(1) + 4 cos x = −x sin x + 5 cos x
22.
dy/dx = x2 (− sin x) + (cos x)(2x) + 4 cos x = −x2 sin x + 2x cos x + 4 cos x,
d2 y/dx2 = −[x2 (cos x) + (sin x)(2x)] + 2[x(− sin x) + cos x] − 4 sin x = (2 − x2 ) cos x − 4(x + 1) sin x
23.
dy/dx = (sin x)(− sin x) + (cos x)(cos x) = cos2 x − sin2 x,
d2 y/dx2 = (cos x)(− sin x) + (cos x)(− sin x) − [(sin x)(cos x) + (sin x)(cos x)] = −4 sin x cos x
24.
dy/dx = sec2 x; d2 y/dx2 = 2 sec2 x tan x
26.
Let f (x) = sin x, then f 0 (x) = cos x.
(a) f (0) = 0 and f 0 (0) = 1 so y − 0 = (1)(x − 0), y = x
(b)
(c)
27.
f (π) = 0 and f 0 (π) = −1 so y − 0 = (−1)(x − π), y = −x + π
³π ´
³π´
1
1
1
1
1 ³
π
1
π´
= √ and f 0
= √ so y − √ = √ x −
, y = √ x− √ + √
f
4
4
4
2
2
2
2
2
4 2
2
Let f (x) = tan x, then f 0 (x) = sec2 x.
f (0) = 0 and f 0 (0) = 1 so y − 0 = (1)(x − 0), y = x.
³
³π ´
³π ´
π´
π
= 1 and f 0
= 2 so y − 1 = 2 x −
, y = 2x − + 1.
(b) f
4
4
4
2
³
³ π´
³ π´
π´
π
0
= −1 and f −
= 2 so y + 1 = 2 x +
, y = 2x + − 1.
(c) f −
4
4
4
2
(a)
28.
29.
(a)
If y = cos x then y 0 = − sin x and y 00 = − cos x so y 00 + y = (− cos x) + (cos x) = 0;
if y = sin x then y 0 = cos x and y 00 = − sin x so y 00 + y = (− sin x) + (sin x) = 0.
(b)
y 0 = A cos x − B sin x, y 00 = −A sin x − B cos x so
y 00 + y = (−A sin x − B cos x) + (A sin x + B cos x) = 0.
(a)
f 0 (x) = cos x = 0 at x = ±π/2, ±3π/2.
(b)
f 0 (x) = 1 − sin x = 0 at x = −3π/2, π/2.
(c)
f 0 (x) = sec2 x ≥ 1 always, so no horizontal tangent line.
(d)
f 0 (x) = sec x tan x = 0 when sin x = 0, x = ±2π, ±π, 0
81
30.
Chapter 3
(a)
0.5
0
2p
-0.5
(b)
y = sin x cos x = (1/2) sin 2x and y 0 = cos 2x. So y 0 = 0 when 2x = (2n + 1)π/2 for n = 0, 1, 2, 3
or x = π/4, 3π/4, 5π/4, 7π/4
31.
x = 10 sin θ, dx/dθ = 10 cos θ; if θ = 60◦ , then
dx/dθ = 10(1/2) = 5 ft/rad = π/36 ft/deg ≈ 0.087 ft/deg
32.
s = 3800 csc θ, ds/dθ = −3800 csc θ cot θ; if θ = 30◦ , then
√
√
√
ds/dθ = −3800(2)( 3) = −7600 3 ft/rad = −380 3π/9 ft/deg ≈ −230 ft/deg
33.
D = 50 tan θ, dD/dθ = 50 sec2 θ; if θ = 45◦ , then
√
dD/dθ = 50( 2)2 = 100 m/rad = 5π/9 m/deg ≈ 1.75 m/deg
34.
(a)
From the right triangle shown, sin θ = r/(r + h) so r + h = r csc θ, h = r(csc θ − 1).
(b)
dh/dθ = −r csc θ cot θ; if θ = 30◦ , then
√
dh/dθ = −6378(2)( 3) ≈ −22, 094 km/rad ≈ −386 km/deg
35.
36.
(a)
d4k
d87
d3 d4·21
d3
d4
sin x = sin x, so
sin x = sin x;
sin x = 3 4·21 sin x = 3 sin x = − cos x
4
4k
87
dx
dx
dx
dx dx
dx
(b)
d100
d4k
cos
x
=
cos x = cos x
dx100
dx4k
d
[x sin x] = x cos x + sin x
dx
d2
[x sin x] = −x sin x + 2 cos x
dx2
d3
[x sin x] = −x cos x − 3 sin x
dx3
d4
[x sin x] = x sin x − 4 cos x
dx4
By mathematical induction one can show
d4k
[x sin x] = x sin x − (4k) cos x;
dx4k
d4k+1
[x sin x] = x cos x + (4k + 1) sin x;
dx4k+1
d4k+3
d4k+2
[x
sin
x]
=
−x
sin
x
+
(4k
+
2)
cos
x;
[x sin x] = −x cos x − (4k + 3) sin x;
dx4k+2
dx4k+3
d17
[x sin x] = x cos x + 17 sin x
Since 17 = 4 · 4 + 1,
dx17
37.
(a)
all x
(b)
all x
(c)
x 6= π/2 + nπ, n = 0, ±1, ±2, . . .
(d) x 6= nπ, n = 0, ±1, ±2, . . .
(e)
x 6= π/2 + nπ, n = 0, ±1, ±2, . . .
(f )
x 6= nπ, n = 0, ±1, ±2, . . .
(g)
x 6= (2n + 1)π, n = 0, ±1, ±2, . . .
(h)
x 6= nπ/2, n = 0, ±1, ±2, . . .
(i)
all x
Exercise Set 3.5
38.
(a)
(b)
(c)
82
d
cos(x + h) − cos x
cos x cos h − sin x sin h − cos x
[cos x] = lim
= lim
h→0
h→0
dx
h
h
¶
µ
¶¸
·
µ
sin h
cos h − 1
− sin x
= (cos x)(0) − (sin x)(1) = − sin x
= lim cos x
h→0
h
h
·
¸
1
d
d
cos x(0) − (1)(− sin x)
sin x
[sec x] =
=
=
= sec x tan x
dx
dx cos x
cos2 x
cos2 x
d
d h cos x i sin x(− sin x) − cos x(cos x)
[cot x] =
=
dx
dx sin x
sin2 x
− sin2 x − cos2 x
−1
=
= − csc2 x
sin2 x
sin2 x
·
¸
1
cos x
d
sin x(0) − (1)(cos x)
d
= − 2 = − csc x cot x
[csc x] =
=
dx
dx sin x
sin2 x
sin x
=
(d)
39. f 0 (x) = − sin x, f 00 (x) = − cos x, f 000 (x) = sin x, and f (4) (x) = cos x with higher order derivatives
repeating this pattern, so f (n) (x) = sin x for n = 3, 7, 11, . . .
µ
40.
(a)
(b)
sin h
cos h
h
¶
µ
sin h
h
cos h
¶
1
=1
h→0
1
tan x + tan h
− tan x
tan(x + h) − tan x
d
[tan x] = lim
= lim 1 − tan x tan h
h→0
h→0
dx
h
h
tan h
= lim
h→0
h→0
h
lim
= lim
=
tan x + tan h − tan x + tan2 x tan h
tan h(1 + tan2 x)
= lim
h→0
h→0 h(1 − tan x tan h)
h(1 − tan x tan h)
= lim
tan h
tan h sec2 x
2
h
= sec x lim
= lim
h→0 h(1 − tan x tan h)
h→0 1 − tan x tan h
tan h
h
= sec2 x
= sec x
lim (1 − tan x tan h)
lim
h→0
2
h→0
41.
d
tan(x + y) − tan y
tan(y + h) − tan y
= lim
=
(tan y) = sec2 y
x→0
h→0
x
h
dy
43.
Let t be the radian measure, then h =
lim
(a)
180
t and cos h = cos t, sin h = sin t.
π
π
cos h − 1
cos t − 1
cos t − 1
= lim
=
lim
=0
lim
t→0 180t/π
h→0
h
180 t→0
t
(b)
sin h
sin t
sin t
π
π
= lim
=
lim
=
t→0 180t/π
h→0 h
180 t→0 t
180
(c)
π
d
cos h − 1
sin h
[sin x] = sin x lim
+ cos x lim
= sin x(0) + cos x(π/180) =
cos x
h→0
h→0 h
dx
h
180
lim
EXERCISE SET 3.5
d 3
(x + 2x) = 37(x3 + 2x)36 (3x2 + 2)
dx
1.
f 0 (x) = 37(x3 + 2x)36
2.
f 0 (x) = 6(3x2 + 2x − 1)5
d
(3x2 + 2x − 1) = 6(3x2 + 2x − 1)5 (6x + 2) = 12(3x2 + 2x − 1)5 (3x + 1)
dx
83
3.
Chapter 3
µ
¶−3
µ
¶
µ
¶−3 µ
¶
7
7
7
7
d
f 0 (x) = −2 x3 −
x3 −
= −2 x3 −
3x2 + 2
x
dx
x
x
x
4. f (x) = (x5 − x + 1)−9 ,
f 0 (x) = −9(x5 − x + 1)−10
5.
9(5x4 − 1)
d 5
(x − x + 1) = −9(x5 − x + 1)−10 (5x4 − 1) = − 5
dx
(x − x + 1)10
f (x) = 4(3x2 − 2x + 1)−3 ,
f 0 (x) = −12(3x2 − 2x + 1)−4
6.
d
24(1 − 3x)
(3x2 − 2x + 1) = −12(3x2 − 2x + 1)−4 (6x − 2) =
dx
(3x2 − 2x + 1)4
d 3
1
3x2 − 2
(x − 2x + 5) = √
f 0 (x) = √
2 x3 − 2x + 5 dx
2 x3 − 2x + 5
√
d
1
3
7. f 0 (x) = p
(4 + 3 x) = √ p
√
√
dx
2 4+3 x
4 x 4+3 x
8. f 0 (x) = 3 sin2 x
9.
10.
f 0 (x) = cos(x3 )
d
(sin x) = 3 sin2 x cos x
dx
d 3
(x ) = 3x2 cos(x3 )
dx
√
√
√
√ d
√
√ d √
3 cos(3 x) sin(3 x)
√
f 0 (x) = 2 cos(3 x) [cos(3 x)] = −2 cos(3 x) sin(3 x) (3 x) = −
dx
dx
x
11. f 0 (x) = sec2 (4x2 )
d
(4x2 ) = 8x sec2 (4x2 )
dx
12.
f 0 (x) = 12 cot3 x
d
(cot x) = 12 cot3 x(− csc2 x) = −12 cot3 x csc2 x
dx
13.
f 0 (x) = 20 cos4 x
d
(cos x) = 20 cos4 x(− sin x) = −20 cos4 x sin x
dx
14.
f 0 (x) = − csc(x3 ) cot(x3 )
15.
f 0 (x) = cos(1/x2 )
16.
f 0 (x) = 4 tan3 (x3 )
17.
f 0 (x) = 4 sec(x7 )
d 3
(x ) = −3x2 csc(x3 ) cot(x3 )
dx
d
2
(1/x2 ) = − 3 cos(1/x2 )
dx
x
d
d
[tan(x3 )] = 4 tan3 (x3 ) sec2 (x3 ) (x3 ) = 12x2 tan3 (x3 ) sec2 (x3 )
dx
dx
18.
d
d
[sec(x7 )] = 4 sec(x7 ) sec(x7 ) tan(x7 ) (x7 ) = 28x6 sec2 (x7 ) tan(x7 )
dx
dx
¶
µ
¶
¶·
µ
¶¸
µ
µ
d
x
x
(x + 1)(1) − x(1)
x
x
cos
= 3 cos2
− sin
f 0 (x) = 3 cos2
x + 1 dx
x+1
x+1
x+1
(x + 1)2
¶
µ
¶
µ
3
x
x
sin
=−
cos2
2
(x + 1)
x+1
x+1
19.
5 sin(5x)
d
1
[cos(5x)] = − p
f 0 (x) = p
2 cos(5x) dx
2 cos(5x)
20.
1
3 − 8 sin(4x) cos(4x)
d
p
[3x − sin2 (4x)] =
f 0 (x) = p
2
dx
2 3x − sin (4x)
2 3x − sin2 (4x)
Exercise Set 3.5
21.
22.
84
£
¤−4 d £
¤
f 0 (x) = −3 x + csc(x3 + 3)
x + csc(x3 + 3)
dx
·
¸
¤−4
£
d 3
3
3
3
1 − csc(x + 3) cot(x + 3) (x + 3)
= −3 x + csc(x + 3)
dx
¤−4 £
¤
£
3
2
3
3
1 − 3x csc(x + 3) cot(x + 3)
= −3 x + csc(x + 3)
£
¤−5 d £ 4
¤
x − sec(4x2 − 2)
f 0 (x) = −4 x4 − sec(4x2 − 2)
dx
·
¸
¤
£ 4
d
−5
2
3
2
2
2
4x − sec(4x − 2) tan(4x − 2) (4x − 2)
= −4 x − sec(4x − 2)
dx
¤
¤
£
£ 4
−5
x2 − 2 sec(4x2 − 2) tan(4x2 − 2)
= −16x x − sec(4x2 − 2)
p
−2x
x(10 − 3x2 )
23. f 0 (x) = x2 · √
+ 2x 5 − x2 = √
2 5 − x2
5 − x2
√
24.
25.
26.
f 0 (x) =
√
1
1 − x2 (1) − x(−x/ 1 − x2 )
=
1 − x2
(1 − x2 )3/2
dy
d
= x3 (2 sin 5x) (sin 5x) + 3x2 sin2 5x = 10x3 sin 5x cos 5x + 3x2 sin2 5x
dx
dx
·
¸
√
√
√
√
√
√
√
1
3
1
dy
1
2
2
= x 3 tan ( x) sec ( x) √ + √ tan3 ( x) = tan2 ( x) sec2 ( x) + √ tan3 ( x)
dx
2
2 x
2 x
2 x
µ ¶
µ ¶
µ ¶
µ ¶µ
µ ¶
µ ¶
µ ¶
¶
1 d 1
1
1
1
1
1
1
tan
+ sec
(5x4 ) = x5 sec
tan
− 2 + 5x4 sec
x
x dx x
x
x
x
x
x
µ ¶
µ ¶
µ ¶
1
1
1
tan
+ 5x4 sec
= −x3 sec
x
x
x
27.
dy
= x5 sec
dx
28.
sec(3x + 1) cos x − 3 sin x sec(3x + 1) tan(3x + 1)
cos x − 3 sin x tan(3x + 1)
dy
=
=
2
dx
sec (3x + 1)
sec(3x + 1)
29.
d
dy
= − sin(cos x) (cos x) = − sin(cos x)(− sin x) = sin(cos x) sin x
dx
dx
30.
d
dy
= cos(tan 3x) (tan 3x) = 3 sec2 3x cos(tan 3x)
dx
dx
31.
d
d
dy
= 3 cos2 (sin 2x) [cos(sin 2x)] = 3 cos2 (sin 2x)[− sin(sin 2x)] (sin 2x)
dx
dx
dx
= −6 cos2 (sin 2x) sin(sin 2x) cos 2x
32.
(1 − cot x2 )(−2x csc x2 cot x2 ) − (1 + csc x2 )(2x csc2 x2 )
1 + cot x2 csc x2
dy
=
= −2x csc x2
2
2
dx
(1 − cot x )
(1 − cot x2 )2
33.
d
d
dy
= (5x + 8)13 12(x3 + 7x)11 (x3 + 7x) + (x3 + 7x)12 13(5x + 8)12 (5x + 8)
dx
dx
dx
= 12(5x + 8)13 (x3 + 7x)11 (3x2 + 7) + 65(x3 + 7x)12 (5x + 8)12
34.
dy
= (2x − 5)2 3(x2 + 4)2 (2x) + (x2 + 4)3 2(2x − 5)(2) = 6x(2x − 5)2 (x2 + 4)2 + 4(2x − 5)(x2 + 4)3
dx
= 2(2x − 5)(x2 + 4)2 (8x2 − 15x + 8)
85
Chapter 3
35.
·
¸2
·
¸
·
¸2
dy
11
33(x − 5)2
x−5
x−5
d x−5
·
=
=3
=3
2
dx
2x + 1 dx 2x + 1
2x + 1
(2x + 1)
(2x + 1)4
36.
dy
= 17
dx
= 17
37.
38.
39.
40.
41.
µ
µ
1 + x2
1 − x2
1 + x2
1 − x2
¶16
¶16
d
dx
µ
1 + x2
1 − x2
¶
µ
= 17
1 + x2
1 − x2
¶16
(1 − x2 )(2x) − (1 + x2 )(−2x)
(1 − x2 )2
4x
68x(1 + x2 )16
=
2
2
(1 − x )
(1 − x2 )18
dy
(4x2 − 1)8 (3)(2x + 3)2 (2) − (2x + 3)3 (8)(4x2 − 1)7 (8x)
=
dx
(4x2 − 1)16
2
2
7
2(2x + 3) (4x − 1) [3(4x2 − 1) − 32x(2x + 3)]
2(2x + 3)2 (52x2 + 96x + 3)
=
=−
2
16
(4x − 1)
(4x2 − 1)9
dy
d
= 12[1 + sin3 (x5 )]11 [1 + sin3 (x5 )]
dx
dx
d
= 12[1 + sin3 (x5 )]11 3 sin2 (x5 ) sin(x5 ) = 180x4 [1 + sin3 (x5 )]11 sin2 (x5 ) cos(x5 )
dx
£
¤4 d £
¤
dy
= 5 x sin 2x + tan4 (x7 )
x sin 2x tan4 (x7 )
dx
dx
¸
·
¤
£
d
d
4
= 5 x sin 2x + tan4 (x7 )
x cos 2x (2x) + sin 2x + 4 tan3 (x7 ) tan(x7 )
dx
dx
¤ £
¤
£
4 7 4
6
3 7
= 5 x sin 2x + tan (x ) 2x cos 2x + sin 2x + 28x tan (x ) sec2 (x7 )
d
dy
= cos(3x2 ) (3x2 ) = 6x cos(3x2 ),
dx
dx
d
d2 y
= 6x(− sin(3x2 )) (3x2 ) + 6 cos(3x2 ) = −36x2 sin(3x2 ) + 6 cos(3x2 )
2
dx
dx
d
d
dy
= x(− sin(5x)) (5x) + cos(5x) − 2 sin x (sin x)
dx
dx
dx
= −5x sin(5x) + cos(5x) − 2 sin x cos x = −5x sin(5x) + cos(5x) − sin(2x),
d
d
d
d2 y
= −5x cos(5x) (5x) − 5 sin(5x) − sin(5x) (5x) − cos(2x) (2x)
dx2
dx
dx
dx
= −25x cos(5x) − 10 sin(5x) − 2 cos(2x)
µ ¶
µ ¶
µ ¶
µ ¶
µ ¶
1
1
1
1 d 1
1
2
+ tan
= − sec
+ tan
,
x dx x
x
x
x
x
µ ¶
µ ¶
µ ¶
µ ¶
µ ¶
µ ¶
µ ¶
1 d
1
1
2
1
1
1 d 1
1
2
d2 y
2
2
2
sec
sec
+
+
sec
=
tan
=
−
sec
sec
dx2
x
x dx
x
x2
x
x dx x
x3
x
x
42.
dy
= x sec2
dx
43.
(1 − x) + (1 + x)
2
d2 y
dy
−2
=
=
=
2(1
−
x)
and
= −2(2)(−1)(1 − x)−3 = 4(1 − x)−3
dx
(1 − x)2
(1 − x)2
dx2
45.
dy
dy
= −3x sin 3x + cos 3x; if x = π then y = −π and
= −1 so y + π = −(x − π), y = −x
dx
dx
46.
dy
dy
= 3x2 cos(1 + x3 ); if x = −3 then y = sin(−26) = − sin 26 and
= 27 cos 26
dx
dx
so y + sin 26 = 27(cos 26)(x + 3)
Exercise Set 3.5
47.
48.
49.
dy
dy
= −3 sec3 (π/2 − x) tan(π/2 − x); if x = −π/2 then y = −1 and
=0
dx
dx
so y + 1 = (0)(x + π/2), y = −1
dy
dy
= 3(x − 1/x)2 (1 + 1/x2 ); if x = 2 then y = 27/8 and
= 135/16
dx
dx
135
135
27
27
=
(x − 2), y =
x−
so y −
8
16
16
2
y = cot3 (π − θ) = − cot3 θ so dy/dx = 3 cot2 θ csc2 θ
µ
50.
51.
86
6
au + b
cu + d
¶5
ad − bc
(cu + d)2
d
[a cos2 πω + b sin2 πω] = −2πa cos πω sin πω + 2πb sin πω cos πω
dω
= π(b − a)(2 sin πω cos πω) = π(b − a) sin 2πω
52.
2 csc2 (π/3 − y) cot(π/3 − y)
53.
(a)
(c)
2
f 0 (x) = x √
p
−x
4 − 2x2
+ 4 − x2 = √
4 − x2
4 − x2
2
-2
2
-2
2
-2
(d)
f (1) =
√
-6
√
2
2
3 and f 0 (1) = √ so the tangent line has the equation y − 3 = √ (x − 1).
3
3
3
0
2
0
54.
(a)
(c)
0.5
f 0 (x) = 2x cos x2 cos x − sin x sin x2
1.2
^
^
0
6
6
-1.2
87
Chapter 3
(d)
f (1) = sin 1 cos 1 and f 0 (1) = 2 cos2 1 − sin2 1, so the tangent line has the equation
y − sin 1 cos 1 = (2 cos2 1 − sin2 1)(x − 1).
0.8
^
55.
0
6
(a)
dy/dt = −Aω sin ωt, d2 y/dt2 = −Aω 2 cos ωt = −ω 2 y
(b)
one complete oscillation occurs when ωt increases over an interval of length 2π, or if t increases
over an interval of length 2π/ω
(c)
f = 1/T
(d)
amplitude = 0.6 cm,
T = 2π/15 s/oscillation,
f = 15/(2π) oscillations/s
56.
dy/dt = 3A cos 3t, d2 y/dt2 = −9A sin 3t, so −9A sin 3t + 2A sin 3t = 4 sin 3t,
−7A sin 3t = 4 sin 3t, −7A = 4, A = −4/7
57.
(a)
p ≈ 10 lb/in2 , dp/dh ≈ −2 lb/in2 /mi
(b)
dp dh
dp
=
≈ (−2)(0.3) = −0.6 lb/in2 /s
dt
dh dt
58.
(a)
(b)
59.
60.
61.
dF dθ
dF
=
≈ (0.18)(−0.5) = −0.09 lb/s
dt
dθ dt
With u = sin x,
½
d
d
du
d
d
cos x, u > 0
(| sin x|) =
(|u|) =
(|u|)
=
(|u|) cos x =
− cos x, u < 0
dx
dx
du
dx
du
½
½
cos x, sin x > 0
cos x,
0 0 an increase in L gives an increase in T , which is the period. To speed up a clock,
Since
dL
decrease the period; to decrease T , decrease L.
√
dT
L
= − 3/2 < 0; a decrease in g will increase T and the clock runs slower
dg
g
√
¶
µ
√
dT
−1
L
−3/2
=2 L
g
= − 3/2
(f ) s3 /m
dg
2
g
97
Chapter 3
(a)
f 0 (x) = 2x, f 0 (1.8) = 3.6
(b)
f 0 (x) = (x2 − 4x)/(x − 2)2 , f 0 (3.5) ≈ −0.777778
24.
(a)
f 0 (x) = 3x2 − 2x, f 0 (2.3) = 11.27
25.
f 0 (x) = 2x ln 2; f 0 (2) ≈ 2.772589
23.
(b)
f 0 (x) = (1 − x2 )/(x2 + 1)2 , f 0 (−0.5) = 0.48
26. f 0 (x) = xsin x (cos x ln x + sin x/x); f 0 (2) = 0.312141
¯
3(h + 1)2.5 + 580h − 3
1 d 2.5 ¯¯
1
= 58 +
3x ¯
= lim
= 58 + (2.5)(3)(1)1.5 = 58.75 ft/s
h→0
10h
10 dx
10
x=1
27.
vinst
28.
164 ft/s
29.
Solve 3x2 − cos x = 0 to get x = ±0.535428.
2500
1
20
0
30.
When x4 − x − 1 > 0, f (x) = x4 − 2x − 1; when x4 − x − 1 < 0,
f (x) = −x4 + 1, and f is differentiable in both cases. The roots of
x4 − x − 1 = 0 are x1 = −0.724492, x2 = 1.220744. So x4 − x − 1 > 0
on (−∞, x1 ) and (x2 , +∞), and x4 − x − 1 < 0 on (x1 , x2 ). Then
lim− f 0 (x) = lim− (4x3 − 2) = 4x31 − 2 and
x→x1
x→x1
x→x+
1
x→x1
1.5
-1.5
2
lim f 0 (x) = lim+ −4x3 = −4x31 which is not equal to 4x31 − 2, so
f is not differentiable at x = x1 ; similarly f is not differentiable at
x = x2 .
31.
(a)
f 0 (x) = 5x4
(b)
(d)
f 0 (x) = −3/(x − 1)2
(e)
0
2
32. f (x) = 2x sin x + x cos x
34.
f 0 (x) =
0
36.
f (x) =
37.
f 0 (x) =
6x2 + 8x − 17
(3x + 2)2
x2 cos
f 0 (x) = −1/x2
√
f 0 (x) = 3x/ 3x2 + 5
33.
√
1 − 2 x sin 2x
√
f (x) =
2 x
35.
f 0 (x) =
-1.5
(c) f 0 (x) = −1/2x3/2
(f )
f 0 (x) = 3 cos 3x
0
(1 + x2 ) sec2 x − 2x tan x
(1 + x2 )2
√
√
x − 2x3/2 sin x
2x7/2
−2x5 sin x − 2x4 cos x + 4x4 + 6x2 sin x + 6x − 3x cos x − 4x sin x + 4 cos x − 8
√
2x2 x4 − 3 + 2(2 − cos x)2
Horizon Module 3
98
CHAPTER 3 HORIZON MODULE
1.
x1 = l1 cos θ1 , x2 = l2 cos(θ1 + θ2 ), so x = x1 + x2 = l1 cos θ1 + l2 cos(θ1 + θ2 ) (see Figure 3 in text);
similarly y1 = l1 sin θ1 + l2 sin(θ1 + θ2 ).
2.
Fix θ1 for the moment and let θ2 vary; then the distance r from (x, y) to the origin (see Figure 3 in
text) is at most l1 + l2 and at least l1 − l2 if l1 ≥ l2 and l2 − l1 otherwise. For any fixed θ2 let θ1 vary
and the point traces out a circle of radius r.
(a) {(x, y) : 0 ≤ x2 + y 2 ≤ 2l1 }
3.
(b)
{(x, y) : l1 − l2 ≤ x2 + y 2 ≤ l1 + l2 }
(c)
{(x, y) : l2 − l1 ≤ x2 + y 2 ≤ l1 + l2 }
(x, y) = (l1 cos θ + l2 cos(θ1 + θ2 ), l1 sin θ1 + l2 sin(θ1 + θ2 ))
= (cos(π/4) + 3 cos(5π/12), sin(π/4) + 3 sin(5π/12)) =
Ã√
√
√
√ !
2+3 6 7 2+3 6
,
4
4
4. x = (1) cos 2t + (1) cos(2t + 3t) = cos 2t + cos 5t,
y = (1) sin 2t + (1) sin(2t + 3t) = sin 2t + sin 5t
y
5.
y
2
-2
-2
6.
2
2
v1 = 3, v2 = 5
y
x
-2
2
2
-2
v1 = 1, v2 = 4
x
x
1
2
-2
v1 = 4, v2 = 1
x = 2 cos t, y = 2 sin t, a circle of radius 2
9 = [3 sin(θ1 + θ2 )]2 + [3 cos(θ1 + θ2 )]2 = [5 − 3 sin θ1 ]2 + [3 − 3 cos θ1 ]2
= 25 − 30 sin θ1 + 9 sin2 θ1 + 9 − 18 cos θ1 + 9 cos2 θ1 = 43 − 30 sin θ1 − 18 cos θ1 ,
so 15 sin θ1 + 9 cos θ1 = 17
¶2
µ
17 − 9 cos θ1
2
2
+ cos θ1 , or 306 cos2 θ1 − 306 cos θ1 = −64
(b) 1 = sin θ1 + cos θ2 =
15
√
³
´
p
1 5 17
2
(c) cos θ1 = 153 ± (153) − 4(153)(32) /306 = ±
2
102
(e) If θ1 = 0.792436 rad, then θ2 = 0.475882 rad ≈ 27.2660◦ ;
if θ1 = 1.26832 rad, then θ2 = −0.475882 rad ≈ −27.2660◦ .
µ
¶
dθ2
dθ2
dθ1
dx
dθ1
dθ1
= −3 sin θ1
− (3 sin(θ1 + θ2 ))
+
= −3
(sin θ1 + sin(θ1 + θ2 )) − 3 (sin(θ1 + θ2 ))
8.
dt
dt
dt
dt
dt
dt
dy
dθ1
dx
dy
dθ2
dθ2
dθ1
− 3(sin(θ1 + θ2 ))
; similarly
=x
+ 3(cos(θ1 + θ2 ))
. Now set
= 0,
= 1.
= −y
dt
dt
dt
dt
dt
dt
dt
7.
9.
(a)
1
x = 3 cos(π/3) + 3 cos(−π/3) = 6 = 3 and y = 3 sin(π/3) − 3 sin(π/3) = 0; equations (4)
2
dθ1
dθ2
dθ2
= 0, 3
+ 3 cos(π/3)
= 1 with solution dθ2 /dt = 0, dθ1 /dt = 1/3.
become 3 sin(π/3)
dt
dt
dt
dθ1
dθ2
dθ1
= 0 and −3
−3
= 1, with solution dθ1 /dt = 0, dθ2 /dt = −1/3.
(b) x = −3, y = 3, so −3
dt
dt
dt
(a)
CHAPTER 4
Logarithmic and Exponential Functions
EXERCISE SET 4.1
1.
(a)
f (g(x)) = 4(x/4) = x, g(f (x)) = (4x)/4 = x, f and g are inverse functions
(b)
f (g(x)) = 3(3x − 1) + 1 = 9x − 2 6= x so f and g are not inverse functions
p
f (g(x)) = 3 (x3 + 2) − 2 = x, g(f (x)) = (x − 2) + 2 = x, f and g are inverse functions
(c)
6 x, f and g are not inverse functions
(d) f (g(x)) = (x1/4 )4 = x, g(f (x)) = (x4 )1/4 = |x| =
2.
(a) They are inverse functions.
2
-2
2
-2
(b) The graphs are not reflections of each other
about the line y = x.
2
-2
2
-2
(c) They are inverse functions provided the domain of
g is restricted to [0, +∞)
5
0
5
0
(d) They are inverse functions provided the domain of f (x)
is restricted to [0, +∞)
2
0
2
0
3.
4.
(a)
yes; all outputs (the elements of row two) are distinct
(b)
no; f (1) = f (6)
(a)
no; it is easy to conceive of, say, 8 people in line at two different times
(b)
no; perhaps your weight remains constant for more than a year
(c)
yes, since the function is increasing, in the sense that the greater the volume, the greater the
weight
99
Exercise Set 4.1
100
5.
(a)
yes
(b)
yes
(c)
6.
(a)
no, the horizontal line test fails
no
(d)
yes
(e)
(b)
yes, horizontal line test
6
no
(f )
no
10
-1
-3
3
3
-2
-10
7.
(a) no, the horizontal line test fails
(b) no, the horizontal line test fails
(c) yes, horizontal line test
9.
(a)
8.
(a) no, the horizontal line test fails
(b) no, the horizontal line test fails
(c) yes, horizontal line test
f has an inverse because the graph passes the horizontal line test. To compute f −1 (2) start at
2 on the y-axis and go to the curve and then down, so f −1 (2) = 8; similarly, f −1 (−1) = −1 and
f −1 (0) = 0.
(b) domain of f −1 is [−2, 2], range is [−8, 8]
y
(c)
8
4
-2
-1
1
2
x
-4
-8
10.
11.
12.
(a)
the horizontal line test fails
(b)
−∞ < x ≤ −1; −1 ≤ x ≤ 2; and 2 ≤ x < +∞.
(a)
f 0 (x) = 2x + 8; f 0 < 0 on (−∞, −4) and f 0 > 0 on (−4, +∞); not one-to-one
(b)
f 0 (x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f 0 (x) is positive for all x, so f is one-to-one
(c)
f 0 (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one
(a)
f 0 (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so f is not one-to-one
(b)
f 0 (x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f 0 is positive for all x, so f is one-to-one
(c)
1
; f is one-to-one because:
(x + 1)2
if x1 < x2 < −1 then f 0 > 0 on [x1 , x2 ], so f (x1 ) 6= f (x2 )
if −1 < x1 < x2 then f 0 > 0 on [x1 , x2 ], so f (x1 ) 6= f (x2 )
if x1 < −1 < x2 then f (x1 ) > 1 > f (x2 ) since f (x) > 1 on (−∞, −1) and f (x) < 1 on (−1, +∞)
f 0 (x) =
13.
y = f −1 (x), x = f (y) = y 5 , y = x1/5 = f −1 (x)
14.
y = f −1 (x), x = f (y) = 6y, y =
15.
y = f −1 (x), x = f (y) = 7y − 6, y =
16.
y = f −1 (x), x = f (y) =
1
x = f −1 (x)
6
1
(x + 6) = f −1 (x)
7
x+1
y+1
, xy − x = y + 1, (x − 1)y = x + 1, y =
= f −1 (x)
y−1
x−1
101
Chapter 4
17.
y = f −1 (x), x = f (y) = 3y 3 − 5, y =
18.
y = f −1 (x), x = f (y) =
√
5
19.
y = f −1 (x), x = f (y) =
√
3
−1
p
3
(x + 5)/3 = f −1 (x)
4y + 2, y =
1 5
(x − 2) = f −1 (x)
4
2y − 1, y = (x3 + 1)/2 = f −1 (x)
5
,y=
(x), x = f (y) = 2
y +1
r
5−x
= f −1 (x)
x
20.
y=f
21.
p
y = f −1 (x), x = f (y) = 3/y 2 , y = − 3/x = f −1 (x)
22.
y = f −1 (x), x = f (y) =
½
23. y = f −1 (x), x = f (y) =
½
2y, y ≤ 0
,
y2 , y > 0
½
x/2,
x≤0
√
x, x > 0
y = f −1 (x) =
5/2 − y, y < 2
, y = f −1 (x) =
1/y, y ≥ 2
½
5/2 − x,
x > 1/2
1/x, 0 < x ≤ 1/2
24.
y = p−1 (x), x = p(y) = y 3 − 3y 2 + 3y − 1 = (y − 1)3 , y = x1/3 + 1 = p−1 (x)
25.
y = f −1 (x), x = f (y) = (y + 2)4 for y ≥ 0, y = f −1 (x) = x1/4 − 2 for x ≥ 16
26. y = f −1 (x), x = f (y) =
27.
√
y + 3 for y ≥ −3, y = f −1 (x) = x2 − 3 for x ≥ 0
√
y = f −1 (x), x = f (y) = − 3 − 2y for y ≤ 3/2, y = f −1 (x) = (3 − x2 )/2 for x ≤ 0
28.
y = f −1 (x), x = f (y) = 3y 2 + 5y − 2 for y ≥ 0, 3y 2 + 5y − 2 − x = 0 for y ≥ 0,
√
y = f −1 (x) = (−5 + 12x + 49)/6 for x ≥ −2
29.
y = f −1 (x), x = f (y) = y − 5y 2 for y ≥ 1, 5y 2 − y + x = 0 for y ≥ 1,
√
y = f −1 (x) = (1 + 1 − 20x)/10 for x ≤ −4
30.
(a)
(b)
31.
32.
(c)
C = −273.15◦ C is equivalent to F = −459.67◦ F, so the domain is F ≥ −459.67, the range is
C ≥ −273.15
(a)
y = f (x) = (6.214 × 10−4 )x
(c)
how many meters in y miles
(b) x = f −1 (y) =
104
y
6.214
f and f −1 are continuous so f (3) = lim f (x) = 7; then f −1 (7) = 3, and
x→3
³
´
−1
−1
−1
lim x = f (7) = 3
lim f (x) = f
x→7
33.
5
(F − 32)
9
how many degrees Celsius given the Fahrenheit temperature
C=
(a)
x→7
√
f (g(x)) = f ( x)
√
= ( x)2 = x, x > 1;
g(f (x)) = g(x2 )
√
= x2 = x, x > 1
(b)
y
y = f(x)
y = g(x)
x
Exercise Set 4.1
(c)
34.
102
no, because f (g(x)) = x for every x in the domain of g is not satisfied (the domain of g is x ≥ 0)
2
2
y = f −1 (x),
p x = f (y) = ay + by + c, ay + by + c − x = 0, use the quadratic formula to get
2
−b ± b − 4a(c − x)
;
y=
2a
p
p
−b + b2 − 4a(c − x)
−b − b2 − 4a(c − x)
(b) f −1 (x) =
(a) f −1 (x) =
2a
2a
3−x
1 − x = 3 − 3x − 3 + x = x so f = f −1
35. (a) f (f (x)) =
3−x
1−x−3+x
1−
1−x
(b) symmetric about the line y = x
3−
36. y = m(x − x0 ) is an equation of the line. The graph of the inverse of f (x) = m(x − x0 ) will be the
reflection of this line about y = x. Solve y = m(x − x0 ) for x to get x = y/m + x0 = f −1 (y) so
y = f −1 (x) = x/m + x0 .
37.
f (x) = x3 − 3x2 + 2x = x(x − 1)(x − 2) so f (0) = f (1) = f (2) = 0 thus f is not one-to-one.
√
√
6 ± 36 − 24
= 1 ± 3/3. f 0 (x) > 0 (f is increasing)
(b) f 0 (x) = 3x2 − 6x + 2, f 0 (x) = 0 when x =
6√
√
√
if x < 1 − 3/3,√f 0 (x) < 0 (f is decreasing) √
if 1 − 3/3 < x√< 1 + 3/3, so f (x) takes on values
less than f (1 − 3/3) on both sides of 1 − 3/3 thus 1 − 3/3 is the largest value of k.
38.
(a)
(a)
f (x) = x3 (x − 2) so f (0) = f (2) = 0 thus f is not one to one.
(b) f 0 (x) = 4x3 − 6x2 = 4x2 (x − 3/2), f 0 (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2]
and increasing on [3/2, +∞) so 3/2 is the smallest value of k.
39.
if f −1 (x) = 1, then x = f (1) = 2(1)3 + 5(1) + 3 = 10
40.
if f −1 (x) = 2, then x = f (2) = (2)3 /[(2)2 + 1] = 8/5
41.
42.
6
-2
10
-5
6
10
-2
43.
-5
44.
3
0
3
0
6
0
6
0
45.
f (f (x)) = x thus f = f −1 so the graph is symmetric about y = x.
46.
(a)
Suppose x1 6= x2 where x1 and x2 are in the domain of g and g(x1 ), g(x2 ) are in the domain of f
then g(x1 ) 6= g(x2 ) because g is one-to-one so f (g(x1 )) 6= f (g(x2 )) because f is one-to-one thus
f ◦ g is one-to-one because (f ◦ g)(x1 ) 6= (f ◦ g)(x2 ) if x1 6= x2 .
103
Chapter 4
(b)
f , g, and f ◦ g all have inverses because they are all one-to-one. Let h = (f ◦ g)−1 then
(f ◦ g)(h(x)) = f [g(h(x))] = x, apply f −1 to both sides to get g(h(x)) = f −1 (x), then apply g −1
to get h(x) = g −1 (f −1 (x)) = (g −1 ◦ f −1 )(x), so h = g −1 ◦ f −1
47.
y
x
48.
Suppose that g and h are both inverses of f then f (g(x)) = x, h[f (g(x))] = h(x), but h[f (g(x))] = g(x)
because h is an inverse of f so g(x) = h(x).
49.
F 0 (x) = 2f 0 (2g(x))g 0 (x) so F 0 (3) = 2f 0 (2g(3))g 0 (3). By inspection f (1) = 3, so g(3) = f −1 (3) = 1
and g 0 (3) = (f −1 )0 (3) = 1/f 0 (f −1 (3)) = 1/f 0 (1) = 1/7 because f 0 (x) = 4x3 + 3x2 . Thus
F 0 (3) = 2f 0 (2)(1/7) = 2(44)(1/7) = 88/7.
EXERCISE SET 4.2
1.
(a)
−4
(b)
4
(c) 1/4
2.
(a)
1/16
(b)
8
(c) 1/3
3.
(a)
2.9690
(b) 0.0341
4.
(a)
1.8882
(b) 0.9381
¶
1
= log2 (2−5 ) = −5
(b) log2
32
(d) log9 3 = log9 (91/2 ) = 1/2
µ
5.
4
(a)
log2 16 = log2 (2 ) = 4
(c)
log4 4 = 1
6.
(a) log10 (0.001) = log10 (10−3 ) = −3
(c) ln(e3 ) = 3
(b) log10 (104 ) = 4
√
(d) ln( e) = ln(e1/2 ) = 1/2
7.
(a)
1.3655
(b) −0.3011
8.
(a)
−0.5229
(b)
9.
(a)
2 ln a +
10.
(a)
1
ln c − ln a − ln b = t/3 − r − s
3
11.
(a)
1 + log x +
12.
(a)
13.
log
1
1
ln b + ln c = 2r + s/2 + t/2
2
2
1.1447
(b) ln b − 3 ln a − ln c = s − 3r − t
(b)
1
(ln a + 3 ln b − 2 ln c) = r/2 + 3s/2 − t
2
1
log(x − 3)
2
(b)
2 ln |x| + 3 ln sin x −
1
log(x + 2) − log cos 5x
3
(b)
1
1
ln(x2 + 1) − ln(x3 + 5)
2
2
24 (16)
= log(256/3)
3
14.
log
1
ln(x2 + 1)
2
√
√
100 x
x − log(sin3 2x) + log 100 = log
sin3 2x
Exercise Set 4.2
√
3
15.
17.
ln
√
104
x(x + 1)2
cos x
16. 1 + x = 103 = 1000, x = 999
x = 10−1 = 0.1, x = 0.01
18. x2 = e4 , x = ±e2
19.
1/x = e−2 , x = e2
20.
x=7
21.
2x = 8, x = 4
22.
log10 x3 = 30, x3 = 1030 , x = 1010
23.
log10 x = 5, x = 105
√
4
4
= ln 2, 5 = 2, x5 = 2, x = 5 2
5
x
x
p
p
25. ln 2x2 = ln 3, 2x2 = 3, x2 = 3/2, x = 3/2 (we discard − 3/2 because it does not satisfy the original
equation)
24.
ln 4x − ln x6 = ln 2, ln
26.
ln 3x = ln 2, x ln 3 = ln 2, x =
27.
ln 5−2x = ln 3, −2x ln 5 = ln 3, x = −
28.
1
e−2x = 5/3, −2x = ln(5/3), x = − ln(5/3)
2
29.
e3x = 7/2, 3x = ln(7/2), x =
30.
ex (1 − 2x) = 0 so ex = 0 (impossible) or 1 − 2x = 0, x = 1/2
31.
e−x (x + 2) = 0 so e−x = 0 (impossible) or x + 2 = 0, x = −2
32.
e2x − ex − 6 = (ex − 3)(ex + 2) = 0 so ex = −2 (impossible) or ex = 3, x = ln 3
33.
e−2x − 3e−x + 2 = (e−x − 2)(e−x − 1) = 0 so e−x = 2, x = − ln 2 or e−x = 1, x = 0
34.
(a)
ln 2
ln 3
ln 3
2 ln 5
1
ln(7/2)
3
(b)
y
y
2
2
x
6
2
-2
-2
35.
(a)
2
(b)
y
x
y
2
6
4
x
-4
2
2
-2
2
4
x
-4
105
36.
Chapter 4
(a)
(b)
y
-1
y
x
x
3
-1
-10
37.
38.
log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈ 2.8777;
log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈ −0.3174
39.
10
2
0
0
-5
40.
41.
3
2
-3
(a)
Let X = logb x and Y = loga x. Then bX = x and aY = x so aY = bX , or aY /X = b, which means
loga x
loga x
= loga b, logb x =
.
loga b = Y /X. Substituting for Y and X yields
logb x
loga b
(b)
Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = 1.
(log2 81)(log3 32) = (log2 [34 ])(log3 [25 ]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20
x = 3.6541, y = 1.2958
2
2
0.6
6
42.
Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)x for x by
taking common logarithms, resulting in 3 = log 0.051517 + x log 1.1306727, which yields x ≈ 77.4, so
the debt first reached one trillion dollars around 1977.
43.
(a) no, the curve passes through the origin
(c) y = 2−x
(b) y = 2x/4
√
(d) y = ( 5)x
5
-1
2
0
Exercise Set 4.2
44.
(a)
106
As x → +∞ the function grows very slowly,
but it is always increasing and tends to +∞.
As x → 1+ the function tends to −∞.
(b)
y
x
150
-2
45.
log(1/2) < 0 so 3 log(1/2) < 2 log(1/2)
46.
Let x = logb a and y = logb c, so a = bx and c = by .
First, ac = bx by = bx+y or equivalently, logb (ac) = x + y = logb a + logb c.
Secondly, a/c = bx /by = bx−y or equivalently, logb (a/c) = x − y = logb a − logb c.
Next, ar = (bx )r = brx or equivalently, logb ar = rx = r logb a.
Finally, 1/c = 1/by = b−y or equivalently, logb (1/c) = −y = − logb c.
47.
75e−t/125 = 15, t = −125 ln(1/5) = 125 ln 5 ≈ 201 days.
48.
(a)
If t = 0, then Q = 12 grams
(c)
12e−0.055t = 6, e−0.055t = 0.5, t = −(ln 0.5)/(0.055) ≈ 12.6 hours
49.
(a)
7.4; basic
50.
(a)
log[H + ] = −2.44, [H + ] = 10−2.44 ≈ 3.6 × 10−3 mol/L
(b)
log[H + ] = −8.06, [H + ] = 10−8.06 ≈ 8.7 × 10−9 mol/L
(b)
(b)
4.2; acidic
Q = 12e−0.055(4) = 12e−0.22 ≈ 9.63 grams
(c) 6.4; acidic
51.
(a) 140 dB; damage
(c) 80 dB; no damage
52.
Suppose that I1 = 3I2 and β1 = 10 log10 I1 /I0 , β2 = 10 log10 I2 /I0 . Then
I1 /I0 = 3I2 /I0 , log10 I1 /I0 = log10 3I2 /I0 = log10 3 + log10 I2 /I0 , β1 = 10 log10 3 + β2 ,
β1 − β2 = 10 log10 3 ≈ 4.8 decibels.
53.
Let IA and IB be the intensities of the automobile and blender, respectively. Then
(d) 5.9; acidic
(b) 120 dB; damage
(d) 75 dB; no damage
log10 IA /I0 = 7 and log10 IB /I0 = 9.3, IA = 107 I0 and IB = 109.3 I0 , so IB /IA = 102.3 ≈ 200.
54.
The decibel level of the nth echo is 120(2/3)n ;
log 12
log(1/12)
=
≈ 6.13 so 6 echoes can be heard.
120(2/3)n < 10 if (2/3)n < 1/12, n >
log(2/3)
log 1.5
55.
(a)
log E = 4.4 + 1.5(8.2) = 16.7, E = 1016.7 ≈ 5 × 1016 J
(b)
Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E,
respectively. Then 1.5(M2 − M1 ) = log(10E) − log E = log 10 = 1,
M2 − M1 = 1/1.5 = 2/3 ≈ 0.67.
56.
Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively. Then
log E2 − log E1 = log(E2 /E1 ) = 1.5, E2 /E1 = 101.5 ≈ 31.6.
57.
If t = −2x, then x = −t/2 and lim(1 − 2x)1/x = lim(1 + t)−2/t = lim[(1 + t)1/t ]−2 = e−2 .
58.
If t = 3/x, then x = 3/t and lim (1 + 3/x)x = lim+ (1 + t)3/t = lim+ [(1 + t)1/t ]3 = e3 .
x→0
x→+∞
t→0
t→0
t→0
t→0
107
Chapter 4
EXERCISE SET 4.3
2
(2x − 5)−2/3
3
1.
y = (2x − 5)1/3 ; dy/dx =
2.
dy/dx =
¤−2/3 2 2
£
¤−2/3
1£
2
sec (x )(2x) = x sec2 (x2 ) 2 + tan(x2 )
2 + tan(x2 )
3
3
3.
dy/dx =
·
¸1/2
·
¸
¸1/2
·
9
d x−1
3 x−1
x−1
=
2 x+2
dx x + 2
2(x + 2)2 x + 2
4.
·
¸−1/2
·
¸
·
¸−1/2
¸−1/2
· 2
d x2 + 1
−12x
1 x2 + 1
6x
1 x2 + 1
x +1
dy/dx =
=− 2
=
2 x2 − 5
dx x2 − 5
2 x2 − 5
(x2 − 5)2
(x − 5)2 x2 − 5
µ
5.
3
dy/dx = x
2
−
3
¶
(5x2 + 1)−5/3 (10x) + 3x2 (5x2 + 1)−2/3 =
1 2
x (5x2 + 1)−5/3 (25x2 + 9)
3
4
x2 (3 − 2x)1/3 (−2) − (3 − 2x)4/3 (2x)
2(3 − 2x)1/3 (2x − 9)
=
6. dy/dx = 3
4
x
3x3
5
15[sin(3/x)]3/2 cos(3/x)
[sin(3/x)]3/2 [cos(3/x)](−3/x2 ) = −
2
2x2
7.
dy/dx =
8.
dy/dx = −
9.
10.
¤−3/2 £
¤
£
¤−3/2
1£
3
cos(x3 )
− sin(x3 ) (3x2 ) = x2 sin(x3 ) cos(x3 )
2
2
dy
dy
2 − 3x2 − y
+ y − 2 = 0,
=
dx
dx
x
dy
1 + 2x − x3
1
1
(b) y =
= + 2 − x2 so
= − 2 − 2x
x
x
dx
x
2 − 3x2 − y
2 − 3x2 − (1/x + 2 − x2 )
1
dy
=
=
= −2x − 2
(c) from part (a),
dx
x
x
x
(a)
3x2 + x
1 −1/2 dy
dy
√
y
− ex = 0 or
= 2ex y
2
dx
dx
dy
(b) y = (2 + ex )2 = 2 + 4ex + e2x so
= 4ex + 2e2x
dx
dy
√
= 2ex y = 2ex (2 + ex ) = 4ex + 2e2x
(c) from part (a),
dx
(a)
dy
x
dy
= 0 so
=−
dx
dx
y
11.
2x + 2y
12.
3x2 − 3y 2
13.
x2
14.
x3 (2y)
dy
x2 − 2y
dy
dy
dy
= 6(x
+ y), −(3y 2 + 6x)
= 6y − 3x2 so
= 2
dx
dx
dx
dx
y + 2x
dy
dy
+ 2xy + 3x(3y 2 )
+ 3y 3 − 1 = 0
dx
dx
1 − 2xy − 3y 3
dy
dy
= 1 − 2xy − 3y 3 so
=
(x2 + 9xy 2 )
dx
dx
x2 + 9xy 2
dy
dy
+ 3x2 y 2 − 5x2
− 10xy + 1 = 0
dx
dx
10xy − 3x2 y 2 − 1
dy
dy
(2x3 y − 5x2 )
= 10xy − 3x2 y 2 − 1 so
=
dx
dx
2x3 y − 5x2
Exercise Set 4.3
108
1 dy
dy
1
y2
− 2 = 0 so
=− 2
2
y dx x
dx
x
15.
−
16.
2x =
(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx)
,
(x − y)2
dy
x(x − y)2 + y
dy
so
=
dx
dx
x
·
¸
dy
1 − 2xy 2 cos(x2 y 2 )
dy
+ 2xy 2 = 1,
=
cos(x2 y 2 ) x2 (2y)
dx
dx
2x2 y cos(x2 y 2 )
2x(x − y)2 = −2y + 2x
17.
18.
19.
(1 + csc y)(− csc2 y)(dy/dx) − (cot y)(− csc y cot y)(dy/dx)
,
(1 + csc y)2
dy
2x(1 + csc y)2 = − csc y(csc y + csc2 y − cot2 y) ,
dx
2x(1 + csc y)
dy
but csc2 y − cot2 y = 1, so
=−
dx
csc y
2x =
¶
µ
dy
dy
+ y2 +
=1
3 tan2 (xy 2 + y) sec2 (xy 2 + y) 2xy
dx
dx
so
20.
21.
dy
1 − 3y 2 tan2 (xy 2 + y) sec2 (xy 2 + y)
=
dx
3(2xy + 1) tan2 (xy 2 + y) sec2 (xy 2 + y)
dy
(1 + sec y)[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y)(dy/dx)
= 4y 3 ,
2
(1 + sec y)
dx
dy
multiply through by (1 + sec y)2 and solve for
dx
y(1 + sec y)
dy
=
to get
dx
4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y
3x d2 y
(4y)(3) − (3x)(4dy/dx)
12y − 12x(3x/(4y))
12y 2 − 9x2
−3(3x2 − 4y 2 )
dy
=
,
=
=
=
=
,
dx
4y dx2
16y 2
16y 2
16y 3
16y 3
but 3x2 − 4y 2 = 7 so
22.
d2 y
−3(7)
21
=
=−
2
3
dx
16y
16y 3
x2 d2 y
dy
y 2 (2x) − x2 (2ydy/dx)
2xy 2 − 2x2 y(−x2 /y 2 )
2x(y 3 + x3 )
= − 2,
=−
=−
=−
,
2
4
4
dx
y dx
y
y
y5
but x3 + y 3 = 1 so
d2 y
2x
=− 5
dx2
y
23.
y d2 y
dy
x(dy/dx) − y(1)
x(−y/x) − y
2y
=− ,
=−
=−
= 2
dx
x dx2
x2
x2
x
24.
y
dy
=
,
dx
y−x
(y − x)(dy/dx) − y(dy/dx − 1)
d2 y
=
=
2
dx
(y − x)2
=
25.
µ
(y − x)
¶
µ
¶
y
y
−y
−1
y−x
y−x
(y − x)2
y 2 − 2xy
d2 y
3
2
but
y
−
2xy
=
−3,
so
=−
(y − x)3
dx2
(y − x)3
sin y
dy
d2 y
dy
= (1 + cos y)−1 ,
=
= −(1 + cos y)−2 (− sin y)
2
dx
dx
dx
(1 + cos y)3
109
26.
Chapter 4
dy
cos y
=
,
dx
1 + x sin y
(1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y]
d2 y
=
2
dx
(1 + x sin y)2
=−
2 sin y cos y + (x cos y)(2 sin2 y + cos2 y)
,
(1 + x sin y)3
but x cos y = y, 2 sin y cos y = sin 2y, and sin2 y + cos2 y = 1 so
sin 2y + y(sin2 y + 1)
d2 y
=−
2
dx
(1 + x sin y)3
27.
28.
√
√
√
√
dy
dy
x
= − ; at (1/ 2, 1/ 2),
= −1; at (1/ 2, −1/ 2),
dx
y
dx
√
−x
dy
dy
2
= +1. Directly, at the upper point y = 1 − x ,
= √
= −1 and at the lower point
dx
dx
1 − x2
√
x
dy
=√
= +1.
y = − 1 − x2 ,
dx
1 − x2
By implicit differentiation, 2x+2y(dy/dx) = 0,
√
√
If y 2 − x + 1 = 0, then y = x − 1 goes through the point (10, 3) so dy/dx = 1/(2 x − 1).
√ By
implicit differentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = − x − 1
√
goes through (10, −3) so dy/dx = −1/(2 x − 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6.
dy
x3
1
dy
= 0, so
= − 3 = − 3/4 ≈ −0.1312.
dx
dx
y
15
29.
4x3 + 4y 3
30.
3y 2
31.
dy
4(x + y ) 2x + 2y
dx
dy
dy
dy
dy
y+1
+ x2
+ 2xy + 2x − 6y
= 0, so
= −2x 2
= 0 at x = 0
dx
dx
dx
dx
3y + x2 − 6y
µ
2
2
¶
µ
¶
dy
= 25 2x − 2y
,
dx
x[25 − 4(x2 + y 2 )]
dy
dy
=
; at (3, 1)
= −9/13
dx
y[25 + 4(x2 + y 2 )]
dx
µ
32.
2
3
34.
(a)
x−1/3 + y −1/3
dy
dx
¶
= 0,
√
√
dy
y 1/3
= − 1/3 = 3 at (−1, 3 3)
dx
x
y
2
0
x
1
2
–2
(b)
(c)
dy
dy
= (x − a)(x − b) + x(x − b) + x(x − a) = 3x2 − 2(a + b)x + ab. If
= 0 then
dx
dx
2
3x − 2(a + b)x + ab = 0. By the quadratic formula
p
i
2(a + b) ± 4(a + b)2 − 4 · 3ab
1h
=
a + b ± (a2 + b2 − ab)1/2 .
x=
6
3
p
y = ± x(x − a)(x − b). The square root is only defined for nonnegative arguments, so it is
necessary that all three of the factors x, x − a, x − b be nonnegative, or that two of them be
nonpositive. If, for example, 0 < a < b then the function is defined on the disjoint intervals
0 < x < a and b < x < +∞, so there are two parts.
2y
Exercise Set 4.3
35.
110
(a)
(b)
y
±1.1547
2
-2
2
x
2
(c)
dy
y − 2x
dy
dy
dy
− y + 2y
= 0. Solve for
=
. If
= 0 then
dx
dx
dx
2y − x
dx
2
y − 2x = 0 or y = 2x. Thus 4 = x2 − xy + y 2 = x2 − 2x2 + 4x2 = 3x2 , x = ± √ .
3
Implicit differentiation yields 2x − x
36.
√
du
1 −1/2 du 1 −1/2
u
u
+ v
= −√
= 0 so
2
dv
2
dv
v
37.
¶
da
da
2t3 + 3a2
da
4a
− 4t = 6 a + 2at
, solve for
to get
= 3
dt
dt
dt
dt
2a − 6at
3 da
38. 1 = (cos x)
µ
3
2
dx
1
dx
so
=
= sec x
dy
dy
cos x
39.
2a2 ω
b2 λ
dω
dω
+ 2b2 λ = 0 so
=− 2
dλ
dλ
aω
40.
Let P (x0 , y0 ) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope of
the tangent to y 2 = 2x3 at P must be −3/4. By implicit differentiation dy/dx = 3x2 /y, so at P ,
3x20 /y0 = −3/4, or y0 = −4x20 . But y02 = 2x30 because P is on the curve y 2 = 2x3 . Elimination of y0
gives 16x40 = 2x30 , x30 (8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x20 it follows that y0 = 0 when
x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solution because
dy/dx = 3x2 /y (the slope of the curve as determined by implicit differentiation) is valid only if y 6= 0.
Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point (1/8, −1/16) is
the only solution.
41.
The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use
2xy
2
4
dy
=− 2
so at (1,1), −
= − , 1 + 2a = 3/2, a = 1/4 and
implicit differentiation to get
dx
x + 2ay
1 + 2a
3
hence b = 1 + 1/4 = 5/4.
42.
Use implicit differentiation to get dy/dx = (y − 3x2 )/(3y 2 − x), so dy/dx
= 0 if y = 3x2 √
. Substitute
√
3
3
3
6
3
3
this into x − xy + y = 0 to obtain 27x − 2x = 0, x = 2/27, x = 2/3 and hence y = 3 4/3.
43.
Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve
x2 − 4x + y 2 + 3 = 0. Implicit differentiation applied to the equation of the curve gives
dy/dx = (2 − x)/y. At P the slope of the curve must equal the slope of the line so
(2 − x0 )/y0 = y0 /x0 , or y02 = 2x0 − x20 . But x20 − 4x0 + y02 + 3 = 0 because (x0 , y0 ) is on the curve,
2
and elimination of y02 in the latter two equations gives x20 − 4x0 + (2x
√ 0 − x0 ) + 3 = 0, x0 = 3/2 which
2
2
2
when
so y0 = ± 3/2. The
√ substituted into
√ y0 = 2x0 − x0 yields y0 = 3/4, √
√ slopes of the lines are
(± 3/2)/(3/2) = ± 3/3 and their equations are y = ( 3/3)x and y = −( 3/3)x.
44.
By implicit differentiation, dy/dx = k/(2y) so the slope of the tangent to y 2 = kx at (x0 , y0 ) is
k
(x − x0 ), or 2y0 y − 2y02 = kx − kx0 .
k/(2y0 ) if y0 6= 0. The tangent line in this case is y − y0 =
2y0
But y02 = kx0 because (x0 , y0 ) is on the curve y 2 = kx, so the equation of the tangent line becomes
2y0 y − 2kx0 = kx − kx0 which gives y0 y = k(x + x0 )/2. If y0 = 0, then x0 = 0; the graph of y 2 = kx
has a vertical tangent at (0, 0) so its equation is x = 0, but y0 y = k(x + x0 )/2 gives the same result
when x0 = y0 = 0.
111
Chapter 4
dy
dy dt
=
. Use implicit differentiation on 2y 3 t + t3 y = 1 to get
dx
dt dx
1
dy
dt
dy
2y 3 + 3t2 y
2y 3 + 3t2 y
=−
=
so
=−
.
, but
2
3
dt
6ty + t
dx
cos t
dx
(6ty 2 + t3 ) cos t
45.
By the chain rule,
46.
2x3 y
47.
2xy
48.
49.
3x2 y 2 dx
dy
dx dy
dy
+ 3x2 y 2
+
= 0,
=− 3
dt
dt
dt
dt
2x y + 1 dt
dx dy
3 cos 3x − y 2 dx
dy
dx
= y2
= 3(cos 3x) ,
=
dt
dt
dt dt
2xy
dt
4 1/3 00
4
x , f (x) = x−2/3
3
9
7
28 1/3 000
28 −2/3
x , f (x) =
x
(b) f 0 (x) = x4/3 , f 00 (x) =
3
9
27
(c) generalize parts (a) and (b) with k = (n − 1) + 1/3 = n − 2/3
(a)
f 0 (x) =
£
¤
¡
¢
y 0 = rxr−1 , y 00 = r(r − 1)xr−2 so 3x2 r(r − 1)xr−2 + 4x rxr−1 − 2xr = 0,
3r(r − 1)xr + 4rxr − 2xr = 0, (3r2 + r − 2)xr = 0,
3r2 + r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/3
50.
£
¤
¡
¢
y 0 = rxr−1 , y 00 = r(r − 1)xr−2 so 16x2 r(r − 1)xr−2 + 24x rxr−1 + xr = 0,
16r(r − 1)xr + 24rxr + xr = 0, (16r2 + 8r + 1)xr = 0,
16r2 + 8r + 1 = 0, (4r + 1)2 = 0; r = −1/4
51.
52.
53.
54.
We shall find when the curves intersect and check that the slopes are negative reciprocals. For the
intersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k)2 + y 2 = k 2 to obtain
x−k
y−c
1
= −
.
cy = kx = (x2 + y 2 ). Thus x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and
2
x
y
dy
x
dy
x−k
Differentiating the two families yields (black)
=−
, and (gray)
=−
. But it was
dx
y−c
dx
y
proven that these quantities are negative reciprocals of each other.
dy
dy
Differentiating, we get the equations (black) x
+ y = 0 and (gray) 2x − 2y
= 0. The first says
dx
dx
y
x
the (black) slope is = − and the second says the (gray) slope is , and these are negative reciprocals
x
y
of each other.
1
dx
dy
= 15y 2 + 1,
=
;
dy
dx
15y 2 + 1
dy dy
1
dy
check: 1 = 15y 2
+
,
=
dx dx dx
15y 2 + 1
y = f −1 (x), x = f (y) = 5y 3 + y − 7,
y = f −1 (x), x = f (y) = 1/y 2 ,
check: 1 = −2y −3
55.
dx
dy
= −2y −3 ,
= −y 3 /2;
dy
dx
dy dy
,
= −y 3 /2
dx dx
1
dx
dy
= 10y 4 + 3y 2 ,
=
;
dy
dx
10y 4 + 3y 2
1
dy
dy dy
check: 1 = 10y 4
+ 3y 2 ,
=
dx
dx dx
10y 4 + 3y 2
y = f −1 (x), x = f (y) = 2y 5 + y 3 + 1,
Exercise Set 4.4
56.
112
dx
dy
1
= 5 − 2 cos 2y,
=
;
dy
dx
5 − 2 cos 2y
1
dy dy
=
check: 1 = (5 − 2 cos 2y) ,
dx dx
5 − 2 cos 2y
y = f −1 (x), x = f (y) = 5y − sin 2y,
EXERCISE SET 4.4
1
(2) = 1/x
2x
µ ¶
2 ln x
1
=
3. 2(ln x)
x
x
1.
2.
1
(3x2 ) = 3/x
x3
4.
1
(cos x) = cot x
sin x
6.
1
√
2+ x
5.
1
sec2 x
(sec2 x) =
tan x
tan x
7.
·
¸
(1 + x2 )(1) − x(2x)
1 − x2
1
8.
=
x/(1 + x2 )
(1 + x2 )2
x(1 + x2 )
1
ln x
9.
3x2 − 14x
x3 − 7x2 − 3
x3
10.
µ ¶
1
1
= √
x
2x ln x
11.
1
(ln x)−1/2
2
13.
1
− sin(ln x)
x
12.
14.
15. 3x2 log2 (3 − 2x) +
−2x3
(ln 2)(3 − 2x)
16.
µ
¶
1
√
2 x
1
√
= √
2 x(2 + x)
µ ¶
1
1
=
x
x ln x
µ ¶
1
+ (3x2 ) ln x = x2 (1 + 3 ln x)
x
1
2 2(ln x)(1/x)
p
2
ln x
= p
1 + ln x
x 1 + ln2 x
2 sin(ln x) cos(ln x)
£
¤3
log2 (x2 − 2x)
sin(ln x2 )
1
=
x
x
£
¤2
+ 3x log2 (x2 − 2x)
17.
2x(1 + log x) − x/(ln 10)
(1 + log x)2
18.
1/[x(ln 10)(1 + log x)2 ]
19.
7e7x
20.
−10xe−5x
21.
x3 ex + 3x2 ex = x2 ex (x + 3)
22.
−
23.
(ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x )
dy
=
dx
(ex + e−x )2
=
2x − 2
(x2 − 2x) ln 2
2
1 1/x
e
x2
(e2x + 2 + e−2x ) − (e2x − 2 + e−2x )
= 4/(ex + e−x )2
(ex + e−x )2
24.
ex cos(ex )
25.
(x sec2 x + tan x)ex tan x
27.
(1 − 3e3x )e(x−e
29.
(x − 1)e−x
x−1
= x
−x
1 − xe
e −x
3x
)
26.
(ln x)ex − ex (1/x)
dy
ex (x ln x − 1)
=
=
dx
(ln x)2
x(ln x)2
28.
p
15 2
x (1 + 5x3 )−1/2 exp( 1 + 5x3 )
2
30.
1
[− sin(ex )]ex = −ex tan(ex )
cos(ex )
113
Chapter 4
µ
¶
31.
dy
1
+
dx xy
32.
1
dy
=
dx
x tan y
33.
¸
·
d
3x
1
2
ln cos x − ln(4 − 3x ) = − tan x +
dx
2
4 − 3x2
34.
d
dx
35.
ln |y| = ln |x| +
36.
1
dy
1
ln |y| = [ln |x − 1| − ln |x + 1|],
=
5
dx
5
µ
x
dy
+y
dx
µ
= 0,
dy
y
=−
dx
x(y + 1)
¶
dy
dy
tan y
x sec y
+ tan y ,
=
dx
dx
x(tan y − sec2 y)
2
1
[ln(x − 1) − ln(x + 1)]
2
¶
1
=
2
µ
1
1
−
x−1 x+1
¶
¸
·
p
2x
dy
1
1
3
ln |1 + x2 |,
= x 1 + x2
+
3
dx
x 3(1 + x2 )
r
5
·
¸
x−1
1
1
−
x+1 x−1 x+1
1
1
ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5|
3
2
√
·
¸
dy
(x2 − 8)1/3 x3 + 1
2x
3x2
6x5 − 7
=
+
−
dx
x6 − 7x + 5
3(x2 − 8) 2(x3 + 1) x6 − 7x + 5
37. ln |y| =
38.
1
ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| − ln |x|
2
¸
·
sin x cos x tan3 x
1
dy
3 sec2 x
√
=
−
cot x − tan x +
dx
tan x
2x
x
1 0
y = ln 2, y 0 = y ln 2 = 2x ln 2
y
39.
f 0 (x) = 2x ln 2; y = 2x , ln y = x ln 2,
40.
f 0 (x) = −3−x ln 3; y = 3−x , ln y = −x ln 3,
41.
f 0 (x) = π sin x (ln π) cos x;
y = π sin x , ln y = (sin x) ln π,
42.
43.
44.
45.
1 0
y = − ln 3, y 0 = −y ln 3 = −3−x ln 3
y
1 0
y = (ln π) cos x, y 0 = π sin x (ln π) cos x
y
f 0 (x) = π x tan x (ln π)(x sec2 x + tan x);
1
y = π x tan x , ln y = (x tan x) ln π, y 0 = (ln π)(x sec2 x + tan x)
y
y 0 = π x tan x (ln π)(x sec2 x + tan x)
1 dy
3x2 − 2
1
= 3
ln x + ln(x3 − 2x),
ln y = (ln x) ln(x3 − 2x),
y dx
x − 2x
x
¸
· 2
1
−
2
dy
3x
= (x3 − 2x)ln x 3
ln x + ln(x3 − 2x)
dx
x − 2x
x
ln y = (sin x) ln x,
¸
·
sin x
dy
sin x
1 dy
=
+ (cos x) ln x,
= xsin x
+ (cos x) ln x
y dx
x
dx
x
1
1 dy
=
tan x + (sec2 x) ln(ln x),
ln y = (tan x) ln(ln x),
y dx
x ln x
·
¸
dy
tan x tan x
2
= (ln x)
+ (sec x) ln(ln x)
dx
x ln x
Exercise Set 4.4
46.
47.
114
1 dy
2x
1
ln y = (ln x) ln(x2 + 3),
= 2
ln x + ln(x2 + 3),
y dx
x +3
x
¸
·
1
2x
dy
2
ln x
2
= (x + 3)
ln x + ln(x + 3)
dx
x2 + 3
x
y = Ae2x + Be−4x , y 0 = 2Ae2x − 4Be−4x , y 00 = 4Ae2x + 16Be−4x so
y 00 + 2y 0 − 8y = (4Ae2x + 16Be−4x ) + 2(2Ae2x − 4Be−4x ) − 8(Ae2x + Be−4x ) = 0
48.
y = Aekt , dy/dt = kAekt = k(Aekt ) = ky
49.
(a)
f 0 (x) = kekx , f 00 (x) = k 2 ekx , f 000 (x) = k 3 ekx , . . . , f (n) (x) = k n ekx
(b)
f 0 (x) = −ke−kx , f 00 (x) = k 2 e−kx , f 000 (x) = −k 3 e−kx , . . . , f (n) (x) = (−1)n k n e−kx
50.
51.
dy
= e−λt (ωA cos ωt − ωB sin ωt) + (−λ)e−λt (A sin ωt + B cos ωt)
dt
= e−λt [(ωA − λB) cos ωt − (ωB + λA) sin ωt]
"
"
µ
¶2 #
µ
¶2 #
1 x−µ
d
1
1 x−µ
−
f (x) = √
exp −
2
σ
dx
2
σ
2πσ
"
µ
¶2 # · µ
¶ µ ¶¸
1
x−µ
1
1 x−µ
−
=√
exp −
2
σ
σ
σ
2πσ
0
"
µ
¶2 #
1 x−µ
1
(x − µ) exp −
= −√
2
σ
2πσ 3
52.
53.
54.
55.
(a)
y 0 = −xe−x + e−x = e−x (1 − x), xy 0 = xe−x (1 − x) = y(1 − x)
(b)
y 0 = −x2 e−x /2 + e−x /2 = e−x /2 (1 − x2 ), xy 0 = xe−x /2 (1 − x2 ) = y(1 − x2 )
(a)
logx e =
1
d
ln e
1
=
, [logx e] = −
ln x
ln x dx
x(ln x)2
(b)
logx 2 =
ln 2 d
ln 2
, [logx 2] = −
ln x dx
x(ln x)2
2
2
2
dβ
10
β = 10 log I − 10 log I0 ,
=
dI
I ln 10
¸
1
dβ
db/W/m2
=
(a)
dI I=10I0
I0 ln 10
¸
dβ
1
db/W/m2
(c)
=
dI I=100I0
100I0 ln 10
2
(b)
dβ
dI
¸
=
I=100I0
¸
¸
·
·
qk0
dk
q(T − T0 ) ³ q ´
q(T − T0 )
= k0 exp −
− 2 = − 2 exp −
dT
2T0 T
2T
2T
2T0 T
56.
because xx is not of the form ax where a is constant
1
(b) y = xx , ln y = x ln x, y 0 = 1 + ln x, y 0 = xx (1 + ln x)
y
57.
f 0 (x) = exe−1
(a)
1
db/W/m2
10I0 ln 10
115
Chapter 4
58.
Let P (x0 , y0 ) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and an equation
of the tangent line at P is y − y0 = 3e3x0 (x − x0 ), y − e3x0 = 3e3x0 (x − x0 ). If the line passes through
the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0 e3x0 which gives x0 = 1/3 and thus
y0 = e. The point is (1/3, e).
59.
(a)
(b)
¯
ln(1 + h) − ln 1
ln(1 + h)
1¯
= lim
= ¯¯
=1
h→0
h→0
h
h
x x=1
¯
¯
¯
¯
d
10h − 1
=
(10x )¯¯
= 10x ln 10¯¯
= ln 10
f (x) = 10x ; f 0 (0) = lim
h→0
h
dx
x=0
x=0
f (x) = ln x; f 0 (1) = lim
¯
¯
¯
ln(e2 + h) − 2
d
1 ¯¯
¯
=
(ln x)¯
= ¯
= e−2
60. (a) f (x) = ln x; f (e ) = lim
h→0
h
dx
x x=e2
x=e2
¯
¯
¯
d x ¯¯
2x − 2
x
x
0
=
(2 )¯
(b) f (x) = 2 ; f (1) = lim
= 2 ln 2¯¯
= 2 ln 2
x→1 x − 1
dx
x=1
x=1
0
61. (b)
2
(c)
y
6
dy
1 1
dy
= − so
< 0 at
dx
2 x
dx
dy
> 0 at x = e
x = 1 and
dx
4
2
1
2
3
4
x
(d) The slope is a continuous function which goes from a negative value to a positive value; therefore
it must take the value zero in between, by the Intermediate Value Theorem.
(e)
62.
dy
= 0 when x = 2
dx
(a)
(c)
100
-5
125
5
-5
5
-100
63.
(d)
x = −2, 5/3
(a)
ex cos πx oscillates between +ex and −ex
as cos πx oscillates between −1 and +1.
-25
y
(b)
20
1
-20
2
3
x
Exercise Set 4.5
64.
(a)
116
12
0
9
0
(b)
P tends to 12 as t gets large; lim P (t) = lim
(c)
the rate of population growth tends to zero
t→+∞
t→+∞
60
60
60
=
=
= 12
5 + 7e−t
5
5 + 7 lim e−t
t→+∞
3.2
0
9
0
65.
(a)
100
0
8
20
(b)
as t tends to +∞, the population tends to 19
95
95
95
= 19
=
=
lim P (t) = lim
−t/4
t→+∞
t→+∞ 5 − 4e−t/4
5
5 − 4 lim e
t→+∞
(c)
the rate of population growth tends to zero
0
0
8
-80
EXERCISE SET 4.5
1.
(a)
−π/2
(b)
π
(c) −π/4
(d) 0
2.
(a)
π/3
(b)
π/3
(c) π/4
(d) 4π/3
3.
√
√
√
θ = −π/3; cos θ = 1/2, tan θ = − 3, cot θ = −1/ 3, sec θ = 2, csc θ = −2/ 3
117
4.
Chapter 4
θ = π/3; sin θ =
√
√
√
√
3/2, tan θ = 3, cot θ = 1/ 3, sec θ = 2, csc θ = 2/ 3
5. tan θ = 4/3, 0 < θ < π/2; use the triangle shown to get sin θ = 4/5,
cos θ = 3/5, cot θ = 3/4, sec θ = 5/3, csc θ = 5/4
5
4
θ
3
6.
domain
range
sin−1
[−1, 1]
[−π/2, π/2]
−1
[−1, 1]
[0, π]
tan−1
(−∞, +∞)
(−π/2, π/2)
−1
(−∞, +∞)
(0, π)
cos
cot
sec−1
csc
7.
−1
(−∞, −1] ∪ [1, +∞) [0, π/2) ∪ [π, 3π/2)
(−∞, −1] ∪ [1, +∞) (0, π/2] ∪ (−π, −π/2]
(b) sin−1 (sin π) = sin−1 (sin 0) = 0
(a)
π/7
(c)
sin−1 (sin(5π/7)) = sin−1 (sin(2π/7)) = 2π/7
(d) Note that π/2 < 630 − 200π < π so
sin(630) = sin(630 − 200π) = sin(π − (630 − 200π)) = sin(201π − 630) where
0 < 201π − 630 < π/2; sin−1 (sin 630) = sin−1 (sin(201π − 630)) = 201π − 630.
8.
9.
10.
(a)
π/7
−1
(b)
π
−1
(c)
cos (cos(12π/7)) = cos (cos(2π/7)) = 2π/7
(d)
Note that −π/2 < 200 − 64π < 0 so cos(200) = cos(200 − 64π) = cos(64π − 200) where
0 < 64π − 200 < π/2; cos−1 (cos 200) = cos−1 (cos(64π − 200)) = 64π − 200.
(a) 0 ≤ x ≤ π
(c) −π/2 < x < π/2
(b) −1 ≤ x ≤ 1
(d) −∞ < x < +∞
Let θ = sin−1 (−3/4) then sin θ = −3/4, −π/2 < θ < 0 and
√
(see figure) sec θ = 4/ 7
÷7
θ
−3
4
11.
Let θ = cos−1 (3/5), sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25
5
4
θ
3
Exercise Set 4.5
12.
(a)
118
sin(cos−1 x) =
1
√
1 − x2
(b)
1 − x2
tan(cos−1 x) =
1
cos−1x
1 − x2
cos−1x
x
√
(c)
csc(tan−1 x) =
1 + x2
x
1+
x
x2
(d)
sin(tan−1 x) = √
1 + x2
x
tan−1x
(a)
cos(tan−1 x) = √
1
1 + x2
(b)
tan(cot−1 x) =
1 + x2
cot−1x
x
1
√
x2 − 1
sin(sec−1 x) =
x
(d)
cot(csc−1 x) =
p
x2 − 1
x
x
1
x2 − 1
csc−1x
sec−1x
x2 − 1
1
(a)
1
x
1
x
tan−1x
14.
x
1
1 + x2
(c)
x
1 + x2
tan−1x
1
13.
√
1 − x2
x
x
−1.00 −0.80 −0.6 −0.40 −0.20 0.00 0.20 0.40 0.60 0.80 1.00
−1
sin x −1.57 −0.93 −0.64 −0.41 −0.20 0.00 0.20 0.41 0.64 0.93 1.57
cos−1 x
3.14
2.50
2.21
1.98
1.77 1.57 1.37 1.16 0.93 0.64 0.00
y
(b)
(c)
y
3
2
1
1
x
x
0.5
1
-1
1
119
15.
Chapter 4
(a)
y
y
π/2
3
x
-5
1
x
-10
10
1
5
-π/2
(b)
The domain of cot−1 x is (−∞, +∞), the range is (0, π); the domain of csc−1 x is
(−∞, −1] ∪ [1, +∞), the range is [−π/2, 0) ∪ (0, π/2].
(a)
y = cot−1 x, x = cot y, tan y = 1/x, y = tan−1 (1/x)
(b)
y = sec−1 x, x = sec y, cos y = 1/x, y = cos−1 (1/x)
(c)
y = csc−1 x, x = csc y, sin y = 1/x, y = sin−1 (1/x)
17.
(a)
55.0◦
18.
(a)
x = π − sin−1 (0.37) ≈ 2.7626 rad
(b)
θ = 180◦ + sin−1 (0.61) ≈ 217.6◦
19.
(a)
x = π + cos−1 (0.85) ≈ 3.6964 rad
(b)
θ = − cos−1 (0.23) ≈ −76.7◦
20.
(a)
x = tan−1 (3.16) − π ≈ −1.8773
(b)
θ = 180◦ − tan−1 (0.45) ≈ 155.8◦
21.
(a)
p
1
p
(1/3) = 1/ 9 − x2
1 − x2 /9
(b)
p
−2/ 1 − (2x + 1)2
22.
(a)
2x/(1 + x4 )
(b) −
23.
(a)
24.
(a)
y = 1/ tan x = cot x, dy/dx = − csc2 x
(b)
y = (tan−1 x)−1 , dy/dx = −(tan−1 x)−2
25.
(a)
1
1
p
(−1/x2 ) = − √
2
|x| x2 − 1
1 − 1/x
26.
(a)
−
27.
(a)
28.
(a)
29.
x3 + x tan−1 y = ey , 3x2 +
16.
x7
√
(b)
33.6◦
(c) 25.8◦
1
1+x
µ
1 −1/2
x
2
¶
=−
1
√
2(1 + x) x
√
(b) −1/ e2x − 1
7
1
(7x6 ) = √
14
x −1
x x14 − 1
µ
1
1 + x2
¶
sin x
sin x
=
=
2
|
sin x|
1 − cos x
½
1, sin x > 0
−1, sin x < 0
(b)
√
(b)
1
− √
−1
2 cot x(1 + x2 )
ex
√
+ ex sec−1 x
x x2 − 1
(b)
3x2 (sin−1 x)2
√
+ 2x(sin−1 x)3
1 − x2
0
(b)
0
(cos−1
1
√
x) 1 − x2
x
(3x2 + tan−1 y)(1 + y 2 )
y 0 + tan−1 y = ey y 0 , y 0 =
2
1+y
(1 + y 2 )ey − x
Exercise Set 4.5
30.
120
sin−1 (xy) = cos−1 (x − y), p
1−
p
y 1 − (x −
+ 1 − x2 y 2
p
y0 = p
1 − x2 y 2 − x 1 − (x − y)2
p
31.
1
x2 y 2
1
(xy 0 + y) = − p
(1 − y 0 ),
2
1 − (x − y)
y)2
(a)
32.
y
(a)
3
y
1
2
1
x
-1
33.
1
(a)
2
x
y
π
−2π
34.
35.
4π
x
(a)
sin−1 0.9 > 1, so it is not in the domain of sin−1 x
(b)
−1 ≤ sin−1 x ≤ 1 is necessary, or −0.841471 ≤ x ≤ 0.841471
(a)
(b)
y
y
6
6
x
-0.5
x
0.5
-6
36.
(a)
x = 2π − cos−1 k
(c)
2x = sin−1 k or 2x = π − sin−1 k so x =
-6
(b) x = π + tan−1 k
1
2
sin−1 k or x = π/2 − 12 sin−1 k
R
6378
= sin−1
≈ 23◦
R+h
16, 378
37. (b)
θ = sin−1
38.
(a)
p
p
If γ = 90◦ , then sin γ = 1, 1 − sin2 φ sin2 γ = 1 − sin2 φ = cos φ,
D = tan φ tan λ = (tan 23.55◦ )(tan 65◦ ) ≈ 0.934684245 so h ≈ 21.2 hours.
(b)
If γ = 270◦ , then sin γ = −1, D = − tan φ tan λ ≈ −0.934684245 so h ≈ 2.8 hours.
39. sin 2θ = gR/v 2 = (9.8)(18)/(14)2 = 0.9, 2θ = sin−1 (0.9) or 2θ = 180◦ − sin−1 (0.9) so
θ=
1
2
sin−1 (0.9) ≈ 32◦ or θ = 90◦ − 12 sin−1 (0.9) ≈ 58◦ . The ball will have a lower
parabolic trajectory for θ = 32◦ and hence will result in the shorter time of flight.
40. 42 = 22 + 32 − 2(2)(3) cos θ, cos θ = −1/4, θ = cos−1 (−1/4) ≈ 104◦
121
Chapter 4
41.
p
√
y = 0 when x2 = 6000v 2 /g, x = 10v 60/g = 1000 30 for v = 400 and g = 32;
√
√
tan θ = 3000/x = 3/ 30, θ = tan−1 (3/ 30) ≈ 29◦ .
42.
θ = α − β, cot α =
θ = cot−1
x
x
and cot β = so
a+b
b
x
−1 x
− cot
a+b
b
a
θ
b
β
α
x
43.
(a)
Let θ = sin−1 (−x) then sin θ = −x, −π/2 ≤ θ ≤ π/2.
But sin(−θ) = − sin θ and
−π/2 ≤ −θ ≤ π/2 so sin(−θ) = −(−x) = x, −θ = sin−1 x, θ = − sin−1 x.
44.
45.
(b)
proof is similar to that in part (a)
(a)
Let θ = cos−1 (−x) then cos θ = −x, 0 ≤ θ ≤ π. But cos(π − θ) = − cos θ and
0 ≤ π − θ ≤ π so cos(π − θ) = x, π − θ = cos−1 x, θ = π − cos−1 x
(b)
Let θ = sec−1 (−x) for x ≥ 1; then sec θ = −x and π/2 < θ ≤ π. So 0 ≤ π − θ < π/2 and
π − θ = sec−1 sec(π − θ) = sec−1 (− sec θ) = sec−1 x, or sec−1 (−x) = π − sec−1 x.
(a)
sin−1 x = tan−1 √
x
1 − x2
1
x
sin−1x
÷1 − x2
(b)
46.
sin−1 x + cos−1 x = π/2; cos−1 x = π/2 − sin−1 x = π/2 − tan−1 √
tan(α + β) =
tan α + tan β
,
1 − tan α tan β
tan(tan−1 x + tan−1 y) =
47.
x
1 − x2
x+y
tan(tan−1 x) + tan(tan−1 y)
=
1 − tan(tan−1 x) tan(tan−1 y)
1 − xy
so
tan−1 x + tan−1 y = tan−1
(a)
tan−1
(b)
2 tan−1
x+y
1 − xy
1
1
1/2 + 1/3
+ tan−1 = tan−1
= tan−1 1 = π/4
2
3
1 − (1/2) (1/3)
2 tan−1
1
1
= tan−1 + tan−1
3
3
1
1
+ tan−1 = tan−1
3
7
1
1/3 + 1/3
3
= tan−1
= tan−1 ,
3
1 − (1/3) (1/3)
4
3
1
3/4
+
1/7
+ tan−1 = tan−1
= tan−1 1 = π/4
4
7
1 − (3/4) (1/7)
s
48. sin(sec−1 x) = sin(cos−1 (1/x)) =
1−
√
µ ¶2
1
x2 − 1
=
x
|x|
Exercise Set 4.6
122
EXERCISE SET 4.6
1. (b)
(d)
2. (b)
(d)
A = x2
Find
dx
dA
= 2x
dt
dt
¯
¯
dA ¯
= 2. From part (c),
= 2(3)(2) = 12 ft2 /min.
dt ¯
(c)
¯
¯
dx ¯¯
dA ¯¯
given
that
dt ¯x=3
dt ¯x=3
A = πr2
x=3
dr
dA
= 2πr
dt
dt
¯
dA ¯¯
= 2. From part (c),
= 2π(5)(2) = 20π cm2 /s.
dt ¯
(c)
¯
¯
dr ¯¯
dA ¯¯
given that
Find
dt ¯r=5
dt ¯r=5
r=5
µ
¶
dr
dV
2 dh
=π r
+ 2rh
.
3. (a) V = πr h, so
dt
dt
dt
¯
¯
¯
dV ¯¯
dh ¯¯
dr ¯¯
(b) Find
given that
= 1 and
= −1. From part (a),
dt ¯h=6,
dt ¯h=6,
dt ¯h=6,
r=10
r=10
r=10
¯
dV ¯¯
= π[102 (1) + 2(10)(6)(−1)] = −20π in3 /s; the volume is decreasing.
dt ¯h=6,
2
r=10
4.
(a)
`2 = x2 + y 2 , so
(b)
Find
1
d`
=
dt
`
µ
x
¶
dx
dy
+y
.
dt
dt
¯
1
dy
1
d` ¯¯
dx
= and
=− .
given that
dt ¯x=3,
dt
2
dt
4
y=4
From part (a) and the fact that ` = 5 when x = 3 and y = 4,
¯
· µ ¶
µ
¶¸
d` ¯¯
1
1
1
1
3
+
4
−
=
ft/s; the diagonal is increasing.
=
¯
dt x=3,
5
2
4
10
y=4
dy
dx
¶
µ
x
−y
y
cos2 θ
dx
dy
dθ
dθ
2
dt
dt
5. (a) tan θ = , so sec θ
=
=
−y
x
,
x
dt
x2
dt
x2
dt
dt
¯
¯
¯
dθ ¯¯
dx ¯¯
dy ¯¯
1
(b) Find
given
that
=
1
and
=− .
dt ¯x=2,
dt ¯x=2,
dt ¯x=2,
4
y=2
y=2
y=2
π
π
1
When x = 2 and y = 2, tan θ = 2/2 = 1 so θ = and cos θ = cos = √ . Thus
4
4
2
√
¯
¶
¸
· µ
5
1
(1/ 2)2
dθ ¯¯
− 2(1) = −
rad/s; θ is decreasing.
2 −
=
from part (a),
dt ¯x=2,
22
4
16
y=2
6.
¯
¯
¯
dz ¯¯
dx ¯¯
dy ¯¯
Find
given that
= −2 and
= 3.
dt ¯x=1,
dt ¯x=1,
dt ¯x=1,
y=2
y=2
y=2
¯
¯
dz
3 dy
2 2 dx dz ¯
= 2x y
+ 3x y
,
= (4)(3) + (12)(−2) = −12 units/s; z is decreasing
dt
dt
dt dt ¯x=1,
y=2
7.
Let A be the area swept out, and θ the angle through which the minute hand has rotated.
dθ
π
1
dθ
4π 2
dA
dA
given that
=
rad/min; A = r2 θ = 8θ, so
=8
=
in /min.
Find
dt
dt
30
2
dt
dt
15
123
8.
Chapter 4
Let r be the radius and A the area enclosed by the ripple. We want
¯
dA ¯¯
dr
given that
= 3. We
dt ¯t=10
dt
dr
dA
= 2πr . Because r is increasing at the constant rate of 3 ft/s, it follows
dt
dt ¯
dA ¯¯
= 2π(30)(3) = 180π ft2 /s.
that r = 30 ft after 10 seconds so
dt ¯t=10
know that A = πr2 , so
¯
dr
dr
1 dA
dA
dA
dr ¯¯
= 6. From A = πr2 we get
= 2πr
so
=
. If A = 9 then
given that
dt ¯A=9
dt
dt
dt
dt
2πr dt
¯
√
√
1
dr ¯¯
2
√ (6) = 1/ π mi/h.
πr = 9, r = 3/ π so
=
dt ¯A=9
2π(3/ π)
9. Find
10.
11.
D
4
where D
The volume V of a sphere of radius r is given by V = πr3 or, because r =
3
2
¯
µ ¶3
D
1
dD ¯¯
dV
4
= 3. From
= πD3 . We want
given that
is the diameter, V = π
3
2
6
dt ¯r=1
dt
¯
dD dD
1
2 dV
dD ¯¯
dV
2
3
1
= πD2
,
=
,
so
ft/min.
=
(3) =
V = πD3 we get
6
dt
2
dt dt
πD2 dt
dt ¯r=1
π(2)2
2π
¯
dr
dV
dr
4
dV ¯¯
= −15. From V = πr3 we get
= 4πr2
so
given that
Find
dt ¯r=9
dt
3
dt
dt
¯
dV ¯¯
= 4π(9)2 (−15) = −4860π. Air must be removed at the rate of 4860π cm3 /min.
dt ¯
r=9
12.
Let¯ x and y be the distances shown in the diagram. We want to find
dx
dy ¯¯
= 5. From x2 + y 2 = 172 we get
given that
¯
dt y=8
dt
dy
dy
x dx
dx
+ 2y
= 0, so
=−
. When y = 8, x2 + 82 = 172 ,
2x
dt
dt
dt
y dt ¯
dy ¯¯
15
75
x2 = 289 − 64 = 225, x = 15 so
= − (5) = −
ft/s; the
dt ¯
8
8
17
y
y=8
x
top of the ladder is moving down the wall at a rate of 75/8 ft/s.
¯
dx ¯¯
dy
= −2. From x2 + y 2 = 132 we get
13. Find
given that
dt ¯y=5
dt
dy
dx
y dy
dx
+ 2y
= 0 so
=−
. Use x2 + y 2 = 169 to find that
2x
dt
dt
dt ¯ x dt
5
5
dx ¯¯
= − (−2) = ft/s.
x = 12 when y = 5 so
dt ¯y=5
12
6
13
y
x
14.
¯
dθ ¯¯
Let θ be the acute angle, and x the distance of the bottom of the plank from the wall. Find
dt ¯x=2
¯
x
dx ¯¯
1
so
given that
= − ft/s. The variables θ and x are related by the equation cos θ =
dt ¯x=2
2
10
√
√
1 dx dθ
1
dθ
dx
=
,
=−
. When x = 2, the top of the plank is 102 − 22 = 96 ft above
dt
10 dt dt
10 sin θ dt
¯
¶
µ
√
1
dθ ¯¯
1
1
= √ ≈ 0.051 rad/s.
= −√
the ground so sin θ = 96/10 and
−
¯
dt x=2
2
96
2 96
− sin θ
Let x denote the distance from first base and y the distance from
dy
dx
home plate. Then x2 + 602 = y 2 and 2x
= 2y . When x = 50
dt
dt
√
x dx
50
dy
125
=
= √ (25) = √ ft/s.
then y = 10 61 so
dt
y dt
10 61
61
x
60
15.
124
ft
Exercise Set 4.6
y
First
Home
16.
17.
18.
¯
¯
dx ¯¯
dy ¯¯
given
that
= 2000. From x2 + 52 = y 2 we get
dt ¯x=4
dt ¯x=4
√
dx
y dy
dx
dy
= 2y
so
=
. Use x2 + 25 = y 2 to find that y = 41
2x
dt
dt
dt
x√
dt
¯
√
dx ¯¯
41
(2000) = 500 41 mi/h.
=
when x = 4 so
¯
dt x=4
4
Find
¯
¯
dy ¯¯
dx ¯¯
given
that
= 880. From y 2 = x2 + 30002
dt ¯x=4000
dt ¯x=4000
dx
dy
x dx
dy
= 2x
so
=
. If x = 4000, then y = 5000 so
we get 2y
dt
dt
dt
y dt
¯
4000
dy ¯¯
(880) = 704 ft/s.
=
dt ¯x=4000
5000
Rocket
y
Radar
station
x
5 mi
Find
Rocket
y
x
Camera
3000 ft
¯
¯
dx ¯¯
dφ ¯¯
Find
given that
= 0.2. But x = 3000 tan φ so
dt ¯φ=π/4
dt ¯φ=π/4
¯
´
³
dφ dx ¯¯
dx
2 π
= 3000(sec2 φ) ,
(0.2) = 1200 ft/s.
=
3000
sec
dt
dt dt ¯
4
φ=π/4
19.
(a)
If x denotes the altitude, then r − x = 3960, the radius of the Earth. θ = 0 at perigee, so
r = 4995/1.12 ≈ 4460; the altitude is x = 4460 − 3960 = 500 miles. θ = π at apogee, so
r = 4995/0.88 ≈ 5676; the altitude is x = 5676 − 3960 = 1716 miles.
(b)
If θ = 120◦ , then r = 4995/0.94 ≈ 5314; the altitude is 5314 − 3960 = 1354 miles. The rate of
change of the altitude is given by
dr
dr dθ
4995(0.12 sin θ) dθ
dx
=
=
=
.
dt
dt
dθ dt
(1 + 0.12 cos θ)2 dt
Use θ = 120◦ and dθ/dt = 2.7◦ /min = (2.7)(π/180) rad/min to get dr/dt ≈ 27.7 mi/min.
20.
(a)
Let x be the horizontal distance shown in the figure. Then x = 4000 cot θ and
dθ
sin2 θ dx
dθ
dx
= −4000 csc2 θ , so
=−
. Use θ = 30◦ and
dt
dt
dt
4000 dt
dx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to get
dθ/dt = −0.0275 rad/s ≈ −1.6◦ /s; θ is decreasing at the rate of 1.6◦ /s.
(b)
Let y be the distance between the observation point and the aircraft. Then y = 4000 csc θ so
dy/dt = −4000(csc θ cot θ)(dθ/dt). Use θ = 30◦ and dθ/dt = −0.0275 rad/s to get dy/dt ≈ 381
ft/s.
125
21.
Chapter 4
¯
dh ¯¯
dV
given that
= 20. The volume of water in the tank
dt ¯h=16
dt
1
at a depth h is V = πr2 h. Use similar triangles (see figure) to get
3
µ
¶2
r
10
5
1
5
25
=
so r =
h thus V = π
h h=
πh3 ,
h
24
12
3
12 ¯
432
25
144 dV dh ¯¯
144
9
dh dh
dV
=
πh2 ;
=
,
=
(20) =
dt
144
dt dt
25πh2 dt dt ¯
25π(16)2
20π
Find
10
r
24
h
h=16
ft/min.
22.
¯
1
dh ¯¯
dV
1
= 8. V = πr2 h, but r = h so
given that
dt ¯h=6
dt
3
2
µ ¶2
1
4 dV
dV
dh dh
h
1
1
πh3 ,
= πh2 ,
=
,
h=
V = π
2
12
dt
4
dt dt
πh2 dt
¯ 3
dh ¯¯
4
8
ft/min.
=
(8) =
dt ¯h=6
π(6)2
9π
Find
h
r
¯
1
dV ¯¯
dh
1
= 5. V = πr2 h, but r = h so
23. Find
given that
dt ¯h=10
dt
3
2
µ ¶2
h
1
1
πh3 ,
h=
V = π
3
2
12¯
1
1
dV
dh dV ¯¯
= πh2 ,
= π(10)2 (5) = 125π ft3 /min.
dt
4
dt dt ¯h=10
4
h
r
24.
Let r and h be as shown in the figure.
¯ If C is the circumference of
dC ¯¯
dV
the base, then we want to find
given that
= 10. It is
¯
dt h=8
dt
1
given that r = h, thus C = 2πr = πh so
2
dh
dC
=π
dt
dt
Use
V =
h
r
(1)
dh
1
1
dV
1 2
πr h =
πh3 to get
= πh2 , so
3
12
dt
4
dt
4 dV
dh
=
dt
πh2 dt
(2)
4 dV
dC
= 2
so
Substitution of (2) into (1) gives
dt
h dt
¯
5
4
dC ¯¯
(10) = ft/min.
=
dt ¯h=8
64
8
25.
dh
With s and h as shown in the figure, we want to find
given that
dt
1
ds
= 500. From the figure, h = s sin 30◦ = s so
dt
2
1 ds
1
dh
=
= (500) = 250 mi/h.
dt
2 dt
2
s
30°
Ground
h
Exercise Set 4.6
26.
27.
28.
¯
dx ¯¯
dy
given that
= −20. From x2 + 102 = y 2 we get
dt ¯y=125
dt
dx
y dy
dy
dx
= 2y
so
=
. Use x2 + 100 = y 2 to find that
2x
dt√
dt
dt
x dt
√
x =¯ 15, 525 = 15 69 when y = 125 so
125
500
dx ¯¯
= √ (−20) = − √ . The boat is approaching the
dt ¯y=125
15 69
3 69
500
dock at the rate of √ ft/min.
3 69
Find
¯
dx ¯¯
dy
given that
= −12. From x2 + 102 = y 2 we get
dt
dt ¯y=125
dy
x dx
dy
dx
= 2y
so
=
. Use x2 + 100 = y 2 to find that
2x
dtp
dt
dt
√ y dt
x = 15,√525 = 15 69 when
√ y = 125 so
15 69
36 69
dy
=
(−12) = −
. The rope must be pulled at the rate
dt √ 125
25
36 69
ft/min.
of
25
Find
(a)
Let x and y be as shown in the figure. It is required to find
dy
x
x+y
dx
, given that
= −3. By similar triangles, =
,
dt
dt
6
18
1
18x = 6x + 6y, 12x = 6y, x = y, so
2
1 dy
1
3
dx
=
= (−3) = − ft/s.
dt
2 dt
2
2
126
Pulley
y
10
Boat
x
Pulley
y
10
Boat
x
Light
18
Man
Shadow
6
x
(b)
29.
y
The tip of the shadow is z = x + y feet from the street light, thus the rate at which it is
dx
dy
dx
3
dy
dz
=
+
. In part (a) we found that
= − when
= −3 so
moving is given by
dt
dt
dt
dt
2
dt
dz
= (−3/2) + (−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/s toward
dt
the street light.
¯
dx ¯¯
dθ
2π
π
Find
given that
=
= rad/s. Then
¯
dt θ=π/4
dt
10
5
dθ
dx
= 4 sec2 θ ,
x = 4 tan θ (see figure) so
dt
dt
¯
³
´ ³π´
dx ¯¯
2 π
= 4 sec
= 8π/5 km/s.
dt ¯θ=π/4
4
5
x
4
Ship
θ
127
30.
Chapter 4
¯
dz ¯¯
If x, y, and z are as shown in the figure, then we want
given
dt ¯x=2,
y=4
¯
dx
dy ¯¯
2
2
2
that
= −600 and
= −1200. But z = x + y so
dt
dt ¯x=2,
y=4
µ
¶
1
dx
dz
dx
dy dz
dy
2z
= 2x
+ 2y ,
=
x
+y
. When x = 2 and
dt
dt
dt dt
z√ dt √ dt
y =¯ 4, z 2 = 22 + 42 = 20, z = 20 = 2 5 so
√
dz ¯¯
1
3000
= √ [2(−600) + 4(−1200)] = − √ = −600 5 mi/h; the
¯
dt x=2,
2 5
5
y=4
distance
between
missile
and
aircraft
is
decreasing
at the rate of
√
600 5 mi/h.
¯
¯
dy ¯¯
dz ¯¯
dx
= −600 and
31. We wish to find
given
= −1200 (see
dt ¯x=2,
dt
dt ¯x=2,
y=4
y=4
figure). From the law of cosines,
z 2 = x2 + y 2 − 2xy cos 120◦ = x2 + y 2 − 2xy(−1/2) = x2 + y 2 + xy, so
dx
dy
dy
dx
dz
= 2x
+ 2y
+x
+y ,
2z
dt
dt
dt
dt
dt ¸
·
1
dx
dy
dz
=
(2x + y)
+ (2y + x)
. When x = 2 and y = 4,
dt
2z
dt
dt√
√
2
2
z2 =
¯ 2 + 4 + (2)(4) = 28, so z = 28 = 2 7, thus
1
dz ¯¯
4200
√ [(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = − √ =
=
dt ¯x=2,
2(2 7)
7
y=4
√
−600 7 mi/h; the distance between
missile and aircraft is
√
decreasing at the rate of 600 7 mi/h.
32.
(a)
Let x, y, and z be the distances shown in the first figure. Find
P
Aircraft
x
y
z
Missile
P
x
Aircraft
120°
y
z
Missile
¯
dz ¯¯
dx
= −75 and
given that
dt ¯x=2,
dt
y=0
dy
= −100. In order to find an equation relating x, y, and z, first draw the line segment that
dt
joins the point P to the car, as shown p
in the second figure. Because triangle OP C is a right
triangle, it follows that P C has length x2 + (1/2)2 ; but triangle HP C is also a right triangle
´2
³p
dx
dy
dz
= 2x
+ 2y
+ 0,
x2 + (1/2)2 + y 2 = x2 + y 2 + 1/4 and 2z
so z 2 =
dt
dt
dt
µ
¶
√
1
dx
dy
dz
=
x
+y
. Now, when x = 2 and y = 0, z 2 = (2)2 + (0)2 + 1/4 = 17/4, z = 17/2
dt
z
dt
dt
¯
√
1
dz ¯¯
= √
so
[2(−75) + 0(−100)] = −300/ 17 mi/h
¯
dt x=2,
( 17/2)
y=0
North
1
2
mi
y
West
1
mi
2
P
x
East
z
Helicopter
(b)
decreasing, because
P
y
Car
x
O
z
H
dz
< 0.
dt
C
Exercise Set 4.6
33.
(a)
We want
128
¯
¯
dy ¯¯
dx ¯¯
given
that
= 6. For convenience, first rewrite the equation as
dt ¯x=1,
dt ¯x=1,
y=2
y=2
8 8
16 dy dy
y3
dy
dx
dx
+ y3
=
y ,
=
so
xy = + y 2 then 3xy 2
16
5 5
dt
dt
5 dt dt
dt
y − 3xy 2
5
¯
23
dy ¯¯
=
(6) = −60/7 units/s.
16
dt ¯x=1,
(2) − 3(1)22
y=2
5
dy
<0
(b) falling, because
dt
¯
¯
dy ¯¯
dx ¯¯
given that
= 2. Square and rearrange to get x3 = y 2 − 17
34. Find
dt ¯(2,5)
dt ¯(2,5)
¯
µ ¶
2y dy dx ¯¯
5
5
dx
dy dx
= 2y ,
= 2 ,
(2) = units/s.
=
so 3x2
dt
dt dt
3x dt dt ¯(2,5)
6
3
3
35.
The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) is
¯
¯
p
√
dD ¯¯
dx ¯¯
2
2
2
given that
= −2.
D = (x − 3) + (2x − 0) = 5x − 6x + 9. Find
dt ¯x=3
dt ¯x=3
¯
12
5x − 3
dD ¯¯
dD
dx
= √ (−2) = −4 units/s.
=√
, so
¯
2
dt
dt x=3
36
5x − 6x + 9 dt
¯
¯
dx ¯¯
dD ¯¯
given
that
= 4.
36. (a) Let D be the distance between P and (2, 0). Find
dt ¯x=3
dt ¯x=3
p
p
√
dD
2x − 3
= √
D = (x − 2)2 + y 2 = (x − 2)2 + x = x2 − 3x + 4 so
;
dt
2 x2 − 3x + 4
¯
3
3
dD ¯¯
= √ = units/s.
dt ¯x=3
4
2 4
¯
¯
dx ¯¯
dθ ¯¯
given that
= 4.
(b) Let θ be the angle of inclination. Find
dt ¯x=3
dt ¯x=3
√
x+2
dθ
x+2
y
x
dx dθ
dx
=
so sec2 θ
=− √
,
= − cos2 θ √
.
tan θ =
x−2
x−2
dt
2 x(x − 2)2 dt dt
2 x(x − 2)2 dt
¯
1
dθ ¯¯
1 5
5
When x = 3, D = 2 so cos θ = and
= − √ (4) = − √ rad/s.
¯
2
dt
42 3
2 3
x=3
37. Solve
dx
dy
dy dx
dx
dy
=3
given y = x ln x. Then
=
= (1 + ln x) , so 1 + ln x = 3, ln x = 2, x = e2 .
dt
dt
dt
dx dt
dt
dy
dy
dx
dx
16
dx
+ 18y
= 0; if
=
6= 0, then (32x + 18y)
= 0, 32x + 18y = 0, y = − x so
dt
dt
dt
dt
dt
9
9
9
16 9
16
256 2
400 2
81
2
2
x = 144,
x = 144, x =
, x = ± . If x = , then y = −
= − . Similarly,
16x + 9
81
9
25
5
5
9 5
5
16
9 16
9 16
9
. The points are ( , − ) and (− , ).
if x = − , then y =
5
5
5
5
5 5
¯
¯
1
1
1 ds
1 dS
ds ¯¯
1
dS ¯¯
+
=
we get − 2
− 2
= 0, so
given that
= −2. From
39. Find
dt ¯s=10
dt ¯s=10
s
S
6
s
dt
S
dt
¯
2
S ds
1
1
1
dS ¯¯
225
dS
= − 2 . If s = 10, then
+ = which gives S = 15. So
(−2) = 4.5 cm/s.
=−
¯
dt
s dt
10 S
6
dt s=10
100
The image is moving away from the lens.
38. 32x
129
Chapter 4
40.
Suppose that the reservoir has height H and that the radius at the top is R. At any instant of time
let h and r be the corresponding dimensions of the cone of water (see figure). We want to show that
dh
dV
is constant and independent of H and R, given that
= −kA where V is the volume of water,
dt
dt
A is the area of a circle of radius r, and k is a positive constant. The volume of a cone of radius r and
µ ¶2
1 2
R
R
1
R
r
height h is V = πr h. By similar triangles = , r = h thus V = π
h3 so
3
h
H
H
3
H
dV
=π
dt
µ
R
H
¶2
h2
dh
dt
(1)
R
µ ¶2
R
dV
2
= −kA or, because A = πr = π
h2 ,
But it is given that
dt
H
µ ¶2
R
dV
= −kπ
h2 , which when substituted into equation (1) gives
dt
H
µ ¶2
µ ¶2
R
R
dh dh
2
= −k.
h =π
h2 ,
−kπ
H
H
dt dt
41.
r
H
h
dr
is a constant
Let r be the radius, V the volume, and A the surface area of a sphere. Show that
dt
4
dV
= −kA, where k is a positive constant. Because V = πr3 ,
given that
dt
3
dV
dr
= 4πr2
(1)
dt
dt
dV
dV
= −kA or, because A = 4πr2 ,
= −4πr2 k which when substituted into
dt
dt
dr dr
= −k.
equation (1) gives −4πr2 k = 4πr2 ,
dt dt
But it is given that
42.
Let x be the distance between the tips of the minute and hour hands, and α and β the angles shown
in the figure. Because the minute hand makes one revolution in 60 minutes,
2π
dα
=
= π/30 rad/min; the hour hand makes one revolution in 12 hours (720 minutes), thus
dt
60
¯
dα
2π
dx ¯¯
dβ
dβ
given that
=
= π/360 rad/min. We want to find
= π/30 and
= π/360.
dt
720
dt ¯α=2π,
dt
dt
β=3π/2
Using the law of cosines on the triangle shown in the figure,
x2 = 32 + 42 − 2(3)(4) cos(α
µ − β) = 25
¶ − 24 cos(α − β), so
dα dβ
dx
= 0 + 24 sin(α − β)
−
,
2x
dt
dt
dt
µ
¶
12 dα dβ
dx
=
−
sin(α − β). When α = 2π and β = 3π/2,
dt
x dt
dt
x2 =
¯ 25 − 24 cos(2π − 3π/2) = 25, x = 5; so
11π
12
dx ¯¯
(π/30 − π/360) sin(2π − 3π/2) =
in/min.
=
¯
dt α=2π,
5
150
4
x
3
b
a
β=3π/2
43.
Extend sides of cup to complete the cone and let V0 be the volume
1
of the portion added, then (see figure) V = πr2 h − V0 where
3
µ ¶2
4
1
1
1
1
r
h
=
= so r = h and V = π
πh3 − V0 ,
h − V0 =
h
12
3
3
3¯ 3
27
1
9 dV dh ¯¯
9
20
dh dh
dV
= πh2 ,
=
,
cm/s.
=
(20) =
¯
2
2
dt
9
dt dt
πh dt dt h=9
π(9)
9π
4
r
6
2
6
h
Exercise Set 4.7
130
EXERCISE SET 4.7
1.
x2 − 4
(x − 2)(x + 2)
x+2
2
= lim
= lim
=
+ 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4
3
5
2 − lim
2
2x − 5
x→+∞ x
lim
=
=
7
x→+∞ 3x + 7
3
3 + lim
x→+∞ x
(a)
lim
x→2 x2
(b)
2.
sin x
cos x
sin x
= sin x
= cos x so lim
=0
x→0
tan x
sin x
tan x
x2 − 1
x2 − 1
(x − 1)(x + 1)
x+1
2
=
=
so
lim
=
x→1 x3 − 1
x3 − 1
(x − 1)(x2 + x + 1)
x2 + x + 1
3
(a)
(b)
3.
lim
1/x
=1
x→1 1
4.
5.
lim
ex
=1
x→0 cos x
6.
7.
sec2 θ
=1
θ→0
1
8.
9.
lim
lim
x→π +
cos x
= −1
1
10.
lim
2 cos 2x
= 2/5
5 cos 5x
lim
1
= 1/5
6x − 13
lim
tet + et
= −1
−et
lim
cos x
= +∞
2x
x→0
x→3
t→0
x→0+
3e3x
9e3x
= lim
= +∞
x→+∞ 2x
x→+∞ 2
11.
1/x
=0
x→+∞ 1
13.
lim+
− csc2 x
−x
−1
= lim+
= lim+
= −∞
2
x→0 sin x
x→0 2 sin x cos x
1/x
lim+
−1/x
x
= lim+ 1/x = 0
2
1/x
x→0 e
(−1/x )e
14.
15.
16.
lim
x→0
x→0
20.
21.
22.
23.
lim
100x99
(100)(99)x98
(100)(99)(98) · · · (1)
= lim
= · · · = lim
=0
x
x→+∞
x→+∞
x→+∞
e
ex
ex
√
cos x/ sin x
2/ 1 − 4x2
2
=
lim
cos
x
=
1
lim+
17. lim
=2
x→0 sec2 x/ tan x
x→0+
x→0
1
lim
1−
18.
12.
lim
x→0
1
1
1
1 + x2 = lim
=
2
2
x→0
3x
3(1 + x )
3
lim (x − π) tan(x/2) = lim
x→π
x→π
19.
x
1
= lim x = 0
x
x→+∞ e
x→+∞ e
lim xe−x = lim
x→+∞
x−π
1
= lim
= −2
cot(x/2) x→π −(1/2) csc2 (x/2)
sin(π/x)
(−π/x2 ) cos(π/x)
= lim
= lim π cos(π/x) = π
x→+∞
x→+∞
x→+∞
1/x
−1/x2
lim x sin(π/x) = lim
x→+∞
lim+ tan x ln x = lim+
x→0
lim
x→(π/2)−
x→0
ln x
1/x
− sin2 x
−2 sin x cos x
= lim+
= lim+
= lim+
=0
2
x→0
cot x x→0 − csc x x→0
x
1
sec 3x cos 5x =
lim
x→(π/2)−
−5(+1)
5
cos 5x
−5 sin 5x
= lim −
=
=−
cos 3x x→(π/2) −3 sin 3x
(−3)(−1)
3
131
24.
Chapter 4
x−π
1
= lim
=1
x→π sec2 x
tan x
lim (x − π) cot x = lim
x→π
x→π
ln(1 − 3/x)
−3
= lim
= −3, lim y = e−3
x→+∞
x→+∞ 1 − 3/x
x→+∞
1/x
25.
y = (1 − 3/x)x , lim ln y = lim
26.
y = (1 + 2x)−3/x , lim ln y = lim −
27.
y = (ex + x)1/x , lim ln y = lim
28.
y = (1 + a/x)bx , lim ln y = lim
29.
y = (2 − x)tan(πx/2) , lim ln y = lim
30.
y = [cos(2/x)]x , lim ln y = lim
x→+∞
x→0
x→+∞
x→+∞
ln cos(2/x)
(−2/x2 )(− tan(2/x))
− tan(2/x)
= lim
= lim
2
x→+∞
x→+∞
x→+∞
1/x
−2/x3
1/x
2
x→+∞
lim
x→0
lim
x→0
b ln(1 + a/x)
ab
= lim
= ab, lim y = eab
x→+∞ 1 + a/x
x→+∞
1/x
ln(2 − x)
2 sin2 (πx/2)
= lim
= 2/π, lim y = e2/π
x→1 cot(πx/2)
x→1
x→1
π(2 − x)
x→1
µ
32.
ln(ex + x)
ex + 1
= lim x
= 2, lim y = e2
x→0
x→0 e + x
x→0
x
x→0
= limx→+∞
31.
x→0
3 ln(1 + 2x)
6
= lim −
= −6, lim y = e−6
x→0
x→0
x
1 + 2x
(2/x2 ) sec2 (2/x)
= −2, lim y = e−2
x→+∞
−1/x2
1
1
−
sin x x
¶
= lim
x→0
x − sin x
1 − cos x
sin x
= lim
= lim
=0
x→0 x cos x + sin x
x→0 2 cos x − x sin x
x sin x
1 − cos 3x
3 sin 3x
9
9
= lim cos 3x =
= lim
x→0
x→0 2
x2
2x
2
33.
(x2 + x) − x2
x
1
lim √
= 1/2
= lim √
= lim p
2
2
x→+∞
x→+∞
x→+∞
x +x+x
x +x+x
1 + 1/x + 1
34.
ex − 1 − x
ex − 1
ex
= lim x
= lim x
= 1/2
x
x
x→0 xe − x
x→0 xe + e − 1
x→0 xe + 2ex
35.
lim
lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln
x→+∞
x
lim
x→+∞
36.
38.
(a)
(b)
39.
40.
x→+∞
x
e
e
e
= lim
= lim
= +∞ so lim [x − ln(x2 + 1)] = +∞
x→+∞
x2 + 1 x→+∞ 2x x→+∞ 2
lim ln
x→+∞
x→+∞
x
ex
,
x2 + 1
x
1
= lim ln
= ln(1) = 0
1 + x x→+∞ 1/x + 1
ln x
1/x
1
= lim
= lim
=0
n
n−1
x→+∞
x→+∞
x
nx
nxn
xn
nxn−1
= lim
= lim nxn = +∞
lim
x→+∞ ln x
x→+∞ 1/x
x→+∞
lim
x→+∞
0
3x2 − 2x + 1
because it is not a form
x→1
3x2 − 2x
0
(a)
L’Hôpital’s Rule does not apply to the problem lim
(b)
3x2 − 2x + 1
=2
x→1
3x2 − 2x
lim
4x3 − 12x2 + 12x − 4
12x2 − 24x + 12
24x − 24
=
lim
= lim
=0
x→1 4x3 − 9x2 + 6x − 1
x→1 12x2 − 18x + 6
x→1 24x − 18
lim
Exercise Set 4.7
41.
lim
x→+∞
132
1/(x ln x)
2
√
= lim √
=0
x→+∞
1/(2 x)
x ln x
0.15
100
10000
0
42.
y = xx , lim+ ln y = lim+
x→0
x→0
ln x
= lim −x = 0, lim+ y = 1
x→0
1/x x→0+
1
0
0.5
0
43.
y = (sin x)3/ ln x , lim+ ln y = lim+
lim y = e3
x→0
x→0
x
3 ln sin x
= lim+ (3 cos x)
= 3,
x→0
ln x
sin x
25
x→0+
0
0.5
19
44.
lim −
x→π/2
4 sec2 x
4
= lim
=4
sec x tan x x→π/2− sin x
4
0.5
p/ 2
0
45.
e−x ln x − 1
;
e−x
e−x
ln
x
1/x
= lim
= 0 by L’Hôpital’s Rule, so
lim e−x ln x = lim
x→+∞
x→+∞ ex
x→+∞ ex
−x
e ln x − 1
lim [ln x − ex ] = lim
= −∞
x→+∞
x→+∞
e−x
ln x − ex = ln x −
1
=
0
0
3
-16
46.
lim [ln ex − ln(1 + 2ex )] = lim ln
x→+∞
x→+∞
1
1
= ln ;
= lim ln −x
x→+∞
e +2
2
horizontal asymptote y = − ln 2
ex
1 + 2ex
-0.6
0
-1.2
12
133
47.
Chapter 4
ln(ln x)
1
= lim
= 0;
x→+∞ x ln x
x
lim y = 1, y = 1 is the horizontal asymptote
y = (ln x)1/x , lim ln y = lim
x→+∞
1.02
x→+∞
x→+∞
100
10000
1
µ
48.
y=
x+1
x+2
¶x
1
,
ln
lim ln y = lim
x→+∞
x→+∞
−1
lim y = e
x→+∞
x+1
−x2
x + 2 = lim
= −1;
x→+∞ (x + 1)(x + 2)
1/x
is the horizontal asymptote
0
50
0
49.
(a)
0
50.
(a)
y = x 1+ln x ; lim+ ln y = lim+
(b)
same as part (a) with x → +∞
52.
+∞
(c)
ln a
x→0
x→0
0
(d) −∞
(e)
+∞
(f )
−∞
(ln a) ln x
(ln a)/x
= lim+
= lim+ ln a = ln a, lim+ y = eln a = a
x→0
x→0
x→0
1 + ln x
1/x
(ln a) ln(x + 1)
ln a
= lim
= ln a, lim y = eln a = a
x→0 x + 1
x→0
x
¶
µ
1 + 2 cos 2x
x + sin 2x
sin 2x
does not exist, nor is it ±∞; lim
= lim 1 +
=1
lim
x→+∞
x→+∞
x→+∞
1
x
x
(c)
51.
(b)
y = (x + 1)
ln a
x
, lim ln y = lim
x→0
x→0
2
2 − cos x
2x − sin x
2 − (sin x)/x
does not exist, nor is it ±∞; lim
= lim
=
x→+∞ 3 + cos x
x→+∞ 3x + sin x
x→+∞ 3 + (sin x)/x
3
lim
x(2 + sin 2x)
2 + sin 2x
= lim
, which
x→+∞
x→+∞
x→+∞ 1 + 1/x
x+1
does not exist because sin 2x oscillates between −1 and 1 as x → +∞
¶
µ
sin x
1 1
+ cos x +
does not exist, nor is it ±∞;
54. lim
x→+∞ x
2
2x
x(2 + sin x)
2 + sin x
= lim
=0
lim
x→+∞
x→+∞ x + 1/x
x2 + 1
53.
lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim
55.
lim+
56.
R→0
(a)
(b)
V t −Rt/L
L e
1
=
Vt
L
π/2 − x
−1
= lim
= lim sin2 x = 1
cot x
x→π/2 − csc2 x
x→π/2
¶
¶
µ
µ
cos x − (π/2 − x) sin x
sin x
1
1
− tan x = lim
−
= lim
lim
cos x
(π/2 − x) cos x
x→π/2 π/2 − x
x→π/2 π/2 − x
x→π/2
lim (π/2 − x) tan x = lim
x→π/2
x→π/2
−(π/2 − x) cos x
−(π/2
− x) sin x − cos x
x→π/2
= lim
= lim
x→π/2
(π/2 − x) sin x + cos x
=0
−(π/2 − x) cos x + 2 sin x
Supplementary Exercises 4
(c)
1/(π/2 − 1.57) ≈ 1255.765849, tan 1.57 ≈ 1255.765592;
1/(π/2 − 1.57) − tan 1.57 ≈ 0.000265
(c)
kt − 1
(ln k)k t
= lim+
= ln k
x→+∞
t→0
t→0
t
1
¢
¡ √
¢
¡ √
ln 0.3 = −1.20397, 1024 1024 0.3 − 1 = −1.20327; ln 2 = 0.69315, 1024 1024 2 − 1 = 0.69338
(a)
No; sin(1/x) oscillates as x → 0.
57. (b)
58.
134
lim x(k 1/x − 1) = lim+
(b)
0.05
-0.35
0.35
-0.05
(c)
For the limit as x → 0+ use the Squeezing Theorem together with the inequalities −x2 ≤
x2 sin(1/x) ≤ x2 . For x → 0− do the same; thus lim f (x) = 0.
x→0
k + cos `x
= ±∞. Hence k = −1, and by the rule
x2
√
−1 + cos `x
−` sin `x
−`2 cos `x
`2
lim
=
lim
=
lim
=
−
= 4 if ` = ±2 2.
x→0
x→0
x→0
x2
2x
2
2
59.
If k 6= −1 then lim (k + cos `x) = k + 1 6= 0, so lim
60.
(a)
x→0
(b)
61.
x→0
− cos(1/x) + 2x sin(1/x)
which does not exist (nor is it ±∞).
cos x
h x i
x
1
[x sin(1/x)], but lim
= lim
= 1 and lim x sin(1/x) = 0, thus
Rewrite as lim
x→0 sin x
x→0 cos x
x→0
h x i x→0 sin x
[x sin(1/x)] = (1)(0) = 0
lim
x→0 sin x
Apply the rule to get lim
x→0
sin(1/x)
sin x
, lim
= 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between −1
x→0
(sin x)/x x→0+ x
x sin(1/x)
does not exist.
and 1 as x → +∞, so lim+
x→0
sin x
lim
x→0+
CHAPTER 4 SUPPLEMENTARY EXERCISES
1.
(a)
f (g(x)) = x for all x in the domain of g, and g(f (x)) = x for all x in the domain of f .
(b) They are reflections of each other through the line y = x.
(c)
The domain of one is the range of the other and vice versa.
(d) The equation y = f (x) can always be solved for x as a function of y. Functions with no inverses
include y = x2 , y = sin x.
2.
(e)
Yes, g is continuous; this is evident from the statement about the graphs in part (b) above.
(f )
Yes, g must be differentiable (where f 0 6= 0); this can be inferred from the graphs. Note that if
f 0 = 0 at a point then g 0 cannot exist (infinite slope).
(a)
For sin x, −π/2 ≤ x ≤ π/2; for cos x, 0 ≤ x ≤ π; for tan x, −π/2 < x < π/2; for sec x,
0 ≤ x < π/2 or π/2 < x ≤ π.
135
Chapter 4
(b)
y
y
y = sin -1 x
1
y = sin x
π/2
y = cos -1 x
x
π
-1
x
y = cos x
-1
y
y
y = tan x
2
y = sec -1 x
y = tan -1 x
y = sec x
2
y = sec -1 x
π/2
- π/2
x
π/2
-1
-2
3.
x
y = sec x
(a)
when the limit takes the form 0/0 or ∞/∞
(b)
Not necessarily; only if lim f (x) = 0. Consider g(x) = x; lim g(x) = 0. For f (x) choose
x→0
cos x
x2
cos x, x2 , and |x|1/2 . There are three possibilities; lim
does not exist; lim
= 0, and
x→0 x
x→0 x
1/2
|x|
lim
= +∞.
x→0 x2
4.
In the case +∞−(−∞) the limit is +∞; in the case −∞−(+∞) the limit is −∞, because large positive
(negative) quantities are being added to large positive (negative) quantities. The cases +∞ − (+∞)
and −∞ − (−∞) are indeterminate; large numbers of opposite sign are being subtracted, and more
information about the sizes is needed.
µ
5.
−1
x+1
8
¶1/3
1
(x + 1)1/3
2
(a)
x = f (y) = 8y − 1; y = f
(b)
(c)
f (x) = (x − 1)2 ; f does not have an inverse because f is not one-to-one, for example
f (0) = f (2) = 1.
√
x = f (y) = (ey )2 + 1; y = f −1 (x) = ln x − 1 = 12 ln(x − 1)
(d)
x = f (y) =
3
(x) =
=
y+2
x+2
; y = f −1 (x) =
y−1
x−1
ad − bc
; if ad−bc = 0 then the function represents a horizontal line, no inverse. If ad−bc 6= 0
(cx + d)2
ay + b
b − xd
then y = f −1 (x) =
.
then f 0 (x) > 0 or f 0 (x) < 0 so f is invertible. If x = f (y) =
cy + d
xc − a
6. f 0 (x) =
7.
(a)
2 −1/3 2 −1/3 0
x
− y
y − y 0 = 0. At x = 1 and y = −1, y 0 = 2. The tangent line is
3
3
y + 1 = 2(x − 1).
(b)
(xy 0 + y) cos xy = y 0 . With x = π/2 and y = 1 this becomes y 0 = 0, so the equation of the
tangent line is y − 1 = 0(x − π/2) or y = 1.
Differentiating,
8. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1 (4/5)] = 3/5,
sin[cos−1 (5/13)] = 12/13, cos[sin−1 (4/5)] = 3/5, cos[sin−1 (5/13)] = 12/13
(a)
cos[cos−1 (4/5) + sin−1 (5/13)] = cos(cos−1 (4/5)) cos(sin−1 (5/13)) − sin(cos−1 (4/5)) sin(sin−1 (5/13))
33
4 12 3 5
−
=
.
=
5 13 5 13
65
Supplementary Exercises 4
(b)
136
sin[sin−1 (4/5) + cos−1 (5/13)] = sin(sin−1 (4/5)) cos(cos−1 (5/13)) + cos(sin−1 (4/5)) sin(cos−1 (5/13))
3 12
56
4 5
+
=
.
=
5 13 5 13
65
¢
¡
9. 3 ln e2x (ex )3 + 2 exp(ln 1) = 3 ln e2x + 3 ln(ex )3 + 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 2
10.
Y = ln(Cekt ) = ln C + ln ekt = ln C + kt, a line with slope k and Y -intercept ln C
11.
(a)
ex
ex
ex
=
lim
= +∞
=
lim
x→+∞
x→+∞
x→+∞ x2
x→+∞ 2x
x→+∞ 2
x
2
2 x
2
so lim (e /x − 1) = +∞ and thus lim x (e /x − 1) = +∞
lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim
x→+∞
(b)
(c)
12.
13.
1
ln x
1/x
= lim 3 = ; lim
lim 4
x→1 x − 1
x→1 4x
4 x→1
r
x→+∞
ln x
=
x4 − 1
r
1
ln x
=
4
x→1 x − 1
2
lim
lim ax ln a = ln a
x→0
y 0 = aeax sin bx + beax cos bx and y 00 = (a2 − b2 )eax sin bx + 2abeax cos bx, so y 00 − 2ay 0 + (a2 + b2 )y
= (a2 − b2 )eax sin bx + 2abeax cos bx − 2a(aeax sin bx + beax cos bx) + (a2 + b2 )eax sin bx = 0.
√
√
sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y 0 =
y 00 + 2 sin y cos3 y =
1
−2x
x
+ 2√
= 0.
2
(1 + x2 )2
(1
+
x2 )3/2
1+x
14.
ln y = 2x ln 3 + 7x ln 5;
15.
Find
1
−2x
, y 00 =
, hence
2
1+x
(1 + x2 )2
dy
dy
/y = 2 ln 3 + 7 ln 5, or
= (2 ln 3 + 7 ln 5)y
dx
dx
¯
dy
dz
dθ ¯¯
= a and
= −b. From the figure
given
dt ¯x=1
dt
dt
y=1
√
sin θ = y/z; when xµ = y = 1, z = ¶ 2. So θ = sin−1 (y/z) and
1
y dz
a
1 dy
dθ
=p
− 2
= −b − √ when x = y = 1.
2
2
dt
z
dt
z
dt
2
1 − y /z
z
y
θ
x
16.
(a)
f 0 (x) = −3/(x + 1)2 . If x = f (y) = 3/(y + 1) then y = f −1 (x) = (3/x) − 1, so
and
(b)
17.
1
f 0 (f −1 (x))
=−
(f −1 (x) + 1)2
(3/x)2
3
=−
= − 2.
3
3
x
d −1
−3
f (x) = 2 ;
dx
x
d −1
2
f (x) = ; and
f (x) = ex/2 , f 0 (x) = 12 ex/2 . If x = f (y) = ey/2 then y = f −1 (x) = 2 ln x, so
dx
x
2
1
−f −1 (x)/2
− ln x
−1
=
2e
=
2e
=
2x
=
f 0 (f −1 (x))
x
y
(a)
(b)
2
2
-2
4
x
The curve y = e−x/2 sin 2x has x-intercepts
at x = 0, π/2. It intersects the curve y =
e−x/2 at x = π/4, and it intersects the curve
y = −e−x/2 at x = −π/4, 3π/4.
137
18.
Chapter 4
(a)
(b)
y
y
π/2
π/2
x
x
1
(c)
1
(d)
y
y
π/2
x
x
1
5
−π/2
19.
(a)
(b)
(c)
y = x3 + 1 so y 0 = 3x2 .
(d)
y0 =
(e)
20.
21.
x/(1 + x) − ln(1 + x)
ln(1 + x)
ln(1 + x) y 0
1
,
=
−
=
,
x
y
x2
x(1 + x)
x2
(1 + x)(1/x)
1
dy
ln(1 + x)
= (1 + x)(1/x)−1 −
dx
x
x2
¶
¶
µ
µ
¤
£ x
dy
y0
1
1
x
x
ex x
x
=e
+ ln x ,
=x e
+ ln x = ex xe −1 + xe ln x
ln y = e ln x,
y
x
dx
x
ln y =
abe−x
(1 + be−x )2
dy
2 −1/3 dy
2
xy
+ y 2/3 + yx−1/3 + x2/3
= 2x. Multiply by 3x1/3 y 1/3 :
3
dx
3
dx
dy
dy
+ 3x1/3 y + 2y 4/3 + 3xy 1/3
= 6x4/3 y 1/3 . Regroup:
2x4/3
dx
dx
´
6x4/3 y 1/3 − 3x1/3 y − 2y 4/3
dy
dy ³ 4/3
2x + 3xy 1/3 = 6x4/3 y 1/3 − 3x1/3 y − 2y 4/3 ,
=
.
dx
dx
2x4/3 + 3xy 1/3
1
1
ln x + ln(x + 1) − ln sin x + ln cos x, so
2
3
1
cos x
sin x
5x + 3
1
0
y =
+
−
−
=
− cot x − tan x.
2x 3(x + 1)
sin x
cos x
6x(x + 1)
(f )
y=
(a)
Find x when y = 5 · 12 = 60 in. Since y = log x, x = 10y = 1060 in. This is approximately
2.68 × 1042 light-years, so even in astronomical terms it is a fabulously long distance.
(b)
Find x when y = 100(5280)(12) in. Since y = 10x , x = log y = 6.80 in or 0.57 ft, approximately.
(a)
The function ln x − x0.2 is negative at x = 1 and positive at x = 4, so it must be zero in between
(IVT).
(b)
x = 3.654
Supplementary Exercises 4
22.
138
ln x
1
= . The steps are reversible.
x
k
(b) By zooming it is seen that the maximum value of y is
approximately 0.368 (actually, 1/e), so there are two distinct
solutions of xk = ex whenever k > 1/0.368 ≈ 2.717.
(a)
If xk = ex then k ln x = x, or
y
x
2
-2
(c)
23.
24.
x = 1.155
1
1
= 1 so x =
. The
x ln b
ln b
curves intersect when (x, x) lies on the graph of y = logb x, so
ln x
from which
x = logb x. From Formula (9), Section 4.2, logb x =
ln b
1/e
ln x = 1, x = e, ln b = 1/e, b = e ≈ 1.4447.
Set y = logb x and solve y 0 = 1: y 0 =
(a)
(b)
√
Find the point of intersection: f (x) = x + k = ln x. The
1
1 √
slopes are equal, so m1 = = m2 = √ , x = 2, x = 4.
x
2
x
√
Then ln 4 = 4 + k, k = ln 4 − 2.
√
k
1
Since the slopes are equal m1 = √ = m2 = , so k x = 2.
x
2
x
√
At the point of intersection k x = ln x, 2 = ln x, x = e2 ,
k = 2/e.
y
2
2
x
y
2
x
2
y
2
0
x
5
CHAPTER 5
Analysis of Functions and Their Graphs
EXERCISE SET 5.1
1.
(a)
f 0 > 0 and f 00 > 0
(b)
y
f 0 > 0 and f 00 < 0
y
x
x
(c)
f 0 < 0 and f 00 > 0
(d)
f 0 < 0 and f 00 < 0
y
y
x
x
2.
(a)
(b)
y
y
x
(c)
x
(d)
y
y
x
3.
A: dy/dx < 0, d2 y/dx2 > 0
B: dy/dx > 0, d2 y/dx2 < 0
C: dy/dx < 0, d2 y/dx2 < 0
x
4. A: dy/dx < 0, d2 y/dx2 < 0
B: dy/dx < 0, d2 y/dx2 > 0
C: dy/dx > 0, d2 y/dx2 < 0
139
Exercise Set 5.1
140
5.
An inflection point occurs when f 00 changes sign: at x = −1, 0, 1 and 2.
6.
(a) f (0) < f (1) since f 0 > 0 on (0, 1).
(c) f 0 (0) > 0 by inspection.
(e) f 00 (0) < 0 since f 0 is decreasing there.
7.
(a) [4, 6]
(d) (2, 3) and (5, 7)
8.
f0
f 00
(1, 2)
−
+
(2, 3)
−
−
(3, 4)
−
+
(b) f (1) > f (2) since f 0 < 0 on (1, 2).
(d) f 0 (1) = 0 by inspection.
(f ) f 00 (2) = 0 since f 0 has a minimum there.
(b) [1, 4] and [6, 7]
(e) x = 2, 3, 5
(4, 5)
+
+
(5, 6)
+
−
(c) (1, 2) and (3, 5)
(6, 7)
−
−
f 0 (x) = 2x − 5
f 00 (x) = 2
(a) [5/2, +∞)
(c) (−∞, +∞)
(e) none
(b) (−∞, 5/2]
(d) none
f 0 (x) = −2(x + 3/2)
f 00 (x) = −2
(a) (−∞, −3/2]
(c) none
(e) none
(b)
(d)
11. f 0 (x) = 3(x + 2)2
f 00 (x) = 6(x + 2)
(a) (−∞, +∞)
(c) (−2, +∞)
(e) −2
(b) none
(d) (−∞, −2)
12. f 0 (x) = 3(4 − x2 )
f 00 (x) = −6x
(a) [−2, 2]
(c) (−∞, 0)
(e) 0
(b) (−∞, −2], [2, +∞)
(d) (0, +∞)
9.
10.
[−3/2, +∞)
(−∞, +∞)
13.
f 0 (x) = 12x2 (x − 1)
f 00 (x) = 36x(x − 2/3)
(a) [1, +∞)
(c) (−∞, 0), (2/3, +∞)
(e) 0, 2/3
(b)
(d)
14.
f 0 (x) = 4x(x2 − 4)
f 00 (x) = 12(x2 − 4/3)
(a) [−2, 0], [2, +∞)
√
√
(c) (−∞, −2/ 3), (2/ 3, +∞)
√
√
(e) −2/ 3, 2/ 3
(b) (−∞, −2], [0, 2]
√
√
(d) (−2/ 3, 2/ 3)
15.
f 0 (x) =
4x
2
(x + 2)2
f 00 (x) = −4
3x2 − 2
(x2 + 2)3
(a) [0, +∞)
p
p
(d) (−∞, − 2/3), (+ 2/3, +∞)
16.
17.
(b) (−∞, 0]
p
p
(e) − 2/3, 2/3
2 − x2
2x(x2 − 6)
f 00 (x) =
2
2
(x + 2)
(x2 + 2)3
√ √
√
√
(a) [− 2, 2]
(b) (−∞, − 2], [ 2, +∞)
√
√
√
√
(d) (−∞, − 6), (0, 6)
(e) − 6, 0, 6
(−∞, 1]
(0, 2/3)
p
p
(c) (− 2/3, + 2/3)
f 0 (x) =
f 0 (x) = 13 (x + 2)−2/3
00
f (x) =
− 29 (x
−5/3
+ 2)
(a) (−∞, +∞)
(c) (−∞, −2)
(e) −2
√
√
(c) (− 6, 0), ( 6, +∞)
(b) none
(d) (−2, +∞)
141
18.
Chapter 5
f 0 (x) = 23 x−1/3
f 00 (x) = − 29 x−4/3
(a) [0, +∞)
(c) none
(e) none
(b) (−∞, 0]
(d) (−∞, 0), (0, +∞)
4(x + 1)
3x2/3
4(x − 2)
f 00 (x) =
9x5/3
(a) [−1, +∞)
(c) (−∞, 0), (2, +∞)
(e) 0, 2
(b) (−∞, −1]
(d) (0, 2)
4(x − 1/4)
3x2/3
4(x
+ 1/2)
f 00 (x) =
9x5/3
(a) [1/4, +∞)
(c) (−∞, −1/2), (0, +∞)
(e) −1/2, 0
(b)
(d)
(a) (−∞, 0]
(c) (−∞, −1), (1, +∞)
(e) −1, 1
(b) [0, +∞)
(d) (−1, 1)
(a) (−∞, +∞)
(c) (0, +∞)
(e) 0
(b)
(d)
19.
f 0 (x) =
20.
f 0 (x) =
21.
f 0 (x) = −xe−x /2
2
00
2
−x2 /2
f (x) = (−1 + x )e
(−∞, 1/4]
(−1/2, 0)
22.
f 0 (x) = (2x2 + 1)ex
2
f 00 (x) = 2x(2x2 + 3)ex
23.
f 0 (x) =
(a) [0, +∞)
(c) (−1, 1)
(e) −1, 1
(b) (−∞, 0]
(d) (−∞, −1), (1, +∞)
24.
f 0 (x) = x(2 ln x + 1)
f 00 (x) = 2 ln x + 3
(a) [e−1/2 , +∞)
(c) (e−3/2 , +∞)
(e) e−3/2
(b) (0, e−1/2 ]
(d) (0, e−3/2 )
25.
f 0 (x) = − sin x
f 00 (x) = − cos x
(a) [π, 2π]
(c) (π/2, 3π/2)
(e) π/2, 3π/2
2
2x
1 + x2
1 − x2
f 00 (x) = 2
(1 + x2 )2
none
(−∞, 0)
1
(b)
(d)
[0, π]
(0, π/2), (3π/2, 2π)
0
2p
-1
26.
f 0 (x) = 2 sin 4x
f 00 (x) = 8 cos 4x
(a) (0, π/4], [π/2, 3π/4]
(b) [π/4, π/2], [3π/4, π]
(c) (0, π/8), (3π/8, 5π/8), (7π/8, π) (d) (π/8, 3π/8), (5π/8, 7π/8)
(e) π/8, 3π/8, 5π/8, 7π/8
1
0
p
0
Exercise Set 5.1
27.
142
f 0 (x) = sec2 x
f 00 (x) = 2 sec2 x tan x
(a) (−π/2, π/2)
(c) (0, π/2)
(e) 0
10
(b) none
(d) (−π/2, 0)
^
6
-10
28.
f 0 (x) = 2 − csc2 x
8
cos x
f (x) = 2 csc x cot x = 2 3
sin x
(a) [π/4, 3π/4]
(c) (0, π/2)
(e) π/2
00
2
(b) (0, π/4], [3π/4, π)
(d) (π/2, π)
0
p
-2
29.
f 0 (x) = cos 2x
f 00 (x) = −2 sin 2x
(a) [0, π/4], [3π/4, π]
(c) (π/2, π)
(e) π/2
0.5
(b) [π/4, 3π/4]
(d) (0, π/2)
0
p
-0.5
30.
f 0 (x) = −2 cos x sin x − 2 cos x = −2 cos x(1 + sin x)
f 00 (x) = 2 sin x (sin x + 1) − 2 cos2 x = 2 sin x(sin x + 1) − 2 + 2 sin2 x = 4(1 + sin x)(sin x − 1/2)
Note: 1 + sin x ≥ 0
2
(a) [π/2, 3π/2]
(b) [0, π/2], [3π/2, 2π]
(c) (π/6, 5π/6)
(d) (0, π/6), (5π/6, 2π)
(e) π/6, 5π/6
0
o
-2
31.
(a)
(b)
y
(c)
y
4
4
x
2
y
4
x
x
2
2
143
32.
Chapter 5
(a)
(b)
y
4
(c)
y
4
4
x
2
33.
34.
y
x
x
2
2
(a)
f 0 (x) = 3(x − a)2 , f 00 (x) = 6(x − a); inflection point is (a, 0)
(b)
f 0 (x) = 4(x − a)3 , f 00 (x) = 12(x − a)2 ; no inflection points
For n ≥ 2, f 00 (x) = n(n − 1)(x − a)n−2 ; there is a sign change of f 00 (point of inflection) at (a, 0) if and
only if n is odd. For n = 1, y = x − a, so there is no point of inflection.
on [0, +∞) thus if
35. f 0 (x) = 1/3 − 1/[3(1 + x)2/3 ] so f is increasing
√
3
1
+
x
> 0,
x
>
0,
then
f
(x)
>
f
(0)
=
0,
1
+
x/3
−
√
3
1 + x < 1 + x/3.
2.5
0
10
0
36.
f 0 (x) = sec2 x − 1 so f is increasing on [0, π/2) thus if 0 < x < π/2,
then f (x) > f (0) = 0, tan x − x > 0, x < tan x.
10
0
0
37.
x ≥ sin x on [0, +∞): let f (x) = x − sin x. Then f (0) = 0 and
f 0 (x) = 1 − cos x ≥ 0, so f (x) is increasing on [0, +∞).
6
4
0
4
-1
38.
(a)
Let h(x) = ex − 1 − x for x ≥ 0. Then h(0) = 0 and h0 (x) = ex − 1 ≥ 0 for x ≥ 0, so h(x) is
increasing.
(b) Let h(x) = ex − 1 − x − 12 x2 . Then h(0) = 0 and h0 (x) = ex − 1 − x. By part (a), ex − 1 − x ≥ 0
for x ≥ 0, so h(x) is increasing.
Exercise Set 5.1
(c)
6
0
6
2
0
39.
144
0
2
0
Points of inflection at x = −2, +2. Concave up on (−5, −2) and
(2, 5); concave down on (−2, 2). Increasing on [−3.5829, 0.2513] and
[3.3316, 5] , and decreasing on [−5, −3.5829] and [0.2513, 3.3316].
250
-5
5
-250
40.
√
√
Points
up on (−5, −1/ 3) and
√
√
√ of inflection at x = ±1/ 3. Concave
(1/ 3, 5), and concave down on (−1/ 3, 1/ 3). Increasing on
[−5, 0] and decreasing on [0, 5].
1
-5
5
-2
41.
Break the interval [−5, 5] into ten subintervals and check f 00 (x) at each endpoint. We find f 00 (−1) > 0
and f 00 (0) < 0. Refine [−1, 0] into ten subintervals; f 00 (−0.2) > 0, f 00 (−0.1) < 0; repeat, f 00 (−0.18) > 0,
f 00 (−0.17) < 0, so x = −0.175 is correct to two decimal places. Note also that f 00 (1) = 0 so there are
two inflection points.
42. Break the interval [−5, 5] into ten subintervals and check f 00 (x) at each endpoint. We discover
f 00 (−1) > 0, f 00 (0) < 0 and f 00 (1) > 0. Refine [−1, 0] into ten subintervals and we see that f 00 (−0.6) > 0,
f 00 (−0.5) < 0. Subdivide [−0.6, −0.5] into 10 subintervals and we see that f 00 (−0.58) > 0 and
f 00 (−0.57) < 0. Thus x = −0.575 is within 0.005 of the true root and is thus correct to two decimal places. For the other root we could proceed in a similar manner, but it easier to note that f 00 (x)
is an even function and thus the other root is x = 0.575 to two decimal places.
43.
44.
45.
90x3 − 81x2 − 585x + 397
. The denominator has complex roots, so is always positive; hence
(3x2 − 5x + 8)3
the x-coordinates of the points of inflection of f (x) are the roots of the numerator (if it changes sign).
A plot of the numerator over [−5, 5] shows roots lying in [−3, −2], [0, 1], and [2, 3]. Breaking each of
these intervals into ten subintervals locates the roots in [−2.5, −2.4], [0.6, 0.7] and [2.7, 2.8]. Thus to
one decimal place the roots are x = −2.45, 0.65, 2.75.
f 00 (x) = 2
2x5 + 5x3 + 14x2 + 30x − 7
. Points of inflection will occur when the numerator changes sign,
(x2 + 1)5/2
since the denominator is always positive. A plot of y = 2x5 + 5x3 + 14x2 + 30x − 7 suggests that there
is only one root and it lies in [0, 1]. Subdivide into ten subintervals and determine that the root lies
between x = 0.2 and x = 0.3. Thus to one decimal place the point of inflection is located at x = 0.25.
f 00 (x) =
f (x1 ) − f (x2 ) = x21 − x22 = (x1 + x2 )(x1 − x2 ) < 0 if x1 < x2 for x1 , x2 in [0, +∞), so f (x1 ) < f (x2 )
and f is thus increasing.
46. f (x1 ) − f (x2 ) =
is decreasing.
1
1
x2 − x1
−
=
> 0 if x1 < x2 for x1 , x2 in (0, +∞), so f (x1 ) > f (x2 ) and thus f
x1 x2
x1 x2
145
47.
Chapter 5
(a)
If x1 < x2 where x1 and x2 are in I, then f (x1 ) < f (x2 ) and g(x1 ) < g(x2 ), so f (x1 ) + g(x1 ) <
f (x2 ) + g(x2 ), (f + g)(x1 ) < (f + g)(x2 ). Thus f + g is increasing on I.
(b)
Case I: If f and g are ≥ 0 on I, and if x1 < x2 where x1 and x2 are in I, then 0 < f (x1 ) < f (x2 )
and 0 < g(x1 ) < g(x2 ), so f (x1 )g(x1 ) < f (x2 )g(x2 ), (f · g)(x1 ) < (f · g)(x2 ). Thus f · g is
increasing on I.
Case II: If f and g are not necessarily positive on I then no conclusion can be drawn: for example,
f (x) = g(x) = x are both increasing on (−∞, 0), but (f · g)(x) = x2 is decreasing there.
48.
(a)
f (x) = x, g(x) = 2x
49.
(a)
(b)
f (x) = x, g(x) = x + 6
(c)
f (x) = 2x, g(x) = x
b
b
f 00 (x) = 6ax + 2b = 6a(x + ), f 00 (x) = 0 when x = − . f changes its direction of concavity
3a
3a
b
b
so −
is an inflection point.
at x = −
3a
3a
(b) If f (x) = ax3 + bx2 + cx + d has three x-intercepts, then it has three roots, say x1 , x2 and x3 , so
we can write f (x) = a(x − x1 )(x − x2 )(x − x3 ) = ax3 + bx2 + cx + d, from which it follows that
1
b
= (x1 + x2 + x3 ), which is the average.
b = −a(x1 + x2 + x3 ). Thus −
3a
3
(c)
f (x) = x(x2 − 3x2 + 2) = x(x − 1)(x − 2) so the intercepts are 0, 1, and 2 and the average is 1.
f 00 (x) = 6x − 6 = 6(x − 1) changes sign at x = 1.
50.
b
f 00 (x) = 6x + 2b, so the point of inflection is at x = − . Thus an increase in b moves the point of
3
inflection to the left.
51.
(a)
Let x1 < x2 belong to (a, b). If both belong to (a, c] or both belong to [c, b) then we have
f (x1 ) < f (x2 ) by hypothesis. So assume x1 < c < x2 . We know by hypothesis that f (x1 ) < f (c),
and f (c) < f (x2 ). We conclude that f (x1 ) < f (x2 ).
(b)
Use the same argument as in part (a), but with inequalities reversed.
52.
By Theorem 5.1.2, f is increasing on any interval ((2n − 1)π, 2(n + 1)π) (n = 0, ±1, ±2, · · ·), because
f 0 (x) = 1 + cos x > 0 on [(2n − 1)π, (2n + 1)π]. By Exercise 51 (a) we can piece these intervals together
to show that f (x) is increasing on (−∞, +∞).
53.
t = 7.67
1000
0
15
0
54.
By zooming on the graph of y 0 (t), maximum increase is at x = −0.577 and maximum decrease is at
x = 0.577.
55.
56.
y
y
4
2
infl pt
3
1
t
infl pts
2
1
x
Exercise Set 5.2
146
EXERCISE SET 5.2
1.
(a)
(b)
y
y
f(x)
f(x)
x
x
(c)
(d)
y
y
f(x)
f(x)
x
2.
(a)
x
(b)
y
y
x
x
(c)
(d)
y
y
x
x
3.
4.
(a)
f 0 (x) = 6x − 6 and f 00 (x) = 6, with f 0 (1) = 0. For the first derivative test, f 0 < 0 for x < 1 and
f 0 > 0 for x > 1. For the second derivative test, f 00 (1) > 0.
(b)
f 0 (x) = 3x2 − 3 and f 00 (x) = 6x. f 0 (x) = 0 at x = ±1. First derivative test: f 0 > 0 for x < −1
and x > 1, and f 0 < 0 for −1 < x < 1, so there is a relative maximum at x = −1, and a relative
minimum at x = 1. Second derivative test: f 00 < 0 at x = −1, a relative maximum; and f 00 > 0
at x = 1, a relative minimum.
(a)
f 0 (x) = 2 sin x cos x = sin 2x (so f 0 (0) = 0) and f 00 (x) = 2 cos 2x. First derivative test: if x is
near 0 then f 0 < 0 for x < 0 and f 0 > 0 for x > 0, so a relative minimum at x = 0. Second
derivative test: f 00 (0) = 2 > 0, so relative minimum at x = 0.
(b) g 0 (x) = 2 tan x sec2 x (so g 0 (0) = 0) and g 00 (x) = 2 sec2 x(sec2 x + 2 tan2 x). First derivative test:
g 0 < 0 for x < 0 and g 0 > 0 for x > 0, so a relative minimum at x = 0. Second derivative test:
g 00 (0) = 2 > 0, relative minimum at x = 0.
5.
(c)
Both functions are squares, and so are positive for values of x near zero; both functions are zero
at x = 0, so that must be a relative minimum.
(a)
f 0 (x) = 4(x − 1)3 , g 0 (x) = 3x2 − 6x + 3 so f 0 (1) = g 0 (1) = 0.
(b)
f 00 (x) = 12(x − 1)2 , g 00 (x) = 6x − 6, so f 00 (1) = g 00 (1) = 0, which yields no information.
(c)
f 0 < 0 for x < 1 and f 0 > 0 for x > 1, so there is a relative minimum at x = 1;
g 0 (x) = 3(x − 1)2 > 0 on both sides of x = 1, so there is no relative extremum at x = 1.
147
6.
Chapter 5
(a)
f 0 (x) = −5x4 , g 0 (x) = 12x3 − 24x2 so f 0 (0) = g 0 (0) = 0.
(b)
f 00 (x) = −20x3 , g 00 (x) = 36x2 − 48x, so f 00 (0) = g 00 (0) = 0, which yields no information.
(c)
f 0 < 0 on both sides of x = 0, so there is no relative extremum there; g 0 (x) = 12x2 (x − 2) < 0
on both sides of x = 0 (for x near 0), so again there is no relative extremum there.
7.
f 0 (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1), f 0 (x) = 0 when x = −3, 1 (stationary points).
√
(b) f 0 (x) = 4x(x2 − 3), f 0 (x) = 0 when x = 0, ± 3 (stationary points).
8.
(a)
f 0 (x) = 6(x2 − 1), f 0 (x) = 0 when x = ±1 (stationary points).
(b)
f 0 (x) = 12x3 − 12x2 = 12x2 (x − 1), f 0 (x) = 0 when x = 0, 1 (stationary points).
(a)
√
f 0 (x) = (2 − x2 )/(x2 + 2)2 , f 0 (x) = 0 when x = ± 2 (stationary points).
(b)
f 0 (x) = 23 x−1/3 = 2/(3x1/3 ), f 0 (x) does not exist when x = 0.
(a)
f 0 (x) = 8x/(x2 + 1)2 , f 0 (x) = 0 when x = 0 (stationary point).
9.
10.
(a)
(b) f 0 (x) = 13 (x + 2)−2/3 , f 0 (x) does not exist when x = −2.
11.
(a)
(b)
12.
13.
4(x + 1) 0
, f (x) = 0 when x = −1 (stationary point), f 0 (x) does not exist when x = 0.
3x2/3
f 0 (x) = −3 sin 3x, f 0 (x) = 0 when sin 3x = 0, 3x = nπ, n = 0, ±1, ±2, · · ·
x = nπ/3, n = 0, ±1, ±2, · · · (stationary points)
f 0 (x) =
4(x − 3/2) 0
, f (x) = 0 when x = 3/2 (stationary point), f 0 (x) does not exist when x = 0.
3x2/3
½
½
cos x, sin x > 0
sin x, sin x ≥ 0
and f 0 (x) does not
(b) f (x) = | sin x| =
so f 0 (x) =
− cos x, sin x < 0
− sin x, sin x < 0
exist when x = nπ, n = 0, ±1, ±2, · · · (sin x = 0) because limx→nπ− f 0 (x) 6= limx→nπ+ f 0 (x) (see
Theorem preceding Exercise 75, Section 3.3). Now f 0 (x) = 0 when ± cos x = 0 provided sin x 6= 0
so x = π/2 + nπ, n = 0, ±1, ±2, · · · are stationary points.
(a)
f 0 (x) =
(a)
x = 2 because f 0 (x) changes sign from − to + there.
(b) x = 0 because f 0 (x) changes sign from + to − there.
(c)
x = 1, 3 because f 00 (x) (the slope of the graph of f 0 (x)) changes sign at these points.
14.
(a)
x=1
15.
(a)
√
critical points x = 0, ± 5; f 0 :
(b)
x=5
(c)
x = −1, 0, 3
− − 0+ + 0 − − 0 + +
− 5
0
5
√
x = 0: relative maximum; x = ± 5: relative minimum
(b)
critical point x = 0; f 0 :
− − −
0
+ + +
0
x = 0: relative minimum
16.
(a)
critical points x = 0, −1/2, 1; f 0 :
+ + + 0 − 0 − − − 0 +
− 12
0
1
x = 0: neither; x = −1/2: relative maximum; x = 1: relative minimum
Exercise Set 5.2
(b)
critical points: x = ±3/2, −1; f 0 :
148
+ + 0 − − ? + + 0 − −
− 32
−1
3
2
x = ±3/2: relative maximum; x = −1: relative minimum
17.
f 0 (x) = −2(x + 2); critical point x = −2; f 0 (x):
+ + + 0 − − −
-2
f 00 (x) = −2; f 00 (−2) < 0, f (−2) = 5; relative max of 5 at x = −2
18.
f 0 (x) = 6(x − 2)(x − 1); critical points x = 1, 2; f 0 (x) :
+ + + 0 − − − 0 + + +
1
2
f 00 (x) = 12x − 18; f 00 (1) < 0, f 00 (2) > 0, f (1) = 5, f (2) = 4; relative min of 4 at x = 2, relative max
of 5 at x = 1
19.
f 0 (x) = 2 sin x cos x = sin 2x; critical points x = π/2, π, 3π/2; f 0 (x) :
+ + 0 − − 0 + + 0 − −
π
2
π
f 00 (x) = 2 cos 2x; f 00 (π/2) < 0, f 00 (π) > 0, f 00 (3π/2) < 0, f (π/2) = f (3π/2) = 1,
f (π) = 0; relative min of 0 at x = π, relative max of 1 at x = π/2, 3π/2
20.
f 0 (x) = 1/2 − cos x; critical points x = π/3, 5π/3; f 0 (x):
− − 0 + + + + + 0 − −
π
3
5π
3
f 00 (x) = − sin x; f 00 (π/3) < 0, f 00 (5π/3) > 0
√
√
f (π/3) = π/6 − 3/2, f (5π/3) = 5π/6 + 3/2;
√
√
relative min of π/6 − 3/2 at x = π/3, relative max of 5π/6 + 3/2 at x = 5π/3
21.
f 0 (x) = 3x2 + 5; no relative extrema because there are no critical points.
22.
f 0 (x) = 4x(x2 − 1); critical points x = 0, 1, −1
f 00 (x) = 12x2 − 4; f 00 (0) < 0, f 00 (1) > 0, f 00 (−1) > 0
relative min of 6 at x = 1, −1, relative max of 7 at x = 0
23.
f 0 (x) = (x − 1)(3x − 1); critical points x = 1, 1/3
f 00 (x) = 6x − 4; f 00 (1) > 0, f 00 (1/3) < 0
relative min of 0 at x = 1, relative max of 4/27 at x = 1/3
24.
f 0 (x) = 2x2 (2x + 3); critical points x = 0, −3/2
relative min of −27/16 at x = −3/2 (first derivative test)
25.
f 0 (x) = 4x(1 − x2 ); critical points x = 0, 1, −1
f 00 (x) = 4 − 12x2 ; f 00 (0) > 0, f 00 (1) < 0, f 00 (−1) < 0
relative min of 0 at x = 0, relative max of 1 at x = 1, −1
26.
f 0 (x) = 10(2x − 1)4 ; critical point x = 1/2; no relative extrema (first derivative test)
27.
f 0 (x) = 45 x−1/5 ; critical point x = 0; relative min of 0 at x = 0 (first derivative test)
28.
f 0 (x) = 2 + 23 x−1/3 ; critical points x = 0, −1/27
relative min of 0 at x = 0, relative max of 1/27 at x = −1/27
29.
f 0 (x) = 2x/(x2 + 1)2 ; critical point x = 0; relative min of 0 at x = 0
3π
2
149
30.
Chapter 5
f 0 (x) = 2/(x + 2)2 ; no critical points (x = −2 is not in the domain of f ) no relative extrema
31. f 0 (x) = 2x/(1 + x2 ); critical point at x = 0; relative min of 0 at x = 0 (first derivative test)
32.
f 0 (x) = x(2 + x)ex ; critical points at x = 0, −2; relative min of 0 at x = 0 and relative max of 4/e2 at
x = −2 (first derivative test)
33. f 0 (x) = 2x if |x| > 2, f 0 (x) = −2x if |x| < 2,
f 0 (x) does not exist when x = ±2; critical points x = 0, 2, −2
relative min of 0 at x = 2, −2, relative max of 4 at x = 0
34.
f 0 (x) = −1 if x < 3, f 0 (x) = 2x if x > 3, f 0 (3) does not exist;
critical point x = 3, relative min of 6 at x = 3
35.
f 0 (x) = 2 cos 2x if sin 2x > 0, f 0 (x) = −2 cos 2x if sin 2x < 0,
f 0 (x) does not exist when x = π/2, π, 3π/2;
critical points x = π/4, 3π/4, 5π/4, 7π/4, π/2, π, 3π/2
relative min of 0 at x = π/2, π, 3π/2; relative max of 1 at
x = π/4, 3π/4, 5π/4, 7π/4
1
0
0
36.
√
f 0 (x) = 3 + 2 cos
√ x; critical points x = 5π/6, 7π/6
relative
min of 7 3π/6 − 1 at x = 7π/6; relative max of
√
5 3π/6 + 1 at x = 5π/6
12
0
0
37.
f 0 (x) = − sin 2x; critical points x = π/2, π, 3π/2
relative min of 0 at x = π/2, 3π/2; relative max of 1 at x = π
0
2
f 0 (x) = (2 cos x −
√ 1)/(2 − cos x) ; critical points x = π/3, 5π/3
relative
max of 3/3 at x = π/3, relative min of
√
− 3/3 at x = 5π/3
o
1
0
38.
o
o
0.8
0
-0.8
o
Exercise Set 5.2
39.
f 0 (x) = ln x + 1, f 00 (x) = 1/x; f 0 (1/e) = 0, f 00 (1/e) > 0;
relative min of −1/e at x = 1/e
150
2.5
0
2.5
-0.5
40.
ex − e−x
= 0 when x = 0. By the first derivative test
(ex + e−x )2
0
f (x) > 0 for x < 0 and f 0 (x) < 0 for x > 0; relative max of 1 at
x=0
f 0 (x) = −2
1
-2
2
0
41.
f 0 (x) = 2x(1 − x)e−2x = 0 at x = 0, 1. f 00 (x) = (4x2 − 8x + 2)e−2x ;
f 00 (0) > 0 and f 00 (1) < 0, so a relative min of 0 at x = 0 and a
relative max of 1/e2 at x = 1.
0.14
-0.3
4
0
42.
f 0 (x) = 10/x − 1 = 0 at x = 10; f 00 (x) = −10/x2 < 0;
relative max of 10(ln(10) − 1) ≈ 13.03 at x = 10
14
0
20
-4
43.
Relative minima at x = −3.58, 3.33; relative max at x = 0.25
250
-5
5
-250
44.
Relative min at x = −0.84; relative max at x = 0.84
1.2
6
-6
-1.2
151
Chapter 5
45.
relative max at x = 0.255
46. relative max at x = 0.845
47.
Relative min at x = −1.20 and a relative max at x = 1.80
y
f''(x)
1
f'(x)
-4
-2
2
4
2
4
x
-1
-2
48.
Relative max at x = −0.78 and a relative min at x = 1.55
y
f''(x)
5
-4
-2
f'(x)
49.
(a)
x
-5
k
k
2x3 − k
Let f (x) = x2 + , then f 0 (x) = 2x− 2 =
. f has a relative extremum when 2x3 −k = 0,
x
x
x2
so k = 2x3 = 2(3)3 = 54.
k − x2
x
0
,
then
f
(x)
=
. f has a relative extremum when k − x2 = 0, so
x2 + k
(x2 + k)2
k = x2 = 32 = 9.
(b) Let f (x) =
50.
(a)
one relative maximum, located at x = n
0.3
0
14
0
51.
(b)
f 0 (x) = cxn−1 (−x + n)e−x = 0 at x = n. Since f 0 (x) > 0 for x < n and f 0 (x) < 0 for
x > n it’s a maximum.
(a)
f 0 (x) = −xf (x). Since f (x) is always
positive, f 0 (x) = 0 at x = 0, f 0 (x) > 0
for x < 0 and f 0 (x) < 0 for x > 0, so
x = 0 is a maximum.
(b)
y
1
2π
( µ,
µ
1
2π
)
x
Exercise Set 5.3
52.
(a)
152
relative minima at x = ±0.6436,
relative max at x = 0
(b)
x = ±0.6436, 0
y
2
1.8
1.6
1.4
1.2
1
x
-1.5 -1 -0.5
0.5
1 1.5
53.
f 0 (x) = 3ax2 + 2bx + c and f 0 (x) has roots at x = 0, 1, so f 0 (x) must be of the form f 0 (x) = 3ax(x − 1);
thus c = 0 and 2b = −3a, b = −3a/2. f 00 (x) = 6ax + 2b = 6ax − 3a, so f 00 (0) > 0 and f 00 (1) < 0
provided a < 0. Finally f (0) = d, so d = 0; and f (1) = a + b + c + d = a + b = −a/2 so a = −2. Thus
f (x) = −2x3 + 3x2 .
54.
(a)
Because h and g have relative maxima at x0 , h(x) ≤ h(x0 ) for all x in I1 and g(x) ≤ g(x0 ) for all
x in I2 , where I1 and I2 are open intervals containing x0 . If x is in I1 ∩ I2 then both inequalities
are true and by addition so is h(x) + g(x) ≤ h(x0 ) + g(x0 ) which shows that h + g has a relative
maximum at x0 .
(b) By counterexample; both h(x) = −x2 and g(x) = −2x2 have relative maxima at x = 0 but
h(x) − g(x) = x2 has a relative minimum at x = 0 so in general h − g does not necessarily have
a relative maximum at x0 .
55.
(a)
(b)
y
(
x0
)
x
f (x0 ) is not an extreme value.
(c)
y
(
x0
)
y
(
x
y = x2 − 2x − 3;
y
y 0 = 2(x − 1);
y 00 = 2
x
1
-4
)
x
f (x0 ) is a relative minimum.
f (x0 ) is a relative maximum.
EXERCISE SET 5.3
1.
x0
(1, -4)
153
2.
Chapter 5
y = 1 + x − x2 ;
y
y 0 = −2(x − 1/2);
y 00 = −2
( 12 , 54 )
1
x
1
3.
y = x3 − 3x + 1;
y
y 0 = 3(x2 − 1);
(-1, 3)
y 00 = 6x
(0, 1)
x
(1, -1)
4.
y = 2x3 − 3x2 + 12x + 9;
0
y
y = 6(x − x + 2);
2
y 00 = 12(x − 1/2)
( 12 , 292 )
10
x
1
5.
y = x4 + 2x3 − 1;
0
y
2
y = 4x (x + 3/2);
y 00 = 12x(x + 1)
x
(0, -1)
(-1, -2)
(- 32 , - 4316 )
6.
y = x4 − 2x2 − 12;
y
y 0 = 4x(x2 − 1);
10
y 00 = 12(x2 − 1/3)
1
(1, -13)
(-1, -13)
(
-
1
3
,-
113
9
x
(0, -12)
)
( 13 , - 1139 )
Exercise Set 5.3
7.
154
y = x3 (3x2 − 5);
y 0 = 15x2 (x2 − 1);
00
y
(- 13 , 783 )
(-1, 2)
y = 30x(2x − 1)
2
(0, 0)
x
(1, -2)
( 13 , - 78 3 )
8.
y = 3x3 (x + 4/3);
y
y 0 = 12x2 (x + 1);
y 00 = 36x(x + 2/3)
(0, 0)
(-1, -1)
9.
y = x(x − 1)3 ;
x
(- 23 , - 1627 )
y
y 0 = (4x − 1)(x − 1)2 ;
y 00 = 6(2x − 1)(x − 1)
0.1
x
(1, 0)
(
-0.1
(
10.
1,
4
1
,- 1
2 16
27
- 256
)
)
y = x4 (x + 5);
y 0 = 5x3 (x + 4);
00
y
(-4, 256)
300
2
y = 20x (x + 3)
x
(-3, 162)
11.
y = 2x/(x − 3);
(0, 0)
y
y 0 = −6/(x − 3)2 ;
y 00 = 12/(x − 3)3
y=2
x
x=3
155
12.
Chapter 5
x
;
x2 − 1
x2 + 1
;
y0 = − 2
(x − 1)2
2x(x2 + 3)
y 00 =
(x2 − 1)3
y
y=
x = -1
1
x
(0, 0)
x=1
13.
x2
;
−1
2x
y0 = − 2
;
(x − 1)2
2
2(3x + 1)
y 00 =
(x2 − 1)3
y=
y
x2
y=1
(0, 0)
x = -1
14.
x2 − 1
;
x2 + 1
4x
y0 = 2
;
(x + 1)2
4(1 − 3x2 )
y 00 =
(x2 + 1)3
x
x=1
y
y=
y=1
x
(- 13 , - 12 )
( 13 , - 12 )
(0, -1)
15.
16.
x3 − 1
1
=
;
x
x
3
2x + 1
y0 =
,
x2
r
3 1
0
y = 0 when x = −
≈ −0.8;
2
3
2(x − 1)
y 00 =
x3
y = x2 −
2x2 − 1
;
x2
2
y0 = 3 ;
x
6
y 00 = − 4
x
y
≈(-0.8, 1.9)
x
(1, 0)
y
y=
y=2
x
Exercise Set 5.3
17.
156
x3 − 1
;
x3 + 1
6x2
;
y0 = 3
(x + 1)2
12x(1 − 2x3 )
y 00 =
(x3 + 1)3
y
y=
y=1
x = -1
18.
19.
8
;
4 − x2
16x
y0 =
;
(4 − x2 )2
16(3x2 + 4)
y 00 =
(4 − x2 )3
y
y=
x−1
;
x2 − 4
2
x − 2x + 4
y0 = −
(x2 − 4)2
x
3
(1/ 2 , -1/3)
(0, -1)
x = -2
x=2
1
x
y
y=
x = -2
4
x
x=2
20.
4
4
− ;
x x2
4(x + 2)
;
y0 =
x3
8(x + 3)
y 00 = −
x4
y =3−
y
(-3, 359 )
(-2, 4)
y=3
x
21.
(x − 1)2
;
x2
2(x − 1)
y0 =
;
x3
2(3
−
2x)
y 00 =
x4
y=
y
y=1
x
(1, 0)
( )
3, 1
2 9
157
22.
Chapter 5
3
1
− ;
x x3
3(1 − x2 )
y0 =
;
x4
2
6(x − 2)
y 00 =
x5
y
y =2+
(1, 4)
(
2, 2 +
5
4
2)
y=2
x
(−
(b)
I
(c)
III
2, 2 –
5
4
2)
(–1, 0)
23.
(a)
VI
(d) V
(e)
IV
(f )
II
24.
(a)
When n is even the function is defined only for x ≥ 0; as n increases the graph approaches the
line y = 1 for x > 0.
y
x
(b)
When n is odd the graph is symmetric with respect to the origin; as n increases the graph
approaches the line y = 1 for x > 0 and the line y = −1 for x < 0.
y
x
25.
26.
√
x2 − 1;
x
y0 = √
;
2
x −1
1
y 00 = − 2
(x − 1)3/2
y
y=
-1
x
1
p
3
x2 − 4;
2x
;
y0 =
2
3(x − 4)2/3
2(3x2 + 4)
y 00 = −
9(x2 − 4)5/3
y=
3
y
3
(-2, 0)
(2, 0)
-2
2
(0, -2)
x
Exercise Set 5.3
27.
158
y
y = 2x + 3x2/3 ;
y 0 = 2 + 2x−1/3 ;
2
y 00 = − x−4/3
3
5
4
(0, 0)
28.
y = 4x − 3x4/3 ;
x
y
y 0 = 4 − 4x1/3 ;
4
y 00 = − x−2/3
3
3
(1, 1)
x
1
29.
30.
y
y = x(3 − x)1/2 ;
3(2 − x)
;
y0 = √
2 3−x
3(x − 4)
y 00 =
4(3 − x)3/2
(2, 2)
x
y = x1/3 (4 − x);
4(1 − x)
;
y0 =
3x2/3
4(x + 2)
y 00 = −
9x5/3
y
10
(1, 3)
x
3
(-2, -6 2 )
-10
√
8( x − 1)
;
31. y =
x√
4(2 − x)
;
y0 =
x2
√
2(3 x − 8)
y 00 =
x3
y
4
(4, 2)
(649 , 158 )
x
15
159
32.
33.
Chapter 5
√
1+ x
√ ;
1− x
1
0
√ ;
y = √
2 x(1−
√ x)
3 x−1
√
y 00 = 3/2
2x (1 − x)3
y
y=
x=1
2
( 19 , 2 )
x
-2
y = -1
y = x + sin x;
0
y
0
y = 1 + cos x, y = 0 when x = π + 2nπ;
y 00 = − sin x; y 00 = 0 when x = nπ
c
n = 0, ±1, ±2, . . .
34.
y = x − cos x;
c
x
c
x
y
y 0 = 1 + sin x;
y 0 = 0 when x = −π/2 + 2nπ;
c
00
y = cos x;
00
y = 0 when x = π/2 + nπ
n = 0, ±1, ±2, . . .
35.
y
y = sin x + cos x;
y 0 = cos x − sin x;
y 0 = 0 when x = π/4 + nπ;
00
y = − sin x − cos x;
2
x
o
-o
-2
00
y = 0 when x = 3π/4 + nπ
36.
√
3 cos x + sin x;
√
0
y = − 3 sin x + cos x;
y
y=
2
0
y = 0 when x = π/6 + nπ;
√
y 00 = − 3 cos x − sin x;
y 00 = 0 when x = 2π/3 + nπ
o
-2
x
Exercise Set 5.3
37.
160
y = sin2 x, 0 ≤ x ≤ 2π;
y
y 0 = 2 sin x cos x = sin 2x;
1
y 00 = 2 cos 2x
c
38.
y = x tan x, −π/2 < x < π/2;
x
o
y
y 0 = x sec2 x + tan x;
y 0 = 0 when x = 0;
y 00 = 2 sec2 x(x tan x + 1), which is always
positive for −π/2 < x < π/2
39.
(a)
x
-6
6
lim xex = +∞, lim xex = 0
x→+∞
y
x→−∞
(b) y = xex ;
y 0 = (x + 1)ex ;
y 00 = (x + 2)ex
1
-5
-3
-1
x
(-2, -0.27)
-1
(-1, -0.37)
40.
(a)
(b)
lim xe−2x = 0, lim xe−2x = −∞
x→−∞
¶
µ
1 −2x 00
−2x
0
y = xe ; y = −2 x −
e ; y = 4(x − 1)e−2x
2
y
x→+∞
0.3
(0.5, 0.18)
(1, 0.14)
0.1
x
-3
-1
-0.1
1
3
-0.3
41.
x2
x2
=
0,
lim
= +∞
x→+∞ e2x
x→−∞ e2x
(b) y = x2 /e2x = x2 e−2x ;
y 0 = 2x(1 − x)e−2x ;
y 00 = 2(2x2 − 4x + 1)e−2x ;
2x2 − 4x + 1 = 0, when
y 00 = 0 if √
√
4 ± 16 − 8
x=
= 1 ± 2/2 ≈ 0.29, 1.71
4
(a)
lim
y
0.3
(1, 0.14)
(1.71, 0.10)
(0, 0)
1
2
(0.29, 0.05)
42.
(a)
lim x2 e2x = +∞, lim x2 e2x = 0.
x→+∞
x→−∞
3
x
161
Chapter 5
(b)
y = x2 e2x ;
y 0 = 2x(x + 1)e2x ;
y 00 = 2(2x2 + 4x + 1)e2x ;
2
y 00 = 0 if 2x
√ + 4x + 1 = 0, when
√
−4 ± 16 − 8
= −1 ± 2/2 ≈ −0.29, −1.71
x=
4
y
0.3
0.2
(-1, 0.14)
(-1.71, 0.10)
x
-3
-2
-1
(0, 0)
(-0.29, 0.05)
43.
(a)
lim f (x) = +∞, lim f (x) = −∞
x→+∞
x→−∞
x2
y
(b) y = xe ;
2
y 0 = (1 + 2x2 )ex ;
00
2 x2
y = 2x(3 + 2x )e
no relative extrema, inflection point at (0, 0)
100
(0,0)
-2
2
x
-100
44.
(a)
lim f (x) = 1
x→±∞
0
.
(b) f (x) = 2x−3 e−1/x so f 0 (x) < 0 for x < 0 and f 0 (x) > 0 for
x > 0. By L’Hôpital’s Rule limx→0 f 0 (x) = 0, so (by the first
derivative test) f (x) has a minimum at x = 0.
2
−4
+ 4x−6 )e−1/x , so f (x) has points of inflection
f 00 (x) = (−6x
p
at x = ± 2/3
2
y
1
0.8
0.4
(−
2/3, e −3/2)
-5 (0, 0)
-10
45.
(a)
lim+ y = lim+ x ln x = lim+
x→0
x→0
x→0
lim y = +∞
(
ln x
1/x
= 0;
= lim+
1/x x→0 −1/x2
2/3, e −3/2)
5
10
x
y
x→+∞
(b) y = x ln x,
y 0 = 1 + ln x, y 00 = 1/x,
y 0 = 0 when x = e−1
x
1
(e-1, -e-1)
46.
(a)
ln x
1/x
= lim+
= 0,
x→0 −2/x3
1/x2
lim y = +∞
lim y = lim+
x→0+
y
x→0
0.2
x→+∞
(b) y = x2 ln x, y 0 = x(1 + 2 ln x),
y 00 = 3 + 2 ln x,
y 0 = 0 if x = e−1/2 ,
y 00 = 0 if x = e−3/2 ,
lim+ y 0 = 0
x→0
0.1
(e -3/2, - 32 e-3)
x
1
-0.1
-0.2
(e -1/2, - 12 e-1)
Exercise Set 5.3
47.
48.
162
ln x
= −∞;
x2
ln x
1/x
=0
lim y = lim
= lim
2
x→+∞
x→+∞ x
x→+∞ 2x
ln x
1 − 2 ln x
(b) y = 2 , y 0 =
,
x
x3
6 ln x − 5
,
y 00 =
x4
y 0 = 0 if x = e1/2 ,
y 00 = 0 if x = e5/6
(a)
(a)
lim y = lim+
x→0+
y
x→0
0.4
0.3
0.2
0.1
x
x→+∞
1
-0.1
-0.2
-0.3
-0.4
√
√
lim (ln x)/ x = −∞ by inspection, lim (ln x)/ x = lim
x→0+
(e1/2, 12 e -1) (e 5/6, 5 e -5/3)
6
x→+∞
2
3
1/x
2
√ = lim √ = 0, L’Hôpital’s
1/2 x x→+∞ x
Rule.
2 − ln x
ln x
(b) y = √ , y 0 =
x
2x3/2
−8 + 3 ln x
y 00 =
4x5/2
y 0 = 0 if x = e2 ,
y 00 = 0 if x = e8/3
y
( e 8/3 , 38 e -4/3 )
(e 2 , 2/e)
0.5
x
2
6
10
14
-1
49.
(a)
lim y = −∞, lim y = +∞;
x→−∞
y
x→+∞
curve crosses x-axis at x = 0, 1, −1
4
2
x
-2
-1
1
-2
-4
-6
(b)
y
lim y = +∞;
x→±∞
curve never crosses x-axis
0.2
0.1
-1
(c)
1
lim y = −∞, lim y = +∞;
x→−∞
x
y
x→+∞
curve crosses x-axis at x = −1
0.4
0.2
x
-1
1
-0.2
163
Chapter 5
(d)
y
lim y = +∞;
x→±∞
curve crosses x-axis at x = 0, 1
0.4
0.2
x
-1
50.
(a)
y
y
y
a
a
b
y
a
(c)
(a)
b
x
x
b
y
a
51.
x
x
b
y
a
b
x
a
(b)
1
b
x
horizontal asymptote y = 3 as x → ±∞, vertical asymptotes of
x = ±2
y
10
5
x
-5
5
-5
Exercise Set 5.3
(b)
164
horizontal asymptote of y = 1 as x → ±∞, vertical asymptotes
at x = ±1
y
10
x
-5
5
-10
(c)
horizontal asymptote of y = −1 as x → ±∞, vertical
asymptotes at x = −2, 1
y
10
x
-5
5
-10
(d)
horizontal asymptote of y = 1 as x → ±∞,
vertical asymptote at x = −1, 2
y
10
x
-5
5
-10
52.
y
a
53.
b
x
0.4
(a)
-0.5
(b)
y 0 = (1 − bx)e−bx , y 00 = b2 (x − 2/b)e−bx ;
relative max at x = 1/b, y = 1/be; point of
inflection at x = 2/b, y = 2/be2 . Increasing
b moves the relative max and the point of
inflection to the left and down, i.e. towards
the origin.
(b)
y 0 = −2bxe−bx , y 00 = 2b(−1 + 2bx2 )e−bx ;
relative max at x = p
0, y = 1; points
√
of inflection at x = ± 1/2b, y = 1/ e.
Increasing b moves the points of inflection
towards the y-axis; the relative max doesn’t
move.
3
-0.2
54.
(a)
1
-2
2
0
2
2
165
55.
Chapter 5
(a)
The oscillations of ex cos x about zero
increase as x → ±∞ so the limit does not
exist.
(b)
y
6
4
(0,1)
(1.52, 0.22)
x
-2
-1
1
2
-2
(c)
The curve y = eax cos bx oscillates between y = eax and y = −eax . The frequency of oscillation
increases when b increases.
y
y
5
-1
10
b=3
5
x
2
b=1
1
a=2
a=1
x
-1
-5
a=3
b=2
0.5
1
b=1
a=1
·
¸
P (x)
− (ax + b)
56. lim
x→±∞ Q(x)
the degree of Q(x).
57.
58.
=
lim
x→±∞
R(x)
Q(x)
=
0 because the degree of R(x) is less than
2
x2 − 2
= x − so
x
x
y = x is an oblique asymptote;
x2 + 2
,
y0 =
x2
4
y 00 = − 3
x
y
y=
5
x2 − 2x − 3
=x−4+
so
x+2
x+2
y = x − 4 is an oblique asymptote;
x2 + 4x − 1 00
10
,y =
y0 =
2
(x + 2)
(x + 2)3
y=x
4
4
x
y
y=
x = -2
-10
10
x
≈(0.24, -1.52)
y=x-4
≈(-4.24, -10.48)
59.
(x − 2)3
12x − 8
=x−6+
so
x2
x2
y = x − 6 is an oblique asymptote;
(x − 2)2 (x + 4)
,
y0 =
x3
24(x − 2)
y 00 =
x4
y
y=
10
-10
(-4, -13.5)
(2, 0)
x
y=x-6
Exercise Set 5.3
60.
166
4 − x3
,
x2
x3 + 8
y0 = −
,
x3
24
y 00 = 4
x
y
y=
(-2, 3)
y = -x
x
61.
1
1
(x − 1)(x + 1)2
− 2 =
,
x x
x2
y = x + 1 is an oblique asymptote;
(x + 1)(x2 − x + 2)
,
y0 =
x3
2(x + 3)
y 00 = −
x4
y =x+1−
y
y = x+1
(-1, 0)
x
(-3, - 169 )
62.
63.
The oblique asymptote is y = 2x so (2x3 − 3x + 4)/x2 = 2x, −3x + 4 = 0, x = 4/3.
lim [f (x) − x2 ] = lim (1/x) = 0
x→±∞
y
x→±∞
x3 + 1 0
1
2x3 − 1
1
=
, y = 2x − 2 =
,
x
x
x
x2
√
2
2(x3 + 1) 0
y 00 = 2 + 3 =
, y = 0 when x = 1/ 3 2 ≈ 0.8,
3
x
√ x
y = 3 3 2/2 ≈ 1.9; y 00 = 0 when x = −1, y = 0
3
y = x2 +
64.
3
lim [f (x) − (3 − x2 )] = lim (2/x) = 0
x→±∞
x→±∞
3
2 + 3x − x
2
2
2(x3 + 1)
=
, y 0 = −2x − 2 = −
,
x
x
x
x2
3
4
2(x − 2) 0
y 00 = −2 + 3 = −
, y = 0 when x = −1, y = 0;
x
x3
√
y 00 = 0 when x = 3 2 ≈ 1.3, y = 3
y = 3 − x2 +
65.
y = x2
x
y
y = 3 - x2
1
3
x
L
Let y be the length of the other side of the rectangle, then
L = 2x + 2y and xy = 400 so y = 400/x and hence L = 2x + 800/x.
L = 2x is an oblique asymptote (see Exercise 48)
800
2(x2 + 400) 0
800
2(x2 − 400)
L = 2x +
=
, L =2− 2 =
,
x
x
x
x2
1600
L00 = 3 , L0 = 0 when x = 20, L = 80
x
100
20
x
167
66.
Chapter 5
Let y be the height of the box, then S = x2 + 4xy and x2 y = 500 so
S
y = 500/x2 and hence S = x2 + 2000/x.
The graph approaches the curve S = x2 asymptotically
1000
(see Exercise 63)
x3 + 2000 0
2000
2000
2(x3 − 1000)
S = x2 +
=
, S = 2x − 2 =
,
x
x
x
x2
3
4000
2(x + 2000) 00
, S = 0 when x = 10, S = 300
S 00 = 2 + 3 =
x
x3
67.
30
y 0 = 0.1x4 (6x − 5);
critical points: x = 0, x = 5/6;
relative minimum at x = 5/6,
y ≈ −6.7 × 10−3
x
y
0.01
-1
68.
1
y 0 = 0.1x4 (x + 1)(7x + 5);
critical points: x = 0, x = −1, x = −5/7,
relative maximum at x = −1, y = 0;
relative minimum at x = −5/7, y ≈ −1.5 × 10−3
x
y
0.001
1
69.
x
kL2 Ae−kLt
kL2 A
0
S,
so
P
(0)
=
(1 + Ae−kLt )2
(1 + A)2
(a)
P 0 (t) =
(b)
The rate of growth increases to its maximum, which occurs when P is halfway between 0 and
1
ln A; it then decreases back towards zero.
L, or when t =
Lk
dP
is maximized when P lies half way between 0 and L, i.e. P = L/2.
From (6) one sees that
dt
This follows since the right side of (6) is a parabola (with P as independent variable) with
1
ln A, from (8).
P -intercepts P = 0, L. The value P = L/2 corresponds to t =
Lk
(c)
70. Since 0 < P < L the right-hand side of (7) can change sign only if the factor L − 2P changes sign,
L
1
L
=
ln A.
, 1 = Ae−kLt , t =
which it does when P = L/2. From (5) we have
2
1 + Ae−kLt
Lk
SUPPLEMENTARY EXERCISES FOR CHAPTER 5
4.
(a)
x3 x2
−
on [−2, 2]; x = 0 is a relative maximum and x = 1 is a relative
3
2
1
minimum, but y = 0 is not the largest value of y on the interval, nor is y = − the smallest.
6
False; an example is y =
Supplementary Exercises
6.
168
(b)
true
(c)
False; for example y = x3 on (−1, 1) which has a critical point but no relative extrema
(a)
(b)
y
4
(c)
y
4
4
x
2
7.
(a)
y
x
x
2
2
7(x − 7)(x − 1)
; critical points at x = 0, 1, 7;
3x2/3
neither at x = 0, relative max at x = 1, relative min at x = 7 (first derivative test)
f 0 (x) =
f 0 (x) = 2 cos x(1 + 2 sin x); critical points at x = π/2, 3π/2, 7π/6, 11π/6;
relative max at x = π/2, 3π/2, relative min at x = 7π/6, 11π/6
√
3 x−1
; critical points at x = 5; relative max at x = 5
(c) f 0 (x) = 3 −
2
(b)
8.
f 0 (x) =
(b)
x3 − 4 00
x3 + 8
,
f
(x)
=
2
;
x2
x3
1/3
00 1/3
critical point at x = 4 , f (4 ) > 0, relative min at x = 41/3
(c)
9.
x−9
27 − x
, f 00 (x) =
; critical point at x = 9; f 00 (9) > 0, relative min at x = 9
18x3/2
36x5/2
(a)
f 0 (x) = 2
f 0 (x) = sin x(2 cos x + 1), f 00 (x) = 2 cos2 x − 2 sin2 x + cos x; critical points at x = 2π/3, π, 4π/3;
f 00 (2π/3) < 0, relative max at x = 2π/3; f 00 (π) > 0, relative min at x = π; f 00 (4π/3) < 0, relative
max at x = 4π/3
y
lim f (x) = +∞, lim f (x) = +∞
x→−∞
0
x→+∞
f (x) = x(4x2 − 9x + 6), f 00 (x) = 6(2x − 1)(x − 1)
relative min at x = 0,
points of inflection when x = 1/2, 1,
no asymptotes
4
3
2
1
(1,2)
(0,1)
( 12 , 2316 )
1
10.
lim f (x) = −∞, lim f (x) = +∞
x→−∞
x→+∞
2
0
f (x) = x3 (x − 2) , f (x) = x2 (5x − 6)(x − 2),
f 00 (x) = 4x(5x2 − 12x + 6) √
8 ± 2 31
critical points at x = 0,
√5
8 − 2 31
= −0.63
relative max at x =
5√
8 + 2 31
= 3.83
relative min at x =
5
√
6 ± 66
= 0, −0.42, 2.82
points of inflection at x = 0,
5
no asymptotes
(-0.42,0.16)
2
x
y
(0,0)
(-0.63,0.27)
x
-1
1
2
3
4
-100
-200
(2.82,-165.00)
(3.83,-261.31)
169
11.
Chapter 5
y
lim f (x) doesn’t exist
x→±∞
0
f (x) = 2x sec2 (x2 + 1),
£
¤
f 00 (x) = 2 sec2 (x2 + 1) 1 + 4x2 tan(x2 + 1)
4
2
critical point at x = 0; relative min at x = 0
2
x
-2
-1
1
2
point of inflection when 1 + 4x tan(x + 1) = 0
q
vertical asymptotes at x = ± π(n + 12 ) − 1, n = 0, 1, 2, . . .
12.
-2
-4
lim f (x) = −∞, lim f (x) = +∞
x→−∞
0
2
y
x→+∞
00
f (x) = 1 + sin x, f (x) = cos x
4
critical points at x = 2nπ + π/2, n = 0, ±1, ±2, . . .,
2
−π
no extrema because f 0 ≥ 0 and by Exercise 51 of Section 5.1,
x
π
-2
f is increasing on (−∞, +∞)
-4
inflections points at x = nπ + π/2, n = 0, ±1, ±2, . . .
-6
no asymptotes
13.
x(x + 5)
2x3 + 15x2 − 25
00
,
f
(x)
=
−2
(x2 + 2x + 5)2
(x2 + 2x + 5)3
critical points at x = −5, 0;
f 0 (x) = 2
y
1
relative max at x = −5,
0.8
relative min at x = 0
0.6
points of inflection at x = −7.26, −1.44, 1.20
0.4
0.2
horizontal asymptote y = 1 as x → ±∞
14.
x
-20
3x2 − 25 00
3x2 − 50
,
f
(x)
=
−6
x4
x5
√
critical points at x = ±5 3/3;
√
relative max at x = −5 3/3,
√
relative min at x = +5 3/3
p
inflection points at x = ±5 2/3
-10
10
f 0 (x) = 3
20
y
5
x
-4
6
-5
horizontal asymptote of y = 0 as x → ±∞,
vertical asymptote x = 0
15.
lim f (x) = +∞, lim f (x) = −∞
x→+∞
½
½
x
x≤0
f 0 (x) =
if
−2x
x>0
critical point at x = 0, no extrema
y
x→−∞
inflection point at x = 0 (f changes concavity)
no asymptotes
2
1
-2
1
-2
x
Supplementary Exercises
16.
170
5 − 3x
,
3(1 + x)1/3 (3 − x)2/3
−32
f 00 (x) =
9(1 + x)4/3 (3 − x)5/3
critical point at x = 5/3;
f 0 (x) =
y
4
2
relative max at x = 5/3
-4
-2
x
2
-1
cusp at x = −1;
-3
point of inflection at x = 3
oblique asymptote y = −x as x → ±∞
17.
y
lim f (x) = +∞
x→+∞
0
f (x) = 1 + ln x, f 00 (x) = 1/x
lim f (x) = 0, lim+ f 0 (x) = −∞
x→0+
1
x→0
critical point at x = 1/e;
relative min at x = 1/e
1
2
x
no points of inflection, no asymptotes
18.
y
lim f (x) = +∞
x→+∞
0
f (x) = x(2 ln x + 1), f 00 (x) = 2 ln x + 3
0.2
0.4
x
0.6
lim f (x) = 0, lim+ f 0 (x) = 0
x→0+
x→0
critical point at x = e−1/2 ;
-0.1
−1/2
relative min at x = e
point of inflection at x = e−3/2
19.
1 − 2 ln x 00
6 ln x − 5
, f (x) =
x3
x4
critical point at x = e1/2 ,
f 0 (x) =
relative max at x = e1/2
y
x
1
2
-1
5/6
point of inflection at x = e
-2
horizontal asymptote y = 0 as x → +∞
20.
lim f (x) = +∞
y
x→±∞
2x
1 − x2
, f 00 (x) = 2 2
+1
(x + 1)2
critical point at x = 0;
f 0 (x) =
2
x2
1
relative min at x = 0
points of inflection at x = ±1
no asymptotes
x
-2
2
171
21.
Chapter 5
y
lim f (x) = +∞
x→+∞
x − 1 00
x2 − 2x + 2
, f (x) = ex
2
x
x3
critical point at x = 1;
f 0 (x) = ex
6
relative min at x = 1
2
no points of inflection
-2
1
-2
x
3
vertical asymptote x = 0,
horizontal asymptote y = 0 for x → −∞
22.
f 0 (x) = (1 − x)e−x , f 00 (x) = (x − 2)e−x
y
critical point at x = 1; relative max at x = 1
0.2
point of inflection at x = 2
1
x
2
horizontal asymptote y = 0 as x → +∞, lim f (x) = −∞
x→−∞
-0.8
23.
f 0 (x) = x(2 − x)e1−x , f 00 (x) = (x2 − 4x + 2)e1−x
y
1.8
critical points at x = 0, 2;
relative min at x = 0,
relative max at x = 2
points of inflection at x = 2 ±
1
√
2
0.6
horizontal asymptote y = 0 as x → +∞,
x
1
2
3
4
lim f (x) = +∞
x→−∞
24.
f 0 (x) = x2 (3 + x)ex−1 , f 00 (x) = x(x2 + 6x + 6)ex−1
y
critical points at x = −3, 0;
1
relative min at x = −3
points of inflection at x = 0, −3 ±
√
3
0.4
horizontal asymptote y = 0 as x → −∞
-4
lim f (x) = +∞
-2
1
-0.4
x→+∞
25.
(a)
(b)
40
-5
5
-40
1
, f 00 (x) = 2x
400
1
critical points at x = ± ;
20
1
relative max at x = − ,
20
1
relative min at x =
20
f 0 (x) = x2 −
x
Supplementary Exercises
172
(c) The finer details can be seen when graphing over
a much smaller x-window.
0.0001
-0.1
0.1
-0.0001
26.
(a)
(b)
200
-5
5
√
critical points at x = ± 2, 32 , 2;
√
relative max at x = − 2,
√
relative min at x = 2,
relative max at x = 32 ,
relative min at x = 2
-200
(c)
10
-2.909
-2.2
3.5
1.3
-2.912
-4
27.
(a)
(b)
6
-5
1.6
Divide y = x2 + 1 into y = x3 − 8 to get the
asymptote ax + b = x
5
-6
28.
29.
(a)
p(x) = x3 − x
(b) p(x) = x4 − x2
(c)
p(x) = x5 − x4 − x3 + x2
(d)
p(x) = x5 − x3
f 0 (x) = 4x3 − 18x2 + 24x − 8, f 00 (x) = 12(x − 1)(x − 2)
f 00 (1) = 0, f 0 (1) = 2, f (1) = 2; f 00 (2) = 0, f 0 (2) = 0, f (2) = 3,
so the tangent lines at the inflection points are y = 2x and y = 3.
30.
dy
dy dy
dy
cos x
=2 ;
= 0 when cos x = 0. Use the first derivative test:
=
and
dx
dx dx
dx
2 + sin y
2 + sin y > 0, so critical points when cos x = 0, relative maxima when x = 2nπ + π/2, relative minima
when x = 2nπ − π/2, n = 0, ±1, ±2, . . .
cos x − (sin y)
173
31.
Chapter 5
(2x − 1)(x2 + x − 7)
x2 + x − 7
= 2
,
2
(2x − 1)(3x + x − 1)
3x + x − 1
horizontal asymptote: y = 1/3, √
vertical asymptotes: x = (−1 ± 13)/6
f (x) =
x 6= 1/2
y
10
5
-4
-2
2
4
x
-5
32.
(a)
(x − 2)(x2 + x + 1)(x2 − 2)
(x − 2)(x2 − 2)2 (x2 + 1)
x2 + x + 1
= 2
(x − 2)(x2 + 1)
f (x) =
y
(b)
4
x
-2
1
3
-2
33.
(a) sin x = −1 yields the smallest values, and
sin x = +1 yields the largest
3
O
0
o
(b)
f 0 (x) = esin x cos x; relative maxima at x = 2nπ + π/2, y = e; relative minima at x = 2nπ − π/2,
y = 1/e; n = 0, ±1, ±2, . . . (first derivative test)
(c)
f 00 (x) = (1 − sin x − sin2 x)esin x ; f 00 (x) = 0 when sin x = t, a root of t2 + t − 1 = 0,
√
√
−1 − 5
−1 ± 5
; sin x =
is impossible. So the points of inflection on 0 < x < 2π occur
t=
2
2
√
−1 + 5
, or x = 0.66624, 2.47535
when sin x =
2
34.
f 0 (x) = 3ax2 + 2bx + c; f 0 (x) > 0 or f 0 (x) < 0 on (−∞, +∞) if f 0 (x) = 0 has no real solutions so
from the quadratic formula (2b)2 − 4(3a)c < 0, 4b2 − 12ac < 0, b2 − 3ac < 0. If b2 − 3ac = 0, then
f 0 (x) = 0 has only one real solution at, say, x = c so f is always increasing or always decreasing on
both (−∞, c] and [c, +∞), and hence on (−∞, +∞) because f is continuous everywhere. Thus f is
always increasing or decreasing if b2 − 3ac ≤ 0.
35.
(a)
relative minimum −0.232466 at x = 0.450184
2
-1
1.5
-0.5
Supplementary Exercises
174
0.2
(b) relative maximum 0 at x = 0;
relative minimum −0.107587 at x = ±0.674841
-1.2
1.2
-0.15
1
(c) relative maximum 0.876839; at x = 0.886352;
relative minimum −0.355977 at x = −1.244155
-1.5
1.5
-0.4
36.
(a)
f 0 (x) = 2+3x2 −4x3 has one real root at x = 1.14, a relative max; so f is one-to-one for x ≤ 1.14
(b) f (1.14) = 3.07 so the domain of f −1 is (−∞, 3.07) and the range is (−∞, 1.14); f −1 (−1) = −0.70
37.
(a)
(b)
0.5
0
y = 0 at x = 0; lim y = 0
x→+∞
5
0
38.
(c)
relative max at x = 1/a, inflection point at x = 2/a
(d)
As a increases, the x-coordinate of the maximum and the inflection point move towards the
origin.
(a)
4
4
-2
2
-2
0
2
0
(b) y = 1 at x = a; lim y = +∞, lim y = 0
x→−∞
0
x→+∞
00
(c) Since y is always negative and y is always positive, there are
no relative extrema and no inflection points.
(d) An increase in b makes the graph flatter.
(e)
An increase in a shifts the graph to the right.
175
39.
Chapter 5
1
1
and f 00 (x) = 2
; so f 00 > 0 if x > 1 and therefore f 0 is increasing on
x
x (x + 1)
[1, +∞). Next, f 0 (1) = ln 2 − 1 < 0. Then by L’Hôpital’s Rule,
f 0 (x) = ln(1 + 1/x) −
ln(1 + 1/x)
−1/x2
= lim
=1
x→+∞
x→+∞ (1 + 1/x)(−1/x2 )
1/x
lim x ln(1 + 1/x) = lim
x→+∞
x ln(1 + 1/x) − 1
is indeterminate.
x
¶
¸
· µ
1
1
0
−
= 0.
By L’Hôpital’s Rule lim f (x) = lim ln 1 +
x→+∞
x→+∞
x
x+1
Thus on [1, +∞) the function f 0 starts negative and increases towards zero, so it is negative
on the whole interval. So f (x) is decreasing, and f (x) > f (x + 1). Set x = n and obtain
ln(1 + 1/n)n+1 > ln(1 + 1/(n + 1))n+2 . Since ln x and its inverse function ex are both increasing,
it follows that (1 + 1/n)n+1 > (1 + 1/(n + 1))n+2 .
and thus lim f 0 (x) = lim
x→+∞
x→+∞
CALCULUS HORIZON MODULE CHAPTER 5
1.
The sum of the squares for the residuals for line I is approximately 12 +12 +12 +02 +22 +12 +12 +12 = 10,
and the same for line II is approximately 02 + (0.4)2 + (1.2)2 + 02 + (2.2)2 + (0.6)2 + (0.2)2 + 02 = 6.84;
line II is the regression line.
2.
(a)
y = 5.035714286x − 4.232142857
(b)
y
14
10
6
2
1
4.
r = 0.9907002406
5.
(a)
2
3
4
x
S = 2.155239850t + 190.3600714; r = 0.9569426456
(b) yes, because r is close to 1
6.
7.
(c)
244.241068 mi/h
(d)
It is assumed that the line still gives a good estimate in the year 2000.
(a)
Y = ln y = bx + ln a has slope b and Y -intercept ln a.
(b)
y = a + bX has slope b and y-intercept a.
(c)
Y = ln y = b ln x + ln a = bX + ln a has slope b and Y -intercept ln a.
(d)
The same algebraic rules hold.
(a)
y = 3.923208367 + e0.2934589528x
(b)
y
30
25
20
15
10
5
1
2 3 4 5 6 7
x
Calculus Horizon Module
8.
9.
176
(a)
It appears that log T = a + b log d, so T = 10a db , an exponential model.
(b)
log T = 1.719666407 × 10−4 + 1.499661719 log d
(c)
T = 1.000396046 d1.499661719
(d)
“The squares of the periods of revolution of the planets are proportional to the cubes of their
mean distances”
(a)
T = 27 + 57.8 e−0.046t
(b)
T0 = 84.9◦ C
(c)
53.19 min
CHAPTER 6
Applications of the Derivative
EXERCISE SET 6.1
1.
relative maxima at x = 2, 6; absolute maximum at x = 6; relative and absolute minimum at x = 4
2.
relative maximum at x = 3; absolute maximum at x = 7; relative minima at x = 1, 5; absolute minima
at x = 1, 5
y
3.
(a)
(b)
y
(c)
y
3
2
10
4.
(a)
5
x
7
x
7
x
(b)
y
(c)
y
y
x
x
-5
5
x
5.
f 0 (x) = 8x − 4, f 0 (x) = 0 when x = 1/2; f (0) = 1, f (1/2) = 0, f (1) = 1 so the maximum value is 1
at x = 0, 1 and the minimum value is 0 at x = 1/2.
6.
f 0 (x) = 8 − 2x, f 0 (x) = 0 when x = 4; f (0) = 0, f (4) = 16, f (6) = 12 so the maximum value is 16 at
x = 4 and the minimum value is 0 at x = 0.
7.
f 0 (x) = 3(x − 1)2 , f 0 (x) = 0 when x = 1; f (0) = −1, f (1) = 0, f (4) = 27 so the maximum value is 27
at x = 4 and the minimum value is −1 at x = 0.
8.
f 0 (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2), f 0 (x) = 0 when x = −1, 2; f (−2) = −4, f (−1) = 7,
f (2) = −20, f (3) = −9 so the maximum value is 7 at x = −1 and the minimum value is −20 at x = 2.
9.
√
√
2
3/2
f 0 (x)
√ = 3/(4x + 1) , no critical points; f (−1)
√ = −3/ 5, f (1) = 3/ 5 so the maximum value is
3/ 5 at x = 1 and the minimum value is −3/ 5 at x = −1.
10.
2(2x + 1)
, f 0 (x) = 0 when x = −1/2 and f 0 (x) does not exist when x = −1, 0;
3(x2 + x)1/3
f (−2) = 22/3 , f (−1) = 0, f (−1/2) = 4−2/3 , f (0) = 0, f (3) = 122/3 so the maximum value is 122/3 at
x = 3 and the minimum value is 0 at x = −1, 0.
f 0 (x) =
11. f 0 (x) = 1 − sec2 x, f 0 (x) = 0 for x in (−π/4, π/4) when x = 0; f (−π/4) = 1 − π/4, f (0) = 0,
f (π/4) = π/4 − 1 so the maximum value is 1 − π/4 at x = −π/4 and the minimum value is π/4 − 1
at x = π/4.
177
Exercise Set 6.1
12.
178
0
f 0 (x) = cos x + sin x, f √
(x) = 0 for x in (0, π) when x = 3π/4; f (0) = −1, f (3π/4) =
the maximum value is 2 at x = 3π/4 and the minimum value is −1 at x = 0.
√
2, f (π) = 1 so
½
13.
½
10 − x2 , |x| ≤ 3
−2x, |x| < 3
0
,
f
(x)
=
thus f 0 (x) = 0 when x = 0,
2x, |x| > 3
−8 + x2 , |x| > 3
f 0 (x) does not exist for x in (−5, 1) when x = −3 because lim − f 0 (x) 6= lim + f 0 (x) (see Theorem
f (x) = 1 + |9 − x2 | =
x→−3
x→−3
preceding Exercise 75, Section 3.3); f (−5) = 17, f (−3) = 1, f (0) = 10, f (1) = 9 so the maximum
value is 17 at x = −5 and the minimum value is 1 at x = −3.
½
½
−4, x < 3/2
6 − 4x, x ≤ 3/2
0
, f 0 (x) does not exist when
14. f (x) = |6 − 4x| =
, f (x) =
4, x > 3/2
−6 + 4x, x > 3/2
x = 3/2 thus 3/2 is the only critical point in (−3, 3); f (−3) = 18, f (3/2) = 0, f (3) = 6 so the
maximum value is 18 at x = −3 and the minimum value is 0 at x = 3/2.
15.
f 0 (x) = 2x − 3; critical point x = 3/2. Minimum value f (3/2) = −13/4, no maximum.
16.
f 0 (x) = −4(x + 1); critical point x = −1. Maximum value f (−1) = 5, no minimum.
17.
f 0 (x) = 12x2 (1 − x); critical points x = 0, 1. Maximum value f (1) = 1, no minimum because
lim f (x) = −∞.
x→+∞
18.
f 0 (x) = 4(x3 + 1); critical point x = −1. Minimum value f (−1) = −3, no maximum.
19.
No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞.
x→+∞
x→−∞
20. No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞.
x→+∞
x→−∞
21. f 0 (x) = x(x + 2)/(x + 1)2 ; critical point x = −2 in (−5, −1). Maximum value f (−2) = −4, no
minimum.
22. f 0 (x) = −6/(x − 3)2 ; no critical points in [−5, 5] (x = 3 is not in the domain of f ). No maximum or
minimum because lim+ f (x) = +∞ and lim− f (x) = −∞.
x→3
23.
x→3
(x2 − 1)2 can never be less than zero because it is the square of
x2 − 1; the minimum value is 0 for x = ±1, no maximum because
lim f (x) = +∞.
10
x→+∞
-2
2
0
24.
(x − 1)2 (x + 2)2 can never be less than zero because it is the product
of two squares; the minimum value is 0 for x = 1 or −2, no
maximum because lim f (x) = +∞.
15
x→+∞
-3
2
0
179
25.
Chapter 6
5(8 − x) 0
, f (x) = 0 when x = 8 and f 0 (x) does not exist
3x1/3
when x = 0; f (−1) = 21, f (0) = 0, f (8) = 48, f (20) = 0 so the
maximum value is 48 at x = 8 and the minimum value is 0 at
x = 0, 20.
f 0 (x) =
50
-1
20
0
26.
f 0 (x) = (2 √
− x2 )/(x2 + 2)2 , f 0 (x) √
= 0 for√x in the interval (−1, 4)
when x = 2; f (−1)
√ = −1/3, f (√ 2) = 2/4, f (4) = 2/9 so the
maximum value is 2/4 at x = 2 and the minimum value is
−1/3 at x = −1.
0.4
-1
4
-0.4
27.
f 0 (x) = −1/x2 ; no maximum or minimum because there are no
critical points in (0, +∞).
25
0
10
0
28.
f 0 (x) = (1 − x2 )/(x2 + 1)2 ; critical point x = 1. Maximum value
f (1) = 1/2, minimum value 0 because f (x) is never less than zero
on [0, +∞) and f (0) = 0.
0.5
0
20
0
29.
f 0 (x) = 2 sec x tan x − sec2 x = (2 sin x − 1)/√
cos2 x, f 0 (x) = √
0 for x in
(0, π/4) when x = π/6; f (0) = 2, f (π/6) = 3, f (π/4) = 2 √
2 − 1 so
the maximum value is 2 at x = 0 and the minimum value is 3 at
x = π/6.
2
3
0
1.5
30.
f 0 (x) = 2 sin x cos x − sin x = sin x(2 cos x − 1), f 0 (x) = 0 for x in
(−π, π) when x = 0, ±π/3; f (−π) = −1, f (−π/3) = 5/4, f (0) = 1,
f (π/3) = 5/4, f (π) = −1 so the maximum value is 5/4 at
x = ±π/3 and the minimum value is −1 at x = ±π.
1.5
C
c
-1.5
Exercise Set 6.1
31.
180
f 0 (x) = x2 (2x − 3)e−2x , f 0 (x) = 0 for x in [1, 4] when x = 3/2;
27
if x = 1, 3/2, 4, then f (x) = e−2 , e−3 , 64e−8 ;
8
27 −3
e at x = 3/2,
critical point at x = 3/2; absolute maximum of
8
absolute minimum of 64e−8 at x = 4
0.2
1
4
0
32.
1 − ln x 0
f 0 (x) =
, f (x) = 0 when x = e;
x2
absolute minimum of 0 at x = 1;
absolute maximum of 1/e at x = e
0.4
1
e
0
33.
f 0 (x) = −[cos(cos x)] sin x; f 0 (x) = 0 if sin x = 0 or if cos(cos x) = 0.
If sin x = 0, then x = π is the critical point in (0, 2π); cos(cos x) = 0
has no solutions because −1 ≤ cos x ≤ 1. Thus f (0) = sin(1),
f (π) = sin(−1) = − sin(1), and f (2π) = sin(1) so the maximum
value is sin(1) ≈ 0.84147 and the minimum value is
− sin(1) ≈ −0.84147.
1
o
0
-1
34.
f 0 (x) = −[sin(sin x)] cos x; f 0 (x) = 0 if cos x = 0 or if sin(sin x) = 0.
If cos x = 0, then x = π/2 is the critical point in (0, π);
sin(sin x) = 0 if sin x = 0, which gives no critical points in (0, π).
Thus f (0) = 1, f (π/2) = cos(1), and f (π) = 1 so the maximum
value is 1 and the minimum value is cos(1) ≈ 0.54030.
1.5
0
0
35.
c
½
4, x < 1
so f 0 (x) = 0 when x = 5/2, and f 0 (x) does not exist when x = 1 because
2x − 5, x > 1
lim− f 0 (x) 6= lim+ f 0 (x) (see Theorem preceding Exercise 75, Section 3.3); f (1/2) = 0, f (1) = 2,
0
f (x) =
x→1
x→1
f (5/2) = −1/4, f (7/2) = 3/4 so the maximum value is 2 and the minimum value is −1/4.
36.
f 0 (x) = 2x + p which exists throughout the interval (0, 2) for all values of p so f 0 (1) = 0 because
f (1) is an extreme value, thus 2 + p = 0, p = −2. f (1) = 3 so 12 + (−2)(1) + q = 3, q = 4 thus
f (x) = x2 − 2x + 4 and f (0) = 4, f (2) = 4 so f (1) is the minimum value.
37.
sin 2x has a period of π, and sin 4x a period of π/2 so f (x) is periodic with period π. Consider the
interval [0, π]. f 0 (x) = 4 cos 2x + 4 cos 4x, f 0 (x) = 0 when cos 2x + cos 4x = 0, but cos 4x = 2 cos2 2x − 1
(trig identity) so
2 cos2 2x + cos 2x − 1 = 0
(2 cos 2x − 1)(cos 2x + 1) = 0
cos 2x = 1/2 or cos 2x = −1.
From cos 2x = 1/2, 2x
5π/6. From cos 2x = −1, 2x = π so x =√π/2.
√ = π/3 or 5π/3 so x = π/6 or √
f (0) = 0, f (π/6) = 3 3/2, f (π/2) = 0, f (5π/6)√= −3 3/2, f (π) = 0. The maximum value is 3 3/2
at x = π/6 + nπ and the minimum value is −3 3/2 at x = 5π/6 + nπ, n = 0, ±1, ±2, · · ·.
181
38.
Chapter 6
x
x
has a period of 6π, and cos a period of 4π, so f (x) has a period of 12π. Consider the interval
3
2
x
x
x
x
[0, 12π]. f 0 (x) = − sin − sin , f 0 (x) = 0 when sin + sin = 0 thus, by use of the trig identity
3
2
3
µ ¶
³ x´ 2
a−b
5x
5x
x
a+b
cos
, 2 sin
cos −
= 0 so sin
= 0 or cos
= 0. Solve
sin a + sin b = 2 sin
2
2
12
12
12
12
5x
x
= 0 to get x = 12π/5, 24π/5, 36π/5, 48π/5 and then solve cos
= 0 to get x = 6π. The
sin
12
12
corresponding values of f (x) are −4.0450, 1.5450, 1.5450, −4.0450, 1, 5, 5 so the maximum value is 5
and the minimum value is −4.0450 (approximately).
cos
39. Let f (x) = x − sin x, then f 0 (x) = 1 − cos x and so f 0 (x) = 0 when cos x = 1 which has no solution
for 0 < x < 2π thus the minimum value of f must occur at 0 or 2π. f (0) = 0, f (2π) = 2π so 0 is the
minimum value on [0, 2π] thus x − sin x ≥ 0, sin x ≤ x for all x in [0, 2π].
40.
Let f (x) = ln x − x + 1, then f 0 (x) = 1/x − 1 and so f 0 (x) = 0 at x = 1. Since lim+ f (x) = −∞ and
x→0
lim f (x) = −∞, f (x) has a maximum of f (1) = 0 at x = 1 and so f (x) ≤ 0 for 0 < x < +∞, so
x→+∞
ln x ≤ x − 1 on (0, +∞).
41.
Let m = slope at x, then m = f 0 (x) = 3x2 − 6x + 5, dm/dx = 6x − 6; critical point for m is x = 1,
minimum value of m is f 0 (1) = 2
42.
(a)
−64 cos3 x + 27 sin3 x 0
64 cos x 27 sin x
+
, f (x) = 0 when
=
2
2
cos x
sin x
sin2 x cos2 x
27 sin3 x = 64 cos3 x, tan3 x = 64/27, tan x = 4/3 so the critical point is x = x0 where
tan x0 = 4/3 and 0 < x0 < π/2. To test x0 first rewrite f 0 (x) as
27 cos x(tan3 x − 64/27)
27 cos3 x(tan3 x − 64/27)
=
;
f 0 (x) =
sin2 x cos2 x
sin2 x
0
if x < x0 then tan x < 4/3 and f (x) < 0, if x > x0 then tan x > 4/3 and f 0 (x) > 0 so f (x0 ) is
the minimum value. f has no maximum because lim+ f (x) = +∞.
f 0 (x) = −
x→0
(b)
If tan x0 = 4/3 then (see figure)
sin x0 = 4/5 and cos x0 = 3/5
so f (x0 ) = 64/ sin x0 + 27/ cos x0
= 64/(4/5) + 27/(3/5)
= 80 + 45 = 125
5
4
x0
3
43.
√
2x(x3 − 24x2 + 192x − 640)
; real root of x3 − 24x2 + 192x − 640 at x = 4(2 + 3 2). Since
3
(x − 8)
lim+ f (x) = lim f (x) = +∞ and there is only one relative extremum, it must be a minimum.
f 0 (x) =
x→+∞
x→8
44.
(a)
¢
K ¡ −at
ln(a/b)
dC
dC
=
ae − be−bt so
= 0 at t =
. This is the only stationary point and
dt
a−b
dt
a−b
C(0) = 0, lim C(t) = 0, C(t) > 0 for 0 < t < +∞, so it is an absolute maximum.
x→+∞
(b)
0.7
0
10
0
Exercise Set 6.2
182
45.
The slope of the line is −1, and the slope of the tangent to y = −x2 is −2x so −2x = −1, x = 1/2.
The line lies above the curve so the vertical distance is given by F (x) = 2 − x + x2 ; F (−1) = 4,
F (1/2) = 7/4, F (3/2) = 11/4. The point (1/2, −1/4) is closest, the point (−1, −1) farthest.
46.
The slope of the line is 4/3; and the slope of the tangent to y = x3 is 3x2 so 3x2 = 4/3, x2 = 4/9,
x = ±2/3. The line lies below the curve so the vertical distance is given by F (x) = x3 − 4x/3 + 1;
F (−1) = 4/3, F (−2/3) = 43/27, F (2/3) = 11/27, F (1) = 2/3. The closest point is (2/3, 8/27), the
farthest is (−2/3, −8/27).
47.
The absolute extrema of y(t) can occur at the endpoints t = 0, 12 or when dy/dt = 2 sin t = 0, i.e.
t = 0, 12, kπ, k = 1, 2, 3; the absolute maximum is y = 4 at t = π, 3π; the absolute minimum is y = 0
at t = 0, 2π.
48.
(a)
The absolute extrema of y(t) can occur at the endpoints t = 0, 2π or when
dy/dt = 2 cos 2t − 4 sin t cos t = 2 cos 2t − 2 sin 2t = 0, t = 0, 2π, π/8, 5π/8, 9π/8, 13π/8;
the absolute maximum is y = 3.4142 at t = π/8, 9π/8; the absolute minimum is y = 0.5859
at t = 5π/8, 13π/8.
(b)
The absolute extrema of x(t) occur at the endpoints t = 0, 2π or when
49.
50.
2 sin t + 1
dx
=−
= 0,
dt
(2 + sin t)2
t = 7π/6, 11π/6. The absolute maximum is x = 0.5774 at t = 11π/6 and the absolute minimum
is x = −0.5774 at t = 7π/6.
b
f 0 (x) = 2ax + b; critical point is x = −
2a
µ
¶
b
is the minimum value of f , but
f 00 (x) = 2a > 0 so f −
2a
¶
µ
¶2
¶
µ
µ
b
−b2 + 4ac
b
b
=a −
+c=
thus f (x) ≥ 0 if and only if
+b −
f −
2a
2a
2a
4a
¶
µ
−b2 + 4ac
b
≥ 0,
≥ 0, −b2 + 4ac ≥ 0, b2 − 4ac ≤ 0
f −
2a
4a
Use the proof given in the text, replacing “maximum” by “minimum” and “largest” by “smallest” and
reversing the order of all inequality symbols.
EXERCISE SET 6.2
1.
Let x = one number, y = the other number, and P = xy where x + y = 10. Thus y = 10 − x so
P = x(10 − x) = 10x − x2 for x in [0, 10]. dP/dx = 10 − 2x, dP/dx = 0 when x = 5. If x = 0, 5, 10
then P = 0, 25, 0 so P is maximum when x = 5 and, from y = 10 − x, when y = 5.
2.
Let x and y be nonnegative numbers and z the sum of their squares, then z = x2 + y 2 . But x + y = 1,
y = 1 − x so z = x2 + (1 − x)2 = 2x2 − 2x + 1 for 0 ≤ x ≤ 1. dz/dx = 4x − 2, dz/dx = 0 when x = 1/2.
If x = 0, 1/2, 1 then z = 1, 1/2, 1 so
(a) z is as large as possible when one number is 0 and the other is 1.
(b)
3.
z is as small as possible when both numbers are 1/2.
If y = x + 1/x for 1/2 ≤ x ≤ 3/2 then dy/dx = 1 − 1/x2 = (x2 − 1)/x2 , dy/dx = 0 when x = 1. If
x = 1/2, 1, 3/2 then y = 5/2, 2, 13/6 so
(a)
y is as small as possible when x = 1.
(b) y is as large as possible when x = 1/2.
183
4.
Chapter 6
A = xy where x + 2y = 1000 so y = 500 − x/2 and A = 500x − x2 /2
for x in [0, 1000]; dA/dx = 500 − x, dA/dx = 0 when x = 500. If
x = 0 or 1000 then A = 0, if x = 500 then A = 125, 000 so the area
is maximum when x = 500 ft and y = 500 − 500/2 = 250 ft.
Stream
y
x
5.
6.
Let x and y be the dimensions shown in the figure and A the area,
then A = xy subject to the cost condition 3(2x) + 2(2y) = 6000, or
y = 1500 − 3x/2. Thus A = x(1500 − 3x/2) = 1500x − 3x2 /2 for x in
[0, 1000]. dA/dx = 1500 − 3x, dA/dx = 0 when x = 500. If x = 0 or
1000 then A = 0, if x = 500 then A = 375, 000 so the area is greatest
when x = 500 ft and (from y = 1500 − 3x/2) when y = 750 ft.
Heavy-duty
y
Standard
x
Let x and y be the dimensions shown in the figure and A the area of
the rectangle, then A = xy and, by similar triangles,
x/6 = (8 − y)/8, y = 8 − 4x/3 so A = x(8 − 4x/3) = 8x − 4x2 /3 for
x in [0, 6]. dA/dx = 8 − 8x/3, dA/dx = 0 when x = 3. If x = 0, 3, 6
then A = 0, 12, 0 so the area is greatest when x = 3 in and (from
y = 8 − 4x/3) y = 4 in.
10
8
x
y
6
7.
8.
Let x, y, and z be as shown in the figure and A the area of the
rectangle, then A = xy and, by similar triangles, z/10 = y/6,
z = 5y/3; also x/10 = (8 − z)/8 = (8 − 5y/3)/8 thus
y = 24/5 − 12x/25 so A = x(24/5 − 12x/25) = 24x/5 − 12x2 /25 for
x in [0, 10]. dA/dx = 24/5 − 24x/25, dA/dx = 0 when x = 5. If
x = 0, 5, 10 then A = 0, 12, 0 so the area is greatest when x = 5 in.
and y = 12/5 in.
z
y
10
8
x
6
A = (2x)y = 2xy where y = 16 − x2 so A =√
32x − 2x3 for 0√
≤ x ≤ 4;
2
dA/dx = 0 when x = 4/ 3. If x = 0, 4/ √3, 4
dA/dx = 32 − 6x ,√
then A = 0, 256/(3 3), 0 so the area is largest when x = 4/ 3 and
y=
√32/3. The dimensions of the rectangle with largest area are
8/ 3 by 32/3.
y
16
y
x
x
-4
9.
10.
√
A = xy√where x2 + y 2 = 202 = 400 so y = 400 − x2 and
√
≤ x ≤ 20; √
dA/dx = 2(200 √
− x2 )/ 400 − x2 ,
A = x 400 − x2 for 0√
dA/dx = 0 when x = 200 = 10 2. If x = 0, 10 2,
√20 then
2 and
A =√
0, 200, 0 so the area
is
maximum
when
x
=
10
√
y = 400 − 200 = 10 2.
Let x and y be the dimensions shown in the figure, then the area of
the rectangle is A = xy.
³ x ´2
p
1p 2
But
+ y 2 = R2 , thus y = R2 − x2 /4 =
4R − x2 so
2p
2
√
1
A = x 4R2 − x2 for 0 ≤ x ≤ 2R. dA/dx = (2R2 − x2 )/ 4R2 − x2 ,
2
√
√
2
dA/dx = 0 when x = 2R. If x = 0,
√ A = 0, R , 0 so
√ 2R, 2R then
the greatest area occurs when x = 2R and y = 2R/2.
10
4
y
x
x
R
y
x
2
Exercise Set 6.2
11.
184
Let x = length of each side that uses the $1 per foot fencing,
y = length of each side that uses the $2 per foot fencing.
The cost is C = (1)(2x) + (2)(2y) = 2x + 4y, but A = xy = 3200 thus y = 3200/x so
C = 2x + 12800/x for x > 0,
dC/dx = 2 − 12800/x2 , dC/dx = 0 when x = 80, d2 C/dx2 > 0 so
C is least when x = 80, y = 40.
12.
A = xy where 2x + 2y = p so y = p/2 − x and A = px/2 − x2 for x
in [0, p/2]; dA/dx = p/2 − 2x, dA/dx = 0 when x = p/4. If x = 0 or
p/2 then A = 0, if x = p/4 then A = p2 /16 so the area is maximum
when x = p/4 and y = p/2 − p/4 = p/4, which is a square.
y
x
13.
Let x and y be the dimensions of a rectangle; the perimeter is p = 2x + 2y. But A = xy thus
√
y = A/x so p = 2x + 2A/x for x > 0, dp/dx = 2 − 2A/x2 = √
2(x2 − A)/x2 ,√dp/dx = 0 when x = A,
d2 p/dx2 = 4A/x3 > 0 if x > 0 so p is a minimum when x = A and y = A and thus the rectangle
is a square.
14.
With x, y, r, and s as shown in the figure, the sum of the enclosed
x
y
x
and s = because x is the
areas is A = πr 2 + s2 where r =
2π
4
circumference of the circle and y is the perimeter of the square, thus
r
x2
y2
A=
+ . But x + y = 12, so y = 12 − x and
4π 16
(12 − x)2
π+4 2 3
x2
A=
+
=
x − x + 9 for 0 ≤ x ≤ 12.
4π
16
16π
2
π+4
3 dA
12π
12π
dA
=
x− ,
= 0 when x =
. If x = 0,
, 12
dx
8π
2 dx
π+4
π+4
36 36
then A = 9,
,
so the sum of the enclosed areas is
π+4 π
(a) a maximum when x = 12 in. (when all of the wire is used for the circle)
(b)
15.
(a)
17.
y
cut
s
a minimum when x = 12π/(π + 4) in.
dN
= 250(20 − t)e−t/20 = 0 at t = 20, N (0) = 125000, N (20) ≈ 161788, and N (100) ≈ 128,369;
dt
the absolute maximum is N = 161788 at t = 20, the absolute minimum is N = 125000 at t = 0.
(b) The absolute minimum of
16.
12
dN
d2 N
occurs when
= 12.5(t − 40)e−t/20 = 0, t = 40.
dt
dt2
The area of the window is A = 2rh + πr2 /2, the perimeter is
1
p = 2r + 2h + πr thus h = [p − (2 + π)r] so
2
A = r[p − (2 + π)r] + πr2 /2
= pr − (2 + π/2)r2 for 0 ≤ r ≤ p/(2 + π),
dA/dr = p − (4 + π)r, dA/dr = 0 when r = p/(4 + π) and
d2 A/dr2 < 0, so A is maximum when r = p/(4 + π).
r
h
2r
V = x(12 − 2x)2 for 0 ≤ x ≤ 6; dV /dx = 12(x − 2)(x − 6),
dV /dx = 0 when x = 2 for 0 < x < 6. If x = 0, 2, 6 then
V = 0, 128, 0 so the volume is largest when x = 2 in.
12
x
x
x
x
12 − 2x
12
x
x
x
x
12 − 2x
185
Chapter 6
18.
The dimensions of the box will be (k − 2x) by (k − 2x) by x so V = (k − 2x)2 x = 4x3 − 4kx2 + k 2 x for
x in [0, k/2]. dV /dx = 12x2 − 8kx + k 2 = (6x − k)(2x − k), dV /dx = 0 for x in (0, k/2) when x = k/6.
If x = 0, k/6, k/2 then V = 0, 2k 3 /27, 0 so V is maximum when x = k/6. The squares should have
dimensions k/6 by k/6.
19.
Let x be the length of each side of a square, then V = x(3 − 2x)(8 − 2x) = 4x3 − 22x2 + 24x for
0 ≤ x ≤ 3/2; dV /dx = 12x2 − 44x + 24 = 4(3x − 2)(x − 3), dV /dx = 0 when x = 2/3 for 0 < x < 3/2.
If x = 0, 2/3, 3/2 then V = 0, 200/27, 0 so the maximum volume is 200/27 ft3 .
20.
Let x = length of each edge of base, y = height. The cost is
C = (cost of top and bottom) + (cost of sides) = (2)(2x2 ) + (3)(4xy) = 4x2 + 12xy, but
V = x2 y = 2250 thus y = 2250/x2 so C = 4x2 + 27000/x for x > 0, dC/dx = 8x − 27000/x2 ,
√
dC/dx = 0 when x = 3 3375 = 15, d2 C/dx2 > 0 so C is least when x = 15, y = 10.
21.
Let x = length of each edge of base, y = height, k = $/cm2 for the sides. The cost is
C = (2k)(2x2 ) + (k)(4xy) = 4k(x2 + xy), but V = x2 y = 2000 thus y = 2000/x2 so
C = 4k(x2 + 2000/x) for x > 0 dC/dx = 4k(2x − 2000/x2 ), dC/dx = 0 when
√
x = 3 1000 = 10, d2 C/dx2 > 0 so C is least when x = 10, y = 20.
22.
Let x and y be the dimensions shown in the figure and V the
volume, then V = x2 y. The amount of material is to be 1000 ft2 ,
thus (area of base) + (area of sides) = 1000, x2 + 4xy = 1000,
1
1000 − x2
1000 − x2
so V = x2
= (1000x − x3 ) for
y=
4x √
4x
4
0 < x ≤ 10 10.
p
p
dV
1
dV
= (1000 − 3x2 ),
= 0 when x = 1000/3 = 10 10/3.
dx
4
dx
p
√
5000 p
10/3, 0;
If x = 0, 10 10/3, 10 10 then V = 0,
3
p
p
the volume is greatest for x = 10 10/3 ft and y = 5 10/3 ft.
y
x
x
23.
Let x = height and width, y = length. The surface area is S = 2x2 + 3xy where
x2 y = V , so y = V /x2
p
3
2
2
2
2
and S = 2x + 3V /x forrx > 0; dS/dx
r = 4x − 3V /x , dS/dx = 0 when x = 3V /4, d S/dx > 0 so S
4 3 3V
3 3V
,y=
.
is minimum when x =
4
3
4
24.
Let r and h be the dimensions shown in the figure, then the volume
of the inscribed cylinder is V = πr2 h. But
µ ¶2
h
h2
2
r +
= R2 thus r2 = R2 −
2 µ
4
h
¶
µ
¶
h2
h3
R
h
2
2
2
h=π R h−
so V = π R −
4
4
µ
¶
dV
dV
3
r
= π R 2 − h2 ,
=0
for 0 ≤ h ≤ 2R.
dh
4
dh
√
√
4π
when h = 2R/ 3. If h = 0, 2R/ 3, 2R then V = 0, √ R3 , 0 so the volume is largest when
3
3
p
√
h = 2R/ 3 and r = 2/3R.
Exercise Set 6.2
186
25.
Let r and h be the dimensions shown in the figure, then the surface
2
area is S µ
= 2πrh
¶2 + 2πr .
√
h
But r2 +
= R2 thus h = 2 R2 − r2 so
h
√ 2
S = 4πr R2 − r2 + 2πr2 for 0 ≤ r ≤ R,
R
h
2
4π(R2 − 2r2 )
dS
dS
= √
= 0 when
+ 4πr;
dr
dr
R2 − r 2
r
R2 − 2r2
√
= −r
(i)
R2 − r 2
√
R2 − 2r2 = −r R2 − r2
R4 − 4R2 r2 + 4r4 = r2 (R2 − r2 )
5r4 − 5R2 r2 + R4 = 0
s
√
√
√
2
4 − 20R4
±
25R
5
5± 5
5R
5
±
2
2
and using the quadratic formula r =
=
R ,r=
R, of which
10
10
10
s
s
√
√
√
5+ 5
5+ 5
R satisfies (i). If r = 0,
R, 0 then S = 0, (5 + 5)πR2 , 2πR2 so the surface
only r =
10
10
s
s
√
√
p
5+ 5
5− 5
R and, from h = 2 R2 − r2 , h = 2
R.
area is greatest when r =
10
10
26.
Let R and H be the radius and height of the cone, and r and h the
radius and height of the cylinder (see figure), then the volume of the
r
H −h
cylinder is V = πr 2 h. By similar triangles (see figure)
=
H
R
thus
H
H
H
h = (R − r) so V = π (R − r)r2 = π (Rr2 − r3 ) for 0 ≤ r ≤ R.
R
R
R
dV
H
dV
H
= π (2Rr − 3r2 ) = π r(2R − 3r),
= 0 for 0 < r < R
dr
R
R
dr
when r = 2R/3. If r = 0, 2R/3, R then V = 0, 4πR2 H/27, 0 so the
4
4 1 2
4πR2 H
=
πR H = · (volume of cone).
maximum volume is
27
9 3
9
r
H
h
R
27.
2
From (13), S = 2πr2 + 2πrh. But V = πr
h thus h = V /(πr2 ) and so S = 2πr2 + 2V /r for r > 0.
p
3
2
if r = V /(2π). Since d2 S/dr2 = 4π + 4V /r3 > 0, the minimum
dS/dr = 4πr − 2V /r , dS/dr = 0 p
surface area is achieved when r = 3 V /2π and so h = V /(πr2 ) = [V /(πr3 )]r = 2r.
28.
V = πr2 h where S = 2πr2 + 2πrh so h =
29.
The surface area is S = πr2 + 2πrh where V = πr2 h = 500 so
h = 500/(πr2 ) and S = πr2 + 1000/r for r > 0;
dS/dr
= 2πr − 1000/r2 = (2πr3 − 1000)/r2 , dS/dr = 0 when
p
3
r = 500/π, d2 S/dr2 > 0 for r > 0 so S is minimum when
p
500
500
500 p
3
500/π
r = 3 500/π and h = 2 = 3 r =
πr
πr
π(500/π)
p
= 3 500/π.
1
S − 2πr2
, V = (Sr − 2πr3 ) for r > 0.
2πr
2
2
p
1
d
V
dV
= (S − 6πr2 ) = 0 if r = S/(6π),
= −6πr < 0 so V is maximum when
dr
2
dr2
p
S − 2πr2
S − 2πr2
S − S/3
r = S/(6π) and h =
=
r = 2r, thus the height is equal to the
r =
2πr
2πr2
S/3
diameter of the base.
r
h
187
30.
Chapter 6
The total area of material used is
A = Atop + Abottom + Aside = (2r)2 + (2r)2 + 2πrh = 8r2 + 2πrh.
The volume is V = πr2 h thus h = V /(πr2 ) so A = 8r2 + 2V /r for r > 0,
√
dA/dr = 16r − 2V /r2 = 2(8r3 − V )/r2 , dA/dr = 0 when r = 3 V /2. This is the only critical point,
√
r
r
π 3
=
=
r and, for
d2 A/dr2 > 0 there so the least material is used when r = 3 V /2,
h
V /(πr2 )
V
√
πV
π
r
3
= .
r = V /2, =
h
V 8
8
31. Let x be the length of each side of the squares and y the height of the frame, then the volume is V = x2 y.
The total length of the wire is L thus 8x+4y = L, y = (L−8x)/4 so V = x2 (L−8x)/4 = (Lx2 −8x3 )/4
for 0 ≤ x ≤ L/8. dV /dx = (2Lx − 24x2 )/4, dV /dx = 0 for 0 < x < L/8 when x = L/12. If
x = 0, L/12, L/8 then V = 0, L3 /1728, 0 so the volume is greatest when x = L/12 and y = L/12.
32.
(a)
Let x = diameter of the sphere, y = length of an edge of the cube. The combined volume is
1
(S − πx2 )1/2
V = πx3 + y 3 and the surface area is S = πx2 + 6y 2 = constant. Thus y =
and
6
61/2
r
2 3/2
S
(S − πx )
π
;
for 0 ≤ x ≤
V = x3 +
6
π
63/2
p
√
dV
π
3π
π
dV
= x2 − 3/2 x(S − πx2 )1/2 = √ x( 6x − S − πx2 ).
= 0 when x = 0, or when
dx
2
dx
6
2 6
r
r
r
√
√
S
S
S
S
, x =
. If x = 0,
,
, then
6x = S − πx2 , 6x2 = S − πx2 , x2 =
6+π
6
+
π
6
+
π
π
r
r
S 3/2
S 3/2
S 3/2
S
S
, √ so that V is smallest when x =
, and hence when y =
,
V = 3/2 , √
6+π
6+π
6
6 6+π 6 π
thus x = y.
(b) From part (a), the sum of the volumes is greatest when there is no cube.
33.
34.
Let h and r be the dimensions shown in the figure, then the volume
1
is V = πr2 h. But r2 + h2 = L2 thus r2 = L2 − h2 so
3
1
1
V = π(L2 − h2 )h = π(L2 h − h3 ) for 0 ≤ h ≤ L.
3
3
√
√
dV
1
dV
2
2
= π(L − 3h ).
= 0 when h = L/ 3. If h = 0, L/ 3, 0
dh
3
dh
2π 3
then V = 0, √ L , 0 so the volume is as large as possible when
9 3 p
√
h = L/ 3 and r = 2/3L.
Let r and h be the radius and height of the cone (see figure). The
slant height of any such cone will be R, the radius of the circular
sheet. Refer to the solution of Exercise 33 to find that the largest
2π
volume is √ R3 .
9 3
h
L
r
h
R
r
Exercise Set 6.2
35.
36.
37.
38.
39.
188
√
The area of the paper is A = πrL = πr r2 + h2 , but
p
1
V = πr2 h = 10 thus h = 30/(πr2 ) so A = πr r2 + 900/(π 2 r4 ).
3
To simplify
let S = A2 ,
µ the computations
¶
900
900
S = π 2 r2 r2 + 2 4 = π 2 r4 + 2 for r > 0,
π r
r
1800
4(π 2 r6 − 450)
dS
2 3
= 4π r − 3 =
, dS/dr = 0 when
dr p
r
r3
r = 6 450/π 2 , d2 S/dr2 > 0, so S and hence A is least when
p
30 p
3
π 2 /450.
r = 6 450/π 2 , h =
π
1
hb. By similar triangles (see figure)
2
R
b/2
2Rh
Rh2
=√
,b= √
so A = √
for h > 2R,
h
h2 − 2Rh
h2 − 2Rh
h2 − 2Rh
2
Rh (h − 3R) dA
dA
= 2
= 0 for h > 2R when h = 3R, by the first
,
dh
(h − 2Rh)3/2 dh
derivative
√ test A is minimum when h = 3R. If h = 3R then
b = 2 3R (the triangle is equilateral).
r
The area of the triangle is A =
1
The volume of the cone is V = πr2 h. By similar triangles (see
3
R
r
Rh
figure) = √
,r=√
so
2
2
h
h − 2Rh
h − 2Rh
1
h3
h2
1
= πR2
for h > 2R,
V = πR2 2
3
h − 2Rh
3
h − 2R
1
dV
h(h − 4R) dV
= πR2
= 0 for h > 2R when h = 4R, by the
,
dh
3
(h − 2R)2 dh
first √
derivative test V is minimum when h = 4R. If h = 4R then
r = 2R.
h−R
dA/dθ = 0 for 0 < θ < π/2 when
√ cos θ = 1/2, θ = π/3. If
θ = 0, π/3, π/2 then A = 0, 75 3/4, 25 so the cross-sectional area is
greatest when θ = π/3.
h2 − 2Rh
h
R
R
b/2
b
h−R
h2 − 2Rh
h
R
R
r
The area is (see figure)
1
A = (2 sin θ)(4 + 4 cos θ)
2
= 4(sin θ + sin θ cos θ)
for 0 ≤ θ ≤ π/2;
dA/dθ = 4(cos θ − sin2 θ + cos2 θ)
= 4(cos θ − [1 − cos2 θ] + cos2 θ)
= 4(2 cos2 θ + cos θ − 1)
= 4(2 cos θ − 1)(cos θ + 1)
dA/dθ =√0 when θ = π/3 for 0 < θ < π/2.
√ If θ = 0, π/3, π/2 then
A = 0, 3 3, 4 so the maximum area is 3 3.
Let b and h be the dimensions shown in the figure, then the
1
cross-sectional area is A = h(5 + b). But h = 5 sin θ and
2
5
b = 5 + 2(5 cos θ) = 5 + 10 cos θ so A = sin θ(10 + 10 cos θ)
2
= 25 sin θ(1 + cos θ) for 0 ≤ θ ≤ π/2.
dA/dθ = −25 sin2 θ + 25 cos θ(1 + cos θ)
= 25(− sin2 θ + cos θ + cos2 θ)
= 25(−1 + cos2 θ + cos θ + cos2 θ)
= 25(2 cos2 θ + cos θ − 1) = 25(2 cos θ − 1)(cos θ + 1).
L
h
4 cos θ
2 sin θ
2
θ
4
5 cos θ
5
2 cos θ
b
h = 5 sin θ
θ
5
5
189
40.
41.
42.
Chapter 6
cos φ
, k the constant of proportionality. If h is the height of the lamp above the table then
`2
√
r2 − 2h2
h
h
dI
dI
cos φ = h/` and ` = h2 + r2 so I = k 3 = k 2
=k 2
=0
for h > 0,
,
2
3/2
2
5/2
`
dh
(h + r )
(h + r )
√
√dh
when h = r/ 2, by the first derivative test I is maximum when h = r/ 2.
I = k
Let L, L1 , and L2 be as shown in the figure, then
L = L1 + L2 = 8 csc θ + sec θ,
dL
= −8 csc θ cot θ + sec θ tan θ, 0 < θ < π/2
dθ
sin θ
8 cos θ
−8 cos3 θ + sin3 θ
+
;
=−
=
2
2
cos θ
sin θ
sin2 θ cos2 θ
dL
= 0 if sin3 θ = 8 cos3 θ, tan3 θ = 8, tan θ = 2 which gives the
dθ
absolute minimum for L because lim+ L = lim − L = +∞. If
θ→0
√θ→π/2
√
5/2
and
sec
θ
=
5 so
tan θ =√2, then csc
θ
=
√
√
L = 8( 5/2) + 5 = 5 5 ft.
L2
L
L1
8
θ
1
Let
x = number of steers per acre
w = average market weight per steer
T = total market weight per acre
then T = xw where w = 2000 − 50(x − 20) = 3000 − 50x
so
T = x(3000 − 50x) = 3000x − 50x2 for 0 ≤ x ≤ 60,
dT /dx = 3000 − 100x and dT /dx = 0 when x = 30. If x = 0, 30, 60 then T = 0, 45000, 0 so the total
market weight per acre is largest when 30 steers per acre are allowed.
43.
44.
(a)
The daily profit is
P = (revenue) − (production cost) = 100x − (100, 000 + 50x + 0.0025x2 )
= −100, 000 + 50x − 0.0025x2
for 0 ≤ x ≤ 7000, so dP/dx = 50 − 0.005x and dP/dx = 0 when x = 10, 000. Because 10,000
is not in the interval [0, 7000], the maximum profit must occur at an endpoint. When x = 0,
P = −100, 000; when x = 7000, P = 127, 500 so 7000 units should be manufactured and sold
daily.
(b)
Yes, because dP/dx > 0 when x = 7000 so profit is increasing at this production level.
(a)
R(x) = px but p = 1000 − x so R(x) = (1000 − x)x
(b)
P (x) = R(x) − C(x) = (1000 − x)x − (3000 + 20x) = −3000 + 980x − x2
(c)
P 0 (x) = 980 − 2x, P 0 (x) = 0 for 0 < x < 500 when x = 490; test the points 0, 490, 500 to find
that the profit is a maximum when x = 490.
(d) P (490) = 237,100
(e)
45.
p = 1000 − x = 1000 − 490 = 510.
The profit is
P = (profit on nondefective) − (loss on defective) = 100(x − y) − 20y = 100x − 120y
but y = 0.01x + 0.00003x2 so P = 100x − 120(0.01x + 0.00003x2 ) = 98.8x − 0.0036x2 for x > 0,
dP/dx = 98.8 − 0.0072x, dP/dx = 0 when x = 98.8/0.0072 ≈ 13, 722, d2 P/dx2 < 0 so the profit is
maximum at a production level of about 13,722 pounds.
Exercise Set 6.2
46.
190
The total cost C is
C = c · (hours to travel 3000 mi at a speed of v mi/h)
3000
3000
= (a + bv n )
= 3000(av −1 + bv n−1 ) for v > 0,
=c·
v
v
dC/dv = 3000[−av −2 + b(n − 1)v n−2 ] = 3000[−a + b(n − 1)v n ]/v 2 ,
¸1/n
·
a
. This is the only critical point and dC/dv changes sign from − to
dC/dv = 0 when v =
b(n − 1)
¸1/n
·
a
mi/h.
+ at this point so the total cost is least when v =
b(n − 1)
p
√
47. The distance between the particles is D = (1 − t − t)2 + (t − 2t)2 = 5t2 − 4t + 1 for t ≥ 0. For
convenience, we minimize D2 instead, so D2 = 5t2 − 4t + 1, dD2 /dt = 10t − 4, which is 0 when t =
√ 2/5.
d2 D2 /dt2 > 0 so D2 and hence D is minimum when t = 2/5. The minimum distance is D = 1/ 5.
p
√
48. The distance between the particles is D = (2t − t)2 + (2 − t2 )2 = t4 − 3t2 + 4 for t ≥ 0. For
2
so D2 = t4 − 3t2 + 4, dD2 /dt
= 4t3 − 6t = 4t(t2 − 3/2), which is
convenience we minimize
p
p D instead
= 12t2 − 6 > 0 when t = 3/2 so D2 and hence D is minimum
0 for t > 0 when t = 3/2. d2 D2 /dt2 √
there. The minimum distance is D = 7/2.
49.
50.
51.
1. The distance between
Let P p
(x, y) be a point on the curve x2 + y 2 = p
√ P (x, y) and P0 (2, 0) is
(x − 2)2 + y 2 , but y 2 = 1 − x2 so D =
(x − 2)2 + 1 − x2 = 5 − 4x for −1 ≤ x ≤ 1,
D =
2
dD
which has no critical points for −1 < x < 1. If x = −1, 1 then D = 3, 1 so the
= −√
dx
5 − 4x
closest point occurs when x = 1 and y = 0.
√
Let P (x, y) be a point on y = x, then the distance D between P and (2, 0) is
p
p
√
D = (x − 2)2 + y 2 = (x − 2)2 + x = x2 − 3x + 4, for 0 ≤ x ≤ 3. For convenience we find the
2
extrema for D2 instead, so D√
= x2 − 3x + 4, dD2 /dx = 2x − 3√= 0 when x = 3/2. If x = 0, 3/2, 3
2
then D
p = 4, 7/4, 4 so D = 2, 7/2, 2. The points (0, 0) and (3, 3) are at the greatest distance, and
(3/2, 3/2) the shortest distance from (2, 0).
Let (x, y) be a point on the curve, then the square of the distance between (x, y) and (0, 2) is
S = x2 + (y − 2)2 where x2 − y 2 = 1, x2 = y 2 + 1 so
S = (y 2 + 1) + (y − 2)2 = 2y 2 − 4y + 5 for any y, dS/dy = 4y − 4, dS/dy = 0 when y = 1,
√
d2 S/dy 2 > 0 so S is least when y = 1 and x = ± 2.
52.
The square of the distance between a point (x, y) on the curve and the point (0, 9) is
S = x2 + (y − 9)2 where x = 2y 2 so S = 4y 4 + (y − 9)2 for any y,
dS/dy = 16y 3 + 2(y − 9) = 2(8y 3 + y − 9), dS/dy = 0 when y = 1 (which is the only real solution),
d2 S/dy 2 > 0 so S is least when y = 1, x = 2.
53.
If P (x0 , y0 ) is on the curve y = 1/x2 , then y0 = 1/x20 . At P the slope of the tangent line is −2/x30 so its
2
2
3
3
1
equation is y − 2 = − 3 (x − x0 ), or y = − 3 x + 2 . The tangent line crosses the y-axis at 2 , and
x0
x0
x0
x0
x0
s
9
3
9
x0 . The length of the segment then is L =
+ x20 for x0 > 0. For convenience,
2
x40
4
9
9
dL2
36 9
9(x60 − 8)
= − 5 + x0 =
, which is 0 when x60 = 8,
we minimize L2 instead, so L2 = 4 + x20 ,
x0
4
dx0
x0
2
2x50
√ d2 L2
√
x0 = 2.
> 0 so L2 and hence L is minimum when x0 = 2, y0 = 1/2.
dx20
the x-axis at
191
Chapter 6
54.
If P (x0 , y0 ) is on the curve y = 1 − x2 , then y0 = 1 − x20 . At P the slope of the tangent line is −2x0 so
its equation is y − (1 − x20 ) = −2x0 (x − x0 ), or y = −2x0 x + x20 + 1. The y-intercept is x20 + 1 and the
1
1
1
x-intercept is (x0 +1/x0 ) so the area A of the triangle is A = (x20 +1)(x0 +1/x0 ) = (x30 +2x0 +1/x0 )
2
4
4
for 0 ≤ x0 ≤ 1.
1
1
dA/dx0 = (3x20 + 2 − 1/x20 ) = (3x40 + 2x20 − 1)/x20 which is 0 when x20 = −1 (reject), or
4
4
√
√
1
2
2
when x0 = 1/3 so x0 = 1/ 3. d A/dx20 = (6x0 + 2/x30 ) > 0 at x0 = 1/ 3 so a relative minimum and
4
hence the absolute minimum occurs there.
55.
At each point (x, y) on the curve the slope of the tangent line is m =
dy
2x
=−
for any x,
dx
(1 + x2 )2
√
2(3x2 − 1) dm
dm
=
=
0
when
x
=
±1/
,
3, by the first derivative test the only relative maximum
dx
(1 + x2 )3 √
dx
occurs at x = −1/ 3, which is the absolute maximum because lim m = 0. The tangent line has
x→±∞
√
greatest slope at the point (−1/ 3, 3/4).
56.
Let x be how far P is upstream from where the man starts
(see figure), then the total time to reach T is
t= √
(time from M to P ) + (time from P to T )
x2 + 1 1 − x
+
for 0 ≤ x ≤ 1,
=
rR
rW
where rR and rW are the rates at which he can row and walk,
respectively.
1
x
P
T
1
M
√
p
x
dt
1
x2 + 1 1 − x dt
+
,
= √
= 0 when 5x = 3 x2 + 1,
− so
3
5
dx
dx
3 x2 + 1 5
√
2
2
2
25x = 9(x + 1), x = 9/16, x = 3/4. If x = 0, 3/4, 1 then t = 8/15, 7/15, 2/3 so the time is a
minimum when x = 3/4 mile.
√
x2 + 1 1 − x dt
x
dt
1
+
,
= √
= 0 when x = 4/3 which is not in the interval
− so
(b) t =
2
4
5
dx
5
dx
4 x +1
[0, 1]. Check the endpoints to find that the time is a minimum when x = 1 (he should row
directly to the town).
(a)
57.
t=
With x and y as shown in the figure, the maximum length of pipe
will be the smallest value of L = x + y. By similar triangles
x
8x
y
=√
,y=√
so
8
x2 − 16
x2 − 16
dL
128
dL
8x
,
=1− 2
= 0 when
for x > 4,
L=x+ √
3/2
2
dx
dx
(x − 16)
x − 16
(x2 − 16)3/2 = 128
x2 − 16 = 1282/3 = 16(22/3 )
x2 = 16(1 + 22/3 )
x = 4(1 + 22/3 )1/2 ,
y
x
x2 − 16
4
d2 L/dx2 = 384x/(x2 − 16)5/2 > 0 if x > 4 so L is smallest when x = 4(1 + 22/3 )1/2 .
For this value of x, L = 4(1 + 22/3 )3/2 ft.
8
Exercise Set 6.2
58.
192
s = (x1 − x̄)2 + (x2 − x̄)2 + · · · + (xn − x̄)2 ,
ds/dx̄ = −2(x1 − x̄) − 2(x2 − x̄) − · · · − 2(xn − x̄),
ds/dx̄ = 0 when
(x1 − x̄) + (x2 − x̄) + · · · + (xn − x̄) = 0
(x1 + x2 + · · · xn ) − (x̄ + x̄ + · · · + x̄) = 0
(x1 + x2 + · · · + xn ) − nx̄ = 0
1
x̄ = (x1 + x2 + · · · + xn ),
n
d2 s/dx̄2 = 2 + 2 + · · · + 2 = 2n > 0, so s is minimum when x̄ =
1
(x1 + x2 + · · · + xn ).
n
59. Let x = distance from the weaker light source, I = the intensity at that point, and k the constant of
proportionality. Then
8kS
kS
if 0 < x < 90;
I= 2 +
x
(90 − x)2
dI
2kS
16kS
2kS[8x3 − (90 − x)3 ]
kS(x − 30)(x2 + 2700)
=− 3 +
=
=
18
,
dx
x
(90 − x)3
x3 (90 − x)3
x3 (x − 90)3
dI
dI
which is 0 when x = 30;
< 0 if x < 30, and
> 0 if x > 30, so the intensity is minimum at a
dx
dx
distance of 30 cm from the weaker source.
60. p
If f (x0 ) ispa maximum then f (x) ≤ f (x0 ) for all x in some
p open interval containingpx0 thus
√
f (x) ≤
f (x0 ) because x is an increasing function, so f (x0 ) is a maximum of f (x) at
x0 . The proof is similar for a minimum value, simply replace ≤ by ≥.
61.
Let v = speed of light in the medium. The total time required for the light to travel from A to P to
B is
p
1 p
t = (total distance from A to P to B)/v = ( (c − x)2 + a2 + x2 + b2 ),
v
"
#
1
c−x
x
dt
=
−p
+√
2
2
2
dx
v
x + b2
(c − x) + a
√
dt
x
c−x
and
= 0 when √
=p
. But x/ x2 + b2 = sin θ2 and
2
2
2
2
dx
x +b
(c − x) + a
p
2
2
(c − x)/ (c − x) + a = sin θ1 thus dt/dx = 0 when sin θ2 = sin θ1 so θ2 = θ1 .
62.
The total time required for the light to travel from A to P to B is
p
√
(c − x)2 + b2
x2 + a2
+
,
t = (time from A to P ) + (time from P to B) =
v1
v2
√
dt
x
c−x
= √
− p
but x/ x2 + a2 = sin θ1 and
dx
v1 x2 + a2
v2 (c − x)2 + b2
p
sin θ1
sin θ1
dt
sin θ2
dt
sin θ2
(c − x)/ (c − x)2 + b2 = sin θ2 thus
=
= 0 when
−
so
=
.
dx
v1
v2
dx
v1
v2
63.
(a)
The rate at which the farmer walks is analogous to the speed of light in Fermat’s principle.
(b)
the best path occurs when θ1 = θ2
(see figure).
Barn
House
θ2
θ1
3
4
1
4
x
1−x
(c)
by similar triangles,
x/(1/4) = (1 − x)/(3/4)
3x = 1 − x
4x = 1
x = 1/4 mi.
193
Chapter 6
EXERCISE SET 6.3
1.
(a) positive, negative, slowing down
(c) negative, positive, slowing down
(b) positive, positive, speeding up
2.
(a) positive, slowing down
(c) positive, speeding up
(b)
3.
(a)
left because v = ds/dt < 0 at t0
(b)
negative because a = d2 s/dt2 and the curve is concave down at t0 (d2 s/dt2 < 0)
(c)
speeding up because v and a have the same sign
(d)
v < 0 and a > 0 at t1 so the particle is slowing down because v and a have opposite signs.
(a)
C
4.
5.
(b)
negative, slowing down
A
(c)
B
s(m)
t(s)
6.
(a) when s ≥ 0, so 0 < t < 2 and 4 < t ≤ 8
(c) when s is decreasing, so 0 ≤ t < 3
7.
s
(b) when the slope is zero, at t = 3
a
25
–15
t
6
t
–15
6
8.
(a) v ≈ (30 − 10)/(15 − 10) = 20/5 = 4 m/s
v
(b)
a
t
t
25
(1)
9.
25
(2)
(a)
At 60 mi/h the slope of the estimated tangent line is about 4.6 mi/h/s. Use 1 mi = 5, 280 ft
and 1 h = 3600 s to get a = dv/dt ≈ 4.6(5,280)/(3600) ≈ 6.7 ft/s2 .
(b)
The slope of the tangent to the curve is maximum at t = 0 s.
Exercise Set 6.3
10.
11.
12.
13.
14.
(a)
194
t
1
2
3
4
5
s 0.71
1.00
0.71
0.00 −0.71
v 0.56
0.00 −0.56 −0.79 −0.56
a −0.44 −0.62 −0.44 0.00
0.44
(b)
to the right at t = 1, stopped at t = 2, otherwise to the left
(c)
speeding up at t = 3; slowing down at t = 1, 5; neither at t = 2, 4
(a)
v(t) = 3t2 − 12t, a(t) = 6t − 12
(b)
s(1) = −5 ft, v(1) = −9 ft/s, speed = 9 ft/s, a(1) = −6 ft/s2
(c)
v = 0 at t = 0, 4
(d)
for t ≥ 0, v(t) changes sign at t = 4, and a(t) changes sign at t = 2; so the particle is speeding
up for 0 < t < 2 and 4 < t and is slowing down for 2 < t < 4
(e)
total distance = |s(4) − s(0)| + |s(5) − s(4)| = | − 32 − 0| + | − 25 − (−32)| = 39 ft
(a)
v(t) = 4t3 − 4, a(t) = 12t2
(b)
s(1) = −1 ft, v(1) = 0 ft/s, speed = 0 ft/s, a(1) = 12 ft/s2
(c)
v = 0 at t = 1
(d)
speeding up for t > 1, slowing down for 0 < t < 1
(e)
total distance = |s(1) − s(0)| + |s(5) − s(1)| = | − 1 − 2| + |607 − (−1)| = 611 ft
(a)
v(t) = −(3π/2) sin(πt/2), a(t) = −(3π 2 /4) cos(πt/2)
(b)
s(1) = 0 ft, v(1) = −3π/2 ft/s, speed = 3π/2 ft/s, a(1) = 0 ft/s2
(c)
v = 0 at t = 0, 2, 4
(d)
v changes sign at t = 0, 2, 4 and a changes sign at t = 1, 3, 5, so the particle is speeding up for
0 < t < 1, 2 < t < 3 and 4 < t < 5, and it is slowing down for 1 < t < 2 and 3 < t < 4
(e)
total distance = |s(2) − s(0)| + |s(4) − s(2)| + |s(5) − s(4)|
= | − 3 − 3| + |3 − (−3)| + |0 − 3| = 15 ft
(a)
v(t) =
(b)
s(1) = 1/5 ft, v(1) = 3/25 ft/s, speed = 3/25 ft/s, a(1) = −22/125 ft/s2
(c)
v = 0 at t = 2
2t(t2 − 12)
4 − t2
,
a(t)
=
(t2 + 4)2
(t2 + 4)3
√
√
a changes sign at t = 2√ 3, so the particle is speeding up for 2 < t < 2 3 and it is slowing down
for 0 < t < 2 and for 2 3 < t
¯
¯ ¯
¯
¯1
¯ ¯5
1 ¯ 19
(e) total distance = |s(2) − s(0)| + |s(5) − s(2)| = ¯¯ − 0¯¯ + ¯¯ − ¯¯ =
ft
4
29 4
58
(d)
15. v(t) =
2t(t2 − 15)
5 − t2
,
a(t)
=
(t2 + 5)2
(t2 + 5)3
0.2
0.25
0.01
0
0
0
10
20
20
0
s(t)
-0.05
v(t)
-0.15
a(t)
195
Chapter 6
(a)
(c)
16.
√
5
√
√
5/10 at t = 5
√
√
√
a changes sign
at
t
=
15,
so
the
particle
is
speeding
up
for
5
<
t
<
15 and slowing down
√
√
for 0 < t < 5 and 15 < t
v = 0 at t =
(b)
s=
v(t) = (1 − t)e−t , a(t) = (t − 2)e−t
1
0.4
0.1
0
0
0
18.
19.
8
10
0
17.
-0.2
s(t)
(b)
v = 0 at t = 1
(c)
a changes sign at t = 2, so the particle is speeding up for 1 < t < 2 and slowing down for
0 < t < 1 and 2 < t
s = −3t + 2
v = −3
a=0
Constant speed
Speeding up
1
½
a=
3
Slowing down
s
5
Speeding up
Slowing down
(Stopped) t = 4
t=3
t=0
9
t+1
(t + 4)(t − 2)
v=
(t + 1)2
18
a=
(t + 1)3
Slowing down
t = 2 (Stopped)
16 18 20
s
Speeding up
(Stopped) t = 2
5
0 ≤ t ≤ 2π
t > 2π
cos t,
1,
t = 1 (Stopped)
t=0
s=t+
½
s
t=0
t=2
(Stopped) t = 3
s = t3 − 9t2 + 24t
v = 3(t − 2)(t − 4)
a = 6(t − 3)
s=
s = 1/e at t = 1
2
s = t3 − 6t2 + 9t + 1
v = 3(t − 1)(t − 3)
a = 6(t − 2)
v=
22.
a(t)
(a)
½
21.
-2
v(t)
0
20.
6
− sin t,
0,
0 ≤ t ≤ 2π
t > 2π
− cos t,
0,
0 ≤ t < 2π
t > 2π
Slowing down
Slowing down
t=π
t = 3π/2
t = π/2
-1
0
9
s
(Stopped
permanently)
t = 2π
t=0
1
Speeding up
s
√
15t2 − 6t − 2
5t2 − 6t + 2
3 + 39
√
is always positive, a(t) =
v(t) =
has a positive root at t =
15
2t3/2
t
Slowing down
0
3 + 39
15
Speeding up
s
Exercise Set 6.3
23.
(a)
196
v = 10t − 22, speed = |v| = |10t − 22|. d|v|/dt does not exist at t = 2.2 which is the only critical
point. If t = 1, 2.2, 3 then |v| = 12, 0, 8. The maximum speed is 12 ft/s.
(b) the distance from the origin is |s| = |5t2 − 22t| = |t(5t − 22)|, but t(5t − 22) < 0 for
1 ≤ t ≤ 3 so |s| = −(5t2 − 22t) = 22t − 5t2 , d|s|/dt = 22 − 10t, thus the only critical point is
t = 2.2. d2 |s|/dt2 < 0 so the particle is farthest from the origin when t = 2.2. Its position is
s = 5(2.2)2 − 22(2.2) = −24.2.
24.
200t
d|v|
600(4 − t2 )
200t
,
speed
=
|v|
=
for
t
≥
0.
= 0 when t = 2, which is the
=
(t2 + 12)2
(t2 + 12)2
dt
(t2 + 12)3
only critical point in (0, +∞). By the first derivative test there is a relative maximum, and hence an
absolute maximum, at t = 2. The maximum speed is 25/16 ft/s to the left.
v=−
25. s(t) = s0 − 12 gt2 = s0 − 4.9t2 m, v = −9.8t m/s, a = −9.8 m/s2
(a)
|s(1.5) − s(0)| = 11.025 m
(b)
v(1.5) = −14.7 m/s
(c)
|v(t)| = 12 when t = 12/9.8 = 1.2245 s
(d) s(t) − s0 = −100 when 4.9t2 = 100, t = 4.5175 s
r
26.
27.
= 800 − 16t ft, s(t) = 0 when t =
√
800
=5 2
16
(a)
s(t) = s0 −
(b)
√
√
v(t) = −32t and v(5 2) = −160 2 ≈ 226.27 ft/s = 154.28 mi/h
1 2
2 gt
2
s(t) = s0 + v0 t − 12 gt2 = 60t − 4.9t2 m and v(t) = v0 − gt = 60 − 9.8t m/s
(a)
v(t) = 0 when t = 60/9.8 ≈ 6.12 s
(b)
s(60/9.8) ≈ 183.67 m
(c)
another 6.12 s; solve for t in s(t) = 0 to get this result, or use the symmetry of the parabola
s = 60t − 4.9t2 about the line t = 6.12 in the t-s plane
(d) also 60 m/s, as seen from the symmetry of the parabola (or compute v(6.12))
28.
(a)
they are the same
(b) s(t) = v0 t − 12 gt2 and v(t) = v0 − gt; s(t) = 0 when t = 0, 2v0 /g;
v(0) = v0 and v(2v0 /g) = v0 − g(2v0 /g) = −v0 so the speed is the same
at launch (t = 0) and at return (t = 2v0 /g).
29.
If g = 32 ft/s2 , s0 = 7 and v0 is unknown, then s(t) = 7 + v0 t − 16t2 and v(t) = v0 − 32t; s = smax when
v = 0, or√
t = v0 /32; and smax = 208 yields 208 = s(v0 /32) = 7 + v0 (v0 /32) − 16(v0 /32)2 = 7 + v02 /64,
so v0 = 8 201 ≈ 113.42 ft/s.
30.
(a)
Use (6) and then (5) to get v 2 = v02 − 2v0 gt + g 2 t2 = v02 − 2g(v0 t − 12 gt2 ) = v02 − 2g(s − s0 ).
(b)
Add v0 to both sides of (6): 2v0 − gt = v0 + v, v0 − 12 gt = 12 (v0 + v);
from (5) s = s0 + t(v0 − 12 gt) = s0 + 12 (v0 + v)t
(c)
Add v to both sides of (6): 2v+gt = v0 +v, v+ 12 gt = 12 (v0 +v); from part (b), s = s0 + 12 (v0 +v)t =
s0 + vt + 12 gt2
31.
v0 = 0 and g = 9.8, so v 2 = −19.6(s − s0 ); since v = 24 when s = 0 it follows that 19.6s0 = 242 or
s0 = 29.39 m.
32.
s = 1000 + vt + 12 (32)t2 = 1000 + vt + 16t2 ; s = 0 when t = 5, so v = −(1000 + 16 · 52 )/5 = −280 ft/s.
197
33.
34.
Chapter 6
(a)
s = smax when v = 0, so 0 = v02 − 2g(smax − s0 ), smax = v02 /2g + s0 .
(b)
s0 = 7,√smax = 208, g = 32 and v0 is unknown, so from part (a) v02 = 2g(208 − 7) = 64 · 201,
v0 = 8 201 ≈ 113.42 ft/s.
s = t3 − 6t2 + 1, v = 3t2 − 12t, a = 6t − 12.
(a)
a = 0 when t = 2; s = −15, v = −12.
(b) v = 0 when 3t2 − 12t = 3t(t − 4) = 0, t = 0 or t = 4. If t = 0, then s = 1 and a = −12; if t = 4,
then s = −31 and a = 12.
35.
(a)
(b)
1.5
0
v=√
√
2t
2
, lim v = √ = 2
2
2t2 + 1 t→+∞
5
0
36.
(a)
(b)
37.
(a)
(b)
(c)
38.
39.
40.
dv ds
dv
ds
dv
=
=v
because v =
dt
ds dt
ds
dt
3 dv
3
3
9
= ;
v= √
= − 2 ; a = − 3 = −9/500
2s ds
2s
4s
2 3t + 7
a=
1 2
1
3
t − t + 3 = − t2 + t + 1, t2 − 2t + 2 = 0 which has no real solution.
2
4
4
¯3 2
¯
3
Find the minimum value of D = |s1 − s2 | = ¯ 4 t − 2t + 2¯. From part (a), t2 − 2t + 2
4
3 2
is never zero, and for t = 0 it is positive, hence it is always positive, so D = t − 2t + 2.
4
4
3
4 d2 D
2
dD
> 0 so D is minimum when t = , D = .
= t − 2 = 0 when t = .
dt
2
3 dt2
3
3
1
v1 = t − 1, v2 = − t + 1. v1 < 0 if 0 ≤ t < 1, v1 > 0 if t > 1; v2 < 0 if t > 2, v2 > 0 if 0 ≤ t < 2.
2
They are moving in opposite directions during the intervals 0 ≤ t < 1 and t > 2.
s1 = s2 if they collide, so
(a)
sA − sB = 20 − 0 = 20 ft
(b)
sA = sB , 15t2 + 10t + 20 = 5t2 + 40t, 10t2 − 30t + 20 = 0, (t − 2)(t − 1) = 0, t = 1 or t = 2 s.
(c)
vA = vB , 30t + 10 = 10t + 40, 20t = 30, t = 3/2 s. When t = 3/2, sA = 275/4 and sB = 285/4
so car B is ahead of car A.
(a)
From the estimated tangent to the graph at the point where v = 2000, dv/ds ≈ −1.25 ft/s/ft.
(b)
a = v dv/ds ≈ (2000)(−1.25) = −2500 ft/s2
p
r0 (t) = 2v(t)v 0 (t)/[2 v 2 (t)] = v(t)a(t)/|v(t)| so r0 (t) > 0 (speed is increasing) if v and a have the same
sign, and r0 (t) < 0 (speed is decreasing) if v and a have opposite signs.
EXERCISE SET 6.4
1.
x2n − 2
2xn
x1 = 1, x2 = 1.5, x3 = 1.416666667, · · ·, x5 = x6 = 1.414213562
f (x) = x2 − 2, f 0 (x) = 2x, xn+1 = xn −
Exercise Set 6.4
198
x2n − 7
2xn
x1 = 3, x2 = 2.666666667, x3 = 2.645833333, · · ·, x5 = x6 = 2.645751311
2.
f (x) = x2 − 7, f 0 (x) = 2x, xn+1 = xn −
3.
f (x) = x3 − 6, f 0 (x) = 3x2 , xn+1 = xn −
4.
xn − a = 0
5.
f (x) = x3 − x + 3, f 0 (x) = 3x2 − 1, xn+1 = xn −
x3n − 6
3x2n
x1 = 2, x2 = 1.833333333, x3 = 1.817263545, · · ·, x5 = x6 = 1.817120593
x3n − xn + 3
3x2n − 1
x1 = −2, x2 = −1.727272727, x3 = −1.673691174, · · · , x5 = x6 = −1.671699882
x3n + xn − 1
3x2n + 1
x1 = 1, x2 = 0.75, x3 = 0.686046512, · · ·, x5 = x6 = 0.682327804
6. f (x) = x3 + x − 1, f 0 (x) = 3x2 + 1, xn+1 = xn −
x5n + x4n − 5
5x4n + 4x3n
x1 = 1, x2 = 1.333333333, x3 = 1.239420573, · · · , x6 = x7 = 1.224439550
7. f (x) = x5 + x4 − 5, f 0 (x) = 5x4 + 4x3 , xn+1 = xn −
8.
9.
x5n − xn + 1
5x4n − 1
x1 = −1, x2 = −1.25, x3 = −1.178459394, · · · , x6 = x7 = −1.167303978
f (x) = x5 − x + 1, f 0 (x) = 5x4 − 1, xn+1 = xn −
f (x) = x4 + x − 3, f 0 (x) = 4x3 + 1, xn+1 = xn −
x1 = −2, x2 = −1.645161290,
x3 = −1.485723955, · · · , x6 = x7 = −1.452626879
x4n + xn − 3
4x3n + 1
15
-2
2
-6
10.
x5n − 5x3n − 2
5x4n − 15x2n
x1 = 2, x2 = 2.5, x3 = 2.327384615, · · · , x7 = x8 = 2.273791732
f (x) = x5 − 5x3 − 2, f 0 (x) = 5x4 − 15x2 , xn+1 = xn −
10
-2.5
2.5
-20
11.
2 sin xn − xn
2 cos xn − 1
x1 = 2, x2 = 1.900995594, x3 = 1.895511645, x4 = x5 = 1.895494267
f (x) = 2 sin x − x, f 0 (x) = 2 cos x − 1, xn+1 = xn −
1
0
–7
6
199
12.
Chapter 6
f (x) = sin x − x2 , f 0 (x) = cos x − 2x, xn+1 = xn −
x1 = 1, x2 = 0.891395995,
x3 = 0.876984845, · · · , x5 = x6 = 0.876726215
sin xn − x2n
cos xn − 2xn
0.3
0
1.5
–1.3
13.
f (x) = x − tan x, f 0 (x) = 1 − sec2 x = − tan2 x,
xn − tan xn
xn+1 = xn +
tan2 xn
x1 = 4.5, x2 = 4.493613903, x3 = 4.493409655,
x4 = x5 = 4.493409458
100
6
i
–100
14.
f (x) = 1 − ex cos x, f 0 (x) = ex (sin x − cos x),
1 − ex cos x
xn+1 = xn + x
e (sin x − cos x)
x1 = 1, x2 = 1.572512605, x3 = 1.363631415, x7 = x8 = 1.292695719
25
c
0
-5
15.
16.
At the point of intersection, x3 = 0.5x − 1, x3 − 0.5x + 1 = 0. Let
f (x) = x3 − 0.5x + 1. By graphing y = x3 and y = 0.5x − 1 it is
evident that there is only one point of intersection and it occurs in
the interval [−2, −1]; note that f (−2) < 0 and f (−1) > 0.
f 0 (x) = 3x2 − 0.5 so
x3 − 0.5x + 1
; x1 = −1, x2 = −1.2,
xn+1 = xn − n 2
3xn − 0.5
x3 = −1.166492147, · · ·,
x5 = x6 = −1.165373043
The graphs of y = e−x and y = ln x intersect near x = 1.3; let
f (x) = e−x − ln x, f 0 (x) = −e−x − 1/x, x1 = 1.3,
e−xn − ln xn
, x2 = 1.309759929, x4 = x5 = 1.309799586
xn+1 = xn + −x
e n + 1/xn
2
-2
2
-2
1
0
2
-4
17.
√
The graphs of y = x2 and y√= 2x + 1 intersect at points near
x = −0.5 and x = 1; x2 = 2x + 1, x4 − 2x − 1 = 0. Let
f (x) = x4 − 2x − 1, then f 0 (x) = 4x3 − 2 so
x4 − 2xn − 1
.
xn+1 = xn − n 3
4xn − 2
If x1 = −0.5, then x2 = −0.475, x3 = −0.474626695,
x4 = x5 = −0.474626618; if x1 = 1, then x2 = 2,
x3 = 1.633333333, · · · , x8 = x9 = 1.395336994.
4
-0.5
2
0
Exercise Set 6.4
18.
200
The graphs of y = x3 /8 + 1 and y = cos 2x intersect at x = 0 and at
a point near x = −2;
x3 /8 + 1 = cos 2x, x3 − 8 cos 2x + 8 = 0. Let f (x) = x3 − 8 cos 2x + 8,
x3 − 8 cos 2xn + 8
then f 0 (x) = 3x2 + 16 sin 2x so xn+1 = xn − n 2
.
3xn + 16 sin 2xn
x1 = −2, x2 = −2.216897577,
x3 = −2.193821581, · · · , x5 = x6 = −2.193618950.
2
-3
2
-2
19.
20.
x2n − a
1
=
2xn
2
µ
a
xn
¶
(a)
f (x) = x2 − a, f 0 (x) = 2x, xn+1 = xn −
(b)
a = 10; x1 = 3, x2 = 3.166666667, x3 = 3.162280702, x4 = x5 = 3.162277660
(a)
f (x) =
(b)
xn +
1
1
− a, f 0 (x) = − 2 , xn+1 = xn (2 − axn )
x
x
a = 17; x1 = 0.05, x2 = 0.0575, x3 = 0.058793750, x5 = x6 = 0.058823529
21.
f 0 (x) = x3 + 2x + 5; solve f 0 (x) = 0 to find the critical points. Graph y = x3 and y = −2x − 5 to see
x3 + 2xn + 5
that they intersect at a point near x = −1; f 00 (x) = 3x2 + 2 so xn+1 = xn − n 2
.
3xn + 2
x1 = −1, x2 = −1.4, x3 = −1.330964467, · · · , x5 = x6 = −1.328268856 so the minimum value of f (x)
occurs at x ≈ −1.328268856 because f 00 (x) > 0; its value is approximately −4.098859132.
22.
From a rough sketch of y = x sin x we see that the maximum occurs at a point near x = 2, which will
be a point where f 0 (x) = x cos x + sin x = 0. f 00 (x) = 2 cos x − x sin x so
xn cos xn + sin xn
xn + tan xn
xn+1 = xn −
= xn −
.
2 cos xn − xn sin xn
2 − xn tan xn
x1 = 2, x2 = 2.029048281, x3 = 2.028757866, x4 = x5 = 2.028757838; the maximum value is approximately 1.819705741.
23.
Let f (x) be the square of the distance between (1, 0) and any point (x, x2 ) on the parabola, then
f (x) = (x − 1)2 + (x2 − 0)2 = x4 + x2 − 2x + 1 and f 0 (x) = 4x3 + 2x − 2. Solve f 0 (x) = 0 to find
4x3 + 2xn − 2
2x3 + xn − 1
the critical points; f 00 (x) = 12x2 + 2 so xn+1 = xn − n 2
= xn − n 2
. x1 = 1,
12xn + 2
6xn + 1
x2 = 0.714285714, x3 = 0.605168701, · · · , x6 = x7 = 0.589754512; the coordinates are approximately
(0.589754512, 0.347810385).
24.
The area is A = xy = x cos x so dA/dx = cos x − x sin x. Find x so that dA/dx = 0;
cos xn − xn sin xn
1 − xn tan xn
= xn +
.
d2 A/dx2 = −2 sin x − x cos x so xn+1 = xn +
2 sin xn + xn cos xn
2 tan xn + xn
x1 = 1, x2 = 0.864536397, x3 = 0.860339078, x4 = x5 = 0.860333589; y ≈ 0.652184624.
25.
(a)
(b)
26.
Let s be the arc length, and L the length of the chord, then s = 1.5L. But s = rθ and
L = 2r sin(θ/2) so rθ = 3r sin(θ/2), θ − 3 sin(θ/2) = 0.
θn − 3 sin(θn /2)
.
1 − 1.5 cos(θn /2)
θ1 = 3, θ2 = 2.991592920, θ3 = 2.991563137, θ4 = θ5 = 2.991563136 rad so θ ≈ 171◦ .
Let f (θ) = θ − 3 sin(θ/2), then f 0 (θ) = 1 − 1.5 cos(θ/2) so θn+1 = θn −
r2 (θ − sin θ)/2 = πr2 /4 so θ − sin θ − π/2 = 0. Let f (θ) = θ − sin θ − π/2, then f 0 (θ) = 1 − cos θ so
θn − sin θn − π/2
.
θn+1 =
1 − cos θn
θ1 = 2, θ2 = 2.339014106, θ3 = 2.310063197, · · · , θ5 = θ6 = 2.309881460 rad; θ ≈ 132◦ .
201
Chapter 6
27.
If x = 1, then y 4 + y = 1, y 4 + y − 1 = 0. Graph z = y 4 and z = 1 − y to see that they intersect near
y 4 + yn − 1
.
y = −1 and y = 1. Let f (y) = y 4 + y − 1, then f 0 (y) = 4y 3 + 1 so yn+1 = yn − n 3
4yn + 1
If y1 = −1, then y2 = −1.333333333, y3 = −1.235807860, · · · , y6 = y7 = −1.220744085;
if y1 = 1, then y2 = 0.8, y3 = 0.731233596, · · · , y6 = y7 = 0.724491959.
28.
If x = 1, then 2y − cos y = 0. Graph z = 2y and z = cos y to see that they intersect near y = 0.5. Let
2yn − cos yn
f (y) = 2y − cos y, then f 0 (y) = 2 + sin y so yn+1 = yn −
.
2 + sin yn
y1 = 0.5, y2 = 0.450626693, y3 = 0.450183648, y4 = y5 = 0.450183611.
29.
30.
¤
5000 £
(1 + i)25 − 1 ; set f (i) = 50i − (1 + i)25 + 1, f 0 (i) = 50 − 25(1 + i)24 ; solve
i
£
¤ £
¤
f (i) = 0. Set i0 = .06 and ik+1 = ik − 50i − (1 + i)25 + 1 / 50 − 25(1 + i)24 . Then i1 = 0.05430,
i2 = 0.05338, i3 = 0.05336, · · ·, i = 0.053362.
S(25) = 250000 =
0.5
(a) x1 = 2, x2 = 5.3333,
x3 = 11.055, x4 = 22.293,
x5 = 44.676
0
15
0
(b)
31.
(a)
(b)
x1 = 0.5, x2 = −0.3333, x3 = 0.0833, x4 = −0.0012, x5 = 0.0000 (and xn = 0 for n ≥ 6)
x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
0.5000 −0.7500 0.2917 −1.5685 −0.4654 0.8415 −0.1734 2.7970 1.2197 0.1999
The sequence xn must diverge, since if it did converge then f (x) = x2 + 1 = 0 would have a
solution. It seems the xn are oscillating back and forth in a quasi-cyclical fashion.
EXERCISE SET 6.5
1.
2. f (−3) = f (3) = 0; f 0 (0) = 0
f (0) = f (4) = 0; f 0 (3) = 0; [0, 4], c = 3
3. f (2) = f (4) = 0, f 0 (x) = 2x − 6, 2c − 6 = 0, c = 3
4.
0
f (0) = f (2) = 0, f (x) = 3x − 6x + 2, 3c − 6c + 2 = 0; c =
2
2
6±
√
√
36 − 24
= 1 ± 3/3
6
5. f (π/2) = f (3π/2) = 0, f 0 (x) = − sin x, − sin c = 0, c = π
x2 − 4x + 1 c2 − 4c + 1
,
= 0, c2 − 4c + 1 = 0
(x − 2)2
(c − 2)2
√
√
√
4 ± 16 − 4
= 2 ± 3, of which only c = 2 − 3 is in (−1, 1)
c=
2
6. f (−1) = f (1) = 0, f 0 (x) =
7. f (0) = f (4) = 0, f 0 (x) =
8.
1
1
1
1
− √ , − √ = 0, c = 1
2 2 x 2 2 c
f (1) = f (3) = 0, f 0 (x) = −
2
4
2
4
+ 2 , − 3 + 2 = 0, −6 + 4c = 0, c = 3/2
3
x
3x
c
3c
Exercise Set 6.5
9.
202
6
3
f (8) − f (0)
= = = f 0 (1.54); c = 1.54
8−0
8
4
10.
f (4) − f (0)
= 1.19 = f 0 (0.77)
4−0
42 − 12
= 3, c = 1
6 − (−4)
11.
f (−4) = 12, f (6) = 42, f 0 (x) = 2x + 1, 2c + 1 =
12.
f (−1) = −6, f (2) = 6, f 0 (x) = 3x2 + 1, 3c2 + 1 =
6 − (−6)
= 4, c2 = 1, c = ±1 of which only c = 1 is
2 − (−1)
in (−1, 2)
13.
1
2−1
1 √
1
, √
=
= , c + 1 = 3/2, c + 1 = 9/4, c = 5/4
f (0) = 1, f (3) = 2, f 0 (x) = √
3−0
3
2 x+1 2 c+1
14. f (3) = 10/3, f (4) = 17/4, f 0 (x) = 1 − 1/x2 , 1 − 1/c2 =
√
which only c = 2 3 is in (3, 4)
15.
f (−5) = 0, f (3) = 4, f 0 (x) = − √
√
17/4 − 10/3
= 11/12, c2 = 12, c = ±2 3 of
4−3
√
1
x
c
4−0
= , −2c = 25 − c2 ,
, −√
=
2
2
3
−
(−5)
2
25 − x
25 − c
√
4c2 = 25 − c2 , c2 = 5, c = − 5
√
√
(we reject c = 5 because it does not satisfy the equation −2c = 25 − c2 )
16. f (2) = 1, f (5) = 1/4, f 0 (x) = −1/(x − 1)2 , −
c = −1 (reject), or c = 3
17.
(a)
1
1
1/4 − 1
= − , (c − 1)2 = 4, c − 1 = ±2,
=
2
(c − 1)
5−2
4
(b) c = −1.29
f (−2) = f (1) = 0
6
-2
1
-2
(c)
18.
x0 = −1, x1 = −1.5, x2 = −1.32, x3 = −1.290, x4 = −1.2885843
f (−2) − f (1)
−16 − 5
=
= 7 so y − 5 = 7(x − 1),
−2 − 1
−3
y = 7x − 2
(a) m =
(b) f 0 (x) = 3x2 + 4 = 7 has solutions x = ±1;
discard x = 1, so c = −1
(d)
5
-2
1
(c) y − f (−1) = 7(x − (−1)) or y = 7x + 2
-20
19.
(a)
f 0 (x) = sec2 x, sec2 c = 0 has no solution
20.
(a)
f (−1) = 1, f (8) = 4, f 0 (x) =
(b) tan x is not continuous on [0, π]
2 −1/3
x
3
4−1
1
2 −1/3
=
c
= , c1/3 = 2, c = 8 which is not in (−1, 8).
3
8 − (−1)
3
(b)
x2/3 is not differentiable at x = 0, which is in (−1, 8).
203
21.
22.
Chapter 6
(a)
Two x-intercepts of f determine two solutions a and b of f (x) = 0; by Rolle’s Theorem there
exists a point c between a and b such that f 0 (c) = 0, i.e. c is an x-intercept for f 0 .
(b)
f (x) = sin x = 0 at x = nπ, and f 0 (x) = cos x = 0 at x = nπ + π/2, which lies between nπ and
(n + 1)π, (n = 0, ±1, ±2, . . .)
f (x1 ) − f (x0 )
is the average rate of change of y with respect to x on the interval [x0 , x1 ]. By the
x1 − x0
Mean-Value Theorem there is a value c in (x0 , x1 ) such that the instantaneous rate of change f 0 (c) =
f (x1 ) − f (x0 )
.
x1 − x0
23.
Let s(t) be the position function of the automobile for 0 ≤ t ≤ 5, then by the Mean-Value Theorem
there is at least one point c in (0, 5) where
s0 (c) = v(c) = [s(5) − s(0)]/(5 − 0) = 4/5 = 0.8 mi/min = 48 mi/h.
24.
Let T (t) denote the temperature at time with t = 0 denoting 11 AM, then T (0) = 76 and T (12) = 52.
(a)
By the Mean-Value Theorem there is a value c between 0 and 12 such that
T 0 (c) = [T (12) − T (0)]/(12 − 0) = (52 − 76)/(12) = −2◦ F/h.
(b)
Assume that T (t1 ) = 88◦ F where 0 < t1 < 12, then there is at least one point c in (t1 , 12) where
T 0 (c) = [T (12) − T (t1 )]/(12 − t1 ) = (52 − 88)/(12 − t1 ) = −36/(12 − t1 ). But 12 − t1 < 12 so
T 0 (c) < −36/12 = −3◦ F/h.
25.
Let f (t) and g(t) denote the distances from the first and second runners to the starting point, and let
h(t) = f (t) − g(t). Since they start (at t = 0) and finish (at t = t1 ) at the same time, h(0) = h(t1 ) = 0,
so by Rolle’s Theorem there is a time t2 for which h0 (t2 ) = 0, i.e. f 0 (t2 ) = g 0 (t2 ); so they have the
same velocity at time t2 .
26.
f (x) = x6 − 2x2 + x satisfies f (0) = f (1) = 0, so by Rolle’s Theorem f 0 (c) = 0 for some c in (0, 1).
27.
(a)
By the Constant Difference Theorem f (x) − g(x) = k for some k; since f (x0 ) = g(x0 ), k = 0, so
f (x) = g(x) for all x.
(b)
Set f (x) = sin2 x + cos2 x, g(x) = 1; then f 0 (x) = 2 sin x cos x − 2 cos x sin x = 0 = g 0 (x). Since
f (0) = 1 = g(0), f (x) = g(x) for all x.
(a)
By the Constant Difference Theorem f (x) − g(x) = k for some k; since f (x0 ) − g(x0 ) = c, k = c,
so f (x) − g(x) = c for all x.
(b)
Set f (x) = (x − 1)3 , g(x) = (x2 + 3)(x − 3). Then
f 0 (x) = 3(x − 1)2 , g 0 (x) = (x2 + 3) + 2x(x − 3) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2 ,
so f 0 (x) = g 0 (x) and hence f (x) − g(x) = k. Expand f (x) and g(x) to get
h(x) = f (x) − g(x) = (x3 − 3x2 + 3x − 1) − (x3 − 3x2 + 3x − 9) = 8.
(c)
h(x) = x3 − 3x2 + 3x − 1 − (x3 − 3x2 + 3x − 9) = 8
28.
29.
30.
f (y) − f (x)
= f 0 (c),
y−x
so |f (x) − f (y)| = |f 0 (c)||x − y| ≤ M |x − y|; if x > y exchange x and y; if x = y the inequality
also holds.
(b) f (x) = sin x, f 0 (x) = cos x, |f 0 (x)| ≤ 1 = M , so |f (x) − f (y)| ≤ |x − y| or | sin x − sin y| ≤ |x − y|.
(a)
If x, y belong to I and x < y then for some c in I,
(a)
If x, y belong to I and x < y then for some c in I,
f (y) − f (x)
= f 0 (c),
y−x
so |f (x) − f (y)| = |f 0 (c)||x − y| ≥ M |x − y|; if x > y exchange x and y; if
x = y the inequality also holds.
(b) If x and y belong to (−π/2, π/2) and f (x) = tan x, then |f 0 (x)| = sec2 x ≥ 1 and
| tan x − tan y| ≥ |x − y|
Exercise Set 6.5
(c)
31.
(a)
(b)
204
y lies in (−π/2, π/2) if and only if −y does; use part (b) and replace y with −y
√
Let f (x) = x. By the Mean-Value Theorem there is a number c between x and y such that
√
√
√
y− x
1
y−x
1
√
= √ < √ for c in (x, y), thus y − x < √
y−x
2 c
2 x
2 x
multiply through and rearrange to get
√
xy <
1
(x + y).
2
32.
Suppose that f (x) has at least two distinct real solutions r1 and r2 in I. Then
f (r1 ) = f (r2 ) = 0 so by Rolle’s Theorem there is at least one number between r1 and r2 where
f 0 (x) = 0, but this contradicts the assumption that f 0 (x) 6= 0, so f (x) = 0 must have fewer than two
distinct solutions in I.
33.
(a)
If f (x) = x3 + 4x − 1 then f 0 (x) = 3x2 + 4 is never zero, so by Exercise 32 f has at most one real
root; since f is a cubic polynomial it has at least one real root, so it has exactly one real root.
(b)
Let f (x) = ax3 + bx2 + cx + d. If f (x) = 0 has at least two distinct real solutions r1 and r2 , then
f (r1 ) = f (r2 ) = 0 and by Rolle’s Theorem there is at least one number between r1 and r2 where
f 0 (x) = 0. But f 0 (x) = 3ax2 + 2bx + c = 0 for
√
√
x = (−2b ± 4b2 − 12ac)/(6a) = (−b ± b2 − 3ac)/(3a), which are not real if b2 − 3ac < 0
so f (x) = 0 must have fewer than two distinct real solutions.
34.
35.
√
√
√
1
1
1
1
1
4− 3
= 2 − 3. But < √ < √ for c in (3, 4) so
f (x) = √ , √ =
4−3
4
2 x 2 c
2 c
2 3
√
√
√
√
1
1
< 2 − 3 < √ , 0.25 < 2 − 3 < 0.29, −1.75 < − 3 < −1.71, 1.71 < 3 < 1.75.
4
2 3
0
(a)
d 2
[f (x) + g 2 (x)] = 2f (x)f 0 (x) + 2g(x)g 0 (x) = 2f (x)g(x) + 2g(x)[−f (x)] = 0,
dx
so f 2 (x) + g 2 (x) is constant.
(b) f (x) = sin x and g(x) = cos x
36.
37.
(a)
d 2
[f (x) − g 2 (x)] = 2f (x)f 0 (x) − 2g(x)g 0 (x) = 2f (x)g(x) − 2g(x)f (x) = 0 so f 2 (x) − g 2 (x)
dx
is constant.
(b)
f 0 (x) = 12 (ex − e−x ) = g(x) and g 0 (x) = 12 (ex + e−x ) = f (x)
If f 0 (x) = g 0 (x), then f (x) = g(x) + k. Let x = 1,
f (1) = g(1) + k = (1)3 − 4(1) + 6 + k = 3 + k = 2, so k = −1. f (x) = x3 − 4x + 5.
38.
39.
Let h = f − g, then h is continuous on [a, b], differentiable on (a, b), and h(a) = f (a) − g(a) = 0,
h(b) = f (b) − g(b) = 0. By Rolle’s Theorem there is some c in (a, b) where h0 (c) = 0. But
h0 (c) = f 0 (c) − g 0 (c) so f 0 (c) − g 0 (c) = 0, f 0 (c) = g 0 (c).
y
c
x
205
40.
Chapter 6
(a)
Suppose f 0 (x) = 0 more than once in (a, b), say at c1 and c2 . Then f 0 (c1 ) = f 0 (c2 ) = 0 and
by using Rolle’s Theorem on f 0 , there is some c between c1 and c2 where f 00 (c) = 0, which
contradicts the fact that f 00 (x) > 0 so f 0 (x) = 0 at most once in (a, b).
(b)
If f 00 (x) > 0 for all x in (a, b), then f is concave up on (a, b) and has at most one relative
extremum, which would be a relative minimum, on (a, b).
41.
similar to the proof of part (a) with f 0 (c) < 0
42.
similar to the proof of part (a) with f 0 (c) = 0
CHAPTER 6 SUPPLEMENTARY EXERCISES
3.
(a)
If f has an absolute extremum at a point of (a, b) then it must, by Theorem 6.1.4, be at a critical
point of f ; since f is differentiable on (a, b) the critical point is a stationary point.
(b)
It could occur at a critical point which is not a stationary point: for example, f (x) = |x| on
[−1, 1] has an absolute minimum at x = 0 but is not differentiable there.
4.
No; speeding up means the velocity and acceleration have the same sign, i.e. av > 0; the velocity is
increasing when the acceleration is positive, i.e. a > 0. These are not the same thing. An example is
s = t − t2 at t = 1, where v = −1 and a = −2, so av > 0 but a < 0.
5.
Yes; by the Mean Value Theorem there is a point c in (a, b) such that f 0 (c) =
7.
(a)
f (b) − f (a)
= 0.
b−a
f 0 (x) = −1/x2 6= 0, no critical points; by inspection M = −1/2 at x = −2; m = −1 at x = −1
(b) f 0 (x) = 3x2 − 4x3 = 0 at x = 0, 3/4; f (−1) = −2, f (0) = 0, f (3/4) = 27/256, f (3/2) = −27/16,
so m = −2 at x = −1, M = 27/256 at x = 3/4
µ
¶1/3
2
x(7x − 12)
144
−
,
critical
points
at
x
=
12/7,
2;
m
=
f
(12/7)
=
≈ −1.9356 at
(c) f 0 (x) =
49
7
3(x − 2)2/3
x = 12/7, M = 9 at x = 3
8.
(d)
ex (x − 2)
, stationary point at x = 2; by Theorem
x→+∞
x→0
x3
6.1.5 f (x) has an absolute minimum at x = 2, and m = e2 /4.
(a)
√
3 − x2
, critical point at x = 3. Since lim+ f (x) = 0, f (x) has no minimum, and
f 0 (x) = 2 2
2
x→0
√ (x + 3) √
M = 3/3 at x = 3.
(b)
f 0 (x) = 10x3 (x−2), critical points at x = 0, 2; lim− f (x) = 88, so f (x) has no maximum; m = −9
(c)
at x = 2
critical point at x = 2; m = −3 at x = 3, M = 0 at x = 2
(d)
f 0 (x) = (1 + ln x)xx , critical point at x = 1/e; lim+ f (x) = lim+ ex ln x = 1, lim f (x) = +∞; no
lim+ f (x) = lim f (x) = +∞ and f 0 (x) =
x→3
x→0
x→0
absolute maximum, absolute minimum m = e−1/e at x = 1/e
9.
x = 2.3561945
10.
x→+∞
x = −2.11491, 0.25410, 1.86081
11.
(a) yes; f 0 (0) = 0
p
(c) yes, f 0 ( π/2) = 0
(b)
no, f is not differentiable on (−1, 1)
12.
(a)
(b)
yes,
no, f is not differentiable on (−2, 2)
√
f (3) − f (2)
= −1 = f 0 (1 + 2)
3−2
Supplementary Exercises 6
lim f (x) = 2, lim+ f (x) = 2 so f is continuous on [0, 2]; lim− f 0 (x) = lim− −2x = −2 and
x→1
x→1
x→1
√
f (2) − f (0)
0
2
= −1 = f 0 ( 2)
lim+ f (x) = lim+ (−2/x ) = −2, so f is differentiable on (0, 2); and
x→1
x→1
2−0
(c)
13.
206
x→1−
Let k be the amount of light admitted per unit area of clear glass. The total amount of light admitted
by the entire window is
1
1
T = k · (area of clear glass) + k · (area of blue glass) = 2krh + πkr2 .
2
4
But P = 2h + 2r + πr which gives 2h = P − 2r − πr so
h
³
1
π ´ 2i
T = kr(P − 2r − πr) + πkr2 = k P r − 2 + π −
r
4
4
¸
·
P
8 + 3π 2
r for 0 < r <
,
= k Pr −
4
2+π
µ
¶
8 + 3π
dT
2P
dT
=k P−
r ,
= 0 when r =
.
dr
2
dr
8 + 3π
This is the only critical point and d2 T /dr2 < 0 there so the most light is admitted when
r = 2P/(8 + 3π) ft.
14.
15.
p
If one corner of the rectangle is at (x, y) with x > 0, y > 0, then A = 4xy, y = 3 1 − (x/4)2 ,
p
√
√
8 − x2
dA
A = 12x 1 − (x/4)2 = 3x 16 − x2 ,
= 6√
, critical point at x = 2 2. Since A = 0 when
dx
16 − x2
√
x = 0, 4 and A > 0 otherwise, there is an absolute maximum A = 24 at x = 2 2.
(a) If a = k, a constant, then v = kt + b where b is constant; so
the velocity changes sign at t = −b/k.
v
t
- b/ k
b
(b) Consider the equation s = 5 − t3 /6, v = −t2 /2, a = −t. Then
for t > 0, a is decreasing and av > 0, so the particle is
speeding up.
16.
v
t
s(t) = t/(t2 + 5), v(t) = (5 − t2 )/(t2 + 5)2 , a(t) = 2t(t2 − 15)/(t2 + 5)3
(a) 0.25
0.01
0.2
0
0
0
20
20
0
(b)
10
-0.05
s(t)
v changes sign at t =
√
5
-0.15
v(t)
(c)
s=
a(t)
√
√
5/10, v = 0, a = − 5/50
207
Chapter 6
(d)
(e)
√
√
√
a changes sign at √
t = 15,
so
the
particle
is
speeding
up
for
5
<
t
<
15, and it is slowing
√
down for 0 < t < 5 and 15 < t
√
v(0) = 1/5, lim v(t) = 0, v(t) has one t-intercept at t = 5 and v(t) has one critical point at
t→+∞
√
the maximum velocity occurs when t = 0 and the minimum velocity
t = 15. Consequently
√
occurs when t = 15.
17.
s(t) = s0 + v0 t − 12 gt2 = v0 t − 4.9t2 , v(t) = v0 − 9.8t; smax occurs when v = 0, i.e. t = v0 /9.8,
√
and then 0.76 = smax = v0 (v0 /9.8) − 4.9(v0 /9.8)2 = v02 /19.6, so v0 = 0.76 · 19.6 = 3.86 m/s and
s(t) = 3.86t − 4.9t2 . Then s(t) = 0 when t = 0, 0.7878, s(t) = 0.15 when t = 0.0410, 0.7468, and
s(t) = 0.76−0.15 = 0.61 when t = 0.2188, 0.5689, so the player spends 0.5689−0.2188 = 0.3501 s
in the top 15.0 cm of the jump and 0.0410 + (0.7878 − 0.7468) = 0.0820 s in the bottom
15.0 cm.
(b) The height vs time plot is a parabola that opens down, and the slope is smallest near the top
of the parabola, so a given change ∆h in height corresponds to a large time change ∆t near the
top of the parabola and a narrower time change at points farther away from the top.
18.
(a)
(a)
s(t) = s0 + v0 − 4.9t2 ; assume s0 = v0 = 0, so s(t) = −4.9t2 , v(t) = −9.8t
t
s
v
19.
0
1
2
3
4
0 −4.9 −19.6 −44.1 −78.4
0 −9.8 −19.6 −29.4 −39.2
(b)
The formula for v is linear (with no constant term).
(c)
The formula for s is quadratic (with no linear or constant term).
(a)
(b)
2.1
-10
minimum: (−2.111985, −0.355116)
maximum: (0.372591, 2.012931)
10
-0.5
20.
(a)
y
0.2
0.6
1
x
-0.5
-1
-1.5
-2
(b)
21.
p
The distance between the boat and the origin is x2 + y 2 , where y = (x10/3 − 1)/(2x2/3 ). The
minimum distance is 0.8247 mi when x = 0.6598 mi. The boat gets swept downstream.
(c)
Use the equation of the path to obtain dy/dt = (dy/dx)(dx/dt), dx/dt = (dy/dt)/(dy/dx). Let
dy/dt = −4 and find the value of dy/dx for the value of x obtained in part (b) to get
dx/dt = −3 mi/h.
(a)
v = −2
3t8 + 10t6 − 12t4 − 6t2 + 1
t(t4 + 2t2 − 1)
,a=2
4
2
(t + 1)
(t4 + 1)3
Supplementary Exercises 6
(b)
208
s
a
v
1.2
2
0.4
0.2
1
0.8
0.6
0.4
0.2
1
2
3
4
5
6
t
-0.2
-0.4
-0.6
-0.8
-1
1
1
2
3
4
5
6
t
1
2
3
4
5
6
t
-1
-2
-3
(c)
It is farthest from the origin at approximately t = 0.64 (when v = 0) and s = 1.2
(d)
Find t so that the velocity v = ds/dt > 0. The particle is moving in the positive direction for
0 ≤ t ≤ 0.64 s.
(e)
It is speeding up when a, v > 0 or a, v < 0, so for 0 ≤ t < 0.36 and 0.64 < t < 1.1, otherwise it
is slowing down.
(f )
Find the maximum value of |v| to obtain: maximum speed = 1.05 m/s when t = 1.10 s.
22.
Find t so that N 0 (t) is maximum. The size of the population is increasing most rapidly when t = 8.4
years.
23.
Solve φ − 0.0167 sin φ = 2π(90)/365 to get φ = 1.565978 so
r = 150 × 106 (1 − 0.0167 cos φ) = 149.988 × 106 km.
24.
Solve φ − 0.0934 sin φ = 2π(1)/1.88 to get φ = 3.325078 so
r = 228 × 106 (1 − 0.0934 cos φ) = 248.938 × 106 km.
CHAPTER 7
Integration
EXERCISE SET 7.1
·
1.
A = 1(1)/2 = 1/2; ∆x = (b − a)/n = 1/n, x∗k = k/n, f (x∗k ) = k/n, An =
n
An
2.
1
2
3
4
5
6
7
8
9
10
0.0000 2.0000 2.6667 3.0000 3.2000 3.3333 3.4286 3.50000 3.5556 3.6000
∗
∗
A = 2(2
¶ = µ16; ∆x ¶= (b − a)/n
µ = 2/n,
¶¸ xk = 2k/n, f (xk ) = 2 + 12k/n = 2(1 + 6k/n),
·µ+ 14)/2
2
12
6n
6
An = 2 1 +
+ 1+
+ ··· + 1 +
n
n
n
n
n
An
4.
1
2
3
4
5
6
7
8
9
10
1.0000 0.7500 0.6666 0.6250 0.6000 0.5833 0.5714 0.5625 0.5556 0.5500
A = 4(2)/2
a)/n = 2/n, x∗k = ¸2k/n, f (x∗k ) = 4 − 2(2k/n) = 4 − 4k/n = 4(1 − k/n),
·µ = 4;¶∆x µ= (b − ¶
³
2
n´ 2
1
An = 4 1 −
+ 1−
+ ··· + 1 −
n
n
n
n
n
An
3.
¸
2
n 1
1
+ + ··· +
n n
n n
1
2
3
4
5
6
7
8
9
10
28.0000 22.0000 20.0000 19.0000 18.4000 18.0000 17.7143 17.5000 17.3333 17.2000
p
A = π(1)2 /4 = π/4; ∆x = (b − a)/n = 1/n, x∗k = k/n, f (x∗k ) = 1 − (k/n)2 ,
i1
hp
p
p
1 + (1/n)2 + 1 + (2/n)2 + · · · + 1 + (n/n)2
An =
n
n
1
2
3
4
5
6
7
8
9
10
An 0.0000 0.4330 0.5627 0.6239 0.6593 0.6822 0.6982 0.7100 0.7190 0.7261
5.
A(1) − A(0) = 1/2
6. A(2) − A(0) = 4
7.
A(2) − A(0) = 16
8. A(1) − A(0) =
9. A(x) = ex , area = A(1) − A(0) = e − 1
10.
1π
= π/4
22
A(x) = − cos x, area = A(π) − A(0) = 2
EXERCISE SET 7.2
Z
1.
(a)
2.
(a)
(b)
√
Z
p
x
dx = 1 + x2 + C
1 + x2
(b)
(x + 1)ex dx = xex + C
d
(sin x − x cos x + C) = cos x − cos x + x sin x = x sin x
dx
√
µ
¶ √
x
1
d
1 − x2 + x2 / 1 − x2
√
+C =
=
2
2
dx
1
−
x
(1
−
x2 )3/2
1−x
3.
i
d hp 3
3x2
x +5 = √
dx
2 x3 + 5
4.
·
¸
x
d
3 − x2
=
dx x2 + 3
(x2 + 3)2
Z
so
Z
so
p
3x2
√
dx = x3 + 5 + C
2 x3 + 5
3 − x2
x
dx = 2
+C
(x2 + 3)2
x +3
209
Exercise Set 7.2
210
5.
√
d £ ¡ √ ¢¤ cos (2 x)
√
sin 2 x =
dx
x
6.
d
[sin x − x cos x] = x sin x
dx
7.
(a)
8.
(a)
9.
(a)
Z
so
√
¡ √ ¢
cos (2 x)
√
dx = sin 2 x + C
x
Z
so
x9 /9 + C
3 5/3
x +C
5
Z
1
1
x−3 dx = − x−2 + C
2
4
x sin x dx = sin x − x cos x + C
(b)
7 12/7
x
+C
12
(c)
2 9/2
x +C
9
(b)
1
1
− x−5 + C = − 5 + C
5
5x
(c)
8x1/8 + C
(b) u4 /4 − u2 + 7u + C
3 5/3
x − 5x4/5 + 4x + C
5
Z
2
12
1
1
11.
(x−3 + x1/2 − 3x1/4 + x2 )dx = − x−2 + x3/2 − x5/4 + x3 + C
2
3
5
3
Z
3
8
12.
(7y −3/4 − y 1/3 + 4y 1/2 )dy = 28y 1/4 − y 4/3 + y 3/2 + C
4
3
Z
13.
(x + x4 )dx = x2 /2 + x5 /5 + C
10.
Z
4
1
(4 + 4y 2 + y 4 )dy = 4y + y 3 + y 5 + C
3
5
Z
Z
12
3
1/3
2
15.
x (4 − 4x + x )dx = (4x1/3 − 4x4/3 + x7/3 )dx = 3x4/3 − x7/3 + x10/3 + C
7
10
Z
1
2
1
16.
(2 − x + 2x2 − x3 )dx = 2x − x2 + x3 − x4 + C
2
3
4
Z
17.
(x + 2x−2 − x−4 )dx = x2 /2 − 2/x + 1/(3x3 ) + C
14.
Z
18.
1
(t−3 − 2)dt = − t−2 − 2t + C
2
Z ·
20.
¸
√
1 −1 √ t
1
t − 2e dt = ln t − 2et + C
2
2
19.
2 ln x + 3ex + C
21. −4 cos x + 2 sin x + C
Z
22.
4 tan x − csc x + C
(sec2 x + sec x tan x)dx = tan x + sec x + C
23.
Z
24.
(sec x tan x + 1)dx = sec x + x + C
25. ln θ − 2eθ + cot θ + C
sin y dy = − cos y + C
27.
Z
26.
Z
Z
28.
sec x tan x dx = sec x + C
Z
(φ + 2 csc φ)dφ = φ /2 − 2 cot φ + C
2
2
29.
(1 + sin θ)dθ = θ − cos θ + C
211
Chapter 7
Z
30.
31.
33.
Z
2 sin x cos x
dx = 2 sin x dx = −2 cos x + C
cos x
Z
Z
Z
¡ 2
¢
1 − sin x
1 − sin x
dx
=
=
sec x − sec x tan x dx = tan x − sec x + C
2
2
cos x
1 − sin x
(a)
y
(b)
f (x) = x2 /2 + 5
(b)
f (x) = ex /2 + 1/2
6
4
2
-1
34.
1
(a)
2
x
y
6
4
2
-1
1
35.
2
x
36.
y
y
5
4
π/4
π/2
2
x
x
1
2
3
4
5
-5
Z
37.
f 0 (x) = m = − sin x so f (x) =
38.
f 0 (x) = m = (x + 1)2 , so f (x) =
39.
Z
(x + 1)2 dx =
1
1
1
(x + 1)3 + C; f (−2) = 8 = (−2 + 1)3 + C = − + C,
3
3
3
C =8+
1
25
(x + 1)3 +
3
3
(a)
1
25
=
, f (x) =
3
3
Z
y(x) = x1/3 dx =
3
3 4/3
3
5
x + C, y(1) = + C = 2, C = 5/4; y(x) = x4/3 +
4
4
4
4
(b)
y(t) =
Z
Z
(c)
y(x) =
(a)
t−1 dt = ln |t| + C, y(−1) = C = 5, C = 5; y(t) = ln |t| + 5
(x1/2 + x−1/2 )dx =
8
2 3/2
8
x + 2x1/2 + C, y(1) = 0 = + C, C = − ,
3
3
3
2 3/2
8
x + 2x1/2 −
3
3
¶
Z µ
1
1
1
1
1
1 −3
x
; y(x) = − x−2 +
dx = − x−2 + C, y(1) = 0 = − + C, C =
y(x) =
8
16
16
16
16
16
y(x) =
40.
(− sin x)dx = cos x + C; f (0) = 2 = 1 + C so C = 1, f (x) = cos x + 1
Exercise Set 7.2
212
√
√
π
2
2
(sec2 t − sin t) dt = tan t + cos t + C, y( ) = 1 = 1 +
+ C, C = −
;
4
2
2
√
2
y(t) = tan t + cos t −
2
Z
2
2
y(x) = x7/2 dx = x9/2 + C, y(0) = 0 = C, C = 0; y(x) = x9/2
9
9
Z
(b)
(c)
y(t) =
2 3/2
4 5/2
x + C1 ; f (x) =
x + C1 x + C2
3
15
41.
f 0 (x) =
42.
f 0 (x) = x2 /2 + sin x + C1 , use f 0 (0) = 2 to get C1 = 2 so f 0 (x) = x2 /2 + sin x + 2,
f (x) = x3 /6 − cos x + 2x + C2 , use f (0) = 1 to get C2 = 2 so f (x) = x3 /6 − cos x + 2x + 2
Z
43.
dy/dx = 2x + 1, y =
(2x + 1)dx = x2 + x + C; y = 0 when x = −3
so (−3)2 + (−3) + C = 0, C = −6 thus y = x2 + x − 6
Z
44.
x2 dx = x3 /3 + C; y = 2 when x = −1 so (−1)3 /3 + C = 2, C = 7/3
dy/dx = x2 , y =
thus y = x3 /3 + 7/3
Z
45.
6xdx = 3x2 + C1 . The slope of the tangent line is −3 so dy/dx = −3 when x = 1. Thus
Z
2
2
3(1) + C1 = −3, C1 = −6 so dy/dx = 3x − 6, y = (3x2 − 6)dx = x3 − 6x + C2 ; If x = 1, then
dy/dx =
y = 5 − 3(1) = 2 so (1)2 − 6(1) + C2 = 2, C2 = 7 thus y = x3 − 6x + 7.
46.
dT /dx = C1 , T = C1 x + C2 ; T = 25 when x = 0 so C2 = 25, T = C1 x + 25. T = 85 when x = 50 so
50C1 + 25 = 85, C1 = 1.2, T = 1.2x + 25
47.
(a)
F 0 (x) = G0 (x) = 3x + 4
(b) F (0) = 16/6 = 8/3, G(0) = 0, so F (0) − G(0) = 8/3
48.
(c)
F (x) = (9x2 + 24x + 16)/6 = 3x2 /2 + 4x + 8/3 = G(x) + 8/3
(a)
F 0 (x) = G0 (x) = 10x/(x2 + 5)2
(b)
F (0) = 0, G(0) = −1, so F (0) − G(0) = 1
(c)
F (x) =
x2
(x2 + 5) − 5
5
=
=1− 2
= G(x) + 1
2
2
x +5
x +5
x +5
Z
49.
51.
Z
(sec x − 1)dx = tan x − x + C
2
(a)
1
2
Z
1
(1 − cos x)dx = (x − sin x) + C
2
(b)
½
52.
0
0
F (x) = G (x) = f (x), where f (x) =
(b)
G(x) − F (x) =
(c)
no, because (−∞, 0) ∪ (0, +∞) is not an interval
1087
v= √
2 273
Z
2,
3,
1
2
Z
(1 + cos x) dx =
1
(x + sin x) + C
2
1, x > 0
−1, x < 0
(a)
½
53.
(csc2 x − 1)dx = − cot x − x + C
50.
x>0
so G(x) 6= F (x) plus a constant
x<0
1087 1/2
1087 1/2
ft/s
T −1/2 dT = √
T + C, v(273) = 1087 = 1087 + C so C = 0, v = √
T
273
273
213
Chapter 7
EXERCISE SET 7.3
Z
1.
2.
3.
u23 du = u24 /24 + C = (x2 + 1)24 /24 + C
Z
(b) − u3 du = −u4 /4 + C = −(cos4 x)/4 + C
Z
√
(c) 2 sin u du = −2 cos u + C = −2 cos x + C
Z
3
3
3p 2
(d)
u−1/2 du = u1/2 + C =
4x + 5 + C
8
4
4
Z
1
1
1
(e)
u−1 du = ln u + C = ln(x3 − 4) + C
3
3
3
(a)
Z
1
1
1
sec2 u du = tan u + C = tan(4x + 1) + C
4
4
4
Z
1
1
1
(b)
u1/2 du = u3/2 + C = (1 + 2y 2 )3/2 + C
4
6
6
Z
1
2
2
(c)
u1/2 du =
u3/2 + C =
sin3/2 πθ + C
π
3π
3π
Z
5
5
(d)
u4/5 du = u9/5 + C = (x2 + 7x + 3)9/5 + C
9
9
Z
du
= ln u + C = ln(1 + ex ) + C
(e)
u
(a)
Z
1
1
− u du = − u2 + C = − cot2 x + C
2
2
Z
1 10
1
(b)
u9 du =
u +C =
(1 + sin t)10 + C
10
10
Z
1
(c)
du = ln |u| + C = ln | ln x| + C
u
Z
1
1
1
(d) −
eu du = − eu + C = − e−5x + C
5
5
5
Z
1
1
1
1
(e) −
du = − ln |u| + C = − ln |(1 + cos 3θ)| + C
3
u
3
3
(a)
Z
4.
Z
2 7/2 4 5/2 2 3/2
u − u + u +C
7
5
3
4
2
2
= (1 + x)7/2 − (1 + x)5/2 + (1 + x)3/2 + C
7
5
3
(u − 1)2 u1/2 du =
(a)
Z
(u5/2 − 2u3/2 + u1/2 )du =
csc2 u du = − cot u + C = − cot(sin x) + C
(b)
Z
eu du = eu + C = etan x + C
Z
1
1
1
(d)
u1/2 du = u3/2 + C = (1 + e2t )3/2 + C
2
3
3
Z
¯
¯
1
(e)
du = ln |u| + C = ln ¯x5 + 1¯ + C
u
(c)
5.
u = 2x, du = 2dx;
1
2
Z
eu du =
1 u
1
e + C = e2x + C
2
2
Exercise Set 7.3
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
214
1
2
Z
1
1
1
du = ln |u| + C = ln |2x| + C
u
2
2
Z
1
u = 2 − x2 , du = −2x dx; −
u3 du = −u4 /8 + C = −(2 − x2 )4 /8 + C
2
Z
1
1 6
1
u = 3x − 1, du = 3dx;
u5 du =
u +C =
(3x − 1)6 + C
3
18
18
Z
1
1
1
u = 8x, du = 8dx;
cos u du = sin u + C = sin 8x + C
8
8
8
Z
1
1
1
u = 3x, du = 3dx;
sin u du = − cos u + C = − cos 3x + C
3
3
3
Z
1
1
1
u = 4x, du = 4dx;
sec u tan u du = sec u + C = sec 4x + C
4
4
4
Z
1
1
1
u = 5x, du = 5dx;
sec2 u du = tan u + C = tan 5x + C
5
5
5
Z
1
1 3/2
1
u = 7t2 + 12, du = 14t dt;
u1/2 du =
u +C =
(7t2 + 12)3/2 + C
14
21
21
Z
1
1
1p
2
u = 4 − 5x , du = −10x dx; −
u−1/2 du = − u1/2 + C = −
4 − 5x2 + C
10
5
5
Z
1
2
2p 3
u = x3 + 1, du = 3x2 dx;
u−1/2 du = u1/2 + C =
x +1+C
3
3
3
Z
1
1
1
u = 1 − 3x, du = −3dx; −
u−2 du = u−1 + C = (1 − 3x)−1 + C
3
3
3
Z
1
1
1
u = 4x2 + 1, du = 8x dx;
u−3 du = − u−2 + C = − (4x2 + 1)−2 + C
8
16
16
Z
1
1
1
u = 3x2 , du = 6x dx;
cos u du = sin u + C = sin(3x2 ) + C
6
6
6
Z
u = sin x, du = cos x dx;
eu du = eu + C = esin x + C
u = 2x, du = 2dx;
1
u = x , du = 4x dx;
4
4
Z
3
eu du =
Z
1
1
3
eu du = − eu + C = − e−2x + C
6
6
Z
¯
¯
1
22. u = ex − e−x , du = (ex + e−x )dx,
du = ln |u| + C = ln ¯ex − e−x ¯ + C
u
Z
1
1
1
23. u = 5/x, du = −(5/x2 )dx; −
sin u du = cos u + C = cos(5/x) + C
5
5
5
Z
√
√
1
24. u = x, du = √ dx; 2 sec2 u du = 2 tan u + C = 2 tan x + C
2 x
Z
1
1
1
sec2 u du = tan u + C = tan(x3 ) + C
25. u = x3 , du = 3x2 dx;
3
3
3
21.
u = −2x3 , du = −6x2 , −
1
6
1 u
1 4
e + C = ex + C
4
4
215
26.
27.
Chapter 7
Z
1
1
1
u3 du = − u4 + C = − cos4 2t + C
2
8
8
Z
Z
e−x dx; u = −x, du = −dx; − eu du = −eu + C = −e−x + C
u = cos 2t, du = −2 sin 2t dt; −
Z
Z
28.
e
x/2
dx; u = x/2, du = dx/2; 2
1
3
√
eu du = 2eu + C = 2ex/2 + C = 2 ex + C
Z
u5 du =
29.
u = sin 3t, du = 3 cos 3t dt;
30.
1
u = 5 + cos 2θ, du = −2 sin 2θ dθ; −
2
31. u = 2 − sin 4θ, du = −4 cos 4θ dθ; −
32.
u = tan 5x, du = 5 sec2 5x dx;
1
5
1
4
Z
Z
1 6
1
u +C =
sin6 3t + C
18
18
u−3 du =
1 −2
1
u + C = (5 + cos 2θ)−2 + C
4
4
1
1
u1/2 du = − u3/2 + C = − (2 − sin 4θ)3/2 + C
6
6
Z
u3 du =
1 4
1
u +C =
tan4 5x + C
20
20
Z
1
1
1
u2 du = u3 + C = sec3 2x + C
33. u = sec 2x, du = 2 sec 2x tan 2x dx;
2
6
6
Z
34. u = sin θ, du = cos θ dθ;
sin u du = − cos u + C = − cos(sin θ) + C
35.
36.
u=
u=
√
1
y, du = √ dy, 2
2 y
√
1
y, du = √ dy, 2
2 y
Z
√
eu du = 2eu + C = 2e
Z
1
du = 2
eu
Z
y
+C
√
e−u du = −2e−u + C = −2e−
y
+C
1
38. u = a + bx, du = b dx, dx = du
b
Z
1
n
n
u1/n du =
u(n+1)/n + C =
(a + bx)(n+1)/n + C
b
b(n + 1)
b(n + 1)
39.
u = sin(a + bx), du = b cos(a + bx)dx
Z
1
1
1
un du =
un+1 + C =
sinn+1 (a + bx) + C
b
b(n + 1)
b(n + 1)
41.
u = x − 3, x = u + 3, dx = du
Z
Z
2
2
(u + 3)u1/2 du = (u3/2 + 3u1/2 )du = u5/2 + 2u3/2 + C = (x − 3)5/2 + 2(x − 3)3/2 + C
5
5
42.
u = y + 1, y = u − 1, dy = du
Z
Z
2
2
u−1
du
=
(u1/2 − u−1/2 )du = u3/2 − 2u1/2 + C = (y + 1)3/2 − 2(y + 1)1/2 + C
1/2
3
3
u
43.
u = 3θ, du = 3 dθ
Z
Z
1
1
1
1
tan2 u du =
(sec2 u − 1)du = (tan u − u) + C = (tan 3θ − 3θ) + C
3
3
3
3
Z
Z
sin2 2θ sin 2θ dθ = (1 − cos2 2θ) sin 2θ dθ; u = cos 2θ, du = −2 sin 2θ dθ,
Z
1
1
1
1
1
−
(1 − u2 )du = − u + u3 + C = − cos 2θ + cos3 2θ + C
2
2
6
2
6
44.
Exercise Set 7.3
Z µ
45.
1+
1
t
216
¶
dt = t + ln |t| + C
Z
2
Z
e2 ln x dx =
46.
e2 ln x = eln x = x2 , x > 0, so
47.
ln(ex ) + ln(e−x ) = ln(ex e−x ) = ln 1 = 0 so
Z
48.
cos x
dx; u = sin x, du = cos xdx;
sin x
Z
x2 dx =
Z
1 3
x +C
3
[ln(ex ) + ln(e−x )]dx = C
1
du = ln |u| + C = ln | sin x| + C
u
Z
49.
(a)
1
1
u du = u2 + C1 = sin2 x + C1 ;
2
2
Z
1 2
1
with u = cos x, du = − sin x dx; − u du = − u + C2 = − cos2 x + C2
2
2
with u = sin x, du = cos x dx;
(b)
because
by a µconstant:
¶
µ they differ ¶
1
1
1
2
sin x + C1 − − cos2 x + C2 = (sin2 x + cos2 x) + C1 − C2 = 1/2 + C1 − C2
2
2
2
(a)
First method:
(25x2 − 10x + 1)dx =
second method:
1
5
Z
50.
Z
u2 du =
25 3
x − 5x2 + x + C1 ;
3
1 3
1
u + C2 =
(5x − 1)3 + C2
15
15
1
1
25 3
1
(5x − 1)3 + C2 =
(125x3 − 75x2 + 15x − 1) + C2 =
x − 5x2 + x −
+ C2 ;
15
15
3
15
the answers differ by a constant.
Z
√
2
3x + 1dx = (3x + 1)3/2 + C,
51. y(x) =
9
29
29
2
16
+ C = 5, C =
so y(x) = (3x + 1)3/2 +
y(1) =
9
9
9
9
Z
5
52. y(x) = (6 − 5 sin 2x)dx = 6x + cos 2x + C,
2
1
5
1
5
y(0) = + C = 3, C = so y(x) = 6x + cos 2x +
2
2
2
2
Z
√
7
2
2
53. f 0 (x) = m = 3x + 1, f (x) = (3x + 1)1/2 dx = (3x + 1)3/2 + C; f (0) = 1 = + C, C = , so
9
9
9
2
7
3/2
f (x) = (3x + 1) +
9
9
(b)
54.
y
4
-4
4
x
217
Chapter 7
Z
55.
56.
8
8
256
(4 + 0.15t)5/2 + C; p(0) = 100, 000 = 45/2 + C =
+ C,
3
3
3
8
8
256
≈ 99, 915, p(t) = (4 + 0.15t)5/2 + 99, 915, p(5) = (4.75)5/2 + 99, 915 ≈ 100, 416
C = 100, 000 −
3
3
3
Z
√
2
dr
= −k t, r = −k t1/2 dt = − kt3/2 + C; r(0) = 10, 000 = C so C = 10, 000 and
dt
3
2
250
2 3/2
k, so
r = − kt + 10, 000; r(25) = 9, 000 = − k(25)3/2 + 10, 000 = 10, 000 −
3
3
3
3/2
k = 12, r = −8t + 10, 000, and r(60) = 6281.94 m
(4 + 0.15t)3/2 dt =
p(t) =
EXERCISE SET 7.4
1.
(a) 1 + 8 + 27 = 36
(c) 20 + 12 + 6 + 2 + 0 + 0 = 40
(e) 1 − 2 + 4 − 8 + 16 = 11
(b) 5 + 8 + 11 + 14 + 17 = 55
(d) 1 + 1 + 1 + 1 + 1 + 1 = 6
(f ) 0 + 0 + 0 + 0 + 0 + 0 = 0
2.
(a) 1 + 0 − 3 + 0 = −2
(c) e2 + e2 + · · · + e2 = 14e2
(14 terms)
(b) 1 − 1 + 1 − 1 + 1 − 1 = 0
(d) 24 + 25 + 26 = 112
3.
(e)
ln 1 + ln 2 + ln 3 + ln 4 + ln 5 + ln 6 = ln(1 · 2 · 3 · 4 · 5 · 6) = ln 720
(f )
1−1+1−1+1−1+1−1+1−1+1=1
10
X
k
4.
k=1
7.
20
X
3k
5.
k=1
10
X
2k
5
X
8.
8
X
(−1)k+1
50
X
(a)
1
k
11.
5
X
9.
16.
(−1)k+1 (2k − 1)
(−1)k
1
k
2k
12.
3
X
k=0
(b)
50
X
cos
kπ
7
(2k − 1)
(−1)k+1 ak
(b)
5
X
(−1)k+1 bk
(c)
k=0
n
X
ak xk
k=0
(d)
5
X
a5−k bk
k=0
1
(100)(100 + 1) = 5050
2
100
X
k−
k=1
17.
6
X
k=1
k=1
k=1
15.
2k
1
5
X
(a)
4
X
k=0
(2k − 1)
1
14.
6.
k=1
k=1
13.
k(k + 1)
k=1
k=1
10.
49
X
2
X
k=
k=1
1
(100)(100 + 1) − (1 + 2) = 5050 − 3 = 5047
2
1
(20)(21)(41) = 2, 870
6
19. 4
6
X
k=1
k3 − 2
6
X
k=1
k+
6
X
k=1
18.
7
100
X
k=1
·
1=4
k+
100
X
1=
k=1
¸
¸
·
1
1 2 2
(6) (7) − 2 (6)(7) + 6 = 1728
4
2
7
(100)(101) + 100 = 35, 450
2
Exercise Set 7.4
20.
20
X
k2 −
3
X
k=1
21.
218
k 2 = 2, 870 − 14 = 2, 856
k=1
30
X
30
X
k(k 2 − 4) =
k=1
22.
k=1
6
X
k−
k=1
23.
n
X
6
X
k3 =
k=1
(4k − 3) = 4
k=1
24.
n−1
X
26.
27.
k=
k=1
1
1
(30)2 (31)2 − 4 · (30)(31) = 214, 365
4
2
1
1
(6)(7) − (6)2 (7)2 = −420
2
4
n
X
k−
n
X
k=1
1
3 = 4 · n(n + 1) − 3n = 2n2 − n
2
n
k=1
n−1 2
X
k
n−1
k=1
k=1
1X 2
1
1 1
=
k = · (n − 1)(n)(2n − 1) = (n − 1)(2n − 1)
n
n
n 6
6
n−1 3
n−1
X
1 X 3
1 1
1
k
= 2
k = 2 · (n − 1)2 n2 = (n − 1)2
n2
n
n 4
4
k=1
n µ
X
5
k=1
30.
30
X
3X
3
3 1
=
k = · n(n + 1) = (n + 1)
n
n
n 2
2
k=1
28.
k3 − 4
1
1
(n − 1)[(n − 1) + 1][2(n − 1) + 1] = n(n − 1)(2n − 1)
6
6
k2 =
n
X
3k
k=1
30
X
k=1
k=1
k=1
25.
(k 3 − 4k) =
n
−
S − rS =
2k
n
¶
n
X
=
5X
2 1
2X
5
1−
k = (n) − · n(n + 1) = 4 − n
n
n
n
n 2
ark −
n
n
k=1
k=1
n
X
k=0
ark+1
k=0
= (a + ar + ar2 + · · · + arn ) − (ar + ar2 + ar3 + · · · + arn+1 )
= a − arn+1 = a(1 − rn+1 )
so (1 − r)S = a(1 − rn+1 ), hence S = a(1 − rn+1 )/(1 − r)
31.
(a)
19
X
3k+1 =
k=0
(b)
2k+5 =
k=0
(c)
k=0
3(3k ) =
k=0
25
X
100
X
19
X
µ
(−1)
25
X
25 2k =
k=0
−1
2
¶k
=
3
3(1 − 320 )
= (320 − 1)
1−3
2
25 (1 − 226 )
= 231 − 25
1−2
(−1)(1 − (−1/2)101 )
2
= − (1 + 1/2101 )
1 − (−1/2)
3
32.
(a)
1.999023438, 1.999999046, 2.000000000; 2
33.
n
n
n+1
1
1 X
1 1
1 + 2 + 3 + ··· + n X k
n+1
; lim
=
=
=
k = 2 · n(n + 1) =
n2
n2
n2
n 2
2n n→+∞ 2n
2
k=1
k=1
(b)
2.831059456, 2.990486364, 2.999998301; 3
219
34.
Chapter 7
n
n
X
12 + 22 + 32 + · · · + n2
1 X 2
1 1
(n + 1)(2n + 1)
k2
=
=
k = 3 · n(n + 1)(2n + 1) =
;
n3
n3
n3
n 6
6n2
k=1
lim
n→+∞
35.
36.
(n + 1)(2n + 1)
1
1
= lim (1 + 1/n)(2 + 1/n) =
n→+∞ 6
6n2
3
n
X
5k
k=1
n2
n−1
X
2k 2
k=1
k=1
n3
=
n
5
5 X
5(n + 1)
5(n + 1)
5 1
; lim
=
k = 2 · n(n + 1) =
n→+∞
n2
n 2
2n
2n
2
k=1
=
n−1
(n − 1)(2n − 1)
2 X 2
2 1
k = 3 · (n − 1)(n)(2n − 1) =
;
n3
n 6
3n2
k=1
(n − 1)(2n − 1)
1
2
= lim (1 − 1/n)(2 − 1/n) =
lim
n→+∞
n→+∞ 3
3n2
3
37.
(a)
5
X
2j
(b)
j=0
38.
(a)
5
X
6
X
2j−1
(c)
j=1
(a)
(k + 4)2k+8
18
X
(b)
13
X
(k − 4)2k
k=9
sin
³π ´
k=1
(b)
k
6
X
ek =
k=0
40. 1 + 3 + 5 + · · · + (2n − 1) =
n
X
(2k − 1) = 2
k=1
41.
2j−2
j=2
k=1
39.
7
X
n
X
k−
k=1
n
X
k=1
e7 − 1
e−1
1
1 = 2 · n(n + 1) − n = n2
2
For 1 ≤ k ≤ n the k-th L-shaped strip consists of the corner square, a strip above and a strip to the
n
X
(2k − 1) = n2 .
right for a combined area of 1 + (k − 1) + (k − 1) = 2k − 1, so the total area is
k=1
42.
n(n + 1)
= 465, n2 + n − 930 = 0, (n + 31)(n − 30) = 0, n = 30.
2
43.
(35 − 34 ) + (36 − 35 ) + · · · + (317 − 316 ) = 317 − 34
µ
1−
44.
µ
45.
1
2
¶
1
1
− 2
22
1
µ
+
1 1
−
2 3
¶
µ
+
¶
µ
+ ··· +
1
1
− 2
32
2
1
1
−
50 51
¶
µ
+ ··· +
¶
=
50
51
1
1
− 2
202
19
¶
=
46.
(22 − 2) + (23 − 22 ) + · · · + (2101 − 2100 ) = 2101 − 2
47.
(a)
(b)
399
1
−1=−
202
400
¶
n µ
1X
1
1
1
=
−
(2k − 1)(2k + 1) 2
2k − 1 2k + 1
k=1
k=1
·µ
¶ µ
¶ µ
¶
µ
¶¸
1
1
1
1 1
1 1
1
1−
+
−
+
−
+ ··· +
−
=
2
3
3 5
5 7
2n − 1 2n + 1
·
¸
1
n
1
1−
=
=
2
2n + 1
2n + 1
1
n
lim
=
n→+∞ 2n + 1
2
n
X
Exercise Set 7.4
48.
n
X
(a)
k=1
220
¶
n µ
X
1
1
1
=
−
k(k + 1)
k k+1
k=1
¶ µ
¶ µ
¶
µ
¶
µ
1 1
1 1
1
1
1
+
−
+
−
+ ··· +
−
= 1−
2
2 3
3 4
n n+1
=1−
(b)
49.
51.
lim
n→+∞
n
=1
n+1
both are valid
n
X
(xi − x̄) =
i=1
n
X
xi = nx̄ so
50.
n
X
n
X
xi −
i=1
n
X
i=1
52.
n
1
=
n+1
n+1
n
X
x̄ =
n
X
i=1
none is valid
1X
xi thus
n i=1
n
xi − nx̄ but x̄ =
i=1
(xi − x̄) = nx̄ − nx̄ = 0
i=1
(ak − bk ) = (a1 − b1 ) + (a2 − b2 ) + · · · + (an − bn )
k=1
= (a1 + a2 + · · · + an ) − (b1 + b2 + · · · + bn ) =
n
X
ak −
k=1
53.
n
X
bk
k=1
n
X
£
¤
(k + 1)4 − k 4 = (n + 1)4 − 1 (telescoping sum), expand the
k=1
quantity in brackets to get
n
X
(4k 3 + 6k 2 + 4k + 1) = (n + 1)4 − 1,
k=1
4
n
X
k3 + 6
k=1
n
X
k=1
54.
n
X
k=1
"
k2 + 4
n
X
k+
k=1
1
(n + 1)4 − 1 − 6
k =
4
3
n
X
1 = (n + 1)4 − 1
k=1
n
X
k −4
2
n
X
k=1
k=1
k−
n
X
#
1
k=1
=
1
[(n + 1)4 − 1 − n(n + 1)(2n + 1) − 2n(n + 1) − n]
4
=
1
(n + 1)[(n + 1)3 − n(2n + 1) − 2n − 1]
4
=
1
1
(n + 1)(n3 + n2 ) = n2 (n + 1)2
4
4
If n = 2m then 2m + 2(m − 1) + · · · + 2 · 2 + 2 = 2
m
X
k =2·
k=1
if n = 2m + 1 then (2m + 1) + (2m − 1) + · · · + 5 + 3 + 1 =
=2
m+1
X
k=1
55.
k−
m+1
X
k=1
n2 + 2n
m(m + 1)
= m(m + 1) =
;
2
4
m+1
X
(2k − 1)
k=1
1=2·
(m + 1)(m + 2)
n2 + 2n + 1
− (m + 1) = (m + 1)2 =
2
4
50 · 30 + 49 · 29 + · · · + 22 · 2 + 21 · 1 =
30
X
k=1
k(k + 20) =
30
X
k=1
k 2 + 20
30
X
k=1
k=
30 · 31
30 · 31 · 61
+ 20
= 18,755
6
2
221
Chapter 7
EXERCISE SET 7.5
1.
(a)
1
1
bh = · 4 · 4 = 8
2
2
y
A
1
x
1
y
(b) The approximation is greater than A,
as the rectangles extend beyond the area.
4
X
f (x∗k )∆x = (4 + 3 + 2 + 1)(1) = 10
k=1
1
x
1
y
(c) The approximation is less than A,
as the rectangles lie inside the area.
4
X
f (x∗k )∆x = (3 + 2 + 1 + 0)(1) = 6
k=1
1
x
1
y
(d) The approximation is equal to A,
as can be seen by measuring congruent triangles.
4
X
f (x∗k )∆x = (3.5 + 2.5 + 1.5 + 0.5)(1) = 8
k=1
1
x
1
2.
(a)
1
1
bh = 4 · 12 = 24
2
2
y
18
9
6
2
3
6
x
Exercise Set 7.5
(b) The approximation is less than A,
as the rectangles lie inside the area.
4
X
f (x∗k )∆x = (6 + 9 + 12 + 15)(1) = 42
222
y
18
k=1
9
6
(c) The approximation is greater than A,
as the rectangles extend beyond the area.
4
X
f (x∗k )∆x = (9 + 12 + 15 + 18)(1) = 54
2
3
6
2
3
6
2
3
6
x
y
18
k=1
9
6
(d) The approximation is equal to A,
as can be seen by measuring congruent triangles.
4
X
f (x∗k )∆x = (7.5 + 10.5 + 13.5 + 16.5)(1) = 48
x
y
18
k=1
9
6
3.
(a)
x∗k = 0, 1, 2, 3, 4
5
X
f (x∗k )∆x = (1 + 2 + 5 + 10 + 17)(1) = 35
k=1
(b)
x∗k = 1, 2, 3, 4, 5
5
X
f (x∗k )∆x = (2 + 5 + 10 + 17 + 26)(1) = 60
k=1
(c)
x∗k = 1/2, 3/2, 5/2, 7/2, 9/2
5
X
f (x∗k )∆x = (5/4 + 13/4 + 29/4 + 53/4 + 85/4)(1) = 185/4 = 46.25
k=1
4.
(a)
x∗k = 1, 2, 3, 4, 5
5
X
f (x∗k )∆x = (1 + 8 + 27 + 64 + 125)(1) = 225
k=1
(b)
x∗k = 2, 3, 4, 5, 6
5
X
f (x∗k )∆x = (8 + 27 + 64 + 125 + 216)(1) = 440
k=1
(c)
x∗k = 3/2, 5/2, 7/2, 9/2, 11/2
5
X
k=1
f (x∗k )∆x = (27/8 + 125/8 + 343/8 + 729/8 + 1331/8)(1) = 2555/8 = 319.375
x
223
5.
Chapter 7
(a)
x∗k = −π/2, −π/4, 0, π/4
4
X
√
√
√
f (x∗k )∆x = (0 + 1/ 2 + 1 + 1/ 2)(π/4) = (1 + 2)π/4 ≈ 1.896
k=1
(b)
x∗k = −π/4, 0, π/4, π/2
4
X
√
√
√
f (x∗k )∆x = (1/ 2 + 1 + 1/ 2 + 0)(π/4) = (1 + 2)π/4 ≈ 1.896
k=1
(c)
x∗k = −3π/8, −π/8, π/8, 3π/8
·
¸
4
X
3π
π´
π
π
3π π
π
π ³ √
f (x∗k )∆x = cos
+ cos + cos + cos
= π cos cos = π 2 cos
/2 ≈ 2.052
8
8
8
8 4
4
8
8
k=1
6.
(a)
x∗k = 0, 1, 2, 3, 4
4
X
¡
¢
f (x∗k )∆x = e0 + e1 + e2 + e3 + e4 (1) = (1 − e5 )/(1 − e) = 85.791
k=1
(b)
x∗k = 1, 2, 3, 4, 5
4
X
¡
¢
f (x∗k )∆x = e1 + e2 + e3 + e4 + e5 (1) = e(1 − e5 )/(1 − e) = 233.204
k=1
(c)
x∗k = 1/2, 3/2, 5/2, 7/2, 9/2
4
´
³
X
f (x∗k )∆x = e1/2 + e3/2 + e5/2 + e7/2 + e9/2 (1) = e1/2 (1 − e5 )/(1 − e) = 141.446
k=1
7. left endpoints: x∗k = 1, 2, 3, 4;
4
X
f (x∗k )∆x = (2 + 3 + 2 + 1)(1) = 8
k=1
right endpoints: x∗k = 2, 3, 4, 5;
4
X
f (x∗k )∆x = (3 + 2 + 1 + 2)(1) = 8
k=1
8.
(a)
A = 12 (h1 + h2 )w = 12 (4 + 16)(4) = 40
y
16
7
4
1
(b)
(left) x∗k = 1, 2, 3, 4;
4
X
2
5
x
f (x∗k )∆x = (4 + 7 + 10 + 13)(1) = 34
k=1
(right) x∗k = 2, 3, 4, 5;
4
X
f (x∗k )∆x = (7 + 10 + 13 + 16)(1) = 46; the average is 12 (34 + 46) = 40
k=1
(c)
9.
The right endpoint approximation exceeds the true area by four triangles; the true area exceeds
the left endpoint approximation by four different, but congruent, triangles.
0.718771403, 0.668771403, 0.692835360
10. 0.761923639, 0.584145862, 0.663501867
11.
0.919403170, 1.07648280, 1.001028825
12. 4.884074732, 5.684074732, 5.347070728
13.
0.351220577, 0.420535296, 0.386502483
14.
1.63379940, 1.805627583, 1.717566087
Exercise Set 7.5
15.
(a)
(b)
(c)
17.
(a)
224
√
n
1/x
1/x2
sin x
x
ln x
ex
25 0.693097198 0.666154270 1.000164512 5.336963538 0.386327689 1.718167282
50 0.693134682 0.666538346 1.000041125 5.334644416 0.386302694 1.718253191
100 0.693144056 0.666634573 1.000010281 5.333803776 0.3862964444 1.718274669
A=
1
(3)(3) = 9/2
2
(b)
1
−A = − (1)(1 + 2) = −3/2
2
y
y
-2
-1
x
A
A
x
3
(c)
1
−A1 + A2 = − + 8 = 15/2
2
(d)
−A1 + A2 = 0
y
y
A2
-5
A2
5
-1
4
A1
18.
(a)
A=
x
1
(1)(2) = 1
2
(b)
A=
1
(2)(3/2 + 1/2) = 2
2
y
y
1
1
A
A
x
2
(c)
x
A1
-1
1
−A = − (1/2)(1) = −1/4
2
(d)
x
1
A1 − A2 = 1 − 1/4 = 3/4
y
y
1
1
A1
A2
A
2
x
x
2
225
19.
Chapter 7
(a)
A = 2(5) = 10
(b)
0; A1 = A2 by symmetry
y
y
2
A1
1
c
A
6
x
A2
x
1
(c)
2
3
5
4
1
1
(5)(5/2) + (1)(1/2)
2
2
= 13/2
A1 + A2 =
(d)
1
[π(1)2 ] = π/2
2
y
1
y
5
A
-1
3
2
-1
(a)
x
2
A = (6)(5) = 30
(b)
y
−A1 + A2 = 0 because
A1 = A2 by symmetry
y
6
A
A2
$
x
-10
(c)
1
1
(2)(2) + (1)(1) = 5/2
2
2
(d)
x
4
A1
-5
A1 + A2 =
x
A2
A1
20.
1
1
π(2)2 = π
4
y
y
2
2
A
A1
A2
2
x
3
2
21.
(a)
0.8
Z
22.
Z
1
0
Z
(b)
¸1
1
2
f (x)dx =
(a)
−2.6
(b)
2xdx = x
0
Z
1
−1
f (x)dx =
=1
0
¸1
1
−1
2xdx = x2
−1
= 12 − (−1)2 = 0
(c) −1.8
x
(d)
−0.3
Exercise Set 7.5
Z
226
Z
10
(c)
1
Z
5
2
f (x)dx + 2
Z
4
3
2
Z
4
f (x)dx −
Z
1
26.
f (x)dx −
Z
f (x)dx = −
Z
Z
0
Z
(a)
−3
(b)
30.
3
Z
dx − 5
2
−2
−2
3
f (x)dx +
¸
f (x)dx = −(2 − 6) = 4
1
p
2dx +
0
−3
Z
dx − 3
3
−1
Z
0
Z
29.
1
Z
1
1 − x2 dx = 1/2 + 2(π/4) = (1 + π)/2
0
−1
Z
28.
1
f (x)dx = 1 − (−2) = 3
f (x)dx = −
xdx + 2
4
1
·Z
3
−2
1
(a)
(b)
1/2
0
3
27.
= 12 − (1/2)2 + 2 · 5 − 2 · 1 = 3/4 + 8 = 35/4
+ 2x
g(x)dx = 3(2) − 10 = −4
0
−2
2dx = x
1
Z
5
f (x)dx =
Z
2
1
5
25.
#5
¸1
5
g(x)dx = 5 + 2(−3) = −1
−1
1
Z
1
1/2
Z
−1
Z
2xdx +
1/2
24.
1
f (x)dx =
(d)
23.
¸10
2dx = 2x
= 18
1
Z
Z
10
f (x)dx =
2
−2
xdx = 4 · 4 − 5(−1/2 + (3 · 3)/2) = −4
p
9 − x2 dx = 2 · 3 + (π(3)2 )/4 = 6 + 9π/4
|x|dx = 4 · 1 − 3(2)(2 · 2)/2 = −8
(a)
√
x > 0, 1 − x < 0 on [2, 3] so the integral is negative
(b)
x2 > 0, 3 − cos x > 0 for all x so the integral is positive
(a)
x4 > 0,
(b)
x − 9 < 0, |x| + 1 > 0 on [−2, 2] so the integral is negative
Z
3 − x > 0 on [−3, −1] so the integral is positive
3
10
31.
√
p
25 − (x − 5)2 dx = π(5)2 /2 = 25π/2
0
Z
p
9 − (x − 3)2 dx = π(3)2 /4 = 9π/4
3
32.
0
Z
33.
(a)
−3
Z
34.
3
Z
4x(1 − 3x)dx
0
Z
2
x3 dx
(a)
Z
0
Z
1
(3x + 1)dx = 5/2
0
37.
(a)
π/2
sin2 xdx
(b)
1
35.
1
ex dx
(b)
lim
max ∆xk →0
n
X
k=1
2x∗k ∆xk ; a = 1, b = 2
36.
2
−2
(b)
p
4 − x2 dx = π(2)2 /2 = 2π
lim
n
X
x∗k
∆xk ; a = 0, b = 1
+1
max ∆xk →0
x∗
k=1 k
227
38.
Chapter 7
(a)
(b)
39.
n
X
lim
max ∆xk →0
k=1
n
X
lim
ln x∗k ∆xk , a = 1, b = 2
max ∆xk →0
(1 + cos x∗k ) ∆xk , a = −π/2, b = π/2
k=1
(a)
1
1
1
1
2
k+1
and x∗1 = 1 + , so x∗2 = x∗1 +
= 1 + , x∗k+1 = x∗k +
= 1+
for
x∗k+1 = x∗k +
n
n
n
n
n
n
k = 2, 3, · · · , n − 1
(b)
f (x∗k ) = 1 +
(c)
(d)
k
1
and ∆x =
n
n
n
n
X
X
1
3
1
1
1 n(n + 1)
1
= +
1+ 2
k = n+ 2
n
n
n
n
2
2 2n
k=1
k=1
¶
µ
1
3
3
+
= which is the area of the trapezoid with base 1 and sides 1 and 2
lim
n→+∞ 2
2n
2
¶
n
n µ
X
1
1 (n − 1)n
3
1
k−1 X
k−1 1
∗
,
= n+ 2
= −
;
=1+
f (xk )∆x =
1+
n
n
n
n
n
2
2 2n
k=1
k=1
¶
µ
1
3
3
−
=
lim
n→+∞ 2
2n
2
40.
f (x∗k )
41.
(a)
1
b−a
= , x∗k =
(right) ∆x =
n µ n¶
n
n
2
X
X
1
k
∗
f (xk )∆x =
=
n
n
k=1
(b)
lim
n→+∞
(a)
k=1
k=1
1
k−1
b−a
= , x∗k =
,
(left) ∆x =
n
n
n
¶
µ
n
n
n
2
X
X k−1
1
1 X
1 (n − 1)n(2n − 1)
= 3
;
f (x∗k )∆x =
(k − 1)2 = 3
n
n
n
n
6
k=1
42.
k=1
k
,
n
n
n
X
1 n(n + 1)(2n + 1)
2
1 X 2
1
k
=
f (x∗k )∆x = =
;
lim
3
3
n→+∞
n
n
6
6
3
n
X
k=1
f (x∗k )∆x =
k=1
k=1
1
2
=
6
3
3
3k
b−a
= , x∗ =
,
n à n kµ ¶
n !
n
n
n
n
2
X
X
12 X
27 X 2
27 n(n + 1)(2n + 1)
1 3k
3
=
;
f (x∗k )∆x =
1− 3
k = 12 − 3
4−
4 n
n
n
4n
4n
6
k=1
k=1
k=1
k=1
n
X
39
27(2)
lim
=
f (x∗k )∆x = 12 −
n→+∞
4(6)
4
(right) ∆x =
k=1
(b)
b−a
3
3(k − 1)
= , x∗k =
,
n
n ¶ !
Ãn
µ
n
n
n
n
2
X
X
3
27 X
27 (n − 1)n(2n − 1)
12 X
1 3(k − 1)
f (x∗k )∆x =
1− 3
(k−1)2 = 12− 3
=
;
4−
4
n
n
n
4n
4n
6
(left) ∆x =
k=1
lim
n→+∞
k=1
n
X
k=1
f (x∗k )∆x = 12 −
k=1
39
27(2)
=
4(6)
4
k=1
Exercise Set 7.5
43.
(a)
228
b−a
4k
4
(right) ∆x =
= , x∗ = 2 +
,
n µ n k¶
n
¶2
µ
n
n
3
X
X
4
192 n(n + 1) 384 n(n + 1)(2n + 1) 256 n(n + 1)
4k
∗
f (xk )∆x =
;
= 32+ 2
+ 3
+ 4
2+
n
n
n
2
n
6
n
2
k=1
k=1
n
X
192 384(2) 256
lim
+
+
= 320
f (x∗k )∆x = 32 +
n→+∞
2
6
4
k=1
(b)
b−a
4
4(k − 1)
(left) ∆x =
= , x∗k = 2 +
,
n
n
¶3 n
¶2
µ
n
n µ
X
X
192 (n − 1)n 384 (n − 1)n(2n − 1) 256 (n − 1)n
4(k − 1)
4
∗
= 32+ 2
+ 3
+ 4
2+
f (xk )∆x =
;
n
n
n
2
n
6
n
2
k=1
k=1
n
X
192 384(2) 256
+
+
= 320
f (x∗k )∆x = 32 +
lim
n→+∞
2
6
4
k=1
44.
(a)
2
2k
b−a
= , x∗k = −3 +
,
n
n
n
Ã
!
¶3
µ
n
n
n
n
n
n
X
X
56 X
108 X
72 X 2 16 X 3
2
2k
−3
=
f (x∗k )∆x =
1− 2
k+ 3
k − 4
k
1−
n
n
n
n
n
n
k=1
k=1
k=1
k=1
k=1
k=1
¶2
µ
108 n(n + 1) 72 n(n + 1)(2n + 1) 16 n(n + 1)
+ 3
− 4
= 56 − 2
;
n
2
n
6
n
2
n
X
72(2) 16
lim
−
= 22
f (x∗k )∆x = 56 − 54 +
n→+∞
6
4
(right) ∆x =
k=1
(b)
b−a
2(k − 1)
2
= , x∗k = −3 +
,
n
n
n
Ã
¶3 !
µ
n
n
X
X
2(k − 1)
2
∗
−3
f (x k)∆x =
1−
n
n
(left) ∆x =
k=1
k=1
56 X
108 X
72 X
16 X
1− 2
(k − 1) + 3
(k − 1)2 − 4
(k − 1)3
n
n
n
n
k=1
k=1
k=1
k=1
¶2
µ
108 (n − 1)n 72 (n − 1)n(2n − 1) 16 (n − 1)n
+ 3
− 4
= 56 − 2
;
n
2
n
6
n
2
n
X
108 72(2) 16
lim
+
−
= 22
f (x∗k )∆x = 56 −
n→+∞
2
6
4
n
n
n
n
=
k=1
45.
(a)
f is continuous on [−1, 1] so f is integrable there by part (a) of Theorem 7.5.8
(b)
|f (x)| ≤ 1 so f is bounded on [−1, 1], and f has one point of discontinuity, so by part (b) of
Theorem 7.5.8 f is integrable on [−1, 1]
(c)
f is not bounded on [-1,1] because lim f (x) = +∞, so f is not integrable on [0,1]
(d)
46.
x→0
1
does not exist. f is continuous
x
elsewhere. −1 ≤ f (x) ≤ 1 for x in [−1, 1] so f is bounded there. By part (b), Theorem 7.5.8, f
is integrable on [−1, 1].
f (x) is discontinuous at the point x = 0 because lim sin
x→0
Each subinterval of a partition of [a, b] contains both rational and irrational numbers. If all x∗k are
chosen to be rational then
n
n
n
n
X
X
X
X
∗
f (xk )∆xk =
(1)∆xk =
∆xk = b − a so
lim
f (x∗k )∆xk = b − a.
k=1
k=1
If all x∗k are irrational then
max ∆xk →0
k=1
lim
max ∆xk →0
preceding limits are not equal.
n
X
k=1
k=1
f (x∗k )∆xk = 0. f is not integrable on [a, b] because the
229
47.
Chapter 7
(a)
Let Sn =
n
X
f (x∗k )∆xk and S =
k=1
that
lim
max ∆xk →0
Z
b
f (x)dx then
a
n
X
cf (x∗k )∆xk = cSn and we want to prove
k=1
cSn = cS. If c = 0 the result follows immediately, so suppose that c 6= 0 then for
any ² > 0, |cSn − cS| = |c||Sn − S| < ² if |Sn − S| < ²/|c|. But because f is integrable on [a, b],
there is a number δ > 0 such that |Sn − S| < ²/|c| whenever max ∆xk < δ so |cSn − cS| < ² and
hence
lim cSn = cS.
max ∆xk →0
(b)
Let Rn =
Z
n
X
f (x∗k )∆xk , Sn =
k=1
n
X
g(x∗k )∆xk , Tn =
k=1
n
X
[f (x∗k ) + g(x∗k )]∆xk , R =
k=1
Z
b
f (x)dx, and
a
b
g(x)dx then Tn = Rn + Sn and we want to prove that
S=
a
lim
max ∆xk →0
Tn = R + S.
|Tn − (R + S)| = |(Rn − R) + (Sn − S)| ≤ |Rn − R| + |Sn − S|
so for any ² > 0 |Tn − (R + S)| < ² if |Rn − R| + |Sn − S| < ².
Because f and g are integrable on [a, b], there are numbers δ1 and δ2 such that
|Rn − R| < ²/2 for max ∆xk < δ1 and |Sn − S| < ²/2 for max ∆xk < δ2 .
If δ = min(δ1 , δ2 ) then |Rn − R| < ²/2 and |Sn − S| < ²/2 for max ∆xk < δ thus
|Rn − R| + |Sn − S| < ² and so |Tn − (R + S)| < ² for max ∆xk < δ which shows that
lim Tn = R + S.
max ∆xk →0
48.
For the smallest, find x∗k so that f (x∗k ) is minimum on each subinterval: x∗1 = 1, x∗2 = 3/2, x∗3 = 3 so
(2)(1) + (7/4)(2) + (4)(1) = 9.5. For the largest, find x∗k so that f (x∗k ) is maximum on each subinterval:
x∗1 = 0, x∗2 = 3, x∗3 = 4 so (4)(1) + (4)(2) + (8)(1) = 20.
EXERCISE SET 7.6
Z
1.
2
(a)
0
Z
(b)
¤2
(2 − x)dx = (2x − x2 /2) 0 = 4 − 4/2 = 2
−1
Z
2dx = 2x
−1
Z
5
Z
−1
(x + 3)dx = (x /2 + 3x)
−1
¸3
x3 dx = x4 /4 = 81/4 − 16/4 = 65/4
5.
1
2
xdx = x3/2
3
x
µ
¸9
1
2
= (27 − 1) = 52/3
3
3
−1
Z
4
6.
¶¸0
1 3
2
x − 2x + 7x
= 48
3
−3
5
8.
1
µ
10.
x4 dx = x5 /5
−1
−3/5
x
1
1
¸1
1
Z
=e −e
x
e dx = e
1
Z
4.
¸3
3
7.
9.
√
3
= 4/2 + 6 − (1/2 − 3) = 21/2
2
9
= 5(9) − 5(3) = 30
5dx = 5x
3
¸2
2
¸9
9
(b)
0
2
Z
Z
2
(c)
Z
= 9/2 + 3 − (1/2 + 1) = 6
¸5
xdx = x /2 = 25/2
0
3
1
2
(a)
Z
i3
(x + 1)dx = (x2 /2 + x)
1
3.
= 2(1) − 2(−1) = 4
3
(c)
2.
¸1
1
5
dx = x2/5
2
= 1/5 − (−1)/5 = 2/5
¸4
=
1
5 2/5
(4 − 1)
2
¸5
1
dx = ln x = ln 5 − ln 1 = ln 5
x
1
1 2 1 5
x + x
2
5
¶¸2
= 81/10
−1
Exercise Set 7.6
Z
3
1
x dx = −
x
−2
11.
1
13.
230
4 5/2
x
5
sin x
19.
5ex
=0
16.
√
2
18.
ln 2
a
1/2
√
3/2
= (ln 2)/2
¶¸9
= 10819/324
4
π/6
´i2
¶ #4a
a
1
= ln 2 +
√ 2
2(e − e) + csc 2 − csc 1
5
= − a3/2
3
Z
(3 − 2x)dx +
Z
π/2
¸3/2
2
(2x − 3)dx = (3x − x2 )
¸π/2
3π/4
cos x dx +
0
¸3π/4
− sin x
(− cos x)dx = sin x
π/2
¸2
+ (x2 − 3x)
= 9/4 + 1/4 = 5/2
3/2
=2−
0
√
2/2
π/2
¸0
¸2
2
2
2 + x dx = − (2 − x)3/2
+ (2 + x)3/2
3
3
−1
0
−1
0
√
√
√
√
√
2
2
2
= − (2 2 − 3 3) + (8 − 2 2) = (8 − 4 2 + 3 3)
3
3
3
¸0
¸1
Z 1
Z 0
(1−ex )dx+ (ex −1)dx = (x−ex )
+(ex −x) = −1−(−1−e−1 )+e−1−1 = e+1/e−2
(b)
Z
0
(a)
√
Z
2 − x dx +
−1
Z
Z
0
−2
Z
1
3
30.
√
0
1
(−x)dx = x3
3
Z
4
x dx +
√
2
−1
0
2
29.
¸0
1
− x2
2
−2
1
2
1
dx = x3/2
2
x
3
Z
3
x dx +
0
31.
1/2
√
= π 2 /9 + 2 3
0
28.
¶¸1
1 2
x − sec x
= 3/2 − sec(1)
2
0
4
2
√
8 y + y 3/2 − 3/2
3
3y
22.
3/2
(b)
= 179/2
1
=1
0
(ln x)/2
1
0
Z
¶¸8
µ
= −55/3
2ex + csc x
2
x − x3/2
3
(a)
= 31/160
1
i1
20.
¶¸4
¶¸π/2
1 2
x − 2 cot x
2
ln x +
tan θ
µ
= 5e3 − 5(2) = 5e3 − 10
Z
27.
=
−π/4
¸2
iπ/4
−π/2
i3
µ
4
x
3x5/3 +
14.
iπ/2
³
26.
1
√
10
2
6 t − t3/2 + √
3
t
µ
12.
1
x dx = − 5
5x
−6
µ
iπ/4
23.
2
4
17.
24.
= 2/3
1
= 844/5
− cos θ
µ
Z
¸9
15.
21.
¸3
0.665867079;
1
Z
32. 1.000257067;
0
1
1
dx = −
2
x
x
π/2
¸1
0
1
−
x
¸3
= −11/6
0
¸4
= 17/12
1
¸3
= 2/3
1
¸π/2
sin x dx = − cos x
=1
0
0
231
Chapter 7
Z
33.
3
1.098242635;
1
Z
35.
¸3
1
dx = ln x
x
µ
3
(x2 + 1)dx =
A=
0
Z
36.
2
A=
= ln 3 ≈ 1.098612289
1
¶¸3
1 3
x +x
3
µ
(−x2 + 3x − 2)dx =
1
Z
37.
¶¸2
1
3
− x3 + x2 − 2x
= 1/6
3
2
1
#2π/3
2π/3
3 sin x dx = −3 cos x
A=
0
A1 =
Z
= 9/2
A=−
38.
−2
−3
Z
A2 = −
Z
µ
(x2 − 3x − 10)dx =
5
−2
8
A3 =
−1
−2
0
Z
39.
= 12
0
1
x3 dx = − x4
4
¶¸−2
1 3 3 2
x − x − 10x
= 23/6,
3
2
−3
¸−1
= 15/4
−2
y
(x2 − 3x − 10)dx = 343/6,
10
A1
(x2 − 3x − 10)dx = 243/6, A = A1 + A2 + A3 = 203/2
5
-3
A3
A2
5
8
x
-2
40.
(a)
the area is positive
¶
µ
¶¸5
Z 5µ
1
1 4
1 3
1 2
1
1 3
1 2 1
343
(b)
x − x − x+
dx =
x − x − x + x
=
100
20
25
5
400
60
50
5
1200
−2
−2
41.
(a)
(b)
the area between the curve and the x-axis breaks into equal parts, one above and one below the
x-axis, so the integral is zero
¸1
Z 1
1
1
x3 dx = x4
= (14 − (−1)4 ) = 0;
4
4
−1
−1
#π/2
Z
π/2
−π/2
(c)
sin xdx = − cos x
The
Z a area on Zthea left side of the y-axis is equal to the area on the right side, so
f (x)dx = 2
f (x)dx
−a
Z
(d)
= − cos(π/2) + cos(−π/2) = 0 + 0 = 0
−π/2
−1
Z
0
1
x2 dx =
π/2
−π/2
1 3
x
3
¸1
=
−1
1 3
2
(1 − (−1)3 ) = = 2
3
3
Z
#π/2
1
x2 dx;
0
Z
= sin(π/2) − sin(−π/2) = 1 + 1 = 2 = 2
cos xdx = sin x
π/2
cos xdx
0
−π/2
42.
The numerator is an odd function and the denominator is an even function, so the integrand is an
odd function and the integral is zero.
43.
(a)
µ
44.
45.
(a)
(a)
x3 + 1
cos 2x
sin
√
x
(b)
(b)
(b)
F (x) =
1
F (x) = sin 2t
2
2
ex
¶¸x
1 4
t +t
4
=
1
¸x
=
π/4
1 4
5
x + x − ; F 0 (x) = x3 + 1
4
4
1
1
sin 2x − , F 0 (x) = cos 2x
2
2
Exercise Set 7.6
1
√
1+ x
46.
(a)
47.
−
49.
F 0 (x) =
51.
(b) ln x
x
cos x
(a)
50.
232
48.
√
3x
3x2 + 1, F 00 (x) = √
3x2 + 1
0
F 0 (x) =
(b)
|u|
√
13
(c)
√
6/ 13
(c)
0
cos x
−(x2 + 3) sin x − 2x cos x
00
,
F
(x)
=
x2 + 3
(x2 + 3)2
(a)
0
(b)
1/3
(a)
F 0 (x) =
(b)
increasing on [3, +∞), decreasing on (−∞, 3]
(c)
7 + 6x − x2
(7 − x)(1 + x)
=
; concave up on (−1, 7), concave down on (−∞, −1) and
(x2 + 7)2
(x2 + 7)2
on (7, +∞)
x−3
= 0 when x = 3, which is a relative minimum, and hence the absolute minimum,
x2 + 7
by the first derivative test.
F 00 (x) =
52.
F
3
2
-20
53.
-10
(a)
10
t
20
(0, +∞) because f is continuous there and 1 is in (0, +∞)
(b) at x = 1 because F (1) = 0
54.
(a)
(−3, 3) because f is continuous there and 1 is in (−3, 3)
(b) at x = 1 because F (1) = 0
55.
(a)
fave =
(b)
fave =
1
9
Z
9
x1/2 dx = 2;
0
1
e−1
Z
1
e
√
x∗ = 2, x∗ = 4
¸e
1
1
1
1
1
dx =
ln x =
; ∗ =
, x∗ = e − 1
x
e−1
e
−
1
x
e
−
1
1
Z
56.
π
1
sin x dx = 0; sin x∗ = 0, x∗ = −π, 0, π
2π −π
Z
√
1
1
1
1 3 1
dx = ; ∗ 2 = , x∗ = 3
fave =
2
2 1 x
3 (x )
3
(a)
fave =
(b)
57.
√
2≤
√
√
√
x3 + 2 ≤ 29, so 3 2 ≤
Z
3
p
√
x3 + 2dx ≤ 3 29
0
58.
Let f (x) = x sin x, f (0) = f (1) = 0, f 0 (x) = sin x + x cos x = 0 when x = − tan x, x ≈ 2.0288,
so fZ has an absolute maximum at x ≈ 2.0288; f (2.0288) ≈ 1.8197, so 0 ≤ x sin x ≤ 1.82 and
π
0≤
x sin xdx ≤ 1.82π = 5.72
0
233
Chapter 7
Z
59.
5
0 ≤ ln x ≤ ln 5 for x in [1, 5], so 0 ≤
ln xdx ≤ 4 ln 5
1
60.
£
¤b
¤b
cF (x) a = cF (b) − cF (a) = c[F (b) − F (a)] = c F (x) a
£
¤b
F (x) + G(x) a = [F (b) + G(b)] − [F (a) + G(a)]
£
(a)
(b)
¤b
= [F (b) − F (a)] + [G(b) − G(a)] = F (x)
£
¤b
F (x) − G(x) a = [F (b) − G(b)] − [F (a) − G(a)]
(c)
a
¤b
= [F (b) − F (a)] − [G(b) − G(a)] = F (x)
a
¤b
+ G(x)
a
¤b
− G(x)
a
EXERCISE SET 7.7
1.
(a)
the increase in height in inches, during the first ten years
(b)
the change in the radius in centimeters, during the time interval t = 1 to t = 2 seconds
the change in the speed of sound in ft/s, during an increase in temperature from t = 32◦ F to
t = 100◦ F
(d) the displacement of the particle in cm, during the time interval t = t1 to t = t2 seconds
(c)
Z
2.
1
V (t)dt gal
(a)
0
3.
(b)
the change f (x1 ) − f (x2 ) in the values of f over the interval
(a)
displ = s(3) − s(0)
¸2
¸3
Z 2
Z 3
Z 3
v(t)dt =
(1 − t)dt +
(t − 3)dt = (t − t2 /2) + (t2 /2 − 3t) = −1/2;
=
Z
0
3
dist =
0
|v(t)|dt = (t − t2 /2)
Z
0
Z
1
0
0
¸1
= t /2 + t
dt +
1
5/2
1
3
¸2
2
Z
0
3
(2t − 5)dt
5/2
¸5/2
¸3
+ (t − 5t)
2
2
= 5/2
5/2
v
1
2
4
6
8
10
Z
t
t
-1
5.
(a)
v(t) = 20 +
= 1/2
2
(5 − 2t)dt = t2 /2
(5 − 2t)dt +
2
2
0
Z
2
+ (5t − t )
2
4.
Z
dt +
1
¸2
1
tdt +
2
tdt +
dist =
Z
1
2
¸3
+ (t2 /2 − 3t)
0
displ = s(3) − s(0)
Z
Z 3
v(t)dt =
=
0
¸2
+ (t2 /2 − t)
0
(b)
2
¸1
a(u)du; add areas of the small blocks to get
0
1
v(5) ≈ 20 + (1.5 + 2.7 + 4.6 + 6.2 + 7.6) = 31.3
2
¸3
¸3
+ t + (5t − t2 ) = 5;
2
2
Exercise Set 7.7
234
Z
(b)
v(10) = v(4) +
5
10
1
a(u)du ≈ 31.3 + (8.6 + 9.3 + 9.7 + 10 + 10.1) = 55.15
2
6. a > 0 and therefore (Theorem 7.5.6(a)) v > 0, so the particle is always speeding up for 0 < t < 10
Z
7.
(t3 − 2t2 + 1)dt =
1 4 2 3
t − t + t + C,
4
3
(a)
s(t) =
(b)
2
1 4 2 3
1
(0) − (0) + 0 + C = 1, C = 1, s(t) = t4 − t3 + t + 1
4
3
4
3
Z
v(t) = 4 cos 2t dt = 2 sin 2t + C1 , v(0) = 2 sin 0 + C1 = −1, C1 = −1,
Z
v(t) = 2 sin 2t − 1, s(t) = (2 sin 2t − 1)dt = − cos 2t − t + C2 ,
s(0) =
s(0) = − cos 0 − 0 + C2 = −3, C2 = −2, s(t) = − cos 2t − t − 2
Z
8.
(a)
(1 + sin t)dt = t − cos t + C, s(0) = 0 − cos 0 + C = −3, C = −2, s(t) = t − cos t − 2
s(t) =
Z
(b)
(t2 − 3t + 1)dt =
v(t) =
1 3 3 2
t − t + t + C1 ,
3
2
1 3 3 2
1
3
(0) − (0) + 0 + C1 = 0, C1 = 0, v(t) = t3 − t2 + t,
3
2
3
2
¶
Z µ
1 4 1 3 1 2
1 3 3 2
t − t + t dt =
t − t + t + C2 ,
s(t) =
3
2
12
2
2
1
1
1 4 1 3 1 2
1
(0)4 − (0)3 + (0)2 + C2 = 0, C2 = 0, s(t) =
t − t + t
s(0) =
12
2
2
12
2
2
Z
(a) s(t) = (2t − 3)dt = t2 − 3t + C, s(1) = (1)2 − 3(1) + C = 5, C = 7, s(t) = t2 − 3t + 7
v(0) =
9.
Z
(b)
v(t) =
Z
cos tdt = sin t + C1 , v(π/2) = 2 = 1 + C1 , C1 = 1, v(t) = sin t + 1,
(sin t+1)dt = − cos t+t+C2 , s(π/2) = 0 = π/2+C2 , C2 = −π/2, s(t) = − cos t+t−π/2
s(t) =
Z
10.
(a)
t2/3 dt =
s(t) =
Z
√
96
3
3 5/3
3
96
t + C, s(8) = 0 = 32 + C, C = − , s(t) = t5/3 −
5
5
5
5
5
2
2 3/2
2
13
13
t + C1 , v(4) = 1 = 8 + C1 , C1 = − , v(t) = t3/2 − ,
3
3
3
3
3
¶
Z µ
2 3/2 13
4 5/2 13
13
4
44
s(t) =
t −
dt =
t − t + C2 , s(4) = −5 =
32 − 4 + C2 = − + C2 ,
3
3
15
3
15
3
5
4 5/2 13
19
19
, s(t) =
t − t+
C2 =
5
15
3
5
(b) v(t) =
tdt =
Z
11.
(a)
displacement = s(π/2) − s(0) =
Z π/2
| sin t|dt = 1
distance =
0
π/2
distance =
π/2
Z
| cos t|dt = −
2π
¸2π
cos tdt = sin t
= −1
π/2
3π/2
Z
π/2
2π
cos tdt +
π/2
=1
0
Z
2π
sin tdt = − cos t
0
(b) displacement = s(2π) − s(π/2) =
Z
¸π/2
cos tdt = 3
3π/2
235
12.
Chapter 7
(a)
¸6
Z 6
displacement = s(6) − s(0) =
(2t − 4)dt = (t2 − 4t) = 12
0
¸2
¸6
Z 2 0
Z 6
Z 6
|2t − 4|dt =
(4 − 2t)dt +
(2t − 4)dt = (4t − t2 ) + (t2 − 4t) = 20
distance
0
(b)
0
Z
5
displacement =
Z
2
Z
3
|t − 3|dt =
0
5
5
−(t − 3)dt +
0
distance =
0
Z
(t − 3)dt = 13/2
3
|t − 3|dt = 13/2
0
13.
(a)
v(t) = t3 − 3t2 + 2t = t(t − 1)(t − 2)
Z 3
(t3 − 3t2 + 2t)dt = 9/4
displacement =
Z
0
3
distance =
Z
|v(t)|dt =
0
3
(b) displacement =
Z
3
1
Z
Z
(a)
v(t)dt = e3 − 9 + 4 ln 2
ln 2
1 1
( − )dt = 1 − ln 3
2
t
1
Z 2
Z
Z 3
|v(t)|dt = −
v(t)dt +
distance =
1
9
displacement =
Z
9
3
v(t)dt = 2 ln 2 − ln 3
2
3t−1/2 dt = 6
4
Z
9
|v(t)|dt =
v(t)dt = 6
4
4
v(t) = −2t + 3
Z 4
(−2t + 3)dt = −6
displacement =
1
Z
distance =
4
Z
| − 2t + 3|dt =
1
17.
3
3
distance =
16.
v(t)dt = 11/4
displacement =
Z
15.
Z
v(t)dt +
0
1
(b)
3
2
ln 2
|v(t)|dt = −
0
14.
−v(t)dt +
(et − 2)dt = e3 − 7
0
distance =
Z
2
v(t)dt +
0
Z
Z
1
Z
3/2
4
(−2t + 3)dt +
1
(2t − 3)dt = 13/2
3/2
1 2
t − 2t
2
¶
Z 5µ
1 2
t − 2t dt = −10/3
displacement =
2
1
¯
¶
¶
Z 4 µ
Z 5µ
Z 5¯
¯
¯1 2
1 2
1 2
¯
¯
t − 2t dt +
t − 2t dt = 17/3
−
distance =
¯ 2 t − 2t¯ dt =
2
2
1
1
4
v(t) =
8
2√
5t + 1 +
5
5
¶
¸3
Z 3µ
4
8
8
2√
dt =
(5t + 1)3/2 + t = 204/25
5t + 1 +
displacement =
5
5
75
5 0
0
Z 3
Z 3
|v(t)|dt =
v(t)dt = 204/25
distance =
v(t) =
0
0
2
Exercise Set 7.7
18.
236
v(t) = − cos t + 2
Z π/2
√
(− cos t + 2)dt = (π + 2 − 2)/2
displacement =
Z
π/4
π/2
distance =
Z
| − cos t + 2|dt =
π/4
π/2
(− cos t + 2)dt = (π +
√
2 − 2)/2
π/4
Z
19.
20.
1
2
1
sin πt dt = − cos πt + C
2
π
2
2
2
1
2
s = 0 when t = 0 which gives C = so s = − cos πt + .
π
π
2
π
π
1
dv
= cos πt. When t = 1 : s = 2/π, v = 1, |v| = 1, a = 0.
a=
dt
2
2
Z
3
3
(b) v = −3 t dt = − t2 + C1 , v = 0 when t = 0 which gives C1 = 0 so v = − t2
2
2
Z
3
1
1
t2 dt = − t3 + C2 , s = 1 when t = 0 which gives C2 = 1 so s = − t3 + 1.
s=−
2
2
2
When t = 1 : s = 1/2, v = −3/2, |v| = 3/2, a = −3.
(a)
s=
(a)
negative, because v is decreasing
(b)
speeding up when av > 0, so 2 < t < 5; slowing down when 1 < t < 2
(c)
negative, because the area between the graph of v(t) and the t-axis appears to be greater where
v < 0 compared to where v > 0
Z
21.
1
A = A1 + A2 =
Z
0
Z
22.
Z
π
0
Z
(1 − ex )dx +
1
A = A1 + A2 =
1/2
25.
s(t) =
1
(ex − 1)dx = 1/e + e − 2
0
1−x
dx +
x
Z
1
2
x−1
dx = −
x
µ
¶
1
− ln 2 + (1 − ln 2) = 1/2
2
20 3
20 3
t − 50t2 + 50t + s0 , s(0) = 0 gives s0 = 0, so s(t) =
t − 50t2 + 50t, a(t) = 40t − 100
3
3
150
0
sin xdx = 2 + 1 = 3
π
−1
Z
24.
3π/2
sin xdx −
0
Z
(x2 − 1)dx = 2/3 + 20/3 = 22/3
1
A = A1 + A2 =
23. A = A1 + A2 =
3
(1 − x2 )dx +
150
6
0
50
6
0
-100
-100
6
0
237
26.
Chapter 7
2
v(t) = 2t2 − 30t + v0 , v(0) = 3 = v0 , so v(t) = 2t2 − 30t + 3, s(t) = t3 − 15t2 + 3t + s0 , s(0) = −5 = s0 ,
3
2
so s(t) = t3 − 15t2 + 3t − 5
3
500
1200
0
70
25
0
-1200
27.
25
-200
s(t)
0
25
-30
v(t)
(a) from the graph the velocity is positive, so the displacement is
always increasing and is therefore positive
a(t)
v
0.5
0.4
0.3
0.2
t
1
(b)
28.
2
3
4
5
s(t) = t/2 + (t + 1)e−t
(a) If t0 < 1 then the area between the velocity curve and the
t-axis, between t = 0 and t = t0 , will always be negative, so
the displacement will be negative.
v
0.1
t
0.2 0.4 0.6 0.8
-0.1
-0.2
µ
(b)
s(t) =
½
29.
(a)
a(t) =
1
t2
−
2
200
0,
−10,
¶
ln(t + 0.1) −
t<4
t>4
t2
t
1
+
−
ln 10
4
20 200
½
(b)
25, t < 4
65 − 10t, t > 4
v(t) =
a
v
2
4
12
t
20
2
-5
4
6
-20
-40
-10
½
25t,
65t − 5t2 − 80,
(c)
x(t) =
(d)
x(6.5) = 131.25
t<4
, so x(8) = 120, x(12) = −20
t>4
8
10 12
t
1
Exercise Set 7.7
238
v 2 − v02
2(s − s0 )
(a)
From exercise 30 part (a), in Section 3 of Chapter 6, v 2 = v02 − 2g(s − s0 ), so a = −g =
(b)
From exercise 30 part (b), in Section 3 of Chapter 6, s = s0 + 12 (v0 + v)t, so t =
(c)
From exercise 30 part (c), in Section 3 of Chapter 6, s = s0 + vt + 12 gt2 = s0 + vt − 12 at2
31.
(a)
a = −1 mi/h/s = −22/15 ft/s2
32.
Take t = 0 when deceleration begins, then a = −10 so v = −10t + C1 , but v = 88 when t = 0 which
gives C1 = 88 thus v = −10t + 88, t ≥ 0
30.
(a)
(b)
2(s − s0 )
v0 + v
a = 30 km/h/min = 1/7200 km/s2
v = 45 mi/h = 66 ft/s, 66 = −10t + 88, t = 2.2 s
(b) v = 0 (the car is stopped) when t = 8.8 s
Z
Z
s =
v dt = (−10t + 88)dt = −5t2 + 88t + C2 , and taking s = 0 when t = 0, C2 = 0 so
s = −5t2 + 88t. At t = 8.8, s = 387.2. The car travels 387.2 ft before coming to a stop.
33.
a = a0 ft/s2 , v = a0 t + v0 = a0 t + 132 ft/s, s = a0 t2 /2 + 132t + s0 = a0 t2 /2 + 132t ft; s = 200 ft
121
20
when t =
, so
when v = 88 ft/s. Solve 88 = a0 t + 132 and 200 = a0 t2 /2 + 132t to get a0 = −
5
11
121
t + 132.
s = −12.1t2 + 132t, v = −
5
70
121
242
ft/s2
ft/s when t =
s
(b) v = 55 mi/h =
(a) a0 = −
5
3
33
60
(c) v = 0 when t =
s
11
34.
dv/dt = 3, v = 3t + C1 , but v = v0 when t = 0 so C1 = v0 , v = 3t + v0 . From ds/dt = v = 3t + v0 we
get s = 3t2 /2 + v0 t + C2 and, with s = 0 when t = 0, C2 = 0 so s = 3t2 /2 + v0 t. s = 40 when t = 4
thus 40 = 3(4)2 /2 + v0 (4), v0 = 4 m/s
35.
Suppose s = s0 = 0, v = v0 = 0 at t = t0 = 0; s = s1 = 120, v = v1 at t = t1 ; and s = s2 , v = v2 = 12
at t = t2 . From Exercise 30(a),
v12 − v02
, v 2 = 2as1 = 5.2(120) = 624. Applying the formula again,
2(s1 − s0 ) 1
v 2 − v12
, v 2 = v12 − 3(s2 − s1 ), so
−1.5 = a = 2
2(s2 − s1 ) 2
s2 = s1 − (v22 − v12 )/3 = 120 − (144 − 624)/3 = 280 m.
2.6 = a =
½
36.
a(t) =
4,
0,
t<2
, so, with v0 = 0, v(t) =
t>2
½
4t, t < 2
and, since s0 = 0, s(t) =
8, t > 2
½
2t2 , t < 2
8t − 8, t > 2
s = 100 when 8t − 8 = 100, t = 108/8 = 13.5 s
37. The truck’s velocity is vT = 50 and its position is sT = 50t + 5000. The car’s acceleration is aC = 2,
so vC = 2t, sC = t2 (initial position and initial velocity of the car are both zero). sT = sC when
50t + 5000 = t2 , t2 − 50t − 5000 = (t + 50)(t − 100) = 0, t = 100 s and sC = sT = t2 = 10, 000 ft
38.
Let t = 0 correspond to the time when the leader is 100 m from the finish line; let s = 0 correspond to
the finish line. Then vC = 12, sC = 12t − 115; aL = 0.5 for√t > 0, vL = 0.5t + 8, sL = 0.25t2 + 8t − 100.
sC = 0 at t = 115/12 ≈ 9.58 s, and sL = 0 at t = −16 + 4 41 ≈ 9.61, so the challenger wins.
39.
s = 0 and v = 112 when t = 0 so v(t) = −32t + 112, s(t) = −16t2 + 112t
(a)
v(3) = 16 ft/s, v(5) = −48 ft/s
(b)
v = 0 when the projectile is at its maximum height so −32t + 112 = 0, t = 7/2 s,
s(7/2) = −16(7/2)2 + 112(7/2) = 196 ft.
239
Chapter 7
(c)
s = 0 when it reaches the ground so −16t2 + 112t = 0, −16t(t − 7) = 0, t = 0, 7 of which t = 7
is when it is at ground level on its way down. v(7) = −112, |v| = 112 ft/s.
40.
s = 112 when t = 0 so s(t) = −16t2 + v0 t + 112. But s = 0 when t = 2 thus
−16(2)2 + v0 (2) + 112 = 0, v0 = −24 ft/s.
41.
(a)
s(t) = 0 when it hits the ground, s(t) = −16t2 + 16t = −16t(t − 1) = 0 when t = 1 s.
(b)
The projectile moves upward until it gets to its highest point where v(t) = 0,
v(t) = −32t + 16 = 0 when t = 1/2 s.
√
555/4 s
42.
s(t) = 0 when the rock hits the ground, s(t) = −16t2 + 555 = 0 when t =
√
√
√
(b) v(t) = −32t, v( 555/4) = −8 555, the speed at impact is 8 555 ft/s
43.
(a)
s(t) = 0 when the package hits the ground,
√
s(t) = −16t2 + 20t + 200 = 0 when t = (5 + 5 33)/8 s
√
√
√
(b) v(t) = −32t + 20, v[(5 + 5 33)/8] = −20 33, the speed at impact is 20 33 ft/s
44.
(a)
s(t) = 0 when the stone hits the ground,
s(t) = −16t2 − 96t + 112 = −16(t2 + 6t − 7) = −16(t + 7)(t − 1) = 0 when t = 1 s
(b)
v(t) = −32t − 96, v(1) = −128, the speed at impact is 128 ft/s
45.
(a)
s(t) = −4.9t2 + 49t + 150 and v(t) = −9.8t + 49
(a)
the projectile reaches its maximum height when v(t) = 0, −9.8t + 49 = 0, t = 5 s
(b)
s(5) = −4.9(5)2 + 49(5) + 150 = 272.5 m
(c)
the projectile reaches its starting point when s(t) = 150, −4.9t2 + 49t + 150 = 150,
−4.9t(t − 10) = 0, t = 10 s
(d)
v(10) = −9.8(10) + 49 = −49 m/s
(e)
s(t) = 0 when the projectile hits the ground, −4.9t2 + 49t + 150 = 0 when (use the quadratic
formula) t ≈ 12.46 s
(f )
v(12.46) = −9.8(12.46) + 49 ≈ −73.1, the speed at impact is about 73.1 m/s
46. take s = 0 at the water level and let h be the height of the bridge, then s = h and v = 0 when t = 0
so s(t) = −16t2 + h
(a)
s = 0 when t = 4 thus −16(4)2 + h = 0, h = 256 ft
(b)
2
First,
√ find how long it takes for the stone to hit the water (find t for s = 0) : −16t + h = 0,
t = h/4. Next, find how long it takes the sound to travel to the bridge: this time is h/1080
because the speed is constant
√ at 1080 ft/s. Finally, use the fact that the total of these two
√
√
h
h
+
= 4, h + 270 h = 4320, h + 270 h − 4320 = 0, and by the
times must be 4 s:
1080
4 p
√
√
−270 ± (270)2 + 4(4320)
, reject the negative value to get h ≈ 15.15,
quadratic formula h =
2
h ≈ 229.5 ft.
p
47.
g = 9.8/6 =p4.9/3 m/s2 , so v = −(4.9/3)t, s = −(4.9/6)t2 + 5,s = 0 when t =
v = −(4.9/3) 30/4.9 ≈ −4.04, so the speed of the module upon landing is 4.04 m/s
48.
s(t) = − 12 gt2 + v0 t; s = 1000 when v = 0, so 0 = v = −gt + v0 , t = v0 /g,
√
1000 = s(v0 /g) = − 12 g(v0 /g)2 + v0 (v0 /g) = 12 v02 /g, so v02 = 2000g, v0 = 2000g.
√
The initial velocity on the Earth would have to be 6 times faster than that on the Moon.
30/4.9 and
Exercise Set 7.7
49.
51.
52.
53.
fave =
240
Z
1
3−1
3x dx =
1
Z
1
fave =
π−0
fave =
fave =
3
Z
Z
e
1
54.
55.
(a)
(c)
50. fave =
1
1
sin x
π
0
1
fave =
ln 5 − (−1)
1
2−0
fave =
=6
1
2 − (−1)
Z
2
−1
x2 dx =
1 3
x
9
¸2
=1
−1
¸π
π
cos x dx =
1
=
e−1
¸3
¸π
1
sin x dx = − cos x = 2/π
π
0
π
0
1
π−0
3 2
x
4
=0
0
1
1
1
dx =
(ln e − ln 1) =
x
1−e
e−1
Z
ln 5
−1
Z
ex dx =
1
5 − e−1
(5 − e−1 ) =
ln 5 + 1
1 + ln 5
√
(b) (x∗ )2 = 4/3,√x∗ = ±2/ 3,
but only 2/ 3 is in [0, 2]
2
x2 dx = 4/3
0
y
4
2
3
56.
(a)
1
fave =
4−0
(c)
x
2
Z
4
2x dx = 4
(b)
2x∗ = 4, x∗ = 2
0
y
8
4
2
57.
58.
(a)
vave =
1
4−1
x
4
Z
4
(3t3 + 2)dt =
1
(b)
vave =
100 − 7
s(4) − s(1)
=
= 31
4−1
3
(a)
aave =
1
5−0
Z
5
(t + 1)dt = 7/2
0
v(π/4) − v(0)
=
(b) aave =
π/4 − 0
59.
263
1 789
=
3 4
4
√
√
2/2 − 1
= (2 2 − 4)/π
π/4
time to fill tank = (volume of tank)/(rate of filling) = [π(3)2 5]/(1) = 45π, weight of water in tank at
time t = (62.4) (rate of filling)(time) = 62.4t,
Z 45π
1
62.4t dt = 1404π lb
weightave =
45π 0
241
60.
61.
Chapter 7
(a)
If x is the distance from the cooler end, then the temperature is T (x) = (15 + 1.5x)◦ C, and
Z 10
1
(15 + 1.5x)dx = 22.5◦ C
Tave =
10 − 0 0
(b)
By the Mean-Value Theorem for Integrals there exists x∗ in [0, 10] such that
Z 10
1
(15 + 1.5x)dx = 22.5, 15 + 1.5x∗ = 22.5, x∗ = 5
f (x∗ ) =
10 − 0 0
(a)
amount of water = (rate of flow)(time) = 4t gal, total amount = 4(30) = 120 gal
Z 60
(4 + t/10)dt = 420 gal
(b) amount of water =
Z
(c)
0
120
amount of water =
(10 +
√
√
t)dt = 1200 + 160 30 ≈ 2076.36 gal
0
62.
(a)
The maximum value of R occurs at 4:30 P.M. when t = 0.
Z 60
100(1 − 0.0001t2 )dt = 5280 cars
(b)
0
Z
63.
Z
b
b
[f (x) − fave ] dx =
(a)
a
Z
f (x)dx −
a
Z
because fave (b − a) =
Z
b
b
f (x)dx − fave (b − a) = 0
fave dx =
a
a
b
f (x)dx
a
(b)
Z b
Z b
no, because if
[f (x) − c]dx = 0 then
f (x)dx − c(b − a) = 0 so
a
a
Z b
1
f (x)dx = fave is the only value
c=
b−a a
EXERCISE SET 7.8
Z
1.
3
7
(a)
u du
(b)
1
2.
(a)
1
2
Z
Z
1
1
−
2
Z
4
u
1/2
7
(c)
Z
−1
eu du
1
u = 2x + 1,
2
1
4. u = 4x − 2,
4
5.
sin u du
−π
udu
1Z
(d)
Z
3
1 5
u du =
u
10
¸3
4
1
Z
1
u = 1 − 2x, −
2
1
6. u = 4 − 3x, −
3
6
1 4
u du =
u
16
1
¸6
3
2
Z
1
2
1
u du = − u4
8
¸1
3
3
Z
1
−2
1
2
3
1
u du = − u9
27
¸−2
1
(u − 3)u1/2 du
¸1
= 121/5
0
¸2
= 80
1
1
= 10, or − (1 − 2x)4
8
8
4
3
1
= 121/5, or
(2x + 1)5
10
1
= 80, or
(4x − 2)4
16
(d)
2
(b)
0
3.
Z
π
Z
1
u2 du
(c)
du
1
π
¸0
= 10
−1
1
= 19, or − (4 − 3x)9
27
¸2
= 19
1
0
−3
(u + 5)u20 du
Exercise Set 7.8
242
Z
7.
9
u = 1 + x,
Z
1
2
2
(1 + x)5/2 − (1 + x)3/2
5
3
or
Z
8.
4
u = 4 − x,
Z
¸8
π/4
u = x/2, 8
11.
Z
2
3
u = 3x,
¸9
4
(u3/2 − 4u1/2 )du =
¸0
9
2 5/2 8 3/2
u − u
5
3
¸π/4
2
sin u
3
0
= −506/15
9
¸π/2
√
= 8 − 4 2, or − 8 cos(x/2)
0
¸π/6
2
= 2/3, or sin 3x
= 2/3
3
0
0
13
1
+4=
when x = − ln 3,
3
3
¸7
Z 7
1
ln 3
= ln(7) − ln(13/3) = ln(21/13)
du = ln u
u = e + 4 = 3 + 4 = 7 when x = ln 3,
13/3 u
13/3
u = ex + 4, du = ex dx, u = e− ln 3 + 4 =
12. u = 3 − 4ex , du = −4ex dx, u = −1 when x = 0, u = −17 when x = ln 5
¸−17
Z
1 −17
1 2
−
u du = − u
= −36
4 −1
8
−1
13.
14.
15.
Z
1
3
5
p
25 −
Z
4
p
Z
0
·
¸
1 1
25
2
π(5) =
π
du =
3 4
12
16 − u2 du =
0
−
1
2
6
−6
Z
1
17.
0
p
1 − u2 du =
1
π/8
0
7
−2
p
1 − u2 du =
0
−1
1 1
· [π(1)2 ] = π/8
2 4
= −(x + 5)
1
=−
A=
0
1
dx
=−
(3x + 1)2
3(3x + 1)
1
fave =
4−0
Z
4
−2x
e
0
√
= 3 2/4
0
¸7
3
Z
21.
1
¸π/8
3
20.
Z
3
3 cos 2x dx = sin 2x
2
(x + 5)
19.
1
2
¸1
1
1
sin πxdx = − cos πx = − (−1 − 1) = 2/π
π
π
0
A=
Z
·
¸
1 1
π(4)2 = 2π
2 4
p
36 − u2 du = π(6)2 /2 = 18π
Z
18.
u2
0
1
2
Z
16.
√
=8−4 2
0
¸π/2
cos u du =
¸4
= −506/15
−5
sin u du = −8 cos u
π/2
= 1192/15,
1
= 1192/15
0
10.
2 5/2 2 3/2
u − u
5
3
0
Z
8
2
or (4 − x)5/2 − (4 − x)3/2
5
3
(u3/2 − u1/2 )du =
1
(u − 4)u1/2 du =
9
9.
9
(u − 1)u1/2 du =
1
1
1
+ =
12 8
24
¸1
=
0
1
dx = − e−2x
8
1
4
¸4
=
0
1 − e−8
8
243
22.
23.
25.
Chapter 7
1
fave =
1/4 − (−1/4)
2
(3x + 1)1/2
3
2 3
(x + 9)1/2
3
#1/4
2
4
sec πxdx = tan πx
=
π
π
−1/4
Z
2
−1/4
¸1
= 2/3
¸1
24.
−1
√
2 √
= ( 10 − 2 2)
3
1
27. u = x + 4x + 7,
2
2
28.
1
29.
32.
Z
28
u
−1/2
¸28
du = u
√
Z
34.
35.
1
3
Z
= 1/2
sin u du = −2 cos u
Z
1
3
−1
u2 du =
0
9
u = 5 + x,
ln(x + e)
¸√π
= 2/3
31.
0
5
sin(x2 )
2
=0
0
= −4
1
tan u
3
1 3
u
9
¸π/3
√
= ( 3 − 1)/3
π/4
¸−1
= −1/9
0
0
u−5
√ du =
u
Z
9
(u
− 5u
1/2
4
= ln(2e) − ln e = ln 2
−1/2
2
)du = u3/2 − 10u1/2
3
38.
1
2
− e−x
2
¸√2
¸9
= 8/3
4
= (e−1 − e−2 )/2
1
p
1
4 − u2 du = [π(2)2 ] = 2π
2
−2
2
(a)
(c)
Z
1 4
f (u)du = 5/3
3 1
Z 0
Z
2
f (u)du = −1/2
u = x , 1/2
(b)
u = 3x + 1,
4
Z
42.
¸π/4
1
1
u = 4 − 3y, y = (4 − u), dy = − du
3
3
Z 1
Z 4
1
1
16 − 8u + u2
du =
(16u−1/2 − 8u1/2 + u3/2 )du
−
27 4
27 1
u1/2
·
¸4
16 3/2 2 5/2
1
1/2
32u − u + u
= 106/405
=
27
3
5
1
Z
41.
2
(tan x)3/2
3
30.
π/4
ie
40.
√
√
√
√
28 − 12 = 2( 7 − 3)
1
π/3
4
37.
= 1/10
−1
π
Z
36.
¸0
¸2
sec2 u du =
u = sin 3θ,
= 38/15
1
¸2π
2π
x, 2
u = 3θ,
¸2
12
π
33.
=
12
1
1
dx = −
(x − 3)2
x−3
1 3
(t + 1)20
10
26.
1/2
¸π/4
1
2
sin x
=0
2
−3π/4
u=
2
(5x − 1)3/2
15
0
2
Z
1/4
1
u = 1 − x,
0
4
u = 3x,
1
3
Z
9
f (u)du = 5/3
0
f (u)du = −1/2
0
Z
xm (1 − x)n dx = −
1
0
Z
(1 − u)m un du =
0
1
Z
un (1 − u)m du =
0
1
xn (1 − x)m dx
Exercise Set 7.8
43.
244
sin x = cos(π/2 − x),
Z π/2
Z
Z π/2
sinn x dx =
cosn (π/2 − x)dx = −
0
0
Z
Z
π/2
cosn u du =
=
44.
0
u = 1 − x, −
(u = π/2 − x)
cosn u du
π/2
π/2
cosn x dx
0
Z
0
(by replacing u by x)
0
Z
(1 − u)u du =
n
1
1
Z
1
(1 − u)u du =
n
0
(un − un+1 )du =
0
1
1
1
−
=
n+1 n+2
(n + 1)(n + 2)
Z
45.
e1.528t dt = 524.959e1.528t + C; y(0) = 750 = 524.959 + C, C = 225.041,
y(t) = (802.137)
y(t) = 524.959e1.528t + 225.041, y(12) = 48, 233, 525, 650
46.
275000
Vave =
10 − 0
Z
47.
s(t) =
−0.17t
e
−0.17t
¸10
dt = −161764.7059e
0
= $132, 212.96
0
(25 + 10e−0.05t )dt = 25t − 200e−0.05t + C
√
s(10) − s(0) = 250 − 200(e−0.5 − 1) = 450 − 200/ e ≈ 328.69 ft
(b)
yes; without it the distance would have been 250 ft
Z
k
e2x dx = 3,
0
2
Vrms
=
(a)
1 2x
e
2
¸k
0
1
1/f − 0
1
1
= 3, (e2k − 1) = 3, e2k = 7, k = ln 7
2
2
Z
1/f
Vp2 sin2 (2πf t)dt =
0
1
fV 2
2 p
Z
1/f
[1 − cos(4πf t)]dt
0
¸1/f
√
1
1
1
2
sin(4πf t)]
= Vp2 , so Vrms = Vp / 2
= f Vp [t −
2
4πf
2
0
√
√
Vp / 2 = 120, Vp = 120 2 ≈ 169.7 V
(b)
50.
10
(a)
48.
49.
Z
Let u = t − x, then du = −dx and
Z 0
Z t
Z t
f (t − x)g(x)dx = −
f (u)g(t − u)du =
f (u)g(t − u)du;
0
t
0
the result follows by replacing u by x in the last integral.
Z
51.
Z a
f (a − u)
f (a − u) + f (u) − f (u)
du =
du
f
(a
−
u)
+
f
(u)
f (a − u) + f (u)
a
0
Z a
Z a
f (u)
du, I = a − I so 2I = a, I = a/2
du −
=
f
(a
−
u) + f (u)
0
0
I=−
(a)
0
(b) 3/2
52.
(c)
π/4
Z 1
Z 1
1
1
1
1
2
, dx = − 2 du, I =
du = −I so I = 0 which is impossible
(−1/u
)du
=
−
2
2
u
u
−1 1 + 1/u
−1 u + 1
1
because
is positive on [−1, 1]. The substitution u = 1/x is not valid because u is not continuous
1 + x2
for all x in [−1, 1].
x=
Z
53.
1
sin πxdx = 2/π
0
245
55.
Chapter 7
(a)
Let u = −x then
Z a
Z
f (x)dx = −
−a
Z
−a
f (−u)du =
a
−a
a
Z
f (−u)du = −
so, replacing u by x in the latter integral,
Z a
Z a
Z
Z a
f (x)dx = −
f (x)dx, 2
f (x)dx = 0,
a
−a
f (u)du
a
f (x)dx = 0
Z 0
Z a
The graph of f is symmetric about the origin so
f (x)dx is the negative of
f (x)dx thus
−a
0
Z a
Z 0
Z a
f (x)dx =
f (x) +
f (x)dx = 0
−a
−a
−a
Z
(b)
−a
Z
a
−a
Z
0
−a
f (x)dx =
0
0
−a
Z
Z
f (−u)du =
Z
0
−a
f (x)dx to get
Z
a
0
a
f (x)dx
0
a
f (x)dx = 2
0
0
f (u)du =
Z
a
f (x)dx +
0
Z
a
f (−u)du =
Z
a
f (x)dx =
−a
f (x)dx, let u = −x in
0
a
−a
Z
0
Z
a
a
f (x)dx +
f (x)dx = −
Z
so
−a
f (x)dx
0
The graph of f (x) is symmetric about the y-axis so there is as much signed area to the left of
the y-axis as there is to the right.
EXERCISE SET 7.9
1.
(a)
(b)
y
(c)
y
3
3
3
2
2
2
1
1
1
t
1
2.
y
2
3
0.5
t
1
1
y
3
2
1
2
3
3
2
1
t
iac
3.
(a)
ln t
(c)
ln t
1
i1/c
= ln(ac) = ln a + ln c = 7
ia/c
1
= ln(a/c) = 2 − 5 = −3
(b)
ln t
(d)
ln t
1
ia3
1
= ln(1/c) = −5
= ln a3 = 3 ln a = 6
e2
t
Exercise Set 7.9
246
i√a
4.
(a)
ln t
(c)
ln t
1
i2/a
1
i2a
= ln a1/2 = 2
(b)
= ln 2 − 4
(d) ln t
ln t
= ln 2 + 4
1
ia
2
= 4 − ln 2
5.
ln 5 ≈ 1.603210678; ln 5 = 1.609437912; magnitude of error is < 0.0063
6.
ln 3 ≈ 1.098242635; ln 3 = 1.098612289; magnitude of error is < 0.0004
7.
(a) x−1 , x > 0
(c) −x2 , −∞ < x < +∞
(e) x3 , x > 0
√
(g) x − 3 x, −∞ < x < +∞
8.
(a)
f (ln 3) = e−2 ln 3 = eln(1/9) = 1/9
(b)
f (ln 2) = eln 2 + 3e− ln 2 = 2 + 3eln(1/2) = 2 + 3/2 = 7/2
9.
(a)
3π = eπ ln 3
(b)
2
10.
(a)
π −x = e−x ln π
(b)
x2x = e2x ln x
·µ
11.
(a)
(b)
1
1+
x
lim
x→+∞
¶x ¸2
lim
x→+∞
1
1+
x
1
1+
y
¶y/3
·µ
y→+∞
x→0
15.
(a)
16.
(a)
17.
F 0 (x) =
y→+∞
0
=
µ
lim
y→+∞
1
1+
y
¶y ¸1/3
= e1/3
x→0
14. g 0 (x) = 1 − cos x
(b)
√
2x x2 + 1
0
·
¶y ¸1/3
3
1
(3x2 ) =
3
x
x
F 0 (x) =
(b)
= lim
1
1+
y
h
i1/3
lim (1 + x)1/3x = lim (1 + x)1/x
= e1/3
(a)
2 ln 2
y→0
(b)
(a)
√
=e
= e2
y = 3x, lim
(a)
2
¶x ¸2
(a)
g 0 (x) = x2 − x
19.
=
y→0
13.
18.
µ
·
√
h
i2
y = 2x, lim (1 + y)2/y = lim (1 + y)1/y = e2
µ
12.
(b) x2 , x 6= 0
(d) −x, −∞ < x < +∞
(f ) ln x + x, x > 0
ex
(h)
,x>0
x
(b)
1
=1
x
µ ¶
µ ¶
1
1
−
sin
2
x
x
eln x
cos x
−(x2 + 3) sin x − 2x cos x
, F 00 (x) =
2
x +3
(x2 + 3)2
(b) 1/3
√
3x
3x2 + 1, F 00 (x) = √
3x2 + 1
(b)
√
13
Z x2
p
p
√
d
t 1 + tdt = x2 1 + x2 (2x) = 2x3 1 + x2
dx 1
√
Z x2
√
2 2
2 2
4 2
3/2
5/2
t 1 + tdt = − (x + 1) + (x + 1) −
3
5
15
1
(c)
0
(c)
√
6/ 13
247
20.
Chapter 7
(a)
(b)
d
dx
d
dx
Z
a
f (t)dt = −
x
Z
a
d
dx
f (t)dt = −
g(x)
Z
d
dx
x
f (t)dt = −f (x)
a
Z
g(x)
f (t)dt = −f (g(x))g 0 (x)
a
tan2 x
sec2 x = − tan2 x
1 + tan2 x
21.
(a)
− sin x2
(b)
−
22.
(a)
−(x2 + 1)40
(b)
− cos3
23.
−3
µ ¶µ
¶
1
1
cos3 (1/x)
− 2 =
x
x
x2
x2 − 1
3x − 1
+
2x
9x2 + 1
x4 + 1
24. If f is continuous on an open interval I and g(x), h(x), and a are in I then
Z g(x)
Z a
Z g(x)
Z h(x)
Z g(x)
f (t)dt =
f (t)dt +
f (t)dt = −
f (t)dt +
f (t)dt
h(x)
so
25.
26.
27.
d
dx
h(x)
Z
g(x)
a
a
a
f (t)dt = −f (h(x))h0 (x) + f (g(x))g 0 (x)
h(x)
sin2 (x3 )(3x2 ) − sin2 (x2 )(2x) = 3x2 sin2 (x3 ) − 2x sin2 (x2 )
1
2
1
(1) −
(−1) =
(b)
1+x
1−x
1 − x2
(a)
1
1
(3) − (1) = 0 so F (x) is constant on (0, +∞). F (1) = ln 3 so F (x) = ln 3 for all x > 0.
3x
x
Z 5
Z 7
Z 10
Z 10
Z 3
f (t)dt = 0,
f (t)dt = 6,
f (t)dt = 0; and
f (t)dt =
(4t − 37)/3dt = −3
from geometry,
F 0 (x) =
0
(a)
3
5
7
7
F (0) = 0, F (3) = 0, F (5) = 6, F (7) = 6, F (10) = 3
(b) F is increasing where F 0 = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2] and
[6, 37/4]
(c)
(d)
critical points when F 0 (x) = f (x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum
−9/4 at x = 3/2
F(x)
6
4
2
2
4
6
8
10
x
-2
28.
fave =
1
10 − 0
Z
f (t)dt =
0
1
F (10) = 0.3
10
¸x
1
1
(−t)dt = − t2
= (1 − x2 ),
2 −1
2
−1
(
Z x
Z 0
(1 − x2 )/2, x < 0
1 1 2
(−t)dt +
t dt = + x ; F (x) =
x ≥ 0 : F (x) =
2 2
(1 + x2 )/2, x ≥ 0
−1
0
Z
29.
10
x < 0 : F (x) =
x
Exercise Set 7.9
248
Z
30.
0 ≤ x ≤ 2 : F (x) =
Z
x
t dt =
0
Z
2
x > 2 : F (x) =
t dt +
0
Z
x
31. y(x) = 2 +
1
Z
32.
x
y(x) =
x
x
¸x
=
2x − 2, x > 2
1
5 3 4/3
+ x
4 4
2 3/2 2
x − + 2x1/2 − 2
3
3
(sec2 t − sin t)dt = tan x + cos x −
y(x) = 1 +
x2 /2, 0 ≤ x ≤ 2
2 dt = 2 + 2(x − 2) = 2x − 2; F (x) =
3
t1/3 dt = 2 + t4/3
4
(t1/2 + t−1/2 )dt =
Z
(
2
1
33.
1 2
x,
2
√
2/2
π/4
Z
34.
x
2
tet dt =
y(x) =
0
Z
36.
Z
1 −x2 1
e
−
2
2
35. P (x) = P0 +
x
r(t)dt individuals
0
T
s(T ) = s1 +
v(t)dt
1
37. II has a minimum at x = 1, and I has a zero there, so I could be the derivative of II; on the other
hand I has a minimum near x = 1/3, but II is not zero there, so II could not be the derivative of I
38. (b)
39.
1
lim (xk − 1) = ln x lim xk = ln x by L’Hôpital’s rule (with respect to k)
k→0
k
k→0
(a)
where f (t) = 0; by the first derivative test, at t = 3
(b)
where f (t) = 0; by the first derivative test, at t = 1
(c)
at t = 0, 1 or 5; from the graph it is evident that it is at t = 5
(d)
at t = 0, 3 or 5; from the graph it is evident that it is at t = 3
(e)
F is concave up when F 00 = f 0 is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4);
it is concave down on (1/2, 2) and (4, 5)
(f )
F(x)
1
0.5
1
2
3
4
2
4
5
x
-0.5
-1
40.
(a)
erf(x)
1
-4
-2
x
-1
(c)
(e)
(g)
erf0 (x) > 0 for all x, so there are no relative extrema
2 √
erf00 (x) = −4xe−x / π changes sign only at x = 0 so that is the only point of inflection
lim erf(x) = +1, lim erf(x) = −1
x→+∞
x→−∞
249
41.
Chapter 7
C 0 (x) = cos(πx2 /2), C 00 (x) = −πx sin(πx2 /2)
cos t goes from negative to positive at 2kπ − π/2, and from positive
√ to negative at t = 2kπ + π/2,
so C(x) has relative minima when πx2 /2 = 2kπ − π/2,√x = ± 4k − 1, k = 1, 2, . . ., and C(x)
has relative maxima when πx2 /2 = (4k + 1)π/2, x = ± 4k + 1, k = 0, 1, . . ..
√
(b) sin t changes sign at t = kπ, so C(x) has inflection points at πx2 /2 = kπ, x = ± 2k, k = 1, 2, . . .;
the case k = 0 is distinct due to the factor of x in C 00 (x), but x changes sign at x = 0 and
sin(πx2 /2) does not, so there is also a point of inflection at x = 0
(a)
Z
42.
Let F (x) =
1
h→0 h
Z
lim
1
x
F (x + h) − F (x)
1
ln tdt, F (x) = lim
= lim
h→0
h→0 h
h
0
ln tdt; but F 0 (x) = ln x so
x
ln tdt = ln x
x
Differentiate: f (x) = 3e3x , so 2 +
Z
x
¸x
x
= 2 + e3x − e3a = e3x provided
3e3t dt = 2 + e3t
f (t)dt = 2 +
a
e3a = 2, a = (ln 2)/3.
44.
x+h
x+h
Z
43.
Z
a
a
The area under 1/t for x ≤ t ≤ x + 1 is less than the area of the rectangle with altitude 1/x and
base 1, but greater than the area of the rectangle with altitude 1/(x + 1) and base 1.
¸x+1
Z x+1
1
(b)
dt = ln t
= ln(x + 1) − ln x = ln(1 + 1/x), so
t
x
x
1/(x + 1) < ln(1 + 1/x) < 1/x for x > 0.
(a)
(c)
from part (b), e1/(x+1) < eln(1+1/x) < e1/x , e1/(x+1) < 1 + 1/x < e1/x ,
ex/(x+1) < (1 + 1/x)x < e; by the Squeezing Theorem, lim (1 + 1/x)x = e.
x→+∞
(d)
45.
Use the inequality ex/(x+1) < (1 + 1/x)x to get e < (1 + 1/x)x+1 so
(1 + 1/x)x < e < (1 + 1/x)x+1 .
¯
µ
¶50 ¯¯
¯
1
¯
¯
From Exercise 44(d) ¯e − 1 +
¯ < y(50), and from the graph y(50) < 0.06
¯
¯
50
0.2
0
100
0
46.
F 0 (x) = f (x), thus F 0 (x) has a value at each x in I because f is continuous on I so F is continuous
on I because a function that is differentiable at a point is also continuous at that point
CHAPTER 7 SUPPLEMENTARY EXERCISES
5.
If the acceleration a = const, then v(t) = at + v0 , s(t) = 12 at2 + v0 t + s0
6.
(a)
Divide the base into n equal subintervals. Above each subinterval choose the lowest and highest
points on the curved top. Draw a rectangle above the subinterval going through the lowest point,
and another through the highest point. Add the rectangles that go through the lowest points to
obtain a lower estimate of the area; add the rectangles through the highest points to obtain an
upper estimate of the area.
Supplementary Exercises 7
7.
(b)
n = 10: 25.0 cm, 22.4 cm
(c)
n = 20: 24.4 cm, 23.1 cm
(a)
(c)
(e)
8.
(a)
(c)
3
1 1
+ =
2 4
4
¶
µ
35
3
=−
5 −1 −
4
4
(b) −1 −
(d)
−2
not enough information
(f )
not enough information
5
1
+2=
2
2
not enough information
Z
9.
(a)
(b)
(c)
10.
250
Z
1
−1
dx +
1
−1
1 2
(x + 1)3/2
3
p
(b)
not enough information
13
1
(d) 4(2) − 3 =
2
2
1 − x2 dx = 2(1) + π(1)2 /2 = 2 + π/2
¸3
1
(103/2 − 1) − 9π/4
3
− π(3)2 /4 =
0
1
u = x , du = 2xdx;
2
Z
2
1
2
1
3
=−
2
2
0
1
p
1 − u2 du =
1
π(1)2 /4 = π/8
2
y
1
0.8
0.6
0.4
0.2
x
0.2
11.
12.
13.
0.6
1
The rectangle with vertices (0, 0), (π, 0), (π, 1) and (0, 1) has area π and is much too large; so is the
triangle with vertices (0, 0), (π, 0) and (π, 1) which has area π/2; 1 − π is negative; so the answer is
35π/128.
e2x
3ex
= ex − x
so
Divide ex + 3 into e2x to get x
e +3
e +3
Z
Z
Z
ex
e2x
x
dx
−
3
dx
=
e
dx = ex − 3 ln(ex + 3) + C
ex + 3
ex + 3
Since y = ex and y = ln x are inverse functions, their graphs are
symmetric with respect to the line y = x; consequently the areas A1
and A3 are equal (see figure). But A1 + A2 = e, so
Z 1
Z e
ln xdx +
ex dx = A2 + A3 = A2 + A1 = e
1
0
y
e
A1
1
A2
A3
1
1X
n
(a)
x
n
X
√
k
=
f (x∗k )∆x where f (x) = x, x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
n
k=1
k=1
Z 1
n r
X
2
1
k
=
x1/2 dx =
lim
n→+∞ n
n
3
0
n
14.
r
e
k=1
251
Chapter 7
µ ¶4
n
X
k
=
f (x∗k )∆x where f (x) = x4 , x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
n
k=1
k=1
Z 1
n µ ¶4
1X k
1
lim
=
x4 dx =
n→+∞ n
n
5
0
1X
n
n
(b)
k=1
(c)
n
X
X
ek/n
=
f (x∗k )∆x where f (x) = ex , x∗k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
n
k=1
k=1
Z 1
n
n
k/n
X
X
e
= lim
lim
f (x∗k )∆x =
ex dx = e − 1.
n→+∞
n→+∞
n
0
n
k=1
15.
k=1
1
is positive and increasing on the interval [1, 2], the left endpoint approximation
x
1
overestimates the integral of and the right endpoint approximation underestimates it.
x
Since f (x) =
(a)
(b)
For n = 5 this becomes
¸ Z 2
·
¸
·
1
1
1
1
1
1
1
1
1
1
1
+
+
+
+
<
dx < 0.2
+
+
+
+
0.2
1.2 1.4 1.6 1.8 2.0
1.0 1.2 1.4 1.6 1.8
1 x
Z 2
1
dx = ln 2 is
For general n the left endpoint approximation to
1 x
n
n
n−1
X
X
1
1
1
1X
=
=
and the right endpoint approximation is
n
1 + (k − 1)/n
n+k−1
n+k
k=1
k=1
k=0
Z 2
n
n
n−1
X
X
X
1
1
1
1
. This yields
<
dx <
which is the desired inequality.
n+k
n+k
x
n
+
k
1
k=1
k=1
k=0
1
1
1
1
=
so
≤ 0.1, n ≥ 5
(c) By telescoping, the difference is −
n 2n
2n
2n
(d) n ≥ 1, 000
16.
The direction field is clearly an even function, which means that the solution is even, its derivative
is odd. Since sin x is periodic and the direction field is not, that eliminates all but x, the solution of
which is the family y = x2 /2 + C.
17.
(a)
1 · 2 + 2 · 3 + · · · + n(n + 1) =
n
X
k=1
k(k + 1) =
n
X
k2 +
k=1
n
X
k
k=1
1
1
1
n(n + 1)(2n + 1) + n(n + 1) = n(n + 1)(n + 2)
6
2
3
µ
¶
¶
n−1 µ
n−1
n−1
X
X
X
k
1 1
17 n − 1
9
9
1
9
−
;
=
1− 2
k = (n − 1) − 2 · (n − 1)(n) =
n n2
n
n
n
n 2
2
n
k=1
k=1
k=1
µ
¶
17
17 n − 1
=
lim
n→+∞ 2
n
2
"
#
¸
3
2
2
3 ·
3
3
X
X
X
X
X
X
1
1
2i + (2)(3) = 2
i+
j =
i+
3 = 2 · (3)(4) + (3)(3) = 21
2
2
i=1 j=1
j=1
i=1
i=1
i=1
=
(b)
(c)
18.
(a)
14
X
k=0
(k + 4)(k + 1)
(b)
19
X
k=5
(k − 1)(k − 4)
Supplementary Exercises 7
19.
(a)
(b)
20.
21.
22.
252
Z
Z
If u = sec x, du = sec x tan xdx,
sec2 x tan xdx = udu = u2 /2 + C1 = (sec2 x)/2 + C1 ;
Z
Z
2
2
if u = tan x, du = sec xdx,
sec x tan xdx = udu = u2 /2 + C2 = (tan2 x)/2 + C2 .
They are equal only if sec2 x and tan2 x differ by a constant, which is true.
iπ/4
iπ/4
1
1
sec2 x
= (2 − 1) = 1/2 and 12 tan2 x
= (1 − 0) = 1/2
0
0
2
2
Z p
Z
p
2
1 + x−2/3 dx = x−1/3 x2/3 + 1dx; u = x2/3 + 1, du = x−1/3 dx
3
Z
3
u1/2 du = u3/2 + C = (x2/3 + 1)3/2 + C
2
1
2
(a)
Z bX
n
fk (x)dx =
a k=1
n Z
X
k=1
b
fk (x)dx
a
Z
(b) yes; substitute ck fk (x) for fk (x) in part (a), and then use
ck fk (x)dx = ck
a
Theorem 7.5.4
Z x
1
dt
23. (a)
1
+
t2
1
Z x
1
(b)
dt
2
tan(π/4−2) 1 + t
24.
Z
b
b
fk (x)dx from
a
(a)
F 0 (x) =
x−3
; increasing on [3, +∞), decreasing on (−∞, 3]
x2 + 7
(b)
F 00 (x) =
7 + 6x − x2
(7 − x)(1 + x)
=
; concave up on (−1, 7), concave down on (−∞, −1) and
(x2 + 7)2
(x2 + 7)2
(7, +∞)
(c)
x−3
= 0 when x = 3, which is a relative minimum, and hence the absolute minimum,
x2 + 7
by the first derivative test.
F 0 (x) =
(d)
F(x)
3
2
1
-10
10
20
x
1
1
+
(−1/x2 ) = 0 so F is constant on (0, +∞).
2
1+x
1 + (1/x)2
25.
F 0 (x) =
26.
(−3, 3) because f is continuous there and 1 is in (−3, 3)
27.
(a)
The domain is (−∞, +∞); F (x) is 0 if x = 1, positive if x > 1, and negative if x < 1, because
the integrand is positive, so the sign of the integral depends on the orientation (forwards or
backwards).
(b)
The domain is [−2, 2]; F (x) is 0 if x = −1, positive if −1 < x ≤ 2, and negative if −2 ≤ x < −1;
same reasons as in part (a).
253
Chapter 7
28.
The left endpoint of the top boundary is ((b − a)/2, h) and the right endpoint of the top boundary is
((b + a)/2, h) so
x < (b − a)/2
2hx/(b − a),
h,
(b − a)/2 < x < (b + a)/2
f (x) =
2h(x − b)/(a − b),
x > (a + b)/2
The area of the trapezoid is given by
Z (b+a)/2
Z b
Z (b−a)/2
2hx
2h(x − b)
dx +
dx = (b − a)h/4 + ah + (b − a)h/4 = h(a + b)/2.
hdx +
b
−
a
a−b
0
(b−a)/2
(b+a)/2
Z
29.
24
³
´
√
2000e−t/48 + 500 sin(πt/12) dt = 96000(1 − 1/ e) ≈ 37, 773.06
(a)
0
(b)
(c)
1
8−0
Z
8
³
´
2000e−t/48 + 500 sin(πt/12)dt = 1125/π + 12000(1 − e−1/6 ) ≈ 2, 200.32
0
2300
0
2000
(d)
30.
31.
32.
8
maximum rate is 2285.32 kW/h at t = 4.8861
1
wave =
52 − 26
Z
52
(t/7)dt = 39/7; t∗ /7 = 39/7, t∗ = 39
26
(a)
no, since the velocity curve is not a straight line
(b)
25 < t < 40
(c)
3.54 ft/s
(d)
141.5 ft
(e)
no since the velocity is positive and the acceleration is never negative
(f )
need the position at any one given time (e.g. s0 )
(a)
x = aekt + be−kt , dx/dt = akekt − bke−kt ,
d2 x/dt2 = ak 2 ekt + bk 2 e−kt = k 2 (aekt + be−kt ) = k 2 x
(b)
At t = 0, v = ak − bk = (a − b)k = v0 so k = v0 /(a − b) and a = k 2 x = v02 x/(a − b)2 .
Z
33.
u = 5 + 2 sin 3x, du = 6 cos 3xdx;
34.
u=3+
35.
u = ax3 + b, du = 3ax2 dx;
√
1
x, du = √ dx;
2 x
Z
Z
1√
1
1
√ du = u1/2 + C =
5 + 2 sin 3x + C
3
3
6 u
√
√
4
4
2 udu = u3/2 + C = (3 + x)3/2 + C
3
3
1
1
1
+C =− 2 3
+C
du = −
3au2
3au
3a x + 3ab
Z
1
1
tan u + C =
tan(ax2 ) + C
2a
2a
Z
x
−x
x −x
37. ln(e ) + ln(e ) = ln(e e ) = ln 1 = 0 so [ln(ex ) + ln(e−x )]dx = C
36. u = ax2 , du = 2axdx;
1
2a
sec2 udu =
Supplementary Exercises 7
µ
−
38.
1
3
1
− + 4
3u3
u 4u
254
¶¸−1
= 389/192
−2
Z
39.
2
u = ln x, du = (1/x)dx;
1
Z
1
40.
¸2
1
du = ln u
u
= ln 2
1
√
e−x/2 dx = 2(1 − 1/ e)
0
41.
u=e
−2x
−2x
, du = −2e
Z
1
dx; −
2
42.
¸1
1
sin3 πx = 0
3π
0
43.
With b = 1.618034, area =
Z
1/4
1
b
3 1
(1 + cos u)du = +
8 2
µ
1
sin 1 − sin
4
¶
(x + x2 − x3 )dx = 1.007514.
0
44.
(a)
(b)
2
2
1 2
x sin 3x −
sin 3x + x cos 3x − 0.251607
3
27
9
p
4
−6
f (x) = 4 + x2 + √
4 + x2
f (x) =
7
1 4
k − k − k 2 + = 0 to get k = 2.073948.
4
4
1
1
1
(b) Solve − cos 2k + k 3 + = 3 to get k = 1.837992.
2
3
2
Z x
x
t
√
46. F (x) =
dt, F 0 (x) = √
, so F is increasing on [1, 3]; Fmax = F (3) ≈ 1.152082854
3
2
+
t
2
+ x3
−1
and Fmin = F (1) ≈ −0.07649493141
45.
(a)
47.
(a)
Solve
(b) 0.7651976866
y
(c)
J0 (x) = 0 if x = 2.404826
1
y = J0(x)
0.5
1 2 3 4 5 6 7 8
x
-0.5
¸1
1 3
x dx = x
= 1/3
48. (a) A =
3
0
0
#
" n
¶2
¶
µ
n µ
n
n
X
X
X
1 X 2
n(n + 1)
1 n(n + 1)(2n + 1)
1
k−1
= 3
−2
+n ,
k −2
k+
1 = 3
(b)
n
n
n
n
6
2
k=1
k=1
k=1
k=1
¶
µ
n
X k−1 2 1
2
1
lim
= =
n→+∞
n
n
6
3
Z
1
2
(c)
n µ
X
k=1
k=1
k
n
¶2
n µ ¶2
n
X
1 X 2
2
1
1 n(n + 1)(2n + 1)
1
1
k
= 3
and lim
= =
k = 3
n→+∞
n
n
n
6
n
n
6
3
k=1
k=1
Z
49.
100,000
100, 000/(ln 100, 000) ≈ 8,686;
2
1
dt ≈ 9,629, so the integral is better
ln t
255
Chapter 7
CHAPTER 7 HORIZON MODULE
1.
2.
3.
4.
vx (0) = 35 cos α, so from Equation (1), x(t) = (35 cos α)t; vy (0) = 35 sin α, so from Equation (2),
y(t) = (35 sin α)t − 4.9t2 .
dx(t)
dy(t)
= 35 cos α, vy (t) =
= 35 sin α − 9.8t
dt
dt
(b) vy (t) = 35 sin α − 9.8t, vy (t) = 0 when t = 35 sin α/9.8;
y = vy (0)t − 4.9t2 = (35 sin α)(35 sin α)/9.8 − 4.9((35 sin α)/9.8)2 = 62.5 sin2 α, so
ymax = 62.5 sin2 α.
(a)
vx (t) =
0.004 2
t = x/(35 cos α) so y = (35 sin α)(x/(35 cos α)) − 4.9(x/(35 cos α))2 = (tan α)x −
x;
cos2 α
the trajectory is a parabola because y is a quadratic function of x.
15◦
no
25◦
yes
35◦
no
45◦
no
55◦
no
65◦
yes
75◦
no
85◦
no
65
0
120
0
5.
y(t) = (35 sin α s)t − 4.9t2 = 0 when t = 35 sin α/4.9, at which time
x = (35 cos α)(35 sin α/4.9) = 125 sin 2α; this is the maximum value of x, so R = 125 sin 2α m.
6.
(a)
R = 95 when sin 2α = 95/125 = 0.76, α = 0.4316565575, 1.139139769 rad ≈ 24.73◦ , 65.27◦ .
(b)
y(t) < 50 is required; but y(1.139) ≈ 51.56 m, so his height would be 56.56 m.
7. 0.4019 < α < 0.4636 (radians), or 23.03◦ < α < 26.57◦
CHAPTER 8
Applications of the Definite Integral
in Geometry, Science, and Engineering
EXERCISE SET 8.1
Z
2
2
(x + 1 − x)dx = (x /3 + x − x /2)
2
1. A =
−1
Z
4
2. A =
3
2
√
( x + x/4)dx = (2x3/2 /3 + x2 /8)
4
0
Z
= 22/3
0
2
2
(y − 1/y 2 )dy = (y 2 /2 + 1/y)
3. A =
1
Z
= 9/2
−1
=1
1
2
2
(2 − y + y)dy = (2y − y /3 + y /2)
2
4. A =
3
0
2
= 10/3
0
Z
Z
4
(4x − x2 )dx = 32/3
5. (a) A =
16
(b) A =
0
√
( y − y/4)dy = 32/3
0
y
(4, 16)
y = 4x
y = x2
5
x
1
6. Eliminate x to get y 2 = 4(y + 4)/2, y 2 − 2y − 8 = 0,
(y − 4)(y + 2) = 0; y = −2, 4 with corresponding
values of x = 1, 4.
Z 1
Z 4
√
√
√
(a) A =
[2 x − (−2 x)]dx +
[2 x − (2x − 4)]dx
1
Z 4
Z0 1
√
√
4 xdx +
(2 x − 2x + 4)dx = 8/3 + 19/3 = 9
=
0
1
Z 4
(b) A =
[(y/2 + 2) − y 2 /4]dy = 9
y
y = 2x – 4
x
(1, -2)
−2
Z
1
7. A =
√
( x − x2 )dx = 49/192
y
1/4
y = √x
(1, 1)
y = x2
x
1
4
256
(4, 4)
y2 = 4x
257
Chapter 8
Z
Z
2
[0 − (x − 4x)]dx
3
8. A =
(0 − cos 2x)dx
Z0 2
π/4
Z
(4x − x3 )dx = 4
=
π/2
9. A =
π/2
=−
cos 2x dx = 1/2
0
π/4
y
y
1
x
y = cos 2x
2
x
3
y = 2x 3 – 4x
6
-1
Z
2
2
10. Equate sec x and 2 to get sec x = 2,
3π/4
11. A =
sin y dy =
2
π/4
y
y
2
(# , 2)
(3 , 2)
y = sec 2 x
1
x = sin y
x
9
√
sec x = ± 2, x = ±π/4
Z π/4
(2 − sec2 x)dx = π − 2
A=
3
x
−π/4
Z
√
Z
2
[(x + 2) − x ]dx = 9/2
2
12. A =
−1
ln 2
13. A =
e
2x
−e
x
dx =
0
= 1/2
y
y
(2, 4)
y = e 2x
4
y=
(–1, 1)
x2
x
2
y = ex
x=y–2
x
ln 2
Z
14. A =
1
e
ie
dy
= ln y = 1
y
1
y
e
1
x
1/e
1
1 2x
e − ex
2
#ln 2
0
Exercise Set 8.1
258
(
3 − x, x ≤ 1
,
1 + x, x ≥ 1
Z 1
1
− x + 7 − (3 − x) dx
A=
5
−5
Z 5
1
+
− x + 7 − (1 + x) dx
5
1
Z 5
Z 1
6
4
x + 4 dx +
=
6 − x dx
5
5
−5
1
y
15. y = 2 + |x − 1| =
(–5, 8)
y = – 15 x + 7
(5, 6)
y = 3–x
y= 1+x
x
= 72/5 + 48/5 = 24
Z
Z
2/5
(4x − x)dx
16. A =
Z
17. A =
0
(x3 − 4x2 + 3x)dx
Z 3
[−(x3 − 4x2 + 3x)]dx
+
0
1
(−x + 2 − x)dx
+
Z
1
2/5
Z
2/5
1
(2 − 2x)dx = 3/5
3x dx +
=
= 5/12 + 32/12 = 37/12
1
0
4
2/5
(
y
2
5
,
8
5
)
-1
y = −x + 2
y = 4x
4
(1, 1)
x
-8
y= x
18. Equate y = x3 − 2x2 and y = 2x2 − 3x
to get x3 − 4x2 + 3x = 0,
x(x − 1)(x − 3) = 0; x = 0, 1, 3
with corresponding values of y = 0, −1.9.
Z 1
A=
[(x3 − 2x2 ) − (2x2 − 3x)]dx
0
Z
3
-2
[(2x3 − 3x) − (x3 − 2x2 )]dx
1
1
Z
(x3 − 4x2 + 3x)dx +
=
-1
3
+
Z
9
0
3
(−x3 + 4x2 − 3x)dx
1
8
37
5
+ =
=
12 3
12
19. From the symmetry of the region
Z 5π/4
√
A=2
(sin x − cos x)dx = 4 2
1
π/4
0
-1
o
259
Chapter 8
Z
20. The region is symmetric about the
origin so
Z 0
A=2
(x3 − 4x)dx = 8
Z
0
21. A =
−1
1
(y 3 − y)dy +
−(y 3 − y)dy
0
= 1/2
1
−2
3.1
-1
-3
1
3
-1
-3.1
Z
y 3 − 4y 2 + 3y − (y 2 − y) dy
1
22. A =
4.1
0
Z
+
4
y 2 − y − (y 3 − 4y 2 + 3y) dy
1
= 7/12 + 45/4 = 71/6
-2.2
12.1
0
23. Solve 3−2x = x6 +2x5 −3x4 +x2 to find the real roots
Z x = −3, 1; from a plot it is seen that the line
1
is above the polynomial when −3 < x < 1, so A =
−3
(3−2x−(x6 +2x5 −3x4 +x2 )) dx = 9152/105
q
√
1
24. Solve x − 2x − 3x = x to find the roots x = 0, ±
6 + 2 21. Thus, by symmetry,
2
Z √(6+2√21)/2
7√
27
+
(x3 − (x5 − 2x3 − 3x)) dx =
21
A=2
4
4
0
5
Z
k
25.
3
√
2 ydy =
Z
9
Z
√
2 ydy
Z
k
y 1/2 dy =
x2 dx
1
1 3
k = (8 − k 3 )
3
3
9
k
k3 = 4
√
k= 34
2
2 3/2
k
= (27 − k 3/2 )
3
3
k 3/2 = 27/2
2/3
k = (27/2)
2
k
0
y 1/2 dy
0
Z
k
x2 dx =
26.
k
0
Z
3
√
3
y
x = √y
= 9/ 4
y
y = 9
y = k
x
2
x=k
x
Exercise Set 8.1
260
Z
2
(2x − x2 )dx = 4/3
27. (a) A =
0
(b) y = mx intersects y = 2x − x2 where mx = 2x − x2 , x2 + (m − 2)x = 0, x(x + m − 2) = 0 so
x = 0 or x = 2 − m. The area below the curve and above the line is
2−m
Z 2−m
Z 2−m
1 3
1
1
2
2
2
(2 − m)x − x
(2x − x − mx)dx =
[(2 − m)x − x ]dx =
= (2 − m)3
2
3
6
0
0
0
√
3
so (2 − m)3 /6 = (1/2)(4/3) = 2/3, (2 − m)3 = 4, m = 2 − 4.
28. The line through (0, 0) and (5π/6, 1/2) is y =
Z
y
3
x;
5π
√
5
3
3
x dx =
− π+1
sin x −
5π
2
24
5π/6
A=
0
y = sin x
1
5c
6
(
,
c
1
2
)
x
29. (a) It gives the area of the region that is between f and g when f (x) > g(x) minus the area of
the region between f and g when f (x) < g(x), for a ≤ x ≤ b.
(b) It gives the area of the region that is between f and g for a ≤ x ≤ b.
Z
30. (b)
n→+∞
1
(x1/n − x) dx = lim
lim
n→+∞
0
x2
n
x(n+1)/n −
n+1
2
1
= lim
0
n→+∞
1
n
−
n+1 2
= 1/2
31. The curves intersect at x = 0 and, by Newton’s Method, at x ≈ 2.595739080 = b, so
Z b
b
A≈
(sin x − 0.2x)dx = − cos x + 0.1x2 0 ≈ 1.180898334
0
32. By Newton’s Method, the points of intersection are at x ≈ ±0.824132312, so with
b
Z b
(cos x − x2 )dx = 2(sin x − x3 /3) ≈ 1.094753609
b = 0.824132312 we have A ≈ 2
0
0
R
|v| dt, so
Z 60
(3t − t2 /20) dt = 1800 ft.
(a) distance =
0
Z T
3
1
(b) If T ≤ 60 then distance =
(3t − t2 /20) dt = T 2 − T 3 ft.
2
60
0
33. distance =
Z
34. Since a1 (0) = a2 (0) = 0, A =
of the two cars at time T .
T
(a2 (t)−a1 (t)) dt = v2 (T )−v1 (T ) is the difference in the velocities
0
y
35. Solve x1/2 + y 1/2 = a1/2 for y to get
y = (a1/2 − x1/2 )2 = a − 2a1/2 x1/2 + x
Z a
A=
(a − 2a1/2 x1/2 + x)dx = a2 /6
a
0
x
a
261
Chapter 8
√
36. Solve for y to get y = (b/a) a2 − x2 for the upper half of the ellipse; make use of symmetry to
Z a p
Z
4b a p 2
4b 1 2
b
2
2
get A = 4
× πa = πab.
a − x dx =
a − x2 dx =
a
a
a
4
0
0
37. Let A be the area between the curve and the x-axis and AR the area of the rectangle, then
b
Z b
k
kbm+1
A=
xm+1 =
, AR = b(kbm ) = kbm+1 , so A/AR = 1/(m + 1).
kxm dx =
m
+
1
m
+
1
0
0
EXERCISE SET 8.2
Z
Z
3
1. V = π
−1
(3 − x)dx = 8π
1
[(2 − x2 )2 − x2 ]dx
2. V = π
0
Z
1
(4 − 5x2 + x4 )dx
=π
0
= 38π/15
Z
2
3. V = π
0
Z
Z
1
(3 − y)2 dy = 13π/6
4
4. V = π
x4 dx = 32π/5
6. V = π
(4 − 1/y 2 )dy = 9π/2
1/2
Z
2
5. V = π
2
√
sec2 x dx = π( 3 − 1)
π/3
π/4
0
y
y
y = sec x
y = x2
2
1
x
x
2
3
-1
4
-2
Z
π/2
cos x dx = (1 −
7. V = π
√
2/2)π
y
y = √cos x
1
π/4
x
3
6
-1
Z
y
1
[(x2 )2 − (x3 )2 ]dx
8. V = π
0
Z
1
(x4 − x6 )dx = 2π/35
=π
1
(1, 1)
y = x2
y = x3
0
1
x
Exercise Set 8.2
Z
262
Z
4
9. V = π
−4
Z
[(25 − x2 ) − 9]dx
−3
Z
4
(16 − x2 )dx = 256π/3
= 2π
y
y=
(9 − x2 )2 dx
3
=π
−3
0
5
3
10. V = π
√25 – x 2
(81 − 18x2 + x4 )dx = 1296π/5
y
9
y = 9 – x2
y= 3
-3
x
Z
ln 3
π
e dx = e2x
2
0
Z
1
−4x
e
12. V = π
0
x
ln 3
2x
11. V = π
3
= 4π
0
π
dx = (1 − e−4 )
4
Z
4
[(4x)2 − (x2 )2 ]dx
13. V = π
0
Z
y
4
(16x2 − x4 )dx = 2048π/15
=π
1
0
y
(4, 16)
16
x
y = 4x
1
y = x2
x
4
-1
Z
Z
π/4
(cos2 x − sin2 x)dx
14. V = π
0
Z
1
y 2/3 dy = 3π/5
15. V = π
0
y
π/4
cos 2x dx = π/2
=π
0
y
1
1
x
y = sin x
3
-1
y = x3
y = cos x
x
-1
1
263
Chapter 8
Z
Z
1
16. V = π
−1
Z
(1 − y 2 )2 dy
3
17. V = π
(1 + y)dy = 8π
−1
1
=π
−1
y
(1 − 2y 2 + y 4 )dy = 16π/15
3
y
1
x = 1 – y2
x=
√1 + y
x
2
x
-1
1
-1
Z
Z
3
[2 − (y + 1)]dy
2
18. V = π
csc2 y dy = 2π
π/4
0
Z
3π/4
19. V = π
3
y
(3 − y)dy = 9π/2
=π
9
0
y = x2 – 1
y
6
x = csc y
(2, 3)
3
3
-2
x
-1
1
2
x
Z
Z
1
(y − y 4 )dy = 3π/10
20. V = π
0
y
2
21. V = π
−1
[(y + 2)2 − y 4 ]dy = 72π/5
y
x = y2
1
-1
(1, 1)
y = x2
x = y2
(4, 2)
x =y+2
x
1
x
-1
Z
1
22. V = π
−1
Z
(1, –1)
(2 + y 2 )2 − (1 − y 2 )2 dy
x = 2 + y2
y
x = 1 – y2
1
1
(3 + 6y 2 )dy = 10π
=π
x
−1
1
-1
2
Exercise Set 8.2
Z
264
a
23. V = π
−a
y
b2 2
(a − x2 )dx = 4πab2 /3
a2
b
y=
b
a
√a 2 – x 2
x
–a
Z
2
1
dx = π(1/b − 1/2); π(1/b − 1/2) = 3, b = 2π/(π + 6)
x2
24. V = π
b
Z
a
Z
0
(x + 1)dx
25. V = π
Z
4
26. V = π
−1
Z
1
[(x + 1) − 2x]dx
0
= π/2 + π/2 = π
4
= 8π + 8π/3 = 32π/3
y
y = √x
y
1
-1
y = √ 2x
4
6
x
1
Z
Z
3
(9 − y ) dy
2 2
27. V = π
Z
9
[32 − (3 −
28. V = π
0
√
x)2 ]dx
0
Z
3
(81 − 18y 2 + y 4 )dy
=π
9
=π
0
√
(6 x − x)dx
0
= 648π/5
= 135π/2
y
3
y =6–x
x
(1, √ 2)
y = √x + 1
(6 − x)2 dx
0
+π
6
x dx + π
y
x = y2
x
y=3
y = √x
x
9
9
Z
1
29. V = π
√
[( x + 1)2 − (x + 1)2 ]dx
0
Z
=π
0
1
√
(2 x − x − x2 )dx = π/2
y
x=y
x = y2
1
x
1
y = −1
265
Chapter 8
Z
y
1
[(y + 1)2 − (y 2 + 1)2 ]dy
30. V = π
x=y
0
Z
1
1
(2y − y 2 − y 4 )dy = 7π/15
=π
0
x
1
x = y2
x = –1
Z
31. A(x) = π(x2 /4)2 = πx4 /16,
1
(x − x4 )dx = 3π/10
32. V = π
0
Z
20
(πx4 /16)dx = 40, 000π ft3
V =
0
Z
2
1
1
1√
34. A(x) = π
x = πx,
2
2
8
Z 4
1
πx dx = π
V =
0 8
1
(x − x ) dx
2 2
33. V =
0
Z
1
(x2 − 2x3 + x4 )dx = 1/30
=
0
Square
y
y
y = √x
y = x (1, 1)
y = x2
1
4
x
35. On the upper half of the circle, y =
x
√
1 − x2 , so:
(a) A(x) is the area of a semicircle of radius y, so
Z
Z 1
π 1
(1 − x2 ) dx = π
(1 − x2 ) dx = 2π/3
A(x) = πy 2 /2 = π(1 − x2 )/2; V =
2 −1
0
y
y
-1
1
x
y = √1 – x 2
(b) A(x) is the area of a square of side 2y, so
Z 1
Z 1
2
2
2
(1 − x ) dx = 8
(1 − x2 ) dx = 16/3
A(x) = 4y = 4(1 − x ); V = 4
−1
0
y
-1
2y
y = √1 – x 2
1
x
Exercise Set 8.2
266
(c) A(x) is the area of an equilateral triangle with sides 2y, so
√
√
√
3
(2y)2 = 3y 2 = 3(1 − x2 );
A(x) =
4
Z 1√
√ Z 1
√
3(1 − x2 ) dx = 2 3
(1 − x2 ) dx = 4 3/3
V =
−1
0
2y
2y
y
-1
2y
y = √1 – x 2
1
x
36. By similar triangles, R/r = y/h so
r
R = ry/h and A(y) = πr2 y 2 /h2 .
Z h
y 2 dy = πr2 h/3
V = (πr2 /h2 )
R
0
h
y
37. The two curves cross at x = b ≈ 1.403288534, so
Z π/2
Z b
((2x/π)2 − sin16 x) dx + π
(sin16 x − (2x/π)2 ) dx = 0.710172176.
V =π
b
0
Z
e
(1 − (ln y)2 ) dy = π
38. V = π
1
Z
r
(r2 − y 2 ) dy = π(rh2 − h3 /3) =
39. (a) V = π
r−h
1 2
πh (3r − h)
3
y
(b) By the Pythagorean Theorem,
r2 = (r − h)2 + ρ2 , 2hr = h2 + ρ2 ; from part (a),
πh 3 2
πh
2
2
2
(3hr − h ) =
(h + ρ ) − h )
V =
3
3 2
1
= πh(h2 + 3ρ2 ).
6
h
r
π
π
dV
dh
V = (30h2 − h3 ),
= (60h − 3h2 ) ,
3
dt
3
dt
π
dh dh
1
= (300 − 75) ,
= 1/(150π) ft/min
2
3
dt dt
x
r
40. Find the volume generated by revolving
the shaded region about the y-axis.
Z −10+h
π
V =π
(100 − y 2 )dy = h2 (30 − h)
3
−10
Find dh/dt when h = 5 given that dV /dt = 1/2.
x2 + y2 = r 2
y
h – 10
10
x
h
−10
x=
√ 100 – y2
267
Chapter 8
5
= 0.5; {y0 , y1 , · · · , y10 } = {0, 2.00, 2.45, 2.45, 2.00, 1.46, 1.26, 1.25, 1.25, 1.25, 1.25};
10
9
X
yi 2
∆x ≈ 11.157;
left = π
2
i=0
41. (b) ∆x =
right = π
10
X
yi 2
2
i=1
∆x ≈ 11.771; V ≈ average = 11.464 cm3
√
42. If x = r/2 then from y 2 = r2 − x2 we get y = ± 3r/2
√
√
as limits of integration; for − 3 ≤ y ≤ 3,
y
√ 3r
2
x=
A(y) = π[(r2 − y 2 ) − r2 /4] = π(3r2 /4 − y 2 ), thus
Z
V =π
√ r 2 – y2
x
√
3r/2
(3r
√
− 3r/2
Z √3r/2
2
/4 − y 2 )dy
r
2
(3r2 /4 − y 2 )dy =
= 2π
√
3πr3 /2.
–
0
y
43. (a)
√ 3r
2
y
(b)
h –4
x
h –4
-2
x
h
h
-4
-4
2≤h≤4
0≤h<2
If the cherry is partially submerged then 0 ≤ h < 2 as shown in Figure (a); if it is totally submerged then 2 ≤ h ≤ 4 as shown in Figure (b). The radius of the glass is 4 cm and that of the
cherry is 1 cm so points on the sections shown in the figures satisfy the equations x2 + y 2 = 16
and x2 + (y + 3)2 = 1. We will find the volumes of the solids that are generated when the shaded
regions are revolved about the y-axis.
For 0 ≤ h < 2,
Z
Z
h−4
[(16 − y ) − (1 − (y + 3) )]dy = 6π
2
V =π
−4
h−4
2
(y + 4)dy = 3πh2 ;
−4
for 2 ≤ h ≤ 4,
Z
−2
V =π
−4
Z
Z
[(16 − y ) − (1 − (y + 3) )]dy + π
2
Z
−2
−4
=
so
h−4
(y + 4)dy + π
= 6π
h−4
2
−2
−2
(16 − y 2 )dy
1
(16 − y 2 )dy = 12π + π(12h2 − h3 − 40)
3
1
π(12h2 − h3 − 4)
3
2
3πh
V =
1 π(12h2 − h3 − 4)
3
if 0 ≤ h < 2
if 2 ≤ h ≤ 4
Exercise Set 8.3
44.
268
p
x = h ± r2 − y2 ,
Z r h
i
p
p
V =π
(h + r2 − y 2 )2 − (h − r2 − y 2 )2 dy
−r
Z
r
= 4πh
−r
= 4πh
y
(x – h 2) + y2 = r 2
p
r2 − y 2 dy
1 2
πr
2
x
= 2π 2 r2 h
45. tan θ = h/x so h = x tan θ,
A(y) =
1
1
1
hx = x2 tan θ = (r2 − y 2 ) tan θ
2
2
2
h
because x2 = r2 − y 2 ,
Z r
1
(r2 − y 2 )dy
V = tan θ
2
−r
Z r
2
= tan θ
(r2 − y 2 )dy = r3 tan θ
3
0
u
x
p
46. A(x) = (x tan θ)(2 r2 − x2 )
p
= 2(tan θ)x r2 − x2 ,
Z r p
V = 2 tan θ
x r2 − x2 dx
47. Each cross section perpendicular to the
y-axis is a square so
A(y) = x2 = r2 − y 2 ,
Z r
1
V =
(r2 − y 2 )dy
8
0
0
2
= r3 tan θ
3
V = 8(2r3 /3) = 16r3 /3
y
x tan u
y
x
x=
√ r 2 – y2
√r 2 – x 2
r
x
48. The regular cylinder of radius r and height h has the same circular cross sections as do those of
the oblique clinder, so by Cavalieri’s Principle, they have the same volume: πr2 h.
EXERCISE SET 8.3
Z
Z
2
2πx(x2 )dx = 2π
1. V =
1
x3 dx = 15π/2
1
√
2
Z
2
2πx(
p
2. V =
Z
4−
x2
− x)dx = 2π
0
Z
0
Z
1
2πy(2y − 2y )dy = 4π
0
p
√
8π
(2 − 2)
(x 4 − x2 − x2 )dx =
3
1
(y 2 − y 3 )dy = π/3
2
3. V =
√
2
0
269
Chapter 8
Z
Z
2
2
2πy[y − (y 2 − 2)]dy = 2π
4. V =
0
Z
(y 2 − y 3 + 2y)dy = 16π/3
0
Z
1
3
2π(x)(x )dx
5. V =
0
9
6. V =
4
Z
Z
1
x4 dx = 2π/5
= 2π
√
2πx( x)dx
9
x3/2 dx = 844π/5
= 2π
0
4
y
y
y = x3
3
1
2
x
-1
y = √x
1
1
x
-1
-9
Z
Z
3
7. V =
Z
3
2πx(1/x)dx = 2π
1
-4
dx = 4π
4
√
π/2
8. V =
1
9
√
2πx cos(x2 )dx = π/ 2
0
y
y
y = cos (x 2)
1
y= x
x
-3
-1
1
x
3
√p
2
Z
Z
2
1
Z
10. V = 2π
2
(x2 − x)dx = 20π/3
= 8π
1
x
dx
2+1
x
0
i1
= π ln(x2 + 1) = π ln 2
2πx[(2x − 1) − (−2x + 3)]dx
9. V =
0
1
y
y
1
(2, 3)
y=
1
x2 + 1
(1, 1)
x
-1
x
1
(2, –1)
√
3
Z
11. V =
x2
2πxe
1
x2
i√3
dx = πe
1
y
= π(e − e)
3
20
y = ex
2
10
x
-√3 -1
1
√3
Exercise Set 8.3
Z
270
y
2
2πx(2x − x2 )dx
12. V =
0
Z
2
(2x2 − x3 )dx =
= 2π
0
y = 2x – x 2
8
π
3
x
2
Z
Z
1
3
13. V =
2πy dy = π/2
14. V =
0
Z
3
3
y 2 dy = 76π/3
2πy(2y)dy = 4π
2
2
y
y
x = y2
1
3
2
x = 2y
x
Z
x
1
2πy(1 −
15. V =
0
√
Z
y)dy
4
2πy(5 − y − 4/y)dy
16. V =
1
Z
1
(y − y 3/2 )dy = π/5
= 2π
Z
0
4
(5y − y 2 − 4)dy = 9π
= 2π
1
y
y
(1, 4)
x = 5–y
y = √x
(4, 1)
x
x = 4/y
1
Z
Z
π
x sin xdx = 2π 2
17. V = 2π
π/2
x cos xdx = π 2 − 2π
18. V = 2π
0
x
0
Z
1
2πx(x3 − 3x2 + 2x)dx = 7π/30
19. (a) V =
0
(b) much easier; the method of slicing would require that x be expressed in terms of y.
y
y = x3 – 3x2 + 2x
x
-1
1
271
Chapter 8
Z
y
2
2π(x + 1)(1/x3 )dx
20. V =
1
Z
x+1
2
= 2π
(x−2 + x−3 )dx = 7π/4
y = 1/x 3
1
x
-1
Z
1x 2
y
1
2π(1 − y)y 1/3 dy
21. V =
0
Z
1
(y 1/3 − y 4/3 )dy = 9π/14
= 2π
0
1
x = y1/3
1–y
Z
Z
b
2πx[f (x) − g(x)]dx
22. (a)
d
2πy[f (y) − g(y)]dy
(b)
a
x
c
y
h
(r − y) is an equation of the line
r
through (0, r) and (h, 0) so
Z r
h
(r − y) dy
2πy
V =
r
0
Z
2πh r
=
(ry − y 2 )dy = πr2 h/3
r 0
23. x =
Z
k/4
24. V =
(0, r)
x
(h, 0)
√
2π(k/2 − x)2 kxdx
y
0
√ Z
= 2π k
k/2 – x
y = √kx
k/4
1/2
(kx
− 2x
3/2
3
)dx = 7πk /60
x
0
y = – √kx
Z
Z a
p
2πx(2 r2 − x2 )dx = 4π
x(r2 − x2 )1/2 dx
0
0
a
i
4π
4π h 3
= − (r2 − x2 )3/2 =
r − (r2 − a2 )3/2
3
3
0
a
x = k/2
x = k/4
y
25. V =
y=
√ r 2 – x2
x
a
y = – √ r 2 – x2
Exercise Set 8.4
Z
272
a
26. V =
−a
p
2π(b − x)(2 a2 − x2 )dx
Z
p
a
= 4πb
−a
Z
a2 − x2 dx − 4π
a
−a
y
b−x
p
x a2 − x2 dx
= 4πb · (area of a semicircle of radius a) − 4π(0)
= 2π 2 a2 b
√ a2 – x2
–a
x
a
– √ a2 – x2
x=b
Z
b
27. Vx = π
1/2
1
dx = π(2 − 1/b), Vy = 2π
x2
Z
b
dx = π(2b − 1);
1/2
Vx = Vy if 2 − 1/b = 2b − 1, 2b2 − 3b + 1 = 0, solve to get b = 1/2 (reject) or b = 1.
EXERCISE SET 8.4
1. (a)
(b)
2.
dy
= 2, L =
dx
1
dx
= ,L =
dy
2
Z
2
√
1 + 4dx =
√
5
1
Z
4
p
√
√
1 + 1/4 dy = 2 5/2 = 5
2
dy
dx
= 1,
= 5, L =
dt
dt
Z
1
p
√
12 + 52 dt = 26
0
9 1/2
81
x , 1 + [f 0 (x)]2 = 1 + x,
2
4
3/2 #1
Z 1p
√
81
8
1+ x
1 + 81x/4dx =
= (85 85 − 8)/243
L=
243
4
0
3. f 0 (x) =
0
4. g 0 (y) = y(y 2 + 2)1/2 , 1 + [g 0 (y)]2 = 1 + y 2 (y 2 + 2) = y 4 + 2y 2 + 1 = (y 2 + 1)2 ,
Z
1
Z
p
(y 2
L=
+
1)2 dy
0
1
(y 2 + 1)dy = 4/3
=
0
2
2 −1/3
4
9x2/3 + 4
dy
dy
= x
, 1+
= 1 + x−2/3 =
,
5.
dx
3
dx
9
9x2/3
Z 40
Z 8 √ 2/3
1
9x + 4
dx =
u1/2 du, u = 9x2/3 + 4
L=
18 13
3x1/3
1
40
√
√
√
√
1
1
1 3/2
u
(40 40 − 13 13) =
(80 10 − 13 13)
=
=
27
27
27
13
or (alternate solution)
2
4 + 9y
9
dx
,
=1+ y =
dy
4
4
Z
Z 40
√
√
1
1
1 4p
(80 10 − 13 13)
4 + 9y dy =
u1/2 du =
L=
2 1
18 13
27
x = y 3/2 ,
3
dx
= y 1/2 , 1 +
dy
2
273
Chapter 8
2
1 3
1 6 1
1 6 1
1 3
−3
0
2
−6
−6
−3
x − +x
x + +x =
x +x
6. f (x) = x − x , 1 + [f (x)] = 1 +
,
=
4
16
2
16
2
4
s
2
Z 3
Z 3
1 3
1 3
x + x−3 dx =
x + x−3 dx = 595/144
L=
4
4
2
2
0
1 2x
1 x
1 2x
e − e−x , 1 + [f 0 (x)]2 = 1 +
e − 2 + e−2x =
e + 2 + e−2x , so
7. f 0 (x) =
2
4
4
Z 3
Z 3p
1
1 + (f 0 (x))2 dx =
ex + e−x dx = (e3 − e−3 )/2
L=
2 0
0
1 3 1 −3
y − y , 1 + [g 0 (y)]2 = 1 +
2
2
Z 4
1 3 1 −3
y + y
dy = 2055/64
L=
2
2
1
8. g 0 (y) =
1 6 1 1 −6
y − + y
4
2 4
Z
2
2
2 2
2
2
1
2
9. (dx/dt) + (dy/dt) = (t ) + (t) = t (t + 1), L =
=
2
1 3 1 −3
y + y
,
2
2
√
t(t2 + 1)1/2 dt = (2 2 − 1)/3
0
10. (dx/dt)2 + (dy/dt)2 = [2(1 + t)]2 + [3(1 + t)2 ]2 = (1 + t)2 [4 + 9(1 + t)2 ],
Z 1
√
√
(1 + t)[4 + 9(1 + t)2 ]1/2 dt = (80 10 − 13 13)/27
L=
0
Z
π/2
11. (dx/dt)2 + (dy/dt)2 = (−2 sin 2t)2 + (2 cos 2t)2 = 4, L =
2 dt = π
0
12. (dx/dt)2 + (dy/dt)2 = (− sin t + sin t + t cos t)2 + (cos t − cos t + t sin t)2 = t2 ,
Z π
L=
t dt = π 2 /2
0
13. (dx/dt)2 + (dy/dt)2 = [et (cos t − sin t)]2 + [et (cos t + sin t)]2 = 2e2t ,
Z π/2 √
√
2et dt = 2(eπ/2 − 1)
L=
0
Z
2
2
t
2
t
2
4
2et dt = 2(e4 − e)
2t
14. (dx/dt) + (dy/dt) = (2e cos t) + (−2e sin t) = 4e , L =
1
√
p
sec x tan x
= tan x, 1 + (y 0 )2 = 1 + tan2 x = sec x when 0 < x < π/4, so
sec x
Z π/4
√
sec x dx = ln(1 + 2)
L=
15. dy/dx =
0
√
p
cos x
= cot x, 1 + (y 0 )2 = 1 + cot2 x = csc x when π/4 < x < π/2, so
sin x
!
√
Z π/2
√
√
2−1 √
csc x dx = − ln( 2 − 1) = − ln √
L=
( 2 + 1) = ln(1 + 2)
2+1
π/4
16. dy/dx =
17. (a) (dx/dθ)2 + (dy/dθ)2 = (a(1 − cos θ))2 + (a sin θ)2 = a2 (2 − 2 cos θ), so
Z 2π p
Z 2π p
2
2
(dx/dθ) + (dy/dθ) dθ = a
2(1 − cos θ) dθ
L=
0
0
Exercise Set 8.4
274
18. (a) Use the interval 0 ≤ φ < 2π.
(b) (dx/dφ)2 + (dy/dφ)2 = (−3a cos2 φ sin φ)2 + (3a sin2 φ cos φ)2
= 9a2 cos2 φ sin2 φ(cos2 φ + sin2 φ) = (9a2 /4) sin2 2φ, so
π/2
Z π/2
Z 2π
| sin 2φ| dφ = 6a
sin 2φ dφ = −3a cos 2φ
= 6a
L = (3a/2)
0
0
y
19. (a)
0
(b) dy/dx does not exist at x = 0.
(8, 4)
(-1, 1)
x
(c)
x = g(y) = y 3/2 , g 0 (y) =
Z
1
p
1 + 9y/4dy
L=
0
Z
+
0
=
8
27
4
p
1 + 9y/4dy
3 1/2
y ,
2
(portion for − 1 ≤ x ≤ 0)
(portion for 0 ≤ x ≤ 8)
√
√
√
8
13 √
13 − 1 + (10 10 − 1) = (13 13 + 80 10 − 16)/27
8
27
20. For (4), express the curve y = f (x) in the parametric form x = t, y = f (t) so dx/dt = 1 and
dy/dt = f 0 (t) = f 0 (x) = dy/dx. For (5), express x = g(y) as x = g(t), y = t so
dx/dt = g 0 (t) = g 0 (y) = dx/dy and dy/dt = 1.
Z
2
Z
p
1 + 4x2 dx ≈ 4.645975301
21. L =
0
22. L =
π
p
1 + cos2 y dy ≈ 3.820197789
0
23. Numerical integration yields: in Exercise 21, L ≈ 4.646783762; in Exercise 22, L ≈ 3.820197788.
24. 0 ≤ m ≤ f 0 (x) ≤ M , so m2 ≤ [f 0 (x)]2 ≤ M 2 , and 1 + m2 ≤ 1 + [f 0 (x)]2 ≤ 1 + M 2 ; thus
p
√
√
1 + m2 ≤ 1 + [f 0 (x)]2 ≤ 1 + M 2 ,
Z bp
Z bp
Z bp
2
0
2
1 + m dx ≤
1 + [f (x)] dx ≤
1 + M 2 dx, and
a
a
a
p
p
(b − a) 1 + m2 ≤ L ≤ (b − a) 1 + M 2
√
25. f 0 (x) = cos x, 2/2 ≤ cos x ≤ 1 for 0 ≤ x ≤ π/4 so
p
√
πp
π√
(π/4) 1 + 1/2 ≤ L ≤ (π/4) 1 + 1,
3/2 ≤ L ≤
2.
4
4
26. (dx/dt)2 + (dy/dt)2 = (−a sin t)2 + (b cos t)2 = a2 sin2 t + b2 cos2 t
= a2 (1 − cos2 t) + b2 cos2 t = a2 − (a2 − b2 ) cos2 t
a2 − b2
2
2
cos t = a2 [1 − k 2 cos2 t],
=a 1−
a2
Z π/2 p
Z 2π p
a 1 − k 2 cos2 t dt = 4a
1 − k 2 cos2 t dt
L=
0
0
275
Chapter 8
27. (a) (dx/dt)2 + (dy/dt)2 = 4 sin2 t + cos2 t = 4 sin2 t + (1 − sin2 t) = 1 + 3 sin2 t,
Z π/2 p
Z 2π p
1 + 3 sin2 t dt = 4
1 + 3 sin2 t dt
L=
0
0
(b) 9.69
Z
4.8
(c) distance traveled =
p
1 + 3 sin2 t dt ≈ 5.16 cm
1.5
Z
4.6
28. The distance is
p
1 + (2.09 − 0.82x)2 dx ≈ 6.65 m
0
Z
π
29. L =
p
1 + (k cos x)2 dx
0
k
1
2
1.84
1.83
1.832
L
3.8202
5.2704
5.0135
4.9977
5.0008
Experimentation yields the values in the table, which by the Intermediate-Value Theorem show
that the true solution k to L = 5 lies between k = 1.83 and k = 1.832, so k = 1.83 to two decimal
places.
EXERCISE SET 8.5
Z
1
1. S =
√
√
2π(7x) 1 + 49dx = 70π 2
0
Z
1
√
x dx = 35π 2
0
1
1
2. f 0 (x) = √ , 1 + [f 0 (x)]2 = 1 +
4x
2 x
r
Z 4
Z 4p
√
√
√
1
S=
dx = 2π
2π x 1 +
x + 1/4dx = π(17 17 − 5 5)/6
4x
1
1
√
3. f 0 (x) = −x/ 4 − x2 , 1 + [f 0 (x)]2 = 1 +
Z
1
S=
−1
x2
4
=
,
4 − x2
4 − x2
Z 1
p
p
2
2
2π 4 − x (2/ 4 − x )dx = 4π
dx = 8π
−1
4. y = f (x) = x3 for 1 ≤ x ≤ 2, f 0 (x) = 3x2 ,
2
Z 2
p
√
√
π
3
4 3/2
4
(1 + 9x )
2πx 1 + 9x dx =
= 5π(29 145 − 2 10)/27
S=
27
1
1
Z
5. S =
2
√
√ Z 2
√
2π(9y + 1) 82dy = 2π 82
(9y + 1)dy = 40π 82
0
6. g 0 (y) = 3y 2 , S =
0
Z
1
2πy 3
p
√
1 + 9y 4 dy = π(10 10 − 1)/27
0
p
7. g 0 (y) = −y/ 9 − y 2 , 1 + [g 0 (y)]2 =
9
,S=
9 − y2
Z
2
2π
−2
Z 2
p
3
9 − y2 · p
dy = 6π
dy = 24π
9 − y2
−2
2−y
,
1−y
√
Z 0
Z 0p
p
√
√
2−y
S=
dy = 4π
2π(2 1 − y) √
2 − y dy = 8π(3 3 − 2 2)/3
1−y
−1
−1
8. g 0 (y) = −(1 − y)−1/2 , 1 + [g 0 (y)]2 =
Exercise Set 8.5
276
2
1 −1/2 1 1/2
1 −1 1 1
1 −1/2 1 1/2
0
2
x
9. f (x) = x
− x , 1 + [f (x)] = 1 + x − + x =
+ x
,
2
2
4
2 4
2
2
Z
Z 3
1 3/2
1 −1/2 1 1/2
π 3
1/2
x
2π x − x
+ x
(3 + 2x − x2 )dx = 16π/9
dx =
S=
3
2
2
3 1
1
0
2
1
1
1
1
10. f 0 (x) = x2 − x−2 , 1 + [f 0 (x)]2 = 1 + x4 − + x−4 = x2 + x−2 ,
4
2 16
4
Z 2
Z 2
1
1
1 5 1
1 3 1 −1
x + x
x + x + x−3 dx = 515π/64
x2 + x−2 dx = 2π
2π
S=
3
4
4
3
3
16
1
1
1
1 4 1 −2 0
y + y , g (y) = y 3 − y −3 ,
4
8
4
2
1 −6
1
1 −3
0
2
6
3
= y + y
,
1 + [g (y)] = 1 + y − + y
2 16
4
Z 2
Z 2
1 −3
π
1 4 1 −2
3
y + y
2π
(8y 7 + 6y + y −5 )dy = 16, 911π/1024
y + y
dy =
S=
4
8
4
16 1
1
11. x = g(y) =
√
1
65 − 4y
,
, 1 + [g 0 (y)]2 =
16 − y; g 0 (y) = − √
4(16 − y)
2 16 − y
s
Z 15 p
Z 15 p
√
√ π
65 − 4y
S=
dy = π
2π 16 − y
65 − 4y dy = (65 65 − 5 5)
4(16 − y)
6
0
0
12. x = g(y) =
13. f 0 (x) = ex ,
1 + [f 0 (x)]2 = 1 + e2x , S =
Z
1
2πex
p
1 + e2x dx ≈ 22.94
0
0
0
Z
2
π
2
14. f (x) = cos x, 1 + [f (x)] = 1 + cos x, S =
p
2π sin x 1 + cos2 x dx ≈ 14.42
0
0
0
Z
2
2
15. x = g(y) = ln y, g (y) = 1/y, 1 + [g (y)] = 1 + 1/y ; S =
e
2π
p
1 + 1/y 2 ln y dy ≈ 7.05
1
16. x = g(y) = tan y, g 0 (y) = sec2 y, 1 + [g 0 (y)]2 = 1 + sec4 y;
Z π/4
p
2π tan y 1 + sec4 y dy ≈ 3.84
S=
0
17. Revolve the line segment joining the points (0, 0) and (h, r) about the x-axis. An equation of the
line segment is y = (r/h)x for 0 ≤ x ≤ h so
Z h
Z h
p
p
2πr p 2
2
2
2
2π(r/h)x 1 + r /h dx = 2 r + h
x dx = πr r2 + h2
S=
h
0
0
√
√
18. f (x) = r2 − x2 , f 0 (x) = −x/ r2 − x2 , 1 + [f 0 (x)]2 = r2 /(r2 − x2 ),
Z r
Z r
p
p
S=
2π r2 − x2 (r/ r2 − x2 )dx = 2πr
dx = 4πr2
−r
−r
p
19. g(y) = r2 − y 2 , g 0 (y) = −y/ r2 − y 2 , 1 + [g 0 (y)]2 = r2 /(r2 − y 2 ),
Z r
Z r
p
p
2π r2 − y 2 r2 /(r2 − y 2 ) dy = 2πr
dy = 2πrh
(a) S =
p
r−h
r−h
(b) From part (a), the surface area common to two polar caps of height h1 > h2 is
2πrh1 − 2πrh2 = 2πr(h1 − h2 ).
277
Chapter 8
20. For (4), express the curve y = f (x) in the parametric form x = t, y = f (t) so dx/dt = 1 and
dy/dt = f 0 (t) = f 0 (x) = dy/dx. For (5), express x = g(y) as x = g(t), y = t so
dx/dt = g 0 (t) = g 0 (y) = dx/dy and dy/dt = 1.
21. x0 = 2t, y 0 = 2, (x0 )2 + (y 0 )2 = 4t2 + 4
Z 4 p
Z 4
p
√
8π
(17 17 − 1)
(2t) 4t2 + 4dt = 8π
t t2 + 1dt =
S = 2π
3
0
0
22. x0 = et (cos t − sin t), y 0 = et (cos t + sin t), (x0 )2 + (y 0 )2 = 2e2t
Z π/2
√
√ Z π/2 2t
t
2t
S = 2π
(e sin t) 2e dt = 2 2π
e sin t dt
0
√
π/2
1
= 2 2π e2t (2 sin t − cos t)
5
0
0
√
2 2
π(2eπ + 1)
=
5
23. x0 = 1, y 0 = 4t, (x0 )2 + (y 0 )2 = 1 + 16t2 , S = 2π
Z
1
t
p
1 + 16t2 dt =
0
√
π
(17 17 − 1)
24
24. x0 = −2 sin t cos t, y 0 = 2 sin t cos t, (x0 )2 + (y 0 )2 = 8 sin2 t cos2 t
Z π/2
p
√ Z π/2
√
2
2
2
cos t 8 sin t cos t dt = 4 2π
cos3 t sin t dt = 2π
S = 2π
0
0
25. x0 = −r sin t, y 0 = r cos t, (x0 )2 + (y 0 )2 = r2 ,
Z π
Z π
√
2
2
S = 2π
r sin t r dt = 2πr
sin t dt = 4πr2
0
0
2 2
dy
dx
dx
dy
= a(1 − cos φ),
= a sin φ,
26.
+
= 2a2 (1 − cos φ)
dφ
dφ
dφ
dφ
Z 2π
Z 2π
p
√
2
2
a(1 − cos φ) 2a (1 − cos φ) dφ = 2 2πa
(1 − cos φ)3/2 dφ,
S = 2π
0
0
√
φ
φ
but 1 − cos φ = 2 sin2 so (1 − cos φ)3/2 = 2 2 sin3 for 0 ≤ φ ≤ π and, taking advantage of the
2
2
Z π
3 φ
2
symmetry of the cycloid, S = 16πa
sin dφ = 64πa2 /3.
2
0
27. (a) length of arc of sector = circumference of base of cone,
1
`θ = 2πr, θ = 2πr/`; S = area of sector = `2 (2πr/`) = πr`
2
(b) S = πr2 `2 − πr1 `1 = πr2 (`1 + `) − πr1 `1 = π[(r2 − r1 )`1 + r2 `];
Using similar triangles `2 /r2 = `1 /r1 , r1 `2 = r2 `1 , r1 (`1 + `) = r2 `1 , (r2 − r1 )`1 = r1 `
so S = π (r1 ` + r2 `) = π (r1 + r2 ) `.
l1
r1
l2
l
r2
Exercise Set 8.6
278
p
p
p
28. 2πk 1 + [f 0 (x)]2 ≤ 2πf (x) 1 + [f 0 (x)]2 ≤ 2πK 1 + [f 0 (x)]2 , so
Z b
Z b
Z b
p
p
p
0
2
0
2
2πk 1 + [f (x)] dx ≤
2πf (x) 1 + [f (x)] dx ≤
2πK 1 + [f 0 (x)]2 dx,
a
a
Z
b
2πk
p
Z
a
b
1 + [f 0 (x)]2 dx ≤ S ≤ 2πK
a
p
1 + [f 0 (x)]2 dx, 2πkL ≤ S ≤ 2πKL
a
p
p
29. (a) 1 ≤ 1 + [f 0 (x)]2 so 2πf (x) ≤ 2πf (x) 1 + [f 0 (x)]2 ,
Z b
Z b
Z b
p
2πf (x)dx ≤
2πf (x) 1 + [f 0 (x)]2 dx, 2π
f (x)dx ≤ S, 2πA ≤ S
a
a
a
0
(b) 2πA = S if f (x) = 0 for all x in [a, b] so f (x) is constant on [a, b].
EXERCISE SET 8.6
1. (a) W = F · d = 30(7) = 210 ft·lb
6
Z 6
Z 6
1
(b) W =
F (x) dx =
x−2 dx = −
= 5/6 ft·lb
x 1
1
1
Z
Z
5
0
Z
2
5
40 dx −
F (x) dx =
2. W =
0
Z
3. distance traveled =
2
40
(x − 5) dx = 80 + 60 = 140 J
3
Z
5
v(t) dt =
0
work done is 10 · 10 = 100 ft·lb.
0
5
2 i5
4t
dt = t2 = 10 ft. The force is a constant 10 lb, so the
5
5 0
4. (a) F (x) = kx, F (0.05) = 0.05k = 45, k = 900 N/m
Z
Z 0.03
900x dx = 0.405 J
(c) W =
(b) W =
0
Z
0.10
900x dx = 3.375 J
0.05
0.8
500xdx = 160 J
5. F (x) = kx, F (0.2) = 0.2k = 100, k = 500 N/m, W =
0
Z
2
12x dx = 24 J
6. F (x) = kx, F (1/2) = k/2 = 6, k = 12 N/m, W =
0
Z
1
kx dx = k/2 = 10, k = 20 lb/ft
7. W =
0
Z
6
(9 − x)62.4(25π)dx
8. W =
0
Z
(9 − x)dx = 56, 160π ft·lb
= 1560π
5
9
6
0
6
x
0
Z
6
(9 − x)ρ(25π)dx = 900πρ ft·lb
9. W =
0
9-x
279
Chapter 8
10. r/10 = x/15, r = 2x/3,
Z 10
(15 − x)62.4(4πx2 /9)dx
W =
0
=
15
15 - x
10
Z
83.2
π
3
10
10
x
(15x2 − x3 )dx
r
0
= 208, 000π/3 ft·lb
0
11. w/4 = x/3, w = 4x/3,
Z 2
(3 − x)(9810)(4x/3)(6)dx
W =
0
4
3
3-x
2
Z
2
x
(3x − x2 )dx
= 78480
w(x)
0
= 261, 600 J
0
12.
p
w = 2 4 − x2
Z 2
p
W =
(3 − x)(50)(2 4 − x2 )(10)dx
−2
Z
2
= 3000
−2
p
Z
4 − x2 dx − 1000
2
−2
3
w(x)
2
p
x 4 − x2 dx
2
= 3000[π(2)2 /2] − 0 = 6000π ft·lb
Z
3-x
x
0
-2
9
(10 − x)62.4(300)dx
13. (a) W =
0
Z
10
9
9
(10 − x)dx
= 18, 720
x
0
0
= 926,640 ft·lb
(b) to empty the pool in one hour would require
926,640/3600 = 257.4 ft·lb of work per second
so hp of motor = 257.4/550 = 0.468
20
14. When the rocket is x ft above the ground
15
3000
total weight = weight of rocket
+ weight of fuel
= 3 + [40 − 2(x/1000)]
= 43 − x/500 tons,
Z
3000
(43 − x/500)dx = 120, 000 ft·tons
W =
0
x
Rocket
0
10 - x
Exercise Set 8.7
Z
280
100
15(100 − x)dx
15. W =
16. Let F (x) be the force needed to hold
charge A at position x, then
c
c
F (x) =
, F (−a) = 2 = k,
2
(a − x)
4a
0
= 75, 000 ft·lb
Pulley
so c = 4a2 k.
Z 0
4a2 k(a − x)−2 dx = 2ak J
W =
100
−a
100 - x
A
−a
x
B
0
a
x
Chain
0
17. (a) 150 = k/(4000)2 , k = 2.4 × 109 , w(x) = k/x2 = 2,400,000,000/x2 lb
(b) 6000 = k/(4000)2 , k = 9.6 × 1010 , w(x) = 9.6 × 1010 /(x + 4000)2 lb
Z 5000
9.6(1010 )x−2 dx = 4,800,000 mi·lb = 2.5344 × 1010 ft·lb
(c) W =
4000
18. (a) 20 = k/(1080)2 , k = 2.3328 × 107 , weight = w(x + 1080) = 2.3328 · 107 /(x + 1080)2 lb
Z 10.8
(b) W =
[2.3328 · 107 /(x + 1080)2 ] dx = 213.86 mi·lb = 1,129,188 ft·lb
0
19. W = F · d = (6.40 × 105 )(3.00 × 103 ) = 1.92 × 109 J; from Theorem 8.6.4,
vf2 = 2W/m + vi2 = 2(1.92 · 109 )/(4 · 105 ) + 202 = 10,000, vf = 100 m/s
20. W = F · d = (2.00 × 105 )(2.00 × 105 ) = 4 × 1010 J; from Theorem 8.6.4,
vf2 = 2W/m + vi2 = 8 · 1010 /(2 · 103 ) + 108 ≈ 11.832 m/s.
21. (a) The kinetic energy would have decreased by
(b) (4.5 × 1014 )/(4.2 × 1015 ) ≈ 0.107
1
1
mv 2 = 4 · 106 (15000)2 = 4.5 × 1014 J
2
2
1000
(0.107) ≈ 8.24 bombs
(c)
13
EXERCISE SET 8.7
1. (a) F = ρhA = 62.4(5)(100) = 31,200 lb
2
P = ρh = 62.4(5) = 312 lb/ft
2. (a) F = P A = 6 · 105 (160) = 9.6 × 107 N
Z
(b) F = P A = 100(60) = 6000 lb
2
62.4x(4)dx
3. F =
(b) F = ρhA = 9810(10)(25) = 2,452,500 N
P = ρh = 9810(10) = 98.1 kPa
0
Z
0
x
2
x dx = 499.2 lb
= 249.6
0
2
4
281
Chapter 8
Z
3
4. F =
9810x(4)dx
1
Z
0
x dx
= 39240
4
1
x
3
1
3
= 156, 960 N
Z
5. F =
5
p
9810x(2 25 − x2 )dx
0
Z
5
x(25 − x )
2 1/2
= 19, 620
0
5
y
x
y=
√25 – x 2
5
2 √25 – x 2
dx
0
= 8.175 × 105 N
4
0
6. By similar triangles
√
2 3−x
2 √
w(x)
√
=
, w(x) = √ (2 3 − x),
4
2 3
3
√
Z 2 3
2 √
62.4x √ (2 3 − x) dx
F =
3
0
Z 2√3 √
124.8
(2 3x − x2 )dx = 499.2 lb
= √
3 0
w(x)
x
4
4
2√ 3
7. By similar triangles
0
10 − x
w(x)
=
6
8
2
6
x
3
w(x) = (10 − x),
4
Z 10
3
9810x (10 − x) dx
F =
4
2
Z 10
= 7357.5
(10x − x2 )dx = 1,098,720 N
w(x)
8
10
2
0
8. w(x) = 16 + 2u(x), but
12 − x
1
u(x)
=
so u(x) = (12 − x),
4
8
2
4
4
x
w(x)
w(x) = 16 + (12 − x) = 28 − x,
Z
u(x)
4
12
16
12
62.4x(28 − x)dx
F =
4
Z
12
(28x − x2 )dx = 77, 209.6 lb.
= 62.4
4
Z
9. Yes: if ρ2 = 2ρ1 then F2 =
Z
b
ρ2 h(x)w(x) dx =
a
Z
b
2ρ1 h(x)w(x) dx = 2
a
b
ρ1 h(x)w(x) dx = 2F1 .
a
Exercise Set 8.7
Z
2
10. F =
0
282
p
50x(2 4 − x2 )dx
Z
0
y
2
2
x(4 − x2 )1/2 dx
= 100
y = √ 4 – x2
0
= 800/3 lb
2√ 4 – x 2
x
11. Find the forces on the upper and lower halves and add them:
w1 (x)
x
√
=√
, w1 (x) = 2x
2a
2a/2
Z √2a/2
Z
F1 =
ρx(2x)dx = 2ρ
0
0
a
x
√
2a/2
√2a / 2
√
x2 dx = 2ρa3 /6,
x
0
√
√
w2 (x)
2a − x
√
= √
, w2 (x) = 2( 2a − x)
2a
2a/2
√
Z 2a
Z √2a √
√
√
ρx[2( 2a − x)]dx = 2ρ √
( 2ax − x2 )dx = 2ρa3 /3,
F2 = √
2a/2
a
√2a
2a/2
√
√
√
F = F1 + F2 = 2ρa3 /6 + 2ρa3 /3 = ρa3 / 2 lb
√
12. h(x) = x sin 60◦ = 3x/2,
Z 100
√
9810( 3x/2)(200)dx
F =
200
0
0
√ Z
= 981000 3
100
60°
100
0
√
= 4,905,000 3 N
13.
h(x)
x
x dx
√
√
√
162 + 42 = 272 = 4 17 is the
other dimension of the bottom.
√
(h(x) − 4)/4 = x/(4 17)
√
h(x) = x/ 17 + 4,
Z
√
4 17
F =
0
√
62.4(x/ 17 + 4)10dx
√
4 17
Z
= 624
16
4
0
h(x)
4
x
√
(x/ 17 + 4)dx
4
4√17
0
√
= 14, 976 17 lb
Z
h+2
14. F =
ρ0 x(2)dx
h
Z
0
h
h+2
x dx
= 2ρ0
h
= 4ρ0 (h + 1)
h
x
h+2
2
2
10
a
w1(x)
a √
w2(x)
2a
283
Chapter 8
15. (a) From Exercise 14, F = 4ρ0 (h + 1) so (assuming that ρ0 is constant) dF/dt = 4ρ0 (dh/dt)
which is a positive constant if dh/dt is a positive constant.
(b) If dh/dt = 20 then dF/dt = 80ρ0 lb/min from part (a).
EXERCISE SET 8.8
1. (a)
(b)
(c)
(d)
(e)
(f )
2. (a) csch(−1) ≈ −0.8509
(b) sech(ln 2) = 0.8
sinh 3 ≈ 10.0179
cosh(−2) ≈ 3.7622
tanh(ln 4) = 15/17 ≈ 0.8824
sinh−1 (−2) ≈ −1.4436
cosh−1 3 ≈ 1.7627
3
tanh−1 ≈ 0.9730
4
(c) coth 1 ≈ 1.3130
1
(d) sech−1 ≈ 1.3170
2
(e) coth−1 3 ≈ 0.3466
√
(f ) csch−1 (− 3) ≈ −0.5493
1
4
3−
=
3
3
5
1
1 1
+2 =
(b) cosh(− ln 2) = (e− ln 2 + eln 2 ) =
2
2 2
4
1
1
3. (a) sinh(ln 3) = (eln 3 − e− ln 3 ) =
2
2
312
25 − 1/25
e2 ln 5 − e−2 ln 5
=
=
2
ln
5
−2
ln
5
e
+e
25 + 1/25
313
63
1 1
1 −3 ln 2
3 ln 2
−8 =−
−e
)=
(d) sinh(−3 ln 2) = (e
2
2 8
16
(c) tanh(2 ln 5) =
4. (a)
1
1 ln x
(e
+ e− ln x ) =
2
2
1
1 ln x
(e
− e− ln x ) =
(b)
2
2
1
x+
x
1
x−
x
=
x2 + 1
,x>0
2x
=
x2 − 1
,x>0
2x
x2 − 1/x2
x4 − 1
e2 ln x − e−2 ln x
,x>0
=
=
e2 ln x + e−2 ln x
x2 + 1/x2
x4 + 1
1 − ln x
1 + x2
1 1
ln x
(d)
(e
+x =
,x>0
+e )=
2
2 x
2x
(c)
5.
sinh x0
cosh x0
tanh x0
coth x0
sech x0
csch x0
(a)
2
√5
2/√ 5
√ 5/2
1/ √ 5
1/ 2
(b)
3/ 4
5/ 4
3/ 5
5/3
4/5
4/3
(c)
4/3
5/ 3
4/5
5/4
3/ 5
3/ 4
(a) cosh2 x0 = 1 + sinh2 x0 = 1 + (2)2 = 5, cosh x0 =
√
5
9
25
3
−1=
, sinh x0 = (because x0 > 0)
16
16
4
2
9
16
3
4
=
, sech x0 = ,
=1−
(c) sech2 x0 = 1 − tanh2 x0 = 1 −
5
25
25
5
sinh x0
4
1
5
4
5
=
cosh x0 =
= , from
= tanh x0 we get sinh x0 =
sech x0
3
cosh x0
3
5
3
(b) sinh2 x0 = cosh2 x0 − 1 =
Exercise Set 8.8
6.
284
d
cosh x
d
1
cschx =
=−
= − coth x csch x for x 6= 0
dx
dx sinh x
sinh2 x
d
1
sinh x
d
sech x =
=−
= − tanh x sech x for all x
dx
dx cosh x
cosh2 x
d cosh x
sinh2 x − cosh2 x
d
coth x =
=
= − csch2 x for x 6= 0
dx
dx sinh x
sinh2 x
dy
dy dx
=
cosh y; so
dx dy
dx
1
1
1
dy
d
=√
[sinh−1 x] =
=
=p
for all x.
2
dx
dx
cosh y
1
+ x2
1 + sinh y
7. (a) y = sinh−1 x if and only if x = sinh y; 1 =
(b) Let x ≥ 1. Then y = cosh−1 x if and only if x = cosh y; 1 =
dy
dy dx
=
sinh y, so
dx dy
dx
1
1
1
dy
d
= 2
[cosh−1 x] =
=
=p
for x ≥ 1.
2
dx
dx
sinh y
x −1
cosh y − 1
(c) Let −1 < x < 1. Then y = tanh−1 x if and only if x = tanh y; thus
dy
1
dy
d
dy
dy dx
=
sech2 y =
(1 − tanh2 y) = 1 − x2 , so
[tanh−1 x] =
=
.
1=
dx dy
dx
dx
dx
dx
1 − x2
9. 4 cosh(4x − 8)
12. 2
15.
1
11. − csch2 (ln x)
x
10. 4x3 sinh(x4 )
sech2 2x
tanh 2x
13.
1
csch(1/x) coth(1/x)
x2
2 + 5 cosh(5x) sinh(5x)
q
4x + cosh2 (5x)
14. −2e2x sech(e2x ) tanh(e2x )
16. 6 sinh2 (2x) cosh(2x)
√
√
√
17. x5/2 tanh( x) sech2 ( x) + 3x2 tanh2 ( x)
18. −3 cosh(cos 3x) sin 3x
20.
19.
q
√
(sinh−1 x)2 − 1 1 + x2
23. −(tanh−1 x)−2 /(1 − x2 )
sinh x
sinh x
p
=
=
2
| sinh x|
cosh x − 1
24. 2(coth−1 x)/(1 − x2 )
25.
p
26. (sech2 x)/ 1 + tanh2 x
27. −
28. 10(1 + x csch
31.
−1
1
sinh7 x + C
7
9
x)
p
1
= 1/ 9 + x2
3
√
21. 1/ (cosh−1 x) x2 − 1
1
1
p
(−1/x2 ) = − √
2
|x| x2 + 1
1 + 1/x
22. 1/
1
p
1 + x2 /9
1, x > 0
−1, x < 0
ex
√
+ ex sech−1 x
2x 1 − x
x
− √
+ csch−1 x
|x| 1 + x2
32.
1
sinh(2x − 3) + C
2
33.
2
(tanh x)3/2 + C
3
285
Chapter 8
1
34. − coth(3x) + C
3
1
36. − coth3 x + C
3
35. ln(cosh x) + C
ln 3
37.
1
− sech3 x
3
40. x =
√
√
0
2
√
du =
2
2u − 2
2u,
Z
√
1
u2
−1
√
du = cosh−1 (x/ 2) + C
1
du = − sech−1 (ex ) + C
u 1 − u2
√
41. u = ex ,
Z
√
42. u = cos θ, −
Z
1
du = − sinh−1 (cos θ) + C
1 + u2
du
= − csch−1 |u| + C = − csch−1 |2x| + C
u 1 + u2
√
43. u = 2x,
Z
√
44. x = 5u/3,
i1/2
45. tanh−1 x
0
i√3
46. sinh−1 t
√
1
u2
= sinh−1
√
du =
1
cosh−1 (3x/5) + C
3
1
1 1 + 1/2
ln
= ln 3
2 1 − 1/2
2
√
3 − sinh−1 0 = ln( 3 + 2)
ln 3
sinh 2x dx =
0
−1
= tanh−1 (1/2) − tanh−1 (0) =
ln 3
49. A =
Z
1
5/3
du =
2
3
25u − 25
0
Z
= ln 5 − ln 3
1
1
du = sinh−1 3x + C
2
3
1+u
√
Z
Z
38. ln(cosh x)
ln 2
Z
1
3
39. u = 3x,
iln 3
= 37/375
1
cosh 2x
2
1
[cosh(2 ln 3) − 1],
2
=
0
1
1
1
but cosh(2 ln 3) = cosh(ln 9) = (eln 9 + e− ln 9 ) = (9 + 1/9) = 41/9 so A = [41/9 − 1] = 16/9.
2
2
2
Z
sech2 x dx = π tanh x
0
Z
0
1
52.
0
= π tanh(ln 2) = 3π/5
0
Z
5
5
(cosh2 2x − sinh2 2x)dx = π
51. V = π
Z
ln 2
ln 2
50. V = π
dx = 5π
0
1
1
1
cosh ax dx = 2, sinh ax = 2, sinh a = 2, sinh a = 2a;
a
a
0
let f (a) = sinh a − 2a, then an+1 = an −
sinh an − 2an
, a1 = 2.2, . . . , a4 = a5 = 2.177318985.
cosh an − 2
53. y 0 = sinh x, 1 + (y 0 )2 = 1 + sinh2 x = cosh2 x
ln 2
Z ln 2
1
3
1 ln 2
1
− ln 2
2−
=
cosh x dx = sinh x
= sinh(ln 2) = (e − e
)=
L=
2
2
2
4
0
0
Exercise Set 8.8
286
54. y 0 = sinh(x/a), 1 + (y 0 )2 = 1 + sinh2 (x/a) = cosh2 (x/a)
x1
Z x1
L=
cosh(x/a)dx = a sinh(x/a)
= a sinh(x1 /a)
0
0
55. sinh(−x) =
cosh(−x) =
1
1 −x
(e − ex ) = − (ex − e−x ) = − sinh x
2
2
1
1 −x
(e + ex ) = (ex + e−x ) = cosh x
2
2
1
1 x
(e + e−x ) + (ex − e−x ) = ex
2
2
1 x
1
(b) cosh x − sinh x = (e + e−x ) − (ex − e−x ) = e−x
2
2
1 x
1
(c) sinh x cosh y + cosh x sinh y = (e − e−x )(ey + e−y ) + (ex + e−x )(ey − e−y )
4
4
56. (a) cosh x + sinh x =
=
1 x+y
[(e
− e−x+y + ex−y − e−x−y ) + (ex+y + e−x+y − ex−y − e−x−y )]
4
1 (x+y)
[e
− e−(x+y) ] = sinh(x + y)
2
Let y = x in part (c).
The proof is similar to part (c), or: treat x as variable and y as constant, and differentiate
the result in part (c) with respect to x.
Let y = x in part (e).
Use cosh2 x = 1 + sinh2 x together with part (f).
Use sinh2 x = cosh2 x − 1 together with part (f).
=
(d)
(e)
(f )
(g)
(h)
57. (a) Divide cosh2 x − sinh2 x = 1 by cosh2 x.
sinh y
sinh x
+
tanh x + tanh y
sinh x cosh y + cosh x sinh y
cosh x cosh y
=
=
(b) tanh(x + y) =
sinh
x
sinh
y
cosh x cosh y + sinh x sinh y
1 + tanh x tanh y
1+
cosh x cosh y
(c) Let y = x in part (b).
1
58. (a) Let y = cosh−1 x; then x = cosh y = (ey + e−y ), ey − 2x + e−y = 0, e2y − 2xey + 1 = 0,
2
√
p
2x ± 4x2 − 4
y
= x ± x2 − 1. To determine which sign to take, note that y ≥ 0
e =
2
√
so e−y ≤ ey , x = (ey + e−y )/2 ≤ (ey + ey )/2 = ey , hence ey ≥ x thus ey = x + x2 − 1,
√
y = cosh−1 x = ln(x + x2 − 1).
ey − e−y
e2y − 1
, xe2y + x = e2y − 1,
=
ey + e−y
e2y + 1
1 1+x
1+x
, y = ln
.
= (1 + x)/(1 − x), 2y = ln
1−x
2 1−x
(b) Let y = tanh−1 x; then x = tanh y =
1 + x = e2y (1 − x), e2y
√
p
d
1 + x/ x2 − 1
−1
√
59. (a)
(cosh x) =
= 1/ x2 − 1
2
dx
x+ x −1
d
1
1
d 1
1
−1
(b)
(tanh x) =
(ln(1 + x) − ln(1 − x)) =
+
= 1/(1 − x2 )
dx
dx 2
2 1+x 1−x
287
Chapter 8
60. Let y = sech−1 x then x = sech y = 1/ cosh y, cosh y = 1/x, y = cosh−1 (1/x); the proofs for the
remaining two are similar.
Z
61. If |u| < 1 then, by Theorem 8.8.6,
Z
For |u| > 1,
62. (a)
du
= tanh−1 u + C.
1 − u2
du
= coth−1 u + C = tanh−1 (1/u) + C.
1 − u2
√
d
1
1
d
x
√ =− √
(sech−1 |x|) =
(sech−1 x2 ) = − √ √
dx
dx
x 1 − x2
x2 1 − x2 x2
(b) Similar to solution of part (a)
63. (a)
(b)
(c)
(d)
(e)
(f )
1 x
(e − e−x ) = +∞ − 0 = +∞
2
1 x
lim sinh x = lim
(e − e−x ) = 0 − ∞ = −∞
x→−∞
x→−∞ 2
ex − e−x
=1
lim tanh x = lim x
x→+∞
x→+∞ e + e−x
ex − e−x
= −1
lim tanh x = lim x
x→−∞
x→−∞ e + e−x
p
lim sinh−1 x = lim ln(x + x2 + 1) = +∞
lim sinh x = lim
x→+∞
x→+∞
x→+∞
x→+∞
lim tanh−1 x = lim
x→1−
x→1−
1
[ln(1 + x) − ln(1 − x)] = +∞
2
p
x2 − 1) − ln x]
x→+∞
x→+∞
√
p
x + x2 − 1
= lim ln(1 + 1 − 1/x2 ) = ln 2
= lim ln
x→+∞
x→+∞
x
cosh x
ex + e−x
1
(1 + e−2x ) = 1/2
= lim
= lim
(b)
lim
x→+∞
x→+∞
x→+∞ 2
ex
2ex
64. (a)
lim (cosh−1 x − ln x) = lim [ln(x +
65. For |x| < 1, y = tanh−1 x is defined and dy/dx = 1/(1 − x2 ) > 0; y 00 = 2x/(1 − x2 )2 changes sign
at x = 0, so there is a point of inflection there.
Z
Z
1
1
√
du = − √
dx = − cosh−1 x + C = − cosh−1 (−u) + C.
66. Let x = −u,
2
2
u −1
x −1
p
1
√
− cosh−1 (−u) = − ln(−u + u2 − 1) = ln
−u + u2 − 1
p
p
= ln(−u − u2 − 1) = ln |u + u2 − 1|
67. Using sinh x + cosh x = ex (Exercise 56a), (sinh x + cosh x)n = (ex )n = enx = sinh nx + cosh nx.
Z
68.
a
1
e dx = etx
t
−a
a
tx
=
−a
1 at
2 sinh at
(e − e−at ) =
for t 6= 0.
t
t
69. (a) y 0 = sinh(x/a), 1 + (y 0 )2 = 1 + sinh2 (x/a) = cosh2 (x/a)
b
Z b
cosh(x/a) dx = 2a sinh(x/a) = 2a sinh(b/a)
L=2
0
0
(b) The highest point is at x = b, the lowest at x = 0,
so S = a cosh(b/a) − a cosh(0) = a cosh(b/a) − a.
Chapter 8 Supplementary Exercises
288
70. From part (a) of Exercise 69, L = 2a sinh(b/a) so 120 = 2a sinh(50/a), a sinh(50/a) = 60. Let
u = 50/a, then a = 50/u so (50/u) sinh u = 60, sinh u = 1.2u. If f (u) = sinh u − 1.2u, then
sinh un − 1.2un
; u1 = 1, . . . , u5 = u6 = 1.064868548 ≈ 50/a so a ≈ 46.95415231.
un+1 = un −
cosh un − 1.2
From part (b), S = a cosh(b/a) − a ≈ 46.95415231[cosh(1.064868548) − 1] ≈ 29.2 ft.
71. From part (b) of Exercise 69, S = a cosh(b/a) − a so 30 = a cosh(200/a) − a. Let u = 200/a,
then a = 200/u so 30 = (200/u)[cosh u − 1], cosh u − 1 = 0.15u. If f (u) = cosh u − 0.15u − 1,
cosh un − 0.15un − 1
; u1 = 0.3, . . . , u4 = u5 = 0.297792782 ≈ 200/a so
then un+1 = un −
sinh un − 0.15
a ≈ 671.6079505. From part (a), L = 2a sinh(b/a) ≈ 2(671.6079505) sinh(0.297792782) ≈ 405.9 ft.
72. (a) When the bow of the boat is at the point (x, y) and the person has walked a distance D,
then the person is located at the point (0, D), the line segment connecting (0, D) and (x, y)
√
has length a; thus a2 = x2 + (D − y)2 , D = y + a2 − x2 = a sech−1 (x/a).
!
p
1 + 5/9
−1
≈ 14.44 m.
(b) Find D when a = 15, x = 10: D = 15 sech (10/15) = 15 ln
2/3
2
1p 2
x
1
a
a2
+√
=√
a − x2 ,
− +x =−
(c) dy/dx = − √
x
x
x a2 − x2
a2 − x2
a2 − x2
15
Z 15
a2 − x2
a2
225
225
=
;
with
a
=
15
and
x
=
5,
L
=
dx
=
−
= 30 m.
1 + [y 0 ]2 = 1 +
x2
x2
x2
x 5
5
CHAPTER 8 SUPPLEMENTARY EXERCISES
Z
Z
2
(2 + x − x2 ) dx
6. (a) A =
2
(b) A =
0
√
Z
0
Z
4
y dy +
√
[( y − (y − 2)] dy
2
2
[(2 + x)2 − x4 ] dx
Z 4
Z 2
√
√
y y dy + 2π
y[ y − (y − 2)] dy
(d) V = 2π
0
2
Z
Z 2
x(2 + x − x2 ) dx
(f ) V = π
(e) V = 2π
(c) V = π
0
0
Z
Z
a
Z
Z
0
(b) A =
−1
Z
(x3 − x) dx +
8/27
8. (a) S =
0
Z
2
(c) S =
Z
c
(f (x) − g(x)) dx +
(g(x) − f (x)) dx +
b
1
Z
(x − x3 ) dx +
0
p
2π(y + 2) 1 + y 4 /81 dy
2
d
(f (x) − g(x)) dx
c
2
11
1 1 9
+ + =
4 4 4
4
(x3 − x) dx =
1
p
2πx 1 + x−4/3 dx
4
π(y − (y − 2)2 ) dy
y dy +
0
b
7. (a) A =
Z
2
Z
(b) S =
2
2π
0
y3 p
1 + y 4 /81 dy
27
0
2
y 1/3
y 2/3
x2/3 + y 2/3
a2/3
dy
dy
=−
, so 1 +
=1+
=
= 2/3 ,
2/3
dx
x
dx
x
x
x
Z −a/8
dx = −a1/3
x−1/3 dx = 9a/8.
9. By implicit differentiation
Z
−a/8
L=
−a
a1/3
(−x1/3 )
−a
289
Chapter 8
10. The base of the dome is a hexagon of side r. An equation of the circle of radius r that lies in a
vertical x-y plane and passes through two opposite vertices of the base hexagon is x2 + y 2 = r2 .
A horizontal,
√ cross section at height y above theZ base
√ hexagonal
√has area
r
√
3 3 2
3 3 2
3
3 2
x =
(r − y 2 ), hence the volume is V =
(r − y 2 ) dy = 3r3 .
A(y) =
2
2
2
0
11. Let the sphere have radius R, the hole radius r. By the Pythagorean Theorem, r2 + (L/2)2 = R2 .
Use cylindrical shells to √
calculate the volume
√ of the solid obtained by rotating about the y-axis
the region r < x < R, − R2 − x2 < y < R2 − x2 :
R
Z R
p
4
4
(2πx)2 R2 − x2 dx = − π(R2 − x2 )3/2
= π(L/2)3 ,
V =
3
3
r
r
so the volume is independent of R.
Z
L/2
12. V = 2
π
0
16R2 2
4π
LR2
(x − L2 /4)2 =
L4
15
13. Set a = 68.7672, b = 0.0100333, c = 693.8597, d = 299.2239.
650
(a)
Z
d
p
1 + a2 b2 sinh2 bx dx
(b) L = 2
0
= 1480.2798 ft
-300
300
0
(d) 82◦
(c) x = 283.6249 ft
Z
b
p
1 + (12.54 − 0.82x)2 dx = 196.306 yd
14. y = 0 at x = b = 30.585; distance =
0
Z
15. Let u = ax then du = a dx,
Z
Z
1
√
du =
2
u − a2
1
1
du =
a2 − u2
a
Z
√
Z
1
x2
−1
√
1
du =
a2 + u2
Z
√
a
dx =
a2 + a2 x2
Z
√
1
dx = sinh−1 (u/a)+C,
1 + x2
dx = cosh−1 (u/a) + C, u > a,
1
tanh−1 (u/a) + C, |u| < a
a
a+u
1
1
ln
+C
=
dx =
1 − x2
2a
a−u
1
−1
coth (u/a) + C, |u| > a
a
(a) sinh−1 (x/2) + C
(b) cosh−1 (x/3) + C
(c)
1
√
√
−1
√
(x/
2)
+
C,
|x|
<
2
tanh
2
√
1
2+x
= √ ln √
+C
2 2
2−x
√
√
1
√ coth−1 (x/ 2) + C, |x| > 2
2
Z
Z
1
1
dx
dx
√
p
=√
dx = √ sinh−1
(d)
5
5
16 + 5x2
16/5 + x2
√ !
5x
+C
4
Chapter 8 Supplementary Exercises
290
16. (a) cosh 3x = cosh(2x + x) = cosh 2x cosh x + sinh 2x sinh x
= (2 cosh2 x − 1) cosh x + (2 sinh x cosh x) sinh x
= 2 cosh3 x − cosh x + 2 sinh2 x cosh x
= 2 cosh3 x − cosh x + 2(cosh2 x − 1) cosh x = 4 cosh3 x − 3 cosh x
x
x
(b) from Theorem 8.8.2 with x replaced by : cosh x = 2 cosh2 − 1,
2
2
1
2 x
2 x
= cosh x + 1, cosh
= (cosh x + 1),
2 cosh
2r
2
2
x
x
1
cosh =
(cosh x + 1) (because cosh > 0)
2
2
2
x
x
(c) from Theorem 8.8.2 with x replaced by : cosh x = 2 sinh2 + 1,
2
2
r
1
x
x
x
1
2 sinh2 = cosh x − 1, sinh2 = (cosh x − 1), sinh = ±
(cosh x − 1)
2
2
2
2
2
1
1
= k , k = 2, W =
2
4
17. (a) F = kx,
Z
Z
1/4
kx dx = 1/16 J
0
L
kx dx = kL2 /2, L = 5 m
(b) 25 =
0
Z
150
(30x + 2000) dx = 15 · 1502 + 2000 · 150 = 637,500 lb·ft
18. F = 30x + 2000, W =
0
Z
1
19. (a) F =
ρx3 dx N
0
x
w(x)
= , w(x) = 2x, so
(b) By similar triangles
4
2
Z 4
2
ρ(1 + x)2x dx lb/ft .
F =
h(x) = 1 + x
0
1
w(x)
x
2
8 2
(c) A formula for the parabola is y =
x − 10, so F =
125
r
0
9810|y|2
−10
125
(y + 10) dy N.
8
(b) r = 1 when t ≈ 0.673080 s.
r
20. (a)
Z
4
(c) dr/dt = 4.48 m/s.
2
1
t
1
291
Chapter 8
y
21. (a)
x
100
200
-0.4
(b) The maximum deflection occurs at
x = 96 inches (the midpoint
of the beam) and is about 1.42 in.
(c) The length of the centerline is
Z 192 p
1 + (dy/dx)2 dx = 192.026 in.
-0.8
0
-1.2
-1.6
22. The x-coordinates of the points of intersection are a ≈ −0.423028 and b ≈ 1.725171; the area is
Z b
(2 sin x − x2 + 1)dx ≈ 2.542696.
a
23. Let (a, k), where π/2 < a < π, be the coordinates of the point of intersection of y = k with
y = sin x. Thus k = sin a and if the shaded areas are equal,
Z a
Z a
(k − sin x)dx =
(sin a − sin x) dx = a sin a + cos a − 1 = 0
0
0
Solve for a to get a ≈ 2.331122, so k = sin a ≈ 0.724611.
Z
0
k = 1.736796.
Z
a+2
25. (a)
a
√
k
x sin x dx = 2π(sin k − k cos k) = 8; solve for k to get
24. The volume is given by 2π
x
dx
1 + x3
(b) Use the result in Exercise 24, Section 7.9, to obtain
Z a+2
a+2
a
x
d
√
dx = p
−√
= 0; solve for a to get a ≈ 0.683772.
3
3
da a
1+x
1 + a3
1 + (a + 2)
Z a+2
x
√
dx ≈ 1.347655 J.
The maximum work is
1
+ x3
a
CHAPTER 9
Principles of Integral Evaluation
EXERCISE SET 9.1
1. u = 3 − 2x, du = −2dx,
1
9
2. u = 4 + 9x, du = 9dx,
3. u = x2 , du = 2xdx,
1
2
4. u = x2 , du = 2xdx,
2
1
2
−
Z
1
1
u3 du = − u4 + C = − (3 − 2x)4 + C
8
8
Z
2 3/2
2
u +C =
(4 + 9x)3/2 + C
3·9
27
u1/2 du =
Z
1
1
tan u + C = tan(x2 ) + C
2
2
sec2 u du =
Z
tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C
5. u = 2 + cos 3x, du = −3 sin 3xdx,
6. u =
3
3x
, du = dx,
2
2
2
3
Z
−
1
3
1
du
=
2
4 + 4u
6
Z
Z
1
1
du
= − ln |u| + C = − ln(2 + cos 3x) + C
u
3
3
1
1
du
= tan−1 u + C = tan−1 (3x/2) + C
2
1+u
6
6
Z
sinh u du = cosh u + C = cosh ex + C
7. u = ex , du = ex dx,
8. u = ln x, du =
1
dx,
x
Z
sec u tan u du = sec u + C = sec(ln x) + C
Z
9. u = cot x, du = − csc2 xdx,
10. u = x2 , du = 2xdx,
1
2
−
Z
√
1
1
du
= sin−1 u + C = sin−1 (x2 ) + C
2
2
2
1−u
11. u = cos 7x, du = −7 sin 7xdx,
Z
12. u = sin x, du = cos x dx,
Z
13. u = ex , du = ex dx,
14. u = tan
15. u =
√
−1
√
1
7
−
Z
u5 du = −
1 6
1
u + C = − cos6 7x + C
42
42
p
√
1 + 1 + sin2 x
1 + 1 + u2
du
√
+ C = − ln
+C
= − ln
u
sin x
u u2 + 1
p
p
du
= ln u + u2 + 4 + C = ln ex + e2x + 4 + C
4 + u2
1
x, du =
dx,
1 + x2
1
dx,
x − 2, du = √
2 x−2
2
eu du = −eu + C = −ecot x + C
16. u = 3x + 2x, du = (6x + 2)dx,
Z
eu du = eu + C = etan
Z
x
+C
√
x−2
eu du = 2eu + C = 2e
2
1
2
−1
Z
cot u du =
292
+C
1
1
ln | sin u| + C = ln sin |3x2 + 2x| + C
2
2
293
Chapter 9
17. u =
√
18. u = ln x, du =
19. u =
√
Z
1
x, du = √ dx,
2 x
Z
dx
,
x
√
2 cosh u du = 2 sinh u + C = 2 sinh x + C
du
= ln |u| + C = ln |ln x| + C
u
Z
1
x, du = √ dx,
2 x
2 du
=2
3u
Z
e−u ln 3 du = −
2 −u ln 3
2 −√x
e
3
+C =−
+C
ln 3
ln 3
Z
20. u = sin θ, du = cos θdθ,
2
2
, du = − 2 dx,
x
x
21. u =
Z
22.
sec u tan u du = sec u + C = sec(sin θ) + C
√
−
1
2
Z
csch2 u du =
1
2
1
coth u + C = coth + C
2
2
x
p
dx
= ln x + x2 − 3 + C
x2 − 3
Z
23. u = e−x , du = −e−x dx,
24. u = ln x, du =
1
dx,
x
cos u du = sin u + C = sin(ln x) + C
x
25. u = e , du = e dx,
26. u = x−1/2 , du = −
ex dx
√
=
1 − e2x
1
dx,
2x3/2
1
2
27. u = x2 , du = 2xdx,
Z
Z
−
1
du
=
sec u
2
Z
28. 2u = ex , 2du = ex dx,
2+u
1
2 + e−x
1
du
ln
+
C
=
−
ln
=
−
+C
4 − u2
4
2−u
4
2 − e−x
Z
Z
x
−
√
Z
√
du
= sin−1 u + C = sin−1 ex + C
1 − u2
2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C
Z
cos u du =
1
1
sin u + C = sin(x2 ) + C
2
2
2du
= sin−1 u + C = sin−1 (ex /2) + C
4 − 4u2
29. 4−x = e−x ln 4 , u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx,
Z
1 u
1 −x2 ln 4
1 −x2
1
eu du = −
e +C =−
e
4
+C =−
+C
−
ln 16
ln 16
ln 16
ln 16
2
2
Z
30. 2
πx
πx ln 2
=e
,
2πx dx =
1 πx ln 2
1
e
2πx + C
+C =
π ln 2
π ln 2
31. (a) With u = x we get
Z
dx
√
= sin−1 x + C (|x| < 1)
from (15),
1 − x2
Z
dx
= tan−1 x + C
from (17),
1 + x2
Z
dx
√
from (19),
= sec−1 x + C (x > 1)
x x2 − 1
Exercise Set 9.2
294
(b) With u = ax, du = adx, we get (for a > 0)
Z
Z
adx
du
√
√
=
= sin−1 x + C = sin−1 (u/a) + C
2
2
a −u
a 1 − x2
Z
Z
1
1
1
adx
du
=
= tan−1 x + C = tan−1 (u/a) + C
2
2
2
2
a +u
a
1+x
a
a
Z
Z
1
1
du
adx
√
√
=
= sec−1 x + C = sec−1 (u/a) + C
2
2
2
2
a
a
u u −a
a x x −1
EXERCISE SET 9.2
−x
1. u = x, dv = e
−x
dx, du = dx, v = −e
Z
;
−x
xe
−x
dx = −xe
Z
+
e−x dx = −xe−x − e−x + C
Z
1 3x 1
1
1
xe −
e3x dx = xe3x − e3x + C
3
3
3
9
Z
Z
3. u = x2 , dv = ex dx, du = 2x dx, v = ex ;
x2 ex dx = x2 ex − 2 xex dx.
2. u = x, dv = e3x dx, du = dx, v =
1 3x
e ;
3
Z
xe3x dx =
Z
xex dx use u = x, dv = ex dx, du = dx, v = ex to get
For
Z
Z
xe dx = xe − e + C1 so
x
x
x
x2 ex dx = x2 ex − 2xex + 2ex + C
1
4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ;
2
Z
For xe−2x dx use u = x, dv = e−2x dx to get
Z
1
x2 e−2x dx = − x2 e−2x +
2
Z
xe−2x dx
Z
Z
1
1
1
1
xe−2x dx = − xe−2x +
e−2x dx = − xe−2x − e−2x + C
2
2
2
4
Z
1
1
1
so
x2 e−2x dx = − x2 e−2x − xe−2x − e−2x + C
2
2
4
1
5. u = x, dv = sin 2x dx, du = dx, v = − cos 2x;
2
Z
Z
1
1
1
1
cos 2x dx = − x cos 2x + sin 2x + C
x sin 2x dx = − x cos 2x +
2
2
2
4
1
6. u = x, dv = cos 3x dx, du = dx, v = sin 3x;
3
Z
Z
1
1
1
1
sin 3x dx = x sin 3x + cos 3x + C
x cos 3x dx = x sin 3x −
3
3
3
9
Z
Z
7. u = x2 , dv = cos x dx, du = 2x dx, v = sin x;
x2 cos x dx = x2 sin x − 2 x sin x dx
Z
For
x sin x dx use u = x, dv = sin x dx to get
Z
Z
x sin x dx = −x cos x + sin x + C1 so
x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C
295
Chapter 9
8. u = x2 , dv = sin x dx, du = 2x dx, v = − cos x;
Z
Z
Z
2
2
x sin x dx = −x cos x + 2 x cos x dx; for x cos x dx use u = x, dv = cos x dx to get
Z
Z
x cos x dx = x sin x + cos x + C1 so
x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C
√
2
1
9. u = ln x, dv = x dx, du = dx, v = x3/2 ;
x
3
Z
Z
√
2 3/2
2
2
4
x1/2 dx = x3/2 ln x − x3/2 + C
x ln x dx = x ln x −
3
3
3
9
1
1
10. u = ln x, dv = x dx, du = dx, v = x2 ;
x
2
Z
1
1
x ln x dx = x2 ln x −
2
2
Z
x dx =
1 2
1
x ln x − x2 + C
2
4
Z
Z
ln x
dx, v = x; (ln x)2 dx = x(ln x)2 − 2 ln x dx.
x
Z
Z
Use u = ln x, dv = dx to get
ln x dx = x ln x − dx = x ln x − x + C1 so
11. u = (ln x)2 , dv = dx, du = 2
Z
(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C
√
1
1
12. u = ln x, dv = √ dx, du = dx, v = 2 x;
x
x
Z
√
ln x
√ dx = 2 x ln x−2
x
Z
√
√
1
√ dx = 2 x ln x−4 x+C
x
Z
Z
2
2x
dx, v = x;
ln(2x + 3)dx = x ln(2x + 3) −
dx
13. u = ln(2x + 3), dv = dx, du =
2x + 3
2x + 3
Z
Z
3
3
2x
dx =
1−
dx = x − ln(2x + 3) + C1 so
but
2x + 3
2x + 3
2
Z
3
ln(2x + 3)dx = x ln(2x + 3) − x + ln(2x + 3) + C
2
Z
Z
2x
x2
2
2
dx,
v
=
x;
ln(x
dx
+
4)dx
=
x
ln(x
+
4)
−
2
x2 + 4
x2 + 4
Z
Z
4
x
x2
dx =
1− 2
dx = x − 2 tan−1 + C1 so
but
x2 + 4
x +4
2
Z
x
ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 + C
2
14. u = ln(x2 + 4), dv = dx, du =
√
15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x;
Z
Z
p
p
−1
−1
sin x dx = x sin x − x/ 1 − x2 dx = x sin−1 x + 1 − x2 + C
16. u = cos−1 (2x), dv = dx, du = − √
Z
cos−1 (2x)dx = x cos−1 (2x) +
Z
2
dx, v = x;
1 − 4x2
√
1p
2x
dx = x cos−1 (2x) −
1 − 4x2 + C
2
2
1 − 4x
Exercise Set 9.2
296
2
17. u = tan−1 (2x), dv = dx, du =
dx, v = x;
1 + 4x2
Z
Z
2x
1
tan−1 (2x)dx = x tan−1 (2x) −
dx = x tan−1 (2x) − ln(1 + 4x2 ) + C
2
1 + 4x
4
Z
Z
x2
1
1 2
1 2
1
−1
−1
x
x
dx,
v
=
;
x
tan
x
dx
=
tan
x
−
dx
18. u = tan x, dv = x dx, du =
1 + x2
2
2
2
1 + x2
Z
Z
1
x2
dx = x − tan−1 x + C1 so
dx
=
1
−
but
1 + x2
1 + x2
Z
1
1
1
x tan−1 x dx = x2 tan−1 x − x + tan−1 x + C
2
2
2
−1
Z
19. u = e , dv = sin x dx, du = e dx, v = − cos x;
x
Z
e sin x dx = −e cos x +
x
x
Z
Z
x
For
x
e cos x = e sin x −
e cos x dx use u = e , dv = cos x dx to get
Z
Z
x
x
ex cos x dx.
x
ex sin x dx so
Z
ex sin x dx = −ex cos x + ex sin x −
ex sin x dx,
Z
Z
ex sin x dx = ex (sin x − cos x) + C1 ,
2
20. u = e2x , dv = cos 3x dx, du = 2e2x dx, v =
Z
e2x cos 3x dx =
Z
Z
Z
1
2
e2x sin 3x dx = − e2x cos 3x +
3
3
2x
e
13
9
1 2x
2
e sin 3x −
3
3
ex sin x dx =
1
sin 3x;
3
e2x sin 3x dx. Use u = e2x , dv = sin 3x dx to get
Z
e2x cos 3x dx so
1
2
4
cos 3x dx = e2x sin 3x + e2x cos 3x −
3
9
9
Z
e2x cos 3x dx =
1 x
e (sin x − cos x) + C
2
Z
e2x cos 3x dx,
1 2x
e (3 sin 3x + 2 cos 3x) + C1 ,
9
Z
e2x cos 3x dx =
1 2x
e (3 sin 3x + 2 cos 3x) + C
13
1
21. u = eax , dv = sin bx dx, du = aeax dx, v = − cos bx (b 6= 0);
b
Z
Z
1
a
eax sin bx dx = − eax cos bx +
eax cos bx dx. Use u = eax , dv = cos bx dx to get
b
b
Z
Z
1
a
eax cos bx dx = eax sin bx −
eax sin bx dx so
b
b
Z
Z
1
a
a2
eax sin bx dx,
eax sin bx dx = − eax cos bx + 2 eax sin bx − 2
b
b
b
Z
eax
eax sin bx dx = 2
(a sin bx − b cos bx) + C
a + b2
e−3θ
22. From Exercise 21 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C
34
297
Chapter 9
cos(ln x)
dx, v = x;
x
Z
Z
sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to get
23. u = sin(ln x), dv = dx, du =
Z
Z
cos(ln x)dx = x cos(ln x) +
sin(ln x)dx so
Z
Z
sin(ln x)dx = x sin(ln x) − x cos(ln x) −
sin(ln x)dx,
Z
sin(ln x)dx = (x/2)[sin(ln x) − cos(ln x)] + C
1
24. u = cos(ln x), dv = dx, du = − sin(ln x)dx, v = x;
x
Z
Z
cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to get
Z
Z
sin(ln x)dx = x sin(ln x) −
cos(ln x)dx so
Z
Z
cos(ln x)dx = x cos(ln x) + x sin(ln x) −
Z
cos(ln x)dx =
cos(ln x)dx,
1
x[cos(ln x) + sin(ln x)] + C
2
25. u = x, dv = sec2 x dx, du = dx, v = tan x;
Z
Z
Z
sin x
x sec2 x dx = x tan x − tan x dx = x tan x −
dx = x tan x + ln | cos x| + C
cos x
26. u = x, dv = tan2 x dx = (sec2 x − 1)dx, du = dx, v = tan x − x;
Z
Z
x tan2 x dx = x tan x − x2 − (tan x − x)dx
1
1
= x tan x − x2 + ln | cos x| + x2 + C = x tan x − x2 + ln | cos x| + C
2
2
2
27. u = x2 , dv = xex dx, du = 2x dx, v =
Z
2
1
x e dx = x2 ex −
2
3 x2
Z
2
xex dx =
1 x2
e ;
2
1 2 x2 1 x2
x e − e +C
2
2
1
1
;
dx, du = (x + 1)ex dx, v = −
(x + 1)2
x+1
Z
Z
xex
xex
ex
xex
x
x
+
e
+
e
+C
dx
=
−
dx
=
−
+
C
=
(x + 1)2
x+1
x+1
x+1
28. u = xex , dv =
Exercise Set 9.2
298
1
29. u = x, dv = e−5x dx, du = dx, v = − e−5x ;
5
1
Z
Z 1
1
1 1 −5x
xe−5x dx = − xe−5x +
e
dx
5
5 0
0
0
1
1
1
1
1
= − e−5 − e−5x = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25
5
25
5
25
0
30. u = x, dv = e2x dx, du = dx, v =
Z
2
xe2x dx =
0
1 2x
xe
2
2
−
0
1
2
Z
2
0
1 2x
e ;
2
1
e2x dx = e4 − e2x
4
2
0
1
= e4 − (e4 − 1) = (3e4 + 1)/4
4
1
1
dx, v = x3 ;
x
3
e
e
Z
Z e
1 3
1 e 2
1 3 1 3
1
1
2
x ln x dx = x ln x −
x dx = e − x
= e3 − (e3 − 1) = (2e3 + 1)/9
3
3
3
9
3
9
1
1
1
1
31. u = ln x, dv = x2 dx, du =
1
1
1
dx, du = dx, v = − ;
2
x
x
x
e
Z e
1
ln x
1
dx = − ln x √ + √ 2 dx
2
x
x
e x
e
e
√
√
1
1
1
1
1
1
1
3 e−4
= − + √ ln e −
=− + √ − +√ =
e
x √e
e 2 e e
2e
e
e
32. u = ln x, dv =
Z
e
√
e
1
dx, v = x;
x+3
2
Z 2
Z 2
Z 2
3
x
dx = 2 ln 5 + 2 ln 1 −
dx
ln(x + 3)dx = x ln(x + 3)
−
1−
x+3
−2
−2 x + 3
−2
−2
33. u = ln(x + 3), dv = dx, du =
2
= 2 ln 5 − [x − 3 ln(x + 3)]−2 = 2 ln 5 − (2 − 3 ln 5) + (−2 − 3 ln 1) = 5 ln 5 − 4
1
dx, v = x;
1 − x2
1/2
1/2 Z 1/2
Z 1/2
1
x
1 p
√
sin−1 x dx = x sin−1 x
−
dx = sin−1 + 1 − x2
2
2
1 − x2
0
0
0
0
r
√
π
1 π
3
3
+
−1=
+
−1
=
2 6
4
12
2
34. u = sin−1 x, dv = dx, du = √
√
1
√
dθ, v = θ;
2θ θ − 1
4
Z 4
√
√ 4 1Z 4
√
√
1
√
dθ = 4 sec−1 2 − 2 sec−1 2 − θ − 1
sec−1 θdθ = θ sec−1 θ −
2 2
θ−1
2
2
2
π √
π
5π √
−2
− 3+1=
− 3+1
=4
3
4
6
35. u = sec−1
θ, dv = dθ, du =
299
Chapter 9
1
1
36. u = sec−1 x, dv = x dx, du = √
dx, v = x2 ;
2
2
x x −1
2
Z
Z 2
1
1 2
x
√
x sec−1 x dx = x2 sec−1 x −
dx
2−1
2
2
x
1
1
1
2
√
1p 2
1
x − 1 = 2π/3 − 3/2
= [(4)(π/3) − (1)(0)] −
2
2
1
1
37. u = x, dv = sin 4x dx, du = dx, v = − cos 4x;
4
π/2
π/2
Z
Z π/2
1
1 π/2
1
sin 4x
x sin 4x dx = − x cos 4x
+
cos 4x dx = −π/8 +
= −π/8
4
4 0
16
0
0
0
Z
π
38.
(x + x cos x)dx =
0
1 2
x
2
π
Z
π
+
x cos x dx =
0
0
π2
+
2
Z
π
x cos x dx;
0
u = x, dv = cos x dx, du = dx, v = sin x
π Z π
π
Z π
Z π
x cos x dx = x sin x −
sin x dx = cos x = −2 so
(x + x cos x)dx = π 2 /2 − 2
0
0
0
√
0
0
√
1
2
xdx, du = √
dx, v = x3/2 ;
3
2 x(1 + x)
3
Z 3
Z 3
√
√
√
2
1
x
dx
x tan−1 xdx = x3/2 tan−1 x −
3
3 1 1+x
1
1
Z
√ 3 1 3
1
2 3/2
−1
dx
1−
x −
= x tan
3
3 1
1+x
1
3
√
√
1
1
2 3/2
−1
x tan
x − x + ln |1 + x| = (2 3π − π/2 − 2 + ln 2)/3
=
3
3
3
1
39. u = tan−1
x, dv =
2x
dx, v = x;
+1
2 Z 2
Z 2
Z 2
1
2x2
dx
=
2
ln
5
−
2
dx
1
−
ln(x2 + 1)dx = x ln(x2 + 1) −
2
x2 + 1
0
0 x +1
0
0
i2
= 2 ln 5 − 2(x − tan−1 x) = 2 ln 5 − 4 + 2 tan−1 2
40. u = ln(x2 + 1), dv = dx, du =
x2
0
41. t =
√
x, t2 = x, dx = 2t dt
Z
(a)
√
x
e
Z
√
x
e
Z
(b)
cos
Z
cos
Z
dx = 2
tet dt; u = t, dv = et dt, du = dt, v = et ,
Z
dx = 2tet − 2
√
√
√
et dt = 2(t − 1)et + C = 2( x − 1)e x + C
Z
x dx = 2
t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,
√
x dx = 2t sin t − 2
Z
√
√
√
sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C
Exercise Set 9.2
300
Z
e
ln x dx = (x ln x − x) = 1
e
43. (a) A =
1
1
Z
e
(b) V = π
h
ie
(ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) = π(e − 2)
1
1
Z
π/2
(x − x sin x)dx =
44. A =
0
Z
1 2
x
2
π/2
Z
π/2
π/2
−
x sin x dx =
0
0
0
= 2π 2
x sin x dx = 2π(−x cos x + sin x)
0
Z
= π 2 /8 − 1
π
π
45. V = 2π
π2
− (−x cos x + sin x)
8
0
π/2
46. V = 2π
π/2
x cos x dx = 2π(cos x + x sin x)
= π(π − 2)
0
0
Z
47. distance =
5
t2 e−t dt; u = t2 , dv = e−t dt, du = 2tdt, v = −e−t ,
0
2 −t
distance = −t e
Z
i5
0
= −25e
te−t dt; u = 2t, dv = e−t dt, du = 2dt, v = −e−t ,
0
distance = −25e−5 − 2te−t
−5
5
+2
−5
− 10e
Z
i5
5
+2
0
e−t dt = −25e−5 − 10e−5 − 2e−t
0
−5
− 2e
−5
+ 2 = −37e
i5
0
+2
1
cos(kωt); the integrand is an even function of t so
kω
π/ω
Z π/ω
Z π/ω
Z π/ω
2
1
cos(kωt) dt
t sin(kωt) dt = 2
t sin(kωt) dt = − t cos(kωt)
+2
kω
kω
−π/ω
0
0
0
π/ω
2
2π(−1)k+1
2π(−1)k+1
+
sin(kωt)
=
=
kω 2
k2 ω2
kω 2
0
48. u = 2t, dv = sin(kωt)dt, du = 2dt, v = −
Z
Z
1
2
sin x dx = − sin2 x cos x − cos x + C
3
3
Z
Z
Z
1
1
3
1
(b)
sin4 x dx = − sin3 x cos x +
sin2 x dx,
sin2 x dx = − sin x cos x + x + C1 so
4
4
2
2
π/4
Z π/4
3
1
3
4
3
sin x dx = − sin x cos x − sin x cos x + x
4
8
8 0
0
1
2
sin x dx = − sin2 x cos x +
3
3
3
49. (a)
√
√
√
√
1
3
= − (1/ 2)3 (1/ 2) − (1/ 2)(1/ 2) + 3π/32 = 3π/32 − 1/4
4
8
Z
50. (a)
1
4
cos x dx = cos4 x sin x +
5
5
5
=
Z
1
4 1
2
4
2
cos x dx = cos x sin x +
cos x sin x + sin x + C
5
5 3
3
3
4
8
1
cos4 x sin x +
cos2 x sin x +
sin x + C
5
15
15
301
Chapter 9
Z
Z
1
5
cos x dx = cos5 x sin x +
6
6
6
(b)
cos4 x dx
Z
5 1
3
1
5
3
2
cos x sin x +
cos x dx
= cos x sin x +
6
6 4
4
1
5
5 1
1
cos3 x sin x +
cos x sin x + x + C,
= cos5 x sin x +
6
24
8 2
2
π/2
5
5
5
1
cos5 x sin x +
cos3 x sin x +
cos x sin x + x
= 5π/32
6
24
16
16 0
51. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x;
Z
Z
sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx
Z
= − sinn−1 x cos x + (n − 1)
sinn−2 x(1 − sin2 x)dx
Z
= − sinn−1 x cos x + (n − 1)
Z
sinn−2 x dx − (n − 1)
Z
n
Z
sinn x dx,
Z
sinn x dx = − sinn−1 x cos x + (n − 1)
1
n−1
sin x dx = − sinn−1 x cos x +
n
n
sinn−2 x dx,
Z
n
sinn−2 x dx
52. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x;
Z
Z
n
n−2
x tan x − (n − 2) secn−2 x tan2 x dx
sec x dx = sec
Z
= secn−2 x tan x − (n − 2)
secn−2 x(sec2 x − 1)dx
Z
= secn−2 x tan x − (n − 2)
Z
secn x dx + (n − 2)
Z
(n − 1)
secn−2 x dx,
Z
secn x dx = secn−2 x tan x + (n − 2)
secn−2 x dx,
Z
Z
1
n−2
n−2
sec x dx =
sec
secn−2 x dx
x tan x +
n−1
n−1
Z
Z
Z
Z
n
n−2
2
n−1
2
(b)
tan x dx = tan
x(sec x − 1) dx = tan
x sec x dx − tann−2 x dx
n
1
tann−1 x −
=
n−1
n
x
n−1
(c) u = x , dv = e dx, du = nx
Z
Z
tann−2 x dx
Z
x
dx, v = e ;
x e dx = x e − n
n x
xn−1 ex dx
Z
1
1
3
53. (a)
tan x dx = tan x − tan x +
dx = tan3 x − tan x + x + C
3
3
Z
Z
1
2
1
2
(b)
sec4 x dx = sec2 x tan x +
sec2 x dx = sec2 x tan x + tan x + C
3
3
3
3
1
tan x dx = tan3 x −
3
4
Z
Z
n x
2
Exercise Set 9.2
302
Z
Z
x3 ex dx = x3 ex − 3
(c)
Z
x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx
e dx = x3 ex − 3x2 ex + 6xex − 6ex + C
Z
= x e − 3x e + 6 xe −
3 x
2 x
x
x
54. (a) u = 3x,
Z
Z
Z
Z
1
1
1 2 u
2
u2 eu du =
u2 eu − 2 ueu du =
u e −
ueu − eu du
x2 e3x dx =
27
27
27
27
2
2
1
2
2
1 2 u
u e − ueu + eu + C = x2 e3x − xe3x + e3x + C
27
27
27
3
9
27
=
√
(b) u = − x,
Z
Z 1
√
xe− x dx = 2
Z
0
−1
0
Z
u e du = u e − 3
3 u
u3 eu du,
3 u
3 u
Z
u e du = u e − 3 u e − 2
2 u
2 u
u
ue du
Z
= u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C,
Z
−1
2
−1
u e du = 2(u − 3u + 6u − 6)e
3 u
3
u
2
0
= 12 − 32e−1
0
55. u = x, dv = f 00 (x)dx, du = dx, v = f 0 (x);
1
Z 1
Z 1
00
0
x f (x)dx = xf (x)
−
f 0 (x)dx
−1
−1
−1
1
= f 0 (1) + f 0 (−1) − f (1) + f (−1)
= f (1) + f (−1) − f (x)
0
0
−1
56. (a) u = f (x), dv = dx, du = f 0 (x), v = x;
b Z b
Z
Z b
f (x) dx = xf (x) −
xf 0 (x) dx = bf (b) − af (a) −
a
a
a
b
xf 0 (x) dx
a
(b) Substitute y = f (x), dy = f 0 (x) dx, x = a when y = f (a), x = b when y = f (b),
Z f (b)
Z f (b)
Z b
0
xf (x) dx =
x dy =
f −1 (y) dy
a
f (a)
f (a)
y
(c) From a = f −1 (α) and b = f −1 (β) we get
b
bf (b) − af (a) = βf −1 (β) − αf −1 (α); then
Z β
Z f (b)
Z β
f −1 (x) dx =
f −1 (y) dy =
f −1 (y) dy,
α
α
a
a = f –1(a)
b
f (x) dx
a
= βf −1 (β) − αf −1 (α) −
A2
x
f (a)
which, by part (b), yields
Z β
Z
f −1 (x) dx = bf (b) − af (a) −
α
A1
Z
f −1 (β)
f (x) dx
f −1 (α)
b = f –1(b)
303
Chapter 9
Z
β
Note from the figure that A1 =
f
−1
Z
(x) dx, A2 =
α
f −1 (β)
f (x) dx, and
f −1 (α)
A1 + A2 = βf −1 (β) − αf −1 (α), a ”picture proof”.
57. (a) Use Exercise 56(c);
Z π/6
Z sin−1 (1/2)
Z 1/2
1
1
1
1
−1
−1
−1
−1
−0·sin 0−
−
sin x dx = sin
sin x dx = sin
sin x dx
2
2
2
2
0
sin−1 (0)
0
(b) Use Exercise 56(b);
Z
Z e2
ln x dx = e2 ln e2 − e ln e −
e
f −1 (y) dy = 2e2 − e −
Z
ln e
Z
Z
2
2
ey dy = 2e2 − e −
1
ex dx
1
Z
xex dx = x(ex + C1 ) −
58. (a)
ln e2
(ex + C1 )dx = xex + C1 x − ex − C1 x + C = xex − ex + C
Z
Z
(b) u(v + C1 ) −
Z
(v + C1 )du = uv + C1 u −
v du − C1 u = uv −
v du
EXERCISE SET 9.3
Z
1
u5 du = − cos6 x + C
6
1. u = cos x, −
3. u = sin ax,
1
a
Z
u du =
Z
cos2 3x dx =
4.
1
2
2. u = sin 3x,
1
3
Z
u4 du =
1
sin5 3x + C
15
1
sin2 ax + C, a 6= 0
2a
Z
(1 + cos 6x)dx =
1
1
x+
sin 6x + C
2
12
Z
Z
1
1
1
5.
sin 5θ dθ =
(1 − cos 10θ)dθ = θ −
sin 10θ + C
2
2
20
Z
Z
3
6.
cos at dt = (1 − sin2 at) cos at dt
2
Z
=
Z
7.
Z
cos at dt −
sin2 at cos at dt =
Z
5
cos θdθ =
1
1
sin at −
sin3 at + C (a 6= 0)
a
3a
Z
(1 − sin θ) cos θdθ =
2
2
(1 − 2 sin2 θ + sin4 θ) cos θdθ
1
2
sin3 θ + sin5 θ + C
3
5
Z
Z
8.
sin3 x cos3 x dx = sin3 x(1 − sin2 x) cos x dx
= sin θ −
Z
1
1
sin4 x − sin6 x + C
4
6
Z
Z
Z
9.
sin2 2t cos3 2t dt = sin2 2t(1 − sin2 2t) cos 2t dt = (sin2 2t − sin4 2t) cos 2t dt
=
=
(sin3 x − sin5 x) cos x dx =
1
1
sin3 2t −
sin5 2t + C
6
10
Exercise Set 9.3
304
Z
Z
3
10.
(1 − cos2 2x) cos2 2x sin 2x dx
2
sin 2x cos 2x dx =
Z
1
1
cos5 2x + C
(cos2 2x − cos4 2x) sin 2x dx = − cos3 2x +
6
10
Z
Z
Z
1
1
1
1
11.
sin2 x cos2 x dx =
sin2 2x dx =
(1 − cos 4x)dx = x −
sin 4x + C
4
8
8
32
Z
Z
Z
1
1
2
4
2
12.
sin x cos x dx =
(1 − cos 2x)(1 + cos 2x) dx =
(1 − cos2 2x)(1 + cos 2x)dx
8
8
Z
Z
Z
1
1
1
1
sin2 2x dx +
sin2 2x cos 2x dx =
(1 − cos 4x)dx +
sin3 2x
=
8
8
16
48
=
1
1
1
x−
sin 4x +
sin3 2x + C
16
64
48
Z
Z
1
1
1
(sin 3x − sin x)dx = − cos 3x + cos x + C
sin x cos 2x dx =
2
6
2
Z
Z
1
1
1
sin 3θ cos 2θdθ =
(sin 5θ + sin θ)dθ = − cos 5θ − cos θ + C
2
10
2
Z
Z
1
1
sin x cos(x/2)dx =
[sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C
2
3
Z
5
u = cos x, − u1/5 du = − cos6/5 x + C
6
=
13.
14.
15.
16.
Z
Z
π/4
π/4
(1 − sin2 x) cos x dx
cos3 x dx =
17.
0
0
π/4
= sin x −
Z
1
sin3 x
3
π/2
1
4
sin2 (x/2) cos2 (x/2)dx =
18.
0
1
8
=
Z
Z
π/3
0
Z
π/2
sin2 x dx =
0
x−
1
8
Z
π/2
1
sin 2x
2
π/2
(1 − cos 2x)dx
0
= π/16
0
sin4 3x(1 − sin2 3x) cos 3x dx =
0
π
cos2 5θ dθ =
20.
−π
Z
Z
π/3
sin4 3x cos3 3x dx =
19.
0
√
√
1 √
= ( 2/2) − ( 2/2)3 = 5 2/12
3
1
2
Z
π
(1 + cos 10θ)dθ =
−π
π/6
21.
sin 2x cos 4x dx =
0
1
2
Z
0
π/6
1
2
θ+
1
sin 10θ
10
π/3
=0
1
1
sin5 3x −
sin7 3x
15
21
0
π
=π
−π
π/6
1
1
(sin 6x − sin 2x)dx = − cos 6x + cos 2x
12
4
0
= [(−1/12)(−1) + (1/4)(1/2)] − [−1/12 + 1/4] = 1/24
Z
2π
sin2 kx dx =
22.
0
1
2
Z
2π
(1 − cos 2kx)dx =
0
1
2
2π
x−
1
sin 2kx
2k
=π−
0
1
sin 4πk (k 6= 0)
4k
305
23.
Chapter 9
−2x
25. u = e
26.
1
24. − ln | cos 5x| + C
5
1
tan(3x + 1) + C
3
−2x
, du = −2e
1
dx; −
2
Z
tan u du =
1
ln | sin 3x| + C
3
28. u =
√
1
x, du = √ dx;
2 x
1
1
ln | cos u| + C = ln | cos(e−2x )| + C
2
2
27.
1
ln | sec 2x + tan 2x| + C
2
Z
2 sec u du = 2 ln | sec u + tan u| + C = 2 ln | sec
√
√
x + tan x| + C
Z
1
tan3 x + C
3
Z
Z
1
1
5
2
2
30.
tan x(1 + tan x) sec x dx = (tan5 x + tan7 x) sec2 x dx = tan6 x + tan8 x + C
6
8
Z
Z
1
1
31.
tan3 4x(1 + tan2 4x) sec2 4x dx = (tan3 4x + tan5 4x) sec2 4x dx =
tan4 4x +
tan6 4x + C
16
24
Z
1
1
32.
tan4 θ(1 + tan2 θ) sec2 θ dθ = tan5 θ + tan7 θ + C
5
7
Z
Z
1
1
33.
sec4 x(sec2 x − 1) sec x tan x dx = (sec6 x − sec4 x) sec x tan x dx = sec7 x − sec5 x + C
7
5
Z
Z
1
2
34.
(sec2 θ − 1)2 sec θ tan θdθ = (sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ = sec5 θ − sec3 θ + sec θ + C
5
3
Z
Z
Z
Z
Z
2
2
5
3
5
3
35.
(sec x − 1) sec x dx = (sec x − 2 sec x + sec x)dx = sec x dx − 2 sec x dx + sec x dx
29. u = tan x,
u2 du =
1
3
= sec3 x tan x +
4
4
Z
Z
sec x dx − 2
3
sec3 x dx + ln | sec x + tan x|
1
5 1
1
3
sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C
= sec x tan x −
4
4 2
2
3
5
1
sec3 x tan x − sec x tan x + ln | sec x + tan x| + C
4
8
8
Z
Z
36.
[sec2 (x/2) − 1] sec3 (x/2)dx = [sec5 (x/2) − sec3 (x/2)]dx
Z
Z
=2
sec5 u du − sec3 u du
=
Z
Z
3
1
3
3
3
sec u tan u +
sec u du − sec u du
=2
4
4
Z
1
1
= sec3 u tan u −
sec3 u du
2
2
(u = x/2)
=
1
1
1
sec3 u tan u − sec u tan u − ln | sec u + tan u| + C
2
4
4
=
x 1
x
x 1
x
x
1
x
sec3 tan − sec tan − ln sec + tan
+C
2
2
2 4
2
2 4
2
2
(equation (20))
(equation (20), (22))
Exercise Set 9.3
Z
306
Z
1
sec 2t(sec 2t tan 2t)dt = sec3 2t + C
6
2
37.
Z
Z
sec4 x dx =
39.
38.
sec4 x(sec x tan x)dx =
1
sec5 x + C
5
Z
(1 + tan2 x) sec2 x dx =
(sec2 x + tan2 x sec2 x)dx = tan x +
1
tan3 x + C
3
40. Using equation (20),
Z
Z
1
3
5
3
sec3 x dx
sec x dx = sec x tan x +
4
4
=
3
3
1
sec3 x tan x + sec x tan x + ln | sec x + tan x| + C
4
8
8
Z
tan4 x dx =
41. Use equation (19) to get
1
tan3 x − tan x + x + C
3
42. u = 4x, use equation (19) to get
Z
1 1
1
1
1
tan3 u du =
tan2 u + ln | cos u| + C = tan2 4x + ln | cos 4x| + C
4
4 2
8
4
43.
Z √
tan x(1 + tan2 x) sec2 x dx =
Z
sec1/2 x(sec x tan x)dx =
44.
Z
π/6
2
2
tan3/2 x + tan7/2 x + C
3
7
2
sec3/2 x + C
3
π/6
√
1
tan 2x − x
(sec 2x − 1)dx =
= 3/2 − π/6
2
0
2
45.
0
Z
π/6
1
sec θ(sec θ tan θ)dθ = sec3 θ
3
π/6
2
46.
0
√
√
= (1/3)(2/ 3)3 − 1/3 = 8 3/27 − 1/3
0
47. u = x/2,
Z
π/4
tan5 u du =
2
0
48. u = πx,
1
π
Z
π/4
√
1
tan4 u − tan2 u − 2 ln | cos u|
= 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2
2
0
π/4
sec u tan u du =
0
π/4
√
1
sec u
= ( 2 − 1)/π
π
0
Z
Z
(csc2 x − 1) csc2 x(csc x cot x)dx =
49.
Z
50.
1
cos2 3t
·
dt =
sin2 3t cos 3t
Z
Z
Z
52.
1
csc 3t cot 3t dt = − csc 3t + C
3
Z
(csc2 x − 1) cot x dx =
51.
1
1
(csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C
5
3
Z
csc x(csc x cot x)dx −
1
(cot2 x + 1) csc2 x dx = − cot3 x − cot x + C
3
1
cos x
dx = − csc2 x − ln | sin x| + C
sin x
2
307
Chapter 9
2π
Z
1 2π
cos(m + n)x cos(m − n)x
−
53. (a)
sin mx cos nx dx =
[sin(m+n)x+sin(m−n)x]dx = −
2 0
2(m + n)
2(m − n) 0
0
2π
2π
but cos(m + n)x
= 0, cos(m − n)x
= 0.
Z
2π
Z
2π
0
Z
0
2π
1
[cos(m + n)x + cos(m − n)x]dx;
2 0
0
since m 6= n, evaluate sin at integer multiples of 2π to get 0.
Z
Z 2π
1 2π
sin mx sin nx dx =
[cos(m − n)x − cos(m + n)x] dx;
(c)
2 0
0
since m 6= n, evaluate sin at integer multiples of 2π to get 0.
cos mx cos nx dx =
(b)
55. y 0 = tan x, 1 + (y 0 )2 = 1 + tan2 x = sec2 x,
Z
π/4
L=
√
Z
sec2
sec x dx = ln | sec x + tan x|
x dx =
0
π/4
π/4
0
Z
0
Z
π/4
π/4
π/4
(1 − tan x)dx = π
(2 − sec x)dx = π(2x − tan x)
2
56. V = π
0
Z
√
= ln( 2 + 1)
2
0
0
Z
π/4
π/4
(cos x − sin x)dx = π
2
2
57. V = π
0
Z
0
π
sin2 x dx =
58. V = π
0
π
2
Z
=
(1 − cos 2x)dx =
0
L
59. With 0 < α < β, D = Dβ −Dα =
2π
Z
β
α
#π/4
1
cos 2x dx = π sin 2x
2
π
1
π(π − 2)
2
π
2
= π/2
0
π
x−
1
sin 2x
2
= π 2 /2
0
L
ln | sec x + tan x|
sec x dx =
2π
#β
=
α
sec β + tan β
L
ln
2π
sec α + tan α
sec 50◦ + tan 50◦
100
100
ln(sec 25◦ + tan 25◦ ) = 7.18 cm
ln
(b) D =
= 7.34 cm
2π
2π
sec 30◦ + tan 30◦
Z
Z
61. (a)
csc x dx = sec(π/2 − x)dx = − ln | sec(π/2 − x) + tan(π/2 − x)| + C
60. (a) D =
= − ln | csc x + cot x| + C
(b) − ln | csc x + cot x| = ln
| csc x − cot x|
1
= ln | csc x − cot x|,
= ln
| csc x + cot x|
| csc2 x − cot2 x|
− ln | csc x + cot x| = − ln
= ln
cos x
sin x
1
+
= ln
sin x
sin x
1 + cos x
2 sin(x/2) cos(x/2)
= ln | tan(x/2)|
2 cos2 (x/2)
√
√
√
2[(1/ 2) sin x + (1/ 2) cos x]
√
√
= 2[sin x cos(π/4) + cos x sin(π/4)] = 2 sin(x + π/4),
Z
Z
1
1
dx
=√
csc(x + π/4)dx = − √ ln | csc(x + π/4) + cot(x + π/4)| + C
sin x + cos x
2
2
√
1
2 + cos x − sin x
+C
= − √ ln
sin x + cos x
2
62. sin x + cos x =
Exercise Set 9.4
308
p
p
b
a
a2 + b2 √
sin x + √
cos x = a2 + b2 (sin x cos θ + cos x sin θ)
a2 + b2
a2 + b2
√
√
√
where cos θ = a/ a2 + b2 and sin θ = b/ a2 + b2 so a sin x + b cos x = a2 + b2 sin(x + θ)
Z
Z
1
1
dx
=√
csc(x + θ)dx = − √
ln | csc(x + θ) + cot(x + θ)| + C
and
2
2
2
a sin x + b cos x
a +b
a + b2
√
a2 + b2 + a cos x − b sin x
1
+C
ln
= −√
a sin x + b cos x
a2 + b2
63. a sin x + b cos x =
Z
π/2
sinn x dx = −
64. (a)
0
π/2
Z
Z
1
n − 1 π/2 n−2
n − 1 π/2 n−2
sinn−1 x cos x
+
sin
x dx =
sin
x dx
n
n
n
0
0
0
(b) By repeated application of the formula in part (a)
Z π/2
Z π/2
n−3
n−1
sinn x dx =
sinn−4 x dx
n
n
−
2
0
0
Z π/2
n−3
n−5
1
n−1
···
dx, n even
n
n−2
n−4
2
0
=
Z π/2
n − 1 n − 3 n − 5
2
·
·
·
sin x dx, n odd
n
n−2
n−4
3
0
1 · 3 · 5 · · · (n − 1) π
· , n even
2 · 4 · 6···n
2
=
2 · 4 · 6 · · · (n − 1) , n odd
3 · 5 · 7···n
Z
π/2
2
sin x dx =
3
3
65. (a)
0
Z
π/2
sin5 x dx =
(c)
0
2·4
= 8/15
3·5
Z
sin4 x dx =
1·3 π
· = 3π/16
2·4 2
sin6 x dx =
1·3·5 π
· = 5π/32
2·4·6 2
π/2
(b)
0
Z
(d)
π/2
0
66. Similar to proof in Exercise 64.
EXERCISE SET 9.4
1. x = 2 sin θ, dx = 2 cos θ dθ,
Z
Z
4 cos2 θ dθ = 2 (1 + cos 2θ)dθ = 2θ + sin 2θ + C
1 p
= 2θ + 2 sin θ cos θ + C = 2 sin−1 (x/2) + x 4 − x2 + C
2
1
1
sin θ, dx = cos θ dθ,
2
2
Z
Z
1
1
1
1
2
cos θ dθ =
(1 + cos 2θ)dθ = θ + sin 2θ + C
2
4
4
8
2. x =
=
1
1
1 p
1
θ + sin θ cos θ + C = sin−1 2x + x 1 − 4x2 + C
4
4
4
2
309
Chapter 9
3. x = 3 sin θ, dx = 3 cos θ dθ,
Z
Z
9
9
9
9
9
(1 − cos 2θ)dθ = θ − sin 2θ + C = θ − sin θ cos θ + C
9 sin2 θ dθ =
2
2
4
2
2
p
1
9
= sin−1 (x/3) − x 9 − x2 + C
2
2
4. x = 4 sin θ, dx = 4 cos θ dθ,
√
Z
Z
1
1
1
16 − x2
1
2
dθ =
csc θ dθ = − cot θ + C = −
+C
2
16
16
16
16x
sin θ
5. x = 2 tan θ, dx = 2 sec2 θ dθ,
Z
Z
Z
1
1
1
1
1
1
2
dθ
=
cos
(1 + cos 2θ)dθ =
θ+
sin 2θ + C
θ
dθ
=
2
8
sec θ
8
16
16
32
=
1
1
x
1
x
θ+
sin θ cos θ + C =
tan−1 +
+C
16
16
16
2 8(4 + x2 )
√
√
6. x = 5 tan θ, dx = 5 sec2 θ dθ,
Z
Z
1
1
5 tan2 θ sec θ dθ = 5 (sec3 θ − sec θ)dθ = 5
sec θ tan θ − ln | sec θ + tan θ| + C1
2
2
√
p
1 p
5
1 p
5
5 + x2 + x
√
= x 5 + x2 − ln
+ C1 = x 5 + x2 − ln( 5 + x2 + x) + C
2
2
2
2
5
7. x = 3 sec θ, dx = 3 sec θ tan θ dθ,
Z
Z
p
x
2
3 tan θ dθ = 3 (sec2 θ − 1)dθ = 3 tan θ − 3θ + C = x2 − 9 − 3 sec−1 + C
3
8. x = 4 sec θ, dx = 4 sec θ tan θ dθ,
√
Z
Z
1
1
x2 − 16
1
1
dθ =
cos θ dθ =
sin θ + C =
+C
16
sec θ
16
16
16x
√
2 cos θ dθ,
Z
√ Z
√
1 − cos2 θ sin θ dθ
2 2 sin3 θ dθ = 2 2
9. x =
√
2 sin θ, dx =
p
1
1
3
= 2 2 − cos θ + cos θ + C = −2 2 − x2 + (2 − x2 )3/2 + C
3
3
√
√
√
5 sin θ, dx = 5 cos θ dθ,
√ Z
√
1
1
5
1
25 5 sin3 θ cos2 θ dθ = 25 5 − cos3 θ + cos5 θ + C = − (5 − x2 )3/2 + (5 − x2 )5/2 + C
3
5
3
5
10. x =
3
2
3
11. x = sec θ, dx = sec θ tan θ dθ,
2
2
9
Z
2
1
dθ =
sec θ
9
Z
2
cos θ dθ = sin θ + C =
9
12. t = tan θ, dt = sec2 θ dθ,
Z
Z
Z
tan2 θ + 1
sec3 θ
dθ =
sec θ dθ = (sec θ tan θ + csc θ)dθ
tan θ
tan θ
√
p
1 + t2 − 1
+C
= sec θ + ln | csc θ − cot θ| + C = 1 + t2 + ln
|t|
√
4x2 − 9
+C
9x
Exercise Set 9.4
310
Z
13. x = sin θ, dx = cos θ dθ,
1
dθ =
cos2 θ
Z
p
sec2 θ dθ = tan θ + C = x/ 1 − x2 + C
√
Z
Z
1
1
1
x2 + 25
sec θ
dθ =
csc θ cot θ dθ = − csc θ+C = −
+C
14. x = 5 tan θ, dx = 5 sec θ dθ,
2
25 tan θ
25
25
25x
Z
p
15. x = sec θ, dx = sec θ tan θ dθ,
sec θ dθ = ln | sec θ + tan θ| + C = ln x + x2 − 1 + C
2
16. 1 + 2x2 + x4 = (1 + x2 )2 , x = tan θ, dx = sec2 θ dθ,
Z
Z
Z
1
1
1
1
2
dθ = cos θ dθ =
(1 + cos 2θ)dθ = θ + sin 2θ + C
sec2 θ
2
2
4
=
1
1
x
1
θ + sin θ cos θ + C = tan−1 x +
+C
2
2
2
2(1 + x2 )
1
1
sec θ, dx = sec θ tan θ dθ,
3
3
Z
Z
p
1
1
1
sec θ
dθ
=
csc θ cot θ dθ = − csc θ + C = −x/ 9x2 − 1 + C
2
3
3
3
tan θ
17. x =
18. x = 5 sec θ, dx = 5 sec θ tan θ dθ,
Z
25
25
sec θ tan θ +
ln | sec θ + tan θ| + C1
25 sec3 θ dθ =
2
2
p
1 p
25
ln |x + x2 − 25| + C
= x x2 − 25 +
2
2
19. ex = sin θ, ex dx = cos θ dθ,
Z
Z
1
1
1
1
1 p
cos2 θ dθ =
(1 + cos 2θ)dθ = θ + sin 2θ + C = sin−1 (ex ) + ex 1 − e2x + C
2
2
4
2
2
Z
20. u = sin θ,
√
1
du = sin−1
2 − u2
sin θ
√
2
+C
21. x = 4 sin θ, dx = 4 cos θ dθ,
π/2
Z π/2
1
1
3
2
3
5
1024
sin θ cos θ dθ = 1024 − cos θ + cos θ
= 1024(1/3 − 1/5) = 2048/15
3
5
0
0
2
2
sin θ, dx = cos θ dθ,
3
3
π/6
Z π/6
Z π/6
1
1
1
1
1
3
dθ =
sec θ tan θ +
ln | sec θ + tan θ|
sec θ dθ =
24 0
cos3 θ
24 0
48
48
0
√
√
√
√
1
1 2 1
[(2/ 3)(1/ 3) + ln |2/ 3 + 1/ 3|] =
+ ln 3
=
48
48 3 2
22. x =
Z
π/3
23. x = sec θ, dx = sec θ tan θ dθ,
π/4
24. x =
√
1
dθ =
sec θ
√
2 sec θ, dx = 2 sec θ tan θ dθ, 2
Z
π/4
Z
π/3
π/3
cos θ dθ = sin θ
π/4
π/4
π/4
tan θ dθ = 2 tan θ − 2θ
= 2 − π/2
2
0
√
√
= ( 3 − 2)/2
0
311
Chapter 9
√
√
25. x = 3 tan θ, dx = 3 sec2 θ dθ,
Z
Z
Z
Z √3/2
1 π/3 1 − sin2 θ
1
1 π/3 cos3 θ
1 − u2
1 π/3 sec θ
dθ
=
cos
θ
dθ
=
dθ
=
du (u = sin θ)
9 π/6 tan4 θ
9 π/6 sin4 θ
9 π/6
9 1/2
u4
sin4 θ
√
√3/2
Z √3/2
1
1
1
10 3 + 18
1
−4
−2
− 3+
(u − u )du =
=
=
9 1/2
9
3u
u 1/2
243
√
√
26. x = 3 tan θ, dx = 3 sec2 θ dθ,
√ Z π/3
√ Z π/3
√ Z π/3
3
tan3 θ
3
3
3
dθ =
sin θ dθ =
1 − cos2 θ sin θ dθ
3
3 0
sec θ
3 0
3 0
√
√
π/3
√
1
1
1
1
3
3
3
− cos θ + cos θ
− +
− −1 +
= 5 3/72
=
=
3
3
3
2 24
3
0
27. u = x2 + 4, du = 2x dx,
Z
1
1
1
1
du = ln |u| + C = ln(x2 + 4) + C; or x = 2 tan θ, dx = 2 sec2 θ dθ,
2
u
2
2
Z
tan θ dθ = ln | sec θ| + C1 = ln
=
√
x2 + 4
+ C1 = ln(x2 + 4)1/2 − ln 2 + C1
2
1
ln(x2 + 4) + C with C = C1 − ln 2
2
1
1
x2 + 1
, 1 + (y 0 )2 = 1 + 2 =
,
x
x
x2
Z 2√ 2
Z 2r 2
x +1
x +1
dx; x = tan θ, dx = sec2 θ dθ,
dx
=
L=
2
x
x
1
1
29. y 0 =
Z
Z tan−1 (2)
tan2 θ + 1
sec θ dθ =
(sec θ tan θ + csc θ)dθ
tan θ
π/4
π/4
π/4
!
√
tan−1 (2)
h√
i
√
√
5 1
−
−
= 5 + ln
2 + ln | 2 − 1|
= sec θ + ln | csc θ − cot θ|
2
2
π/4
√
√
√
2+2 2
√
= 5 − 2 + ln
1+ 5
tan−1 (2)
L=
sec3 θ
dθ =
tan θ
Z
tan−1 (2)
30. y 0 = 2x, 1 + (y 0 )2 = 1 + 4x2 ,
Z 1p
1
1
1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ,
L=
2
2
0
1
L=
2
=
Z
tan−1 2
1
sec θ dθ =
2
3
0
tan−1 2
1
1
sec θ tan θ + ln | sec θ + tan θ|
2
2
0
√
√
1 √
1
1√
1
( 5)(2) + ln | 5 + 2| =
5 + ln(2 + 5)
4
4
2
4
31. y 0 = 2x, 1 + (y 0 )2 = 1 + 4x2 ,
Z 1 p
1
1
x2 1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ,
S = 2π
2
2
0
Exercise Set 9.4
−1
Z
π tan 2
(sec θ − 1) sec θ dθ =
(sec5 θ − sec3 θ)dθ
4 0
0
0
tan−1 2
√
√
1
1
π
π 1
sec3 θ tan θ − sec θ tan θ − ln | sec θ + tan θ|
[18 5 − ln(2 + 5)]
=
=
4 4
8
8
32
0
π
S=
4
Z
312
tan−1 2
Z
1
y2
32. V = π
Z
π
tan θ sec θ dθ =
4
2
tan−1 2
3
2
3
p
1 − y 2 dy; y = sin θ, dy = cos θ dθ,
0
Z
π/2
π
sin θ cos θ dθ =
4
2
V =π
0
Z
π/2
π
sin 2θ dθ =
8
Z
π/2
π
(1 − cos 4θ)dθ =
8
2
2
0
Z
0
1
θ − sin 4θ
4
π/2
π2
=
16
0
du = u + C = sinh−1 (x/3) + C
33. (a) x = 3 sinh u, dx = 3 cosh u du,
(b) x = 3 tan θ, dx = 3 sec2 θ dθ,
Z
p
sec θ dθ = ln | sec θ + tan θ| + C = ln( x2 + 9/3 + x/3) + C
but sinh−1 (x/3) = ln(x/3 +
p
√
x2 /9 + 1) = ln(x/3 + x2 + 9/3) so the results agree.
(c) x = cosh u, dx = sinh u du,
Z
Z
1
1
1
2
(cosh 2u − 1)du = sinh 2u − u + C
sinh u du =
2
4
2
1
1 p 2
1
1
= sinh u cosh u − u + C = x x − 1 − cosh−1 x + C
2
2
2
2
p
√
because cosh u = x, and sinh u = cosh2 u − 1 = x2 − 1
34. A =
4b
a
Z
4b
A=−
a
Z
35.
Z
37.
Z
38.
Z
39.
a
p
a2 − x2 dx; x = a cos θ, dx = −a sin θ dθ,
0
Z
Z
0
2
1
p
dx = sin−1
9 − (x − 1)2
1
1
dx =
2
16(x + 1/2) + 1
16
Z
(1 − cos 2θ) dθ = πab
sin θ dθ = 2ab
0
1
1
dx = tan−1
2
(x − 2) + 9
3
π/2
2
a sin θ dθ = 4ab
π/2
Z
π/2
2
0
x−2
3
x−1
3
Z
+C
36.
1
p
dx = sin−1 (x − 1) + C
1 − (x − 1)2
+C
1
1
dx = tan−1 (4x + 2) + C
2
(x + 1/2) + 1/16
4
p
1
p
dx = ln x − 3 + (x − 3)2 + 1 + C
(x − 3)2 + 1
Z
40.
x
dx, let u = x + 3,
(x + 3)2 + 1
Z
Z
3
1
u
u−3
du =
−
du = ln(u2 + 1) − 3 tan−1 u + C
u2 + 1
u2 + 1 u2 + 1
2
=
1
ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C
2
313
Chapter 9
Z p
41.
4 − (x + 1)2 dx, let x + 1 = 2 sin θ,
Z
cos2 θ dθ = 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C
4
= 2 sin−1
Z
x+1
2
p
1
+ (x + 1) 3 − 2x − x2 + C
2
ex
p
dx, let u = ex + 1/2,
(ex + 1/2)2 + 3/4
42.
x
√
2e + 1
1
p
√
du = sinh−1 (2u/ 3) + C = sinh−1
+C
3
u2 + 3/4
√
3
tan θ,
Alternate solution: let ex + 1/2 =
2
!
√
Z
2 e2x + ex + 1 2ex + 1
√
sec θ dθ = ln | sec θ + tan θ| + C = ln
+ √
+ C1
3
3
p
= ln(2 e2x + ex + 1 + 2ex + 1) + C
Z
Z
1
1
dx =
2(x + 1)2 + 5
2
43.
Z
p
1
1
−1
√
dx
=
2/5(x + 1) + C
tan
(x + 1)2 + 5/2
10
Z
44.
2x + 3
dx, let u = x + 1/2,
4(x + 1/2)2 + 4
Z
Z
1
1
1
1
u
2u + 2
du =
+
du = ln(u2 + 1) + tan−1 u + C
4u2 + 4
2
u2 + 1 u2 + 1
4
2
=
Z
2
45.
1
Z
1
1
1
ln(x2 + x + 5/4) + tan−1 (x + 1/2) + C
4
2
1
√
dx =
4x − x2
Z
1
Z
p
4x − x2 dx =
46.
0
2
1
1
−1
p
dx = sin
4 − (x − 2)2
p
x−2
2
2
= π/6
1
4 − (x − 2)2 dx, let x − 2 = 2 sin θ,
0
Z
−π/6
−π/6
2
cos θ dθ = 2θ + sin 2θ
4
−π/2
−π/2
√
2π
3
−
=
3
2
48. u = x sin x, du = (x cos x + sin x) dx;
Z p
p
1 p
1
1
1
1 + u2 du = u 1 + u2 + sinh−1 u + C = x sin x 1 + x2 sin2 x + sinh−1 (x sin x) + C
2
2
2
2
49. u = sin2 x, du = 2 sin x cos x dx;
Z
i
i
1h p
1h 2 p
1 p
u 1 − u2 + sin−1 u + C =
sin x 1 − sin4 x + sin−1 (sin2 x) + C
1 − u2 du =
2
4
4
Exercise Set 9.5
314
50. u = 3x = ex ln 3 , du = (ln 3)3x dx;
√
√
p
3
Z 3p
p
1
6 2 − ln(3 + 2 2)
1
2
2
2
u u − 1 − ln u + u − 1
u − 1 du =
=
ln 3 1
2 ln 3
2 ln 3
1
EXERCISE SET 9.5
1.
B
A
+
(x − 2) (x + 5)
2.
5
A
B
C
= +
+
x(x − 3)(x + 3)
x
x−3 x+3
3.
A
B
C
2x − 3
= + 2+
x2 (x − 1)
x
x
x−1
4.
A
B
C
+
+
x + 2 (x + 2)2
(x + 2)3
5.
B
A
C
Dx + E
+ 2+ 3+ 2
x
x
x
x +1
6.
A
Bx + C
+ 2
x−1
x +5
7.
Cx + D
Ax + B
+ 2
x2 + 5
(x + 5)2
8.
Bx + C
Dx + E
A
+ 2
+ 2
x−2
x +1
(x + 1)2
9.
10.
11.
12.
13.
A
B
1
1
1
=
+
; A = − , B = so
(x + 4)(x − 1)
x+4 x−1
5
5
Z
Z
1
1
1
1
x−1
1
1
1
dx +
dx = − ln |x + 4| + ln |x − 1| + C = ln
+C
−
5
x+4
5
x−1
5
5
5
x+4
A
B
1
1
1
=
+
; A = , B = − so
(x + 1)(x + 7)
x+1 x+7
6
6
Z
Z
1
1
1
1
x+1
1
1
1
dx −
dx = ln |x + 1| − ln |x + 7| + C = ln
+C
6
x+1
6
x+7
6
6
6
x+7
A
B
11x + 17
=
+
; A = 5, B = 3 so
(2x − 1)(x + 4)
2x − 1 x + 4
Z
Z
5
1
1
dx + 3
dx = ln |2x − 1| + 3 ln |x + 4| + C
5
2x − 1
x+4
2
A
B
5x − 5
=
+
; A = 1, B = 2 so
(x − 3)(3x + 1)
x − 3 3x + 1
Z
Z
2
1
1
dx + 2
dx = ln |x − 3| + ln |3x + 1| + C
x−3
3x + 1
3
A
B
C
2x2 − 9x − 9
= +
+
; A = 1, B = 2, C = −1 so
x(x + 3)(x − 3)
x
x+3 x−3
Z
Z
Z
x(x + 3)2
1
1
1
dx + 2
dx −
dx = ln |x| + 2 ln |x + 3| − ln |x − 3| + C = ln
+C
x
x+3
x−3
x−3
Note that the symbol C has been recycled; to save space this recycling is usually not mentioned.
315
14.
Chapter 9
A
B
C
1
1
1
= +
+
; A = −1, B = , C = so
x(x + 1)(x − 1)
x
x+1 x−1
2
2
Z
Z
Z
1
1
1
1
1
1
1
−
dx +
dx +
dx = − ln |x| + ln |x + 1| + ln |x − 1| + C
x
2
x+1
2
x−1
2
2
=
15.
6
x2 + 2
=x−2+
,
x+2
x+2
16.
3
x2 − 4
=x+1−
,
x−1
x−1
17.
18.
19.
Z
6
x−2+
x+2
Z
x+1−
3
x−1
(x + 1)(x − 1)
1 |x2 − 1|
1
ln
+C
+ C = ln
2
2
x
2
x2
dx =
1 2
x − 2x + 6 ln |x + 2| + C
2
dx =
1 2
x + x − 3 ln |x − 1| + C
2
12x − 22 12x − 22
B
A
3x2 − 10
=3+ 2
,
+
=
; A = 12, B = 2 so
− 4x + 4
x − 4x + 4 (x − 2)2
x − 2 (x − 2)2
Z
Z
Z
1
1
dx + 2
3dx + 12
dx = 3x + 12 ln |x − 2| − 2/(x − 2) + C
x−2
(x − 2)2
x2
x2
3x − 2
3x − 2
A
B
=1+ 2
,
=
+
; A = −1, B = 4 so
x2 − 3x + 2
x − 3x + 2 (x − 1)(x − 2)
x−1 x−2
Z
Z
Z
1
1
dx + 4
dx = x − ln |x − 1| + 4 ln |x − 2| + C
dx −
x−1
x−2
2x2 + x + 1
x5 + 2x2 + 1
2
=
x
,
+
1
+
x3 − x
x3 − x
A
B
C
2x2 + x + 1
= +
+
; A = −1, B = 1, C = 2 so
x(x + 1)(x − 1)
x
x+1 x−1
Z
Z
Z
Z
1
1
1
dx +
dx + 2
dx
(x2 + 1)dx −
x
x+1
x−1
=
20.
1
(x + 1)(x − 1)2
1 3
x + x − ln |x| + ln |x + 1| + 2 ln |x − 1| + C = x3 + x + ln
+C
3
3
x
28x − 1
2x5 − x3 − 1
= 2x2 + 7 + 3
,
3
x − 4x
x − 4x
A
B
C
1
57
55
28x − 1
= +
+
;A= ,B=− ,C=
so
x(x + 2)(x − 2)
x
x+2 x−2
4
8
8
Z
Z
Z
Z
57
55
1
1
1
1
dx −
dx +
dx
(2x2 + 7)dx +
4
x
8
x+2
8
x−2
=
21.
2 3
57
55
1
x + 7x + ln |x| −
ln |x + 2| +
ln |x − 2| + C
3
4
8
8
2x2 + 3
B
C
A
+
= +
; A = 3, B = −1, C = 5 so
x(x − 1)2
x
x − 1 (x − 1)2
Z
Z
Z
1
1
1
dx −
dx + 5
dx = 3 ln |x| − ln |x − 1| − 5/(x − 1) + C
3
x
x−1
(x − 1)2
Exercise Set 9.5
22.
23.
A
B
3x2 − x + 1
C
= + 2+
; A = 0, B = −1, C = 3 so
2
x (x − 1)
x
x
x−1
Z
Z
1
1
−
dx = 1/x + 3 ln |x − 1| + C
dx
+
3
2
x
x−1
B
C
A
x2 + x − 16
+
+
=
; A = −1, B = 2, C = −1 so
2
(x + 1)(x − 3)
x + 1 x − 3 (x − 3)2
Z
Z
Z
1
1
1
dx + 2
dx −
−
dx
x+1
x−3
(x − 3)2
= − ln |x + 1| + 2 ln |x − 3| +
24.
25.
26.
27.
28.
B
A
C
x2
+
=
+
; A = 1, B = −4, C = 4 so
3
2
(x + 2)
x + 2 (x + 2)
(x + 2)3
Z
Z
Z
2
4
1
1
1
dx − 4
−
dx
+
4
dx = ln |x + 2| +
+C
2
3
x+2
(x + 2)
(x + 2)
x + 2 (x + 2)2
2x2 + 3x + 3
B
A
C
+
=
+
; A = 2, B = −1, C = 2 so
3
2
(x + 1)
x + 1 (x + 1)
(x + 1)3
Z
Z
Z
1
1
1
1
1
dx −
−
dx
+
2
dx = 2 ln |x + 1| +
+C
2
2
3
x+1
(x + 1)
(x + 1)
x + 1 (x + 1)2
2x2 − 1
A
Bx + C
=
+ 2
; A = −14/17, B = 12/17, C = 3/17 so
2
(4x − 1)(x + 1)
4x − 1
x +1
Z
7
6
3
2x2 − 1
dx = − ln |4x − 1| +
ln(x2 + 1) +
tan−1 x + C
(4x − 1)(x2 + 1)
34
17
17
A Bx + C
1
= + 2
; A = 1, B = −1, C = 0 so
+ 1)
x
x +1
x(x2
x3
30.
1
(x − 3)2
1
+ C = ln
+
+C
x−3
|x + 1|
x−3
2x2 − 2x − 1
A
B
C
= + 2+
; A = 3, B = 1, C = −1 so
x2 (x − 1)
x
x
x−1
Z
Z
Z
1
1
1
1
dx +
dx = 3 ln |x| − − ln |x − 1| + C
3
dx
−
x
x2
x−1
x
Z
29.
316
1
x2
1
1
dx = ln |x| − ln(x2 + 1) + C = ln 2
+C
+x
2
2 x +1
Ax + B
Cx + D
x3 + 3x2 + x + 9
= 2
+ 2
; A = 0, B = 3, C = 1, D = 0 so
2
2
(x + 1)(x + 3)
x +1
x +3
Z 3
1
x + 3x2 + x + 9
dx = 3 tan−1 x + ln(x2 + 3) + C
(x2 + 1)(x2 + 3)
2
x3 + x2 + x + 2
Ax + B
Cx + D
= 2
+ 2
; A = D = 0, B = C = 1 so
(x2 + 1)(x2 + 2)
x +1
x +2
Z 3
1
x + x2 + x + 2
dx = tan−1 x + ln(x2 + 2) + C
(x2 + 1)(x2 + 2)
2
317
31.
32.
Chapter 9
x
x3 − 3x2 + 2x − 3
=x−3+ 2
,
2
x +1
x +1
Z 3
1
1
x − 3x2 + 2x − 3
dx = x2 − 3x + ln(x2 + 1) + C
2
x +1
2
2
x
x4 + 6x3 + 10x2 + x
= x2 + 2
,
2
x + 6x + 10
x + 6x + 10
Z
Z
Z
x
x
u−3
dx =
dx =
du,
x2 + 6x + 10
(x + 3)2 + 1
u2 + 1
=
Z
u=x+3
1
ln(u2 + 1) − 3 tan−1 u + C1
2
1
1
x4 + 6x3 + 10x2 + x
dx = x3 + ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C
x2 + 6x + 10
3
2
Z
1
A
B
1
dx, and
=
+
; A = −1/6,
33. Let x = sin θ to get
x2 + 4x − 5
(x + 5)(x − 1)
x+5 x−1
Z
Z
1
1
x−1
1
1
1
1
1 − sin θ
dx +
dx = ln
+ C = ln
+ C.
B = 1/6 so we get −
6
x+5
6
x−1
6
x+5
6
5 + sin θ
so
Z
et
dt =
e2t − 4
t
34. Let x = e ; then
Z
1
dx,
x2 − 4
A
B
1
=
+
; A = −1/4, B = 1/4 so
(x + 2)(x − 2)
x+2 x−2
Z
Z
1
1
x−2
1
et − 2
1
1
1
dx +
dx = ln
+ C = ln t
+ C.
−
4
x+2
4
x−2
4
x+2
4
e +2
Z
2
35. V = π
0
18x2 − 81
x4
x4
=1+ 4
,
dx, 4
2
2
2
(9 − x )
x − 18x + 81
x − 18x2 + 81
B
D
18x2 − 81
A
C
18x − 81
+
+
=
=
+
;
(9 − x2 )2
(x + 3)2 (x − 3)2
x + 3 (x + 3)2
x − 3 (x − 3)2
2
9
9
9
9
A = − , B = , C = , D = so
4
4
4
4
2
9/4
9
9/4
9
19 9
+ ln |x − 3| −
− ln 5
=π
V = π x − ln |x + 3| −
4
x+3 4
x−3 0
5
4
Z
ln 5
36. Let u = ex to get
− ln 5
dx
=
1 + ex
Z
ln 5
− ln 5
A
B
1
= +
; A = 1, B = −1;
u(1 + u)
u
1+u
37.
(x2
Z
ex dx
=
x
e (1 + ex )
Z
5
1/5
Z
5
1/5
du
,
u(1 + u)
5
du
= (ln u − ln(1 + u))
u(1 + u)
= ln 5
1/5
Cx + D
Ax + B
x2 + 1
+ 2
= 2
; A = 0, B = 1, C = D = −2 so
2
+ 2x + 3)
x + 2x + 3 (x + 2x + 3)2
x2 + 1
dx =
2
(x + 2x + 3)2
Z
1
dx −
(x + 1)2 + 2
Z
2x + 2
dx
(x2 + 2x + 3)2
1
x+1
= √ tan−1 √ + 1/(x2 + 2x + 3) + C
2
2
Exercise Set 9.5
38.
318
Cx + D
x5 + x4 + 4x3 + 4x2 + 4x + 4
Ax + B
Ex + F
+ 2
= 2
+ 2
;
2
3
2
(x + 2)
x +2
(x + 2)
(x + 2)3
A = B = 1, C = D = E = F = 0 so
Z
√
1
1
x+1
2
−1
√
dx
=
ln(x
+
2)
+
(x/
2) + C
tan
x2 + 2
2
2
39. x4 − 3x3 − 7x2 + 27x − 18 = (x − 1)(x − 2)(x − 3)(x + 3),
A
B
C
D
1
=
+
+
+
;
(x − 1)(x − 2)(x − 3)(x + 3)
x−1 x−2 x−3 x+3
A = 1/8, B = −1/5, C = 1/12, D = −1/120 so
Z
1
1
1
1
dx
= ln |x − 1| − ln |x − 2| +
ln |x − 3| −
ln |x + 3| + C
x4 − 3x3 − 7x2 + 27x − 18
8
5
12
120
40. 16x3 − 4x2 + 4x − 1 = (4x − 1)(4x2 + 1),
A
Bx + C
1
=
+ 2
; A = 4/5, B = −4/5, C = −1/5 so
(4x − 1)(4x2 + 1)
4x − 1
4x + 1
Z
1
1
1
dx
= ln |4x − 1| −
ln(4x2 + 1) −
tan−1 (2x) + C
3
2
16x − 4x + 4x − 1
5
10
10
41. (a) x4 + 1 = (x4 + 2x2 + 1) − 2x2 = (x2 + 1)2 − 2x2
√
√
= [(x2 + 1) + 2x][(x2 + 1) − 2x]
√
√
√
√
= (x2 + 2x + 1)(x2 − 2x + 1); a = 2, b = − 2
(b)
Ax + B
Cx + D
x
√
√
√
=
+
;
2x + 1)(x2 − 2x + 1)
x2 + 2x + 1 x2 − 2x + 1
√
√
2
2
, C = 0, D =
so
A = 0, B = −
4
4
√
√ Z 1
Z
Z 1
x
2 1
1
2
1
√
√
dx = −
dx +
dx
4
2
2
4 0 x + 2x + 1
4 0 x − 2x + 1
0 x +1
√ Z 1
√ Z 1
2
1
2
1
√
√
dx +
dx
=−
2
4 0 (x + 2/2) + 1/2
4 0 (x − 2/2)2 + 1/2
√ Z 1−√2/2
√ Z 1+√2/2
2
1
2
1
du +
du
=−
√
√
2
2
4
u + 1/2
4 − 2/2 u + 1/2
2/2
1+√2/2
1−√2/2
√
√
1
1
+ tan−1 2u √
= − tan−1 2u √
2
2
2/2
− 2/2
1
√
√
1
1
1 π
π
+ tan−1 ( 2 − 1) −
−
= − tan−1 ( 2 + 1) +
2
2 4
2
2
4
√
√
π 1
= − [tan−1 ( 2 + 1) − tan−1 ( 2 − 1)]
4
2
√
√
π 1
= − [tan−1 (1 + 2) + tan−1 (1 − 2)]
4
2
"
√ #
√
π 1
(1 + 2) + (1 − 2)
−1
√
√
= − tan
(Exercise 46, Section 4.5)
4
2
1 − (1 + 2)(1 − 2)
π 1 π π
π 1
=
= − tan−1 1 = −
4
2
4
2 4
8
(x2 +
√
319
42.
Chapter 9
B
1
1
1
A
+
;A =
,B =
so
=
2
−x
a−x a+x
2a
2a
Z
1
1
1
a+x
1
1
+
dx =
(− ln |a − x| + ln |a + x| ) + C =
ln
+C
2a
a−x a+x
2a
2a
a−x
a2
EXERCISE SET 9.6
1
2
+ ln |2 − 3x| + C
9 2 − 3x
1. Formula (60):
i
3h
4x + ln |−1 + 4x| + C
16
2. Formula (62):
3. Formula (65):
x
1
ln
+C
5
5 + 2x
4. Formula (66): −
5. Formula (102):
1
(x + 1)(−3 + 2x)3/2 + C
5
6. Formula (105):
√
1
4 − 3x − 2
+C
7. Formula (108): ln √
2
4 − 3x + 2
1
1 − 5x
− 5 ln
+C
x
x
√
2
(−x − 4) 2 − x + C
3
√
−1
8. Formula (108): tan
√
1
x+ 5
√ +C
9. Formula (69): √ ln
2 5
x− 5
10. Formula (70):
3x − 4
+C
2
x−3
1
ln
+C
6
x+3
p
3
xp 2
x − 3 − ln x + x2 − 3 + C
2
2
√
p
x2 + 5
+ ln(x + x2 + 5) + C
12. Formula (93): −
x
p
xp 2
x + 4 − 2 ln(x + x2 + 4) + C
13. Formula (95):
2
√
xp
9
x2 − 2
x
+C
15. Formula (74):
9 − x2 + sin−1 + C
14. Formula (90): −
2x
2
2
3
11. Formula (73):
√
16. Formula (80): −
17. Formula (79):
√
4 − x2
x
− sin−1 + C
x
2
3 − x2 −
√
18. Formula (117): −
20. Formula (40): −
√
√
3 ln
3+
√
9 − x2
x
6x − x2
+C
3x
+C
19. Formula (38): −
1
1
sin(5x) + sin x + C
10
2
1
1
cos(7x) + cos(3x) + C
14
6
√ 1
22. Formula (50): 4 x
ln x − 1 + C
2
21. Formula (50):
x4
[4 ln x − 1] + C
16
23. Formula (42):
e−2x
(−2 sin(3x) − 3 cos(3x)) + C
13
Exercise Set 9.6
24. Formula (43):
320
ex
(cos(2x) + 2 sin(2x)) + C
5
4
1
u du
2x
=
+ ln 4 − 3e
+C
(4 − 3u)2
18 4 − 3e2x
Z
1
25. u = e , du = 2e dx, Formula (62):
2
2x
2x
Z
1
sin 2x
du
= ln
+C
2u(3 − u)
6
3 − sin 2x
26. u = sin 2x, du = 2 cos 2xdx, Formula (116):
√
1
du
−1 3 x
= tan
+C
u2 + 4
3
2
Z
√
3
2
27. u = 3 x, du = √ dx, Formula (68):
3
2 x
1
4
28. u = sin 4x, du = 4 cos 4xdx, Formula (68):
29. u = 3x, du = 3dx, Formula (76):
Z
1
3
√
Z
1
sin 4x
du
tan−1
+C
=
2
9+u
12
3
p
1
du
= ln 3x + 9x2 − 4 + C
2
3
u −4
√
√
2x2 , du = 2 2xdx, Formula (72):
Z p
√
p
x2 p 4
3
1
√
u2 + 3 du =
2x + 3 + √ ln
2x2 + 2x4 + 3 + C
4
2 2
4 2
30. u =
31. u = 3x2 , du = 6xdx, u2 du = 54x5 dx, Formula (81):
Z
x2 p
5
3x2
1
u2 du
√
sin−1 √ + C
=−
5 − 9x4 +
54
36
108
5
5 − u2
Z
1 p
du
√
=−
3 − 4x2 + C
3x
u2 3 − u 2
32. u = 2x, du = 2dx, Formula (83): 2
Z
sin2 u du =
33. u = ln x, du = dx/x, Formula (26):
34. u = e−2x , du = −2e−2x , Formula (27): −
35. u = −2x, du = −2dx, Formula (51):
1
4
36. u = 5x − 1, du = 5dx, Formula (50):
1
5
Z
1
2
ueu du =
1
(−2x − 1)e−2x + C
4
ln u du =
1
1
(u ln u − u) + C = (5x − 1)[ln(5x − 1) − 1] + C
5
5
Z
Z
39. u = 4x2 , du = 8xdx, Formula (70):
Z
√
1
8
Z
1
40. u = 2e , du = 2e dx, Formula (69):
2
x
x
1
1
cos2 u du = − e−2x − sin 2e−2x + C
4
8
Z
37. u = cos 3x, du = −3 sin 3x, Formula (67): −
1
38. u = lnx, du = dx, Formula (105):
x
1
1
ln x + sin(2 ln x) + C
2
4
cos 3x
1
1
du
+
ln
+C
=
−
u(u + 1)2
3 1 + cos 3x
1 + cos 3x
√
1
u du
=
(2 ln x + 1) 4 ln x − 1 + C
12
4u − 1
1
4x2 − 1
du
=
ln
+C
−1
16
4x2 + 1
u2
Z
√
1
2ex + 3
du
√ +C
= √ ln
3 − u2
4 3
2ex − 3
321
Chapter 9
41. u = 2ex , du = 2ex dx, Formula (74):
Z
√
√
1 p
3
1 p
3
1 p
3 − u2 du = u 3 − u2 + sin−1 (u/ 3) + C = ex 3 − 4e2x + sin−1 (2ex / 3) + C
2
4
4
2
4
42. u = 3x, du = 3dx, Formula (80):
√
√
Z √
4 − u2 du
4 − u2
4 − 9x2
−1
−
3
sin
− 3 sin−1 (3x/2) + C
=
−3
(u/2)
+
C
=
−
3
2
u
u
x
43. u = 3x, du = 3dx, Formula (112):
r
Z r
5
1
25
5
5
u−5
1
2
u − u du =
u−
u − u2 +
sin−1
+C
3
3
6
6
3
216
5
25
18x − 5 p
18x − 5
sin−1
+C
5x − 9x2 +
=
36
216
5
√
√
5 dx, Formula (117):
q √
√
Z
(u/ 5) − u2
x − 5x2
du
q √
√
+C
=−
+ C = −2
x
u/(2 5)
u (u/ 5) − u2
44. u =
5x, du =
45. u = 3x, du = 3dx, Formula (44):
Z
1
1
1
u sin u du = (sin u − u cos u) + C = (sin 3x − 3x cos 3x) + C
9
9
9
Z
√
√
√
√
2
46. u = x, u = x, 2udu = dx, Formula (45): 2 u cos u du = 2 cos x + 2 x sin x + C
√
Z
47. u = − x, u = x, 2udu = dx, Formula (51): 2
2
√
√
ueu du = −2( x + 1)e− x + C
48. u = 2 − 3x2 , du = −6xdx, Formula (50):
Z
1
1
1
1
−
ln u du = − (u ln u − u) + C = − ((2 − 3x2 ) ln(2 − 3x2 ) + (2 − 3x2 ) + C
6
6
6
6
49. x2 + 4x − 5 = (x + 2)2 − 9; u = x + 2, du = dx, Formula (70):
Z
1
u−3
1
x−1
du
= ln
+ C = ln
+C
2
u −9
6
u+3
6
x+5
50. x2 + 2x − 3 = (x + 1)2 − 4, u = x + 1, du = dx, Formula (77):
Z p
1 p
4 − u2 du = u 4 − u2 + 2 sin−1 (u/2) + C
2
p
1
= (x + 1) 3 − 2x − x2 + 2 sin−1 ((x + 1)/2) + C
2
51. x2 − 4x − 5 = (x − 2)2 − 9, u = x − 2, du = dx, Formula (77):
Z
Z
Z
p
u
u du
du
u+2
√
√
du =
+2 √
= − 9 − u2 + 2 sin−1 + C
3
9 − u2
9 − u2
9 − u2
p
x−2
+C
= − 5 + 4x − x2 + 2 sin−1
3
Exercise Set 9.6
322
52. x2 + 6x + 13 = (x + 3)2 + 4, u = x + 3, du = dx, Formula (71):
Z
1
3
1
3
(u − 3) du
= ln(u2 + 4) − tan−1 (u/2) + C = ln(x2 + 6x + 13) − tan−1 ((x + 3)/2) + C
u2 + 4
2
2
2
2
√
53. u = x − 2, x = u2 + 2, dx = 2u du;
Z
Z
2
4
2
4
2u2 (u2 + 2)du = 2 (u4 + 2u2 )du = u5 + u3 + C = (x − 2)5/2 + (x − 2)3/2 + C
5
3
5
3
√
54. u = x + 1, x = u2 − 1, dx = 2u du;
Z
√
2
2
2 (u2 − 1)du = u3 − 2u + C = (x + 1)3/2 − 2 x + 1 + C
3
3
√
55. u = x3 + 1, x3 = u2 − 1, 3x2 dx = 2u du;
Z
Z
2
2 5 2 3
2 3
2
2
u2 (u2 − 1)du =
(u4 − u2 )du =
u − u +C =
(x + 1)5/2 − (x3 + 1)3/2 + C
3
3
15
9
15
9
√
56. u = x3 − 1, x3 = u2 + 1, 3x2 dx = 2u du;
Z
p
2
2
2
1
du = tan−1 u + C = tan−1 x3 − 1 + C
2
3
u +1
3
3
57. u = x1/6 , x = u6 , dx = 6u5 du;
Z
Z
Z
1
u3
6u5
2
du
=
6
u
du
du
=
6
−
u
+
1
−
u3 + u2
u+1
u+1
= 2x1/2 − 3x1/3 + 6x1/6 − 6 ln(x1/6 + 1) + C
Z
58. u = x
1/5
59. u = x
1/4
5
4
4
3
5u4
du = 5
u5 − u3
, x = u , dx = 5u du;
Z
, x = u , dx = 4u du; 4
Z
60. u = x1/3 , x = u3 , dx = 3u2 du; 3
Z
1
du = 4
u(1 − u)
u4
du = 3
3
u +1
5
u
du = ln |x2/5 − 1| + C
u2 − 1
2
Z
1
x1/4
1
+C
+
du = 4 ln
u 1−u
|1 − x1/4 |
Z
(u −
u3
u
)du,
+1
u
−1/3 (1/3)u + 1/3
u
=
=
+ 2
so
2
+1
(u + 1)(u − u + 1)
u+1
u −u+1
Z
Z
1
u+1
u
du =
3u +
− 2
du
3
u− 3
u +1
u+1 u −u+1
u3
√
1
3 2
2u − 1
u + ln |u + 1| − ln(u2 − u + 1) − 3 tan−1 √
+C
2
2
3
√
3
1
2x1/3 − 1
√
= x2/3 + ln |x1/3 + 1| − ln(x2/3 − x1/3 + 1) − 3 tan−1
+C
2
2
3
=
61. u = x1/6 , x = u6 , dx = 6u5 du;
Z
Z
1
u3
2
du = 6
u +u+1+
du = 2x1/2 + 3x1/3 + 6x1/6 + 6 ln |x1/6 − 1| + C
6
u−1
u−1
323
Chapter 9
√
62. u = x, x = u2 , dx = 2u du;
Z 2
Z
√
√
2
u +u
−2
du = −2
u+2+
du = −x − 4 x − 4 ln | x − 1| + C
u−1
u−1
√
63. u = 1 + x2 , x2 = u2 − 1, 2x dx = 2u du, x dx = u du;
Z
1
(u2 − 1)du = (1 + x2 )3/2 − (1 + x2 )1/2 + C
3
64. u = (x + 3)1/5 , x = u5 − 3, dx = 5u4 du;
Z
5
15
5 (u8 − 3u3 )du = (x + 3)9/5 − (x + 3)4/5 + C
9
4
65. u =
66. u =
√
√
Z
2
x, x = u , dx = 2u du, Formula (44): 2
Z
x, x = u2 , dx = 2u du, Formula (51): 2
Z
1
67.
2u
1−u
+
1 + u2
1 + u2
2
1+
Z
68.
1
2u
2+
1 + u2
2
du =
1 + u2
2
du =
1 + u2
Z
Z
=
Z
69. u = tan(θ/2),
dθ
=
1 − cos θ
Z
√
√
√
u sin u du = 2 sin x − 2 x cos x + C
√
√ √
ueu du = 2 xe x − 2e x + C
1
du = ln | tan(x/2) + 1| + C
u+1
1
du
u2 + u + 1
2
1
du = √ tan−1
2
(u + 1/2) + 3/4
3
Z
2 tan(x/2) + 1
√
3
Z
71. u = tan(x/2), 2
(3u2
Z
+C
1
1
du = − + C = − cot(θ/2) + C
u2
u
70. u = tan(x/2),
Z
Z
Z
2
2
2
1
1
du
=
du
=
dz
3u2 + 8u − 3
3
(u + 4/3)2 − 25/9
3
z 2 − 25/9
=
(z = u + 4/3)
z − 5/3
1
tan(x/2) − 1/3
1
ln
+ C = ln
+C
5
z + 5/3
5
tan(x/2) + 3
1 − u2
du;
(3u2 + 1)(u2 + 1)
(0)u + 2 (0)u − 1
2
1
1 − u2
=
+ 2
= 2
−
so
+ 1)(u2 + 1)
3u2 + 1
u +1
3u + 1 u2 + 1
√
4
cos x
dx = √ tan−1 [ 3 tan(x/2)] − x + C
2 − cos x
3
1
72. u = tan(x/2),
2
Z
1
1 − u2
du =
u
2
Z
(1/u − u)du =
1
1
ln | tan(x/2)| − tan2 (x/2) + C
2
4
Exercise Set 9.6
Z
x
73.
2
324
x
1
t
1
dt = ln
(Formula (65), a = 4, b = −1)
t(4 − t)
4 4−t 2
x
1
x
1
x
x
1
ln
− ln 1 = ln
, ln
= 0.5, ln
= 2,
=
4
4−x
4 4−x 4 4−x
4−x
x
= e2 , x = 4e2 − e2 x, x(1 + e2 ) = 4e2 , x = 4e2 /(1 + e2 ) ≈ 3.523188312
4−x
x
Z x
√
1
√
dt = 2 tan−1 2t − 1
74.
(Formula (108), a = −1, b = 2)
1 t 2t − 1
1
√
√
= 2 tan−1 2x − 1 − tan−1 1 = 2 tan−1 2x − 1 − π/4 ,
√
√
√
2(tan−1 2x − 1 − π/4) = 1, tan−1 2x − 1 = 1/2 + π/4, 2x − 1 = tan(1/2 + π/4),
x = [1 + tan2 (1/2 + π/4)]/2 ≈ 6.307993516
Z
p
4
75. A =
25 −
x2
dx =
0
25
x
1 p
x 25 − x2 +
sin−1
2
2
5
=6+
Z
p
2
76. A =
4
(Formula (74), a = 5)
0
25
4
sin−1 ≈ 17.59119023
2
5
9x2 − 4 dx; u = 3x,
2/3
6
p
p
1 1 p 2
2
2
u u − 4 − 2 ln u + u − 4
u − 4 du =
(Formula (73), a2 = 4)
3 2
2
2
√
√
√
2
1 √
3 32 − 2 ln(6 + 32) + 2 ln 2 = 4 2 − ln(3 + 2 2) ≈ 4.481689467
=
3
3
Z
1
A=
3
Z
6
1
1
dx; u = 4x,
25 − 16x2
77. A =
0
A=
1
4
Z
Z
4
u+5
1
1
ln
du =
25 − u2
40
u−5
0
4
78. A =
√
x ln x dx =
1
=
Z
π/2
79. V = 2π
4 3/2
x
9
8
4
4
(Formula (50), n = 1/2)
1
4
(12 ln 4 − 7) ≈ 4.282458815
9
π/2
x cos x dx = 2π(cos x + x sin x)
= π(π − 2) ≈ 3.586419094
√
4π
(3x + 8)(x − 4)3/2
x x − 4 dx =
15
=
Z
0
1
ln 9 ≈ 0.054930614 (Formula (69), a = 5)
40
(Formula (45))
0
80. V = 2π
81. V = 2π
=
3
ln x − 1
2
0
Z
4
3
8
(Formula (102), a = −4, b = 1)
4
1024
π ≈ 214.4660585
15
xe−x dx; u = −x,
0
Z
V = 2π
0
−3
−3
ueu du = 2πeu (u − 1)
= 2π(1 − 4e−3 ) ≈ 5.031899801
0
(Formula (51))
325
Chapter 9
Z
5
π
x ln x dx = x2 (2 ln x − 1)
2
82. V = 2π
1
= π(25 ln 5 − 12) ≈ 88.70584621
Z
5
1
(Formula (50), n = 1)
p
2
1 + 16x2 dx; u = 4x,
83. L =
0
1
L=
4
Z
8
0
Z
3
8
p
p
1 up
1
2
2
2
1 + u du =
1 + u + ln u + 1 + u
4 2
2
0
√
√
1
= 65 + ln(8 + 65) ≈ 8.409316783
8
Z
p
3
1 + 9/x2 dx =
84. L =
1
1
√
x2 + 9
dx =
x
p
3+
x2 + 9 − 3 ln
√
√
√
√
3 + 10
√ ≈ 3.891581644
= 3 2 − 10 + 3 ln
1+ 2
Z
π
85. S = 2π
(Formula (72), a2 = 1)
x2 + 9
x
!#3
1
(Formula (89), a = 3)
p
(sin x) 1 + cos2 x dx; u = cos x,
0
Z
−1p
S = −2π
1+
u2
1
Z
1
1p
1 + 1/x4 dx = 2π
x
16
√
4
86. S = 2π
Z
p
1
p
1
u
2
2
du = 4π
1 + du = 4π
1 + u + ln u + 1 + u
a2 = 1)
2
2
0
0
√
√
= 2π[ 2 + ln(1 + 2)] ≈ 14.42359945 (Formula (72)
Z 1p
S=π
1
Z
4
u2
√
1
x4 + 1
dx; u = x2 ,
x3
!#
√
16
p
u2 + 1
u2 + 1
+ ln u + u2 + 1
du = π −
u2
u
1
!
√
√
√
16 + 257
257
√
+ ln
=π
2−
≈ 9.417237485 (Formula (93), a2 = 1)
16
1+ 2
Z
t
20 cos6 u sin3 u du
87. (a) s(t) = 2 +
=−
s(t)
(b)
4
0
40
166
20
sin2 t cos7 t −
cos7 t +
9
63
63
3
2
1
t
3
Z
t
a(u) du = −
88. (a) v(t) =
0
Z
6
9
12
15
1
1 −t
1
1
3
1
e cos 2t + e−t sin 2t + e−t cos 6t − e−t sin 6t +
−
10
5
74
37
10 74
t
v(u) du
s(t) = 10 +
0
=−
343866
2
35 −t
6 −t
16
3 −t
e cos 2t − e−t sin 2t +
e cos 6t +
e sin 6t +
t+
50
25
2738
1369
185
34225
Exercise Set 9.7
(b)
12
10
8
6
4
2
326
s(t)
t
2
6
14
18
Z
1 + tan(x/2)
1+u
2
1
dx =
+ C = ln
+C
du = ln
2
cos x
1−u
1−u
1 − tan(x/2)
1 + sin x
cos(x/2) + sin(x/2) cos(x/2) + sin(x/2)
+ C = ln
+C
= ln
cos(x/2) − sin(x/2) cos(x/2) + sin(x/2)
cos x
Z
89. (a)
10
Z
sec x dx =
= ln |sec x + tan x| + C
x
π
x
1 + tan
tan + tan
x
4
2
2
+
=
(b) tan
x =
π
x
4
2
1 − tan tan
1 − tan
4
2
2
Z
Z
Z
1
dx = 1/u du = ln | tan(x/2)| + C but
90.
csc x dx =
sin x
π
ln | tan(x/2)| =
1 sin2 (x/2)
1 (1 − cos x)/2
1 1 − cos x
ln
= ln
= ln
; also,
2
2 cos (x/2)
2 (1 + cos x)/2
2 1 + cos x
1 − cos2 x
1
1 1 − cos x
1 − cos x
=
= − ln | csc x + cot x|
=
so ln
1 + cos x
(1 + cos x)2
(csc x + cot x)2
2 1 + cos x
q
√
91. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/ 1 − tanh2 (x/2) = 1/ 1 − u2 ,
√
sinh(x/2) = tanh(x/2) cosh(x/2) = u/ 1 − u2 , so sinh x = 2 sinh(x/2) cosh(x/2) = 2u/(1 − u2 ),
cosh x = cosh2 (x/2) + sinh2 (x/2) = (1 + u2 )/(1 − u2 ), x = 2 tanh−1 u, dx = [2/(1 − u2 )]du;
Z
Z
2
2
2u + 1
2 tanh(x/2) + 1
dx
1
√
=
du = √ tan−1 √
+ C = √ tan−1
+ C.
2
2 cosh x + sinh x
u +u+1
3
3
3
3
EXERCISE SET 9.7
1. exact value = 14/3 ≈ 4.666666667
(a) 4.667600663, |EM | ≈ 0.000933996
(b) 4.664795679, |ET | ≈ 0.001870988
(c) 4.666651630, |ES | ≈ 0.000015037
3. exact value = 2
(a) 2.008248408, |EM | ≈ 0.008248408
(b) 1.983523538, |ET | ≈ 0.016476462
(c) 2.000109517, |ES | ≈ 0.000109517
5. exact value = e−1 − e−3 ≈ 0.318092373
(a) 0.317562837, |EM | ≈ 0.000529536
(b) 0.319151975, |ET | ≈ 0.001059602
(c) 0.318095187, |ES | ≈ 0.000002814
2. exact value = 2
(a) 1.998377048, |EM | ≈ 0.001622952
(b) 2.003260982, |ET | ≈ 0.003260982
(c) 2.000072698, |ES | ≈ 0.000072698
4. exact value = sin(1) ≈ 0.841470985
(a) 0.841821700, |EM | ≈ 0.000350715
(b) 0.840769642, |ET | ≈ 0.000701343
(c) 0.841471453, |ES | ≈ 0.000000468
1
ln 5 ≈ 0.804718956
2
(a) 0.801605339, |EM | ≈ 0.003113617
(b) 0.811019505, |ET | ≈ 0.006300549
(c) 0.805041497, |ES | ≈ 0.000322541
6. exact value =
327
Chapter 9
7. f (x) =
√
1
15
x + 1, f 00 (x) = − (x + 1)−3/2 , f (4) (x) = − (x + 1)−7/2 ; K2 = 1/4, K4 = 15/16
4
16
27
(1/4) = 0.002812500
2400
243
(15/16) ≈ 0.000126563
(c) |ES | ≤
180 × 104
(a) |EM | ≤
(b) |ET | ≤
27
(1/4) = 0.005625000
1200
√
3
105 −9/2
x
; K2 = 3/4, K4 = 105/16
8. f (x) = 1/ x, f 00 (x) = x−5/2 , f (4) (x) =
4
16
27
(3/4) = 0.008437500
2400
243
(105/16) ≈ 0.000885938
(c) |ES | ≤
180 × 104
(a) |EM | ≤
(b) |ET | ≤
27
(3/4) = 0.016875000
1200
9. f (x) = sin x, f 00 (x) = − sin x, f (4) (x) = sin x; K2 = K4 = 1
π3
(1) ≈ 0.012919282
2400
π5
(1) ≈ 0.000170011
(c) |ES | ≤
180 × 104
(a) |EM | ≤
(b) |ET | ≤
π3
(1) ≈ 0.025838564
1200
10. f (x) = cos x, f 00 (x) = − cos x, f (4) (x) = cos x; K2 = K4 = 1
1
(1) ≈ 0.000416667
2400
1
(1) ≈ 0.000000556
(c) |ES | ≤
180 × 104
(a) |EM | ≤
(b) |ET | ≤
1
(1) ≈ 0.000833333
1200
11. f (x) = e−x , f 00 (x) = f (4) (x) = e−x ; K2 = K4 = e−1
8
(e−1 ) ≈ 0.001226265
2400
32
(e−1 ) ≈ 0.000006540
(c) |ES | ≤
180 × 104
(a) |EM | ≤
(b) |ET | ≤
8
(e−1 ) ≈ 0.002452530
1200
12. f (x) = 1/(2x + 3), f 00 (x) = 8(2x + 3)−3 , f (4) (x) = 384(2x + 3)−5 ; K2 = 8, K4 = 384
8
(8) ≈ 0.026666667
2400
32
(384) ≈ 0.006826667
(c) |ES | ≤
180 × 104
(a) |EM | ≤
13. (a) n >
(27)(1/4)
(24)(5 × 10−4 )
1/2
(243)(15/16)
(c) n >
(180)(5 × 10−4 )
≈ 23.7; n = 24
(b) n >
1/4
8
(8) ≈ 0.053333333
1200
(27)(1/4)
(12)(5 × 10−4 )
1/2
≈ 33.5; n = 34
≈ 7.1; n = 8
1/2
(27)(3/4)
≈ 41.1; n = 42
14. (a) n >
(24)(5 × 10−4 )
1/4
(243)(105/16)
≈ 11.5; n = 12
(c) n >
(180)(5 × 10−4 )
(b) |ET | ≤
(27)(3/4)
(b) n >
(12)(5 × 10−4 )
1/2
≈ 58.1; n = 59
Exercise Set 9.7
328
1/2
(π 3 )(1)
15. (a) n >
≈ 35.9; n = 36
(24)(10−3 )
1/4
(π 5 )(1)
(c) n >
≈ 6.4; n = 8
(180)(10−3 )
16. (a) n >
(1)(1)
(24)(10−3 )
(π 3 )(1)
(b) n >
(12)(10−3 )
1/2
≈ 6.5; n = 7
(1)(1)
(c) n >
(180)(10−3 )
(b) n >
(8)(8)
(24)(10−6 )
(8)(e−1 )
(b) n >
(12)(10−6 )
1/2
≈ 1632.99; n = 1633
(32)(384)
(c) n >
(180)(10−6 )
1/2
≈ 9.1; n = 10
≈ 1.5; n = 2
1/2
(8)(e−1 )
17. (a) n >
≈ 350.2; n = 351
(24)(10−6 )
1/4
(32)(e−1 )
(c) n >
≈ 15.99; n = 16
(180)(10−6 )
≈ 50.8; n = 51
1/4
18. (a) n >
(1)(1)
(12)(10−3 )
1/2
(b) n >
(8)(8)
(12)(10−6 )
1/2
≈ 495.2; n = 496
1/2
≈ 2309.4; n = 2310
1/4
≈ 90.9; n = 92
19. 0.746824948,
0.746824133
20. 1.137631378,
1.137630147
21. 2.129861595,
2.129861293
22. 2.418388347,
2.418399152
23. 0.805376152,
0.804776489
24. 1.536963087,
1.544294774
25. (a) 3.142425985, |EM | ≈ 0.000833331
(b) 3.139925989, |ET | ≈ 0.001666665
(c) 3.141592614, |ES | ≈ 0.000000040
26. (a) 3.152411433, |EM | ≈ 0.010818779
(b) 3.104518326, |ET | ≈ 0.037074328
(c) 3.127008159, |ES | ≈ 0.014584495
27. S14 = 0.693147984, |ES | ≈ 0.000000803 = 8.03 × 10−7 ; the method used in Example 5 results
in a value of n which ensures that the magnitude of the error will be less than 10−6 , this is not
necessarily the smallest value of n.
28. (a) greater, because the graph of e−x is concave up on the interval (1, 2)
2
(b) less, because the graph of e−x is concave down on the interval (0, 0.5)
2
29. f (x) = x sin x, f 00 (x) = 2 cos x − x sin x, |f 00 (x)| ≤ 2| cos x| + |x| | sin x| ≤ 2 + 2 = 4 so K2 ≤ 4,
(8)(4)
n>
(24)(10−4 )
1/2
≈ 115.5; n = 116 (a smaller n might suffice)
30. f (x) = ecos x , f 00 (x) = (sin2 x)ecos x − (cos x)ecos x , |f 00 (x)| ≤ ecos x (sin2 x + | cos x|) ≤ 2e so
K2 ≤ 2e, n >
31. f (x) =
√
(1)(2e)
(24)(10−4 )
x, f 00 (x) = −
1/2
≈ 47.6; n = 48 (a smaller n might suffice)
1
, lim |f 00 (x)| = +∞
4x3/2 x→0+
329
Chapter 9
√
√
√
√
x sin x + cos x
32. f (x) = sin x, f 00 (x) = −
, lim+ |f 00 (x)| = +∞
x→0
4x3/2
Z
π
33. L =
Z
p
1 + cos2 x dx ≈ 3.820187623
34. L =
0
35.
0
5
10
15
20
v (mi/hr)
0
40
60
73
84
v (ft/s)
0
58.67
88
107.07
123.2
20
v dt ≈
0
36.
Z
p
1 + 1/x4 dx ≈ 2.146822803
1
t (s)
Z
3
20
[0 + 4(58.67) + 2(88) + 4(107.07) + 123.2] ≈ 1604 ft
(3)(4)
t
0
1
2
3
4
5
6
7
8
a
0
0.02
0.08
0.20
0.40
0.60
0.70
0.60
0
8
8
[0 + 4(0.02) + 2(0.08) + 4(0.20) + 2(0.40) + 4(0.60) + 2(0.70) + 4(0.60) + 0]
(3)(8)
≈ 2.7 cm/s
a dt ≈
0
Z
180
v dt ≈
37.
0
Z
1800
38.
0
180
[0.00 + 4(0.03) + 2(0.08) + 4(0.16) + 2(0.27) + 4(0.42) + 0.65] = 37.9 mi
(3)(6)
4
2
4
2
4
1
1800
1
+
+
+
+
+
+
≈ 0.71 s
(1/v)dx ≈
(3)(6) 3100 2908 2725 2549 2379 2216 2059
Z
Z
16
0
16
r2 dy ≈ π
πr2 dy = π
39. V =
0
16
[(8.5)2 + 4(11.5)2 + 2(13.8)2 + 4(15.4)2 + (16.8)2 ]
(3)(4)
≈ 9270 cm3 ≈ 9.3 L
Z
600
600
[0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2 ,
(3)(6)
0
V = 75A ≈ 75(9000) = 675, 000 ft3
40. A =
Z
h dx ≈
b
f (x) dx ≈ A1 + A2 + · · · + An =
41.
a
=
b−a 1
1
1
(y0 + y1 ) + (y1 + y2 ) + · · · + (yn−1 + yn )
n
2
2
2
b−a
[y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn ]
2n
42. right endpoint, trapezoidal, midpoint, left endpoint
43. (a) The maximum value of |f 00 (x)| is approximately 3.844880.
(b) n = 18
(c) 0.904741
44. (a) The maximum value of |f 00 (x)| is approximately 1.467890.
(b) n = 12
(c) 1.112062
Exercise Set 9.8
330
45. (a) The maximum value of |f (4) (x)| is approximately 42.551816.
(b) n = 8
(c) 0.904524
46. (a) The maximum value of |f (4) (x)| is approximately 7.022710.
(b) n = 8
(c) 1.111443
EXERCISE SET 9.8
1. (a)
(b)
(c)
(d)
(e)
(f )
improper; infinite discontinuity at x = 3
continuous integrand, not improper
improper; infinite discontinuity at x = 0
improper; infinite interval of integration
improper; infinite interval of integration and infinite discontinuity at x = 1
continuous integrand, not improper
2. (a) improper if p > 0
(b) improper if 1 < p < 2
(c) integrand is continuous for all p, not improper
3.
−x
lim (−e
`→+∞
`
)
= lim (−e−` + 1) = 1
`→+∞
0
`
4.
5.
6.
7.
8.
9.
10.
11.
1
ln(1 + x2 )
lim
`→+∞ 2
x−1
lim ln
`→+∞
x+1
2
1
lim − e−x
`→+∞
2
`→−∞
0
3
5
= ln
5
3
1
1
1
−
2 + 2 = 2
2 ln `
`
= lim
`→+∞
e
√
√
= lim (2 ln ` − 2 ln 2) = +∞, divergent
`→+∞
1
x
tan−1
`→−∞ 2
2
0
= lim
`→−∞
`
2
1
[−1 + 1/(2` − 1)2 ] = −1/4
4
1
1 π
`
− tan−1
= [π/4 − (−π/2)] = 3π/8
`→−∞ 2 4
2
2
= lim
`
0
= lim
`→−∞
0
12.
= − ln
1 −`2
−e
+ 1 = 1/2
`→+∞ 2
lim
`
= lim
1
4(2x − 1)2
1 3x
e
`→−∞ 3
3
`−1
− ln
ln
`+1
5
`→+∞
`
2
lim
1
[ln(1 + `2 ) − ln 2] = +∞, divergent
2
4
lim 2 ln x
lim −
−1
= lim
`
`→+∞
`→+∞
`
1
lim −
`→+∞
2 ln2 x
√
= lim
1
lim − ln(3 − 2ex )
`→−∞
2
1
1 1 3`
− e
=
3 3
3
1
1
ln(3 − 2e` ) = ln 3
`→−∞ 2
2
= lim
`
331
Chapter 9
Z
Z
+∞
Z
0
x3 dx converges if
13.
−∞
Z
−∞
+∞
1 4
x
`→+∞ 4
0
Z
+∞
14.
0
Z
√
∞
so
−∞
x
x2 + 2
√
dx = lim
`→+∞
x
x2
+2
x3 dx both converge; it diverges if either (or both)
0
`
x3 dx = lim
diverge.
+∞
x3 dx and
1 4
` = +∞ so
`→+∞ 4
Z
+∞
x3 dx is divergent.
= lim
0
−∞
`
p
p
√
x2 + 2 = lim ( `2 + 2 − 2) = +∞
0
`→+∞
dx is divergent.
`
1
1
x
1
dx = lim −
= lim [−1/(`2 + 3) + 1/3] = ,
15.
2
2
2
`→+∞
(x + 3)
2(x + 3) 0 `→+∞ 2
6
0
Z ∞
Z 0
x
x
dx = −1/6 so
dx = 1/6 + (−1/6) = 0
similarly
2
2
2
2
−∞ (x + 3)
−∞ (x + 3)
Z
+∞
Z
+∞
16.
Z
Z
0
0
−∞
`
h
πi π
e−t
−1 −t
−1 −`
= ,
−
tan
dt
=
lim
−
tan
(e
)
=
lim
(e
)
+
`→+∞
`→+∞
1 + e−2t
4
4
0
0
h π
i π
e−t
dt = lim − tan−1 (e−t ) = lim − + tan−1 (e−` ) = ,
−2t
`→−∞
`→−∞
1+e
4
4
`
+∞
−∞
17.
π
π π
e−t
dt = + =
1 + e−2t
4
4
2
lim −
+
`→3
18.
1
x−3
3
lim+ x2/3
`→0 2
4
= lim+ −1 +
`→3
`
8
= lim
`→0+
`
1
= +∞, divergent
`−3
3
(4 − `2/3 ) = 6
2
`
lim − − ln(cos x)
19.
=
`→π/2
`
20.
√
lim− −2 9 − x
`→9
21.
−1
lim sin
`→1−
`
x
0
lim − ln(cos `) = +∞, divergent
`→π/2−
√
= lim 2(− 9 − ` + 3) = 6
`→9−
0
= lim sin−1 ` = π/2
`→1−
1
p
p
√
√
2
= lim + (− 8 + 9 − `2 ) = − 8
lim + − 9 − x
22.
`→−3
23.
0
`→−3
`
√
`
lim − 1 − 2 sin x
=
`→π/6−
0
√
lim (− 1 − 2 sin ` + 1) = 1
`→π/6−
`
24.
lim − ln(1 − tan x)
`→π/4−
Z
2
25.
0
=
0
lim − ln(1 − tan `) = +∞, divergent
`→π/4−
`
dx
= lim ln |x − 2|
x−2
`→2−
0
= lim− (ln |` − 2| − ln 2) = −∞, divergent
`→2
Exercise Set 9.8
Z
2
26.
0
Z
8
27.
0
Z
332
2
dx
= lim −1/x
x2
`→0+
Z
`→0+
`
3
x−1/3 dx = lim+ x2/3
`→0 2
0
x−1/3 dx = lim
`→0−
−1
Z
8
2
= lim (−1/2 + 1/`) = +∞ so
8
`
3 2/3
x
2
`
−2
dx
is divergent
x2
3
= lim+ (4 − `2/3 ) = 6,
`→0 2
= lim
`→0−
−1
3 2/3
(` − 1) = −3/2
2
x−1/3 dx = 6 + (−3/2) = 9/2
so
−1
`
√
dx
3
1/3
=
lim
3(x
−
2)
= lim− 3[(` − 2)1/3 − (−2)1/3 ] = 3 2,
2/3
−
`→2
`→2
0 (x − 2)
0
4
Z 4
Z 4
√
√
dx
dx
3
3
1/3
=
lim
3(x
−
2)
=
3
2
so
=6 2
similarly
2/3
2/3
+
`→2
(x
−
2)
2 (x − 2)
0
`
Z
28.
2
Z
+∞
1
dx =
x2
Z
+∞
dx
√
=
x x2 − 1
Z +∞
1
1
29. Define
dx +
dx where a > 0; take a = 1 for convenience,
2
x2
0
0 x
a
1
Z +∞
Z 1
1
1
dx = lim (−1/x) = lim (1/` − 1) = +∞ so
dx is divergent.
2
2
+
+
x
x
`→0
`→0
0
0
`
30. Define
1
Z
a
Z
a
1
dx
√
+
x x2 − 1
Z
+∞
a
dx
√
where a > 1,
x x2 − 1
take a = 2 for convenience to get
2
Z 2
dx
−1
√
= lim+ sec x = lim+ (π/3 − sec−1 `) = π/3,
2
`→1
`→1
1 x x −1
`
`
Z +∞
Z +∞
dx
dx
√
√
= lim sec−1 x = π/2 − π/3 so
= π/2.
2
`→+∞
x x −1
x x2 − 1
2
1
2
Z
+∞
31.
0
Z
+∞
32.
√
0
Z
+∞
33.
√
0
Z
+∞
34.
√
0
Z
x
Z
lim
`
−e−u
`→+∞
Z
+∞
0
Z
1
0
e−x
dx = −
1 − e−2x
+∞
e−u du = 2 lim
0
e−x
dx =
1 − e−x
0
`→+∞
+∞
dx = 2
dx
=2
x(x + 4)
36. A =
37.
Z
√
x
e−
√
`→+∞
du
1
u
= 2 lim
tan−1
`→+∞ 2
u2 + 4
2
`
Z
0
√
1
du
=
1 − u2
Z
= lim+ 2(1 −
√
π
`
=
2
2
`) = 2
`
√
0
du
= lim sin−1 u
`→1
1 − u2
0
= lim sin−1 ` =
`→1
π
2
`
= 1/3
0
`
1 −x
e (sin x − cos x)
`→+∞ 2
(b) 2.804364
`→+∞
`→0
`
1
1 − e−` = 2
= lim tan−1
0
1
√
du
√ = lim 2 u
u
`→0+
e−x cos x dx = lim
39. (a) 2.726585
= 2 lim
0
1
xe−3x dx = lim − (3x + 1)e−3x
`→+∞
9
0
`
= 1/2
0
(c) 0.219384
(d) 0.504067
333
Chapter 9
40. 1 +
dy
dx
2
16x2
9 + 12x2
=1+
=
; the arc length is
2
9 − 4x
9 − 4x2
Z
3/2
s
0
9 + 12x2
dx ≈ 3.633168
9 − 4x2
Z
ln x dx = x ln x − x + C,
41.
Z
0
Z
1
ln x dx = lim+
`→0
`
but lim+ ` ln ` = lim+
`→0
`→0
1
1
ln x dx = lim+ (x ln x − x)
`→0
ln `
= lim (−`) = 0 so
1/` `→0+
`
1
Z
= lim+ (−1 − ` ln ` + `),
`→0
ln x dx = −1
0
Z
42.
ln x 1
ln x
− + C,
dx = −
2
x
x
x
`
Z `
Z +∞
ln x 1
ln ` 1
ln x
ln x
−
−
+
1
,
dx
=
lim
dx
=
lim
=
lim
−
−
`→+∞ 1 x2
`→+∞
x2
x
x 1 `→+∞
`
`
1
Z +∞
ln `
1
ln x
but lim
= lim
= 0 so
=1
`→+∞ `
`→+∞ `
x2
1
Z
43.
1
1
xe−3x dx = − xe−3x − e−3x + C,
3
9
`
Z `
Z +∞
1
1
xe−3x dx = lim
xe−3x dx = lim
− xe−3x − e−3x
`→+∞ 0
`→+∞
3
9
0
0
1 −3` 1 −3` 1
= lim − `e
− e
+
`→+∞
3
9
9
Z +∞
`
1
but lim `e−3` = lim 3` = lim
=
0
so
xe−3x dx = 1/9
`→+∞
`→+∞ e
`→+∞ 3e3`
0
Z
44. A =
3
+∞
x−2
8
dx = lim 2 ln
2
`→+∞
x −4
x+2
Z
45. (a) V = π
+∞
e−2x dx = −
0
Z
π
2
`
3
1
`−2
− ln
= 2 ln 5
= lim 2 ln
`→+∞
`+2
5
lim e−2x
`→+∞
`
= π/2
0
p
1 + e−2x dx, let u = e−x to get
0
p
1
Z 0p
h√
p
√ i
1
u
2
2
2
1 + u du = 2π
1 + u + ln | u + 1 + u | = π
2 + ln(1 + 2)
S = −2π
2
2
1
0
(b) S = 2π
+∞
e−x
47. (a) For x ≥ 1, x2 ≥ x, e−x ≤ e−x
Z `
Z +∞
i`
(b)
e−x dx = lim
e−x dx = lim −e−x = lim (e−1 − e−` ) = 1/e
`→+∞ 1
`→+∞
`→+∞
1
1
Z +∞
2
(c) By parts (a) and (b) and Exercise 46(b),
e−x dx is convergent and is ≤ 1/e.
2
1
ex
1
≤
48. (a) If x ≥ 0 then ex ≥ 1,
2x + 1
2x + 1
`
Z `
dx
1
(b) lim
= lim
ln(2x + 1) = +∞
`→+∞ 0 2x + 1
`→+∞ 2
0
Z +∞
(c) By parts (a) and (b) and Exercise 46(a),
0
ex
dx is divergent.
2x + 1
Exercise Set 9.8
334
Z
`→+∞
i`
`
(π/x2 ) dx = lim −(π/x)
49. V = lim
1
Z `
A = lim
`→+∞
`→+∞
1
= lim (π − π/`) = π
`→+∞
p
p
2π(1/x) 1 + 1/x4 dx; use Exercise 46(a) with f (x) = 2π/x, g(x) = (2π/x) 1 + 1/x4
1
and a = 1 to see that the area is infinite.
√
Z +∞
x3 + 1
for x ≥ 2,
1dx = +∞
50. (a) 1 ≤
x
2
`
Z +∞
Z +∞
1
x
dx
dx
≤
=
lim
−
= 1/24
(b)
`→+∞
x5 + 1
x4
3x3 2
2
2
`
Z +∞
ex
1
1
1
≤
for x ≥ 0,
dx = lim
ln(2x + 1) = +∞
(c)
`→+∞ 2
2x + 1
2x + 1
2x + 1
0
0
Z
2x
51.
Z
p
3
1 + t dt ≥
0
t
3/2
0
Z
2x
t3/2 dt = lim
lim
x→+∞
2x
x→+∞
0
Z
2x
p
1 + t3 dt
0
lim
x5/2
x→+∞
52. (b) u =
√
Z
+∞
x,
0
2
dt = t5/2
5
2
(2x)5/2
5
2x
2
(2x)5/2 ,
5
0
Z +∞ p
= +∞ so
1 + t3 dt = +∞; by L’Hôpital’s Rule
=
0
p
p
√
2 1 + (2x)3
2 1/x3 + 8
= 8 2/5
= lim
= lim
x→+∞ (5/2)x3/2
x→+∞
5/2
√
Z +∞
Z +∞
cos x
√ dx = 2
cos u du;
cos u du diverges by part (a).
x
0
0
Z
1
1
x
dx
cos θ dθ = 2 sin θ + C = √
=
+C
r2
r
(r2 + x2 )3/2
r2 r2 + x2
`
p
p
2πN I
2πN Ir
x
√
lim
lim (`/ r2 + `2 − a/ r2 + a2 )
=
so u =
2
2
2
`→+∞ r
k
kr `→+∞
r +x a
p
2πN I
(1 − a/ r2 + a2 ).
=
kr
Z
53. Let x = r tan θ to get
M
to get
2RT
r
r
3/2
−2
2
4
1
2RT
8RT
M
M
=
=√
(a) v̄ = √
2 2RT
M
πM
π 2RT
π
r
3/2 √
−5/2
4
3RT
3 π
3RT
M
M
2
so vrms =
=
(b) vrms = √
8
2RT
M
M
π 2RT
54. Let a2 =
55. (a) Satellite’s weight = w(x) = k/x2 lb when x = distance from center of Earth; w(4000) = 6000
Z 4000+`
10
9.6 × 1010 x−2 dx mi·lb.
so k = 9.6 × 10 and W =
Z
4000
+∞
(b)
`
9.6 × 1010 x−2 dx = lim −9.6 × 1010 /x
`→+∞
4000
= 2.4 × 107 mi·lb
4000
`
1
1
e−st dt = lim − e−st =
`→+∞
s
s
0
0
`
Z +∞
Z +∞
1 −(s−2)t
1
−st 2t
−(s−2)t
2t
e
e e dt =
e
dt = lim −
=
(b) L{e } =
`→+∞
s
−
2
s
−
2
0
0
0
Z
56. (a) L{1} =
+∞
335
Chapter 9
Z
+∞
(c) L{sin t} =
−st
e
0
Z
+∞
(d) L{cos t} =
0
Z
+∞
57. (a) L{f (t)} =
`
e−st
(−s sin t − cos t)
sin t dt = lim 2
`→+∞ s + 1
te−st dt = lim −(t/s + 1/s2 )e−st
2 −st
t e
+∞
(c) L{f (t)} =
3
58.
=
0
1
s2
dt = lim −(t /s + 2t/s + 2/s3 )e−st
2
1
e−st dt = lim − e−st
`→+∞
s
10
100
1000
10,000
0.8862269
0.8862269
0.8862269
0.8862269
Z
`
2
`→+∞
0
Z
`
`→+∞
+∞
(b) L{f (t)} =
0
`
s
e−st
(−s cos t + sin t) = 2
e−st cos t dt = lim 2
`→+∞ s + 1
s +1
0
0
Z
1
s2 + 1
=
`
3
=
0
2
s3
e−3s
=
s
Z
p
2
e−u du = π/a
0
0
Z +∞
Z +∞
√
√
2
2
2
−x2 /2σ 2
e
dx = √
e−u du = 1
(b) x = 2σu, dx = 2σ du, √
π 0
2πσ 0
59. (a) u =
Z
√
ax, du =
√
+∞
−ax2
a dx, 2
e
2
dx = √
a
√
2
e−x dx ≈ 0.8862; π/2 ≈ 0.8862
Z 3
Z +∞
Z
Z0 +∞
−x2
−x2
−x2
e
dx = e
dx +
e
dx so E =
(b)
+∞
3
60. (a)
0
Z
0
3
+∞
−x2
e
Z
+∞
xe−x dx =
dx <
3
2
3
1 −9
e < 7 × 10−5
2
4
1
dx ≈ 1.047; π/3 ≈ 1.047
+1
0
Z +∞
Z 4
Z +∞
1
1
1
(b)
dx =
dx +
dx so
6+1
6+1
6+1
x
x
x
0
0
4
Z +∞
Z +∞
1
1
1
E=
dx <
dx =
< 2 × 10−4
6
6
x +1
x
5(4)5
4
4
61. (a)
x6
Z
`
+∞
62. If p = 0, then
(1)dx = lim x
Z
= +∞,
`→+∞
0
+∞
if p 6= 0, then
0
1
e dx = lim epx
`→+∞ p
`
px
0
Z
1
63. If p = 1, then
Z
0
1
if p 6= 1, then
0
0
1
= lim (ep` − 1) =
`→+∞ p
−1/p, p < 0
.
+∞,
p>0
1
dx
= lim ln x
x
`→0+
= +∞;
`
dx
x1−p
= lim
p
x
`→0+ 1 − p
1
= lim [(1 − `1−p )/(1 − p)] =
`
`→0+
1/(1 − p), p < 1
.
+∞,
p>1
√
64. u = 1 − x, u2 = 1 − x, 2u du = −dx;
Z 0p
Z 1p
h p
√ i1 √
−2
2 − u2 du = 2
2 − u2 du = u 2 − u2 + 2 sin−1 (u/ 2) = 2 + π/2
1
Z
0
0
Z
1
cos(u )du ≈ 1.809
2
65. 2
0
66. −2
Z
0
sin(1 − u )du = 2
1
1
sin(1 − u2 )du ≈ 1.187
2
0
Chapter 9 Supplementary Exercises
336
CHAPTER 9 SUPPLEMENTARY EXERCISES
1. (a)
(c)
(e)
(g)
(i)
integration by parts, u = x, dv = sin x dx
reduction formula
u-substitution: u = x3 + 1
integration by parts: dv = dx, u = tan−1 x
u-substitution: u = 4 − x2
(b)
(d)
(f )
(h)
u-substitution: u = sin x
u-substitution: u = tan x
u-substitution: u = x + 1
trigonometric substitution: x = 2 sin θ
1
3
1
9
2. (a) x = 3 tan θ
(d) x = 3 sec θ
(b) x = 3 sin θ
√
(e) x = 3 tan θ
(c) x =
(f ) x =
5. (a) #40
(d) #108
(b) #57
(e) #52
(c) #113
(f ) #71
6. (a) u = x2 , dv = √
Z
1
0
√
3
x
x2 + 1
x
dx, du = 2x dx, v =
x2 + 1
Z
i1
p
2
2
dx = x x + 1 − 2
0
=
√
2
2 − (x2 + 1)3/2
3
sin θ
tan θ
√
x2 + 1;
1
x(x2 + 1)1/2 dx
0
1
=
0
√
√
2 √
2 − [2 2 − 1] = (2 − 2)/3
3
(b) u2 = x2 + 1, x2 = u2 − 1, 2x dx = 2u du, x dx = u du;
Z 1
Z 1
Z √2 2
x3
x2
u −1
√
√
u du
dx =
x dx =
2
2
u
x +1
x +1
0
0
1
√2
Z √2
√
1 3
2
u −u
(u − 1)du =
= (2 − 2)/3
=
3
1
1
7. (a) u = 2x,
Z
Z
1
1
1
4
4
sin u du =
− sin3 u cos u +
sin 2x dx =
2
2
4
1
3
1
− sin u cos u +
= − sin3 u cos u +
8
8
2
3
4
1
2
Z
2
sin u du
Z
du
3
3
1
sin u cos u + u + C
= − sin3 u cos u −
8
16
16
3
3
1
sin 2x cos 2x + x + C
= − sin3 2x cos 2x −
8
16
8
(b) u = x2 ,
Z
Z
Z
1
1 1
3
4 2
4
3
2
cos u du =
cos u sin u +
cos u du
x cos (x )dx =
2
2 4
4
Z
1
3 1
1
cos u sin u +
du
= cos3 u sin u +
8
8 2
2
=
3
3
1
cos3 u sin u +
cos u sin u + u + C
8
16
16
=
1
3
3
cos3 (x2 ) sin(x2 ) +
cos(x2 ) sin(x2 ) + x2 + C
8
16
16
8. (a) With x = sec θ:
√
Z
Z
x2 − 1
1
dx
=
cot
θ
dθ
=
ln
|
sin
θ|
+
C
=
ln
+ C; valid for |x| > 1.
3
x −x
|x|
337
Chapter 9
(b) With x = sin θ:
Z
Z
Z
1
1
dx
=
−
dθ
=
−
2 csc 2θ dθ
x3 − x
sin θ cos θ
√
= − ln | csc 2θ − cot 2θ| + C = ln | cot θ| + C = ln
1 − x2
+ C, 0 < |x| < 1.
|x|
(c) By partial fractions:
Z
1/2
1/2
1
1
1
+
dx = − ln |x| + ln |x + 1| + ln |x − 1| + C
− +
x x+1 x−1
2
2
p
2
|x − 1|
+ C; valid for all x except x = 0, ±1.
= ln
|x|
√
9. (a) With u = x:
Z
Z
p
√
1
1
dx = 2 √
du = 2 sin−1 (u/ 2) + C = 2 sin−1 ( x/2) + C;
√ √
x 2−x
2 − u2
√
with u = 2 − x:
Z
Z
√
√
√
1
1
dx = −2 √
du = −2 sin−1 (u/ 2) + C = −2 sin−1 ( 2 − x/ 2) + C;
√ √
2
x 2−x
2−u
completing the square:
Z
1
p
dx = sin−1 (x − 1) + C.
1 − (x − 1)2
(b) In the three results in part (a) the antiderivatives differ by a constant, in particular
p
√
√
2 sin−1 ( x/2) = π − 2 sin−1 ( 2 − x/ 2) = π/2 + sin−1 (x − 1).
Z
A
B
3−x
C
3−x
= + 2+
; A = −4, B = 3, C = 4
dx, 2
3
2
x (x + 1)
x
x
x+1
1 x +x
2
3
A = −4 ln |x| − + 4 ln |x + 1|
x
1
2
10. A =
= (−4 ln 2 −
3
3
3
3
+ 4 ln 3) − (−4 ln 1 − 3 + 4 ln 2) = − 8 ln 2 + 4 ln 3 = + 4 ln
2
2
2
4
11. Solve y = 1/(1 + x2 ) for x getting
r
1−y
x=
and integrate with respect to
y
Z 1r
1−y
dy (see figure)
to y to get A =
y
0
Z
+∞
12. A =
e
Z
13. V = 2π
0
Z
14.
0
+∞
= 1/e
e
`
`→+∞
`→+∞
y = 1 / (1 + x2)
x
xe−x dx = 2π lim −e−x (x + 1)
but lim e−` (` + 1) = lim
`→+∞
1
`
ln x
ln x − 1
dx = lim −
`→+∞
x2
x
+∞
y
= 2π lim
0
`→+∞
1 − e−` (` + 1)
`+1
1
= lim ` = 0 so V = 2π
`→+∞ e
e`
`
dx
1
tan−1 (x/a)
= lim
`→+∞ a
x2 + a2
= lim
0
`→+∞
π
1
tan−1 (`/a) =
= 1, a = π/2
a
2a
Chapter 9 Supplementary Exercises
338
Z
2
u1/2 du = − cos3/2 θ + C
3
Z
1
1
1
16. Use Endpaper Formula (31) to get tan7 θdθ = tan6 θ − tan4 θ + tan2 θ + ln | cos θ| + C.
6
4
2
Z
1
1
17. u = tan(x2 ),
u2 du = tan3 (x2 ) + C
2
6
15. u = cos θ, −
√
√
18. x = (1/ 2) sin θ, dx = (1/ 2) cos θ dθ,
1
√
2
Z
π/2
)
π/2
Z
3 π/2
1
3
2
cos θ sin θ
+
cos θ dθ
4
4 −π/2
−π/2
)
(
π/2
Z
3π
1 π/2
3 1
3
1
cos θ sin θ
+
dθ = √ π = √
= √
2
2
4 2 2
4
2
8
2
−π/2
−π/2
1
cos θ dθ = √
2
−π/2
4
(
√
√
19. x = 3 tan θ, dx = 3 sec2 θ dθ,
Z
Z
1
1
x
1
1
dθ =
cos θ dθ = sin θ + C = √
+C
3
sec θ
3
3
3 3 + x2
Z
Z
√
1
1
cos θ
dθ, let u = sin θ − 3,
du = √ tan−1 [(sin θ − 3)/ 3] + C
20.
(sin θ − 3)2 + 3
u2 + 3
3
Z
x+3
p
dx, let u = x + 1,
21.
(x + 1)2 + 1
Z
Z
p
2
u+2
2
−1/2
√
+√
du =
u(u + 1)
du = u2 + 1 + 2 sinh−1 u + C
u2 + 1
u2 + 1
p
= x2 + 2x + 2 + 2 sinh−1 (x + 1) + C
Alternate solution: let x + 1 = tan θ,
Z
Z
Z
(tan θ + 2) sec θ dθ = sec θ tan θ dθ + 2 sec θ dθ = sec θ + 2 ln | sec θ + tan θ| + C
=
p
p
x2 + 2x + 2 + 2 ln( x2 + 2x + 2 + x + 1) + C.
Z
22. Let x = tan θ to get
x3
1
dx.
− x2
A
B
C
1
= + 2+
; A = −1, B = −1, C = 1 so
x2 (x − 1)
x
x
x−1
Z
Z
Z
1
1
1
1
−
dx −
dx = − ln |x| + + ln |x − 1| + C
dx +
x
x2
x−1
x
=
23.
x−1
tan θ − 1
1
+ ln
+ C = cot θ + ln
+ C = cot θ + ln |1 − cot θ| + C
x
x
tan θ
A
B
C
1
1
1
1
=
+
+
;A=− ,B=
,C=
so
(x − 1)(x + 2)(x − 3)
x−1 x+2 x−3
6
15
10
Z
Z
Z
1
1
1
1
1
1
dx +
dx +
dx
−
6
x−1
15
x+2
10
x−3
1
1
1
ln |x + 2| +
ln |x − 3| + C
= − ln |x − 1| +
6
15
10
339
24.
Chapter 9
x(x2
Z
A
Bx + C
1
= + 2
; A = 1, B = C = −1 so
+ x + 1)
x
x +x+1
−x − 1
dx = −
2
x +x+1
Z
Z
x+1
dx = −
(x + 1/2)2 + 3/4
u + 1/2
du,
u2 + 3/4
u = x + 1/2
√
1
1
= − ln(u2 + 3/4) − √ tan−1 2u/ 3 + C1
2
3
Z
1
1
2x + 1
dx
= ln |x| − ln(x2 + x + 1) − √ tan−1 √
+C
x(x2 + x + 1)
2
3
3
so
25. u =
Z 2
0
√
x − 4, x = u2 + 4, dx = 2u du,
2
Z 2
4
2u2
−1
du
=
2
du
=
2u
−
4
tan
1
−
(u/2)
=4−π
u2 + 4
u2 + 4
0
0
√
26. u = x, x = u2 , dx = 2u du,
Z 3
Z 3
3
9
3
u2
−1 u
du = 2
du = 2u − 6 tan
1− 2
=6− π
2
2+9
u
u
+
9
3
2
0
0
0
27. u =
Z
Z
1
0
30.
ex + 1, ex = u2 − 1, x = ln(u2 − 1), dx =
2
du =
u2 − 1
28. u =
29.
√
√
Z
√ x
1
e +1−1
1
+C
−
du = ln |u − 1| − ln |u + 1| + C = ln √ x
u−1 u+1
e +1+1
2u
du,
u2 + 1
1
Z 1
π
1
2u2
−1
du
=
2
du
=
(2u
−
2
tan
u)
=2−
1
−
2
2
u +1
u +1
2
0
0
ex − 1, ex = u2 + 1, x = ln(u2 + 1), dx =
lim −
`→+∞
1
2(x2 + 1)
`
1
bx
tan−1
lim
`→+∞ ab
a
1
1
1
+
=
2(`2 + 1) 2(a2 + 1)
2(a2 + 1)
= lim
−
= lim
π
1
b`
tan−1
=
ab
a
2ab
`→+∞
a
`
1
31. Let u = x to get
4
4
0
`→+∞
Z
√
1
1
1
du = sin−1 u + C = sin−1 (x4 ) + C.
2
4
4
1−u
Z
Z
(cos32 x sin30 x − cos30 x sin32 x)dx =
32.
cos30 x sin30 x(cos2 x − sin2 x)dx
1
= 30
2
Z q
33.
Z
34.
2u
du,
−1
u2
Z
sin30 2x cos 2x dx =
sin31 2x
+C
31(231 )
√
Z
p
√
√
1
2
2
√
[(x + 2)3/2 − (x − 2)3/2 ] + C
x − x − 4dx =
( x + 2 − x − 2)dx =
3
2
1
1
dx = −
x10 (1 + x−9 )
9
Z
1
1
1
du = − ln |u| + C = − ln |1 + x−9 | + C
u
9
9
Chapter 9 Supplementary Exercises
340
35. (a) (x + 4)(x − 5)(x2 + 1)2 ;
(b) −
B
Cx + D
Ex + F
A
+
+ 2
+ 2
x+4 x−5
x +1
(x + 1)2
2
x−2
3
3
+
−
−
x + 4 x − 5 x2 + 1 (x2 + 1)2
−1
(c) −3 ln |x + 4| + 2 ln |x − 5| + 2 tan
Z
+∞
−t
−t
`
e dt = lim −e
36. (a) Γ(1) =
`→+∞
0
Z
+∞
(b) Γ(x + 1) =
x
+ tan−1 x
2
x +1
= lim (−e−` + 1) = 1
`→+∞
tx e−t dt; let u = tx , dv = e−t dt to get
0
Γ(x + 1) =
0
1
3
x − ln(x2 + 1) −
2
2
+∞
−tx e−t 0
Z
+x
+∞
tx−1 e−t dt = −tx e−t
0
+∞
0
+ xΓ(x)
tx
= 0 (by multiple applications of L’Hôpital’s rule)
t→+∞
t→+∞ et
so Γ(x + 1) = xΓ(x)
lim tx e−t = lim
(c) Γ(2) = (1)Γ(1) = (1)(1) = 1, Γ(3) = 2Γ(2) = (2)(1) = 2, Γ(4) = 3Γ(3) = (3)(2) = 6
It appears that Γ(n) = (n − 1)! if n is a positive integer.
Z +∞
Z +∞
√
√
√
2
1
(d) Γ
=
t−1/2 e−t dt = 2
e−u du (with u = t) = 2( π/2) = π
2
0
0
1
1√
3
3√
1
3
5
3
= Γ
=
= Γ
=
π, Γ
π
(e) Γ
2
2
2
2
2
2
2
4
37. (a) t = − ln x, x = e−t , dx = −e−t dt,
Z
Z 0
Z 1
(ln x)n dx = −
(−t)n e−t dt = (−1)n
0
+∞
+∞
tn e−t dt = (−1)n Γ(n + 1)
0
(b) t = xn , x = t1/n , dx = (1/n)t1/n−1 dt,
Z +∞
Z +∞
−xn
e
dx = (1/n)
t1/n−1 e−t dt = (1/n)Γ(1/n) = Γ(1/n + 1)
0
0
q
p
q
2
2
cos θ − cos θ0 = 2 sin (θ0 /2) − sin (θ/2) = 2(k 2 − k 2 sin2 φ) = 2k 2 cos2 φ
38. (a)
q
√
1
1
1 − sin2 (θ/2) dθ
= 2 k cos φ; k sin φ = sin(θ/2) so k cos φ dφ = cos(θ/2) dθ =
2
2
q
1
2k cos φ
dφ and hence
=
1 − k 2 sin2 φ dθ, thus dθ = p
2
1 − k 2 sin2 φ
s Z
s
Z
2k cos φ
8L π/2
1
L π/2
1
√
p
dφ
=
4
dφ
T =
·p
2
2
g 0
g
2k cos φ
0
1 − k sin φ
1 − k 2 sin2 φ
p
(b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, then
√ Z π/2
3
dφ
q
≈ 1.37 s.
T =
2 0
1 − sin2 (π/18) sin2 φ
341
Chapter 9
CHAPTER 9 HORIZON MODULE
1. The depth of the cut equals the terrain elevation minus the track elevation. From Figure 2, the
1
cross sectional area of a cut of depth D meters is 10D + 2 · D2 = D2 + 10D square meters.
2
Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area
town A (m)
(m)
(m)
(m)
f (x) of cut (m2 )
0
2000
4000
6000
8000
10,000
12,000
14,000
16,000
18,000
20,000
100
105
108
110
104
106
120
122
124
128
130
100
101
102
103
104
105
106
107
108
109
110
0
4
6
7
0
1
14
15
16
19
20
0
56
96
119
0
11
336
375
416
551
600
Z
2000
The total volume of dirt to be excavated, in cubic meters, is
f (x) dx.
0
By Simpson’s Rule, this is approximately
20,000 − 0
[0 + 4 · 56 + 2 · 96 + 4 · 119 + 2 · 0 + 4 · 11 + 2 · 336 + 4 · 375 + 2 · 416 + 4 · 551 + 600]
3 · 10
= 4,496,000 m3 .
Excavation costs $4 per m3 , so the ttal cost of the railroad from kA to M is about
4 · 4,496,000 = 17,984,000 dollars.
2. (a)
Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area
town A (m)
(m)
(m)
(m)
f (x) of cut (m2 )
20,000
20,100
20,200
20,300
20,400
20,500
20,600
20,700
20,800
20,900
21,000
130
135
139
142
145
147
148
146
143
139
133
110
109.8
109.6
109.4
109.2
109
108.8
108.6
108.4
108.2
108
20
25.2
29.4
32.6
35.8
38
39.2
37.4
34.6
30.8
25
300
887.04
1158.36
1388.76
1639.64
1824
1928.64
1772.76
1543.16
1256.64
875
Z
21,000
The total volume of dirt to be excavated, in cubic meters, is
f (x) dx.
20,000
By Simpson’s Rule this is approximately
21,000 − 20,000
[600 + 4 · 887.04 + 2 · 1158.36 + . . . + 4 · 1256.64 + 875] = 1,417,713.33 m3 .
3 · 10
The total cost of a trench from M to N is about 4 · 1,417,713.33 ≈ 5,670,853 dollars.
Chapter 9 Horizon Module
(b)
342
Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area
town A (m)
(m)
(m)
(m)
f (x) of cut (m2 )
21,000
22,000
23,000
24,000
25,000
26,000
27,000
28,000
29,000
30,000
31,000
133
120
106
108
106
98
100
102
96
91
88
108
106
104
102
100
98
96
94
92
90
88
25
14
2
6
6
0
4
8
4
1
0
875
336
24
96
96
0
56
144
56
11
0
Z
31,000
The total volume of dirt to be excavated, in cubic meters, is
f (x) dx. By Simpson’s
21,000
Rule this is approximately
31,000 − 21,000
[875 + 4 · 336 + 2 · 24 + . . . + 4 · 11 + 0] = 1,229,000 m3 .
3 · 10
The total cost of the railroad from N to B is about 4 · 1,229,000 ≈ 4,916,000 dollars.
3. The total cost if trenches are used everywhere is about
17,984,000 + 5,670,853 + 4,916,000 = 28,570,853 dollars.
1 2
π5 ≈ 119.27 m2 . The length of the
2
tunnel is 1000 m, so the volume of dirt to be removed is about 1000AT ≈ 1,119,269.91 m3 ,
and the drilling and dirt-piling costs are 8 · 1000AT ≈ 954,159 dollars.
(b) To extend the tunnel from a length of x meters to a length of x + dx meters, we must move
a volume of AT dx cubic meters of dirt a distance of about x meters. So the cost of this
extension is about 0.06 × AT dx dollars. The cost of moving all of the dirt in the tunnel is
therefore
1000
Z 1000
x2
0.06 × AT dx = 0.06AT
= 30,000AT ≈ 3,578,097 dollars.
2 0
0
4. (a) The cross-sectional area of a tunnel is AT = 80 +
(c) The total cost of the tunnel is about 954,159 + 3,578,097 ≈ 4,532,257 dollars.
5. The total cost of the railroad, using a tunnel, is 17,894,000 + 4,532,257 + 4,916,000 + 27,432,257
dollars, which is smaller than the cost found in Exercise 3. It will be cheaper to build the railroad
if a tunnel is used.
CHAPTER 10
Mathematical Modeling with Differential
Equations
EXERCISE SET 10.1
1. y 0 = 2x2 ex
3
/3
= x2 y and y(0) = 2 by inspection.
2. y 0 = x3 − 2 sin x, y(0) = 3 by inspection.
dy
dy
= c; (1 + x)
= (1 + x)c = y
dx
dx
0
(b) second order; y = c1 cos t − c2 sin t, y 00 + y = −c1 sin t − c2 cos t + (c1 sin t + c2 cos t) = 0
3. (a) first order;
c
dy
+ y = 2 − e−x/2 + 1 + ce−x/2 + x − 3 = x − 1
dx
2
(b) second order; y 0 = c1 et − c2 e−t , y 00 − y = c1 et + c2 e−t − c1 et + c2 e−t = 0
4. (a) first order; 2
5.
dy
dy
y2
dy
1 dy
=x
+ y,
(1 − xy) = y 2 ,
=
y dx
dx
dx
dx
1 − xy
6. 2x + y 2 + 2xy
dy
= 0, by inspection.
dx
d 3x
ye
= 0, ye3x = C, y = Ce−3x
dx
dy
= −3dx, ln |y| = −3x + C1 , y = ±e−3x eC1 = Ce−3x
separation of variables:
y
R
d
(b) IF: µ = e−2 dt = e−2t , [ye−2t ] = 0, ye−2t = C, y = Ce2t
dt
dy
= 2dt, ln |y| = 2t + C1 , y = ±eC1 e2t = Ce2t
separation of variables:
y
7. (a) IF: µ = e3
R
8. (a) IF: µ = e−4
dx
R
= e3x ,
x dx
= e−2x ,
2
2
d h −2x2 i
ye
= 0, y = Ce2x
dx
2
2
dy
= 4x dx, ln |y| = 2x2 + C1 , y = ±eC1 e2x = Ce2x
y
R
d t
t
dt
ye = 0, y = Ce−t
(b) IF: µ = e
=e,
dt
dy
= −dt, ln |y| = −t + C1 , y = ±eC1 e−t = Ce−t
separation of variables:
y
separation of variables:
9.
1
1
y
y
dy = dx, ln |y| = ln |x| + C1 , ln
= C1 , = ±eC1 = C, y = Cx
y
x
x
x
10.
dy
1
= x2 dx, tan−1 y = x3 + C, y = tan
2
1+y
3
11.
√
√
p
x
dy
2
2
= −√
dx, ln |1 + y| = − 1 + x2 + C1 , 1 + y = ±e− 1+x eC1 = Ce− 1+x ,
2
1+y
1+x
√
1+x2
y = Ce−
1 3
x +C
3
−1
343
Exercise Set 10.1
p
1
x3 dx y 2
= ln(1 + x4 ) + C1 , 2y 2 = ln(1 + x4 ) + C, y = ± [ln(1 + x4 ) + C]/2
,
4
1+x 2
4
12. y dy =
13.
14.
1
+y
y
17.
18.
dy = ex dx, ln |y| + y 2 /2 = ex + C; by inspection, y = 0
2
2
dy
= −x dx, ln |y| = −x2 /2 + C1 , y = ±eC1 e−x /2 = Ce−x /2
y
15. ey dy =
16.
344
sin x
dx = sec x tan x dx, ey = sec x + C, y = ln(sec x + C)
cos2 x
dy
x2
+ C, y = tan(x + x2 /2 + C)
= (1 + x) dx, tan−1 y = x +
2
1+y
2
Z
Z
dy
dx
1
1
y−1
=
,
−
+
dy
=
csc x dx, ln
= ln | csc x − cot x| + C1 ,
y2 − y
sin x
y y−1
y
1
y−1
= ±eC1 (csc x − cot x) = C(csc x − cot x), y =
;
y
1 − C(csc x − cot x)
by inspection, y = 0 is also a solution.
1
3
cos y
dy =
dx,
dy = 3 cos x dx, ln | sin y| = 3 sin x + C1 ,
tan y
sec x
sin y
sin y = ±e3 sin x+C1 = ±eC1 e3 sin x = Ce3 sin x
R
19. µ = e
20. µ = e2
R
21. µ = e
22.
23.
24.
25.
Z
3dx
R
= e3x , e3x y =
x dx
2
= ex ,
2
2
2
d h x2 i
1 2
1
ye
= xex , yex = ex + C, y = + Ce−x
dx
2
2
Z
dx
ex dx = ex + C, y = e−2x + Ce−3x
= ex , ex y =
ex cos(ex )dx = sin(ex ) + C, y = e−x sin(ex ) + Ce−x
R
1
dy
+ 2y = , µ = e 2dx = e2x , e2x y =
dx
2
Z
1
1
1 2x
e dx = e2x + C, y = + Ce−2x
2
4
4
p
R
2
2
1
x
dy
+ 2
y = 0, µ = e (x/(x +1))dx = e 2 ln(x +1) = x2 + 1,
dx x + 1
i
p
C
d h p 2
y x + 1 = 0, y x2 + 1 = C, y = √
dx
x2 + 1
R
1
dy
+y =
, µ = e dx = ex , ex y =
x
dx
1+e
Z
ex
dx = ln(1 + ex ) + C, y = e−x ln(1 + ex ) + Ce−x
1 + ex
R
1
1
dy
d
+ y = 1, µ = e (1/x)dx = eln x = x,
[xy] = x, xy = x2 + C, y = x/2 + C/x
dx x
dx
2
3
1
(a) 2 = y(1) = + C, C = , y = x/2 + 3/(2x)
2
2
(b) 2 = y(−1) = −1/2 − C, C = −5/2, y = x/2 − 5/(2x)
345
26.
Chapter 10
2
2
x2
dy
= x dx, ln |y| =
+ C1 , y = ±eC1 ex /2 = Cex /2
y
2
2
(a) 1 = y(0) = C so C = 1, y = ex /2
1 2
1
= y(0) = C, so y = ex /2
(b)
2
2
Z
R
2
2
−x2 /2
− x dx
−x2 /2
27. µ = e
=e
,e
y = xe−x /2 dx = −e−x /2 + C,
2
R
28. µ = e
dt
/2
29. (y + cos y) dy = 4x2 dx,
C=
30.
2
, 3 = −1 + C, C = 4, y = −1 + 4ex /2
Z
t
t
= e , e y = 2et dt = 2et + C, y = 2 + Ce−t , 1 = 2 + C, C = −1, y = 2 − e−t
y = −1 + Cex
4
4
4
y2
π2
π2
+ sin y = x3 + C,
+ sin π = (1)3 + C,
= + C,
2
3
2
3
2
3
π2
4
− , 3y 2 + 6 sin y = 8x3 + 3π 2 − 8
2
3
1
dy
= (x + 2)ey , e−y dy = (x + 2)dx, −e−y = x2 + 2x + C, −1 = C,
dx
2
1 2
1 2
1 2
−y
−y
−e = x + 2x − 1, e = − x − 2x + 1, y = − ln 1 − 2x − x
2
2
2
31. 2(y − 1) dy = (2t + 1) dt, y 2 − 2y = t2 + t + C, 1 + 2 = C, C = 3, y 2 − 2y = t2 + t + 3
R
sinh x
y = cosh x, µ = e (sinh x/ cosh x)dx = eln cosh x = cosh x,
cosh x Z
Z
1
1
1
2
(cosh 2x + 1)dx = sinh 2x + x + C =
(cosh x)y = cosh x dx =
2
4
2
1
1
1
1
1
y = sinh x + x sech x + C sech x, = C, y = sinh x + x sech x +
2
2
4
2
2
32. y 0 +
33. (a)
dx
1
dy
=
, ln |y| = ln |x| + C1 , |y| = C|x|1/2 ;
y
2x
2
1
1
sinh x cosh x + x + C,
2
2
1
sech x
4
y
y = 2.5x 2
y = 2x 2
2
y = x2
by inspection y = 0 is also a solution.
(b) 2 = C(1)2 , C = 2, x = 2y 2
y=0
-2
x
2
y = –0.5x 2
-2
y = –3x 2
p
x2
y2
= − + C1 , y = ± C 2 − x2
34. (a) y dy = −x dx,
2
2
√
(b) y = 25 − x2
3
y
y = –1.5x 2
y = √ 9 – x2
y = √ 2.25 – x 2
y = √ 0.25 – x 2
x
-3
3
y = – √ 1 – x2
y = – √ 4 – x2
-3
y = – √ 6.25 – x 2
Exercise Set 10.1
35.
346
x dx
dy
=− 2
,
y
x +4
1
ln |y| = − ln(x2 + 4) + C1 ,
2
C
y=√
x2 + 4
36. y 0 + 2y = 3et , µ = e2
= e2t ,
100
C=2
C=1
C=0
-2
C=2
C=1
-2
dt
d 2t
ye = 3e3t , ye2t = e3t + C,
dt
y = et + Ce−2t
1.5
C=0
R
2
C = –1
C = –2
2
C = –1
C = –2
-100
-1
37. (1 − y ) dy = x dx,
2
y−
2
38.
x3
y3
=
+ C1 , x3 + y 3 − 3y = C
3
3
1
+y
y
yey
2
/2
dy = dx, ln |y| +
y2
= x + C1 ,
2
= ±eC1 ex = Cex
y
y
3
2
x
-5
5
x
-2
-3
2
-2
41.
dy
x2
= xey , e−y dy = x dx, −e−y =
+ C, x = 2 when y = 0 so −1 = 2 + C, C = −3, x2 + 2e−y = 6
dx
2
42.
3x2
dy
=
, 2y dy = 3x2 dx, y 2 = x3 + C, 1 = 1 + C, C = 0,
dx
2y
2
y 2 = x3 , y = x3/2 passes through (1, 1).
0
1.6
0
43.
dy
= rate in − rate out, where y is the amount of salt at time t,
dt
y
1
dy
1
dy
= (4)(2) −
(2) = 8 − y, so
+ y = 8 and y(0) = 25.
dt
50
25
dt
25
Z
R
t/25
t/25
(1/25)dt
µ=e
=e
,e
y = 8et/25 dt = 200et/25 + C,
y = 200 + Ce−t/25 , 25 = 200 + C, C = −175,
(a) y = 200 − 175e−t/25 oz
(b) when t = 25, y = 200 − 175e−1 ≈ 136 oz
347
44.
Chapter 10
y
1
dy
1
dy
= (5)(10) −
(10) = 50 − y, so
+ y = 50 and y(0) = 0.
dt
200
20
dt
20
Z
R 1
dt
t/20
t/20
t/20
µ = e 20 = e
,e
y = 50e
dt = 1000et/20 + C,
y = 1000 + Ce−t/20 , 0 = 1000 + C, C = −1000;
(a)
y = 1000 − 1000e−t/20 lb
(b)
when t = 30, y = 1000 − 1000e−1.5 ≈ 777 lb
45. The volume V of the (polluted) water is V (t) = 500 + (20 − 10)t = 500 + 10t;
if y(t) is the number of pounds of particulate matter in the water,
R dt
y
1
dy
1
dy
= 0 − 10 = −
y,
+
y = 0; µ = e 50+t = 50 + t;
then y(0) = 50, and
dt
V
50 + t
dt
50 + t
d
[(50 + t)y] = 0, (50 + t)y = C, 2500 = 50y(0) = C, y(t) = 2500/(50 + t).
dt
The tank reaches the point of overflowing when V = 500 + 10t = 1000, t = 50 min, so
y = 2500/(50 + 50) = 25 lb.
46. The volume of the lake (in gallons) is V = 246πr2 h = 246π(15)2 3 = 166, 050π. Let y(t) denote
y
dy
y
= 0 − 103
= −
lb/h and
the number of pounds of mercury salts at time t, then
dt
V
166.05π
dt
t
dy
= −
, ln y = −
+ C1 , y = Ce−t/(166.05π) , and
y0 = 10−5 V = 1.6605π lb;
y
166.05π
166.05π
C = y(0) = y0 = 1.6605π, y = 1.6605πe−t/(166.05π) .
t
1
2
3
4
5
6
7
8
9
10
11
12
y(t) 5.2066 5.1967 5.1867 5.1768 5.1669 5.1570 5.1471 5.1372 5.1274 5.1176 5.1078 5.0980
R
c
dv
d h ct/m i
gm ct/m
+ v = −g, µ = e(c/m) dt = ect/m ,
ve
e
+ C,
= −gect/m , vect/m = −
dt
m
dt
c
gm
gm
gm
gm −ct/m
gm
+Ce−ct/m , but v0 = v(0) = −
+C, C = v0 +
,v = −
+ v0 +
e
v=−
c
c
c
c
c
mg
with vτ and −ct/m with −gt/vτ in (23).
(b) Replace
c
vτ
(c) From part (b), s(t) = C − vτ t − (v0 + vτ ) e−gt/vτ ;
g
vτ
vτ
vτ
s0 = s(0) = C −(v0 +vτ ) , C = s0 +(v0 +vτ ) , s(t) = s0 −vτ t+ (v0 +vτ ) 1 − e−gt/vτ
g
g
g
47. (a)
48. Given m = 240, g = 32, vτ = mg/c: with a closed parachute vτ = 120 so c = 64, and with an open
parachute vτ = 24, c = 320.
(a) Let t denote time elapsed in seconds after the moment of the drop. From Exercise 47(b),
while the parachute is closed
v(t) = e−gt/vτ (v0 + vτ ) − vτ = e−32t/120 (0 + 120) − 120 = 120 e−4t/15 − 1 and thus
v(25) = 120 e−20/3 − 1 ≈ −119.85, so the parachutist is falling at a speed of 119.85 ft/s
120
120 1 − e−4t/15 ,
when the parachute opens. From Exercise 47(c), s(t) = s0 − 120t +
32
s(25) = 10000 − 120 · 25 + 450 1 − e−20/3 ≈ 7449.43 ft.
(b) If t denotes time elapsed after the parachute opens, then, by Exercise 47(c),
24
(−119.85 + 24) 1 − e−32t/24 = 0, with the solution (Newton’s
s(t) = 7449.43 − 24t +
32
Method) t = 310.42 s, so the sky diver is in the air for about 25 + 310 = 335 s.
Exercise Set 10.1
49.
348
R
R
V (t)
dI
d
V (t) Rt/L
+ I=
, µ = e(R/L) dt = eRt/L , (eRt/L I) =
e
,
dt
L
L
dt
L
Z
Z t
1 t
1
IeRt/L = I(0) +
V (u)eRu/L du, I(t) = I(0)e−Rt/L + e−Rt/L
V (u)eRu/L du.
L 0
L
0
t
Z t
1
6
6
(a) I(t) = e−5t/2
1 − e−5t/2 A.
12e5u/2 du = e−5t/2 e5u/2 =
4
5
5
0
0
(b)
lim I(t) =
t→+∞
6
A
5
50. From Exercise 49 and Endpaper Table #42,
t
Z t
1
e2u
(2 sin u − cos u)
3e2u sin u du = 15e−2t + e−2t
I(t) = 15e−2t + e−2t
3
5
0
0
1
1
= 15e−2t + (2 sin t − cos t) + e−2t .
5
5
ck
dv
=
− g, v = −c ln(m0 − kt) − gt + C; v = 0 when t = 0 so 0 = −c ln m0 + C,
dt
m0 − kt
m0
− gt.
C = c ln m0 , v = c ln m0 − c ln(m0 − kt) − gt = c ln
m0 − kt
(b) m0 − kt = 0.2m0 when t = 100 so
m0
v = 2500 ln
− 9.8(100) = 2500 ln 5 − 980 ≈ 3044 m/s.
0.2m0
51. (a)
dv dx
dv
dv
dv
dv
=
=
v so m
= mv .
dt
dx dt
dx
dt
dx
m
mv
2
dv = −dx,
ln(kv + mg) = −x + C; v = v0 when x = 0 so
(b)
kv 2 + mg
2k
52. (a) By the chain rule,
C=
m
m
m kv02 + mg
m
ln(kv02 + mg),
ln(kv 2 + mg) = −x +
ln(kv02 + mg), x =
ln
.
2k
2k
2k
2k kv 2 + mg
(c) x = xmax when v = 0 so
xmax =
3.56 × 10−3
(7.3 × 10−6 )(988)2 + (3.56 × 10−3 )(9.8)
m kv02 + mg
ln
=
ln
≈ 1298 m
2k
mg
2(7.3 × 10−6 )
(3.56 × 10−3 )(9.8)
√
√
dh
π
= −0.025 h, √ dh = −0.025dt, 2π h = −0.025t + C; h = 4 when
53. (a) A(h) = π(1)2 = π, π
dt
h
√
√
0.025
t, h ≈ (2 − 0.003979t)2 .
t = 0, so 4π = C, 2π h = −0.025t + 4π, h = 2 −
2π
(b) h = 0 when t ≈ 2/0.003979 ≈ 502.6 s ≈ 8.4 min.
h p
i
p
54. (a) A(h) = 6 2 4 − (h − 2)2 = 12 4h − h2 ,
p
√
√
dh
= −0.025 h, 12 4 − h dh = −0.025dt,
12 4h − h2
dt
−8(4 − h)3/2 = −0.025t + C; h = 4 when t = 0 so C = 0,
(4 − h)3/2 = (0.025/8)t, 4 − h = (0.025/8)2/3 t2/3 ,
h ≈ 4 − 0.021375t2/3 ft
(b) h = 0 when t =
8
(4 − 0)3/2 = 2560 s ≈ 42.7 min
0.025
2√4 − (h − 2)2
h−2
2
h
349
55.
56.
Chapter 10
1
dv
1
1
= −0.04v 2 , 2 dv = −0.04dt, − = −0.04t + C; v = 50 when t = 0 so −
= C,
dt
v
v
50
1
50
dx
dx
50
1
cm/s. But v =
so
=
, x = 25 ln(2t + 1) + C1 ; x = 0
− = −0.04t − , v =
v
50
2t + 1
dt
dt
2t + 1
when t = 0 so C1 = 0, x = 25 ln(2t + 1) cm.
√
√
1
dv
= −0.02 v, √ dv = −0.02dt, 2 v = −0.02t + C; v = 9 when t = 0 so 6 = C,
dt
v
√
dx
dx
so
= (3 − 0.01t)2 ,
2 v = −0.02t + 6, v = (3 − 0.01t)2 cm/s. But v =
dt
dt
100
100
(3 − 0.01t)3 + C1 ; x = 0 when t = 0 so C1 = 900, x = 900 −
(3 − 0.01t)3 cm.
x=−
3
3
2
dy
= − sin x + e−x , y(0) = 1.
dx
57. Differentiate to get
Z
58.
Z
h(y) dy =
h(y(x))y 0 (x) dx; since h(y)
dy
= g(x) it follows that
dx
Z
Z
h(y) dy =
g(x) dx.
EXERCISE SET 10.2
y
1.
y
2.
4
4
3
3
2
2
1
1
y
3.
y(0) = 2
2
y(0) = 1
x
x
1
4.
2
3
4
1
R
dy
+ y = 1, µ = e dx = ex ,
dx
d
[yex ] = ex ,
dx
yex = ex + C, y = 1 + Ce−x
R
dy
− 2y = −x, µ = e−2
dx
1
y = (2x + 1) + Ce2x
4
dx
2
3
4
y(0) = –1
-1
y
5.
10
y(1) = 1
y(–1) = 0
x
-2
(a) −1 = 1 + C, C = −2, y = 1 − 2e−x
(b) 1 = 1 + C, C = 0, y = 1
(c) 2 = 1 + C, C = 1, y = 1 + e−x
6.
5
x
= e−2x ,
y(0) = –1
-10
d −2x
1
ye
= −xe−2x , ye−2x = (2x + 1)e−2x + C,
dx
4
1
1
(2x + 1) + e2x−2
4
4
5
1
(b) −1 = 1/4 + C, C = −5/4, y = (2x + 1) − e2x
4
4
1
1
(c) 0 = −1/4 + Ce−2 , C = e2 /4, y = (2x + 1) + e2x+2
4
4
(a) 1 = 3/4 + Ce2 , C = 1/4e2 , y =
2
Exercise Set 10.2
350
(
7.
lim y = 1
8.
x→+∞
lim y =
x→+∞
+∞
if y0 ≥ 1/4
−∞,
if y0 < 1/4
9. (a) IV, since the slope is positive for x > 0 and negative for x < 0.
(b) VI, since the slope is positive for y > 0 and negative for y < 0.
(c) V, since the slope is always positive.
(d) II, since the slope changes sign when crossing the lines y = ±1.
(e) I, since the slope can be positive or negative in each quadrant but is not periodic.
(f )
III, since the slope is periodic in both x and y.
11. (a) y0 = 1,
yn+1 = yn + (xn + yn )(0.2) = (xn + 6yn )/5
n
xn
1
0.2
1.20
2
3
4
5
0.4
1.48
0.6
1.86
0.8
2.35
1.0
2.98
xn
0
0.2
y(xn)
abs. error
perc. error
1
0
0
1.24
0.04
0.4
1.58
0.10
0.6
2.04
0.19
0.8
2.65
0.30
1.0
3.44
0.46
3
7
9
11
13
yn
d −x
ye
= xe−x ,
dx
ye−x = −(x + 1)e−x + C, 1 = −1 + C,
C = 2, y = −(x + 1) + 2ex
(b) y 0 − y = x, µ = e−x ,
0
0
1
y
(c)
3
x
0.2
0.4
0.6
0.8
1
12. h = 0.1, yn+1 = (xn + 11yn )/10
n
xn
yn
0
0
1.00
1
0.1
1.10
2
3
4
5
6
7
8
0.2
1.22
0.3
1.36
0.4
1.53
0.5
1.72
0.6
1.94
0.7
2.20
0.8
2.49
9
0.9
10
1.0
2.82
3.19
In Exercise 11, y(1) ≈ 2.98; in Exercise 12, y(1) ≈ 3.19; the true solution is y(1) ≈ 3.44; so the
absolute errors are approximately 0.46 and 0.25 respectively.
13. y0 = 1, yn+1 = yn +
√
yn /2
y
9
n
xn
yn
0
0
1
1
0.5
1.50
2
1
2.11
3
1.5
2.84
4
2
3.68
5
2.5
4.64
6
3
5.72
7
3.5
6.91
8
4
8.23
x
1
2
3
4
351
Chapter 10
y
14. y0 = 1, yn+1 = yn + (xn − yn2 )/4
n
xn
yn
0
0
1
1
0.25
0.75
2
2
3
4
5
6
7
8
0.50
0.67
0.75
0.68
1.00
0.75
1.25
0.86
1.50
0.99
1.75
1.12
2.00
1.24
1.5
1
0.5
x
0.5
15. y0 = 1, yn+1 = yn +
n
tn
yn
0
0
1
1
0.5
1.42
2
1
1.92
3
1.5
2
y
1
sin yn
2
1.5
2.39
1
3
4
2
2.73
t
3
16. y0 = 0, yn+1 = yn + e−yn /10
n
tn
yn
0
0
0
1
0.1
0.10
y
1
2
3
4
5
6
7
8
9
10
0.2
0.19
0.3
0.27
0.4
0.35
0.5
0.42
0.6
0.49
0.7
0.55
0.8
0.60
0.9
0.66
1.0
0.71
t
1
17. h = 1/5, y0 = 1, yn+1 = yn +
n
tn
yn
0
0
1.00
1
0.2
1.06
2
0.4
0.90
18. (a) By inspection,
1
cos(2πn/5)
5
3
4
5
0.6
0.74
0.8
0.80
1.0
1.00
2
dy
= e−x and y(0) = 0.
dx
(b) yn+1 = yn + e−xn /20 = yn + e−(n/20) /20 and y20 = 0.7625. From a CAS, y(1) = 0.7468.
2
2
19. (b) y dy = −x dx, y 2 /2 = −x2 /2 + C1 , x2 + y 2 = C; if y(0) = 1 then C = 1 so y(1/2) =
√
3/2.
dy
1
√
√
(c) √ = dx, 2 y = x/2 + C, 2 = C,
20. (a) y0 = 1, yn+1 = yn + ( yn /2)h
y
2
√
h = 0.2 : yn+1 = yn + yn /10; y5 ≈ 1.5489
√
√
y = x/4 + 1, y = (x/4 + 1)2 ,
h = 0.1 : yn+1 = yn + yn /20; y10 ≈ 1.5556
√
y(1) = 25/16 = 1.5625
h = 0.05 : yn+1 = yn + yn /40; y20 ≈ 1.5590
Exercise Set 10.3
352
EXERCISE SET 10.3
1. (a)
dy
= ky 2 , y(0) = y0 , k > 0
dt
(b)
dy
= −ky 2 , y(0) = y0 , k > 0
dt
3. (a)
1
ds
= s
dt
2
(b)
d2 s
ds
=2
2
dt
dt
4. (a)
dv
= −2v 2
dt
(b)
d2 s
= −2
dt2
dy
= 0.01y, y0 = 10,000
dt
1
1
ln 2 ≈ 69.31 hr
(c) T = ln 2 =
k
0.01
ds
dt
2
(b) y = 10,000et/100
5. (a)
(d) 45,000 = 10,000et/100 ,
45,000
≈ 150.41 hr
t = 100 ln
10,000
1
1
ln 2 =
ln 2
T
20
dy
= ((ln 2)/20)y, y(0) = 1
(a)
dt
6. k =
(b) y(t) = et(ln 2)/20 = 2t/20
(d) 1,000,000 = 2t/20 ,
(c) y(120) = 26 = 64
t = 20
7. (a)
ln 106
≈ 398.63 min
ln 2
ln 2
1
dy
= −ky, y(0) = 5.0 × 107 ; 3.83 = T = ln 2, so k =
≈ 0.1810
dt
k
3.83
(b) y = 5.0 × 107 e−0.181t
(c) y(30) = 5.0 × 107 e−0.1810(30) ≈ 219,297
(d) y(t) = 0.1y0 = y0 e−kt , −kt = ln 0.1, t = −
8. (a) k =
ln 0.1
= 12.72 days
0.1810
1
dy
1
ln 2 =
ln 2 ≈ 0.0050, so
= −0.0050y, y0 = 10.
T
140
dt
(b) y = 10e−0.0050t
(c) 10 weeks = 70 days so y = 10e−0.35 ≈ 7 mg.
(d) 0.3y0 = y0 e−kt , t = −
ln 0.3
≈ 243.2 days
0.0050
9. 100e0.02t = 5000, e0.02t = 50, t =
1
ln 50 ≈ 196 days
0.02
1
ln 1.2. y = 20,000 when
10. y = 10,000ekt , but y = 12,000 when t = 10 so 10,000e10k = 12,000, k =
10
ln 2
ln 2
= 10
≈ 38, in the year 2025.
2 = ekt , t =
k
ln 1.2
3.5
1
1
≈ 0.2100, T = ln 2 ≈ 3.30 days
11. y(t) = y0 e−kt = 10.0e−kt , 3.5 = 10.0e−k(5) , k = − ln
5 10.0
k
353
12.
Chapter 10
dy
1
= y0 e−kt , 0.6y0 = y0 e−5k , k = − ln 0.6 ≈ 0.10
dt
5
ln 2
≈ 6.8 yr
(a) T =
k
y
(b) y(t) ≈ y0 e−0.10t ,
≈ e−0.10t , so e−0.10t × 100 percent will remain.
y0
13. (a) k =
ln 2
≈ 0.1386; y ≈ 2e0.1386t
5
(b) y(t) = 5e0.015t
1
ln 100 ≈ 0.5117,
9
≈ 0.5995, y ≈ 0.5995e0.5117t .
(c) y = y0 ekt , 1 = y0 ek , 100 = y0 e10k . Divide: 100 = e9k , k =
y ≈ y0 e0.5117t ; also y(1) = 1, so y0 = e−0.5117
(d) k =
14. (a) k =
ln 2
≈ 0.1386, 1 = y(1) ≈ y0 e0.1386 , y0 ≈ e−0.1386 ≈ 0.8706, y ≈ 0.8706e0.1386t
T
ln 2
≈ 0.1386, y ≈ 10e−0.1386t
T
(b) y = 10e−0.015t
(c) 100 = y0 e−k , 1 = y0 e−10k . Divide: e9k = 100, k =
y0 = e10k ≈ e5.117 ≈ 166.81, y = 166.81e−0.5117t .
(d) k =
1
ln 100 ≈ 0.5117;
9
ln 2
≈ 0.1386, 10 = y(1) ≈ y0 e−0.1386 , y0 ≈ 10e0.1386 ≈ 11.4866, y ≈ 11.4866e−0.1386t
T
16. (a) None; the half-life is independent of the initial amount.
(b) kT = ln 2, so T is inversely proportional to k.
ln 2
; and ln 2 ≈ 0.6931. If k is measured in percent, k 0 = 100k,
k
69.31
70
ln 2
≈
≈ 0.
then T =
0
k
k
k
17. (a) T =
(b) 70 yr
(c) 20 yr
(d) 7%
18. Let y = y0 ekt with y = y1 when t = t1 and y = 3y1 when t = t1 + T ; then y0 ekt1 = y1 (i) and
1
y0 ek(t1 +T ) = 3y1 (ii). Divide (ii) by (i) to get ekT = 3, T = ln 3.
k
19. From (12), y(t) = y0 e−0.000121t . If 0.27 =
and if 0.30 =
ln 0.30
y(t)
≈ 9950, or roughly between 9000 B.C. and 8000 B.C.
then t = −
y0
0.000121
1
20. (a)
y(t)
ln 0.27
≈ 10,820 yrs,
= e−0.000121t then t = −
y0
0.000121
(b) t = 1988 yields
y/y0 = e−0.000121(1988) ≈ 79%.
0
50000
0
21. y0 ≈ 2, L ≈ 8; since the curve y =
6e−2k = 2, k =
1
ln 3 ≈ 0.5493.
2
2·8
16
passes through the point (2, 4), 4 =
,
2 + 6e−kt
2 + 6e−2k
Exercise Set 10.3
354
400,000
passes through the point (200, 600),
400 + 600e−kt
1
800
, k=
ln 2.25 ≈ 0.00405.
=
3
200
22. y0 ≈ 400, L ≈ 1000; since the curve y =
600 =
400,000
, 600e−200k
400 + 600e−200k
23. (a) y0 = 5
(b) L = 12
(d) L/2 = 6 =
(e)
60
, 5 + 7e−t = 10, t = − ln(5/7) ≈ 0.3365
5 + 7e−t
1
dy
=
y(12 − y), y(0) = 5
dt
12
24. (a) y0 = 1
(d) 750 =
(e)
(c) k = 1
(b) L = 1000
(c) k = 0.9
1
1000
ln(3 · 999) ≈ 8.8949
, 3(1 + 999e−0.9t ) = 4, t =
1 + 999e−0.9t
0.9
0.9
dy
=
y(1000 − y), y(0) = 1
dt
1000
25. See (13):
(a) L = 10
(b) k = 10
dy
= 10(1 − 0.1y)y = 25 − (y − 5)2 is maximized when y = 5.
dt
1
dy
= 50y 1 −
y ; from (13), k = 50, L = 50,000.
dt
50,000
(c)
26.
(a) L = 50,000
(c)
d
dy
is maximized when 0 =
dt
dy
dy
dt
(b) k = 50
= 50 − y/500, y = 25,000
27. Assume y(t) students have had the flu t days after semester break. Then y(0) = 20, y(5) = 35.
dy
= ky(L − y) = ky(1000 − y), y0 = 20
dt
1000
20000
=
;
(b) Part (a) has solution y =
−1000kt
20 + 980e
1 + 49e−1000kt
1000
1000
35 =
, k = 0.000115, y ≈
.
1 + 49e−5000k
1 + 49e−0.115t
(a)
(c)
t 0 1 2 3
y(t) 20 22 25 28
4 5 6 7 8 9 10 11 12 13 14
31 35 39 44 49 54 61 67 75 83 93
y
(d)
100
75
50
25
t
3
6
9
12
355
Chapter 10
28. (a)
dp
= −kh, p(0) = p0
dh
(b) p0 = 1, so p = e−kh , but p = 0.83 when h = 5000 thus e−5000k = 0.83,
k=−
29. (a)
ln 0.83
≈ 0.0000373, p ≈ e−0.0000373h atm.
5000
dT
dT
= −k(T − 21), T (0) = 95,
= −k dt, ln(T − 21) = −kt + C1 ,
dt
T − 21
T = 21 + eC1 e−kt = 21 + Ce−kt , 95 = T (0) = 21 + C, C = 74, T = 21 + 74e−kt
32
64
= − ln , T = 21 + 74et ln(32/37) = 21 + 74
(b) 85 = T (1) = 21 + 74e−k , k = − ln
74
37
t
30
ln(30/74)
32
T = 51 when
=
≈ 6.22 min
, t=
74
37
ln(32/37)
30.
32
37
t
,
dT
= k(70 − T ), T (0) = 40; − ln(70 − T ) = kt + C, 70 − T = e−kt e−C , T = 40 when t = 0, so
dt
5
70 − 52
= ln ≈ 0.5,
30 = e−C , T = 70 − 30e−kt ; 52 = T (1) = 70 − 30e−k , k = − ln
30
3
T ≈ 70 − 30e−0.5t
31. Let T denote the body temperature of McHam’s body at time t, the number of hours elapsed after
dT
dT
10:06 P.M.; then
= −k(T − 72),
= −kdt, ln(T − 72) = −kt + C, T = 72 + eC e−kt ,
dt
T − 72
3.6
77.9 = 72 + eC , eC = 5.9, T = 72 + 5.9e−kt , 75.6 = 72 + 5.9e−k , k = − ln
≈ 0.4940,
5.9
ln(26.6/5.9)
T = 72+5.9e−0.4940t . McHam’s body temperature was last 98.6◦ when t = −
≈ −3.05,
0.4940
so around 3 hours and 3 minutes before 10:06; the death took place at approximately 7:03 P.M.,
while Moore was on stage.
dT
dT
= k(Ta − T ) where k > 0. If T0 > Ta then
= −k(T − Ta ) where k > 0;
dt
dt
−kt
with k > 0.
both cases yield T (t) = Ta + (T0 − Ta )e
32. If T0 < Ta then
33. k/m = 0.25/1 = 0.25
(a) From (21), y = 0.3 cos(t/2)
y
(c)
(b) T = 2π · 2 = 4π s, f = 1/T = 1/(4π) Hz
(d) y = 0 at the equilibrium position,
so t/2 = π/2, t = π s.
0.3
(e)
t
6
c
i
t/2 = π at the maximum position below
the equlibrium position, so t = 2π s.
o
-0.3
34. 64 = w = −mg, m = 2, k/m = 0.25/2 = 1/8,
√
(a) From (21), y = cos(t/(2 2))
p
√
k/m = 1/(2 2)
r
√
√
m
= 2π(2 2) = 4π 2 s,
(b) T = 2π
k
√
f = 1/T = 1/(4π 2) Hz
Exercise Set 10.3
356
y
(c)
(d) y = 0 at
√ the equilibrium√position,
so t/(2 2) = π/2, t = π 2 s
√
√
(e) t/(2 2) = π, t = 2π 2 s
1
t
2p
6p
10p
-1
35. l = 0.05, k/m = g/l = 9.8/0.05 = 196 s−2
p
(b) T = 2π m/k = 2π/14 = π/7 s,
f = 7/π Hz
(d) 14t = π/2, t = π/28 s
(a) From (21), y = −0.12 cos 14t.
y
(c)
0.15
(e)
14t = π, t = π/14 s
t
π
7
2π
7
-0.15
36. l = 0.5, k/m = g/l = 32/0.5 = 64,
p
k/m = 8
p
(b) T = 2π m/k = 2π/8 = π/4 s;
f = 1/T = 4/π Hz
(a) From (21), y = −1.5 cos 8t.
(c)
y
(d) 8t = π/2, t = π/16 s
2
(e)
8t = π, t = π/8 s
t
3
6
-2
r
37. Assume y = y0 cos
dy
k
t, so v =
= −y0
m
dt
r
r
k
sin
m
r
k
t
m
r
k
k
(a) The maximum speed occurs when sin
t = ±1,
t = nπ + π/2,
m
m
r
k
t = 0, y = 0.
so cos
m
r
r
r
k
k
k
t = 0,
t = nπ, so cos
t = ±1, y = ±y0 .
(b) The minimum speed occurs when sin
m
m
m
r
4π 2
4π 2 w
4π 2 w + 4
4π 2 w
m
, k = 2 m = 2 , so k =
=
, 25w = 9(w + 4),
k
T
T g
g 9
g
25
4π 2 1
π2
4π 2 w
9
=
=
25w = 9w + 36, w = , k =
g 9
32 4
32
4
38. (a) T = 2π
(b) From part (a), w =
1
4
357
Chapter 10
39. By Hooke’s Law, F (t) = −kx(t), since the only force is the restoring force of the spring. Newton’s
Second Law gives F (t) = mx00 (t), so mx00 (t) + kx(t) = 0, x(0) = x0 , x0 (0) = 0.
r
0
40. 0 = v(0) = y (0) = c2
k
, so c2 = 0; y0 = y(0) = c1 , so y = y0 cos
m
r
k
t.
m
41. (a) y = y0 bt = y0 et ln b = y0 ekt with k = ln b > 0 since b > 1.
(b) y = y0 bt = y0 et ln b = y0 e−kt with k = − ln b > 0 since 0 < b < 1.
(d) y = 4(0.5t ) = 4et ln 0.5 = 4e−t ln 2
(c) y = 4(2t ) = 4et ln 2
42. If y = y0 ekt and y = y1 = y0 ekt1 then y1 /y0 = ekt1 , k =
y = y1 = e−kt1 then y1 /y0 = e−kt1 , k = −
ln(y1 /y0 )
; if y = y0 e−kt and
t1
ln(y1 /y0 )
.
t1
CHAPTER 10 SUPPLEMENTARY EXERCISES
4. The differential equation in part (c) is not separable; the others are.
5. (a)
linear
(b)
6. IF: µ = e−2x ,
2
Sep of var:
linear and separable
(c)
separable
(d)
neither
2
2
2
2
d h −2x2 i
1
1
ye
= xe−2x , ye−2x = − e−2x + C, y = − + Ce2x
dx
4
4
2
2
2
1
x2
dy
1
= x dx, ln |4y + 1| =
+ C1 , 4y + 1 = ±e4C1 e2x = C2 e2x ; y = − + Ce2x
4y + 1
4
2
4
7. The parabola ky(L − y) opens down and has its maximum midway between the y-intercepts, that
dy
1
= k(L/2)2 = kL2 /4.
is, at the point y = (0 + L) = L/2, where
2
dt
8. (a) If y = y0 ekt , then y1 = y0 ekt1 , y2 = y0 ekt2 , divide: y2 /y1 = ek(t2 −t1 ) , k =
T =
ln 2
(t2 − t1 ) ln 2
=
. If y = y0 e−kt , then y1 = y0 e−kt1 , y2 = y0 e−kt2 ,
k
ln(y2 /y1 )
y2 /y1 = e−k(t2 −t1 ) , k = −
(t2 − t1 ) ln 2
1
ln 2
=−
.
ln(y2 /y1 ), T =
t2 − t1
k
ln(y2 /y1 )
In either case, T is positive, so T =
(t2 − t1 ) ln 2
.
ln(y2 /y1 )
(b) In part (a) assume t2 = t1 + 1 and y2 = 1.25y1 . Then T =
9.
1
ln(y2 /y1 ),
t2 − t1
ln 2
≈ 3.1 h.
ln 1.25
4π 3 dV
dV
dr
= −kS; but V =
r ,
= 4πr2 , S = 4πr2 , so dr/dt = −k, r = −kt + C, 4 = C,
dt
3
dt
dt
r = −kt + 4, 3 = −k + 4, k = 1, r = 4 − t m.
Chapter 10 Supplementary Exercises
358
10. Assume the tank contains y(t) oz of salt at time t. Then y0 = 0 and for 0 < t < 15,
y
dy
= 5 · 10 −
10 = (50 − y/100) oz/min, with solution y = 5000 + Ce−t/100 . But y(0) = 0 so
dt
1000
C = −5000, y = 5000(1 − e−t/100 ) for 0 ≤ t ≤ 15, and y(15) = 5000(1 − e−0.15 ). For 15 < t < 30,
y
dy
= 0−
5, y = C1 e−t/200 , C1 e−0.075 = y(15) = 5000(1−e−0.15 ), C1 = 5000(e0.075 −e−0.075 ),
dt
1000
y = 5000(e0.075 − e−0.075 )e−t/100 , y(30) = 5000(e0.075 − e−0.075 )e−0.3 ≈ 556.13 oz.
11. (a) Assume the air contains y(t) ft3 of carbon monoxide at time t. Then y0 = 0 and for
y
d h t/12000 i
1 t/12000
dy
= 0.04(0.1) −
(0.1) = 1/250 − y/12000,
ye
e
,
=
t > 0,
dt
1200
dt
250
yet/12000 = 48et/12000 + C, y(0) = 0, C = −48; y = 48(1 − e−t/12000 ). Thus the percentage
y
100 = 4(1 − e−t/12000 ) percent.
of carbon monoxide is P =
1200
(b) 0.012 = 4(1 − e−t/12000 ), t = 35.95 min
12.
dy
= dx, tan−1 y = x + C, π/4 = C; y = tan(x + π/4)
y2 + 1
13.
14.
1
1
+
y5
y
dy =
dx
1
1
, − y −4 + ln |y| = ln |x| + C; − = C, y −4 + 4 ln(x/y) = 1
x
4
4
R
2
d 2
dy
+ y = 4x, µ = e (2/x)dx = x2 ,
yx = 4x3 , yx2 = x4 + C, y = x2 + Cx−2 ,
dx x
dx
2 = y(1) = 1 + C, C = 1, y = x2 + 1/x2
π
π
dy
1
= 4 sec2 2x dx, − = 2 tan 2x + C, −1 = 2 tan 2 + C = 2 tan + C = 2 + C, C = −3,
y2
y
8
4
1
y=
3 − 2 tan 2x
dy
dy
1
y−3
1
16.
= dx,
= dx,
−
dy = dx, ln
= x + C1 ,
2
y − 5y + 6
(y − 3)(y − 2)
y−3 y−2
y−2
15.
3 − 2Cex
ln 2 − 3
3 ln 2 − 6 − (2 ln 2 − 6)ex
y−3
= Cex ; y = ln 2 if x = 0, so C =
; y=
=
x
y−2
ln 2 − 2
1 − ce
ln 2 − 2 − (ln 2 − 3)ex
d −x
ye
= xe−x sin 3x,
dx
Z
3
2
3
1
−x
−x
−x
e cos 3x + − x +
e−x sin 3x + C;
ye = xe sin 3x dx = − x −
10
50
10
25
53
3
3
2
1
53
3
, y = − x−
cos 3x + − x +
sin 3x + ex
1 = y(0) = − + C, C =
50
50
10
50
10
25
50
17. (a) µ = e−
R
dx
= e−x ,
y
(c)
4
x
-10
-2
-2
359
Chapter 10
ln 2
≈ 0.00012182; T2 = 5730 + 40 = 5770, k2 ≈ 0.00012013.
T1
1
y
1
= 684.5, 595.7; t2 = − ln(y/y0 ) = 694.1, 604.1; in
With y/y0 = 0.92, 0.93, t1 = − ln
k1 y0
k2
1988 the shroud was at most 695 years old, which places its creation in or after the year 1293.
19. (a) Let T1 = 5730 − 40 = 5690, k1 =
(b) If the true half-life is T with decay rate k and solution y(t) = y0 e−kt , and if the half-life is
taken to be T1 = T (1 + r/100) with decay rate k1 and solution y1 (t) = y0 e−k1 t , then
k
100k
r
ln 2
ln 2
100k
=
=
; k − k1 = k −
=k
and the
=
k1 =
T1
T (1 + r/100)
1 + r/100
100 + r
100 + r
100 + r
y1 − y
= 100 e(k−k1 )t − 1 = 100 ektr/(100+r) − 1 percent.
y
percentage error is given by 100
20. (a) yn+1 = yn + 0.1(1 + 5tn − yn ), y0 = 5
n
tn
yn
0
1
5.00
1
1.1
5.10
2
3
4
5
6
7
8
9
1.2
5.24
1.3
5.42
1.4
5.62
1.5
5.86
1.6
6.13
1.7
6.41
1.8
6.72
1.9
7.05
10
2
7.39
(b) The true solution is y(t) = 5t − 4 + 4e1−t , so the percentage errors are given by
tn
yn
y(tn)
abs. error
rel. error (%)
1
5.00
5.00
0.00
0.00
1.1
5.10
5.12
0.02
0.38
1.2
5.24
5.27
0.03
0.66
1.3
5.42
5.46
0.05
0.87
1.4
5.62
5.68
0.06
1.00
1.5
5.86
5.93
0.06
1.08
1.6
6.13
6.20
0.07
1.12
1.7
6.41
6.49
0.07
1.13
1.8
6.72
6.80
0.08
1.11
1.9
7.05
7.13
0.08
1.07
2
7.39
7.47
0.08
1.03
21. (b) y = C1 ex + C2 e−x
(c) 1 = y(0) = C1 + C2 , 1 = y 0 (0) = C1 − C2 ; C2 = 0, C1 = 1, y = ex
22. (a) 2ydy = dx, y 2 = x + C; if y(0) = 1 then C = 1, y 2 = x + 1, y =
√
C = 1, y 2 = x + 1, y = − x + 1.
y
1
x
1
-1
(b)
x + 1; if y(0) = −1 then
y
1
-1
√
x
-1
1
-1
dy
1
= −2x dx, − = −x2 + C, −1 = C, y = 1/(x2 + 1)
y2
y
y
1
x
-1
1
1805
1805
, 25 = y(1) =
,
−kt
19 + 76e
19 + 76e−k
y0 L
5
95
, t ≈ 7.77 yr.
k ≈ 0.3567; when 0.8L = y(t) =
, 19 + 76e−kt = y0 =
19 + 76e−kt
4
4
23. (a) Use (15) in Section 10.3 with y0 = 19, L = 95: y(t) =
Chapter 10 Supplementary Exercises
360
y
dy
=k 1−
y, y(0) = 19.
dt
95
r
r
r
r
r
k
m
k
m
k
0
24. (a) y0 = y(0) = c1 , v0 = y (0) = c2
, c2 =
v0 , y = y0 cos
t + v0
sin
t
m
k
m
k
m
(b) From (13),
(b) l = 0.5, k/m = g/l = 9.8/0.5 = 19.6,
√
√
1
y = − cos( 19.6 t) + 0.25 √
sin( 19.6 t)
19.6
1.1
0
3.1
-1.1
√
√
√
0.25
19.6 sin 19.6 t + 0.25 cos 19.6 t = 0 when tan 19.6 t = − √
, so
19.6
√
√
√
0.25
19.6
, cos 19.6 t = ∓ √
,
sin 19.6 t = ± √
19.6625
19.6625
√
√
0.25
0.25
19.6
19.6625
√
+√
= √
|y(t)| = √
19.6625
19.6 19.6625
19.6
(c) v = y 0 (t) =
√
≈ 1.0016 m is the maximum displacement.
r
25. y = y0 cos
k
t, T = 2π
m
r
2πt
m
, y = y0 cos
k
T
2π
2πt
y0 sin
has maximum magnitude 2π|y0 |/T and occurs when
T
T
2πt/T = nπ + π/2, y = y0 cos(nπ + π/2) = 0.
(a) v = y 0 (t) = −
4π 2
2πt
has maximum magnitude 4π 2 |y0 |/T 2 and occurs when
y0 cos
2
T
T
2πt/T = jπ, y = y0 cos jπ = ±y0 .
(b) a = h00 (t) = −
26. (a) In t years the interest will be compounded nt times at an interest rate of r/n each time. The
value at the end of 1 interval is P + (r/n)P = P (1 + r/n), at the end of 2 intervals it is
P (1 + r/n) + (r/n)P (1 + r/n) = P (1 + r/n)2 , and continuing in this fashion the value at the
end of nt intervals is P (1 + r/n)nt .
(b) Let x = r/n, then n = r/x and
lim P (1 + r/n)nt = lim+ P (1 + x)rt/x = lim+ P [(1 + x)1/x ]rt = P ert .
n→+∞
x→0
x→0
(c) The rate of increase is dA/dt = rP ert = rA.
27. (a) A = 1000e(0.08)(5) = 1000e0.4 ≈ $1, 491.82
(b) P e(0.08)(10) = 10, 000, P e0.8 = 10, 000, P = 10, 000e−0.8 ≈ $4, 493.29
(c) From (11) with k = r = 0.08, T = (ln 2)/0.08 ≈ 8.7 years.
CHAPTER 11
Infinite Series
EXERCISE SET 11.1
1. (a)
1
(b)
3n−1
(−1)n−1
3n−1
(c)
2. (a) (−r)n−1 ; (−r)n
2n − 1
2n
(d)
n2
π 1/(n+1)
(b) (−1)n+1 rn ; (−1)n rn+1
(b) 1, −1, 1, −1
3. (a) 2, 0, 2, 0
(c) 2(1 + (−1)n ); 2 + 2 cos nπ
(b) (2n − 1)!
4. (a) (2n)!
5. 1/3, 2/4, 3/5, 4/6, 5/7, . . .; lim
n→+∞
n
= 1, converges
n+2
n2
= +∞, diverges
n→+∞ 2n + 1
6. 1/3, 4/5, 9/7, 16/9, 25/11, . . .; lim
7. 2, 2, 2, 2, 2, . . .; lim 2 = 2, converges
n→+∞
1
1
1
1
8. ln 1, ln , ln , ln , ln , . . .; lim ln(1/n) = −∞, diverges
n→+∞
2
3
4
5
9.
ln 1 ln 2 ln 3 ln 4 ln 5
,
,
,
,
, . . .;
1
2
3
4
5
ln x
ln n
1
= lim
= 0 apply L’Hôpital’s Rule to
, converges
lim
n→+∞ n
n→+∞ n
x
10. sin π, 2 sin(π/2), 3 sin(π/3), 4 sin(π/4), 5 sin(π/5), . . .;
sin(π/n)
(−π/n2 ) cos(π/n)
= lim
= π, converges
n→+∞
n→+∞
1/n
−1/n2
lim n sin(π/n) = lim
n→+∞
11. 0, 2, 0, 2, 0, . . .; diverges
(−1)n+1
= 0, converges
n→+∞
n2
12. 1, −1/4, 1/9, −1/16, 1/25, . . .; lim
13. −1, 16/9, −54/28, 128/65, −250/126, . . .; diverges because odd-numbered terms approach
−2, even-numbered terms approach 2.
14. 1/2, 2/4, 3/8, 4/16, 5/32, . . .; lim
n→+∞
n
1
= 0, converges
= lim n
n
n→+∞
2
2 ln 2
15. 6/2, 12/8, 20/18, 30/32, 42/50, . . .; lim
n→+∞
1
(1 + 1/n)(1 + 2/n) = 1/2, converges
2
16. π/4, π 2 /42 , π 3 /43 , π 4 /44 , π 5 /45 , . . .; lim (π/4)n = 0, converges
n→+∞
17. cos(3), cos(3/2), cos(1), cos(3/4), cos(3/5), . . .; lim cos(3/n) = 1, converges
n→+∞
18. 0, −1, 0, 1, 0, . . .; diverges
361
Exercise Set 11.1
362
x2
= 0, so lim n2 e−n = 0, converges
x→+∞ ex
n→+∞
19. e−1 , 4e−2 , 9e−3 , 16e−4 , 25e−5 , . . .; lim x2 e−x = lim
x→+∞
√
√
√
√
10 − 2, 18 − 3, 28 − 4, 40 − 5, . . .;
p
3
3n
3
lim ( n2 + 3n − n) = lim √
= , converges
= lim p
2
n→+∞
n→+∞
n→+∞
2
n + 3n + n
1 + 3/n + 1
20. 1,
x+3
21. 2, (5/3) , (6/4) , (7/5) , (8/6) , . . .; let y =
x+1
2
3
4
ln
lim ln y = lim
x→+∞
x→+∞
x
5
, converges because
x+3
n
2x2
n+3
x + 1 = lim
= 2, so lim
= e2
x→+∞ (x + 1)(x + 3)
n→+∞ n + 1
1/x
22. −1, 0, (1/3)3 , (2/4)4 , (3/5)5 , . . .; let y = (1 − 2/x)x , converges because
lim ln y = lim
x→+∞
23.
24.
x→+∞
2n − 1
2n
n−1
n2
ln(1 − 2/x)
−2
= lim
= −2, lim (1 − 2/n)n = lim y = e−2
x→+∞
n→+∞
x→+∞
1/x
1 − 2/x
+∞
; lim
n→+∞
n=1
+∞
2n − 1
= 1, converges
2n
n−1
= 0, converges
n→+∞ n2
; lim
n=1
25.
1
3n
+∞
; lim
n=1
n→+∞
1
= 0, converges
3n
+∞
26. {(−1)n n}n=1 ; diverges because odd-numbered terms tend toward −∞,
even-numbered terms tend toward +∞.
27.
28.
29.
30.
32.
1
1
−
n n+1
3/2n−1
+∞
; lim
n=1
n→+∞
1
1
−
n n+1
+∞
; lim 3/2n−1
n=1 n→+∞
= 0, converges
= 0, converges
√
+∞
n + 1 − n + 2 n=1 ; converges because
√
√
(n + 1) − (n + 2)
−1
√
√
= lim √
=0
lim ( n + 1 − n + 2) = lim √
n→+∞
n→+∞
n + 1 + n + 2 n→+∞ n + 1 + n + 2
√
(−1)n+1 /3n+4
lim
n→+∞
√
n
+∞
; lim (−1)n+1 /3n+4
n=1 n→+∞
n = 1, so lim
n→+∞
√
n
n3 = 13 = 1
(
33. (a) 1, 2, 1, 4, 1, 6
= 0, converges
(b)
an =
n,
(
n odd
n
1/2 , n even
(c)
an =
1/n,
n odd
1/(n + 1), n even
(d) In part (a) the sequence diverges, since the even terms diverge to +∞ and the odd terms
equal 1; in part (b) the sequence diverges, since the odd terms diverge to +∞ and the even
terms tend to zero; in part (c) lim an = 0.
n→+∞
34. The even terms are zero, so the odd terms must converge to zero, and this is true if and only if
lim bn = 0, or −1 < b < 1.
n→+∞
363
35.
Chapter 11
1
1
√
(yn + p/yn ), L = (L + p/L), L2 = p, L = ± p;
lim yn+1 = lim
n→+∞ 2
2
√
√
√
L = − p (reject, because the terms in the sequence are positive) or L = p; lim yn = p.
n→+∞
n→+∞
36. (a) an+1 =
(b)
√
6 + an
lim an+1 = lim
n→+∞
√
n→+∞
√
6 + an , L =
6 + L, L2 − L − 6 = 0, (L − 3)(L + 2) = 0,
L = −2 (reject, because the terms in the sequence are positive) or L = 3; lim an = 3.
n→+∞
1 2 1 2 3 1
+ , + + ,
+
4 4 9 9 9 16
1
(c) an = 2 (1 + 2 + · · · + n) =
n
37. (a) 1,
2
3
4
3 2 5
+
+
= 1, , ,
16 16 16
4 3 8
1n+1
1 1
n(n + 1) =
, lim an = 1/2
n2 2
2 n n→+∞
1 4 1
4
9 1
4
9
16
5 14 15
+ ,
+
+ ,
+
+
+
= 1, , ,
8 8 27 27 27 64 64 64 64
8 27 32
1 (n + 1)(2n + 1)
1
1 1
,
(c) an = 3 (12 + 22 + · · · + n2 ) = 3 n(n + 1)(2n + 1) =
n
n 6
6
n2
1
(1 + 1/n)(2 + 1/n) = 1/3
lim an = lim
n→+∞
n→+∞ 6
38. (a) 1,
sin2 n
1
, cn = ; then an ≤ bn ≤ cn , lim an = lim cn = 0, so lim bn = 0.
n→+∞
n→+∞
n→+∞
n
n
n
n
n
1+n
3
n/2 + n
, cn =
; then (for n ≥ 2), an ≤ bn ≤
= cn ,
40. Let an = 0, bn =
2n
4
2n
lim an = lim cn = 0, so lim bn = 0.
39. Let an = 0, bn =
n→+∞
n→+∞
n→+∞
41. (a) a1 = (0.5)2 , a2 = a21 = (0.5)4 , . . . , an = (0.5)2
(c)
lim an = lim e2
n→+∞
n
ln(0.5)
n→+∞
n
= 0, since ln(0.5) < 0.
(d) Replace 0.5 in part (a) with a0 ; then the sequence converges for −1 ≤ a0 ≤ 1, because if
a0 = ±1, then an = 1 for n ≥ 1; if a0 = 0 then an = 0 for n ≥ 1; and if 0 < |a0 | < 1 then
n−1
ln a1
= 0 since 0 < a1 < 1.
a1 = a20 > 0 and lim an = lim e2
n→+∞
n→+∞
42. f (0.2) = 0.4, f (0.4) = 0.8, f (0.8) = 0.6, f (0.6) = 0.2 and then the cycle repeats, so the sequence
does not converge.
30
43. (a)
0
5
0
ln(2x + 3x )
2x ln 2 + 3x ln 3
= lim
x→+∞
x→+∞
x
2x + 3x
(b) Let y = (2x + 3x )1/x , lim ln y = lim
x→+∞
(2/3)x ln 2 + ln 3
= ln 3, so lim (2n + 3n )1/n = eln 3 = 3
x→+∞
n→+∞
(2/3)x + 1
= lim
Exercise Set 11.2
364
44. Let f (x) = 1/(1 + x), 0 ≤ x ≤ 1. Take ∆xk = 1/n and x∗k = k/n then
1
Z 1
n
n
X
X
1
1
1
(1/n) =
dx
=
ln(1
+
x)
∆x
so
lim
a
=
= ln 2
an =
k
n
n→+∞
1 + (k/n)
1 + x∗k
0 1+x
0
k=1
k=1
Z n
ln n
1
1
ln n
1
dx =
, lim an = lim
= lim
= 0,
45. an =
n→+∞
n→+∞
n→+∞
n−1 1 x
n−1
n−1
n
ln n
apply L’Hôpital’s Rule to
, converges
n−1
46. (a) If n ≥ 1, then an+2 = an+1 + an , so
an+2
an
=1+
.
an+1
an+1
(c) With L = lim (an+2 /an+1 ) = lim (an+1 /an ), L = 1 + 1/L, L2 − L − 1 = 0,
n→+∞
n→+∞
√
√
L = (1 ± 5)/2, so L = (1 + 5)/2 because the limit cannot be negative.
47.
1
1
− 0 = < if n > 1/
n
n
(a) 1/ = 1/0.5 = 2, N = 3
(b) 1/ = 1/0.1 = 10, N = 11
(c) 1/ = 1/0.001 = 1000, N = 1001
48.
1
n
−1 =
< if n + 1 > 1/, n > 1/ − 1
n+1
n+1
(a) 1/ − 1 = 1/0.25 − 1 = 3, N = 4
(b) 1/ − 1 = 1/0.1 − 1 = 9, N = 10
(c) 1/ − 1 = 1/0.001 − 1 = 999, N = 1000
49. (a)
(b)
1
1
− 0 = < if n > 1/, choose any N > 1/.
n
n
1
n
−1 =
< if n > 1/ − 1, choose any N > 1/ − 1.
n+1
n+1
50. If |r| < 1 then
lim rn = 0; if r > 1 then
n→+∞
lim rn = +∞, if r < −1 then rn oscillates between
n→+∞
positive and negative values that grow in magnitude so lim rn does not exist for |r| > 1; if r = 1
n→+∞
then lim 1n = 1; if r = −1 then (−1)n oscillates between −1 and 1 so lim (−1)n does not exist.
n→+∞
n→+∞
EXERCISE SET 11.2
1. an+1 − an =
1
1
1
− =−
< 0 for n ≥ 1, so strictly decreasing.
n+1 n
n(n + 1)
2. an+1 − an = (1 −
3. an+1 − an =
1
1
1
) − (1 − ) =
> 0 for n ≥ 1, so strictly increasing.
n+1
n
n(n + 1)
n
1
n+1
−
=
> 0 for n ≥ 1, so strictly increasing.
2n + 3 2n + 1
(2n + 1)(2n + 3)
365
Chapter 11
4. an+1 − an =
n
1
n+1
−
=−
< 0 for n ≥ 1, so strictly decreasing.
4n + 3 4n − 1
(4n − 1)(4n + 3)
5. an+1 − an = (n + 1 − 2n+1 ) − (n − 2n ) = 1 − 2n < 0 for n ≥ 1, so strictly decreasing.
6. an+1 − an = [(n + 1) − (n + 1)2 ] − (n − n2 ) = −2n < 0 for n ≥ 1, so strictly decreasing.
7.
(n + 1)(2n + 1)
2n2 + 3n + 1
(n + 1)/(2n + 3)
an+1
=
=
> 1 for n ≥ 1, so strictly increasing.
=
an
n/(2n + 1)
n(2n + 3)
2n2 + 3n
8.
an+1
2n+1
1 + 2n
2 + 2n+1
1
=
·
=
=1+
> 1 for n ≥ 1, so strictly increasing.
n+1
n
an
1+2
2
1 + 2n+1
1 + 2n+1
9.
an+1
(n + 1)e−(n+1)
=
= (1 + 1/n)e−1 < 1 for n ≥ 1, so strictly decreasing.
an
ne−n
10.
an+1
(2n)!
10n+1
10
·
< 1 for n ≥ 1, so strictly decreasing.
=
=
an
(2n + 2)! 10n
(2n + 2)(2n + 1)
11.
an+1
(n + 1)n+1 n!
(n + 1)n
· n =
=
= (1 + 1/n)n > 1 for n ≥ 1, so strictly increasing.
an
(n + 1)!
n
nn
12.
an+1
5n+1
2n
5
= (n+1)2 · n = 2n+1 < 1 for n ≥ 1, so strictly decreasing.
an
5
2
2
2
13. f (x) = x/(2x + 1), f 0 (x) = 1/(2x + 1)2 > 0 for x ≥ 1, so strictly increasing.
14. f (x) = 3 − 1/x, f 0 (x) = 1/x2 > 0 for x ≥ 1, so strictly increasing.
15. f (x) = 1/(x + ln x), f 0 (x) = −
1 + 1/x
< 0 for x ≥ 1, so strictly decreasing.
(x + ln x)2
16. f (x) = xe−2x , f 0 (x) = (1 − 2x)e−2x < 0 for x ≥ 1, so strictly decreasing.
17. f (x) =
1 − ln(x + 2)
ln(x + 2) 0
, f (x) =
< 0 for x ≥ 1, so strictly decreasing.
x+2
(x + 2)2
18. f (x) = tan−1 x, f 0 (x) = 1/(1 + x2 ) > 0 for x ≥ 1, so strictly increasing.
19. f (x) = 2x2 − 7x, f 0 (x) = 4x − 7 > 0 for x ≥ 2, so eventually strictly increasing.
20. f (x) = x3 − 4x2 , f 0 (x) = 3x2 − 8x = x(3x − 8) > 0 for x ≥ 3, so eventually strictly increasing.
21. f (x) =
10 − x2
x
0
,
f
(x)
=
< 0 for x ≥ 4, so eventually strictly decreasing.
x2 + 10
(x2 + 10)2
22. f (x) = x +
23.
17 0
x2 − 17
, f (x) =
> 0 for x ≥ 5, so eventually strictly increasing.
x
x2
an+1
n+1
(n + 1)! 3n
=
> 1 for n ≥ 3, so eventually strictly increasing.
=
·
an
3n+1
n!
3
24. f (x) = x5 e−x , f 0 (x) = x4 (5 − x)e−x < 0 for x ≥ 6, so eventually strictly decreasing.
Exercise Set 11.2
366
25. (a) Yes: a monotone sequence is increasing or decreasing; if it is increasing, then it is increasing and bounded above, so by Theorem 11.2.3 it converges; if decreasing, then use Theorem 11.2.4. The limit lies in the interval [1, 2].
(b) Such a sequence may converge, in which case, by the argument in Part (a), its limit is ≤ 2.
But convergence may not happen: for example, the sequence {−n}+∞
n=1 diverges.
26. (a) an+1 =
|x| |x|n
|x|
|x|n+1
=
=
an
(n + 1)!
n + 1 n!
n+1
(b) an+1 /an = |x|/(n + 1) < 1 if n > |x| − 1.
(c) From Part (b) the sequence is eventually decreasing, and it is bounded below by 0, so by
Theorem 11.2.4 it converges.
|x|
L = 0.
(d) If lim an = L then from Part (a), L =
n→+∞
lim (n + 1)
n→+∞
|x|
= lim an = 0
n→+∞
n!
r
q
q
√
√
√
2, 2 + 2, 2 + 2 + 2
27. (a)
√
√
√
√
√
(b) a1 = 2 < 2 so a2 = 2 + a1 < 2 + 2 = 2, a3 = 2 + a2 < 2 + 2 = 2, and so on
indefinitely.
n
(e)
lim
n→+∞
(c) a2n+1 − a2n = (2 + an ) − a2n = 2 + an − a2n = (2 − an )(1 + an )
(d) an > 0 and, from Part (b), an < 2 so 2 − an > 0 and 1 + an > 0 thus, from Part (c),
a2n+1 − a2n > 0, an+1 − an > 0, an+1 > an ; {an } is a strictly increasing sequence.
(e) The sequence is increasing and has 2 as an upper bound so it must converge to a limit L,
√
√
2 + an , L =
2 + L, L2 − L − 2 = 0, (L − 2)(L + 1) = 0
lim an+1 = lim
n→+∞
n→+∞
thus lim an = 2.
n→+∞
√
28. (a) If f (x) = 12 (x + 3/x), then f 0 (x) = (x2 − 3)/(2x2 ) and f 0 (x) = 0 for x = 3; the minimum
√
√
√
√
value of f (x) for x > 0 is f ( 3) = 3. Thus f (x) ≥ 3 for x > 0 and hence an ≥ 3 for
n ≥ 2.
√
(b) an+1 − an = (3 − a2n )/(2an ) ≤ 0 for n ≥ 2 since an ≥ 3 for n ≥ 2; {an } is eventually
decreasing.
√
3 is a lower bound for an so {an } converges; lim an+1 = lim 12 (an + 3/an ),
(c)
n→+∞
n→+∞
√
L = 12 (L + 3/L), L2 − 3 = 0, L = 3.
29. (a) The altitudes of the rectangles are ln k for k = 2 to n, and their bases all have length 1 so
the sum of their areas is ln 2 + ln 3 + · · · + ln n = ln(2 · 3 · · · n) = ln n!. The area under the
Z n+1
Z n
ln x dx, and
ln x dx is the area for x in
curve y = ln x for x in the interval [1, n] is
1Z
Z n1
n+1
the interval [1, n + 1] so, from the figure,
ln x dx < ln n! <
ln x dx.
1
Z
n
n
ln x dx = (x ln x − x)
(b)
1
1
Z
= n ln n − n + 1 and
1
n+1
ln x dx = (n + 1) ln(n + 1) − n so from
1
Part (a), n ln n − n + 1 < ln n! < (n + 1) ln(n + 1) − n, en ln n−n+1 < n! < e(n+1) ln(n+1)−n ,
nn
(n + 1)n+1
en ln n e1−n < n! < e(n+1) ln(n+1) e−n , n−1 < n! <
e
en
367
Chapter 11
1/n
(n + 1)n+1
< n! <
,
en
√
n
√
1
(1 + 1/n)(n + 1)1/n
(n + 1)1+1/n
n!
n
, 1−1/n <
<
,
< n! <
e
n
e
e
nn
(c) From part (b), n−1
e
n
e1−1/n
but
30. n! >
1
e1−1/n
→
1/n
√
n
√
n
(1 + 1/n)(n + 1)1/n
1
1
1
n!
and
→ as n → +∞ (why?), so lim
= .
n→+∞ n
e
e
e
e
√
nn √
n
n
n
n
,
n!
>
,
lim
=
+∞
so
lim
n! = +∞.
n→+∞
en−1
e1−1/n n→+∞ e1−1/n
EXERCISE SET 11.3
1. (a) s1 = 2, s2 = 12/5, s3 =
lim sn =
n→+∞
5 5
62
312
2 − 2(1/5)n
, s4 =
sn =
= − (1/5)n ,
25
125
1 − 1/5
2 2
5
, converges
2
1 1
1
3
7
15
(1/4) − (1/4)2n
, s2 = , s3 = , s4 =
sn =
= − + (2n ),
4
4
4
4
1−2
4 4
lim sn = +∞, diverges
(b) s1 =
n→+∞
(c)
1
1
1
1
1
3
1
=
−
, s1 = , s2 = , s3 =
, s4 = ;
(k + 1)(k + 2)
k+1 k+2
6
4
10
3
1
1
1
, lim sn = , converges
sn = −
2 n + 2 n→+∞
2
2. (a) s1 = 1/4, s2 = 5/16, s3 = 21/64, s4 = 85/256
n−1 !
n
1
1
1
1 1 − (1/4)n
1
1
1 + + ··· +
=
1−
;
sn =
=
4
4
4
4 1 − 1/4
3
4
(b) s1 = 1, s2 = 5, s3 = 21, s4 = 85; sn =
4n − 1
, diverges
3
(c) s1 = 1/20, s2 = 1/12, s3 = 3/28, s4 = 1/8;
n
X
1
1
1
1
sn =
−
= −
, lim sn = 1/4
k+3 k+4
4 n + 4 n→+∞
k=1
3. geometric, a = 1, r = −3/4, sum =
1
= 4/7
1 − (−3/4)
4. geometric, a = (2/3)3 , r = 2/3, sum =
5. geometric, a = 7, r = −1/6, sum =
(2/3)3
= 8/9
1 − 2/3
7
=6
1 + 1/6
6. geometric, r = −3/2, diverges
7. sn =
n
X
k=1
1
1
−
k+2 k+3
=
1
1
−
, lim sn = 1/3
3 n + 3 n→+∞
lim sn =
n→+∞
1
3
Exercise Set 11.3
368
n
X
1
1
1
1
8. sn =
− k+1 = − n+1 , lim sn = 1/2
n→+∞
2k
2
2 2
k=1
n
X
1/3
1
1/3
1/3
9. sn =
−
= −
, lim sn = 1/6
3k − 1 3k + 2
6 3n + 2 n→+∞
k=1
10. sn =
n+1
X
k=2
=
11.
∞
X
k=3
"n+1
#
n+1
X 1
1/2
1 X 1
1/2
−
=
−
k−1 k+1
2
k−1
k+1
"n+1
1 X
2
k=2
#
k=2
k=2
n+3
X 1
1
1
1
1
1
−
=
1+ −
−
;
k−1
k−1
2
2 n+1 n+2
k=4
∞
k=1
∞
X
4k+2
k=1
3
4
X
1
=
1/k, the harmonic series, so the series diverges.
k−2
12. geometric, a = (e/π)4 , r = e/π < 1, sum =
13.
lim sn =
n→+∞
7k−1
=
∞
X
k=1
e4
(e/π)4
= 3
1 − e/π
π (π − e)
k−1
64
4
= 448/3
64
; geometric, a = 64, r = 4/7, sum =
7
1 − 4/7
14. geometric, a = 125, r = 125/7, diverges
15. 0.4444 · · · = 0.4 + 0.04 + 0.004 + · · · =
0.4
= 4/9
1 − 0.1
16. 0.9999 · · · = 0.9 + 0.09 + 0.009 + · · · =
0.9
=1
1 − 0.1
17. 5.373737 · · · = 5 + 0.37 + 0.0037 + 0.000037 + · · · = 5 +
0.37
= 5 + 37/99 = 532/99
1 − 0.01
18. 0.159159159 · · · = 0.159 + 0.000159 + 0.000000159 + · · · =
0.159
= 159/999 = 53/333
1 − 0.001
19. 0.782178217821 · · · = 0.7821 + 0.00007821 + 0.000000007821 + · · · =
0.7821
7821
869
=
=
1 − 0.0001
9999
1111
20. 0.451141414 · · · = 0.451 + 0.00014 + 0.0000014 + 0.000000014 + · · · = 0.451 +
22. (a)
geometric; 18/5
(b)
geometric; diverges
(c)
∞
X
1
k=1
2
0.00014
44663
=
1 − 0.01
99000
1
1
−
2k − 1 2k + 1
3 3
3 3 3
3
· 10 + 2 · · · 10 + 2 · · · · 10 + · · ·
4
4 4
4 4 4
2
3
20(3/4)
3
3
3
+ 20
= 10 + 60 = 70 meters
+ 20
+ · · · = 10 +
= 10 + 20
4
4
4
1 − 3/4
23. d = 10 + 2 ·
= 1/2
369
Chapter 11
2
3 3
3
n
1
1
1
1
1
1
24. volume = 1 +
+
+ ··· +
+ ··· = 1 + +
+ ··· +
+ ···
2
4
2n
8
8
8
1
= 8/7
=
1 − (1/8)
3
2
3
n
1
= ln
25. (a) sn = ln + ln + ln + · · · + ln
2
3
4
n+1
lim sn = −∞, series diverges.
n
1 2 3
· · ···
2 3 4
n+1
= ln
1
= − ln(n + 1),
n+1
n→+∞
k+1
k−1
k
k2 − 1
(k − 1)(k + 1)
k−1
+ ln
= ln
− ln
,
= ln
= ln
2
2
k
k
k
k
k
k+1
n+1
X k−1
k
sn =
− ln
ln
k
k+1
k=2
2
2
3
3
4
n
n+1
1
+ ln − ln
+ ln − ln
+ · · · + ln
− ln
= ln − ln
2
3
3
4
4
5
n+1
n+2
(b) ln(1 − 1/k 2 ) = ln
= ln
26. (a)
∞
X
n+1
1
1
− ln
, lim sn = ln = − ln 2
2
n + 2 n→+∞
2
(−1)k xk = 1 − x + x2 − x3 + · · · =
k=0
(b)
∞
X
1
1
=
if | − x| < 1, |x| < 1, −1 < x < 1.
1 − (−x)
1+x
(x − 3)k = 1 + (x − 3) + (x − 3)2 + · · · =
k=0
(c)
∞
X
(−1)k x2k = 1−x2 +x4 −x6 +· · · =
k=0
1
1
=
if |x − 3| < 1, 2 < x < 4.
1 − (x − 3)
4−x
1
1
=
if |−x2 | < 1, |x| < 1, −1 < x < 1.
1 − (−x2 )
1 + x2
27. (a) Geometric series, a = x, r = −x2 . Converges for | − x2 | < 1, |x| < 1;
x
x
=
.
S=
2
1 − (−x )
1 + x2
(b) Geometric series, a = 1/x2 , r = 2/x. Converges for |2/x| < 1, |x| > 2;
S=
1
1/x2
= 2
.
1 − 2/x
x − 2x
(c) Geometric series, a = e−x , r = e−x . Converges for |e−x | < 1, e−x < 1, ex > 1, x > 0;
1
e−x
.
= x
S=
−x
1−e
e −1
√
√
√
√
1
1
k+1− k
k+1− k
√
,
28.
=√ −√
= √ √
k+1
k2 + k
k k+1
k
n
X
1
1
1
1
1
1
1
1
√ −√
= √ −√
sn =
+ √ −√
+ √ −√
k+1
1
2
2
3
3
4
k
k=1
1
1
1
=1− √
; lim sn = 1
+··· + √ − √
n
n+1
n + 1 n→+∞
29. sn = (1 − 1/3) + (1/2 − 1/4) + (1/3 − 1/5) + (1/4 − 1/6) + · · · + [1/n − 1/(n + 2)]
= (1 + 1/2 + 1/3 + · · · + 1/n) − (1/3 + 1/4 + 1/5 + · · · + 1/(n + 2))
= 3/2 − 1/(n + 1) − 1/(n + 2), lim sn = 3/2
n→+∞
Exercise Set 11.3
370
" n
#
n
n
X
1/2
1 X1 X 1
1/2
1
=
−
=
−
30. sn =
k(k + 2)
k
k+2
2
k
k+2
k=1
k=1
k=1
k=1
" n
#
n+2
1
1
1
1
1 X1 X1
3
−
=
1+ −
−
; lim sn =
=
n→+∞
2
k
k
2
2 n+1 n+2
4
n
X
k=1
k=3
" n
#
n
n
X
X
1/2
1 X 1
1
1
1/2
=
−
=
−
31. sn =
(2k − 1)(2k + 1)
2k − 1 2k + 1
2
2k − 1
2k + 1
k=1
k=1
k=1
k=1
" n
#
n+1
X 1
1
1
1 X 1
1
−
=
1−
; lim sn =
=
n→+∞
2
2k − 1
2k − 1
2
2n + 1
2
n
X
k=1
k=2
1
1
32. Geometric series, a = sin x, r = − sin x. Converges for | − sin x| < 1, | sin x| < 2,
2
2
sin x
2 sin x
so converges for all values of x. S =
=
.
1
2 + sin x
1 + sin x
2
33. a2 =
1
1
1
1
1
1
1
1
1
1
1
1
1
a1 + , a3 = a2 + = 2 a1 + 2 + , a4 = a3 + = 3 a1 + 3 + 2 + ,
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
1
1
a4 + = 4 a1 + 4 + 3 + 2 + , . . . , an = n−1 a1 + n−1 + n−2 + · · · + ,
2
2
2
2
2
2
2
2
2
2
2
∞
n
X 1
1/2
a1
lim an = lim n−1 +
=1
=0+
n→+∞
n→+∞ 2
2
1 − 1/2
n=1
a5 =
34. 0.a1 a2 · · · an 9999 · · · = 0.a1 a2 · · · an + 0.9 (10−n ) + 0.09 (10−n ) + · · ·
= 0.a1 a2 · · · an +
0.9 (10−n )
= 0.a1 a2 · · · an + 10−n
1 − 0.1
= 0.a1 a2 · · · (an + 1) = 0.a1 a2 · · · (an + 1) 0000 · · ·
35. The series converges to 1/(1 − x) only if −1 < x < 1.
36. P0 P1 = a sin θ,
P1 P2 = a sin θ cos θ,
P2 P3 = a sin θ cos2 θ,
P3 P4 = a sin θ cos3 θ, . . .
(see figure)
Each sum is a geometric series.
P1
a sin u cos u
P3
a sin u
u
a sin u cos2 u
a sin u cos3 u
u
u
P0
P2
P4
a
(a) P0 P1 + P1 P2 + P2 P3 + · · · = a sin θ + a sin θ cos θ + a sin θ cos2 θ + · · · =
(b) P0 P1 + P2 P3 + P4 P5 + · · · = a sin θ + a sin θ cos2 θ + a sin θ cos4 θ + · · ·
a sin θ
a sin θ
= a csc θ
=
=
2
1 − cos θ
sin2 θ
(c) P1 P2 + P3 P4 + P5 P6 + · · · = a sin θ cos θ + a sin θ cos3 θ + · · ·
a sin θ cos θ
a sin θ cos θ
= a cot θ
=
=
2
1 − cos θ
sin2 θ
P
a sin θ
1 − cos θ
371
Chapter 11
37. By inspection,
θ
θ/2
θ θ θ
− + −
+ ··· =
= θ/3
2 4 8 16
1 − (−1/2)
38. A1 + A2 + A3 + · · · = 1 + 1/2 + 1/4 + · · · =
39. (b)
1
=2
1 − (1/2)
2k 3k+1 − 2k+1 A + 2k 3k − 2k B
2k B
2k A
+
=
3k − 2k
3k+1 − 2k+1
(3k − 2k ) (3k+1 − 2k+1 )
3 · 6k − 2 · 22k A + 6k − 22k B
(3A + B)6k − (2A + B)22k
=
=
(3k − 2k ) (3k+1 − 2k+1 )
(3k − 2k ) (3k+1 − 2k+1 )
so 3A + B = 1 and 2A + B = 0, A = 1 and B = −2.
X
n
n
X
2k+1
2k
2k
(c) sn =
−
(ak − ak+1 ) where ak = k
.
=
k
k
k+1
k+1
3 −2
3
−2
3 − 2k
k=1
k=1
But sn = (a1 − a2 ) + (a2 − a3 ) + (a3 − a4 ) + · · · + (an − an+1 ) which is a telescoping sum,
2n+1
(2/3)n+1
,
lim
s
=
lim
2
−
= 2.
sn = a1 − an+1 = 2 − n+1
n
n→+∞
3
− 2n+1 n→+∞
1 − (2/3)n+1
EXERCISE SET 11.4
∞
X
1/2
1
= 1;
1. (a)
=
2k
1 − 1/2
∞
X
1/4
1
= 1/3;
=
4k
1 − 1/4
k=1
(b)
∞
X
k=1
k=1
1/5
1
= 1/4;
=
5k
1 − 1/5
∞
X
k=1
1
=1
k(k + 1)
∞
X
1
1
+ k = 1 + 1/3 = 4/3
2k
4
k=1
(Example 5, Section 11.3);
∞
X
1
1
= 1/4 − 1 = −3/4
−
5k
k(k + 1)
k=1
2. (a)
∞
X
k=2
so
∞
X
7/10
1
7
=
3/4
(Exercise
10,
Section
11.3);
= 7/9;
=
k2 − 1
10k−1
1 − 1/10
∞
X
k=2
k=2
7
1
−
= 3/4 − 7/9 = −1/36
k 2 − 1 10k−1
(b) with a = 9/7, r = 3/7, geometric,
with a = 4/5, r = 2/5, geometric,
∞
X
7−k 3k+1 −
k+1
2
k=1
5k
∞
X
k=1
∞
X
k=1
7−k 3k+1 =
9/7
= 9/4;
1 − (3/7)
4/5
2k+1
= 4/3;
=
k
5
1 − (2/5)
= 9/4 − 4/3 = 11/12
3. (a) p = 3, converges
(b) p = 1/2, diverges
(c) p = 1, diverges
(d) p = 2/3, diverges
4. (a) p = 4/3, converges
(b) p = 1/4, diverges
(c) p = 5/3, converges
(d) p = π, converges
Exercise Set 11.4
372
1
k2 + k + 3
= ; the series diverges.
5. (a)
lim
k→+∞ 2k 2 + 1
2
(b)
lim cos kπ does not exist;
(d)
(c)
k→+∞
lim
k→+∞
1
1+
k
k
= e; the series diverges.
1
= 0; no information
k→+∞ k!
lim
the series diverges.
k
= 0; no information
k→+∞ ek
6. (a)
(b)
lim
1
(c)
lim √ = 0; no information
k→+∞
k
Z
√
lim √
k→+∞
k
= 1; the series diverges.
k+3
`
+∞
1
1
= lim
ln(5x + 2)
`→+∞
5x + 2
5
7. (a)
1
Z
(d)
lim ln k = +∞; the series diverges.
k→+∞
= +∞, the series diverges by the Integral Test.
1
`
+∞
1
1
tan−1 3x
dx = lim
2
`→+∞ 3
1 + 9x
(b)
1
=
1
1
π/2 − tan−1 3 ,
3
the series converges by the Integral Test.
Z
`
+∞
x
1
ln(1 + x2 )
dx = lim
`→+∞ 2
1 + x2
8. (a)
1
Z
+∞
(b)
−3/2
(4 + 2x)
1
= +∞, the series diverges by the Integral Test.
1
`
√
dx = lim −1/ 4 + 2x
`→+∞
√
= 1/ 6,
1
the series converges by the Integral Test.
9.
∞
X
k=1
∞
X1
1
=
, diverges because the harmonic series diverges.
k+6
k
k=7
∞
∞
X
X
3
3 1
10.
=
, diverges because the harmonic series diverges.
5k
5 k
k=1
11.
∞
X
k=1
∞
√
k=1
12.
X 1
1
√ , diverges because the p-series with p = 1/2 ≤ 1 diverges.
=
k + 5 k=6 k
1
lim
k→+∞ e1/k
Z
= 1, the series diverges because lim uk = 1 6= 0.
+∞
−1/3
(2x − 1)
13.
1
14.
k→+∞
3
dx = lim (2x − 1)2/3
`→+∞ 4
ln x
is decreasing for x ≥ e, and
x
Z
+∞
3
`
= +∞, the series diverges by the Integral Test.
1
ln x
1
= lim (ln x)2
`→+∞ 2
x
`
= +∞,
3
so the series diverges by the Integral Test.
15.
lim
k→+∞
Z
+∞
k
1
= lim
= +∞, the series diverges because lim uk 6= 0.
k→+∞
ln(k + 1) k→+∞ 1/(k + 1)
−x2
xe
16.
1
2
1
dx = lim − e−x
`→+∞
2
`
1
= e−1 /2, the series converges by the Integral Test.
373
17.
18.
Chapter 11
lim (1 + 1/k)−k = 1/e 6= 0, the series diverges.
k→+∞
k2 + 1
= 1 6= 0, the series diverges.
k→+∞ k 2 + 3
lim
Z
2
tan−1 x
1
tan−1 x
dx = lim
`→+∞ 2
1 + x2
+∞
19.
1
`
= 3π 2 /32, the series converges by the Integral Test, since
1
1 − 2x tan−1 x
d tan−1 x
=
< 0 for x ≥ 1.
2
dx 1 + x
1 + x2
Z
+∞
√
20.
1
21.
`
1
x2 + 1
dx = lim sinh−1 x
`→+∞
= +∞, the series diverges by the Integral Test.
1
lim k 2 sin2 (1/k) = 1 6= 0, the series diverges.
k→+∞
Z
+∞
22.
2 −x3
x e
1
3
1
dx = lim − e−x
`→+∞
3
`
= e−1 /3,
1
the series converges by the Integral Test (x2 e−x is decreasing for x ≥ 1).
3
23. 7
∞
X
k −1.01 , p-series with p > 1, converges
k=5
Z
`
+∞
`→+∞
1
25.
= 1 − tanh(1), the series converges by the Integral Test.
sech2 x dx = lim tanh x
24.
1
Z +∞
1
dx
−p
is
decreasing
for
x
≥
e
,
so
use
the
Integral
Test
with
to get
p
p
x(ln x)
e−p x(ln x)
if p < 1
`
+∞
1−p `
(ln x)
= +∞ if p = 1,
lim
=
lim ln(ln x)
−1
`→+∞
`→+∞
1 − p e−p
if p > 1
e−p
(−1)p pp (p − 1)
Thus the series converges for p > 1.
26. Set g(x) = x(ln x)[ln(ln x)]p , g 0 (x) = (1 + ln x) ln(ln x) + p, so for fixed p there exists A > 0 such
Z +∞
dx
that g 0 (x) > 0, 1/g(x) is decreasing for x > A; use the Integral Test with
x(ln
x)[ln(ln
x)]p
A
to get
if p < 1,
`
+∞
1−p `
[ln(ln x)]
lim ln[ln(ln x)] = +∞ if p = 1, lim
=
1
`→+∞
`→+∞
1−p
if p > 1
A
A
(p − 1)[ln(ln A)]p−1
Thus the series converges for p > 1.
27. (a) 3
∞
∞
X
X
1
1
−
= π 2 /2 − π 4 /90
2
k
k4
k=1
(c)
∞
X
k=2
k=1
1
=
(k − 1)4
∞
X
k=1
1
= π 4 /90
k4
(b)
∞
X
1
1
− 1 − 2 = π 2 /6 − 5/4
2
k
2
k=1
Exercise Set 11.4
374
28. (a) Suppose Σ(uk + vk ) converges; then so does Σ[(uk + vk ) − uk ], but Σ[(uk + vk ) − uk ] = Σvk ,
so Σvk converges which contradicts the assumption that Σvk diverges. Suppose Σ(uk − vk )
converges; then so does Σ[uk − (uk − vk )] = Σvk which leads to the same contradiction as
before.
(b) Let uk = 2/k and vk = 1/k; then both Σ(uk + vk ) and Σ(uk − vk ) diverge; let uk = 1/k and
vk = −1/k then Σ(uk + vk ) converges; let uk = vk = 1/k then Σ(uk − vk ) converges.
29. (a) diverges because
∞
X
∞
X
(2/3)k−1 converges and
k=1
(b) diverges because
∞
X
∞
X
1/(3k + 2) diverges and
k=1
∞
X
k=2
∞
X
1/k 3/2 converges.
k=1
(c) converges because both
30. (a) If S =
1/k diverges.
k=1
uk and sn =
k=1
n
X
∞
X
1
(Exercise
25)
and
1/k 2 converge.
k(ln k)2
k=2
∞
X
uk , then S − sn =
k=1
uk . Interpret uk , k = n + 1, n + 2, . . ., as
k=n+1
the areas of inscribed or circumscribed rectangles with height uk and base of length one for
the curve y = f (x) to obtain the result.
n
X
(b) Add sn =
Z
uk to each term in the conclusion of part (a) to get the desired result:
k=1
+∞
sn +
f (x) dx <
n+1
+∞
X
Z
+∞
uk < sn +
31. (a) In Exercise 30 above let f (x) =
f (x) dx
n
k=1
1
. Then
x2
Z
+∞
f (x) dx = −
n
1
x
+∞
=
n
1
;
n
use this result and the same result with n + 1 replacing n to obtain the desired result.
(b) s3 = 1 + 1/4 + 1/9 = 49/36; 58/36 = s3 +
(d) 1/11 <
1
1
1
< π 2 < s3 + = 61/36
4
6
3
1 2
π − s10 < 1/10
6
1
32. Apply Exercise 30 with f (x) =
,
(2x + 1)2
∞
X
1
1
1
(a)
<
− s10 <
46
(2k + 1)2
42
Z
+∞
f (x) dx =
n
1
:
2(2n + 1)
k=1
(b) π/2 − tan−1 (11) <
∞
X
k=1
(c) 12e−11 <
∞
X
k=1
Z
+∞
33. (a)
n
(b)
1
− s10 < π/2 − tan−1 (10)
k2 + 1
k
< 11e−10
ek
1
1
dx = 2 ; use Exercise 30(b)
3
x
2n
1
1
−
< 0.01 for n = 5.
2
2n
2(n + 1)2
(c) From Part (a) with n = 5 obtain 1.1995 < S < 1.2057, so S ≈ 1.203.
375
Chapter 11
Z
+∞
34. (a)
n
1
1
1
1
dx = 3 ; choose n so that
−
< 0.005, n = 4; S ≈ 1.084
x4
3n
3n3
3(n + 1)3
Z n
Z n+1
1
1
1
dx = ln n and
dx = ln(n + 1), u1 = 1 so
35. (a) Let F (x) = , then
x
x
x
1
1
ln(n + 1) < sn < 1 + ln n.
(b) ln(1, 000, 001) < s1,000,000 < 1 + ln(1, 000, 000), 13 < s1,000,000 < 15
(c) s109 < 1 + ln 109 = 1 + 9 ln 10 < 22
(d) sn > ln(n + 1) ≥ 100, n ≥ e100 − 1 ≈ 2.688 × 1043 ; n = 2.69 × 1043
36. p-series with p = ln a; convergence for p > 1, a > e
37. x2 e−x is decreasing and positive for x > 2 so the Integral Test applies:
∞
Z ∞
x2 e−x dx = (x2 + 2x + 2)e−x
= 5e−1 so the series converges.
1
1
38. (a) f (x) = 1/(x3 + 1) is decreasing and continuous on the interval [1, +∞], so the Integral Test
applies.
(c)
n
10
20
30
40
50
sn 0.681980 0.685314 0.685966 0.686199 0.686307
n
60
70
80
90
100
sn 0.686367 0.686403 0.686426 0.686442 0.686454
Z
+∞
(e) Set g(n) =
n
√
√
1 n3 + 1
1
3
3
2n − 1
−1
√
dx
=
π
+
tan
−
; for n ≥ 10,
x3 + 1
6
6 (n + 1)3
3
3
g(n) − g(n + 1) ≤ 0.001; s10 + (g(10) + g(11))/2 ≈ 0.6865, so the sum ≈ 0.6865 to three
decimal places.
EXERCISE SET 11.5
1. (a) f (k) (x) = (−1)k e−x , f (k) (0) = (−1)k ; e−x ≈ 1 − x + x2 /2 (quadratic), e−x ≈ 1 − x (linear)
(b) f 0 (x) = − sin x, f 00 (x) = − cos x, f (0) = 1, f 0 (0) = 0, f 00 (0) = −1,
cos x ≈ 1 − x2 /2 (quadratic), cos x ≈ 1 (linear)
(c) f 0 (x) = cos x, f 00 (x) = − sin x, f (π/2) = 1, f 0 (π/2) = 0, f 00 (π/2) = −1,
sin x ≈ 1 − (x − π/2)2 /2 (quadratic), sin x ≈ 1 (linear)
(d) f (1) = 1, f 0 (1) = 1/2, f 00 (1) = −1/4;
√
√
1
1
1
x = 1 + (x − 1) − (x − 1)2 (quadratic), x ≈ 1 + (x − 1) (linear)
2
8
2
2. (a) p2 (x) = 1 + x + x2 /2, p1 (x) = 1 + x
1
1
1
(x − 9)2 , p1 (x) = 3 + (x − 9)
(b) p2 (x) = 3 + (x − 9) −
6
216
6
√
√
√
7
π
π
3
3
(x − 2) −
(x − 2)
3(x − 2)2 , p1 (x) = +
(c) p2 (x) = +
3
6
72
3
6
(d) p2 (x) = x, p1 (x) = x
Exercise Set 11.5
376
1 −1/2 00
1
1
1
x
, f (x) = − x−3/2 ; f (1) = 1, f 0 (1) = , f 00 (1) = − ;
2
4
2
4
√
1
1
x ≈ 1 + (x − 1) − (x − 1)2
2
8
√
1
1
(b) x = 1.1, x0 = 1, 1.1 ≈ 1 + (0.1) − (0.1)2 = 1.04875, calculator value ≈ 1.0488088
2
8
3. (a) f 0 (x) =
4. (a) cos x ≈ 1 − x2 /2
(b) 2◦ = π/90 rad, cos 2◦ = cos(π/90) ≈ 1 −
π2
≈ 0.99939077, calculator value ≈ 0.99939083
2 · 902
5. f (x) = tan x, 61◦ = π/3 + π/180 rad; x0 = π/3, f 0 (x) = sec2 x, f 00 (x) = 2 sec2 x tan x;
√
√
√
√
f (π/3) = 3, f 0 (π/3) = 4, f 00 (x) = 8 3; tan x ≈ 3 + 4(x − π/3) + 4 3(x − π/3)2 ,
√
√
tan 61◦ = tan(π/3 + π/180) ≈ 3 + 4π/180 + 4 3(π/180)2 ≈ 1.80397443,
calculator value ≈ 1.80404776
√
1 −1/2 00
1
x
, f (x) = − x−3/2 ;
2
4
1
1 00
1 √
1
, f (36) = −
; x ≈ 6 + (x − 36) −
(x − 36)2 ;
f (36) = 6, f 0 (36) =
12
864
12
1728
√
0.03 (0.03)2
−
≈ 6.00249947917, calculator value ≈ 6.00249947938
36.03 ≈ 6 +
12
1728
6. f (x) =
x, x0 = 36, f 0 (x) =
1
7. f (k) (x) = (−1)k e−x , f (k) (0) = (−1)k ; p0 (x) = 1, p1 (x) = 1 − x, p2 (x) = 1 − x + x2 ,
2
∞
X
1
1
1
1
1
(−1)k k
p3 (x) = 1 − x + x2 − x3 , p4 (x) = 1 − x + x2 − x3 + x4 ;
x
2
3!
2
3!
4!
k!
k=0
a2 2
x ,
2
∞
X
a2
a3
a2
a3
a4
ak k
p3 (x) = 1 + ax + x2 + x3 , p4 (x) = 1 + ax + x2 + x3 + x4 ;
x
2
3!
2
3!
4!
k!
8. f (k) (x) = ak eax , f (k) (0) = ak ; p0 (x) = 1, p1 (x) = 1 + ax, p2 (x) = 1 + ax +
k=0
9. f (k) (0) = 0 if k is odd, f (k) (0) is alternately π k and −π k if k is even; p0 (x) = 1, p1 (x) = 1,
∞
π2 2
π2 2
π 2 2 π 4 4 X (−1)k π 2k 2k
p2 (x) = 1 −
x ; p3 (x) = 1 −
x , p4 (x) = 1 −
x +
x ;
x
2!
2!
2!
4!
(2k)!
k=0
10. f (k) (0) = 0 if k is even, f (k) (0) is alternately π k and −π k if k is odd; p0 (x) = 0, p1 (x) = πx,
∞
π3 3
π 3 3 X (−1)k π 2k+1 2k+1
p2 (x) = πx; p3 (x) = πx −
x , p4 (x) = πx −
x ;
x
3!
3!
(2k + 1)!
k=0
(−1)k+1 (k − 1)! (k)
, f (0) = (−1)k+1 (k − 1)!; p0 (x) = 0,
(1 + x)k
∞
X
1
1
1
1
1
1
(−1)k+1 k
p1 (x) = x, p2 (x) = x − x2 , p3 (x) = x− x2 + x3 , p4 (x) = x− x2 + x3 − x4 ;
x
2
2
3
2
3
4
k
11. f (0) (0) = 0; for k ≥ 1, f (k) (x) =
k=1
12. f (k) (x) = (−1)k
k!
; f (k) (0) = (−1)k k!; p0 (x) = 1, p1 (x) = 1 − x,
(1 + x)k+1
p2 (x) = 1 − x + x2 , p3 (x) = 1 − x + x2 − x3 , p4 (x) = 1 − x + x2 − x3 + x4 ;
∞
X
(−1)k xk
k=0
377
Chapter 11
13. f (k) (0) = 0 if k is odd, f (k) (0) = 1 if k is even; p0 (x) = 1, p1 (x) = 1,
p2 (x) = 1 + x2 /2, p3 (x) = 1 + x2 /2, p4 (x) = 1 + x2 /2 + x4 /4!;
∞
X
k=0
1 2k
x
(2k)!
14. f (k) (0) = 0 if k is even, f (k) (0) = 1 if k is odd; p0 (x) = 0, p1 (x) = x, p2 (x) = x,
∞
X
1
p3 (x) = x + x3 /3!, p4 (x) = x + x3 /3!;
x2k+1
(2k + 1)!
k=0
(
15. f
(k)
(x) =
(−1)k/2 (x sin x − k cos x)
(
k even
(−1)(k−1)/2 (x cos x + k sin x) k odd
,
f
(k)
(0) =
(−1)1+k/2 k
k even
0
k odd
∞
X
1
(−1)k 2k+2
p0 (x) = 0, p1 (x) = 0, p2 (x) = x2 , p3 (x) = x2 , p4 (x) = x2 − x4 ;
x
6
(2k + 1)!
k=0
16. f (k) (x) = (k + x)ex , f (k) (0) = k; p0 (x) = 0, p1 (x) = x, p2 (x) = x + x2 ,
∞
X
1
1
1
1
p3 (x) = x + x2 + x3 , p4 (x) = x + x2 + x3 + x4 ;
xk
2
2
3!
(k − 1)!
k=1
17. (a) f (0) = 1, f 0 (0) = 2, f 00 (0) = −2, f 000 (0) = 6, f (k) (0) = 0 for k ≥ 4;
the Maclaurin series for f (x) is f (x).
(b) f (k) (0) = k!ck for k ≤ n, and f (k) (0) = 0 for k > n; the Maclaurin series for f (x) is f (x).
19. f (k) (x0 ) = e; p0 (x) = e, p1 (x) = e + e(x − 1),
e
e
e
p2 (x) = e + e(x − 1) + (x − 1)2 , p3 (x) = e + e(x − 1) + (x − 1)2 + (x − 1)3 ,
2
2
3!
∞
X
e
e
e
e
p4 (x) = e + e(x − 1) + (x − 1)2 + (x − 1)3 + (x − 1)4 ;
(x − 1)k
2
3!
4!
k!
k=0
1
1 1
1
20. f (k) (x) = (−1)k e−x , f (k) (ln 2) = (−1)k ; p0 (x) = , p1 (x) = − (x − ln 2),
2
2
2 2
p2 (x) =
1
1
1 1
1 1
1
− (x−ln 2)+
(x−ln 2)2 , p3 (x) = − (x−ln 2)+
(x−ln 2)2 −
(x−ln 2)3 ,
2 2
2·2
2 2
2·2
2 · 3!
p4 (x) =
1
1 1
1
1
− (x − ln 2) +
(x − ln 2)2 −
(x − ln 2)3 +
(x − ln 2)4 ;
2 2
2·2
2 · 3!
2 · 4!
∞
X
(−1)k
k=0
2 · k!
21. f (k) (x) =
(x − ln 2)k
(−1)k k! (k)
, f (−1) = −k!; p0 (x) = −1; p1 (x) = −1 − (x + 1);
xk+1
p2 (x) = −1 − (x + 1) − (x + 1)2 ; p3 (x) = −1 − (x + 1) − (x + 1)2 − (x + 1)3 ;
p4 (x) = −1 − (x + 1) − (x + 1)2 − (x + 1)3 − (x + 1)4 ;
∞
X
(−1)(x + 1)k
k=0
Exercise Set 11.5
22. f (k) (x) =
378
1
(−1)k k!
(−1)k k!
1
1
, f (k) (3) =
; p0 (x) = ; p1 (x) = − (x − 3);
k+1
(x + 2)
5k+1
5
5 25
p2 (x) =
1
1
1
1
1
1
1
− (x − 3) +
(x − 3)2 ; p3 (x) = − (x − 3) +
(x − 3)2 −
(x − 3)3 ;
5 25
125
5 25
125
625
p4 (x) =
∞
X
1
1
1
1
1
(−1)k
− (x − 3) +
(x − 3)2 −
(x − 3)3 +
(x − 3)4 ;
(x − 3)k
5 25
125
625
3025
5k+1
k=0
23. f (k) (1/2) = 0 if k is odd, f (k) (1/2) is alternately π k and −π k if k is even;
π2
(x − 1/2)2 ,
2
∞
X
π2
π4
(−1)k π 2k
p4 (x) = 1 −
(x − 1/2)2 +
(x − 1/2)4 ;
(x − 1/2)2k
2
4!
(2k)!
p0 (x) = p1 (x) = 1, p2 (x) = p3 (x) = 1 −
k=0
24. f (k) (π/2) = 0 if k is even, f (k) (π/2) is alternately −1 and 1 if k is odd; p0 (x) = 0,
p1 (x) = −(x − π/2), p2 (x) = −(x − π/2), p3 (x) = −(x − π/2) +
p4 (x) = −(x − π/2) +
1
(x − π/2)3 ,
3!
∞
X
1
(−1)k+1
(x − π/2)3 ;
(x − π/2)2k+1
3!
(2k + 1)!
k=0
(−1)k−1 (k − 1)! (k)
; f (1) = (−1)k−1 (k − 1)!;
xk
1
1
1
p0 (x) = 0, p1 (x) = (x − 1); p2 (x) = (x − 1) − (x − 1)2 ; p3 (x) = (x − 1) − (x − 1)2 + (x − 1)3 ,
2
2
3
∞
k−1
X
1
1
1
(−1)
p4 (x) = (x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 ;
(x − 1)k
2
3
4
k
25. f (1) = 0, for k ≥ 1, f (k) (x) =
k=1
(−1)k−1 (k − 1)! (k)
(−1)k−1 (k − 1)!
; f (e) =
;
k
x
ek
1
1
1
p0 (x) = 1, p1 (x) = 1 + (x − e); p2 (x) = 1 + (x − e) − 2 (x − e)2 ;
e
e
2e
1
1
1
p3 (x) = 1 + (x − e) − 2 (x − e)2 + 3 (x − e)3 ,
e
2e
3e
∞
X
1
1
1
1
(−1)k−e
p4 (x) = 1 + (x − e) − 2 (x − e)2 + 3 (x − e)3 − 4 (x − e)4 ; 1 +
(x − e)k
e
2e
3e
4e
kek
26. f (e) = 1, for k ≥ 1, f (k) (x) =
k=1
27. (a) f (1) = 1, f 0 (1) = 2, f 00 (1) = −2, f 000 (1) = 6, f (k) (1) = 0 for k ≥ 4;
the Taylor series for f (x) is f (x).
(b) f (k) (x0 ) = k!ck for k ≤ n; for k > n, f (k) (x0 ) = 0; the Taylor series for f (x) is f (x).
379
Chapter 11
29. f (k) (0) = (−2)k ; p0 (x) = 1, p1 (x) = 1 − 2x,
4
4
p2 (x) = 1 − 2x + 2x2 , p3 (x) = 1 − 2x + 2x2 − x3
3
-0.6
0.6
-1
30. f (k) (π/2) = 0 if k is odd, f (k) (π/2) is alternately 1
1
and −1 if k is even; p0 (x) = 1, p2 (x) = 1 − (x − π/2)2 ,
2
1
1
p4 (x) = 1 − (x − π/2)2 + (x − π/2)4 ,
2
24
1
1
1
(x − π/2)6
p6 (x) = 1 − (x − π/2)2 + (x − π/2)4 −
2
24
720
31. f (k) (π) = 0 if k is odd, f (k) (π/2) is alternately −1
1
and 1 if k is even; p0 (x) = −1, p2 (x) = −1 + (x − π)2 ,
2
1
1
p4 (x) = −1 + (x − π)2 − (x − π)4 ,
2
24
1
1
1
(x − π)6
p6 (x) = −1 + (x − π)2 − (x − π)4 +
2
24
720
32. f (0) = 0; for k ≥ 1, f (k) (x) =
1.25
^
i
-1.25
1.25
o
0
-1.25
(−1)k−1 (k − 1)!
,
(x + 1)k
1.5
f (k) (0) = (−1)k−1 (k − 1)!; p0 (x) = 0, p1 (x) = x,
1
1
1
p2 (x) = x − x2 , p3 (x) = x − x2 + x3
2
2
3
-1
1
-1.5
33. p(0) = 1, p(x) has slope −1 at x = 0, and p(x) is concave up at x = 0, eliminating I, II and III
respectively and leaving IV.
34. Let p0 (x) = 2, p1 (x) = p2 (x) = 1 − 3x, p3 (x) = 1 − 3x + x3 , and, for any arbitrary integer n ≥ 4
n
X
3
and constants c4 , c5 , . . . , cn , let pn (x) = 2 − 3(x − 1) + (x − 1) +
ck (x − 1)k ; then any of the
k=4
polynomials p0 , p1 , . . . , pn is a possible Taylor polynomial for f about x = 1.
35. f (k) (ln 4) = 15/8 for k even, f (k) (ln 4) = 17/8 for k odd, which can be written as
f (k) (ln 4) =
∞
16 − (−1)k X 16 − (−1)k
;
(x − ln 4)k
8
8k!
k=0
Exercise Set 11.6
380
36. (a) cos α ≈ 1 − α2 /2; x = r − r cos α = r(1 − cos α) ≈ rα2 /2
(b) In Figure Ex-36 let r = 4000 mi and α = 1/80 so that the arc has length 2rα = 100 mi.
4000
= 5/16 mi.
Then x ≈ rα2 /2 =
2 · 802
37. From Exercise 2(a), p1 (x) = 1 + x, p2 (x) = 1 + x + x2 /2
3
(a)
-1
1
-1
(b)
x –1.000 –0.750 –0.500 –0.250 0.000 0.250 0.500
f (x)
0.431 0.506 0.619 0.781 1.000 1.281 1.615
p1(x) 0.000 0.250 0.500 0.750 1.000 1.250 1.500
p2(x) 0.500 0.531 0.625 0.781 1.000 1.281 1.625
(c)
|esin x − (1 + x)| < 0.01
for − 0.14 < x < 0.14
0.750
1.977
1.750
2.031
(d)
1.000
2.320
2.000
2.500
|esin x − (1 + x + x2 /2)| < 0.01
for − 0.50 < x < 0.50
0.01
0.015
-0.15
0.15
-0.6
0
0.6
0
EXERCISE SET 11.6
1. (a)
∞
1
1
1 X 1
≤
=
,
converges
5k 2 − k
5k 2 − k 2
4k 2
4k 2
k=1
∞
3 X
3
> ,
(b)
3/k diverges
k − 1/4
k
k=1
2. (a)
∞
k
k+1
1 X
>
,
=
1/k diverges
k2 − k
k2
k
(b)
∞
1
1 X 1
<
,
converges
3k + 5
3k
3k
(b)
k=2
3. (a)
k=1
4. (a)
∞
X
ln k
1
1
> for k ≥ 3,
diverges
k
k
k
k=1
∞
k
k
1 X 1
√
√ diverges
(b)
>
=
;
k 3/2 − 1/2
k 3/2
k k=1 k
∞
2 X 2
2
<
,
converges
k4 + k
k4
k4
k=1
∞
5 X 5
5 sin2 k
< ,
converges
k!
k!
k!
k=1
381
Chapter 11
5. compare with the convergent series
∞
X
4k 7 − 2k 6 + 6k 5
= 1/2, converges
k→+∞
8k 7 + k − 8
1/k 5 , ρ = lim
k=1
6. compare with the divergent series
∞
X
k
= 1/9, diverges
k→+∞ 9k + 6
1/k, ρ = lim
k=1
7. compare with the convergent series
∞
X
3k
= 1, converges
k→+∞ 3k + 1
5/3k , ρ = lim
k=1
8. compare with the divergent series
∞
X
k 2 (k + 3)
= 1, diverges
k→+∞ (k + 1)(k + 2)(k + 5)
1/k, ρ = lim
k=1
9. compare with the divergent series
ρ = lim
k→+∞ (8k 2
∞
X
1
,
2/3
k
k=1
k 2/3
1
= lim
= 1/2, diverges
k→+∞ (8 − 3/k)1/3
− 3k)1/3
10. compare with the convergent series
∞
X
1/k 17 ,
k=1
17
ρ = lim
k→+∞
k
1
= lim
= 1/217 , converges
k→+∞ (2 + 3/k)17
(2k + 3)17
3k+1 /(k + 1)!
3
= lim
= 0, the series converges
k
k→+∞
k→+∞ k + 1
3 /k!
11. ρ = lim
4k+1 /(k + 1)2
4k 2
=
lim
= 4, the series diverges
k→+∞
k→+∞ (k + 1)2
4k /k 2
12. ρ = lim
13. ρ = lim
k→+∞
k
= 1, the result is inconclusive
k+1
(k + 1)(1/2)k+1
k+1
= 1/2, the series converges
= lim
k
k→+∞
k→+∞ 2k
k(1/2)
14. ρ = lim
(k + 1)!/(k + 1)3
k3
=
lim
= +∞, the series diverges
k→+∞
k→+∞ (k + 1)2
k!/k 3
15. ρ = lim
(k + 1)/[(k + 1)2 + 1]
(k + 1)(k 2 + 1)
= lim
= 1, the result is inconclusive.
2
k→+∞
k→+∞ k(k 2 + 2k + 2)
k/(k + 1)
16. ρ = lim
17. ρ = lim
k→+∞
3k + 2
= 3/2, the series diverges
2k − 1
18. ρ = lim k/100 = +∞, the series diverges
k→+∞
k 1/k
= 1/5, the series converges
k→+∞ 5
19. ρ = lim
20. ρ = lim (1 − e−k ) = 1, the result is inconclusive
k→+∞
Exercise Set 11.6
382
21. Ratio Test, ρ = lim 7/(k + 1) = 0, converges
k→+∞
22. Limit Comparison Test, compare with the divergent series
∞
X
1/k
k=1
(k + 1)2
= 1/5, converges
k→+∞
5k 2
23. Ratio Test, ρ = lim
24. Ratio Test, ρ = lim (10/3)(k + 1) = +∞, diverges
k→+∞
25. Ratio Test, ρ = lim e−1 (k + 1)50 /k 50 = e−1 < 1, converges
k→+∞
26. Limit Comparison Test, compare with the divergent series
∞
X
1/k
k=1
27. Limit Comparison Test, compare with the convergent series
converges
28.
1/k 5/2 , ρ =
k=1
k3
= 1,
+1
k→+∞ k 3
k=1
29. Limit Comparison Test, compare with the divergent series
diverges
∞
X
1/k, ρ =
k=1
∞
∞
X
3 X
2 + (−1)k
2 + (−1)k
k
≤
,
3/5
converges
so
converges
5k
5k
5k
k=1
k=1
31. Limit Comparison Test, compare with the convergent series
∞
X
1/k 5/2 ,
k=1
k 3 + 2k 5/2
ρ = lim 3
= 1, converges
k→+∞ k + 3k 2 + 3k
32.
lim
∞
∞
X
4 X 4
4
4
< k ,
converges (Ratio Test) so
converges by the Comparison Test
2 + 3k k
3 k
3k k
2 + k3k
k=1
30.
∞
X
∞
∞
X
5 X
4 + | cos k|
4 + | cos k|
3
<
,
5/k
converges
so
converges
3
3
k
k
k3
k=1
k=1
33. Limit Comparison Test, compare with
∞
X
√
1/ k
k=1
34. Ratio Test, ρ = lim (1 + 1/k)−k = 1/e < 1, converges
k→+∞
ln(k + 1)
k
= lim
= 1/e < 1, converges
k→+∞
k→+∞ e(k + 1)
e ln k
35. Ratio Test, ρ = lim
k+1
1
= lim
= 0, converges
2k+1
2k+1
k→+∞ e
k→+∞ 2e
36. Ratio Test, ρ = lim
37. Ratio Test, ρ = lim
k→+∞
k+5
= 1/4, converges
4(k + 1)
lim √
k→+∞
k
= 1,
k2 + k
383
Chapter 11
38. Root Test, ρ = lim (
k→+∞
39. diverges because lim
k k
1
) = lim
= 1/e, converges
k→+∞ (1 + 1/k)k
k+1
k→+∞
40.
∞
X
k=1
Z
√
∞ √
k ln k
ln k
k ln k X k ln k
k ln k
=
because ln 1 = 0, 3
<
= 2 ,
3
3
3
k +1
k +1
k +1
k
k
√
k=2
+∞
2
41.
1
= 1/4 6= 0
4 + 2−k
`
∞
∞ √
X
X
1
ln x 1
ln x
ln k
k ln k
−
.
dx = lim −
= (ln 2 + 1) so
converges and so does
`→+∞
x2
x
x 2
2
k2
k3 + 1
k=2
∞
∞
X
tan−1 k
π/2 X π/2
tan−1 k
<
,
converges
so
converges
2
2
2
k
k
k
k2
k=1
42.
k=1
k=1
∞
∞
X
2 5k X
5k + 5k
5k + k
5k + k
5k
<
=
,
converges (Ratio Test) so
converges
2
k! + 3
k!
k!
k!
k! + 3
k=1
k=1
(k + 1)2
= 1/4, converges
k→+∞ (2k + 2)(2k + 1)
43. Ratio Test, ρ = lim
2(k + 1)2
= 1/2, converges
k→+∞ (2k + 4)(2k + 3)
44. Ratio Test, ρ = lim
45. uk =
k!
k+1
, by the Ratio Test ρ = lim
= 1/2; converges
k→+∞ 2k + 1
1 · 3 · 5 · · · (2k − 1)
46. uk =
1 · 3 · 5 · · · (2k − 1)
1
, by the Ratio Test ρ = lim
= 0; converges
k→+∞ 2k
(2k − 1)!
47. Root Test: ρ = lim
k→+∞
1
(ln k)1/k = 1/3, converges
3
k+1
π(k + 1)
= π, diverges
= lim π
1+1/k
k→+∞ k
k→+∞
k
48. Root Test: ρ = lim
∞
X
sin(π/k)
= 1 and
π/k diverges
49. (b) ρ = lim
k→+∞
π/k
k=1
1
1
2
≈ 2
50. (a) cos x ≈ 1 − x /2, 1 − cos
k
2k
√
(b) ρ = lim
k→+∞
1 − cos(1/k)
= 2, converges
1/k 2
1
d
1
g(x) = √ − = 0 when x = 4. Since lim g(x) = lim g(x) = +∞
x→+∞
x→0+
dx
2 x x
√
√
it follows that g(x) has its minimum at x = 4, g(4) = 4 − ln 4 > 0, and thus x − ln x > 0 for
x > 0.
√
∞
∞
X
ln k
1 X 1
k
ln k
(a)
< 2 = 3/2 ,
converges so
converges.
k2
k
k2
k
k 3/2
51. Set g(x) =
x − ln x;
k=1
(b)
1
1
> ,
2
(ln k)
k
∞
X
k=2
1
diverges so
k
k=1
∞
X
k=2
1
diverges.
(ln k)2
Exercise Set 11.7
384
α
α
= α = α, the series converges if α < 1 and diverges
1
(k 1/k )α
∞
X
if α > 1. If α = 1 then the series is
1/k which diverges.
52. By the Root Test, ρ = lim
k→+∞
k=1
P
bk . Then
53. (a) If
bk converges, then set M =
Pa1 + a2 + · · · + an ≤ b1 + b2 + · · · + bn ≤ M ;
apply Theorem 11.4.6 to get convergence of
ak .
P
(b) Assume
the contrary, that
bk converges; then use part (a) of the Theorem to show that
P
ak converges, a contradiction.
P
54. (a) If lim (ak /bk ) = 0 then for k ≥ K, ak /bk < 1, ak < bk so
k→+∞
P
ak converges by the Comparison
Test.
(b) If
lim (ak /bk ) = +∞ then for k ≥ K, ak /bk > 1, ak > bk so
k→+∞
Comparison Test.
EXERCISE SET 11.7
1. ak+1 < ak , lim ak = 0, ak > 0
k→+∞
2.
1
k
ak+1
< for k > 0, so {ak } is decreasing and tends to zero.
=
ak
3(k + 1)
3
3. diverges because lim ak = lim
k→+∞
k→+∞
k+1
= 1/3 6= 0
3k + 1
k+1
4. diverges because lim ak = lim √
= +∞ 6= 0
k→+∞
k→+∞
k+1
5. {e−k } is decreasing and lim e−k = 0, converges
k→+∞
6.
ln k
k
ln k
= 0, converges
k→+∞ k
is decreasing and lim
(3/5)k+1
= 3/5, converges absolutely
k→+∞ (3/5)k
7. ρ = lim
8. ρ = lim
k→+∞
2
= 0, converges absolutely
k+1
3k 2
= 3, diverges
k→+∞ (k + 1)2
9. ρ = lim
10. ρ = lim
k→+∞
k+1
= 1/5, converges absolutely
5k
(k + 1)3
= 1/e, converges absolutely
k→+∞
ek 3
11. ρ = lim
(k + 1)k+1 k!
= lim (1 + 1/k)k = e, diverges
k→+∞ (k + 1)!k k
k→+∞
12. ρ = lim
P
ak diverges by the
385
Chapter 11
∞
X
(−1)k+1
13. conditionally convergent,
3k
k=1
14. absolutely convergent,
∞
X
1
converges by the Alternating Series Test but
diverges
3k
k=1
∞
X
1
converges
4/3
k
k=1
15. divergent, lim ak 6= 0
k→+∞
16. absolutely convergent, Ratio Test for absolute convergence
17.
∞
X
cos kπ
k
k=1
Test but
=
∞
X
(−1)k
k=1
∞
X
k
is conditionally convergent,
∞
X
(−1)k
k=1
k
converges by the Alternating Series
1/k diverges.
k=1
∞
X
(−1)k ln k
18. conditionally convergent,
k
k=3
diverges (Limit Comparison Test with
19. conditionally convergent,
∞
X
(−1)k+1
k=1
∞
X
Alternating Series Test but
k=1
20. conditionally convergent,
∞
X
k=1
21.
∞
X
P
∞
X
ln k
k=3
k
1/k).
k+2
converges by the
k(k + 3)
P
k+2
diverges (Limit Comparison Test with
1/k)
k(k + 3)
∞
X
(−1)k+1 k 2
k3 + 1
k=1
converges by the Alternating Series Test but
converges by the Alternating Series Test but
P
k2
diverges (Limit Comparison Test with (1/k))
+1
k3
sin(kπ/2) = 1 + 0 − 1 + 0 + 1 + 0 − 1 + 0 + · · ·, divergent ( lim sin(kπ/2) does not exist)
k→+∞
k=1
22. absolutely convergent,
∞
X
| sin k|
k=1
23. conditionally convergent,
k3
converges (compare with
∞
X
(−1)k
k=2
k ln k
P
1/k 3 )
converges by the Alternating Series Test but
∞
X
k=2
1
diverges
k ln k
(Integral Test)
24. conditionally convergent,
∞
X
∞
X
k=1
(−1)k
p
k(k + 1)
converges by the Alternating Series Test but
P
1
p
diverges (Limit Comparison Test with
1/k)
k(k + 1)
k=1
25. absolutely convergent,
∞
X
(1/ ln k)k converges by the Root Test
k=2
Exercise Set 11.7
386
∞
X
(−1)k+1
√ converges by the Alternating Series Test but
k+1+ k
k=1
P √
diverges (Limit Comparison Test with
1/ k)
26. conditionally convergent,
∞
X
√
k=1
1
√
k+1+ k
√
27. conditionally convergent, let f (x) =
+∞
{ak }k=2 =
k2 + 1
k3 + 2
+∞
is nonincreasing, lim ak = 0; the series converges by the
k→+∞
k=2
Alternating Series Test but
∞
X
k=2
28.
∞
X
k cos kπ
k=1
k2 + 1
=
∞
X
(−1)k k
k=1
x2 + 1
x(4 − 3x − x3 )
then f 0 (x) =
≤ 0 for x ≥ 2 so
3
x +2
(x3 + 2)2
k2 + 1
P
k +1
diverges (Limit Comparison Test with
1/k)
3
k +2
2
is conditionally convergent,
Alternating Series Test but
∞
X
(−1)k k
k=1
∞
X
k=1
k2
k2 + 1
converges by the
k
diverges
+1
k+1
=0
k→+∞ (2k + 1)(2k)
29. absolutely convergent by the Ratio Test, ρ = lim
30. divergent, lim ak = +∞
31. |error| < a8 = 1/8 = 0.125
32. |error| < a6 = 1/6! < 0.0014
√
33. |error| < a100 = 1/ 100 = 0.1
k→+∞
34. |error| < a4 = 1/(5 ln 5) < 0.125
35. |error| < 0.0001 if an+1 ≤ 0.0001, 1/(n + 1) ≤ 0.0001, n + 1 ≥ 10, 000, n ≥ 9, 999, n = 9, 999
36. |error| < 0.00001 if an+1 ≤ 0.00001, 1/(n + 1)! ≤ 0.00001, (n + 1)! ≥ 100, 000. But 8! = 40, 320,
9! = 362, 880 so (n + 1)! ≥ 100, 000 if n + 1 ≥ 9, n ≥ 8, n = 8
√
√
37. |error| < 0.005 if an+1 ≤ 0.005, 1/ n + 1 ≤ 0.005, n + 1 ≥ 200, n + 1 ≥ 40, 000, n ≥ 39, 999,
n = 39, 999
38. |error| < 0.05 if an+1 ≤ 0.05, 1/[(n + 2) ln(n + 2)] ≤ 0.05, (n + 2) ln(n + 2) ≥ 20. But 9 ln 9 ≈ 19.8
and 10 ln 10 ≈ 23.0 so (n + 2) ln(n + 2) ≥ 20 if n + 2 ≥ 10, n ≥ 8, n = 8
39. ak =
3
3
3/4
= 0.5
, |error| < a11 = 12 < 0.00074; s10 ≈ 0.4995; S =
2k+1
2
1 − (−1/2)
40. ak =
k−1
10
1
2
2
= 0.6
, |error| < a11 =
< 0.01735; s10 ≈ 0.5896; S =
3
3
1 − (−2/3)
1
1
, an+1 =
≤ 0.005, (2n + 1)! ≥ 200, 2n + 1 ≥ 6, n ≥ 2.5; n = 3,
(2k − 1)!
(2n + 1)!
s3 = 1 − 1/6 + 1/120 ≈ 0.84
41. ak =
42. ak =
1
1
, an+1 =
≤ 0.005, (2n)! ≥ 200, 2n ≥ 6, n ≥ 3; n = 3, s3 ≈ 0.54
(2k − 2)!
(2n)!
387
Chapter 11
43. ak =
44. ak =
1
1
, an+1 =
≤ 0.005, (n + 1)2n+1 ≥ 200, n + 1 ≥ 6, n ≥ 5; n = 5, s5 ≈ 0.41
k2k
(n + 1)2n+1
(2k −
1)5
1
1
, an+1 =
≤ 0.005,
5
+ 4(2k − 1)
(2n + 1) + 4(2n + 1)
(2n + 1)5 + 4(2n + 1) ≥ 200, 2n + 1 ≥ 3, n ≥ 1; n = 1, s1 = 0.20
45. (c) ak =
46.
P
1
1
, an+1 =
≤ 10−2 , 2n + 1 ≥ 100, n ≥ 49.5; n = 50
2k − 1
2n + 1
(1/k p ) converges if p > 1 and diverges if p ≤ 1, so
∞
X
(−1)k
k=1
1
converges absolutely if p > 1,
kp
and converges conditionally if 0 < p ≤ 1 since it satisfies the Alternating Series Test; it diverges
for p ≤ 0 since lim ak 6= 0.
k→+∞
1
1
1
1
1
1
1
47. 1 + 2 + 2 + · · · = 1 + 2 + 2 + · · · − 2 + 2 + 2 + · · ·
3
5
2
3
2
4
6
2
2
1
1 π2
π2
1
1
π
π
− 2 1 + 2 + 2 + ··· =
−
=
=
6
2
2
3
6
4 6
8
1
1
1
1
1
1
1
48. 1 + 4 + 4 + · · · = 1 + 4 + 4 + · · · − 4 + 4 + 4 + · · ·
3
5
2
3
2
4
6
1
1 π4
π4
1
1
π4
π4
− 4 1 + 4 + 4 + ··· =
−
=
=
90 2
2
3
90 16 90
96
49. Every positive integer can be written in exactly one of the three forms 2k − 1 or 4k − 2 or 4k,
so a rearrangement is
1
1
1
1
1 1
1 1 1
1
1
+
− −
+
−
−
+ ··· +
−
−
+ ···
1− −
2 4
3 6 8
5 10 12
2k − 1 4k − 2 4k
1
1
1
1 1
1
1
1 1
−
+
−
+
−
+ ··· +
−
+ · · · = ln 2
=
2 4
6 8
10 12
4k − 2 4k
2
50. (a)
1.5
(b)
Yes; since f (x) is decreasing for x ≥ 1
and lim f (x) = 0, so the series
x→+∞
satisfies the Alternating Series Test.
0
10
0
51. (a) The distance d from the starting point is
180 180
180
1 1
1
+
− ··· −
= 180 1 − + − · · · −
.
2
3
1000
2 3
1000
1
1 1
differs from ln 2 by less than 1/1001 so
From Theorem 11.7.2, 1 − + − · · · −
2 3
1000
180(ln 2 − 1/1001) < d < 180 ln 2, 124.58 < d < 124.77.
d = 180 −
Exercise Set 11.8
388
180
180 180
+
+ ··· +
, and from inequality (2) in
(b) The total distance traveled is s = 180 +
2
3
1000
Section 11.4,
Z 1001
Z 1000
180
180
dx < s < 180 +
dx
x
x
1
1
180 ln 1001 < s < 180(1 + ln 1000)
1243 < s < 1424
52. (a) Suppose Σ|ak | converges, then
lim |ak | = 0 so |ak | < 1 for k ≥ K and thus |ak |2 < |ak |,
k→+∞
a2k < |ak | hence Σa2k converges by the Comparison Test.
(b) Let ak =
P 2
P
1
, then
ak converges but
ak diverges.
k
EXERCISE SET 11.8
uk+1
= |x|, so the interval of convergence is −1 < x < 1, converges
uk
1. geometric series, ρ = lim
k→+∞
there to
1
(the series diverges for x = ±1)
1+x
uk+1
= |x|2 , so the interval of convergence is −1 < x < 1, converges
uk
2. geometric series, ρ = lim
k→+∞
there to
1
(the series diverges for x = ±1)
1 − x2
3. geometric series, ρ = lim
k→+∞
there to
uk+1
= |x − 2|, so the interval of convergence is 1 < x < 3, converges
uk
1
1
=
(the series diverges for x = 1, 3)
1 − (x − 2)
3−x
4. geometric series, ρ =
converges there to
lim
k→+∞
uk+1
= |x + 3|, so the interval of convergence is −4 < x < −2,
uk
1
1
=
(the series diverges for x = −4, −2)
1 + (x + 3)
4+x
5. (a) geometric series, ρ =
converges there to
lim
k→+∞
uk+1
= |x/2|, so the interval of convergence is −2 < x < 2,
uk
2
1
=
; (the series diverges for x = −2, 2)
1 + x/2
2+x
(b) f (0) = 1; f (1) = 2/3
6. (a) geometric series, ρ = lim
k→+∞
converges to
uk+1
x−5
, so the interval of convergence is 2 < x < 8,
=
uk
3
3
1
=
(the series diverges for x = 2, 8)
1 + (x − 5)/3
x−2
(b) f (3) = 3, f (6) = 3/4
389
Chapter 11
7. ρ =
k+1
|x| = |x|, the series converges if |x| < 1 and diverges if |x| > 1. If x = −1,
k→+∞ k + 2
lim
∞
X
(−1)k
k=0
k+1
converges by the Alternating Series Test; if x = 1,
∞
X
k=0
1
diverges. The radius of
k+1
convergence is 1, the interval of convergence is [−1, 1).
8. ρ = lim 3|x| = 3|x|, the series converges if 3|x| < 1 or |x| < 1/3 and diverges if |x| > 1/3. If
k→+∞
x = −1/3,
∞
X
(−1)k diverges, if x = 1/3,
k=0
∞
X
(1) diverges. The radius of convergence is 1/3, the
k=0
interval of convergence is (−1/3, 1/3).
|x|
= 0, the radius of convergence is +∞, the interval is (−∞, +∞).
k→+∞ k + 1
9. ρ = lim
10. ρ = lim
k→+∞
k+1
|x| = +∞, the radius of convergence is 0, the series converges only if x = 0.
2
∞
X
5k 2 |x|
(−1)k
=
5|x|,
converges
if
|x|
<
1/5
and
diverges
if
|x|
>
1/5.
If
x
=
−1/5,
2
k→+∞ (k + 1)
k2
k=1
∞
X
converges; if x = 1/5,
1/k 2 converges. Radius of convergence is 1/5, interval of convergence is
11. ρ = lim
k=1
[−1/5, 1/5].
12. ρ =
ln k
|x| = |x|, the series converges if |x| < 1 and diverges if |x| > 1. If x = −1,
k→+∞ ln(k + 1)
lim
∞
X
(−1)k
k=2
ln k
converges; if x = 1,
∞
X
1/(ln k) diverges (compare to
P
(1/k)). Radius of convergence
k=2
is 1, interval of convergence is [−1, 1).
∞
X
k|x|
(−1)k
= |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1,
converges;
13. ρ = lim
k→+∞ k + 2
k(k + 1)
k=1
∞
X
1
if x = 1,
converges. Radius of convergence is 1, interval of convergence is [−1, 1].
k(k + 1)
k=1
∞
X
k+1
−1
|x| = 2|x|, converges if |x| < 1/2, diverges if |x| > 1/2. If x = −1/2,
k→+∞ k + 2
2(k + 1)
k=0
∞
X
(−1)k
diverges; if x = 1/2,
converges. Radius of convergence is 1/2, interval of convergence
2(k + 1)
k=0
is (−1/2, 1/2].
14. ρ = lim 2
√
15. ρ = lim √
k→+∞
x = 1,
∞
X
k
−1
√ diverges; if
|x| = |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1,
k+1
k
k=1
∞
X
(−1)k−1
√
converges. Radius of convergence is 1, interval of convergence is (−1, 1].
k
k=1
Exercise Set 11.8
390
|x|2
= 0, radius of convergence is +∞, interval of convergence is (−∞, +∞).
k→+∞ (2k + 2)(2k + 1)
16. ρ = lim
|x|2
= 0, radius of convergence is +∞, interval of convergence is (−∞, +∞).
k→+∞ (2k + 3)(2k + 2)
17. ρ = lim
∞
X
k 3/2 |x|3
1
3
=
|x|
,
converges
if
|x|
<
1,
diverges
if
|x|
>
1.
If
x
=
−1,
converges;
3/2
3/2
k→+∞ (k + 1)
k
k=0
∞
X
(−1)k
if x = 1,
converges. Radius of convergence is 1, interval of convergence is [−1, 1].
k 3/2
k=0
18. ρ = lim
19. ρ = lim
k→+∞
3|x|
= 0, radius of convergence is +∞, interval of convergence is (−∞, +∞).
k+1
k(ln k)2 |x|
= |x|, converges if |x| < 1, diverges if |x| > 1. If x = −1, then, by
k→+∞ (k + 1)[ln(k + 1)]2
20. ρ = lim
Exercise 11.4.25,
∞
X
k=2
∞
X
−1
(−1)k+1
converges;
if
x
=
1,
converges. Radius of convergence
k(ln k)2
k(ln k)2
k=2
is 1, interval of convergence is [−1, 1].
∞
X
1 + k2
(−1)k
|x|
=
|x|,
converges
if
|x|
<
1,
diverges
if
|x|
>
1.
If
x
=
−1,
2
k→+∞ 1 + (k + 1)
1 + k2
k=0
∞
X
1
converges; if x = 1,
converges. Radius of convergence is 1, interval of convergence is
1 + k2
k=0
[−1, 1].
21. ρ = lim
∞
X
1
1
|x − 3| = |x − 3|, converges if |x − 3| < 2, diverges if |x − 3| > 2. If x = 1,
(−1)k
k→+∞ 2
2
k=0
∞
X
diverges; if x = 5,
1 diverges. Radius of convergence is 2, interval of convergence is (1, 5).
22. ρ = lim
k=0
∞
X
k|x + 1|
−1
= |x + 1|, converges if |x + 1| < 1, diverges if |x + 1| > 1. If x = −2,
k→+∞ k + 1
k
k=1
∞
k+1
X (−1)
converges. Radius of convergence is 1, interval of convergence is
diverges; if x = 0,
k
k=1
(−2, 0].
23. ρ = lim
24. ρ =
∞
X
(k + 1)2
|x − 4| = |x − 4|, converges if |x − 4| < 1, diverges if |x − 4| > 1. If x = 3,
k→+∞ (k + 2)2
lim
1/(k +1)2 converges; if x = 5,
k=0
of convergence is [3, 5].
∞
X
(−1)k /(k +1)2 converges. Radius of convergence is 1, interval
k=0
391
Chapter 11
25. ρ =
lim (3/4)|x + 5| =
k→+∞
x = −19/3,
∞
X
3
|x + 5|, converges if |x + 5| < 4/3, diverges if |x + 5| > 4/3. If
4
(−1)k diverges; if x = −11/3,
k=0
∞
X
1 diverges. Radius of convergence is 4/3, interval
k=0
of convergence is (−19/3, −11/3).
(2k + 3)(2k + 2)k 3
|x − 2| = +∞, radius of convergence is 0,
k→+∞
(k + 1)3
series converges only at x = 2.
26. ρ = lim
k2 + 4
|x + 1|2 = |x + 1|2 , converges if |x + 1| < 1, diverges if |x + 1| > 1. If x = −2,
k→+∞ (k + 1)2 + 4
27. ρ = lim
∞
X
(−1)3k+1
k=1
k2 + 4
converges; if x = 0,
∞
X
(−1)k
converges. Radius of convergence is 1, interval of
k2 + 4
k=1
convergence is [−2, 0].
28. ρ =
lim
k→+∞
x = 2,
k ln(k + 1)
|x − 3| = |x − 3|, converges if |x − 3| < 1, diverges if |x − 3| > 1. If
(k + 1) ln k
∞
X
(−1)k ln k
k
k=1
converges; if x = 4,
∞
X
ln k
k=1
k
diverges. Radius of convergence is 1, interval of
convergence is [2, 4).
π|x − 1|2
= 0, radius of convergence +∞, interval of convergence (−∞, +∞).
k→+∞ (2k + 3)(2k + 2)
29. ρ = lim
30. ρ =
lim
k→+∞
1
1
1
|2x − 3| =
|2x − 3|, converges if
|2x − 3| < 1 or |x − 3/2| < 8, diverges if
16
16
16
|x − 3/2| > 8. If x = −13/2,
∞
X
(−1)k diverges; if x = 19/2,
k=0
∞
X
1 diverges. Radius of convergence
k=0
is 8, interval of convergence is (−13/2, 19/2).
31. ρ =
lim
k→+∞
p
k
|uk | =
|x|
= 0, the series converges absolutely for all x so the interval of
k→+∞ ln k
lim
convergence is (−∞, +∞).
32.
ρ = lim
k→+∞
2k + 1
|x| = 0
(2k)(2k − 1)
10
33. (a)
so R = +∞.
-1
1
-1
Exercise Set 11.8
392
|x|2
= 0, R = +∞
k→+∞ 4(k + 1)(k + 2)
34. Ratio Test: ρ = lim
35. By the Ratio Test for absolute convergence,
(pk + p)!(k!)p
(pk + p)(pk + p − 1)(pk + p − 2) · · · (pk + p − [p − 1])
|x| = lim
|x|
k→+∞ (pk)![(k + 1)!]p
k→+∞
(k + 1)p
2
p−1
1
p−
··· p −
|x| = pp |x|,
= lim p p −
k→+∞
k+1
k+1
k+1
ρ = lim
converges if |x| < 1/pp , diverges if |x| > 1/pp . Radius of convergence is 1/pp .
36. By the Ratio Test for absolute convergence,
ρ = lim
k→+∞
(k + 1 + p)!k!(k + q)!
k+1+p
|x| = lim
|x| = 0,
k→+∞ (k + 1)(k + 1 + q)
(k + p)!(k + 1)!(k + 1 + q)!
radius of convergence is +∞.
37. (a) By Theorem 11.4.3(b) both series converge or diverge together, so they have the same radius
of convergence.
P
(b) By Theorem
11.4.3(a) the series (ck + dk )(x − x0 )k converges
P if |x − x0 |k < R; if |x − x0 | > R
P
then (ck + dk )(x − x0 )k cannot converge, as otherwise
P ck (x − x0 ) would converge by
the same Theorem. Hence the radius of convergence of (ck + dk )(x − x0 )k is R.
P
k
(c) Let rPbe the radius of convergence
of
(ck + dk )(x
0 | < min(R1 , R2 )
P
P− x0 ) . If |x − x
k
k
then
ck (x − x0 ) and
dk (x − x0 ) converge, so (ck + dk )(x − x0 )k converges. Hence
r ≥ min(R1 , R2 ) (to see that r > min(R1 , R2 ) is possible consider the case ck = −dk = 1).
If in addition R1 6= R2 , and R1 < |x − x0 | < R2 (or R2 < |x − x0 | < R1 ) then
P
(ck + dk )(x − x0 )k cannot converge, as otherwise all three series would converge. Thus
in this case r = min(R1 , R2 ).
38. By the Root Test for absolute convergence,
ρ = lim |ck |1/k |x| = L|x|, L|x| < 1 if |x| < 1/L so the radius of convergence is 1/L.
k→+∞
39. By assumption
|x| <
√
∞
X
ck x converges if |x| < R so
k=0
R. Moreover,
k
∞
X
2k
ck x
=
k=0
has radius of convergence
40. The assumption is that
√
∞
X
∞
X
∞
X
2k
ck x
=
k=0
∞
X
ck (x2 )k converges if |x2 | < R,
k=0
ck (x ) diverges if |x | > R, |x| >
2 k
2
k=0
k=0
R.
ck Rk is convergent and
k=0
is absolutely convergent then
∞
X
∞
X
ck (−R)k is divergent. Suppose that
k=0
because |ck Rk | = |ck (−R)k |, which contradicts the assumption that
k=0
ck Rk
k=0
∞
X
k=0
k
∞
X
ck (−R)k is also absolutely convergent and hence convergent
k=0
∞
X
∞
X
√
R. Thus
ck x2k
ck R must be conditionally convergent.
ck (−R)k is divergent so
393
Chapter 11
EXERCISE SET 11.9
1. sin 4◦ = sin
π
π
(π/45)3
(π/45)5
=
−
+
− ···
45
45
3!
5!
(a) Method 1: |Rn (π/45)| ≤
sin 4◦ ≈
(π/45)n+1
< 0.000005 for n + 1 = 4, n = 3;
(n + 1)!
(π/45)3
π
−
≈ 0.069756
45
3!
(b) Method 2: The first term in the alternating series that is less than 0.000005 is
the result is the same as in part (a).
2. cos 3◦ = cos
(π/45)5
, so
5!
π
(π/60)2
(π/60)4
=1−
+
− ···
60
2
4!
(π/60)n+1
(π/60)2
< 0.0005 for n = 2; cos 3◦ ≈ 1 −
≈ 0.9986.
(n + 1)!
2
(π/60)4
(b) Method 2: The first term in the alternating series that is less than 0.0005 is
, so the
4!
result is the same as in part (a).
(a) Method 1: |Rn (π/60)| ≤
3. f (k) (x) = ex , |f (k) (x)| ≤ e1/2 < 2 on [0, 1/2], let M = 2, e1/2 = 1 +
|Rn (1/2)| ≤
e1/2 ≈ 1 +
1
1
1 1
+ +
+
+ ···;
2 8 48 24 · 16
M
2
(1/2)n+1 ≤
(1/2)n+1 ≤ 0.00005 for n = 5;
(n + 1)!
(n + 1)!
1
1
1 1
+ +
+
≈ 1.64844, calculator value 1.64872
2 8 48 24 · 16
4. f (x) = ex , f (k) (x) = ex , |f (k) (x)| ≤ 1 on [−1, 0], |Rn (x)| ≤
if n = 6, so e−1 ≈ 1 − 1 +
5. |Rn (0.1)| ≤
1
1
(1)n+1 =
< 0.5 × 10−3
(n + 1)!
(n + 1)!
1
1
1
1
1
− + − +
≈ 0.3681, calculator value 0.3679
2! 3! 4! 5! 6!
(0.1)n+1
≤ 0.000005 for n = 3; cos 0.1 ≈ 1 − (0.1)2 /2 = 0.99500, calculator value
(n + 1)!
0.995004 . . .
6. (0.1)3 /3 < 0.5 × 10−3 so tan−1 (0.1) ≈ 0.100, calculator value ≈ 0.0997
7. Expand about π/2 to get sin x = 1 −
|Rn (x)| ≤
1
1
(x − π/2)2 + (x − π/2)4 − · · ·, 85◦ = 17π/36 radians,
2!
4!
(π/36)n+1
|x − π/2|n+1
|17π/36 − π/2|n+1
, |Rn (17π/36)| ≤
=
< 0.5 × 10−4
(n + 1)!
(n + 1)!
(n + 1)!
1
if n = 3, sin 85◦ ≈ 1 − (−π/36)2 ≈ 0.99619, calculator value 0.99619 . . .
2
8. −175◦ = −π + π/36 rad; x0 = −π, x = −π + π/36, cos x = −1 +
|Rn | ≤
(x + π)4
(x + π)2
−
− ···;
2
4!
(π/36)2
(π/36)n+1
≤ 0.00005 for n = 3; cos(−π + π/36) = −1 +
≈ −0.99619,
(n + 1)!
2
calculator value −0.99619 . . .
Exercise Set 11.9
394
9. f (k) (x) = cosh x or sinh x, |f (k) (x)| ≤ cosh x ≤ cosh 0.5 =
so |Rn (x)| <
1 0.5
1
e + e−0.5 < (2 + 1) = 1.5
2
2
1.5(0.5)n+1
(0.5)3
≤ 0.5 × 10−3 if n = 4, sinh 0.5 ≈ 0.5 +
≈ 0.5208, calculator
(n + 1)!
3!
value 0.52109 . . .
10. f (k) (x) = cosh x or sinh x, |f (k) (x)| ≤ cosh x ≤ cosh 0.1 =
1 0.1
e + e−0.1 < 1.06 so |Rn (x)| <
2
(0.1)2
1.06(0.1)n+1
≤ 0.5 × 10−3 for n = 2, cosh 0.1 ≈ 1 +
= 1.005, calculator value 1.0050 . . .
(n + 1)!
2!
11. f (x) = sin x, f (n+1) (x) = ± sin x or ± cos x, |f (n+1) (x)| ≤ 1, |Rn (x)| ≤
|x − π/4|n+1
,
(n + 1)!
|x − π/4|n+1
= 0; by the Squeezing Theorem, lim |Rn (x)| = 0
n→+∞
n→+∞
(n + 1)!
lim
so lim Rn (x) = 0 for all x.
n→+∞
12. f (x) = ex , f (n+1) (x) = ex ; if x > 1 then |Rn (x)| ≤
|Rn (x)| ≤
ex
|x − 1|n+1 ; if x < 1 then
(n + 1)!
e
|x − 1|n+1
|x − 1|n+1 . But lim
= 0 so lim Rn (x) = 0.
n→+∞ (n + 1)!
n→+∞
(n + 1)!
13. (a) Let x = 1/9 in series (17).
(1/9)3
= 2(1/9 + 1/37 ) ≈ 0.223, which agrees with the calculator value
(b) ln 1.25 ≈ 2 1/9 +
3
0.22314 . . . to three decimal places.
14. (a) Let x = 1/2 in series (17).
(1/2)3
= 2(1/2 + 1/24) = 13/12 ≈ 1.083; the calculator value is 1.099 to
(b) ln 3 ≈ 2 1/2 +
3
three decimal places.
15. (a) (1/2)9 /9 < 0.5 × 10−3 and (1/3)7 /7 < 0.5 × 10−3 so
tan−1 1/2 ≈ 1/2 −
(1/2)5
(1/2)7
(1/2)3
+
−
≈ 0.4635
3
5
7
tan−1 1/3 ≈ 1/3 −
(1/3)5
(1/3)3
+
≈ 0.3218
3
5
(b) From Formula (21), π ≈ 4(0.4635 + 0.3218) = 3.1412
1
1
(c) Let a = tan−1 , b = tan−1 ; then |a − 0.4635| < 0.0005 and |b − 0.3218| < 0.0005, so
2
3
|4(a + b) − 3.140| ≤ 4|a − 0.4635| + 4|b − 0.3218| < 0.004, so two decimal-place accuracy is
guaranteed, but not three.
1
1·2
1·2·5
16. (27+x)1/3 = 3(1+x/33 )1/3 = 3 1 + 4 x − 8 x2 + 12 x3 + . . . , alternates after first term,
3
3 2
3 3!
√
1
3·2
< 0.0005, 28 ≈ 3 1 + 4 ≈ 3.0370
8
3 2
3
395
Chapter 11
17. (a)
sin x = x −
|R4 (x)| ≤
x3
+ 0 · x4 + R4 (x),
3!
0.0001
(b)
|x|5
< 0.5 × 10−3 if |x|5 < 0.06,
5!
|x| < (0.06)1/5 ≈ 0.569, (−0.569, 0.569)
-0.7
0.7
0
18. (a)
f (k) (x) = ex ≤ eb ,
|R2 (x)| ≤
0.0004
(b)
eb b3
< 0.0005,
3!
eb b3 < 0.003 if b < 0.137 (by trial and
error with a hand calculator), so [0, 0.136].
-0.2
0.2
0
19. (a)
cos x = 1 −
|R5 (x)| ≤
x4
x2
+
+ (0)x5 + R5 (x),
2!
4!
0.000000005
(b)
(0.2)6
|x|6
≤
< 9 × 10−8
6!
6!
-0.2
0.2
0
20. (a)
f 00 (x) = −1/(1 + x)2 ,
0.00005
(b)
f 00 (x) < 1/(0.99)2 ≤ 1.03,
|R1 (x)| ≤
1.03(0.01)2
1.03|x|2
≤
2
2
≤ 5.11 × 10−5 for − 0.01 ≤ x ≤ 0.01
-0.01
0.01
0
21. (a) (1 + x)−1 = 1 − x +
=
∞
X
−1(−2) 2 −1(−2)(−3) 3
−1(−2)(−3) · · · (−k) k
x +
x + ··· +
x + ···
2!
3!
k!
(−1)k xk
k=0
(b) (1 + x)1/3 = 1 + (1/3)x +
+
(1/3)(−2/3) 2 (1/3)(−2/3)(−5/3) 3
x +
x + ···
2!
3!
∞
X
(1/3)(−2/3) · · · (4 − 3k)/3 k
2 · 5 · · · (3k − 4) k
x + · · · = 1+x/3+ (−1)k−1
x
k!
3k k!
k=2
(c) (1 + x)−3 = 1 − 3x +
=
∞
X
(−3)(−4) 2 (−3)(−4)(−5) 3
(−3)(−4) · · · (−2 − k) k
x +
x + ··· +
x + ···
2!
3!
k!
∞
(−1)k
k=0
(k + 2)! k X
(k + 2)(k + 1) k
x =
x
(−1)k
2 · k!
2
k=0
Exercise Set 11.9
396
m
22. (1 + x)
23. (a)
(b)
(c)
(d)
(e)
=
m
0
+
∞
X
m
k=1
k
k
x =
∞
X
m
k=0
k
xk
(k − 1)!
1
dk
(k − 1)!
d
d
ln(1 + x) =
, k ln(1 + x) = (−1)k−1
ln(1 − x) = −
; similarly
,
k
dx
1 + x dx
(1 + x)
dx
(1 − x)k
1
(−1)n
(n+1)
(x) = n!
+
.
so f
(1 + x)n+1
(1 − x)n+1
(−1)n
1
1
1
+
+
n!
=
n!
f (n+1) (x) ≤ n!
(1 + x)n+1
(1 − x)n+1
(1 + x)n+1
(1 − x)n+1
n+1
M
1
(n+1)
(x) ≤ M on the interval [0, 1/3] then |Rn (1/3)| ≤
.
If f
(n + 1)! 3
1
.
If 0 ≤ x ≤ 1/3 then 1 + x ≥ 1, 1 − x ≥ 2/3, f (n+1) (x) ≤ M = n! 1 +
(2/3)n+1
"
"
#
#
n+1
n+1
n+1 n+1
1
(1/3)n+1
M
1
1
1
1
1
=
+
+
0.000005 ≥
=
(n + 1)! 3
n+1
3
(2/3)n+1
n+1
3
2
24. Set x = 1/4 in Formula (17). Follow the argument of Exercise 23: parts (a) and (b) remain
n+1
M
1
≤ 0.000005 for x in the interval
unchanged; in part (c) replace (1/3) with (1/4):
(n + 1)! 4
From part (b), together with 0 ≤ x ≤ 1/4, 1 + x ≥ 1, 1 − x ≥ 3/4, follows
n+1
M
1
1
.
Part
(e)
now
becomes
0.000005
≥
part (d): M = n! 1 +
(3/4)n+1
(n + 1)! 4
"
n+1 #
n+1
1
1
1
=
+
, which is true for n = 9.
n+1
4
3
[0, 1/4].
25. f (x) = cos x, f (n+1) (x) = ± sin x or ± cos x, |f (n+1) (x)| ≤ 1, set M = 1,
1
|x − a|n+1
|x − a|n+1 , lim
= 0 so lim Rn (x) = 0 for all x.
n→+∞ (n + 1)!
n→+∞
(n + 1)!
|Rn (x)| ≤
26. f (x) = sin x, f (n+1) (x) = ± sin x or ± cos x, |f (n+1) (x)| ≤ 1, follow Exercise 25.
π
27. (a) From Machin’s formula and a CAS, ≈ 0.7853981633974483096156609, accurate to the 25th
4
decimal place.
(b)
n
0
1
2
3
1/p
sn
0.3183098 78 . . .
0.3183098 861837906 067 . . .
0.3183098 861837906 7153776 695 . . .
0.3183098 861837906 7153776 752674502 34 . . .
0.3183098 861837906 7153776 752674502 87 . . .
f (h) − f (0)
e−1/h
= lim
, let t = 1/h then h = 1/t and
h→0
h→0
h
h
2
28. (a) f 0 (0) = lim
2
e−1/h
t
1
e−1/h
= lim te−t = lim t2 = lim
= 0 so
lim
2 = 0, similarly
t
−
t→+∞
t→+∞ e
t→+∞ 2te
h
h
h→0
2
lim
h→0+
f 0 (0) = 0.
2
397
Chapter 11
(b) The Maclaurin series is 0 + 0 · x + 0 · x2 + · · · = 0, but f (0) = 0 and f (x) > 0 if x 6= 0 so the
series converges to f (x) only at the point x = 0.
EXERCISE SET 11.10
1
= 1 − x + x2 − · · · + (−1)k xk + · · · ; R = 1.
1+x
1
(b) Replace x with x2 :
= 1 + x2 + x4 + · · · + x2k + · · · ; R = 1.
1 − x2
1
(c) Replace x with 2x :
= 1 + 2x + 4x2 + · · · + 2k xk + · · · ; R = 1/2.
1 − 2x
1. (a) Replace x with −x :
(d)
1/2
1
1
1
1
1
1
=
; replace x with x/2 :
= + 2 x + 3 x2 + · · · + k+1 xk + · · · ; R = 2.
2−x
1 − x/2
2−x
2 2
2
2
2. (a) Replace x with −x : ln(1 − x) = −x − x2 /2 − x3 /3 − · · · − xk /k − · · · ; R = 1.
(b) Replace x with x2 : ln(1 + x2 ) = x2 − x4 /2 + x6 /3 − · · · + (−1)k−1 x2k /k + · · · ; R = 1.
(c) Replace x with 2x : ln(1+2x) = 2x−(2x)2 /2+(2x)3 /3−· · ·+(−1)k−1 (2x)k /k+· · · ; R = 1/2.
(d) ln(2 + x) = ln 2 + ln(1 + x/2); replace x with x/2 :
ln(2 + x) = ln 2 + x/2 − (x/2)2 /2 + (x/2)3 /3 + · · · + (−1)k−1 (x/2)k /k + · · · ; R = 2.
1
1·3 2 1·3·5 3
1
=1− x+ 2
x − 3
x + · · ·, so
2
2 · 2!
2 · 3!
1+x
1
1
1
1·3 2
1·3·5 3
x − 13/2
x + ···
= 1/2 − 5/2 x + 9/2
=√ p
2
2
2
·
2!
2
· 3!
2 1 + x/2
3. (a) From Section 11.9, Example 5(b), √
(2 + x)−1/2
(b) Example 5(a):
1
1
= 1 − 2x + 3x2 − 4x3 + · · ·, so
= 1 + 2x2 + 3x4 + 4x6 + · · ·
(1 + x)2
(1 − x2 )2
1
1/a
=
= 1/a + x/a2 + x2 /a3 + · · · + xk /ak+1 + · · · ; R = |a|.
a−x
1 − x/a
1
1
1
= 2 1 − 2(x/a) + 3(x/a)2 − 4(x/a)3 + · · ·
(b) 1/(a + x)2 = 2
2
a (1 + x/a)
a
4. (a)
=
2
3
4
1
− 3 x + 4 x2 − 5 x3 + · · ·
a2
a
a
a
23 3 25 5 27 7
x + x − x + · · ·; R = +∞
3!
5!
7!
1
1
(c) 1 + x2 + x4 + x6 + · · ·; R = +∞
2!
3!
5. (a) 2x −
4
(b) 1 − 2x + 2x2 − x3 + · · ·; R = +∞
3
π2 4 π4 6 π6 8
x +
x −
x + · · · ; R = +∞
(d) x2 −
2
4!
6!
24
26
22
6. (a) 1 − x2 + x4 − x6 + · · ·; R = +∞
2!
4!
6!
1
1
1
1
(b) x2 1 + x + x2 + x3 + · · · = x2 + x3 + x4 + x5 + · · ·; R = +∞
2!
3!
2!
3!
1
1
1
1
(c) x 1 − x + x2 − x3 + · · · = x − x2 + x3 − x4 + · · ·; R = +∞
2!
3!
2!
3!
(d) x2 −
1 6
1
1
x + x10 − x14 + · · ·; R = +∞
3!
5!
7!
Exercise Set 11.10
398
7. (a) x2 1 − 3x + 9x2 − 27x3 + · · · = x2 − 3x3 + 9x4 − 27x5 + · · ·; R = 1/3
23
25
27
23 3 25 5 27 7
(b) x 2x + x + x + x + · · · = 2x2 + x4 + x6 + x8 + · · ·; R = +∞
3!
5!
7!
3!
5!
7!
(c) Substitute
3
x − x3 +
2
8. (a)
3/2 for m and −x2 for x in Equation (22) of Section 11.9, then multiply by x:
3 5
1
x + x7 + · · ·; R = 1
8
16
−x
x
=
= −x 1 + x + x2 + x3 + · · · = −x − x2 − x3 − x4 − · · · ; R = 1.
x−1
1−x
(b) 3 +
3 4
3
3
x + x8 + x12 + · · ·; R = +∞
2!
4!
6!
(c) From Table 11.9 with m = −3, (1 + x)−3 = 1 − 3x + 6x2 − 10x3 + · · ·, so
x(1 + 2x)−3 = x − 6x2 + 24x3 − 80x4 + · · ·; R = 1/2
1
22 2 24 4 26 6
1
1 − 1 − x + x − x + ···
9. (a) sin x = (1 − cos 2x) =
2
2
2!
4!
6!
2
23 4 25 6 27 8
x + x − x + ···
4!
6!
8!
(b) ln (1 + x3 )12 = 12 ln(1 + x3 ) = 12x3 − 6x6 + 4x9 − 3x12 + · · ·
= x2 −
1
22 2 24 4 26 6
1
1 + 1 − x + x − x + ···
10. (a) cos x = (1 + cos 2x) =
2
2
2!
4!
6!
2
= 1 − x2 +
23 4 25 6
x − x + ···
4!
6!
(b) In Equation (17) of Section 11.9 replace x with −x : ln
11. (a)
1−x
1+x
1 3 1 5
= −2 x + x + x + · · ·
3
5
1
1
=
= 1 + (1 − x) + (1 − x)2 + · · · + (1 − x)k + · · ·
x
1 − (1 − x)
= 1 − (x − 1) + (x − 1)2 − · · · + (−1)k (x − 1)k + · · ·
(b) (0, 2)
12. (a)
1/x0
1
=
= 1/x0 − (x − x0 )/x20 + (x − x0 )2 /x30 − · · · + (−1)k (x − x0 )k /xk+1
+···
0
x
1 + (x − x0 )/x0
(b) (0, 2x0 )
13. (a) (1 + x + x2 /2 + x3 /3! + x4 /4! + · · ·)(x − x3 /3! + x5 /5! − · · ·) = x + x2 + x3 /3 − x5 /30 + · · ·
(b) (1 + x/2 − x2 /8 + x3 /16 − (5/128)x4 + · · ·)(x − x2 /2 + x3 /3 − x4 /4 + x5 /5 − · · ·)
= x − x3 /24 + x4 /24 − (71/1920)x5 + · · ·
1
1
1 6
3
25
331 6
x · · · = 1 − x2 + x4 −
x +···
14. (a) (1 − x2 + x4 /2 − x6 /6 + · · ·) 1 − x2 + x4 −
2
24
720
2
24
720
1 2
11
1
5 3
1
41
4 2
1 + x − x + x − · · · = 1 + x + x2 + x3 + · · ·
(b)
1 + x + ···
3
3
9
81
3
9
81
399
Chapter 11
15. (a)
(b)
16. (a)
(b)
1
1
1
1
1
5
61 6
=1
1 − x2 + x4 − x6 + · · · = 1 + x2 + x4 +
x + ···
cos x
2!
4!
6!
2
24
720
x5
x2
x3
x4
x3
1
1
sin x
+
−
·
·
·
1
+
x
+
+
+
+
·
·
·
= x − x2 + x3 − x5 + · · ·
=
x
−
ex
3!
5!
2!
3!
4!
3
30
tan−1 x
2
2
= x − x3 /3 + x5 /5 − · · · / (1 + x) = x − x2 + x3 − x4 · · ·
1+x
3
3
1
5
7
ln(1 + x)
= x − x2 /2 + x3 /3 − x4 /4 + · · · / (1 − x) = x + x2 + x3 + x4 + · · ·
1−x
2
6
12
17. ex = 1 + x + x2 /2 + x3 /3! + · · · + xk /k! + · · · , e−x = 1 − x + x2 /2 − x3 /3! + · · · + (−1)k xk /k! + · · · ;
1 x
e − e−x = x + x3 /3! + x5 /5! + · · · + x2k+1 /(2k + 1)! + · · · , R = +∞
2
1 x
e + e−x = 1 + x2 /2 + x4 /4! + · · · + x2k /(2k)! + · · · , R = +∞
cosh x =
2
sinh x =
18. tanh x =
19.
x + x3 /3! + x5 /5! + · · ·
1
2
17 7
= x − x3 + x5 −
x + ···
1 + x2 /2 + x4 /4! + · · ·
3
15
315
−1
3
4x − 2
=
+
= − 1 + x + x2 + x3 + x4 + · · · + 3 1 − x + x2 − x3 + x4 + · · ·
x2 − 1
1−x 1+x
= 2 − 4x + 2x2 − 4x3 + 2x4 + · · ·
20.
1
2
x3 + x2 + 2x − 2
=x+1−
+
x2 − 1
1−x 1+x
= x + 1 − 1 + x + x2 + x3 + x4 + · · · + 2 1 − x + x2 − x3 + x4 + · · ·
= 2 − 2x + x2 − 3x3 + x4 − · · ·
d
1 − x2 /2! + x4 /4! − x6 /6! + · · · = −x + x3 /3! − x5 /5! + · · · = − sin x
dx
d
x − x2 /2 + x3 /3 − · · · = 1 − x + x2 − · · · = 1/(1 + x)
(b)
dx
21. (a)
d
x + x3 /3! + x5 /5! + · · · = 1 + x2 /2! + x4 /4! + · · · = cosh x
dx
1
d
x − x3 /3 + x5 /5 − x7 /7 + · · · = 1 − x2 + x4 − x6 + · · · =
(b)
dx
1 + x2
22. (a)
Z
23. (a)
Z
(b)
1 + x + x2 /2! + · · · dx = (x + x2 /2! + x3 /3! + · · ·) + C1
= 1 + x + x2 /2! + x3 /3! + · · · + C1 − 1 = ex + C
x + x3 /3! + x5 /5! + · · · = x2 /2! + x4 /4! + · · · + C1
= 1 + x2 /2! + x4 /4! + · · · + C1 − 1 = cosh x + C
Exercise Set 11.10
Z
24. (a)
Z
(b)
400
x − x3 /3! + x5 /5! − · · · dx = x2 /2! − x4 /4! + x6 /6! − · · · + C1
= − 1 − x2 /2! + x4 /4! − x6 /6! + · · · + C1 + 1
= − cos x + C
1 − x + x2 − · · · dx = x − x2 /2 + x3 /3 − · · · + C = ln(1 + x) + C
(Note: −1 < x < 1, so |1 + x| = 1 + x)
25. (a) Substitute x2 for x in the Maclaurin Series for 1/(1 − x) (Table 11.9.1)
and then multiply by x:
∞
∞
k=0
k=0
X
X
x
=x
(x2 )k =
x2k+1
2
1−x
(b) f (5) (0) = 5!c5 = 5, f (6) (0) = 6!c6 = 0
26. x2 cos 2x =
∞
X
(−1)k 22k
(2k)!
k=0
27. (a)
lim
x→0
(c) f (n) (0) = n!cn =
(
n!
if n odd
0
if n even
x2k+2 ; f (99) (0) = 0 because c99 = 0.
sin x
= lim 1 − x2 /3! + x4 /5! − · · · = 1
x→0
x
x − x3 /3 + x5 /5 − x7 /7 + · · · − x
tan−1 x − x
= lim
= −1/3
(b) lim
x→0
x→0
x3
x3
28. (a)
x2 /2! − x4 /4! + x6 /6! − · · ·
1 − cos x 1 − 1 − x2 /2! + x4 /4! − x6 /6! + · · ·
=
=
3
5
sin x
x − x /3! + x /5! − · · ·
x − x3 /3! + x5 /5! − · · ·
0
1 − cos x
x/2! − x3 /4! + x5 /6! − · · ·
, x 6= 0; lim
= =0
2
4
x→0
1 − x /3! + x /5! − · · ·
sin x
1
1 √
1 1
ln 1 + x − sin 2x = lim
ln(1 + x) − sin 2x
(b) lim
x→0 x
x→0 x 2
1 2 1 3
4 3
4 5
1 1
x − x + x − · · · − 2x − x + x − · · ·
= lim
x→0 x 2
2
3
3
15
3
3 1
= lim − − x + x2 + · · · = −3/2
x→0
2 4
2
=
Z
1
sin x
29.
2
Z
1 6
1 10
1 14
x − x + x − x + · · · dx
3!
5!
7!
2
dx =
0
1
0
=
1
1 7
1
1
1 3
x −
x +
x11 −
x15 + · · ·
3
7 · 3!
11 · 5!
15 · 7!
0
=
1
1
1
1
−
+
−
+ ···,
3 7 · 3! 11 · 5! 15 · 7!
1
< 0.5 × 10−3 so
but
15 · 7!
Z
1
sin(x2 )dx ≈
0
1
1
1
−
+
≈ 0.3103
3 7 · 3! 11 · 5!
401
Chapter 11
Z
1/2
30.
−1
tan
2
2x
0
Z
1/2
2
=
but
2 1
8 1
32 1
128 1
−
+
−
− ···,
3
7
11
32
21 2
55 2
105 215
128
< 0.5 × 10−4 so
105 · 215
0.2
1 + x4
31.
1/3
Z
0.2
but
(1 + x2 )−1/4 dx =
0
Z
1/2
Z
0.2
(1 + x4 )1/3 dx ≈ 0.200
0
5
15 6
1
x + · · · dx
1 − x2 + x4 −
4
32
128
1/2
1
15 7
1 3
x + x5 −
x + ···
12
32
896
0
= 1/2 −
but
0
0
=x−
2
8
32
−
+
≈ 0.0806
3 · 23
21 · 27
55 · 211
tan−1 (2x2 )dx ≈
0.2
1
1
1
1 5
x − x9 + · · ·
= 0.2 + (0.2)5 − (0.2)9 + · · · ,
15
81
15
81
0
1
(0.2)5 < 0.5 × 10−3 so
15
1/2
32.
1/2
1
1
1 + x4 − x8 + · · · dx
3
9
0
=x+
Z
dx =
0
Z
8 6 32 10 128 14
x + · · · dx
2x − x + x −
dx =
3
5
7
0
1/2
8
32
128 15
2
x + ···
= x3 − x7 + x11 −
3
21
55
105
0
Z
1
1
15
(1/2)3 + (1/2)5 −
(1/2)7 + · · · ,
12
32
896
15
(1/2)7 < 0.5 × 10−3 so
896
Z
1/2
(1 + x2 )−1/4 dx ≈ 1/2 −
0
1
1
(1/2)3 + (1/2)5 ≈ 0.4906
12
32
"∞
#
"∞
#
∞
X
X
d X k
d
x
1
k−1
=
x
=
x
x
kx
kxk
33. (a)
=
x
=
(1 − x)2
dx 1 − x
dx
k=0
k=1
k=1
#
Z "X
Z
∞
1
dx − C =
(b) − ln(1 − x) =
xk dx − C
1−x
k=0
=
∞
X
k=0
xk+1
−C =
k+1
∞
X
k=1
xk
− C, − ln(1 − 0) = 0 so C = 0.
k
(c) Replace x with −x in part (b): ln(1 + x) = −
+∞
X
(−1)k
k=1
(d)
+∞
X
(−1)k+1
k=1
k
k
k
x =
+∞
X
(−1)k+1
k=1
k
xk
converges by the Alternating Series Test.
(e) By parts (c) and (d) and the remark,
+∞
X
(−1)k+1
k=1
k
xk converges to ln(1 + x) for −1 < x ≤ 1.
Exercise Set 11.10
402
34. (a) In Exercise 33(a), set x =
1/3
1
3
,S=
=
2
3
(1 − 1/3)
4
(b) In part (b) set x = 1/4, S = ln(4/3)
35. (a) sinh
(b)
−1
(c) In part (e) set x = 1, S = ln 2
1 2 3 4
5 6
x=
1+x
dx − C =
1 − x + x − x + · · · dx − C
2
8
16
3
5 7
1
x + · · · − C; sinh−1 0 = 0 so C = 0.
= x − x3 + x5 −
6
40
112
1 + x2
Z
−1/2
Z
2 −1/2
=1+
=1+
∞
X
(−1/2)(−3/2)(−5/2) · · · (−1/2 − k + 1)
k!
k=1
∞
X
(−1)k
k=1
sinh−1 x = x +
∞
X
(x2 )k
1 · 3 · 5 · · · (2k − 1) 2k
x ,
2k k!
1 · 3 · 5 · · · (2k − 1) 2k+1
x
2k k!(2k + 1)
(−1)k
k=1
(c) R = 1
−1
36. (a) sin
(b)
1 2 3 4
5 6
x = (1 − x )
dx − C =
1 + x + x + x + · · · dx − C
2
8
16
3
5 7
1
x + · · · − C, sin−1 0 = 0 so C = 0
= x + x3 + x5 +
6
40
112
Z
1 − x2
−1/2
Z
2 −1/2
=1+
=1+
=1+
sin−1 x = x +
∞
X
(−1/2)(−3/2)(−5/2) · · · (−1/2 − k + 1)
k=1
∞
X
k=1
∞
X
k=1
∞
X
k=1
k!
−x2
k
(−1)k (1/2)k (1)(3)(5) · · · (2k − 1)
(−1)k x2k
k!
1 · 3 · 5 · · · (2k − 1) 2k
x
2k k!
1 · 3 · 5 · · · (2k − 1) 2k+1
x
2k k!(2k + 1)
(c) R = 1
37. (a) y(t) = y0
∞
X
(−1)k (0.000121)k tk
k=0
k!
i
(b) y(1) ≈ y0 (1 − 0.000121t)
t=1
= 0.999879y0
(c) y0 e−0.000121 ≈ 0.9998790073y0
38. (a) If
ct
ct
≈ 0 then e−ct/m ≈ 1 − , and v(t) ≈
m
m
1−
ct
m
v0 +
cv
mg mg
0
−
= v0 −
+ g t.
c
c
m
(b) The quadratic approximation is
cv
(ct)2
c2
ct
mg mg
mg 2
0
+
−
= v0 −
+g t+
t .
v0 +
v0 +
v0 ≈ 1 −
2
2
m
2m
c
c
m
2m
c
403
Chapter 11
39. θ0 = 5◦ = π/36 rad, k = sin(π/72)
r
p
L
= 2π 1/9.8 ≈ 2.00709
(a) T ≈ 2π
g
r
(b) T ≈ 2π
L
g
1+
k2
4
≈ 2.008044621
(c) 2.008045644
40. The third order model gives the same result as the second, because there is no term of degree three
Z π/2
1·3π
in (5). By the Wallis sine formula,
, and
sin4 φ dφ =
2·4 2
0
s
r Z π/2
1·3 4 4
1 2 2
L
L π k 2 π 3k 4 3π
+
+
1 + k sin φ + 2 k sin φ dφ = 4
T ≈4
g 0
2
2 2!
g 2
2 4
8 16
s
k2
9k 4
L
1+
+
= 2π
g
4
64
41. (a) F =
mg
mgR2
=
= mg 1 − 2h/R + 3h2 /R2 − 4h3 /R3 + · · ·
2
2
(R + h)
(1 + h/R)
(b) If h = 0, then the binomial series converges to 1 and F = mg.
(c) Sum the series to the linear term, F ≈ mg − 2mgh/R.
(d)
2h
2 · 29,028
mg − 2mgh/R
=1−
=1−
≈ 0.9973, so about 0.27% less.
mg
R
4000 · 5280
42. (a) We can differentiate term-by-term:
0
y =
∞
X
k=1
xy 00 + y 0 + xy =
xy 00 + y 0 + xy =
k=0
k=0
k=0
22k+1 (k + 1)!k!
∞
X
(−1)k+1 x2k+1
k=0
(b) y 0 =
∞
∞
X
(−1)k+1 (2k + 1)x2k+1
k=0
∞
X
∞
X (−1)k+1 x2k+1
X (−1)k+1 (2k + 1)x2k
(−1)k x2k−1
00
=
,
y
, and
=
22k−1 k!(k − 1)!
22k+1 (k + 1)!k!
22k+1 (k + 1)!k!
22k (k!)2
+
∞
∞
X
X
(−1)k+1 x2k+1
(−1)k x2k+1
+
,
22k+1 (k + 1)!k!
22k (k!)2
k=0
k=0
1
2k + 1
+
−1 =0
2(k + 1) 2(k + 1)
∞
(−1)k (2k + 1)x2k 00 X (−1)k (2k + 1)x2k−1
, y =
.
22k+1 k!(k + 1)!
22k (k − 1)!(k + 1)!
k=1
Since J1 (x) =
∞
X
k=0
∞
X (−1)k−1 x2k+1
(−1)k x2k+1
and x2 J1 (x) =
, it follows that
2k+1
2
k!(k + 1)!
22k−1 (k − 1)!k!
k=1
x2 y 00 + xy 0 + (x2 − 1)y
=
∞
X
(−1)k (2k + 1)x2k+1
k=1
22k (k − 1)!(k + 1)!
+
∞
X
(−1)k (2k + 1)x2k+1
−
∞
X
k=0
∞
=
x x X (−1)k x2k+1
− +
2
2
22k−1 (k − 1)!k!
k=1
22k+1 (k!)(k + 1)!
k=0
+
∞
X
(−1)k−1 x2k+1
22k−1 (k − 1)!k!
k=1
(−1)k x2k+1
22k+1 k!(k + 1)!
2k + 1
1
2k + 1
+
−1−
2(k + 1) 4k(k + 1)
4k(k + 1)
= 0.
Chapter 11 Supplementary Exercises
(c) From part (a),
J00 (x)
404
∞
X
(−1)k+1 x2k+1
= −J1 (x).
=
22k+1 (k + 1)!k!
k=0
43. If
∞
X
ak xk =
k=0
∞
X
bk xk for x in (−r, r) then
k=0
∞
X
(ak −bk )xk = 0 so by Theorem 11.10.6
k=0
∞
X
(ak −bk )xk
k=0
is the Taylor series for f (x) = 0 about 0 and hence ak − bk = 0, ak = bk for all k.
CHAPTER 11 SUPPLEMENTARY EXERCISES
4. (a)
∞
X
f (k) (0)
k=0
9. (a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
(j)
(k)
(l)
k!
xk
(b)
∞
X
f (k) (x0 )
k=0
k!
(x − x0 )k
always true by Theorem 11.4.2
P
sometimes false, for example the harmonic series diverges but (1/k2 ) converges
sometimes false, for example f (x) = sin πx, ak = 0, L = 0
always true by Example 3(d) of Section 11.1
1
1
sometimes false, for example an = + (−1)n
2
4
sometimes false, for example uk = 1/2
always false by Theorem 11.4.3
sometimes false, for example uk = vk = (2/k)
always true by the Comparison Test
always true by the Comparison Test
P
sometimes false, for example (−1)k /k
P
sometimes false, for example (−1)k /k
10. (a) false, f (x) is not differentiable at x = 0, Definition 11.5.4
(b) true: sn = 1 if n is odd and s2n = 1 + 1/(n + 1);
lim sn = 1
n→+∞
(c) false, lim ak 6= 0
(b) 1/(5k + 1) < 1/5k , converges
11. (a) geometric, r = 1/5, converges
(c)
∞
9
9
9 X 9
√
√ = √ ,
√ diverges
≥√
k+1
k+ k
2 k k=1 2 k
12. (a) converges by Alternating Series Test
k
∞
X
k+2
converges by the Root Test.
(b) absolutely convergent,
3k − 1
k=1
(c)
13. (a)
∞
X
k −1
1
k −1/2
1
>
=
,
diverges
2
2+1
3k
3k
2 + sin k
k=1
∞
∞
X
1
1 X
1
3
<
converges by the Comparison Test
,
1/k
converges,
so
k 3 + 2k + 1
k3
k 3 + 2k + 1
k=1
k=1
(b) Limit Comparison Test, compare with the divergent series
∞
X
1
, diverges
2/5
k
k=1
405
Chapter 11
∞
∞
X
cos(1/k)
1 X 1
cos(1/k)
converges, so
converges absolutely
< 2,
k2
k
k2
k2
(c)
k=1
k=1
∞
∞
X
X
ln k
ln k
√ =
√ because ln 1 = 0,
k k k=2 k k
k=1
14. (a)
Z
+∞
2
`
∞
X
√
4
2 ln x
ln x
ln k
dx
=
lim
−
=
2(ln
2
+
2)
so
converges
−
3/2
1/2
1/2
`→+∞
x
x
x
k 3/2
2
k=2
∞
X
k 4/3
k 4/3
1
1
≥
=
,
diverges
8k 2 + 5k + 1
8k 2 + 5k 2 + k 2
14k 2/3 k=1 14k 2/3
(b)
(c) absolutely convergent,
∞
X
k=1
k2
P
1
converges (compare with
1/k 2 )
+1
∞
∞
99
∞
X
X
X
1 X 1
1
1
1
1
−
=
= 100
=
15.
5k
5k
5k
5
5k
4 · 599
k=0
k=0
16. no, lim ak =
k→+∞
k=100
k=0
1
6= 0 (Divergence Test)
2
17. (a) p0 (x) = 1, p1 (x) = 1 − 7x, p2 (x) = 1 − 7x + 5x2 , p3 (x) = 1 − 7x + 5x2 − 4x3 ,
p4 (x) = 1 − 7x + 5x2 − 4x3
(b) If f (x) is a polynomial of degree n and k ≥ n then the Maclaurin polynomial of degree k is
the polynomial itself; if k < n then it is the truncated polynomial.
18. ln(1 + x) = x − x2 /2 + · · · ; so |ln(1 + x) − x| ≤ x2 /2 by Theorem 11.7.2.
19. sin x = x − x3 /3! + x5 /5! − x7 /7! + · · · is an alternating series, so
| sin x − x + x3 /3! − x5 /5!| ≤ x7 /7! ≤ π 7 /(47 7!) ≤ 0.00005
Z
1
20.
0
2
1
x4
x6
1
1
1
1
1 − cos x
x
dx =
−
+
− ··· =
−
+
− · · ·, and
< 0.0005,
x
2 · 2! 4 · 4! 6 · 6!
2 · 2! 4 · 4! 6 · 6!
6 · 6!
0
Z
so
0
1
1
1
1 − cos x
dx =
−
= 0.2396 to three decimal-place accuracy.
x
2 · 2! 4 · 4!
21. (a) ρ = lim
k→+∞
2k
k!
1/k
(b) ρ = lim uk
k→+∞
1/k
= lim
k→+∞
= lim
k→+∞
2
√
= 0, converges
k
k!
k
√
= e, diverges
k
k!
22. (a) 11 ≥ 1!; if k k ≥ k!, then (k +1)k+1 ≥ (k +1)k k ≥ (k +1)k! = (k +1)!; mathematical induction
(b)
X 1
X 1
, converges
≤
kk
k!
Chapter 11 Supplementary Exercises
(c)
lim
k→+∞
1
kk
406
1/k
= lim
k→+∞
1
= 0, converges
k
1
1
1
− 2−
=
23. (a) u100 =
uk −
uk = 2 −
100
99
9900
k=1
k=1
1
1
1
− 2−
=
, lim uk = 0
(b) u1 = 1; for k ≥ 2, uk = 2 −
k
k−1
k(k − 1) k→+∞
n
∞
X
X
1
(c)
=2
2−
uk = lim
uk = lim
n→+∞
n→+∞
n
100
X
99
X
k=1
k=1
X
∞
∞
∞
X
2
3 X 2
1
2
1
3
3
−
= 2 (geometric series)
24. (a)
− k =
−
=
2k
3
2k
3k
2 1 − (1/2)
3 1 − (1/3)
(b)
k=1
n
X
k=1
[ln(k + 1) − ln k] = ln(n + 1), so
k=1
(c)
(d)
lim
n→+∞
lim
n→+∞
n
X
1
k=1
n
X
k=1
2
1
1
−
k k+2
∞
X
[ln(k + 1) − ln k] = lim ln(n + 1) = +∞, diverges
n→+∞
k=1
1
= lim
n→+∞ 2
1
1
1
1+ −
−
2 n+1 n+2
=
3
4
π π
π
tan−1 (k + 1) − tan−1 k = lim tan−1 (n + 1) − tan−1 (1) = − =
n→+∞
2
4
4
25. (a) e2 − 1
26. ak+1 =
k=1
(b) sin π = 0
(d) e− ln 3 = 1/3
(c) cos e
k+1
√
1/2
1/4
1/2k
ak = ak = ak−1 = · · · = a1
= c1/2
(a) If c = 1/2 then lim ak = 1.
(b) if c = 3/2 then lim ak = 1.
k→+∞
k→+∞
(x − 100)2
(x − 100)4
+
, so
162
2 · 164
Z 110
(x − 100)4
105
(x − 100)2
1
103
1
√
+
+
1−
dx
=
10
−
p≈ √
162
2 · 164
3 · 162
2 · 5 · 164
16 2π 100
16 2π
27. Expand in a Maclaurin Series about x = 100 : e−[(x−100)/16] = 1 −
2
≈ 0.220678, or about 22.07%.
∞
28. f (x) = xex = x + x2 +
X xk+1
x4
x3
+
+ ··· =
,
2!
3!
k!
k=0
f 0 (x) = (x + 1)ex = 1 + 2x +
∞
∞
X
4x
3x
k+1 k X k+1
+
+ ··· =
x ;
= f 0 (1) = 2e.
2!
3!
k!
k!
2
3
k=0
k=0
1
1
1
+ 2 − 2 + · · · ; since the series all converge absolutely,
2
2
3
4
2
1
1
1 π2
π2
1
1
1
1
1 π2
π
− A = 2 2 + 2 2 + 2 2 + ··· =
1 + 2 + 2 + ··· =
, so A =
=
.
6
2
4
6
2
2
3
2 6
2 6
12
29. Let A = 1 −
30. Compare with 1/k p : converges if p > 1, diverges otherwise.
407
Chapter 11
1
3
3
k+1
|x| =
31. (a) x + x2 + x3 + x4 + · · ·; ρ = lim
k→+∞ 3k + 1
2
14
35
1
converges if |x| < 1, |x| < 3 so R = 3.
3
2 5 2 7
8
k+1 2
3
|x|
(b) −x + x − x + x9 − · · ·; ρ = lim
k→+∞
3
5
35
2k + 1
√
√
1
converges if |x|2 < 1, |x|2 < 2, |x| < 2 so R = 2.
2
32. By the Ratio Test for absolute convergence, ρ =
|x − x0 | < b, diverges if |x − x0 | > b.
∞
X
1
|x|,
3
=
|x − x0 |
|x − x0 |
=
; converges if
b
b
lim
k→+∞
If x = x0 − b,
1 2
|x| ,
2
∞
X
(−1)k diverges; if x = x0 + b,
k=0
1 diverges. The interval of convergence is (x0 − b, x0 + b).
k=0
√
√
√
x x2 x3
( x)2 ( x)4 ( x)6
+
−
+··· = 1− +
−
+ · · ·; if x ≤ 0, then
2!
4!
6!
2! 4!
6!
√
√
√
√
( −x)4
( −x)6
x
x2
x3
( −x)2
cosh( −x) = 1 +
+
+
+ ··· = 1 − +
−
+ · · ·.
2!
4!
6!
2!
4!
6!
33. If x ≥ 0, then cos
√
x = 1−
34. By Exercise 70 of Section 3.5, the derivative of an odd (even) function is even (odd); hence all
odd-numbered derivatives of an odd function are even, all even-numbered derivatives of an odd
function are odd; a similar statement holds for an even function.
(a) If f (x) is an even function, then
2f 0 (0) = lim+
h→0
f (0 + h) − f (0)
f (0) − f (0 − h)
f (0 + h) − f (0 − h)
+ lim
= lim
= 0, so
+
+
h
h
h
h→0
h→0
0
f (0) = 0. If f (x) is even then so is f (2k) , thus f (2k+1) (0) = 0, k = 0, 1, 2, . . . . Hence
c2k+1 = f (2k+1) (0)/(2k + 1)! = 0.
(b) If f (x) is an odd function, then f (2k−1) is even (k = 1, 2, . . .), and thus by Part (a),
f (2k) (0) = 0, c2k = f (2k) (0)/(2k)! = 0.
35.
v2
1− 2
c
Z
−1/2
+∞
36. (a)
n
#
"
v2
1
2 p
≈ 1 + 2 , so K = m0 c
− 1 ≈ m0 c2 (v 2 /(2c2 ) = m0 v 2 /2
2c
1 − v 2 /c2
1
dx < (0.5)10−5 if n > 63.7; let n = 64.
x3.7
(b) sn ≈ 1.10628342; CAS: 1.10628824
CHAPTER 12
Analytic Geometry in Calculus
EXERCISE SET 12.1
1.
(5, 8)
(1, 6 )
(–3, i)
(–5, @)
(3, 3)
(–1, r)
(4, e)
p/ 2
2.
p/ 2
( 32 , L)
(0, c)
0
(–6, –p)
√
3. (a) (3 3, 3)
(2, g)
0
(2, $)
√
(b) (−7/2, 7 3/2)
√
(e) (−7 3/2, 7/2)
√
(c) (3 3, 3)
√
√
4. (a) (−4 2, −4 2)
(d) (5, 0)
√
√
(b) (7 2/2, −7 2/2)
(e) (0, −2)
√ √
(c) (4 2, 4 2)
(f ) (0, 0)
5. (a) both (5, π)
√
√
(d) (8 2, 5π/4), (8 2, −3π/4)
(b) (4, 11π/6), (4, −π/6)
(c) (2, 3π/2), (2, −π/2)
√
(f ) both ( 2, π/4)
(d) (0, 0)
6. (a) (2, 5π/6)
(f ) (−5, 0)
(e) both (6, 2π/3)
(c) (2, −7π/6)
(b) (−2, 11π/6)
(d) (−2, −π/6)
7. (a) (5, 0.6435)
√
(b) ( 29, 5.0929)
(c) (1.2716, 0.6658)
8. (a) (5, 2.2143)
(b) (3.4482, 2.6260)
p
(c) ( 4 + π 2 /36, 0.2561)
9. (a) r2 = x2 + y 2 = 4; circle
(b) y = 4; horizontal line
(c) r2 = 3r cos θ, x2 + y 2 = 3x, (x − 3/2)2 + y 2 = 9/4; circle
(d) 3r cos θ + 2r sin θ = 6, 3x + 2y = 6; line
10. (a) r cos θ = 5, x = 5; vertical line
(b) r2 = 2r sin θ, x2 + y 2 = 2y, x2 + (y − 1)2 = 1; circle
(c) r2 = 4r cos θ + 4r sin θ, x2 + y 2 = 4x + 4y, (x − 2)2 + (y − 2)2 = 8; circle
1 sin θ
, r cos2 θ = sin θ, r2 cos2 θ = r sin θ, x2 = y; parabola
(d) r =
cos θ cos θ
11. (a) r cos θ = 7
(b) r = 3
(c) r − 6r sin θ = 0, r = 6 sin θ
2
(d) 4(r cos θ)(r sin θ) = 9, 4r2 sin θ cos θ = 9, r2 sin 2θ = 9/2
12. (a) r sin θ = −3
(b) r =
(c) r2 + 4r cos θ = 0, r = −4 cos θ
(d) r4 cos2 θ = r2 sin2 θ, r2 = tan2 θ, r = tan θ
408
√
5
409
Chapter 12
p/ 2
13.
2
14.
3
-2
-3
2
0
3
-2
-3
r = 2 cos 3θ
r = 3 sin 2θ
15.
16.
p/ 2
p/2
0
-4
-1
4
0
r = 3 − 4 sin 3θ
r = 2 + 2 sin θ
17. (a) r = 5
(b) (x − 3)2 + y 2 = 9, r = 6 cos θ
(c) Example 6, r = 1 − cos θ
18. (a) From (8-9), r = a ± b sin θ or r = a ± b cos θ. The curve is not symmetric about the y-axis,
so Theorem 12.2.1(a) eliminates the sine function, thus r = a ± b cos θ. The cartesian point
(−3, 0) is either the polar point (3, π) or (−3, 0), and the cartesian point (−1, 0) is either
the polar point (1, π) or (−1, 0). A solution is a = 1, b = −2; we may take the equation as
r = 1 − 2 cos θ.
(b) x2 + (y + 3/2)2 = 9/4, r = −3 sin θ
(c) Figure 12.1.18, a = 1, n = 3, r = sin 3θ
19. (a) Figure 12.1.18, a = 3, n = 2, r = 3 sin 2θ
(b) From (8-9), symmetry about the y-axis and Theorem 12.1.1(b), the equation is of the form
r = a ± b sin θ. The cartesian points (3, 0) and (0, 5) give a = 3 and 5 = a + b, so b = 2 and
r = 3 + 2 sin θ.
(c) Example 8, r2 = 9 cos 2θ
20. (a) Example 6 rotated through π/2 radian: a = 3, r = 3 − 3 sin θ
(b) Figure 12.1.18, a = 1, r = cos 5θ
(c) x2 + (y − 2)2 = 4, r = 4 sin θ
Exercise Set 12.1
410
21.
22.
23.
2
3
(
Line
Line
24.
Circle
25.
26.
4
2
6
Circle
1
Circle
27.
Cardioid
28.
29.
2
1
2
3
4
6
30.
Cardioid
Cardioid
Circle
31.
10
4
32.
3
1
5
8
1
Cardioid
Limaçon
Cardioid
33.
34.
35.
4
1
3
1
2
7
2
1
Cardioid
36.
Limaçon
Limaçon
37.
3
38.
8
3
4
2
1
7
5
2
Limaçon
Limaçon
Limaçon
411
Chapter 12
39.
40.
5
41.
1
3
3
3
7
Lemniscate
7
Limaçon
Limaçon
42.
43.
1
44.
4
2p
4p
8p
6p
45.
Spiral
Lemniscate
Lemniscate
46.
2p
47.
2p
4p
8p
4p
1
6p
6p
Spiral
Four-petal rose
Spiral
48.
49.
50.
3
9
2
Eight-petal rose
Four-petal rose
1
52.
-1
Three-petal rose
3
53.
1
-3
-1
-3
2
54.
-2
1
55.
2
-2
3
-1
1
-1
Exercise Set 12.1
412
56. 0 ≤ θ ≤ 8π
57. (a) −4π < θ < 4π
58. In I, along the
√x-axis, x = r grows ever slower with θ. In II x = r grows linearly with θ.
Hence I: r = θ; II: r = θ.
59. (a) r = a/ cos θ, x = r cos θ = a, a family of vertical lines
(b) r = b/ sin θ, y = r sin θ = b, a family of horizontal lines
60. The image of (r0 , θ0 ) under a rotation through an angle α is (r0 , θ0 + α). Hence (f (θ), θ) lies on
the original curve if and only if (f (θ), θ + α) lies on the rotated curve, i.e. (r, θ) lies on the rotated
curve if and only if r = f (θ − α).
√
2
(cos θ + sin θ)
2
(b) r = 1 + cos(θ − π/2) = 1 + sin θ
61. (a) r = 1 + cos(θ − π/4) = 1 +
(c) r = 1 + cos(θ − π) = 1 − cos θ
√
(d) r = 1 + cos(θ − 5π/4) = 1 −
2
(cos θ + sin θ)
2
62. r2 = 4 cos 2(θ − π/2) = −4 cos 2θ
63. Either r − 1 = 0 or θ − 1 = 0,
so the graph consists of the
circle r = 1 and the line θ = 1.
p/2
u=1
r=1
0
64. (a) r2 = Ar sin θ + Br cos θ, x2 + y 2 = Ay + Bx, (x − B/2)2 + (y − A/2)2 = (A2 + B 2 )/4, which
1p 2
A + B2.
is a circle of radius
4
(b) Formula (4) follows by setting A = 0, B = 2a, (x − a)2 + y 2 = a2 , the circle of radius a about
(a, 0). Formula (5) is derived in a similar fashion.
65. y = r sin θ = (1 + cos θ) sin θ = sin θ + sin θ cos θ,
dy/dθ = cos θ − sin2 θ + cos2 θ = 2 cos2 θ + cos θ − 1 = (2 cos θ − 1)(cos θ + 1);
dy/dθ = 0 if cos θ = 1/2 or if cos θ = −1; θ = π/3 or π.
√
√
If θ = π/3, π, then y = 3 3/4, 0 so the maximum value of y is 3 3/4 and the polar coordinates
of the highest point are (3/2, π/3).
66. x = r cos θ = (1 + cos θ) cos θ = cos θ + cos2 θ, dx/dθ = − sin θ − 2 sin θ cos θ = − sin θ(1 + 2 cos θ),
dx/dθ = 0 if sin θ = 0 or if cos θ = −1/2; θ = 0, 2π/3, or π. If θ = 0, 2π/3, π, then x = 2, −1/4, 0
so the minimum value of x is −1/4. The leftmost point has polar coordinates (1/2, 2π/3).
67. (a) Let (x1 , y1 ) and (x2 , y2 ) be the rectangular coordinates of the points (r1 , θ1 ) and (r2 , θ2 ) then
p
p
d = (x2 − x1 )2 + (y2 − y1 )2 = (r2 cos θ2 − r1 cos θ1 )2 + (r2 sin θ2 − r1 sin θ1 )2
p
p
= r12 + r22 − 2r1 r2 (cos θ1 cos θ2 + sin θ1 sin θ2 ) = r12 + r22 − 2r1 r2 cos(θ1 − θ2 ).
(b) Let P and Q have polar coordinates (r1 , θ1 ), (r2 , θ2 ), respectively, then the perpendicular
from OQ to OP has length h = r2 sin(θ2 − θ1 ) and A = 12 hr1 = 12 r1 r2 sin(θ2 − θ1 ).
413
Chapter 12
p
p
√
(c) From Part (a), d = 9 + 4 − 2 · 3 · 2 cos(π/6 − π/3) = 13 − 6 3 ≈ 1.615
1
(d) A = 2 sin(5π/6 − π/3) = 1
2
68. (a) 0 = (x2 + y 2 + a2 )2 − a4 − 4a2 x2 = x4 + y 4 + a4 + 2x2 y 2 + 2x2 a2 + 2y 2 a2 − a4 − 4a2 x2
= x4 + y 4 + 2x2 y 2 − 2x2 a2 + 2y 2 a2 = (x2 + y 2 )2 + 2a2 (y 2 − x2 )
= r4 + 2a2 r2 (sin2 θ − cos2 θ) = r4 − 2a2 r2 cos 2θ, so r2 = 2a2 cos 2θ
(b) (x2 + a2 + y 2 )2 − 4x2 a2 = a4 ; (x − a)2 + y 2 (x + a)2 + y 2 = a4 ;
p
p
(x + a)2 + y 2 (x − a) + y 2 = a2
69.
70.
lim y = lim r sin θ = lim
sin θ
=1
θ
lim y = lim± r sin θ = lim±
sin θ
sin θ
1
1
lim
= 1 · lim , so lim y does not exist.
= lim
θ2
θ→0± θ θ→0± θ
θ→0± θ
θ→0±
θ→0+
θ→0+
θ→0±
θ→0+
θ→0
θ→0
71. Note that r → ±∞ as θ approaches odd multiples of π/2;
x = r cos θ = 4 tan θ cos θ = 4 sin θ,
y = r sin θ = 4 tan θ sin θ
so x → ±4 and y → ±∞ as θ approaches
odd multiples of π/2.
72.
lim
θ→(π/2)−
x=
lim
θ→(π/2)−
r cos θ =
lim
θ→(π/2)−
r
u
-4
4
2 sin2 θ = 2, x = 2 is a vertical asymptote.
73. Let r = a sin nθ (the proof for r = a cos nθ is similar). If θ starts at 0, then θ would have to increase
by some positive integer multiple of π radians in order to reach the starting point and begin to
retrace the curve. Let (r, θ) be the coordinates of a point P on the curve for 0 ≤ θ < 2π. Now
a sin n(θ + 2π) = a sin(nθ + 2πn) = a sin nθ = r so P is reached again with coordinates (r, θ + 2π)
thus the curve is traced out either exactly once or exactly twice for 0 ≤ θ < 2π. If for 0 ≤ θ < π,
P (r, θ) is reached again with coordinates (−r, θ + π) then the curve is traced out exactly once for
0 ≤ θ < π, otherwise exactly once for 0 ≤ θ < 2π. But
a sin nθ, n even
a sin n(θ + π) = a sin(nθ + nπ) =
−a sin nθ, n odd
so the curve is traced out exactly once for 0 ≤ θ < 2π if n is even, and exactly once for 0 ≤ θ < π
if n is odd.
EXERCISE SET 12.2
1. (a) dy/dx =
1/2
= 1/(4t); dy/dx
2t
t=−1
(b) x = (2y)2 + 1, dx/dy = 8y, dy/dx
= −1/4; dy/dx
y=±(1/2)
t=1
= 1/4
= ±1/4
2. (a) dy/dx = (4 cos t)/(−3 sin t) = −(4/3) cot t; dy/dx
t=π/4
= −4/3, dy/dx
(b) (x/3)2 + (y/4)2 = 1, 2x/9 + (2y/16)(dy/dx) = 0, dy/dx = −16x/9y,
dy/dx
√
x=3/ 2
√
y=4/ 2
= −4/3; dy/dx
√
x=3/ 2
√
y=−4/ 2
= 4/3
t=7π/4
= 4/3
Exercise Set 12.2
414
3.
d dy dt
1
d2 y
d dy
=
= − 2 (1/2t) = −1/(8t3 ); positive when t = −1,
=
2
dx
dx dx
dt dx dx
4t
negative when t = 1
4.
d2 y
d
=
2
dx
dt
dy
dx
dt
−(4/3)(− csc2 t)
4
=
= − csc3 t; negative at t = π/4, positive at t = 7π/4.
dx
−3 sin t
9
5. dy/dx =
√
√
2/ t
2
√ = 4 t, d2 y/dx2 =
√ = 4, dy/dx
1/(2 t)
1/(2 t)
6. dy/dx =
1
t2
= t, d2 y/dx2 = , dy/dx
t
t
t=2
= 2, d2 y/dx2
t=1
t=2
= 4, d2 y/dx2
t=1
=4
= 1/2
sec2 t
− csc t cot t
= csc t, d2 y/dx2 =
= − cot3 t,
sec t tan t
sec t tan t
√
√
dy/dx t=π/3 = 2/ 3, d2 y/dx2 t=π/3 = −1/(3 3)
7. dy/dx =
d2 y
sinh t
= tanh t,
= sech2 t/ cosh t = sech3 t, dy/dx t=0 = 0, d2 y/dx2 t=0 = 1
cosh t
dx2
dy
dy/dθ
− cos θ d2 y
dx
2
1
d dy
1
=
=
;
/
=
=
=
;
2
2
dx
dx/dθ
2 − sin θ dx
dθ dx
dθ
(2 − sin θ) 2 − sin θ
(2 − sin θ)3
8. dy/dx =
9.
dy
dx
10.
=
θ=π/3
−1/2
−1
d2 y
√
√ ;
=
2 − 3/2
4 − 3 dx2
=
θ=π/3
(2 −
1
√
3/2)3
=
(4 −
8
√
3)3
3 cos φ
d2 y
dφ
dy
d
=
= −3 cot φ;
(−3 cot φ)
= −3(− csc2 φ)(− csc φ) = −3 csc3 φ;
=
2
dx
− sin φ
dx
dφ
dx
dy
dx
φ=5π/6
√ d2 y
= 3 3;
dx2
= −24
φ=5π/6
−e−t
= −e−2t ; for t = 1, dy/dx = −e−2 , (x, y) = (e, e−1 ); y − e−1 = −e−2 (x − e),
et
y = −e−2 x + 2e−1
11. (a) dy/dx =
(b) y = 1/x, dy/dx = −1/x2 , m = −1/e2 , y − e−1 = −
1
1
2
(x − e), y = − 2 x +
e2
e
e
12. dy/dx =
16t − 2
= 8t − 1; for t = 1, dy/dx = 7, (x, y) = (6, 10); y − 10 = 7(x − 6), y = 7x − 32
2
13. dy/dx =
4 cos t
= −2 cot t
−2 sin t
(a) dy/dx = 0 if cot t = 0, t = π/2 + nπ for n = 0, ±1, · · ·
1
(b) dx/dy = − tan t = 0 if tan t = 0, t = nπ for n = 0, ±1, · · ·
2
14. dy/dx =
2t + 1
2t + 1
=
6t2 − 30t + 24
6(t − 1)(t − 4)
(a) dy/dx = 0 if t = −1/2
(b) dx/dy =
6(t − 1)(t − 4)
= 0 if t = 1, 4
2t + 1
415
Chapter 12
15. x = y = 0 when t = 0, π;
lines are y = −2x, y = 2x.
2 cos 2t dy
dy
=
;
dx
cos t
dx
= 2,
t=0
dy
dx
= −2, the equations of the tangent
t=π
16. y(t) = 0 has three solutions, t = 0, ±π/2; the last two correspond to the crossing point.
dy
2
2
For t = ±π/2, m =
=
; the tangent lines are given by y = ± (x − 2).
dx
±π
π
17. If y = 4 then t2 = 4, t = ±2, x = 0 for t = ±2 so (0, 4) is reached when t = ±2.
dy/dx = 2t/(3t2 − 4). For t = 2, dy/dx = 1/2 and for t = −2, dy/dx = −1/2. The tangent lines
are y = ±x/2 + 4.
18. If x = 3 then t2 − 3t + 5 = 3, t2 − 3t + 2 = 0, (t − 1)(t − 2) = 0, t = 1 or 2. If t = 1 or 2 then
y = 1 so (3, 1) is reached when t = 1 or 2. dy/dx = (3t2 + 2t − 10)/(2t − 3). For t = 1, dy/dx = 5,
the tangent line is y − 1 = 5(x − 3), y = 5x − 14. For t = 2, dy/dx = 6, the tangent line is
y − 1 = 6(x − 3), y = 6x − 17.
19. (a)
1
-1
1
-1
(b)
dx
dy
= −3 cos2 t sin t and
= 3 sin2 t cos t are both zero when t = 0, π/2, π, 3π/2, 2π,
dt
dt
so singular points occur at these values of t.
20. (a) when y = 0
(b)
dy
a sin θ
=
= 0 when θ = 2nπ + π/2, n = 0, 1, . . . (which is when y = 0).
dx
a − a cos θ
√
√
21. Substitute θ = π/3, r = 1, and dr/dθ = − 3 in equation (7) gives slope m = 1/ 3.
22. As in Exercise 21, θ = π/4, dr/dθ =
√
2/2, r = 1 +
√
√
2/2, m = −1 − 2
23. As in Exercise 21, θ = 2, dr/dθ = −1/4, r = 1/2, m =
tan 2 − 2
2 tan 2 + 1
√
√
24. As in Exercise 21, θ = π/6, dr/dθ = 4 3a, r = 2a, m = 3 3/5
√
√
25. As in Exercise 21, θ = 3π/4, dr/dθ = −3 2/2, r = 2/2, m = −2
26. As in Exercise 21, θ = π, dr/dθ = 3, r = 4, m = 4/3
r cos θ + (sin θ)(dr/dθ)
cos θ + 2 sin θ cos θ
dy
; if θ = 0, π/2, π,
=
=
dx
−r sin θ + (cos θ)(dr/dθ)
− sin θ + cos2 θ − sin2 θ
then m = 1, 0, −1.
27. m =
28. m =
dy
cos θ(4 sin θ − 1)
=
; if θ = 0, π/2, π then m = −1/2, 0, 1/2.
dx
4 cos2 θ + sin θ − 2
Exercise Set 12.2
416
29. dx/dθ = −a sin θ(1 + 2 cos θ), dy/dθ = a(2 cos θ − 1)(cos θ + 1)
(a) horizontal if dy/dθ = 0 and dx/dθ 6= 0. dy/dθ = 0 when cos θ = 1/2 or cos θ = −1 so θ = π/3,
5π/3, or π; dx/dθ 6= 0 for θ = π/3 and 5π/3. For the singular point θ = π we find that
lim dy/dx = 0. There is a horizontal tangent line at (3a/2, π/3), (0, π), and (3a/2, 5π/3).
θ→π
(b) vertical if dy/dθ 6= 0 and dx/dθ = 0. dx/dθ = 0 when sin θ = 0 or cos θ = −1/2 so θ = 0, π,
2π/3, or 4π/3; dy/dθ 6= 0 for θ = 0, 2π/3, and 4π/3. The singular point θ = π was discussed
in part (a). There is a vertical tangent line at (2a, 0), (a/2, 2π/3), and (a/2, 4π/3).
30. dx/dθ = a(cos2 θ − sin2 θ) = a cos 2θ, dy/dθ = 2a sin θ cos θ = a sin 2θ
(a) horizontal if dy/dθ = 0 and dx/dθ 6= 0. dy/dθ = 0 when θ = 0, π/2, π, 3π/2, 2π; dx/dθ 6= 0
for (0, 0), (a, π/2), (0, π), (−a, 3π/2), (0, 2π); in reality only two distinct points
(b) vertical if dy/dθ 6= 0 and dx/dθ = 0. √
dx/dθ = 0 when
dy/dθ 6= 0
√ θ = π/4, 3π/4,
√ 5π/4, 7π/4; √
there, so vertical tangent line at (a/ 2, π/4), (a/ 2, 3π/4), (−a/ 2, 5π/4), (−a/ 2, 7π/4),
only two distinct points
31. dy/dθ = (d/dθ)(sin2 θ cos2 θ) = (sin 4θ)/2 = 0 at θ = 0, π/4, π/2, 3π/4, π; at the same points,
dx
= 0 at θ = π/2, a singular point; and
dx/dθ = (d/dθ)(sin θ cos3 θ) = cos2 θ(4 cos2 θ − 3). Next,
dθ
θ = 0, π both give the same point, so there are just three points with a horizontal tangent.
32. dx/dθ = 4 sin2 θ − sin θ − 2, dy/dθ = cos θ(1 − 4 sin θ). dy/dθ = 0 when cos θ = 0 or sin θ = 1/4 so
θ = π/2, 3π/2, sin−1 (1/4), or π − sin−1 (1/4); dx/dθ 6= 0 at these points, so there is a horizontal
tangent at each one.
33.
p/2
34.
p/2
p/2
35.
0
0
4
2
36.
θ0 = ±π/4
θ0 = π/2
θ0 = π/6, π/2, 5π/6
p/2
p/2
37.
0
θ0 = 0, π/2
θ0 = 2π/3, 4π/3
Z
2π
adθ = 2πa
0
p/2
38.
3
39. r2 + (dr/dθ)2 = a2 + 02 = a2 , L =
0
0
0
θ0 = 0
417
Chapter 12
Z
2
2
2
2
π/2
2
40. r + (dr/dθ) = (2a cos θ) + (−2a sin θ) = 4a , L =
2adθ = 2πa
−π/2
Z
π
41. r2 + (dr/dθ)2 = [a(1 − cos θ)]2 + [a sin θ]2 = 4a2 sin2 (θ/2), L = 2
2a sin(θ/2)dθ = 8a
0
Z
π
42. r2 + (dr/dθ)2 = [sin2 (θ/2)]2 + [sin(θ/2) cos(θ/2)]2 = sin2 (θ/2), L =
sin(θ/2)dθ = 2
0
Z
43. r2 + (dr/dθ)2 = (e3θ )2 + (3e3θ )2 = 10e6θ , L =
2
√
10e3θ dθ =
√
10(e6 − 1)/3
0
44. r2 + (dr/dθ)2 = [sin3 (θ/3)]2 + [sin2 (θ/3) cos(θ/3)]2 = sin4 (θ/3),
Z π/2
√
sin2 (θ/3)dθ = (2π − 3 3)/8
L=
0
3 sin t
dy
=
dx
1 − 3 cos t
dy
3 sin 10
(b) At t = 10,
=
≈ −0.4640, θ ≈ tan−1 (−0.4640) = −0.4344
dx
1 − 3 cos 10
45. (a) From (3),
dy
dy
= 0 when
= 2 sin t = 0, t = 0, π, 2π, 3π
dx
dt
dx
= 0 when 1 − 2 cos t = 0, cos t = 1/2, t = π/3, 5π/3, 7π/3
(b)
dt
46. (a)
47. (a) r2 + (dr/dθ)2 = (cos nθ)2 + (−n sin nθ)2 = cos2 nθ + n2 sin2 nθ
= (1 − sin2 nθ) + n2 sin2 nθ = 1 + (n2 − 1) sin2 nθ,
Z π/(2n) q
1 + (n2 − 1) sin2 nθdθ
L=2
Z
0
π/4
p
1 + 3 sin2 2θdθ ≈ 2.42
(b) L = 2
0
(c)
n
L
2
3
4
5
6
7
8
9
10
11
2.42211 2.22748 2.14461 2.10100 2.07501 2.05816 2.04656 2.03821 2.03199 2.02721
n
L
13
14
15
16
17
18
19
20
12
2.02346 2.02046 2.01802 2.01600 2.01431 2.01288 2.01167 2.01062 2.00971
48. (a)
p/2
0
2
2
−θ 2
−θ 2
−2θ
Z
+∞
(b) r + (dr/dθ) = (e ) + (−e ) = 2e , L =
0
Z θ0 √
√
√
(c) L = lim
2e−θ dθ = lim
2(1 − e−θ0 ) = 2
θ0 →+∞
0
θ0 →+∞
√
2e−θ dθ
Exercise Set 12.2
418
49. x0 = 2t, y 0 = 2, (x0 )2 + (y 0 )2 = 4t2 + 4
4
Z 4 p
Z 4
p
√
8π 2
8π
7/2
2
2
(t + 1)
(17 17 − 1)
(2t) 4t + 4dt = 8π
t t + 1dt =
=
S = 2π
3
3
0
0
0
50. x0 = et (cos t − sin t), y 0 = et (cos t + sin t), (x0 )2 + (y 0 )2 = 2e2t
Z π/2
√
√ Z π/2 2t
t
2t
S = 2π
(e sin t) 2e dt = 2 2π
e sin t dt
0
√
= 2 2π
π/2
1 2t
e (2 sin t − cos t)
5
0
0
√
2 2
π(2eπ + 1)
=
5
51. x0 = −2 sin t cos t, y 0 = 2 sin t cos t, (x0 )2 + (y 0 )2 = 8 sin2 t cos2 t
π/2
Z π/2
p
√ Z π/2
√
√
2
2
3
4
2
cos t 8 sin t cos t dt = 4 2π
cos t sin t dt = − 2π cos t
= 2π
S = 2π
0
0
52. x0 = 1, y 0 = 4t, (x0 )2 + (y 0 )2 = 1 + 16t2 , S = 2π
0
Z
p
1
t
1 + 16t2 dt =
0
53. x0 = −r sin t, y 0 = r cos t, (x0 )2 + (y 0 )2 = r2 , S = 2π
Z
π
√
π
(17 17 − 1)
24
Z
√
r sin t r2 dt = 2πr2
0
π
sin t dt = 4πr2
0
2 2
dy
dx
dx
dy
= a(1 − cos φ),
= a sin φ,
54.
+
= 2a2 (1 − cos φ)
dφ
dφ
dφ
dφ
Z 2π
Z 2π
p
√
a(1 − cos φ) 2a2 (1 − cos φ) dφ = 2 2πa2
(1 − cos φ)3/2 dφ,
S = 2π
0
0
√
φ
φ
so (1 − cos φ)3/2 = 2 2 sin3 for 0 ≤ φ ≤ π and, taking advantage of the
but 1 − cos φ = 2 sin
2
2
Z π
3 φ
2
symmetry of the cycloid, S = 16πa
sin dφ = 64πa2 /3
2
0
2
55. (a)
dθ
dr
dr/dt
2
dr
= 2 and
= 1 so
=
=
= 2, r = 2θ + C, r = 10 when θ = 0 so
dt
dt
dθ
dθ/dt
1
10 = C, r = 2θ + 10.
(b) r2 + (dr/dθ)2 = (2θ + 10)2 + 4, during the first 5 seconds the rod rotates through an angle
Z 5p
(2θ + 10)2 + 4dθ, let u = 2θ + 10 to get
of (1)(5) = 5 radians so L =
L=
=
1
2
Z
20
10
"
p
0
u2 + 4du =
1 hup
2 2
u2 + 4 + 2 ln |u +
#
i20
p
u2 + 4|
10
√
√
√
1
20 + 404
√
10 404 − 5 104 + 2 ln
≈ 75.7 mm
2
10 + 104
dr
dy
dr
dx
=
cos θ − r sin θ,
= r cos θ +
sin θ,
56. x = r cos θ, y = r sin θ,
dθ
dθ
dθ
dθ
2
2 2
dy
dr
dx
+
= r2 +
, and Formula (6) of Section 8.4 becomes
dθ
dθ
dθ
s
2
Z β
dr
2
r +
dθ
L=
dθ
α
419
Chapter 12
EXERCISE SET 12.3
Z
π
1. (a)
Z
π/2
2π
(d)
0
1
(1 − cos θ)2 dθ
2
(d) 4π /3
0
Z
π/2
−π/2
3
2π
0
π/2
−π/2
3π/4
−π/4
Z
π
5. A = 2
0
Z
π/6
0
Z
0
2π/3
1
sin2 2θ dθ
2
π/4
1
cos2 2θ dθ
2
1 2 2
4a sin θ dθ = πa2
2
1 2
4a cos2 θ dθ = πa2
2
x−
1
2
2
2
1
1
+ y−
=
2
2
1
(sin θ + cos θ)2 dθ = π/2
2
Z
π/2
6. A =
0
1
(1 + sin θ)2 dθ = 3π/8 + 1
2
1
(16 cos2 3θ)dθ = 4π
2
π/2
1
4 sin2 2θ dθ = π/2, so total area = 2π.
2
0
9. A = 2
π
(b) A =
Z
π
0
(f ) π/8
8. The petal in the first quadrant has area
Z
(f ) 2
(e) 3π/4
1
(2 + 2 cos θ)2 dθ = 6π
2
7. A = 6
Z
(c) π/8
(b) A =
0
(b) π/2
4. (a) r2 = r sin θ + r cos θ, x2 + y 2 − y − x = 0,
Z
π/2
(c)
1
(1 − sin θ)2 dθ
2
1 2
a dθ = πa2
2
(c) A =
Z
1
4 cos2 θ dθ
2
(e)
2. (a) 3π/8 + 1
Z
π/2
Z
1 2
θ dθ
2
3. (a) A =
Z
(b)
Z
√
1
(1 + 2 cos θ)2 dθ = π − 3 3/2
2
Z
π/2
11. area = A1 − A2 =
0
Z
12. area = A1 − A2 =
0
π
1
4 cos2 θ dθ −
2
Z
1
π/4
0
1
(1 + cos θ)2 dθ −
2
10. A =
Z
0
3
2
dθ = 4/3
θ2
1
1
cos 2θ dθ = π/2 −
2
4
π/2
1
cos2 θ dθ = 5π/8
2
√
√
13. The circles intersect when cos t = 3 sin t, tan t = 1/ 3, t = π/6, so
Z π/6
Z π/2
√
√
√
1 √
1
(4 3 sin t)2 dt+
(4 cos t)2 dt = 2π −3 3+4π/3− 3 = 10π/3−4 3.
A = A1 +A2 =
2
0
π/6 2
14. The curves intersect when 1 + cos t = 3 cos t, cos t = 1/2, t = ±π/3, and hence
Z π/2
Z π/3
√
√
1
1
(1+cos t)2 dt+2
9 cos2 t dt = 2(π/4+9 3/16+3π/8−9 3/16) = 5π/4.
total area = 2
2
0
π/3 2
Exercise Set 12.3
Z
420
π/2
15. A = 2
π/6
Z
π
1
[16 − (2 − 2 cos θ)2 ]dθ = 10π
2
16. A = 2
0
Z
π/3
√
1
[(2 + 2 cos θ)2 − 9]dθ = 9 3/2 − π
2
π/4
1
(16 sin2 θ)dθ = 2π − 4
2
17. A = 2
0
Z
√
1
[25 sin2 θ − (2 + sin θ)2 ]dθ = 8π/3 + 3
2
18. A = 2
0
"Z
2π/3
19. A = 2
0
Z
π/3
20. A = 2
0
Z
1
(1/2 + cos θ)2 dθ −
2
cos−1 (3/5)
0
Z
π/8
0
π
2π/3
#
√
1
2
(1/2 + cos θ) dθ = (π + 3 3)/4
2
9
9√
1
(2 + 2 cos θ)2 − sec2 θ dθ = 2π +
3
2
4
4
21. A = 2
22. A = 8
Z
1
(100 − 36 sec2 θ)dθ = 100 cos−1 (3/5) − 48
2
1
(4a2 cos2 2θ − 2a2 )dθ = 2a2
2
23. (a) r is not real for π/4 < θ < 3π/4 and 5π/4 < θ < 7π/4
Z π/4
1 2
a cos 2θ dθ = a2
(b) A = 4
2
0
Z π/6 h
i
√
2π
1
4 cos 2θ − 2 dθ = 2 3 −
(c) A = 4
2
3
0
Z
π/2
24. A = 2
0
1
sin 2θ dθ = 1
2
Z
4π
25. A =
2π
1 2 2
a θ dθ −
2
Z
0
2π
1 2 2
a θ dθ = 8π 3 a2
2
26. (a) x = r cos θ, y = r sin θ,
(dx/dθ)2 + (dy/dθ)2 = (f 0 (θ) cos θ − f (θ) sin θ)2 + (f 0 (θ) sin θ + f (θ) cos θ)2 = f 0 (θ)2 + f (θ)2 ;
Z β
p
2πf (θ) sin θ f 0 (θ)2 + f (θ)2 dθ if about θ = 0; similarly for θ = π/2
S=
α
0
0
(b) f , g are continuous and no segment of the curve is traced more than once.
2
dr
= cos2 θ + sin2 θ = 1,
27. r +
dθ
Z π/2
2π cos2 θ dθ = π 2 .
so S =
2
−π/2
421
Chapter 12
Z
π/2
28. S =
√
2πeθ cos θ 2e2θ dθ
0
√ Z
= 2 2π
π/2
e2θ cos θ dθ =
0
Z
π
2π(1 − cos θ) sin θ
29. S =
0
√
Z
p
1 − 2 cos θ + cos2 θ + sin2 θ dθ
π
sin θ(1 − cos θ)3/2 dθ =
= 2 2π
0
Z
√
2 2π π
(e − 2)
5
2 √
2 2π(1 − cos θ)5/2
5
π
= 32π/5
0
π
2πa sin(θ)a dθ = 4πa2
30. S =
0
31. (a) r3 cos3 θ − 3r2 cos θ sin θ + r3 sin3 θ = 0, r =
Z
π/(2n)
32. (a) A = 2
0
πa2
1 2
a cos2 nθ dθ =
2
4n
3 cos θ sin θ
cos3 θ + sin3 θ
Z
(b) A = 2
0
2
(c)
πa
1
× total area =
2n
4n
(d)
π/(2n)
πa2
1 2
a cos2 nθ dθ =
2
4n
1
πa2
× total area =
n
4n
33. If the upper right corner of the square is the point (a, a) then the large circle has equation r =
and the small circle has equation (x − a)2 + y 2 = a2 , r = 2a cos θ, so
Z π/4 h
i
√
1
(2a cos θ)2 − ( 2a)2 dθ = a2 = area of square.
area of crescent = 2
2
0
√
2a
Exercise Set 12.4
Z
2π
34. A =
0
422
1
(cos 3θ + 2)2 dθ = 9π/2
2
Z
π/2
35. A =
0
1
4 cos2 θ sin4 θ dθ = π/16
2
1
3
-3
3
0
1
–1
-3
EXERCISE SET 12.4
1. (a)
(c)
4px = y 2 , point (1, 1), 4p = 1, x = y 2
a = 3, b = 2,
y2
x2
+
=1
9
4
(b)
−4py = x2 , point (3, −3), 12p = 9, −3y = x2
(d)
a = 3, b = 2,
x2
y2
+
=1
4
9
(e) asymptotes: y = ±x, so a = b; point (0, 1), so y 2 − x2 = 1
(f )
asymptotes: y = ±x, so b = a; point (2, 0), so
x2
y2
−
=1
4
4
2. (a) part (a), vertex (0, 0), p = 1/4; focus (1/4, 0), directrix: x = −1/4
part (b), vertex (0, 0), p = 3/4; focus (0, −3/4), directrix: y = 3/4
√
√
√
(b) part (c), c = a2 − b2 = 5, foci (± 5, 0)
√
√
√
part (d), c = a2 − b2 = 5, foci (0, ± 5)
√
√
√
(c) part (e), c = a2 + b2 = 2, foci at (0, ± 2); asymptotes: y 2 − x2 = 0, y = ±x
√
√
√
√
y2
x2
−
= 0, y = ±x
part (f), c = a2 + b2 = 8 = 2 2, foci at (±2 2, 0); asymptotes:
4
4
y
3. (a)
y
(b)
5
3
y=
F
( 32 , 0)
-3
x
x
-5
3
x=–
3 -3
2
5
-5
y
4. (a)
9
4
(
9
F 0, – 4
)
y
(b)
F(0, 1)
(
5
2
F – ,0
)
x
x
y = –1
x=
5
2
423
Chapter 12
y
5. (a)
x=
6
y
(b)
1
2
x
-4
V(–2, –2)
V(2, 3)
( )
7
F 2, 3
y=–
7
4
(
F –2, – 9
x
4
)
-4
6
y
6. (a)
x=
y
(b)
23
4
F
x
y=
F
( 94 , –1 )
( 12 , 32 )
1
2
x
V(4, –1)
V
y
7. (a)
4
(b)
( 2)
x=–
( 12 , 1 )
y
9
4
V 2, 5
4
V(–2, 2)
(
)
7
F –4, 2
y=3
F(2, 2)
x
x
2
4
8. (a)
y
x = –9
y
(b)
2
V(–4, 3)
(
7
2
F – ,3
(
F –1,
)
V (–1, 1)
17
16
x
y=
15
16
x
9. (a) c2 = 16 − 9 = 7, c =
√
7
(b)
c2 = 9 − 1 = 8, c =
y
(0, 3)
(–4, 0)
y2
x2
+
=1
1
9
(√7, 0)
√
8
y
(0, 3)
(0, √ 8)
x
(4, 0)
x
(– √7, 0)
(0, –3)
(–1, 0)
(1, 0)
(0, – √ 8)
(0, –3)
)
Exercise Set 12.4
424
10. (a) c2 = 25 − 4 = 21, c =
(0, 5)
√
21
(b)
y
y2
x2
+
=1
9
4
√
c2 = 9 − 4 = 5, c = 5
(0, √ 21)
y
(0, 2)
x
(–2, 0)
(√ 5, 0)
(–3, 0)
(2, 0)
x
(0, – √ 21)
(0, –5)
11. (a)
(3, 0)
(– √ 5, 0)
(y − 3)2
(x − 1)2
+
=1
16
9
√
c2 = 16 − 9 = 7, c = 7
(b)
(0, –2)
(y + 1)2
(x + 2)2
+
=1
4
3
c2 = 4 − 3 = 1, c = 1
y
y
(1 – √7, 3)
(1, 6)
(–2, –1 + √ 3)
(1 + √7, 3)
x
(–3, –1)
(–3, 3)
(0, –1)
(–4, –1)
(5, 3)
(–1, –1)
x
(–2, –1 – √ 3)
(1, 0)
12. (a)
(y − 5)2
(x + 3)2
+
=1
16
4
√
c2 = 16 − 4 = 12, c = 2 3
(b)
(y + 2)2
x2
+
=1
4
9
√
c2 = 9 − 4 = 5, c = 5
y
(–3 –2 √ 3, 5)
y
(0, 1)
(–3, 7)
(0, –2 + √ 5)
x
(–7, 5)
(1, 5)
(–3, 3)
(–3 +2 √ 3, 5)
(–2, –2)
(2, –2)
x
(0, –5)
(0, –2 – √ 5)
425
13. (a)
Chapter 12
(y − 1)2
(x + 1)2
+
=1
9
1
√
c2 = 9 − 1 = 8, c = 8
(b)
(y − 5)2
(x + 1)2
+
=1
4
16
√
c2 = 16 − 4 = 12, c = 2 3
y
y
(–1, 9)
(–1 – √8, 1)
(–1, 2)
(–1, 5 + 2 √3)
(2, 1)
(–3, 5)
x
(–4, 1)
(–1, 0)
(1, 5)
(–1, 5 – 2 √3)
(–1 + √8, 1)
x
(–1, 1)
14. (a)
(y − 3)2
(x + 1)2
+
=1
4
9
√
c2 = 9 − 4 = 5, c = 5
(b)
(y + 3)2
(x − 2)2
+
=1
9
5
c2 = 9 − 5 = 4, c = 2
y
(–1, 6)
(2, –3 + √5)
y
(–3, 3)
(1, 3)
(4, –3)
(–1, –3)
(0, –3)
x
(2, –3 – √5)
(–1, 3 – √5)
(–1, 0)
√
15. (a) c2 = a2 + b2 = 16 + 4 = 20, c = 2 5
y
(0, √ 13)
y = –2 x
3
(b) y 2 /4 − x2 /9 = 1
√
c2 = 4 + 9 = 13, c = 13
y
(0, 2)
y= 2x
3
x
y = –1 x
2
(–4, 0)
y= 1x
(4, 0)
2
x
(0, – √ 13)
x
(–1, 3 + √5)
(0, –2)
(–2 √ 5, 0)
(2 √ 5, 0)
(5, –3)
Exercise Set 12.4
426
16. (a) c2 = a2 + b2 = 9 + 25 = 34, c =
√
34
y
(0, √ 34)
(0, 3)
3
5
y=– x
y=
(b) x2 /25 − y 2 /16 = 1
√
c2 = 25 + 16 = 41, c = 41
y
4
5
y=– x
3
x
5 x
y=
4
x
5
(–5, 0) (5, 0)
x
(0, – √ 34)
(0, –3)
(– √41, 0)
17. (a) c2 = 9 + 4 = 13, c =
y
√
13
(√41, 0)
(b) (y + 3)2 /36 − (x + 2)2 /4 = 1
√
c2 = 36 + 4 = 40, c = 2 10
y – 4 = 2 (x – 2)
3
(2 – √13, 4)
(–2, –3 + 2 √10)
(2 + √13, 4)
(5, 4)
y
(–2, 3)
(–1, 4)
x
y + 3 = –3(x + 2)
x
y + 3 = 3(x + 2)
(–2, –9)
y – 4 = – 2 (x – 2)
3
(–2, –3 – 2 √10)
√
18. (a) c2 = 3 + 5 = 8, c = 2 2
y+4=
y
√35 (x – 2)
(2, – 4 + 2 √ 2)
(b) (x + 1)2 /1 − (y − 3)2 /2 = 1
√
c2 = 1 + 2 = 3, c = 3
y − 3 = −√2(x + 1)
y
x
(–2, 3)
(2, – 4 + √ 3)
(-1 − √3, 3)
(2, – 4 – √ 3)
√35 (x – 2)
(-1 + √3, 3)
x
(2, – 4 – 2 √ 2)
y+4=–
(0, 3)
y − 3 = √2(x + 1)
427
Chapter 12
19. (a) (x + 1)2 /4 − (y − 1)2 /1 = 1
√
c2 = 4 + 1 = 5, c = 5
(b) (x − 1)2 /4 − (y + 3)2 /64 = 1
√
c2 = 4 + 64 = 68, c = 2 17
y
y + 3 = 4(x –1)
y – 1 = 1 (x + 1)
y
2
(–3, 1)
(–1 + √5, 1)
(3, –3)
(–1, –3)
x
x
(1, 1)
(–1 – √5, 1)
y – 1 = – 1 (x + 1)
(–1 – 2 √17, –3)
(1 + 2 √17, –3)
2
y + 3 = –4(x –1)
20. (a) (y − 3)2 /4 − (x + 2)2 /9 = 1
√
c2 = 4 + 9 = 13, c = 13
(b) (y + 5)2 /9 − (x + 2)2 /36 = 1
√
c2 = 9 + 36 = 45, c = 3 5
y
y
(–2, 3 + √ 13)
y + 5 = 1 (x + 2)
(-2, 5)
(–2, –5 + 3 √ 5)
y – 3 = 2 (x + 2)
y – 3 = – 2 (x + 2)
3
2
(–2, –2)
x
3
x
(-2, 1)
(–2, 3 – √ 13)
(–2, –8)
(–2, –5 – 3 √ 5)
y + 5 = – 1 (x + 2)
2
21. (a) y 2 = 4px, p = 3, y 2 = 12x
(b) y 2 = −4px, p = 7, y 2 = −28x
22. (a) x2 = −4py, p = 4, x2 = −16y
(b) x2 = −4py, p = 1/2, x2 = −2y
23. (a) x2 = −4py, p = 3, x2 = −12y
(b) The vertex is 3 units above the directrix so p = 3, (x − 1)2 = 12(y − 1).
24. (a) y 2 = 4px, p = 6, y 2 = 24x
(b) The vertex is half way between the focus and directrix so the vertex is at (2, 4), the focus is 3
units to the left of the vertex so p = 3, (y − 4)2 = −12(x − 2)
25. y 2 = a(x − h), 4 = a(3 − h) and 9 = a(2 − h), solve simultaneously to get h = 19/5, a = −5 so
y 2 = −5(x − 19/5)
26. (x − 5)2 = a(y + 3), (9 − 5)2 = a(5 + 3) so a = 2, (x − 5)2 = 2(y + 3)
27. (a) x2 /9 + y 2 /4 = 1
(b) a = 26/2 = 13, c = 5, b2 = a2 − c2 = 169 − 25 = 144; x2 /169 + y 2 /144 = 1
Exercise Set 12.4
428
28. (a) x2 + y 2 /5 = 1
(b) b = 8, c = 6, a2 = b2 + c2 = 64 + 36 = 100; x2 /64 + y 2 /100 = 1
29. (a) c = 1, a2 = b2 + c2 = 2 + 1 = 3; x2 /3 + y 2 /2 = 1
(b) b2 = 16 − 12 = 4; x2 /16 + y 2 /4 = 1 and x2 /4 + y 2 /16 = 1
30. (a) c = 3, b2 = a2 − c2 = 16 − 9 = 7; x2 /16 + y 2 /7 = 1
(b) a2 = 9 + 16 = 25; x2 /25 + y 2 /9 = 1 and x2 /9 + y 2 /25 = 1
31. (a) a = 6, (2, 3) satisfies x2 /36 + y 2 /b2 = 1 so 4/36 + 9/b2 = 1, b2 = 81/8; x2 /36 + y 2 /(81/8) = 1
(b) The center is midway between the foci so it is at (1, 3), thus c = 1, b = 1, a2 = 1 + 1 = 2;
(x − 1)2 + (y − 3)2 /2 = 1
32. (a) Substitute (3, 2) and (1, 6) into x2 /A + y 2 /B = 1 to get 9/A + 4/B = 1 and 1/A + 36/B = 1
which yields A = 10, B = 40; x2 /10 + y 2 /40 = 1
(b) The center is at (2, −1) thus c = 2, a = 3, b2 = 9 − 4 = 5; (x − 2)2 /5 + (y + 1)2 /9 = 1
33. (a) a = 2, c = 3, b2 = 9 − 4 = 5; x2 /4 − y 2 /5 = 1
(b) a = 1, b/a = 2, b = 2; x2 − y 2 /4 = 1
34. (a) a = 3, c = 5, b2 = 25 − 9 = 16; y 2 /9 − x2 /16 = 1
(b) a = 3, a/b = 1, b = 3; y 2 /9 − x2 /9 = 1
35. (a) vertices along x-axis: b/a = 3/2 so a = 8/3; x2 /(64/9) − y 2 /16 = 1
vertices along y-axis: a/b = 3/2 so a = 6; y 2 /36 − x2 /16 = 1
(b) c = 5, a/b = 2 and a2 + b2 = 25, solve to get a2 = 20, b2 = 5; y 2 /20 − x2 /5 = 1
36. (a) foci along the x-axis: b/a = 3/4 and a2 + b2 = 25, solve to get a2 = 16, b2 = 9;
x2 /16 − y 2 /9 = 1
foci along the y-axis: a/b = 3/4 and a2 + b2 = 25 which results in y 2 /9 − x2 /16 = 1
(b) c = 3, b/a = 2 and a2 + b2 = 9 so a2 = 9/5, b2 = 36/5; x2 /(9/5) − y 2 /(36/5) = 1
37. (a) the center is at (6, 4), a = 4, c = 5, b2 = 25 − 16 = 9; (x − 6)2 /16 − (y − 4)2 /9 = 1
(b) The asymptotes intersect at (1/2, 2) which is the center, (y − 2)2 /a2 − (x − 1/2)2 /b2 = 1 is
the form of the equation because (0, 0) is below both asymptotes, 4/a2 − (1/4)/b2 = 1 and
a/b = 2 which yields a2 = 3, b2 = 3/4; (y − 2)2 /3 − (x − 1/2)2 /(3/4) = 1.
38. (a) the center is at (1, −2); a = 2, c = 10, b2 = 100 − 4 = 96; (y + 2)2 /4 − (x − 1)2 /96 = 1
(b) the center is at (1, −1); 2a = 5 − (−3) = 8, a = 4,
(y + 1)2
(x − 1)2
−
=1
16
16
39. (a) y = ax2 + b, (20, 0) and (10, 12) are on the curve so
400a + b = 0 and 100a + b = 12. Solve for b to get
b = 16 ft = height of arch.
(b)
y
y2
100 144
x2
+ 2 = 1,
+ 2 = 1, 400 = a2 , a = 20;
2
a
b
400
b
√
b = 8 3 ft = height of arch.
(10, 12)
x
-20
-10
10
20
429
Chapter 12
40. (a) (x − b/2)2 = a(y − h), but (0, 0) is on the parabola so b2 /4 = −ah, a = −
(x − b/2)2 = −
b2
(y − h)
4h
(b) As in part (a), y = −
4h
(x − b/2)2 + h, A =
b2
Z
b
−
0
b2
,
4h
2
4h
2
(x
−
b/2)
+
h
dx = bh
b2
3
41. We may assume that the vertex is (0, 0) and the parabola opens to the right. Let P (x0 , y0 ) be a
point on the parabola y 2 = 4px, then by the definition of a parabola, P F = distance from P to
directrix x = −p, so P F = x0 + p where x0 ≥ 0 and P F is a minimum when x0 = 0 (the vertex).
42. Let p = distance (in millions of miles) between
the vertex (closest point) and the focus F ,
then P D = P F , 2p + 20 = 40, p = 10 million miles.
P
D
40
p
60°
40 cos 60° = 20
p
Directrix
43. Use an xy-coordinate system so that y 2 = 4px is an equation of the parabola, then (1, 1/2) is a
point on the curve so (1/2)2 = 4p(1), p = 1/16. The light source should be placed at the focus
which is 1/16 ft. from the vertex.
44. (a) Substitute x2 = y/2 into y 2 − 8x2 = 5 to get y 2 − 4y − 5 = 0;
y = −1, 5. Use x2 = y/2 to find that there is no solution if
p
y = −1 and that x = ± 5/2 if y = 5. The curves intersect
p
p
at ( 5/2, 5) and (− 5/2, 5), and thus the area is
Z √5/2 p
√
A=2
( 5 + 8x2 − 2x) dx
y
(– √52 , 5)
( √52 , 5)
x
0
√
√
5√
5 2 √
( 5 − 1) +
2 ln(2 + 5)
=
2
4
(b) Eliminate x to get y 2 = 1, y = ±1. Use either equation
to find that x = ±2 if y = 1 or if y = −1. The curves
intersect at (2, 1), (2, −1), (−2, 1), and (−2, −1),
and thus the area is
Z √5/3 p
1
A=4
1 + 2x2 dx
3
0
Z 2 p
1 p 2
1
1 + 2x2 − √
3x − 5 dx
+4 √
7
5/3 3
=
√
√
√
10 √
5
1√
ln 5
2 ln(2 2 + 3) +
21 ln(2 3 + 7) −
3
21
21
y
(–2, 1)
(2, 1)
x
(–2, –1)
(2, –1)
Exercise Set 12.4
430
(c) Add both equations to get x2 = 4, x = ±2.
√
Use either equation to find that y = ± 3 if x = 2
or if x = −2. The curves intersect at
√
√
√
√
(2, 3), (2, − 3), (−2, 3), (−2, − 3) and thus
Z 2 hp
Z 1p
i
p
7 − x2 dx + 4
7 − x2 − x2 − 1 dx
A=4
0
√
= 4 3 + 14 sin−1
1
y
(–2, √ 3)
(2, √ 3)
x
(–2, – √ 3)
√
√
2√
7 − 4 3 + 2 ln(2 + 3)
7
(2, – √ 3)
45. (a) P : (b cos t, b sin t); Q : (a cos t, a sin t); R : (a cos t, b sin t)
(b) For a circle, t measures the angle between the positive x-axis and the line segment joining
the origin to the point. For an ellipse, t measures the angle between the x-axis and OP Q,
not OR.
46. (a) For any point (x, y), the equation
y = b sinh t has a unique solution t,
3
(b)
−∞ < t < +∞. On the hyperbola,
-3
y2
x2
= 1 + 2 = 1 + sinh2 t
2
a
b
= cosh2 t, so x = ±a cosh t.
3
-3
47. (a) For any point (x, y), the equation y = b tan t has a unique solution t where −π/2 < t < π/2.
y2
x2
On the hyperbola, 2 = 1 + 2 = 1 + tan2 t = sec2 t, so x = ±a sec t.
a
b
3
(b)
-3
3
-3
48. By Definition 12.4.1, (x + 1)2 + (y − 4)2 = (y − 1)2 , (x + 1)2 = 6y − 15, (x + 1)2 = 6(y − 5/2)
49. (4, 1) and (4, 5) are the foci so the center is at (4, 3) thus c = 2, a = 12/2 = 6, b2 = 36 − 4 = 32;
(x − 4)2 /32 + (y − 3)2 /36 = 1
p
p
(x − 1)2 + (y − 1)2 − x2 + y 2 = 1,
50. From the definition of a hyperbola,
p
p
(x − 1)2 + (y − 1)2 − x2 + y 2 = ±1, transpose the second radical to the right hand side of the
p
equation and square and simplify to get ±2 x2 + y 2 = −2x − 2y + 1, square and simplify again
to get 8xy − 4x − 4y + 1 = 0.
4x2
4 2 y2
x +
= 1, then A(x) = (2y)2 = 16 1 −
,
51. Let the ellipse have equation
81
4
81
Z 9/2
4x2
dx = 96
16 1 −
V =2
81
0
431
Chapter 12
52. See Exercise 51, A(y) =
√
y2
x2
53. Assume 2 + 2 = 1, A = 4
a
b
54. (a) Assume
√ 81
3
4
3x2 =
Z
p
a
b
1−
y2
4
,V =
√ 81 Z 2
√
y2
dy = 54 3
1−
3
2 0
4
1 − x2 /a2 dx = πab
0
y2
x2
+
= 1, V = 2
a2
b2
Z
a
0
4
πb2 1 − x2 /a2 dx = πab2
3
(b) In Part (a) interchange a and b to obtain the result.
2
bx
y2
dy
a4 − (a2 − b2 )x2
x2
dy
=− √
,
,1 +
=
55. Assume 2 + 2 = 1,
a
b
dx
dx
a2 (a2 − x2 )
a a2 − x2
s
Z a
p
a
2πb p
a4 − (a2 − b2 )x2
b
−1 c
2
2
S=2
+
sin
,
c
=
1 − x /a
dx
=
2πab
a2 − b2
a
a2 − x2
a
c
a
0
56. As in Exercise 55, 1 +
Z
b
S=2
0
dx
dy
s
p
2πa 1 − y 2 /b2
2
=
b4 + (a2 − b2 )y 2
,
b2 (b2 − y 2 )
b4 + (a2 − b2 )y 2
dy = 2πab
b2 (b2 − y 2 )
a b a+c
+ ln
b
c
b
,c =
p
a2 − b2
57. Open the compass to the length of half the major axis, place the point of the compass at an end
of the minor axis and draw arcs that cross the major axis to both sides of the center of the ellipse.
Place the tacks where the arcs intersect the major axis.
58. Let P denote the pencil tip, and let R(x, 0) be the point below Q and P which lies on the line L.
Then QP + P F is the length of the string and QR = QP + P R is the length of the side of the
triangle. These two are equal, so P F = P R. But this is the definition of a parabola according to
Definition 12.4.1.
59. Let P denote the pencil tip, and let k be the difference between the length of the ruler and that
of the string. Then QP + P F2 + k = QF1 , and hence P F2 + k = P F1 , P F1 − P F2 = k. But this
is the definition of a hyperbola according to Definition 12.4.3.
60. In the x0 y 0 -plane an equation of the circle is x02 + y 02 = r2 where r is the radius of the cylinder. Let
P (x, y) be a point on the curve in the xy-plane, then x0 = x cos θ and y 0 = y so x2 cos2 θ + y 2 = r2
which is an equation of an ellipse in the xy-plane.
p
p
1 p
1
D2 + p2 D2 = D 1 + p2 (see figure), so a = D 1 + p2 , but b = D,
2
2
r
p
1
1
1 2
T = c = a2 − b2 =
D (1 + p2 ) − D2 = pD.
4
4
2
61. L = 2a =
pD
D
62. y =
1 2
1
x , dy/dx =
x, dy/dx
4p
2p
=
x=x0
1
x0 , the tangent line at (x0 , y0 ) has the formula
2p
x2
x2
x0
x0
1 2
(x − x0 ) =
x − 0 , but 0 = 2y0 because (x0 , y0 ) is on the parabola y =
x .
2p
2p
2p
2p
4p
x0
x0
Thus the tangent line is y − y0 =
x − 2y0 , y =
x − y0 .
2p
2p
y − y0 =
Exercise Set 12.4
432
dy
dx
63. By implicit differentiation,
=−
(x0 ,y0 )
b2 x0
if y0 6= 0, the tangent line is
a2 y0
b2 x0
y − y0 = − 2 (x − x0 ), a2 y0 y − a2 y02 = −b2 x0 x + b2 x20 , b2 x0 x + a2 y0 y = b2 x20 + a2 y02 ,
a y0
but (x0 , y0 ) is on the ellipse so b2 x20 + a2 y02 = a2 b2 ; thus the tangent line is b2 x0 x + a2 y0 y = a2 b2 ,
x0 x/a2 + y0 y/b2 = 1. If y0 = 0 then x0 = ±a and the tangent lines are x = ±a which also follows
from x0 x/a2 + y0 y/b2 = 1.
64. By implicit differentiation,
dy
dx
=
(x0 ,y0 )
b2 x0
b2 x0
if
y
=
6
0,
the
tangent
line
is
y
−
y
=
(x − x0 ),
0
0
a2 y0
a2 y0
= a2 b2 , x0 x/a2 − y0 y/b2 = 1. If y0 = 0 then x0 = ±a and the tangent
b x0 x − a y0 y =
lines are x = ±a which also follow from x0 x/a2 − y0 y/b2 = 1.
2
2
b2 x20 − a2 y02
y2
x2
y2
x2
+
=
1
and
−
= 1 as the equations of the ellipse and hyperbola. If (x0 , y0 ) is
a2
b2
A2
B2
x20
y02
x20
y02
1
1
1
1
2
2
a point of intersection then 2 + 2 = 1 = 2 − 2 , so x0
− 2 = y0
+ 2 and
a
b
A
B
A2
a
B2
b
65. Use
a2 A2 y02 (b2 + B 2 ) = b2 B 2 x20 (a2 − A) . Since the conics have the same foci, a2 − b2 = c2 = A2 + B 2 ,
so a2 − A2 = b2 + B 2 . Hence a2 A2 y02 = b2 B 2 x20 . From Exercises 63 and 64, the slopes of the
tangent lines are −
perpendicular.
B 2 x0
b2 B 2 x20
b2 x0
and
,
whose
product
is
−
= −1. Hence the tangent lines are
a2 y0
A2 y0
a2 A2 y02
dy
x0
=−
where (x0 , y0 ) is the point
dx (x0 ,y0 )
4y0
of tangency, but −x0 /(4y0 ) = −1/2 because the slope of the line is −1/2 so x0 = 2y0 . (x0 , y0 ) is
on the ellipse so x20 + 4y02 = 8 which when solved with x0 = 2y0 yields the points of tangency (2, 1)
and (−2, −1). Substitute these into the equation of the line to get k = ±4.
66. Use implicit differentiation on x2 + 4y 2 = 8 to get
√
√
67. Let (x0 , y0 ) be such a point. The foci are at (− 5, 0) and ( 5, 0), the lines are perpendicular if
y0
y0
√ ·
√ = −1, y02 = 5 − x20 and 4x20 − y02 = 4. Solve
the product of their slopes is −1 so
x0 + 5 x0 − 5
√
√
√
√
√
√
to get x0 = ±3/ 5, y0 = ±4/ 5. The coordinates are (±3/ 5, 4/ 5), (±3/ 5, −4/ 5).
68. Let (x0 , y0 ) be one of the points; then dy/dx
(x0 ,y0 )
= 4x0 /y0 , the tangent line is y = (4x0 /y0 )x+4,
2
2
2
2
but (x0 , y0 ) is on both
√ the line and the curve which leads to 4x0 − y0 + 4y0 = 0 and 4x0 − y0 = 36,
solve to get x0 = ±3 13/2, y0 = −9.
69. Let d1 and d2 be the distances of the first and second observers, respectively, from the point
where the gun was fired. Then t = (time for sound to reach the second observer) − (time for
sound to reach the first observer) = d2 /v − d1 /v so d2 − d1 = vt. For constant v and t the
difference of distances, d2 and d1 is constant so the gun was fired somewhere on a branch of a
vt
v 2 t2
, and
hyperbola whose foci are where the observers are. Since d2 − d1 = 2a, a = , b2 = c2 −
2
4
2
2
y
x
− 2
= 1.
2
2
v t /4 c − (v 2 t2 /4)
70. As in Exercise 69, d2 − d1 = 2a = vt = 299,792,458 × 10−7 , a2 = (vt/2)2 ≈ 224.6888; c2 = (50)2
y2
x2
−
= 1. But y = 200 km, so x ≈ 64.6 km.
= 2500, b2 = c2 − a2 ≈ 2275.3112,
224.6888 2275.3112
433
Chapter 12
y2
3p
x2
+
= 1, x =
4 − y2 ,
9
4
2
Z
Z −2+h
p
2
(2)(3/2) 4 − y (18)dy = 54
V=
71. (a) Use
−2
−2+h
−2
p
4 − y 2 dy
−2+h
p
p
y
−1 y
−1 h − 2
2
2
+ (h − 2) 4h − h + 2π ft3
4 − y + 2 sin
= 27 4 sin
= 54
2
2 −2
2
(b) When h = 4 ft, Vfull = 108 sin−1 1 + 54π = 108π ft3 , so solve for h when V = (k/4)Vfull ,
k = 1, 2, 3, to get h = 1.19205, 2, 2.80795 ft or 14.30465, 24, 33.69535 in.
72. We may assume A > 0, since if A < 0 then one can multiply the equation by −1, and if A = 0
then one can exchange A with C (C cannot be zero simultaneously with A). Then
2
2
E2
D
E
D2
−
= 0.
+C y+
+F −
Ax2 + Cy 2 + Dx + Ey + F = A x +
2A
2C
4A 4C
D2
E2
+
the equation represents an ellipse (a circle if A = C);
4A
4C
D2 E 2
D2 E 2
if F =
+
, the point x = −D/(2A), y = −E/(2C); and if F >
+
then there is
4A 4C
4A 4C
no graph.
(a) Let AC > 0. If F <
D2
E2
(b) If AC < 0 and F =
+
, then
4A 4C
√
√
√
√
D
E
D
E
+ −C y +
− −C y +
= 0, a pair of lines;
A x+
A x+
2A
2C
2A
2C
otherwise a hyperbola
(c) Assume C = 0, so Ax2 +Dx+Ey+F = 0. If E 6= 0, parabola; if E = 0 then Ax2 +Dx+F = 0.
If this polynomial has roots x = x1 , x2 with x1 6= x2 then a pair of parallel lines; if x1 = x2
then one line; if no roots, then no graph. If A = 0, C 6= 0 then a similar argument applies.
73. (a) (x − 1)2 − 5(y + 1)2 = 5, hyperbola
√
(b) x2 − 3(y + 1)2 = 0, x = ± 3(y + 1), two lines
(c) 4(x + 2)2 + 8(y + 1)2 = 4, ellipse
(d) 3(x + 2)2 + (y + 1)2 = 0, the point (−2, −1) (degenerate case)
(e) (x + 4)2 + 2y = 2, parabola
(f )
5(x + 4)2 + 2y = −14, parabola
74. distance from the point (x, y) to the focus (0, p) = distance to the directrix y = −p, so x2 + (y − p)2
= (y + p)2 , x2 = 4py
75. distance from the point (x, y) to the focus (0, −c) plus distance to the focus (0, c) = const = 2a,
p
p
p
x2 + (y + c)2 + x2 + (y − c)2 = 2a, x2 + (y + c)2 = 4a2 + x2 + (y − c)2 − 4a x2 + (y − c)2 ,
p
c
x2
y2
x2 + (y − c)2 = a − y, and since a2 − c2 = b2 , 2 + 2 = 1
a
b
a
76. distance from the point (x, y) to the focus (−c, 0) less distance to the focus (c, 0) is equal to 2a,
p
p
p
(x + c)2 + y 2 − (x − c)2 + y 2 = ±2a, (x + c)2 + y 2 = (x − c)2 + y 2 + 4a2 ± 4a (x − c)2 + y 2 ,
cx
p
x2
y2
− a , and, since c2 − a2 = b2 , 2 − 2 = 1
(x − c)2 + y 2 = ±
a
a
b
Exercise Set 12.5
434
EXERCISE SET 12.5
1. (a) r =
3/2
, e = 1, d = 3/2
1 − cos θ
(b) r =
3/2
, e = 1/2, d = 3
1 + 12 sin θ
p/ 2
p/ 2
2
-2
–2
0
2
0
2
–2
-2
(c) r =
2
3
2
1+
cos θ
, e = 3/2, d = 4/3
(d) r =
5/3
, e = 1, d = 5/3
1 + sin θ
p/ 2
p/ 2
7
3
-7
-5
10
7
0
-11
-7
2. (a) r =
1−
0
4/3
, e = 2/3, d = 2
2
3 cos θ
(b) r =
1
1−
4
3
sin θ
, e = 4/3, d = 3/4
p/ 2
p/ 2
0
0
(c) r =
1/3
, e = 1, d = 1/3
1 + sin θ
(d) r =
1/2
, e = 3, d = 1/6
1 + 3 sin θ
p/ 2
p/ 2
0
0
435
Chapter 12
3. (a) e = 1, d = 8, parabola, opens up
10
4
, e = 3/4, d = 16/3,
1 + sin θ
ellipse, directrix 16/3 units
above the pole
(b) r =
3
4
5
-15
15
-8
8
-10
-20
2
, e = 3/2, d = 4/3,
1 − 32 sin θ
hyperbola, directrix 4/3 units
below the pole
(c) r =
3
, e = 1/4, d = 12,
1 + 14 cos θ
ellipse, directrix 12 units
to the right of the pole
(d) r =
4
4
-6
6
-5
3
-4
-8
4. (a) e = 1, d = 15, parabola, opens left
20
2/3
, e = 1,
1 + cos θ
d = 2/3, parabola, opens left
(b) r =
10
-20
20
-15
5
-20
-10
64/7
, e = 12/7, d = 16/3,
1 − 12
7 sin θ
hyperbola, directrix 16/3 units below pole
(c) r =
, e = 2/3, d = 6,
1 − cos θ
ellipse, directrix 6 units left of the pole
20
-30
4
(d) r =
2
3
6
30
-3
-40
13
-6
Exercise Set 12.5
436
5. (a) d = 1, r =
2
2/3
ed
=
=
2
1 + e cos θ
3 + 2 cos θ
1 + 3 cos θ
(b) e = 1, d = 1, r =
1
ed
=
1 − e cos θ
1 − cos θ
(c) e = 3/2, d = 1, r =
3
3/2
ed
=
=
3
1 + e sin θ
2 + 3 sin θ
1 + 2 sin θ
6. (a) e = 2/3, d = 1, r =
2
ed
2/3
=
=
1 − e sin θ
3 − 2 sin θ
1 − 23 sin θ
(b) e = 1, d = 1, r =
1
ed
=
1 + e sin θ
1 + sin θ
(c) e = 4/3, d = 1, r =
7. (a) r =
4
4/3
ed
=
=
1 − e cos θ
3 − 4 cos θ
1 − 43 cos θ
ed
ed
ed
,θ = 0 : 6 =
,θ = π : 4 =
, 6 ± 6e = 4 ∓ 4e, 2 = ∓10e, use bottom
1 ± e cos θ
1±e
1∓e
sign to get e = 1/5, d = 24, r =
24
24/5
=
1 − cos θ
5 − 5 cos θ
d
d
2
, 1 = , d = 2, r =
1 − sin θ
2
1 − sin θ
ed
ed
ed
(c) r =
, θ = π/2 : 3 =
, θ = 3π/2 : −7 =
, ed = 3 ± 3e = −7 ± 7e, 10 = ±4e,
1 ± e sin θ
1±e
1∓e
(b) e = 1, r =
e = 5/2, d = 21/5, r =
21
21/2
=
1 + (5/2) sin θ
2 + 5 sin θ
ed
ed
ed
,1 =
,4 =
, 1 ± e = 4 ∓ 4e, upper sign yields e = 3/5, d = 8/3,
1 ± e sin θ
1±e
1∓e
8
8/5
=
r=
5 + 3 sin θ
1 + 35 sin θ
8. (a) r =
d
6
d
, 3 = , d = 6, r =
1 − cos θ
2
1 − cos θ
√
√
√
5
2d
√
(c) a = b = 5, e = c/a = 50/5 = 2, r =
; r = 5 when θ = 0, so d = 5 + √ ,
1 + 2 cos θ
2
√
5 2+5
√
.
r=
1 + 2 cos θ
(b) e = 1, r =
9. (a) r =
3
1+
1
2
sin θ
, e = 1/2, d = 6, directrix 6 units above pole; if θ = π/2 : r0 = 2;
if θ = 3π/2 : r1 = 6, a = (r0 + r1 )/2 = 4, b =
coordinates),
√
√
r0 r1 = 2 3, center (0, −2) (rectangular
x2
(y + 2)2
+
=1
12
16
1/2
1/2
, e = 1/2, d = 1, directrix 1/2 unit left of pole; if θ = π : r0 =
= 1/3;
1
3/2
1 − 2 cos θ
√
if θ = 0 : r1 = 1, a = 2/3, b = 1/ 3, center = (1/3, 0) (rectangular coordinates),
9
(x − 1/3)2 + 3y 2 = 1
4
(b) r =
437
Chapter 12
6/5
, e = 2/5, d = 3, directrix 3 units right of pole, if θ = 0 : r0 = 6/7,
1 + 25 cos θ
√ √
if θ = π : r1 = 2, a = 10/7, b = 2 3/ 7, center (−4/7, 0) (rectangular coordinates),
7
49
(x + 4/7)2 + y 2 = 1
100
12
10. (a) r =
(b) r =
2
1−
3
4
sin θ
, e = 3/4, d = 8/3, directrix 8/3 units below pole, if θ = 3π/2 : r0 = 8/7,
√
if θ = π/2; r1 = 8, a = 32/7, b = 8/ 7, center: (0, 24/7) (rectangular coordinates),
2
24
49
7 2
x +
y−
=1
64
1024
7
2
, e = 3, d = 2/3, hyperbola, directrix 2/3 units above pole, if θ = π/2 :
1 + 3 sin θ
2
√
3
2
r0 = 1/2; θ = 3π/2 : r1 = 1, center (0, 3/4), a = 1/4, b = 1/ 2, −2x + 16 y −
=1
4
11. (a) r =
5/3
, e = 3/2, d = 10/9, hyperbola, directrix 3/2 units left of pole, if θ = π :
1 − 32 cos θ
p
5/3
9
9
r0 = 2/3; θ = 0 : r1 =
= 10/3, center (−2, 0), a = 4/3, b = 20/9, (x+2)2 − y 2 = 1
1/2
16
20
(b) r =
4
, e = 2, d = 2, hyperbola, directrix 2 units below pole, if θ = 3π/2 : r0 = 4/3;
1 − 2 sin θ
2
√
8
4
9
3
= 4, center (0, −8/3), a = 4/3, b = 4/ 3,
y+
θ = π/2 : r1 =
− x2 = 1
1−2
16
3
16
12. (a) r =
15/2
, e = 4, d = 15/8, hyperbola, directrix 15/8 units right of pole, if θ = 0 :
1 + 4 cos θ
√
5
4
15
, center (2, 0), 4(x − 2)2 − y 2 = 1
r0 = 3/2; θ = π : r1 = − = 5/2, a = 1/2, b =
2
2
15
(b) r =
13. (a) r =
1+
8=a=
(b) r =
1−
4=a=
(c) r =
1−
1
2d
1
2 cos θ
d
, if θ = 0 : r0 = d/3; θ = π, r1 = d,
2 + cos θ
12
2
1
(r1 + r0 ) = d, d = 12, r =
2
3
2 + cos θ
3
5d
3
5 sin θ
=
3d
3
3
, if θ = 3π/2 : r0 = d; θ = π/2, r1 = d,
5 − 3 sin θ
8
2
1
64
3(64/15)
64
15
(r1 + r0 ) =
d, d =
,r =
=
2
16
15
5 − 3 sin θ
25 − 15 sin θ
3
5d
3
5 cos θ
d = 16/3, r =
(d)
=
1
5d
1
5 sin θ
=
3d
3
3
3
, if θ = π : r0 = d; θ = 0, r1 = d, 4 = b = d,
5 − 3 cos θ
8
2
4
16
5 − 3 cos θ
d
=
, if θ = π/2 : r0 = d/6; θ = 3π/2, r1 = d/4,
5 + sin θ
1
120
1
1 1
−
d, d = 120, r =
=
5=c= d
2
4 6
24
5 + sin θ
1+
Exercise Set 12.5
14. (a) r =
438
1+
1
2d
1
2 sin θ
10 = a =
=
d
, if θ = π/2 : r0 = d/3; θ = 3π/2 : r1 = d,
2 + sin θ
15
2
1
(r0 + r1 ) = d, d = 15, r =
2
3
2 + sin θ
1
5d
1
5 cos θ
d
, if θ = π : r0 = d/6, θ = 0 : r1 = d/4,
5 − cos θ
5
144/5
144
1
1
1 1
+
=
d, d = 144/5, r =
=
6 = a = (r1 + r0 ) = d
2
2
4 6
24
5 − cos θ
25 − 5 cos θ
(b) r =
1−
=
√
3d
3
, if θ = 3π/2 : r0 = d, θ = π/2 : r1 = 3d, 4 = b = 3d/ 7,
4 − 3 sin θ
7
1−
√
4√
4 7
d=
7, r =
3
4 − 3 sin θ
(c) r =
3
4d
3
4 sin θ
=
4
5d
4
5 cos θ
=
4d
4
, if θ = 0 : r0 = d; θ = π : r1 = 4d,
5 + 4 cos θ
9
1+
4
16
45
45/2
45
1
1
=
d, d =
, r=
=
c = 10 = (r1 − r0 ) = d 4 −
2
2
9
9
8
5 + 4 cos θ
10 + 8 cos θ
(d) r =
15. (a) e = c/a =
(b) e =
− r0 )
r1 − r0
=
r 1 + r0
+ r0 )
r1 /r0 − 1
r1
1+e
, e(r1 /r0 + 1) = r1 /r0 − 1,
=
r1 /r0 + 1
r0
1−e
16. (a) e = c/a =
(b) e =
1
2 (r1
1
2 (r1
1
2 (r1
1
2 (r1
+ r0 )
r1 + r0
=
r 1 − r0
− r0 )
r1 /r0 + 1
r1
e+1
, e(r1 /r0 − 1) = r1 /r0 + 1,
=
r1 /r0 − 1
r0
e−1
17. (a) T = a3/2 = 39.51.5 ≈ 248 yr
(b) r0 = a(−e) = 39.5(1 − 0.249) = 29.6645 AU ≈ 4,449,675,000 km
r1 = a(1 + e) = 39.5(1 + 0.249) = 49.3355 AU ≈ 7,400,325,000 km
(c) r =
a(1 − e2 )
39.5(1 − (0.249)2 )
37.05
≈
≈
AU
1 + e cos θ
1 + 0.249 cos θ
1 + 0.249 cos θ
p/ 2
(d)
50
-30
0
20
-50
T
=
18. (a) In yr and AU, T = a ; in days and km,
365
a 3/2
days.
so T = 365 × 10−9
150
3/2
a
150 × 106
3/2
,
439
Chapter 12
−9
(b) T = 365 × 10
57.95 × 106
150
3/2
≈ 87.6 days
(c) From (17) the polar equation of the orbit has the form r =
(d)
37.82
a(1 − e2 )
=
1 + e cos θ
1 + 0.205 cos θ
p/2
40
-50
0
20
-40
19. (a) a = T 2/3 = 23802/3 ≈ 178.26 AU
(b) r0 = a(1 − e) ≈ 0.8735 AU, r1 = a(1 + e) ≈ 355.64 AU
(c) r =
1.74
a(1 − e2 )
≈
AU
1 + e cos θ
1 + 0.9951 cos θ
p/ 2
(d)
20
0
-400
-20
20. (a) By Exercise 15(a), e =
r 1 − r0
≈ 0.093
r1 + r0
(b) r = 12 (a0 + a1 ) = 225,400,000 km ≈ 1.503 AU, so T = a3/2 ≈ 1.84 yr
(c) r =
a(1 − e2 )
1.49
≈
AU
1 + e cos θ
1 + 0.093 cos θ
(d)
p/ 2
1.49
1.6428
1.3632
0
1.49
21. r0 = a(1 − e) ≈ 7003 km, hmin ≈ 7003 − 6440 = 563 km,
r1 = a(1 + e) ≈ 10,726 km, hmax ≈ 10,726 − 6440 = 4286 km
Chapter 12 Supplementary Exercises
440
22. r0 = a(1 − e) ≈ 651,736 km, hmin ≈ 581,736 km; r1 = a(1 + e) ≈ 6,378,102 km,
hmax ≈ 6,308,102 km
a
→ +∞.
e
1
− e and lim d = +∞;
Let d be the distance between the directrix and the focus, then d = a
e→0
e
but the center and the focus come together, so distance between the directrix and the center also
tends to +∞.
23. Since the foci are fixed, a is constant; since e → 0, the distance
2
c
x2
y2
x2
y2
x−a ,
24. (a) From Figure 12.4.22, 2 − 2 = 1, 2 − 2
= 1, (x − c)2 + y 2 =
2
a
b
a
c −a
a
p
c
(x − c)2 + y 2 = x − a for x > 0.
a
(b) From Part (a), P F =
PF
c
P D,
= c/a.
a
PD
CHAPTER 12 SUPPLEMENTARY EXERCISES
√
2. (a) ( 2, 3π/4)
√
(b) (− 2, 7π/4)
√
(c) ( 2, 3π/4)
√
(d) (− 2, −π/4)
3. (a) circle
(e) limaçon
(b) rose
(f ) none
(c) line
(g) none
(d) limaçon
(h) spiral
4. (a) r =
1/3
, ellipse, right of pole, distance = 1
1 + 13 cos θ
(b) hyperbola, left of pole, distance = 1/3
(c) r =
1/3
, parabola, above pole, distance = 1/3
1 + sin θ
(d) parabola, below pole, distance = 3
5. (a)
(b)
p/ 2
p/ 2
(1, 9)
0
0
(c)
p/ 2
(d)
(e)
p/ 2
0
p/ 2
(2, 3)
0
(1, 3)
(1, #)
(–1, 3)
0
441
Chapter 12
6. Family I: x2 + (y − b)2 = b2 , b < 0, or r = 2b sin θ; Family II: (x − a)2 + y 2 = a2 , a < 0, or
r = 2a cos θ
7. (a) r = 2a/(1 + cos θ), r + x = 2a, x2 + y 2 = (2a − x)2 , y 2 = −4ax + 4a2 , parabola
(b) r2 (cos2 θ − sin2 θ) = x2 − y 2 = a2 , hyperbola
√
√
(c) r sin(θ − π/4) = ( 2/2)r(sin θ − cos θ) = 4, y − x = 4 2, line
(d) r2 = 4r cos θ + 8r sin θ, x2 + y 2 = 4x + 8y, (x − 2)2 + (y − 4)2 = 20, circle
9. (a)
2
4
45
7
5 2 1 2
c
= e = and 2b = 6, b = 3, a2 = b2 + c2 = 9 + a2 , a2 = 9, a = √ ,
x + y =1
a
7
49
49
49
9
5
(b) x2 = −4py, directrix y = 4, focus (−4, 0), 2p = 8, x2 = −16y
√
√
(c) For the ellipse, a = 4, b = 3, c2 = a2 − b2 = 16 − 3 = 13, foci (± 13, 0);
√
4
13 2
a ,
for the hyperbola, c = 13, b/a = 2/3, b = 2a/3, 13 = c2 = a2 + b2 = a2 + a2 =
9
9
x2
y2
a = 3, b = 2,
−
=1
9
4
10. (a) e = 4/5 = c/a, c = 4a/5, but a = 5 so c = 4, b = 3,
(x + 3)2
(y − 2)2
+
=1
25
9
(b) directrix y = 2, p = 2, (x + 2)2 = −8y
(c) center (−1, 5), vertices (−1, 7) and (−1, 3), a = 2, a/b = 8, b = 1/4,
y
11. (a)
(b)
2
y
(y − 5)2
−16(x+1)2 = 1
4
y
(c)
4
4
x
-4
x
8
x
2
-8
8
10
-3
-12
-10
470
13. (a) The equation of the parabola is y = ax2 and it passes through (2100, 470), thus a =
,
21002
470 2
x .
y=
21002
s
2
Z 2100
470
1+ 2
x dx ≈ 4336.3 ft
(b) L = 2
21002
0
14. (a) As t runs from 0 to π, the upper portion of the curve is traced out from right to left; as t
runs from π to 2π the bottom portion of the curve is traced out from right to left. The loop
1
1
occurs for π + sin−1 < t < 2π − sin−1 .
4
4
(b) lim x = +∞, lim y = 1; lim x = −∞, lim y = 1; lim x = +∞, lim y = 1;
t→0+
t→0+
t→π −
t→π −
t→π +
t→π +
lim x = −∞, lim y = 1; the horizontal asymptote is y = 1.
t→2π −
t→2π −
(c) horizontal tangent line when dy/dx = 0, or dy/dt = 0, so cos t = 0, t = π/2, 3π/2;
1
1
, 2π−sin−1 √
,
vertical tangent line when dx/dt = 0, so − csc2 t−4 sin t = 0, t = π+sin−1 √
3
3
4
4
t = 3.823, 5.602
Chapter 12 Supplementary Exercises
442
(d) r2 = x2 + y 2 = (cot t + 4 cos t)2 + (1 + 4 sin t)2 = (4 + csc t)2 , r = 4 + csc t; with t = θ,
f (θ) = 4 + csc θ; m = dy/dx = (f (θ) cos θ + f 0 (θ) sin θ)/(−f (θ) sin θ + f 0 (θ) cos θ); when
√
√
θ = π + sin−1 (1/4), m = 15/15, when θ = 2π − sin−1 (1/4), m = − 15/15, so the tangent
√
lines to the conchoid at the pole have polar equations θ = ± 15/15.
15.
A
=
2
Z
π/6
0
1
(2 sin θ)2 dθ +
2
Z
π/4
π/6
√
π/6
5π
1
π
1 2
3
1 , dθ = 2 θ − sin 2θ
−
+ ,A =
2
4
24
12
2
0
16. The circle has radius a/2 and lies entirely inside the cardioid, so
Z 2π
a2
5a2
3a2
1 2
a (1 + sin θ)2 dθ − πa2 /4 =
π− π=
π
A=
2
2
4
4
0
Z
π/2
17. (a) r = 1/θ, dr/dθ = −1/θ , r + (dr/dθ) = 1/θ + 1/θ , L =
2
2
2
2
4
π/4
1p
1 + θ2 dθ ≈ 0.9457 by
θ2
Endpaper Table Formula 93.
Z +∞ p
1
1 + θ2 dθ diverges by the comparison test (with 1/θ), and thus the
(b) The integral
θ2
1
arc length is infinite.
18. (a) When the point of departure of the thread from the circle has traversed an angle θ, the
amount of thread that has been unwound is equal to the arc length traversed by the point of
departure, namely aθ. The point of departure is then located at (a cos θ, a sin θ), and the tip of
the string, located at (x, y), satisfies the equations x−a cos θ = aθ sin θ, y−a sin θ = −aθ cos θ;
hence x = a(cos θ + θ sin θ), y = a(sin θ − θ cos θ).
(b) Assume for simplicity that a = 1. Then dx/dθ = θ cos θ, dy/dθ = θ sin θ; dx/dθ = 0 has
solutions θ = 0, π/2, 3π/2; and dy/dθ = 0 has solutions θ = 0, π, 2π. At θ = π/2, dy/dθ > 0,
so the direction is North; at θ = π, dx/dθ < 0, so West; at θ = 3π/2, dy/dθ < 0, so South; at
θ = 2π, dx/dθ > 0, so East. Finally, lim+ dy/dx = lim+ tan θ = 0, so East.
θ→0
(c)
u
x
y
0
1
0
p/ 2
p/ 2
1
p
–1
p
3p/ 2
–3p/ 2
–1
θ→0
2p
1
–2p
Note that the parameter θ in these equations does not satisfy equations (1) and (2) of Section
12.1, since it measures the angle of the point of departure and not the angle of the tip of the
thread.
4
-5
2
-8
443
Chapter 12
Z
√
a2 +b2
19. (a) V =
π b2 x2 /a2 − b2 dx
y
a
=
p
2
πb2 2
(b − 2a2 ) a2 + b2 + ab2 π
2
3a
3
x
√
a2 +b2
Z
(b) V = 2π
p
x b2 x2 /a2 − b2 dx = (2b4 /3a)π
y
a
x
p/ 2
20. (a)
(b) θ = π/2, 3π/2, r = 1
5
-5
5
0
-5
r cos θ + (dr/dθ) sin θ
; at θ = π/2, m1 = (−1)/(−1) = 1, m2 = 1/(−1) = −1,
−r sin θ + (dr/dθ) cos θ
m1 m2 = −1; and at θ = 3π/2, m1 = −1, m2 = 1, m1 m2 = −1
(c) dy/dx =
22. The tips are located at r = 1, θ = π/6, 5π/6, 3π/2 and, for example,
p
p
√
d = 1 + 1 − 2 cos(5π/6 − π/6) = 2(1 − cos(2π/3)) = 3
23. (a) x = r cos θ = cos θ + cos2 θ, dx/dθ = − sin θ − 2 sin θ cos θ = − sin θ(1 + 2 cos θ) = 0 if sin θ = 0
or cos θ = −1/2, so θ = 0, π, 2π/3, 4π/3; maximum x = 2 at θ = 0, minimum x = −1/4 at
θ=π
2
(b) y = r sin θ = sin θ + sin θ cos θ, dy/dθ
√ = 2 cos θ + cos θ − 1 = 0 at cos θ√= 1/2, −1, so
θ = π/3, 5π/3, π; maximum y = 3 3/4 at θ = π/3, minimum y = −3 3/4 at θ = 5π/3
√
2θ cos θ − sin θ
= 0 if 2θ cos θ = sin θ, tan θ = 2θ which
24. (a) y = r sin θ = (sin θ)/ θ, dy/dθ =
2θ3/2
only happens once on (0, π]. Since lim y = 0 and y = 0 at θ = π, y has a maximum when
θ→0+
tan θ = 2θ.
(b) θ ≈ 1.16556
√
(c) ymax = (sin θ)/ θ ≈ 0.85124
Chapter 12 Supplementary Exercises
444
25. The width is twice the maximum value of y for 0 ≤ θ ≤ π/4:
y = r sin θ√ = sin θ cos
sin3 θ, dy/dθ = cos θ − 6 sin2 θ cos θ √= 0 when cos θ = 0 or
√ 2θ = sin√θ − 2 √
sin θ = 1/ 6, y = 1/ 6 − 2/(6 6) = 6/9, so the width of the petal is 2 6/9.
25
y2
dy =
πh3 + 225πh ft3 .
225π 1 +
26. (a)
1521
2028
0
v
2 !
Z h/2 u
Z h/2
u
p
225
225
t
2
2
+
dy
2πx 1 + (dx/dy) dy = 4π
225 + y
(b) S = 2
1521
1521
0
0
y2
x2
−
= 1, so V = 2
225 1521
Z
h/2
#
"
√
p
2
5
6084
+
h
h
+
πh 6084 + h2 + 1170π ln
ft2
=
26
78
27. (a) The end of the inner arm traces out the circle x1 = cos t, y1 = sin t. Relative to the end of
the inner arm, the outer arm traces out the circle x2 = cos 2t, y2 = − sin 2t. Add to get the
motion of the center of the rider cage relative to the center of the inner arm:
x = cos t + cos 2t, y = sin t − sin 2t.
(b) Same as part (a), except x2 = cos 2t, y2 = sin 2t, so x = cos t + cos 2t, y = sin t + sin 2t
Z 2π " 2 2 #1/2
Z 2π
√
dy
dx
+
dt =
5 − 4 cos 3t dt ≈ 13.36489321,
(c) L1 =
dt
dt
0
0
Z 2π
√
5 + 4 cos t dt ≈ 13.36489322; L1 and L2 appear to be equal, and indeed, with the
L2 =
0
substitution u = 3t − π and the periodicity of cos u,
Z
Z 2π
√
1 5π p
5 − 4 cos(u + π) du =
5 + 4 cos u du = L2 .
L1 =
3 −π
0
Z
π/2
"
29. C = 4
0
Z
dx
dt
2
+
dy
dt
2 #1/2
Z
π/2
(a2 sin2 t + b2 cos2 t)1/2 dt
dt = 4
0
Z
π/2
(a sin t + (a − c ) cos t)
2
=4
2
2
2
2
1/2
0
30. a = 3, b = 2, c =
π/2
(1 − e2 cos2 t)1/2 dt
dt = 4a
0
√
Z
π/2
p
1 − (5/9) cos2 u du ≈ 15.86543959
5, C = 4(3)
0
31. (a)
1−e
1
59
r0
=
,e =
=
r1
61
1+e
60
59
93 × 106 = 91,450,000 mi
60
2 #1/2
Z π/2 "
cos
θ
dθ ≈ 584,295,652.5 mi
1−
(c) C = 4 × 93 × 106
60
0
(b) a = 93 × 106 , r0 = a(1 − e) =
32. (a) y = y0 + (v0 sin α)
(b)
g
x
−
v0 cos α 2
x
v0 cos α
2
= y0 + x tan α −
2v02
g
x2
cos2 α
g
v2
dy
= tan α − 2
x, dy/dx = 0 at x = 0 sin α cos α,
2
dx
v0 cos α
g
2
v2
g
v2
v02 sin α cos α
= y0 + 0 sin2 α
y = y0 + 0 sin2 α − 2
2
g
2v0 cos α
g
2g
445
Chapter 12
√
√
33. α = π/4, y0 = 3, x = v0 t/ 2, y = 3 + v0 t/ 2 − 16t2
√
(a) Assume the ball passes through x = 391, y = 50, then 391 = v0 t/ 2, 50 = 3 + 391 − 16t2 ,
√
√
16t2 = 344, t = 21.5, v0 = 2x/t ≈ 119.3 ft/s
v0
v0 v0
v2
v0
v02
dy
√ − 16 11
= √ − 32t = 0 when t = √ , ymax = 3 + √
= 3 + 0 ≈ 114.2 ft
dt
2
128
2
32 2
2 32 2
p
√
−v0 / 2 ± v02 /2 + 192
, t ≈ −0.04 (discard) and 5.31, dist = 447.9 ft
(c) y = 0 when t =
−32
(b)
y
34. (a)
1
x
-1
1
-1
Z
1
cos2
(c) L =
−1
πt2
2
+ sin2
πt2
2
dt = 2
y
dy
−
tan φ − tan θ
35. tan ψ = tan(φ − θ) =
= dx x
y dy
1 + tan φ tan θ
1+
x dx
sin θ
r cos θ + (dr/dθ) sin θ
−
r
−r sin θ + (dr/dθ) cos θ cos θ
=
=
sin θ
r cos θ + (dr/dθ) sin θ)
dr/dθ
1+
−r sin θ + (dr/dθ) cos θ)
cos θ
36. (a) From Exercise 35,
1 − cos θ
θ
r
=
= tan ,
tan ψ =
dr/dθ
sin θ
2
(b)
p/ 2
so ψ = θ/2.
0
(c) At θ = π/2, ψ = θ/2 = π/4. At θ = 3π/2, ψ = θ/2 = 3π/4.
37. tan ψ =
aebθ
1
r
=
= is constant, so ψ is constant.
dr/dθ
abebθ
b
Chapter 12 Horizon Module
446
CHAPTER 12 HORIZON MODULE
1. For the Earth, aE (1 − e2E ) = 1(1 − 0.0172 ) = 0.999711, so the polar equation is
r=
0.999711
aE (1 − e2E )
=
.
1 − eE cos θ
1 − 0.017 cos θ
For Rogue 2000, aR (1 − e2R ) = 5(1 − 0.982 ) = 0.198, so the polar equation is
r=
0.198
aR (1 − e2R )
=
.
1 − eR cos θ
1 − 0.98 cos θ
1
2.
-1
10
-1
kR
kE
=
, so kE − kE eR cos θ = kR − kR eE cos θ.
1 − eE cos θ
1 − eR cos θ
kE − kR
.
Solving for cos θ gives cos θ =
kE eR − kR eE
3. At the intersection point A,
4. From Exercise 1, kE = 0.999711 and kR = 0.198, so
cos θ =
kE − kR
0.999711 − 0.198
≈ 0.821130
=
kE eR − kR eE
0.999711(0.98) − 0.198(0.017)
and θ = cos−1 0.821130 ≈ 0.607408 radian.
5. Substituting cos θ ≈ 0.821130 into the polar equation for the Earth gives
r≈
0.999711
≈ 1.013864,
1 − 0.017(0.821130)
so the polar coordinates of intersection A are approximately (1.013864, 0.607408).
Z
θF
1 2
r dθ. By Exercise 12.4.53 the area of
θI 2
the entire ellipse is πab, where a is the semimajor axis and b is the semiminor axis. But
p
√
√
b = a2 − c2 = a2 − (ea)2 = a 1 − e2 ,
Z θF
r2 dθ
t
θI
√
=
, which implies Formula (2).
so Formula (1) becomes
T
2πa2 1 − e2
6. By Theorem 12.3.2 the area of the elliptic sector is
7. In Formula (2) substitute T = 1, θI = 0, and θF ≈ 0.607408, and use the polar equation of the
Earth’s orbit found in Exercise 1:
Z
t=
0
θF
kE
1 − eE cos θ
p
2π 1 − e2E
2
Z
dθ
≈
0
0.607408
0.999711
1 − 0.017 cos θ
√
2π 0.999711
2
dθ
≈ 0.099792 yr.
447
Chapter 12
Note: This calculation can be done either by numerical integration or by using the integration
formula
r
θ
1+e
−1
Z
tan
2 tan
e sin θ
dθ
1−e
2
+ C,
=
+
2
2
2
3/2
(1 − e cos θ)
(1 − e )(1 − e cos θ)
(1 − e )
obtained by using a CAS or by the substitution u = tan(θ/2).
√
8. In Formula (2) we substitute T = 5 5 and θI = 0.45, and use the polar equation of Rogue 2000’s
orbit found in Exercise 1:
Z
θF
T
θI
t=
aR (1 − e2R )
1 − eR cos θ
p
2πa2R 1 − e2R
2
√ Z
5 5
dθ
=
aR (1 − e2R )
1 − eR cos θ
0.45
p
2
2πaR 1 − e2R
θF
2
dθ
,
so
Z
θF
0.45
aR (1 − e2R )
1 − eR cos θ
2
dθ =
p
2tπa2R 1 − e2R
√
.
5 5
9. (a) A CAS shows that
2
q
r
Z
eR (1 − e2R ) sin θ
θ
1 + eR
aR (1 − e2R )
+
+C
dθ = a2R 2 1 − e2R tan−1
tan
1 − eR cos θ
1 − eR
2
1 − eR cos θ
(b) Evaluating the integral above from θ = 0.45 to θ = θF , setting the result equal to the right
side of (3), and simplifying gives
p
θ
r
eR 1 − e2R sin θ F
θ
tπ
1+R
+
tan
= √ .
tan−1
1 − eR
2
2(1 − eR cos θ) 0.45
5 5
Using the known values of eR and t, and solving numerically, θF ≈ 0.611346.
10. Substituting θF ≈ 0.611346 in the equation for Rogue 2000’s orbit gives r ≈ 1.002525 AU. So the
polar coordinates of Rogue 2000 when the Earth is at intersection A are about (1.002525, 0.611346).
11. Substituting the values
pfound in Exercises 5 and 10 into the distance formula in Supplementary
Exercise 22 gives d = r12 + r22 − 2r1 r2 cos(θ1 − θ2 ) ≈ 0.012013 AU ≈ 1.797201 × 106 km.
Since this is less than 4 million kilometers, a notification should be issued. (Incidentally, Rogue
2000’s closest approach to the Earth does not occur when the Earth is at A, but about 9 hours
earlier, at t ≈ 0.098768 yr, at which time the distance is about 1.219435 million kilometers.)
CHAPTER 13
Three-Dimensional Space; Vectors
EXERCISE SET 13.1
1. (a) (0, 0, 0), (3, 0, 0), (3, 5, 0), (0, 5, 0), (0, 0, 4), (3, 0, 4), (3, 5, 4), (0, 5, 4)
(b) (0, 1, 0), (4, 1, 0), (4, 6, 0), (0, 6, 0), (0, 1, −2), (4, 1, −2), (4, 6, −2), (0, 6, −2)
2. corners: (2, 2, ±2), (2, −2, ±2),
(−2, 2, ±2), (−2, −2, ±2)
3. corners: (4, 2, −2), (4,2,1), (4,1,1), (4, 1, −2),
(−6, 1, 1), (−6, 2, 1), (−6, 2, −2), (−6, 1, −2)
z
(–2, –2, 2)
z
(–6, 2, 1)
(–2, 2, 2)
(–6, 1, –2)
(2, –2, 2)
(2, 2, 2)
(–2, –2, –2)
y
(–2, 2, –2)
(4, 1, 1)
(2, –2, –2)
x
(–6, 2, –2)
y
(4, 2, 1)
(2, 2, –2)
(4, 1, –2)
x
4. (a) (x2 , y1 , z1 ), (x2 , y2 , z1 ), (x1 , y2 , z1 )(x1 , y1 , z2 ), (x2 , y1 , z2 ), (x1 , y2 , z2 )
(b) The midpoint of the diagonal has coordinates which are the coordinates of the midpoints
1
(x1 + x2 ), y1 , z1 ;
of the edges. The midpoint of the edge (x1 , y1 , z1 ) and (x2 , y1 , z1 ) is
2
1
the midpoint of the edge (x2 , y1 , z1 ) and (x2 , y2 , z1 ) is x2 , (y1 + y2 ), z1 ; the midpoint
2
1
of the edge (x2 , y2 , z1 ) and (x2 , y2 , z2 )) is x2 , y2 , (z1 + z2 ) . Thus the coordinates of the
2
1
1
1
midpoint of the diagonal are (x1 + x2 ), (y1 + y2 ), (z1 + z2 ).
2
2
2
p
√
√
√
5. The diameter is d = (1 − 3)2 + (−2 − 4)2 + (4 + 12)2 = 296, so the radius is 296/2 = 74.
The midpoint (2, 1, −4) of the endpoints of the diameter is the center of the sphere.
6. Each side has length
√
14 so the triangle is equilateral.
√
7. (a) The sides have lengths 7,√ 14, and 7 5; it is a right triangle because the sides satisfy the
Pythagorean theorem, (7 5)2 = 72 + 142 .
(b) (2,1,6) is the vertex of the 90◦ angle because it is opposite the longest side
(the hypotenuse).
(c) area = (1/2)(altitude)(base) = (1/2)(7)(14) = 49
8. (a) 3
p
√
(2)2 + (−3)2 = 13
(d)
(b) 2
p
√
(e)
(−5)2 + (−3)2 = 34
(c) 5
p
√
(f )
(−5)2 + (2)2 = 29
9. (a) (x − 1)2 + y 2 + (z + 1)2 = 16
p
√
(b) r = (−1 − 0)2 + (3 − 0)2 + (2 − 0)2 = 14, (x + 1)2 + (y − 3)2 + (z − 2)2 = 14
448
449
Chapter 13
1p
1√
(−1 − 0)2 + (2 − 2)2 + (1 − 3)2 =
5, center (−1/2, 2, 2),
2
2
(x + 1/2)2 + (y − 2)2 + (z − 2)2 = 5/4
(c) r =
10. r = |[distance between (0,0,0) and (3, −2, 4)] ± 1| =
√
√
2
x2 + y 2 + z 2 = r 2 =
29 ± 1 = 30 ± 2 29
√
29 ± 1,
11. (x − 2)2 + (y + 1)2 + (z + 3)2 = r2 ,
(a) r2 = 32 = 9
(b) r2 = 12 = 1
12. (a) The sides have length 1, so the radius is
(b) The diagonal has length
√
1+1+1=
(c) r2 = 22 = 4
1
1
; hence (x + 2)2 + (y − 1)2 + (z − 3)2 =
2
4
√
3
3 and is a diameter, so (x+2)2 +(y−1)2 +(z−3)2 = .
4
13. (x + 5)2 + (y + 2)2 + (z + 1)2 = 49; sphere, C(−5, −2, −1), r = 7
14. x2 + (y − 1/2)2 + z 2 = 1/4; sphere, C(0, 1/2, 0), r = 1/2
√
15. (x − 1/2)2 + (y − 3/4)2 + (z + 5/4)2 = 54/16; sphere, C(1/2, 3/4, −5/4), r = 3 6/4
16. (x + 1)2 + (y − 1)2 + (z + 1)2 = 0; the point (−1, 1, −1)
17. (x − 3/2)2 + (y + 2)2 + (z − 4)2 = −11/4; no graph
18. (x − 1)2 + (y − 3)2 + (z − 4)2 = 25; sphere, C(1, 3, 4), r = 5
z
19. (a)
z
(b)
y
y
y
x
x
x
z
20. (a)
z
(c)
(b)
z
(c)
z
y=1
x=1
z=1
y
y
y
x
x
x
z
21. (a)
z
(b)
z
(c)
y
y
5
y
5
5
x
x
x
Exercise Set 13.1
450
z
22. (a)
z
(b)
z
(c)
y
y
y
x
x
x
23. (a) −2y + z = 0
(b) −2x + z = 0
(c) (x − 1) + (y − 1) = 1
2
(d) (x − 1)2 + (z − 1)2 = 1
2
24. (a) (x − a)2 + (z − a)2 = a2
(b) (x − a)2 + (y − a)2 = a2
(c) (y − a)2 + (z − a)2 = a2
z
25.
y
x
z
26.
z
27.
1
1
y
x
y
x
z
28.
z
29.
3
y
3
2
y
x
x
z
30.
z
31.
2
3
y
2
z
32.
√3
-3
y
y
x
3
3
x
x
451
Chapter 13
z
33.
-2
z
34.
2
y
x
y
x
1.4
35. (a)
z
(b)
-1.4
1.4
x
y
-1.4
36. (a)
-2
z
(b)
1
2
x
y
-1
37. Complete the square to get (x + 1)2 + (y − √
1)2 + (z − 2)2 = 9; center (−1, 1, 2), radius 3. The
distance between
√ is 6 < 3 so the origin is inside the sphere. The largest
√ the origin and the center
distance is 3 + 6, the smallest is 3 − 6.
38. (x − 1)2 + y 2 + (z + 4)2 ≤ 25; all points on and inside the sphere of radius 5 with center
at (1, 0, −4).
39. (y + 3)2 + (z − 2)2 > 16; all points outside the circular cylinder (y + 3)2 + (z − 2)2 = 16.
40.
p
(x − 1)2 + (y + 2)2 + z 2 = 2 x2 + (y − 1)2 + (z − 1)2 , square and simplify to get
3x2 + 3y 2 + 3z 2 + 2x − 12y − 8z + 3 = 0, then complete the square to get
√
(x + 1/3)2 + (y − 2)2 + (z − 4/3)2 = 44/9; center (−1/3, 2, 4/3), radius 2 11/3.
p
41. Let r be the radius of a styrofoam sphere. The distance from the origin to the center of the bowling
ball is equal to the sum of the distance from the origin to the center of the styrofoam sphere nearest
the origin and the distance between the center of this
√ sphere and the center of the bowling ball so
√
√
√
√
√
3−1
3R = 3r + r + R, ( 3 + 1)r = ( 3 − 1)R, r = √
R = (2 − 3)R.
3+1
42. (a) Complete the square to get (x + G/2)2 + (y + H/2)2 + (z + I/2)2 = K/4, so the equation
represents a sphere when K > 0, a point when K = 0, and no graph when K < 0.
√
(b) C(−G/2, −H/2, −I/2), r = K/2
Exercise Set 13.2
452
43. (a sin φ cos θ)2 + (a sin φ sin θ)2 + (a cos φ)2 = a2 sin2 φ cos2 θ + a2 sin2 φ sin2 θ + a2 cos2 φ
= a2 sin2 φ(cos2 θ + sin2 θ) + a2 cos2 φ
= a2 sin2 φ + a2 cos2 φ = a2 (sin2 φ + cos2 φ) = a2
EXERCISE SET 13.2
y
1. (a–c)
y
(d–f )
-5i + 3j
〈2, 5〉
x
x
〈2, 0〉
3i - 2j
〈–5, –4〉
-6j
y
2. (a–c)
y
(d–f )
〈-3, 7〉
4i + 2 j
x
x
〈6, -2〉
4i
-2 i - j
〈0, -8〉
z
3. (a–b)
z
(c–d)
–i + 2j + 3k
〈1, -2, 2〉
y
y
〈2, 2, –1〉
x
x
z
4. (a–b)
2i + 3j – k
z
(c–d)
〈-1, 3, 2 〉
i - j + 2k
y
y
〈3, 4, 2 〉
2j - k
x
x
453
Chapter 13
5. (a) h4 − 1, 1 − 5i = h3, −4i
(b) h0 − 2, 0 − 3, 4 − 0i = h−2, −3, 4i
y
z
–2i – 3j + 4k
x
3i – 4j
y
x
6. (a) h−3 − 2, 3 − 3i = h−5, 0i
(b) h0 − 3, 4 − 0, 4 − 4i = h−3, 4, 0i
y
z
– 3i + 4j
x
y
–5i
x
7. (a) h2 − 3, 8 − 5i = h−1, 3i
8. (a) h−4 − (−6), −1 − (−2)i = h2, 1i
(b) h0 − 7, 0 − (−2)i = h−7, 2i
(b) h−1, 6, 1i
(c) h−3, 6, 1i
(c) h5, 0, 0i
9. (a) Let (x, y) be the terminal point, then x − 1 = 3, x = 4 and y − (−2) = −2, y = −4.
The terminal point is (4, −4).
(b) Let (x, y, z) be the initial point, then 5 − x = −3, −y = 1, and −1 − z = 2 so x = 8,
y = −1, and z = −3. The initial point is (8, −1, −3).
10. (a) Let (x, y) be the terminal point, then x − 2 = 7, x = 9 and y − (−1) = 6, y = 5.
The terminal point is (9,5).
(b) Let (x, y, z) be the terminal point, then x + 2 = 1, y − 1 = 2, and z − 4 = −3 so x = −1,
y = 3, and z = 1. The terminal point is (−1, 3, 1).
11. (a) −i + 4j − 2k
(d) 40i − 4j − 4k
12. (a)
(c)
(e)
(f )
(b) 18i + 12j − 6k
(e) −2i − 16j − 18k
(c) −i − 5j − 2k
(f ) −i + 13j − 2k
h1, −2, 0i
(b) h28, 0, −14i + h3, 3, 9i = h31, 3, −5i
h3, −1, −5i
(d) 3(h2, −1, 3i − h28, 0, −14i) = 3h−26, −1, 17i = h−78, −3, 51i
h−12, 0, 6i − h8, 8, 24i = h−20, −8, −18i
h8, 0, −4i − h3, 0, 6i = h5, 0, −10i
√
√
13. (a) kvk = 1 + 1 = 2
√
(c) kvk = 21
√
√
(b) kvk = 1 + 49 = 5 2
√
(d) kvk = 14
√
14. (a) kvk = 9 + 16 = 5
(c) kvk = 3
√
(b) kvk = 2 + 7 = 3
√
(d) kvk = 3
Exercise Set 13.2
454
√
15. (a) ku + vk = k2i − 2j + 2kk = 2 3
√
√
(c) k − 2uk + 2kvk = 2 14 + 2 2
√
√
√
(e) (1/ 6)i + (1/ 6)j − (2/ 6)k
(b) kuk + kvk =
√
√
14 + 2
√
(d) k3u − 5v + wk = k − 12j + 2kk = 2 37
(f ) 1
16. If one vector is a positive multiple of the other, say u = αv with α > 0, then u, v and u + v are
parallel and ku + vk = (1 + α)kvk = kuk + kvk.
√
√
17 so the required vector is −1/ 17 i + 4/ 17 j
√
√
(b) k6i − 4j + 2kk = 2 14 so the required vector is (−3i + 2j − k)/ 14
−→
−→
√
√
(c) AB= 4i + j − k, k AB k = 3 2 so the required vector is (4i + j − k)/ 3 2
17. (a) k − i + 4jk =
√
3
4
1
18. (a) k3i − 4jk = 5 so the required vector is − (3i − 4j) = − i + j
5
5
5
1
2
2
(b) k2i − j − 2kk = 3 so the required vector is i − j − k
3
3
3
−→
−→
3
4
(c) AB = 4i − 3j, k AB k = 5 so the required vector is i − j
5
5
1
19. (a) − v = h−3/2, 2i
2
√
√
√
1
17
(b) kvk = 85, so √ v = √ h7, 0, −6i has length 17
85
5
20. (a) 3v = −6i + 9j
(b)
−
6
2
8
2
v = √ i− √ j− √ k
kvk
26
26
26
√
√
21. (a) v = kvkhcos π/4, sin π/4i = h3 2/2, 3 2/2i
(b) v = kvkhcos 90◦ , sin 90◦ i = h0, 2i
√
(c) v = kvkhcos 120◦ , sin 120◦ i = h−5/2, 5 3/2i
(d) v = kvkhcos π, sin πi = h−1, 0i
√
√
√
22. From (12), v =
π/6, sin π/6i√= h 3/2, 1/2i and
√w =√hcos 3π/4, sin
√ 3π/4i = h− 2/2, 2/2i,
√ hcos√
so v + w = (( 3 − 2)/2, (1 + 2)/2, v − w = (( 3 + 2)/2, (1 − 2)/2)
√
√
√
◦
◦
◦
◦
23. From (12), √
v = hcos
√ 30 , sin 30
√ i = h 3/2, 1/2i and w = hcos 135 , sin 135 i = h− 2/2, 2/2i, so
v + w = (( 3 − 2)/2, (1 + 2)/2)
√
√
24. w = h1, 0i, and from (12), v = hcos 120◦ , sin 120◦ i = h−1/2, 3/2i, so v + w = h1/2, 3/2i
25. (a) The initial point of u + v + w
is the origin and the endpoint
is (−2, 5), so u + v + w = h−2, 5i.
–2i + 5j
(b) The initial point of u + v + w
is (−5, 4) and the endpoint
is (−2, −4), so u + v + w = h3, −8i.
y
y
5
2
x
-5
5
x
-5
5
3i – 8j
-5
-8
455
Chapter 13
26. (a) v = h−10, 2i by inspection, so u − v + w = u + v + w − 2v = h−2, 5i + h20, −4i = h18, 1i.
(b) v = h−3, 8i by inspection, so u − v + w = u + v + w − 2v = h3, −8i + h6, −16i = h9, −24i.
27. 6x = 2u − v − w = h−4, 6i, x = h−2/3, 1i
28. u − 2x = x − w + 3v, 3x = u + w − 3v, x =
1
(u + w − 3v) = h2/3, 2/3i
3
2
1
8
1
4
5
i + j + k, v = i − j − k
30. u = h−5, 8i, v = h7, −11i
7
7
7
7
7
7
√
31. k(i + j) + (i − 2j)k = k2i − jk = 5, k(i + j − (i − 2j)k = k3jk = 3
29. u =
32. Let A, B, C be the vertices (0,0), (1,3), (2,4) and D the fourth vertex (x, y). For the parallelogram
−→
−→
ABCD, AD = BC, hx, yi = h1, 1i so x = 1, y = 1 and D is at (1,1). For the parallelogram ACBD,
−→
−→
AD = CB, hx, yi = h−1, −1i so x = −1, y = −1 and D is at (−1, −1). For the parallelogram
−→
−→
ABDC, AC=BD, hx − 1, y − 3i = h2, 4i, so x = 3, y = 7 and D is at (3, 7).
33. (a) 5 = kkvk = |k|kvk = ±3k, so k = ±5/3
(b) 6 = kkvk = |k|kvk = 2kvk, so kvk = 3
34. If kkvk = 0 then |k|kvk = 0 so either k = 0 or kvk = 0; in the latter case, by (9) or (10), v = 0.
−→
35. (a) Choose two points on the line, √for example√P1 (0, 2)
P2 = h1, 3i is
√ and P2 (1, 5);√then P1√
parallel to the line, kh1, 3ik = 10, so h1/ 10, 3/ 10i and h−1/ 10, −3/ 10i are unit
vectors parallel to the line.
−→
4) and P2 (1, 3);
(b) Choose two points on the line, √
for example
√ P1 (0,√
√ then
√ P1 P2 = h1, −1i is
parallel to the line, kh1, −1ik = 2 so h1/ 2, −1/ 2i and h−1/ 2, 1/ 2i are unit vectors
parallel to the line.
(c) Pick any line that is perpendicular to the line x + y = 4, for example y = x/5; then P1 (0, 0)
−→
1
and P2 (5, 1) are on the line, so P1 P2 = h5, 1i is perpendicular to the line, so ± √ h5, 1i are
26
unit vectors perpendicular to the line.
36. (a) ±k
(b) ±j
(c) ±i
37. (a) the circle of radius 1 about the origin
(b) the solid disk of radius 1 about the origin
(c) all points outside the solid disk of radius 1 about the origin
38. (a) the circle of radius 1 about the tip of r0
(b) the solid disk of radius 1 about the tip of r0
(c) all points outside the solid disk of radius 1 about the tip of r0
39. (a) the (hollow) sphere of radius 1 about the origin
(b) the solid ball of radius 1 about the origin
(c) all points outside the solid ball of radius 1 about the origin
40. The sum of the distances between (x, y) and the points (x1 , y1 ), (x2 , y2 ) is the constant k, so
the set consists of all points on the ellipse with foci at (x1 , y1 ) and (x2 , y2 ), and major axis of
length k.
Exercise Set 13.2
456
41. Since φ = π/2, from (13) we get kF1 + F2 k2 = kF1 k2 + kF2 k2 = 3600 + 900,
√
30
kF2 k
sin φ = √ , α ≈ 26.57◦ , θ = α ≈ 26.57◦ .
so kF1 + F2 k = 30 5 lb, and sin α =
kF1 + F2 k
30 5
1
42. kF1 + F2 k2 = kF1 k2 + kF2 k2 + 2kF1 kkF2 k cos φ = 14,400 + 10,000 + 2(120)(100) = 36,400, so
2
√
√
100
kF2 k
5 3
sin φ = √ sin 60◦ = √ , α ≈ 27.16◦ ,
kF1 + F2 k = 20 91 N, sin α =
kF1 + F2 k
20 91
2 91
θ = α ≈ 27.16◦ .
√
3
2
2
2
,
43. kF1 + F2 k = kF1 k + kF2 k + 2kF1 kkF2 k cos φ = 160,000 + 160,000 − 2(400)(400)
2
400
kF2 k
1
sin φ ≈
, α = 75.00◦ ,
so kF1 + F2 k ≈ 207.06 N, and sin α =
kF1 + F2 k
207.06 2
θ = α − 30◦ = 45.00◦ .
44. kF1 + F2 k2 = kF1 k2 + kF2 k2 + 2kF1 kkF2 k cos φ = 16 + 4 + 2(4)(2) cos 77◦ , so
2
kF2 k
sin φ =
sin 77◦ , α ≈ 23.64◦ , θ = α−27◦ ≈ −3.36◦ .
kF1 +F2 k ≈ 4.86 lb, and sin α =
kF1 + F2 k
4.86
45. Let F1 , F2 , F3 be the forces in the diagram with magnitudes 40, 50, 75 respectively. Then
F1 + F2 + F3 = (F1 + F2 ) + F3 . Following the examples, F1 + F2 has magnitude 45.83 N and
makes an angle 79.11◦ with the positive x-axis. Then
k(F1 +F2 )+F3 k2 ≈ 45.832 +752 +2(45.83)(75) cos 79.11◦ , so F1 +F2 +F3 has magnitude ≈ 94.995
N and makes an angle θ = α ≈ 28.28◦ with the positive x-axis.
46. Let F1 , F2 , F3 be the forces in the diagram with magnitudes 150, 200, 100 respectively. Then
F1 + F2 + F3 = (F1 + F2 ) + F3 . Following the examples, F1 + F2 has magnitude 279.34 N and
makes an angle 91.24◦ with the positive x-axis. Then
kF1 + F2 + F3 k2 ≈ 279.342 + 1002 + 2(279.34)(100) cos(270 − 91.24)◦ , and F1 + F2 + F3 has
magnitude ≈ 179.37 N and makes an angle 91.94◦ with the positive x-axis.
be the forces in the diagram with√magnitudes 8, 10 respectively. Then kF1 + F2 k has
47. Let F1 , F2 √
magnitude 82 + 102 + 2 · 8 · 10 cos 120◦ = 2 21 ≈ 9.165 lb, and makes an angle
kF1 k
sin 120 ≈ 109.11◦ with the positive x-axis, so F has magnitude 9.165 lb and
60◦ + sin−1
kF1 + F2 k
makes an angle −70.89◦ with the positive x-axis.
√
48. kF1 + F2 k = 1202 + 1502 + 2 · 120 · 150 cos 75◦ = 214.98 N and makes an angle 92.63◦ with the
positive x-axis, and kF1 +F2 +F3 k = 232.90 N and makes an angle 67.23◦ with the positive x-axis,
hence F has magnitude 232.90 N and makes an angle −112.77◦ with the positive x-axis.
49. F1 + F2 + F = 0, where F has magnitude 250 and makes an angle −90◦ with the positive x-axis.
Thus kF1 + F2 k2 = kF1 k2 + kF2 k2 + 2kF1 kkF2 k cos 105◦ = 2502 and
√
kF2 k
2
kF2 k
−1
◦
◦
sin 105 , so
≈
0.9659, kF2 k ≈ 183.02 lb,
45 = α = sin
250
2
250
kF1 k2 + 2(183.02)(−0.2588)kF1 k + (183.02)2 = 62,500, kF1 k = 224.13 lb.
√
50. Similar to Exercise 49, kF1 k = 100 3 N, kF2 k = 100 N
51. (a) c1 v1 + c2 v2 = (2c1 + 4c2 ) i + (−c1 + 2c2 ) j = −4j, so 2c1 + 4c2 = 0 and −c1 + 2c2 = −4
which gives c1 = 2, c2 = −1.
(b) c1 v1 + c2 v2 = hc1 − 2c2 , −3c1 + 6c2 i = h3, 5i, so c1 − 2c2 = 3 and −3c1 + 6c2 = 5
which has no solution.
457
Chapter 13
52. (a) Equate corresponding components to get the system of equations c1 + 3c2 = −1,
2c2 + c3 = 1, and c1 + c3 = 5. Solve to get c1 = 2, c2 = −1, and c3 = 3.
(b) Equate corresponding components to get the system of equations c1 + 3c2 + 4c3 = 2,
−c1 − c3 = 1, and c2 + c3 = −1. From the second and third equations, c1 = −1 − c3 and
c2 = −1 − c3 ; substitute these into the first equation to get −4 = 2, which is nonsense so the
system has no solution.
53. Place u and v tip to tail so that u + v is the vector from the initial point of u to the terminal
point of v. The shortest distance between two points is along the line joining these points so
ku + vk ≤ kuk + kvk.
54. (a): u + v = (u1 i + u2 j) + (v1 i + v2 j) = (v1 i + v2 j) + u1 i + u2 j = v + u
(c): u + 0 = u1 i + u2 j + 0i + 0j = u1 i + u2 j = u
(e): k(lu) = k(l(u1 i + u2 j)) = k(lu1 i + lu2 j) = klu1 i + klu2 j = (kl)u
55. (d): u + (−u) = u1 i + u2 j + (−u1 i − u2 j) = (u1 − u1 )i + (u1 − u1 ) j = 0
(g): (k + l)u = (k + l)(u1 i + u2 j) = ku1 i + ku2 j + lu1 i + lu2 j = ku + lu
(h): 1u = 1(u1 i + u2 j) = 1u1 i + 1u2 j = u1 i + u2 j = u
56. Draw the triangles with sides formed by the vectors u, v, u + v and ku, kv, ku + kv. By similar
triangles, k(u + v) = ku + kv.
57. Let a, b, c be vectors along the sides of the
1
1
midpoints of a and b, then u = a − b =
2
2
parallel to c and half as long.
triangle and A,B the
1
1
(a − b) = c so u is
2
2
a
A
c
u
b
B
58. Let a, b, c, d be vectors along the sides of the quadrilateral
and A, B, C, D the corresponding midpoints, then
1
1
1
1
u = b + c and v = d − a but d = a + b + c so
2
2
2
2
1
1
1
1
v = (a + b + c) − a = b + c = u thus ABCD
2
2
2
2
is a parallelogram because sides AD and BC are equal
and parallel.
B
a
b
u
v
D
EXERCISE SET 13.3
√
√
1. (a) (1)(6) + (2)(−8) = −10; cos θ = (−10)/[( 5)(10)] = −1/ 5
√
√
(b) (−7)(0) + (−3)(1) = −3; cos θ = (−3)/[( 58)(1)] = −3/ 58
(c) (1)(8) + (−3)(−2) + (7)(−2) = 0; cos θ = 0
√
√
√
(d) (−3)(4) + (1)(2) + (2)(−5) = −20; cos θ = (−20)/[( 14)( 45)] = −20/(3 70)
2. (a) u · v = 1(2) cos(π/6) =
√
3. (a) u · v = −34 < 0, obtuse
(c) u · v = −1 < 0, obtuse
3
C
c
A
√
(b) u · v = 2(3) cos 135◦ = −3 2
(b) u · v = 6 > 0, acute
(d) u · v = 0, orthogonal
d
Exercise Set 13.3
458
−→
4. Let the points be P, Q, R in order, then P Q= h2 − (−1), −2 − 2, 0 − 3i = h3, −4, −3i,
−→
−→
QR= h3 − 2, 1 − (−2), −4 − 0i = h1, 3, −4i, RP = h−1 − 3, 2 − 1, 3 − (−4)i = h−4, 1, 7i;
−→
−→
−→
−→
−→
−→
since QP · QR= −3(1) + 4(3) + 3(−4) = −3 < 0, 6 P QR is obtuse;
since RP · RQ= −4(−1) + (−3) + 7(4) = 29 > 0, 6 P RQ is acute;
since P R · P Q= 4(3) − 1(−4) − 7(−3) = 37 > 0, 6 RP Q is acute
5. Since v1 · vi = cos φi , the answers are, in order,
√
√
√
√
2/2, 0, − 2/2, −1, − 2/2, 0, 2/2
6. Proceed as in Exercise 5; 25/2, −25/2, −25, −25/2, 25/2
−→
−→
−→
−→
−→
−→
7. (a) AB = h1, 3, −2i, BC = h4, −2, −1i, AB · BC = 0 so AB and BC are orthogonal; it is a right
triangle with the right angle at vertex B.
(b) Let A, B, and C be the vertices (−1, 0), (2, −1), and (1,4) with corresponding interior angles
α, β, and γ, then
cos α =
cos β =
cos γ =
−→
−→
−→
−→
−→
−→
−→
−→
−→
−→
−→
−→
−→
AB · AC
k AB k k AC k
BA · BC
k BA k k BC k
CA · CB
k CA k k CB k
=
√
h3, −1i · h2, 4i
√ √
= 1/(5 2), α ≈ 82◦
10 20
=
√
h−3, 1i · h−1, 5i
√ √
= 4/ 65, β ≈ 60◦
10 26
=
√
h−2, −4i · h1, −5i
√ √
= 9/ 130, γ ≈ 38◦
20 26
−→
8. AB · AP = [2i + j + 2k] · [(k − 1)i + (k + 1)j + (k − 3)k]
= 2(k − 1) + (k + 1) + 2(k − 3) = 5k − 7 = 0, k = 7/5.
9. (a) v · v1 = −ab + ba = 0; v · v2 = ab + b(−a) = 0
(b) Let v1 = 2i + 3j, v2 = −2i − 3j;
2
v1
3
= √ i + √ j, u2 = −u1 .
take u1 =
kv1 k
13
13
y
3
u1
x
3
-3
u2
-3
v
10. By inspection, 3i − 4j is orthogonal to and has the same length as 4i + 3j
so u1 = (4i + 3j) + (3i − 4j) = 7i − j and u2 = (4i + 3j) + (−1)(3i − 4j) = i +√
7j each make an angle
√
of 45◦ with 4i + 3j; unit vectors in the directions of u1 and u2 are (7i − j)/ 50 and (i + 7j)/ 50.
11. (a) The dot product of a vector u and a scalar v · w is not defined.
(b) The sum of a scalar u · v and a vector w is not defined.
(c) u · v is not a vector.
(d) The dot product of a scalar k and a vector u + v is not defined.
459
Chapter 13
12. (b): u · (v + w) = (6i − j + 2k) · ((2i + 7j + 4k) + (i + j − 3k)) = (6i − j + 2k) · (3i + 8j + k) = 12;
u · v + u · w = (6i − j + 2k) · (2i + 7j + 4k) + (6i − j + 2k) · (i + j − 3k) = 13 − 1 = 12
(c): k(u · v) = −5(13) = −65; (ku) · v = (−30i + 5j − 10k) · (2i + 7j + 4k) = −65;
u · (kv) = (6i − j + 2k) · (−10i − 35j − 20k) = −65
13. (a) h1, 2i · (h28, −14i + h6, 0i) = h1, 2i · h34, −14i = 6
√
(b) k6wk = 6kwk = 36
(c) 24 5
√
(d) 24 5
14. false, for example a = h1, 2i, b = h−1, 0i, c = h5, −3i
15. (a) kvk =
√
√
√
3 so cos α = cos β = 1/ 3, cos γ = −1/ 3, α = β ≈ 55◦ , γ ≈ 125◦
(b) kvk = 3 so cos α = 2/3, cos β = −2/3, cos γ = 1/3, α ≈ 48◦ , β ≈ 132◦ , γ ≈ 71◦
16. (a) kvk = 7 so cos α = 3/7, cos β = −2/7, cos γ = −6/7, α ≈ 65◦ , β ≈ 107◦ , γ ≈ 149◦
(b) kvk = 5, cos α = 3/5, cos β = 0, cos γ = −4/5, α ≈ 53◦ , β ≈ 90◦ , γ ≈ 143◦
17. cos2 α + cos2 β + cos2 γ =
v12
v22
v32
+
+
= v12 + v22 + v32 /kvk2 = kvk2 /kvk2 = 1
2
2
2
kvk
kvk
kvk
p
p
p
18. Let v = hx, y, zi, then x = x2 + y 2 cos θ, y = x2 + y 2 sin θ, x2 + y 2 = kvk cos λ, and
z = kvk sin λ, so x/kvk = cos θ cos λ, y/kvk = sin θ cos λ, and z/kvk = sin λ.
√
√ √
√
3
1
31
3
3 3
=
, cos β =
= , cos γ = ; α ≈ 64◦ , β ≈ 41◦ , γ = 60◦
19. cos α =
2 2
4
2 2
4
2
20. Let u1 = ku1 khcos α1 , cos β1 , cos γ1 i, u2 = ku2 khcos α2 , cos β2 , cos γ2 i, u1 and u2 are perpendicular
if and only if u1 · u2 = 0 so ku1 k ku2 k(cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 ) = 0,
cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 = 0.
21. (a)
b
= h3/5, 4/5i, so projb v = h6/25, 8/25i
kbk
and v − projb v = h44/25, −33/25i
y
2
projbv
x
2
v
-2
-2
(b)
√
√
b
= h1/ 5, −2/ 5i, so projb v = h−6/5, 12/5i
kbk
and v − projb v = h26/5, 13/5i
5
v – projbv
y v – projbv
v
projbv
-5
x
5
-5
Exercise Set 13.3
(c)
460
√
√
b
= h2/ 5, 1/ 5i, so projb v = h−16/5, −8/5i
kbk
and v − projb v = h1/5, −2/5i
y
x
-4
v – projbv
projbv
v
-4
22. (a)
(b)
b
= h1/3, 2/3, 2/3i, so projb v = h2/3, 4/3, 4/3i and v − projb v = h4/3, −7/3, 5/3i
kbk
b
= h2/7, 3/7, −6/7i, so projb v = h−74/49, −111/49, 222/49i
kbk
and v − projb v = h270/49, 62/49, 121/49i
23. (a) projb v = h−1, −1i, so v = h−1, −1i + h3, −3i
(b) projb v = h16/5, 0, −8/5i, so v = h16/5, 0, −8/5i + h−1/5, 1, −2/5i
24. (a) projb v = h1, 1i, so v = h1, 1i + h−4, 4i
(b) projb v = h0, −8/5, 4/5i, so v = h0, −8/5, 4/5i + h−2, 13/5, 26/5i
−→
−→
−→
−→
−→
−→
25. AP = −i + 3j, AB= 3i + 4j, kproj −→ AP k = | AP · AB |/k AB k = 9/5
AB
−→
√ p
k AP k = 10, 10 − 81/25 = 13/5
−→
−→
−→
−→
−→
−→
√
26. AP = −4i + 2k, AB = −3i + 2j − 4k, kproj −→ AP k = | AP · AB |/k AB k = 4/ 29.
AB
−→
p
p
√
k AP k = 20, 20 − 16/29 = 564/29
27. Let F be the downward force of gravity on the block, then kFk = 10(9.8) = 98 N, and
√ if F1 and
perpendicular
to
the
ramp,
then
kF
k
=
kF
k
=
49
2 N. Thus
F2 are the forces parallel to and
1
2 √
√
the block exerts a force of 49 2 N against the ramp and it requires a force of 49 2 N to prevent
the block from sliding down the ramp.
28. Let x denote the magnitude of the force in the direction of Q. Then the force F acting on the
1
1
block is F = xi − 10j. Let u = − √ (i + j) and v = √ (i − j) be the unit vectors in the directions
2
2
x + 10
x − 10
along and against the ramp. Then F decomposes as F = − √ u + √ v, and thus the block
2
2
will not slide down the ramp provided x ≥ 10 N.
29. Three forces act on the block: its weight −300j; the tension √
in cable A, which has the form
j), where a, b are√ positive
a(−i + j); and the tension in cable B, which has the form b( 3i −√
constants. The sum of these forces is zero, which yields a = 450 + 150 3, b = 150 + 150 3. Thus
the forces√along cables A and
respectively,
√ B are, √
√
√
√
k150(3 + 3)(i − j)k = 450 2 + 150 6 lb, and k150( 3 + 1)( 3i − j)k = 300 + 300 3 lb.
461
Chapter 13
30. (a) Let TA and TB be the forces exerted on the block by cables A and B. Then
TA = a(−10i + dj) and TB = b(20i + dj) for some positive a, b. Since TA + TB − 100j = 0, we
100
200
100
2000
2000
200
,b =
, TA = −
i+
j, and TB =
i+
j.
find a =
3d
3d
3d
3
3d
3
500
-20
100
-100
(b) An increase in d will decrease both forces.
500
-20
100
-100
40
(c) The inequality kTA k ≤ 150 is equivalent to d ≥ √ , and kTB k ≤ 150 is equivalent to
65
40
40
.
d ≥ √ . Hence we must have d ≥
65
77
−→
−→
31. Let P and Q be the points (1,3) and (4,7) then P Q = 3i + 4j so W = F · P Q = −12 ft · lb.
−→
−→
32. W = F · P Q= kFk k P Q k cos 45◦ = (500)(100)
√
√
2/2 = 25,000 2 N · m
33. W = F ·15i = 15 · 50 cos 60◦ = 375 ft · lb.
√
√
34. W = F ·(15/ 3)(i + j + k) = −15/ 3 N · m
35. With the cube as shown in the diagram,
and a the length of each edge,
d1 = ai + aj + ak, d2 = ai + aj − ak,
cos θ = (d1 · d2 ) / (kd1 k kd2 k) = 1/3, θ ≈ 71◦
z
d2
y
d1
x
36. Take i, j, and k along adjacent edges of the box, then 10i + 15j + 25k is along a diagonal, and a
√
2
3
5
unit vector in this direction is √ i + √ j + √ k. The direction cosines are cos α = 2/ 38,
38
38
38
√
√
◦
cos β = 3/ 38, and cos γ = 5/ 38 so α ≈ 71 , β ≈ 61◦ , and γ ≈ 36◦ .
Exercise Set 13.3
462
37. u + v and u − v are vectors along the diagonals,
(u + v) · (u − v) = u · u − u · v + v · u − v · v = kuk2 − kvk2 so (u + v) · (u − v) = 0
if and only if kuk = kvk.
38. The diagonals have lengths ku + vk and ku − vk but
ku + vk2 = (u + v) · (u + v) = kuk2 + 2u · v + kvk2 , and
ku − vk2 = (u − v) · (u − v) = kuk2 − 2u · v + kvk2 . If the parallelogram is a rectangle then
u · v = 0 so ku + vk2 = ku − vk2 ; the diagonals are equal. If the diagonals are equal, then
4u · v = 0, u · v = 0 so u is perpendicular to v and hence the parallelogram is a rectangle.
39. ku + vk2 = (u + v) · (u + v) = kuk2 + 2u · v + kvk2 and
ku − vk2 = (u − v) · (u − v) = kuk2 − 2u · v + kvk2 , add to get
ku + vk2 + ku − vk2 = 2kuk2 + 2kvk2
The sum of the squares of the lengths of the diagonals of a parallelogram is equal to twice the sum
of the squares of the lengths of the sides.
40. ku + vk2 = (u + v) · (u + v) = kuk2 + 2u · v + kvk2 and
ku − vk2 = (u − v) · (u − v) = kuk2 − 2u · v + kvk2 , subtract to get
ku + vk2 − ku − vk2 = 4u · v, the result follows by dividing both sides by 4.
41. v = c1 v1 + c2 v2 + c3 v3 so v · vi = ci vi · vi because vi · vj = 0 if i 6= j,
thus v · vi = ci kvi k2 , ci = v · vi /kvi k2 for i = 1, 2, 3.
42. v1 · v2 = v1 · v3 = v2 · v3 = 0 so they are mutually perpendicular. Let v = i − j + k, then
v · v1
3
v · v2
1
v · v3
1
c1 =
.
= , c2 =
= − , and c3 =
=
kv1 k2
7
kv2 k2
3
kv3 k2
21
43. (a) u = xi + (x2 + 1)j, v = xi − (x + 1)j, θ = cos−1 [(u · v)/(kukkvk)].
Solve dθ/dx = 0 to find that the minimum value of θ occurs when x ≈ −3.136742 so the
minimum angle is about 40◦ .
(b) Solve u · v = 0 for x to get x ≈ −0.682328.
44. (a) u = cos θ1 i ± sin θ1 j, v = ± sin θ2 j + cos θ2 k, cos θ = u · v = ± sin θ1 sin θ2
(b) cos θ = ± sin2 45◦ = ±1/2, θ = 60◦
(c) Let θ(t) = cos−1 (sin t sin 2t); solve θ0 (t) = 0 for t to find that θmax ≈ 140◦ (reject, since θ
is acute) when t ≈ 2.186276 and that θmin ≈ 40◦ when t ≈ 0.955317; for θmax check the
endpoints t = 0, π/2 to obtain θmax = cos−1 (0) = π/2.
45. Let u = hu1 , u2 , u3 i, v = hv1 , v2 , v3 i, w = hw1 , w2 , w3 i. Then
u · (v + w) = hu1 (v1 + w1 ), u2 (v2 + w2 ), u3 (v3 + w3 )i = hu1 v1 + u1 w1 , u2 v2 + u2 w2 , u3 v3 + u3 w3 i
= hu1 v1 , u2 v2 , u3 v3 i + hu1 w1 , u2 w2 , u3 w3 i = u · v + u · w
0 · v = 0 · v1 + 0 · v2 + 0 · v3 = 0
463
Chapter 13
EXERCISE SET 13.4
1. (a) i × (i + j + k) =
i
1
1
j k
0 0
1 1
= −j + k
(b) i × (i + j + k) = (i × i) + (i × j) + (i × k) = −j + k
2. (a) j × (i + j + k) =
i
0
1
j k
1 0
1 1
=i−k
j × (i + j + k) = (j × i) + (j × j) + (j × k) = i − k
(b) k × (i + j + k) =
i
0
1
j k
0 1
1 1
= −i + j
k × (i + j + k) = (k × i) + (k × j) + (k × k) = j − i + 0 = −i + j
3. h7, 10, 9i
4. −i − 2j − 7k
5. h−4, −6, −3i
6. i + 2j − 4k
7. (a) v × w = h−23, 7, −1i, u × (v × w) = h−20, −67, −9i
(b) u × v = h−10, −14, 2i, (u × v) × w = h−78, 52, −26i
(c) (u × v) × (v × w) = h−10, −14, 2i × h−23, 7, −1i = h0, −56, −392i
(d) (v × w) × (u × v) = h0, 56, 392i
1
1
9. u × v = (i + j) × (i + j + k) = k − j − k + i = i − j, the direction cosines are √ , − √ , 0
2
2
!
√
1
5
2
10. u × v = 12i + 30j − 6k, so ± √ i + √ j − √ k
30
6
30
−→
−→
1
11. n = AB × AC = h1, 1, −3i × h−1, 3, −1i = h8, 4, 4i, unit vectors are ± √ h2, 1, 1i
6
12. A vector parallel to the yz-plane must be perpendicular to i;
√
√
i × (3i − j + 2k) = −2j − k, k − 2j − kk = 5, the unit vectors are ±(2j + k)/ 5.
13. A = ku × vk = k − 7i − j + 3kk =
√
59
14. A = ku × vk = k − 6i + 4j + 7kk =
15. A =
−→
√
1
1
1 −→
k P Q × P R k = kh−1, −5, 2i × h2, 0, 3ik = kh−15, 7, 10ik = 374/2
2
2
2
16. A =
−→
√
1
1
1 −→
k P Q × P R k = kh−1, 4, 8i × h5, 2, 12ik = kh32, 52, −22ik = 9 13
2
2
2
17. 80
18. 29
21. V = |u · (v × w)| = | − 16| = 16
23. (a) u · (v × w) = 0, yes
19. −3
√
101
20. 1
22. V = |u · (v × w)| = |45| = 45
(b) u · (v × w) = 0, yes
(c) u · (v × w) = 245, no
Exercise Set 13.4
464
24. (a) u · (w × v) = −u · (v × w) = −3
(c) w · (u × v) = u · (v × w) = 3
(e) (u × w) · v = u · (w × v) = −3
(b) (v × w) · u = u · (v × w) = 3
(d) v · (u × w) = u · (w × v) = −3
(f ) v · (w × w) = 0 because w × w = 0
√
25. (a) V = |u · (v × w)| = | − 9| = 9
(b) A = ku × wk = k3i − 8j + 7kk = 122
(c) v × w = −3i − j + 2k is perpendicular to the plane determined by v and w; let θ be the
angle between u and v × w then
cos θ =
−9
u · (v × w)
= √ √ = −9/14
kuk kv × wk
14 14
so the acute angle φ that u makes with the plane determined by v and w is
φ = θ − π/2 = sin−1 (9/14).
26. From the diagram,
P
ku × vk
kukkvk sin θ
=
d = kuk sin θ =
kvk
kvk
u
d
θ
v
−→
−→
27. (a) u = AP = −4i + 2k, v = ABp
= −3i + 2j − 4k, u × v = −4i − 22j − 8k;
distance = ku × vk/kvk = 2 141/29
−→
−→
√
(b) u = AP = 2i + 2j, v = AB = −2i + j, u × v = 6k; distance = ku × vk/kvk = 6/ 5
1
kv × wk and
2
1
1
|u · (v × w)|
height = kprojv×w uk =
so V = (area of base) (height) = |u · (v × w)|
kv × wk
3
6
28. Take v and w as sides of the (triangular) base, then area of base =
−→
−→
−→
29. P Q = h3, −1, −3i, P R = h2, −2, 1i, P S = h4, −4, 3i,
V =
−→
−→
1
1 −→
| P Q · (P R × P S)| = |−4| = 2/3
6
6
√
ku × vk
k36i − 24jk
12 13
(b) sin θ =
=
=
kukkvk
49
49
23
u·v
=−
30. (a) cos θ =
kukkvk
49
(c)
232
144 · 13
2401
+
=
=1
2
2
49
49
492
31. From Theorems 13.3.3 and 13.4.5a it follows that sin θ = cos θ, so θ = π/4.
32. ku × vk2 = kuk2 kvk2 sin2 θ = kuk2 kvk2 (1 − cos2 θ) = kuk2 kvk2 − (u · v)2
−→
33. (a) F = 10j and P Q= i + j + k, so the vector moment of F about P is
−→
PQ ×F =
i
1
0
j k
1 1
10 0
√
= −10i + 10k, and the scalar moment is 10 2 lb·ft.
The direction of rotation of the cube about P is counterclockwise looking along
−→
P Q × F = −10i + 10k toward its initial point.
465
Chapter 13
−→
(b) F = 10j and P Q= j + k, so the vector moment of F about P is
−→
PQ ×F =
i
0
0
j k
1 1
10 0
= −10i, and the scalar moment is 10 lb·ft. The direction of rotation
of the cube about P is counterclockwise looking along −10i toward its initial point.
−→
(c) F = 10j and P Q= j, so the vector moment of F about P is
−→
PQ ×F =
i
0
0
j k
1 0
10 0
= 0, and the scalar moment is 0 lb·ft. Since the force is parallel to
the direction of motion, there is no rotation about P .
−→
1000
34. (a) F = √ (−i + k) and P Q= 2j − k, so the vector moment of F about P is
2
i
j k
−→
√
√
P Q × F = 500 2 0 2 −1 = 500 2(2i + j + 2k), and the scalar moment is
−1 0 1
√
1500 2 N·m.
(b) The direction angles of the vector moment of F about the point P are
cos−1 (2/3) ≈ 48◦ , cos−1 (1/3) ≈ 71◦ , and cos−1 (2/3) ≈ 48◦ .
35. F makes an angle 72◦ with the positive x-axis, and P~Q = 0.2i + 0.03j makes an angle
α = tan−1 (0.03/0.2) with the x-axis, so φ ≈ 72◦ − 8.53◦ = 63.47◦ and kF k = 200 N, so
−→
k P Q ×Fk = kP~QkkFk| sin 63.47◦ | = 200k(0.2i + 0.030j)k| sin 63.47◦ | ≈ 36.19 N·m.
36.
Part (b) :
let u = hu1 , u2 , u3 i , v = hv1 , v2 , v3 i , and w = hw1 , w2 , w3 i ; show that
u × (v + w) and (u × v) + (u × w) are the same.
Part (c) :
(u + v) × w = −[w × (u + v)] from part (a)
= −[(w × u) + (w × v)] from part (b)
= (u × w) + (v × w) from part (a)
37. Let u = hu1 , u2 , u3 i and v = hv1 , v2 , v3 i; show that k(u × v), (ku) × v, and u × (kv) are all the
same; Part (e) is proved in a similar fashion.
38. Suppose the first two rows are interchanged. Then by definition,
b1
a1
c1
b2
a2
c2
b3
a3
c3
= b1
a2
c2
a3
c3
− b2
a1
c1
a3
c3
+ b3
a1
c1
a2
c2
= b1 (a2 c3 − a3 c2 ) − b2 (a1 c3 − a3 c1 ) + b3 (a1 c2 − a2 c1 ),
which is the negative of the right hand side of (2) after expansion. If two other rows were to be
exchanged, a similar proof would hold. Finally, suppose ∆ were a determinant with two identical
rows. Then the value is unchanged if we interchange those two rows, yet ∆ = −∆ by Part (b) of
Theorem 13.4.1. Hence ∆ = −∆, ∆ = 0.
39. −8i − 20j + 2k, −8i − 8k
40. (a) From the first formula in Exercise 39 it follows that u × (v × w) is a linear combination of
v and w and hence lies in the plane determined by them.
(b) Similar to (a).
Exercise Set 13.5
466
41. If a, b, c, and d lie in the same plane then a × b and c × d are parallel so (a × b) × (c × d) = 0
42. Let u and v be the vectors from a point on the curve to the points (2, −1, 0) and (3, 2, 2), respectively. Then u = (2 − x)i + (−1 − lnx)j and v = (3 − x)i + (2 − lnx)j + 2k. The area of the triangle
is given by A = (1/2)ku × vk; solve dA/dx = 0 for x to get x = 2.091581. The minimum area is
1.887850.
−→
−→
−→
−→
−→
43. P Q0 × F =P Q × F+ QQ0 × F =P Q × F, since F and QQ0 are parallel.
EXERCISE SET 13.5
In many of the Exercises in this section other answers are also possible.
1. (a) L1 : P (1, 0), v = j, x = 1, y = t
L2 : P (0, 1), v = i, x = t, y = 1
L3 : P (0, 0), v = i + j, x = t, y = t
(b) L1 :
L2 :
L3 :
L4 :
P (1, 1, 0), v = k, x = 1, y = 1, z = t
P (0, 1, 1), v = i, x = t, y = 1, z = 1
P (1, 0, 1), v = j, x = 1, y = t, z = 1
P (0, 0, 0), v = i + j + k, x = t,
y = t, z = t
2. (a) L1 : x = t, y = 1, 0 ≤ t ≤ 1
L2 : x = 1, y = t, 0 ≤ t ≤ 1
L3 : x = t, y = t, 0 ≤ t ≤ 1
(b) L1 :
L2 :
L3 :
L4 :
x = 1, y = 1, z = t, 0 ≤ t ≤ 1
x = t, y = 1, z = 1, 0 ≤ t ≤ 1
x = 1, y = t, z = 1, 0 ≤ t ≤ 1
x = t, y = t, z = t, 0 ≤ t ≤ 1
−→
3. (a) P1 P2 = h2, 3i so x = 3 + 2t, y = −2 + 3t for the line; for the line segment add the condition
0 ≤ t ≤ 1.
−→
(b) P1 P2 = h−3, 6, 1i so x = 5 − 3t, y = −2 + 6t, z = 1 + t for the line; for the line segment add
the condition 0 ≤ t ≤ 1.
−→
4. (a) P1 P2 = h−3, −5i so x = −3t, y = 1 − 5t for the line; for the line segment add the condition
0 ≤ t ≤ 1.
−→
(b) P1 P2 = h0, 0, −3i so x = −1, y = 3,z = 5 − 3t for the line; for the line segment add the
condition 0 ≤ t ≤ 1.
5. (a) x = 2 + t, y = −3 − 4t
(b) x = t, y = −t, z = 1 + t
6. (a) x = 3 + 2t, y = −4 + t
(b) x = −1 − t, y = 3t, z = 2
7. (a) r0 = 2i − j so P (2, −1) is on the line, and v = 4i − j is parallel to the line.
(b) At t = 0, P (−1, 2, 4) is on the line, and v = 5i + 7j − 8k is parallel to the line.
8. (a) At t = 0, P (−1, 5) is on the line, and v = 2i + 3j is parallel to the line.
(b) r0 = i + j − 2k so P (1, 1, −2) is on the line, and v = j is parallel to the line.
9. (a) hx, yi = h−3, 4i + th1, 5i; r = −3i + 4j + t(i + 5j)
(b) hx, y, zi = h2, −3, 0i + th−1, 5, 1i; r = 2i − 3j + t(−i + 5j + k)
10. (a) hx, yi = h0, −2i + th1, 1i; r = −2j + t(i + j)
(b) hx, y, zi = h1, −7, 4i + th1, 3, −5i; r = i − 7j + 4k + t(i + 3j − 5k)
11. x = −5 + 2t, y = 2 − 3t
12. x = t, y = 3 − 2t
467
Chapter 13
13. 2x + 2yy 0 = 0, y 0 = −x/y = −(3)/(−4) = 3/4, v = 4i + 3j; x = 3 + 4t, y = −4 + 3t
14. y 0 = 2x = 2(−2) = −4, v = i − 4j; x = −2 + t, y = 4 − 4t
15. x = −1 + 3t, y = 2 − 4t, z = 4 + t
16. x = 2 − t, y = −1 + 2t, z = 5 + 7t
17. The line is parallel to the vector h2, −1, 2i so x = −2 + 2t, y = −t, z = 5 + 2t.
18. The line is parallel to the vector h1, 1, 0i so x = t, y = t, z = 0.
19. (a) y = 0, 2 − t = 0, t = 2, x = 7
(b) x = 0, 1 + 3t = 0, t = −1/3, y = 7/3
√
√
√
−1 ± 85
43 ∓ 85
−7 ± 85
2
2
2
,x =
,y =
(c) y = x , 2 − t = (1 + 3t) , 9t + 7t − 1 = 0, t =
18
6
18
20. (4t)2 + (3t)2 = 25, 25t2 = 25, t = ±1, the line intersects the circle at ±h4, 3i
21. (a) z = 0 when t = 3 so the point is (−2, 10, 0)
(b) y = 0 when t = −2 so the point is (−2, 0, −5)
(c) x is always −2 so the line does not intersect the yz-plane
22. (a) z = 0 when t = 4 so the point is (7,7,0)
(b) y = 0 when t = −3 so the point is (−7, 0, 7)
(c) x = 0 when t = 1/2 so the point is (0, 7/2, 7/2)
23. (1 + t)2 + (3 − t)2 = 16, t2 − 2t − 3 = 0, (t + 1)(t − 3) = 0; t = −1, 3. The points of intersection
are (0, 4, −2) and (4,0,6).
24. 2(3t) + 3(−1 + 2t) = 6, 12t = 9; t = 3/4. The point of intersection is (5/4, 9/4, 1/2).
25. The lines intersect if we can find values of t1 and t2 that satisfy the equations 2 + t1 = 2 + t2 ,
2 + 3t1 = 3 + 4t2 , and 3 + t1 = 4 + 2t2 . Solutions of the first two of these equations are t1 = −1,
t2 = −1 which also satisfy the third equation so the lines intersect at (1, −1, 2).
26. Solve the equations −1 + 4t1 = −13 + 12t2 , 3 + t1 = 1 + 6t2 , and 1 = 2 + 3t2 . The third equation
yields t2 = −1/3 which when substituted into the first and second equations gives t1 = −4 in both
cases; the lines intersect at (−17, −1, 1).
27. The lines are parallel, respectively, to the vectors h7, 1, −3i and h−1, 0, 2i. These vectors are not
parallel so the lines are not parallel. The system of equations 1 + 7t1 = 4 − t2 , 3 + t1 = 6, and
5 − 3t1 = 7 + 2t2 has no solution so the lines do not intersect.
28. The vectors h8, −8, 10i and h8, −3, 1i are not parallel so the lines are not parallel. The lines do not
intersect because the system of equations 2 + 8t1 = 3 + 8t2 , 6 − 8t1 = 5 − 3t2 , 10t1 = 6 + t2 has no
solution.
29. The lines are parallel, respectively, to the vectors v1 = h−2, 1, −1i and v2 = h−4, 2, −2i;
v2 = 2v1 , v1 and v2 are parallel so the lines are parallel.
30. The lines are not parallel because the vectors h3, −2, 3i and h9, −6, 8i are not parallel.
−→
−→
31. P1 P2 = h3, −7, −7i, P2 P3 = h−9, −7, −3i; these vectors are not parallel so the points do not lie on
the same line.
Exercise Set 13.5
468
−→
−→
−→
−→
32. P1 P2 = h2, −4, −4i, P2 P3 = h1, −2, −2i; P1 P2 = 2 P2 P3 so the vectors are parallel and the points
lie on the same line.
33. If t2 gives the point h−1 + 3t2 , 9 − 6t2 i on the second line, then t1 = 4 − 3t2 yields the point
h3 − (4 − 3t2 ), 1 + 2(4 − 3t2 )i = h−1 + 3t2 , 9 − 6t2 i on the first line, so each point of L2 is a point
of L1 ; the converse is shown with t2 = (4 − t1 )/3.
34. If t1 gives the point h1 + 3t1 , −2 + t1 , 2t1 i on L1 , then t2 = (1 − t1 )/2 gives the point
h4 − 6(1 − t1 )/2, −1 − 2(1 − t1 )/2, 2 − 4(1 − t1 )/2i = h1 + 3t1 , −2 + t1 , 2t1 i on L2 , so each point of
L1 is a point of L2 ; the converse is shown with t1 = 1 − 2t2 .
35. The line segment joining the points (1,0) and (−3, 6).
36. The line segment joining the points (−2, 1, 4) and (7,1,1).
37. A(3, 0, 1) and B(2, 1, 3) are on the line, and (method of Exercise 25)
−→
−→
−→
−→
−→
−→
AP = −5i + j, AB= −i + j + 2k, kproj −→ AP k = | AP · AB |/k AB k =
AB
so distance =
√
√
−→
6 and k AP k =
−→
√
26,
−→
√
√
k AP × AB k
26 − 6 = 4 5. Using the method of Exercise 26, distance =
= 2 5.
−→
k AB k
38. A(2, −1, 0) and B(3, −2, 3) are on the line, and (method of Exercise 25)
−→
−→
−→
−→
AP = −i + 5j − 3k, AB= i − j + 3k, kproj −→ AP k = | AP
AB
·
−→
−→
15
AB |/k AB k = √
and
11
p
p
√
k AP k = 35, so distance = 35 − 225/11 = 4 10/11. Using the method of Exercise 26,
−→
−→
distance =
−→
k AP × AB k
−→
k AB k
p
= 4 10/11.
39. The vectors v1 = −i + 2j + k and v2 = 2i − 4j − 2k are parallel to the lines, v2 = −2v1 so v1 and
v2 are parallel. Let t = 0 to get the points P (2, 0, 1) and Q(1, 3, 5) on the first and second lines,
−→
respectively. Let u = P Q = −i + 3j + 4k, v = 12 v2 = i − 2j − k; u × v = 5i + 3j − k; by the method
p
of Exercise 26 of Section 13.4, distance = ku × vk/kvk = 35/6.
40. The vectors v1 = 2i + 4j − 6k and v2 = 3i + 6j − 9k are parallel to the lines, v2 = (3/2)v1 so v1
and v2 are parallel. Let t = 0 to get the points P (0, 3, 2) and Q(1, 0, 0) on the first and second
−→
lines, respectively. Let u = P Q = i − 3j − 2k, v = 12 v1 = i + 2j − 3k; u × v = 13i + j + 5k,
p
distance = ku × vk/kvk = 195/14 (Exer. 26, Section 13.4).
41. (a) The line is parallel to the vector hx1 − x0 , y1 − y0 , z1 − z0 i so
x = x0 + (x1 − x0 ) t, y = y0 + (y1 − y0 ) t, z = z0 + (z1 − z0 ) t
(b) The line is parallel to the vector ha, b, ci so x = x1 + at, y = y1 + bt, z = z1 + ct
42. Solve each of the given parametric equations (2) for t to get t = (x − x0 ) /a, t = (y − y0 ) /b,
t = (z − z0 ) /c, so (x, y, z) is on the line if and only if (x − x0 ) /a = (y − y0 ) /b = (z − z0 ) /c.
43. (a) It passes through the point (1, −3, 5) and is parallel to v = 2i + 4j + k
(b) hx, y, zi = h1 + 2t, −3 + 4t, 5 + ti
−→
−→
44. Let the desired point be P (x0 , y0 , z0 ), then P1 P = (2/3) P1 P2 ,
hx0 − 1, y0 − 4, z0 + 3i = (2/3)h0, 1, 2i = h0, 2/3, 4/3i; equate corresponding components to get
x0 = 1, y0 = 14/3, z0 = −5/3.
469
Chapter 13
45. (a) Let t = 3 and t = −2, respectively, in the equations for L1 and L2 .
(b) u = 2i − j − 2k and v = i + 3j √
− k are parallel to L1 and L2 ,
cos θ = u · v/(kuk kvk) = 1/(3 11), θ ≈ 84◦ .
(c) u × v = 7i + 7k is perpendicular to both L1 and L2 , and hence so is i + k, thus x = 7 + t,
y = −1, z = −2 + t.
46. (a) Let t = 1/2 and t = 1, respectively, in the equations for L1 and L2 .
(b) u = 4i − 2j + 2k and v = i − j√+ 4k are parallel to L1 and L2 ,
cos θ = u · v/(kuk kvk) = 14/ 432, θ ≈ 48◦ .
(c) u × v = −6i − 14j − 2k is perpendicular to both L1 and L2 , and hence so is 3i + 7j + k,
thus x = 2 + 3t, y = 7t, z = 3 + t.
47. (0,1,2) is on the given line (t = 0) so u = j − k is a vector from this point to the point (0,2,1),
v = 2i − j + k is parallel to the given line. u × v = −2j−2k, and hence w = j + k, is perpendicular
to both lines so v × w = −2i − 2j + 2k, and hence i + j − k, is parallel to the line we seek. Thus
x = t, y = 2 + t, z = 1 − t are parametric equations of the line.
48. (−2, 4, 2) is on the given line (t = 0) so u = 5i − 3j − 4k is a vector from this point to the point
(3, 1, −2), v = 2i + 2j + k is parallel to the given line. u × v = 5i − 13j + 16k is perpendicular to
both lines so v × (u × v) = 45i − 27j − 36k, and hence 5i − 3j − 4k is parallel to the line we seek.
Thus x = 3 + 5t, y = 1 − 3t, z = −2 − 4t are parametric equations of the line.
49. (a) When t = 0 the bugs are at (4, 1, 2) and (0,1,1) so the distance between them is
√
√
42 + 02 + 12 = 17 cm.
10
(b)
0
(c) The distance has a minimum value.
5
0
(d) Minimize D2 instead of D (the distance between the bugs).
D2 = [t − (4 − t)]2 + [(1 + t) − (1 + 2t)]2 + [(1 + 2t) − (2 + t)]2 = 6t2 − 18t + 17,
d(D2 )/dt = 12t − 18 = 0 when t = 3/2; the minimum
p
√
distance is 6(3/2)2 − 18(3/2) + 17 = 14/2 cm.
50. The line intersects the xz-plane when t = −1, the xy-plane when t = 3/2. Along the line,
T = 25t2 (1 + t)(3 − 2t) for −1 ≤ t ≤ 3/2. Solve dT /dt = 0 for t to find that the maximum value
of T is about 50.96 when t ≈ 1.073590.
EXERCISE SET 13.6
1. x = 3, y = 4, z = 5
2. x = x0 , y = y0 , z = z0
3. (x − 2) + 4(y − 6) + 2(z − 1) = 0, x + 4y + 2z = 28
4. −(x + 1) + 7(y + 1) + 6(z − 2) = 0, −x + 7y + 6z = 6
Exercise Set 13.6
470
6. 2x − 3y − 4z = 0
5. z = 0
7. n = i − j, x − y = 0
8. n = i + j, P (1, 0, 0), (x − 1) + y = 0, x + y = 1
9. n = j + k, P (0, 1, 0), (y − 1) + z = 0, y + z = 1
−→
−→
−→
−→
10. n = j − k, y − z = 0
11. P1 P2 × P1 P3 = h2, 1, 2i × h3, −1, −2i = h0, 10, −5i,for convenience choose h0, 2, −1i which is also
normal to the plane. Use any of the given points to get 2y − z = 1
12. P1 P2 × P1 P3 = h−1, −1, −2i × h−4, 1, 1i = h1, 9, −5i, x + 9y − 5z = 16
13. (a) parallel, because h2, −8, −6i and h−1, 4, 3i are parallel
(b) perpendicular, because h3, −2, 1i and h4, 5, −2i are orthogonal
(c) neither, because h1, −1, 3i and h2, 0, 1i are neither parallel nor orthogonal
14. (a) neither, because h3, −2, 1i and h6, −4, 3i are neither parallel nor orthogonal
(b) parallel, because h4, −1, −2i and h1, −1/4, −1/2i are parallel
(c) perpendicular, because h1, 4, 7i and h5, −3, 1i are orthogonal
15. (a) parallel, because h2, −1, −4i and h3, 2, 1i are orthogonal
(b) neither, because h1, 2, 3i and h1, −1, 2i are neither parallel nor orthogonal
(c) perpendicular, because h2, 1, −1i and h4, 2, −2i are parallel
16. (a) parallel, because h−1, 1, −3i and h2, 2, 0i are orthogonal
(b) perpendicular, because h−2, 1, −1i and h6, −3, 3i are parallel
(c) neither, because h1, −1, 1i and h1, 1, 1i are neither parallel nor orthogonal
17. (a) 3t − 2t + t − 5 = 0, t = 5/2 so x = y = z = 5/2, the point of intersection is (5/2, 5/2, 5/2)
(b) 2(2 − t) + (3 + t) + t = 1 has no solution so the line and plane do not intersect
18. (a) 2(3t) − 5t + (−t) + 1 = 0, 1 = 0 has no solution so the line and the plane do not intersect.
(b) (1 + t) − (−1 + 3t) + 4(2 + 4t) = 7, t = −3/14 so x = 1 − 3/14 = 11/14,
y = −1 − 9/14 = −23/14, z = 2 − 12/14 = 8/7, the point is (11/14, −23/14, 8/7)
19. n 1 = h1, 0, 0i, n2 = h2, −1, 1i, n1 · n2 = 2 so
√
√
2
n 1 · n2
= √ √ = 2/ 6, θ = cos−1 (2/ 6) ≈ 35◦
cos θ =
kn1 k kn2 k
1 6
20. n1 = h1, 2, −2i, n2 = h6, −3, 2i, n1 · n2 = −4 so
4
(−n1 ) · n2
=
= 4/21, θ = cos−1 (4/21) ≈ 79◦
cos θ =
k − n1 k kn2 k
(3)(7)
(Note: −n1 is used instead of n1 to get a value of θ in the range [0, π/2])
21. h4, −2, 7i is normal to the desired plane and (0,0,0) is a point on it; 4x − 2y + 7z = 0
22. v = h3, 2, −1i is parallel to the line and n = h1, −2, 1i is normal to the given plane so
v × n = h0, −4, −8i is normal to the desired plane. Let t = 0 in the line to get (−2, 4, 3) which is
also a point on the desired plane, use this point and (for convenience) the normal h0, 1, 2i to find
that y + 2z = 10.
471
Chapter 13
23. Find two points P1 and P2 on the line of intersection of the given planes and then find an equation
of the plane that contains P1 , P2 , and the given point P0 (−1, 4, 2). Let (x0 , y0 , z0 ) be on the
line of intersection of the given planes; then 4x0 − y0 + z0 − 2 = 0 and 2x0 + y0 − 2z0 − 3 = 0,
eliminate y0 by addition of the equations to get 6x0 − z0 − 5 = 0; if x0 = 0 then z0 = −5, if x0 = 1
then z0 = 1. Substitution of these values of x0 and z0 into either of the equations of the planes
gives the corresponding values y0 = −7 and y0 = 3 so P1 (0, −7, −5) and P2 (1, 3, 1) are on the
−→
−→
line of intersection of the planes. P0 P1 × P0 P2 = h4, −13, 21i is normal to the desired plane whose
equation is 4x − 13y + 21z = −14.
24. h1, 2, −1i is parallel to the line and hence normal to the plane x + 2y − z = 10
25. n1 = h2, 1, 1i and n2 = h1, 2, 1i are normals to the given planes, n1 × n2 = h−1, −1, 3i so h1, 1, −3i
is normal to the desired plane whose equation is x + y − 3z = 6.
−→
26. n = h4, −1, 3i is normal to the given plane, P1 P2 = h3, −1, −1i is parallel to the line,
−→
n × P1 P2 = h4, 13, −1i is normal to the desired plane whose equation is 4x + 13y − z = 1.
27. n1 = h2, −1, 1i and n2 = h1, 1, −2i are normals to the given planes,
n1 × n2 = h1, 5, 3i is normal to the desired plane whose equation is x + 5y + 3z = −6.
28. Let t = 0 and t = 1 to get the points P1 (−1, 0, −4) and P2 (0, 1, −2) that lie on the line. Denote the
−→
−→
given point by P0 , then P0 P1 × P0 P2 = h7, −1, −3i is normal to the desired plane whose equation
is 7x − y − 3z = 5.
29. The plane is the perpendicular bisector of the line segment that joins P1 (2, −1, 1) and P2 (3, 1, 5).
−→
The midpoint of the line segment is (5/2, 0, 3) and P1 P2 = h1, 2, 4i is normal to the plane so an
equation is x + 2y + 4z = 29/2.
30. n1 = h2, −1, 1i and n2 = h0, 1, 1i are normals to the given planes, n1 × n2 = h−2, −2, 2i so
n = h1, 1, −1i is parallel to the line of intersection of the planes. v = h3, 1, 2i is parallel to the
given line, v × n = h−3, 5, 2i so h3, −5, −2i is normal to the desired plane. Let t = 0 to find the
point (0,1,0) that lies on the given line and hence on the desired plane. An equation of the plane
is 3x − 5y − 2z = −5.
31. The line is parallel to the line of intersection of the planes if it is parallel to both planes. Normals
to the given planes are n1 = h1, −4, 2i and n2 = h2, 3, −1i so n1 × n2 = h−2, 5, 11i is parallel to
the line of intersection of the planes and hence parallel to the desired line whose equations are
x = 5 − 2t, y = 5t, z = −2 + 11t.
−→
−→
32. Denote the points by A, B, C, and D, respectively. The points lie in the same plane if AB × AC
−→
−→
−→
−→
−→
−→
and AB × AD are parallel (method 1). AB × AC = h0, −10, 5i, AB × AD = h0, 16, −8i, these
vectors are parallel because h0, −10, 5i = (−10/16)h0, 16, −8i. The points lie in the same plane
−→
−→
if D lies in the plane determined by A, B, C (method 2), and since AB × AC = h0, −10, 5i, an
equation of the plane is −2y + z + 1 = 0, 2y − z = 1 which is satisfied by the coordinates of D.
33. v = h0, 1, 1i is parallel to the line.
(a) For any t, 6·0 + 4t − 4t = 0, so (0, t, t) is in the plane.
(b) n = h5, −3, 3i is normal to the plane, v · n = 0 so the line is parallel to the plane. (0,0,0) is
on the line, (0, 0, 1/3) is on the plane. The line is below the plane because (0,0,0) is below
(0, 0, 1/3).
Exercise Set 13.6
472
(c) n = h6, 2, −2i, v · n = 0 so the line is parallel to the plane. (0,0,0) is on the line, (0, 0, −3/2)
is on the plane. The line is above the plane because (0,0,0) is above (0, 0, −3/2).
−→
−→
34. The intercepts correspond to the points A(a, 0, 0), B(0, b, 0), and C(0, 0, c). AB × AC = hbc, ac, abi
is normal to the plane so bcx + acy + abz = abc or x/a + y/b + z/c = 1.
35. v1 = h1, 2, −1i and v2 = h−1, −2, 1i are parallel, respectively, to the given lines and to each
other so the lines are parallel. Let t = 0 to find the points P1 (−2, 3, 4) and P2 (3, 4, 0) that lie,
−→
respectively, on the given lines. v1 × P1 P2 = h−7, −1, −9i so h7, 1, 9i is normal to the desired plane
whose equation is 7x + y + 9z = 25.
36. The system 4t1 − 1 = 12t2 − 13, t1 + 3 = 6t2 + 1, 1 = 3t2 + 2 has the solution (Exercise 26,
Section 13.5) t1 = −4, t2 = −1/3 so (−17, −1, 1) is the point of intersection. v1 = h4, 1, 0i and
v2 = h12, 6, 3i are (respectively) parallel to the lines, v1 × v2 = h3, −12, 12i so h1, −4, 4i is normal
to the desired plane whose equation is x − 4y + 4z = −9.
37. n1 = h−2, 3, 7i and n2 = h1, 2, −3i are normals to the planes, n1 × n2 = h−23, 1, −7i is parallel
to the line of intersection. Let z = 0 in both equations and solve for x and y to get x = −11/7,
y = −12/7 so (−11/7, −12/7, 0) is on the line, a parametrization of which is
x = −11/7 − 23t, y = −12/7 + t, z = −7t.
38. Similar to Exercise 37 with n1 = h3, −5, 2i, n2 = h0, 0, 1i, n1 × n2 = h−5, −3, 0i. z = 0 so
3x − 5y = 0, let x = 0 then y = 0 and (0,0,0) is on the line, a parametrization of which is
x = −5t, y = −3t, z = 0.
√
39. D = |2(1) − 2(−2) + (3) − 4|/ 4 + 4 + 1 = 5/3
√
40. D = |3(0) + 6(1) − 2(5) − 5|/ 9 + 36 + 4 = 9/7
√
√
41. (0,0,0) is on the first plane so D = |6(0) − 3(0) − 3(0) − 5|/ 36 + 9 + 9 = 5/ 54.
√
√
42. (0,0,1) is on the first plane so D = |(0) + (0) + (1) + 1|/ 1 + 1 + 1 = 2/ 3.
43. (1,3,5) and (4,6,7) are on L1 and L2 , respectively. v1 = h7, 1, −3i and v2 = h−1, 0, 2i are,
respectively, parallel to L1 and L2 , v1 × v2 = h2, −11, 1i so the
√z + 51 = 0
√ plane 2x − 11y +
contains L2 and is parallel to L1 , D = |2(1) − 11(3) + (5) + 51|/ 4 + 121 + 1 = 25/ 126.
44. (3,4,1) and (0,3,0) are on L1 and L2 , respectively. v1 = h−1, 4, 2i and v2 = h1, 0, 2i are parallel to
L1 and L2 , v1 × v2 = h8, 4, −4i = 4h2, 1, −1i so 2x + y − z − 3 = 0 contains L2 and is parallel to
√
√
L1 , D = |2(3) + (4) − (1) − 3|/ 4 + 1 + 1 = 6.
√
√
45. The distance between (2, 1, −3) and the plane is |2 − 3(1) + 2(−3) − 4|/ 1 + 9 + 4 = 11/ 14 which
is the radius of the sphere; an equation is (x − 2)2 + (y − 1)2 + (z + 3)2 = 121/14.
46. The vector 2i + j − k is normal to the plane and hence parallel to the line so parametric equations
of the line are x = 3 + 2t, y = 1 + t, z = −t. Substitution into the equation of the plane yields
2(3 + 2t) + (1 + t) − (−t) = 0, t = −7/6; the point of intersection is (2/3, −1/6, 7/6).
47. v = h1, 2, −1i is parallel to the line, n = h2, −2, −2i is normal to the plane, v · n = 0 so v is
parallel to the plane because v and n are perpendicular. (−1, 3, 0) is on the line so
√
√
D = |2(−1) − 2(3) − 2(0) + 3|/ 4 + 4 + 4 = 5/ 12
473
Chapter 13
48. (a)
(b)
n
r – r0
P(x0, y0)
n · (r − r0 ) = a(x − x0 ) + b(y − y0 ) = 0
P(x, y)
r
r0
O
(c) See the proof of Theorem 13.6.1. Since a and b are not both zero, there is at least one point
(x0 , y0 ) that satisfies ax+by +d = 0, so ax0 +bx0 +d = 0. If (x, y) also satisfies ax+by +d = 0
then, subtracting, a(x − x0 ) + b(y − y0 ) = 0, which is the equation of a line with n = ha, bi
as normal.
√
(d) Let Q(x1 , y1 ) be a point on the line, and position the normal n = ha, bi, with length a2 + b2 ,
so that its initial point is at Q. The distance is the orthogonal projection of
−→
QP0 = hx0 − x1 , y0 − y1 i onto n. Then
−→
|ax0 + by0 + d|
QP0 · n
√
n =
.
D = kprojn QP 0 k =
2
knk
a2 + b2
−→
√
√
49. D = |2(−3) + (5) − 1|/ 4 + 1 = 2/ 5
50. (a) If hx0 , y0 , z0 i lies on the second plane, so that ax0 + by0 + cz0 + d2 = 0, then by Theorem
| − d2 + d1 |
|ax0 + by0 + cz0 + d1 |
√
=√
13.6.2, the distance between the planes is D =
2
2
2
a +b +c
a2 + b2 + c2
5
(b) The distance between the planes −2x + y + z = 0 and −2x + y + z + = 0 is
3
5
|0 − 5/3|
= √ .
D= √
4+1+1
3 6
EXERCISE SET 13.7
1. (a) elliptic paraboloid, a = 2, b = 3
(b) hyperbolic paraboloid, a = 1, b = 5
(c) hyperboloid of one sheet, a = b = c = 4
(d) circular cone, a = b = 1
(e) elliptic paraboloid, a = 2, b = 1
(f ) hyperboloid of two sheets, a = b = c = 1
√
√
2. (a) ellipsoid, a = 2/2, b = 1/2, e = 3/3
(b) hyperbolic paraboloid, a = b = 1
(c) hyperboloid of one sheet, a = 1, b = 3, c = 1
(d) hyperboloid of two sheets, a = 1, b = 2, c = 1
√
√
(e) elliptic paraboloid, a = 2, b = 2/2
√
(f ) elliptic cone, a = 2, b = 3
Exercise Set 13.7
474
3. (a) −z = x2 + y 2 , circular paraboloid
opening down the negative z-axis
z
x
y
(b) z = x2 + y 2 , circular paraboloid, no change
(c) z = x2 + y 2 , circular paraboloid, no change
(d) z = x2 + y 2 , circular paraboloid, no change
z
x
(e) x = y 2 + z 2 , circular paraboloid
opening along the positive x-axis
y
(f ) y = x2 + z 2 , circular paraboloid
opening along the positive y-axis
z
z
y
x
x
y
z
4. (a)
(b)
(c)
(d)
x +y −z
x2 + y 2 − z 2
x2 + y 2 − z 2
x2 + y 2 − z 2
2
2
2
= 1,
= 1,
= 1,
= 1,
no
no
no
no
change
change
change
change
x
y
475
Chapter 13
(e) −x2 + y 2 + z 2 = 1, hyperboloid
of one sheet with x-axis as axis
(f ) x2 − y 2 + z 2 = 1, hyperboloid
of one sheet with y-axis as axis
z
z
x
x
5. (a)
(b)
(c)
(d)
(e)
(f )
y
y
hyperboloid of one sheet, axis is y-axis
hyperboloid of two sheets separated by yz-plane
elliptic paraboloid opening along the positive x-axis
elliptic cone with x-axis as axis
hyperbolic paraboloid straddling x- and z-axes
paraboloid opening along the negative y-axis
6. (a) same
(b) same
(c) same
2
(d) same
7. (a) x = 0 :
(e) y =
2
z
x
− 2
2
a
c
(f ) y =
z2
x2
z2
y2
+
= 1; y = 0 :
+
= 1;
25
4
9
4
x2
z2
+
a2
c2
z
x2
z2
+
=1
9
4
x2
y2
z=0:
+
=1
9
25
y2
z2
+
=1
4
25
y
x2
y2
+
=1
9
25
x
(b) x = 0 : z = 4y 2 ; y = 0 : z = x2 ;
z
z=0:x=y=0
z = 4y2
z = x2
x2 + 4y2 = 0
(0, 0, 0)
x
y
Exercise Set 13.7
(c) x = 0 :
476
z2
x2
z2
y2
−
= 1; y = 0 :
−
= 1;
16
4
9
4
z
y2
z2
–
=1
16
4
x2
y2
z=0:
+
=1
9
16
y
y2
x2
+
=1
9
16
x
x2
z2
–
=1
9
4
8. (a) x = 0 : y = z = 0; y = 0 : x = 9z 2 ; z = 0 : x = y 2
z
x = 9z2
y
x
(b) x = 0 : −y 2 + 4z 2 = 4; y = 0 : x2 + z 2 = 1;
x = y2
z=0
z
z = 0 : 4x2 − y 2 = 4
y=0
x=0
y
x
y
(c) x = 0 : z = ± ; y = 0 : z = ±x; z = 0 : x = y = 0
2
x=0
z
z=0
x
y
y=0
9. (a) 4x2 + z 2 = 3; ellipse
(b) y 2 + z 2 = 3; circle
(c) y 2 + z 2 = 20; circle
(d) 9x2 − y 2 = 20; hyperbola
(e) z = 9x2 + 16; parabola
(f ) 9x2 + 4y 2 = 4; ellipse
10. (a) y 2 − 4z 2 = 27; hyperbola
(b) 9x2 + 4z 2 = 25; ellipse
(c) 9z 2 − x2 = 4; hyperbola
(d) x2 + 4y 2 = 9; ellipse
(e) z = 1 − 4y 2 ; parabola
(f ) x2 − 4y 2 = 4; hyperbola
477
Chapter 13
z
11.
z
12.
z
13.
(0, 0, 3)
(0, 0, 2)
(1, 0, 0)
x
x
y
(0, 3, 0)
(2, 0, 0)
(0, 3, 0)
(0, 2, 0)
y
x
y
(6, 0, 0)
Ellipsoid
Ellipsoid
Hyperboloid
of one sheet
z
14.
z
15.
z
16.
(0, 3, 0)
(3, 0, 0)
x
y
x
Hyperboloid
of one sheet
(0, 0, 2)
Elliptic cone
Elliptic cone
z
17.
z
18.
y
x
y
z
19.
(0, 0, –2)
x
y
x
y
Hyperboloid
of two sheets
x
y
Hyperboloid
of two sheets
Hyperbolic paraboloid
z
20.
z
21.
z
22.
y
x
Hyperbolic
paraboloid
y
x
y
x
Elliptic paraboloid
Circular paraboloid
Exercise Set 13.7
478
z
23.
z
24.
z
25.
(0, 0, 2)
y
x
x
y
x
(0, 2, 0)
Elliptic paraboloid
Hyperboloid
of one sheet
Circular cone
z
26.
z
27.
y
z
28.
(0, 0, 2)
(-3, 0, 0)
x
(3, 0, 0)
y
Hyperboloid
of two sheets
(2, 0, 0)
Hyperboloid
of one sheet
Hyperbolic
paraboloid
z
29.
x
y
x
30.
z
(0, 0, 1)
(0, 1, 0)
y
x
y
x
z
31.
(1, 0, 0)
z
32.
(0, 0, 1)
x
x
(1, 0, 0)
(0, 1, 0)
y
y
y
479
Chapter 13
z
33.
z
34.
(0, 0, 2)
x
y
x
y
Hyperboloid
of one sheet
(–2, 3, –9)
Circular paraboloid
z
35.
z
36.
(-1, 1, 2)
(1, –1, –2)
y
y
x
x
Ellipsoid
Hyperboloid
of one sheet
y2
x2
+
=1
(b) 6, 4
9
4
(d) The focal axis is parallel to the x-axis.
√
√
(c) (± 5, 0, 2)
√
z2
y2
+
=1
(b) 4, 2 2
4
2
(d) The focal axis is parallel to the y-axis.
√
(c) (3, ± 2, 0)
x2
y2
−
=1
(b) (0, ±2, 4)
4
4
(d) The focal axis is parallel to the y-axis.
√
(c) (0, ±2 2, 4)
37. (a)
38. (a)
39. (a)
y2
x2
−
=1
(b) (±2, 0, −4)
40. (a)
4
4
(e) The focal axis is parallel to the x-axis.
41. (a) z + 4 = y 2
(d)
(c) (2, 0, −15/4)
The focal axis is parallel to the z-axis.
42. (a) z − 4 = −x2
(d)
(b) (2, 0, −4)
√
(c) (±2 2, 0, −4)
(b) (0, 2, 4)
The focal axis is parallel to the z-axis.
(c) (0, 2, 15/4)
Exercise Set 13.7
480
43. x2 + y 2 = 4 − x2 − y 2 , x2 + y 2 = 2; circle of radius
in the plane z = 2, centered at (0, 0, 1)
√
2
z
4
2
2
x +y = 2
(z = 2)
x
44. y 2 + z = 4 − 2(y 2 + z), y 2 + z = √
4/3;
parabolas in the planes x = ±2/ 3
which open in direction of the negative y-axis
y
z
x
y
z = 4 – y2, x = 4
3
3
45. y = 4(x2 + z 2 )
47. |z − (−1)| =
paraboloid
46. y 2 = 4(x2 + z 2 )
p
x2 + y 2 + (z − 1)2 , z 2 + 2z + 1 = x2 + y 2 + z 2 − 2z + 1, z = (x2 + y 2 )/4; circular
p
48. |z + 1| = 2 x2 + y 2 + (z − 1)2 , z 2 + 2z + 1 = 4 x2 + y 2 + z 2 − 2z + 1 ,
4x2 + 4y 2 + 3z 2 − 10z + 3 = 0,
y2
(z − 5/3)2
x2
+
+
= 1; ellipsoid, center at (0, 0, 5/3).
4/3 4/3
16/9
x2
y2
x2
z2
+ 2 = 1; if y = 0 then 2 + 2 = 1; since c < a the major axis has length 2a, the
2
a
a
a
c
minor axis length 2c.
49. If z = 0,
50.
x2
y2
z2
+ 2 + 2 = 1, where a = 6378.1370, b = 6356.5231.
2
a
a
b
51. Each slice perpendicular to the z-axis for |z| < c is an ellipse whose equation is
y2
y2
c2 − z 2
x2
x2
+
= 1, the area of which is
+
=
,
or
a2
b2
c2
(a2 /c2 )(c2 − z 2 ) (b2 /c2 )(c2 − z 2 )
Z c
bp
ap
ab
ab
4
c2 − z 2
c2 − z 2 = π 2 c2 − z 2 so V = 2
π 2 c2 − z 2 dz = πabc.
π
c
c
c
c
3
0
481
Chapter 13
EXERCISE SET 13.8
√
5 2, 3π/4, 6
1. (a) (8, π/6, −4)
(b)
2. (a) (2, 7π/4, 1)
(b) (1, π/2, 1)
(c) (2, π/2, 0)
(d) (8, 5π/3, 6)
√
(c) (4 2, 3π/4, −7)
√
(d) (2 2, 7π/4, −2)
(c) (5, 0, 4)
(d) (−7, 0, −9)
3. (a)
√
2 3, 2, 3
(b)
4. (a)
√
3, −3 3, 7
(b) (0, 1, 0)
(c) (0, 3, 5)
(d) (0, 4, −1)
5. (a)
√
2 2, π/3, 3π/4
(b) (2, 7π/4, π/4)
(c) (6, π/2, π/3)
(d) (10, 5π/6, π/2)
6. (a)
√
8 2, π/4, π/6
(b)
(c) (2, 0, π/2)
(d) (4, π/6, π/6)
√ √
−4 2, 4 2, −2
√
2 2, 5π/3, 3π/4
√
√
√
7. (a) (5 6/4, 5 2/4, 5 2/2)
(c) (0,0,1)
(b) (7,0,0)
(d) (0, −2, 0)
√
√
√
− 2/4, 6/4, − 2/2
√ √ √
(c) (2 6, 2 2, 4 2)
√
√
√
3 2/4, −3 2/4, −3 3/2
√
(d) 0, 2 3, 2
8. (a)
9. (a)
(b)
√
2 3, π/6, π/6
(c) (2, 3π/4, π/2)
10. (a)
√
(b)
2, π/4, 3π/4
√
4 3, 1, 2π/3
(d)
√
4 2, 5π/6, π/4
√
2 2, 0, 3π/4
√
(d) (2 10, π, tan−1 3)
(b)
(c) (5, π/2, tan−1 (4/3))
√
11. (a) 5 3/2, π/4, −5/2
(c) (0, 0, 3)
(b) (0, 7π/6, −1)
(d) (4, π/6, 0)
12. (a) (0, π/2, −5/2)
√
√
(b) (3 2, 0, −3 2)
√
(d) (5/2, 2π/3, −5 3/2)
(c) (0, 3π/4, −3)
z
15.
z
16.
z
17.
y
y
(3, 0, 0)
x
y
2
z= x +y
x
2
2
x +y =9
x
y = x, x ≥ 0
2
Exercise Set 13.8
482
z
18.
z
19.
z
20.
y
(0, 4, 0)
y
y
(2, 0, 0)
x
x=2
x
z=x
x
2
z
21.
2
x + (y – 2) = 4
z
22.
y
x
(1, 0, 0)
x
2
2
y
x2 – y2 = z
2
x +y +z =1
z
23.
z
24.
y
y
y = √3x
x
x
(3, 0, 0)
2
2
2
x +y +z =9
z
25.
z
26.
(0, 0, 2)
y
y
x
z = √x2 + y2
x
z=2
483
Chapter 13
z
27.
z
28.
(0, 0, 2)
y
y
(1, 0, 0)
x
2
2
x
2
x2 + y2 = 1
x + y + (z – 2) = 4
29.
z
z
30.
y
y
(1, 0, 0)
(1, 0, 0)
x
(x – 1)2 + y2 + z2 = 1
x
(x – 1)2 + y2 = 1
31. (a) z = 3
(b) ρ cos φ = 3, ρ = 3 sec φ
32. (a) r sin θ = 2, r = 2 csc θ
(b) ρ sin φ sin θ = 2, ρ = 2 csc φ csc θ
33. (a) z = 3r2
(b) ρ cos φ = 3ρ2 sin2 φ, ρ =
34. (a) z =
√
3r
(b) ρ cos φ =
1
csc φ cot φ
3
√
1
π
3ρ sin φ, tan φ = √ , φ =
6
3
35. (a) r = 2
(b) ρ sin φ = 2, ρ = 2 csc φ
36. (a) r2 − 6r sin θ = 0, r = 6 sin θ
(b) ρ sin φ = 6 sin θ, ρ = 6 sin θ csc φ
37. (a) r2 + z 2 = 9
(b) ρ = 3
38. (a) z 2 = r2 cos2 θ − r2 sin2 θ = r2 (cos2 θ − sin2 θ), z 2 = r2 cos 2θ
(b) Use the result in part (a) with r = ρ sin φ, z = ρ cos φ to get ρ2 cos2 φ = ρ2 sin2 φ cos 2θ,
cot2 φ = cos 2θ
39. (a) 2r cos θ + 3r sin θ + 4z = 1
(b) 2ρ sin φ cos θ + 3ρ sin φ sin θ + 4ρ cos φ = 1
40. (a) r2 − z 2 = 1
(b) Use the result of part (a) with r = ρ sin φ, z = ρ cos φ to get ρ2 sin2 φ − ρ2 cos2 φ = 1,
ρ2 cos 2φ = −1
Exercise Set 13.8
484
41. (a) r2 cos2 θ = 16 − z 2
(b) x2 = 16 − z 2 , x2 + y 2 + z 2 = 16 + y 2 , ρ2 = 16 + ρ2 sin2 φ sin2 θ, ρ2 1 − sin2 φ sin2 θ = 16
42. (a) r2 + z 2 = 2z
(b) ρ2 = 2ρ cos φ, ρ = 2 cos φ
43. all points on or above the paraboloid z = x2 + y 2 , that are also on or below the plane z = 4
44. a right circular cylindrical solid of height 3 and radius 1 whose axis is the line x = 0, y = 1
45. all points on or between concentric spheres of radii 1 and 3
46. all points on or above the cone φ = π/6, that are also on or below the sphere ρ = 2
√
√
47. θ = π/6, φ = π/6, spherical (4000, π/6, π/6), rectangular 1000 3, 1000, 2000 3
48. (a) y = r sin θ = a sin θ but az = a sin θ so y = az, which is a plane that contains the curve of
intersection of z = sin θ and the circular cylinder r = a. From Exercise 60, Section 12.4, the
curve of intersection of a plane and a circular cylinder is an ellipse.
z
(b)
z = sin θ
y
x
49. (a) (10, π/2, 1)
50.
(b) (0, 10, 1)
√
(c) ( 101, π/2, tan−1 10)
20
0
30
0
51. Using spherical coordinates: for point A, θA = 360◦ − 60◦ = 300◦ , φA = 90◦ − 40◦ = 50◦ ; for point
B, θB = 360◦ − 40◦ = 320◦ , φB = 90◦ − 20◦ = 70◦ . Unit vectors directed from the origin to the
points A and B, respectively, are
uA = sin 50◦ cos 300◦ i + sin 50◦ sin 300◦ j + cos 50◦ k,
uB = sin 70◦ cos 320◦ i + sin 70◦ sin 320◦ j + cos 70◦ k
The angle α between uA and uB is α = cos−1 (uA · uB ) ≈ 0.459486 so the shortest distance is
6370α ≈ 2,927 km.
485
Chapter 13
CHAPTER 13 SUPPLEMENTARY EXERCISES
2. (c) F = −i − j
(d) kh1, −2, 2ik = 3, so kr − h1, −2, 2ik = 3, or (x − 1)2 + (y + 2)2 + (z − 2)2 = 9
√
3. (b) x = cos 120◦ = −1/2, y = ± sin 120◦ = ± 3/2
(d) true: ku × vk = kukkvk| sin(θ)| = 1
4. (d)
x + 2y − z = 0
5. (b) (y, x, z), (x, z, y), (z, y, x)
(c) circle of radius 5 in plane z = 1 with center at (0, 0, 1) (rectangular coordinates)
(d) the two half-lines z = ±x, x ≥ 0 in the xz-plane
6. (x + 3)2 + (y − 5)2 + (z + 4)2 = r2 ,
(b) r2 = 52 = 25
(a) r2 = 42 = 16
−→
−→
−→
(c) r2 = 32 = 9
−→
7. (a) AB = −i + 2j + 2k, AC = i + j − k, AB × AC= −4i + j − 3k, area =
(b) area =
√
1√
3
1 −→
hk AB k = h =
26, h = 26/3
2
2
2
−→
√
1 −→
k AB × AC k = 26/2
2
8. The sphere x2 + (y − 1)2 + (z + 3)2 = 16 has center Q(0, 1, −3) and radius 4, and
−→
√
√
√
√
k P Q k = 12 + 42 = 17, so minimum distance is 17 − 4, maximum distance is 17 + 4.
9. (a) a · b = 0, 4c + 3 = 0, c = −3/4
p
p
√
(b) Use a · b = kak kbk cos θ to get 4c + 3 = c2 + 1(5) cos(π/4), 4c + 3 = 5 c2 + 1/ 2
Square both sides and rearrange to get 7c2 + 48c − 7 = 0, (7c − 1)(c + 7) = 0 so c = −7
(invalid) or c = 1/7.
2
(c) Proceed as in (b)√with
θ = π/6 to get 11c − 96c + 39 = 0 and use the quadratic formula to
get c = 48 ± 25 3 /11.
(d) a must be a scalar multiple of b, so ci + j = k(4i + 3j), k = 1/3, c = 4/3.
−→
−→
−→
−→
10. OS = OP + P S = 3i + 4j+ QR = 3i + 4j + (4i + j) = 7i + 5j
11. (a) the plane through the origin which is perpendicular to r0
(b) the plane through the tip of r0 which is perpendicular to r0
12. The normals to the planes are given by ha1 , b1 , c1 i and ha2 , b2 , c2 i, so the condition is
a1 a2 + b1 b2 + c1 c2 = 0.
−→
−→
−→
−→
−→
−→
−→
−→
−→
13. Since AC · (AB × AD) =AC · (AB × CD) + AC · (AB × AC) = 0 + 0 = 0, the volume of the
−→
−→
−→
parallelopiped determined by AB, AC, and AD is zero, thus A, B, C, and D are coplanar (lie in
−→
−→
the same plane). Since AB × CD6= 0, the lines are not parallel. Hence they must intersect.
14. The points P lie on the plane determined by A, B and C.
15. (a) false, for example i · j = 0
(b) false, for example i × i = 0
(c) true; 0 = kuk · kvk cos θ = kuk · kvk sin θ, so either u = 0 or v = 0 since cos θ = sin θ = 0 is
impossible.
Chapter 13 Supplementary Exercises
486
16. (a) Replace u with a × b, v with c, and w with d in the first formula of Exercise 39.
(b) From the second formula of Exercise 39,
(a × b) × c + (b × c) × a + (c × a) × b
= (c · a)b − (c · b)a + (a · b)c − (a · c)b + (b · c)a − (b · a)c = 0
17. ku − vk2 = (u − v) · (u − v) = kuk2 + kvk2 − 2kukkvk cos θ = 2(1 − cos θ) = 4 sin2 (θ/2), so
ku − vk = 2 sin(θ/2)
−→
−→
−→
18. AB= i − 2j − 2k, AC= −2i − j − 2k, AD= i + 2j − 3k
1 −2 −2
−2 −1 −2
1
2 −3
(a) From Theorem 13.4.6 and formula (9) of Section 13.4,
−→
= 29, so V = 29.
−→
(b) The plane containing A, B, and C has normal AB × AC= 2i + 6j − 5k, so the equation of
the plane is 2(x − 1) + 6(y + 1) − 5(z − 2) = 0, 2x + 6y − 5z = −14. From Theorem 13.6.2,
15
|2(2) + 6(1) − 5(−1)|
√
=√ .
D=
65
65
19. (a) h2, 1, −1i × h1, 2, 1i = h3, −3, 3i, so the line is parallel to i − j + k. By inspection, (0, 2, −1)
lies on both planes, so the line has an equation r = 2j − k + t(i − j + k), that is,
x = t, y = 2 − t, z = −1 + t.
h2, 1, −1i · h1, 2, 1i
= 1/2, so θ = π/3
(b) cos θ =
kh2, 1, −1ikkh1, 2, 1ik
20. Let α = 50◦ , β = 70◦ , then γ = cos−1
p
1 − cos2 α − cos2 β ≈ 47◦ .
√
21. 5hcos 60◦ , cos 120◦ , cos 135◦ i = h5/2, −5/2, −5 2/2i
22. (a) Let k be the length of an edge and introduce a coordinate system as shown in the figure,
√
2k2
d·u
= √ √ = 2/ 6
then d = hk, k, ki, u = hk, k, 0i, cos θ =
kdk kuk
k 3 k 2
√
so θ = cos−1 (2/ 6) ≈ 35◦
z
d
θ
y
u
x
(b) v = h−k, 0, ki, cos θ =
d·v
= 0 so θ = π/2 radians.
kdk kvk
23. (a) (x − 3)2 + 4(y + 1)2 − (z − 2)2 = 9, hyperboloid of one sheet
(b) (x + 3)2 + (y − 2)2 + (z + 6)2 = 49, sphere
(c) (x − 1)2 + (y + 2)2 − z 2 = 0, circular cone
24. (a) perpendicular, since h2, 1, 2i · h−1, −2, 2i = 0
3
(b) L1 : hx, y, zi = h1 + 2t, − + t, −1 + 2ti; L2 : hx, y, zi = h4 − t, 3 − 2t, −4 + 2ti
2
487
Chapter 13
3
(c) Solve simultaneously 1 + 2t1 = 4 − t2 , − + t1 = 3 − 2t2 , −1 + 2t1 = −4 + 2t2 , solution
2
1
t1 = , t2 = 2, x = 2, y = −1, z = 0
2
25. (a) r2 = z; ρ2 sin2 φ = ρ cos φ, ρ = cot φ csc φ
(b) r2 (cos2 θ − sin2 θ) − z 2 = 0, z 2 = r2 cos 2θ;
ρ2 sin2 φ cos2 θ − ρ2 sin2 φ sin2 θ − ρ2 cos2 φ = 0, cos 2θ = cot2 φ
26. (a) z = r2 cos2 θ − r2 sin2 θ = x2 − y 2
z
27. (a)
(b) (ρ sin φ cos θ)(ρ cos φ) = 1, xz = 1
z
(b)
(0, 0, 2)
p/6
(2, 0, 0)
x
z
(c)
(0, 2, 0)
x
y
y
(0, 0, 2)
p/6
y
x
z
28. (a)
z
(b)
5
z=2
f
5
5
x
y
z
(c)
2
2
x
y
x
y
Chapter 13 Supplementary Exercises
z
29. (a)
r=2
488
z
(b)
r=1
z=3
y
z=2
x
x
y
z
(c)
z
(d)
r=2
r=1
z=3
z=2
x
y
u = p/6
u = p/6
x
u = p/3
z
30. (a)
z
1
(b)
2
2
u = p/3
y
y
z
(c)
1
2
2y
y
2
x
2
x
x
31. (a) At x = c the trace of the surface is the circle y 2 + z 2 = [f (c)]2 , so the
y 2 + z 2 = [f (x)]2
r
3 2
2
2
2x
2
2
(b) y + z = e
(c) y + z = 4 − x , so let f (x) = 4 −
4
surface is given by
3 2
x
4
32. (a) Permute x and y in Exercise 31a: x2 + z 2 = [f (y)]2
(b) Permute x and z in Exercise 31a: x2 + y 2 = [f (z)]2
(c) Permute y and z in Exercise 31a: y 2 + z 2 = [f (x)]2
−→
−→
34. P Q= h1, −1, 6i, and W = F · PQ = 13 lb·ft
z
33.
x
y
489
Chapter 13
−→
−→
35. F = F1 + F2 = 2i − j + 3k, P Q= i + 4j − 3k, W = F· P Q= −11 N·m
36. F1 = 250 cos 38◦ i + 250 sin 38◦ j, F = 1000i, F2 = F − F1 = (1000 − 250 cos 38◦ )i − 250 sin 38◦ j;
r
17 1
250 sin 38◦
− cos 38◦ ≈ 817.62 N·m, θ = tan−1
≈ −11◦
kF2 k = 1000
16 2
250 cos 38◦ − 1000
37. (a) F = −6i + 3j − 6k
−→
−→
(b) OA= h5, 0, 2i, so the vector moment is OA ×F = −6i + 18j + 15k
CHAPTER 14
Vector-Valued Functions
EXERCISE SET 14.1
1. (−∞, +∞); r(π) = −i − 3πj
2. [−1/3, +∞); r(1) = h2, 1i
3. [2, +∞); r(3) = −i − ln 3j + k
4. [−1, 1); r(0) = h2, 0, 0i
5. r = 3 cos ti + (t + sin t)j
6. r = (t2 + 1)i + e−2t j
7. r = 2ti + 2 sin 3tj + 5 cos 3tk
8. r = t sin ti + ln tj + cos2 tk
9. x = 3t2 , y = −2, z = 0
10. x = sin2 t, y = 1 − cos 2t, z = 0
√
11. x = 2t − 1, y = −3 t, z = sin 3t
12. x = te−t , y = 0, z = −5t2
13. the line in 2-space through the point (2, 0) and parallel to the vector −3i − 4j
14. the circle of radius 3 in the xy-plane, with center at the origin
15. the line in 3-space through the point (0, −3, 1) and parallel to the vector 2i + 3k
16. the circle of radius 2 in the plane x = 3, with center at (3, 0, 0)
17. an ellipse in the plane z = −1, center at (0, 0, −1), major axis of length 6 parallel to x-axis, minor
axis of length 4 parallel to y-axis
18. a parabola in the plane x = −2, vertex at (−2, 0, −1), opening upward
19. (a) The line is parallel to the vector −2i + 3j; the slope is −3/2.
(b) y = 0 in the xz-plane so 1 − 2t = 0, t = 1/2 thus x = 2 + 1/2 = 5/2 and z = 3(1/2) = 3/2;
the coordinates are (5/2, 0, 3/2).
20. (a) x = 3 + 2t = 0, t = −3/2 so y = 5(−3/2) = −15/2
(b) x = t, y = 1 + 2t, z = −3t so 3(t) − (1 + 2t) − (−3t) = 2, t = 3/4; the point of intersection is
(3/4, 5/2, −9/4).
21. (a)
y
(b)
(0, 1)
y
(1, 1)
x
x
(1, 0)
(1, -1)
490
491
Chapter 14
z
22. (a)
z
(b)
(0, 0, 1)
(1, 1, 1)
y
(1, 1, 0)
y
(1, 1, 0)
x
x
23. r = (1 − t)(3i + 4j), 0 ≤ t ≤ 1
24. r = (1 − t)4k + t(2i + 3j), 0 ≤ t ≤ 1
25. x = 2
26. y = 2x + 10
y
y
10
x
x
2
-5
27. (x − 1)2 + (y − 3)2 = 1
28. x2 /4 + y 2 /25 = 1
y
y
5
3
x
2
x
1
29. x2 − y 2 = 1, x ≥ 1
30. y = 2x2 + 4, x ≥ 0
y
y
2
x
1
4
x
1
Exercise Set 14.1
492
z
31.
z
32.
(0, 2, π /2)
(0, 4, π /2)
y
(2, 0, 0)
y
(9, 0, 0)
x
x
z
33.
z
34.
2
y
c
o
y
x
x
36. x = t, y = −t, z =
35. x = t, y = t, z = 2t2
z
z
√ √
2 1 − t2
y+x=0
y
z=
√ 2 – x 2 – y2
x
x
y
37. r = ti + t2 j ±
1p
81 − 9t2 − t4 k
3
38. r = ti + tj + (1 − 2t)ks
z
z
y=x
x
y
x
x+y+z=1
y
493
Chapter 14
39. x2 + y 2 = (t sin t)2 + (t cos t)2 = t2 (sin2 t + cos2 t) = t2 = z
40. x − y + z + 1 = t − (1 + t)/t + (1 − t2 )/t + 1 = [t2 − (1 + t) + (1 − t2 ) + t]/t = 0
√
√
41. x = sin t, y = 2 cos t, z = 3 sin t so x2 + y 2 + z 2 = sin2 t + 4 cos2 t + 3 sin2 t√= 4 and z = 3x; it
is the curve of intersection of the sphere x2 + y 2 + z 2 = 4 and the plane z = 3x, which is a circle
with center at (0, 0, 0) and radius 2.
42. x = 3 cos t, y = 3 sin t, z = 3 sin t so x2 + y 2 = 9 cos2 t + 9 sin2 t = 9 and z = y; it is the curve
of intersection of the circular
cylinder x2 + y 2 = 9 and the plane z = y, which is an ellipse with
√
major axis of length 6 2 and minor axis of length 6.
43. The helix makes one turn as t varies from 0 to 2π so z = c(2π) = 3, c = 3/(2π).
44. 0.2t = 10, t = 50; the helix has made one revolution when t = 2π so when t = 50 it has made
50/(2π) = 25/π ≈ 7.96 revolutions.
45. x2 + y 2 = t2 cos2 t + t2 sin2 t = t2 ,
p
x2 + y 2 = t = z; a conical helix.
46. The curve wraps around an elliptic cylinder with axis along the z-axis; an elliptical helix.
47. (a)
(b)
(c)
(d)
III, since the curve is a subset of the plane y = −x
IV, since only x is periodic in t, and y, z increase without bound
II, since all three components are periodic in t
I, since the projection onto the yz-plane is a circle and the curve increases without bound in
the x-direction
49. (a) Let x = 3 cos t and y = 3 sin t, then z = 9 cos2 t.
z
(b)
x
50. The plane is parallel to a line on the surface of
the cone and does not go through the vertex so
the curve of intersection
is a parabola. Eliminate
p
z to get y + 2 = x2 + y 2 , (y + 2)2 = x2 + y 2 ,
y = x2 /4 − 1; let x = t, then y = t2 /4 − 1
and z = t2 /4 + 1.
y
z
x
y
Exercise Set 14.2
494
EXERCISE SET 14.2
y
1.
y
2.
2
r(2p) – r(3p/2)
r''(p)
x
-2
x
2
-2
r'(p/4)
3. r0 (t) = 5i + (1 − 2t)j
6. r0 (t) =
7.
4. r0 (t) = sin tj
5. r0 (t) = −
1
i + sec2 tj + 2e2t k
t2
1
1
i + (cos t − t sin t)j − √ k
2
1+t
2 t
r0 (t) = h1, 2ti,
8.
r0 (t) = 3t2 i + 2tj,
r0 (2) = h1, 4i,
r0 (1) = 3i + 2j
r(2) = h2, 4i
r(1) = i + j
y
y
3
〈1, 4 〉
2
4
1
x
x
1
2
9.
r0 (t) = sec t tan ti + sec2 tj,
r0 (0) = j
r(0) = i
y
1
10.
2
3
4
r0 (t) = 2 cos ti − 3 sin tj,
π √
3
= 3i − j
r0
6
2
√
π
3 3
=i+
j
r
6
2
y
3
x
2
1.5
1
x
-1
-2 -1
1
-1
-2
-3
2
495
Chapter 14
r0 (t) = 2 cos ti − 2 sin tk,
11.
r0 (t) = − sin ti + cos tj + k,
12.
r0 (π/2) = −2k,
1
1
r0 (π/4) = − √ i + √ j + k,
2
2
r(π/2) = 2i + j
1
π
1
r(π/4) = √ i + √ j + k
4
2
2
z
z
( ) = − √12 i + √12 j + k
r′ 3
y
(2, 1, 0)
1 1
,
,3
√2 √2
(
x
r′ 6 = -2k
()
y
)
x
13. 9i + 6j
√
√
14. h 2/2, 2/2i
15. h1/3, 0i
16. j
17. 2i − 3j + 4k
18. h3, 1/2, sin 2i
19. (a) continuous, lim r(t) = 0 = r(0)
(b) not continuous, lim r(t) does not exist
20. (a) not continuous, lim r(t) does not exist.
(b) continuous, lim r(t) = 5i − j + k = r(0)
t→0
t→0
t→0
21. (a)
t→0
lim (r(t) − r0 (t)) = i − j + k
t→0
(b) lim (r(t) × r0 (t)) = lim (− cos ti − sin tj + k) = −i + k
t→0
(c)
t→0
0
lim (r(t) · r (t)) = 0
t→0
0
00
22. r(t) · (r (t) × r (t)) =
t
1
0
t2
2t
2
t3
3t2
6t
= 2t3 , so lim r(t) · (r0 (t) × r00 (t)) = 2
t→1
1
23. r0 (t) = 2ti − j, r0 (1) = 2i − j, r(1) = i + 2j; x = 1 + 2t, y = 2 − t, z = 0
t
24. r0 (t) = 2e2t i + 6 sin 3tj, r0 (0) = 2i, r(0) = i − 2j; x = 1 + 2t, y = −2, z = 0
√
25. r0 (t) = −2π sin πti + 2π cos πtj + 3k, r0 (1/3) = − 3 πi + πj + 3k,
√
√
√
r(1/3) = i + 3 j + k; x = 1 − 3 πt, y = 3 + πt, z = 1 + 3t
1
1
i − e−t j + 3t2 k, r0 (2) = i − e−2 j + 12k,
t
2
1
r(2) = ln 2i + e−2 j + 8k; x = ln 2 + t, y = e−2 − e−2 t, z = 8 + 12t
2
26. r0 (t) =
Exercise Set 14.2
496
3
3
j, t = 0 at P0 so r0 (0) = 2i + j,
27. r0 (t) = 2i + √
4
2 3t + 4
3
r(0) = −i + 2j; r = (−i + 2j) + t 2i + j
4
√
28. r0 (t) = −4 sin ti − 3j, t = π/3 at P0 so r0 (π/3) = −2 3i − 3j,
√
r(π/3) = 2i − πj; r = (2i − πj) + t(−2 3i − 3j)
29. r0 (t) = 2ti +
1
j − 2tk, t = −2 at P0 so r0 (−2) = −4i + j + 4k,
(t + 1)2
r(−2) = 4i + j; r = (4i + j) + t(−4i + j + 4k)
30. r0 (t) = cos ti + sinh tj +
1
k, t = 0 at P0 so r0 (0) = i + k, r(0) = j; r = ti + j + tk
1 + t2
31. 3ti + 2t2 j + C
32. (sin t)i − (cos t)j + C
33. (−t cos t + sin t)i + tj + C
34. h(t − 1)et , t(ln t − 1)i + C
35. (t3 /3)i − t2 j + ln |t|k + C
36. h−e−t , et , t3 i + C
π/3
1
1
sin 3t, cos 3t
3
3
37.
Z
2
40.
2
t(1 + t2 )1/2 dt =
0
3
2
2
− (3 − t)5/2 , (3 + t)5/2 , t
5
5
9
2 3/2
t i + 2t1/2 j
3
Z
43. y(t) =
Z
44. y(t) =
38.
0
Z
p
0
41.
= h0, −2/3i
t2 + t4 dt =
39.
=
1
−3
3/2
1
1 + t2
3
2
1
1
1 3
t i + t4 j
3
4
=
0
1
1
i+ j
3
4
√
= (5 5 − 1)/3
0
√
√
= h72 6/5, 72 6/5, 6i
52
i + 4j
3
42.
1 2
1
(e − 1)i + (1 − e−1 )j + k
2
2
y0 (t)dt = 13 t3 i + t2 j + C, y(0) = C = i + j, y(t) = ( 13 t3 + 1)i + (t2 + 1)j
y0 (t)dt = (sin t)i − (cos t)j + C,
y(0) = −j + C = i − j so C = i and y(t) = (1 + sin t)i − (cos t)j.
Z
45. y0 (t) = y00 (t)dt = ti + et j + C1 , y0 (0) = j + C1 = j so C1 = 0 and y0 (t) = ti + et j.
Z
1
y(t) = y0 (t)dt = t2 i + et j + C2 , y(0) = j + C2 = 2i so C2 = 2i − j and
2
1 2
t + 2 i + (et − 1)j
y(t) =
2
497
Chapter 14
0
Z
46. y (t) = y00 (t)dt = 4t3 i − t2 j + C1 , y0 (0) = C1 = 0, y0 (t) = 4t3 i − t2 j
Z
1
1
y(t) = y0 (t)dt = t4 i − t3 j + C2 , y(0) = C2 = 2i − 4j, y(t) = (t4 + 2)i − ( t3 + 4)j
3
3
47.
0
0
6
48.
1.5
0
1.5
5
0
p
p
49. r0 (t) = −4 sin ti + 3 cos tj, kr(t)k = 16 cos2 t + 9 sin2 t, kr0 (t)k = 16 sin2 t + 9 cos2 t,
p
−7 sin t cos t
krkkr0 k = 144 + 49 sin2 t cos2 t, θ = cos−1 p
144 + 49 sin2 t cos2 t
0
r and r are parallel when θ = 0, so t = 0, π/2, π, 3π/2, 2π; the angle between r and r0 is greatest at
θmax = 0.28 (t ≈ 2.23, 5.53) and θmin = −0.28, (t ≈ 0.77, 3.95), so r and r0 are never perpendicular.
3
0
0
o
√
√
2 + 3t2
√
50. r0 (t) = 2ti + 3t2 j, kr(t)k = t2 1 + t2 , kr0 (t)k = t 4 + 9t2 , cos θ = √
1 + t2 4 + 9t2
√
cos θ = 1 when t = 0, ±23/4 / 3 and r and r0 are parallel; cos θ > 0, so they are never perpendicular.
0.3
0
1
0
51. (a) 2t − t2 − 3t = −2, t2 + t − 2 = 0, (t + 2)(t − 1) = 0 so t = −2, 1. The points of intersection
are (−2, 4, 6) and (1, 1, −3).
(b) r0 = i + 2tj − 3k; r0 (−2) = i − 4j − 3k, r0 (1) = i + 2j − 3k, and n = 2i − j + k is normal to
the plane. Let θ be the acute angle, then
√
for t = −2: cos θ = |n · r0 |/(knk kr0 k) = 3/ 156, θ ≈ 76◦ ;
√
for t = 1: cos θ = |n · r0 |/(knk kr0 k) = 3/ 84, θ ≈ 71◦ .
Exercise Set 14.2
498
52. r0 = −2e−2t i − sin tj + 3 cos tk, t = 0 at the point (1, 1, 0) so r0 (0) = −2i + 3k and hence the tangent
line is x = 1 − 2t, y = 1, z = 3t. But x = 0 in the yz-plane so 1 − 2t = 0, t = 1/2. The point of
intersection is (0, 1, 3/2).
53. r1 (1) = r2 (2) = i + j + 3k so the graphs intersect at P; r01 (t) = 2ti + j + 9t2 k and
1
r02 (t) = i + tj − k so r01 (1) = 2i + j + 9k and r02 (2) = i + j − k are tangent to the graphs at P,
2
√
r0 (1) · r02 (2)
6
thus cos θ = 01
= − √ √ , θ = cos−1 (6/ 258) ≈ 68◦ .
kr1 (1)k kr02 (2)k
86 3
54. r1 (0) = r2 (−1) = 2i + j + 3k so the graphs intersect at P; r01 (t) = −2e−t i − (sin t)j + 2tk and
r02 (t) = −i + 2tj + 3t2 k so r01 (0) = −2i and r02 (−1) = −i − 2j + 3k are tangent to the graphs at P,
thus cos θ =
r01 (0) · r02 (−1)
1
= √ , θ ≈ 74◦ .
0
0
kr1 (0)k kr2 (−1)k
14
55. (a) r01 = 2i + 6tj + 3t2 k, r02 = 4t3 k, r1 · r2 = t7 ;
(b) r1 × r2 = 3t6 i − 2t5 j,
d
(r1 · r2 ) = 7t6 = r1 · r02 + r01 · r2
dt
d
(r1 × r2 ) = 18t5 i − 10t4 j = r1 × r02 + r01 × r2
dt
56. (a) r01 = − sin ti+cos tj+k, r02 = k, r1 · r2 = cos t+t2 ;
d
(r1 · r2 ) = − sin t+2t = r1 · r02 +r01 · r2
dt
(b) r1 × r2 = t sin ti + t(1 − cos t)j − sin tk,
d
(r1 × r2 ) = (sin t + t cos t)i + (1 + t sin t − cos t)j − cos tk = r1 × r02 + r01 × r2
dt
57.
d
[r(t) × r0 (t)] = r(t) × r00 (t) + r0 (t) × r0 (t) = r(t) × r00 (t) + 0 = r(t) × r00 (t)
dt
58.
d
du
d
[u · (v × w)] = u · [v × w] +
· [v × w] = u ·
dt
dt
dt
v×
dw dv
du
+
×w +
· [v × w]
dt
dt
dt
du
dw
dv
+u·
×w +
· [v × w]
=u· v×
dt
dt
dt
59. In Exercise 60, write each scalar triple product as a determinant.
60. Let c = c1 i + c2 j, r(t) = x(t)i + y(t)j, r1 (t) = x1 (t)i + y1 (t)j, r2 (t) = x2 (t)i + y2 (t)j and use
properties of derivatives.
61. Let r1 (t) = x1 (t)i + y1 (t)j + z1 (t)k and r2 (t) = x2 (t)i + y2 (t)j + z2 (t)k, in both (8) and (9); show
that the left and right members of the equalities are the same.
Z
62. (a)
Z
kr(t) dt =
k(x(t)i + y(t)j + z(t)k) dt
Z
Z
Z
Z
= k x(t) dt i + k y(t) dt j + k z(t) dt k = k r(t) dt
(b) Similar to Part (a)
(c)
Use Part (a) on Part (b) with k = −1
499
Chapter 14
EXERCISE SET 14.3
1. (a) The tangent vector reverses direction at the four cusps.
(b) r0 (t) = −3 cos2 t sin ti + 3 sin2 t cos tj = 0 when t = 0, π/2, π, 3π/2, 2π.
2. r0 (t) = cos ti + 2 sin t cos tj = 0 when t = π/2, 3π/2. The tangent vector reverses direction at (1, 1)
and (−1, 1).
3. r0 (t) = 3t2 i + (6t − 2)j + 2tk; smooth
4. r0 (t) = −2t sin(t2 )i + 2t cos(t2 )j − e−t k; smooth
5. r0 (t) = (1 − t)e−t i + (2t − 2)j − π sin(πt)k; not smooth, r0 (1) = 0
6. r0 (t) = π cos(πt)i + (2 − 1/t)j + (2t − 1)k; not smooth, r0 (1/2) = 0
7. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = (−3 cos2 t sin t)2 + (3 sin2 t cos t)2 + 02 = 9 sin2 t cos2 t,
Z π/2
3 sin t cos t dt = 3/2
L=
0
Z
π
8. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = (−3 sin t)2 + (3 cos t)2 + 16 = 25, L =
5dt = 5π
0
0
−t
√
0
Z
−t
1
9. r (t) = he , −e , 2i, kr (t)k = e + e , L =
t
t
(et + e−t )dt = e − e−1
0
Z
10. (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = 1/4 + (1 − t)/4 + (1 + t)/4 = 3/4, L =
√
11. r (t) = 3t i + j + 6 tk, kr0 (t)k = 3t2 + 1, L =
0
Z
3
(3t2 + 1)dt = 28
2
1
12. r0 (t) = 3i − 2j + k, kr0 (t)k =
√
14, L =
Z
4
√
√
14
14 dt =
3
Z
√
13. r (t) = −3 sin ti + 3 cos tj + k, kr (t)k = 10, L =
0
0
√
2π
√
10 dt = 2π 10
0
14. r0 (t) = 2ti + t cos tj + t sin tk, kr0 (t)k =
√
5t, L =
Z
π
√
√
5t dt = π 2 5/2
0
15. (dr/dt)(dt/dτ ) = (i + 2tj)(4) = 4i + 8tj = 4i + 8(4τ + 1)j;
r(τ ) = (4τ + 1)i + (4τ + 1)2 j, r0 (τ ) = 4i + 2(4)(4τ + 1)
16. (dr/dt)(dt/dτ ) = h−3 sin t, 3 cos ti(π) = h−3π sin πτ , 3π cos πτ i;
r(τ ) = h3 cos πτ, 3 sin πτ i, r0 (τ ) = h−3π sin πτ, 3π cos πτ i
17. (dr/dt)(dt/dτ ) = (et i − 4e−t j)(2τ ) = 2τ eτ i − 8τ e−τ j;
2
2
r(τ ) = eτ i + 4e−τ j, r0 (τ ) = 2τ eτ i − 4(2)τ e−τ j
2
2
2
2
1
−1
√
√
( 3/2)dt = 3
Exercise Set 14.3
500
18. (dr/dt)(dt/dτ ) =
9
1
9 1/2
t j + k (−1/τ 2 ) = − 5/2 j − 2 k;
2
τ
2τ
1
9
1
k, r0 (τ ) = − τ −5/2 j − 2 k
τ
2
τ
Z t√
√
√
s
s
s
s
19. (a) kr0 (t)k = 2, s =
2 dt = 2t; r = √ i + √ j, x = √ , y = √
2
2
2
2
0
s
(b) Similar to Part (a), x = y = z = √
3
r(τ ) = i + 3τ −3/2 j +
s
s
20. (a) x = − √ , y = − √
2
2
s
s
s
(b) x = − √ , y = − √ , z = − √
3
3
3
21. (a) r(t) = h1, 3, 4i when t = 0,
Z t
√
so s =
1 + 4 + 4 du = 3t, x = 1 + s/3, y = 3 − 2s/3, z = 4 + 2s/3
0
(b) r
= h28/3, −41/3, 62/3i
s=25
Z
t√
√
22. (a) r(t) = h−5, 0, 1i when t = 0, so s =
9 + 4 + 1 du = 14t,
0
√
√
√
x = −5 + 3s/ 14, y = 2s/ 14, z = 5 + s/ 14
√
√
√
= h−5 + 30/ 14, 20/ 14, 5 + 10/ 14i
(b) r(s)
s=10
23. x = 3 + cos t, y = 2 + sin t, (dx/dt)2 + (dy/dt)2 = 1,
Z t
s=
du = t so t = s, x = 3 + cos s, y = 2 + sin s for 0 ≤ s ≤ 2π.
0
24. x = cos3 t, y = sin3 t, (dx/dt)2 + (dy/dt)2 = 9 sin2 t cos2 t,
Z t
3
s=
3 sin u cos u du = sin2 t so sin t = (2s/3)1/2 , cos t = (1 − 2s/3)1/2 ,
2
0
x = (1 − 2s/3)3/2 , y = (2s/3)3/2 for 0 ≤ s ≤ 3/2
25. x = t3 /3, y = t2 /2, (dx/dt)2 + (dy/dt)2 = t2 (t2 + 1),
Z t
1
s=
u(u2 + 1)1/2 du = [(t2 + 1)3/2 − 1] so t = [(3s + 1)2/3 − 1]1/2 ,
3
0
x=
1
1
[(3s + 1)2/3 − 1]3/2 , y = [(3s + 1)2/3 − 1] for s ≥ 0
3
2
26. x = (1 + t)2 , y = (1 + t)3 , (dx/dt)2 + (dy/dt)2 = (1 + t)2 [4 + 9(1 + t)2 ],
Z t
√
1
s=
([4 + 9(1 + t)2 ]3/2 − 13 13) so
(1 + u)[4 + 9(1 + u)2 ]1/2 du =
27
0
√
√
1
1
1 + t = [(27s + 13 13)2/3 − 4]1/2 , x = [(27s + 13 13)2/3 − 4],
3
9
√ 2/3
√
√
1
[(27s + 13 13) − 4]3/2 for 0 ≤ s ≤ (80 10 − 13 13)/27
y=
27
501
Chapter 14
Z t√
√
27. x = et cos t, y = et sin t, (dx/dt)2 + (dy/dt)2 = 2e2t , s =
2 eu du = 2(et − 1) so
0
√
√
√
√
√
t = ln(s/ 2 + 1), x = (s/ 2 + 1) cos[ln(s/ 2 + 1)], y = (s/ 2 + 1) sin[ln(s/ 2 + 1)]
√
for 0 ≤ s ≤ 2(eπ/2 − 1)
28. x = sin(et ), y = cos(et ), z =
√ t
3e ,
Z
t
2eu du = 2(et − 1) so
√
et = 1 + s/2; x = sin(1 + s/2), y = cos(1 + s/2), z = 3(1 + s/2) for s ≥ 0
2
2
2
2t
(dx/dt) + (dy/dt) + (dz/dt) = 4e , s =
0
29. dx/dt = −a sin t, dy/dt = a cos t, dz/dt = c,
Z t0 p
Z t0 p
p
L=
a2 sin2 t + a2 cos2 t + c2 dt =
a2 + c2 dt = t0 a2 + c2
0
0
30. x = a cos t, y = a sin t, z = ct, (dx/dt)2 + (dy/dt)2 + (dz/dt)2 = a2 + c2 = w2 ,
Z t
s=
w du = wt so t = s/w; x = a cos(s/w), y = a sin(s/w), z = cs/w for s ≥ 0
0
31. x = at − a sin t, y = a − a cos t, (dx/dt)2 + (dy/dt)2 = 4a2 sin2 (t/2),
Z t
s=
2a sin(u/2)du = 4a[1 − cos(t/2)] so cos(t/2) = 1 − s/(4a), t = 2 cos−1 [1 − s/(4a)],
0
cos t = 2 cos2 (t/2) − 1 = 2[1 − s/(4a)]2 − 1,
sin t = 2 sin(t/2) cos(t/2) = 2(1 − [1 − s/(4a)]2 )1/2 (2[1 − s/(4a)]2 − 1),
x = 2a cos−1 [1 − s/(4a)] − 2a(1 − [1 − s/(4a)]2 )1/2 (2[1 − s/(4a)]2 − 1),
y=
32.
s(8a − s)
for 0 ≤ s ≤ 8a
8a
dr
dθ dy
dr
dθ
dx
= cos θ − r sin θ ,
= sin θ + r cos θ ,
dt
dt
dt dt
dt
dt
2 2
2 2 2 2
dθ
dy
dz
dr
dz
dx
+
+
=
+ r2
+
dt
dt
dt
dt
dt
dt
Z
2
2
2
2
ln 2
3
3e dt = e2t
2
ln 2
2t
4t
33. (a) (dr/dt) + r (dθ/dt) + (dz/dt) = 9e , L =
0
= 9/2
0
(b) (dr/dt)2 + r2 (dθ/dt)2 + (dz/dt)2 = 5t2 + t4 = t2 (5 + t2 ),
Z 2
√
t(5 + t2 )1/2 dt = 9 − 2 6
L=
1
34.
dρ
dφ
dθ
dx
= sin φ cos θ
+ ρ cos φ cos θ
− ρ sin φ sin θ ,
dt
dt
dt
dt
dφ
dθ dz
dρ
dφ
dρ
dy
+ ρ cos φ sin θ
+ ρ sin φ cos θ ,
= cos φ
− ρ sin φ ,
= sin φ sin θ
dt
dt
dt dt
dt
dt
dt
2
2
2 2 2 2
dy
dz
dρ
dθ
dφ
dx
+
+
=
+ ρ2 sin2 φ
+ ρ2
dt
dt
dt
dt
dt
dt
Exercise Set 14.4
502
Z
35. (a) (dρ/dt)2 + ρ2 sin2 φ(dθ/dt)2 + ρ2 (dφ/dt)2 = 3e−2t , L =
Z
(b) (dρ/dt)2 + ρ2 sin2 φ(dθ/dt)2 + ρ2 (dφ/dt)2 = 5, L =
2
√
3e−t dt =
√
3(1 − e−2 )
0
5
√
√
5dt = 4 5
1
d 3
d 3
d
r(t) = i + 2tj is never zero, but
r (τ ) =
(τ i + τ 6 j) = 3τ 2 i + 6τ 5 j is zero at τ = 0.
dt
dτ
dτ
dr dt
dt
dr
=
, and since t = τ 3 ,
= 0 when τ = 0.
(b)
dτ
dt dτ
dτ
36. (a)
37. (a)
g(τ ) = πτ
(b)
g(τ ) = π(1 − τ )
38.
t=1−τ
39. Represent the helix by x = a cos t, y = a sin t, z = ct with a = 6.25 and c = 10/π, so that the
radius of the helix is the distance from the axis of the cylinder to the center of the copper cable,
and the helix makes one turn in a distance of 20 in. (t = 2π). From Exercise 29 the length of the
p
helix is 2π 6.252 + (10/π)2 ≈ 44 in.
3
40. r(t) = cos ti + sin tj + t3/2 k, r0 (t) = − sin ti + cos tj + t1/2 k
2
q
1√
4 + 9t
(a) kr0 (t)k = sin2 t + cos2 t + 9t/4 =
2
Z 2
√
1√
2
ds
1√
=
(11 22 − 4)
4 + 9t
(c)
4 + 9t dt =
(b)
dt
2
27
0 2
41. r0 (t) = (1/t)i + 2j + 2tk
p
p
(a) kr0 (t)k = 1/t2 + 4 + 4t2 = (2t + 1/t)2 = 2t + 1/t
Z 3
ds
= 2t + 1/t
(c)
(2t + 1/t)dt = 8 + ln 3
(b)
dt
1
42. If r(t) = x(t)i + y(t)j + z(t)k is smooth, then kr0 (t)k is continuous and nonzero. Thus the angle
between r0 (t) and i, given by cos−1 (x0 (t)/kr0 (t)k), is a continuous function of t. Similarly, the
angles between r0 (t) and the vectors j and k are continuous functions of t.
43. Let r(t) = x(t)i + y(t)j and use the chain rule.
EXERCISE SET 14.4
1. (a)
y
(b)
y
x
x
503
Chapter 14
y
2.
x
3. r0 (t) = 2ti + j, kr0 (t)k =
√
4t2 + 1, T(t) = (4t2 + 1)−1/2 (2ti + j),
T0 (t) = (4t2 + 1)−1/2 (2i) − 4t(4t2 + 1)−3/2 (2ti + j);
2
2
1
1
2
T(1) = √ i + √ j, T0 (1) = √ (i − 2j), N(1) = √ i − √ j.
5
5
5 5
5
5
4. r0 (t) = ti + t2 j, T(t) = (t2 + t4 )−1/2 (ti + t2 j),
T0 (t) = (t2 + t4 )−1/2 (i + 2tj) − (t + 2t3 )(t2 + t4 )−3/2 (ti + t2 j);
1
1
1
1
1
T(1) = √ i + √ j, T0 (1) = √ (−i + j), N(1) = − √ i + √ j
2
2
2 2
2
2
5. r0 (t) = −5 sin ti + 5 cos tj, kr0 (t)k = 5, T(t) = − sin ti + cos tj, T0 (t) = − cos ti − sin tj;
√
√
√
1
1
1
3
3
3
0
i + j, T (π/3) = − i −
j, N(π/3) = − i −
j
T(π/3) = −
2
2
2
2
2
2
√
1 + t2
, T(t) = (1 + t2 )−1/2 (i + tj),
t
1
e
i+ √
j,
T0 (t) = (1 + t2 )−1/2 (j) − t(1 + t2 )−3/2 (i + tj); T(e) = √
1 + e2
1 + e2
1
e
1
(−ei + j), N(e) = − √
i+ √
j
T0 (e) =
(1 + e2 )3/2
1 + e2
1 + e2
1
6. r (t) = i + j, kr0 (t)k =
t
0
1
7. r0 (t) = −4 sin ti + 4 cos tj + k, T(t) = √ (−4 sin ti + 4 cos tj + k),
17
1
4
1
T0 (t) = √ (−4 cos ti − 4 sin tj), T(π/2) = − √ i + √ k
17
17
17
4
T0 (π/2) = − √ j, N(π/2) = −j
17
8. r0 (t) = i + tj + t2 k, T(t) = (1 + t2 + t4 )−1/2 (i + tj + t2 k),
T0 (t) = (1 + t2 + t4 )−1/2 (j + 2tk) − (t + 2t3 )(1 + t2 + t4 )−3/2 (i + tj + t2 k),
T(0) = i, T0 (0) = j = N(0)
1
9. r0 (t) = et [(cos t − sin t)i + (cos t + sin t)j + k], T(t) = √ [(cos t − sin t)i + (cos t + sin t)j + k],
3
1
T0 (t) = √ [(− sin t − cos t)i + (− sin t + cos t)j],
3
1
1
1
1
1
1
T(0) = √ i + √ j + √ k, T0 (0) = √ (−i + j), N(0) = − √ i + √ j
3
3
3
3
2
2
Exercise Set 14.4
504
p
√
10. r0 (t) = sinh ti + cosh tj + k, kr0 (t)k = sinh2 t + cosh2 t + 1 = 2 cosh t,
1
1
T(t) = √ (tanh ti + j + sech tk), T0 (t) = √ (sech2 ti − sech t tanh tk), at t = ln 2,
2
2
4
3
3
1
4
tanh(ln 2) = and sech(ln 2) = so T(ln 2) = √ i + √ j + √ k,
5
5
5 2
2
5 2
3
4
4
T0 (ln 2) = √ (4i − 3k), N(ln 2) = i − k
5
5
25 2
11. From the remark, the line is parametrized by normalizing v, but T(t0 ) = v/kvk, so r = r(t0 ) + tv
becomes r = r(t0 ) + sT(t0 ).
i
i
1
2
= h1, 2i, and T(1) = h √ , √ i, so the tangent line can be parametrized as
5
5
2
s
2s
1
r = h1, 1i + s √ , √ , so x = 1 + √ , y = 1 + √ .
5
5
5
5
12. r0 (t)
t=1
= h1, 2ti
t=1
13. r0 (t) = cos ti − sin tj + tk, r0 (0) = i, r(0) = j, T(0) = i, so the tangent line has the parametrization
x = s, y = 1.
√
t
1
17
k, r0 (1) = i + j − √ k, kr0 (1)k = √ , so the tangent
8
8
9 − t2
√
√
√
1
s 8
1
i+j− √ k .
line has parametrizations r = i + j + 8k + t i + j − √ k = i + j + 8k + √
8
17
8
14. r(1) = i + j +
15. T =
√
8k, r0 (t) = i + j − √
3
4
4
4
3
3
cos ti − sin tj + k, N = − sin ti − cos tj, B = T × N = cos ti − sin tj − k
5
5
5
5
5
5
1
1
16. T0 (t) = √ [(cos t + sin t)i + (− sin t + cos t)j ], N = √ [(− sin t + cos t)i − (cos t + sin t)j],
2
2
B = T × N = −k
17. r0 (t) = t sin ti + t cos tj, kr0 k = t, T = sin ti + cos tj, N = cos ti − sin tj, B = T × N = −k
√
18. T = (−a sin ti + a cos tj + bk)/ a2 + b2 , N = − cos ti − sin tj,
√
B = T × N = (b sin ti − b cos tj + ak)/ a2 + b2
√
√
√
2
2
2
2
i+
j + k, T = − sin ti + cos tj =
(−i + j), N = −(cos ti + sin tj) = −
(i + j),
19. r(π/4) =
2
2
2
2
√
B = k; the rectifying, osculating, and normal planes are given (respectively) by x + y = 2,
z = 1, −x + y = 0.
√
1
1
1
20. r(0) = i + j, T = √ (i + j + k), N = √ (−j + k), B = √ (2i − j − k); the rectifying, osculating,
3
2
6
and normal planes are given (respectively) by −y + z = −1, 2x − y − z = 1, x + y + z = 2.
21. (a) By formulae (1) and (11), N(t) = B(t) × T(t) =
r0 (t)
r0 (t) × r00 (t)
× 0
.
0
00
kr (t) × r (t)k kr (t)k
(b) Since r0 is perpendicular to r0 × r00 it follows from Lagrange’s Identity (Exercise 32 of Section
13.4) that k(r0 (t) × r00 (t)) × r0 (t)k = kr0 (t) × r00 (t)kkr0 (t)k, and the result follows.
(c) From Exercise 39 of Section 13.4, (r0 (t) × r00 (t)) × r0 (t) = kr0 (t)k2 r00 (t) − (r0 (t) · r00 (t))r0 (t) =
u(t), so N(t) = u(t)/ku(t)k
505
Chapter 14
1
2
22. (a) r0 (t) = 2ti + j, r0 (1) = 2i + j, r00 (t) = 2i, u = 2i − 4j, N = √ i − √ j
5
5
π
(b) r0 (t) = −4 sin ti + 4 cos tj + k, r0 ( ) = −4i + k, r00 (t) = −4 cos ti − 4 sin tj,
2
π
r00 ( ) = −4j, u = 17(−4j), N = −j
2
23. r0 (t) = cos ti−sin tj+k, r00 (t) = − sin ti−cos tj, u = −2(sin ti+cos tj), kuk = 2, N = − sin ti−cos tj
24. r0 (t) = i + 2tj + 3t2 k, r00 (t) = 2j + 6tk, u(t) = −(4t + 18t3 )i + (2 − 18t4 )j + (6t + 12t3 )k,
1
N= √
−(4t + 18t3 )i + (2 − 18t4 )j + (6t + 12t3 )k
2 81t8 + 117t6 + 54t4 + 13t2 + 1
EXERCISE SET 14.5
1. κ ≈
1
=2
0.5
2. κ ≈
1
3
=
4/3
4
3. r0 (t) = 2ti + 3t2 j, r00 (t) = 2i + 6tj, κ = kr0 (t) × r00 (t)k/kr0 (t)k3 =
6
t(4 + 9t2 )3/2
4. r0 (t) = −4 sin ti+cos tj, r00 (t) = −4 cos ti−sin tj, κ = kr0 (t)×r00 (t)k/kr0 (t)k3 =
4
(16 sin t + cos2 t)3/2
5. r0 (t) = 3e3t i − e−t j, r00 (t) = 9e3t i + e−t j, κ = kr0 (t) × r00 (t)k/kr0 (t)k3 =
12e2t
2
3/2
(9e6t + e−2t )
p
6 t2 (t − 1)2
6. r (t) = −3t i + (1 − 2t)j, r (t) = −6ti − 2j, κ = kr (t) × r (t)k/kr (t)k =
(9t4 + 4t2 − 4t + 1)3/2
0
00
2
0
00
0
3
7. r0 (t) = −4 sin ti + 4 cos tj + k, r00 (t) = −4 cos ti − 4 sin tj,
κ = kr0 (t) × r00 (t)k/kr0 (t)k3 = 4/17
√
0
00
0
00
0
8. r (t) = i + tj + t k, r (t) = j + 2tk, κ = kr (t) × r (t)k/kr (t)k =
2
3
t4 + 4t2 + 1
(t4 + t2 + 1)3/2
9. r0 (t) = sinh ti + cosh tj + k, r00 (t) = cosh ti + sinh tj, κ = kr0 (t) × r00 (t)k/kr0 (t)k3 =
10. r0 (t) = j + 2tk, r00 (t) = 2k, κ = kr0 (t) × r00 (t)k/kr0 (t)k3 =
2
(4t2 + 1)3/2
11. r0 (t) = −3 sin ti + 4 cos tj + k, r00 (t) = −3 cos ti − 4 sin tj,
r0 (π/2) = −3i + k, r00 (π/2) = −4j; κ = k4i + 12kk/k − 3i + kk3 = 2/5, ρ = 5/2
12. r0 (t) = et i − e−t j + k, r00 (t) = et i + e−t j,
√
√
r0 (0) = i − j + k, r00 (0) = i + j; κ = k − i + j + 2kk/ki − j + kk3 = 2/3, ρ = 3/ 2
13. r0 (t) = et (cos t − sin t)i + et (cos t + sin t)j + et k,
r00 (t) = −2et sin ti + 2et cos tj + et k, r0 (0) = i + j + k,
√
√
r00 (0) = 2j + k; κ = k − i − j + 2kk/ki + j + kk3 = 2/3, ρ = 3 2/2
1
2 cosh2 t
Exercise Set 14.5
506
14. r0 (t) = cos ti − sin tj + tk, r00 (t) = − sin ti − cos tj + k,
√
√
r0 (0) = i, r00 (0) = −j + k; κ = k − j − kk/kik3 = 2, ρ = 2/2
√
s
1
s
1
3
i − sin 1 +
j+
k, kr0 (s)k = 1, so
15. r (s) = cos 1 +
2
2
2
2
2
1
s
1
s
dT
1
dT
= − sin 1 +
i − cos 1 +
j, κ =
=
ds
4
2
4
2
ds
16
0
r
3 − 2s
si +
3
r
2s
j, kr0 (s)k = 1, so
3
s
r
1
1
1
dT
dT
1
3
i + √ j, κ =
=√
=
+
=
ds
ds
9
−
6s
6s
2s(9
− 6s)
9 − 6s
6s
0
16. r (s) = −
17. (a) r0 = x0 i + y 0 j, r00 = x00 i + y 00 j, kr0 × r00 k = |x0 y 00 − x00 y 0 |, κ =
|x0 y 00 − y 0 x00 |
(x02 + y 02 )3/2
(b) r = xi + y(x)j, r0 (x) = i + (dy/dx)j, r00 (x) = (d2 y/dx2 )j, so
κ(x) = k(d2 y/dx2 )kk/ki + (dy/dx)jk3 = |d2 y/dx2 |/[1 + (dy/dx)2 ]3/2 .
18.
dy
|y 00 |
= tan φ, (1 + tan2 φ)3/2 = (sec2 φ)3/2 = | sec φ|3 , κ(x) =
= |y 00 cos3 φ|
dx
| sec φ|3
19. κ(x) =
| sin x|
, κ(π/2) = 1
(1 + cos2 x)3/2
20. κ(x) =
2|x|
, κ(0) = 0
(1 + x4 )3/2
21. κ(x) =
√
2|x|3
, κ(1) = 1/ 2
4
3/2
(x + 1)
22. κ(x) =
e−x
e−1
,
κ(1)
=
(1 + e−2x )3/2
(1 + e−2 )3/2
23. κ(x) =
√
2 sec2 x| tan x|
, κ(π/4) = 4/(5 5)
4
3/2
(1 + sec x)
24. By implicit differentiation, dy/dx = 4x/y, d2 y/dx2 = 36/y 3 so κ =
if (x, y) = (2, 5) then κ =
36/125
36
= √
(1 + 64/25)3/2
89 89
25. x0 (t) = 2t, y 0 (t) = 3t2 , x00 (t) = 2, y 00 (t) = 6t,
x0 (1/2) = 1, y 0 (1/2) = 3/4, x00 (1/2) = 2, y 00 (1/2) = 3; κ = 96/125
26. x0 (t) = −4 sin t, y 0 (t) = cos t, x00 (t) = −4 cos t, y 00 (t) = − sin t,
x0 (π/2) = −4, y 0 (π/2) = 0, x00 (π/2) = 0, y 00 (π/2) = −1; κ = 1/16
27. x0 (t) = 3e3t , y 0 (t) = −e−t , x00 (t) = 9e3t , y 00 (t) = e−t ,
√
x0 (0) = 3, y 0 (0) = −1, x00 (0) = 9, y 00 (0) = 1; κ = 6/(5 10)
28. x0 (t) = −3t2 , y 0 (t) = 1 − 2t, x00 (t) = −6t, y 00 (t) = −2,
x0 (1) = −3, y 0 (1) = −1, x00 (1) = −6, y 00 (1) = −2; κ = 0
36/|y|3
;
(1 + 16x2 /y 2 )3/2
507
Chapter 14
29. x0 (t) = 1, y 0 (t) = −1/t2 , x00 (t) = 0, y 00 (t) = 2/t3
√
x0 (1) = 1, y 0 (1) = −1, x00 (1) = 0, y 00 (1) = 2; κ = 1/ 2
30. x0 (t) = 4 cos 2t, y 0 (t) = 3 cos t, x00 (t) = −8 sin 2t, y 00 (t) = −3 sin t,
x0 (π/2) = −4, y 0 (π/2) = 0, x00 (π/2) = 0, y 00 (π/2) = −3, κ = 12/43/2 = 3/2
31. (a) κ(x) =
ρ(x) =
| cos x|
,
(1 + sin2 x)3/2
(b) κ(t) =
(1 + sin2 x)3/2
| cos x|
ρ(t) =
2
,
(4 sin2 t + cos2 t)3/2
1
(4 sin2 t + cos2 t)3/2 ,
2
ρ(0) = 1/2, ρ(π/2) = 4
ρ(0) = ρ(π) = 1.
y
y
c
r (0) = r (c) = 1
x
1
ρ 6 =4
()
x
2
1
ρ (0) = 2
32. x0 (t) = −e−t (cos t + sin t),
0
κ
−t
y (t) = e (cos t − sin t),
00
−t
x (t) = 2e
00
sin t,
−t
y (t) = −2e
6
cos t;
using the formula of Exercise 17(a),
1
κ = √ et .
2
t
-3
3
33. (a) At x = 0 the curvature of I has a large value, yet the value of II there is zero, so II is not the
curvature of I; hence I is the curvature of II.
(b) I has points of inflection where the curvature is zero, but II is not zero there, and hence is
not the curvature of I; so I is the curvature of II.
34. (a) II takes the value zero at x = 0, yet the curvature of I is large there; hence I is the curvature
of II.
(b) I has constant zero curvature; II has constant, positive curvature; hence I is the curvature
of II.
35. (a)
0
0
1
(b)
1
5
0
5
0
Exercise Set 14.5
508
4
36. (a)
4
(b)
-1
1
-1
1
-4
-4
|12x2 − 4|
37. (a) κ =
(1 +
(4x3
−
y
(b)
3/2
4x)2 )
8
k
f(x)
x
-2
2
(c) f 0 (x) = 4x3 − 4x = 0 at x = 0, ±1, f 00 (x) = 12x2 − 4, so extrema at x = 0, ±1, and ρ = 1/4
for x = 0 and ρ = 1/8 when x = ±1.
y
38. (a)
(c) κ(t) =
30
t2 + 2
(t2 + 1)3/2
(d)
lim κ(t) = 0
t→+∞
x
-30
30
-30
dr
dr
i + r cos θ + sin θ
j;
39. r (θ) = −r sin θ + cos θ
dθ
dθ
d2 r
d2 r
dr
dr
+ cos θ 2 i + −r sin θ + 2 cos θ
+ sin θ 2 j;
r00 (θ) = −r cos θ − 2 sin θ
dθ
dθ
dθ
dθ
0
2
d2 r
dr
−r 2
dθ
dθ
.
"
2 #3/2
dr
r2 +
dθ
r2 + 2
κ=
40. Let r = a be the circle, so that dr/dθ = 0, and κ(θ) =
3
3
, κ(π/2) = √
41. κ(θ) = √
1/2
2 2(1 + cos θ)
2 2
43. κ(θ) =
10 + 8 cos2 3θ
2
, κ(0) =
2
3/2
3
(1 + 8 cos θ)
1
1
=
r
a
42. κ(θ) = √
44. κ(θ) =
1
1
, κ(1) = √
2θ
5e
5e2
θ2 + 2
3
, κ(1) = √
2
3/2
(θ + 1)
2 2
509
Chapter 14
45. The radius of curvature is zero when θ = π, so there is a cusp there.
46.
d2 r
3
dr
= − sin θ, 2 = − cos θ, κ(θ) = 3/2 √
dθ
dθ
2
1 + cos θ
47. Let y = t, then x =
1/|2p|
t2
and κ(t) = 2
;
4p
[t /(4p2 ) + 1]3/2
t = 0 when (x, y) = (0, 0) so κ(0) = 1/|2p|, ρ = 2|p|.
ex (1 − 2e2x ) 0
ex
, κ0 (x) =
; κ (x) = 0 when e2x = 1/2, x = −(ln 2)/2. By the first
2x
3/2
(1 + e )
(1 + e2x )5/2
√
1
1
derivative test, κ(− ln 2) is maximum so the point is (− ln 2, 1/ 2).
2
2
48. κ(x) =
49. Let x = 3 cos t, y = 2 sin t for 0 ≤ t < 2π, κ(t) =
6
so
(9 sin t + 4 cos2 t)3/2
2
1
1
(9 sin2 t + 4 cos2 t)3/2 = (5 sin2 t + 4)3/2 which, by inspection, is minimum when
6
6
t = 0 or π. The radius of curvature is minimum at (3, 0) and (−3, 0).
ρ(t) =
6(1 − 45x4 )
6x
for x > 0, κ0 (x) =
; κ0 (x) = 0 when x = 45−1/4 which, by the
4
3/2
(1 + 9x )
(1 + 9x4 )5/2
first derivative test, yields the maximum.
50. κ(x) =
51. r0 (t) = − sin ti + cos tj − sin tk, r00 (t) = − cos ti − sin tj − cos tk,
√
√
kr0 (t) × r00 (t)k = k − i + kk = 2, kr0 (t)k = (1 + sin2 t)1/2 ; κ(t) = 2/(1 + sin2 t)3/2 ,
√
√
ρ(t) = (1 + sin2 t)3/2 / 2. The minimum value of ρ is 1/ 2; the maximum value is 2.
√
52. r0 (t) = et i − e−t j + 2k, r00 (t) = et i + e−t j;
√
√
√
1
2
, ρ(t) = √ (et + e−t )2 = 2 2 cosh2 t. The minimum value of ρ is 2 2.
κ(t) = 2t
−2t
e +e
+2
2
p
53. From Exercise 39: dr/dθ = aeaθ = ar, d2 r/dθ2 = a2 eaθ = a2 r; κ = 1/[ 1 + a2 r].
dr
dr
= −2a2 sin 2θ, r
= −a2 sin 2θ, and
dθ
dθ
2
2
2
d2 r
d2 r
dr
dr
dr
= −2a2 cos 2θ so r 2 = −
− 2a2 cos 2θ = −
− 2r2 , thus
again to get r 2 +
dθ
dθ
dθ
dθ
dθ
"
2
2 #
a2 sin 2θ
d2 r
dr
3
dr
dr
2
2
=−
so
−r 2 =3 r +
;
,κ= 2
r +2
2
1/2
dθ
dθ
dθ
dθ
r
[r + (dr/dθ) ]
54. Use implicit differentiation on r2 = a2 cos 2θ to get 2r
2
r +
dr
dθ
2
= r2 +
a4 sin2 2θ
r4 + a4 sin2 2θ
a4 cos2 2θ + a4 sin2 2θ
a4
3r
=
=
=
, hence κ = 2 .
r2
r2
r2
r2
a
Exercise Set 14.5
510
55. (a) d2 y/dx2 = 2, κ(φ) = |2 cos3 φ|
√
√
(b) dy/dx = tan φ = 1, φ = π/4, κ(π/4) = |2 cos3 (π/4)| = 1/ 2, ρ = 2
y
(c)
3
x
-2
56. (a)
1
5
5
, 0 , 0, −
3
2
(b)
clockwise
(c)
it is a point, namely the center of the circle
57. κ = 0 along y = 0; along y = x2 , κ(x) = 2/(1 + 4x2 )3/2 , κ(0) = 2. Along y = x3 ,
κ(x) = 6|x|/(1 + 9x4 )3/2 , κ(0) = 0.
y
58. (a)
(b) For y = x2 , κ(x) =
2
(1 + 4x2 )3/2
4
so κ(0) = 2; for y = x4 ,
κ(x) =
x
-2
12x2
so κ(0) = 0.
(1 + 16x6 )3/2
κ is not continuous at x = 0.
2
59. κ = 1/r along the circle; along y = ax2 , κ(x) = 2a/(1 + 4a2 x2 )3/2 , κ(0) = 2a so 2a = 1/r,
a = 1/(2r).
|y 00 |
so the transition will be smooth if the values of y are equal, the values of y 0
(1 + y 02 )3/2
are equal, and the values of y 00 are equal at x = 0. If y = ex , then y 0 = y 00 = ex ; if y = ax2 + bx + c,
then y 0 = 2ax + b and y 00 = 2a. Equate y, y 0 , and y 00 at x = 0 to get c = 1, b = 1, and a = 1/2.
60. κ(x) =
61. The result follows from the definitions N =
T0 (s)
and κ = kT0 (s)k.
kT0 (s)k
dB
dB
= 0 because kB(s)k = 1 so
is perpendicular to B(s).
ds
ds
dB
dT
dB
dT
+
· T(s) = 0, but
= κN(s) so κB(s) · N(s) +
·
(b) B(s) · T(s) = 0, B(s) ·
ds
ds
ds
ds
dB
dB
· T(s) = 0 because B(s) · N(s) = 0; thus
is perpendicular to T(s).
T(s) = 0,
ds
ds
dB
dB
is perpendicular to both B(s) and T(s) but so is N(s), thus
is parallel to N(s) and
(c)
ds
ds
hence a scalar multiple of N(s).
62. (a) B ·
(d) If C lies in a plane, then T(s) and N(s) also lie in the plane; B(s) = T(s) × N(s) so B(s) is
always perpendicular to the plane and hence dB/ds = 0, thus τ = 0.
511
63.
Chapter 14
dT dB
dN
= B×
+
× T = B × (κN) + (−τ N) × T = κB × N − τ N × T, but B × N = −T and
ds
ds
ds
dN
= −κT + τ B
N × T = −B so
ds
64. r00 (s) = dT/ds = κN so r000 (s) = κdN/ds + (dκ/ds)N but dN/ds = −κT + τ B so
r000 (s) = −κ2 T + (dκ/ds)N + κτ B, r0 (s) × r00 (s) = T × (κN) = κT × N = κB,
[r0 (s) × r00 (s)] · r000 (s) = −κ3 B · T + κ(dκ/ds)B · N + κ2 τ B · B = κ2 τ ,
τ = [r0 (s) × r00 (s)] · r000 (s)/κ2 = [r0 (s) × r00 (s)] · r000 (s)/kr00 (s)k2 and
B = T× N = [r0 (s) × r00 (s)]/kr0 (s)k
65. r = a cos(s/w)i + a sin(s/w)j + (cs/w)k, r0 = −(a/w) sin(s/w)i + (a/w) cos(s/w)j + (c/w)k,
r00 = −(a/w2 ) cos(s/w)i − (a/w2 ) sin(s/w)j, r000 = (a/w3 ) sin(s/w)i − (a/w3 ) cos(s/w)j,
r0 × r00 = (ac/w3 ) sin(s/w)i − (ac/w3 ) cos(s/w)j + (a2 /w3 )k, (r0 × r00 ) · r000 = a2 c/w6 ,
kr00 (s)k = a/w2 , so τ = c/w2 and B = (c/w) sin(s/w)i − (c/w) cos(s/w)j + (a/w)k
dT ds
dT
=
= (κN)s0 = κs0 N,
dt
ds dt
dN ds
dN
=
= (−κT + τ B)s0 = −κs0 T + τ s0 B.
N0 =
dt
ds dt
66. (a) T0 =
(b) kr0 (t)k = s0 so r0 (t) = s0 T and r00 (t) = s00 T + s0 T0 = s00 T + s0 (κs0 N) = s00 T + κ(s0 )2 N.
(c) r000 (t) = s00 T0 + s000 T + κ(s0 )2 N0 + [2κs0 s00 + κ0 (s0 )2 ]N
= s00 (κs0 N) + s000 T + κ(s0 )2 (−κs0 T + τ s0 B) + [2κs0 s00 + κ0 (s0 )2 ]N
= [s000 − κ2 (s0 )3 ]T + [3κs0 s00 + κ0 (s0 )2 ]N + κτ (s0 )3 B.
(d) r0 (t) × r00 (t) = s0 s00 T × T + κ(s0 )3 T × N = κ(s0 )3 B, [r0 (t) × r00 (t)] · r000 (t) = κ2 τ (s0 )6 so
τ=
[r0 (t) × r00 (t)] · r000 (t)
[r0 (t) × r00 (t)] · r000 (t)
=
κ2 (s0 )6
kr0 (t) × r00 (t)k2
67. r0 = 2i + 2tj + t2 k, r00 = 2j + 2tk, r000 = 2k, r0 × r00 = 2t2 i − 4tj + 4k, kr0 × r00 k = 2(t2 + 2),
τ = 8/[2(t2 + 2)]2 = 2/(t2 + 2)2
68. r0 = −a sin ti + a cos tj + bk, r00 = −a cos ti − a sin tj, r000 = a sin ti − a cos tj,
p
r0 × r00 = ab sin ti − ab cos tj + a2 k, kr0 × r00 k = a2 (a2 + b2 ),
τ = a2 b/[a2 (a2 + b2 )] = b/(a2 + b2 )
√
√
√
69. r0 = et i − e−t j + 2k, r00 = et i + e−t j, r000 = et i − e−t j, r0 × r00 = − 2e−t i + 2et j + 2k,
√
√
√
kr0 × r00 k = 2(et + e−t ), τ = (−2 2)/[2(et + e−t )2 ] = − 2/(et + e−t )2
70. r0 = (1 − cos t)i + sin tj + k, r00 = sin ti + cos tj, r000 = cos ti − sin tj,
r0 × r00 = − cos ti + sin tj + (cos t − 1)k,
q
q
kr0 × r00 k = cos2 t + sin2 t + (cos t − 1)2 = 1 + 4 sin4 (t/2), τ = −1/[1 + 4 sin4 (t/2)]
Exercise Set 14.6
512
EXERCISE SET 14.6
1. v(t) = −3 sin ti + 3 cos tj
a(t) = −3 cos ti − 3 sin tj
p
kv(t)k = 9 sin2 t + 9 cos2 t = 3
√
r(π/3) = (3/2)i + (3 3/2)j
√
v(π/3) = −(3 3/2)i + (3/2)j
√
a(π/3) = −(3/2)i − (3 3/2)j
2. v(t) = i + 2tj
a(t) = 2j
kv(t)k =
√
1 + 4t2
r(2) = 2i + 4j
v(2) = i + 4j
a(2) = 2j
y
y v = − 3√3i + 3 j
2
2
v = i + 4j
8
a = 2j
(32 , 3 √2 3)
(2, 4)
x
3
a=−
3
3√3
j
i−
2
2
3. v(t) = et i − e−t j
4. v(t) = 4i − j
−t
t
x
4
a(t) = e i + e j
√
kv(t)k = e2t + e−2t
a(t) = 0
r(0) = i + j
r(1) = 6i
v(0) = i − j
v(1) = 4i − j
a(0) = i + j
a(1) = 0
kv(t)k =
√
17
y
y
a= i + j
(6, 0)
(1, 1)
x
x
v = 4i − j
a=0
v = i − j
5. v = i + tj + t2 k, a = j + 2tk; at t = 1, v = i + j + k, kvk =
√
3, a = j + 2k
6. r = (1 + 3t)i + (2 − 4t)j + (7 + t)k, v = 3i − 4j + k,
√
a = 0; at t = 2, v = 3i − 4j + k, kvk = 26, a = 0
7. v = −2 sin ti + 2 cos tj + k, a = −2 cos ti − 2 sin tj;
√
√
√
√
√
at t = π/4, v = − 2i + 2j + k, kvk = 5, a = − 2i − 2j
8. v = et (cos t + sin t)i + et (cos t − sin t)j + k, a = 2et cos ti − 2et sin tj; at t = π/2,
v = eπ/2 i − eπ/2 j + k, kvk = (1 + 2eπ )1/2 , a = −2eπ/2 j
513
Chapter 14
9. (a) v = −aω sin ωti + bω cos ωtj, a = −aω 2 cos ωti − bω 2 sin ωtj = −ω 2 r
(b) From Part (a), kak = ω 2 krk
10. (a) v = 16π cos πti − 8π sin 2πtj, a = −16π 2 sin πti − 16π 2 cos 2πtj;
at t = 1, v = −16πi, kvk = 16π, a = −16π 2 j
(b) x = 16 sin πt, y = 4 cos 2πt = 4 cos2 πt − 4 sin2 πt = 4 − 8 sin2 πt, y = 4 − x2 /32
(c) Both x(t) and y(t) are periodic and have period 2, so after 2 s the particle retraces its path.
p
p
√
11. v = (6/ t)i + (3/2)t1/2 j, kvk = 36/t + 9t/4, dkvk/dt = (−36/t2 + 9/4)/(2 36/t + 9t/4) = 0
√
if t = 4 which yields a minimum by the first derivative test. The minimum speed is 3 2 when
r = 24i + 8j.
12. v = (1 − 2t)i − 2tj, kvk =
p
√
(1 − 2t)2 + 4t2 = 8t2 − 4t + 1,
8t − 2
d
1
kvk = √
which yields a minimum by the first derivative test. The
= 0 if t =
dt
4
8t2 − 4t + 1
√
1
3
i − j.
minimum speed is 1/ 2 when the particle is at r =
16
16
13. (a)
6
0
3
8
p
p
(b) v = 3 cos 3ti + 6 sin 3tj, kvk = 9 cos2 3t + 36 sin2 3t = 3 1 + 3 sin2 3t; by inspection, maximum speed is 6 and minimum speed is 4
(d)
d
9 sin 6t
kvk = p
= 0 when t = 0, π/6, π/3, π/2, 2π/3; the maximum speed is 6 which
dt
2 1 + 3 sin2 3t
occurs first when sin 3t = 1, t = π/6.
8
14. (a)
0
0
c
p
p
(d) v = −6 sin 2ti + 2 cos 2tj + 4k, kvk = 36 sin2 2t + 4 cos2 2t + 16 = 2 8 sin2 t + 5; by inspec√
√
tion the maximum speed is 2 13 when t = π, the minimum speed is 2 5 when t = π/2.
15. v(t) = − sin ti + cos tj + C1 , v(0) = j + C1 = i, C1 = i − j, v(t) = (1 − sin t)i + (cos t − 1)j;
r(t) = (t + cos t)i + (sin t − t)j + C2 , r(0) = i + C2 = j,
C2 = −i + j so r(t) = (t + cos t − 1)i + (sin t − t + 1)j
Exercise Set 14.6
514
16. v(t) = ti − e−t j + C1 , v(0) = −j + C1 = 2i + j; C1 = 2i + 2j so
v(t) = (t + 2)i + (2 − e−t )j; r(t) = (t2 /2 + 2t)i + (2t + e−t )j + C2
r(0) = j + C2 = i − j, C2 = i − 2j so r(t) = (t2 /2 + 2t + 1)i + (2t + e−t − 2)j
17. v(t) = − cos ti + sin tj + et k + C1 , v(0) = −i + k + C1 = k so
C1 = i, v(t) = (1 − cos t)i + sin tj + et k; r(t) = (t − sin t)i − cos tj + et k + C2 ,
r(0) = −j + k + C2 = −i + k so C2 = −i + j, r(t) = (t − sin t − 1)i + (1 − cos t)j + et k.
1
1
1
j + e−2t k + C1 , v(0) = −j + k + C1 = 3i − j so
t+1
2
2
1
1
1 −2t 1
C1 = 3i − k, v(t) = 3i −
j+
e
k;
−
2
t+1
2
2
1 −2t 1
e
+ t k + C2 ,
r(t) = 3ti − ln(t + 1)j −
4
2
9
1
9 1 −2t 1
− e
− t k.
r(0) = − k + C2 = 2k so C2 = k, r(t) = 3ti − ln(t + 1)j +
4
4
4 4
2
18. v(t) = −
19. If a = 0 then x00 (t) = y 00 (t) = z 00 (t) = 0, so x(t) = x1 t + x0 , y(t) = y1 t + y0 , z(t) = z1 t + z0 , the
motion is along a straight line and has constant speed.
20. (a) If krk is constant then so is krk2 , but then x2 + y 2 = c2 (2-space) or x2 + y 2 + z 2 = c2
(3-space), so the motion is along a circle or a sphere of radius c centered at the origin, and
the velocity vector is always perpendicular to the locating vector.
(b) If kvk is constant then by the Theorem, v(t) · a(t) = 0, so the velocity is always perpendicular
to the acceleration.
21. v = 3t2 i + 2tj, a = 6ti + 2j; v = 3i + 2j and a = 6i + 2j when t = 1 so
√
cos θ = (v · a)/(kvk kak) = 11/ 130, θ ≈ 15◦ .
22. v = et (cos t − sin t)i + et (cos t + sin t)j, a = −2et sin ti + 2et cos tj, v · a = 2e2t , kvk =
√
kak = 2et , cos θ = (v · a)/(kvk kak) = 1/ 2, θ = 45◦ .
23. (a) displacement = r1 − r0 = 0.7i + 2.7j − 3.4k
(b) ∆r = r1 − r0 , so r0 = r1 − ∆r = −0.7i − 2.9j + 4.8k.
y
24. (a)
(b) distance =
p
50(1 − cos 2πt)
4
2
x
-4 -2
2
4
-2
-4
Z
25. ∆r = r(3) − r(1) = 8i + 26/3j; v = 2ti + t2 j, s =
p
3
t
√
√
4 + t2 dt = (13 13 − 5 5)/3.
1
Z
26. ∆r = r(3π/2) − r(0) = 3i − 3j; v = 3 cos ti − 3 sin tj, s =
3π/2
3dt = 9π/2.
0
√ t
2e ,
515
Chapter 14
27. ∆r = r(ln 3) − r(0) = 2i − 2/3j +
√
√
2 ln 3k; v = et i − e−t j + 2 k, s =
Z
ln 3
(et + e−t )dt = 8/3.
0
28. ∆r = r(π) − r(0) = 0; v = −2 sin 2ti + 2 sin 2tj − sin 2tk,
Z π/2
Z π
3| sin 2t|dt = 6
sin 2t dt = 6.
kvk = 3| sin 2t|, s =
0
0
29. In both cases, the equation of the path in rectangular coordinates is x2 + y 2 = 4, the particles
move counterclockwise around this circle; v1 = −6 sin 3ti + 6 cos 3tj and
v2 = −4t sin(t2 )i + 4t cos(t2 )j so kv1 k = 6 and kv2 k = 4t.
30. Let u = 1 − t3 in r2 to get r2 (u) = (3 + 2u)i + uj + (1 − u)k so both particles move along the same
√
√
line; v1 = 2i + j − k and v2 = −6t2 i − 3t2 j + 3t2 k so kv1 k = 6 and kv2 k = 3 6t2 .
31. (a) v = −e−t i + et j, a = e−t i + et j; when t = 0, v = −i + j, a = i + j, kvk =
√
v × a = −2k so aT = 0, aN = 2.
√
(c) κ = 1/ 2
(b) aT T = 0, aN N = a − aT T = i + j
√
2, v · a = 0,
32. (a) v = −2t sin(t2 )i + 2t cos(t2 )j, a = [−4t2 cos(t2 ) − 2 sin(t2 )]i + [−4t2 sin(t2 ) + 2 cos(t2 )]j; when
p
p
√
√
√
√
√
√
t = π/2, v = − π/2i + π/2j, a = (−π/ 2 − 2)i + (−π/ 2 + 2)j, kvk = π,
√
v · a = 2 π, v × a = π 3/2 k so aT = 2, aN = π
√
√
(b) aT T = − 2(i − j), aN N = a − aT T = −(π/ 2)(i + j)
(c) κ = 1
33. (a) v = (3t2 − 2)i + 2tj, a = 6ti + 2j; when t = 1, v = i + 2j, a = 6i + 2j, kvk =
√
√
v × a = −10k so aT = 2 5, aN = 2 5
√
2 5
(b) aT T = √ (i + 2j) = 2i + 4j, aN N = a − aT T = 4i − 2j
5
√
(c) κ = 2/ 5
√
5, v · a = 10,
√
34. (a) v = et (− sin t+cos t)i+et (cos t+sin t)j, a = −2et sin ti+2et cos tj; when t = π/4, v = 2eπ/4 j,
√
√
√
√
a = − 2eπ/4 i + 2eπ/4 j, kvk = 2eπ/4 , v · a = 2eπ/2 , v × a = 2eπ/2 k so aT = 2eπ/4 ,
√
aN = 2eπ/4
√
√
(b) aT T = 2eπ/4 j, aN N = a − aT T = − 2eπ/4 i
(c) κ = √
1
2eπ/4
35. (a) v = i + 2tj + 3t2 k, a = 2j + 6tk; when t = 1, v = i + 2j + 3k, a = 2j + 6k, kvk =
p
√
√ √
v · a = 22, v × a = 6i − 6j + 2k so aT = 22/ 14, aN = 76/ 14 = 38/7
(b) aT T =
√
14,
22
33
8
9
11
11
i + j + k, aN N = a − aT T = − i − j + k
7
7
7
7
7
7
√
19
(c) κ = √
7 14
36. (a) v = et i − 2e−2t j + k, a = et i + 4e−2t j; when t = 0, v = i − 2j + k, a = i + 4j, kvk =
p
√
v · a = −7, v × a = −4i + j + 6k so aT = −7/ 6, aN = 53/6
√
6,
Exercise Set 14.6
516
19
7
7
13
i+ j+ k
(b) aT T = − (i − 2j + k), aN N = a − aT T =
6
6
3
6
√
53
(c) κ = √
6 6
37. (a) v = 3 cos ti−2 sin tj−2 cos 2tk, a = −3 sin ti−2 cos tj+4 sin 2tk; when t = π/2, v = −2j+2k,
√
a = −3i, kvk = 2 2, v · a = 0, v × a = −6j − 6k so aT = 0, aN = 3
(b) aT T = 0, aN N = a = −3i
(c) κ =
3
8
38. (a) v = 3t2 j − (16/t)k, a = 6tj + (16/t2 )k; when t = 1, v = 3j − 16k, a = 6j + 16k, kvk =
√
√
v · a = −238, v × a = 144i so aT = −238/ 265, aN = 144/ 265
(b) aT T = −
(c) κ =
√
265,
3808
432
714
2304
j+
k, aN N = a − aT T =
j+
k
265
265
265
265
144
2653/2
39. kvk = 4, v · a = −12, v × a = 8k so aT = −3, aN = 2, T = −j, N = (a − aT T)/aN = i
√
√
√
5, v · a = 3, v × a = −6k so aT = 3/ 5, aN = 6/ 5, T = (1/ 5)(i + 2j),
√
N = (a − aT T)/aN = (1/ 5)(2i − j)
40. kvk =
√
41. kvk = 3, v · a = 4, v × a = 4i − 3j − 2k so aT = 4/3, aN =
√
N = (a − aT T)/aN = (i − 8j + 14k)/(3 29)
√
29/3, T = (1/3)(2i + 2j + k),
42. kvk = 5, v · a = −5, v × a = −4i − 10j − 3k so aT = −1, aN =
√
N = (a − aT T)/aN = (8i − 5j + 6k)/(5 5)
√
5, T = (1/5)(3i − 4k),
43. aT =
p
d2 s
dp 2
=
3t
+
4
=
3t/
3t2 + 4 so when t = 2, aT = 3/2.
dt2
dt
44. aT =
p
d2 s
dp2
=
t + e−3t = (2t − 3e−3t )/[2 t2 + e−3t ] so when t = 0, aT = −3/2.
2
dt
dt
p
d2 s
dp
=
(4t − 1)2 + cos2 πt = [4t − 1 − π cos πt sin πt]/ (4t − 1)2 + cos2 πt so when
2
dt
dt
√
t = 1/4, aT = −π/ 2.
45. aT =
46. aT =
p
d2 s
dp4
2 + 3 = (2t3 + 5t)/ t4 + 5t2 + 3 so when t = 1, a = 7/3.
=
t
+
5t
T
dt2
dt
47. aN = κ(ds/dt)2 = (1/ρ)(ds/dt)2 = (1/1)(2.9 × 105 )2 = 8.41 × 1010 km/s2
48. a = (d2 s/dt2 )T + κ(ds/dt)2 N where κ =
|d2 y/dx2 |
. If d2 y/dx2 = 0, then κ = 0 and
[1 + (dy/dx)2 ]3/2
a = (d2 s/dt2 )T so a is tangent to the curve.
49. aN = κ(ds/dt)2 = [2/(1 + 4x2 )3/2 ](3)2 = 18/(1 + 4x2 )3/2
50. y = ex , aN = κ(ds/dt)2 = [ex /(1 + e2x )3/2 ](2)2 = 4ex /(1 + e2x )3/2
517
Chapter 14
51. a = aT T + aN N; by the Pythagorean Theorem aN =
p
√
kak2 − a2T = 9 − 9 = 0
52. As in Exercise 51, kak2 = a2T + a2N , 81 = 9 + a2N , aN =
√
√
72 = 6 2.
2
√ √
1 2
c , so c2 = 1000aN , c ≤ 10 10 1.5 ≈ 38.73 m/s.
1000
53. Let c = ds/dt, aN = κ
ds
dt
54. 10 km/h is the same as
1
100
m/s, so kFk = 500
36
15
, aN =
100
36
2
≈ 257.20 N.
√
55. (a) v0 = 320, α = 60◦ , s0 = 0 so x = 160t, y = 160 3t − 16t2 .
√
√
(b) dy/dt = 160 3 − 32t, dy/dt = 0 when t = 5 3 so
√ √
√ 2
ymax = 160 3(5 3) − 16(5 3) = 1200 ft.
√
√
√
√
(c) y = 16t(10 3 − t), y = 0 when t = 0 or 10 3 so xmax = 160(10 3) = 1600 3 ft.
√
√
√
√
(d) v(t) = 160i + (160 3 − 32t)j, v(10 3) = 160(i − 3j), kv(10 3)k = 320 ft/s.
√
√
56. (a) v0 = 980, α = 45◦ , s0 = 0 so x = 490 2 t, y = 490 2t − 4.9t2
√
√
(b) dy/dt = 490 2 − 9.8t, dy/dt = 0 when t = 50 2 so
√
√
√ 2
ymax = 490 2(50 2) − 4.9(50 2) = 24, 500 m.
√
√
(c) y = 4.9t(100 2 − t), y = 0 when t = 0 or 100 2 so
√
√
xmax = 490 2(100 2) = 98, 000 m.
√
√
√
√
√
(d) v(t) = 490 2 i + (490 2 − 9.8t)j, v(100 2) = 490 2(i − j), kv(100 2)k = 980 m/s.
√
57. v0 = 80, α = −60◦ , s0 = 168 so x = 40t, y = 168 − 40 3 t − 16t2 ; y = 0 when
√
√
√
√
t = −7 3/2 (invalid) or t = 3 so x( 3) = 40 3 ft.
√
58. v0 = 80, α = 0◦ , s0 = 168 so x = 80t, y = 168 − 16t2 ; y = 0 when t = − 42/2 (invalid) or
√
√
√
t = 42/2 so x( 42/2) = 40 42 ft.
√
59. α = 30◦ , s0 = 0 so x = 3v0 t/2, y = v0 t/2 − 16t2 ; dy/dt = v0 /2 − 32t, dy/dt = 0 when t = v0 /64
so ymax = v02 /256 = 2500, v0 = 800 ft/s.
60. α = 45◦ , s0 = 0 so x =
√
√
√
2 v0 t/2, y = 2v0 t/2 − 4.9t2 ; y = 0 when t = 0 or 2v0 /9.8 so
xmax = v02 /9.8 = 24, 500, v0 = 490 m/s.
61. v0 = 800, s0 = 0 so x = (800 cos α)t, y = (800 sin α)t − 16t2 = 16t(50 sin α − t); y = 0 when t = 0
or 50 sin α so xmax = 40, 000 sin α cos α = 20, 000 sin 2α = 10, 000, 2α = 30◦ or 150◦ , α = 15◦
or 75◦ .
62. (a) v0 = 5, α = 0◦ , s0 = 4 so x = 5t, y = 4 − 16t2 ; y = 0 when t = −1/2 (invalid) or 1/2 so it
takes the ball 1/2 s to hit the floor.
√
(b) v(t) = 5i − 32tj, v(1/2) = 5i − 16j, kv(1/2)k = 281 so the ball hits the floor with a speed
√
of 281 ft/s.
(c) v0 = 0, α = −90◦ , s0 = 4 so x = 0, y = 4 − 16t2 ; y = 0 when t = 1/2 so both balls would hit
the ground at the same instant.
Exercise Set 14.6
518
√
3
63. (a) v0 = 40, α = 60, s0 = 4, so x = 20t, y = 4 + 20 3t − 16t2 ; when x = 15, t = ,
4
2
√ 3
3
≈ 20.98 ft, so the water clears the corner point A with 0.98 ft to
y = 4 + 20 3 − 16
4
4
spare.
√
(b) y = 20 when −16t2 + 25 3t − 16 = 0, t = 0.668 (reject) or 1.500, x(1.500) ≈ 30 ft, so the
water hits the roof.
(c) about 15 ft
√
64. x = (v0 /2)t, y = 4 + (v0 3/2)t − 16t2 , solve x = 15, y = 20 simultaneously for v0 and t,
15 √
3 − 1, t ≈ 0.7898, v0 ≈ 30/0.7898 ≈ 37.98 ft/s.
v0 /2 = 15/t, t2 =
16
√
√
65. (a) x = (35 2/2)t, y = (35 2/2)t − 4.9t2 , from Exercise 17a in Section 14.5
κ=
√
|x0 y 00 − x00 y 0 |
9.8
√ = 0.004 2 ≈ 0.00566
, κ(0) =
2
[(x0 )2 + (y 0 )2 ]3/2
35 2
(b) y 0 (t) = 0 when t =
25 √
125
m
2, y =
14
4
2
d2 s
1
1322
ds
2
ft/s2 ,
= (132)2 =
66. (a) a = aT T + aN N, aT = 2 = −7.5 ft/s , aN = κ
dt
dt
ρ
3000
s
2
p
1322
2
2
2
kak = aT + aN = (7.5) +
≈ 9.49 ft/s2
3000
(b) cos θ =
aT
7.5
a·T
=
≈−
≈ −0.79, θ ≈ 2.48 radians ≈ 142◦
kakkTk
kak
9.49
67. s0 = 0 so x = (v0 cos α)t, y = (v0 sin α)t − gt2 /2
(a) dy/dt = v0 sin α − gt so dy/dt = 0 when t = (v0 sin α)/g, ymax = (v0 sin α)2 /(2g)
(b) y = 0 when t = 0 or (2v0 sin α)/g, so x = R = (2v02 sin α cos α)/g = (v02 sin 2α)/g when
t = (2v0 sin α)/g; R is maximum when 2α = 90◦ , α = 45◦ , and the maximum value of R
is v02 /g.
68. The range is (v02 sin 2α)/g and the maximum range is v02 /g so (v02 sin 2α)/g = (3/4)v02 /g,
sin 2α = 3/4, α = (1/2) sin−1 (3/4) ≈ 24.3◦ or α = (1/2)[180◦ − sin−1 (3/4)] ≈ 65.7◦ .
√
69. v0 = 80, α = 30◦ , s0 = 5 so x = 40 3t, y = 5 + 40t − 16t2
p
√
√
(a) y = 0 when t = (−40 ± (40)2 − 4(−16)(5))/(−32) = (5 ± 30)/4, reject (5 − 30)/4 to get
√
t = (5 + 30)/4 ≈ 2.62 s.
√
(b) x ≈ 40 3(2.62) ≈ 181.5 ft.
1
70. (a) v0 = v, s0 = h so x = (v cos α)t, y = h + (v sin α)t − gt2 . If x = R, then (v cos α)t = R,
2
1
R
but y = 0 for this value of t so h + (v sin α)[R/(v cos α)] − g[R/(v cos α)]2 = 0,
t=
v cos α
2
h + (tan α)R − g(sec2 α)R2 /(2v 2 ) = 0, g(sec2 α)R2 − 2v 2 (tan α)R − 2v 2 h = 0.
519
Chapter 14
dR
dR
dR
− 2v 2 sec2 αR − 2v 2 tan α
= 0; if
= 0 and α = α0
dα
dα
dα
2
2
2
2
2
when R = R0 , then 2g sec α0 tan α0 R0 − 2v sec α0 R0 = 0, g tan α0 R0 − v = 0,
tan α0 = v 2 /(gR0 ).
(b) 2g sec2 α tan αR2 + 2g sec2 αR
(c) If α = α0 and R = R0 , then from part (a) g(sec2 α0 )R02 − 2v 2 (tan α0 )R0 − 2v 2 h = 0, but from
part (b) tan α0 = v 2 /(gR0 ) so sec2 α0 = 1 + tan2 α0 = 1 + v 4 /(gR0 )2 thus
g[1 + v 4 /(gR0 )2 ]R02 − 2v 2 [v 2 /(gR0 )]R0 − 2v 2 h = 0, gR02 − v 4 /g − 2v 2 h = 0,
p
R02 = v 2 (v 2 + 2gh)/g 2 , R0 = (v/g) v 2 + 2gh and
p
p
p
tan α0 = v 2 /(v v 2 + 2gh) = v/ v 2 + 2gh, α0 = tan−1 (v/ v 2 + 2gh).
71. (a) v0 (cos α)(2.9) = 259 cos 23◦ so v0 cos α ≈ 82.21061, v0 (sin α)(2.9) − 16(2.9)2 = −259 sin 23◦
so v0 sin α ≈ 11.50367; divide v0 sin α by v0 cos α to get tan α ≈ 0.139929, thus α ≈ 8◦
and v0 ≈ 82.21061/ cos 8◦ ≈ 83 ft/s.
(b) From part (a), x ≈ 82.21061t and y ≈ 11.50367t − 16t2 for 0 ≤ t ≤ 2.9; the distance traveled
Z 2.9 p
(dx/dt)2 + (dy/dt)2 dt ≈ 268.76 ft.
is
0
EXERCISE SET 14.7
1. The results follow from formulae (1) and (7) of Section 12.5.
2. (a) (rmax − rmin )/(rmax + rmin ) = 2ae/(2a) = e
(b) rmax /rmin = (1 + e)/(1 − e), and the result follows.
3. (a) From (15) and (6), at t = 0,
C = v0 × b0 − GM u = v0 j × r0 v0 k − GM u = r0 v02 i − GM i = (r0 v02 − GM )i
(b) From (22), r0 v02 − GM = GM e, so from (7) and (17), v × b = GM (cos θi + sin θj) + GM ei,
and the result follows.
(c) From (10) it follows that b is perpendicular to v, and the result follows.
(d) From Part (c) and (10), kv × bk = kvkkbk = vr0 v0 . From Part (b),
q
√
kv × bk = GM (e + cos θ)2 + sin2 θ = GM e2 + 2e cos θ + 1. By (10) and
GM p 2
e + 2e cos θ + 1. From (22),
Part (d), kv × bk = kvkkbk = v(r0 v0 ) thus v =
r0 v0
v0 p 2
e + 2e cos θ + 1.
r0 v02 /(GM ) = 1 + e, GM/(r0 v0 ) = v0 /(1 + e) so v =
1+e
4. At the end of the minor axis, cos θ = −c/a = −e so
r
v0 p
v0 p 2
1−e
.
e + 2e(−e) + 1 =
1 − e2 = v0
v=
1+e
1+e
1+e
5. vmax occurs when θ = 0 so vmax = v0 ; vmin occurs when θ = π so
v0 p 2
1−e
1+e
vmin =
, thus vmax = vmin
.
e − 2e + 1 = vmax
1+e
1+e
1−e
6. If the orbit is a circle then e = 0 so from Part (e) of Exercise 3, v = v0 at all points on the orbit.
p
p
Use (22) with e = 0 to get v0 = GM/r0 so v = GM/r0 .
Chapter 14 Supplementary Exercises
520
7. r0 = 6440 + 200 = 6640 km so v =
p
3.99 × 105 /6640 ≈ 7.75 km/s.
s
8. From Example 1, the orbit is 22,352 mi above the Earth, thus v ≈
r
9. From (23) with r0 = 6440 + 300 = 6740 km, vesc =
1.24 × 1012
≈ 6859.68 mi/h.
26,352
2(3.99) × 105
≈ 10.88 km/s.
6740
2π 3/2
4π 2 a3
a . But T = 1 yr = 365 · 24 · 3600 s, thus M =
10. From (29), T = √
≈ 1.99 × 1030 kg.
GT 2
GM
11. (a) At perigee, r = rmin = a(1 − e) = 238,900 (1 − 0.055) ≈ 225,760 mi; at apogee,
r = rmax = a(1 + e) = 238,900(1 + 0.055) ≈ 252,040 mi. Subtract the sum
of the radius of the moon and the radius of the Earth to get
minimum distance = 225,760 − 5080 = 220,680 mi,
and maximum distance = 252,040 − 5080 = 246,960 mi.
p
p
(b) T = 2π a3 /(GM ) = 2π (238,900)3 /(1.24 × 1012 ) ≈ 659 hr ≈ 27.5 days.
12. (a) rmin = 6440 + 649 = 7,089 km, rmax = 6440 + 4,340 = 10,780 km so
a = (rmin + rmax )/2 = 8934.5 km.
(b) e = (10,780 − 7,089)/(10,780 + 7,089) ≈ 0.207.
p
p
(c) T = 2π a3 /(GM ) = 2π (8934.5)3 /(3.99 × 105 ) ≈ 8400 s ≈ 140 min
13. (a) r0 = 4000 + 180 = 4180 mi, v =
p
1.24 × 1012 /4180 ≈ 17,224 mph
(4180)(17,824)2
r0 v02
−1 =
−1 ≈ 0.07094.
GM
1.24 × 1012
rmax = 4180(1 + 0.07094)/(1 − 0.07094) ≈ 4818 mi; the apogee altitude
is 4818 − 4000 = 818 mi.
(b) r0 = 4180 mi, v0 = 17,224+600 = 17,824 mi/h; e =
14. By equation (20), r =
hence e ≥ 0.
k
, where k > 0. By assumption, r is minimal when θ = 0,
1 + e cos θ
CHAPTER 14 SUPPLEMENTARY EXERCISES
2. (a) the line through the tips of r0 and r1
(b) the line segment connecting the tips of r0 and r1
(c) the line through the tip of r0 which is parallel to r0 (t0 )
4. (a)
speed
(b)
distance traveled
distance of the particle from the origin
πu2
7. (a) r(t) =
du i +
du j;
cos
sin
2
0
0
2
2
2
dr
πt
πt
+ sin2
= 1 and r(0) = 0
= x0 (t)2 + y 0 (t)2 = cos2
dt
2
2
Z
t
πu2
2
(c)
Z
t
521
Chapter 14
2
2
2
πs
πs
πs2
πs
i + sin
j, r00 (s) = −πs sin
i + πs cos
j,
2
2
2
2
κ = kr00 (s)k = π|s|
(b) r0 (s) = cos
(c) κ(s) → +∞, so the spiral winds ever tighter.
8. (a) The tangent vector to the curve is always tangent to the sphere.
(b) kvk = const, so v · a = 0; the acceleration vector is always perpendicular to the velocity
vector
1
1
2
2
(c) kr(t)k = 1 − cos t (cos2 t + sin2 t) + cos2 t = 1
4
4
9. (a) kr(t)k = 1, so by Theorem 14.2.7, r0 (t) is always perpendicular to the vector r(t). Then
v(t) = Rω(− sin ωti + cos ωtj), v = kv(t)k = Rω
(b) a = −Rω 2 (cos ωti + sin ωtj), a = kak = Rω 2 , and a = −ω 2 r is directed toward the origin.
(c) The smallest value of t for which r(t) = r(0) satisfies ωt = 2π, so T = t =
10. (a) F = kFk = mkak = mRω 2 = mR
2π
.
ω
v2
mv 2
=
R2
R
13
6.5
=
,
9600
19,200
r
13
r
13t
13t
a = −a
= −Rω 2 = −Rω 2 (cos ωti+sin ωtj) = −
cos
i + sin
j
krk
R
2
19,200
19,200
(b) R = 6440 + 3200 = 9600 km, 6.5 = v = Rω = 9600ω, ω =
(c) F = ma = 70
2
kg · km/s2 ≈ 10.77 N
13
11. (a) Let r = xi + yj + zk, then x2 + z 2 = t2 (sin2 πt + cos2 πt) = t2 = y 2
z
y
x
p
(b) Let x = t, then y = t2 , z = ± 4 − t2 /3 − t4 /6
z
x
y
Chapter 14 Supplementary Exercises
12.
522
y
t=
2
3
t=0
x
t=1
t=
1
3
13. (a) ker (t)k2 = cos2 θ + sin2 θ = 1, so er (t) is a unit vector; r(t) = r(t)e(t), so they have the same
direction if r(t) > 0, opposite if r(t) < 0. eθ (t) is perpendicular to er (t) since er (t) · eθ (t) = 0,
and it will result from a counterclockwise rotation of er (t) provided e(t) × eθ (t) = k, which
is true.
(b)
dθ
dθ
d
dθ
d
dθ
er (t) =
(− sin θi + cos θj) =
eθ (t) and eθ (t) = − (cos θi + sin θj) = − er (t), so
dt
dt
dt
dt
dt
dt
d
dθ
d
0
v(t) = r(t) = (r(t)er (t)) = r (t)er (t) + r(t) eθ (t)
dt
dt
dt
(c) From Part (b), a =
d
v(t)
dt
dθ
dθ
d2 θ
= r (t)er (t) + r (t) eθ (t) + r0 (t) eθ (t) + r(t) 2 eθ (t) − r(t)
dt
dt
dt
#
"
2
d2 θ
dr dθ
dθ
d2 r
eθ (t)
−r
er (t) + r 2 + 2
=
2
dt
dt
dt
dt dt
00
0
14. The height y(t) of the rocket satisfies tan θ = y/b, y = b tan θ, v =
−→
15. r = r0 + t P Q= (t − 1)i + (4 − 2t)j + (3 + 2t)k;
dθ
dt
2
er (t)
dy dθ
dθ
dy
=
= b sec2 θ .
dt
dθ dt
dt
s−3
12 − 2s
9 + 2s
dr
= 3, r(s) =
i+
j+
k
dt
3
3
3
16. By equation (26) of Section 14.6, r(t) = (60 cos α)ti + ((60 sin α)t − 16t2 + 4)j, and the maximum
15
sin α, so the ball clears the
height of the baseball occurs when y 0 (t) = 0, 60 sin α = 32t, t =
8
2
152
15
152 sin α
28
ceiling if ymax = (60 sin α) sin α − 16 2 sin2 α + 4 ≤ 25,
≤ 21, sin2 α ≤
. The ball
8
8
4
75
2
hits the wall when x = 60, t = sec α, and y(sec α) = 60 sin α sec α − 16 sec α + 4. Maximize the
28
. Then
height h(α) = y(sec α) = 60 tan α − 16 sec2 α + 4, subject to the constraint sin2 α ≤
75
15
15
60
15
=
, so sin α = √
, but for
=
h0 (α) = 60 sec2 α − 32 sec2 α tan α = 0, tan α =
2
2
32
8
17
8 + 15
this value of α the constraint is not satisfied (the ball hits the ceiling). Hence the maximum
hvalue of hpoccursiat one of the endpoints of the α-interval on which the ball clears the ceiling, i.e.
0, sin−1 28/75 . Since h0 (0) = 60, it follows that h is increasing throughout the interval, since
28
, hmax = 60 tan α − 16 sec2 α + 4 =
h0 > 0 inside the interval. Thus hmax occurs when sin2 α =
75
√
√
120 329 − 1012
75
28
≈ 24.78 ft. Note: the possibility that the baseball keeps
60 √ − 16 + 4 =
47
47
47
climbing until it hits the wall can be rejected as follows: if so, then y 0 (t) = 0 after the ball hits
15
15
sin α occurs after t = sec α, hence
sin α ≥ sec α, 15 sin α cos α ≥ 8,
the wall, i.e. t =
8
8
15 sin 2α ≥ 16, impossible.
523
Chapter 14
17. r0 (1) = 3i + 10j + 10k, so if r0 (t) = 3t2 i + 10j + 10tk is perpendicular to r0 (1), then
9t2 + 100 + 100t = 0, t = −10, −10/9,
so r = −1000i − 100j + 500k, −(1000/729)i − (100/9)j + (500/81)k.
dy
dx
= x(t),
= y(t), x(0) = x0 , y(0) = y0 , so
dt
dt
t
t
t
x(t) = x0 e , y(t) = y0 e , r(t) = e r0
18. Let r(t) = x(t)i + y(t)j, then
dv
2
1
= 2t2 i + j + cos 2tk, v0 = i + 2j − k, so x0 (t) = t3 + 1, y 0 (t) = t + 2, z 0 (t) = sin 2t − 1,
dt
3
2
1
1
1
1
x(t) = t4 + t, y(t) = t2 + 2t, z(t) = − cos 2t − t + , since r(0) = 0. Hence
6
2
4
4
1
1 2
1
1 4
t +t i+
t + 2t j −
cos 2t + t −
k
r(t) =
6
2
4
4
i p
ds i
= kr0 (t)k
(5/3)2 + 9 + (1 − (sin 2)/2)2 ≈ 3.475
(b)
dt t=1
t=1
19. (a)
20. kvk2 = v(t) · v(t), 2kvk
d
1
d
kvk = 2v · a,
(kvk) =
(v · a)
dt
dt
kvk
CHAPTER 15
Partial Derivatives
EXERCISE SET 15.1
1. (a) f (2, 1) = (2)2 (1) + 1 = 5
(c) f (0, 0) = (0)2 (0) + 1 = 1
(e) f (3a, a) = (3a)2 (a) + 1 = 9a3 + 1
2. (a) 2t
(b) f (1, 2) = (1)2 (2) + 1 = 3
(d) f (1, −3) = (1)2 (−3) + 1 = −2
( f ) f (ab, a − b) = (ab)2 (a − b) + 1 = a3 b2 − a2 b3 + 1
(c) 2y 2 + 2y
(b) 2x
3. (a) f (x + y, x − y) = (x + y)(x − y) + 3 = x2 − y 2 + 3
(b) f xy, 3x2 y 3 = (xy) 3x2 y 3 + 3 = 3x3 y 4 + 3
4. (a) (x/y) sin(x/y)
(c) (x − y) sin(x − y)
(b) xy sin(xy)
3
5. F (g(x), h(y)) = F x3 , 3y + 1 = x3 ex (3y+1)
h
i
2
6. g(u(x, y), v(x, y)) = g x2 y 3 , πxy = πxy sin x2 y 3 (πxy) = πxy sin πx5 y 7
7. (a) t2 + 3t10
8.
√
(b) 0
√
−3 ln(t2 +1)
te
=
(t2
(c) 3076
t
3
+ 1)
9. (a) 19
6
(b) −9
(c) 3
(e) −t + 3
(f ) (a + b)(a − b)2 b3 + 3
8
(d) a + 3
10. (a) x2 (x + y)(x − y) + (x + y) = x2 x2 − y 2 + (x + y) = x4 − x2 y 2 + x + y
(b) (xz)(xy)(y/x) + xy = xy 2 z + xy
2
2
11. F x2 , y + 1, z 2 = (y + 1)ex (y+1)z
12. g x2 z 3 , πxyz, xy/z = (xy/z) sin πx3 yz 4
√
√
13. (a) f ( 5, 2, π, −3π) = 80 π
(b) f (1, 1, . . . , 1) =
n
X
k = n(n + 1)/2
k=1
14. (a) f (−2, 2, 0, π/4) = 1
(b) f (1, 2, . . . , n) = n(n + 1)(2n + 1)/6, see (Theorem 2(b), Section 7.4)
15.
y
16.
y
x
x
1
2
524
525
Chapter 15
y
17.
y
18.
x
x
19. (a) all points above or on the line y = −2
(b) all points on or within the sphere x2 + y 2 + z 2 = 25
(c) all points in 3-space
20. (a) all points on or between the vertical lines x = ±2.
(b) all points above the line y = 2x
(c) all points not on the plane x + y + z = 0
z
21.
z
22.
3
(0, 3, 0)
y
y
x
x
z
23.
z
24.
x
x
y
z
25.
y
z
26.
(0, 0, 4)
(0, 2, 0)
x
x
y
(2, 0, 0)
y
Exercise Set 15.1
526
z
27.
z
28.
(0, 0, 1)
x
(0, 1, 0)
y
(1, 0, 0)
y
x
z
29.
z
30.
(0, 0, 1)
(0, –1, 0)
y
y
x
x
31.
y
k = -2
32.
k= 0 1234
y
k=2
k = -1
k=1
k=0 x
x
33.
y
34.
y
k=0 1234
x
k
k
k
k
k
=
=
=
=
=
2
1
0
–1
–2
x
527
Chapter 15
35.
k = –1
y
k = –2
2
y
36.
k=0
k=1
k=2
k = -2
2
c
o
k=1
k=2
x
-2
k = -1
k=0
x
2
-2
k=2
k=1
k = –2
k=0
k = –1
z
37.
z
38.
(0, 0, 2)
y
x
(0, 4, 0)
x
(2, 0, 0)
y
z
39.
z
40.
(0, 0, 1)
(0, − 12 , 0 )
(0, 0, 3)
x
y
y
( 14 , 0, 0 )
x
41. concentric spheres, common center at (2,0,0)
42. parallel planes, common normal 3i − j + 2k
43. concentric cylinders, common axis the y-axis
44. circular paraboloids, common axis the z-axis, all the same shape but with different vertices along
z-axis.
45. (a) f (−1, 1) = 0; x2 − 2x3 + 3xy = 0
(c) f (2, −1) = −18; x2 − 2x3 + 3xy = −18
(b) f (0, 0) = 0; x2 − 2x3 + 3xy = 0
Exercise Set 15.1
528
46. (a) f (ln 2, 1) = 2; yex = 2
(c) f (1, −2) = −2e; yex = −2e
(b) f (0, 3) = 3; yex = 3
47. (a) f (1, −2, 0) = 5; x2 + y 2 − z = 5
(c) f (0, 0, 0) = 0; x2 + y 2 − z = 0
(b) f (1, 0, 3) = −2; x2 + y 2 − z = −2
48. (a) f (1, 0, 2) = 3; xyz + 3 = 3, xyz = 0
(c) f (0, 0, 0) = 3; xyz = 0
(b) f (−2, 4, 1) = −5; xyz + 3 = −5, xyz = −8
y
49. (a)
(b) At (1, 4) the temperature is
T (1, 4) = 4 so the temperature
will remain constant along
the path xy = 4.
4
T=1
T=2
T=3
x
4
p
50. V = 8/
16 + x2 + y 2
p
16 + x2 + y 2
y
= 8/V x + y = 64/V − 16
2
2
2
20
the equipotential curves are circles.
V = 2.0
V = 1.0
V = 0.5
x
10 20
√
51. (a) f (x, y) = 1 − x2 − y 2 , because f = c is a circle of radius 1 − c (provided c ≤ 1), and the
radii in (a) decrease as c increases.
p
(b) f (x, y) = x2 + y 2 because f = c is a circle of radius c, and the radii increase uniformly.
√
(a) is the contour plot of f (x, y) = 1 − x2 − y 2 , because f = c is a circle of radius 1 − c
(provided c ≤ 1), and the radii in (a) decrease as c increases.
√
(c) f (x, y) = x2 + y 2 because f = c is a circle of radius c and the radii in the plot grow like
the square root function.
52. (a) III
53. (a) A
(d) decrease
(b) IV
(c) I
(b) B
(e) increase
(d) II
(c) increase
(f ) decrease
54. (a) Calgary, since the contour lines are closer together near Calgary than they are near Chicago.
(b) The change in atmospheric pressure is about ∆p ≈ 1012 − 999 = 13, so the average rate of
change is ∆p/1600 ≈ 0.0081.
529
Chapter 15
55. (a)
(b)
2
1
-4
-5
4
5
-3
56. (a)
-3
(b)
10
-10
10
10
-10
10
-10
-10
z
57. (a)
(b)
2
1
0
x
y
-1
-2
-2
y
0 6
58. (a)
c i
5
(b)
o
-1
0
1
2
o
l
i
f
z 0
c
-5
4
3
2
1
x
0
9
6
3
0
0
1
2
3
4
59. (a) The graph of g is the graph of f shifted one unit in the positive x-direction.
(b) The graph of g is the graph of f shifted one unit up the z-axis.
(c) The graph of g is the graph of f shifted one unit down the y-axis and then inverted with
respect to the plane z = 0.
Exercise Set 15.2
530
z
60. (a)
x
y
(b) If a is positive and increasing then the graph of g is more pointed, and in the limit as a → +∞
the graph approaches a ’spike’ on the z-axis of height 1. As a decreases to zero the graph of
g gets flatter until it finally approaches the plane z = 1.
EXERCISE SET 15.2
y
1.
y
2.
y
3.
x
x
x
5
-1
y=x
y
4.
5.
1
y
x
y
6.
x
x
y = 2x + 1
y
7.
xy = –1
8.
xy = 1
x
xy = 1
9. all of 3-space
y
xy = –1
x
531
Chapter 15
10. all points inside the sphere with radius 2 and center at the origin
11. all points not on the cylinder x2 + z 2 = 1
12.
all of 3-space
13. 35
14. π 2 /2
15. −8
16. e−7
17. 0
18. 0
19. (a) Along x = 0
lim
(x,y)→(0,0)
(b) Along x = 0,
3
3
= lim 2 does not exist.
y→0 2y
x2 + 2y 2
lim
(x,y)→(0,0)
x+y
1
= lim does not exist.
y→0 y
x + y2
1
x
1
→ +∞ as x → 0 so the original
= lim does not exist because
x→0 x
x2
x
limit does not exist.
20. (a) Along y = 0 : lim
x→0
(b) Along y = 0 : lim
x→0
1
does not exist, so the original limit does not exist.
x
sin x2 + y 2
sin z
=1
lim
= lim+
21. Let z = x + y , then
x2 + y 2
z
(x,y)→(0,0)
z→0
2
2
22. Let z = x2 + y 2 , then
1 − cos x2 + y 2
1 − cos z
sin z
= lim
=0
= lim+
x2 + y 2
z
(x,y)→(0,0)
z→0
z→0+ 1
lim
23. Let z = x2 + y 2 , then
lim
(x,y)→(0,0)
24. With z =
25.
lim
(x,y)→(0,0)
26.
lim
(x,y)→(0,0)
e−1/(x
2
+y 2 )
= lim+ e−1/z = 0
z→0
1
1
1 −1/√z
w
√
,
lim
; let w = √ , lim w = 0
e
x2 + y 2 z→+∞ z
z w→+∞ e
x2 + y 2 x2 − y 2
=
lim
x2 − y 2 = 0
2
2
x +y
(x,y)→(0,0)
x2 + 4y 2 x2 − 4y 2
=
lim
x2 − 4y 2 = 0
2
2
x + 4y
(x,y)→(0,0)
0
x2
=
lim
0
=
0;
along
y
=
x
:
lim
= lim 1/5 = 1/5
x→0 3x2
x→0
x→0 5x2
x→0
so the limit does not exist.
27. along y = 0 : lim
28. Let z = x2 + y 2 , then
1 − x2 − y 2
1 − z2
=
lim
= +∞ so the limit does not exist.
x2 + y 2
z2
(x,y)→(0,0)
z→0+
29. 8/3
31. Let t =
lim
30. ln 5
p
x2
+
y2
+
z2,
sin x2 + y 2 + z 2
sin t2
p
=0
then
lim
= lim
t
(x,y,z)→(0,0,0)
t→0+
x2 + y 2 + z 2
Exercise Set 15.2
32. With t =
532
p
sin t
cos t
= +∞ so the limit does not exist.
x2 + y 2 + z 2 , lim 2 = lim
+
+
t
2t
t→0
t→0
33. y ln(x2 + y 2 ) = r sin θ ln r2 = 2r(ln r) sin θ, so
lim
(x,y)→(0,0)
34.
35.
(r2 cos2 θ)(r2 sin2 θ)
x2 y 2
p
= r3 cos2 θ sin2 θ, so
=
r
x2 + y 2
√ 2 2 2
e x +y +z
eρ
p
= , so
ρ
x2 + y 2 + z 2
36.
lim
−1
tan
(x,y,z)→(0,0,0)
y ln(x2 + y 2 ) = lim+ 2r(ln r) sin θ = 0
r→0
lim
(x,y)→(0,0)
x2 y 2
p
x2 + y 2
=0
√ 2 2 2
e x +y +z
eρ
p
lim
does not exist.
= lim+
ρ
(x,y,z)→(0,0,0)
ρ→0
x2 + y 2 + z 2
π
1
1
= lim+ tan−1 2 =
2
2
2
x +y +z
ρ
2
ρ→0
37. (a) No, since there seem to be points near (0, 0) with z = 0 and other points near (0, 0) with
z ≈ 1/2.
mx3
mx
x4
=
lim
=
0
(c)
lim
= lim 1/2 = 1/2
(b) lim 4
x→0 x + m2 x2
x→0 x2 + m2
x→0 2x4
x→0
(d) A limit must be unique if it exists, so f (x, y) cannot have a limit as (x, y) → (0, 0).
38. (a)
Along y = mx :
mx4
mx2
=
lim
= 0;
x→0 2x6 + m2 x2
x→0 2x4 + m2
along y = kx2 :
lim
lim
x→0 2x6
kx5
kx
= lim 2
= 0.
x→0 2x + k 2
+ k 2 x4
1
x6
1
= lim = 6= 0
x→0 2x6 + x6
x→0 3
3
(b) lim
39. (a)
abct3
abct
= lim 2
=0
t→0 a2 t2 + b4 t4 + c4 t4
t→0 a + b4 t2 + c4 t2
lim
t4
= lim 1/3 = 1/3
t→0 t4 + t4 + t4
t→0
(b) lim
40. π/2 because
x2 + 1
→ +∞ as (x, y) → (0, 1)
x2 + (y − 1)2
41. −π/2 because
x2 − 1
→ −∞ as (x, y) → (0, 1)
x2 + (y − 1)2
42. with z = x2 + y 2 , lim
+
z→0
sin z
= 1 = f (0, 0)
z
x2
does not exist.
(x,y)→(0,0) x2 + y 2
Along x = 0 : lim 0/y 2 = lim 0 = 0; along y = 0 : lim x2 /x2 = lim 1 = 1.
43. No, because
lim
y→0
y→0
x→0
x→0
44. Using polar coordinates with r > 0, xy = r2 sin θ cos θ and x2 + y 2 = r2 so
|xy ln x2 + y 2 | = |r2 sin θ cos θ ln r2 | ≤ |2r2 ln r|, but lim+ 2r2 ln r = 0 thus
r→0
xy ln x2 + y 2 = 0; f (x, y) will be continuous at (0,0) if we define f (0, 0) = 0.
lim
(x,y)→(0,0)
533
Chapter 15
EXERCISE SET 15.3
1. (a) 9x2 y 2
(e) 6y
(b) 6x3 y
(f ) 6x3
(c) 9y 2
(g) 36
(d) 9x2
(h) 12
2. (a) 2e2x sin y
(e) cos y
(b) e2x cos y
(f ) e2x
(c) 2 sin y
(g) 0
(d) 0
(h) 4
√
(b) − x cos y
sin y
(c) − √
2 x
sin y
(d) − √
2 x
(b) 140x4 y 3
(c) 140x3 y 4
(d) 140x3 y 4
3. (a) −
1
cos y
4x3/2
4. (a) 8 + 84x2 y 5
5. (a)
3
3
∂z
= √
; slope =
∂x
8
2 3x + 2y
(b)
∂z
1
1
=√
; slope =
∂y
4
3x + 2y
6. (a)
∂z
= e−y ; slope = 1
∂x
(b)
∂z
= −xe−y + 5; slope = 2
∂y
∂z
= −4 cos(y 2 − 4x); rate of change = −4 cos 7
∂x
∂z
= 2y cos(y 2 − 4x); rate of change = 2 cos 7
(b)
∂y
7. (a)
8. (a)
∂z
1
1
=−
; rate of change = −
∂x
(x + y)2
4
(b)
∂z
1
1
=−
; rate of change = −
∂y
(x + y)2
4
9. ∂z/∂x = slope of line parallel to xz-plane = −4; ∂z/∂y = slope of line parallel to yz-plane = 1/2
10. The slope at P in the positive x-direction is negative, the slope in the positive y-direction is
negative, thus ∂z/∂x < 0, ∂z/∂y < 0; the curve through P which is parallel to the x-axis is
concave down, so ∂ 2 z/∂x2 < 0; the curve parallel to the y-axis is concave down, so ∂ 2 z/∂y 2 < 0.
2 3
11. ∂z/∂x = 8xy 3 ex
y
2 3
, ∂z/∂y = 12x2 y 2 ex
y
12. ∂z/∂x = −5x4 y 4 sin(x5 y 4 ), ∂z/∂y = −4x5 y 3 sin x5 y 4
13. ∂z/∂x = x3 /(y 3/5 + x) + 3x2 ln(1 + xy −3/5 ), ∂z/∂y = −(3/5)x4 /(y 8/5 + xy)
14. ∂z/∂x = yexy sin(4y 2 ), ∂z/∂y = 8yexy cos(4y 2 ) + xexy sin(4y 2 )
15.
y(x2 − y 2 ) ∂z
x(x2 − y 2 )
∂z
=− 2
= 2
,
2
2
∂x
(x + y ) ∂y
(x + y 2 )2
16.
xy 3 (3x + 4y) ∂z
x2 y 2 (6x + 5y)
∂z
=
=
,
3/2
∂x
∂y
2(x + y)
2(x + y)3/2
−1/2
17. fx (x, y) = (3/2)x2 y 5x2 − 7 3x5 y − 7x3 y
−1/2
fy (x, y) = (1/2)x3 3x2 − 7 3x5 y − 7x3 y
18. fx (x, y) = −2y/(x − y)2 , fy (x, y) = 2x/(x − y)2
19. fx (x, y) =
y −1/2
xy −3/2
3
, fy (x, y) = − 2
− y −5/2 tan−1 (x/y)
2
2
2
y +x
y +x
2
Exercise Set 15.3
534
20. fx (x, y) = 3x2 e−y + (1/2)x−1/2 y 3 sec
√
√
√
x tan x, fy (x, y) = −x3 e−y + 3y 2 sec x
−7/3
−7/3
, fy (x, y) = −(8/3)y tan x y 2 tan x
21. fx (x, y) = −(4/3)y 2 sec2 x y 2 tan x
1
√
x sinh xy 2 cosh xy 2 + x−1/2 sinh x sinh2 xy 2
2
√
fy (x, y) = 4xy cosh x sinh xy 2 cosh xy 2
22. fx (x, y) = 2y 2 cosh
√
23. fx (x, y) = −2x, fx (3, 1) = −6; fy (x, y) = −21y 2 , fy (3, 1) = −21
24. ∂f /∂x = x2 y 2 exy + 2xyexy , ∂f /∂x
25. ∂z/∂x = x(x2 +4y 2 )−1/2 , ∂z/∂x
26. ∂w/∂x = −x2 y sin xy + 2x cos xy,
(1,1)
(1,2)
= 3e; ∂f /∂y = x3 yexy + x2 exy , ∂f /∂y
(1,1)
√
= 1/ 17 ; ∂z/∂y = 4y(x2 +4y 2 )−1/2 , ∂z/∂y
= 2e
(1,2)
√
= 8/ 17
∂w
∂w
(1/2, π) = −π/4; ∂w/∂y = −x3 sin x,
(1/2, π) = −1/8
∂x
∂y
27. fx = 8x − 8y 4 , fy = −32xy 3 + 35y 4 , fxy = fyx = −32y 3
p
p
28. fx = x/ x2 + y 2 , fy = y/ x2 + y 2 , fxy = fyx = −xy(x2 + y 2 )−3/2
29. fx = ex cos y, fy = −ex sin y, fxy = fyx = −ex sin y
2
2
30. fx = ex−y , fy = −2yex−y , fxy = fyx = −2yex−y
2
31. fx = 4/(4x − 5y), fy = −5/(4x − 5y), fxy = fyx = 20/(4x − 5y)2
32. fx = 4xy 2 /(x2 + y 2 )2 , fy = −4x2 y/(x2 + y 2 )2 , fxy = fyx = 8xy(x2 − y 2 )/(x2 + y 2 )3
√
√
√
35. (a) 2x − 2z(∂z/∂x) = 0, ∂z/∂x = x/z = ±3/(2 6) = ± 6/4, ∂z/∂x = ± 6/4
p
p
√
(b) z = ± x2 + y 2 − 1, ∂z/∂x = ±x/ x2 + y 2 − 1 = ± 6/4
√
√
36. (a) 2y − 2z(∂z/∂y) = 0, ∂z/∂y = y/z = ±4/(2 6) = ± 6/3
p
p
√
(b) z = ± x2 + y 2 − 1, ∂z/∂y = ±y/ x2 + y 2 − 1 = ± 6/3
37.
1/2
3 2
∂z
x + y2 + z2
= 0, ∂z/∂x = −x/z; similarly, ∂z/∂y = −y/z
2x + 2z
2
∂x
38.
4x − 2x2 − y + z 3 1 − 3z 2 (∂z/∂y)
1 − 2x2 − y + z 3
∂z
∂z
4x − 3z 2 (∂z/∂x)
=
=
=
1,
;
=
1,
2x2 + y − z 3
∂x
3z 2
2x2 + y − z 3
∂y
3z 2
∂z
∂z
2x + yz 2 cos xyz
∂z
+ yz cos xyz +
sin xyz = 0,
=−
;
39. 2x + z xy
∂x
∂x
∂x
xyz cos xyz + sin xyz
∂z
∂z
xz 2 cos xyz
∂z
+ xz cos xyz +
sin xyz = 0,
=−
z xy
∂y
∂y
∂y
xyz cos xyz + sin xyz
40. exy (cosh z)
exy (cosh z)
∂z
z 2 − yexy sinh z
∂z
∂z
+ yexy sinh z − z 2 − 2xz
= 0,
= xy
;
∂x
∂x
∂x
e cosh z − 2xz
∂z
xexy sinh z
∂z
∂z
+ xexy sinh z − 2xz
= 0,
= − xy
∂y
∂y
∂y
e cosh z − 2xz
535
Chapter 15
41. III is a plane, and its partial derivatives are constants, so III cannot be f (x, y). If I is the graph
of z = f (x, y) then (by inspection) fy is constant as y varies, but neither II nor III is constant as
y varies. Hence z = f (x, y) has II as its graph, and as II seems to be an odd function of x and an
even function of y, fx has I as its graph and fy has III as its graph.
42. Moving to the right from (x0 , y0 ) decreases f (x, y), so fx < 0; moving up increases f , so fy > 0.
43. (a) 30xy 4 − 4
(b) 60x2 y 3
(c) 60x3 y 2
44. (a) 120(2x − y)2
(b) −240(2x − y)2
(c) 480(2x − y)
45. (a) fxyy (0, 1) = −30
(b) fxxx (0, 1) = −125
(c) fyyxx (0, 1) = 150
46. (a)
∂3w
∂3w
= −ey sin x,
2
∂y ∂x
∂y 2 ∂x
3
(b)
47. (a)
√
= −1/ 2
(π/4,0)
√
= −1/ 2
3
∂ w
∂ w
= −ey cos x,
2
∂x ∂y
∂x2 ∂y
∂3f
∂x3
(b)
(π/4,0)
∂3f
∂y 2 ∂x
(b) fxxxx
48. (a) fxyy
(c)
∂4f
∂x2 ∂y 2
(d)
(c) fxxyy
∂4f
∂y 3 ∂x
(d) fyyyxx
49. (a) 2xy 4 z 3 + y
(d) 2y 4 z 3 + y
(b) 4x2 y 3 z 3 + x
(e) 32z 3 + 1
(c) 3x2 y 4 z 2 + 2z
(f ) 438
50. (a) 2xy cos z
(d) 4y cos z
(b) x2 cos z
(e) 4 cos z
(c) −x2 y sin z
(f ) 0
51. fx = 2z/x, fy = z/y, fz = ln(x2 y cos z) − z tan z
52. fx = y −5/2 z sec(xz/y) tan(xz/y), fy = −xy −7/2 z sec(xz/y) tan(xz/y) − (3/2)y −5/2 sec(xz/y),
fz = xy −5/2 sec(xz/y) tan(xz/y)
53. fx = −y 2 z 3 / 1 + x2 y 4 z 6 , fy = −2xyz 3 / 1 + x2 y 4 z 6 , fz = −3xy 2 z 2 / 1 + x2 y 4 z 6
√
z sinh x2 yz cosh x2 yz , fy = 2x2 z cosh z sinh x2 yz cosh x2 yz ,
√
√
fz = 2x2 y cosh z sinh x2 yz cosh x2 yz + (1/2)z −1/2 sinh z sinh2 x2 yz
54. fx = 4xyz cosh
√
55. ∂w/∂x = yzez cos xz, ∂w/∂y = ez sin xz, ∂w/∂z = yez (sin xz + x cos xz)
2
2
56. ∂w/∂x = 2x/ y 2 + z 2 , ∂w/∂y = −2y x2 + z 2 / y 2 + z 2 , ∂w/∂z = 2z y 2 − x2 / y 2 + z 2
p
p
p
57. ∂w/∂x = x/ x2 + y 2 + z 2 , ∂w/∂y = y/ x2 + y 2 + z 2 , ∂w/∂z = z/ x2 + y 2 + z 2
58. ∂w/∂x = 2y 3 e2x+3z , ∂w/∂y = 3y 2 e2x+3z , ∂w/∂z = 3y 3 e2x+3z
59. (a) e
(b) 2e
(c) e
Exercise Set 15.3
536
√
60. (a) 2/ 7
√
(b) 4/ 7
62.
-2
-1
0
√
(c) 1/ 7
-2
y
1
1
z 0
-1
2
2
2
1
2
63. (3/2) x + y + z + w
0
2
z
-1
0
x
-1
and ∂w/∂z = −z/w
1
2
6
0
-2
2 1/2
y
2
∂w
2x + 2w
∂x
1
0
-1
x
-2
= 0, ∂w/∂x = −x/w; similarly, ∂w/∂y = −y/w
64. ∂w/∂x = −4x/3, ∂w/∂y = −1/3, ∂w/∂z = (2x2 + y − z 3 + 3z 2 + 3w)/3
65.
yzw cos xyz ∂w
xzw cos xyz ∂w
xyw cos xyz
∂w
=−
,
=−
,
=−
∂x
2w + sin xyz ∂y
2w + sin xyz ∂z
2w + sin xyz
66.
∂w
yexy sinh w
∂w
xexy sinh w
∂w
2zw
= 2
,
= 2
,
= xy
xy
∂x
z − e cosh w ∂y
z − exy cosh w ∂z
e cosh w − z 2
67. (a) fxy = 15x2 y 4 z 7 + 2y
(b) fyz = 35x3 y 4 z 6 + 3y 2
(c) fxz = 21x2 y 5 z 6
(d) fzz = 42x3 y 5 z 5
(e) fzyy = 140x3 y 3 z 6 + 6y
(f ) fxxy = 30xy 4 z 7
(g) fzyx = 105x2 y 4 z 6
(h) fxxyz = 210xy 4 z 6
68. (a) 160(4x − 3y + 2z)3
2
69. fx = ex , fy = −ey
(b) −1440(4x − 3y + 2z)2
2
2 2
70. fx = yex
71. ∂w/∂xi = −i sin(x1 + 2x2 + . . . + nxn )
y
(c) −5760(4x − 3y + 2z)
2 2
, fy = xex
1
72. ∂w/∂xi =
n
n
X
y
!(1/n)−1
xk
k=1
73. (a) fx = 2x + 2y, fxx = 2, fy = −2y + 2x, fyy = −2; fxx + fyy = 2 − 2 = 0
(b) zx = ex sin y − ey sin x, zxx = ex sin y − ey cos x, zy = ex cos y + ey cos x,
zyy = −ex sin y + ey cos x; zxx + zyy = ex sin y − ey cos x − ex sin y + ey cos x = 0
(c) zx =
zy =
2x
y
2x − 2y
x2 − y 2 − 2xy
1
−
2
=
,
z
=
−2
,
xx
x2 + y 2
x2 1 + (y/x)2
x2 + y 2
(x2 + y 2 )2
2y
1
2y + 2x
y 2 − x2 + 2xy
1
+
2
=
,
z
=
−2
;
yy
x2 + y 2
x 1 + (y/x)2
x2 + y 2
(x2 + y 2 )2
zxx + zyy = −2
x2 − y 2 − 2xy
y 2 − x2 + 2xy
−2
=0
2
2
2
(x + y )
(x2 + y 2 )2
537
Chapter 15
74. (a) zt = −e−t sin(x/c), zx = (1/c)e−t cos(x/c), zxx = −(1/c2 )e−t sin(x/c);
zt − c2 zxx = −e−t sin(x/c) − c2 (−(1/c2 )e−t sin(x/c)) = 0
(b) zt = −e−t cos(x/c), zx = −(1/c)e−t sin(x/c), zxx = −(1/c2 )e−t cos(x/c);
zt − c2 zxx = −e−t cos(x/c) − c2 (−(1/c2 )e−t cos(x/c)) = 0
75. ux = ω sin c ωt cos ωx, uxx = −ω 2 sin c ωt sin ωx, ut = c ω cos c ωt sin ωx, utt = −c2 ω 2 sin c ωt sin ωx;
1
1
uxx − 2 utt = −ω 2 sin c ωt sin ωx − 2 (−c2 )ω 2 sin c ωt sin ωx = 0
c
c
76. (a) ∂u/∂x = ∂v/∂y = 2x, ∂u/∂y = −∂v/∂x = −2y
(b) ∂u/∂x = ∂v/∂y = ex cos y, ∂u/∂y = −∂v/∂x = −ex sin y
(c) ∂u/∂x = ∂v/∂y = 2x/(x2 + y 2 ), ∂u/∂y = −∂v/∂x = 2y/(x2 + y 2 )
77. ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x so ∂ 2 u/∂x2 = ∂ 2 v/∂x∂y, and ∂ 2 u/∂y 2 = −∂ 2 v/∂y∂x,
∂ 2 u/∂x2 + ∂ 2 u/∂y 2 = ∂ 2 v/∂x∂y − ∂ 2 v/∂y∂x, if ∂ 2 v/∂x∂y = ∂ 2 v/∂y∂x then
∂ 2 u/∂x2 + ∂ 2 u/∂y 2 = 0; thus u satisfies Laplace’s equation. The proof that v satisfies Laplace’s
equation is similar. Adding Laplace’s equations for u and v gives Laplaces’ equation for u + v.
78. ∂z/∂y = 6y, ∂z/∂y
(2,1)
=6
79. ∂z/∂x = −x 29 − x2 − y 2
−1/2
, ∂z/∂x](4,3) = −2
80. (a) ∂z/∂y = 8y, ∂z/∂y](−1,1) = 8
(b) ∂z/∂x = 2x, ∂z/∂x](−1,1) = −2
81. (a) ∂V /∂r = 2πrh
(b) ∂V /∂h = πr2
(c)
∂V /∂r]r=6, h=4 = 48π
πsd2
82. (a) ∂V /∂s = √
6 4s2 − d2
(c)
∂V /∂s]s=10, d=16 = 320π/9
(d)
∂V /∂h]r=8, h=10 = 64π
(b) ∂V /∂d =
(d)
πd(8s2 − 3d2 )
√
24 4s2 − d2
∂V /∂d]s=10, d=16 = 16π/9
83. (a) P = 10T /V , ∂P/∂T = 10/V , ∂P/∂T ]T =80, V =50 = 1/5 lb/(in2 K)
(b) V = 10T /P, ∂V /∂P = −10T /P 2 , if V = 50 and T = 80 then
P = 10(80)/(50) = 16, ∂V /∂P ]T =80, P =16 = −25/8(in5 /lb3 )
84. (a) ∂z/∂y = x2 , ∂z/∂y](1,3) = 1, j + k is parallel to the tangent line so x = 1, y = 3 + t,
z =3+t
(b) ∂z/∂x = 2xy, ∂z/∂x](1,3) = 6, i + 6k is parallel to the tangent line so x = 1 + t, y = 3,
z = 3 + 6t
∂z
∂z
cos(x − y) ∂z
85.
1+
cos(x + z) + cos(x − y) = 0,
= −1 −
;
cos(x + z) − cos(x − y) = 0,
∂x
∂x
cos(x + z) ∂y
cos(x − y) ∂ 2 z
− cos(x + z) sin(x − y) + cos(x − y) sin(x + z)(1 + ∂z/∂x)
∂z
=
;
=
,
∂y
cos(x + z) ∂x∂y
cos2 (x + z)
substitute for ∂z/∂x and simplify to get
∂2z
cos2 (x + z) sin(x − y) + cos2 (x − y) sin(x + z)
=−
.
∂x∂y
cos3 (x + z)
Exercise Set 15.4
86. ∂V /∂r =
538
2 1
2
πrh = ( πr2 h) = 2V /r
3
r 3
87. (a) ∂T /∂x = 3x2 + 1, ∂T /∂x](1,2) = 4
(b) ∂T /∂y = 4y, ∂T /∂y](1,2) = 8
88. ∂ 2 R/∂R12 = −2R22 /(R1 + R2 )3 , ∂ 2 R/∂R22 = −2R12 /(R1 + R2 )3 ,
i
h
6
4
2
∂ 2 R/∂R12 ∂ 2 R/∂R22 = 4R12 R22 / (R1 + R2 ) = 4/ (R1 + R2 ) [R1 R2 / (R1 + R2 )]
4
= 4R2 / (R1 + R2 )
89.
2(x + ∆x)2 − 3(x + ∆x)y + y 2 − (2x2 − 3xy + y 2 )
f (x + ∆x, y) − f (x, y)
=
= 4x + 2∆x − 3y,
∆x
∆x
fx = lim
∆x→0
f (x + ∆x, y) − f (x, y)
= lim (4x + 2∆x − 3y) = 4x − 3y; fx (2, −1) = 11
∆x→0
∆x
2x2 − 3x(y + ∆y) + (y + ∆y)2 − (2x2 − 3xy + y 2 )
f (x, y + ∆y) − f (x, y)
=
= −3x + 2y + ∆y,
∆y
∆y
fy = lim
∆y→0
f (x, y + ∆y) − f (x, y)
= −3x + 2y; fy (2, −1) = −8
∆y
2 2
4x
(x + y 2 )−1/3 (2x) =
, (x, y) 6= (0, 0);
3
3(x2 + y 2 )1/3
d
d 4/3
4
[f (x, 0)]
[x ]
=
= x1/3
= 0.
fx (0, 0) =
dx
dx
3
x=0
x=0
x=0
90. fx (x, y) =
d
[f (0, y)]
91. (a) fy (0, 0) =
dy
y=0
d
[y]
=
dy
=1
y=0
1 3
y2
(x + y 3 )−2/3 (3y 2 ) = 3
;
3
(x + y 3 )2/3
fy (x, y) does not exist where y = −x, x 6= 0.
(b) If (x, y) 6= (0, 0), then fy (x, y) =
EXERCISE SET 15.4
1. 42t13
2.
2(3 + t−1/3 )
3(2t + t2/3 )
3. 3t−2 sin(1/t)
4.
1 − 2t4 − 8t4 ln t
√
2t 1 + ln t − 2t4 ln t
5. −
10 7/3 1−t10/3
t e
3
6. (1 + t)et cosh (tet /2) sinh (tet /2)
7. ∂z/∂u = 24u2 v 2 − 16uv 3 − 2v + 3, ∂z/∂v = 16u3 v − 24u2 v 2 − 2u − 3
8. ∂z/∂u = 2u/v 2 − u2 v sec2 (u/v) − 2uv 2 tan(u/v)
∂z/∂v = −2u2 /v 3 + u3 sec2 (u/v) − 2u2 v tan(u/v)
9. ∂z/∂u = −
2 sin u
2 cos u cos v
, ∂z/∂v = −
3 sin v
3 sin2 v
10. ∂z/∂u = 3 + 3v/u − 4u, ∂z/∂v = 2 + 3 ln u + 2 ln v
539
Chapter 15
11. ∂z/∂u = eu , ∂z/∂v = 0
12. ∂z/∂u = − sin(u − v) sin u2 + v 2 + 2u cos(u − v) cos u2 + v 2
∂z/∂v = sin(u − v) sin u2 + v 2 + 2v cos(u − v) cos u2 + v 2
13. ∂T /∂r = 3r2 sin θ cos2 θ − 4r3 sin3 θ cos θ
∂T /∂θ = −2r3 sin2 θ cos θ + r4 sin4 θ + r3 cos3 θ − 3r4 sin2 θ cos2 θ
14. dR/dφ = 5e5φ
15. ∂t/∂x = x2 + y 2 / 4x2 y 3 , ∂t/∂y = y 2 − 3x2 / 4xy 4
16. ∂w/∂u =
2v 2 u2 v 2 − (u − 2v)2
2
[u2 v 2 + (u − 2v)2 ]
, ∂w/∂v =
u2 (u − 2v)2 − u2 v 2
17. −π
19.
√
2
[u2 v 2 + (u − 2v)2 ]
18. 351/2, −168
√
3
3e
√ √
, 2−4 3 e 3
21. F (x, y) = x2 y 3 + cos y,
20. 1161
2xy 3
dy
=− 2 2
dx
3x y − sin y
22. F (x, y) = x3 − 3xy 2 + y 3 − 5,
23. F (x, y) = exy + yey − 1,
24. F (x, y) = x − (xy)
1/2
3x2 − 3y 2
dy
x2 − y 2
=−
=
2
dx
−6xy + 3y
2xy − y 2
yexy
dy
= − xy
dx
xe + yey + ey
√
2 xy − y
1 − (1/2)(xy)−1/2 y
dy
=
=−
+ 3y − 4,
√
dx
x − 6 xy
−(1/2)(xy)−1/2 x + 3
1/2
where x and y are the distances of cars A and B, respectively, from the
25. D = x2 + y 2
intersection and D is the distance between them.
h
h
1/2 i
1/2 i
(dx/dt) + y/ x2 + y 2
(dy/dt), dx/dt = −25 and dy/dt = −30
dD/dt = x/ x2 + y 2
when x = 0.3 and y = 0.4 so dD/dt = (0.3/0.5)(−25) + (0.4/0.5)(−30) = −39 mph.
26. T = (1/10)P V , dT /dt = (V /10)(dP/dt) + (P/10)(dV /dt), dV /dt = 4 and dP/dt = −1 when
V = 200 and P = 5 so dT /dt = (20)(−1) + (1/2)(4) = −18 K/s.
1
1
ab sin θ but θ = π/6 when a = 4 and b = 3 so A = (4)(3) sin(π/6) = 3.
2
2
1
6
−1
, 0 ≤ θ ≤ π/2.
Solve ab sin θ = 3 for θ to get θ = sin
2
ab
1
1
∂θ da ∂θ db
6
6
da
db
dθ
+
=r
+r
=
− 2
− 2
∂a dt
∂b dt
a b dt
ab
dt
dt
36
36
1− 2 2
1− 2 2
a b
a b
da
db
6
1 da 1 db
+
,
= 1 and
=1
= −√
2
2
a
dt
b
dt
dt
dt
a b − 36
6
7
dθ
7√
1 1
= −√
+
=− √ =−
3 radians/s
when a = 4 and b = 3 so
dt
36
144 − 36 4 3
12 3
27. A =
Exercise Set 15.4
540
√
28. From the law of cosines, c = a2 + b2 − 2ab cos θ where c is the length of the third side.
√
θ = π/3 so c = a2 + b2 − ab,
−1/2
∂c da ∂c db
1
da 1 2
db
dc
=
+
= (a2 + b2 − ab)−1/2 (2a − b)
+
a + b2 − ab
(2b − a)
dt ∂a dt
∂b dt
2
dt
2
dt
db da
db
1
da
+ (2b − a)
,
= 2 and
= 1 when a = 5 and b = 10
= √
(2a − b)
dt
dt dt
dt
2 a2 + b2 − ab
√
1
dc
= √ [(0)(2) + (15)(1)] = 3/2 cm/s. The third side is increasing.
so
dt
2 75
29. V = (π/4)D2 h where D is the diameter and h is the height, both measured in inches,
dV /dt = (π/2)Dh(dD/dt) + (π/4)D2 (dh/dt), dD/dt = 3 and dh/dt = 24 when D = 30 and
h = 240, so dV /dt = (π/2)(30)(240)(3) + (π/4)(30)2 (24) = 16,200π in3 /year.
30.
∂T dx ∂T dy
y 2 dx
dy
dT
=
+
=
+ 2y ln x , dx/dt = 1 and dy/dt = −4 at (3,2) so
dt
∂x dt
∂y dt
x dt
dt
dT /dt = (4/3)(1) + (4 ln 3)(−4) = 4/3 − 16 ln 3◦ C/s.
31. (a) xy-plane, fx = 12x2 y + 6xy, fy = 4x3 + 3x2 , fxy = fyx = 12x2 + 6x
(b) y 6= 0, fx = 3x2 /y, fy = −x3 /y 2 , fxy = fyx = −3x2 /y 2
32. (a) x2 + y 2 > 1, (the exterior of the circle of radius 1 about the origin);
p
p
−3/2
fx = x/ x2 + y 2 − 1, fy = y/ x2 + y 2 − 1, fxy = fyx = −xy x2 + y 2 − 1
(b) xy-plane, fx = 2x cos(x2 + y 3 ), fy = 3y 2 cos(x2 + y 3 ), fxy = fyx = −6xy 2 sin x2 + y 3
33. (a)
4:
fxxx , fxxy = fxyx = fyxx , fxyy = fyxy = fyyx , fyyy
(b)
5:
fxxxx , fxxxy = fxxyx = fxyxx = fyxxx ,
fxxyy = fxyxy = fxyyx = fyxyx = fyyxx = fyxxy ,
fxyyy = fyxyy = fyyxy = fyyyx , fyyyy
34. (a) Since ew has infinitely many continuous derivatives, as does xy 2 , by the Chain Rule the
2
function exy has infinitely many continuous derivatives, hence by Theorem 15.4.6,
fxyx = fxxy ; since fxy = fyx if follows that fxyx = fyxx .
2
(b) fxyx = fxxy = fyxx = 2xy 5 exy + 4y 3 exy
2
35. (a) f (tx, ty) = 3t2 x2 + t2 y 2 = t2 f (x, y); n = 2
p
(b) f (tx, ty) = t2 x2 + t2 y 2 = tf (x, y); n = 1
(c) f (tx, ty) = t3 x2 y − 2t3 y 3 = t3 f (x, y); n = 3
2
(d) f (tx, ty) = 5/ t2 x2 + 2t2 y 2 = t−4 f (x, y); n = −4
36. (a) If f (u, v) = tn f (x, y), then
let t = 1 to get x
∂f
∂f du ∂f dv
∂f
+
= ntn−1 f (x, y), x
+y
= ntn−1 f (x, y);
∂u dt
∂v dt
∂u
∂v
∂f
∂f
+y
= nf (x, y).
∂x
∂y
541
Chapter 15
(b) If f (x, y) = 3x2 + y 2 then xfx + yfy = 6x2 + 2y 2 = 2f (x, y);
p
p
p
p
If f (x, y) = x2 + y 2 then xfx + yfy = x2 / x2 + y 2 + y 2 / x2 + y 2 = x2 + y 2 = f (x, y);
If f (x, y) = x2 y − 2y 3 then xfx + yfy = 3x2 y − 6y 3 = 3f (x, y);
5(−2)2x
5(−2)4y
5
then xfx + yfy = x 2
+y 2
= −4f (x, y)
If f (x, y) = 2
2
2
2
3
(x + 2y )
(x + 2y )
(x + 2y 2 )3
dz ∂u ∂z
dz ∂u
∂z
=
,
=
∂x
du ∂x ∂y
du ∂y
2
∂2z
dz ∂ 2 u d2 z ∂u
dz ∂ 2 u
∂
dz ∂u
(b)
=
=
+
+
;
∂x2
du ∂x2
∂x du ∂x
du ∂x2
du2 ∂x
dz ∂ 2 u
∂ dz ∂u
dz ∂ 2 u
d2 z ∂u ∂u
∂2z
=
+
=
+ 2
∂y∂x
du ∂y∂x ∂y du ∂x
du ∂y∂x du ∂x ∂y
37. (a)
∂z ∂ 2 u
∂
∂2z
=
+
∂y 2
∂u ∂y 2
∂y
∂z
∂u
dz ∂ 2 u d2 z
∂u
=
+ 2
∂y
du ∂y 2
du
∂u
∂z
2
38. (a) z = f (u), u = x2 − y 2 ; ∂z/∂x = (dz/du)(∂u/∂x) = 2xdz/du
∂z/∂y = (dz/du)(∂u/∂y) = −2ydz/du, y∂z/∂x + x∂z/∂y = 2xydz/du − 2xydz/du = 0
dz ∂u
dz ∂z
dz ∂u
dz
∂z
=
=y ,
=
=x ,
∂x
du ∂x
du ∂y
du ∂y
du
∂z
∂z
dz
dz
x
−y
= xy
− xy
= 0.
∂x
∂y
du
du
(b) z = f (u), u = xy;
(c) yzx + xzy = y(2x cos(x2 − y 2 )) − x(2y cos(x2 − y 2 )) = 0
(d) xzx − yzy = xyexy − yxexy = 0
∂r
∂θ
∂r
∂θ
+ cos θ
and 0 = r cos θ
+ sin θ ; solve for ∂r/∂x and ∂θ/∂x.
∂x
∂x
∂x
∂x
∂r
∂θ
∂r
∂θ
+ cos θ
and 1 = r cos θ
+ sin θ ; solve for ∂r/∂y and ∂θ/∂y.
(b) 0 = −r sin θ
∂y
∂y
∂y
∂y
∂z
∂z ∂r
∂z ∂θ
∂z
1 ∂z
(c)
=
+
=
cos θ −
sin θ.
∂x
∂r ∂x ∂θ ∂x
∂r
r ∂θ
∂z ∂r
∂z ∂θ
∂z
1 ∂z
∂z
=
+
=
sin θ +
cos θ.
∂y
∂r ∂y
∂θ ∂y
∂r
r ∂θ
39. (a) 1 = −r sin θ
(d) Square and add the results of parts (a) and (b).
(e) From Part (c),
1 ∂z
∂ ∂z
1 ∂z
∂ ∂z
∂2z
∂r
∂θ
cos
θ
−
sin
θ
+
cos
θ
−
sin
θ
=
2
∂x
∂r ∂r
r ∂θ
∂x ∂θ ∂r
r ∂θ
∂x
2
2
1 ∂ z
1 ∂z
∂ z
sin θ −
sin θ cos θ
cos θ + 2
=
2
∂r
r ∂θ
r ∂r∂θ
2
∂z
1 ∂2z
sin θ
1 ∂z
∂ z
cos θ −
sin θ −
cos θ
−
sin θ −
+
∂θ∂r
∂r
r ∂θ2
r ∂θ
r
=
2 ∂2z
1 ∂2z
2 ∂z
1 ∂z
∂2z
2
sin
θ
cos
θ
−
sin
θ
cos
θ
+
sin2 θ.
cos
θ
+
sin2 θ +
∂r2
r2 ∂θ
r ∂θ∂r
r2 ∂θ2
r ∂r
Exercise Set 15.4
542
Similarly, from Part (c),
2 ∂2z
1 ∂2z
∂2z
2 ∂z
1 ∂z
∂2z
2
sin
θ
cos
θ
+
sin
θ
cos
θ
+
cos2 θ.
=
sin
θ
−
cos2 θ +
2
2
2
2
2
∂y
∂r
r ∂θ
r ∂θ∂r
r ∂θ
r ∂r
Add to get
40. zx =
∂2z
∂2z
1 ∂2z
1 ∂z
∂2z
.
+
=
+
+
∂x2
∂y 2
∂r2
r2 ∂θ2
r ∂r
−2y
4xy
2x
4xy
, zxx = 2
, zy = 2
, zyy = − 2
, zxx + zyy = 0;
x2 + y 2
(x + y 2 )2
x + y2
(x + y 2 )2
z = tan−1
2r2 cos θ sin θ
= tan−1 tan 2θ = 2θ, zr = 0, zθθ = 0
r2 (cos2 θ − sin2 θ)
41. (a) By the chain rule,
∂u
∂u
∂v
∂v
∂v
∂u
=
cos θ +
sin θ and
= − r sin θ +
r cos θ, use the
∂r
∂x
∂y
∂θ
∂x
∂y
Cauchy-Riemann conditions
∂u
∂v
∂u
∂v
∂u
=
and
=−
in the equation for
to get
∂x
∂y
∂y
∂x
∂r
∂v
∂v
∂v
∂u
1 ∂v
∂v
1 ∂u
∂u
=
cos θ −
sin θ and compare to
to see that
=
. The result
=−
∂r
∂y
∂x
∂θ
∂r
r ∂θ
∂r
r ∂θ
can be obtained by considering
(b) ux =
uy =
x2
∂v
∂u
and
.
∂r
∂θ
2x
1
2x
1
, vy = 2
= 2
= ux ;
2
2
+y
x 1 + (y/x)
x + y2
2y
y
2y
1
, vx = −2 2
=− 2
= −uy ;
x2 + y 2
x 1 + (y/x)2
x + y2
u = ln r2 , v = 2θ, ur = 2/r, vθ = 2, so ur =
1
1
vθ , uθ = 0, vr = 0, so vr = − uθ
r
r
42. z = f (u, v) where u = x − y and v = y − x,
∂z ∂u ∂z ∂v
∂z
∂z
∂z
∂z ∂u ∂z ∂v
∂z
∂z
∂z
∂z
∂z
=
+
=
−
and
=
+
=−
+
so
+
=0
∂x
∂u ∂x ∂v ∂x
∂u ∂v
∂y
∂u ∂y
∂v ∂y
∂u ∂v
∂x ∂y
43. (a) ux = f 0 (x + ct), uxx = f 00 (x + ct), ut = cf 0 (x + ct), utt = c2 f 00 (x + ct); utt = c2 uxx
(b) Substitute g for f and −c for c in Part (a).
(c) Since the sum of derivatives equals the derivative of the sum, the result follows from Parts
(a) and (b).
1
(d) sin t sin x = (− cos(x + t) + cos(x − t))
2
44. fx (x0 , y0 ) = y0 , fy (x0 , y0 ) = x0 ,
∆f = (x0 + ∆x)(y0 + ∆y) − x0 y0 = y0 ∆x + x0 ∆y + ∆x∆y
= fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (0)∆x + (∆x)∆y where 1 = 0 and 2 = ∆x.
45. fx (x0 , y0 ) = 2x0 , fy (x0 , y0 ) = 2y0 ,
∆f = (x0 + ∆x)2 + (y0 + ∆y)2 − x20 + y02 = 2x0 ∆x + (∆x)2 + 2y0 ∆y + (∆y)2
= fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (∆x)∆x + (∆y)∆y where 1 = ∆x and 2 = ∆y.
543
Chapter 15
46. fx (x0 , y0 ) = 2x0 y0 , fy (x0 , y0 ) = x20 ,
∆f = (x0 + ∆x)2 (y0 + ∆y) − x20 y0 = 2x0 y0 ∆x + x20 ∆y + y0 (∆x)2 + (∆x)2 ∆y + 2x0 ∆x∆y
= fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (y0 ∆x + ∆x∆y)∆x + (2x0 ∆x)∆y
where 1 = y0 ∆x + ∆x∆y and 2 = 2x0 ∆x.
47. fx (x0 , y0 ) = 3, fy (x0 , y0 ) = 2y0 ,
∆f = 3(x0 + ∆x) + (y0 + ∆y)2 − 3x0 − y02 = 3∆x + 2y0 ∆y + (∆y)2
= fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + (0)∆x + (∆y)∆y where 1 = 0 and 2 = ∆y.
48. (a)
lim
f (x, y) = 0 = f (0, 0)
(x,y)→(0,0)
(b)
lim
∆x→0
f (0 + ∆x, 0) − f (0, 0)
f (∆x, 0)
|∆x|
= lim
= lim
, which does not exist because
∆x→0
∆x→0 ∆x
∆x
∆x
lim |∆x|/∆x = 1 and
∆x→0+
lim |∆x|/∆x = −1.
∆x→0−
f (∆x, 0) − f (0, 0)
−3∆x
= lim
= −3,
∆x→0 ∆x
∆x
f (0, ∆y) − f (0, 0)
−2∆y
= lim
= −2;
fy (0, 0) = lim
∆y→0
∆y→0 ∆y
∆y
lim
f (x, y) does not exist because f (x, y) → 5 if (x, y) → (0, 0) where x ≥ 0 or y ≥ 0, but
49. fx (0, 0) = lim
∆x→0
(x,y)→(0,0)
f (x, y) → 0 if (x, y) → (0, 0) where x < 0 and y < 0, so f is not continuous at (0,0).
50. fx (0, 0) = lim
∆x→0
f (∆x, 0) − f (0, 0)
0−0
= lim
= lim 0 = 0
∆x→0 ∆x
∆x→0
∆x
f (0, ∆y) − f (0, 0)
0−0
0
= lim
= 0; along y = 0, lim 2 = lim 0 = 0,
x→0 x
x→0
∆y→0 ∆y
∆y
2
x
lim
f (x, y) does not exist.
along y = x, lim 2 = lim 1/2 = 1/2 so
x→0 2x
x→0
(x,y)→(0,0)
fy (0, 0) = lim
∆y→0
51. (a) fx (0, 0) = lim
∆x→0
f (∆x, 0) − f (0, 0)
= 0; similarly, fy (0, 0) = 0
∆x
f (∆x, y) − f (0, y)
y(∆x2 − y 2 )
= lim
= −y
∆x→0
∆x→0 ∆x2 + y 2
∆x
(b) fx (0, y) = lim
f (x, ∆y) − f (x, 0)
x(x2 − ∆y 2 )
= lim
=x
∆y→0
∆y→0 x2 + ∆y 2
∆y
fy (x, 0) = lim
fx (0, ∆y) − fx (0, 0)
−∆y
= lim
= −1
∆y→0
∆y→0 ∆y
∆y
fy (∆x, 0) − fy (0, 0)
∆x
= lim
=1
fyx (0, 0) = lim
∆x→0
∆x→0 ∆x
∆x
(c) fxy (0, 0) = lim
(d) No, since fxy and fyx are not continuous.
52. Represent the line segment C that joins A and B by x = x0 + (x1 − x0 )t, y = y0 + (y1 − y0 )t
for 0 ≤ t ≤ 1. Let F (t) = f (x0 + (x1 − x0 )t, y0 + (y1 − y0 )t) for 0 ≤ t ≤ 1; then
f (x1 , y1 ) − f (x0 , y0 ) = F (1) − F (0). Apply the Mean Value Theorem to F (t) on the interval [0,1]
to get [F (1) − F (0)]/(1 − 0) = F 0 (t∗ ), F (1) − F (0) = F 0 (t∗ ) for some t∗ in (0,1) so
f (x1 , y1 ) − f (x0 , y0 ) = F 0 (t∗ ). By the chain rule, F 0 (t) = fx (x, y)(dx/dt) + fy (x, y)(dy/dt) =
fx (x, y)(x1 − x0 ) + fy (x, y)(y1 − y0 ). Let (x∗ , y ∗ ) be the point on C for t = t∗ then
f (x1 , y1 ) − f (x0 , y0 ) = F 0 (t∗ ) = fx (x∗ , y ∗ ) (x1 − x0 ) + fy (x∗ , y ∗ ) (y1 − y0 ).
Exercise Set 15.5
544
53. Let (a, b) be any point in the region, if (x, y) is in the region then by the result of Exercise 52
f (x, y) − f (a, b) = fx (x∗ , y ∗ )(x − a) + fy (x∗ , y ∗ )(y − b) where (x∗ , y ∗ ) is on the line segment joining
(a, b) and (x, y). If fx (x, y) = fy (x, y) = 0 throughout the region then
f (x, y) − f (a, b) = (0)(x − a) + (0)(y − b) = 0, f (x, y) = f (a, b) so f (x, y) is constant on the region.
EXERCISE SET 15.5
1. At P , ∂z/∂x = 48 and ∂z/∂y = −14, tangent plane 48x − 14y − z = 64, normal line x = 1 + 48t,
y = −2 − 14t, z = 12 − t.
2. At P , ∂z/∂x = 14 and ∂z/∂y = −2, tangent plane 14x − 2y − z = 16, normal line x = 2 + 14t,
y = 4 − 2t, z = 4 − t.
3. At P , ∂z/∂x = 1 and ∂z/∂y = −1, tangent plane x − y − z = 0, normal line x = 1 + t, y = −t,
z = 1 − t.
4. At P , ∂z/∂x = −1 and ∂z/∂y = 0, tangent plane x + z = −1, normal line x = −1 − t, y = 0,
z = −t.
5. At P , ∂z/∂x = 0 and ∂z/∂y = 3, tangent plane 3y − z = −1, normal line x = π/6, y = 3t,
z = 1 − t.
6. At P , ∂z/∂x = 1/4 and ∂z/∂y = 1/6, tangent plane 3x + 2y − 12z = −30, normal line x = 4 + t/4,
y = 9 + t/6, z = 5 − t.
7. By implicit differentiation ∂z/∂x = −x/z, ∂z/∂y = −y/z so at P , ∂z/∂x = 3/4 and
∂z/∂y = 0, tangent plane 3x − 4z = −25, normal line x = −3 + 3t/4, y = 0, z = 4 − t.
8. By implicit differentiation ∂z/∂x = (xy)/(4z), ∂z/∂y = x2 /(8z) so at P , ∂z/∂x = 3/8 and
∂z/∂y = −9/16, tangent plane 6x − 9y − 16z = 5, normal line x = −3 + 3t/8, y = 1 − 9t/16,
z = −2 − t.
9. The tangent plane is horizontal if the normal ∂z/∂xi + ∂z/∂yj − k is parallel to k which occurs
when ∂z/∂x = ∂z/∂y = 0.
(a) ∂z/∂x = 3x2 y 2 , ∂z/∂y = 2x3 y; 3x2 y 2 = 0 and 2x3 y = 0 for all (x, y) on the x-axis or y-axis,
and z = 0 for these points, the tangent plane is horizontal at all points on the x-axis or
y-axis.
(b) ∂z/∂x = 2x − y − 2, ∂z/∂y = −x + 2y + 4; solve the system 2x − y − 2 = 0, −x + 2y + 4 = 0,
to get x = 0, y = −2. z = −4 at (0, −2), the tangent plane is horizontal at (0, −2, −4).
10. ∂z/∂x = 6x, ∂z/∂y = −2y, so 6x0 i − 2y0 j − k is normal to the surface at a point (x0 , y0 , z0 ) on
the surface. 6i + 4j − k is normal to the given plane. The tangent plane and the given plane are
parallel if their normals are parallel so 6x0 = 6, x0 = 1 and −2y0 = 4, y0 = −2. z = −1 at (1, −2),
the point on the surface is (1, −2, −1).
11. ∂z/∂x = −6x, ∂z/∂y = −4y so −6x0 i − 4y0 j − k is normal to the surface at a point (x0 , y0 , z0 ) on
the surface. This normal must be parallel to the given line and hence to the vector
−3i + 8j − k which is parallel to the line so −6x0 = −3, x0 = 1/2 and −4y0 = 8, y0 = −2.
z = −3/4 at (1/2, −2). The point on the surface is (1/2, −2, −3/4).
12. (3,4,5) is a point of intersection because it satisfies both equations. Both surfaces have
(3/5)i + (4/5)j − k as a normal so they have a common tangent plane at (3,4,5).
545
Chapter 15
13. df = 2xydx + x2 dy = 0.6 + 0.2 = 0.8, ∆f = (x + ∆x)2 (y + ∆y) − x2 y = (1.1)2 (3.2) − 12 · 3 = 0.872
14. dz = 6xdx − 2dy = −12(0.02) − 2(−0.03) = −0.18,
∆z = 3(−2 + 0.02)2 − 2(4 − 0.03) − (3(−2)2 − 2(4)) = −0.1788
16. 2(4)(−2) − (−2)3 − (0 − 13 ) = −7
15. 3/1 − (−1)/2 = 7/2
17. dz = 3x2 y 2 dx + 2x3 ydy, ∆z = (x + ∆x)3 (y + ∆y)2 − x3 y 2
18. dz = yexy dx + xexy dy, ∆z = e(x+∆x)(y+∆y) − exy
19. dz = 7dx − 2dy
20. dz = (10xy 5 − 2)dx + (25x2 y 4 + 4)dy
21. dz = y/ 1 + x2 y 2 dx + x/ 1 + x2 y 2 dy
22. dz = 2 sec2 (x − 3y) tan(x − 3y)dx − 6 sec2 (x − 3y) tan(x − 3y)dy
23. (a) Let f (x, y) = ex sin y; f (0, 0) = 0, fx (0, 0) = 0, fy (0, 0) = 1, so ex sin y ≈ y
(b) Let f (x, y) =
2x + 1
2x + 1
; f (0, 0) = 1, fx (0, 0) = 2, fy (0, 0) = −1, so
≈ 1 + 2x − y
y+1
y+1
24. f (1, 1) = 1, fx (x, y) = αxα−1 y β , fx (1, 1) = α, fy (x, y) = βxα y β−1 , fy (1, 1) = β, so
xα y β ≈ 1 + α(x − 1) + β(y − 1)
25. dT = Tx dx + Ty dy ≈ 2(−0.02) − (0.02) = −0.06, T ≈ T (1, 3) + dT ≈ 93 − 0.06 = 92.94◦
26. p(104, 103) ≈ p(100, 98) − px (100, 98)(104 − 100) − py (100, 98)(103 − 98)
= 1008 + (−2)4 + (1)5 = 1005 mb
27. f (x, y) = (x2 + y 2 )−1/2 , fx (4, 3) =
−x
4
−y
3
, fy (4, 3) = 2
,
=−
=−
125
125
(x2 + y 2 )3/2
(x + y 2 )3/2
3
1
4
1
p
(−0.08) −
(0.01) = 0.20232; actual value ≈ 0.202334.
≈√
−
2
2
2
2
125
125
4 +3
(3.92) + (3.01)
28. From Exercise 24, x0.5 y 0.3 ≈ 1 + 0.5(x − 1) + 0.3(y − 1), so
(1.05)0.5 (0.97)0.3 ≈ 1 + 0.5(0.05) + 0.3(−0.03) = 1.016, actual value ≈ 1.01537
29. df = (2x + 2y − 4)dx + 2xdy; x = 1, y = 2, dx = 0.01, dy = 0.04 so df = 0.10
30. df = (1/3)x−2/3 y 1/2 dx + (1/2)x1/3 y −1/2 dy; x = 8, y = 9, dx = −0.02, dy = 0.03 so df = 0.005
31. df = −x−2 dx − y −2 dy; x = −1, y = −2, dx = −0.02, dy = −0.04 so df = 0.03
32. df =
33. z =
x
y
dx +
dy; x = 0, y = 2, dx = −0.09, dy = −0.02 so df = −0.09
2(1 + xy)
2(1 + xy)
p
−1/2
−1/2
x2 + y 2 , dz = x x2 + y 2
dx + y x2 + y 2
dy; x = 3, y = 4, dx = 0.2,
dy = −0.04 so dz = 0.088 cm.
34. dV = (2/3)πrhdr + (1/3)πr2 dh; r = 4, h = 20, dr = 0.05, dh = −0.05 so dV = 2.4π ≈ 7.54 in3 .
Exercise Set 15.5
546
35. A = xy, dA = ydx + xdy, dA/A = dx/x + dy/y, |dx/x| ≤ 0.03 and |dy/y| ≤ 0.05,
|dA/A| ≤ |dx/x| + |dy/y| ≤ 0.08 = 8%
36. V = (1/3)πr2 h, dV = (2/3)πrhdr + (1/3)πr2 dh, dV /V = 2(dr/r) + dh/h, |dr/r| ≤ 0.01 and
|dh/h| ≤ 0.04, |dV /V | ≤ 2|dr/r| + |dh/h| ≤ 0.06 = 6%.
p
x
y
x2 + y 2 , dz = p
dx + p
dy,
2
2
2
x +y
x + y2
x
y2
y
x2
dz
dx
dy
= 2
+ 2
,
dx + 2
dy = 2
2
2
2
2
z
x +y
x +y
x +y
x
x +y
y
37. z =
x2
y2
dx
dy
dx
dy
dz
≤ 2
+ 2
, if
≤ r/100 and
≤ r/100 then
2
2
z
x +y x
x +y y
x
y
x2
y2
r
dz
≤ 2
so the percentage error in z is at most about r%.
(r/100) + 2
(r/100) =
2
z
x +y
x + y2
100
p
−1/2
−1/2
x2 + y 2 , dz = x x2 + y 2
dx + y x2 + y 2
dy,
2
2 −1/2
2
2 −1/2
|dx| + y x + y
|dy|; if x = 3, y = 4, |dx| ≤ 0.05, and
|dz| ≤ x x + y
|dy| ≤ 0.05 then |dz| ≤ (3/5)(0.05) + (4/5)(0.05) = 0.07 cm
38. (a) z =
(b) A = (1/2)xy, dA = (1/2)ydx + (1/2)xdy,
|dA| ≤ (1/2)y|dx| + (1/2)x|dy| ≤ 2(0.05) + (3/2)(0.05) = 0.175 cm2 .
R22
R12
R2
dR
39. dR =
2 dR1 +
2 dR2 , R = R + R
1
2
(R1 + R2 )
(R1 + R2 )
dR1
R1
R1
+
R1 + R2
dR2
,
R2
R2
dR1
R1
dR2
dR
≤
+
; if R1 = 200, R2 = 400, |dR1 /R1 | ≤ 0.02, and
R
R1 + R 2 R1
R1 + R2 R 2
|dR2 /R2 | ≤ 0.02 then |dR/R| ≤ (400/600)(0.02) + (200/600)(0.02) = 0.02 = 2%.
40. dP = (k/V )dT − (kT /V 2 )dV , dP/P = dT /T − dV /V ; if dT /T = 0.03 and dV /V = 0.05 then
dP/P = −0.02 so there is about a 2% decrease in pressure.
a
1
da − √
dc; if a = 3, c = 5, |da| ≤ 0.01, and |dc| ≤ 0.01 then
2
2
−a
c c − a2
|dθ| ≤ (1/4)(0.01) + (3/20)(0.01) = 0.004 radians.
41. dθ = √
c2
42. V = πr2 h, dV = 2πrhdr + πr2 dh; r = 2, h = 5, dr = 0.01, and dh = 0.01 so
dV = (20π)(0.01) + (4π)(0.01) = 0.24π, or about 0.754 cm3 .
1 dL 1 dg
πL
dT
π
=
−
; |dL/L| ≤ 0.005 and |dg/g| ≤ 0.001 so
dL − p
dg,
43. dT = p
2
T
2
L
2 g
g L/g
g L/g
|dT /T | ≤ (1/2)(0.005) + (1/2)(0.001) = 0.003 = 0.3%
44. Let h be the height of the building, x the distance to the building, and θ the angle of elevation,
◦
then h = x tan θ, dh = tan θdx + x sec2 θdθ; if x = 100,
√ θ = 60 , |dx| ≤ 1/6 ft, and
3 (1/6) + (100)(4)(π/900) < 1.7 ft.
|dθ| ≤ (0.2)(π/180) = π/900 radians, then |dh| ≤
45. (a) z = xy, dz = ydx + xdy, dz/z = dx/x + dy/y; (r + s)%.
(b) z = x/y, dz = dx/y − xdy/y 2 , dz/z = dx/x − dy/y; (r + s)%.
547
Chapter 15
(c) z = x2 y 3 , dz = 2xy 3 dx + 3x2 y 2 dy, dz/z = 2dx/x + 3dy/y; (2r + 3s)%.
(d) z = x3 y 1/2 , dz = 3x2 y 1/2 dx + x3 dy/ 2y 1/2 , dz/z = 3dx/x + (1/2)dy/y; (3r + s/2)%.
k
k
k
k
; at a point a, b,
on the surface, − 2 , − 2 , −1 and hence bk, ak, a2 b2 is
46. z =
xy
ab
a b ab
normal to the surface so the tangent plane is bkx + aky + a2 b2 z = 3abk. The plane cuts the x,
3k
y, and z-axes at the points 3a, 3b, and
, respectively, so the volume of the tetrahedron that is
ab
9
1 3k
1
(3a)(3b) = k, which does not depend on a and b.
formed is V =
3 ab
2
2
47. (a) 2t + 7 = (−1 + t)2 + (2 + t)2 , t2 = 1, t = ±1 so the points of intersection are (−2, 1, 5) and
(0, 3, 9).
(b) ∂z/∂x = 2x, ∂z/∂y = 2y so at (−2, 1, 5) the vector n = −4i + 2j − k is normal to the surface.
v = i + j+2k is parallel to the line; n · v = −4 so the cosine of the acute angle is
√
√ √
[n · (−v)]/(knk k − vk) = 4/ 21 6 = 4/ 3 14 . Similarly, at (0,3,9) the vector
n = 6j − k is normal to the surface, n · v = 4 so the cosine of the acute angle is
√
√ √
4/ 37 6 = 4/ 222.
48. z = xf (u) where u = x/y, ∂z/∂x = xf 0 (u)∂u/∂x + f (u) = (x/y)f 0 (u) + f (u) = uf 0 (u) + f (u),
∂z/∂y = xf 0 (u)∂u/∂y = −(x2 /y 2 )f 0 (u) = −u2 f 0 (u). If (x0 , y0 , z0 ) is on the surface then, with
u0 = x0 /y0 , [u0 f 0 (u0 ) + f (u0 )] i − u20 f 0 (u0 ) j − k is normal to the surface so the tangent plane is
[u0 f 0 (u0 ) + f (u0 )] x − u20 f 0 (u0 )y − z = [u0 f 0 (u0 ) + f (u0 )]x0 − u20 f 0 (u0 )y0 − z0
x2
x0 0
=
f (u0 ) + f (u0 ) x0 − 20 f 0 (u0 ) y0 − z0
y0
y0
= x0 f (u0 ) − z0 = 0
so all tangent planes pass through the origin.
49. Use implicit differentiation to get ∂z/∂x = −c2 x/ a2 z , ∂z/∂y = −c2 y/ b2 z . At (x0 , y0 , z0 ),
z0 6= 0, a normal to the surface is − c2 x0 / a2 z0 i − c2 y0 / b2 z0 j − k so the tangent plane is
c2 y0
c2 x20
c2 y02
x0 x y0 y z0 z
x20
y02
z02
c2 x0
x
−
y
−
z
=
−
−
−
z
,
+
+
=
+
+
=1
0
a2 z0
b 2 z0
a2 z0
b 2 z0
a2
b2
c2
a2
b2
c2
50. ∂z/∂x = 2x/a2 , ∂z/∂y = 2y/b2 . At (x0 , y0 , z0 ) the vector 2x0 /a2 i + 2y0 /b2 j − k is normal
to the surface so the tangent plane is 2x0 /a2 x + 2y0 /b2 y − z = 2x20 /a2 + 2y02 /b2 − z0 , but
z0 = x20 /a2 + y02 /b2 so 2x0 /a2 x + 2y0 /b2 y − z = 2z0 − z0 = z0 , 2x0 x/a2 + 2y0 y/b2 = z + z0
−
51. n1 = fx (x0 , y0 ) i+fy (x0 , y0 ) j − k and n2 = gx (x0 , y0 ) i+gy (x0 , y0 ) j − k are normal, respectively,
to z = f (x, y) and z = g(x, y) at P ; n1 and n2 are perpendicular if and only if n1 · n2 = 0,
fx (x0 , y0 ) gx (x0 , y0 ) + fy (x0 , y0 ) gy (x0 , y0 ) + 1 = 0,
fx (x0 , y0 ) gx (x0 , y0 ) + fy (x0 , y0 ) gy (x0 , y0 ) = −1.
x0
x0
y0
j − k; similarly n2 = − p 2
i− p 2
j − k;
2
+
+
x0 + y0
x0 + y02
p
since a normal to the sphere is N = x0 i + y0 j + z0 k, and n1 · N = x20 + y02 − z0 = 0,
p
n2 · N = − x20 + y02 − z0 = 0, the result follows.
52. n1 = fx i + fy j − k = p
x20
y02
y0
i+ p
x20
y02
Exercise Set 15.6
548
EXERCISE SET 15.6
1. f increases the most in the direction of III.
2. The contour lines are closer at P , so the function is increasing more rapidly there, hence ∇f is
larger at P .
4. ∇z = −4e−3y sin 4xi − 3e−3y cos 4xj
3. ∇z = 4i − 8j
5. ∇z =
y
x
i+ 2
j
x2 + y 2
x + y2
6. ∇z = e−5x sec x2 y
2xy tan x2 y − 5 i + x2 tan x2 yj
7. ∇f (x, y) = 3(2x + y) x2 + xy
8. ∇f (x, y) = −x x2 + y 2
−3/2
2
i + 3x x2 + xy
i − y x2 + y 2
2
−3/2
j, ∇f (−1, −1) = −36i − 12j
j, ∇f (3, 4) = −(3/125)i − (4/125)j
9. ∇f (x, y) = [y/(x + y)]i + [y/(x + y) + ln(x + y)]j, ∇f (−3, 4) = 4i + 4j
10. ∇f (x, y) = 3y 2 tan2 x sec2 xi + 2y tan3 xj, ∇f (π/4, −3) = 54i − 6j
11. ∇f (x, y) = (3y/2)(1 + xy)1/2 i + (3x/2)(1 + xy)1/2 j, ∇f (3, 1) = 3i + 9j,
√
√
Du f = ∇f · u = 12/ 2 = 6 2
12. ∇f (x, y) = 2ye2xy i + 2xe2xy j, ∇f (4, 0) = 8j, Du f = ∇f · u = 32/5
√
13. ∇f (x, y) = 2x/ 1 + x2 + y i + 1/ 1 + x2 + y j, ∇f (0, 0) = j, Du f = −3/ 10
14. ∇f (x, y) = − (c + d)y/(x − y)2 i + (c + d)x/(x − y)2 j,
∇f (3, 4) = −4(c + d)i + 3(c + d)j, Du f = −(7/5)(c + d)
15. ∇f (x, y) = 12x2 y 2 i + 8x3 yj, ∇f (2, 1) = 48i + 64j, u = (4/5)i − (3/5)j, Du f = ∇f · u = 0
√
√
16. ∇f (x, y) = (2x − 3y)i + −3x + 12y 2 j, ∇f (−2, 0) = −4i + 6j, u = (i + 2j)/ 5, Du f = 8/ 5
√
√
17. ∇f (x, y) = y 2 /x i + 2y ln xj, ∇f (1, 4) = 16i, u = (−i + j)/ 2, Du f = −8 2
√
√
√
18. ∇f (x, y) = ex cos yi − ex sin yj, ∇f (0, π/4) = (i − j)/ 2, u = (5i − 2j)/ 29, Du f = 7/ 58
19. ∇f (x, y) = − y/ x2 + y 2 i + x/ x2 + y 2 j,
√
√
∇f (−2, 2) = −(i + j)/4, u = −(i + j)/ 2, Du f = 2/4
√
√
20. ∇f (x, y) = (ey − yex ) i + (xey − ex )j, ∇f (0, 0) = i − j, u = (5i − 2j)/ 29, Du f = 7/ 29
21. ∇f (x, y) = (y/2)(xy)−1/2 i + (x/2)(xy)−1/2 j, ∇f (1, 4) = i + (1/4)j,
√
√
u = cos θi + sin θj = (1/2)i +
3/2 j, Du f = 1/2 + 3/8
22. ∇f (x, y) = [2y/(x + y)2 ]i − [2x/(x + y)2 ]j, ∇f (−1, −2) = −(4/9)i + (2/9)j, u = j, Du f = 2/9
549
Chapter 15
√
√
23. ∇f (x, y) = 2 sec2 (2x + y)i + sec2 (2x + y)j, ∇f (π/6, π/3) = 8i + 4j, u = (i − j)/ 2, Du f = 2 2
24. ∇f (x, y) = cosh x cosh yi + sinh x sinh yj, ∇f (0, 0) = i, u = −i, Du f = −1
25. f (1, 2) = 3, level curve 4x − 2y + 3 = 3, 2x − y = 0;
∇f (x, y) = 4i − 2j
y
∇f (1, 2) = 4i − 2j
(1, 2)
4i – 2j
26. f (−2, 2) = 1/2, level curve y/x2 = 1/2, y = x2 /2 for x 6= 0.
∇f (x, y) = − 2y/x3 i + 1/x2 j
∇f (−2, 2) = (1/2)i + (1/4)j
x
y
1
1
i+ j
2
4
(−2, 2)
x
27. f (−2, 0) = 4, level curve x2 + 4y 2 = 4, x2 /4 + y 2 = 1.
∇f (x, y) = 2xi + 8yj
∇f (−2, 0) = −4i
y
1
x
2
-4i
28. f (2, −1) = 3, level curve x2 − y 2 = 3.
y
∇f (x, y) = 2xi − 2yj
∇f (2, −1) = 4i + 2j
4i + 2j
x
(2, −1)
√
√
29. ∇f (x, y) = 12x2 y 2 i + 8x3 yj, ∇f (−1, 1) = 12i − 8j, u = (3i − 2j)/ 13, k∇f (−1, 1)k = 4 13
Exercise Set 15.6
550
√
√
30. ∇f (x, y) = 3i − (1/y)j, ∇f (2, 4) = 3i − (1/4)j, u = (12i − j)/ 145, k∇f (2, 4)k = 145/4
31. ∇f (x, y) = x x2 + y 2
−1/2
i + y x2 + y 2
−1/2
j,
∇f (4, −3) = (4i − 3j)/5, u = (4i − 3j)/5, k∇f (4, −3)k = 1
32. ∇f (x, y) = y(x + y)−2 i − x(x + y)−2 j, ∇f (0, 2) = (1/2)i, u = i, k∇f (0, 2)k = 1/2
√
√
33. ∇f (x, y) = −2xi − 2yj, ∇f (−1, −3) = 2i + 6j, u = −(i + 3j)/ 10, −k∇f (−1, −3)k = −2 10
√
√
34. ∇f (x, y) = yexy i + xexy j; ∇f (2, 3) = e6 (3i + 2j), u = −(3i + 2j)/ 13, −k∇f (2, 3)k = − 13e6
35. ∇f (x, y) = −3 sin(3x − y)i + sin(3x − y)j,
√
√
√
∇f (π/6, π/4) = (−3i + j)/ 2, u = (3i − j)/ 10, −k∇f (π/6, π/4)k = − 5
r
r
√
x
x+y
x+y
y
i
−
j, ∇f (3, 1) = ( 2/16)(i − 3j),
2
2
(x + y)
x−y
(x + y)
x−y
√
√
u = −(i − 3j)/ 10, −k∇f (3, 1)k = − 5/8
36. ∇f (x, y) =
−→
√
37. ∇f (x, y) = y(x + y)−2 i − x(x + y)−2 j, ∇f (1, 0) = −j, P Q = −2i − j, u = (−2i − j)/ 5,
√
Du f = 1/ 5
38. ∇f (x, y) = −e−x sec yi + e−x sec y tan yj,
−→
√
√
∇f (0, π/4) = 2(−i + j), P O = −(π/4)j, u = −j, Du f = − 2
xey
yey
√
j, ∇f (1, 1) = (e/2)(i + 3j), u = −j, Du f = −3e/2
xyey + √
39. ∇f (x, y) = √ i +
2 xy
2 xy
40. ∇f (x, y) = −y(x + y)−2 i + x(x + y)−2 j, ∇f (2, 3) = (−3i + 2j)/25, if Du f = 0√then u and ∇f are
orthogonal, by inspection 2i + 3j is orthogonal to ∇f (2, 3) so u = ±(2i + 3j)/ 13.
41. ∇f (x, y) = 8xyi + 4x2 j, ∇f (1, −2) = −16i + 4j is normal to the level curve through P so
√
u = ±(−4i + j)/ 17.
42. ∇f (x, y) = (6xy − y)i + 3x2 − x j, ∇f (2, −3) = −33i + 10j is normal to the level curve through
√
P so u = ±(−33i + 10j)/ 1189.
43. Solve the system (3/5)fx (1, 2) − (4/5)fy (1, 2) = −5, (4/5)fx (1, 2) + (3/5)fy (1, 2) = 10 for
fx (1, 2) and fy (1, 2) to get fx (1, 2) = 5, fy (1, 2) = 10. For (c), ∇f (1, 2) = 5i + 10j,
√
√
u = (−i − 2j)/ 5, Du f = −5 5.
−→
√
√
44. ∇f (−5, 1) = −3i + 2j, P Q = i + 2j, u = (i + 2j)/ 5, Du f = 1/ 5
√
√
45. ∇f (4, −5) = 2i − j, u = (5i + 2j)/ 29, Du f = 8/ 29
46. Let u = u1 i + u2 j where u21 + u22 = 1, but Du f = ∇f · u = u1 − 2u2 = −2 so u1 = 2u2 − 2,
2
(2u2 − 2) + u22 = 1, 5u22 − 8u2 + 3 = 0, u2 = 1 or u2 = 3/5 thus u1 = 0 or u1 = −4/5; u = j or
3
4
u = − i + j.
5
5
551
Chapter 15
47. (a) At (1, 2) the steepest ascent seems to be in the direction i + j and the slope in that direction
√
√
1
1
seems to be 0.5/( 2/2) = 1/ 2, so ∇f ≈ i + j, which has the required direction and
2
2
magnitude.
y
(b)
5
−∇f (4, 4)
x
5
48. (a)
200
300 400
500
100
0 ft
P
Depart from each contour line in a direction orthogonal to that contour line, as an approximation to the optimal path.
(b)
500
200
300 400
100
0 ft
P
At the top there is no contour line, so head for the nearest contour line. From then on depart
from each contour line in a direction orthogonal to that countour line, as in Part (a).
49. ∇z = 6xi − 2yj, k∇zk =
50. ∇z = 3i + 2yj, k∇zk =
p
36x2 + 4y 2 = 6 if 36x2 + 4y 2 = 36; all points on the ellipse 9x2 + y 2 = 9.
p
8
4y
8
=
9 + 4y 2 , so ∇k∇zk = p
=√
2
5
9
+
16
9 + 4y
√
51. r = ti − t2 j, dr/dt = i − 2tj = i − 4j at the point (2, −4), u = (i − 4j)/ 17;
√
∇z = 2xi + 2yj = 4i − 8j at (2, −4), hence dz/ds = Du z = ∇z · u = 36/ 17.
52. (a) ∇T (x, y) =
y 1 − x2 + y 2
2
y2 )
i+
x 1 + x2 − y 2
(1 + x2 +
(1 + x2 +
√
Du T = 1/ 9 5
√
(b) u = −(i + j)/ 2, opposite to ∇T (1, 1)
2
y2 )
√
j, ∇T (1, 1) = (i + j)/9, u = (2i − j)/ 5,
Exercise Set 15.6
552
53. (a) ∇V (x, y) = −2e−2x cos 2yi − 2e−2x sin 2yj, E = −∇V (π/4, 0) = 2e−π/2 i
(b) V (x, y) decreases most rapidly in the direction of −∇V (x, y) which is E.
54. ∇z = −4xi − 8yj, if x = −20 and y = 5 then ∇z = 80i − 40j.
(a) u = −i points due west, Du z = −80, the climber will descend because z is decreasing.
√
√
√
(b) u = (i + j)/ 2 points northeast, Du z = 20 2, the climber will ascend at the rate of 20 2
ft per ft of travel in the xy−plane.
(c) The climber will travel
√ a level path in a direction perpendicular to√∇z = 80i − 40j, by
(i + 2j)/ 5 makes an angle of
inspection ±(i + 2j)/ 5 are unit vectors in these directions;
√
tan−1 (1/2) ≈ 27◦ with the positive y-axis so −(i+2j)/ 5 makes the same angle with the
negative y-axis. The compass direction should be N 27◦ E or S 27◦ W.
55. (a) ∇r = p
(b) ∇f (r) =
x
x2
+
y2
y
i+ p
j = r/r
2
x + y2
∂f (r)
∂r
∂r
∂f (r)
i+
j = f 0 (r) i + f 0 (r) j = f 0 (r)∇r
∂x
∂y
∂x
∂y
(1 − 3r) −3r
e r
56. (a) ∇ re−3r =
r
f 0 (r)
3
3
(b) 3r2 r =
r so f 0 (r) = 3r3 , f (r) = r4 + C, f (2) = 12 + C = 1, C = −11; f (r) = r4 − 11
r
4
4
57. ur = cos θi + sin θj, uθ = − sin θi + cos θj,
∂z
∂z
1 ∂z
1 ∂z
∂z
∂z
∇z =
i+
j=
cos θ −
sin θ i +
sin θ +
cos θ j
∂x
∂y
∂r
r ∂θ
∂r
r ∂θ
=
1 ∂z
∂z
1 ∂z
∂z
(cos θi + sin θj) +
(− sin θi + cos θj) =
ur +
uθ
∂r
r ∂θ
∂r
r ∂θ
58. (a) ∇(f + g) = (fx + gx ) i + (fy + gy ) j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g
(b) ∇(cf ) = (cfx ) i + (cfy ) j = c (fx i + fy j) = c∇f
(c) ∇(f g) = (f gx + gfx ) i + (f gy + gfy ) j = f (gx i + gy j) + g (fx i + fy j) = f ∇g + g∇f
gfy − f gy
g (fx i + fy j) − f (gx i + gy j)
g∇f − f ∇g
gfx − f gx
i+
j=
=
2
2
2
g
g
g
g2
(e) ∇ (f n ) = nf n−1 fx i + nf n−1 fy j = nf n−1 (fx i + fy j) = nf n−1 ∇f
(d) ∇(f /g) =
dy
dx
= −8kx,
= −2ky. Divide and solve
dt
dt
4
−8t
−2t
to get y = 256x; one parametrization is x(t) = e , y(t) = 4e .
59. r0 (t) = v(t) = k(x, y)∇T = −8k(x, y)xi − 2k(x, y)yj;
60. r0 (t) = v(t) = k∇T = −2k(x, y)xi−4k(x, y)yj. Divide and solve to get y =
tion is x(t) = 5e−2t , y(t) = 3e−4t .
3 2
x ; one parametriza25
553
Chapter 15
y
61.
62.
5
4
C = –10
C = –5
(5, 3)
T = 80
T = 95 T = 90
C = –15
-6
C=0
6
T = 97
x
-3
-4
3
-5
z
63. (a)
y
x
(c) ∇f = [2x − 2x(x2 + 3y 2 )]e−(x
2
+y 2 )
i + [6y − 2y(x2 + 3y 2 )]e−(x
2
+y 2 )
j
(d) ∇f = 0 if x = y = 0 or x = 0, y = ±1 or x = ±1, y = 0.
64. dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)
= (∂z/∂xi + ∂z/∂yj) · (dx/dti + dy/dtj) = ∇z · r0 (t)
65. ∇f (x, y) = fx (x, y)i + fy (x, y)j, if ∇f (x, y) = 0 throughout the region then
fx (x, y) = fy (x, y) = 0 throughout the region, the result follows from Exercise 51, Section 15.4.
66. Let u1 and u2 be nonparallel unit vectors for which the directional derivative is zero. Let u be
any other unit vector, then u = c1 u1 + c2 u2 for some choice of scalars c1 and c2 ,
Du f (x, y) = ∇f (x, y) · u = c1 ∇f (x, y) · u1 + c2 ∇f (x, y) · u2
= c1 Du1 f (x, y) + c2 Du2 f (x, y) = 0.
EXERCISE SET 15.7
1. 165t32
4.
1 − 512t5 − 2560t5 ln t
√
2t 1 + ln t − 512t5 ln t
2.
3 − (4/3)t−1/3 − 24t−7
3t − 2t2/3 + 4t−6
5. 3264
3. −2t cos t2
6. 0
7. ∇f (x, y, z) = 20x4 y 2 z 3 i + 8x5 yz 3 j + 12x5 y 2 z 2 k, ∇f (2, −1, 1) = 320i − 256j + 384k, Du f = −320
Exercise Set 15.7
554
8. ∇f (x, y, z) = yzexz i + exz j + (xyexz + 2z) k, ∇f (0, 2, 3) = 6i + j + 6k, Du f = 45/7
4y
6z
2x
i+ 2
j+ 2
k,
2
2
2
2
+ 2y + 3z
x + 2y + 3z
x + 2y 2 + 3z 2
∇f (−1, 2, 4) = (−2/57)i + (8/57)j + (24/57)k, Du f = −314/741
9. ∇f (x, y, z) =
x2
10. ∇f (x, y, z) = yz cos xyzi + xz cos xyzj + xy cos xyzk,
√
√
√
∇f (1/2, 1/3, π) = (π 3/6)i + (π 3/4)j + ( 3/12)k, Du f = (1 − π)/12
11. ∇f (x, y, z) = 3x2 z − 2xy i − x2 j + x3 + 2z k, ∇f (2, −1, 1) = 16i − 4j + 10k,
√
√
u = (3i − j + 2k)/ 14, Du f = 72/ 14
−1/2
−1/2
i + j − z x2 + z 2
k, ∇f (−3, 1, 4) = (3/5)i + j−(4/5)k,
12. ∇f (x, y, z) = −x x2 + z 2
u = (2i − 2j − k)/3, Du f = 0
z−x
y+x
1
i−
j+
k, ∇f (1, 0, −3) = (1/3)i + (4/9)j + (1/9)k,
2
z+y
(z + y)
(z + y)2
u = (−6i + 3j − 2k)/7, Du f = −8/63
13. ∇f (x, y, z) = −
14. ∇f (x, y, z) = ex+y+3z (i + j+3k), ∇f (−2, 2, −1) = e−3 (i + j+3k), u = (20i − 4j + 5k)/21,
Du f = (31/21)e−3
√
√
15. ∇f (1, 1, −1) = 3i − 3j, u = (i − j)/ 2, k∇f (1, 1, −1)k = 3 2
√
√
16. ∇f (0, −3, 0) = (i−3j+4k)/6, u = (i−3j+4k)/ 26, k∇f (0, −3, 0)k = 26/6
√
√
17. ∇f (1, 2, −2) = (−i + j)/2, u = (−i + j)/ 2, k∇f (1, 2, −2)k = 1/ 2
√
√
18. ∇f (4, 2, 2) = (i − j − k)/8, u = (i − j − k)/ 3, k∇f (4, 2, 2)k = 3/8
√
√
19. ∇f (5, 7, 6) = −i + 11j − 12k, u = (i − 11j + 12k)/ 266, −k∇f (5, 7, 6)k = − 266
√
√
20. ∇f (0, 1, π/4) = 2 2(i − k), u = −(i − k)/ 2, −k∇f (0, 1, π/4)k = −4
−→
√
√
21. ∇f (2, 1, −1) = −i + j − k. P Q = −3i + j + k, u = (−3i + j + k)/ 11, Du f = 3/ 11
22. ∇f (−1, −2, 1) = 13i + 5j − 20k, u = −k, Du f = 20
23. Let u be the unit vector in the direction of a, then
Du f (3, −2, 1) = ∇f (3, −2, 1) · u = k∇f (3, −2, 1)k cos θ = 5 cos θ = −5, cos θ = −1, θ = π so
∇f (3, −2, 1) is oppositely directed to u; ∇f (3, −2, 1) = −5u = −10/3i + 5/3j + 10/3k.
√
√
24. (a) ∇T (1, 1, 1) = (i + j + k)/8, u = −(i + j + k)/ 3, Du T = − 3/8
√
√
(c)
3/8
(b) (i + j + k)/ 3
25. (a) f (x, y, z) = x2 + y 2 + 4z 2 , ∇f = 2xi + 2yj + 8zk, ∇f (2, 2, 1) = 4i + 4j + 8k,
n = i + j + 2k, x + y + 2z = 6
(b) r(t) = 2i + 2j + k + t(i + j + 2k), x(t) = 2 + t, y(t) = 2 + t, z(t) = 1 + 2t
√
n·k
2
= √ , θ ≈ 35.26◦
(c) cos θ =
knk
3
555
Chapter 15
26. (a) f (x, y, z) = xz − yz 3 + yz 2 , n = ∇f (2, −1, 1) = i + 3k; tangent plane x + 3z = 5
(b) normal line x = 2 + t, y = −1, z = 1 + 3t
(c) cos θ =
3
n·k
= √ , θ ≈ 18.43◦
knk
10
27. Set f (x, y) = z + x − z 4 (y − 1), then f (x, y, z) = 0, n = ±∇f (3, 5, 1) = ±(i − j − 19k),
1
(i − j − 19k)
unit vectors ±
363
28. f (x, y, z) = sin xz − 4 cos yz, ∇f (π, π, 1) = −i − πk; unit vectors ± √
1
(i + πk)
1 + π2
29. f (x, y, z) = x2 + y 2 + z 2 , if (x0 , y0 , z0 ) is on the sphere then ∇f (x0 , y0 , z0 ) = 2 (x0 i + y0 j + z0 k)
is normal to the sphere at (x0 , y0 , z0 ), the normal line is x = x0 + x0 t, y = y0 + y0 t, z = z0 + z0 t
which passes through the origin when t = −1.
30. f (x, y, z) = 2x2 + 3y 2 + 4z 2 , if (x0 , y0 , z0 ) is on the ellipsoid then
∇f (x0 , y0 , z0 ) = 2 (2x0 i + 3y0 j + 4z0 k) is normal there and hence so is n1 = 2x0 i + 3y0 j + 4z0 k;
n1 must be parallel to n2 = i − 2j + 3k which is normal to the given plane so n1 = cn2
for some constant c. Equate corresponding components to get x0 = c/2, y0 = −2c/3, and
z0 = 3c/4; substitute into the equation of the ellipsoid yields 2 c2 /4 + 3 4c2 /9 + 4 9c2 /16 = 9,
√
√
√
√
c2 = 108/49, c = ±6 3/7. The points on the ellipsoid are 3 3/7, −4 3/7, 9 3/14 and
√
√
√
−3 3/7, 4 3/7, −9 3/14 .
31. f (x, y, z) = x2 + y 2 − z 2 , if (x0 , y0 , z0 ) is on the surface then ∇f (x0 , y0 , z0 ) = 2 (x0 i + y0 j − z0 k)
−→
is normal there and hence so is n1 = x0 i + y0 j − z0 k; n1 must be parallel to P Q = 3i + 2j − 2k so
−→
n1 = c P Q for some constant c. Equate components to get x0 = 3c, y0 = 2c and z0 = 2c which
when substituted into the equation of the surface yields 9c2 + 4c2 − 4c2 = 1, c2 = 1/9, c = ±1/3
so the points are (1, 2/3, 2/3) and (−1, −2/3, −2/3).
32. f1 (x, y, z) = 2x2 + 3y 2 + z 2 , f2 (x, y, z) = x2 + y 2 + z 2 − 6x − 8y − 8z + 24,
n1 = ∇f1 (1, 1, 2) = 4i + 6j + 4k, n2 = ∇f2 (1, 1, 2) = −4i − 6j − 4k, n1 = −n2 so n1 and n2 are
parallel.
33. n1 = 2i − 2j − k, n2 = 2i − 8j + 4k, n1 × n2 = 16i − 10j − 12k is tangent to the line, so
x(t) = 1 + 8t, y(t) = −1 + 5t, z(t) = 2 + 6t
p
4 3
x2 + y 2 −z, n1 = ∇f (4, 3, 5) = i+ j−k, n2 = i+2j+2k, n1 ×n2 = (16i−13j+5k)/5
5 5
is tangent to the line, x(t) = 4 + 16t, y(t) = 3 − 13t, z(t) = 5 + 5t
34. f (x, y, z) =
35. f (x, y, z) = x2 + z 2 − 25, g(x, y, z) = y 2 + z 2 − 25, n1 = ∇f (3, −3, 4) = 6i + 8k,
n2 = ∇g(3, −3, 4) = −6j + 8k, n1 × n2 = 48i − 48j − 36k is tangent to the line,
x(t) = 3 + 4t, y(t) = −3 − 4t, z(t) = 4 − 3t
36. (a) f (x, y, z) = z − 8 + x2 + y 2 , g(x, y, z) = 4x + 2y − z, n1 = 4j + k, n2 = 4i + 2j − k,
n1 × n2 = −6i + 4j − 16k is tangent to the line, x(t) = 3t, y(t) = 2 − 2t, z(t) = 4 + 8t
37. dw = 3x2 y 2 zdx + 2x3 yzdy + x3 y 2 dz, ∆w = (x + ∆x)3 (y + ∆y)2 (z + ∆z) − x3 y 2 z
38. dw = yzexyz dx + xzexyz dy + xyexyz dz, ∆w = e(x+∆x)(y+∆y)(z+∆z) − exyz
Exercise Set 15.7
556
39. dw = 8dx − 3dy + 4dz
40. dw = 8xy 3 z 7 − 3y dx + 12x2 y 2 z 7 − 3x dy + 28x2 y 3 z 6 + 1 dz
41. dw =
yz
xz
xy
dx +
dy +
dz
2
2
2
2
2
2
1+x y z
1+x y z
1 + x2 y 2 z 2
1
1
1
42. dw = √ dx + √ dy + √ dz
2 y
2 x
2 z
43. df = 2y 2 z 3 dx + 4xyz 3 dy + 6xy 2 z 2 dz
= 2(−1)2 (2)3 (−0.01) + 4(1)(−1)(2)3 (−0.02) + 6(1)(−1)2 (2)2 (0.02) = 0.96
44. df =
xz(x + z)
xy(x + y)
yz(y + z)
dx +
dy +
dz
2
2
(x + y + z)
(x + y + z)
(x + y + z)2
= (−16)(−0.04) + (−12)(0.02) + (−6)(−0.03) = 0.58
45. V = `wh, dV = wh d` + `h dw + `w dh,
|dV | ≤ wh|d`| + `h|dw| + `w|dh| ≤ (4)(5)(0.05) + (3)(5)(0.05) + (3)(4)(0.05) = 2.35 cm3
1
= R2 /R12 , similarly
R12 (1/R1 + 1/R2 + 1/R3 )2
dR
dR1
dR2
dR3
= (R/R1 )
+ (R/R2 )
+ (R/R3 )
,
∂R/∂R2 = R2 /R22 and ∂R/∂R3 = R2 /R32 so
R
R1
R2
R3
dR1
dR2
dR3
dR
≤ (R/R1 )
+ (R/R2 )
+ (R/R3 )
R
R1
R2
R3
46. R = 1/ (1/R1 + 1/R2 + 1/R3 ), ∂R/∂R1 =
≤ (R/R1 ) (0.10) + (R/R2 ) (0.10) + (R/R3 ) (0.10)
= R (1/R1 + 1/R2 + 1/R3 ) (0.10) = (1)(0.10) = 0.10 = 10%
1
1
1
b sin θda + a sin θdb + ab cos θdθ,
2
2
2
1
1
1
|dA| ≤ b sin θ|da| + a sin θ|db| + ab cos θ|dθ|
2
2
2
47. dA =
√
1
1
1
(50)(1/2)(1/2) + (40)(1/2)(1/4) + (40)(50)
3/2 (π/90)
2
2
2
√
= 35/4 + 50π 3/9 ≈ 39 ft2
≤
48. V = `wh, dV = whd` + `hdw + `wdh, |dV /V | ≤ |d`/`| + |dw/w| + |dh/h| ≤ 3(r/100) = 3r%
49. ∂f /∂v = 8vw3 x4 y 5 , ∂f /∂w = 12v 2 w2 x4 y 5 , ∂f /∂x = 16v 2 w3 x3 y 5 , ∂f /∂y = 20v 2 w3 x4 y 4
50. ∂w/∂r = cos st + ueu cos ur, ∂w/∂s = −rt sin st,
∂w/∂t = −rs sin st, ∂w/∂u = reu cos ur + eu sin ur
2
51. ∂f /∂v1 = 2v1 / v32 + v42 , ∂f /∂v2 = −2v2 / v32 + v42 , ∂f /∂v3 = −2v3 v12 − v22 / v32 + v42 ,
2
∂f /∂v4 = −2v4 v12 − v22 / v32 + v42
52.
∂V
∂V
∂V
∂V
= 2xe2x−y + e2x−y ,
= −xe2x−y + w,
= w2 ezw ,
= wzezw + ezw + y
∂x
∂y
∂z
∂w
557
Chapter 15
53. (a) 0
(b) 0
yw
(e) 2(yw + 1)e
(c) 0
(d) 0
yw
sin z cos z
(f ) 2xw(yw + 2)e
sin z cos z
54. 128, −512, 32, 64/3
55. ∂z/∂r = (dz/dx)(∂x/∂r) = 2r cos2 θ/ r2 cos2 θ + 1 ,
∂z/∂θ = (dz/dx)(∂x/∂θ) = −2r2 sin θ cos θ/ r2 cos2 θ + 1
56. ∂u/∂x = (∂u/∂r)(dr/dx) + (∂u/∂t)(∂t/∂x)
= s2 ln t (2x) + rs2 /t y 3 = x(4y + 1)2 1 + 2 ln xy 3
∂u/∂y = (∂u/∂s)(ds/dy) + (∂u/∂t)(∂t/∂y)
= (2rs ln t)(4) + rs2 /t 3xy 2 = 8x2 (4y + 1) ln xy 3 + 3x2 (4y + 1)2 /y
57. ∂w/∂ρ = 2ρ 4 sin2 φ + cos2 φ , ∂w/∂φ = 6ρ2 sin φ cos φ, ∂w/∂θ = 0
58.
∂w ∂w dy
∂w dz
dw
1
=
+
+
= 3y 2 z 3 + (6xyz 3 )(6x) + 9xy 2 z 2 √
dx
∂x
∂y dx
∂z dx
2 x−1
√
9
2
2
3/2
2
2
3/2
= 3(3x + 2) (x − 1) + 36x (3x + 2)(x − 1) + x(3x2 + 2)2 x − 1
2
√
3
= (3x2 + 2)(39x3 − 30x2 + 10x − 4) x − 1
2
59. (a) V = `wh,
∂V d` ∂V dw ∂V dh
d`
dw
dh
dV
=
+
+
= wh + `h
+ `w
dt
∂` dt
∂w dt
∂h dt
dt
dt
dt
= (3)(6)(1) + (2)(6)(2) + (2)(3)(3) = 60 in3 /s
p
(b) D = `2 + w2 + h2 ; dD/dt = (`/D)d`/dt + (w/D)dw/dt + (h/D)dh/dt
= (2/7)(1) + (3/7)(2) + (6/7)(3) = 26/7 in/s
60. (a) ∂A/∂a = (1/2)b sin θ = (1/2)(10)
√
√
3/2 = 5 3/2
(b) ∂A/∂θ = (1/2)ab cos θ = (1/2)(5)(10)(1/2) = 25/2
(c) b = (2A csc θ)/a, ∂b/∂a = −(2A csc θ)/a2 = −b/a = −2
61. Let z = f (u) where u = x + 2y; then ∂z/∂x = (dz/du)(∂u/∂x) = dz/du,
∂z/∂y = (dz/du)(∂u/∂y) = 2dz/du so 2∂z/∂x − ∂z/∂y = 2dz/du − 2dz/du = 0
62. Let z = f (u) where u = x2 + y 2 ; then ∂z/∂x = (dz/du)(∂u/∂x) = 2x dz/du,
∂z/∂y = (dz/du)(∂u/∂y) = 2ydz/du so y ∂z/∂x − x∂z/∂y = 2xydz/du − 2xydz/du = 0
63. ∂w/∂x = (dw/dρ)(∂ρ/∂x) = (x/ρ)dw/dρ, similarly ∂w/∂y = (y/ρ)dw/dρ and
∂w/∂z = (z/ρ)dw/dρ so (∂w/∂x)2 + (∂w/∂y)2 + (∂w/∂z)2 = (dw/dρ)2
64. Let w = f (r, s, t) where r = x − y, s = y − z, t = z − x;
∂w/∂x = (∂w/∂r)(∂r/∂x) + (∂w/∂t)(∂t/∂x) = ∂w/∂r − ∂w/∂t, similarly
∂w/∂y = −∂w/∂r + ∂w/∂s and ∂w/∂z = −∂w/∂s + ∂w/∂t so ∂w/∂x + ∂w/∂y + ∂w/∂z = 0
65. ∂w/∂ρ = sin φ cos θ∂w/∂x + sin φ sin θ∂w/∂y + cos φ∂w/∂z
∂w/∂φ = ρ cos φ cos θ∂w/∂x + ρ cos φ sin θ∂w/∂y − ρ sin φ∂w/∂z
∂w/∂θ = −ρ sin φ sin θ∂w/∂x + ρ sin φ cos θ∂w/∂y
Exercise Set 15.7
558
66.
∂F ∂z
∂z
∂F/∂x ∂F
∂F ∂z
∂z
∂F/∂y
∂F
+
= 0 so
=−
,
+
= 0 so
=−
.
∂x
∂z ∂x
∂x
∂F/∂z ∂y
∂z ∂y
∂y
∂F/∂z
67.
∂z
2x + yz ∂z
xz − 3z 2
=
,
=
∂x
6yz − xy ∂y
6yz − xy
68. ln(1 + z) + xy 2 + z − 1 = 0;
69. yex − 5 sin 3z − 3z = 0;
70.
y 2 (1 + z) ∂z
2xy(1 + z)
∂z
=−
,
=−
∂x
2+z
∂y
2+z
yex
yex
∂z
ex
∂z
=−
=
,
=
∂x
−15 cos 3z − 3
15 cos 3z + 3 ∂y
15 cos 3z + 3
∂z
zeyz cos xz − yexy cos yz
∂z
zexy sin yz − xexy cos yz + zeyz sin xz
= − xy
,
=
−
∂x
ye sin yz + xeyz cos xz + yeyz sin xz ∂y
yexy sin yz + xeyz cos xz + yeyz sin xz
∂f
∂f
∂f
i+
j+
k
∂x
∂y
∂z
∂f ∂w
∂f ∂w
∂f ∂u ∂f ∂v
∂f ∂u ∂f ∂v
+
+
i+
+
+
j
=
∂u ∂x ∂v ∂x ∂w ∂x
∂u ∂y
∂v ∂y ∂w ∂y
∂f ∂w
∂f
∂f
∂f
∂f ∂u ∂f ∂v
+
+
k=
∇u +
∇v +
∇w
+
∂u ∂z
∂v ∂z
∂w ∂z
∂u
∂v
∂w
71. ∇f (u, v, w) =
72. (a)
∂f
∂f ∂z
∂w
=
+
∂x
∂x
∂z ∂x
(b)
∂w
∂f
∂f ∂z
=
+
∂y
∂y
∂z ∂y
2
73. wr = er / (er + es + et + eu ), wrs = −er es / (er + es + et + eu ) ,
3
wrst = 2er es et / (er + es + et + eu ) ,
4
wrstu = −6er es et eu / (er + es + et + eu ) = −6er+s+t+u /e4w = −6er+s+t+u−4w
74. ∂w/∂y1 = a1 ∂w/∂x1 + a2 ∂w/∂x2 + a3 ∂w/∂x3 ,
∂w/∂y2 = b1 ∂w/∂x1 + b2 ∂w/∂x2 + b3 ∂w/∂x3
75. (a) dw/dt =
4
X
(∂w/∂xi ) (dxi /dt)
i=1
(b) ∂w/∂vj =
4
X
(∂w/∂xi ) (∂xi /∂vj ) for j = 1, 2, 3
i=1
76. Let u = x21 + x22 + ... +x2n ; then w = uk , ∂w/∂xi = kuk−1 (2xi ) = 2k xi uk−1 ,
∂ 2 w/∂x2i = 2k(k − 1)xi uk−2 (2xi ) + 2kuk−1 = 4k(k − 1)x2i uk−2 + 2kuk−1 for i = 1, 2, . . . , n
so
n
X
i=1
∂ 2 w/∂x2i = 4k (k − 1) uk−2
n
X
x2i + 2kn uk−1
i=1
= 4k(k − 1)uk−2 u + 2kn uk−1 = 2kuk−1 [2(k − 1) + n]
which is 0 if k = 0 or if 2(k − 1) + n = 0, k = 1 − n/2.
77. dF/dx = (∂F/∂u)(du/dx) + (∂F/∂v)(dv/dx)
= f (u)g 0 (x) − f (v)h0 (x) = f (g(x))g 0 (x) − f (h(x))h0 (x)
559
Chapter 15
78. ∇f = fx i + fy j + fz k and ∇g = gx i + gy j + gz k evaluated at (x0 , y0 , z0 ) are normal, respectively,
to the surfaces f (x, y, z) = 0 and g(x, y, z) = 0 at (x0 , y0 , z0 ). The surfaces are orthogonal at
(x0 , y0 , z0 ) if and only if ∇f · ∇g = 0 so fx gx + fy gy + fz gz = 0.
79. f (x, y, z) = x2 + y 2 + z 2 − a2 = 0, g(x, y, z) = z 2 − x2 − y 2 = 0,
fx gx + fy gy + fz gz = −4x2 − 4y 2 + 4z 2 = 4g(x, y, z) = 0
EXERCISE SET 15.8
1. (a) minimum at (2, −1), no maxima
(b) maximum at (0, 0), no minima
(c) no maxima or minima
2. (a) maximum at (−1, 5), no minima
(b) no maxima or minima
(c) no maxima or minima
3. f (x, y) = (x − 3)2 + (y + 2)2 , minimum at (3, −2), no maxima
4. f (x, y) = −(x + 1)2 − 2(y − 1)2 + 4, maximum at (−1, 1), no minima
5. fx = 6x + 2y = 0, fy = 2x + 2y = 0; critical point (0,0); D = 8 > 0 and fxx = 6 > 0 at (0,0),
relative minimum.
6. fx = 3x2 − 3y = 0, fy = −3x − 3y 2 = 0; critical points (0,0) and (−1, 1); D = −9 < 0 at (0,0),
saddle point; D = 27 > 0 and fxx = −6 < 0 at (−1, 1), relative maximum.
7. fx = 2x − 2xy = 0, fy = 4y − x2 = 0; critical points (0,0) and (±2, 1); D = 8 > 0 and fxx = 2 > 0
at (0,0), relative minimum; D = −16 < 0 at (±2, 1), saddle points.
8. fx = 3x2 − 3 = 0, fy = 3y 2 − 3 = 0; critical points (−1, ±1) and (1, ±1); D = −36 < 0 at (−1, 1)
and (1, −1), saddle points; D = 36 > 0 and fxx = 6 > 0 at (1,1), relative minimum; D = 36 > 0
and fxx = −36 < 0 at (−1, −1), relative maximum.
9. fx = y + 2 = 0, fy = 2y + x + 3 = 0; critical point (1, −2); D = −1 < 0 at (1, −2), saddle point.
10. fx = 2x + y − 2 = 0, fy = x − 2 = 0; critical point (2, −2); D = −1 < 0 at (2, −2), saddle point.
11. fx = 2x + y − 3 = 0, fy = x + 2y = 0; critical point (2, −1); D = 3 > 0 and fxx = 2 > 0 at (2, −1),
relative minimum.
12. fx = y − 3x2 = 0, fy = x − 2y = 0; critical points (0,0) and (1/6, 1/12); D = −1 < 0 at (0,0),
saddle point; D = 1 > 0 and fxx = −1 < 0 at (1/6, 1/12), relative maximum.
13. fx = 2x − 2/ x2 y = 0, fy = 2y − 2/ xy 2 = 0; critical points (−1, −1) and (1, 1); D = 32 > 0
and fxx = 6 > 0 at (−1, −1) and (1, 1), relative minima.
14. fx = ey = 0 is impossible, no critical points.
15. fx = 2x = 0, fy = 1 − ey = 0; critical point (0, 0); D = −2 < 0 at (0, 0), saddle point.
16. fx = y − 2/x2 = 0, fy = x − 4/y 2 = 0; critical point (1,2); D = 3 > 0 and fxx = −4 > 0 at (1, 2),
relative minimum.
Exercise Set 15.8
560
17. fx = ex sin y = 0, fy = ex cos y = 0, sin y = cos y = 0 is impossible, no critical points.
18. fx = y cos x = 0, fy = sin x = 0; sin x = 0 if x = nπ for n = 0, ±1, ±2, . . . and cos x 6= 0 for these
values of x so y = 0; critical points (nπ, 0) for n = 0, ±1, ±2, . . .; D = −1 < 0 at (nπ, 0), saddle
points.
2
2
2
2
19. fx = −2(x + 1)e−(x +y +2x) = 0, fy = −2ye−(x +y +2x) = 0; critical point (−1, 0); D = 4e2 > 0
and fxx = −2e < 0 at (−1, 0), relative maximum.
20. fx = y − a3 /x2 = 0, fy = x − b3 /y 2 = 0; critical point a2 /b, b2 /a ; if ab > 0 then D = 3 > 0 and
fxx = 2b3 /a3 > 0 at a2 /b, b2 /a , relative minimum; if ab < 0 then D = 3 > 0 and
fxx = 2b3 /a3 < 0 at a2 /b, b2 /a , relative maximum.
21.
2
1
0
-1
-2
-2
-1
0
1
2
∇f = (4x − 4y)i − (4x − 4y 3 )j = 0 when x = y, x = y 3 , so x = y = 0 or x = y = ±1. At
(0, 0), D = −16, a saddle point; at (1, 1) and (−1, −1), D = 32 > 0, fxx = 4, a relative minimum.
22.
5
140
4
100
-100
60
3
-60
20
2
-20
1
0
0
0
-1
-2 40 20
20
-3
0
60100
-20
-4
140
-40
-5
-5 -4 -3 -2 -1 0 1 2 3 4 5
∇f = (2y 2 − 2xy + 4y)i + (4xy − x2 + 4x)j = 0 when 2y 2 − 2xy + 4y = 0, 4xy − x2 + 4x = 0, with
solutions (0, 0), (0, −2), (4, 0), (4/3, −2/3). At (0, 0), D = −16, a saddle point. At (0, −2),
D = −16, a saddle point. At (4, 0), D = −16, a saddle point. At (4/3, −2/3), D = 16/3,
fxx = 4/3 > 0, a relative minimum.
23. (a) critical point (0,0); D = 0
(b) f (0, 0) = 0, x4 + y 4 ≥ 0 so f (x, y) ≥ f (0, 0), relative minimum.
24. (a) critical point (0,0); D = 0
(b) f (0, 0) = 0, inside any circle centered at (0, 0) there are points where f (x, y) > 0 (along the
x-axis) and points where f (x, y) < 0 (along the y-axis) so (0, 0) is a saddle point.
561
Chapter 15
25. (a) fx = 3ey − 3x2 = 3 ey − x2 = 0, fy = 3xey − 3e3y = 3ey x − e2y = 0, ey = x2 and
e2y = x, x4 = x, x x3 − 1 = 0 so x = 0, 1; critical point (1, 0); D = 27 > 0 and fxx = −6 < 0
at (1, 0), relative maximum.
(b)
lim f (x, 0) = lim 3x − x3 − 1 = +∞ so no absolute maximum.
x→−∞
x→−∞
26. fx = 8xey − 8x3 = 8x(ey − x2 ) = 0, fy = 4x2 ey − 4e4y = 4ey (x2 − e3y ) = 0, x2 = ey and
x2 = e3y , e3y = ey , e2y = 1, so y = 0 and x = ±1; critical points (1,0) and (−1, 0). D = 128 > 0
and fxx = −16 < 0 at both points so a relative maximum occurs at each one.
27. fx = y − 1 = 0, fy = x − 3 = 0; critical point (3,1).
Along y = 0 : u(x) = −x; no critical points,
along x = 0 : v(y) = −3y; no critical points,
4
along y = − x + 4 :
5
4
27
w(x) = − x2 + x − 12; critical point (27/8, 13/10).
5
5
(x, y) (3, 1)
f (x, y) −3
(0, 0)
0
(5, 0)
−5
(0, 4)
−12
(27/8, 13/10)
−231/80
Absolute maximum value is 0, absolute minimum value is −12.
28. fx = y − 2 = 0, fy = x = 0; critical point (0,2), but (0,2) is not in the interior of R.
Along y = 0 : u(x) = −2x; no critical points,
along x = 0 : v(y) = 0; no critical points,
along y = 4 − x :
w(x) = 2x − x2 ; critical point (1, 3).
(x, y) (0, 0)
f (x, y)
0
(0, 4)
0
(4, 0)
−8
(1, 3)
1
Absolute maximum value is 1, absolute minimum value is −8.
29. fx = 2x − 2 = 0, fy = −6y + 6 = 0; critical point (1,1).
Along y = 0 :
along y = 2 :
along x = 0 :
along x = 2 :
u1 (x) = x2 − 2x; critical point (1, 0),
u2 (x) = x2 − 2x; critical point (1, 2)
v1 (y) = −3y 2 + 6y; critical point (0, 1),
v2 (y) = −3y 2 + 6y; critical point (2, 1)
(x, y) (1, 1)
f (x, y)
2
(1, 0)
−1
(1, 2)
−1
(0, 1)
3
(2, 1)
3
(0, 0)
0
(0, 2)
0
(2, 0)
0
(2, 2)
0
Absolute maximum value is 3, absolute minimum value is −1.
30. fx = ey − 2x = 0, fy = xey − ey = ey (x − 1) = 0; critical point (1, ln 2).
Along y = 0 :
along y = 1 :
along x = 0 :
along x = 2 :
u1 (x) = x − x2 − 1; critical point (1/2, 0),
u2 (x) = ex − x2 − e; critical point (e/2, 1),
v1 (y) = −ey ; no critical points,
v2 (y) = ey − 4; no critical points (for 0 < y < 1).
(x, y) (0, 0)
f (x, y) −1
(0, 1)
−e
(2, 1)
e−4
(2, 0)
−3
(1, ln 2) (1/2, 0)
−1
−3/4
Absolute maximum value is −3/4, absolute minimum value is −3.
(e/2, 1)
e(e − 4)/4 ≈ −0.87
Exercise Set 15.8
562
31. fx = 2x − 1 = 0, fy = 4y = 0; critical point (1/2, 0).
√
Along x2 + y 2 = 4 : y 2 = 4 − x2 , u(x) = 8 − x − x2 for −2 ≤ x ≤ 2; critical points (−1/2, ± 15/2).
(x, y) (1/2, 0)
f (x, y) −1/4
√
−1/2, 15/2
33/4
√
−1/2, − 15/2
33/4
(−2, 0) (2, 0)
6
2
Absolute maximum value is 33/4, absolute minimum value is −1/4.
32. fx = y 2 = 0, fy = 2xy = 0; no critical points in the interior of R.
Along y = 0 : u(x) = 0; no critical points,
along x = 0 : v(y) = 0; no critical points
along x2 + y 2 = 1 : w(x) = x − x3 for 0 ≤ x ≤ 1; critical point
√ p
1/ 3, 2/3 .
(x, y)
(0, 0)
(0, 1)
(1, 0)
f (x, y)
0
0
0
Absolute maximum value is
√ p
1/ 3, 2/3
√
2 3/9
2√
3, absolute minimum value is 0.
9
33. Maximize P = xyz subject to x + y + z = 48, x > 0, y > 0, z > 0. z = 48 − x − y so
P = xy(48 − x − y) = 48xy − x2 y − xy 2 , Px = 48y − 2xy − y 2 = 0, Py = 48x − x2 − 2xy = 0. But
2
>0
x 6= 0 and y 6= 0 so 48 − 2x − y = 0 and 48 − x − 2y = 0; critical point (16,16). Pxx Pyy − Pxy
and Pxx < 0 at (16,16), relative maximum. z = 16 when x = y = 16, the product is maximum for
the numbers 16,16,16.
34. Minimize S = x2 + y 2 + z 2 subject to x + y + z = 27, x > 0, y > 0, z > 0. z = 27 − x − y so
S = x2 + y 2 + (27 − x − y)2 , Sx = 4x + 2y − 54 = 0, Sy = 2x + 4y − 54 = 0; critical point (9,9);
2
= 12 > 0 and Sxx = 4 > 0 at (9,9), relative minimum. z = 9 when x = y = 9, the
Sxx Syy − Sxy
sum of the squares is minimum for the numbers 9,9,9.
35. Maximize w = xy 2 z 2 subject to x + y + z = 5, x > 0, y > 0, z > 0. x = 5 − y − z so
w = (5 − y − z)y 2 z 2 = 5y 2 z 2 − y 3 z 2 − y 2 z 3 , wy = 10yz 2 − 3y 2 z 2 − 2yz 3 = yz 2 (10 − 3y − 2z) = 0,
wz = 10y 2 z − 2y 3 z − 3y 2 z 2 = y 2 z(10 − 2y − 3z) = 0, 10 − 3y − 2z = 0 and 10 − 2y − 3z = 0; critical
2
= 320 > 0 and wyy = −24 < 0 when y = z = 2, relative
point when y = z = 2; wyy wzz − wyz
maximum. x = 1 when y = z = 2, xy 2 z 2 is maximum at (1,2,2).
36. Minimize w = D2 = x2 + y 2 + z 2 subject to x2 − yz = 5. x2 = 5 + yz so w = 5 + yz + y 2 + z 2 ,
2
= 3 > 0 and
wy = z + 2y = 0, wz = y + 2z = 0; critical point when y = z = 0; wyy wzz − wyz
√
2
wyy√= 2 > 0 when y = z = 0, relative minimum. x = 5, x = ± 5 when y = z = 0. The points
± 5, 0, 0 are closest to the origin.
37. The diagonal of the box must equal the diameter of the sphere, thus we maximize V = xyz or, for
2
2
convenience, w = V 2 = x2 y 2 z 2 subject to x2 +y 2 +z 2 = 4a2 , x > 0, y > 0, z > 0; z 2 = 4a2 −x
−y
2 2 2
4 2
2 4
2
2
2
2
2
2
2
2
hence w = 4a x y −x y −x y , wx = 2xy (4a −2x −y ) = 0, wy = 2x y 4a − x − 2y = 0,
√
√
4a2 − 2x2 − y 2 = 0 and 4a2 − x2 − 2y 2 = 0; critical point 2a/ 3, 2a/ 3 ;
√
√
4096 8
128 4
2
a > 0 and wxx = −
a < 0 at 2a/ 3, 2a/ 3 , relative maximum.
=
wxx wyy − wxy
27
9
√
√
z =√
2a/ 3 √
when x√= y = 2a/ 3, the dimensions of the box of maximum volume are
2a/ 3, 2a/ 3, 2a/ 3.
38. Maximize V = xyz subject to x+y+z = 1, x > 0, y > 0, z > 0. z = 1−x−y so V = xy−x2 y−xy 2 ,
Vx = y(1 − 2x − y) = 0, Vy = x(1 − x − 2y) = 0, 1 − 2x − y = 0 and 1 − x − 2y = 0; critical point
2
= 1/3 > 0 and Vxx = −2/3 < 0 at (1/3, 1/3), relative maximum. The
(1/3, 1/3); Vxx Vyy − Vxy
maximum volume is V = (1/3)(1/3)(1/3) = 1/27.
563
Chapter 15
39. Let x, y, and z be, respectively, the length, width, and height of the box. Minimize
C = 10(2xy) + 5(2xz + 2yz) = 10(2xy + xz + yz) subject to xyz = 16. z = 16/(xy)
so C = 20(xy + 8/y + 8/x), Cx = 20(y − 8/x2 ) = 0, Cy = 20(x − 8/y 2 ) = 0;
2
= 1200 > 0
critical point (2,2); Cxx Cyy − Cxy
and Cxx = 40 > 0 at (2,2), relative minimum. z = 4 when x = y = 2. The cost of materials is
minimum if the length and width are 2 ft and the height is 4 ft.
40. Maximize the profit P = 500(y − x)(x − 40) + [45, 000 + 500(x − 2y)](y − 60)
= 500(−x2 − 2y 2 + 2xy − 20x + 170y − 5400).
Px = 1000(−x + y − 10) = 0, Py = 1000(−2y + x + 85) = 0; critical point (65,75);
2
= 1,000,000 > 0 and Pxx = −1000 < 0 at (65,75), relative maximum. The profit
Pxx Pyy − Pxy
will be maximum when x = 65 and y = 75.
41. (a) x = 0 : f (0, y) = −3y 2 , minimum −3, maximum 0;
∂f
(1, y) = −6y + 2 = 0 at y = 1/3, minimum 3,
x = 1, f (1, y) = 4 − 3y 2 + 2y,
∂y
maximum 13/3;
y = 0, f (x, 0) = 4x2 , minimum 0, maximum 4;
∂f
(x, 1) = 8x + 2 6= 0 for 0 < x < 1, minimum −3, maximum 3
y = 1, f (x, 1) = 4x2 + 2x − 3,
∂x
d
(b) f (x, x) = 3x2 , minimum 0, maximum 3; f (x, 1−x) = −x2 +8x−3, f (x, 1−x) = −2x+8 6= 0
dx
for 0 < x < 1, maximum 4, minimum −3
(c) fx (x, y) = 8x + 2y = 0, fy (x, y) = −3y + 2x = 0, solution is (0, 0), which is not an interior
point of the square, so check the sides: minimum −3, maximum 14/3.
42. Maximize A = ab sin α subject to 2a + 2b = `, a > 0, b > 0, 0 < α < π. b = (` − 2a)/2 so
A = (1/2)(`a − 2a2 ) sin α, Aa = (1/2)(` − 4a) sin α, Aα = (a/2)(` − 2a) cos α; sin α 6= 0 so from
Aa = 0 we get a = `/4 and then from Aα = 0 we get cos α = 0, α = π/2. Aaa Aαα −A2aα = `2 /8 > 0
and Aaa = −2 < 0 when a = `/4 and α = π/2, the area is maximum.
43. Minimize S = xy + 2xz + 2yz subject to xyz = V , x > 0, y > 0, z > 0 where x, y, and z are,
respectively, the length, width, and height of the box. z√= V /(xy)
so S = xy + 2V /y + 2V /x,
√
2
= 3 > 0 and
Sx = y − 2V /x2 = 0, Sy = x − 2V /y 2 = 0; critical point ( 3 2V , 3 2V ); Sxx Syy − Sxy
Sxx = 2 > 0 at this point so there is a relative minimum there. The length and width are each
√
√
3
2V , the height is z = 3 2V /2.
44. The altitude of the trapezoid is x sin φ and the lengths of the lower and upper bases are, respectively,
27 − 2x and 27 − 2x + 2x cos φ so we want to maximize
A = (1/2)(x sin φ)[(27 − 2x) + (27 − 2x + 2x cos φ)] = 27x sin φ − 2x2 sin φ + x2 sin φ cos φ.
Ax = sin φ(27 − 4x + 2x cos φ),
Aφ = x(27 cos φ − 2x cos φ − x sin2 φ + x cos2 φ) = x(27 cos φ − 2x cos φ + 2x cos2 φ − x).
sin φ 6= 0 so from Ax = 0 we get cos φ = (4x − 27)/(2x), x 6= 0 so from Aφ = 0 we get
(27 − 2x + 2x cos φ) cos φ − x = 0 which, for cos φ = (4x − 27)/(2x), yields 4x − 27 − x = 0,
x = 9. If x = 9 then cos φ = 1/2, φ = π/3.
√ The critical point occurs when x = 9 and φ = π/3;
Axx Aφφ − A2xφ = 729/2 > 0 and Axx = −3 3/2 < 0 there, the area is maximum when x = 9 and
φ = π/3.
Exercise Set 15.8
45. (a)
564
n
n
n
n
X
X
X
X
∂g
=
2 (mxi + b − yi ) xi = 2 m
x2i + b
xi −
xi yi
∂m
i=1
i=1
i=1
i=1
!
!
n
n
n
X
X
X
2
xi m +
xi b =
xi yi ,
i=1
∂g
=
∂b
i=1
n
X
i=1
2 (mxi + b − yi ) = 2 m
n
X
i=1
xi + bn −
i=1
n
X
!
= 0 if
yi
i=1
!
= 0 if
n
X
!
xi
i=1
m + nb =
n
X
yi
i=1
(b) The function z = g(m, b), as a function of m and b, has only one critical point, found in part
(a), and tends to +∞ as either |m| or |b| tends to infinity. Thus the only critical point must
be a minimum.
46. (a)
n
X
2
(xi − x̄) =
i=1
n
X
n
n
X
X
x2i − 2x̄
xi + nx̄2
x2i − 2x̄xi + x̄2 =
i=1
=
n
X
x2i −
i=1
=
n
X
x2i
i=1
(b) gmm = 2
n
X
2
n
1
−
n
n
X
xi
i=1
n
X
i=1
!2
+
1
n
!2
xi
i=1
n
X
xi
i=1
> 0 so n
i=1
!2
n
X
x2i
−
i=1
x2i , gbb = 2n, gmb = 2
i=1
n
X
n
X
!2
xi
>0
i=1
xi ,
i=1
2
= 4 n
D = gmm gbb − gmb
n
X
n
X
x2i −
i=1
!2
xi > 0 and gmm > 0
i=1
(c) g(m, b) is of the second-degree in m and b so the graph of z = g(m, b) is a quadric surface.
(d) The only quadric surface of this form having a relative minimum is a paraboloid that opens
upward where the relative minimum is also the absolute minimum.
47. n = 3,
3
X
xi = 3,
i=1
48. n = 4,
4
X
4
X
xi = 7,
xi = 10,
i=1
50.
5
X
i=1
yi = 7,
i=1
i=1
49.
3
X
4
X
5
X
i=1
xi yi = 13,
i=1
yi = 4,
i=1
i=1
xi = 15,
4
X
3
X
yi = 8.2,
4
X
i=1
x2i = 21,
i=1
4
X
x2i = 30,
i=1
yi = 15.1,
5
X
i=1
3
X
4
X
i=1
4
X
x2i = 11, y =
19
3
x+
4
12
xi yi = −2, y = −
14
36
x+
35
5
xi yi = 23, n = 4; m = 0.5, b = 0.8, y = 0.5x + 0.8.
i=1
x2i = 55,
5
X
i=1
xi yi = 39.8, n = 5; m = −0.55, b = 4.67, y = 4.67 − 0.55x
565
Chapter 15
51. (a) y =
57
8843
+
t ≈ 63.1643 + 0.285t
140
200
(b)
80
0(1930)
(c) y =
60
60
2909
≈ 83.1143
35
52. (a) y ≈ 119.84 − 1.13x
90
(b)
35
50
60
(c) about 52 units
53. (a) P =
2798 171
+
T ≈ 133.2381 + 0.4886T
21
350
(b)
190
0
130
(c) T ≈ −
120
139,900
=≈ −272.7096◦ C
513
54. (a) for example, z = y
(b) For example, on 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 let z =
(
y if 0 < x < 1, 0 < y < 1
1/2 if x = 0, 1 or y = 0, 1
55. f (x0 , y0 ) ≥ f (x, y) for all (x, y) inside a circle centered at (x0 , y0 ) by virtue of Definition 15.8.1.
If r is the radius of the circle, then in particular f (x0 , y0 ) ≥ f (x, y0 ) for all x satisfying
|x − x0 | < r so f (x, y0 ) has a relative maximum at x0 . The proof is similar for the function
f (x0 , y).
EXERCISE SET 15.9
1. (a) xy = 4 is tangent to the line, so the maximum value of f is 4.
(b) xy = 2 intersects the curve and so gives a smaller value of f .
(c) Maximize f (x, y) = xy subject to the constraint g(x, y) = x + y − 4 = 0, ∇f = λ∇g,
yi + xj = λ(i + j), so solve the equations y = λ, x = λ with solution x = y = λ, but x + y = 4,
so x = y = 2, and the maximum value of f is f = xy = 4.
Exercise Set 15.9
566
2. (a) x2 + y 2 = 25 is tangent to the line at (3, 4), so the minimum value of f is 25.
(b) A smaller value of f yields a circle of a smaller radius, and hence does not intersect the line.
(c) Minimize f (x, y) = x2 + y 2 subject to the constraint g(x, y) = 3x + 4y − 25 = 0, ∇f = λ∇g,
2xi + 2yj = 3λi + 4λj, so solve 2x = 3λ, 2y = 4λ; solution is x = 3, y = 4, minimum = 25.
3. y = 8xλ, x = 16yλ; y/(8x) = x/(16y), x2 = 2y 2 so 4 2y 2 + 8y 2 = 16, y 2 = 1, y = ±1. Test
√
√
√
√
√
√
√
√
± 2, −1 and (± 2, 1). f − 2, −1 = f
2, 1 = 2, f − 2, 1 = f
2, −1 = − 2.
√
√
√
√
√
√
Maximum 2 at − 2, −1 and
2, 1 , minimum − 2 at − 2, 1 and
2, −1 .
2
2
4. 2x = 2xλ, −1 = 2yλ.
√ If x 6= 0 then λ = 1 2and 2y = −1/2 so x + (−1/2)√= 25,
2
x = 99/4, x = ±3 11/2. If x = 0 then 0 + y = 25, y = ±5. Test ±3 11/2, −1/2 and (0, ±5).
√
√
f ±3 11/2, −1/2 = 101/4, f (0, −5) = 5, f (0, 5) = −5. Maximum 101/4 at ±3 11/2, −1/2 ,
minimum −5 at (0,5).
5. 12x2 = 4xλ, 2y = 2yλ. If y 6= 0 then λ = 1 and 12x2 = 4x, 12x(x√− 1/3) = 0, x = 0 or x = 1/3
that y = ±1 when x = 0,√y = ± 7/3 when
so from 2x2 + y 2 = 1 we find √
√ x = 1/3. If y = 0
then 2x2 + (0)2 = 1, x = ±1/ 2. Test (0, ±1), 1/3, ± 7/3 , and ±1/ 2, 0 . f (0, ±1) = 1,
√
√
√
√
√
√
√
f 1/3, ± 7/3 = 25/27, f 1/ 2, 0 = 2, f −1/ 2, 0 = − 2. Maximum 2 at 1/ 2, 0 ,
√
√
minimum − 2 at −1/ 2, 0 .
6. 1 = 2xλ, −3 = 6yλ; 1/(2x) = −1/(2y), y = −x so x2 + 3(−x)2 = 16, x = ±2. Test (−2, 2) and
(2, −2). f (−2, 2) = −9, f (2, −2) = 7. Maximum 7 at (2, −2), minimum −9 at (−2, 2).
7. 2 = 2xλ, 1 = 2yλ, −2 = 2zλ; 1/x = 1/(2y) = −1/z thus x = 2y, z = −2y so
(2y)2 + y 2 + (−2y)2 = 4, y 2 = 4/9, y = ±2/3. Test (−4/3, −2/3, 4/3) and (4/3, 2/3, −4/3).
f (−4/3, −2/3, 4/3) = −6, f (4/3, 2/3, −4/3) = 6. Maximum 6 at (4/3, 2/3, −4/3), minimum −6
at (−4/3, −2/3, 4/3).
8. 3 = 4xλ, 6 = 8yλ, 2 = 2zλ; 3/(4x) = 3/(4y) = 1/z thus y = x, z = 4x/3, so
2x2 + 4(x)2 + (4x/3)2 = 70, x2 = 9, x = ±3. Test (−3, −3, −4) and (3,3,4).
f (−3, −3, −4) = −35, f (3, 3, 4) = 35. Maximum 35 at (3, 3, 4), minimum −35 at (−3, −3, −4).
2
2
2
2
9. yz = 2xλ, xz = 2yλ, xy =
√ 2zλ; yz/(2x) = xz/(2y) = xy/(2z) thus y√ = x , z =
√ x so
2
2
2
with x√=±1/ √3, y = √
±1/ 3,√and
x + x +√x = 1, x = ±1/ 3. Test the eight
√ possibilities
√
√
z = ±1/ 3 to find the maximum is 1/ 3 3 at 1/ 3, 1/ 3, 1/ 3 , 1/ 3, −1/ 3, −1/ 3 ,
√
√
√
√
√
√
√
−1/ 3, 1/ 3, −1/ 3 , and −1/ 3, −1/ 3, 1/ 3 ; the minimum is −1/ 3 3 at
√
√
√
√
√
√
√
√
√
√
√
√
1/ 3, 1/ 3, −1/ 3 , 1/ 3, −1/ 3, 1/ 3 , −1/ 3, 1/ 3, 1/ 3 , and −1/ 3, −1/ 3, −1/ 3 .
10. 4x3 = λ, 4y 3 = λ, 4z 3 = λ, so x = y = z; x + y + z = 3x = 1, x = y = z = 1/3, minimum value
1/27, no maximum
11. f (x, y) = x2 + y 2 ; 2x = 2λ, 2y = −4λ; y = −2x so 2x − 4(−2x) = 3, x = 3/10. The point is
(3/10, −3/5).
12. f (x, y) = (x − 4)2 + (y − 2)2 , g(x, y) = y − 2x − 3; 2(x − 4) = −2λ, 2(y − 2) = λ; x − 4 = −2(y − 2),
x = −2y + 8 so y = 2(−2y + 8) + 3, y = 19/5. The point is (2/5, 19/5).
13. f (x, y, z) = x2 + y 2 + z 2 ; 2x = λ, 2y = 2λ, 2z = λ; y = 2x, z = x so x + 2(2x) + x = 1, x = 1/6.
The point is (1/6, 1/3, 1/6).
14. f (x, y, z) = (x − 1)2 + (y + 1)2 + (z − 1)2 ; 2(x − 1) = 4λ, 2(y + 1) = 3λ, 2(z − 1) = λ; x = 4z − 3,
y = 3z − 4 so 4(4z − 3) + 3(3z − 4) + z = 2, z = 1. The point is (1, −1, 1).
567
Chapter 15
15. f (x, y) = (x − 1)2 + (y − 2)2 ; 2(x − 1) = 2xλ, 2(y − 2) = 2yλ; (x − 1)/x = (y − 2)/y, y = 2x
so x2 + (2x)2 = 45, x = ±3. f (−3, −6) = 80 and f (3, 6) = 20 so (3,6) is closest and (−3, −6) is
farthest.
16. f (x, y, z) = x2 + y 2 + z 2 ; 2x = yλ, 2y = xλ, 2z = −2zλ. If z 6= 0 then λ = −1 so 2x = −y and
2y = −x, x = y = 0; substitute into xy − z 2 = 1 to get z 2 = −1 which has no real solution. If
z = 0 then xy − (0)2 = 1, y = 1/x, and also (from 2x = yλ and 2y = xλ), 2x/y = 2y/x, y 2 = x2
so (1/x)2 = x2 , x4 = 1, x = ±1. Test (1,1,0) and (−1, −1, 0) to see that they are both closest to
the origin.
17. f (x, y, z) = x + y + z, x2 + y 2 + z 2 = 25 where x, y, and z are the components of the vector;
2
2
2
1 = 2xλ,√1 = 2yλ, √
1 = 2zλ;
√ 1/(2x)√ = 1/(2y)
√ = 1/(2z); y√= x,√z = x
√ so x +
√ x + x = 25,
x = ±5/ 3. f −5/ 3, −5/ 3, −5/ 3 = −5 3 and f 5/ 3, 5/ 3, 5/ 3 = 5 3 so the vector
√
is 5(i + j + k)/ 3.
18. x2 + y 2 = 25 is the constraint; 8x − 4y = 2xλ, −4x + 2y = 2yλ; (4x − 2y)/x = (−2x + y)/y,
2
2x2 + √
3xy − 2y 2 = 0, (2x − y)(x + 2y) = 0, y = 2x or
= 2x then√x2 +√(2x)
√ x = −2y.
√ If y √
= 25,
2
2
5, 2 5 = 0 and
x = ± 5. If x = −2y then −2y + y = 25, y = ± 5. T − 5, −2 5 = T
√
√
√ √
T 2 5, − 5 = T −2 5, 5 = 125. The highest temperature is 125 and the lowest is 0.
19. (a)
(b) minimum value −5,
maximum value 101/4
15
31.5
-31.5
-27
(c) Find the minimum and maximum values of f (x, y) = x2 − y subject to the constraint
g(x, y) = x2 + y 2 − 25 = 0, ∇f = λ∇g, 2xi − j = 2λxi + 2λyj, so solve 2x = 2λx, −1 = 2λy.
If x = 0 then y = ±5, f = ∓5, and if x 6= 0 then λ = 1, y = −1/2, x2 =√25 − 1/4 = 99/4,
f = 99/4 + 1/2 = 101/4, so the maximum value of f is 101/4 at (±3 11/2, −1/2) and the
minimum value of f is −5 at (0, 5).
20. (a)
(b) f ≈ 15
6
5
4
3
2
1
0
0
1
2
3
4
5
6
(d) Set f (x, y) = x3 + y 3 − 3xy, g(x, y) = (x − 4)2 + (y − 4)2 − 25; minimize f subject to the
constraint g = 0 : ∇f = λg, (3x2 − 3y)i + (3y 2 − 3x)j = 2λ(x − 4)i + 2λ(y − 4)j, so solve
(use a CAS) 3x2 − 3y = 2λ(x − 4), 3y 2 − 3x = 2λ(y − 4); minimum value f = 14.52 at
(2.5858, 2.5858)
21. Minimize f = x2 + y 2 + z 2 subject to g(x, y, z) = x + y + z − 27 = 0. ∇f = λ∇g,
2xi + 2yj + 2zk = λi + λj + λk, solution x = y = z = 9, minimum value 243
Exercise Set 15.9
568
22. Maximize f (x, y, z) = xy 2 z 2 subject to g(x, y, z) = x + y + z − 5 = 0, ∇f = λ∇g = λ(i + j + k),
λ = y 2 z 2 = 2xyz 2 = 2xy 2 z, λ = 0 is impossible, hence x, y, z 6= 0, and z = y = 2x, 5x − 5 = 0,
x = 1, y = z = 2, maximum value 8 at (1, 2, 2)
23. Minimize f = x2 + y 2 + z 2 subject to√x2 − yz = 5, ∇f = λ∇g, 2x = 2xλ, 2y = −zλ, 2z = −yλ.
If λ 6= ±2, √then y = z = 0, x = ± 5, f = 5; if λ = ±2
√ then x = 0, and since −yz = 5,
y = −z = ± 5, f = 10, thus the minimum value is 5 at (± 5, 0, 0).
24. The diagonal of the box must equal the diameter of the sphere so maximize V = xyz or, for
convenience, maximize f = V 2 = x2 y 2 z 2 subject to g(x, y, z) = x2 + y 2 + z 2 − 4a2 = 0, ∇f = λ∇g,
2xy 2 z 2 = 2λx, 2x2 yz 2 = 2λy, 2x2 y 2√
z = 2λz. Since V 6= 0 it follows
√ that x, y, z 6= 0, hence
2
2
3
x = ±y = ±z, 3x = 4a , x = ±2a/ 3, maximum volume 8a /(3 3).
25. Let x, y, and z be, respectively, the length, width, and height of the box. Minimize
f (x, y, z) = 10(2xy) + 5(2xz + 2yz) = 10(2xy + xz + yz) subject to g(x, y, z) = xyz − 16 = 0,
∇f = λ∇g, 20y + 10z = λyz, 20x + 10z = λxz, 10x + 10y = λz. Since V = xyz = 16, x, y, z 6= 0,
thus λz = 20 + 10(z/y) = 20 + 10(z/x) = 10x + 10y, so x = y. Then z = 16/x2 , thus
20 + 10(16/x3 ) = 20x, x3 + 8 = x4 , the only real solution of which is x = 2,thus x = y = 2, z = 4,
minimum value 240.
26. Minimize f (p, q, r) = 2pq +2pr +2qr, subject to g(p, q, r) = p+q +r −1 = 0, ∇(p,q,r) f = λ∇(p,q,r) g,
2(q + r) = λ, 2(p + r) = λ, 2(p + q) = λ, solution p = q = r = 1/3, minimum value 2/3.
27. Maximize A(a, b, α) = ab sin α subject to g(a, b, α) = 2a + 2b − ` = 0, ∇(a,b,α) f = λ∇(a,b,α) g,
b sin α = 2λ, a sin α = 2λ, ab cos α = 0 with solution a = b, α = π/2 maximum value if parallelogram
is a square.
28. Minimize f (x, y, z) = xy + 2xz + 2yz subject to g(x, y, z) = xyz − V = 0, ∇f = λ∇g,
y + 2z = λyz, x + 2z = λxz, 2x + 2y = λxy; λ = 0 leads to x = y = z = 0, impossible, so solve for
λ = 1/z + 2/x = 1/z + 2/y = 2/y + 2/x, so x = y = 2z, x3 = 2V , minimum value 3(2V )2/3
29. (a) Maximize f (α, β, γ) = cos α cos β cos γ subject to g(α, β, γ) = α + β + γ − π = 0,
∇f = λ∇g, − sin α cos β cos γ = λ, −√
cos α sin β cos γ = λ, − cos α cos β sin γ = λ with solution
α = β = γ = π/3, maximum value 3 3/8
(b) for example, f (α, β) = cos α cos β cos(π − α − β)
z
x
y
30. Find maxima and minima z = x2 + 4y 2 subject to the constraint g(x, y) = x2 + y 2 − 1 = 0,
∇z = λ∇g, 2xi + 8yj = 2λxi + 2λyj, solve 2x = 2λx, 8y = 2λy. If y 6= 0 then λ = 4, x = 0, y 2 = 1
and z = x2 + 4y 2 = 4. If y = 0 then x2 = 1 and z = 1, so the maximum height is obtained for
(x, y) = (0, ±1), z = 4 and the minimum height is z = 1 at (±1, 0).
569
Chapter 15
CHAPTER 15 SUPPLEMENTARY EXERCISES
1. (a) They approximate the profit per unit of any additional sales of the standard or high-resolution
monitors, respectively.
(b) The rates of change with respect to the two directions x and y, and with respect to time.
p
3. z = x2 + y 2 = c implies x2 + y 2 = c2 , which is the equation of a circle; x2 + y 2 = c is also the
equation of a circle (for c > 0).
y
y
3
3
x
x
-3
-3
3
3
-3
-3
z = x2 + y2
5. (b)
z=
√x2 + y2
f (x, y, z) = z − x2 − y 2
7. (a) f (ln y, ex ) = eln y ln ex = xy
(b) er+s ln rs
8. (a)
(b)
y
1
y= x
y
x
x
–1
1
√
1
y, wxy = 8xy sec2 (x2 + y 2 ) tan(x2 + y 2 ) + y −1/2 ,
2
1 −1/2
1
2 2
2
2 2
2
2
, wyx = 8xy sec (x + y ) tan(x + y 2 ) + y −1/2
wy = 2y sec (x + y ) + xy
2
2
9. wx = 2x sec2 (x2 + y 2 ) +
10. ∂w/∂x =
1
1
− sin(x + y), ∂ 2 w/∂x2 = −
− cos(x + y),
x−y
(x − y)2
∂w/∂y = −
1
1
− sin(x + y), ∂ 2 w/∂y 2 = −
− cos(x + y) = ∂ 2 w/∂x2
x−y
(x − y)2
11. Fx = −6xz, Fxx = −6z, Fy = −6yz, Fyy = −6z, Fz = 6z 2 − 3x2 − 3y 2 ,
Fzz = 12z, Fxx + Fyy + Fzz = −6z − 6z + 12z = 0
12. fx = yz + 2x, fxy = z, fxyz = 1, fxyzx = 0; fz = xy − (1/z), fzx = y, fzxx = 0, fzxxy = 0
Chapter 15 Supplementary Exercises
13. (a) P =
10 dT
10T dP
,
=
= 12 K/(m2 min)
V dT
V dt
14. (a) z = 1 − y 2 , slope =
∂z
= −2y = 4
∂y
570
(b)
P dV
dP
= −10 2
= 240 K/(m2 min)
dt
V dt
(b) z = 1 − 4x2 ,
∂z
= −8x = −8
∂x
15. x4 − x + y − x3 y = (x3 − 1)(x − y), limit = −1, continuous
16.
x4 − y 4
= x2 − y 2 , limit =
lim (x2 − y 2 ) = 0, continuous
x2 + y 2
(x,y)→(0,0)
√
1
1
1
2
1
2
√
√
√
√
√
17. Use the unit vectors u = h
,
i, v = h0, −1i, w = h−
,−
i=−
u + √ v, so that
2
2
5
5
5
5
√
√
1
1
7
2
2 √
Dw f = − √ Duf + √ Dvf = − √ 2 2 + √ (−3) = − √
5
5
5
5
5
18. (a) n = zx i + zy j − k = 8i + 8j − k, tangent plane 8x + 8y − z = 4 + 8 ln 2, normal line
x(t) = 1 + 8t, y(t) = ln 2 + 8t, z(t) = 4 − t
(b) n = 3i + 10j − 14k, tangent plane 3x + 10y − 14z = 30, normal line
x(t) = 2 + 3t, y(t) = 1 + 10t, z(t) = −1 − 14t
19. The origin is not such a point, so assume that the normal line at (x0 , y0 , z0 ) 6= (0, 0, 0) passes
through the origin, then n = zx i + zy j − k = −y0 i − x0 j − k; the line passes through the origin and
is normal to the surface if it has the form r(t) = −y0 ti−x0 tj−tk and (x0 , y0 , z0 ) = (x0 , y0 , 2−x0 y0 )
lies on the line if −y0 t = x0 , −x0 t = y0 , −t = 2 − x0 y0 , with solutions x0 = y0 = −1,
x0 = y0 = 1, x0 = y0 = 0; thus the points are (0, 0, 2), (1, 1, 1), (−1, −1, 1).
2 −1/3 2 −1/3 2 −1/3
−1/3
−1/3
−1/3
2/3
2/3
2/3
x
i+ y0
j+ z0
k, tangent plane x0 x+y0
y +z0
z = x0 +y0 +z0 = 1;
3 0
3
3
1/3
1/3
1/3
2/3
2/3
2/3
intercepts are x = x0 , y = y0 , z = z0 , sum of squares of intercepts is x0 + y0 + z0 = 1.
20. n =
21. A tangent to the line is 6i + 4j + k, a normal to the surface is n = 18xi + 8yj − k, so solve
18x = 6k, 8y = 4k, −1 = k; k = −1, x = −1/3, y = −1/2, z = 2
22. ∆w = (1.1)2 (−0.1) − 2(1.1)(−0.1) + (−0.1)2 (1.1) − 0 = 0.11,
dw = (2xy − 2y + y 2 )dx + (x2 − 2x + 2yx)dy = −(−0.1) = 0.1
23. dV =
1
1
2
2
xhdx + x2 dh = 2(−0.1) + (0.2) = −0.06667 m3 ; ∆V = −0.07267 m3
3
3
3
3
24. ∇f = (2x + 3y − 6)i + (3x + 6y + 3)j = 0 if 2x + 3y = 6, x + 2y = −1, x = 15, y = −8, D = 3 > 0,
fxx = 2 > 0, so f has a relative minimum at (15, −8).
25. ∇f = (2xy − 6x)i + (x2 − 12y)j = 0 if 2xy − 6x = 0, x2 − 12y = 0; if x = 0 then y = 0, and
if x 6= 0 then y = 3, x = ±6, thus the gradient vanishes at (0, 0), (−6, 3), (6, 3); fxx = 0 at all
three points, fyy = −12 < 0, D = −4x2 , so (±6, 3) are saddle points, and near the origin we write
f (x, y) = (y − 3)x2 − 6y 2 ; since y − 3 < 0 when |y| < 3, f has a maximum by inspection.
26. ∇f = (3x2 − 3y)i − (3x − y)j = 0 if y = x2 , 3x = y, so x = y = 0 or x = 3, y = 9; at
x = y = 0, D = −9, saddle point; at x = 3, y = 9, D = 9, fxx = 18 > 0, relative minimum
571
Chapter 15
27. (a)
(b) 5
K
P=3
5
4
P=2
4
3
3
P=1
2
2
1
L
1
2
3
4
5
1
0
0
1
2
3
4
5
28. (a) ∂P/∂L = cαLα−1 K β , ∂P/∂K = cβLα K β−1
(b) the rates of change of output with respect to labor and capital equipment, respectively
(c) K(∂P/∂K) + L(∂P/∂L) = cβLα K β + cαLα K β = (α + β)P = P
29. (a) L + K = 200,000, P = 1000L0.6 (200,000 − L)0.4 ,
L6
(200,000 − L)0.4
−
400
= 0 when L = 120,000,
L4
(200,000 − L)0.6
P = 102,033,960.1, which is a maximum because P = 0 at L = 0, 200,000, P > 0 in between,
and dP/dL = 0 has only the one solution.
dP/dL = 600
(b) Since L + K = 200,000 and L = 120,000, K = 80,000
√
√
30. (a) y 2 = 8 − 4x2 , find extrema of f (x) = x2 (8 − 4x2 ) = −4x4 + 8x2 defined for − 2 ≤ x ≤ 2.
Then f 0 (x) = −16x3 + 16x = 0 when x = 0, ±1, f 00 (x) = −48x2 + 16,
√ so f has a relative
maximum
√ at x = ±1, y = ±2 and a relative minimum at x = 0, y = ±2 2. At the endpoints
x = ± 2, y = 0 we obtain the minimum f = 0 again.
(b) f (x, y) = x2 y 2 , g(x, y) = 4x2 + y 2 − 8 = 0, ∇f = 2xy 2√i + 2x2 yj = λ∇g = 8λxi + 2λyj,
√ so
solve 2xy 2 = λ8x, 2x2 y = λ2y. If x = 0 then y = ±2 2, and if y = 0 then x = ± 2. In
either case f has a relative and absolute minimum. Assume x, y 6= 0, then y 2 = 4λ, x2 = λ,
use g = 0 to obtain x2 = 1, x = ±1, y = ±2, and f = 4 is a relative and absolute maximum
at (±1, ±2).
31. Let a corner of the box be at (x, y, z), so that (x/a)2 + (y/b)2 + (z/c)2 = 1. Maximize V = xyz
subject to g(x, y, z) = (x/a)2 + (y/b)2 + (z/c)2 = 1, solve ∇V = λ∇g, or
yzi + xzj + xyk = (2λx/a2 )i + (2λy/b2 )j + (2λz/c2 )k, a2 yz = 2λx, b2 xz = 2λy, c2 xy = 2λz. For the
maximum volume, x, y, z 6= 0; divide the first and second equations to obtain a2 y 2 = b2 x2 ; the first
√
and third to obtain a2 z 2 = c2 x2 , and finally b2 z 2 = c2 y 2 . From g = 0 get 3(x/a)2 = 1, x = ±a/ 3,
√
√
2a
2b
2c
and then y = ±b/ 3, z = ±c/ 3. The dimensions of the box are √ × √ × √ , and the
3
3
3
√
maximum volume is 8abc/(3 3).
32. (a) Let f (x, y) = 3x2 − 5xy + tan xy = 0. Then
dy
6x − 5y + y sec2 xy
dy
df
= 6x − 5y + y sec2 xy + (−5x + x sec2 xy)
= 0, so
=
.
dx
dx
dx
5x − x sec2 xy
dg
x
dy
(b) Let g(x, y) = x ln y + sin(x − y) = π,
= ln y + cos(x − y) +
− cos(x − y)
= 0,
dx
y
dx
ln y + cos(x − y)
dy
=
dx
−x/y + cos(x − y)
Chapter 15 Supplementary Exercises
33. F (x, y) = 0, Fx + Fy
572
dy
dy
dy
d2 y
= 0, Fxx + Fxy
+ Fyx
+ Fy 2 = 0, thus
dx
dx
dx
dx
Fx d2 y
Fxx + 2Fxy (dy/dx)
Fxx Fy − 2Fx Fxy
dy
=− , 2 =−
=−
dx
Fy dx
Fy
Fy2
34. Denote the currents I1 , I2 , I3 by x, y, z respectively. Then minimize F (x, y, z) = x2 R1 +y 2 R2 +z 2 R3
subject to g(x, y, z) = x+y +z −I = 0, so solve ∇F = λ∇g, 2xR1 i+2yR2 j+2zR3 k = λ(i + j + k),
1
1
1
:
:
.
λ = 2xR1 = 2yR2 = 2zR3 , so the minimum value of F occurs when I1 : I2 : I3 =
R1 R2 R3
√
35. Solve (t − 1)2 /4 + 16e−2t + (2 − t)2 = 1 for t to get t = 1.8332, 2.83984; the particle strikes
the surface at the points P1 (0.8332, 0.63959, 0.64603), P2 (1.83984, 0.23374, 0.31482). The velocity
√
dy
dz
dx
i + j + k = i − 4e−t j − 1/(2 t)k, and a normal to the surface is
vectors are given by v =
dt
dt
dt
n = ∇(x2 /4 + y 2 + z 2 ) = x/2i + 2yj + 2zk. At the points Pi these are
v1 = i − 0.639589j − 0.369286k, v2 = i − 0.23374j + 0.29670k;
n1 = 0.41661i + 1.27918j + 1.29207k and n2 = 0.91992i + 0.46748j + 0.62963k so
cos−1 [(vi · ni )/(kvi k kni k)] = 112.3◦ , 61.1◦ ; the acute angles are 67.7◦ , 61.1◦ .
36. (a) F 0 (x) =
Z
1
ey cos(xey )dy =
0
sin(ex) − sin x
x
π
so the maximum value of F (x) is
e+1
Z 1
π
π y
F
=
e
dy ≈ 0.909026.
sin
e+1
e+1
0
(b) Use a CAS to get x =
37. Let x, y, z be the lengths of the sides opposite angles α, β, γ, located at A,√B, C respectively. Then
x2 = y 2 + z 2 − 2yz cos α and x2 = 100 + 400 − 2(10)(20)/2 = 300, x = 10 3 and
dy
dz
dz
dy
dα
dx
= 2y
+ 2z
− 2 y cos α + z cos α − yz(sin α)
2x
dt
dt
dt
dt
dt
dt
!
√
1
10π
1
3 π
= 60 + √
= 2(10)(4) + 2(20)(2) − 2 10(2) + 20(4) − 10(20)
2
2
2 60
3
π
dx √
= 3 + , the length of BC is increasing.
dt
6
d ∂z
∂
∂ ∂z dy
∂ 2 z dx
∂ 2 z dy
∂z dx
38. (a)
=
+
=
+
by the Chain Rule, and
2
dt ∂x
∂x ∂x dt
∂y ∂x dt
∂x dt
∂y∂x dt
∂ ∂z dy
∂ 2 z dx ∂ 2 z dy
∂ ∂z dx
d ∂z
=
+
=
+ 2
dt ∂y
∂x ∂y dt
∂y ∂y dt
∂x∂y dt
∂y dt
so
(b)
∂z dx ∂z dy
dz
=
+
,
dt
∂x dt
∂y dt
2
∂ 2 z dy
∂z d2 x dy
∂z d2 y
dx ∂ 2 z dx
d2 z
∂ z dx ∂ 2 z dy
+
+
+
+
=
+
dt2
dt ∂x2 dt
∂y∂x dt
∂x dt2
dt ∂x∂y dt
∂y 2 dt
∂y dt2
CHAPTER 16
Multiple Integrals
EXERCISE SET 16.1
Z
1
Z
Z
2
1.
Z
1
(x + 3)dy dx =
0
Z
(2x + 6)dx = 7
0
4
Z
0
Z
1
4
x y dx dy =
3.
2
Z
0
2
Z
ln 3
Z
ln 2
0
Z
2
Z
0
Z
1
0
2
0
Z
5
7.
Z
1
1
Z
0
Z
π
Z
1
0
Z
Z
Z
1
2
Z
2
ln 2
Z
1
1
Z
Z
2
Z
Z
1
−2
Z
1
Z
Z
Z
2
π/2
4
3
dx = 1 − ln 2
1
1
−
x+1 x+2
dx = ln(25/24)
Z
1
dy dx =
√
√
[x(x2 + 2)1/2 − x(x2 + 1)1/2 ]dx = (3 3 − 4 2 + 1)/3
0
1
x(1 − x2 )1/2 dx = 1/3
0
Z
Z
π/3
(x sin y − y sin x)dy dx =
16.
0
0 dx = 0
Z
p
2
x 1 − x dy dx =
3
15.
0
xy
x2 + y 2 + 1
0
1
10dx = 20
4
−1
p
14.
0
−3
1
4xy 3 dy dx =
13.
−1
6
dy dx =
1 x
(e − 1)dx = (1 − ln 2)/2
2
0
1
dy dx =
(x + y)2
12.
3
1
1−
x+1
Z
7
(sin 2x − sin x)dx = −2
0
4
Z
π
xy ey x dy dx =
Z
(3 + 3y 2 )dy = 14
−2
π/2
11.
0
6
4
1
ln 2
−1
8.
x cos xy dy dx =
Z
−2
3 dy = 3
2
10.
π/2
0
2
(x + y )dx dy =
Z
x
dy dx =
(xy + 1)2
9.
0
2
−1
2
Z
Z
2
4.
0
dx dy =
−1
Z
1
sin x dx = (1 − cos 2)/2
2
0
Z
0
4x dx = 16
1
0
y sin x dy dx =
Z
−1
3
(2x − 4y)dy dx =
ex dx = 2
0
6.
Z
Z
1
ln 3
ex+y dy dx =
5.
Z
1
1
y dy = 2
3
2
3
2.
0
Z
5
0
Z
Z
2
(2x + y)dy dx =
19. V =
3
Z
1
3
Z
1
0
x π2
−
sin x dx = π 2 /144
2
18
5
(2x + 3/2)dx = 19
3
Z
2
(3x3 + 3x2 y)dy dx =
20. V =
π/2
3
(6x3 + 6x2 )dx = 172
1
573
Exercise Set 16.1
Z
2
574
Z
Z
3
x2 dy dx =
21. V =
0
Z
0
3
Z
2
3x2 dx = 8
0
Z
4
3
5(1 − x/3)dy dx =
22. V =
0
0
5(4 − 4x/3)dx = 30
0
z
23. (a)
z
(b)
(0, 0, 5)
(1, 0, 4)
y
y
(2, 5, 0)
(3, 4, 0)
x
x
z
24. (a)
z
(b)
(2, 2, 8)
(0, 0, 2)
y
y
(2, 2, 0)
(1, 1, 0)
x
Z
1/2
x
Z
Z
π
0
0
iπ
1/2
x cos(xy) cos2 πx dy dx =
25.
cos2 πx sin(xy)
Z
dx
0
0
1/2
=
0
i1/2
1
1
cos2 πx sin πx dx = − cos3 πx
=
3
3π
0
z
26. (a)
y
3
5
x
(b) The projection onto the xy-plane consists of R1 : [0, 5] × [0, 1], over which lie each of the two
skew planes, and R2 : [0, 5] × [1, 3], over which is only the plane z = −2y + 6, so
Z 5Z 3
Z 5Z 1
85
45
+ 20 =
((−2y + 6) − y) dy dx +
(−2y + 6) dy dx =
V =
2
2
0
0
0
1
27. fave =
2
π
Z
π/2
Z
1
y sin xy dy dx =
0
0
2
π
Z
π/2
− cos xy
0
i1
dx =
0
2
π
Z
π/2
(1 − cos x) dx = 1 −
0
2
π
575
Chapter 16
1
29. Tave =
2
Z
1
3
28. average =
Z
Z
Z
1
√
1
[(1 + y)3/2 − y 3/2 ]dy = 2(31 − 9 3)/45
9
3
x(x2 + y)1/2 dx dy =
0
0
1Z
0
2
10 − 8x − 2y
2
0
2
0
1
A(R)
30. fave =
3
Z
b
Z
d
44
− 16x2
3
0
c
31. 0.6211310829
dx =
14
3
◦
32. 2.230985141
Z
ZZ
b
"Z
#
d
f (x, y)dA =
33.
1
1
(b − a)(d − c)k = k
A(R)
k dy dx =
a
Z
1
dy dx =
2
g(x)h(y)dy dx =
a
c
"Z
R
Z
# "Z
b
g(x)dx
=
#
d
"Z
b
#
d
g(x)
h(y)dy dx
a
c
h(y)dy
a
c
34. The integral of tan x (an odd function) over the interval [−1, 1] is zero.
35. The first integral equals 1/2, the second equals −1/2. No, because the integrand is not continuous.
16
X
36. (a)
4 X
4
X
i
f (x∗k , yk∗ )∆A∗k =
i=1 j=1
k=1
Z
2
Z
Z
2
0
1
4
−2
1
j
−
2 4
2
1
= −4
2
2
(x − 2y) dy dx =
(b)
2
−
(2x − 4) dx = −4
0
0
EXERCISE SET 16.2
Z
1
Z
Z
x
x2
0
Z
1 4
(x − x7 )dx = 1/40
3
0
Z
3/2
Z
3−y
3/2
(3y − 2y 2 )dy = 7/24
y dx dy =
2.
y
1
Z
3
1
Z √9−y2
Z
3
y dx dy =
3.
0
0
Z
1
Z
x
√
2π
Z
√
Z
Z
p
x/y dy dx =
x2
1/4
y
p
Z
Z
1
1
Z
x3
Z
x2
(x − y)dy dx =
−1
−x2
Z
2(x − x3/2 )dx = 13/80
[−x cos(x2 ) + x]dx = π/2
π
Z
4
2x dx = 4/5
−1
1
1/4
1
2
6.
x1/2 y −1/2 dy dx =
√
2π
√
0
Z
x
x2
1/4
sin(y/x)dy dx =
π
9 − y 2 dy = 9
0
4.
5.
1
xy 2 dy dx =
1.
π
Z
7.
π/2
0
x2
1
cos(y/x)dy dx =
x
Z
π
sin x dx = 1
π/2
Exercise Set 16.2
Z
1
Z
x
576
Z
2
1
0
Z
0
2
Z
y2
e
Z
2
Z
3
Z
(y+7)/2
Z
1
1
Z
√
x
Z
x2
√
1 Z
1−x2
(b)
√
− 1−x2
−1
Z
Z
Z
8
2 −1/2
Z
0
Z
Z
x
Z
17.
5
√
1−x2
Z
(3x − 2y)dy dx =
√
25−x2
Z
Z
Z
Z
Z
6−y
2
y2
0
Z
π/4
√
1/ 2
Z
π/4
Z
sin y
0
Z
0
Z
x3
x
Z
x
√
1/ 2
Z
x dy dx +
x
1
(−x4 + x3 + x2 − x)dx = −1/2
0
Z
2
Z
(x4 − x3 − x2 + x)dx +
2x
22.
(x − 1)dy dx
x3
0
−1
0
Z
0
=
Z
1
(x − 1)dy dx +
21.
−1
−1
p
6x 1 − x2 dx = 0
1
cos 2y dy = 1/8
4
x dx dy =
20.
0
1
1
(36y − 12y 2 + y 3 − y 5 )dy = 50/3
2
xy dx dy =
Z
√
1
y(1 + y 2 )−1/2 dy = ( 17 − 1)/2
2
0
19.
0
1
1 4
y dy = 31/10
2
(5x − x2 )dx = 125/6
5−x
2
0
2
5
y dy dx =
18.
0
xy 2 dx dy =
0
√
− 1−x2
−1
Z
y
x sin x dx = π
0
1
Z
π
x cos y dy dx =
0
4
dx dy =
0
π
2
p
2x 1 − x2 dx + 0 = 0
14.
1
x(1 + y )
16.
1
−1
4
√
y
15.
Z
y dy dx =
(x3 − 16x)dx = 576
16/x
Z
√
− 1−x2
−1
Z
√
1−x2
Z
1
x dy dx +
x2 dy dx =
Z
(x3/2 + x/2 − x3 − x4 /2)dx = 3/10
0
x
13.
Z
1
(x + y)dy dx =
0
Z
0
2
−(y−5)/2
12. (a)
0
0
1 3
x dx = 1/12
3
16
1 5
x dx =
3
0 2
Z 3
xy dx dy =
(3y 2 + 3y)dy = 38
0
0
1
4
1
(e − 1)y 2 dy = 7(e − 1)/3
Z
x2
(b)
Z
Z
x2 − y 2 dy dx =
xy dy dx =
Z
4
p
1
11. (a)
8
x
2
dx dy =
0
Z
Z
Z
y
0
Z
x/y 2
1
9.
0
10.
1
Z
2
xex dx = (e − 1)/2
ex dy dx =
8.
1
√
1/ 2
Z
Z
1/x
2
Z
3
x dy dx =
x
√
1/ 2
x dx +
0
1
√
1/ 2
(x − x3 )dx = 1/8
577
Chapter 16
y
24. (a)
(b)
(1, 3), (3, 27)
25
R
15
5
x
1
Z
3
Z
2
Z
4x3 −x4
3
x[(4x3 − x4 ) − (3 − 4x + 4x2 )] dx =
x dy dx =
(c)
1
3−4x+4x2
π/4
Z
Z
1
Z
cos x
25. A =
π/4
(cos x − sin x)dx =
dy dx =
sin x
0
Z
Z
1
Z
Z
−4
3y−4
Z
3
(−y 2 − 3y + 4)dy = 125/6
Z
9−y 2
27. A =
3
8(1 − y 2 /9)dy = 32
dx dy =
−3
Z
1
−3
1−y 2 /9
Z
Z
cosh x
28. A =
1
dy dx =
sinh x
0
4Z
(cosh x − sinh x)dx = 1 − e−1
0
Z
6−3x/2
4
(3 − 3x/4 − y/2) dy dx =
29.
0
Z
0
2
Z
[(3 − 3x/4)(6 − 3x/2) − (6 − 3x/2)2 /4] dx = 12
0
√
4−x2
Z 2
p
2
4 − x dy dx =
(4 − x2 ) dx = 16/3
30.
0
√
2−1
1
dx dy =
−4
224
15
0
−y 2
26. A =
Z
3
0
0
Z
3
Z
(3
√
− 9−x2
31. V =
−3
Z
1
√
9−x2
Z
Z
− x)dy dx =
−3
Z
x
(2x3 − x4 − x6 )dx = 11/70
x2
0
Z
3
Z
0
Z
2
3
(9x2 + y 2 )dy dx =
33. V =
0
Z
Z
0
(1 − x)dx dy =
−1
y2
3/2
Z
35. V =
3
Z
1
(1/2 − y 2 + y 4 /2)dy = 8/15
−1
√
9−4x2
√
− 9−4x2
−3/2
Z
Z
1
34. V =
Z
(18x2 + 8/3)dx = 170
0
1
Z
−3/2
Z
(9 − x2 )dx dy =
y 2 /3
3/2
(y + 3)dy dx =
3
36. V =
0
p
p
(6 9 − x2 − 2x 9 − x2 )dx = 27π
1
(x2 + 3y 2 )dy dx =
32. V =
3
p
6 9 − 4x2 dx = 27π/2
3
(18 − 3y 2 + y 6 /81)dy = 216/7
0
Exercise Set 16.2
Z
578
√
25−x2
Z
5
Z
p
25 −
37. V = 8
x2 dy dx
(25 − x2 )dx = 2000/3
0
0
5
=8
0
Z √1−(y−1)2
1
2 3/2
2
2 1/2
[1 − (y − 1) ] + y [1 − (y − 1) ]
dy,
(x + y )dx dy = 2
38. V = 2
3
0
0
0
Z π/2
1
cos3 θ + (1 + sin θ)2 cos θ cos θ dθ which eventually yields
let y − 1 = sin θ to get V = 2
−π/2 3
V = 3π/2
Z
2
2
Z
√
1−x2
Z
1
0
0
Z
√
4−x2
Z
2
Z
2
√
2
Z
0
Z
2
f (x, y)dx dy
41.
1
Z
0
Z
f (x, y)dx dy
Z
4
Z
−y 2
e
47.
0
Z
Z
Z
4
dx dy =
0
1
0
Z
2x
Z
x2
Z
3
2
0
ln 3
Z
3
1
2
x dx dy =
ey
0
2
Z
(9 − e2y )dy =
0
Z
0
1
Z
1
Z
Z
e
f (x, y)dy dx
0
x2
1
(9 ln 3 − 4)
2
2
y 2 sin(y 3 )dy = (1 − cos 8)/3
ex
1
(ex − xex )dx = e/2 − 1
x dy dx =
0
Z
4
53. (a)
2
0
2
√
x
0
Z
Z
Z
0
y2
sin πy 3 dy dx; the inner integral is non-elementary.
sin πy 3 dx dy =
Z
0
2
√
x
46.
0
0
52.
0
Z
sin x
f (x, y)dy dx
ln 3
sin(y 3 )dx dy =
Z
f (x, y)dy dx
3
Z
y2
51.
0
Z
2
ln x
1
0
50.
Z
Z
43.
0
π/2
e2
x2 ex dx = (e8 − 1)/3
ex dy dx =
Z
Z
x/2
0
49.
0
Z
2x cos(x2 )dx = sin 1
0
2
8
1
cos(x )dy dx =
48.
Z
p
1
2 3/2
2
x 4 − x + (4 − x )
dx = 2π
3
1 −y2
ye dy = (1 − e−16 )/8
4
2
0
0
y/4
(1 − x2 )3/2 dx = π/2
0
45.
ey
0
1
f (x, y)dy dx
e
44.
Z
42.
y2
0
2
2
0
0
Z
2
2
(x + y )dy dx =
40. V =
Z
8
3
(1 − x2 − y 2 )dy dx =
39. V = 4
Z
2
1
cos πy 3
y 2 sin πy 3 dy = −
3π
2
=0
0
579
Chapter 16
Z
1
Z
π/2
sec2 (cos x)dx dy ; the inner integral is non-elementary.
(b)
sin−1 y
0
Z
π/2
Z
Z
sin x
π/2
2
sec2 (cos x) sin x dx = tan 1
sec (cos x)dy dx =
0
0
Z
2
0
√
4−x2
Z
π/2
128
64 64
+
sin2 θ −
sin4 θ
3
3
3
0
Z
Z
2
Z
1
4−
0
x2
Z
−1
−2
Z
2
Z
1
8
x dx +
3
Z
7
x dx +
3
−1
Z
1
Z
1
x
0
Z
Z
0
1
2
8
x dx = 0
3
1
dy dx = 2
1 + x2
Z
0
p
1 − x2 − y 2 , so 1/8 of the volume of
1
x
π
1
−
dx = − ln 2
1 + x2
1 + x2
2
(3x − x2 − x) dx = 4/3, so
0
Z
2
Z
3x−x2
(x2 − xy)dy dx =
0
x
3
4
Z
2
(−2x3 + 2x4 − x5 /2)dx = −
0
59. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so
Z a Z sin x p
1 + x + y dy dx = 0.676089
V =
0
x/2
EXERCISE SET 16.3
Z
π/2
Z
Z
sin θ
1.
π/2
r cos θdr dθ =
0
Z
0
πZ
0
Z
1+cos θ
2.
π
r dr dθ =
0
Z
0
π/2
0
Z
Z
a sin θ
Z
0
π/2
2
a3
sin3 θ dθ = a3
3
9
π/3
Z
0
Z
cos 3θ
4.
π/3
r dr dθ =
0
Z
0
π
Z
0
1
cos2 3θ dθ = π/12
2
Z
1−sin θ
2
5.
r cos θ dr dθ =
0
0
1
sin2 θ cos θ dθ = 1/6
2
1
(1 + cos θ)2 dθ = 3π/4
2
r2 dr dθ =
3.
0
(x = 2 sin θ)
2
58. Area =
3
f¯ =
4
dx
2
56. This is the volume in the first octant under the surface z =
π
the sphere of radius 1, thus .
6
57. Area of triangle is 1/2, so f¯ = 2
xy 2 dy dx
1
1
1
+ (4 − x2 )3/2
3
64 π 64 π 128 π 1 · 3
+
−
= 8π
3 2
3 4
3 2 2·4
dθ =
xy 2 dy dx +
−1
=
2
xy 2 dy dx +
55.
x
p
0
=
−2
2
0
0
Z
−1
2
2
(x + y ) dy dx = 4
54. V = 4
Z
Z
2
0
π
1
(1 − sin θ)3 cos θ dθ = 0
3
2
3 8
=−
4 15
5
Exercise Set 16.3
Z
π
Z
580
Z
cos θ
π
r3 dr dθ =
6.
0
0
0
Z
2π
Z
Z
1−cos θ
7. A =
0
Z
Z
π/2
Z
sin 2θ
Z
π/2
Z
0
Z
1
π/2
1
(1 − sin2 2θ)dθ = π/16
2
r dr dθ =
π/4
sin 2θ
Z
Z
π/3
π/4
Z
2
π/3
(4 − sec2 θ)dθ = 4π/3 −
r dr dθ =
10. A = 2
sec θ
0
Z
Z
π/2
Z
4 sin θ
Z
π/6
2
Z
π
Z
1
π
(−2 cos θ − cos2 θ)dθ = 2 − π/4
r dr dθ =
12. A = 2
π/2
Z
1+cos θ
Z
π/2
3
13. V = 8
0
Z
1
Z
π/2
π/2
Z π/2
p
128 √
64 √
2
r 9 − r dr dθ =
2
dθ =
2π
3
3
0
2 sin θ
r2 dr dθ =
14. V = 2
0
Z
0
Z
π/2
cos θ
Z
16
3
π/2
sin3 θ dθ = 32/9
0
Z
1
(1 − r )r dr dθ =
2
15. V = 2
Z
0
Z
π/2
Z
3
(1 − sin4 θ)dθ = 5π/32
0
Z
1
π/2
Z
0
π/2
dr dθ = 8
16. V = 4
dθ = 4π
0
Z
3 sin θ
π/2
sin4 θ dθ = 27π/16
r2 sin θ dr dθ = 9
17. V =
0
0
Z
π/2
0
Z
2 cos θ
18. V = 4
0
Z
1
19.
0
0
Z
2
3
20.
0
π/4
Z
0
0
π/2
Z
1
(1 − e−1 )
2
Z
p
2
r 9 − r dr dθ = 9
0
21.
Z
p
32 π/2
16
2
(3π − 4)
r 4 − r dr dθ =
(1 − sin3 θ)dθ =
3 0
9
e−r r dr dθ =
0
π/2
Z
2π
22.
0
dθ = (1 − e−1 )π
0
π/2
dθ = 9π/2
0
2
1
1
r dr dθ = ln 5
2
1+r
2
2 cos θ
Z
16
2r sin θ dr dθ =
3
π/4
dθ =
0
Z
π
ln 5
8
π/2
2
π/4
π/2
2
0
3
√
(16 sin2 θ − 4)dθ = 4π/3 + 2 3
π/2
r dr dθ =
π/6
√
0
11. A = 2
Z
sin2 2θ dθ = π/2
0
9. A =
Z
π/2
r dr dθ = 2
0
Z
1
(1 − cos θ)2 dθ = 3π/2
2
0
8. A = 4
2π
2π
r dr dθ =
0
Z
1
cos4 θ dθ = 3π/32
4
cos3 θ sin θ dθ = 1/3
π/4
581
Chapter 16
Z
π/2
Z
1
1
r dr dθ =
4
Z
π/2
3
23.
0
Z
0
2π
Z
2
−r 2
e
24.
0
Z
0
π/2
Z
dθ = π/8
0
1
r dr dθ = (1 − e−4 )
2
2 cos θ
0
Z
0
π/2
Z
1
Z
8
3
r2 dr dθ =
25.
Z
2π
π/2
cos3 θ dθ = 16/9
0
1
cos(r )r dr dθ = sin 1
2
Z
π/2
2
26.
0
Z
0
π/2
Z
Z
0
π/4
Z
dθ =
0
sec θ tan θ
1
r dr dθ =
3
Z
π/4
2
28.
0
Z
0
π/4
Z
2
29.
0
Z
√
0
5
1
r dr dθ =
2
3 csc θ
30.
tan−1 (3/4)
=
2π
Z
Z
a
0
Z
0
π/2
Z
a
32. (a) V = 8
0
0
(25 − 9 csc2 θ)dθ
i
25
25 h π
− tan−1 (3/4) − 6 =
tan−1 (4/3) − 6
2 2
2
2π
h
0
Z
π/2
tan−1 (3/4)
hr dr dθ =
31. V =
0
√
sec3 θ tan3 θ dθ = 2( 2 + 1)/45
π √
r
dr dθ = ( 5 − 1)
4
1 + r2
Z
π/2
Z
π
sin 1
4
p
π
r
2
1
−
1/
dr
dθ
=
1
+
a
2
(1 + r2 )3/2
a
27.
0
dθ = (1 − e−4 )π
0
a2
dθ = πa2 h
2
4c
c 2
(a − r2 )1/2 r dr dθ = − π(a2 − r2 )3/2
a
3a
a
=
0
4 2
πa c
3
4
(b) V ≈ π(6378.1370)2 6356.5231 ≈ 1,083,168,200,000 km3
3
Z
π/2
Z
a sin θ
33. V = 2
0
Z
0
π/4
Z
√
a 2 cos 2θ
Z
π/4
Z
4 sin θ
π/2
Z
4 sin θ
r dr dθ +
r dr dθ
π/4
0
Z
π/4
π/2
(8 sin θ − 4 cos 2θ)dθ +
2
=
π/6
Z
φ
√
8 sin2 θ dθ = 4π/3 + 2 3 − 2
π/4
Z
Z
2a sin θ
2
r dr dθ = 2a
36. A =
0
(1 − cos3 θ)dθ = (3π − 4)a2 c/9
0
0
√
8 cos 2θ
π/6
π/2
cos 2θ dθ = 2a2
0
35. A =
Z
π/4
r dr dθ = 4a
0
Z
Z
2
34. A = 4
Z
2
c 2
(a − r2 )1/2 r dr dθ = a2 c
a
3
0
0
φ
1
sin2 θ dθ = a2 φ − a2 sin 2φ
2
Exercise Set 16.4
582
Z
+∞
37. (a) I 2 =
0
Z
+∞
2
Z
+∞
Z
2
e−y dy =
0
Z
+∞
=
0
Z
e−x dx
2
2
0
π/2
Z
+∞
2
(b) I =
−r 2
e
0
0
1
r dr dθ =
2
Z
+∞
0
+∞
2π
Z
R
D(r)r dr dθ =
0
Z
Z
tan−1 (2)
0
Z
−r
tan−1 (1/3)
(c)
Z
π
Z
1
re−r dr dθ = π
4
0
Z
1
re−r dr ≈ 1.173108605
4
0
−r
R
= 2πk[1 − (R + 1)e−R ]
0
tan−1 (2)
2
cos2 θ dθ =
tan−1 (1/3)
0
√
π/2
I=
r dr dθ = −2πk(1 + r)e
r cos θ dr dθ = 4
40.
dx dy
0
2
3
+y 2 )
0
ke
0
2
π/2
(b)
R
39. V =
e−(x
0
0
Z
2
2
e−x dx e−y dy
0
Z
Z
2π
+∞
dθ = π/4
38. (a) 1.173108605
Z
Z
0
Z
e−x e−y dx dy =
+∞
1
1 π
+ 2[tan−1 (2) − tan−1 (1/3)] = +
5
5
2
EXERCISE SET 16.4
z
1. (a)
z
(b)
y
x
x
y
z
(c)
x
y
z
2. (a)
z
(b)
x
y
y
x
583
Chapter 16
z
(c)
y
x
3. (a) x = u, y = v, z =
5 3
+ u − 2v
2 2
(b) x = u, y = v, z = u2
4. (a) x = u, y = v, z =
v
1 + u2
(b) x = u, y = v, z =
1 2 5
v −
3
3
5. (a) x = 5 cos u, y = 5 sin u, z = v; 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1
(b) x = 2 cos u, y = v, z = 2 sin u; 0 ≤ u ≤ 2π, 1 ≤ v ≤ 3
6. (a) x = u, y = 1 − u, z = v; −1 ≤ v ≤ 1
7. x = u, y = sin u cos v, z = sin u sin v
9. x = r cos θ, y = r sin θ, z =
1
1 + r2
(b) x = u, y = 5 + 2v, z = v; 0 ≤ u ≤ 3
8. x = u, y = eu cos v, z = eu sin v
10. x = r cos θ, y = r sin θ, z = e−r
2
11. x = r cos θ, y = r sin θ, z = 2r2 cos θ sin θ
12. x = r cos θ, y = r sin θ, z = r2 (cos2 θ − sin2 θ)
13. x = r cos θ, y = r sin θ, z =
√
√
9 − r2 ; r ≤ 5
14. x = r cos θ, y = r sin θ, z = r; r ≤ 3
√
1
1
3
ρ
15. x = ρ cos θ, y = ρ sin θ, z =
2
2
2
16. x = 3 cos θ, y = 3 sin θ, z = 3 cot φ
17. z = x − 2y; a plane
18. y = x2 + z 2 , 0 ≤ y ≤ 4; part of a circular paraboloid
19. (x/3)2 + (y/2)2 = 1; 2 ≤ z ≤ 4; part of an elliptic cylinder
20. z = x2 + y 2 ; 0 ≤ z ≤ 4; part of a circular paraboloid
21. (x/3)2 + (y/4)2 = z 2 ; 0 ≤ z ≤ 1; part of an elliptic cone
22. x2 + (y/2)2 + (z/3)2 = 1; part of an ellipsoid
23. (a) x = r cos θ, y = r sin θ, z = r; x = u, y = v, z =
√
u2 + v 2 ; 0 ≤ z ≤ 2
24. (a) I: x = r cos θ, y = r sin θ, z = r2 ; II: x = u, y = v, z = u2 + v 2 ; u2 + v 2 ≤ 2
25. (a) 0 ≤ u ≤ 3, 0 ≤ v ≤ π
(b) 0 ≤ u ≤ 4, −π/2 ≤ v ≤ π/2
Exercise Set 16.4
584
26. (a) 0 ≤ u ≤ 6, −π ≤ v ≤ 0
(b) 0 ≤ u ≤ 5, π/2 ≤ v ≤ 3π/2
27. (a) 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π
(b) 0 ≤ φ ≤ π, 0 ≤ θ ≤ π
28. (a) π/2 ≤ φ ≤ π
(b) 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/2
29. u = 1, v = 2, ru × rv = −2i − 4j + k; 2x + 4y − z = 5
30. u = 1, v = 2, ru × rv = −4i − 2j + 8k; 2x + y − 4z = −6
31. u = 0, v = 1, ru × rv = 6k; z = 0
32. ru × rv = 2i − j − 3k; 2x − y − 3z = −4
√
√
33. ru × rv = ( 2/2)i − ( 2/2)j + (1/2)k; x − y +
√
√
π 2
2
z=
2
8
34. ru × rv = 2i − ln 2k; 2x − (ln 2)z = 0
p
p
9 − y 2 , zx = 0, zy = −y/ 9 − y 2 , zx2 + zy2 + 1 = 9/(9 − y 2 ),
Z 2Z 3
Z 2
3
p
S=
dy dx =
3π dx = 6π
9 − y2
0
−3
0
35. z =
Z
36. z = 8 − 2x − 2y,
zx2
+
zy2
4
Z
+ 1 = 4 + 4 + 1 = 9, S =
Z
4−x
0
0
4
3(4 − x)dx = 24
3 dy dx =
0
37. z 2 = 4x2 + 4y 2 , 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus
Z 1Z x√
√ Z 1
√
zx2 + zy2 + 1 = (16x2 + 16y 2 )/z 2 + 1 = 5, S =
5 dy dx = 5
(x − x2 )dx = 5/6
0
x2
0
38. z 2 = x2 + y 2 , zx = x/z, zy = y/z, zx2 + zy2 + 1 = (z 2 + y 2 )/z 2 + 1 = 2,
ZZ √
Z π/2 Z 2 cos θ √
√ Z π/2
√
S=
2 dA = 2
2 r dr dθ = 4 2
cos2 θ dθ = 2π
0
R
0
0
39. zx = −2x, zy = −2y, zx2 + zy2 + 1 = 4x2 + 4y 2 + 1,
ZZ p
Z 2π Z 1 p
S=
4x2 + 4y 2 + 1 dA =
r 4r2 + 1 dr dθ
0
R
0
√
1
(5 5 − 1)
=
12
Z
2π
√
dθ = (5 5 − 1)π/6
0
40. zx = 2, zy = 2y, zx2 + zy2 + 1 = 5 + 4y 2 ,
Z 1Z yp
Z 1 p
√
2
S=
5 + 4y dx dy =
y 5 + 4y 2 dy = (27 − 5 5)/12
0
0
0
41. ∂r/∂u = cos vi + sin vj + 2uk, ∂r/∂v = −u sin vi + u cos vj,
Z 2π Z 2 p
√
√
√
u 4u2 + 1 du dv = (17 17 − 5 5)π/6
k∂r/∂u × ∂r/∂vk = u 4u2 + 1; S =
0
1
42. ∂r/∂u = cos vi + sin vj + k, ∂r/∂v = −u sin vi + u cos vj,
√
Z π/2 Z 2v √
√
2 3
π
2 u du dv =
k∂r/∂u × ∂r/∂vk = 2u; S =
12
0
0
585
Chapter 16
43. zx = y, zy = x, zx2 + zy2 + 1 = x2 + y 2 + 1,
ZZ p
Z π/6 Z 3 p
Z π/6
√
√
1
S=
x2 + y 2 + 1 dA =
r r2 + 1 dr dθ = (10 10 − 1)
dθ = (10 10 − 1)π/18
3
0
0
0
R
44. zx = x, zy = y, zx2 + zy2 + 1 = x2 + y 2 + 1,
ZZ p
Z
Z 2π Z √8 p
26 2π
2
2
S=
x + y + 1 dA =
r r2 + 1 dr dθ =
dθ = 52π/3
3 0
0
0
R
45. On the sphere, zx = −x/z and zy = −y/z so zx2 + zy2 + 1 = (x2 + y 2 + z 2 )/z 2 = 16/(16 − x2 − y 2 );
the planes z = 1 and z = 2 intersect the sphere along the circles x2 + y 2 = 15 and x2 + y 2 = 12;
ZZ
Z 2π Z √15
Z 2π
4
4r
p
√
S=
dA =
dr dθ = 4
dθ = 8π
√
16 − r2
16 − x2 − y 2
12
0
0
R
46. On the sphere, zx = −x/z and zy = −y/z so zx2 + zy2 + 1 = (x2 + y 2 + z 2 )/z 2 = 8/(8 − x2 − y 2 );
the cone cuts the sphere in the circle x2 + y 2 = 4;
Z 2π Z 2 √
√ Z 2π
√
2 2r
√
S=
dr dθ = (8 − 4 2)
dθ = 8(2 − 2)π
2
8−r
0
0
0
47. r(u, v) = a cos u sin vi + a sin u sin vj + a cos vk, kru × rv k = a2 sin v,
Z π
Z π Z 2π
a2 sin v du dv = 2πa2
sin v dv = 4πa2
S=
0
0
0
Z
h
Z
48. r = r cos ui + r sin uj + vk, kru × rv k = r; S =
2π
r du dv = 2πrh
0
0
h
h
h2 x2 + h2 y 2
x
y
p
+ 1 = (a2 + h2 )/a2 ,
, zy = p
, zx2 + zy2 + 1 = 2 2
a x2 + y 2
a x2 + y 2
a (x + y 2 )
Z 2π
Z 2π Z a √ 2
p
1 p
a + h2
r dr dθ = a a2 + h2
dθ = πa a2 + h2
S=
a
2
0
0
0
49. zx =
50. Revolving a point (a0 , 0, b0 ) of the xz-plane around the z-axis generates a circle, an equation of
which is r = a0 cos ui + a0 sin uj + b0 k, 0 ≤ u ≤ 2π. A point on the circle (x − a)2 + z 2 = b2 which
generates the torus can be written r = (a + b cos v)i + b sin vk, 0 ≤ v ≤ 2π. Set a0 = a + b cos v
and b0 = a + b sin v and use the first result: any point on the torus can thus be written in the form
r = (a + b cos v) cos ui + (a + b cos v) sin uj + b sin vk, which yields the result.
51. ∂r/∂u = −(a + b cos v) sin ui + (a + b cos v) cos uj,
∂r/∂v = −b sin v cos ui − b sin v sin uj + b cos vk, k∂r/∂u × ∂r/∂vk = b(a + b cos v);
Z 2π Z 2π
b(a + b cos v)du dv = 4π 2 ab
S=
0
0
√
52. kru × rv k = u2 + 1; S =
Z
0
4π
Z
0
5
Z
p
2
u + 1 du dv = 4π
5
p
u2 + 1 du = 174.7199011
0
53. z = −1 when v ≈ 0.27955, z = 1 when v ≈ 2.86204, kru × rv k = | cos v|;
Z 2π Z 2.86204
| cos v| dv du ≈ 9.099
S=
0
0.27955
Exercise Set 16.5
586
54. (a) x = v cos u, y = v sin u, z = f (v), for example
x = v cos u, y = v sin u, z = 1/v 2
(b)
z
(c)
x
y
55. (x/a)2 + (y/b)2 + (z/c)2 = cos2 v(cos2 u + sin2 u) + sin2 v = 1, ellipsoid
56. (x/a)2 + (y/b)2 − (z/c)2 = cos2 u cosh2 v + sin2 u cosh2 v − sinh2 v = 1, hyperboloid of one sheet
57. (x/a)2 + (y/b)2 − (z/c)2 = sinh2 v + cosh2 v(sinh2 u − cosh2 u) = −1, hyperboloid of two sheets
EXERCISE SET 16.5
Z
Z
1
Z
2
Z
1
2
1.
2
Z
1
2
(x + y + z )dx dy dz =
−1
Z
0
1/2
Z
π
−1
Z
Z
1
Z
2
0
Z
y2
0
Z
1/2
Z
1/3
Z
z
2
Z
π
−1
0
Z
π/4
−1
Z
1
Z
Z
−1
Z
x2
Z
π/4
Z
3
Z
√
9−z 2
0
Z
Z
x
3
0
√
9−z 2
Z
0
0
Z
3
Z
0
x2
Z
0
0
Z
ln z
3
Z
x
1
Z
2
Z
0
√
4−x2
x
1
Z
3−x2 −y 2
2
Z
−5+x2 +y 2
0
0
Z
2
0
2
Z
2
Z
8.
1
z
0
√
3y
y
dx dy dz =
2
x + y2
Z
1
√
1
cos y dy = 2/8
4
1
(81 − 18z 2 + z 4 )dz = 81/5
8
1 5 3 3
x − x + x2 dx = 118/3
2
2
√
4−x2
0
0
=
Z
3
1/3
1
1
(1 − cos πx)dx =
+
2
12
[2x(4 − x2 ) − 2xy 2 ]dy dx
x dz dy dx =
7.
3
0
(xz − x)dz dx =
1
Z
Z
Z
x2
xey dy dz dx =
6.
1 3
x dx dz =
2
1/2
π/4
0
xy dy dx dz =
5.
Z
1
Z
47
1 7 1 5 1
y + y − y dy =
3
2
6
3
x3 cos y dx dy =
0
0
0
x cos y dz dx dy =
0
2
(yz 2 + yz)dz dy =
0
4.
0
1
x sin xy dy dx =
2
y2
yz dx dz dy =
3.
(10/3 + 2z 2 )dz = 8
−1
0
zx sin xy dz dy dx =
1/3
1
2
(1/3 + y + z )dy dz =
0
2.
Z
2
2
2
Z
z
2
4
x(4 − x2 )3/2 dx = 128/15
3
π
dy dz =
3
Z
1
2
π
(2 − z)dz = π/6
3
√
3−2
4π
587
Chapter 16
Z
π
Z
Z
1
Z
π/6
9.
Z
π
Z
0
1
0
Z
0
Z
1−x2
Z
y
Z
1
0
−1
0
Z
√
2
Z
Z
Z
x
y dy dx =
−1
−1
0
√
2
Z
2−x2
11.
Z
x
0
0
Z
π/2
Z
π/2
0
Z
0
Z
xy
π/2
Z
π/6
Z
1
y
π/6
Z √1−x2 −y2
14. 8
0
0
Z
e−x
2
−y 2 −z 2
Z
4
y
Z
0
(4−x)/2
0
4
dz dy dx ≈ 2.381
Z
Z
(12−3x−6y)/4
4
Z
(4−x)/2
0
1
0
0
Z
1−x
Z
Z
√
y
Z
1
1−x
dz dy dx =
16. V =
0
0
Z
0
Z
2
Z
4
0
0
Z
4−y
2
x2
0
Z
y
Z
0
18. V =
Z
0
0
1
y dy dx =
2
(1 − x)3/2 dx = 4/15
3
Z
4
2
1
x2
0
y
Z
p
1 − y 2 dx dy =
Z
dz dx dy =
0
Z
1
8 − 4x2 + x4 dx = 256/15
2
(4 − y)dy dx = 2
0
Z √1−y2
√
0
dz dy dx = 2
17. V = 2
1
0
1
(12 − 3x − 6y)dy dx
4
3
(4 − x)2 dx = 4
16
=
Z
√
y cos y dy = (5π − 6 3)/12
π/6
dz dy dx =
Z
π/2
1 3
x (2 − x2 )2 dx = 1/6
4
0
15. V =
Z
0
y sin x dx dy =
0
√
1−x2
Z
√
2
Z
π/2
cos(z/y)dz dx dy =
12.
1
(1 − x2 )3 dx = 32/105
3
1
xy(2 − x2 )2 dy dx =
2
xyz dz dy dx =
0
1
2
0
(1 − 3/π)x dx = π(π − 3)/2
0
1−x2
y dz dy dx =
10.
π
x[1 − cos(πy/6)]dy dx =
xy sin yz dz dy dx =
0
Z
1
0
0
1
y
p
1 − y 2 dy = 1/3
0
19. The projection of the curve of intersection onto the xy-plane is x2 + y 2 = 1,
Z 1 Z √1−x2 Z 4−3y2
V =4
dz dy dx
4x2 +y 2
0
0
20. The projection of the curve of intersection onto the xy-plane is 2x2 + y 2 = 4,
Z √2 Z √4−2x2 Z 8−x2 −y2
V =4
dz dy dx
Z
3x2 +y 2
0
0
3
Z
√
9−x2 /3
Z
Z
x+3
dz dy dx
21. V = 2
−3
0
0
1
Z
√
1−x2
Z
22. V = 8
√
1−x2
dz dy dx
0
0
0
Exercise Set 16.5
588
z
23. (a)
z
(b)
(0, 9, 9)
(0, 0, 1)
y
x
(1, 0, 0)
(3, 9, 0)
y
x
z
(c)
(0, 0, 1)
y
(1, 2, 0)
x
z
24. (a)
z
(b)
(0, 0, 2)
(0, 0, 2)
(0, 2, 0)
y
y
(3, 9, 0)
(2, 0, 0)
x
x
z
(c)
(0, 0, 4)
y
(2, 2, 0)
x
Z
1
Z
1−x
Z
25. V =
Z
1−x−y
1
Z
1−x
Z
dz dy dx = 1/6, fave = 6
0
0
0
1−x−y
(x + y + z) dz dy dx =
0
0
0
26. The integrand is an odd function of each of x, y, and z, so the answer is zero.
3
4
589
Chapter 16
Z
a
Z
b(1−x/a)
Z
Z
c(1−x/a−y/b)
27. (a)
Z
b
Z
a(1−y/b)
c(1−x/a−y/b)
dz dy dx,
0
Z
0
Z
c
Z
a(1−z/c)
0
dz dx dy,
Z
b(1−x/a−z/c)
0
Z
a
0
Z
c(1−x/a)
0
b(1−x/a−z/c)
dy dx dz,
Z
0
Z
c
0
Z
b(1−z/c)
0
dy dz dx,
Z
a(1−y/b−z/c)
0
Z
b
0
c(1−y/b)
0
Z
a(1−y/b−z/c)
dx dy dz,
0
0
dx dz dy
0
0
0
0
(b) Use the first integral in part (a) to get
Z a
Z a Z b(1−x/a)
x 2
x y
1
1
dy dx =
bc 1 −
c 1− −
dx = abc
a
b
2
a
6
0
0
0
Z
√
9−x2
Z
3
Z √9−x2 −y2
f (x, y, z)dz dy dx
28. (a)
Z
0
Z
4
0
x/2
Z
0
Z
2
f (x, y, z)dz dy dx
(b)
0
0
Z
a
Z
d
4−y
x2
0
0
dz dy dx
0
Z
0
Z
`
b
Z
c
k
"Z
d
f (x)g(y)h(z)dz dy dx =
30.
a
Z
Z b√1−x2 /a2 Z c√1−x2 /a2 −y2 /b2
0
b
4−x2
f (x, y, z)dz dy dx
0
29. V = 8
Z
Z
2
(c)
f (x)g(y)
a
k
"Z
b
f (x)
=
"Z
31. (a)
−1
Z
f (x)dx
y dy
ey dy
e2x dx
0
# "Z
ln 3
0
# "Z
d
g(y)dy
c
#
`
h(z)dz
k
sin z dz = (0)(1/3)(1) = 0
0
"Z
h(z)dz
#
π/2
2
0
1
(b)
"Z
1
#
`
k
# "Z
a
Z
x dx
# "Z
c
b
=
1
#
d
g(y)dy dx
a
Z
h(z)dz dy dx
c
"Z
#
`
ln 2
#
e−z dz = [(e2 − 1)/2](2)(1/2) = (e2 − 1)/2
0
32. (a) At any point outside the closed sphere {x2 + y 2 + z 2 ≤ 1} the integrand is negative, so to
maximize the integral it suffices to include all points inside the sphere; hence the maximum
value is taken on the region G = {x2 + y 2 + z 2 ≤ 1}.
(b)
8π
15
EXERCISE SET 16.6
1. Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum is
5(0 − a) + 10(5 − a) + 20(10 − a) = 0 for equilibrium, so 250 − 35a = 0, a = 50/7. The fulcrum
should be placed 50/7 ft to the right of m1 .
Exercise Set 16.6
590
2. At equilibrium, 10(0 − 4) + 3(2 − 4) + 4(3 − 4) + m(6 − 4) = 0, m = 25
Z
Z
1
1
1
x dy dx = , ȳ =
2
3. A = 1, x̄ =
0
1
2
4. A = 2, x̄ =
0
Z
Z
0
Z
1
1
y dy dx =
1
0
0
Z
1−x
x dy dx +
−1
−1−x
ZZ
5. A = 1/2,
Z
1+x
Z
Z
x dy dx = 0, similarly ȳ = 0.
0
1
Z
−1+x
ZZ
x
x dA =
x dy dx = 1/3,
0
R
1
2
Z
1
Z
x
y dA =
0
y dy dx = 1/6;
0
R
0
centroid (2/3, 1/3)
Z
1
Z
ZZ
x2
dy dx = 1/3,
6. A =
0
0
Z
ZZ
1
x2
ZZ
2−x2
dy dx = 7/6,
7. A =
x
0
Z
ZZ
1
Z
Z
1
Z
2−x2
x dA =
x dy dx = 5/12,
x
0
R
2−x2
y dA =
y dy dx = 19/15; centroid (5/14, 38/35)
x
0
R
ZZ
π
8. A = ,
4
x dy dx = 1/4,
0
y dy dx = 1/10; centroid (3/4, 3/10)
R
Z
x2
0
0
1
Z
1
0
R
Z
y dA =
Z
Z
x dA =
Z
1
Z
x dA =
x dy dx =
0
R
√
1−x2
0
4
4
1
, x̄ =
, ȳ =
by symmetry
3
3π
3π
9. x̄ = 0 from the symmetry of the region,
ZZ
Z πZ b
2
4(b3 − a3 )
1
.
y dA =
r2 sin θ dr dθ = (b3 − a3 ); centroid x̄ = 0, ȳ =
A = π(b2 − a2 ),
2
3
3π(b2 − a2 )
0
a
R
10. ȳ = 0 from the symmetry of the region, A = πa2 /2,
Z π/2 Z a
ZZ
4a
2
3
,0
x dA =
r cos θ dr dθ = 2a /3; centroid
3π
−π/2 0
R
Z
1
Z
Z
√
x
11. M =
1
Z
(x + y)dy dx = 13/20, Mx =
0
0
Z
1
Z
My =
(x + y)y dy dx = 3/10,
0
√
x
√
x
0
(x + y)x dy dx = 19/42, x̄ = My /M = 190/273, ȳ = Mx /M = 6/13;
0
0
the mass is 13/20 and the center of gravity is at (190/273, 6/13).
Z
π
Z
sin x
y dy dx = π/4, x̄ = π/2 from the symmetry of the density and the region,
12. M =
0
Z
0
π
Z
sin x
y 2 dy dx = 4/9, ȳ = Mx /M =
Mx =
0
0
16
; mass π/4, center of gravity
9π
π 16
,
.
2 9π
591
Chapter 16
Z
Z
π/2
13. M =
0
a
r3 sin θ cos θ dr dθ = a4 /8, x̄ = ȳ from the symmetry of the density and the
Z π/2 Z a
=
r4 sin θ cos2 θ dr dθ = a5 /15, x̄ = 8a/15; mass a4 /8, center of gravity
0
region, My
0
0
(8a/15, 8a/15).
Z
π
Z
1
r3 dr dθ = π/4, x̄ = 0 from the symmetry of density and region,
0
0
Z πZ 1
8
8
Mx =
; mass π/4, center of gravity 0,
.
r4 sin θ dr dθ = 2/5, ȳ =
5π
5π
0
0
14. M =
Z
1
Z
1
Z
1
x dz dy dx =
15. V = 1, x̄ =
0
0
0
1
1
, similarly ȳ = z̄ = ; centroid
2
2
ZZZ
16. V = πr2 h = 2π, x̄ = ȳ = 0 by symmetry,
2
Z
2π
1 1 1
, ,
2 2 2
Z
z dz dy dx =
G
centroid = (0, 0, 1)
Z
1
rz dr dθ dz = 2π,
0
0
0
17. x̄ = ȳ = z̄ from the symmetry of the region, V = 1/6,
Z Z
Z
1 1 1−x 1−x−y
x dz dy dx = (6)(1/24) = 1/4; centroid (1/4, 1/4, 1/4)
x̄ =
V 0 0
0
18. The solid is described by −1 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y 2 , 0 ≤ x ≤ 1 − z;
2 Z
Z 1 Z 1−y2 Z 1−z
Z Z
1−z
1 1 1−y
4
5
V =
, ȳ = 0 by symmetry,
dx dz dy = , x̄ =
x dx dz dy =
5
V
14
−1 0
0
−1 0
0
2 Z
Z Z
1−z
1 1 1−y
2
2
5
z̄ =
, 0,
.
z dx dz dy = ; the centroid is
V −1 0
7
14
7
0
19. x̄ = 1/2 and ȳ = 0 from the symmetry of the region,
Z 1Z 1 Z 1
ZZZ
1
V =
dz dy dx = 4/3, z̄ =
z dV = (3/4)(4/5) = 3/5; centroid (1/2, 0, 3/5)
V
0
−1 y 2
G
20. x̄ = ȳ from the symmetry of the region,
ZZZ
Z 2 Z 2 Z xy
1
dz dy dx = 4, x̄ =
x dV = (1/4)(16/3) = 4/3,
V =
V
0
0
0
z̄ =
1
V
ZZZ
G
z dV = (1/4)(32/9) = 8/9; centroid (4/3, 4/3, 8/9)
G
21. x̄ = ȳ = z̄ from the symmetry of the region, V = πa3 /6,
Z Z √a2 −x2 Z √a2 −x2 −y2
Z Z √a2 −x2 p
1 a
1 a
x̄ =
x dz dy dx =
x a2 − x2 − y 2 dy dx
V 0 0
V 0 0
0
Z
Z
6
1 π/2 a 2 p 2
r a − r2 cos θ dr dθ =
(πa4 /16) = 3a/8; centroid (3a/8, 3a/8, 3a/8)
=
3
V 0
πa
0
Exercise Set 16.6
592
22. x̄ = ȳ = 0 from the symmetry of the region, V = 2πa3 /3
Z Z √a2 −x2 Z √a2 −x2 −y2
Z Z √a2 −x2
1 a
1 a
1 2
z̄ =
(a − x2 − y 2 )dy dx
z dz dy dx =
√
√
V −a − a2 −x2 0
V −a − a2 −x2 2
Z
Z
3
1 2π a 1 2
(a − r2 )r dr dθ =
(πa4 /4) = 3a/8; centroid (0, 0, 3a/8)
=
V 0
2πa3
0 2
Z
Z
a
a
Z
a
(a − x)dz dy dx = a4 /2, ȳ = z̄ = a/2 from the symmetry of density and
23. M =
0
0
0
1
region, x̄ =
M
Z
Z
a
a
Z
a
x(a − x)dz dy dx = (2/a4 )(a5 /6) = a/3;
0
0
0
mass a4 /2, center of gravity (a/3, a/2, a/2)
Z
√
a2 −x2
Z
a
24. M =
√
− a2 −x2
−a
1
M
and region, z̄ =
Z
h
(h − z)dz dy dx =
0
ZZZ
z(h − z)dV =
1 2 2
πa h , x̄ = ȳ = 0 from the symmetry of density
2
2
(πa2 h3 /6) = h/3;
πa2 h2
G
2 2
mass πa h /2, center of gravity (0, 0, h/3)
Z
Z
1
1
Z
1−y 2
yz dz dy dx = 1/6, x̄ = 0 by the symmetry of density and region,
25. M =
−1
1
ȳ =
M
0
0
ZZZ
1
y z dV = (6)(8/105) = 16/35, z̄ =
M
ZZZ
2
yz 2 dV = (6)(1/12) = 1/2;
G
G
mass 1/6, center of gravity (0, 16/35, 1/2)
Z
Z
3
Z
9−x2
1
26. M =
0
0
ȳ =
1
M
0
1
xz dz dy dx = 81/8, x̄ =
M
ZZZ
x2 z dV = (8/81)(81/5) = 8/5,
G
ZZZ
xyz dV = (8/81)(243/8) = 3, z̄ =
G
1
M
ZZZ
xz 2 dV = (8/81)(27/4) = 2/3;
G
mass 81/8, center of gravity (8/5, 3, 2/3)
Z
1
Z
0
x̄ =
1
k(x2 + y 2 )dy dx = 2k/3, x̄ = ȳ from the symmetry of density and region,
27. (a) M =
1
M
0
ZZ
kx(x2 + y 2 )dA =
3
(5k/12) = 5/8; center of gravity (5/8, 5/8)
2k
R
(b) ȳ = 1/2 from the symmetry of density and region,
ZZ
Z 1Z 1
1
M=
kx dy dx = k/2, x̄ =
kx2 dA = (2/k)(k/3) = 2/3,
M
0
0
center of gravity (2/3, 1/2)
R
593
Chapter 16
28. (a) x̄ = ȳ = z̄ from the symmetry of density and region,
Z 1Z 1Z 1
k(x2 + y 2 + z 2 )dz dy dx = k,
M=
0
1
x̄ =
M
0
0
ZZZ
kx(x2 + y 2 + z 2 )dV = (1/k)(7k/12) = 7/12; center of gravity (7/12, 7/12, 7/12)
G
(b) x̄ = ȳ = z̄ from the symmetry of density and region,
Z 1Z 1Z 1
k(x + y + z)dz dy dx = 3k/2,
M=
0
1
x̄ =
M
0
0
ZZZ
kx(x + y + z)dV =
2
(5k/6) = 5/9; center of gravity (5/9, 5/9, 5/9)
3k
G
Z
ZZZ
π
Z
sin x
Z
1/(1+x2 +y 2 )
dV =
29. V =
G
1
x=
V
dz dy dx = 0.666633,
0
ZZZ
0
0
1
xdV = 1.177406, y =
V
ZZZ
G
1
ydV = 0.353554, z =
V
ZZZ
G
zdV = 0.231557
G
30. (b) Use cylindrical coordinates to get
Z 2π Z a Z 1/(1+r2 )
ZZZ
dV =
r dz dr dθ = π ln(1 + a2 ),
V =
0
G
ZZZ
1
V
z=
zdV =
0
0
a2
2(1 + a2 ) ln(1 + a2 )
G
lim+ z̄ =
(c)
a→0
1
; lim z̄ = 0; solve z = 1/4 for a to obtain a ≈ 1.980291.
2 a→+∞
31. Let x = r cos θ, y = r sin θ, and dA = r dr dθ in formulas (10) and (11).
Z
2π
Z
a(1+sin θ)
r dr dθ = 3πa2 /2,
32. x̄ = 0 from the symmetry of the region, A =
1
ȳ =
A
Z
2π
Z
0
a(1+sin θ)
r2 sin θ dr dθ =
0
0
0
2
(5πa3 /4) = 5a/6; centroid (0, 5a/6)
3πa2
Z
π/2
Z
sin 2θ
r dr dθ = π/8,
33. x̄ = ȳ from the symmetry of the region, A =
x̄ =
1
A
Z
π/2
0
Z
sin 2θ
r2 cos θ dr dθ = (8/π)(16/105) =
0
0
0
128
; centroid
105π
128 128
,
105π 105π
34. x̄ = 3/2 and ȳ = 1 from the symmetry of the region,
ZZ
ZZ
x dA = x̄A = (3/2)(6) = 9,
y dA = ȳA = (1)(6) = 6
R
R
35. x̄ = 0 from the symmetry of the region, πa2 /2 is the area of the semicircle, 2π ȳ is the distance
traveled by the centroid to generate the sphere so 4πa3 /3 = (πa2 /2)(2π ȳ), ȳ = 4a/(3π)
Exercise Set 16.7
594
36. (a) V =
1 2
πa
2
1
4a
= π(3π + 4)a3
2π a +
3π
3
√
4a
2
a+
so
(b) the distance between the centroid and the line is
2
3π
#
" √
4a
1√
2
1 2
πa
a+
=
V =
2π
2π(3π + 4)a3
2
2
3π
6
37. x̄ = k so V = (πab)(2πk) = 2π 2 abk
38. ȳ = 4 from the symmetry of the region,
Z 2 Z 8−x2
A=
dy dx = 64/3 so V = (64/3)[2π(4)] = 512π/3
−2
x2
1
39. The region generates a cone of volume πab2 when it is revolved about the x-axis, the area of the
3
1
1
1
1
2
ab (2π ȳ), ȳ = b/3. A cone of volume πa2 b is generated when the
region is ab so πab =
2
3
2
3
1 2
1
ab (2πx̄), x̄ = a/3. The centroid is (a/3, b/3).
region is revolved about the y-axis so πa b =
3
2
Z
a
Z
b
y 2 δ dy dx =
40. Ix =
Z
0
a
Z
0
b
Z
2π
Z
a
b
x2 δ dy dx =
0
0
0
1
(x2 + y 2 )δ dy dx = δab(a2 + b2 )
3
Z
a
Iz =
0
Z
1
δab3 , Iy =
3
Z
2
3
41. Ix =
2π
Z
4
a
r3 cos2 θ δ dr dθ = δπa4 /4 = Ix ;
r sin θ δ dr dθ = δπa /4; Iy =
0
1 3
δa b,
3
0
0
0
Iz = Ix + Iy = δπa4 /2
EXERCISE SET 16.7
Z
2π
Z
1
Z
√
1−r 2
1.
Z
Z
2π
1
0
1
(1 − r2 )r dr dθ =
2
π/2
Z
zr dz dr dθ =
0
Z
π/2
0
0
Z
cos θ
0
Z
Z
r2
0
Z
0
0
π/2
Z
π/2
Z
0
0
2π Z
r sin θ dr dθ =
0
Z
1
0
π/4 Z
0
π/2
0
Z
a sec φ
Z
π/2
0
2π Z
ρ2 sin φ dρ dφ dθ =
4.
0
0
Z
0
2π
Z
3
0
Z
Z
9
2π
Z
0
0
r2
0
π/4
0
0
π/2
1
cos4 θ sin θ dθ = 1/20
4
1
sin φ cos φ dφ dθ =
4
1 3
a sec3 φ sin φ dφ dθ =
3
Z
3
r(9 − r2 )dr dθ =
r dz dr dθ =
5. V =
0
Z
ρ3 sin φ cos φ dρ dφ dθ =
Z
1
dθ = π/4
8
3
0
3.
2π
cos θ
r sin θ dz dr dθ =
2.
Z
0
2π
Z
π/2
0
Z
2π
0
81
dθ = 81π/2
4
1
dθ = π/16
8
1 3
a dθ = πa3 /3
6
595
Chapter 16
Z
Z
2π
√
9−r 2
Z
2
6. V = 2
Z
2π
Z
2
r dz dr dθ = 2
0
r
0
0
0
p
9 − r2 dr dθ
0
Z
√
2
= (27 − 5 5)
3
√
dθ = 4(27 − 5 5)π/3
2π
0
7. r2 + z 2 = 20 intersects z = r2 in a circle of radius 2,
Z 2π Z 2 Z √20−r2
Z 2π Z 2 p
V =
r dz dr dθ =
(r 20 − r2 − r3 )dr dθ
0
r2
0
0
0
Z
√
4
(10 5 − 19)
3
=
2π
√
dθ = 8(10 5 − 19)π/3
0
8. z = hr/a intersects z = h in a circle of radius a,
Z 2π Z a
Z 2π
Z 2π Z a Z h
h
1 2
2
(ar − r )dr dθ =
a h dθ = πa2 h/3
r dz dr dθ =
V =
a
6
0
0
hr/a
0
0
0
Z
2π
Z
π/3
Z
Z
4
2π
Z
π/3
2
ρ sin φ dρ dφ dθ =
9. V =
0
Z
0
2π
Z
0
π/4
Z
0
Z
2
0
2π
Z
π/4
ρ2 sin φ dρ dφ dθ =
10. V =
0
0
1
0
0
32
64
sin φ dφ dθ =
3
3
Z
2π
dθ = 64π/3
0
√
7
7
sin φ dφ dθ = (2 − 2)
3
6
Z
2π
√
dθ = 7(2 − 2)π/3
0
11. In spherical coordinates the sphere and the plane z = a are ρ = 2a and ρ = a sec φ, respectively.
They intersect at φ = π/3,
Z 2π Z π/2 Z 2a
Z 2π Z π/3 Z a sec φ
2
ρ sin φ dρ dφ dθ +
ρ2 sin φ dρ dφ dθ
V=
Z
0
0
2π
Z
0
Z
π/3
π/3
0
Z
2π
1 3
a sec3 φ sin φ dφ dθ +
3
0
0
0
Z 2π
Z 2π
4
1
dθ + a3
dθ = 11πa3 /3
= a3
2
3
0
0
π/2
8 3
a sin φ dφ dθ
3
=
Z
2π
Z
π/2
Z
π/2
Z
a
Z
0
0
a2 −r 2
13.
Z
Z
π/2
Z
1
Z
π/2
e−ρ ρ2 sin φ dρ dφ dθ =
π/4
Z
15.
0
Z
0
2π
3
0
Z
Z
π
√
8
3
ρ3 sin φ dρ dφ dθ = 81π
16.
0
0
0
a
1 6
a
12
Z
0
π/2
cos2 θ dθ = πa6 /48
0
1
(1 − e−1 )
3
Z
0
π
Z
π/2
sin φ dφ dθ = (1 − e−1 )π/3
0
√
ρ4 cos2 φ sin φ dρ dφ dθ = 32(2 2 − 1)π/15
0
Z
Z
√ Z
√
9 2 2π
9 sin φ dφ dθ =
dθ = 9 2π
2
0
(a2 r3 − r5 ) cos2 θ dr dθ
0
14.
0
π/2
r3 cos2 θ dz dr dθ =
=
0
π/2
π/4
0
0
0
π
Z
ρ sin φ dρ dφ dθ =
π/4
0
Z
2π
π/3
2
12. V =
Z
Z
3
0
Exercise Set 16.7
Z
π/2
Z
596
√ Z π/2
4,294,967,296 √
1
2
37
cos θ cos φ dφ dθ =
cos37 θ dθ =
2 ≈ 0.008040
18
36 0
755,505,013,725
π/4
18.
0
0
Z
Z
2π
Z
a
√
a2 −r 2
19. (a) V = 2
0
Z
0
Z
2π
π
0
Z
a
ρ2 sin φ dρ dφ dθ = 4πa3 /3
(b) V =
0
Z
0
0
Z √4−x2 −y2
√
4−x2
Z
2
r dz dr dθ = 4πa3 /3
xyz dz dy dx
20. (a)
0
0
0
Z
2
√
4−x2
Z
=
0
0
Z
π/2
Z
2
1
1
xy(4 − x2 − y 2 )dy dx =
2
8
Z
2
x(4 − x2 )2 dx = 4/3
0
√
4−r 2
Z
r3 z sin θ cos θ dz dr dθ
(b)
0
0
0
Z
π/2
Z
=
0
Z
π/2
Z
Z
π/2
0
2
8
1 3
(4r − r5 ) sin θ cos θ dr dθ =
2
3
Z
π/2
sin θ cos θ dθ = 4/3
0
2
ρ5 sin3 φ cos φ sin θ cos θ dρ dφ dθ
(c)
0
0
0
Z
π/2
Z
π/2
=
0
Z
2π
Z
3
Z
0
8
32
sin3 φ cos φ sin θ cos θ dφ dθ =
3
3
Z
3
2π
Z
3
27
1
r(3 − r)2 dr dθ =
2
8
(3 − z)r dz dr dθ =
21. M =
0
Z
r
0
2π
Z
a
Z
0
Z
h
22. M =
0
Z
2π
a
1
1 2
kh r dr dθ = ka2 h2
2
4
k zr dz dr dθ =
0
Z
0
2π
Z
0
π
Z
0
0
Z
a
2π
Z
π
kρ3 sin φ dρ dφ dθ =
23. M =
0
0
2π Z
Z
0
πZ
0
2
24. M =
0
2π Z
Z
ρ sin φ dρ dφ dθ =
0
0
1
0
0
π
Z
Z
π/2
sin θ cos θ dθ = 4/3
0
Z
2π
dθ = 27π/4
0
2π
dθ = πka2 h2 /2
0
1
1 4
ka sin φ dφ dθ = ka4
4
2
3
sin φ dφ dθ = 3
2
Z
Z
2π
dθ = πka4
0
2π
dθ = 6π
0
25. x̄ = ȳ = 0 from the symmetry of the region,
Z 2π Z 1 p
Z 2π Z 1 Z √2−r2
√
r dz dr dθ =
(r 2 − r2 − r3 )dr dθ = (8 2 − 7)π/6,
V =
0
z̄ =
1
V
centroid
r2
0
2π Z
Z
1
Z
0
√
2−r 2
zr dz dr dθ =
0
0
0, 0,
r2
7
√
16 2 − 14
0
√
6
(7π/12) = 7/(16 2 − 14);
(8 2 − 7)π
√
26. x̄ = ȳ = 0 from the symmetry of the region, V = 8π/3,
Z
Z Z
3
1 2π 2 2
(4π) = 3/2; centroid (0, 0, 3/2)
zr dz dr dθ =
z̄ =
V 0
8π
0
r
597
Chapter 16
27. x̄ = ȳ = z̄ from the symmetry of the region, V = πa3 /6,
Z
Z
Z
6
1 π/2 π/2 a 3
ρ cos φ sin φ dρ dφ dθ =
(πa4 /16) = 3a/8;
z̄ =
V 0
πa3
0
0
centroid (3a/8, 3a/8, 3a/8)
Z
2π
Z
π/3
Z
4
ρ2 sin φ dρ dφ dθ = 64π/3,
28. x̄ = ȳ = 0 from the symmetry of the region, V =
Z
1
z̄ =
V
Z
2π
0
Z
π/3
4
0
0
3
(48π) = 9/4; centroid (0, 0, 9/4)
64π
ρ3 cos φ sin φ dρ dφ dθ =
0
0
0
Z
π/2
Z
2 cos θ
Z
r2
r dz dr dθ = 3π/2,
29. ȳ = 0 from the symmetry of the region, V = 2
Z
2
x̄ =
V
Z
π/2
0
Z
2 cos θ
r2
r2 cos θ dz dr dθ =
Z
2
z̄ =
V
0
Z
π/2
0
Z
2 cos θ
0
r2
Z
0
0
Z
π/2
Z
2 cos θ
Z
4−r 2
Z
π/2
2 cos θ
zr dz dr dθ =
30. M =
0
=
4
(π) = 4/3,
3π
4
(5π/6) = 10/9; centroid (4/3, 0, 10/9)
3π
rz dz dr dθ =
0
16
3
Z
0
0
Z
0
0
1
r(4 − r2 )2 dr dθ
2
π/2
(1 − sin6 θ)dθ = (16/3)(11π/32) = 11π/6
0
π/2
Z
Z
π/3
Z
2
π/2
Z
π/3
ρ2 sin φ dρ dφ dθ =
31. V =
π/6
0
0
2π
Z
π/4
Z
π/6
0
√
= 2( 3 − 1)π/3
Z
0
0
Z
1
2π
Z
π/4
3
ρ sin φ dρ dφ dθ =
32. M =
0
0
0
0
0
Z
4 √
8
sin φ dφ dθ = ( 3 − 1)
3
3
√
1
1
sin φ dφ dθ = (2 − 2)
4
8
π/2
dθ
0
Z
2π
dθ = (2 −
√
2)π/4
0
33. x̄ = ȳ = 0 from the symmetry of density and region,
Z 2π Z 1 Z 1−r2
(r2 + z 2 )r dz dr dθ = π/4,
M=
0
z̄ =
1
M
Z
0
2π
Z
0
1
Z
1−r 2
z(r2 +z 2 )r dz dr dθ = (4/π)(11π/120) = 11/30; center of gravity (0, 0, 11/30)
0
0
0
Z
2π
Z
1
Z
34. x̄ = ȳ = 0 from the symmetry of density and region, M =
1
z̄ =
M
Z
2π
Z
1
zr dz dr dθ = π/4,
0
Z
r
0
0
r
z 2 r dz dr dθ = (4/π)(2π/15) = 8/15; center of gravity (0, 0, 8/15)
0
0
0
35. x̄ = ȳ = 0 from the symmetry of density and region,
Z 2π Z π/2 Z a
kρ3 sin φ dρ dφ dθ = πka4 /2,
M=
0
1
z̄ =
M
Z
0
2π
Z
0
π/2
Z
a
kρ4 sin φ cos φ dρ dφ dθ =
0
0
0
2
(πka5 /5) = 2a/5; center of gravity (0, 0, 2a/5)
πka4
Exercise Set 16.7
598
36. x̄ = z̄ = 0 from the symmetry of the region, V = 54π/3 − 16π/3 = 38π/3,
Z Z Z
Z Z
1 π π 65
1 π π 3 3 2
sin2 φ sin θ dφ dθ
ρ sin φ sin θ dρ dφ dθ =
ȳ =
V 0 0 2
V 0 0 4
Z
1 π 65π
3
=
sin θ dθ =
(65π/4) = 195/152; centroid (0, 195/152, 0)
V 0 8
38π
Z
2π
Z
Z
π
R
37. M =
δ0 e
0
0
Z
(ρ/R)3 2
2π
Z
π
ρ sin φ dρ dφ dθ =
0
0
0
4
1
(e − 1)R3 δ0 sin φ dφ dθ = π(e − 1)δ0 R3
3
3
38. (a) The sphere and cone intersect in a circle of radius ρ0 sin φ0 ,
Z θ2 Z ρ0 sin φ0 q
Z θ2 Z ρ0 sin φ0 Z √ρ20 −r2
2
2
2
r ρ0 − r − r cot φ0 dr dθ
r dz dr dθ =
V=
θ1
Z
θ2
θ1
0
1
1 3
ρ0 (1 − cos3 φ0 − sin3 φ0 cot φ0 )dθ = ρ30 (1 − cos3 φ0 − sin2 φ0 cos φ0 )(θ2 − θ1 )
3
3
=
θ1
=
r cot φ0
0
1 3
ρ (1 − cos φ0 )(θ2 − θ1 ).
3 0
(b) From part (a), the volume of the solid bounded by θ = θ1 , θ = θ2 , φ = φ1 , φ = φ2 , and
1
1
1
ρ = ρ0 is ρ30 (1 − cos φ2 )(θ2 − θ1 ) − ρ30 (1 − cos φ1 )(θ2 − θ1 ) = ρ30 (cos φ1 − cos φ2 )(θ2 − θ1 )
3
3
3
so the volume of the spherical wedge between ρ = ρ1 and ρ = ρ2 is
1
1
∆V = ρ32 (cos φ1 − cos φ2 )(θ2 − θ1 ) − ρ31 (cos φ1 − cos φ2 )(θ2 − θ1 )
3
3
1 3
(ρ − ρ31 )(cos φ1 − cos φ2 )(θ2 − θ1 )
3 2
=
(c)
d
cos φ = − sin φ so from the Mean-Value Theorem cos φ2 −cos φ1 = −(φ2 −φ1 ) sin φ∗ where
dφ
d 3
φ∗ is between φ1 and φ2 . Similarly
ρ = 3ρ2 so ρ32 −ρ31 = 3ρ∗2 (ρ2 −ρ1 ) where ρ∗ is between
dρ
ρ1 and ρ2 . Thus cos φ1 −cos φ2 = sin φ∗ ∆φ and ρ32 −ρ31 = 3ρ∗2 ∆ρ so ∆V = ρ∗2 sin φ∗ ∆ρ∆φ∆θ.
Z
2π
Z
a
Z
Z
h
2π
Z
a
Z
h
r2 δ r dz dr dθ = δ
39. Iz =
0
Z
0
2π
Z
0
Z
a
r3 dz dr dθ =
0
0
0
Z
h
2π
Z
a
1
(hr3 cos2 θ + h3 r)dr dθ
3
(r2 cos2 θ + z 2 )δr dz dr dθ = δ
40. Iy =
0
0
Z
0
2π
=δ
0
Z
2π
Z
0
π
1
π
1 4
a h cos2 θ + a2 h3 dθ = δ
a4 h + a2 h3
4
6
4
3
a2
Z
Z
h
2π
Z
a2
Z
h
r2 δ r dz dr dθ = δ
Z
a1
2π
Z
0
41. Iz =
0
π
0
Z
r3 dz dr dθ =
0
a1
0
Z
a
2π
Z
π
Z
(ρ2 sin2 φ)δ ρ2 sin φ dρ dφ dθ = δ
42. Iz =
0
0
1
δπa4 h
2
0
1
δπh(a42 − a41 )
2
a
ρ4 sin3 φ dρ dφ dθ =
0
0
0
8
δπa5
15
599
Chapter 16
EXERCISE SET 16.8
1.
∂(x, y)
=
∂(u, v)
1
3
3.
∂(x, y)
=
∂(u, v)
cos u
sin u
4.
∂(x, y)
=
∂(u, v)
5. x =
4
−5
2.
− sin v
cos v
2(v 2 − u2 )
(u2 + v 2 )2
−
4uv
(u2 + v 2 )2
∂(x, y)
=
∂(u, v)
1/u
v
0
u
2/9
−1/9
8. x = u3/2 /v 1/2 , y = v 1/2 /u1/2 ;
∂(x, y, z)
=
∂(u, v, w)
3
1
0
1
0
1
10.
∂(x, y, z)
=
∂(u, v, w)
1−v
v − vw
vw
0
−2
1
= −1 − 16uv
5/9
2/9
=
1
9
=1
1
√ √
2 2 u+v
1
− √ √
2 2 v−u
√
√
√
∂(x, y)
=
u + v/ 2, y = v − u/ 2;
∂(u, v)
9.
1 4v
4u −1
= 4/(u2 + v 2 )2
5
1
2
∂(x, y)
2
u + v, y = − u + v;
=
9
9
9
9
∂(u, v)
√
∂(x, y)
=
∂(u, v)
= cos u cos v + sin u sin v = cos(u − v)
2(v 2 − u2 )
(u2 + v 2 )2
4uv
2
(u + v 2 )2
6. x = ln u, y = uv;
7. x =
= −17
∂(x, y)
=
∂(u, v)
3u1/2
2v 1/2
−
−
1/2
v
2u3/2
1
√ √
2 2 u+v
1
√ √
2 2 v−u
u3/2
2v 3/2
1
=
1
= √
2
4 v − u2
1
2v
2u1/2 v 1/2
=5
−u
u − uw
uw
0
−uv
uv
= u2 v
∂(x, y, z)
=
11. y = v, x = u/y = u/v, z = w − x = w − u/v;
∂(u, v, w)
12. x = (v + w)/2, y = (u − w)/2, z = (u − v)/2,
∂(x, y, z)
=
∂(u, v, w)
1/v
0
−1/v
0
1/2
1/2
−u/v 2
1
u/v 2
1/2
0
−1/2
0
0
1
= 1/v
1/2
−1/2
0
=−
1
4
Exercise Set 16.8
600
y
13.
y
14.
(3, 4)
4
(0, 2)
3
2
1
x
x
(0, 0)
(–1, 0)
(0, 0)
2
(4, 0)
3
(1, 0)
y
15.
1
3
y
16.
(0, 3)
2
(2, 0)
-3
x
1
3
x
-3
17. x =
1
2
2
1 ∂(x, y)
1 1
1
u + v, y = − u + v,
= ;
5
5
5
5 ∂(u, v)
5 5
ZZ
1
u
dAuv =
v
5
Z
1
1
1 ∂(x, y)
1 1
1
u + v, y = u − v,
=− ;
2
2
2
2 ∂(u, v)
2 2
ZZ
veuv dAuv =
3
Z
1
S
18. x =
2
1
2
S
Z
4
4
1
3
u
du dv = ln 3
v
2
Z
1
veuv du dv =
1
0
1 4
(e − e − 3)
2
∂(x, y)
= −2; the boundary curves of the region S in the uv-plane are
∂(u, v)
Z 1Z u
ZZ
1
sin u cos vdAuv = 2
sin u cos v dv du = 1 − sin 2
v = 0, v = u, and u = 1 so 2
2
0
0
19. x = u + v, y = u − v,
S
p
√
1
∂(x, y)
= − ; the boundary curves of the region S in
∂(u, v)
2u
Z Z
ZZ
1 4 3 2
1
2
dAuv =
uv
v du dv = 21
the uv-plane are u = 1, u = 3, v = 1, and v = 4 so
2u
2 1 1
20. x =
v/u, y =
uv so, from Example 3,
S
∂(x, y)
= 12; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1.
∂(u, v)
Z 2π Z 1
ZZ
p
12 u2 + v 2 (12) dAuv = 144
r2 dr dθ = 96π
Use polar coordinates to obtain
21. x = 3u, y = 4v,
0
S
0
∂(x, y)
= 2; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1. Use
∂(u, v)
Z 2π Z 1
ZZ
2
2
2
e−(4u +4v ) (2) dAuv = 2
re−4r dr dθ = (1 − e−4 )π/2
polar coordinates to obtain
22. x = 2u, y = v,
S
0
0
601
Chapter 16
23. Let S be the region in the uv-plane bounded by u2 + v 2 = 1, so u = 2x, v = 3y,
∂(x, y)
=
∂(u, v)
x = u/2, y = v/3,
1
6
ZZ
1/2
0
0 1/3
Z
1
6
sin(u2 + v 2 )du dv =
π/2
1
r sin r2 dr dθ =
0
S
0
24. u = x/a, v = y/b, x = au, y = bv;
25. x = u/3, y = v/2, z = w,
Z
= 1/6, use polar coordinates to get
i1
π
π
(− cos r2 ) =
(1 − cos 1)
24
24
0
∂(x, y)
= ab; A = ab
∂(u, v)
Z
2π
Z
1
r dr dθ = πab
0
0
∂(x, y, z)
= 1/6; S is the region in uvw-space enclosed by the sphere
∂(u, v, w)
u2 + v 2 + w2 = 36 so
ZZZ 2
Z 2π Z π Z 6
1
u 1
dVuvw =
(ρ sin φ cos θ)2 ρ2 sin φ dρ dφ dθ
9 6
54 0
0
0
S
Z 2π Z π Z 6
192
1
π
ρ4 sin3 φ cos2 θdρ dφ dθ =
=
54 0
5
0
0
26. Let G1 be the region u2 + v 2 + w2 ≤ 1, let x = au, y = bv, z = cw,
ZZZ
ZZZ
(y 2 + z 2 )dx dy dz =
Ix =
Z
G
2π
Z
π
Z
(b2 v 2 + c2 w2 )du dv dw
G1
1
abc(b2 sin2 φ sin2 θ + c2 cos2 φ)ρ4 sin φ dρ dφ dθ
=
Z
0
0
2π
=
0
0
4
abc 2 2
(4b sin θ + 2c2 )dθ =
πabc(b2 + c2 )
15
15
27. Let u = y − 4x, v = y + 4x, then x =
1
8
ZZ
1
u
dAuv =
v
8
S
Z
5
Z
2
2
0
1
2
ZZ
uv dAuv = −
S
1
2
Z
1
∂(x, y)
1
1
(v − u), y = (v + u) so
=− ;
8
2
∂(u, v)
8
1 5
u
du dv = ln
v
4 2
28. Let u = y + x, v = y − x, then x =
−
2
Z
1
∂(x, y)
1
1
(u − v), y = (u + v) so
= ;
2
2
∂(u, v)
2
1
uv du dv = −
0
0
29. Let u = x − y, v = x + y, then x =
1
2
1
∂(x, y)
1
1
(v + u), y = (v − u) so
= ; the boundary curves of
2
2
∂(u, v)
2
the region S in the uv-plane are u = 0, v = u, and v = π/4; thus
ZZ
Z
Z
√
1
1 π/4 v sin u
1
sin u
dAuv =
du dv = [ln( 2 + 1) − π/4]
2
cos v
2 0
cos
v
2
0
S
∂(x, y, z)
= abc;
∂(u, v, w)
Exercise Set 16.8
602
1
∂(x, y)
1
1
(v − u), y = (u + v) so
= − ; the boundary
2
2
∂(u, v)
2
curves of the region S in the uv-plane are v = −u, v = u, v = 1, and v = 4; thus
ZZ
Z Z
1
1 4 v u/v
15
(e − e−1 )
eu/v dAuv =
e du dv =
2
2 1 −v
4
30. Let u = y − x, v = y + x, then x =
S
31. Let u = y/x, v = x/y 2 , then x = 1/(u2 v), y = 1/(uv) so
ZZ
1
dAuv =
u4 v 3
Z
S
4
Z
1
2
1
1
∂(x, y)
= 4 3;
∂(u, v)
u v
1
du dv = 35/256
u4 v 3
∂(x, y)
= 6; S is the region in the uv-plane enclosed by the circle u2 + v 2 = 1
∂(u, v)
ZZ
Z 2π Z 1
ZZ
(9 − x − y)dA =
6(9 − 3u − 2v)dAuv = 6
(9 − 3r cos θ − 2r sin θ)r dr dθ = 54π
so
32. Let x = 3u, y = 2v,
R
0
S
33. x = u, y = w/u, z = v + w/u,
ZZZ
v2 w
dVuvw =
u
S
Z
4
2
Z
1
Z
0
0
1
∂(x, y, z)
=− ;
∂(u, v, w)
u
1
3
v2 w
du dv dw = 2 ln 3
u
34. u = xy, v = yz, w = xz, 1 ≤ u ≤ 2, 1 ≤ v ≤ 3, 1 ≤ w ≤ 4,
p
p
p
1
∂(x, y, z)
= √
x = uw/v, y = uv/w, z = vw/u,
∂(u, v, w)
2 uvw
Z 2Z 3Z 4
ZZZ
√
√
1
√
dV =
dw dv du = 4( 2 − 1)( 3 − 1)
V =
1
1
1 2 uvw
G
35. (b) If x = x(u, v), y = y(u, v) where u = u(x, y), v = v(x, y), then by the chain rule
∂x
∂x ∂u ∂x ∂v
∂x
∂x ∂u ∂x ∂v
+
=
= 1,
+
=
=0
∂u ∂x ∂v ∂x
∂x
∂u ∂y
∂v ∂y
∂y
∂y
∂y ∂u ∂y ∂v
∂y
∂y ∂u ∂y ∂v
+
=
= 0,
+
=
=1
∂u ∂x ∂v ∂x
∂x
∂u ∂y
∂v ∂y
∂y
36. (a)
∂(x, y)
=
∂(u, v)
1−v
v
−u
u
∂(u, v)
=
∂(x, y)
1
−y/(x + y)2
= u; u = x + y, v =
1
x/(x + y)2
=
y
,
x+y
1
y
1
x
= ;
+
=
(x + y)2
(x + y)2
x+y
u
∂(u, v) ∂(x, y)
=1
∂(x, y) ∂(u, v)
(b)
∂(x, y)
=
∂(u, v)
∂(u, v)
=
∂(x, y)
√
√
u
= 2v 2 ; u = x/ y, v = y,
2v
√
1 ∂(u, v) ∂(x, y)
1
1/ y −x/(2y −3/2 )
√
= 2;
=1
=
0
1/(2 y)
2y
2v ∂(x, y) ∂(u, v)
v
0
603
Chapter 16
(c)
∂(x, y)
=
∂(u, v)
∂(u, v)
=
∂(x, y)
37.
√
√
v
= −2uv; u = x + y, v = x − y,
−v
√
√
1/(2 x + y) 1/(2 x + y)
1 ∂(u, v) ∂(x, y)
1
;
=1
=−
=− p
√
√
2
2
2uv ∂(x, y) ∂(u, v)
1/(2 x − y) −1/(2 x − y)
2 x −y
u
u
1 1
∂(x, y)
∂(u, v)
= 3xy 4 = 3v so
=
;
∂(x, y)
∂(u, v)
3v 3
ZZ
1
sin u
dAuv =
v
3
Z
1
S
38.
∂(x, y)
1
∂(x, y)
∂(u, v)
= 8xy so
=
; xy
= xy
∂(x, y)
∂(u, v)
8xy
∂(u, v)
ZZ
Z Z
1 16 4
1
dAuv =
du dv = 21/8
8
8 9
1
1
8xy
2
=
Z
2π
π
2
sin u
du dv = − ln 2
v
3
1
so
8
S
39.
1
∂(x, y)
∂(u, v)
= −2(x2 + y 2 ) so
=−
;
∂(x, y)
∂(u, v)
2(x2 + y 2 )
x4 − y 4 xy
1
1
∂(x, y)
=
e = (x2 − y 2 )exy = veu so
∂(u, v)
2(x2 + y 2 )
2
2
ZZ
Z Z
1 4 3 u
7
1
veu dAuv =
ve du dv = (e3 − e)
2
2 3 1
4
(x4 − y 4 )exy
S
40. Set u = x + y + 2z, v = x − 2y + z, w = 4x + y + z, then
Z
ZZZ
6
Z
Z
2
3
dx dy dz =
V =
−6
R
−2
−3
∂(u, v, w)
=
∂(x, y, z)
1
1 2
1 −2 1
4
1 1
= 18, and
1
∂(x, y, z)
du dv dw = 6(4)(12)
= 16
∂(u, v, w)
18
41. (a) Let u = x + y, v = y, then the triangle R with vertices (0, 0), (1, 0) and (0, 1) becomes the
triangle in the uv-plane with vertices (0, 0), (1, 0), (1, 1), and
Z 1
Z 1Z u
ZZ
∂(x, y)
dv du =
f (x + y)dA =
f (u)
uf (u) du
∂(u, v)
0
0
0
Z
R
1
ueu du = (u − 1)eu
(b)
0
42. (a)
(b)
∂(x, y)
=
∂(r, θ)
∂(x, y, z)
=
∂(ρ, θ, φ)
cos θ
sin θ
i1
=1
0
−r sin θ
r cos θ
sin φ cos θ
sin φ sin θ
cos φ
= r,
∂(x, y)
=r
∂(r, θ)
−ρ sin φ sin θ
ρ sin φ cos θ
0
ρ cos φ cos θ
ρ cos φ sin θ
−ρ sin φ
= −ρ2 sin φ;
∂(x, y, z)
= ρ2 sin φ
∂(ρ, θ, φ)
Chapter 16 Supplementary Exercises
604
CHAPTER 16 SUPPLEMENTARY EXERCISES
ZZ
ZZZ
3. (a)
dA
(b)
R
ZZ
dV
s
(c)
G
2
∂z
∂x
+
∂z
∂y
2
dA
R
4. (a) x = a sin φ cos θ, y = a sin φ sin θ, z = ρ cos φ, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π
(b) x = a cos θ, y = a sin θ, z = z, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h
Z
1
Z
1+
7.
1−
0
√
√
Z
1−y 2
f (x, y) dx dy
2
Z
Z
2x
8.
3
Z
6−x
f (x, y) dy dx +
1−y 2
x
0
f (x, y) dy dx
2
x
9. (a) (1, 2) = (b, d), (2, 1) = (a, c), so a = 2, b = 1, c = 1, d = 2
Z 1Z 1
Z 1Z 1
ZZ
∂(x, y)
du dv =
dA =
3du dv = 3
(b)
0
0 ∂(u, v)
0
0
R
√
10. 0 < sin xy < 1 for 0 < x,y < π, so
Z πZ π
Z πZ π
Z
√
0=
0 dy dx <
sin xy dy dx <
0
Z
0
0
1
2x cos(πx2 ) dx =
11.
1/2
Z
2
x2 y3
e
2
12.
0
Z
1
Z
x=2y
dy =
x=−y
0
π
0
i1
√
1
sin(πx2 )
= −1/( 2π)
π
1/2
3
2
Z
2
3
y 2 ey dy =
0
1 y3
e
2
2
=
0
1 8
e −1
2
Z
ex ey dx dy
π
Z
x
sin x
dy dx
x
14.
2y
0
y
15.
Z
1 dy dx = π 2
0
2
13.
0
π
0
16. p /2
1
r = a(1 + cos u)
y = sin x
r=a
y = tan (x/2)
x
p /6
6
Z
8
Z
y 1/3
x2 sin y 2 dx dy =
17. 2
0
0
π/2 Z
Z
2
3
2
(4 − r2 )r dr dθ = 2π
18.
0
0
Z
0
8
1
y sin y 2 dy = − cos y 2
3
8
=
0
0
1
(1 − cos 64) ≈ 0.20271
3
605
Chapter 16
2xy
, and r = 2a sin θ is the circle x2 + (y − a)2 = a2 , so
x2 + y 2
Z a h
i
p
p
2xy
2 − x2 − ln a −
2 − x2
dx = a2
dy
dx
=
x
ln
a
+
a
a
x2 + y 2
0
19. sin 2θ = 2 sin θ cos θ =
Z
√
a− a2 −x2
0
Z
√
a+ a2 −x2
Z
a
π/2
Z
π/2
2
4r (cos θ sin θ) r dr dθ = −4 cos 2θ
2
20.
π/4
Z
2π
Z
π/4
Z
2
Z
16
Z
2π
r2 cos2 θ r dz dr dθ =
21.
0
r4
Z
π/2
0
Z
π/2
Z
1
0
0
Z
0
2π
Z
r3 (16 − r4 ) dr = 32π
0
Z
π π π/2
1
2
ρ sin φ dρ dφ dθ = 1 −
sin φ dφ
1 + ρ2
4 2 0
iπ/2
π π
π π
(− cos φ)
= 1−
= 1−
4 2
4 2
0
π/3
Z
Z
a
2π
Z
π/3
Z
(ρ2 sin2 φ)ρ2 sin φ dρ dφ dθ =
23. (a)
Z
0
0
2π
Z
0
√
√
3a/2 Z
a2 −r 2
(b)
0
Z
(c)
√
r/ 3
0
√
3a/2
Z √(3a2 /4)−x
√
− 3a/2
Z
0
Z
√
−
Z
r2 dz rdr dθ =
√ 2 2 2
2 Z
a −x −y
√
(3a2 /4)−x2
√
4x−x2
Z
4
24. (a)
√
− 4x−x2
Z
π/2
4 cos θ
Z
2π
0
a
ρ4 sin3 φ dρ dφ dθ
0
0
0
√
√
3a/2 Z
a2 −r 2
Z
√
r/ 3
0
r3 dz dr dθ
(x2 + y 2 ) dz dy dx
√
x2 +y 2 / 3
4x
dz dy dx
x2 +y 2
Z 4r cos θ
(b)
r dz dr dθ
−π/2
2
2
cos2 θ dθ
0
22.
Z
=4
0
Z
r2
0
Z
2−y/2
2
dx dy =
25.
(y/2)1/3
0
Z
π/6
0
Z
2
3 y 4/3
y y 1/3
y2
3
−
2− −
dy = 2y −
=
2
2
4
2 2
2
0
Z
cos 3θ
π/6
cos2 3θ = π/4
r dr dθ = 3
26. A = 6
0
Z
0
2π
Z
0
√
a/ 3
27. V =
0
Z
√
3r
0
√
a/ 3
Z
a
r(a −
r dz dr dθ = 2π
0
√
πa3
3r) dr =
9
28. The intersection of the two surfaces projects onto the yz-plane as 2y 2 + z 2 = 1, so
Z 1/√2 Z √1−2y2 Z 1−y2
dx dz dy
V =4
0
0
√
Z
1/
√
2Z
1−2y 2
y 2 +z 2
Z
(1 − 2y − z ) dz dy = 4
2
=4
0
2
0
0
√
29. kru × rv k = 2u2 + 2v 2 + 4,
Z
ZZ p
2u2 + 2v 2 + 4 dA =
S=
u2 +v 2 ≤4
0
2π
Z
0
2
√
1/ 2
8
(1 − 2x2 )3/2 dx =
3
√
2π
4
√ p
8π √
(3 3 − 1)
2 r2 + 2 r dr dθ =
3
Chapter 16 Supplementary Exercises
30. kru × rv k =
√
1 + u2 , S =
606
Z
2
Z
31. (ru × rv )
u=1
v=2
0
i
4
Z
p
2
3u
1 + u2 du = 53/2 − 1
0
= h−2, −4, 1i, tangent plane 2x + 4y − z = 5
32. u = −3, v = 0, (ru × rv )
Z
Z
p
1 + u2 dv du =
0
i
3u
u=−3
v=0
Z
2+y 2 /8
4
= h−18, 0, −3i, tangent plane 6x + z = −9
y2
2−
8
32
; ȳ = 0 by symmetry;
3
−4
4
Z 4
Z 4 Z 2+y2 /8
3 256
8
3 4
1
256
8
y dy =
, x̄ =
= ; centroid
,0
x dx dy =
2 + y2 −
4
128
15
32 15
5
5
−4 y 2 /4
−4
dx dy =
33. A =
y 2 /4
dy =
34. A = πab/2, x̄ = 0 by symmetry,
Z a Z b√1−x2 /a2
Z
1 a 2
4b
y dy dx =
b (1 − x2 /a2 )dx = 2ab2 /3, centroid 0,
2 −a
3π
−a 0
1
35. V = πa2 h, x̄ = ȳ = 0 by symmetry,
3
Z a
Z 2π Z a Z h−rh/a
r 2
rz dz dr dθ = π
rh2 1 −
dr = πa2 h2 /12, centroid (0, 0, h/4)
a
0
0
0
0
1 4
256
,
dz dy dx =
(4 − y)dy dx =
36. V =
8 − 4x + x dx =
2
15
−2 x2 0
−2 x2
−2
Z 2 Z 4 Z 4−y
Z 2Z 4
Z 2
1024
32
1 6
2
4
x − 2x +
dx =
y dz dy dx =
(4y − y ) dy dx =
3
3
35
−2 x2 0
−2 x2
−2
Z 2 Z 4 Z 4−y
Z 2Z 4
Z 2 6
2048
32
x
1
(4 − y)2 dy dx =
dx =
z dz dy dx =
− + 2x4 − 8x2 +
2
6
3
105
2
2
−2 x
0
−2 x
−2
8
x̄ = 0 by symmetry, centroid 0, , 4
7
Z
2
Z
4
Z
4−y
Z
2
Z
4
Z
2
2
√
37. The two quarter-circles with center at the origin and of radius A and 2A lie inside and outside
of the square with corners (0, 0), (A, 0), (A, A), (0, A), so the following inequalities hold:
Z AZ A
Z π/2 Z √2A
Z π/2 Z A
1
1
1
rdr dθ ≤
dx dy ≤
rdr dθ
2
2
2
2
2
(1 + r2 )2
0
0 (1 + r )
0
0 (1 + x + y )
0
0
πA2
and the integral on the right equals
4(1 + A2 )
2
π
2πA
. Since both of these quantities tend to , it follows by sandwiching that
2
4(1 + 2A )
4
Z +∞ Z +∞
π
1
dx dy = .
2 + y 2 )2
(1
+
x
4
0
0
The integral on the left can be evaluated as
38. The centroid of the circle which generates the tube travels a distance
Z 4π q
√
√
√
s=
sin2 t + cos2 t + 1/16 dt = 17π, so V = π(1/2)2 17π = 17π 2 /4.
0
607
Chapter 16
39. (a) The values of x in the formula of the astroidal sphere lie between −a and a; and the same is
true for x = (a cos u cos v)3 ; similarly for y and z. Moreover,
x2/3 + y 2/3 + z 2/3 = a2/3 cos2 u cos2 v + a2/3 sin2 u cos2 v + a2/3 sin2 v
= a2/3 cos2 v + a2/3 sin2 v = a2/3
Z
Z
π/2
π/2
kru × rv kdu dv
(b) S = 8
0
0
Z
π/2
Z
π/2
sin u cos u sin v cos4 v
= 72
Z
0
0
(1−x2/3 )3/2
Z
1
(c) 8
0
p
sin2 v + sin2 u cos2 u cos2 vdu dv ≈ 4.451
p
1 − x2/3 − y 2/3 dy dx ≈ 0.3590
0
(d) Let x = t cos3 u, y = t sin3 u, 0 ≤ t ≤ 1, 0 ≤ u ≤ π/2, then J =
Z
1
Z
π/2
(1 − t2/3 )3/2 3t sin2 u cos2 u du dt =
V =8
0
0
4
3
40. V = πa3 , d¯ =
3
4πa3
ZZZ
ρ≤a
3
ρdV =
4πa3
Z
π
Z
2π
Z
3π
2
Z
∂(x, y)
= 3t sin2 u cos2 u,
∂(t, u)
1
t(1 − t2/3 )3/2 dt =
0
a
ρ3 sin φ dρ dθ dφ =
0
0
0
4π
35
3
3
a4
= a
2π(2)
3
4πa
4
4
41. (a) (x/a)2 +(y/b)2 +(z/c)2 = sin2 φ cos2 θ+sin2 φ sin2 θ+cos2 φ = sin2 φ+cos2 φ = 1, an ellipsoid
(b) r(φ, θ) = h2 sin φ cos θ, 3 sin φ sin θ, 4 cos φi; rφ ×rθ = 2h6 sin2 φ cos θ, 4 sin2 φ sin θ, 3 cos φ sin φi,
p
krφ × rθ k = 2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ,
Z 2π Z π q
S=
2 16 sin4 φ + 20 sin4 φ cos2 θ + 9 sin2 φ cos2 φ dφ dθ ≈ 111.5457699
0
0
CHAPTER 17
Topics in Vector Calculus
EXERCISE SET 17.1
1. (a) III because the vector field is independent of y and the direction is that of the negative x-axis
for negative x, and positive for positive
(b) IV, because the y-component is always positive, and the x-component is positive for positive
x, negative for negative x
2. (a) I, since the vector field is constant
(b) II, since the vector field points away from the origin
3. (a) true
(b) true
(c) true
4. (a) false, the lengths are equal to 1
(c) false, the x-component is then zero
5.
y
6.
(b) false, the y-component is then zero
y
7.
y
x
x
x
8.
y
9.
y
y
x
x
11. (a) ∇φ = φx i + φy j =
10.
x
y
x
i+
j = F, so F is conservative for all x, y
1 + x2 y 2
1 + x2 y 2
(b) ∇φ = φx i + φy j = 2xi − 6yj + 8zk = F so F is conservative for all x, y
12. (a) ∇φ = φx i + φy j = (6xy − y 3 )i + (4y + 3x2 − 3xy 2 )j = F, so F is conservative for all x, y
(b) ∇φ = φx i + φy j + φz k = (sin z + y cos x)i + (sin x + z cos y)j + (x cos z + sin y)k = F, so F is
conservative for all x, y
608
609
Chapter 17
13. div F = 2x + y, curl F = zi
14. div F = z 3 + 8y 3 x2 + 10zy, curl F = 5z 2 i + 3xz 2 j + 4xy 4 k
15. div F = 0, curl F = (40x2 z 4 − 12xy 3 )i + (14y 3 z + 3y 4 )j − (16xz 5 + 21y 2 z 2 )k
16. div F = yexy + sin y + 2 sin z cos z, curl F = −xexy k
2
17. div F = p
x2
18. div F =
+ y2 + z2
, curl F = 0
x
z
1
+ xzexyz + 2
, curl F = −xyexyz i + 2
j + yzexyz k
x
x + z2
x + z2
19. ∇ · (F × G) = ∇ · (−(z + 4y 2 )i + (4xy + 2xz)j + (2xy − x)k) = 4x
20. ∇ · (F × G) = ∇ · ((x2 yz 2 − x2 y 2 )i − xy 2 z 2 j + xy 2 zk) = −xy 2
21. ∇ · (∇ × F) = ∇ · (− sin(x − y)k) = 0
22. ∇ · (∇ × F) = ∇ · (−zeyz i + xexz j + 3ey k) = 0
23. ∇ × (∇ × F) = ∇ × (xzi − yzj + yk) = (1 + y)i + xj
24. ∇ × (∇ × F) = ∇ × ((x + 3y)i − yj − 2xyk) = −2xi + 2yj − 3k
25. div (kF) = k
∂q
∂h
∂f
+k
+k
= k div F
∂x
∂y
∂z
26. curl (kF) = k
∂h ∂g
−
∂y
∂z
i+k
∂h
∂f
−
∂z
∂x
j+k
∂f
∂g
−
∂x ∂y
k = k curl F
27. Let F = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, then
∂P
∂Q
∂g
∂h ∂R
∂f
+
+
+
+
+
div (F + G) =
∂x
∂x
∂y
∂y
∂z
∂z
∂g
∂h
∂Q ∂R
∂P
∂f
+
+
+
+
+
= div F + div G
=
∂x ∂y
∂z
∂x
∂y
∂z
28. Let F = f (x, y, z)i + g(x, y, z)j + h(x, y, z)k and G = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k, then
∂
∂
∂
∂
(h + R) −
(g + Q) i +
(f + P ) −
(h + R) j
curl (F + G) =
∂y
∂z
∂z
∂x
∂
∂
(g + Q) −
(f + P ) k;
+
∂x
∂y
expand and rearrange terms to get curl F + curl G.
∂f
∂φ
∂g
∂φ
∂h ∂φ
29. div (φF) = φ
+
f + φ
+
g + φ
+
h
∂x ∂x
∂y
∂y
∂z
∂z
∂g
∂h
∂φ
∂φ
∂φ
∂f
+
+
+
f+
g+
h
=φ
∂x ∂y
∂z
∂x
∂y
∂z
= φ div F + ∇φ · F
Exercise Set 17.1
610
∂
∂
∂
∂
∂
∂
(φh) −
(φg) i+
(φf ) −
(φh) j+
(φg) −
(φf ) k; use the product
∂y
∂z
∂z
∂x
∂x
∂y
rule to expand each of the partial derivatives, rearrange to get φ curl F + ∇φ × F
30. curl (φF) =
∂
31. div(curl F) =
∂x
∂h ∂g
−
∂y
∂z
∂
+
∂y
∂f
∂h
−
∂z
∂x
∂
+
∂z
∂g
∂f
−
∂x ∂y
∂2g
∂2f
∂2h
∂2g
∂2f
∂2h
−
+
−
+
−
= 0,
∂x∂y ∂x∂z
∂y∂z
∂y∂x ∂z∂x ∂z∂y
assuming equality of mixed second partial derivatives
=
2
2
∂2φ
∂2φ
∂2φ
∂ φ
∂ φ
∂2φ
−
i+
−
j+
−
k = 0, assuming equality
∂y∂z
∂z∂y
∂z∂x ∂x∂z
∂x∂y ∂y∂x
of mixed second partial derivatives
32. curl (∇φ) =
33. ∇ · (kF) = k∇ · F, ∇ · (F + G) = ∇ · F + ∇ · G, ∇ · (φF) = φ∇ · F + ∇φ · F, ∇ · (∇ × F) = 0
34. ∇ × (kF) = k∇ × F, ∇ × (F + G) = ∇ × F + ∇ × G, ∇ × (φF) = φ∇ × F + ∇φ × F, ∇ × (∇φ) = 0
37. (a) curl r = 0i + 0j + 0k = 0
p
(b) ∇krk = ∇ x2 + y 2 + z 2 = p
x
x2
+
y2
+
z2
y
i+ p
x2
+
y2
+
z2
z
r
j+ p
k=
2
2
2
krk
x +y +z
38. (a) div r = 1 + 1 + 1 = 3
(b) ∇
xi + yj + zk
r
1
= ∇(x2 + y 2 + z 2 )−1/2 = − 2
=−
krk
krk3
(x + y 2 + z 2 )3/2
39. (a) ∇f (r) = f 0 (r)
∂r
∂r
∂r
f 0 (r)
i + f 0 (r) j + f 0 (r) k = f 0 (r)∇r =
r
∂x
∂y
∂z
r
(b) div[f (r)r] = f (r)div r + ∇f (r) · r = 3f (r) +
f 0 (r)
r · r = 3f (r) + rf 0 (r)
r
40. (a) curl[f (r)r] = f (r)curl r + ∇f (r) × r = f (r)0 +
f 0 (r)
r×r=0+0=0
r
f 0 (r)
f 0 (r)
f 0 (r)
r =
div r + ∇
·r
(b) ∇ f (r) = div[∇f (r)] = div
r
r
r
2
=3
f 0 (r) rf 00 (r) − f 0 (r)
f 0 (r)
+
+ f 00 (r)
r·r=2
3
r
r
r
41. f (r) = 1/r3 , f 0 (r) = −3/r4 , div(r/r3 ) = 3(1/r3 ) + r(−3/r4 ) = 0
42. Multiply 3f (r) + rf 0 (r) = 0 through by r2 to obtain 3r2 f (r) + r3 f 0 (r) = 0,
d[r3 f (r)]/dr = 0, r3 f (r) = C, f (r) = C/r3 , so F = Cr/r3 (an inverse-square field).
43. (a) At the point (x, y) the slope of the line along which the vector −yi + xj lies is −x/y; the
slope of the tangent line to C at (x, y) is dy/dx, so dy/dx = −x/y.
(b) ydy = −xdx, y 2 /2 = −x2 /2 + K1 , x2 + y 2 = K
611
Chapter 17
44. dy/dx = x, y = x2 /2 + K
45. dy/dx = 1/x, y = ln x + K
y
y
x
x
46. dy/dx = −y/x, (1/y)dy = (−1/x)dx, ln y = − ln x + K1 ,
y
y = eK1 e− ln x = K/x
x
EXERCISE SET 17.2
Z
1
1. (a)
dy = 1 because s = y is arclength measured from (0, 0)
0
(b) 0, because sin xy = 0 along C
Z
ds = length of line segment = 2
2. (a)
C
s
3. (a) ds =
Z
dx
dt
2
+
dy
dt
2
Z
dt, so
0
1
(2t − 3t2 )2 dt = 0
(b)
0
Z
1
t(3t2 )(6t3 )2
4. (a)
Z
0
t(3t2 )(6t3 )2 6t dt =
0
p
√
11 √
1
4
ln( 10 − 3) −
(2t − 3t2 ) 4 + 36t2 dt = −
10 −
108
36
27
Z 1
1
(c)
(2t − 3t2 )6t dt = −
2
0
p
864
1 + 36t2 + 324t4 dt =
5
1
(c)
1
(b) 0, because x is constant and dx = 0
Z
t(3t2 )(6t3 )2 dt =
0
Z
648
11
0
1
5. (a) C : x = t, y = t, 0 ≤ t ≤ 1;
6t dt = 3
0
Z
(b) C : x = t, y = t2 , 0 ≤ t ≤ 1;
1
(3t + 6t2 − 2t3 )dt = 3
0
54
5
1
t(3t2 )(6t3 )2 18t2 dt = 162
(d)
Z
1
(b)
Exercise Set 17.2
612
(c) C : x = t, y = sin(πt/2), 0 ≤ t ≤ 1;
Z 1
[3t + 2 sin(πt/2) + πt cos(πt/2) − (π/2) sin(πt/2) cos(πt/2)]dt = 3
0
Z
1
(d) C : x = t3 , y = t, 0 ≤ t ≤ 1;
(9t5 + 8t3 − t)dt = 3
0
Z
1
6. (a) C : x = t, y = t, z = t, 0 ≤ t ≤ 1;
1
2
(t + t − t) dt =
0
Z
1
(b) C : x = t, y = t2 , z = t3 , 0 ≤ t ≤ 1;
(t2 + t3 (2t) − t(3t2 )) dt = −
0
Z
(c) C : x = cos πt, y = sin πt, z = t, 0 ≤ t ≤ 1;
1
60
1
(−π sin2 πt + πt cos πt − cos πt) dt = −
0
Z
√
3
1+t
dt =
1+t
7.
0
Z
Z
3
−1/2
(1 + t)
dt = 2
1
5
0
1
0
√
1 + 2t
dt = 5(π/4 + ln 2)
1 + t2
1
3(t2 )(t2 )(2t3 /3)(1 + 2t2 ) dt = 2
t7 (1 + 2t2 ) dt = 13/20
0
√ Z 2π
√
5
e−t dt = 5(1 − e−2π )/4
4 0
Z
Z
0
Z
9.
10.
8.
√
1
12.
−1
Z
π/4
(8 cos2 t−16 sin2 t−20 sin t cos t)dt = 1−π
11.
0
2
2
t − t5/3 + t2/3 dt = 6/5
3
3
Z
3
13. C : x = (3 − t)2 /3, y = 3 − t, 0 ≤ t ≤ 3;
0
Z
14. C : x = t2/3 , y = t, −1 ≤ t ≤ 1;
1
−1
1
(3 − t)2 dt = 3
3
2 2/3 2 1/3
t − t + t7/3 dt = 4/5
3
3
Z
π/2
15. C : x = cos t, y = sin t, 0 ≤ t ≤ π/2;
(− sin t − cos2 t)dt = −1 − π/4
0
Z
1
(−37 + 41t − 9t2 )dt = −39/2
16. C : x = 3 − t, y = 4 − 3t, 0 ≤ t ≤ 1;
0
Z
1
(−3)e3t dt = 1 − e3
17.
0
Z
2
π
−
2
π
π/2
(sin2 t cos t − sin2 t cos t + t4 (2t)) dt =
18.
0
Z
π6
192
q
cos21 t sin9 t (−3 cos2 t sin t)2 + (3 sin2 t cos t)2 dt
0
Z π/2
61,047
π
cos22 t sin10 t dt =
= 3
4,294,967,296
0
Z e
5 6 59 4 491
1
5
2
4
(b)
dt =
e + e −
t ln t + 7t (2t) + t (ln t)
t
36
16
144
1
20. (a)
π/2
613
Chapter 17
21. (a) C1 : (0, 0) to (1, 0); x = t, y = 0, 0 ≤ t ≤ 1
C2 : (1, 0) to (0, 1); x = 1 − t, y = t, 0 ≤ t ≤ 1
C3 : (0, 1) to (0, 0); x = 0, y = 1 − t, 0 ≤ t ≤ 1
Z 1
Z 1
Z 1
(0)dt +
(−1)dt +
(0)dt = −1
0
0
0
(b) C1 : (0, 0) to (1, 0); x = t, y = 0, 0 ≤ t ≤ 1
C2 : (1, 0) to (1, 1); x = 1, y = t, 0 ≤ t ≤ 1
C3 : (1, 1) to (0, 1); x = 1 − t, y = 1, 0 ≤ t ≤ 1
C4 : (0, 1) to (0, 0); x = 0, y = 1 − t, 0 ≤ t ≤ 1
Z 1
Z 1
Z 1
Z 1
(0)dt +
(−1)dt +
(−1)dt +
(0)dt = −2
0
0
0
0
22. (a) C1 : (0, 0) to (1, 1); x = t, y = t, 0 ≤ t ≤ 1
C2 : (1, 1) to (2, 0); x = 1 + t, y = 1 − t, 0 ≤ t ≤ 1
C3 : (2, 0) to (0, 0); x = 2 − 2t, y = 0, 0 ≤ t ≤ 1
Z 1
Z 1
Z 1
(0)dt +
2dt +
(0)dt = 2
0
0
0
(b) C1 : (−5, 0) to (5, 0); x = t, y = 0, −5 ≤ t ≤ 5
C2 : x = 5 cos t, y = 5 sin t, 0 ≤ t ≤ π
Z π
Z 5
(0)dt +
(−25)dt = −25π
−5
0
Z
Z
1
23. C1 : x = t, y = z = 0, 0 ≤ t ≤ 1,
0 dt = 0; C2 : x = 1, y = t, z = 0, 0 ≤ t ≤ 1,
0
Z
C3 : x = 1, y = 1, z = t, 0 ≤ t ≤ 1,
x2 z dx − yx2 dy + 3 dz = 0 −
3 dt = 3;
C
0
(−t) dt = −
0
Z
1
1
5
1
+3=
2
2
24. C1 : (0, 0, 0) to (1, 1, 0); x = t, y = t, z = 0, 0 ≤ t ≤ 1
C2 : (1, 1, 0) to (1, 1, 1); x = 1, y = 1, z = t, 0 ≤ t ≤ 1
C3 : (1, 1, 1) to (0, 0, 0); x = 1 − t, y = 1 − t, z = 1 − t, 0 ≤ t ≤ 1
Z 1
Z 1
Z 1
(−t3 )dt +
3 dt +
−3dt = −1/4
0
Z
0
0
Z
π
25.
(0)dt = 0
26.
0
Z
1
(e2t − 4e−t )dt = e2 /2 + 4e−1 − 9/2
0
1
27.
Z
e−t dt = 1 − e−1
0
π/2
(7 sin2 t cos t + 3 sin t cos t)dt = 23/6
28.
0
29. Represent the circular arc by x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π/2.
Z
√ Z π/2 √
√
√
x yds = 9 3
sin t cos t dt = 6 3
C
0
p
30. δ(x, y) = k x2 + y 2 where k is the constant of proportionality,
Z p
Z 1 √
√ Z 1 2t
√
t
t
2
2
k x + y ds = k
e ( 2e ) dt = 2k
e dt = (e2 − 1)k/ 2
C
0
0
1
2
Exercise Set 17.2
Z
31.
C
614
kx
ds = 15k
1 + y2
Z
0
π/2
cos t
dt = 5k tan−1 3
1 + 9 sin2 t
32. δ(x, y, z) = kz where k is the constant of proportionality,
Z 4
Z
√
kzds =
k(4 t)(2 + 1/t) dt = 136k/3
C
1
Z
1
33. C : x = t2 , y = t, 0 ≤ t ≤ 1; W =
3t4 dt = 3/5
0
Z
Z
3
(t2 + 1 − 1/t3 + 1/t)dt = 92/9 + ln 3
34. W =
35.
1
(t3 + 5t6 )dt = 27/28
W =
1
0
36. C1 : (0, 0, 0) to (1, 3, 1); x = t, y = 3t, z = t, 0 ≤ t ≤ 1
C2 : (1, 3, 1) to (2, −1, 4); x = 1 + t, y = 3 − 4t, z = 1 + 3t, 0 ≤ t ≤ 1
Z 1
Z 1
(4t + 8t2 )dt +
(−11 − 17t − 11t2 )dt = −37/2
W =
0
0
37. Since F and r are parallel, F · r = kFkkrk, and since F is constant,
Z
Z
Z
√ Z 4√
2dt = 16
F · dr = d(F · r) = d(kFkkrk) = 2
Z
C
F·r=
√
√
2(4 2)
−4
C
C
Z
F · r = 0, since F is perpendicular to the curve
38.
C
39. C : x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ π/2
Z π/2
1
− sin t + cos t dt = 3/4
4
0
40. C1 : (0, 3) to (6, 3); x = 6t, y = 3, 0 ≤ t ≤ 1
C2 : (6, 3) to (6, 0); x = 6, y = 3 − 3t, 0 ≤ t ≤ 1
Z 1
Z 1
1
2
6
−12
dt
+
dt = tan−1 2 − tan−1 (1/2)
2+9
2
36t
36
+
9(1
−
t)
3
3
0
0
41. Represent the parabola by x = t, y = t2 , 0 ≤ t ≤ 2.
Z
Z 2 p
√
3xds =
3t 1 + 4t2 dt = (17 17 − 1)/4
C
0
42. Represent the semicircle by x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π.
Z π
Z
x2 yds =
16 cos2 t sin t dt = 32/3
C
0
Z
43. (a) 2πrh = 2π(1)2 = 4π
(b) S =
Z
(c) C : x = cos t, y = sin t, 0 ≤ t ≤ 2π; S =
2π
z(t) dt
C
(2 + (1/2) sin 3t) dt = 4π
0
615
Chapter 17
Z
44. C : x = a cos t, y = a sin t, 0 ≤ t ≤ 2π,
Z
Z
−C
dt = −
Z
dt = −
C
2π
dt = −2π
0
1
λ[(1 − λ)t + (3λ − 1)t2 − (1 + 2λ)t3 ]dt = −λ/12, W = 1 when λ = −12
45. W =
0
3
1
z k = 170 −
t k, so
10
4π
Z 60
1
1
170 − z dz = 10,020. Note that the functions
F · dr = 170 − z dz, and W =
10
10
0
x(z), y(z) are irrelevant.
46. The force exerted by the farmer is F =
150 + 20 −
EXERCISE SET 17.3
1. ∂x/∂y = 0 = ∂y/∂x, conservative so ∂φ/∂x = x and ∂φ/∂y = y, φ = x2 /2 + k(y), k 0 (y) = y,
k(y) = y 2 /2 + K, φ = x2 /2 + y 2 /2 + K
2. ∂(3y 2 )/∂y = 6y = ∂(6xy)/∂x, conservative so ∂φ/∂x = 3y 2 and ∂φ/∂y = 6xy,
φ = 3xy 2 + k(y), 6xy + k 0 (y) = 6xy, k 0 (y) = 0, k(y) = K, φ = 3xy 2 + K
3. ∂(x2 y)/∂y = x2 and ∂(5xy 2 )/∂x = 5y 2 , not conservative
4. ∂(ex cos y)/∂y = −ex sin y = ∂(−ex sin y)/∂x, conservative so ∂φ/∂x = ex cos y and
∂φ/∂y = −ex sin y, φ = ex cos y + k(y), −ex sin y + k 0 (y) = −ex sin y,
k 0 (y) = 0, k(y) = K, φ = ex cos y + K
5. ∂(cos y + y cos x)/∂y = − sin y + cos x = ∂(sin x − x sin y)/∂x, conservative so
∂φ/∂x = cos y + y cos x and ∂φ/∂y = sin x − x sin y, φ = x cos y + y sin x + k(y),
−x sin y + sin x + k 0 (y) = sin x − x sin y, k 0 (y) = 0, k(y) = K, φ = x cos y + y sin x + K
6. ∂(x ln y)/∂y = x/y and ∂(y ln x)/∂x = y/x, not conservative
7. (a) ∂(y 2 )/∂y = 2y = ∂(2xy)/∂x, independent of path
Z 1
(4 + 14t + 6t2 )dt = 13
(b) C : x = −1 + 2t, y = 2 + t, 0 ≤ t ≤ 1;
0
(c) ∂φ/∂x = y 2 and ∂φ/∂y = 2xy, φ = xy 2 + k(y), 2xy + k 0 (y) = 2xy, k 0 (y) = 0, k(y) = K,
φ = xy 2 + K. Let K = 0 to get φ(1, 3) − φ(−1, 2) = 9 − (−4) = 13
8. (a) ∂(y sin x)/∂y = sin x = ∂(− cos x)/∂x, independent of path
Z 1
(π sin πt − 2πt sin πt + 2 cos πt)dt = 0
(b) C1 : x = πt, y = 1 − 2t, 0 ≤ t ≤ 1;
0
(c) ∂φ/∂x = y sin x and ∂φ/∂y = − cos x, φ = −y cos x + k(y), − cos x + k 0 (y) = − cos x,
k 0 (y) = 0, k(y) = K, φ = −y cos x+K. Let K = 0 to get φ(π, −1)−φ(0, 1) = (−1)−(−1) = 0
9. ∂(3y)/∂y = 3 = ∂(3x)/∂x, φ = 3xy, φ(4, 0) − φ(1, 2) = −6
10. ∂(ex sin y)/∂y = ex cos y = ∂(ex cos y)/∂x, φ = ex sin y, φ(1, π/2) − φ(0, 0) = e
11. ∂(2xey )/∂y = 2xey = ∂(x2 ey )/∂x, φ = x2 ey , φ(3, 2) − φ(0, 0) = 9e2
Exercise Set 17.3
616
12. ∂(3x − y + 1)/∂y = −1 = ∂[−(x + 4y + 2)]/∂x,
φ = 3x2 /2 − xy + x − 2y 2 − 2y, φ(0, 1) − φ(−1, 2) = 11/2
13. ∂(2xy 3 )/∂y = 6xy 2 = ∂(3x2 y 2 )/∂x, φ = x2 y 3 , φ(−1, 0) − φ(2, −2) = 32
14. ∂(ex ln y − ey /x)/∂y = ex /y − ey /x = ∂(ex /y − ey ln x)/∂x,
φ = ex ln y − ey ln x, φ(3, 3) − φ(1, 1) = 0
15. φ = x2 y 2 /2, W = φ(0, 0) − φ(1, 1) = −1/2
16. φ = x2 y 3 , W = φ(4, 1) − φ(−3, 0) = 16
17. φ = exy , W = φ(2, 0) − φ(−1, 1) = 1 − e−1
18. φ = e−y sin x, W = φ(−π/2, 0) − φ(π/2, 1) = −1 − 1/e
19. ∂(ey + yex )/∂y = ey + ex = ∂(xey + ex )/∂x so F is conservative, φ(x, y) = xey + yex so
Z
F · dr = φ(0, ln 2) − φ(1, 0) = ln 2 − 1
C
20. ∂(2xy)/∂y = 2x = ∂(x2 + cos y)/∂x so F is conservative, φ(x, y) = x2 y + sin y so
Z
F · dr = φ(π, π/2) − φ(0, 0) = π 3 /2 + 1
C
21. F · dr = [(ey + yex )i + (xey + ex )j] · [(π/2) cos(πt/2)i + (1/t)j]dt
π
cos(πt/2)(ey + yex ) + (xey + ex )/t dt,
=
2
Z 2
Z
π
cos(πt/2) eln t + (ln t)esin(πt/2) + sin(πt/2)eln t + esin(πt/2)
F · dr =
dt ≈ −0.307
so
2
C
1
22. F · dr = 2t2 cos(t/3) + [t2 + cos(t cos(t/3))](cos(t/3) − (t/3) sin(t/3)) dt, so
Z
Z π
F · dr =
2t2 cos(t/3) + [t2 + cos(t cos(t/3))](cos(t/3) − (t/3) sin(t/3)) dt ≈ 16.503
C
0
23. No; a closed loop can beZ found whose tangent everywhere makes an angle < π with the vector
F · dr > 0, and by Theorem 17.3.2 the vector field is not conservative.
field, so the line integral
C
24. The vector field is constant, say F = ai + bj, so let φ(x, y) = ax + by and F is conservative.
25. If F is conservative, then F = ∇φ =
Thus
∂φ
∂φ
∂φ
∂φ
∂φ
∂φ
i+
j+
k and hence f =
,g =
, and h =
.
∂x
∂y
∂z
∂x
∂y
∂z
∂2φ
∂g
∂ 2 φ ∂f
∂2φ
∂h
∂ 2 φ ∂g
∂2φ
∂h
∂2φ
∂f
=
and
=
,
=
and
=
,
=
and
=
.
∂y
∂y∂x
∂x
∂x∂y ∂z
∂z∂x
∂x
∂x∂z ∂z
∂z∂y
∂y
∂y∂z
The result follows from the equality of mixed second partial derivatives.
26. Let f (x, y, z) = yz, g(x, y, z) = xz, h(x, y, z) = yx2 , then ∂f /∂z = y, ∂h/∂x = 2xy 6= ∂f /∂z, thus
Z
by Exercise 25, F = f i+gj+hk is not conservative, and by Theorem 17.3.2,
yz dx+xz dy+yx2 dz
C
is not independent of the path.
617
27.
Chapter 17
∂
(h(x)[x sin y + y cos y]) = h(x)[x cos y − y sin y + cos y]
∂y
∂
(h(x)[x cos y − y sin y]) = h(x) cos y + h0 (x)[x cos y − y sin y],
∂x
equate these two partial derivatives to get (x cos y − y sin y)(h0 (x) − h(x)) = 0 which holds for all
x and y if h0 (x) = h(x), h(x) = Cex where C is an arbitrary constant.
28. (a)
3cxy
∂
∂
cx
cy
=− 2
=
when (x, y) 6= (0, 0),
2
2
3/2
2
−5/2
2
∂y (x + y )
∂x (x + y 2 )3/2
(x + y )
so by Theorem 17.3.3, F is conservative. Set ∂φ/∂x = cx/(x2 + y 2 )−3/2 ,
then φ(x, y) = −c(x2 + y 2 )−1/2 + k(y), ∂φ/∂y = cy/(x2 + y 2 )−3/2 + k 0 (y), so k 0 (y) = 0.
c
is a potential function.
Thus φ(x, y) = − 2
(x + y 2 )1/2
(b) curl F = 0 is similar to part (a), so F is conservative. Let
Z
cx
dx = −c(x2 + y 2 + z 2 )−1/2 + k(y, z). As in part (a),
φ(x, y, z) =
(x2 + y 2 + z 2 )3/2
∂k/∂y = ∂k/∂z = 0, so φ(x, y, z) = −c/(x2 + y 2 + z 2 )1/2 is a potential function for F.
Z
Q
1
1
F · dr = φ(3, 2, 1) − φ(1, 1, 2) = − √ + √
14
6
P
1
1
(b) C begins at P (1, 1, 2) and ends at Q(3, 2, 1) so the answer is again W = − √ + √ .
14
6
29. (a) See Exercise 28, c = 1; W =
(c) C begins at, say, (3, 0) and ends at the same point so W = 0.
dy
dx
−x
dt for points on the circle x2 + y 2 = 1, so
y
dt
dt
Z
Z π
C1 : x = cos t, y = sin t, 0 ≤ t ≤ π,
F · dr =
(− sin2 t − cos2 t) dt = −π
30. (a) F · dr =
C1
0
Z
C2 : x = cos t, y = − sin t, 0 ≤ t ≤ π,
Z
C2
(b)
π
F · dr =
(sin2 t + cos2 t) dt = π
0
x2 − y 2 ∂g
y 2 − x2
∂f
∂f
= 2
=− 2
,
=
2
2
2
2
∂y
(x + y ) ∂x
(x + y )
∂y
(c) The circle about the origin of radius 1, which is formed by traversing C1 and then traversing
C2 in the reverse direction, does not lie in an open simply connected region inside which F
is continuous, since F is not defined at the origin, nor can it be defined there in such a way
as to make the resulting function continuous there.
31. If C is composed of smooth curves C1 , C2 , . . . , Cn and curve Ci extends from (xi−1 , yi−1 ) to (xi , yi )
Z
n Z
n
X
X
then
F · dr =
F · dr =
[φ(xi , yi ) − φ(xi−1 , yi−1 )] = φ(xn , yn ) − φ(x0 , y0 )
C
i=1
Ci
i=1
where (x0 , y0 ) and (xn , yn ) are the endpoints of C.
Z
Z
F · dr +
32.
Z
C1
−C2
Z
F · dr = 0, but
F · dr is independent of path.
C
−C2
Z
Z
F · dr = −
F · dr so
C2
Z
F · dr =
C1
F · dr, thus
C2
Exercise Set 17.4
618
33. Let C1 be an arbitrary piecewise smooth curve from (a, b) to a point (x, y1 ) in the disk, and C2
the vertical line segment from (x, y1 ) to (x, y). Then
Z
Z (x,y1 )
Z
Z
F · dr +
F · dr =
F · dr +
F · dr.
φ(x, y) =
C1
C2
C2
(a,b)
The first term does not depend on y;
Z
Z
∂
∂
∂φ
=
F · dr =
f (x, y)dx + g(x, y)dy.
hence
∂y
∂y C2
∂y C2
Z
∂
∂φ
=
g(x, y) dy.
∂y
∂y C2
Z y
∂
∂φ
Express C2 as x = x, y = t where t varies from y1 to y, then
=
g(x, t) dt = g(x, y).
∂y
∂y y1
However, the line integral with respect to x is zero along C2 , so
EXERCISE SET 17.4
ZZ
Z
1
Z
1
(2x − 2y)dA =
1.
(2x − 2y)dy dx = 0; for the line integral, on x = 0, y 2 dx = 0, x2 dy = 0;
0
R
0
on y = 0, y 2 dx = x2 dy = 0; on x = 1, y 2 dx + x2 dy = dy; and on y = 1, y 2 dx + x2 dy = dx,
I
Z 1
Z 0
2
2
hence
y dx + x dy =
dy +
dx = 1 − 1 = 0
0
C
1
ZZ
(1 − 1)dA = 0; for the line integral let x = cos t, y = sin t,
2.
R
Z
I
2π
(− sin2 t + cos2 t)dt = 0
y dx + x dy =
0
C
Z
4
Z
(2y − 3x)dy dx = 0
−2
Z
Z
2
3.
1
π/2
Z
ZZ
[1 − (−1)]dA = 2
R
Z
dA = 8π
1
y
−
−
1+y 1+y
ZZ
dA = −
R
Z
dA = −4
R
π/2
Z
4
(−r2 )r dr dθ = −32π
0
0
ZZ
1
y2
−
−
2
1+y
1 + y2
11.
R
1
Z
x
(2x − 2y)dy dx = 1/30
0
ZZ
dA = −1
dA = −
R
dA = π
R
8.
R
10.
ZZ
(sec2 x − tan2 x)dA =
R
ZZ
9.
(1 + 2r sin θ)r dr dθ = 9π
0
6.
0
ZZ
3
ZZ
π/2
(−y cos x + x sin y)dy dx = 0
7.
Z
0
5.
0
2π
4.
x2
619
Chapter 17
ZZ
Z
(cos x cos y − cos x cos y)dA = 0
12.
1
Z
√
x
(y 2 − x2 )dy dx = 0
13.
x2
0
R
15. (a) C : x = cos t, y = sin t, 0 ≤ t ≤ 2π;
Z 2π
I
=
esin t (− sin t) + sin t cos tecos t dt ≈ −3.550999378;
C
ZZ
R
0
ZZ
∂ y
∂
(yex ) −
e dA =
[yex − ey ] dA
∂x
∂y
Z
Z
R
r sin θer cos θ − er sin θ r dr dθ ≈ −3.550999378
0
0
Z
Z 1h
i
2
y
x
2
(b) C1 : x = t, y = t , 0 ≤ t ≤ 1; [e dx + ye dy] =
et + 2t3 et dt ≈ 2.589524432
2π
1
=
0
C1
Z
C2 : x = t2 , y = t, 0 ≤ t ≤ 1;
Z
C1
ZZ
≈ −0.269616482;
C
17. A =
1
2
1
Z
ab cos t dt = πab
Z
−y dx =
(b)
C
0
−y dx + x dy =
C
=
1
2
Z
dt =
e+3
≈ 2.859140914
2
x2
I
I
i
√
x
2π
2
x dy =
16. (a)
2
2tet + tet
[yex − ey ] dy dx ≈ −0.269616482
0
R
Z
I
Z
=
C2
h
1
0
C2
Z
−
Z
[ey dx + yex dy] =
2π
ab sin2 t dt = πab
0
2π
(3a2 sin4 φ cos2 φ + 3a2 cos4 φ sin2 φ)dφ
0
3 2
a
2
Z
2π
sin2 φ cos2 φ dφ =
0
3 2
a
8
Z
2π
sin2 2φ dφ = 3πa2 /8
0
18. C1 : (0, 0) to (a, 0); x = at,
y = 0,
0≤t≤1
0≤t≤1
C2 : (a, 0) to (0, b); x = a − at, y = bt,
y = b − bt, 0 ≤ t ≤ 1
C3 : (0, b) to (0, 0); x = 0,
Z 1
Z 1
Z 1
I
1
x dy =
(0)dt +
ab(1 − t)dt +
(0)dt = ab
A=
2
C
0
0
0
19. C1 : (0, 0) to (a, 0); x = at, y = 0, 0 ≤ t ≤ 1
C2 : (a, 0) to (a cos t0 , b sin t0 ); x = a cos t, y = b sin t, 0 ≤ t ≤ t0
C3 : (a cos t0 , b sin t0 ) to (0, 0); x = −a(cos t0 )t, y = −b(sin t0 )t, −1 ≤ t ≤ 0
I
Z
Z
Z
1 1
1 t0
1 0
1
1
−y dx + x dy =
(0) dt +
ab dt +
(0) dt = ab t0
A=
2 C
2 0
2 0
2 −1
2
20. C1 : (0, 0) to (a, 0); x = at, y = 0, 0 ≤ t ≤ 1
C2 : (a, 0) to (a cosh t0 , b sinh t0 ); x = a cosh t, y = b sinh t, 0 ≤ t ≤ t0
C3 : (a cosh t0 , b sinh t0 ) to (0, 0); x = −a(cosh t0 )t, y = −b(sinh t0 )t, −1 ≤ t ≤ 0
I
Z
Z
Z
1 1
1 t0
1 0
1
1
−y dx + x dy =
(0) dt +
ab dt +
(0) dt = ab t0
A=
2 C
2 0
2 0
2 −1
2
Exercise Set 17.4
620
ZZ
Z
21. W =
π
Z
5
r2 sin θ dr dθ = 250/3
y dA =
0
R
0
22. We cannot apply Green’s Theorem on the region enclosed by the closed curve C, since F does not
have first order partial derivatives at the origin. However, the curve C , consisting of
y = x30 /4, x0 ≤ x ≤ 2; x = 2, x0 ≤ y ≤ 2; and y = x3 /4, x0 ≤ x ≤ 2 encloses a region R in which
Green’s Theorem does hold, and
I
I
ZZ
Z 2 Z x3 /4
1 −1/2 1 −1/2
W = F · dr = lim
x
F · dr = lim+
∇ · F dA =
− y
dy dx
2
2
→0+
→0
x0 x30 /4
C
C
R
√
3 7/2
3 5/2
18 √
2 3
3/2
x + x0 + x0 − x0
2−
−
35
4 0
14
10
= lim+
→0
ZZ
I
Z
y dx − x dy =
23.
C
1
A
ZZ
I
1 2
x dy =
2
x dA, but
C
I
C
1
1 2
x dy =
2
2A
Z
x
18 √
2
35
a(1+cos θ)
0
ZZ
x dA from Green’s Theorem so
R
I
=−
r dr dθ = −3πa2
0
R
1
x̄ =
A
Z
(−2)dA = −2
R
24. x̄ =
2π
!
1
x dy. Similarly, ȳ = −
2A
C
I
2
y 2 dx.
C
Z
Z 1
1
3
; C1 : x = t, y = t3 , 0 ≤ t ≤ 1,
x2 dy =
t2 (3t2 ) dt =
4
5
C1
0
x3
0
I
Z
Z 1
Z
Z
4
8
1
3 1
, x̄ =
C2 : x = t, y = t, 0 ≤ t ≤ 1;
x2 dy =
t2 dt = ,
x2 dy =
−
= − =
3
5
3
15
15
C2
C
C1
C2
0
Z
Z 1
Z 1
4
8
1 1
8 8
2
6
2
, centroid
,
y dx =
t dt −
t dt = − = − , ȳ =
7 3
21
21
15 21
C
0
0
Z
25. A =
1
dy dx =
a2
26. A = ; C1 : x = t, y = 0, 0 ≤ t ≤ a, C2 : x = a − t, y = t, 0 ≤ t ≤ a; C3 : x = 0, y = a − t, 0 ≤ t ≤ a;
2
Z
Z
Z
Z a
I
Z
Z
Z
a
a3
a3
2
x dy = 0, x2 dy =
(a − t)2 dt = , x2 dy = 0, x2 dy = + + = , x̄ = ;
3
3
3
0
C1
Z
C2
Z
y 2 dx = 0 −
C
0
C3
a
C
C1
C2
a a
a
a
,
t2 dt + 0 = − , ȳ = , centroid
3
3
3 3
3
27. x̄ = 0 from the symmetry of the region,
√
C1 : (a, 0) to (−a, 0) along y = a2 − x2 ; x = a cos t, y = a sin t, 0 ≤ t ≤ π
C2 : (−a, 0) to (a, 0); x = t, y = 0, −a ≤ t ≤ a
Z π
Z a
1
−a3 sin3 t dt +
(0)dt
A = πa2 /2, ȳ = −
2A 0
−a
=−
1
πa2
−
4a3
3
=
4a
; centroid
3π
0,
4a
3π
C3
621
Chapter 17
ab
; C1 : x = t, y = 0, 0 ≤ t ≤ a, C2 : x = a, y = t, 0 ≤ t ≤ b;
2
C3 : x = a − t, y = b − bt/a, 0 ≤ t ≤ a;
Z
Z b
Z
Z a
Z
ba2
2
2
2
2
2
,
x dy = 0,
x dy =
a dt = ba ,
x dy =
(a − t)2 (−b/a) dt = −
3
C1
C2
C3
0
0
Z
Z
Z
I
2a
2ba2
, x̄ =
;
x2 dy =
+
+
=
3
3
C
C1
C2
C3
Z a
Z
ab2
b
t
2a b
2
2
dt = −
, ȳ = , centroid
,
y dx = 0 + 0 −
b a−
a
3
3
3 3
C
0
28. A =
ZZ
(1−x2 −y 2 )dA where R is the region enclosed
29. From Green’s Theorem, the given integral equals
R
by C. The value of this integral is maximum if the integration extends over the largest region for
which the integrand 1 − x2 − y 2 is nonnegative so we want 1 − x2 − y 2 ≥ 0, x2 + y 2 ≤ 1. The
largest region is that bounded by the circle x2 + y 2 = 1 which is the desired curve C.
30. (a) C : x = a + (c − a)t, y = b + (d − b)t, 0 ≤ t ≤ 1
Z 1
Z
−y dx + x dy =
(ad − bc)dt = ad − bc
C
0
(b) Let C1 , C2 , and C3 be the line segments from (x1 , y1 ) to (x2 , y2 ), (x2 , y2 ) to (x3 , y3 ), and
(x3 , y3 ) to (x1 , y1 ), then if C is the entire boundary consisting of C1 , C2 , and C3
Z
3 Z
1
1X
A=
−y dx + x dy =
−y dx + x dy
2 C
2 i=1 Ci
=
(c) A =
1
[(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + (x3 y1 − x1 y3 )]
2
1
[(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + · · · + (xn y1 − x1 yn )]
2
1
[(0 − 0) + (6 + 8) + (0 + 2) + (0 − 0)] = 8
2
Z
Z
ZZ
31.
F · dr =
(x2 + y) dx + (4x − cos y) dy = 3
dA = 3(25 − 2) = 69
(d) A =
C
C
Z
Z
F · dr =
32.
C
R
(e−x + 3y) dx + x dy = −2
C
ZZ
dA = −2[π(4)2 − π(2)2 ] = −24π
R
EXERCISE SET 17.5
1. R is the annular region between x2 + y 2 = 1 and x2 + y 2 = 4;
s
ZZ
ZZ
y2
x2
z 2 dS =
(x2 + y 2 )
+ 2
+ 1 dA
2
2
x +y
x + y2
σ
R
√ Z 2π Z 2 3
√ ZZ 2
15 √
2
π 2.
(x + y )dA = 2
r dr dθ =
= 2
2
0
1
R
Exercise Set 17.5
622
2. z = 1 − x − y, R is the triangular region enclosed by x + y = 1, x = 0 and y = 0;
√
ZZ
ZZ
√
√ Z 1 Z 1−x
3
.
xy dS =
xy 3 dA = 3
xy dy dx =
24
0
0
σ
R
3. Let r(u, v) = cos ui + vj + sin uk, 0 ≤ u ≤ π, 0 ≤ v ≤ 1. Then ru = − sin ui + cos uk, rv = j,
ZZ
Z 1Z π
2
x y dS =
v cos2 u du dv = π/4
ru × rv = − cos ui + sin uk, kru × rv k = 1,
σ
0
0
p
4 − x2 − y 2 , R is the circular region enclosed by x2 + y 2 = 3;
s
ZZ
ZZ
p
y2
x2
(x2 + y 2 )z dS =
(x2 + y 2 ) 4 − x2 − y 2
+
+ 1 dA
4 − x2 − y 2
4 − x2 − y 2
4. z =
σ
R
Z
ZZ
2
2π
√
3
Z
2
r3 dr dθ = 9π.
2(x + y )dA = 2
=
0
R
0
5. If we use the projection of σ onto the xz-plane then y = 1 − x and R is the rectangular region in
the xz-plane enclosed by x = 0, x = 1, z = 0 and z = 1;
ZZ
ZZ
√
√ Z 1Z 1
√
(x − y − z)dS =
(2x − 1 − z) 2dA = 2
(2x − 1 − z)dz dx = − 2/2
σ
0
R
0
6. R is the triangular region enclosed by 2x + 3y = 6, x = 0, and y = 0;
ZZ
ZZ
√
√ Z 3 Z (6−2x)/3
√
(x + y)dS =
(x + y) 14 dA = 14
(x + y)dy dx = 5 14.
σ
0
R
0
7. There are six surfaces, parametrized by projecting onto planes:
σ1 : z = 0; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (onto xy-plane), σ2 : x = 0; 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 (onto yz-plane),
σ3 : y = 0; 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 (onto xz-plane), σ4 : z = 1; 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 (onto xy-plane),
σ5 : x = 1; 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 (onto yz-plane), σ6 : y = 1; 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 (onto xz-plane).
By symmetry the integrals over σ1 , σ2 and σ3 are equal, as are those over σ4 , σ5 and σ6 , and
ZZ
Z 1Z 1
ZZ
Z 1Z 1
(x + y + z)dS =
(x + y)dx dy = 1;
(x + y + z)dS =
(x + y + 1)dx dy = 2,
σ1
0
0
σ4
ZZ
(x + y + z)dS = 3 · 1 + 3 · 2 = 9.
thus,
σ
8. Let r(φ, θ) = sin φ cos θi + sin φ sin θj + cos φk,
p
0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2; krφ × rθ k = 1 − cos2 φ = sin φ,
Z 2π Z π/2
ZZ
q
p
(1 + 1 − x2 − y 2 ) dS =
(1 + 1 − sin2 φ) sin φ dφ dθ
σ
0
Z
0
π/2
(1 + cos φ) sin φ dφ = 3π
= 2π
0
0
0
623
Chapter 17
9. R is the circular region enclosed by x2 + y 2 = 1;
s
ZZ p
ZZ p
y2
x2
x2 + y 2 + z 2 dS =
2(x2 + y 2 )
+
+ 1 dA
x2 + y 2
x2 + y 2
σ
R
ZZ p
= lim+ 2
x2 + y 2 dA
r0 →0
R0
0
where R is the annular region enclosed by x2 + y 2 = 1 and x2 + y 2 = r02 with r0 slightly larger
s
y2
x2
+
+ 1 is not defined for x2 + y 2 = 0, so
x2 + y 2
x2 + y 2
ZZ p
Z 2π Z 1
4π
4π
(1 − r03 ) =
.
x2 + y 2 + z 2 dS = lim 2
r2 dr dθ = lim
+
+
3
3
r0 →0
r0 →0
0
r0
than 0 because
σ
10. Let r(φ, θ) = a sin φ cos θi + a sin φ sin θj + a cos φk,
p
0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/2; krφ × rθ k = a2 1 − cos2 φ,
ZZ
Z 2π Z π
p
8
f (x, y, z) = a2
a2 sin2 φ 1 − cos2 φ dφ dθ = πa4
3
0
0
σ
√
Z Z
29 6 (12−2x)/3
xy(12 − 2x − 3y)dy dx
16 0 0
√ Z 3 Z (12−4z)/3
29
yz(12 − 3y − 4z)dy dz
(b)
4
0
0
√ Z 3 Z 6−2z
29
xz(12 − 2x − 4z)dx dz
(c)
9
0
0
11. (a)
Z
a
√
a2 −x2
Z
Z
x dy dx
12. (a) a
Z
a
(c) a
√
0
0
a2
Z
4
Z
2
y3 z
0
1
Z
9
16. a
0
Z
0
xz
dx dz
− x2 − z 2
√
13. 18 29/5
15.
√
a2 −z 2
z dy dz
0
√
a2 −z 2
Z
Z
(b) a
0
0
a
14. a4 /3
Z Z
p
1 4 4 √
4y 2 + 1 dy dz;
xz 1 + 4x dx dz
2 0 1
√
a/ 2
√
a/ 5
Z
p
x2 y
a2
−
y2
dy dx, a
√
2a/ 5
√
a/ 2
Z
9
x2 dx dz
17.
√
√
391 17/15 − 5 5/3
0
18. The region R : 3x2 + 2y 2 = 5 is symmetric in y. The integrand is
ZZZ
p
x2 yz dS = 0.
x2 yz dS = x2 y(5 − 3x2 − 2y 2 ) 1 + 36x2 + 16y 2 dy dx, which is odd in y, hence
σ
Exercise Set 17.5
624
√
x
∂z
∂z
= −√
= 0;
4 − x2 ,
,
2
∂x
4 − x ∂y
ZZ
ZZ r
Z 4Z 1
4
x2
1
√
δ0 dS = δ0
+ 1 dA = 2δ0
dx dy = πδ0 .
2
2
4−x
3
4−x
0
0
19. z =
σ
R
1 2
(x + y 2 ), R is the circular region enclosed by x2 + y 2 = 8;
2
ZZ p
Z 2π Z √8 p
ZZ
52
2
2
πδ0 .
δ0 dS = δ0
x + y + 1 dA = δ0
r2 + 1 r dr dθ =
3
0
0
20. z =
σ
R
21. z = 4 − y 2 , R is the rectangular region enclosed by x = 0, x = 3, y = 0 and y = 3;
ZZ
ZZ p
Z 3Z 3 p
√
1
y dS =
y 4y 2 + 1 dA =
y 4y 2 + 1 dy dx = (37 37 − 1).
4
0
0
σ
R
22. R is the annular region enclosed by x2 + y 2 = 1 and x2 + y 2 = 16;
s
ZZ
ZZ
p
y2
x2
2
2
2
2
x z dS =
x x +y
+ 2
+ 1 dA
2
2
x +y
x + y2
σ
R
√
√ Z 2π Z 4 4
√ Z Z 2p
1023 2
2
2
2
π.
x x + y dA = 2
r cos θ dr dθ =
= 2
5
0
1
R
ZZ
ZZ
ZZ
23. M =
δ(x, y, z)dS =
σ
δ0 dS = δ0
σ
dS = δ0 S
σ
p
24. δ(x, y, z) = |z|; use z = a2 −x2 −y 2 , let R be the circular region enclosed by x2 +y 2 = a2 , and
σ the hemisphere above R. By the symmetry of both the surface and the density function with
respect to the xy-plane we have
s
ZZ
ZZ p
ZZ
y2
x2
2
2
2
M =2
z dS = 2
a −x −y
+ 2
+ 1 dA = lim− 2a
dA
a2 − x2 − y 2
a − x2 − y 2
r0 →a
σ
R
ZZ
where Rr0 is the circular region with radius r0 that is slightly less than a. But
dA is simply
Rr0
the area of the circle with radius r0 so M = lim− 2a(πr02 ) = 2πa3 .
r0 →a
25. By symmetry x̄ = ȳ = 0.
ZZ
ZZ p
Z
dS =
x2 + y 2 + 1 dA =
σ
0
R
ZZ
√
8
p
r2 + 1 r dr dθ =
0
52π
,
3
p
p
1
x2 + y 2 + 1 dA =
(x2 + y 2 ) x2 + y 2 + 1 dA
2
R
R
√
Z
Z 8 p
596π
1 2π
r3 r2 + 1 dr dθ =
=
2 0
15
0
z dS =
so z̄ =
Z
ZZ
ZZ
σ
2π
z
149
596π/15
=
. The centroid is (x̄, ȳ, z̄) = (0, 0, 149/65).
52π/3
65
Rr0
625
Chapter 17
26. By symmetry x̄ = ȳ = 0.
ZZ
ZZ
Z 2π Z √3
2
r
p
√
dS =
dA = 2
dr dθ = 4π,
4
− r2
4 − x2 − y 2
0
0
σ
R
ZZ
ZZ
√
z dS =
2 dA = (2)(area of circle of radius 3) = 6π
σ
R
3
6π
= . The centroid is (x̄, ȳ, z̄) = (0, 0, 3/2).
so z̄ =
4π
2
27. ∂r/∂u = cos vi + sin vj + 3k, ∂r/∂v = −u sin vi + u cos vj, k∂r/∂u × ∂r/∂vk =
√ Z π/2 Z 2 4
√
√ ZZ 4
u sin v cos v dA = 3 10
u sin v cos v du dv = 93/ 10
3 10
0
R
√
10u;
1
28. ∂r/∂u = j, ∂r/∂v = −2 sin vi + 2 cos vk, k∂r/∂u × ∂r/∂vk = 2;
Z 2π Z 3
ZZ
1
1
dA = 8
du dv = 16π ln 3
8
u
u
0
1
R
√
29. ∂r/∂u = cos vi + sin vj + 2uk, ∂r/∂v = −u sin vi + u cos vj, k∂r/∂u × ∂r/∂vk = u 4u2 + 1;
Z π Z sin v
ZZ
u dA =
u du dv = π/4
0
R
0
30. ∂r/∂u = 2 cos u cos vi + 2 cos u sin vj − 2 sin uk, ∂r/∂v = −2 sin u sin vi + 2 sin u cos vj;
k∂r/∂u × ∂r/∂vk = 4 sin u;
Z 2π Z π/2
ZZ
e−2 cos u sin u dA = 4
e−2 cos u sin u du dv = 4π(1 − e−2 )
4
0
R
31. ∂z/∂x = −2xe−x
2
2
−y 2
0
, ∂z/∂y = −2ye−x
2
2
2
2
−y 2
,
−2(x2 +y 2 )
+ 1; use polar coordinates to get
(∂z/∂x) + (∂z/∂y) + 1 = 4(x + y )e
Z 2π Z 3 p
r2 4r2 e−2r2 + 1 dr dθ ≈ 57.895751
M=
0
0
Z
ZZ
2π
Z
1
dS =
32. (b) A =
σ
0
−1
1p
40u cos(v/2) + u2 + 4u2 cos2 (v/2) + 100du dv ≈ 62.93768644;
2
x̄ ≈ 0.01663836266; ȳ = z̄ = 0 by symmetry
EXERCISE SET 17.6
1. (a) zero
(d) negative
2. (a) positive
(d) zero
(b) zero
(c) positive
(e) zero
(f ) zero
(b) zero
(e) negative
(c) zero
(f ) zero
Exercise Set 17.6
626
3. (a) positive
(d) zero
(b) zero
(e) positive
(c) positive
(f ) zero
4. 0; the flux is zero on the faces y = 0, 1 and z = 0, 1; it is 1 on x = 1 and −1 on x = 0
5. (a) n = − cos vi − sin vj
(b) inward, by inspection
6. (a) −r cos θi − r sin θj + rk
(b) inward, by inspection
ZZ
7. n = −zx i − zy j + k,
ZZ
F · n dS =
R
Z
2π
Z
(2x + 2y + 2(1 − x − y )) dS =
2
2
2
2
2r dr dθ = 2π
0
R
1
0
8. With z = 1 − x − y, R is the triangular region enclosed by x + y = 1, x = 0 and y = 0; use upward
normals to get
ZZ
ZZ
ZZ
F · n dS = 2
(x + y + z)dA = 2
dA = (2)(area of R) = 1.
σ
R
R
9. R is the annular region enclosed by x2 + y 2 = 1 and x2 + y 2 = 4;
!
ZZ
ZZ
y2
x2
F · n dS =
−p
+ 2z dA
−p
x2 + y 2
x2 + y 2
σ
R
Z
ZZ p
2
2
x + y dA =
=
2π
Z
r2 dr dθ =
0
R
2
1
14π
.
3
10. R is the circular region enclosed by x2 + y 2 = 4;
ZZ
ZZ
Z 2π Z 2
F · n dS =
(2y 2 − 1)dA =
(2r2 sin2 θ − 1)r dr dθ = 4π.
σ
0
R
0
ZZ
11. R is the circular region enclosed by x + y − y = 0;
2
ZZ
F · n dS =
2
σ
(−x)dA = 0 since the
R
region R is symmetric across the y-axis.
1
12. With z = (6 − 6x − 3y), R is the triangular region enclosed by 2x + y = 2, x = 0, and y = 0;
2
ZZ
ZZ
Z 1 Z 2−2x
ZZ
3
2
F · n dS =
x dA = 3
x dy dx = 1.
3x + yx + zx dA = 3
2
0
0
σ
R
R
13. ∂r/∂u = cos vi + sin vj − 2uk, ∂r/∂v = −u sin vi + u cos vj,
∂r/∂u × ∂r/∂v = 2u2 cos vi + 2u2 sin vj + uk;
Z 2π Z 2
ZZ
3
(2u + u) dA =
(2u3 + u)du dv = 18π
R
0
1
14. ∂r/∂u = k, ∂r/∂v = −2 sin vi + cos vj, ∂r/∂u × ∂r/∂v = − cos vi − 2 sin vj;
Z 2π Z 5
ZZ
2
− sin v
(2 sin v − e
cos v) dA =
(2 sin2 v − e− sin v cos v)du dv = 10π
R
0
0
627
Chapter 17
15. ∂r/∂u = cos vi + sin vj + 2k, ∂r/∂v = −u sin vi + u cos vj,
∂r/∂u × ∂r/∂v = −2u cos vi − 2u sin vj + uk;
Z π Z sin v
ZZ
2
u dA =
u2 du dv = 4/9
0
R
0
16. ∂r/∂u = 2 cos u cos vi + 2 cos u sin vj − 2 sin uk, ∂r/∂v = −2 sin u sin vi + 2 sin u cos vj;
∂r/∂u × ∂r/∂v = 4 sin2 u cos vi + 4 sin2 u sin vj + 4 sin u cos uk;
Z 2π Z π/3
ZZ
8 sin u dA = 8
sin u du dv = 8π
0
R
0
17. In each part, divide σ into the six surfaces
σ1 : x = −1 with |y| ≤ 1, |z| ≤ 1, and n = −i, σ2 : x = 1 with |y| ≤ 1, |z| ≤ 1, and n = i,
σ3 : y = −1 with |x| ≤ 1, |z| ≤ 1, and n = −j, σ4 : y = 1 with |x| ≤ 1, |z| ≤ 1, and n = j,
σ5 : z = −1 with |x| ≤ 1, |y| ≤ 1, and n = −k, σ6 : z = 1 with |x| ≤ 1, |y| ≤ 1, and n = k,
ZZ
ZZ
F · n dS =
(a)
σ1
ZZ
F · n dS =
dS = 4,
σ1
ZZ
σ2
ZZ
σ
ZZ
F · n dS =
(b)
σ1
σi
ZZ
F · n dS = 4 for i = 2, 3, 4, 5, 6 so
dS = 4, similarly
σ1
ZZ
F · n dS = 0 for
dS = 4, and
σ2
F · n dS = 4 + 4 + 0 + 0 + 0 + 0 = 8.
i = 3, 4, 5, 6 so
ZZ
ZZ
σi
F · n dS = 4 + 4 + 4 + 4 + 4 + 4 = 24.
σ
ZZ
ZZ
ZZ
F · n dS = −
(c)
ZZ
σ1
ZZ
dS = −4,
σ1
F · n dS = 4, similarly
σ2
ZZ
F · n dS = 4 for i = 4, 6 so
and
σi
F · n dS = −4 for i = 3, 5
σi
F · n dS = −4 + 4 − 4 + 4 − 4 + 4 = 0.
σ
18. Decompose σ into a top σ1 (the disk) and a bottom σ2 (the portion of the paraboloid). Then
ZZ
ZZ
Z 2π Z 1
F · n1 dS = −
y dS = −
r2 sin θ dr dθ = 0,
n1 = k,
σ1
0
σ1
p
n2 = (2xi + 2yj − k)/ 1 + 4x2 + 4y 2 ,
0
ZZ
ZZ
F · n2 dS =
σ2
σ2
y(2x2 + 2y 2 + 1)
p
dS = 0,
1 + 4x2 + 4y 2
because the surface σ2 is symmetric with respect to the xy-plane and the integrand is an odd
function of y. Thus the flux is 0.
19. R is the circular region enclosed by x2 + y 2 = 1;
!
p
ZZ
ZZ
ZZ
y
x + y − x2 + y 2
x
p
p
F · n dS =
+p
− 1 dA =
dA
x2 + y 2
x2 + y 2
x2 + y 2
σ
R
= lim+
r0 →0
Z
0
2π
Z
R
1
r0
(r cos θ + r sin θ − r)dr dθ = lim+ π(r02 − 1) = −π.
r0 →0
Exercise Set 17.6
628
20. Let r = cos vi + uj + sin vk, −2 ≤ u ≤ 1, 0 ≤ v ≤ 2π; ru × rv = cos vi + sin vk,
ZZ
ZZ
F · n dS =
(cos2 v + sin2 v) dA = area of R = 3 · 2π = 6π
σ
R
ZZ
ZZ
F · ndS =
21. (a)
F·
σ
R
ZZ
ZZ
F · ndS =
F·
σ
∂x
∂x
j−
k dA, if σ is oriented by front normals, and
i−
∂y
∂z
−i +
∂x
∂x
j+
k dA, if σ is oriented by back normals,
∂y
∂z
R
where R is the projection of σ onto the yz-plane.
p
(b) R is the semicircular region in the yz-plane enclosed by z = 1 − y 2 and z = 0;
ZZ
Z 1 Z √1−y2
ZZ
32
.
F · n dS =
(−y − 2yz + 16z)dA =
(−y − 2yz + 16z)dz dy =
3
−1 0
σ
R
ZZ
F·
22. (a)
R
∂y
∂y
i−j+
k dA, σ oriented by right normals,
∂x
∂z
ZZ
F·
and
−
∂y
∂y
i+j−
k dA, σ oriented by left normals,
∂x
∂z
R
where R is the projection of σ onto the xz-plane.
√
(b) R is the semicircular region in the xz-plane enclosed by z = 1 − x2 and z = 0;
ZZ
Z 1 Z √1−x2
ZZ
π
2
2
2
2
F · n dS =
(2x − (x + z ) + z )dA =
(x2 + z 2 )dz dx = .
4
−1 0
σ
R
23. (a) On the sphere, krk = a so F = ak r and F · n = ak r · (r/a) = ak−1 krk2 = ak−1 a2 = ak+1 ,
ZZ
ZZ
k+1
hence
F · n dS = a
dS = ak+1 (4πa2 ) = 4πak+3 .
σ
σ
ZZ
(b) If k = −3, then
F · n dS = 4π.
σ
24. Let r = sin u cos vi + sin u sin vj + cos uk, ru × rv = sin2 u cos vi + sin2 u sin vj + sin u cos uk,
F · (ru × rv ) = a2 sin3 u cos2 v +
ZZ
Z
2π
Z
π
1
sin3 u sin2 v + a sin u cos3 u) du dv
9
0
0
Z π
Z
Z 2π
1 π 3
3
2
2
=a
sin u du
cos v dv +
sin u du
a 0
0
0
Z π
Z 2π
sin2 v dv + 2πa
sin u cos3 u du
+
F · n dS =
σ
1
sin3 u sin2 v + a sin u cos3 u,
9
(a2 sin3 u cos2 v +
0
=
4π
3
a2 +
1
a
0
= 10 if a ≈ −1.722730, 0.459525, 1.263205
629
Chapter 17
EXERCISE SET 17.7
ZZ
ZZ
1. σ1 : x = 0, F · n = −x = 0,
σ2 : x = 1, F · n = x = 1,
(0)dA = 0
σ1
σ2
ZZ
σ3 : y = 0, F · n = −y = 0,
ZZ
σ4 : y = 1, F · n = y = 1,
(0)dA = 0
σ3
ZZ
F · n = 3;
σ
ZZ
σ6 : z = 1, F · n = z = 1,
(0)dA = 0
σ5
ZZZ
(1)dA = 1
σ4
ZZ
σ5 : z = 0, F · n = −z = 0,
(1)dA = 1
(1)dA = 1
σ6
ZZZ
div FdV =
3dV = 3
G
G
2. For any point r = xi + yj + zk on σ let n = xi + yj + zk; then F · n = x2 + y 2 + z 2 = 1, so
ZZ
ZZ
ZZZ
ZZZ
F · n dS =
dS = 4π; also
div FdV =
3dV = 3(4π/3) = 4π
σ
σ
G
G
ZZ
3. σ1 : z = 1, n = k, F · n = z 2 = 1,
(1)dS = π,
σ1
σ2 : n = 2xi + 2yj − k, F · n = 4x2 − 4x2 y 2 − x4 − 3y 4 ,
ZZ
Z 2π Z 1
2
π
F · n dS =
4r cos2 θ − 4r4 cos2 θ sin2 θ − r4 cos4 θ − 3r4 sin4 θ r dr dθ = ;
3
0
0
σ2
ZZ
=
4π
3
σ
ZZZ
ZZZ
div FdV =
G
Z
2π
Z
1
Z
1
(2 + z)dV =
(2 + z)dz r dr dθ = 4π/3
0
G
r2
0
ZZ
4. σ1 : x = 0, F · n = −xy = 0,
ZZ
σ2 : x = 2, F · n = xy = 2y,
(0)dA = 0
σ1
ZZ
σ3 : y = 0, F · n = −yz = 0,
ZZ
σ4 : y = 2, F · n = yz = 2z,
(0)dA = 0
σ3
ZZ
ZZZ
F · n = 24; also
σ
(2z)dA = 8
σ4
ZZ
σ5 : z = 0, F · n = −xz = 0,
(2y)dA = 8
σ2
ZZ
σ6 : z = 2, F · n = xz = 2x,
(0)dA = 0
σ5
ZZZ
div FdV =
G
(2x)dA = 8
σ6
(y + z + x)dV = 24
G
ZZZ
5. G is the rectangular solid;
Z
2
Z
1
Z
0
G
3
(2x − 1) dx dy dz = 12.
div F dV =
0
0
ZZZ
6. G is the spherical solid enclosed by σ;
ZZZ
div F dV =
G
ZZZ
0 dV = 0
G
dV = 0.
G
Exercise Set 17.7
630
7. G is the cylindrical solid;
ZZZ
ZZZ
div F dV = 3
dV = (3)(volume of cylinder) = (3)[πa2 (1)] = 3πa2 .
G
G
8. G is the solid bounded by z = 1 − x2 − y 2 and the xy-plane;
ZZZ
ZZZ
Z 2π Z 1 Z 1−r2
3π
.
div F dV = 3
dV = 3
r dz dr dθ =
2
0
0
0
G
G
9. G is the cylindrical solid;
ZZZ
Z
ZZZ
div F dV = 3
(x2 + y 2 + z 2 )dV = 3
G
Z
div F dV =
Z
0
ZZZ
ZZZ
2
3
(r2 + z 2 )r dz dr dθ = 180π.
0
G
10. G is the tetrahedron;
2π
0
Z
1
Z
1−x
Z
1−x−y
x dV =
G
x dz dy dx =
0
G
0
0
1
.
24
p
11. G is the hemispherical solid bounded by z = 4 − x2 − y 2 and the xy-plane;
ZZZ
ZZZ
Z 2π Z π/2 Z 2
192π
.
div F dV = 3
(x2 + y 2 + z 2 )dV = 3
ρ4 sin φ dρ dφ dθ =
5
0
0
0
G
G
12. G is the hemispherical solid;
ZZZ
Z
ZZZ
div F dV = 5
z dV = 5
G
2π
0
G
Z
Z
π/2
ρ3 sin φ cos φ dρ dφ dθ =
0
0
13. G is the conical solid;
ZZZ
Z
ZZZ
div F dV = 2
(x + y + z)dV = 2
G
2π
0
G
a
Z
1
Z
5πa4
.
4
1
(r cos θ + r sin θ + z)r dz dr dθ =
0
r
14. G is the solid bounded by z = 2x and z = x2 + y 2 ;
ZZZ
ZZZ
Z π/2 Z 2 cos θ Z 2r cos θ
π
div F dV =
dV = 2
r dz dr dθ = .
2
2
0
0
r
G
G
15. G is the solid bounded by z = 4 − x2 , y + z = 5, and the coordinate planes;
ZZZ
ZZZ
Z 2 Z 4−x2 Z 5−z
4608
2
.
div F dV = 4
x dV = 4
x2 dy dz dx =
35
−2 0
0
G
G
ZZ
ZZZ
F · n dS =
16.
σ
ZZZ
div F dV =
G
0 dV = 0;
G
since the vector field is constant, the same amount enters as leaves.
ZZZ
ZZ
r · n dS =
17.
σ
ZZZ
div r dV = 3
G
dV = 3vol(G)
G
π
.
2
631
Chapter 17
ZZ
F · n dS = 3[π(32 )(5)] = 135π
18.
σ
ZZ
ZZZ
ZZZ
curl F · n dS =
19.
div(curl F)dV =
σ
G
ZZ
ZZZ
σ
G
ZZZ
G
ZZZ
(f ∇g) · n =
σ
div (f ∇g)dV =
G
ZZ
(f ∇2 g + ∇f · ∇g)dV by Exercise 29, Section 17.1.
G
ZZZ
(f ∇g) · n dS =
22.
∇2 f dV
div (∇f )dV =
ZZ
21.
G
ZZZ
∇f · n dS =
20.
(0)dV = 0
σ
(f ∇2 g + ∇f · ∇g)dV by Exercise 29, Section 17.1;
G
ZZ
ZZZ
(g∇f ) · n dS =
σ
(g∇2 f + ∇g · ∇f )dV by interchanging f and g;
G
subtract to obtain the result.
23. Let the normal n = n1 i + n2 j + n3 k, then we have to show that
ZZ
ZZZ
(f n1 i + f n2 j + f n3 k) dS =
(fx i + fy j + fz k) dV . We shall show that
σ
ZZ
G
ZZZ
f n1 i dS =
ZZ
fx i dV , i.e. that
σ
ZZZ
f n1 dS =
σ
G
fx dV ; the other parts will follow in a
G
similar manner. Let F(x, y, z) = f (x, y, z)i, then by the Divergence Theorem
ZZ
ZZ
ZZZ
ZZZ
f n1 dS =
F · n dS =
∇ · F dV =
fx dV .
σ
σ
G
G
∂
∂
x
y
z
∂
+
+
2
2
2
3/2
2
2
2
3/2
2
2
∂x (x + y + z )
∂y (x + y + z )
∂z (x + y + z 2 )3/2
(−2x2 + y 2 + z 2 ) + (−2y 2 + x2 + z 2 ) + (−2z 2 + x2 + y 2 )
=0
=c
(x2 + y 2 + z 2 )5/2
24. div r = c
25. (a) The flux through any cylinder whose axis is the z-axis is positive by inspection; by the
Divergence Theorem, this says that the divergence cannot be negative at the origin, else
the flux through a small enough cylinder would also be negative (impossible), hence the
divergence at the origin must be ≥ 0.
(b) Similar to Part (a), ≤ 0.
26. (a) F = xi + yj + zk, div F = 3
(b) F = −xi − yj − zk, div F = −3
27. div F = 0; no sources or sinks.
28. div F = y − x; sources where y > x, sinks where y < x.
29. div F = 3x2 + 3y 2 + 3z 2 ; sources at all points except the origin, no sinks.
Exercise Set 17.8
632
30. div F = 3(x2 + y 2 + z 2 − 1); sources outside the sphere x2 + y 2 + z 2 = 1, sinks inside the sphere
x2 + y 2 + z 2 = 1.
31. Let σ1 be the portion of the paraboloid z = 1 − x2 − y 2 for z ≥ 0, and σ2 the portion of the plane
z = 0 for x2 + y 2 ≤ 1. Then
ZZ
ZZ
F · n dS =
F · (2xi + 2yj + k) dA
σ1
R
Z
Z
1
√
1−x2
(2x[x2 y
√
− 1−x2
=
−1
− (1 − x2 − y 2 )2 ] + 2y(y 3 − x) + (2x + 2 − 3x2 − 3y 2 )) dy dx
= 3π/4;
ZZ
z = 0 and n = −k on σ2 so F · n = 1 − 2x,
F · n dS =
σ2
ZZ
ZZ
(1 − 2x)dS = π. Thus
σ2
F · n dS = 3π/4 + π = 7π/4. But div F = 2xy + 3y 2 + 3 so
σ
Z
ZZZ
1
div F dV =
G
−1
Z
√
1−x2
√
− 1−x2
Z
1−x2 −y 2
(2xy + 3y 2 + 3) dz dy dx = 7π/4.
0
EXERCISE SET 17.8
1. (a) The flow is independent of z and has no component in the direction of k, and so by inspection
the only nonzero component of the curl is in the direction of k. However both sides of (9)
are zero, as the flow is orthogonal to the curve Ca . Thus the curl is zero.
(b) Since the flow appears to be tangential to the curve Ca , it seems that the right hand side of
(9) is nonzero, and thus the curl is nonzero, and points in the positive z-direction.
2. (a) The only nonzero vector component of the vector field is in the direction of i, and it increases
with y and is independent of x. Thus the curl of F is nonzero, and points in the negative
z-direction.
(b) By inspection the vector field is constant, and thus its curl is zero.
3. If σ is oriented with upward normals then C consists of three parts parametrized as
C1 : r(t) = (1 − t)i + tj for 0 ≤ t ≤ 1, C2 : r(t) = (1 − t)j + tk for 0 ≤ t ≤ 1,
C3 : r(t) = ti + (1 − t)k for 0 ≤ t ≤ 1.
Z
Z
Z 1
Z
1
F · dr =
F · dr =
F · dr =
(3t − 1)dt = so
2
C1
C2
C3
0
I
1
1
3
1
F · dr = + + = . curl F = i + j + k, z = 1 − x − y, R is the triangular region in
2
2
2
2
C
the xy-plane enclosed by x + y = 1, x = 0, and y = 0;
ZZ
ZZ
3
1
(curl F) · n dS = 3
dA = (3)(area of R) = (3) (1)(1) = .
2
2
σ
R
633
Chapter 17
4. If σ is oriented with upward normals then C can be parametrized as r(t) = cos ti + sin tj + k for
0 ≤ t ≤ 2π.
Z 2π
I
F · dr =
(sin2 t cos t − cos2 t sin t)dt = 0;
C
0
ZZ
ZZ
(curl F) · n dS =
curl F = 0 so
σ
0 dS = 0.
σ
5. If σ is oriented with upward normals then C can be parametrized as r(t) = a cos ti + a sin tj for
0 ≤ t ≤ 2π.
Z 2π
ZZ
ZZ
I
F · dr =
0 dt = 0; curl F = 0 so
(curl F) · n dS =
0 dS = 0.
C
0
σ
σ
6. If σ is oriented with upward normals then C can be parametrized as r(t) = 3 cos ti + 3 sin tj for
0 ≤ t ≤ 2π.
Z 2π
Z 2π
I
2
2
F · dr =
(9 sin t + 9 cos t)dt = 9
dt = 18π.
C
0
0
curl F = −2i + 2j + 2k, R is the circular region in the xy-plane enclosed by x2 + y 2 = 9;
ZZ
ZZ
Z 2π Z 3
(curl F) · n dS =
(−4x + 4y + 2)dA =
(−4r cos θ + 4r sin θ + 2)r dr dθ = 18π.
σ
0
R
0
7. Take σ as the part of the plane z = 0 for x2 + y 2 ≤ 1 with n = k; curl F = −3y 2 i + 2zj + 2k,
ZZ
ZZ
(curl F) · n dS = 2
dS = (2)(area of circle) = (2)[π(1)2 ] = 2π.
σ
σ
8. curl F = xi + (x − y)j + 6xy 2 k;
ZZ
Z
ZZ
(curl F) · n dS =
(x − y − 6xy 2 )dA =
σ
1
0
R
Z
3
(x − y − 6xy 2 )dy dx = −30.
0
9. C is the boundary of R and curl F = 2i + 3j + 4k, so
ZZ
I
ZZ
curl F · n dS =
4 dA = 4(area of R) = 16π
F·r=
R
R
10. curl F = −4i − 6j + 6yk, z = y/2 oriented with upward normals, R is the triangular region in the
xy-plane enclosed by x + y = 2, x = 0, and y = 0;
ZZ
ZZ
Z 2 Z 2−x
(curl F) · n dS =
(3 + 6y)dA =
(3 + 6y)dy dx = 14.
σ
0
R
0
11. curl F = xk, take σ as part of the plane z = y oriented with upward normals, R is the circular
region in the xy-plane enclosed by x2 + y 2 − y = 0;
ZZ
ZZ
Z π Z sin θ
(curl F) · n dS =
x dA =
r2 cos θ dr dθ = 0.
σ
R
0
0
12. curl F = −yi − zj − xk, z = 1 − x − y oriented with upward normals, R is the triangular region in
the xy-plane enclosed by x + y = 1, x = 0 and y = 0;
ZZ
ZZ
ZZ
1
1
(curl F) · n dS =
(−y − z − x)dA = −
dA = − (1)(1) = − .
2
2
σ
R
R
Exercise Set 17.8
634
13. curl F = i + j + k, take σ as the part of the plane z = 0 with x2 + y 2 ≤ a2 and n = k;
ZZ
ZZ
(curl F) · n dS =
dS = area of circle = πa2 .
σ
σ
√
14. curl F = i + j + k, take σ as the part of the plane z = 1/ 2 with x2 + y 2 ≤ 1/2 and n = k.
ZZ
ZZ
π
(curl F) · n dS =
dS = area of circle = .
2
σ
σ
15. (a) Take σ as the part of the plane 2x + y + 2z = 2 in the first octant, oriented with downward
normals; curl F = −xi + (y − 1)j − k,
ZZ
I
F · T ds =
(curl F) · n dS
C
ZσZ
=
3
1
x− y+
2
2
Z
1
Z
2−2x
dA =
0
R
0
3
1
x− y+
2
2
dy dx =
3
.
2
(b) At the origin curl F = −j − k and with n = k, curl F(0, 0, 0) · n = (−j − k) · k = −1.
(c) The rotation of F has its maximum value at the origin about the unit vector in the same
1
1
direction as curl F(0, 0, 0) so n = − √ j − √ k.
2
2
16. (a) div(curl F) =
=
∂
∂x
∂h ∂g
−
∂y
∂z
+
∂
∂y
∂h
∂f
−
∂z
∂x
+
∂
∂z
∂f
∂g
−
∂x ∂y
∂2g
∂2f
∂2h
∂2g
∂2f
∂2h
−
+
−
+
−
= 0,
∂x∂y ∂x∂z
∂y∂z
∂y∂x ∂z∂x ∂z∂y
assuming equality of mixed second partial derivatives
(b) By the Divergence Theorem
ZZZ
ZZZ
ZZ
(curl F) · n dS =
div(curl F)dV =
0 dV = 0.
σ
G
G
(c) The flux of the curl field through the boundary of a solid is zero.
ZZ
I
E · rdr =
17. Since
C
ZZ
curl E · n dS, it follows that
σ
ZZ
curl E · ndS = −
σ
∂B
· ndS. This
∂t
σ
∂B
.
relationship holds for any volume σ, hence curl E = −
∂t
18. Parametrize C by x = cos t, y = sin t, 0 ≤ t ≤ 2π. But F = x2 yi + (y 3 − x)j + (2x − 1)k along C
I
so
F · dr = −5π/4. Since curl F = (−2z − 2)j + (−1 − x2 )k,
C
ZZ
ZZ
(curl F) · n dS =
σ
(curl F) · (2xi + 2yj + k) dA
R
Z
1
=
−1
Z
√
1−x2
[2y(2x2
√
− 1−x2
+ 2y 2 − 4) − 1 − x2 ] dy dx = −5π/4
635
Chapter 17
CHAPTER 17 SUPPLEMENTARY EXERCISES
2. (b)
c
(r − r0 )
kr − r0 k3
Z
b
3. (a)
f (x(t), y(t))
Z
a
b
(b)
xi + yj + zk
(c) c p
x2 + y 2 + z 2
dy
dx
+ g(x(t), y(t))
dt
dt
dt
p
f (x(t), y(t)) x0 (t)2 + y 0 (t)2 dt
a
Z
Z
4. (a) M =
δ(x, y, z) ds
C
I
I
x dy = −
(d) A =
C
C
ZZ
11.
(b) L =
1
y dx =
2
ZZ
ds
(c) S =
σ
C
I
dS
−y dx + x dy
C
ZZ
f (x(u, v), y(u, v), z(u, v))kru × rv k du dv
f (x, y, z)dS =
σ
R
13. C1 : (0, 0) to (1, 0); x = t, y = 0, 0 ≤ t ≤ 1
C2 : (1, 0) to (cosh t0 , sinh t0 ); x = cosh t, y = sinh t, 0 ≤ t ≤ t0
C3 : (cosh t0 , sinh t0 ) to (0, 0); x = −(cosh t0 )t, y = −(sinh t0 )t, −1 ≤ t ≤ 0
I
Z
Z
Z
1 1
1 t0
1 0
1
1
−y dx + x dy =
(0) dt +
dt +
(0) dt = t0
A=
2 C
2 0
2 0
2 −1
2
qQ(xi + yj + zk)
4π0 (x2 + y 2 + z 2 )3/2
1
qQ
qQ 1
√
−
(b) F = ∇φ, where φ = −
,
so
W
=
φ(3,
1,
5)−φ(3,
0,
0)
=
.
4π0 3
4π0 (x2 + y 2 + z 2 )1/2
35
14. (a) F(x, y, z) =
C : x = 3, y = t, z = 5t, 0 ≤ t ≤ 1; F · dr =
Z
W =
0
1
qQ
26qQt dt
=
2
3/2
4π0
4π0 (26t + 9)
qQ[0 + t + 25t] dt
4π0 (9 + t2 + 25t2 )3/2
1
1
√ −
35 3
15. (a) Assume the mass M is located at the origin and the mass m at (x, y, z), then
GmM
r,
+ y 2 + z 2 )3/2
Z t2
dy
dz
dx
GmM
W =−
+
y
+
z
dt
x
2
2
2 3/2
dt
dt
dt
t1 (x + y + z )
it2
1
1
2
2
2 −1/2
= GmM
−
= GmM (x + y + z )
r2
r1
t1
1
1
−
≈ −1.597 × 109 J
(b) W = 3.99 × 105 × 103
7200 7000
r = xi + yj + zk, F(x, y, z) = −
(x2
Chapter 17 Supplementary Exercises
636
Z
2
16. Let g(x, y) = x , then
ZZ
g dy =
C
1
2x dA, so x̄ =
A
ZZ
R
1
x dA =
2A
R
Z
C
1
g dy =
2A
Z
x2 dy,
C
similarly for ȳ.
17. x̄ = 0 by symmetry; by Exercise 16, ȳ =
1
2A
Z
y 2 dx; C1 : y = 0, −a ≤ x ≤ a, y 2 dx = 0;
C
C2 : x = a cos θ, y = a sin θ, 0 ≤ θ ≤ π, so
Z π
4a
1
a2 sin2 θ(−a sin θ) dθ =
ȳ = −
2
2(πa /2) 0
3π
1
18. ȳ = x̄ by symmetry; by Exercise 16, x̄ =
2A
Z
x2 dy; C1 : y = 0, 0 ≤ x ≤ a, x2 dy = 0;
C
C2 : x = a cos θ, y = a sin θ, 0 ≤ θ ≤ π/2; C3 : x = 0, x2 dy = 0;
Z π/2
4a
1
a2 (cos2 θ)a cos θ dθ =
x̄ =
2(πa2 /4) 0
3π
1
19. ȳ = 0 by symmetry; x̄ =
2A
Z
x2 dy; A = αa2 ; C1 : x = t cos α, y = −t sin α, 0 ≤ t ≤ a;
C
C2 : x = a cos θ, y = a sin θ, −α ≤ θ ≤ α; C3 : x = t cos α,
y = t sin α, 0 ≤ t ≤ a (reverse orientation);
Z
Z
a
2Ax̄ = −
0
2a3
cos2 α sin α + 2a3
=−
3
x̄ =
Z
Z
α
t2 cos2 α sin α dt +
−α
α
cos3 θ dθ = −
0
a
a3 cos3 θ dθ −
t2 cos2 α sin α dt,
0
2a3
2a3
4
cos2 α sin α +
cos2 α sin α = a3 sin α;
3
3
3
2a sin α
3 α
Z
a
b−
20. A =
0
b 2
x
a2
dx =
2ab
, C1 : x = t, y = bt2 /a2 , 0 ≤ t ≤ a;
3
C2 : x = a − t, y = b, 0 ≤ t ≤ a, x2 dy = 0; C3 : x = 0, y = b − t, 0 ≤ t ≤ b, x2 dy = y 2 dx = 0;
Z a
3a
a2 b
, x̄ =
;
t2 (2bt/a2 ) dt =
2Ax̄ =
2
8
0
Z a
Z a
3b
ab2
4ab2
+ ab2 =
, ȳ =
(bt2 /a2 )2 dt +
b2 dt = −
2Aȳ = −
5
5
5
0
0
ZZ
Z
f (x) dx + g(y) dy =
21. (a)
C
Z
Z
F · dr =
(b) W =
C
∂
∂
g(y) −
f (x)
∂x
∂y
dA = 0
R
f (x) dx + g(y) dy = 0, so the work done by the vector field around any
C
simple closed curve is zero. The field is conservative.
637
Chapter 17
22. (a) Let r = d cos θi + d sin θj + zk in cylindrical coordinates, so
dr dθ
dr
dr
=
= ω(−d sin θi + d cos θj), v =
= ωk × r = ω × r.
dt
dθ dt
dt
(b) From Part (a), v = ωd(− sin θi + cos θj) = −ωyi + ωxj
(c) From Part (b), curl v = 2ωk = 2ω
(d) No; from Exercise 32 in Section 17.1, if φ were a potential function for v, then
curl (∇φ) = curl v = 0, contradicting Part (c) above.
23. Yes; by imagining a normal vector sliding around the surface it is evident that the surface has two
sides.
ZZ
ZZ
ZZZ
24. Dn φ = n · ∇φ, so
Dn φ dS =
n · ∇φ dS =
∇ · (∇φ) dV
ZZZ
=
σ
2
2
2
σ
G
∂ φ ∂ φ ∂ φ
+ 2 + 2 dV
∂x2
∂y
∂z
G
ZZ
ZZZ
Dn f dS = −
25. By Exercise 24,
σ
ZZZ
[fxx + fyy + fzz ] dV = −6
G
dV = −6vol(G) = −8π
G
26. (a) fy − gx = exy + xyexy − exy − xyexy = 0 so the vector field is conservative.
(b) φx = yexy − 1, φ = exy − x + k(x), φy = xexy , let k(x) = 0; φ(x, y) = exy − x
Z
(c) W = F · dr = φ(x(8π), y(8π)) − φ(x(0), y(0)) = φ(8π, 0) − φ(0, 0) = −8π
C
∂
∂
(yh(x)) =
(−2xh(x)), or h(x) = −2h(x) − 2xh0 (x) which
∂y
∂x
y
2
has the general solution x3 h(x)2 = C1 , h(x) = Cx−3/2 , so C 3/2 i − C 1/2 j is conservative,
x
x
√
with potential function φ = −2Cy/ x.
27. (a) If h(x)F is conservative, then
∂
∂
(yg(y)) =
(−2xg(y)), or g(y) + yg 0 (y) = −2g(y),
∂y
∂x
1
2x
with general solution g(y) = C/y 3 , so F = C 2 i − C 3 j is conservative, with potential
y
y
function Cx/y 2 .
(b) If g(y)F(x, y) is conservative then
28. A computation of curl F shows that curl F = 0 if and only if the three given equations hold.
Moreover the equations hold if F is conservative, so it remains to show that F is conservative if
curl F = 0. Let C by any simple closed curve in the region. Since the region is simply connected,
there
is a piecewis
smooth, oriented
I
ZZ
Z Z surface σ in the region with boundary C. By Stokes’ Theorem,
F·r=
C
(curl F) · n dS =
σ
0 dS = 0.
σ
By the 3-space analog of Theorem 17.3.2, F is conservative.
29. (a) conservative, φ(x, y, z) = xz 2 − e−y
(b) not conservative, fy 6= gx (for example)
30. (a) conservative, φ(x, y, z) = − cos x + yz
(b) not conservative, fz 6= hx
Chapter 17 Horizon Module
638
CHAPTER 17 HORIZON MODULE
1. (a) If r = xi + yj denotes the position vector, then F1 · r = 0 by inspection, so the velocity field
k
is tangent to the circle. The relationship F1 × r = − k indicates that r, F1 , k is a right2π
k
(sin θi − cos θj)
handed system, so the flow is counterclockwise. The polar form F1 = −
2πr
k
on a circle of radius r; and it also shows that the
shows that the speed is the constant
2πr
1
speed is proportional to with constant of proportionality 2kπ.
r
(b) Since kF1 k =
k
, when r = 1 we get k = 2πkF1 k
2πr
y
2.
1
x
–1
1
–1
q
q
is constant for constant
r so F2 is directed toward the origin, and kF2 k =
2
2πkrk
2πr
r, and the speed is proportional to the distance from the origin (constant of proportionality
q
). Since the velocity vector is directed toward the origin, the fluid flows towards the origin,
2π
which must therefore be a sink.
3. (a) F2 = −
(b) From Part (a) when r = 1, q = 2πkF2 k.
y
4.
1
x
–1
1
–1
5. (b) The magnitudes of the field vectors increase, and their directions become more tangent to
circles about the origin.
(c) The magnitudes of the field vectors increase, and their directions tend more towards the
origin.
639
Chapter 17
6. (a) The inward component is F2 , so at r = 20, 15 = kF2 k =
component is F1 , so at r = 20, 45 = kF1 k =
q
, so q = 600π; the tangential
2π(20)
k
, so k = 1800π.
2π(20)
1
[(300x + 900y)i + (300y − 900x)j]
+ y2
√
√
300 10
≤ 5 km/hr if r ≥ 60 10 ≈ 189.7 km.
(c) kFk =
r
(b) F = −
x2
q
k
1
1
[(q cos θ + k sin θ)i + (q sin θ − k cos θ)j] = −
ur +
uθ = −
(qur − kuθ )
2πr
2πr
2πr
2πr
7. F = −
1
1
1
(qur − kuθ ) · (− (kur + quθ )) =
(qk − kq) = 0, since ur and uθ are unit
2πr
2π
(2πr)2
orthogonal vectors.
8. F · ∇ψ = −
9. From the hypotheses of Exercise 8, ψ = −
α=−
1
∂
q
k ln r + α(θ), ψ = α0 (θ) = − ,
2π
∂θ
2π
1
q
θ, ψ = − (k ln r + qθ)
2π
2π
10. The streamline ψ = c becomes k ln r + qθ = −2πc, ln r = −qθ/k − 2πc/k,
r = e−qθ/k e−2πc/k = κe−qθ , where κ > 0.
y
11.
1
x
1
2
3
12. q = 600π, k = 1800π, r = κe−θ/3 ; at r = 20, θ = π/4, κ = reθ/3 = 20eπ/12 ≈ 25.985; the desired
streamline has the polar equation r = 25.985e−θ/3 .
y
20
10
x
25
APPENDIX A
Real Numbers, Intervals, and Inequalities
EXERCISE SET A
1. (a) rational
(d) rational
(g) rational
(b) integer, rational
(e) integer, rational
(h) integer, rational
2. (a) irrational
(b) rational
(c) rational
(c) integer, rational
(f ) irrational
(d) rational
3. (a) x = 0.123123123 . . ., 1000x = 123 + x, x = 123/999 = 41/333
(b) x = 12.7777 . . ., 10(x − 12) = 7 + (x − 12), 9x = 115, x = 115/9
(c) x = 38.07818181 . . ., 100x = 3807.81818181 . . ., 99x = 100x − x = 3769.74,
376974
20943
3769.74
=
=
x=
99
9900
550
4296
(d)
10000
4. x = 0.99999 . . ., 10x = 9 + x, 9x = 9, x = 1
2
2
256 2
8
16
5. (a) If r is the radius, then D = 2r so
D =
r =
r . The area of a circle of radius
9
9
81
r is πr2 so 256/81 was the approximation used for π.
(b) 22/7 ≈ 3.1429 is better than 256/81 ≈ 3.1605.
√ !
333
63 17 + 15 5
22
223
355
√
<
<
<
6. (a)
<
71
106
25 7 + 15 5
113
7
(b) Ramanujan’s
(c) Athoniszoon’s
7.
Line
Blocks
8.
Line
1
2
3
4
5
Blocks
all blocks
none
2, 4
2
2, 3
9. (a)
(b)
(c)
(d)
(e)
(f )
2
3, 4
3
1, 2
4
3, 4
5
2, 4, 5
6
1, 2
(d) Ramanujan’s
7
3, 4
always correct (add −3 to both sides of a ≤ b)
not always correct (correct only if a = b)
not always correct (correct only if a = b)
always correct (multiply both sides of a ≤ b by 6)
not always correct (correct only if a ≥ 0)
always correct (multiply both sides of a ≤ b by the nonnegative quantity a2 )
10. (a) always correct
(b) not always correct (for example let a = b = 0, c = 1, d = 2)
(c) not always correct (for example let a = 1, b = 2, c = d = 0)
11. (a) all values because a = a is always valid
(b) none
12. a = b, because if a 6= b then a < b and b < a are contradictory
640
641
Appendix A
13. (a) yes, because a ≤ b is true if a < b
(b) no, because a < b is false if a = b is true
14. (a) x2 − 5x = 0, x(x − 5) = 0 so x = 0 or x = 5
(b) −1, 0, 1, 2 are the only integers that satisfy −2 < x < 3
15. (a) {x : x is a positive odd integer}
(c) {x : x is irrational}
(b) {x : x is an even integer}
(d) {x : x is an integer and 7 ≤ x ≤ 10}
16. (a) not equal to A because 0 is not in A
(b) equal to A
(c) equal to A because (x − 3)(x − 3x + 2) = 0, (x − 3)(x − 2)(x − 1) = 0 so x = 1, 2, or 3
2
17. (a) false, there are points inside the triangle that are not inside the circle
(b) true, all points inside the triangle are also inside the square
(c) true
(d) false
(f ) true, a is inside the circle
(g) true
18. (a)
(e) true
∅, {a1 }, {a2 }, {a3 }, {a1 , a2 }, {a1 , a3 }, {a2 , a3 }, {a1 , a2 , a3 }
(b)
19. (a)
4
(c)
∅
(b)
−3
(d)
−1
7
−3
3
(e)
(f )
20. (a)
−3
3
−3
3
(b)
4
8
(c)
2
5
(d) none
3
(b) (−∞, −2) ∪ (2, +∞)
21. (a) [−2, 2]
(b)
22. (a)
−3
4
−5
1
0
4
(c)
6
8
11
(d)
(e)
(g)
4
2
4
7
(f )
1
2.3
0
5
(h)
Exercise Set A
642
23. 3x < 10; (−∞, 10/3)
10
3
25. 2x ≤ −11; (−∞, −11/2]
24.
1
5x
≥ 8; [40, +∞)
40
26. 9x < −10; (−∞, −10/9)
− 11
−
2
27. 2x ≤ 1 and 2x > −3; (−3/2, 1/2]
3
−2
29.
1
2
12 − 3x
4−x
x
− 4 < 0,
< 0,
< 0;
x−3
x−3
x−3
(−∞, 3) ∪ (4, +∞)
10
9
28. 8x ≥ 5 and 8x ≤ 14; [ 58 , 74 ]
5
8
7
4
+ + + + + + + 0 − − −
4
− − − 0 + + + + + + +
3
− − −
30.
16 − x
x
+2=
≥ 0;
8−x
8−x
(−∞, 8) ∪ [16, +∞)
+ + + 0 − − −
3
4
3
4
+ + + + + + + 0 − − −
16
+ + + 0 − − − − − − −
8
+ + +
31.
2x + 3
x + 3/2
3x + 1
−1=
< 0,
< 0;
x−2
x−2
x−2
(− 32 , 2)
− − − 0 + + +
8
16
8
16
− − − 0 + + + + + + +
3
−2
− − − − − − − 0 + + +
2
+ + + 0 − − −
+ + +
− 32
2
− 32
2
4−x
x−3
4−x
x−3
16 − x
8−x
16 − x
8−x
x+
3
2
x−2
x+
3
2
x−2
643
32.
Appendix A
x−6
x + 14
x/2 − 3
− 1 > 0,
− 2 > 0,
< 0;
4+x
4+x
x+4
(−14, −4)
− − − 0 + + + + + + +
−14
− − − − − − − 0 + + +
−4
+ + + 0 − − −
33.
x+2
4
−1=
≤ 0; (−∞, −2] ∪ (2, +∞)
2−x
2−x
+ + +
−14
−4
−14
−4
− − − 0 + + + + + + +
−2
+ + + + + + + 0 − − −
2
− − − 0 + + +
34.
13 − 2x
13/2 − x
3
−2=
≤ 0,
≤ 0;
x−5
x−5
x−5
(−∞, 5) ∪ [
13
, +∞)
2
− − −
−2
2
−2
2
+ + + + + + + 0 − − −
13
2
− − − 0 + + + + + + +
5
− − −
35.
x2 − 9 = (x + 3)(x − 3) > 0;
(−∞, −3) ∪ (3, +∞)
+ + + 0 − − −
5
13
2
5
13
2
− − − 0 + + + + + + +
−3
− − − − − − − 0 + + +
3
+ + + 0 − − − 0 + + +
−3
3
−3
3
x + 14
x+4
x + 14
x+4
x+2
2−x
x+2
2−x
13
2
−x
x−5
13
2
−x
x−5
x+3
x−3
(x + 3)(x − 3)
Exercise Set A
36. x2 − 5 = (x −
644
√
5)(x +
√
√ √
5) ≤ 0; [− 5, 5]
− − − 0 + + + + + + +
− 5
− − − − − − − 0 + + +
x+ 5
x− 5
5
+ + + 0 − − − 0 + + +
37. (x − 4)(x + 2) > 0; (−∞, −2) ∪ (4, +∞)
− 5
5
− 5
5
− − − − − − − 0 + + +
4
− − − 0 + + + + + + +
−2
+ + + 0 − − − 0 + + +
38. (x − 3)(x + 4) < 0; (−4, 3)
− − − − − − − 0 + + +
3
− − − 0 + + + + + + +
−4
+ + + 0 − − − 0 + + +
−4
3
−4
3
x−3
x+4
(x − 3)(x + 4)
40. (x − 2)(x − 1) ≥ 0; (−∞, 1] ∪ [2, +∞)
−2
4
−2
4
( x + 5 )( x − 5 )
x−4
x+2
(x − 4)(x + 2)
39. (x − 4)(x − 5) ≤ 0; [4, 5]
− − − 0 + + + + + + +
4
− − − − − − − 0 + + +
5
+ + + 0 − − − 0 + + +
4
5
4
5
− − − 0 + + + + + + +
1
− − − − − − − 0 + + +
2
+ + + 0 − − − 0 + + +
1
2
1
2
x−4
x−5
(x − 4)(x − 5)
x−1
x−2
(x − 1)(x − 2)
645
41.
Appendix A
2
x+8
3
− =
> 0; (−8, 0) ∪ (4, +∞)
x−4 x
x(x − 4)
− − 0 + + + + + + + +
−8
x+8
− − − − − − 0 + + + +
x
0
− − − − − − − − 0 + +
4
− − 0 + + +
42.
− −
+ +
−8
0
4
−8
0
4
3
−2x − 5
1
−
=
≥ 0,
x+1 x−2
(x + 1)(x − 2)
− − 0 + + + + + + + +
x + 5/2
≤ 0;
(x + 1)(x − 2)
− − − − 0 + + + + + +
(−∞, − 52 ] ∪ (−1, 2)
5
−2
−1
− − − − − − − − 0 + +
2
− − 0 +
− − −
+ +
5
−1
2
5
−1
2
−2
−2
x−4
(x + 8)
x(x − 4)
x+
5
2
x+1
x−2
x+
5
2
(x + 1)(x − 2)
43. By trial-and-error we find that x = 2 is a root of the equation x3 − x2 − x − 2 = 0 so x − 2
is a factor of x3 − x2 − x − 2. By long division we find that x2 + x + 1 is another factor so
x3 − x2 − x − 2 = (x − 2)(x2 + x + 1). The linear factors of x2 + x + 1 can be determined by first
finding the roots of x2 + x + 1 = 0 by the quadratic formula. These roots are complex numbers
so x2 + x + 1 6= 0 for all real x; thus x2 + x + 1 must be always positive or always negative.
Since x2 + x + 1 is positive when x = 0, it follows that x2 + x + 1 > 0 for all real x. Hence
x3 − x2 − x − 2 > 0, (x − 2)(x2 + x + 1) > 0, x − 2 > 0, x > 2, so S = (2, +∞).
44. By trial-and-error we find that x = 1 is a root of the equation x3 − 3x + 2 = 0 so x − 1
is a factor of x3 − 3x + 2. By long division we find that x2 + x − 2 is another factor so
x3 − 3x + 2 = (x − 1)(x2 + x − 2) = (x − 1)(x − 1)(x + 2) = (x − 1)2 (x + 2). Therefore we
want to solve (x − 1)2 (x + 2) ≤ 0. Now if x 6= 1, then (x − 1)2 > 0 and so x + 2 ≤ 0, x ≤ −2. By
inspection, x = 1 is also a solution so S = (−∞, −2] ∪ {1}.
45.
46.
√
x2 + x − 6 is real if x2 + x − 6 ≥ 0. Factor to get (x + 3)(x − 2) ≥ 0 which has as its solution
x ≤ −3 or x ≥ 2.
x+2
≥ 0; (−∞, −2] ∪ (1, +∞)
x−1
47. 25 ≤
5
(F − 32) ≤ 40, 45 ≤ F − 32 ≤ 72, 77 ≤ F ≤ 104
9
48. (a) n = 2k, n2 = 4k 2 = 2(2k 2 ) where 2k 2 is an integer.
(b) n = 2k + 1, n2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 where 2k 2 + 2k is an integer.
Exercise Set A
646
r
p
49. (a) Assume m and n are rational, then m = and n = where p, q, r, and s are integers so
q
s
p r
ps + rq
m+n= + =
which is rational because ps + rq and qs are integers.
q
s
qs
p
(b) (proof by contradiction) Assume m is rational and n is irrational, then m = where p and
q
r
q are integers. Suppose that m + n is rational, then m + n = where r and s are integers
s
r
p
rq − ps
r
. But rq − ps and sq are integers, so n is rational which
so n = − m = − =
s
s
q
sq
contradicts the assumption that n is irrational.
r
p
50. (a) Assume m and n are rational, then m = and n = where p, q, r, and s are integers so
q
s
p r
pr
mn = · =
which is rational because pr and qs are integers.
q s
qs
(b) (proof by contradiction) Assume m is rational and nonzero and that n is irrational, then
r
p
m = where p and q are integers and p 6= 0. Suppose that mn is rational, then mn =
q
s
r/s
rq
r/s
=
=
. But rq and ps are integers, so n is
where r and s are integers so n =
m
p/q
ps
rational which contradicts the assumption that n is irrational.
√
√
√
√
√
51. a = 2, b = 3, c = 6, d = − 2 are irrational, and a + d = 0, a rational; a + a = 2 2, an
irrational; ad = −2, a rational; and ab = c, an irrational.
52. (a) irrational (Exercise 49(b))
(b) irrational (Exercise 50(b))
(c) rational by inspection; Exercise 51 gives no information
√
√ 2
(d)
π must be irrational, for if it were rational, then so would be π = ( π) by Exercise 50(a);
but π is known to be irrational.
53. The average of a and b is 12 (a + b); if a and b are rational then so is the average, by Exercise 49(a)
√
and Exercise 50(a). On the other hand if a = b = 2 then the average of a and b is irrational, but
the average of a and −b is rational.
54. If 10x = 3, then x > 0 because 10x ≤ 1 for x ≤ 0. If 10p/q = 3 with p, q integers, then 10p = 3q ,
so 3q must be even (as well as 10p odd), a contradiction to Exercise 48(b).
55. 8x3 − 4x2 − 2x + 1 can be factored by grouping terms:
(8x3 −4x2 )−(2x−1) = 4x2 (2x−1)−(2x−1) = (2x−1)(4x2 −1) = (2x−1)2 (2x+1). The problem,
then, is to solve (2x − 1)2 (2x + 1) < 0. By inspection, x = 1/2 is not a solution. If x 6= 1/2, then
(2x − 1)2 > 0 and it follows that 2x + 1 < 0, 2x < −1, x < −1/2, so S = (−∞, −1/2).
56. First, rewrite the inequality as 12x3 − 20x2 + 11x − 2 ≥ 0. Now, if a polynomial in x with integer
p
coefficients has a rational zero , then p will be a factor of the constant term and q will be a factor
q
of the coefficient of the highest power of x. By trial-and-error we find that x = 1/2 is a zero, thus
(x − 1/2) is a factor so
12x3 − 20x2 + 11x − 2 = (x − 1/2) 12x2 − 14x + 4
= 2(x − 1/2) 6x2 − 7x + 2
= 2(x − 1/2)(2x − 1)(3x − 2) = (2x − 1)2 (3x − 2).
Now to solve (2x − 1)2 (3x − 2) ≥ 0 we first note that x = 1/2 is a solution. If x 6= 1/2 then
(2x − 1)2 > 0 and 3x − 2 ≥ 0, 3x ≥ 2, x ≥ 2/3 so S = [2/3, +∞) ∪ {1/2}.
57. If a < b, then ac < bc because c is positive; if c < d, then bc < bd because b is positive, so ac < bd
(Theorem A.1(a)).
58. no, since the decimal representation is not repeating (the string of zeros does not have constant
length)
APPENDIX B
Absolute Value
EXERCISE SET B
1. (a) 7
2.
p
(b)
(x − 6)2 = x − 6 if x ≥ 6,
√
2
(c) k 2
(d) k 2
p
(x − 6)2 = −(x − 6) = −x + 6 if x < 6
3. |x − 3| = |3 − x| = 3 − x if 3 − x ≥ 0, which is true if x ≤ 3
4. |x + 2| = x + 2 if x + 2 ≥ 0 so x ≥ −2.
5. All real values of x because x2 + 9 > 0.
6. |x2 + 5x| = x2 + 5x if x2 + 5x ≥ 0 so x(x + 5) ≥ 0 which is true for x ≤ −5 or x ≥ 0.
7. |3x2 + 2x| = |x(3x + 2)| = |x||3x + 2|. If |x||3x + 2| = x|3x + 2|, then |x||3x + 2| − x|3x + 2| = 0,
(|x| − x)|3x + 2| = 0, so either |x| − x = 0 or |3x + 2| = 0. If |x| − x = 0, then |x| = x, which is
true for x ≥ 0. If |3x + 2| = 0, then x = −2/3. The statement is true for x ≥ 0 or x = −2/3.
8. |6 − 2x| = |2(3 − x)| = |2||3 − x| = 2|x − 3| for all real values of x.
9.
10.
p
(x + 5)2 = |x + 5| = x + 5 if x + 5 ≥ 0, which is true if x ≥ −5.
p
(3x − 2)2 = |3x − 2| = |2 − 3x| = 2 − 3x if 2 − 3x ≥ 0 so x ≤ 2/3.
13. (a) |7 − 9| = | − 2| = 2
(c) |6 − (−8)| = |14| = 14
(e) | − 4 − (−11)| = |7| = 7
14.
(b) |3 − 2| = |1| = 1
√
√
√
(d) | − 3 − 2| = | − (3 + 2)| = 3 + 2
(f ) | − 5 − 0| = | − 5| = 5
√
p
a4 = (a2 )2 = |a2 |, but |a2 | = a2 because a2 ≥ 0 so it is valid for all values of a.
15. (a) B is 6 units to the left of A; b = a − 6 = −3 − 6 = −9.
(b) B is 9 units to the right of A; b = a + 9 = −2 + 9 = 7.
(c) B is 7 units from A; either b = a + 7 = 5 + 7 = 12 or b = a − 7 = 5 − 7 = −2. Since it is
given that b > 0, it follows that b = 12.
16. In each case we solve for e in terms of f :
(a) e = f − 4; e is to the left of f .
(c) e = f + 6; e is to the right of f .
18. |3 + 2x| = 11
17. |6x − 2| = 7
Case 1:
6x − 2 = 7
6x = 9
x = 3/2
Case 2:
Case 1:
Case 2:
6x − 2 = −7
6x = −5
x = −5/6
3 + 2x = 11
2x = 8
x= 4
3 + 2x = −11
2x = −14
x = −7
20. |4x + 5| = |8x − 3|
19. |6x − 7| = |3 + 2x|
Case 1:
6x − 7 = 3 + 2x
4x = 10
x = 5/2
(b) e = f + 4; e is to the right of f .
(d) e = f − 7; e is to the left of f .
Case 2:
6x − 7 = −(3 + 2x)
8x = 4
x = 1/2
Case 1:
4x + 5 = 8x − 3
−4x = −8
x= 2
647
Case 2:
4x + 5 = −(8x − 3)
12x = −2
x = −1/6
Exercise Set B
648
22. 2x − 7 = |x + 1|
21. |9x| − 11 = x
Case 1:
2x − 7 = x + 1
x= 8
Case 2:
9x − 11 = x
−9x − 11 = x
8x = 11
−10x = 11
x = 11/8
x = −11/10
23.
x+5
=6
2−x
24.
Case 2:
2x − 7 = −(x + 1)
3x = 6
x = 2; not a solution
because x must also
satisfy x < −1
x−3
=5
x+4
Case 1:
Case 2:
Case 1:
Case 2:
x+5
=6
2−x
x + 5 = 12 − 6x
7x = 7
x= 1
x+5
= −6
2−x
x + 5 = −12 + 6x
−5x = −17
x = 17/5
x−3
=5
x+4
x − 3 = 5x + 20
−4x = 23
x = −23/4
x−3
= −5
x+4
x − 3 = −5x − 20
6x = −17
x = −17/6
25.
|x + 6| < 3
−3 < x + 6 < 3
−9 < x < −3
S = (−9, −3)
26.
|7 − x| ≤ 5
−5 ≤ 7 − x ≤ 5
−12 ≤ −x ≤ −2
12 ≥ x ≥ 2
S = [2, 12]
27.
|2x − 3| ≤ 6
−6 ≤ 2x − 3 ≤ 6
−3 ≤ 2x ≤ 9
−3/2 ≤ x ≤ 9/2
S = [−3/2, 9/2]
28.
|3x + 1| < 4
−4 < 3x + 1 < 4
−5 < 3x < 3
−5/3 < x < 1
S = (−5/3, 1)
29. |x + 2| > 1
Case 1:
30.
Case 2:
x + 2> 1
x + 2 < −1
x > −1
x < −3
S = (−∞, −3) ∪ (−1, +∞)
1
x−1 ≥2
2
Case 1:
Case 2:
1
1
x − 1≥ 2
x − 1 ≤ −2
2
2
1
1
x≥ 3
x ≤ −1
2
2
x≥ 6
x ≤ −2
S = (−∞, −2] ∪ [6, +∞)
32. |7x + 1| > 3
31. |5 − 2x| ≥ 4
Case 1:
Case 2:
Case 1:
Case 2:
5 − 2x ≥ 4
−2x ≥ −1
5 − 2x ≤ −4
−2x ≤ −9
7x + 1 > 3
7x > 2
7x + 1 < −3
7x < −4
x ≤ 1/2
x ≥ 9/2
S = (−∞, 1/2] ∪ [9/2, +∞)
x > 2/7
x < −4/7
S = (−∞, −4/7) ∪ (2/7, +∞)
649
33.
Appendix B
1
< 2, x 6= 1
|x − 1|
34.
|x − 1| > 1/2
Case 1:
|3x + 1| ≤ 1/5
−1/5 ≤ 3x + 1 ≤ 1/5
Case 2:
−6/5 ≤ 3x ≤ −4/5
−2/5 ≤ x ≤ −4/15
x − 1 > 1/2 x − 1 < −1/2
x > 3/2
x < 1/2
S = [−2/5, −1/3) ∪ (−1/3, −4/15]
S = (−∞, 1/2) ∪ (3/2, +∞)
35.
3
≥ 4, x 6= 1/2
|2x − 1|
36.
|2x − 1| 1
≤
3
4
|2x − 1| ≤ 3/4
2
< 1, x 6= −3
|x + 3|
|x + 3|
>1
2
|x + 3| > 2
−3/4 ≤ 2x − 1 ≤ 3/4
1/4 ≤ 2x ≤ 7/4
1/8 ≤ x ≤ 7/8
S = [1/8, 1/2) ∪ (1/2, 7/8]
37.
1
≥ 5, x 6= −1/3
|3x + 1|
Case 1:
Case 2:
x + 3> 2
x + 3 < −2
x > −1
x < −5
S = (−∞, −5) ∪ (−1, +∞)
p
(x2 − 5x + 6)2 = x2 − 5x + 6 if x2 − 5x + 6 ≥ 0 or, equivalently, if (x − 2)(x − 3) ≥ 0;
x ∈ (−∞, 2] ∪ [3, +∞).
38. If x ≥ 2 then 3 ≤ x − 2 ≤ 7 so 5 ≤ x ≤ 9; if x < 2 then 3 ≤ 2 − x ≤ 7 so −5 ≤ x ≤ −1.
S = [−5, −1] ∪ [5, 9].
39. If u = |x − 3| then u2 − 4u = 12, u2 − 4u − 12 = 0, (u − 6)(u + 2) = 0, so u = 6 or u = −2. If
u = 6 then |x − 3| = 6, so x = 9 or x = −3. If u = −2 then |x − 3| = −2 which is impossible. The
solutions are −3 and 9.
41. |a − b| = |a + (−b)|
≤ |a| + | − b| (triangle inequality)
= |a| + |b|.
42.
a = (a − b) + b
|a| = |(a − b) + b|
|a| ≤ |a − b| + |b| (triangle inequality)
|a| − |b| ≤ |a − b|.
43. From Exercise 42
(i) |a| − |b| ≤ |a − b|; but |b| − |a| ≤ |b − a| = |a − b|, so (ii) |a| − |b| ≥ −|a − b|.
Combining (i) and (ii): −|a − b| ≤ |a| − |b| ≤ |a − b|, so ||a| − |b|| ≤ |a − b|.
APPENDIX C
Coordinate Planes and Lines
EXERCISE SET C
2. area = 12 bh = 12 (5 − (−3))(1) = 4
y
1.
(-4, 7)
(6, 7)
(-4, 1)
(6, 1)
x
y
(4, 3)
(5, 2)
(-3, 2)
x
(b) y = −3
3. (a) x = 2
y
y
x
x
2
-3
(c) x ≥ 0
(d) y = x
y
y
x
x
(e) y ≥ x
(f ) |x| ≥ 1
y
y
x
x
-1
650
1
651
Appendix C
4. (a) x = 0
(b) y = 0
y
y
x
x
(d) x ≥ 1 and y ≤ 2
(c) y < 0
y
y
2
x
x
1
(f ) |x| = 5
(e) x = 3
y
y
x
3
-5
5. y = 4 − x2
5
6. y = 1 + x2
y
y
5
4
x
-5
5
-5
5
x
x
Exercise Set C
√
7. y =
652
√
8. y = − x + 1
x−4
y
y
5
5
x
4
x
-5
9. x2 − x + y = 0
5
10. x = y 3 − y 2
y
y
1
x
-1
1
2
x
-1
-2
-1
1
2
-1
-2
-3
12. xy = −1
11. x2 y = 2
y
y
x
-1
1
2
-1
1
1
-2
x
-3
1
4−2
=
3 − (−1)
2
√
√
2− 2
=0
(c) m =
−3 − 4
13. (a) m =
14. m1 =
(b) m =
1−3
= −1
7−5
(d) m =
12 − (−6)
18
=
, not defined
−2 − (−2)
0
3
5
1
5−2
7−2
7−5
= , m2 =
= , m3 =
=−
6 − (−1)
7
2 − (−1)
3
2−6
2
15. (a) The line through (1, 1) and (−2, −5) has slope m1 =
(0, −1) has slope m2 =
−5 − 1
= 2, the line through (1, 1) and
−2 − 1
−1 − 1
= 2. The given points lie on a line because m1 = m2 .
0−1
653
Appendix C
(b) The line through (−2, 4) and (0, 2) has slope m1 =
(1, 5) has slope m2 =
1
5−4
= . The given points do not lie on a line because m1 6= m2 .
1+2
3
y
16.
2−4
= −1, the line through (−2, 4) and
0+2
y
17.
(b)
(c)
(a)
(b)
(4, 2)
x
x
(-1, -2)
(a)
(c)
18. The triangle is equiangular because it is equilateral. The angles of
◦
(see figure), thus the√
inclination of the sides are 0◦ , 60◦ , and 120√
◦
◦
slopes of its sides are tan 0 = 0, tan 60 = 3, and tan 120◦ = − 3.
y
60°
19. III < II < IV < I
120°
x
20. III < IV < I < II
21. Use the points (1, 2) and (x, y) to calculate the slope: (y − 2)/(x − 1) = 3
(a) if x = 5, then (y − 2)/(5 − 1) = 3, y − 2 = 12, y = 14
(b) if y = −2, then (−2 − 2)/(x − 1) = 3, x − 1 = −4/3, x = −1/3
22. Use (7, 5) and (x, y) to calculate the slope: (y − 5)/(x − 7) = −2
(a) if x = 9, then (y − 5)/(9 − 7) = −2, y − 5 = −4, y = 1
(b) if y = 12, then (12 − 5)/(x − 7) = −2, x − 7 = −7/2, x = 7/2
23. Using (3, k) and (−2, 4) to calculate the slope, we find
k−4
= 5, k − 4 = 25, k = 29.
3 − (−2)
24. The slope obtained by using the points (1, 5) and (k, 4) must be the same as that obtained from
−3 − 5
1
4−5
=
,−
= −8, k − 1 = 1/8, k = 9/8.
the points (1, 5) and (2, −3) so
k−1
2−1
k−1
25.
0−2
0−5
=−
, −2x + 8 = 5x − 5, 7x = 13, x = 13/7
x−1
x−4
y−0
1
1
y−5
= , y = x. Use (7, 5) and (x, y) to get
= 2,
x−0
2
2
x−7
1
y − 5 = 2(x − 7), y = 2x − 9. Solve the system of equations y = x and y = 2x − 9 to get
2
x = 6, y = 3.
26. Use (0, 0) and (x, y) to get
27. Show that opposite sides are parallel by showing that they have the same slope:
using (3, −1) and (6, 4), m1 = 5/3; using (6, 4) and (−3, 2), m2 = 2/9;
using (−3, 2) and (−6, −3), m3 = 5/3; using (−6, −3) and (3, −1), m4 = 2/9.
Opposite sides are parallel because m1 = m3 and m2 = m4 .
Exercise Set C
654
28. The line through (3, 1) and (6, 3) has slope m1 = 2/3, the line through (3, 1) and (2, 9) has slope
m2 = −8, the line through (6, 3) and (2, 9) has slope m3 = −3/2. Because m1 m3 = −1, the
corresponding lines are perpendicular so the given points are vertices of a right triangle.
y
29. (a)
y
(b)
3
x
x
3
8
y
(c)
y
(c)
x
4
x
-2
-7
y
30. (a)
y
(b)
x
x
3
-8
-4
(c)
y
y
(d)
5
x
x
2
655
Appendix C
y
31. (a)
y
(b)
y
(c)
5
3
x
-5
-5
y
32. (a)
x
5
-1
x
5
-5
y
(b)
5
y
(c)
5
5
2
-5
5
x
-5
5
x
x
-5
5
1
(b) m = − , b = 3
4
(d) m = 0, b = 1
33. (a) m = 3, b = 2
8
3
8
3
(c) y = − x + so m = − , b =
5
5
5
5
b
b
(e) y = − x + b so m = − , y-intercept b
a
a
2
1
2
1
x − so m = , b = −
3
3
3
3
(d) y = 3 so m = 0, b = 3
34. (a) m = −4, b = 2
(b) y =
3
3
(c) y = − x + 3 so m = − , b = 3
2
2
a0
a0
(e) y = − x so m = − , b = 0
a1
a1
35. (a) m = (0 − (−3))/(2 − 0)) = 3/2 so y = 3x/2 − 3
(b) m = (−3 − 0)/(4 − 0) = −3/4 so y = −3x/4
36. (a) m = (0 − 2)/(2 − 0)) = −1 so y = −x + 2
(b) m = (2 − 0)/(3 − 0) = 2/3 so y = 2x/3
37. y = −2x + 4
38. y = 5x − 3
39. The slope m of the line must equal the slope of y = 4x−2, thus m = 4 so the equation is y = 4x+7.
40. The slope of the line 3x + 2y = 5 is −3/2 so the line through (−1, 2) with this slope
3
1
3
is y − 2 = − (x + 1); y = − x + .
2
2
2
41. The slope m of the line must be the negative reciprocal of the slope of y = 5x + 9, thus m = −1/5
and the equation is y = −x/5 + 6.
Exercise Set C
656
42. The slope of the line x − 4y = 7 is 1/4 so a line perpendicular to it must have a slope of −4;
y + 4 = −4(x − 3); y = −4x + 8.
43. y − 4 =
−7 − 4
(x − 2) = 11(x − 2), y = 11x − 18.
1−2
44. y − 6 =
1−6
(x − (−3)), y − 6 = −5(x + 3), y = −5x − 9.
−2 − (−3)
45. The line passes through (0, 2) and (−4, 0), thus m =
1
1
0−2
= so y = x + 2.
−4 − 0
2
2
46. The line passes through (0, b) and (a, 0), thus m =
b
y = − x + b.
a
47. y = 1
0−b
b
= − , so the equation is
a−0
a
48. y = −8
m1 = 4, m2 = 4; parallel because m1 = m2
m1 = 2, m2 = −1/2; perpendicular because m1 m2 = −1
m1 = 5/3, m2 = 5/3; parallel because m1 = m2
If A 6= 0 and B 6= 0, then m1 = −A/B, m2 = B/A and the lines are perpendicular because
m1 m2 = −1. If either A or B (but not both) is zero, then the lines are perpendicular because
one is horizontal and the other is vertical.
(e) m1 = 4, m2 = 1/4; neither
49. (a)
(b)
(c)
(d)
m1 = −5, m2 = −5; parallel because m1 = m2
m1 = 2, m2 = −1/2; perpendicular because m1 m2 = −1.
m1 = −4/5, m2 = 5/4; perpendicular because m1 m2 = −1.
If B 6= 0, then m1 = m2 = −A/B and the lines are parallel because m1 = m2 . If B = 0 (and
A 6= 0), then the lines are parallel because they are both perpendicular to the x-axis.
(e) m1 = 1/2, m2 = 2; neither
50. (a)
(b)
(c)
(d)
51. y = (−3/k)x + 4/k, k 6= 0
(a) −3/k = 2, k = −3/2
(b) 4/k = 5, k = 4/5
(c) 3(−2) + k(4) = 4, k = 5/2
(d) The slope of 2x − 5y = 1 is 2/5 so −3/k = 2/5, k = −15/2.
(e) The slope of 4x + 3y = 2 is −4/3 so the slope of the line perpendicular to it is 3/4;
−3/k = 3/4, k = −4.
52. y 2 = 3x: the union of the graphs of y =
√
√
3x and y = − 3x
y
5
x
-5
5
657
Appendix C
53. (x − y)(x + y) = 0: the union of the graphs of
x − y = 0 and x + y = 0
y
5
-5
54. F =
9
C + 32
5
55. u = 3v 2
x
5
56. Y = 4X + 5
v
X
F
5
50
u
8
-30
30
Y
5
-1
C
-30
57. Solve x = 5t + 2 for t to get t =
2
1
x − , so y =
5
5
58. Solve x = 1 + 3t2 for t2 to get t2 =
2
1
x−
5
5
1
1
x − , so y = 2 −
3
3
−3=
1
1
x−
3
3
1
17
x − , which is a line.
5
5
1
7
= − x + , which is a line;
3
3
1 + 3t2 ≥ 1 for all t so x ≥ 1.
59. An equation of the line through (1, 4) and (2, 1) is y = −3x + 7. It crosses the y-axis at y = 7, and
1
the x-axis at x = 7/3, so the area of the triangle is (7)(7/3) = 49/6.
2
2
60. (2x − 3y)(2x + 3y) = 0, so 2x − 3y = 0, y = x or 2x + 3y = 0,
3
2
2
y = − x. The graph consists of the lines y = ± x.
3
3
y
5
x
-5
61. (a)
yes
(b)
yes
(c)
no
(d)
yes
(e)
yes
(f )
5
yes
(g)
no
APPENDIX D
Distance, Circles, and Quadratic Equations
EXERCISE SET D
1. in the proof of Theorem D.1
p
√
√
2. (a) d = (−1 − 2)2 + (1 − 5)2 = 9 + 16 = 25 = 5
2 + (−1) 5 + 1
,
= (1/2, 3)
(b)
2
2
p
√
√
3. (a) d = (1 − 7)2 + (9 − 1)2 = 36 + 64 = 100 = 10
7+1 1+9
,
= (4, 5)
(b)
2
2
p
√
√
4. (a) d = (−3 − 2)2 + (6 − 0)2 = 25 + 36 = 61
2 + (−3) 0 + 6
,
= (−1/2, 3)
(b)
2
2
p
√
√
5. (a) d = [−7 − (−2)]2 + [−4 − (−6)]2 = 25 + 4 = 29
−2 + (−7) −6 + (−4)
,
= (−9/2, −5)
(b)
2
2
6. Let A(1, 1), B(−2, −8), and C(4, 10) be the given points
(see diagram). A, B, and C lie on a straight line if and only if
d1 + d2 = d3 , where d1 , d2 , and d3 are the lengths of the line
segments AB, AC, and BC. But
p
√
d1 = (−2 − 1)2 + (−8 − 1)2 = 3 10,
p
√
d2 = (4 − 1)2 + (10 − 1)2 = 3 10,
p
√
d3 = (4 + 2)2 + (10 + 8)2 = 6 10; because d1 + d2 = d3 ,
it follows that A, B, and C lie on a straight line.
y
C(4, 10)
d2
d3
A(1, 1)
x
d1
B(-2, -8)
7. Let A(5, −2), B(6, 5), and C(2, 2) be the given vertices and a, b, and c the lengths of the sides
opposite these vertices; then
p
p
√
√
a = (2 − 6)2 + (2 − 5)2 = 25 = 5 and b = (2 − 5)2 + (2 + 2)2 = 25 = 5.
Triangle ABC is isosceles because it has two equal sides (a = b).
8. A triangle is a right triangle if and only if the square of the longest side is equal to the sum of the
squares of the other two sides (Pythagorean theorem). With A(1, 3), B(4, 2), and C(−2, −6) as vertices and s1 , s2 , and s3 the lengths of the sides opposite these vertices we find that
s21 = (−2 − 4)2 + (−6 − 2)2 = 100, s22 = (−2 − 1)2 + (−6 − 3)2 = 90, s23 = (4 − 1)2 + (2 − 3)2 = 10,
and that s21 = s22 + s23 , so ABC is a right triangle.
9. P1 (0, −2), P2 (−4, 8), and P3 (3, 1) all lie on a circle whose center is C(−2, 3) if the points P1 ,
P2 and P3 are equidistantpfrom C. Denoting the distances
between
P1 , P2 , P3 and C √
by d1 , d2
p
√
and dp
(0 + 2)2 + (−2 − 3)2 = 29, d2 = (−4 + 2)2 + (8 − 3)2 = 29, and
3 we find that d1 =
√
d3 = (3 + 2)2 + (1 − 3)2 = 29, so P1 , P2 and P3 lie on a circle whose center is C(−2, 3) because
d 1 = d2 = d3 .
658
659
Appendix D
10. The distance between (t, 2t − 6) and (0, 4) is
p
p
√
(t − 0)2 + (2t − 6 − 4)2 = t2 + (2t − 10)2 = 5t2 − 40t + 100;
p
√
the distance between (t, 2t − 6) and (8, 0) is (t − 8)2 + (2t − 6)2 = 5t2 − 40t + 100,
so (t, 2t − 6) is equidistant from (0, 4) and (8, 0).
11. If (2, k) is equidistant from (3,7) and (9,1), then
p
p
(2 − 3)2 + (k − 7)2 = (2 − 9)2 + (k − 1)2 , 1 + (k − 7)2 = 49 + (k − 1)2 ,
1 + k 2 − 14k + 49 = 49 + k 2 − 2k + 1, −12k = 0, k = 0.
12. (x − 3)/2 = 4 and (y + 2)/2 = −5 so x = 11 and y = −12.
1
6−8
= so the slope of the perpen−4 − 2
3
dicular bisector is −3. The midpoint of the line segment is (−1, 7) so an equation of the bisector is
y − 7 = −3(x + 1); y = −3x + 4.
13. The slope of the line segment joining (2, 8) and (−4, 6) is
14. The slope of the line segment joining (5, −1) and (4, 8) is
8 − (−1)
= −9 so the slope of the perpen4−5
1
dicular bisector is . The midpoint of the line segment is (9/2, 7/2) so an equation of the bisector is
9
1
9
1
7
x−
; y = x + 3.
y− =
2
9
2
9
15. Method (see figure): Find an equation of the perpendicular bisector of the line segment joining
A(3, 3) and B(7, −3). All points on this perpendicular bisector are equidistant from A and B, thus
find where it intersects the given line.
The midpoint of AB is (5, 0), the slope of AB is −3/2 thus the
slope of the perpendicular bisector is 2/3 so an equation is
y
y−0=
2
(x − 5)
3
4x - 2y + 3 = 0
A(3, 3)
3y = 2x − 10
x
2x − 3y − 10 = 0.
The solution of the system
(
B(7, -3)
4x − 2y + 3 = 0
2x − 3y − 10 = 0
gives the point (−29/8, −23/4).
16. (a) y = 4 is a horizontal line, so the vertical distance is |4 − (−2)| = |6| = 6.
(b) x = −1 is a vertical line, so the horizontal distance is | − 1 − 3| = | − 4| = 4.
17. Method (see figure): write an equation of the line that goes through
the given point and that is perpendicular to the given line; find the
point P where this line intersects the given line; find the
distance between P and the given point.
The slope of the given line is 4/3, so the slope of a line
perpendicular to it is −3/4.
y
4x – 3y + 10 = 0
P
(2, 1)
x
Exercise Set D
660
3
The line through (2, 1) having a slope of −3/4 is y − 1 = − (x − 2) or, after simplification,
4
3x + 4y = 10 which when solved simultaneously with 4x − 3y + 10 = 0 yields (−2/5, 14/5) as the
point of intersection. The distance d between (−2/5, 14/5) and (2, 1) is
p
d = (2 + 2/5)2 + (1 − 14/5)2 = 3.
18. (See the solution to Exercise 17 for a description of the method.) The slope of the line
5x + 12y − 36 = 0 is −5/12. The line through (8, 4) and perpendicular to the given line is
12
(x − 8) or, after simplification, 12x − 5y = 76. The point of intersection of this line with
y−4 =
5
84 4
,
and the distance between it and (8, 4) is 4.
the given line is found to be
13 13
19. If B = 0, then the line Ax + C = 0 is vertical and x = −C/A for each point on the line. The line
through (x0 , y0 ) and perpendicular to the given line is horizontal and intersects the given line at
the point (−C/A, y0 ). The distance d between (−C/A, y0 ) and (x0 , y0 ) is
r
p
|Ax0 + C|
(Ax0 + C)2
√
d = (x0 + C/A)2 + (y0 − y0 )2 =
=
A2
A2
|Ax0 + By0 + C|
√
which is the value of
for B = 0.
A2 + B 2
If B 6= 0, then the slope of the given line is −A/B and the line through (x0 , y0 ) and perpendicular
to the given line is
B
(x − x0 ), Ay − Ay0 = Bx − Bx0 , Bx − Ay = Bx0 − Ay0 .
A
The point of intersection of this line and the given line is obtained by solving
Ax + By = −C and Bx − Ay = Bx0 − Ay0 .
Multiply the first equation through by A and the second by B and add the results to get
y − y0 =
(A2 + B 2 )x = B 2 x0 − ABy0 − AC so x =
B 2 x0 − ABy0 − AC
A2 + B 2
Similarly, by multiplying by B and −A, we get y =
−ABx0 + A2 y0 − BC
.
A2 + B 2
The square of the distance d between (x, y) and (x0 , y0 ) is
B 2 x0 − ABy0 − AC
d = x0 −
A2 + B 2
2
so d =
20. d =
2
2
−ABx0 + A2 y0 − BC
+ y0 −
A2 + B 2
=
(A2 x0 + ABy0 + AC)2
(ABx0 + B 2 y0 + BC)2
+
2
2
2
(A + B )
(A2 + B 2 )2
=
A2 (Ax0 + By0 + C)2 + B 2 (Ax0 + By0 + C)2
(A2 + B 2 )2
=
(Ax0 + By0 + C)2 (A2 + B 2 )
(Ax0 + By0 + C)2
=
2
2
2
(A + B )
A2 + B 2
|Ax0 + By0 + C|
√
.
A2 + B 2
|15|
15
|4(2) − 3(1) + 10|
p
= 3.
=√ =
2
2
5
25
4 + (−3)
21. d =
|5(8) + 12(4) − 36|
|52|
52
√
= 4.
=√
=
2
2
13
169
5 + 12
661
Appendix D
22. Method (see figure): Let A(0, a), B(b, 0), and C(c, 0) be the given
vertices; find equations for the perpendicular bisectors L1 , L2 , and
L3 and show that they all intersect at the same point.
line L1 : The midpoint of BC is
an equation for L1 is x =
b+c
;
2
line L2 : The midpoint of AB is
(if b 6= 0) so the slope of
A(0, a)
b+c
,0
2
y
and since L1 is vertical,
L3
L2
x
B(b, 0)
C(c, 0)
L1
a
b a
,
; the slope of AB is −
2 2
b
b
b
b
a
x−
;
L2 is (even if b = 0) and an equation of L2 is y − =
a
2
a
2
c a
a
,
; the slope of AC is − (if c 6= 0) so the slope of
line L3 : The midpoint of AC is
2 2
c
c
c
c
a
x−
.
L3 is (even if c = 0) and an equation of L3 is y − =
a
2
a
2
a
b
b
b+c
and y − =
x−
.
For the point of intersection of L1 and L2 , solve x =
2
2
a
2
b + c a2 + bc
,
. The point of intersection of L1 and L3 is obtained by
The point is found to be
2
2
a
c
c
b+c
b + c a2 + bc
and y− =
x−
, its solution yields the point
,
.
solving the system x =
2
2
a
2
2
2
So L1 L2 , and L3 all intersect at the same point.
23. (a) center (0,0), radius 5
√
(c) center (−1, −3), radius 5
(b) center (1,4), radius 4
(d) center (0, −2), radius 1
24. (a) center (0, 0), radius 3
√
(c) center (−4, −1), radius 8
(b) center (3, 5), radius 6
(d) center (−1, 0), radius 1
25. (x − 3)2 + (y − (−2))2 = 42 , (x − 3)2 + (y + 2)2 = 16
√
26. (x − 1)2 + (y − 0)2 = ( 8/2)2 , (x − 1)2 + y 2 = 2
27. r = 8 because the circle is tangent to the x-axis, so (x + 4)2 + (y − 8)2 = 64.
28. r = 5 because the circle is tangent to the y-axis, so (x − 5)2 + (y − 8)2 = 25.
29. (0, 0) is on the circle, so r =
30. r =
p
p
(4 − 1)2 + (−5 − 3)2 =
(−3 − 0)2 + (−4 − 0)2 = 5; (x + 3)2 + (y + 4)2 = 25.
√
73; (x − 4)2 + (y + 5)2 = 73.
31. The center is the p
midpoint of the line segment
joining (2, 0) and (0, 2) so the center is at (1, 1).
√
The radius is r = (2 − 1)2 + (0 − 1)2 = 2, so (x − 1)2 + (y − 1)2 = 2.
32. The center is the midpoint
of the line segment
p
√ joining (6, 1) and (−2, 3), so the center is at (2, 2).
The radius is r = (6 − 2)2 + (1 − 2)2 = 17, so (x − 2)2 + (y − 2)2 = 17.
33. (x2 − 2x) + (y 2 − 4y) = 11, (x2 − 2x + 1) + (y 2 − 4y + 4) = 11 + 1 + 4, (x − 1)2 + (y − 2)2 = 16;
center (1,2) and radius 4
Exercise Set D
662
34. (x√2 + 8x) + y 2 = −8, (x2 + 8x + 16) + y 2 = −8 + 16, (x + 4)2 + y 2 = 8; center (−4, 0) and radius
2 2
35. 2(x2 + 2x) + 2(y 2 − 2y) =√0, 2(x2 + 2x + 1) + 2(y 2 − 2y + 1) = 2 + 2, (x + 1)2 + (y − 1)2 = 2;
center (−1, 1) and radius 2
36. 6(x2 − x) + 6(y 2 + y) = 3, 6(x2 − x + 1/4) + 6(y 2 + y + 1/4) = 3 + 6/4 + 6/4,
(x − 1/2)2 + (y + 1/2)2 = 1; center (1/2, −1/2) and radius 1
37. (x2 + 2x) + (y 2 + 2y) = −2, (x2 + 2x + 1) + (y 2 + 2y + 1) = −2 + 1 + 1, (x + 1)2 + (y + 1)2 = 0;
the point (−1, −1)
38. (x2 − 4x) + (y 2 − 6y) = −13, (x2 − 4x + 4) + (y 2 − 6y + 9) = −13 + 4 + 9, (x − 2)2 + (y − 3)2 = 0;
the point (2, 3)
39. x2 + y 2 = 1/9; center (0, 0) and radius 1/3
40. x2 + y 2 = 4; center (0,0) and radius 2
41. x2 + (y 2 + 10y) = −26, x2 + (y 2 + 10y + 25) = −26 + 25, x2 + (y + 5)2 = −1; no graph
42. (x2 − 10x) + (y 2 − 2y) = −29, (x2 − 10x + 25) + (y 2 − 2y + 1) = −29 + 25 + 1,
(x − 5)2 + (y − 1)2 = −3; no graph
25
5
5
1
+ 16 y 2 + y +
= 7 + 25 + 4,
43. 16 x2 + x + 16(y 2 + y) = 7, 16 x2 + x +
2
2
16
4
(x + 5/4)2 + (y + 1/2)2 = 9/4; center (−5/4, −1/2) and radius 3/2
44. 4(x2 − 4x) + 4(y 2 − 6y) = 9, 4(x2 − 4x + 4) + 4(y 2 −
√6y + 9) = 9 + 16 + 36,
(x − 2)2 + (y − 3)2 = 61/4; center (2, 3) and radius 61/2
√
√
45. (a) y 2 = 16 − x2 , so y = ± 16 − x2 . The bottom half is y = − 16 − x2 .
√
2
− 2x − x2 , so y − 2 = ± 3 − 2x − x2 ,
(b) Complete the
√ square in y to get (y − 2) = 3 √
or y = 2 ± 3 − 2x − x2 . The top half is y = 2 + 3 − 2x − x2 .
p
p
46. (a) x2 = 9 − y 2 so x = ± 9 − y 2 . The right half is x = 9 − y 2 .
p
p
(b) Complete the square in p
x to get (x − 2)2 = 1 − y 2 so x − 2 = ± 1 − y 2 , x = 2 ± 1 − y 2 .
The left half is x = 2 − 1 − y 2 .
y
47. (a)
(b)
5
√
5 + 4x − x2
p
= 5 − (x2 − 4x)
p
= 5 + 4 − (x2 − 4x + 4)
p
= 9 − (x − 2)2
y=
x
-5
5
y
3
-1
5
x
663
Appendix D
y
48. (a)
y
(b)
2
2
x
3
-2
-2
5
x
-2
49. The tangent line is perpendicular to the radius at the point. The slope of the radius is 4/3, so
3
the slope of the perpendicular is −3/4. An equation of the tangent line is y − 4 = − (x − 3), or
4
25
3
y =− x+ .
4
4
50. (a) (x + 1)2 + y 2 = 10, center at C(−1, 0). The slope of CP is −1/3 so the slope of the tangent
is 3; y + 1 = 3(x − 2), y = 3x − 7.
(b) (x − 3)2 + (y + 2)2 = 26, center at C(3, −2). The slope of CP is 5 so the slope of the tangent
1
1
19
1
is − ; y − 3 = − (x − 4), y = − x + .
5
5
5
5
√
√
51. (a) The center ofp
the circle is at (0,0) and its radius is 20 = 2 5. The distance between P and
√
√
the center is (−1)2 + (2)2 = 5 which is less than 2 5, so P is inside the circle.
(b) Draw the diameter of the circle that passes through P , then the shorter segment of the
diameter is the shortest line that can be drawn from P to the circle, and the longer segment
is the longest line that
from P to the circle
prove it?). Thus, the
√
√ (can
√ you √
√ can√be drawn
smallest distance is 2 5 − 5 = 5, and the largest is 2 5 + 5 = 3 5.
√
√
52. (a) x2 + (y − 1)2 = 5, center at C(0, 1) and radius 5. The distance between P and C is 3 5/2
so P is outside the circle.
√
√
1√
3√
5√
3√
5− 5=
5, the largest distance is
5+ 5=
5.
(b) The smallest distance is
2
2
2
2
53. Let (a, b) be the coordinates of T (or T 0 ). The radius from (0, 0) to T (or T 0 ) will be perpendicular
2
2
to L (or L0 ) so, using slopes, b/a = −(a−3)/b, a2 +b2 = 3a. But (a, b) is on
√ the circle so a +b = 1,
2
2
2
thus 3a = 1, a = 1/3.
√ Let a = 1/3 in a√ + b = 1 to get b = 8/9, b = ± 8/3. The coordinates of
T and T 0 are (1/3, 8/3) and (1/3, − 8/3).
54. (a)
√ p
(x − 2)2 + (y − 0)2 = 2 (x − 0)2 + (y − 1)2 ; square both sides and expand to get
x2 − 4x + 4 + y 2 = 2(x2 + y 2 − 2y + 1), x2 + y 2 + 4x − 4y − 2 = 0, which is a circle.
p
(b) (x2 + 4x) + (y 2 − 4y) = 2, (x2 + 4x + 4) + (y 2 − 4y + 4) = 2 + 4 + 4, (x + 2)2 + (y − 2)2 = 10;
√
center (−2, 2), radius 10.
55. (a) [(x − 4)2 + (y − 1)2 ] + [(x − 2)2 + (y + 5)2 ] = 45
x2 − 8x + 16 + y 2 − 2y + 1 + x2 − 4x + 4 + y 2 + 10y + 25 = 45
2x2 + 2y 2 − 12x + 8y + 1 = 0, which is a circle.
(b) 2(x2 − 6x) + 2(y 2 + 4y) = −1, 2(x2 − 6x + 9) + 2(y 2 + 4y + 4) = −1 + 18 + 8,
√
(x − 3)2 + (y + 2)2 = 25/2; center (3, −2), radius 5/ 2.
Exercise Set D
664
56. If x2 − y 2 = 0, then y 2 = x2 so y = x or y = −x. The graph of x2 − y 2 = 0 consists of the graphs
of the two lines y = ±x. The graph of (x − c)2 + y 2 = 1 is a circle of radius 1 with center at (c, 0).
Examine
√ the system cannot have just one solution, and
√ has 0 solutions if
√ the figure to see that
|c| > 2, 2 solutions if |c| = 2, 3 solutions if |c| = 1, and 4 solutions if |c| < 2, |c| 6= 1.
y
y
1
x
x
1
2
2 solutions
3 solutions
y
y
x
4 solutions
x
0 solutions
58. y = x2 − 3
57. y = x2 + 2
y
y
x
– 3
3
(0, 2)
(0, -3)
x
59. y = x2 + 2x − 3
60. y = x2 − 3x − 4
y
61. y = −x2 + 4x + 5
y
y
(2, 9)
x
-3
-3
(-1, -4)
x
-1
4
5
1
-4
x
( 32 , – 254 )
-1
5
665
Appendix D
62. y = −x2 + x
63. y = (x − 2)2
64. y = (3 + x)2
y
y
y
9
4
( 12 , 14 )
x
1
x
(2, 0)
x
(-3, 0)
65. x2 − 2x + y = 0
67. y = 3x2 − 2x + 1
66. x2 + 8x + 8y = 0
y
y
y
(1, 1)
(-4, 2)
x
2
x
-8
1
( 13 , 23 )
x
69. x = −y 2 + 2y + 2
68. y = x2 + x + 2
70. x = y 2 − 4y + 5
y
y
y
1+ 3
(1, 2)
2
(3, 1)
(– 12 , 74 )
x
x
5
2
x
1– 3
√
√
71. (a) x2 = 3 − y, x = ± 3 − y. The right half is x = 3 − y.
√
√
(b) Complete the square in x to get (x−1)2 = y+1, x = 1± y + 1. The left half is x = 1− y + 1.
√
√
72. (a) y 2 = x + 5, y = ± x + 5. The upper half is y = x + 5.
p
(b) Complete the square in y to get (y − 1/2)2 = x + 9/4, y − 1/2 = ± x + 9/4,
p
p
y = 1/2 ± x + 9/4. The lower half is y = 1/2 − x + 9/4.
Exercise Set D
666
y
73. (a)
y
(b)
5
4
x
-5
5
y
74. (a)
y
(b)
5
5
x
x
4
3
s
75. (a)
x
(b) The ball will be at its highest point
when t = 1 sec; it will rise 16 ft.
16
8
t
1
2
76. (a) 2x + y = 500, y = 500 − 2x.
(b) A = xy = x(500 − 2x) = 500x − 2x2 .
(c) The graph of A versus x is a parabola with its vertex (high point) at
x = −b/(2a) = −500/(−4) = 125, so the maximum value of A is
A = 500(125) − 2(125)2 = 31,250 ft2 .
77. (a) (3)(2x) + (2)(2y) = 600, 6x + 4y = 600, y = 150 − 3x/2
(b) A = xy = x(150 − 3x/2) = 150x − 3x2 /2
(c) The graph of A versus x is a parabola with its vertex (high point) at
x = −b/(2a) = −150/(−3) = 50, so the maximum value of A is
A = 150(50) − 3(50)2 /2 = 3,750 ft2 .
667
Appendix D
b
78. (a) y = ax2 + bx + c = a x2 + x + c
a
2
2
b
b
b
b2
b2
=a x+
+ c−
= a x2 + x + 2 + c −
a
4a
4a
2a
4a
b
b2
except when x = − , so the graph has its high
4a
2a
b2
b
point there. If a > 0 then y is always greater than c −
except when x = − , so the
4a
2a
graph has its low point there.
(b) If a < 0 then y is always less than c −
79. (a) The parabola y = 2x2 + 5x√− 1 opens upward and
√ has x-intercepts of x = (−5 ±
2x2 + 5x − 1 < 0 if (−5 − 33)/4 < x < (−5 + 33)/4.
√
33)/4, so
(b) The parabola y = x2 − 2x + 3 opens upward and has no x-intercepts, so x2 − 2x + 3 > 0 if
−∞ < x < +∞.
1 opens upward and√has x-intercepts of x = (−1 ±
80. (a) The parabola y = x2 + x − √
x2 + x − 1 > 0 if x < (−1 − 5 )/2 or x > (−1 + 5 )/2.
√
5 )/2, so
(b) The parabola y = x2 − 4x + 6 opens upward and has no x-intercepts, so x2 − 4x + 6 < 0 has
no solution.
81. (a) The t-coordinate of the vertex is t = −40/[(2)(−16)] = 5/4, so the maximum height is
s = 5 + 40(5/4) − 16(5/4)2 = 30 ft.
(b) s = 5 + 40t − 16t2 = 0 if t ≈ 2.6 s
√
√
+ 7 < 0, which √
is true if (5√− 3 2 )/4 < t < (5 + 3 2 )/4.
(c) s = 5 + 40t − 16t2 > 12 if 16t2 − 40t√
The length of this interval is (5 + 3 2 )/4 − (5 − 3 2 )/4 = 3 2/2 ≈ 2.1 s.
82. x + 3 − x2 > 0, x2 − x − 3 < 0, (1 −
√
√
13)/2 < x < (1 + 13)/2
APPENDIX E
Trigonometry Review
EXERCISE SET E
1. (a) 5π/12
(b) 13π/6
(c) π/9
(d) 23π/30
2. (a) 7π/3
(b) π/12
(c) 5π/4
(d) 11π/12
3. (a) 12◦
(b) (270/π)◦
(c) 288◦
(d) 540◦
4. (a) 18◦
(b) (360/π)◦
(c) 72◦
(d) 210◦
5.
(c)
3/4
√
3/ 10
tan θ
csc θ
sec θ
cot θ
√
√
√
2/5
21/2 5/ 21 5/2 2/ 21
√
√
√
√
7/4 3/ 7
4/3
4/ 7
7/3
√
√
√
1/ 10
3
10/3
10
1/3
(a)
sin θ
√
1/ 2
cos θ
√
1/ 2
(b)
3/5
(c)
1/4
(a)
(b)
6.
sin θ
√
21/5
cos θ
tan θ
1
4/5
3/4
√
√
15/4 1/ 15
csc θ
√
2
sec θ
√
2
cot θ
5/3
5/4
√
4/ 15
4/3
√
15
4
1
√
√
7. sin θ = 3/ 10, cos θ = 1/ 10
8. sin θ =
√
√
5/3, tan θ = 5/2
√
21/2, csc θ = 5/ 21
10. cot θ =
√
√
15, sec θ = 4/ 15
9. tan θ =
√
11. Let x be the length of the side adjacent to θ, then cos θ = x/6 = 0.3, x = 1.8.
12. Let x be the length of the hypotenuse, then sin θ = 2.4/x = 0.8, x = 2.4/0.8 = 3.
13.
θ
◦
sin θ
cos θ
tan θ
csc θ
sec θ
cot θ
√
√
√
√
−1/ 2 −1/ 2
1
− 2
− 2
1
√
√
√
√
1/2
− 3/2 −1/ 3
2
−2/ 3 − 3
√
√
√
√
− 3/2
1/2
− 3 −2/ 3
2
−1/ 3
(a)
225
(b)
−210◦
(c)
5π/3
(d)
−3π/2
1
θ
sin θ
14.
(a)
330◦
(b)
−120◦
(c)
9π/4
(d)
−3π
0
—
1
—
0
cos θ
tan θ
csc θ
sec θ cot θ
√
√
√
√
−1/2
3/2 −1/ 3
−2
2/ 3 − 3
√
√
√
√
− 3/2 −1/2
3
−2/ 3
−2
1/ 3
√
√
√
√
1/ 2 1/ 2
1
2
2
1
0
−1
0
—
668
−1
—
669
Appendix E
15.
sin θ
cos θ
tan θ
csc θ
sec θ
cot θ
(a)
4/5
3/5
4/3
5/4
5/3
3/4
(b)
−4/5
(c)
(d)
(e)
(f )
16.
3/5
−4/3 −5/4
5/3
−3/4
√
√
√
√
1/2 − 3/2 −1/ 3
2
−2 3 − 3
√
√
√
√
−1/2
3/2 −1/ 3 −2
2/ 3 − 3
√
√
√
√
1/ 2 1/ 2
1
1
2
2
√
√
√
√
1/ 2 −1/ 2
−1
2
− 2
−1
sin θ
(a)
(b)
(c)
(d)
(e)
(f )
cos θ
tan θ
csc θ
sec θ
cot θ
√
√
√
√
1/4
15/4
1/ 15
4
4/ 15
15
√
√
√
√
1/4
− 15/4 −1/ 15
4
−4/ 15 − 15
√
√
√
√
3/ 10
1/ 10
3
10/3
10
1/3
√
√
√
√
−3/ 10 −1/ 10
3
− 10/3 − 10
1/3
√
√
√
√
21/5
−2/5
− 21/2 5/ 21
−5/2
−2/ 21
√
√
√
√
− 21/5
−2/5
21/2 −5/ 21
−5/2
2/ 21
17. (a) x = 3 sin 25◦ ≈ 1.2679
(b) x = 3/ tan(2π/9) ≈ 3.5753
18. (a) x = 2/ sin 20◦ ≈ 5.8476
(b) x = 3/ cos(3π/11) ≈ 4.5811
19.
sin θ
(a)
(b)
(c)
√
cos θ
a/3
9−
√
√
a/ a2 + 25 5/ a2 + 25
√
a2 − 1/a
1/a
a2 /3
tan θ
√
a/ 9 − a2
√
a/5
a2 − 1
csc θ
sec θ
cot θ
√
√
2
3/a
3/ 9 − a
9 − a2 /a
√
√
a2 + 25/a
a2 + 25/5
5/a
√
√
2
a/ a − 1
a
1/ a2 − 1
20. (a) θ = 3π/4 ± 2nπ and θ = 5π/4 ± 2nπ, n = 0, 1, 2, . . .
(b) θ = 5π/4 ± 2nπ and θ = 7π/4 ± 2nπ, n = 0, 1, 2, . . .
21. (a) θ = 3π/4 ± nπ, n = 0, 1, 2, . . .
(b) θ = π/3 ± 2nπ and θ = 5π/3 ± 2nπ, n = 0, 1, 2, . . .
22. (a) θ = 7π/6 ± 2nπ and θ = 11π/6 ± 2nπ, n = 0, 1, 2, . . .
(b) θ = π/3 ± nπ, n = 0, 1, 2, . . .
23. (a) θ = π/6 ± nπ, n = 0, 1, 2, . . .
(b) θ = 4π/3 ± 2nπ and θ = 5π/3 ± 2nπ, n = 0, 1, 2, . . .
24. (a) θ = 3π/2 ± nπ, n = 0, 1, 2, . . .
(b) θ = π ± 2nπ, n = 0, 1, 2, . . .
25. (a) θ = 3π/4 ± nπ, n = 0, 1, 2, . . .
(b) θ = π/6 ± nπ, n = 0, 1, 2, . . .
26. (a) θ = 2π/3 ± 2nπ and θ = 4π/3 ± 2nπ, n = 0, 1, 2, . . .
(b) θ = 7π/6 ± 2nπ and θ = 11π/6 ± 2nπ, n = 0, 1, 2, . . .
Exercise Set E
670
27. (a) θ = π/3 ± 2nπ and θ = 2π/3 ± 2nπ, n = 0, 1, 2, . . .
(b) θ = π/6 ± 2nπ and θ = 11π/6 ± 2nπ, n = 0, 1, 2, . . .
28. sin θ = −3/5, cos θ = −4/5, tan θ = 3/4, csc θ = −5/3, sec θ = −5/4, cot θ = 4/3
√
√
√
√
29. sin θ = 2/5, cos θ = − 21/5, tan θ = −2/ 21, csc θ = 5/2, sec θ = −5/ 21, cot θ = − 21/2
30. (a) θ = π/2 ± 2nπ, n = 0, 1, 2, . . .
(c) θ = π/4 ± nπ, n = 0, 1, 2, . . .
(e) θ = ±2nπ, n = 0, 1, 2, . . .
(b) θ = ±2nπ, n = 0, 1, 2, . . .
(d) θ = π/2 ± 2nπ, n = 0, 1, 2, . . .
(f ) θ = π/4 ± nπ, n = 0, 1, 2, . . .
31. (a) θ = ±nπ, n = 0, 1, 2, . . .
(c) θ = ±nπ, n = 0, 1, 2, . . .
(e) θ = π/2 ± nπ, n = 0, 1, 2, . . .
(b) θ = π/2 ± nπ, n = 0, 1, 2, . . .
(d) θ = ±nπ, n = 0, 1, 2, . . .
(f ) θ = ±nπ, n = 0, 1, 2, . . .
32. Construct a right triangle with one angle equal to 17◦ , measure the lengths of the sides and
hypotenuse and use formula (6) for sin θ and cos θ to approximate sin 17◦ and cos 17◦ .
33. (a) s = rθ = 4(π/6) = 2π/3 cm
34. r = s/θ = 7/(π/3) = 21/π
36. θ = s/r so A =
35. θ = s/r = 2/5
1
1
1 2
r θ = r2 (s/r) = rs
2
2
2
37. (a) 2πr = R(2π − θ), r =
(b) h =
(b) s = rθ = 4(5π/6) = 10π/3 cm
2π − θ
R
2π
p
p
R2 − r2 = R2 − (2π − θ)2 R2 /(4π 2 ) =
√
4πθ − θ2
R
2π
38. The circumference of the circular base is 2πr. When cut and flattened, the cone becomes a circular
sector of radius L. If θ is the central angle that subtends the arc of length 2πr, then θ = (2πr)/L
so the area S of the sector is S = (1/2)L2 (2πr/L) = πrL which is the lateral surface area of the
cone.
39. Let h be the altitude as shown in the figure, then
√
√
1 √
h = 3 sin 60◦ = 3 3/2 so A = (3 3/2)(7) = 21 3/4.
2
3
h
60°
7
40. Draw the perpendicular from vertex C√as shown in the figure, then
h = 9 sin 30◦ = 9/2, a = h/ sin 45◦ = 9 2/2,
√
c1 = 9 cos 30◦ = 9 3/2, c2 = a cos 45◦ = 9/2,
√
c1 + c2 = 9( 3 + 1)/2, angle C = 180◦ − (30◦ + 45◦ ) = 105◦
C
9
A
41. Let x be the distance above the ground, then x = 10 sin 67◦ ≈ 9.2 ft.
42. Let x be the height of the building, then x = 120 tan 76◦ ≈ 481 ft.
a
h
45°
30°
c1
c2
B
671
Appendix E
43. From the figure, h = x − y but x = d tan β,
y = d tan α so h = d(tan β − tan α).
h
x
y
β α
d
44. From the figure, d = x − y but x = h cot α,
y = h cot β so d = h(cot α − cot β),
d
.
h=
cot α − cot β
h
α
β
d
y
x
√
√
45. (a) sin 2θ = 2 sin θ cos θ = 2( 5/3)(2/3) = 4 5/9
(b) cos 2θ = 2 cos2 θ − 1 = 2(2/3)2 − 1 = −1/9
√
√
√
46. (a) sin(α − β) = sin α cos β − cos α sin β = (3/5)(1/ 5) − (4/5)(2/ 5) = −1/ 5
√
√
√
(b) cos(α + β) = cos α cos β − sin α sin β = (4/5)(1/ 5) − (3/5)(2/ 5) = −2/(5 5)
47. sin 3θ = sin(2θ + θ) = sin 2θ cos θ + cos 2θ sin θ = (2 sin θ cos θ) cos θ + (cos2 θ − sin2 θ) sin θ
= 2 sin θ cos2 θ + sin θ cos2 θ − sin3 θ = 3 sin θ cos2 θ − sin3 θ; similarly, cos 3θ = cos3 θ − 3 sin2 θ cos θ
48.
cos θ sec θ
cos θ sec θ
cos θ
cos θ
=
=
=
= cos2 θ
2
2
sec θ
sec θ
(1/ cos θ)
1 + tan θ
49.
cos θ tan θ + sin θ
cos θ(sin θ/ cos θ) + sin θ
=
= 2 cos θ
tan θ
sin θ/ cos θ
50. 2 csc 2θ =
2
2
=
=
sin 2θ
2 sin θ cos θ
51. tan θ + cot θ =
1
sin θ
1
cos θ
= csc θ sec θ
sin θ
cos θ
sin2 θ + cos2 θ
1
2
2
+
=
=
=
=
= 2 csc 2θ
cos θ
sin θ
sin θ cos θ
sin θ cos θ
2 sin θ cos θ
sin 2θ
52.
sin 2θ cos θ − cos 2θ sin θ
sin θ
sin 2θ cos 2θ
−
=
=
= sec θ
sin θ
cos θ
sin θ cos θ
sin θ cos θ
53.
sin θ + cos 2θ − 1
sin θ + (1 − 2 sin2 θ) − 1
sin θ(1 − 2 sin θ)
=
=
= tan θ
cos θ − sin 2θ
cos θ − 2 sin θ cos θ
cos θ(1 − 2 sin θ)
54. Using (47), 2 sin 2θ cos θ = 2(1/2)(sin θ + sin 3θ) = sin θ + sin 3θ
55. Using (47), 2 cos 2θ sin θ = 2(1/2)[sin(−θ) + sin 3θ] = sin 3θ − sin θ
Exercise Set E
672
56. tan(θ/2) =
2 sin2 (θ/2)
1 − cos θ
sin(θ/2)
=
=
cos(θ/2)
2 sin(θ/2) cos(θ/2)
sin θ
57. tan(θ/2) =
2 sin(θ/2) cos(θ/2)
sin θ
sin(θ/2)
=
=
cos(θ/2)
2 cos2 (θ/2)
1 + cos θ
58. From (52), cos(π/3 + θ) + cos(π/3 − θ) = 2 cos(π/3) cos θ = 2(1/2) cos θ = cos θ
1
59. From the figure, area = hc but h = b sin A
2
1
so area = bc sin A. The formulas
2
1
1
area = ac sin B and area = ab sin C
2
2
follow by drawing altitudes from vertices B and C, respectively.
C
b
a
h
A
60. From right triangles ADC and BDC,
h1 = b sin A = a sin B so a/ sin A = b/ sin B.
From right triangles AEB and CEB,
h2 = c sin A = a sin C so a/ sin A = c/ sin C
thus a/ sin A = b/ sin B = c/ sin C.
B
c
C
h1
b
h2
E
A
a
D
c
61. (a) sin(π/2 + θ) = sin(π/2) cos θ + cos(π/2) sin θ = (1) cos θ + (0) sin θ = cos θ
(b) cos(π/2 + θ) = cos(π/2) cos θ − sin(π/2) sin θ = (0) cos θ − (1) sin θ = − sin θ
(c) sin(3π/2 − θ) = sin(3π/2) cos θ − cos(3π/2) sin θ = (−1) cos θ − (0) sin θ = − cos θ
(d) cos(3π/2 + θ) = cos(3π/2) cos θ − sin(3π/2) sin θ = (0) cos θ − (−1) sin θ = sin θ
sin α cos β + cos α sin β
sin(α + β)
=
, divide numerator and denominator by
cos(α + β)
cos α cos β − sin α sin β
sin β
sin α
and tan β =
to get (38);
cos α cos β and use tan α =
cos α
cos β
tan α − tan β
tan α + tan(−β)
=
because
tan(α − β) = tan(α + (−β)) =
1 − tan α tan(−β)
1 + tan α tan β
tan(−β) = − tan β.
62. tan(α + β) =
63. (a) Add (34) and (36) to get sin(α − β) + sin(α + β) = 2 sin α cos β so
sin α cos β = (1/2)[sin(α − β) + sin(α + β)].
(b) Subtract (35) from (37).
(c) Add (35) and (37).
A−B
1
A+B
cos
= (sin B + sin A) so
2
2
2
A−B
A+B
cos
.
sin A + sin B = 2 sin
2
2
64. (a) From (47), sin
(b) Use (49)
(c) Use (48)
B
673
Appendix E
α+β
α−β
cos
, but sin(−β) = − sin β so
2
2
α−β
α+β
sin
.
sin α − sin β = 2 cos
2
2
65. sin α + sin(−β) = 2 sin
66. (a) From (34), C sin(α + φ) = C sin α cos φ + C cos α sin φ so C cos φ = 3√and C sin φ = 5, √
square and add√to get C 2 (cos2 φ + sin2 φ) = 9 + 25, C 2 = 34. If C = 34 then cos φ = 3/ 34
and sin φ = 5/ 34√so φ is the first-quadrant angle for which tan φ = 5/3.
3 sin α + 5 cos α = 34 sin(α + φ).
√
(b) Follow the procedure of part (a) to get C cos φ = A and C sin φ = B, C = A2 + B 2 ,
tan φ = B/A where the quadrant in which φ lies is determined
by the signs of A and B because
√
cos φ = A/C and sin φ = B/C, so A sin α + B cos α = A2 + B 2 sin(α + φ).
67. Consider the triangle having a, b, and d as sides. The angle formed by sides√a and b is π − θ so
from the law of cosines, d2 = a2 + b2 − 2ab cos(π − θ) = a2 + b2 + 2ab cos θ, d = a2 + b2 + 2ab cos θ.
APPENDIX F
Solving Polynomial Equations
EXERCISE SET F
1. (a) q(x) = x2 + 4x + 2, r(x) = −11x + 6
(c) q(x) = x3 − x2 + 2x − 2, r(x) = 2x + 1
(b) q(x) = 2x2 + 4, r(x) = 9
2. (a) q(x) = 2x2 − x + 2, r(x) = 5x + 5
(c) q(x) = 5x3 − 5, r(x) = 4x2 + 10
(b) q(x) = x3 + 3x2 − x + 2, r(x) = 3x − 1
3. (a) q(x) = 3x2 + 6x + 8, r(x) = 15
(c) q(x) = x4 + x3 + x2 + x + 1, r(x) = 0
(b) q(x) = x3 − 5x2 + 20x − 100, r(x) = 504
4. (a) q(x) = 2x2 + x − 1, r(x) = 0
(b) q(x) = 2x3 − 5x2 + 3x − 39, r(x) = 147
(c) q(x) = x6 + x5 + x4 + x3 + x2 + x + 1, r(x) = 2
5.
x
0
1 −3
7
p(x) −4 −3 101 5001
6.
x
1 −1 3 −3
7
−7
21
−21
p(x) −24 −12 12
0 420 −168 10416 −7812
7. (a) q(x) = x2 + 6x + 13, r = 20
(b) q(x) = x2 + 3x − 2, r = −4
8. (a) q(x) = x4 − x3 + x2 − x + 1, r = −2
(b) q(x) = x4 + x3 + x2 + x + 1, r = 0
9. Assume r = a/b a and b integers with a > 0:
(a) b divides 1, b = ±1; a divides 24, a = 1, 2, 3, 4, 6, 8, 12, 24;
the possible candidates are {±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24}
(b) b divides 3 so b = ±1, ±3; a divides −10 so a = 1, 2, 5, 10;
the possible candidates are {±1, ±2, ±5, ±10, ±1/3, ±2/3, ±5/3, ±10/3}
(c) b divides 1 so b = ±1; a divides 17 so a = 1, 17;
the possible candidates are {±1, ±17}
10. An integer zero c divides −21, so c = ±1, ±3, ±7, ±21 are the only possibilities; substitution of
these candidates shows that the integer zeros are −7, −1, 3
11. (x + 1)(x − 1)(x − 2)
12. (x + 2)(3x + 1)(x − 2)
13. (x + 3)3 (x + 1)
14. 2x4 + x3 − 19x2 + 9
15. (x + 3)(x + 2)(x + 1)2 (x − 3)
17. −3 is the only real root.
18. x = −3/2 is the only real root.
19. x = −2, −2/3 are the only real roots.
674
675
Appendix F
20. −2, −1, 1/2, 3
21. −2, 2, 3 are the only real roots.
23. If x − 1 is a factor then p(1) = 0, so k 2 − 7k + 10 = 0, k 2 − 7k + 10 = (k − 2)(k − 5), so k = 2, 5.
24. (−3)7 = −2187, so −3 is a root and thus by Theorem F.4(a), x + 3 is a factor of x7 + 2187.
25. If the side of the cube is x then x2 (x − 3) = 196; the only real root of this equation is x = 7 cm.
a 3
a
>
+ 1. The polynomial p(x) = x3 − x + 1 has only one real root
b
b
c ≈ −1.325, and p(0) = 1 so p(x) > 0 for all x > c; hence there is no positive rational
a a 3
+ 1.
solution of >
b
b
26. (a) Try to solve
(b) From part (a), any real x < c is a solution.
27. Use the Factor Theorem with x as the variable and y as the constant c.
(a) For any positive integer n the polynomial xn − y n has x = y as a root.
(b) For any positive even integer n the polynomial xn − y n has x = −y as a root.
(c) For any positive odd integer n the polynomial xn + y n has x = −y as a root.
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