ISMT11_C01_A Thomas' Calculus 11th Edition Solution Manual
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CHAPTER 1 PRELIMINARIES
1.1 REAL NUMBERS AND THE REAL LINE
1. Executing long division,
"
9
2. Executing long division,
"
11
œ 0.1,
2
9
œ 0.2,
œ 0.09,
2
11
3
9
œ 0.3,
œ 0.18,
3
11
8
9
œ 0.8,
œ 0.27,
9
11
9
9
œ 0.9
œ 0.81,
11
11
œ 0.99
3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6.
a) NNT. 5 is a counter example.
b) NT. 2 < x < 6 Ê 2 2 < x 2 < 6 2 Ê 0 < x 2 < 2.
c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3.
d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2.
e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3.
f) NT. 2 < x < 6 Ê x < 6 Ê (x 4) < 2 and 2 < x < 6 Ê x > 2 Ê x < 2 Ê x + 4 < 2 Ê (x 4) < 2.
The pair of inequalities (x 4) < 2 and (x 4) < 2 Ê | x 4 | < 2.
g) NT. 2 < x < 6 Ê 2 > x > 6 Ê 6 < x < 2. But 2 < 2. So 6 < x < 2 < 2 or 6 < x < 2.
h) NT. 2 < x < 6 Ê 1(2) > 1(x) < 1(6) Ê 6 < x < 2
4. NT = necessarily true, NNT = Not necessarily true. Given: 1 < y 5 < 1.
a) NT. 1 < y 5 < 1 Ê 1 + 5 < y 5 + 5 < 1 + 5 Ê 4 < y < 6.
b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6)
c) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y > 4.
d) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y < 6.
e) NT. 1 < y 5 < 1 Ê 1 + 1 < y 5 + 1 < 1 + 1 Ê 0 < y 4 < 2.
f) NT. 1 < y 5 < 1 Ê (1/2)(1 + 5) < (1/2)(y 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3.
g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4.
h) NT. 1 < y 5 < 1 Ê y 5 > 1 Ê y > 4 Ê y < 4 Ê y + 5 < 1 Ê (y 5) < 1.
Also, 1 < y 5 < 1 Ê y 5 < 1. The pair of inequalities (y 5) < 1 and (y 5) < 1 Ê | y 5 | < 1.
5. 2x 4 Ê x 2
6. 8 3x
5 Ê 3x
3 Ê x Ÿ 1
7. 5x $ Ÿ ( 3x Ê 8x Ÿ 10 Ê x Ÿ
ïïïïïïïïïñqqqqqqqqp x
1
5
4
8. 3(2 x) 2(3 x) Ê 6 3x 6 2x
Ê 0 5x Ê 0 x
9. 2x
10.
"
#
Ê
"
5
6 x
4
7x
ˆ
10 ‰
6
3x4
2
7
6
Ê "#
x or
"
3
7
6
ïïïïïïïïïðqqqqqqqqp x
0
5x
x
Ê 12 2x 12x 16
Ê 28 14x Ê 2 x
qqqqqqqqqðïïïïïïïïî x
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
11.
Chapter 1 Preliminaries
4
5
"
3
(x 2)
(x 6) Ê 12(x 2) 5(x 6)
Ê 12x 24 5x 30 Ê 7x 6 or x 67
12. x2 5 Ÿ
123x
4
Ê (4x 20) Ÿ 24 6x
Ê 44 Ÿ 10x Ê 22
5 Ÿ x
qqqqqqqqqñïïïïïïïïî x
22/5
13. y œ 3 or y œ 3
14. y 3 œ 7 or y 3 œ 7 Ê y œ 10 or y œ 4
15. 2t 5 œ 4 or 2t & œ 4 Ê 2t œ 1 or 2t œ 9 Ê t œ "# or t œ 9#
16. 1 t œ 1 or 1 t œ 1 Ê t œ ! or t œ 2 Ê t œ 0 or t œ 2
17. 8 3s œ
18.
s
#
9
2
or 8 3s œ #9 Ê 3s œ 7# or 3s œ 25
# Ê sœ
1 œ 1 or
s
#
1 œ 1 Ê
s
#
œ 2 or
s
#
7
6
or s œ
25
6
œ ! Ê s œ 4 or s œ 0
19. 2 x 2; solution interval (2ß 2)
20. 2 Ÿ x Ÿ 2; solution interval [2ß 2]
qqqqñïïïïïïïïñqqqqp x
2
2
21. 3 Ÿ t 1 Ÿ 3 Ê 2 Ÿ t Ÿ 4; solution interval [2ß 4]
22. 1 t 2 1 Ê 3 t 1;
solution interval (3ß 1)
qqqqðïïïïïïïïðqqqqp t
3
1
23. % 3y 7 4 Ê 3 3y 11 Ê 1 y
solution interval ˆ1ß
11
3
;
11 ‰
3
24. 1 2y 5 " Ê 6 2y 4 Ê 3 y 2;
solution interval (3ß 2)
25. 1 Ÿ
z
5
1Ÿ1 Ê 0Ÿ
z
5
qqqqðïïïïïïïïðqqqqp y
3
2
Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10]
26. 2 Ÿ
1 Ÿ 2 Ê 1 Ÿ
solution interval 23 ß 2‘
3z
#
27. "# 3
Ê
2
7
28. 3
"
x
x
2
x
2
5
"
#
2
7
Ÿ 3 Ê 32 Ÿ z Ÿ 2;
qqqqñïïïïïïïïñqqqqp z
2
2/3
Ê 7# x" 5# Ê
7
#
"
x
5
#
; solution interval ˆ 27 ß 25 ‰
43 Ê 1
Ê 2x
3z
#
Ê
2
7
2
x
( Ê 1
x
#
"
7
x 2; solution interval ˆ 27 ß 2‰
qqqqðïïïïïïïïðqqqqp x
2
2/7
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.1 Real Numbers and the Real Line
4 or 2s
29. 2s
4 Ê s
2 or s Ÿ 2;
solution intervals (_ß 2] [2ß _)
"
#
30. s 3
or (s 3)
"
#
Ê s
5# or s
7
#
Ê s
5# or s Ÿ 7# ;
solution intervals ˆ_ß 7# ‘ 5# ß _‰
ïïïïïïñqqqqqqñïïïïïïî s
7/2
5/2
31. 1 x 1 or (" x) 1 Ê x 0 or x 2
Ê x 0 or x 2; solution intervals (_ß !) (2ß _)
32. 2 3x 5 or (2 3x) 5 Ê 3x 3 or 3x 7
Ê x 1 or x 73 ;
solution intervals (_ß 1) ˆ 73 ß _‰
33.
1 or ˆ r# 1 ‰
r"
#
Ê r
34.
3r
5
2 or r 1 Ÿ 2
1 or r Ÿ 3; solution intervals (_ß 3] [1ß _)
"
Ê
1 Ê r1
ïïïïïïðqqqqqqðïïïïïïî x
1
7/3
or ˆ 3r5 "‰
2
5
or 3r5 53 Ê r 37 or r 1
solution intervals (_ß ") ˆ 73 ß _‰
3r
5
2
5
7
5
ïïïïïïðqqqqqqðïïïïïïî r
1
7/3
35. x# # Ê kxk È2 Ê È2 x È2 ;
solution interval ŠÈ2ß È2‹
qqqqqqðïïïïïïðqqqqqqp x
È#
È #
36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ 2;
solution interval (_ß 2] [2ß _)
ïïïïïïñqqqqqqñïïïïïïî r
2
2
37. 4 x# 9 Ê 2 kxk 3 Ê 2 x 3 or 2 x 3
Ê 2 x 3 or 3 x 2;
solution intervals (3ß 2) (2ß 3)
38.
"
9
x#
Ê
x
"
#
"
3
kxk
"
#
Ê
"
3
x
or #" x 3" ;
solution intervals ˆ "# ß 3" ‰ ˆ 3" ß #" ‰
Ê
"
3
"
4
"
#
or
"
3
x
39. (x 1)# 4 Ê kx 1k 2 Ê 2 x 1 2
Ê 1 x 3; solution interval ("ß $)
qqqqðïïïïðqqqqðïïïïðqqqp x
3
2
2
3
"
#
qqqqðïïïïðqqqqðïïïïðqqqp x
1/2 1/3
1/3
1/2
qqqqqqðïïïïïïïïðqqqqp x
1
3
40. (x 3)# # Ê kx 3k È2
Ê È2 x 3 È2 or 3 È2 x 3 È2 ;
solution interval Š3 È2ß 3 È2‹
qqqqqqðïïïïïïïïðqqqqp x
3 È #
3 È #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3
4
Chapter 1 Preliminaries
41. x# x 0 Ê x# x +
1
4
<
1
4
2
Ê ˆx 12 ‰ <
ʹx
1
4
1
2
¹<
1
2
Ê 12 < x
1
2
<
1
2
Ê 0 < x < 1.
So the solution is the interval (0ß 1)
42. x# x 2
0 Ê x# x +
1
4
9
4
Ê ¹x
1
2
¹
3
2
Ê x
1
2
3
2
or ˆx 12 ‰
3
2
Ê x
2 or x Ÿ 1.
The solution interval is (_ß 1] [2ß _)
43. True if a
0; False if a 0.
44. kx 1k œ 1 x Í k(x 1)k œ 1 x Í 1 x
0 Í xŸ1
45. (1) ka bk œ (a b) or ka bk œ (a b);
both squared equal (a b)#
(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ a, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka bk and
y œ kak kbk so that ka bk# Ÿ akak kbkb# Ê ka bk Ÿ kak kbk .
46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk .
If a 0 and b 0, then ab 0 and kabk œ ab œ (a)(b) œ kak kbk .
If a 0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk .
If a 0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk .
47. 3 Ÿ x Ÿ 3 and x "# Ê
"
#
x Ÿ 3.
48. Graph of kxk kyk Ÿ 1 is the interior
of “diamond-shaped" region.
49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x1 | < $ . Then | x1 | < $ Ê 2| x1 | < 2$ Ê
| 2x # | < 2$ Ê | (2x + 1) 3 | < 2$ Ê | f(x) f(1) | < 2$
50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x 0 | < % /2. Then 2| x 0 | < % and
| 2x + 3 3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) f(0) | < %.
51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê a < 0. Let a = b. By definition, | b | œ b. Since b = a,
| a | œ (a) œ a and | a | œ | a | œ a.
ii) For a < 0, | a | œ a. Now, a < 0 Ê a > 0. Let a œ b. By definition, | b | œ b and thus |a| œ a. So again
| a | œ |a|.
iii) By definition | 0 | œ 0 and since 0 œ 0, | 0 | œ 0. Thus, by i), ii), and iii) | a | œ | a | for any real number.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.2 Lines, Circles and Parabolas
Prove | x | > 0 Ê x > a or x < a for any positive number, a.
For x
0, | x | œ x. | x | > a Ê x > a.
For x < 0, | x | œ x. | x | > a Ê x > a Ê x < a.
ii) Prove x > a or x < a Ê | x | > 0 for any positive number, a.
a > 0 and x > a Ê | x | œ x. So x > a Ê | x | > a.
For a > 0, a < 0 and x < a Ê x < 0 Ê | x | œ x. So x < a Ê x > a Ê | x | > a.
52. i)
53. a)
1=1 Ê |1|=1 ʹb
b)
lal
lbl
œ ¹a
†
"
b
¹ œ ¹ a¹
† "b ¹ œ
†¹
"
b
l bl
lbl
¹ œ ¹ a¹
Ê ¹ b¹
† l bl
"
† ¹ b" ¹ œ
œ
lbl
lbl
Ê
†
¹ b ¹ ¹ "b ¹
¹ b¹
œ
¹ b¹
†
¹ b¹ ¹ b¹
Ê ¹ b" ¹ œ "
¹ b¹
lal
lbl
54. Prove Sn œ kan k œ kakn for any real number a and any positive integer n.
ka" k œ kak " œ a, so S" is true. Now, assume that Sk œ ¸ak ¸ œ kak k is true form some positive integer 5 .
Since ka" k œ kak " and ¸ak ¸ œ kak k , we have ¸ak" ¸ œ ¸ak † a" ¸ œ ¸ak ¸ka" k œ kak k kak " œ kak k+" . Thus,
Sk" œ ¸ak" ¸ œ kak k+" is also true. Thus by the Principle of Mathematical Induction, Sn œ l an l œ l a ln
is true for all n positive integers.
1.2 LINES, CIRCLES, AND PARABOLAS
1. ?x œ 1 (3) œ 2, ?y œ 2 2 œ 4; d œ È(?x)# (?y)# œ È4 16 œ 2È5
2. ?x œ $ (1) œ 2, ?y œ 2 (2) œ 4; d œ È(2)# 4# œ 2È5
3. ?x œ 8.1 (3.2) œ 4.9, ?y œ 2 (2) œ 0; d œ È(4.9)# 0# œ 4.9
#
4. ?x œ 0 È2 œ È2, ?y œ 1.5 4 œ 2.5; d œ ÊŠÈ2‹ (2.5)# œ È8.25
5. Circle with center (!ß !) and radius 1.
6. Circle with center (!ß !) and radius È2.
7. Disk (i.e., circle together with its interior points) with center (!ß !) and radius È3.
8. The origin (a single point).
9. m œ
?y
?x
œ
1 2
2 (1)
œ3
perpendicular slope œ "3
10. m œ
?y
?x
œ
# "
2 (2)
œ 34
perpendicular slope œ
4
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5
6
Chapter 1 Preliminaries
11. m œ
?y
?x
œ
33
1 2
œ0
12. m œ
14. (a) x œ È2
(b) y œ
# 0
# (#)
; no slope
15. (a) x œ 0
16. (a) x œ 1
(b) y œ È2
(b) y œ 1.3
4
3
œ
perpendicular slope œ 0
perpendicular slope does not exist
13. (a) x œ 1
?y
?x
(b) y œ 0
17. P(1ß 1), m œ 1 Ê y 1 œ 1ax (1)b Ê y œ x
18. P(2ß 3), m œ
"
#
Ê y (3) œ
19. P(3ß 4), Q(2ß 5) Ê m œ
?y
?x
20. P(8ß 0), Q(1ß 3) Ê m œ
œ
?y
?x
"
#
(x 2) Ê y œ
54
2 3
œ
"
#
x4
œ "5 Ê y 4 œ "5 (x 3) Ê y œ "5 x
30
1 (8)
œ
3
7
Ê y0œ
3
7
ax (8)b Ê y œ
3
7
23
5
x
21. m œ 54 , b œ 6 Ê y œ 54 x 6
22. m œ "# , b œ 3 Ê y œ
"
#
23. m œ 0, P(12ß 9) Ê y œ 9
24. No slope, P ˆ "3 ß %‰ Ê x œ
24
7
x3
"
3
25. a œ 1, b œ 4 Ê (0ß 4) and ("ß 0) are on the line Ê m œ
?y
?x
œ
04
1 0
œ 4 Ê y œ 4x 4
26. a œ 2, b œ 6 Ê (2ß 0) and (!ß 6) are on the line Ê m œ
?y
?x
œ
6 0
02
œ 3 Ê y œ 3x 6
27. P(5ß 1), L: 2x 5y œ 15 Ê mL œ 25 Ê parallel line is y (1) œ 25 (x 5) Ê y œ 25 x 1
È
È
È
28. P ŠÈ2ß 2‹ , L: È2x 5y œ È3 Ê mL œ 52 Ê parallel line is y 2 œ 52 Šx ŠÈ2‹‹ Ê y œ 52 x
8
5
29. P(4ß 10), L: 6x 3y œ 5 Ê mL œ 2 Ê m¼ œ "# Ê perpendicular line is y 10 œ "# (x 4) Ê y œ "# x 12
30. P(!ß 1), L: 8x 13y œ 13 Ê mL œ
8
13
13
Ê m¼ œ 13
8 Ê perpendicular line is y œ 8 x 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.2 Lines, Circles and Parabolas
31. x-intercept œ 4, y-intercept œ 3
32. x-intercept œ 4, y-intercept œ 2
33. x-intercept œ È3, y-intercept œ È2
34. x-intercept œ 2, y-intercept œ 3
35. Ax By œ C" Í y œ AB x
C"
B
and Bx Ay œ C# Í y œ
B
A
x
C#
A.
Since ˆ AB ‰ ˆ AB ‰ œ 1 is the
product of the slopes, the lines are perpendicular.
36. Ax By œ C" Í y œ AB x
slope
AB ,
C"
B
and Ax By œ C# Í y œ AB x
C#
B.
Since the lines have the same
they are parallel.
37. New position œ axold ?xß yold ?yb œ (# &ß 3 (6)) œ ($ß 3).
38. New position œ axold ?xß yold ?yb œ (6 (6)ß 0 0) œ (0ß 0).
39. ?x œ 5, ?y œ 6, B(3ß 3). Let A œ (xß y). Then ?x œ x# x" Ê 5 œ 3 x Ê x œ 2 and
?y œ y# y" Ê 6 œ 3 y Ê y œ 9. Therefore, A œ (#ß 9).
40. ?x œ " " œ !, ?y œ ! ! œ !
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
7
8
Chapter 1 Preliminaries
41. C(!ß 2), a œ 2 Ê x# (y 2)# œ 4
42. C($ß 0), a œ 3 Ê (x 3)# y# œ 9
43. C(1ß 5), a œ È10 Ê (x 1)# (y 5)# œ 10
44. C("ß "), a œ È2 Ê (x 1)# (y 1)# œ 2
x œ 0 Ê (0 1)# (y 1)# œ 2 Ê (y 1)# œ 1
Ê y 1 œ „ 1 Ê y œ 0 or y œ 2.
Similarly, y œ 0 Ê x œ 0 or x œ 2
#
45. C ŠÈ3ß 2‹ , a œ 2 Ê Šx È3‹ (y 2)# œ 4,
#
x œ 0 Ê Š0 È3‹ (y 2)# œ 4 Ê (y 2)# œ 1
Ê y 2 œ „ 1 Ê y œ 1 or y œ 3. Also, y œ 0
#
#
Ê Šx È3‹ (0 2)# œ 4 Ê Šx È3‹ œ 0
Ê x œ È 3
#
46. C ˆ3ß "# ‰, a œ 5 Ê (x 3)# ˆy "# ‰ œ 25, so
#
x œ 0 Ê (0 3)# ˆy "# ‰ œ 25
#
Ê ˆy "# ‰ œ 16 Ê y
"
#
œ „4 Ê yœ
9
#
#
or y œ 7# . Also, y œ 0 Ê (x 3)# ˆ0 "# ‰ œ 25
Ê (x 3)# œ
Ê xœ3„
99
4
3È11
#
Ê x3œ „
3È11
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.2 Lines, Circles and Parabolas
47. x# y# 4x 4y % œ 0
Ê x# %B y# 4y œ 4
Ê x# 4x 4 y# 4y 4 œ 4
Ê (x 2)# (y 2)# œ 4 Ê C œ (2ß 2), a œ 2.
48. x# y# 8x 4y 16 œ 0
Ê x# 8x y# 4y œ 16
Ê x# 8x 16 y# 4y 4 œ 4
Ê (x 4)# (y 2)# œ 4
Ê C œ (%ß 2), a œ 2.
49. x# y# 3y 4 œ 0 Ê x# y# 3y œ 4
Ê x# y# 3y 94 œ 25
4
#
Ê x# ˆy 3# ‰ œ
25
4
Ê C œ ˆ0ß 3# ‰ ,
a œ 5# .
50. x# y# 4x
#
9
4
#
œ0
Ê x 4x y œ
9
4
#
Ê x# 4x 4 y œ
Ê (x 2)# y# œ
25
4
25
4
Ê C œ (2ß 0), a œ 5# .
51. x# y# 4x 4y œ 0
Ê x# 4x y# 4y œ 0
Ê x# 4x 4 y# 4y 4 œ 8
Ê (x 2)# (y 2)# œ 8
Ê C(2ß 2), a œ È8.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
9
10
Chapter 1 Preliminaries
52. x# y# 2x œ 3
Ê x# 2x 1 y# œ 4
Ê (x 1)# y# œ 4
Ê C œ (1ß 0), a œ 2.
2
53. x œ #ba œ 2(1)
œ1
Ê y œ (1)# 2(1) 3 œ 4
Ê V œ ("ß 4). If x œ 0 then y œ 3.
Also, y œ 0 Ê x# 2x 3 œ 0
Ê (x 3)(x 1) œ 0 Ê x œ 3 or
x œ 1. Axis of parabola is x œ 1.
4
54. x œ #ba œ 2(1)
œ 2
Ê y œ (2)# 4(2) 3 œ 1
Ê V œ (2ß 1). If x œ 0 then y œ 3.
Also, y œ 0 Ê x# 4x 3 œ 0
Ê (x 1)(x 3) œ 0 Ê x œ 1 or
x œ 3. Axis of parabola is x œ 2.
55. x œ #ba œ 2(4 1) œ 2
Ê y œ (2)# 4(2) œ 4
Ê V œ (2ß 4). If x œ 0 then y œ 0.
Also, y œ 0 Ê x# 4x œ 0
Ê x(x 4) œ 0 Ê x œ 4 or x œ 0.
Axis of parabola is x œ 2.
56. x œ #ba œ 2(4 1) œ 2
Ê y œ (2)# 4(2) 5 œ 1
Ê V œ (2ß 1). If x œ 0 then y œ 5.
Also, y œ 0 Ê x# 4x 5 œ 0
Ê x# 4x 5 œ 0 Ê x œ
4 „È 4
#
Ê no x intercepts. Axis of parabola is x œ 2.
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Section 1.2 Lines, Circles and Parabolas
57. x œ #ba œ 2(61) œ 3
Ê y œ (3)# 6(3) 5 œ 4
Ê V œ (3ß %). If x œ 0 then y œ 5.
Also, y œ 0 Ê x# 6x 5 œ 0
Ê (x 5)(x 1) œ 0 Ê x œ 5 or
x œ 1. Axis of parabola is x œ 3.
1
58. x œ #ba œ 2(2)
œ
"
4
#
Ê y œ 2 ˆ "4 ‰ 4" 3 œ 23
8
‰
Ê V œ ˆ "4 ß 23
.
If
x
œ
0
then
y œ 3.
8
Also, y œ 0 Ê 2x# x 3 œ 0
Ê xœ
1„È23
4
Ê no x intercepts.
Axis of parabola is x œ "4 .
1
59. x œ #ba œ 2(1/2)
œ 1
"
#
(1)# (1) 4 œ 72
Ê V œ ˆ"ß 72 ‰ . If x œ 0 then y œ 4.
Ê yœ
Also, y œ 0 Ê
Ê xœ
1 „ È 7
1
"
#
x# x 4 œ 0
Ê no x intercepts.
Axis of parabola is x œ 1.
60. x œ #ba œ 2(21/4) œ 4
Ê y œ "4 (4)# 2(4) 4 œ 8
Ê V œ (4ß 8) . If x œ 0 then y œ 4.
Also, y œ 0 Ê "4 x# 2x 4 œ 0
Ê xœ
2 „ È 8
1/2
œ 4 „ 4È2.
Axis of parabola is x œ 4.
61. The points that lie outside the circle with center (!ß 0) and radius È7.
62. The points that lie inside the circle with center (!ß 0) and radius È5.
63. The points that lie on or inside the circle with center ("ß 0) and radius 2.
64. The points lying on or outside the circle with center (!ß 2) and radius 2.
65. The points lying outside the circle with center (!ß 0) and radius 1, but inside the circle with center (!ß 0),
and radius 2 (i.e., a washer).
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11
12
Chapter 1 Preliminaries
66. The points on or inside the circle centered
at (!ß !) with radius 2 and on or inside the
circle centered at (2ß 0) with radius 2.
67. x# y# 6y 0 Ê x# (y 3)# 9.
The interior points of the circle centered at
(!ß 3) with radius 3, but above the line
y œ 3.
68. x# y# 4x 2y 4 Ê (x 2)# (y 1)# 9.
The points exterior to the circle centered at
(2ß 1) with radius 3 and to the right of the
line x œ 2.
69. (x 2)# (y 1)# 6
70. (x 4)# (y 2)# 16
71. x# y# Ÿ 2, x
72. x# y# 4, (x 1)# (y 3)# 10
1
73. x# y# œ 1 and y œ 2x Ê 1 œ x# 4x# œ 5x#
Ê Šx œ
"
È5
and y œ
2
È5 ‹
or Šx œ È"5 and y œ È25 ‹ .
Thus, A Š È"5 ß È25 ‹ , B Š È"5 ß È25 ‹ are the
points of intersection.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.2 Lines, Circles, and Parabolas
74. x y œ 1 and (x 1)# y# œ 1
Ê 1 œ (y)# y# œ 2y#
Ê Šy œ
"
È2
and x œ "
Šy œ È"2 and x œ 1
A Š"
"
È2
"
È2 ‹
"
È2 ‹ .
ß È"2 ‹ and B Š1
or
Thus,
"
È2
ß È"2 ‹
are intersection points.
75. y x œ 1 and y œ x# Ê x# x œ 1
1 „È 5
.
#
1 È 5
3 È 5
If x œ # , then y œ x 1 œ # .
È
È
If x œ 1# 5 , then y œ x 1 œ 3# 5 .
È
È
È
È
Thus, A Š 1# 5 ß 3# 5 ‹ and B Š 1# 5 ß 3# 5 ‹
Ê x# x 1 œ 0 Ê x œ
are the intersection points.
76. y œ x and C œ (x 1)# Ê (x 1)# œ x
3 „È 5
.
#
È 5 3
3 È 5
x œ # , then y œ x œ # . If
È
È
x œ 3# 5 , then y œ x œ 3# 5 .
È
È
È
Thus, A Š 3# 5 ß 5#3 ‹ and B Š 3# 5
Ê x# 3x " œ 0 Ê x œ
If
È
ß 3# 5 ‹
are the intersection points.
77. y œ 2x# 1 œ x# Ê 3x# œ 1
Ê x œ È"3 and y œ 3" or x œ È"3 and y œ 3" .
Thus, A Š È"3 ß 3" ‹ and B Š È"3 ß 3" ‹ are the
intersection points.
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13
14
Chapter 1 Preliminaries
78. y œ
x#
4
œ (x 1)# Ê 0 œ
#
3x#
4
2x 1
Ê 0 œ 3x 8x 4 œ (3x 2)(x 2)
Ê x œ 2 and y œ
yœ
#
x
4
x#
4
œ 1, or x œ
œ 9" . Thus, A(2ß 1) and
2
3 and
2
B ˆ 3 ß 9" ‰
are the intersection points.
79. x# y# œ 1 œ (x 1)# y#
Ê x# œ (x 1)# œ x# 2x 1
Ê 0 œ 2x 1 Ê x œ "# . Hence
y# œ " x # œ
A Š "# ß
È3
# ‹
and
3
4
or y œ „
È3
#
È
B Š "# ß #3 ‹
. Thus,
are the
intersection points.
80. x# y# œ 1 œ x# y Ê y# œ y
Ê y(y 1) œ 0 Ê y œ 0 or y œ 1.
If y œ 1, then x# œ " y# œ 0 or x œ 0.
If y œ 0, then x# œ 1 y# œ 1 or x œ „ 1.
Thus, A(0ß 1), B("ß 0), and C(1ß 0) are the
intersection points.
81. (a) A ¸ (69°ß 0 in), B ¸ (68°ß .4 in) Ê m œ
(b) A ¸ (68°ß .4 in), B ¸ (10°ß 4 in) Ê m œ
(c) A ¸ (10°ß 4 in), B ¸ (5°ß 4.6 in) Ê m œ
82. The time rate of heat transfer across a material,
to the temperature gradient across the material,
of the material.
?U
?>
œ
X
-kA ?
?B
Ê
?U ÎA
k = ??> X .
?B
68° 69°
.4 0 ¸ 2.5°/in.
10° 68°
4 .4 ¸ 16.1°/in.
5° 10°
4.6 4 ¸ 8.3°/in.
?U
?> , is directly
?X
?B (the slopes
Note that
?U
?>
proportional to the cross-sectional area, A, of the material,
from the previous problem), and to a constant characteristic
and
?X
?B
are of opposite sign because heat flow is toward lower
temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good
insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are
X
not changing), we may define another constant, K, characteristics of the material: K œ ?"X Þ Using the values of ?
?B from
?B
the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the
poorest insulator, with K œ 0.4.
83. p œ kd 1 and p œ 10.94 at d œ 100 Ê k œ
10.94"
100
œ 0.0994. Then p œ 0.0994d 1 is the diver's
pressure equation so that d œ 50 Ê p œ (0.0994)(50) 1 œ 5.97 atmospheres.
84. The line of incidence passes through (!ß 1) and ("ß 0) Ê The line of reflection passes through ("ß 0) and (#ß ")
0
Ê m œ 1#
1 œ 1 Ê y 0 œ 1(x 1) Ê y œ x 1 is the line of reflection.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.2 Lines, Circles, and Parabolas
85. C œ
5
9
(F 32) and C œ F Ê F œ
86. m œ
37.1
100
œ
14
?x
Ê ?x œ
14
.371 .
5
9
F
160
9
Ê
4
9
15
F œ 160
9 or F œ 40° gives the same numerical reading.
#
14 ‰
Therefore, distance between first and last rows is É(14)# ˆ .371
¸ 40.25 ft.
87. length AB œ È(5 1)# (5 2)# œ È16 9 œ 5
length AC œ È(4 1)# (# #)# œ È9 16 œ 5
length BC œ È(4 5)# (# 5)# œ È1 49 œ È50 œ 5È2 Á 5
#
88. length AB œ Ê(1 0)# ŠÈ3 0‹ œ È1 3 œ 2
length AC œ È(2 0)# (0 0)# œ È4 0 œ 2
#
length BC œ Ê(2 1)# Š0 È3‹ œ È1 3 œ 2
89. Length AB œ È(?x)# (?y)# œ È1# 4# œ È17 and length BC œ È(?x)# (?y)# œ È4# 1# œ È17.
Also, slope AB œ 41 and slope BC œ "4 , so AB ¼ BC. Thus, the points are vertices of a square. The coordinate
increments from the fourth vertex D(xß y) to A must equal the increments from C to B Ê 2 x œ ?x œ 4 and
1 y œ ?y œ " Ê x œ 2 and y œ 2. Thus D(#ß 2) is the fourth vertex.
90. Let A œ (xß 2) and C œ (9ß y) Ê B œ (xß y). Then 9 x œ kADk and 2 y œ kDCk Ê 2(9 x) 2(2 y) œ 56
and 9 x œ 3(2 y) Ê 2(3(2 y)) 2(2 y) œ 56 Ê y œ 5 Ê 9 x œ 3(2 (5)) Ê x œ 12.
Therefore, A œ (12ß 2), C œ (9ß 5), and B œ (12ß 5).
91. Let A("ß "), B(#ß $), and C(2ß !) denote the points.
Since BC is vertical and has length kBCk œ 3, let
D" ("ß 4) be located vertically upward from A and
D# ("ß 2) be located vertically downward from A so
that kBCk œ kAD" k œ kAD# k œ 3. Denote the point
D$ (xß y). Since the slope of AB equals the slope of
3
"
CD$ we have yx
2 œ 3 Ê 3y 9 œ x 2 or
x 3y œ 11. Likewise, the slope of AC equals the slope
0
2
of BD$ so that yx
2 œ 3 Ê 3y œ 2x 4 or 2x 3y œ 4.
Solving the system of equations
x 3y œ ""
we find x œ 5 and y œ 2 yielding the vertex D$ (5ß #).
2x 3y œ 4
92. Let ax, yb, x Á ! and/or y Á ! be a point on the coordinate plane. The slope, m, of the segment a!ß !b to ax, yb is yx . A 90‰
rotation gives a segment with slope mw œ m" œ xy . If this segment has length equal to the original segment, its endpoint
will be ay, xb or ay, xb, the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise
rotation.
(a) ("ß 4);
(b) (3ß 2);
(c) (5ß 2);
(d) (0ß x);
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16
Chapter 1 Preliminaries
(e) (yß 0);
(f) (yß x);
(g) (3ß 10)
93. 2x ky œ 3 has slope 2k and 4x y œ 1 has slope 4. The lines are perpendicular when 2k (4) œ 1 or
k œ 8 and parallel when 2k œ 4 or k œ "# .
94. At the point of intersection, 2x 4y œ 6 and 2x 3y œ 1. Subtracting these equations we find 7y œ 7 or
y œ 1. Substitution into either equation gives x œ 1 Ê (1ß 1) is the intersection point. The line through (1ß 1)
and ("ß #) is vertical with equation x œ 1.
95. Let M(aß b) be the midpoint. Since the two triangles
shown in the figure are congruent, the value a must
lie midway between x" and x# , so a œ x" #x# .
Similarly, b œ
y " y #
# .
96. (a) L has slope 1 so M is the line through P(2ß 1) with slope 1; or the line y œ x 3. At the intersection
point, Q, we have equal y-values, y œ x 2 œ x 3. Thus, 2x œ 1 or x œ "# . Hence Q has coordinates
ˆ "# ß 5# ‰ . The distance from P to L œ the distance from P to Q œ Ɉ #3 ‰# ˆ 3# ‰# œ É 18
4 œ
(b) L has slope 43 so M has slope
3
4
3È 2
# .
and M has the equation 4y 3x œ 12. We can rewrite the equations of
84
the lines as L: x y œ 3 and M: B 43 y œ 4. Adding these we get 25
12 y œ 7 so y œ 25 . Substitution
12
‰
ˆ 12 84 ‰
into either equation gives x œ 43 ˆ 84
25 4 œ 25 so that Q 25 ß 25 is the point of intersection. The distance
3
4
from P to L œ Ɉ4
12 ‰#
25
ˆ6
84 ‰#
25
œ
22
5 .
(c) M is a horizontal line with equation y œ b. The intersection point of L and M is Q("ß b). Thus, the
distance from P to L is È(a 1)# 0# œ ka 1k .
(d) If B œ 0 and A Á 0, then the distance from P to L is ¸ AC x! ¸ as in (c). Similarly, if A œ 0 and B Á 0, the
distance is ¸ CB y! ¸ . If both A and B are Á 0 then L has slope AB so M has slope AB . Thus,
L: Ax By œ C and M: Bx Ay œ Bx! Ay! . Solving these equations simultaneously we find the
point of intersection Q(xß y) with x œ
ACB aAy! Bx! b
A# B#
P to Q equals È(?x)# (?y)# , where (?x)# œ
œ
A# aAx! By! Cb#
aA# B# b#
#
#
BCA aAy! Bx! b
.
A# B#
#
#
#
#
ABy! B x!
Š x! aA B bAAC
‹
# B#
and y œ
#
#
A y! ABx!
, and (?y)# œ Š y! aA B bABC
‹ œ
# B#
#
! Cb
Thus, È(?x)# (?y)# œ É aAx!A#By
œ
B#
kAx! By! Ck
ÈA# B#
The distance from
B# aAx! By! Cb#
.
aA# B# b#
.
1.3 FUNCTIONS AND THEIR GRAPHS
1. domain œ (_ß _); range œ [1ß _)
3. domain œ (!ß _); y in range Ê y œ
2. domain œ [0ß _); range œ (_ß 1]
"
Èt
, t 0 Ê y# œ
"
t
and y ! Ê y can be any positive real number
Ê range œ (!ß _).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.3 Functions and Their Graphs
4. domain œ [0ß _); y in range Ê y œ
"
1 È t
17
, t 0. If t œ 0, then y œ 1 and as t increases, y becomes a smaller
and smaller positive real number Ê range œ (0ß 1].
5. 4 z# œ (2 z)(2 z) 0 Í z − [2ß 2] œ domain. Largest value is g(0) œ È4 œ 2 and smallest value is
g(2) œ g(2) œ È0 œ 0 Ê range œ [0ß 2].
6. domain œ (2ß 2) from Exercise 5; smallest value is g(0) œ "# and as 0 z increases to 2, g(z) gets larger and
larger (also true as z 0 decreases to 2) Ê range œ "# ß _‰ .
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
9. y œ Ɉ "x ‰ " Ê
(a) No (x !Ñ;
(c) No; if x ",
"
x
"
x
"
! Ê x Ÿ 1 and x !. So,
"Ê
10. y œ É# Èx Ê # Èx
"
x
(b) No; division by ! undefined;
(d) Ð!ß "Ó
" !;
! Ê Èx
! and Èx Ÿ #. Èx
! and Èx Ÿ # Ê x Ÿ %Þ So, ! Ÿ x Ÿ %.
!Êx
(a) No; (b) No; (c) Ò!ß %Ó
#
11. base œ x; (height)# ˆ #x ‰ œ x# Ê height œ
È3
#
x; area is a(x) œ
"
#
(base)(height) œ
"
#
(x) Š
È3
# x‹
œ
È3
4
x# ;
perimeter is p(x) œ x x x œ 3x.
12. s œ side length Ê s# s# œ d# Ê s œ
d
È2
; and area is a œ s# Ê a œ
"
#
d#
13. Let D œ diagonal of a face of the cube and j œ the length of an edge. Then j# D# œ d# and (by Exercise 10)
D# œ 2j# Ê 3j# œ d# Ê j œ
d
È3
. The surface area is 6j# œ
6d#
3
14. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ
ˆx, Èx‰ œ ˆ m"# ,
#
œ 2d# and the volume is j$ œ Š d3 ‹
Èx
x
œ
"
Èx
"‰
m .
15. The domain is a_ß _b.
16. The domain is a_ß _b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
$Î#
œ
d$
3È 3
(x 0). Thus,
.
18
Chapter 1 Preliminaries
17. The domain is a_ß _b.
18. The domain is Ð_ß !Ó.
19. The domain is a_ß !b a!ß _b.
20. The domain is a_ß !b a!ß _b.
21. Neither graph passes the vertical line test
(a)
(b)
22. Neither graph passes the vertical line test
(a)
(b)
Ú xyœ" Þ
Ú yœ1x Þ
or
or
kx yk œ 1 Í Û
Í Û
ß
ß
Ü x y œ " à
Ü y œ " x à
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.3 Functions and Their Graphs
23.
x
y
0
0
25. y œ œ
1
1
2
0
24.
x
y
0
1
1
0
2
0
"
, x0
26. y œ œ x
x, 0 Ÿ x
3 x, x Ÿ 1
2x, 1 x
27. (a) Line through a!ß !b and a"ß "b: y œ x
Line through a"ß "b and a#ß !b: y œ x 2
x, 0 Ÿ x Ÿ 1
f(x) œ œ
x 2, 1 x Ÿ 2
Ú
Ý
Ý 2, ! Ÿ x "
!ß " Ÿ x #
(b) f(x) œ Û
Ý
Ý 2ß # Ÿ x $
Ü !ß $ Ÿ x Ÿ %
28. (a) Line through a!ß 2b and a#ß !b: y œ x 2
"
Line through a2ß "b and a&ß !b: m œ !&
# œ
x #, 0 x Ÿ #
f(x) œ œ "
$ x &$ , # x Ÿ &
"
$
$ !
! Ð"Ñ œ
" $
%
#! œ #
(b) Line through a"ß !b and a!ß $b: m œ
Line through a!ß $b and a#ß "b: m œ
f(x) œ œ
œ "$ , so y œ "$ ax 2b " œ "$ x
&
$
$, so y œ $x $
œ #, so y œ #x $
$x $, " x Ÿ !
#x $, ! x Ÿ #
29. (a) Line through a"ß "b and a!ß !b: y œ x
Line through a!ß "b and a"ß "b: y œ "
Line through a"ß "b and a$ß !b: m œ !"
$" œ
Ú
x
" Ÿ x !
"
!xŸ"
f(x) œ Û
Ü "# x $#
"x$
"
#
œ "# , so y œ "# ax "b " œ "# x
$
#
(b) Line through a#ß "b and a!ß !b: y œ "# x
Line through a!ß #b and a"ß !b: y œ #x #
Line through a"ß "b and a$ß "b: y œ "
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19
20
Chapter 1 Preliminaries
Ú
"
#x
f(x) œ Û #x #
Ü "
# Ÿ x Ÿ !
!xŸ"
"xŸ$
30. (a) Line through ˆ T# ß !‰ and aTß "b: m œ
Ú
A,
Ý
Ý
Ý
Aß
f(x) œ Û
Aß
Ý
Ý
Ý
Ü Aß
! Ÿ x T#
T
# Ÿ x T
T Ÿ x $#T
$T
# Ÿ x Ÿ #T
x
#
31. (a) From the graph,
(b)
x
#
1
x 0:
x
#
x 0:
x
2
œ T# , so y œ T# ˆx T# ‰ 0 œ T# x "
!, 0 Ÿ x Ÿ T#
#
T
T x ", # x Ÿ T
f(x) œ
(b)
"!
TaTÎ#b
4
x
1
x
#
Ê
4
x
1
4
x
Ê x − (2ß 0) (%ß _)
1 4x 0
#
2x8
0 Ê x 2x
0 Ê
(x4)(x2)
#x
0
(x4)(x2)
#x
0
Ê x 4 since x is positive;
1
4
x
0 Ê
x# 2x8
2x
0 Ê
Ê x 2 since x is negative;
sign of (x 4)(x 2)
ïïïïïðïïïïïðïïïïî
2
%
Solution interval: (#ß 0) (%ß _)
3
2
x 1 x 1
3
2
x 1 x 1
32. (a) From the graph,
(b) Case x 1:
Ê x − (_ß 5) (1ß 1)
Ê
3(x1)
x 1
2
Ê 3x 3 2x 2 Ê x 5.
Thus, x − (_ß 5) solves the inequality.
Case 1 x 1:
3
x 1
2
x 1
Ê
3(x1)
x 1
2
Ê 3x 3 2x 2 Ê x 5 which is true
if x 1. Thus, x − (1ß 1) solves the
inequality.
Case 1 x: x3 1 x2 1 Ê 3x 3 2x 2 Ê x 5
which is never true if 1 x, so no solution
here.
In conclusion, x − (_ß 5) (1ß 1).
33. (a) ÚxÛ œ 0 for x − [0ß 1)
(b) ÜxÝ œ 0 for x − (1ß 0]
34. ÚxÛ œ ÜxÝ only when x is an integer.
35. For any real number x, n Ÿ x Ÿ n ", where n is an integer. Now: n Ÿ x Ÿ n " Ê Ðn "Ñ Ÿ x Ÿ n. By
definition: ÜxÝ œ n and ÚxÛ œ n Ê ÚxÛ œ n. So ÜxÝ œ ÚxÛ for all x − d .
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Section 1.3 Functions and Their Graphs
21
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.
37. v œ f(x) œ xÐ"% 2xÑÐ22 2xÑ œ %x$ 72x# $!)x; ! x 7Þ
38. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # AB # œ 2# Ê AB œ È2Þ So,
#
h# "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ " Ê The equation of AB is
y œ f(x) œ B "; x − Ò!ß "Ó.
(b) AÐxÑ œ 2x y œ 2xÐx "Ñ œ 2x# #x; x − Ò!ß "Ó.
39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed.
x
x
(b) r œ )1#
1 œ % #1
(c) h œ È"' r# œ É"' ˆ%
#
x‰
(d) V œ "$ 1 r# h œ "$ 1ˆ )1#
†
1
x ‰#
#1
œ É"' ˆ16
È"'1x x#
#1
œ
4x
1
x# ‰
%1#
œ É 4x
1
x#
%1#
œ É "'%11#x
x#
%1#
œ
È"'1xx#
#1
a)1 xb# È"'1x x#
#%1#
40. (a) Note that 2 mi = 10,560 ft, so there are È)!!# x# feet of river cable at $180 per foot and a"!ß &'! xb feet of land
cable at $100 per foot. The cost is Caxb œ ")!È)!!# x# "!!a"!ß &'! xb.
(b) Ca!b œ $"ß #!!ß !!!
Ca&!!b ¸ $"ß "(&ß )"#
Ca"!!!b ¸ $"ß ")'ß &"#
Ca"&!!b ¸ $"ß #"#ß !!!
Ca#!!!b ¸ $"ß #%$ß ($#
Ca#&!!b ¸ $"ß #()ß %(*
Ca$!!!b ¸ $"ß $"%ß )(!
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the
point P.
41. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, yb lie on the same
vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !,
for any x.
42. Pick 11, for example: "" & œ "' Ä # † "' œ $# Ä $# ' œ #' Ä
faxb œ
#ax&b'
#
#'
#
œ "$ Ä "$ # œ "", the original number.
# œ x, the number you started with.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
22
Chapter 1 Preliminaries
1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS
1. (a) linear, polynomial of degree 1, algebraic.
(c) rational, algebraic.
(b) power, algebraic.
(d) exponential.
2. (a) polynomial of degree 4, algebraic.
(c) algebraic.
(b) exponential.
(d) power, algebraic.
3. (a) rational, algebraic.
(c) trigonometric.
(b) algebraic.
(d) logarithmic.
4. (a) logarithmic.
(c) exponential.
(b) algebraic.
(d) trigonometric.
5. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
6. (a) Graph f because it is linear.
(b) Graph g because it contains a!ß "b.
(c) Graph h because it is a nonlinear odd function.
7. Symmetric about the origin
Dec: _ x _
Inc: nowhere
8. Symmetric about the y-axis
Dec: _ x !
Inc: ! x _
9. Symmetric about the origin
Dec: nowhere
Inc: _ x !
!x_
10. Symmetric about the y-axis
Dec: ! x _
Inc: _ x !
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.4 Identifying Functions; Mathematical Models
11. Symmetric about the y-axis
Dec: _ x Ÿ !
Inc: ! x _
12. No symmetry
Dec: _ x Ÿ !
Inc: nowhere
13. Symmetric about the origin
Dec: nowhere
Inc: _ x _
14. No symmetry
Dec: ! Ÿ x _
Inc: nowhere
15. No symmetry
Dec: ! Ÿ x _
Inc: nowhere
16. No symmetry
Dec: _ x Ÿ !
Inc: nowhere
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
23
24
Chapter 1 Preliminaries
17. Symmetric about the y-axis
Dec: _ x Ÿ !
Inc: ! x _
18. Symmetric about the y-axis
Dec: ! Ÿ x _
Inc: _ x !
19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the
function is even.
20. faxb œ x& œ
"
x&
and faxb œ axb& œ
"
a x b&
œ ˆ x"& ‰ œ faxb. Thus the function is odd.
21. Since faxb œ x# " œ axb# " œ faxb. The function is even.
22. Since Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ and Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ the function is neither even nor
odd.
23. Since gaxb œ x$ x, gaxb œ x$ x œ ax$ xb œ gaxb. So the function is odd.
24. gaxb œ x% $x# " œ axb% $axb# " œ gaxbß thus the function is even.
25. gaxb œ
"
x# "
26. gaxb œ
x
x# " ;
27. hatb œ
"
t ";
œ
"
axb# "
œ gaxb. Thus the function is even.
gaxb œ x#x" œ gaxb. So the function is odd.
h a t b œ
"
t " ;
h at b œ
"
" t.
Since hatb Á hatb and hatb Á hatb, the function is neither even nor odd.
28. Since l t$ | œ l atb$ |, hatb œ hatb and the function is even.
29. hatb œ 2t ", hatb œ 2t ". So hatb Á hatb. hatb œ 2t ", so hatb Á hatb. The function is neither even nor
odd.
30. hatb œ 2l t l " and hatb œ 2l t l " œ 2l t l ". So hatb œ hatb and the function is even.
31. (a)
The graph supports the assumption that y is proportional to x. The
constant of proportionality is estimated from the slope of the
regression line, which is 0.166.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.4 Identifying Functions; Mathematical Models
(b)
25
The graph supports the assumption that y is proportional to x"Î# .
The constant of proportionality is estimated from the slope of the
regression line, which is 2.03.
32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the
regression line.
The graphs support the assumption that y is proportional to $x . The constant of proportionality is estimated from the
slope of the regression line, which is 5.00.
(b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from
the slope of the regression line, which is 2.99.
33. (a) The scatterplot of y œ reaction distance versus x œ speed is
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is
approximately 1.1.
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26
Chapter 1 Preliminaries
(b) Calculate x w œ speed squared. The scatterplot of x w versus y œ braking distance is:
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which
is approximately 0.059.
34. Kepler's 3rd Law is Tadaysb œ !Þ%"R$Î# , R in millions of miles. "Quaoar" is 4 ‚ "!* miles from Earth, or about
4 ‚ "!* *$ ‚ "!' ¸ % ‚ "!* miles from the sun. Let R œ 4000 (millions of miles) and
T œ a!Þ%"ba%!!!b$Î# days ¸ "!$ß (#$ days.
35. (a)
The hypothesis is reasonable.
(b) The constant of proportionality is the slope of the line ¸
(c) y(in.) œ a!Þ)( in./unit massba"$ unit massb œ ""Þ$" in.
36. (a)
)Þ(%" !
"! !
in./unit mass œ !Þ)(% in./unit mass.
(b)
Graph (b) suggests that y œ k x$ is the better model. This graph is more linear than is graph (a).
1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. Df : _ x _, Dg : x
2. Df : x 1
Rf œ Rg : y
0 Ê x
1 Ê Dfbg œ Dfg : x
1, Dg : x 1
0, Rfbg : y È2, Rfg : y
0 Ê x
1. Rf : _ y _, Rg : y
1. Therefore Dfbg œ Dfg : x
0, Rfbg : y
1.
0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1, Rfg : y
0
Section 1.5 Combining Functions; Shifting and Scaling Graphs
3. Df : _ x _, Dg : _ x _ Ê DfÎg : _ x _ since g(x) Á 0 for any x; DgÎf : _ x _
since f(x) Á 0 for any x. Rf : y œ 2, Rg : y 1, RfÎg : 0 y Ÿ 2, RgÎf : y "#
4. Df : _ x _, Dg : x 0 Ê DfÎg : x 0 since g(x) Á 0 for any x
for any x 0. Rf : y œ 1, Rg : y 1, RfÎg : 0 y Ÿ 1, RgÎf : y "
5. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
0; DgÎf : x
0 since f(x) Á 0
f(g(0)) œ f(3) œ 2
g(f(0)) œ g(5) œ 22
f(g(x)) œ f(x# 3) œ x# 3 5 œ x# 2
g(f(x)) œ g(x 5) œ (x 5)# 3 œ x# 10x 22
f(f(5)) œ f(0) œ 5
g(g(2)) œ g(1) œ 2
f(f(x)) œ f(x 5) œ (x 5) 5 œ x 10
g(g(x)) œ g(x# 3) œ (x# 3)# 3 œ x% 6x# 6
6. (a) f ˆg ˆ "# ‰‰ œ f ˆ 23 ‰ œ 3"
(b) g ˆf ˆ "# ‰‰ œ g ˆ "# ‰ œ 2
(c) f(g(x)) œ f ˆ x " 1 ‰ œ
"
x 1
1œ
(d) g(f(x)) œ g(x 1) œ
"
(x1) 1
(e) f(f(2)) œ f(1) œ 0
(f) g(g(2)) œ g ˆ "3 ‰ œ
œ
"
4
3
œ
x
x1
"
x
3
4
(g) f(f(x)) œ f(x 1) œ (x 1) 1 œ x 2
"
(h) g(g(x)) œ g ˆ x " 1 ‰ œ " " 1 œ xx
# (x Á 1 and x Á 2)
x1
#
7. (a) u(v(f(x))) œ u ˆv ˆ "x ‰‰ œ u ˆ x"# ‰ œ 4 ˆ x" ‰ 5 œ x4# 5
(b) u(f(v(x))) œ u af ax# bb œ u ˆ x"# ‰ œ 4 ˆ x"# ‰ 5 œ x4# 5
#
(c) v(u(f(x))) œ v ˆu ˆ "x ‰‰ œ v ˆ4 ˆ x" ‰ 5‰ œ ˆ 4x 5‰
(d) v(f(u(x))) œ v(f(4x 5)) œ v ˆ 4x " 5 ‰ œ ˆ 4x " 5 ‰
(e) f(u(v(x))) œ f au ax# bb œ f a4 ax# b 5b œ
"
4x# 5
(f) f(v(u(x))) œ f(v(4x 5)) œ f a(4x 5)# b œ
8. (a) h(g(f(x))) œ h ˆg ˆÈx‰‰ œ h Š
Èx
4 ‹
#
œ 4Š
"
(4x 5)#
Èx
4 ‹
8 œ Èx 8
(b) h(f(g(x))) œ h ˆf ˆ x4 ‰‰ œ h ˆÈ x4 ‰ œ 4È x4 8 œ 2Èx 8
4È x 8
œ Èx 2
4
È4x 8
È
œ 4
œ x# 2
(c) g(h(f(x))) œ g ˆh ˆÈx‰‰ œ g ˆ4Èx 8‰ œ
(d) g(f(h(x))) œ g(f(4x 8)) œ g ŠÈ4x 8‹
(e) f(g(h(x))) œ f(g(4x 8)) œ f ˆ 4x 4 8 ‰ œ f(x 2) œ Èx 2
(f) f(h(g(x))) œ f ˆh ˆ x ‰‰ œ f ˆ4 ˆ x ‰ 8‰ œ f(x 8) œ Èx 8
4
4
9. (a) y œ f(g(x))
(c) y œ g(g(x))
(e) y œ g(h(f(x)))
(b) y œ j(g(x))
(d) y œ j(j(x))
(f) y œ h(j(f(x)))
10. (a) y œ f(j(x))
(c) y œ h(h(x))
(e) y œ j(g(f(x)))
(b) y œ h(g(x)) œ g(h(x))
(d) y œ f(f(x))
(f) y œ g(f(h(x)))
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27
28
Chapter 1 Preliminaries
g(x)
f(x)
(f ‰ g)(x)
(a)
x7
Èx
Èx 7
(b)
x2
3x
3(x 2) œ 3x 6
(c)
x#
Èx 5
Èx# 5
(d)
x
x1
x
x1
"
x1
"
x
1
11.
(e)
(f)
x
xc1
x
xc1 1
"
x
(b) af‰gbaxb œ
gaxb"
g ax b
œ
x
x (x1)
œx
x
"
x
12. (a) af‰gbaxb œ lgaxbl œ
œ
x
"
lx "l .
x
x"
Ê"
"
g ax b
œ
x
x"
Ê"
x
x"
œ
"
g ax b
Ê
"
x"
œ
"
gaxb ß so
gaxb œ x ".
(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x .
(d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.)
#
The completed table is shown. Note that the absolute value sign in part (d) is optional.
gaxb
faxb
af‰gbaxb
"
"
lxl
x"
lx "l
x"
x
x"
Èx
x#
Èx
x#
x
x"
lxl
lxl
13. (a) fagaxbb œ É 1x 1 œ É 1x x
gafaxbb œ
1
Èx 1
(b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð1, _Ñ
(c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ
14. (a) fagaxbb œ 1 2Èx x
gafaxbb œ 1 kxk
(b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð0, _Ñ
(c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð_, 1Ñ
15. (a) y œ (x 7)#
(b) y œ (x 4)#
16. (a) y œ x# 3
(b) y œ x# 5
17. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
18. (a) y œ (x 1)# 4
(b) y œ (x 2)# 3
(c) y œ (x 4)# 1
(d) y œ (x 2)#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs
19.
20.
21.
22.
23.
24.
25.
26.
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29
30
Chapter 1 Preliminaries
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs
37.
38.
39.
40.
41.
42.
43.
44.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
31
32
Chapter 1 Preliminaries
45.
46.
47.
48.
49. (a) domain: [0ß 2]; range: [#ß $]
(b) domain: [0ß 2]; range: [1ß 0]
(c) domain: [0ß 2]; range: [0ß 2]
(d) domain: [0ß 2]; range: [1ß 0]
(e) domain: [2ß 0]; range: [!ß 1]
(f) domain: [1ß 3]; range: [!ß "]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs
(g) domain: [2ß 0]; range: [!ß "]
(h) domain: [1ß 1]; range: [!ß "]
50. (a) domain: [0ß 4]; range: [3ß 0]
(b) domain: [4ß 0]; range: [!ß $]
(c) domain: [4ß 0]; range: [!ß $]
(d) domain: [4ß 0]; range: ["ß %]
(e) domain: [#ß 4]; range: [3ß 0]
(f) domain: [2ß 2]; range: [3ß 0]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
33
34
Chapter 1 Preliminaries
(g) domain: ["ß 5]; range: [3ß 0]
(h) domain: [0ß 4]; range: [0ß 3]
51. y œ 3x# 3
52. y œ a2xb# 1 œ %x# 1
53. y œ "# ˆ"
54. y œ 1
"‰
x#
"
axÎ$b#
œ
"
#
œ1
"
#x#
*
x#
55. y œ È%x 1
56. y œ 3Èx 1
#
57. y œ É% ˆ x# ‰ œ "# È16 x#
58. y œ "$ È% x#
59. y œ " a3xb$ œ " 27x$
$
60. y œ " ˆ x# ‰ œ "
x$
)
"Î#
"Î#
61. Let y œ È#x " œ faxb and let gaxb œ x"Î# , haxb œ ˆx "# ‰ , iaxb œ È#ˆx "# ‰ , and
"Î#
jaxb œ ’È#ˆx "# ‰ “ œ faBb. The graph of haxb is the graph of gaxb shifted left
"
#
unit; the graph of iaxb is the graph
of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs
62. Let y œ È"
x
#
œ faxbÞ Let gaxb œ axb"Î# , haxb œ ax #b"Î# , and iaxb œ
"
È # a x
#b"Î# œ È"
x
#
35
œ faxbÞ
The graph of gaxb is the graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right
two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#.
63. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax "b3 #.
64. y œ a" Bb$ # œ Òax "b$ a#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax "b$ , iaxb œ ax "b$ a#b, and
jaxb œ Òax "b$ a#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of
haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
36
Chapter 1 Preliminaries
65. Compress the graph of faxb œ
get haxb œ
"
#x
".
66. Let faxb œ
"
x#
and gaxb œ
#
x#
"
x
horizontally by a factor of 2 to get gaxb œ
"œ
"
#
Š B# ‹
"œ
"
#
ŠxÎÈ#‹
"œ
"
#
’Š"ÎÈ#‹B“
"
#x .
Then shift gaxb vertically down 1 unit to
"Þ Since È# ¸ "Þ%, we see that the graph of
faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.
$
$
67. Reflect the graph of y œ faxb œ È
x across the x-axis to get gaxb œ È
x.
68. y œ faxb œ a#xb#Î$ œ Òa"ba#bxÓ#Î$ œ a"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$
compressed horizontally by a factor of 2.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.5 Combining Functions; Shifting and Scaling Graphs
69.
70.
71. *x# #&y# œ ##& Ê
x#
&#
73. $x# ay #b# œ $ Ê
x#
"#
75. $ax "b# #ay #b# œ '
Ê
ax " b #
#
ŠÈ#‹
y a#b‘#
#
ŠÈ$‹
y#
$#
a y #b #
#
ŠÈ$‹
œ"
74. ax "b# #y# œ % Ê
y#
%#
x a"b‘#
##
œ"
#
#
76. 'ˆx $# ‰ *ˆy "# ‰ œ &%
#
œ"
x#
#
È
Š (‹
72. "'x# (y# œ ""# Ê
œ"
Ê
’xˆ $# ‰“
$#
ˆy "# ‰#
#
ŠÈ'‹
œ"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y#
#
ŠÈ#‹
œ"
37
38
77.
Chapter 1 Preliminaries
x#
"'
y#
*
œ " has its center at a!ß !b. Shiftinig 4 units left and 3 units up gives the center at ah, kb œ a%ß $b. So the
equation is
x a4b‘#
4#
ay 3 b #
3#
œ"Ê
ax %b #
4#
a y $b #
3#
œ ". Center, C, is a%ß $b, and major axis, AB, is the segment
from a)ß $b to a!ß $b.
78. The ellipse
x#
%
y#
#&
œ " has center ah, kb œ a!ß !b. Shifting the ellipse 3 units right and 2 units down produces an ellipse
with center at ah, kb œ a$ß #b and an equation
a$ß $b to a$ß (b is the major axis.
ax 3 b#
%
y a#b‘#
#&
œ ". Center, C, is a3ß #b, and AB, the segment from
79. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd
(b) Š gf ‹ (x) œ
(c) ˆ gf ‰ (x) œ
(d)
(e)
(f)
(g)
(h)
(i)
f(x)
g(x)
g(x)
f(x)
œ
œ
f(x)
g(x)
g(x)
f(x)
œ Š gf ‹ (x), odd
œ ˆ gf ‰ (x), odd
f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even
g# (x) œ (g(x))# œ (g(x))# œ g# (x), even
(f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even
(g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even
(f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even
(g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd
80. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.6 Trigonometric Functions
81. (a)
(b)
(c)
(d)
82.
1.6 TRIGONOMETRIC FUNCTIONS
1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m
radians and
51
4
1 ‰
3. ) œ 80° Ê ) œ 80° ˆ 180°
œ
41
9
2. ) œ
s
r
œ
101
8
œ
51
4
1 ‰
(b) s œ r) œ (10)(110°) ˆ 180°
œ
1101
18
œ
551
9
m
ˆ 180°
‰ œ 225°
1
Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
39
40
Chapter 1 Preliminaries
4. d œ 1 meter Ê r œ 50 cm Ê ) œ
5.
1
)
231
0
1
#
s
r
œ
30
50
31
4
"
È2
È"
2
sin )
0
cos )
1
È
#3
"#
tan )
0
È3
0
und.
"
und.
"
È3
und.
0
1
und.
È 2
cot )
1
#
und.
È23
sec )
csc )
0
"
"
0
"
und.
7. cos x œ 45 , tan x œ 34
9. sin x œ
È8
3
, tan x œ È8
"
‰ ¸ 34°
œ 0.6 rad or 0.6 ˆ 180°
1
6.
È2
3#1
)
sin )
"
cos )
!
"
#
tan )
und.
È 3
cot )
!
È"3
sec )
und.
#
csc )
"
È23
2
È5
10. sin x œ
12
13
11. sin x œ È"5 , cos x œ È25
12. cos x œ
13.
14.
15.
1'
È
#3
8. sin x œ
period œ 1
13
, cos x œ
"
È2
&1
'
"
#
È
#3
È"3
"
È"3
È 3
"
È 3
2
È3
È2
È23
#
È2
#
"#
È3
#
"
È5
, tan x œ 12
5
È3
#
, tan x œ
"
È3
period œ 41
16.
period œ 2
17.
period œ 4
18.
period œ 6
period œ 1
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1
%
"
È2
Section 1.6 Trigonometric Functions
19.
20.
period œ 21
period œ 21
21.
22.
period œ 21
period œ 21
23. period œ 1# , symmetric about the origin
24. period œ 1, symmetric about the origin
25. period œ 4, symmetric about the y-axis
26. period œ 41, symmetric about the origin
27. (a) Cos x and sec x are positive in QI and QIV and
negative in QII and QIII. Sec x is undefined when
cos x is 0. The range of sec x is (_ß 1] ["ß _);
the range of cos x is ["ß 1].
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41
42
Chapter 1 Preliminaries
(b) Sin x and csc x are positive in QI and QII and
negative in QIII and QIV. Csc x is undefined when
sin x is 0. The range of csc x is (_ß 1] [1ß _);
the range of sin x is ["ß "].
28. Since cot x œ
"
tan x
, cot x is undefined when tan x œ 0
and is zero when tan x is undefined. As tan x approaches
zero through positive values, cot x approaches infinity.
Also, cot x approaches negative infinity as tan x
approaches zero through negative values.
29. D: _ x _; R: y œ 1, 0, 1
30. D: _ x _; R: y œ 1, 0, 1
31. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x
32. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x
33. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x
34. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x
35. cos (A B) œ cos (A (B)) œ cos A cos (B) sin A sin (B) œ cos A cos B sin A (sin B)
œ cos A cos B sin A sin B
36. sin (A B) œ sin (A (B)) œ sin A cos (B) cos A sin (B) œ sin A cos B cos A (sin B)
œ sin A cos B cos A sin B
37. If B œ A, A B œ 0 Ê cos (A B) œ cos 0 œ 1. Also cos (A B) œ cos (A A) œ cos A cos A sin A sin A
œ cos# A sin# A. Therefore, cos# A sin# A œ 1.
38. If B œ 21, then cos (A 21) œ cos A cos 21 sin A sin 21 œ (cos A)(1) (sin A)(0) œ cos A and
sin (A 21) œ sin A cos 21 cos A sin 21 œ (sin A)(1) (cos A)(0) œ sin A. The result agrees with the
fact that the cosine and sine functions have period 21.
39. cos (1 x) œ cos 1 cos B sin 1 sin x œ (1)(cos x) (0)(sin x) œ cos x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.6 Trigonometric Functions
40. sin (21 x) œ sin 21 cos (x) cos (21) sin (x) œ (0)(cos (x)) (1)(sin (x)) œ sin x
41. sin ˆ 3#1 x‰ œ sin ˆ 3#1 ‰ cos (x) cos ˆ 3#1 ‰ sin (x) œ (1)(cos x) (0)(sin (x)) œ cos x
42. cos ˆ 3#1 x‰ œ cos ˆ 3#1 ‰ cos x sin ˆ 3#1 ‰ sin x œ (0)(cos x) (1)(sin x) œ sin x
œ sin ˆ 14 13 ‰ œ sin
44. cos
111
1#
45. cos
1
12
œ cos ˆ 13 14 ‰ œ cos
46. sin
51
1#
œ sin ˆ 231 14 ‰ œ sin ˆ 231 ‰ cos ˆ 14 ‰ cos ˆ 231 ‰ sin ˆ 14 ‰ œ Š
21 ‰
3
cos
œ cos
È
47. cos#
1
8
œ
1cos ˆ 281 ‰
#
œ
1 # 2
#
49. sin#
1
1#
œ
1cos ˆ 211# ‰
#
œ
1 # 3
#
È
1
3
1
4
1
3
cos
cos
21
3
1
4
1
3
È2
È3
# ‹Š # ‹
71
1#
œ cos ˆ 14
1
4
È2
ˆ"‰
# ‹ #
43. sin
sin
sin
1
4
cos ˆ 14 ‰ sin
œŠ
sin
1
3
21
3
œŠ
Š
È2
ˆ "‰
# ‹ #
sin ˆ 14 ‰ œ ˆ "# ‰ Š
œ
2 È 2
4
48. cos#
1
1#
œ
2 È 3
4
50. sin#
1
8
Š
È2
# ‹
œ
È2
È3
# ‹Š # ‹
Š
1cos ˆ 211# ‰
#
1cos ˆ 281 ‰
#
51. tan (A B) œ
sin (AB)
cos (AB)
œ
sin A cos Bcos A cos B
cos A cos Bsin A sin B
œ
sin A cos B
cos A sin B
cos A cos B cos A cos B
sin A sin B
cos A cos B
cos A cos B cos A cos B
œ
tan Atan B
1tan A tan B
52. tan (A B) œ
sin (AB)
cos (AB)
œ
sin A cos Bcos A cos B
cos A cos Bsin A sin B
œ
sin A cos B
cos A sin B
cos A cos B cos A cos B
sin A sin B
cos A cos B
cos A cos B cos A cos B
œ
tan Atan B
1tan A tan B
È 2 È 6
4
œ
È3
È2
# ‹ Š # ‹
È3
È2
# ‹Š # ‹
œ
È 6 È 2
4
œ
ˆ "# ‰ Š
œ
œ
œ
È
1 # 3
#
È
1 # 2
#
œ
œ
1 È 3
2È 2
È2
# ‹
œ
1 È 3
2È 2
2 È 3
4
2 È 2
4
53. According to the figure in the text, we have the following: By the law of cosines, c# œ a# b# 2ab cos )
œ 1# 1# 2 cos (A B) œ 2 2 cos (A B). By distance formula, c# œ (cos A cos B)# (sin A sin B)#
œ cos# A 2 cos A cos B cos# B sin# A 2 sin A sin B sin# B œ 2 2(cos A cos B sin A sin B). Thus
c# œ 2 2 cos (A B) œ 2 2(cos A cos B sin A sin B) Ê cos (A B) œ cos A cos B sin A sin B.
54. (a) cosaA Bb œ cos A cos B sin A sin B
sin ) œ cosˆ 1# )‰ and cos ) œ sinˆ 1# )‰
Let ) œ A B
sinaA Bb œ cos’ 1# aA Bb“ œ cos’ˆ 1# A‰ B“ œ cos ˆ 1# A‰ cos B sin ˆ 1# A‰ sin B
œ sin A cos B cos A sin B
(b) cosaA Bb œ cos A cos B sin A sin B
cosaA aBbb œ cos A cos aBb sin A sin aBb
Ê cosaA Bb œ cos A cos aBb sin A sin aBb œ cos A cos B sin A asin Bb
œ cos A cos B sin A sin B
Because the cosine function is even and the sine functions is odd.
55. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (60°) œ 4 9 12 cos (60°) œ 13 12 ˆ "# ‰ œ 7.
Thus, c œ È7 ¸ 2.65.
56. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (40°) œ 13 12 cos (40°). Thus, c œ È13 12 cos 40° ¸ 1.951.
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44
Chapter 1 Preliminaries
57. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand,
if C is obtuse (as in the figure on the right), then sin C œ sin (1 C) œ hb . Thus, in either case,
h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B.
a # b # c #
2ab
By the law of cosines, cos C œ
and cos B œ
a # c # b #
.
2ac
Moreover, since the sum of the
interior angles of a triangle is 1, we have sin A œ sin (1 (B C)) œ sin (B C) œ sin B cos C cos B sin C
#
#
#
#
#
#
b c
c b ˆ h ‰
h ‰
œ ˆ hc ‰ ’ a 2ab
a2a# b# c# c# b# b œ
“ ’ a 2ac
“ b œ ˆ 2abc
ah
bc
Ê ah œ bc sin A.
Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives
h
sin A
sin C
sin B
bc œ ðóóóóóóóñóóóóóóóò
a œ c œ b .
law of sines
58. By the law of sines,
Thus sin B œ
3È 3
2È 7
sin A
#
œ
sin B
3
œ
È3/2
c .
By Exercise 55 we know that c œ È7.
¶ 0.982.
59. From the figure at the right and the law of cosines,
b# œ a# 2# 2(2a) cos B
œ a# 4 4a ˆ "# ‰ œ a# 2a 4.
Applying the law of sines to the figure,
Ê
È2/2
a
œ
È3/2
b
sin A
a
œ
sin B
b
Ê b œ É 3# a. Thus, combining results,
a# 2a 4 œ b# œ
3
#
a# Ê 0 œ
"
#
a# 2a 4
Ê 0 œ a# 4a 8. From the quadratic formula and the fact that a 0, we have
aœ
4È4# 4(1)(8)
#
œ
4 È 3 4
#
¶ 1.464.
60. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator
is in radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The
curves look like intersecting straight lines near the origin when the calculator is in degree mode.
61. A œ 2, B œ 21, C œ 1, D œ 1
62. A œ "# , B œ 2, C œ 1, D œ
"
#
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Section 1.6 Trigonometric Functions
63. A œ 12 , B œ 4, C œ 0, D œ
64. A œ
L
21 ,
"
1
B œ L, C œ 0, D œ 0
65. (a) amplitude œ kAk œ 37
(c) right horizontal shift œ C œ 101
(b) period œ kBk œ 365
(d) upward vertical shift œ D œ 25
66. (a) It is highest when the value of the sine is 1 at f(101) œ 37 sin (0) 25 œ 62° F.
The lowest mean daily temp is 37(1) 25 œ 12° F.
(b) The average of the highest and lowest mean daily temperatures œ
The average of the sine function is its horizontal axis, y œ 25.
62°(12)°
#
œ 25° F.
67-70. Example CAS commands:
Maple
f := x -> A*sin((2*Pi/B)*(x-C))+D1;
A:=3; C:=0; D1:=0;
f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )];
plot( f_list, x=-4*Pi..4*Pi, scaling=constrained,
color=[red,blue,green,cyan], linestyle=[1,3,4,7],
legend=["B=1","B=3","B=2*Pi","B=3*Pi"],
title="#67 (Section 1.6)" );
Mathematica
Clear[a, b, c, d, f, x]
f[x_]:=a Sin[21/b (x c)] + d
Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x, 41, 41 }]
67. (a) The graph stretches horizontally.
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45
46
Chapter 1 Preliminaries
(b) The period remains the same: period œ l B l. The graph has a horizontal shift of
"
#
period.
68. (a) The graph is shifted right C units.
(b) The graph is shifted left C units.
(c) A shift of „ one period will produce no apparent shift. l C l œ '
69. The graph shifts upwards l D lunits for D ! and down l D lunits for D !Þ
70. (a) The graph stretches l A l units.
(b) For A !, the graph is inverted.
1.7 GRAPHING WITH CALCULATORS AND COMPUTERS
1-4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and
has little unused space.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.7 Graphing with Calculators and Computers
1. d.
2. c.
3. d.
4. b.
5-30.
For any display there are many appropriate display widows. The graphs given as answers in Exercises 530
are not unique in appearance.
5. Ò2ß 5Ó by Ò15ß 40Ó
6. Ò4ß 4Ó by Ò4ß 4Ó
7. Ò2ß 6Ó by Ò250ß 50Ó
8. Ò1ß 5Ó by Ò5ß 30Ó
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47
48
Chapter 1 Preliminaries
9. Ò4ß 4Ó by Ò5ß 5Ó
10. Ò2ß 2Ó by Ò2ß 8Ó
11. Ò2ß 6Ó by Ò5ß 4Ó
12. Ò4ß 4Ó by Ò8ß 8Ó
13. Ò"ß 'Ó by Ò"ß %Ó
14. Ò"ß 'Ó by Ò"ß &Ó
15. Ò3ß 3Ó by Ò!ß "!Ó
16. Ò"ß #Ó by Ò!ß "Ó
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Section 1.7 Graphing with Calculators and Computers
17. Ò&ß "Ó by Ò&ß &Ó
18. Ò&ß "Ó by Ò#ß %Ó
19. Ò%ß %Ó by Ò!ß $Ó
20. Ò&ß &Ó by Ò#ß #Ó
21. Ò"!ß "!Ó by Ò'ß 'Ó
22. Ò&ß &Ó by Ò#ß #Ó
23. Ò'ß "!Ó by Ò'ß 'Ó
24. Ò$ß &Ó by Ò#ß "!Ó
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49
50
Chapter 1 Preliminaries
25. Ò!Þ!$ß !Þ!$Ó by Ò"Þ#&ß "Þ#&Ó
26. Ò!Þ"ß !Þ"Ó by Ò$ß $Ó
27. Ò$!!ß $!!Ó by Ò"Þ#&ß "Þ#&Ó
28. Ò&!ß &!Ó by Ò!Þ"ß !Þ"Ó
29. Ò!Þ#&ß !Þ#&Ó by Ò!Þ$ß !Þ$Ó
30. Ò!Þ"&ß !Þ"&Ó by Ò!Þ!#ß !Þ!&Ó
31. x# #x œ % %y y# Ê y œ # „ Èx# #x ).
The lower half is produced by graphing
y œ # Èx# #x ).
32. y# "'x# œ " Ê y œ „ È" "'x# . The upper branch
is produced by graphing y œ È" "'x# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 1.7 Graphing with Calculators and Computers
33.
34.
35.
36.
37.
38Þ
39.
40.
41. (a) y œ "!&*Þ"%x #!(%*(#
(b) m œ "!&*Þ"% dollars/year, which is the yearly increase in compensation.
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51
52
Chapter 1 Preliminaries
(c)
(d) Answers may vary slightly. y œ a"!&*Þ14ba#!"!b #!(%*(# œ $&$ß 899
42. (a) Let C œ cost and x œ year.
C œ a(*'!Þ("bx "Þ' ‚ "!(
(b) Slope represents increase in cost per year
(c) C œ a#'$(Þ"%bx &Þ# ‚ "!'
(d) The median price is rising faster in the northeast (the slope is larger).
43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is
d œ !Þ!)''x# "Þ*(x &!Þ".
(b)
(c) From the graph in part (b), the stopping distance is about $(! feet when the vehicle is (# mph and it is about &#& feet
when the speed is )& mph.
Algebraically: dquadratic a(#b œ !Þ!)''a(#b# "Þ*(a(#b &!Þ" œ $'(Þ' ft.
dquadratic a)&b œ !Þ!)''a)&b# "Þ*(a)&b &!Þ" œ &##Þ) ft.
(d) The linear regression function is d œ 'Þ)*x "%!Þ% Ê dlinear a(#b œ 'Þ)*a(#b "%!Þ% œ $&&Þ( ft and
dlinear a)&b œ 'Þ)*a)&b "%!Þ% œ %%&Þ# ft. The linear regression line is shown on the graph in part (b). The quadratic
regression curve clearly gives the better fit.
44. (a) The power regression function is y œ %Þ%%'%(x!Þ&""%"% .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 1 Practice Exercises
(b)
(c) 15Þ2 km/h
(d) The linear regression function is y œ !Þ*"$'(&x %Þ")**(' and it is shown on the graph in part (b). The linear
regession function gives a speed of "%Þ# km/h when y œ "" m. The power regression curve in part (a) better fits the
data.
CHAPTER 1 PRACTICE EXERCISES
1. ( 2x
$ Ê #x
% Ê x
2.
3x "! Ê x "!
$
3.
"
& ax
#
qqqqqqqqðïïïïïïïî
x
"!
$
"b "% ax #b Ê %ax "b &ax #b
Ê %x % &x "! Ê ' x
4.
x$
#
%$ x Ê $ax $b
Ê $x *
5.
#a% xb
) #x Ê &x
"Êx
qqqqqqqqñïïïïïïïî
x
"
&
"
&
lx " l œ ( Ê x " œ ( or ax "b œ ( Ê x œ ' or x œ )
6. ly $ l % Ê % y $ % Ê " y (
7. ¹" x# ¹
$
#
Ê "
x
#
$# or "
x
#
$
#
Ê x# &# or x#
"
#
Ê x & or x "
Ê x & or x "
8. ¹ #x$( ¹ Ÿ & Ê & Ÿ
#x(
$
Ÿ & Ê 1& Ÿ #x ( Ÿ 1& Ê 22 Ÿ #x Ÿ 8 Ê "" Ÿ x Ÿ %
9. Since the particle moved to the y-axis, # ?x œ ! Ê ?x œ 2. Since ?y œ 3?x œ 6, the new coordinates
are (x ?xß y ?y) œ (# #ß & ') œ (0ß 11).
10. (a)
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53
54
Chapter 1 Preliminaries
(b)
line
AB
slope
10 1
9
3
2 8 œ 6 œ #
10 6
4
2
2 (4) œ 6 œ 3
6 (3)
9
3
% 2 œ 6 œ #
1 (3)
4
2
82 œ 6 œ 3
66
œ0
% 14
3
BC
CD
DA
CE
BD
is vertical and has no slope
(c) Yes; A, B, C and D form a parallelogram.
3 ˆ 14
‰
(d) Yes. The line AB has equation y 1 œ 3# (x 8). Replacing x by 14
3 gives y œ # 3 8 "
3 ˆ
10 ‰
14
œ # 3 1 œ 5 1 œ 6. Thus, E ˆ 3 ß 6‰ lies on the line AB and the points A, B and E are collinear.
(e) The line CD has equation y 3 œ 3# (x 2) or y œ 3# x. Thus the line passes through the origin.
11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are
È53, È72 and È65, respectively. The slopes of AB, BC and AC are 7 , 1 and " , respectively.
#
8
12. P(xß 3x 1) is a point on the line y œ 3x 1. If the distance from P to (!ß 0) equals the distance from P to
($ß %), then x# (3x 1)# œ (x 3)# (3 3x)# Ê x# 9x# 6x 1 œ x# 6x 9 9 18x 9x#
23
ˆ 17 ‰
ˆ 17 23 ‰
Ê 18x œ 17 or x œ 17
18 Ê y œ 3x 1 œ 3 18 1 œ 6 . Thus the point is P 18 ß 6 .
13. y œ $ax "b a'b Ê y œ $x *
14. y œ "# ax "b # Ê y œ "# x
$
#
15. x œ !
16. m œ
# '
" a$b
œ
)
%
œ # Ê y œ #ax $b ' Ê y œ #x
17. y œ #
18. m œ
&$
# $
œ
#
&
œ &# Ê y œ &# ax $b $ Ê y œ &# x
#"
&
19. y œ $x $
20. Since #x y œ # is equivalent to y œ #x #, the slope of the given line (and hence the slope of the desired line) is 2.
y œ #a x "b " Ê y œ # x &
21. Since %x $y œ "# is equivalent to y œ %$ x %, the slope of the given line (and hence the slope of the desired line) is
%$ . y œ %$ ax 4b "2 Ê y œ %$ x
#!
$
22. Since $x &y œ " is equivalent to y œ $& x "& , the slope of the given line is
5$ .
yœ
5$ ax
#b $ Ê y œ
5$ x
"*
$
$
&
and the slope of the perpendicular line is
23. Since "# x "$ y œ " is equivalent to y œ $# x $, the slope of the given line is $# and the slope of the perpendicular line
is #$ . y œ #$ ax "b # Ê y œ #$ x
)
$
24. The line passes through a!ß &b and a$ß !b. m œ
! a&b
$!
œ
&
$
Ê y œ $& x &
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Chapter 1 Practice Exercises
25. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ
26. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰
surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰
"Î#
#Î$
C
#1
#
Ê A œ 1ˆ #C1 ‰ œ
C#
%1 .
$ $V
. The volume is V œ %$ 1 r$ Ê r œ É
%1 . Substitution into the formula for
.
27. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies
the equation tan ) œ
28. tan ) œ
rise
run
œ
h
&!!
x#
x
œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.
Ê h œ &!! tan ) ft.
29.
30.
Symmetric about the origin.
Symmetric about the y-axis.
31.
32.
Neither
Symmetric about the y-axis.
33. yaxb œ axb# " œ x# " œ yaxb. Even.
34. yaxb œ axb& axb$ axb œ x& x$ x œ yaxb. Odd.
35. yaxb œ " cosaxb œ " cos x œ yaxb. Even.
36. yaxb œ secaxb tanaxb œ
37. yaxb œ
axb% "
axb$ #axb
œ
x% "
x$ #x
sinaxb
cos# axb
œ
sin x
cos# x
œ sec x tan x œ yaxb. Odd.
%
"
œ xx$ #
x œ yaxb. Odd.
38. yaxb œ " sinaxb œ " sin x. Neither even nor odd.
39. yaxb œ x cosaxb œ x cos x. Neither even nor odd.
40. yaxb œ Éaxb% " œ Èx% " œ yaxb. Even.
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55
56
Chapter 1 Preliminaries
41. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since l x l attains all nonnegative values, the range is Ò#ß _Ñ.
42. (a) Since the square root requires " x !, the domain is Ð_ß "Ó.
(b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ.
43. (a) Since the square root requires "' x#
!, the domain is Ò%ß %Ó.
(b) For values of x in the domain, ! Ÿ "' x# Ÿ "', so ! Ÿ È"' x# Ÿ %. The range is Ò!ß %Ó.
44. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since $#x attains all positive values, the range is a"ß _b.
45. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since #ex attains all positive values, the range is a$ß _b.
46. (a) The function is equivalent to y œ tan #x, so we require #x Á
k1
#
for odd integers k. The domain is given by x Á
k1
%
for
odd integers k.
(b) Since the tangent function attains all values, the range is a_ß _b.
47. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) The sine function attains values from " to ", so # Ÿ #sina$x 1b Ÿ # and hence $ Ÿ #sina$x 1b " Ÿ ". The
range is Ò3ß 1Ó.
48. (a) The function is defined for all values of x, so the domain is a_ß _b.
&
(b) The function is equivalent to y œ È
x# , which attains all nonnegative values. The range is Ò!ß _Ñ.
49. (a) The logarithm requires x $ !, so the domain is a$ß _b.
(b) The logarithm attains all real values, so the range is a_ß _b.
50. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) The cube root attains all real values, so the range is a_ß _b.
51. (a) The function is defined for % Ÿ x Ÿ %, so the domain is Ò%ß %Ó.
(b) The function is equivalent to y œ Èl x l, % Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The
range is Ò!ß #Ó.
52. (a) The function is defined for # Ÿ x Ÿ #, so the domain is Ò#ß #Ó.
(b) The range is Ò"ß "Ó.
53. First piece: Line through a!ß "b and a"ß !b. m œ
Second piece: Line through a"ß "b and a#ß !b. m
faxb œ œ
" x, ! Ÿ x "
# x, " Ÿ x Ÿ #
54. First piece: Line through a!ß !b and a2ß 5b. m œ
Second piece: Line through a2ß 5b and a4ß !b. m
faxb œ
10
5
2 x,
5x
2 ,
!"
"
"! œ " œ
"
"
œ !#
" œ "
" Ê y œ x " œ " x
œ " Ê y œ ax "b " œ x # œ # x
5!
5
5
2! œ 2 Ê y œ 2x
5
5
5
œ !4
2 œ 2 œ 2 Ê
y œ 52 ax 2b 5 œ 52 x 10 œ 10
!Ÿx2
(Note: x œ 2 can be included on either piece.)
2ŸxŸ4
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5x
2
Chapter 1 Practice Exercises
55. (a) af‰gba"b œ faga"bb œ fŠ È"" # ‹ œ fa"b œ
(b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ
(c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ
"
É "# #
"
"Îx
œ
"
È#Þ&
"
"
œ"
or É &#
œ x, x Á !
(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ
"
"
É Èx # #
œ
% x#
È
É " #È x #
$
56. (a) af‰gba"b œ faga"bb œ fˆÈ
" "‰ œ fa!b œ # ! œ #
$
(b) ag‰f ba#b œ faga#bb œ ga# #b œ ga!b œ È
!"œ"
(c) af‰f baxb œ fafaxbb œ fa# xb œ # a# xb œ x
$
$
$
È
(d) ag‰gbaxb œ gagaxbb œ gˆÈ
x "‰ œ É
x""
#
57. (a) af‰gbaxb œ fagaxbb œ fˆÈx #‰ œ # ˆÈx #‰ œ x, x #.
ag‰f baxb œ fagaxbb œ ga# x# b œ Èa# x# b # œ È% x#
(b) Domain of f‰g: Ò#ß _ÑÞ
Domain of g‰f: Ò#ß #ÓÞ
(c) Range of f‰g: Ð_ß #ÓÞ
Range of g‰f: Ò!ß #ÓÞ
%
58. (a) af‰gbaxb œ fagaxbb œ fŠÈ" x‹ œ ÉÈ" x œ È
" x.
ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É" Èx
(b) Domain of f‰g: Ð_ß "ÓÞ
Domain of g‰f: Ò!ß "ÓÞ
(c) Range of f‰g: Ò!ß _ÑÞ
Range of g‰f: Ò!ß "ÓÞ
59.
60.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
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58
Chapter 1 Preliminaries
61.
62.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
It does not change the graph.
63.
64.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
65.
66.
Whenever g" (x) is positive, the graph of y œ g# (x)
œ kg" (x)k is the same as the graph of y œ g" (x).
When g" (x) is negative, the graph of y œ g# (x) is
the reflection of the graph of y œ g" (x) across the
x-axis.
67.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
It does not change the graph.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is
the same as the graph of y œ g" (x). When g" (x) is negative, the
graph of y œ g# (x) is the reflection of the graph of y œ g" (x)
across the x-axis.
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Chapter 1 Practice Exercises
59
68.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is
the same as the graph of y œ g" (x). When g" (x) is negative, the
graph of y œ g# (x) is the reflection of the graph of y œ g" (x)
across the x-axis.
69.
70.
period œ 1
period œ 41
71.
72.
period œ 2
period œ 4
73.
74.
period œ 21
period œ 21
75. (a) sin B œ sin
1
3
œ
b
c
œ
b
#
Ê b œ 2 sin
1
3
œ 2Š
È3
# ‹
œ È3. By the theorem of Pythagoras,
a# b# œ c# Ê a œ Èc# b# œ È4 3 œ 1.
(b) sin B œ sin
1
3
œ
b
c
œ
2
c
Ê cœ
2
sin 13
œ È23 œ
Š ‹
#
4
È3
#
. Thus, a œ Èc# b# œ ÊŠ È43 ‹ (2)# œ É 43 œ
76. (a) sin A œ
a
c
Ê a œ c sin A
(b) tan A œ
a
b
Ê a œ b tan A
77. (a) tan B œ
b
a
Ê aœ
(b) sin A œ
a
c
Ê cœ
b
tan B
a
sin A
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2
È3
.
60
Chapter 1 Preliminaries
78. (a) sin A œ
(c) sin A œ
a
c
a
c
œ
È c # b #
c
79. Let h œ height of vertical pole, and let b and c denote the
distances of points B and C from the base of the pole,
measured along the flatground, respectively. Then,
tan 50° œ hc , tan 35° œ hb , and b c œ 10.
Thus, h œ c tan 50° and h œ b tan 35° œ (c 10) tan 35°
Ê c tan 50° œ (c 10) tan 35°
Ê c (tan 50° tan 35°) œ 10 tan 35°
tan 35°
Ê c œ tan10
50°tan 35° Ê h œ c tan 50°
œ
10 tan 35° tan 50°
tan 50°tan 35°
¸ 16.98 m.
80. Let h œ height of balloon above ground. From the figure at
the right, tan 40° œ ha , tan 70° œ hb , and a b œ 2. Thus,
h œ b tan 70° Ê h œ (2 a) tan 70° and h œ a tan 40°
Ê (2 a) tan 70° œ a tan 40° Ê a(tan 40° tan 70°)
70°
œ 2 tan 70° Ê a œ tan 240°tantan
70° Ê h œ a tan 40°
œ
2 tan 70° tan 40°
tan 40°tan 70°
¸ 1.3 km.
81. (a)
(b) The period appears to be 41.
(c) f(x 41) œ sin (x 41) cos ˆ x#41 ‰ œ sin (x 21) cos ˆ x# 21‰ œ sin x cos
since the period of sine and cosine is 21. Thus, f(x) has period 41.
x
#
82. (a)
(b) D œ (_ß 0) (!ß _); R œ [1ß 1]
(c) f is not periodic. For suppose f has period p. Then f ˆ #"1 kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all
integers k. Choose k so large that
"
#1
kp
"
1
Ê 0
"
(1/21)kp
1. But then
f ˆ #"1 kp‰ œ sin Š (1/#1")kp ‹ 0 which is a contradiction. Thus f has no period, as claimed.
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Chapter 1 Additional and Advanced Exercises
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. (a) The given graph is reflected about the y-axis.
(b) The given graph is reflected about the x-axis.
(c) The given graph is shifted left 1 unit, stretched
vertically by a factor of 2, reflected about the
x-axis, and then shifted upward 1 unit.
2. (a)
(d) The given graph is shifted right 2 units, stretched
vertically by a factor of 3, and then shifted
downward 2 units.
(b)
3. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy
f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)).
4. Yes, there are many such function pairs. For example, if g(x) œ (2x 3)$ and f(x) œ x"Î$ , then
(f ‰ g)(x) œ f(g(x)) œ f a(2x 3)$ b œ a(2x 3)$ b
"Î$
œ 2x 3.
5. If f is odd and defined at x, then f(x) œ f(x). Thus g(x) œ f(x) 2 œ f(x) 2 whereas
g(x) œ (f(x) 2) œ f(x) 2. Then g cannot be odd because g(x) œ g(x) Ê f(x) 2 œ f(x) 2
Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is
even, then g(x) œ f(x) 2 is also even: g(x) œ f(x) 2 œ f(x) 2 œ g(x).
6. If g is odd and g(0) is defined, then g(0) œ g(0) œ g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0.
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62
Chapter 1 Preliminaries
7. For (xß y) in the 1st quadrant, kxk kyk œ 1 x
Í x y œ 1 x Í y œ 1. For (xß y) in the 2nd
quadrant, kxk kyk œ x 1 Í x y œ x 1
Í y œ 2x 1. In the 3rd quadrant, kxk kyk œ x 1
Í x y œ x 1 Í y œ 2x 1. In the 4th
quadrant, kxk kyk œ x 1 Í x (y) œ x 1
Í y œ 1. The graph is given at the right.
8. We use reasoning similar to Exercise 7.
(1) 1st quadrant: y kyk œ x kxk
Í 2y œ 2x Í y œ x.
(2) 2nd quadrant: y kyk œ x kxk
Í 2y œ x (x) œ 0 Í y œ 0.
(3) 3rd quadrant: y kyk œ x kxk
Í y (y) œ x (x) Í 0 œ 0
Ê all points in the 3rd quadrant
satisfy the equation.
(4) 4th quadrant: y kyk œ x kxk
Í y (y) œ 2x Í 0 œ x. Combining
these results we have the graph given at the
right:
9. By the law of sines,
10. By the law of sines,
sin 13
È3
œ
sin 14
4
œ
sin A
a
œ
sin A
a
œ
sin B
b
œ
sin B
b
œ
sin 14
b
Ê bœ
sin B
3
Ê sin B œ
È3 sin (1/4)
sin (1/3)
3
4
11. By the law of cosines, a# œ b# c# 2bc cos A Ê cos A œ
sin
œ
È3 Š È2 ‹
È3
#
œ È2.
#
1
4
œ
b # c # a#
2bc
3
4
Š
œ
12. By the law of cosines, c# œ a# b# 2ab cos C œ 2# 3# (2)(2)(3) cos
È2
# ‹
œ
3È 2
8
2# 3# 2#
2(2)(3)
1
4
.
œ 34 .
œ 4 9 12 Š
È2
# ‹
œ 13 6È2 Ê c œ É13 6È2 , since c 0.
#
a # c # b #
4 # 3 #
œ 2(2)(2)(4)
#ac
È135
3È15
121
256 œ 16 œ 16 .
œ
4169
16
#
4 # 5 #
a # b # c #
œ 2(2)(2)(4)
2ab
È231
25
256 œ 16 .
œ
41625
16
13. By the law of cosines, b# œ a# c# 2ac cos B Ê cos B œ
œ
11
16 .
Since 0 B 1, sin B œ È1 cos# B œ É1
14. By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
5
œ 16
. Since 0 C 1, sin C œ È1 cos# C œ É1
15. (a) sin# x cos# x œ 1 Ê sin# x œ 1 cos# x œ (1 cos x)(1 cos x) Ê (1 cos x) œ
Ê
1cos x
sin x
œ
sin# x
1cos x
sin x
1cos x
(b) Using the definition of the tangent function and the double angle formulas, we have
tan# ˆ x# ‰ œ
sin# ˆ x# ‰
cos# ˆ #x ‰
œ
"cos Š2 Š #x ‹‹
#
"cos Š2 Š #x ‹‹
#
œ
1cos x
1cos x
.
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Chapter 1 Additional and Advanced Exercises
16. The angles labeled # in the accompanying figure are
equal since both angles subtend arc CD. Similarly, the
two angles labeled ! are equal since they both subtend
arc AB. Thus, triangles AED and BEC are similar which
) b
implies ab c œ 2a cos
a c
Ê (a c)(a c) œ b(2a cos ) b)
Ê a# c# œ 2ab cos ) b#
Ê c# œ a# b# 2ab cos ).
17. As in the proof of the law of sines of Section P.5, Exercise 57, ah œ bc sin A œ ab sin C œ ac sin B
Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B.
18. As in Section P.5, Exercise 57, (Area of ABC)# œ
œ
"
4
(base)# (height)# œ
"
4
"
4
a# h # œ
a# b# sin# C
a# b# a" cos# Cb . By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
Thus, (area of ABC)# œ
œ
"
4
"
16
"
4
a# b# a" cos# Cb œ
#
Š4a# b# aa# b# c# b ‹ œ
"
16
"
4
a# b# Œ" Š a
#
b # c#
‹
#ab
#
œ
a# b#
4
a # b # c#
2ab
Š"
#
.
#
aa b c # b
4a# b#
#
‹
ca2ab aa# b# c# bb a2ab aa# b# c# bbd
"
ca(a b)# c# b ac# (a b)# bd œ 16
c((a b) c)((a b) c)(c (a b))(c (a b))d
a
b
c
a
b
c
a
b
c
a
b
c
œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s a)(s b)(s c), where s œ a#bc .
œ
"
16
Therefore, the area of ABC equals Ès(s a)(s b)(s c) .
19. 1.
2.
3.
4.
5.
6.
b c (a c) œ b a, which is positive since a b. Thus, a c b c.
b c (a c) œ b a, which is positive since a b. Thus, a c b c.
c 0 and a b Ê c 0 œ c and b a are positive Ê (b a)c œ bc ac is positive Ê ac bc.
a b and c 0 Ê b a and c are positive Ê (b a)(c) œ ac bc is positive Ê bc ac.
Since a 0, a and "a are positive Ê "a 0.
Since 0 a b, both "a and b" are positive. By (3), a b and "a 0 Ê a ˆ "a ‰ b ˆ "a ‰ or 1 ba
Ê 1 ˆ "b ‰
7.
b
a
b
a
"
a
"
b
0 Ê
"
b
"
a .
"
a
and b" are both negative, i.e.,
0 and b" 0. By (4), a b and
1 Ê 1 ˆ b" ‰ ba ˆ b" ‰ by (4) since b" 0 Ê b" "a .
ab0 Ê
Ê
ˆ b" ‰ by (3) since
"
a
0 Ê b ˆ "a ‰ a ˆ "a ‰
20. (a) If a œ 0, then 0 œ kak kbk Í b Á 0 Í 0 œ kak# kbk# . Since kak# œ kak kak œ ka# k œ a# and
kbk# œ b# we obtain a# b# . If a Á 0 then kak 0 and kak kbk Ê a# b# . On the other hand,
if a# b# then a# œ kak# kbk# œ b# Ê 0 kbk# kak# œ akbk kakb akbk kakb . Since akbk kakb 0
and the product akbk kakb akbk kakb is positive, we must have akbk kakb 0 Ê kbk kak . Thus
kak kbk Í a# b# .
(b) ab Ÿ kabk Ê ab
#
#
2 kabk by Exercise 19(4) above Ê a# 2ab b#
kak œ a# and kbk œ b# . Factoring both sides, (a b)#
#
kak# 2 kak kbk kbk# , since
akak kbkb Ê ka bk
kkak kbkk , by part (a).
21. The fact that ka" a# á an k Ÿ ka" k ka# k á kan k holds for n œ 1 is obvious. It also holds for
n œ 2 by the triangle inequality. We now show it holds for all positive integers n, by induction.
Suppose it holds for n œ k 1: ka" a# á ak k Ÿ ka" k ka# k á kak k (this is the induction
hypothesis). Then ka" a# á ak akb1 k œ kaa" a# á ak b akb1 k Ÿ ka" a# á ak k kakb1 k
(by the triangle inequality) Ÿ ka" k ka# k á kak k kakb1 k (by the induction hypothesis) and the
inequality holds for n œ k 1. Hence it holds for all n by induction.
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Chapter 1 Preliminaries
22. The fact that ka" a# á an k ka" k ka# k á kan k holds for n œ 1 is obvious. It holds for n œ 2
by Exercise 21(b), since ka" a# k œ ka" (a# )k kka" k ka# kk œ kka" k ka# kk ka" k ka# k .
We now show it holds for all positive integers n by induction.
Suppose the inequality holds for n œ k 1. Then ka" a# á ak k ka" k ka# k á kak k (this is
the induction hypothesis). Thus ka" á ak akb1 k œ kaa" á ak b aakb1 bk
kkaa" á ak bk kakb1 kk (by Exercise 21(b)) œ kka" á ak k kakb1 kk ka" á ak k kakb1 k
ka" k ka# k á kak k kakb1 k (by the induction hypothesis). Hence the inequality holds for all
n by induction.
23. If f is even and odd, then f(x) œ f(x) and f(x) œ f(x) Ê f(x) œ f(x) for all x in the domain of f.
Thus 2f(x) œ 0 Ê f(x) œ 0.
f(x) f((x))
œ f(x) #f(x) œ E(x) Ê E
#
even function. Define O(x) œ f(x) E(x) œ f(x) f(x) #f(x) œ f(x) #f(x) . Then
O(x) œ f(x) #f((x)) œ f(x)# f(x) œ Š f(x) #f(x) ‹ œ O(x) Ê O is an odd function
24. (a) As suggested, let E(x) œ
f(x) f(x)
#
Ê E(x) œ
is an
Ê f(x) œ E(x) O(x) is the sum of an even and an odd function.
(b) Part (a) shows that f(x) œ E(x) O(x) is the sum of an even and an odd function. If also
f(x) œ E" (x) O" (x), where E" is even and O" is odd, then f(x) f(x) œ 0 œ aE" (x) O" (x)b
(E(x) O(x)). Thus, E(x) E" (x) œ O" (x) O(x) for all x in the domain of f (which is the same as the
domain of E E" and O O" ). Now (E E" )(x) œ E(x) E" (x) œ E(x) E" (x) (since E and E" are
even) œ (E E" )(x) Ê E E" is even. Likewise, (O" O)(x) œ O" (x) O(x) œ O" (x) (O(x))
(since O and O" are odd) œ (O" (x) O(x)) œ (O" O)(x) Ê O" O is odd. Therefore, E E" and
O" O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is,
E" œ E and O" œ O, so the decomposition of f found in part (a) is unique.
25. y œ ax# bx c œ a Šx# ba x
b#
4a# ‹
b#
4a
c œ a ˆx
b ‰#
2a
b#
4a
c
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift
of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward.
Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the
graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the
right.
If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward
to the right. If b 0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward ?c
units if ?c 0.
26. (a) If a 0, the graph rises to the right of the vertical line x œ b and falls to the left. If a 0, the graph
falls to the right of the line x œ b and rises to the left. If a œ 0, the graph reduces to the horizontal
line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph
becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls
more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right.
(c) Increasing c shifts the graph upward; decreasing c shifts it downward.
27. If m 0, the x-intercept of y œ mx 2 must be negative. If m 0, then the x-intercept exceeds
Ê 0 œ mx 2 and x
"
#
Ê x œ m2
"
#
Ê 0 m 4.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
#
Chapter 1 Additional and Advanced Exercises
28. Each of the triangles pictured has the same base
b œ v?t œ v(1 sec). Moreover, the height of each
triangle is the same value h. Thus "# (base)(height) œ
"
#
bh
œ A" œ A# œ A$ œ á . In conclusion, the object sweeps
out equal areas in each one second interval.
29. (a) By Exercise #95 of Section 1.2, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope
of OP œ
(b)
?y
?x
œ
b/2
a/2
œ
b
a .
b 0
The slope of AB œ 0a œ ba . The line
#
of their slopes is " œ ˆ ba ‰ ˆ ba ‰ œ ba#
segments AB and OP are perpendicular when the product
. Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB
is perpendicular to OP when a œ b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
65
66
Chapter 1 Preliminaries
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 2 LIMITS AND CONTINUITY
2.1 RATES OF CHANGE AND LIMITS
1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x)
approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1.
(b) 1
(c) 0
2. (a) 0
(b) 1
(c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t)
approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0.
3. (a) True
(d) False
(b) True
(e) False
(c) False
(f) True
4. (a) False
(d) True
(b) False
(e) True
(c) True
5.
x
lim
x Ä 0 kx k
x
kx k
does not exist because
x
kx k
œ
x
x
œ 1 if x 0 and
approaches 1. As x approaches 0 from the right,
x
kx k
x
kxk
œ
x
x
œ 1 if x 0. As x approaches 0 from the left,
approaches 1. There is no single number L that all
the function values get arbitrarily close to as x Ä 0.
6. As x approaches 1 from the left, the values of
"
x 1
become increasingly large and negative. As x approaches 1
from the right, the values become increasingly large and positive. There is no one number L that all the
function values get arbitrarily close to as x Ä 1, so lim x" 1 does not exist.
xÄ1
7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function
is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when
x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the
definition of f(x) at x! itself.
8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of
xÄ0
the value f(0) itself.
9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1)
is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5.
xÄ1
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If
lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x),
xÄ1
xÄ1
whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
68
Chapter 2 Limits and Continuity
11. (a) f(x) œ ax# *b/(x 3)
x
3.1
f(x)
6.1
2.9
5.9
x
f(x)
3.01
6.01
3.001
6.001
3.0001
6.0001
3.00001
6.00001
3.000001
6.000001
2.99
5.99
2.999
5.999
2.9999
5.9999
2.99999
5.99999
2.999999
5.999999
The estimate is lim f(x) œ 6.
x Ä $
(b)
(c) f(x) œ
x# 9
x3
œ
(x 3)(x 3)
x3
œ x 3 if x Á 3, and lim (x 3) œ 3 3 œ 6.
x Ä $
12. (a) g(x) œ ax# #b/ Šx È2‹
x
g(x)
1.4
2.81421
1.41
2.82421
1.414
2.82821
1.4142
2.828413
1.41421
2.828423
1.414213
2.828426
(b)
(c) g(x) œ
x# 2
x È2
œ
Šx È2‹ Šx È2‹
Šx È2‹
œ x È2 if x Á È2, and
13. (a) G(x) œ (x 6)/ ax# 4x 12b
x
5.9
5.99
G(x)
.126582 .1251564
x
G(x)
6.1
.123456
6.01
.124843
5.999
.1250156
6.001
.124984
lim
x Ä È#
5.9999
.1250015
6.0001
.124998
Šx È2‹ œ È2 È2 œ 2È2.
5.99999
.1250001
6.00001
.124999
5.999999
.1250000
6.000001
.124999
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.1 Rates of Change and Limits
(b)
(c) G(x) œ
x6
ax# 4x 12b
œ
x6
(x 6)(x 2)
œ
"
x#
14. (a) h(x) œ ax# 2x 3b / ax# 4x 3b
x
2.9
2.99
h(x)
2.052631
2.005025
x
h(x)
3.1
1.952380
3.01
1.995024
"
if x Á 6, and lim
x Ä ' x 2
œ
"
' 2
œ "8 œ 0.125.
2.999
2.000500
2.9999
2.000050
2.99999
2.000005
2.999999
2.0000005
3.001
1.999500
3.0001
1.999950
3.00001
1.999995
3.000001
1.999999
(b)
(c) h(x) œ
x# 2x 3
x# 4x 3
œ
(x 3)(x 1)
(x 3)(x 1)
œ
x1
x1
15. (a) f(x) œ ax# 1b / akxk 1b
x
1.1
1.01
f(x)
2.1
2.01
x
f(x)
.9
1.9
.99
1.99
if x Á 3, and lim
x1
x Ä $ x1
œ
31
31
œ
4
#
œ 2.
1.001
2.001
1.0001
2.0001
1.00001
2.00001
1.000001
2.000001
.999
1.999
.9999
1.9999
.99999
1.99999
.999999
1.999999
(b)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
69
70
Chapter 2 Limits and Continuity
(c) f(x) œ
x# "
kx k 1
(x 1)(x 1)
1
œ (x x1)(x
1)
(x 1)
œ x 1, x 0 and x Á 1
, and lim (1 x) œ 1 (1) œ 2.
x Ä 1
œ 1 x, x 0 and x Á 1
16. (a) F(x) œ ax# 3x 2b / a2 kxkb
x
2.1
2.01
F(x)
1.1
1.01
1.9
.9
x
F(x)
1.99
.99
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999
(b)
(c) F(x) œ
x# 3x 2
2 kx k
(x 2)(x 1)
œ (x 2)(x# x")
2x
17. (a) g()) œ (sin ))/)
)
.1
g())
.998334
,
x 0
, and lim (x 1) œ 2 1 œ 1.
x Ä #
œ x 1, x 0 and x Á 2
.01
.999983
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
.1
.998334
.01
.999983
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
18. (a) G(t) œ (1 cos t)/t#
t
.1
G(t)
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
.1
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
)
g())
lim g()) œ 1
)Ä!
(b)
t
G(t)
lim G(t) œ 0.5
tÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.1 Rates of Change and Limits
(b)
Graph is NOT TO SCALE
19. (a) f(x) œ x"ÎÐ"xÑ
x
.9
f(x)
.348678
x
f(x)
1.1
.385543
.99
.366032
.999
.367695
.9999
.367861
.99999
.367877
.999999
.367879
1.01
.369711
1.001
.368063
1.0001
.367897
1.00001
.367881
1.000001
.367878
lim f(x) ¸ 0.36788
xÄ1
(b)
Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point
(1ß 2.71820).
20. (a) f(x) œ a3x 1b /x
x
.1
f(x)
1.161231
.01
1.104669
.001
1.099215
.0001
1.098672
.00001
1.098618
.000001
1.098612
.1
1.040415
.01
1.092599
.001
1.098009
.0001
1.098551
.00001
1.098606
.000001
1.098611
x
f(x)
lim f(x) ¸ 1.0986
xÄ!
(b)
21. lim 2x œ 2(2) œ 4
22. lim 2x œ 2(0) œ 0
23. lim" (3x 1) œ 3 ˆ "3 ‰ 1 œ 0
24. lim
xÄ#
xÄ
$
xÄ!
1
x Ä 1 3x1
œ
"
3(1) 1
œ #"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
71
72
25.
Chapter 2 Limits and Continuity
lim 3x(2x 1) œ 3(1)(2(1) 1) œ 9
26.
x Ä "
1
#
27. lim1 x sin x œ
xÄ
#
1
#
sin
œ
1
#
28. xlim
Ä1
29. (a)
?f
?x
œ
f(3) f(2)
3#
30. (a)
?g
?x
œ
g(1) g(1)
1 (1)
31. (a)
?h
?t
œ
h ˆ 341 ‰ h ˆ 14 ‰
1
31
4 4
œ
?g
?t
œ
g(1) g(0)
10
(2 1) (2 1)
10
32. (a)
33.
?R
?)
œ
R(2) R(0)
20
34.
?P
?)
œ
P(2) P(1)
21
35. (a)
œ
œ
28 9
1
œ
œ
œ
1 1
2
Q% (18ß 550)
œ
3
3
œ 1
"
1 1
œ
"
1 1
œ
f(1) f(")
1 (1)
œ
20
#
œ1
œ0
(b)
?g
?x
œ
g(0)g(2)
0(2)
œ
04
#
œ 2
(b)
?h
?t
œ
h ˆ 1# ‰ h ˆ 16 ‰
11
#
6
œ
?g
?t
œ
g(1) g(1)
1 (1)
œ
1 1
1
#
œ 14
œ
650 225
20 10
650 375
20 14
650 475
20 16.5
650 550
20 18
Q$ (16.5ß 475)
cos 1
1 1
œ
?f
?x
œ 12
3"
#
(b)
0 È3
1
3
œ
3 È 3
1
(2 1) (2 ")
#1
œ0
œ1
œ22œ0
Slope of PQ œ
Q# (14ß 375)
œ
3(1)#
2(1)1
(b)
(8 16 10)(" % &)
1
Q" (10ß 225)
cos x
1 1
œ
œ 19
È 8 1 È 1
#
Q
3x#
lim
x Ä 1 2x1
?p
?t
œ 42.5 m/sec
œ 45.83 m/sec
œ 50.00 m/sec
œ 50.00 m/sec
(b) At t œ 20, the Cobra was traveling approximately 50 m/sec or 180 km/h.
36. (a)
Slope of PQ œ
Q
Q" (5ß 20)
Q# (7ß 39)
Q$ (8.5ß 58)
Q% (9.5ß 72)
80 20
10 5
80 39
10 7
80 58
10 8.5
80 72
10 9.5
?p
?t
œ 12 m/sec
œ 13.7 m/sec
œ 14.7 m/sec
œ 16 m/sec
(b) Approximately 16 m/sec
37. (a)
(b)
?p
?t
œ
174 62
1994 1992
œ
112
#
œ 56 thousand dollars per year
(c) The average rate of change from 1991 to 1992 is ??pt œ
The average rate of change from 1992 to 1993
is ??pt
œ
62 27
1992 1991
111 62
1993 1992
œ 35 thousand dollars per year.
œ 49 thousand dollars per year.
So, the rate at which profits were changing in 1992 is approximatley "# a35 49b œ 42 thousand dollars per year.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.1 Rates of Change and Limits
38. (a) F(x) œ (x 2)/(x 2)
x
1.2
F(x)
4.0
?F
?x
?F
?x
?F
?x
œ
?g
?x
?g
?x
œ
œ
œ
1.1
3.4
1.01
3.04
1.001
3.004
1.0001
3.0004
1
3
4.0 (3)
œ 5.0;
1.2 1
3.04 (3)
œ 4.04;
1.01 1
3.!!!% (3)
œ 4.!!!%;
1.0001 1
?F
?x
?F
?x
œ
œ
3.4 (3)
œ 4.4;
1.1 1
3.004 (3)
œ 4.!!%;
1.001 1
È
g(2) g(1)
œ #21" ¸ 0.414213
21
È1 h"
g(1 h) g(1)
(1 h) 1 œ
h
?g
?x
œ
g(1.5) g(1)
1.5 1
(b) The rate of change of F(x) at x œ 1 is 4.
39. (a)
œ
œ
È1.5 "
0.5
¸ 0.449489
(b) g(x) œ Èx
1h
È1 h
1.1
1.04880
1.01
1.004987
1.001
1.0004998
1.0001
1.0000499
1.00001
1.000005
1.000001
1.0000005
ŠÈ1 h 1‹ /h
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g(x) at x œ 1 is 0.5.
(d) The calculator gives lim
hÄ!
40. (a) i)
ii)
(b)
f(3) f(2)
32
f(T) f(2)
T#
œ
œ
""
3
#
1
" "
T #
T#
T
f(T)
af(T) f(2)b/aT 2b
œ
œ
"
6
1
È1 h"
h
œ 6"
#TT
T#
2
#T
œ "# .
œ
2.1
0.476190
0.2381
2T
#T(T 2)
œ
2T
#T(2 T)
2.01
0.497512
0.2488
œ #"T , T Á 2
2.001
0.499750
0.2500
2.0001
0.4999750
0.2500
2.00001
0.499997
0.2500
2.000001
0.499999
0.2500
(c) The table indicates the rate of change is 0.25 at t œ 2.
" ‰
(d) lim ˆ #T
œ 4"
TÄ#
41-46. Example CAS commands:
Maple:
f := x -> (x^4 16)/(x 2);
x0 := 2;
plot( f(x), x œ x0-1..x0+1, color œ black,
title œ "Section 2.1, #41(a)" );
limit( f(x), x œ x0 );
In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be
overcome in Maple by entering the function as f := x -> (surd(x+1, 3) 1)/x.
Mathematica: (assigned function and values for x0 and h may vary)
Clear[f, x]
f[x_]:=(x3 x2 5x 3)/(x 1)2
x0= 1; h = 0.1;
Plot[f[x],{x, x0 h, x0 h}]
Limit[f[x], x Ä x0]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
73
74
Chapter 2 Limits and Continuity
2.2 CALCULATING LIMITS USING THE LIMIT LAWS
1.
3.
4.
5.
lim (2x 5) œ 2(7) 5 œ 14 5 œ 9
lim ax# 5x 2b œ (2)# 5(2) 2 œ 4 10 2 œ 4
lim ax$ 2x# 4x 8b œ (2)$ 2(2)# 4(2) 8 œ 8 8 8 8 œ 16
x Ä #
lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8
x3
œ
9.
lim
y Ä & 5 y
y#
23
26
#
y Ä # y 5y 6
œ
5
8
(5)#
5 (5)
œ
y2
10. lim
13.
6.
tÄ'
lim
x Ä # x6
12.
lim (10 3x) œ 10 3(12) œ 10 36 œ 26
x Ä 1#
xÄ#
7.
11.
2.
x Ä (
œ
8.
œ
25
10
œ
22
(2)# 5(#) 6
lim# 3s(2s 1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰ 1‘ œ 2 ˆ 43 1‰ œ
sÄ
$
4
lim
x Ä & x7
œ
4
57
œ
4
#
œ 2
5
#
œ
4
4 10 6
œ
4
#0
œ
"
5
lim 3(2x 1)# œ 3(2(1) 1)# œ 3(3)# œ 27
x Ä "
lim (x 3)"*)% œ (4 3)"*)% œ (1)"*)% œ 1
x Ä %
%
lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16
y Ä $
14. lim (2z 8)"Î$ œ (2(0) 8)"Î$ œ (8)"Î$ œ 2
zÄ!
15. lim
3
œ
3
È3(0) 1 1
œ
3
È1 1
œ
3
2
16. lim
5
œ
5
È5(0) 4 2
œ
5
È4 #
œ
5
4
17. lim
È3h 1 "
h
h Ä ! È3h 1 1
h Ä ! È5h 4 2
hÄ0
œ
3
È" "
œ
È5h 4 2
h
hÄ0
5
È4 2
19. lim
œ
x5
#
x Ä & x 25
20.
21.
œ lim
a3h "b 1
È5h 4 2
h
hÄ0
†
È5h 4 2
È5h 4 2
œ lim
a5h 4b 4
h Ä 0 hŠÈ3h 1 "‹
œ lim
3h
œ lim
5h
h Ä 0 hŠÈ3h 1 "‹
œ lim
3
h Ä 0 È3h1"
h Ä 0 hŠÈ5h 4 2‹
h Ä 0 hŠÈ5h 4 2‹
œ lim
5
4
x5
œ lim
x Ä & (x 5)(x 5)
x3
lim
È3h 1 1
È3h 1 1
œ lim
lim
#
x Ä $ x 4x 3
x Ä &
†
hÄ0
$
#
18. lim
œ
È3h 1 "
h
œ lim
x# 3x "0
x5
œ lim
œ lim
x3
x Ä $ (x 3)(x 1)
œ lim
x Ä &
1
x Ä & x5
(x 5)(x 2)
x5
œ
œ lim
"
55
œ
"
10
1
œ
"
3 1
x Ä $ x 1
œ "2
œ lim (x 2) œ & # œ 7
x Ä &
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5
h Ä 0 È5h 4 2
2
3
Section 2.2 Calculating Limits Using the Limit Laws
(x 5)(x 2)
x2
22. lim
x# 7x "0
x#
œ lim
23. lim
t# t 2
t# 1
t Ä " (t 1)(t 1)
xÄ#
tÄ"
t# 3t 2
lim
#
t Ä " t t 2
25.
lim
$
#
x Ä # x 2x
2x 4
5y$ 8y#
u% "
$
u Ä 1 u 1
œ lim
y# (5y 8)
œ lim
œ lim
4x x#
œ lim
x Ä 1 Èx 3 2
lim
x Ä "
œ lim
xÄ%
œ
x Ä "
35.
x2
x Ä 2 È x # 5 3
œ lim
œ
lim
"
œ lim
x ˆ2 È x ‰ ˆ 2 È x ‰
2 Èx
œ lim
xÄ1
x1
œ lim
x2
x Ä 2 Èx# 12 4
œ lim
x Ä 2
(x 3) Š2 Èx# 5‹
9 x#
œ
lim
12
32
œ
3
8
"
6
œ lim x ˆ2 Èx‰ œ 4(2 2) œ 16
xÄ%
(x 1) ˆÈx 3 #‰
(x 3) 4
2
33
œ lim ŠÈx 3 #‹
xÄ1
œ "3
ax# 12b 16
x Ä 2 (x 2) ŠÈx# 12 4‹
œ lim
4
È16 4
œ
œ lim
"
2
ax 2b ŠÈx# 5 3‹
ax # 5 b 9
x Ä 2
œ
Š2 Èx# 5‹ Š2 Èx# 5‹
x Ä 3
x Ä 3 (x 3) Š2 Èx# 5‹
È x# 5 3
x2
œ
œ
4
3
ax # 8 b *
x Ä 1 (x 1) ŠÈx# 8 $‹
ax 2b ŠÈx# 5 3‹
œ lim
444
(4)(8)
œ
œ
œ lim
œ
œ
(1 1)(1 1)
111
"
È9 3
œ
x Ä * Èx 3
x Ä 2 ŠÈx# 5 3‹ ŠÈx# 5 3‹
(x 2)(x 2)
œ
v# 2v 4
(v
2) av# 4b
vÄ#
(x 2) ŠÈx# 12 4‹
œ lim
œ #"
œ lim
ŠÈx# 12 4‹ ŠÈx# 12 4‹
xÄ2
œ 13
au# "b (u 1)
u# u 1
x Ä 1 È x # ) $
ax 2b ŠÈx# 5 3‹
2 È x# 5
x3
x Ä 3
lim
uÄ1
œ lim
(x 2)(x 2)
lim
x Ä 2
8
16
(x 1) ŠÈx# 8 $‹
x Ä 2 (x 2) ŠÈx# 12 4‹
lim
œ
ŠÈx# 8 $‹ ŠÈx# 8 $‹
lim
(x 1)(x 1)
Èx# 12 4
x2
xÄ2
œ
5y 8
œ È4 2 œ 4
33. lim
34.
œ 21
x Ä 1 ˆÈ x 3 # ‰ ˆ È x 3 # ‰
x Ä 1 (x 1) ŠÈx# ) $‹
œ lim
2
4
(x 1) ˆÈx 3 2‰
œ lim
È x# 8 3
x1
œ lim
x(4 x)
x Ä % 2 Èx
x1
31. lim
Èx 3
x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰
x Ä % 2 Èx
œ
2
œ lim
(v 2) av# 2v 4b
(v
2)(v 2) av# 4b
vÄ#
3
#
1 2
1 2
#
y Ä ! 3y 16
au# "b (u 1)(u 1)
au# u 1b (u 1)
œ
œ
#
x Ä # x
#
#
y Ä ! y a3y 16b
Èx 3
x9
t2
t Ä " t 2
œ lim
uÄ1
12
11
œ
œ lim
2(x 2)
œ lim
30. lim
t2
œ lim
v$ 8
% 16
v
vÄ#
xÄ*
32.
t Ä " (t 2)(t 1)
œ lim
28. lim
29. lim
xÄ#
t Ä " t1
(t 2)(t 1)
œ lim
œ lim (x 5) œ 2 5 œ 3
œ lim
#
x Ä # x (x 2)
%
#
y Ä 0 3y 16y
27. lim
(t 2)(t 1)
œ lim
24.
26. lim
xÄ#
È9 3
4
œ 23
œ lim
(3 x)(3 x)
4 ax # 5 b
x Ä 3 (x 3) Š2 Èx# 5‹
x Ä 3 (x 3) Š2 Èx# 5‹
œ lim
3x
x Ä 3 2 È x # 5
œ
6
2 È4
œ
3
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
75
76
Chapter 2 Limits and Continuity
4x
x Ä 4 5 È x# 9
œ lim
a4 xb Š5 Èx# 9‹
œ lim
36. lim
x Ä 4 Š5 Èx# 9‹ Š5 Èx# 9‹
a4 xb Š5 Èx# 9‹
xÄ4
16 x#
œ lim
(4 x)(4 x)
xÄ4
25 ax# 9b
xÄ4
a4 xb Š5 Èx# 9‹
œ lim
a4 xb Š5 Èx# 9‹
5 È x# 9
4x
œ lim
xÄ4
œ
5 È25
8
œ
5
4
37. (a) quotient rule
(b) difference and power rules
(c) sum and constant multiple rules
38. (a) quotient rule
(b) power and product rules
(c) difference and constant multiple rules
39. (a) xlim
f(x) g(x) œ ’xlim
f(x)“ ’ x lim
g(x)“ œ (5)(2) œ 10
Äc
Äc
Äc
(b) xlim
2f(x) g(x) œ 2 ’xlim
f(x)“ ’ xlim
g(x)“ œ 2(5)(2) œ 20
Äc
Äc
Äc
(c) xlim
[f(x) 3g(x)] œ xlim
f(x) 3 xlim
g(x) œ 5 3(2) œ 1
Äc
Äc
Äc
lim
f(x)
f(x)
5
5
xÄc
(d) xlim
œ lim f(x)
lim g(x) œ 5(2) œ 7
Ä c f(x) g(x)
x
40. (a)
(b)
(c)
(d)
41. (a)
(b)
(c)
(d)
42. (a)
(b)
(c)
Äc
Äc
lim [g(x) 3] œ lim g(x) lim 3 œ $ $ œ !
xÄ%
xÄ%
xÄ%
lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0
xÄ%
xÄ%
#
xÄ%
#
lim [g(x)] œ ’ lim g(x)“ œ [3]# œ 9
xÄ%
g(x)
x Ä % f(x) 1
lim
xÄ%
œ
Ä%
lim g(x)
x
lim f(x) lim 1
xÄ%
xÄ%
œ
3
01
œ3
lim [f(x) g(x)] œ lim f(x) lim g(x) œ 7 (3) œ 4
xÄb
xÄb
xÄb
lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(3) œ 21
xÄb
xÄb
xÄb
lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(3) œ 12
xÄb
xÄb
xÄb
lim f(x)/g(x) œ lim f(x)/ lim g(x) œ
xÄb
xÄb
xÄb
7
3
œ 73
lim [p(x) r(x) s(x)] œ lim p(x) lim r(x) lim s(x) œ 4 0 (3) œ 1
x Ä #
x Ä #
x Ä #
x Ä #
lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0
x Ä #
x Ä #
(1 h)# 1#
h
hÄ!
œ lim
hÄ!
(2 h)# (2)#
h
45. lim
[3(2 h) 4] [3(2) 4]
h
hÄ!
x Ä #
x Ä #
44. lim
hÄ!
x Ä #
lim [4p(x) 5r(x)]/s(x) œ ’4 lim p(x) 5 lim r(x)“ ‚ lim s(x) œ [4(4) 5(0)]/3 œ
x Ä #
43. lim
"‰
ˆ #" h ‰ ˆ #
h
hÄ!
46. lim
x
1 2h h# 1
h
œ lim
hÄ!
œ lim
hÄ!
44hh# 4
h
œ lim
hÄ!
œ lim
3h
hÄ! h
2
2 h "
2h
œ lim
x Ä #
h(2 h)
h
hÄ!
x Ä #
œ lim (2 h) œ 2
h(h 4)
h
hÄ!
œ lim (h 4) œ 4
hÄ!
œ3
œ lim
hÄ!
2 (2 h)
2h(# h)
œ lim
h
h Ä ! h(4 2h)
œ "4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"6
3
Section 2.2 Calculating Limits Using the Limit Laws
È7 h È7
h
hÄ!
47. lim
œ lim
ŠÈ7 h È7‹ ŠÈ7 h È7‹
œ lim
h ŠÈ7 h È7‹
hÄ!
h
h Ä ! h ŠÈ7hÈ7‹
È3(0 h) 1 È3(0) 1
h
hÄ!
3h
h Ä ! h ŠÈ3h 1 "‹
œ
h Ä ! È 7 h È 7
48. lim
œ lim
"
œ lim
œ lim
"
#È 7
ŠÈ3h 1 "‹ ŠÈ3h 1 "‹
œ lim
h ŠÈ3h 1 "‹
hÄ!
œ lim
œ
3
h Ä ! È3h 1 1
(7 h) 7
h Ä ! h ŠÈ7 h È7‹
(3h 1) "
œ lim
h Ä ! h ŠÈ3h 1 1 ‹
3
#
49. lim È5 2x# œ È5 2(0)# œ È5 and lim È5 x# œ È5 (0)# œ È5; by the sandwich theorem,
xÄ!
xÄ!
lim f(x) œ È5
xÄ!
50. lim a2 x# b œ 2 0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2
xÄ!
51. (a)
xÄ!
lim Š1
xÄ!
x#
6‹
œ1
0
6
xÄ!
œ 1 and lim 1 œ 1; by the sandwich theorem, lim
(b) For x Á 0, y œ (x sin x)/(2 2 cos x)
lies between the other two graphs in the
figure, and the graphs converge as x Ä 0.
52. (a)
lim Š "#
xÄ!
lim
xÄ!
1cos x
x#
x#
24 ‹
œ lim
1
xÄ! #
lim
x#
x Ä ! #4
œ
"
#
x sin x
x Ä ! 22 cos x
xÄ!
0œ
"
#
and lim
"
xÄ! #
œ1
œ "# ; by the sandwich theorem,
œ "# .
(b) For all x Á 0, the graph of f(x) œ (1 cos x)/x#
lies between the line y œ "# and the parabola
yœ
"
#
x# /24, and the graphs converge as x Ä 0.
53. xlim
f(x) exists at those points c where xlim
x% œ xlim
x# . Thus, c% œ c# Ê c# a1 c# b œ 0
Äc
Äc
Äc
Ê c œ 0, 1, or 1. Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1.
xÄ!
xÄ!
x Ä 1
xÄ1
54. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the
conditions of the sandwich theorem are satisfied, lim f(x) œ 5 Á 0.
xÄ#
55. 1 œ lim
xÄ%
f(x)5
x 2
lim f(x) lim 5
%
xÄ%
œ xÄlim
x lim 2 œ
xÄ%
xÄ%
lim f(x) 5
xÄ%
%#
Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7.
xÄ%
xÄ%
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
77
78
Chapter 2 Limits and Continuity
56. (a) 1 œ lim
f(x)
x#
lim f(x)
lim f(x)
#
xÄ#
œ xÄlim
Ê
%
x# œ
(b) 1 œ lim
f(x)
x#
œ ’ lim
x Ä #
x Ä #
xÄ#
f(x)
lim x" “
x “ ’x Ä
#
x Ä #
57. (a) 0 œ 3 † 0 œ ’ lim
xÄ#
lim f(x) œ 4.
x Ä #
œ ’ lim
x Ä #
f(x) ˆ " ‰
x “ #
Ê
lim
x Ä #
f(x)
x
œ 2.
f(x) 5
x # “ ’xlim
Ä#
5
(x 2)“ œ lim ’Š f(x)
x # ‹ (x 2)“ œ lim [f(x) 5] œ lim f(x) 5
f(x) 5
x # “ ’xlim
Ä#
(x 2)“ Ê lim f(x) œ 5 as in part (a).
xÄ#
Ê lim f(x) œ 5.
xÄ#
xÄ#
xÄ#
(b) 0 œ 4 † 0 œ ’ lim
xÄ#
58. (a) 0 œ 1 † 0 œ ’ lim
f(x)
# “ ’ lim
xÄ! x
xÄ!
(b) 0 œ 1 † 0 œ
59. (a)
lim x sin
xÄ!
(b) 1 Ÿ sin
60. (a)
"
x
’ lim f(x)
# “ ’ lim
xÄ! x
xÄ!
"
x
xÄ#
#
x“ œ ’ lim
f(x)
#
xÄ! x
x“ œ
lim ’ f(x)
x#
xÄ!
#
“ ’ lim x# “ œ lim ’ f(x)
x# † x “ œ lim f(x). That is, lim f(x) œ 0.
xÄ!
† x“ œ
xÄ!
lim f(x) .
xÄ! x
That is,
xÄ!
lim f(x)
xÄ! x
œ 0.
œ0
Ÿ 1 for x Á 0:
x 0 Ê x Ÿ x sin
"
x
Ÿ x Ê lim x sin
"
x
œ 0 by the sandwich theorem;
x 0 Ê x
"
x
x Ê lim x sin
"
x
œ 0 by the sandwich theorem.
x sin
xÄ!
xÄ!
lim x# cos ˆ x"$ ‰ œ 0
xÄ!
(b) 1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich
theorem since lim x# œ 0.
xÄ!
xÄ!
2.3 PRECISE DEFINITION OF A LIMIT
1.
Step 1:
Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5
$ 5 œ 7 Ê $ œ 2, or $ 5 œ 1 Ê $ œ 4.
The value of $ which assures kx 5k $ Ê 1 x 7 is the smaller value, $ œ 2.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
xÄ!
Section 2.3 Precise Definition of a Limit
2.
Step 1:
Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ 2
$ 2 œ 1 Ê $ œ 1, or $ 2 œ 7 Ê $ œ 5.
The value of $ which assures kx 2k $ Ê 1 x 7 is the smaller value, $ œ 1.
Step 1:
Step 2:
kx (3)k $ Ê $ x $ $ Ê $ 3 x $ 3
$ 3 œ 7# Ê $ œ "# , or $ $ œ "# Ê $ œ 5# .
3.
The value of $ which assures kx (3)k $ Ê 7# x "# is the smaller value, $ œ "# .
4.
Step 1:
¸x ˆ 3# ‰¸ $ Ê $ x
Step 2:
$
Step 1:
¸x "# ¸ $ Ê $ x
Step 2:
$
œ
3
#
$ Ê $
"
#
3
#
x$
3
#
Ê $ œ #, or $ œ Ê $ œ 1.
The value of $ which assures ¸x ˆ 3# ‰¸ $ Ê 7# x "# is the smaller value, $ œ ".
3
#
7
#
3
#
5.
"
#
$ Ê $
"
#
x$
"
or $ #" œ 47 Ê $ œ 14
.
"
4
The value of $ which assures ¸x # ¸ $ Ê 9 x
œ
4
9
Ê $œ
"
18 ,
"
#
4
7
"
#
is the smaller value, $ œ
"
18 .
6.
Step 1:
Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3
$ $ œ 2.7591 Ê $ œ 0.2409, or $ $ œ 3.2391 Ê $ œ 0.2391.
The value of $ which assures kx 3k $ Ê 2.7591 x 3.2391 is the smaller value, $ œ 0.2391.
7. Step 1:
Step 2:
kx 5k $ Ê $ x 5 $ Ê $ 5 x $ 5
From the graph, $ 5 œ 4.9 Ê $ œ 0.1, or $ 5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.
8. Step 1:
Step 2:
kx (3)k $ Ê $ x 3 $ Ê $ 3 x $ 3
From the graph, $ 3 œ 3.1 Ê $ œ 0.1, or $ 3 œ 2.9 Ê $ œ 0.1; thus $ œ 0.1.
9. Step 1:
Step 2:
kx 1k $ Ê $ x 1 $ Ê $ 1 x $ 1
9
7
From the graph, $ 1 œ 16
Ê $ œ 16
, or $ 1 œ 25
16 Ê $ œ
10. Step 1:
Step 2:
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3
From the graph, $ 3 œ 2.61 Ê $ œ 0.39, or $ 3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.
11. Step 1:
kx 2k $ Ê $ x 2 $ Ê $ 2 x $ 2
From the graph, $ 2 œ È3 Ê $ œ 2 È3 ¸ 0.2679, or $ 2 œ È5 Ê $ œ È5 2 ¸ 0.2361;
thus $ œ È5 2.
Step 2:
9
16 ;
thus $ œ
7
16 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
79
80
Chapter 2 Limits and Continuity
12. Step 1:
Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1
From the graph, $ 1 œ
thus $ œ
È5 2
# .
È5
#
Ê $œ
È5 2
#
¸ 0.1180, or $ 1 œ
13. Step 1:
Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1
7
16
From the graph, $ 1 œ 16
9 Ê $ œ 9 ¸ 0.77, or $ 1 œ 25 Ê
14. Step 1:
¸x "# ¸ $ Ê $ x
Step 2:
From the graph, $
thus $ œ 0.00248.
"
#
œ
"
#
1
2.01
$ Ê $
Ê $œ
1
2
"
#
x$
"
#.01
"
#
¸ 0.00248, or $
"
#
œ
È3
#
9
25
Ê $œ
2 È3
#
œ 0.36; thus $ œ
1
1.99
Ê $œ
1
1.99
¸ 0.1340;
9
25
"
#
œ 0.36.
¸ 0.00251;
15. Step 1:
Step 2:
k(x 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4 Ê $ œ 0.01.
16. Step 1:
k(2x 2) (6)k 0.02 Ê k2x 4k 0.02 Ê 0.02 2x 4 0.02 Ê 4.02 2x 3.98
Ê 2.01 x 1.99
kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2 Ê $ œ 0.01.
Step 2:
17. Step 1:
Step 2:
18. Step 1:
Step 2:
¹Èx 1 "¹ 0.1 Ê 0.1 Èx 1 " 0.1 Ê 0.9 Èx 1 1.1 Ê 0.81 x 1 1.21
Ê 0.19 x 0.21
kx 0k $ Ê $ x $ . Then, $ œ !Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19.
¸Èx "# ¸ 0.1 Ê 0.1 Èx "# 0.1 Ê 0.4 Èx 0.6 Ê 0.16 x 0.36
¸x "4 ¸ $ Ê $ x 4" $ Ê $ 4" B $ 4" .
Then, $
19. Step 1:
Step 2:
20. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
Step 2:
"
4
œ 0.16 Ê $ œ 0.09 or $
"
4
œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.
¹È19 x $¹ " Ê " È19 x $ 1 Ê 2 È19 x % Ê 4 19 x 16
Ê % x 19 16 Ê 15 x 3 or 3 x 15
kx 10k $ Ê $ x 10 $ Ê $ 10 x $ 10.
Then $ 10 œ 3 Ê $ œ 7, or $ 10 œ 15 Ê $ œ 5; thus $ œ 5.
¹Èx 7 4¹ 1 Ê " Èx 7 % 1 Ê 3 Èx 7 5 Ê 9 x 7 25 Ê 16 x 32
kx 23k $ Ê $ x 23 $ Ê $ 23 x $ 23.
Then $ 23 œ 16 Ê $ œ 7, or $ 23 œ 32 Ê $ œ 9; thus $ œ 7.
¸ "x 4" ¸ 0.05 Ê 0.05
"
x
"
4
0.05 Ê 0.2
"
x
0.3 Ê
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4.
2
2
Then $ % œ 10
3 or $ œ 3 , or $ 4 œ 5 or $ œ 1; thus $ œ 3 .
10
#
x
10
3
or
10
3
x 5.
kx# 3k !.1 Ê 0.1 x# 3 0.1 Ê 2.9 x# 3.1 Ê È2.9 x È3.1
¹x È3¹ $ Ê $ x È3 $ Ê $ È3 x $ È3.
Then $ È3 œ È2.9 Ê $ œ È3 È2.9 ¸ 0.0291, or $ È3 œ È3.1 Ê $ œ È3.1 È3 ¸ 0.0286;
thus $ œ 0.0286.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.3 Precise Definition of a Limit
23. Step 1:
Step 2:
81
kx# 4k 0.5 Ê 0.5 x# 4 0.5 Ê 3.5 x# 4.5 Ê È3.5 kxk È4.5 Ê È4.5 x È3.5,
for x near 2.
kx (2)k $ Ê $ x 2 $ Ê $ # x $ 2.
Then $ # œ È4.5 Ê $ œ È4.5 # ¸ 0.1213, or $ # œ È3.5 Ê $ œ # È3.5 ¸ 0.1292;
thus $ œ È4.5 2 ¸ 0.12.
24. Step 1:
Step 2:
25. Step 1:
Step 2:
¸ "x (1)¸ 0.1 Ê 0.1
"
x
11
1 0.1 Ê 10
"
x
9
10
10
10
10
Ê 10
11 x 9 or 9 x 11 .
kx (1)k $ Ê $ x 1 $ Ê $ " x $ ".
"
10
"
Then $ " œ 10
9 Ê $ œ 9 , or $ " œ 11 Ê $ œ 11 ; thus $ œ
"
11 .
kax# 5b 11k " Ê kx# 16k 1 Ê " x# 16 1 Ê 15 x# 17 Ê È15 x È17.
kx 4k $ Ê $ x 4 $ Ê $ % x $ %.
Then $ % œ È15 Ê $ œ % È15 ¸ 0.1270, or $ % œ È17 Ê $ œ È17 % ¸ 0.1231;
thus $ œ È17 4 ¸ 0.12.
26. Step 1:
Step 2:
27. Step 1:
Step 2:
28. Step 1:
Step 2:
29. Step 1:
Step 2:
¸ 120
¸
x 5 " Ê "
Step 2:
&1 Ê 4
120
x
6 Ê
"
4
x
120
"
6
Ê 30 x 20 or 20 x 30.
kx 24k $ Ê $ x 24 $ Ê $ 24 x $ 24.
Then $ 24 œ 20 Ê $ œ 4, or $ 24 œ 30 Ê $ œ 6; thus Ê $ œ 4.
kmx 2mk 0.03 Ê 0.03 mx 2m 0.03 Ê 0.03 2m mx 0.03 2m Ê
0.03
2 0.03
m x2 m .
kx 2k $ Ê $ x 2 $ Ê $ # x $ #.
0.03
0.03
Then $ # œ # 0.03
m Ê $ œ m , or $ # œ # m Ê $ œ
0.03
m .
In either case, $ œ
kmx 3mk c Ê c mx 3m c Ê c 3m mx c 3m Ê 3
kx 3k $ Ê $ x 3 $ Ê $ $ B $ $.
Then $ $ œ $ mc Ê $ œ mc , or $ $ œ $ mc Ê $ œ
¸(mx b) ˆ m# b‰¸ - Ê c mx m# c Ê c
¸x "# ¸ $ Ê $ x "# $ Ê $ "# x $ "# .
Then $
30. Step 1:
120
x
"
#
œ
"
#
c
m
Ê $œ
c
m,
or $
"
#
œ
"
#
c
m
c
m.
m
#
Ê $œ
c
m
x 3
In either case, $ œ
c
m.
In either case, $ œ
c
m.
m
#
Ê
c
m
c
m.
"
#
mx c
0.03
m .
c
m
x
"
#
c
m.
k(mx b) (m b)k 0.05 Ê 0.05 mx m 0.05 Ê 0.05 m mx 0.05 m
0.05
Ê 1 0.05
m x" m .
kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
0.05
0.05
Then $ " œ " 0.05
m Ê $ œ m , or $ " œ " m Ê $ œ
0.05
m .
In either case, $ œ
0.05
m .
31. lim (3 2x) œ 3 2(3) œ 3
xÄ3
Step 1:
Step 2:
32.
ka3 2xb (3)k 0.02 Ê 0.02 6 2x 0.02 Ê 6.02 2x 5.98 Ê 3.01 x 2.99 or
2.99 x 3.01.
0 k x 3k $ Ê $ x 3 $ Ê $ $ x $ $ .
Then $ $ œ 2.99 Ê $ œ 0.01, or $ $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.
lim (3x #) œ (3)(1) 2 œ 1
x Ä 1
Step 1:
k(3x 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
82
Chapter 2 Limits and Continuity
kx (1)k $ Ê $ x 1 $ Ê $ " x $ 1.
Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.
Step 2:
33. lim
x# 4
x Ä # x#
34.
35.
œ lim
xÄ#
#
(x 2)(x 2)
(x 2)
œ lim (x 2) œ # # œ 4, x Á 2
xÄ#
(x 2)(x 2)
(x 2)
Step 1:
¹Š xx 24 ‹
Step 2:
Ê 1.95 x 2.05, x Á 2.
kx 2k $ Ê $ x 2 $ Ê $ # x $ 2.
Then $ # œ 1.95 Ê $ œ 0.05, or $ # œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.
lim
x Ä &
x# 6x 5
x5
4¹ 0.05 Ê 0.05
œ lim
x Ä &
(x 5)(x 1)
(x 5)
% 0.05 Ê 3.95 x 2 4.05, x Á 2
œ lim (x 1) œ 4, x Á 5.
x Ä &
(x 5)(x ")
(x 5)
Step 1:
#
¹Š x x 6x5 5 ‹
Step 2:
Ê 5.05 x 4.95, x Á 5.
kx (5)k $ Ê $ x 5 $ Ê $ & x $ &.
Then $ & œ 5.05 Ê $ œ 0.05, or $ & œ 4.95 Ê $ œ 0.05; thus $ œ 0.05.
(4)¹ 0.05 Ê 0.05
4 0.05 Ê 4.05 x 1 3.95, x Á 5
lim È1 5x œ È1 5(3) œ È16 œ 4
x Ä $
Step 1:
¹È1 5x 4¹ 0.5 Ê 0.5 È1 5x 4 0.5 Ê 3.5 È1 5x 4.5 Ê 12.25 1 5x 20.25
Step 2:
Ê 11.25 5x 19.25 Ê 3.85 x 2.25.
kx (3)k $ Ê $ x 3 $ Ê $ $ x $ $.
Then $ $ œ 3.85 Ê $ œ 0.85, or $ $ œ 2.25 Ê 0.75; thus $ œ 0.75.
36. lim
4
xÄ# x
œ
4
#
œ2
Step 1:
¸ 4x 2¸ 0.4 Ê 0.4
Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ #.
Then $ # œ 53 Ê $ œ "3 , or $ # œ 5# Ê $ œ "# ; thus $ œ 3" .
4
x
2 0.4 Ê 1.6
4
x
2.4 Ê
10
16
x
4
10
24
Ê
10
4
x
10
6
or
5
3
x 25 .
37. Step 1:
Step 2:
k(9 x) 5k % Ê % 4 x % Ê % 4 x % 4 Ê % % x 4 % Ê % % x 4 %.
kx 4k $ Ê $ x 4 $ Ê $ % x $ %.
Then $ 4 œ % 4 Ê $ œ %, or $ % œ % % Ê $ œ %. Thus choose $ œ %.
38. Step 1:
k(3x 7) 2k % Ê % 3x 9 % Ê 9 % 3x * % Ê 3
Step 2:
39. Step 1:
Step 2:
40. Step 1:
%
3
x 3 3% .
kx 3k $ Ê $ x 3 $ Ê $ 3 x $ 3.
Then $ 3 œ $ 3% Ê $ œ 3% , or $ 3 œ 3 3% Ê $ œ 3% . Thus choose $ œ 3% .
¹Èx 5 2¹ % Ê % Èx 5 # % Ê # % Èx 5 # % Ê (# %)# x 5 (# %)#
Ê (# %)# & x (# %)# 5.
kx 9k $ Ê $ x 9 $ Ê $ 9 x $ 9.
Then $ * œ %# %% * Ê $ œ %% %# , or $ * œ %# %% * Ê $ œ %% %# . Thus choose
the smaller distance, $ œ %% %# .
¹È4 x 2¹ % Ê % È4 x # % Ê # % È4 x # % Ê (# %)# % x (# %)#
Ê (# %)# x 4 (# %)# Ê (# %)# % x (# %)# %.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.3 Precise Definition of a Limit
Step 2:
41. Step 1:
Step 2:
42. Step 1:
Step 2:
43. Step 1:
Step 2:
kx 0k $ Ê $ x $ .
Then $ œ (# %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (# %)# 4 œ 4% %# . Thus choose
the smaller distance, $ œ 4% %# .
For x Á 1, kx# 1k % Ê % x# " % Ê " % x# " % Ê È1 % kxk È1 %
Ê È" % x È1 % near B œ ".
kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ È1 % Ê $ œ " È1 %, or $ 1 œ È" % Ê $ œ È" % 1. Choose
$ œ min š" È1 %ß È1 % "›, that is, the smaller of the two distances.
For x Á 2, kx# 4k % Ê % x# 4 % Ê 4 % x# 4 % Ê È4 % kxk È4 %
Ê È4 % x È4 % near B œ 2.
kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2.
Then $ 2 œ È% % Ê $ œ È% % #, or $ # œ È% % Ê $ œ # È% %. Choose
$ œ min šÈ% % #ß # È% %› .
¸ "x 1¸ % Ê %
"
x
"% Ê "%
"
x
"% Ê
"
1%
%
"%,
"
1%.
x
kx 1k $ Ê $ x 1 $ Ê " $ x " $ .
Then " $ œ " " % Ê $ œ " " " % œ " % % , or " $ œ " " % Ê $ œ
Choose $ œ
44. Step 1:
83
"
"%
"œ
%
"%.
the smaller of the two distances.
¸ x"# "3 ¸ % Ê %
"
x#
"
3
% Ê
"
3
%
"
x#
"
3
% Ê
1 3%
3
"
x#
1 $%
3
Ê
3
" $%
x#
3
" $%
3
È $.
Ê É 1 3 $% kxk É " 3 $% , or É " 3 $% x É "$
% for x near
Step 2:
¹x È3¹ $ Ê $ x È3 $ Ê È3 $ x È3 $ .
Then È3 $ œ É " 3 $% Ê $ œ È3 É " 3 $% , or È3 $ œ É " 3 $% Ê $ œ É " 3 $% È3.
Choose $ œ min šÈ3 É " 3 $% ß É " 3 $% È3›.
45. Step 1:
Step 2:
46. Step 1:
Step 2:
47. Step 1:
#
¹Š xx*3 ‹ (6)¹ % Ê % (x 3) 6 %, x Á 3 Ê % x 3 % Ê % $ x % $.
kx (3)k $ Ê $ x 3 $ Ê $ $ x $ 3.
Then $ $ œ % $ Ê $ œ %, or $ $ œ % $ Ê $ œ %. Choose $ œ %.
#
¹Š xx11 ‹ 2¹ % Ê % (x 1) 2 %, x Á 1 Ê " % x " %.
kx 1k $ Ê $ x 1 $ Ê " $ x " $ .
Then " $ œ " % Ê $ œ %, or " $ œ " % Ê $ œ %. Choose $ œ %.
x 1: l(4 2x) 2l % Ê ! 2 2x % since x 1Þ Thus, 1
x
Step 2:
48. Step 1:
1: l(6x 4) 2l % Ê ! Ÿ 6x 6 % since x
%
#
x !;
1. Thus, " Ÿ x 1 6% .
kx 1k $ Ê $ x 1 $ Ê " $ x 1 $ .
Then 1 $ œ " #% Ê $ œ #% , or " $ œ 1 6% Ê $ œ 6% . Choose $ œ 6% .
x !: k2x 0k % Ê % 2x ! Ê #% x 0;
x 0: ¸ x# !¸ % Ê ! Ÿ x #%.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
84
Chapter 2 Limits and Continuity
Step 2:
kx 0k $ Ê $ x $ .
Then $ œ #% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .
49. By the figure, x Ÿ x sin
"
x
Ÿ x for all x 0 and x
x sin
then by the sandwich theorem, in either case, lim x sin
xÄ!
50. By the figure, x# Ÿ x# sin
"
x
"
x
"
x
x for x 0. Since lim (x) œ lim x œ 0,
xÄ!
œ 0.
xÄ!
Ÿ x# for all x except possibly at x œ 0. Since lim ax# b œ lim x# œ 0, then
by the sandwich theorem, lim x# sin
xÄ!
"
x
xÄ!
œ 0.
xÄ!
51. As x approaches the value 0, the values of g(x) approach k. Thus for every number % 0, there exists a $ !
such that ! kx 0k $ Ê kg(x) kk %.
52. Write x œ h c. Then ! lx cl $ Í $ x c $ , x Á c Í $ ah cb c $ , h c Á c
Í $ h $ , h Á ! Í ! lh !l $ .
Thus, limfaxb œ L Í for any % !, there exists $ ! such that lfaxb Ll % whenever ! lx cl $
xÄc
Í lfah cb Ll % whenever ! lh !l $ Í limfah cb œ L.
hÄ!
53. Let f(x) œ x# . The function values do get closer to 1 as x approaches 0, but lim f(x) œ 0, not 1. The
xÄ!
function f(x) œ x# never gets arbitrarily close to 1 for x near 0.
54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x "# ¸ % for any
given % 0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to
xÄ!
x! . As another example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0
such that sin
"
x
œ
"
#
as you can see from the accompanying figure. However, lim sin
xÄ!
"
x
fails to exist. The
wrong statement does not require all values of x arbitrarily close to x! œ 0 to lie within % 0 of L œ "# . Again
you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we
choose % 4" we cannot satisfy the inequality ¸sin x" #" ¸ % for all values of x sufficiently near x! œ 0.
#
55. kA *k Ÿ 0.01 Ê 0.01 Ÿ 1 ˆ x# ‰ 9 Ÿ 0.01 Ê 8.99 Ÿ
Ê
2É 8.99
1
ŸxŸ
2É 9.01
1
1 x#
4
Ÿ 9.01 Ê
4
1
(8.99) Ÿ x# Ÿ
4
1
(9.01)
or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right
endpoint was rounded down.
56. V œ RI Ê
V
R
œ I Ê ¸ VR 5¸ Ÿ 0.1 Ê 0.1 Ÿ
120
R
5 Ÿ 0.1 Ê 4.9 Ÿ
120
R
Ÿ 5.1 Ê
10
49
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
R
1#0
10
51
Ê
Section 2.3 Precise Definition of a Limit
(120)(10)
51
ŸRŸ
(120)(10)
49
85
Ê 23.53 Ÿ R Ÿ 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) $ x 1 0 Ê " $ x 1 Ê f(x) œ x. Then kf(x) 2k œ kx 2k œ 2 x 2 1 œ 1. That is,
kf(x) 2k 1 "# no matter how small $ is taken when " $ x 1 Ê lim f(x) Á 2.
xÄ1
(b) 0 x 1 $ Ê " x " $ Ê f(x) œ x 1. Then kf(x) 1k œ k(x 1) 1k œ kxk œ x 1. That is,
kf(x) 1k 1 no matter how small $ is taken when " x " $ Ê lim f(x) Á 1.
xÄ1
(c) $ x 1 ! Ê " $ x 1 Ê f(x) œ x. Then kf(x) 1.5k œ kx 1.5k œ 1.5 x 1.5 1 œ 0.5.
Also, ! x 1 $ Ê 1 x " $ Ê f(x) œ x 1. Then kf(x) 1.5k œ k(x 1) 1.5k œ kx 0.5k
œ x 0.5 " 0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that
$ x 1 $ but kf(x) 1.5k "# Ê lim f(x) Á 1.5.
xÄ1
58. (a) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 4k œ 2. Thus for % 2, kh(x) 4k
matter how small we choose $ 0 Ê lim h(x) Á 4.
% whenever 2 x 2 $ no
(b) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 3k œ 1. Thus for % 1, kh(x) 3k
matter how small we choose $ 0 Ê lim h(x) Á 3.
% whenever 2 x 2 $ no
xÄ#
xÄ#
(c) For 2 $ x 2 Ê h(x) œ x# so kh(x) 2k œ kx# 2k . No matter how small $ 0 is chosen, x# is close to 4
when x is near 2 and to the left on the real line Ê kx# 2k will be close to 2. Thus if % 1, kh(x) 2k %
whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim h(x) Á 2.
xÄ#
59. (a) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 4k 0.8. Thus for % 0.8, kf(x) 4k
3 $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 4.
xÄ$
% whenever
(b) For 3 x 3 $ Ê f(x) 3 Ê kf(x) 4.8k 1.8. Thus for % 1.8, kf(x) 4.8k
no matter how small we choose $ 0 Ê lim f(x) Á 4.8.
% whenever 3 x 3 $
(c) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 3k 1.8. Again, for % 1.8, kf(x) 3k
no matter how small we choose $ 0 Ê lim f(x) Á 3.
% whenever $ $ x 3
xÄ$
xÄ$
60. (a) No matter how small we choose $ 0, for x near 1 satisfying " $ x " $ , the values of g(x) are
near 1 Ê kg(x) 2k is near 1. Then, for % œ "# we have kg(x) 2k "# for some x satisfying
" $ x " $ , or ! kx 1k $ Ê
lim g(x) Á 2.
x Ä 1
(b) Yes, lim g(x) œ 1 because from the graph we can find a $ ! such that kg(x) 1k % if ! kx (1)k $ .
x Ä 1
61-66. Example CAS commands (values of del may vary for a specified eps):
Maple:
f := x -> (x^4-81)/(x-3);x0 := 3;
plot( f(x), x=x0-1..x0+1, color=black,
# (a)
title="Section 2.3, #61(a)" );
L := limit( f(x), x=x0 );
# (b)
epsilon := 0.2;
# (c)
plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01,
color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" );
q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d)
delta := abs(x0-q);
plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" );
for eps in [0.1, 0.005, 0.001 ] do
# (e)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
86
Chapter 2 Limits and Continuity
q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 );
delta := abs(x0-q);
head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta );
print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta,
color=black, linestyle=[1,3,3], title=head ));
end do:
Mathematica (assigned function and values for x0, eps and del may vary):
Clear[f, x]
y1: œ L eps; y2: œ L eps; x0 œ 1;
f[x_]: œ (3x2 (7x 1)Sqrt[x] 5)/(x 1)
Plot[f[x], {x, x0 0.2, x0 0.2}]
L: œ Limit[f[x], x Ä x0]
eps œ 0.1; del œ 0.2;
Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange Ä {L 2eps, L 2eps}]
2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY
1. (a) True
(e) True
(i) False
(b) True
(f) True
(j) False
(c) False
(g) False
(k) True
(d) True
(h) False
(l) False
2. (a) True
(e) True
(i) True
(b) False
(f) True
(j) False
(c) False
(g) True
(k) True
(d) True
(h) True
3. (a)
lim f(x) œ
x Ä #b
2
#
" œ #, lim c f(x) œ $ # œ "
xÄ#
(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x)
xÄ#
xÄ#
xÄ#
(c) lim c f(x) œ 4# 1 œ 3, lim b f(x) œ 4# " œ $
xÄ%
xÄ%
(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x)
xÄ%
xÄ%
xÄ%
4. (a)
lim f(x) œ
x Ä #b
2
#
œ 1, lim c f(x) œ $ # œ ", f(2) œ 2
xÄ#
(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x)
xÄ#
xÄ#
xÄ#
(c)
lim c f(x) œ 3 (1) œ 4, lim b f(x) œ 3 (1) œ 4
x Ä "
x Ä "
(d) Yes, lim f(x) œ 4 because 4 œ
x Ä "
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x)
5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0
xÄ!
(b) lim c f(x) œ lim c 0 œ 0
xÄ!
(c)
xÄ!
lim f(x) does not exist because lim b f(x) does not exist
xÄ!
xÄ!
6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0
xÄ!
(b) No, lim c g(x) does not exist since Èx is not defined for x 0
xÄ!
(c) No, lim g(x) does not exist since lim c g(x) does not exist
xÄ!
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.4 One-Sided Limits and Limits at Infinity
7. (aÑ
lim f(x) œ " œ lim b f(x)
xÄ1
(c) Yes, lim f(x) œ 1 since the right-hand and left-hand
(b)
x Ä 1c
xÄ1
limits exist and equal 1
8. (a)
(b)
lim f(x) œ 0 œ lim c f(x)
xÄ1
x Ä 1b
(c) Yes, lim f(x) œ 0 since the right-hand and left-hand
xÄ1
limits exist and equal 0
9. (a) domain: 0 Ÿ x Ÿ 2
range: 0 y Ÿ 1 and y œ 2
(b) xlim
f(x) exists for c belonging to
Äc
(0ß 1) ("ß #)
(c) x œ 2
(d) x œ 0
10. (a) domain: _ x _
range: " Ÿ y Ÿ 1
(b) xlim
f(x) exists for c belonging to
Äc
(_ß 1) ("ß ") ("ß _)
(c) none
(d) none
11.
x Ä !Þ&c
lim
13.
x Ä #b
14.
x Ä 1c
15.
lim b
lim
lim
hÄ!
2
0.5 2
È3
É 3/2
É xx
É
1 œ
0.5 1 œ
1/2 œ
12.
lim
x Ä 1b
"
1
È0 œ !
É "1
É xx
# œ
# œ
5‰
ˆ x x 1 ‰ ˆ 2x
ˆ 2 ‰ 2(2) 5
ˆ"‰
x# x œ # " Š (#)# (2) ‹ œ (2) # œ 1
ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3 7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3 7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1
Èh# 4h 5 È5
h
œ lim b
hÄ!
œ lim b Š
hÄ!
ah# 4h 5b 5
h ŠÈh# 4h 5 È5‹
Èh# 4h 5 È5
È # 4h 5 È5
‹ Š Èhh#
‹
h
4h 5 È5
œ lim b
hÄ!
h(h 4)
h ŠÈh# 4h 5 È5‹
œ
04
È5 È5
œ
2
È5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
87
88
16.
Chapter 2 Limits and Continuity
lim
h Ä !c
È6 È5h# 11h 6
h
6 a5h# 11h 6b
œ lim c
hÄ!
17. (a)
19. (a)
) Ä $b
20. (a)
t Ä %b
(x2)
(x#)
Ú) Û
)
lim
akx 2k œ x 2 for x 2b
(x 3) ’ (x(x#2)
) “
lim
x Ä #c
akx 2k œ (x 2) for x 2b
(x 3)(1) œ (2 3) œ 1
È2x (x 1)
(x 1)
akx 1k œ x 1 for x 1b
œ lim b È2x œ È2
xÄ1
œ lim c
xÄ1
È2x (x 1)
(x 1)
akx 1k œ (x 1) for x 1b
œ
œ1
3
3
lim at ÚtÛb œ 4 4 œ 0
sin È2)
È 2)
22. lim
sin kt
t
23. lim
sin 3y
4y
)Ä!
tÄ!
yÄ!
œ
26. lim
2t
t Ä ! tan t
27. lim
xÄ!
)Ä!
3 sin 3y
"
4 ylim
3y
Ä!
sin 2x ‰
ˆ cos
2x
x
œ lim
xÄ!
œ 2 lim
t
sin t
t Ä ! ˆ cos t ‰
x csc 2x
cos 5x
œ
œ
œ lim
tÄ!
" ‰
cos 5x
29. lim
x x cos x
) Ä $c
(b)
t Ä %c
Ú) Û
)
lim
œ
2
3
lim at ÚtÛb œ 4 3 œ 1
t cos t
sin t
œ lim ˆ sin xxcos x
xÄ!
œk†1œk
3
sin )
4 )lim
Ä! )
œ
"
3
Œ
œ
"
lim
)Ä!c
(where ) œ kt)
(where ) œ 3y)
3
4
œ
sin )
)
"
œ Š lim
‹ Š lim
x Ä ! cos 2x
xÄ!
sin 2x
"
3
†1œ
2 sin 2x
#x ‹
"
3
(where ) œ 3h)
œ1†2œ2
œ 2 Š lim cos t‹ Œ lim" sin t œ 2 † " † " œ 2
tÄ!
œ Š #" lim
t
Ä!
t
"
‹ Š lim cos 5x ‹
x Ä ! sin 2x
xÄ!
6x# cos x
sin
x sin 2x
xÄ!
x Ä ! sin x cos x
œ
"
"
sin 3h
3 h lim
Ä !c ˆ 3h ‰
28. lim 6x# (cot x)(csc 2x) œ lim
xÄ!
)Ä!
x Ä ! x cos 2x
œ 2 lim
sin )
)
œ k lim
sin 3y
3
4 ylim
Ä ! 3y
3h ‰
sin 3h
œ lim ˆ sinx2x †
xÄ!
k sin )
)
œ lim
œ lim c ˆ "3 †
hÄ!
h
tan 2x
x
k sin kt
kt
(b)
(where x œ È2))
œ1
sin x
x
xÄ!
tÄ!
lim
h Ä !c sin 3h
xÄ!
œ lim
œ lim
25. lim
œ 211
È6
œ lim c È2x œ È2
xÄ1
21. lim
24.
lim
(0 11)
È6 È6
œ
(x 3) œ (2) 3 œ 1
lim
x Ä #b
x Ä #c
œ lim b
xÄ1
È2x (x 1)
kx 1 k
lim
x Ä 1c
œ
h(5h 11)
h ŠÈ6 È5h# 11h 6‹
(x 3)
lim
x Ä #b
œ
È2x (x 1)
kx 1 k
lim
(b)
kx 2 k
x2
(x 3)
lim
x Ä 1b
œ
œ
x Ä #c
18. (a)
kx 2 k
x 2
(x 3)
lim
È5h# 11h 6
È6 È5h# 11h 6
È
‹ Š È66
‹
h
È5h# 11h 6
œ lim c
hÄ!
h ŠÈ6 È5h# 11h 6‹
x Ä #b
(b)
œ lim c Š
hÄ!
2x
œ lim ˆ3 cos x †
xÄ!
x cos x ‰
sin x cos x
œ lim ˆ sinx x †
xÄ!
x
sin x
†
" ‰
cos x
œ ˆ #" † 1‰ (1) œ
2x ‰
sin 2x
"
#
œ3†"†1œ3
lim
x
x Ä ! sin x
œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2
xÄ!
30. lim
xÄ!
x
x# x sin x
#x
xÄ!
œ lim ˆ #x
xÄ!
xÄ!
"
#
x
"# ˆ sinx x ‰‰ œ 0
"
#
"# (1) œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.4 One-Sided Limits and Limits at Infinity
31. lim
sin(1 cos t)
1cos t
32. lim
sin (sin h)
sin h
tÄ!
hÄ!
sin )
33. lim
) Ä ! sin 2)
34. lim
sin 5x
35. lim
tan 3x
œ
3
8 xlim
Ä!
36. lim
yÄ!
œ
)Ä!
sin )
)
sin )
)
œ lim
)Ä!
œ 1 since ) œ 1 cos t Ä 0 as t Ä 0
œ 1 since ) œ sin h Ä 0 as h Ä 0
sin )
œ lim ˆ sin
2) †
2) ‰
#)
5x
œ lim ˆ sin
sin 4x †
4x
5x
sin 3x
œ lim ˆ cos
3x †
" ‰
sin 8x
)Ä!
x Ä ! sin 4x
x Ä ! sin 8x
œ lim
xÄ!
xÄ!
"
# )lim
Ä!
œ
† 54 ‰ œ
œ lim
yÄ!
2) ‰
sin 2)
ˆ sin5x5x †
5
4 xlim
Ä!
sin 3x
œ lim ˆ cos
3x †
3
8
†1†1†1œ
4x ‰
sin 4x
†
8x
3x
†1†1œ
œ
5
4
3
8
"
lim
xÄ „_
12
5
œ
12
5
œ 0 whenever
m
n
0. This result follows immediately from
ˆ xm"În ‰ œ
lim
xÄ „_
37. (a) 3
(b) 3
38. (a) 1
(b) 1
39. (a)
"
#
(b)
"
#
40. (a)
"
8
(b)
"
8
41. (a) 53
45.
lim
tÄ_
46. r Ä
lim_
ˆ x" ‰mÎn œ Š
"
lim
‹
xÄ „_ x
mÎn
(b) 53
3
4
(b)
44. 3") Ÿ
5
4
† 83 ‰
œ1†1†1†1†
lim
mÎn
xÄ „_ x
Example 6 and the power rule in Theorem 8:
43. "x Ÿ
†1†1œ
yÄ!
cos 5y ˆ 3†4 ‰
lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos
4y ‹ 5
yÄ!
42. (a)
"
#
sin 4y
cos 5y
3†4†5y
œ lim Š siny3y ‹ Š cos
4y ‹ Š sin 5y ‹ Š 3†4†5y ‹
sin 3y sin 4y cos 5y
y cos 4y sin 5y
Note: In these exercises we use the result
"
#
œ
"
sin 8x
xÄ!
ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ
sin 3y cot 5y
y cot 4y
ˆ sin) ) †
sin 2x
x
Ÿ
"
x
cos )
3)
Ÿ
"
3)
2 t sin t
t cos t
Ê x lim
Ä_
Ê
47. (a) x lim
Ä_
lim
) Ä _
œ lim
2
t
tÄ_
r sin r
2r 7 5 sin r
2x 3
5x 7
$
sin 2x
x
œrÄ
lim_
œ x lim
Ä_
œ 0 by the Sandwich Theorem
cos )
3)
œ 0 by the Sandwich Theorem
1 ˆ sint t ‰
1 ˆ cost t ‰
œ
1 ˆ sinr r ‰
2 7r 5 ˆ sinr r ‰
2 3x
5 7x
2x 7
48. (a) x lim
œ x lim
Ä _ x$ x# x 7
Ä_
(b) 2 (same process as part (a))
3
4
œ
010
10
œ 1
œrÄ
lim_
2
5
2 Š x7$ ‹
1 "x x"# x7$
10
200
œ
(b)
"
#
2
5
(same process as part (a))
œ2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 0mÎn œ 0.
89
90
Chapter 2 Limits and Continuity
"
x
x"#
49. (a) x lim
Ä_
x1
x# 3
œ x lim
Ä_
1 x3#
50. (a) x lim
Ä_
3x 7
x# 2
œ x lim
Ä_
1 x2#
51. (a) x lim
Ä_
7x$
x$ 3x# 6x
52. (a) x lim
Ä_
"
x$ 4x 1
&
3
x
x7#
œ x lim
Ä_
(b)
9
#
2x%
œ0
(b) 0 (same process as part (a))
"
x$
œ x lim
Ä_
10
x
œ x lim
Ä_
œ(
x"# x31'
1
2
(b) 7 (same process as part (a))
œ!
1 x4# x"$
%
9x% x
5x# x 6
(b) 0 (same process as part (a))
7
1 3x x6#
10x x 31
53. (a) x lim
œ x lim
x'
Ä_
Ä_
(b) 0 (same process as part (a))
54. (a) x lim
Ä_
œ0
(b) 0 (same process as part (a))
œ0
9 x"$
5
x#
x"$ x6%
œ
9
#
(same process as part (a))
55. (a) x lim
Ä_
2x$ 2x 3
3x$ 3x# 5x
œ x lim
Ä_
2 x2# x3$
3 3x x5#
œ 23
(b) 23 (same process as part (a))
%
x
56. (a) x lim
œ x lim
Ä _ x% 7x$ 7x# 9
Ä_
(b) 1 (same process as part (a))
57. x lim
Ä_
2Èx x"
3x 7
59. x Ä
lim
_
œ x lim
Ä_
$
& x
È
xÈ
$ xÈ
& x
È
x" x%
x# x$
61. x lim
Ä_
2x&Î$ x"Î$ 7
x)Î& 3x Èx
62. x Ä
lim
_
2
Š "Î#
‹ Š x"# ‹
x
3 7x
œxÄ
lim
_
60. x lim
Ä_
œ x lim
Ä_
$
È
x 5x 3
2x x#Î$ 4
"
1 7x x7# x9%
œ0
1 xÐ"Î&Ñ Ð"Î$Ñ
1 xÐ"Î&Ñ Ð"Î$Ñ
x x"#
1 x"
œ x lim
Ä_
œxÄ
lim
_
œ 1
58. x lim
Ä_
œxÄ
lim
_
" ‹
1 Š #Î"&
x
" ‹
1 Š #Î"&
2 Èx
2 Èx
œ x lim
Ä_
2
Š "Î#
‹"
x
2
Š "Î#
‹1
x
œ 1
œ1
x
œ_
" 7
2x"Î"& "*Î"&
x
x)Î&
"
3
1 $Î&
""Î"!
x
"
x#Î$
2
5 3x
"
x"Î$
œ_
x
4x
œ 5#
63. Yes. If lim b f(x) œ L œ lim c f(x), then xlim
f(x) œ L. If lim b f(x) Á lim c f(x), then xlim
f(x) does not exist.
Äa
Äa
xÄa
xÄa
xÄa
xÄa
64. Since xlim
f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim
f(x) can be found by calculating
Äc
Äc
xÄc
xÄc
lim b f(x).
xÄc
65. If f is an odd function of x, then f(x) œ f(x). Given lim b f(x) œ 3, then lim c f(x) œ $.
xÄ!
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.4 One-Sided Limits and Limits at Infinity
91
66. If f is an even function of x, then f(x) œ f(x). Given lim c f(x) œ 7 then lim b f(x) œ 7. However, nothing
xÄ#
x Ä #
can be said about
67. Yes. If x lim
Ä_
lim
x Ä #c
f(x)
g(x)
f(x) because we don't know lim b f(x).
xÄ#
œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä
lim
_
f(x)
g(x)
œ 2 as well.
68. Yes, it can have a horizontal or oblique asymptote.
69. At most 1 horizontal asymptote: If x lim
Ä_
f(x)
lim
x Ä _ g(x)
f(x)
g(x)
œ L, then the ratio of the polynomials' leading coefficients is L, so
œ L as well.
Èx# x Èx# x œ lim ’Èx# x Èx# x“ † ’ Èx# x Èx# x “ œ lim
70. x lim
È x# x È x# x
Ä_
xÄ_
xÄ_
2x
2
2
œ x lim
œ
lim
œ
œ
1
È #
1 1
"
"
Ä_ È #
xÄ_
x x
x x
ax # x b a x # x b
È x# x È x# x
É1 x É1 x
71. For any % 0, take N œ 1. Then for all x N we have that kf(x) kk œ kk kk œ 0 %.
72. For any % 0, take N œ 1. Then for all y N we have that kf(x) kk œ kk kk œ 0 %.
73. I œ (5ß 5 $ ) Ê 5 x & $ . Also, Èx 5 % Ê x 5 %# Ê x & %# . Choose $ œ %#
Ê lim Èx 5 œ 0.
x Ä &b
74. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %#
Ê lim È% x œ 0.
x Ä %c
75. As x Ä 0 the number x is always negative. Thus, ¹ kxxk (1)¹ % Ê ¸ xx 1¸ % Ê 0 % which is always
true independent of the value of x. Hence we can choose any $ 0 with $ x ! Ê
x
lim
x Ä ! c kx k
œ 1.
2
¸ x 2
¸
76. Since x Ä # we have x 2 and kx 2k œ x 2. Then, ¹ kxx
2 k " ¹ œ x 2 " % Ê 0 %
which is always true so long as x #. Hence we can choose any $ !, and thus # x # $
2
Ê ¹ kxx
2k "¹ % . Thus,
77. (a)
lim
x Ä %!!b
x 2
lim
x Ä #b kx2k
œ 1.
ÚxÛ œ 400. Just observe that if 400 x 401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any
number % ! that 400 x 400 $ Ê lÚxÛ 400l œ l400 400l œ ! %.
(b)
lim c ÚxÛ œ 399. Just observe that if 399 x 400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any
x Ä %!!
number % ! that 400 $ x 400 Ê lÚxÛ 399l œ l399 399l œ ! %.
(c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist.
x Ä %!!
78. (a)
x Ä %!!
x Ä %!!
lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %#
xÄ!
Ê lim b f(x) œ 0.
x Ä !b
xÄ!
(b)
lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0.
x Ä !c
xÄ!
Since kx# 0k œ kx# 0k œ x# % whenever kxk È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰ 0¸ %
if $ x 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
92
Chapter 2 Limits and Continuity
(c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0.
79.
81.
82.
83.
84.
lim
"
x
x sin
xÄ „_
3x 4
x Ä „ _ 2x 5
œ lim
œ
lim
"
)Ä0 )
sin ) œ 1, ˆ) œ x" ‰
3 4x
5
x Ä „ _ 2 x
lim
œ lim
3 4t
t Ä 0 2 5t
œ
80.
3
#
cos
lim
"
x
x Ä _ 1 x"
œ lim c
)Ä!
cos )
1)
œ
"
1
œ 1, ˆ) œ x" ‰
, ˆt œ "x ‰
"Îx
lim ˆ "x ‰ œ lim b zz œ 1, ˆz œ x" ‰
zÄ!
xÄ_
ˆ3 2x ‰ ˆcos "x ‰ œ lim (3 2))(cos )) œ (3)(1) œ 3, ˆ) œ x" ‰
lim
xÄ „_
)Ä0
lim ˆ x3# cos x" ‰ ˆ1 sin x" ‰ œ lim b a3)# cos )b (1 sin )) œ (0 1)(1 0) œ 1, ˆ) œ x" ‰
)Ä!
xÄ_
2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES
"
œ_
1.
lim
x Ä !b 3x
3.
lim
x Ä #c x 2
5.
lim
x Ä )b x8
7.
lim
#
x Ä ( (x7)
3
2x
4
œ _
œ _
œ_
lim
"Î$
x Ä !b 3x
10. (a)
lim
"Î&
x Ä !b x
4
11. lim
#Î&
xÄ! x
13.
œ lim
4
#
x Ä ! ax"Î& b
œ_
Š positive
positive ‹
lim
x Ä !c 2x
positive
Š negative
‹
4.
lim
x Ä $b x 3
Š negative
positive ‹
6.
lim
x Ä &c 2x10
3x
œ_
positive
Š positive
‹
8.
lim #
x Ä ! x (x1)
œ _
(b)
lim
"Î$
x Ä !c 3x
(b)
lim
"Î&
x Ä !c x
œ_
2
positive
Š negative
‹
2.
œ_
2
9. (a)
œ _
Š positive
positive ‹
œ_
5
"
"
2
2
12. lim
"
#Î$
xÄ! x
lim tan x œ _
14.
x Ä ˆ 1# ‰
Š negative
negative ‹
negative
Š positive
†positive ‹
œ _
œ _
œ lim
"
#
x Ä ! ax"Î$ b
œ_
lim sec x œ _
x Ä ˆ #1 ‰
lim (1 csc )) œ _
15.
) Ä !
16.
) Ä !b
lim (2 cot )) œ _ and lim c (2 cot )) œ _, so the limit does not exist
)Ä!
"
œ lim b
xÄ#
"
(x2)(x2)
œ_
Š positive"†positive ‹
"
œ lim c
xÄ#
"
(x2)(x2)
œ _
Š positive†"negative ‹
17. (a)
lim
#
x Ä # b x 4
(b)
lim
#
x Ä # c x 4
(c)
lim
#
x Ä #b x 4
(d)
lim
#
x Ä #c x 4
"
œ
lim
x Ä #b (x2)(x2)
"
œ _
Š positive†"negative ‹
"
œ
lim
x Ä #c (x2)(x2)
"
œ_
Š negative"†negative ‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.5 Infinite Limits and Vertical Asymptotes
x
œ lim b
xÄ"
x
(x1)(x1)
œ_
positive
Š positive
†positive ‹
x
œ lim c
xÄ"
x
(x1)(x1)
œ _
positive
Š positive
†negative ‹
18. (a)
lim
#
x Ä "b x 1
(b)
lim
#
x Ä "c x 1
(c)
lim
#
x Ä "b x 1
(d)
lim
#
x Ä "c x 1
x
œ
lim
x Ä "b (x1)(x1)
x
œ_
negative
Š positive
†negative ‹
x
œ
lim
x Ä "c (x1)(x1)
x
œ _
negative
Š negative
†negative ‹
19. (a)
lim
x Ä !b #
x#
"
x
œ 0 lim b
xÄ!
"
x
œ _
"
Š negative
‹
(b)
lim
x Ä !c #
x#
"
x
œ 0 lim c
xÄ!
"
x
œ_
"
Š positive
‹
(c)
lim
#
x Ä $È2
(d)
lim
x Ä 1 #
20. (a)
x#
x#
lim
x Ä #b
"
x
"
x
œ
x# 1
2x 4
x# 1
2#Î$
#
œ
"
#
(d)
lim
x Ä !c 2x 4
œ
(b)
(c)
(d)
(e)
22. (a)
x# 3x 2
x$ 2x#
lim b
x# 3x 2
x$ 2x#
#
x 3x 2
x$ 2x#
lim
x Ä #c
lim
xÄ#
#
x 3x 2
x$ 2x#
#
x 3x 2
x$ 2x#
lim
xÄ!
lim
x Ä #b
(c)
x Ä 0c
œ lim b
xÄ#
œ lim c
xÄ#
(d)
x Ä "b
(e)
lim
x Ä !b x(x #)
x# 1
2x 4
lim
x Ä #c
œ _
positive
Š negative
‹
œ0
(x 2)(x 1)
x# (x 2)
œ _
(x 2)(x 1)
x# (x 2)
œ lim b
xÄ#
(x 2)(x 1)
x# (x 2)
œ lim c
xÄ#
œ lim
œ lim
(x 2)(x 1)
x# (x 2)
œ _
xÄ!
x# 3x 2
x$ 4x
lim
x# 3x 2
x$ 4x
x"
2†0
#4
(x 2)(x 1)
x# (x 2)
xÄ#
lim
and
œ lim b
xÄ!
œ lim b
xÄ#
x# 3x 2
x$ 4x
(b)
œ
œ lim
x# 3x 2
x$ 4x
lim
x Ä #b
(x 1)(x 1)
2x 4
(b)
"
4
lim b
xÄ#
3
#
Š positive
positive ‹
œ lim b
xÄ"
xÄ!
œ 2"Î$ 2"Î$ œ 0
œ_
lim
x Ä "b 2x 4
21. (a)
"
#"Î$
ˆ "1 ‰ œ
(c)
x# 1
œ
xÄ#
(x 2)(x ")
x(x #)(x 2)
(x 2)(x ")
œ lim c
xÄ!
œ lim b
xÄ"
(x 2)(x ")
x(x #)(x 2)
(x 2)(x ")
x(x #)(x 2)
œ
x1
x#
œ
"
4
,xÁ2
x1
x#
œ
"
4
,xÁ2
x1
x#
œ
"
4
,xÁ2
†negative
Š negative
positive†negative ‹
œ lim b
xÄ#
lim
x Ä #b x(x #)(x 2)
†negative
Š negative
positive†negative ‹
(x 1)
x(x #)
œ
(x 1)
lim
x Ä #b x(x #)
œ lim c
xÄ!
œ lim b
xÄ"
œ_
(x 1)
x(x #)
œ
negative
Š positive
†positive ‹
x"
negative
Š negative
†positive ‹
œ_
œ
"
8
œ_
(x 1)
x(x #)
œ _
lim
x Ä !c x(x #)
"
#(4)
0
(1)(3)
negative
Š negative
†positive ‹
negative
Š negative
†positive ‹
œ0
so the function has no limit as x Ä 0.
lim 2
23. (a)
t Ä !b
24. (a)
t Ä !b
25. (a)
x Ä !b
(c)
x Ä "b
3 ‘
t"Î$
œ _
"
lim t$Î&
7‘ œ _
lim
2
lim
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(b)
t Ä !c
(b)
t Ä !c
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(b)
x Ä !c
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(d)
x Ä "c
"
t$Î&
3 ‘
t"Î$
œ_
7‘ œ _
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
93
94
Chapter 2 Limits and Continuity
26. (a)
x Ä !b
(c)
x Ä "b
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ_
(b)
x Ä !c
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
(d)
x Ä "c
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
27. y œ
"
x1
28. y œ
"
x1
29. y œ
"
#x 4
30. y œ
3
x3
31. y œ
x3
x2
32. y œ
2x
x1
œ1
"
x#
œ#
2
x1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.5 Infinite Limits and Vertical Asymptotes
33. y œ
x#
x"
35. y œ
x# %
x"
œx"
37. y œ
x# 1
x
œx
œx1
"
x"
$
x"
"
x
39. Here is one possibility.
34. y œ
x# "
x1
œx"
36. y œ
x2 "
#x %
œ #" x "
38. y œ
x$ 1
x#
œx
#
x1
$
#x %
"
x#
40. Here is one possibility.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
95
96
Chapter 2 Limits and Continuity
41. Here is one possibility.
42. Here is one possibility.
43. Here is one possibility.
44. Here is one possibility.
45. Here is one possibility.
46. Here is one possibility.
"
x#
47. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 0k $ Ê
"
x#
Ê
B ! Í
"
x#
"
x#
#
B0 Í x
"
B
"
ÈB
Í kxk
. Choose $ œ
"
ÈB
, then 0 kxk $ Ê kxk
xÄ!
"
B.
B ! Í lxl
"
B.
Choose $ œ
Then ! kx 0k $ Ê lxl
"
B
Ê
"
lx l
"
lx l
2
(x 3)#
B ! Í
2
(x 3)#
$ œ É B2 , then 0 kx 3k $ Ê
B0 Í
2
(x 3)#
(x 3)#
2
"
B
Í (x 3)#
B 0 so that lim
2
#
x Ä $ (x 3)
2
B
B. Now,
x Ä ! lx l
2
(x 3)#
Now,
#
B ! Í (x 5)
Ê kx 5k
"
ÈB
Ê
"
(x 5)#
"
B
Í kx 5k
B so that lim
"
"
ÈB
#
x Ä & (x 5)
. Choose $ œ
œ _.
B.
Í ! kB $k É B2 . Choose
œ _.
50. For every real number B 0, we must find a $ 0 such that for all x, 0 kx (5)k $ Ê
1
(x 5)#
"
B so that lim
49. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 3k $ Ê
Now,
"
ÈB
B so that lim x"# œ _.
48. For every real number B 0, we must find a $ 0 such that for all x, ! kx 0k $ Ê
"
lx l
B. Now,
"
ÈB
1
(x 5)#
B.
. Then 0 kx (5)k $
œ _.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.5 Infinite Limits and Vertical Asymptotes
97
51. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim f(x) œ _, if
x Ä x!
for every positive number B, there exists a corresponding number $ 0 such that for all x,
x! $ x x! Ê f(x) B.
(b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim f(x) œ _,
x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such
that for all x, x! x x! $ Ê f(x) B.
(c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ _,
x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such
that for all x, x! $ x x! Ê f(x) B.
52. For B 0,
"
x
B 0 Í x B" . Choose $ œ B" . Then ! x $ Ê 0 x
53. For B 0,
"
x
B 0 Í x" B 0 Í x
Ê B" x Ê
54. For B !,
"
x#
"
x
B so that lim c
xÄ!
"
x
"
B
Ê
"
x#
"
x#
Ê
"
x#
"
1 x#
"
x
œ _.
"
B
Í x 2 B" Í x 2 B" . Choose $ œ B" . Then
"
x#
B 0 so that lim c
xÄ#
"
x#
œ _.
œ _.
B Í 1 x#
"
#B . Then " $ x " Ê
"
1 x# B for ! x 1 and
"
x
B so that lim b
xÄ!
B Í ! x 2 B" . Choose $ œ B" . Then # x # $ Ê ! x # $ Ê ! x 2
B ! so that lim b
xÄ#
57. y œ sec x
"
x
œ _.
B Í x " # B Í (x 2)
56. For B 0 and ! x 1,
$
Ê
Í B" x. Choose $ œ B" . Then $ x !
2 $ x 2 Ê $ x 2 ! Ê B" x 2 0 Ê
55. For B 0,
"
B
"
B
Í (" x)(" x) B" . Now
$ x 1 0 Ê " x $
x near 1 Ê
"
lim
#
x Ä "c " x
"
#B
1x
#
1 since x 1. Choose
Ê (" x)(" x) B" ˆ 1 # x ‰ B"
œ _.
58. y œ sec x
"
x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
B
98
Chapter 2 Limits and Continuity
59. y œ tan x
61. y œ
"
x#
x
È 4 x#
63. y œ x#Î$
60. y œ
"
x
62. y œ
"
È 4 x#
tan x
64. y œ sin ˆ x# 1 1 ‰
"
x"Î$
2.6 CONTINUITY
1. No, discontinuous at x œ 2, not defined at x œ 2
2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5
xÄ$
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.6 Continuity
3. Continuous on [1ß 3]
4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ !
xÄ"
xÄ"
5. (a) Yes
(b) Yes,
(c) Yes
(d) Yes
6. (a) Yes, f(1) œ 1
lim
x Ä "b
f(x) œ 0
(b) Yes, lim f(x) œ 2
xÄ1
(c) No
(d) No
7. (a) No
(b) No
8. ["ß !) (!ß ") ("ß #) (#ß $)
9. f(2) œ 0, since lim c f(x) œ 2(2) 4 œ 0 œ lim b f(x)
xÄ#
xÄ#
10. f(1) should be changed to 2 œ lim f(x)
xÄ1
11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0).
xÄ"
xÄ1
xÄ"
Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than
xÄ!
f(0) œ 1.
12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1).
xÄ"
xÄ1
xÄ"
Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than
xÄ#
f(2) œ 2.
13. Discontinuous only when x 2 œ 0 Ê x œ 2
14. Discontinuous only when (x 2)# œ 0 Ê x œ 2
15. Discontinuous only when x# %x $ œ ! Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1
16. Discontinuous only when x# 3x 10 œ 0 Ê (x 5)(x 2) œ 0 Ê x œ 5 or x œ 2
17. Continuous everywhere. ( kx 1k sin x defined for all x; limits exist and are equal to function values.)
18. Continuous everywhere. ( kxk " Á 0 for all x; limits exist and are equal to function values.)
19. Discontinuous only at x œ 0
20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n ") 1# , n an integer, but continuous at all other x.
21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ
n1
# ,
n an integer, but
continuous at all other x.
22. Discontinuous when
1x
#
is an odd integer multiple of 1# , i.e.,
1x
#
œ (2n 1) 1# , n an integer Ê x œ 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
99
100
Chapter 2 Limits and Continuity
23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n 1) 1# , n an integer, but continuous at all other x.
24. Continuous everywhere since x% 1
and are equal to the function values.
1 and " Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1 sin# x
1; limits exist
25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval 3# ß _‰ .
26. Discontinuous when 3x 1 0 or x
"
3
Ê continuous on the interval 3" ß _‰ .
27. Continuous everywhere: (2x 1)"Î$ is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2 x)"Î& is defined for all x; limits exist and are equal to function values.
29. xlim
sin (x sin x) œ sin (1 sin 1) œ sin (1 0) œ sin 1 œ 0, and function continuous at x œ 1.
Ä1
30. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !.
tÄ!
31. lim sec ay sec# y tan# y 1b œ lim sec ay sec# y sec# yb œ lim sec a(y 1) sec# yb œ sec a(" ") sec# 1b
yÄ1
yÄ1
yÄ1
œ sec 0 œ 1, , and function continuous at y œ ".
32. lim tan 14 cos ˆsin x"Î$ ‰‘ œ tan 14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !.
xÄ!
33. lim cos ’ È19 13 sec 2t “ œ cos ’ È19 13 sec 0 “ œ cos
tÄ!
1
È16
œ cos
1
4
œ
È2
# ,
and function continuous at t œ !.
34. lim1 Écsc# x 5È3 tan x œ Écsc# ˆ 16 ‰ 5È3 tan ˆ 16 ‰ œ Ê4 5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at
xÄ
'
x œ 1' .
35. g(x) œ
x# 9
x3
(x 3)(x 3)
(x 3)
œ
36. h(t) œ
t# 3t 10
t#
37. f(s) œ
s$ "
s# 1
38. g(x) œ
œ
œ
œ x 3, x Á 3 Ê g(3) œ lim (x 3) œ 6
(t 5)(t 2)
t#
as# s 1b (s 1)
(s 1)(s 1)
x# 16
x# 3x 4
œ
xÄ$
œ t 5, t Á # Ê h(2) œ lim (t 5) œ 7
tÄ#
œ
(x 4)(x 4)
(x 4)(x 1)
s# s "
s1 ,
œ
x4
x1
s Á 1 Ê f(1) œ lim Š s
sÄ1
#
s1
s1 ‹
4‰
, x Á 4 Ê g(4) œ lim ˆ xx
1 œ
xÄ%
œ
3
#
8
5
39. As defined, lim c f(x) œ (3)# 1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have
xÄ$
xÄ$
6a œ 8 Ê a œ 43 .
40. As defined,
lim
x Ä #c
g(x) œ 2 and
4b œ 2 Ê b œ "# .
lim
x Ä #b
g(x) œ b(2)# œ 4b. For g(x) to be continuous we must have
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.6 Continuity
41. The function can be extended: f(0) ¸ 2.3.
42. The function cannot be extended to be continuous at
x œ 0. If f(0) ¸ 2.3, it will be continuous from the
right. Or if f(0) ¸ 2.3, it will be continuous from the
left.
43. The function cannot be extended to be continuous
at x œ 0. If f(0) œ 1, it will be continuous from
the right. Or if f(0) œ 1, it will be continuous
from the left.
44. The function can be extended: f(0) ¸ 7.39.
101
45. f(x) is continuous on [!ß "] and f(0) 0, f(1) 0
Ê by the Intermediate Value Theorem f(x) takes
on every value between f(0) and f(1) Ê the
equation f(x) œ 0 has at least one solution between
x œ 0 and x œ 1.
46. cos x œ x Ê (cos x) x œ 0. If x œ 1# , cos ˆ 1# ‰ ˆ 1# ‰ 0. If x œ 1# , cos ˆ 1# ‰
for some x between
1
#
and
1
#
1
#
0. Thus cos x x œ 0
according to the Intermediate Value Theorem.
47. Let f(x) œ x$ 15x 1 which is continuous on [4ß 4]. Then f(4) œ 3, f(1) œ 15, f(1) œ 13, and f(4) œ 5.
By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals % x 1, " x 1, and
" x 4. That is, x$ 15x 1 œ 0 has three solutions in [%ß 4]. Since a polynomial of degree 3 can have at most 3
solutions, these are the only solutions.
48. Without loss of generality, assume that a b. Then F(x) œ (x a)# (x b)# x is continuous for all values of
x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value
Theorem, since a a # b b, there is a number c between a and b such that F(x) œ a # b .
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102
Chapter 2 Limits and Continuity
49. Answers may vary. Note that f is continuous for every value of x.
(a) f(0) œ 10, f(1) œ 1$ 8(1) 10 œ 3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c
so that ! c 1 and f(c) œ 1.
(b) f(0) œ 10, f(4) œ (4)$ 8(4) 10 œ 22. Since 22 È3 10, by the Intermediate Value
Theorem, there exists a c so that 4 c 0 and f(c) œ È3.
(c) f(0) œ 10, f(1000) œ (1000)$ 8(1000) 10 œ 999,992,010. Since 10 5,000,000 999,992,010, by the
Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c) œ 5,000,000.
50. All five statements ask for the same information because of the intermediate value property of continuous
functions.
(a) A root of f(x) œ x$ 3x 1 is a point c where f(c) œ 0.
(b) The points where y œ x$ crosses y œ 3x 1 have the same y-coordinate, or y œ x$ œ 3x 1
Ê f(x) œ x$ 3x 1 œ 0.
(c) x$ 3x œ 1 Ê x$ 3x 1 œ 0. The solutions to the equation are the roots of f(x) œ x$ 3x 1.
(d) The points where y œ x$ 3x crosses y œ 1 have common y-coordinates, or y œ x$ 3x œ 1
Ê f(x) œ x$ 3x 1 œ !.
(e) The solutions of x$ 3x 1 œ 0 are those points where f(x) œ x$ 3x 1 has value 0.
51. Answers may vary. For example, f(x) œ
sin (x 2)
x2
is discontinuous at x œ 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2.
52. Answers may vary. For example, g(x) œ
"
x1
has a discontinuity at x œ 1 because lim g(x) does not exist.
x Ä "
Š lim c g(x) œ _ and lim b g(x) œ _.‹
x Ä "
x Ä "
53. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $ 0 there is an irrational number x (actually
infinitely many) in the interval (x! $ ß x! $ ) Ê f(x) œ 0. Then 0 kx x! k $ but kf(x) f(x! )k
œ 1 "# œ %, so x lim
f(x) fails to exist Ê f is discontinuous at x! rational.
Äx
!
On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x! $ ß x! $ ) Ê f(x)
œ 1. Again x lim
f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at
Äx
!
every point.
(b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x! $ ß x! ) or
(x! ß x! $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and
x Ä x!
lim f(x) exist by the same arguments used in part (a).
x Ä x!
54. Yes. Both f(x) œ x and g(x) œ x
g ˆ "# ‰ œ 0 Ê
f(x)
g(x)
"
#
are continuous on [!ß "]. However
f(x)
g(x)
is undefined at x œ
"
#
since
is discontinuous at x œ "# .
55. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not.
56. Let f(x) œ
œ
"
x1
"
(x 1) 1
œ
and g(x) œ x 1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x))
"
x
is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be
continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1.
57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to
equal zero at some point between a and b since f is continuous on [aß b].
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives
58. Let f(x) be the new position of point x and let d(x) œ f(x) x. The displacement function d is negative if x is
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is
then in its original position.
59. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a 0 and f(1) œ b 1
because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x) x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0) 0 œ a 0 and
g(1) œ f(1) 1 œ b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that
g(c) œ 0 Ê f(c) c œ 0 or f(c) œ c.
60. Let % œ
kf(c)k
#
0. Since f is continuous at x œ c there is a $ 0 such that kx ck $ Ê kf(x) f(c)k %
Ê f(c) % f(x) f(c) %.
If f(c) 0, then % œ "# f(c) Ê
"
#
"
#
If f(c) 0, then % œ f(c) Ê
f(c) f(x)
3
#
3
#
f(c) f(x)
f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
"
#
f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
61. By Exercises 52 in Section 2.3, we have xlim
faxb œ L Í lim fac hb œ L.
Äc
hÄ0
Thus, faxb is continuous at x œ c Í xlim
faxb œ facb Í lim fac hb œ facb.
Äc
hÄ0
62. By Exercise 61, it suffices to show that lim sinac hb œ sin c and lim cosac hb œ cos c.
hÄ0
hÄ0
Now lim sinac hb œ lim asin cbacos hb acos cbasin hb‘ œ asin cbŠ lim cos h‹ acos cbŠ lim sin h‹
hÄ0
hÄ0
hÄ0
hÄ0
By Example 6 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac hb œ sin c and thus faxb œ sin x is
hÄ0
continuous at x œ c. Similarly,
hÄ0
hÄ0
lim cosac hb œ lim acos cbacos hb asin cbasin hb‘ œ acos cbŠ lim cos h‹ asin cbŠ lim sin h‹ œ cos c.
hÄ0
hÄ0
Thus, gaxb œ cos x is continuous at x œ c.
hÄ0
63. x ¸ 1.8794, 1.5321, 0.3473
64. x ¸ 1.4516, 0.8547, 0.4030
65. x ¸ 1.7549
66. x ¸ 1.5596
67. x ¸ 3.5156
68. x ¸ 3.9058, 3.8392, 0.0667
69. x ¸ 0.7391
70. x ¸ 1.8955, 0, 1.8955
hÄ0
2.7 TANGENTS AND DERIVATIVES
1. P" : m" œ 1, P# : m# œ 5
2. P" : m" œ 2, P# : m# œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 2 Limits and Continuity
3. P" : m" œ 5# , P# : m# œ "#
5. m œ lim
hÄ!
4. P" : m" œ 3, P# : m# œ 3
c4 (" h)# d a4 (1)# b
h
a1 2h h# b1
h
hÄ!
œ lim
œ lim
hÄ!
h(# h)
h
œ 2;
at ("ß $): y œ $ #(x (1)) Ê y œ 2x 5,
tangent line
6. m œ lim
hÄ!
c(1 h 1)# 1d c(" ")# 1d
h
h#
œ lim
hÄ! h
œ lim h œ 0; at ("ß "): y œ 1 0(x 1) Ê y œ 1,
hÄ!
tangent line
È
2È 1 h 2È 1
œ lim 2 1 h h 2
h
hÄ!
hÄ!
4(1 h) 4
œ lim
œ lim È1 2h 1
h Ä ! 2h ŠÈ1 h 1‹
hÄ!
7. m œ lim
†
2È 1 h 2
2È 1 h #
œ 1;
at ("ß #): y œ 2 1(x 1) Ê y œ x 1, tangent line
8. m œ lim
hÄ!
"
(1 h)#
("" )#
h
a2h h# b
#
h Ä ! h(1 h)
œ lim
1 (1 h)#
#
h Ä ! h(1h)
2h
lim
# œ 2;
h Ä ! (1 h)
œ lim
œ
at ("ß "): y œ 1 2(x (1)) Ê y œ 2x 3,
tangent line
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives
(2 h)$ (2)$
h
9. m œ lim
hÄ!
8 12h 6h# h$ 8
h
œ lim
hÄ!
œ lim a12 6h h# b œ 12;
hÄ!
at (2ß 8): y œ 8 12(x (2)) Ê y œ 12x 16,
tangent line
"
(# h)$
10. m œ lim
h
hÄ!
œ
"2
8(8)
hÄ!
hÄ!
at ˆ#ß "8 ‰ : y œ 8"
Ê yœ
11. m œ lim
hÄ!
x
"
#,
12 6h h#
8(2 h)$
œ lim
3
œ 16
;
3
16
8 (# h)$
8h(# h)$
œ lim
a12h 6h# h$ b
8h(# h)$
œ lim
hÄ!
(#" )$
3
16 (x
(2))
tangent line
c(2 h)# 1d 5
h
œ lim
hÄ!
a5 4h h# b 5
h
hÄ!
at (2ß 5): y 5 œ 4(x 2), tangent line
12. m œ lim
hÄ!
c(" h) 2(1 h)# d (1)
h
œ lim
hÄ!
h(4 h)
h
œ lim
a1 h 2 4h 2h# b 1
h
hÄ!
3h
(3 h) 2
3
h
œ lim
hÄ!
(3 h) 3(h 1)
h(h 1)
h Ä ! h(h 1)
at ($ß $): y 3 œ 2(x 3), tangent line
14. m œ lim
hÄ!
8
(2 h)#
2
h
hÄ!
hÄ!
(2 h)$ 8
h
œ lim
hÄ!
œ lim
a8 12h 6h# h$ b 8
h
hÄ!
at (2ß )): y 8 œ 12(t 2), tangent line
16. m œ lim
hÄ!
c(1 h)$ 3(1 h)d 4
h
a1 3h 3h# h$ 3 3hb 4
h
œ lim
hÄ!
at ("ß %): y 4 œ 6(t 1), tangent line
È4 h 2
h
hÄ!
17. m œ lim
È4 h 2
h
hÄ!
œ lim
œ "4 ; at (%ß #): y 2 œ
18. m œ lim
hÄ!
œ
"
È9 3
È(8 h) 1 3
h
"
4
†
È4 h 2
È4 h 2
œ lim
hÄ!
h a12 6h h# b
h
œ lim
œ 3;
œ 2;
8 2 a4 4h h# b
h(2 h)#
hÄ!
8 2(2 h)#
#
h Ä ! h(2 h)
œ lim
at (2ß 2): y 2 œ 2(x 2)
15. m œ lim
2h
œ lim
h(3 2h)
h
œ lim
at ("ß "): y 1 œ 3(x 1), tangent line
13. m œ lim
œ %;
œ lim
2h(4 h)
h(2 h)#
œ
8
4
œ 2;
œ 12;
œ lim
hÄ!
(4 h) 4
h Ä ! h ŠÈ4 h #‹
h a6 3h h# b
h
œ lim
œ 6;
h
h Ä ! h ŠÈ4 h #‹
œ
"
È4 #
(x 4), tangent line
œ lim
hÄ!
È9 h 3
h
œ 6" ; at (8ß 3): y 3 œ
19. At x œ 1, y œ 5 Ê m œ lim
hÄ!
"
6
†
È9 h 3
È9 h 3
œ lim
(9 h) 9
h Ä ! h ŠÈ9 h 3‹
œ lim
h
h Ä ! h ŠÈ9 h 3‹
(x 8), tangent line
5(" h)# 5
h
œ lim
hÄ!
5 a1 2h h# b 5
h
œ lim
hÄ!
5h(2 h)
h
œ 10, slope
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
105
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Chapter 2 Limits and Continuity
c1 (2 h)# d (3)
h
20. At x œ 2, y œ 3 Ê m œ lim
hÄ!
21. At x œ 3, y œ
"
#
"
(3 h) 1
Ê m œ lim
#"
22. At x œ 0, y œ 1 Ê m œ lim
hÄ!
h1
h1
hÄ!
hÄ!
(1)
h
a1 4 4h h# b 3
h
2 (2 h)
2h(2 h)
œ lim
h
hÄ!
œ lim
hÄ!
hÄ!
h
œ lim
h Ä ! 2h(2 h)
(h 1) (h ")
h(h 1)
œ lim
œ lim
œ lim
h(4 h)
h
œ 4, slope
œ "4 , slope
2h
h Ä ! h(h 1)
œ 2, slope
c(x h)# 4(x h) 1d ax# 4x 1b
h
hÄ!
a2xh h# 4hb
lim
œ lim (2x h 4) œ 2x
h
hÄ!
hÄ!
23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim
ax# 2xh h# 4x 4h 1b ax# 4x 1b
h
hÄ!
œ lim
œ
4;
2x 4 œ 0 Ê x œ 2. Then f(2) œ 4 8 1 œ 5 Ê (2ß 5) is the point on the graph where there is a
horizontal tangent.
24. 0 œ m œ lim
hÄ!
c(x h)$ 3(x h)d ax$ 3xb
h
3x# h 3xh# h$ 3h
h
œ lim
hÄ!
œ lim
hÄ!
ax$ 3x# h 3xh# h$ 3x 3hb ax$ 3xb
h
œ lim a3x# 3xh h# 3b œ 3x# 3; 3x# 3 œ 0 Ê x œ 1 or x œ 1. Then
hÄ!
f(1) œ 2 and f(1) œ 2 Ê ("ß 2) and ("ß 2) are the points on the graph where a horizontal tangent exists.
"
(x h) 1
25. 1 œ m œ lim
x " 1
h
hÄ!
(x 1) (x h 1)
h(x 1)(x h 1)
œ lim
hÄ!
h
œ lim
h Ä ! h(x 1)(x h 1)
œ (x " 1)#
Ê (x 1)# œ 1 Ê x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ 1 and m œ 1
Ê y œ 1 (x 0) œ (x 1). If x œ 2, then y œ 1 and m œ 1 Ê y œ 1 (x 2) œ (x 3).
26.
"
4
œ m œ lim
Èx h Èx
œ lim
h
y œ 2 "4 (x 4) œ
hÄ!
f(2 h) f(2)
h
x
4
Èx h Èx
h
hÄ!
h Ä ! h ŠÈx h Èx‹
27. lim
œ lim
h
hÄ!
œ
"
#È x
. Thus,
"
4
œ
†
Èx h Èx
Èx h Èx
"
#Èx
(x h) x
œ lim
h Ä ! h ŠÈx h Èx‹
Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is
1.
œ lim
hÄ!
a100 4.9(# h)# b a100 4.9(2)# b
h
4.9 a4 4h h# b 4.9(4)
h
œ lim
hÄ!
œ lim (19.6 4.9h) œ 19.6. The minus sign indicates the object is falling downward at a speed of
hÄ!
19.6 m/sec.
f(10 h) f(10)
h
hÄ!
28. lim
3(10 h)# 3(10)#
h
hÄ!
œ lim
29. lim
f(3 h) f(3)
h
œ lim
30. lim
f(2 h) f(2)
h
œ lim
hÄ!
hÄ!
hÄ!
hÄ!
1(3 h)# 1(3)#
h
41
3
œ lim
(2 h)$ 431 (2)$
h
f(0 h) f(0)
h
hÄ!
31. Slope at origin œ lim
3 a20h h# b
h
hÄ!
œ lim
hÄ!
œ lim
1 c9 6h h# 9d
h
41
3
hÄ!
h# sin ˆ "h ‰
h
hÄ!
œ lim
œ 60 ft/sec.
œ lim 1(6 h) œ 61
c12h 6h# h$ d
h
hÄ!
œ lim
hÄ!
41
3
c12 6h h# d œ 161
œ lim h sin ˆ h" ‰ œ 0 Ê yes, f(x) does have a tangent at
hÄ!
the origin with slope 0.
32. lim
hÄ!
g(0 h) g(0)
h
œ lim
hÄ!
h sin ˆ "h ‰
h
œ lim sin h" . Since lim sin
hÄ!
hÄ!
"
h
does not exist, f(x) has no tangent at
the origin.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives
33.
lim
h Ä !c
lim
hÄ!
34.
f(0 h) f(0)
h
f(0 h) f(0)
h
œ lim c
hÄ!
1 0
h
œ _, and lim b
hÄ!
f(0 h) f(0)
h
10
h
œ lim b
hÄ!
œ _ Ê yes, the graph of f has a vertical tangent at the origin.
œ _, and lim b U(0 h)h U(0) œ lim b
hÄ!
hÄ!
does not have a vertical tangent at (!ß ") because the limit does not exist.
lim
h Ä !c
œ _. Therefore,
U(0 h) U(0)
h
œ lim c
hÄ!
01
h
11
h
œ 0 Ê no, the graph of f
35. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
h#Î& 0
h
œ lim c
hÄ!
"
h$Î&
œ _ and lim b
hÄ!
"
h$Î&
œ _ Ê limit does not exist
Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.
36. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
h%Î& 0
h
œ lim c
hÄ!
"
h"Î&
œ _ and lim b
hÄ!
"
h"Î&
œ _ Ê limit does not exist
Ê y œ x%Î& does not have a vertical tangent at x œ 0.
37. (a) The graph appears to have a vertical tangent at x œ !.
(b)
f(0 h) f(0)
h
hÄ!
lim
h"Î& 0
h
hÄ!
œ lim
œ lim
"
%Î&
hÄ! h
œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.
38. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0)
h
at x œ 0.
œ lim
hÄ!
h$Î& 0
h
œ lim
"
#Î&
hÄ! h
œ _ Ê the graph of y œ x$Î& has a vertical tangent
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
107
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Chapter 2 Limits and Continuity
39. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
4h#Î& 2h
h
œ lim c
hÄ!
4
h$Î&
2 œ _ and lim b
hÄ!
4
h$Î&
#œ_
Ê limit does not exist Ê the graph of y œ 4x#Î& 2x does not have a vertical tangent at x œ 0.
40. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h&Î$ 5h#Î$
h
œ lim h#Î$
hÄ!
5
h"Î$
œ 0 lim
5
"Î$
hÄ! h
y œ x&Î$ 5x#Î$ does not have a vertical tangent at x œ !.
does not exist Ê the graph of
41. (a) The graph appears to have a vertical tangent at x œ 1
and a cusp at x œ 0.
(b) x œ 1:
(1 h)#Î$ (1 h 1)"Î$ "
h
hÄ!
#Î$
"Î$
lim
Ê yœx
x œ 0:
(x 1)
lim f(0 h)h f(0)
hÄ!
(1 h)#Î$ h"Î$ "
h
hÄ!
œ lim
has a vertical tangent at x œ 1;
h#Î$ (h 1)"Î$ (1)"Î$
h
hÄ!
#Î$
"Î$
œ lim
does not exist Ê y œ x
œ _
(x 1)
"
œ lim ’ h"Î$
hÄ!
(h ")"Î$
h
h" “
does not have a vertical tangent at x œ 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 2.7 Tangents and Derivatives
42. (a) The graph appears to have vertical tangents at x œ 0 and
x œ 1.
(b) x œ 0:
h"Î$ (h 1)"Î$ (")"Î$
h
hÄ!
f(0 h) f(0)
h
hÄ!
œ lim
f(1 h) f(1)
h
œ lim
lim
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ 0;
x œ 1:
lim
hÄ!
hÄ!
(1 h)"Î$ (" h 1)"Î$ 1
h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ ".
43. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
h Ä !b
f(0 h) f(0)
h
œ lim b
xÄ!
Èh 0
h
œ lim
È kh k 0
f(0 h) f(0)
h
"
h Ä ! Èh
œ lim c
œ lim c
h
hÄ!
hÄ!
Ê y has a vertical tangent at x œ 0.
lim
h Ä !c
œ _;
È kh k
kh k
œ lim c
hÄ!
"
È kh k
œ_
44. (a) The graph appears to have a cusp at x œ 4.
(b)
lim
f(4 h) f(4)
h
œ lim b
hÄ!
Èk4 (4 h)k 0
h
lim
f(4 h) f(4)
h
œ lim c
hÄ!
Èk4 (4 h)k
h
h Ä !b
h Ä !c
œ lim b
hÄ!
œ lim c
hÄ!
È kh k
h
È kh k
lhl
œ lim b
hÄ!
œ lim c
hÄ!
"
Èh
"
È kh k
œ _;
œ _
Ê y œ È% x does not have a vertical tangent at x œ 4.
45-48. Example CAS commands:
Maple:
f := x -> x^3 + 2*x;x0 := 0;
plot( f(x), x=x0-1/2..x0+3, color=black,
# part (a)
title="Section 2.7, #45(a)" );
q := unapply( (f(x0+h)-f(x0))/h, h );
# part (b)
L := limit( q(h), h=0 );
# part (c)
sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 );
# part (d)
tan_line := f(x0) + L*(x-x0);
plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
109
110
Chapter 2 Limits and Continuity
linestyle=[1,2,5,6,7], title="Section 2.7, #45(d)",
legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)",
"Secant line (h=2)","Secant line (h=3)"] );
Mathematica: (function and value for x0 may change)
Clear[f, m, x, h]
x0 œ p;
f[x_]: œ Cos[x] 4Sin[2x]
Plot[f[x], {x, x0 1, x0 3}]
dq[h_]: œ (f[x0+h] f[x0])/h
m œ Limit[dq[h], h Ä 0]
ytan: œ f[x0] m(x x0)
y1: œ f[x0] dq[1](x x0)
y2: œ f[x0] dq[2](x x0)
y3: œ f[x0] dq[3](x x0)
Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}]
CHAPTER 2 PRACTICE EXERCISES
1. At x œ 1:
Ê
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x) œ 1
lim f(x) œ 1 œ f(1)
x Ä 1
Ê f is continuous at x œ 1.
At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0.
xÄ!
xÄ!
xÄ!
But f(0) œ 1 Á lim f(x)
xÄ!
Ê f is discontinuous at x œ 0.
If we define fa!b œ !, then the discontinuity at x œ ! is
removable.
At x œ 1: lim c f(x) œ 1 and lim b f(x) œ 1
xÄ"
Ê lim f(x) does not exist
xÄ"
xÄ1
Ê f is discontinuous at x œ 1.
2. At x œ 1:
Ê
lim
x Ä "c
f(x) œ 0 and
lim
x Ä "b
f(x) œ 1
lim f(x) does not exist
x Ä "
Ê f is discontinuous at x œ 1.
At x œ 0: lim c f(x) œ _ and lim b f(x) œ _
xÄ!
Ê lim f(x) does not exist
xÄ!
xÄ!
Ê f is discontinuous at x œ 0.
At x œ 1: lim c f(x) œ lim b f(x) œ 1 Ê lim f(x) œ 1.
xÄ"
xÄ1
xÄ"
But f(1) œ 0 Á lim f(x)
xÄ1
Ê f is discontinuous at x œ 1.
If we define fa"b œ ", then the discontinuity at x œ " is
removable.
3. (a)
(b)
lim a3fatbb œ 3 lim fatb œ 3(7) œ 21
t Ä t!
t Ä t!
#
#
lim afatbb œ Š lim fatb‹ œ a(b# œ 49
t Ä t!
t Ä t!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 2 Practice Exercises
(c)
(d)
(e)
(f)
111
lim afatb † gatbb œ lim fatb † lim gatb œ (7)(0) œ 0
t Ä t!
t Ä t!
lim fatb
t Ä t! g(t)7
Ät
t Ä t!
lim fatb
œ
Ät
t
t
Ät
t
!
Ät
t
!
7
07
œ
lim gatb lim 7
t
!
Ät
lim fatb
œ
!
lim agatb 7b
!
œ1
lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1
t Ä t!
t Ä t!
lim kfatbk œ ¹ lim fatb¹ œ k7k œ 7
t Ä t!
t Ä t!
(g) lim afatb gatbb œ lim fatb lim gatb œ 7 0 œ 7
t Ä t!
(h)
4. (a)
(b)
(c)
(d)
(e)
(f)
t Ä t!
lim Š " ‹
t Ä t! fatb
œ
"
lim fatb
t
Ät
t Ä t!
"
7
œ
!
œ 71
lim g(x) œ lim g(x) œ È2
xÄ!
xÄ!
lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ
xÄ!
xÄ!
xÄ!
lim af(x) g(x)b œ lim f(x) lim g(x) œ
xÄ!
"
lim
x Ä ! f(x)
œ
xÄ!
"
lim f(x)
œ
xÄ!
"
"
#
xÄ!
œ2
"
#
lim ax f(x)b œ lim x lim f(x) œ 0
xÄ!
xÄ!
f(x)†cos x
x 1
xÄ!
lim
xÄ!
lim f(x)† lim cos x
œ
xÄ!
xÄ!
lim x lim 1
xÄ!
xÄ!
œ
ˆ "# ‰ (1)
01
"
#
È2
#
È2
œ
"
#
œ #"
5. Since lim x œ 0 we must have that lim (4 g(x)) œ 0. Otherwise, if lim (% g(x)) is a finite positive
xÄ!
xÄ!
xÄ!
’ 4xg(x) “
’ 4xg(x) “
œ _ and lim b
œ _ so the limit could not equal 1 as
xÄ!
x Ä 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) œ 4.
number, we would have lim c
xÄ!
xÄ!
6. 2 œ lim
x Ä %
xÄ!
’x lim g(x)“ œ lim x † lim
xÄ!
x Ä %
x Ä %
’ lim g(x)“ œ 4 lim
xÄ!
(since lim g(x) is a constant) Ê lim g(x) œ
xÄ!
xÄ!
2
%
x Ä %
œ #" .
’ lim g(x)“ œ 4 lim g(x)
xÄ!
xÄ!
7. (a) xlim
faxb œ xlim
x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a_ß _b.
Äc
Äc
(b) xlim
gaxb œ xlim
x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ.
Äc
Äc
(c) xlim
haxb œ xlim
x#Î$ œ
Äc
Äc
(d) xlim
kaxb œ xlim
x"Î' œ
Äc
Äc
"
c#Î$
"
c"Î'
œ hacb for every nonzero real number c Ê h is continuous on a_ß !b and a_ß _b.
œ kacb for every positive real number c Ê k is continuous on a!ß _b
8. (a) - ˆˆn "# ‰1ß ˆn "# ‰1‰, where I œ the set of all integers.
n−I
(b) - an1ß an 1b1b, where I œ the set of all integers.
n−I
(c) a_ß 1b a1ß _b
(d) a_ß !b a!ß _b
9.
(a)
(b)
10. (a)
(x 2)(x 2)
x# 4x 4
$ 5x# 14x œ lim
x
xÄ!
x Ä ! x(x 7)(x 2)
x2
x2
lim
œ _ and lim b x(x
7)
x Ä !c x(x 7)
œ lim
lim (x 2)(x 2)
x Ä # x(x 7)(x #)
œ lim
lim
#
x 4x 4
xÄ!
lim $
#
x Ä # x 5x 14x
x# x
lim &
%
$
x Ä ! x 2x x
Now lim c
xÄ!
œ
œ lim
x(x 1)
$
#
x Ä ! x ax 2x 1b
1
x# (x 1)
œ _ and lim b
xÄ!
x2
, x Á 2; the limit does not exist because
x2
, x Á 2, and lim
x Ä ! x(x 7)
œ _
x Ä # x(x 7)
œ lim
x1
#
x Ä ! x (x 1)(x 1)
1
x# (x 1)
x2
x Ä # x(x 7)
œ lim
œ _ Ê lim
"
#
x Ä 0 x (x 1)
#
x x
&
%
$
x Ä ! x 2x x
œ
0
2(9)
œ0
, x Á 0 and x Á 1.
œ _.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
112
Chapter 2 Limits and Continuity
(b)
x# x
exist because
$
#
x Ä " x ax 2x 1b
"
1 Èx
1x
œ lim
12. xlim
Äa
x # a#
x % a%
œ xlim
Äa
13. lim
(x h)# x#
h
œ lim
(x h)# x#
h
xÄ!
œ lim
hÄ!
"
15. lim
#x #
16. lim
(# x)$ 8
x
xÄ!
xÄ!
17.
18.
xÄ!
xÄ!
lim
xÄ1
"
x g(x)
3x# 1
g(x)
œ2 Ê
xÄ!
"
x Ä ! 4 #x
xÄ!
xÄ1
5 x#
œ0 Ê
"
#
lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2.
xÄ!
x Ä !b
Ê È5 lim
x Ä È5
g(x) œ
"
#
Ê
lim
x Ä È5
g(x) œ
%x )
$x $
x Ä #
#!
&!
ˆ"
œxÄ
lim
_ $x
%
$x#
"
#
#
#
&
œ
"
x#
24. x lim
œ x lim
œ
Ä _ x # (x "
Ä _ " (x x"#
#x $
##. x Ä
lim
œxÄ
lim
_ &x# (
_ &
) ‰
$x$
!
"!!
$
x#
(
x#
œ
#!
&!
œ!!!œ!
œ!
#
%
x (x
x(
25. x Ä
lim
œxÄ
lim
œ _
_ x 1
_ " "x
$
x x
x"
#'. x lim
œxÄ
lim
œ_
Ä _ "#x$ "#)
_ "# "#)
x$
lsin xl
lsin xl
"
27. x lim
Ÿ x lim
œ ! since int x Ä _ as x Ä _ Êx lim
œ !.
Ä _ gx h
Ä _ gx h
Ä _ gx h
lim
)Ä_
29. x lim
Ä_
lcos ) "l
)
Ÿ lim
"
l#l
)Ä_ )
x sin x #Èx
x sin x
#Î$
È5
lim g(x) œ _ since lim a5 x# b œ 1
# $
28.
"
#
xÄ1
x Ä #
#x $
x
21. x lim
œ x lim
œ
Ä _ &x (
Ä _ & (x
#
œ2 Ê
œ _ Ê lim g(x) œ 0 since lim a3x# 1b œ 4
lim
x Ä # Èg(x)
x
23. x Ä
lim
_
œ "4
œ lim ax# 6x 12b œ 12
(x g(x)) œ
lim
"
#a#
œ lim (2x h) œ h
œ lim
x Ä È&
œ
"
#
hÄ!
"Î$
x Ä È&
œ
œ _.
œ lim (2x h) œ 2x
ax$ 6x# 12x 8b 8
x
œ lim
"
x # a#
œ xlim
Äa
ax# 2hx h# b x#
h
2 (2 x)
2x(# x)
œ lim
"
lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“
x Ä !b
xÄ!
19. lim
20.
x
ax # a # b
ax # a # b a x # a # b
"
lim
#
x Ä "b x (x 1)
x Ä 1 1 Èx
, x Á 0 and x Á 1. The limit does not
1
#
x Ä " x (x 1)
œ lim
ax# 2hx h# b x#
h
xÄ!
14. lim
"
" Èx
x Ä 1 ˆ1 È x ‰ ˆ 1 È x ‰
hÄ!
œ lim
œ _ and
lim
#
x Ä "c x (x 1)
11. lim
xÄ1
x(x 1)
œ lim
lim
&
%
$
x Ä " x 2x x
œ ! Ê lim
œ x lim
Ä_
)Ä_
" sinx x È#x
" sinx x
&Î$
x x
" x
30. x lim
œ x lim
#x œ
Ä _ x#Î$ cos# x
Ä _Œ " cos#Î$
lcos ) "l
)
œ !.
"!!
"!
œ"
œ
"!
"!
œ"
x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
#
&
Chapter 2 Practice Exercises
31. At x œ 1:
œ
lim
x Ä "c
lim
x Ä "c
lim
x Ä "b
f(x) œ
x ax # 1 b
x# 1
f(x) œ
œ
lim
x ax # 1 b
kx # 1 k
lim
x Ä "c
lim
x Ä "c
x Ä "b
x œ 1, and
x ax # 1 b
kx # 1 k
œ
x ax # 1 b
lim
#
x Ä "b ax "b
œ lim (x) œ (1) œ 1. Since
x Ä 1
lim f(x) Á lim b f(x)
x Ä "c
x Ä "
Ê
lim f(x) does not exist, the function f cannot be
x Ä 1
extended to a continuous function at x œ 1.
At x œ 1:
lim f(x) œ lim c
xÄ"
x Ä "c
#
x ax # 1 b
kx # 1 k
œ lim c
xÄ"
#
x ax # 1 b
ax # 1 b
œ lim c (x) œ 1, and
xÄ"
lim f(x) œ lim b xkaxx# 11k b œ lim b x axx# "1b œ lim b x œ 1. Again lim f(x) does not exist so f
xÄ1
xÄ"
xÄ"
xÄ1
cannot be extended to a continuous function at x œ 1 either.
x Ä "b
32. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin
xÄ!
"
x
does not exist.
33. Yes, f does have a continuous extension to a œ 1:
"
define f(1) œ lim xxÈ
œ 43 .
%
x
xÄ1
34. Yes, g does have a continuous extension to a œ 1# :
)
5
g ˆ 1# ‰ œ lim1 45)cos
#1 œ 4 .
)Ä #
35. From the graph we see that lim h(t) Á lim h(t)
tÄ!
tÄ!
so h cannot be extended to a continuous function
at a œ 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
113
114
Chapter 2 Limits and Continuity
36. From the graph we see that lim c k(x) Á lim b k(x)
xÄ!
xÄ!
so k cannot be extended to a continuous function
at a œ 0.
37. (a) f(1) œ 1 and f(2) œ 5 Ê f has a root between 1 and 2 by the Intermediate Value Theorem.
(b), (c) root is 1.32471795724
38. (a) f(2) œ 2 and f(0) œ 2 Ê f has a root between 2 and 0 by the Intermediate Value Theorem.
(b), (c) root is 1.76929235424
CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
1. (a)
x
xx
0.1
0.7943
0.01
0.9550
0.001
0.9931
10
100
1000
0.3679
0.3679
0.3679
0.0001
0.9991
0.00001
0.9999
Apparently, lim b xx œ 1
xÄ!
(b)
2. (a)
x
ˆ "x ‰"ÎÐln xÑ
Apparently,
"ÎÐln xÑ
lim ˆ " ‰
xÄ_ x
œ 0.3678 œ
"
e
(b)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 2 Additional and Advanced Exercises 115
3.
lim L œ lim c L! É" vc# œ L! É1 vÄcc# œ L! É1 cc# œ 0
vÄc
The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster
than the speed of light).
lim v#
#
v Ä cc
#
4. ¹
Èx
#
1¹ 0.2 Ê 0.2
Èx
#
1 0.2 Ê 0.8
Èx
#
1.2 Ê 1.6 Èx 2.4 Ê 2.56 x 5.76.
¹
Èx
#
1¹ 0.1 Ê 0.1
Èx
#
1 0.1 Ê 0.9
Èx
#
1.1 Ê 1.8 Èx 2.2 Ê 3.24 x 4.84.
5. k10 (t 70) ‚ 10% 10k 0.0005 Ê k(t 70) ‚ 10% k 0.0005 Ê 0.0005 (t 70) ‚ 10% 0.0005
Ê 5 t 70 5 Ê 65° t 75° Ê Within 5° F.
6. We want to know in what interval to hold values of h to make V satisfy the inequality
lV "!!!l œ l$'1h "!!!l Ÿ "!. To find out, we solve the inequality:
**!
l$'1h "!!!l Ÿ "! Ê "! Ÿ $'1h "!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $'
1 Ÿ hŸ
"!"!
$'1
Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe.
The interval in which we should hold h is about )Þ* )Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect
to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f(x) œ lim ax# 7b œ ' œ f(1).
xÄ1
xÄ1
Step 1: kax# 7b 6k % Ê % x# 1 % Ê 1 % x# 1 % Ê È1 % x È1 %.
Step 2: kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ È1 % or $ " œ È1 %. Choose $ œ min š1 È1 %ß È1 % 1› , then
0 kx 1k $ Ê kax# (b 6k % and lim f(x) œ 6. By the continuity test, f(x) is continuous at x œ 1.
xÄ1
8. Show lim" g(x) œ lim"
xÄ
xÄ
%
"
2x
œ 2 œ g ˆ "4 ‰ .
%
Step 1: ¸ #"x 2¸ % Ê % #"x # % Ê # % #"x # % Ê
Step 2: ¸B "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" .
Then $
Choose $ œ
"
4
œ
"
4 #%
%
4(#%)
Ê $œ
"
4
"
4 #%
œ
%
4(2 %)
, or $
"
4
œ
, the smaller of the two values. Then 0 ¸x
By the continuity test, g(x) is continuous at x œ
"
4
"
4 #% Ê
4" ¸ $
"
4#%
x
"
4 #%
¸ #"x
"
4#%
.
"
4
%
4(2 %)
$œ
œ
Ê
2¸ % and lim"
.
xÄ
%
"
#x
œ 2.
.
9. Show lim h(x) œ lim È2x 3 œ " œ h(2).
xÄ#
xÄ#
Step 1: ¹È2x 3 1¹ % Ê % È2x 3 " % Ê " % È2x 3 " % Ê
(1 %)# $
#
x
(" %)# 3
.
#
Step 2: kx 2k $ Ê $ x 2 $ or $ # x $ #.
(" % )# $
Ê $œ
#
(" % Ñ # $
(" %Ñ# "
#œ
#
#
Then $ # œ
#
Ê $œ
œ%
#
(" %)# $
œ " (1# %)
#
#
%# . Choose $ œ %
œ%
%#
#,
%#
#
, or $ # œ
(" %)# $
#
the smaller of the two values . Then,
! kx 2k $ Ê ¹È2x 3 "¹ %, so lim È2x 3 œ 1. By the continuity test, h(x) is continuous at x œ 2.
xÄ#
10. Show lim F(x) œ lim È9 x œ # œ F(5).
xÄ&
xÄ&
Step 1: ¹È9 x 2¹ % Ê % È9 x # % Ê 9 (2 %)# x * (# %)# .
Step 2: 0 kx 5k $ Ê $ x & $ Ê $ & x $ &.
Then $ & œ * (# %)# Ê $ œ (# %)# % œ %# #%, or $ & œ * (# %)# Ê $ œ % (# %)# œ %# #%.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
116
Chapter 2 Limits and Continuity
Choose $ œ %# #%, the smaller of the two values. Then, ! kx 5k $ Ê ¹È9 x #¹ %, so
lim È9 x œ #. By the continuity test, F(x) is continuous at x œ 5.
xÄ&
11. Suppose L" and L# are two different limits. Without loss of generality assume L# L" . Let % œ
"
3
(L# L" ).
Since x lim
f(x) œ L" there is a $" 0 such that 0 kx x! k $" Ê kf(x) L" k % Ê % f(x) L" %
Äx
!
Ê "3 (L# L" ) L" f(x)
"
3
(L# L" ) L" Ê 4L" L# 3f(x) 2L" L# . Likewise, x lim
f(x) œ L#
Ä x!
so there is a $# such that 0 kx x! k $# Ê kf(x) L# k % Ê % f(x) L# %
Ê "3 (L# L" ) L# f(x) 3" (L# L" ) L# Ê 2L# L" 3f(x) 4L# L"
Ê L" 4L# 3f(x) 2L# L" . If $ œ min e$" ß $# f both inequalities must hold for 0 kx x! k $ :
4L" L# 3f(x) 2L" L#
Ê 5(L" L# ) 0 L" L# . That is, L" L# 0 and L" L# 0,
L" %L# 3f(x) 2L# L"
a contradiction.
12. Suppose xlim
f(x) œ L. If k œ !, then xlim
kf(x) œ xlim
0 œ ! œ ! † xlim
f(x) and we are done.
Äc
Äc
Äc
Äc
%
If k Á 0, then given any % !, there is a $ ! so that ! lx cl $ Ê lfaxb Ll l5l
Ê lkllfaxb Ll %
Ê lkafaxb Lb| % Ê lakfaxbb akLbl %. Thus, xlim
kf(x) œ kL œ kŠxlim
f(x)‹.
Äc
Äc
13. (a) Since x Ä 0 , 0 x$ x 1 Ê ax$ xb Ä 0 Ê
lim f ax$ xb œ lim c f(y) œ B where y œ x$ x.
yÄ!
x Ä !b
(b) Since x Ä 0 , 1 x x$ 0 Ê ax$ xb Ä 0 Ê
(c) Since x Ä 0 , 0 x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax$ xb œ lim b f(y) œ A where y œ x$ x.
yÄ!
x Ä !c
lim f ax# x% b œ lim b f(y) œ A where y œ x# x% .
yÄ!
x Ä !b
(d) Since x Ä 0 , 1 x 0 Ê ! x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax# x% b œ A as in part (c).
x Ä !b
14. (a) True, because if xlim
(f(x) g(x)) exists then xlim
(f(x) g(x)) xlim
f(x) œ xlim
[(f(x) g(x)) f(x)]
Äa
Äa
Äa
Äa
œ xlim
g(x) exists, contrary to assumption.
Äa
(b) False; for example take f(x) œ
"
x
and g(x) œ x" . Then neither lim f(x) nor lim g(x) exists, but
lim (f(x) g(x)) œ lim ˆ "x x" ‰ œ lim 0 œ 0 exists.
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous
functions).
1, x Ÿ 0
Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is
(d) False; for example let f(x) œ œ
1, x 0
continuous at x œ 0.
15. Show lim f(x) œ lim
x Ä 1
x# "
x Ä 1 x 1
œ lim
x Ä 1
(x 1)(x ")
(x 1)
Define the continuous extension of f(x) as F(x) œ œ
œ #, x Á 1.
x# 1
x1 ,
2
x Á "
. We now prove the limit of f(x) as x Ä 1
, x œ 1
exists and has the correct value.
#
Step 1: ¹ xx 1" (#)¹ % Ê %
(x 1)(x ")
(x 1)
# % Ê % (x 1) # %, x Á " Ê % " x % ".
Step 2: kx (1)k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ % " Ê $ œ %, or $ " œ % " Ê $ œ %. Choose $ œ %. Then ! kx (1)k $
#
Ê ¹ xx 1" a#b¹ % Ê
lim F(x) œ 2. Since the conditions of the continuity test are met by F(x), then f(x) has a
x Ä 1
continuous extension to F(x) at x œ 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 2 Additional and Advanced Exercises 117
16. Show lim g(x) œ lim
xÄ$
xÄ$
x# 2x 3
2x 6
œ lim
xÄ$
(x 3)(x ")
2(x 3)
œ #, x Á 3.
#
x 2x 3
2x 6 ,
Define the continuous extension of g(x) as G(x) œ œ
xÁ3
. We now prove the limit of g(x) as
, xœ3
2
x Ä 3 exists and has the correct value.
Step 1: ¹ x
#
2x 3
#x 6
2¹ % Ê %
(x 3)(x ")
2(x 3)
# % Ê %
x"
#
# % , x Á $ Ê $ #% x $ #% .
Step 2: kx 3k $ Ê $ x 3 $ Ê $ $ x $ $.
Then, $ $ œ $ #% Ê $ œ #%, or $ $ œ $ #% Ê $ œ #%. Choose $ œ #%. Then ! kx 3k $
Ê ¹x
#
2x 3
2x 6
2¹ % Ê lim
xÄ$
(x 3)(x ")
#(x 3)
œ 2. Since the conditions of the continuity test hold for G(x),
g(x) can be continuously extended to G(x) at B œ 3.
17. (a) Let % ! be given. If x is rational, then f(x) œ x Ê kf(x) 0k œ kx 0k % Í kx 0k %; i.e., choose
$ œ %. Then kx 0k $ Ê kf(x) 0k % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x) 0k %
Í ! % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case,
given % ! there is a $ œ % ! such that ! kx 0k $ Ê kf(x) 0k %. Therefore, f is continuous at
x œ 0.
(b) Choose x œ c !. Then within any interval (c $ ß c $ ) there are both rational and irrational numbers.
If c is rational, pick % œ #c . No matter how small we choose $ ! there is an irrational number x in
(c $ ß c $ ) Ê kf(x) f(c)k œ k0 ck œ c
c
#
œ %. That is, f is not continuous at any rational c 0. On
the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $ !
there is a rational number x in (c $ ß c $ ) with kx ck
œ kxk
c
#
œ% Í
œ % Ê f is not continuous at any irrational c 0.
If x œ c 0, repeat the argument picking % œ
nonzero value x œ c.
18. (a) Let c œ
c
#
kc k
#
œ
c
# .
x
c
#
Then kf(x) f(c)k œ kx 0k
3c
#.
Therefore f fails to be continuous at any
m
n
be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ
"
#n
œ %. Therefore f is discontinuous at x œ c, a rational number.
"
#n .
No matter how
small $ ! is taken, there is an irrational number x in the interval (c $ ß c $ ) Ê kf(x) f(c)k œ ¸0 "n ¸
œ
"
n
(b) Now suppose c is an irrational number Ê f(c) œ 0. Let % 0 be given. Notice that
number reduced to lowest terms with denominator 2 and belonging to [0ß 1];
denominator 3 belonging to [0ß 1];
"
4
and
[0ß 1]; etc. In general, choose N so that
"
N
3
4
with denominator 4 in [0ß 1];
"
3
and
" 2 3
5, 5, 5
2
3
and
"
#
is the only rational
the only rationals with
4
5
with denominator 5 in
% Ê there exist only finitely many rationals in [!ß "] having
denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc ri k : i œ 1ß á ß pf . Then the interval (c $ ß c $ )
contains no rational numbers with denominator Ÿ N. Thus, 0 kx ck $ Ê kf(x) f(c)k œ kf(x) 0k
œ kf(x)k Ÿ N" % Ê f is continuous at x œ c irrational.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
118
Chapter 2 Limits and Continuity
(c) The graph looks like the markings on a typical ruler
when the points (xß f(x)) on the graph of f(x) are
connected to the x-axis with vertical lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the
zero point, 0, on the equator Ê 0 1R represents the midnight point (at the same exact time). Suppose x"
is a point on the equator “just after" noon Ê x" 1R is simultaneously “just after" midnight. It seems
reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically
opposite point just after midnight: That is, T(x" ) T(x" 1R) 0. At exactly the same moment in time
pick x# to be a point just before midnight Ê x# 1R is just before noon. Then T(x# ) T(x# 1R) 0.
Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate
Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that
T(c) T(c 1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the
temperatures are the same.
#
#
#
#
"
20. xlim
f(x)g(x) œ xlim
af(x) g(x)b‹ Šxlim
af(x) g(x)b‹ “
’af(x) g(x)b af(x) g(x)b “ œ "% ’Šxlim
Äc
Äc %
Äc
Äc
œ "% ˆ$# a"b# ‰ œ #.
21. (a) At x œ 0: lim r (a) œ lim
aÄ!
œ lim
1 (" a)
aÄ!
a Ä ! a ˆ" È1 a‰
At x œ 1:
(b) At x œ 0:
lim
a Ä "b
œ
r (a) œ
" È1 a
a
1
" È1 0
œ lim c
aÄ!
1 (" a)
a ˆ" È1 a‰
œ
"
#
aÄ!
1 (1 a)
lim
a Ä "b a ˆ1 È1 a‰
lim r (a) œ lim c
aÄ!
a Ä !c
È1 a
œ lim Š " a
" È1 a
a
œ lim c
aÄ!
" È1 a
‹ Š " È1 a ‹
a
œ lim
a Ä 1 a ˆ" È1 a‰
È1 a
œ lim c Š " a
aÄ!
a
a ˆ 1 È 1 a ‰
œ lim c
aÄ!
œ
"
" È0
œ1
" È1 a
‹ Š " È1 a ‹
"
œ _ (because the
" È1 a
"
œ _ (because the
" È1 a
denominator is always negative); lim b r (a) œ lim b
aÄ!
aÄ!
is always positive). Therefore, lim r (a) does not exist.
aÄ!
At x œ 1:
lim r (a) œ lim b
a Ä "b
a Ä "
1 È 1 a
a
œ
lim
"
a Ä 1b " È1 a
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
denominator
Chapter 2 Additional and Advanced Exercises 119
(c)
(d)
22. f(x) œ x 2 cos x Ê f(0) œ 0 2 cos 0 œ 2 0 and f(1) œ 1 2 cos (1) œ 1 # 0. Since f(x) is
continuous on [1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [1 #ß #].
Thus there is some number c in [1ß !] such that f(c) œ 0; i.e., c is a solution to x 2 cos x œ 0.
23. (a) The function f is bounded on D if f(x) M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x
in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then
B Ÿ f(x) Ÿ B Ê f(x) B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and
M œ B a lower bound.
(b) Assume f(x) Ÿ N for all x and that L N. Let % œ L # N . Since x lim
f(x) œ L there is a $ ! such that
Äx
!
0 kx x! k $ Ê kf(x) Lk % Í L % f(x) L % Í L
Í
LN
#
f(x)
3L N
# .
But L N Ê
LN
#
Ê L
f(x) L
b, then a b
24. (a) If a
ML
#
Í
3L M
#
f(x)
ML
# . As in part (b), 0 kx
L
M
M, a contradiction.
#
0 Ê ka bk œ a b Ê max (aß b) œ
ab
#
ka b k
#
If a Ÿ b, then a b Ÿ 0 Ê ka bk œ (a b) œ b a Ê max (aß b) œ
œ
2b
#
f(x) L
x! k $
ab
ab
2a
# # œ # œ a.
ka b k
ab
œ a # b b # a
#
#
œ
œ b.
(b) Let min (aß b) œ
ab
#
ka b k
#
LN
#
N Ê N f(x) contrary to the boundedness assumption
f(x) Ÿ N. This contradiction proves L Ÿ N.
(c) Assume M Ÿ f(x) for all x and that L M. Let % œ
ML
#
LN
#
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
120
Chapter 2 Limits and Continuity
25. lim œ
xÄ0
sina" cos xb
x
œ lim
œ lim
xÄ0
†
sin x
xÄ0 x
26.
lim
sin x
x Ä 0b sin Èx
œ
sina" cos xb
" cos x
sin x
" cos x
sin x
lim
x Ä 0b B
†
†
" cos x
x
Èx
sin Èx
†
†
œ lim
x
Èx
œ lim
sinasin xb
sin x
28. lim
sinax# xb
x
œ lim
sinax# xb
x# x
† ax "b œ lim
sinax# %b
x Ä 2 x2
œ lim
sinax# %b
#
x Ä 2 x %
† ax 2b œ lim
xÄ0
29. lim
xÄ0
sinˆÈx $‰
x9
xÄ9
30. lim
sin x
x
xÄ0
sinasin xb
sin x
†
sina" cos xb
" cos x
† lim
" cos# x
x Ä 0 xa" cos xb
"
Èx $
† lim
sin x
xÄ0 x
œ " † lim
œ " † " œ ".
sinax# xb
x# x
† lim ax "b œ " † " œ "
sinax# %b
#
x Ä 2 x %
† lim ax 2b œ " † % œ %
xÄ0
sinˆÈx $‰
x Ä 9 Èx $
œ lim
xÄ0
œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !.
x Ä 0 Š Èx ‹ x Ä 0
sinasin xb
x
xÄ0
œ lim
œ " † ˆ #! ‰ œ !.
27. lim
xÄ0
" cos x
" cos x
†
xÄ0
xÄ2
sinˆÈx $‰
x Ä 9 Èx $
œ lim
† lim
"
x Ä 9 Èx $
œ"†
"
'
œ
"
'
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
sin# x
x Ä 0 xa" cos xb
CHAPTER 3 DIFFERENTIATION
3.1 THE DERIVATIVE OF A FUNCTION
1. Step 1: f(x) œ 4 x# and f(x h) œ 4 (x h)#
f(x h) f(x)
h
Step 2:
œ
c4 (x h)# d a4 x# b
h
œ
a4 x# 2xh h# b 4 x#
h
œ
2xh h#
h
œ
h(2x h)
h
œ 2x h
Step 3: f w (x) œ lim (2x h) œ 2x; f w ($) œ 6, f w (0) œ 0, f w (1) œ 2
hÄ!
c(x h 1)# 1d c(x 1)# 1d
h
hÄ!
2xh h# 2h
lim
œ lim (2x h 2)
h
hÄ!
hÄ!
2. F(x) œ (x 1)# 1 and F(x h) œ (x h 1)# " Ê Fw (x) œ lim
ax# 2xh h# 2x 2h 1 1b ax# 2x 1 1b
h
œ lim
hÄ!
œ
œ 2(x 1); Fw (1) œ 4, Fw (0) œ 2, Fw (2) œ 2
3. Step 1: g(t) œ
"
t#
and g(t h) œ
"
"
# #
g(t h) g(t)
œ (t h)h t
h
2t h)
2t h
œ h(
(t h)# t# h œ (t h)# t#
Step 2:
2t h
Step 3: gw (t) œ lim
1 z
#z
4. k(z) œ
and k(z h) œ
Œ
œ
œ
# #
h Ä ! (t h) t
"
(t h)#
h
2t
t# †t#
1 (z h)
2(z h)
œ
(" z)(z h)
lim (1 z h)z
#(z h)zh
hÄ!
œ
"
2z#
œ
t# (t h)#
(t h)# †t#
œ
2
t$
t# at# 2th h# b
(t h)# †t# †h
œ
œ
2th h#
(t h)# t# h
2
; gw (1) œ 2, gw (2) œ "4 , gw ŠÈ3‹ œ 3È
3
Ê kw (z) œ lim
Š
" (z h) " z ‹
#(z h)
#z
h
hÄ!
#
z h z# zh
lim z z zh
2(z h)zh
hÄ!
h
œ lim
h Ä ! 2(z h)zh
œ lim
"
h Ä ! #(z h)z
; kw (") œ "# , kw (1) œ "# , kw ŠÈ2‹ œ "4
5. Step 1: p()) œ È3) and p() h) œ È3() h)
Step 2:
p() h) p())
h
œ
œ
È3() h) È3)
h
3h
h ŠÈ3) 3h È3)‹
Step 3: pw ()) œ lim
œ
ŠÈ3) 3h È3)‹
œ
3
È3) 3h È3)
3
œ
h Ä ! È3) 3h È3)
†
h
3
È 3) È 3)
œ
3
2È 3 )
ŠÈ3) 3h È3)‹
ŠÈ3) 3h È3)‹
; pw (1) œ
6. r(s) œ È2s 1 and r(s h) œ È2(s h) 1 Ê rw (s) œ lim
hÄ!
œ lim
hÄ!
œ lim
ŠÈ2s h 1 È2s 1‹
h
†
2h
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
œ
"
È2s 1
; rw (0) œ 1, rw (1) œ
ŠÈ2s 2h 1 È2s 1‹
ŠÈ2s 2h 1 È2s 1‹
œ lim
"
È3
6x# h 6xh# 2h$
h
hÄ!
3
#È2
È2s 2h 1 È2s 1
h
œ
2
È2s 1 È2s 1
œ
2
2È2s 1
"
È2
dy
dx
h a6x# 6xh 2h# b
h
hÄ!
œ lim
, pw (3) œ "# , pw ˆ 32 ‰ œ
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
2
7. y œ f(x) œ 2x$ and f(x h) œ 2(x h)$ Ê
œ lim
(3) 3h) 3)
h ŠÈ3) 3h È3)‹
(2s 2h 1) (2s 1)
œ lim
h Ä ! È2s 2h 1 È2s 1
, rw ˆ #" ‰ œ
3
2È 3
œ
2(x h)$ 2x$
h
hÄ!
œ lim
2 ax$ 3x# h 3xh# h$ b 2x$
h
hÄ!
œ lim
œ lim a6x# 6xh 2h# b œ 6x#
hÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
122
Chapter 3 Differentiation
8. r œ
œ
s$
#
1 Ê
”
œ lim
dr
ds
(s h)$
#
t
2t1
Š
œ lim
(t b h)(2t b 1) c t(2t b 2h b 1)
‹
(2t b 2h b 1)(2t b 1)
h
œ
2t# t 2ht h 2t# 2ht t
(2t 2h 1)(2t 1)h
hÄ!
"
"
(2t 1)(2t 1) œ (2t 1)#
dv
dt
œ lim
œ
"
th“
ˆt "t ‰
#
#
th
lim ht h(th h)t
hÄ!
"
Èq 1
11. p œ f(q) œ
œ
œ lim
h
" "
t
h
th
œ
t# 1
t#
"
È(q h) 1
Ê
h Ä ! (2t 2h 1)(2t 1)
Š
h(t h)t t (t h)
‹
(t h)t
h
hÄ!
œ1
"
t#
Š È(q "h) 1 ‹ Š Èq" 1 ‹
œ lim
dp
dq
"
œ lim
œ lim
h
hÄ!
h
hÄ!
Èq 1 Èq h 1
œ lim
h Ä ! hÈ q h 1 È q 1
h
hÄ!
3 #
# s
h
hÄ!
h Ä ! (2t 2h 1)(2t 1)h
and f(q h) œ
3sh h# b œ
t ‰
Š 2(t bt bh)hb 1 ‹ ˆ 2t b
1
œ lim
ds
dt
œ lim
#
1
lim t (thth)t
hÄ!
Èq b 1 c Èq b h b 1
Œ Èq b h b 1 Èq b 1
œ
(t h)(2t 1) t(2t 2h 1)
(2t 2h 1)(2t 1)h
hÄ!
œ lim
h
hÄ!
Ê
œ lim
œ lim
’(t h)
œ
th
2(th)1
and r(t h) œ
hÄ!
c(s h)$ 2d cs$ 2d
"
h
# hlim
Ä!
h c3s# 3sh h# d
"
"
œ # lim a3s#
# hlim
h
Ä!
hÄ!
h
hÄ!
"
s$ 3s# h 3sh# h$ 2 s$ 2
# hlim
h
Ä!
9. s œ r(t) œ
10.
$
1• ’ s# 1“
œ
ˆÈ q 1 È q h 1 ‰ ˆ È q 1 È q h 1 ‰
1) (q h 1)
† ˆÈq 1 Èq h 1‰ œ lim hÈq h 1(qÈq
1 ˆÈ q 1 È q h 1 ‰
h Ä ! h Èq h 1 Èq 1
hÄ!
h
"
lim
œ lim Èq h 1 Èq 1 ˆÈq 1 Èq h 1‰
h Ä ! h È q h 1 È q 1 ˆÈ q 1 È q h 1 ‰
hÄ!
"
"
œ
È q 1 È q 1 ˆÈ q 1 È q 1 ‰
2(q 1) Èq 1
dz
dw
œ lim
œ lim
œ
12.
"
Š È3(w
h) 2
h
hÄ!
"
È3w 2 ‹
ŠÈ3w 2 È3w 3h 2‹
œ lim
hÈ3w 3h 2 È3w 2
hÄ!
È3w 2 È3w 3h 2
œ lim
h Ä ! hÈ3w 3h 2 È3w 2
†
ŠÈ3w2È3w3h2‹
ŠÈ3w 2 È3w 3h 2‹
œ lim
(3w 2) (3w 3h 2)
œ lim
3
h Ä ! hÈ3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
h Ä ! È3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
œ
9
x
and f(x h) œ (x h)
œ
x(x h)# 9x x# (x h) 9(x h)
x(x h)h
œ
h(x# xh 9)
x(x h)h
14. k(x) œ
"
#x
œ lim
hÄ!
œ lim
(# x) (2 x h)
h(2 x)(2 x h)
hÄ!
œ lim
hÄ!
œ
x# xh 9
x(x h)
œ
w
œ
9
(x h)
Ê
f(x h) f(x)
h
œ
’(x h)
x$ 2x# h xh# 9x x$ x# h 9x 9h
x(x h)h
; f (x) œ
and k(x h) œ
"
kw (2) œ 16
ds
dt
3
È3w 2 È3w 2 ŠÈ3w 2 È3w 2‹
3
2(3w 2) È3w 2
13. f(x) œ x
15.
œ
#
lim x xh 9
h Ä ! x(x h)
œ
x# 9
x#
9
9
(x b h) “ ’x x “
h
œ
œ1
x# h xh# 9h
x(x h)h
9
x#
; m œ f w (3) œ 0
Š # "x h
k(x h) k(x)
œ
lim
h
h
hÄ!
hÄ!
h
"
"
lim
œ lim (2 x)(# x h) œ (2 x)# ;
h Ä ! h(2 x)(2 x h)
hÄ!
"
2 (x h)
c(t h)$ (t h)# d at$ t# b
h
3t# h 3th# h$ 2th h#
h
Ê kw (x) œ lim
œ lim
hÄ!
œ lim
hÄ!
"
#x‹
at$ 3t# h 3th# h$ b at# 2th h# b t$ t#
h
h a3t# 3th h# 2t hb
h
œ lim a3t# 3th h# 2t hb
hÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.1 The Derivative of a Function
œ 3t# 2t; m œ
16.
dy
dx
ds ¸
dt tœ"
œ5
(x h 1)$ (x 1)$
h
œ lim
hÄ!
(x 1)$ 3(x ")# h 3(x 1)h# h$ (x 1)$
h
œ lim
hÄ!
œ lim c3(x 1)# 3(x 1)h h# d œ 3(x 1)# ; m œ
hÄ!
17. f(x) œ
œ
8
Èx 2
and f(x h) œ
8 ŠÈx 2 Èx h 2‹
hÈ x h 2 È x 2
†
8
È(x h) 2
f(x h) f(x)
h
Ê
ŠÈx 2 Èx h 2‹
œ
ŠÈx 2 Èx h 2‹
œ
8h
hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
œ
8
Èx 2 Èx 2 ŠÈx 2 Èx 2‹
œ
œ3
dy
dx ¹ x=#
È(x b h) c 2 Èx c 2
8
œ
8
h
8[(x 2) (x h 2)]
hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
8
Ê f w (x) œ lim
h Ä ! Èx h 2 Èx 2 ŠÈx 2 Èx h 2‹
4
(x 2)Èx 2
; m œ f w (6) œ
4
4È 4
œ "# Ê the equation of the tangent
line at (6ß 4) is y 4 œ "# (x 6) Ê y œ "# x $ % Ê y œ "# x (.
18. gw (z) œ lim
ˆ1 È4 (z h)‰ Š1 È4 z‹
h
hÄ!
œ
h
lim
h Ä ! h ŠÈ4 z h È4 z‹
œ
†
h
hÄ!
(4 z h) (4 z)
lim
h Ä ! h ŠÈ4 z h È4 z‹
"
œ "#
2È 4 3
"# z $# # Ê w
ŠÈ4 z h È4 z‹
œ lim
ŠÈ4 z h È4 z‹
ŠÈ4 z h È4 z‹
"
œ lim
h Ä ! ŠÈ4 z h È4 z‹
œ
"
2È 4 z
m œ gw (3) œ
Ê the equation of the tangent line at ($ß #) is w 2 œ "# (z 3)
Êwœ
œ "# z (# .
19. s œ f(t) œ 1 3t# and f(t h) œ 1 3(t h)# œ 1 3t# 6th 3h# Ê
a1 3t# 6th 3h# b a1 3t# b
h
hÄ!
œ lim
20. y œ f(x) œ "
œ lim
hÄ!
h
h Ä ! x(x h)h
hÄ!
2È 4 ) 2È 4 ) h
hÈ 4 ) È 4 ) h
œ
œ
"
xh
Ê
"
lim
h Ä ! x(x h)
2
È4 () h)
4(% )) 4(% ) h)
œ lim
h Ä ! 2hÈ4 ) È4 ) h ŠÈ4 ) È4 ) h‹
œ
2
(4 )) Š2È4 )‹
œ
dy
dx
œ
œ lim
"
x#
"
3
Ê
"
(4 ))È4 )
Ê
Šz h Èz h‹ ˆz Èz‰
hÄ!
h
œ 1 lim
(z h) z
h Ä ! h ŠÈz h Èz‹
hÄ!
Š1
dr ¸
d) )œ!
Ê
dr
d)
È œ
È4 c ) c h È4 c )
2
2
h
2
h Ä ! È4 ) È4 ) h ŠÈ4 ) È% ) h‹
œ
"
8
h Èz h Èz
h
hÄ!
œ lim
œ 1 lim
"
h
œ lim
œ lim
"
x h ‹ Š1 x ‹
hÄ!
dy
dx ¹x= 3
f(t h) f(t)
h
œ6
f(x h) f(x)
h
hÄ!
œ lim
22. w œ f(z) œ z Èz and f(z h) œ (z h) Èz h Ê
œ lim
ds ¸
dt t="
œ lim
f() h) f())
œ lim
h
hÄ!
hÄ!
È
È
2È4 ) #È% ) h Š2 % ) 2 4 ) h‹
lim
† È
Š2 4 ) #È4 ) h‹
h Ä ! hÈ 4 ) È 4 ) h
and f() h) œ
2
È4 )
hÄ!
and f(x h) œ 1
" "
x
xh
œ lim
h
21. r œ f()) œ
œ lim
"
x
œ lim (6t 3h) œ 6t Ê
ds
dt
;
"
h Ä ! Èz h Èz
dw
dz
œ lim
hÄ!
œ lim –1
hÄ!
œ"
"
2È z
Ê
f(z h) f(z)
h
Èz h Èz
dw ¸
dz zœ4
h
œ
†
ŠÈz h Èz‹
ŠÈz h Èz‹ —
5
4
" "
fazb faxb
a x #b a z # b
xz
"
z#
x#
23. f w axb œ zlim
œ zlim
œ zlim
œ zlim
œ zlim
Äx zx
Ä x zx
Ä x az xbaz #bax #b
Ä x az xbaz #bax #b
Ä x az #bax #b
œ ax "
#b#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
123
124
Chapter 3 Differentiation
"
"
#
#
#
#
Òax "b az "bÓÒax "b az "bÓ
ax "b az"b
fazb faxb
az"b
ax"b
24. f w axb œ zlim
œ zlim
œ zlim
œ zlim
zx
Ä x zx
Äx
Ä x az xbaz "b# ax "b#
Äx
az xbaz "b# ax "b#
ax zbax z 2b
"ax z 2b
œ zlim
œ zlim
œ
Ä x az xbaz "b# ax "b#
Ä x a z " b # a x "b#
z
"a#x #b
a x "b %
œ
#ax "b
a x "b %
œ
#
a x "b $
x
gazb gaxb
z a x "b x a z " b
z x
"
zc"
x"
25. gw axb œ zlim
œ zlim
œ zlim
œ zlim
œ zlim
Äx zx
Äx zx
Ä x az xbaz "bax "b
Ä x az xbaz "bax "b
Ä x az "bax "b
œ ax "
"b#
gazb gaxb
26. gw axb œ zlim
œ zlim
Äx zx
Äx
"
"
œ zlim
œ
#È x
Ä x Èz Èx
ˆ" Èz‰ˆ" Èx‰
zx
œ zlim
Äx
Èz Èx
zx
†
Èz Èx
Èz Èx
zx
œ zlim
Ä x az x bˆÈ z È x ‰
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0),
then positive Ê the slope is always increasing which matches (b).
28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x
increases. When x œ 0, the slope of the tangent line to x is 0. For x 0, f#w (x) is positive and increasing. This
graph matches (a).
29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we
expect f$w to be zero, and (d) matches this condition.
30. The graph matches with (c).
31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree.
For example, lim c
xÄ!
f(x) f(0)
x0
œ slope of line joining (%ß 0) and (!ß #) œ
"
#
but lim b
xÄ!
line joining (0ß 2) and ("ß 2) œ 4. Since these values are not equal, f w (0) œ lim
xÄ!
f(x) f(0)
x0
f(x) f(0)
x0
(b)
32. (a)
(b) Shift the graph in (a) down 3 units
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ slope of
does not exist.
Section 3.1 The Derivative of a Function
33.
(b) The fastest is between the 20th and 30th days;
slowest is between the 40th and 50th days.
34. (a)
35. Left-hand derivative: For h 0, f(0 h) œ f(h) œ h# (using y œ x# curve) Ê
œ lim c
hÄ!
h# 0
h
œ lim c h œ 0;
hÄ!
Right-hand derivative: For h 0, f(0 h) œ f(h) œ h (using y œ x curve) Ê
œ lim b
hÄ!
Then lim c
hÄ!
h0
h
lim
h Ä !c
œ lim b 1 œ 1;
hÄ!
f(0 h) f(0)
h
Á lim b
hÄ!
f(0 h) f(0)
h
lim
h Ä !b
œ lim c 0 œ 0;
hÄ!
f(1 h) f(1)
h
lim
h Ä !c
Right-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2(1 h) œ 2 2h Ê
Then lim c
hÄ!
(2 2h)2
h
œ lim b
hÄ!
f(1 h) f(1)
h
2h
h
È1 h "
h
œ lim c
hÄ!
lim
h Ä !b
ŠÈ1 h "‹
h
†
ŠÈ1 h "‹
ŠÈ1 h 1‹
œ lim c
hÄ!
lim
h Ä !c
Then lim c
hÄ!
(2h 1) "
h
f(1 h) f(1)
h
38. Left-hand derivative:
lim
h Ä !c
Right-hand derivative:
œ lim b
hÄ!
Then lim c
hÄ!
h
h(1 h)
œ lim b 2 œ 2;
hÄ!
Á lim b
hÄ!
lim b
hÄ!
œ lim b
hÄ!
f(1 h) f(1)
h
f(1 h) f(1)
h
f(1 h) f(")
h
(1 h) "
h ŠÈ1 h "‹
œ lim c
hÄ!
Á lim b
hÄ!
"
È1 h 1
lim
h Ä !b
Ê the derivative f w (1) does not exist.
œ lim c
f(1 h) f(")
h
"
1h
f(1 h) f(1)
h
f(" h) f(1)
h
Right-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ 2(1 h) 1 œ 2h 1 Ê
œ lim b
hÄ!
22
h
Ê the derivative f w (1) does not exist.
37. Left-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ È1 h Ê
œ lim c
hÄ!
œ lim c
hÄ!
œ lim b 2 œ 2;
hÄ!
f(1 h) f(1)
h
Á lim b
hÄ!
f(0 h) f(0)
h
Ê the derivative f w (0) does not exist.
36. Left-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2 Ê
œ lim b
hÄ!
f(0 h) f(0)
h
hÄ!
œ lim b
hÄ!
(1 h) "
h
œ lim c 1 œ 1;
" "‹
Š1
h
h
hÄ!
œ lim b
hÄ!
Š
1 (1 h)
1h ‹
h
œ 1;
f(1 h) f(1)
h
Ê the derivative f w (1) does not exist.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ #" ;
f("h)f(1)
h
125
126
Chapter 3 Differentiation
39. (a) The function is differentiable on its domain $ Ÿ x Ÿ 2 (it is smooth)
(b) none
(c) none
40. (a) The function is differentiable on its domain # Ÿ x Ÿ 3 (it is smooth)
(b) none
(c) none
41. (a) The function is differentiable on $ Ÿ x 0 and ! x Ÿ 3
(b) none
(c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x)
xÄ!
xÄ!
42. (a) f is differentiable on # Ÿ x 1, " x 0, 0 x 2, and 2 x Ÿ 3
(b) f is continuous but not differentiable at x œ 1: lim f(x) œ 0 exists but there is a corner at x œ 1 since
x Ä 1
œ 3 and lim b f(" h)h f(1) œ 3 Ê f w (1) does not exist
hÄ!
hÄ!
(c) f is neither continuous nor differentiable at x œ 0 and x œ 2:
at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist;
lim c
f(1 h) f(")
h
xÄ!
xÄ0
xÄ!
at x œ 2, lim f(x) exists but lim f(x) Á f(2)
xÄ#
xÄ#
43. (a) f is differentiable on " Ÿ x 0 and 0 x Ÿ 2
(b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so
f(0 h) f(0)
h
hÄ!
f w (0) œ lim
xÄ!
does not exist
(c) none
44. (a) f is differentiable on $ Ÿ x 2, 2 x 2, and 2 x Ÿ 3
(b) f is continuous but not differentiable at x œ 2 and x œ 2: there are corners at those points
(c) none
45. (a) f w (x) œ lim
hÄ!
f(x h) f(x)
h
œ lim
hÄ!
(x h)# ax# b
h
œ lim
hÄ!
x# 2xh h# x#
h
œ lim (2x h) œ 2x
hÄ!
(b)
(c) yw œ 2x is positive for x 0, yw is zero when x œ 0, yw is negative when x 0
(d) y œ x# is increasing for _ x 0 and decreasing for ! x _; the function is increasing on intervals
where yw 0 and decreasing on intervals where yw 0
f(x h) f(x)
h
hÄ!
46. (a) f w (x) œ lim
œ lim
hÄ!
Š xc"
h
h
1 ‹
x
œ lim
hÄ!
x (x h)
x(x h)h
œ lim
"
h Ä ! x(x h)
œ
"
x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.1 The Derivative of a Function
(b)
(c) yw is positive for all x Á 0, yw is never 0, yw is never negative
(d) y œ "x is increasing for _ x 0 and ! x _
w
47. (a) Using the alternate formula for calculating derivatives: f (x) œ
œ
$
$
lim z x
z Ä x 3(z x)
œ
az# zx x# b
lim (z x)3(z
x)
zÄx
œ
#
#
lim z zx3 x
zÄx
f(x)
lim f(z)z
x
zÄx
#
w
$
Š z3
œ zlim
Äx
x$
3 ‹
zx
œ x Ê f (x) œ x#
(b)
(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative
(d) y œ
x$
3
is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw 0; y is
never decreasing
%
48. (a) Using the alternate
œ zlim
Äx
z% x%
4(z x)
œ
%
z
x
Œ4 4
f(z) f(x)
form for calculating derivatives: f (x) œ zlim
œ
lim
zx
zx
Äx
zÄx
(z x) az$ xz# x# z x$ b
z$ xz# x# z x$
$
w
lim
œ zlim
œ x Ê f (x) œ x$
4(z x)
4
zÄx
Äx
w
(b)
(c) yw is positive for x 0, yw is zero for x œ 0, yw is negative for x 0
(d) y œ
x%
4
is increasing on 0 x _ and decreasing on _ x 0
#
#
(xc) ax xc c b
f(x) f(c)
x c
49. yw œ xlim
œ xlim
œ xlim
œ xlim
ax# xc c# b œ 3c# .
xc
xc
Äc
Ä c xc
Äc
Äc
The slope of the curve y œ x$ at x œ c is yw œ 3c# . Notice that 3c# 0 for all c Ê y œ x$ never has a negative
slope.
$
$
50. Horizontal tangents occur where yw œ 0. Thus, yw œ lim
hÄ!
œ lim
hÄ!
2 ŠÈx h Èx‹
h
†
ŠÈx h Èx‹
ŠÈx h Èx‹
œ lim
2È x h 2È x
h
2((x h) x))
h Ä ! h ŠÈx h Èx‹
œ lim
2
h Ä ! Èx h Èx
œ
"
Èx
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
127
128
Chapter 3 Differentiation
Then yw œ 0 when
51. yw œ lim
hÄ!
œ lim
hÄ!
"
Èx
œ 0 which is never true Ê the curve has no horizontal tangents.
a2(x h)# 13(x h) 5b a2x# 13x 5b
h
4xh 2h# 13h
h
œ lim
hÄ!
2x# 4xh 2h# 13x 13h 5 2x# 13x 5
h
œ lim (4x 2h 13) œ 4x 13, slope at x. The slope is 1 when 4x 13 œ "
hÄ!
Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3# 13 † 3 5 œ 16. Thus the tangent line is y 16 œ (1)(x 3)
Ê y œ x "$ and the point of tangency is (3ß 16).
52. For the curve y œ Èx, we have yw œ lim
ŠÈx h Èx‹
hÄ!
"
œ lim
h Ä ! Èx h Èx
œ
"
#Èx
h
†
ŠÈx h Èx‹
ŠÈx h Èx‹
h Ä ! ŠÈx h Èx‹ h
. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and ("ß !) is the point
on the line where it crosses the x-axis. Then the slope of the line is
"
;
2È a
(x h) x
œ lim
using the derivative formula at x œ a Ê
exist: its point of tangency is ("ß "), its slope is
Èa
a1
œ
"
#È a
œ
Èa 0
a (1)
œ
Èa
a1
which must also equal
"
Ê 2a œ a 1 Ê a œ 1.
#Èa
"
# ; and an equation of the line is
Thus such a line does
y1œ
"
#
(x 1)
Ê y œ "# x "# .
53. No. Derivatives of functions have the intermediate value property. The function f(x) œ ÚxÛ satisfies f(0) œ 0
and f(1) œ 1 but does not take on the value "# anywhere in [!ß "] Ê f does not have the intermediate value
property. Thus f cannot be the derivative of any function on [!ß "] Ê f cannot be the derivative of any function
on (_ß _).
54. The graphs are the same. So we know that
for f(x) œ kxk , we have f w (x) œ
kx k
x
.
55. Yes; the derivative of f is f w so that f w (x! ) exists Ê f w (x! ) exists as well.
56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well.
57. Yes, lim
g(t)
t Ä ! h(t)
can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)
œ 0, but lim
g(t)
t Ä ! h(t)
œ lim
tÄ!
mt
t
œ lim m œ m, which need not be zero.
tÄ!
58. (a) Suppose kf(x)k Ÿ x# for " Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim
œ lim
hÄ!
f(h) 0
h
œ lim
hÄ!
f(h)
h .
For khk Ÿ 1, h# Ÿ f(h) Ÿ h# Ê h Ÿ
hÄ!
f(h)
h
f(0 h) f(0)
h
Ÿ h Ê f w (0) œ lim
hÄ!
f(h)
h
œ0
by the Sandwich Theorem for limits.
(b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since " Ÿ sin x Ÿ 1). By part (a),
f is differentiable at x œ 0 and f w (0) œ 0.
59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ
y œ Èx so that
"
#È x
œ lim
hÄ!
Èx h Èx
h
"
2È x
. The graphs reveal that y œ
is the derivative of the function
Èx h Èx
h
gets closer to y œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
#È x
Section 3.1 The Derivative of a Function
as h gets smaller and smaller.
60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so
that 3x# œ lim
hÄ!
(xh)$ x$
h
. The graphs reveal that y œ
(xh)$ x$
h
gets closer to y œ 3x# as h
gets smaller and smaller.
61. Weierstrass's nowhere differentiable continuous function.
62-67. Example CAS commands:
Maple:
f := x -> x^3 + x^2 - x;
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
129
130
Chapter 3 Differentiation
x0 := 1;
plot( f(x), x=x0-5..x0+2, color=black,
title="Section 3_1, #62(a)" );
q := unapply( (f(x+h)-f(x))/h, (x,h) );
# (b)
L := limit( q(x,h), h=0 );
# (c)
m := eval( L, x=x0 );
tan_line := f(x0) + m*(x-x0);
plot( [f(x),tan_line], x=x0-2..x0+3, color=black,
linestyle=[1,7], title="Section 3.1 #62(d)",
legend=["y=f(x)","Tangent line at x=1"] );
Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ):
# (e)
Yvals := map( f, Xvals ):
evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" );
Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
< Ê kv(0)k œ l 'l œ ' m/sec and kv(')k œ l'l œ ' m/sec;
Ê a(0) œ # m/sec# and a(') œ # m/sec#
(c) v œ 0 Ê ' #t œ 0 Ê t œ $. v is positive in the interval ! t $ and v is negative when $ t ' Ê the body
changes direction at t œ $.
3. s œ t$ 3t# 3t, 0 Ÿ t Ÿ 3
(a) displacement œ ?s œ s(3) s(0) œ 9 m, vav œ
(b) v œ
ds
dt
?s
?t
œ
9
3
œ 3 m/sec
œ 3t# 6t 3 Ê kv(0)k œ k3k œ 3 m/sec and kv(3)k œ k12k œ 12 m/sec; a œ
#
#
d# s
dt#
œ 6t 6
Ê a(0) œ 6 m/sec and a(3) œ 12 m/sec
(c) v œ 0 Ê 3t# 6t 3 œ 0 Ê t# 2t 1 œ 0 Ê (t 1)# œ 0 Ê t œ 1. For all other values of t in the
interval the velocity v is negative (the graph of v œ 3t# 6t 3 is a parabola with vertex at t œ 1 which
opens downward Ê the body never changes direction).
4. s œ
t%
4
t$ t# , 0 Ÿ t Ÿ $
(a) ?s œ s($) s(0) œ
*
%
m, vav œ
?s
?t
œ
*
%
$
œ
$
%
m/sec
(b) v œ t$ 3t# 2t Ê kv(0)k œ 0 m/sec and kv($)k œ ' m/sec; a œ 3t# 6t 2 Ê a(0) œ 2 m/sec# and
a($) œ "" m/sec#
(c) v œ 0 Ê t$ 3t# 2t œ 0 Ê t(t 2)(t 1) œ 0 Ê t œ 0, 1, 2 Ê v œ t(t 2)(t 1) is positive in the
interval for 0 t 1 and v is negative for 1 t 2 and v is positive for # t $ Ê the body changes direction at
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
136
Chapter 3 Differentiation
t œ 1 and at t œ #.
5. s œ
25
t#
5t , 1 Ÿ t Ÿ 5
(a) ?s œ s(5) s(1) œ 20 m, vav œ
(b) v œ
50
t$
5
t#
m/sec#
(c) v œ 0 Ê
505t
t$
6. s œ
25
t5
œ 5 m/sec
Ê kv(1)k œ 45 m/sec and kv(5)k œ
4
25
a(5) œ
20
4
"
5
m/sec; a œ
150
t%
10
t$
Ê a(1) œ 140 m/sec# and
œ 0 Ê 50 5t œ 0 Ê t œ 10 Ê the body does not change direction in the interval
, % Ÿ t Ÿ 0
(a) ?s œ s(0) s(4) œ 20 m, vav œ 20
4 œ 5 m/sec
(b) v œ
a(0)
(c) v œ
25
(t5)# Ê kv(4)k œ
œ 25 m/sec#
0 Ê (t25
5)# œ 0 Ê
25 m/sec and kv(0)k œ " m/sec; a œ
50
(t5)$
Ê a(4) œ 50 m/sec# and
v is never 0 Ê the body never changes direction
7. s œ t$ 6t# 9t and let the positive direction be to the right on the s-axis.
(a) v œ 3t# 12t 9 so that v œ 0 Ê t# 4t 3 œ (t 3)(t 1) œ 0 Ê t œ 1 or 3; a œ 6t 12 Ê a(1)
œ 6 m/sec# and a(3) œ 6 m/sec# . Thus the body is motionless but being accelerated left when t œ 1, and
motionless but being accelerated right when t œ 3.
(b) a œ 0 Ê 6t 12 œ 0 Ê t œ 2 with speed kv(2)k œ k12 24 9k œ 3 m/sec
(c) The body moves to the right or forward on 0 Ÿ t 1, and to the left or backward on 1 t 2. The
positions are s(0) œ 0, s(1) œ 4 and s(2) œ 2 Ê total distance œ ks(1) s(0)k ks(2) s(1)k œ k4k k2k
œ 6 m.
8. v œ t# 4t 3 Ê a œ 2t 4
(a) v œ 0 Ê t# 4t 3 œ 0 Ê t œ 1 or 3 Ê a(1) œ 2 m/sec# and a(3) œ 2 m/sec#
(b) v 0 Ê (t 3)(t 1) 0 Ê 0 Ÿ t 1 or t 3 and the body is moving forward; v 0 Ê (t 3)(t 1) 0
Ê " t 3 and the body is moving backward
(c) velocity increasing Ê a 0 Ê 2t 4 0 Ê t 2; velocity decreasing Ê a 0 Ê 2t 4 0 Ê ! Ÿ t 2
9. sm œ 1.86t# Ê vm œ 3.72t and solving 3.72t œ 27.8 Ê t ¸ 7.5 sec on Mars; sj œ 11.44t# Ê vj œ 22.88t and
solving 22.88t œ 27.8 Ê t ¸ 1.2 sec on Jupiter.
10. (a) v(t) œ sw (t) œ 24 1.6t m/sec, and a(t) œ vw (t) œ sw w (t) œ 1.6 m/sec#
(b) Solve v(t) œ 0 Ê 24 1.6t œ 0 Ê t œ 15 sec
(c) s(15) œ 24(15) .8(15)# œ 180 m
(d) Solve s(t) œ 90 Ê 24t .8t# œ 90 Ê t œ
30„15È2
#
¸ 4.39 sec going up and 25.6 sec going down
(e) Twice the time it took to reach its highest point or 30 sec
11. s œ 15t "# gs t# Ê v œ 15 gs t so that v œ 0 Ê 15 gs t œ 0 Ê gs œ
15
t
. Therefore gs œ
15
20
œ
3
4
œ 0.75 m/sec#
12. Solving sm œ 832t 2.6t# œ 0 Ê t(832 2.6t) œ 0 Ê t œ 0 or 320 Ê 320 sec on the moon; solving
se œ 832t 16t# œ 0 Ê t(832 16t) œ 0 Ê t œ 0 or 52 Ê 52 sec on the earth. Also, vm œ 832 5.2t œ 0
Ê t œ 160 and sm (160) œ 66,560 ft, the height it reaches above the moon's surface; ve œ 832 32t œ 0
Ê t œ 26 and se (26) œ 10,816 ft, the height it reaches above the earth's surface.
13. (a) s œ 179 16t# Ê v œ 32t Ê speed œ kvk œ 32t ft/sec and a œ 32 ft/sec#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.3 The Derivative as a Rate of Change
(b) s œ 0 Ê 179 16t# œ 0 Ê t œ É 179
16 ¸ 3.3 sec
È
É 179
(c) When t œ É 179
16 , v œ 32
16 œ 8 179 ¸ 107.0 ft/sec
14. (a)
lim1 v œ lim1 9.8(sin ))t œ 9.8t so we expect v œ 9.8t m/sec in free fall
)Ä
(b) a œ
#
dv
dt
)Ä
#
œ 9.8 m/sec#
(b) between 3 and 6 seconds: $ Ÿ t Ÿ 6
(d)
15. (a) at 2 and 7 seconds
(c)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing
still when 1 t 2 or 3 t 5
(b)
17. (a)
(c)
(e)
(f)
190 ft/sec
at 8 sec, 0 ft/sec
From t œ 8 until t œ 10.8 sec, a total of 2.8 sec
Greatest acceleration happens 2 sec after launch
(g) From t œ 2 to t œ 10.8 sec; during this period, a œ
(b) 2 sec
(d) 10.8 sec, 90 ft/sec
v(10.8)v(2)
10.82
¸ 32 ft/sec#
18. (a) Forward: 0 Ÿ t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;
Slows down: 0 Ÿ t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0 Ÿ t 2 and 6 t 7; zero: 2 t 3 and 7 t 9
(c) t œ 0 and 2 Ÿ t Ÿ 3
(d) 7 Ÿ t Ÿ 9
19. s œ 490t# Ê v œ 980t Ê a œ 980
(a) Solving 160 œ 490t# Ê t œ
4
7
sec. The average velocity was
s(4/7)s(0)
4/7
œ 280 cm/sec.
(b) At the 160 cm mark the balls are falling at v(4/7) œ 560 cm/sec. The acceleration at the 160 cm mark
was 980 cm/sec# .
17
(c) The light was flashing at a rate of 4/7
œ 29.75 flashes per second.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
137
138
Chapter 3 Differentiation
20. (a)
(b)
21. C œ position, A œ velocity, and B œ acceleration. Neither A nor C can be the derivative of B because B's
derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while
C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position.
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That
leaves B for acceleration.
22. C œ position, B œ velocity, and A œ acceleration. Curve C cannot be the derivative of either A or B because
C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is
negative where B has negative slopes and positive where B has positive slopes.
23. (a) c(100) œ 11,000 Ê cav œ
#
11,000
100
œ $110
w
(b) c(x) œ 2000 100x .1x Ê c (x) œ 100 .2x. Marginal cost œ cw (x) Ê the marginal cost of producing 100
machines is cw (100) œ $80
(c) The cost of producing the 101st machine is c(101) c(100) œ 100 201
10 œ $79.90
24. (a) r(x) œ 20000 ˆ1 "x ‰ Ê rw (x) œ
(b) rw a"!!b œ
20000
100#
20000
x#
, which is marginal revenue.
œ $#Þ
(c) x lim
rw (x) œ x lim
Ä_
Ä_
will approach zero.
20000
x#
œ 0. The increase in revenue as the number of items increases without bound
25. b(t) œ 10' 10% t 10$ t# Ê bw (t) œ 10% (2) a10$ tb œ 10$ (10 2t)
(a) bw (0) œ 10% bacteria/hr
(b) bw (5) œ 0 bacteria/hr
w
%
(c) b (10) œ 10 bacteria/hr
26. Q(t) œ 200(30 t)# œ 200 a900 60t t# b Ê Qw (t) œ 200(60 2t) Ê Qw (10) œ 8,000 gallons/min is the rate
the water is running at the end of 10 min. Then
Q(10)Q(0)
10
œ 10,000 gallons/min is the average rate the
water flows during the first 10 min. The negative signs indicate water is leaving the tank.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.3 The Derivative as a Rate of Change
27. (a) y œ 6 ˆ1
t ‰#
1#
œ 6 Š1
(b) The largest value of
(c)
dy
dt
t
6
t#
144 ‹
Ê
dy
dt
œ
t
12
139
1
is 0 m/h when t œ 12 and the fluid level is falling the slowest at that time. The
smallest value of dy
dt is 1 m/h, when t œ 0, and
dy
In this situation, dt Ÿ 0 Ê the graph of y is
always decreasing. As dy
dt increases in value,
the fluid level is falling the fastest at that time.
the slope of the graph of y increases from 1
to 0 over the interval 0 Ÿ t Ÿ 12.
28. (a) V œ
4
3
1 r$ Ê
(b) When r œ 2,
dV
dr
dV
dr
œ 41 r # Ê
dV ¸
dr r=2
œ 41(2)# œ 161 ft$ /ft
œ 161 so that when r changes by 1 unit, we expect V to change by approximately 161.
Therefore when r changes by 0.2 units V changes by approximately (161)(0.2) œ 3.21 ¸ 10.05 ft$ . Note
that V(2.2) V(2) ¸ 11.09 ft$ .
29. 200 km/hr œ 55 59 m/sec œ
t œ 25, D œ
10
9
#
(25) œ
500
9 m/sec,
6250
9 m
and D œ
10 #
9 t
30. s œ v! t 16t# Ê v œ v! 32t; v œ 0 Ê t œ
v!
32
Ê Vœ
20
9
t. Thus V œ
500
9
Ê
; 1900 œ v! t 16t# so that t œ
Ê v! œ È(64)(1900) œ 80È19 ft/sec and, finally,
80È19 ft
sec
†
60 sec
1 min
†
60 min
1 hr
†
1 mi
5280 ft
v!
32
20
9
tœ
500
9
Ê t œ 25 sec. When
Ê 1900 œ
v!#
3#
¸ 238 mph.
31.
v œ 0 when t œ 6.25 sec
v 0 when 0 Ÿ t 6.25 Ê body moves up; v 0 when 6.25 t Ÿ 12.5 Ê body moves down
body changes direction at t œ 6.25 sec
body speeds up on (6.25ß 12.5] and slows down on [0ß 6.25)
The body is moving fastest at the endpoints t œ 0 and t œ 12.5 when it is traveling 200 ft/sec. It's
moving slowest at t œ 6.25 when the speed is 0.
(f) When t œ 6.25 the body is s œ 625 m from the origin and farthest away.
(a)
(b)
(c)
(d)
(e)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
v!#
64
140
Chapter 3 Differentiation
32.
(a) v œ 0 when t œ
3
#
sec
(b) v 0 when 0 Ÿ t 1.5 Ê body moves down; v 0 when 1.5 t Ÿ 5 Ê body moves up
(c) body changes direction at t œ 3# sec
(d) body speeds up on ˆ 3# ß &‘ and slows down on !ß 3# ‰
(e) body is moving fastest at t œ 5 when the speed œ kv(5)k œ 7 units/sec; it is moving slowest at
t œ 3# when the speed is 0
(f) When t œ 5 the body is s œ 12 units from the origin and farthest away.
33.
6 „ È15
3
6 È15
t
3
(a) v œ 0 when t œ
sec
(b) v 0 when
6 È15
3
Ê body moves left; v 0 when 0 Ÿ t
6 È15
3
or
6 È15
3
tŸ4
Ê body moves right
6 „ È15
sec
3
È
È
Š 6 3 15 ß #‹ Š 6 3 15 ß %“
(c) body changes direction at t œ
(d) body speeds up on
È15
and slows down on ’0ß 6 3
È15
‹ Š#ß 6 3
‹.
(e) The body is moving fastest at t œ 0 and t œ 4 when it is moving 7 units/sec and slowest at t œ
(f) When t œ
6È15
3
the body is at position s ¸ 6.303 units and farthest from the origin.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
6„È15
3
sec
Section 3.4 Derivatives of Trigonometric Functions
141
34.
6 „ È15
3
È
È
v 0 when 0 Ÿ t 6 3 15 or 6 3 15 t Ÿ 4 Ê body is moving left; v 0 when
È
6 È15
t 6 3 15 Ê body is moving right
3
È
body changes direction at t œ 6 „ 3 15 sec
È
È
È
È
body speeds up on Š 6 3 15 ß #‹ Š 6 3 15 ß %“ and slows down on ’!ß 6 3 15 ‹ Š#ß 6 3 15 ‹
(a) v œ 0 when t œ
(b)
(c)
(d)
(e) The body is moving fastest at 7 units/sec when t œ 0 and t œ 4; it is moving slowest and stationary at
tœ
6 „ È15
3
(f) When t œ
6 È15
3
the position is s ¸ 10.303 units and the body is farthest from the origin.
35. (a) It takes 135 seconds.
(b) Average speed œ ??Ft œ
&!
($ !
œ
&
($
¸ !Þ!') furlongs/sec.
(c) Using a symmetric difference quotient, the horse's speed is approximately
?F
?t
œ
%#
&* $$
œ
#
#'
¸ !Þ!(( furlongs/sec.
(d) The horse is running the fastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes
only 11 seconds to run, which is the least amount of time for a furlong.
(e) The horse accelerates the fastest during the first furlong (between markers 0 and 1).
3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
1. y œ 10x 3 cos x Ê
dy
dx
œ 10 3
œ
3
x#
3. y œ csc x 4Èx 7 Ê
dy
dx
2. y œ
3
x
5 sin x Ê
4. y œ x# cot x
"
x#
dy
dx
Ê
dy
dx
5
(cos x) œ 10 3 sin x
(sin x) œ
3
x#
œ csc x cot x
œ x#
œ x# csc# x 2x cot x
d
dx
d
dx
d
dx
5 cos x
4
#È x
0 œ csc x cot x
ax# b
2
x$
œ (sec x tan x)
d
dx
(cot x) cot x †
d
dx
2
Èx
œ x# csc# x (cot x)(2x)
2
x$
2
x$
5. y œ (sec x tan x)(sec x tan x) Ê
#
dy
dx
(sec x tan x) (sec x tan x)
#
d
dx
(sec x tan x)
œ (sec x tan x) asec x tan x sec xb (sec x tan x) asec x tan x sec xb
œ asec# x tan x sec x tan# x sec$ x sec# x tan xb asec# x tan x sec x tan# x sec$ x tan x sec# xb œ 0.
ŠNote also that y œ sec# x tan# x œ atan# x 1b tan# x œ 1 Ê
dy
dx
œ 0.‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
142
Chapter 3 Differentiation
6. y œ (sin x cos x) sec x Ê
œ (sin x cos x)
dy
dx
d
dx
(sec x) sec x
œ (sin x cos x)(sec x tan x) (sec x)(cos x sin x) œ
œ
sin# x cos x sin x cos# x cos x sin x
cos# x
œ
"
cos# x
d
dx (sin x cos x)
(sin x cos x) sin x
x sin x
cos cos
cos# x
x
œ sec# x
ŠNote also that y œ sin x sec x cos x sec x œ tan x 1 Ê
7. y œ
œ
(1 cot x)
œ
dy
dx
csc# x csc# x cot x csc# x cot x
(1 cot x)#
8. y œ
œ
Ê
cot x
1 cot x
Ê
cos x
1 sin x
sin x sin# x cos# x
(1 sin x)#
9. y œ
4
cos x
"
tan x
10. y œ
cos x
x
x
cos x
œ
(cot x) (cot x)
(1 cot x)#
œ
(1 sin x)
œ
dy
dx
d
dx
œ
(1 cot x) acsc# xb (cot x) acsc# xb
(1 cot x)#
d
(cos x) (cos x) dx
(1 sin x)
b (cos x) acos xb
œ (1 sin x) a(1sin xsin
(1 sin x)#
x)#
(1 sin x)
sin x 1
"
(1 sin x)# œ (1 sin x)# œ 1 sin x
œ
dy
dx
(1 cot x)
œ sec# x.‹
csc# x
(1 cot x)#
d
dx
œ 4 sec x cot x Ê
Ê
d
dx
dy
dx
œ 4 sec x tan x csc# x
dy
dx
x(sin x) (cos x)(1)
x#
11. y œ x# sin x 2x cos x 2 sin x Ê
#
(cos x)(1) x(sin x)
cos# x
œ
x sin x cos x
x#
cos x x sin x
cos# x
dy
dx
œ ax# cos x (sin x)(2x)b a(2x)(sin x) (cos x)(2)b 2 cos x
dy
dx
œ ax# (sin x) (cos x)(2x)b a2x cos x (sin x)(2)b 2(sin x)
œ x cos x 2x sin x 2x sin x 2 cos x 2 cos x œ x# cos x
12. y œ x# cos x 2x sin x 2 cos x Ê
#
œ x sin x 2x cos x 2x cos x 2 sin x 2 sin x œ x# sin x
13. s œ tan t t Ê
ds
dt
œ
14. s œ t# sec t 1 Ê
15. s œ
œ
16. s œ
œ
1 csc t
1 csc t
Ê
ds
dt
œ
d
dt
(tan t) 1 œ sec# t 1 œ tan# t
ds
dt
œ 2t
d
dt
(sec t) œ 2t sec t tan t
(1 csc t)(csc t cot t) (" csc t)(csc t cot t)
(1 csc t)#
csc t cot t csc# t cot t csc t cot t csc# t cot t
(1 csc t)#
sin t
1 cos t
"
cos t 1
Ê
ds
dt
œ
17. r œ 4 )# sin ) Ê
œ
2 csc t cot t
(1 csc t)#
(1 cos t)(cos t) (sin t)(sin t)
(1 cos t)#
dr
d)
18. r œ ) sin ) cos ) Ê
œ ˆ) #
dr
d)
d
d)
œ
cos t cos# t sin# t
(1 cos t)#
œ
cos t "
(1 cos t)#
œ 1 "cos t
(sin )) (sin ))(2))‰ œ a)# cos ) 2) sin )b œ )() cos ) # sin ))
œ () cos ) (sin ))(1)) sin ) œ ) cos )
dr
19. r œ sec ) csc ) Ê d)
œ (sec ))(csc ) cot )) (csc ))(sec ) tan ))
"
"
"
"
cos
)
sin ) ‰
#
#
œ ˆ cos ) ‰ ˆ sin ) ‰ ˆ sin ) ‰ ˆ sin" ) ‰ ˆ cos" ) ‰ ˆ cos
) œ sin# ) cos# ) œ sec ) csc )
20. r œ (1 sec )) sin ) Ê
21. p œ &
"
cot q
dr
d)
œ (" sec )) cos ) (sin ))(sec ) tan )) œ (cos ) ") tan# ) œ cos ) sec# )
œ 5 tan q Ê
22. p œ (1 csc q) cos q Ê
dp
dq
dp
dq
œ sec# q
œ (1 csc q)(sin q) (cos q)(csc q cot q) œ (sin q 1) cot# q œ sin q csc# q
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.4 Derivatives of Trigonometric Functions
23. p œ
œ
sin q cos q
cos q
Ê
œ
dp
dq
(cos q)(cos q sin q) (sin q cos q)(sin q)
cos# q
cos# q cos q sin q sin# q cos q sin q
cos# q
24. p œ
tan q
1 tan q
Ê
dp
dq
œ
œ
"
cos# q
œ sec# q
(1 tan q) asec# qb (tan q) asec# qb
(1 tan q)#
œ
sec# q tan q sec# q tan q sec# q
(1 tan q)#
œ
sec# q
(1 tan q)#
25. (a) y œ csc x Ê yw œ csc x cot x Ê yww œ a(csc x) acsc# xb (cot x)(csc x cot x)b œ csc$ x csc x cot# x
œ (csc x) acsc# x cot# xb œ (csc x) acsc# x csc# x 1b œ 2 csc$ x csc x
(b) y œ sec x Ê yw œ sec x tan x Ê yww œ (sec x) asec# xb (tan x)(sec x tan x) œ sec$ x sec x tan# x
œ (sec x) asec# x tan# xb œ (sec x) asec# x sec# x 1b œ 2 sec$ x sec x
26. (a) y œ 2 sin x Ê yw œ 2 cos x Ê yww œ 2(sin x) œ 2 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ 2 sin x
(b) y œ 9 cos x Ê yw œ 9 sin x Ê yww œ 9 cos x Ê ywww œ 9(sin x) œ 9 sin x Ê yÐ%Ñ œ 9 cos x
27. y œ sin x Ê yw œ cos x Ê slope of tangent at
x œ 1 is yw (1) œ cos (1) œ "; slope of
tangent at x œ 0 is yw (0) œ cos (0) œ 1; and
slope of tangent at x œ 3#1 is yw ˆ 3#1 ‰ œ cos 3#1
œ 0. The tangent at (1ß !) is y 0 œ 1(x 1),
or y œ x 1; the tangent at (0ß 0) is
y 0 œ 1(x 0), or y œ x; and the tangent at
ˆ 31
‰
# ß 1 is y œ 1.
28. y œ tan x Ê yw œ sec# x Ê slope of tangent at x œ 13
is sec# ˆ 13 ‰ œ 4; slope of tangent at x œ 0 is sec# (0) œ 1;
and slope of tangent at x œ
1
3
is sec# ˆ 13 ‰ œ 4. The tangent
at ˆ 13 ß tanˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4ˆx 13 ‰ ;
the tangent at (0ß 0) is y œ x; and the tangent at ˆ 13 ß tan ˆ 13 ‰‰
œ Š 13 ß È3‹ is y È3 œ 4 ˆx 13 ‰ .
29. y œ sec x Ê yw œ sec x tan x Ê slope of tangent at
x œ 13 is sec ˆ 13 ‰ tan ˆ 13 ‰ œ 2È3 ; slope of tangent
is sec ˆ 14 ‰ tan ˆ 14 ‰ œ È2 . The tangent at the point
ˆ 1 ß sec ˆ 1 ‰‰ œ ˆ 1 ß #‰ is y 2 œ #È3 ˆx 1 ‰ ;
at x œ
1
4
3
3
3
3
the tangent at the point ˆ 14 ß sec ˆ 14 ‰‰ œ Š 14 ß È2‹ is y È2
œ È2 ˆx 14 ‰ .
30. y œ 1 cos x Ê yw œ sin x Ê slope of tangent at
È
x œ 13 is sin ˆ 13 ‰ œ #3 ; slope of tangent at x œ
‰ œ 1. The tangent at the point
is sin ˆ 31
#
31
#
ˆ 13 ß " cos ˆ 13 ‰‰ œ ˆ 13 ß 3# ‰
È
is y 3# œ #3 ˆx 13 ‰ ; the tangent at the point
ˆ 3#1 ß " cos ˆ 3#1 ‰‰ œ ˆ 3#1 ß 1‰ is y 1 œ x 3#1
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143
144
Chapter 3 Differentiation
31. Yes, y œ x sin x Ê yw œ " cos x; horizontal tangent occurs where 1 cos x œ 0 Ê cos x œ 1
Ê xœ1
32. No, y œ 2x sin x Ê yw œ 2 cos x; horizontal tangent occurs where 2 cos x œ 0 Ê cos x œ #. But there
are no x-values for which cos x œ #.
33. No, y œ x cot x Ê yw œ 1 csc# x; horizontal tangent occurs where 1 csc# x œ 0 Ê csc# x œ 1. But there
are no x-values for which csc# x œ 1.
34. Yes, y œ x 2 cos x Ê yw œ 1 2 sin x; horizontal tangent occurs where 1 2 sin x œ 0 Ê 1 œ 2 sin x
Ê "# œ sin x Ê x œ 16 or x œ 561
35. We want all points on the curve where the tangent
line has slope 2. Thus, y œ tan x Ê yw œ sec# x so
that yw œ 2 Ê sec# x œ 2 Ê sec x œ „ È2
Ê x œ „ 14 . Then the tangent line at ˆ 14 ß "‰ has
equation y 1 œ 2 ˆx 14 ‰ ; the tangent line at
ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ .
36. We want all points on the curve y œ cot x where
the tangent line has slope 1. Thus y œ cot x
Ê yw œ csc# x so that yw œ 1 Ê csc# x œ 1
Ê csc# x œ 1 Ê csc x œ „ 1 Ê x œ 1# . The
tangent line at ˆ 1# ß !‰ is y œ x 12 .
2 cos x ‰
37. y œ 4 cot x 2 csc x Ê yw œ csc# x 2 csc x cot x œ ˆ sin" x ‰ ˆ 1 sin
x
(a) When x œ 1# , then yw œ 1; the tangent line is y œ x
w
1
#
2.
(b) To find the location of the horizontal tangent set y œ 0 Ê 1 2 cos x œ 0 Ê x œ
then y œ % È3 is the horizontal tangent.
38. y œ 1 È2 csc x cot x Ê yw œ È2 csc x cot x csc# x œ ˆ sin" x ‰ Š
1
3
È2 cos x 1
‹
sin x
(a) If x œ 14 , then yw œ 4; the tangent line is y œ 4x 1 4.
(b) To find the location of the horizontal tangent set yw œ 0 Ê È2 cos x 1 œ 0 Ê x œ
xœ
31
4 ,
radians. When x œ 13 ,
31
4
radians. When
then y œ 2 is the horizontal tangent.
39. lim sin ˆ "x #" ‰ œ sin ˆ #" #" ‰ œ sin 0 œ 0
xÄ2
40.
lim
x Ä 16
È1 cos (1 csc x) œ É1 cos ˆ1 csc ˆ 1 ‰‰ œ È1 cos a1 † a2bb œ È2
6
1 ‰
‘
ˆ 1 ‰
‘
ˆ1‰
‘
41. lim sec cos x 1 tan ˆ 4 sec
x 1 œ sec cos 0 1 tan 4 sec 0 1 œ sec 1 1 tan 4 1 œ sec 1 œ 1
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.4 Derivatives of Trigonometric Functions
x ‰
ˆ 1tan 0 ‰
ˆ 1‰
42. lim sin ˆ tan1xtan
2 sec x œ sin tan 02 sec 0 œ sin # œ 1
xÄ!
43. lim tan ˆ1
tÄ!
sin t ‰
t
œ tan Š1 lim
tÄ!
1) ‰
44. lim cos ˆ sin
) œ cos Š1 lim
)
sin t
t ‹
‹
) Ä ! sin )
)Ä!
œ tan (1 1) œ 0
"
œ cos Œ1 †
lim
"
œ cos ˆ1 † 1 ‰ œ 1
sin )
)Ä!
)
dv
da
ˆ1‰
45. s œ # # sin t Ê v œ ds
dt œ 2 cos t Ê a œ dt œ 2 sin t Ê j œ dt œ 2 cos t. Therefore, velocity œ v 4
œ È2 m/sec; speed œ ¸v ˆ 1 ‰¸ œ È2 m/sec; acceleration œ a ˆ 1 ‰ œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ È2 m/sec$ .
4
46. s œ sin t cos t Ê v œ
4
œ cos t sin t Ê a œ
ds
dt
v ˆ 14 ‰
velocity œ
œ 0 m/sec; speed œ
1‰
ˆ
jerk œ j 4 œ 0 m/sec$ .
47. lim f(x) œ lim
xÄ!
Ê 9 œ c.
48.
xÄ!
sin# 3x
x#
¸v ˆ 14 ‰¸
dv
dt
4
œ sin t cos t Ê j œ
œ 0 m/sec; acceleration œ
a ˆ 14 ‰
œ cos t sin t. Therefore
È
œ 2 m/sec# ;
da
dt
œ lim 9 ˆ sin3x3x ‰ ˆ sin3x3x ‰ œ 9 so that f is continuous at x œ 0 Ê lim f(x) œ f(0)
xÄ!
xÄ!
lim g(x) œ lim c (x b) œ b and lim b g(x) œ lim b cos x œ 1 so that g is continuous at x œ 0 Ê lim c g(x)
xÄ!
xÄ!
xÄ!
xÄ!
œ lim b g(x) Ê b œ 1. Now g is not differentiable at x œ 0: At x œ 0, the left-hand derivative is
x Ä !c
xÄ!
d
dx
(x b)¸ x=0 œ 1, but the right-hand derivative is
d
dx
(cos x)¸ x=0 œ sin 0 œ 0. The left- and right-hand
derivatives can never agree at x œ 0, so g is not differentiable at x œ 0 for any value of b (including b œ 1).
49.
d***
dx***
d%
dx%
(cos x) œ sin x because
(cos x) œ cos x Ê the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 œ 249 † 4 3 Ê
œ
d$
dx$
d#%*†%
’ dx
#%*†%
(cos x)“ œ
50. (a) y œ sec x œ
Ê
d
dx
"
cos x
d$
dx$
Ê
d***
dx***
(cos x)
(cos x) œ sin x.
dy
dx
œ
(cos x)(0) (1)(sin x)
(cos x)#
œ
(sin x)(0) (1)(cos x)
(sin x)#
œ
sin x
cos# x
sin x ‰
œ ˆ cos" x ‰ ˆ cos
x œ sec x tan x
(sec x) œ sec x tan x
(b) y œ csc x œ
Ê
d
dx
d
dx
Ê
dy
dx
œ
cos x
sin# x
" ‰ ˆ cos x ‰
œ ˆ sin
x
sin x œ csc x cot x
(csc x) œ csc x cot x
(c) y œ cot x œ
Ê
"
sin x
cos x
sin x
Ê
dy
dx
#
œ
(sin x)(sin x) (cos x)(cos x)
(sin x)#
œ
sin# xcos# x
sin# x
œ
"
sin# x
œ csc# x
(cot x) œ csc x
51.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y œ
closer and closer to the black curve y œ cos x because
d
dx
(sin x) œ
is true as h takes on the values of 1, 0.5, 0.3 and 0.1.
sin x
lim sin (x h)
h
hÄ!
sin (x h) sin x
h
get
œ cos x. The same
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
145
146
Chapter 3 Differentiation
52.
cos (x h) cos x
h
cos x
lim cos (x h)
œ
sin x.
h
hÄ!
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y œ
get
closer and closer to the black curve y œ sin x because
The
d
dx
(cos x) œ
same is true as h takes on the values of 1, 0.5, 0.3, and 0.1.
53. (a)
The dashed curves of y œ
sinax hb sinax hb
#h
are closer to the black curve y œ cos x than the corresponding dashed
curves in Exercise 51 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
(b)
The dashed curves of y œ
cosax hb cosax hb
#h
are closer to the black curve y œ sin x than the corresponding dashed
curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
54. lim
hÄ!
k0 h k k 0 h k
2h
œ lim
xÄ!
k h k k hk
2h
œ lim 0 œ 0 Ê the limits of the centered difference quotient exists even
hÄ!
though the derivative of f(x) œ kxk does not exist at x œ 0.
55. y œ tan x Ê yw œ sec# x, so the smallest value
yw œ sec# x takes on is yw œ 1 when x œ 0;
yw has no maximum value since sec# x has no
largest value on ˆ 1# ß 1# ‰ ; yw is never negative
since sec# x
1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.4 Derivatives of Trigonometric Functions
56. y œ cot x Ê yw œ csc# x so yw has no smallest
value since csc# x has no minimum value on
(!ß 1); the largest value of yw is 1, when x œ 1# ;
the slope is never positive since the largest
value yw œ csc2 x takes on is 1.
57. y œ
sin x
x appears to cross the y-axis at y œ 1, since
lim sin x œ 1; y œ sinx2x appears to cross the y-axis
xÄ! x
at y œ 2, since lim sinx2x œ 2; y œ sinx4x appears to
xÄ!
cross the y-axis at y œ 4, since lim sinx4x œ 4.
xÄ!
However, none of these graphs actually cross the y-axis
since x œ 0 is not in the domain of the functions. Also,
lim
xÄ!
sin 5x
x
sin (3x)
x
œ 5, lim
xÄ!
œ k Ê the graphs of y œ
yœ
sin kx
x
œ 3, and lim
sin kx
x
xÄ!
yœ
sin 5x
x ,
sin (3x)
,
x
and
approach 5, 3, and k, respectively, as
x Ä 0. However, the graphs do not actually cross the
y-axis.
58. (a)
sin h
h
h
1
0.01
0.001
0.0001
ˆ sinh h ‰ ˆ 180
‰
1
.99994923
1
1
1
.017452406
.017453292
.017453292
.017453292
1 ‰
sin ˆh† 180
h
xÄ!
œ lim
sin h°
h
hÄ!
lim
œ lim
1
180
hÄ!
1 ‰
sin ˆh† 180
1 †h
180
œ lim
1 sin )
180
)Ä!
)
œ
1
180
() œ h †
1
180 )
(converting to radians)
(b)
cos h1
h
h
1
0.01
0.001
0.0001
lim
hÄ!
0.0001523
0.0000015
0.0000001
0
cos h1
h
(c) In degrees,
œ 0, whether h is measured in degrees or radians.
d
dx
(sin x) œ lim
hÄ!
œ lim ˆsin x †
hÄ!
cos h 1 ‰
h
sin (x h) sin x
h
lim ˆcos x †
hÄ!
œ lim
hÄ!
sin h ‰
h
(sin x cos h cos x sin h) sin x
h
œ (sin x) † lim ˆ cos hh 1 ‰ (cos x) † lim ˆ sinh h ‰
hÄ!
1 ‰
œ (sin x)(0) (cos x) ˆ 180
œ
(d)
1
180 cos x
cos x
d
In degrees, dx
(cos x) œ lim cos (x h)
œ lim (cos x cos h sinh x sin h) cos x
h
hÄ!
hÄ!
(cos x)(cos h 1) sin x sin h
ˆ
œ lim
œ lim cos x † cos hh 1 ‰ lim ˆsin x † sinh h ‰
h
hÄ!
hÄ!
hÄ!
œ (cos x) lim ˆ
hÄ!
(e)
d#
dx#
d#
dx#
cos h 1 ‰
h
hÄ!
1 ‰
1
(sin x) lim ˆ sinh h ‰ œ (cos x)(0) (sin x) ˆ 180
œ 180
sin x
hÄ!
(sin x) œ
d
dx
1
1 ‰#
ˆ 180
cos x‰ œ ˆ 180
sin x;
(cos x) œ
d
dx
1
1 ‰#
ˆ 180
sin x‰ œ ˆ 180
cos x;
d$
dx$
(sin x) œ
d$
dx$
d
dx
(cos x) œ
#
$
1 ‰
1 ‰
sin x‹ œ ˆ 180
cos x;
Š ˆ 180
d
dx
#
$
1 ‰
1 ‰
cos x‹ œ ˆ 180
sin x
Š ˆ 180
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
147
148
Chapter 3 Differentiation
3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS
1. f(u) œ 6u 9 Ê f w (u) œ 6 Ê f w (g(x)) œ 6; g(x) œ
œ 6 † 2x$ œ 12x$
"
#
x% Ê gw (x) œ 2x$ ; therefore
dy
dx
œ f w (g(x))gw (x)
2. f(u) œ 2u$ Ê f w (u) œ 6u# Ê f w (g(x)) œ 6(8x 1)# ; g(x) œ 8x 1 Ê gw (x) œ 8; therefore
œ 6(8x 1)# † 8 œ 48(8x 1)#
dy
dx
œ f w (g(x))gw (x)
3. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (3x 1); g(x) œ 3x 1 Ê gw (x) œ 3; therefore
œ f w (g(x))gw (x)
dy
dx
œ (cos (3x 1))(3) œ 3 cos (3x 1)
4. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin ˆ 3x ‰ ; g(x) œ
‰ œ "3 sin ˆ 3x ‰
œ sin ˆ 3x ‰ † ˆ "
3
x
3
Ê gw (x) œ "3 ; therefore
dy
dx
œ f w (g(x))gw (x)
5. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin (sin x); g(x) œ sin x Ê gw (x) œ cos x; therefore
dy
dx
œ f w (g(x))gw (x) œ (sin (sin x)) cos x
6. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (x cos x); g(x) œ x cos x Ê gw (x) œ 1 sin x; therefore
dy
dx
œ f w (g(x))gw (x) œ (cos (x cos x))(1 sin x)
7. f(u) œ tan u Ê f w (u) œ sec# u Ê f w (g(x)) œ sec# (10x 5); g(x) œ 10x 5 Ê gw (x) œ 10; therefore
dy
dx
œ f w (g(x))gw (x) œ asec# (10x 5)b (10) œ 10 sec# (10x 5)
8. f(u) œ sec u Ê f w (u) œ sec u tan u Ê f w (g(x)) œ sec ax# 7xb tan ax# 7xb ; g(x) œ x# 7x
Ê gw (x) œ 2x 7; therefore
œ f w (g(x))gw (x) œ (2x 7) sec ax# 7xb tan ax# 7xb
dy
dx
9. With u œ (2x 1), y œ u& :
dy
dx
œ
dy du
du dx
œ 5u% † 2 œ 10(2x 1)%
10. With u œ (4 3x), y œ u* :
dy
dx
œ
dy du
du dx
œ 9u) † (3) œ 27(4 3x))
11. With u œ ˆ1 x7 ‰ , y œ ?( :
œ
dy
dx
12. With u œ ˆ x# 1‰ , y œ ?"! :
dy
dx
œ
#
13. With u œ Š x8 x "x ‹ , y œ ?% :
14. With u œ ˆ x5
" ‰
5x ,
y œ ?& :
15. With u œ tan x, y œ sec u:
16. With u œ 1 "x , y œ cot u:
17. With u œ sin x, y œ u$ :
dy
dx
dy
dx
dy
dx
œ
œ
dy du
du dx
dy
dx
œ
œ
dy du
du dx
œ
dy du
du dx
18. With u œ cos x, y œ 5u% :
dy
dx
œ
œ 10u"" † ˆ "# ‰ œ 5 ˆ x# 1‰
dy du
du dx
dy du
du dx
dy du
du dx
dy
dx
)
œ 7u) † ˆ "7 ‰ œ ˆ" x7 ‰
dy du
du dx
œ 4u$ † ˆ x4 1
œ 5u% † ˆ 15
" ‰
5x#
"‰
x#
œ ˆ x5
""
$
#
œ 4 Š x8 x x" ‹ ˆ x4 1
" ‰%
5x
ˆ1
"‰
x#
œ (sec u tan u) asec# xb œ (sec (tan x) tan (tan x)) sec# x
œ acsc# ub ˆ x"# ‰ œ x"# csc# ˆ1 x" ‰
œ 3u# cos x œ 3 asin# xb (cos x)
dy du
du dx
œ a20u& b (sin x) œ 20 acos& xb (sin x)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"‰
x#
Section 3.5 The Chain Rule and Parametric Equations
19. p œ È3 t œ (3 t)"Î# Ê
20. q œ È2r r# œ a2r r# b
21. s œ
œ
4
31
4
1
sin 3t
4
51
dp
dt
"Î#
Ê
cos 5t Ê
23. r œ (csc ) cot ))" Ê
dr
d)
dq
dr
(3 t)"Î# †
œ
"
#
d
dt
a2r r# b
(3 t) œ "# (3 t)"Î# œ
"Î#
†
d
dr
ds
dt
œ
cos 3t †
d
dt
(3t)
ds
dt
œ cos ˆ 3#1t ‰ †
d
dt
ˆ 3#1t ‰ sin ˆ 3#1t ‰ †
4
31
4
51
"
#
a2r r# b œ
(sin 5t) †
a2r r# b
(5t) œ
d
dt
"
2È 3 t
4
1
"Î#
"r
È2r r#
(2 2r) œ
cos 3t
4
1
sin 5t
d
dt
ˆ 3#1t ‰ œ
31
2
cos ˆ 3#1t ‰
31
2
sin ˆ 3#1t ‰
œ (csc ) cot ))#
d
d)
(csc ) cot )) œ
csc ) cot ) csc# )
(csc ) cot ))#
œ
csc ) (cot ) csc ))
(csc ) cot ))#
œ (sec ) tan ))#
d
d)
(sec ) tan )) œ
sec ) tan ) sec# )
(sec ) tan ))#
œ
sec ) (tan ) sec ))
(sec ) tan ))#
csc )
csc ) cot )
24. r œ (sec ) tan ))" Ê
œ
"
#
(cos 3t sin 5t)
22. s œ sin ˆ 3#1t ‰ cos ˆ 3#1t ‰ Ê
œ 321 ˆcos 3#1t sin 3#1t ‰
œ
œ
dr
d)
sec )
sec ) tan )
d
d
# d
%
%
#
#
25. y œ x# sin% x x cos# x Ê dy
xb cos# x †
dx œ x dx asin xb sin x † dx ax b x dx acos
d
d
œ x# ˆ4 sin$ x dx
(sin x)‰ 2x sin% x x ˆ2 cos$ x † dx
(cos x)‰ cos# x
d
dx
(x)
œ x# a4 sin$ x cos xb 2x sin% x xa a2 cos$ xb (sin x)b cos# x
œ 4x# sin$ x cos x 2x sin% x 2x sin x cos$ x cos# x
d ˆ"‰
x d
$
$
asin& xb sin& x † dx
x 3 dx acos xb cos x †
œ "x a5 sin' x cos xb asin& xb ˆ x"# ‰ 3x a a3 cos# xb (sin x)b acos$ xb ˆ 3" ‰
26. y œ
"
x
sin& x
x
3
cos$ x Ê yw œ
œ 5x sin' x cos x
"
x#
" d
x dx
sin& x x cos# x sin x
"
3
ˆ x3 ‰
cos$ x
"
" ‰"
7
d
(
'
ˆ
Ê dy
21 (3x 2) 4 #x#
dx œ 21 (3x 2) † dx (3x 2)
7
" ‰# ˆ " ‰
"
'
'
ˆ
#
21 (3x 2) † 3 (1) 4 #x#
x$ œ (3x 2) $
x Š4 "# ‹
27. y œ
œ
d
dx
(1) ˆ4
" ‰#
#x #
†
d
dx
ˆ4
" ‰
#x #
#x
%
28. y œ (5 2x)$ "8 ˆ 2x 1‰ Ê
dy
dx
$
$
œ 3(5 2x)% (2) 84 ˆ x2 1‰ ˆ x2# ‰ œ 6(5 2x)% ˆ x"# ‰ ˆ 2x 1‰
$
œ
6
(5 2x)%
Š 2x 1‹
x#
29. y œ (4x 3)% (x 1)$ Ê
%
dy
dx
œ (4x 3)% (3)(x 1)% †
%
$
d
dx
(x 1) (x 1)$ (4)(4x 3)$ †
$
%
œ (4x 3) (3)(x 1) (1) (x 1) (4)(4x 3) (4) œ 3(4x 3) (x 1)
œ
$
(4x 3)
(x 1)%
c3(4x 3) 16(x 1)d œ
'
30. y œ (2x 5)" ax# 5xb Ê
&
œ 6 ax# 5xb
2 ax# 5xb
(2x 5)#
dy
dx
%
d
dx
$
(4x 3)
16(4x 3) (x 1)$
$
(4x 3) (4x 7)
(x 1)%
&
'
œ (2x 5)" (6) ax# 5xb (2x 5) ax# 5xb (1)(2x 5)# (2)
'
d ˆ
d
31. h(x) œ x tan ˆ2Èx‰ 7 Ê hw (x) œ x dx
tan ˆ2x"Î# ‰‰ tan ˆ2x"Î# ‰ † dx
(x) 0
"
d ˆ "Î# ‰
# ˆ "Î# ‰
"Î#
#
† dx 2x
œ x sec 2x
tan ˆ2x ‰ œ x sec ˆ2Èx‰ † È tan ˆ2Èx‰ œ Èx sec# ˆ2Èx‰ tan ˆ2Èx‰
x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
149
150
Chapter 3 Differentiation
d ˆ
d
32. k(x) œ x# sec ˆ "x ‰ Ê kw (x) œ x# dx
sec x" ‰ sec ˆ x" ‰ † dx
ax# b œ x# sec ˆ x" ‰ tan ˆ x" ‰ †
œ x# sec ˆ "x ‰ tan ˆ x" ‰ † ˆ x"# ‰ 2x sec ˆ x" ‰ œ 2x sec ˆ x" ‰ sec ˆ x" ‰ tan ˆ x" ‰
#
33. f()) œ ˆ 1 sincos) ) ‰ Ê f w ()) œ 2 ˆ 1 sincos) ) ‰ †
œ
(2 sin )) acos ) cos# ) sin# )b
(1 cos ))$
34. g(t) œ ˆ 1 sincost t ‰
œ
"
(2 sin )) (cos ) 1)
(1 cos ))$
œ
Ê gw (t) œ ˆ 1 sincost t ‰
asin# t cos t cos# tb
(1 cos t)#
#
†
(1 cos ))(cos )) (sin ))(sin ))
(1 cos ))#
2 sin )
(1 cos ))#
œ
†
2 sin )
1 cos )
d
dt
#
ˆ 1 sincost t ‰ œ (1 sincost t)# †
(sin t)(sin t) (" cos t)(cos t)
(sin t)#
"
1 cos t
œ
35. r œ sin a)# b cos (2)) Ê
ˆ 1 sincos) ) ‰ œ
d
d)
ˆ x" ‰ 2x sec ˆ x" ‰
d
dx
œ sin a)# b (sin 2))
dr
d)
(2)) cos (2)) acos a)# bb †
d
d)
d
d)
a) # b
œ sin a)# b (sin 2))(2) (cos 2)) acos a)# bb (2)) œ 2 sin a)# b sin (#)) 2) cos (2)) cos a)# b
36. r œ Šsec È)‹ tan ˆ ") ‰ Ê
dr
d)
œ )"# sec È) sec# ˆ ") ‰
37. q œ sin Š Ètt 1 ‹ Ê
œ cos Š Ètt 1 ‹ †
39. y œ sin# (1t 2) Ê
Èt 1
t
2
t1
dq
dt
"
#È )
tan ˆ ") ‰ sec È) tan È) œ Šsec È)‹ ”
œ cos Š Ètt 1 ‹ †
dq
dt
Èt 1
38. q œ cot ˆ sint t ‰ Ê
œ Šsec È)‹ ˆ sec# ") ‰ ˆ )"# ‰ tan ˆ ") ‰ Šsec È) tan È)‹ Š
d
dt
Èt 1 (1)t †
d
dt
sec# ˆ )" ‰
)# •
ˆÈ t 1 ‰
ˆÈ t 1 ‰
#
1) t
œ cos Š Ètt 1 ‹ Š 2(t
‹ œ Š 2(tt1)2$Î# ‹ cos Š Ètt 1 ‹
2(t 1)$Î#
œ csc# ˆ sint t ‰ †
d
dt
ˆ sint t ‰ œ ˆcsc# ˆ sint t ‰‰ ˆ t cos tt# sin t ‰
œ 2 sin (1t 2) †
dy
dt
Š Ètt 1 ‹ œ cos Š Ètt 1 ‹ †
tan È) tan ˆ ") ‰
#È )
"
‹
#È )
d
dt
sin (1t 2) œ 2 sin (1t 2) † cos (1t 2) †
d
dt
(1t 2)
œ 21 sin (1t 2) cos (1t 2)
40. y œ sec# 1t Ê
dy
dt
œ (2 sec 1t) †
41. y œ (1 cos 2t)% Ê
42. y œ ˆ1 cot ˆ #t ‰‰
œ
#
dy
dt
Ê
(sec 1t) œ (2 sec 1t)(sec 1t tan 1t) †
œ 4(1 cos 2t)& †
dy
dt
ˆ #t ‰
$
ˆ1 cot ˆ t ‰‰
#
csc#
43. y œ sin acos (2t 5)b Ê
d
dt
dy
dt
œ 2 ˆ1 cot ˆ #t ‰‰
(1t) œ 21 sec# 1t tan 1t
(1 cos 2t) œ 4(1 cos 2t)& (sin 2t) †
d
dt
$
œ cos (cos (2t 5)) †
d
dt
cot ˆ #t ‰‰ œ 2 ˆ1 cot ˆ #t ‰‰
$
d
dt
(2t) œ
8 sin 2t
(1 cos 2t)&
† ˆcsc# ˆ #t ‰‰ †
†
dˆ
dt 1
d
dt
cos (2t 5) œ cos (cos (2t 5)) † (sin (2t 5)) †
d
dt
dˆt‰
dt #
(2t 5)
œ 2 cos (cos (2t 5))(sin (2t 5))
ˆ
ˆ t ‰‰ †
44. y œ cos ˆ5 sin ˆ 3t ‰‰ Ê dy
dt œ sin 5 sin 3
5
t
t
œ 3 sin ˆ5 sin ˆ 3 ‰‰ ˆcos ˆ 3 ‰‰
d
dt
$
dy
% ˆ t ‰‘# d
† dt 1
dt œ 3 1 tan
1#
#
tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰ † 1"# ‘ œ 1
45. y œ 1 tan% ˆ 1t# ‰‘ Ê
œ 12 1
46. y œ
"
6
ˆ5 sin ˆ 3t ‰‰ œ sin ˆ5 sin ˆ 3t ‰‰ ˆ5 cos ˆ 3t ‰‰ †
$
c1 cos# (7t)d Ê
dy
dt
œ
3
6
#
d
dt
ˆ 3t ‰
#
tan% ˆ 1t# ‰‘ œ 3 1 tan% ˆ 1t# ‰‘ 4 tan$ ˆ 1t# ‰ †
#
tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰‘
#
d
dt
tan ˆ 1t# ‰‘
c1 cos# (7t)d † 2 cos (7t)(sin (7t))(7) œ 7 c1 cos# (7t)d (cos (7t) sin (7t))
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.5 The Chain Rule and Parametric Equations
"Î#
Ê
œ "# a1 cos at# bb
"Î#
47. y œ a1 cos at# bb
dy
dt
2 cos ŒÉ1 Èt
É1 Èt†2Èt
œ
a1 cos at# bb
"Î#
†
d
dt
a1 cos at# bb œ
#
"
#
a1 cos at# bb
œ 4 cos ŒÉ1 Èt †
dy
dt
" ‰#
x
œ
6
x%
œ
"
#x
51. y œ
"
9
œ
"
ˆ1 Èx‰$ Š
ˆ1 x" ‰
"
#
6
x$
ˆ1
" ‰#
x
œ
d
dx
6
x$
ˆ1 x" ‰# ˆ1 x" ‰# †
d
dt
ˆ1 Èt‰
d
dx
ˆ x3# ‰
ˆ1 x" ‰ ˆ x" 1 x" ‰
ˆ1 Èx‰# x"Î#
"
#È x
"
#
1‹ œ
"
#x
"
#
$
x" ˆ1 Èx‰ "# x"Î# ˆ1 Èx‰ 1‘
ˆ1 Èx‰$ Š 3#
"
‹
#Èx
cot (3x 1) Ê yw œ 9" csc# (3x 1)(3) œ 3" csc# (3x 1) Ê yww œ ˆ 32 ‰ (csc (3x 1) †
2
3
†
# É 1 È t
#
$
’ˆ1 Èx‰ ˆ "# x$Î# ‰ x"Î# (2) ˆ1 Èx‰ ˆ "# x"Î# ‰“
#
$
$Î# ˆ
1 Èx‰ x" ˆ1 Èx‰ “ œ
’ "
# x
"
#
a t# b ‰
É t Èt
"
#
50. y œ ˆ1 Èx‰ Ê yw œ ˆ1 Èx‰ ˆ "# x"Î# ‰ œ
œ
d
dt
cos ŒÉ1 Èt
œ ˆ x3# ‰ ˆ2 ˆ1 x" ‰ ˆ x"# ‰‰ ˆ x6$ ‰ ˆ1
œ x6$ ˆ1 x" ‰ ˆ1 2x ‰
"
#
ˆsin at# b †
ŒÉ1 Èt œ 4 cos ŒÉ1 Èt †
d
dt
$
#
#
49. y œ ˆ1 "x ‰ Ê yw œ 3 ˆ1 x" ‰ ˆ x"# ‰ œ x3# ˆ1 x" ‰ Ê yww œ ˆ x3# ‰ †
Ê yww œ
"Î#
at b
asin at# bb † 2t œ È1t sin
cos at# b
48. y œ 4 sin ŒÉ1 Èt Ê
œ
"
#
œ
151
csc (3x 1)(csc (3x 1) cot (3x 1) †
d
dx
csc (3x 1))
d
dx
#
(3x 1)) œ 2 csc (3x 1) cot (3x 1)
52. y œ 9 tan ˆ x3 ‰ Ê yw œ 9 ˆsec# ˆ x3 ‰‰ ˆ "3 ‰ œ 3 sec# ˆ x3 ‰ Ê yww œ 3 † 2 sec ˆ x3 ‰ ˆsec ˆ x3 ‰ tan ˆ x3 ‰‰ ˆ "3 ‰ œ 2 sec# ˆ 3x ‰ tan ˆ 3x ‰
53. g(x) œ Èx Ê gw (x) œ
"
#È x
Ê g(1) œ 1 and gw (1) œ
therefore, (f ‰ g)w (1) œ f w (g(1)) † gw (1) œ 5 †
"
#
œ
"
u#
1
10
14
=
"
(1x)#
Ê g(1) œ
"
#
w
and gw (1) œ
"
4
; f(u) œ 1
Ê f w (g(1)) œ f w ˆ #" ‰ œ 4; therefore, (f ‰ g)w (1) œ f (g(1))gw (1) œ 4 †
55. g(x) œ 5Èx Ê gw (x) œ
œ
; f(u) œ u& 1 Ê f w (u) œ 5u% Ê f w (g(1)) œ f w (1) œ 5;
5
#
54. g(x) œ (1 x)" Ê gw (x) œ (1 x)# (1) œ
Ê f w (u) œ
"
#
5
#Èx
Ê g(1) œ 5 and gw (1) œ
5
#
"
4
"
u
œ1
1‰
; f(u) œ cot ˆ 110u ‰ Ê f w (u) œ csc# ˆ 110u ‰ ˆ 10
1
1
1
csc# ˆ 110u ‰ Ê f w (g(1)) œ f w (5) œ 10
csc# ˆ 1# ‰ œ 10
; therefore, (f ‰ g)w (1) œ f w (g(1))gw (1) œ 10
†
5
#
56. g(x) œ 1x Ê gw (x) œ 1 Ê g ˆ "4 ‰ œ 14 and gw ˆ 4" ‰ œ 1; f(u) œ u sec# u Ê f w (u) œ 1 2 sec u † sec u tan u
œ 1 2 sec# u tan u Ê f w ˆg ˆ "4 ‰‰ œ f w ˆ 14 ‰ œ 1 2 sec# 14 tan 14 œ 5; therefore, (f ‰ g)w ˆ 4" ‰ œ f w ˆg ˆ 4" ‰‰ gw ˆ 4" ‰ œ 51
57. g(x) œ 10x# x 1 Ê gw (x) œ 20x 1 Ê g(0) œ 1 and gw (0) œ 1; f(u) œ
œ
2u# 2
au # 1 b #
Ê f w (u) œ
au# 1b(2) (2u)(2u)
au # 1 b #
Ê f w (g(0)) œ f w (1) œ 0; therefore, (f ‰ g)w (0) œ f w (g(0))gw (0) œ 0 † 1 œ 0
"
2
w
x# 1 Ê g (x) œ x$ Ê g(1) œ 0 and
4(u 1)
1 ‰ (u 1)(1) (u 1)(1)
2 ˆ uu
œ 2(u(u1)(2)
1 †
(u 1)#
1)$ œ (u 1)$
w
w
w
58. g(x) œ
œ
2u
u # 1
#
1‰
1‰
gw (1) œ 2; f(u) œ ˆ uu
Ê f w (u) œ 2 ˆ uu
1
1
Ê f w (g(1)) œ f w (0) œ 4; therefore,
(f ‰ g) (1) œ f (g(1))g (1) œ (4)(2) œ 8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
d
du
1‰
ˆ uu
1
152
Chapter 3 Differentiation
59. (a) y œ 2f(x) Ê
œ 2f w (x) Ê
dy
dx
(b) y œ f(x) g(x) Ê
(c) y œ f(x) † g(x) Ê
œ 2f w (2) œ 2 ˆ "3 ‰ œ
dy
dx ¹ x=2
œ f w (x) gw (x) Ê
dy
dx
dy
dx
œ
g(x)f w (x) f(x)gw (x)
[g(x)]#
(e) y œ f(g(x)) Ê
dy
dx
œ f w (g(x))gw (x) Ê
(d) y œ
f(x)
g(x)
Ê
"
#
Ê
dy
dx ¹ x=2
œ
(g) y œ (g(x))# Ê
dy
dx
œ 2(g(x))$ † gw (x) Ê
(h) y œ a(f(x))# (g(x))# b
dy
dx ¹ x=2
œ
"
#
(f(x))"Î# † f w (x) œ
"Î#
Ê
"
#
# a(f(x))
# "Î#
w
a(f(2))# (g(2)) b
Ê
dy
dx ¹ x=3
œ
dy
dx
(2) ˆ "3 ‰ (8)(3)
##
œ
"
3
œ f w (g(2))gw (2) œ f w (2)(3) œ
f w (x)
#Èf(x)
dy
dx
œ f(3)gw (3) g(3)f w (3) œ 3 † 5 (4)(21) œ 15 81
dy
dx ¹ x=3
g(2)f w (2) f(2)gw (2)
[g(2)]#
œ
dy
dx ¹ x=2
(f) y œ (f(x))"Î# Ê
Ê
œ f w (3) gw (3) œ 21 5
dy
dx ¹ x=3
œ f(x)gw (x) g(x)f w (x) Ê
dy
dx
2
3
dy
dx ¹ x=2
œ
f w (2)
#Èf(2)
ˆ "3 ‰
œ
œ
(3) œ 1
œ
#È 8
37
6
"
6È 8
œ
"
1 #È 2
œ 2(g(3))$ gw (3) œ 2(4)$ † 5 œ
(g(x))# b
"Î#
œ
È2
24
5
3#
a2f(x) † f w (x) 2g(x) † gw (x)b
a2f(2)f (2) 2g(2)gw (2)b œ
"
#
a8# 2# b
"Î#
ˆ2 † 8 †
"
3
2 † 2 † (3)‰
œ 3È517
60. (a) y œ 5f(x) g(x) Ê
(b) y œ f(x)(g(x))$ Ê
œ 5f w (x) gw (x) Ê
dy
dx
dy
dx ¹ x=1
œ 5f w (1) gw (1) œ 5 ˆ "3 ‰ ˆ 38 ‰ œ 1
œ f(x) a3(g(x))# gw (x)b (g(x))$ f w (x) Ê
dy
dx
dy
dx ¹ x=0
œ $f(0)(g(0))# gw (0) (g(0))$ f w (0)
œ 3(1)(1)# ˆ 3" ‰ (1)$ (5) œ 6
(c) y œ
œ
f(x)
g(x) 1
Ê
(g(x) 1)f w (x) f(x) gw (x)
(g(x) 1)#
œ
dy
dx
(4") ˆ "3 ‰(3) ˆ 83 ‰
(41)#
Ê
dy
dx ¹ x=1
œ
(g(1) 1)f w (1) f(1)gw (1)
(g(1) 1)#
œ1
(d) y œ f(g(x)) Ê
dy
dx
œ f w (g(x))gw (x) Ê
dy
dx ¹ x=0
œ f w (g(0))gw (0) œ f w (1) ˆ "3 ‰ œ ˆ "3 ‰ ˆ 3" ‰ œ 9"
(e) y œ g(f(x)) Ê
dy
dx
œ gw (f(x))f w (x) Ê
dy
dx ¹ x=0
œ gw (f(0))f w (0) œ gw (1)(5) œ ˆ 83 ‰ (5) œ 40
3
(f) y œ ax"" f(x)b
#
Ê
œ 2 ax"" f(x)b
dy
dx
$
a11x"! f w (x)b Ê
dy
dx ¹ x=1
œ 2(1 f(1))$ a11 f w (1)b
"
‰
œ 2(1 3)$ ˆ11 "3 ‰ œ ˆ 42$ ‰ ˆ 32
3 œ 3
(g) y œ f(x g(x)) Ê
œ f w (x g(x)) a1 gw (x)b Ê
dy
dx
dy
dx ¹ x=0
œ f w (0 g(0)) a1 gw (0)b œ f w (1) ˆ1 "3 ‰
œ ˆ "3 ‰ ˆ 43 ‰ œ 49
61.
ds
dt
œ
ds
d)
†
d)
dt :
s œ cos ) Ê
62.
dy
dt
œ
dy
dx
†
dx
dt :
y œ x# 7x 5 Ê
63. With y œ x, we should get
(a) y œ
(b)
u
5
7 Ê
y œ 1 "u Ê
"
œ "
u# † (x 1)#
œ
dy
du
dy
du
œ
dy
dx
"
5
dy
du
dy
dx
œ
"
u#
ds ¸
d) )= 321
œ sin ˆ 3#1 ‰ œ 1 so that
œ 2x 7 Ê
dy
dx ¹ x=1
œ 9 so that
dy
dt
œ
ds
dt
dy
dx
œ
†
ds
d)
dx
dt
†
d)
dt
œ9†
œ 1†5œ 5
"
3
œ3
œ 1 for both (a) and (b):
; u œ 5x 35 Ê
; u œ (x 1)
"
a(x 1)" b#
64. With y œ x$Î# , we should get
(a) y œ u$ Ê
œ sin ) Ê
ds
d)
†
dy
dx
"
(x 1)#
œ
3
#
"
du
dx
œ 5; therefore,
Ê
du
dx
œ (x 1)# †
dy
dy
dx œ du
#
†
œ (x 1) (1) œ
"
(x 1)#
du
"
dx œ 5
"
(x 1)# ;
† 5 œ 1, as expected
therefore
dy
dx
œ
dy
du
†
du
dx
œ 1, again as expected
x"Î# for both (a) and (b):
œ 3u# ; u œ Èx Ê
du
dx
"
#È x
; therefore,
dy
dx
œ
dy
du
†
du
dx
œ 3u# †
œ 3x# ; therefore,
dy
dx
œ
dy
du
†
du
dx
œ
œ
"
#È x
#
œ 3 ˆÈx‰ †
"
#Èx
œ
3
#
Èx,
as expected.
(b) y œ Èu Ê
dy
du
œ
"
#È u
; u œ x$ Ê
du
dx
"
#Èu
† 3x# œ
"
#È x $
again as expected.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
† 3x# œ
3
#
x"Î# ,
Section 3.5 The Chain Rule and Parametric Equations
65. y œ 2 tan ˆ 14x ‰ Ê
(a)
dy
dx ¹ x=1
œ
1
#
dy
dx
œ ˆ2 sec#
1x ‰ ˆ 1 ‰
4
4
œ
1
#
sec#
1x
4
sec# ˆ 14 ‰ œ 1 Ê slope of tangent is 2; thus, y(1) œ 2 tan ˆ 14 ‰ œ 2 and yw (1) œ 1 Ê tangent line is
given by y 2 œ 1(x 1) Ê y œ 1x 2 1
(b) yw œ 1# sec# ˆ 14x ‰ and the smallest value the secant function can have in # x 2 is 1 Ê the minimum
value of yw is 1# and that occurs when 1# œ 1# sec# ˆ 14x ‰ Ê 1 œ sec# ˆ 14x ‰ Ê „ 1 œ sec ˆ 14x ‰ Ê x œ 0.
66. (a) y œ sin 2x Ê yw œ 2 cos 2x Ê yw (0) œ 2 cos (0) œ 2 Ê tangent to y œ sin 2x at the origin is y œ 2x;
y œ sin ˆ x# ‰ Ê yw œ "# cos ˆ x# ‰ Ê yw (0) œ "# cos 0 œ "# Ê tangent to y œ sin ˆ x# ‰ at the origin is
y œ "# x. The tangents are perpendicular to each other at the origin since the product of their slopes is
1.
(b) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m cos 0 œ m; y œ sin ˆ mx ‰ Ê yw œ m" cos ˆ mx ‰
Ê yw (0) œ m" cos (0) œ m" . Since m † ˆ m" ‰ œ 1, the tangent lines are perpendicular at the origin.
(c) y œ sin (mx) Ê yw œ m cos (mx). The largest value cos (mx) can attain is 1 at x œ 0 Ê the largest value
yw can attain is kmk because kyw k œ km cos (mx)k œ kmk kcos mxk Ÿ kmk † 1 œ kmk . Also, y œ sin ˆ mx ‰
ˆ x ‰¸ Ÿ ¸ m" ¸ ¸cos ˆ mx ‰¸ Ÿ km" k Ê the largest value yw can attain is ¸ m" ¸ .
Ê yw œ m" cos ˆ mx ‰ Ê kyw k œ ¸ "
m cos m
(d) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m Ê slope of curve at the origin is m. Also, sin (mx) completes
m periods on [0ß 21]. Therefore the slope of the curve y œ sin (mx) at the origin is the same as the number
of periods it completes on [0ß 21]. In particular, for large m, we can think of “compressing" the graph of
y œ sin x horizontally which gives more periods completed on [0ß 21], but also increases the slope of the
graph at the origin.
67. x œ cos 2t, y œ sin 2t, 0 Ÿ t Ÿ 1
Ê cos# 2t sin# 2t œ 1 Ê x# y# œ 1
68. x œ cos (1 t), y œ sin (1 t), 0 Ÿ t Ÿ 1
Ê cos# (1 t) sin# (1 t) œ 1
Ê x# y# œ 1, y !
69. x œ 4 cos t, y œ 2 sin t, 0 Ÿ t Ÿ 21
70. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21
Ê
16 cos# t
16
4 sin# t
4
œ1 Ê
x#
16
y#
4
œ1
Ê
16 sin# t
16
25 cos# t
25
œ1 Ê
x#
16
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y#
#5
œ1
153
154
Chapter 3 Differentiation
71. x œ 3t, y œ 9t# , _ t _ Ê y œ x#
72. x œ Èt , y œ t, t
0 Ê x œ È y
#
or y œ x , x Ÿ 0
73. x œ 2t 5, y œ 4t 7, _ t _
Ê x 5 œ 2t Ê 2(x 5) œ 4t
Ê y œ 2(x 5) 7 Ê y œ 2x 3
75. x œ t, y œ È1 t# , 1 Ÿ t Ÿ 0
Ê y œ È1 x#
#
Ê y œ 2 23 x, ! Ÿ x Ÿ $
76. x œ Èt 1, y œ Èt, t 0
Ê y# œ t Ê x œ Èy# 1, y
77. x œ sec# t 1, y œ tan t, 1# t
#
74. x œ 3 3t, y œ 2t, 0 Ÿ t Ÿ 1 Ê y# œ t
Ê x œ 3 3 ˆ y# ‰ Ê 2x œ 6 3y
#
Ê sec t 1 œ tan t Ê x œ y
1
#
78. x œ sec t, y œ tan t, 1# t
#
#
#
0
1
#
#
Ê sec t tan t œ 1 Ê x y œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.5 The Chain Rule and Parametric Equations
79. (a) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
80. (a) x œ a sin t, y œ b cos t,
(b) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
(c) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
1
#
ŸtŸ
155
51
#
(b) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 21
(c) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 9#1
(d) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
(d) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 41
81. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes
through a"ß $b at t œ !. We determine a and b so that the line goes through a%ß "b when t œ ".
Since % œ " a Ê a œ &. Since " œ $ b Ê b œ %. Therefore, one possible parameterization is x œ " &t,
y œ $ %t, 0 Ÿ t Ÿ ".
82. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through
a"ß $b at t œ !. We determine a and b so that the line goes through a$ß #b when t œ ". Since $ œ " a Ê a œ %.
Since # œ $ b Ê b œ &. Therefore, one possible parameterization is x œ " %t, y œ $ &t, 0 Ÿ t Ÿ ".
83. The lower half of the parabola is given by x œ y# " for y Ÿ !. Substituting t for y, we obtain one possible
parameterization x œ t# ", y œ t, t Ÿ 0Þ
84. The vertex of the parabola is at a"ß "b, so the left half of the parabola is given by y œ x# #x for x Ÿ ". Substituting
t for x, we obtain one possible parametrization: x œ t, y œ t# #t, t Ÿ ".
85. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a#ß $b for t œ ! and passes
through a"ß "b at t œ ". Then x œ fatb, where fa!b œ # and fa"b œ ".
Since slope œ ??xt œ "#
"! œ $, x œ fatb œ $t # œ # $t. Also, y œ gatb, where ga!b œ $ and ga"b œ ".
Since slope œ
?y
?t
"3
"!
œ
œ 4. y œ gatb œ %t $ œ $ %t.
One possible parameterization is: x œ # $t, y œ $ %t, t
!.
86. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a"ß #b for t œ ! and
passes through a!ß !b at t œ ". Then x œ fatb, where fa!b œ " and fa"b œ !.
Since slope œ
Since slope œ
?x
?t
?y
?t
! a"b
"!
!#
"! œ
œ
œ
œ ", x œ fatb œ "t a"b œ " t. Also, y œ gatb, where ga!b œ # and ga"b œ !.
#. y œ gatb œ #t # œ # #t.
One possible parameterization is: x œ " t, y œ # #t, t
87. t œ
Ê
1
4 Ê
dy
dx ¹ tœ 1
x œ 2 cos
d# y
dx#
dyw /dt
dx/dt
œ cot
1
4
1
4
œ È2, y œ 2 sin
1
4
œ È2;
dx
dt
!.
œ 2 sin t,
dy
dt
œ 2 cos t Ê
dy
dx
œ 1; tangent line is y È2 œ 1 Šx È2‹ or y œ x
œ
dy/dt
dx/dt
w
È
2 2 ; dy
dt
œ
2 cos t
2 sin t
œ cot t
œ csc# t
4
Ê
88. t œ
Ê
œ
21
3 Ê
dy
dx ¹ tœ 21
œ
csc# t
2 sin t
x œ cos
21
3
"
œ 2 sin
$t Ê
d# y
dx# ¹ tœ 21
œ È 2
4
È3 dx
21
3 œ # ; dt
È3
È 3 x
# ‹œ
œ "# , y œ È3 cos
œ È3 ; tangent line is y Š
3
Ê
d# y
dx# ¹ tœ 1
œ sin t,
dy
dt
œ È3 sin t Ê
ˆ "# ‰‘ or y œ È3 x;
dy
dx
œ
È3 sin t
sin t
œ È3
d# y
dx#
œ
œ0
dyw
dt
œ0 Ê
"
œ 1; tangent line is
0
sin t
œ0
3
89. t œ
y
1
4
"
#
Ê xœ
1
4
,yœ
"
#
;
dx
dt
œ 1,
dy
dt
œ 1 † ˆx 4" ‰ or y œ x 4" ;
œ
dyw
dt
"
#Èt
Ê
dy
dx
œ
œ 4" t$Î# Ê
dy/dt
dx/dt
d# y
dx#
œ
œ
1
2È t
dyw /dt
dx/dt
Ê
dy
dx ¹ tœ 1
4
œ
#É "4
œ 4" t$Î# Ê
d# y
dx# ¹ tœ 1
œ 2
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
156
Chapter 3 Differentiation
90. t œ 3 Ê x œ È3 1 œ 2, y œ È3(3) œ 3;
È
œ 3 Èt3t1 œ
dyw
dt
Ê
œ
dy
dx ¹ tœ3
3 È 3 1
È3(3)
œ
d# y
dx# ¹ tœ3
#
3t
œ 4t,
dx
dt
y 1 œ 1 † (x 5) or y œ x 4;
œ
1
3
1
3
Ê xœ
Ê
sin t
1cos t
93. t œ
Ê
1
3
sin
dy
dx ¹ tœ 1
œ
3
Ê y œ È3x
œ
1
(1 cos t)#
1
2 Ê
dy
dx ¹ tœ 1
1È3
3
œ
1
3
sin ˆ 13 ‰
1cos ˆ 13 ‰
2;
d# y
dx# ¹ tœ 1
Ê
œ
Ê
3
2tÈ3t Èt 1
d# y
dx#
dyw
dt
œ 4t$ Ê
dy
dt
dyw
dt
œ 2t Ê
dy
dx
d# y
dx#
œ
œ
(3t)"Î# Ê
dy
dx
ˆ 3# ‰ (3t)"Î#
ˆ "# ‰ (t 1)"Î#
œ
Š 2tÈ3t3Èt b 1 ‹
œ tÈ33t
Š 2Èct1b 1 ‹
œ
dyw /dt
dx/dt
4t$
4t
œ
œ t# Ê
œ
2t
4t
"
#
dy
dx ¹ tœc1
d# y
dx# ¹ tœc1
Ê
œ (1)# œ 1; tangent line is
œ
dy
, y œ 1 cos 13 œ 1 "# œ "# ; dx
dt œ 1 cos t, dt œ sin t Ê
È
Š #3 ‹
È
œ ˆ " ‰ œ È3 ; tangent line is y "# œ È3 Šx 13 #3 ‹
#
œ
"
#
È3
#
(1 cos t)(cos t) (sin t)(sin t)
(1cos t)#
œ
1
1cos t
d# y
dx#
Ê
dyw /dt
dx/dt
œ
œ
dy
dx
œ
dy/dt
dx/dt
1 ‰
ˆ 1 cos
t
1 cos t
œ 4
x œ cos
œ cot
1
2
œ 0, y œ 1 sin
1
#
1
2
œ 2;
œ sin t,
dx
dt
œ 0; tangent line is y œ 2;
sec# t
2 sec# t tan t
œ
"
2 tan t
œ
"
#
cot t Ê
w
dy
dt
dy
dx ¹ tœc 1
4
y (1) œ "# (x 1) or y œ "# x "# ;
Ê
œ
œ
dy/dt
dx/dt
d# y
dx# ¹ tœc 1
œ
dyw
dt
œ
dy
dt
œ cos t Ê
œ csc# t Ê
94. t œ 14 Ê x œ sec# ˆ 14 ‰ 1 œ 1, y œ tan ˆ 14 ‰ œ 1;
dy
dx
3
#
3
2
Ê
œ
œ "3
91. t œ 1 Ê x œ 5, y œ 1;
92. t œ
dy
dt
œ 2; tangent line is y 3 œ 2[x (2)] or y œ 2x 1;
È3t 3 (t 1)"Î# ‘ 3Èt 1 3 (3t)"Î# ‘
#
œ "# (t 1)"Î# ,
dx
dt
"
#
dx
dt
#
d y
dx#
œ
dy
dx
#
csc t
sin t
œ
cos t
sin t
œ cot t
œ csc$ t Ê
d# y
dx# ¹ tœ 1
œ 1
2
œ 2 sec# t tan t,
dy
dt
œ sec# t
cot ˆ 14 ‰ œ #" ; tangent line is
œ "# csc# t Ê
d# y
dx#
œ
"# csc# t
2 sec# t tan t
œ "4 cot$ t
"
4
4
95. s œ A cos (21bt) Ê v œ
ds
dt
œ A sin (21bt)(21b) œ 21bA sin (21bt). If we replace b with 2b to double the
frequency, the velocity formula gives v œ 41bA sin (41bt) Ê doubling the frequency causes the velocity to
# #
double. Also v œ #1bA sin (21bt) Ê a œ dv
dt œ 41 b A cos (21bt). If we replace b with 2b in the
acceleration formula, we get a œ 161# b# A cos (41bt) Ê doubling the frequency causes the acceleration to
$ $
quadruple. Finally, a œ 41# b# A cos (21bt) Ê j œ da
dt œ 81 b A sin (21bt). If we replace b with 2b in the jerk
formula, we get j œ 641$ b$ A sin (41bt) Ê doubling the frequency multiplies the jerk by a factor of 8.
21
21
21 ‰
96. (a) y œ 37 sin 365
(x 101)‘ 25 Ê yw œ 37 cos 365
(x 101)‘ ˆ 365
œ
741
365
21
cos 365
(x 101)‘ .
The temperature is increasing the fastest when yw is as large as possible. The largest value of
21
21
cos 365
(x 101)‘ is 1 and occurs when 365
(x 101) œ 0 Ê x œ 101 Ê on day 101 of the year
( µ April 11), the temperature is increasing the fastest.
1
741
21
‘ 741
(b) yw (101) œ 74
365 cos 365 (101 101) œ 365 cos (0) œ 365 ¸ 0.64 °F/day
97. s œ (" 4t)"Î# Ê v œ
ds
dt
œ
v œ 2(" 4t)"Î# Ê a œ
dv
dt
"
#
(1 4t)"Î# (4) œ 2(1 4t)"Î# Ê v(6) œ 2(" % † 6)"Î# œ
"
#
2
5
m/sec;
œ † 2(1 4t)$Î# (4) œ 4(1 4t)$Î# Ê a(6) œ 4(1 4 † 6)$Î# œ 14#5 m/sec#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.5 The Chain Rule and Parametric Equations
98. We need to show a œ
œ
k
2È s
† kÈs œ
99. v proportional to
œ 2sk$Î# †
dx
dt
100. Let
k
Ès
kT
2
k
#
is constant: a œ
dv
dt
œ
dv
ds
†
ds
dt
and
dv
ds
œ
d
ds
ˆkÈs‰ œ
Ê aœ
k
2È s
dv
ds
†
†
ds
dt
ds
dt
œ
dv
ds
†v
which is a constant.
"
Ès
Ê vœ
k
Ès
for some constant k Ê
#
dv
ds
œ 2sk$Î# . Thus, a œ
œ k# ˆ s"# ‰ Ê acceleration is a constant times
œ f(x). Then, a œ
101. T œ 21É Lg Ê
œ
#
dv
dt
dT
dL
dv
dt
œ 21 †
œ
"
#É Lg
dv
dx
†
dx
dt
œ
†
"
g
œ
1
gÉ Lg
dv
dx
† f(x) œ
œ
1
ÈgL
d
dx
"
s#
œ
dv
ds
œ
dv
ds
†v
so a is inversely proportional to s# .
ˆ dx
‰
dt † f(x) œ
. Therefore,
dv
dt
dT
du
œ
d
dx
(f(x)) † f(x) œ f w (x)f(x), as required.
dT
dL
†
dL
du
œ
1
ÈgL
† kL œ
1 kÈ L
Èg
œ
"
#
† 21kÉ Lg
, as required.
102. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f ‰ g is
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so
there is no contradiction.
103. The graph of y œ (f ‰ g)(x) has a horizontal tangent at x œ 1 provided that (f ‰ g)w (1) œ 0 Ê f w (g(1))gw (1) œ 0
Ê either f w (g(1)) œ 0 or gw (1) œ 0 (or both) Ê either the graph of f has a horizontal tangent at u œ g(1), or the
graph of g has a horizontal tangent at x œ 1 (or both).
104. (f ‰ g)w (5) 0 Ê f w (g(5)) † gw (5) 0 Ê f w (g(5)) and gw (5) are both nonzero and have opposite signs.
That is, either cf w (g(5)) 0 and gw (5) 0d or cf w (g(5)) 0 and gw (5) 0d .
105. As h Ä 0, the graph of y œ
sin 2(xh)sin 2x
h
approaches the graph of y œ 2 cos 2x because
lim
hÄ!
sin 2(xh)sin 2x
h
œ
d
dx
(sin 2x) œ 2 cos 2x.
106. As h Ä 0, the graph of y œ
cos c(x h)# dcos ax# b
h
#
approaches the graph of y œ 2x sin ax b because
lim
hÄ!
cos c(x h)# dcos ax# b
h
œ
d
dx
ccos ax# bd œ 2x sin ax# b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
157
158
107.
Chapter 3 Differentiation
dx
dt
œ cos t and
œ 2 cos 2t Ê
dy
dt
dy
dx
Ê 2 cos# t 1 œ 0 Ê cos t œ „
y œ sin 2 ˆ 14 ‰ œ 1 Ê Š
È2
# ß 1‹
œ
dy/dt
dx/dt
"
È2
œ
œ
2 cos 2t
cos t
Ê tœ
1
4
31
4
,
2 a2 cos# t 1b
cos t
51
4
,
108.
dx
dt
œ 2 cos 2t and
dy
dx ¹ tœ0
3 ca2 cos# t 1b (cos t) 2 sin t cos t sin td
2 a2 cos# t1b
œ
dy
dx
œ 3 cos 3t Ê
dy
dt
(3 cos t) a4 cos# t 3b
2 a2 cos# t 1b
œ0 Ê
and y œ sin 3 ˆ 16 ‰ œ 1 Ê Š
œ
œ 2 Ê y œ 2x and
dy
dx
œ
œ
dy/dt
dx/dt
3 cos 3t
2 cos 2t
È3
#
È3
# ß 1‹
1
6
Ê tœ
,
51
6
,
71
6
œ
"
#É È x
†
d
dx
ˆÈx‰ œ
"
#ÉÈx
†
"
#Èx
œ
dy
dx
œ
"
#É x È x
3È x
4È x É È x
œ
3
4
†
d
dx
ˆxÈx‰ Ê
dy
dx
œ
dy
dx
œ
110. From the power rule, with y œ x$Î% , we get
Ê
œ
df
dt
œ
(3 cos t) a4 cos# t 3b
2 a2 cos# t1b
,
111
6
. In the 1st quadrant: t œ
1
6
1
#
,
31
#
and
and
Ê x œ sin 2 ˆ 16 ‰ œ
1
#
, 1,
31
#
dy
dx ¹ tœ0
and t œ 0,
œ
3 cos 0
2 cos 0
1
3
œ
,
21
3
3
#
, 1,
41
3
Ê yœ
,
51
3
3
#
È3
#
x, and
Ê t œ 0 and t œ 1 give
dy
dx ¹ tœ1
"
4
dy
dx
"
#ÉxÈx
œ
"
4
x$Î% . From the chain rule, y œ ÉÈx
x$Î% , in agreement.
œ
3
4
x"Î% . From the chain rule, y œ ÉxÈx
† Šx †
"
#È x
È x‹ œ
"
#ÉxÈx
† ˆ 3# Èx‰ œ
3È x
4É xÈ x
œ 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14t
df
dt
È2
#
; then
x"Î% , in agreement.
(c) The curve of y œ
œ
3(cos 2t cos t sin 2t sin t)
2 a2 cos# t1b
111. (a)
(b)
1
4
1 give the tangent lines at
œ 3# Ê y œ 3# x
109. From the power rule, with y œ x"Î% , we get
dy
dx
Ê x œ sin
is the point where the graph has a horizontal tangent. At the origin: x œ 0
the tangent lines at the origin. Tangents at the origin:
Ê
1
31
# , 1, # ; thus t œ 0 and t œ
dy
dx ¹ tœ1 œ 2 Ê y œ 2x
(3 cos t) a2 cos# t 1 2 sin# tb
2 a2 cos# t1b
and y œ 0 Ê sin 2t œ 0 and sin 3t œ 0 Ê t œ 0,
3 cos (31)
2 cos (21)
1
4
. In the 1st quadrant: t œ
œ 0 Ê 3 cos t œ 0 or 4 cos# t 3 œ 0: 3 cos t œ 0 Ê t œ
4 cos# t 3 œ 0 Ê cos t œ „
œ
71
4
œ0
is the point where the tangent line is horizontal. At the origin: x œ 0 and y œ 0
Ê sin t œ 0 Ê t œ 0 or t œ 1 and sin 2t œ 0 Ê t œ 0,
the origin. Tangents at origin:
,
2 a2 cos# t 1b
cos t
œ0 Ê
dy
dx
; then
approximates y œ
dg
dt
the best when t is not 1, 1# , 0, 1# , nor 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 3.5 The Chain Rule and Parametric Equations
112. (a)
(b)
dh
dt
œ 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)
(c)
111-116. Example CAS commands:
Maple:
f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t)
- 0.02546*cos(10*t) - 0.01299*cos(14*t);
g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t x • 3 x • 2 2*x;
plot(f(x), x=1..2);
diff(f(x),x);
fp := unapply (ww ,x);
L:=x -> f(a) fp(a)*(x a);
plot({f(x), L(x)}, x=1..2);
err:=x -> abs(f(x) L(x));
plot(err(x), x=1..2, title = #absolute error function#);
err(1);
Mathematica: (function, x1, x2, and a may vary):
Clear[f, x]
{x1, x2} = {1, 2}; a = 1;
f[x_]:=x3 x2 2x
Plot[f[x], {x, x1, x2}]
lin[x_]=f[a] f'[a](x a)
Plot[{f[x], lin[x]}, {x, x1, x2}]
err[x_]=Abs[f[x] lin[x]]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
180
Chapter 3 Differentiation
Plot[err[x], {x, x1,x 2}]
err//N
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del)
eps = 0.5; del = 0.4
Plot[{err[x], eps},{x, a del, a del}]
CHAPTER 3 PRACTICE EXERCISES
1. y œ x& 0.125x# 0.25x Ê
2. y œ 3 0.7x$ 0.3x( Ê
4. y œ x( È7x
"
1 1
Ê
œ 5x% 0.25x 0.25
œ 2.1x# 2.1x'
dy
dx
3. y œ x$ 3 ax# 1# b Ê
dy
dx
dy
dx
œ 3x# 3(2x 0) œ 3x# 6x œ 3x(x 2)
dy
dx
œ 7x' È7
5. y œ (x 1)# ax# 2xb Ê
dy
dx
œ (x 1)# (2x 2) ax# 2xb (2(x 1)) œ 2(x 1) c(x 1)# x(x 2)d
dy
dx
œ (2x 5)(1)(4 x)# (1) (4 x)" (2) œ (4 x)# c(2x 5) 2(4 x)d
#
œ 2(x 1) a2x 4x 1b
6. y œ (2x 5)(4 x)" Ê
œ 3(4 x)
#
$
7. y œ a)# sec ) 1b Ê
8. y œ Š1
csc )
#
)#
4‹
9. s œ
Èt
1 Èt
Ê
ds
dt
œ
10. s œ
"
Èt 1
Ê
ds
dt
œ
#
Ê
ˆ1 Èt‰†
"
sin# x
2
sin x
œ 2 Š1
"
"
Èt Èt Š #Èt ‹
#
ˆÈ t 1 ‰
dy
dx
#
"
Èt ‹
#
œ
ds
dt
)# ˆ csc ) cot )
4‹
#
ˆ1 Èt‰ Èt
2Èt ˆ1 Èt‰
#
œ
#) ‰ œ Š1
csc )
#
)#
4 ‹ (csc
) cot ) ))
"
#
#Èt ˆ1 Èt‰
"
#
2 È t ˆÈ t 1 ‰
dy
dx
œ (2 csc x)(csc x cot x) 2( csc x cot x) œ (2 csc x cot x)(1 csc x)
œ 4 cos$ (1 2t)(sin (1 2t))(2) œ 8 cos$ (1 2t) sin (1 2t)
œ 3 cot# ˆ 2t ‰ ˆcsc# ˆ 2t ‰‰ ˆ t#2 ‰ œ
15. s œ (sec t tan t)& Ê
ds
dt
16. s œ csc& a1 t 3t# b Ê
&
csc )
#
œ (4 tan x) asec# xb (2 sec x)(sec x tan x) œ 2 sec# x tan x
œ csc# x 2 csc x Ê
ds
dt
œ
#
ˆÈt 1‰ (0) 1 Š
13. s œ cos% (1 2t) Ê
14. s œ cot$ ˆ 2t ‰ Ê
dy
d)
ˆ1 Èt‰
11. y œ 2 tan# x sec# x Ê
12. y œ
#
œ 3 a)# sec ) 1b (2) sec ) tan ))
dy
d)
6
t#
cot# ˆ 2t ‰ csc# ˆ 2t ‰
œ 5(sec t tan t)% asec t tan t sec# tb œ 5(sec t)(sec t tan t)&
ds
dt
œ 5 csc% a1 t 3t# b acsc a1 t 3t# b cot a1 t 3t# bb (1 6t)
œ 5(6t 1) csc a1 t 3t# b cot a1 t 3t# b
17. r œ È2) sin ) œ (2) sin ))"Î# Ê
dr
d)
œ
"
#
(2) sin ))"Î# (#) cos ) 2 sin )) œ
) cos ) sin )
È2) sin )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Practice Exercises
18. r œ 2)Ècos ) œ 2) (cos ))"Î# Ê
œ 2) ˆ "# ‰ (cos ))"Î# (sin )) 2(cos ))"Î# œ
dr
d)
) sin )
Ècos )
181
2Ècos )
2 cos ) ) sin )
Ècos )
œ
19. r œ sin È2) œ sin (2))"Î# Ê
20. r œ sin Š) È) 1‹ Ê
cos È2)
È 2)
œ cos (2))"Î# ˆ "# (2))"Î# (2)‰ œ
œ cos Š) È) 1‹ Š1
"
‹
2È ) 1
2È)"1
#È ) "
œ
œ
"
#
22. y œ 2Èx sin Èx Ê
dy
dx
"
2
œ 2Èx ˆcos Èx‰ Š 2È
‹ ˆsin Èx‰ Š 2È
‹ œ cos Èx
x
x
x# csc
Ê
2
x
x# ˆcsc
2
x
cot x2 ‰ ˆ x#2 ‰ ˆcsc x2 ‰ ˆ "# † 2x‰ œ csc
cos Š) È) 1‹
dy
dx
21. y œ
"
#
dr
d)
dr
d)
2
x
cot
2
x
x csc
2
x
sin Èx
Èx
dy
"Î#
sec (2x)# tan (2x)# (2(2x) † 2) sec (2x)# ˆ "# x$Î# ‰
dx œ x
8x"Î# sec (2x)# tan (2x)# "# x$Î# sec (2x)# œ "# x"Î# sec (2x)# c16 tan (2x)# x# d or #x"$Î# seca#xb2 16x# tana2xb#
23. y œ x"Î# sec (2x)# Ê
œ
"‘
24. y œ Èx csc (x 1)$ œ x"Î# csc (x 1)$
Ê
dy
dx
œ x"Î# acsc (x 1)$ cot (x 1)$ b a3(x 1)# b csc (x 1)$ ˆ "# x"Î# ‰
œ 3Èx (x 1)# csc (x 1)$ cot (x 1)$
or
"
csc(x
#È x
csc (x 1)$
2È x
œ
"
#
Èx csc (x 1)$ x" 6(x 1)# cot (x 1)$ ‘
1)$ c1 6x(x 1)# cot (x 1)$ d
25. y œ 5 cot x# Ê
26. y œ x# cot 5x Ê
dy
dx
œ 5 acsc# x# b (2x) œ 10x csc# ax# b
œ x# acsc# 5xb (5) (cot 5x)(2x) œ 5x# csc# 5x 2x cot 5x
dy
dx
27. y œ x# sin# a2x# b Ê
dy
dx
œ x# a2 sin a2x# bb acos a2x# bb (4x) sin# a2x# b (2x) œ 8x$ sin a2x# b cos a2x# b 2x sin# a2x# b
28. y œ x# sin# ax$ b Ê
dy
dx
œ x# a2 sin ax$ bb acos ax$ bb a3x# b sin# ax$ b a2x$ b œ 6 sin ax$ b cos ax$ b 2x$ sin# ax$ b
29. s œ ˆ t 4t 1 ‰
30. s œ
#
"
15(15t 1)$
Èx
Ê
œ 2 ˆ t 4t 1 ‰
$
(4t)(1)
Š (t 1)(4)
‹ œ 2 ˆ t 4t 1 ‰
(t 1)#
"
œ 15
(15t 1)$ Ê
#
31. y œ Š x 1 ‹ Ê
2È x
ds
dt
#
dy
dx
32. y œ Š 2Èx 1 ‹ Ê
dy
dx
(x 1)#
2È x
œ 2 Š 2È x 1 ‹
#
"Î#
33. y œ É x x# x œ ˆ1 "x ‰
Ê
dy
dx
œ
34. y œ 4xÉx Èx œ 4x ˆx x"Î# ‰
"Î#
œ ˆx Èx‰
’2x Š1
"
‹
#È x
"Î#
"
#
4
(t 1)#
"
œ 15
(3)(15t 1)% (15) œ
ds
dt
"
(x 1) Š #È
‹ ˆÈx‰ (1)
x
Èx
œ 2 Šx1‹ †
$
œ
(x 1) 2x
(x 1)$
ˆ2Èx 1‰ Š È" ‹ ˆ2Èx‰ Š È" ‹
x
x
ˆ2 È x 1 ‰
#
ˆ1 "x ‰"Î# ˆ x"# ‰ œ
Ê
dy
dx
œ
œ
"
"Î#
3
(15t 1)%
1x
(x 1)$
#x # É 1
œ 4x ˆ "# ‰ ˆx x"Î# ‰
œ (t 8t$1)
4Èx Š È"x ‹
ˆ2 È x 1 ‰$
œ
4
ˆ2 È x 1 ‰$
"
x
ˆ1 "# x"Î# ‰ ˆx x"Î# ‰"Î# (4)
"Î#
ˆ2x Èx 4x 4Èx‰ œ
4 ˆx Èx‰“ œ ˆx Èx‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
6x 5Èx
É x Èx
182
Chapter 3 Differentiation
#
35. r œ ˆ cossin) ) 1 ‰ Ê
œ 2 ˆ cossin) ) " ‰ Š cos
#
#
1 ‰
36. r œ ˆ 1sin )cos
Ê
)
œ
2(sin ) ")
(1 cos ))$
)) (sin ))(sin ))
œ 2 ˆ cossin) ) 1 ‰ ’ (cos ) 1)(cos
“
(cos ) 1)#
dr
d)
) cos ) sin# )
‹
(cos ) ")#
œ
(2 sin )) (1 cos ))
(cos ) 1)$
1 ‰ (1 cos ))(cos )) (sin ) ")(sin ))
œ 2 ˆ 1sin )cos
“
) ’
(1 cos ))#
dr
d)
2(sin ) 1)(cos ) sin ) 1)
(1 c os ))$
acos ) cos# ) sin# ) sin )b œ
37. y œ (2x 1) È2x 1 œ (2x 1)$Î# Ê
œ
dy
dx
3
#
(2x 1)"Î# (2) œ 3È2x 1
38. y œ 20(3x 4)"Î% (3x 4)"Î& œ 20(3x 4)"Î#! Ê
39. y œ 3 a5x# sin 2xb
40. y œ a3 cos$ 3xb
2 sin )
(cos ) ")#
œ
$Î#
"Î$
Ê
Ê
" ‰
œ 20 ˆ 20
(3x 4)"*Î#! (3) œ
œ 3 ˆ 3# ‰ a5x# sin 2xb
dy
dx
œ "3 a3 cos$ 3xb
dy
dx
dy
dx
%Î$
&Î#
[10x (cos 2x)(2)] œ
a3 cos# 3xb (sin 3x)(3) œ
3
(3x 4)"*Î#!
9(5x cos 2x)
a5x# sin 2xb&Î#
3 cos# 3x sin 3x
a3 cos$ 3xb%Î$
2
41. xy 2x 3y œ 1 Ê axyw yb 2 3yw œ 0 Ê xyw 3yw œ 2 y Ê yw (x 3) œ 2 y Ê yw œ yx
3
42. x# xy y# 5x œ 2 Ê 2x Šx
œ 5 2x y Ê
œ
dy
dx
dy
dx
dy
dx
dy
dx
ˆ4x 4y"Î$ ‰ œ 2 3x# 4y Ê
44. 5x%Î& 10y'Î& œ 15 Ê 4x"Î& 12y"Î&
"
#
45. (xy)"Î# œ 1 Ê
(xy)"Î# Šx
46. x# y# œ 1 Ê x# Š2y
47. y# œ
x
x 1
Ê 2y
x‰
48. y# œ ˆ 11
x
"Î#
dy
dx
dy
dx ‹
œ
dp
dq
œ
dp
dq
dy
dx
2y
dy
dx
œ 5 2x y Ê
dy
dx
(x 2y)
"x
1x
Ê 4y$
dp
dq
dy
dx
œ 2 Ê 4x
œ
dy
dx
œ 0 Ê 12y"Î&
dy
dx
dy
dx
4y"Î$
œ 4x"Î& Ê
œ x"Î# y"Î# Ê
œ 2xy# Ê
dy
dx
œ yx
Ê
dy
dx
œ
6q œ 0 Ê 3p#
dp
dq
dy
dx
dy
dx
œ 2 3x# 4y
dy
dx
dy
dx
"
œ "3 x"Î& y"Î& œ 3(xy)
"Î&
œ x" y Ê
dy
dx
œ yx
"
#y(x 1)#
dy
dx
œ
œ
(1 x)(1) (1 x)Ð")
(" x)#
4 Šp q
dy
dx
2 3x# 4y
4x 4y"Î$
dy
dx
y‹ œ 0 Ê x"Î# y"Î#
Ê
dy
dx
dp
dq ‹
"
2y$ (1 x)#
4q
dp
dq
œ 6q 4p Ê
dp
dq
a3p# 4qb œ 6q 4p
6q 4p
3p# 4q
50. q œ a5p# 2pb
Ê
&œ! Ê x
4y‹ 4y"Î$
y# (2x) œ 0 Ê 2x# y
49. p$ 4pq 3q# œ 2 Ê 3p#
Ê
dy
dx
(x 1)(1) (x)(1)
(x 1)#
Ê y% œ
dy
dx
5 2x y
x 2y
43. x$ 4xy 3y%Î$ œ 2x Ê 3x# Š4x
Ê
y‹ 2y
$Î#
#
Ê 1 œ 3# a5p# 2pb
&Î#
Š10p
dp
dq
2
dp
dq ‹
Ê 23 a5p# 2pb
&Î#
œ
dp
dq
(10p 2)
&Î#
œ a5p3(5p 2p1)b
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Practice Exercises
dr ‰
51. r cos 2s sin# s œ 1 Ê r(sin 2s)(2) (cos 2s) ˆ ds
2 sin s cos s œ 0 Ê
Ê
dr
ds
2r sin 2s sin 2s
cos 2s
œ
œ
(2r 1)(sin 2s)
cos 2s
52. 2rs r s s# œ 3 Ê 2 ˆr s
53. (a) x$ y$ œ 1 Ê 3x# 3y#
Ê
d# y
dx#
(b) y# œ 1
Ê
d# y
dx#
x#
‹
y#
2xy# a2yx# b Š
œ
y%
Ê 2y
2
x
2xy x# Š
œ
œ
dy
dx
y# x%
"
‹
yx#
œ
54. (a) x# y# œ 1 Ê 2x 2y
(b)
dy
dx
œ
x
y
d# y
dx#
Ê
œ
y(1) x
y#
dr
ds
d# y
dx#
œ
#
œ xy# Ê
dy
dx
2x%
y
y%
œ
"
yx#
œ
dy
dx
1 2s œ 0 Ê
(2s 1) œ 1 2s 2r Ê
y# (2x) ax# b Š2y
dr
ds
" 2s 2r
2s 1
œ
dy
dx ‹
y%
2xy$ 2x%
y&
Ê
dy
dx
œ ayx# b
"
Ê
dy
dx
œ
d# y
dx#
œ ayx# b
#
’y(2x) x#
dy
dx “
2xy# 1
y$ x%
œ 0 Ê 2y
dy
dx
dy
dx
dr
ds
2xy#
Ê
2
x#
œ0 Ê
œ
(cos 2s) œ 2r sin 2s 2 sin s cos s
œ (2r 1)(tan 2s)
dr ‰
ds
dy
dx
dr
ds
y x Š xy ‹
œ
œ
y#
dy
dx
œ 2x Ê
y# x#
y$
œ
"
y$
x
y
(since y# x# œ 1)
55. (a) Let h(x) œ 6f(x) g(x) Ê hw (x) œ 6f w (x) gw (x) Ê hw (1) œ 6f w (1) gw (1) œ 6 ˆ "# ‰ a%b œ (
(b) Let h(x) œ f(x)g# (x) Ê hw (x) œ f(x) a#g(x)b gw (x) g# (x)f w (x) Ê hw (0) œ #f(0)g(0)gw (0) g# (0)f w (0)
œ #(1)(1) ˆ "# ‰ (1)# ($) œ #
(c) Let h(x) œ
œ
f(x)
g(x) 1
(& 1) ˆ "# ‰ 3 a%b
(& 1)#
(g(x) 1)f (x) f(x)g (x)
(g(x) 1)#
Ê hw (x) œ
œ
w
w
Ê hw (1) œ
(g(1) ")f (1) f(1)g (1)
(g(1) 1)#
w
w
&
"#
(d) Let h(x) œ f(g(x)) Ê hw (x) œ f w (g(x))gw (x) Ê hw (0) œ f w (g(0))gw (0) œ f w (1) ˆ "# ‰ œ ˆ "# ‰ ˆ "# ‰ œ
"
%
(e) Let h(x) œ g(f(x)) Ê hw (x) œ gw (f(x))f w (x) Ê hw (0) œ gw (f(0))f w (0) œ gw (1)f w (0) œ a%b ($) œ "#
(f) Let h(x) œ (x f(x))$Î# Ê hw (x) œ 3# (x f(x))"Î# a1 f w (x)b Ê hw (1) œ 3# (1 f(1))"Î# a1 f w (1)b
œ 3# (1 3)"Î# ˆ1 "# ‰ œ *#
(g) Let h(x) œ f(x g(x)) Ê hw (x) œ f w (x g(x)) a1 gw (x)b Ê hw (0) œ f w (g(0)) a1 gw (0)b
œ f w (1) ˆ1 "# ‰ œ ˆ "# ‰ ˆ $# ‰ œ $%
56. (a) Let h(x) œ Èx f(x) Ê hw (x) œ Èx f w (x) f(x) †
(b) Let h(x) œ (f(x))"Î# Ê hw (x) œ
"
#
"
#È x
(f(x))"Î# af w (x)b Ê hw (0) œ
(c) Let h(x) œ f ˆÈx‰ Ê hw (x) œ f w ˆÈx‰ †
"
#È x
"
œ 5" (3) ˆ #" ‰
#È 1
"
"Î#
(2) œ 3"
# (9)
Ê hw (1) œ È1 f w (1) f(1) †
"
#
(f(0))"Î# f w (0) œ
Ê hw (1) œ f w ŠÈ1‹ †
"
#È 1
w
œ
"
5
†
"
#
œ
œ 13
10
"
10
(d) Let h(x) œ f(1 5 tan x) Ê hw (x) œ f w (1 5 tan x) a5 sec# xb Ê h (0) œ f w (1 5 tan 0) a5 sec# 0b
œ f w (1)(5) œ "5 (5) œ 1
(2 cos x)f (x) f(x)(sin x)
f(0)(0)
Ê hw (0) œ (2 1)f(2(0)
œ 3(9 2) œ
(2 cos x)#
1)#
h(x) œ 10 sin ˆ 1#x ‰ f # (x) Ê hw (x) œ 10 sin ˆ 1#x ‰ a2f(x)f w (x)b f # (x) ˆ10 cos ˆ 1#x ‰‰ ˆ 1# ‰
hw (1) œ 10 sin ˆ 1# ‰ a2f(1)f w (1)b f # (1) ˆ10 cos ˆ 1# ‰‰ ˆ 1# ‰ œ 20(3) ˆ "5 ‰ ! œ 12
(e) Let h(x) œ
(f) Let
Ê
57. x œ t# 1 Ê
dy
dt
œ
dy
dx
†
dx
dt
f(x)
2 cos x
dx
dt
Ê hw (x) œ
œ 2t; y œ 3 sin 2x Ê
œ 6 cos a2t# b † 2t Ê
58. t œ au# 2ub
"Î$
œ 2 au# 2ub
"Î$
w
Ê
dt
du
œ
5; thus
"
3
dy
dt ¹ t=0
au# 2ub
ds
du
œ
ds
dt
†
dy
dx
32
œ 3(cos 2x)(2) œ 6 cos 2x œ 6 cos a2t# 21b œ 6 cos a2t# b ; thus,
œ 6 cos (0) † 0 œ 0
#Î$
dt
du
w
(2u 2) œ
2
#
3 au
"Î$
œ ’2 au# 2ub
2ub
#Î$
(u 1); s œ t# 5t Ê
5“ ˆ 32 ‰ au#
2ub
#Î$
(u 1)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
ds
dt
œ 2t 5
183
184
Chapter 3 Differentiation
Ê
ds ¸
du u=2
œ ’2 a2# 2(2)b
59. r œ 8 sin ˆs 16 ‰ Ê
œ
; thus,
2É8 sin ˆs 16 ‰
Ê
dw ¸
ds s=0
œ
#Î$
(2 1) œ 2 ˆ2 † 8"Î$ 5‰ ˆ8#Î$ ‰ œ 2(2 † 2 5) ˆ 4" ‰ œ
dw
dr
†
œ
2É8 sin ˆ 16 ‰
d ) ‰‰
dt
d)
dt
œ
dr
ds
cos ŠÉ8 sin ˆs 16 ‰ 2‹
# É8 sinˆ s 16 ‰
(cos 0)(8) Š
È3 ‹
2È4
œ0 Ê
d)
dt
#
and
61. y$ y œ 2 cos x Ê 3y#
2 sin (0)
3 1
œ 0;
d# y
dx# ¹ (0ß1)
Ê
œ
d# y
dx#
d# y
dx#
œ
"3
8#Î$
œ
"
3
2
3
dy
dx
œ 2 sin x Ê
a3y# 1b (2 cos x) (2 sin x) Š6y
"
3
œ
"
œ cos ˆÈr 2‰ Š #È
‹
r
† 8 cos ˆs 16 ‰‘
(2)t 1) œ )# Ê
dy
dx
d)
dt
œ
) #
2)t1
; r œ a)# 7b
0 and )# t ) œ 1 Ê ) œ 1 so that
œ
ˆ 6" ‰ (1)
œ
"Î$
d) ¸
dt t=0, )=1
œ
1
1
œ 1
"
6
a3y# 1b œ 2 sin x Ê
dy
dx
œ
2 sin x
3y# 1
Ê
dy
dx ¹ (0ß1)
dy
dx ‹
a3y# 1b#
x#Î$ 3" y#Î$
ˆx#Î$ ‰ Š 23 y"Î$
4
(3 1)(2 cos 0) (2 sin 0)(6†0)
(3 1)#
62. x"Î$ y"Î$ œ 4 Ê
Ê
œ
dy
dx
dw
dr
9
#
œ È3
#Î$
#Î$
dr
"
#
(2)) œ 32 ) a)# 7b
; now t œ
d) œ 3 a) 7b
dr ¸
2
dr ¸
dr ¸
"
#Î$
œ 6 Ê dt t=0 œ d) t=0 † ddt) ¸ t=0
d) )=1 œ 3 (1 7)
Ê
œ
œ
dw
ds
cos ŠÉ8 sin ˆ 16 ‰ 2‹†8 cos ˆ 16 ‰
60. )# t ) œ 1 Ê ˆ)# t ˆ2)
œ
5“ ˆ 23 ‰ a2# 2(2)b
œ 8 cos ˆs 16 ‰ ; w œ sin ˆÈr 2‰ Ê
dr
ds
cos É8 sin ˆs 16 ‰ 2
"Î$
dy
ˆ #Î$ ‰ ˆ 23
dx ‹ y
œ #"
dy
dx
x"Î$ ‰
#
ax#Î$ b
œ0 Ê
Ê
dy
dx
#Î$
œ yx#Î$ Ê
d# y
dx# ¹ (8ß8)
œ
dy
dx ¹ (8ß8)
œ 1;
dy
dx
œ
y#Î$
x#Î$
ˆ8#Î$ ‰ 23 †8"Î$ †(1)‘ ˆ8#Î$ ‰ ˆ 23 †8"Î$ ‰
8%Î$
"
6
"
"
f(t h) f(t)
#(th)1 #t1 œ 2t 1 (2t 2h 1)
"
"
and
f(t
h)
œ
Ê
œ
2t 1
#(t h) 1
h
h
(2t 2h 1)(2t 1)h
f(t h) f(t)
2h
2
2
w
œ
Ê
f
(t)
œ
lim
œ
lim
(2t 2h 1)(2t 1)h
(2t 2h 1)(2t 1)
h
hÄ!
h Ä ! (2t 2h 1)(#t 1)
#
(2t 1)#
63. f(t) œ
œ
œ
g(x h) g(x)
h
lim g(x h)h g(x) œ lim
hÄ!
hÄ!
64. g(x) œ 2x# 1 and g(x h) œ 2(x h)# 1 œ 2x# 4xh 2h# 1 Ê
œ
#
#
#
a2x 4xh 2h 1b a2x 1b
h
œ
4xh 2h#
h
œ 4x 2h Ê gw (x) œ
œ 4x
(4x 2h)
65. (a)
lim f(x) œ lim c x# œ 0 and lim b f(x) œ lim b x# œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0) it
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
follows that f is continuous at x œ 0.
(c) lim c f w (x) œ lim c (2x) œ 0 and lim b f w (x) œ lim b (2x) œ 0 Ê lim f w (x) œ 0. Since this limit exists, it
(b)
x Ä !c
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Practice Exercises
185
follows that f is differentiable at x œ 0.
66. (a)
lim f(x) œ lim c x œ 0 and lim b f(x) œ lim b tan x œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0), it
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
follows that f is continuous at x œ 0.
(c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b sec# x œ 1 Ê lim f w (x) œ 1. Since this limit exists it
(b)
x Ä !c
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0.
67. (a)
lim f(x) œ lim c x œ 1 and lim b f(x) œ lim b (2 x) œ 1 Ê lim f(x) œ 1. Since lim f(x) œ 1 œ f(1), it
xÄ"
xÄ"
xÄ"
xÄ"
xÄ"
follows that f is continuous at x œ 1.
(c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b 1 œ 1 Ê lim c f w (x) Á lim b f w (x), so lim f w (x) does
(b)
x Ä "c
xÄ"
xÄ"
xÄ"
not exist Ê f is not differentiable at x œ 1.
xÄ"
xÄ"
xÄ1
xÄ"
lim f(x) œ lim c sin 2x œ 0 and lim b f(x) œ lim b mx œ 0 Ê lim f(x) œ 0, independent of m; since
xÄ!
xÄ!
xÄ!
xÄ!
f(0) œ 0 œ lim f(x) it follows that f is continuous at x œ 0 for all values of m.
68. (a)
x Ä !c
xÄ!
lim f w (x) œ lim c (sin 2x)w œ lim c 2 cos 2x œ 2 and lim b f w (x) œ lim b (mx)w œ lim b m œ m Ê f is
x Ä !c
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
differentiable at x œ 0 provided that lim c f w (x) œ lim b f w (x) Ê m œ 2.
(b)
xÄ!
69. y œ
œ
"
#
x
#
"
#x 4
œ
2(2x 4)
"
# x
#
(2x 4)" Ê
dy
dx
œ
"
#
xÄ!
2(2x 4)# ; the slope of the tangent is 3# Ê 3#
Ê 2 œ 2(2x 4)# Ê 1 œ
"
(2x 4)#
Ê 4x# 16x 15 œ 0 Ê (2x 5)(2x 3) œ 0 Ê x œ
Ê (2x 4)# œ 1 Ê 4x# 16x 16 œ 1
5
#
or x œ
3
#
Ê ˆ 5# ß 94 ‰ and ˆ 3# ß "4 ‰ are points on the
curve where the slope is .
3
#
70. y œ x
"
2x
Ê xœ „
Ê
"
#
dy
dx
œ1
2
(2x)#
Ê ˆ "# ß "# ‰ and ˆ
71. y œ 2x$ 3x# 12x 20 Ê
#
#
"
"
#x# ; the slope of the tangent is 3 Ê 3 œ 1 #x#
" "‰
# ß # are points on the curve where the slope is 3.
œ1
dy
dx
Ê 2œ
œ 6x# 6x 12; the tangent is parallel to the x-axis when
dy
dx
"
#x #
Ê x# œ
"
4
œ0
Ê 6x 6x 12 œ 0 Ê x x 2 œ 0 Ê (x 2)(x 1) œ 0 Ê x œ 2 or x œ 1 Ê (#ß !) and ("ß #7) are
points on the curve where the tangent is parallel to the x-axis.
72. y œ x$ Ê
dy
dx
œ 3x# Ê
dy
dx ¹ (2ß8)
œ 12; an equation of the tangent line at (#ß )) is y 8 œ 12(x 2)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
186
Chapter 3 Differentiation
Ê y œ 12x 16; x-intercept: 0 œ 12x 16 Ê x œ 43 Ê ˆ 43 ß !‰ ; y-intercept: y œ 12(0) 16 œ 16 Ê (0ß 16)
73. y œ 2x$ 3x# 12x 20 Ê
dy
dx
œ 6x# 6x 12
(a) The tangent is perpendicular to the line y œ 1
x
24
when
dy
dx
œ Š ˆ" " ‰ ‹ œ 24; 6x# 6x 12 œ 24
#4
Ê x# x 2 œ 4 Ê x# x 6 œ 0 Ê (x 3)(x 2) œ 0 Ê x œ 2 or x œ 3 Ê (#ß 16) and ($ß 11) are
x
points where the tangent is perpendicular to y œ 1 24
.
dy
(b) The tangent is parallel to the line y œ È2 12x when dx œ 12 Ê 6x# 6x 12 œ 12 Ê x# x œ 0
Ê x(x 1) œ 0 Ê x œ 0 or x œ 1 Ê (!ß 20) and ("ß () are points where the tangent is parallel to
y œ È2 12x.
74. y œ
1 sin x
x
Ê
dy
dx
œ
x(1 cos x) (1 sin x)(1)
x#
Ê m" œ
dy
dx ¹ x=1
œ
1 #
1#
œ 1 and m# œ
Since m" œ m"# the tangents intersect at right angles.
75. y œ tan x, 1# x
1
#
Ê
dy
dx
dy
1#
dx ¹ x=c1 1#
œ 1.
œ sec# x; now the slope
of y œ x# is "# Ê the normal line is parallel to
y œ x# when
#
Ê cos x œ
dy
dx
"
#
œ 2. Thus, sec# x œ 2 Ê
Ê cos x œ
for 1# x
1
#
„"
È2
Ê xœ
1
4
"
cos# x
œ2
and x œ
1
4
Ê ˆ 14 ß 1‰ and ˆ 14 ß "‰ are points
where the normal is parallel to y œ x# .
76. y œ 1 cos x Ê
œ sin x Ê
dy
dx
dy
dx ¹ ˆ 1 ß1‰
œ 1
2
Ê the tangent at ˆ 1# ß 1‰ is the line y 1 œ ˆx 1# ‰
Ê y œ x 1# 1; the normal at ˆ 1# ß 1‰ is
y 1 œ (1) ˆx 1# ‰ Ê y œ x
77. y œ x# C Ê
thus,
"
#
œ
ˆ "# ‰#
78. y œ x$ Ê
dy
dx
dy
dx
1
#
1
œ 2x and y œ x Ê
C Ê Cœ
œ 3x# Ê
dy
dx
œ 1; the parabola is tangent to y œ x when 2x œ 1 Ê x œ
"
#
Ê yœ
"
#
;
"
4
dy
dx ¹ x=a
œ 3a# Ê the tangent line at aaß a$ b is y a$ œ 3a# (x a). The tangent line
intersects y œ x$ when x$ a$ œ 3a# (x a) Ê (x a) ax# xa a# b œ 3a# (x a) Ê (x a) ax# xa 2a# b œ 0
Ê (x a)# (x 2a) œ 0 Ê x œ a or x œ 2a. Now
dy
dx ¹ x=c2a
œ 3(2a)# œ 12a# œ 4 a3a# b, so the slope at
x œ 2a is 4 times as large as the slope at aaß a$ b where x œ a.
79. The line through (!ß $) and (5ß 2) has slope m œ
y œ x 3; y œ
c
x1
Ê
dy
dx
œ
c
(x 1)# ,
3 (2)
05
œ 1 Ê the line through (!ß $) and (&ß 2) is
so the curve is tangent to y œ x 3 Ê
Ê (x 1)# œ c, x Á 1. Moreover, y œ
c
x1
intersects y œ x 3 Ê
#
c
x 1
dy
dx
œ 1 œ
c
(x 1)#
œ x 3, x Á 1
Ê c œ (x 1)(x 3), x Á 1. Thus c œ c Ê (x 1) œ (x 1)(x 3) Ê (x 1)[x 1 (x 3)]
œ !, x Á 1 Ê (x 1)(2x 2) œ 0 Ê x œ 1 (since x Á 1) Ê c œ 4.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Practice Exercises
80. Let Šbß „ Èa# b# ‹ be a point on the circle x# y# œ a# . Then x# y# œ a# Ê 2x 2y
Ê
dy
dx ¹ x=b
œ
b
„È a # b #
y Š „ Èa# b# ‹ œ
Ê normal line through Šbß „ Èa# b# ‹ has slope
„È a # b #
b
(x b) Ê y … Èa# b# œ
„È a # b #
b
„È a # b #
b
œ0 Ê
dy
dx
dy
dx
œ xy
Ê normal line is
x … Èa# b# Ê y œ „
È a# b #
b
x
which passes through the origin.
81. x# 2y# œ 9 Ê 2x 4y
œ "4 x
9
4
5
#
x
œ 2y
Ê
dy
dx
œ "4 Ê the tangent line is y œ 2 "4 (x 1)
dy
dx ¹ (1ß2)
and the normal line is y œ 2 4(x 1) œ 4x 2.
82. x$ y# œ 2 Ê 3x# 2y
œ 3# x
œ0 Ê
dy
dx
œ0 Ê
dy
dx
œ
dy
dx
3x#
2y
Ê
dy
dx ¹ (1ß1)
and the normal line is y œ 1 23 (x 1) œ
83. xy 2x 5y œ 2 Ê Šx
y‹ 2 5
dy
dx
œ0 Ê
dy
dx
(x 5) œ y 2 Ê
Ê the tangent line is y œ 2 2(x 3) œ 2x 4 and the normal line is y œ 2
84. (y x)# œ 2x 4 Ê 2(y x) Š dy
dx 1‹ œ 2 Ê (y x)
Ê the tangent line is y œ 2 34 (x 6) œ
85. x Èxy œ 6 Ê 1
"
#Èxy
dy
dx
Šx
3
4
x
dy
dx
œ 1 (y x) Ê
y 2
x 5
dy
dx
œ
Ê
1
#
(x 3) œ "# x 7# .
dy
dx
œ
1yx
yx
3
2
x"Î# 3y"Î#
y œ 4 "4 (x 1) œ 4" x
dy
dx
dy
dx ‹
dy
dx
y œ 2Èxy Ê
dy
dx
œ
y$ a3x# b“ 2y
a3x$ y# 2y 1b œ 1 3x# y$ Ê
dy
dx
2Èxy y
x
œ
x"Î#
2y"Î#
Ê
dy
dx ¹ (1ß4)
dy
dx
dy
dx ¹ (6ß2)
œ
3
4
Ê
œ
dy
dx ¹ (4ß1)
4
5
x
5
4
11
5
.
œ "4 Ê the tangent line is
dy
dx
œ
dy
dx
œ1
1 3x# y$
3x$ y# 2y 1
Ê
dy
dx
Ê 3x$ y#
dy
dx ¹ (1ß1)
dy
dx
2y
œ 24 , but
Therefore, the curve has slope "# at ("ß ") but the slope is undefined at ("ß 1).
88. y œ sin (x sin x) Ê
œ2
and the normal line is y œ 4 4(x 1) œ 4x.
17
4
87. x$ y$ y# œ x y Ê ’x$ Š3y#
Ê
œ0 Ê
dy
dx
dy
dx ¹ (3ß2)
Ê
Ê the tangent line is y œ 1 54 (x 4) = 54 x 6 and the normal line is y œ " 45 (x 4) œ
86. x$Î# 2y$Î# œ 17 Ê
(x 1)
and the normal line is y œ 2 43 (x 6) œ 43 x 10.
5
#
y‹ œ 0 Ê x
3
#
x "3 .
2
3
dy
dx
œ #3 Ê the tangent line is y œ 1
dy
dx
dy
dx ¹ (1ßc1) is
dy
dx
œ " 3x# y$
undefined.
œ [cos (x sin x)](1 cos x); y œ 0 Ê sin (x sin x) œ 0 Ê x sin x œ k1,
k œ 2, 1, 0, 1, 2 (for our interval) Ê cos (x sin x) œ cos (k1) œ „ 1. Therefore,
dy
dx
œ 0 and y œ 0 when
1 cos x œ 0 and x œ k1. For #1 Ÿ x Ÿ 21, these equations hold when k œ 2, 0, and 2 (since
cos (1) œ cos 1 œ 1). Thus the curve has horizontal tangents at the x-axis for the x-values 21, 0, and 21
(which are even integer multiples of 1) Ê the curve has an infinite number of horizontal tangents.
89. x œ
"
#
tan t, y œ
Ê xœ
"
#
tan
1
3
œ 2 cos$ ˆ 13 ‰ œ
90. x œ "
"
t#
"
#
sec t Ê
œ
È3
#
dy
dx
œ
dy/dt
dx/dt
œ
"
#
sec
1
3
and y œ
"
#
sec t tan t
"
#
# sec t
œ
tan t
sec t
œ sin t Ê
œ1 Ê yœ
È3
#
x 4" ;
d# y
dx#
"
4
,yœ"
3
t
Ê
dy
dx
œ
dy/dt
dx/dt
œ
Š t3# ‹
Š t2$ ‹
œ 32 t Ê
dy
dx ¹ tœ2
dy
dx ¹ tœ1Î3
œ
dyw /dt
dx/dt
œ sin
œ
"
#
1
3
cos t
sec# t
œ
È3
#
;tœ
1
3
œ 2 cos$ t Ê
d# y ¸
dx# tœ1Î3
œ 3# (2) œ 3; t œ 2 Ê x œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
##
œ
5
4
and
187
188
Chapter 3 Differentiation
yœ1
3
#
œ "# Ê y œ 3x
"3
4
d# y
dx#
;
œ
dyw /dt
dx/dt
œ
ˆ #3 ‰
Š t2$ ‹
œ
3 $
4 t
d# y
dx# ¹ tœ2
Ê
œ
3
4
(2)$ œ 6
91. B œ graph of f, A œ graph of f w . Curve B cannot be the derivative of A because A has only negative slopes
while some of B's values are positive.
92. A œ graph of f, B œ graph of f w . Curve A cannot be the derivative of B because B has only negative slopes
while A has positive values for x 0.
93.
94.
95. (a) 0, 0
(b) largest 1700, smallest about 1400
96. rabbits/day and foxes/day
sin x
97. lim
#
x Ä ! 2x x
98. lim
3x tan 7x
#x
99. lim
sin r
xÄ!
r Ä ! tan 2r
100.
sin 7x ‰
2x cos 7x
xÄ!
2r
tan 2r
103.
104.
105.
)Ä!
xÄ!
4 tan# ) tan ) 1
tan# ) &
œ
lim c
) Ä ˆ1‰
lim
tan x
x
œ lim
)Ä!
2 sin# ˆ #) ‰
)#
)Ä!
œ lim ˆ cos" x †
tan )
)
xÄ!
sin (sin ))
sin )
Š" tan5# ) ‹
Š5 cot7 ) cot8# ) ‹
œ
œ
‹œ
3
#
ˆ1 † 1 † 27 ‰ œ 2
"
#
. Let x œ sin ). Then x Ä 0 as ) Ä 0
(4 0 0)
(1 0)
(0 2)
(5 0 0)
œ lim
x sin x
# x
x Ä ! 2 ˆ2 sin ˆ # ‰‰
œ4
œ 52
†
x x
œ lim ’ sin## ˆ# x ‰ †
xÄ!
#
sin x
x “
(1)(1)(1) œ 1
œ lim ’
sin x ‰
x
"
ˆ 27 ‰
†
œ ˆ "# ‰ (1) ˆ 1" ‰ œ
cos 2r
Š4 tan" ) tan"# ) ‹
Š cot"# ) 2‹
œ lim b
)Ä!
œ lim
sin 7x
7x
œ1
sin x
x
x sin x
œ lim 2(1xsincosx x)
x Ä ! 2 2 cos x
xÄ!
ˆ x# ‰
ˆx‰
œ lim ’ sin ˆ x ‰ † sin #ˆ x ‰ † sinx x “ œ
xÄ!
#
#
xÄ!
xÄ!
sin 2r
r Ä ! ˆ 2r ‰
lim
1cos )
)#
)Ä!
lim Š cos"7x †
† "# ‰ œ ˆ "# ‰ (1) lim
2
lim
3
#
)Ä!
œ lim
1 2 cot# )
5 cot# ) 7 cot ) 8
lim b
œ
)Ä!
2
102.
œ (1) ˆ "1 ‰ œ 1
(sin )) ˆ sin ) ‰
œ lim Š sinsin
œ lim
) ‹
)
sin (sin ))
sin )
)Ä!
lim c
) Ä ˆ1‰
œ lim ˆ 3x
2x
rÄ!
Ê lim
101.
"
(#x 1) “
œ lim ˆ sinr r †
sin (sin ))
)
lim
)Ä!
xÄ!
œ lim ’ˆ sinx x ‰ †
)Ä!
sin ˆ #) ‰
ˆ #) ‰
†
sin ˆ #) ‰
ˆ #) ‰
† "# “ œ (1)(1) ˆ "# ‰ œ
"
#
œ 1; let ) œ tan x Ê ) Ä 0 as x Ä 0 Ê lim g(x) œ lim
xÄ!
xÄ!
œ 1. Therefore, to make g continuous at the origin, define g(0) œ 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
tan (tan x)
tan x
Chapter 3 Practice Exercises
106.
lim f(x) œ lim
xÄ!
(tan x)
œ lim ’ tantan
†
x
tan (tan x)
x Ä ! sin (sin x)
sin x
sin (sin x)
xÄ!
#105); let ) œ sin x Ê ) Ä 0 as x Ä 0 Ê
(b) S œ 21r# 21rh and r constant Ê
(c) S œ 21r# 21rh Ê
(d) S constant Ê
dh
dt
(b) r constant Ê
dr
dt
109. A œ 1r# Ê
110. V œ s$ Ê
111.
dR"
dt
dV
dt
œ0 Ê
dh ‰
ˆr dr
dt h dt
È r# h #
dr
dt
œ 3s# †
ds
dt
œ 1 ohm/sec,
dR#
dt
ds
dt
œ
œ
(using the result of
œ 1. Therefore, to make f
œ (41r 21h)
dr
dt
Ê (2r
dh
dt
dr
dt
dr
dt 21r
h) dr
dt œ r
dh
dt
dh
dt
Ê
dr
dt
œ
r dh
2rh dt
;
1Èr# h#
dr
1 r#
Èr# h# “ dt
; so r œ 10 and
Ê
dr
dt
lim )
) Ä ! sin )
œ (41r 21h)
21r
dr
dt
sin x
x Ä ! sin (sin x)
œ ’1Èr# h#
dr
dt
1 r#
dr
Èr# h# “ dt
1rh
dh
Èr# h# dt
œ
dS
dt
dr ‰
dt
1Èr# h#
1r# dr
dt
È r# h #
œ
dS
dt
œ0 Ê
œ 21 r
dA
dt
dr
dt
œ ’1Èr# h#
dS
dt
(c) In general,
œ 1r †
dS
dt
(a) h constant Ê
œ 21r dh
dt
#1 ˆr dh
h
dt
œ 0 Ê 0 œ (41r 21h)
dS
dt
108. S œ 1rÈr# h# Ê
œ 41r
dS
dt
œ 41r dr
dt 21 h
dS
dt
dS
dt
œ 1 † lim
sin x
lim
x Ä ! sin (sin x)
continuous at the origin, define f(0) œ 1.
107. (a) S œ 21r# 21rh and h constant Ê
"
cos x “
†
dr
dt
" dV
3s# dt
dh
1rh
Èr# h# dt
œ 12 m/sec Ê
; so s œ 20 and
œ 0.5 ohm/sec; and
"
R
œ
"
R"
"
R#
dV
dt
dA
dt
œ (21)(10) ˆ 12 ‰ œ 40 m# /sec
œ 1200 cm$ /min Ê
" dR
R# dt
Ê
œ
" dR"
R"# dt
ds
dt
œ
" dR#
R## dt
"
3(20)#
(1200) œ 1 cm/min
. Also,
"
"
R" œ 75 ohms and R# œ 50 ohms Ê R" œ 75
50
Ê R œ 30 ohms. Therefore, from the derivative
9(625)
" dR
"
"
"
"
"
ˆ
‰ Ê dR
ˆ 50005625 ‰
(30)# dt œ (75)# (1) (50)# (0.5) œ 5625 5000
dt œ (900) 5625†5000 œ 50(5625) œ 50
equation,
œ 0.02 ohm/sec.
112.
dR
dt
œ 3 ohms/sec and
X œ 20 ohms Ê
dZ
dt
dX
dt
œ 2 ohms/sec; Z œ ÈR# X# Ê
œ
(10)(3)(20)(2)
È10# 20#
113. Given
dx
dt
œ 10 m/sec and
œ 2x
dx
dt
2y
&
dD
dt
dy
dt
Ê D
"
È5
œ
dX
R dR
dt X dt
È R # X#
so that R œ 10 ohms and
¸ 0.45 ohm/sec.
œ 5 m/sec, let D be the distance from the origin Ê D# œ x# y# Ê 2D
dy
dt
dD
dt
œ
dZ
dt
œx
œ (5)(10) (12)(5) Ê
dD
dt
y
dx
dt
œ
110
5
dy
dt
dD
dt
. When (xß y) œ ($ß %), D œ É$# a%b# œ & and
œ 22. Therefore, the particle is moving away from the origin at 22 m/sec
(because the distance D is increasing).
114. Let D be the distance from the origin. We are given that
œ x# ˆ x
$Î# ‰#
œ x# x$ Ê 2D
œ 2x
dD
dt
3x#
dx
dt
dx
dt
dD
dt
œ 11 units/sec. Then D# œ x# y#
œ x(2 3x)
dx
dt
and substitution in the derivative equation gives (2)(6)(11) œ (3)(2 9)
115. (a) From the diagram we have
(b) V œ
"
3
1 r# h œ
"
3
#
10
h
1 ˆ 25 h‰ h œ
œ
4
r
41 h$
75
116. From the sketch in the text, s œ r) Ê
Ê
ds
dt
œr
d)
dt
œ (1.2)
d)
dt
. Therefore,
Ê rœ
Ê
dV
dt
2
5
œ
; x œ 3 Ê D œ È 3# 3$ œ 6
dx
dt
Ê
dx
dt
œ 4 units/sec.
h.
41h# dh
25 dt
ds
d)
dr
dt œ r dt ) dt .
ds
dt œ 6 ft/sec and r
œ 5 and h œ 6 Ê
dh
dt
125
œ 144
1 ft/min.
Also r œ 1.2 is constant Ê
dr
dt
œ0
, so
dV
dt
œ 1.2 ft Ê
d)
dt
œ 5 rad/sec
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
189
190
Chapter 3 Differentiation
117. (a) From the sketch in the text,
d)
dt
point A, x œ 0 Ê ) œ 0 Ê
œ 0.6 rad/sec and x œ tan ). Also x œ tan ) Ê
dx
dt
dx
dt
œ sec# )
d)
dt ;
at
#
œ asec 0b (0.6) œ 0.6. Therefore the speed of the light is 0.6 œ
3
5
km/sec
when it reaches point A.
(3/5) rad
sec
(b)
†
1 rev
21 rad
118. From the figure,
a
r
†
60 sec
min
œ
b
BC
œ
18
1
Ê
a
r
revs/min
œ
b
Èb# r#
. We are given
that r is constant. Differentiation gives,
"
r
†
da
dt
‰
ŠÈb# r# ‹ ˆ db
dt (b) Š È
œ
b# r#
b œ 2r and
Ê
œ
da
dt
db
dt
b
‰
‹ ˆ db
dt
b# r#
. Then,
œ 0.3r
Ô È(2r)# r# (0.3r) (2r) É2r(#0.3r)# ×
(2r) r
Ù
œ rÖ
(2r)# r#
Õ
Ø
È3r# (0.3r) 4r# (0.3r)
È3r#
3r
œ
a3r# b (0.3r) a4r# b (0.3r)
3 È 3 r#
œ
0.3r
3È 3
œ
r
10È3
m/sec. Since
da
dt
is positive,
the distance OA is increasing when OB œ 2r, and B is moving toward O at the rate of 0.3r m/sec.
119. (a) If f(x) œ tan x and x œ 14 , then f w (x) œ sec# x,
f ˆ 14 ‰ œ 1 and f w ˆ 14 ‰ œ 2. The linearization of
f(x) is L(x) œ 2 ˆx 14 ‰ (1) œ 2x
1 2
#
.
(b) If f(x) œ sec x and x œ 14 , then f w (x) œ sec x tan x,
f ˆ 1 ‰ œ È2 and f w ˆ 1 ‰ œ È2. The linearization
4
4
of f(x) is L(x) œ È2 ˆx 14 ‰ È2
œ È2x
120. f(x) œ
"
1 tan x
È2(% 1)
.
4
Ê f w (x) œ
sec# x
(1 tan x)#
. The linearization at x œ 0 is L(x) œ f w (0)(x 0) f(0) œ 1 x.
121. f(x) œ Èx 1 sin x 0.5 œ (x 1)"Î# sin x 0.5 Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x
Ê L(x) œ f w (0)(x 0) f(0) œ 1.5(x 0) 0.5 Ê L(x) œ 1.5x 0.5, the linearization of f(x).
122. f(x) œ
œ
2
1 x
2
(1 x)#
È1 x 3.1 œ 2(1 x)" (1 x)"Î# 3.1 Ê f w (x) œ 2(1 x)# (1) "# (1 x)"Î#
"
2È 1 x
Ê L(x) œ f w (0)(x 0) f(0) œ 2.5x 0.1, the linearization of f(x).
123. S œ 1 rÈr# h# , r constant Ê dS œ 1 r † "# ar# h# b
Ê dS œ
"Î#
#h dh œ
1rh
Èr# h# dh.
Height changes from h! to h! dh
1 r h! adhb
Ér# h#!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Additional and Advanced Exercises
124. (a) S œ 6r# Ê dS œ 12r dr. We want kdSk Ÿ (2%) S Ê k12r drk Ÿ
12r#
100
Ê kdrk Ÿ
r
100
191
. The measurement of the
edge r must have an error less than 1%.
#
3r dr
‰
(b) When V œ r$ , then dV œ 3r# dr. The accuracy of the volume is ˆ dV
V (100%) œ Š r$ ‹ (100%)
r ‰
œ ˆ 3r ‰ (dr)(100%) œ ˆ 3r ‰ ˆ 100
(100%) œ 3%
125. C œ 21r Ê r œ
dV œ
#
C
21
, S œ 41 r # œ
C#
1
, and V œ
4
3
1 r$ œ
C$
61 #
. It also follows that dr œ
"
#1
dC, dS œ
2C
1
dC and
dC. Recall that C œ 10 cm and dC œ 0.4 cm.
0.2
ˆ drr ‰ (100%) œ ˆ 0.2
‰ ˆ 2101 ‰ (100%) œ (.04)(100%) œ 4%
(a) dr œ 0.4
21 œ 1 cm Ê
1
8
1 ‰
ˆ dS
‰
ˆ 8 ‰ ˆ 100
(b) dS œ 20
(100%) œ 8%
1 (0.4) œ 1 cm Ê
S (100%) œ 1
C
21 #
10#
21 #
#
(0.4) œ
20
1#
‰
ˆ 20 ‰ 61
cm Ê ˆ dV
V (100%) œ 1# Š 1000 ‹ (100%) œ 12%
126. Similar triangles yield
35
h
œ
(c) dV œ
Ê dh œ 120a# da œ
15
6
120
a#
Ê h œ 14 ft. The same triangles imply that 20h a œ 6a Ê h œ 120a" 6
" ‰
2
‰ ˆ „ 1"# ‰ œ ˆ "#!
‰ˆ „ "#
da œ ˆ 120
œ „ 45
¸ „ .0444 ft œ „ 0.53 inches.
a#
"&#
CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2) œ 2 sin ) cos ) Ê
#
#
d
d)
Ê cos 2) œ cos ) sin )
(b) cos 2) œ cos# ) sin# ) Ê
(sin 2)) œ
d
d)
d
d)
(cos 2)) œ
(2 sin ) cos )) Ê 2 cos 2) œ 2[(sin ))(sin )) (cos ))(cos ))]
d
d)
acos# ) sin# )b Ê 2 sin 2) œ (2 cos ))(sin )) (2 sin ))(cos ))
Ê sin 2) œ cos ) sin ) sin ) cos ) Ê sin 2) œ 2 sin ) cos )
2. The derivative of sin (x a) œ sin x cos a cos x sin a with respect to x is
cos (x a) œ cos x cos a sin x sin a, which is also an identity. This principle does not apply to the
equation x# 2x 8 œ 0, since x# 2x 8 œ 0 is not an identity: it holds for 2 values of x (2 and 4), but not
for all x.
3. (a) f(x) œ cos x Ê f w (x) œ sin x Ê f ww (x) œ cos x, and g(x) œ a bx cx# Ê gw (x) œ b 2cx Ê gww (x) œ 2c;
also, f(0) œ g(0) Ê cos (0) œ a Ê a œ 1; f w (0) œ gw (0) Ê sin (0) œ b Ê b œ 0; f ww (0) œ gww (0)
Ê cos (0) œ 2c Ê c œ "# . Therefore, g(x) œ 1 "# x# .
(b) f(x) œ sin (x a) Ê f w (x) œ cos (x a), and g(x) œ b sin x c cos x Ê gw (x) œ b cos x c sin x; also,
f(0) œ g(0) Ê sin (a) œ b sin (0) c cos (0) Ê c œ sin a; f w (0) œ gw (0) Ê cos (a) œ b cos (0) c sin (0)
Ê b œ cos a. Therefore, g(x) œ sin x cos a cos x sin a.
(c) When f(x) œ cos x, f www (x) œ sin x and f Ð%Ñ (x) œ cos x; when g(x) œ 1 "# x# , gwww (x) œ 0 and gÐ%Ñ (x) œ 0.
Thus f www (0) œ 0 œ gwww (0) so the third derivatives agree at x œ 0. However, the fourth derivatives do not
agree since f Ð%Ñ (0) œ 1 but gÐ%Ñ (0) œ 0. In case (b), when f(x) œ sin (x a) and g(x)
œ sin x cos a cos x sin a, notice that f(x) œ g(x) for all x, not just x œ 0. Since this is an identity, we
have f ÐnÑ (x) œ gÐnÑ (x) for any x and any positive integer n.
4. (a) y œ sin x Ê yw œ cos x Ê yww œ sin x Ê yww y œ sin x sin x œ 0; y œ cos x Ê yw œ sin x
Ê yww œ cos x Ê yww y œ cos x cos x œ 0; y œ a cos x b sin x Ê yw œ a sin x b cos x
Ê yww œ a cos x b sin x Ê yww y œ (a cos x b sin x) (a cos x b sin x) œ 0
(b) y œ sin (2x) Ê yw œ 2 cos (2x) Ê yww œ 4 sin (2x) Ê yww 4y œ 4 sin (2x) 4 sin (2x) œ 0. Similarly,
y œ cos (2x) and y œ a cos (2x) b sin (2x) satisfy the differential equation yw w 4y œ 0. In general,
y œ cos (mx), y œ sin (mx) and y œ a cos (mx) b sin (mx) satisfy the differential equation yww m# y œ 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
192
Chapter 3 Differentiation
5. If the circle (x h)# (y k)# œ a# and y œ x# 1 are tangent at ("ß #), then the slope of this tangent is
m œ 2xk (1 2) œ 2 and the tangent line is y œ 2x. The line containing (hß k) and ("ß #) is perpendicular to
ß
y œ 2x Ê
k2
h1
œ "# Ê h œ 5 2k Ê the location of the center is (5 2kß k). Also, (x h)# (y k)# œ a#
Ê x h (y k)yw œ 0 Ê 1 ayw b# (y k)yw w œ 0 Ê yww œ
w
1 ay b#
ky
w
. At the point ("ß #) we know
ww
y œ 2 from the tangent line and that y œ 2 from the parabola. Since the second derivatives are equal at ("ß #)
we obtain 2 œ
1 (2)#
k#
Ê kœ
9
#
#
. Then h œ 5 2k œ 4 Ê the circle is (x 4)# ˆy 9# ‰ œ a# . Since ("ß #)
lies on the circle we have that a œ
5È 5
2
.
6. The total revenue is the number of people times the price of the fare: r(x) œ xp œ x ˆ3
x ‰#
, where
40
x ‰ ˆ
x ‰
40
3 40
" ‰
dr
x ‰#
x ‰ˆ
dr
‘
0 Ÿ x Ÿ 60. The marginal revenue is dx
œ ˆ3 40
2x ˆ3 40
40
Ê dx
œ ˆ3
2x
40
x
x
dr
œ 3 ˆ3 40 ‰ ˆ1 40 ‰ . Then dx œ 0 Ê x œ 40 (since x œ 120 does not belong to the domain). When 40 people
are on the bus the marginal revenue is zero and the fare is p(40) œ ˆ3
7. (a) y œ uv Ê
dy
dt
œ
du
dt
x ‰#
40 ¹ x=40
œ $4.00.
v u dv
dt œ (0.04u)v u(0.05v) œ 0.09uv œ 0.09y Ê the rate of growth of the total production is
9% per year.
(b) If
œ 0.02u and
du
dt
dv
dt
œ 0.03v, then
dy
dt
œ (0.02u)v (0.03v)u œ 0.01uv œ 0.01y, increasing at 1% per
year.
8. When x# y# œ 225, then yw œ xy . The tangent
line to the balloon at (12ß 9) is y 9 œ
Ê yœ
4
3
4
3
(x 12)
x 25. The top of the gondola is 15 8
œ 23 ft below the center of the balloon. The intersection of y œ 23 and y œ 43 x 25 is at the far
right edge of the gondola Ê 23 œ
Ê xœ
3
#
4
3
x 25
. Thus the gondola is 2x œ 3 ft wide.
9. Answers will vary. Here is one possibility.
10. s(t) œ 10 cos ˆt 14 ‰ Ê v(t) œ
10
(a) s(0) œ 10 cos ˆ 14 ‰ œ È
ds
dt
œ 10 sin ˆt 14 ‰ Ê a(t) œ
dv
dt
œ
d# s
dt#
œ 10 cos ˆt 14 ‰
2
(b) Left: 10, Right: 10
(c) Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 341 when the particle is farthest to the left.
Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 14 , but t 0 Ê t œ 21 41 œ 741 when the particle
is farthest to the right. Thus, v ˆ 341 ‰ œ 0, v ˆ 741 ‰ œ 0, a ˆ 341 ‰ œ 10, and a ˆ 741 ‰ œ 10.
(d) Solving 10 cos ˆt 14 ‰ œ 0 Ê t œ
1
4
Ê v ˆ 14 ‰ œ 10, ¸v ˆ 14 ‰¸ œ 10 and a ˆ 14 ‰ œ !.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Additional and Advanced Exercises
11. (a) s(t) œ 64t 16t# Ê v(t) œ
ds
dt
193
œ 64 32t œ 32(2 t). The maximum height is reached when v(t) œ 0
Ê t œ 2 sec. The velocity when it leaves the hand is v(0) œ 64 ft/sec.
(b) s(t) œ 64t 2.6t# Ê v(t) œ ds
dt œ 64 5.2t. The maximum height is reached when v(t) œ 0 Ê t ¸ 12.31 sec.
The maximum height is about s(12.31) œ 393.85 ft.
12. s" œ 3t$ 12t# 18t 5 and s# œ t$ 9t# 12t Ê v" œ 9t# 24t 18 and v# œ 3t# 18t 12; v" œ v#
Ê 9t# 24t 18 œ 3t# 18t 12 Ê 2t# 7t 5 œ 0 Ê (t 1)(2t 5) œ 0 Ê t œ 1 sec and t œ 2.5 sec.
13. m av# v#! b œ k ax#! x# b Ê m ˆ2v
substituting
dx
dt
œv Ê m
dv
dt
dv ‰
dt
œ k ˆ2x
dx ‰
dt
Ê m
dv
dt
2x ‰
œ k ˆ 2v
dx
dt
Ê m
dv
dt
œ kx ˆ "v ‰
dx
dt
œ 2At B Ê v ˆ t" # t# ‰ œ 2A ˆ t" # t# ‰ B œ A at" t# b B is the
instantaneous velocity at the midpoint. The average velocity over the time interval is vav œ
œ
Bt# Cb aAt#"
t# t"
. Then
œ kx, as claimed.
14. (a) x œ At# Bt C on ct" ß t# d Ê v œ
aAt##
dx
dt
Bt" Cb
œ
at# t" b cA at# t" b Bd
t# t"
#
?x
?t
œ A at# t" b B.
(b) On the graph of the parabola x œ At Bt C, the slope of the curve at the midpoint of the interval
ct" ß t# d is the same as the average slope of the curve over the interval.
15. (a) To be continuous at x œ 1 requires that lim c sin x œ lim b (mx b) Ê 0 œ m1 b Ê m œ 1b ;
xÄ1
xÄ1
(b) If yw œ œ
cos x, x 1
is differentiable at x œ 1, then lim c cos x œ m Ê m œ 1 and b œ 1.
xÄ1
m, x 1
16. faxb is continuous at ! because lim
xÄ!
œ
x ‰ ˆ 1 cos x ‰
lim ˆ 1 xcos
#
1 cos x
xÄ!
œ
" cos x
x
œ ! œ fa!b. f w (0) œ lim
f(x) f(0)
x0
ˆ 1 "cos x ‰
w
#
lim ˆ sinx x ‰
xÄ!
xÄ!
œ
"
#
œ lim
xÄ!
1 c cos x
0
x
x
. Therefore f (0) exists with value
"
#
.
17. (a) For all a, b and for all x Á 2, f is differentiable at x. Next, f differentiable at x œ 2 Ê f continuous at x œ 2
Ê lim c f(x) œ f(2) Ê 2a œ 4a 2b 3 Ê 2a 2b 3 œ 0. Also, f differentiable at x Á 2
xÄ2
Ê f w (x) œ œ
a, x 2
. In order that f w (2) exist we must have a œ 2a(2) b Ê a œ 4a b Ê 3a œ b.
2ax b, x 2
Then 2a 2b 3 œ 0 and 3a œ b Ê a œ
3
4
and b œ
9
4
.
(b) For x #, the graph of f is a straight line having a slope of
$
%
and passing through the origin; for x
is a parabola. At x œ #, the value of the y-coordinate on the parabola is
$
#
#, the graph of f
which matches the y-coordinate of the point
on the straight line at x œ #. In addition, the slope of the parabola at the match up point is
$
%
which is equal to the
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.
18. (a) For any a, b and for any x Á 1, g is differentiable at x. Next, g differentiable at x œ 1 Ê g continuous at
x œ 1 Ê lim b g(x) œ g(1) Ê a 1 2b œ a b Ê b œ 1. Also, g differentiable at x Á 1
x Ä "
Ê gw (x) œ œ
a, x 1
. In order that gw (1) exist we must have a œ 3a(1)# 1 Ê a œ 3a 1
3ax# 1, x 1
Ê a œ "# .
(b) For x Ÿ ", the graph of f is a straight line having a slope of
"
#
and a y-intercept of ". For x ", the graph of f is
a parabola. At x œ ", the value of the y-coordinate on the parabola is
$
#
which matches the y-coordinate of the point
on the straight line at x œ ". In addition, the slope of the parabola at the match up point is "# which is equal to the
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.
19. f odd Ê f(x) œ f(x) Ê
d
dx
(f(x)) œ
d
dx
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is even.
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194
Chapter 3 Differentiation
20. f even Ê f(x) œ f(x) Ê
d
dx
(f(x)) œ
d
dx
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is odd.
21. Let h(x) œ (fg)(x) œ f(x) g(x) Ê hw (x) œ x lim
Äx
œ x lim
Äx
œ
f(x) g(x) f(x) g(x! ) f(x) g(x! ) f(x! ) g(x! )
x x!
!
g(x! )
f(x! ) x lim
’ g(x)x
x! “
Ä x!
!
h(x) h(x! )
x x!
œ x lim
Äx
!
f(x) g(x) f(x! ) g(x! )
x x!
f(x! )
!)
œ x lim
’f(x) ’ g(x)x xg(x
““ x lim
’g(x! ) ’ f(x)x
x! ““
Äx
Äx
!
!
w
g(x! ) f (x! ) œ 0 †
g(x! )
lim ’ g(x)x
x! “
x Ä x!
!
w
g(x! ) f (x! ) œ g(x! ) f w (x! ), if g is
continuous at x! . Therefore (fg)(x) is differentiable at x! if f(x! ) œ 0, and (fg)w (x! ) œ g(x! ) f w (x! ).
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f(0) œ 0 and g is continuous
at 0.
(a) If f(x) œ sin x and g(x) œ kxk , then kxk sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0
and g(x) œ kxk is continuous at x œ 0.
(b) If f(x) œ sin x and g(x) œ x#Î$ , then x#Î$ sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0
and g(x) œ x#Î$ is continuous at x œ 0.
(c) If f(x) œ 1 cos x and g(x) œ $Èx, then $Èx (1 cos x) is differentiable because f w (0) œ sin (0) œ 0,
f(0) œ 1 cos (0) œ 0 and g(x) œ x"Î$ is continuous at x œ 0.
(d) If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable because f w (0) œ 1, f(0) œ 0 and
sin ˆ "x ‰
lim x sin ˆ "x ‰ œ lim
xÄ!
"
x
xÄ!
œ lim
tÄ_
sin t
t
œ 0 (so g is continuous at x œ 0).
23. If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable at x œ 0 because f w (0) œ 1, f(0) œ 0 and
lim x sin ˆ "x ‰ œ lim
xÄ!
sin ˆ "x ‰
"
x
xÄ!
œ lim
tÄ_
sin t
t
œ 0 (so g is continuous at x œ 0). In fact, from Exercise 21,
h (0) œ g(0) f (0) œ 0. However, for x Á 0, hw (x) œ x# cos ˆ "x ‰‘ ˆ x"# ‰ 2x sin ˆ x" ‰ . But
lim hw (x) œ lim cos ˆ "x ‰ 2x sin ˆ x" ‰‘ does not exist because cos ˆ x" ‰ has no limit as x Ä 0. Therefore,
w
w
xÄ!
xÄ!
the derivative is not continuous at x œ 0 because it has no limit there.
24. From the given conditions we have f(x h) œ f(x) f(h), f(h) 1 œ hg(h) and lim g(h) œ 1. Therefore,
hÄ!
w
f (x) œ
lim f(xh)h f(x)
hÄ!
w
œ
lim f(x) f(h)h f(x)
hÄ!
œ
lim f(x) ’ f(h)h 1 “
hÄ!
œ f(x) ’ lim g(h)“ œ f(x) † 1 œ f(x)
Ê f w (x) œ f(x) and f axbexists at every value of x.
hÄ!
25. Step 1: The formula holds for n œ 2 (a single product) since y œ u" u# Ê
dy
dx
œ
du"
dx
u# u"
du#
dx
.
Step 2: Assume the formula holds for n œ k:
y œ u" u# âuk Ê
du#
duk
dx u$ âuk á u" u# âuk-1 dx
d(u" u# âuk )
If y œ u" u# âuk ukb1 œ au" u# âuk b ukb1 , then dy
ukb1 u" u# âuk dudxkb1
dx œ
dx
dukb1
du#
duk ‰
"
œ ˆ du
dx u# u$ âuk u" dx u$ âuk â u" u# âukc1 dx ukb1 u" u# âuk dx
dukb1
du#
duk
"
œ du
dx u# u$ âukb1 u" dx u$ â ukb1 â u" u# âukc1 dx ukb1 u" u# âuk dx .
dy
dx
œ
du"
dx
u# u$ âuk u"
.
Thus the original formula holds for n œ (k1) whenever it holds for n œ k.
26. Recall ˆ mk ‰ œ
œ
m!
m!
m!
m!
ˆm‰
ˆm‰ ˆ m ‰
k! (m k)! . Then 1 œ 1! (m 1)! œ m and k k 1 œ k! (m k)! (k 1)! (m k 1)!
m! (k 1) m! (m k)
(m 1)!
ˆm1‰
œ (k m!1)!(m(m 1)k)! œ (k 1)! ((m
(k 1)! (m k)!
1) (k 1))! œ k 1 . Now, we prove
Leibniz's rule by mathematical induction.
Step 1: If n œ 1, then
d(uv)
dv
du
dx œ u dx v dx . Assume that the statement is true for n œ k, that is:
"
#
k
k#
k"
d (uv)
du
d u dv
dk v
ˆk‰ d u d v
ˆ k ‰ du d v
dxk œ dxk v k dxk" dx 2 dxk# dx# á k 1 dv dxk" u dxk .
kb"
k
k"
k
(uv)
d
dk u dv
dk" u d# v
ddxk"u v ddxuk dv
‘
If n œ k 1, then d dx(uv)
œ dx
Š d dx
k"
k ‹ œ
dx ’k dxk dx k dxk" dx# “
k
Step 2:
k
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 3 Additional and Advanced Exercises
’ˆ k2 ‰
du
dx
dk" u d# v
dxk" dx#
ˆ k2 ‰
kb"
dk# u d$ v
dxk# dx$ “
á ’ˆ k k 1 ‰
k"
d# u dk" v
dx# dxk"
dv
d u‘
d u
d u dv
dxk u dxk" œ dxk" v (k 1) dxk dx
k
kb"
k"
du d v
d v
d u
ˆ k k 1 ‰ ˆ kk ‰‘ dx
dxk u dxk" œ dxk" v (k
k
kb"
dv
d v
ˆ k k 1 ‰ du
dx dxk u dxk" .
k
ˆ k1 ‰
k
1)
ˆ kk 1 ‰
du dk u
dx dxk
v“
k"
#
ˆ k2 ‰‘ ddxk"u ddxv# á
dk u dv
dk" u d# v
ˆ k 2 1 ‰ dx
k"
dxk dx
dx#
á
Therefore the formula (c) holds for n œ (k 1) whenever it holds for n œ k.
27. (a) T# œ
(b) T# œ
41 # L
g
#
41 L
g
ÊLœ
T# g
41 #
ÊTœ
#1 È
L;
Èg
ÊLœ
a1 sec# ba32.2 ft/sec# b
41 #
dT œ
#1
Èg
†
"
dL
#È L
Ê L ¸ 0.8156 ft
œ
1
ÈLg dL;
dT œ
1
Èa!Þ)"&' ftba32.2 ft/sec# b a!Þ!"
ftb ¸ 0.00613 sec.
(c) Since there are 86,400 sec in a day, we have a0.00613 secba86,400 sec/dayb ¸ 529.6 sec/day, or 8.83 min/day; the
clock will lose about 8.83 min/day.
28. v œ s$ Ê
dv
dt
#
œ $s# ds
dt œ ka's b Ê
ds
dt
œ #k. If s! œ the initial length of the cube's side, then s" œ s! #k
Ê #k œ s! s" . Let t œ the time it will take the ice cube to melt. Now, t œ
œ
"
"Î$
" ˆ $% ‰
s!
#k
œ
s!
s ! s "
œ
av! b"Î$
"Î$
av! b ˆ $% v! ‰
"Î$
¸ "" hr.
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195
196
Chapter 3 Differentiation
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 4 APPLICATIONS OF DERIVATIVES
4.1 EXTREME VALUES OF FUNCTIONS
1. An absolute minimum at x œ c# , an absolute maximum at x œ b. Theorem 1 guarantees the existence of such
extreme values because h is continuous on [aß b].
2. An absolute minimum at x œ b, an absolute maximum at x œ c. Theorem 1 guarantees the existence of such
extreme values because f is continuous on [aß b].
3. No absolute minimum. An absolute maximum at x œ c. Since the function's domain is an open interval, the
function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values.
4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill
the conclusions of Theorem 1.
5. An absolute minimum at x œ a and an absolute maximum at x œ c. Note that y œ g(x) is not continuous but
still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
6. Absolute minimum at x œ c and an absolute maximum at x œ a. Note that y œ g(x) is not continuous but still
has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when
the hypothesis is not satisfied, absolute extrema may or may not occur.
7. Local minimum at a"ß !b, local maximum at a"ß !b
8. Minima at a#ß !b and a#ß !b, maximum at a!ß #b
9. Maximum at a!ß &b. Note that there is no minimum since the endpoint a#ß !b is excluded from the graph.
10. Local maximum at a$ß !b, local minimum at a#ß !b, maximum at a"ß #b, minimum at a!ß "b
11. Graph (c), since this the only graph that has positive slope at c.
12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c.
13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a.
14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.
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Chapter 4 Applications of Derivatives
15. f(x) œ
2
3
x 5 Ê f w (x) œ
f(2) œ
19
3 ,
2
3
Ê no critical points;
f(3) œ 3 Ê the absolute maximum
is 3 at x œ 3 and the absolute minimum is 19
3 at
x œ 2
16. f(x) œ x 4 Ê f w (x) œ 1 Ê no critical points;
f(4) œ 0, f(1) œ 5 Ê the absolute maximum is 0
at x œ 4 and the absolute minimum is 5 at x œ "
17. f(x) œ x# 1 Ê f w (x) œ 2x Ê a critical point at
x œ 0; f(1) œ 0, f(0) œ 1, f(2) œ 3 Ê the absolute
maximum is 3 at x œ 2 and the absolute minimum is 1
at x œ 0
18. f(x) œ % x# Ê f w (x) œ 2x Ê a critical point at
x œ 0; f(3) œ 5, f(0) œ 4, f(1) œ 3 Ê the absolute
maximum is 4 at x œ 0 and the absolute minimum is 5
at x œ 3
19. F(x) œ x"# œ x# Ê Fw (x) œ 2x$ œ
2
x$
, however
x œ 0 is not a critical point since 0 is not in the domain;
F(0.5) œ 4, F(2) œ 0.25 Ê the absolute maximum is
0.25 at x œ 2 and the absolute minimum is 4 at
x œ 0.5
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Section 4.1 Extreme Values of Functions
20. F(x) œ "x œ x" Ê Fw (x) œ x# œ
"
x#
, however
x œ 0 is not a critical point since 0 is not in the domain;
F(2) œ "# , F(1) œ 1 Ê the absolute maximum is 1 at
x œ 1 and the absolute minimum is
21. h(x) œ $Èx œ x"Î$ Ê hw (x) œ
"
3
"
#
at x œ 2
x#Î$ Ê a critical point
at x œ 0; h(1) œ 1, h(0) œ 0, h(8) œ 2 Ê the absolute
maximum is 2 at x œ 8 and the absolute minimum is 1
at x œ 1
22. h(x) œ 3x#Î$ Ê hw (x) œ #x"Î$ Ê a critical point at
x œ 0; h(1) œ 3, h(0) œ 0, h(1) œ 3 Ê the absolute
maximum is 0 at x œ 0 and the absolute minimum is 3
at x œ 1 and at x œ 1
23. g(x) œ È4 x# œ a4 x# b
Ê gw (x) œ
"
#
a4 x# b
"Î#
"Î#
(2x) œ
x
È 4 x#
Ê critical points at x œ 2 and x œ 0, but not at x œ 2
because 2 is not in the domain; g(2) œ 0, g(0) œ 2,
g(1) œ È3 Ê the absolute maximum is 2 at x œ 0 and the
absolute minimum is 0 at x œ 2
24. g(x) œ È5 x# œ a& x# b
a5 x# b
(2x)
x
"‰
w
ˆ
Ê g (x) œ # œ È # Ê critical points at x œ È5
"Î#
"Î#
&x
and x œ 0, but not at x œ È5 because È5 is not in the
domain; f ŠÈ5‹ œ 0, f(0) œ È5
Ê the absolute maximum is 0 at x œ È5 and the absolute
minimum is È5 at x œ 0
25. f()) œ sin ) Ê f w ()) œ cos ) Ê ) œ
1
#
1
#
is a critical point,
but ) œ
is not a critical point because #1 is not interior
the domain; f ˆ #1 ‰ œ 1, f ˆ 1# ‰ œ 1, f ˆ 561 ‰ œ "#
Ê the absolute maximum is 1 at ) œ 1# and the absolute
minimum is 1 at ) œ #1
to
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Chapter 4 Applications of Derivatives
26. f()) œ tan ) Ê f w ()) œ sec# ) Ê f has no critical points in
1‰
ˆ 1
3 ß 4 . The extreme values therefore occur at the
‰ œ È3 and f ˆ 14 ‰ œ 1 Ê the absolute
endpoints: f ˆ 1
3
maximum is 1 at ) œ 14 and the absolute
minimum is È3 at ) œ 1
3
27. g(x) œ csc x Ê gw (x) œ (csc x)(cot x) Ê a critical point
at x œ 1# ; g ˆ 13 ‰ œ È23 , g ˆ 1# ‰ œ 1, g ˆ 231 ‰ œ È23 Ê the
absolute maximum is
at x œ
2
È3
absolute minimum is 1 at x œ
1
3
and x œ
21
3 ,
and the
1
#
28. g(x) œ sec x Ê gw (x) œ (sec x)(tan x) Ê a critical point at
x œ 0; g ˆ 13 ‰ œ 2, g(0) œ 1, g ˆ 16 ‰ œ È23 Ê the absolute
maximum is 2 at x œ 13 and the absolute minimum is 1
at x œ 0
29. f(t) œ 2 ktk œ # Èt# œ # at# b
Ê f w (t) œ "# at# b
"Î#
"Î#
(2t) œ Èt # œ kttk
t
Ê a critical point at t œ 0; f(1) œ 1,
f(0) œ 2, f(3) œ 1 Ê the absolute maximum is 2 at t œ 0
and the absolute minimum is 1 at t œ 3
30. f(t) œ kt 5k œ È(t 5)# œ a(t 5)# b
œ
"
#
a(t 5)# b
"Î#
(2(t 5)) œ
t5
È(t 5)#
"Î#
œ
Ê f w (t)
t5
kt 5 k
Ê a critical point at t œ 5; f(4) œ 1, f(5) œ 0, f(7) œ 2
Ê the absolute maximum is 2 at t œ 7 and the absolute
minimum is 0 at t œ 5
31. f(x) œ x%Î$ Ê f w (x) œ
4
3
x"Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 16 Ê the absolute
maximum is 16 at x œ 8 and the absolute minimum is 0 at x œ 0
32. f(x) œ x&Î$ Ê f w (x) œ
5
3
x#Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 32 Ê the absolute
maximum is 32 at x œ 8 and the absolute minimum is 1 at x œ 1
33. g()) œ )$Î& Ê gw ()) œ
3
5
)#Î& Ê a critical point at ) œ 0; g(32) œ 8, g(0) œ 0, g(1) œ 1 Ê the absolute
maximum is 1 at ) œ 1 and the absolute minimum is 8 at ) œ 32
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Section 4.1 Extreme Values of Functions
34. h()) œ 3)#Î$ Ê hw ()) œ 2)"Î$ Ê a critical point at ) œ 0; h(27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute
maximum is 27 at ) œ 27 and the absolute minimum is 0 at ) œ 0
35. Minimum value is 1 at x œ #.
36. To find the exact values, note that yw œ $x# #,
which is zero when x œ „ É #$ . Local maximum at
ŠÉ #$ ß %
%È '
* ‹
¸ a!Þ)"'ß &Þ!)*b; local
minimum at ŠÉ #$ ß %
%È '
* ‹
¸ a!Þ)"'ß #Þ*""b
37. To find the exact values, note that that yw œ $x# #x )
œ a$x %bax #b, which is zero when x œ # or x œ %$ .
‰
Local maximum at a#ß "(b; local minimum at ˆ %$ ß %"
#(
38. Note that yw œ $x# 'x $ œ $ax "b# , which is zero at
x œ ". The graph shows that the function assumes lower
values to the left and higher values to the right of this point,
so the function has no local or global extreme values.
39. Minimum value is 0 when x œ " or x œ ".
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Chapter 4 Applications of Derivatives
40. The minimum value is 1 at x œ !.
41. The actual graph of the function has asymptotes at x œ „ ",
so there are no extrema near these values. (This is an
example of grapher failure.) There is a local minimum at
a!ß "b.
42. Maximum value is 2 at x œ ";
minimum value is 0 at x œ " and x œ $.
"
# at x œ "à
"# as x œ ".
43. Maximum value is
minimum value is
"
# at x œ 0à
"# as x œ 2.
44. Maximum value is
minimum value is
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Section 4.1 Extreme Values of Functions
45. yw œ x#Î$ a"b #$ x"Î$ ax #b œ
crit. pt.
x œ %&
xœ!
derivative
!
undefined
&x %
$ x
$È
extremum
local max
local min
46. yw œ x#Î$ a#xb #$ x"Î$ ax# %b œ
crit. pt.
x œ "
xœ!
xœ"
derivative
!
undefined
!
"
a #xb a"bÈ%
#È % x #
x# a% x# b
% #x #
œÈ
È % x#
% x#
crit. pt.
x œ #
x œ È #
x œ È#
xœ#
)x# )
$ x
$È
extremum
minimum
local max
minimum
47. yw œ x
œ
value
"#
"Î$
œ "Þ!$%
#& "!
0
derivative
undefined
!
!
undefined
value
$
0
$
x#
extremum
local max
minimum
maximum
local min
value
!
#
#
!
48. yw œ x# #È$" x a 1b #xÈ$ x
œ
x# a%xba$ xb
#È $ x
crit. pt.
xœ0
x œ "#
&
xœ$
_5x# "#x
#È $ x
derivative
!
!
undefined
#,
49. yw œ œ
",
crit. pt.
xœ"
œ
extremum
minimum
local max
minimum
value
!
"%%
"Î#
¸ %Þ%'#
"#& "&
!
extremum
minimum
value
#
x"
x"
derivative
undefined
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Chapter 4 Applications of Derivatives
", x !
50. yw œ œ
# #x, x !
crit. pt.
xœ!
xœ"
51. yw œ œ
derivative
undefined
!
2x 2,
2x 6,
crit. pt.
x œ 1
xœ1
xœ3
extremum
local min
local max
value
$
%
x1
x1
derivative
!
undefined
!
extremum
maximum
local min
maximum
value
5
1
5
"% x# "# x "&
% , xŸ"
x$ 'x# )x,
x"
w
w
#
if x ", and limc f a" hb œ ". Also, f axb œ $x "#x ) if x ", and
52. We begin by determining whether f w axb is defined at x œ ", where faxb œ œ
Clearly, f w axb œ "# x
"
#
hÄ!
limb f w a" hb œ ". Since f is continuous at x œ ", we have that f w a"b œ ". Thus,
hÄ!
f w axb œ œ
"# x "# ,
$x "#x ) ,
#
Note that "# x
But #
#È $
$
crit. pt.
x œ "
x ¸ $Þ"&&
"
#
xŸ"
x"
œ ! when x œ ", and $x# "#x ) œ ! when x œ
¸ !Þ)%& ", so the critical points occur at x œ " and x œ
derivative
!
!
extremum
local max
local min
È
"# „ È"## %a$ba)b
œ "# „' %)
#a$b
È
# # $ $ ¸ $Þ"&&.
œ#„
#È$
$ .
value
4
¸ $Þ!(*
53. (a) No, since f w axb œ #$ ax #b"Î$ , which is undefined at x œ #.
(b) The derivative is defined and nonzero for all x Á #. Also, fa#b œ ! and faxb ! for all x Á #.
(c) No, faxb need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed
interval of the form Òa, bÓ would have both a maximum value and minimum value on the interval.
(d) The answers are the same as (a) and (b) with 2 replaced by a.
x$ *x, x Ÿ $ or ! Ÿ x $
$x$ *, x $ or ! x $
. Therefore, f w axb œ œ
.
$
x *x, $ x ! or x $
$x$ *, $ x ! or x $
(a) No, since the left- and right-hand derivatives at x œ !, are * and *, respectively.
(b) No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively.
54. Note that faxb œ œ
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Section 4.1 Extreme Values of Functions
205
(c) No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively.
(d) The critical points occur when f w axb œ ! (at x œ „ È$) and when f w axb is undefined (at x œ ! and x œ „ $). The
minimum value is ! at x œ $, at x œ !, and at x œ $; local maxima occur at ŠÈ$ß 'È$‹ and ŠÈ$ß 'È$‹.
55.
(a) The construction cost is Caxb œ !Þ$È"' x# !Þ#a* xb million dollars, where ! Ÿ x Ÿ * miles. The following is
a graph of Caxb.
Solving Cw axb œ
!Þ$x
È"' x#
!Þ# œ ! gives x œ „
)È &
&
¸ „ $Þ&) miles, but only x œ $Þ&) miles is a critical point is
È
the specified domain. Evaluating the costs at the critical and endpoints gives Ca!b œ $3 million, CŠ ) & & ‹ ¸ $2.694
million, and Ca*b ¸ $2.955 million. Therefore, to minimize the cost of construction, the pipeline should be placed
from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the
refinery.
(b) If the per mile cost of underwater construction is p, then Caxb œ pÈ"' x# !Þ#a* xb and
Cw axb œ È !Þ$x x# !Þ# œ ! gives xc œ Èp#!Þ)
, which minimizes the construction cost provided xc Ÿ *. The value
!Þ!%
"'
of p that gives xc œ * miles is !Þ#"))'%. Consequently, if the underwater construction costs $218,864 per mile or less,
then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost
of construction.
In theory, p would have to be infinite to justify running the pipe directly from the docking facility to point A (i.e., for
xc to be zero). For all values of p !Þ#"))'% there is always an xc − Ð!ß *Ñ that will give a minimum value for C.
This is proved by looking at Cww axc b œ
"'p
a"' x#c b$Î#
which is always positive for p !.
56. There are two options to consider. The first is to build a new road straight from Village A to Village B. The second is to
build a new highway segment from Village A to the Old Road, reconstruct a segment of Old Road, and build a new
highway segment from Old Road to Village B, as shown in the figure. The cost of the first option is C" œ !Þ&a"&!b million
dollars œ 75 million dollars.
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Chapter 4 Applications of Derivatives
The construction cost for the second option is C# axb œ !Þ&Š#È#&!! x# ‹ !Þ$a"&! #xb million dollars for
! Ÿ x Ÿ (& miles. The following is a graph of C# axb.
Solving Cw# axb œ
x
È#&!! x#
!Þ' œ ! give x œ „ $(Þ& miles, but only x œ $(Þ& miles is in the specified domain. In
summary, C" œ $75 million, C# a!b œ $95 million, C# a$(Þ&b œ $85 million, and C# a(&b œ $90.139 million. Consequently,
a new road straight from village A to village B is the least expensive option.
57.
The length of pipeline is Laxb œ È% x# É#& a"! xb# for ! Ÿ x Ÿ "!. The following is a graph of Laxb.
Setting the derivative of Laxb equal to zero gives Lw axb œ
"! x
É#& a"! xb#
x
È % x#
a"! xb
É#& a"! xb#
œ !. Note that
x
È % x#
œ cos )A and
œ cos )B , therefore, Lw axb œ ! when cos )A œ cos )B , or )A œ )B and ˜ACP is similar to ˜BDP. Use
simple proportions to determine x as follows:
x
2
œ
"!x
&
Êxœ
#!
(
¸ #Þ)&( miles along the coast from town A to town B.
If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight
line (the shortest distance) between two towns, again forcing )A œ )B . The shortest length of pipe is the same regardless of
whether the towns are on thee same or opposite sides of the river.
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Section 4.1 Extreme Values of Functions
207
58.
(a) The length of guy wire is Laxb œ È*!! x# É#&!! a"&! xb# for ! Ÿ x Ÿ "&!. The following is a graph of
Laxb.
Setting Lw axb equal to zero gives Lw axb œ
a"&! xb
É#&!! a"&! xb#
x
È*!! x#
a"&! xb
É#&!! a"&! xb#
œ !. Note that
x
È*!! x#
œ cos )A and
œ cos )B . Therefore, Lw axb œ ! when cos )A œ cos )B , or )A œ )B and ˜ACE is similar to ˜ABD.
Use simple proportions to determine x:
x
$!
œ
"&! x
&!
Êxœ
##&
%
œ &'Þ#& feet.
(b) If the heights of the towers are hB and hC , and the horizontal distance between them is s, then
Laxb œ Éh#C x# Éh#B as xb# and Lw axb œ
as x b
É h B as x b #
x
Éh#C x#
as x b
É h B as x b #
. However,
x
Éh#C x#
œ cos )G and
œ cos )B . Therefore, Lw axb œ ! when cos )C œ cos )B , or )C œ )B and ˜ACE is similar to ˜ABD.
Simple proportions can again be used to determine the optimum x: hxc œ
sx
hB
Ê x œ Š hB hc hc ‹s.
59. (a) Vaxb œ "'!x &#x# %x$
Vw axb œ "'! "!%x "#x# œ %ax #ba$x #!b
The only critical point in the interval a!ß &b is at x œ #. The maximum value of Vaxb is 144 at x œ #.
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x œ # units.
60. (a) Pw axb œ # #!!x#
The only critical point in the interval a!ß _b is at x œ "!. The minimum value of Paxb is %! at x œ "!.
(b) The smallest possible perimeter of the rectangel is 40 units and it occurs at x œ "! units which makes the rectangle a
10 by 10 square.
61. Let x represent the length of the base and È#& x# the height of the triangle. The area of the triangle is represented by
#
Aaxb œ x È#& x# where ! Ÿ x Ÿ &. Consequently, solving Aw axb œ ! Ê #& #x œ ! Ê x œ & . Since
#È#& x#
#
Aa!b œ Aa&b œ !, Aaxb is maximized at x œ
&
È# .
The largest possible area is AŠ È&# ‹ œ
È#
#&
%
cm# .
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Chapter 4 Applications of Derivatives
62. (a) From the diagram the perimeter P œ #x #1r œ %!!
Ê x œ #!! 1r. The area A is 2rx
Ê Aarb œ %!!r #1r# where ! Ÿ r Ÿ #!!
1 .
(b) Aw arb œ %!! %1r so the only critical point is r œ
"!!
1 .
Since Aarb œ ! if r œ ! and x œ #!! 1r œ !, the
values r œ "!!
1 ¸ 31.83 m and x œ "!! m maximize the
area over the interval ! Ÿ r Ÿ
63. s œ "# gt# v! t s! Ê
ds
dt
#!!
1 .
œ gt v! œ ! Ê t œ
2
Thus sŠ vg! ‹ œ "# gŠ vg! ‹ v0 Š vg! ‹ s0 œ
64.
Now satb œ s0 Í tˆ gt2 v0 ‰ œ 0 Í t œ 0 or t œ
s0 s0 is the maximum height over the interval 0 Ÿ t Ÿ
œ ! Ê tan t œ " Ê t œ
never negative) Ê the peak current is #È# amps.
dI
dt
œ #sin t #cos t, solving
v!2
2g
v!
g.
dI
dt
65. Yes, since f(x) œ kxk œ Èx# œ ax# b
"Î#
Ê f w (x) œ
"
#
ax# b
1
%
2v0
g .
2v0
g .
n1 where n is a nonnegative integer (in this exercise t is
"Î#
(2x) œ
x
ax# b"Î#
œ
x
kx k
is not defined at x œ 0. Thus it
is not required that f w be zero at a local extreme point since f w may be undefined there.
66. If f(c) is a local maximum value of f, then f(x) Ÿ f(c) for all x in some open interval (aß b) containing c. Since
f is even, f(x) œ f(x) Ÿ f(c) œ f(c) for all x in the open interval (bß a) containing c. That is, f assumes
a local maximum at the point c. This is also clear from the graph of f because the graph of an even function
is symmetric about the y-axis.
67. If g(c) is a local minimum value of g, then g(x) g(c) for all x in some open interval (aß b) containing c. Since
g is odd, g(x) œ g(x) Ÿ g(c) œ g(c) for all x in the open interval (bß a) containing c. That is, g
assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd
function is symmetric about the origin.
68. If there are no boundary points or critical points the function will have no extreme values in its domain. Such
functions do indeed exist, for example f(x) œ x for _ x _. (Any other linear function f(x) œ mx b
with m Á 0 will do as well.)
69. (a) f w axb œ $ax# #bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The
function faxb œ x$ $x has two critical points at x œ " and x œ ". The function faxb œ x$ " has one critical point
at x œ !Þ The function faxb œ x$ x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the
cubic function has no extreme values.)
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Section 4.1 Extreme Values of Functions
209
70. (a)
fa!b œ ! is not a local extreme value because in any open interval containing x œ !, there are infinitely many points
where faxb œ " and where faxb œ ".
(b) One possible answer, on the interval Ò!ß "Ó:
"
a" xbcos "x
,
!Ÿx"
faxb œ œ
!, x œ "
This function has no local extreme value at x œ ". Note that it is continuous on Ò!ß "Ó.
71. Maximum value is 11 at x œ &;
minimum value is 5 on the interval Ò$ß #Ó;
local maximum at a&ß *b
72. Maximum value is 4 on the interval Ò&ß (Ó;
minimum value is % on the interval Ò#ß "Ó.
73. Maximum value is & on the interval Ò$ß _Ñ;
minimum value is & on the interval Ð_ß #Ó.
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Chapter 4 Applications of Derivatives
74. Minimum value is 4 on the interval Ò"ß $Ó
75-80. Example CAS commands:
Maple:
with(student):
f := x -> x^4 - 8*x^2 + 4*x + 2;
domain := x=-20/25..64/25;
plot( f(x), domain, color=black, title="Section 4.1 #75(a)" );
Df := D(f);
plot( Df(x), domain, color=black, title="Section 4.1 # 75(b)" )
StatPt := fsolve( Df(x)=0, domain )
SingPt := NULL;
EndPt := op(rhs(domain));
Pts :=evalf([EndPt,StatPt,SingPt]);
Values := [seq( f(x), x=Pts )];
Maximum value is 2.7608 and occurs at x=2.56 (right endpoint).
%
Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point).
Mathematica: (functions may vary) (see section 2.5 re. RealsOnly ):
< cos(x)^2 + sin(x);
ic := [x=Pi,y=1];
F := unapply( int( f(x), x ) + C, x );
eq := eval( y=F(x), ic );
solnC := solve( eq, {C} );
Y := unapply( eval( F(x), solnC ), x );
DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]],
color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" );
Mathematica: (functions and values may vary)
The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution
of the initial value problems for exercises 103 - 105.
Clear[x, y, yprime]
yprime[x_] = Cos[x]2 Sin[x];
initxvalue = 1; inityvalue = 1;
y[x_] = Integrate[yprime[t], {t, initxvalue, x}] inityvalue
If the solution satisfies the differential equation and initial condition, the following yield True
yprime[x]==D[y[x], x] //Simplify
y[initxvalue]==inityvalue
Since exercise 106 is a second order differential equation, two integrations will be required.
Clear[x, y, yprime]
y2prime[x_] = 3 Exp[x/2] 1;
initxval = 0; inityval = 4; inityprimeval = 1;
yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] inityprimeval
y[x_] = Integrate[yprime[t], {t, initxval, x}] inityval
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue).
y2prime[x]==D[y[x], {x, 2}]//Simplify
y[initxval]==inityval
yprime[initxval]==inityprimeval
Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,0,1]}]
CHAPTER 4 PRACTICE EXERCISES
1. No, since f(x) œ x$ 2x tan x Ê f w (x) œ 3x# 2 sec# x 0 Ê f(x) is always increasing on its domain
cos x
2. No, since g(x) œ csc x 2 cot x Ê gw (x) œ csc x cot x 2 csc# x œ sin
#x
2
sin# x
œ sin"# x (cos x 2) 0
Ê g(x) is always decreasing on its domain
3. No absolute minimum because x lim
(7 x)(11 3x)"Î$ œ _. Next f w (x) œ
Ä_
(11 3x)"Î$ (7 x)(11 3x)#Î$ œ
w
(11 3x) (7 x)
(11 3x)#Î$
œ
4(1 x)
(11 3x)#Î$
Ê x œ 1 and x œ
11
3
are critical points.
w
Since f 0 if x 1 and f 0 if x 1, f(1) œ 16 is the absolute maximum.
4. f(x) œ
ax b
x# 1
Ê f w (x) œ
We require also that f(3)
w
#a$x "bax $b
ax # 1 b #
#
a ax# 1b 2x(ax b)
ab
œ aaxax#2bx
1 b#
ax # 1 b#
œ 1. Thus " œ 3a8b Ê 3a b œ
w
"
; f w (3) œ 0 Ê '%
(*a 'b a) œ ! Ê &a $b œ !.
). Solving both equations yields a œ 6 and b œ 10. Now,
so that f œ ± ± ± ± . Thus f w changes sign at x œ $ from
1
1
3
1/3
positive to negative so there is a local maximum at x œ $ which has a value f(3) œ 1.
f (x) œ
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Chapter 4 Practice Exercises
275
5. Yes, because at each point of [!ß "Ñ except x œ 0, the function's value is a local minimum value as well as a
local maximum value. At x œ 0 the function's value, 0, is not a local minimum value because each open
interval around x œ 0 on the x-axis contains points to the left of 0 where f equals 1.
6. (a) The first derivative of the function f(x) œ x$ is zero at x œ 0 even though f has no local extreme value at
x œ 0.
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x œ c then f w (c) œ 0. It does not
assert the (false) reverse implication f w (c) œ 0 Ê f has a local extreme at x œ c.
7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is
continuous throughout a finite closed interval a Ÿ x Ÿ b then the existence of absolute extrema is guaranteed on
that interval.
8. The absolute maximum is k1k œ 1 and the absolute minimum is k0k œ 0. This is not inconsistent with the Extreme Value
Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that
interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half
closed, such as Ò"ß "Ñ, so there is nothing to contradict.
9. (a) There appear to be local minima at x œ 1.75
and 1.8. Points of inflection are indicated at
approximately x œ 0 and x œ „ 1.
(b) f w (x) œ x( 3x& 5x% 15x# œ x# ax# 3b ax$ 5b. The pattern yw œ ± ± ± ±
$È
!
È$
È $
&
$È
indicates a local maximum at x œ
5 and local minima at x œ „ È3 .
(c)
10. (a) The graph does not indicate any local
extremum. Points of inflection are indicated at
approximately x œ $% and x œ ".
(b) f w (x) œ x( 2x% 5
10
x$
œ x$ ax$ 2b ax( 5b . The pattern f w œ )( ± ± indicates
(È
$È
!
&
#
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276
Chapter 4 Applications of Derivatives
a local maximum at x œ (È5 and a local minimum at x œ $È2 .
(c)
11. (a) g(t) œ sin# t 3t Ê gw (t) œ 2 sin t cos t 3 œ sin (2t) 3 Ê gw 0 Ê g(t) is always falling and hence must
decrease on every interval in its domain.
(b) One, since sin# t 3t 5 œ 0 and sin# t 3t œ 5 have the same solutions: f(t) œ sin# t 3t 5 has the same
derivative as g(t) in part (a) and is always decreasing with f(3) 0 and f(0) 0. The Intermediate Value
Theorem guarantees the continuous function f has a root in [$ß 0].
12. (a) y œ tan ) Ê
dy
d)
œ sec# ) 0 Ê y œ tan ) is always rising on its domain Ê y œ tan ) increases on every
interval in its domain
(b) The interval 14 ß 1‘ is not in the tangent's domain because tan ) is undefined at ) œ
1
#
. Thus the tangent
need not increase on this interval.
13. (a) f(x) œ x% 2x# 2 Ê f w (x) œ 4x$ 4x. Since f(0) œ 2 0, f(1) œ 1 0 and f w (x) 0 for 0 Ÿ x Ÿ 1, we
may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 Ÿ x Ÿ 1.
È
(b) x# œ 2 „ 4 8 0 Ê x# œ È3 1 and x 0 Ê x ¸ È.7320508076 ¸ .8555996772
#
14. (a) y œ
x
x1
Ê yw œ
"
(x 1)#
0, for all x in the domain of
x
x1
Ê yœ
x
x1
is increasing in every interval in
its domain
(b) y œ x$ 2x Ê yw œ 3x# 2 0 for all x Ê the graph of y œ x$ 2x is always increasing and can never
have a local maximum or minimum
15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) œ a! be the initial
amount and V(1440) œ a! (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir
after the rain, where 24 hr œ 1440 min. Assume that V(t) is continuous on [!ß 1440] and differentiable on
(!ß 1440). The Mean Value Theorem says that for some t! in (!ß 1440) we have Vw (t! ) œ
œ
a! (1400)(43,560)(7.48) a!
1440
œ
456,160,320 gal
1440 min
V(1440) V(0)
1440 0
œ 316,778 gal/min. Therefore at t! the reservoir's volume
was increasing at a rate in excess of 225,000 gal/min.
16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the
d
difference 3x g(x) is a constant K because gw (x) œ 3 œ dx
(3x). Thus g(x) œ 3x K, the same form as F(x).
x
1
x
1
x 1 œ 1 x 1 Ê x 1 differs from x 1
(x 1) x(1)
d ˆ x ‰
d ˆ " ‰
œ (x " 1)# œ dx
dx x 1 œ
(x 1)#
x1 .
17. No,
18. f w (x) œ gw (x) œ
2x
ax # 1 b #
by the constant 1. Both functions have the same derivative
Ê f(x) g(x) œ C for some constant C Ê the graphs differ by a vertical shift.
19. The global minimum value of
"
#
occurs at x œ #.
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Chapter 4 Practice Exercises
277
20. (a) The function is increasing on the intervals Ò$ß #Ó and Ò"ß #Ó.
(b) The function is decreasing on the intervals Ò#ß !Ñ and Ð!ß "Ó.
(c) The local maximum values occur only at x œ #, and at x œ #; local minimum values occur at x œ $ and at x œ "
provided f is continuous at x œ !.
21. (a) t œ 0, 6, 12
(b) t œ 3, 9
(c) 6 t 12
(d) 0 t 6, 12 t 14
22. (a) t œ 4
(b) at no time
(c) 0 t 4
(d) 4 t 8
23.
24.
25.
26.
27.
28.
29.
30.
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278
Chapter 4 Applications of Derivatives
31.
32.
33. (a) yw œ 16 x# Ê yw œ ± ± Ê the curve is rising on (%ß %), falling on (_ß 4) and (%ß _)
%
%
Ê a local maximum at x œ 4 and a local minimum at x œ 4; yww œ 2x Ê yww œ ± Ê the curve
!
is concave up on (_ß !), concave down on (!ß _) Ê a point of inflection at x œ 0
(b)
34. (a) yw œ x# x 6 œ (x $)(x 2) Ê yw œ ± ± Ê the curve is rising on (_ß 2) and ($ß _),
#
$
falling on (#ß $) Ê local maximum at x œ 2 and a local minimum at x œ 3; yww œ 2x 1
Ê yww œ ± Ê concave up on ˆ "# ß _‰ , concave down on ˆ_ß "# ‰ Ê a point of inflection at x œ "#
"Î#
(b)
35. (a) yw œ 6x(x 1)(x 2) œ 6x$ 6x# 12x Ê yw œ ± ± ± Ê the graph is rising on ("ß !)
"
!
#
and (#ß _), falling on (_ß 1) and (!ß #) Ê a local maximum at x œ 0, local minima at x œ 1 and
x œ 2; yww œ 18x# 12x 12 œ 6 a3x# 2x 2b œ 6 Šx
yww œ ±
on
±
"È(
$
È
È
Š 1 3 7 ß 1 3 7 ‹
"È(
$
1 È7
3 ‹ Šx
Ê the curve is concave up on Š_ß
Ê points of inflection at x œ
1 È7
3 ‹
1 È7
3 ‹
Ê
È7
and Š 1 3
1 „ È7
3
(b)
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ß _‹ , concave down
Chapter 4 Practice Exercises
279
36. (a) yw œ x# (6 4x) œ 6x# 4x$ Ê yw œ ± ± Ê the curve is rising on ˆ_ß #3 ‰, falling on ˆ #3 ß _‰
!
$Î#
3
ww
Ê a local maximum at x œ # ; y œ 12x 12x# œ 12x(" x) Ê yww œ ± ± Ê concave up on
!
"
(!ß "), concave down on (_ß !) and ("ß _) Ê points of inflection at x œ 0 and x œ 1
(b)
37. (a) yw œ x% 2x# œ x# ax# 2b Ê yw œ ± ± ± Ê the curve is rising on Š_ß È2‹ and
!
È#
È #
ŠÈ2ß _‹ , falling on ŠÈ2ß È2‹ Ê a local maximum at x œ È2 and a local minimum at x œ È2 ;
yww œ 4x$ 4x œ 4x(x 1)(x 1) Ê yww œ ± ± ± Ê concave up on ("ß 0) and ("ß _),
"
!
"
concave down on (_ß 1) and (0ß 1) Ê points of inflection at x œ 0 and x œ „ 1
(b)
38. (a) yw œ 4x# x% œ x# a4 x# b Ê yw œ ± ± ± Ê the curve is rising on (2ß 0) and (0ß 2),
#
!
#
falling on (_ß 2) and (#ß _) Ê a local maximum at x œ 2, a local minimum at x œ 2; yww œ 8x 4x$
œ 4x a2 x# b Ê yww œ ± ± ± Ê concave up on Š_ß È2‹ and Š0ß È2‹ , concave
!
È#
È #
down on ŠÈ2ß 0‹ and ŠÈ2ß _‹ Ê points of inflection at x œ 0 and x œ „ È2
(b)
39. The values of the first derivative indicate that the curve is rising on (!ß _) and falling on (_ß 0). The slope
of the curve approaches _ as x Ä ! , and approaches _ as x Ä 0 and x Ä 1. The curve should therefore
have a cusp and local minimum at x œ 0, and a vertical tangent at x œ 1.
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280
Chapter 4 Applications of Derivatives
40. The values of the first derivative indicate that the curve is rising on ˆ!ß "# ‰ and ("ß _), and falling on (_ß !)
and ˆ "# ß "‰ . The derivative changes from positive to negative at x œ "# , indicating a local maximum there. The
slope of the curve approaches _ as x Ä 0 and x Ä 1 , and approaches _ as x Ä 0 and as x Ä 1 ,
indicating cusps and local minima at both x œ 0 and x œ 1.
41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _
as x Ä 0 and as x Ä 1, indicating vertical tangents at both x œ 0 and x œ 1.
È33
42. The graph of the first derivative indicates that the curve is rising on Š!ß 17 16
on (_ß !) and
xœ
17 È33
16
È
È
Š 17 16 33 ß 17 16 33 ‹
Ê a local maximum at x œ
17 È33
16
È33
‹ and Š 17 16
ß _‹ , falling
, a local minimum at
. The derivative approaches _ as x Ä 0 and x Ä 1, and approaches _ as x Ä 0 ,
indicating a cusp and local minimum at x œ 0 and a vertical tangent at x œ 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Practice Exercises
43. y œ
x1
x3
45. y œ
x# 1
x
œ1
œx
4
x3
"
x
44. y œ
2x
x5
œ2
46. y œ
x# x 1
x
10
x5
œx1
"
x
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281
282
Chapter 4 Applications of Derivatives
47. y œ
x$ 2
#x
œ
49. y œ
x# 4
x# 3
œ1
51. lim
xÄ"
52. lim
54. lim
tan x
x
œ
tan x
sin# x
sinamxb
x Ä 1Î#c
58.
xÄ"
axa"
b"
x Ä " bx
x Ä ! sinanxb
57. lim
œ lim
œ lim
#
x Ä ! tanax b
56. lim
"
x
"
x# 3
xa "
x Ä ! x sin x
55. lim
x # $x %
x"
b
x Ä " x "
53. xlim
Ä1
x#
#
tan 1
1
#x $
"
œ
x% 1
x#
œ x#
50. y œ
x#
x# 4
œ1
"
x#
4
x# 4
œ&
a
b
œ!
œ lim
sec# x
x Ä ! " cos x
œ lim
#sin x†cos x
# #
x Ä ! #x sec ax b
œ lim
xÄ!
m cosamxb
n cosanxb
œ
"
""
œ
"
#
sina#xb
œ lim
# #
x Ä ! #x sec ax b
œ
cosa$xb
x Ä 1Î#c cosa(xb
Èx
cos x
59. lim acsc x cot xb œ lim
xÄ!
#cosa#xb
œ lim
# #
#
# #
x Ä ! #x a#sec ax btanax b†#xb #sec ax b
œ
#
! #†"
œ"
m
n
seca(xbcosa$xb œ lim
lim Èx sec x œ lim b
x Ä !b
xÄ!
xÄ!
48. y œ
œ
!
"
$sina$xb
œ lim
x Ä 1Î#c (sina(xb
œ
$
(
œ!
" cos x
sin x
œ lim
sin x
x Ä ! cos x
œ
!
"
œ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Practice Exercises
60. lim ˆ x"%
xÄ!
61.
#
œ lim Š " x%x ‹ œ lim a" x# b †
xÄ!
xÄ!
"
x%
œ lim a" x# b œ lim
ŠÈx# x " Èx# x‹ œ lim ŠÈx# x " Èx# x‹ †
xÄ_
œ lim È # #x " È #
xÄ_
x x" x x
Notice that x œ Èx# for x ! so this is equivalent to
lim
#x "
x
#
x
x
" É x # x
É
#
x
x#
$
œ lim
xÄ_
$
lim Š x#x " x#x " ‹ œ lim
xÄ_
xÄ_
"#
"
œ lim #%
œ
lim
œ!
x
xÄ_
x Ä _ #x
"
%
xÄ! x
xÄ!
xÄ_
œ lim
xÄ_
62.
"‰
x#
œ"†_œ_
Èx# x "Èx# x
È x# x " È x# x
# x"
œ È"# È" œ "
"
É" x x"# É" x"
x $ a x # "b x $ a x # " b
ax# "bax# "b
œ lim
xÄ_
#x $
x% "
œ lim
xÄ_
'x#
%x $
œ lim
xÄ_
"#x
"#x#
63. (a) Maximize f(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36
Ê f w (x) œ
"
#
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx
#Èx È36 x
Ê derivative fails to exist at 0 and 36; f(0) œ 6,
and f(36) œ 6 Ê the numbers are 0 and 36
(b) Maximize g(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36
Ê gw (x) œ
"
#
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx
#Èx È36 x
Ê critical points at 0, 18 and 36; g(0) œ 6,
g(18) œ 2È18 œ 6È2 and g(36) œ 6 Ê the numbers are 18 and 18
64. (a) Maximize f(x) œ Èx (20 x) œ 20x"Î# x$Î# where 0 Ÿ x Ÿ 20 Ê f w (x) œ 10x"Î# 3# x"Î#
œ
20 3x
#È x
œ 0 Ê x œ 0 and x œ
œ
40È20
3È 3
Ê the numbers are
20
3
20
3
‰ É 20
ˆ
are critical points; f(0) œ f(20) œ 0 and f ˆ 20
3 œ
3 20
and
40
3
.
(b) Maximize g(x) œ x È20 x œ x (20 x)"Î# where 0 Ÿ x Ÿ 20 Ê gw (x) œ
Ê È20 x œ
"
#
Ê xœ
the numbers must be
65. A(x) œ
"
#
79
4
and
79
4 .
"
4 .
The critical points are x œ
79
4
2È20 x 1
#È20 x
‰
and x œ 20. Since g ˆ 79
4 œ
(2x) a27 x# b for 0 Ÿ x Ÿ È27
Ê Aw (x) œ 3(3 x)(3 x) and Aw w (x) œ 6x.
The critical points are 3 and 3, but 3 is not in the
domain. Since Aw w (3) œ 18 0 and A ŠÈ27‹ œ 0,
the maximum occurs at x œ 3 Ê the largest area is
A(3) œ 54 sq units.
66. The volume is V œ x# h œ 32 Ê h œ 32
x# . The
32 ‰
#
ˆ
surface area is S(x) œ x 4x x# œ x# 128
x ,
where x 0 Ê Sw (x) œ
20 ‰
3
2(x 4) ax# 4x 16b
x#
Ê the critical points are 0 and 4, but 0 is not in the
domain. Now Sw w (4) œ 2 256
4$ 0 Ê at x œ 4 there
is a minimum. The dimensions 4 ft by 4 ft by 2 ft
minimize the surface area.
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81
4
œ0
and g(20) œ 20,
283
284
Chapter 4 Applications of Derivatives
#
67. From the diagram we have ˆ h# ‰ r# œ ŠÈ3‹
Ê r# œ
12h#
4
#
. The volume of the cylinder is
#
V œ 1r# h œ 1 Š 12 4 h ‹ h œ
1
4
0 Ÿ h Ÿ 2È3 . Then Vw (h) œ
a12h h$ b , where
31
4
(2 h)(2 h)
Ê the critical points are 2 and 2, but 2 is not in
the domain. At h œ 2 there is a maximum since
Vw w (2) œ 31 0. The dimensions of the largest
cylinder are radius œ È2 and height œ 2.
68. From the diagram we have x œ radius and
y œ height œ 12 2x and V(x) œ "3 1x# (12 2x), where
0 Ÿ x Ÿ 6 Ê Vw (x) œ 21x(4 x) and Vw w (4) œ 81. The
critical points are 0 and 4; V(0) œ V(6) œ 0 Ê x œ 4
gives the maximum. Thus the values of r œ 4 and
h œ 4 yield the largest volume for the smaller cone.
‰ , where p is the profit on grade B tires and 0 Ÿ x Ÿ 4. Thus
69. The profit P œ 2px py œ 2px p ˆ 40510x
x
Pw (x) œ
2p
(5 x)#
ax# 10x 20b Ê the critical points are Š5 È5‹, 5, and Š5 È5‹ , but only Š5 È5‹ is in
the domain. Now Pw (x) 0 for 0 x Š5 È5‹ and Pw (x) 0 for Š5 È5‹ x 4 Ê at x œ Š5 È5‹ there
is a local maximum. Also P(0) œ 8p, P Š5 È5‹ œ 4p Š5 È5‹ ¸ 11p, and P(4) œ 8p Ê at x œ Š5 È5‹ there
is an absolute maximum. The maximum occurs when x œ Š5 È5‹ and y œ 2 Š5 È5‹ , the units are
hundreds of tires, i.e., x ¸ 276 tires and y ¸ 553 tires.
70. (a) The distance between the particles is lfatbl where fatb œ cos t cosˆt 1% ‰. Then, f w atb œ sin t sinˆt 1% ‰.
Solving f w atb œ ! graphically, we obtain t ¸ "Þ"(), t ¸ %Þ$#!, and so on.
Alternatively, f w atb œ ! may be solved analytically as follows. f w atb œ sin’ˆt 1) ‰ 1) “ sin’ˆt 1) ‰ 1) “
œ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ œ #sin 1) cosˆt 1) ‰
so the critical points occur when cosˆt 1) ‰ œ !, or t œ
$1
)
k1. At each of these values, fatb œ „ cos $)1
¸ „ !Þ('& units, so the maximum distance between the particles is !Þ('& units.
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Chapter 4 Practice Exercises
285
(b) Solving cos t œ cos ˆt 1% ‰ graphically, we obtain t ¸ #Þ(%*, t ¸ &Þ)*!, and so on.
Alternatively, this problem can be solved analytically as follows.
cos t œ cos ˆt 1% ‰
cos’ˆt 1) ‰ 1) “ œ cos’ˆt 1) ‰ 1) “
cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) œ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1)
#sin ˆt 1) ‰sin 1) œ !
sin ˆt 1) ‰ œ !
tœ
The particles collide when t œ
(1
)
(1
)
k1
¸ #Þ(%*. (plus multiples of 1 if they keep going.)
71. The dimensions will be x in. by "! #x in. by "' #x in., so Vaxb œ xa"! #xba"' #xb œ %x$ &#x# "'!x for
! x &. Then Vw axb œ "#x# "!%x "'! œ %ax #ba$x #!b , so the critical point in the correct domain is x œ #.
This critical point corresponds to the maximum possible volume because Vw axb ! for ! x # and Vw axb ! for
2 x &. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.$
Graphical support:
72. The length of the ladder is d" d# œ 8 sec ) 6 csc ). We
wish to maximize I()) œ 8 sec ) 6 csc ) Ê Iw ())
œ 8 sec ) tan ) 6 csc ) cot ). Then Iw ()) œ 0
Ê 8 sin$ ) 6 cos$ ) œ 0 Ê tan ) œ
$
È
6
#
Ê
d" œ 4 É4 $È36 and d# œ $È36 É4 $È36
Ê the length of the ladder is about
Š4 $È36‹ É4 $È36 œ Š4 $È36‹
$Î#
¸ "*Þ( ft.
73. g(x) œ 3x x$ 4 Ê g(2) œ 2 0 and g(3) œ 14 0 Ê g(x) œ 0 in the interval [#ß 3] by the Intermediate
Value Theorem. Then gw (x) œ 3 3x# Ê xnb1 œ xn
3xn x$n 4
33xn#
; x! œ 2 Ê x" œ 2.22 Ê x# œ 2.196215, and
so forth to x& œ 2.195823345.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
286
Chapter 4 Applications of Derivatives
74. g(x) œ x% x$ 75 Ê g(3) œ 21 0 and g(4) œ 117 0 Ê g(x) œ 0 in the interval [$ß %] by the Intermediate
Value Theorem. Then gw (x) œ 4x$ 3x# Ê xnb1 œ xn
x%n x$n 75
4xn$ 3xn#
; x! œ 3 Ê x" œ 3.259259
Ê x# œ 3.229050, and so forth to x& œ 3.22857729.
75.
' ax$ 5x 7b dx œ
76.
' Š8t$ t# t‹ dt œ 8t4% t6$ t## C œ 2t% t6$ t## C
77.
' ˆ3Èt t4# ‰ dt œ ' ˆ3t"Î# 4t# ‰ dt œ 3t$Î# 4t"1 C œ 2t$Î# 4t C
78.
' Š #È" t t3% ‹ dt œ ' ˆ #" t"Î# 3t% ‰ dt œ #" Œ t"Î# (3t$3) C œ Èt t"$ C
x%
4
5x#
#
7x C
#
Š 3# ‹
"
#
79. Let u œ r 5 Ê du œ dr
' ar dr5b
œ'
#
du
u#
u"
1
œ ' u# du œ
C œ u" C œ ar " 5b C
80. Let u œ r È2 Ê du œ dr
'
6 dr
$
Šr È2‹
œ 6'
dr
$
Šr È2‹
œ 6'
du
u$
81. Let u œ )# 1 Ê du œ 2) d) Ê
'
"
#
3)È)# 1 d) œ ' Èu ˆ #3 du‰ œ
82. Let u œ 7 )2 Ê du œ 2) d) Ê
'È)
d) œ '
7 ) 2
"
Èu
ˆ #" du‰ œ
"
#
x$ a 1 x % b
"Î%
#
C
du œ ) d)
3
#
"
#
3
ŠrÈ2‹
$Î#
$Î#
' u"Î# du œ 3# Œ u$Î#
C œ a ) # 1b C
3 C œ u
#
du œ ) d)
"Î#
' u"Î# du œ #" Œ u"Î#
C œ È7 )2 C
" C œ u
#
83. Let u œ 1 x% Ê du œ 4x$ dx Ê
'
#
œ 6' u$ du œ 6 Š u# ‹ C œ 3u# C œ
"
4
du œ x$ dx
dx œ ' u"Î% ˆ "4 du‰ œ
"
4
$Î%
" $Î%
' u"Î% du œ 4" Œ u$Î%
C œ 3" a1 x% b C
3 C œ 3 u
4
84. Let u œ 2 x Ê du œ dx Ê du œ dx
' (2 x)$Î& dx œ ' u$Î& ( du) œ ' u$Î& du œ u
)Î&
Š 85 ‹
85. Let u œ
'
s
10
sec# 10s
Ê du œ
"
10
C œ 58 u)Î& C œ 58 (2 x))Î& C
ds Ê 10 du œ ds
ds œ ' asec# ub (10 du) œ 10 ' sec# u du œ 10 tan u C œ 10 tan
86. Let u œ 1s Ê du œ 1 ds Ê
"
1
s
10
C
du œ ds
' csc# 1s ds œ ' acsc# ub ˆ 1" du‰ œ 1" ' csc# u du œ 1" cot u C œ 1" cot 1s C
87. Let u œ È2 ) Ê du œ È2 d) Ê
' csc È2) cot È2) d) œ '
"
È2
du œ d)
(csc u cot u) Š È"2 du‹ œ
"
È2
(csc u) C œ È"2 csc È2) C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Additional and Advanced Exercises
)
3
88. Let u œ
'
sec
)
3
tan
89. Let u œ
'
Ê du œ
x
4
)
3
"
3
d) Ê 3 du œ d)
d) œ ' (sec u tan u)(3 du) œ 3 sec u C œ 3 sec
Ê du œ
"
4
)
3
C
dx Ê 4 du œ dx
2u ‰
dx œ ' asin# ub (4 du) œ ' 4 ˆ 1 cos
du œ 2' (1 cos 2u) du œ 2 ˆu
#
œ 2u sin 2u C œ 2 ˆ x4 ‰ sin 2 ˆ x4 ‰ C œ x# sin x# C
sin#
x
4
90. Let u œ
'
cos#
œ
x
#
91. y œ '
x
#
x
#
"
#
Ê du œ
"
#
2u ‰
dx œ ' acos# ub (2 du) œ ' 2 ˆ 1 cos
du œ ' (1 cos 2u) du œ u
#
x# "
x#
dx œ ' a1 x# b dx œ x x" C œ x
y œ 1 when x œ 1 Ê
"
x
œ ' Š15Èt
3
Èt ‹
"
3
2
1
1
"‰
x#
Ê
"
x
C; y œ 1 when x œ 1 Ê 1
dx œ ' ax# 2 x# b dx œ
C œ 1 Ê C œ 3" Ê y œ
$
x
3
x$
3
2x x" C œ
2x
dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
dr
dt œ
&Î#
œ 4t&Î# 4t$Î# 8t C; r œ 0 when t œ 1 Ê 4(1)
r œ 4t&Î# 4t$Î# 8t
d# r
dt#
C
dr
dt
"
x
C œ 1
x$
3
2x
"
x
C;
"
3
œ 8 when t œ 1
10t$Î# 6t"Î# 8 Ê r œ ' ˆ10t$Î# 6t"Î# 8‰ dt
4(1)$Î# 8(1) C" œ 0 Ê C" œ 0. Therefore,
œ ' cos t dt œ sin t C; rw w œ 0 when t œ 0 Ê sin 0 C œ 0 Ê C œ 0. Thus,
dr
dt
1
1
1
Ê 10(1)$Î# 6(1)"Î# C œ 8 Ê C œ 8. Thus
94.
sin 2u
#
sin x C
#
92. y œ ' ˆx x" ‰ dx œ ' ˆx# 2
dr
dt
C
dx Ê 2 du œ dx
Ê C œ 1 Ê y œ x
93.
sin 2u ‰
#
œ ' sin t dt œ cos t C" ; rw œ 0 when t œ 0 Ê 1 C" œ 0 Ê C" œ 1. Then
d# r
dt# œ sin t
dr
dt œ cos t
1
Ê r œ ' (cos t 1) dt œ sin t t C# ; r œ 1 when t œ 0 Ê 0 0 C# œ 1 Ê C# œ 1. Therefore,
r œ sin t t 1
CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m Ÿ f(x) Ÿ M for all x − I. If m œ M
then f is constant on I.
3x 6, 2 Ÿ x 0
has an absolute minimum value of 0 at x œ 2 and an absolute
9 x# , 0 Ÿ x Ÿ 2
maximum value of 9 at x œ 0, but it is discontinuous at x œ 0.
2. No, the function f(x) œ œ
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical
point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the
closed endpoint. Extreme values occur only where f w œ 0, f w does not exist, or at the endpoints of the interval.
Thus the extreme points will not be at the ends of an open interval.
4. The pattern f w œ ± ± ± ± indicates a local maximum at x œ 1 and a local
"
#
$
%
minimum at x œ 3.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
287
288
Chapter 4 Applications of Derivatives
5. (a) If yw œ 6(x 1)(x 2)# , then yw 0 for x 1 and yw 0 for x 1. The sign pattern is
f w œ ± ± Ê f has a local minimum at x œ 1. Also yww œ 6(x 2)# 12(x 1)(x 2)
"
#
œ 6(x 2)(3x) Ê yw w 0 for x 0 or x 2, while yww 0 for 0 x 2. Therefore f has points of inflection
at x œ 0 and x œ 2. There is no local maximum.
(b) If yw œ 6x(x 1)(x 2), then yw 0 for x 1 and 0 x 2; yw 0 for " x 0 and x 2. The sign
sign pattern is yw œ ± ± ± . Therefore f has a local maximum at x œ 0 and
"
!
#
È7
local minima at x œ 1 and x œ 2. Also, yww œ ") ’x Š 1 $
1 È 7
$
x
1 È 7
$
È7
‹“ ’x Š 1 $
‹“ , so yww 0 for
and yww 0 for all other x Ê f has points of inflection at x œ
6. The Mean Value Theorem indicates that
f(6) f(0)
60
1 „È 7
$
.
œ f w (c) Ÿ 2 for some c in (0ß 6). Then f(6) f(0) Ÿ 12
indicates the most that f can increase is 12.
7. If f is continuous on [aß c) and f w (x) Ÿ 0 on [aß c), then by the Mean Value Theorem for all x − [aß c) we have
f(c) f(x)
cx
f(c). Also if f is continuous on (cß b] and f w (x)
Ÿ 0 Ê f(c) f(x) Ÿ 0 Ê f(x)
f(x) f(c)
xc
all x − (cß b] we have
0 Ê f(x) f(c)
0 Ê f(x)
f(c). Therefore f(x)
8. (a) For all x, (x 1)# Ÿ 0 Ÿ (x 1)# Ê a1 x# b Ÿ 2x Ÿ a1 x# b Ê "# Ÿ
(b) There exists c − (aß b) such that
Ê kf(b) f(a)k Ÿ
"
#
c
1 c#
œ
f(b) f(a)
ba
f(a)
¸ c ¸
Ê ¹ f(b)b
a ¹ œ 1 c# Ÿ
"
#
0 on (cß b], then for
f(c) for all x − [aß b].
x
1 x#
Ÿ
"
#
.
, from part (a)
kb ak .
9. No. Corollary 1 requires that f w (x) œ 0 for all x in some interval I, not f w (x) œ 0 at a single point in I.
10. (a) h(x) œ f(x)g(x) Ê hw (x) œ f w (x)g(x) f(x)gw (x) which changes signs at x œ a since f w (x), gw (x) 0 when
x a, f w (x), gw (x) 0 when x a and f(x), g(x) 0 for all x. Therefore h(x) does have a local maximum
at x œ a.
(b) No, let f(x) œ g(x) œ x$ which have points of inflection at x œ 0, but h(x) œ x' has no point of inflection
(it has a local minimum at x œ 0).
11. From (ii), f(1) œ
lim
xÄ „_
f(x) œ
12.
dy
dx
œ 0 Ê a œ 1; from (iii), either 1 œ x lim
f(x) or 1 œ x Ä
lim
f(x). In either case,
Ä_
_
x"
#
x Ä „ _ bx cx #
1 "x
x
c 2x
xÄ „_
lim
" a
bc#
œ ! and if c œ 0,
œ 3x# 2kx 3 œ 0 Ê x œ
1 "x
bx
c 2x
xÄ „_
1 x"
then lim
bx
2x
xÄ „_
œ
lim
lim
2k „ È4k# 36
6
œ " Ê b œ 0 and c œ ". For if b œ ", then
œ
lim
xÄ „_
1 x"
2
x
œ „ _. Thus a œ 1, b œ 0, and c œ 1.
Ê x has only one value when 4k# 36 œ 0 Ê k# œ 9 or
k œ „ 3.
13. The area of the ?ABC is A(x) œ
w
where 0 Ÿ x Ÿ 1. Thus A (x) œ
"
#
(2) È1 x# œ a1 x# b
x
È 1 x#
"Î#
,
Ê 0 and „ 1 are
critical points. Also A a „ 1b œ 0 so A(0) œ 1 is the
maximum. When x œ 0 the ?ABC is isosceles since
AC œ BC œ È2 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Additional and Advanced Exercises
f (c h) f (c)
œ f ww (c) Ê for % œ "# kf ww (c)k 0
h
hÄ0
Ê ¹ f (ch)h f (c) f ww (c)¹ "# kf ww (c)k . Then f w (c) œ
w
14. lim
w
w
w
3
#
f ww (c)
0 Ê "# kf ww (c)k
f (c h)
f ww (c) "# kf ww (c)k . If f ww (c) 0, then
h
f (c h)
"# f ww (c) 0; likewise if f ww (c) 0, then 0 "#
h
w
Ê f ww (c) "# kf ww (c)k
Ê
there exists a $ 0 such that 0 khk $
w
w
f (c h)
h
"
#
f ww (c)
w
kf ww (c)k
kf ww (c)k œ f ww (c)
f ww (c)
f (c h)
h
w
3
#
f ww (c).
(a) If f ww (c) 0, then $ h 0 Ê f (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local
maximum.
(b) If f ww (c) 0, then $ h 0 Ê f w (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local
minimum.
15. The time it would take the water to hit the ground from height y is É 2y
g , where g is the acceleration of
gravity. The product of time and exit velocity (rate) yields the distance the water travels:
È64(h y) œ 8 É 2 ahy y# b
D(y) œ É 2y
g
g
are critical points. Now D(0) œ 0,
D ˆ h# ‰
16. From the figure in the text, tan (" )) œ
give
ba
h
œ
tan "
1 ha tan "
a
h
h tan " a
h a tan "
œ
œ
"Î#
, 0 Ÿ y Ÿ h Ê Dw (y) œ 4 É g2 ahy y# b
8h
Èg
ba
h ;
"Î#
(h 2y) Ê 0,
and D(h) œ 0 Ê the best place to drill the hole is at y œ
tan (" )) œ
tan " tan )
1 tan " tan )
. Solving for tan " gives tan " œ
; and tan ) œ
bh
h# a(b a)
a
h
h
#
h
#
and h
.
. These equations
or
#
ah a(b a)b tan " œ bh. Differentiating both sides with respect to h gives
2h tan " ah# a(b a)b sec# "
d"
dh
œ b. Then
d"
dh
bh
œ 0 Ê 2h tan " œ b Ê 2h Š h# a(b
a) ‹ œ b
Ê 2bh# œ bh# ab(b a) Ê h# œ a(b a) Ê h œ Èa(a b) .
17. The surface area of the cylinder is S œ 21r# 21rh. From
the diagram we have Rr œ H H h Ê h œ RH R rH and
S(r) œ 21r(r h) œ 21r ˆr H r HR ‰
œ 21 ˆ1 HR ‰ r# 21Hr, where 0 Ÿ r Ÿ R.
Case 1: H R Ê S(r) is a quadratic equation containing
the origin and concave upward Ê S(r) is maximum at
r œ R.
Case 2: H œ R Ê S(r) is a linear equation containing the
origin with a positive slope Ê S(r) is maximum at
r œ R.
Case 3: H R Ê S(r) is a quadratic equation containing the origin and concave downward. Then
dS
H‰
dS
H‰
RH
ˆ
ˆ
dr œ 41 1 R r 21H and dr œ 0 Ê 41 1 R r 21H œ 0 Ê r œ 2(H R) . For simplification
we let r‡ œ
RH
2(H R)
.
(a) If R H 2R, then 0 H 2R Ê H 2(H R) Ê
(b) If H œ 2R, then r‡ œ
#
2R
2R
RH
2(H R)
R which is impossible.
œ R Ê S(r) is maximum at r œ R.
(c) If H 2R, then 2R H 2H Ê H 2(H R) Ê
S(r) is a maximum at r œ r‡ œ
RH
2(H R)
H
2(H R)
1 Ê
RH
2(H R)
R Ê r‡ R. Therefore,
.
Conclusion: If H − (0ß R] or H œ 2R, then the maximum surface area is at r œ R. If H − (Rß 2R), then r R
which is not possible. If H − (2Rß _), then the maximum is at r œ r‡ œ 2(HRH
R) .
18. f(x) œ mx 1
"
x
Ê f w (x) œ m
minimum. If f Š È"m ‹
"
x#
and f w w (x) œ
2
x$
0 when x 0. Then f w (x) œ 0 Ê x œ
0, then Èm 1 Èm œ 2Èm 1
0 Ê m
"
4
"
Èm
yields a
. Thus the smallest acceptable value
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
289
290
Chapter 4 Applications of Derivatives
for m is
19. (a)
(b)
(c)
"
4
.
lim
xÄ!
#sina&xb
$x
œ lim
xÄ!
x
x Ä ! sin# È#x
lim
x Ä 1/2
"
xÄ!
(g)
(h)
cos$ x
#
sinax# b
lim
x Ä ! xsin x
lim
xÄ!
x$ )
20. (a) x lim
Ä_
(b) x lim
Ä_
"!
$
$sina&xbsina$xb &cosa&xbcosa$xb
$cosa$xb
xÄ!
È #x
"
œ lim
#sinÈ#x cosÈ#x
xÄ!
È#x
xÄ!
sinŠ#È#x‹
œ
&
$
"
È#x
œ lim
x Ä ! cosŠ#È#x‹ È#
#x
"sin x
cos x
lim " cos#x
x Ä ! " sec x
œ
cos x
œ lim
œ !Þ
x Ä 1/2 sin x
lim " cos# x
x Ä ! tan x
œ
lim cos x# "
x Ä ! tan x
œ lim
sin x
#
x Ä ! # tan x sec x
œ lim
sin x
xÄ!
# sin x
cos$ x
œ
œ #"
#x cosax# b
œ lim
x Ä ! xcos xsin x
sec x "
x#
lim #
x Ä # x %
†"œ
"
#
œ
x Ä 1/2
œ
"!
$
œ
œ lim
œ lim
asec x tan xb œ lim
lim x sin x
x Ä ! x tan x
œ lim
(f)
sina&xbcosa$xb
sina$xb
lim x csc# È#x œ lim
x Ä ! cosŠ#È#x‹†#
(e)
xÄ! $
xÄ!
xÄ!
"! sina&xb
a&xb
œ lim
lim sina&xbcota$xb œ lim
xÄ!
œ lim
(d)
#sina&xb
$
& a &x b
œ lim
xÄ!
œ lim
xÄ#
sec x tan x
#x
xÄ!
ax #bax# #x %b
ax #bax #b
œ x lim
Ä_
#
sec x tan x sec x
#
xÄ!
x # #x %
x#
œ lim
xÄ#
œ x lim
œ x lim
Ä _ Èx &
Ä_
#x
x (È x
$
œ lim
Èx &
Èx
Èx
Èx &
Èx &
a#x# bsinax# b #cosax# b
xsin x#cos x
œ lim
É" x&
" È&x
#x
x
x( x
x
È œ x lim
Ä_
œ
#
" (É x"
œ
œ
"
"
œ"
œ
#
"!
œ
"!
#
#
#
œ"
œ
%%%
%
"
#
œ$
œ#
21. (a) The profit function is Paxb œ ac exbx aa bxb œ ex# ac bbx a. Pw axb œ #ex c b œ !
Ê x œ c#eb . Pww axb œ #e ! if e ! so that the profit function is maximized at x œ c #e b .
(b) The price therefore that corresponds to a production level yeilding a maximum profit is
p¹
xœ c#eb
œ c eˆ c #e b ‰ œ
c b
#
dollars.
#
(c) The weekly profit at this production level is Paxb œ eˆ c #e b ‰ ac bbˆ c #e b ‰ a œ
ac b b #
%e
#
a.
(d) The tax increases cost to the new profit function is Faxb œ ac exbx aa bx txb œ ex ac b tbx a.
bc
cbt
ww
Now Fw axb œ #ex c b t œ ! when x œ t #
e œ
#e . Since F axb œ #e ! if e !, F is maximized
when x œ c #be t units per week. Thus the price per unit is p œ c eˆ c #be t ‰ œ c #b t dollars. Thus, such a tax
increases the cost per unit by
cbt
#
The x-intercept occurs when
"
x
cb
#
œ
t
#
dollars if units are priced to maximize profit.
22. (a)
$œ!Ê
(b) By Newton's method, xn" œ xn
œ xn xn
$x#n
œ #xn
$xn#
faxn b
f ax n b .
w
"
x
œ $ Ê x œ $" .
Here f w axn b œ x#
n œ
"
x#n .
" $
So xn" œ xn xn" œ xn Š x"n $‹x#n
x#
n
œ xn a# $xn b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 4 Additional and Advanced Exercises
23. x" œ x!
and
a
q"
x!
fax! b
f w ax ! b
q
qx!q xq! a
qxq!"
x a
œ x! ! q" œ
qx
!
with weights m! œ
In the case where x! œ
a
xq!"
q"
q
q
x aq "b a
œ ! q"
œ x! Š q q " ‹
a
Š"‹
xq!" q
qx!
and m" œ "q .
we have xq! œ a and x" œ
q"
a
a
"
q" Š q ‹ q" Š q ‹
x!
x!
œ
291
so that x" is a weighted average of x!
q"
a
q" Š q
x!
q" ‹ œ
a
q" .
x!
#
dy
d y
24. We have that ax hb# ay hb# œ r# and so #ax hb #ay hb dy
dx œ ! and # # dx #ay hb dx# œ ! hold.
dy
x y dx
dy .
" dx
dy
Thus #x #y dy
dx œ #h #h dx , by the former. Solving for h, we obtain h œ
#
d y
equation yields # # dy
dx #y dx# #Œ
dy
x y dx
dy
" dx
œ !. Dividing by 2 results in "
Substituting this into the second
dy
dx
#
y ddxy# Œ
dy
x y dx
dy
" dx
œ !.
25. (a) aatb œ sww atb œ k ak !b Ê sw atb œ kt C" , where sw a!b œ )) Ê C" œ )) Ê sw atb œ kt )). So
satb œ
kt#
#
kt#
#
))t C# where sa!b œ ! Ê C# œ ! so satb œ
))t œ "!!. Solving for t we obtain t œ
kŠ ))
È))# #!!k
‹
k
))#
#!!
so that k œ
)) œ ! or kŠ ))
)) „ È))# #!!k
.
k
È))# #!!k
‹
k
kt#
#
))t. Now satb œ "!! when
At such t we want sw atb œ !, thus
)) œ !. In either case we obtain ))# #!!k œ !
¸ $)Þ(# ft/sec# .
(b) The initial condition that sw a!b œ %% ft/sec implies that sw atb œ kt %% and satb œ
w
The car is stopped at a time t such that s atb œ kt %% œ ! Ê t œ
‰
sˆ %%
k
œ
k ˆ %% ‰#
#
k
‰
%%ˆ %%
k
œ
%%#
#k
*')
k
œ
œ
‰
*')ˆ #!!
))#
%%
k .
kt#
#
%%t where k is as above.
At this time the car has traveled a distance
œ #& feet. Thus halving the initial velocity quarters
stopping distance.
26. haxb œ f # axb g# axb Ê hw axb œ #faxbf w axb #gaxbgw axb œ #faxbf w axb gaxbgw axb‘ œ #faxbgaxb gaxbafaxbb‘
œ # † ! œ !. Thus haxb œ c, a constant. Since ha!b œ &, haxb œ & for all x in the domain of h. Thus ha"!b œ &.
27. Yes. The curve y œ x satisfies all three conditions since
dy
dx
œ " everywhere, when x œ !, y œ !, and
d# y
dx#
œ ! everywhere.
28. yw œ $x# # for all x Ê y œ x$ #x C where " œ "$ # † " C Ê C œ % Ê y œ x$ #x %.
29. sww atb œ a œ t# Ê v œ sw atb œ
maximum for this t‡ . Since satb
t‡ œ a$Cb"Î$ . So
ÊCœ
a%bb$Î%
$ .
a$Cb"Î$ ‘%
12
t$
w
‡
‡
$ C. We seek v! œ s a!b œ C. We know that sat b œ b for some t and s is at a
%
%
œ 12t Ct k and sa!b œ ! we have that satb œ 12t Ct and also sw at‡ b œ ! so that
Ca$Cb"Î$ œ b Ê a$Cb"Î$ ˆC
Thus v! œ sw a!b œ
a%bb$Î%
$
œ
#È# $Î%
.
$ b
$C ‰
"#
œ b Ê a$Cb"Î$ ˆ $%C ‰ œ b Ê $"Î$ C%Î$ œ
30. (a) sww atb œ t"Î# t"Î# Ê vatb œ sw atb œ #$ t$Î# #t"Î# k where va!b œ k œ
(b) satb œ
% &Î#
"& t
%
%$ t$Î# %$ t k# where sa!b œ k# œ "&
. Thus satb œ
%
# $Î#
#t"Î#
$ Ê vatb œ $ t
% &Î#
%
%$ t$Î# %$ t "&
.
"& t
31. The graph of faxb œ ax# bx c with a ! is a parabola opening upwards. Thus faxb
one real value of x. The solutions to faxb œ ! are, by the quadratic equation
#
Thus we require
#
a#bb %ac Ÿ ! Ê b ac Ÿ !.
#
Thus aa" b" a# b# Þ Þ Þ an bn b
aa#"
a##
ÞÞÞ
an# bab"#
b##
ÞÞÞ
bn# b
%$ Þ
! for all x if faxb œ ! for at most
#b „ Éa#bb# %ac
.
#a
32. (a) Clearly faxb œ aa" x b" b# Þ Þ Þ aan x bn b# ! for all x. Expanding we see
faxb œ aa#" x# #a" b" x b"# b Þ Þ Þ aan# x# #an bn x bn# b
œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b
%b
$
!.
Ÿ ! by Exercise 31.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
292
Chapter 4 Applications of Derivatives
Thus aa" b" a# b# Þ Þ Þ an bn b# Ÿ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b.
(b) Referring to Exercise 31: It is clear that faxb œ ! for some real x Í b# %ac œ !, by quadratic formula.
Now notice that this implies that
faxb œ aa" x b" b# Þ Þ Þ aan x bn b#
œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# œ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b
But now faxb œ ! Í ai x bi œ ! for all i œ "ß #ß Þ Þ Þ ß n Í ai x œ bi œ ! for all i œ "ß #ß Þ Þ Þ ß n.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 5 INTEGRATION
5.1 ESTIMATING WITH FINITE SUMS
1. faxb œ x#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
"
#
iœ!
$
"
#
œ "# Š!# ˆ "# ‰ ‹ œ
#
"
4
œ 4" Š!# ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê an upper sum is !ˆ 4i ‰ †
2. faxb œ x$
iœ!
2
iœ1
%
#
"
)
#
#
#
#
"
#
#
œ "# Šˆ "# ‰ +1# ‹ œ
#
"
4
#
#
#
œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1# ‹ œ
iœ"
"
%
†
(
)
"
%
$! ‰
† ˆ "'
œ
œ
(
$#
&
)
"&
$#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
"
$
iœ!
$
"
#
œ "# Š!$ ˆ "# ‰ ‹ œ
$
"
4
œ 4" Š!$ ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê an upper sum is !ˆ 4i ‰ †
iœ!
2
iœ1
%
iœ"
$
"
"'
$
$
$
$'
#&'
$
"
#
$
œ "# Šˆ "# ‰ +1$ ‹ œ
$
"
4
$
$
$
œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1$ ‹ œ œ
"
#
†
*
)
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
*
'%
"!!
#&'
œ
*
"'
#&
'%
294
Chapter 5 Integration
3. faxb œ
"
x
Since f is decreasing on Ò!ß "Ó, we use left endpoints to obtain
upper sums and right endpoints to obtain lower sums.
#
(a) ˜x œ
&"
#
œ # and xi œ " i˜x œ " #i Ê a lower sum is ! x"i † # œ #ˆ $" &" ‰ œ
(b) ˜x œ
&"
%
œ 1 and xi œ " i˜x œ " i Ê a lower sum is
(c) ˜x œ
&"
#
œ # and xi œ " i˜x œ " #i Ê an upper sum is ! x"i † # œ #ˆ" $" ‰ œ
(d) ˜x œ
&"
%
œ 1 and xi œ " i˜x œ " i Ê an upper sum is
4. faxb œ % x#
iœ"
%
!"
xi
iœ"
"
† " œ "ˆ #"
iœ!
$
!"
xi
iœ!
"
$
† " œ "ˆ"
"
#
"
%
"'
"&
&" ‰ œ
"
$
((
'!
)
$
"% ‰ œ
#&
"#
Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use
left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Ó to obtain
lower sums and use right endpoints on Ò#ß !Ó and left endpoints
on Ò!ß #Ó to obtain upper sums.
(a) ˜x œ
# a#b
#
œ # and xi œ # i˜x œ # #i Ê a lower sum is # † ˆ% a#b# ‰ # † a% ## b œ !
(b) ˜x œ
# a#b
%
œ " and xi œ # i˜x œ # i Ê a lower sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
"
%
iœ!
iœ$
œ "ˆˆ% a#b# ‰ ˆ% a"b# ‰ a% "# b a% ## b‰ œ '
(c) ˜x œ
# a#b
#
œ # and xi œ # i˜x œ # #i Ê a upper sum is # † ˆ% a!b# ‰ # † a% !# b œ "'
(d) ˜x œ
# a#b
%
œ " and xi œ # i˜x œ # i Ê a upper sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
#
$
iœ"
iœ#
œ "ˆˆ% a"b# ‰ a% !# b a% !# b a% "# b‰ œ "%
5. faxb œ x#
œ
"
#
Using 4 rectangles Ê ˜x œ " % ! œ
Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
"
%
Using 2 rectangles Ê ˜x œ
#
#
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
#
#
"!
$#
#
œ
"!
#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
&
"'
#
œ "% Šˆ ") ‰ ˆ $) ‰ ˆ &) ‰ ˆ () ‰ ‹ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
#"
'%
Section 5.1 Estimating with Finite Sums
6. faxb œ x$
œ
"
#
Using 4 rectangles Ê ˜x œ " % ! œ
Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
"
%
$
$
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
$
$
$
$
(
œ "% Š " $ )&
‹œ
$
7. faxb œ
"
x
"!
#
Using 2 rectangles Ê ˜x œ
#)
# † '%
%*'
% † )$
œ
œ
Using 2 rectangles Ê ˜x œ
œ #ˆ "# "% ‰ œ $#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
(
$#
"#%
)$
&"
#
œ
$"
"#)
œ # Ê #afa#b fa%bb
Using 4 rectangles Ê ˜x œ & % " œ "
Ê "ˆfˆ $# ‰ fˆ &# ‰ fˆ (# ‰ fˆ *# ‰‰
œ "ˆ #$
8. faxb œ % x#
#
&
#
(
#* ‰ œ
"%))
$†&†(†*
Using 2 rectangles Ê ˜x œ
œ #a$ $b œ "#
295
œ
# a#b
#
%*'
&†(†*
œ
%*'
$"&
œ # Ê #afa"b fa"bb
Using 4 rectangles Ê ˜x œ # %a#b œ "
Ê "ˆfˆ $# ‰ fˆ "# ‰ fˆ "# ‰ fˆ $# ‰‰
#
#
#
#
œ "ŠŠ% ˆ $# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ $# ‰ ‹‹
œ "' ˆ *% † # "% † #‰ œ "' "!
# œ ""
9. (a) D ¸ (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) œ 87 inches
(b) D ¸ (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) œ 87 inches
10. (a) D ¸ (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300)
(1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) œ 5220 meters (NOTE: 5 minutes œ 300 seconds)
(b) D ¸ (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300)
(1.8)(300) (1.5)(300) (1.2)(300) (0)(300) œ 4920 meters (NOTE: 5 minutes œ 300 seconds)
11. (a) D ¸ (0)(10) (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10)
(35)(10) (44)(10) (30)(10) œ 3490 feet ¸ 0.66 miles
(b) D ¸ (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10)
(44)(10) (30)(10) (35)(10) œ 3840 feet ¸ 0.73 miles
12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the
midpoints of each time interval to approximate this area using rectangles. Thus,
D ¸ (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001)
(128)(0.001) (134)(0.001) (139)(0.001) ¸ 0.967 miles
(b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours œ 22.7 sec. At 22.7
sec, the velocity was approximately 120 mi/hr.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
296
Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing
acceleration † ?t. Thus, ?t œ 1 and speed ¸ [32.00 19.41 11.77 7.14 4.33](1) œ 74.65 ft/sec
(b) Using right end-points we obtain a lower estimate: speed ¸ [19.41 11.77 7.14 4.33 2.63](1)
œ 45.28 ft/sec
(c) Upper estimates for the speed at each second are:
t
0
1
2
3
4
5
v
0
32.00 51.41 63.18 70.32
74.65
Thus, the distance fallen when t œ 3 seconds is s ¸ [32.00 51.41 63.18](1) œ 146.59 ft.
14. (a) The speed is a decreasing function of time Ê right end-points give an lower estimate for the height (distance)
attained. Also
t
0
1
2
3
4
5
v
400
368
336
304
272
240
gives the time-velocity table by subtracting the constant g œ 32 from the speed at each time increment
?t œ 1 sec. Thus, the speed ¸ 240 ft/sec after 5 seconds.
(b) A lower estimate for height attained is h ¸ [368 336 304 272 240](1) œ 1520 ft.
15. Partition [!ß #] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of these
subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating
1
125
343
$
$
rectangles are f(m" ) œ (0.25)$ œ 64
, f(m# ) œ (0.75)$ œ 27
64 , f(m$ ) œ (1.25) œ 64 , and f(m% ) œ (1.75) œ 64
Notice that the average value is approximated by
œ
"
length of [!ß#]
†”
"
#
$
$
$
$
’ˆ 4" ‰ ˆ #" ‰ ˆ 34 ‰ ˆ #" ‰ ˆ 54 ‰ ˆ #" ‰ ˆ 74 ‰ ˆ #" ‰“ œ
$"
"'
approximate area under
• . We use this observation in solving the next several exercises.
curve f(x) œ x$
16. Partition [1ß 9] into the four subintervals ["ß $], [3ß &], [&ß (], and [(ß *]. The midpoints of these subintervals are
m" œ 2, m# œ 4, m$ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m" ) œ "# ,
f(m# ) œ "4 , f(m$ ) œ 6" , and f(m% ) œ 8" . The width of each rectangle is ?x œ 2. Thus,
Area ¸ 2 ˆ "# ‰ 2 ˆ 4" ‰ 2 ˆ 6" ‰ 2 ˆ 8" ‰ œ
Ê average value ¸
25
1#
area
length of ["ß*]
œ
ˆ 25
‰
12
8
œ
25
96 .
17. Partition [0ß 2] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of the subintervals
are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating rectangles are
"
#
f(m" ) œ
œ
"
#
"
#
sin#
1
4
œ
"
#
"
#
œ 1, and f(m% ) œ
œ 1, f(m# ) œ
"
2
sin#
71
4
œ
sin#
"
#
Š È"2 ‹ œ 1. The width of each rectangle is ?x œ #" . Thus,
31
4
œ
"
#
"
#
œ 1, f(m$ ) œ
"
2
sin#
51
4
œ
"
#
Š È"2 ‹
#
"
2
#
Area ¸ (1 1 1 1) ˆ "# ‰ œ 2 Ê average value ¸
area
length of [0ß2]
œ
2
#
œ 1.
18. Partition [0ß 4] into the four subintervals [0ß 1], [1ß 2ß ], [2ß 3], and [3ß 4]. The midpoints of the subintervals
are m" œ "# , m# œ #3 , m$ œ 5# , and m% œ 7# . The heights of the four approximating rectangles are
f(m" ) œ 1 Šcos Š
%
1 ˆ "# ‰
4 ‹‹
œ 1 ˆcos ˆ 18 ‰‰ œ 0.27145 (to 5 decimal places),
f(m# ) œ 1 Šcos Š
%
1 ˆ 3# ‰
4 ‹‹
œ 1 ˆcos ˆ 381 ‰‰ œ 0.97855, f(m3 ) œ 1 Šcos Š
%
œ 0.97855, and f(m% ) œ 1 Šcos Š
%
%
1 ˆ 7# ‰
4 ‹‹
%
1 ˆ #5 ‰
4 ‹‹
œ 1 ˆcos ˆ 581 ‰‰
%
%
œ 1 ˆcos ˆ 781 ‰‰ œ 0.27145. The width of each rectangle is
?x œ ". Thus, Area ¸ (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) œ 2.5 Ê average
2.5
5
value ¸ lengtharea
of [0ß4] œ 4 œ 8 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.1 Estimating with Finite Sums
297
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left
endpoints:
(a) upper estimate œ (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) œ 758 gal,
lower estimate œ (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) œ 543 gal.
(b) upper estimate œ (70 97 136 190 265 369 516 720) œ 2363 gal,
lower estimate œ (50 70 97 136 190 265 369 516) œ 1693 gal.
(c) worst case: 2363 720t œ 25,000 Ê t ¸ 31.4 hrs;
best case: 1693 720t œ 25,000 Ê t ¸ 32.4 hrs
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints:
(a) upper estimate œ (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) œ 60.9 tons
lower estimate œ (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) œ 46.8 tons
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) œ 126.6 tons,
so near the end of September 125 tons of pollutants will have been released.
#
21. (a) The diagonal of the square has length 2, so the side length is È#. Area œ ŠÈ#‹ œ #
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring
#1
1
"' œ ) .
Area œ "'ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ % sin 1 œ #È# ¸ #Þ)#)
#
)
)
%
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring
#1
1
$# œ "' .
Area œ $#ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ ) sin 1 œ #È# ¸ $Þ!'"
#
"'
"'
)
(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1.
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring
#1
1
1 ‰ˆ
ˆ " ‰ˆ
cos 1n ‰ œ "# sin #n1 .
#n œ n . The area of each isosceles triangle is AT œ # # sin n
(b) The area of the polygon is AP œ nAT œ
n
#
sin
#1
n ,
n
nÄ_ #
so lim
sin
#1
n
œ lim 1 †
nÄ_
sin #n1
ˆ #n1 ‰
œ1
(c) Multiply each area by r# .
AT œ "# r# sin #n1
AP œ n# r# sin
lim AP œ 1r
#
#1
n
nÄ_
23-26. Example CAS commands:
Maple:
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do
# (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
298
Chapter 5 Integration
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot);
# (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
Mathematica: (assigned function and values for a and b may vary):
Symbols for 1, Ä , powers, roots, fractions, etc. are available in Palettes (under File).
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_]:=x Sin[1/x]
{a,b}={1/4, 1}
Plot[f[x],{x, a, b}]
The following code computes the value of the function for each interval midpoint and then finds the average. Each
sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell.
n =100; dx = (b a) /n;
values = Table[N[f[x]], {x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =200; dx = (b a) /n;
values = Table[N[f[x]],{x, a + dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =1000; dx = (b a) /n;
values = Table[N[f[x]],{x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
FindRoot[f[x] == average,{x, a}]
5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS
2
1. !
kœ1
3
2. !
kœ1
6k
k1
œ
6(1)
11
6(2)
21
œ
6
2
k1
k
œ
11
1
21
2
31
3
12
3
œ7
œ0
1
2
2
3
œ
7
6
4
3. ! cos k1 œ cos (11) cos (21) cos (31) cos (41) œ 1 1 1 1 œ 0
kœ1
5
4. ! sin k1 œ sin (11) sin (21) sin (31) sin (41) sin (51) œ 0 0 0 0 0 œ 0
kœ1
3
5. ! (1)kb1 sin
kœ1
1
k
œ (1)"" sin
1
1
(1)#" sin
1
#
(")$" sin
1
3
œ 01
È3
#
œ
È3 2
#
4
6. ! (1)k cos k1 œ (1)" cos (11) (1)# cos (21) (1)$ cos (31) (1)% cos (41)
kœ1
œ (1) 1 (1) 1 œ 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.2 Sigma Notation and Limits of Finite Sums
6
7. (a) ! 2kc1 œ 2"" 2#" 2$" 2%" 2&" 2'" œ 1 2 4 8 16 32
kœ1
5
(b) ! 2k œ 2! 2" 2# 2$ 2% 2& œ 1 2 4 8 16 32
kœ0
4
(c) ! 2k1 œ 2"" 2!" 2"" 2#" 2$" 2%" œ 1 2 4 8 16 32
kœ"
All of them represent 1 2 4 8 16 32
6
8. (a) ! (2)k1 œ (2)"" (2)#" (2)$" (2)%" (2)&" (2)'" œ 1 2 4 8 16 32
kœ1
5
(b) ! (1)k 2k œ (1)! 2! Ð")" 2" (1)# 2# (1)$ 2$ (1)% 2% (1)& 2& œ 1 2 4 8 16 32
kœ0
3
(c) ! (1)k1 2k2 œ Ð")#" 2## (")"" 2"# (")!" 2!# (1)"" 2"# (")#" 2##
kœ2
(1)$" 2$# œ 1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
4
9. (a) !
kœ2
2
(b) !
kœ0
1
(c) !
kœ"
(")k"
k1
(1)#"
21
œ
(")k
k1
œ
(1)!
01
(")k
k2
œ
(1)"
1 2
(")$"
31
(")"
11
(")!
02
(")#
21
(")%"
41
œ 1
œ1
(")"
12
"
#
œ 1
"
#
"
#
"
3
"
3
"
3
(a) and (c) are equivalent; (b) is not equivalent to the other two.
4
10. (a) ! (k 1)# œ (1 1)# (2 1)# (3 1)# (4 1)# œ 0 1 4 9
kœ1
3
(b) ! (k 1)# œ (1 1)# (0 1)# (1 1)# (2 1)# (3 1)# œ 0 1 4 9 16
kœ1
"
(c) ! k# œ (3)# (2)# (1)# œ 9 4 1
kœ3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
6
4
4
12. ! k#
11. ! k
kœ1
13. !
kœ1
5
5
15. ! (1)k1
14. ! 2k
kœ1
kœ1
n
kœ1
"
k
5
16. ! (1)k
kœ1
n
17. (a) ! 3ak œ 3 ! ak œ 3(5) œ 15
kœ1
n
(b) !
kœ1
n
bk
6
œ
"
6
kœ1
n
! bk œ
kœ1
"
6
(6) œ 1
n
n
kœ1
n
kœ1
n
kœ1
n
kœ1
n
kœ1
n
kœ1
(c) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 1
(d) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 11
n
(e) ! (bk 2ak ) œ ! bk 2 ! ak œ 6 2(5) œ 16
kœ1
kœ1
"
#k
kœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
k
5
299
300
Chapter 5 Integration
n
n
kœ1
n
kœ1
n
18. (a) ! 8ak œ 8 ! ak œ 8(0) œ 0
n
n
kœ1
kœ1
(c) ! (ak 1) œ ! ak ! 1 œ 0 n œ n
kœ1
10
19. (a) ! k œ
kœ1
10(10 1)
#
n
(b) ! 250bk œ 250 ! bk œ 250(1) œ 250
kœ1
n
n
kœ1
n
kœ1
kœ1
kœ1
(d) ! (bk 1) œ ! bk ! 1 œ " n
10
(b) ! k# œ
œ 55
kœ1
10(10 1)(2(10) 1)
6
œ 385
13(13 1)(2(13) 1)
6
œ 819
#
10
(c) ! k$ œ ’ 10(10# 1) “ œ 55# œ 3025
kœ1
13
20. (a) ! k œ
kœ1
13(13 1)
#
13
(b) ! k# œ
œ 91
kœ1
#
13
(c) ! k$ œ ’ 13(13# 1) “ œ 91# œ 8281
kœ1
7
7
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
5
21. ! 2k œ 2 ! k œ 2 Š 7(7 # ") ‹ œ 56
23. ! a3 k# b œ ! 3 ! k# œ 3(6)
24. ! ak# 5b œ ! k# ! 5 œ
22. !
kœ1
6(6 ")(2(6) 1)
6
6(6 ")(2(6) 1)
6
1k
15
œ
1
15
5
!kœ
kœ1
1
15
Š 5(5 # 1) ‹ œ 1
œ 73
5(6) œ 61
5
5
5
5
kœ1
kœ1
kœ1
kœ1
7
7
7
7
kœ1
kœ1
kœ1
kœ1
1)
25. ! k(3k 5) œ ! a3k# 5kb œ 3 ! k# 5 ! k œ 3 Š 5(5 1)(2(5)
‹ 5 Š 5(5 # 1) ‹ œ 240
6
1)
26. ! k(2k 1) œ ! a2k# kb œ 2 ! k# ! k œ 2 Š 7(7 1)(2(7)
‹
6
5
k$
225
27. !
kœ1
7
kœ1
#
7
28. Œ! k !
kœ1
29. (a)
$
5
Œ! k œ
kœ1
k$
4
"
2 #5
7
5
5
kœ1
kœ1
#
œ Œ! k
kœ1
$
! k $ Œ! k œ
"
4
"
#25
#
! k$ œ Š 7(7 1) ‹
#
kœ1
(b)
œ 308
$
Š 5(5 # 1) ‹ Š 5(5 # 1) ‹ œ 3376
#
7
7(7 1)
#
"
4
#
Š 7(7 # 1) ‹ œ 588
(c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.2 Sigma Notation and Limits of Finite Sums
30. (a)
(b)
(c)
31. (a)
(b)
(c)
32. (a)
(b)
(c)
301
33. kx" x! k œ k1.2 0k œ 1.2, kx# x" k œ k1.5 1.2k œ 0.3, kx$ x# k œ k2.3 1.5k œ 0.8, kx% x$ k œ k2.6 2.3k œ 0.3,
and kx& x% k œ k3 2.6k œ 0.4; the largest is lPl œ 1.2.
34. kx" x! k œ k1.6 (2)k œ 0.4, kx# x" k œ k0.5 (1.6)k œ 1.1, kx$ x# k œ k0 (0.5)k œ 0.5,
kx% x$ k œ k0.8 0k œ 0.8, and kx& x% k œ k1 0.8k œ 0.2; the largest is lPl œ 1.1.
35. faxb œ " x#
Since f is decreasing on Ò !, 1Ó we use left endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n"
is ! a" x#i b "n œ
iœ!
œ
n$
n$
n
"! #
i
n$
iœ!
"
n
n"
! Š" ˆ i ‰# ‹ œ
n
iœ!
œ"
an " b n a# an " b " b
'n $
# $n n"#
. Thus,
'
n"
lim ! a" x#i b "n œ lim Œ"
nÄ_ iœ!
nÄ_
"
n$
n"
! an# i# b
iœ!
œ"
#n$ $n# n
'n $
œ"
# $n n"#
'
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ"
"
$
œ
#
$
302
Chapter 5 Integration
Since f is increasing on Ò !, $Ó we use right endpoints to obtain
upper sums. ˜x œ $ n ! œ $n and xi œ i˜x œ $ni . So an upper
36. faxb œ #x
n
n
sum is !#xi ˆ n$ ‰ œ ! 'ni †
iœ"
iœ"
n
Thus,
37. faxb œ x# "
lim ! 'i
nÄ_ iœ" n
†
$
n
œ
#
œ lim *n n# *n
nÄ_
$
n
n
")
n#
")
n#
!i œ
iœ"
n an " b
#
†
œ
*n# *n
n#
œ lim ˆ* n* ‰ œ *.
nÄ_
Since f is increasing on Ò !, $Ó we use right endpoints to obtain
upper sums. ˜x œ $ n ! œ $n and xi œ i˜x œ $ni . So an upper
n
n
#
sum is !ax#i "b $n œ !Šˆ $ni ‰ "‹ n$ œ
œ
iœ"
n
#( ! #
i n$
n
iœ"
iœ"
†nœ
#( nan "ba#n "b
‹
n$ Š
'
n
!Š *i## "‹
n
$
n
iœ"
$
*
") #(
*a#n$ $n# nb
n n#
$œ
$Þ Thus,
#n $
#
n
#(
") n n*#
lim !ax#i "b $n œ lim Œ
$ œ
#
nÄ_ iœ"
nÄ_
œ
38. faxb œ $x#
Since f is increasing on Ò !, "Ó we use right endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n
n
iœ"
iœ"
#
is !$x#i ˆ "n ‰ œ !$ˆ ni ‰ ˆ n" ‰ œ
œ
#n$ $n# n
#n $
œ lim Œ
nÄ_
39. faxb œ x x# œ xa" xb
œ
# $n n"#
#
# $n n"#
#
#
#
œ
n
$
n$
! i# œ
iœ"
$
n$
† Š nan "ba' #n "b ‹
n
. Thus, lim !$x#i ˆ "n ‰
nÄ_ iœ"
œ ".
Since f is increasing on Ò !, "Ó we use right endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n
n
#
is !axi xi# b n" œ !Š ni ˆ ni ‰ ‹ n" œ
iœ"
iœ"
œ
" n a n "b
‹
n# Š
#
œ
" "n
#
n"$ Š nan "ba' #n "b ‹
n
# $ "
œ lim ”Š
nÄ_
40. faxb œ $x #x#
* $ œ "#.
n
'
"
#
"
n
n#
n
"!
i
n#
iœ"
n
"! #
i
n$
iœ"
n# n
#n#
#n $ $ n # n
'n$
œ
. Thus, lim !axi x#i b "n
nÄ_ iœ"
‹Œ
$
n
# n"#
'
• œ
"
#
#
'
œ &' .
Since f is increasing on Ò !, "Ó we use right endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n
n
#
is !a$xi #x#i b "n œ !Š $ni #ˆ ni ‰ ‹ n" œ
iœ"
iœ"
œ
$ n a n "b
‹
n# Š
#
œ
$ $n
#
œ lim ”Š
nÄ_
n#$ Š nan "ba' #n "b ‹
n
# $ "
$
#
n
$
$
n
n#
œ
n
$!
i
n#
iœ"
$n# $n
#n#
. Thus, lim !a$xi #x#i b "n
‹Œ
nÄ_ iœ"
$
n
# n"#
$
• œ
$
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
#
$
œ
"$
' .
n
#! #
i
n$
iœ"
#n# $n "
$n#
Section 5.3 The Definite Integral
303
5.3 THE DEFINITE INTEGRAL
1.
'02 x# dx
2.
'"! 2x$ dx
3.
'(& ax# 3xb dx
4.
'"% "x dx
5.
'#$ 1 " x dx
6.
'0" È4 x# dx
7.
'! Î% (sec x) dx
8.
'0 Î% (tan x) dx
1
1
9. (a)
(c)
(e)
(f)
10. (a)
(b)
(c)
(d)
(e)
(f)
11. (a)
(c)
12. (a)
(c)
13. (a)
(b)
14. (a)
(b)
"
&
'#2 g(x) dx œ 0
(b) ' g(x) dx œ ' g(x) dx œ 8
&
"
2
2
&
&
2
'" 3f(x) dx œ 3'" f(x) dx œ 3(4) œ 12
(d) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 (4) œ 10
#
"
"
&
&
&
'" [f(x) g(x)] dx œ '" f(x) dx '" g(x) dx œ 6 8 œ 2
'"& [4f(x) g(x)] dx œ 4 '"& f(x) dx '"& g(x) dx œ 4(6) 8 œ 16
'"* 2f(x) dx œ 2 '"* f(x) dx œ 2(1) œ 2
'(* [f(x) h(x)] dx œ '(*f(x) dx '(* h(x) dx œ 5 4 œ 9
'(* [2f(x) 3h(x)] dx œ 2 '(* f(x) dx 3 '(* h(x) dx œ 2(5) 3(4) œ 2
'*"f(x) dx œ '"* f(x) dx œ (1) œ 1
'"( f(x) dx œ '"* f(x) dx '(* f(x) dx œ 1 5 œ 6
'*( [h(x) f(x)] dx œ '(* [f(x) h(x)] dx œ '(* f(x) dx '(* h(x) dx œ 5 4 œ 1
'"2 f(u) du œ '"2 f(x) dx œ 5
'#" f(t) dt œ '"2 f(t) dt œ 5
'!$ g(t) dt œ '$! g(t) dt œ È2
'$! [g(x)] dx œ '$! g(x) dx œ È2
(b)
(d)
(b)
(d)
'"2 È3 f(z) dz œ È3 '"2 f(z) dz œ 5È3
'"2 [f(x)] dx œ '"2 f(x) dx œ 5
'$! g(u) du œ '$! g(t) dt œ È2
'$! Èg(r)2 dr œ È"2 '$! g(t) dt œ Š È"2 ‹ ŠÈ2‹ œ 1
'$% f(z) dz œ '!% f(z) dz '!$ f(z) dz œ 7 3 œ 4
'%$ f(t) dt œ '$% f(t) dt œ 4
'"$ h(r) dr œ '"$ h(r) dr '"" h(r) dr œ 6 0 œ 6
"
$
$
' h(u) du œ Œ ' h(u) du œ ' h(u) du œ 6
$
"
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
304
Chapter 5 Integration
15. The area of the trapezoid is A œ
œ
"
#
(5 2)(6) œ 21 Ê
œ 21 square units
"
#
(3 1)(1) œ 2 Ê
(B b)h
"
#
(B b)h
'# ˆ #x 3‰ dx
%
16. The area of the trapezoid is A œ
œ
"
#
'"Î#
$Î#
(2x 4) dx
œ 2 square units
17. The area of the semicircle is A œ
œ
9
#
1 Ê
"
#
1r# œ
1(3)#
'$$ È9 x# dx œ 9# 1 square units
18. The graph of the quarter circle is A œ
œ 41 Ê
"
#
"
4
1 r# œ
"
4
1(4)#
'%! È16 x# dx œ 41 square units
19. The area of the triangle on the left is A œ
"
#
bh œ
œ 2. The area of the triangle on the right is A œ
œ
Ê
"
#
(1)(1) œ
"
#.
"
#
"
#
(2)(2)
bh
Then, the total area is 2.5
'# kxk dx œ 2.5 square units
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.3 The Definite Integral
20. The area of the triangle is A œ
Ê
"
#
bh œ
'" a1 kxkb dx œ 1 square unit
"
21. The area of the triangular peak is A œ
"
#
(2)(1) œ 1
"
#
bh œ
"
#
(2)(1) œ 1.
The area of the rectangular base is S œ jw œ (2)(1) œ 2.
Then the total area is 3 Ê
'"" a2 kxkb dx œ 3 square units
22. y œ 1 È1 x# Ê y 1 œ È1 x#
Ê (y 1)# œ 1 x# Ê x# (y 1)# œ 1, a circle with
center (!ß ") and radius of 1 Ê y œ 1 È1 x# is the
upper semicircle. The area of this semicircle is
A œ "# 1r# œ "# 1(1)# œ 1# . The area of the rectangular base
is A œ jw œ (2)(1) œ 2. Then the total area is 2
Ê
23.
'"" Š1 È1 x# ‹ dx œ 2 1# square units
'!b x2 dx œ "# (b)( b2 ) œ b4
#
1
#
24.
'!b 4x dx œ "# b(4b) œ 2b#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
305
306
Chapter 5 Integration
25.
'ab 2s ds œ "# b(2b) "# a(2a) œ b# a#
27.
'"
29.
'1#1 ) d) œ (2#1)
31.
'0
33.
'!"Î# t# dt œ ˆ 3‰
œ
35.
'a#a x dx œ (2a)#
37.
'!
39.
'$" 7 dx œ 7(1 3) œ 14
41.
'!2 5x dx œ 5 '!2 x dx œ 5 ’ 2#
43.
'!2 (2t 3) dt œ 2 '"" t dt '!2 3 dt œ 2 ’ 2#
44.
'!
45.
'#" ˆ1 #z ‰ dz œ '#" 1 dz '#" #z dz œ '#" 1 dz "# '"# z dz œ 1[1 2] "# ’ 2# 1# “ œ " "# ˆ 3# ‰ œ 74
46.
'$! (2z 3) dz œ '$! 2z dz '$! 3 dz œ 2 '!$ z dz '$! 3 dz œ 2 ’ 3#
47.
'"# 3u# du œ 3 '"# u# du œ 3 ”'!# u# du '!" u# du• œ 3 Š’ 23
È#
3
È
7
ŠÈ2‹
#
#
(1)#
#
œ
1#
#
œ
31 #
#
"
#
28.
'!Þ&#Þ& x dx œ (2.5)#
30.
'È& # # r dr œ Š5È#2‹
x dx œ
3
7‹
ŠÈ
œ
32.
'!!Þ$ s# ds œ (0.3)3
3
34.
'!1Î# )# d) œ ˆ 3‰
7
3
"
24
a#
#
œ
36.
'a
$
b‹
ŠÈ
3
œ
b
3
38.
'!$b x# dx œ (3b)3
40.
'!2 È2 dx œ È2 (# !) œ 2È2
#
0#
#“
œ 10
42.
'$& 8x dx œ "8 '$& x dx œ 8" ’ 5#
#
Št È2‹ dt œ
È2
'!
È2
t dt '
!
0#
#“
(0.5)#
#
#
$
1 $
#
È$a
3a#
#
$
x# dx œ
#
È
$
#
#
È2
'ab 3t dt œ "# b(3b) "# a(3a) œ 3# ab# a# b
#
x dx œ
" $
#
$
È
b
26.
œ3
#
ŠÈ2‹
#
œ 24
œ 0.009
œ
1$
#4
#
ŠÈ3a‹
x dx œ
#
$
a#
#
œ a#
œ 9b$
#
3#
#“
œ
16
16
œ1
3(2 0) œ 4 6 œ 2
#
È2 dt œ
ŠÈ2‹
–
#
0#
#—
È2 ’È2 0“ œ 1 2 œ 1
#
#
$
0$
3“
$
’ "3
0#
#“
#
3[0 3] œ 9 9 œ 0
0$
3 “‹
$
œ 3 ’ 23
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1$
3“
œ 3 ˆ 73 ‰ œ 7
Section 5.3 The Definite Integral
48.
'"Î#" 24u# du œ 24 '"Î#"
49.
'!# a3x# x 5b dx œ 3 '!# x# dx '!# x dx '!# 5 dx œ 3 ’ 23
50.
'"! a3x# x 5b dx œ '!" a3x# x 5b dx œ ”3 '!" x# dx '!" x dx '!" 5 dx•
u# du œ 24 –
'!"
u# du
'!"Î#
$
u# du— œ 24 ” 13
$
$
0$
3‹
b0
n
œ
œ ’3 Š 13
51. Let ?x œ
#
Š 1#
0#
#‹
5(1 0)“ œ ˆ 3# 5‰ œ
0$
3“
ˆ "# ‰$
3 •
#
’ 2#
œ 24 ’
0#
#“
ˆ 78 ‰
3
“œ7
5[2 0] œ (8 2) 10 œ 0
7
#
and let x! œ 0, x" œ ?x,
b
n
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 3(?x)# ?x œ 3(?x)$
f(c# ) ?x œ f(2?x) ?x œ 3(2?x)# ?x œ 3(2)# (?x)$
f(c$ ) ?x œ f(3?x) ?x œ 3(3?x)# ?x œ 3(3)# (?x)$
ã
f(cn ) ?x œ f(n?x) ?x œ 3(n?x)# ?x œ 3(n)# (?x)$
n
n
kœ1
n
kœ1
Then Sn œ ! f(ck ) ?x œ ! 3k# (?x)$
$
1)
œ 3(?x) ! k# œ 3 Š bn$ ‹ Š n(n 1)(2n
‹
6
$
kœ1
œ
$
b
#
ˆ2
52. Let ?x œ
3
n
b0
n
"‰
n#
œ
Ê
b
n
'!b 3x# dx œ n lim
Ä_
b$
#
ˆ2
3
n
"‰
n#
œ b$ .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 1(?x)# ?x œ 1(?x)$
f(c# ) ?x œ f(2?x) ?x œ 1(2?x)# ?x œ 1(2)# (?x)$
f(c$ ) ?x œ f(3?x) ?x œ 1(3?x)# ?x œ 1(3)# (?x)$
ã
f(cn ) ?x œ f(n?x) ?x œ 1(n?x)# ?x œ 1(n)# (?x)$
n
n
Then Sn œ ! f(ck ) ?x œ ! 1k# (?x)$
kœ1
n
kœ1
1)
œ 1(?x)$ ! k# œ 1 Š bn$ ‹ Š n(n 1)(2n
‹
6
$
kœ1
œ
1b
6
$
ˆ2
3
n
"‰
n#
Ê
'!b 1x# dx œ n lim
Ä_
1 b$
6
ˆ2
3
n
"‰
n#
œ
307
1 b$
3 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
308
Chapter 5 Integration
b0
n
53. Let ?x œ
œ
b
n
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 2(?x)(?x) œ 2(?x)#
f(c# ) ?x œ f(2?x) ?x œ 2(2?x)(?x) œ 2(2)(?x)#
f(c$ ) ?x œ f(3?x) ?x œ 2(3?x)(?x) œ 2(3)(?x)#
ã
f(cn ) ?x œ f(n?x) ?x œ 2(n?x)(?x) œ 2(n)(?x)#
n
n
Then Sn œ ! f(ck ) ?x œ ! 2k(?x)#
kœ1
n
kœ1
œ 2(?x)# ! k œ 2 Š bn# ‹ Š n(n 2 1) ‹
#
kœ1
œ b# ˆ1 "n ‰ Ê
b0
n
54. Let ?x œ
œ
'!b 2x dx œ n lim
Ä_
b
n
b# ˆ1 n" ‰ œ b# .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
"
#
‰
f(c" ) ?x œ f(?x) ?x œ ˆ ?x
# 1 (?x) œ # (?x) ?x
2
?
x
"
f(c# ) ?x œ f(2?x) ?x œ ˆ # 1‰ (?x) œ # (2)(?x)# ?x
f(c$ ) ?x œ f(3?x) ?x œ ˆ 3?# x 1‰ (?x) œ
"
#
(3)(?x)# ?x
f(cn ) ?x œ f(n?x) ?x œ ˆ n?# x 1‰ (?x) œ
"
#
(n)(?x)# ?x
ã
n
n
kœ1
kœ1
Then Sn œ ! f(ck ) ?x œ ! ˆ "# k(?x)# ?x‰ œ
œ
"
4
b# ˆ1 n1 ‰ b Ê
55. av(f) œ Š È3" 0 ‹
œ
'! ˆ x# 1‰ dx œ n lim
Ä_
b
"
È3
È$
'!
È$
'!
x# dx
"
#
n
n
kœ1
kœ1
(?x)# ! k ?x ! 1 œ
ˆ 4" b# ˆ1 n" ‰ b‰ œ
"
4
"
#
Š bn# ‹ Š n(n 2 1) ‹ ˆ bn ‰ (n)
#
b# b.
ax# 1b dx
È$
'!
"
È3
1 dx
$
œ
"
È3
ŠÈ3‹
3
56. av(f) œ ˆ 3 " 0 ‰
$
"
È3
ŠÈ3 0‹ œ 1 1 œ 0.
'!$ Š x# ‹ dx œ 3" ˆ #" ‰ '!$ x# dx
#
#
œ "6 Š 33 ‹ œ 3# ; x# œ 3# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.3 The Definite Integral
'!" a3x# 1b dx œ
"
"
œ 3 ' x# dx ' 1 dx œ 3 Š 13 ‹ (1 0)
!
!
57. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!" a3x# 3b dx œ
"
"
œ 3 ' x# dx ' 3 dx œ 3 Š 13 ‹ 3(1 0)
!
!
58. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!$ (t 1)# dt
$
$
$
œ 3" ' t# dt 23 ' t dt 3" ' 1 dt
!
!
!
59. av(f) œ ˆ 3 " 0 ‰
œ
"
3
$
#
Š 33 ‹ 32 Š 3#
60. av(f) œ Š 1 1(2) ‹
0#
#‹
3" (3 0) œ 1.
'#" at# tb dt
'#" t# dt 3" '#" t dt
"
#
œ "3 ' t# dt 3" '
t# dt 3" Š 1#
!
!
œ
"
3
#
œ
"
3
$
Š 13 ‹ 3" Š (32) ‹
$
61. (a) av(g) œ Š 1 "(1) ‹
"
#
œ
3
#
(2)#
# ‹
.
'"" akxk 1b dx
'"! (x 1) dx "# '!" (x 1) dx
!
!
"
"
œ "# ' x dx "# ' 1 dx "# ' x dx "# ' 1 dx
"
"
!
!
œ
"
#
#
œ "# Š 0#
(1)#
# ‹
#
"# (0 (1)) "# Š 1#
0#
#‹
"# (1 0)
œ "# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
309
310
Chapter 5 Integration
'"$ akxk 1b dx œ #" '"$ (x 1) dx
$
$
œ "# ' x dx "# ' 1 dx œ "# Š 3# 1# ‹ "# (3 1)
"
"
(b) av(g) œ ˆ 3 " 1 ‰
#
#
œ 1.
(c) av(g) œ Š 3 "(1) ‹
œ
"
4
"
4
'"$ akxk 1b dx
'"" akxk 1b dx 4" '"$ akxk 1b dx
"
4
(see parts (a) and (b) above).
62. (a) av(h) œ Š 0 "(1) ‹
'"0 kxk dx œ '"0 (x) dx
œ
œ
(1 2) œ
'"0 x dx œ 0#
#
(b) av(h) œ ˆ 1 " 0 ‰
#
œ Š "#
(1)#
#
œ "# .
'0" kxk dx œ '0" x dx
0#
#‹
œ "# .
(c) av(h) œ Š 1 "(1) ‹
'"" kxk dx
'"0 kxk dx '0" kxk dx
œ
"
#
Œ
œ
"
#
ˆ "# ˆ "# ‰‰ œ "# (see parts (a) and (b)
above).
63. To find where x x# 0, let x x# œ 0 Ê x(1 x) œ 0 Ê x œ 0 or x œ 1. If 0 x 1, then 0 x x# Ê a œ 0
and b œ 1 maximize the integral.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.3 The Definite Integral
311
64. To find where x% 2x# Ÿ 0, let x% 2x# œ 0 Ê x# ax# 2b œ 0 Ê x œ 0 or x œ „ È2. By the sign graph,
0 0 0 , we can see that x% 2x# Ÿ 0 on ’È2ß È2“ Ê a œ È2 and b œ È2
!
È#
È #
minimize the integral.
"
1 x #
65. f(x) œ
is decreasing on [0ß 1] Ê maximum value of f occurs at 0 Ê max f œ f(0) œ 1; minimum value of f
occurs at 1 Ê min f œ f(1) œ
Ê
"
#
Ÿ
'0" 1 " x
"
1 1#
œ
"
#
. Therefore, (1 0) min f Ÿ
dx Ÿ 1. That is, an upper bound œ 1 and a lower bound œ
#
66. See Exercise 65 above. On [0ß 0.5], max f œ
'0
0.5
(0.5 0) min f Ÿ
"
1 1#
min f œ
Then
"
4
2
5
"
1 0#
'0
0.5
"
1 x#
dx
'0.5 1 " x
"
#
dx Ÿ
67. 1 Ÿ sin ax# b Ÿ 1 for all x Ê (1 0)(1) Ÿ
"
#
"
1 (0.5)#
œ 1, min f œ
f(x) dx Ÿ (0.5 0) max f Ê
Ÿ
2
5
'0
'0.5 1 " x
"
œ 0.5. Therefore (1 0.5) min f Ÿ
Ÿ
'0" 1 " x
2
5
Ê
0.5
#
"
1 x#
#
"
#
dx Ÿ (1 0) max f
.
œ 0.8. Therefore
dx Ÿ
"
#
. On [0.5ß 1], max f œ
dx Ÿ (1 0.5) max f Ê
13
20
Ÿ
'0 1 " x
"
#
dx Ÿ
9
10
"
4
Ÿ
"
1 (0.5)#
'0.5 1 1 x
"
#
dx Ÿ
œ 0.8 and
2
5
.
.
'0" sin ax# b dx Ÿ (1 0)(1) or '0"sin x# dx Ÿ 1
Ê
'0"sin x# dx cannot
equal 2.
68. f(x) œ Èx 8 is increasing on [!ß "] Ê max f œ f(1) œ È1 8 œ 3 and min f œ f(0) œ È0 8 œ 2È2 .
Therefore, (1 0) min f Ÿ
'0" Èx 8 dx Ÿ (1 0) max f
0 on [aß b], then min f
69. If f(x)
a Ê ba
Then b
Ê 2È 2 Ÿ
'0" Èx 8 dx Ÿ 3.
0 on [aß b]. Now, (b a) min f Ÿ
0 and max f
0 Ê (b a) min f
0 Ê
'ab f(x) dx
0.
70. If f(x) Ÿ 0 on [aß b], then min f Ÿ 0 and max f Ÿ 0. Now, (b a) min f Ÿ
b
a Ê ba
71. sin x Ÿ x for x
Ÿ0 Ê
72. sec x
0 Ê (b a) max f Ÿ 0 Ê
1
x#
#
Ê
'ab f(x) dx Ÿ 0.
x#
#‹
Exercise 69) since [0ß 1] is contained in ˆ 1# ß 1# ‰ Ê
#
Ê
#
0 on ˆ 1# ß 1# ‰ Ê
'0" ’sec x Š1 x# ‹“ dx
#
'0"sec x dx '0" Š1 x# ‹ dx
'0" sec x dx '0" 1 dx "# '0" x# dx
#
Ê
'0" sec x dx
0 Ê
$
(1 0) "# Š 13 ‹ Ê
'ab f(x) dx is a constant K. Thus'ab av(f) dx œ 'ab K dx
'ab av(f) dx œ (b a)K œ (b a) † b " a 'ab f(x) dx œ 'ab f(x) dx.
73. Yes, for the following reasons: av(f) œ
0 (see
'0" sec x dx
Thus a lower bound is 76 .
œ K(b a) Ê
Then
0 Ê
#
on ˆ 1# ß 1# ‰ Ê sec x Š1
'0" Š1 x# ‹ dx
'ab f(x) dx Ÿ (b a) max f.
'0" (sin x x) dx Ÿ 0 (see Exercise 70) Ê '0" sin x dx '0" x dx
'0" sin x dx Ÿ Š 1# 0# ‹ Ê '0" sin x dx Ÿ "# . Thus an upper bound is "# .
0 Ê sin x x Ÿ 0 for x
'0" sin x dx Ÿ '0" x dx
'ab f(x) dx Ÿ (b a) max f.
"
ba
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
'0" sec x dx
7
6.
312
Chapter 5 Integration
74. All three rules hold. The reasons: On any interval [aß b] on which f and g are integrable, we have:
(a) av(f g) œ
"
ba
'ab [f(x) g(x)] dx œ b " a ”'ab f(x) dx 'ab g(x) dx• œ b " a 'ab f(x) dx b " a 'ab g(x) dx
œ av(f) av(g)
(b) av(kf) œ
(c) av(f) œ
"
ba
"
ba
'ab kf(x) dx œ b " a ”k 'ab f(x) dx• œ k ” b " a 'ab f(x) dx• œ k av(f)
'ab f(x) dx Ÿ b " a 'ab g(x) dx since f(x) Ÿ g(x) on [aß b], and b " a 'ab g(x) dx œ av(g).
Therefore, av(f) Ÿ av(g).
ba
n and let ck be the right
n ab a b
× and ck œ a kabn ab .
n
75. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
endpoint of each subinterval. So the partition is P œ Öa, a
n
n
kœ"
kœ"
b
We get the Riemann sum ! fack b˜x œ ! c †
this expression remains cab ab. Thus,
ba
n
œ
ba
n ,
n
c ab a b !
"
n
kœ"
a
œ
#a b a b
,
n
c ab a b
n
...,a
† n œ cab ab. As n Ä _ and mPm Ä !
'a c dx œ cab ab.
76. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
endpoint of each subinterval. So the partition is P œ
n
n
We get the Riemann sum ! fack b˜x œ ! ck# ˆ b n a ‰
œ
n
ba ! #
a
n Œ
kœ"
kœ"
n
#a a b a b !
k
n
kœ"
œ ab aba# aab ab# †
n"
n
kœ"
n
ab a b # ! #
k
n#
kœ"
ab a b $
'
†
œ
b a
n
ba
n
and let ck be the right
Öa, a b n a , a #abn ab , . . ., a nabn ab × and ck œ a kabn ab .
n
n
#
#
#
œ b n a ! Ša kabn ab ‹ œ bn a ! Ša# #akabn ab k abn# ab ‹
kœ"
kœ"
† na#
an "ba#n "b
n#
#a a b a b#
n#
†
n a n "b
#
ab a b $
n$
†
nan "ba#n "b
'
$
"
" n"
ab ab$ # n n#
†
"
'
"
ab a b $
†#
'
b
$
$
x# dx œ b$ a$ .
a
œ ab aba# aab ab# †
As n Ä _ and mPm Ä ! this expression has value ab aba# aab ab# † "
œ ba# a$ ab# #a# b a$ "$ ab$ $b# a $ba# a$ b œ
b$
$
a$
$.
Thus,
'
77. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is
increasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x! ), min# œ f(x" ), á ,
minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore
U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x
œ (f(x" ) f(x! )) ?x (f(x# ) f(x" ))?x á (f(xn ) f(xnc1 )) ?x œ (f(xn ) f(x! )) ?x œ (f(b) f(a)) ?x.
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f
is increasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where
min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore
U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn
œ (f(x" ) f(x! )) ?x" (f(x# ) f(x" ))?x# á (f(xn ) f(xnc1 )) ?xn
Ÿ (f(x" ) f(x! )) ?xmax (f(x# ) f(x" )) ?xmax á (f(xn ) f(xnc1 )) ?xmax . Then
U L Ÿ (f(xn ) f(x! )) ?xmax œ (f(b) f(a)) ?xmax œ kf(b) f(a)k ?xmax since f(b) f(a). Thus
lim (U L) œ lim (f(b) f(a)) ?xmax œ 0, since ?xmax œ lPl .
lPl Ä 0
lPl Ä 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.3 The Definite Integral
313
78. (a) U œ max" ?x max# ?x á maxn ?x where
max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xnc" )
since f is decreasing on [aß b];
L œ min" ?x min# ?x á minn ?x where
min" œ f(x" ), min# œ f(x# )ß á , minn œ f(xn )
since f is decreasing on [aß b]. Therefore
U L œ (max" min" ) ?x (max# min# ) ?x
á (maxn minn ) ?x
œ (f(x! ) f(x" )) ?x (f(x" ) f(x# ))?x
á (f(xn" ) f(xn )) ?x œ (f(x! ) f(xn )) ?x
œ (f(a) f(b)) ?x.
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xn" ) since
f is decreasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where
min" œ f(x" ), min# œ f(x# ), á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore
U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn
œ (f(x! ) f(x" )) ?x" (f(x" ) f(x# ))?x# á (f(xn" ) f(xn )) ?xn
Ÿ (f(x! ) f(xn )) ?xmax œ (f(a) f(b) ?xmax œ kf(b) f(a)k ?xmax since f(b) Ÿ f(a). Thus
lim (U L) œ lim kf(b) f(a)k ?xmax œ 0, since ?xmax œ lPl .
lPl Ä 0
lPl Ä 0
79. (a) Partition 0ß 1# ‘ into n subintervals, each of length ?x œ
x# œ 2?x, á , xn œ n?x œ
1
#.
1
#n
with points x! œ 0, x" œ ?x,
Since sin x is increasing on 0ß 1# ‘ , the upper sum U is the sum of the areas
of the circumscribed rectangles of areas f(x" ) ?x œ (sin ?x)?x, f(x# ) ?x œ (sin 2?x) ?x, á , f(xn ) ?x
œ (sin n?x) ?x. Then U œ (sin ?x sin 2?x á sin n?x) ?x œ ”
œ”
1 cos ˆˆn " ‰ 1 ‰
cos 4n
1
# 2n
1
• ˆ #n ‰
# sin 4n
'!
1 cos ˆ 1 1 ‰‰
1 ˆcos 4n
#
4n
1
4n sin 4n
œ
1 cos ˆ 1 1 ‰
cos 4n
#
4n
sin 1
Š 14n ‹
4n
1Î#
(b) The area is
œ
cos ?#x cosˆ ˆn #" ‰ ?x‰
• ?x
# sin ?#x
sin x dx œ n lim
Ä_
1 cos ˆ 1 1 ‰
cos 4n
#
4n
sin 1
Š 14n ‹
œ
1 cos 1#
1
œ 1.
4n
n
80. (a) The area of the shaded region is !˜xi † mi which is equal to L.
iœ"
n
(b) The area of the shaded region is !˜xi † Mi which is equal to U.
iœ"
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure
and the first part of the figure. Thus this area is U L.
n
n
iœ"
iœ"
81. By Exercise 80, U L œ !˜xi † Mi !˜xi † mi where Mi œ maxÖfaxb on the ith subinterval× and
n
n
iœ"
iœ"
mi œ minÖfaxb on the ith subinterval×. Thus U L œ !aMi mi b˜xi !% † ˜xi provided ˜xi $ for each
n
n
iœ"
iœ"
i œ "ß Þ Þ Þ , n. Since !% † ˜xi œ % !˜xi œ %ab ab the result, U L %ab ab follows.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
314
Chapter 5 Integration
82. The car drove the first 150 miles in 5 hours and the
second 150 miles in 3 hours, which means it drove 300
miles in 8 hours, for an average of 300
8 mi/hr
œ 37.5 mi/hr. In terms of average values of functions,
the function whose average value we seek is
30, 0 Ÿ t Ÿ 5
v(t) œ œ
, and the average value is
50, 5 1 Ÿ 8
(30)(5) (50)(3)
8
œ 37.5.
83-88. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
f := x -> 1-x;
a := 0;
b := 1;
N :=[ 4, 10, 20, 50 ];
P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]:
display( P, insequence=true );
89-92. Example CAS commands:
Maple:
with( Student[Calculus1] );
f := x -> sin(x);
a := 0;
b := Pi;
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do
# (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot);
# (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
83-92. Example CAS commands:
Mathematica: (assigned function and values for a, b, and n may vary)
Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands
Clear[x, f, a, b, n]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.4 The Fundamental Theorem of Calculus
{a, b}={0, 1}; n =10; dx = (b a)/n;
f = Sin[x]2 ;
xvals =Table[N[x], {x, a, b dx, dx}];
yvals = f /.x Ä xvals;
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}];
Plot[f, {x, a, b}, Epilog Ä boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}={0, 1}; n =10; dx = (b a)/n;
f = Sin[x]2 ;
xvals =Table[N[x], {x, a dx, b, dx}];
yvals = f /.x Ä xvals;
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx,xvals, yvals}];
Plot[f, {x, a, b}, Epilog Ä boxes];
Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N
Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}={0, 1}; n =10; dx = (b a)/n;
f = Sin[x]2 ;
xvals =Table[N[x], {x, a dx/2, b dx/2, dx}];
yvals = f /.x Ä xvals;
boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals dx/2, xvals dx/2, yvals}];
Plot[f, {x, a, b},Epilog Ä boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
5.4 THE FUNDAMENTAL THEOREM OF CALCULUS
1.
'c (2x 5) dx œ cx# 5xd#! œ a0# 5(0)b a(2)# 5(2)b œ 6
2.
'c ˆ5 x# ‰ dx œ ’5x x4 “ %
0
2
4
3.
#
$
3
'
4
0
Š3x
x$
4‹
'c ax$ 2x 3b dx œ ’ x4
%
'
6.
'
7.
'
1
#
#
Š5(3)
4%
16 ‹
#
Š 3(0)
#
(3)#
4 ‹
(0)%
16 ‹
œ
133
4
œ8
%
œ Š 24 2# 3(2)‹ Š (42) (2)# 3(2)‹ œ 12
%
"
$
ˆx# Èx‰ dx œ ’ x3 23 x$Î# “ œ ˆ "3 23 ‰ 0 œ 1
!
0
0
#
4#
4‹
œ Š 3(4)
#
x# 3x“
2
5.
5
&
x$Î# dx œ 25 x&Î# ‘ ! œ
32
1
(5)&Î# 0 œ 2(5)$Î# œ 10È5
$#
'cc x2
2
2
5
x'Î& dx œ 5x"Î& ‘ " œ ˆ #5 ‰ (5) œ
1
8.
%
x%
16 “ !
#
dx œ ’ 3x#
2
4.
œ Š5(4)
#
dx œ
5
#
'cc 2x# dx œ c2x" d "
ˆ 2 ‰ ˆ 2 ‰
# œ 1 # œ 1
1
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
315
316
Chapter 5 Integration
'
1
9.
'
1
10.
'
1Î3
11.
'
51Î6
12.
'
31Î4
13.
'
1Î3
14.
15.
'
0
0
sin x dx œ [cos x]1! œ (cos 1) (cos 0) œ (1) (1) œ 2
(1 cos x) dx œ [x sin x]1! œ (1 sin 1) (0 sin 0) œ 1
0
1Î$
1Î6
1Î4
0
œ ˆ2 tan ˆ 13 ‰‰ (2 tan 0) œ 2È3 0 œ 2È3
2 sec# x dx œ [2 tan x]!
&1Î'
csc# x dx œ [cot x]1Î' œ ˆcot ˆ 561 ‰‰ ˆcot ˆ 16 ‰‰ œ ŠÈ3‹ ŠÈ3‹ œ 2È3
$1Î%
csc ) cot ) d) œ [csc )]1Î% œ ˆcsc ˆ 341 ‰‰ ˆcsc ˆ 14 ‰‰ œ È2 ŠÈ2‹ œ 0
1Î$
4 sec u tan u du œ [4 sec u]!
0
" cos 2t
#
1Î2
dt œ
'
0
ˆ"
1Î2 #
"
#
œ 4 sec ˆ 13 ‰ 4 sec 0 œ 4(2) 4(1) œ 4
cos 2t‰ dt œ "# t
"
4
!
sin 2t‘ 1Î# œ ˆ "# (0)
"
4
sin 2(0)‰ ˆ "# ˆ 1# ‰
"
4
sin 2 ˆ 1# ‰‰
œ 14
Î
1Î$
2t
'c ÎÎ " cos
dt œ ' ˆ "# "# cos 2t‰ dt œ "# t 4" sin 2t‘ 1Î$
#
Î
1 3
16.
1 3
1 3
1 3
œ ˆ "# ˆ 13 ‰
"
4
sin 2 ˆ 13 ‰‰ ˆ #" ˆ 13 ‰
"
4
sin 2 ˆ 13 ‰‰ œ
1
6
"
4
sin 231
17.
'c
18.
'cÎ Î% ˆ4 sec# t t1 ‰ dt œ 'Î Î% a4 sec# t 1t# b dt œ 4 tan t 1t ‘ 11Î%Î$
1Î#
1Î#
$
a8y# sin yb dy œ ’ 8y3 cos y“
1Î#
1Î#
1
œŒ
8 ˆ 1# ‰
3
$
8 ˆ 1# ‰
3
cos 1# Œ
1
6
$
œ Š4 tan ˆ
cos ˆ 1# ‰ œ
1‰
4
Š4 tan ˆ 13 ‰
1
ˆ 13 ‰ ‹
È3
4
21 $
3
œ (4(1) 4) Š4 ŠÈ3‹ 3‹ œ 4È3 3
'"" (r 1)# dr œ '"" ar# 2r 1b dr œ ’ r3 r# r“ " œ Š (31)
20.
'È (t 1) at# 4b dt œ 'È at$ t# 4t 4b dt œ ’ t4 t3 2t# 4t“ÈÈ$
$
"
È3
È3
3
œ
%
$
$
(1)# (1)‹ Š 13 1# 1‹ œ 38
$
3
%
ŠÈ3‹
$
ŠÈ3‹
4
3
21.
'È" Š u#
"
u& ‹
22.
' " ˆ v"
"‰
v%
23.
'
(
2
1
1
3
1 3
1
ˆ 14 ‰ ‹
19.
È2
sin ˆ 321 ‰ œ
1
#
1 3
1Î2
"
4
$
s# È s
s#
(
2
dv œ
ds œ
'
1
È 3 ‹
#
Š
2 ŠÈ3‹ 4È3
"
du œ 'È Š u#
%
&
u ‹ du œ
u)
’ 16
4
$
"
"
4u% “È#
ŠÈ3‹
È2
ˆ1 s$Î# ‰ ds œ ’s
#
)
œ
1)
Š 16
$
È#
2
“
Ès
"
#
2 ŠÈ3‹ 4 ŠÈ3‹ œ 10È3
3
"
' " av$ v% b dv œ 2v1 3v" ‘ ""Î# œ Š 2(1)
1Î2
$
#
œ È 2
"
4(1)% ‹
"
3(1)$ ‹
2
É È2
ŠÈ2‹
16
"
%
4 ŠÈ2‹
"
$
3 ˆ "# ‰
œ 34
Œ
"
#
2 ˆ "# ‰
Š1
2
È1 ‹
œ È2 2$Î% 1
œ È2 %È8 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 56
Section 5.4 The Fundamental Theorem of Calculus
24.
'
4
1 Èu
Èu
9
du œ
'
4
9
ˆu"Î# 1‰ du œ 2Èu u‘ % œ Š2È4 4‹ Š2È9 9‹ œ 3
*
'c% kxk dx œ '%! kxk dx '!
4
25.
4
kxk dx œ
'%! x dx '!
4
#
x dx œ ’ x# “
œ 16
26.
'
1
"
! #
acos x kcos xk b dx œ
1
#
œ sin
27. (a)
!
d
dx
28. (a)
'
(b)
d
dx
29. (a)
'
(b)
d
dt
30. (a)
'
!
"
# (cos
x cos x) dx
'
1
"
1Î# #
#
’ x# “ œ Š 0#
!
(cos x cos x) dx œ
'
1Î#
!
(4)#
# ‹
#
Š 4#
Èx
cos t dt œ [sin t]! œ sin Èx sin 0 œ sin Èx Ê
Èx
'
Œ
sin x
1
!
d
dx
Œ
'
Èx
!
cos t dt œ
sin x
'
!
d ˆÈ ‰‰
cos t dt œ ˆcos Èx‰ ˆ dx
x œ ˆcos Èx ‰ ˆ "# x"Î# ‰ œ
3t# dt œ ct$ d "
Œ
t%
1Î#
d
dx
ˆsin Èx‰ œ cos Èx ˆ "# x"Î# ‰
sin x
Èu du œ
t%
!
tan )
!
d
dx
Œ
'
sin x
1
3t# dt œ
d
dx
asin$ x 1b œ 3 sin# x cos x
d
3t# dt œ a3 sin# xb ˆ dx
(sin x)‰ œ 3 sin# x cos x
1
Œ'
œ sin$ x 1 Ê
cos Èx
2È x
'
t%
!
t%
u"Î# du œ 23 u$Î# ‘ ! œ
2
3
at% b
$Î#
0œ
2 '
3 t
Ê
d
dt
Œ'
Œ
'
t%
!
Èu du œ
d
dt
ˆ 23 t' ‰ œ 4t&
Èu du œ Èt% ˆ dtd at% b‰ œ t# a4t$ b œ 4t&
)
sec# y dy œ [tan y]tan
œ tan (tan )) 0 œ tan (tan )) Ê
!
d
d)
tan )
!
sec# y dy œ
d
d)
(tan (tan )))
œ asec# (tan ))b sec# )
(b)
d
d)
'
Œ
tan )
!
sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )
31. y œ
'
33. y œ
'È sin t# dt œ '
34. y œ
'
35. y œ
'
36. y œ
'
x
!
È1 t# dt Ê
Èx
!
x
x#
cos Èt dt Ê
sin x
dt
È1 t#
!
!
tan x
dt
1 t#
, kxk
Ê
dy
dx
œ È1 x#
dy
dx
!
!
0#
#‹
cos x dx œ [sin x]!
cos Èx
2È x
œ
(b)
1Î#
%
#
%
sin 0 œ 1
Èx
'
'
!
dy
dx
1
#
sin t# dt Ê
32. y œ
dy
dx
'
1
x
"
t
dt Ê
dy
dx
œ
"
x
,x0
#
d ˆÈ ‰‰
œ Šsin ˆÈx‰ ‹ ˆ dx
x œ (sin x) ˆ "# x"Î# ‰ œ 2sinÈxx
d
œ Šcos Èx# ‹ ˆ dx
ax# b‰ œ 2x cos kxk
Ê
dy
dx
œ
"
È1 sin# x
d
ˆ dx
(sin x)‰ œ
"
Ècos# x
(cos x) œ
cos x
kcos xk
œ
cos x
cos x
œ 1 since kxk
"
d
‰ ˆ dx
œ ˆ 1 tan
(tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1
#x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
#
317
318
Chapter 5 Integration
37. x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2; Area
œ
'$# ax# 2xbdx '#! ax# 2xbdx '!# ax# 2xbdx
$
œ ’ x3 x# “
œ ŠŠ
(2)$
3
#
$
$
’ x3 x# “
#
(2) ‹ Š
!
$
#
(3)$
3
’ x3 x# “
#
!
#
(3) ‹‹
$
ŠŠ 03 0# ‹ Š (32) (2)# ‹‹
$
$
$
ŠŠ 23 2# ‹ Š 03 0# ‹‹ œ
28
3
38. 3x# 3 œ 0 Ê x# œ 1 Ê x œ „ 1; because of symmetry about
the y-axis, Area œ 2 Œ
'!" a3x# 3bdx '"# a3x# 3bdx
"
#
2 Š cx$ 3xd ! cx$ 3xd " ‹ œ 2 c aa1$ 3(1)b a0$ 3(0)bb
aa2$ 3(2)b a1$ 3(1)bd œ 2(6) œ 12
39. x$ 3x# 2x œ 0 Ê x ax# 3x 2b œ 0
Ê x(x 2)(x 1) œ 0 Ê x œ 0, 1, or 2;
Area œ
'!" ax$ 3x# 2xbdx '"# ax$ 3x# 2xbdx
"
%
%
œ ’ x4 x$ x# “ ’ x4 x$ x# “
!
%
#
"
%
œ Š 14 1$ 1# ‹ Š 04 0$ 0# ‹
%
%
’Š 24 2$ 2# ‹ Š 14 1$ 1# ‹“ œ
"
#
40. x$ 4x œ 0 Ê x ax# 4b œ 0 Ê x(x 2)(x 2) œ 0
Ê x œ 0, 2, or 2. Area œ
œ
%
’ x4
#
2x “
%
!
#
%
’ x4
'c! ax$ 4xbdx '!# ax$ 4xbdx
2
#
#
%
2x “ œ Š 04 2(0)# ‹
!
Š (42) 2(2)# ‹ ’Š 24 2(2)# ‹ Š 04 2(0)# ‹“ œ 8
%
41. x"Î$ œ 0 Ê x œ 0; Area œ
%
'c! x"Î$ dx '!) x"Î$ dx
"
!
)
œ 34 x%Î$ ‘ " 34 x%Î$ ‘ !
œ ˆ 34 (0)%Î$ ‰ ˆ 34 (1)%Î$ ‰ ˆ 34 (8)%Î$ ‰ ˆ 34 (0)%Î$ ‰
œ
51
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.4 The Fundamental Theorem of Calculus
42. x"Î$ x œ 0 Ê x"Î$ ˆ1 x#Î$ ‰ œ 0 Ê x"Î$ œ 0 or
1 x#Î$ œ 0 Ê x œ 0 or 1 œ x#Î$ Ê x œ 0 or
1 œ x# Ê x œ 0 or „ 1;
Area œ
œ
'c! ˆx"Î$ x‰dx '!" ˆx"Î$ x‰dx '") ˆx"Î$ x‰dx
"
%Î$
’ 34
x
!
x#
# “ "
œ ’Š 34 (0)%Î$
’ 34 x%Î$
0#
#‹
"
x#
# “!
’ 43 x%Î$
(1)#
# ‹“
Š 34 (1)%Î$
’Š 34 (1)%Î$
1#
#‹
Š 34 (0)%Î$
0#
# ‹“
’Š 34 (8)%Î$
8#
#‹
Š 34 (1)%Î$
1#
# ‹“
œ
"
4
"
4
ˆ2!
$
4
#" ‰ œ
)
x#
# “"
83
4
43. The area of the rectangle bounded by the lines y œ 2, y œ 0, x œ 1, and x œ 0 is 21. The area under the curve
y œ 1 cos x on [0ß 1] is
'!
1
(1 cos x) dx œ [x sin x]!1 œ (1 sin 1) (0 sin 0) œ 1. Therefore the area of
the shaded region is 21 1 œ 1.
44. The area of the rectangle bounded by the lines x œ 16 , x œ
"
#
51
6 ,
y œ sin
ˆ 561 16 ‰ œ 13 . The area under the curve y œ sin x on 16 ß 561 ‘ is
œ ˆcos
51 ‰
6
È3
# ‹
ˆcos 16 ‰ œ Š
È3
#
'
1
6
œ
51Î6
1Î6
"
#
œ sin
51
6
, and y œ 0 is
&1Î'
sin x dx œ [cos x]1Î'
œ È3. Therefore the area of the shaded region is È3 13 .
45. On 14 ß 0‘ : The area of the rectangle bounded by the lines y œ È2, y œ 0, ) œ 0, and ) œ 14 is È2 ˆ 14 ‰
œ
1È2
4
. The area between the curve y œ sec ) tan ) and y œ 0 is
'c
!
1Î4
sec ) tan ) d) œ [sec )]!1Î%
œ (sec 0) ˆsec ˆ 14 ‰‰ œ È2 1. Therefore the area of the shaded region on 14 ß !‘ is
1È2
4
On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ
under the curve y œ sec ) tan ) is
of the shaded region on !ß 14 ‘ is
È
'
1Î4
!
1È2
4
1Î%
sec ) tan ) d) œ [sec )]!
œ sec
1
4
ŠÈ2 1‹ .
1È2
4
. The area
sec 0 œ È2 1. Therefore the area
ŠÈ2 1‹ . Thus, the area of the total shaded region is
È
1È2
#
Š 1 4 2 È2 1‹ Š 1 4 2 È2 1‹ œ
.
46. The area of the rectangle bounded by the lines y œ 2, y œ 0, t œ 14 , and t œ 1 is 2 ˆ1 ˆ 14 ‰‰ œ 2
area under the curve y œ sec# t on 14 ß !‘ is
under the curve y œ 1 t# on [!ß "] is
'c
!
1Î4
"
$
œ
dt 3 œ 0 3 œ 3 Ê (d) is a solution to this problem.
dy
dx
œ
48. y œ
'c sec t dt 4
Ê
dy
dx
œ sec x and y(1) œ
49. y œ
'
sec t dt 4 Ê
dy
dx
œ sec x and y(0) œ
x
1
!
x
and y(1) œ
'
"
t
dt 3 Ê
"
x
1
1
Thus, the total
. Therefore the area of the shaded region is ˆ2 1# ‰
'
"
1 t
$
5
3
47. y œ
x
$
!
2
3
'
!
. The
!
ˆ 1‰
sec# t dt œ [tan t]
1Î% œ tan 0 tan 4 œ 1. The area
'! a1 t# b dt œ ’t t3 “ " œ Š1 13 ‹ Š0 03 ‹ œ 32 .
area under the curves on 14 ß "‘ is 1
1
#
'cc sec t dt 4 œ 0 4 œ 4
1
1
!
5
3
œ
"
3
1
#
.
Ê (c) is a solution to this problem.
sec t dt 4 œ 0 4 œ 4 Ê (b) is a solution to this problem.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
319
320
Chapter 5 Integration
50. y œ
'
51. y œ
'
53. s œ
'
x
"
" t
"
"
t
dt 3 œ 0 3 œ 3 Ê (a) is a solution to this problem.
'
'c ÎÎ
b 2
b 2
$
ˆ bh
#
bh ‰
6
Š2
b
Œh ˆ # ‰
2
(x 1)# ‹
bh ‰
6
dx œ 2
'
$
!
œ
bh
3
t
t!
È1 t# dt 2
g(x) dx v!
$
2
3
bh
"
(x 1)# ‹
Š1
"
bÎ2
4h ˆ #b ‰
3b#
œ bh
x
4hx$
3b# “ bÎ2
ˆh ˆ 4h
‰ # ‰ dx œ ’hx
b# x
œ ˆ bh
#
!
"
54. v œ
4h ˆ b# ‰
3b#
$
'
f(x) dx s!
œ Œhˆ #b ‰
'
and y(1) œ
'
55. Area œ
56. r œ
"
x
52. y œ
#
t!
œ
dy
dx
sec t dt 3
x
t
dt 3 Ê
$
dx œ 2 x ˆ x11 ‰‘ ! œ 2 ’Š3
"
(3 1) ‹
Š0
"
(0 1) ‹“
œ 2 3 "4 1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500
57.
dc
dx
œ
"
#È x
œ
"
#
x"Î# Ê c œ
'
x
!
" "Î#
dt
# t
œ t"Î# ‘ 0 œ Èx
x
c(100) c(1) œ È100 È1 œ $9.00
58. By Exercise 57, c(400) c(100) œ È400 È100 œ 20 10 œ $10.00
59. (a) v œ
(b) a œ
(c) s œ
(d)
(e)
(f)
(g)
ds
dt
df
dt
'
!
œ
d
dt
'
t
!
f(x) dx œ f(t) Ê v(5) œ f(5) œ 2 m/sec
is negative since the slope of the tangent line at t œ 5 is negative
3
f(x) dx œ
"
#
(3)(3) œ
9
#
m since the integral is the area of the triangle formed by y œ f(x), the x-axis,
and x œ 3
t œ 6 since from t œ 6 to t œ 9, the region lies below the x-axis
At t œ 4 and t œ 7, since there are horizontal tangents there
Toward the origin between t œ 6 and t œ 9 since the velocity is negative on this interval. Away from the
origin between t œ 0 and t œ 6 since the velocity is positive there.
Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the
x-axis than below it.
60. (a) v œ
(b) a œ
dg
dt
df
dt
œ
d
dt
'
!
t
g(x) dx œ g(t) Ê v(3) œ g(3) œ 0 m/sec.
is positive, since the slope of the tangent line at t œ 3 is positive
(c) At t œ 3, the particle's position is
'
!
$
g(x) dx œ
"
#
(3)(6) œ 9
(d) The particle passes through the origin at t œ 6 because s(6) œ
'
!
'
g(x) dx œ 0
(e) At t œ 7, since there is a horizontal tangent there
(f) The particle starts at the origin and moves away to the left for 0 t 3. It moves back toward the origin
for 3 t 6, passes through the origin at t œ 6, and moves away to the right for t 6.
(g) Right side, since its position at t œ 9 is positive, there being more area above the x-axis than below it at t œ *.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.4 The Fundamental Theorem of Calculus
61. k 0 Ê one arch of y œ sin kx will occur over the interval 0ß 1k ‘ Ê the area œ
œ "k cos ˆk ˆ 1k ‰‰ ˆ k" cos (0)‰ œ
62. lim x"$
xÄ!
63.
'
64.
'
x
1
x
!
'
!
x
t%
t#
dt
"
œ lim
' x %t# dt
! t "
x$
xÄ!
x
66.
1
x#
"
k
x#
d
dx
xÄ!
'
d
dx
'
!
1
x
x
f(t) dt œ
d
dx
ax# 2x 1b œ 2x 2
f(t) dt œ cos 1x 1x sin 1x Ê f(4) œ cos 1(4) 1(4) sin 1(4) œ 1
Ê f w (x) œ 1 (x9 1) œ
9
x 2
Ê f w (1) œ 3; f(1) œ 2
'#"" 1 9 t dt œ 2 0 œ 2;
sec (t 1) dt Ê gw (x) œ asec ax# 1bb (2x) œ 2x sec ax# 1b Ê gw (1) œ 2(1) sec a(1)# 1b
#
a"b
"
œ 2; g(1) œ 3 '
sec (t 1) dt œ 3 ' sec (t 1) dt œ 3 0 œ 3; L(x) œ 2(x (1)) g(1)
1
1
œ 2(x 1) 3 œ 2x 1
67. (a)
(b)
(c)
(d)
(e)
(f)
(g)
True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus.
True: g is continuous because it is differentiable.
True, since gw (1) œ f(1) œ 0.
False, since gww (1) œ f w (1) 0.
True, since gw (1) œ 0 and gww (1) œ f w (1) 0.
False: gww (x) œ f w (x) 0, so gww never changes sign.
True, since gw (1) œ f(1) œ 0 and gw (x) œ f(x) is an increasing function of x (because f w (x) 0).
68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, hw (x) œ f(x). Since f is differentiable for all x,
h has a second derivative for all x.
(b) True: they are continuous because they are differentiable.
(c) True, since hw (1) œ f(1) œ 0.
(d) True, since hw (1) œ 0 and hww (1) œ f w (1) 0.
(e) False, since hww (1) œ f w (1) 0.
(f) False, since hww (x) œ f w (x) 0 never changes sign.
(g) True, since hw (1) œ f(1) œ 0 and hw (x) œ f(x) is a decreasing function of x (because f w (x) 0).
69.
1 Îk
cos kx‘ !
%
L(x) œ 3(x 1) f(1) œ 3(x 1) 2 œ 3x 5
g(x) œ 3 '
sin kx dx œ
2
k
xÄ!
f(t) dt œ x cos 1x Ê f(x) œ
'# " 1 9 t dt
!
1Îk
œ lim x$x#" œ lim $ax%" "b œ _.
f(t) dt œ x# 2x 1 Ê f(x) œ
65. f(x) œ 2
'
321
70. The limit is 3x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
322
Chapter 5 Integration
71-74. Example CAS commands:
Maple:
with( plots );
f := x -> x^3-4*x^2+3*x;
a := 0;
b := 4;
F := unapply( int(f(t),t=a..x), x );
# (a)
p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#71(a) (Section 5.4)" ):
p1;
dF := D(F);
# (b)
q1 := solve( dF(x)=0, x );
pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ];
p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ):
display( [p1,p2], title="71(b) (Section 5.4)" );
incr := solve( dF(x)>0, x );
# (c)
decr := solve( dF(x)<0, x );
df := D(f);
# (d)
p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#71(d) (Section 5.4)" ):
p3;
q2 := solve( df(x)=0, x );
pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ];
p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '(x)=0" ):
display( [p3,p4], title="71(d) (Section 5.4)" );
75-78. Example CAS commands:
Maple:
a := 1;
u := x -> x^2;
f := x -> sqrt(1-x^2);
F := unapply( int( f(t), t=a..u(x) ), x );
dF := D(F);
# (b)
cp := solve( dF(x)=0, x );
solve( dF(x)>0, x );
solve( dF(x)<0, x );
d2F := D(dF);
# (c)
solve( d2F(x)=0, x );
plot( F(x), x=-1..1, title="#75(d) (Section 5.4)" );
79.
Example CAS commands:
Maple:
f := `f`;
q1 := Diff( Int( f(t), t=a..u(x) ), x );
d1 := value( q1 );
80.
Example CAS commands:
Maple:
f := `f`;
q2 := Diff( Int( f(t), t=a..u(x) ), x,x );
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.5 Indefinite Integrals and the Substitution Rule
value( q2 );
71-80. Example CAS commands:
Mathematica: (assigned function and values for a, and b may vary)
For transcendental functions the FindRoot is needed instead of the Solve command.
The Map command executes FindRoot over a set of initial guesses
Initial guesses will vary as the functions vary.
Clear[x, f, F]
{a, b}= {0, 21}; f[x_] = Sin[2x] Cos[x/3]
F[x_] = Integrate[f[t], {t, a, x}]
Plot[{f[x], F[x]},{x, a, b}]
x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}]
x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}]
Slightly alter above commands for 75 - 80.
Clear[x, f, F, u]
a=0; f[x_] = x2 2x 3
u[x_] = 1 x2
F[x_] = Integrate[f[t], {t, a, u(x)}]
x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}]
x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}]
After determining an appropriate value for b, the following can be entered
b = 4;
Plot[{F[x], {x, a, b}]
5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE
"
3
1. Let u œ 3x Ê du œ 3 dx Ê
'
sin 3x dx œ '
"
3
du œ dx
sin u du œ 3" cos u C œ 3" cos 3x C
"
4
2. Let u œ 2x# Ê du œ 4x dx Ê
'
x sin a2x b dx œ '
"
4
#
sin u du œ 4" cos u C œ 4" cos 2x# C
"
#
3. Let u œ 2t Ê du œ 2 dt Ê
'
sec 2t tan 2t dt œ '
"
#
'
ˆ1 cos
du œ dt
sec u tan u du œ
4. Let u œ 1 cos 2t Ê du œ
t ‰#
#
"
#
sin
28(7x 2)& dx œ '
"
7
"
7
x$ ax% 1b dx œ '
#
"
4
2
3
"
#
sec u C œ
u$ C œ
t
2
2
3
sec 2t C
dt
ˆ1 cos #t ‰$ C
du œ dx
(28)u& du œ ' 4u& du œ u% C œ (7x 2)% C
6. Let u œ x% " Ê du œ 4x$ dx Ê
'
"
#
dt Ê 2 du œ sin
t
#
ˆsin #t ‰ dt œ ' 2u# du œ
5. Let u œ 7x 2 Ê du œ 7 dx Ê
'
du œ x dx
u# du œ
u$
1#
"
4
du œ x$ dx
Cœ
"
1#
$
ax% 1b C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
323
324
Chapter 5 Integration
7. Let u œ 1 r$ Ê du œ 3r# dr Ê 3 du œ 9r# dr
' È9r
#
dr
1 r$
œ ' 3u"Î# du œ 3(2)u"Î# C œ 6 a1 r$ b
"Î#
C
8. Let u œ y% 4y# 1 Ê du œ a4y$ 8yb dy Ê 3 du œ 12 ay$ 2yb dy
'
12 ay% 4y# 1b ay$ 2yb dy œ ' 3u# du œ u$ C œ ay% 4y# 1b C
#
$
9. Let u œ x$Î# 1 Ê du œ
'
x"Î# dx Ê
Èx sin# ˆx$Î# 1‰ dx œ '
10. Let u œ "x Ê du œ
'
3
#
"
x#
2
3
2
3
du œ Èx dx
sin# u du œ
2
3
"
4
ˆ #u
sin 2u‰ C œ
"
3
ˆx$Î# 1‰
"
6
sin ˆ2x$Î# 2‰ C
dx
cos# ˆ x" ‰ dx œ ' cos# aub du œ
"
œ 2x
"4 sin ˆ 2x ‰ C
"
x#
'
cos# aub du œ ˆ u#
"
4
"
sin 2u‰ C œ 2x
"
4
sin ˆ x2 ‰ C
11. (a) Let u œ cot 2) Ê du œ 2 csc# 2) d) Ê "# du œ csc# 2) d)
'
csc# 2) cot 2) d) œ '
"
#
#
#
u du œ "# Š u# ‹ C œ u4 C œ 4" cot# 2) C
(b) Let u œ csc 2) Ê du œ 2 csc 2) cot 2) d) Ê "# du œ csc 2) cot 2) d)
'
csc# 2) cot 2) d) œ ' "# u du œ "# Š u# ‹ C œ u4 C œ 4" csc# 2) C
#
"
5
12. (a) Let u œ 5x 8 Ê du œ 5 dx Ê
'
dx
È5x8
œ'
"
5
Š È"u ‹ du œ
"
#
(b) Let u œ È5x 8 Ê du œ
'
dx
È5x8
œ'
du œ
2
5
2
5
'
"
5
#
du œ dx
u"Î# du œ
"
5
ˆ2u"Î# ‰ C œ
(5x 8)"Î# (5) dx Ê
uCœ
2
5
2
5
du œ
2
5
u"Î# C œ
2
5
È5x 8 C
dx
È5x8
È5x 8 C
13. Let u œ 3 2s Ê du œ 2 ds Ê "# du œ ds
'
È3 2s ds œ ' Èu ˆ " du‰ œ " ' u"Î# du œ ˆ " ‰ ˆ 2 u$Î# ‰ C œ " (3 2s)$Î# C
#
#
#
3
3
"
#
14. Let u œ 2x 1 Ê du œ 2 dx Ê
'
(2x 1) dx œ ' u
$
$
ˆ "#
du‰ œ
15. Let u œ 5s 4 Ê du œ 5 ds Ê
'
"
È5s 4
ds œ '
"
Èu
ˆ 5" du‰ œ
"
5
"
#
'
"
5
'
du œ dx
%
u$ du œ ˆ "# ‰ Š u4 ‹ C œ
"
8
(2x 1)% C
du œ ds
u"Î# du œ ˆ 5" ‰ ˆ2u"Î# ‰ C œ
2
5
È5s 4 C
16. Let u œ 2 x Ê du œ dx Ê du œ dx
'
3
(2 x)#
dx œ '
3(du)
u#
œ 3 ' u# du œ 3 Š u1 ‹ C œ
"
3
2 x
C
17. Let u œ 1 )# Ê du œ 2) d) Ê "# du œ ) d)
'
%
&Î%
) È1 )# d) œ ' %Èu ˆ "# du‰ œ "# ' u"Î% du œ ˆ "# ‰ ˆ 45 u&Î% ‰ C œ 25 a1 )# b C
18. Let u œ )# 1 Ê du œ 2) d) Ê 4 du œ 8) d)
' 8) $È)# 1 d) œ ' $Èu (4 du) œ 4 ' u"Î$ du œ 4 ˆ 3 u%Î$ ‰ C œ 3 a)# 1b%Î$ C
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.5 Indefinite Integrals and the Substitution Rule
19. Let u œ 7 3y# Ê du œ 6y dy Ê "# du œ 3y dy
'
$Î#
3yÈ7 3y# dy œ ' Èu ˆ "# du‰ œ "# ' u"Î# du œ ˆ "# ‰ ˆ 23 u$Î# ‰ C œ 3" a7 3y# b C
20. Let u œ 2y# 1 Ê du œ 4y dy
'
4y dy
È2y# 1
œ'
"
Èu
du œ ' u"Î# du œ 2u"Î# C œ 2È2y# 1 C
"
2È x
21. Let u œ 1 Èx Ê du œ
'
dx œ '
"
È x ˆ" È x ‰ #
œ 2u C œ
2 du
u#
"
2È x
22. Let u œ 1 Èx Ê du œ
'
ˆ1 È x ‰ $
Èx
dx Ê 2 du œ
"
3
cos (3z 4) dz œ ' (cos
u) ˆ "3
sin (8z 5) dz œ ' (sin
u) ˆ "8
'
"
3
ˆ1 Èx‰% C
"
3
' cos u du œ 3" sin u C œ 3" sin (3z 4) C
du œ dz
du‰ œ
25. Let u œ 3x 2 Ê du œ 3 dx Ê
"
#
dx
du œ dz
du‰ œ
"
8
dx
C
"
Èx
dx œ ' u$ (2 du) œ 2 ˆ 4" u% ‰ C œ
24. Let u œ 8z 5 Ê du œ 8 dz Ê
'
2
1 È x
dx Ê 2 du œ
23. Let u œ 3z 4 Ê du œ 3 dz Ê
'
"
Èx
"
8
'
"
8
sin u du œ
(cos u) C œ 8" cos (8z 5) C
du œ dx
sec# (3x 2) dx œ ' asec# ub ˆ "3 du‰ œ
"
3
'
"
3
sec# u du œ
tan u C œ
"
3
tan (3x 2) C
26. Let u œ tan x Ê du œ sec# x dx
'
tan# x sec# x dx œ ' u# du œ
27. Let u œ sin ˆ x3 ‰ Ê du œ
'
r$
18
1 Ê du œ
"
#
r#
6
$
cos ˆ x3 ‰ dx Ê 3 du œ cos ˆ x3 ‰ dx
"
#
sin' ˆ x3 ‰ C
sec# ˆ x# ‰ dx Ê 2 du œ sec# ˆ x# ‰ dx
"
4
tan) ˆ x# ‰ C
dr Ê 6 du œ r# dr
r % Š7
r&
10
$
r&
10 ‹
'
$
'
Ê du œ "# r% dr Ê 2 du œ r% dr
dr œ ' u$ (2 du) œ 2 ' u$ du œ 2 Š u4 ‹ C œ "# Š7
%
31. Let u œ x$Î# 1 Ê du œ
'
tan$ x C
r
r
r# Š 18
1‹ dr œ ' u& (6 du) œ 6 ' u& du œ 6 Š u6 ‹ C œ Š 18
1‹ C
&
30. Let u œ 7
'
"
3
tan( ˆ x# ‰ sec# ˆ x# ‰ dx œ ' u( (2 du) œ 2 ˆ 8" u) ‰ C œ
29. Let u œ
'
u$ C œ
sin& ˆ x3 ‰ cos ˆ x3 ‰ dx œ ' u& (3 du) œ 3 ˆ 6" u' ‰ C œ
28. Let u œ tan ˆ x# ‰ Ê du œ
'
"
3
"
3
3
#
x"Î# dx Ê
2
3
r&
10 ‹
%
C
du œ x"Î# dx
x"Î# sin ˆx$Î# 1‰ dx œ ' (sin u) ˆ 23 du‰ œ
2
3
'
sin u du œ
2
3
(cos u) C œ 23 cos ˆx$Î# 1‰ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
325
326
Chapter 5 Integration
32. Let u œ x%Î$ 8 Ê du œ
'
4
3
x"Î$ dx Ê
3
4
du œ x"Î$ dx
x"Î$ sin ˆx%Î$ 8‰ dx œ ' (sin u) ˆ 34 du‰ œ
3
4
'
sin u du œ
3
4
(cos u) C œ 34 cos ˆx%Î$ 8‰ C
33. Let u œ sec ˆv 1# ‰ Ê du œ sec ˆv 1# ‰ tan ˆv 1# ‰ dv
'
sec ˆv 1# ‰ tan ˆv 1# ‰ dv œ ' du œ u C œ sec ˆv 1# ‰ C
34. Let u œ csc ˆ v # 1 ‰ Ê du œ "# csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv Ê 2 du œ csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv
'
csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv œ ' 2 du œ 2u C œ 2 csc ˆ v # 1 ‰ C
35. Let u œ cos (2t 1) Ê du œ 2 sin (2t 1) dt Ê "# du œ sin (2t 1) dt
'
sin (2t 1)
cos# (2t 1)
dt œ ' #"
du
u#
œ
"
#u
Cœ
"
# cos (2t 1)
C
36. Let u œ 2 sin t Ê du œ cos t dt
'
dt œ '
6 cos t
(2 sin t)$
6
u$
du œ 6 ' u$ du œ 6 Š u# ‹ C œ 3(2 sin t)# C
#
37. Let u œ cot y Ê du œ csc# y dy Ê du œ csc# y dy
'
Ècot y csc# y dy œ ' Èu (du) œ ' u"Î# du œ 23 u$Î# C œ 23 (cot y)$Î# C œ 23 acot$ yb"Î# C
38. Let u œ sec z Ê du œ sec z tan z dz
'
sec z tan z
Èsec z
39. Let u œ
'
"
t#
"
t
dz œ '
"
Èu
du œ ' u"Î# du œ 2u"Î# C œ 2Èsec z C
"
Èt
"
)#
" "Î#
# t
dt Ê 2 du œ
sin
"
)
"
)
cos
Ê du œ ˆcos ") ‰ ˆ )"# ‰ d) Ê du œ
"
)
cos È)
È) sin# È)
"
)#
cos
d) œ ' u du œ #" u# C œ #" sin#
42. Let u œ csc È) Ê du œ Šcsc È) cot È)‹ Š
'
"
Èt
dt
cos ˆÈt 3‰ dt œ ' (cos u)(2 du) œ 2 ' cos u du œ 2 sin u C œ 2 sin ˆÈt 3‰ C
41. Let u œ sin
'
dt
cos ˆ "t 1‰ dt œ ' (cos u)(du) œ ' cos u du œ sin u C œ sin ˆ "t 1‰ C
40. Let u œ Èt 3 œ t"Î# 3 Ê du œ
'
"
t#
1 œ t" 1 Ê du œ t# dt Ê du œ
d) œ '
"
È)
"
‹
#È )
"
)
"
)
d)
C
d) Ê 2 du œ
"
È)
cot È) csc È) d)
cot È) csc È) d) œ ' 2 du œ 2u C œ 2 csc È) C œ
2
sin È)
C
43. Let u œ s$ 2s# 5s 5 Ê du œ a3s# 4s 5b ds
' as$ 2s# 5s 5b a3s# 4s 5b ds œ '
44.
u du œ
u#
#
Let u œ )% 2)# 8) 2 Ê du œ a4)$ 4) 8b d) Ê
'
as$ 2s# 5s 5b
#
Cœ
a)% 2)# 8) 2b a)$ ) 2b d) œ ' u ˆ "4 du‰ œ
"
4
'
"
4
#
C
du œ a)$ ) 2b d)
u du œ
"
4
#
Š u# ‹ C œ
ˆ) % 2 ) # 8 ) 2 ‰ #
8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
C
Section 5.5 Indefinite Integrals and the Substitution Rule
"
4
45. Let u œ 1 t% Ê du œ 4t$ dt Ê
'
t$ a1 t% b dt œ ' u$ ˆ "4 du‰ œ
$
46. Let u œ 1
'
"
x
Ê du œ
É x x& 1 dx œ '
"
x#
"
x#
"
4
du œ t$ dt
ˆ 4" u% ‰ C œ
"
16
%
a 1 t% b C
dx
É x x 1 dx œ '
"
x#
"
x
É1
dx œ ' Èu du œ ' u"Î# du œ
2
3
u$Î# C œ
2
3
ˆ1 "x ‰$Î# C
47. Let u œ x# ". Then du œ #xdx and "# du œ xdx and x# œ u ". Thus ' x$ Èx# " dx œ ' au "b "# Èu du
œ
"
#
' au$Î# u"Î# bdu œ "# ’ #& u&Î# #$ u$Î# “ C œ "& u&Î# "$ u$Î# C œ "& ax# "b&Î# "$ ax# "b$Î# C
48. Let u œ x$ " Ê du œ $x# dx and x$ œ u ". So ' $B& Èx$ " dx œ ' au "bÈu du œ ' au$Î# u"Î# bdu
œ #& u&Î# #$ u$Î# C œ #& ax$ "b
&Î#
#$ ax$ "b
$Î#
C
49. (a) Let u œ tan x Ê du œ sec# x dx; v œ u$ Ê dv œ 3u# du Ê 6 dv œ 18u# du; w œ 2 v Ê dw œ dv
'
18 tan# x sec# x
dx œ
a2 tan$ xb#
6
œ 2 u$ C
$
'
'
18u#
du œ
a 2 u $ b#
6
2 tan$ x C
#
#
œ
6 dv
(2 v)#
œ'
6 dw
w#
œ 6 ' w# dw œ 6w" C œ # 6 v C
(b) Let u œ tan x Ê du œ 3 tan x sec x dx Ê 6 du œ 18 tan# x sec# x dx; v œ 2 u Ê dv œ du
'
18 tan# x sec# x
a2 tan$ xb#
dx œ '
œ'
6 du
(2 u)#
6 dv
v#
6
œ v6 C œ 2 6 u C œ # tan
$x C
(c) Let u œ 2 tan$ x Ê du œ 3 tan# x sec# x dx Ê 6 du œ 18 tan# x sec# x dx
'
18 tan# x sec# x
a2 tan$ xb#
dx œ '
6 du
u#
6
œ u6 C œ 2 tan
$x C
50. (a) Let u œ x 1 Ê du œ dx; v œ sin u Ê dv œ cos u du; w œ 1 v# Ê dw œ 2v dv Ê
'
"
#
Èw dw œ
"
3
w$Î# C œ
"
3
a 1 v# b
$Î#
Cœ
"
3
a1 sin# ub
$Î#
#
Cœ
(b) Let u œ sin (x 1) Ê du œ cos (x 1) dx; v œ 1 u Ê dv œ 2u du Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' u È1 u# du œ '
œ ˆ "# ˆ 23 ‰ v$Î# ‰ C œ
"
3
v$Î# C œ
"
3
a1 u # b
$Î#
Cœ
"
3
(c) Let u œ 1 sin (x 1) Ê du œ 2 sin (x 1) cos (x 1) dx Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ '
œ
"
3
a1 sin# (x 1)b
$Î#
"
6
dr œ ' Š
cos Èu
ˆ"
È u ‹ 1#
sin È)
É) cos$ È)
d) œ
'
$Î#
dv œ u du
Èv dv œ '
$Î#
"
#
v"Î# dv
C
"
#
du œ sin (x 1) cos (x 1) dx
"
#
u"Î# du œ
"
#
ˆ 23 u$Î# ‰ C
"
1#
du œ (2r 1) dr; v œ Èu Ê dv œ
"
#È u
du Ê
du‰ œ ' (cos v) ˆ 6" dv‰ œ
"
6
sin v C œ
"
6
sin Èu C
sin È3(2r 1)# 6 C
52. Let u œ cos È) Ê du œ Šsin È)‹ Š
'
a1 sin# (x 1)b
du
(2r 1) cos È3(2r 1)# 6
È3(2r 1)# 6
œ
Èu du œ '
"
3
C
51. Let u œ 3(2r 1)# 6 Ê du œ 6(2r 1)(2) dr Ê
"
1#Èu
"
#
"
#
"
#
a1 sin# (x 1)b
#
'
dw œ v dv
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' È1 sin# u sin u cos u du œ ' vÈ1 v# dv
œ'
œ
"
#
sin È)
È) Écos$ È)
"
‹
#È )
d) œ '
d) Ê 2 du œ
2 du
u$Î#
sin È)
È)
d)
œ 2 ' u$Î# du œ 2 ˆ2u"Î# ‰ C œ
4
Èu
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
C
"
6
dv
C
327
328
Chapter 5 Integration
œ
4
Écos È)
C
53. Let u œ 3t# 1 Ê du œ 6t dt Ê 2 du œ 12t dt
s œ ' 12t a3t# 1b dt œ ' u$ (2 du) œ 2 ˆ "4 u% ‰ C œ
$
s œ 3 when t œ 1 Ê 3 œ
"
#
"
#
u% C œ
%
"
#
a3t# 1b C;
(3 1)% C Ê 3 œ 8 C Ê C œ 5 Ê s œ
"
#
%
a3t# 1b 5
54. Let u œ x# 8 Ê du œ 2x dx Ê 2 du œ 4x dx
y œ ' 4x ax# 8b
"Î$
dx œ ' u"Î$ (2 du) œ 2 ˆ 3# u#Î$ ‰ C œ 3u#Î$ C œ 3 ax# 8b
y œ 0 when x œ 0 Ê 0 œ 3(8)#Î$ C Ê C œ 12 Ê y œ 3 ax# 8b
55. Let u œ t
1
1#
#Î$
#Î$
C;
12
Ê du œ dt
s œ ' 8 sin# ˆt
dt œ ' 8 sin# u du œ 8 ˆ u# "4 sin 2u‰ C œ 4 ˆt 11# ‰ 2 sin ˆ2t 16 ‰ C;
s œ 8 when t œ 0 Ê 8 œ 4 ˆ 11# ‰ 2 sin ˆ 16 ‰ C Ê C œ 8 13 1 œ 9 13
Ê s œ 4ˆt 11# ‰ 2 sin ˆ2t 16 ‰ 9 13 œ 4t 2 sin ˆ2t 16 ‰ 9
56. Let u œ
1
4
1‰
1#
) Ê du œ d)
r œ ' 3 cos# ˆ 14 )‰ d) œ ' 3 cos# u du œ 3 ˆ u#
sin 2u‰ C œ 3# ˆ 14 )‰ 43 sin ˆ 1# 2)‰ C;
C Ê C œ 1# 43 Ê r œ 3# ˆ 14 )‰ 43 sin ˆ 1# 2)‰ 1#
when ) œ 0 Ê 18 œ 381 43 sin 1#
Ê r œ 3# ) 34 sin ˆ 1# 2)‰ 18 43 Ê r œ
rœ
1
8
57. Let u œ 2t
ds
dt
1
#
3
2
)
3
4
"
4
cos 2)
1
8
3
4
3
4
Ê du œ 2 dt Ê 2 du œ 4 dt
œ ' 4 sin ˆ2t 1# ‰ dt œ ' (sin u)(2 du) œ 2 cos u C" œ 2 cos ˆ2t 1# ‰ C" ;
at t œ 0 and
ds
dt
œ 100 we have 100 œ 2 cos ˆ 1# ‰ C" Ê C" œ 100 Ê
œ 2 cos ˆ2t 1# ‰ 100
ds
dt
Ê s œ ' ˆ2 cos ˆ2t 1# ‰ 100‰ dt œ ' (cos u 50) du œ sin u 50u C# œ sin ˆ2t 1# ‰ 50 ˆ2t 1# ‰ C# ;
at t œ 0 and s œ 0 we have 0 œ sin ˆ 1# ‰ 50 ˆ 1# ‰ C# Ê C# œ 1 251
Ê s œ sin ˆ2t 1# ‰ 100t 251 (1 251) Ê s œ sin ˆ2t 1# ‰ 100t 1
58. Let u œ tan 2x Ê du œ 2 sec# 2x dx Ê 2 du œ 4 sec# 2x dx; v œ 2x Ê dv œ 2 dx Ê
dy
dx
œ ' 4 sec# 2x tan 2x dx œ ' u(2 du) œ u# C" œ tan# 2x C" ;
at x œ 0 and
dy
dx
œ 4 we have 4 œ 0 C" Ê C" œ 4 Ê
Ê y œ ' asec# 2x 3b dx œ ' asec# v 3b ˆ "# dv‰ œ
at x œ 0 and y œ 1 we have 1 œ
"
#
dy
dx
"
#
"
#
dv œ dx
œ tan# 2x 4 œ asec# 2x 1b 4 œ sec# 2x 3
tan v 3# v C# œ
(0) 0 C# Ê C# œ 1 Ê y œ
"
#
"
#
tan 2x 3x C# ;
tan 2x 3x 1
59. Let u œ 2t Ê du œ 2 dt Ê 3 du œ 6 dt
s œ ' 6 sin 2t dt œ ' (sin u)(3 du) œ 3 cos u C œ 3 cos 2t C;
at t œ 0 and s œ 0 we have 0 œ 3 cos 0 C Ê C œ 3 Ê s œ 3 3 cos 2t Ê s ˆ 1# ‰ œ 3 3 cos (1) œ 6 m
60. Let u œ 1t Ê du œ 1 dt Ê 1 du œ 1# dt
v œ ' 1# cos 1t dt œ ' (cos u)(1 du) œ 1 sin u C" œ 1 sin (1t) C" ;
at t œ 0 and v œ 8 we have 8 œ 1(0) C" Ê C" œ 8 Ê v œ
ds
dt
œ 1 sin (1t) 8 Ê s œ ' (1 sin (1t) 8) dt
œ ' sin u du 8t C# œ cos (1t) 8t C# ; at t œ 0 and s œ 0 we have 0 œ 1 C# Ê C# œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.6 Substitution and Area Between Curves
329
Ê s œ 8t cos (1t) 1 Ê s(1) œ 8 cos 1 1 œ 10 m
61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on
the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
sin# x C" œ 1 cos# x C" Ê C# œ 1 C" ; also cos# x C# œ cos#2x "# C# Ê C$ œ C# "# œ C" "# .
62. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the
left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
#
tan# x
sec# x1
ˆC "# ‰
C œ sec# x ðñò
# Cœ
#
a constant
63. (a) Š
"
60
"
‹
0
'
0
1Î60
"Î'!
Vmax sin 1201t dt œ 60 Vmax ˆ 120" 1 ‰ cos (1201t)‘ !
œ V#1max [1 1] œ 0
(b) Vmax œ È2 Vrms œ È2 (240) ¸ 339 volts
(c)
'
1Î60
0
aVmax b# sin# 1201t dt œ aVmax b#
aVmax b
#
œ
#
"Î'!
2401t‘ !
t ˆ 240" 1 ‰ sin
'
1Î60
0
œ
ˆ 1 cos# 2401t ‰ dt œ
aVmax b
#
#
aVmax b#
#
œ V#max
1 [cos 21 cos 0]
'
1Î60
0
(1 cos 2401t) dt
"
ˆ 60
ˆ 240" 1 ‰ sin (41)‰ ˆ0 ˆ #40" 1 ‰ sin (0)‰‘ œ
5.6 SUBSTITUTION AND AREA BETWEEN CURVES
1. (a) Let u œ y 1 Ê du œ dy; y œ 0 Ê u œ 1, y œ 3 Ê u œ 4
'
3
0
Èy 1 dy œ
'
4
1
%
u"Î# du œ 23 u$Î# ‘ " œ ˆ 23 ‰ (4)$Î# ˆ 23 ‰ (1)$Î# œ ˆ 23 ‰ (8) ˆ 23 ‰ (1) œ
14
3
(b) Use the same substitution for u as in part (a); y œ 1 Ê u œ 0, y œ 0 Ê u œ 1
'c Èy 1 dy œ '
0
1
1
0
"
u"Î# du œ 23 u$Î# ‘ ! œ ˆ 23 ‰ (1)$Î# 0 œ
2
3
2. (a) Let u œ 1 r# Ê du œ 2r dr Ê "# du œ r dr; r œ 0 Ê u œ 1, r œ 1 Ê u œ 0
'
1
0
r È1 r# dr œ
'
0
!
"# Èu du œ "3 u$Î# ‘ " œ 0 ˆ 3" ‰ (1)$Î# œ
1
"
3
(b) Use the same substitution for u as in part (a); r œ 1 Ê u œ 0, r œ 1 Ê u œ 0
'c r È1 r# dr œ '
1
1
0
0
"# Èu du œ 0
3. (a) Let u œ tan x Ê du œ sec# x dx; x œ 0 Ê u œ 0, x œ
'
1Î4
0
tan x sec# x dx œ
'
1
0
"
#
u du œ ’ u# “ œ
!
1#
#
0œ
1
4
Ê uœ1
"
#
(b) Use the same substitution as in part (a); x œ 14 Ê u œ 1, x œ 0 Ê u œ 0
'c Î
0
1 4
tan x sec# x dx œ
' u du œ ’ u# “ !
0
#
"
1
œ0
"
#
œ "#
4. (a) Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 1
'
1
0
3 cos# x sin x dx œ
$
$
' 3u# du œ cu$ d "
" œ (1) a(1) b œ 2
1
1
(b) Use the same substitution as in part (a); x œ 21 Ê u œ 1, x œ 31 Ê u œ 1
'
31
21
3 cos# x sin x dx œ
'
1
1
3u# du œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
aVmax b#
1#0
330
Chapter 5 Integration
"
4
5. (a) u œ 1 t% Ê du œ 4t$ dt Ê
'
1
0
'
$
t$ a1 t% b dt œ
2
#
%
"
4
1
du œ t$ dt; t œ 0 Ê u œ 1, t œ 1 Ê u œ 2
2%
16
u
u$ du œ ’ 16
“ œ
"
1%
16
œ
15
16
(b) Use the same substitution as in part (a); t œ 1 Ê u œ 2, t œ 1 Ê u œ 2
'c
1
'
$
t$ a1 t% b dt œ
1
2
"
4
2
u$ du œ 0
"
#
6. (a) Let u œ t# 1 Ê du œ 2t dt Ê
'
È7
t at# 1b
0
"Î$
'
dt œ
8
)
"
#
1
du œ t dt; t œ 0 Ê u œ 1, t œ È7 Ê u œ 8
u"Î$ du œ ˆ "# ‰ ˆ 34 ‰ u%Î$ ‘ " œ ˆ 38 ‰ (8)%Î$ ˆ 38 ‰ (1)%Î$ œ
45
8
(b) Use the same substitution as in part (a); t œ È7 Ê u œ 8, t œ 0 Ê u œ 1
'cÈ
0
t at# 1b
7
"Î$
dt œ
'
1
"
8 #
"
#
7. (a) Let u œ 4 r# Ê du œ 2r dr Ê
' a4 5rr b
1
dr œ 5
# #
1
'
5
"
#
5
'
u"Î$ du œ
8
1
"
#
u"Î$ du œ 45
8
du œ r dr; r œ 1 Ê u œ 5, r œ 1 Ê u œ 5
u# du œ 0
(b) Use the same substitution as in part (a); r œ 0 Ê u œ 4, r œ 1 Ê u œ 5
'
1
5r
#
0 a4 r# b
dr œ 5
'
5
"
#
4
&
u# du œ 5 "# u" ‘ % œ 5 ˆ "# (5)" ‰ 5 ˆ "# (4)" ‰ œ
8. (a) Let u œ 1 v$Î# Ê du œ
'
1
10Èv
a1 v$Î# b
0
#
dv œ
'
2
"
u#
1
3
#
v"Î# dv Ê
ˆ 20
‰
3 du œ
'
20
3
20
3
2
1
"
8
du œ 10Èv dv; v œ 0 Ê u œ 1, v œ 1 Ê u œ 2
20 "
1‘
"‘#
u# du œ 20
3 u " œ 3 # 1 œ
10
3
(b) Use the same substitution as in part (a); v œ 1 Ê u œ 2, v œ 4 Ê u œ 1 4$Î# œ 9
'
4
10Èv
#
1 a1 v$Î# b
dv œ
'
9
"
u#
2
20 " ‘ *
20 ˆ "
1‰
20 ˆ
7 ‰
ˆ 20
‰
3 du œ 3 u # œ 3 9 2 œ 3 18 œ
70
#7
9. (a) Let u œ x# 1 Ê du œ 2x dx Ê 2 du œ 4x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4
'
È3
0
4x
È x# 1
dx œ
'
4
2
1 Èu
'
du œ
4
1
%
2u"Î# du œ 4u"Î# ‘ " œ 4(4)"Î# 4(1)"Î# œ 4
(b) Use the same substitution as in part (a); x œ È3 Ê u œ 4, x œ È3 Ê u œ 4
È3
'cÈ
4x
3 È x# 1
dx œ
'
4
4
2
Èu
du œ 0
10. (a) Let u œ x% 9 Ê du œ 4x$ dx Ê
'
1
0
x$
È x% 9
dx œ
'
10
9
"
4
"
4
du œ x$ dx; x œ 0 Ê u œ 9, x œ 1 Ê u œ 10
"!
u"Î# du œ 4" (2)u"Î# ‘ * œ
"
#
(10)"Î# #" (9)"Î# œ
È10 3
#
(b) Use the same substitution as in part (a); x œ 1 Ê u œ 10, x œ 0 Ê u œ 9
'c
0
x$
1 È x% 9
dx œ
'
9
"
10 4
u"Î# du œ
'
9
10
"
4
11. (a) Let u œ 1 cos 3t Ê du œ 3 sin 3t dt Ê
'
1Î6
0
(1 cos 3t) sin 3t dt œ
'
1
0
"
3
'
1Î6
(1 cos 3t) sin 3t dt œ
'
1
2
"
3
"
3
3 È10
#
du œ sin 3t dt; t œ 0 Ê u œ 0, t œ
#
"
u du œ ’ 3" Š u# ‹ “ œ
(b) Use the same substitution as in part (a); t œ
1Î3
u"Î# du œ
1
6
!
"
6
(1)# 6" (0)# œ
Ê u œ 1, t œ
#
#
u du œ ’ 3" Š u# ‹ “ œ
"
"
6
1
3
1
6
Ê u œ 1 cos
"
6
Ê u œ 1 cos 1 œ 2
(2)# 6" (1)# œ
"
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
#
œ1
Section 5.6 Substitution and Area Between Curves
12. (a) Let u œ 2 tan
'c
0
1Î2
t
#
"
#
Ê du œ
ˆ2 tan #t ‰ sec#
'
dt œ
t
#
sec#
2
1
t
#
dt Ê 2 du œ sec#
'c
1Î2
ˆ2 tan #t ‰ sec#
'
dt œ 2
t
#
1
'
0
cos z
È4 3 sin z
dz œ
'
4
"
Èu
4
1
#
Ê u œ 2 tan ˆ 41 ‰ œ 1, t œ 0 Ê u œ 2
#
3
1
#
Ê u œ 1, t œ
1
#
Ê uœ3
$
u du œ cu# d " œ 3# 1# œ 8
"
3
13. (a) Let u œ 4 3 sin z Ê du œ 3 cos z dz Ê
21
dt; t œ
u (2 du) œ cu# d " œ 2# 1# œ 3
(b) Use the same substitution as in part (a); t œ
1Î2
t
#
331
du œ cos z dz; z œ 0 Ê u œ 4, z œ 21 Ê u œ 4
ˆ 3" du‰ œ 0
(b) Use the same substitution as in part (a); z œ 1 Ê u œ 4 3 sin (1) œ 4, z œ 1 Ê u œ 4
'c
1
cos z
1 È4 3 sin z
dz œ
'
4
"
Èu
4
ˆ 3" du‰ œ 0
14. (a) Let u œ 3 2 cos w Ê du œ 2 sin w dw Ê "# du œ sin w dw; w œ 1# Ê u œ 3, w œ 0 Ê u œ 5
'c
0
'
dw œ
sin w
#
1Î2 (3 2 cos w)
5
u# ˆ #" du‰ œ
3
"
#
&
cu" d $ œ
"
#
"
ˆ "5 "3 ‰ œ 15
(b) Use the same substitution as in part (a); w œ 0 Ê u œ 5, w œ
'
1Î2
!
sin w
(3 2 cos w)#
'
dw œ
3
5
u# ˆ #" du‰ œ
"
#
5
' u# du œ
1
#
Ê uœ3
"
15
3
15. Let u œ t& 2t Ê du œ a5t% 2b dt; t œ 0 Ê u œ 0, t œ 1 Ê u œ 3
'
1
0
16. Let u œ 1 Èy Ê du œ
'
'
Èt& 2t a5t% 2b dt œ
4
dy
#
1 2 È y ˆ1 È y ‰
œ
'
3
"
#
2 u
3
0
$
u"Î# du œ 23 u$Î# ‘ ! œ
; y œ 1 Ê u œ 2, y œ 4 Ê u œ 3
dy
2È y
du œ
(3)$Î# 23 (0)$Î# œ 2È3
2
3
'
3
2
$
u# du œ cu" d # œ ˆ 13 ‰ ˆ 12 ‰ œ
"
6
17. Let u œ cos 2) Ê du œ 2 sin 2) d) Ê "# du œ sin 2) d); ) œ 0 Ê u œ 1, ) œ
'
1Î6
!
cos$ 2) sin 2) d) œ
18. Let u œ tan ˆ 6) ‰ Ê du œ
u œ tan
'
31Î2
1
1
4
'
1Î2
1
"
6
u$ ˆ "# du‰ œ "#
!
#
u$ du œ ’ 2" Š u# ‹“
"Î#
"
œ
cot& ˆ 6) ‰ sec# ˆ 6) ‰ d) œ
'
"
#
4 ˆ 1# ‰
1
!
"
4(1)#
"
È3
œ
,)œ
%
"
"
u
3
3
3
&
È3 u (6 du) œ ’6 Š 4 ‹“ "ÎÈ$ œ 2u% ‘ "ÎÈ$ œ 2(1)% # Š "
"
#
3
4
31
#
Ê
(1 sin 2t)$Î# cos 2t dt œ
%
È3 ‹
'
9
1
"
4
5u"Î% ˆ "4 du‰ œ
'
1
0
œ 12
du œ sin t dt; t œ 0 Ê u œ 5 4 cos 0 œ 1, t œ 1 Ê
5
4
'
1
9
*
u"Î% du œ 54 ˆ 45 u&Î% ‰‘ " œ 9&Î% 1 œ $&Î# "
20. Let u œ 1 sin 2t Ê du œ 2 cos 2t dt Ê "# du œ cos 2t dt; t œ 0 Ê u œ 1, t œ
'
1Î
5 (5 4 cos t)"Î% sin t dt œ
1Î4
Ê u œ cos 2 ˆ 16 ‰ œ
œ1
u œ 5 4 cos 1 œ 9
1
1
1
6
sec# ˆ 6) ‰ d) Ê 6 du œ sec# ˆ 6) ‰ d); ) œ 1 Ê u œ tan ˆ 16 ‰ œ
19. Let u œ 5 4 cos t Ê du œ 4 sin t dt Ê
'
'
1Î2
1
4
Ê uœ0
!
"# u$Î# du œ "2 ˆ 25 u&Î# ‰‘ " œ ˆ 15 (0)&Î# ‰ ˆ 15 (1)&Î# ‰ œ
"
5
21. Let u œ 4y y# 4y$ 1 Ê du œ a4 2y 12y# b dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 4(1) (1)# 4(1)$ 1 œ 8
'
!
1
a4y y# 4y$ 1b
#Î$
a12y# 2y 4b dy œ
'
1
8
)
u#Î$ du œ 3u"Î$ ‘ " œ 3(8)"Î$ 3(1)"Î$ œ 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
332
Chapter 5 Integration
"
3
22. Let u œ y$ 6y# 12y 9 Ê du œ a3y# 12y 12b dy Ê
Ê uœ4
'
1
!
ay$ 6y# 12y 9b
œ
'
$
)"Î# d) Ê
È) cos# ˆ)$Î# ‰ d) œ
24. Let u œ 1
'c Î
1 2
1
"
#
3
#
#
!
œ
ay# 4y 4b dy œ
'
4
9
%
"
3
u"Î# du œ 3" ˆ2u"Î# ‰‘ * œ
2
3
(4)"Î# 32 (9)"Î# œ
2
3
(2 3)
2
3
23. Let u œ )$Î# Ê du œ
È1
"Î#
du œ ay# 4y 4b dy; y œ 0 Ê u œ 9, y œ 1
"
t
'
"
4
!
du œ È) d); ) œ 0 Ê u œ 0, ) œ $È1# Ê u œ 1
cos# u ˆ 23 du‰ œ 23 ˆ #u
"
4
1
sin 2u‰‘ ! œ
ˆ 1#
2
3
"
4
sin 21‰ 32 (0) œ
1
3
Ê du œ t# dt; t œ 1 Ê u œ 0, t œ #" Ê u œ 1
t# sin# ˆ1 "t ‰ dt œ
1
2
3
'! sin# u du œ ˆ u2 4" sin 2u‰‘ "
œ ’ˆ #" 4" sin (2)‰ ˆ #0 4" sin 0‰“
!
1
sin 2
25. Let u œ 4 x# Ê du œ 2x dx Ê "# du œ x dx; x œ 2 Ê u œ 0, x œ 0 Ê u œ 4, x œ 2 Ê u œ 0
Aœ
'c
0
2
xÈ4 x# dx
%
œ 23 u$Î# ‘ ! œ
2
3
'
2
!
xÈ4 x# dx œ
(4)$Î# 23 (0)$Î# œ
'
4
!
"# u"Î# du
'
0
4
'
"# u"Î# du œ 2
4
"
! #
u"Î# du œ
'
!
4
u"Î# du
16
3
26. Let u œ 1 cos x Ê du œ sin x dx; x œ 0 Ê u œ 0, x œ 1 Ê u œ 2
'
!
1
(1 cos x) sin x dx œ
'
2
!
#
#
u du œ ’ u2 “ œ
!
2#
#
0#
#
œ2
27. Let u œ 1 cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 1 Ê u œ 1 cos (1) œ 0, x œ 0
Ê u œ 1 cos 0 œ 2
Aœ
'c
0
1
3 (sin x) È1 cos x dx œ
'
2
!
28. Let u œ 1 1 sin x Ê du œ 1 cos x dx Ê
Because of symmetry about x œ 1# , A œ 2
œ
'
!
1
3u"Î# (du) œ 3
"
1
'c
'
!
2
#
u"Î# du œ 2u$Î# ‘ ! œ 2(2)$Î# 2(0)$Î# œ 2&Î#
du œ cos x dx; x œ 1# Ê u œ 1 1 sin ˆ 1# ‰ œ 0, x œ 0 Ê u œ 1
0
1
1Î2 #
(cos x) (sin (1 1 sin x)) dx œ 2
'
1
(" cos 2x)
#
!
dx œ
"
#
'
1
!
(1 cos 2x) dx œ
"
#
30. For the sketch given, a œ 13 , b œ 13 ; f(t) g(t) œ
Aœ
œ
"
#
1
!
1
#
(sin u) ˆ 1" du‰
sin u du œ [cos u]1! œ (cos 1) (cos 0) œ 2
29. For the sketch given, a œ 0, b œ 1; f(x) g(x) œ 1 cos# x œ sin# x œ
Aœ
'
x
"
#
sin 2x ‘ 1
#
!
œ
"
#
1 cos 2x
;
#
[(1 0) (0 0)] œ
sec# t a4 sin# tb œ
"
#
1
#
sec# t 4 sin# t;
2t)
'c ÎÎ ˆ "# sec# t 4 sin# t‰ dt œ "# ' ÎÎ sec# t dt 4' ÎÎ sin# t dt œ "# ' ÎÎ sec# t dt 4' ÎÎ (" cos
dt
#
1 3
1 3
1 3
1Î3
1Î3
1 3
1 3
1 3
1 3
'c Î sec# t dt 2' Î (1 cos 2t) dt œ
1 3
"
#
[tan
1Î$
t]1Î$
2[t
1 3
1 3
1 3
1Î$
sin 2t
# ]1Î$
1 3
œ È3 4 †
1
3
È3 œ
31. For the sketch given, a œ 2, b œ 2; f(x) g(x) œ 2x# ax% 2x# b œ 4x# x% ;
Aœ
'c
2
2
$
a4x# x% b dx œ ’ 4x3
#
x&
5 “ #
œ ˆ 32
3
32 ‰
5
ˆ 32 ‰‘ œ
32
3 5
64
3
64
5
œ
320192
15
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
128
15
41
3
Section 5.6 Substitution and Area Between Curves
333
32. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ y# y$ ;
Aœ
'
1
!
ay# y$ b dy œ
'
1
!
y# dy
'
!
1
$
"
%
"
(" 0)
3
y$ dy œ ’ y3 “ ’ y4 “ œ
!
!
(" 0)
4
"
3
œ
"
4
œ
"
1#
33. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ a12y# 12y$ b a2y# 2yb œ 10y# 12y$ 2y;
Aœ
'
1
!
a10y# 12y$ 2yb dy œ
'
‰
œ ˆ 10
3 0 (3 0) (1 0) œ
!
1
10y# dy
'
1
!
12y$ dy
'
1
!
"
"
"
$‘
12 % ‘
2 #‘
2y dy œ 10
3 y ! 4 y ! # y !
4
3
34. For the sketch given, a œ 1, b œ 1; f(x) g(x) œ x# a2x% b œ x# 2x% ;
Aœ
'c ax# 2x% b dx œ ’ x3
1
$
1
"
2x&
5 “ "
œ ˆ "3 25 ‰ 3" ˆ 52 ‰‘ œ
35. We want the area between the line y œ 1, 0 Ÿ x Ÿ 2, and the curve y œ
'
(formed by y œ x and y œ 1) with base 1 and height 1. Thus, A œ
œ ˆ2
8 ‰
1#
"
#
œ2
2
3
"
#
œ
!
2
2
3
x#
4,
4
5
œ
1012
15
œ
22
15
738?= the area of a triangle
Š1
x#
4‹
dx "# (1)(1) œ ’x
#
x$
1# “ !
"
#
5
6
36. We want the area between the x-axis and the curve y œ x# , 0 Ÿ x Ÿ 1 :6?= the area of a triangle (formed by x œ 1,
x y œ 2, and the x-axis) with base 1 and height 1. Thus, A œ
'
1
!
$
"
x# dx "# (1)(1) œ ’ x3 “
!
"
#
œ
"
3
"
#
œ
5
6
37. AREA œ A1 A2
A1: For the sketch given, a œ 3 and we find b by solving the equations y œ x# 4 and y œ x# 2x
simultaneously for x: x# 4 œ x# 2x Ê 2x# 2x 4 œ 0 Ê 2(x 2)(x 1) Ê x œ 2 or x œ 1 so
b œ 2: f(x) g(x) œ ax# 4b ax# 2xb œ 2x# 2x 4 Ê A1 œ
$
œ ’ 2x3
2x
#
#
4x“
#
$
‰
œ ˆ 16
3 4 8 (18 9 12) œ 9
16
3
œ
'cc a2x# 2x 4b dx
2
3
11
3 ;
A2: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ ax# 2xb ax# 4b œ 2x# 2x 4
'c a2x# 2x 4b dx œ ’ 2x3
1
Ê A2 œ
$
2
œ 23 1 4
16
3
x# 4x“
"
#
‰
œ ˆ 23 1 4‰ ˆ 16
3 48
4 8 œ 9;
Therefore, AREA œ A1 A2 œ
11
3
9œ
38
3
38. AREA œ A1 A2
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ a2x$ x# 5xb ax# 3xb œ 2x$ 8x
Ê A1 œ
'c a2x$ 8xb dx œ ’ 2x4
0
%
2
!
8x#
# “ #
œ 0 (8 16) œ 8;
A2: For the sketch given, a œ 0 and b œ 2: f(x) g(x) œ ax# 3xb a2x$ x# 5xb œ 8x 2x$
Ê A2 œ
'
2
0
#
a8x 2x$ b dx œ ’ 8x2
Therefore, AREA œ A1 A2 œ 16
#
2x%
4 “!
œ (16 8) œ 8;
39. AREA œ A1 A2 A3
A1: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ (x 2) a4 x# b œ x# x 2
Ê A1 œ
'cc ax# x 2b dx œ ’ x3
1
$
2
x#
#
2x“
"
#
œ ˆ "3
"
#
2‰ ˆ 83
4
2
4‰ œ
7
3
"
#
œ
143
6
œ
1"
6 ;
"
#
œ 9# ;
A2: For the sketch given, a œ 1 and b œ 2: f(x) g(x) œ a4 x# b (x 2) œ ax# x 2b
Ê A2 œ
'c
2
1
$
ax# x 2b dx œ ’ x3
x#
#
2x“
#
"
œ ˆ 83
4
#
4‰ ˆ 13
1
2
2‰ œ 3 8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
334
Chapter 5 Integration
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ (x 2) a4 x# b œ x# x 2
Ê A3 œ
'
3
$
x#
#
ax# x 2b dx œ ’ x3
2
Therefore, AREA œ A1 A2 A3 œ
11
6
9
#
$
2x“ œ ˆ 27
3
#
9
#
ˆ9
9
#
6‰ ˆ 38
83 ‰ œ 9
5
6
œ
4
2
4‰ œ 9
9
#
38 ;
49
6
40. AREA œ A1 A2 A3
$
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ Š x3 x‹
Ê A1 œ
"
3
'c
0
2
ax$ 4xb dx œ
"
3
%
’ x4 2x# “
!
x
3
for x:
xœ
f(x) g(x) œ
"
3
œ
Ê
x
3
x$
3
xœ0 Ê
4
3
x
3
"
3
ax$ 4xb
x$
3
x and y œ
x
3
$
'
0
2
ax$ 4xb dx œ
"
3
'
0
2
a4x x$ b œ
$
Ê A3 œ
"
3
'
3
2
ax$ 4xb dx œ
"
3
Therefore, AREA œ A1 A2 A3 œ
$
%
’ x4 2x# “ œ
#
4
3
4
3
25
12
œ
"
3
x
3
œ
"
3
ax$ 4xb
ˆ 81
‰ ˆ 16
‰‘ œ
4 2†9 4 8
3225
1#
œ
"
3
ˆ 81
‰
4 14 œ
19
4
41. a œ 2, b œ 2;
f(x) g(x) œ 2 ax# 2b œ 4 x#
'c a4 x# bdx œ ’4x x3 “ #
2
$
#
2
8‰
œ 2 † ˆ 24
3 3 œ
œ ˆ8 83 ‰ ˆ8 83 ‰
32
3
42. a œ 1, b œ 3;
f(x) g(x) œ a2x x# b (3) œ 2x x# 3
'c a2x x# 3b dx œ ’x# x3
3
Ê Aœ
œ ˆ9
$
1
27
3
9‰ ˆ1
1
3
3‰ œ 11
43. a œ 0, b œ 2;
f(x) g(x) œ 8x x% Ê A œ
#
œ ’ 8x2
#
x&
“
5 !
œ 16
32
5
œ
'
2
0
"
3
3x“
œ
$
"
32
3
a8x x% b dx
80 32
5
œ
48
5
44. Limits of integration: x# 2x œ x Ê x# œ 3x
Ê x(x 3) œ 0 Ê a œ 0 and b œ 3;
f(x) g(x) œ x ax# 2xb œ 3x x#
Ê Aœ
œ
27
#
"
3
(8 4) œ 43 ;
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ Š x3 x‹
Ê Aœ
simultaneously
(x 2)(x 2) œ 0 Ê x œ 2, x œ 0, or x œ 2 so b œ 2:
Š x3 x‹ œ 3" ax$ 4xb Ê A2 œ 3"
x
3
43 x œ
œ 0 3" (4 8) œ 43 ;
#
A2: For the sketch given, a œ 0 and we find b by solving the equations y œ
x$
3
x$
3
œ
'
0
9œ
3
#
a3x x# b dx œ ’ 3x2
27 18
#
œ
$
x$
3 “!
9
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
25
12 ;
’2x#
#
x%
4 “!
Section 5.6 Substitution and Area Between Curves
45. Limits of integration: x# œ x# 4x Ê 2x# 4x œ 0
Ê 2x(x 2) œ 0 Ê a œ 0 and b œ 2;
f(x) g(x) œ ax# 4xb x# œ 2x# 4x
Ê Aœ
'
2
0
œ 16
3
#
4x#
2 “!
$
a2x# 4xb dx œ ’ 2x
3
œ
16
#
32 48
6
œ
8
3
46. Limits of integration: 7 2x# œ x# 4 Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê a œ 1 and b œ 1;
f(x) g(x) œ a7 2x# b ax# 4b œ 3 3x#
Ê Aœ
'c a3 3x# b dx œ 3 ’x x3 “ "
1
$
"
1
œ 3 ˆ1 "3 ‰ ˆ1 3" ‰‘ œ 6 ˆ 23 ‰ œ 4
47. Limits of integration: x% 4x# 4 œ x#
Ê x% 5x# 4 œ 0 Ê ax# 4b ax# 1b œ 0
Ê (x 2)(x 2)(x 1)(x 1) œ ! Ê x œ 2, 1, 1, 2;
f(x) g(x) œ ax% 4x# 4b x# œ x% 5x# 4 and
g(x) f(x) œ x# ax% 4x# 4b œ x% 5x# 4
Ê Aœ
'
2
1
'cc ax% 5x# 4bdx 'c ax% 5x# 4bdx
1
2
1
%
#
ax 5x 4bdx
&
œ ’ x5
œ ˆ "5
œ
1
5
3
60
5
5x$
3
4x“
"
&
’ x5
#
4‰ ˆ 32
5
60
3
œ
300180
15
40
3
5x$
3
4x“
8‰ ˆ 5"
5
3
"
&
"
’ 5x
5x$
3
4‰ ˆ 5"
4x“
5
3
#
"
4‰ ˆ 32
5
40
3
8‰ ˆ 5"
œ8
48. Limits of integration: xÈa# x# œ 0 Ê x œ 0 or
Èa# x# œ 0 Ê x œ 0 or a# x# œ 0 Ê x œ a, 0, a;
Aœ
œ
œ
"
#
"
3
'c xÈa# x# dx '
0
a
0
’ 23 aa# x# b
# $Î#
aa b
a
$Î# !
“
"
3
a
"# ’ 23 aa# x# b
# $Î#
’ aa b
xÈa# x# dx
“œ
2a
3
$Î# a
“
!
$
49. Limits of integration: y œ Èkxk œ
Èx, x Ÿ 0
and
Èx, x 0
5y œ x 6 or y œ x5 65 ; for x Ÿ 0: Èx œ x5 65
Ê 5Èx œ x 6 Ê 25(x) œ x# 12x 36
Ê x# 37x 36 œ 0 Ê (x 1)(x 36) œ 0
Ê x œ 1, 36 (but x œ 36 is not a solution);
for x 0: 5Èx œ x 6 Ê 25x œ x# 12x 36
Ê x# 13x 36 œ 0 Ê (x 4)(x 9) œ 0
Ê x œ 4, 9; there are three intersection points and
Aœ
'c ˆ x 5 6 Èx‰dx '
0
1
0
4
ˆ x 5 6 Èx‰dx
'
4
9
ˆÈ x
x6‰
5
dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5
3
4‰
335
336
Chapter 5 Integration
#
œ ’ (x 106) 23 (x)$Î# “
œ ˆ 36
10
25
10
!
%
#
’ (x 106) 23 x$Î# “ ’ 23 x$Î#
"
23 ‰ ˆ 100
10
2
3
† 4$Î#
!
36
10
0‰ ˆ 32 † 9$Î#
50. Limits of integration:
x# 4, x Ÿ 2 or x
y œ kx# 4k œ œ
4 x# , 2 Ÿ x Ÿ 2
for x Ÿ 2 and x
#
2: x# 4 œ
#
#
x#
2
*
(x 6)#
10 “ %
225
10
2
3
† 4$Î#
100 ‰
10
œ 50
10
20
3
2
4
Ê 2x 8 œ x 8 Ê x œ 16 Ê x œ „ 4;
x#
#
for 2 Ÿ x Ÿ 2: 4 x# œ
#
4 Ê 8 2x# œ x# 8
Ê x œ 0 Ê x œ 0; by symmetry of the graph,
Aœ2
'
2
0
#
’Š x2 4‹ a4 x# b“dx 2
œ 2 ˆ 8# 0‰ 2 ˆ32
'
4
2
#
#
!
16 68 ‰ œ 40
64
6
$
’Š x2 4‹ ax# 4b“dx œ 2 ’ x2 “ 2 ’8x
56
3
œ
%
x$
6 “#
64
3
51. Limits of integration: c œ 0 and d œ 3;
f(y) g(y) œ 2y# 0 œ 2y#
'
Ê Aœ
0
3
$
$
2y# dy œ ’ 2y3 “ œ 2 † 9 œ 18
!
52. Limits of integration: y# œ y 2 Ê (y 1)(y 2) œ 0
Ê c œ 1 and d œ 2; f(y) g(y) œ (y 2) y#
Ê Aœ
'c ay 2 y# b dy œ ’ y#
2
#
1
2y
œ ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰ œ 6
8
3
"
#
#
y$
3 “ "
2
"
3
œ
9
#
53. Limits of integration: 4x œ y# 4 and 4x œ 16 y
Ê y# 4 œ 16 y Ê y# y 20 œ 0 Ê
(y 5)(y 4) œ 0 Ê c œ 4 and d œ 5;
#
f(y) g(y) œ ˆ 164y ‰ Š y 44 ‹ œ
Ê Aœ
"
4
y$
3
'c ay# y 20b dy
5
4
y#
#
œ
"
4
’
œ
"
4
"
4
ˆ 125
3
189
ˆ 3
œ
y# y20
4
20y“
&
%
25
‰
100
4" ˆ 64
2
3
9
243
‰
180
œ
2
8
16
#
80‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
5
3
Section 5.6 Substitution and Area Between Curves
54. Limits of integration: x œ y# and x œ 3 2y#
Ê y# œ 3 2y# Ê 3y# œ 3 Ê 3(y 1)(y 1) œ 0
Ê c œ 1 and d œ 1; f(y) g(y) œ a3 2y# b y#
œ 3 3y# œ 3 a1 y# b Ê A œ 3
œ 3 ’y
"
y$
3 “ "
œ 3 ˆ1
"‰
3
'c a1 y# b dy
1
1
3 ˆ1 3" ‰
œ 3 † 2 ˆ1 "3 ‰ œ 4
55. Limits of integration: x œ y# and x œ 2 3y#
Ê y# œ 2 3y# Ê 2y# 2 œ 0
Ê 2(y 1)(y 1) œ 0 Ê c œ 1 and d œ 1;
f(y) g(y) œ a2 3y# b ay# b œ 2 2y# œ 2 a1 y# b
Ê Aœ2
'c a1 y# b dy œ 2 ’y y3 “ "
1
$
"
1
œ 2 ˆ1 "3 ‰ 2 ˆ1 3" ‰ œ 4 ˆ 23 ‰ œ
8
3
56. Limits of integration: x œ y#Î$ and x œ 2 y%
Ê y#Î$ œ 2 y% Ê c œ 1 and d œ 1;
f(y) g(y) œ a2 y% b y#Î$
'c ˆ2 y% y#Î$ ‰ dy
1
Ê Aœ
1
œ ’2y
y&
5
35 y&Î$ “
"
"
œ ˆ2 "5 35 ‰ ˆ2
œ 2 ˆ2 "5 35 ‰ œ 12
5
"
5
35 ‰
57. Limits of integration: x œ y# 1 and x œ kyk È1 y#
Ê y# 1 œ kyk È1 y# Ê y% 2y# 1 œ y# a1 y# b
Ê y% 2y# 1 œ y# y% Ê 2y% 3y# 1 œ 0
Ê a2y# 1b ay# 1b œ 0 Ê 2y# 1 œ 0 or y# 1 œ 0
Ê y# œ
"
#
or y# œ 1 Ê y œ „
„È 2
#
È2
#
or y œ „ 1.
are not solutions Ê y œ „ 1;
for 1 Ÿ y Ÿ 0, f(x) g(x) œ yÈ1 y# ay# 1b
Substitution shows that
œ 1 y# y a1 y# b
"Î#
, and by symmetry of the graph,
'c ’1 y# y a1 y# b"Î# “ dy
"Î#
œ 2' a1 y# b dy 2 ' y a1 y# b dy
c
c
0
Aœ2
1
0
0
1
œ 2 ’y
1
$
y
3
“
!
"
# $Î#
2 ˆ "# ‰ ” 2 a1 3y b •
!
"
œ 2 (! 0) ˆ1 3" ‰‘ ˆ 23 0‰ œ 2
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337
338
Chapter 5 Integration
58. AREA œ A1 A2
Limits of integration: x œ 2y and x œ y$ y# Ê
y$ y# œ 2y Ê y ay# y 2b œ y(y 1)(y 2) œ 0
Ê y œ 1, 0, 2:
for 1 Ÿ y Ÿ 0, f(y) g(y) œ y$ y# 2y
'c
0
Ê A1 œ
1
œ 0 ˆ "4
"
3
%
y$
3
ay$ y# 2yb dy œ ’ y4
1‰ œ
y# “
!
"
5
12 ;
for 0 Ÿ y Ÿ 2, f(y) g(y) œ 2y y$ y#
'! a2y y$ y# b dy œ ’y# y4
2
Ê A2 œ
Ê ˆ4
%
#
y$
3 “!
38 ‰ 0 œ 38 ;
16
4
Therefore, A1 A2 œ
5
12
8
3
œ
37
12
59. Limits of integration: y œ 4x# 4 and y œ x% 1
Ê x% 1 œ 4x# 4 Ê x% 4x# 5 œ 0
Ê ax# 5b (x 1)(x 1) œ 0 Ê a œ 1 and b œ 1;
f(x) g(x) œ 4x# 4 x% 1 œ 4x# x% 5
Ê Aœ
'c a4x# x% 5b dx œ ’ 4x3
1
$
1
œ ˆ 43
"
5
"
5
5‰ ˆ 43
5‰ œ 2 ˆ 43
"
x&
5
5x“
"
5
5‰ œ
"
104
15
60. Limits of integration: y œ x$ and y œ 3x# 4
Ê x$ 3x# 4 œ 0 Ê ax# x 2b (x 2) œ 0
Ê (x 1)(x 2)# œ 0 Ê a œ 1 and b œ 2;
f(x) g(x) œ x$ a3x# 4b œ x$ 3x# 4
Ê Aœ
'c ax$ 3x# 4b dx œ ’ x4
œ ˆ 16
4
24
3
2
%
1
8‰ ˆ 41 " 4‰ œ
3x$
3
4x“
#
"
27
4
61. Limits of integration: x œ 4 4y# and x œ 1 y%
Ê 4 4y# œ 1 y% Ê y% 4y# 3 œ 0
Ê Šy È3‹ Šy È3‹ (y 1)(y 1) œ 0 Ê c œ 1
0; f(y) g(y) œ a4 4y# b a1 y% b
and d œ 1 since x
'c a3 4y# y% b dy
1
œ 3 4y# y% Ê A œ
œ ’3y
4y$
3
"
y&
5 “ "
1
œ 2ˆ3
4
3
5" ‰ œ
56
15
#
62. Limits of integration: x œ 3 y# and x œ y4
#
Ê 3 y# œ y4 Ê
3y#
4
3œ0 Ê
3
4
(y 2)(y 2) œ 0
#
Ê c œ 2 and d œ 2; f(y) g(y) œ a3 y# b Š y4 ‹
œ 3 Š1
œ 3 ˆ2
y#
4‹
8 ‰
12
'c Š1 y4 ‹ dy œ 3 ’y 1y# “ #
Ê Aœ3
ˆ 2
2
#
$
#
2
8 ‰‘
12
œ 3 ˆ4
16 ‰
12
œ 12 4 œ 8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.6 Substitution and Area Between Curves
63. a œ 0, b œ 1; f(x) g(x) œ 2 sin x sin 2x
Ê Aœ
'
1
0
(2 sin x sin 2x) dx œ 2 cos x
cos 2x ‘ 1
2
!
œ 2(1) "# ‘ ˆ2 † 1 "# ‰ œ 4
64. a œ 13 , b œ 13 ; f(x) g(x) œ 8 cos x sec# x
Ê Aœ
œ Š8 †
'c
1Î3
1Î3
È3
#
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x] 1Î$
È3
#
È3‹ Š8 †
È3‹ œ 6È3
‰
65. a œ 1, b œ 1; f(x) g(x) œ a1 x# b cos ˆ 1x
#
Ê Aœ
œ ˆ1
"
3
'c 1 x# cos ˆ 1#x ‰‘ dx œ ’x x3
1
$
1
12 ‰ ˆ1
"
3
12 ‰ œ 2 ˆ 23 12 ‰ œ
2
1
4
3
sin ˆ 1#x ‰“
"
"
4
1
66. A œ A1 A2
a" œ 1, b" œ 0 and a# œ 0, b# œ 1;
f" (x) g" (x) œ x sin ˆ 1#x ‰ and f# (x) g# (x) œ sin ˆ 1#x ‰ x
Ê by symmetry about the origin,
A" A# œ 2A" Ê A œ 2
œ 2 ’ 12 cos ˆ 1#x ‰
"
x#
# “!
œ 2 ˆ 12 "# ‰ œ 2 ˆ 4211 ‰ œ
'
1
0
sin ˆ 1x
‰
‘
# x dx
œ 2 ˆ 12 † 0 "# ‰ ˆ 12 † 1 0‰‘
4 1
1
67. a œ 14 , b œ 14 ; f(x) g(x) œ sec# x tan# x
Ê Aœ
œ
'c
1Î4
1Î4
'c
1Î4
1Î4
asec# x tan# xb dx
csec# x asec# x 1bd dx
1Î4
1Î%
œ ' 1 † dx œ [x]1Î% œ
1Î4
1
4
ˆ 14 ‰ œ
1
#
68. c œ 14 , d œ 14 ; f(y) g(y) œ tan# y a tan# yb œ 2 tan# y
œ 2 asec# y 1b Ê A œ
'c
1Î4
1Î4
2 asec# y 1b dy
1Î%
œ 2[tan y y]1Î% œ 2 ˆ1 14 ‰ ˆ1 14 ‰‘
œ 4 ˆ1 14 ‰ œ 4 1
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339
340
Chapter 5 Integration
69. c œ 0, d œ 1# ; f(y) g(y) œ 3 sin yÈcos y 0 œ 3 sin yÈcos y
Ê Aœ3
'
1Î2
0
1Î#
sin yÈcos y dy œ 3 23 (cos y)$Î# ‘ !
œ 2(0 1) œ 2
"Î$
‰
70. a œ 1, b œ 1; f(x) g(x) œ sec# ˆ 1x
3 x
Ê Aœ
œ
Š 13
'c sec# ˆ 13x ‰ x"Î$ ‘ dx œ 13 tan ˆ 13x ‰ 43 x%Î$ ‘ ""
1
1
È3 3 ‹ ’ 3 ŠÈ3‹ 3 “ œ
4
1
4
6È 3
1
71. A œ A" A#
Limits of integration: x œ y$ and x œ y Ê y œ y$
Ê y$ y œ 0 Ê y(y 1)(y 1) œ 0 Ê c" œ 1, d" œ 0
and c# œ 0, d# œ 1; f" (y) g" (y) œ y$ y and
f# (y) g# (y) œ y y$ Ê by symmetry about the origin,
A" A# œ 2A# Ê A œ 2
œ 2 ˆ "# 4" ‰ œ
'
1
0
#
ay y$ b dy œ 2 ’ y#
"
#
"
y%
4 “!
72. A œ A" A#
Limits of integration: y œ x$ and y œ x& Ê x$ œ x&
Ê x& x$ œ 0 Ê x$ (x 1)(x 1) œ 0 Ê a" œ 1, b" œ 0
and a# œ 0, b# œ 1; f" (x) g" (x) œ x$ x& and
f# (x) g# (x) œ x& x$ Ê by symmetry about the origin,
A" A# œ 2A# Ê A œ 2
œ 2 ˆ "4 6" ‰ œ
'
0
1
%
ax$ x& b dx œ 2 ’ x4
"
6
73. A œ A" A#
Limits of integration: y œ x and y œ
$
"
x#
Ê xœ
"
x# ,
"
x'
6 “!
xÁ0
Ê x œ 1 Ê x œ 1 , f" (x) g" (x) œ x 0 œ x
Ê A" œ
'
0
1
#
"
x dx œ ’ x2 “ œ "# ; f# (x) g# (x) œ
œ x# Ê A# œ
A œ A" A# œ
'
"
#
!
2
1
"
x#
0
#
"
"
‘
x# dx œ "
x " œ # 1 œ #;
"
#
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.6 Substitution and Area Between Curves
74. Limits of integration: sin x œ cos x Ê x œ
1
4;
and b œ
œŠ
È2
#
1Î4
0
1Î%
(cos x sin x) dx œ [sin x cos x]!
È2
# ‹
Ê aœ0
f(x) g(x) œ cos x sin x
'
Ê Aœ
1
4
341
(0 1) œ È2 1
75. (a) The coordinates of the points of intersection of the
line and parabola are c œ x# Ê x œ „ Èc and y œ c
(b) f(y) g(y) œ Èy ˆÈy‰ œ 2Èy Ê the area of the
lower section is, AL œ
œ2
'
c
0
'
0
c
[f(y) g(y)] dy
Èy dy œ 2 23 y$Î# ‘ ! œ
c
c$Î# . The area of the
4
3
entire shaded region can be found by setting c œ 4: A œ ˆ 43 ‰ 4$Î# œ 43†8 œ 32
3 . Since we want c to divide the region
32
4
$Î#
into subsections of equal area we have A œ 2AL Ê 3 œ 2 ˆ 3 c ‰ Ê c œ 4#Î$
(c) f(x) g(x) œ c x# Ê AL œ
œ
4
3
Èc
Èc
c
c
'cÈ [f(x) g(x)] dx œ 'cÈ ac x# b dx œ ’cx x3 “ È
$
c
cÈc
œ 2 ’c$Î#
c$Î# . Again, the area of the whole shaded region can be found by setting c œ 4 Ê A œ
condition A œ 2AL , we get
4
3
$Î#
c
œ
#Î$
Ê cœ4
32
3
32
3 .
c$Î#
3 “
From the
as in part (b).
76. (a) Limits of integration: y œ 3 x# and y œ 1
Ê 3 x# œ 1 Ê x# œ 4 Ê a œ 2 and b œ 2;
f(x) g(x) œ a3 x# b (1) œ 4 x#
Ê Aœ
'c a4 x# b dx œ ’4x x3 “ #
2
$
#
32
3
1
œ ˆ8 83 ‰ ˆ8 83 ‰ œ 16
16
3
œ
(b) Limits of integration: let x œ 0 in y œ 3 x#
Ê y œ 3; f(y) g(y) œ È3 y ˆÈ3 y‰
œ 2(3 y)"Î#
Ê Aœ2
'c (3 y)"Î# dy œ 2 'c (3 y)"Î# (1) dy œ (2) ’ 2(3 3y)
œ ˆ 43 ‰ (8) œ
3
3
1
1
“
$
"
œ ˆ 43 ‰ 0 (3 1)$Î# ‘
32
3
77. Limits of integration: y œ 1 Èx and y œ
Ê 1 Èx œ
$Î#
2
Èx
2
Èx
, x Á 0 Ê Èx x œ 2 Ê x œ (2 x)#
Ê x œ 4 4x x# Ê x# 5x 4 œ 0
Ê (x 4)(x 1) œ 0 Ê x œ 1, 4 (but x œ 4 does not
satisfy the equation); y œ È2x and y œ x4 Ê È2x œ x4
Ê 8 œ xÈx Ê 64 œ x$ Ê x œ 4.
Therefore, AREA œ A" A# : f" (x) g" (x) œ ˆ1 x"Î# ‰
Ê A" œ
'
œ ˆ1
"8 ‰ 0 œ
2
3
0
œ ˆ4 † 2
1
ˆ1 x"Î# x4 ‰ dx œ ’x 23 x$Î#
16 ‰
8
37
24 ; f# (x)
"
x#
8 “!
g# (x) œ 2x"Î#
ˆ4 "8 ‰ œ 4
15
8
œ
17
8 ;
x
4
x
4
Ê A# œ
'
1
4
ˆ2x"Î# 4x ‰ dx œ ’4x"Î#
Therefore, AREA œ A" A# œ
37
24
17
8
œ
3751
24
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
88
24
%
x#
8 “"
œ
11
3
342
Chapter 5 Integration
78. Limits of integration: (y 1)# œ 3 y Ê y# 2y 1
œ 3 y Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0
Ê y œ 2 since y 0; also, 2Èy œ 3 y
Ê 4y œ 9 6y y# Ê y# 10y 9 œ 0
Ê (y 9)(y 1) œ 0 Ê y œ 1 since y œ 9 does not
satisfy the equation;
AREA œ A" A#
f" (y) g" (y) œ 2Èy 0 œ 2y"Î#
Ê A" œ 2
Ê A# œ
'
1
0
'
2
1
"
$Î#
y"Î# dy œ 2 ’ 2y3 “ œ 43 ; f# (y) g# (y) œ (3 y) (y 1)#
!
#
c3 y (y 1) d dy œ 3y "# y# "3 (y 1)$ ‘ " œ ˆ6 2 3" ‰ ˆ3
#
Therefore, A" A# œ
4
3
7
6
œ
œ
15
6
"
#
80. A œ
'
b
a
2f(x) dx
'
a
b
'
0
a
'
b
a
a$
3‹
aa# x# b dx œ 2 a# x "3 x$ ‘ ! œ 2 Ša$
a
a$
$
Š 4a3 ‹
(2a) aa# b œ a$ ; limit of ratio œ lim b
aÄ!
f(x) dx œ 2
0‰ œ 1
"
3
"
#
œ 67 ;
5
2
79. Area between parabola and y œ a# : A œ 2
Area of triangle AOC:
"
#
f(x) dx
'
b
a
f(x) dx œ
'
b
a
œ
3
4
0œ
4a$
3 ;
which is independent of a.
f(x) dx œ 4
81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the
region's upper and lower bounding curves at x œ 0. The area of the shaded region is actually
Aœ
'c [x (x)] dx '
0
1
0
1
[x (x)] dx œ
'c 2x dx '
0
1
0
1
2x dx œ #.
82. It is sometimes true. It is true if f(x) g(x) for all x between a and b. Otherwise it is false. If the graph of f
lies below the graph of g for a portion of the interval of integration, the integral over that portion will be
negative and the integral over [aß b] will be less than the area between the curves (see Exercise 53).
83. Let u œ 2x Ê du œ 2 dx Ê
'
3
sin 2x
x
1
dx œ
'
6
2
sin u
ˆ u# ‰
"
#
ˆ "# du‰ œ
du œ dx; x œ 1 Ê u œ 2, x œ 3 Ê u œ 6
'
6
du œ cF(u)d '# œ F(6) F(2)
sin u
u
2
84. Let u œ 1 x Ê du œ dx Ê du œ dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 0
'
0
1
f(1 x) dx œ
'
1
0
f(u) ( du) œ
'
0
1
f(u) du œ
'
1
f(u) du œ
0
'
0
1
f(x) dx
85. (a) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
f odd Ê f(x) œ f(x). Then
'c f(x) dx œ '
0
1
0
1
f(u) ( du) œ
'
1
œ 3
(b) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
f even Ê f(x) œ f(x). Then
'c f(x) dx œ '
0
1
1
0
0
f(u) ( du) œ
'
f(u) ( du) œ
1
0
f(u) du œ
'
0
1
'
1
0
f(u) du œ
'
0
1
f(u) du
f(u) du œ 3
'c f(x) dx when f is odd. Let u œ x Ê du œ dx Ê du œ dx and x œ a Ê u œ a and x œ !
Ê u œ !. Thus ' f(x) dx œ ' f(u) du œ ' f(u) du œ ' f(u) du œ ' f(x) dx.
c
Thus ' f(x) dx œ ' f(x) dx ' f(x) dx œ ' f(x) dx ' f(x) dx œ !.
c
c
0
86. (a) Consider
a
0
0
a
0
a
a
0
a
a
a
a
0
a
0
a
0
a
0
a
0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 5.6 Substitution and Area Between Curves
'c
1/2
(b)
1/2
1Î#
sin x dx œ [cos x]1Î# œ cos ˆ 1# ‰ cos ˆ 1# ‰ œ ! ! œ !.
87. Let u œ a x Ê du œ dx; x œ 0 Ê u œ a, x œ a Ê u œ 0
Iœ
'
' f(u)f(af(au) duu) œ ' f(x)f(af(ax) dxx)
dx
f(x)f(ax)
' f(x)f(af(ax) dxx) œ ' f(x)
' dx œ [x]! œ a 0 œ a.
I I œ ' f(x)f(x)
f(ax)
f(ax) dx œ
a
f(x) dx
0 f(x)f(ax)
œ
'
0
f(au)
a f(au)f(u)
a
Ê
a
0
0
a
a
a
a
0
0
Therefore, 2I œ a Ê I œ
88. Let u œ
'
a
( du) œ
xy
x
"
t
xy
t
dt œ
a
#
0
.
t
Ê du œ xy
t# dt Ê xy du œ
'
y
1
u" du œ
0
'
y
1
"
u
du œ
'
y
1
"
t
"
u
dt Ê u" du œ
du œ
'
1
y
"
t
"
t
dt; t œ x Ê u œ y, t œ xy Ê u œ 1. Therefore,
dt
89. Let u œ x c Ê du œ dx; x œ a c Ê u œ a, x œ b c Ê u œ b
' cc
b c
a c
90. (a)
f(x c) dx œ
'
a
b
f(u) du œ
'
a
b
f(x) dx
(b)
(c)
91-94. Example CAS commands:
Maple:
f := x -> x^3/3-x^2/2-2*x+1/3;
g := x -> x-1;
plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" );
q1 := [ -5, -2, 1, 4 ];
# (b)
q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )];
for i from 1 to nops(q2)-1 do
# (c)
area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] );
end do;
add( area[i], i=1..nops(q2)-1 );
# (d)
Mathematica: (assigned functions may vary)
Clear[x, f, g]
f[x_] = x2 Cos[x]
g[x_] = x3 x
Plot[{f[x], g[x]}, {x, 2, 2}]
After examining the plots, the initial guesses for FindRoot can be determined.
pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {1, 0, 1}]
i1=NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}]
i2=NIntegrate[f[x] g[x], {x, pts[[2]], pts[[3]]}]
i1 i2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
343
344
Chapter 5 Integration
CHAPTER 5 PRACTICE EXERCISES
1. (a) Each time subinterval is of length ?t œ 0.4 sec. The distance traveled over each subinterval, using the
midpoint rule, is ?h œ "# avi vib1 b ?t, where vi is the velocity at the left endpoint and vib1 the velocity at
the right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to
arrive at the height associated with velocity vib1 at the right endpoint. Using this methodology we build
the following table based on the figure in the text:
t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6
6.0
v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90
76
65
h (ft)
0 2
9
25 56 114 188 257 320 378 432 481 525 564 592 620.2
t (sec)
v (fps)
h (ft)
6.4
50
643.2
6.8
37
660.6
7.2
25
672
7.6
12
679.4
8.0
0
681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph.
Remember that some shifting of the graph occurs in the printing process.
The total height attained is about 680 ft.
(b) The graph is based on the table in part (a).
2. (a) Each time subinterval is of length ?t œ 1 sec. The distance traveled over each subinterval, using the
midpoint rule, is ?s œ "# avi vib1 b ?t, where vi is the velocity at the left, and vib1 the velocity at the
right, endpoint of the subinterval. We then add ?s to the distance attained so far at the left endpoint vi
to arrive at the distance associated with velocity vib1 at the right endpoint. Using this methodology we
build the table given below based on the figure in the text, obtaining approximately 26 m for the total
distance traveled:
t (sec)
0
1
2
3
4
5
6
7
8
9
10
v (m/sec)
0
0.5
1.2
2
3.4
4.5
4.8
4.5
3.5
2
0
s (m)
0
0.25
1.1
2.7
5.4
9.35
14
18.65 22.65 25.4 26.4
(b) The graph shows the distance traveled by the
moving body as a function of time for
0 Ÿ t Ÿ 10.
3. (a)
(c)
10
!
kœ1
10
ak
4
œ
"
4
10
! ak œ
kœ1
"
4
(2) œ #"
(b)
10
10
10
kœ1
kœ1
kœ1
10
10
10
kœ1
kœ1
kœ1
! (bk 3ak ) œ ! bk 3 ! ak œ 25 3(2) œ 31
! (ak bk 1) œ ! ak ! bk ! " œ 2 25 (1)(10) œ 13
kœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Practice Exercises
10
10
kœ1
kœ1
! ˆ 5 bk ‰ œ !
#
(d)
20
5
#
10
! bk œ
kœ1
20
kœ1
20
(b)
kœ1
! ˆ"
#
(c)
kœ1
20
2bk ‰
7
20
œ !
kœ1
20
"
#
20
! bk œ
2
7
kœ1
20
"
#
20
20
20
kœ1
kœ1
kœ1
! (ak bk ) œ ! ak ! bk œ 0 7 œ 7
(20) 27 (7) œ 8
! aak 2b œ ! ak ! 2 œ 0 2(20) œ 40
(d)
kœ1
kœ1
kœ1
"
#
5. Let u œ 2x 1 Ê du œ 2 dx Ê
5
1
'
(2x 1)"Î# dx œ
9
1
3
1
x ax# 1b
7. Let u œ
"Î$
'
dx œ
8
0
du œ dx; x œ 1 Ê u œ 1, x œ 5 Ê u œ 9
*
u"Î# ˆ "# du‰ œ u"Î# ‘ " œ 3 1 œ 2
6. Let u œ x# 1 Ê du œ 2x dx Ê
'
(10) 25 œ 0
! 3ak œ 3 ! ak œ 3(0) œ 0
4. (a)
'
5
#
"
#
du œ x dx; x œ 1 Ê u œ 0, x œ 3 Ê u œ 8
)
u"Î$ ˆ "# du‰ œ 38 u%Î$ ‘ ! œ
3
8
(16 0) œ 6
Ê 2 du œ dx; x œ 1 Ê u œ 1# , x œ 0 Ê u œ 0
x
2
' cos ˆ x# ‰ dx œ ' Î (cos u)(2 du) œ [2 sin u]!1Î# œ 2 sin 0 2 sin ˆ 1# ‰ œ 2(0 (1)) œ 2
0
0
1
1 2
8. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
(sin x)(cos x) dx œ
(e)
10. (a)
(c)
(e)
#
u du œ ’ u2 “ œ
!
Ê uœ1
"
#
2
2
5
2
2
(c)
0
"
'c f(x) dx œ "3 'c 3 f(x) dx œ 3" (12) œ 4
(b) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 4 œ 2
c
c
c
' g(x) dx œ 'c g(x) dx œ 2
(d) ' (1 g(x)) dx œ 1 ' g(x) dx œ 1(2) œ 21
c
c
'c Š f(x) 5 g(x) ‹ dx œ 5" 'c f(x) dx 5" 'c g(x) dx œ 5" (6) 5" (2) œ 85
2
9. (a)
'
1
1
#
'
'
'
2
5
5
2
5
5
5
2
2
2
2
0
0
2
' 7 g(x) dx œ "7 (7) œ 1
f(x) dx œ ' f(x) dx œ 1
[g(x) 3 f(x)] dx œ ' g(x) dx 3'
g(x) dx œ
2
"
7
(b)
0
2
(d)
0
2
0
5
2
0
0
2
'
'
2
2
2
5
5
2
2
2
1
2
0
g(x) dx œ
'
0
2
g(x) dx
È2 f(x) dx œ È2
'
0
2
'
0
1
g(x) dx œ 1 2 œ 1
f(x) dx œ È2 (1) œ 1È2
f(x) dx œ 1 31
11. x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1;
Area œ
'
0
1
ax# 4x 3b dx
"
$
'
1
$
3
ax# 4x 3b dx
œ ’ x3 2x# 3x“ ’ x3 2x# 3x“
!
$
"
$
œ ’Š "3 2(1)# 3(1)‹ 0“
$
$
’Š 33 2(3)# 3(3)‹ Š 13 2(1)# 3(1)‹“
œ ˆ "3 1‰ 0 ˆ 3" 1‰‘ œ
8
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
345
346
Chapter 5 Integration
12. 1
x#
4
œ 0 Ê 4 x# 0 Ê x œ „ 2;
Area œ
'c Š1 x4 ‹ dx '
2
2
œ ’x
2
#
x$
12 “ #
2$
12 ‹
œ ’Š2
3
#
’x
Š1
x#
4‹
dx
$
x$
12 “ #
Š2
(2)$
12 ‹“
œ 43 ˆ 43 ‰‘ ˆ 34 43 ‰ œ
’Š3
3$
12 ‹
2$
12 ‹“
Š2
13
4
13. 5 5x#Î$ œ 0 Ê 1 x#Î$ œ 0 Ê x œ „ 1;
Area œ
'c ˆ5 5x#Î$ ‰ dx '
1
1
1
8
ˆ5 5x#Î$ ‰ dx
"
)
œ 5x 3x&Î$ ‘ " 5x 3x&Î$ ‘ "
œ ˆ5(1) 3(1)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
ˆ5(8) 3(8)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
œ [2 (2)] [(40 96) 2] œ 62
14. 1 Èx œ 0 Ê x œ 1;
Area œ
œ
œ
œ
'
0
1
ˆ1 Èx‰ dx
4
1
ˆ1 Èx‰ dx
x 23 x$Î# ‘ " x 23 x$Î# ‘ %
!
"
ˆ1 23 (1)$Î# ‰ 0‘ ˆ4 23
"
ˆ4 16
‰ "‘
3
3 3 œ 2
15. f(x) œ x, g(x) œ
œ
'
'
2
1
ˆx
"‰
x#
œ
'
œ
Š 42
1
Šx
a œ 1, b œ 2 Ê A œ
'
b
[f(x) g(x)] dx
a
#
#
dx œ ’ x# x" “ œ ˆ 4# "# ‰ ˆ "# 1‰ œ 1
16. f(x) œ x, g(x) œ
2
"
x# ,
(4)$Î# ‰ ˆ1 23 (1)$Î# ‰‘
"
"
Èx
"
Èx ‹dx
, a œ 1, b œ 2 Ê A œ
#
œ ’ x# 2Èx“
2È2‹ ˆ "# 2‰ œ
'
a
b
[f(x) g(x)] dx
#
"
7 4 È 2
#
'
#
17. f(x) œ ˆ1 Èx‰ , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ
'
0
1
ˆ1 2x"Î# x‰ dx œ ’x 43 x$Î#
#
"
x#
# “!
x%
#
"
x(
7 “!
œ1
"
#
"
7
œ
a
œ1
18. f(x) œ a1 x$ b , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ ’x
b
'
a
b
'
[f(x) g(x)] dx œ
4
3
"
#
œ
"
6
1
0
ˆ1 Èx‰# dx œ
(6 8 3) œ
[f(x) g(x)] dx œ
'
0
1
'
0
1
ˆ1 2Èx x‰ dx
"
6
#
a1 x$ b dx œ
'
0
9
14
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
a1 2x$ x' b dx
Chapter 5 Practice Exercises
19. f(y) œ 2y# , g(y) œ 0, c œ 0, d œ 3
Ê Aœ
œ2
'
3
'
d
c
y# dy œ
0
'
[f(y) g(y)] dy œ
2
3
3
0
a2y# 0b dy
$
cy$ d ! œ 18
20. f(y) œ 4 y# , g(y) œ 0, c œ 2, d œ 2
Ê Aœ
'
d
c
#
y$
3 “ #
œ ’4y
'c a4 y# b dy
2
[f(y) g(y)] dy œ
œ 2 ˆ8
8‰
3
2
œ
32
3
y#
4
21. Let us find the intersection points:
y 2
4
œ
Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0 Ê y œ 1
or y œ 2 Ê c œ 1, d œ 2; f(y) œ
Ê Aœ
'
d
c
y 2
4
2
#
1
œ
"
4
'c ay 2 y# b dy œ 4" ’ y#
œ
"
4
ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰‘ œ
#
1
y#
4
'c Š y4 2 y4 ‹ dy
[f(y) g(y)] dy œ
2
, g(y) œ
2y
9
8
y# 4
4
22. Let us find the intersection points:
#
y$
3 “ "
œ
y 16
4
Ê y# y 20 œ 0 Ê (y 5)(y 4) œ 0 Ê y œ 4
or y œ 5 Ê c œ 4, d œ 5; f(y) œ
Ê Aœ
'
d
c
[f(y) g(y)] dy œ
y 16
4
, g(y) œ
y# 4
4
'c Š y 416 y 4 4 ‹ dy
5
#
4
œ
"
4
'c ay 20 y# b dy œ "4 ’ y#
œ
"
4
"
4
125 ‰
ˆ 25
‰‘
ˆ "#6 80 64
# 100 3
3
9
"
9
"
ˆ # 180 63‰ œ 4 ˆ # 117‰ œ 8 (9 234) œ
œ
5
#
20y
4
23. f(x) œ x, g(x) œ sin x, a œ 0, b œ
Ê Aœ
'
b
a
#
[f(x) g(x)] dx œ
œ ’ x# cos x“
1Î%
!
#
œ Š 31#
'
1Î4
(x sin x) dx
1
24. f(x) œ 1, g(x) œ ksin xk , a œ 1# , b œ
Ê Aœ
œ
'c
œ2
0
'
a
b
[f(x) g(x)] dx œ
(1 sin x) dx
1Î2
1Î2
'
0
'
0
1Î2
243
8
1
4
0
È2
# ‹
&
y$
3 “ %
'c
1Î2
1Î2
1
2
a1 ksin xkb dx
(1 sin x) dx
1Î#
(1 sin x) dx œ 2[x cos x]!
œ 2 ˆ 1# 1‰ œ 1 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
347
348
Chapter 5 Integration
25. a œ 0, b œ 1, f(x) g(x) œ 2 sin x sin 2x
'
Ê Aœ
1
0
(2 sin x sin 2x) dx œ 2 cos x
cos 2x ‘ 1
#
!
œ 2 † (1) "# ‘ ˆ2 † 1 "# ‰ œ 4
26. a œ 13 , b œ 13 , f(x) g(x) œ 8 cos x sec# x
'c
1Î3
Ê Aœ
œ Š8 †
1Î3
È3
#
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x]1Î$
È3‹ Š8 †
È3
#
È3‹ œ 6È3
27. f(y) œ Èy, g(y) œ 2 y, c œ 1, d œ 2
'
Ê Aœ
œ
'
1
2
d
c
[f(y) g(y)] dy œ
'
2
1
Èy (2 y)‘ dy
ˆÈy 2 y‰ dy œ ’ 23 y$Î# 2y
œ Š 43 È2 4 2‹ ˆ 23 2 "# ‰ œ
4
3
#
y#
# “"
È2
7
6
œ
8 È 2 7
6
28. f(y) œ 6 y, g(y) œ y# , c œ 1, d œ 2
Ê Aœ
'
œ ’6y
y#
#
œ4
c
7
3
d
[f(y) g(y)] dy œ
"
#
#
y$
3 “"
œ
'
2
1
a6 y y# b dy
œ ˆ12 2 83 ‰ ˆ6
24143
6
œ
"
#
3" ‰
13
6
29. f(x) œ x$ 3x# œ x# (x 3) Ê f w (x) œ 3x# 6x œ 3x(x 2) Ê f w œ ± ±
!
#
Ê f(0) œ 0 is a maximum and f(2) œ 4 is a minimum. A œ
‰
œ ˆ 81
4 27 œ
30. A œ
'
a
0
4
3
a#
6
"# ‰ œ
'
a
&Î$
A# œ
"
y#
# “!
œ
0
(6 8 3) œ
"
10
'
0
1
a
x#
# “0
$
!
œ a# 34 Èa † aÈa
a#
6
ˆy#Î$ y‰ dy
; the area below the x-axis is
'c ˆy#Î$ y‰ dy œ ’ 3y5
0
%
ax$ 3x# b dx œ ’ x4 x$ “
ˆa 2Èa x"Î# x‰ dx œ ’ax 43 Èa x$Î#
31. The area above the x-axis is A" œ
œ ’ 3y5
0
3
27
4
ˆa"Î# x"Î# ‰# dx œ
œ a# ˆ1
'
&Î$
1
Ê the total area is A" A# œ
!
y#
# “ "
œ
11
10
6
5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
a#
#
Chapter 5 Practice Exercises
'
32. A œ
1Î4
0
'
31Î2
51Î4
(cos x sin x) dx
'
51Î4
1Î4
(sin x cos x) dx
1Î%
(cos x sin x) dx œ [sin x cos x]!
&1Î%
$1Î#
[ cos x sin x]1Î% [sin x cos x]&1Î%
œ ’Š
È2
#
È2
# ‹
(0 1)“ ’Š
’(1 0) Š
33. y œ x#
34. y œ
'
x
0
'
x
"
1 t
È2
#
dt Ê
dy
dx
È2
# ‹“
xœ0 Ê yœ
'
0
0
È2
# ‹
È2
#
Š
œ
8È 2
#
2 œ 4È2 2
"
x
Ê
d# y
dx#
œ 2x
ˆ1 2Èsec t‰ dt Ê
È2
#
œ2
"
x#
; y(1) œ 1
œ 1 2Èsec x Ê
dy
dx
d# y
dx#
ˆ1 2Èsec t‰ dt œ 0 and x œ 0 Ê
dy
dx
36. y œ
'c È2 sin# t dt 2 so that dydx œ È2 sin# x; x œ 1
sin t
t
dt 3 Ê
dy
dx
œ
;xœ5 Ê yœ
sin x
x
5
5
sin t
t
1
1
"
t
dt œ 1 and yw (1) œ 2 1 œ 3
œ 1 2Èsec 0 œ 3
'
x
'
œ 2 ˆ "# ‰ (sec x)"Î# (sec x tan x) œ Èsec x (tan x);
35. y œ
5
'
È2
# ‹“
dt 3 œ 3
x
Ê yœ
1
'cc È2 sin# t dt 2 œ 2
1
1
37. Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx
' 2(cos x)"Î# sin x dx œ ' 2u"Î# ( du) œ 2 ' u"Î# du œ 2 Š u"Î#" ‹ C œ 4u"Î# C
#
œ 4(cos x)"Î# C
38. Let u œ tan x Ê du œ sec# x dx
' (tan x)$Î# sec# x dx œ ' u$Î# du œ ˆu"Î#"‰ C œ 2u"Î# C œ (tanx)2 "Î# C
#
39. Let u œ 2) 1 Ê du œ 2 d) Ê
"
#
du œ d)
' [2) 1 2 cos (2) 1)] d) œ ' (u 2 cos u) ˆ "# du‰ œ u4
#
œ )# ) sin (2) 1) C, where C œ C"
40. Let u œ 2) 1 Ê du œ 2 d) Ê
'Š
œ
41.
42.
"
#
"
È 2 ) 1
"
#
2 sec# (#) 1)‹ d) œ
"
4
sin u C" œ
(2)1)#
4
sin (2) 1) C"
is still an arbitrary constant
du œ d)
' Š È"u 2 sec# u‹ ˆ #" du‰ œ #" ' ˆu"Î# 2 sec# u‰ du
"Î#
Š u " ‹ "# (2 tan u) C œ u"Î# tan u C œ (2) 1)"Î# tan (2) 1) C
#
' ˆt 2t ‰ ˆt 2t ‰ dt œ ' ˆt# t4 ‰ dt œ ' at# 4t# b dt œ t3$ 4 Š t"1 ‹ C œ t3$ 4t C
#
t#
' (t1)t%#1 dt œ ' t#t%2t dt œ ' ˆ t"# t2$ ‰ dt œ ' at# 2t$ b dt œ (t"1) 2 Š #
‹ C œ "t t"# C
43. Let u œ #t$Î# Ê du œ $Èt dt Ê "$ du œ Èt dt
' Èt sin ˆ#t$Î# ‰dt œ "$ ' sin u du œ "$ cos u C œ "$ cosˆ#t$Î# ‰ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
349
350
Chapter 5 Integration
44. Let u œ " sec ) Ê du œ sec ) tan) d) Ê
' sec ) tan) È" sec ) d) œ ' u"Î# du œ #$ u$Î# C
œ #$ a" sec )b$Î# C
'c a3x# 4x 7b dx œ cx$ 2x# 7xd "" œ c1$ 2(1)# 7(1)d c(1)$ 2(1)# 7(1)d œ 6 (10) œ 16
1
45.
1
46.
'
47.
'
48.
'
49.
'
1
0
"
a8s$ 12s# 5b ds œ c2s% 4s$ 5sd ! œ c2(1)% 4(1)$ 5(1)d 0 œ 3
2
4
#
1 v
27
1
'
dv œ
2
#
4v# dv œ c4v" d " œ ˆ #4 ‰ ˆ 14 ‰ œ 2
1
#(
x%Î$ dx œ 3x"Î$ ‘ " œ 3(27)"Î$ ˆ3(1)"Î$ ‰ œ 3 ˆ "3 ‰ 3(1) œ 2
4
dt
1 tÈt
œ
'
4
œ
dt
t$Î#
1
'
4
1
%
50. Let x œ 1 Èu Ê dx œ
'
4
ˆ1 Èu‰"Î#
Èu
1
du œ
'
3
2
2
È4
t$Î# dt œ 2t"Î# ‘ " œ
"
#
u"Î# du Ê 2 dx œ
du
Èu
(2)
È1
œ1
; u œ 1 Ê x œ 2, u œ 4 Ê x œ 3
$
x"Î# (2 dx) œ 2 ˆ 23 ‰ x$Î# ‘ # œ
4
3
ˆ3$Î# ‰ 43 ˆ2$Î# ‰ œ 4È3 83 È2 œ
4
3
Š3È3 2È2‹
51. Let u œ 2x 1 Ê du œ 2 dx Ê 18 du œ 36 dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 3
'
1
36 dx
$
0 (2x1)
œ
'
3
# $
$
9 ‘
ˆ 9 ‰ ˆ 9 ‰
18u$ du œ ’ "8u
2 “ œ u # " œ 3 # 1 # œ 8
"
1
52. Let u œ 7 5r Ê du œ 5 dr Ê "5 du œ dr; r œ 0 Ê u œ 7, r œ 1 Ê u œ 2
'
1
dr
$
È
(7 5r)#
0
œ
'
'
1
(7 5r)#Î$ dr œ
0
2
7
#
u#Î$ ˆ 5" du‰ œ 5" 3u"Î$ ‘ ( œ
53. Let u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 3# du œ x"Î$ dx; x œ
x œ 1 Ê u œ 1 1#Î$ œ 0
'
1
1Î8
œ
x"Î$ ˆ1 x#Î$ ‰
$Î#
'
dx œ
0
1Î2
0
x$ a1 9x% b
" ˆ 25 ‰
œ 18
16
"Î#
$Î#
dx œ
'
0
1
sin# 5r dr œ
56. Let u œ 4t
'
1Î%
0
œ
1
8
1
4
$Î%
"
36
#Î$
œ
3
4
,
!
&Î#
œ 35 u&Î# ‘ $Î% œ 35 (0)&Î# ˆ 35 ‰ ˆ 34 ‰
'
51
0
"
16
"
16
"
5
"Î#
#
1
8
' Î acos
1 4
"
%
Ê u œ 1 9 ˆ #" ‰ œ
25
16
#&Î"'
" "Î# ‘
œ 18
u
"
du œ dr; r œ 0 Ê u œ 0, r œ 1 Ê u œ 51
Ê du œ 4 dt Ê
œ
#&Î"'
"
#
"
90
asin# ub ˆ "5 du‰ œ
31Î4
du œ x$ dx; x œ 0 Ê u œ 1, x œ
"
"
u$Î# ˆ 36
du‰ œ ’ 36
Š u " ‹“
"
ˆ 18
(1)"Î# ‰ œ
cos# ˆ4t 14 ‰ dt œ
25Î16
1
55. Let u œ 5r Ê du œ 5 dr Ê
'
!
#
Ê u œ 1 ˆ 8" ‰
27È3
160
54. Let u œ 1 9x% Ê du œ 36x$ dx Ê
'
&Î#
u$Î# ˆ #3 du‰ œ ’ˆ 32 ‰ Š u 5 ‹“
3Î4
"
8
Š $È7 $È2‹
3
5
"
4
#
"
5
u2
sin 2u ‘ &1
4
!
œ ˆ 1#
sin 101 ‰
#0
du œ dt; t œ 0 Ê u œ 14 , t œ
ub ˆ "4 du‰ œ
"
4
u2
sin 2u ‘ $1Î%
4
1Î%
œ
ˆ0
1
4
"
4
sin 0 ‰
20
Ê uœ
Š 381
œ
1
#
31
4
sin ˆ 3#1 ‰
‹
4
4" Š 18
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
sin ˆ 1# ‰
‹
4
Chapter 5 Practice Exercises
'
1Î$
57.
'
31Î4
58.
0
1Î$
sec# ) d) œ [tan )]!
1Î4
31
1
1
3
tan 0 œ È3
$1Î%
csc# x dx œ [cot x]1Î% œ ˆ cot
59. Let u œ
'
œ tan
cot#
"
6
Ê du œ
x
6
dx œ
x
6
'
1Î6
31 ‰
4
ˆ cot 14 ‰ œ 2
dx Ê 6 du œ dx; x œ 1 Ê u œ 16 , x œ 31 Ê u œ
1Î2
351
'
6 cot# u du œ 6
1Î2
1
#
1Î#
acsc# u 1b du œ [6(cot u u)]1Î' œ 6 ˆ cot
1Î6
1
#
1# ‰ 6 ˆcot
1
6
16 ‰
œ 6È3 21
60. Let u œ
'
1
0
tan#
)
3
)
3
œ 3 tan
"
3
Ê du œ
d) œ
1
3
'
0
1
d) Ê 3 du œ d); ) œ 0 Ê u œ 0, ) œ 1 Ê u œ
)
3
ˆsec#
1‰ d) œ
'
1Î3
0
1
3
1Î$
3 asec# u 1b du œ [3 tan u 3u]!
3 ˆ 13 ‰‘ (3 tan 0 0) œ 3È3 1
'c
sec x tan x dx œ [sec x]!1Î$ œ sec 0 sec ˆ 13 ‰ œ 1 2 œ 1
'
csc z cot z dz œ [csc z]1Î% œ ˆ csc
0
61.
62.
1Î3
31Î4
1Î4
$1Î%
31 ‰
4
ˆ csc 14 ‰ œ È2 È2 œ 0
63. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
'
5(sin x)$Î# cos x dx œ
1
0
1
#
Ê uœ1
"
"
5u$Î# du œ 5 ˆ 25 ‰ u&Î# ‘ ! œ 2u&Î# ‘ ! œ 2(1)&Î# 2(0)&Î# œ 2
64. Let u œ 1 x# Ê du œ 2x dx Ê du œ 2x dx; x œ 1 Ê u œ 0, x œ 1 Ê u œ 0
'c
1
1
'
2x sin a1 x# b dx œ
0
sin u du œ 0
0
"
3
65. Let u œ sin 3x Ê du œ 3 cos 3x dx Ê
œ 1
'c ÎÎ
1 2
15 sin% 3x cos 3x dx œ
1 2
du œ cos 3x dx; x œ 1# Ê u œ sin ˆ 3#1 ‰ œ 1, x œ
1
#
&
&
' 15u% ˆ "3 du‰ œ ' 5u% du œ cu& d "
" œ (1) (1) œ 2
1
1
1
1
66. Let u œ cos ˆ x# ‰ Ê du œ "# sin ˆ x# ‰ dx Ê 2 du œ sin ˆ x# ‰ dx; x œ 0 Ê u œ cos ˆ 0# ‰ œ 1, x œ
œ
'
"
#
21Î3
cos% ˆ x# ‰ sin ˆ x# ‰ dx œ
0
'
1
1Î2
$
u% (2 du) œ ’2 Š u3 ‹“
67. Let u œ 1 3 sin# x Ê du œ 6 sin x cos x dx Ê
Ê u œ 1 3 sin#
'
1Î2
0
3 sin x cos x
È1 3 sin# x
1
#
œ4
dx œ
'
4
"
Èu
1
ˆ #" du‰ œ
'
4
1
"
#
'
0
1Î4
1
4
Ê u œ 1 7 tan
sec# x
(1 7 tan x)#Î$
dx œ
'
1
1
4
8
"
#
"Î#
"
œ
2
3
ˆ "# ‰$ 32 (1)$ œ
2
3
(8 1) œ
du œ 3 sin x cos x dx; x œ 0 Ê u œ 1, x œ
"Î#
%
#
"
21
3
21
Ê u œ cos Š #3 ‹
14
3
1
#
%
u"Î# du œ ’ 2" Š u " ‹“ œ u"Î# ‘ " œ 4"Î# 1"Î# œ 1
"
7
68. Let u œ 1 7 tan x Ê du œ 7 sec# x dx Ê
xœ
Ê u œ sin ˆ 3#1 ‰
du œ sec# x dx; x œ 0 Ê u œ 1 7 tan 0 œ 1,
œ8
"
u#Î$
ˆ 7" du‰ œ
'
1
8
"
7
"Î$
)
3
"
)
u#Î$ du œ ’ 7" Š u " ‹“ œ 37 u"Î$ ‘ " œ
3
7
(8)"Î$ 37 (1)"Î$ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3
7
352
Chapter 5 Integration
69. Let u œ sec ) Ê du œ sec ) tan ) d); ) œ 0 Ê u œ sec 0 œ 1, ) œ
'
1Î3
0
tan )
È2 sec )
"
È2
œ
'
d) œ
"Î#
#
#
"
'
1Î3
0
’ ˆu " ‰ “ œ ’
1Î3
sec ) tan )
sec ) tan )
d) œ
È2 (sec ))$Î#
sec ) È2 sec )
0
#
2
2
2
È2u “ œ È2(2) Š È2(1) ‹ œ
"
cos Èt
2È t
70. Let u œ sin Èt Ê du œ ˆcos Èt‰ ˆ "# t"Î# ‰ dt œ
#
1
4
tœ
'
1 Î4
#
Ê u œ sin
71. (a) av(f) œ
'
"
1 (1)
'c
"
k (k)
(b) av(f) œ
"
#k
œ
1
"
1Î2 Èu
1
1
'c
k
k
(2 du) œ 2
73. favw œ
'
a
b
"
È2 u$Î#
du œ
"
È2
'
1
2
œ2
u$Î# du
È2 1
dt Ê 2 du œ
cos Èt
Èt
dt; t œ
1#
36
Ê u œ sin
1
6
œ
"
#
,
"
"
"
#k
’ mx2 bx“
"
#
’ mx2 bx“
(2bk) œ b
"
30
1
1
3
u"Î# du œ 4Èu‘ "Î# œ 4È1 4É #" œ 2 Š2 È2‹
#
(mx b) dx œ
3
0
3
0
0
"
ba
1
1Î2
"
#
a
(b) yav
'
(mx b) dx œ
' È3x dx œ "3 '
œ a " 0 ' Èax dx œ "a '
72. (a) yav œ
'
Ê u œ sec
œ1
dt œ
cos Èt
Ét sin Èt
1# Î36
1
#
d) œ
2
1
3
È3 x"Î# dx œ
a
Èa x"Î# dx œ
0
"
b a
Èaxf w (x) dx œ
[f(x)]ab œ
"
b a
k
ck
#
#
m(1)
b(1)‹“ œ
’Š m(1)
2 b(1)‹ Š #
œ
"
#
œ
"
#k
#
"
#
(2b) œ b
#
m(k)
b(k)‹“
’Š m(k)
2 b(k)‹ Š #
È3
3
23 x$Î# ‘ $ œ
!
È3
3
23 (3)$Î# 23 (0)$Î# ‘ œ
È3
3
Š2È3‹ œ 2
Èa
a
23 x$Î# ‘ a œ
!
Èa
a
ˆ 23 (a)$Î# 23 (0)$Î# ‰ œ
Èa
a
ˆ 32 aÈa‰ œ
[f(b) f(a)] œ
f(b) f(a)
ba
2
3
a
so the average value of f w over [aß b] is the
slope of the secant line joining the points (aß f(a)) and (bß f(b)), which is the average rate of change of f over [aß b].
74. Yes, because the average value of f on [aß b] is
and the average value of the function is
"
#
'
a
"
ba
'
a
b
f(x) dx. If the length of the interval is 2, then b a œ 2
b
f(x) dx.
75. We want to evaluate
"
$'& !
'
$'&
!
f(x) dx œ
"
$'&
'
$'&
!
#1
Œ$(sin” $'& ax "!"b• #&dx œ
#1
Notice that the period of y œ sin” $'&
ax "!"b• is
length 365. Thus the value of
76.
"
'(&#!
œ
'#
'(&
!
$(
$'&
'
$'&
!
"
'&& Œ”)Þ#(a'(&b
#'a'(&b
#†"!&
#1
$'&
"Þ)(a'(&b
$†"!&
$
'
!
$'&
#1
sin” $'&
ax "!"b•dx
#&
$'&
'
$'&
!
dx
œ $'& and that we are integrating this function over an iterval of
#1
ax "!"b•dx
sin” $'&
a)Þ#( "!& a#'T "Þ)(T# bbdT œ
#
#1
$(
$'&
"
'&& ”)Þ#(T
• ”)Þ#(a#!b
#&
$'&
#'T#
#†"!&
#'a#!b
#†"!&
#
'
!
$'&
dx is
"Þ)(T$
$†"!& •
"Þ)(a#!b
$†"!&
$
$(
$'&
†!
#&
$'&
† $'& œ #&.
'(&
#!
• ¸
"
'&& a$(#%Þ%%
"'&Þ%!b
œ &Þ%$ œ the average value of Cv on [20, 675]. To find the temperature T at which Cv œ &Þ%$, solve
&Þ%$ œ )Þ#( "!& a#'T "Þ)(T# b for T. We obtain "Þ)(T# #'T #)%!!! œ !
ÊTœ
#' „ Éa#'b# %a"Þ)(ba#)%!!!b
È#"#%**'
œ #' „ $Þ(%
.
#a"Þ)(b
‰
So T œ $)#Þ)# or T œ $*'Þ(#. Only T œ $*'Þ(# lies in the
interval [20, 675], so T œ $*'Þ(# C.
77.
dy
dx
œ È# cos$ x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Practice Exercises
78.
dy
dx
œ È# cos$ a(x# b †
79.
dy
dx
œ
d
dx Œ
80.
dy
dx
œ
d
dx Œ
d
#
dx a(x b
œ "%xÈ# cos$ a(x# b
' x $ ' t dt œ $'x
%
1
353
%
'sec# x t " " dt œ dxd Œ'#sec x t " " dt œ sec "x " dxd asec xb œ sec" xsectan xx
#
#
#
#
81. Yes. The function f, being differentiable on [aß b], is then continuous on [aß b]. The Fundamental Theorem of
Calculus says that every continuous function on [aß b] is the derivative of a function on [aß b].
82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on
[aß b], then
'
0
1
'
b
a
f(x) dx œ F(b) F(a). In particular, if F(x) is an antiderivaitve of È1 x% on [0ß 1], then
È1 x% dx œ F(1) F(0).
83. y œ
' x È1 t# dt œ ' x È1 t# dt
84. y œ
'
1
1
0
"
#
cos x 1 t
dt œ
'
0
cos x
"
1 t#
dt Ê
Ê
dy
dx
œ
d
dx
dy
dx
œ
d
dx
”
' x È1 t# dt• œ dxd ”' x È1 t# dt• œ È1 x#
”
'
1
cos x
0
"
d
‰ ˆ dx
œ ˆ 1 cos
(cos x)‰ œ ˆ sin"# x ‰ ( sin x) œ
#x
1
"
1 t#
"
sin x
d
dt• œ dx
”
'
0
cos x
"
1 t#
dt•
œ csc x
85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each
interval by averaging the widths at top and bottom. This gives the estimate
A ¸ "& † ˆ ! # $' $' # &% &% # &" &" #%*Þ& %*Þ&# &% &% #'%Þ% '%Þ% # '(Þ& '(Þ&# %# ‰
A ¸ &*'" ft# . The cost is Area † ($2.10/ft# ) ¸ a5961 ft# b a$2.10/ft# b œ $12,518.10 Ê the job cannot be done for $11,000.
86. (a) Before the chute opens for A, a œ 32 ft/sec# . Since the helicopter is hovering, v! œ 0 ft/sec
Ê v œ ' 32 dt œ 32t v! œ 32t. Then s! œ 6400 ft Ê s œ ' 32t dt œ 16t# s! œ 16t# 6400.
At t œ 4 sec, s œ 16(4)# 6400 œ 6144 ft when A's chute opens;
(b) For B, s! œ 7000 ft, v! œ 0, a œ 32 ft/sec# Ê v œ ' 32 dt œ 32t v! œ 32t Ê s œ ' 32t dt
œ 16t# s! œ 16t# 7000. At t œ 13 sec, s œ 16(13)# 7000 œ 4296 ft when B's chute opens;
(c) After the chutes open, v œ 16 ft/sec Ê s œ ' 16 dt œ 16t s! . For A, s! œ 6144 ft and for B,
s! œ 4296 ft. Therefore, for A, s œ 16t 6144 and for B, s œ 16t 4296. When they hit the ground,
4296
s œ 0 Ê for A, 0 œ 16t 6144 Ê t œ 6144
16 œ 384 seconds, and for B, 0 œ 16t 4296 Ê t œ 16
œ 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens
Ê B hits the ground first.
87. av(I) œ
œ
"
30
"
30
'
30
0
(1200 40t) dt œ
"
30
$!
c1200t 20t# d! œ
"
30
ca(1200(30) 20(30)# b a1200(0) 20(0)# bd
(18,000) œ 600; Average Daily Holding Cost œ (600)($0.03) œ $18
88. av(I) œ
"
14
'
0
14
(600 600t) dt œ
"
14
"%
c600t 300t# d! œ
"
14
c600(14) 300(14)# 0d œ 4800; Average Daily
Holding Cost œ (4800)($0.04) œ $192
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
354
Chapter 5 Integration
'
"
30
89. av(I) œ
30
#
0
$
"
30
Š450 t# ‹ dt œ
’450t t6 “
œ (300)($0.02) œ $6
œ
"
60
'
"
60
90. av(I) œ
60
0
40È15
3
’600(60)
(60)
$Î#
'
"
60
Š600 20È15t‹ dt œ
0
"
60
0“ œ
60
$!
!
"
30
œ
30$
6
’450(30)
0“ œ 300; Average Daily Holding Cost
Š600 20È15 t"Î# ‹ dt œ
"
60
’600t 20È15 ˆ 23 ‰ t$Î# “
'!
!
ˆ36,000 ˆ 320
‰ 15# ‰ œ 200; Average Daily Holding Cost
3
œ (200)($0.005) œ $1.00
CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES
'
1. (a) Yes, because
1
0
(b) No. For example,
4È 2
3
œ
'
1
0
'
"
7
f(x) dx œ
1
0
"
7
7f(x) dx œ
(7) œ 1
'
"
8x dx œ c4x# d ! œ 4, but
1
0
"
È8x dx œ ’2È2 Š x$Î#
œ
3 ‹“
2
2
5
5
5
2
2
2
2
2
œ432œ9
'c f(x) dx œ 4 3 œ 7 2 œ 'c g(x) dx
(c) False:
5
5
2
2
œ
'
x
0
sin ax
a
sin ax
a
'
0
f(t) cos at dt
'
d
Πdx
œ cos ax
x
0
'
x
0
f(x) dx
5
2
5
0
2
x
0
0
x
x
0
0
'
f(t) cos at dt
sin ax
a
x
0
f(t) sin at dt
(f(x) cos ax) sin ax
x
cos ax
a
'
x
0
d
Πdx
'
0
x
f(t) sin at dt
f(t) sin at dt
cos ax
a
(f(x) sin ax)
x
0
x
#
#
0
x
0
'
d
(sin ax) Πdx
a cos ax
'
0
x
0
x
0
f(t) sin at dt œ a sin ax
4. x œ
'
"
#
x
0
y
"
0 È1 4t#
Ê 1œ
œ
'
'
f(t) cos at dt
dt Ê
"
È14y#
a1 4y# b
'
0
x
d
dx
Š dy
dx ‹ Ê
"Î#
(x) œ
dy
dx
0
x
'
d
dx
0
'
0
y
x
'
'
"
È1 4t#
dy
4y Š dx
‹
È1 4y#
0
x
0
x
f(t) cos at dt a cos ax
'
0
x
f(t) sin at dt f(x).
f(t) cos at dt f(x)
f(t) sin at dt œ f(x). Note also that yw (0) œ y(0) œ 0.
dt œ
œ È1 4y# . Then
(8y) Š dy
dx ‹ œ
f(t) sin at dt
f(t) cos at dt (cos ax)f(x) cos ax
f(t) sin at dt a sin ax
cos ax
a
x
0
f(t) sin at dt (sin ax)f(x) sin ax œ a sin ax
Therefore, yww a# y œ a cos ax
a# Πsinaax
2
2
x
"
a
f(t) cos at dt sin ax
œ cos ax
dy
dx
5
5
' f(t) cos at dt sin ax ' f(t) sin at dt. Next,
d y
' f(t) cos at dt (cos ax) Πdxd ' f(t) cos at dt a cos ax '
dx œ a sin ax
Ê
'c g(x) dx
' f(t) sin ax cos at dt "a ' f(t) cos ax sin at dt
cos ax '
' f(t) cos at dt
f(t) sin at dt Ê dy
a
dx œ cos ax Œ
f(t) sin a(x t) dt œ
x
2
5
'c [f(x) g(x)] dx 0 Ê 'c [g(x) f(x)] dx 0.
Ê ' [g(x) f(x)] dx 0 which is a contradiction.
c
Ê
On the other hand, f(x) Ÿ g(x) Ê [g(x) f(x)]
"
a
ˆ1$Î# 0$Î# ‰
5
5
(b) True:
4È 2
3
Á È4
' f(x) dx œ ' f(x) dx œ 3
'c [f(x) g(x)] dx œ 'c f(x) dx 'c g(x) dx œ 'c f(x) dx '
2. (a) True:
3. y œ
!
#
œ
'
d
dy
”
d# y
dx#
œ
4y ˆÈ1 4y# ‰
È1 4y#
y
0
"
È1 4t#
d
dx
ˆÈ1 4y# ‰ œ
œ 4y. Thus
dt• Š dy
dx ‹ from the chain rule
d# y
dx#
d
dy
ˆÈ1 4y# ‰ Š dy
dx ‹
œ 4y, and the constant of
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Additional and Advanced Exercises
355
proportionality is 4.
'
5. (a)
x#
f(t) dt œ x cos 1x Ê
0
cos 1x 1x sin 1x
.
2x
Ê f ax# b œ
'
(b)
f(x)
0
d
dx
$
t# dt œ ’ t3 “
f(x)
!
œ
"
3
'
x#
f(t) dt œ cos 1x 1x sin 1x Ê f ax# b (2x) œ cos 1x 1x sin 1x
0
cos 21 21 sin 21
4
Thus, x œ 2 Ê f(4) œ
"
3
(f(x))$ Ê
"
4
œ
(f(x))$ œ x cos 1x Ê (f(x))$ œ 3x cos 1x Ê f(x) œ $È3x cos 1x
Ê f(4) œ $È3(4) cos 41 œ $È12
6.
'
a
f(x) dx œ
0
a#
#
a
#
Ê f(a) œ Fw (a) œ a
7.
'
b
1
1
#
sin a
"
#
cos a. Let F(a) œ
sin a
a
#
1
#
cos a
f(x) dx œ Èb# 1 È2 Ê f(b) œ
d
db
'
a
0
f(t) dt Ê f(a) œ Fw (a). Now F(a) œ
sin a Ê f ˆ 1# ‰ œ
'
b
1
f(x) dx œ
"
#
side of the equation is:
œ
d
dx
œ
'
0
'
”x
0
x
f(u) du•
'
d
dx
”
d
dx
'
x
0
x
0
f(u)(x u) du• œ
u f(u) du œ
'
x
0
' ”'
dy
dx
d
dx
0
'
x
0
0
u
"
#
1
#
sin
"Î#
'
0
x
d
dx
(2b) œ
f(t) dt• du• œ
f(u) x du
d
f(u) du x ” dx
'
'
0
cos
1
#
1
#
a
#
sin a
sin
1
#
Ê f(x) œ
b
È b# 1
œ
1
#
1
#
cos a
"
#
1
#
œ
"
#
x
È x# 1
x
f(t) dt; the derivative of the right
x
0
u f(u) du
f(u) du• xf(x) œ
'
0
x
f(u) du xf(x) xf(x)
x
f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
when x œ 0, the constant must be 0. Therefore,
9.
”
ab# 1b
x
d
dx
8. The derivative of the left side of the equation is:
1
#
ˆ 1# ‰
#
a#
#
' ”'
x
0
0
u
f(t) dt• du œ
'
0
x
f(u)(x u) du.
œ 3x# 2 Ê y œ ' a3x# 2b dx œ x$ 2x C. Then (1ß 1) on the curve Ê 1$ 2(1) C œ 1 Ê C œ 4
Ê y œ x$ 2x 4
10. The acceleration due to gravity downward is 32 ft/sec# Ê v œ ' 32 dt œ 32t v! , where v! is the initial
velocity Ê v œ 32t 32 Ê s œ ' (32t 32) dt œ 16t# 32t C. If the release point, at t œ !, is s œ 0, then
C œ 0 Ê s œ 16t# 32t. Then s œ 17 Ê 17 œ 16t# 32t Ê 16t# 32t 17 œ 0. The discriminant of this
quadratic equation is 64 which says there is no real time when s œ 17 ft. You had better duck.
11.
'c f(x) dx œ 'c x#Î$ dx '
œ
œ
œ
12.
3
0
8
8
3
4 dx
0
35 x&Î$ ‘ ! [4x]!$
)
ˆ0 35 (8)&Î$ ‰ (4(3)
36
5
0) œ
'c f(x) dx œ 'c Èx dx '
3
0
4
4
!
3
0
$
œ 23 (x)$Î# ‘ % ’ x3 4x“
96
5
12
ax# 4b dx
$
!
$
œ 0 ˆ 23 (4)$Î# ‰‘ ’ Š 33 4(3)‹ 0 “
œ
16
3
3œ
7
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
356
13.
Chapter 5 Integration
'
2
0
'
g(t) dt œ
1
t dt
0
"
#
'
2
1
sin 1t dt
#
œ ’ t2 “ 1" cos 1t‘ "
!
œ ˆ "# 0‰ 1" cos 21 ˆ 1" cos 1‰‘
"
#
œ
14.
'
2
0
2
1
h(z) dz œ
'
1
0
'
È1 z dz
1
2
(7z 6)"Î$ dz
"
#
3
œ 23 (1 z)$Î# ‘ ! 14
(7z 6)#Î$ ‘ "
œ 23 (1 1)$Î# ˆ 23 (1 0)$Î# ‰‘
3
14
(7(2) 6)#Î$
6
3 ‰
55
œ ˆ 7 14
œ 42
3
14
(7(1) 6)#Î$ ‘
2
3
'c f(x) dx œ 'cc dx 'c a1 x# b dx '
2
15.
1
2
1
2
1
"
x$
3 “ "
œ [x]"
# ’x
1$
3‹
16.
ˆ 23 ‰ 4 2 œ
2
3
'c h(r) dr œ 'c r dr '
2
0
1
1
#
œ ’ r2 “
!
"
2
3
Š Š1
1œ
a1 r# b dr
"
ba
1$
3‹
'
2
1
7
6
'
b
a
f(x) dx œ
"
#0
'
2
0
f(x) dx œ
#
’Š 1# 0‹ Š 2# 2‹ Š 1# 1‹“ œ
'
20. f(x) œ
'
x
"
t
1/x
"
ba
sin x
" "
t 1 t#
"
sin x
'
a
'ÈÈ sin t# dt
22. f(x) œ
'
y
x
xb3
f(x) dx œ
y
"
x
"
30
'
3
0
"
#
”
'
1
0
x dx
'
2
1
(x 1) dx• œ
"
#
#
"
#
’ x2 “ #" ’ x2 x“
!
#
"
"
#
f(x) dx œ
"
3
”
'
dx ‰
d ˆ " ‰‰
ˆ dx
Š "" ‹ ˆ dx
œ
x
x
0
"
x
1
dx
'
1
2
0 dx
x ˆ x"# ‰ œ
"
x
'
3
2
dx• œ
"
x
œ
#
#
"
3
[1 0 0 3 2] œ
2
3
2
x
"
d
"
d
‰ ˆ dx
‰ ˆ dx
dt Ê f w (x) œ ˆ 1 sin
(sin x)‰ ˆ 1 cos
(cos x)‰ œ
#x
#x
21. g(y) œ
2
b
dt Ê f w (x) œ
cos x
"
cos x
dr
0‹ a2 1b
#
18. Ave. value œ
œ
13
3
"
#
19. f(x) œ
’2(2) 2(1)“
!
(1)#
# ‹
17. Ave. value œ
"
#
(1)$
3 ‹•
Š1
’r r3 “ [r]#"
œ "#
œ
1
0
$
œ Š0
2 dx
[2x]#"
œ a1 (2)b ”Š1
œ1
2
1
cos x
cos# x
sin x
sin# x
d ˆ
d ˆ
Èy‰‹ œ
Ê gw (y) œ Šsin ˆ2Èy‰ ‹ Š dy
2Èy‰‹ Šsin ˆÈy‰ ‹ Š dy
sin 4y
Èy
sin y
2È y
d
‰
t(5 t) dt Ê f w (x) œ (x 3)(& (x 3)) ˆ dx
(x 3)‰ x(5 x) ˆ dx
dx œ (x 3)(2 x) x(5 x)
œ 6 x x# 5x x# œ 6 6x. Thus f w (x) œ 0 Ê 6 6x œ 0 Ê x œ 1. Also, f ww (x) œ 6 ! Ê x œ 1 gives a
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Additional and Advanced Exercises
maximum.
23. Let f(x) œ x& on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , n2 , á ,
_
n
n
&
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for
_
œ
j 1
&
! Š j ‹ ˆ"‰ œ
f(x) œ x& on [0ß 1] Ê n lim
n
n
Ä_
jœ1
œ
'
1
'
"
x& dx œ ’ x6 “ œ
!
0
&
lim " ’ˆ n" ‰
nÄ_ n
ˆ n2 ‰&
&
á ˆ nn ‰ “ œ n lim
’1
Ä_
&
2& á n&
“
n'
"
6
24. Let f(x) œ x$ on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , n2 , á ,
_
n
n
$
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for
_
œ
j 1
$
! Š j ‹ ˆ " ‰ œ lim
f(x) œ x$ on [0ß 1] Ê n lim
n
n
Ä_
nÄ_
jœ1
œ
'
0
1
%
"
x$ dx œ ’ x4 “ œ
!
"
n
$
$
$
’ˆ n" ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ n lim
’1
Ä_
$
2$ á n$
“
n%
"
4
25. Let y œ f(x) on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , 2n , á ,
_
n
n
are the
right-hand endpoints of the subintervals. Since f is continuous on [!ß 1], ! f Š nj ‹ ˆ "n ‰ is a Riemann sum of
œ
j 1
_
! f Š j ‹ ˆ " ‰ œ lim
y œ f(x) on [0ß 1] Ê n lim
n
n
Ä_
nÄ_
jœ1
"
n
'
1
"
26. (a) n lim
[2 4 6 á 2n] œ n lim
Ä _ n#
Ä_
on [0ß 1] (see Exercise 25)
"
n
n2
"
(b) n lim
c1"& 2"& á n"& d œ n lim
Ä _ n"'
Ä_
"&
f(x) œ x on [0ß 1] (see Exercise 25)
"
n
"&
"&
"&
’ˆ 1n ‰ ˆ 2n ‰ á ˆ nn ‰ “ œ
'
'
f ˆ n" ‰ f ˆ n2 ‰ á f ˆ nn ‰‘ œ
4
n
6
n
á
œ
2n ‘
n
0
1
0
f(x) dx
"
2x dx œ cx# d! œ 1, where f(x) œ 2x
'
0
1
"'
"
"
16 ,
x"& dx œ ’ x16 “ œ
!
where
1
"
"
(c) n lim
sin 1n sin 2n1 á sin nn1 ‘ œ
sin n1 dx œ 1" cos 1x‘ ! œ 1" cos 1 ˆ 1" cos 0‰
Ä_ n
0
œ 12 , where f(x) œ sin 1x on [0ß 1] (see Exercise 25)
"
(d) n lim
c1"& 2"& á n"& d œ Šn lim
Ä _ n"(
Ä_
" ‰
ˆ
œ 0 16 œ 0 (see part (b) above)
"
n"&
(e) n lim
Ä_
c1"& 2"& á n"& d œ n lim
Ä_
œ Šn lim
n‹ Šn lim
Ä_
Ä_
"
n"'
"
"
n ‹ Šn lim
Ä _ n"'
n
n"'
c1"& 2"& á n"& d‹ œ Šn lim
Ä_
"
n‹
'
1
0
x"& dx
c1"& 2"& á n"& d
c1"& 2"& á n"& d‹ œ Šn lim
n‹
Ä_
'
0
1
x"& dx œ _ (see part (b) above)
27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal
to r, the radius of the circle) and a vertex angle of )n where )n œ 2n1 . The area of each triangle is
An œ
" #
# r
sin )n Ê the area of the polygon is A œ nAn œ
(b) n lim
A œ n lim
Ä_
Ä_
nr#
#
sin
21
n
œ n lim
Ä_
n1r#
21
sin
21
n
nr#
#
nr#
21
# sin n .
sin ˆ 2n1 ‰
#
ˆ 2n1 ‰ œ a1r b
sin )n œ
œ n lim
a1 r # b
Ä_
lim
2 1 În Ä 0
sin ˆ 2n1 ‰
ˆ 2n1 ‰
œ 1 r#
'x cos 2t dt " œ sin x ' x cos 2t dt " Ê yw œ cos x cosa2xb; when x œ 1 we have
1
yw œ cos 1 cosa21b œ " " œ #. And yww œ sin x 2sina2xb; when x œ 1, y œ sin 1 ' cos 2t dt "
x
28. y œ sin x
1
1
œ ! ! " œ ".
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357
358
Chapter 5 Integration
'
ga$b œ '
29. (a) ga"b œ
(b)
1
1
$
1
(c) ga"b œ
fatb dt œ !
fatb dt œ "# a#ba"b œ "
' fatb dt œ ' fatb dt œ "% a1 ## b œ 1
1
1
1
1
(d) gw axb œ faxb œ ! Ê x œ $, ", $ and the sign chart for gw axb œ faxb is
relative maximum at x œ ".
(e) gw a"b œ fa"b œ # is the slope and ga"b œ
± ± ± . So g has a
3
1
3
' fatb dt œ 1, by (c). Thus the equation is y 1 œ #ax "b
1
1
y œ #x # 1 .
(f) gww axb œ f w axb œ ! at x œ " and gww axb œ f w axb is negative on a$ß "b and positive on a"ß "b so there is an
inflection point for g at x œ ". We notice that gww axb œ f w axb ! for x on a"ß #b and gww axb œ f w axb ! for x on
a#ß %b, even though gww a#b does not exist, g has a tangent line at x œ #, so there is an inflection point at x œ #.
(g) g is continuous on Ò$ß %Ó and so it attains its absolute maximum and minimum values on this interval. We saw in (d)
that gw axb œ ! Ê x œ $, ", $. We have that
ga$b œ
' $ fatb dt œ '$" fatb dt œ 1##
1
#
œ #1
' fatb dt œ !
$
ga$b œ ' fatb dt œ "
%
ga%b œ ' fatb dt œ " "# † " † " œ "#
ga"b œ
1
1
1
1
Thus, the absolute minimum is #1 and the absolute maximum is !. Thus, the range is Ò#1ß !Ó.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 5 Additional and Advanced Exercises
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
359
360
Chapter 5 Integration
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS
1. (a) A œ 1(radius)# and radius œ È1 x# Ê A(x) œ 1 a1 x# b
(b) A œ width † height, width œ height œ 2È1 x# Ê A(x) œ 4 a1 x# b
(diagonal)#
;
#
(c) A œ (side)# and diagonal œ È2(side) Ê A œ
(d) A œ
È3
4
diagonal œ 2È1 x# Ê A(x) œ 2 a1 x# b
(side)# and side œ 2È1 x# Ê A(x) œ È3 a1 x# b
2. (a) A œ 1(radius)# and radius œ Èx Ê A(x) œ 1x
(b) A œ width † height, width œ height œ 2Èx Ê A(x) œ 4x
(diagonal)#
;
#
(c) A œ (side)# and diagonal œ È2(side) Ê A œ
(d) A œ
È3
4
(side)# and side œ 2Èx Ê A(x) œ È3x
(diagonal)#
#
3. A(x) œ
diagonal œ 2Èx Ê A(x) œ 2x
œ
ˆ È x ˆ È x ‰ ‰ #
#
œ 2x (see Exercise 1c); a œ 0, b œ 4;
V œ 'a A(x) dx œ '0 2x dx œ cx# d ! œ 16
b
4
1(diameter)#
4
4. A(x) œ
œ
%
1 c a2 x # b x # d
4
#
œ
1c2 a1 x# bd
4
#
œ 1 a1 2x# x% b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 1 a1 2x# x% b dx œ 1 ’x 23 x$
b
1
"
x&
5 “ "
#
œ 21 ˆ1
2
3
5" ‰ œ
161
15
#
5. A(x) œ (edge)# œ ’È1 x# ŠÈ1 x# ‹“ œ Š2È1 x# ‹ œ 4 a1 x# b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 4a1 x# b dx œ 4 ’x
b
1
#
(diagonal)#
#
6. A(x) œ
œ
œ
#
Š2È1 x# ‹
V œ 'a A(x) dx œ 2'c1 a1 x# b dx œ 2 ’x
1
"
#
7. (a) STEP 1) A(x) œ
#
"
x$
3 “ "
(side) † (side) † ˆsin 13 ‰ œ
STEP 2) a œ 0, b œ 1
œ 8 ˆ1 "3 ‰ œ
16
3
#
’È1 x# ŠÈ1 x# ‹“
b
"
x$
3 “ "
"
#
œ 2 a1 x# b (see Exercise 1c); a œ 1, b œ 1;
œ 4 ˆ1 "3 ‰ œ
8
3
† Š2Èsin x‹ † Š2Èsin x‹ ˆsin 13 ‰ œ È3 sin x
STEP 3) V œ 'a A(x) dx œ È3 '0 sin x dx œ ’È3 cos x“ œ È3(1 1) œ 2È3
1
1
b
!
#
(b) STEP 1) A(x) œ (side) œ Š2Èsin x‹ Š2Èsin x‹ œ 4 sin x
STEP 2) a œ 0, b œ 1
STEP 3) V œ 'a A(x) dx œ '0 4 sin x dx œ c4 cos xd 1! œ 8
1
b
#
8. (a) STEP 1) A(x) œ 1(diameter)
œ 14 (sec x tan x)# œ
4
sin x ‘
œ 14 sec# x asec# x 1b 2 cos
#x
STEP 2) a œ 13 , b œ
asec# x tan# x 2 sec x tan xb
1
3
STEP 3) V œ 'a A(x) dx œ 'c1Î3
b
1
4
1Î3
1
4
ˆ2 sec# x 1
2 sin x ‰
cos# x
dx œ
1
4
2 tan x x 2 ˆ cos" x ‰‘1Î$
1Î$
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
362
Chapter 6 Applications of Definite Integrals
œ
1
4
’2È3
1
3
2 Š ˆ "" ‰ ‹ Š2È3
#
1
3
2 Š ˆ "" ‰ ‹‹“ œ
#
(b) STEP 1) A(x) œ (edge)# œ (sec x tan x)# œ ˆ2 sec# x 1 2
STEP 2) a œ 13 , b œ
1
3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 ˆ2 sec# x 1
1Î3
b
1
4
9. A(y) œ
(diameter)# œ
1
4
#
ŠÈ5y# 0‹ œ
c œ 0, d œ 2; V œ 'c A(y) dy œ '0
d
#
&
œ ’ˆ 541 ‰ Š y5 ‹“ œ
!
"
#
10. A(y) œ
2
1
4
51
4
51
4
Š4È3
21
3 ‹
sin x ‰
cos# x
dx œ 2 Š2È3 13 ‹ œ 4È3
21
3
y% ;
y% dy
a2& 0b œ 81
"
#
(leg)(leg) œ
#
È1 y# ˆÈ1 y# ‰‘ œ
V œ 'c A(y) dy œ 'c1 2a1 y# b dy œ 2 ’y
d
2 sin x ‰
cos# x
1
4
1
"
y$
3 “ "
"
#
#
ˆ2È1 y# ‰ œ 2 a1 y# b ; c œ 1, d œ 1;
œ 4 ˆ1 "3 ‰ œ
8
3
11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) œ (side length)# œ s# ;
STEP 2) a œ 0, b œ h; STEP 3) V œ 'a A(x) dx œ '0 s# dx œ s# h
b
h
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the
prism described above, regardless of the number of turns Ê V œ s# h
12. 1) The solid and the cone have the same altitude of 12.
2) The cross sections of the solid are disks of diameter
x ˆ x# ‰ œ x# . If we place the vertex of the cone at the
origin of the coordinate system and make its axis of
symmetry coincide with the x-axis then the cone's cross
sections will be circular disks of diameter
x
ˆ x‰ x
4 4 œ # (see accompanying figure).
3) The solid and the cone have equal altitudes and identical
parallel cross sections. From Cavalieri's Principle we
conclude that the solid and the cone have the same
volume.
13. R(x) œ y œ 1
œ 1 ˆ2
4
2
14. R(y) œ x œ
3y
#
x
#
8 ‰
12
#
Ê V œ '0 1[R(x)]# dx œ 1'0 ˆ1 x# ‰ dx œ 1'0 Š1 x
2
œ
2
2
x#
4‹
dx œ 1 ’x
x#
#
21
3
‰ dy œ 1'
Ê V œ '0 1[R(y)]# dy œ 1'0 ˆ 3y
#
0
2
15. R(x) œ tan ˆ 14 y‰ ; u œ
1
4
y Ê du œ
2
1
4
#
2
9
4
#
y# dy œ 1 34 y$ ‘ ! œ 1 †
3
4
dy Ê 4 du œ 1 dy; y œ 0 Ê u œ 0, y œ 1 Ê u œ
#
x$
12 “ !
† 8 œ 61
1
4
;
1Î%
V œ '0 1[R(y)]# dy œ 1'0 tan ˆ 14 y‰‘ dy œ 4 '0 tan# u du œ 4 '0 a1 sec# ub du œ 4[u tan u]!
1
1
#
1Î4
1Î4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.1 Volumes by Slicing and Rotation About an Axis
œ 4 ˆ 14 1 0‰ œ 4 1
1
#
16. R(x) œ sin x cos x; R(x) œ 0 Ê a œ 0 and b œ
œ 1'0
1Î2
xœ
1
#
(sin x cos x)# dx œ 1 '0
1Î2
Ê u œ 1‘ Ä V œ 1'0
1
"
8
(sin 2x)#
4
are the limits of integration; V œ '0
1Î2
dx; u œ 2x Ê du œ 2 dx Ê
sin# u du œ
1
8
#u
"
4
sin
1
2u‘ !
œ
1
8
ˆ 1#
du
8
œ
dx
4
1[R(x)]# dx
; x œ 0 Ê u œ 0,
0‰ 0‘ œ
1#
16
17. R(x) œ x# Ê V œ '0 1[R(x)]# dx œ 1 '0 ax# b dx
2
2
œ 1 '0 x% dx œ 1 ’ x5 “ œ
2
#
&
#
321
5
!
18. R(x) œ x$ Ê V œ '0 1[R(x)]# dx œ 1'0 ax$ b dx
2
2
œ 1 '0 x' dx œ 1 ’ x7 “ œ
2
(
#
!
#
1281
7
19. R(x) œ È9 x# Ê V œ 'c3 1[R(x)]# dx œ 1 'c3 a9 x# b dx
3
$
x$
3 “ $
œ 1 ’9x
3
œ 21 9(3)
27 ‘
3
œ 2 † 1 † 18 œ 361
20. R(x) œ x x# Ê V œ '0 1[R(x)]# dx œ 1'0 ax x# b dx
1
1
œ 1'0 ax# 2x$ x% b dx œ 1 ’ x3
1
œ 1 ˆ 13
$
"
#
5" ‰ œ
1
30
(10 15 6) œ
21. R(x) œ Ècos x Ê V œ '0
1Î2
1Î#
œ 1 csin xd !
2x%
4
1
30
#
"
x&
5 “!
1[R(x)]# dx œ 1'0 cos x dx
1Î2
œ 1(1 0) œ 1
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363
364
Chapter 6 Applications of Definite Integrals
1Î4
1Î4
22. R(x) œ sec x Ê V œ 'c1Î4 1[R(x)]# dx œ 1 '1Î4 sec# x dx
1Î%
œ 1 ctan xd 1Î% œ 1[1 (1)] œ 21
23. R(x) œ È2 sec x tan x Ê V œ
œ1
'01Î4 1[R(x)]# dx
'01Î4 ŠÈ2 sec x tan x‹# dx
œ 1 '0 Š2 2È2 sec x tan x sec# x tan# x‹ dx
1Î4
œ 1 Œ'0 2 dx 2È2 '0 sec x tan x dx
1Î4
1Î%
œ 1 Œ[2x]!
'01Î4 (tan x)# sec# x dx
1Î4
1Î%
2È2 [sec x]!
$
’ tan3 x “
1Î%
!
œ 1 ’ˆ 1# 0‰ 2È2 ŠÈ2 1‹ "3 a1$ 0b“ œ 1 Š 1# 2È2
11
3 ‹
24. R(x) œ 2 2 sin x œ 2(1 sin x) Ê V œ '0 1[R(x)]# dx
1Î2
œ 1 '0 4(1 sin x)# dx œ 41 '0 a1 sin# x 2 sin xb dx
1Î2
1Î2
œ 41'0 1 "# (1 cos 2x) 2 sin x‘ dx
1Î2
œ 41'0 ˆ 3#
1Î2
2 sin x‰
cos 2x
2
1Î#
œ 41 3# x sin42x 2 cos x‘ !
œ 41 ˆ 341 0 0‰ (0 0 2)‘ œ 1(31 8)
25. R(y) œ È5 † y# Ê V œ 'c1 1[R(y)]# dy œ 1 'c1 5y% dy
1
1
"
œ 1 cy& d " œ 1[1 (1)] œ 21
26. R(y) œ y$Î# Ê V œ '0 1[R(y)]# dy œ 1'0 y$ dy
2
%
2
#
œ 1 ’ y4 “ œ 41
!
27. R(y) œ È2 sin 2y Ê V œ '0 1[R(y)]# dy
1Î2
œ 1'0 2 sin 2y dy œ 1 c cos 2yd !
1Î2
1Î#
œ 1[1 (1)] œ 21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.1 Volumes by Slicing and Rotation About an Axis
28. R(y) œ Écos
1y
4
Ê V œ 'c2 1[R(y)]# dy
0
œ 1 'c2 cos ˆ 14y ‰ dy œ 4 sin
0
29. R(y) œ
2
y1
1y ‘ !
4 #
œ 4[0 (1)] œ 4
Ê V œ '0 1[R(y)]# dy œ 41 '0
3
3
"
(y 1)#
dy
$
"
‘
œ 41 ’ y"
1 “ œ 41 4 (1) œ 31
!
30. R(y) œ
È2y
y # 1
Ê V œ '0 1[R(y)]# dy œ 1'0 2y ay# 1b
1
1
#
dy;
#
cu œ y 1 Ê du œ 2y dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 2d
Ä V œ 1'1 u# du œ 1 "u ‘ " œ 1 #" (1)‘ œ
2
#
1
#
31. For the sketch given, a œ 1# , b œ 1# ; R(x) œ 1, r(x) œ Ècos x; V œ 'a 1 a[R(x)]# [r(x)]# b dx
b
œ 'c1Î2 1(1 cos x) dx œ 21'0 (1 cos x) dx œ 21[x sin x]!
1Î2
1Î2
1Î#
œ 21 ˆ 1# 1‰ œ 1# 21
32. For the sketch given, c œ 0, d œ 14 ; R(y) œ 1, r(y) œ tan y; V œ 'c 1 a[R(y)]# [r(y)]# b dy
d
œ 1'0 a1 tan# yb dy œ 1 '0 a2 sec# yb dy œ 1[2y tan y]!
1Î4
1Î4
33. r(x) œ x and R(x) œ 1 Ê V œ
œ '0 1 a1 x# b dx œ 1 ’x
1
1Î%
œ 1 ˆ 1# 1‰ œ
1#
#
1
'01 1 a[R(x)]# [r(x)]# b dx
"
x$
3 “!
œ 1 ˆ1 "3 ‰ 0‘ œ
21
3
34. r(x) œ 2Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
1
œ 1'0 (4 4x) dx œ 41’x
1
"
x#
# “!
œ 41 ˆ1 "# ‰ œ 21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
365
366
Chapter 6 Applications of Definite Integrals
35. r(x) œ x# 1 and R(x) œ x 3
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
2
œ 1'c1 ’(x 3)# ax# 1b “ dx
2
#
œ 1 'c1 cax# 6x 9b ax% 2x# 1bd dx
2
œ 1 'c1 ax% x# 6x 8b dx
2
&
œ 1 ’ x5
x$
3
œ 1 ˆ 32
5
8
3
#
6x#
#
8x“
24
#
16‰ ˆ 5"
"
"
3
6
#
‰
ˆ 5†30533 ‰ œ
8‰‘ œ 1 ˆ 33
5 3 28 3 8 œ 1
36. r(x) œ 2 x and R(x) œ 4 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
2
œ 1'c1 ’a4 x# b (2 x)# “ dx
2
#
œ 1 'c1 ca16 8x# x% b a4 4x x# bd dx
2
œ 1'c1 a12 4x 9x# x% b dx
2
œ 1 ’12x 2x# 3x$
œ 1 ˆ24 8 24
#
x&
5 “ "
32 ‰
5
ˆ12 2 3 "5 ‰‘ œ 1 ˆ15
33 ‰
5
œ
1081
5
37. r(x) œ sec x and R(x) œ È2
Ê V œ 'c1Î4 1 a[R(x)]# [r(x)]# b dx
1Î4
œ 1 'c1Î4 a2 sec# xb dx œ 1[2x tan x]1Î%
1Î4
1Î%
œ 1 ˆ 1# 1‰ ˆ 1# 1‰‘ œ 1(1 2)
38. R(x) œ sec x and r(x) œ tan x
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
1
œ 1 '0 asec# x tan# xb dx œ 1 '0 1 dx œ 1[x]!" œ 1
1
1
39. r(y) œ 1 and R(y) œ 1 y
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 c(1 y)# 1d dy œ 1 '0 a1 2y y# 1b dy
1
1
œ 1 '0 a2y y# b dy œ 1 ’y#
1
"
y$
3 “!
œ 1 ˆ1 3" ‰ œ
41
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1171
5
Section 6.1 Volumes by Slicing and Rotation About an Axis
40. R(y) œ 1 and r(y) œ 1 y Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 c1 (1 y)# d dy œ 1'0 c1 a1 2y y# bd dy
1
1
œ 1'0 a2y y# b dy œ 1 ’y#
1
"
y$
3 “!
œ 1 ˆ1 "3 ‰ œ
21
3
41. R(y) œ 2 and r(y) œ Èy
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
4
œ 1'0 (4 y) dy œ 1 ’4y
4
%
y#
2 “!
œ 1(16 8) œ 81
42. R(y) œ È3 and r(y) œ È3 y#
È3
Ê V œ '0
È3
œ 1 '0
$
œ 1 ’ y3 “
1 a[R(y)]# [r(y)]# b dy
È3
c3 a3 y# bd dy œ 1'0
È$
!
y# dy
œ 1È3
43. R(y) œ 2 and r(y) œ 1 Èy
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 ’4 ˆ1 Èy‰ “ dy
1
#
œ 1 '0 ˆ4 1 2Èy y‰ dy
1
œ 1 '0 ˆ3 2Èy y‰ dy
1
œ 1 ’3y 43 y$Î#
œ 1 ˆ3
"
y#
# “!
"# ‰ œ 1 ˆ 18683 ‰ œ
4
3
71
6
44. R(y) œ 2 y"Î$ and r(y) œ 1
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
#
œ 1'0 ’ˆ2 y"Î$ ‰ 1“ dy
1
œ 1'0 ˆ4 4y"Î$ y#Î$ 1‰ dy
1
œ 1 '0 ˆ3 4y"Î$ y#Î$ ‰ dy
1
œ 1 ’3y 3y%Î$
"
3y&Î$
5 “!
œ 1 ˆ3 3 53 ‰ œ
31
5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
367
368
Chapter 6 Applications of Definite Integrals
45. (a) r(x) œ Èx and R(x) œ 2
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
4
œ 1'0 (4 x) dx œ 1 ’4x
4
(b) r(y) œ 0 and R(y) œ y#
%
x#
# “!
œ 1(16 8) œ 81
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
2
œ 1'0 y% dy œ 1 ’ y5 “ œ
2
&
#
!
321
5
#
(c) r(x) œ 0 and R(x) œ 2 Èx Ê V œ '0 1 a[R(x)]# [r(x)]# b dx œ 1'0 ˆ2 Èx‰ dx
4
œ 1'0 ˆ4 4Èx x‰ dx œ 1 ’4x
4
4
8x$Î#
3
%
x#
# “!
œ 1 ˆ16
64
3
16 ‰
#
œ
81
3
(d) r(y) œ 4 y# and R(y) œ 4 Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’16 a4 y# b “ dy
2
2
œ 1 '0 a16 16 8y# y% b dy œ 1 '0 a8y# y% b dy œ 1 ’ 83 y$
2
2
46. (a) r(y) œ 0 and R(y) œ 1
#
y&
5 “!
#
œ 1 ˆ 64
3
32 ‰
5
œ
2241
15
y
#
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
2
#
œ 1'0 ˆ1 y# ‰ dy œ 1'0 Š1 y
2
œ 1 ’y
2
y#
#
#
y$
12 “ !
œ 1 ˆ#
(b) r(y) œ 1 and R(y) œ 2
4
2
8 ‰
12
y#
4‹
œ
dy
21
3
y
#
#
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’ˆ2 y# ‰ 1“ dy œ 1 '0 Š4 2y
2
2
œ 1'0 Š3 2y
2
y#
4‹
dy œ 1 ’3y y#
#
y$
12 “ !
2
œ 1 ˆ6 4
8 ‰
12
œ 1 ˆ2 23 ‰ œ
y#
4
1‹ dy
81
3
47. (a) r(x) œ 0 and R(x) œ 1 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
1
œ 1 'c1 a1 x# b dx œ 1 'c1 a1 2x# x% b dx
1
œ 1 ’x
1
#
2x$
3
"
x&
5 “ "
103 ‰
œ 21 ˆ 1515
œ
œ 21 ˆ1
2
3
15 ‰
161
15
(b) r(x) œ 1 and R(x) œ 2 x# Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’a2 x# b 1“ dx
1
1
œ 1 'c1 a4 4x# x% 1b dx œ 1'c1 a3 4x# x% b dx œ 1 ’3x 43 x$
1
œ
21
15
1
(45 20 3) œ
561
15
#
"
x&
5 “ "
œ 21 ˆ3
4
3
15 ‰
2
3
15 ‰
(c) r(x) œ 1 x# and R(x) œ 2 Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’4 a1 x# b “ dx
1
1
œ 1 'c1 a4 1 2x# x% b dx œ 1'c1 a3 2x# x% b dx œ 1 ’3x 23 x$
1
œ
21
15
1
(45 10 3) œ
641
15
#
"
x&
5 “ "
œ 21 ˆ3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.1 Volumes by Slicing and Rotation About an Axis
369
48. (a) r(x) œ 0 and R(x) œ hb x h
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
b
#
œ 1 '0 ˆ hb x h‰ dx
b
œ 1'0 Š hb# x#
b
#
$
x
œ 1h# ’ 3b
#
x#
b
2h#
b
x h# ‹ dx
b
x“ œ 1h# ˆ b3 b b‰ œ
!
1 h# b
3
#
(b) r(y) œ 0 and R(y) œ b ˆ1 yh ‰ Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1b# '0 ˆ1 yh ‰ dy
h
œ 1b# '0 Š1
h
2y
h
y#
h# ‹
dy œ 1b# ’y
y#
h
h
h
y$
3h# “ !
1 b# h
3
œ 1b# ˆh h 3h ‰ œ
49. R(y) œ b Èa# y# and r(y) œ b Èa# y#
Ê V œ 'ca 1 a[R(y)]# [r(y)]# b dy
a
œ 1 'ca ’ˆb Èa# y# ‰ ˆb Èa# y# ‰ “ dy
#
a
#
œ 1 'ca 4bÈa# y# dy œ 4b1'ca Èa# y# dy
a
a
1a#
#
œ 4b1 † area of semicircle of radius a œ 4b1 †
œ 2a# b1#
50. (a) A cross section has radius r œ È#y and area 1r# œ #1y. The volume is '0 #1ydy œ 1 cy# d ! œ #&1.
&
(b) Vahb œ ' Aahbdh, so
dV
dh
œ Aahb. Therefore
For h œ %, the area is #1a%b œ )1, so
dh
dt
œ
dV
dt
"
)1
œ
dV
dh
†
œ Aahb †
dh
dt
$
$
)1
† $ units
sec œ
hca
51. (a) R(y) œ Èa# y# Ê V œ 1'ca aa# y# b dy œ 1 ’a# y
œ 1 ’a# h "3 ah$ 3h# a 3ha# a$ b
dV
$
dt œ 0.2 m /sec
dV
#
dh œ 101h 1h
(b) Given
Ê
and a œ 5 m, find
Ê
dV
dt
œ
dV
dh
†
dh
dt
a$
3“
œ 1 Ša# h
†
so
dh
dt
œ
"
A ah b
†
dV
dt .
units$
sec .
hca
y$
3 “ ca
h$
3
dh
dt ,
&
œ 1 ’a# h a$
h# a ha# ‹ œ
(h a)$
3
Ša$
a$
3 ‹“
1h# (3a h)
3
#
From part (a), V(h) œ 1h (153 h) œ 51h# 13h
dh ¸
0.2
"
"
œ 1h(10 h) dh
dt Ê dt hœ4 œ 41(10 4) œ (201)(6) œ 1#01 m/sec.
dh ¸
dt hœ4 .
$
52. Suppose the solid is produced by revolving y œ 2 x about
the y-axis. Cast a shadow of the solid on a plane parallel to
the xy-plane.
Use an approximation such as the Trapezoid Rule, to
#
estimate 'a 1cRaybd# dy ¸ ! 1Œ #k ˜y.
b
n
d^
kœ"
53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius
h has been removed. Thus its area is A" œ 1R# 1h# œ 1 aR# h# b . The cross section of the hemisphere is a disk of
#
radius ÈR# h# . Therefore its area is A# œ 1 ŠÈR# h# ‹ œ 1 aR# h# b . We can see that A" œ A# . The altitudes of
both solids are R. Applying Cavalieri's Principle we find
Volume of Hemisphere œ (Volume of Cylinder) (Volume of Cone) œ a1R# b R "3 1 aR# b R œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
3
1 R$ .
370
Chapter 6 Applications of Definite Integrals
54. R(x) œ
rx
h
Ê V œ '0 1[R(x)]# dx œ 1'0
h
h # #
r x
dx œ
h#
1r#
h#
h
$
#
$
’ x3 “ œ Š 1hr# ‹ Š h3 ‹ œ
!
"
3
1r# h, the volume of
a cone of radius r and height h.
c7
c7
55. R(y) œ È256 y# Ê V œ 'c16 1[R(y)]# dy œ 1'c16 a256 y# b dy œ 1 ’256y
œ 1 ’(256)(7)
56. R(x) œ
œ
1
144
x
1#
7$
3
Š(256)(16)
16$
3 ‹“
$
È36 x# Ê V œ ' 1[R(x)]# dx œ 1'
0
0
6
'
x&
5 “!
’12x$
œ
1
144
6
Š12 † 6$
6&
5‹
œ
16$
3 ‹
œ 1 Š 73 256(16 7)
1 †6 $
144
x#
144
ˆ12
a36 x# b dx œ
36 ‰
5
1
144
(
y$
3 “ "'
œ 10531 cm$ ¸ 3308 cm$
'06 a36x# x% b dx
1 ‰ ˆ 6036 ‰
œ ˆ 196
œ
144
5
361
5
cm$ . The plumb bob will
weigh about W œ (8.5) ˆ 3651 ‰ ¸ 192 gm, to the nearest gram.
57. (a) R(x) œ kc sin xk , so V œ 1'0 [R(x)]# dx œ 1'0 (c sin x)# dx œ 1'0 ac# 2c sin x sin# xb dx
1
1
œ 1'0 ˆc# 2c sin x
'1
1
(b)
1
1cos 2x ‰
dx œ 1 0 ˆc# "# 2c sin x cos#2x ‰ dx
#
1
œ 1 ˆc# "# ‰ x 2c cos x sin42x ‘ ! œ 1 ˆc# 1 1# 2c 0‰ (0 2c 0)‘ œ 1 ˆc# 1 1# 4c‰ . Let
2
V(c) œ 1 ˆc# 1 1# 4c‰ . We find the extreme values of V(c): dV
dc œ 1(2c1 4) œ 0 Ê c œ 1 is a critical
#
#
point, and V ˆ 12 ‰ œ 1 ˆ 14 1# 18 ‰ œ 1 ˆ 1# 14 ‰ œ 1# 4; Evaluate V at the endpoints: V(0) œ 1# and
#
#
V(1) œ 1 ˆ 3# 1 4‰ œ 1# (4 1)1. Now we see that the function's absolute minimum value is 1# 4,
taken on at the critical point c œ 12 . (See also the accompanying graph.)
#
From the discussion in part (a) we conclude that the function's absolute maximum value is 1# , taken on at
the endpoint c œ 0.
(c) The graph of the solid's volume as a function of c for
0 Ÿ c Ÿ 1 is given at the right. As c moves away from
[!ß "] the volume of the solid increases without bound.
If we approximate the solid as a set of solid disks, we
can see that the radius of a typical disk increases without
bounds as c moves away from [0ß 1].
58. (a) R(x) œ 1
œ 1 'c4 Š1
4
œ 1 ’x
x$
24
œ 21 ˆ4
Ê V œ 'c4 1[R(x)]# dx
4
x#
16
8
3
x#
16 ‹
#
dx œ 1'c4 Š1
4
%
x&
5†16# “ %
45 ‰ œ
21
15
œ 21 Š4
x#
8
x%
16# ‹
4$
24
4&
5†16# ‹
(60 40 12) œ
641
15
dx
ft$
(b) The helicopter will be able to fly ˆ 64151 ‰ (7.481)(2) ¸ 201 additional miles.
6.2 VOLUME BY CYLINDRICAL SHELLS
1. For the sketch given, a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š1
b
2
x#
4‹
dx œ 21'0 Šx
x#
4‹
dx œ 21'0 Š2x
2
x$
4‹
œ 21 † 3 œ 61
2. For the sketch given, a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š2
b
2
2
#
dx œ 21 ’ x#
x$
4‹
#
x%
16 “ !
dx œ 21 ’x#
œ 21 ˆ 4#
#
x%
16 “ !
16 ‰
16
œ 21(4 1) œ 61
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.2 Volume by Cylindrical Shells
3. For the sketch given, c œ 0, d œ È2;
È2
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0
d
È2
21y † ay# b dy œ 21'0
%
y$ dy œ 21 ’ y4 “
4. For the sketch given, c œ 0, d œ È3;
È3
È3
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y † c3 a3 y# bd dy œ 21 '0
d
È#
!
œ 21
%
y$ dy œ 21 ’ y4 “
È3
!
œ
5. For the sketch given, a œ 0, b œ È3;
È3
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † ŠÈx# 1‹ dx;
b
’u œ x# 1 Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4“
Ä V œ 1'1 u"Î# du œ 1 23 u$Î# ‘ " œ
%
4
21
3
ˆ4$Î# 1‰ œ ˆ 231 ‰ (8 1) œ
141
3
6. For the sketch given, a œ 0, b œ 3;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š Èx9x
‹ dx;
$9
b
3
cu œ x$ 9 Ê du œ 3x# dx Ê 3 du œ 9x# dx; x œ 0 Ê u œ 9, x œ 3 Ê u œ 36d
Ä V œ 21 '9 3u"Î# du œ 61 2u"Î# ‘ * œ 121 ŠÈ36 È9‹ œ 361
$'
36
7. a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x x ˆ x2 ‰‘ dx
b
2
œ '0 21x# †
2
3
#
dx œ 1 '0 3x# dx œ 1 cx$ d ! œ 81
2
#
8. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ˆ2x x2 ‰ dx
b
1
œ 1 '0 2 Š 3x# ‹ dx œ 1 ' 3x# dx œ 1 cx$ d ! œ 1
1
1
#
"
0
9. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x c(2 x) x# d dx
b
1
œ 21'0 a2x x# x$ b dx œ 21 ’x#
1
œ 21 ˆ1
"
3
4" ‰ œ 21 ˆ 12 124 3 ‰ œ
x$
3
101
12
œ
"
x%
4 “!
51
6
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
91
#
371
372
Chapter 6 Applications of Definite Integrals
10. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ca2 x# b x# d dx
b
1
œ 21'0 x a2 2x# b dx œ 41'0 ax x$ b dx
1
1
"
x%
4 “!
#
œ 41 ’ x#
œ 41 ˆ "2 4" ‰ œ 1
11. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Èx (2x 1)‘ dx
b
1
"
œ 21'0 ˆx$Î# 2x# x‰ dx œ 21 25 x&Î# 23 x$ "# x# ‘ !
1
œ 21 ˆ 25
2
3
15 ‰
"# ‰ œ 21 ˆ 12 20
œ
30
71
15
12. a œ ", b œ 4;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21x ˆ 32 x"Î# ‰ dx
b
4
œ 31'1 x"Î# dx œ 31 23 x$Î# ‘ " œ 21 ˆ4$Î# "‰
%
4
œ 21(8 1) œ 141
13. (a) xf(x) œ œ
xf(x) œ œ
x†
sin x, 0 x Ÿ 1
0xŸ1
Ê xf(x) œ œ
; since sin 0 œ 0 we have
0, x œ 0
x, x œ 0
sin x
x ,
sin x, 0 x Ÿ 1
Ê xf(x) œ sin x, 0 Ÿ x Ÿ 1
sin x, x œ 0
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † f(x) dx and x † f(x) œ sin x, 0 Ÿ x Ÿ 1 by part (a)
1
b
Ê V œ 21'0 sin x dx œ 21[ cos x]1! œ 21( cos 1 cos 0) œ 41
1
tan# x
x ,
tan# x, 0 x Ÿ 1/4
0 x Ÿ 14
Ê xg(x) œ œ
; since tan 0 œ 0 we have
0, x œ 0
x † 0, x œ 0
tan# x, 0 x Ÿ 1/4
Ê xg(x) œ tan# x, 0 Ÿ x Ÿ 1/4
xg(x) œ œ
tan# x, x œ 0
14. (a) xg(x) œ œ
x†
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † g(x) dx and x † g(x) œ tan# x, 0 Ÿ x Ÿ 1/4 by part (a)
1Î4
b
Ê V œ 21'0 tan# x dx œ 21'0 asec# x 1b dx œ 21[tan x x]!
1Î4
1Î4
1Î%
œ 21 ˆ1 14 ‰ œ
41 1 #
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.2 Volume by Cylindrical Shells
15. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y Èy (y)‘ dy
d
2
œ 21'0 ˆy$Î# y# ‰ dy œ 21 ’ 2y5
2
&Î#
&
œ 21 ” 25 ŠÈ2‹
œ
161
15
2$
3•
#
y$
3 “!
È
œ 21 Š 8 5 2 83 ‹ œ 161 Š
È2
5
"3 ‹
Š3È2 5‹
16. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y cy# (y)ddy
d
2
œ 21'0 ay$ y# b dy œ 21 ’ y4
2
%
œ 161 ˆ 56 ‰ œ
401
3
#
y$
3 “!
œ 161 ˆ 24 "3 ‰
17. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y a2y y# bdy
d
2
œ 21'0 a2y# y$ b dy œ 21 ’ 2y3
2
$
œ 321 ˆ "3 4" ‰ œ
321
12
œ
81
3
#
y%
4 “!
œ 21 ˆ 16
3
"6 ‰
4
18. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y a2y y# ybdy
d
1
œ 21'0 y ay y# b dy œ 21'0 ay# y$ b dy
1
1
$
œ 21 ’ y3
"
y%
“
4 !
œ 21 ˆ 13 "4 ‰ œ
1
6
19. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ 21'0 y[y (y)]dy
d
1
œ 21'0 2y# dy œ
1
41
3
"
cy$ d ! œ
41
3
20. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 yˆy y2 ‰dy
d
2
œ 21 '0
2
y2
2 dy
1
œ 13 c y3 d ! œ
81
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
373
374
Chapter 6 Applications of Definite Integrals
21. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y c(2 y) y# d dy
d
2
œ 21 '0 a2y y# y$ b dy œ 21 ’y#
2
œ 21 ˆ4
8
3
16 ‰
4
1
6
œ
y$
3
(48 32 48) œ
#
y%
4 “!
161
3
22. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y c(2 y) y# d dy
d
1
œ 21'0 a2y y# y$ b dy œ 21 ’y#
1
œ 21 ˆ1
14 ‰ œ
1
3
1
6
(12 4 3) œ
y$
3
51
6
"
y%
4 “!
shell ‰
shell
23. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y † 12 ay# y$ b dy œ 241 '0 ay$ y% b dy œ 241 ’ y4
d
1
œ 241 ˆ 14 15 ‰ œ
241
20
œ
1
"
y&
5 “!
%
61
5
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(1 y) c12 ay# y$ bd dy œ 241'0 (1 y) ay# y$ b dy
d
1
1
œ 241'0 ay# 2y$ y% b dy œ 241 ’ y3
1
$
y%
2
"
y&
5 “!
œ 241 ˆ "3
1
2
" ‰
51 ‰ œ 241 ˆ 30
œ
41
5
shell ‰
shell
(c) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆ 85 y‰ c12 ay# y$ bd dy œ 241 '0 ˆ 85 y‰ ay# y$ b dy
d
1
œ 241'0 ˆ 85 y#
1
œ
241
12
13
5
1
8 $
y$ y% ‰ dy œ 241 ’ 15
y
13
20
y%
"
y&
5 “!
8
œ 241 ˆ 15
13
20
241
60
15 ‰ œ
(32 39 12)
œ 21
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆy 25 ‰ c12 ay# y$ bd dy œ 241'0 ˆy 25 ‰ ay# y$ b dy
d
1
1
2 $
œ 241'0 ˆy$ y% 25 y# 25 y$ ‰ dy œ 241'0 ˆ 25 y# 35 y$ y% ‰ dy œ 241 ’ 15
y
1
1
2
œ 241 ˆ 15
3
20
15 ‰ œ
241
60
(8 9 12) œ
241
12
2
%
œ 21 ’ y4
#
y'
24 “ !
%
2'
24 ‹
œ 21 Š 24
#
%
œ 321 ˆ 4"
4 ‰
24
dy œ '0 21y Šy#
2
y#
# ‹“
2
œ 21 '0 Š2y#
2
y%
2
y$
y&
4‹
#
$
dy œ 21 ’ 2y3
y%
4
2
œ 21'0 Š5y# 54 y% y$
2
y&
4‹
#
$
dy œ 21 ’ 5y3
y%
4
#
y'
#4 “ !
œ 21 ˆ 16
3
œ
2
œ 21'0 Šy$
2
y&
4
58 y#
5
32
#
%
y% ‹ dy œ 21 ’ y4
%
y'
#4
5y$
#4
2
y#
# ‹“
16
4
64 ‰
24
2
y'
#4 “ !
y#
# ‹“
32
10
dy œ '0 21(5 y) Šy#
#
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆy 58 ‰ ’ y# Š y4
d
81
3
y%
4‹
%
5y&
20
2
dy œ '0 21(2 y) Šy#
shell ‰
shell
(c) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(5 y) ’ y# Š y4
d
dy œ 21'0 Šy$
y#
# ‹“
%
y&
10
y%
4‹
2 ‰
œ 321 ˆ 4" 6" ‰ œ 321 ˆ 24
œ
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(2 y) ’ y# Š y4
d
"
y&
5 “!
y%
œ 21
shell ‰
shell
24. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y ’ y# Š y4
d
3
20
œ 21 ˆ 40
3
160
20
16
4
dy œ '0 21 ˆy 58 ‰ Šy#
#
5y&
160 “ !
2
œ 21 ˆ 16
4
64
24
40
24
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
81
5
y%
4‹
64 ‰
24
dy
dy
œ 81
y%
4‹
160 ‰
160
dy
œ 41
y&
4‹
dy
Section 6.2 Volume by Cylindrical Shells
375
shell ‰
shell
25. (a) About x-axis: V œ 'c 21 ˆ radius
Š height
‹dy
d
œ '0 21yˆÈy y‰dy œ 21'0 ˆy$Î# y# ‰dy
1
1
"
œ 21 #& y&Î# "$ y$ ‘ ! œ 21ˆ #& "$ ‰ œ
#1
"&
shell ‰
shell
About y-axis: V œ 'a 21 ˆ radius
Š height
‹dx
b
œ '0 21xax x# bdx œ 21'0 ax2 x3 bdx
1
1
$
œ 21’ x$
"
x%
% “!
œ 21ˆ "$ "% ‰ œ
1
'
(b) About x-axis: Raxb œ x and raxb œ x# Ê V œ 'a 1Raxb# raxb# ‘dx œ '0 1cx# x% ddx
b
$
œ 1’ x$
"
x&
& “!
œ 1ˆ "$ "& ‰ œ
1
#1
"&
About y-axis: Rayb œ Èy and rayb œ y Ê V œ 'c 1Rayb# rayb# ‘dy œ '0 1cy y2 ddy
d
#
œ 1’ y#
"
y$
$ “!
œ 1ˆ "# "$ ‰ œ
1
1
'
#
26. (a) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’ˆ #x #‰ x# “dx
%
b
œ 1'0 ˆ $% x# #x %‰dx œ 1’ x% x# %x“
%
$
œ 1a"' "' "'b œ "'1
%
!
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹dx œ '0 #1xˆ x# # x‰dx
%
b
œ '0 #1xˆ# x# ‰dx œ #1'0 Š#x
%
%
œ #1’x#
%
x$
' “!
'% ‰
'
œ #1ˆ"'
œ
x#
# ‹dx
$#1
$
shell ‰
shell
(c) V œ 'a 21 ˆ radius
Š height
‹dx œ '0 #1a% xbˆ x# # x‰dx œ '0 #1a% xbˆ# x# ‰dx œ #1'0 Š) %x
%
b
œ #1’)x #x#
%
x$
“
' !
œ #1ˆ$# $#
%
'% ‰
'
%
x#
# ‹dx
'%1
$
œ
#
(d) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’a) xb# ˆ' #x ‰ “dx œ 1'0 ’a'% "'x x# b Š$' 'x x% ‹“dx
%
b
%
#
1'0 ˆ $% x# "!x #)‰dx œ 1’ x% &x# #)x“ œ 1"' a&ba"'b a(ba"'b‘ œ 1a$ba"'b œ %)1
%
%
$
!
shell ‰
shell
27. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '1 21y(y 1) dy
d
2
œ 21'1 ay# yb dy œ 21 ’ y3
2
$
#
y#
# “"
œ 21 ˆ 83 42 ‰ ˆ "3 #" ‰‘
œ 21 ˆ 73 2 "# ‰ œ 13 (14 12 3) œ
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx
51
3
b
œ '1 21x(2 x) dx œ 21'1 a2x x# b dx œ 21 ’x#
2
2
œ 21 ˆ 12 3 8 ‰ ˆ 3 3 " ‰‘ œ 21 ˆ 34 32 ‰ œ
41
3
#
x$
3 “"
œ 21 ˆ4 83 ‰ ˆ1 "3 ‰‘
shell ‰
shell
' ˆ 203
‰
(c) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21 ˆ 10
3 x (2 x) dx œ 21 1
b
2
#
" $‘
8 #
ˆ 40
œ 21 20
3 x 3 x 3 x " œ 21
3
2
32
3
38 ‰ ˆ 20
3
8
3
16
3
x x# ‰ dx
3" ‰‘ œ 21 ˆ 33 ‰ œ 21
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '1 21(y 1)(y 1) dy œ 21'1 (y 1)# œ 21 ’ (y31) “ œ
d
2
2
$
#
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
21
3
376
Chapter 6 Applications of Definite Integrals
shell ‰
shell
28. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21yay# 0b dy
d
2
œ 21'0 y$ dy œ 21 ’ y4 “ œ 21 Š 24 ‹ œ 81
2
%
#
%
!
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx
b
œ '0 21x ˆ2 Èx‰ dx œ 21'0 ˆ2x x$Î# ‰ dx
4
4
%
2 †2 &
5 ‹
œ 21 x# 25 x&Î# ‘ ! œ 21 Š16
œ 21 ˆ16
64 ‰
5
21
5
œ
321
5
(80 64) œ
shell ‰
shell
(c) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(4 x) ˆ2 Èx‰ dx œ 21'0 ˆ8 4x"Î# 2x x$Î# ‰ dx
b
4
4
%
œ 21 8x 83 x$Î# x# 25 x&Î# ‘ ! œ 21 ˆ32
64
3
16
64 ‰
5
œ
21
15
(240 320 192) œ
21
15
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(2 y) ay# b dy œ 21 '0 a2y# y$ b dy œ 21 ’ 23 y$
d
2
œ 21 ˆ 16
3
16 ‰
4
œ
321
12
2
81
3
(4 3) œ
(112) œ
2241
15
#
y%
4 “!
shell ‰
shell
29. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21yay y$ b dy
d
1
œ '0 21 ay# y% b dy œ 21 ’ y3
1
œ
$
41
15
"
y&
“
5 !
œ 21 ˆ "3 "5 ‰
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy
d
œ '0 21(1 y) ay y$ b dy
1
œ 21 '0 ay y# y$ y% b dy œ 21 ’ y#
1
#
y$
3
y%
4
"
y&
5 “!
œ 21 ˆ "#
"
3
"
4
5" ‰ œ
21
60
(30 20 15 12) œ
71
30
shell ‰
shell
30. (a) V œ 'c 21 ˆ radius
Š height
‹dy
d
œ '0 21y c1 ay y$ bddy
1
œ 21 '0 ay y# y% b dy œ 21 ’ y#
1
#
œ 21 ˆ "#
œ
"
3
5" ‰ œ
21
30
y$
3
"
y&
5 “!
(15 10 6)
111
15
(b) Use the washer method:
V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’1# ay y$ b “ dy œ 1 '0 a1 y# y' 2y% b dy œ 1 ’y
d
1
œ 1 ˆ1
"
3
"
7
25 ‰ œ
1
105
1
#
(105 35 15 42) œ
y$
3
y(
7
971
105
"
2y&
5 “!
(c) Use the washer method:
V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’c1 ay y$ bd 0“ dy œ 1'0 ’1 2 ay y$ b ay y$ b “ dy
d
1
œ 1'0 a1 y# y' 2y 2y$ 2y% b dy œ 1 ’y
1
œ
1
210
(70 30 105 2 † 42) œ
1
#
y$
3
y(
7
y#
#
y%
#
1211
210
"
2y&
5 “!
œ 1 ˆ1
"
3
"
7
1
"
#
25 ‰
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(1 y) c1 ay y$ bd dy œ 21 '0 (1 y) a1 y y$ b dy
d
1
1
œ 21'0 a1 y y$ y y# y% b dy œ 21'0 a1 2y y# y$ y% b dy œ 21 ’y y#
1
œ 21 ˆ1 1
1
"
3
"
4
5" ‰ œ
21
60
(20 15 12) œ
231
30
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y$
3
y%
4
"
y&
5 “!
Section 6.2 Volume by Cylindrical Shells
shell ‰
shell
31. (a) V œ 'c 21 ˆ radius
Š height
‹dy œ '0 21y ˆÈ8y y# ‰ dy
d
2
œ 21'0 Š2È2 y$Î# y$ ‹ dy œ 21 ’ 4 5 2 y&Î#
È
2
#
y%
4 “!
&
œ 21
4È2†ŠÈ2‹
2%
4
5
œ 21 † 4 ˆ 85 1‰ œ
81
5
$
œ 21 Š 4†52
(8 5) œ
4 †4
4 ‹
241
5
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ŠÈx
b
4
&
œ 21 Š 2†52
4%
3# ‹
'
œ 21 Š 25
2)
32 ‹
œ
1†2(
160
x#
8‹
dx œ 21'0 Šx$Î#
(32 20) œ
4
1†2* †3
160
œ
1†2% †3
5
œ
x$
8‹
dx œ 21 ’ 25 x&Î#
481
5
shell ‰
shell
32. (a) V œ 'a 21 ˆ radius
Š height
‹ dx
b
œ '0 21x ca2x x# b xd dx
1
œ 21 '0 x ax x# b dx œ 21'0 ax# x$ b dx
1
$
1
œ 21 ’ x3
"
x%
4 “!
œ 21 ˆ "3 4" ‰ œ
1
6
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(1 x) ca2x x# b xd dx œ 21'0 (1 x) ax x# b dx
b
1
1
œ 21 '0 ax 2x# x$ b dx œ 21 ’ x2 23 x$
1
"
x%
4 “!
#
œ 21 ˆ 12
2
3
"4 ‰ œ
21
1#
(6 8 3) œ
1
6
33. (a) V œ 'a 1 cR# (x) r# (x)d dx œ 1 '1Î16 ˆx"Î# 1‰ dx
b
1
"
œ 1 2x"Î# x‘"Î"' œ 1 (2 1) ˆ2 †
œ 1 ˆ1
7 ‰
16
œ
"
4
" ‰‘
16
91
16
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dy œ '1 21y Š y"%
b
2
œ 21'1 ˆy$
2
y ‰
16
dy œ 21 ’ 12 y#
œ 21 ˆ "8 8" ‰ ˆ #"
œ
21
32
(8 1) œ
91
16
" ‰‘
3#
d
2
œ 1 "3 y$
œ
1
48
y ‘#
16 "
y#
32 “ "
" ‰
32
"
16 ‹
dy
"
œ 1 ˆ 24
8" ‰ ˆ 3"
(2 6 16 3) œ
dy
#
œ 21 ˆ 4"
34. (a) V œ 'c 1 cR# (y) r# (y)d dy œ '1 1 Š y"%
"
16 ‹
" ‰‘
16
111
48
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1Î4 21x Š È"x "‹ dx
b
1
œ 21 '1Î4 ˆx"Î# x‰ dx œ 21 ’ 23 x$Î#
1
œ 21 ˆ 23 "# ‰ ˆ 23 †
"
8
" ‰‘
3#
"
x#
2 “ "Î%
œ 1 ˆ 43 1
"
6
" ‰
16
œ
1
48
(4 † 16 48 8 3) œ
111
48
35. (a) H3=k: V œ V" V#
V" œ 'a 1[R" (x)]# dx and V# œ 'a 1[R# (x)]# with R" (x) œ É x 3 2 and R# (x) œ Èx,
b"
b#
"
#
a" œ 2, b" œ 1; a# œ 0, b# œ 1 Ê two integrals are required
(b) [ +=2/<: V œ V" V#
V" œ 'a 1 a[R" (x)]# [r" (x)]# b dx with R" (x) œ É x 3 2 and r" (x) œ 0; a" œ 2 and b" œ 0;
b"
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
%
x%
32 “ !
377
378
Chapter 6 Applications of Definite Integrals
V# œ 'a 1 a[R# (x)]# [r# (x)]# b dx with R# (x) œ É x 3 2 and r# (x) œ Èx; a# œ 0 and b# œ 1
b#
#
Ê two integrals are required
shell ‰
shell
shell
(c) W2/66: V œ 'c 21 ˆ radius
Š height
‹ dy œ 'c 21y Š height
‹ dy where shell height œ y# a3y# 2b œ 2 2y# ;
d
d
c œ 0 and d œ 1. Only 98/ integral is required. It is, therefore preferable to use the =2/66 method.
However, whichever method you use, you will get V œ 1.
36. (a) H3=k: V œ V" V# V$
Vi œ 'c 1[Ri (y)]# dy, i œ 1, 2, 3 with R" (y) œ 1 and c" œ 1, d" œ 1; R# (y) œ Èy and c# œ 0 and d# œ 1;
di
i
R$ (y) œ (y)"Î% and c$ œ 1, d$ œ 0 Ê three integrals are required
(b) [ +=2/<: V œ V" V#
Vi œ 'c 1a[Ri (y)]# [ri (y)]# b dy, i œ 1, 2 with R" (y) œ 1, r" (y) œ Èy, c" œ 0 and d" œ 1;
di
i
R# (y) œ 1, r# (y) œ (y)"Î% , c# œ 1 and d# œ 0 Ê two integrals are required
shell ‰
shell
shell
(c) W2/66: V œ 'a 21 ˆ radius
Š height
‹dx œ 'a 21xŠ height
‹dx, where shell height œ x# ax% b œ x# x% ,
b
b
a œ 0 and b œ 1 Ê only one integral is required. It is, therefore preferable to use the =2/66 method.
However, whichever method you use, you will get V œ 561 .
6.3 LENGTHS OF PLANE CURVES
1.
dx
dt
œ 1 and
dy
dt
#
#
È(1)# (3)# œ È10
‰ Š dy
œ 3 Ê Êˆ dx
dt
dt ‹ œ
Ê Length œ '2/3 È10 dt œ È10 ctd 12/3 œ È10 Š 23 È10‹ œ
1
2.
dx
dt
œ sin t and
dy
dt
5È10
3
#
#
È( sin t)# (1 cos t)# œ È2 2 cos t
‰ Š dy
œ 1 cos t Ê Êˆ dx
dt
dt ‹ œ
cos t ‰
È2 ' É sin# t dt
Ê Length œ '! È2 2 cos t dt œ È2 '! Ɉ 11
cos t (1 cos t) dt œ
1 cos t
!
1
œ È2'!
1
sin t
È1 cos t
1
1
0 on [0ß 1]); [u œ 1 cos t Ê du œ sin t dt; t œ 0 Ê u œ 0,
dt (since sin t
#
t œ 1 Ê u œ 2] Ä È2 '! u"Î# du œ È2 2u"Î# ‘ ! œ 4
2
3.
dx
dt
œ 3t# and
dy
dt
#
#
‰ Š dy
Éa3t# b# (3t)# œ È9t% 9t# œ 3tÈt# 1 Šsince t
œ 3t Ê Êˆ dx
dt
dt ‹ œ
È3
Ê Length œ '! 3tÈt# 1 dt; ’u œ t# 1 Ê
Ä
4.
dx
dt
3
#
0 on ’0ß È3“‹
du œ 3t dt; t œ 0 Ê u œ 1, t œ È3 Ê u œ 4“
'14 3# u"Î# du œ u$Î# ‘ %" œ (8 1) œ 7
œ t and
dy
dt
#
#
Èt# (2t 1) œ È(t 1)# œ kt 1k œ t 1 since 0 Ÿ t Ÿ 4
‰ Š dy
œ (2t 1)"Î# Ê Êˆ dx
dt
dt ‹ œ
Ê Length œ '0 (t 1) dt œ ’ t2 t“ œ (8 4) œ 12
4
%
#
!
5.
dx
dt
œ (2t 3)"Î# and
dy
dt
#
#
È(2t 3) (1 t)# œ Èt# 4t 4 œ kt 2k œ t 2
‰ Š dy
œ 1 t Ê Êˆ dx
dt
dt ‹ œ
since 0 Ÿ t Ÿ 3 Ê Length œ '0 (t 2) dt œ ’ t2 2t“ œ
$
#
$
!
21
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.3 Lengths of Plane Curves
6.
dx
dt
œ 8t cos t and
dy
dt
œ k8tk œ 8t since 0 Ÿ t Ÿ
7.
dy
dx
œ
"
3
† 3# ax# 2b
#
#
È(8t cos t)# (8t sin t)# œ È64t# cos# t 64t# sin# t
‰ Š dy
œ 8t sin t Ê Êˆ dx
dt
dt ‹ œ
"Î#
Ê Length œ '0 8t dt œ c4t# d !
1Î2
1
#
1Î#
œ 1#
† 2x œ Èax# 2b † x
Ê L œ '0 È1 ax# 2b x# dx œ '0 È1 2x# x% dx
$
3
œ '0 Éa1 x# b# dx œ '0 a1 x# b dx œ ’x
$
œ3
8.
dy
dx
œ
3
#
œ 12
27
3
Èx Ê L œ ' É1 94 x dx; u œ 1 94 x
0
4
Ê du œ
dx Ê
9
4
du œ dx; x œ 0 Ê u œ 1; x œ 4
4
9
Ê u œ 10d Ä L œ '1 u"Î# ˆ 49 du‰ œ
10
œ
9.
dx
dy
8
27
œ y#
#
"
4y#
%
Ê Š dx
dy ‹ œ y
3
œ '1 Éy%
3
"
#
œ '1 ÊŠy#
3
$
œ ’ y3
dx
dy
œ
4
9
23 u$Î# ‘ "!
"
Š10È10 1‹
Ê L œ '1 É1 y%
10.
$
x$
3 “!
$
"
#
"
16y%
"
4y# ‹
$
y"
4 “"
#
"
#
"
16y%
"
#
"
16y%
dy
dy
dy œ '1 Šy#
3
" ‰
1#
œ ˆ 27
3
"
4y# ‹
dy
ˆ 3" 4" ‰ œ 9
#
y"Î# "# y"Î# Ê Š dx
dy ‹ œ
"
4
"
1#
"
3
"
4
œ9
(1 4 3)
1#
œ9
(2)
1#
œ
53
6
Šy 2 y" ‹
Ê L œ '1 Ê1 "4 Šy 2 y" ‹ dy
9
œ '1 Ê "4 Šy 2 y" ‹ dy œ '1
9
œ
"
#
9
"
#
"
Èy ‹
Ê ŠÈ y
*
$Î#
$
"
dx
dy
dy
'19 ˆy"Î# y"Î# ‰ dy œ "# 23 y$Î# 2y"Î# ‘ *"
œ ’ y 3 y"Î# “ œ Š 33 3‹ ˆ "3 1‰ œ 11
11.
#
œ y$
"
4y$
#
'
Ê Š dx
dy ‹ œ y
Ê L œ '1 É1 y'
2
œ '1 Éy'
2
œ '1 Šy$
2
œ Š 16
4
"
2
y$
4 ‹
"
(16)(2) ‹
"
16y'
"
2
"
16y'
"
#
32
3
dy
2
%
œ
"
16y'
dy œ '1 ÊŠy$
dy œ ’ y4
"
3
y$
4 ‹
#
dy
#
y#
8 “"
ˆ "4 "8 ‰ œ 4
"
32
"
4
"
8
œ
128184
32
œ
123
32
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
379
380
12.
Chapter 6 Applications of Definite Integrals
dx
dy
œ
y#
#
"
#y #
#
Ê Š dx
dy ‹ œ
"
4
ay% 2 y% b
Ê L œ '2 É1 "4 ay% 2 y% b dy
3
œ '2 É "4 ay% 2 y% b dy
3
13.
œ
"
#
'23 Éay# y# b# dy œ "# '23 ay# y# b dy
œ
"
#
’ y3 y" “ œ
dy
dx
$
$
"
#
#
"‰
ˆ 27
ˆ 8 " ‰‘ œ
3 3 3 #
#
#Î$
œ x"Î$ "4 x"Î$ Ê Š dy
dx ‹ œ x
Ê L œ '1 É1 x#Î$
8
œ '1 Éx#Î$
8
"
#
x#Î$
16
"
#
x#Î$
16
"
#
"
#
ˆ 26
3
8
3
#" ‰ œ
"
#
ˆ6 #" ‰ œ
13
4
x#Î$
16
dx
dx
#
œ '1 Ɉx"Î$ "4 x"Î$ ‰ dx œ '1 ˆx"Î$ "4 x"Î$ ‰ dx
8
8
)
œ 34 x%Î$ 38 x#Î$ ‘ " œ
œ
14.
dy
dx
3
8
2x%Î$ x#Î$ ‘ )
"
3
8
ca2 † 2% 2# b (2 1)d œ
œ x# 2x 1
œ (1 x)#
#
2
œ '0 É(1 x)%
"
#
œ '0 Ê’(1 x)#
2
œ '0 ’(1 x)#
2
"
#
(1x)%
16
#
(1x)#
“
4
(1x)#
“
4
(1x)%
16
99
8
"
"
4 (1x)#
%
Ê Š dy
dx ‹ œ (1 x)
Ê L œ '0 É1 (1 x)%
2
(32 4 3) œ
œ x# 2x 1
4
(4x4)#
"
"
4 (1x)#
3
8
"
#
"
16(1x)%
dx
dx
dx
dx; cu œ 1 x Ê du œ dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 3d
Ä L œ '1 ˆu# "4 u# ‰ du œ ’ u3 "4 u" “ œ ˆ9
3
$
$
"
15.
dx
dy
" ‰
1#
ˆ 3" 4" ‰ œ
108143
12
œ
106
12
œ
53
6
#
%
œ Èsec% y 1 Ê Š dx
dy ‹ œ sec y 1
1Î4
1Î4
Ê L œ '1Î4 È1 asec% y 1b dy œ '1Î4 sec# y dy
1Î%
œ ctan yd 1Î% œ 1 (1) œ 2
16.
dy
dx
#
%
œ È3x% 1 Ê Š dy
dx ‹ œ 3x 1
c1
c1
Ê L œ 'c2 È1 a3x% 1b dx œ 'c2 È3 x# dx
$
œ È3 ’ x3 “
"
#
œ
È3
3
c1 (2)$ d œ
È3
3
(" 8) œ
7È 3
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.3 Lengths of Plane Curves
17. (a)
dy
dx
#
(b)
#
œ 2x Ê Š dy
dx ‹ œ 4x
Ê L œ 'c1 Ê1 Š dy
dx ‹ dx
#
2
œ 'c1 È1 4x# dx
2
(c) L ¸ 6.13
18. (a)
dy
dx
#
(b)
%
œ sec# x Ê Š dy
dx ‹ œ sec x
Ê L œ 'c1Î3 È1 sec% x dx
0
(c) L ¸ 2.06
19. (a)
dx
dy
#
(b)
#
œ cos y Ê Š dx
dy ‹ œ cos y
Ê L œ '0 È1 cos# y dy
1
(c) L ¸ 3.82
20. (a)
dx
dy
#
œ È1y y# Ê Š dx
dy ‹ œ
1Î2
Ê L œ 'c1Î2 É1
œ 'c1Î2 a1 y# b
1Î2
"Î#
y#
a1 y # b
y#
1 y#
(b)
1Î2
dy œ '1Î2 É 1 " y# dy
dy
(c) L ¸ 1.05
21. (a) 2y 2 œ 2
dx
dy
#
#
Ê Š dx
dy ‹ œ (y 1)
(b)
Ê L œ 'c1 È1 (y 1)# dy
3
(c) L ¸ 9.29
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
381
382
Chapter 6 Applications of Definite Integrals
22. (a)
dy
dx
#
(b)
#
#
œ cos x - cos x + x sin x Ê Š dy
dx ‹ œ x sin x
Ê L œ '0 È1 x# sin# x dx
1
(c) L ¸ 4.70
23. (a)
dy
dx
#
(b)
#
œ tan x Ê Š dy
dx ‹ œ tan x
# x cos# x
Ê L œ '0 È1 tan# x dx œ '0 É sin cos
dx
#x
1Î6
œ '0
1Î6
1Î6
œ '0 sec x dx
1Î6
dx
cos x
(c) L ¸ 0.55
24. (a)
dx
dy
#
(b)
#
œ Èsec# y 1 Ê Š dx
dy ‹ œ sec y 1
Ê L œ 'c1Î3 È1 asec# y 1b dy
1Î4
1Î4
1Î4
œ 'c1Î3 ksec yk dy œ '1Î3 sec y dy
(c) L ¸ 2.20
25. È2 x œ '0 Ê1 Š dy
dt ‹ dt, x
#
x
#
0 Ê È2 œ Ê1 Š dy
dx ‹ Ê
dy
dx
œ „ 1 Ê y œ f(x) œ „ x C where C is any
real number.
26. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see that
dy œ f w (xkc1 ) ˜ xk Ê length of kth tangent fin is
È( ˜ xk )# (dy)# œ È( ˜ xk )# [f w (xkc1 ) ˜ xk ]# .
n
n
! (length of kth tangent fin) œ lim ! È( ˜ xk )# [f w (xk1 ) ˜ xk ]#
(b) Length of curve œ n lim
Ä_
nÄ_
kœ1
! È1 [f w (xk1 )]# ˜ xk œ ' È1 [f w (x)]# dx
œ n lim
Ä_
a
n
kœ1
b
kœ1
#
"
27. (a) Š dy
dx ‹ correspondes to 4x here, so take
So y œ Èx from ("ß ") to (4ß 2).
dy
dx
as
"
.
#È x
Then y œ Èx C and since ("ß ") lies on the curve, C œ 0.
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.3 Lengths of Plane Curves
#
28. (a) Š dx
dy ‹ correspondes to
So y œ
"
y%
here, so take
dy
dx
as
"
y# .
Then x œ y" C and, since (!ß ") lies on the curve, C œ 1
"
"x.
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
29. (a)
dx
dt
œ 2 sin 2t and
dy
dt
Ê Length œ '0 2 dt œ c2td !
1Î2
(b)
dx
dt
œ 1 cos 1t and
1Î#
dy
dt
#
#
È(2 sin 2t)# (2 cos 2t)# œ 2
‰ Š dy
œ 2 cos 2t Ê Êˆ dx
dt
dt ‹ œ
œ1
#
#
È(1 cos 1t)# (1 sin 1t)# œ 1
‰ Š dy
œ 1 sin 1t œ ʈ dx
dt
dt ‹ œ
Ê Length œ 'c1Î2 1 dt œ c1td "Î# œ 1
1Î2
30. x œ a() sin )) Ê
Ê
dy
d)
dx
d)
"Î#
‰# œ a# a1 2 cos ) cos# )b and y œ a(1 cos ))
œ a(1 cos )) Ê ˆ dx
d)
#
#
' ˆ dx ‰# Š dyd) ‹ d) œ '0 È2a# (1 cos )) d)
œ a sin ) Ê Š dy
d) ‹ œ a sin ) Ê Length œ 0 Ê d)
#
œ aÈ2'0 È2 É 1 #cos ) d) œ 2a '0 ¸sin #) ¸ d) œ 2a '0 sin
21
#
21
21
21
)
#
21
#1
d) œ 4a cos 2) ‘ ! œ 8a
31-36. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
with( student );
f := x -> sqrt(1-x^2);a := -1;
b := 1;
N := [2, 4, 8 ];
for n in N do
xx := [seq( a+i*(b-a)/n, i=0..n )];
pts := [seq([x,f(x)],x=xx)];
L := simplify(add( distance(pts[i+1],pts[i]), i=1..n ));
T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L );
P[n] := plot( [f(x),pts], x=a..b, title=T ):
end do:
display( [seq(P[n],n=N)], insequence=true, scaling=constrained );
L := ArcLength( f(x), x=a..b, output=integral ):
L = evalf( L );
# (b)
# (a)
# (c)
37-40. Example CAS commands:
Maple:
with( plots );
with( student );
x := t -> t^3/3;
y := t -> t^2/2;
a := 0;
b := 1;
N := [2, 4, 8 ];
for n in N do
tt := [seq( a+i*(b-a)/n, i=0..n )];
pts := [seq([x(t),y(t)],t=tt)];
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
383
384
Chapter 6 Applications of Definite Integrals
L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n ));
T := sprintf("#37(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L );
P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ):
end do:
display( [seq(P[n],n=N)], insequence=true );
ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ):
L := Int( ds(t), t=a..b ):
L = evalf(L);
# (b)
# (a)
# (c)
31-40. Example CAS commands:
Mathematica: (assigned function and values for a, b, and n may vary)
Clear[x, f]
{a, b} = {1, 1}; f[x_] = Sqrt[1 x2 ]
p1 = Plot[f[x], {x, a, b}]
n = 8;
pts = Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N
Show[{p1,Graphics[{Line[pts]}]}]
Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]])2 (pts[[i 1, 2]] pts[[i, 2]])2 ], {i, 1, n}]
NIntegrate[ Sqrt[ 1 f'[x]2 ],{x, a, b}]
6.4 MOMENTS AND CENTERS OF MASS
1. Because the children are balanced, the moment of the system about the origin must be equal to zero:
5 † 80 œ x † 100 Ê x œ 4 ft, the distance of the 100-lb child from the fulcrum.
2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x œ a and the
200-lb end at x œ a. Then the center of mass x satisfies x œ
at a distance a
or
2
3
a
3
œ
2a
3
œ
"
3
(2a) which is
"
3
100(a) 200(a)
300
Ê x œ 3a . That is, x is located
of the length of the log from the 200-lb (heavier) end (see figure)
of the way from the lighter end toward the heavier end.
"
3
(2a)
èëëéëëê
100 lbs.
ñïïïïïïïïïïïïïïñïïïïñïïïïïïñ
a 200 lbs
a
x œ a/3
!
3. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point
m
masses located at the centers of the rods at coordinates ˆ L# ß !‰ and ˆ0ß L# ‰. Therefore x œ y
m
œ
x" m" x# m#
m" m#
œ
L
# †m0
mm
œ
L
4
and y œ
mx
m
œ
y" m# y# m#
m" m#
œ
0 L2 †m
mm
œ
L
4
Ê
ˆ L4 ß L4 ‰
is the center of
mass location.
4. Let the rods have lengths x œ L and y œ 2L. The center of mass of each rod is in its center (see Example 1).
The rod system is equivalent to two point masses located at the centers of the rods at coordinates ˆ L# ß !‰ and
(!ß L). Therefore x œ
L
# †m0†2m
m2m
œ
5. M! œ '0 x † 4 dx œ ’4 x# “ œ 4 †
2
#
#
!
6. M! œ '1 x † 4 dx œ ’4 x# “ œ
3
#
$
"
4
#
!†mL†2m
m2m
L
6
and y œ
œ
4
#
œ 8; M œ '0 4 dx œ [4x]#! œ 4 † 2 œ 8 Ê x œ
2
2L
3
‰
Ê ˆ L6 ß 2L
3 is the center of mass location.
M!
M
œ1
(9 1) œ 16; M œ '1 4 dx œ [4x]$" œ 12 4 œ 8 Ê x œ
3
M!
M
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
16
8
œ2
Section 6.4 Moments and Centers of Mass
7. M! œ '0 x ˆ1 x3 ‰ dx œ '0 Šx
3
œ3
3
œ
9
6
9
#
Ê xœ
ˆ 15
‰
2
ˆ 92 ‰
œ
M!
M
x#
3‹
œ
8. M! œ '0 x ˆ2 x4 ‰ dx œ '0 Š2x
4
4
M œ '0 ˆ2 x4 ‰ dx œ ’2x
4
9. M! œ '1 x Š1
4
"
Èx ‹
%
x#
8 “!
15
9
$
x$
9 “!
#
dx œ ’ x#
œ
x#
4‹
œ
27 ‰
9
M œ '0 ˆ1 3x ‰ dx œ ’x
3
15
# ;
16
8
%
x$
12 “ !
œ ˆ16
64 ‰
12
œ
œ
œ6 Ê xœ
dx œ '1 ˆx x"Î# ‰ dx œ ’ x#
4
#
M!
M
%
2x$Î#
3 “"
32
3 †6
œ ˆ8
œ 16
16 ‰
3
M!
M
10. M! œ '1Î4 x † 3 ˆx$Î# x&Î# ‰ dx œ 3'1Î4 ˆx"Î# x$Î# ‰ dx œ 3 2x"Î#
1
1
œ 6 14 œ 20 Ê x œ
M!
M
œ
2 ‘"
3x$Î# "Î%
œ
9
3
1
2
œ 3; M œ '0 (2 x) dx '1 x dx œ ’2x
1
2
"
x#
# “!
#
# #
’ x# “
"
œ ˆ2
12. M! œ '0 x(x 1) dx '1 2x dx œ '0 ax# xb dx '1 2x dx œ ’ x3
œ3
œ
32
3 ;
ˆ 73
‰
6
5
œ
2 ‘"
x"Î# "Î%
œ
15
#
14
3
2
1
1
2
2
œ
23
6 ;
Ê xœ
M!
M
‰ ˆ 72 ‰ œ
œ ˆ 23
6
œ
73
6
;
73
30
œ 3 ’(2 2) Š2 †
16 ‰‘
3
$
"‰
#
"
x$
3 “!
"
#
2
ˆ "# ‰ ‹“
œ 3 ˆ2
14 ‰
3
"
x#
2 “!
ˆ 4#
#
$
’ x3 “ œ ˆ1 "3 ‰ ˆ 83 "3 ‰
"
"‰
#
œ3 Ê xœ
#
"
M!
M
œ1
#
cx# d " œ ˆ "3 2" ‰ (4 1)
M œ '0 (x 1) dx '1 2 dx œ ’ x# x“ c2xd #" œ ˆ "# 1‰ (4 #) œ 2
5
6
4528
6
œ
9
20
2
1
2
3
œ 3 ˆ2 32 ‰ ˆ4
11. M! œ '0 x(2 x) dx '1 x † x dx œ '0 a2x x# b dx '1 x# dx œ ’ 2x#
1
œ 16 †
ˆ "# 32 ‰ œ
4
œ 3(4 1) œ 9; M œ 3'1Î4 ˆx$Î# x&Î# ‰ dx œ 3 x"Î#2
16
3
16
9
%
M œ '1 ˆ1 x"Î# ‰ dx œ x 2x"Î# ‘ " œ (4 4) (1 2) œ 5 Ê x œ
1
$
x#
6 “!
5
3
dx œ ’x#
œ8
œ ˆ 92
385
!
3
#
œ
7
#
23
21
13. Since the plate is symmetric about the y-axis and its density is
constant, the distribution of mass is symmetric about the y-axis
and the center of mass lies on the y-axis. This means that
x œ 0. It remains to find y œ MMx . We model the distribution of
mass with @/<>3-+6 strips. The typical strip has center of mass:
#
(µ
x ßµ
y ) œ Šxß x 4 ‹ , length: 4 x# , width: dx, area:
#
#
dA œ a4 x b dx, mass: dm œ $ dA œ $ a4 x# b dx. The moment of the strip about the x-axis is
#
µ
C dm œ Š x #4 ‹ $ a4 x# b dx œ #$ a16 x% b dx. The moment of the plate about the x-axis is Mx œ ' µ
C dm
œ 'c2 #$ a16 x% b dx œ
2
$
#
’16x
#
x&
5 “ #
plate is M œ ' $ a4 x# b dx œ $ ’4x
œ
$
#
’Š16 † 2
#
x$
3 “ #
2&
5‹
œ 2$ ˆ8 83 ‰ œ
32$
3 .
2&
5 ‹“
œ
$ †2
#
ˆ32
Therefore y œ
Mx
M
œ
Š16 † 2
32 ‰
5
$
Š 128
5 ‹
Š 323$ ‹
‰
mass is the point (xß y) œ ˆ!ß 12
5 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
œ
128$
5 .
12
5 .
The mass of the
The plate's center of
386
Chapter 6 Applications of Definite Integrals
14. Applying the symmetry argument analogous to the one in
Exercise 13, we find x œ 0. To find y œ MMx , we use the
@/<>3-+6 strips technique. The typical strip has center of
#
mass: (µ
x ßµ
y ) œ Šxß 25 x ‹ , length: 25 x# , width: dx,
#
#
area: dA œ a25 x bdx, mass: dm œ $ dA œ $ a25 x# b dx.
The moment of the strip about the x-axis is
#
µ
y dm œ Š 25 # x ‹ $ a25 x# b dx œ
œ 'c5 #$ a25 x# b dx œ
5
#
œ $ † 625 ˆ5
œ 2$ Š5$
10
3
5$
3‹
$
#
'c55
a25 x# b dx. The moment of the plate about the x-axis is Mx œ ' µ
y dm
#
$
#
$
#
a625 50x# x% b dx œ
’625x
50
3
x$
&
x&
5 “ &
œ 2 † #$ Š625 † 5
50
3
† 5$
1‰ œ $ † 625 † ˆ 38 ‰ . The mass of the plate is M œ ' dm œ 'c5 $ a25 x# b dx œ $ ’25x
5
œ
$ † 5$ . Therefore y œ
4
3
Mx
M
œ
$ †5% † ˆ 83 ‰
$ †5$ †ˆ 43 ‰
5&
5‹
&
x$
3 “ &
œ 10. The plate's center of mass is the point (xß y) œ (!ß 10).
15. Intersection points: x x# œ x Ê 2x x# œ 0
Ê x(2 x) œ 0 Ê x œ 0 or x œ 2. The typical @/<>3-+6
#
strip has center of mass: (µ
x ßµ
y ) œ Šxß ax x b (x) ‹
#
œ Šxß
x#
# ‹,
#
length: ax x b (x) œ 2x x# , width: dx,
area: dA œ a2x x# b dx, mass: dm œ $ dA œ $ a2x x# b dx.
The moment of the strip about the x-axis is
#
µ
y dm œ Š x# ‹ $ a2x x# b dx; about the y-axis it is µ
x dm œ x † $ a2x x# b dx. Thus, Mx œ ' µ
y dm
œ '0 ˆ #$ x# ‰ a2x x# b dx œ #$ '0 a2x$ x% b dx œ #$ ’ x#
2
2
%
#
x&
5 “!
œ #$ Š2$
œ 45$ ; My œ ' µ
x dm œ '0 x † $ a2x x# b dx œ $ '0 a2x# x$ b œ $ ’ 23 x$
2
2
M œ ' dm œ '0 $ a2x x# b dx œ $ '0 a2x x# b dx œ $ ’x#
2
2
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1 and y œ
Mx
M
#
x$
3 “!
#
x%
4 “!
2&
5‹
œ #$ † 2$ ˆ1 45 ‰
œ $ Š2 †
œ $ ˆ4 83 ‰ œ
4$
3
2$
3
2%
4‹
œ
. Therefore, x œ
$ †2%
1#
œ
My
M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ 35 Ê (xß y) œ ˆ1ß 35 ‰ is the center of mass.
16. Intersection points: x# 3 œ 2x# Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Applying the
symmetry argument analogous to the one in Exercise 13, we
find x œ 0. The typical @/<>3-+6 strip has center of mass:
#
#
#
(µ
x ßµ
y ) œ Šxß 2x ax 3b ‹ œ Šxß x 3 ‹ ,
#
#
#
#
#
length: 2x ax 3b œ 3 a1 x b, width: dx,
area: dA œ 3 a1 x# b dx, mass: dm œ $ dA œ 3$ a1 x# b dx.
The moment of the strip about the x-axis is
µ
y dm œ 3 $ ax# 3b a1 x# b dx œ 3 $ ax% 3x# x# 3b dx œ
#
œ
3
#
$ 'c1 ax% 2x# 3b dx œ
1
#
3
#
&
$ ’ x5
2x$
3
M œ ' dm œ 3$ 'c1 a1 x# b dx œ 3$ ’x
1
3x“
"
x$
3 “ "
"
"
œ
3
#
3
#
$ ax% 2x# 3b dx; Mx œ ' µ
y dm
† $ † 2 ˆ 5"
2
3
45 ‰
3‰ œ 3$ ˆ 310
œ 325$ ;
15
œ 3$ † 2 ˆ1 3" ‰ œ 4$ . Therefore, y œ
Mx
M
œ 5$††$32†4 œ 58
Ê (xß y) œ ˆ0ß 85 ‰ is the center of mass.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
4$
3
;
Section 6.4 Moments and Centers of Mass
387
17. The typical 29<3D98>+6 strip has center of mass:
$
(µ
x ßµ
y ) œ Š y y ß y‹ , length: y y$ , width: dy,
#
area: dA œ ay y$ b dy, mass: dm œ $ dA œ $ ay y$ b dy.
The moment of the strip about the y-axis is
$
#
µ
x dm œ $ Š y y ‹ ay y$ b dy œ $ ay y$ b dy
#
œ
$
#
#
#
%
'
ay 2y y b dy; the moment about the x-axis is
1
$
µ
y dm œ $ y ay y$ b dy œ $ ay# y% b dy. Thus, Mx œ ' µ
y dm œ $ '0 ay# y% b dy œ $ ’ y3
My œ ' µ
x dm œ
$
#
'01 ay# 2y% y' b dy œ #$ ’ y3
$
œ $ '0 ay y$ b dy œ $ ’ y#
1
œ
#
"
y%
4 “!
2y&
5
œ $ ˆ "# 4" ‰ œ
$
4
"
y(
7 “!
œ
$
#
ˆ "3
. Therefore, x œ
2
5
7" ‰ œ
$
#
œ $ ˆ "3 "5 ‰ œ
15 ‰
ˆ 35 3†42
œ
5†7
4$ ‰ ˆ 4 ‰
œ ˆ 105
$ œ
My
M
"
y&
5 “!
16
105
2$
15
;
4$
105
; M œ ' dm
Mx
M
2$ ‰ ˆ 4 ‰
œ ˆ 15
$
and y œ
16 8 ‰
Ê (xß y) œ ˆ 105
ß 15 is the center of mass.
8
15
18. Intersection points: y œ y# y Ê y# 2y œ 0
Ê y(y 2) œ 0 Ê y œ 0 or y œ 2. The typical
29<3D98>+6 strip has center of mass:
#
#
(µ
x ßµ
y ) œ Š ay yby ß y‹ œ Š y ß y‹ ,
#
2
#
length: y ay yb œ 2y y# , width: dy,
area: dA œ a2y y# b dy, mass: dm œ $ dA œ $ a2y y# b dy.
The moment about the y-axis is µ
x dm œ #$ † y# a2y y# b dy
œ #$ a2y$ y% b dy; the moment about the x-axis is µ
y dm œ $ y a2y y# b dy œ $ a2y# y$ b dy. Thus,
Mx œ ' µ
y dm œ '0 $ a2y# y$ b dy œ $ ’ 2y3
2
œ '0
2
$
#
$
a2y$ y% b dy œ
œ $ ’y#
#
y$
3 “!
$
#
%
’ y2
œ $ ˆ4 83 ‰ œ
#
y&
5 “!
4$
3
œ
$
#
ˆ8
#
y%
4 “!
16$
1#
ˆ 405 32 ‰ œ
4$
5
; M œ ' dm œ '0 $ a2y y# b dy
œ
$
#
My
M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ
32 ‰
5
. Therefore, x œ
œ
(4 3) œ
4$
3
; My œ ' µ
x dm
16 ‰
4
œ $ ˆ 16
3
2
3
5
and y œ
Mx
M
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1
Ê (xß y) œ ˆ 35 ß "‰ is the center of mass.
19. Applying the symmetry argument analogous to the one used
in Exercise 13, we find x œ 0. The typical @/<>3-+6 strip has
center of mass: (µ
x ßµ
y ) œ ˆxß cos# x ‰ , length: cos x, width: dx,
area: dA œ cos x dx, mass: dm œ $ dA œ $ cos x dx. The
moment of the strip about the x-axis is µ
y dm œ $ † cos# x † cos x dx
2x ‰
œ #$ cos# x dx œ #$ ˆ 1 cos
dx œ 4$ (1 cos 2x) dx; thus,
#
1Î2
Mx œ ' µ
y dm œ 'c1Î2 4$ (1 cos 2x) dx œ
1Î#
œ $ [sin x]1Î# œ 2$ . Therefore, y œ
Mx
M
œ
$
4
x
$1
4 †# $
œ
sin 2x ‘ 1Î#
#
1Î#
1
8
œ
$
4
ˆ 1# 0‰ ˆ 1# ‰‘ œ
$1
4
Ê (xß y) œ ˆ!ß 18 ‰ is the center of mass.
20. Applying the symmetry argument analogous to the one used
in Exercise 13, we find x œ 0. The typical vertical strip has
#
center of mass: (µ
x ßµ
y ) œ Šxß sec# x ‹ , length: sec# x, width: dx,
area: dA œ sec# x dx, mass: dm œ $ dA œ $ sec# x dx. The
#
moment about the x-axis is µ
y dm œ Š sec x ‹ a$ sec# xb dx
œ
$
#
1Î4
sec% x dx. Mx œ 'c1Î4 µ
y dm œ
#
$
#
1Î2
; M œ ' dm œ $ '1Î2 cos x dx
'11ÎÎ44 sec% x dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
388
Chapter 6 Applications of Definite Integrals
œ
$
#
'c11ÎÎ44 atan# x 1b asec# xb dx œ #$ '11ÎÎ44 (tan x)# asec# xb dx #$ '11ÎÎ44 sec# x dx œ 2$ ’ (tan3x) “ 1Î4
œ
$
2
3" ˆ 3" ‰‘ #$ [1 (1)] œ
$
1Î%
1 Î4
Therefore, y œ
Mx
M
œ ˆ 43$ ‰ ˆ 2"$ ‰ œ
2
3
$
3
$ œ
4$
3
#$ [tan x]1Î%
; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]1Î4 œ $ [1 (1)] œ 2$ .
1Î4
1Î4
Ê (xß y) œ ˆ!ß 32 ‰ is the center of mass.
21. Since the plate is symmetric about the line x œ 1 and its
density is constant, the distribution of mass is symmetric
about this line and the center of mass lies on it. This means
that x œ 1. The typical @/<>3-+6 strip has center of mass:
#
#
#
(µ
x ßµ
y ) œ Šxß a2x x ba2x 4xb ‹ œ Šxß x 2x ‹ ,
#
#
#
#
#
length: a2x x b a2x 4xb œ 3x 6x œ 3 a2x x# b ,
width: dx, area: dA œ 3 a2x x# b dx, mass: dm œ $ dA
œ 3$ a2x x# b dx. The moment about the x-axis is
#
µ
y dm œ 3 $ ax# 2xb a2x x# b dx œ 3 $ ax# 2xb dx
#
#
œ 3# $ ax% 4x$ 4x# b dx. Thus, Mx œ ' µ
y dm œ '0
2
œ $
3
2
&
Š 25
%
2
4
3
$
%
†2 ‹œ $†2
3
#
œ '0 3$ a2x x# b dx œ 3$ ’x#
2
#
x$
3 “!
ˆ 25
1
2‰
3
3
2
&
$ ax% 4x$ 4x# b dx œ 32 $ ’ x5 x% 34 x$ “
%
œ $ †2
3
#
10 ‰
ˆ 6 15
15
œ 3$ ˆ4 83 ‰ œ 4$ . Therefore, y œ
Mx
M
œ
8$
5
; M œ ' dm
#
!
œ ˆ 85$ ‰ ˆ 4"$ ‰ œ 25
Ê (xß y) œ ˆ1ß 25 ‰ is the center of mass.
22. (a) Since the plate is symmetric about the line x œ y and
its density is constant, the distribution of mass is
symmetric about this line. This means that x œ y. The
typical @/<>3-+6 strip has center of mass:
È
#
(µ
x ßµ
y ) œ Šxß 9 x ‹ , length: È9 x# , width: dx,
#
area: dA œ È9 x# dx,
mass: dm œ $ dA œ $ È9 x# dx.
The moment about the x-axis is
È
#
µ
y dm œ $ Š 9# x ‹ È9 x# dx œ
$
#
a9 x# b dx. Thus, Mx œ ' µ
y dm œ '0
3
$
#
a9 x# b dx œ
$
#
’9x
$
x$
3 “!
(27 9) œ 9$ ; M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a quarter of a circle of radius 3) œ $ ˆ 941 ‰ œ
4 ‰
Therefore, y œ MMx œ (9$ ) ˆ 91$
œ 14 Ê (xß y) œ ˆ 14 ß 14 ‰ is the center of mass.
œ
$
#
(b) Applying the symmetry argument analogous to the one
used in Exercise 13, we find that x œ 0. The typical
vertical strip has the same parameters as in part (a).
3
Thus, M œ ' µ
y dm œ ' $ a9 x# b dx
x
œ #'0
3
$
#
c3 #
a9 x# b dx œ 2(9$ ) œ 18$ ;
M œ ' dm œ ' $ dA œ $ ' dA
œ $ (Area of a semi-circle of radius 3) œ $ ˆ 921 ‰ œ 91$
2 . Therefore, y œ
4
as in part (a) Ê (xß y) œ ˆ0ß 1 ‰ is the center of mass.
Mx
M
2 ‰
œ (18$ ) ˆ 91$
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
4
1
, the same y
91$
4
.
Section 6.4 Moments and Centers of Mass
389
23. Since the plate is symmetric about the line x œ y and its
density is constant, the distribution of mass is symmetric
about this line. This means that x œ y. The typical @/<>3-+6
strip has
È
#
center of mass: (µ
x ßµ
y ) œ Šxß 3 9 x ‹ ,
#
length: 3 È9 x# , width: dx,
area: dA œ Š3 È9 x# ‹ dx,
mass: dm œ $ dA œ $ Š3 È9 x# ‹ dx.
The moment about the x-axis is
µ
y dm œ $
Š3 È9 x# ‹ Š3 È9 x# ‹
#
dx œ
$
#
c9 a9 x# bd dx œ
$ x#
#
dx. Thus, Mx œ '0
3
$ x#
#
dx œ
$
6
$
cx$ d ! œ
#
9$
#
equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 Ê A œ 3
œ
9
4
9$
4
(4 1) Ê M œ $ A œ
(4 1). Therefore, y œ
Mx
M
œ ˆ 9#$ ‰ ’ 9$(44 1) “ œ
2
41
. The area
19
4
Ê (xß y) œ ˆ 4 2 1 ß 4 2 1 ‰ is the
center of mass.
24. Applying the symmetry argument analogous to the one used
in Exercise 13, we find that y œ 0. The typical @/<>3-+6 strip
has center of mass: (µ
x ßµ
y ) œ Œxß
"
x$
length:
ˆ x"$ ‰ œ
2
x$
"
x$
x"$
#
œ (xß 0),
, width: dx, area: dA œ
2
x$
dx,
2$
x$
mass: dm œ $ dA œ dx. The moment about the y-axis is
a
µ
x dm œ x † 2x$$ dx œ 2x$# dx. Thus, My œ ' µ
x dm œ '1 2x$# dx
œ 2$ x" ‘ " œ 2$ ˆ "a 1‰ œ
a
xœ
My
M
œ
’ 2$(aa 1) “
25. Mx œ ' µ
y dm œ '1
2
#
’ $ aa#a 1b “
Š x2# ‹
#
2$ (a1)
a
œ
2a
a1
; M œ ' dm œ '1
a
Ê (xß y) œ
2$
x$
ˆ a 2a
‰
1ß 0 .
dx œ $ x"# ‘ " œ $ ˆ a"# 1‰ œ
a
$ aa# 1b
a#
. Therefore,
Also, a lim
x œ 2.
Ä_
† $ † ˆ x2# ‰ dx
œ '1 ˆ x"# ‰ ax# b ˆ x2# ‰ dx œ '1
2
2
2
x#
dx œ 2'1 x# dx
2
#
œ 2 cx" d " œ 2 ˆ "# ‰ (1)‘ œ 2 ˆ "# ‰ œ 1;
My œ ' µ
x dm œ '1 x † $ † ˆ x2# ‰ dx
2
œ '1 x ax# b ˆ x2# ‰ dx œ 2'1 x dx œ 2 ’ x# “
2
2
#
#
"
œ 2 ˆ2 "# ‰ œ 4 1 œ 3; M œ ' dm œ '1 $ ˆ x2# ‰ dx œ '1 x# ˆ x2# ‰ dx œ 2'1 dx œ 2[x]"# œ 2(2 1) œ 2. So
xœ
My
M
œ
3
#
and y œ
Mx
M
œ
"
#
2
2
2
Ê (xß y) œ ˆ 3# ß "# ‰ is the center of mass.
26. We use the @/<>3-+6 strip approach:
1
#
M œ'µ
y dm œ ' ax x b ax x# b † $ dx
x
œ
0
"
#
#
'0 ax# x% b † 12x dx
1
œ 6'0 ax$ x& b dx œ 6 ’ x4
1
œ 6 ˆ "4 6" ‰ œ
%
6
4
1œ
"
#
"
x'
6 “!
;
My œ ' µ
x dm œ '0 x ax x# b † $ dx œ '0 ax# x$ b † 12x dx œ 12'0 ax$ x% b dx œ 12 ’ x4
1
1
1
%
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
x&
5 “!
œ 12 ˆ "4 5" ‰
390
Chapter 6 Applications of Definite Integrals
œ
12
#0
xœ
œ
; M œ ' dm œ ' ax x# b † $ dx œ 12'0 ax# x$ b dx œ 12 ’ x3
1
3
5
1
$
0
My
M
œ
3
5
and y œ
Mx
M
œ
"
#
"
x%
4 “!
œ 12 ˆ "3 4" ‰ œ
12
12
œ 1. So
Ê ˆ 35 ß "# ‰ is the center of mass.
shell ‰
shell
27. (a) We use the shell method: V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21x ’ È4x Š È4x ‹“ dx
b
œ 161'1
4
x
Èx
4
%
dx œ 161'1 x"Î# dx œ 161 32 x$Î# ‘ " œ 161 ˆ 32 † 8 32 ‰ œ
4
(b) Since the plate is symmetric about the x-axis and its density $ (x) œ
"
x
321
3
(8 1) œ
2241
3
is a function of x alone, the
distribution of its mass is symmetric about the x-axis. This means that y œ 0. We use the vertical strip
4
4
4
approach to find x: My œ ' µ
x dm œ '1 x † ’ È4x Š È4x ‹“ † $ dx œ '1 x † È8x † x" dx œ 8'1 x"Î# dx
4
œ 8 2x"Î# ‘ " œ 8(2 † 2 2) œ 16; M œ ' dm œ '1 ’ È4x Š È
‹“ † $ dx œ 8'1 Š È"x ‹ ˆ "x ‰ dx œ 8'1 x$Î# dx
x
%
4
%
œ 8 2x"Î# ‘ " œ 8[1 (2)] œ 8. So x œ
My
M
4
œ
4
œ 2 Ê (xß y) œ (2ß 0) is the center of mass.
16
8
(c)
28. (a) We use the disk method: V œ 'a 1R# (x) dx œ '1 1 ˆ x4# ‰ dx œ 41'1 x# dx œ 41 x" ‘ "
b
4
4
%
‘
œ 41 "
4 (1) œ 1[1 4] œ 31
(b) We model the distribution of mass with vertical strips: Mx œ ' µ
y dm œ '1
4
2
œ 2'1 x$Î# dx œ 2 ’ È
x dm œ '1 x †
“ œ 2[1 (2)] œ 2; My œ ' µ
x
%
4
$Î#
4
"
%
2‘
œ 2 ’ 2x3 “ œ 2 16
3 3 œ
"
œ 2(4 2) œ 4. So x œ
My
M
28
3
œ
; M œ ' dm œ '1
4
ˆ 28
‰
3
4
œ
7
3
and y œ
2
x
Mx
M
† $ dx œ 2'1
4
œ
2
4
œ
"
#
Èx
x
2
x
ˆ 2x ‰
2
† ˆ 2x ‰ † $ dx œ '1
4
2
x#
† Èx dx
† $ dx œ 2'1 x"Î# dx
4
dx œ 2'1 x"Î# dx œ 2 2x"Î# ‘ "
%
4
Ê (xß y) œ ˆ 73 ß "# ‰ is the center of mass.
(c)
29. The mass of a horizontal strip is dm œ $ dA œ $ L dy, where L is the width of the triangle at a distance of y above
its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have
Ê Lœ
b
h
(h y). Thus, Mx œ ' µ
y dm œ '0 $ y ˆ bh ‰ (h y) dy œ
h
œ
$b
h
Š h#
$
h$
3‹
œ
$b
h
#
h#
2‹
Šh
œ $ bh# ˆ "# 3" ‰ œ
œ
$ bh
2
. So y œ
Mx
M
$ bh#
6
œ
$b
h
#
ˆ 2 ‰
Š $bh
6 ‹ $ bh
œ
h
3
œ
hy
h
'0h ahy y# b dy œ $hb ’ hy#
; M œ ' dm œ '0 $ ˆ hb ‰ (h y) dy œ
h
L
b
$b
h
#
h
y$
3 “!
'0h ah yb dy œ $hb ’hy y2 “ h
Ê the center of mass lies above the base of the
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
#
!
Section 6.4 Moments and Centers of Mass
triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be
placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the
medians, as claimed.
30. From the symmetry about the y-axis it follows that x œ 0.
It also follows that the line through the points (!ß !) and
(!ß $) is a median Ê y œ "3 (3 0) œ 1 Ê (xß y) œ (!ß ").
31. From the symmetry about the line x œ y it follows that
x œ y. It also follows that the line through the points (!ß !)
and ˆ "# ß "# ‰ is a median Ê y œ x œ 23 † ˆ "# 0‰ œ 3"
Ê (xß y) œ ˆ "3 ß 3" ‰ .
32. From the symmetry about the line x œ y it follows that
x œ y. It also follows that the line through the point (!ß !)
and ˆ #a ß #a ‰ is a median Ê y œ x œ 32 ˆ #a 0‰ œ 3" a
Ê (xß y) œ ˆ 3a ß 3a ‰ .
33. The point of intersection of the median from the vertex (0ß b)
to the opposite side has coordinates ˆ!ß #a ‰
Ê y œ (b 0) † "3 œ b3 and x œ ˆ #a !‰ † 32 œ 3a
Ê (xß y) œ ˆ 3a ß b3 ‰ .
34. From the symmetry about the line x œ
a
#
it follows that
xœ
It also follows that the line through the points
a
ˆ # ß !‰ and ˆ #a ß b‰ is a median Ê y œ "3 (b 0) œ b3
a
#.
Ê (xß y) œ ˆ #a ß b3 ‰ .
35. y œ x"Î# Ê dy œ
"
#
x"Î# dx
Ê ds œ È(dx)# (dy)# œ É1
Mx œ $ '0 Èx É1
2
œ $ '0 Éx
2
dx œ
" ‰$Î#
4
œ
2$
3
œ
2$ ˆ 9 ‰$Î#
3 ’ 4
’ˆ2
"
4
ˆ 4" ‰
"
4x
"
4x
dx
2$
3
$Î#
’ˆx 4" ‰ “
dx ;
#
!
ˆ 4" ‰$Î# “
$Î#
“œ
2$
3
"‰
ˆ 27
8 8 œ
13$
6
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
391
392
Chapter 6 Applications of Definite Integrals
36. y œ x$ Ê dy œ 3x# dx
Ê dx œ É(dx)# a3x# dxb# œ È1 9x% dx;
Mx œ $ '0 x$ È1 9x% dx;
1
"
36
[u œ 1 9x% Ê du œ 36x$ dx Ê
du œ x$ dx;
x œ 0 Ê u œ 1, x œ 1 Ê u œ 10]
Ä Mx œ $ '1
10
"
36
u"Î# du œ
$
36
23 u$Î# ‘ "! œ
"
$
54
ˆ10$Î# 1‰
37. From Example 6 we have Mx œ '0 a(a sin ))(k sin )) d) œ a# k'0 sin# ) d) œ
1
œ
a# k
#
)
sin 2) ‘ 1
#
!
œ
a# k 1
#
1
'01 (1 cos 2)) d)
; My œ '0 a(a cos ))(k sin )) d) œ a# k '0 sin ) cos ) d) œ
1
1
M œ '0 ak sin ) d) œ ak[ cos )]1! œ 2ak. Therefore, x œ
1
a# k
#
My
M
œ 0 and y œ
Mx
M
a# k
#
1
csin# )d ! œ 0;
#
" ‰
œ Š a 2k1 ‹ ˆ 2ak
œ
a1
4
Ê ˆ!ß a41 ‰
is the center of mass.
38. Mx œ ' µ
y dm œ '0 (a sin )) † $ † a d)
1
œ '0 aa# sin )b a1 k kcos )kb d)
1
œ a# '0 (sin ))(1 k cos )) d)
1Î2
a# '1Î2 (sin ))(1 k cos )) d)
1
œ a# '0 sin ) d) a# k'0 sin ) cos ) d) a# '1Î2 sin ) d) a# k '1Î2 sin ) cos ) d)
1Î2
1Î2
1Î#
#
1
a# k ’ sin# ) “
œ a# [ cos )]!
1Î#
!
1
#
a# [ cos )]11Î# a# k ’ sin# ) “
1
1Î#
œ a [0 (1)] a k ˆ "# 0‰ a# [(1) 0] a# k ˆ0 "# ‰ œ a#
#
#
a# k
#
a#
œ 2a# a# k œ a# (2 k);
1
1
M œ'µ
x dm œ ' (a cos )) † $ † a d) œ ' aa# cos )b a1 k kcos )kb d)
y
0
0
œa
'0
œ a#
'01Î2 cos ) d) a# k '
#
1Î2
#
(cos ))(1 k cos )) d) a
1Î2
0
#
œ a [sin
1Î#
) ]!
œ a# (1 0)
a# k
#
a# k
#
a# k
#
)
'1Î2 (cos ))(1 k cos )) d)
1
2) ‰
2) ‰
ˆ 1 cos
d) a# '1Î2 cos ) d) a# k'1Î2 ˆ 1 cos
d)
#
#
sin 2) ‘ 1Î#
#
!
1
a# [sin )]11Î#
1
a# k
#
ˆ 1# 0‰ (! 0)‘ a# (0 1)
)
a# k
#
sin 2) ‘ 1
#
1Î#
(1 0) ˆ 1# 0‰‘ œ a#
a# k 1
4
a#
M œ '0 $ † a d) œ a'0 (1 k kcos )k) d) œ a '0 (1 k cos )) d) a'1Î2 (1 k cos )) d)
1
1
1Î#
œ a[) k sin )]!
œ
a1
#
1Î2
a# k 1
4
œ 0;
1
a[) k sin )]11Î# œ a ˆ 1# k‰ 0‘ a (1 0) ˆ 1# k‰‘
ak a ˆ 1# k‰ œ a1 2ak œ a(1 2k). So x œ
My
M
œ 0 and y œ
Mx
M
œ
a# (2 k)
a(1 #k)
œ
a(2 k)
1 #k
ka ‰
Ê ˆ0ß 2a
1 #k is the center of mass.
39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc
length we have that the length of a particular segment is ds œ È(dx)# (dy)# . This implies that
Mx œ ' $ y ds, My œ ' $ x ds and M œ ' $ ds. If $ is constant, then x œ
yœ
Mx
M
' y ds
œ ' ds œ
' y ds
length
My
M
' x ds
œ ' ds œ
' x ds
length
and
.
40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus
x#
a
vertical strip has center of mass: (µ
x ßµ
y ) œ Œxß 2 4p , length: a
mass: dm œ $ dA œ $ Ša
œ
$
#
2
%
&
#
œ $ ’ax
8a$ Èpa
3
2
Èpa
œ 2 † $ ’ax
È
Mx
M
œŠ
Èpa
2
Èpa
x$
12p “ !
œ
3
5
Èpa
8a# $Èpa
5
2$ paÈpa
12p ‹
œ 2$ Š2aÈpa
8a# $ Èpa
3
‹ Š 8a$È
5
pa ‹
2
x&
80p# “ 0
œ 2 † #$ ’a# x
"6 ‰
‰
œ 2a# $ Èpa ˆ 8080
œ 2a# $ Èpa ˆ 64
80 œ
x$
12p “ c2 pa
. So y œ
c2Èpa
#
16 ‰
80
width: dx, area: dA œ Ša
dx. Thus, Mx œ ' µ
y dm œ 'c2Èpa "# Ša
#Èpa
x
x
'c22ÈÈpapa Ša# 16p
‹ dx œ #$ ’a# x 80p
“
œ 2a# $ Èpa ˆ1
œ
x#
4p ‹
x#
4p ,
x#
4p ‹ Ša
x#
4p ‹ $
œ 4a$ Èpa ˆ1
dx,
dx
2& p# a# Èpa
‹
80p#
œ $ Š2a# Èpa
; M œ ' dm œ $
x#
4p ‹
2
Èpa
'
c2Èpa
4 ‰
12
Ša
x#
4p ‹
dx
œ 4a$ Èpa ˆ 121#4 ‰
a, as claimed.
41. Since the density is constant, its value will not affect our answers, so we can set $ œ ".
1Î2 !
A generalization of Example 6 yields M œ ' µ
y dm œ '
a# sin ) d) œ a# [ cos )]1Î2 !
1Î2 !
1Î# !
x
1Î2 !
œ a# cos ˆ 1# !‰ cos ˆ 1# !‰‘ œ a# (sin ! sin !) œ 2a# sin !; M œ ' dm œ '1Î# ! a d) œ a[)]11ÎÎ22 !!
œ a ˆ 1# !‰ ˆ 1# !‰‘ œ 2a!. Thus, y œ
Ê c œ 2a sin !. Then y œ
a(2a sin !)
2a!
œ
ac
s ,
Mx
M
œ
2a# sin !
2a!
œ
a sin !
!
lim
! Ä !b
(b)
sin ! ! cos !
! ! cos !
!
f(!)
¸
d
h
œ
a sin !
!
sin ! ! cos !
! ! cos ! .
a(sin ! ! cos !)
a(! ! cos !)
œ
œ a cos ! d Ê d œ
0.4
0.664879
0.6
0.662615
0.8
0.659389
1.0
0.655145
6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS
1. (a)
dy
dx
#
%
œ sec# x Ê Š dy
dx ‹ œ sec x
Ê S œ 21'0
1Î4
a(sin ! ! cos !)
.
!
The graphs below suggest that
2
3.
0.2
0.666222
c
#
as claimed.
42. (a) First, we note that y œ (distance from origin to AB) d Ê
Moreover, h œ a a cos ! Ê
. Now s œ a(2!) and a sin ! œ
(b)
(tan x) È1 sec% x dx
(c) S ¸ 3.84
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
393
394
2. (a)
Chapter 6 Applications of Definite Integrals
dy
dx
#
(b)
2
œ 2x Ê Š dy
dx ‹ œ 4x
Ê S œ 21'0 x# È1 4x# dx
2
(c) S ¸ 53.23
3. (a) xy œ 1 Ê x œ
Ê S œ 21'1
2
"
y
"
y
Ê
dx
dy
#
œ y"# Ê Š dx
dy ‹ œ
"
y%
(b)
È1 y% dy
(c) S ¸ 5.02
4. (a)
dx
dy
#
#
œ cos y Ê Š dx
dy ‹ œ cos y
(b)
Ê S œ 21'0 (sin y) È1 cos# y dy
1
(c) S ¸ 14.42
#
5. (a) x"Î# y"Î# œ 3 Ê y œ ˆ3 x"Î# ‰
"Î# ‰ ˆ
ˆ
Ê dy
"# x"Î# ‰
dx œ 2 3 x
#
"Î# ‰
ˆ
Ê Š dy
dx ‹ œ 1 3x
(b)
#
#
#
Ê S œ 21'1 ˆ3 x"Î# ‰ É1 a1 3x"Î# b dx
4
(c) S ¸ 63.37
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus
dx
dy
6. (a)
#
"Î# ‰
ˆ
œ 1 y"Î# Ê Š dx
dy ‹ œ 1 y
#
395
(b)
#
Ê S œ 21 '1 ˆy 2Èy‰ É1 a1 y"Î# b dx
2
(c) S ¸ 51.33
dx
dy
7. (a)
#
(b)
#
œ tan y Ê Š dx
dy ‹ œ tan y
Ê S œ 21'0 Š'0 tan t dt‹ È1 tan# y dy
1Î3
y
œ 21'0 Š'0 tan t dt‹ sec y dy
1Î3
y
(c) S ¸ 2.08
dy
dx
8. (a)
#
(b)
#
œ Èx# 1 Ê Š dy
dx ‹ œ x 1
È5
Ê S œ 21'1 Š'1 Èt# 1 dt‹ È1 ax# 1b dx
È5
x
œ 21'1 Š'1 Èt# 1 dt‹ x dx
x
(c) S ¸ 8.55
9. y œ
œ
x
#
1È5
#
Ê
dy
dx
' ˆ x ‰ É1
œ "# ; S œ 'a 21y Ê1 Š dy
dx ‹ dx Ê S œ 0 21 #
x
#
4
"
4
dx œ
1È5
#
'04 x dx
%
#
’ x# “ œ 41È5; Geometry formula: base circumference œ 21(2), slant height œ È4# 2# œ 2È5
!
Ê Lateral surface area œ
10. y œ
#
b
Ê x œ 2y Ê
dx
dy
"
#
(41) Š2È5‹ œ 41È5 in agreement with the integral value
#
È
È '
È #
'
œ 2; S œ 'c 21x Ê1 Š dx
dy ‹ dy œ 0 21 † 2y 1 2 dy œ 41 5 0 y dy œ 21 5 cy d !
#
d
2
2
#
œ 21È5 † 4 œ 81È5; Geometry formula: base circumference œ 21(4), slant height œ È4# 2# œ 2È5
Ê Lateral surface area œ " (81) Š2È5‹ œ 81È5 in agreement with the integral value
#
11.
dy
dx
'
œ "# ; S œ 'a 21yÊ1 Š dy
dx ‹ dx œ 1 21
b
#
3
(x 1)
#
É1 ˆ "# ‰# dx œ
1È5
#
'13 (x 1) dx
œ
1È5
#
1È5
#
#
’ x# x“
$
"
È
ˆ 9# 3‰ ˆ "# 1‰‘ œ 1 # 5 (4 2) œ 31È5; Geometry formula: r" œ "# "# œ 1, r# œ 3# "# œ 2,
œ
slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(r" r# ) ‚ slant height œ 1(1 2)È5
œ 31È5 in agreement with the integral value
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
396
Chapter 6 Applications of Definite Integrals
12. y œ
x
#
"
#
Ê x œ 2y 1 Ê
dx
œ 2; S œ 'c 21x Ê1 Š dy
‹ dy œ '1 21(2y 1)È1 4 dy œ 21È5 '1 (2y 1) dy
#
d
dx
dy
2
#
2
œ 21È5 cy# yd " œ 21È5 [(4 2) (1 1)] œ 41È5; Geometry formula: r" œ 1, r# œ 3,
slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(1 3)È5 œ 41È5 in agreement with
the integral value
13.
dy
dx
#
x#
3
œ
’u œ 1
x%
9
Ê S œ '0
2
x%
9
Ê Š dy
dx ‹ œ
Ê du œ
4
9
x œ 0 Ê u œ 1, x œ 2 Ê u œ
Ä S œ 21 '1
25Î9
14.
œ
1
3
dy
dx
œ
"
1 2
4 du œ # 3
1 ˆ 12527 ‰
œ 98811
3
27
#
Ê S œ '3Î4 21Èx É1
œ 21'3Î4 Éx
15Î4
15.
œ
’ˆ 15
4
œ
41
3
(8 1) œ
dy
dx
" (2 2x)
# È2x x#
œ
ˆ 43
dx;
dx;
#&Î*
u$Î# ‘ "
"
4x
dx
$Î#
dx œ 21 ’ 23 ˆx 4" ‰ “
"
4
" ‰$Î#
4
41
3
x$
9
du œ
x%
9
"
4x
x"Î# Ê Š dy
dx ‹ œ
15Î4
É1
25 ‘
9
u"Î# †
ˆ 125
‰
27 1 œ
"
#
"
4
x$ dx Ê
21 x$
9
" ‰$Î#
“
4
"&Î%
$Î%
41
3
$
’ˆ 24 ‰
Ê Š dy
dx ‹ œ
(1 x)#
2x x#
œ
1“
281
3
œ
#
1x
È2x x#
Ê S œ '0 5 21È2x x# É1
1Þ5
Þ
œ 21'0 5 È2x
1Þ5
Þ
(1 x)#
2x x#
È
x# 1 2x x#
x# 2x È
2x x#
dx
dx
œ 21'0 5 dx œ 21[x]"Þ&
!Þ& œ 21
1Þ5
Þ
16.
dy
dx
"
2È x 1
œ
#
dy
Ê Š dx
‹ œ
"
4(x 1)
Ê S œ '1 21Èx 1 É1
5
œ 21'1 É(x 1)
5
"
4
"
4(x 1)
dx
dx œ 21'1 Éx
5
&
$Î#
œ 21 ’ 23 ˆx 54 ‰ “ œ
17.
œ
41
3
œ
1
6
dx
dy
‰$Î#
’ˆ 25
4
5
4
41 ˆ
5 ‰$Î#
3 ’ 5 4
"
$
$
ˆ 94 ‰$Î# “ œ 431 Š 52$ 32$ ‹
(125 27) œ
981
6
œ
dx
ˆ1 45 ‰$Î# “
491
3
%
'
œ y# Ê Š dx
dy ‹ œ y Ê S œ 0
#
1
u œ 1 y% Ê du œ 4y$ dy Ê
"
4
21 y$
3
È1 y% dy;
du œ y$ dy; y œ 0
Ê u œ 1, y œ 1 Ê u œ 2d Ä S œ '1 21 ˆ "3 ‰ u"Î# ˆ 4" du‰
2
œ
1
6
'12 u"Î# du œ 16 32 u$Î# ‘ #" œ 19 ŠÈ8 1‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus
18. x œ ˆ "3 y$Î# y"Î# ‰ Ÿ 0, when 1 Ÿ y Ÿ 3. To get positive
area, we take x œ ˆ "3 y$Î# y"Î# ‰
Ê
dx
dy
#
œ "# ˆy"Î# y"Î# ‰ Ê Š dx
dy ‹ œ
"
4
ay 2 y" b
Ê S œ '1 21 ˆ "3 y$Î# y"Î# ‰ É1 4" ay 2 y" b dy
3
œ 21'1 ˆ 3" y$Î# y"Î# ‰ É 4" ay 2 y" b dy
3
Éay"Î# y"Î# b#
œ 21'1 ˆ "3 y$Î# y"Î# ‰
3
œ 1'1 ˆ "3 y#
3
2
3
dx
dy
œ
"
È 4 y
œ 41 '0
15Î4
œ
20.
dx
dy
81
3
œ
3
$
y 1‰ dy œ 1 ’ y9
œ 19 (18 1 3) œ
19.
dy œ 1'1 y"Î# ˆ 3" y 1‰ Šy"Î#
#
#
Ê Š dx
dy ‹ œ
Ê S œ '0
15Î4
"
4 y
5È 5
8 ‹
œ
81
3
#
"
È2y 1
Ê Š dx
dy ‹ œ
Š 40
È 5 5 È 5
‹
8
œ
È
È5
1
12
‹œ
#
41 È 2
3
"
2y 1
$Î#
’1$Î# ˆ 58 ‰
“œ
#
1 dy œ ÊŠy'
"
#
"
16y' ‹
2
È2
È2
dy
dx
œ
"
#
aa# x# b
Ê S œ 21'ca Èa# x# É1
a
$Î#
œ
21 r
h
dy
dx
#
#
É h h# r
25. y œ cos x Ê
œ
41 È 2
3
5È 5
‹
8È 8
Š1
21
40
"
#
"
16y' ‹
dy
dy œ 21'1 ˆy% "4 y# ‰ dy
2
"
4y$ ‹
(8 † 31 5) œ
2531
20
È2
x ax# 1b dx œ 21'0 ax$ xb dx œ 21 ’ x4
"Î#
x#
aa # x # b
(2x) œ
x
È a# x#
%
#
Ê Š dy
dx ‹ œ
È#
x#
# “!
œ 21 ˆ 44 22 ‰ œ 41
x#
aa # x # b
dx œ 21'ca Èaa# x# b x# dx œ 21'ca a dx œ 21a[x]ca a
a
a
r
h
#
Ê Š dy
dx ‹ œ
r#
h#
Ê S œ 21 '0
h
r
h
x É1
r#
h#
dx œ 21'0
h
r
h
#
#
x É h h# r dx
'0h x dx œ 2h1r Èh# r# ’ x# “ h œ 2h1r Èh# r# Š h# ‹ œ 1rÈh# r#
#
#
dy
dx
5$Î# “
1
œ 21a[a (a)] œ (21a)(2a) œ 41a#
x Ê
1‰
È2
23. y œ Èa# x# Ê
r
h
"
3
Ê dy œ xÈx# 2 dx Ê ds œ È1 a2x# x% b dx Ê S œ 21'0 x È1 2x# x% dx
$Î#
œ 21'0 xÉax# 1b# dx œ 21'0
24. y œ
dy œ 21'5Î8 È(2y 1) 1 dy
2
#
ax# 2b
"
9
È(4 y) 1 dy
5$Î# “ œ 831 ’ˆ 45 ‰
dy; S œ '1 21y ds œ 21'1 y Šy$
"
4y$ ‹
"
"
3
1‰‘ œ 1 ˆ3
1 dy œ ÊŠy'
"‰
"‰
ˆ " " ‰‘ œ 21 ˆ 31
œ 21 ’ y5 4" y" “ œ 21 ˆ 32
5 8 5 4
5 8 œ
22. y œ
15Î4
15 ‰$Î#
4
1
"
4y$ ‹
dy œ Šy$
&
dy œ 41'0
"
3
Š16È2 5È5‹
21. ds œ Èdx# dy# œ ÊŠy$
"
4y$ ‹
3
351È5
3
"
5
Š 8†2 8†22È
2
dy œ 1 '1 ˆ 3" y 1‰ (y 1) dy
3‰ ˆ "9
9
3
"
4y
21 † 2È4 y É1
Ê S œ '5Î8 21È2y 1 É1
"
2y1
1
œ ÊŠy$
"
È5 y dy œ 41 23 (5 y)$Î# ‘ "&Î% œ 831 ’ˆ5
!
Š 5È 5
41 È 2
3
y“ œ 1 ˆ 27
9
161
9
œ 21'5Î8 È2 y"Î# dy œ 21È2 23 y$Î# ‘ &Î) œ
œ
$
y#
3
"
‹
y"Î#
#
0
#
#
È1 sin# x dx
'
œ sin x Ê Š dy
dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x)
#
1Î2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
397
398
Chapter 6 Applications of Definite Integrals
26. y œ ˆ1 x#Î$ ‰
$Î#
Ê
dy
dx
œ
Ê S œ 2'0 21 ˆ1 x#Î$ ‰
1
3
#
ˆ1 x#Î$ ‰"Î# ˆ 23 x"Î$ ‰ œ
$Î#
#
ˆ1x#Î$ ‰"Î#
x"Î$
Ê Š dy
dx ‹ œ
1x#Î$
x#Î$
œ
"
x#Î$
1
$Î#
"
É1 ˆ x#Î$
1‰ dx œ 41'0 ˆ1 x#Î$ ‰ Èx#Î$ dx
1
$Î#
œ 41'0 ˆ1 x#Î$ ‰ x"Î$ dx; u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 32 du œ x"Î$ dx;
1
!
x œ 0 Ê u œ 1, x œ 1 Ê u œ 0d Ä S œ 41'1 u$Î# ˆ 3# du‰ œ 61 25 u&Î# ‘ " œ 61 ˆ0 25 ‰ œ
0
121
5
#
#
#
È16# y#
27. The area of the surface of one wok is S œ 'c 21x Ê1 Š dx
dy ‹ dy. Now, x y œ 16 Ê x œ
#
d
Ê
dx
dy
œ
c7
#
y
È16# y#
Ê Š dx
dy ‹ œ
c7
; S œ 'c16 21È16# y# É1
y#
16# y#
y#
16# y#
c7
dy œ 21'c16 Èa16# y# b y# dy
œ 21'c16 16 dy œ 321 † 9 œ 2881 ¸ 904.78 cm# . The enamel needed to cover one surface of one wok is
V œ S † 0.5 mm œ S † 0.05 cm œ (904.78)(0.05) cm$ œ 45.24 cm$ . For 5000 woks, we need
5000 † V œ 5000 † 45.24 cm$ œ (5)(45.24)L œ 226.2L Ê 226.2 liters of each color are needed.
28. y œ Èr# x# Ê
œ 21'a
abh
œ 21'a
2x
È r# x #
œ
Èar# x# b x# dx œ 21r'
a
29. y œ ÈR# x# Ê
abh
œ "#
dy
dx
dy
dx
œ "#
2x
È R # x#
x
Èr# x#
abh
œ
abh
30. (a) x# y# œ 45# Ê x œ È45# y# Ê
45
x#
r# x # ;
S œ 21 'a
abh
Èr# x# É1
x#
r# x#
dx
dx œ 21rh, which is independent of a.
#
x
È R # x#
ÈaR# x# b x# dx œ 21R '
a
S œ 'c22Þ5 21 È45# y# É1
#
Ê Š dx
dy ‹ œ
y#
45# y#
dx
Ê Š dy
‹ œ
x#
R # x# ;
S œ 21'a
abh
ÈR# x# É1
x#
R # x#
dx
dx œ 21Rh
dx
dy
œ
y
È45# y#
#
Ê Š dx
dy ‹ œ
y#
45# y#
;
dy œ 21 '22Þ5 Èa45# y# b y# dy œ 21 † 45'22Þ5 dy
45
45
œ (21)(45)(67.5) œ 60751 square feet
(b) 19,085 square feet
dy
'
È1 1 dx œ 21 ' (x)È2 dx 21' xÈ2 dx
31. y œ x Ê Š dy
dx ‹ œ 1 Ê Š dx ‹ œ 1 Ê S œ 21 c1 kxk
c1
0
#
#
œ 2È21 ’ x# “
32.
dy
dx
œ
x#
3
!
2
#
"
!
#
4
9
0
#
2È21 ’ x# “ œ 2È21 ˆ0 "# ‰ 2È21(2 0) œ 5È21
Ê Š dy
dx ‹ œ
Ê du œ
2
x%
9
Ê by symmetry of the graph that S œ 2 'cÈ3 21 Š x9 ‹ É1
0
$
x%
9
dx; ’u œ 1
x$ dx Ê "4 du œ x9 dx; x œ È3 Ê u œ 2, x œ 0 Ê u œ 1“ Ä S œ 41'2 u"Î# ˆ "4 ‰ du
1
$
"
œ 1'2 u"Î# du œ 1 23 u$Î# ‘ # œ 1 Š 23 23 È8‹ œ
1
È3
È3
21
3
ŠÈ8 1‹ . If the absolute value bars are dropped the
integral for S œ 'cÈ3 21f(x) ds will equal zero since 'cÈ3 21 Š x9 ‹ É1
$
x%
9
dx is the integral of an odd function
over the symmetric interval È3 Ÿ x Ÿ È3.
33.
dx
dt
x%
9
œ sin t and
dy
dt
#
È( sin t)# (cos t)# œ 1 Ê S œ ' 21y ds
‰ Š dy
œ cos t Ê Êˆ dx
dt
dt ‹ œ
#
œ '0 21(2 sin t)(1) dt œ 21 c2t cos td #!1 œ 21[(41 1) (0 1)] œ 81#
21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus
34.
dx
dt
œ t"Î# and
È3
dy
dt
œ '0 21 ˆ 23 t$Î# ‰ É t
#
"
t
dt œ
È3
'0
#
21 ˆ 23 t$Î# ‰ É t
#
f(t) œ 21 ˆ 23 t$Î# ‰ É t
Ê
35.
dx
dt
È3
'0
F(t) dt œ
œ 1 and
È2
dy
dt
#
281
9
"
t
41
3
È3
'0
1
t
Ê S œ ' 21x ds
tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1,
'14 231 Èu du œ 491 u$Î# ‘ %" œ 2891
’t œ È3 Ê u œ 4“ Ä
Note:
#
#
Èt t" œ É t
‰ Š dy
œ t"Î# Ê Êˆ dx
dt
dt ‹ œ
1
t
dt is an improper integral but limb f(t) exists and is equal to 0, where
tÄ!
. Thus the discontinuity is removable: define F(t) œ f(t) for t 0 and F(0) œ 0
.
#
#
#
È2‹ œ Ét# 2È2 t 3 Ê S œ ' 21x ds
‰ Š dy
œ t È2 Ê Êˆ dx
dt
dt ‹ œ Ê1 Št
#
œ 'cÈ2 21 Št È2‹ Ét# 2È2 t 3 dt; ’u œ t# 2È2 t 3 Ê du œ Š2t 2È2‹ dt; t œ È2 Ê u œ 1,
*
t œ È2 Ê u œ 9“ Ä '1 1Èu du œ 23 1u$Î# ‘ " œ
9
36.
dx
dt
œ aa1 cos tb and
dy
dt
21
3
(27 1) œ
521
3
#
#
‰ Š dy
Éc aa1 cos tb d# aa sin tb#
œ a sin t Ê Êˆ dx
dt
dt ‹ œ
œ Èa2 2 a2 cos t a2 cos2 t a2 sin2 t œ È2a2 2a2 cos t œ aÈ2È1 cos t Ê S œ ' 21y ds
œ '0 21 aa1 cos tb † aÈ2È1 cos t dt œ 2È2 1 a2 '0 a1 cos tb3/2 dt
21
37.
dx
dt
œ 2 and
21
dy
dt
È2# 1# œ È5 Ê S œ ' 21y ds œ ' 21(t 1)È5 dt
‰ Š dy
œ 1 Ê Êˆ dx
dt
dt ‹ œ
0
#
#
1
"
#
œ 21È5 ’ t2 t“ œ 31È5. Check: slant height is È5 Ê Area is 1(1 2)È5 œ 31È5 .
!
38.
dx
dt
œ h and
dy
dt
Èh# r# Ê S œ ' 21y ds œ ' 21rtÈh# r# dt
‰ Š dy
œ r Ê Êˆ dx
dt
dt ‹ œ
0
œ 21rÈh# r#
#
#
1
'01 t dt œ 21rÈh# r# ’ t2 “ " œ 1rÈh# r# .
#
!
Check: slant height is Èh# r# Ê Area is
1rÈh# r# .
39. (a) An equation of the tangent line segment is
(see figure) y œ f(mk ) f w (mk )(x mk ).
When x œ xkc1 we have
r" œ f(mk ) f w (mk )(x51 mk )
œ f(mk ) f w (mk ) ˆ ?#xk ‰ œ f(mk ) f w (mk )
when x œ xk we have
r# œ f(mk ) f w (mk )(x5 mk )
k
œ f(mk ) f w (mk ) ?x
# ;
(b) L#k œ (?xk )# (r# r" )#
;
#
ˆf w (mk ) ?#xk ‰‘
œ (?xk )# [f w (mk )?xk ]# Ê Lk œ È(?xk )# [f w (mk )?xk ]# , as claimed
(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent
œ (?xk )# f w (mk )
?x k
#
?x k
#
line segment about the x-axis is given by ?Sk œ 1(r" r# )Lk œ 1[2f(mk )] Éa?xk b# [f w (mk )?xk ]#
using parts (a) and (b) above. Thus, ?Sk œ 21f(mk ) È1 [f w (mk )]# ?xk .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
399
400
Chapter 6 Applications of Definite Integrals
! ?Sk œ lim ! 21f(mk ) È1 [f w (mk )]# ?xk œ ' 21f(x) È1 [f w (x)]# dx
(d) S œ n lim
Ä_
nÄ_
a
kœ1
kœ1
n
n
È3
40. S œ 'a 21f(x) dx œ '0
b
21 †
x
È3
b
dx œ
1
È3
È3
c x# d ! œ
31
È3
œ È31
41. The centroid of the square is located at (#ß #). The volume is V œ (21) ayb (A) œ (21)(2)(8) œ 321 and the
surface area is S œ (21) ayb (L) œ (21)(2) Š4È8‹ œ 32È21 (where È8 is the length of a side).
42. The midpoint of the hypotenuse of the triangle is ˆ 3# ß 3‰
Ê y œ 2x is an equation of the median Ê the line
y œ 2x contains the centroid. The point ˆ 3# ß $‰ is
3È 5
#
units from the origin Ê the x-coordinate of the
#
centroid solves the equation Ɉx 3# ‰ (2x 3)#
œ
È5
#
Ê ˆx# 3x 94 ‰ a4x# 12x 9b œ
5
4
Ê 5x# 15x 9 œ 1
Ê x# 3x 2 œ (x 2)(x 1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the
Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5 xb "# (3)(6)‘
œ (21)(4)(9) œ 721
43. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41#
44. We create the cone by revolving the triangle with vertices
(0ß 0), (hß r) and (hß 0) about the x-axis (see the accompanying
figure). Thus, the cone has height h and base radius r. By
Theorem of Pappus, the lateral surface area swept out by the
hypotenuse L is given by S œ 21yL œ 21 ˆ r ‰ Èh# r#
#
œ 1rÈr# h# . To calculate the volume we need the position
of the centroid of the triangle. From the diagram we see that
the centroid lies on the line y œ
œ
"
3
Éh#
r#
4
#
r
2h
#
#
x. The x-coordinate of the centroid solves the equation É(x h)# ˆ 2hr x #r ‰
#
#
Ê Š 4h4h# r ‹ x# Š 4h 2h r ‹ x
inside the triangle Ê y œ
r
2h
47. V œ 21 yA Ê
4
3
2 ar# 4h# b
9
œ0 Ê xœ
2h
3
or
4h
3
Ê xœ
x œ 3r . By the Theorem of Pappus, V œ 21 ˆ 3r ‰‘ ˆ "# hr‰ œ
45. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ
46. S œ 213 L Ê 21 ˆa
r#
4
2a ‰‘
(1a)
1
2a
1,
since the centroid must lie
1r# h.
and by symmetry x œ 0
œ 21a# (1 2)
1ab# œ a21yb ˆ 1#ab ‰ Ê y œ
48. V œ 213A Ê V œ 21 ˆa
"
3
2h
3 ,
4a ‰‘ 1a#
Š # ‹
31
œ
4b
31
and by symmetry x œ 0
1a$ (31 4)
3
49. V œ 213 A œ (21)(area of the region) † (distance from the centroid to the line y œ x a). We must find the
4a ‰
distance from ˆ0ß 31
to y œ x a. The line containing the centroid and perpendicular to y œ x a has slope
1 and contains the point ˆ!ß 34a1 ‰ . This line is y œ x 34a1 . The intersection of y œ x a and y œ x 34a1 is
the point ˆ 4a 613a1 ß 4a 613a1 ‰ . Thus, the distance from the centroid to the line y œ x a is
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.6 Work
Ɉ 4a 613a1 ‰# ˆ 34a1
4a
61
3a1 ‰#
61
œ
È2 (4a 3a1)
61
Ê V œ (21) Š
È2 (4a 3a1)
#
‹ Š 1#a ‹
61
œ
È2 1a$ (4 31)
6
‰
50. The line perpendicular to y œ x a and passing through the centroid ˆ!ß 2a
1 has equation y œ x
intersection of the two perpendicular lines occurs when x a œ x
Ê xœ
2a
1
2a a1
21
Ê yœ
2a ‰#
#
a(21)
È 21
#
the distance from the centroid to the line y œ x a is Ɉ 2a 2 1a 0‰ ˆ 2a 2 1a
œ
2a
1 . The
2a a1
21 . Thus
.
1 )
Therefore, by the Theorem of Pappus the surface area is S œ 21 ’ a(2
“ (1a) œ È21a# (2 1).
È
21
51. From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is Mx œ y M
#
œ ˆ 34a1 ‰ Š 1#a ‹ œ
2a$
3
.
6.6 WORK
1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) œ kx. The
work done by F is W œ '0 F(x) dx œ k '0 x dx œ
3
3
k
#
$
cx# d ! œ
9k
# .
This work is equal to 1800 J Ê
k œ 1800
9
#
Ê k œ 400 N/m
2. (a) We find the force constant from Hooke's Law: F œ kx Ê k œ
Ê kœ
F
x
800
4
œ 200 lb/in.
(b) The work done to stretch the spring 2 inches beyond its natural length is W œ '0 kx dx
2
œ 200 '0 x dx œ 200 ’ x# “ œ 200(2 0) œ 400 in † lb œ 33.3 ft † lb
2
#
#
!
(c) We substitute F œ 1600 into the equation F œ 200x to find 1600 œ 200x Ê x œ 8 in.
3. We find the force constant from Hooke's law: F œ kx. A force of 2 N stretches the spring to 0.02 m
N
Ê 2 œ k † (0.02) Ê k œ 100 m
. The force of 4 N will stretch the rubber band y m, where F œ ky Ê y œ
Ê yœ
4N
N
100 m
œ 100 '0
0Þ04
Ê y œ 0.04 m œ 4 cm. The work done to stretch the rubber band 0.04 m is W œ '0
F
k
0Þ04
#
x dx œ 100 ’ x# “
!Þ!%
œ
!
(100)(0.04)#
#
kx dx
œ 0.08 J
4. We find the force constant from Hooke's law: F œ kx Ê k œ
F
x
Ê kœ
90
1
Ê k œ 90
N
m.
&
The work done to
‰
stretch the spring 5 m beyond its natural length is W œ '0 kx dx œ 90 '0 x dx œ 90 ’ x# “ œ (90) ˆ 25
# œ 1125 J
5
5
#
!
5. (a) We find the spring's constant from Hooke's law: F œ kx Ê k œ
F
x
œ
21,714
8 5
œ
21,714
3
Ê k œ 7238
(b) The work done to compress the assembly the first half inch is W œ '0 kx dx œ 7238 '0
0Þ5
#
œ 7238 ’ x# “
!Þ&
!
#
œ (7238) (0.5)
# œ
(7238)(0.25)
#
1Þ0
1Þ0
Þ
Þ
#
¸ 2714 in † lb
6. First, we find the force constant from Hooke's law: F œ kx Ê k œ
compresses the scale x œ
scale this far is W œ '0
1Î8
in, he/she must weigh F œ kx œ
#
kx dx œ 2400 ’ x# “
"Î)
!
x dx
¸ 905 in † lb. The work done to compress the assembly the
second half inch is: W œ '0 5 kx dx œ 7238 '0 5 x dx œ 7238 ’ x# “
"
8
lb
in
0Þ5
œ
2400
2†64
F
x
2,400 ˆ 8" ‰
"Þ!
!Þ&
œ
œ
150
" ‰
ˆ 16
7238
#
c1 (0.5)# d œ
(7238)(0.75)
#
œ 16 † 150 œ 2,400
lb
in .
If someone
œ 300 lb. The work done to compress the
œ 18.75 lb † in. œ
25
16
ft † lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
401
402
Chapter 6 Applications of Definite Integrals
7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to
x, the length of the rope still hanging: F(x) œ 0.624x. The work done is: W œ '0 F(x) dx œ '0 0.624x dx
50
#
œ 0.624 ’ x# “
&!
!
50
œ 780 J
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the
ground is Faxb œ "%% %x. The work done is: W œ 'a F(x) dx œ '0 a"%% %xbdx œ c144x 2x# d ! œ 1944 ft † lb
b
18
")
9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) œ (4.5)(180 x) where x
is the position of the car off the first floor. The work done is: W œ '0
180
œ 4.5 ’180x
")!
x#
# “!
180#
# ‹
œ 4.5 Š180#
œ
4.5†180#
#
F(x) dx œ 4.5'0
180
(180 x) dx
œ 72,900 ft † lb
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) œ xk# . The
b
work done is W œ 'a xk# dx œ k 'a x"# dx œ k x" ‘ a œ k ˆ b" "a ‰ œ
b
b
k(a b)
ab
11. The force against the piston is F œ pA. If V œ Ax, where x is the height of the cylinder, then dV œ A dx
Ê Work œ ' F dx œ ' pA dx œ 'ap
ap# ßV# b
" ßV" b
p dV.
12. pV"Þ% œ c, a constant Ê p œ cV"Þ% . If V" œ 243 in$ and p" œ 50 lb/in$ , then c œ (50)(243)"Þ% œ 109,350 lb.
‘
ˆ 3#"!Þ%
Thus W œ '243 109,350V"Þ% dV œ 109,350
œ 109,350
0.4
0.4V!Þ% #%$
$#
32
" ‰
#43!Þ%
ˆ 4" 9" ‰
œ 109,350
0.4
œ (109,350)(5)
(0.4)(36) œ 37,968.75 in † lb. Note that when a system is compressed, the work done by the system is negative.
13. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate,
the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of
the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#! xb. So:
W œ '0 0.8a#! xb dx œ 0.8 ’20x
20
#!
x#
# “!
œ 160 ft † lb.
14. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate,
the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of
the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#! xb. So:
W œ '0 2a#! xb dx œ 2 ’20x
20
#!
x#
# “!
œ 400 ft † lb.
Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5
times as great.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.6 Work
403
15. We will use the coordinate system given.
(a) The typical slab between the planes at y and y ?y has
a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . The force
F required to lift the slab is equal to its weight:
F œ 62.4 ?V œ 62.4 † 120 ?y lb. The distance through
which F must act is about y ft, so the work done lifting
the slab is about ?W œ force ‚ distance
œ 62.4 † 120 † y † ?y ft † lb. The work it takes to lift all
20
the water is approximately W ¸ ! ?W
0
20
œ ! 62.4 † 120y † ?y ft † lb. This is a Riemann sum for
0
the function 62.4 † 120y over the interval 0 Ÿ y Ÿ 20. The work of pumping the tank empty is the limit of these sums:
W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “
20
#
#!
!
‰ œ (62.4)(120)(200) œ 1,497,600 ft † lb
œ (62.4)(120) ˆ 400
#
5 ‰
(b) The time t it takes to empty the full tank with ˆ 11
–hp motor is t œ
W
†lb
250 ftsec
œ
1,497,600 ft†lb
†lb
250 ftsec
œ 5990.4 sec
œ 1.664 hr Ê t ¸ 1 hr and 40 min
(c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is
W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “
10
#
œ 1497.6 sec œ 0.416 hr ¸ 25 min
(d) In a location where water weighs 62.26
"!
!
‰ œ 374,400 ft † lb and the time is t œ
œ (62.4)(120) ˆ 100
#
W
†lb
250 ftsec
lb
ft$ :
a) W œ (62.26)(24,000) œ 1,494,240 ft † lb.
b) t œ 1,494,240
œ 5976.96 sec ¸ 1.660 hr Ê t ¸ 1 hr and 40 min
250
In a location where water weighs 62.59
lb
ft$
a) W œ (62.59)(24,000) œ 1,502,160 ft † lb
b) t œ 1,502,160
œ 6008.64 sec ¸ 1.669 hr Ê t ¸ 1 hr and 40.1 min
250
16. We will use the coordinate system given.
(a) The typical slab between the planes at y and y ?y has
a volume of ?V œ (20)(12) ?y œ 240 ?y ft$ . The force
F required to lift the slab is equal to its weight:
F œ 62.4 ?V œ 62.4 † 240 ?y lb. The distance through
which F must act is about y ft, so the work done lifting
the slab is about ?W œ force ‚ distance
20
œ 62.4 † 240 † y † ?y ft † lb. The work it takes to lift all the water is approximately W ¸ ! ?W
10
20
œ ! 62.4 † 240y † ?y ft † lb. This is a Riemann sum for the function 62.4 † 240y over the interval
10
10 Ÿ y Ÿ 20. The work it takes to empty the cistern is the limit of these sums: W œ '10 62.4 † 240y dy
20
#
œ (62.4)(240) ’ y# “
(b) t œ
W
†lb
275 ftsec
œ
#!
œ (62.4)(240)(200 50) œ (62.4)(240)(150) œ 2,246,400 ft † lb
"!
2,246,400 ft†lb
275
¸ 8168.73 sec ¸ 2.27 hours ¸ 2 hr and 16.1 min
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is
W œ '10 62.4 † 240y dy œ (62.4)(240) ’ y# “
15
#
Then the time is t œ
W
†lb
275 ftsec
œ
936,000
#75
"&
"!
œ (62.4)(240) ˆ 225
#
100 ‰
#
‰ œ 936,000 ft.
œ (62.4)(240) ˆ 125
#
¸ 3403.64 sec ¸ 56.7 min
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
404
Chapter 6 Applications of Definite Integrals
lb
ft$ :
(d) In a location where water weighs 62.26
a) W œ (62.26)(240)(150) œ 2,241,360 ft † lb.
b) t œ 2,241,360
œ 8150.40 sec œ 2.264 hours ¸ 2 hr and 15.8 min
275
‰ œ 933,900 ft † lb; t œ 933,900
c) W œ (62.26)(240) ˆ 125
#
#75 œ 3396 sec ¸ 0.94 hours ¸ 56.6 min
lb
ft$
In a location where water weighs 62.59
a) W œ (62.59)(240)(150) œ 2,253,240 ft † lb.
b) t œ 2,253,240
œ 8193.60 sec œ 2.276 hours ¸ 2 hr and 16.56 min
275
‰ œ 938,850 ft † lb; t œ 938,850
c) W œ (62.59)(240) ˆ 125
#
275 ¸ 3414 sec ¸ 0.95 hours ¸ 56.9 min
#
17. The slab is a disk of area 1x# œ 1ˆ y# ‰ , thickness ˜y, and height below the top of the tank a"! yb. So the work to pump
#
the oil in this slab, ˜W, is 57a"! yb1ˆ y# ‰ . The work to pump all the oil to the top of the tank is
W œ '0
10
571
#
4 a"!y
y$ bdy œ
571
4
$
’ "!$y
"!
y%
% “!
œ 11,8751 ft † lb ¸ 37,306 ft † lb.
#
18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"% yba1bˆ y# ‰ and since the tank is
half full and the volume of the original cone is V œ "$ 1r# h œ "$ 1a&# ba"!b œ
with half the volume the cone is filled to a height y,
œ
571 "%y$
4 ’ $
$
È
&!!
y%
“
% !
#&!1
'
#&!1
$
ft3 , half the volume œ
$
È
&!!
$
œ $" 1 y% y Ê y œ È
&!! ft. So W œ '0
#
#&!1
'
ft3 , and
571
#
4 a"%y
y$ b dy
¸ 60,042 ft † lb.
#
‰ ?y
19. The typical slab between the planes at y and and y ?y has a volume of ?V œ 1(radius)# (thickness) œ 1 ˆ 20
#
œ 1 † 100 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 51.2 ?V œ 51.2 † 1001 ?y lb
Ê F œ 51201 ?y lb. The distance through which F must act is about (30 y) ft. The work it takes to lift all the
30
30
kerosene is approximately W ¸ ! ?W œ ! 51201(30 y) ?y ft † lb which is a Riemann sum. The work to pump the
0
0
tank dry is the limit of these sums: W œ '0 51201(30 y) dy œ 51201 ’30y
30
¸ 7,238,229.48 ft † lb
$!
y#
# “!
‰ œ (5120)(4501)
œ 51201 ˆ 900
#
20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires
all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3
additional feet. Thus pumping through the valve requires È$ fta%1b6 ft3 a'#Þ% lb/ft3 b ¸ 14,115 ft † lb less work and thus
less time.
21. (a) Follow all the steps of Example 5 but make the substitution of 64.5
W œ '0
8
œ
64.51
4
64.51†8$
3
(10 y)y# dy œ
64.51
4
$
’ 10y
3
%
y
4
)
“ œ
!
64.51
4
$
Š 103†8
lb
ft$
8%
4‹
for 57
lb
ft$ .
Then,
1‰
‰
œ ˆ 64.5
a8$ b ˆ 10
4
3 2
œ 21.51 † 8$ ¸ 34,582.65 ft † lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance ¸ (13 y) ft.
Then W œ '0
8
571
4
(13 y)y# dy œ
571
4
œ (191) a8# b (7)(2) ¸ 53.482.5 ft † lb
$
’ 13y
3
)
y%
4 “!
œ
571
4
$
Š 133†8
8%
4‹
‰
œ ˆ 5741 ‰ a8$ b ˆ 13
3 2 œ
571†8$ †7
3 †4
22. The typical slab between the planes of y and y?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆÈy‰ ?y œ xy ?y m$ . The force F(y) is equal to the slab's weight: F(y) œ 10,000 mN$ † ?V
œ 110,000y ?y N. The height of the tank is 4# œ 16 m. The distance through which F(y) must act to lift
the slab to the level of the top of the tank is about (16 y) m, so the work done lifting the slab is about
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.6 Work
405
?W œ 10,0001y(16 y) ?y N † m. The work done lifting all the slabs from y œ 0 to y œ 16 to the top is
16
approximately W ¸ ! 10,0001y(16 y)?y. Taking the limit of these Riemann sums, we get
0
W œ '0 10,0001y(16 y) dy œ 10,0001'0 a16y y# b dy œ 10,0001 ’ 16y
#
16
œ
16
10,000†1†16$
6
#
"'
y$
3 “!
$
œ 10,0001 Š 16#
16$
3 ‹
¸ 21,446,605.9 J
23. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆÈ25 y# ‰ ?y m$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 9800 † ?V
#
œ 98001 ˆÈ25 y# ‰ ?y œ 98001 a25 y# b ?y N. The distance through which F(y) must act to lift the
slab to the level of 4 m above the top of the reservoir is about (4 y) m, so the work done is approximately
?W ¸ 98001 a25 y# b (4 y) ?y N † m. The work done lifting all the slabs from y œ 5 m to y œ 0 m is
0
approximately W ¸ ! 98001 a25 y# b (4 y) ?y N † m. Taking the limit of these Riemann sums, we get
c5
W œ 'c5 98001 a25 y# b (4 y) dy œ 98001 'c5 a100 25y 4y# y$ b dy œ 98001 ’100C
0
0
25†25
#
œ 98001 ˆ500
† 125
4
3
625 ‰
4
25
#
y# 34 y$
¸ 15,073,099.75 J
24. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆÈ100 y# ‰ ?y œ 1 a100 y# b ?y ft$ . The force is F(y) œ 56ft$lb † ?V œ 561 a100 y# b ?y lb. The
distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about
(12 y) ft, so the work done is ?W ¸ 561 a100 y# b (12 y) ?y lb † ft. The work done lifting all the slabs
10
from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100 y# b (12 y) ?y lb † ft. Taking the limit of these
0
Riemann sums, we get W œ '0 561 a100 y b (12 y) dy œ 561'0 a100 y# b (12 y) dy
10
10
#
œ 561'0 a1200 100y 12y# y$ b dy œ 561 ’1200C
10
œ 561 ˆ12,000
10,000
#
4 † 1000
10,000 ‰
4
100y#
#
12y$
3
"!
y%
4 “!
œ (561) ˆ12 5 4 5# ‰ (1000) ¸ 967,611 ft † lb.
It would cost (0.5)(967,611) œ 483,805¢ œ $4838.05. Yes, you can afford to hire the firm.
25. F œ m
œ
"
#
dv
dt
œ mv
by the chain rule Ê W œ 'x mv
x#
dv
dx
#
"
m cv# (x# ) v (x" )d œ
26. weight œ 2 oz œ
"
#
weight
32
œ
"
8
3#
œ
"
#56
28. weight œ 1.6 oz œ 0.1 lb Ê m œ
"
8
x#
"
dv ‰
dx
dx œ m "# v# (x)‘ x"
x#
"
slugs; W œ ˆ "# ‰ ˆ #56
slugs‰ (160 ft/sec)# ¸ 50 ft † lb
hr
1 min
5280 ft
27. 90 mph œ 901 hrmi † 601 min
† 60
sec † 1 mi œ 132 ft/sec; m œ
0.3125 lb ‰
#
W œ ˆ "# ‰ ˆ 32
ft/sec# (132 ft/sec) ¸ 85.1 ft † lb
29. weight œ 2 oz œ
dx œ m'x ˆv
mv## "# mv"# , as claimed.
lb; mass œ
2
16
dv
dx
lb Ê m œ
"
8
32
0.1 lb
32 ft/sec#
slugs œ
œ
"
#56
"
3 #0
0.3125 lb
32 ft/sec#
œ
0.3125
32
slugs;
slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb
slugs; 124 mph œ
(124)(5280)
(60)(60)
¸ 181.87 ft/sec;
"
W œ ˆ "# ‰ ˆ 256
slugs‰ (181.87 ft/sec)# ¸ 64.6 ft † lb
30. weight œ 14.5 oz œ
31. weight œ 6.5 oz œ
14.5
16
6.5
16
lb Ê m œ
lb Ê m œ
14.5
(16)(32)
6.5
(16)(32)
"4.5
slugs; W œ ˆ "# ‰ Š (16)(32)
slugs‹ (88 ft/sec)# ¸ 109.7 ft † lb
6.5
slugs; W œ ˆ "# ‰ Š (16)(32)
slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
!
y%
4 “ &
406
Chapter 6 Applications of Definite Integrals
32. F œ (18 lb/ft)x Ê W œ '0 18x dx œ c9x# d !
1Î6
mœ
"
8
32
œ
"
#56
"Î6
slugs and v" œ 0 ft/sec. Thus,
1
4
œ
1
4
ft † lb. Now W œ
È2
4
È2
4 ‹
(16) Š
È2
4 ‹
1
4
ft † lb,
sec when the bearing is at the top of its path.
È2
4
The height the bearing reaches is s œ 8È2 t 16t# Ê at t œ
Š8È2‹ Š
mv# "# mv"# , where W œ
"
ft † lb. œ ˆ #" ‰ ˆ #56
slugs‰ v# Ê v œ 8È2 ft/sec. With v œ 0
at the top of the bearing's path and v œ 8È2 32t Ê t œ
#
"
#
the bearing reaches a height of
œ 2 ft
33. (a) From the diagram,
rayb œ '! x œ '! É&!# ay 325b#
for 325 Ÿ y Ÿ 375 ft.
(b) The volume of a horizontal slice of the funnel
#
is ˜V ¸ 1rayb‘ ˜y
#
œ 1”'! É&!# ay 325b# • ˜y
(c) The work required to lift the single slice of
water is ˜W ¸ 62.4˜Va$(& yb
#
œ 62.4a$(& yb1”'! É&!# ay 325b# • ˜y.
The total work to pump our the funnel is W
#
œ '325 62.4a375 yb1”'! É50# ay 325b# • dy
375
¸ 6.3358 † 10( ft † lb.
34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft † lb. Therefor, the total work
required to pump out the throat and the funnel is 1,353,869,354 63,358,000 œ 1,417227,354 ft † lb.
(b) In horsepower-hours, the work required to pump out the glory hole is 1,417227,354
œ 715.8. Therefore, it would take
1.98†106
715.8 hp†h
1000 hp
œ 0.7158 hours ¸ 43 minutes.
35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a
partition of the interval [!ß (]. The typical slab between the planes at y and y ?y has a volume of about
#
17.5 ‰
?V œ 1(radius)# (thickness) œ 1 ˆ y14
?y in$ . The force F(y) required to lift this slab is equal to its
weight: F(y) œ
4
9
?V œ
41
9
#
17.5 ‰
ˆ y14
?y oz. The distance through which F(y) must act to lift this slab to
the level of 1 inch above the top is about (8 y) in. The work done lifting the slab is about
17.5)#
?W œ ˆ 491 ‰ (y14
(8 y) ?y in † oz. The work done lifting all the slabs from y œ 0 to y œ 7 is
#
7
approximately W œ !
0
41
9†14#
(y 17.5)# (8 y) ?y in † oz which is a Riemann sum. The work is the limit of
these sums as the norm of the partition goes to zero: W œ '0
7
œ
41
9†14#
œ
41
9†14#
'07 a2450 26.25y 27y# y$ b dy œ 9†4141
’
7%
4
$
9†7
26.25
#
41
9†14#
%
#
(y 17.5)# (8 y) dy
’ y4 9y$
26.25
#
y# 2450y“
(
!
#
† 7 2450 † 7“ ¸ 91.32 in † oz
36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius œ 10 ft. Then
?V œ 1 † 100 ?y ft$ . The force required will be F œ 62.4 † ?V œ 62.4 † 1001 ?y œ 62401 ?y lb. The distance through
which F must act is y so the work done lifting the slab is about ?W" œ 62401 † y † ?y lb † ft. The work it takes to
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 6.7 Fluid Pressures and Forces
385
385
360
360
lift all the water into the tank is: W" ¸ ! ?W" œ ! 62401 † y † ?y lb † ft. Taking the limit we end up with
W" œ '360 62401y dy œ 62401 ’ y# “
385
#
$)&
$'!
62401
#
œ
c385# 360# d ¸ 182,557,949 ft † lb
4
#
To find the work required to fill the pipe, do as above, but take the radius to be
Then ?V œ 1 †
"
36
$
?y ft and F œ 62.4 † ?V œ
integration: W# ¸ ! ?W# Ê W# œ '0
360
360
0
62.4
36
62.41
36
"
6
in œ
ft.
?y. Also take different limits of summation and
1y dy œ
62.41
36
#
$'!
#
1 ‰ 360
œ ˆ 62.4
Š # ‹ ¸ 352,864 ft † lb.
36
’ y# “
!
The total work is W œ W" W# ¸ 182,557,949 352,864 ¸ 182,910,813 ft † lb. The time it takes to fill the
W
tank and the pipe is Time œ 1650
¸ 182,910,813
¸ 110,855 sec ¸ 31 hr
1650
37. Work œ '6 370 000
35ß780ß000
ß
1000 MG
r#
ß
dr œ 1000 MG '6 370 000
35ß780ß000
ß
ß
"
œ (1000) a5.975 † 10#% b a6.672 † 10"" b Š 6,370,000
$&ß()!ß!!!
œ 1000 MG "r ‘ 'ß$(!ß!!!
dr
r#
"
35,780,000 ‹
¸ 5.144 ‚ 10"! J
38. (a) Let 3 be the x-coordinate of the second electron. Then r# œ (3 1)# Ê W œ 'c1 F(3) d3
0
œ 'c1 a23(3‚101)# b d3 œ ’ 233‚10" “
#* !
#*
0
"
œ a23 ‚ 10#* b ˆ1 #" ‰ œ 11.5 ‚ 10#*
(b) W œ W" W# where W" is the work done against the field of the first electron and W# is the work done
against the field of the second electron. Let 3 be the x-coordinate of the third electron. Then r#" œ (3 1)#
and r## œ (3 1)# Ê W" œ '3
5
œ a23 ‚ 10
b ˆ "4
#*
"‰
#
œ
23
4
23‚10#*
r#"
‚ 10
d3 œ '3
#*
5
23‚10#*
(3 ")#
, and W# œ '
d3 œ 23 ‚ 10#* ’ 3 " " “
23‚10#*
r##
3
&
œ 23 ‚ 10#* ’ 3 " " “ œ a23 ‚ 10#* b ˆ 6" 4" ‰ œ
$
#* ‰
#* ‰
W œ W" W# œ ˆ 23
ˆ 23
œ
4 ‚ 10
12 ‚ 10
23
3
5
23‚10#*
12
d3 œ '
23‚10#*
#
3 (3 " )
5
(3 2) œ
23
12
&
$
d3
‚ 10#* . Therefore
‚ 10#* ¸ 7.67 ‚ 10#* J
6.7 FLUID PRESSURES AND FORCES
1. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's
right-hand edge: y œ x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then
x œ 5 y and the total width is L(y) œ 2x œ 2(5 y). The depth of the strip is (y). The force exerted by the
c2
c2
water against one side of the plate is therefore F œ 'c5 w(y) † L(y) dy œ 'c5 62.4 † (y) † 2(5 y) dy
c2
œ 124.8 'c5 a5y y# b dy œ 124.8 5# y# "3 y$ ‘ & œ 124.8 ˆ 5# † 4
#
œ (124.8) ˆ 105
#
117 ‰
3
"
3
† 8‰ ˆ 5# † 25
"
3
† 125‰‘
œ (124.8) ˆ 315 6 234 ‰ œ 1684.8 lb
2. An equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is
L(y) œ 2x œ 2(y 3). The depth of the strip is (2 y). The force exerted by the water is
F œ 'c3 w(2 y)L(y) dy œ 'c3 62.4 † (2 y) † 2(3 y) dy œ 124.8'c3 a6 y y# b dy œ 124.8 ’6y
0
0
œ (124.8) ˆ18
9
#
0
y#
#
!
y$
3 “ $
‰
9‰ œ (124.8) ˆ 27
# œ 1684.8 lb
3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is
y œ x 3 Ê x œ y 3. Thus the total width is L(y) œ 2x œ 2(y 3). The depth of the strip changes to (4 y)
Ê F œ 'c3 w(4 y)L(y) dy œ 'c3 62.4 † (4 y) † 2(y 3) dy œ 124.8'c3 a12 y y# b dy
0
œ 124.8 ’12y
0
y#
#
!
y$
“
3 $
œ (124.8) ˆ36
0
9
#
‰
9‰ œ (124.8) ˆ 45
# œ 2808 lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
407
408
Chapter 6 Applications of Definite Integrals
4. Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge
remains the same: y œ x 3 Ê x œ 3 y and L(y) œ 2x œ 2(y 3). The depth of the strip changes to (y)
Ê F œ 'c3 w(y)L(y) dy œ 'c3 62.4 † (y) † 2(y 3) dy œ 124.8'c3 ay# 3yb dy œ 124.8 ’ y3 3# y# “
0
0
œ (124.8) ˆ 27
3
œ
27 ‰
#
0
(124.8)(27)(2 3)
6
$
!
$
œ 561.6 lb
5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be
y œ 2x 4 Ê x œ y # 4 and L(y) œ 2x œ y 4. The depth of the strip is (1 y).
(a) F œ 'c4 w(1 y)L(y) dy œ 'c4 62.4 † (1 y)(y 4) dy œ 62.4 'c4 a4 3y y# b dy œ 62.4 ’4y
0
0
œ (62.4) ’(4)(4)
(3)(16)
#
(b) F œ (64.0) ’(4)(4)
0
(3)(16)
#
64
3 “
œ (62.4) ˆ16 24
64
3 “
œ
(64.0)(120 64)
3
64 ‰
3
œ
(62.4)(120 64)
3
3y#
#
!
y$
3 “ %
œ 1164.8 lb
¸ 1194.7 lb
6. Using the coordinate system given, we find an equation for
the line of the plate's right-hand edge to be y œ 2x 4
Ê x œ 4#y and L(y) œ 2x œ 4 y. The depth of the
strip is (1 y) Ê F œ '0 w(1 y)(4 y) dy
1
œ 62.4'0 ay# 5y 4b dy œ 62.4 ’ y3
1
$
œ (62.4) ˆ "3
5
#
5y#
#
4y“
4‰ œ (62.4) ˆ 2 156 24 ‰ œ
"
!
(62.4)(11)
6
œ 114.4 lb
7. Using the coordinate system given in the accompanying
figure, we see that the total width is L(y) œ 63 and the depth
of the strip is (33.5 y) Ê F œ '0 w(33.5 y)L(y) dy
33
œ '0
33
64
1 #$
64 ‰
† (33.5 y) † 63 dy œ ˆ 12
(63)'0 (33.5 y) dy
$
33
$$
y#
# “!
64 ‰
œ ˆ 12
(63) ’33.5y
$
œ
(64)(63)(33)(67 33)
(#) a12$ b
‰ ’(33.5)(33)
œ ˆ 641#†63
$
33#
# “
œ 1309 lb
8. (a) Use the coordinate system given in the accompanying
‰
figure. The depth of the strip is ˆ 11
6 y ft
Ê F œ '0
11Î6
‰
w ˆ 11
6 y (width) dy
œ (62.4)(width)'0
11Î6
ˆ 11
‰
6 y dy
œ (62.4)(width) ’ 11
6 y
""Î'
y#
# “!
#
‰ † "# “ Ê Fend œ (62.4)(2) ˆ 121
‰ ˆ "# ‰ ¸ 209.73 lb and Fside œ (62.4)(4) ˆ 121
‰ ˆ "# ‰ ¸ 419.47 lb
œ (62.4)(width) ’ˆ 11
6
36
36
(b) Use the coordinate system given in the accompanying
figure. Find Y from the condition that the entire volume
of the water is conserved (no spilling): 11
6 †2†4œ 2†2†Y
11
‰
Ê Y œ 3 ft. The depth of a typical strip is ˆ 11
3 y ft
and the total width is L(y) œ 2 ft. Thus,
F œ '0
113
‰
w ˆ 11
3 y L(y) dy
11
‰
œ '0 (62.4) ˆ 11
3 y † 2 dy œ (62.4)(2) ’ 3 y
113
""Î$
y#
# “!
‰# “ œ
œ (62.4)(2) ’ˆ "# ‰ ˆ 11
3
(62.4)(12")
9
force doubles.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
¸ 838.93 lb Ê the fluid
Section 6.7 Fluid Pressures and Forces
9. Using the coordinate system given in the accompanying
figure, we see that the right-hand edge is x œ È1 y#
so the total width is L(y) œ 2x œ 2È1 y# and the depth
of the strip is (y). The force exerted by the water is
therefore F œ 'c1 w † (y) † 2È1 y# dy
0
œ 62.4'c1 È1 y# d a1 y# b œ 62.4 ’ 23 a1 y# b
0
$Î# !
“
"
œ (62.4) ˆ 23 ‰ (1 0) œ 416 lb
10. Using the same coordinate system as in Exercise 15, the right-hand edge is x œ È3# y# and the total width is
L(y) œ 2x œ 2È9 y# . The depth of the strip is (y). The force exerted by the milk is therefore
F œ 'c3 w † (y) † 2È9 y# dy œ 64.5'c3 È9 y# d a9 y# b œ 64.5 ’ 23 a9 y# b
0
0
$Î# !
“
œ (64.5)(18) œ 1161 lb
$
œ (64.5) ˆ 23 ‰ (27 0)
11. The coordinate system is given in the text. The right-hand edge is x œ Èy and the total width is L(y) œ 2x œ 2Èy.
(a) The depth of the strip is (2 y) so the force exerted by the liquid on the gate is F œ '0 w(2 y)L(y) dy
1
"
œ '0 50(2 y) † 2Èy dy œ 100 '0 (2 y)Èy dy œ 100'0 ˆ2y"Î# y$Î# ‰ dy œ 100 43 y$Î# 25 y&Î# ‘ !
1
1
1
‰
œ 100 ˆ 43 25 ‰ œ ˆ 100
15 (20 6) œ 93.33 lb
2‰
(b) We need to solve 160 œ '0 w(H y) † 2Èy dy for h. 160 œ 100 ˆ 2H
3 5 Ê H œ 3 ftÞ
1
12. Use the coordinate system given in the accompanying figure. The total width is L(y) œ 1.
(a) The depth of the strip is (3 1) y œ (2 y) ft. The force exerted by the fluid in the window is
F œ '0 w(2 y)L(y) dy œ 62.4 '0 (2 y) † 1 dy œ (62.4) ’2y
1
1
"
y#
# “!
œ (62.4) ˆ2 "# ‰ œ
(62.4)(3)
#
œ 93.6 lb
(b) Suppose that H is the maximum height to which the
tank can be filled without exceeding its design
limitation. This means that the depth of a typical
strip is (H 1) y and the force is
F œ '0 w[(H 1) y]L(y) dy œ Fmax , where
1
Fmax œ 312 lb. Thus, Fmax œ w'0 [(H 1) y] † 1 dy œ (62.4) ’(H 1)y
"
y#
# “!
1
œ (62.4) ˆH 3# ‰
‰ (2H 3) œ 93.6 62.4H. Then Fmax œ 93.6 62.4H Ê 312 œ 93.6 62.4H Ê H œ
œ ˆ 62.4
#
405.6
62.4
œ 6.5 ft
13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for
the line of the end plate's right-hand edge is y œ 5# x Ê x œ 25 y. The total width is L(y) œ 2x œ 45 y and the
depth of the typical horizontal strip at level y is (h y). Then the force is F œ '0 w(h y)L(y) dy œ Fmax ,
h
where Fmax œ 6667 lb. Hence, Fmax œ w'0 (h y) † 45 y dy œ (62.4) ˆ 45 ‰ ' ahy y# b dy
h
h
0
œ
#
(62.4) ˆ 45 ‰ ’ hy#
h
y$
3 “0
$
œ (62.4) ˆ 45 ‰ Š h#
$
h
3
‰
œ $Ɉ 54 ‰ ˆ 6667
10.4 ¸ 9.288 ft. The volume of water which the tank can hold is V œ
Height œ h and
"
#
(Base) œ
2
5
$
max ‰
‹ œ (62.4) ˆ 45 ‰ ˆ "6 ‰ h$ œ (10.4) ˆ 45 ‰ h$ Ê h œ Ɉ 54 ‰ ˆ F10.4
"
#
(Base)(Height) † 30, where
h Ê V œ ˆ 25 h# ‰ (30) œ 12h# ¸ 12(9.288)# ¸ 1035 ft$ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
409
410
Chapter 6 Applications of Definite Integrals
14. (a) After 9 hours of filling there are V œ 1000 † 9 œ 9000 cubic feet of water in the pool. The level of the water
V
is h œ Area
, where Area œ 50 † 30 œ 1500 Ê h œ 9000
1500 œ 6 ft. The depth of the typical horizontal strip at
level y is then (6 y) for the coordinate system given in the text. An equation for the drain plate's
right-hand edge is y œ x Ê total width is L(y) œ 2x œ 2y. Thus the force against the drain plate is
F œ '0 w(6 y)L(y) dy œ 62.4 '0 (6 y) † 2y dy œ (62.4)(2)'0 a6y y# b œ (62.4)(2) ’ 6y#
1
1
1
#
œ (124.8) ˆ3 "3 ‰ œ (124.8) ˆ 83 ‰ œ 332.8 lb
"
y$
3 “!
(b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h y) and the force
F œ '0 w(h y)L(y) dy œ Fmax , where Fmax œ 520 lb. Hence, Fmax œ (62.4)'0 (h y) † 2y dy
1
1
œ 124.8'0 ahy y# b dy œ (124.8) ’ hy#
1
Êhœ
27
3
#
"
y$
3 “!
œ (124.8) ˆ h# "3 ‰ œ (20.8)(3h 2) Ê
520
20.8
œ 3h 2
œ 9 ft
15. The pressure at level y is p(y) œ w † y Ê the average
pressure is p œ
œ
#
ˆ wb ‰ Š b# ‹
œ
"
b
'0b p(y) dy œ b" '0b w † y dy œ b" w ’ y# “ b
#
0
wb
#
. This is the pressure at level
b
#
, which
is the pressure at the middle of the plate.
16. The force exerted by the fluid is F œ '0 w(depth)(length) dy œ '0 w † y † a dy œ (w † a)'0 y dy œ (w † a) ’ y# “
b
œ
#
w Š ab# ‹
œ
ˆ wb
‰
# (ab)
b
b
#
b
0
œ p † Area, where p is the average value of the pressure (see Exercise 21).
17. When the water reaches the top of the tank the force on the movable side is 'c2 (62.4) ˆ2È4 y# ‰ (y) dy
0
œ (62.4)'c2 a4 y# b
0
"Î#
(2y) dy œ (62.4) ’ 23 a4 y# b
$Î# !
“
#
œ (62.4) ˆ 23 ‰ ˆ4$Î# ‰ œ 332.8 ft † lb. The force
compressing the spring is F œ 100x, so when the tank is full we have 332.8 œ 100x Ê x ¸ 3.33 ft. Therefore
the movable end does not reach the required 5 ft to allow drainage Ê the tank will overflow.
18. (a) Using the given coordinate system we see that the total
width is L(y) œ 3 and the depth of the strip is (3 y).
Thus, F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) † 3 dy
3
3
œ (62.4)(3)'0 (3 y) dy œ (62.4)(3) ’3y
3
$
y#
# “!
œ (62.4)(3) ˆ9 9# ‰ œ (62.4)(3) ˆ 9# ‰ œ 842.4 lb
(b) Find a new water level Y such that FY œ (0.75)(842.4 lb) œ 631.8 lb. The new depth of the strip is
(Y y) and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy
Y
œ 62.4'0 (Y y) † 3 dy œ (62.4)(3)'0 (Y y) dy œ (62.4)(3) ’Yy
Y
Y
Y
y#
# “0
œ (62.4)(3) ŠY#
Y#
# ‹
#
2FY
È6.75 ¸ 2.598 ft. So, ?Y œ 3 Y
œ (62.4)(3) Š Y# ‹ . Therefore, Y œ É (62.4)(3)
œ É 1263.6
187.2 œ
¸ 3 2.598 ¸ 0.402 ft ¸ 4.8 in
19. Use a coordinate system with y œ 0 at the bottom of the carton and with L(y) œ 3.75 and the depth of a typical strip being
(7.75 y). Then F œ '0
7Þ75
'
‰
w(7.75 y)L(y) dy œ ˆ 64.5
12$ (3.75) 0
7Þ75
‰
(7.75 y) dy œ ˆ 64.5
12$ (3.75) ’7.75y
#
(7.75)
‰
œ ˆ 64.5
¸ 4.2 lb
12$ (3.75)
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
(Þ(&
y#
# “!
Chapter 6 Practice Exercises
411
57 ‰
20. The force against the base is Fbase œ pA œ whA œ w † h † (length)(width) œ ˆ 12
(10)(5.75)(3.5) ¸ 6.64 lb.
$
To find the fluid force against each side, use a coordinate system with y œ 0 at the bottom of the can, so that the depth of a
of ‰
of ‰
ˆ 57 ‰ ˆ width
typical strip is (10 y): F œ '0 w(10 y) ˆ width
the side dy œ 12$
the side ’10y
10
"!
y#
# “!
57 ‰ ˆ width of ‰ ˆ 100 ‰
57 ‰
57 ‰
œ ˆ 12
Ê Fend œ ˆ 12
(50)(3.5) ¸ 5.773 lb and Fside œ ˆ 12
(50)(5.75) ¸ 9.484 lb
$
$
$
the side
#
21. (a) An equation of the right-hand edge is y œ
x Ê xœ
3
#
2
3
y and L(y) œ 2x œ
4y
3
. The depth of the strip
is (3 y) Ê F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) ˆ 43 y‰ dy œ (62.4) † ˆ 43 ‰'0 a3y y# b dy
3
3
$
y$
3 “!
œ (62.4) ˆ 43 ‰ ’ 3# y#
œ (62.4) ˆ 43 ‰ 27
#
3
27 ‘
3
‰
œ (62.4) ˆ 34 ‰ ˆ 27
6 œ 374.4 lb
(b) We want to find a new water level Y such that FY œ
"
#
(374.4) œ 187.2 lb. The new depth of the strip is
(Y y), and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy
Y
œ 62.4'0 (Y y) ˆ 43 y‰ dy œ (62.4) ˆ 43 ‰'0 aYy y# b dy œ (62.4) ˆ 43 ‰ ’Y †
Y
Y
œ (62.4) ˆ 29 ‰ Y$ . Therefore Y$ œ
9FY
2†(62.4)
œ
(9)(187.2)
124.8
y#
#
Y
y$
3 “!
$
œ (62.4) ˆ 34 ‰ Š Y2
Y$
3 ‹
$
$È
Ê Y œ É (9)(187.2)
13.5 ¸ 2.3811 ft. So,
124.8 œ
?Y œ 3 Y ¸ 3 2.3811 ¸ 0.6189 ft ¸ 7.5 in. to the nearest half inch.
(c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depth of the
water.
‰
22. The area of a strip of the face of height ?y and parallel to the base is 100ˆ 26
24 † ?y, where the factor of
26
24
inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then:
‰
F œ '0 w(24 y)a100bˆ 26
24 dy œ '('! ’#%y
24
#%
y#
# “!
œ '('!Š#%#
#%#
# ‹
œ 1,946,880 lb.
CHAPTER 6 PRACTICE EXERCISES
#
1. A(x) œ 14 (diameter)# œ 14 ˆÈx x# ‰
œ 14 ˆx 2Èx † x# x% ‰ ; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ
b
œ
1
4
œ
1
4†70
#
’ x# 74 x(Î#
È3
4
x&
5 “!
(35 40 14) œ
2. A(x) œ
œ
1
4
"
"
#
'01 ˆx 2x&Î# x% ‰ dx
1
4
œ
ˆ "#
4
7
5" ‰
91
280
(side)# ˆsin 13 ‰ œ
È3
4
ˆ2Èx x‰#
ˆ4x 4xÈx x# ‰ ; a œ 0, b œ 4
Ê V œ 'a A(x) dx œ
È3
4
b
œ
È3
4
œ
32È3
4
’2x# 85 x&Î#
ˆ1
8
5
'04 ˆ4x 4x$Î# x# ‰ dx
%
x$
3 “!
32 ‰ œ
8È 3
15
œ
È3
4
ˆ32
8†32
5
(15 24 10) œ
64 ‰
3
8È 3
15
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
accounts for the
412
Chapter 6 Applications of Definite Integrals
3. A(x) œ
œ
1
4
1
4
(diameter)# œ
1
4
(2 sin x 2 cos x)#
† 4 asin# x 2 sin x cos x cos# xb
œ 1(1 sin 2x); a œ
1
4
,bœ
51
4
Ê V œ 'a A(x) dx œ 1 '1Î4 (1 sin 2x) dx
51Î4
b
œ 1 x
cos 2x ‘ &1Î%
#
1Î%
œ 1 ’Š 541
cos 5#1
#
cos 1#
#
‹ Š 14
‹“ œ 1 #
#
#
%
4. A(x) œ (edge)# œ ŒŠÈ6 Èx‹ 0 œ ŠÈ6 Èx‹ œ 36 24È6 Èx 36x 4È6 x$Î# x# ;
a œ 0, b œ 6 Ê V œ 'a A(x) dx œ '0 Š36 24È6 Èx 36x 4È6 x$Î# x# ‹ dx
b
6
œ ’36x 24È6 † 23 x$Î# 18x# 4È6 † 25 x&Î#
œ 216 576 648
5. A(x) œ
(diameter)# œ
1
4
72 œ 360
Š2Èx
x#
4‹
#
œ
1728
5
œ
1
4
œ 216 16 † È6 È6 † 6 18 † 6# 58 È6 È6 † 6#
18001728
5
Š4x x&Î#
œ
6$
3
72
5
x%
16 ‹ ;
a œ 0, b œ 4 Ê V œ 'a A(x) dx
b
'04 Š4x x&Î# 16x ‹ dx œ 14 ’2x# 27 x(Î# 5x†16 “ % œ 14 ˆ32 32 † 87 25 † 32‰
%
œ
1
4
œ
321
4
ˆ1
6. A(x) œ
œ
1
4
1728
5
'
x$
3 “!
È3
4
"
#
8
7
25 ‰ œ
81
35
&
(35 40 14) œ
(edge)# sin ˆ 13 ‰ œ
È3
4
!
721
35
2Èx ˆ2Èx‰‘#
ˆ4Èx‰# œ 4È3 x; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ '0 4È3 x dx œ ’2È3 x# “
b
1
"
!
œ 2È3
7. (a) .3=5 7/>29. :
V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx
b
1
1
#
"
œ 1 cx* d " œ 21
(b) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1
b
1
1
'
!
Note: The lower limit of integration is 0 rather than 1.
(c) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ 21'c1 (1 x) a3x% b dx œ 21 ’ 3x5
b
1
"
&
(d) A+=2/< 7/>29. :
"
x'
2 “ "
œ 21 ˆ 35 "# ‰ ˆ 35 "# ‰‘ œ
R(x) œ 3, r(x) œ 3 3x% œ 3 a1 x% b Ê V œ 'a 1 cR# (x) r# (x)d dx œ 'c1 1 ’9 9 a1 x% b “ dx
b
1
œ 91 'c1 c1 a1 2x% x) bd dx œ 91 'c1 a2x% x) b dx œ 91 ’ 2x5
1
1
&
"
x*
9 “ "
#
œ 181 25 9" ‘ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
21†13
5
œ
261
5
121
5
Chapter 6 Practice Exercises
8. (a) A+=2/< 7/>29. :
R(x) œ
, r(x) œ
4
x$
"
#
#
#
#
&
Ê V œ 'a 1cR# (x) r# (x)d dx œ '1 1 ’ˆ x4$ ‰ ˆ "# ‰ “ dx œ 1 16
x4 ‘ "
5 x
b
2
"‰
"
ˆ 16 " ‰‘ œ 1 ˆ 10
œ 1 ˆ 5†16
32 # 5 4
(b) =2/66 7/>29. :
V œ 21'1 x ˆ x4$ "# ‰ dx œ 21 ’4x"
2
(c) =2/66 7/>29. :
"
#
#
x#
4 “"
4" ‰ œ
16
5
1
20
(2 10 64 5) œ
b
2
4
x
x
(d) A+=2/< 7/>29. :
#
x#
4 “"
571
#0
œ 21 ˆ 4# 1‰ ˆ4 4" ‰‘ œ 21 ˆ 54 ‰ œ
shell ‰
shell
V œ 21'a ˆ radius
Š height
‹ dx œ 21'1 (2 x) ˆ x4$ "# ‰ dx œ 21'1 ˆ x8$
œ 21 ’ x4#
413
2
4
x#
51
#
1 x# ‰ dx
œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ
31
#
V œ 'a 1cR# (x) r# (x)d dx
b
#
œ 1 '1 ’ˆ 7# ‰ ˆ4
2
dx
œ
491
4
161'1 a1 2x$ x' b dx
œ
491
4
161 ’x x#
œ
491
4
491
4
491
4
161 ˆ2 4" 5†"3# ‰ ˆ1 1 5" ‰‘
"
161 ˆ 4" 160
5" ‰
œ
œ
9.
4 ‰#
x$ “
2
161
160
#
x&
5 “"
(40 1 32) œ
491
4
711
10
1031
20
œ
(a) .3=5 7/>29. :
V œ 1 '1 ŠÈx 1‹ dx œ 1'1 (x 1) dx œ 1 ’ x# x“
#
5
5
#
‰ ˆ"
‰‘ œ 1 ˆ 24
‰
œ 1 ˆ 25
# 5 # 1
# 4 œ 81
&
"
(b) A+=2/< 7/>29. :
R(y) œ 5, r(y) œ y# 1 Ê V œ 'c 1 cR# (y) r# (y)d dy œ 1 'c2 ’25 ay# 1b “ dy
d
2
œ 1'c2 a25 y% 2y# 1b dy œ 1 'c2 a24 y% 2y# b dy œ 1 ’24y
2
2
œ 321 ˆ3
2
5
"3 ‰ œ
321
15
(45 6 5) œ
10881
15
#
y&
5
23 y$ “
#
#
œ 21 ˆ24 † 2
(c) .3=5 7/>29. :
R(y) œ 5 ay# 1b œ 4 y#
Ê V œ 'c 1R# (y) dy œ 'c2 1 a4 y# b dy
d
2
#
œ 1 'c2 a16 8y# y% b dy
2
œ 1 ’16y
8y$
3
œ 641 ˆ1
2
3
#
y&
5 “ #
"5 ‰ œ
œ 21 ˆ32
641
15
64
3
(15 10 3) œ
32 ‰
5
5121
15
10. (a) =2/66 7/>29. :
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y Šy
d
4
œ 21'0 Šy#
4
œ
21
1#
† 64 œ
y$
4‹
321
3
$
dy œ 21 ’ y3
%
y%
16 “ !
y#
4‹
dy
œ 21 ˆ 64
3
64 ‰
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
32
5
2
3
† 8‰
414
Chapter 6 Applications of Definite Integrals
(b) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ˆ2Èx x‰ dx œ 21'0 ˆ2x$Î# x# ‰ dx œ 21 ’ 45 x&Î#
b
4
œ 21 ˆ 45 † 32
64 ‰
3
œ
4
1281
15
%
x$
3 “!
(c) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(4 x) ˆ2Èx x‰ dx œ 21'0 ˆ8x"Î# 4x 2x$Î# x# ‰ dx
b
4
$Î#
œ 21 ’ 16
2x# 54 x&Î#
3 x
œ 641 ˆ1 45 ‰ œ
641
5
4
%
x$
3 “!
œ 21 ˆ 16
3 † 8 32
(d) =2/66 7/>29. :
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(4 y) Šy
d
4
œ 21'0 Š4y 2y#
4
y$
4‹
y#
4‹
%
y%
16 “ !
dy œ 21 ’2y# 23 y$
4
5
† 32
64 ‰
3
œ 641 ˆ 34 1
dy œ 21'0 Š4y y# y#
4
œ 21 ˆ32
2
3
4
5
y$
4‹
32 ‰
dy
† 64 16‰ œ 321 ˆ2
321
3
1‰ œ
8
3
11. .3=5 7/>29. :
R(x) œ tan x, a œ 0, b œ
1
3
Ê V œ 1 '0 tan# x dx œ 1'0 asec# x 1b dx œ 1[tan x x]!
1Î3
12. .3=5 7/>29. :
1Î3
1Î$
V œ 1'0 (2 sin x)# dx œ 1 '0 a4 4 sin x sin# xb dx œ 1'0 ˆ4 4 sin x
1
œ 1 4x 4 cos x
1
x
#
sin 2x ‘ 1
4
!
1
œ 1 ˆ41 4
1
#
0‰ (0 4 0 0)‘ œ
œ
1cos 2x ‰
dx
#
9
1
1 ˆ # 8‰ œ 1#
1 Š3È31‹
3
(91 16)
13. (a) .3=5 7/>29. :
V œ 1'0 ax# 2xb dx œ 1'0 ax% 4x$ 4x# b dx œ 1 ’ x5 x% 43 x$ “ œ 1 ˆ 32
5 16
2
œ
161
15
2
#
(6 15 10) œ
#
&
!
161
15
32 ‰
3
(b) A+=2/< 7/>29. :
V œ '0 1’1# ax# 2x "b “ dx œ '0 1 dx '0 1 ax "b% dx œ #1 ’1
2
2
#
2
(c) =2/66 7/>29. :
#
ax"b&
& “!
œ #1 1 †
#
&
œ
)1
&
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ 21'0 (2 x) c ax# 2xbd dx œ 21'0 (2 x) a2x x# b dx
b
2
2
œ 21'0 a4x 2x# 2x# x$ b dx œ 21'0 ax$ 4x# 4xb dx œ 21 ’ x4 43 x$ 2x# “ œ 21 ˆ4
2
œ
21
3
2
(36 32) œ
#
%
!
81
3
32
3
8‰
(d) A+=2/< 7/>29. :
V œ 1 '0 c2 ax# 2xbd dx 1'0 2# dx œ 1'0 ’4 4 ax# 2xb ax# 2xb “ dx 81
2
2
#
2
#
œ 1'0 a4 4x# 8x x% 4x$ 4x# b dx 81 œ 1'0 ax% 4x$ 8x 4b dx 81
2
&
2
#
‰
œ 1 ’ x5 x% 4x# 4x“ 81 œ 1 ˆ 32
5 16 16 8 81 œ
!
1
5
(32 40) 81 œ
721
5
14. .3=5 7/>29. :
V œ 21'0 4 tan# x dx œ 81'0 asec# x 1b dx œ 81[tan x x]!
1Î4
1Î4
1Î%
œ 21(4 1)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
401
5
œ
321
5
Chapter 6 Practice Exercises
15. The material removed from the sphere consists of a cylinder
and two "caps." From the diagram, the height of the cylinder
#
is 2h, where h# ŠÈ$‹ œ ## , i.e. h œ ". Thus
#
Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap,
use the disk method and x# y# œ ## : Vcap œ '" 1x# dy
2
œ '" 1a% y# bdy œ 1’%y
2
œ 1ˆ8 83 ‰ ˆ% "3 ‰‘ œ
#
y3
3 “"
&1
3
ft$ . Therefore,
Vremoved œ Vcyl #Vcap œ '1
"!1
3
#)1
3
œ
ft$ .
16. We rotate the region enclosed by the curve y œ É12 ˆ1
4x# ‰
121
and the x-axis around the x-axis. To find the
11Î2
volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1
b
œ 121'c11Î2 Š1
11Î2
4x#
121 ‹
œ 1321 ˆ1 "3 ‰ œ
17. y œ x"Î#
x$Î#
3
Ê
ˆ4
"
#
œ
#
x"Î# "# x"Î# Ê Š dy
dx ‹ œ
"
4
œ '1
8
dx
dy
È9x#Î$ 4
3x"Î$
œ
5
12
2
3
#
4
4
ˆ2
14 ‰
3
#
x"Î$ Ê Š dx
dy ‹ œ
œ
4x#Î$
9
"
#
ˆx"Î# x"Î# ‰ dx œ
"
#
2x"Î# 23 x$Î# ‘ %
"
10
3
dx
Ê L œ '1 Ê1 Š dy
‹ dy œ '1 É1
#
8
8
4
9x#Î$
dy
'1 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4 Ê du œ 6y"Î$ dy; x œ 1
40
" '
" 2 $Î# ‘ %!
Ä L œ 18
u"Î# du œ 18
œ #"7 40$Î# 13$Î# ‘ ¸ 7.634
3 u
"$
13
dx œ
x œ 8 Ê u œ 40d
19. y œ
"
#
† 8‰ ˆ2 23 ‰‘ œ
18. x œ y#Î$ Ê
8
"
3
x'Î& 58 x%Î& Ê
dx
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx
1
4
2
3
4x#
121 ‹
œ 881 ¸ 276 in$
4
"
#
11Î2
dx œ 1 '11Î2 12 Š1
4 ‰ 11
ˆ 4 ‰ ˆ 11
‰ “ œ 1321 ’1 ˆ 363
œ 241 ’ 11
Š 4 ‹“
2 363
#
#
Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1
œ
#
$
4x$
363 “ ""Î#
dx œ 121 ’x
2641
3
dy
dx
""Î#
4x# ‰
121 ‹
#
"
#
œ
dy
dx
x"Î& "# x"Î& Ê Š dy
dx ‹ œ
"
4
Ê u œ 13,
ˆx#Î& 2 x#Î& ‰
#
Ê L œ '1 É1 4" ax#Î& 2 x#Î& b dx Ê L œ '1 É 4" ax#Î& 2 x#Î& b dx œ ' É 4" ax"Î& x"Î& b dx
32
32
32
1
œ '1
32
œ
"
48
20. x œ
"
#
ˆx
"Î&
x
"Î& ‰
(1260 450) œ
"
1#
y$
"
y
Ê
" %
œ '1 É 16
y
2
"
#
dx œ
œ
1710
48
dx
dt
5
4
$#
x%Î& ‘ "
#
œ
"
4
"
y%
dy œ '1 ÊŠ 4" y#
y#
"
y#
x
'Î&
dx
dy
8
œ ˆ 12
"# ‰ ˆ 1"# 1‰ œ
21.
" 5
# 6
285
8
dx
Ê Š dy
‹ œ
2
7
1#
œ 5 sin t 5 sin 5t and
dy
dt
"
#
œ
"
y# ‹
"
16
#
œ
"
#
ˆ 65
y%
"
#
'
†2
"
y%
5
4
ˆ 56
5 ‰‘
4
œ
"
#
ˆ 315
6
" %
Ê L œ '1 Ê1 Š 16
y
dy œ '1 Š 4" y#
2
†2
%‰
2
"
y# ‹
dy œ ’ 1"# y$ y" “
"
#
75 ‰
4
"
y% ‹
dy
#
"
13
12
#
‰ Š dy
œ 5 cos t 5 cos 5t Ê Êˆ dx
dt
dt ‹
#
œ Éa5 sin t 5 sin 5tb# a5 cos t 5 cos 5tb#
œ 5Èsin# 5t #sin t sin 5t sin# t cos# t #cos t cos 5t cos# 5t œ &È# #asin t sin 5t cos t cos 5 tb
œ 5È#a" cos %tb œ 5É%ˆ "# ‰a" cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
415
416
Chapter 6 Applications of Definite Integrals
Ê Length œ '!
1 Î2
22.
dx
dt
œ 3t2 12t and
1Î#
"!sin #t dt œ c5 cos #td !
dy
dt
œ a&ba"b a&ba"b œ "!
#
#
‰ Š dy
Éa3t2 12tb# a3t2 12tb# œ È288t# "8t4
œ 3t2 12t Ê Êˆ dx
dt
dt ‹ œ
œ 3È2 ktkÈ16 t2 Ê Length œ '! 3È2 ktkÈ16 t2 dt œ 3È2'! t È16 t2 dt; ’u œ 16 t2 Ê du œ 2t dt
"
"
3È 2
2
Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“;
œ
23.
dx
d)
3È 2
2
† 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617.
œ $ sin ) and
Ê Length œ '!
dy
d)
$1Î2
#
24. x œ t and y œ
œ'
'16"7 Èu du œ 3È2 2 23 u3/2 ‘1617 œ 3È2 2 Š 23 a17b3/2 23 a16b3/2 ‹
t$
3
$ d) œ $'!
$1Î2
d) œ $ˆ $#1 !‰ œ
t, È3 Ÿ t Ÿ È3 Ê
È3
È 3
#
#
‰ Š dy
Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $
œ $ cos ) Ê Êˆ dx
d)
d) ‹ œ
Èt% #t# " dt œ
'
dx
dt
*1
#
œ 2t and
dy
dt
È3
È 3
Èt% 2t# " dt œ
œt
'
#
" Ê Length œ '
È3
È 3
È3
È 3
Éat# "b# dt œ
Éa2tb# at# "b# dt
È
'È33 at# "b dt œ ’ t3 t“
3
È3
È 3
œ 4È3
25. Intersection points: 3 x# œ 2x# Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Symmetry
suggests that x œ 0. The typical @/<>3-+6 strip has
#
#
#
center of mass: (µ
x ßµ
y ) œ Šxß 2x a3 x b ‹ œ Šxß x 3 ‹ ,
#
#
#
#
#
length: a3 x b 2x œ 3 a1 x b, width: dx,
area: dA œ 3 a1 x# b dx, and mass: dm œ $ † dA
œ 3$ a1 x# b dx Ê the moment about the x-axis is
µ
y dm œ
œ
3
#
3
#
$ ax# 3b a1 x# b dx œ
&
$ ’ x5
œ 3$ ’x
2x$
3
"
x$
3 “ "
3x“
"
"
3
#
$ ax% 2x# 3b dx Ê Mx œ ' µ
y dm œ
œ 3$ ˆ 5"
2
3
3‰ œ
œ 6$ ˆ1 "3 ‰ œ 4$ Ê y œ
Mx
M
œ
3$
15
(3 10 45) œ
32$
5 †4 $
œ
8
5
32$
5
3
#
$ 'c1 ax% 2x# 3b dx
"
; M œ ' dm œ 3$ 'c1 a1 x# b dx
"
. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .
26. Symmetry suggests that x œ 0. The typical @/<>3-+6
#
strip has center of mass: (µ
x ßµ
y ) œ Šxß x# ‹ , length: x# ,
width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx
Ê the moment about the x-axis is µ
y dm œ #$ x# † x# dx
x% dx Ê Mx œ ' µ
y dm œ
œ
$
#
œ
2$
10
a2& b œ
32$
5
$
#
; M œ ' dm œ $
'c22 x% dx œ 10$ cx& d ##
'c22 x# dx œ $ ’ x3 “ #
$
#
œ
2$
3
a2$ b œ
16$
3
Ê yœ
Mx
M
œ
32†$ †3
5†16†$
œ
centroid is(xß y) œ ˆ!ß 65 ‰ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
6
5
. Therefore, the
Chapter 6 Practice Exercises
417
27. The typical @/<>3-+6 strip has: center of mass: (µ
x ßµ
y )
œ Œxß
#
4 x4
#
, length: 4
area: dA œ Š4
œ $ Š4
x#
4‹
x#
4 ‹dx,
width: dx,
mass: dm œ $ † dA
dx Ê the moment about the x-axis is
#
Š4 x4 ‹
µ
y dm œ $ †
x#
4,
x#
4‹
Š4
#
$
#
dx œ
moment about the y-axis is µ
x dm œ $ Š4
œ
$
2
’16x
%
x&
5†16 “ !
$
#
œ
64
#
œ
16†$ †3
32†$
œ
Mx
M
and y œ
3
2
œ
dx. Thus, Mx œ ' µ
y dm œ
4
4
My
M
x$
4‹
‹ † x dx œ $ Š4x
x
4
œ $ (32 16) œ 16$ ; M œ ' dm œ $ '0 Š4
Ê xœ
dx; the
; My œ ' µ
x dm œ $ '0 Š4x
128$
5
œ
64 ‘
5
x%
16 ‹
Š16
128†$ †3
5†32†$
x#
4‹
œ
dx œ $ ’4x
12
5
%
x$
12 “ !
x$
4‹
dx œ $ ’2x#
œ $ ˆ16
64 ‰
1#
$
#
'04 Š16 16x ‹ dx
%
%
x%
16 “ !
32$
3
œ
‰
. Therefore, the centroid is (xß y) œ ˆ 3# ß 12
5 .
28. A typical 29<3D98>+6 strip has:
#
center of mass: (µ
x ßµ
y ) œ Š y # 2y ß y‹ , length: 2y y# ,
width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA
œ $ a2y y# b dy; the moment about the x-axis is
µ
y dm œ $ † y † a2y y# b dy œ $ a2y# y$ b ; the moment
#
about the y-axis is µ
x dm œ $ † ay 2yb † a2y y# b dy
#
œ a4y y b dy Ê Mx œ ' µ
y dm œ $ '0 a2y# y$ b dy
$
#
#
œ $ ’ 23 y$
œ
$
#
ˆ 43†8
yœ
Mx
M
2
%
œ
#
y%
4 “!
32 ‰
5
4†$ †3
3†4†$
œ
œ $ ˆ 23 † 8
32$
15
16 ‰
4
œ $ ˆ 16
3
16 ‰
4
œ
$ †16
12
œ
4$
3
; My œ ' µ
x dm œ
$
#
'02 a4y# y% b dy œ #$ ’ 34 y$ y5 “ #
&
$
; M œ ' dm œ $ '0 a2y y# b dy œ $ ’y# y3 “ œ $ ˆ4 83 ‰ œ
#
2
!
!
4$
3
Ê xœ
My
M
œ
$ †32†3
15†$ †4
œ
œ 1. Therefore, the centroid is (xß y) œ ˆ 85 ß 1‰ .
29. A typical horizontal strip has: center of mass: (µ
x ßµ
y )
œ Šy
#
2y
# ß y‹ ,
length: 2y y# , width: dy,
area: dA œ a2y y# b dy, mass: dm œ $ † dA
œ (1 y) a2y y# b dy Ê the moment about the
x-axis is µ
y dm œ y(1 y) a2y y# b dy
#
œ a2y 2y$ y$ y% b dy
œ a2y# y$ y% b dy; the moment about the y-axis is
#
µ
x dm œ Š y 2y ‹ (1 y) a2y y# b dy œ " a4y# y% b (1 y) dy œ
#
Ê Mx œ ' µ
y dm œ '0 a2y# y$ y% b dy œ ’ 23 y$
2
œ
16
60
œ
"
#
"
#
#
(20 15 24) œ
$
Š 4†32 2%
2&
5
4
15
(11) œ
2'
6‹
‰ ˆ 38 ‰ œ
œ ˆ 44
15
44
40
œ
11
10
y$
3
; My œ ' µ
x dm œ '0
2
œ 4 ˆ 43 2
œ '0 a2y y# y$ b dy œ ’y#
2
44
15
y%
4
4
5
"
#
#
y&
5 “!
œ ˆ4
8
3
16 ‰
4
œ ˆ 16
3
œ
8
3
24
5
area: dA œ
x$Î#
32 ‰
5
œ 16 ˆ "3
"
#
dx, mass: dm œ $ † dA œ $ †
x$Î#
25 ‰
’ 43 y$ y%
y&
5
2
Ê xœ
My
M
‰ ˆ 83 ‰ œ
œ ˆ 24
5
9
5
‰
. Therefore, the center of mass is (xß y) œ ˆ 95 ß 11
10 .
3
"
4
; M œ ' dm œ '0 (1 y) a2y y# b dy
30. A typical vertical strip has: center of mass: (µ
x ßµ
y ) œ ˆxß 2x3$Î# ‰ , length:
3
16
4
a4y# 4y$ y% y& b dy œ
86 ‰ œ 4 ˆ2 45 ‰ œ
#
y%
4 “!
a4y# 4y$ y% y& b dy
3
x$Î#
, width: dx,
dx Ê the moment about the x-axis is
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
and y œ
Mx
M
#
y'
6 “!
8
5
and
418
Chapter 6 Applications of Definite Integrals
µ
y dm œ
9$
2x$
3
† $ x$Î#
dx œ
3
#x$Î#
(a) Mx œ $ '1
9
M œ $ '1
9
(b) Mx œ '1
9
"
#
9$
#
ˆ x9$ ‰ dx œ
#
*
20$
9
’ x# “ œ
"
*
ˆ x9$ ‰ dx œ
9
#
*
dx.
*
My
M
œ
12$
4$
œ 3 and y œ
Mx
M
œ
ˆ 209$ ‰
4$
œ
5
9
*
3 ‰
3 ‰
"x ‘ * œ 4; My œ ' x# ˆ $Î#
dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î#
dx
"
x
1
9
œ 6 x"Î# ‘ " œ 12 Ê x œ
31. S œ 'a 21y Ê1 Š dy
dx ‹ dx;
#
b
3$
x"Î#
3 ‰
; My œ $ '1 x ˆ x$Î#
dx œ 3$ 2x"Î# ‘ " œ 12$ ;
9
dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ
3
x$Î#
x
#
3
dx; the moment about the y-axis is µ
x dm œ x † $ x$Î#
dx œ
My
M
dy
dx
œ
œ
13
3
and y œ
Mx
M
9
œ
"
3
#
"
È2x 1
Ê Š dy
dx ‹ œ
Ê S œ '0 21È2x 1 É1
3
"
#x 1
"
#x 1
dx
2
È ' Èx 1 dx œ 2È21 2 (x 1)$Î# ‘ $ œ 2È21 † 2 (8 1) œ
œ 21'0 È2x 1 É 2x
2x1 dx œ 2 21 0
3
3
!
3
3
32. S œ 'a 21y Ê1 Š dy
dx ‹ dx;
#
b
œ
1
6
dy
dx
%
'
œ x# Ê Š dy
dx ‹ œ x Ê S œ 0 21 †
#
1
x$
3
È1 x% dx œ
1
6
281È2
3
'01 È1 x% a4x$ b dx
'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“
!
33. S œ 'c 21x Ê1 Š dx
dy ‹ dy;
#
d
dx
dy
œ
ˆ "# ‰ (4 2y)
È4y y#
2y
È4y y#
œ
#
Ê 1 Š dx
dy ‹ œ
4y y# 4 4y y#
4y y#
œ
4
4y y#
Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41
2
2
34. S œ 'c 21x Ê1 Š dx
dy ‹ dy;
#
d
œ 1'2 È4y 1 dy œ
6
35. x œ
t#
#
1
4
dx
dy
œ
*
36. x œ t#
"
2t
761
3
"
È2
1
œ 21 Š2
(125 27) œ
œ t and
ŸtŸ1 Ê
Ê Surface Area œ '1ÎÈ2 21 ˆt#
1
dx
dt
1
6
dy
dt
1
6
"
4y
œ
Ê S œ '2 21Èy †
6
4y 1
4y
(98) œ
È4y 1
È4y
dy
491
3
È5
œ 2 Ê Surface Area œ '0 21(2t)Èt# 4 dt œ '4 21u"Î# du
9
, where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9
and y œ 4Èt ,
œ 21 '1ÎÈ2 ˆt#
#
Ê 1 Š dx
dy ‹ œ 1
23 (4y 1)$Î# ‘ ' œ
#
and y œ 2t, 0 Ÿ t Ÿ È5 Ê
œ 21 23 u$Î# ‘ % œ
1
2È y
" ‰ˆ
2t
2t
" ‰
2t#
"‰
#t
dx
dt
œ 2t
ʈ2t
" ‰#
2t#
dt œ 21 '1ÎÈ2 ˆ2t$
1
"
2t#
and
dy
dt
œ
2
Èt
Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt#
#
3
#
1
" ‰ Ɉ
2t
#t
" ‰#
#t#
dt
"
4" t$ ‰ dt œ 21 2" t% 3# t 8" t# ‘ "ÎÈ#
3È 2
4 ‹
37. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.
The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%!
! œ 4000 J; the rope alone: the force required
b
40
to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40 x). The work
done is W# œ 'a F# (x) dx œ '0 0.8(40 x) dx œ 0.8 ’40x
b
40
the total work is W œ W" W# œ 4000 640 œ 4640 J
%!
x#
# “!
œ 0.8 Š40#
40#
# ‹
œ
(0.8)(1600)
#
œ 640 J;
38. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to
8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is
x‰
F(x) œ 8 † 800 † ˆ 2†24750
œ (6400) ˆ1
†4750
x ‰
9500
lb. The work done is W œ 'a F(x) dx
b
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 6 Practice Exercises
œ '0
4750
6400 ˆ1
x ‰
9500
dx œ 6400 ’x
œ 22,800,000 ft † lb
%(&!
x#
2†9500 “ !
œ 6400 Š4750
4750#
4†4750 ‹
419
œ ˆ 34 ‰ (6400)(4750)
39. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is
W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is
1
1
#
!
#
x#
20 ’ # “
"
W œ '1 kx dx œ k '1 x dx œ
2
"
2
œ 20 ˆ 4# "# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb
40. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ
œ
300
250
F
k
œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx
1Þ2
1Þ2
œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J
41. We imagine the water divided into thin slabs by planes
perpendicular to the y-axis at the points of a partition of the
interval [0ß 8]. The typical slab between the planes at y and
y ?y has a volume of about ?V œ 1(radius)# (thickness)
#
œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to
lift this slab is equal to its weight: F(y) œ 62.4 ?V
œ
(62.4)(25)
16
1y# ?y lb. The distance through which F(y)
must act to lift this slab to the level 6 ft above the top is
about (6 8 y) ft, so the work done lifting the slab is about ?W œ
(62.4)(25)
16
1y# (14 y) ?y ft † lb. The work done
lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately
8
W¸!
!
(62.4)(25)
16
1y# (14 y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of
the partition goes to zero: W œ '0
8
œ
(62.4) ˆ 25161 ‰ Š 14
3
$
†8
8%
4‹
(62.4)(25)
(16)
1y# (14 y) dy œ
(62.4)(25)1
16
'08 a14y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$ y4 “ )
%
!
¸ 418,208.81 ft † lb
42. The same as in Exercise 41, but change the distance through which F(y) must act to (8 y) rather than
(6 8 y). Also change the upper limit of integration from 8 to 5. The integral is:
W œ '0
5
(62.4)(25)1
16
y# (8 y) dy œ (62.4) ˆ 25161 ‰'0 a8y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$
5
œ (62.4) ˆ 25161 ‰ Š 38 † 5$
5%
4‹
&
y%
4 “!
¸ 54,241.56 ft † lb
43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x œ
#
horizontal slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ
slab is its weight: F(y) œ 60 †
1
4
1
4
10
22,500 ft†lb
275 ft†lb/sec
y œ y# . A typical
y# ?y. The force required to lift this
y# ?y. The distance through which F(y) must act is (2 10 y) ft, so the
work to pump the liquid is W œ 60'0 1(12 y) Š y4 ‹ dy œ 151 ’ 12y
3
to empty the tank is
5
10
#
$
"!
y%
4 “!
œ 22,5001 ft † lb; the time needed
¸ 257 sec
44. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16 y# ‰ ?y and the force required to
lift this slab is its weight F(y) œ (57)(20) ˆ2È16 y# ‰ ?y. The distance through which F(y) must act is
(6 4 y) ft, so the work to pump the olive oil from the half-full tank is
W œ 57'c4 (10 y)(20) ˆ2È16 y# ‰ dy œ 2880 'c4 10È16 y# dy 1140'c4 a16 y# b
0
0
0
"Î#
(2y) dy
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
420
Chapter 6 Applications of Definite Integrals
œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 y# b
$Î# !
“
œ 335,153.25 ft † lb
%
œ (22,800)(41) 48,640
strip
45. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ 2 '0 (62.4)(2 y)(2y) dy œ 249.6'0 a2y y# b dy œ 249.6 ’y#
b
2
2
#
y$
3 “!
œ (249.6) ˆ4 83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb
strip
46. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ '0 75 ˆ 56 y‰ (2y 4) dy œ 75'0 ˆ 53 y
5Î6
b
5Î6
10
3
2y# 4y‰ dy
7
7 #
2 $ ‘ &Î'
50 ‰
25 ‰
125 ‰‘
#‰
ˆ 18
œ 75 '0 ˆ 10
dy œ 75 10
ˆ 67 ‰ ˆ 36
ˆ 32 ‰ ˆ 216
3 3 y 2y
3 y 6 y 3 y ! œ (75)
5Î6
œ (75) ˆ 25
9
175
216
250 ‰
3†#16
‰
œ ˆ 9†75
#16 (25 † 216 175 † 9 250 † 3) œ
strip
47. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ 62.4'0 (9 y) Š2 †
b
4
%
œ 62.4 6y$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8
2
5
Èy
2 ‹
(75)(3075)
9†#16
¸ 118.63 lb.
dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy
4
‰ (48 † 5 64) œ
† 32‰ œ ˆ 62.4
5
(62.4)(176)
5
œ 2196.48 lb
strip
48. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth
‹ † L(y) dy, h œ the height of the mercury column,
h
strip depth œ h y, L(y) œ 1 Ê F œ '0 849(h y) " dy œ (849)'0 (h y) dy œ 849’hy
h
œ
849 #
2 h .
Now solve
849 #
2 h
h
h
y#
# “!
œ 849 Šh#
h#
#‹
œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ
49. F œ w" '0 (8 y)(2)(6 y) dy w# 'c6 (8 y)(2)(y 6) dy œ 2w" '0 a48 14y y# b dy 2w# '6 a48 2y y# b dy
6
0
œ 2w" ’48y 7y#
'
y$
3 “!
6
2w# ’48y y#
!
y$
3 “ '
0
œ 216w" 360w#
50. (a) F œ 62.4'0 (10 y) ˆ8 y6 ‰ ˆ y6 ‰‘ dy
6
œ
6
62.4
3
' a240 34y y# b dy
0
œ
62.4
3
’240y 17y#
œ 18,720 lb.
'
y$
3 “!
œ
62.4
3
(1440 612 72)
(b) The centroid ˆ 72 ß 3‰ of the parallelogram is located at the intersection of y œ
6
7
x and y œ 65 x
36
5 .
The centroid of
the triangle is located at (7ß 2). Therefore, F œ (62.4)(7)(36) (62.4)(8)(6) œ (300)(62.4) œ 18,720 lb
CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES
1. V œ 1 'a cf(x)d# dx œ b# ab Ê 1'a cf(t)d# dt œ x# ax for all x a Ê 1 [f(x)]# œ 2x a Ê f(x) œ „ É 2x1 a
b
x
2. V œ 1 '0 [f(x)]# dx œ a# a Ê 1 '0 [f(t)]# dt œ x# x for all x a Ê 1[f(x)]# œ 2x 1 Ê f(x) œ „ É 2x1 1
a
x
3. s(x) œ Cx Ê '0 È1 [f w (t)]# dt œ Cx Ê È1 [f w (x)]# œ C Ê f w (x) œ ÈC# 1 for C
x
1
Ê f(x) œ '0 ÈC# 1 dt k. Then f(0) œ a Ê a œ 0 k Ê f(x) œ '0 ÈC# 1 dt a Ê f(x) œ xÈC# 1 a,
x
where C
x
1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 6 Additional and Advanced Exercises
421
4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1 cos# ) d).
!
The line segment from (0ß 0) to (!ß sin !) has length È(! 0)# (sin ! 0)# œ È!# sin# !. Since the
shortest distance between two points is the length of the straight line segment joining them, we have
!
immediately that ' È1 cos# ) d) È!# sin# ! if 0 ! Ÿ 1 .
#
0
(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1 [f w (t)]# dt È!# f # (!)
!
for ! 0.
5. From the symmetry of y œ 1 xn , n even, about the y-axis for 1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we
n
use the vertical strips technique. The typical strip has center of mass: (µ
x ßµ
y ) œ ˆxß 1 2 x ‰ , length: 1 xn ,
width: dx, area: dA œ a1 xn b dx, mass: dm œ 1 † dA œ a1 xn b dx. The moment of the strip about the
1
1
n #
n #
"
nb1
2n b 1
x-axis is µ
y dm œ a1 x b dx Ê M œ ' a1 x b dx œ 2' " a1 2xn x2n b dx œ x 2x x ‘
#
œ1
2
n1
"
#n 1
œ
x
c1 #
(n 1)(2n 1) 2(2n ") (n 1)
(n 1)(#n 1)
œ
0 #
2n# 3n 1 4n 2 n 1
(n 1)(#n 1)
Also, M œ 'c1 dA œ 'c1 a1 xn b dx œ 2 '0 a1 xn b dx œ 2 x
1
yœ
Mx
M
œ
1
#
2n
(n 1)(2n 1)
†
1
(n 1)
2n
œ
n
2n 1
xn b 1 ‘ "
n1 !
n1
œ
2n#
(n 1)(#n 1)
œ 2 ˆ1
" ‰
n1
#n 1 !
.
œ
2n
n1.
Therefore,
Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä
"
#
so
the limiting position of the centroid is ˆ!ß "# ‰ .
6. Align the telephone pole along the x-axis as shown in the
accompanying figure. The slope of the top length of pole is
9 ‰
ˆ 14.5
"
81 81
œ 8"1 † 40
† (14.5 9) œ 815.5
†40
40
11 ‰
y œ 891 8111†80 x œ 8"1 ˆ9 80
x is an
œ
11
81†80 .
Thus,
equation of the
line representing the top of the pole. Then,
My œ 'a x † 1y# dx œ 1 '0 x 8"1 ˆ9
b
40
11
80
#
x‰‘ dx
b
11 ‰#
'040 x ˆ9 80
x dx; M œ 'a 1y# dx
40
40
‰‘# dx œ 64"1 ' ˆ9 11
‰# dx.
œ 1 '0 8"1 ˆ9 11
80 x
80 x
0
œ
"
641
My
M
Thus, x œ
¸
129,700
5623.3
¸ 23.06 (using a calculator to compute
the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole.
7. (a) Consider a single vertical strip with center of mass (µ
x ßµ
y ). If the plate lies to the right of the line, then
µ
µ b) $ dA Ê the plate's first moment
the moment of this strip about the line x œ b is (x b) dm œ (x
about x œ b is the integral ' (x b)$ dA œ ' $ x dA ' $ b dA œ My b$ A.
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is
ab µ
x b dm œ ab µ
x b $ dA Ê the plate's first moment about x œ b is ' (b x)$ dA œ ' b$ dA ' $ x dA
œ b$ A My .
8. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass:
(µ
x ßµ
y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some
a
proportionality constant k. The moment of the strip about the y-axis is M œ ' µ
x dm œ ' 4kx# Èax dx
y
œ 4kÈa'0 x&Î# dx œ 4kÈa 27 x(Î# ‘ 0 œ 4ka"Î# † 27 a(Î# œ
a
a
œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ
a
a
8ka
7
%
8ka$
5
0
. Also, M œ ' dm œ '0 4kxÈax dx
a
. Thus, x œ
My
M
œ
8ka%
7
†
5
8ka$
œ
5
7
a
‰
Ê (xß y) œ ˆ 5a
7 ß 0 is the center of mass.
y#
#
#
a
(b) A typical horizontal strip has center of mass: (µ
x ßµ
y ) œ Œ 4a # ß y œ Š y 8a4a ß y‹ , length: a
width: dy, area: Ša
y#
4a ‹
dy, mass: dm œ $ dA œ kyk Ša
y#
4a ‹
dy. Thus, Mx œ ' µ
y dm
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y#
4a
,
422
Chapter 6 Applications of Definite Integrals
œ 'c2a y kyk Ša
2a
œ 'c2a Šay#
0
%
32a&
20a
œ 8a3
y#
4a ‹
0
dy '0 Šay#
2a
y%
4a ‹
8a%
3
dy œ 'c2a y# Ša
#
œ
%
&
'c2a a16a y y b dy
œ
"
32a#
’8a% † 4a#
64a'
6 “
"
32a#
'0
2a
"
32a#
2a
#
c2a
"
3 #a #
a16a% y y& b dy œ
64a'
6 “
"
16a#
œ
’8a% y#
32a'
3 ‹
Š32a'
œ
œ2†
16a%
4 ‹
#
Š2a † 4a
"
#a
œ
!
y'
6 “ #a
1
3#a#
’8a% y#
† 32 a32a' b œ
4
3
#a
y'
6 “!
a% ;
'c2a2a kyk a4a# y# b dy
"
4a
#a
#
dy
"
16a#
%
"
4a
#a
y&
“
#0a !
y#
4a ‹
kyk Ša
'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ !
"
4a
yœ
#
4a#
8a ‹
’ 3a y$
' kyk a16a% y% b dy
#
’8a% † 4a#
M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ
y&
#0a “ #a
dy
2a
#
"
3 #a #
#
y#
4a ‹
!
dy œ ’ 3a y$
'c2a2a kyk ay# 4a# b Š 4a 4a y ‹ dy œ 32a"
2a
2a
"
8a
0
dy '0 y# Ša
œ 0; My œ ' µ
x dm œ 'c2a Š y
32a&
#0a
œ
œ
y%
4a ‹
y#
4a ‹
%
%
$
a8a 4a b œ 2a . Therefore, x œ
"
4a
’2a# y#
œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ
My
M
#a
y%
4 “!
2a
3
and
œ 0 is the center of mass.
Mx
M
9. (a) On [0ß a] a typical @/<>3-+6 strip has center of mass: (µ
x ßµ
y ) œ Šx,
È b # x # È a# x#
‹,
#
length: Èb# x# Èa# x# , width: dx, area: dA œ ŠÈb# x# Èa# x# ‹ dx, mass: dm œ $ dA
œ $ ŠÈb# x# Èa# x# ‹ dx. On [aß b] a typical @/<>3-+6 strip has center of mass:
È #
#
(µ
x ßµ
y ) œ Šxß b # x ‹ , length: Èb# x# , width: dx, area: dA œ Èb# x# dx,
mass: dm œ $ dA œ $ Èb# x# dx. Thus, Mx œ ' µ
y dm
œ '0
a
"
#
ŠÈb# x# Èa# x# ‹ $ ŠÈb# x# Èa# x# ‹ dx 'a
b
"
#
Èb# x# $ Èb# x# dx
œ
$
#
'0a cab# x# b aa# x# bd dx #$ 'ab ab# x# b dx œ #$ '0a ab# a# b dx #$ 'ab ab# x# b dx
œ
$
#
cab# a# b xd ! #$ ’b# x
œ
$
#
aab# a$ b #$ Š 23 b$ ab#
a
b
x$
3 “a
œ
a$
3‹
$
#
b$
3‹
cab# a# b ad #$ ’Šb$
œ
$ b$
3
$ a$
3
œ $ Šb
$
a$
3 ‹;
a$
3 ‹“
Š b# a
My œ ' µ
x dm
œ '0 x$ ŠÈb# x# Èa# x# ‹ dx 'a x$ Èb# x# dx
a
b
œ $ '0 x ab# x# b
a
œ
$
#
”
2 ab # x # b
3
$Î#
dx $ '0 x aa# x# b
a
"Î#
a
$ 2 aa
• #”
#
$Î#
x# b
3
# $Î#
#
œ ’ab a b
# $Î#
ab b
a
dx $ 'a x ab# x# b
$ 2 ab
• #”
b
#
!
0
$
3
"Î#
# $Î#
$
3
“ ’0 aa b
•
a
$
3
“ ’0 ab# a# b
#
#
$Î#
We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹ $ Š 14a ‹ œ
œ
$ ab $ a $ b
3
yœ
(b) lim
œ
Mx
M
4
b Ä a 31
†
4
$1 ab# a# b
4 aa# abb# b
31(ab)
Ša
#
ab b#
‹
ab
œ
4
31
$
$
a
Š bb#
a# ‹ œ
dx
b
$Î#
x# b
3
"Î#
4 (b a) aa# ab b# b
31
(b a)(b a)
“œ
$1
4
$ b$
3
$ a$
3
œ
$ ab $ a $ b
3
ab# a# b . Thus, x œ
œ Mx ;
My
M
œ
4 aa# ab b# b
31(a b)
2a
1
2a ‰
Ê (xß y) œ ˆ 2a
1 ß 1 is the limiting
; likewise
.
œ ˆ 341 ‰ Š a
#
a# a#
‹
aa
#
œ ˆ 341 ‰ Š 3a
2a ‹ œ
position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles
coincide when b œ a).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 6 Additional and Advanced Exercises
10. Since the area of the traingle is 36, the diagram may be
labeled as shown at the right. The centroid of the triangle is
ˆ 3a , 24
‰
a . The shaded portion is 144 36 œ 108. Write
ax, yb for the centroid of the remaining region. The centroid
of the whole square is obviously a6, 6b. Think of the square
as a sheet of uniform density, so that the centroid of the
square is the average of the centroids of the two regions,
weighted by area:
'œ
$'ˆ 3a ‰ "!)axb
"%%
and ' œ
‰
$'ˆ 24
a "!)ayb
"%%
which we solve to get x œ )
a
*
and y œ
)a a " b
.
a
Set
x œ 7 in. (Given). It follows that a œ *, whence y œ
œ
7 "*
'%
*
in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.
above, we will find x œ 7 "* .)
11. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ
3
4
3
(1 x)$Î# ‘ $ œ
!
28
3
12. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is
"
# #
# (21a)(21ak) œ 21 a k.
d# x
dt#
13. F œ ma œ t# Ê
œaœ
t#
m
Ê vœ
x œ 0 when t œ 0 Ê C" œ 0 Ê x œ
W œ ' F dx œ '0
Ð12mhÑ"Î%
œ
(12mh)$Î#
18m
œ
F(t) †
12mh†È12mh
18m
œ
2h
3
dx
dt
dx
t$
dt œ 3m C; v œ 0 when t œ 0 Ê
t%
"Î%
.
12m . Then x œ h Ê t œ (12mh)
dt œ '0
Ð12mhÑ"Î%
† 2È3mh œ
14. Converting to pounds and feet, 2 lb/in œ
t# †
t$
3m
dt œ
4h
3
È3mh
†
12 in
1 ft
2 lb
1 in
"
3m
'
’ t6 “
Ð12mh)"Î%
0
Cœ0 Ê
dx
dt
œ
t$
3m
Ê xœ
The work done is
" ‰
œ ˆ 18m
(12mh)'Î%
œ 24 lb/ft. Thus, F œ 24x Ê W œ '0
1Î2
24x dx
"Î#
"
"
‰
œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10
lb‰ ˆ 3# ft/sec
#
"
œ 320
slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height,
s œ 16t# v! t (since s œ 0 at t œ 0) Ê ds
dt œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê
#
and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ
v!#
64
œ
3†640
64
œ 30 ft.
15. The submerged triangular plate is depicted in the figure
at the right. The hypotenuse of the triangle has slope 1
Ê y (2) œ (x 0) Ê x œ (y 2) is an equation
of the hypotenuse. Using a typical horizontal strip, the fluid
strip
strip
pressure is F œ ' (62.4) † Š depth
‹ † Š length
‹ dy
c2
c2
œ 'c6 (62.4)(y)[(y 2)] dy œ 62.4 'c6 ay# 2yb dy
$
œ 62.4 ’ y3 y# “
#
'
‰‘
œ (62.4) ˆ 83 4‰ ˆ 216
3 36
‰
œ (62.4) ˆ 208
3 32 œ
(62.4)(112)
3
t%
12m
¸ 2329.6 lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
tœ
v!
3#
C" ;
423
424
Chapter 6 Applications of Definite Integrals
16. Consider a rectangular plate of length j and width w.
The length is parallel with the surface of the fluid of
weight density =. The force on one side of the plate is
F œ ='cw (y)(j) dy œ =j ’ y# “
0
#
!
œ
w
=jw#
#
. The
=
w
average force on one side of the plate is Fav œ
œ
=
w
#
’ y# “
!
=w
#
œ
w
. Therefore the force
'c0w (y)dy
=jw#
#
‰
œ ˆ =w
# (jw) œ (the average pressure up and down) † (the area of the plate).
17. (a) We establish a coordinate system as shown. A typical
horizontal strip has: center of pressure: (µ
x ßµ
y )
b
ˆ
‰
œ # ß y , length: L(y) œ b, width: dy, area: dA
œ b dy, pressure: dp œ = kyk dA œ =b kyk dy
0
0
Ê Fx œ ' µ
y dp œ 'ch y † =b kyk dy œ =b 'ch y# dy
$
œ =b ’ y3 “
!
œ =b ’0 Š 3h ‹“ œ
=bh$
3
'c0h = kyk L(y) dy œ =b 'c0h
F œ ' dp œ
œ
$
h
# !
=b ’ y# “
h
œ =b ’0
h#
#“
=bh#
#
œ
;
y dy
$
. Thus, y œ
œ
Fx
F
Š =3bh ‹
#
Š =bh
# ‹
œ
2h
3
Ê the distance below the surface is
(b) A typical horizontal strip has length L(y). By similar
triangles from the figure at the right,
L(y)
b
œ
y a
h
Ê L(y) œ bh (y a). Thus, a typical strip has center
of pressure: (µ
x ßµ
y ) œ (µ
x ß y), length: L(y)
œ bh (y a), width: dy, area: dA œ bh (y a) dy,
pressure: dp œ = kyk dA œ =(y) ˆ bh ‰ (y a) dy
œ =b ay# ayb dy Ê F œ ' µ
y dp
h
x
a
œ 'cÐahÑ y †
%
=b
h
#
ay ayb dy œ
'ÐaahÑ
=b
h
a
ay$
3 “ cÐahÑ
ay$ ay# b dy
œ
=b
h
’ y4
œ
=b
h
’Š a4
œ
œ
=b
12h
=b
12h
œ
=bh
12
œ
=b
h
’Š 3a
œ
=b
h
’a
œ
=b
6h
a6a# h 6ah# 2h$ 6a# h 3ah# b œ
œ
%
a%
3‹
%
Š (a 4 h)
a(a h)$
‹“
3
œ
=b
h
’a
%
(a h)%
4
a% a(a h)$
“
3
c3 aa% aa% 4a$ h 6a# h# 4ah$ h% bb 4 aa% a aa$ 3a# h 3ah# h$ bbd
a12a$ h 12a# h# 4ah$ 12a$ h 18a# h# 12ah$ 3h% b œ
a6a# 8ah 3h# b ; F œ ' dp œ ' = kyk L(y) dy œ
$
$
a$
#‹
$
Š (a 3 h)
3a# h 3ah# h$ a$
3
ˆ 1=#bh ‰ a6a# 8ah 3h# b
ˆ =6bh ‰ (3a 2h)
6a# 8ah 3h#
6a 4h
a(a h)#
‹“
#
œ
=b
h
a$ aa$ 2a# h ah# b
“
#
‰ 6a
œ ˆ "
# Š
#
=b
6h
8ah 3h#
‹
3a 2h
$
’ (a h)3
œ
=b
6h
a$
a
=b
12h
a6a# h# 8ah$ 3h% b
=b
h
'cÐahÑ
a$ a(a h)#
“
2
ay# ayb dy œ
=b
h
$
’ y3
c2 a3a# h 3ah# h$ b 3 a2a# h ah# bd
a3ah# 2h$ b œ
=bh
6
(3a 2h). Thus, y œ
Ê the distance below the surface is
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Fx
F
a
ay#
2 “ ÐahÑ
2
3
h.
CHAPTER 7 TRANSCENDENTAL FUNCTIONS
7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES
1. Yes one-to-one, the graph passes the horizontal test.
2. Not one-to-one, the graph fails the horizontal test.
3. Not one-to-one since (for example) the horizontal line y œ # intersects the graph twice.
4. Not one-to-one, the graph fails the horizontal test.
5. Yes one-to-one, the graph passes the horizontal test
6. Yes one-to-one, the graph passes the horizontal test
7. Domain: 0 x Ÿ 1, Range: 0 Ÿ y
9. Domain: 1 Ÿ x Ÿ 1, Range: 1# Ÿ y Ÿ
8. Domain: x 1, Range: y 0
1
#
10. Domain: _ x _, Range: 1# y Ÿ
11. The graph is symmetric about y œ x.
(b) y œ È1 x# Ê y# œ 1 x# Ê x# œ 1 y# Ê x œ È1 y# Ê y œ È1 x# œ f " (x)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
#
426
Chapter 7 Transcendental Functions
12. The graph is symmetric about y œ x.
yœ
"
x
Ê xœ
"
y
Ê yœ
"
x
œ f " (x)
13. Step 1: y œ x# 1 Ê x# œ y 1 Ê x œ Èy 1
Step 2: y œ Èx 1 œ f " (x)
14. Step 1: y œ x# Ê x œ Èy, since x Ÿ !.
Step 2: y œ Èx œ f " (x)
15. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$
Step 2: y œ $Èx 1 œ f " (x)
16. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x
1 Ê x œ 1 Èy
Step 2: y œ 1 Èx œ f " (x)
17. Step 1: y œ (x 1)# Ê Èy œ x 1, since x
Step 2: y œ Èx 1 œ f
"
1 Ê x œ È y 1
(x)
18. Step 1: y œ x#Î$ Ê x œ y$Î#
Step 2: y œ x$Î# œ f " (x)
19. Step 1: y œ x& Ê x œ y"Î&
Step 2: y œ &Èx œ f " (x);
Domain and Range of f " : all reals;
&
f af " (x)b œ ˆx"Î& ‰ œ x and f " (f(x)) œ ax& b
"Î&
œx
"Î%
œx
20. Step 1: y œ x% Ê x œ y"Î%
Step 2: y œ %Èx œ f " (x);
Domain of f " : x
f af
"
(x)b œ ˆx
"Î% ‰%
0, Range of f " : y
œ x and f
"
0;
(f(x)) œ ax% b
21. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$
Step 2: y œ $Èx 1 œ f " (x);
Domain and Range of f " : all reals;
$
f af " (x)b œ ˆ(x 1)"Î$ ‰ 1 œ (x 1) 1 œ x and f " (f(x)) œ aax$ 1b 1b
"Î$
œ ax$ b
"Î$
œx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 7.1 Inverse Functions and Their Derivatives
22. Step 1: y œ
"
#
x
"
#
Ê
7
#
"
xœy
7
#
Ê x œ 2y 7
Step 2: y œ 2x 7 œ f (x);
Domain and Range of f " : all reals;
f af " (x)b œ "# (2x 7) 7# œ ˆx 7# ‰
23. Step 1: y œ
Step 2: y œ
"
x#
Ê x# œ
"
y
"
Èx
œ f " (x)
Ê xœ
7
#
œ x and f " (f(x)) œ 2 ˆ "# x 7# ‰ 7 œ (x 7) 7 œ x
"
Èy
Domain of f " : x 0, Range of f " : y 0;
f af " (x)b œ "" # œ "" œ x and f " (f(x)) œ
Š Èx ‹
24. Step 1: y œ
"
x$
Ê x$ œ
"
x"Î$
"
Step 2: y œ
Domain of f
f af " (x)b œ
Šx‹
"
y
Ê xœ
(c)
26. (a) y œ
"
5
"
$
ax"Î$ b
"
x"
œ
œ 2,
df "
dx ¹ xœ1
x7 Ê
df ¸
dx xœ1
(c)
œ x since x 0
"
y"Î$
œ x and f " (f(x)) œ ˆ x"$ ‰
"
5
œ
" df "
œ 5,
dx
¹
œ 4,
df "
dx ¹ xœ3
œ ˆ x" ‰
"
œx
(b)
x
#
3
#
xœy7
xœ$%Î&
"Î$
"
#
"
(b)
(x) œ 5x 35
œ5
27. (a) y œ 5 4x Ê 4x œ 5 y
Ê x œ 54 y4 Ê f " (x) œ
df ¸
dx xœ1Î#
"
Š "x ‹
: x Á 0, Range of f " : y Á 0;
Ê x œ 5y 35 Ê f
(c)
œ
œ $É x" œ f " (x);
25. (a) y œ 2x 3 Ê 2x œ y 3
Ê x œ y# 3# Ê f " (x) œ
df ¸
dx xœ1
"
É x"#
œ
(b)
5
4
x
4
"
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
427
428
Chapter 7 Transcendental Functions
"
#
28. (a) y œ 2x# Ê x# œ
Ê xœ
(c)
df ¸
dx xœ&
"
È2
(b)
y
Èy Ê f
"
(x) œ
È x#
œ 4xk xœ5 œ 20,
df "
dx ¹ xœ&0
œ
"
#È 2
x"Î# ¹
xœ50
"
#0
œ
$
$
29. (a) f(g(x)) œ ˆ $Èx‰ œ x, g(f(x)) œ Èx$ œ x
w
#
w
(b)
w
(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3;
gw (x) œ 3" x#Î$ Ê gw (1) œ 3" , gw (1) œ
"
3
(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !);
the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)
30. (a) h(k(x)) œ
"
4
ˆ(4x)"Î$ ‰$ œ x,
k(h(x)) œ Š4 †
(c) hw (x) œ
w
k (x) œ
x$
4‹
"Î$
(b)
œx
#
3x
w
w
4 Ê h (2) œ 3, h (2)
4
#Î$
Ê kw (2) œ "3 ,
3 (4x)
œ 3;
kw (2) œ
(d) The line y œ 0 is tangent to h(x) œ
x$
4
"
3
at (!ß !);
the line x œ 0 is tangent to k(x) œ (4x)"Î$ at
(!ß !)
œ 3x# 6x Ê
31.
df
dx
33.
df "
dx ¹ x œ 4
df "
dx ¹ x œ f(3)
df "
dx ¹ x œ f(2)
œ
35. (a) y œ mx Ê x œ
"
m
œ
(b) The graph of y œ f
36. y œ mx b Ê x œ
y
m
"
df
dx
º
œ
xœ2
"
df
dx
œ
º
xœ3
"
ˆ 3" ‰
œ3
y Ê f " (x) œ
"
"
9
œ
"
m
œ 2x 4 Ê
32.
df
dx
34.
dg"
dx ¹x œ 0
b
m
dg"
dx ¹ x œ f(0)
œ
"
dg
dx
º
œ
xœ0
"
df
dx
º
œ
xœ5
œ
"
6
"
2
x
(x) is a line through the origin with slope
œ
df "
dx ¹ x œ f(5)
Ê f " (x) œ
"
m
x
b
m;
"
m.
the graph of f " (x) is a line with slope
"
m
and y-intercept mb .
37. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ x 1
(b) y œ x b Ê x œ y b Ê f " (x) œ x b
(c) Their graphs will be parallel to one another and lie on
opposite sides of the line y œ x equidistant from that
line.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 7.1 Inverse Functions and Their Derivatives
38. (a) y œ x 1 Ê x œ y 1 Ê f " (x) œ 1 x;
the lines intersect at a right angle
(b) y œ x b Ê x œ y b Ê f " (x) œ b x;
the lines intersect at a right angle
(c) Such a function is its own inverse.
39. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x" x# or
x" x# which implies f(x" ) f(x# ) or f(x" ) f(x# ), since f(x) is increasing. In either case,
f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing.
40. f(x) is increasing since x# x" Ê
"
3
x#
5
6
"
3
x" 56 ;
df
dx
œ
"
3
41. f(x) is increasing since x# x" Ê 27x$# 27x"$ ; y œ 27x$ Ê x œ
df
dx
œ 81x# Ê
df "
dx
œ
" ¸
81x# 13 x"Î$
œ
"
9x#Î$
œ
"
9
df "
dx
Ê
"
3
œ
df
dx
œ 24x# Ê
dx
œ
" ¸
24x# 12 Ð1xÑ"Î$
œ
œ3
y"Î$ Ê f " (x) œ
"
3
x"Î$ ;
x#Î$
42. f(x) is decreasing since x# x" Ê 1 8x$# 1 8x"$ ; y œ 1 8x$ Ê x œ
df "
"
ˆ "3 ‰
"
6(" x)#Î$
"
#
(1 y)"Î$ Ê f " (x) œ
"
#
(1 x)"Î$ ;
œ "6 (1 x)#Î$
43. f(x) is decreasing since x# x" Ê (1 x# )$ (1 x" )$ ; y œ (1 x)$ Ê x œ 1 y"Î$ Ê f " (x) œ 1 x"Î$ ;
df
dx
œ 3(1 x)# Ê
df "
dx
œ
"
3(1 x)# ¹ 1cx"Î$
&Î$
44. f(x) is increasing since x# x" Ê x#
df
dx
œ
5
3
x#Î$ Ê
df "
dx
œ
5
3
"
¹
x#Î$ x$Î&
œ
3
5x#Î&
œ
"
3x#Î$
œ "3 x#Î$
&Î$
x" ; y œ x&Î$ Ê x œ y$Î& Ê f " (x) œ x$Î& ;
œ
3
5
x#Î&
45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then
f(x" ) Á f(x# ), so f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well.
46. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then
f(x" ) Á f(x# ), so f(x"" ) Á f(x"# ) , and therefore h(x" ) Á h(x# ).
47. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since
g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because
x" Á x# Ê f(g(x" )) Á f(g(x# )).
48. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g
with g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption
that f ‰ g is one-to-one.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
429
430
Chapter 7 Transcendental Functions
49. The first integral is the area between f(x) and the x-axis
over a Ÿ x Ÿ b. The second integral is the area between
f(x) and the y-axis for f(a) Ÿ y Ÿ f(b). The sum of the
integrals is the area of the larger rectangle with corners
at (0ß 0), (bß 0), (bß f(b)) and (0ß f(b)) minus the area of the
smaller rectangle with vertices at (0ß 0), (aß 0), (aß f(a)) and
(0ß f(a)). That is, the sum of the integrals is bf(b) af(a).
50. f w axb œ
acx dba aax bbc
acx db#
œ
ad bc
.
acx db#
Thus if ad bc Á !, f w axb is either always positive or always negative. Hence faxb is
either always increasing or always decreasing. If follows that faxb is one-to-one if ad bc Á !.
51. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1
52. W(a) œ 'f(a) 1 ’af " (y)b a# “ dy œ 0 œ 'a 21x[f(a) f(x)] dx œ S(a); Ww (t) œ 1’af " (f(t))b a# “ f w (t)
f(a)
a
#
#
œ 1 at# a# b f w (t); also S(t) œ 21f(t)'a x dx 21'a xf(x) dx œ c1f(t)t# 1f(t)a# d 21'a xf(x) dx
t
t
t
Ê Sw (t) œ 1t# f w (t) 21tf(t) 1a# f w (t) 21tf(t) œ 1 at# a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t)
for all t − [aß b].
53-60. Example CAS commands:
Maple:
with( plots );#53
f := x -> sqrt(3*x-2);
domain := 2/3 .. 4;
x0 := 3;
Df := D(f);
# (a)
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"],
title="#53(a) (Section 7.1)" );
q1 := solve( y=f(x), x );
# (b)
g := unapply( q1, y );
m1 := Df(x0);
# (c)
t1 := f(x0)+m1*(x-x0);
y=t1;
m2 := 1/Df(x0);
# (d)
t2 := g(f(x0)) + m2*(x-f(x0));
y=t2;
domaing := map(f,domain);
# (e)
p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ):
p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ):
p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ):
p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ):
p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ):
display( [p1,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" );
Mathematica: (assigned function and values for a, b, and x0 may vary)
If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica
to do this. See section 2.5 for details.
< int((sin(t))/t, t=0..infinity); ˆanswer is 1# ‰
76. (a)
(b) > f:= 2*exp(t^2)/sqrt(Pi);
> int(f, t=0..infinity); (answer is 1)
77. (a) faxb œ
1
x2 /2
È 21 e
f is increasing on (_, 0]. f is decreasing on [0, _). f has a local maximum at a0, fa0bb œ Š!ß
(b) Maple commands:
>f: œ exp(x^2/2)(sqrt(2*pi);
>int(f, x œ 1..1);
>int(f, x œ 2..2);
>int(f, x œ 3..3);
1
È 21 ‹
¸ 0.683
¸ 0.954
¸ 0.997
(c) Part (b) suggests that as n increases, the integral approaches 1. We can take 'cn f(x) dx as close to 1 as we want by
n
_
n
choosing n 1 large enough. Also, we can make 'n f(x) dx and '_ f(x) dx as small as we want by choosing n large
enough. This is because 0 faxb ex/2 for x 1. (Likewise, 0 faxb ex/2 for x 1.)
_
_
Thus, 'n f(x) dx 'n ex/2 dx.
'n_ ex/2 dx œ lim 'nc ex/2 dx œ lim c 2ex/2 dcn œ lim c 2ec/2 2en/2 d œ 2en/2
cÄ_
cÄ_
_
cÄ_
As n Ä _, 2en/2 Ä 0, for large enough n, 'n f(x) dx is as small as we want. Likewise for large enough n,
n
'_
f(x) dx is as small as we want.
78.
'3_ ˆ x " # x" ‰ dx Á '3_ x dx # '3_ dxx , since the left hand integral converges but both of the right hand
integrals diverge.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 8.8 Improper Integrals
_
_
79. (a) The statement is true since 'c_ f(x) dx œ '_ f(x) dx 'a f(x) dx, 'b f(x) dx œ 'a f(x) dx 'a f(x) dx
b
a
b
b
and 'a f(x) dx exists since f(x) is integrable on every interval [aß b].
b
(b)
80. (a)
(b)
81.
a
a
b
b
_
_
'_
f(x) dx 'a f(x) dx œ '_ f(x) dx 'a f(x) dx 'a f(x) dx 'a f(x) dx
b
a
b
_
_
œ '_ f(x) dx 'b f(x) dx 'a f(x) dx œ '_ f(x) dx 'b f(x) dx
0
0
_
_
_
'_
f(x) dx œ '_ f(x) dx '0 f(x) dx œ '_ f(u) du '0 f(x) dx
_
_
_
œ '0 f(u) du '0 f(x) dx œ 2 '0 f(x) dx, where u œ x
0
0
_
_
_
'_
f(x) dx œ '_ f(x) dx '0 f(x) dx œ '_ f(u) du '0 f(x) dx
_
_
_
_
œ '0 f(u) du '0 f(x) dx œ '0 f(x) dx '0 f(x) dx œ 0, where u œ x
_
dx
'_
È
x#
1
1
È x# 1
x
œ x lim
Ä_
82.
_
œ '_ Èxdx
'1
#1
_
"
'_
È
dx
È x# 1
É1
œ x lim
Ä_
"
x#
_
1 x'
dx converges, since '_
_
; '1
dx
È x# 1
diverges because x lim
Ä_
_ dx
œ 1 and '1
"
È 1 x'
x
_
diverges; therefore, '_
_
dx œ 2 '0
"
È 1 x'
Š "x ‹
ŠÈ
"
‹
x# 1
dx
È x# 1
diverges
dx which was shown to converge in
Exercise 51
83.
_
dx
"
'c_
' _ e dx e
e ec œ _ e 1 ; e 1 œ e ec
_
_
Ê 'c_ ee dx1 œ 2 '0 e dxec converges
x
x
x
x
2x
2x
x
x
"
ex
_
and '0
œ c lim
cex d c0 œ c lim
aec 1b œ 1
Ä_
Ä_
dx
ex
x
2x
84.
x
x
_ c
e dx
e
'_
' 1 ec dx ' _ ec dx ' 1 ec dx ' _ e du
x 1 œ _ x 1 1 x 1 ; _ x 1 œ 1 1 u , where u œ x, and since 1 u
_ ec dx
'1_ duu diverges, the integral '1_ 1eduu diverges Ê '_
x 1 diverges
x
x
#
x
#
x
#
u
#
u
#
u
86.
_
'c_
e x
k k
_
dx œ 2 '0 ex dx œ 2 lim
bÄ_
_
dx
'c_
' #
(x 1) œ _
b
lim c 'c# (x dx1)
b Ä 1
87.
dx
(x 1)#
#
#
œ
1
'#
lim
b Ä 1 c
dx
(x 1)#
'0b ex dx œ 2
_
'1 (x dx1)# '2
_
xk
'c__ ksin xkxk kkcos
dx œ 2 '0 ksin xxk k1cos xk dx
1
2
dx
(x 1)#
;
_ sin# x cos# x
2'0
x1
_
dx
'_
(x 1)
dx œ 2 lim
bÄ_
#
diverges
'0b x dx 1 dx
_ ksin xk kcos xk
'c_
dx diverges
kx k 1
_
x
'_
ax 1b ax 2b dx œ 0 by Exercise 80(b) because the integrand is odd and the integral
'0_ ax 1xbdxax 2b Ÿ '0_ dxx converges
#
#
#
#
(u 1) and
lim cex d b0 œ 2, so the integral converges.
bÄ_
x " 1 ‘ b œ _, which diverges Ê
#
bÄ_
89.
"
u
#
œ 2 lim cln kx 1kd b0 œ _, which diverges Ê
88.
x
#
85.
#
$
Example CAS commands:
Maple:
f := (x,p) -> x^p*ln(x);
domain := 0..exp(1);
fn_list := [seq( f(x,p), p=-2..2 )];
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
557
558
Chapter 8 Techniques of Integration
plot( fn_list, x=domain, y=-50..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0],
legend=["p= -2","p = -1","p = 0","p = 1","p = 2"], title="#89 (Section 8.8)" );
q1 := Int( f(x,p), x=domain );
q2 := value( q1 );
q3 := simplify( q2 ) assuming p>-1;
q4 := simplify( q2 ) assuming p<-1;
q5 := value( eval( q1, p=-1 ) );
i1 := q1 = piecewise( p<-1, q4, p=-1, q5, p>-1, q3 );
90.
Example CAS commands:
Maple:
f := (x,p) -> x^p*ln(x);
domain := exp(1)..infinity;
fn_list := [seq( f(x,p), p=-2..2 )];
plot( fn_list, x=exp(1)..10, y=0..100, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0],
legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#90 (Section 8.8)" );
q6 := Int( f(x,p), x=domain );
q7 := value( q6 );
q8 := simplify( q7 ) assuming p>-1;
q9 := simplify( q7 ) assuming p<-1;
q10 := value( eval( q6, p=-1 ) );
i2 := q6 = piecewise( p<-1, q9, p=-1, q10, p>-1, q8 );
91.
Example CAS commands:
Maple:
f := (x,p) -> x^p*ln(x);
domain := 0..infinity;
fn_list := [seq( f(x,p), p=-2..2 )];
plot( fn_list, x=0..10, y=-50..50, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0],
legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#91 (Section 8.8)" );
q11 := Int( f(x,p), x=domain ):
q11 = lhs(i1+i2);
`` = rhs(i1+i2);
`` = piecewise( p<-1, q4+q9, p=-1, q5+q10, p>-1, q3+q8 );
`` = piecewise( p<-1, -infinity, p=-1, undefined, p>-1, infinity );
92.
Example CAS commands:
Maple:
f := (x,p) -> x^p*ln(abs(x));
domain := -infinity..infinity;
fn_list := [seq( f(x,p), p=-2..2 )];
plot( fn_list, x=-4..4, y=-20..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9],
legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#92 (Section 8.8)" );
q12 := Int( f(x,p), x=domain );
q12p := Int( f(x,p), x=0..infinity );
q12n := Int( f(x,p), x=-infinity..0 );
q12 = q12p + q12n;
`` = simplify( q12p+q12n );
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
89-92. Example CAS commands:
Mathematica: (functions and domains may vary)
Clear[x, f, p]
f[x_]:= xp Log[Abs[x]]
int = Integrate[f[x], {x, e, 100)]
int /. p Ä 2.5
In order to plot the function, a value for p must be selected.
p = 3;
Plot[f[x], {x, 2.72, 10}]
CHAPTER 8 PRACTICE EXERCISES
1.
#
' xÈ4x# 9 dx; ” u œ 4x 9 • Ä
du œ 8x dx
2.
#
' 6xÈ3x# 5 dx; ” u œ 3x 5 • Ä ' Èu du œ 23 u$Î# C œ 23 a3x# 5b$Î# C
du œ 6x dx
3.
' x(2x 1)"Î# dx; ” u œ 2x 1 • Ä
du œ 2 dx
œ
4.
(2x 1)&Î#
10
' Èx
dx; ”
1x
œ
2
3
(2x 1)$Î#
6
"
#
"
1#
a4x# 9b
$Î#
C
' ˆ u # 1 ‰ Èu du œ "4 Š' u$Î# du ' u"Î# du‹ œ "4 ˆ 25 u&Î# 23 u$Î# ‰ C
C
uœ1x
Ä '
du œ dx •
(1 u)
Èu
du œ ' ŠÈu
"
Èu ‹
du œ
2
3
u$Î# 2u"Î# C
(1 x)$Î# 2(1 x)"Î# C
5.
' È x dx
6.
' È x dx
7.
' 25ydyy
8.
' 4ydyy ; ”
9.
' Èt
10.
'
11.
' Èu du œ "8 † 23 u$Î# C œ
"
8
8x#
;
1 ”
;”
9 4x#
#
;”
$
%
$
dt
9 4t%
2t dt
t% 1
"
16
' Èduu œ 16" † 2u"Î# C œ È8x8 1 C
u œ 9 4x#
Ä "8 '
du œ 8x dx •
u œ 25 y#
Ä
du œ 2y dy •
u œ 4 y%
• Ä
du œ 4y$ dy
;”
;”
u œ 8x# 1
Ä
du œ 16x dx •
"
#
"
4
#
du
Èu
È9 4x#
4
C
' duu œ #" ln kuk C œ #" ln a25 y# b C
' duu œ 4" ln kuk C œ 4" ln a4 y% b C
u œ 9 4t%
"
• Ä 16 '
du œ 16t$ dt
u œ t#
Ä
du œ 2t dt •
œ 8" † 2u"Î# C œ
du
Èu
"
œ 16
† 2u"Î# C œ
È9 4t%
8
C
' u du1 œ tan" u C œ tan" t# C
#
&Î$
"
' z#Î$ ˆz&Î$ 1‰#Î$ dz; ” u œ z5 #Î$
• Ä
du œ z dz
3
3
5
' u#Î$ du œ 35 † 35 u&Î$ C œ
9
#5
ˆz&Î$ 1‰&Î$ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
559
560
12.
Chapter 8 Techniques of Integration
z%Î&
' z"Î& ˆ1 z%Î& ‰"Î# dz; ” u œ 41 "Î&
• Ä
du œ z
dz
5
5
4
' u"Î# du œ 54 † 2Èu C œ #5 ˆ1 z%Î& ‰"Î# C
13.
2) d)
' (1sin cos
2) )
14.
) d)
' (1 cos
sin ))
15.
u œ 3 4 cos t
dt
' 3 sin4tcos
•
t ;”
16.
' 1cos sin2t dt2t ; ”
17.
' (sin 2x) ecos 2x dx; ” u œ cos 2x • Ä "# ' eu du œ "# eu C œ "# ecos 2x C
du œ 2 sin 2x dx
18.
u œ sec x
' (sec x tan x) esec x dx; ”
Ä ' eu du œ eu C œ esec x C
du œ sec x tan x dx •
19.
' e) sin ae) b cos# ae) b d); ”
20.
)
' e) sec# ae) b d); ” u œ e) • Ä ' sec# u du œ tan u C œ tan ae) b C
du œ e d)
21.
' 2xc1 dx œ
23.
' v dvln v ; ”
24.
' v(2 dvln v) ; ” u œ 2 " ln v •
#
;”
u œ 1 cos 2)
Ä
du œ # sin 2) d) •
"Î#
;”
u œ 1 sin )
Ä
du œ cos ) d) •
u œ 1 sin 2t
Ä
du œ 2 cos 2t dt •
v
dv
œ 4" ln kuk C œ 4" ln k3 4 cos tk C
' duu œ 2" ln kuk C œ 2" ln k1 sin 2tk C
26.
' sinÈ" x dx# ; – u œ sin dx x —
27.
' È 2 dx
;”
28.
' È dx
œ
du œ
"
du œ
' u# du œ "3 u$ C œ "3 cos$ ae) b C
È 1 x#
u œ 2x
Ä
du œ 2 dx •
dx
É1ˆ x7 ‰#
;”
' 5xÈ2 dx œ
"
È2
È2
x
Š 5ln 5 ‹ C
' duu œ ln kuk C œ ln k2 ln vk C
Ä
"
#
'
"
2
du
u
' duu œ ln kuk C œ ln kln vk C
' ax 1b adx2tan" xb ; ” u œ 2 tan
dx
"
7
œ 2u"Î# C œ 2È1 sin ) C
22.
25.
49x#
"Î#
C
du œ
1 4x#
' udu
#
u œ cos ˆe) ‰
• Ä
du œ sin ˆe) ‰ † e) d)
u œ ln v
Ä
du œ v" dv •
1x
' duu œ 2u" C œ #(1 "cos 2)) C
Ä 4" '
du œ 4 sin t dt
2xc1
ln 2
"
#
x # 1
Ä
x
• Ä
' duu œ ln kuk C œ ln k2 tan" xk C
' u du œ "# u# C œ "# asin" xb# C
' È du
1 u#
œ sin" u C œ sin" (2x) C
u œ x7
Ä
du œ "7 dx •
' È du
1 u#
œ sin" u C œ sin" ˆ x7 ‰ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
29.
30.
31.
'È
' È dt
9 4t#
' 9 dt t
#
"
4
œ
dt
16 9t#
œ
'
"
9
'
"
3
œ
'
dt
#
Ê1 Š 3t
4‹
dt
;–
#
Ê1 Š 2t
3‹
dt
#
1ˆt‰
;–
3
uœ
du œ
;–
uœ
du œ
uœ
du œ
"
3
"
3
3
4
3
4
2
3
2
3
t
Ä
dt —
t
Ä
dt —
t
Ä
dt —
"
2
34.
'
35.
' È dx
36.
'È
37.
2)
"
" ˆ y 2 ‰
' y dy4y 8 œ ' (yd(y
C
2) 4 œ # tan
#
38.
2)
"
' t dt4t 5 œ ' (t d(t
(t 2) C
2) 1 œ tan
4 dx
5xÈ25x# 16
œ 3'
6 dx
xÈ4x# 9
4x x#
œ
œ'
dx
4xx# 3
4
25
'
dx
xÉx# 94
œ'
d(x2)
È1(x2)#
40.
'
"
5
dv
(v 1)Èv# 2v
œ'
d(x1)
(x 1)È(x 1)# 1
œ sec" kx 1k C
œ'
d(v 1)
(v 1)È(v 1)# 1
œ sec" kv 1k C
6x
' cos# 3x dx œ ' " cos
dx œ x# sin126x C
#
43.
' sin$ #) d) œ ' ˆ1 cos# #) ‰ ˆsin #) ‰ d); –
tan" (5t) C
cos$
)
#
2 cos
)
#
u œ cos #)
Ä 2 ' a1 u# b du œ
du œ "# sin #) d) —
u$
3
Cœ
2u$
3
2u C
C
' sin$ ) cos# ) d) œ ' a1 cos# )b (sin )) acos# )b d); ”
u&
5
tan" ˆ 3t ‰ C
œ sin" (x 2) C
42.
œ
"
5
"
3
sin" ˆ 2t3 ‰ C
œ sin" ˆ x # 2 ‰ C
2x
' sin# x dx œ ' 1 cos
dx œ #x sin42x C
#
44.
tan" u C œ
"
2
sin" ˆ 3t4 ‰ C
¸
sec" ¸ 5x
4 C
41.
2
3
"
3
sin" u C œ
"
3
#
dx
(x 1)Èx# 2x
œ
œ
"
2
sin" u C œ
#
#
'
œ
œ
#
œ
"
3
¸
œ 2 sec" ¸ 2x
3 C
d(x 2)
È4 (x 2)#
#
39.
dx
xÉx# 16
25
#
1 u#
tan" u C œ
'
"
5
' È du
"
5
33.
u œ 5t
Ä
du œ 5 dt •
œ
1 u#
' 1 duu
' 1 dt25t
;”
' È du
"
3
32.
#
' 1 duu
"
3
cos& )
5
cos$ )
3
u œ cos )
Ä ' a1 u# b u# du œ ' au% u# b du
du œ sin ) d) •
C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
561
562
45.
Chapter 8 Techniques of Integration
' tan$ 2t dt œ '
Ä
œ
46.
"
4
"
#
(tan 2t) asec# 2t 1b dt œ ' tan 2t sec# 2t dt ' tan 2t dt; ”
u œ 2t
du œ 2 dt •
' tan u sec# u du "# ' tan u du œ 4" tan# u #" ln kcos uk C œ 4" tan# 2t #" ln kcos 2tk C
tan# 2t
"
#
ln ksec 2tk C
' 6 sec% t dt œ 6' atan# t 1b asec# tb dt; ”
u œ tan t
Ä 6 ' au# 1b du œ 2u$ 6u C
du œ sec# t dt •
œ 2 tan$ t 6 tan t C
47.
' 2 sin dxx cos x œ ' sindx2x œ '
48.
' cos 2xdxsin x œ ' cos2 dx2x ; ”
#
#
csc 2x dx œ "# ln kcsc 2x cot 2xk C
u œ 2x
Ä
du œ 2 dx •
'
du
cos u
œ ' sec u du œ ln ksec u tan uk C
œ ln ksec 2x tan 2xk C
49.
'11ÎÎ42 Ècsc# y 1 dy œ '11ÎÎ42 cot y dy œ cln ksin ykd 11ÎÎ24 œ ln 1 ln È"
50.
'13Î14Î4 Ècot# t 1 dt œ '13Î14Î4 csc t dt œ c ln kcsc t cot tkd 311Î4Î4 œ ln ¸csc 341 cot 341 ¸ ln ¸csc 14 cot 14 ¸
È
œ ln ¹È2 1¹ ln ¹È2 1¹ œ ln ¹ È2 " ¹ œ ln »
œ ln È2
ŠÈ2 "‹ ŠÈ2 1‹
#1
21
51.
2
» œ ln Š3 2
È2‹
'01 È1 cos# 2x dx œ '01 ksin 2xk dx œ '01Î2 sin 2x dx '11Î2 sin 2x dx œ cos#2x ‘ 10 Î2 cos#2x ‘ 11Î2
œ ˆ "# "# ‰ "# ˆ "# ‰‘ œ 2
52.
'021 È1 sin# x# dx œ '021 ¸cos x# ¸ dx œ '01 cos x# dx '121 cos x# dx œ 2 sin x# ‘ 10 2 sin x# ‘ 1#1
œ (2 0) (0 2) œ 4
53.
'11ÎÎ22 È1 cos 2t dt œ È2 '11ÎÎ22 ksin tk dt œ 2È2 '01Î2 sin t dt œ ’2È2 cos t“ 1Î2 œ 2È2 [0 (1)] œ 2È2
0
54.
'121 È1 cos 2t dt œ È2 '121 kcos tk dt œ È2 '131Î2 cos t dt È2 '3211Î2 cos t dt
œ È2 csin td $11Î2 È2 csin td #$11Î# œ È2 (1 0) È2 [0 (1)] œ 2È2
55.
' xx dx4 œ x ' x4 dx4 œ x 2 tan" ˆ #x ‰ C
56.
' 9xdxx
57.
' 4x2x 13 dx œ ' (2x 1) #x 4 1 ‘ dx œ x x# 2 ln k2x 1k C
58.
' 2xx dx4 œ ' ˆ2 x 8 4 ‰ dx œ 2x 8 ln kx 4k C
59.
' 2yy 41 dy œ '
#
#
#
$
#
œ ' ’ x ax x# 9b9 9x “ dx œ ' ˆx
#
9x ‰
x# 9
dx œ
x#
#
9
#
ln a9 x# b C
#
#
2y dy
y# 4
'
dy
y# 4
œ ln ay# 4b
"
#
tan" ˆ #y ‰ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
60.
' yy 41 dy œ ' yy dy1 4 ' y dy 1 œ #" ln ay# 1b 4 tan" y C
61.
' Èt 2
62.
' 2t È È1 t
63.
#
#
4 t#
dt œ '
#
#
1 t#
t
#
t dt
È4 t#
dt œ '
2t dt
È1 t#
'
dt
t
œ È4 t# 2 sin" ˆ #t ‰ C
œ 2È1 t# ln ktk C
#
#
#
#
cos x)
dx
cos x dx
' cotcotx x csc
' cos
' (cos1x)(1
dx œ ' cos x sin1 x sin x dx
x œ
x1 œ
cos x
x)
' sindx x ' dx œ sin" x cot x x C œ x cot x csc x C
œ ' d(sin
sin x
#
#
#
65.
dt
È4 t#
sin x)
dx
' tantanx x sec
' sinsin xxdx1 œ ' (sin1x)(1
dx œ ' sin x cos1 xcos x dx
x œ
sin x
x)
' cosdx x ' dx œ cos" x tan x x C œ x tan x sec x C
œ ' d(cos
cos x
#
64.
2'
#
#
' sec (5 3x) dx; ” y œ 5 3x •
Ä
dy œ 3 dx
' sec y † Š dy3 ‹ œ "3 ' sec y dy œ 3" ln ksec y tan yk C
œ "3 ln ksec (5 3x) tan (5 3x)k C
66.
' x csc ax# 3b dx œ "# ' csc ax# 3b d ax# 3b œ "# ln kcsc ax# 3b cot ax# 3bk C
67.
' cot ˆ x4 ‰ dx œ 4 ' cot ˆ x4 ‰ d ˆ x4 ‰ œ 4 ln ¸sin ˆ x4 ‰¸ C
68.
' tan (2x 7) dx œ "# ' tan (2x 7) d(2x 7) œ "# ln kcos (2x 7)k C œ "# ln ksec (2x 7)k C
69.
' xÈ1 x dx; ” u œ 1 x •
Ä ' (1 u)Èu du œ ' ˆu$Î# u"Î# ‰ du œ
du œ dx
$
œ
70.
72.
(1 x)
&Î#
(1 x)
2
3
$Î#
C œ 2 –
' 3xÈ2x 1 dx; ” u œ 2x 1 •
du œ 2 dx
œ
71.
2
5
3
10
œ
sec ) tan )
31
œ
sin )
2 cos# )
z œ tan )
Ä
dz œ sec# ) d) •
"
#
32
31
&
ŠÈ1 x‹
3
3 ˆÈ2x 1‰
10
ŠÈ1 x‹
—C
5
&
ˆÈ2x 1‰
#
$
C
' Ètan# ) 1 † sec# ) d) œ ' sec$ ) d)
(FORMULA 92)
ln ksec ) tan )k C œ
' a16 z# b$Î# dz; ”
z
16 a16 z# b"Î#
' sec ) d)
u&Î# 23 u$Î# C
' 3 ˆ u # 1 ‰ Èu † "# du œ 34 ' ˆu$Î# u"Î# ‰ du œ 34 † 25 u&Î# 34 † 23 u$Î# C
(2x 1)&Î# "# (2x 1)$Î# C œ
' Èz# 1 dz; ”
œ
Ä
2
5
z œ 4 tan )
Ä
dz œ 4 sec# ) d) •
zÈ z# 1
#
"
#
ln ¹z È1 z# ¹ C
) d)
" '
"
' 644 sec
cos ) d) œ 16
sin ) C œ
sec ) d) œ 16
#
$
z
16È16 z#
C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
C
563
564
73.
Chapter 8 Techniques of Integration
' È25dy y
œ
#
"
5
'
dy
É1ˆ 5y ‰#
œ'
du
È 1 u#
u œ tan )
, u œ y5 ‘ ; ”
Ä
du œ sec# ) d) •
' Èsec ) d)
#
1 tan# )
œ ' sec ) d)
#
œ ln ksec ) tan )k C" œ ln ¹È1 u# u¹ C" œ ln ºÉ1 ˆ y5 ‰ y5 º C" œ ln ¹
È25 y# y
¹
5
C"
œ ln ¸y È25 y# ¸ C
74.
' È25dy 9y
"
3
Ä
75.
'
76.
' Èx
#
œ
"
5
'
dy
Ê1 Š
3y #
5 ‹
œ
"
3
' È du
dx
x# È 1 x#
$
dx
1 x#
;”
;”
x œ sin )
Ä
dx œ cos ) d) •
x œ sin )
Ä
dx œ cos ) d) •
Note: Ans ´
' Èx
#
dx
1 x#
œ
78.
"
#
"
3
ln ¹È1 u# u¹ C" from Exercise 73
ln ¸È25 9y# 3y¸ C
' sincos))cosd) ) œ ' csc# ) d) œ cot ) C œ È1x x
)
;”
"
#
x
#È
1 x#
3
' sin )coscos) ) d) œ ' sin$ ) d) œ ' a1 cos# )b (sin )) d);
$
u$
3
C œ cos )
"
3
cos$ ) œ È1 x# 3" a1 x# b
x# 9
sin ) cos ) œ
;”
sin" x
#
xÈ 1 x#
#
x œ 2 sin )
Ä
dx œ 2 cos ) d) •
C
' 2 cos ) † 2 cos ) d) œ 2 ' (1 cos 2)) d) œ 2 ˆ) "# sin 2)‰ C
x œ 3 sec )
Ä'
dx œ 3 sec ) tan ) d) •
3 sec ) tan ) d)
È9 sec# ) 9
#
œ'
3 sec ) tan ) d)
3 tan )
œ ln ksec ) tan )k C" œ ln º x3 Ɉ x3 ‰ 1º C" œ ln ¹ x
80.
'
12 dx
ax# 1b$Î#
;”
C
#
#
' È dx
$Î#
2)
' sin )coscos) ) d) œ ' sin# ) d) œ ' 1 cos
d) œ "# ) 4" sin 2) C
#
œ 2) 2 sin ) cos ) C œ 2 sin" ˆ x# ‰ xÉ1 ˆ x# ‰ C œ 2 sin" ˆ x# ‰
79.
C
23 È1 x# C by another method
x œ sin )
Ä
dx œ cos ) d) •
' È4 x# dx; ”
#
#
cu œ cos )d Ä ' a1 u# b du œ u
77.
œ
1 u#
x œ sec )
Ä
dx œ sec ) tan ) d) •
È x# 9
¹
3
#
C
œ ' sec ) d)
cos ) d)
' 12 sectan) tan) ) d) œ ' 12 sin
;”
)
$
xÈ 4 x#
2
C" œ ln ¹x Èx# 9¹ C
u œ sin )
Ä
du œ cos ) d) •
' 12u du
#
12
12 x
œ 12
C
u C œ sin ) C œ È #
x 1
81.
' Èww 1 dw; ”
w œ sec )
tan ) ‰
' tan# ) d) œ ' asec# ) 1b d)
Ä ' ˆ sec
) † sec ) tan ) d) œ
dw œ sec ) tan ) d) •
œ tan ) ) C œ Èw# 1 sec" w C
82.
' Èz z 16 dz; ”
#
#
z œ 4 sec )
Ä
dz œ 4 sec ) tan ) d) •
sec ) tan ) d)
' 4 tan )†44sec
œ 4'
)
tan# ) d) œ 4(tan ) )) C
œ Èz# 16 4 sec" ˆ 4z ‰ C
83. u œ ln (x 1), du œ
dx
x1
; dv œ dx, v œ x;
' ln (x 1) dx œ x ln (x 1) ' x x 1 dx œ x ln (x 1) ' dx ' x dx 1 œ x ln (x 1) x ln (x 1) C"
œ (x 1) ln (x 1) x C" œ (x 1) ln (x 1) (x 1) C, where C œ C" 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
84. u œ ln x, du œ
dx
x
; dv œ x# dx, v œ
"
3
x$ ;
' x# ln x dx œ 3" x$ ln x ' 3" x$ ˆ x" ‰ dx œ x3
$
85. u œ tan" 3x, du œ
3 dx
1 9x#
#
"
6
x$
9
C
; dv œ dx, v œ x;
' tan" 3x dx œ x tan" 3x ' 13x 9xdx
œ x tan" (3x)
ln x
;”
y œ 1 9x#
Ä x tan" 3x
dy œ 18x dx •
"
6
' dyy
ln a1 9x# b C
86. u œ cos" ˆ x# ‰ , du œ
dx
È 4 x#
; dv œ dx, v œ x;
' cos" ˆ x# ‰ dx œ x cos" ˆ x# ‰ ' Èx dx
4 x #
;”
y œ 4 x#
Ä x cos" ˆ x# ‰
dy œ 2x dx •
"
#
' Èdyy
#
œ x cos" ˆ x# ‰ È4 x# C œ x cos" ˆ x# ‰ 2É1 ˆ x# ‰ C
ex
87.
ÐÑ
(x 1)# ïïïïî ex
ÐÑ
2(x 1) ïïïïî ex
ÐÑ
2
ïïïïî ex
Ê
0
' (x 1)# ex dx œ c(x 1)# 2(x 1) 2d ex C
sin (1 x)
88.
ÐÑ
x# ïïïïî
ÐÑ
2x ïïïïî
ÐÑ
2 ïïïïî
cos (1 x)
sin (1 x)
cos (1 x)
Ê
0
' x# sin (1 x) dx œ x# cos (1 x) 2x sin (1 x) 2 cos (1 x) C
89. u œ cos 2x, du œ 2 sin 2x dx; dv œ ex dx, v œ ex ;
I œ ' ex cos 2x dx œ ex cos 2x 2 ' ex sin 2x dx;
u œ sin 2x, du œ 2 cos 2x dx; dv œ ex dx, v œ ex ;
I œ ex cos 2x 2 ’ex sin 2x 2 ' ex cos 2x dx“ œ ex cos 2x 2ex sin 2x 4I Ê I œ
ex cos 2x
5
2ex sin 2x
5
90. u œ sin 3x, du œ 3 cos 3x dx; dv œ ec2x dx, v œ "# ec2x ;
I œ ' ec2x sin 3x dx œ "# ec2x sin 3x
3
2
' ec2x cos 3x dx;
u œ cos 3x, du œ 3 sin 3x dx; dv œ ec2x dx, v œ "# ec2x ;
I œ "# ec2x sin 3x 3# ’ "# ec2x cos 3x
Ê Iœ
4
13
3
#
' ec2x sin 3x dx“ œ "# ec2x sin 3x 34 ec2x cos 3x 94 I
2 c2x
ˆ "# ec2x sin 3x 34 ec2x cos 3x‰ C œ 13
e sin 3x
91.
' x x 3xdx 2 œ ' x2dx2 ' x dx 1 œ 2 ln kx 2k ln kx 1k C
92.
' x x 4xdx 3 œ #3 ' xdx3 #" ' x dx 1 œ #3 ln kx 3k #" ln kx 1k C
3
13
ec2x cos 3x C
#
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
C
565
566
Chapter 8 Techniques of Integration
93.
' x(xdx 1)
œ ' Š "x
94.
' x x(x11) dx œ ' ˆ x2 1 2x x" ‰ dx œ 2 ln ¸ x"
¸ "x C œ 2 ln kxk "x 2 ln kx 1k C
x
95.
' cos )sin )cosd)) 2 ; ccos ) œ yd
#
1
(x 1)# ‹
1
x1
#
Ä '
"
3
œ "3 '
dy
y# y 2
C
dy
y1
"
3
' y dy # œ 3" ln ¹ yy 12 ¹ C
)2¸
"
¸ cos ) 1 ¸
ln ¸ cos
cos ) 1 C œ 3 ln cos ) 2 C
96.
) d)
' sin )cos sin
) 6 ; csin ) œ xd
97.
' 3x x 4xx 4 dx œ '
98.
' x4xdx4x œ ' x4 dx4 œ 2 tan" ˆ #x ‰ C
99.
' (v2v3)8vdv œ #" '
#
4
x
$
$
'
Ä
#
dx '
dx
x# x 6
x4
x# 1
œ
"
5
)2¸
' xdx 2 5" ' xdx 3 œ 5" ln ¸ sin
sin ) 3 C
dx œ 4 ln kxk
"
#
ln ax# 1b 4 tan" x C
#
$
œ
"
x1
#
#
œ
dx œ ln kxk ln kx 1k
"
16
3
Š 4v
5
8(v 2)
"
8(v #) ‹
dv œ 38 ln kvk
5
16
ln kv 2k
&
7) dv
' (v (3v
' (v 2) 1dv '
1)(v 2)(v 3) œ
101.
' t 4tdt 3 œ "# ' t dt 1 #" ' t dt 3 œ #" tan" t
102.
' t t tdt 2 œ "3 ' t t dt2 3" ' t t dt1 œ 6" ln kt# 2k 6" ln at# 1b C
103.
' x xx x 2 dx œ ' ˆx x 2xx 2 ‰ dx œ ' x dx 32 ' x dx 1 34 ' x dx 2
%
#
#
%
#
#
$
dv
v2
'
dv
v3
#
3)
œ ln ¹ (v (v2)(v
1)# ¹ C
"
#È 3
tan" Š Èt 3 ‹ C œ
"
#
tan" t
ln kx 2k
2
3
105.
' x x4x4x 3 dx œ ' ˆx x 3x4x 3 ‰ dx œ '
$
107.
$
$
#
#
'
#
#
x
#
9
#
ln kx 3k
3
#
2x$ x# 21x 24
dx œ
x# 2x 8
2
#
x 3x 3 ln kx 4k
dx
x ˆ3 È x 1 ‰
œ
"
3
x dx
3
#
ln kx 1k ln kxk C
' x dx 1 #9 ' xdx3
ln kx 1k C
'
œ
C
ln kx 1k C
$
'
t
È3
#
4
3
' xx 1x dx œ ' ˆ1 xx "x ‰ dx œ ' ’1 x(x " 1) “ dx œ ' dx ' x dx 1 ' dxx œ x
106.
tan"
#
104.
œ
È3
6
#
#
œ
ln kv 2k C
ln ¹ (v 2)v'(v 2) ¹ C
100.
x#
#
"
16
(2x 3)
"
3
x
‘
x# 2x 8
dx œ ' (2x 3) dx
"
3
' x dx # 32 ' x dx 4
ln kx 2k C
Ô u œ Èx 1 ×
Ù Ä
; Ö du œ 2Èdx
x1
Õ dx œ 2u du Ø
2
3
' au udu1b u œ 3" ' u du 1 3" ' u du 1 œ 3" ln ku 1k 3" ln ku 1k C
#
È
1"
ln ¹ Èxx
¹C
11
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
$
Ô u œ Èx ×
dx
Ö du œ dx Ù Ä '
$ x‰ ;
x ˆ1 È
3x#Î$
Õ dx œ 3u# du Ø
108.
'
109.
' e ds 1 ; Ô du œ es ds ×
u œ es 1
110.
s
'
Õ ds œ
È es 1 "
¹ È es 1 1 ¹
' È16y dy y
111. (a)
' È16y dy y
(b)
Ø
du
u1
#
; cy œ 4 sin xd Ä 4 '
d a16 y# b
È16 y#
œ
(b)
' Èx dx
; cx œ 2 tan yd Ä
(b)
#
#
#
C
' Èx dx
' 4xdxx
' 4xdxx
du
' u a2uu du 1b œ 2 ' (u 1)(u
' u du 1 ' u du 1 œ ln ¸ uu "1 ¸ C
1) œ
u 1
112. (a)
113. (a)
$ x
È
œ 3 ln ¸ u u 1 ¸ C œ 3 ln ¹ 1È
$ x¹ C
s
œ "# '
4 x#
du
u(1 u)
s
#
4 x#
œ 3'
' u(udu 1) œ ' udu1 ' duu œ ln ¸ u u 1 ¸ C œ ln ¸ e e " ¸ C œ ln k1 ecs k C
È s
Ôu œ es 1×
e ds
Ö
; du œ 2Èes 1 Ù Ä
Õ ds œ 2u# du Ø
ds
È es 1
œ ln
Ä
3u# du
u$ (1 u)
"
#
œ È16 y# C
sin x cos x dx
cos x
œ 4 cos x C œ
4È16 y#
4
C œ È16 y# C
' dÈa4 x b œ È4 x# C
#
4 x#
œ #" '
d a4 x # b
4 x#
; cx œ 2 sin )d Ä
' 2 tan y2†2secsecy y dy œ 2 ' sec y tan y dy œ 2 sec y C œ È4 x# C
#
œ #" ln k4 x# k C
È
2 cos ) d)
' 2 sin4)†cos
œ ' tan ) d) œ ln kcos )k C œ ln Š 4 2 x ‹ C
)
#
#
œ "# ln k4 x# k C
114. (a)
' È t dt
œ
(b)
' È t dt
; t œ
4t# 1
4t# 1
"
8
' dÈa4t
#
1b
4t# 1
"
#
œ
"
4
sec )‘ Ä
È4t# 1 C
'
u œ 9 x#
Ä #" '
du œ 2x dx •
"
#
sec ) tan )† "# sec ) d)
tan )
œ
"
4
' sec# ) d) œ tan4 ) C œ È4t4 1 C
#
115.
' 9xdxx
116.
' x a9dx x b œ 9" ' dxx 18" ' 3 dx x 18" ' 3 dx x œ 9" ln kxk 18" ln k3 xk 18" ln k3 xk C
#
;”
du
u
œ #" ln kuk C œ ln
"
Èu
C œ ln
"
È 9 x#
C
#
œ
"
9
ln kxk
"
18
ln k9 x# k C
117.
' 9 dxx
118.
' È dx
119.
' sin3 x cos4 x dx œ ' cos4 xa1 cos2 xbsin x dx œ ' cos4 x sin x dx ' cos6 x sin x dx œ cos5 x cos7 x C
120.
#
œ
9 x#
"
6
;”
' 3 dx x 6" ' 3dxx œ 6" ln k3 xk 6" ln k3 xk C œ 6" ln ¸ xx 33 ¸ C
x œ 3 sin )
Ä
dx œ 3 cos ) d) •
)
' 33 cos
' d) œ ) C œ sin" x C
cos ) d) œ
3
5
' cos5 x sin5 x dx œ ' sin5 x cos4 x cos x dx œ ' sin5 x a1 sin2 xb2 cos x dx
œ ' sin5 x cos x dx 2' sin7 x cos x dx ' sin9 x cos x dx œ sin6 x 2sin8 x sin10 x C
6
8
10
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
7
567
568
Chapter 8 Techniques of Integration
121.
' tan4 x sec2 x dx œ tan5 x C
122.
' tan3 x sec3 x dx œ ' asec2 x 1b sec2 x † sec x † tan x dx œ ' sec4 x † sec x † tan x dx ' sec2 x † sec x † tan x dx
5
œ
123.
sec5 x
5
sec3 x
3
C
' sin 5) cos 6) d) œ "# ' asina)b sina11)bb d) œ "# ' sina)b d) "# ' sina11)b d) œ "# cosa)b ##" cos 11) C
œ "# cos )
"
## cos
11) C
124.
' cos 3) cos 3) d) œ "# ' acos 0 cos 6)b d) œ "# '
d)
125.
' É1 cosˆ 2t ‰ dt œ ' È2¸ cos
t¸
4
126.
' et Ètan2 et 1 dt œ ' k sec et k et dt œ lnk sec et tan et k C
3"
180
Ð%Ñ
127. kEs k Ÿ
t¸
4
dt œ 4È2 ¸ sin
3"
n
(˜x)% M where ˜x œ
œ
2
n
; f(x) œ
"
x
"
#
' cos 6) d) œ "# ) 121 sin 6) C
C
œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f'''(x) œ 6x%
(x) œ 24x& which is decreasing on [1ß 3] Ê maximum of f Ð%Ñ (x) on [1ß 3] is f Ð%Ñ (1) œ 24 Ê M œ 24. Then
" ‰ ˆ 2 ‰%
ˆ 768
‰ ˆ n"% ‰ Ÿ 0.0001 Ê n"% Ÿ (0.0001) ˆ 180
‰ Ê n% 10,000 ˆ 768
‰
kEs k Ÿ 0.0001 Ê ˆ 3180
n (24) Ÿ 0.0001 Ê
180
768
180
Ê f
Ê n
14.37 Ê n
10
12
128. kET k Ÿ
Ê
2
3n#
129. ˜x œ
(˜x)# M where ˜x œ
Ÿ 10$ Ê
ba
n
16 (n must be even)
œ
10
6
3n#
#
œ
1
6
10
n
#
œ
1000 Ê n
Ê
˜x
#
œ
1
1#
"
n ;0
2000
3
! mf(xi ) œ 12 Ê T œ ˆ 1 ‰ (12) œ 1 ;
12
iœ0
6
iœ0
˜x
3
1
18
Ê
2"
n
œ
œ
1‰
S œ ˆ 18
(18) œ 1 .
130. ¸f Ð%Ñ (x)¸ Ÿ 3 Ê M œ 3; ˜x œ
Ê n
131. yav œ
œ
œ
6.38 Ê n
"
365 0
Ê n
25.82 Ê n
"
n
ˆ "n ‰# (8) Ÿ 10$
"
12
26
x!
x"
x#
x$
x%
x&
x'
xi
0
1/6
1/3
1/2
21/3
51/6
1
f(xi )
0
1/2
3/2
2
3/2
1/2
0
m
1
2
2
2
2
2
1
mf(xi )
0
1
3
4
3
1
0
x!
x"
x#
x$
x%
x&
x'
xi
0
1/6
1/3
1/2
21/3
51/6
1
f(xi )
0
1/2
3/2
2
3/2
1/2
0
m
1
4
2
4
2
4
1
mf(xi )
0
2
3
8
3
2
0
;
6
! mf(xi ) œ 18 and
Ÿ f ww (x) Ÿ 8 Ê M œ 8. Then kET k Ÿ 10$ Ê
%
"‰ ˆ"‰
&
. Hence kEs k Ÿ 10& Ê ˆ 2180
Ê
n (3) Ÿ 10
"
60n%
Ÿ 10& Ê n%
8 (n must be even)
21
"
'0365 37 sin ˆ 365
ˆ 21
‰
‰‘ $'&
(x 101)‰ 25‘ dx œ 365
37 ˆ 365
21 cos 365 (x 101) 25x !
" ˆ
‰
21
‘
‰ ˆ
ˆ 365 ‰
21
‘
‰‘
37 ˆ 365
365
21 cos 365 (365 101) 25(365) 37 21 cos 365 (0 101) 25(0)
21
21
21
21
2371 cos ˆ 365
(264)‰ 25 2371 cos ˆ 365
(101)‰ œ 2371 ˆcos ˆ 365
(264)‰ cos ˆ 365
(101)‰‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
25
10&
60
Chapter 8 Practice Exercises
569
37
¸ #1
(0.16705 0.16705) 25 œ 25° F
"
67520
132. av(Cv ) œ
¸
"
655
"
"3
'20675 c8.27 10& a26T 1.87T# bd dT œ 655
8.27T 10
&
T#
0.62333
10&
'(&
T$ ‘ #!
[(5582.25 59.23125 1917.03194) (165.4 0.052 0.04987)] ¸ 5.434;
8.27 10& a26T 1.87T# b œ 5.434 Ê 1.87T# 26T 283,600 œ 0 Ê T ¸
26È676 4(1.87)(283,600)
#(1.87)
¸ 396.45° C
133. (a) Each interval is 5 min œ
1
12
hour.
2a2.4b 2a2.3b Þ Þ Þ 2a2.4b 2.3 d œ
‰
(b) a60 mphbˆ 12
29 hours/gal ¸ 24.83 mi/gal
1
2.5
24 c
29
12
¸ 2.42 gal
134. Using the Simpson's rule, ˜x œ 15 Ê ˜x
3 œ 5;
! mf(xi ) œ 1211.8 Ê Area ¸ a1211.8ba5b œ 6059 ft2 ;
x!
x"
x#
x$
x%
x5
x'
x(
x)
The cost is Area † a$2.10/ft2 b ¸ a6059 ft2 ba$2.10/ft2 b
œ $12,723.90 Ê the job cannot be done for $11,000.
135.
'03 È dx
œ lim c '0
bÄ3
136.
'01 ln x dx œ
b
9 x#
xi
0
15
30
45
60
75
90
105
120
f(xi )
0
36
54
51
49.5
54
64.4
67.5
42
b
œ lim c sin" ˆ x3 ‰‘ 0 œ lim c sin" ˆ 3b ‰ sin" ˆ 30 ‰ œ
bÄ3
bÄ3
dx
È 9 x#
lim cx ln x xd 1b œ (1 † ln 1 1) lim b cb ln b bd œ 1 lim b
b Ä !b
bÄ!
bÄ!
m
1
4
2
4
2
4
2
4
1
1
#
mf(xi )
0
144
108
204
99
216
128.8
270
42
1
#
0œ
ln b
Š "b ‹
œ 1 lim b
bÄ!
œ 1 0 œ 1
137.
'c11 ydy
138.
'c_2 () d1))
#Î$
œ 'c1
0
dy
y#Î$
'0
1
œ 'c2
$Î&
)$Î&
lim
$Î&
) Ä _ ()1)
1
dy
y#Î$
d)
() 1)$Î&
1
_
d)
)$Î&
diverges Ê
'3_ u 2du2u œ '3_ u du 2 '3_ duu œ
140.
'1_ 4v3v v1
#
_
#
dv œ '1 ˆ "v
1
œ 2 † 3 lim b y"Î$ ‘ b œ 6 Š1 lim b b"Î$ ‹ œ 6
bÄ!
bÄ!
2
139.
$
dy
y#Î$
)
'2
'c1 () d1)
$Î&
_
œ 1 and '2
œ 2 '0
"
v#
d)
() 1)$Î&
'2 () d1))
$Î&
diverges
b
lim ln ¸ u u 2 ¸‘ 3 œ lim ln ¸ b b 2 ¸‘ ln ¸ 3 3 2 ¸ œ 0 ln ˆ "3 ‰ œ ln 3
bÄ_
4 ‰
4v 1
bÄ_
dv œ lim ln v
bÄ_
œ lim ln ˆ 4b b 1 ‰ b" ‘ (ln 1 1 ln 3) œ ln
bÄ_
141.
'0_ x# ecx dx œ
142.
'c0_ xe3x dx œ
143.
'c__ 4xdx 9 œ 2 '0_ 4xdx 9 œ #" '0_ x dx
#
converges if each integral converges, but
_
"
4
"
v
b
ln (4v 1)‘ 1
1 ln 3 œ 1 ln
3
4
lim cx# ecx 2xecx 2ecx d 0 œ lim ab# ecb 2becb 2ecb b (2) œ 0 2 œ 2
b
bÄ_
lim
b Ä _
#
bÄ_
x3 e3x 9" e3x ‘ 0 œ 9"
b
#
9
4
œ
lim
b Ä _
"
# b lim
Ä_
ˆ b3 e3b 9" e3b ‰ œ 9" 0 œ 9"
32 tan" ˆ 2x
‰‘ b œ
3
0
"
# b lim
Ä_
32 tan" ˆ 2b
‰‘
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
3
tan" (0)
Š b" ‹
Š
"
‹
b#
570
Chapter 8 Techniques of Integration
"
#
œ
ˆ 23 † 1# ‰ 0 œ
1
6
144.
'c__ x 4dx16 œ 2'0_ x 4dx16 œ 2
145.
)
) Ä _ È)# 1
#
#
b
lim tan" ˆ x4 ‰‘ 0 œ 2 Š lim tan" ˆ b4 ‰‘ tan" (0)‹ œ 2 ˆ 1# ‰ 0 œ 1
bÄ_
_ d)
œ 1 and '6
lim
bÄ_
_
diverges Ê '6
)
d)
È)# 1
_
diverges
_
_
146. I œ '0 ecu cos u du œ lim cecu cos ud b0 '0 ecu sin u du œ 1 lim cecu sin ud b0 '0 aecu b cos u du
bÄ_
Ê I œ 1 0 I Ê 2I œ 1 Ê I œ
147.
bÄ_
"
#
converges
'1_ lnz z dz œ '1e lnz z dz 'e_ lnz z dz œ ’ (ln2z) “ e
#
1
œ _ Ê diverges
148. 0
ect
Èt
_
1 and '1 ect dt converges Ê
Ÿ ect for t
149.
_
2 dx
'_
e ec
_
150.
_
dx
'c_
' 1 dx
' 0 dx
'1
x a1 e b œ _ x a1 e b 1 x a1 e b 0
œ 2'0
x
x
#
Š x"# ‹
x Ä 0 ’ x# a1" ex b “
#
' 1 x dxÈx ; –
œ
œ lim
xÄ0
2x$Î#
3
converges Ê
4 dx
ex
#
x
x # a1 e x b
x#
x
#
#
bÄ_
e
'1_ Èect dt converges
t
_
2 dx
'_
e ec
x
dx
x # a1 e x b
x
converges
_
'1
œ lim a1 ex b œ 2 and '0
'c__ x a1dx e b diverges
Ê
151.
_
'0
#
x
lim
2 dx
ex ecx
b
#
lim ’ (ln#z) “ œ Š 12 0‹ lim ’ (ln2b) "# “
bÄ_
1
xÄ0
dx
x#
dx
x # a1 e x b
;
diverges Ê
'01 x a1dx e b diverges
#
x
x
u œ Èx
— Ä
du œ #dx
Èx
' u 1†2u udu œ ' ˆ2u# 2u 2 1 2 u ‰ du œ 32 u$ u# 2u 2 ln k1 uk C
#
x 2Èx 2 ln ˆ1 Èx‰ C
152.
' x4 x2 dx œ ' ˆx 4xx 42 ‰ dx œ '
153.
'
$
#
#
dx
x ax # 1 b #
;”
x œ tan )
Ä
dx œ sec# ) d) •
œ ln ksin )k
154.
' cosÈÈx x dx; –
155.
'È
156.
' (tÈ 1) dt ' ”
157.
' È du
dx
2x x#
t# 2t
1 u#
"
#
3
#
' x dx 2 #5 ' x dx 2 œ x#
#
3
#
ln kx 2k
) d)
' tansec) sec
œ ' cossin))d) œ ' Š 1 sinsin) ) ‹ d(sin ))
)
#
$
#
%
#
sin# ) C œ ln ¹ Èxx# 1 ¹ "# Š Èx#x 1 ‹ C
u œ Èx
— Ä
du œ #dx
Èx
œ'
x dx
d(x 1)
È1 (x 1)#
' cos uu†2u du œ 2' cos u du œ 2 sin u C œ 2 sin Èx C
œ sin" (x 1) C
u œ t# 2t
Ä
du œ (2t 2) dt œ 2(t 1) dt •
; cu œ tan )d Ä
"
#
' Èduu œ Èu C œ Èt# 2t C
' secsec))d) œ ln ksec ) tan )k C œ ln ¹È1 u# u¹ C
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5
#
ln kx 2k C
Chapter 8 Practice Exercises
158.
' et cos et dt œ sin et C
159.
' 2 cossinx x sin x dx œ '
#
2 csc# x dx '
cos x dx
sin# x
' csc x dx œ 2 cot x
"
sin x
ln kcsc x cot xk C
œ 2 cot x csc x ln kcsc x cot xk C
160.
sin )
' cos
' 1 coscos) ) d) œ ' sec# ) d) ' d) œ tan ) ) C
) d) œ
161.
' 819dvv
162.
x)
"
' 1cos sinx dxx œ ' 1d(sin
(sin x) C
sin x œ tan
163.
#
#
#
#
%
œ
"
#
' v dv 9 12" ' 3 dv v 1"# ' 3 dv v œ 12" ln ¸ 33 vv ¸ 6" tan" v3 C
#
#
#
cos (2) 1)
ÐÑ
"
) ïïïïî
# sin (2) 1)
ÐÑ
1 ïïïïî "4 cos (2) 1)
Ê
0
' ) cos (2) 1) d) œ #) sin (2) 1) "4 cos (2) 1) C
164.
'2_ (x dx1)
165.
' x x 2xdx 1 œ ' ˆx 2 x 3x 2x 2 1 ‰ dx œ '
b
œ lim 1 " x ‘ 2 œ lim 1 " b (1)‘ œ 0 1 œ 1
bÄ_
bÄ_
$
#
œ
166.
#
'
#
x#
#
2x 3 ln kx 1k
d)
É 1 È)
"
x1
×
Ù Ä
2 )
Õ d) œ 2(x 1) dx Ø
œ
Š1 È ) ‹
$Î#
4 Š1 È ) ‹
y œ Èx
— Ä
dy œ 2dx
Èx
"Î#
Cœ4
Ô ŒÉ1 È)
3
Õ
' 2Èsinx secÈxÈdxx ; –
168.
' xx dx16 œ ' ˆx x 16x
‰ dx œ x#
16
169.
'
170.
' ) d2)) 4 œ ' () d1)) 3 œ È33 tan" Š )È " ‹ C
171.
tan x
' cos
' tan x sec# x dx œ ' tan x † d(tan x) œ "# tan# x C
x dx œ
172.
&
y†2y dy
' 2 siny sec
œ ' 2 sin 2y dy œ cos (2y) C œ cos ˆ2Èx‰ C
y
#
%
dy
sin y cos y
œ'
' ˆ x#2x 4
2x ‰
x# 4
dx œ
x#
#
#
4
ln ¹ xx#
4¹ C
œ ' 2 csc (2y) dy œ ln kcsc (2y) cot (2y)k C
2 dy
sin 2y
#
dx
(x 1)#
×
É1 È) C
Ø
167.
%
'
' 2(x È1)x dx œ 2 ' Èx dx 2 ' Èdxx œ 34 x$Î# 4x"Î# C
$
4
3
dx
x1
C
x œ 1 È)
d)
dx œ È
Ô
;Ö
(x 2) dx 3 '
#
3
#
'
dr
(r 1)Èr# 2r
œ'
d(r 1)
(r 1)È(r 1)# 1
œ sec" kr 1k C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
571
572
Chapter 8 Techniques of Integration
173.
' È(r 2) dr
œ'
174.
' 4ydyy
"
#
175.
2) d)
' (1sin cos
2) )
r#
%
4r
œ
#
'
(r 2) dr
È4 (r 2)#
d ay # b
4 ay # b #
œ
œ #" '
u œ % (r 2)#
Ä '
du œ 2(r 2) dr •
du
2È u
œ Èu C œ È4 (r 2)# C
#
tan" Š y# ‹ C
d(1 cos 2))
(1 cos 2))#
œ
"
#(1 cos 2))
Cœ
"
4
sec# ) C
176.
'
177.
'11ÎÎ42 È1 cos 4x dx œ È2 '11ÎÎ42 cos 2x dx œ ’ È#2 sin 2x“ 1Î2 œ È#2
178.
' (15)2x1 dx œ "# ' (15)2x1 d(2x 1) œ "# Š 15ln 15b ‹ C
179.
' Èx dx
dx
ax # 1 b #
œ'
"
4
;”
œ
dx
a 1 x # b#
x
2 a1 x # b
"
4
"¸
ln ¸ xx
1 C (FORMULA 19)
1Î4
2x 1
;”
2x
yœ2x
Ä '
dy œ dx •
(2 y) dy
Èy
œ
y$Î# 4y"Î# C œ
2
3
2
3
(2 x)$Î# 4(2 x)"Î# C
$
ŠÈ2 x‹
œ 2–
180.
3
' È1v v
#
#
2È2 x— C
' cos )sin†cos)) d) œ ' a1 sinsin ))b d) œ ' csc# ) d) ' d) œ cot ) ) C
#
dv; cv œ sin )d Ä
œ sin" v
È 1 v#
v
#
#
C
181.
1)
"
' y dy2y 2 œ ' (yd(y
(y 1) C
1) 1 œ tan
182.
' ln Èx 1 dx; – y œ
#
#
Èx 1
dy œ
Ê
œ
"
#
dx
2È x 1
— Ä
ln y † 2y dy; u œ ln y, du œ
dy
y
; dv œ 2y dy, v œ y#
' 2y ln y dy œ y# ln y ' y dy œ y# ln y "# y# C œ (x 1) ln Èx 1 "# (x 1) C"
c(x 1) ln kx 1k xd ˆC" "# ‰ œ
183.
'
184.
'È
185.
' z azz1 4b dz œ "4 ' ˆ "z z"
186.
' x$ ex
187.
' È t dt
188.
'01Î10 È1 cos 5) d) œ È2 '01Î10
)# tan a)$ b d) œ
#
'
x dx
8 2x# x%
œ
"
#
"
3
'
d ax # 1 b
É 9 ax # 1 b #
#
dx œ
9 4t#
"
#
' x# ex
œ "8 '
cx ln kx 1k x ln kx 1kd C
' tan a)$ b d a)$ b œ 3" ln ksec )$ k C
#
#
"
#
#
d a x# b œ
d a9 4t# b
È9 4t#
œ
"
#
z1 ‰
z# 4
"
#
sin" Š x
dz œ
"
4
#
1
3 ‹
C
ln kzk
"
4z
ˆx# ex# ex# ‰ C œ
"
8
ln az# 4b
#
ax # 1 b e x
#
"
8
tan"
z
#
C
C
œ 4" È9 4t# C
cos ˆ 5#) ‰ d) œ
2È 2
5
sin ˆ 5#) ‰‘ 1Î"! œ
!
2È 2
5
ˆsin
1
4
0‰ œ
2
5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
189.
) d)
' 1cot sin) d)) œ ' (sin )cos
) a1 sin )b ; ”
#
#
' x a1dx x b œ ' dxx ' xx dx1
x œ sin )
Ä
dx œ cos ) d) •
#
#
œ ln ksin )k "# ln a1 sin# )b C
190. u œ tan" x, du œ
dx
1 x#
; dv œ
dx
x#
, v œ x" ;
' tan"x#x dx œ "x tan" x ' x a1dx x#b œ x" tan" x ' dxx ' 1xdxx#
œ x" tan" x ln kxk
191.
' tan2ÈÈyy dy ; Èy œ x‘
192.
'e
œ
193.
" x
"
#
ln a1 x# b C œ tanx
' tan x2x†2x dx œ ln ksec xk C œ ln ¸sec Èy¸ C
Ä
'
et dt
dx
t
(x 1)(x 2)
3et 2 ; ce œ xd Ä
t
"¸
ˆe "‰
ln ¸ xx
# C œ ln et # C
2t
' 4)d))
œ ' ˆ1
#
#
œ ) ln
2¸
¸ ))
2
4 ‰
4 )#
ln kxk ln È1 x# C
œ'
dx
x1
d) œ ' d ) '
'
d)
)#
'
dx
x#
œ ln kx 1k ln kx 2k C
d)
)#
œ ) ln k) 2k ln k) 2k C
C
194.
2x
' 11 cos
'
cos 2x dx œ
tan# x dx œ ' asec# x 1b dx œ tan x x C
195.
' cos Èasin" x#b dx ; – u œ sin dx x —
196.
' sincosx x dxsin x œ ' (sin x)cosa1xdxsin xb œ ' (sincosx) axcosdx xb œ ' sin2 dx2x œ 2' csc 2x dx
"
du œ
1x
È 1 x#
$
Ä
' cos u du œ sin u C œ sin asin" xb C œ x C
#
#
œ ln kcsc (2x) cot (2x)k C
197.
'
198.
' x x 2 dx œ ' x dx 2 '
sin
x
#
cos
x
#
dx œ '
"
#
sin ˆ x# x# ‰ dx œ
#
ax # 2 b#
œ
"
È2
#
tan" Š Èx2 ‹
199.
'
200.
' tan$ t dt œ '
201.
'1_ ln yy dy ; Ö
et dt
1 et
x dx
a x # 2 b#
"
# ax # 2 b
œ
"
È2
"
#
' sin x dx œ "# cos x C
tan" Š Èx2 ‹ "# ax# 2b
"
C
C
œ ln a1 et b C
(tan t) asec# t 1b dt œ
Ô x œ ln y ×
Ù Ä
dx œ dy
y
x
Õ dy œ e dx Ø
$
b
œ lim ˆ 2e
2b
bÄ_
" ‰
4e2b
'0_ xe†e
x
3x
ˆ0 "4 ‰ œ
tan# t
#
' tan t dt œ
ln ksec tk C
_
b
dx œ '0 xec2x dx œ lim x# ec2x 4" ec2x ‘ 0
bÄ_
"
4
202.
' 3 sectanxx sin x dx œ 3 ' cot x dx ' sectanxxdx '
203.
u œ ln (sin v)
v dv
' lncot(sinv dvv) œ ' (sincos
cos v dv •
v) ln (sin v) ; ”
#
#
du œ
tan# t
#
sin v
Ä
cos x dx œ 3 ln ksin xk ln ktan xk sin x C
' duu œ ln kuk C œ ln kln (sin v)k C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
573
574
Chapter 8 Techniques of Integration
'
204.
œ'
dx
(2x 1)Èx# x
œ'
2 dx
(2x 1)È4x# 4x
2 dx
(2x 1)È(2x 1)# 1
;”
u œ 2x 1
Ä
du œ 2 dx •
'
du
uÈ u# 1
œ sec" kuk C œ sec" k2x 1k C
205.
' eln Èx dx œ ' Èx dx œ 23 x$Î# C
206.
' e) È3 4e) d); ”
207.
' 1 sin(cos5t dt5t)
208.
' È dv
209.
'
e2v 1
#
;”
;”
5x%
20x$
60x#
120x
120
'
x œ ev
Ä
dx œ ev dv •
"
3
' È3 u du œ 4" † 23 (3 u)$Î# C œ 6" a3 4e) b$Î# C
dx
xÈ x# 1
du
1 u#
œ 5" tan" u C œ 5" tan" (cos 5t) C
œ sec" x C œ sec" aev b C
' (27)3)1 d(3) 1) œ
sin x
ÐÑ
ïïïïî cos x
ÐÑ
ïïïïî sin x
ÐÑ
ïïïïî cos x
ÐÑ
ïïïïî sin x
ÐÑ
ïïïïî cos x
ÐÑ
ïïïïî sin x
x&
"
4
u œ cos 5t
Ä "5 '
du œ 5 sin 5t dt •
(27)3)1 d) œ
210.
u œ 4e)
Ä
du œ 4e) d) •
"
3 ln 27
(27)3)" C œ
"
3
3)b1
Š 27
ln 27 ‹ C
Ê ' x& sin x dx œ x& cos x 5x% sin x 20x$ cos x 60x# sin x 120x cos x
0
120 sin x C
u œ Èr
dr — Ä
du œ 2È
r
211.
' 1 drÈr ; –
212.
d ˆx 10x 9‰
20x
' x 4x 10x
dx œ ' x 10x 9 œ ln kx% 10x# 9k C
9
213.
' y (y8 dy 2) œ ' dyy ' 2ydy ' 4ydy ' (y dy 2) œ ln ¹ y y 2 ¹ 2y y2
214.
'
215.
'
216.
' t(1 ln t)Èdt(ln t)(2 ln t) ; ” u œ lndtt •
%
$
%
' 2u1 duu œ ' ˆ2 1 2 u ‰ du œ 2u 2 ln k1 uk C œ 2Èr 2 ln ˆ1 Èr‰ C
#
%
$
(t 1) dt
at# 2tb#Î$
#
#
#
;”
8 dm
mÈ49m# 4
$
#
u œ t# 2t
Ä
du œ 2(t 1) dt •
œ
8
7
'
' udu
#Î$
œ
"
#
† 3u"Î$ C œ
3
#
at# 2tb
"Î$
C
¸C
œ 4 sec" ¸ 7m
#
dm
#
mÉm# ˆ 27 ‰
du œ
"
#
C
t
Ä
' (1 u)Èduu(2 u) œ ' (u 1)Èdu
(u 1) 1
#
œ sec" ku 1k C œ sec" kln t 1k C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Practice Exercises
'0x È1 (t 1)% dt and dv œ 3(x 1)# dx, then du œ È1 (x 1)% dx, and v œ (x 1)$ so integration
217. If u œ
by parts Ê '0 3(x 1)# ’'0 È1 (t 1)% dt“ dx œ ’(x 1)$
1
'0x È1 (t 1)% dt“ "
x
1
È
$Î# "
'0 (x 1)$ È1 (x 1)% dx œ ’ "6 a1 (x 1)% b “ œ 86 "
!
!
4v$ v 1
v# (v 1) av# 1b
218.
œ
A
v
#
B
v#
C
v1
Dv E
v# 1
#
Ê 4v$ v 1
œ Av(v 1) av 1b B(v 1) av 1b Cv# av# 1b (Dv E) av# b (v 1)
v œ 0: 1 œ B Ê B œ 1;
v œ 1: 4 œ 2C Ê C œ 2;
coefficient of v% : 0 œ A C D Ê A D œ 2;
coefficient of v$ : 4 œ A B E D
coefficient of v# : 0 œ A B C E Ê C D œ 4 Ê D œ 2 (summing with previous equation);
coefficient of v: 1 œ A B Ê A œ 0;
in summary: A œ 0, B œ 1, C œ 2, D œ 2 and E œ 1
'2_ v (v4v 1) vav 1 1b dv œ
$
Ê
#
#
œ lim ln (v 1)#
bÄ_
#
"
v
1)
œ lim ’ln Š (b1
b# ‹
œ
bÄ_
1
#
ln (5)
"
#
lim
bÄ_
'2b ˆ v 2 1 v# 1 " v
#
2v ‰
1 v#
dv
tan" v ln a1 v# b‘ 2
b
"
b
tan" b“ ˆln 1
"
#
tan" 2 ln 5‰ œ ˆ0 0 1# ‰ ˆ0
"
#
tan" 2 ln 5‰
1
#
tan" a;
tan" 2
219. u œ f(x), du œ f w (x) dx; dv œ dx, v œ x;
'13Î12Î2 f(x) dx œ cx f(x)d 311Î2Î2 '13Î12Î2 xf w (x) dx œ 3#1 f ˆ 3#1 ‰ 1# f ˆ 1# ‰‘ '13Î12Î2 cos x dx
œ ˆ 31#b
'0a 1dxx
220.
1a ‰
#
csin xd 311Î2Î2 œ 1# a3b ab c(1) 1d œ 1# a3b ab 2
œ ctan" xd 0 œ tan" a; 'a
a
#
therefore, tan" a œ
1
#
_
dx
1 x #
œ lim ctan" xd a œ lim atan" b tan" ab œ
b
bÄ_
tan" a Ê tan" a œ
1
4
bÄ_
Ê a œ 1 since a 0.
CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES
#
1. u œ asin" xb , du œ
'
#
'
2.
"
x
; dv œ dx, v œ x;
asin" xb dx œ x asin" xb '
#
u œ sin" x, du œ
'
2 sin" x dx
È 1 x#
2x sin" x dx
È 1 x#
dx
È 1 x#
2x sin" x dx
È 1 x#
;
; dv œ È2x dx # , v œ 2È1 x# ;
1x
œ 2 asin" xb È1 x# ' 2 dx œ 2 asin" xb È1 x# 2x C; therefore
#
#
asin" xb dx œ x asin" xb 2 asin" xb È1 x# 2x C
œ
"
x
,
"
"
"
x(x 1) œ x x 1 ,
"
"
"
"
x(x 1)(x 2) œ 2x x 1 #(x 2) ,
"
"
"
"
"
x(x 1)(x 2)(x 3) œ 6x #(x 1) #(x 2) 6(x 3) ,
"
"
"
"
"
x(x 1)(x 2)(x 3)(x 4) œ #4x 6(x 1) 4(x #) 6(x 3)
"
x(x 1)(x 2) â (x m)
m
œ!
kœ0
(")
(k!)(m k)!(x k) ;
k
therefore '
"
24(x 4)
Ê the following pattern:
dx
x(x 1)(x 2) â (x m)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
575
576
Chapter 8 Techniques of Integration
m
(")
œ ! ’ (k!)(m
k)! ln kx kk“ C
k
kœ0
3. u œ sin" x, du œ
dx
È 1 x#
' x sin" x dx œ x#
#
4.
œ
x#
#
sin" x
œ
x#
#
sin" x
sin" x '
'
z œ Èy
— Ä
dy
2È y
x#
#
;
x œ sin )
Ä
dx œ cos ) d) •
;”
x "# ˆ #)
sin 2) ‰
4
'
Cœ
x sin" x dx œ
x#
#
sin" x
C
" Èy
C
)
' 1 dtan) ) œ ' cos cos
' 12coscos#2)) d) œ #" ' (sec 2) 1) d) œ ln ksec 2) 4tan 2)k 2) C
) sin ) d) œ
#
#
#
#
x "# œ "# sec )
– dx œ " sec ) tan ) d) — Ä
#
tan ) ln ksec ) tan )k
#
œ
"
‹
2È 1 x
'
Cœ
"
4
œ
dx
2È x È 1 x
dt
t È1 t#
Õ d) œ
' u du 1 #" ' u du 1 #" ' uu du1 œ #" ln ¹ Èu 1
œ
"
#
ln Št È1 t# ‹
3e2x
ex b dx
6ex 1
u1œ
Ô
Õ du œ
2
È3
#
' È(2u 1) du
2
È3
sec )
' Š È4
"
È3
œ
2
3
È3u# 6u 1
œ
"
È3
"
16
Ä
3u# 6u 1
"
È3
3
"
È3
"
3
"
#
tan" u C œ
"
#
œ
"
È3
'
(2u 1) du
É(u 1)# 43
Ä
' (u 1)duau 1b
#
ln ¸ tansec) ) " ¸ #" ) C
sec ) 1‹ (sec )) d) œ
4
3
;
' sec# ) d) È" ' sec ) d)
3
† É 34 (u 1)# 1
"
È3
ln ¹
ln ¹u 1 É(u 1)# 43 ¹ ŠC"
"
È3
ln
ln ksec ) tan )k C" œ
’2Ée2x 2ex
' x " 4 dx œ '
œ
×
sec ) tan ) d) Ø
tan )
¹
Ø
du
u# 1
C
sin" t C
u œ ex
Ä
du œ ex dx •
4
3
%
u# 1
;”
œ
4
u œ tan )
"
#
2x
Èx# x ln ¹2x 1 2Èx# x¹
d)
' sincos) ) cos
' tan d)) 1 ; Ô du œ sec# ) d) ×
) œ
œ
' Èa2e
;
C
#
t œ sin )
Ä
;”
dt œ cos ) d) •
"
#
x dx
Ɉx "# ‰# "4
"
#
2Èx# x ln ¹2x 1 2Èx# x¹
#
; dv œ dx, v œ x;
sec ) tan ) d)
' (sec ) ˆ1)†tan
œ #" ' asec# ) sec )b d)
)‰
' ln ŠÈx È1 x‹ dx œ x ln ŠÈx È1 x‹ 2
Ê
9.
sin# ) cos ) d)
2 cos )
C
' ln ŠÈx È1 x‹ dx œ x ln ŠÈx È1 x‹ "# ' ÈxxÈdx1 x ; "# '
8.
sin" x '
sin ) cos ) )
4
' sin" Èy dy œ y sin" Èy Èy È1 y# sin
"
6. u œ ln ŠÈx È1 x‹ , du œ Š Èx dx
È1 x ‹ Š #Èx
7.
x#
#
' 2z sin" z dz; from Exercise 3, ' z sin" z dz
zÈ1 z# sin" z
C Ê
4
" Èy
#
È
sin
y
y
sin" Èy #
#
z# sin" z
#
œy
x# dx
2È 1 x#
#
"
sin# ) d) œ x# sin"
#
xÈ1 x# sin" x
C
4
' sin" Èy dy; –
dz œ
œ
5.
; dv œ x dx, v œ
4
3
È3
#
(u 1) É 34 (u 1)# 1¹ C"
È3
# ‹
ln ¹ex 1 Ée2x 2ex 3" ¹“ C
"
ax# 2b# 4x#
dx œ '
"
ax# 2x 2b ax# 2x 2b
dx
' ’ x 2x 2x 2 2 (x 1)2 1 x 2x 2x 2# (x 1)2 1 “ dx
#
#
#
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Additional and Advanced Exercises
10.
#
"
16
œ
2x 2
"
"
ln ¹ xx#
(x 1) tan" (x 1)d C
2x # ¹ 8 ctan
' x " 1 dx œ 6" '
ˆ x " 1
'
"
x1
x2
x# x 1
x2 ‰
x# x 1
dx
œ
"
6
1¸
ln ¸ xx
1
"
1#
' ” x 2x x " 1 ˆ
œ
"
6
"¸
ln ¸ xx
1
"
1#
x"
" 2x 1
È
’ln ¹ xx#
Š È3 ‹ 2È3 tan" Š 2xÈ3 1 ‹“ C
x 1 ¹ 2 3 tan
#
3
#
x "# ‰ 34
2x 1
x# x 1
3
ˆx "# ‰# 34 •
dx
#
11. x lim
Ä_
'cxx sin t dt œ x lim
c cos td xcx œ x lim
c cos x cos (x)d œ x lim
a cos x cos xb œ x lim
0œ0
Ä_
Ä_
Ä_
Ä_
12.
'x1 cost t dt;
lim b
xÄ!
#
lim b x 'x
1
xÄ!
lim b
xÄ!
cos t
t#
limb
tÄ!
œ '0
1
"
È n# k#
'1
cos t
t#
k
n
dt
"
x
xÄ!
lim b 'x
1
œ1 Ê
xÄ!
cos t
t#
dt diverges since '0
1
œ lim b
xÄ!
Š cos# x ‹
Š
dt
t#
diverges; thus
œ lim b cos x œ 1
xÄ!
x
"
‹
x#
! ln ˆ1 k ˆ " ‰‰ ˆ " ‰ œ ' ln (1 x) dx; ”
œ n lim
n
n
Ä_
0
n
1
kœ1
nc1
! ŠÈ
œ n lim
Ä_
kœ0
"
"
È 1 x #
dx œ csin" xd ! œ
n
‹ ˆ n" ‰
n# k#
nc1
!
œ n lim
Ä_
kœ0
Î
"
u œ 1 x, du œ dx
x œ 0 Ê u œ 1, x œ 1 Ê u œ 2 •
œ
Ê 1 Š dy
dx ‹ œ
1Î2
#
#
a1 x# b 4x#
a1 x # b #
x
' ˆ1
œ '0 Š "1
x# ‹ dx œ 0
1Î2
1Î4
Ê1 ŠÈcos 2t‹ dt œ È2 '0
#
1Î4
œ1
#
2x
1 x#
ˆ n" ‰
1
#
#
1Î%
Ñ
#
Ï Ê1 ’k Š "n ‹“ Ò
#
'
œ Ècos 2x Ê 1 Š dy
dx ‹ œ 1 cos 2x œ 2 cos x; L œ 0
œ È2 csin td !
dy
dx
"
cos t
'12 ln u du œ cu ln u ud #" œ (2 ln 2 2) (ln 1 1) œ 2 ln 2 1 œ ln 4 1
nc1
16.
lim b
#
!
14. n lim
Ä_
kœ0
dy
dx
œ limb
tÄ!
x
1
t
x 'x cos
dt œ
t
n
15.
t
Š cos# t ‹
t
^
dt is an indeterminate 0 † _ form and we apply l'Hopital's
rule:
! ln nÉ1
13. n lim
Ä_
kœ1
Ä
Š "# ‹
2 ‰
1 x#
œ ˆ "# ln 3‰ (0 ln 1) œ ln 3
œ
1 2x# x%
a 1 x # b#
x
'
œ Š 11
x# ‹ ; L œ 0
dx œ '0 ˆ1
1Î2
#
"
1x
#
1Î2
" ‰
1x
#
Ê1 Š dy
dx ‹ dx
"Î#
x ¸‘
dx œ x ln ¸ 11
x !
"
#
shell ‰
shell
17. V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21xy dx
b
1
œ 61 '0 x# È1 x dx;
1
Ô uœ1x ×
du œ dx
Õ x# œ (1 u)# Ø
Ä 61 '1 (1 u)# Èu du
0
œ 61 '1 ˆu"Î# 2u$Î# u&Î# ‰ du
0
!
œ 61 23 u$Î# 45 u&Î# 27 u(Î# ‘ " œ 61 ˆ 23
84 30 ‰
16 ‰
œ 61 ˆ 70 105
œ 61 ˆ 105
œ 32351
4
5
27 ‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Ècos# t dt
577
578
Chapter 8 Techniques of Integration
18. V œ 'a 1y# dx œ 1 '1
b
4
œ 1 '1 ˆ dx
x
4
5 dx
x#
25 dx
x# (5 x)
dx ‰
5x
%
œ 1 ln ¸ 5 x x ¸ 5x ‘ " œ 1 ˆln 4 54 ‰ 1 ˆln
œ
151
4
"
4
5‰
21 ln 4
shell ‰
shell
19. V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21xex dx
b
1
œ 21 cxex ex d "! œ 21
20. V œ '0
ln 2
21(ln 2 x) aex 1b dx
œ 21 '0
ln 2
c(ln 2) ex ln 2 xex xd dx
ln 2
œ 21 ’(ln 2) ex (ln 2)x xex ex
x#
# “0
œ 21 ’2 ln 2 (ln 2)# 2 ln 2 2
(ln 2)#
# “
21(ln 2 1)
#
œ 21 ’ (ln#2) ln 2 1“
21. (a) V œ '1 1 c1 (ln x)# d dx
e
œ 1 cx x(ln x)# d 1 21'1 ln x dx
e
e
(FORMULA 110)
e
œ 1 cx x(ln x)# 2(x ln x x)d 1
e
œ 1 cx x(ln x)# 2x ln xd 1
œ 1 ce e 2e (1)d œ 1
(b) V œ '1 1(1 ln x)# dx œ 1'1 c1 2 ln x (ln x)# d dx
e
e
œ 1 cx 2(x ln x x) x(ln x)# d 1 21'1 ln x dx
e
e
œ 1 cx 2(x ln x x) x(ln x)# 2(x ln x x)d 1
e
œ 1 c5x 4x ln x x(ln x)# d 1
œ 1 c(5e 4e e) (5)d œ 1(2e 5)
e
22. (a) V œ 1 '0 aey b# 1‘ dy œ 1'0 ae2y 1b dy œ 1 e# y‘ ! œ 1 ’ e# 1 ˆ "# ‰“ œ
1
1
"
2y
#
1 ae # 3 b
#
(b) V œ 1'0 aey 1b# dy œ 1'0 ae2y 2ey 1b dy œ 1 e# 2ey y‘ ! œ 1 ’Š e# 2e 1‹ ˆ "# 2‰“
1
1
#
œ 1 Š e# 2e 5# ‹ œ
23. (a)
lim
x Ä !b
x ln x œ 0 Ê
2y
"
#
1 ae# 4e 5b
#
lim
x Ä !b
f(x) œ 0 œ f(0) Ê f is continuous
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Additional and Advanced Exercises
#
Ô u œ (ln x)
×
Ö du œ (2 ln x) dx Ù
#
2
2
x
#
#
Ö
Ù Ä 1 Œ lim ’ x$ (ln x)# “ ' Š x$ ‹ (2 ln x)
(b) V œ '0 1x (ln x) dx; Ö
#
Ù
3
3
b
0
dv œ x dx
b
bÄ!
$
x
Õ
Ø
vœ
579
dx
x
3
$
œ 1 ”ˆ 83 ‰ (ln 2)# ˆ 23 ‰ lim b ’ x3 ln x
bÄ!
2
x$
9 “ b•
#
œ 1 ’ 8(ln3 2)
16(ln 2)
9
16
27 “
24. V œ '0 1( ln x)# dx
1
œ 1 Œ lim b cx(ln x)# d b 2'0 ln x dx
bÄ!
1
1
œ 21 lim b cx ln x xd 1b œ 21
bÄ!
25. M œ '1 ln x dx œ cx ln x xd e1 œ (e e) (0 1) œ 1;
e
Mx œ '1 (ln x) ˆ ln#x ‰ dx œ
e
œ
"
#
e
e
My œ '1 x ln x dx œ ’ x
"
#
'1e (ln x)# dx
Šcx(ln x)# d 1 2 '1 ln x dx‹ œ
e
œ
"
#
’x# ln x
My
M
therefore, x œ
26. M œ '0
1
e
x#
# “1
2 dx
È 1 x#
œ
œ
#
e
ln x
# “1
"
#
’Še#
e# 1
4
"
#
"
#
'1
e#
#‹
and y œ
(e 2);
e
x dx
"# “ œ
Mx
M
œ
"
4
ae# 1b ;
e2
#
"
œ 2 csin" xd ! œ 1;
My œ '0
"
2x dx
È
#
È1 x# œ 2 ’ 1 x “ œ 2;
!
M
therefore, x œ My œ 12 and y œ 0 by symmetry
1
27. L œ '1 É1
e
tan" e
œ '1Î4
"
x#
dx œ '1
e
(sec )) atan# ) 1b
tan )
œ ŠÈ1 e# ln ¹
È x# 1
x
tan" e
d) œ '1Î4
È 1 e#
e
dx; ”
tan" e
x œ tan )
'
Ä
L
œ
1Î4
dx œ sec# ) d) •
sec )†sec# ) d)
tan )
" e
(tan ) sec ) csc )) d) œ csec ) ln kcsc ) cot )kd 1tanÎ4
#
È
"e ¹‹ ’È2 ln Š1 È2‹“ œ È1 e# ln Š 1e e "e ‹ È2 ln Š1 È2‹
#
#
' È
' yÈ1 e2y dy; ”
28. y œ ln x Ê 1 Š dx
dy ‹ œ 1 x Ê S œ 21 c x 1 x dy Ê S œ 21 0 e
#
d
1
tan" e
u œ tan )
Ä S œ 21'1 È1 u# du; ”
Ä 21 '1Î4
du œ sec# ) d) •
e
" e
œ 21 ˆ "# ‰ csec ) tan ) ln ksec ) tan )kd tan
1Î%
u œ ey
du œ ey dy •
sec ) † sec# ) d)
œ 1 ’ŠÈ1 e# ‹ e ln ¹È1 e# e¹“ 1 ’È2 † 1 ln ŠÈ2 1‹“
È 1 e# e
œ 1 ’eÈ1 e# ln Š È
‹ È 2“
21
#Î$
29. L œ 4 '0 Ê1 Š dy
y#Î$ œ 1 Ê y œ ˆ1 x#Î$ ‰
dx ‹ dx; x
1
#
$Î#
Ê
dy
dx
œ 3# ˆ1 x#Î$ ‰
"Î#
ˆx"Î$ ‰ ˆ 23 ‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
580
Chapter 8 Techniques of Integration
#
Ê Š dy
dx ‹ œ
x
Ê L œ 4 '0 Ê1 Š 1 x#Î$
‹ dx œ 4'0
1
1 x#Î$
x#Î$
1
#Î$
30. S œ 21 'c1 f(x) É1 cf w (x)d# dx; f(x) œ ˆ1 x#Î$ ‰
1
$Î# "
‰ dx; –
œ 41'0 ˆ1 x#Î$ ‰ ˆ x"Î$
1
"
œ 61 † 25 (1 u)&Î# ‘ ! œ
#
31. Š dy
dx ‹ œ
"
4x
Ê
dy
dx
„"
#È x
œ
$Î#
"
œ 6 x#Î$ ‘ ! œ 6
dx
x"Î$
Ê cf w (x)d# 1 œ
Ê S œ 21 'c1 ˆ1 x#Î$ ‰
1
"
x#Î$
$Î#
†
dx
Èx#Î$
1
u œ x#Î$
3
$Î#
'1
du œ 61'0 (1 u)$Î# d(1 u)
2 dx — Ä 4 † # 1 0 (1 u)
du œ 3 x"Î$
121
5
Ê y œ Èx or y œ Èx, 0 Ÿ x Ÿ 4
32. The integral 'c1 È1 x# dx is the area enclosed by the x-axis and the semicircle y œ È1 x# . This area is half
1
the circle's area, or
1
#
and multiplying by 2 gives 1. The length of the circular arc y œ È1 x# from x œ 1 to
x
'
x œ 1 is L œ 'c1 Ê1 Š dy
‹ dx œ 'c1 È dx
#
dx ‹ dx œ c1 Ê1 Š È
#
1
#
1
1
1 x#
1x
circle's circumference. In conclusion, 2 'c1 È1 x# dx œ 'c1 È dx
1
_
'c_
e xe
a
x
b
_
dx œ '_ eae b ex dx
'a0 e ce
œaÄ
lim
_
a
x
b
(21) œ 1 since L is half the
.
(a)
x
ex dx
"
#
1
1 x#
33. (b)
œ
lim
b Ä _
'0b e ce
a
x
b
ex dx;
u œ ex
” du œ ex dx • Ä
lim ' ecu du
a Ä _ ea
'1e
b
1
lim
b Ä _
œaÄ
lim
cecu d 1ea
_
ecu du
cecu d e1
b
lim
b Ä _
"e ec aea b ‘
œaÄ
lim
_
ec ˆeb ‰ "e ‘
lim
b Ä _
œ ˆ "e e! ‰ ˆ0 "e ‰ œ 1
34. u œ
"
1y
lim
nÄ_
, du œ (1 dyy)# ; dv œ nync1 dy, v œ yn ;
'01 ny1 y
n 1
dy œ n lim
’ y “ '0
Ä _ Œ 1y
1
"
#
0œ
1
!
'
Ê 0 Ÿ n lim
Ä_ 0
œ
"
n
yn
1 y#
dy Ÿ n lim
Ä_
36.
1
6
uÐn2ÑÎ2
n#
œ
"
È2
"
#
"
#
n lim
Ä_
ˆÈu‰ n2
n#
ŠÈx# a# ‹
Cœ
n#
"
œ sin" x# ‘ ! œ '0
1
È2
sin" u# ‘
0
œ
'01
yn
1 y#
n 1
!
"
#
Cœ
œ sin"
dy œ
b "
'01 yn dy œ n lim
’ y “ œ n lim
Ä _ n1
Ä_
35. u œ x# a# Ê du œ 2x dx;
' x ŠÈx# a# ‹ n dx œ "# ' ˆÈu‰ n du œ
œ
yn
1 y#
"
#
n2
"
n 1
dy. Now, 0 Ÿ
yn
1 y#
'
œ 0 Ê n lim
Ä_ 0
1
' unÎ2 du œ "# Š u Î1 ‹ C, n Á 2
n 2
n
1
#
C
'1
dx
dx
È 4 x# 0 È 4 x# x$
1È2
"
" ˆ1‰
" È2
È2 sin
# œ È2 4 œ 8
'0
1
dx
È4 2x#
œ
"
È2
È2
'0
du
È 4 u#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Ÿ yn
nync1
1y
dy
Chapter 8 Additional and Advanced Exercises
37.
'1_ ˆ x ax 1 #"x ‰ dx œ
lim
#
"
bÄ_ #
œ lim
ab # 1 b
b
’ln
bÄ_
a
integral diverges if a
"
#
œ
Ê
_
'1
ˆln 1
"
#
lim ln
bÄ_
ˆ x#ax 1
bÄ_
ab # 1 b
b
bÄ_
ln 2a “ ; lim
"
#
ab # 1 b
b
a
_ dx
xp
a
"
#
p
"
#
bÄ_
b2a
bÄ_ b
"
b#
"
#
'0b ecxt dt œ
bÄ_
"
#
"
#
"
bÄ_ #
lim
ax # 1 b
x
b
a
“
1
Ê the improper
”ln
ab # 1 b
b
"Î#
ln 2"Î# •
œ lim (b 1)2ac1 œ 0
bÄ_
; in summary, the improper integral
and has the value ln42
cxb
b
lim "x ecxt ‘ 0 œ lim Š 1 xe ‹ œ
bÄ_
bÄ_
œ1 Ê
(b1)2a
b 1
lim
œ _ Ê the improper integral diverges if a
dx converges only when a œ
b
bÄ_
bÄ_
a
ln x‘ 1 œ lim ’ #" ln
1
œ lim É1
ab # 1 b
b
bÄ_
"
#
œ lim b2 ˆa c 2 ‰ œ _ if a
lim
: 0 Ÿ lim
bÄ_
10
x
œ
"
x
if x 0 Ê xG(x) œ x ˆ "x ‰
converges if p 1 and diverges if p Ÿ 1. Thus, p Ÿ 1 for infinite area. The volume of the solid of revolution
_
_ dx
about the x-axis is V œ '1 1 ˆ x"p ‰ dx œ 1 '1
"
#
lim #a ln ax# 1b
È b# 1
"
# : b lim
b
Ä_
; for a œ
œ 1 if x 0
39. A œ '1
#
ln 2‰ œ ln42 ; if a
" ‰
#x
38. G(x) œ lim
'1b ˆ x ax 1 #"x ‰ dx œ
#
x2p
which converges if 2p 1 and diverges if 2p Ÿ 1. Thus we want
for finite volume. In conclusion, the curve y œ xcp gives infinite area and finite volume for values of p satisfying
p Ÿ 1.
40. The area is given by the integral A œ '0
1
dx
xp
;
p œ 1: A œ lim b cln xd œ lim b ln b œ _, diverges;
bÄ!
bÄ!
1
b
p 1: A œ lim b cx1cp d 1b œ 1 lim b b1cp œ _, diverges;
bÄ!
bÄ!
p 1: A œ lim b cx1cp d 1b œ " lim b b1cp œ 1 0, converges; thus, p
bÄ!
bÄ!
The volume of the solid of revolution about the x-axis is Vx œ 1'0
1
p
"
#
"
#
, and diverges if p
dx
x2p
1 for infinite area.
which converges if 2p 1 or
. Thus, Vx is infinite whenever the area is infinite (p
_
1).
_
The volume of the solid of revolution about the y-axis is Vy œ 1 '1 cR(y)d dy œ 1'1
converges if
2
p
#
dy
y2Îp
which
1 Í p 2 (see Exercise 39). In conclusion, the curve y œ xcp gives infinite area and finite
volume for values of p satisfying 1 Ÿ p 2, as described above.
ÐÑ
cos 3x
2e2x
ÐÑ
1
3 sin
4e2x
ÐÑ
19 cos 3x
41. e2x
Iœ
e2x
3
581
sin 3x
2e2x
9
cos 3x 49 I Ê
ÐÑ
42. e3x
3x
ÐÑ
4" cos 4x
9e3x
ÐÑ
"
16
sin 4x
3x
Iœ
e2x
9
(3 sin 3x 2 cos 3x) Ê I œ
e2x
13
(3 sin 3x 2 cos 3x) C
sin 4x
3e3x
I œ e4 cos 4x
13
9
3e3x
16
sin 4x
9
16
I Ê
25
16
Iœ
e3x
16
(3 sin 4x 4 cos 4x) Ê I œ
e3x
25
(3 sin 4x 4 cos 4x) C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
582
Chapter 8 Techniques of Integration
sin 3x
ÐÑ
sin x
3 cos 3x
ÐÑ
cos x
9 sin 3x
ÐÑ
sin x
43.
I œ sin 3x cos x 3 cos 3x sin x 9I Ê 8I œ sin 3x cos x 3 cos 3x sin x
Ê I œ sin 3x cos x83 cos 3x sin x C
cos 5x
ÐÑ
sin 4x
sin 5x
ÐÑ
"4 cos 4x
25cos 5x
ÐÑ
"
16
sin 4
44.
I œ "4 cos 5x cos 4x
Ê Iœ
"
9
ÐÑ
"b cos bx
a# eax
ÐÑ
b"# sin bx
ax
I œ eb cos bx
eax
a# b#
aeax
b#
sin bx
a#
b#
I Ê Ša
ÐÑ
"
b
a# eax
ÐÑ
b"# cos bx
eax
b
sin bx
Ê Iœ
sin 5x sin 4x
eax
a# b#
#
b#
b# ‹ I
œ
eax
b#
(a sin bx b cos bx)
cos bx
aeax
Iœ
5
16
(a sin bx b cos bx) C
ÐÑ
46. eax
9
I Ê 16
I œ "4 cos 5x cos 4x
sin bx
aeax
Ê Iœ
25
16
(4 cos 5x cos 4x 5 sin 5x sin 4x) C
ÐÑ
45. eax
sin 5x sin 4x
5
16
aeax
b#
sin bx
cos bx
a#
b#
I Ê Ša
#
b#
b# ‹ I
œ
eax
b#
(a cos bx b sin bx)
(a cos bx b sin bx) C
47. ln (ax)
ÐÑ
1
"
x
ÐÑ
x
I œ x ln (ax) ' ˆ "x ‰ x dx œ x ln (ax) x C
48. ln (ax)
ÐÑ
x#
"
x
ÐÑ
"
3
Iœ
"
3
x$
x$ ln (ax) ' ˆ "x ‰ Š x3 ‹ dx œ
$
_
49. (a) >(1) œ '0 ect dt œ lim
bÄ_
"
3
x$ ln (ax) 9" x$ C
'0b ect dt œ
lim cect d b0 œ lim e"b (1)‘ œ 0 1 œ 1
bÄ_
bÄ_
(b) u œ tx , du œ xtxc1 dt; dv œ ect dt, v œ ect ; x œ fixed positive real
_
_
Ê >(x 1) œ '0 tx et dt œ lim ctx et d b0 x '0 tx1 et dt œ lim ˆ beb 0x e! ‰ x>(x) œ x>(x)
bÄ_
x
bÄ_
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 8 Additional and Advanced Exercises
(c) >(n 1) œ n>(n) œ n!:
n œ 0: >(0 1) œ >(1) œ 0!;
n œ k: Assume >(k 1) œ k!
n œ k 1: >(k 1 1) œ (k 1) >(k 1)
œ (k 1)k!
œ (k 1)!
Thus, >(n 1) œ n>(n) œ n! for every positive integer n.
for some k 0;
from part (b)
induction hypothesis
definition of factorial
x
n
n
50. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1
(b)
n
10
20
30
40
50
60
ˆ ne ‰n È2n1
3598695.619
2.4227868 ‚ 10")
2.6451710 ‚ 10$#
8.1421726 ‚ 10%(
3.0363446 ‚ 10'%
8.3094383 ‚ 10)"
calculator
3628800
2.432902 ‚ 10")
2.652528 ‚ 10$#
8.1591528 ‚ 10%(
3.0414093 ‚ 10'%
8.3209871 ‚ 10)"
(c)
n
10
ˆ ne ‰n È2n1
3598695.619
ˆ ne ‰n È2n1 e1Î12n
3628810.051
calculator
3628800
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
583
584
Chapter 8 Techniques of Integration
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION
9.1 SLOPE FIELDS AND SEPARABLE DIFFERENTIAL EQUATIONS
1. (a) y œ ex Ê y w œ ex Ê 2y w 3y œ 2 aex b 3ex œ ex
(b) y œ ex e3xÎ2 Ê y w œ ex 3# e3xÎ2 Ê 2y w 3y œ 2 ˆex 3# e3xÎ2 ‰ 3 aex e3xÎ2 b œ ex
(c) y œ ex Ce3xÎ2 Ê y w œ ex 3# Ce3xÎ2 Ê 2y w 3y œ 2 ˆex 3# Ce3xÎ2 ‰ 3 aex Ce3xÎ2 b œ ex
2. (a) y œ "x Ê y w œ
"
x#
(b) y œ x " 3 Ê y w œ
(c) y œ
3. y œ
"
xC
'1x
"
x
et
t
Ê yw œ
#
œ ˆ x" ‰ œ y#
"
(x 3)#
"
(x C)#
#
œ ’ (x " 3) “ œ y#
#
œ x " C ‘ œ y#
dt Ê yw œ x"# '1
x t
e
t
dt ˆ x" ‰ ˆ ex ‰ Ê x# y w œ '1
x t
e
x
t
dt ex œ x Š x"
'1x et
t
dt‹ ex œ xy ex
Ê x# y w xy œ ex
4. y œ
"
È 1 x%
'1x È1 t% dt
$
Ê y w œ Š 12xx% ‹ Š È
"
1 x%
Ê y w œ #" –
' x È1 t% dt È "
4x$
$—
1
È
Š 1 x% ‹
'1x È1 t% dt‹ 1
1 x%
ŠÈ 1 x% ‹
$
Ê y w œ Š 12xx% ‹ y 1 Ê y w
2x$
1 x%
†yœ1
5. y œ ecx tan" a2ex b Ê y w œ ecx tan" a2ex b ecx ’ 1 a"2ex b# “ a2ex b œ ecx tan" a2ex b
Ê y w œ y
2
1 4e2x
Ê yw y œ
2
1 4e2x
2
1 4e2x
; y( ln 2) œ ecÐ ln 2Ñ tan" a2e ln 2 b œ 2 tan" 1 œ 2 ˆ 14 ‰ œ
1
#
#
#
#
#
#
6. y œ (x 2) ex Ê y w œ ex ˆ2xex ‰ (x 2) Ê y w œ ex 2xy; y(2) œ (2 2) e2 œ 0
7. y œ
Ê xy y œ
8. y œ
x
ln x
dy
dx
"Î#
9. 2Èxy
œ 2x
10.
dy
dx
x sin x cos x
Ê y w œ sinx x
x#
1/2)
sin x; y ˆ 1# ‰ œ cos(1(/2)
œ0
Ê yw œ
cos x
x
w
Ê yw œ
ln x x Š "x ‹
(ln x)#
Ê yw œ
"
ln x
"x ˆ cosx x ‰ Ê y w œ sinx x
"
(ln x)#
Ê x# y w œ
œ 1 Ê 2x"Î# y"Î# dy œ dx Ê 2y"Î# dy œ x"Î# dx Ê
C" Ê
2
3
y$Î# x"Î# œ C, where C œ
"
#
"Î#
"
3
x#
(ln x)#
Ê xy w œ sin x y
Ê x# y w œ xy y# ; y(e) œ
' 2y"Î# dy œ ' x"Î# dx
e
ln e
Ê 2 ˆ 23 y$Î# ‰
C"
œ x# Èy Ê dy œ x# y"Î# dx Ê y"Î# dy œ x# dx Ê
Ê 2y
x#
ln x
y
x
' y"Î# dy œ ' x# dx
Ê 2y"Î# œ
x$
3
C
$
x œC
' ey dy œ ' ex dx
11.
dy
dx
œ excy Ê dy œ ex ecy dx Ê ey dy œ ex dx Ê
12.
dy
dx
œ 3x# ey Ê dy œ 3x# ey dx Ê ey dy œ 3x# dx Ê ' ey dy œ ' 3x# dx Ê ey œ x3 C Ê ey x3 œ C
Ê ey œ ex C Ê ey ex œ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ e.
586
13.
Chapter 9 Further Applications of Integration
dy
dx
sec# Èy
Èy dy
œ Èy cos# Èy Ê dy œ ˆÈy cos# Èy‰ dx Ê
side, substitute u œ Èy Ê du œ
1
2Èy dy
Ê 2 du œ
1
Èy dy,
œ dx Ê '
sec# Èy
Èy dy
œ ' dx. In the integral on the left-hand
and we have ' sec# u du œ ' dx Ê 2 tan u œ x C
Ê x 2 tan Èy œ C
14. È2xy dy
dx œ 1 Ê dy œ
Ê È2
15. Èx
dy
dx
y3/2
3
2
dy œ
x1/2
"
#
œ eyÈx Ê
1
È2xy
dx Ê È2Èydy œ
1
Èx
dx Ê È2 y1/2 dy œ x1/2 dx Ê È2 ' y1/2 dy œ ' x1/2 dx
3
C1 Ê È2 y3/2 œ 3Èx 32 C1 Ê È2 ˆÈy‰ 3Èx œ C, where C œ 32 C1
dy
dx
œ
ey eÈ x
Èx
Ê dy œ
ey eÈ x
Èx dx
right-hand side, substitute u œ Èx Ê du œ
"
#È x
eÈ x
Ê ecy dy œ È
dx Ê
x
dx Ê 2 du œ
Ê ecy œ 2eu C1 Ê ecy œ 2eÈx C, where C œ C1
16. asec xb
dy
dx
œ eysin x Ê
dy
dx
"
Èx
' ecy dy œ ' ÈeÈx dxÞ In the integral on the
dx, and we have
x
' ecy dy œ 2 ' eu du
œ eysin x cos x Ê dy œ aey esin x cos xbdx Ê ey dy œ esin x cos x dx
Ê ' ecy dy œ ' esin x cos x dx Ê ecy œ esin x C1 Ê ecy esin x œ C, where C œ C1
17.
dy
dx
œ 2xÈ1 y2 Ê dy œ 2xÈ1 y2 dx Ê
dy
È 1 y2
œ 2x dx Ê '
dy
È 1 y2
œ ' 2x dx Ê sin" y œ x# C since kyk "
Ê y œ sinax2 Cb
18.
dy
dx
œ
e2xcy
exby
2y
Ê dy œ
e2xcy
exby dx
Ê dy œ
e2x ecy
ex ey dx
œ
ex
e2y dx
Ê e2y dy œ ex dx Ê ' e2y dy œ ' ex dx Ê
Ê e 2ex œ C where C œ 2C1
19. y w œ x y Ê slope of 0 for the line y œ x.
For x, y 0, y w œ x y Ê slope 0 in Quadrant I.
For x, y 0, y w œ x y Ê slope 0 in Quadrant III.
For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in
Quadrant II above y œ x.
For kyk kxk, y 0, x 0, y w œ x y Ê slope 0 in
Quadrant II below y œ x.
For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in
Quadrant IV above y œ x.
For kyk kxk, x 0, y 0, y w œ x y Ê slope 0 in
Quadrant IV below y œ x.
All of the conditions are seen in slope field (d).
20. y w œ y 1 Ê slope is constant for a given value of y, slope
is 0 for y œ 1, slope is positive for y 1 and negative for
y 1. These characteristics are evident in slope field (c).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
e2y
#
œ ex C1
Section 9.1 Slope Fields and Separable Differential Equations
21. y w œ xy Ê slope œ 1 on y œ x and 1 on y œ x.
y w œ xy Ê slope œ 0 on the y-axis, excluding a0, 0b,
and is undefined on the x-axis. Slopes are positive for
x 0, y 0 and x 0, y 0 (Quadrants II and IV),
otherwise negative. Field (a) is consistent with these
conditions.
22. y w œ y2 x2 Ê slope is 0 for y œ x and for y œ x.
For kyk kxk slope is positive and for kyk kxk slope is
negative. Field (b) has these characteristics.
23.
24.
25-36. Example CAS commands:
Maple:
ode := diff( y(x), x ) = y(x);
icA := [0, 1];
icB := [0, 2];
icC := [0,-1];
DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#25 (Section 9.1)" );
Mathematica:
To plot vector fields, you must begin by loading a graphics package.
< y: 0 STO > x: y (enter)
y 0.2*x*e^(x^2) STO > y: x 0.1 STO > x: y (enter, 10 times)
The last value displayed gives yEuler a1b ¸ 3.45835
2
2
2
The exact solution: dy œ 2xex dx Ê y œ ex C; ya0b œ 2 œ e0 C Ê C œ 1 Ê y œ 1 ex
Ê yexact a1b œ 1 e ¸ 3.71828
14.
dy
dx
œ y ex 2, ya0b œ 2 Ê yn1 œ yn ayn exn 2bdx œ yn 0.5ayn exn 2b
On a TI-92 Plus calculator home screen, type the following commands:
2 STO > y: 0 STO > x: y (enter)
y 0.5*(y ex 2) STO > y: x 0.5 STO > x: y (enter, 4 times)
The last value displayed gives yEuler a2b ¸ 9.82187
The exact solution:
Êyœ
dy
dx
y œ ex 2 Ê Paxb œ 1, Qaxb œ ex 2 Ê ' P(x) dx œ x Ê vaxb œ ex
' ex aex 2bdx œ ex ax 2ex Cb;ya0b œ 2 Ê 2 œ 2 C Ê C œ 0
1
ecx
x
Ê y œ xe 2 Ê yexact a2b œ 2e2 2 ¸ 16.7781
15.
dy
dx
œ
Èx
y ,y
0ß ya0b œ " Ê yn1 œ yn
È xn
yn dx
œ yn
È xn
yn a0.1b
œ yn 0."
È xn
yn
On a TI-92 Plus calculator home screen, type the following commands:
1 STO > y: 0 STO > x: y (enter)
y 0.1*(Èx /y) STO > y: x 0.1 STO > x: y (enter, 10 times)
The last value displayed gives yEuler a1b ¸ 1.5000
The exact solution: dy œ
Ê
16.
dy
dx
y2
2
œ 23 x3/2
1
2
Èx
y dx
Ê y dy œ Èx dx Ê
y2
2
2
œ 32 x3/2 C; aya20bb œ
12
2
œ
1
2
œ 23 a0b3/2 C Ê C œ
1
2
Ê y œ É 43 x3/2 1 Ê yexact a"b œ É 43 a1b3/2 1 ¸ 1.5275
œ 1 y2 , ya0b œ 0 Ê yn1 œ yn a1 y2n bdx œ yn a1 y2n ba0.1b œ yn 0.1a1 y2n b
On a TI-92 Plus calculator home screen, type the following commands:
0 STO > y: 0 STO > x: y (enter)
y 0.1*(1 y2 ) STO > y: x 0.1 STO > x: y (enter, 10 times)
The last value displayed gives yEuler a1b ¸ 1.3964
The exact solution: dy œ a1 y2 bdx Ê
dy
1 y2
œ dx Ê tan1 y œ x C; tan1 ya0b œ tan1 0 œ 0 œ 0 C Ê C œ 0
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Chapter 9 Further Applications of Integration
Ê tan1 y œ x Ê y œ tan x Ê yexact a"b œ tan 1 ¸ 1.5574
17. (a)
dy
dx
œ 2y2 ax 1b Ê
dy
y2
œ 2ax 1bdx Ê ' y2 dy œ ' a2x 2bdx Ê y" œ x2 2x C
Initial value: ya2b œ "# Ê 2 œ 22 2a2b C Ê C œ 2
1
Solution: y" œ x2 2x 2 or y œ x2 2x
2
ya3b œ 32 21a3b 2 œ 15 œ 0.2
(b) To find the approximation, set y1 œ 2y2 ax 1b and use EULERT with initial values x œ 2 and y œ "# and step size
0.2 for 5 Points. This gives ya3b ¸ 0.1851; error ¸ 0.0149.
(c) Use step size 0.1 for 10 points. This gives ya3b ¸ 0.1929; error ¸ 0.0071.
(d) Use step size 0.05 for 20 points. This gives ya3b ¸ 0.1965; error ¸ 0.0035.
18. (a)
dy
dx
œy1Ê'
dy
y1
œ ' dx Ê ln ky 1k œ x C Ê ky 1k œ exC Ê y 1 œ „ eC ex Ê y œ Aex 1
Initial value: ya0b œ 3 Ê 3 œ Ae0 1 Ê A œ 2
Solution: y œ 2ex 1
ya1b œ 2e 1 ¸ 6.4366
(b) To find the approximation, set y1 œ y 1 and use a graphing calculator or CAS with initial values x œ 0 and y œ 3
and step size 0.2 for 5 Points. This gives ya1b ¸ 5.9766; error ¸ 0.4599
(c) Use step size 0.1 for 10 points. This gives ya1b ¸ 6.1875; error ¸ 0.2491.
(d) Use step size 0.05 for 20 points. This gives ya1b ¸ 6.3066; error ¸ 0.1300.
1
2
x2 2x 2 , so ya3b œ 0.2. To find the approximation, let zn œ yn1 2yn1 axn1 1bdx
ay2n1 axn1 1b z2n ax2n 1bbdx with initial values x0 œ 2 and y0 œ "# . Use a spreadsheet, graphing
19. The exact solution is y œ
yn œ y n 1
and
calculator, or CAS as indicated in parts (a) through (d).
(a) Use dx œ 0.2 with 5 steps to obtain ya3b ¸ 0.2024 Ê error ¸ 0.0024.
(b) Use dx œ 0.1 with 10 steps to obtain ya3b ¸ 0.2005 Ê error ¸ 0.0005.
(c) Use dx œ 0.05 with 20 steps to obtain ya3b ¸ 0.2001 Ê error ¸ 0.0001.
(d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step
size.
20. The exact solution is y œ 2ex 1, so ya1b œ 2e 1 ¸ 6.4366. To find the approximation, let zn œ yn1 ayn1 1bdx
and yn œ yn1 ˆ ync1 2zn 2 ‰dx with initial value yn œ 3. Use a spreadsheet, graphing calculator, or CAS as indicated in
parts (a) through (d).
(a) Use dx œ 0.2 with 5 steps to obtain ya1b ¸ 6.4054 Ê error ¸ 0.0311.
(b) Use dx œ 0.1 with 10 steps to obtain ya1b ¸ 6.4282 Ê error ¸ 0.0084
(c) Use dx œ 0.05 with 20 steps to obtain ya1b ¸ 6.4344 Ê error ¸ 0.0022
(d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step
size.
13-16. Example CAS commands:
Maple:
ode := diff( y(x), x ) = 2*x*exp(x^2);ic := y(0)=2;
xstar := 1;
dx := 0.1;
approx := dsolve( {ode,ic}, y(x), numeric, method=classical[foreuler], stepsize=dx ):
approx(xstar);
exact := dsolve( {ode,ic}, y(x) );
eval( exact, x=xstar );
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Section 9.3 Euler's Method
evalf( % );
17.
Example CAS commands:
Maple:
ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2;
xstar := 3;
exact := dsolve( {ode,ic}, y(x) );
# (a)
eval( exact, x=xstar );
evalf( % );
approx1 := dsolve( {ode,ic}, y(x),
# (b)
numeric, method=classical[foreuler], stepsize=0.2 ):
approx1(xstar);
approx2 := dsolve( {ode,ic}, y(x),
# (c)
numeric, method=classical[foreuler], stepsize=0.1 ):
approx2(xstar);
approx3 := dsolve( {ode,ic}, y(x),
# (d)
numeric, method=classical[foreuler], stepsize=0.05 ):
approx3(xstar);
19.
Example CAS commands:
Maple:
ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2;
xstar := 3;
approx1 := dsolve( {ode,ic}, y(x),
# (a)
numeric, method=classical[heunform], stepsize=0.2 ):
approx1(xstar);
approx2 := dsolve( {ode,ic}, y(x),
# (b)
numeric, method=classical[heunform], stepsize=0.1 ):
approx2(xstar);
approx3 := dsolve( {ode,ic}, y(x),
# (c)
numeric, method=classical[heunform], stepsize=0.05 ):
approx3(xstar);
21.
Example CAS commands:
Maple:
ode := diff( y(x), x ) = x + y(x);ic := y(0)=-7/10;
x0 := -4;x1 := 4;y0 := -4; y1 := 4;
b := 1;
P1 := DEplot( ode, y(x), x=x0..x1, y=y0..y1, arrows=thin, title="#21(a) (Section 9.3)" ):
P1;
Ygen := unapply( rhs(dsolve( ode, y(x) )), x,_C1 );
# (b)
P2 := seq( plot( Ygen(x,c), x=x0..x1, y=y0..y1, color=blue ), c=-2..2 ):
# (c)
display( [P1,P2], title="#21(c) (Section 9.3)" );
CC := solve( Ygen(0,C)=rhs(ic), C );
# (d)
Ypart := Ygen(x,CC);
P3 := plot( Ypart, x=0..b, title="#21(d) (Section 9.3)" ):
P3;
euler4 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/4 ): # (e)
P4 := odeplot( euler4, [x,y(x)], x=0..b, numpoints=4, color=blue ):
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Chapter 9 Further Applications of Integration
display( [P3,P4], title="#21(e) (Section 9.3)" );
euler8 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/8 ): # (f)
P5 := odeplot( euler8, [x,y(x)], x=0..b, numpoints=8, color=green ):
euler16 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/16 ):
P6 := odeplot( euler16, [x,y(x)], x=0..b, numpoints=16, color=pink ):
euler32 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/32 ):
P7 := odeplot( euler32, [x,y(x)], x=0..b, numpoints=32, color=cyan ):
display( [P3,P4,P5,P6,P7], title="#21(f) (Section 9.3)" );
<< N|h
| `percent error` >,
# (g)
< 4 | (x1-x0)/ 4 | evalf[5]( abs(1-eval(y(x),euler4(b))/eval(Ypart,x=b))*100 ) >,
< 8 | (x1-x0)/ 8 | evalf[5]( abs(1-eval(y(x),euler8(b))/eval(Ypart,x=b))*100 ) >,
< 16 | (x1-x0)/16 | evalf[5]( abs(1-eval(y(x),euler16(b))/eval(Ypart,x=b))*100 ) >,
< 32 | (x1-x0)/32 | evalf[5]( abs(1-eval(y(x),euler32(b))/eval(Ypart,x=b))*100 ) > >;
13-24. Example CAS commands:
Mathematica: (assigned functions, step sizes, and values for initial conditions may vary)
For exercises 13 - 20, find the exact solution as follows. Set up two error lists.
Clear[x, y, f]
f[x_,y_]:= 2 y2 (x 1)
a = 2; b = 1/2;
xstar = 3;
desol=DSolve[{y'[x] == f[x, y[x]], y[a] == b}, y[x], x] //Simplify
actual[x_] = desol[[1, 1, 2]];
{xstar, actual[xstar]}
errorlisteuler = { };
errorlisteulerimp = { };
pa = Plot[actual[x], {x, a, xstar}]
Euler's method with error at x*. The Do command is used with a sequence of commands that are repeated n times.
a = 2; b = -1/2;
dx = 0.2;
xstar = 3; n = (xstar a) /dx;
solnslist = {{a,b}};
Do[ {new = b + f[a,b] dx, a = a + dx, b = new, AppendTo[solnslist, {a,b}]},{n}]
solnslist
error= actual[xstar] solnslist[[n, 2]]
relativeerror= error / actual[xstar]
AppendTo[errorlisteuler, error]
pe = ListPlot[solnslist, PlotStyle Ä {Hue[.4], PointSize[0.02]}]
Show[pa, pe]
Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the
error as the step size decreases by entering the input command: errorlisteuler
Improved Euler's method. with error at x*
a = 2; b = 1/2;
dx = 0.2;
xstar = 3; n = (xstar a) /dx;
solnslist = {{a,b}};
Do[{new1 = b f[a,b] dx, new2 = b + (f[a, b] f[a+dx, new1])/2 dx, a = a dx, b = new2,
AppendTo[solnslist, {a,b}]},{n}]
solnslist
error= actual[xstar] solnslist[[n, 2]
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Section 9.4 Graphical Solutions of Autonomous Differential Equations
relativeerror= error / actual[xstar]
AppendTo[errorlisteulerimp, error]
peimp = ListPlot[solnslist, PlotStyle Ä {Hue[.8], PointSize[0.02]}]
Show[pa, peimp]
Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the
error as the step size decreases by entering the input command: errorlisteulerimp
You can also type Show[pa, pe, peimp]. This would be appropriate for a fixed value of dx with each method.
You can also make a list of relative errors.
Problems 21 - 24 involve use of code from section 9.1 together with the above code for Euler's method.
9.4 GRAPHICAL SOUTIONS OF AUTONOMOUS DIFFERENTIAL EQUATIONS
1. y w œ ay 2bay 3b
(a) y œ 2 is a stable equilibrium value and y œ 3 is an unstable equilibrium.
(b) yww œ a2y 1by w œ 2ay 2bˆy 12 ‰ay 3b
(c)
2. y w œ ay 2bay 2b
(a) y œ 2 is a stable equilibrium value and y œ 2 is an unstable equilibrium.
(b) yww œ 2yy w œ 2ay 2byay 2b
(c)
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Chapter 9 Further Applications of Integration
3. y w œ y3 y œ ay 1byay 1b
(a) y œ 1 and y œ 1 is an unstable equilibrium and y œ 0 is a stable equilibrium value.
(b) yww œ a3y2 1by w œ 3ay 1bŠy
1
È3 ‹yŠy
1
È3 ‹ay
1b
(c)
4. y w œ yay 2b
(a) y œ 0 is a stable equilibrium value and y œ 2 is an unstable equilibrium.
(b) yww œ a2y 2by w œ 2yay 1bay 2b
(c)
5. y w œ Èy, y 0
(a) There are no equilibrium values.
1
1
w
Èy œ "#
(b) yww œ 2È
y y œ 2È y
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.4 Graphical Solutions of Autonomous Differential Equations
(c)
6. y w œ y Èy, y 0
(a) y œ 1 is an unstable equilibrium.
(b) yww œ Š1
1
2È y ‹
y w œ Š1
1
ˆ
2È y ‹ y
Èy‰ œ ˆÈy "# ‰ˆÈy 1‰
(c)
7. y w œ ay 1bay 2bay 3b
(a) y œ 1 and y œ 3 is an unstable equilibrium and y œ 2 is a stable equilibrium value.
(b) yww œ a3y2 12y 11bay 1bay 2bay 3b œ 3ay 1bŠy
6 È3
‹ay
3
2bŠy
6 È3
3 ‹ay
(c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3b
601
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Chapter 9 Further Applications of Integration
8. y w œ y3 y2 œ y2 ay 1b
(a) y œ 0 and y œ 1 is an unstable equilibrium.
(b) yww œ a3y2 2ybay3 y2 b œ y3 a3y 2bay 1b
(c)
9.
10.
dP
dt
œ 1 2P has a stable equilibrium at P œ "# .
dP
dt œ Pa1 2Pb has an unstable equilibrium
d2 P
dP
dt2 œ a1 4Pb dt œ Pa1 4Pba1 2Pb
d2 P
dt2
œ 2 dP
dt œ 2a1 2Pb
at P œ 0 and a stable equilibrium at P œ "# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.4 Graphical Solutions of Autonomous Differential Equations
11.
12.
dP
dt œ 2PaP 3b has a
d2 P
dP
dt2 œ 2a2P 3b dt œ
dP
dt
d2 P
dt2
stable equilibrium at P œ 0 and an unstable equilibrium at P œ 3.
4Pa2P 3baP 3b
œ 3Pa1 PbˆP "# ‰ has a stable equilibria at P œ 0 and P œ 1 an unstable equilibrium at P œ "# .
3
œ #3 a6P2 6P+1b dP
dt œ # PŠP
3 È3 ˆ
‹ P
6
"# ‰ŠP
3 È3
‹aP
6
1b
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604
Chapter 9 Further Applications of Integration
13.
Before the catastrophe, the population exhibits logistic growth and Patb Ä M0 , the stable equilibrium. After the
catastrophe, the population declines logistically and Patb Ä M1 , the new stable equilibrium.
14.
dP
dt
œ rPaM PbaP mb, r, M, m 0
The model has 3 equilibrium points. The rest point P œ 0, P œ M are asymptotically stable while P œ m is unstable. For
initial populations greater than m, the model predicts P approaches M for large t. For initial populations less than m, the
model predicts extinction. Points of inflection occur at P œ a and P œ b where a œ "3 M m ÈM2 mM m2 ‘ and
b œ "3 M m ÈM2 mM m2 ‘.
(a) The model is reasonable in the sense that if P m, then P Ä 0 as t Ä _; if m P M, then P Ä M as t Ä _; if
P M, then P Ä M as t Ä _.
(b) It is different if the population falls below m, for then P Ä 0 as t Ä _ (extinction). If is probably a more realistic
model for that reason because we know some populations have become extinct after the population level became too
low.
(c) For P M we see that dP
dt œ rPaM PbaP mb is negative. Thus the curve is everywhere decreasing. Moreover,
P ´ M is a solution to the differential equation. Since the equation satisfies the existence and uniqueness conditions,
solution trajectories cannot cross. Thus, P Ä M as t Ä _.
(d) See the initial discussion above.
(e) See the initial discussion above.
15.
dv
dt
œg
k 2
mv ,
Equilibrium:
Concavity:
g, k, m 0 and vatb
dv
dt
d2 v
dt2
œg
k 2
mv
0
œ 0 Ê v œ É mg
k
ˆ k ‰ˆ
œ 2ˆ mk v‰ dv
dt œ 2 m v g
k 2‰
mv
(a)
(b)
160
(c) vterminal œ É 0.005
œ 178.9
ft
s
œ 122 mph
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Section 9.4 Graphical Solutions of Autonomous Differential Equations
605
16. F œ Fp Fr
ma œ mg kÈv
dv
kÈ
v, va0b œ v0
dt œ g m
Thus,
dv
dt
‰ , the terminal velocity. If v0 ˆ mg
‰ , the object will fall faster and faster, approaching the
œ 0 implies v œ ˆ mg
k
k
2
2
‰ , the object will slow down to the terminal velocity.
terminal velocity; if v0 ˆ mg
k
2
17. F œ Fp Fr
ma œ 50 5kvk
dv
1
dt œ m a50 5kvkb
The maximum velocity occurs when
dv
dt
œ 0 or v œ 10
ft
sec .
18. (a) The model seems reasonable because the rate of spread of a piece of information, an innovation, or a cultural fad is
proportional to the product of the number of individuals who have it (X) and those who do not (N X). When X is
small, there are only a few individuals to spread the item so the rate of spread is slow. On the other hand, when
(N X) is small the rate of spread will be slow because there are only a few indiciduals who can receive it during the
interval of time. The rate of spread will be fastest when both X and (N X) are large because then there are a lot of
individuals to spread the item and a lot of individuals to receive it.
(b) There is a stable equilibrium at X œ N and an unstable equilibrium at X œ 0.
d2 X
dt2
dX
2
œ k dX
dt aN Xb kX dt œ k XaN XbaN 2Xb Ê inflection points at X œ 0, X œ
N
2,
(c)
(d) The spread rate is most rapid when x œ
Eventually all of the people will receive the item.
œ VL RL i œ RL ˆ VR i‰, V, L, R 0
œ RL ˆ VR i‰ œ 0 Ê i œ VR
19. L di
dt Ri œ V Ê
di
dt
d i
2
dt œ
Equilibrium:
Concavity:
N
2.
2
di
dt
ˆ R ‰2 ˆ VR i‰
ˆ RL ‰ di
dt œ L
Phase Line:
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and X œ N.
606
Chapter 9 Further Applications of Integration
If the switch is closed at t œ 0, then ia0b œ 0, and the graph of the solution looks like this:
As t Ä _, it Ä isteady state œ
V
R.
(In the steady state condition, the self-inductance acts like a simple wire connector and, as
a result, the current throught the resistor can be calculated using the familiar version of Ohm's Law.)
20. (a) Free body diagram of he pearl:
(b) Use Newton's Second Law, summing forces in the direction of the acceleration:
ˆ m m P ‰g mk v.
mg Pg kv œ ma Ê dv
dt œ
(c) Equilibrium:
Ê vterminal œ
Concavity:
dv
dt
œ
k amPbg
mŠ
k
v‹ œ 0
am Pbg
k
d2 v
dt2
ˆ k ‰ am k Pbg v‹
œ mk dv
dt œ m Š
2
(d)
(e) The terminal velocity of the pearl is
am Pbg
.
k
9.5 APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS
1. Note that the total mass is 66 7 œ 73 kg, therefore, v œ v0 eakÎmbt Ê v œ 9e3.9tÎ73
3.9tÎ73
(a) satb œ ' 9e3.9tÎ73 dt œ 2190
C
13 e
Since sa0b œ 0 we have C œ
2190
13
3.9tÎ73 ‰
ˆ
and lim satb œ lim 2190
œ
13 1 e
tÄ_
tÄ_
2190
13
¸ 168.5
The cyclist will coast about 168.5 meters.
73 ln 9
(b) 1 œ 9e3.9tÎ73 Ê 3.9t
73 œ ln 9 Ê t œ 3.9 ¸ 41.13 sec
It will take about 41.13 seconds.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.5 Applications of First-Order Differential Equations
2. v œ v0 eakÎmbt Ê v œ 9ea59,000Î51,000,000bt Ê v œ 9e59tÎ51,000
(a) satb œ ' 9e59tÎ51,000 dt œ 459,0000
e59tÎ51,000 C
59
Since sa0b œ 0 we have C œ
459,0000
59
ˆ1 e59tÎ51,000 ‰ œ
and lim satb œ lim 459,0000
59
tÄ_
tÄ_
459,0000
59
¸ 7780 m
The ship will coast about 7780 m, or 7.78 km.
ln 9
59t
(b) 1 œ 9e59tÎ51,000 Ê 51,000
œ ln 9 Ê t œ 51,000
¸ 1899.3 sec
59
It will take about 31.65 minutes.
3. The total distance traveled œ v0km Ê a2.75bak39.92b œ 4.91 Ê k œ 22.36. Therefore, the distance traveled is given by the
function satb œ 4.91ˆ1 ea22.36/39.92bt ‰. The graph shows satb and the data points.
4.
a0.80ba49.90b
œ 1.32 Ê k œ 998
k
33
v0 m
k
998
We know that k œ 1.32 and m œ 33a49.9b œ 20
.
33
v0 m ˆ
ak/mbt ‰
Using Equation 3, we have: satb œ k 1 e
œ 1.32ˆ1
v0 m
k
5. (a)
œ coasting distance Ê
dP
dt
œ 0.0015Pa150 Pb œ
0.255
150 Pa150
Pb œ
Thus, k œ 0.255 and M œ 150, and P œ
Initial condition: Pa0b œ 6 Ê 6 œ
Formula: P œ
150
0.255t
1 24ec0.255t Ê 1 24e
ln 48
Ê t œ 0.255
¸ 17.21 weeks
150
125 œ 1 24ec0.255t Ê 1 24e0.255t
120
Ê t œ ln0.255
¸ 21.28
(b) 100 œ
œ
M
1 Aeckt
150
1 Ae0
150
1 24ec0.255t
k
M PaM
e20t/33 ‰ ¸ 1.32a1 e0.606t b
Pb
150
1 Aec0.255t
Ê 1 A œ 25 Ê A œ 24
œ
3
2
Ê 24e0.255t œ
"
#
Ê e0.255t œ
"
48
œ
6
5
Ê 24e0.255t œ
"
5
Ê e0.255t œ
"
120
Ê 0.255t œ ln 48
Ê 0.255t œ ln 120
It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies.
6. (a)
dP
dt
œ 0.0004Pa250 Pb œ
0.1
250 Pa150
Pb œ
Thus, k œ 0.1 and M œ 250, and P œ
k
M PaM Pb
M
œ 1 250
Aec0.1t
1 Aeckt
Initial condition: Pa0b œ 28, where t œ 0 represents the year 1970
250
111
28 œ 1 250
Ae0 Ê 28a1 Ab œ 250 Ê A œ 28 1 œ 14 ¸ 7.9286
Formula: P œ
250
c0.1t
1 111
14 e
or approximately P œ
250
1 7.9286ec0.1t
(b) The population Patb will round to 250 when Patb
Ê
a249.5bˆ111ec0.1t ‰
14
œ 0.5 Ê e0.1t œ
14
55,389
249.5 Ê 249.5 œ
Ê 0.1t œ ln
14
55,389
250
c0.1t
1 111
14 e
Ê 249.5ˆ1
Ê t œ 10 aln 55,389 ln 14b ¸ 82.8.
It will take about 83 years.
7. (a) Using the general solution form Example 2, part (c),
dy
dt
œ a0.08875 ‚ 107 ba8 ‚ 107 yby Ê yatb œ
M
1 AecrMt
œ
111 0.1t ‰
14 e
8‚107
1 Aeca!Þ!))(&ba)bt
œ
8‚107
1 Aec0.71t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 250
607
608
Chapter 9 Further Applications of Integration
Apply the initial condition:
ya0b œ 1.6 ‚ 107 œ
(b) yatb œ 4 ‚ 107 œ
8‚107
1A
Ê
8‚107
1 4ec0.71t
8
1.6
8‚107
1 4ec0.71 ¸ 2.69671 ‚
lnˆ 1 ‰
0.714 ¸ 1.95253 years.
1 œ 4 Ê ya1b œ
Ê 4e0.71t œ 1 Ê t œ
107 kg.
8. (a) If a part of the population leaves or is removed from the environment (e.g., a preserve or a region) each year, then c
would represent the rate of reduction of the population due to this removal and/or migration. When grizzly bears
become a nuisance (e.g., feeding on livestock) or threaten human safety, they are often relocated to other areas or even
eliminated, but only after relocation efforts fail. In addition, bears are killed, sometimes accidently and sometimes
maliciously. For an environment that has a capacity of about 100 bears, a realistic value for c would probably be
between 0 and 4.
(b)
Equilibrium solutions:
dP
dt
œ 0 œ 0.001a100 PbP 1 Ê P2 100P 1000 œ 0 Ê Peq ¸ 11.27 (unstable) and
Peq ¸ 88.73 (stable)
(c)
For 0 Pa0b Ÿ 11, the bear population will eventually disappear, for 12 Ÿ Pa0b Ÿ 88, the population will grow to
about 89, the population will remain at about 89, and for Pa0b 89, the population will decrease to about 89 bears.
9. (a)
dy
dt
œ 1 y Ê dy œ a1 ybdt Ê
dy
1y
œ dt Ê ln k1 yk œ t C1 Ê eln k1yk œ etC1 Ê k1 yk œ et eC1
1 y œ „ C2 et Ê y œ Cet 1, where C2 œ eC1 and C œ „ C2 . Apply the initial condition: ya0b œ 1 œ Ce0 1
Ê C œ 2 Ê y œ 2et 1.
(b)
dy
dt
œ 0.5a400 yby Ê dy œ 0.5a400 yby dt Ê
Example 2, part (c), we obtain
Ê ' Š 1y
ln¹ y cy400 ¹
Êe
Ê
y
y400
Êyœ
1
400 y ‹dy
1
1
400 Š y
1
400 y ‹dy
dy
a400 yby
œ 0.5 dt. Using the partial fraction decomposition in
œ 0.5 dt Ê Š 1y
1
400 y ‹dy
œ 200 dt
œ ' 200 dt Ê lnkyk lnky 400k œ 200t C1 Ê ln¹ y y400 ¹ œ 200t C1
œ e200tC1 œ e200t eC1 Ê ¹ y y400 ¹ œ C2 e200t (where C2 œ eC1 ) Ê
y
y 400
œ „ C2 e200t
œ Ce200t (where C œ „ C2 ) Ê y œ Ce200t y 400 Ce200t Ê a1 Ce200t by œ 400 Ce200t
400 Ce200t
Ce200t 1
ya0b œ 2 œ
Êyœ
400
1 Ae0
400
1 C1 ec200t
œ
400
1 Aec200t ,
Ê A œ 199 Ê yatb œ
where A œ C1 . Apply the initial condition:
400
1 199ec200t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.5 Applications of First-Order Differential Equations
10.
dP
dt
œ raM PbP Ê dP œ raM PbP dt Ê
we obtain
1 ˆ1
M P
1 ‰
M P dP
dP
aM PbP
œ r dt. Using the partial fraction decomposition in Example 6, part (c),
'
œ r dt Ê ˆ P1
Ê lnkPk lnkP Mk œ arMb
Ê ¸ P P M ¸ œ C2 earMbt (where
'
1 ‰
ˆ P1 P 1 M ‰dP œ rM dt
M P dP œ rM dt Ê
P
t C1 Ê ln¸ P P M ¸ œ arMb t C1 Ê eln¸ P c M ¸ œ earMbtC1 œ earMbt eC1
C2 œ eC1 ) Ê PPM œ „ C2 earMbt Ê PPM œ CearMbt (where C œ „ C2 )
Ê P œ CearMbt P M CearMbt Ê ˆ1 CearMbt ‰P œ M CearMbt Ê P œ
ÊPœ
11. (a)
dP
dt
M
,
1 AecarMbt
œ kP2 Ê ' P2 dP œ ' k dt Ê P" œ kt C Ê P œ
Initial condition: Pa0b œ P0 Ê P0 œ C1 Ê C œ
dP
dt
œ raM PbaP mb Ê
Ê
ÊPœ
M
1 C1 ecarMbt
aP 100ba1200 Pb dP
a1200 PbaP 100b dt
1
P0
"
kt C
P0
1 kP0 t
(b) There is a vertical asymptote at t œ
12. (a)
M CearMbt
CearMbt 1
where A œ C1 .
Solution: P œ kt a11/P0 b œ
dP
dt
1
kpO
œ ra1200 PbaP 100b Ê
œ 1100 r Ê
ˆ 12001 P
1
‰ dP
P 100 dt
1
dP
a1200 PbaP 100b dt
œ 1100 r Ê
œrÊ
ˆ 12001 P
1100
dP
a1200 PbaP 100b dt
1
‰
P 100 dP
P 100
1200 P
1100 r t
where C œ „ eC1 Ê P 100 œ 1200Ce1100 r t CPe1100 r t Ê Pa1 Ce1100 r t b œ 1200Ce
ÊPœ
1200Ce
100
Ce1100 r t 1
œ
c1100 r t
1200 100
C e
1 C1 ec1100 r t
(b) Apply the initial condition: 300 œ
1200 100Aec1100 r t
1 Aec1100 r t
ÊPœ
1200 100A
1A
œ 1100 r
œ 1100 r dt
Ê ' ˆ 12001 P P1100 ‰dP œ ' 1100 r dt Ê ln a1200 Pb ln aP 100b œ 1100 r t C1
P 100 ¸
P 100
C1 1100 r t
¸ P 100 ¸
Ê ln ¸ 1200
Ê
P œ 1100 r t C1 Ê ln 1200 P œ 1100 r t C1 Ê 1200 P œ „ e e
1100 r t
œ Ce1100 r t
100
where A œ C1 .
Ê 300 300A œ 1200 100A Ê A œ
9
2
ÊPœ
2400 900Aec1100 r t
Þ
2 9ec1100 r t
(Note that P Ä 1200 as t Ä _.)
(c)
dP
dt
œ raM PbaP mb Ê
Ê ˆ M 1 P
1 ‰ dP
P m dt
1
dP
aM PbaP mb dt
œrÊ
œ raM mb Ê ' ˆ M 1 P
Ê ln aM Pb ln aP mb œ aM mb r t
Ê
Pm
MP
Mm
dP
aM PbaP mb dt
Ê Pˆ1 Ce
aMmb r t
œ MCe
œ raM mb Ê
'
1 ‰
raM mbdt
P m dP œ
Pm ¸
¸
C1 Ê ln M P œ aM mb r t
aMmb r t
aMmb r t
œ CeaMmb r t where C œ „ eC1 Ê P m œ MCe
aMmb r t ‰
mÊPœ
aP mb aM Pb dP
aM PbaP mb dt
C1 Ê
Pm
MP
Apply the initial condition Pa0b œ P0
MmA
1 A
Ê P0 P0 A œ M mA Ê A œ
M P0
P0 m
ÊPœ
ÊPœ
caMcmb r t
M m
Ce
1 C1 ecaMcmb r t
ÊPœ
MaP0 mb maM P0 becaMcmb r t
aP0 mb aM P0 becaMcmb r t
(Note that P Ä M as t Ä _ provided P0 m.)
13. y œ mx Ê
y
x
orthogonals:
œmÊ
dy
dx
xy y
x2
w
œ 0 Ê y w œ yx . So for
œ xy Ê y dy œ x dx Ê
y2
2
x2
2
œC
Ê x y œ C1
2
œ raM mb
œ „ eC1 eaMmb r t
CPe
MCeaMcmb r t m
CeaMcmb r t 1
A œ C1 .
P0 œ
609
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
M mAecaMcmb r t
1 AecaMcmb r t
610
Chapter 9 Further Applications of Integration
14. y œ cx2 Ê
Ê yw œ
y
x2
2y
x .
œcÊ
x2 y 2xy
x4
œ 0 Ê x2 y w œ 2xy
w
So for the orthogonals:
dy
dx
x
œ 2y
2
Ê 2ydy œ xdx Ê y2 œ x2 C Ê y œ „ É x2 C,
2
C0
15. kx2 y2 œ 1 Ê 1 y2 œ kx2 Ê
1 y2
x2
œk
x a2yby ˆ1 y2 ‰2x
Ê
œ 0 Ê 2yx2 y w œ a1 y2 ba2xb
x%
ˆ1 y2 ‰a2xb
ˆ1 y 2 ‰
Ê y w œ 2xy2 œ xy . So for the orthogonals:
2
ˆ1 y 2 ‰
2
dy
xy
dy œ x dx Ê ln y y2 œ x2 C
dx œ 1y2 Ê
y
2
w
2x
16. 2x2 y2 œ c2 Ê 4x 2yy w œ 0 Ê y w œ 4x
2y œ y . For
orthogonals:
dy
dx
œ
y
2x
Ê
œ
dy
y
dx
2x
Ê ln y œ "# ln x C
Ê ln y œ ln x1/2 ln C1 Ê y œ C1 kxk1/2
17. y œ cex Ê
y
ecx
œcÊ
ex y w c yaex bac1b
aex b2
œ!
Ê ex y w œ yex Ê y w œ y. So for the orthogonals:
dy
dx
œ
1
y
2
Ê y dy œ dx Ê
y2
2
œxC
Ê y œ 2x C1 Ê y œ „ È2x C1
xŠ 1y ‹y c ln y
w
18. y œ ekx Ê ln y œ kx Ê
ln y
x
œkÊ
Ê Š xy ‹ y w ln y œ 0 Ê y w œ
dy
dx
y ln y
x .
x
y ln y Ê y ln y dy œ x dx
1 2
" 2
ˆ " 2‰
# y ln y 4 ay b œ # x
2
y2 ln y y2 œ x2 C1
x2
œ0
So for the orthogonals:
œ
Ê
Ê
C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 9.5 Applications of First-Order Differential Equations
2
w
19. 2x2 3y2 œ 5 and y2 œ x3 intersect at a1, 1b. Also, 2x2 3y2 œ 5 Ê 4x 6y y w œ 0 Ê y w œ 4x
6y Ê y a1, 1b œ 3
y21 œ x3 Ê 2y1 y1w œ 3x2 Ê y1w œ
x2
2
20. (a) x dx y dy œ 0 Ê
y2
2
w
3x2
2y1
Ê y1w a1, 1b œ 32 . Since y w † y1w œ ˆ 23 ‰ˆ 32 ‰ œ 1, the curves are orthogonal.
œ C is the general equation
of the family with slope y œ xy . For the orthogonals:
yw œ
y
x
Ê
dy
y
œ
dx
x
Ê ln y œ ln x C or y œ C1 x
(where C1 œ e Ñ is the general equation of the
orthogonals.
C
(b) x dy 2y dx œ 0 Ê 2y dx œ x dy Ê
Ê "# Š dy
y ‹œ
dx
x
dy
2y
œ
dx
x
Ê "# ln y œ ln x C Ê y œ C1 x2 is
the equation for the solution family.
"
# ln
y ln x œ C Ê
"y
# y
w
Ê slope of orthogonals is
1
x
dy
dx
œ 0 Ê yw œ
2y
x
x
œ 2y
2
Ê 2y dy œ x dx Ê y2 œ x2 C is the general
equation of the orthogonals.
2". y2 œ 4a2 4ax and y2 œ 4b2 4bx Ê (at intersection) 4a2 4ax œ 4b2 4bx Ê a2 b2 œ xaa bb
Ê aa bbaa bb œ aa bbx Ê x œ a b. Now, y2 œ 4a2 4aaa bb œ 4a2 4a2 4ab œ 4ab Ê y œ „ 2Èab.
4a
Thus the intersections are at Ša b, „ 2Èab‹. So, y2 œ 4a2 4ax Ê y1w œ 2y
which are equal to 4a
and
È
2Š2
4a
2Š2Èab‹
4b
2Š2Èab‹
œ
È ba
and
È ba
at the intersections. Also, y œ 4b 4bx Ê
2
2
y2w
œ
4b
2y
which are equal to
œ É ba and É ba at the intersections. ay1w b † ay2w b œ ". Thus the curves are orthogonal.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
ab‹
4b
2Š2Èab‹
and
611
612
Chapter 9 Further Applications of Integration
CHAPTER 9 PRACTICE EXERCISES
".
dy
dx
œ Èy cos2 Èy Ê
2. y w œ
3yax1b2
y 1
Ê
œ dx Ê 2tanÈy œ x C Ê y œ ˆtan1 ˆ x 2 C ‰‰
dy
Èy cos2 Èy
ay 1 b
y dy
3. yy w œ secay2 bsec2 x Ê
œ 3ax 1b2 dx Ê y ln y œ ax 1b3 C
y dy
secay2 b
sinˆy2 ‰
2
œ sec2 x dx Ê
œ tan x C Ê sinay2 b œ 2tan x C1
sin x
4. y cos2 axb dy sin x dx œ 0 Ê y dy œ cos
2 axb dx Ê
y2
2
œ cos1axb C Ê y œ „ É cosa2xb C1
2ax 2b3/2 a3x 4b
15
2ax 2b3/2 a3x 4b
C“
15
5. y w œ xey Èx 2 Ê ey dy œ xÈx 2 dx Ê ey œ
2ax 2b3/2 a3x 4b
15
Ê y œ ln’
6. y w œ xyex Ê
2
dy
y
2
C“ Ê y œ ln’
C Ê ey œ
2ax 2b3/2 a3x 4b
15
C
œ ex x dx Ê ln y œ "# ex C
2
2
7. sec x dy x cos2 y dx œ 0 Ê
dy
cos2 y
x dx
œ sec
x Ê tan y œ cos x x sin x C
8. 2x2 dx 3Èy csc x dy œ 0 Ê 3Èy dy œ
2x2
csc x dx
Ê 2y3/2 œ 2a2 x2 bcos x 4x sin x C
Ê y3/2 œ a2 x2 bcos x 2x sin x C1
9. y w œ
ey
xy
Ê yey dy œ
Ê ay 1bey œ ln kxk C
dx
x
10. y w œ xexy csc y Ê y w œ
x ex
ey csc
yÊ
ey
csc y dy
œ x ex dx Ê
11. xax 1bdy y dx œ 0 Ê xax 1bdy œ y dx Ê
dy
y
œ
ey
2 asin
dx
x ax 1 b
y cos yb œ ax 1bex C
Ê ln y œ lnax 1b lnaxb C
Ê ln y œ lnax 1b lnaxb ln C1 Ê ln y œ lnŠ C1 axx 1b ‹ Ê y œ
12. y w œ ay2 1bax1 b Ê
dy
y 2 1
œ
Ê
dx
x
1
lnŠ yy c
b1‹
2
C1 ax 1b
x
1
œ ln x C Ê lnŠ yy
1 ‹ œ 2ln x ln C1 Ê
y1
y1
œ C1 x2
13. 2y w y œ xex/2 Ê y w "# y œ x2 ex/2 .
' ˆ "‰
paxb œ "# , vaxb œ e c # dx œ ex/2 .
ex/2 y w "# ex/2 y œ ˆex/2 ‰ˆ x2 ‰ˆex/2 ‰ œ
14.
w
y
2
x
2
Ê
d ˆ x/2
dx e
y‰ œ
x
2
Ê ex/2 y œ
x2
4
2
C Ê y œ ex/2 Š x4 C‹
y œ ex sin x Ê y w 2y œ 2ex sin x.
paxb œ 2, vaxb œ e' 2dx œ e2x .
e2x y w 2e2x y œ 2e2x ex sin x œ 2ex sin x Ê
x
d
2x
dx ae
yb œ 2ex sin x Ê e2x y œ ex asin x cos xb C
2x
Ê y œ e asin x cos xb Ce
15. xy w 2y œ 1 x1 Ê y w ˆ 2x ‰y œ
vaxb œ e2'
dx
x
1
x
1
x2 .
2
œ e2ln x œ eln x œ x2 .
x2 y w 2xy œ x 1 Ê
d
2
dx ax yb
œ x 1 Ê x2 y œ
x2
2
xCÊyœ
"
#
1
x
C
x2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 9 Practice Exercises
613
16. xy w y œ 2x ln x Ê y w ˆ 1x ‰y œ 2 ln x.
vaxb œ e
d ˆ1
dx x
' dxx
2
œ eln x œ 1x . ˆ 1x ‰y w ˆ 1x ‰ y œ 2x ln x Ê
† y‰ œ 2x ln x Ê
† y œ c ln x d2 C Ê y œ xc ln x d2 Cx
1
x
17. a1 ex bdy ayex ex bdx œ 0 Ê a1 ex by w ex y œ ex Ê y w œ
' a exdx b
vaxb œ e 1 b ex œ elnae 1b œ ex 1.
x
cx
aex 1by w aex 1bˆ 1 e ex ‰y œ a1e ex b aex 1b Ê
Êyœ
18.
dx
dy
ecx C
ex 1
ex
1 ex y
œ
ecx
a1 e x b .
x
œ
e cx C
1by d œ ex Ê aex 1by œ ex C
d
aex
dx c
1 ex
x 4yey œ 0 Ê x w x œ 4yey . Let vayb œ e
' dy
œ ey . Then ey x w xey œ 4ye2y Ê
d
y
dy axe b
œ 4ye2y
Ê xey œ a2y 1be2y C Ê x œ a2y 1bey Cey
19. ax 3y2 b dy y dx œ 0 Ê x dy y dx œ 3y2 dy Ê
d
dx axyb
œ 3y2 dy Ê xy œ y3 C
'
20. y dx a3x y2 cos yb dx œ 0 Ê x w Š 3y ‹x œ y3 cos y. Let vayb œ e
3dy
y
3
œ e3ln y œ eln y œ y3 .
Then y3 x w 3y2 x œ cos y and y3 x œ ' cos y dy œ sin y C. So x œ y3 asin y Cb
21.
œ exy2 Ê ey dy œ eax2b dx Ê ey œ eax2b C. We have ya0b œ 2, so e2 œ e2 C Ê C œ 2e2 and
e œ eax2b 2e2 Ê y œ lnˆeax2b 2e2 ‰
22.
dy
dx
dy
dx
y
œ
y ln y
1 x2
Ê etan
Ê
c1 a0bC
dy
y ln y
œ
dx
1 x2
Ê lnaln yb œ tan1 axb C Ê y œ ee
So y w ax 1b2
Ê yax 1b2 œ
x
x1.
' x b2 1 dx
Let vaxb œ e
2
ax 1 b a x
1b2 y œ
x3
3
C Ê y œ ax 1b2 Š x3
y œ ax 1b2 Š x3
3
x2
2
2
x
2
x
ax 1 b a x
1b2 Ê
d
dx yax
3
x2
2
dy
dx
tanc1 axbbln 2
2
œ e2lnax1b œ elnax1b œ ax 1b2 .
2
1b2 ‘ œ xax 1b Ê yax 1b œ ' xax 1bdx
1‹
' ˆ 2 ‰dx
25.
tanc1 a0bbC
C‹. We have ya0b œ 1 Ê 1 œ C. So
1
2
w
ˆ2‰
24. x dy
dx 2y œ x 1 Ê y x y œ x x . Let vaxb œ e
So
. We have ya0b œ e2 Ê e2 œ ee
œ 2 Ê tan1 a0b C œ ln 2 Ê 0 C œ ln 2 Ê C œ ln 2 Ê y œ ee
w
ˆ 2 ‰
23. ax 1b dy
dx 2y œ x Ê y x 1 y œ
Ê
tanc1 axbbC
d
x4
2
3
2
dx ax yb œ x x Ê x y œ 4
2
4
2x2 1
y œ x4 4x1 2 "# œ x 4x
2
x2
2
CÊyœ
x2
4
x
C
x2
œ eln x œ x2 . So x2 y w 2xy œ x3 x
2
"# . We have ya1b œ 1 Ê 1 œ
3x2 y œ x2 . Let vaxb œ e' 3x dx œ ex . So ex y w 3x2 ex y œ x2 ex Ê
2
3
3
3
3
We have ya0b œ 1 Ê e0 a1b œ 13 e0 C Ê 1 œ
3
1
3
d
dx axyb
dy
ˆy È y ‰
3
"
#
3
Ê C œ 14 .
3
œ x2 ex Ê ex y œ 13 ex C.
3
cos x
x .
C
3
4
3
Êyœ
1
3
43 ex
3
' 1 dx
x
œ eln x œ x.
Let vaxb œ e
œ cos x Ê xy œ ' cos x dx Ê xy œ sin x C. We have yˆ 12 ‰ œ 0 Ê ˆ 12 ‰0 œ 1 C
Ê C œ 1. So xy œ 1 sin x Ê y œ
27. x dy ˆy Èy‰dx œ 0 Ê
d
x3
dx Še y‹
C Ê C œ 43 and ex y œ 13 ex
26. xdy ay cos xbdx œ 0 Ê xy w y cos x œ 0 Ê y w ˆ 1x ‰y œ
So xy w xˆ 1x ‰y œ cos x Ê
3
1
4
œ
dx
x
1 sin x
x
Ê 2lnˆÈy 1‰ œ ln x C. We have ya1b œ 1 Ê 2 lnŠÈ1 1‹ œ ln 1 C
Ê 2 ln 2 œ C œ ln 22 œ ln 4. So 2 lnˆÈy 1‰ œ ln x ln 4 œ lna4xb Ê lnˆÈy 1‰ œ "# lna4xb œ lna4xb1/2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
614
Chapter 9 Further Applications of Integration
Ê eln
ˆÈ y 1 ‰
28. y2 dx
dy œ
So
y3
3
1/2
œ elna4xb Ê Èy 1 œ 2Èx Ê y œ ˆ2Èx 1‰
ex
e2x 1
Ê
e2x 1
ex dx
œ ex ex
1
3
œ
dy
yc2
Ê
y3
3
2
œ ex ex C. We have ya0b œ 1 Ê
a1 b 3
3
œ e0 e0 C Ê C œ 13 .
Ê y3 œ 3aex ex b 1 Ê y œ c3aex ex b 1 d1/3
xc2
29. xy w ax 2by œ 3x3 ex Ê y w ˆ x x 2 ‰y œ 3x2 ex . Let vaxb œ e' ˆ x ‰dx œ ex2ln x œ xe2 . So
x
x
ex w
ex ˆ x 2 ‰
d ˆ
y œ 3 Ê dx
y † xe2 ‰ œ 3 Ê y † xe2 œ 3x C. We have ya1b œ 0 Ê 0 œ 3a1b C Ê C œ 3
x2 y x2
x
Êy†
ex
x2
œ 3x 3 Ê y œ x2 ex a3x 3b
30. y dx a3x xy 2bdy œ 0 Ê
Payb œ
3
y
x
dx
dy
3x xy 2
y
œ0Ê
dx
dy
3x
y
x œ 2y Ê
dx
dy
Š 3y 1‹x œ 2y .
1 Ê ' Paybdy œ 3ln y y Ê vayb œ e3ln yy œ y3 ey
y3 ey x w y3 ey Š 3y 1‹x œ 2y2 ey Ê y3 ey x œ ' 2y2 ey dy œ 2ey ay2 2y 2b C
Ê y3 œ
Ê y3 œ
2ˆy2 2y 2‰ Cey
.
x
2
yb1
ˆ
‰
2 y 2y 2 4e
x
We have ya2b œ 1 Ê 1 œ
2a1 2 2b Cec1
2
Ê C œ 4e and
31. To find the approximate values let yn œ yn1 ayn1 cos xn1 ba0.1b with x0 œ 0, y0 œ 0, and 20 steps. Use a
spreadsheet, graphing calculator, or CAS to obtain the values in the following table.
x
y
x
y
1.1
1.6241
0
0
1.2
1.8319
0.1
0.1000
1.3
2.0513
0.2
0.2095
1.4
2.2832
0.3
0.3285
1.5
2.5285
0.4
0.4568
1.6
2.7884
0.5
0.5946
1.7
3.0643
0.6
0.7418
1.8
3.3579
0.7
0.8986
1.9
3.6709
0.8
1.0649
2.0
4.0057
0.9
1.2411
1.0
1.4273
32. To find the approximate values let zn œ yn1 aa2 yn1 ba2 xn1 3bba0.1b and
yn œ yn1 Š a2 ync1 ba2 xnc1 32b a2 zn ba2 xn 3b ‹a0.1b with initial values x0 œ 3, y0 œ 1, and 20 steps. Use a
spreadsheet, graphing calculator, or CAS to obtain the values in the following table.
x
y
x
y
1.9 5.9686
3
1
1.8 6.5456
2.9
0.6680
1.7 6.9831
2.8
0.2599
1.6 7.2562
2.7
0.2294
1.5 7.3488
2.6
0.8011
1.4 7.2553
2.5
1.4509
1.3 6.9813
2.4
2.1687
1.2 6.5430
2.3
2.9374
1.1 5.9655
2.2
3.7333
1.0 5.2805
2.1
4.5268
2.0
5.2840
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 9 Practice Exercises
2ync1
" xnc1 2ync1
33. To estimate ya3b, let zn œ yn1 Š xncx1nc1
1 ‹a0.05b and yn œ yn1 # Š xnc1 1
xn 2zn
xn 1 ‹a0.05b
615
with initial values
x0 œ 0, y0 œ 1, and 60 steps. Use a spreadsheet, graphing calculator, or CAS to obtain ya3b ¸ 0.9063.
34. To estimate ya4b, let zn œ yn1 Š
x2nc1 2ync1 1
‹a0.05b
xnc1
with initial values x0 œ 1, y0 œ 1, and 60 steps. Use a
spreadsheet, graphing calculator, or CAS to obtain ya4b ¸ 4.4974.
35. Let yn œ yn1 ˆ exnc1 b1ync1 b 2 ‰adxb with starting values x0 œ 0 and y0 œ 2, and steps of 0.1 and 0.1. Use a spreadsheet,
programmable calculator, or CAS to generate the following graphs.
(a)
(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot
handle the calculations for x Ÿ 1. (This occurs because the analytic solution is y œ 2 lna2 ex b, which has an
asymptote at x œ ln 2 ¸ 0.69. Obviously, the Euler approximations are misleading for x Ÿ 0.7.)
y
y
36. Let zn œ yn1 Š eynncc11 xnncc11 ‹adxb and yn œ yn1 #" Š eynncc11 xnncc11
x2
x2
xn2 zn
ezn xn ‹adxb
with starting values x0 œ 0 and y0 œ 0,
and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs.
(a)
(b)
37.
x
y
dy
dx
1
1.2
1.4
1.6
1 0.8 0.56 0.28
œ x Ê dy œ x dx Ê y œ
Ê 1 œ
"
#
Ê ya2b œ
CÊCœ
2
2
2
3
2
œ
"
#
32
x2
2
1.8
0.04
2.0
0.4
C; x œ 1 and y œ 1
Ê yaexactb œ
x2
2
3
2
is the exact value.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
616
Chapter 9 Further Applications of Integration
38.
x
y
œ
dy
dx
1
1.2
1.4
1.6
1.8
2.0
1 0.8 0.6333 0.4904 0.3654 0.2544
1
x
Ê dy œ x1 dx Ê y œ lnkxk C; x œ 1 and y œ 1
Ê 1 œ ln 1 C Ê C œ 1 Ê yaexactb œ lnkxk 1
Ê ya2b œ ln 2 1 ¸ 0.3069 is the exact value.
39.
x
y
1
1.2
1.4
1.6
1.8
2.0
1 1.2 0.488 1.9046 2.5141 3.4192
œ xy Ê
dy
dx
Êyœe
dy
y
x2
2 C
œ x dx Ê lnkyk œ
x2
2
x2
2
C
x2
2
œ e † eC œ C1 e ; x œ 1 and y œ 1
x2
Ê 1 œ C1 e1/2 Ê C1 œ e1/2 yaexactb œ e1/2 † e 2
œ eˆx 1‰/2 Ê ya2b œ e3/2 ¸ 4.4817 is the
exact value.
2
40.
x
y
1
1.2
1.4
1.6
1.8
2.0
1 1.2 1.3667 1.5130 1.6452 1.7688
dy
y2
1
dx œ y Ê y dy œ dx Ê 2 œ x
"
"
2
# œ 1 C Ê C œ # Ê y œ
C; x œ 1 and y œ 1
2x 1
Ê yaexactb œ È2x 1 Ê ya2b œ È3 ¸ 1.7321 is the
exact value.
41.
dy
dx
œ y2 1 Ê y w œ ay 1bay 1b. We have y w œ 0 Ê ay 1b œ 0, ay 1b œ 0 Ê y œ 1, 1.
(a) Equilibrium points are 1 (stable) and 1 (unstable)
(b) y w œ y2 1 Ê y ww œ 2yy w Ê y ww œ 2yay2 1b œ 2yay 1bay 1b. So y ww œ 0 Ê y œ 0, y œ 1, y œ 1.
(c)
42.
dy
dx
œ y y2 Ê y w œ ya1 yb. We have y w œ 0 Ê ya1 yb œ 0 Ê y œ 0, 1 y œ 0 Ê y œ 0, 1.
(a) The equilibrium points are 0 and 1. So, 0 is unstable and 1 is stable.
(b) Let ïî œ increasing, íï œ decreasing.
yw !
yw !
yw !
qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqpy
0
1
y w œ y y2 Ê y ww œ y w 2yy w Ê y ww œ ay y2 b 2yay y2 b œ y y2 2y2 2y3 Ê y ww œ 2y3 3y2 y
œ ya2y2 3y 1b Ê y ww œ ya2y 1bay 1b. So, y ww œ 0 Ê y œ 0, 2y 1 œ 0, y 1 œ 0 Ê y œ 0, y œ "# ,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 9 Additional and Advanced Exercises
617
y œ 1.
Let ïî œ concave up, íï œ concave down.
y ww !
y ww !
y ww !
y ww !
qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqpy
0
1
1/2
(c)
43. (a) Force œ Mass times Acceleration (Newton's Second Law) or F œ ma. Let a œ
dv
dt
œ
dv
ds
†
ds
dt
œ v dv
ds . Then
2 2
ma œ mgR2 s2 Ê a œ gR2 s2 Ê v dv
Ê v dv œ gR2 s2 ds Ê ' v dv œ ' gR2 s2 ds
ds œ gR s
Ê
v2
2
œ
ÊCœ
gR2
s C1
2
v0 2gR
Ê v2 œ
Êv œ
2
(b) If v0 œ È2gR, then v2 œ
2gR2
s
2gR2
s
2gR
s
2
2C1 œ
v20
2gR2
s
C. When t œ 0, v œ v0 and s œ R Ê v20 œ
2gR2
R
C
2gR
2
Ê v œ É 2gR
s , since v
È2gR. Then
0 if v0
ds
dt
œ
È2gR2
Ès
Ê Ès ds œ È2gR2 dt
Ê ' s1/2 ds œ ' È2gR2 dt Ê 23 s3/2 œ È2gR2 t C1 Ê s3/2 œ ˆ 32 È2gR2 ‰t C; t œ 0 and s œ R
Ê R3/2 œ ˆ 32 È2gR2 ‰a0b C Ê C œ R3/2 Ê s3/2 œ ˆ 32 È2gR2 ‰t R3/2 œ ˆ 32 RÈ2g‰t R3/2
3
œ R3/2 ˆ 32 R1/2 È2g‰t 1 ‘ œ R3/2 ’ Š
44.
v0 m
k
a0.86ba30.84b
k
0.8866t
œ coasting distance Ê
Ê satb œ 0.97a1 e
È2gR
2R ‹t
2/3
0‰
0‰ ‘
‘ Ê s œ R 1 ˆ 3v
1 “ œ R3/2 ˆ 3v
2R t 1
2R t
œ 0.97 Ê k ¸ 27.343. satb œ
v0 m ˆ
k 1
eak/mbt ‰ Ê satb œ 0.97ˆ1 ea27.343/30.84bt ‰
b. A graph of the model is shown superimposed on a graph of the data.
CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES
1. (a)
dy
dt
A
œ kA
V ac yb Ê dy œ k V ay cbdt Ê
dy
yc
'
œ k A
V dt Ê
dy
yc
A
œ ' k A
V dt Ê lnky ck œ k V t C1
Ê y c œ „ eC1 ek V t . Apply the initial condition, ya0b œ y0 Ê y0 œ c C Ê C œ y0 c
A
Ê y œ c ay0 cbek V t .
A
(b) Steady state solution: y_ œ lim yatb œ lim c ay0 cbek V t ‘ œ c ay0 cba0b œ c
A
tÄ_
tÄ_
2. Measure the amounts of oxygen involved in mL. Then the inflow of oxygen is 1000 mL/min (Assumed: it will take 5
minutes to deliver the 5L œ 5000mL); the amount of oxygen at t œ 0 is 210 mL; letting A œ the amount of oxygen in the
flask, the concentration at time t is A mL/L; the outflow rate of oxygen is A mL/L (lb/sec). The rate of change in A, dA
dt ,
equals the rate of gain (1000 mL/min) minus rate of loss (A mL/min). Thus:
dA
dA
t
dt œ 1000 A Ê 1000 A œ dt Ê lnaA 1000b œ t C Ê A 1000 œ Ce . At t œ 0, A œ 210, so C œ 790 and
A œ 1000 790et . Thus, Aa5b œ 1000 790e5 ¸ 994.7 mL. The concentration is
994.7 mL
1000 mL
œ 99.47%.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
618
Chapter 9 Further Applications of Integration
3. The amount of CO2 in the room at time t is Aatb. The rate of change in the amount of CO2 ,
dA
dt
is the rate of internal
production (R1 ) plus the inflow rate (R2 ) minus the outflow rate (R3 ).
R1 œ ˆ20
breaths/min ‰
a30
student
R2 œ Š1000
ft3 CO2
ft3
min ‹Š0.0004 min ‹
A
R3 œ Š 10,000
‹1000 œ 0.1A
dA
dt
3
100
2
studentsbˆ 1728
ft3 ‰Š0.04 ft ftCO
‹ ¸ 1.39
3
œ 0.4
ft3 CO2
min
ft3 CO2
min
ft3 CO2
min
œ 1.39 0.4 0.1A œ 1.79 0.1A Ê Aw 0.1A œ 1.79. Let vatb œ e
' 0.1dt
. We have
' 0.1dt
d
‹
dt ŠAe
' 0.1dt
œ 1.79e
Ê Ae0.1t œ ' 1.79e0.1t dt œ 17.9e0.1t C. At t œ 0, A œ a10,000ba0.0004b œ 4 ft3 CO2 Ê C œ 13.9
Ê A œ 17.9 13.9e0.1t . So Aa60b œ 17.9 13.9e0.1a60b ¸ 17.87 ft3 of CO2 in the 10,000 ft3 room. The percent of
17.87
CO2 is 10,000
‚ 100 œ 0.18%
4.
damvb
damvb
dm
dm
dv
dm
dm
dm
dv
dm
dt œ F av ub dt Ê F œ dt av ub dt Ê F œ m dt v dt v dt u dt Ê F œ m dt u dt .
dm
dt œ b Ê m œ kbkt C. At t œ 0, m œ m0 , so C œ m0 and m œ m0 kbkt.
u kb k
m0 kbkt
dv
Thus, F œ am0 kbktb dv
dt ukbk œ am0 kbktbkgk Ê dt œ g m0 kbkt Ê v œ gt u lnŠ m0 ‹ C1
v œ 0 at t œ 0 Ê C1 œ 0. So v œ gt u lnŠ m0 m0kbkt ‹ œ
t œ 0 Ê y œ "# gt2 c’ t Š
m0 kbkt
m0 kbkt
kbk ‹ lnŠ m0 ‹
dy
dt
Ê y œ ' ’ gt u lnŠ
m0 kbkt
m0 ‹
“dt and u œ c, y œ 0 at
“
'
5. (a) Let y be any function such that vaxby œ ' vaxbQaxb dx C, vaxb œ e Paxb dx . Then
' Paxb dx
'
d
w
w
Ê v w axb œ œ e Paxb dx Paxb œ vaxbPaxb.
dx avaxb † yb œ vaxb † y y † v axb œ vaxbQaxb. We have vaxb œ e
Thus vaxb † y w y † vaxb Paxb œ vaxbQaxb Ê y w y Paxb œ Qaxb Ê the given y is a solution.
(b) If v and Q are continuous on c a, b d and x − aa, bb, then
Ê
d
dx ’
'xx vatbQatb dt“ œ vaxbQaxb
0
'xx vatbQatb dt œ ' vaxbQaxb dx. So C œ y0 vax0 b ' vaxbQaxb dx. From part (a), vaxby œ ' vaxbQaxb dx C.
0
Substituting for C: vaxby œ ' vaxbQaxb dx y0 vax0 b ' vaxbQaxb dx Ê vaxby œ y0 vax0 b when x œ x0 .
6. (a) y w Paxby œ 0, yax0 b œ 0. Use vaxb œ e' Paxb dx as an integrating factor. Then
d
dx avaxbyb
œ 0 Ê vaxby œ C
Ê y œ Ce' Paxb dx and y1 œ C1 e' Paxb dx , y2 œ C# e' Paxb dx , y1 ax0 b œ y2 ax0 b œ 0, y1 y2 œ aC1 C2 be' Paxb dx
œ C3 e
(b)
' Paxb dx
d
y axb
dx avaxbc 1
and y1 y2 œ 0 0 œ 0. So y1 y2 is a solution to y w Paxby œ 0 with yax0 b œ 0.
y2 axb db œ
' Paxb dx ' Paxb dx
d
e
aC1
dx Še
C2 b ‘‹ œ
d
dx aC1
C2 b œ
d
dx aC3 b
œ !.
' dxd avaxbc y1 axb y2 axb dbdx œ avaxbc y1 axb y2 axb db œ ' ! dx œ C
'
'
'
'
(c) y1 œ C1 e Paxb dx , y2 œ C# e Paxb dx , y œ y1 y2 . So yax0 b œ 0 Ê C1 e Paxb dx C# e Paxb dx œ !
Ê C1 C2 œ 0 Ê C1 œ C2 Ê y1 axb œ y2 axb for a x b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 10 CONIC SECTIONS AND POLAR COORDINATES
10.1 CONIC SECTIONS AND QUADRATIC EQUATIONS
1. x œ
y#
8
Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ 2
#
2. x œ y4 Ê 4p œ 4 Ê p œ 1; focus is (1ß 0), directrix is x œ 1
#
3. y œ x6 Ê 4p œ 6 Ê p œ
4. y œ
x#
2
Ê 4p œ 2 Ê p œ
1
#
3
#
; focus is ˆ!ß 3# ‰ , directrix is y œ
3
#
; focus is ˆ!ß 1# ‰ , directrix is y œ 1#
5.
x#
4
y#
9
œ 1 Ê c œ È4 9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x
6.
x#
4
y#
9
œ 1 Ê c œ È9 4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b
7.
x#
2
y# œ 1 Ê c œ È2 1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹
8.
y#
4
x# œ 1 Ê c œ È4 1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x
9. y# œ 12x Ê x œ
y#
1#
Ê 4p œ 12 Ê p œ 3;
focus is ($ß !), directrix is x œ 3
11. x# œ 8y Ê y œ
x#
8
Ê 4p œ 8 Ê p œ 2;
focus is (!ß 2), directrix is y œ 2
#
10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ
focus is ˆ!ß 3# ‰ , directrix is y œ 3#
#
3
#
;
y
12. y# œ 2x Ê x œ #
Ê 4p œ 2 Ê p œ
"
focus is ˆ # ß !‰ , directrix is x œ "#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
#
;
620
Chapter 10 Conic Sections and Polar Coordinates
13. y œ 4x# Ê y œ
x#
ˆ "4 ‰
"
4
Ê 4p œ
Ê pœ
"
16
#
14. y œ 8x# Ê y œ ˆx" ‰ Ê 4p œ
;
8
" ‰
"
focus is ˆ!ß 16
, directrix is y œ 16
#
15. x œ 3y# Ê x œ ˆy" ‰ Ê 4p œ
3
focus is ˆ 1"# ß !‰ , directrix is x œ
#
#
"
3
"
1#
y
17. 16x# 25y# œ 400 Ê #x5 16
œ1
Ê c œ Èa# b# œ È25 16 œ 3
#
19. 2x# y# œ 2 Ê x# y# œ 1
Ê c œ Èa# b# œ È2 1 œ 1
" ‰
focus is ˆ!ß 32
, directrix is y œ
Ê pœ
"
1#
;
16. x œ 2y# Ê x œ
y#
ˆ "# ‰
Ê 4p œ
"
#
"
8
"
3#
Ê pœ
focus is ˆ 8" ß !‰ , directrix is x œ 8"
#
#
x
18. 7x# 16y# œ 112 Ê 16
y7 œ 1
Ê c œ Èa# b# œ È16 7 œ 3
#
#
20. 2x# y# œ 4 Ê x# y4 œ 1
Ê c œ Èa# b# œ È4 2 œ È2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Ê pœ
"
8
;
"
32
;
Section 10.1 Conic Sections and Quadratic Equations
#
#
21. 3x# 2y# œ 6 Ê x# y3 œ 1
Ê c œ Èa# b# œ È3 2 œ 1
#
#
23. 6x# 9y# œ 54 Ê x9 y6 œ 1
Ê c œ Èa# b# œ È9 6 œ È3
#
#
x
22. 9x# 10y# œ 90 Ê 10
y9 œ 1
Ê c œ Èa# b# œ È10 9 œ 1
#
#
y
x
24. 169x# 25y# œ 4225 Ê 25
169
œ1
Ê c œ Èa# b# œ È169 25 œ 12
#
25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a# c# œ 4 ŠÈ2‹ œ 2 Ê
26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25 16 œ 9 Ê
27. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 ;
asymptotes are y œ „ x
x#
9
#
y#
#5
œ1
#
x
28. 9x# 16y# œ 144 Ê 16
y9 œ 1
Ê c œ Èa# b# œ È16 9 œ 5;
asymptotes are y œ „ 34 x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x#
4
y#
#
œ1
621
622
Chapter 10 Conic Sections and Polar Coordinates
29. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b#
œ È8 8 œ 4; asymptotes are y œ „ x
#
#
30. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b#
œ È4 4 œ 2È2; asymptotes are y œ „ x
31. 8x# 2y# œ 16 Ê x# y8 œ 1 Ê c œ Èa# b#
œ È2 8 œ È10 ; asymptotes are y œ „ 2x
32. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b#
œ È3 1 œ 2; asymptotes are y œ „ È3x
#
#
33. 8y# 2x# œ 16 Ê y# x8 œ 1 Ê c œ Èa# b#
œ È2 8 œ È10 ; asymptotes are y œ „ x
y
x
34. 64x# 36y# œ 2304 Ê 36
64
œ 1 Ê c œ Èa# b#
œ È36 64 œ 10; asymptotes are y œ „ 4
#
#
#
#
#
#
#
#
3
35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and
a
b
œ 1 Ê a œ b Ê c# œ a# b# œ 2a# Ê 2 œ 2a#
Ê a œ 1 Ê b œ 1 Ê y# x# œ 1
36. Foci: a „ 2ß !b , Asymptotes: y œ „
Ê 4œ
4a#
3
"
È3
x Ê c œ 2 and
Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê
x#
3
b
a
œ
"
È3
Ê bœ
a
È3
4
3
Ê c# œ a# b# œ a#
y# œ 1
37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and
b
a
œ
4
3
Ê bœ
(3) œ 4 Ê
38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and
a
b
œ
1
2
Ê b œ 2(2) œ 4 Ê
x#
9
y#
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y#
16
x#
16
œ1
œ1
a#
3
œ
4a#
3
Section 10.1 Conic Sections and Quadratic Equations
39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ 2,
focus is (#ß !), and vertex is (!ß 0); therefore the new
directrix is x œ 1, the new focus is (3ß 2), and the
new vertex is (1ß 2)
40. (a) x# œ 4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1,
focus is (!ß 1), and vertex is (!ß 0); therefore the new
directrix is y œ 4, the new focus is (1ß 2), and the
new vertex is (1ß 3)
41. (a)
x#
16
y#
9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
and (%ß !); c œ Èa# b# œ È7 Ê foci are ŠÈ7ß 0‹
and ŠÈ7ß !‹ ; therefore the new center is (%ß $), the
new vertices are (!ß 3) and (8ß 3), and the new foci are
Š4 „ È7ß $‹
42. (a)
x#
9
y#
25
œ 1 Ê center is (!ß 0), vertices are (0ß 5)
and (0ß 5); c œ Èa# b# œ È16 œ 4 Ê foci are
(!ß 4) and (!ß 4) ; therefore the new center is (3ß 2),
the new vertices are (3ß 3) and (3ß 7), and the new
foci are (3ß 2) and (3ß 6)
43. (a)
x#
16
y#
9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
and (4ß 0), and the asymptotes are x4 œ „ y3 or
Èa# b# œ È25 œ 5 Ê foci are
y œ „ 3x
4 ;cœ
(5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the
new vertices are (2ß 0) and (6ß 0), the new foci
are (3ß 0) and (7ß 0), and the new asymptotes are
yœ „
3(x 2)
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
623
624
Chapter 10 Conic Sections and Polar Coordinates
44. (a)
y#
4
x#
5
œ 1 Ê center is (!ß 0), vertices are (0ß 2)
and (0ß 2), and the asymptotes are
yœ „
2x
È5
y
2
œ „
x
È5
or
; c œ Èa# b# œ È9 œ 3 Ê foci are
(0ß 3) and (0ß 3) ; therefore the new center is (0ß 2),
the new vertices are (0ß 4) and (0ß 0), the new foci
are (0ß 1) and (0ß 5), and the new asymptotes are
2x
y2œ „ È
5
45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ 1, and vertex is (0ß 0); therefore the new
vertex is (2ß 3), the new focus is (1ß 3), and the new directrix is x œ 3; the new equation is
(y 3)# œ 4(x 2)
46. y# œ 12x Ê 4p œ 12 Ê p œ 3 Ê focus is (3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new
vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y 3)# œ 12(x 4)
47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ 2, and vertex is (0ß 0); therefore the new
vertex is (1ß 7), the new focus is (1ß 5), and the new directrix is y œ 9; the new equation is
(x 1)# œ 8(y 7)
Ê focus is ˆ!ß #3 ‰ , directrix is y œ 3# , and vertex is (0ß 0); therefore the new
vertex is (3ß 2), the new focus is ˆ3ß "# ‰ , and the new directrix is y œ 7# ; the new equation is
48. x# œ 6y Ê 4p œ 6 Ê p œ
3
#
(x 3)# œ 6(y 2)
49.
x#
6
y#
9
œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß 3); c œ Èa# b# œ È9 6 œ È3 Ê foci are Š!ß È3‹
and Š!ß È3‹ ; therefore the new center is (#ß 1), the new vertices are (2ß 2) and (#ß 4), and the new foci
are Š#ß 1 „ È3‹ ; the new equation is
50.
x#
2
(x 2)#
6
(y 1)#
9
œ1
y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and ŠÈ2ß !‹ ; c œ Èa# b# œ È2 1 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci
are (2ß 4) and (4ß 4); the new equation is
51.
x#
3
y#
#
(x 3)#
#
(y 4)# œ 1
œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and ŠÈ3ß !‹ ; c œ Èa# b# œ È3 2 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci
are (1ß 3) and (3ß 3); the new equation is
52.
x#
16
y#
#5
(x 2)#
3
(y 3)#
#
œ1
œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß 5); c œ Èa# b# œ È25 16 œ 3 Ê foci are
(0ß 3) and (0ß 3); therefore the new center is (4ß 5), the new vertices are (4ß 0) and (4ß 10), and the new
foci are (4ß 2) and (4ß 8); the new equation is
53.
x#
4
y#
5
(x 4)#
16
(y 5)#
#5
œ1
œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (2ß 0); c œ Èa# b# œ È4 5 œ 3 Ê foci are ($ß !) and
(3ß 0); the asymptotes are „
x
#
œ
y
È5
Ê yœ „
È5x
#
; therefore the new center is (2ß 2), the new vertices are
(4ß 2) and (0ß 2), and the new foci are (5ß 2) and (1ß 2); the new asymptotes are y 2 œ „
È5 (x 2)
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
; the new
Section 10.1 Conic Sections and Quadratic Equations
equation is
54.
x#
16
y#
9
(x 2)#
4
(y 2)#
5
625
œ1
œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (4ß 0); c œ Èa# b# œ È16 9 œ 5 Ê foci are (5ß !)
and (5ß 0); the asymptotes are „
x
4
œ
Ê yœ „
y
3
3x
4
; therefore the new center is (5ß 1), the new vertices are
(1ß 1) and (9ß 1), and the new foci are (10ß 1) and (0ß 1); the new asymptotes are y 1 œ „
the new equation is
(x 5)#
16
(y 1)#
9
3(x 5)
4
;
œ1
55. y# x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß 1); c œ Èa# b# œ È1 1 œ È2 Ê foci are
Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (1ß 1), the new vertices are (1ß 0) and
(1ß 2), and the new foci are Š1ß 1 „ È2‹ ; the new asymptotes are y 1 œ „ (x 1); the new equation is
(y 1)# (x 1)# œ 1
56.
y#
3
x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß È3‹ ; c œ Èa# b# œ È3 1 œ 2 Ê foci are (!ß #)
and (!ß 2); the asymptotes are „ x œ
y
È3
Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices
are Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y 3 œ „ È3 (x 1); the new
equation is
(y 3)#
3
(x 1)# œ 1
57. x# 4x y# œ 12 Ê x# 4x 4 y# œ 12 4 Ê (x 2)# y# œ 16; this is a circle: center at
C(2ß 0), a œ 4
58. 2x# 2y# 28x 12y 114 œ 0 Ê x# 14x 49 y# 6y 9 œ 57 49 9 Ê (x 7)# (y 3)# œ 1;
this is a circle: center at C(7ß 3), a œ 1
59. x# 2x 4y 3 œ 0 Ê x# 2x 1 œ 4y 3 1 Ê (x 1)# œ 4(y 1); this is a parabola:
V(1ß 1), F(1ß 0)
60. y# 4y 8x 12 œ 0 Ê y# 4y 4 œ 8x 12 4 Ê (y 2)# œ 8(x 2); this is a parabola:
V(#ß 2), F(!ß #)
61. x# 5y# 4x œ 1 Ê x# 4x 4 5y# œ 5 Ê (x 2)# 5y# œ 5 Ê
(x 2)#
5
y# œ 1; this is an ellipse: the
center is (2ß 0), the vertices are Š2 „ È5ß 0‹ ; c œ Èa# b# œ È5 1 œ 2 Ê the foci are (4ß 0) and (!ß 0)
#
62. 9x# 6y# 36y œ 0 Ê 9x# 6 ay# 6y 9b œ 54 Ê 9x# 6(y 3)# œ 54 Ê x6 (y 9 3) œ 1; this is an ellipse:
the center is (0ß 3), the vertices are (!ß 0) and (!ß 6); c œ Èa# b# œ È9 6 œ È3 Ê the foci are
Š0ß 3 „ È3‹
#
63. x# 2y# 2x 4y œ 1 Ê x# 2x 1 2 ay# 2y 1b œ 2 Ê (x 1)# 2(y 1)# œ 2
#
Ê (x1) (y 1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ;
2
c œ Èa# b# œ È2 1 œ 1 Ê the foci are (2ß 1) and (0ß 1)
64. 4x# y# 8x 2y œ 1 Ê 4 ax# 2x 1b y# 2y 1 œ 4 Ê 4(x 1)# (y 1)# œ 4
Ê (x 1)#
(y1)#
4
œ 1; this is an ellipse: the center is (1ß 1), the vertices are (1ß 3) and
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626
Chapter 10 Conic Sections and Polar Coordinates
(1ß 1); c œ Èa# b# œ È4 1 œ È3 Ê the foci are Š1ß " „ È3‹
65. x# y# 2x 4y œ 4 Ê x# 2x 1 ay# 4y 4b œ 1 Ê (x 1)# (y 2)# œ 1; this is a hyperbola:
the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š1 „ È2ß #‹ ;
the asymptotes are y 2 œ „ (x 1)
66. x# y# 4x 6y œ 6 Ê x# 4x 4 ay# 6y 9b œ 1 Ê (x 2)# (y 3)# œ 1; this is a hyperbola:
the center is (2ß 3), the vertices are (1ß 3) and (3ß 3); c œ Èa# b# œ È1 1 œ È2 Ê the foci are
Š2 „ È2ß 3‹ ; the asymptotes are y 3 œ „ (x 2)
67. 2x# y# 6y œ 3 Ê 2x# ay# 6y 9b œ 6 Ê
(y 3)#
6
x#
3
œ 1; this is a hyperbola: the center is (!ß $),
the vertices are Š!ß 3 „ È6‹ ; c œ Èa# b# œ È6 3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are
y 3
È6
œ „
x
È3
Ê y œ „ È2x 3
68. y# 4x# 16x œ 24 Ê y# 4 ax# 4x 4b œ 8 Ê
y#
8
(x 2)#
2
œ 1; this is a hyperbola: the center is (2ß 0),
the vertices are Š2ß „ È8‹ ; c œ Èa# b# œ È8 2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are
y
È8
œ „
x 2
È2
Ê y œ „ 2(x 2)
69.
70.
71.
72.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.1 Conic Sections and Quadratic Equations
627
74. kx# y# k Ÿ 1 Ê 1 Ÿ x# y# Ÿ 1 Ê 1 Ÿ x# y# and
x# y# Ÿ 1 Ê 1 y# x# and x# y# Ÿ 1
73.
75. Volume of the Parabolic Solid: V" œ '0 21x ˆh
bÎ2
œ
1hb#
8
76. y œ '
; Volume of the Cone: V# œ
w
H
x dx œ
w
H
#
Š x# ‹ C œ
wx#
2H
"
3
#
1 ˆ b# ‰ h œ
"
3
4h
b#
x# ‰ dx œ 21h '0 Šx
bÎ2
#
1 Š b4 ‹ h œ
1hb#
12
4x$
b# ‹
; therefore V" œ
C; y œ 0 when x œ 0 Ê 0 œ
w(0)#
2H
#
dx œ 21h ’ x2
3
#
bÎ2
x%
b# “ !
V#
C Ê C œ 0; therefore y œ
wx#
2H
is the
equation of the cable's curve
77. A general equation of the circle is x# y# ax by c œ 0, so we will substitute the three given points into
a
c œ 1 Þ
b c œ 1 ß
Ê c œ 43 and a œ b œ 73 ; therefore
this equation and solve the resulting system:
2a 2b c œ 8 à
3x# 3y# 7x 7y 4 œ 0 represents the circle
78. A general equation of the circle is x# y# ax by c œ 0, so we will substitute each of the three given points
2a 3b c œ 13 Þ
into this equation and solve the resulting system:
3a 2b c œ 13 ß
Ê a œ 2, b œ 2, and c œ 23;
4a 3b c œ 25 à
therefore x# y# 2x 2y 23 œ 0 represents the circle
79. r# œ (2 1)# (1 3)# œ 13 Ê (x 2)# (y 1)# œ 13 is an equation of the circle; the distance from the
center to (1.1ß 2.8) is È(# 1.1)# (1 2.8)# œ È12.85 È13 , the radius Ê the point is inside the circle
80. (x 2)# (y 1)# œ 5 Ê 2(x 2) 2(y 1)
dy
dx
œ0 Ê
dy
dx
2
#
#
œ yx
1 ; y œ 0 Ê (x 2) (0 1) œ 5
Ê (x 2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0
Ê (0 2)# (y 1)# œ 5 Ê (y 1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !).
At (4ß 0):
At (!ß !):
At (!ß #):
dy
dx
dy
dx
dy
dx
2
œ 40
1 œ 2 Ê the tangent line is y œ 2(x 4) or y œ 2x 8
2
œ 00
1 œ 2 Ê the tangent line is y œ 2x
2
œ 02
1 œ 2 Ê the tangent line is y 2 œ 2x or y œ 2x 2
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628
Chapter 10 Conic Sections and Polar Coordinates
81. (a) y# œ kx Ê x œ
y#
k
; the volume of the solid formed by
Èkx
revolving R" about the y-axis is V" œ '0
œ
1
k#
Èkx
'0
y% dy œ
1x# Èkx
5
#
#
1 Š yk ‹ dy
; the volume of the right
circular cylinder formed by revolving PQ about the
y-axis is V# œ 1x# Èkx Ê the volume of the solid
formed by revolving R# about the y-axis is
V$ œ V# V" œ
41x# Èkx
5
. Therefore we can see the
ratio of V$ to V" is 4:1.
(b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt
#
x
œ
1kx#
#
x
. The volume of the right circular cylinder formed by revolving PS about the x-axis is
#
V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is
1kx#
#
V$ œ V# V" œ 1kx#
œ
1kx#
#
. Therefore the ratio of V$ to V" is 1:1.
82. Let P" (pß y" ) be any point on x œ p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now
y# œ 4px Ê 2y
dy
dx
œ 4p Ê
dy
dx
œ
2p
y
Ê y# yy" œ 2px 2p# . Since x œ
Ê
"
#
tangents from P" are m" œ
y#
4p
2p
y" Èy#" 4p#
œ
dy
dx
œ
#
y
, we have y# yy" œ 2p Š 4p
‹ 2p# Ê y# yy" œ
2y" „ È4y#" 16p#
#
y# yy" 2p# œ 0 Ê y œ
y y"
x (p)
; then the slope of a tangent line from P" is
and m# œ
"
#
2p
y
y# 2p#
œ y" „ Èy#" 4p# . Therefore the slopes of the two
2p
y" Èy#" 4p#
Ê m" m# œ
4p#
y#" ay#" 4p# b
œ 1
Ê the lines are perpendicular
83. Let y œ É1
x#
4
on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by
A(x) œ 2x Š2É1
Ê Aw (x) œ 4É1
x#
4‹
x#
4
œ 4xÉ1
x#
É1 x4#
x#
4
(since the length is 2x and the height is 2y)
. Thus Aw (x) œ 0 Ê 4É1
x#
4
x#
É1 x4#
œ 0 Ê 4 Š1
x#
4‹
x# œ 0 Ê x# œ 2
Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4
is the maximum area when the length is 2È2 and the height is È2.
84. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root
#
Ê V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241
2
2
#
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root
#
Ê V œ 2'0 1 ŠÉ4 49 y# ‹ dy œ 2 '0 1 ˆ4 49 y# ‰ dy œ 21 4y
3
85. 9x# 4y# œ 36 Ê y# œ
œ
91
4
9x# 36
4
'24 ax# 4b dx œ 941 ’ x3
$
3
4
27
$
y$ ‘ ! œ 161
Ê y œ „ #3 Èx# 4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx# 4‹ dx
#
4
%
4x“ œ
#
91
4
ˆ 64
‰ ˆ8
‰‘ œ
3 16 3 8
91
4
ˆ 56
‰
3 8 œ
31
4
(56 24) œ 241
86. x# y# œ 1 Ê x œ „ È1 y# on the interval 3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1 y# ‰ dy œ 2'0 1 ˆÈ1 y# ‰ dy
3
œ 21'0 a1 y# b dy œ 21 ’y
3
$
y$
3 “!
#
œ 241
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3
#
Section 10.1 Conic Sections and Quadratic Equations
87. Let y œ É16
x# on the interval 3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a
16
9
É16
aµ
x ßµ
y b œ xß
#
vertical strip:
Ê mass œ dm œ $ dA œ $É16
#
É16 16
9 x
µ
y dm œ
#
Š$ É16
16
9
16
9
16
9
x#
, length œ É16
16
9
x# , width œ dx Ê area œ dA œ É16
16
9
x# dx
x# ‹ dx œ $ ˆ8 98 x# ‰ dx so the moment of the plate about the x-axis is
3
3
16
9
x# dx. Moment of the strip about the x-axis:
Mx œ ' µ
y dm œ 'c3 $ ˆ8 89 x# ‰ dx œ $ 8x
M œ 'c3 $ É16
629
8
27
$
x$ ‘ $ œ 32$ ; also the mass of the plate is
#
x# dx œ 'c3 4$ É1 ˆ "3 x‰ dx œ 4$ 'c1 3È1 u# du where u œ
3
1
x
3
Ê 3 du œ dx; x œ 3
Ê u œ 1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1 u# du œ 12$ 'c1 È1 u# du
1
œ 12$ ’ "2 ŠuÈ1 u# sin" u‹“
88. y œ Èx# 1 Ê
dy
dx
"
#
œ
È2
1
'
œ É 2x
x# 1 Ê S œ 0
#
–
89.
u œ È2x
— Ä
du œ È2 dx
drA
dt
œ
drB
dt
Ê
d
dt
21
È2
ax# 1b
"
"
1
œ 61$ Ê y œ
"Î#
(2x) œ
x
È x# 1
Mx
M
œ
32$
61$
#
œ
Ê Š dy
dx ‹ œ
È2
16
31
. Therefore the center of mass is ˆ!ß 3161 ‰ .
x#
x # 1
#
dy
Ê Ê1 Š dx
‹ œ É1
È2
dy
1
È #
'
21yÊ1 Š dx
‹ dx œ '0 21Èx# 1 É 2x
x# 1 dx œ 0 21 2x 1 dx ;
#
'02 Èu# 1 du œ È21
#
#
’ " ŠuÈu# 1 ln Šu Èu# 1‹‹“ œ
2 2
!
1
È2
90. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ;
œ
2p
y!
"
#
(4px)"Î# (4p) œ
(b) tan 9 œ mFP œ
œ
2p
È4px
Ê f w (x! ) œ
2p
È4px!
Ê tan " œ
(c) tan ! œ
’2È5 ln Š2 È5‹“
(rA rB ) œ 0 Ê rA rB œ C, a constant Ê the points P(t) lie on a hyperbola with foci at A
and B
f w (x) œ
x#
x# 1
2p
y! .
y! 0
y!
x! p œ x! p
tan 9 tan "
1 tan 9 tan "
y#! 2p(x! p)
y! (x! p 2p)
œ
œ
y!
2p
Šx
p c y ‹
!
!
y!
2p
1 b Šx
p‹ Šy ‹
!
4px! 2px! 2p#
y! (x! p)
!
œ
2p(x! p)
y! (x! p)
œ
2p
y!
91. PF will always equal PB because the string has constant length AB œ FP PA œ AP PB.
92. (a) In the labeling of the accompanying figure we have
y
1 œ tan t so the coordinates of A are (1ß tan t). The
coordinates of P are therefore (1 rß tan t). Since
1# y# œ (OA)# , we have 1# tan# t œ (1 r)#
Ê 1 r œ È1 tan# t œ sec t Ê r œ sec t 1.
The coordinates of P are therefore (xß y) œ (sec tß tan t)
Ê x# y# œ sec# t tan# t œ 1
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630
Chapter 10 Conic Sections and Polar Coordinates
(b) In the labeling of the accompany figure the coordinates
of A are (cos tß sin t), the coordinates of C are (1ß tan t),
and the coordinates of P are (1 dß tan t). By similar
triangles,
d
AB
œ
Ê
OC
OA
d
1 cos t
œ
È1 tan# t
1
Ê d œ (1 cos t)(sec t) œ sec t 1. The coordinates
of P are therefore (sec tß tan t) and P moves on the
hyperbola x# y# œ 1 as in part (a).
93. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (2pß p) and
(2pß p), and these points are È[2p (2p)]# (p p)# œ 4p units apart.
94. x lim
Š b x ba Èx# a# ‹ œ
Ä_ a
œ
b
a x lim
Ä_
’
x # ax # a # b
“
x È x # a#
œ
b
a x lim
Ä_
b
a x lim
Ä_
’
Šx Èx# a# ‹ œ
a#
“
x È x # a#
b
a x lim
Ä_
–
Šx Èx# a# ‹ Šx Èx# a# ‹
x È x # a#
œ0
10.2 CLASSIFYING CONIC SECTIONS BY ECCENTRICITY
#
y#
1. 16x# 25y# œ 400 Ê #x5 16
œ 1 Ê c œ Èa# b#
œ È25 16 œ 3 Ê e œ ca œ 35 ; F a „ 3ß 0b ;
directrices are x œ 0 „
œ „
a
e
5
ˆ 35 ‰
œ „
25
3
#
x#
2. 7x# 16y# œ 112 Ê 16
y7 œ 1 Ê c œ Èa# b#
œ È16 7 œ 3 Ê e œ ca œ 34 ; F a „ 3ß 0b ;
directrices are x œ 0 „
œ „
a
e
4
ˆ 34 ‰
œ „
16
3
3. 2x# y# œ 2 Ê x# y2 œ 1 Ê c œ Èa# b#
œ È2 1 œ 1 Ê e œ ca œ È12 ; F a0ß „ 1b ;
#
directrices are y œ 0 „
a
e
œ „
È2
Š È12 ‹
œ „2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
—
Section 10.2 Classifying Conic Sections by Eccentricity
4. 2x# y# œ 4 Ê
x#
#
œ 1 Ê c œ Èa# b#
y#
4
œ È4 2 œ È2 Ê e œ
directrices are y œ 0 „
a
e
c
a
œ
È2
2
; F Š0ß „ È2‹ ;
œ „ È22 œ „ 2È2
Š ‹
2
#
#
5. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b#
œ È3 2 œ 1 Ê e œ ca œ È13 ; F a0ß „ 1b ;
directrices are y œ 0 „
a
e
œ „
È3
œ „3
Š È13 ‹
#
x#
6. 9x# 10y# œ 90 Ê 10
y9 œ 1 Ê c œ Èa# b#
œ È10 9 œ 1 Ê e œ ca œ È110 ; F a „ 1ß 0b ;
directrices are x œ 0 „
7. 6x# 9y# œ 54 Ê
x#
9
a
e
œ „
y#
6
œ È9 6 œ È3 Ê e œ
directrices are x œ 0 „
a
e
È10
Š È110 ‹
œ „ 10
œ 1 Ê c œ Èa# b#
c
a
œ
È3
3
; F Š „ È3ß 0‹ ;
œ „ È33 œ „ 3È3
Š ‹
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
631
632
Chapter 10 Conic Sections and Polar Coordinates
y#
x#
8. 169x# 25y# œ 4225 Ê 25
169
œ 1 Ê c œ Èa# b#
œ È169 25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ;
a
directrices are y œ 0 „
a
e
œ „
13
13
ˆ 12
‰
13
œ „
169
12
x#
#7
y#
36
9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ
c
e
œ
3
0.5
œ 6 Ê b# œ 36 9 œ 27 Ê
10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ
c
e
œ
8
0.#
œ 40 Ê b# œ 1600 64 œ 1536 Ê
œ1
x#
1600
y#
1536
11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900 49 œ 4851 Ê
œ1
x#
4851
y#
4900
œ1
Šx
9
È5 ‹
12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100 5.76 œ 94.24
x#
100
Ê
y#
94.24
œ1
13. Focus: ŠÈ5ß !‹ , Directrix: x œ
Ê eœ
È5
3
. Then PF œ
PF
œ
Ê È(x
x
256 ‰
9
Ê
a
e
œ
5
9
Šx#
16
3
x
Ê
81
5 ‹
Ê c œ ae œ 4 and
0)#
œ
x# y# œ
16
3
(y
"
4
18
È5
È3
#
¸x
Ê
x#
ˆ 64
‰
3
4
9
a
e
È5
3
16
3
Ê
œ
ae
e#
#
œ
ae
e#
¹x
Ê
16
3
#
Ê (x 4) y œ
y#
ˆ 16
‰
3
9
È5
9
È5 ¹
x#
9
x# y# œ 4 Ê
œ
16 ¸
3
Ê
9
È5
PD Ê ÊŠx È5‹ (y 0)# œ
14. Focus: (%ß 0), Directrix: x œ
4)#
Ê c œ ae œ È5 and
#
È5
3
Ê x# 2È5 x 5 y# œ
È
œ #3 PD
3 ˆ #
32
4 x 3
9
È5
y#
4
4
e#
3
4
Ê
È5
e#
œ
Ê e# œ
9
È5
#
Ê Šx È5‹ y# œ
5
9
5
9
œ1
œ
16
3
ˆx
Ê e# œ
16 ‰#
3
Ê eœ
3
4
È3
#
. Then
#
Ê x 8x 16 y#
œ1
4
"
#
15. Focus: (%ß 0), Directrix: x œ 16 Ê c œ ae œ 4 and ae œ 16 Ê ae
e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ
PF œ 1 PD Ê È(x 4)# (y 0)# œ 1 kx 16k Ê (x 4)# y# œ 1 (x 16)# Ê x# 8x 16 y#
#
œ
1
4
#
#
ax 32x 256b Ê
3
4
#
#
x y œ 48 Ê
œ
"
#
1
È2
. Then PF œ
#
1
È2
y#
48
. Then
œ1
#
PD Ê ÊŠx È2‹ (y 0)# œ
Šx 2È2‹ Ê x# 2È2 x 2 y# œ
1
#
4
x#
64
16. Focus: ŠÈ2ß !‹ , Directrix: x œ 2È2 Ê c œ ae œ È2 and
Ê eœ
#
"
#
a
e
œ 2È 2 Ê
1
È2
ae
e#
œ 2È 2 Ê
È2
e#
œ 2 È 2 Ê e# œ
#
¹x 2È2¹ Ê Šx È2‹ y#
Šx# 4È2 x 8‹ Ê
"
#
x# y# œ 2 Ê
x#
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
y#
#
œ1
"
#
Section 10.2 Classifying Conic Sections by Eccentricity
17. e œ
Ê take c œ 4 and a œ 5; c# œ a# b#
4
5
Ê 16 œ 25 b# Ê b# œ 9 Ê b œ 3; therefore
x#
#5
y#
9
œ1
18. The eccentricity e for Pluto is 0.25 Ê e œ
c
a
œ 0.25 œ
"
4
#
Ê take c œ 1 and a œ 4; c# œ a# b# Ê 1 œ 16 b
#
#
Ê b# œ 15 Ê b œ È15 ; therefore, x y œ 1 is a
16
15
model of Pluto's orbit.
19. One axis is from A("ß ") to B("ß 7) and is 6 units long; the
other axis is from C($ß %) to D(1ß 4) and is 4 units long.
Therefore a œ 3, b œ 2 and the major axis is vertical. The
center is the point C("ß 4) and the ellipse is given by
(x1)#
4
(y4)#
9
œ 1; c# œ a# b# œ 3# 2# œ 5
Ê c œ È5 ; therefore the foci are F Š1ß 4 „ È5‹ , the
eccentricity is e œ
yœ4„
a
e
œ
c
a
È5
3
, and the directrices are
œ 4 „ È5 œ 4 „
Š ‹
3
9È 5
5
.
3
20. Using PF œ e † PD, we have È(x 4)# y# œ
œ
4
9
ax# 18x 81b Ê
5
9
2
3
kx 9k Ê (x 4)# y# œ
x# y# œ 20 Ê 5x# 9y# œ 180 or
x#
36
#
y
20
4
9
(x 9)# Ê x# 8x 16 y#
œ 1.
21. The ellipse must pass through (!ß 0) Ê c œ 0; the point (1ß 2) lies on the ellipse Ê a 2b œ 8. The ellipse
is tangent to the x-axis Ê its center is on the y-axis, so a œ 0 and b œ 4 Ê the equation is 4x# y# 4y œ 0.
Next, 4x# y# 4y 4 œ 4 Ê 4x# (y 24)# œ 4 Ê x#
(y 2)#
4
standard symbols) Ê c# œ a# b# œ 4 1 œ 3 Ê c œ È3 Ê e œ
œ 1 Ê a œ 2 and b œ 1 (now using the
c
a
œ
È3
#
.
22. We first prove a result which we will use: let m" , and
m# be two nonparallel, nonperpendicular lines. Let ! be
the acute angle between the lines. Then tan ! œ 1m" m"mm## .
To see this result, let )" be the angle of inclination of the
line with slope m" , and )# be the angle of inclination of the
line with slope m# . Assume m" m# . Then )" )# and we
have ! œ )" )# . Then tan ! œ tan ()" )# )
)" tan )#
m" m#
œ 1tan
tan )" tan )# œ 1 m" m# , since m" œ tan )" and and
m# œ tan )# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
633
634
Chapter 10 Conic Sections and Polar Coordinates
Now we prove the reflective property of ellipses (see the
x#
a#
accompanying figure): If
# #
# #
# #
b x a y œ a b and y œ
b
a
y#
b#
œ 1, then
Èa# x# Ê yw œ
bx
aÈ a# x#
.
Let P(x! ß y! ) be any point on the ellipse
Ê yw (x! ) œ
bx!
œ
aÉa# x#!
b # x !
a# y!
be the foci. Then mPF" œ
. Let F" (cß 0) and F# (cß 0)
y!
x! c
and mPF# œ
y!
x! c
. Let ! and
" be the angles between the tangent line and PF" and PF# ,
respectively. Then
b# x
tan ! œ
!
Œc a# y! c x!
c
!
y
b# x y
Š1 c a# y (x! ! c) ‹
! !
Similarly, tan " œ
b#
cy!
œ
b# x#! b# x! c a# y#!
a # y ! x ! a# y! c b# x! y!
œ
b# x! c ab# x#! a# y#! b
a # y ! c aa # b # b x ! y!
È2
1
b # x! c a# b #
a # y ! c c # x ! y !
œ
b#
cy!
.
. Since tan ! œ tan " , and ! and " are both less than 90°, we have ! œ " .
23. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 Ê e œ
œ
œ
c
a
œ È2 ; asymptotes are y œ „ x; F Š „ È2 ß !‹ ;
directrices are x œ 0 „
a
e
œ „
"
È2
#
x#
24. 9x# 16y# œ 144 Ê 16
y9 œ 1 Ê c œ Èa# b#
œ È16 9 œ 5 Ê e œ ca œ 54 ; asymptotes are
y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „
œ „
a
e
"6
5
#
#
25. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b#
œ È8 8 œ 4 Ê e œ ca œ È48 œ È2 ; asymptotes are
y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „
œ „
È8
È2
a
e
œ „2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.2 Classifying Conic Sections by Eccentricity
y#
4
26. y# x# œ 4 Ê
x#
4
œ 1 Ê c œ Èa# b#
œ È4 4 œ 2È2 Ê e œ
c
a
œ
2È 2
2
œ È2 ; asymptotes
are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „
œ „
2
È2
a
e
œ „ È2
27. 8x# 2y# œ 16 Ê
x#
2
y#
8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b#
c
a
œ
È10
È2
œ È5 ; asymptotes
are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „
œ „
È2
È5
635
œ „
a
e
2
È10
#
28. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b#
œ È3 1 œ 2 Ê e œ ca œ È23 ; asymptotes are
y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „
œ „
È3
Š È23 ‹
œ „
a
e
3
#
29. 8y# 2x# œ 16 Ê
y#
2
x#
8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b#
c
a
œ
È10
È2
œ È5 ; asymptotes
are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „
œ „
È2
È5
œ „
a
e
2
È10
y#
x#
30. 64x# 36y# œ 2304 Ê 36
64
œ 1 Ê c œ Èa# b#
5
œ È36 64 œ 10 Ê e œ ca œ 10
6 œ 3 ; asymptotes are
y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „
œ „
6
ˆ 53 ‰
œ „
a
e
18
5
31. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ
c
a
œ 3 Ê c œ 3a œ 3 Ê b# œ c# a# œ 9 1 œ 8 Ê y#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x#
8
œ1
636
Chapter 10 Conic Sections and Polar Coordinates
y#
1#
œ1
œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c# a# œ 9 1 œ 8 Ê x#
y#
8
œ1
32. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ
33. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ
c
a
34. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ
œ 25 16 œ 9 Ê
#
y
16
#
x
9
œ 2 Ê c œ 2a œ 4 Ê b# œ c# a# œ 16 4 œ 12 Ê
c
a
c
a
œ 1.25 œ
5
4
Ê cœ
a Ê 5œ
5
4
5
4
x#
4
a Ê a œ 4 Ê b# œ c# a#
œ1
4
#
È2 . Then
35. Focus (4ß 0) and Directrix x œ 2 Ê c œ ae œ 4 and ae œ 2 Ê ae
e# œ 2 Ê e# œ 2 Ê e œ # Ê e œ
PF œ È2 PD Ê È(x 4)# (y 0)# œ È2 kx 2k Ê (x 4)# y# œ 2(x 2)# Ê x# 8x 16 y#
œ 2 ax# 4x 4b Ê x# y# œ 8 Ê
x#
8
y#
8
œ1
36. Focus ŠÈ10ß !‹ and Directrix x œ È2 Ê c œ ae œ È10 and
a
e
œ È2 Ê
ae
e#
œ È2 Ê
È10
e#
œ È 2 Ê e# œ È 5
#
#
Ê e œ %È5 . Then PF œ %È5 PD Ê ÊŠx È10‹ (y 0)# œ %È5 ¹x È2¹ Ê Šx È10‹ y#
#
œ È5 Šx È2‹ Ê x# 2È10 x 10 y# œ È5 Šx# 2È2 x 2‹ Ê Š1 È5‹ x# y# œ 2È5 10
Ê
Š1 È5‹ x#
#È5 10
y#
2È5 10
œ1 Ê
x#
2È 5
y#
10 2È5
œ1
37. Focus (2ß 0) and Directrix x œ "# Ê c œ ae œ 2 and
a
e
œ
"
#
#
Ê
ae
e#
œ
"
#
Ê
PF œ 2PD Ê È(x 2)# (y 0)# œ 2 ¸x "# ¸ Ê (x 2) y# œ 4 ˆx
œ 4 ˆx# x "4 ‰ Ê 3x# y# œ 3 Ê x#
#
y
3
2
e#
" ‰#
#
œ
"
#
Ê e# œ 4 Ê e œ 2. Then
Ê x# 4x 4 y#
œ1
6
#
È3. Then
38. Focus (6ß 0) and Directrix x œ # Ê c œ ae œ 6 and ae œ # Ê ae
e# œ # Ê e# œ # Ê e œ 3 Ê e œ
PF œ È3 PD Ê È(x 6)# (y 0)# œ È3 kx 2k Ê (x 6)# y# œ 3(x 2)# Ê x# 12x 36 y#
x#
1#
œ 3 ax# 4x 4b Ê 2x# y# œ 24 Ê
39. È(x 1)# (y 3)# œ
3
#
y#
24
œ1
ky 2k Ê x# 2x 1 y# 6y 9 œ
Ê 4 ax# 2x 1b 5 ay# 12y 36b œ 4 4 180 Ê
40. c# œ a# b# Ê b# œ c# a# ; e œ
x#
a#
#
y
b#
œ" Ê
x#
a#
#
y
a # ae # 1 b
c
a
(y6)#
36
9
4
ay# 4y 4b Ê 4x# 5y# 8x 60y 4 œ 0
(x1)#
45
œ1
Ê c œ ea Ê c# œ e# a# Ê b# œ e# a# a# œ a# ae# 1b ; thus,
œ 1; the asymptotes of this hyperbola are y œ „ ae# 1bx Ê as e increases, the
absolute values of the slopes of the asymptotes increase and the hyperbola approaches a straight line.
41. To prove the reflective property for hyperbolas:
x#
a#
y#
b#
œ 1 Ê a# y# œ b# x# a# b# and
dy
dx
œ
xb#
ya#
.
Let P(x! ß y! ) be a point of tangency (see the accompanying
figure). The slope from P to F(cß 0) is x!y! c and from
P to F# (cß 0) it is
y!
x ! c
. Let the tangent through P meet
the x-axis in point A, and define the angles nF" PA œ !
and nF# PA œ " . We will show that tan ! œ tan " . From
the preliminary result in Exercise 22,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.3 Quadratic Equations and Rotations
x b#
tan ! œ
!
Πy! a# c x!
c
!
x b#
y!
1 b Š y! a# ‹ Š x
! c‹
!
y
tan " œ
y
!
Πx!
c
x! b#
y! a#
y!
x! b#
1 Šx
c ‹ Š y a# ‹
!
!
œ
œ
x#! b# x! b# cy#! a#
x ! y ! a # y ! a # c x! y! b #
b#
y! c
œ
a# b# x! b# c
x ! y ! c # y! a# c
œ
b#
y! c
. In a similar manner,
. Since tan ! œ tan " , and ! and " are acute angles, we have ! œ " .
42. From the accompanying figure, a ray of light emanating from
the focus A that met the parabola at P would be reflected
from the hyperbola as if it came directly from B
(Exercise 41). The same light ray would be reflected off the
ellipse to pass through B. Thus BPC is a straight line.
Let " be the angle of incidence of the light ray on the
hyperbola. Let ! be the angle of incidence of the light ray
on the ellipse. Note that ! " is the angle between the
tangent lines to the ellipse and hyperbola at P. Since BPC is
a straight line, 2! 2" œ 180°. Thus ! " œ 90°.
10.3 QUADRATIC EQUATIONS AND ROTATIONS
1. x# 3xy y# x œ 0 Ê B# 4AC œ (3)# 4(1)(1) œ 5 0 Ê Hyperbola
2. 3x# 18xy 27y# 5x 7y œ 4 Ê B# 4AC œ (18)# 4(3)(27) œ 0 Ê Parabola
3. 3x# 7xy È17y# œ 1 Ê B# 4AC œ (7)# 4(3) È17 ¸ 0.477 0 Ê Ellipse
#
4. 2x# È15 xy 2y# x y œ 0 Ê B# 4AC œ ŠÈ15‹ 4(2)(2) œ 1 0 Ê Ellipse
5. x# 2xy y# 2x y 2 œ 0 Ê B# 4AC œ 2# 4(1)(1) œ 0 Ê Parabola
6. 2x# y# 4xy 2x 3y œ 6 Ê B# 4AC œ 4# 4(2)(1) œ 24 0 Ê Hyperbola
7. x# 4xy 4y# 3x œ 6 Ê B# 4AC œ 4# 4(1)(4) œ 0 Ê Parabola
8. x# y# 3x 2y œ 10 Ê B# 4AC œ 0# 4(1)(1) œ 4 0 Ê Ellipse (circle)
9. xy y# 3x œ 5 Ê B# 4AC œ 1# 4(0)(1) œ 1 0 Ê Hyperbola
10. 3x# 6xy 3y# 4x 5y œ 12 Ê B# 4AC œ 6# 4(3)(3) œ 0 Ê Parabola
11. 3x# 5xy 2y# 7x 14y œ 1 Ê B# 4AC œ (5)# 4(3)(2) œ 1 0 Ê Hyperbola
12. 2x# 4.9xy 3y# 4x œ 7 Ê B# 4AC œ (4.9)# 4(2)(3) œ 0.01 0 Ê Hyperbola
13. x# 3xy 3y# 6y œ 7 Ê B# 4AC œ (3)# 4(1)(3) œ 3 0 Ê Ellipse
14. 25x# 21xy 4y# 350x œ 0 Ê B# 4AC œ 21# 4(25)(4) œ 41 0 Ê Hyperbola
15. 6x# 3xy 2y# 17y 2 œ 0 Ê B# 4AC œ 3# 4(6)(2) œ 39 0 Ê Ellipse
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
637
638
Chapter 10 Conic Sections and Polar Coordinates
16. 3x# 12xy 12y# 435x 9y 72 œ 0 Ê B# 4AC œ 12# 4(3)(12) œ 0 Ê Parabola
1
1
# Ê !œ 4 ;
È
È
xw sin ! yw cos ! Ê x œ xw #2 yw #2 , y
È
È
È
È
Š #2 xw #2 yw ‹ Š #2 xw #2 yw ‹ œ 2 Ê "#
17. cot 2! œ
yœ
Ê
18. cot 2! œ
AC
B
AC
B
œ
11
1
œ
therefore x œ xw cos ! yw sin !,
œ 0 Ê 2! œ
0
1
œ 0 Ê 2! œ
1
#
Ê !œ
È2
#
œ xw
xw # "# yw # œ 2 Ê xw # yw # œ 4 Ê Hyperbola
1
4
; therefore x œ xw cos ! yw sin !,
È2
w È2
w È2
# y # ,yœ x #
È
È
È
È
Š #2 xw #2 yw ‹ Š #2 xw #2 yw ‹
y œ xw sin ! yw cos ! Ê x œ xw
Ê Š
Ê
"
#
È2
#
xw
#
È2
#
yw ‹
È2
#
yw
yw
È2
#
È
Š #2
xw
xw # xw yw "# yw # "# xw # "# yw # "# xw # xw yw "# yw # œ 1 Ê
19. cot 2! œ
AC
B
31
2È 3
œ
œ
"
È3
1
3
Ê 2! œ
1
6
Ê !œ
È2
#
#
yw ‹ œ 1
xw # "# yw # œ 1 Ê 3xw # yw # œ 2 Ê Ellipse
3
#
; therefore x œ xw cos ! yw sin !,
È3 w
È3 w
" w
1 w
# x # y,yœ # x # y
È
È
Š #3 xw 1# yw ‹ Š 1# xw #3 yw ‹
y œ xw sin ! yw cos ! Ê x œ
Ê 3Š
È3
#
#
xw 1# yw ‹ 2È3
8È3 Š "# xw
20. cot 2! œ
AC
B
È3
#
œ
È3
#
È3
#
#
yw ‹ 8 Š
È3
#
xw "# yw ‹
yw ‹ œ 0 Ê 4xw # 16yw œ 0 Ê Parabola
12
È 3
œ
"
È3
#
1
3
Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
Ê Š
Š 1# xw
xw 1# yw ‹ È3 Š
È3
#
È3
#
1
6
Ê !œ
xw #1 yw , y œ
"
#
; therefore x œ xw cos ! yw sin !,
xw
È3
#
xw 1# yw ‹ Š 1# xw
È3
#
yw
yw ‹ 2 Š 1# xw
È3
#
#
yw ‹ œ 1 Ê
"
#
xw # 5# yw # œ 1
Ê xw # 5yw # œ 2 Ê Ellipse
21. cot 2! œ
AC
B
œ
11
2
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
Ê Š
È2
#
xw
#
È2
#
yw ‹ 2 Š
È2
#
È2
#
1
2
Ê !œ
È2
#
; therefore x œ xw cos ! yw sin !,
È2 w
È2 w
# x # y
È2 w
È2 w
È2 w
È2
# y ‹Š # x # y ‹ Š #
xw
xw
1
4
yw , y œ
xw
È2
#
#
yw ‹ œ 2 Ê yw # œ 1
Ê Parallel horizontal lines
22. cot 2! œ
AC
B
œ
31
2 È 3
œ È"3 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
Ê 3 Š 1# xw
È3
#
#
1
#
xw
yw ‹ 2È3 Š 1# xw
È3
#
È3
#
21
3
Ê !œ
yw , y œ
yw ‹ Š
È3
#
È3
#
1
3
; therefore x œ xw cos ! yw sin !,
xw 1# yw
xw 1# yw ‹ Š
È3
#
#
xw 1# yw ‹ œ 1 Ê 4yw # œ 1
Ê Parallel horizontal lines
23. cot 2! œ
AC
B
œ
È2 È2
2È 2
y œ xw sin ! yw cos ! Ê x œ
È
Ê È 2 Š # 2 xw
È2
#
Ê !œ
#
È2
#
xw
È2
#
yw , y œ
È2
#
xw
È2
#
yw ‹ œ 0 Ê 2È2xw # 8È2 yw œ 0 Ê Parabola
yw ‹ 2È2 Š
È2
#
xw
È2
#
yw ‹ 8 Š
24. cot 2! œ
AC
B
œ
00
1
8Š
1
2
œ 0 Ê 2! œ
È2
#
xw
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
È2
#
1
2
xw
; therefore x œ xw cos ! yw sin !,
È2 w
È2 w
# x # y
È2 w
È2 w
È2 w
# y ‹Š # x # y ‹
Ê !œ
È2
#
1
4
1
4
È2 Š
È2
#
xw
È2
#
yw ‹
#
; therefore x œ xw cos ! yw sin !,
yw , y œ
È2
#
xw
È2
#
yw
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.3 Quadratic Equations and Rotations
Ê Š
È2
#
È2
#
xw
yw ‹ Š
È2
#
xw
È2
#
yw ‹ Š
È2
#
xw
È2
#
yw ‹ Š
È2
#
xw
È2
#
639
yw ‹ 1 œ 0 Ê xw # yw # 2È2 xw 2
œ 0 Ê Hyperbola
25. cot 2! œ
AC
B
œ
33
2
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
Ê 3Š
È2
#
xw
È2
#
#
yw ‹ 2 Š
È2
#
È2
#
1
2
xw
xw
1
4
Ê !œ
È2
#
È2
#
; therefore x œ xw cos ! yw sin !,
È2
#
yw , y œ
yw ‹ Š
È2
#
xw +
xw
È2
#
È2
#
yw
yw ‹ 3 Š
È2
#
xw
È2
#
#
yw ‹ œ 19 Ê 4xw # 2yw # œ 19
Ê Ellipse
26. cot 2! œ
AC
B
œ
3 (1)
4È 3
œ
"
È3
Ê 2! œ
È3
#
È3
È
4 3Š #
y œ xw sin ! yw cos ! Ê x œ
Ê 3Š
È3
#
#
xw 1# yw ‹
1
3
Ê !œ
xw #1 yw , y œ
1
#
1
6
; therefore x œ xw cos ! yw sin !,
xw
xw 1# yw ‹ Š 1# xw
È3
#
È3
#
yw
yw ‹ Š 1# xw
È3
#
#
yw ‹ œ 7 Ê 5xw # 3yw # œ 7
Ê Hyperbola
27. cot 2! œ
14 2
16
œ
3
4
Ê cos 2! œ
2!
and cos ! œ É 1 cos
σ
#
28. cot 2! œ
σ
AC
B
1 ˆ 35 ‰
#
29. tan 2! œ
œ
œ
"
13
w
4"
4
2
È5
1ˆ 35 ‰
#
3
5
2!
(if we choose 2! in Quadrant I); thus sin ! œ É 1 cos
σ
2
œ
2
È5
(or sin ! œ
2
È5
and cos ! œ
1 ˆ 35 ‰
#
œ
"
È5
"
È5 )
2!
œ 34 Ê cos 2! œ 35 (if we choose 2! in Quadrant II); thus sin ! œ É 1 cos
2
2!
and cos ! œ É 1 cos
σ
#
1 ˆ 35 ‰
#
œ
1
È5
(or sin ! œ
1
È5
and cos ! œ
2
È5 )
"
#
Ê 2! ¸ 26.57° Ê ! ¸ 13.28° Ê sin ! ¸ 0.23, cos ! ¸ 0.97; then Aw ¸ 0.9, Bw ¸ 0.0,
œ
"
5
œ
Cw ¸ 3.1, D ¸ 0.7, Ew ¸ 1.2, and Fw œ 3 Ê 0.9 xw # 3.1 yw # 0.7xw 1.2yw 3 œ 0, an ellipse
30. tan 2! œ
"
2 (3)
Ê 2! ¸ 11.31° Ê ! ¸ 5.65° Ê sin ! ¸ 0.10, cos ! ¸ 1.00; then Aw ¸ 2.1, Bw ¸ 0.0,
Cw ¸ 3.1, Dw ¸ 3.0, Ew ¸ 0.3, and Fw œ 7 Ê 2.1 xw # 3.1 yw # 3.0xw 0.3yw 7 œ 0, a hyperbola
31. tan 2! œ
4
14
w
œ
Ê 2! ¸ 53.13° Ê ! ¸ 26.5(° Ê sin ! ¸ 0.45, cos ! ¸ 0.89; then Aw ¸ 0.0, Bw ¸ 0.0,
4
3
Cw ¸ 5.0, D ¸ 0, Ew ¸ 0, and Fw œ 5 Ê 5.0 yw # 5 œ 0 or yw œ „ 1.0, parallel lines
32. tan 2! œ
12
2 18
w
œ
3
4
Ê 2! ¸ 36.87° Ê ! ¸ 18.43° Ê sin ! ¸ 0.32, cos ! ¸ 0.95; then Aw ¸ 0.0, Bw ¸ 0.0,
Cw ¸ 20.1, D ¸ 0, Ew ¸ 0, and Fw œ 49 Ê 20.1 yw # 49 œ 0, parallel lines
33. tan 2! œ
œ 5 Ê 2! ¸ 78.69° Ê ! ¸ 39.35° Ê sin ! ¸ 0.63, cos ! ¸ 0.77; then Aw ¸ 5.0, Bw ¸ 0.0,
5
3 2
Cw ¸ 0.05, Dw ¸ 5.0, Ew ¸ 6.2, and Fw œ 1 Ê 5.0 xw # 0.05 yw # 5.0xw 6.2yw 1 œ 0, a hyperbola
34. tan 2! œ
7
29
w
œ 1 Ê 2! ¸ 45.00° Ê ! ¸ 22.5° Ê sin ! ¸ 0.38, cos ! ¸ 0.92; then Aw ¸ 0.5, Bw ¸ 0.0,
Cw ¸ 10.4, D ¸ 18.4, Ew ¸ 7.6, and Fw œ 86 Ê 0.5 xw # 10.4ayw b# 18.4xw 7.6yw 86 œ 0, an ellipse
35. ! œ 90° Ê x œ xw cos 90° yw sin 90° œ yw and y œ xw sin 90° yw cos 90° œ xw
(a)
xw #
b#
yw #
a#
œ1
(b)
yw #
a#
xw #
b#
œ1
(c) xw # yw # œ a#
(d) y œ mx Ê y mx œ 0 Ê D œ m and E œ 1; ! œ 90° Ê Dw œ 1 and Ew œ m Ê myw xw œ 0 Ê yw œ m" xw
(e) y œ mx b Ê y mx b œ 0 Ê D œ m and E œ 1; ! œ 90° Ê Dw œ 1, Ew œ m and Fw œ b
Ê myw xw b œ 0 Ê yw œ m" xw mb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
640
Chapter 10 Conic Sections and Polar Coordinates
36. ! œ 180° Ê x œ xw cos 180° yw sin 180° œ xw and y œ xw sin 180° yw cos 180° œ yw
(a)
xw #
a#
yw #
b#
œ1
(b)
xw #
a#
yw #
b#
(c) xw # yw # œ a#
œ1
(d) y œ mx Ê y mx œ 0 Ê D œ m and E œ 1; ! œ 180° Ê Dw œ m and Ew œ 1 Ê yw mxw œ 0 Ê
yw œ mxw
(e) y œ mx b Ê y mx b œ 0 Ê D œ m and E œ 1; ! œ 180° Ê Dw œ m, Ew œ 1 and Fw œ b
Ê yw mxw b œ 0 Ê yw œ mxw b
37. (a) Aw œ cos 45° sin 45° œ Š
" w#
"
# x #
Aw œ "# , Cw œ
Ê
(b)
c
a
œ
2È 2
#
œ
"
#
, Bw œ 0, Cw œ cos 45° sin 45° œ "# , Fw œ 1
yw # œ 1 Ê xw # yw # œ 2
"# (see part (a) above), Dw œ Ew œ Bw œ 0, Fw œ a Ê
38. xy œ 2 Ê xw # yw # œ 4 Ê
Ê eœ
È2
È2
# ‹Š # ‹
xw #
4
yw #
4
"
#
xw # "# yw # œ a Ê xw # yw # œ 2a
œ 1 (see Exercise 37(b)) Ê a œ 2 and b œ 2 Ê c œ È4 4 œ 2È2
œ È2
39. Yes, the graph is a hyperbola: with AC 0 we have 4AC 0 and B# 4AC 0.
40. The one curve that meets all three of the stated criteria is the ellipse x# 4xy 5y# 1 œ 0. The reasoning:
The symmetry about the origin means that (xß y) lies on the graph whenever (xß y) does. Adding
Ax# Bxy Cy# Dx Ey F œ 0 and A(x)# B(x)(y) C(y)# D(x) E(y) F œ 0 and dividing
the result by 2 produces the equivalent equation Ax# Bxy Cy# F œ 0. Substituting x œ 1, y œ 0 (because
the point (1ß 0) lies on the curve) shows further that A œ F. Then Fx# Bxy Cy# F œ 0. By implicit
differentiation, 2Fx By Bxyw 2Cyyw œ 0, so substituting x œ 2, y œ 1, and yw œ 0 (from Property 3)
gives 4F B œ 0 Ê B œ 4F Ê the conic is Fx# 4Fxy Cy# F œ 0. Now substituting x œ 2 and y œ 1
again gives 4F 8F C F œ 0 Ê C œ 5F Ê the equation is now Fx# 4Fxy 5Fy# F œ 0. Finally,
dividing through by F gives the equation x# 4xy 5y# 1 œ 0.
41. Let ! be any angle. Then Aw œ cos# ! sin# ! œ 1, Bw œ 0, Cw œ sin# ! cos# ! œ 1, Dw œ Ew œ 0 and Fw œ a#
Ê xw # yw # œ a# .
42. If A œ C, then Bw œ B cos 2! (C A) sin 2! œ B cos 2!. Then ! œ
1
4
Ê 2! œ
1
#
Ê Bw œ B cos
1
#
œ 0 so the
xy-term is eliminated.
43. (a) B# 4AC œ 4# 4(1)(4) œ 0, so the discriminant indicates this conic is a parabola
(b) The left-hand side of x# 4xy 4y# 6x 12y 9 œ 0 factors as a perfect square: (x 2y 3)# œ 0
Ê x 2y 3 œ 0 Ê 2y œ x 3; thus the curve is a degenerate parabola (i.e., a straight line).
44. (a) B# 4AC œ 6# 4(9)(1) œ 0, so the discriminant indicates this conic is a parabola
(b) The left-hand side of 9x# 6xy y# 12x 4y 4 œ 0 factors as a perfect square: (3x y 2)# œ 0
Ê 3x y 2 œ 0 Ê y œ 3x 2; thus the curve is a degenerate parabola (i.e., a straight line).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.3 Quadratic Equations and Rotations
45. (a) B# 4AC œ 1 4(0)(0) œ 1 Ê hyperbola
(b) xy 2x y œ 0 Ê y(x 1) œ 2x Ê y œ
(c) y œ
2x
x1
Ê
dy
dx
œ
2
(x 1)#
and we want
the slope of y œ 2x Ê 2 œ (x#1)
1
dy
Š dx
‹
2x
x1
œ 2,
#
Ê (x 1)# œ 4 Ê x œ 3 or x œ 1; x œ 3
Ê y œ 3 Ê (3ß 3) is a point on the hyperbola
where the line with slope m œ 2 is normal
Ê the line is y 3 œ 2(x 3) or y œ 2x 3;
x œ 1 Ê y œ 1 Ê (1ß 1) is a point on the
hyperbola where the line with slope m œ 2 is
normal Ê the line is y 1 œ 2(x 1) or
y œ 2x 3
46. (a) False: let A œ C œ 1, B œ 2 Ê B# 4AC œ 0 Ê parabola
(b) False: see part (a) above
(c) True: AC 0 Ê 4AC 0 Ê B# 4AC 0 Ê hyperbola
47. Assume the ellipse has been rotated to eliminate the xy-term Ê the new equation is Aw xw # Cw yw # œ 1 Ê the
semi-axes are É A" and É C" Ê the area is 1 ŠÉ A" ‹ ŠÉ C" ‹ œ
w
w
w
w
1
ÈA C
œ Bw # 4Aw Cw œ 4Aw Cw (because Bw œ 0) we find that the area is
w
w
œ
21
È4A C
21
È4AC B#
w
w
. Since B# 4AC
as claimed.
48. (a) Aw Cw œ aA cos# ! B cos ! sin ! C sin# !b aA sin# ! B cos ! sin ! C sin# !b
œ A acos# ! sin# !b C asin# ! cos# !b œ A C
(b) Dw # Ew # œ (D cos ! E sin !)# (D sin ! E cos !)# œ D# cos# ! 2DE cos ! sin ! E# sin# !
D# sin# ! 2DE sin ! cos ! E# cos# ! œ D# acos# ! sin# !b E# asin# ! cos# !b œ D# E#
49. Bw # 4Aw Cw
œ aB cos 2! (C A) sin 2!b# 4 aA cos# ! B cos ! sin ! C sin# !b aA sin# ! B cos ! sin ! C cos# !b
œ B# cos# 2! 2B(C A) sin 2! cos 2! (C A)# sin# 2! 4A# cos# ! sin# ! 4AB cos$ ! sin !
4AC cos% ! 4AB cos ! sin$ ! 4B# cos# ! sin# ! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ !
4C# cos# ! sin# !
#
œ B cos# 2! 2BC sin 2! cos 2! 2AB sin 2! cos 2! C# sin# 2! 2AC sin# 2! A# sin# 2!
4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! B# sin# 2! 4BC cos$ ! sin !
4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# !
#
œ B 2BC(2 sin ! cos !) acos# ! sin# !b 2AB(2 sin ! cos !) acos# ! sin# !b C# a4 sin# ! cos# !b
2AC a4 sin# ! cos# !b A# a4 sin# ! cos# !b 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% !
4AB cos ! sin$ ! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# !
#
œ B 8AC sin# ! cos# ! 4AC cos% ! 4AC sin% !
œ B# 4AC acos% ! 2 sin# ! cos# ! sin% !b
œ B# 4AC acos# ! sin# !b
œ B# 4AC
#
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Chapter 10 Conic Sections and Polar Coordinates
10.4 CONICS AND PARAMETRIC EQUATIONS; THE CYCLOID
1. x œ cos t, y œ sin t, 0 Ÿ t Ÿ 1
Ê cos# t sin# t œ 1 Ê x# y# œ 1
2. x œ sin (21(1 t)), y œ cos (21(1 t)), 0 Ÿ t Ÿ 1
Ê sin# (21(1 t)) cos# (21(1 t)) œ 1
Ê x# y# œ 1
3. x œ 4 cos t, y œ 5 sin t, 0 Ÿ t Ÿ 1
4. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21
Ê
16 cos# t
16
25 sin# t
25
5. x œ t, y œ Èt, t
x#
16
œ1 Ê
y#
25
œ1
0 Ê y œ Èx
Ê
16 sin# t
16
25 cos# t
25
œ1 Ê
x#
16
6. x œ sec# t 1, y œ tan t, 1# t
Ê sec# t 1 œ tan# t Ê x œ y#
7. x œ sec t, y œ tan t, 1# t
#
#
#
1
#
#
Ê sec t tan t œ 1 Ê x y œ 1
y#
#5
œ1
1
#
8. x œ csc t, y œ cot t, 0 t 1
Ê 1 cot# t œ csc# t Ê 1 y# œ x# Ê x# y# œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.4 Conics and Parametric Equations; The Cycloid
9. x œ t, y œ È4 t# , 0 Ÿ t Ÿ 2
Ê y œ È4 x#
10. x œ t# , y œ Èt% 1, t 0
Ê y œ Èx# 1, x 0
11. x œ cosh t, y œ sinh t, _ 1 _
Ê cosh# t sinh# t œ 1 Ê x# y# œ 1
12. x œ 2 sinh t, y œ 2 cosh t, _ t _
Ê 4 cosh# t 4 sinh# t œ 4 Ê y# x# œ 4
13. Arc PF œ Arc AF since each is the distance rolled and
Arc PF
œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ )
b
Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ
nOCG œ
1
#
a
b
);
); nOCG œ nOCP nPCE
œ nOCP ˆ 1# !‰ . Now nOCP œ 1 nFCP
œ 1 ba ). Thus nOCG œ 1 ba )
œ 1 ba )
1
#
1
#
! Ê
! Ê ! œ 1 ba ) ) œ 1
1
# )
ˆ ab b )‰ .
Then x œ OG BG œ OG PE œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1
œ (a b) cos ) b cos ˆ a b b )‰ . Also y œ EG œ CG CE œ (a b) sin ) b sin !
ab
b
)‰
œ (a b) sin ) b sin ˆ1 a b b )‰ œ (a b) sin ) b sin ˆ a b b )‰ . Therefore
x œ (a b) cos ) b cos ˆ a b b )‰ and y œ (a b) sin ) b sin ˆ a b b )‰ .
If b œ 4a , then x œ ˆa 4a ‰ cos )
œ
œ
œ
œ
3a
4
3a
4
3a
4
3a
4
cos )
œ
œ
œ
3a
4
3a
4
3a
4
3a
4
cos 3) œ
3a
4
cos Š
a ˆ 4a ‰
ˆ 4a ‰
)‹
cos ) 4a (cos ) cos 2) sin ) sin 2))
cos ) a(cos )) acos# ) sin# )b (sin ))(2 sin ) cos ))b
a
2a
#
#
4 cos ) sin ) 4 sin ) cos )
#
$
) cos$ ) 3a
4 (cos )) a1 cos )b œ a cos );
a ˆ 4a ‰
a‰
a
3a
a
3a
4 sin ) 4 sin Š ˆ 4a ‰ )‹ œ 4 sin ) 4 sin 3) œ 4
cos )
cos
y œ ˆa
œ
a
4
a
4
a
4
a
4
a
4
cos$ )
sin ) 4a (sin ) cos 2) cos ) sin 2))
sin ) 4a a(sin )) acos# ) sin# )b (cos ))(2 sin ) cos ))b
sin )
sin )
sin )
a
4
3a
4
3a
4
sin ) cos# )
sin ) cos# )
a
4
a
4
#
sin$ )
2a
4
cos# ) sin )
sin$ )
(sin )) a1 sin )b
a
4
sin$ ) œ a sin$ ).
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Chapter 10 Conic Sections and Polar Coordinates
14. P traces a hypocycloid where the larger radius is 2a and the smaller is a Ê x œ (2a a) cos ) a cos ˆ 2a a a )‰
œ 2a cos ), 0 Ÿ ) Ÿ 21, and y œ (2a a) sin ) a sin ˆ 2a a a )‰ œ a sin ) a sin ) œ 0. Therefore P traces the
diameter of the circle back and forth as ) goes from 0 to 21.
15. Draw line AM in the figure and note that nAMO is a right
angle since it is an inscribed angle which spans the diameter
of a circle. Then AN# œ MN# AM# . Now, OA œ a,
AN
AM
a œ tan t, and a œ sin t. Next MN œ OP
Ê OP# œ AN# AM# œ a# tan# t a# sin# t
Ê OP œ Èa# tan# t a# sin# t
œ (a sin t)Èsec# t 1 œ
a sin$ t
cos t œ
#
x œ OP sin t œ
a sin# t
cos t
#
. In triangle BPO,
a sin t tan t and
y œ OP cos t œ a sin t Ê x œ a sin# t tan t and y œ a sin# t.
16. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid
(see the accompanying figure).
Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes
parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and
yw œ b sin !, where ! œ 3#1 ). It follows that xw œ b cos ˆ 3#1 )‰ œ b sin ) and yw œ b sin ˆ 3#1 )‰
œ b cos ) Ê x œ h xw œ a) b sin ) and y œ k yw œ a b cos ) are parametric equations of the trochoid.
#
#
#
17. D œ É(x 2)# ˆy "# ‰ Ê D# œ (x 2)# ˆy "# ‰ œ (t 2)# ˆt# "# ‰ Ê D# œ t% 4t
Ê
d aD # b
dt
17
4
œ 4t$ 4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local
minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest
point on the parabola is (1ß 1).
#
#
18. D œ Ɉ2 cos t 34 ‰ (sin t 0)# Ê D# œ ˆ2 cos t 34 ‰ sin# t Ê
d aD # b
dt
œ 2 ˆ2 cos t 34 ‰ (2 sin t) 2 sin t cos t œ (2 sin t) ˆ3 cos t 3# ‰ œ 0 Ê 2 sin t œ 0 or 3 cos t
Ê t œ 0, 1 or t œ
#
#
1
3
,
51
3
. Now
#
#
d aD b
dt#
œ 6 cos# t 3 cos t 6 sin# t so that
#
#
#
d aD b
dt#
3
#
œ0
(0) œ 3 Ê relative
#
maximum, d dtaD# b (1) œ 9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and
d # aD # b ˆ 5 1 ‰
œ 9# Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse
dt#
3
È
È
the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß #3 ‹ are the desired points.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
closest to
Section 10.4 Conics and Parametric Equations; The Cycloid
19. (a)
(b)
(c)
20. (a)
(b)
(c)
(b)
(c)
21.
22. (a)
23. (a)
(b)
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Chapter 10 Conic Sections and Polar Coordinates
24. (a)
25. (a)
(b)
(b)
(c)
26. (a)
(b)
(c)
(d)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.5 Polar Coordinates
10.5 POLAR COORDINATES
1. a, e; b, g; c, h; d, f
2. a, f; b, h; c, g; d, e
3. (a) ˆ2ß 1# 2n1‰ and ˆ2ß 1# (2n 1)1‰ , n an integer
(b) (#ß 2n1) and (#ß (2n 1)1), n an integer
(c) ˆ2ß 3#1 2n1‰ and ˆ2ß 3#1 (2n 1)1‰ , n an integer
(d) (#ß (2n 1)1) and (#ß 2n1), n an integer
4. (a) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer
(b) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer
(c) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
(d) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0)
(b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0)
(c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹
3
(d) x œ r cos ) œ 2 cos
71
3
3
œ 1, y œ r sin ) œ 2 sin
71
3
œ È3 Ê Cartesian coordinates are Š1ß È3‹
(e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0)
(f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹
3
3
(g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0)
(h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹
3
6. (a) x œ È2 cos
1
4
œ 1, y œ È2 sin
3
1
4
œ 1 Ê Cartesian coordinates are (1ß 1)
(b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0)
(c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0)
(d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1)
4
(e) x œ 3 cos
51
6
œ
4
3È 3
2
, y œ 3 sin
51
6
È
œ 3# Ê Cartesian coordinates are Š 3 # 3 ß 3# ‹
(f) x œ 5 cos ˆtan" 43 ‰ œ 3, y œ 5 sin ˆtan" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4)
(g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0)
(h) x œ 2È3 cos 231 œ È3, y œ 2È3 sin 231 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹
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Chapter 10 Conic Sections and Polar Coordinates
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
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Section 10.5 Polar Coordinates
649
22.
23. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0)
24. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1)
25. r sin ) œ 0 Ê y œ 0, the x-axis
26. r cos ) œ 0 Ê x œ 0, the y-axis
27. r œ 4 csc ) Ê r œ
4
sin )
28. r œ 3 sec ) Ê r œ
Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4)
3
cos )
Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0)
29. r cos ) r sin ) œ 1 Ê x y œ 1, line with slope m œ 1 and intercept b œ 1
30. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0
31. r# œ 1 Ê x# y# œ 1, circle with center C œ (!ß 0) and radius 1
32. r# œ 4r sin ) Ê x# y# œ 4y Ê x# y# 4y 4 œ 4 Ê x# (y 2)# œ 4, circle with center C œ (0ß 2) and radius 2
33. r œ
5
sin )2 cos )
Ê r sin ) 2r cos ) œ 5 Ê y 2x œ 5, line with slope m œ 2 and intercept b œ 5
34. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x
)‰ˆ " ‰
35. r œ cot ) csc ) œ ˆ cos
Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0)
sin )
sin )
which opens to the right
sin ) ‰
36. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos
Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with
#)
vertex œ (!ß 0) which opens upward
37. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function
38. r sin ) œ ln r ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function
39. r# 2r# cos ) sin ) œ 1 Ê x# y# 2xy œ 1 Ê x# 2xy y# œ 1 Ê (x y)# œ 1 Ê x y œ „ 1, two parallel
straight lines of slope 1 and y-intercepts b œ „ 1
40. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular
lines through the origin with slopes 1 and 1, respectively.
41. r# œ 4r cos ) Ê x# y# œ 4x Ê x# 4x y# œ 0 Ê x# 4x 4 y# œ 4 Ê (x 2)# y# œ 4, a circle with
center C(2ß 0) and radius 2
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Chapter 10 Conic Sections and Polar Coordinates
42. r# œ 6r sin ) Ê x# y# œ 6y Ê x# y# 6y œ 0 Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9, a circle with
center C(0ß 3) and radius 3
43. r œ 8 sin ) Ê r# œ 8r sin ) Ê x# y# œ 8y Ê x# y# 8y œ 0 Ê x# y# 8y 16 œ 16
Ê x# (y 4)# œ 16, a circle with center C(0ß 4) and radius 4
44. r œ 3 cos ) Ê r# œ 3r cos ) Ê x# y# œ 3x Ê x# y# 3x œ 0 Ê x# 3x
#
Ê ˆx 3# ‰ y# œ
9
4
, a circle with center C ˆ 3# ß !‰ and radius
9
4
y# œ
9
4
3
#
45. r œ 2 cos ) 2 sin ) Ê r# œ 2r cos ) 2r sin ) Ê x# y# œ 2x 2y Ê x# 2x y# 2y œ 0
Ê (x 1)# (y 1)# œ 2, a circle with center C(1ß 1) and radius È2
46. r œ 2 cos ) sin ) Ê r# œ 2r cos ) r sin ) Ê x# y# œ 2x y Ê x# 2x y# y œ 0
#
Ê (x 1)# ˆy "# ‰ œ 54 , a circle with center C ˆ1ß "# ‰ and radius
È5
#
È
47. r sin ˆ) 16 ‰ œ 2 Ê r ˆsin ) cos 16 cos ) sin 16 ‰ œ 2 Ê #3 r sin ) "# r cos ) œ 2 Ê
Ê È3 y x œ 4, line with slope m œ " and intercept b œ 4
È3
È3
#
È3
È
48. r sin ˆ 231 )‰ œ 5 Ê r ˆsin 231 cos ) cos 231 sin )‰ œ 5 Ê #3 r cos ) "# r sin ) œ 5 Ê
Ê È3 x y œ 10, line with slope m œ È3 and intercept b œ 10
49. x œ 7 Ê r cos ) œ 7
51. x œ y Ê r cos ) œ r sin ) Ê ) œ
y "# x œ 2
È3
#
x "# y œ 5
50. y œ 1 Ê r sin ) œ 1
1
4
52. x y œ 3 Ê r cos ) r sin ) œ 3
53. x# y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2
54. x# y# œ 1 Ê r# cos# ) r# sin# ) œ 1 Ê r# acos# ) sin# )b œ 1 Ê r# cos 2) œ 1
55.
x#
9
y#
4
œ 1 Ê 4x# 9y# œ 36 Ê 4r# cos# ) 9r# sin# ) œ 36
56. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4
57. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos )
58. x# xy y# œ 1 Ê x# y# xy œ 1 Ê r# r# sin ) cos ) œ 1 Ê r# (1 sin ) cos )) œ 1
59. x# (y 2)# œ 4 Ê x# y# 4y 4 œ 4 Ê x# y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin )
60. (x 5)# y# œ 25 Ê x# 10x 25 y# œ 25 Ê x# y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos )
61. (x 3)# (y 1)# œ 4 Ê x# 6x 9 y# 2y 1 œ 4 Ê x# y# œ 6x 2y 6 Ê r# œ 6r cos ) 2r sin ) 6
62. (x 2)# (y 5)# œ 16 Ê x# 4x 4 y# 10y 25 œ 16 Ê x# y# œ 4x 10y 13 Ê r#
œ 4r cos ) 10r sin ) 13
63. (!ß )) where ) is any angle
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Section 10.6 Graphing in Polar Coordinates
64. (a) x œ a Ê r cos ) œ a Ê r œ
(b) y œ b Ê r sin ) œ b Ê r œ
a
cos )
b
sin )
Ê r œ a sec )
Ê r œ b csc )
10.6 GRAPHING IN POLAR COORDINATES
1. 1 cos ()) œ 1 cos ) œ r Ê symmetric about the
x-axis; 1 cos ()) Á r and 1 cos (1 ))
œ 1 cos ) Á r Ê not symmetric about the y-axis;
therefore not symmetric about the origin
2. 2 2 cos ()) œ 2 2 cos ) œ r Ê symmetric about the
x-axis; 2 # cos ()) Á r and 2 2 cos (1 ))
œ 2 2 cos ) Á r Ê not symmetric about the y-axis;
therefore not symmetric about the origin
3. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 ))
œ 1 sin ) Á r Ê not symmetric about the x-axis;
1 sin (1 )) œ 1 sin ) œ r Ê symmetric about
the y-axis; therefore not symmetric about the origin
4. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 ))
œ 1 sin ) Á r Ê not symmetric about the x-axis;
1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the
y-axis; therefore not symmetric about the origin
5. 2 sin ()) œ 2 sin ) Á r and 2 sin (1 ))
œ 2 sin ) Á r Ê not symmetric about the x-axis;
2 sin (1 )) œ 2 sin ) œ r Ê symmetric about the
y-axis; therefore not symmetric about the origin
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Chapter 10 Conic Sections and Polar Coordinates
6. 1 2 sin ()) œ 1 2 sin ) Á r and 1 2 sin (1 ))
œ 1 2 sin ) Á r Ê not symmetric about the x-axis;
1 2 sin (1 )) œ 1 2 sin ) œ r Ê symmetric about the
y-axis; therefore not symmetric about the origin
7. sin ˆ #) ‰ œ sin ˆ #) ‰ œ r Ê symmetric about the y-axis;
sin ˆ 21#) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the
x-axis, and hence the origin.
8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis;
cos ˆ 21#) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the
y-axis, and hence the origin.
9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the
graph when (rß )) is on the graph Ê symmetric about the
x-axis and the y-axis; therefore symmetric about the origin
10. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on
the graph when (rß )) is on the graph Ê symmetric about
the y-axis and the x-axis; therefore symmetric about the
origin
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Section 10.6 Graphing in Polar Coordinates
11. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 ))
are on the graph when (rß )) is on the graph Ê symmetric
about the y-axis and the x-axis; therefore symmetric about
the origin
12. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on
the graph when (rß )) is on the graph Ê symmetric about
the x-axis and the y-axis; therefore symmetric about the
origin
13. Since a „ rß )b are on the graph when (rß )) is on the graph
ˆa „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is
symmetric about the x-axis and the y-axis Ê the graph is
symmetric about the origin
14. Since (rß )) on the graph Ê (rß )) is on the graph
ˆa „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is
symmetric about the origin. But 4 sin 2()) œ 4 sin 2)
Á r# and 4 sin 2(1 )) œ 4 sin (21 2)) œ 4 sin (2))
œ 4 sin 2) Á r# Ê the graph is not symmetric about
the x-axis; therefore the graph is not symmetric about
the y-axis
15. Since (rß )) on the graph Ê (rß )) is on the graph
ˆa „ rb# œ sin 2) Ê r# œ sin 2)‰ , the graph is
symmetric about the origin. But sin 2()) œ ( sin 2))
sin 2) Á r# and sin 2(1 )) œ sin (21 2))
œ sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph
is not symmetric about the x-axis; therefore the graph is
not symmetric about the y-axis
16. Sincea „ rß )b are on the graph when (rß )) is on the
graph ˆa „ rb# œ cos 2()) Ê r# œ cos 2)‰, the
graph is symmetric about the x-axis and the y-axis Ê the
graph is symmetric about the origin.
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Chapter 10 Conic Sections and Polar Coordinates
Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ 1# Ê r œ 1
w
)r cos )
Ê ˆ1ß 1# ‰ ; rw œ ddr) œ sin ); Slope œ rrw sin
cos )r sin )
17. ) œ
1
#
sin# )r cos )
sin ) cos )r sin )
sin# ˆ 1# ‰(1) cos 1#
sin 1# cos 1# (1) sin 1#
œ
Ê Slope at ˆ1ß 1# ‰ is
œ 1; Slope at ˆ1ß 1# ‰ is
sin# ˆ 1# ‰(1) cos ˆ 1# ‰
sin ˆ 1# ‰ cos ˆ 1# ‰(1) sin ˆ 1# ‰
œ1
18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1
dr
Ê ("ß 1); rw œ d)
œ cos );
rw sin )r cos )
cos ) sin )r cos )
rw cos )r sin ) œ cos ) cos )r sin )
0 sin 0(1) cos 0
cos ) sin )r cos )
Ê Slope at ("ß 0) is coscos
# 0(1) sin 0
cos# )r sin )
cos 1 sin 1(1) cos 1
1; Slope at ("ß 1) is cos# 1(1) sin 1 œ 1
Slope œ
œ
œ
Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ 14 Ê r œ 1
Ê ˆ1ß 14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ;
) œ 341 Ê r œ 1 Ê ˆ1ß 341 ‰ ;
19. ) œ
rw œ
1
4
dr
d)
œ 2 cos 2);
Slope œ
r sin )r cos )
r cos )r sin )
w
w
Ê Slope at ˆ1ß 14 ‰ is
Slope at ˆ1ß 14 ‰ is
Slope at ˆ1ß 341 ‰ is
Slope at ˆ1ß 341 ‰ is
2 cos 2) sin )r cos )
2 cos 2) cos )r sin )
2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰
2 cos ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰
œ
#
4
4
œ 1;
2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰
2 cos ˆ 1# ‰ cos ˆ 14 ‰(1) sin ˆ 14 ‰
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹
2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1;
œ 1;
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹
2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1
20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ;
) œ 1# Ê r œ 1 Ê ˆ"ß 12 ‰ ; ) œ 1 Ê r œ 1
Ê (1ß 1); rw œ
dr
d) œ 2 sin 2);
)r cos )
2 sin 2) sin )r cos )
Slope œ rr sin
cos )r sin ) œ 2 sin 2) cos )r sin )
2 sin 0 sin 0cos 0
Ê Slope at (1ß 0) is
2 sin 0 cos 0sin 0 , which is undefined;
2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰
Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 12 ‰ cos ˆ21 ‰(1) sin ˆ 21 ‰ œ 0;
w
w
2
Slope at ˆ1ß 12 ‰ is
Slope at ("ß 1) is
2
2
2 sin 2 ˆ 1# ‰ sin ˆ 1# ‰(1) cos ˆ 1# ‰
2 sin 2 ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰
#
2 sin 21 sin 1cos 1
2 sin 21 cos 1sin 1
#
#
œ 0;
, which is undefined
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Section 10.6 Graphing in Polar Coordinates
21. (a)
(b)
22. (a)
(b)
23. (a)
(b)
24. (a)
(b)
25.
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655
656
Chapter 10 Conic Sections and Polar Coordinates
26. r œ 2 sec ) Ê r œ
2
cos )
Ê r cos ) œ 2 Ê x œ 2
27.
28.
29. ˆ#ß 341 ‰ is the same point as ˆ2ß 14 ‰ ; r œ 2 sin 2 ˆ 14 ‰ œ 2 sin ˆ 1# ‰ œ 2 Ê ˆ2ß 14 ‰ is on the graph
Ê ˆ#ß 341 ‰ is on the graph
30. ˆ "# ß 321 ‰ is the same point as ˆ "# ß 12 ‰ ; r œ sin Š
ˆ 1# ‰
3 ‹
œ sin
1
6
œ "# Ê ˆ "# ß 1# ‰ is on the graph Ê ˆ "# ß 3#1 ‰
is on the graph
31. 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1
Ê r œ 1; points of intersection are ˆ"ß 1# ‰ and ˆ"ß 3#1 ‰ .
The point of intersection (!ß 0) is found by graphing.
32. 1 sin ) œ 1 sin ) Ê sin ) œ 0 Ê ) œ 0, 1 Ê r œ 1;
points of intersection are (1ß 0) and (1ß 1). The point of
intersection (!ß 0) is found by graphing.
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Section 10.6 Graphing in Polar Coordinates
33. 2 sin ) œ 2 sin 2) Ê sin ) œ sin 2) Ê sin )
œ 2 sin ) cos ) Ê sin ) 2 sin ) cos ) œ 0
Ê (sin ))(1 2 cos )) œ 0 Ê sin ) œ 0 or cos ) œ
, or 13 ; ) œ 0 or 1 Ê r œ 0,
Ê r œ È3 , and ) œ 1 Ê r œ È3 ; points of
Ê ) œ 0, 1,
)œ
1
3
1
3
"
#
3
intersection are (!ß 0), ŠÈ3 ß 13 ‹, and ŠÈ3 ß 13 ‹
34. cos ) œ 1 cos ) Ê 2 cos ) œ 1 Ê cos ) œ
"
#
Ê ) œ 13 , 13 Ê r œ "# ; points of intersection are
ˆ "# ß 13 ‰ and ˆ "# , 13 ‰ . The point (0ß 0) is found by
graphing.
#
35. ŠÈ2‹ œ 4 sin ) Ê
"
#
œ sin ) Ê ) œ
1
6
,
51
6
; points
of intersection are ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ . The
points ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ are found by
graphing.
36. È2 sin ) œ È2 cos ) Ê sin ) œ cos ) Ê ) œ
)œ
1
4
#
Ê r œ 1 Ê r œ „ 1 and ) œ
51
4
1
4
,
51
4
;
#
Ê r œ 1
Ê no solution for r; points of intersection are ˆ „ 1ß 14 ‰ .
The points (!ß 0) and ˆ „ 1ß 341 ‰ are found by graphing.
37. 1 œ 2 sin 2) Ê sin 2) œ
"
#
Ê 2) œ
1
6
,
51
6
,
131
6
,
171
6
1
Ê ) œ 12
, 5121 , 13121 , 17121 ; points of intersection are
1‰ ˆ
ˆ1ß 12
, 1ß 5121 ‰ , ˆ1ß 131#1 ‰, and ˆ1ß 171#1 ‰ . No other
points are found by graphing.
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Chapter 10 Conic Sections and Polar Coordinates
38. È2 cos 2) œ È2 sin 2) Ê cos 2) œ sin 2)
Ê 2) œ 14 , 541 , 941 , 1341 Ê ) œ 18 , 581 , 981 ,
)œ
1 91
8 , 8
#
#
Ê r œ 1 Ê r œ „ 1; ) œ
51
8
,
131
8
131
8
;
Ê r œ 1 Ê no solution for r; points of intersection are
ˆ1ß 18 ‰ and ˆ1ß 981 ‰ . The point of intersection (!ß 0) is found
by graphing.
39. r# œ sin 2) and r# œ cos 2) are generated completely for
0 Ÿ ) Ÿ 1# . Then sin 2) œ cos 2) Ê 2) œ 14 is the only
solution on that interval Ê ) œ 18 Ê r# œ sin 2 ˆ 18 ‰ œ È"
Ê rœ „
"
%
È
2
2
"
ß 1‹.
%
È
2 8
; points of intersection are Š „
The point of intersection (!ß 0) is found by graphing.
40. 1 sin
)
#
31
#
Ê )œ
)œ
71
#
œ 1 cos
,
71
#
)
#
;)œ
Ê sin
31
#
Ê r œ 1 cos
intersection are Š"
)
#
œ cos
Ê r œ 1 cos
71
4
œ1
È 2 31
# ß # ‹
È2
#
)
#
Ê
)
#
31
4
œ1
31 71
4 , 4
È2
# ;
œ
; points of
and Š1
È 2 71
# ß # ‹.
three points of intersection (0ß 0) and Š1 „
È2
#
The
ß 1# ‹ are
found by graphing and symmetry.
41. 1 œ 2 sin 2) Ê sin 2) œ
"
#
Ê 2) œ
1
6
,
51
6
,
131
6
,
171
6
Ê ) œ 11# , 511# , 131#1 , 171#1 ; points of intersection are
ˆ"ß 11# ‰ , ˆ"ß 511# ‰ , ˆ1ß 131#1 ‰ , and ˆ"ß 17121 ‰ . The points
of intersection ˆ1ß 711# ‰ , ˆ"ß 111#1 ‰ , ˆ"ß 191#1 ‰ and
ˆ"ß 231#1 ‰ are found by graphing and symmetry.
42. r# œ 2 sin 2) is completely generated on 0 Ÿ ) Ÿ
"
#
1
6
1
#
so
1
that 1 œ 2 sin 2) Ê sin 2) œ Ê 2) œ , 561 Ê ) œ 12
51
1
5
1
ˆ
‰
ˆ
‰
1# ; points of intersection are 1ß 1# and "ß 1# . The
1
5
1
points of intersection ˆ"ß 1# ‰ and ˆ1ß 1# ‰ are found
,
by graphing.
43. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ 1 cos ) Í 1 cos () 1)
œ 1 (cos ) cos 1 sin ) sin 1) œ 1 cos ) œ (1 cos )) œ r; therefore (rß )) is on the graph of
r œ 1 cos ) Í (rß ) 1) is on the graph of r œ 1 cos ) Ê the answer is (a).
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Section 10.6 Graphing in Polar Coordinates
44. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ cos 2) Í sin ˆ2() 1)) 1# ‰
œ sin ˆ2) 5#1 ‰ œ sin (2)) cos ˆ 5#1 ‰ cos (2)) sin ˆ 5#1 ‰ œ cos 2) œ r; therefore (rß )) is on the graph of
r œ sin ˆ2) 1# ‰ Ê the answer is (a).
45.
47. (a)
46.
(b)
(c)
(d)
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659
660
Chapter 10 Conic Sections and Polar Coordinates
48. (a)
(b)
(d)
(c)
(e)
#
#
49. (a) r# œ 4 cos ) Ê cos ) œ r4 ; r œ 1 cos ) Ê r œ 1 Š r4 ‹ Ê 0 œ r# 4r 4 Ê (r 2)# œ 0
#
Ê r œ 2; therefore cos ) œ 24 œ 1 Ê ) œ 1 Ê (2ß 1) is a point of intersection
(b) r œ 0 Ê 0# œ 4 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê ˆ!ß 1# ‰ or ˆ!ß 3#1 ‰ is on the graph; r œ 0 Ê 0 œ 1 cos )
Ê cos ) œ 1 Ê ) œ 0 Ê (0ß 0) is on the graph. Since (!ß 0) œ ˆ!ß 1# ‰ for polar coordinates, the graphs
intersect at the origin.
50. (a) Let r œ f()) be symmetric about the x-axis and the y-axis. Then (rß )) on the graph Ê (rß )) is on the
graph because of symmetry about the x-axis. Then (rß ())) œ (rß )) is on the graph because of
symmetry about the y-axis. Therefore r œ f()) is symmetric about the origin.
(b) Let r œ f()) be symmetric about the x-axis and the origin. Then (rß )) on the graph Ê (rß )) is on the
graph because of symmetry about the x-axis. Then (rß )) is on the graph because of symmetry about
the origin. Therefore r œ f()) is symmetric about the y-axis.
(c) Let r œ f()) be symmetric about the y-axis and the origin. Then (rß )) on the graph Ê (rß )) is on the
graph because of symmetry about the y-axis. Then ((r)ß )) œ (rß )) is on the graph because of
symmetry about the origin. Therefore r œ f()) is symmetric about the x-axis.
51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve
on the interval 0 Ÿ ) Ÿ 14 . So we wish to maximize 2y œ 2r sin ) œ 2 cos 2) sin ) on 0 Ÿ ) Ÿ 14 . Let
f()) œ 2 cos 2) sin ) œ 2 a1 2 sin# )b (sin )) œ 2 sin ) 4 sin$ ) Ê f w ()) œ 2 cos ) 12 sin# ) cos ). Then
f w ()) œ 0 Ê 2 cos ) 12 sin# ) cos ) œ 0 Ê (cos )) a1 6 sin# )b œ 0 Ê cos ) œ 0 or 1 6 sin# ) œ 0 Ê ) œ
sin ) œ
„1
È6 .
Since we want 0 Ÿ ) Ÿ
œ 2 Š È"6 ‹ 4 †
interval 0 Ÿ ) Ÿ
is
2È 6
9
"
œ
. We
6È 6
1
4 . Therefore the
1
4
"
, we choose ) œ sin
Š È"6 ‹
$
Ê f()) œ 2 sin ) 4 sin )
can see from the graph of r œ cos 2) that a maximum does occur in the
maximum width occurs at ) œ sin" Š È"6 ‹ , and the maximum width
2È 6
9 .
52. We wish to maximize y œ r sin ) œ 2(1 cos ))(sin )) œ 2 sin ) 2 sin ) cos ). Then
dy
#
#
#
d) œ 2 cos ) 2(sin ))( sin )) 2 cos ) cos ) œ 2 cos ) 2 sin ) 2 cos ) œ 2 cos ) 4 cos ) 2; thus
dy
#
#
d) œ 0 Ê 4 cos ) 2 cos ) 2 = 0 Ê 2 cos ) cos ) 1 œ 0 Ê (2 cos ) 1)(cos ) 1) œ 0 Ê cos ) œ
or cos ) œ 1 Ê ) œ 13 , 531 , 1. From the graph, we can see that the maximum occurs in the first quadrant so
È
we choose ) œ 13 . Then y œ 2 sin 13 2 sin 13 cos 13 œ 3 # 3 . The x-coordinate of this point is x œ r cos
È
œ 2 ˆ1 cos 13 ‰ ˆcos 13 ‰ œ 3# . Thus the maximum height is h œ 3 # 3 occurring at x œ 3# .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
3
"
#
1
#
or
Section 10.7 Area and Lengths in Polar Coordinates
10.7 AREA AND LENGTHS IN POLAR COORDINATES
1. A œ '0
21
"
#
(4 2 cos ))# d) œ '0
21
"
#
2) ‰‘
a16 16 cos ) 4 cos# )b d) œ '0 8 8 cos ) 2 ˆ 1 cos
d)
#
21
œ '0 (9 8 cos ) cos 2)) d) œ 9) 8 sin )
21
2. A œ '0
21
œ
"
#
a#
"
#
[a(1 cos ))]# d) œ '0
21
"
#
#1
"
2
sin 2)‘ ! œ 181
a# a1 2 cos ) cos# )b d) œ
"
#
a#
2) ‰
'021 ˆ1 2 cos ) 1 cos
d)
#
'021 ˆ 3# 2 cos ) "# cos 2)‰ d) œ "# a# 3# ) 2 sin ) 4" sin 2)‘ #!1 œ 3# 1a#
3. A œ 2 '0
1Î4
"
#
cos# 2) d) œ '0
"
#
a2a# cos 2)b d) œ 2a# '1Î4 cos 2) d) œ 2a# sin22) ‘ 1Î% œ 2a#
1Î4
4. A œ 2 '1Î4
5. A œ '0
1Î2
"
#
1Î4
1 cos 4)
#
d) œ
"
#
)
sin 4) ‘ 1Î%
4
!
œ
1Î4
(4 sin 2)) d) œ '0
1Î2
6. A œ (6)(2)'0
1Î6
1
8
1Î%
1Î#
2 sin 2) d) œ c cos 2)d !
œ2
(2 sin 3)) d) œ 12 '0 sin 3) d) œ 12 cos3 3) ‘ !
1Î6
"
#
1Î'
œ4
7. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin )
Ê cos ) œ sin ) Ê ) œ 14 ; therefore
A œ 2 '0
1Î4
œ '0
1Î4
(2 sin ))# d) œ '0
1Î4
"
#
2) ‰
4 ˆ 1 cos
d) œ '0
#
1Î4
œ c2) sin
1Î%
2) d !
œ
1
#
4 sin# ) d)
(2 2 cos 2)) d)
1
8. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ
Ê )œ
1
6
or
51
6
"
#
; therefore
A œ 1(1)# '1Î6
51Î6
"
#
c(2 sin ))# 1# d d)
œ 1 '1Î6 ˆ2 sin# ) "# ‰ d)
51Î6
œ 1 '1Î6 ˆ1 cos 2) "# ‰ d)
51Î6
œ 1 '1Î6 ˆ "# cos 2)‰ d) œ 1 "2 )
sin 2) ‘ &1Î'
#
1Î'
œ 1 ˆ 511#
41 3È 3
6
51Î6
"
#
sin
51 ‰
3
1
ˆ 12
"
#
sin 13 ‰ œ
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662
Chapter 10 Conic Sections and Polar Coordinates
9. r œ 2 and r œ 2(1 cos )) Ê 2 œ 2(1 cos ))
Ê cos ) œ 0 Ê ) œ „ 1# ; therefore
A œ 2 '0
1Î2
œ '0
1Î2
œ '0
1Î2
œ '0
1Î2
"
#
[2(1 cos ))]# d) "# area of the circle
4 a1 2 cos ) cos# )b d) ˆ "# 1‰ (2)#
4 ˆ1 2 cos )
1 cos 2) ‰
#
d) 2 1
(4 8 cos ) 2 2 cos 2)) d) 21
1Î#
œ c6) 8 sin ) sin 2)d !
21 œ 51 8
10. r œ 2(1 cos )) and r œ 2(1 cos )) Ê 1 cos )
œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also
gives the point of intersection (0ß 0); therefore
A œ 2 '0
1Î2
"
#
[2(1 cos ))]# d) 2 '1Î2 "# [2(1 cos ))]# d)
1
œ '0
4 a1 2 cos ) cos# )b d)
œ '0
4 ˆ1 2 cos )
œ '0
(6 8 cos ) 2 cos 2)) d) '1Î2 (6 8 cos ) 2 cos 2)) d)
1Î2
'1Î2 4 a1 2 cos ) cos# )b d)
1
1Î2
1Î2
1 cos 2) ‰
#
d) '1Î2 4 ˆ1 2 cos )
1
1 cos 2) ‰
#
d)
1
1Î#
œ c6) 8 sin ) sin 2)d !
c6) 8 sin ) sin 2)d 11Î# œ 61 16
11. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ
1
6
Ê )œ
"
#
(in the 1st quadrant); we use symmetry of the
graph to find the area, so
A œ 4 '0 ” "# (6 cos 2)) "# ŠÈ3‹ • d)
#
1Î6
œ 2 '0 (6 cos 2) 3) d) œ 2 c3 sin 2) 3)d !
1Î6
1Î'
œ 3È3 1
12. r œ 3a cos ) and r œ a(1 cos )) Ê 3a cos ) œ a(1 cos ))
Ê 3 cos ) œ 1 cos ) Ê cos ) œ "# Ê ) œ 13 or 13 ;
the graph also gives the point of intersection (0ß 0); therefore
A œ 2 '0
1Î3
"
#
c(3a cos ))# a# (1 cos ))# d d)
œ '0 a9a# cos# ) a# 2a# cos ) a# cos# )b d)
1Î3
œ '0
1Î3
a8a# cos# ) 2a# cos ) a# b d)
œ '0 c4a# (1 cos 2)) 2a# cos ) a# d d)
1Î3
œ '0 a3a# 4a# cos 2) 2a# cos )b d)
1Î3
1Î$
œ c3a# ) 2a# sin 2) 2a# sin )d !
œ 1a# 2a# ˆ "# ‰ 2a# Š
È3
# ‹
œ a# Š1 1 È3‹
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Section 10.7 Area and Lengths in Polar Coordinates
663
13. r œ 1 and r œ 2 cos ) Ê 1 œ 2 cos ) Ê cos ) œ "#
Ê )œ
A œ 2'
1
21
3
in quadrant II; therefore
c(2 cos ))# 1# d d) œ '21Î3 a4 cos# ) 1b d)
1
"
21Î3 #
œ '21Î3 [2(1 cos 2)) 1] d) œ '21Î3 (1 2 cos 2)) d)
1
1
œ c) sin 2)d 1#1Î$ œ
14. (a) A œ 2 '0
21Î3
œ '0
21Î3
"
#
1
3
È3
#
(2 cos ) 1)# d) œ '0
21Î3
a4 cos# ) 4 cos ) 1b d) œ '0
21Î3
#1Î$
(3 2 cos 2) 4 cos )) d) œ c3) sin 2) 4 sin )d !
3È 3
# ‹
(b) A œ Š21
Š1
3È 3
# ‹
1
6
; therefore A œ '1Î6
51Î6
51
6
or
œ '1Î6 ˆ18
51Î6
9
#
È3
#
4È 3
#
œ 21
3È 3
#
œ 1 3È3 (from 14(a) above and Example 2 in the text)
"
#
15. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ
Ê )œ
œ 21
[2(1 cos 2)) 4 cos ) 1] d)
csc# )‰ d) œ 18)
"
#
a6# 9 csc# )b d)
9
#
cot )‘ 1Î'
&1Î'
œ Š151 9# È3‹ Š31 9# È3‹ œ 121 9È3
16. r# œ 6 cos 2) and r œ
Ê
sec ) Ê
3
#
#
9
4
%
sec# ) œ 6 cos 2) Ê
œ 2 cos% ) cos ) Ê 2 cos ) cos# )
3
8
1
6
(in the first quadrant); thus A œ 2 '0
œ 3 sin 2)
9
4
tan )‘ !
1Î6
1Î'
17. (a) r œ tan ) and r œ Š
œ 3Š
È2
# ‹
È3
# ‹
9
4È 3
œ
3È 3
#
csc ) Ê tan ) œ Š
" ˆ
# 6 cos 2)
3È 3
3È 3
4 œ 4
È2
# ‹
È2
# ‹
cos ) Ê 1 cos# ) œ Š
Ê cos# ) Š
È2
# ‹
cos ) 1 œ 0 Ê cos ) œ È2 or
1
4
œ acos# )b a2 cos# ) 1b
9
4
È3
#
(the second equation has no real
sec# )‰ d) œ '0 ˆ6 cos 2)
1Î6
9
4
sec# )‰ d)
csc )
Ê sin# ) œ Š
(use the quadratic formula) Ê ) œ
3
8
or cos# ) œ "4 Ê cos ) œ „
3
4
roots) Ê ) œ
È2
#
œ cos# ) cos 2) Ê
œ 0 Ê 16 cos% ) 8 cos# ) 3 œ 0
3
8
Ê a4 cos# ) 1ba4 cos# ) 3b œ 0 Ê cos# ) œ
9
24
È2
# ‹
cos )
(the solution
in the first quadrant); therefore the area of R" is
A" œ '0
1Î4
AO œ Š
"
#
È2
# ‹
tan# ) d) œ
1
#
csc
œ
È2
#
"
#
and OB œ Š
Ê the area of R# is A# œ
2 ˆ "#
1
8
4" ‰ œ
3
#
1
4
'01Î4 asec# ) 1b d)
"
#
Š
rœ
(b)
lim
lim
œ
"
4
1
4
1Î%
ctan ) )d !
œ
"
#
ˆtan
œ 1 Ê AB œ Ê1# Š
1
4
14 ‰ œ
È2
# ‹
#
œ
) Ä 1 Î2 c
r œ sec ) as ) Ä
" ‰
cos )
1c
#
1
8
;
; therefore the area of the region shaded in the text is
1
4
generates the arc OB of r œ tan ) but does not generate the segment AB of the line
tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then
sin )
ˆ cos
)
"
#
È2
#
csc ). Instead the interval generates the half-line from B to _ on the line r œ
) Ä 1 Î2
=
È2
È2
# ‹Š # ‹
csc
"
#
. Note: The area must be found this way since no common interval generates the region. For
example, the interval 0 Ÿ ) Ÿ
È2
#
È2
# ‹
œ
œ
lim
) Ä 1 Î2 c
ˆ sincos) ) 1 ‰ œ
lim
) Ä 1 Î2 c
lim
) Ä 1 Î2 c
È2
#
csc ).
(tan ) sec ))
) ‰
ˆ cos
sin ) œ 0 Ê r œ tan ) approaches
Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ sec )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
664
Chapter 10 Conic Sections and Polar Coordinates
(or x œ 1) is a vertical asymptote of r œ tan ).
18. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is
A œ 2 '0
1
"
#
2)
(cos ) 1)# d) œ '0 acos# ) 2 cos ) 1b d) œ '0 ˆ 1 cos
2 cos ) 1‰ d)
#
1
œ 32)
31
#
sin 2)
4
1
51
œ
4
4
1
2 sin )‘ ! œ
19. r œ )# , 0 Ÿ ) Ÿ È5 Ê
#
È5
È5
œ '0 k)k È)# 4 d) œ (since )
) œ È5 Ê u œ 9“ Ä '4
9
20. r œ
e)
È2
,0Ÿ)Ÿ1 Ê
dr
d)
"
#
œ
1
4
. The area of the circle is A œ 1 ˆ "# ‰ œ
œ 2); therefore Length œ '0
dr
d)
È5
31
#
1
0) '0
Èu du œ
e)
È2
"
#
Ê the area requested is actually
Éa)# b# (2))# d) œ '
0
) È ) # 4 d ) ; u œ ) # 4 Ê
23 u$Î# ‘ * œ
%
È5
"
#
È ) % 4) # d)
du œ ) d); ) œ 0 Ê u œ 4,
19
3
; therefore Length œ '0 ÊŠ Èe 2 ‹ Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d)
1
#
)
#
)
1
2)
œ '0 e) d) œ e) ‘ ! œ e1 1
1
1
21. r œ 1 cos ) Ê
dr
d)
œ sin ); therefore Length œ '0 È(1 cos ))# ( sin ))# d)
21
1
œ 2 '0 È2 2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1 #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4 2 sin 2) ‘ ! œ 8
1
1
22. r œ a sin#
)
#
1
, 0 Ÿ ) Ÿ 1, a 0 Ê
œ '0 Éa# sin%
1
)
#
a# sin#
)
#
dr
d)
)
#
cos#
œ a sin
)
#
cos
)
#
1
#
; therefore Length œ '0 Ɉa sin# #) ‰ ˆa sin
1
d) œ '0 a ¸sin #) ¸ Ésin#
1
)
#
)
#
cos#
1
6
1 cos )
œ '0
1Î2
,0Ÿ)Ÿ
1
#
É (1 36
cos ))#
œ ˆsince
"
1 cos )
Ê
dr
d)
œ
; therefore Length œ '0
1Î2
6 sin )
(1 cos ))#
d) œ 6 '0
1Î2
36 sin# )
a1 cos )b%
"
¸ 1cos
¸
) É1
0
1Î2
1Î2
1Î2
1Î2
)
#
d)
cos# ) sin# )
0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos
d)
) )#
cos )
È '
œ 6 '0 ˆ 1 "cos ) ‰ É (12 2cos
) )# d) œ 6 2 0
œ 3'0 sec$
#
#
6 sin )
ʈ 1 6cos ) ‰ Š (1
cos ))# ‹ d)
sin# )
(1 cos ))#
d) œ 6'0
1Î4
d)
(1 cos ))$Î#
œ 6È2 '0
1Î2
1Î%
sec$ u du œ (use tables) 6 Œ sec u2tan u ‘ !
d)
ˆ2 cos# #) ‰$Î#
"
#
'01Î4
œ 3'0
1Î2
¸sec$ #) ¸ d)
sec u du
1Î%
œ 6 Š È"2 2" ln ksec u tan uk‘ ! ‹ œ 3 ’È2 ln Š1 È2‹“
24. r œ
2
1 cos )
,
1
#
Ÿ)Ÿ1 Ê
4
œ '1Î2 Ê (1 cos
) ) # Š1
1
œ ˆsince 1 cos )
dr
d)
œ
2 sin )
(1 cos ))#
sin# )
‹
a1 cos )b#
0 on
1
#
sin )
; therefore Length œ '1Î2 ʈ 1 2cos ) ‰ Š (12cos
))# ‹ d)
1
1
#
1
œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc
1
)
#
cos )
d)
È '
È '
œ 2 '1Î2 ˆ 1 "cos ) ‰ É (12 2cos
))# d) œ 2 2 1Î2 (1 cos ))$Î# œ 2 2 1Î2
1
#
d)
cos ) sin
Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos
) )#
1
)
#
0 on
1
#
1
#
#
) sin
d) œ '1Î2 ¸ 1 2cos ) ¸ É (1 (1cos )cos
) )#
#
d)
ˆ2 sin# )# ‰$Î#
)
d)
œ '1Î2 ¸csc$ #) ¸ d)
1
Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables)
1Î2
#
cos #) ‰ d)
d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d)
1
œ 2a cos 2) ‘ ! œ 2a
23. r œ
)
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.7 Area and Lengths in Polar Coordinates
1Î#
2Œ csc u2cot u ‘ 1Î%
'11ÎÎ42
"
#
1Î#
csc u du œ 2 Š È"2 2" ln kcsc u cot uk‘ 1Î% ‹ œ 2 ’ È"2
"
#
ln ŠÈ2 1‹“
œ È2 ln Š1 È2‹
)
3
25. r œ cos$
œ '0
Ê
dr
d)
)
3
œ sin
; therefore Length œ '0
1Î4
)
3
cos#
Écos' ˆ 3) ‰ sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '
1Î4
1Î4 1cos ˆ 2) ‰
3
#
d) œ
"
#
)
3
2
2) ‘ 1Î%
3 !
sin
26. r œ È1 sin 2) , 0 Ÿ ) Ÿ 1È2 Ê
1È 2
Length œ '0
œ '0
È
1 2
É(1 sin 2))
sin 2)
'
É 212sin
2) d) œ 0
27. r œ È1 cos 2) Ê
1È 2
œ '0
È
1 2
œ
dr
d)
"
#
cos# 2)
(1 sin 2))
È
d) œ '0
1 2
1È 2
d) œ '0
2)
1È#
!
#
sin 2) cos
É 1 2 sin 2)1
sin 2)
#
2)
d)
œ 21
È
1È 2
cos 2)
'
É 212cos
2 ) d) œ 0
21
É(1 cos 2))
È2 d) œ ’È2 )“
œ 0; Length œ '0 Èa# 0# d) œ '0 kak d) œ ca)d #!1 œ 21a
1È#
!
sin# 2)
(1 cos 2))
21
œ a sin ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
dr
d)
œ a cos ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
1
œ '0 kak d) œ ca)d 1! œ 1a
1
d)
œ 21
dr
d)
(b) r œ a cos ) Ê
(c) r œ a sin ) Ê
cos# ˆ 3) ‰ d)
(1 sin 2))"Î# (2 cos 2)) œ (cos 2))(1 sin 2))"Î# ; therefore
È2 d) œ ’È2 )“
#
1Î4
3
8
1 2
#
dr
d)
"
#
œ
#
cos# 3) ‰ d)
0
(1 cos 2))"Î# (2 sin 2)); therefore Length œ '0
cos 2) sin
É 1 2 cos 21) cos
2)
28. (a) r œ a Ê
dr
d)
1
8
œ
)
3
ˆcos# 3) ‰ Écos# ˆ 3) ‰ sin# ˆ 3) ‰ d) œ '
1Î4
0
œ '0
Ɉcos$ 3) ‰# ˆ sin
1
1
1
œ '0 kak d) œ ca)d 1! œ 1a
1
29. r œ Ècos 2) , 0 Ÿ ) Ÿ
œ '0
1Î4
1
4
Ê
dr
d)
œ
"
#
(cos 2))"Î# ( sin 2))(2) œ
sin 2)
Ècos 2)
; therefore Surface Area
sin 2)
(21r cos )) ÊŠÈcos 2)‹ Š È
‹ d) œ '0 Š21Ècos 2)‹ (cos ))Écos 2)
cos 2)
#
#
œ '0 Š21Ècos 2)‹ (cos ))É cos" 2) d) œ '0
1Î4
1Î4
30. r œ È2e)Î2 , 0 Ÿ ) Ÿ
1
#
Ê
dr
d)
œ È2 ˆ "# ‰ e)Î2 œ
œ '0 Š21È2 e)Î2 ‹ (sin )) ÊŠÈ2 e)Î2 ‹ Š
#
1Î2
1Î4
1Î%
21 cos ) d) œ c21 sin )d !
È2
#
È2
#
sin# 2)
cos 2)
d)
œ 1È2
e)Î2 ; therefore Surface Area
e)Î2 ‹ d) œ '0 Š21È2 e)Î2 ‹ (sin )) É2e) "# e) d)
#
1Î2
œ '0 Š21È2 e)Î2 ‹ (sin )) É 5# e) d) œ '0 Š21È2 e)Î2 ‹ (sin )) Š È52 e)Î2 ‹ d) œ 21È5 '0 e) sin ) d)
1Î2
1Î2
)
1Î#
œ 21È5 e2 (sin ) cos ))‘ !
œ 1È5 ae1Î2 1b where we integrated by parts
31. r# œ cos 2) Ê r œ „ Ècos 2) ; use r œ Ècos 2) on 0ß 14 ‘ Ê
therefore Surface Area œ 2 '0 Š21Ècos 2)‹ (sin )) Écos 2)
1Î4
œ 41 '0 sin ) d) œ 41 c cos )d !
1Î4
1Î2
È
1Î%
œ 41 ’
È2
#
dr
d)
œ
sin# 2)
cos 2)
"
#
(cos 2))"Î# ( sin 2))(2) œ
d) œ 41 '0
1Î4
sin 2)
Ècos 2)
Ècos 2) (sin )) É
(1)“ œ 21 Š2 È2‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
;
"
cos 2)
d)
665
666
Chapter 10 Conic Sections and Polar Coordinates
32. r œ 2a cos ) Ê
dr
d)
œ 2a sin ); therefore Surface Area œ '0 21(2a cos ))(cos ))È(2a cos ))# (2a sin ))# d)
1
œ 4a1 '0 acos# )b È4a# acos# ) sin# )b d) œ 8a1 '0 acos# )b kak d) œ 8a# 1 '0 cos# ) d)
1
1
1
2) ‰
œ 8a# 1 '0 ˆ 1 cos
d) œ 4a# 1 '0 (1 cos 2)) d) œ 4a# 1 )
#
1
1
33. Let r œ f()). Then x œ f()) cos ) Ê
"
2
1
sin 2)‘ ! œ 4a# 1#
‰# œ cf w ()) cos ) f()) sin )d#
œ f w ()) cos ) f()) sin ) Ê ˆ dx
d)
dx
d)
œ cf w ())d# cos# ) 2f w ()) f()) sin ) cos ) [f())]# sin# ); y œ f()) sin ) Ê
dy
d)
#
œ f w ()) sin ) f()) cos )
#
#
w
w
#
w
#
#
Ê Š dy
d) ‹ œ cf ()) sin ) f()) cos )d œ cf ())d sin ) 2f ())f()) sin ) cos ) [f())] cos ). Therefore
#
#
#
w
#
#
#
#
#
w
#
#
ˆ dx
‰# Š dy
ˆ dr ‰#
d)
d) ‹ œ cf ())d acos ) sin )b [f())] acos ) sin )b œ cf ())d [f())] œ r d) .
' Ér# ˆ ddr) ‰# d).
‰# Š dy
Thus, L œ '! ʈ dx
d)
d) ‹ d) œ !
"
"
#
'021 a(1 cos )) d) œ 2a1 c) sin )d #!1 œ a
21
rav œ 21"0 '0 a d) œ #"1 ca)d #!1 œ a
1Î2
1Î#
rav œ ˆ 1 ‰"ˆ 1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d 1Î# œ 2a
1
34. (a) rav œ
(b)
(c)
"
2 1 0
#
#
35. r œ 2f()), ! Ÿ ) Ÿ " Ê
dr
d)
œ 2f w ()) Ê r# ˆ ddr) ‰ œ [2f())]# c2f w ())d# Ê Length œ '! É4[f())]# 4 cf w ())d# d)
"
#
œ 2 '! É[f())]# cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " .
"
36. Again r œ 2f()) Ê r# ˆ ddr) ‰ œ [2f())]2 c2f w ())d# Ê Surface Area œ '! 21[2f()) sin )] É4[f())]# 4 cf w ())d# d)
"
#
œ 4 '! 21[f()) sin )] É[f())]# cf w ())d# d) which is four times the area of the surface generated by revolving
"
r œ f()) about the x-axis for ! Ÿ ) Ÿ " .
'021 r$ cos ) d)
37. x œ
œ
'021 r# d)
2
3
œ
2
3
2
3
'021 [a(1 cos ))]$ (cos )) d)
œ
'021 [a(1 cos ))]# d)
2
3
a$
'021 a1 3 cos ) 3 cos# ) cos$ )b (cos )) d)
21
a# '0 a1 2 cos ) cos# )b d)
#
2)
1 cos 2)
#
a '0 ’cos ) 3 ˆ 1 cos
# ‰ 3 a1 sin )b (cos )) ˆ # ‰ “ d)
21
'0
21
2)
1 2 cos ) ˆ 1 cos
# ‰‘ d )
œ (After considerable algebra using
"
4
#
' 15 8
1 cos 2A ‰ a 0 ˆ 12 3 cos ) 3 cos 2) 2 cos ) sin ) 12 cos 4)‰ d)
21
#
"
3
' ˆ # 2 cos ) # cos 2)‰ d)
21
the identity cos# A œ
0
œ
#1
"
8
2
2
$
‘
a 15
12 ) 3 sin ) 3 sin 2) 3 sin ) 48 sin 4) !
#3 ) 2 sin ) "4 sin 2)‘ #1
yœ
!
2
3
2
3
'2a2a "a u$ du
31
œ
0
31
'01 r# d) œ '01 a# d) œ ca# )d !1 œ a# 1; x œ
yœ
2
3
œ
5
6
a;
21
2'
$
'021 r$ sin ) d)
3 0 [a(1 cos ))] (sin )) d)
œ
; u œ a(1 cos )) Ê "a du œ sin ) d); ) œ 0 Ê u œ 2a;
21
31
'0 r # d )
) œ 21 Ê u œ 2ad Ä
38.
œ
‰
a ˆ 15
6 1
31
2'
$
'0 r$ sin ) d)
3 0 a sin ) d)
œ
a# 1
'01 r# d) œ
1
1
2
3
œ 0. Therefore the centroid is aBß yb œ ˆ 56 aß 0‰
2
3
'01 r$ cos ) d)
œ
'01 r# d)
a$ c cos )d 1!
a# 1
œ
ˆ 43 ‰ a$
a# 1
œ
2
3
'01 a$ cos ) d)
4a
31 .
a# 1
œ
2
3
a$ c sin )d 1!
a# 1
œ
0
a# 1
œ 0;
Therefore the centroid is axß yb œ ˆ0ß 34a1 ‰ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.8 Conic Sections in Polar Coordinates
667
10.8 CONIC SECTIONS IN POLAR COORDINATES
1. r cos ˆ) 16 ‰ œ 5 Ê r ˆcos ) cos
œ 10 Ê y œ È3 x 10
2. r cos ˆ)
Ê
È2
#
3. r cos ˆ)
Ê 1# x
31 ‰
4
œ 2 Ê r ˆcos ) cos
x
È2
#
1
6
sin ) sin 16 ‰ œ 5 Ê
31
4
31 ‰
4
sin ) sin
È3
#
œ2 Ê
È2
#
41 ‰
œ3 Ê
3
È3
# yœ3
r cos )
È2
#
r cos )
È2
#
È3
#
r ˆcos ) cos
41
3
41 ‰
3
sin ) sin
œ 3 Ê #1 r cos )
È3
3
Ê x È 3 y œ 6 Ê y œ
È2
#
È2
#
r sin ) œ 4 Ê
x
È2
#
r sin ) œ 2
È2
È2
È3
#
r sin ) œ 3
x 2È3
1
4
sin ) sin 14 ‰ œ 4
y œ 4 Ê È2 x È2 y œ 8 Ê y œ x 4È2
5. r cos ˆ) 14 ‰ œ È2 Ê r ˆcos ) cos 14 sin ) sin 14 ‰
œ È2 Ê " r cos ) " r sin ) œ È2 Ê " x
È2
"
È2
y
œ È2 Ê x y œ 2 Ê y œ 2 x
6. r cos ˆ)
Ê
31 ‰
4
œ 1 Ê r ˆcos ) cos
È2
2
r cos )
Ê y œ x È 2
7. r cos ˆ)
21 ‰
3
È2
2
È3
2
31
4
sin ) sin
31 ‰
4
œ1
r sin ) œ 1 Ê x y œ È2
œ 3 Ê r ˆcos ) cos
Ê r cos )
1
2
x "# y œ 5 Ê È3 x y
y œ 2 Ê È2 x È2 y œ 4 Ê y œ x 2È2
4. r cos ˆ) ˆ 14 ‰‰ œ 4 Ê r cos ˆ) 14 ‰ œ 4 Ê r ˆcos ) cos
Ê
r cos ) "# r sin ) œ 5 Ê
21
3
sin ) sin
"
#
r sin ) œ 3 Ê x
Ê x È 3 y œ 6 Ê y œ
È3
3
È3
#
21 ‰
3
œ3
yœ3
x 2È 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
668
Chapter 10 Conic Sections and Polar Coordinates
8. r cos ˆ) 13 ‰ œ 2 Ê r ˆcos ) cos
Ê
1
2
r cos )
È3
2
r sin ) œ 2 Ê
Ê x È3 y œ 4 Ê y œ
È3
3
1
3
sin ) sin 13 ‰ œ 2
"
#
x
x
È3
#
yœ2
4È 3
3
È
9. È2 x È2 y œ 6 Ê È2 r cos ) È2 r sin ) œ 6 Ê r Š #2 cos )
È2
#
sin )‹ œ 3 Ê r ˆcos
1
4
cos ) sin
œ 3 Ê r cos ˆ) 14 ‰ œ 3
È
10. È3 x y œ 1 Ê È3 r cos ) r sin ) œ 1 Ê r Š #3 cos )
œ
"
#
Ê r cos ˆ) 16 ‰ œ
1
#
sin )‹ œ
"
#
Ê r ˆcos
1
6
cos ) sin
1
6
sin )‰
"
#
11. y œ 5 Ê r sin ) œ 5 Ê r sin ) œ 5 Ê r sin ()) œ 5 Ê r cos ˆ 1# ())‰ œ 5 Ê r cos ˆ) 1# ‰ œ 5
12. x œ 4 Ê r cos ) œ 4 Ê r cos ) œ 4 Ê r cos () 1) œ 4
13. r œ 2(4) cos ) œ 8 cos )
14. r œ 2(1) sin ) œ 2 sin )
15. r œ 2È2 sin )
16. r œ 2 ˆ "# ‰ cos ) œ cos )
17.
18.
19.
20.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
4
sin )‰
Section 10.8 Conic Sections in Polar Coordinates
21. (x 6)# y# œ 36 Ê C œ (6ß 0), a œ 6
Ê r œ 12 cos ) is the polar equation
22. (x 2)# y# œ 4 Ê C œ (2ß 0), a œ 2
Ê r œ 4 cos ) is the polar equation
23. x# (y 5)# œ 25 Ê C œ (!ß 5), a œ 5
Ê r œ 10 sin ) is the polar equation
24. x# (y 7)# œ 49 Ê C œ (!ß 7), a œ 7
Ê r œ 14 sin ) is the polar equation
25. x# 2x y# œ 0 Ê (x 1)# y# œ 1
Ê C œ (1ß 0), a œ 1 Ê r œ 2 cos ) is
the polar equation
26. x# 16x y# œ 0 Ê (x 8)# y# œ 64
Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the
polar equation
#
27. x# y# y œ 0 Ê x# ˆy "# ‰ œ 4"
Ê C œ ˆ!ß "# ‰ , a œ "# Ê r œ sin ) is the
#
28. x# y# 43 y œ 0 Ê x# ˆy 23 ‰ œ 49
Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the
polar equation
polar equation
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
669
670
Chapter 10 Conic Sections and Polar Coordinates
29. e œ 1, x œ 2 Ê k œ 2 Ê r œ
2(1)
1 (1) cos )
œ
2
1cos )
30. e œ 1, y œ 2 Ê k œ 2 Ê r œ
2(1)
1 (1) sin )
œ
2
1sin )
31. e œ 5, y œ 6 Ê k œ 6 Ê r œ
6(5)
1 5 sin )
32. e œ 2, x œ 4 Ê k œ 4 Ê r œ
4(2)
1 2 cos )
33. e œ "# , x œ 1 Ê k œ 1 Ê r œ
ˆ "# ‰ (1)
1 ˆ "# ‰ cos )
35. e œ "5 , x œ 10 Ê k œ 10 Ê r œ
37. r œ
"
1 cos )
38. r œ
6
2 cos )
œ
30
15 sin )
8
12 cos )
œ
ˆ "4 ‰ (2)
1 ˆ "4 ‰ cos )
34. e œ 4" , x œ 2 Ê k œ 2 Ê r œ
36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ
œ
1
2cos )
œ
ˆ "5 ‰ (10)
1 ˆ "5 ‰ sin )
ˆ "3 ‰ (6)
1 ˆ "3 ‰ sin )
œ
2
4cos )
œ
10
5sin )
6
3sin )
Ê e œ 1, k œ 1 Ê x œ 1
œ
3
1 ˆ "# ‰ cos )
Ê eœ
"
#
, k œ 6 Ê x œ 6;
#
a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 3 Ê
3
4
aœ3
Ê a œ 4 Ê ea œ 2
39. r œ
25
10 5 cos )
Ê eœ
"
#
Ê rœ
œ
ˆ #5 ‰
1 ˆ "# ‰ cos )
, k œ 5 Ê x œ 5; a a1 e# b œ ke
#
Ê a ’1 ˆ "# ‰ “ œ
40. r œ
ˆ 25
‰
10
5 ‰
1 ˆ 10
cos )
4
22 cos )
Ê rœ
5
#
Ê
2
1cos )
3
4
aœ
5
#
Ê aœ
10
3
Ê ea œ
5
3
Ê e œ 1, k œ 2 Ê x œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.8 Conic Sections in Polar Coordinates
41. r œ
eœ
400
16 8 sin )
"
#
Ê rœ
ˆ 400
‰
16
8 ‰
1 ˆ 16
sin )
25
1 ˆ "# ‰ sin )
, k œ 50 Ê y œ 50; a a1 e# b œ ke
#
Ê a ’1 ˆ "# ‰ “ œ 25 Ê
Ê ea œ
42. r œ
Ê rœ
a œ 25 Ê a œ
3
4
100
3
50
3
12
3 3 sin )
Ê rœ
4
1 sin )
Ê e œ 1,
43. r œ
kœ4 Ê yœ4
44. r œ
4
2 sin )
Ê rœ
8
2 2 sin )
Ê rœ
4
1 sin )
Ê e œ 1,
k œ 4 Ê y œ 4
2
1 ˆ "# ‰ sin )
Ê eœ
"
#
,kœ4
#
Ê y œ 4; a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 2
Ê
3
4
aœ2 Ê aœ
8
3
Ê ea œ
4
3
45.
46.
47.
48.
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671
672
Chapter 10 Conic Sections and Polar Coordinates
49.
50.
51.
52.
53.
54.
55.
56.
57. (a) Perihelion œ a ae œ a(1 e), Aphelion œ ea a œ a(1 e)
(b)
Planet
Perihelion
Aphelion
Mercury
0.3075 AU
0.4667 AU
Venus
0.7184 AU
0.7282 AU
Earth
0.9833 AU
1.0167 AU
Mars
1.3817 AU
1.6663 AU
Jupiter
4.9512 AU
5.4548 AU
Saturn
9.0210 AU
10.0570 AU
Uranus
18.2977 AU
20.0623 AU
Neptune
29.8135 AU
30.3065 AU
Pluto
29.6549 AU
49.2251 AU
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 10.8 Conic Sections in Polar Coordinates
(0.3871) a1 0.2056# b
0.3707
œ 1 0.2056
1 0.2056 cos )
cos )
(0.7233) a1 0.0068# b
0.7233
Venus: r œ 1 0.0068 cos ) œ 1 0.0068 cos )
0.0167# b
0.9997
Earth: r œ 11a10.0167
cos ) œ 1 0.0617 cos )
a1 0.0934# b
1.511
Mars: r œ (1.524)
œ 1 0.0934
1 0.0934 cos )
cos )
(5.203) a1 0.0484# b
5.191
Jupiter: r œ 1 0.0484 cos ) œ 1 0.0484 cos )
a1 0.0543# b
9.511
Saturn: r œ (9.539)
œ 1 0.0543
1 0.0543 cos )
cos )
(19.18) a1 0.0460# b
19.14
Uranus: r œ 1 0.0460 cos ) œ 1 0.0460 cos )
a1 0.0082# b
30.06
Neptune: r œ (30.06)
œ 1 0.0082
1 0.0082 cos )
cos )
58. Mercury: r œ
59. (a) r œ 4 sin ) Ê r# œ 4r sin ) Ê x# y# œ 4y;
È
r œ È3 sec ) Ê r œ cos3) Ê r cos ) œ È3
(b)
#
Ê x œ È3 ; x œ È3 Ê ŠÈ3‹ y# œ 4y
Ê y# 4y 3 œ 0 Ê (y 3)(y 1) œ 0 Ê y œ 3
or y œ 1. Therefore in Cartesian coordinates, the points
of intersection are ŠÈ3ß 3‹ and ŠÈ3ß 1‹. In polar
coordinates, 4 sin ) œ È3 sec ) Ê 4 sin ) cos ) œ È3
Ê 2 sin ) cos ) œ
21
3
Ê )œ
1
6
1
3
or
È3
#
;)œ
Ê sin 2) œ
1
6
È3
#
Ê 2) œ
Ê r œ 2, and ) œ
1
3
or
1
3
Ê r œ 2È3 Ê ˆ2ß 16 ‰ and Š2È3ß 13 ‹ are the points
of intersection in polar coordinates.
60. (a) r œ 8 cos ) Ê r# œ 8r cos ) Ê x# y# œ 8x
Ê x# 8x y# œ 0 Ê (x 4)# y# œ 16;
r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2
(b)
Ê x œ 2; x œ 2 Ê 2# 8(2) y# œ 0
Ê y# œ 12 Ê y œ „ 2È3. Therefore Š2ß „ 2È3‹
are the points of intersection in Cartesian coordinates.
In polar coordinates, 8 cos ) œ 2 sec ) Ê 8 cos# ) œ 2
Ê cos# ) œ "4 Ê cos ) œ „ #" Ê ) œ 13 , 231 , 431 , or
51
3
Ê r œ 4, and ) œ 231 and 431
Ê r œ 4 Ê ˆ4ß 13 ‰ and ˆ4ß 531 ‰ are the points of
intersection in polar coordinates. The points ˆ4ß 231 ‰ and ˆ4ß 431 ‰ are the same points.
;)œ
1
3
and
51
3
61. r cos ) œ 4 Ê x œ 4 Ê k œ 4: parabola Ê e œ 1 Ê r œ
62. r cos ˆ) 1# ‰ œ 2 Ê r ˆcos ) cos
Ê rœ
1
#
4
1 cos )
sin ) sin 1# ‰ œ 2 Ê r sin ) œ 2 Ê y œ 2 Ê k œ 2: parabola Ê e œ 1
2
1 sin )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
673
674
Chapter 10 Conic Sections and Polar Coordinates
63. (a) Let the ellipse be the orbit, with the Sun at one focus.
rmin
Then rmax œ a c and rmin œ a c Ê rrmax
max rmin
œ
(a c) (a c)
(a c) (a c)
œ
2c
2a
œ
c
a
œe
(b) Let F" , F# be the foci. Then PF" PF# œ 10 where
P is any point on the ellipse. If P is a vertex, then
PF" œ a c and PF# œ a c
Ê (a c) (a c) œ 10
Ê 2a œ 10 Ê a œ 5. Since e œ ca we have 0.2 œ
c
5
Ê c œ 1.0 Ê the pins should be 2 inches apart.
64. e œ 0.97, Major axis œ 36.18 AU Ê a œ 18.09, Minor axis œ 9.12 AU Ê b œ 4.56 (1 AU ¸ 1.49 ‚ 10) km)
(a) r œ
ke
1e cos )
œ
(b) ) œ 0 Ê r œ
(c) ) œ 1 Ê r œ
(18.09) c1(0.97)# d
a a1 e # b
1.07
œ 10.97
1e cos ) œ
10.97 cos )
cos ) AU
1.07
(
10.97 ¸ 0.5431 AU ¸ 8.09 ‚ 10 km
1.07
*
10.97 ¸ 35.7 AU ¸ 5.32 ‚ 10 km
65. x# y# 2ay œ 0 Ê (r cos ))# (r sin ))# 2ar sin ) œ 0
Ê r# cos# ) r# sin# ) 2ar sin ) œ 0 Ê r# œ 2ar sin )
Ê r œ 2a sin )
66. y# œ 4ax 4a# Ê (r sin ))# œ 4ar cos ) 4a# Ê r# sin# )
œ 4ar cos ) 4a# Ê r# a1 cos# )b œ 4ar cos ) 4a#
Ê r# r# cos# ) œ 4ar cos ) 4a# Ê r#
œ r# cos# ) 4ar cos ) 4a# Ê r# œ (r cos ) 2a)#
Ê r œ „ (r cos ) 2a) Ê r r cos ) œ 2a or
2a
r r cos ) œ 2a Ê r œ 12a
cos ) or r œ 1cos ) ;
the equations have the same graph, which is a parabola
opening to the right
67. x cos ! y sin ! œ p Ê r cos ) cos ! r sin ) sin ! œ p
Ê r(cos ) cos ! sin ) sin !) œ p Ê r cos () !) œ p
#
68. ax# y# b 2ax ax# y# b a# y# œ 0
Ê
Ê
Ê
Ê
Ê
Ê
Ê
#
ar# b 2a(r cos )) ar# b a# (r sin ))# œ 0
r% 2ar$ cos ) a# r# sin# ) œ 0
r# cr# 2ar cos ) a# a1 cos# )bd œ 0 (assume r Á 0)
r# 2ar cos ) a# a# cos# ) œ 0
ar# 2ar cos ) a# cos# )b a# œ 0
(r a cos ))# œ a# Ê r a cos ) œ „ a
r œ a(1 cos )) or r œ a(1 cos ));
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 10 Practice Exercises
the equations have the same graph, which is a cardioid
69 - 70. Example CAS commands:
Maple:
with( plots );#69
f := (r,k,e) -> k*e/(1+e*cos(theta));
elist := [3/4,1,5/4];
# (a)
P1 := seq( plot( f(r,-2,e), theta=-Pi..Pi, coords=polar ), e=elist ):
display( [P1], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=-2" );
P2 := seq( plot( f(r,2,e), theta=-Pi..Pi, coords=polar ), e=elist ):
display( [P2], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=2" );
elist2 := [7/6,5/4,4/3,3/2,2,3,5,10,20];
# (b)
P3 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist2 ):
display( [P3], insequence=true, view=[-20..20,-20..20], title="#69(b) (Section 10.8)\nk=-1, e>1" );
elist3 := [1/2,1/3,1/4,1/10,1/20];
P4 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist3 ):
display( [P4], insequence=true, title="#69(b) (Section 10.8)\nk=-1, e<1" );
klist := -5..-1;
# (c)
P5 := seq( plot( f(r,k,1/2), theta=-Pi..Pi, coords=polar ), k=klist ):
display( [P5], insequence=true, title="#69(c) (Section 10.8)\ne=1/2, k<0" );
P6 := seq( plot( f(r,k,1), theta=-Pi..Pi, coords=polar ), k=klist ):
display( [P6], insequence=true, view=[-4..50,-50..50], title="#69(c) (Section 10.8)\ne=1, k<0" );
P7 := seq( plot( f(r,k,2), theta=-Pi..Pi, coords=polar ), k=klist ):
display( [P5], insequence=true, title="#69(c) (Section 10.8)\ne=2, k<0" );
Mathematica: (assigned function and values for parameters and bounds may vary):
To do polar plots in Mathematica, it is necessary to first load a graphics package
In the PolarPlot command, it is assumed that the variable r is given as a function of the variable ).
< sin(x);
a := 0;
b := Pi;
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.1 Sequences
705
plot( f(x), x=a..b, title="#23(a) (Section 5.1)" );
N := [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist := [ a+1.*(b-a)/n*i $ i=0..n ];
Ylist := map( f, Xlist );
end do:
for n in N do
# (c)
Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist));
end do;
avg := FunctionAverage( f(x), x=a..b, output=value );
evalf( avg );
FunctionAverage(f(x),x=a..b,output=plot);
# (d)
fsolve( f(x)=avg, x=0.5 );
fsolve( f(x)=avg, x=2.5 );
fsolve( f(x)=Avg[1000], x=0.5 );
fsolve( f(x)=Avg[1000], x=2.5 );
Mathematica: (sequence functions may vary):
Clear[a, n]
a[n_]; = n1 / n
first25= Table[N[a[n]],{n, 1, 25}]
Limit[a[n], n Ä 8]
The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table
to more than the first 25 values.
If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the
limit, do the following.
Clear[minN, lim]
lim= 1
Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}]
minN
For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores
the elements of the sequence and helps to streamline computation.
Clear[a, n]
a[1]= 1;
a[n_]; = a[n]= a[n 1] (1/5)(n1)
first25= Table[N[a[n]], {n, 1, 25}]
The limit command does not work in this case, but the limit can be observed as 1.25.
Clear[minN, lim]
lim= 1.25
Do[{diff=Abs[a[n] lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}]
minN
141. Example CAS Commands:
Maple:
with( Student[Calculus1] );
A := n->(1+r/m)*A(n-1) + b;
A(0) := A0;
A(0) := 1000; r := 0.02015; m := 12; b := 50;
pts1 := [seq( [n,A(n)], n=0..99 )]:
plot( pts1, style=point, title="#141(a) (Section 11.1)");
# (a)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
706
Chapter 11 Infinite Sequences and Series
A(60);
The sequence { A[n] } is not unbounded;
limit( A[n], n=infinity ) = infinity.
A(0) := 5000; r := 0.0589; m := 12; b := -50;
# (b)
pts1 := [seq( [n,A(n)], n=0..99 )]:
plot( pts1, style=point, title="#141(b) (Section 11.1)");
A(60);
pts1 := [seq( [n,A(n)], n=0..199 )]:
plot( pts1, style=point, title="#141(b) (Section 11.1)");
# This sequence is not bounded, and diverges to -infinity:
limit( A[n], n=infinity ) = -infinity.
A(0) := 5000; r := 0.045; m := 4; b := 0;
# (c)
for n from 1 while A(n)<20000 do end do; n;
It takes 31 years (124 quarters) for the investment to grow to $20,000 when the interest rate is 4.5%, compounded
quarterly.
r := 0.0625;
for n from 1 while A(n)<20000 do end do; n;
When the interest rate increases to 6.25% (compounded quarterly), it takes only 22.5 years for the balance to reach
$20,000.
B := k -> (1+r/m)^k * (A(0)+m*b/r) - m*b/r;
# (d)
A(0) := 1000.; r := 0.02015; m := 12; b := 50;
for k from 0 to 49 do
printf( "%5d %9.2f %9.2f %9.2f\n", k, A(k), B(k), B(k)-A(k) );
end do;
A(0) := 'A(0)'; r := 'r'; m := 'm'; b := 'b'; n := 'n';
eval( AA(n+1) - ((1+r/m)*AA(n) + b), AA=B );
simplify( % );
142. Example CAS Commands:
Maple:
r := 3/4.;
# (a)
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 99 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title="#142(a) (Section 11.1)" );
R1 := [1.1, 1.2, 1.5, 2.5, 2.8, 2.9];
# (b)
for r in R1 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 99 do
A := r*A*(1-A);
L := L, [n,A];
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.1 Sequences
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(b) (Section 11.1)\nr = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R1)], insequence=true );
R2 := [3.05, 3.1, 3.2, 3.3, 3.35, 3.4];
# (c)
for r in R2 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 99 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(c) (Section 11.1)\nr = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R2)], insequence=true );
R3 := [3.46, 3.47, 3.48, 3.49, 3.5, 3.51, 3.52, 3.53, 3.542, 3.544, 3.546, 3.548];
for r in R3 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 199 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(d) (Section 11.1)\nr = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R3)], insequence=true );
R4 := [3.5695];
# (e)
for r in R4 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(e) (Section 11.1)\nr = %f", r);
# (d)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
707
708
Chapter 11 Infinite Sequences and Series
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R4)], insequence=true );
R5 := [3.65];
# (f)
for r in R5 do
for k in $1..9 do
A := k/10.;
L := [0,A];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,k/10] := [L];
end do:
t := sprintf("#142(f) (Section 11.1)\nr = %f", r);
P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t );
end do:
display( [seq(P[r], r=R5)], insequence=true );
R6 := [3.65, 3.75];
# (g)
for r in R6 do
for a in [0.300, 0.301, 0.600, 0.601 ] do
A := a;
L := [0,a];
for n from 1 to 299 do
A := r*A*(1-A);
L := L, [n,A];
end do;
pt[r,a] := [L];
end do:
t := sprintf("#142(g) (Section 11.1)\nr = %f", r);
P[r] := plot( [seq( pt[r,a], a=[0.300, 0.301, 0.600, 0.601] )], style=point, title=t );
end do:
display( [seq(P[r], r=R6)], insequence=true );
11.2 INFINITE SERIES
1. sn œ
a a1 r n b
(1 r)
œ
n
2 ˆ1 ˆ "3 ‰ ‰
"
1 ˆ3‰
2. sn œ
a a1 r n b
(1 r)
œ
9 ‰ˆ
" ‰n ‰
ˆ 100
1 ˆ 100
"
1 ˆ 100 ‰
3. sn œ
a a1 r n b
(1 r)
œ
1 ˆ "# ‰
1 ˆ "# ‰
4. sn œ
1 (2)n
1 (2)
, a geometric series where krk 1 Ê divergence
5.
"
(n 1)(n #)
œ
"
n1
n
Ê n lim
s œ
Ä_ n
Ê n lim
s œ
Ä_ n
Ê n lim
s œ
Ä_ n
"
n#
2
1 ˆ "3 ‰
"
ˆ #3 ‰
œ3
9 ‰
ˆ 100
" ‰
1 ˆ 100
œ
œ
"
11
2
3
Ê sn œ ˆ #" 3" ‰ ˆ 3" 4" ‰ á ˆ n " 1
" ‰
n#
œ
"
#
"
n#
Ê n lim
s œ
Ä_ n
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
#
Section 11.2 Infinite Series
6.
5
n(n 1)
œ
5
n
5
n1
Ê sn œ ˆ5 52 ‰ ˆ 52 53 ‰ ˆ 53 54 ‰ á ˆ n 5 1 n5 ‰ ˆ n5
5 ‰
n1
œ 5
709
5
n1
Ê n lim
s œ5
Ä_ n
7. 1
8.
"
16
9.
7
4
"
4
10. 5
"
16
"
64
7
16
5
4
"
256
7
64
5
16
"
64
á , the sum of this geometric series is
"
1 ˆ "3 ‰
œ 10
3
#
œ
"
1 ˆ "3 ‰
œ 10
3
#
œ
13. (1 1) ˆ 1# "5 ‰ ˆ 14
1
1 ˆ "# ‰
14. 2
15.
16.
4
5
"
1 ˆ "5 ‰
8
25
œ2
5
6
œ
"
1 ˆ "4 ‰
œ
A
2n 1
" ‰
25
œ 17
6
B
2n 1
4
5
"
1#
7
3
5
1 ˆ "4 ‰
œ4
" ‰
#7
á , is the sum of two geometric series; the sum is
" ‰
#7
á , is the difference of two geometric series; the sum is
ˆ 18
2
5
" ‰
1#5
4
25
á , is the sum of two geometric series; the sum is
8
125
á ‰ ; the sum of this geometric series is 2 Š 1 "ˆ 2 ‰ ‹ œ
5
4
"
"
"‰
" ‰
ˆ
ˆ" "‰ ˆ"
(4n 3)(4n 1) œ 4n 3 4n 1 Ê sn œ 1 5 5 9 9 13
ˆ1 4n " 1 ‰ œ 1
ˆ 4n " 3 4n " 1 ‰ œ 1 4n " 1 Ê n lim
s œ n lim
Ä_ n
Ä_
6
(2n 1)(2n 1)
œ
17
#
á œ 2 ˆ1
16
125
œ
œ
23
#
12. (5 1) ˆ 5# "3 ‰ ˆ 54 9" ‰ ˆ 85
5
1 ˆ "# ‰
ˆ 74 ‰
1 ˆ "4 ‰
á , the sum of this geometric series is
5
64
11. (5 1) ˆ 5# "3 ‰ ˆ 54 9" ‰ ˆ 85
5
1 ˆ "# ‰
" ‰
ˆ 16
1 ˆ 4" ‰
á , the sum of this geometric series is
á , the sum of this geometric series is
"
1 ˆ "4 ‰
œ
A(2n 1) B(2n 1)
(2n 1)(2n 1)
á ˆ 4n " 7
10
3
" ‰
4n 3
Ê A(2n 1) B(2n 1) œ 6
2A 2B œ 0
ABœ0
Ê (2A 2B)n (A B) œ 6 Ê œ
Ê œ
Ê 2A œ 6 Ê A œ 3 and B œ 3. Hence,
A Bœ6
ABœ6
k
!
nœ1
œ 3 ˆ1
17.
nœ1
" ‰
#k 1
40n
(2n 1)# (2n 1)#
œ
k
œ 3 ! ˆ #n " 1
6
(2n 1)(2n 1)
œ
œ 3 Š 1"
Ê the sum is lim 3 ˆ1
kÄ_
A
(2n 1)
#
" ‰
#n 1
B
(2n 1)#
#
C
(2n 1)
"
3
" ‰
#k 1
"
3
"
5
"
5
"
7
á
"
#(k 1) 1
"
2k 1
"
#k 1 ‹
œ3
D
(2n 1)#
#
A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1) D(2n 1)#
(2n 1)# (2n 1)#
#
#
Ê A(2n 1)(2n 1) B(2n 1) C(2n 1)(2n 1)# D(2n 1)# œ 40n
Ê A a8n$ 4n# 2n 1b B a4n# 4n 1b C a8n$ 4n# 2n 1b œ D a4n# 4n 1b œ 40n
Ê (8A 8C)n$ (4A 4B 4C 4D)n# (2A 4B 2C 4D)n (A B C D) œ 40n
Ú
Ú
8A 8C œ 0
8A 8C œ 0
Ý
Ý
Ý
Ý
4A 4B 4C 4D œ 0
A BC Dœ 0
B Dœ 0
Ê Û
Ê Û
Ê œ
Ê 4B œ 20 Ê B œ 5
œ
2A
4B
2C
4D
40
A
2
œ
2D œ 20
B
C
2D
20
2B
Ý
Ý
Ý
Ý
Ü A B C D œ 0
Ü A B C D œ 0
k
ACœ0
Ê C œ 0 and A œ 0. Hence, ! ’ (#n 1)40n
and D œ 5 Ê œ
# (2n 1)# “
A 5 C 5 œ 0
nœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
710
Chapter 11 Infinite Sequences and Series
k
œ 5 ! ’ (#n " 1)#
nœ1
18.
"
(#n 1)# “
œ 5 Š1
"
(2k 1)# ‹
2n 1
n# (n 1)#
"
n#
œ
"
È2 ‹
Š È"
2
" ‰
#"Î#
"
ˆ #"Î#
Ê n lim
s œ
Ä_ n
21. sn œ ˆ ln"3
œ ln"#
" ‰
ln #
"
#
"
1
œ
"
#5
"
#5
Š È"3
"
(2k 1)# ‹
"
(#k 1)#
"
(#k 1)# ‹
œ5
" ‰
16
á ’ (n " 1)#
"
n# “
’ n"#
"
(n 1)# “
"
È4 ‹
á Š È "
n
1
"
Èn ‹
Š È"n
"
Èn 1 ‹
œ1
"
Èn 1
œ1
"
ˆ #"Î$
" ‰
ln 3
"
(2(k 1) 1)#
á
œ1
"
Èn 1 ‹
" ‰
#"Î$
#"
ˆ ln"4
"
ln (n 2)
"
(n 1)# “
"
È3 ‹
Ê n lim
s œ n lim
Š1
Ä_ n
Ä_
20. sn œ ˆ "#
"
9
Ê sn œ ˆ1 4" ‰ ˆ 4" 9" ‰ ˆ 9"
Ê n lim
s œ n lim
’1
Ä_ n
Ä_
19. sn œ Š1
"
9
Ê the sum is n lim
5 Š1
Ä_
"
(n 1)#
œ 5 Š 1"
" ‰
#"Î%
ˆ ln"5
á ˆ #1ÎÐ"n1Ñ
" ‰
ln 4
" ‰
#1În
á Š ln (n" 1)
ˆ #1"În
"
ln n ‹
" ‰
#1ÎÐn1Ñ
Š ln (n" 2)
œ
"
#
"
#1ÎÐn1Ñ
"
ln (n 1) ‹
Ê n lim
s œ ln"#
Ä_ n
22. sn œ ctan" (1) tan" (2)d ctan" (2) tan" (3)d á ctan" (n 1) tan" (n)d
ctan" (n) tan" (n 1)d œ tan" (1) tan" (n 1) Ê n lim
s œ tan" (1)
Ä_ n
23. convergent geometric series with sum
"
1 Š È" ‹
œ
2
È2
È 2 1
1
#
œ
1
4
1
#
œ 14
œ 2 È2
Š 3# ‹
24. divergent geometric series with krk œ È2 1
25. convergent geometric series with sum
26. n lim
(1)n1 n Á 0 Ê diverges
Ä_
27. n lim
cos (n1) œ n lim
(1)n Á 0 Ê diverges
Ä_
Ä_
28. cos (n1) œ (1)n Ê convergent geometric series with sum
29. convergent geometric series with sum
30. n lim
ln
Ä_
"
n
"
1Š
"
‹
e#
œ
"
1 Š "5 ‹
œ
5
6
e#
e # 1
œ _ Á 0 Ê diverges
31. convergent geometric series with sum
2
"
1 Š 10
‹
32. convergent geometric series with sum
"
1 Š "x ‹
33. difference of two geometric series with sum
ˆ1 "n ‰n œ lim ˆ1
34. n lim
Ä_
nÄ_
" ‰n
n
2œ
œ
20
9
18
9
œ
2
9
x
x1
"
1 Š 23 ‹
"
1 Š 3" ‹
œ3
3
#
œ
3
#
œ e" Á 0 Ê diverges
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1 Š "# ‹
œ1
Section 11.2 Infinite Series
35. n lim
Ä_
n!
1000n
œ _ Á 0 Ê diverges
_
_
nœ1
nœ1
nn
n!
36. n lim
Ä_
œ n lim
Ä_
n†nân
1†#ân
n lim
n œ _ Ê diverges
Ä_
37. ! ln ˆ n n 1 ‰ œ ! cln (n) ln (n 1)d Ê sn œ cln (1) ln (2)d cln (2) ln (3)d cln (3) ln (4)d á
cln (n 1) ln (n)d cln (n) ln (n 1)d œ ln (1) ln (n 1) œ ln (n 1) Ê n lim
s œ _, Ê diverges
Ä_ n
38. n lim
a œ n lim
ln ˆ 2n n 1 ‰ œ ln ˆ #" ‰ Á 0 Ê diverges
Ä_ n
Ä_
"
1 ˆ 1e ‰
39. convergent geometric series with sum
40. divergent geometric series with krk œ
_
_
nœ0
nœ0
e1
1e
¸
1
1e
œ
23.141
22.459
1
41. ! (1)n xn œ ! (x)n ; a œ 1, r œ x; converges to
_
_
nœ0
nœ0
"
1 (x)
n
42. ! (1)n x2n œ ! ax# b ; a œ 1, r œ x# ; converges to
43. a œ 3, r œ
_
44. !
nœ0
œ
x1
#
_
(1)n
#
ˆ 3 "sin x ‰n œ !
nœ0
3 sin x
2(4 sin x)
3 sin x
8 2 sin x
œ
3
"
1 Šx
# ‹
; converges to
"
#
6
3x
ˆ 3 "sin x ‰n ; a œ
"
1 2x
46. a œ 1, r œ x"# ; converges to
,rœ
for k2xk 1 or kxk
"
#
œ
Ÿ
"
3 sin x
x#
x# 1
"
1 x#
x"
#
"
3 sin x
"
#
"
1 Š "
#‹
"
4
"
#
for 1
Ÿ
for all x ˆsince
45. a œ 1, r œ 2x; converges to
œ
œ
"
1x
for kxk 1
for kxk 1
1 or 1 x 3
; converges to
ˆ "# ‰
1 Š 3 "
sin x ‹
for all x‰
for ¸ x1# ¸ 1 or kxk 1.
x
47. a œ 1, r œ (x 1)n ; converges to
48. a œ 1, r œ
3x
#
"
x
1 Š3
# ‹
; converges to
"
1 sin x
49. a œ 1, r œ sin x; converges to
50. a œ 1, r œ ln x; converges to
_
51. 0.23 œ !
nœ0
_
53. 0.7 œ !
nœ0
23
100
7
10
ˆ 10" # ‰n œ
" ‰n
ˆ 10
œ
"
1 ln x
23
Š 100
‹
" ‰
1 ˆ 100
7
Š 10
‹
1
"
Š 10
‹
"
1 (x 1)
œ
œ
2
x1
for kx 1k 1 or 2 x 0
for ¸ 3 # x ¸ 1 or 1 x 5
for x Á (2k 1) 1# , k an integer
for kln xk 1 or e" x e
_
52. 0.234 œ !
23
99
nœ0
_
54. 0.d œ !
7
9
_
6
Š 100
‹
nœ0
"
1 Š 10
‹
1 ‰ ˆ 6 ‰ ˆ " ‰n
55. 0.06 œ ! ˆ 10
œ
10
10
œ
"
#x
œ
nœ0
œ
6
90
œ
d
10
234
1000
ˆ 10" $ ‰n œ
" ‰n
ˆ 10
œ
234
Š 1000
‹
"
1 Š 1000
‹
d
Š 10
‹
"
1 Š 10
‹
œ
d
9
"
15
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
234
999
711
712
Chapter 11 Infinite Sequences and Series
_
56. 1.414 œ 1 !
nœ0
57. 1.24123 œ
124
100
414
1000
_
!
nœ0
_
58. 3.142857 œ 3 !
nœ0
_
59. (a) !
nœ2
_
60. (a) !
nœ1
414
Š 1000
‹
ˆ 10" $ ‰n œ 1
123
10&
ˆ 10" $ ‰n œ
142,857
10'
œ1
"
1 Š 1000
‹
124
100
ˆ 10" ' ‰n œ 3
Š 123& ‹
10
1Š
"
‹
10$
Š 142,857
' ‹
10
1Š
"
‹
10'
414
999
œ
_
(b) !
5
(n 2)(n 3)
(b) !
nœ0
_
nœ3
"
#
"
4
(b) one example is 3#
(c) one example is 1
"
#
"
8
"
16
á œ
3
4
3
8
3
16
"
4
"
8
Š "# ‹
1 Š "# ‹
á œ
"
16
œ3
"
(n 4)(n 5)
61. (a) one example is
124
100
"413
999
œ
œ
123
10& 10#
œ
142,857
10' 1
124
100
3,142,854
999,999
123
99,900
œ
œ
123,999
99,900
œ
41,333
33,300
116,402
37,037
_
"
(n 2)(n 3)
(c) !
5
(n 2)(n 1)
(c) !
nœ5
_
nœ20
"
(n 3)(n #)
5
(n 19)(n 18)
œ1
Š 3# ‹
1 Š "# ‹
œ 3
á ; the series
k
#
k
4
k
8
á œ
Š k# ‹
1 Š "# ‹
œ k where k is any positive or
negative number.
_
Š k# ‹
nœ0
1 Š "# ‹
n 1
62. The series ! kˆ 12 ‰
is a geometric series whose sum is
œ k where k can be any positive or negative number.
_
_
_
_
_
nœ1
nœ1
nœ1
nœ1
nœ1
_
_
_
_
_
nœ1
nœ1
nœ1
nœ1
nœ1
n
n
63. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! Š bann ‹ œ ! (1) diverges.
n
n
n
64. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! aan bn b œ ! ˆ 4" ‰ œ
n
n
_
65. Let an œ ˆ "4 ‰ and bn œ ˆ #" ‰ . Then A œ ! an œ
nœ1
"
3
_
_
_
nœ1
nœ1
nœ1
"
3
Á AB.
n
, B œ ! bn œ 1 and ! Š bann ‹ œ ! ˆ #" ‰ œ 1 Á
66. Yes: ! Š a"n ‹ diverges. The reasoning: ! an converges Ê an Ä 0 Ê
"
an
A
B
.
Ä _ Ê ! Š a"n ‹ diverges by the
nth-Term Test.
67. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series
that diverges does not change the divergence of the series.
68. Let An œ a" a# á an and n lim
A œ A. Assume ! aan bn b converges to S. Let
Ä_ n
Sn œ (a" b" ) (a# b# ) á (an bn ) Ê Sn œ (a" a# á an ) (b" b# á bn )
Ê b" b# á bn œ Sn An Ê n lim
ab" b# á bn b œ S A Ê ! bn converges. This
Ä_
contradicts the assumption that ! bn diverges; therefore, ! aan bn b diverges.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.3 The Integral Test
69. (a)
(b)
œ5 Ê
2
1r
Š 13
2 ‹
1r
œ1r Ê rœ
2
5
œ5 Ê
13
10
70. 1 eb e2b á œ
3
5
#
; 2 2 ˆ 35 ‰ 2 ˆ 35 ‰ á
3
œ 1 r Ê r œ 10
;
"
1 e b
"
9
œ9 Ê
13
2
13
#
3 ‰
ˆ 10
œ 1 eb Ê eb œ
3 ‰#
ˆ 10
13
#
3 ‰$
ˆ 10
á
13
#
Ê b œ ln ˆ 89 ‰
8
9
71. sn œ 1 2r r# 2r$ r% 2r& á r2n 2r2n1 , n œ 0, 1, á
Ê sn œ a1 r# r% á r2n b a2r 2r$ 2r& á 2r2n1 b Ê n lim
s œ
Ä_ n
1 2r
œ 1 r# , if kr# k 1 or krk 1
72. L sn œ
a
1r
a a1 r n b
1r
"
1 r#
2r
1 r#
arn
1 r
œ
#
73. distance œ 4 2 ’(4) ˆ 34 ‰ (4) ˆ 34 ‰ á “ œ 4 2
3
1 Š 34 ‹
#
œ 28 m
$
#
4
4 ‰ ˆ3‰
4 ‰ ˆ3‰
4 ‰ ˆ3‰
4
4
3
3
Ɉ 4.9
74. time œ É 4.9
2Ɉ 4.9
2Ɉ 4.9
á œ É 4.9
2É 4.9
”É 4 É ˆ 4 ‰ á •
4 2
4
4
œ
2
È4.9
Š È44.9 ‹ –
É 34
1 É 34
—œ
2
È4.9
#
Š4 2È3‹ 4È3
È
3
Š È44.9 ‹ Š 2 È
‹œ
3
#
75. area œ 2# ŠÈ2‹ (1)# Š È" ‹ á œ 4 2 1
2
#
76. area œ 2 –
1 Š "# ‹
#
#
— 4–
1 Š 4" ‹
#
œ
È4.9 Š2 È3‹
"
#
á œ
4
1
— 8–
"
— á œ 1 ˆ4
#
#
nc1
"
8
በœ 1
1
16
È3
ˆ " ‰2
4 ‹ 3
A% œ A$ 3a4b2 Š
An œ
lim
È3
4
nÄ_
œ
È3
ˆ " ‰2
4 ‹ 33
n
! 3a4bk2 Š
kœ2
An œ
È3
4
_
nœ1
"
n#
È3
1#
, A$ œ A# 3a4bŠ
, A5 œ A4 3a4b3 Š
È3
ˆ " ‰ k 1
4 ‹ 32
È3
lim
nÄ_ Œ 4
78. Each term of the series !
œ
n
3È3Œ!
kœ2
È3
4
È3
ˆ " ‰2
4 ‹ 32
È3
ˆ " ‰2
4 ‹ 34 ,
œ
kœ2
œ
nc1
œ
1
#
œ_
A" œ
È3
4 ,
...,
n
È3
4
1 Š "# ‹
È3 2
4 s , we see that
È3
È3
È3
4 12 #7 ,
k 1
! 3È3a4bk$ ˆ 9" ‰
œ
4kc$
9k1
Š 4" ‹
Ê n lim
L œ n lim
3 ˆ 43 ‰
Ä_ n
Ä_
(b) Using the fact that the area of an equilateral triangle of side length s is
¸ 12.58 sec
œ 8 m#
"
#
#
1 Š 8" ‹
77. (a) L" œ 3, L# œ 3 ˆ 43 ‰ , L$ œ 3 ˆ 43 ‰ , á , Ln œ 3 ˆ 43 ‰
A# œ A" 3Š
4 2È 3
È4.9 Š2 È3‹
1
36
3È 3 Œ 1 4 œ
9
n
4kc$
.
9k1
È3
4
3È3Œ!
È3
4
1 ‰
3È3ˆ 20
œ
kœ2
2È 3
5
represents the area of one of the squares shown in the figure, and all of the
_
n
squares lie inside the rectangle of width 1 and length ! ˆ "# ‰ œ
œ
n 0
"
1
_
rectangle completely, and the area of the rectangle is 2, we have !
nœ1
"
#
"
n#
œ 2. Since the squares do not fill the
2.
11.3 THE INTEGRAL TEST
1. converges; a geometric series with r œ
"
10
1
2. converges; a geometric series with r œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
e
1
713
714
Chapter 11 Infinite Sequences and Series
3. diverges; by the nth-Term Test for Divergence, n lim
Ä_
4. diverges by the Integral Test; '1
n
_
5. diverges; !
nœ1
3
Èn
_
6. converges; !
nœ1
_
"
Èn
œ3!
nœ1
2
nÈ n
_
dx œ 5 ln (n 1) 5 ln 2 Ê '1
5
x1
_
nœ1
8
n
_
œ 2 !
nœ1
_
œ 8 !
nœ1
5
x1
dx Ä _
, which is a divergent p-series (p œ #" )
"
n$Î#
, which is a convergent p-series (p œ 3# )
7. converges; a geometric series with r œ
8. diverges; !
œ1Á0
n
n1
"
8
1
_
and since !
1
n
nœ1
9. diverges by the Integral Test:
"
n
_
diverges, 8 !
nœ1
'2n lnxx dx œ "# aln# n ln 2b
Ê
t œ ln x ×
dt œ dx
Ä
10. diverges by the Integral Test:
x
Õ dx œ et dt Ø
œ lim 2ebÎ2 (b 2) 2eÐln 2ÑÎ2 (ln 2 2)‘ œ _
'2_ lnÈxx dx; Ô
1
n
diverges
'2_ lnxx dx
Ä _
'ln_2 tetÎ2 dt œ
b
lim 2tetÎ2 4etÎ2 ‘ ln 2
bÄ_
bÄ_
11. converges; a geometric series with r œ
12. diverges; n lim
Ä_
_
13. diverges; !
nœ0
2
n 1
5n
4n 3
_
œ 2 !
nœ0
"
n1
1
ˆ ln 5 ‰ ˆ 54 ‰n œ _ Á 0
œ n lim
Ä _ ln 4
5n ln 5
4n ln 4
œ n lim
Ä_
2
3
, which diverges by the Integral Test
14. diverges by the Integral Test:
'1n 2xdx 1 œ #" ln (2n 1)
15. diverges; n lim
a œ n lim
Ä_ n
Ä_
2n
n1
16. diverges by the Integral Test:
'1n Èx ˆÈdxx 1‰ ; – u œ
œ n lim
Ä_
2n ln 2
1
Ä _ as n Ä _
œ_Á0
Èx "
du œ
dx
Èx
— Ä
Ènb1 du
'2
u
œ ln ˆÈn 1‰ ln 2
Ä _ as n Ä _
17. diverges; n lim
Ä_
Èn
ln n
œ n lim
Ä_
"
Š 2È
‹
n
Š "n ‹
œ n lim
Ä_
Èn
#
œ_Á0
ˆ1 n" ‰n œ e Á 0
18. diverges; n lim
a œ n lim
Ä_ n
Ä_
19. diverges; a geometric series with r œ
"
ln #
20. converges; a geometric series with r œ
¸ 1.44 1
"
ln 3
¸ 0.91 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.3 The Integral Test
21. converges by the Integral Test:
'3_ (ln x) ÈŠ(ln‹x) 1 dx; ”
"
x
#
u œ ln x
Ä
du œ x" dx •
'ln_3
"
uÈ u# 1
du
b
œ lim csec" kukd ln 3 œ lim csec" b sec" (ln 3)d œ lim cos" ˆ "b ‰ sec" (ln 3)‘
bÄ_
bÄ_
œ cos" (0) sec" (ln 3) œ
1
#
22. converges by the Integral Test:
'1_ x a1 "ln xb dx œ '1_ 1 Š(ln‹x)
"
x
#
œ lim ctan" ud 0 œ lim atan" b tan" 0b œ
b
bÄ_
bÄ_
sec" (ln 3) ¸ 1.1439
bÄ_
1
#
0œ
dx; ”
#
u œ ln x
Ä
du œ "x dx •
'0_ 1"u
#
du
1
#
23. diverges by the nth-Term Test for divergence; n lim
n sin ˆ "n ‰ œ n lim
Ä_
Ä_
sin ˆ "n ‰
ˆ "n ‰
œ lim
24. diverges by the nth-Term Test for divergence; n lim
n tan ˆ "n ‰ œ n lim
Ä_
Ä_
tan ˆ "n ‰
ˆ "n ‰
œ n lim
Ä_
xÄ0
œ1Á0
sin x
x
Š n"# ‹ sec# ˆ n" ‰
Š n"# ‹
œ n lim
sec# ˆ "n ‰ œ sec# 0 œ 1 Á 0
Ä_
25. converges by the Integral Test:
œ lim atan" b tan" eb œ
bÄ_
26. converges by the Integral Test:
œ lim 2 ln
bÄ_
u ‘b
u1 e
"
1
#
dx; ”
2x
'e_
u œ ex
Ä
du œ ex dx •
"
1 u#
ctan" ud e
du œ n lim
Ä_
b
tan" e ¸ 0.35
_
'1
u œ ex ×
_
_
dx; du œ ex dx Ä 'e u(1 2 u) du œ 'e ˆ 2u
Õ dx œ " du Ø
u
Ô
2
1 ex
2 ‰
u1
du
bÄ_
28. diverges by the Integral Test:
bÄ_ #
x
œ lim 2 ln ˆ b b 1 ‰ 2 ln ˆ e e 1 ‰ œ 2 ln 1 2 ln ˆ e e 1 ‰ œ 2 ln ˆ e e 1 ‰ ¸ 0.63
27. converges by the Integral Test:
œ lim
'1_ 1 e e
'1_ 81tancx x dx; ” u œ tan dx x •
"
"
#
du œ
1 x#
'1_ x x1 dx; ” u œ x
#
#
1
Ä
du œ 2x dx •
Ä
'11ÎÎ42 8u du œ c4u# d 11ÎÎ24 œ 4 Š 14
#
'2_ du4 œ
"
#
1#
16 ‹
œ
31 #
4
b
lim #" ln u‘ 2
bÄ_
(ln b ln 2) œ _
29. converges by the Integral Test:
'1_ sech x dx œ 2
'1b 1 eae b
x
lim
x #
bÄ_
dx œ 2 lim ctan" ex d 1
b
bÄ_
œ 2 lim atan" eb tan" eb œ 1 2 tan" e ¸ 0.71
bÄ_
30. converges by the Integral Test:
'1_ sech# x dx œ
œ 1 tanh 1 ¸ 0.76
31.
'1_ ˆ x a 2 x " 4 ‰ dx œ
lim
bÄ_
(b 2)a
b4
lim
bÄ_
'1b sech# x dx œ
lim ca ln kx 2k ln kx 4kd 1 œ lim ln
b
bÄ_
œ a lim (b 2)a1 œ œ
bÄ_
bÄ_
lim ctanh xd b1 œ lim (tanh b tanh 1)
bÄ_
(b 2)a
b4
bÄ_
ln ˆ 35 ‰ ;
a
_, a 1
Ê the series converges to ln ˆ 53 ‰ if a œ 1 and diverges to _ if
1, a œ 1
a 1. If a 1, the terms of the series eventually become negative and the Integral Test does not apply. From
that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
715
716
32.
Chapter 11 Infinite Sequences and Series
'3_ ˆ x " 1 x 2a 1 ‰ dx œ
"
2ac1
b Ä _ #a(b 1)
œ lim
if a
"
#
"
#
. If a
b
lim ’ln ¹ (xx1)12a ¹“ œ lim ln
bÄ_
œ
3
bÄ_
b1
(b 1)2a
b"
ln ˆ 422a ‰ ; lim
2a
b Ä _ (b 1)
1, a œ "#
Ê the series converges to ln ˆ #4 ‰ œ ln 2 if a œ
_, a "#
"
#
and diverges to _ if
, the terms of the series eventually become negative and the Integral Test does not apply.
From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it
diverges.
33. (a)
(b) There are (13)(365)(24)(60)(60) a10* b seconds in 13 billion years; by part (a) sn Ÿ 1 ln n where
n œ (13)(365)(24)(60)(60) a10* b Ê sn Ÿ 1 ln a(13)(365)(24)(60)(60) a10* bb
œ 1 ln (13) ln (365) ln (24) 2 ln (60) 9 ln (10) ¸ 41.55
_
34. No, because !
nœ1
"
nx
œ
"
x
_
"
n
!
nœ1
_
and !
nœ1
"
n
diverges
_
_
_
nœ1
nœ1
nœ1
35. Yes. If ! an is a divergent series of positive numbers, then ˆ "# ‰ ! an œ ! ˆ a#n ‰ also diverges and
an
#
an .
_
There is no “smallest" divergent series of positive numbers: for any divergent series ! an of positive
nœ1
_
numbers !
nœ1
ˆ a#n ‰
has smaller terms and still diverges.
_
_
_
nœ1
nœ1
nœ1
36. No, if ! an is a convergent series of positive numbers, then 2 ! an œ ! 2an also converges, and 2an
an .
There is no “largest" convergent series of positive numbers.
n
n
kœ1
kœ1
37. Let An œ ! ak and Bn œ ! 2k aa2k b , where {ak } is a nonincreasing sequence of positive terms converging to
0. Note that {An } and {Bn } are nondecreasing sequences of positive terms. Now,
Bn œ 2a# 4a% 8a) á 2n aa2n b œ 2a# a2a% 2a% b a2a) 2a) 2a) 2a) b á
ˆ2aa2n b 2aa2n b á 2aa2n b ‰ Ÿ 2a" 2a# a2a$ 2a% b a2a& 2a' 2a( 2a) b á
ðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóò
2n1 terms
_
ˆ2aa2nc1 b 2aa2nc1 1b á 2aa2n b ‰ œ 2Aa2n b Ÿ 2 ! ak . Therefore if ! ak converges,
kœ1
then {Bn } is bounded above Ê ! 2k aa2k b converges. Conversely,
_
An œ a" aa# a$ b aa% a& a' a( b á an a" 2a# 4a% á 2n aa2n b œ a" Bn a" ! 2k aa2k b .
kœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.4 Comparison Tests
717
_
Therefore, if ! 2k aa2k b converges, then {An } is bounded above and hence converges.
kœ1
38. (a) aa2n b œ
"
2n ln a2n b
_
"
n ln n
Ê !
nœ2
(b) aa2n b œ
39. (a)
"
#np
œ
"
2n †n(ln 2)
nœ2
_
_
"
nœ1
nœ1
u œ ln x
• Ä
du œ dx
x
;”
"
#n †n(ln 2)
œ
_
"
ln #
"
n
!
nœ2
, which diverges
_
"
#np
"
a2n bpc1
œ!
nœ1
_
n
œ ! ˆ #p"c1 ‰ , a geometric series that
nœ1
1 or p 1, but diverges if p Ÿ 1.
'2_ x(lndxx)
œœ
nœ2
Ê ! 2 n a a2 n b œ ! 2 n †
#pc1
(ln 2)cpb1 , p 1
"
p1
_
diverges.
converges if
p
_
Ê ! 2 n a a2 n b œ ! 2 n
_, p "
if p 1. For p œ 1:
'ln_2 ucp du œ
cpb1
lim ’ up 1 “
bÄ_
b
œ lim Š 1 " p ‹ cbp1 (ln 2)p1 d
bÄ_
ln 2
Ê the improper integral converges if p 1 and diverges
'2_ x dxln x œ
integral diverges if p œ 1.
lim cln (ln x)d b2 œ lim cln (ln b) ln (ln 2)d œ _, so the improper
bÄ_
bÄ_
_
(b) Since the series and the integral converge or diverge together, !
nœ2
"
n(ln n)p
converges if and only if p 1.
40. (a) p œ 1 Ê the series diverges
(b) p œ 1.01 Ê the series converges
_
(c) !
nœ2
"
n aln n$ b
œ
"
3
_
!
nœ2
"
n(ln n)
; p œ 1 Ê the series diverges
(d) p œ 3 Ê the series converges
41. (a) From Fig. 11.8 in the text with f(x) œ
"
x
Ÿ 1 '1 f(x) dx Ê ln (n 1) Ÿ 1
n
Ÿ ˆ1
"
#
"
3
á
"‰
n
and ak œ
"
#
"
3
"
k
á
cln (n 1) ln nd œ ˆ1
If we define an œ 1
"
#
œ
dx Ÿ 1
"
3
"
n
"
#
"
3
á
"
n
Ÿ 1 ln n Ê 0 Ÿ ln (n 1) ln n
"
#
"
3
á n" ‰ ln n™ is bounded above
' nb1
"
"
"
x , n1 n
x dx œ ln (n 1) ln n
#" 3" á n" 1 ln (n 1)‰ ˆ1 #"
(b) From the graph in Fig. 11.8(a) with f(x) œ
"
n1
"
n
"
x
ln n Ÿ 1. Therefore the sequence ˜ˆ1
by 1 and below by 0.
Ê 0
nb1
, we have '1
"
3
á
"
n
ln n‰ .
ln n, then 0 an1 an Ê an1 an Ê {an } is a decreasing sequence of
nonnegative terms.
#
42. ex Ÿ ex for x
_
_
#
b
1, and '1 ecx dx œ lim cex d " œ lim ˆeb e1 ‰ œ ec1 Ê '1 ecx dx converges by
bÄ_
_
bÄ_
n #
the Comparison Test for improper integrals Ê ! e
nœ0
_
#
œ 1 ! en converges by the Integral Test.
nœ1
11.4 COMPARISON TESTS
_
1. diverges by the Limit Comparison Test (part 1) when compared with !
nœ1
"
Èn
, a divergent p-series:
"
lim
nÄ_
Œ #Èn È
$ n
Š È"n ‹
œ n lim
Ä_
Èn
$ n
2È n È
ˆ " ‰œ
œ n lim
Ä _ # n1Î6
"
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
718
Chapter 11 Infinite Sequences and Series
2. diverges by the Direct Comparison Test since n n n n Èn 0 Ê
_
"
n
term of the divergent series !
nœ1
3
n Èn
"
n
, which is the nth
"
n
or use Limit Comparison Test with bn œ
3. converges by the Direct Comparison Test;
sin# n
2n
Ÿ
"
#n
, which is the nth term of a convergent geometric series
4. converges by the Direct Comparison Test;
1 cos n
n#
Ÿ
2
n#
5. diverges since n lim
Ä_
2n
3n 1
œ
2
3
lim
È
"
n#
converges
Á0
6. converges by the Limit Comparison Test (part 1) with
Š nn# "n ‹
and the p-series !
"
n$Î#
, the nth term of a convergent p-series:
ˆ n n " ‰ œ 1
œ n lim
Ä_
nÄ_ Š " ‹
n$Î#
n
n
n
n ‰
7. converges by the Direct Comparison Test; ˆ 3n n 1 ‰ ˆ 3n
œ ˆ "3 ‰ , the nth term of a convergent geometric
series
8. converges by the Limit Comparison Test (part 1) with
"
Š $Î# ‹
n
lim
nÄ_ Š "
È$
n
2
$
‹
É n n$ 2 œ lim É1
œ n lim
Ä_
nÄ_
"
n$Î#
, the nth term of a convergent p-series:
œ1
2
n$
9. diverges by the Direct Comparison Test; n ln n Ê ln n ln ln n Ê
"
n
"
ln n
"
ln (ln n)
_
and !
nœ3
"
n
diverges
_
10. diverges by the Limit Comparison Test (part 3) when compared with !
nœ2
lim
nÄ_
Š (ln"n)# ‹
ˆ n" ‰
œ n lim
Ä_
n
(ln n)#
œ n lim
Ä_
"
#(ln n) Š n" ‹
œ
"
n
# n lim
Ä _ ln n
œ
"
n
"
"
# n lim
Ä _ Š"‹
n
_
11. converges by the Limit Comparison Test (part 2) when compared with !
nœ1
#
lim
nÄ_
’ (lnn$n) “
Š n"# ‹
œ n lim
Ä_
(ln n)#
n
œ n lim
Ä_
2(ln n) Š n" ‹
1
œ 2 n lim
Ä_
ln n
n
_
nœ1
$
lim
nÄ_
Š n"# ‹
œ n lim
Ä_
(ln n)$
n
œ n lim
Ä_
3(ln n)# Š n" ‹
1
œ 3 n lim
Ä_
œ
"
# n lim
Ä_
nœ_
"
n#
, a convergent p-series:
"
n#
, a convergent p-series:
œ0
12. converges by the Limit Comparison Test (part 2) when compared with !
’ (lnn$n) “
, a divergent p-series:
(ln n)#
n
œ 3 n lim
Ä_
2(ln n) Š n" ‹
1
œ 6 n lim
Ä_
œ6†0œ0
13. diverges by the Limit Comparison Test (part 3) with
lim
nÄ_
’È
1
“
n ln n
ˆ n" ‰
œ n lim
Ä_
Èn
ln n
"
n
, the nth term of the divergent harmonic series:
"
œ n lim
Ä_
Š 2È n ‹
ˆ n" ‰
œ n lim
Ä_
Èn
2
œ_
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
ln n
n
Section 11.4 Comparison Tests
"
n&Î%
14. converges by the Limit Comparison Test (part 2) with
lim
n)#
’ (ln$Î#
“
n
nÄ_ Š
"
‹
n&Î%
(ln n)#
n"Î%
œ n lim
Ä_
œ n lim
Ä_
ˆ 2 lnn n ‰
œ 8 n lim
Ä_
"
Š $Î%
‹
4n
15. diverges by the Limit Comparison Test (part 3) with
lim
nÄ_
ˆ 1 "ln n ‰
ˆ "n ‰
œ n lim
Ä_
"
Š n" ‹
œ n lim
Ä_
n
1 ln n
lim
nÄ_
ˆ n" ‰
œ n lim
Ä_
n
(1 ln n)#
17. diverges by the Integral Test:
lim
nÄ_
ˆ "n ‰
œ n lim
Ä_
"
n
"
ˆ 2 lnn n ‰
œ n lim
Ä_
ˆ n" ‰
Š
"
‹
4n$Î%
œ 32 n lÄ
im_
"
n"Î%
œ 32 † 0 œ 0
, the nth term of the divergent harmonic series:
œ n lim
Ä_
'2_ lnx(x11) dx œ 'ln_3 u du œ
n
1 ln# n
œ 8 n lim
Ä_
, the nth term of the divergent harmonic series:
"
’ 2(1 n ln n) “
œ n lim
Ä_
18. diverges by the Limit Comparison Test (part 3) with
Š 1 "ln# n ‹
"
n
ln n
n"Î%
œ n lim
nœ_
Ä_
16. diverges by the Limit Comparison Test (part 3) with
Š (1 "ln n)# ‹
, the nth term of a convergent p-series:
"
n
œ n lim
Ä_
n
#(1 ln n)
"
Š 2n ‹
b
lim 2" u# ‘ ln 3 œ lim
"
bÄ_ #
bÄ_
œ n lim
Ä_
n
#
œ_
ab# ln# 3b œ _
, the nth term of the divergent harmonic series:
œ n lim
Ä_
n
# ln n
œ n lim
Ä_
"
Š 2n ‹
œ n lim
Ä_
n
#
œ_
"
19. converges by the Direct Comparison Test with n$Î#
, the nth term of a convergent p-series: n# 1 n for
"
"
n 2 Ê n# an# 1b n$ Ê nÈn# 1 n$Î# Ê $Î#
or use Limit Comparison Test with
nÈ n# 1
n
"
n$Î#
Èn
n# 1
20. converges by the Direct Comparison Test with
n# 1
Èn
Ê n# 1 Ènn$Î# Ê
_
21. converges because !
nœ1
_
!
nœ1
"
n2n
"n
n2n
n$Î# Ê
_
œ!
nœ1
"
n2n
_
!
nœ1
"
#n
1
n# .
, the nth term of a convergent p-series: n# 1 n#
"
n$Î#
or use Limit Comparison Test with
"
.
n$Î#
which is the sum of two convergent series:
converges by the Direct Comparison Test since
"
n #n
"
#n
_
, and !
nœ1
"
2n
is a convergent geometric
series
_
22. converges by the Direct Comparison Test: !
nœ1
n 2n
n# 2n
_
œ ! ˆ n2" n
nœ1
"‰
n#
and
"
n2n
"
n#
Ÿ
"
#n
"
n#
, the sum of
the nth terms of a convergent geometric series and a convergent p-series
23. converges by the Direct Comparison Test:
"
3nc1 1
"
3nc1
, which is the nth term of a convergent geometric
series
24. diverges; n lim
Š3
Ä_
nc1
"
3n ‹
ˆ"
œ n lim
Ä_ 3
"‰
3n
œ
"
3
Á0
25. diverges by the Limit Comparison Test (part 1) with
"‰
n
ˆsin
lim
n Ä _ ˆ "n ‰
œ lim
xÄ0
sin x
x
"
n
, the nth term of the divergent harmonic series:
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
719
720
Chapter 11 Infinite Sequences and Series
26. diverges by the Limit Comparison Test (part 1) with
ˆtan "n ‰
lim
n Ä _ ˆ "n ‰
œ n lim
Š " ‹
Ä _ cos "
n
ˆsin n" ‰
ˆ n" ‰
"
n
, the nth term of the divergent harmonic series:
œ lim ˆ cos" x ‰ ˆ sinx x ‰ œ 1 † 1 œ 1
xÄ0
"
n#
27. converges by the Limit Comparison Test (part 1) with
lim
nÄ_
"
Š n(n 10n
1)(n 2) ‹
Š n"# ‹
œ n lim
Ä_
10n# n
n# 3n 2
20n 1
2n 3
œ n lim
Ä_
lim
5n$ 3n
n# (n 2) Šn# 5‹
Š n"# ‹
nÄ_
œ n lim
Ä_
5n$ 3n
n$ 2n# 5n 10
œ n lim
Ä_
tanc" n
n1Þ1
29. converges by the Direct Comparison Test:
œ n lim
Ä_
"
n#
28. converges by the Limit Comparison Test (part 1) with
, the nth term of a convergent p-series:
, the nth term of a convergent p-series:
15n# 3
3n# 4n 5
_
1
#
n1Þ1
œ 10
20
2
œ n lim
Ä_
1
and !
œ
#
n1Þ1
nœ1
1
#
_
!
nœ1
30n
6n 4
"
n1Þ1
œ5
is the product of a
convergent p-series and a nonzero constant
30. converges by the Direct Comparison Test: sec" n
1
#
Ê
secc" n
n1 3
Þ
ˆ 1# ‰
n1 3
Þ
_
and !
nœ1
ˆ 1# ‰
n1 3
Þ
œ
1
#
_
!
nœ1
"
n1 3
Þ
is the
product of a convergent p-series and a nonzero constant
31. converges by the Limit Comparison Test (part 1) with
œ n lim
Ä_
" ec2n
1 ec2n
" ec2n
1 ec2n
: n lim
Ä_
"
n#
: n lim
Ä_
34. converges by the Limit Comparison Test (part 1) with
"
123án
lim
nÄ_
36.
"
ˆ n(n # 1) ‰
œ
Š nan 2b 1b ‹
Š n"# ‹
"
1 2# 3# á n#
Š n"# ‹
œ n lim
coth n œ n lim
Ä_
Ä_
en ecn
en ecn
n
Š tanh
‹
n#
Š n"# ‹
œ n lim
tanh n œ n lim
Ä_
Ä_
en en
en en
œ1
33. diverges by the Limit Comparison Test (part 1) with 1n : n lim
Ä_
35.
n
Š coth
‹
n#
œ1
32. converges by the Limit Comparison Test (part 1) with
œ n lim
Ä_
"
n#
œ
œ n lim
Ä_
œ
"
2
n(n 1) .
2n#
n# n
n(n b 1)(2n b 1)
6
œ
1
Š nÈ
n n‹
ˆ 1n ‰
"
n# : n lim
Ä_
Š
œ n lim
Ä_
Èn n ‹
n#
Š n"# ‹
1
n n
È
œ 1.
n
È
œ n lim
nœ1
Ä_
The series converges by the Limit Comparison Test (part 1) with
œ n lim
Ä_
4n
2n 1
6
n(n 1)(2n 1)
Ÿ
œ n lim
Ä_
6
n$
4
2
"
n# :
œ 2.
Ê the series converges by the Direct
Comparison Test
an
37. (a) If n lim
œ 0, then there exists an integer N such that for all n N, ¹ bann 0¹ 1 Ê 1
Ä _ bn
Ê an bn . Thus, if ! bn converges, then ! an converges by the Direct Comparison Test.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
an
bn
1
Section 11.4 Comparison Tests
an
(b) If n lim
œ _, then there exists an integer N such that for all n N,
Ä _ bn
! bn diverges, then ! an diverges by the Direct Comparison Test.
_
38. Yes, !
nœ1
an
n
converges by the Direct Comparison Test because
an
n
an
bn
1 Ê an bn . Thus, if
an
an
39. n lim
œ _ Ê there exists an integer N such that for all n N,
Ä _ bn
then ! bn converges by the Direct Comparison Test
an
bn
1 Ê an bn . If ! an converges,
40. ! an converges Ê n lim
a œ 0 Ê there exists an integer N such that for all n N, 0 Ÿ an 1 Ê an# an
Ä_ n
Ê ! a#n converges by the Direct Comparison Test
41. Example CAS commands:
Maple:
a := n -> 1./n^3/sin(n)^2;
s := k -> sum( a(n), n=1..k );
# (a)]
limit( s(k), k=infinity );
pts := [seq( [k,s(k)], k=1..100 )]:
# (b)
plot( pts, style=point, title="#41(b) (Section 11.4)" );
pts := [seq( [k,s(k)], k=1..200 )]:
# (c)
plot( pts, style=point, title="#41(c) (Section 11.4)" );
pts := [seq( [k,s(k)], k=1..400 )]:
# (d)
plot( pts, style=point, title="#41(d) (Section 11.4)" );
evalf( 355/113 );
Mathematica:
Clear[a, n, s, k, p]
a[n_]:= 1 / ( n3 Sin[n]2 )
s[k_]= Sum[ a[n], {n, 1, k}]
points[p_]:= Table[{k, N[s[k]]}, {k, 1, p}]
points[100]
ListPlot[points[100]]
points[200]
ListPlot[points[200]
points[400]
ListPlot[points[400], PlotRange Ä All]
To investigate what is happening around k = 355, you could do the following.
N[355/113]
N[1 355/113]
Sin[355]//N
a[355]//N
N[s[354]]
N[s[355]]
N[s[356]]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
721
722
Chapter 11 Infinite Sequences and Series
11.5 THE RATIO AND ROOT TESTS
1. converges by the Ratio Test:
ˆ1 "n ‰
œ n lim
Ä_
È2
ˆ #" ‰ œ
lim anb1
n Ä _ an
"
#
”
œ n lim
Ä_
È
(n b 1) 2
2nb1 •
”
È
n 2
#n
œ n lim
Ä_
•
È2
(n 1)
#nb1
n
† 2È2
n
1
Š (nenbb1)1 ‹
2
2. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
3. diverges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
4. diverges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
œ n lim
Ä_
#
Š nen ‹
Š (nenbb1)!
1 ‹
ˆ en!n ‰
b 1)! ‹
Š (n
10nb1
ˆ 10n!n ‰
œ
"
10
anb1
an
(n ")!
enb1
†
en
n!
œ n lim
Ä_
(n ")!
10nb1
†
10n
n!
Š (n10bn1)
1 ‹
œ n lim
Ä_
en
lim
n2 œ n Ä
_
†
œ n lim
Ä_
"!
5. converges by the Ratio Test: n lim
Ä_
(n 1)2
enb1
œ n lim
Ä_
"!
Š n10n ‹
(n ")"!
10n1
œ n lim
Ä_
†
œ n lim
Ä_
10n
n"!
ˆ1 n" ‰# ˆ "e ‰ œ
n"
e
n
10
"
e
1
œ_
œ_
" ‰
ˆ1 n" ‰"! ˆ 10
œ n lim
Ä_
1
ˆ nn 2 ‰n œ lim ˆ1
6. diverges; n lim
a œ n lim
Ä_ n
Ä_
nÄ_
7. converges by the Direct Comparison Test:
2(1)n
(1.25)n
2 ‰ n
n
œ e# Á 0
n
n
œ ˆ 45 ‰ c2 (1)n d Ÿ ˆ 45 ‰ (3) which is the nth term of a convergent
geometric series
8. converges; a geometric series with krk œ ¸ 23 ¸ 1
ˆ1 3n ‰n œ lim ˆ1
9. diverges; n lim
a œ n lim
Ä_ n
Ä_
nÄ_
ˆ1
10. diverges; n lim
a œ n lim
Ä_ n
Ä_
" ‰n
3n
3 ‰ n
n
œ n lim
1
Ä_
11. converges by the Direct Comparison Test:
ln n
n$
n
n$
œ
œ e$ ¸ 0.05 Á 0
Š "3 ‹
n
"
n#
n
"Î$
¸ 0.72 Á 0
œe
for n
2, the nth term of a convergent p-series.
n
n (ln n)
n
È
É
12. converges by the nth-Root Test: n lim
an œ n lim
nn œ n lim
Ä_
Ä_
Ä_
œ n lim
Ä_
Š "n ‹
1
œ n lim
Ä_
ln n
n
œ01
13. diverges by the Direct Comparison Test:
with "n .
a(ln n)n b1În
ann b1În
"
n
"
n#
œ
n1
n#
"
#
ˆ "n ‰ for n 2 or by the Limit Comparison Test (part 1)
n
n
ˆ n"
È
É
14. converges by the nth-Root Test: n lim
an œ n lim
Ä_
Ä_
ˆ " n"# ‰ œ 0 1
œ n lim
Ä_ n
" ‰n
n#
ˆˆ n"
œ n lim
Ä_
" ‰n ‰1În
n#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.5 The Ratio and Root Tests
15. diverges by the Direct Comparison Test:
ln n
n
"
n
for n
3
16. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
(n 1) ln (n 1)
#nb1
†
17. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
(n 2)(n 3)
(n 1)!
†
n!
(n 1)(n 2)
18. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
(n 1)$
en1
œ
19. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
(n 4)!
3! (n 1)! 3nb1
anb1
20. converges by the Ratio Test: n lim
œ n lim
Ä _ an
Ä_
n
1
2
n
2
2
ˆ n ‰ ˆ3‰ ˆn1‰ œ 3 1
œ n lim
Ä_
†
anb1
an
œ n lim
Ä_
(n 1)!
(2n 3)!
22. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
(n 1)!
(n 1)nb1
"
ˆ1 "n ‰n
œ
"
e
†
"
e
†
(2n 1)!
n!
†
†
nn
n!
"
#
œ
1
œ01
1
3! n! 3n
(n 3)!
(n 1)2nb1 (n 2)!
3nb1 (n 1)!
21. converges by the Ratio Test: n lim
Ä_
œ n lim
Ä_
en
n$
2n
n ln (n)
œ n lim
Ä_
n4
3(n 1)
"
3
œ
1
3n n!
n2n (n 1)!
n"
(2n 3)(2n 2)
œ n lim
Ä_
œ01
ˆ n ‰n œ lim
œ n lim
Ä _ n1
nÄ_
"
ˆ n bn " ‰n
1
n n
È
ln n
n
n
n
È
23. converges by the Root Test: n lim
an œ n lim
œ n lim
Ä_
Ä _ É (ln n)n
Ä_
n
n
n
È
24. converges by the Root Test: n lim
an œ n lim
œ n lim
Ä_
Ä _ É (ln n)nÎ2
Ä_
"
ln n
œ n lim
Ä_
n n
È
Èln n
œ
œ01
Ä_
Ä_
n n
lim È
n
Èln n
lim
n
œ01
n
È
n œ 1‹
Šn lim
Ä_
25. converges by the Direct Comparison Test:
œ
n! ln n
n(n 2)!
ln n
n(n 1)(n 2)
n
n(n 1)(n 2)
œ
"
(n 1)(n #)
"
n#
which is the nth-term of a convergent p-series
26. diverges by the Ratio Test: n lim
Ä_
an1
an
27. converges by the Ratio Test: n lim
Ä_
28. converges by the Ratio Test:
approaches 1
1
#
œ n lim
Ä_
anb1
an
lim anb1
n Ä _ an
3n1
(n 1)$ 2n1
œ n lim
Ä_
œ n lim
Ä_
ˆ 1 b nsin n ‰ an
an
Š 1 b tan
n
" n
an
while the denominator tends to _
29. diverges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
†
c1‰
ˆ 3n
2n b 1 an
an
n$ 2n
3n
œ n lim
Ä_
n3
(n 1)3
ˆ #3 ‰ œ
3
#
1
œ01
‹ an
œ n lim
Ä_
œ n lim
Ä_
3n 1
2n 1
" tan" n
n
œ
3
#
œ 0 since the numerator
1
2
‰
30. diverges; an1 œ n n 1 an Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 an1 ‰ Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 ‰ ˆ nn
1 an2
a"
2‰
3
ˆ"‰
Ê an1 œ ˆ n n 1 ‰ ˆ n n 1 ‰ ˆ nn
1 â # a" Ê an1 œ n 1 Ê an1 œ n 1 , which is a constant times the
general term of the diverging harmonic series
31. converges by the Ratio Test: n lim
Ä_
anb1
an
œ n lim
Ä_
Š 2n ‹ an
an
œ n lim
Ä_
2
n
œ01
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
723
724
Chapter 11 Infinite Sequences and Series
32. converges by the Ratio Test:
lim anb1
n Ä _ an
œ n lim
Ä_
anb1
an
œ n lim
Ä_
33. converges by the Ratio Test: n lim
Ä_
34.
n ln n
n 10
0 and a" œ
Ê an1 œ
n ln n
n 10
"
#
Œ
Èn n
#
an
an
Š 1 bnln n ‹ an
an
n n
È
œ n lim
Ä_
n
œ
"ln n
n
œ n lim
Ä_
"
#
1
œ n lim
Ä_
an an ; thus an1 an
"
3
35. diverges by the nth-Term Test: a" œ
œ01
n ln n
n 10
Ê an 0; ln n 10 for n e"! Ê n ln n n 10 Ê
"
#
"
n
1
Ê n lim
a Á 0, so the series diverges by the nth-Term Test
Ä_ n
3
3
6 "
%! "
2 "
2 "
2 "
É
É
É
É
, a# œ É
3 , a$ œ Ê
3 œ
3 , a% œ ËÊ
3 œ
3 ,á ,
%
n! "
n! "
n "
É
an œ É
a œ 1 because šÉ
3 Ê n lim
3 › is a subsequence of š
3 › whose limit is 1 by Table 8.1
Ä_ n
36. converges by the Direct Comparison Test: a" œ
n!
"
#
# $
#
'
' %
#%
, a# œ ˆ "# ‰ , a$ œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , a% œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , á
n
Ê an œ ˆ "# ‰ ˆ "# ‰ which is the nth-term of a convergent geometric series
anb1
an
37. converges by the Ratio Test: n lim
Ä_
n"
"
œ n lim
œ
1
#
Ä _ 2n 1
œ n lim
Ä_
2nb1 (n 1)! (n 1)!
(2n 2)!
†
(2n)!
2n n! n!
œ n lim
Ä_
2(n 1)(n 1)
(2n #)(2n 1)
(3n 3)!
1)! (n 2)!
anb1
38. diverges by the Ratio Test: n lim
œ n lim
† n! (n (3n)!
Ä _ an
Ä _ (n 1)! (n 2)! (n 3)!
(3n 3)(3 2)(3n 1)
2 ‰ ˆ 3n 1 ‰
œ n lim
œ n lim
3 ˆ 3n
n#
n 3 œ 3 † 3 † 3 œ 27 1
Ä _ (n 1)(n 2)(n 3)
Ä_
n
n (n!)
n
È
39. diverges by the Root Test: n lim
an ´ n lim
œ n lim
Ä_
Ä _ É an n b #
Ä_
n
œ_1
n!
n#
n
n (n!)
n (n!)
É
40. converges by the Root Test: n lim
œ n lim
œ n lim
É
an n b n
Ä_
Ä_
Ä_
nn#
"
Ÿ n lim
œ
0
1
Ä_ n
n!
nn
ˆ " ‰ ˆ 2n ‰ ˆ 3n ‰ â ˆ n n 1 ‰ ˆ nn ‰
œ n lim
Ä_ n
n n
n
È
41. converges by the Root Test: n lim
an œ n lim
œ n lim
Ä_
Ä _ É 2n#
Ä_
n
#n
œ n lim
Ä_
n
n
n
È
42. diverges by the Root Test: n lim
an œ n lim
œ n lim
Ä_
Ä _ É a#n b #
Ä_
n
4
œ_1
n
n
anb1
an
43. converges by the Ratio Test: n lim
Ä_
2n "
"
œ n lim
œ
1
4
Ä _ (4†#)(n 1)
44. converges by the Ratio Test: an œ
Ê n lim
Ä_
(2n 2)!
c2nb1 (n 1)!d# a3nb1 1b
#
œ n lim
Š 4n 6n 2 ‹
Ä _ 4n# 8n 4
45. Ratio: n lim
Ä_
anb1
an
†
a1 3cn b
a3 3cn b
œ n lim
Ä_
œ n lim
Ä_
1†3 â (2n 1)
(2†4 â #n) a3n 1b
a2n n!b# a3n 1b
(2n)!
œ1†
"
(n 1)p
†
"
3
np
1
œ
"
3
1†3† â †(2n 1)(2n 1)
4nb1 2nb1 (n 1)!
œ
1†2†3†4 â (2n 1)(2n)
(2†4 â 2n)# a3n 1b
œ n lim
Ä_
†
"
#n ln 2
œ01
4n 2n n!
1†3† â †(2n 1)
œ
(2n)!
a2n n!b# a3n 1b
(2n ")(2n 2) a3n 1b
2# (n 1)# a3n1 1b
1
ˆ n ‰p œ 1p œ 1 Ê no conclusion
œ n lim
Ä_ n1
n "
n
È
É
Root: n lim
an œ n lim
np œ n lim
Ä_
Ä_
Ä_
"
n n ‰p
ˆÈ
œ
"
(1)p
œ 1 Ê no conclusion
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.6 Alternating Series, Absolute and Conditional Convergence
46. Ratio: n lim
Ä_
anb1
an
œ n lim
Ä_
"
(ln (n 1))p
†
(ln n)p
1
œ ’n lim
Ä_
ln n
ln (n 1) “
p
ˆ "n ‰
ˆ n b" 1 ‰ •
œ ”n lim
Ä_
p
œ Šn lim
Ä_
n"
n ‹
725
p
œ (1)p œ 1 Ê no conclusion
"
n
n
È
Root: n lim
an œ n lim
É
(ln n)p œ
Ä_
Ä_
"
p
lim (ln n)1În ‹
ŠnÄ_
ˆ
"
; let f(n) œ (ln n)1În , then ln f(n) œ
‰
"
n ln n
ln (ln n)
n ln n
Ê n lim
ln f(n) œ n lim
œ n lim
œ n lim
n
1
Ä_
Ä_
Ä_
Ä_
"
ln fÐnÑ
!
n
È an œ
œ n lim
e
œ
e
œ
1;
therefore
lim
Ä_
nÄ_
p
lim (ln n)1În ‹
ŠnÄ_
47. an Ÿ
n
2n
_
for every n and the series !
nœ1
_
n
#n
ln (ln n)
n
œ 0 Ê n lim
(ln n)1În
Ä_
œ (1)" p œ 1 Ê no conclusion
converges by the Ratio Test since n lim
Ä_
(n ")
2nb1
†
2n
n
œ
"
#
"
n#
which is a
1
Ê ! an converges by the Direct Comparison Test
nœ1
11.6 ALTERNATING SERIES, ABSOLUTE AND CONDITIONAL CONVERGENCE
1.
_
_
nœ1
nœ1
_
_
nœ1
nœ1
converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ !
convergent p-series
2. converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ !
"
n$Î#
which is a
convergent p-series
3. diverges by the nth-Term Test since for n 10 Ê
4. diverges by the nth-Term Test since n lim
Ä_
"0 n
n"!
n
10
_
n ‰n
ˆ n ‰n Á 0 Ê ! (1)n1 ˆ 10
1 Ê n lim
diverges
Ä _ 10
nœ1
œ n lim
Ä_
"0n (ln 10)"!
10!
^
œ _ (after 10 applications of L'Hopital's
rule)
5. converges by the Alternating Series Test because f(x) œ ln x is an increasing function of x Ê
Ê un
un1 for n
1; also un
0 for n
1 and
"
lim
n Ä _ ln n
6. converges by the Alternating Series Test since f(x) œ
decreasing Ê un
un1 ; also un
ln n
ln n#
œ n lim
Ä_
Ê f w (x) œ
ln n
2 ln n
1 ln x
x#
œ n lim
Ä_
ln n
n
"
#
0 when x e Ê f(x) is
œ n lim
Ä_
œ
8. converges by the Alternating Series Test since f(x) œ ln a1 x" b Ê f w (x) œ
decreasing Ê un
un1 ; also un
0 for n
unb1 ; also un
0 for n
1 and n lim
u œ
Ä_ n
10. diverges by the nth-Term Test since n lim
Ä_
"
#
Š "n ‹
1
œ0
Á0
"
x(x 1)
0 for x 0 Ê f(x) is
ˆ1 n" ‰‹ œ ln 1 œ 0
1 and n lim
u œ n lim
ln ˆ1 "n ‰ œ ln Šn lim
Ä_ n
Ä_
Ä_
9. converges by the Alternating Series Test since f(x) œ
Ê un
is decreasing
œ0
1 and n lim
u œ n lim
Ä_ n
Ä_
0 for n
7. diverges by the nth-Term Test since n lim
Ä_
ln x
x
"
ln x
3È n 1
Èn 1
Èx "
x1
Ê f w (x) œ
Èn "
lim
n Ä _ n1
œ n lim
Ä_
3É 1
"
1 x 2È x
2Èx (x 1)#
0 Ê f(x) is decreasing
œ0
"
n
1 Š Èn ‹
œ3Á0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
726
Chapter 11 Infinite Sequences and Series
_
_
nœ1
nœ1
" ‰n
11. converges absolutely since ! kan k œ ! ˆ 10
a convergent geometric series
12. converges absolutely by the Direct Comparison Test since ¹ (1)
nb1
(0.1)n
n
¹œ
"
(10)n n
n
" ‰
ˆ 10
which is the nth
term of a convergent geometric series
13. converges conditionally since
"
Èn
"
Èn 1
"
Èn
0 and n lim
Ä_
_
_
nœ1
nœ1
œ 0 Ê convergence; but ! kan k œ !
"
n"Î#
is a divergent p-series
14. converges conditionally since
_
_
! kan k œ !
nœ1
nœ1
"
1 Èn
"
1 Èn
"
1 Èn 1
is a divergent series since
_
_
15. converges absolutely since ! kan k œ !
nœ1
n
n $ 1
nœ1
17. converges conditionally since
_
œ!
nœ1
"
n3
diverges because
"
n3
"
n3
"
1 È n
and
"
(n 1) 3
n
n $ 1
and !
"
n#
0 and n lim
Ä_
_
"
4n
"
#È n
_
nœ1
"
n"Î#
œ 0 Ê convergence; but
is a divergent p-series
which is the nth-term of a converging p-series
œ_
n!
#n
16. diverges by the nth-Term Test since n lim
Ä_
"
1 Èn
0 and n lim
Ä_
"
n
and !
nœ1
"
n 3
_
œ 0 Ê convergence; but ! kan k
nœ1
is a divergent series
_
18. converges absolutely because the series ! ¸ sinn# n ¸ converges by the Direct Comparison Test since ¸ sinn# n ¸ Ÿ
nœ1
3n
5n
19. diverges by the nth-Term Test since n lim
Ä_
œ1Á0
"
3 ln x
20. converges conditionally since f(x) œ ln x is an increasing function of x Ê
Ê
"
3 ln n
_
œ!
nœ2
"
3 ln n
"
3 ln (n1)
0 for n
diverges because
2 and n lim
Ä_
"
3 ln n
21. converges conditionally since f(x) œ
un unb1 0 for n
_
œ!
nœ1
"
n#
_
!
nœ1
"
n
"
3n
"
x#
"
3 ln n
_
and !
nœ2
"
x
"
n
"
ln ax$ b
œ
is decreasing
_
_
nœ2
nœ2
œ 0 Ê convergence; but ! kan k œ !
"
ln an$ b
diverges
Ê f w (x) œ ˆ x2$
"‰
x#
0 Ê f(x) is decreasing and hence
_
_
nœ1
nœ1
ˆ " n" ‰ œ 0 Ê convergence; but ! kan k œ !
1 and n lim
Ä _ n#
1 n
n#
is the sum of a convergent and divergent series, and hence diverges
nb1
22. converges absolutely by the Direct Comparison Test since ¹ (n2)5n ¹ œ
2nb1
n 5 n
n
2 ˆ 25 ‰ which is the nth term
of a convergent geometric series
23. converges absolutely by the Ratio Test: n lim
Š uunbn 1 ‹ œ n lim
Ä_
Ä_ ”
(n")# ˆ 23 ‰
n
n# ˆ 23 ‰
n1
•œ
2
3
1
24. diverges by the nth-Term Test since n lim
a œ n lim
101În œ 1 Á 0
Ä_ n
Ä_
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
n#
Section 11.6 Alternating Series, Absolute and Conditional Convergence
_
" xb#
25. converges absolutely by the Integral Test since '1 atan" xb ˆ 1 " x# ‰ dx œ lim ’ atan #
bÄ_
#
#
"
#
œ lim ’atan" bb atan" 1b “ œ
bÄ_
26. converges conditionally since f(x) œ
Ê un unb1 0 for n
'2_ x dxln x œ
lim
"
x
bÄ_
_
_
nœ1
nœ1
Ê ! kan k œ !
"
n ln n
"
n ln n
lim cln (ln x)d b2 œ lim cln (ln b) ln (ln 2)d œ _
œ
(x ln x)#
œ n lim
Ä_
Š "n ‹
1 Š n" ‹
_
_
nœ1
nœ1
! kan k œ !
bÄ_
diverges
28. converges conditionally since f(x) œ
x Š lnxx ‹
œ 0 Ê convergence; but by the Integral Test,
bÄ_
n
n1
27. diverges by the nth-Term Test since n lim
Ä_
1 Š lnxx ‹ ln
œ
1 ln x
(x ln x)#
ln x
x ln x
œ1Á0
Ê f w (x) œ
0 Ê un
Š "x ‹ (x ln x) (ln x) Š1 x" ‹
(x ln x)#
un1 0 when n e and n lim
Ä_
œ 0 Ê convergence; but n ln n n Ê
ln n
n ln n
1
31 #
32
1d
Ê f w (x) œ cln(x(x)
ln x)# 0 Ê f(x) is decreasing
"
x ln x
2 and n lim
Ä_
'2b Šln x‹ dx œ
#
#
’ˆ 1# ‰ ˆ 14 ‰ “ œ
b
“
"
nln n
"
n
Ê
ln n
n ln n
ln n
nln n
"
n
so that
diverges by the Direct Comparison Test
29. converges absolutely by the Ratio Test: n lim
Š uunbn 1 ‹ œ n lim
Ä_
Ä_
_
_
nœ1
nœ1
("00)nb1
(n1)!
†
œ n lim
Ä_
n!
(100)n
"00
n1
œ01
n
30. converges absolutely since ! kan k œ ! ˆ 5" ‰ is a convergent geometric series
_
_
nœ1
nœ1
31. converges absolutely by the Direct Comparison Test since ! kan k œ !
"
n# 2n 1
"
n#
"
n# 2n 1
and
which is the nth-term of a convergent p-series
_
_
_
_
nœ1
nœ1
nœ1
nœ1
_
_
nœ1
nœ1
n
n
n
32. converges absolutely since ! kan k œ ! ˆ lnln nn# ‰ œ ! ˆ 2lnlnnn ‰ œ ! ˆ "# ‰ is a convergent
geometric series
_
33. converges absolutely since ! kan k œ ! ¹ (nÈ1)n ¹ œ !
_
34. converges conditionally since !
nœ1
_
_
nœ1
nœ1
! kan k œ !
"
n
cos n1
n
n
nœ1
_
œ!
nœ1
(1)n
n
"
n$Î#
is a convergent p-series
is the convergent alternating harmonic series, but
diverges
1)
n
È
kan k œ n lim
35. converges absolutely by the Root Test: n lim
Š (n(2n)
n ‹
Ä_
Ä_
n
36. converges absolutely by the Ratio Test: n lim
¹ anb1 ¹ œ n lim
Ä _ an
Ä_
a(n 1)!b#
((2n 2)!)
†
1În
œ n lim
Ä_
(2n)!
(n!)#
œ n lim
Ä_
n"
#n
œ
"
#
1
(n 1)#
(2n 2)(2n 1)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
"
4
1
727
728
Chapter 11 Infinite Sequences and Series
37. diverges by the nth-Term Test since n lim
kan k œ n lim
Ä_
Ä_
ˆ n # 1 ‰n1 œ _ Á 0
n lim
Ä_
(n 1)(n 2)â(n (n 1))
#nc1
œ n lim
Ä_
(2n)!
2n n! n
(n 1)! (n 1)! 3nb1
(2n 3)!
38. converges absolutely by the Ratio Test: n lim
¹ anabn 1 ¹ œ n lim
Ä_
Ä_
(n 1)# 3
(2n 2)(2n 3)
œ n lim
Ä_
œ
Èn 1 Èn
1
†
Èn 1 Èn
Èn 1 Èn
œ
"
Èn 1 Èn
_
(")n
Èn 1 Èn
nœ1
_
nœ1
nœ1
"
Èn 1 Èn
diverges by the Limit Comparison Test (part 1) with
"
Èn 1 Èn
œ lim
"
Èn n Ä _
lim
nÄ_
(2n 1)!
n! n! 3n
and š Èn 1" Èn › is a
decreasing sequence of positive terms which converges to 0 Ê !
_
†
1
3
4
39. converges conditionally since
! kan k œ !
(n ")(n 2)â(2n)
2n n
œ n lim
Ä_
Èn
Èn 1 Èn
œ n lim
Ä_
1
É1 1n 1
œ
converges; but
"
Èn ;
a divergent p-series:
"
#
È
#
n n
40. diverges by the nth-Term Test since n lim
ŠÈn# n n‹ œ n lim
ŠÈn# n n‹ † Š Ènn#
‹
Ä_
Ä_
n n
œ n lim
Ä_
n
È n # n n
œ n lim
Ä_
"
É1 "n 1
œ
"
#
Á0
É n Èn Èn
41. diverges by the nth-Term Test since n lim
ŠÉn Èn Èn‹ œ n lim
ŠÉn Èn Èn‹
Ä_
Ä_ –
É n È n È n —
Èn
œ n lim
Ä_
É n Èn Èn
œ n lim
Ä_
"
É1
"
Èn 1
œ
"
#
Á0
42. converges conditionally since š Èn "Èn 1 › is a decreasing sequence of positive terms converging to 0
_
Ê !
nœ1
(")n
Èn Èn 1
_
so that !
nœ1
converges; but n lim
Ä_
"
Èn Èn 1
Š Èn "Èn 1 ‹
Š È"n ‹
Èn
È n È n 1
œ n lim
Ä_
_
"
Èn
diverges by the Limit Comparison Test with !
nœ1
43. converges absolutely by the Direct Comparison Test since sech (n) œ
œ n lim
Ä_
"
1É1 "n
"
#
œ
which is a divergent p-series
2
en ecn
œ
2en
e2n 1
2en
e2n
œ
2
en
nth term of a convergent geometric series
_
_
nœ1
nœ1
44. converges absolutely by the Limit Comparison Test (part 1): ! kan k œ !
Apply the Limit Comparison Test with
lim
nÄ_
Œ
2
en c ecn
1
en
2en
en ecn
œ n lim
Ä_
1
en ,
the n-th term of a convergent geometric series:
œ n lim
Ä_
45. kerrork ¸(1)' ˆ "5 ‰¸ œ 0.2
47. kerrork ¹(1)'
49.
"
(2n)!
5
10'
(0.01)&
5 ¹
Ê (2n)!
œ 2 ‚ 10""
10'
5
2
en ecn
œ 200,000 Ê n
2
1 ec2n
œ2
46. kerrork ¸(1)' ˆ 10" & ‰¸ œ 0.00001
48. kerrork k(1)% t% k œ t% 1
5 Ê 1
"
#!
"
4!
"
6!
"
8!
¸ 0.54030
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
which is the
Section 11.6 Alternating Series, Absolute and Conditional Convergence
50.
"
n!
10'
5
Ê
5
10'
51. (a) an
n! Ê n
an1 fails since
_
_
nœ1
nœ1
"
3
"
#!
9 Ê 11
"
#
"
3!
"
4!
_
_
nœ1
nœ1
"
5!
"
6!
"
7!
"
8!
¸ 0.367881944
n
n
n
n
(b) Since ! kan k œ ! ˆ 3" ‰ ˆ #" ‰ ‘ œ ! ˆ 3" ‰ ! ˆ #" ‰ is the sum of two absolutely convergent
series, we can rearrange the terms of the original series to find its sum:
ˆ 3"
"
9
"
27
52. s#! œ 1
"
#
"
3
á ‰ ˆ #"
"
4
á
"
19
"
4
"
20
"
8
በœ
ˆ "3 ‰
1 ˆ "3 ‰
ˆ "# ‰
1 ˆ "# ‰
œ
"
#
1 œ #"
"
#
†
"
#1
¸ 0.6687714032 Ê s#!
¸ 0.692580927
_
53. The unused terms are ! (1)j1 aj œ (1)n1 aan1 an2 b (1)n3 aan3 an4 b á
jœn1
œ (1)n1 caan1 an2 b aan3 an4 b á d . Each grouped term is positive, so the remainder
has the same sign as (1)n1 , which is the sign of the first unused term.
54. sn œ
"
1 †2
"
#†3
"
3†4
"
n(n 1)
á
n
œ!
kœ1
"
k(k 1)
n
œ ! ˆ k"
kœ1
œ ˆ1 "# ‰ ˆ "# 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n"
" ‰
k1
" ‰
n1
which are the first 2n terms
of the first series, hence the two series are the same. Yes, for
n
sn œ ! ˆ k"
kœ1
" ‰
k 1
œ ˆ1 #" ‰ ˆ #" 3" ‰ ˆ 3" 4" ‰ ˆ 4" 5" ‰ á ˆ n " 1 n" ‰ ˆ n"
" ‰
n1
œ 1
"
n1
ˆ1 n " 1 ‰ œ 1 Ê both series converge to 1. The sum of the first 2n 1 terms of the first
Ê n lim
s œ n lim
Ä_ n
Ä_
ˆ1 n " 1 ‰ œ 1.
series is ˆ1 n " 1 ‰ n " 1 œ 1. Their sum is n lim
s œ n lim
Ä_ n
Ä_
_
_
_
_
nœ1
nœ1
nœ1
nœ1
55. Theorem 16 states that ! kan k converges Ê ! an converges. But this is equivalent to ! an diverges Ê ! kan k
diverges.
_
_
nœ1
nœ1
56. ka" a# á an k Ÿ ka" k ka# k á kan k for all n; then ! kan k converges Ê ! an converges and these
_
_
nœ1
nœ1
imply that º ! an º Ÿ ! kan k
_
57. (a) ! kan bn k converges by the Direct Comparison Test since kan bn k Ÿ kan k kbn k and hence
nœ1
_
! aan bn b converges absolutely
nœ1
_
_
_
(b) ! kbn k converges Ê ! bn converges absolutely; since ! an converges absolutely and
nœ1
_
nœ1
nœ1
_
_
! bn converges absolutely, we have ! can (bn )d œ ! aan bn b converges absolutely by part (a)
nœ1
_
nœ1
_
_
nœ1
_
nœ1
nœ1
_
(c) ! kan k converges Ê kkk ! kan k œ ! kkan k converges Ê ! kan converges absolutely
nœ1
58. If an œ bn œ (1)n
"
Èn
, then ! (1)n
nœ1
59. s" œ "# , s# œ "# 1 œ
"
#
s$ œ 1
"
4
"
6
"
8
"
#
"
Èn
nœ1
_
_
nœ1
nœ1
converges, but ! an bn œ !
"
n
diverges
,
"
10
"
1#
"
14
"
16
"
18
"
#0
"
2#
¸ 0.5099,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
729
730
Chapter 11 Infinite Sequences and Series
s% œ s$
s& œ s%
s' œ s&
s( œ s'
"
3 ¸ 0.1766,
"
"
"
#4 #6 #8
"
5 ¸ 0.312,
"
"
"
46 48 50
"
30
"
3#
"
34
"
36
"
38
"
40
"
42
"
44
¸ 0.512,
"
52
"
54
"
56
"
58
"
60
"
62
"
64
"
66
¸ 0.51106
N" 1
60. (a) Since ! kan k converges, say to M, for % 0 there is an integer N" such that º ! kan k Mº
nœ1
N" 1
N" 1
_
nœ1
nœ1
nœN"
_
%
#
Í » ! kan k ! kan k ! kan k »
Í » ! kan k»
nœN"
_
%
#
Í ! kan k
nœN"
%
#
%
#
. Also, ! an
converges to L Í for % 0 there is an integer N# (which we can choose greater than or equal to N" ) such
that ksN# Lk
%
#
_
. Therefore, ! kan k
nœN"
%
#
%
#
and ksN# Lk
.
_
k
nœ1
nœ1
(b) The series ! kan k converges absolutely, say to M. Thus, there exists N" such that º ! kan k Mº %
whenever k N" . Now all of the terms in the sequence ekbn kf appear in ekan kf. Sum together all of the
N
terms in ekbn kf, in order, until you include all of the terms ekan kf nœ" 1 , and let N# be the largest index in the
N#
N#
_
nœ1
nœ1
nœ1
sum ! kbn k so obtained. Then º ! kbn k Mº % as well Ê ! kbn k converges to M.
_
_
61. (a) If ! kan k converges, then ! an converges and
nœ1
nœ1
converges where bn œ
a n ka n k
#
_
_
nœ1
nœ1
a n ka n k
#
_
! an
nœ1
"
#
_
_
nœ1
nœ1
! kan k œ !
a n ka n k
#
a , if an 0
œœ n
.
0, if an 0
(b) If ! kan k converges, then ! an converges and
converges where cn œ
"
#
œœ
"
#
_
! an
nœ1
"
#
_
_
nœ1
nœ1
! kan k œ !
a n ka n k
#
0, if an 0
.
an , if an 0
62. The terms in this conditionally convergent series were not added in the order given.
63. Here is an example figure when N œ 5. Notice that u$ u# u" and u$ u& u% , but un
n 5.
un1 for
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.7 Power Series
731
11.7 POWER SERIES
_
nb1
1. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ xxn ¹ 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have ! (1)n , a divergent
Ä_
Ä_
nœ1
_
series; when x œ 1 we have ! 1, a divergent series
nœ1
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
nb1
2. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x(x5)5)n ¹ 1 Ê kx 5k 1 Ê 6 x 4; when x œ 6 we have
Ä_
Ä_
_
_
nœ1
nœ1
! (1)n , a divergent series; when x œ 4 we have ! 1, a divergent series
(a) the radius is 1; the interval of convergence is 6 x 4
(b) the interval of absolute convergence is 6 x 4
(c) there are no values for which the series converges conditionally
nb1
1)
"
"
3. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (4x
(4x 1)n ¹ 1 Ê k4x 1k 1 Ê 1 4x 1 1 Ê # x 0; when x œ # we
Ä_
Ä_
_
_
_
_
nœ1
nœ1
nœ1
have ! (1)n (1)n œ ! (1)2n œ ! 1n , a divergent series; when x œ 0 we have ! (1)n (1)n
nœ1
_
œ ! (1)n , a divergent series
nœ1
(a) the radius is "4 ; the interval of convergence is #" x 0
(b) the interval of absolute convergence is "# x 0
(c) there are no values for which the series converges conditionally
nb1
4. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (3xn2)1
Ä_
Ä_
Ê 1 3x 2 1 Ê
"
3
†
n
(3x 2)n ¹
ˆ n ‰ 1 Ê k3x 2k 1
1 Ê k3x 2k n lim
Ä _ n1
x 1; when x œ
"
3
_
nœ1
(b) the interval of absolute convergence is
"
3
(c) the series converges conditionally at x œ
nœ1
"
n
conditionally convergent; when x œ 1 we have !
(a) the radius is "3 ; the interval of convergence is
_
we have !
"
3
(")n
n
which is the alternating harmonic series and is
, the divergent harmonic series
Ÿx1
x1
"
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
732
Chapter 11 Infinite Sequences and Series
nb1
2)
5. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 10
nb1
Ä_
Ä_
10n
(x 2)n ¹
1 Ê
kx 2 k
10
1 Ê kx 2k 10 Ê 10 x 2 10
_
_
nœ1
nœ1
Ê 8 x 12; when x œ 8 we have ! (")n , a divergent series; when x œ 12 we have ! 1, a divergent
series
(a) the radius is "0; the interval of convergence is 8 x 12
(b) the interval of absolute convergence is 8 x 12
(c) there are no values for which the series converges conditionally
nb1
6. n lim
k2xk 1 Ê k2xk 1 Ê "# x
¹ uunbn 1 ¹ 1 Ê n lim
¹ (2x)
(2x)n ¹ 1 Ê n lim
Ä_
Ä_
Ä_
_
! (")n , a divergent series; when x œ
nœ1
"
#
"
#
; when x œ "# we have
_
we have ! 1, a divergent series
nœ1
(a) the radius is "# ; the interval of convergence is "# x
(b) the interval of absolute convergence is "# x
"
#
"
#
(c) there are no values for which the series converges conditionally
nb1
1)x
7. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n (n
3) †
Ä_
Ä_
(n 2)
nxn ¹
1 Ê kxk n lim
Ä_
_
Ê 1 x 1; when x œ 1 we have ! (")n
nœ1
_
have !
nœ1
n
n#,
n
n#
(n 1)(n 2)
(n 3)(n)
1 Ê kxk 1
, a divergent series by the nth-term Test; when x œ " we
a divergent series
(a) the radius is "; the interval of convergence is " x "
(b) the interval of absolute convergence is " x "
(c) there are no values for which the series converges conditionally
nb1
8. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x n2)1
Ä_
Ä_
†
n
(x 2)n ¹
ˆ
1 Ê kx 2k n lim
Ä_
_
Ê 1 x 2 1 Ê 3 x 1; when x œ 3 we have !
nœ1
_
!
nœ1
(1)n
n ,
"
n,
n ‰
n1
1 Ê kx 2k 1
a divergent series; when x œ " we have
a convergent series
(a) the radius is "; the interval of convergence is 3 x Ÿ "
(b) the interval of absolute convergence is 3 x "
(c) the series converges conditionally at x œ 1
nb1
x
9. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ (n 1)Èn 1 3nb1
Ê
kx k
3
nÈ n 3n
xn ¹
1 Ê
kx k
3
n
n
‹
n 1 ‹ ŠÉ n lim
Ä _ n1
Šn lim
Ä_
_
(1)(1) 1 Ê kxk 3 Ê 3 x 3; when x œ 3 we have !
nœ1
_
when x œ 3 we have !
nœ1
1
,
n$Î#
(")n
,
n$Î#
1
an absolutely convergent series;
a convergent p-series
(a) the radius is 3; the interval of convergence is 3 Ÿ x Ÿ 3
(b) the interval of absolute convergence is 3 Ÿ x Ÿ 3
(c) there are no values for which the series converges conditionally
nb1
10. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (xÈn1) 1 †
Ä_
Ä_
Èn
(x 1)n ¹
1 Ê kx 1k Én lim
Ä_
_
Ê 1 x 1 1 Ê 0 x 2; when x œ 0 we have !
nœ1
(")n
,
n"Î#
n
n1
1 Ê kx 1k 1
a conditionally convergent series; when x œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.7 Power Series
_
we have !
nœ1
1
,
n"Î#
a divergent series
(a) the radius is 1; the interval of convergence is 0 Ÿ x 2
(b) the interval of absolute convergence is 0 x 2
(c) the series converges conditionally at x œ 0
nb1
ˆ " ‰ 1 for all x
11. n lim
† n! ¹ 1 Ê kxk n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ (n 1)! xn
Ä _ n1
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1
nb1
ˆ " ‰ 1 for all x
12. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ 3 x † 3nn!xn ¹ 1 Ê 3 kxk n lim
Ä_
Ä _ (n 1)!
Ä _ n1
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
2nb3
ˆ " ‰ 1 for all x
13. n lim
† n! ¹ 1 Ê x# n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ (n 1)! x2nb1
Ä _ n1
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
2nb3
ˆ " ‰ 1 for all x
14. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (2x(n3)1)! † (2x n!3)2nb1 ¹ 1 Ê (2x 3)# n lim
Ä_
Ä_
Ä _ n 1
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1
x
15. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ È(n 1)# 3
È n# 3
¹
xn
_
Ê 1 x 1; when x œ 1 we have !
nœ1
_
!
nœ1
"
È n# 3
1 Ê kxk Én lim
Ä_
(")n
È n# 3
n# 3
n# 2n 4
" Ê kxk 1
, a conditionally convergent series; when x œ 1 we have
, a divergent series
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
n1
x
16. n lim
†
¹ uunn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ È(n 1)# 3
È n# 3
¹
xn
_
Ê 1 x 1; when x œ 1 we have !
nœ1
1 Ê kxk Én lim
Ä_
"
È n# 3
n# 3
n# 2n 4
" Ê kxk 1
_
, a divergent series; when x œ 1 we have !
nœ1
(")n
È n# 3
a conditionally convergent series
(a) the radius is 1; the interval of convergence is 1 x Ÿ 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
3)
17. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 1)(x
5nb1
Ä_
Ä_
nb1
†
5n
n(x 3)n ¹
1 Ê
kx 3 k
lim
5
nÄ_
ˆ n n " ‰ 1 Ê
kx 3 k
5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
,
733
734
Chapter 11 Infinite Sequences and Series
_
Ê kx 3k 5 Ê 5 x 3 5 Ê 8 x 2; when x œ 8 we have !
nœ1
_
series; when x œ 2 we have !
nœ1
n
n5
5n
n(5)n
5n
_
œ ! (1)n n, a divergent
nœ1
_
œ ! n, a divergent series
nœ1
(a) the radius is 5; the interval of convergence is 8 x 2
(b) the interval of absolute convergence is 8 x 2
(c) there are no values for which the series converges conditionally
nb1
18. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 1)x
Ä_
Ä _ 4nb1 an# 2n 2b
_
Ê 4 x 4; when x œ 4 we have !
nœ1
_
!
nœ1
n
n# 1
4 n an # 1 b
¹
nxn
n(1)n
n# 1
1 Ê
kx k
4 n lim
Ä_
#
(n 1) n
1
¹ n an# a2n 2bb ¹ 1 Ê kxk 4
, a conditionally convergent series; when x œ 4 we have
, a divergent series
(a) the radius is 4; the interval of convergence is 4 Ÿ x 4
(b) the interval of absolute convergence is 4 x 4
(c) the series converges conditionally at x œ 4
19. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä_
Èn 1 xnb1
3nb1
3n
È n xn ¹
†
1 Ê
kx k
3
ˆ n n 1 ‰ 1 Ê
Én lim
Ä_
kx k
3
1 Ê kxk 3
_
Ê 3 x 3; when x œ 3 we have ! (1)n Èn , a divergent series; when x œ 3 we have
nœ1
_
! Èn, a divergent series
nœ1
(a) the radius is 3; the interval of convergence is 3 x 3
(b) the interval of absolute convergence is 3 x 3
(c) there are no values for which the series converges conditionally
20. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä_
nbÈ
1
n 1 (2x5)nb1
¹
n n (2x5)n
È
1 Ê k2x 5k n lim
Š
Ä_
nbÈ
1
n1
‹
n n
È
1
t
È
lim
t
Ä_
Ê k2x 5k Œ tlim
n n 1 Ê k2x 5k 1 Ê 1 2x 5 1 Ê 3 x 2; when x œ 3 we have
È
n
_
Ä_
_
n
! (1) nÈn, a divergent series since lim nÈn œ 1; when x œ 2 we have ! È
n, a divergent series
nÄ_
nœ1
nœ1
(a) the radius is "# ; the interval of convergence is 3 x 2
(b) the interval of absolute convergence is 3 x 2
(c) there are no values for which the series converges conditionally
21.
lim ¹ uunbn 1 ¹
nÄ_
1 Ê n lim
Ä_ »
Š1
nb1
"
n1‹
xnb1
Š1 "n ‹ xn
n
"
t
lim Š1 t ‹
e
Ä_
» 1 Ê kxk lim Š1 " ‹n 1 Ê kxk ˆ e ‰ 1 Ê kxk 1
n
nÄ_
t
_
n
Ê 1 x 1; when x œ 1 we have ! (1)n ˆ1 "n ‰ , a divergent series by the nth-Term Test since
nœ1
lim ˆ1
nÄ_
" ‰n
n
_
n
œ e Á 0; when x œ 1 we have ! ˆ1 n" ‰ , a divergent series
nœ1
(a) the radius is "; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
22. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ ln (nxnln1)xn
Ä_
Ä_
nb1
ˆn1‰
ˆ n ‰ 1 Ê kxk 1
¹ 1 Ê kxk n lim
º
ˆ " ‰ º 1 Ê kxk n lim
Ä_
Ä _ n1
"
n
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.7 Power Series
735
_
Ê 1 x 1; when x œ 1 we have ! (1)n ln n, a divergent series by the nth-Term Test since
nœ1
_
lim ln n Á 0; when x œ 1 we have ! ln n, a divergent series
nÄ_
nœ1
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
(c) there are no values for which the series converges conditionally
nb1 nb1
x
ˆ1 n" ‰n ‹ Š lim (n 1)‹ 1
23. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 1)
¹ 1 Ê kxk Šn lim
nn xn
Ä_
Ä_
Ä_
nÄ_
Ê e kxk n lim
(n
1)
1
Ê
only
x
œ
0
satisfies
this
inequality
Ä_
(a) the radius is 0; the series converges only for x œ 0
(b) the series converges absolutely only for x œ 0
(c) there are no values for which the series converges conditionally
nb1
24. n lim
(n 1) 1 Ê only x œ 4 satisfies this
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n n!1)!(x(x4)4)n ¹ 1 Ê kx 4k n lim
Ä_
Ä_
Ä_
inequality
(a) the radius is 0; the series converges only for x œ 4
(b) the series converges absolutely only for x œ 4
(c) there are no values for which the series converges conditionally
nb1
25. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 2)
Ä_
Ä _ (n 1) 2nb1
n2n
(x 2)n ¹
1 Ê
kx 2 k
lim
#
nÄ_
ˆ n n 1 ‰ 1 Ê
kx 2 k
#
1 Ê kx 2k 2
_
Ê 2 x 2 2 Ê 4 x 0; when x œ 4 we have ! "
n , a divergent series; when x œ 0 we have
nœ1
_
! (1)
nœ1
nb1
, the alternating harmonic series which converges conditionally
n
(a) the radius is 2; the interval of convergence is 4 x Ÿ 0
(b) the interval of absolute convergence is 4 x 0
(c) the series converges conditionally at x œ 0
nb1
nb1
(n 2)(x 1)
ˆ n 2 ‰ 1 Ê 2 kx 1k 1
26. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ ((2)2)n (n
1)(x 1)n ¹ 1 Ê 2 kx 1k n lim
Ä_
Ä_
Ä _ n1
Ê kx 1k
"
#
Ê "# x 1
"
#
Ê
"
#
x 3# ; when x œ
"
#
_
we have ! (n 1) , a divergent series; when x œ
nœ1
_
we have ! (1)n (n 1), a divergent series
n œ1
(a) the radius is "# ; the interval of convergence is
(b) the interval of absolute convergence is
"
#
"
#
x
x
3
#
3
#
(c) there are no values for which the series converges conditionally
nb1
x
27. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ (n 1) aln (n 1)b#
Ê kxk (1) Œn lim
Ä_
_
!
nœ1
(1)n
n(ln n)#
#
ˆ "n ‰
ˆ nb" 1 ‰
n(ln n)#
xn ¹
1 Ê kxk Šn lim
Ä_
1 Ê kxk Šn lim
Ä_
n1
n ‹
#
n
ln n
‹
n 1 ‹ Šn lim
Ä _ ln (n 1)
#
1
1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have
_
which converges absolutely; when x œ 1 we have !
nœ1
"
n(ln n)#
which converges
(a) the radius is "; the interval of convergence is 1 Ÿ x Ÿ 1
(b) the interval of absolute convergence is 1 Ÿ x Ÿ 1
(c) there are no values for which the series converges conditionally
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
3
#
736
Chapter 11 Infinite Sequences and Series
nb1
x
28. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹
Ä_
Ä _ (n 1) ln (n 1)
n ln (n)
xn ¹
1 Ê kxk Šn lim
Ä_
ln (n)
n
‹
n 1 ‹ Šn lim
Ä _ ln (n 1)
_
(1)n
n ln n
Ê kxk (1)(1) 1 Ê kxk 1 Ê 1 x 1; when x œ 1 we have !
nœ2
_
"
n ln n
when x œ 1 we have !
nœ2
1
, a convergent alternating series;
which diverges by Exercise 38, Section 11.3
(a) the radius is "; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
2nb3
5)
29. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ (4x
(n 1)$Î#
Ä_
Ä_
n$Î#
(4x 5)2n1 ¹
1 Ê (4x 5)# Šn lim
Ä_
Ê k4x 5k 1 Ê 1 4x 5 1 Ê 1 x
absolutely convergent; when x œ
3
#
_
we have !
nœ1
(")2nb1
n$Î#
_
; when x œ 1 we have !
3
#
nœ1
$Î#
1 Ê (4x 5)# 1
(1)2nb1
n$Î#
_
œ!
nœ1
"
n$Î#
which is
, a convergent p-series
(a) the radius is "4 ; the interval of convergence is 1 Ÿ x Ÿ
(b) the interval of absolute convergence is 1 Ÿ x Ÿ
n
n1‹
3
#
3
#
(c) there are no values for which the series converges conditionally
nb2
30. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (3x2n1)4
Ä_
Ä_
2n 2
(3x 1)nb1 ¹
†
ˆ 2n 2 ‰ 1 Ê k3x 1k 1
1 Ê k3x 1k n lim
Ä _ 2n 4
_
Ê 1 3x 1 1 Ê 23 x 0; when x œ 23 we have !
nœ1
_
(")nb1
2n 1
when x œ 0 we have !
nœ1
_
"
#n 1
œ!
nœ1
(1)nb1
2n 1
, a conditionally convergent series;
, a divergent series
(a) the radius is "3 ; the interval of convergence is 23 Ÿ x 0
(b) the interval of absolute convergence is 23 x 0
(c) the series converges conditionally at x œ 23
nb1
31. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (xÈn1) 1 †
Ä_
Ä_
Èn
(x 1)n ¹
n
1 Ê kx 1k n lim
¹
¹ 1
Ä _ Én1
ˆ n ‰ 1 Ê kx 1 k 1 Ê 1 x 1 1 Ê 1 1 x 1 1 ;
Ê kx 1k Én lim
Ä _ n1
_
(1)n
Èn
when x œ 1 1 we have !
nœ1
_
!
nœ1
"n
Èn
_
œ!
nœ1
"
n"Î#
_
œ!
nœ1
(")n
n"Î#
, a conditionally convergent series; when x œ 1 1 we have
, a divergent p-series
(a) the radius is "; the interval of convergence is (1 1) Ÿ x (1 1)
(b) the interval of absolute convergence is 1 1 x 1 1
(c) the series converges conditionally at x œ 1 1
2nb3
32.
lim ¹ uunbn 1 ¹
nÄ_
1 Ê n lim
Ä_ »
Šx È2‹
#
Ê
Šx È2‹
#
2nb1
nœ1
nœ1
1 Ê
#
lim
nÄ_
k1k 1
#
_ ŠÈ2‹
we have !
†
2n1
Šx È2‹
1 Ê Šx È2‹ 2 Ê ¹x È2¹ È2 Ê È2 x È2 È2 Ê 0 x 2È2 ; when
x œ 0 we have !
_
#
2n
2n1 »
È
Šx 2‹
2nb1
ŠÈ2‹
2n
2n
_
œ !
nœ1
_
œ!
nœ1
2
nb1Î2
2n
2nb1Î2
#n
_
œ ! È2 which diverges since n lim
a Á 0; when x œ 2È2
Ä_ n
nœ1
_
œ ! È2, a divergent series
nœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.7 Power Series
(a) the radius is È2; the interval of convergence is 0 x 2È2
(b) the interval of absolute convergence is 0 x 2È2
(c) there are no values for which the series converges conditionally
2nb2
33. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 4n1)1
Ä_
Ä_
4n
(x 1)2n ¹
†
1 Ê
(x 1)#
lim
4
nÄ_
_
Ê 2 x 1 2 Ê 1 x 3; at x œ 1 we have !
nœ0
_
we have !
nœ0
_
!
nœ0
(x ")2n
4n
"
#
n
œ!
2
4n
4
4n
nœ0
_
œ!
nœ0
œ
"
1 Šxc
# ‹
_
2n
4 (x ")#
“
4
œ
n
4
4 x# 2x 1
œ
4
3 2x x#
9n
(x 1)2n ¹
†
1 Ê
(x 1)#
lim
9
nÄ_
_
nœ0
!
!
nœ0
k1k 1 Ê (x 1)# 9 Ê kx 1k 3
(3)2n
9n
_
œ ! 1 which diverges; at x œ 2 we have
nœ0
_
œ ! " which also diverges; the interval of convergence is 4 x 2; the series
(x 1)
9n
"
nœ0
is a convergent geometric series when 1 x 3 and the sum is
Ê 3 x 1 3 Ê 4 x 2; when x œ 4 we have !
nœ0
_
nœ0
nœ0
2nb2
32n
9n
_
œ ! 1, a divergent series; the interval of convergence is 1 x 3; the series
34. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 9n1)1
Ä_
Ä_
_
_
n
œ ! 44n œ ! 1, which diverges; at x œ 3
_
#
Šˆ x # 1 ‰ ‹
"
’
(2)2n
4n
k1k 1 Ê (x 1)# 4 Ê kx 1k 2
nœ0
2n
_
n
#
œ ! Šˆ x3 1 ‰ ‹ is a convergent geometric series when 4 x 2 and the sum is
nœ0
1
1 Šxb
3 ‹
#
œ
"
’
9 (x 1)#
“
9
œ
9
9 x# 2x 1
35. n lim
¹ uunbn 1 ¹ 1 Ê n lim
Ä_
Ä_ º
œ
9
8 2x x#
ˆÈx 2‰nb1
2nb1
2n
ˆÈ x 2 ‰ n º
†
1 Ê ¸È x 2 ¸ 2 Ê 2 È x 2 2 Ê 0 È x 4
_
_
Ê 0 x 16; when x œ 0 we have ! (1)n , a divergent series; when x œ 16 we have ! (1)n , a divergent
nœ0
nœ0
_
series; the interval of convergence is 0 x 16; the series !
nœ0
0 x 16 and its sum is
1Œ
"
Èx c 2 œ
#
Œ
2c
"
Èx 2 œ
#
Èx 2 n
Š # ‹
is a convergent geometric series when
2
4 Èx
nb1
36. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (ln(lnx)x)n ¹ 1 Ê kln xk 1 Ê 1 ln x 1 Ê e" x e; when x œ e" or e we
Ä_
Ä_
_
_
nœ0
nœ0
obtain the series ! 1n and ! (1)n which both diverge; the interval of convergence is e" x e;
_
! (ln x)n œ
nœ0
"
1 ln x
when e" x e
37. n lim
¹ uunbn 1 ¹ 1 Ê n lim
Šx
Ä_
Ä_ º
#
1
3 ‹
n 1
n
† ˆ x# 3 1 ‰ º 1 Ê
ax # 1 b
lim
3
nÄ_
k1k 1 Ê
x# "
3
1 Ê x# 2
_
Ê kxk È2 Ê È2 x È2 ; at x œ „ È2 we have ! (1)n which diverges; the interval of convergence is
nœ0
_
È2 x È2 ; the series !
nœ0
"
#
1 Š x 3b 1 ‹
œ
"
#
Š 3 c x3 c 1 ‹
œ
#
Š x 3 1 ‹
n
is a convergent geometric series when È2 x È2 and its sum is
3
# x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
737
738
Chapter 11 Infinite Sequences and Series
a
38. n lim
¹ uunn 1 ¹ 1 Ê n lim
¹x
Ä_
Ä_
# 1bn1
2n1
2n
¹
ax # 1 b n
†
1 Ê kx# 1k 2 Ê È3 x È3 ; when x œ „ È3 we
_
_
nœ0
nœ0
have ! 1n , a divergent series; the interval of convergence is È3 x È3 ; the series ! Š x
"
#
1 Š x 2 1 ‹
convergent geometric series when È3 x È3 and its sum is
nb1
39. n lim
¹ (x #n3)
b1
Ä_
2n
(x 3)n ¹
†
"
œ
2 Šx# 1 ‹
#
œ
#
1
2 ‹
n
is a
2
3 x#
_
1 Ê kx 3k 2 Ê 1 x 5; when x œ 1 we have ! (1)n which diverges;
nœ1
_
when x œ 5 we have ! (1) which also diverges; the interval of convergence is 1 x 5; the sum of this
n
nœ1
"
3
1 Šxc
# ‹
convergent geometric series is
œ
2
x1
œ
2
x 1
n
. If f(x) œ 1 #" (x 3) 4" (x 3)# á ˆ #" ‰ (x 3)n á
n
then f w (x) œ #" #" (x 3) á ˆ #" ‰ n(x 3)n1 á is convergent when 1 x 5, and diverges
2
(x 1)#
when x œ 1 or 5. The sum for f w (x) is
, the derivative of
2
x1
.
n
40. If f(x) œ 1 "# (x 3) 4" (x 3)# á ˆ #" ‰ (x 3)n á œ
œx
(x 3)#
4
(x 3)$
12
_
the series
n
! (1) 2
n 1
á ˆ "# ‰
n (x 3)n1
n 1
2 ln kx 1k (3 ln 4), since '
œ
_
á . At x œ 1 the series ! n21 diverges; at x œ 5
nœ1
dx œ 2 ln kx 1k C, where C œ 3 ln 4 when x œ 3.
2
x1
41. (a) Differentiate the series for sin x to get cos x œ 1
x%
4!
then ' f(x) dx
converges. Therefore the interval of convergence is 1 x Ÿ 5 and the sum is
nœ1
x#
#!
2
x1
x'
6!
)
x"!
1 x8! 10!
á .
a#nb!
x2nb2
2
lim ¹
† x#8 ¹ œ x n lim
n Ä _ (2n 2)!
Ä_
(b) sin 2x œ 2x
2$ x$
3!
2& x&
5!
2( x(
7!
"
6!
œ 2x
42. (a)
(b)
d
x
5x%
5!
7x'
7!
9x)
9!
11x"!
11!
á
The series converges for all values of x since
Š a2n 1ba" 2n 2b ‹ œ 0 1 for all x.
1†00†
"
4!
$ $
( (
2 x
3!
& &
2 x
5!
2x
2!
3x#
3!
aex b œ 1
0†
2 x
7!
4x$
4!
' ex dx œ ex C œ x x#
#
#
(c) ex œ 1 x x#!
ˆ1 † 3!" 1 † #"!
ˆ1 † 5!" 1 † 4!"
x$
3!
"
#!
"
#!
2* x*
9!
2"" x""
11!
"
3!
* *
2 x
9!
"
#
0†
"" ""
á
0†
5x%
5!
2 x
11!
œ
x
#
%
x
1#
'
x
45
)
&
(
*
""
(b) sec# x œ
when
d(tan x)
dx
1
#
œ
x
d
dx
"
5!
0 † 1‰ x' á ‘ œ 2 ’x
x#
#!
x$
3!
x%
4!
4x$
3!
16x&
5!
á“
á œ ex ; thus the derivative of ex is ex itself
x$
x%
x&
x
3! 4! 5! á C, which is the general antiderivative of e
%
&
x4! x5! á ; ecx † ex œ 1 † 1 (1 † 1 1 † 1)x ˆ1 † #"! 1 † 1 #"!
† 1 3!" † 1‰ x$ ˆ1 † 4!" 1 † 3!" #"! † #"! 3!" † 1 4!" † 1‰ x%
† 3!" 3!" † #"! 4!" † 1 5!" † 1‰ x& á œ 1 0 0 0 0 0 á
17x
2520
1
#
converges when
á œ 2x
á œ1x
43. (a) ln ksec xk C œ ' tan x dx œ ' Šx
#
8x$
3!
128x
512x
2048x
32x
5! 7! 9! 11! á
"
"‰ $
‰ # ˆ
(c) 2 sin x cos x œ 2 (0 † 1) (0 † 0 1 † 1)x ˆ0 † "
# 1 † 0 0 † 1 x 0 † 0 1 † # 0 † 0 1 † 3! x
ˆ0 † 4!" 1 † 0 0 † #" 0 † 3!" 0 † 1‰ x% ˆ0 † 0 1 † 4!" 0 † 0 #" † 3!" 0 † 0 1 † 5!" ‰ x&
ˆ0 †
3x#
3!
x
x$
3
2x&
15
17x(
315
62x*
2835
á ‹ dx
"!
á C; x œ 0 Ê C œ 0 Ê ln ksec xk œ
2x&
15
31x
14,175
1#
Šx
x$
3
17x(
315
62x*
2835
† 1‰ x#
á ‹ œ 1 x#
2x%
3
x#
#
17x'
45
x%
12
62x)
315
x'
45
17x)
2520
31x"!
14,175
á , converges
1
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
á ,
Section 11.8 Taylor and Maclaurin Series
x#
#
(c) sec# x œ (sec x)(sec x) œ Š1
œ1
ˆ "#
#
5
"# ‰ x# ˆ 24
4"
%
'
62x)
2x3 17x
45 315
œ1x
5x%
24
5 ‰ %
24 x
61x'
720
1
#
á ,
61
ˆ 720
á ‹ Š1
x
44. (a) ln ksec x tan xk C œ ' sec x dx œ ' Š1
$
œx
x
6
œx
$
x
6
&
x
24
&
x
24
(b) sec x tan x œ
when
1
#
(
61x
5040
(
œ
x#
#
1
#
x
61x'
720
61x'
720
á‹
á
á ‹ dx
á C; x œ 0 Ê C œ 0 Ê ln ksec x tan xk
á , converges when 1# x
Š1
2
œ x ˆ "3 #" ‰ x$ ˆ 15
61 ‰ '
720 x
x#
#
5x%
24
61x'
720
5x$
6
á‹ œ x
1
#
61x&
120
277x(
1008
á , converges
1
#
(c) (sec x)(tan x) œ Š1
1
#
5x%
24
5x%
24
*
277x
72,576
d
dx
x#
2
5
48
*
277x
72,576
61x
5040
d(sec x)
dx
x
5
48
1
#
x#
#
5x%
24
"
6
_
61x'
720
á ‹ Šx
5 ‰ &
24 x
17
ˆ 315
"
15
x$
3
2x&
15
5
72
17x(
315
61 ‰ (
720 x
á‹
á œ x
5x$
6
61x&
120
277x(
1008
á ,
_
45. (a) If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n (k 1)) an xnk and f ÐkÑ (0) œ k!ak
nœ0
Ê ak œ
f
ÐkÑ
(0)
k!
nœk
_
; likewise if f(x) œ ! bn xn , then bk œ
nœ0
f ÐkÑ (0)
k!
Ê ak œ bk for every nonnegative integer k
_
(b) If f(x) œ ! an xn œ 0 for all x, then f ÐkÑ (x) œ 0 for all x Ê from part (a) that ak œ 0 for every
nœ0
nonnegative integer k
46.
"
1x
œ 1 x x# x$ x% á Ê x ’ (1 " x)# “ œ x a1 2x 3x# 4x$ á b Ê
œ x 2x# 3x$ 4x% á Ê x ’ (11x)x $ “ œ x a1 4x 9x# 16x$ á b Ê
œ x 4x# 9x$ 16x% á Ê
_
ˆ "# 4" ‰
ˆ "8 ‰
œ
"
#
4
4
9
8
16
16
_
á Ê !
nœ1
n#
2n
x
(1 x)#
x x#
(1 x)$
œ6
47. The series ! xn converges conditionally at the left-hand endpoint of its interval of convergence [1ß 1Ñ; the
n
nœ1
_
series !
nœ1
xn
an # b
converges absolutely at the left-hand endpoint of its interval of convergence [1ß 1]
48. Answers will vary. For instance:
_
_
n
(a) ! ˆ x3 ‰
(b) ! (x 1)n
nœ1
nœ1
_
n
(c) ! ˆ x # 3 ‰
nœ1
11.8 TAYLOR AND MACLAURIN SERIES
1. f(x) œ ln x, f w (x) œ
"
x
, f ww (x) œ x"# , f www (x) œ
2
x$ ;
f(1) œ ln 1 œ 0, f w (1) œ 1, f ww (1) œ 1, f www (1) œ 2 Ê P! (x) œ 0,
P" (x) œ (x 1), P# (x) œ (x 1) "# (x 1)# , P$ (x) œ (x 1) "# (x 1)# "3 (x 1)$
2. f(x) œ ln (1 x), f w (x) œ
f w (0) œ
œx
1
1
x#
#
"
1x œ
#
œ 1, f ww (0) œ (1)
(1 x)" , f ww (x) œ (1 x)# , f www (x) œ 2(1 x)$ ; f(0) œ ln 1 œ 0,
œ 1, f www (0) œ 2(1)$ œ 2 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x
x$
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x#
#,
P$ (x)
739
740
Chapter 11 Infinite Sequences and Series
3. f(x) œ
"
x
œ x" , f w (x) œ x# , f ww (x) œ 2x$ , f www (x) œ 6x% ; f(2) œ
Ê P! (x) œ
"
#
P$ (x) œ
"
# , P" (x) œ
"
4 (x 2)
"
#
"
8
4" (x 2), P# (x) œ
"
16
(x 2)#
(x 2)
"
#
$
"
#
, f w (2) œ 4" , f ww (2) œ 4" , f www (x) œ 38
4" (x 2) 8" (x 2)# ,
4. f(x) œ (x 2)" , f w (x) œ (x 2)# , f ww (x) œ 2(x 2)$ , f www (x) œ 6(x 2)% ; f(0) œ (2)" œ
œ 4" , f ww (0) œ 2(2)$ œ
"
#
P$ (x) œ
x
4
x#
8
"
4
, f www (0) œ 6(2)% œ 38 Ê P! (x) œ
x$
16
"
#
f ww ˆ 14 ‰ œ sin
È2
#
P# (x) œ
1
4 œ
È2
ˆx
#
È
, f www ˆ 14 ‰ œ cos 14 œ #2 Ê
È2
È
1‰
ˆx 14 ‰# , P$ (x) œ #2
4 4
6. f(x) œ cos x, f w (x) œ sin x, f ww (x) œ cos x, f www (x) œ sin x; f ˆ 14 ‰ œ cos
f w ˆ 14 ‰ œ sin
1
4
œ È"2 , f ww ˆ 14 ‰ œ cos
P" (x) œ
"
È2
"
È2
ˆx 14 ‰ , P# (x) œ
P$ (x) œ
"
È2
"
È2
ˆx 14 ‰
"
#È 2
"
È2
1
4
œ È"2 , f www ˆ 14 ‰ œ sin
"
È2
ˆx 14 ‰#
"
#È 2
1 ‰$
4
ˆx 14 ‰
"
6È 2
ˆx
x4 , P# (x) œ
"
#
, f w (0) œ (2)#
x
4
È2
È2
1
w ˆ1‰
4 œ cos 4 œ #
# ,f
È
È
È
P! œ #2 , P" (x) œ #2 #2 ˆx 14 ‰ ,
È2
È
È
ˆx 14 ‰ 42 ˆx 14 ‰# 1#2 ˆx 14 ‰$
#
5. f(x) œ sin x, f w (x) œ cos x, f ww (x) œ sin x, f www (x) œ cos x; f ˆ 14 ‰ œ sin
È2
#
"
#
, P" (x) œ
"
#
ˆx
1
4
œ
1
4
œ
1
4
œ
"
È2
"
È2
x#
8
,
,
,
Ê P! (x) œ
"
È2
,
1 ‰#
4 ,
7. f(x) œ Èx œ x"Î# , f w (x) œ ˆ "# ‰ x"Î# , f ww (x) œ ˆ 4" ‰ x$Î# , f www (x) œ ˆ 38 ‰ x&Î# ; f(4) œ È4 œ 2,
"
3
f w (4) œ ˆ "# ‰ 4"Î# œ 4" , f ww (4) œ ˆ 4" ‰ 4$Î# œ 32
,f www (4) œ ˆ 38 ‰ 4&Î# œ 256
Ê P! (x) œ 2, P" (x) œ 2 "4 (x 4),
P# (x) œ 2 4" (x 4)
"
64
(x 4)# , P$ (x) œ 2 4" (x 4)
"
64
(x 4)#
"
51#
(x 4)$
8. f(x) œ (x 4)"Î# , f w (x) œ ˆ "# ‰ (x 4)"Î# , f ww (x) œ ˆ 4" ‰ (x 4)$Î# , f www (x) œ ˆ 38 ‰ (x 4)&Î# ; f(0) œ (4)"Î# œ 2,
3
"
f w (0) œ ˆ "# ‰ (4)"Î# œ 4" , f ww (0) œ ˆ 4" ‰ (4)$Î# œ 32
, f www (0) œ ˆ 38 ‰ (4)&Î# œ 256
Ê P! (x) œ 2,
P" (x) œ 2 4" x, P# (x) œ 2 4" x
_
xn
n!
9. ex œ !
nœ0
_
xn
n!
10. ex œ !
nœ0
_
Ê ex œ !
nœ0
(x)n
n!
_ ˆ x ‰n
Ê exÎ2 œ !
nœ0
2
n!
"
64
x# , P$ (x) œ 2 4" x
x#
#!
x$
3!
x#
4†2!
x$
2$ †3!
œ1x
œ1
x
#
x%
4!
"
64
x#
"
512
x$
á
x%
2% †4!
á
11. f(x) œ (1 x)" Ê f w (x) œ (1 x)# , f ww (x) œ 2(1 x)$ , f www (x) œ 3!(1 x)% Ê á f ÐkÑ (x)
œ (1)k k!(1 x)k1 ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3!, á ß f ÐkÑ (0) œ (1)k k!
Ê
"
1x
_
_
nœ0
nœ0
œ 1 x x# x$ á œ ! (x)n œ ! (1)n xn
12. f(x) œ (1 x)" Ê f w (x) œ (1 x)# , f ww (x) œ 2(1 x)$ , f www (x) œ 3!(1 x)% Ê á f ÐkÑ (x)
œ k!(1 x)k1 ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3!, á ß f ÐkÑ (0) œ k!
Ê
"
1x
_
œ 1 x x# x$ á œ ! xn
_
13. sin x œ !
nœ0
nœ0
(")n x2nb1
(#n1)!
_
Ê sin 3x œ !
nœ0
(")n (3x)2nb1
(#n1)!
_
œ!
nœ0
(")n 32nb1 x2nb1
(#n1)!
œ 3x
3$ x$
3!
3& x&
5!
á
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.8 Taylor and Maclaurin Series
_
14. sin x œ !
nœ0
(")n x2nb1
(#n1)!
Ê sin
_
15. 7 cos (x) œ 7 cos x œ 7 !
nœ0
_
16. cos x œ !
nœ0
17. cosh x œ
_
œ!
nœ0
_
nœ0
(")n ˆ #x ‰
(#n1)!
nœ0
(")n x2n
(2n)!
œ7
Ê 5 cos 1x œ 5 !
nœ0
œ
"
#
’Š1 x#
œ
"
#
’Š1 x
x#
#!
x$
3!
_
œ!
nœ0
7x#
#!
7x%
4!
(1)n (1x)2n
(#n)!
x%
4!
(")n x2nb1
#2n1 (2n1)!
7x'
6!
œ5
œ
x$
2$ †3!
x
#
x&
2& †5!
á
á , since the cosine is an even function
51 # x#
2!
51 % x%
4!
á ‹ Š1 x
x#
#!
51 ' x'
6!
x$
3!
á
x%
4!
á ‹“ œ 1
x#
#!
x%
4!
x'
6!
á
x2n
(2n)!
18. sinh x œ
œ!
2n1
_
œ!
_
(1)n x2n
(2n)!
ex ecx
#
x
#
741
ex ecx
#
x#
#!
x$
3!
x%
4!
á ‹ Š1 x
x#
#!
x$
3!
x%
4!
á ‹“ œ x
x$
3!
x&
5!
x'
6!
á
x2n1
(2n 1)!
19. f(x) œ x% 2x$ 5x 4 Ê f w (x) œ 4x$ 6x# 5, f ww (x) œ 12x# 12x, f www (x) œ 24x 12, f Ð4Ñ (x) œ 24
Ê f ÐnÑ (x) œ 0 if n 5; f(0) œ 4, f w (0) œ 5, f ww (0) œ 0, f www (0) œ 12, f Ð4Ñ (0) œ 24, f ÐnÑ (0) œ 0 if n 5
24 %
$
%
$
Ê x% 2x$ 5x 4 œ 4 5x 12
3! x 4! x œ x 2x 5x 4 itself
20. f(x) œ (x 1)# Ê f w (x) œ 2(x 1); f ww (x) œ 2 Ê f ÐnÑ (x) œ 0 if n
n 3 Ê (x 1)# œ 1 2x #2! x# œ 1 2x x#
3; f(0) œ 1, f w (0) œ 2, f ww (0) œ 2, f ÐnÑ (0) œ 0 if
21. f(x) œ x$ 2x 4 Ê f w (x) œ 3x# 2, f ww (x) œ 6x, f www (x) œ 6 Ê f ÐnÑ (x) œ 0 if n 4; f(2) œ 8, f w (2) œ 10,
6
#
$
f ww (2) œ 12, f www (2) œ 6, f ÐnÑ (2) œ 0 if n 4 Ê x$ 2x 4 œ 8 10(x 2) 12
2! (x 2) 3! (x 2)
œ 8 10(x 2) 6(x 2)# (x 2)$
22. f(x) œ 2x$ x# 3x 8 Ê f w (x) œ 6x# 2x 3, f ww (x) œ 12x 2, f www (x) œ 12 Ê f ÐnÑ (x) œ 0 if n
f w (1) œ 11, f ww (1) œ 14, f www (1) œ 12, f ÐnÑ (1) œ 0 if n 4 Ê 2x$ x# 3x 8
12
#
$
#
$
œ 2 11(x 1) 14
2! (x 1) 3! (x 1) œ 2 11(x 1) 7(x 1) 2(x 1)
4; f(1) œ 2,
23. f(x) œ x% x# 1 Ê f w (x) œ 4x$ 2x, f ww (x) œ 12x# 2, f www (x) œ 24x, f Ð4Ñ (x) œ 24, f ÐnÑ (x) œ 0 if n 5;
f(2) œ 21, f w (2) œ 36, f ww (2) œ 50, f www (2) œ 48, f Ð4Ñ (2) œ 24, f ÐnÑ (2) œ 0 if n 5 Ê x% x# 1
48
24
#
$
%
#
$
%
œ 21 36(x 2) 50
2! (x 2) 3! (x 2) 4! (x 2) œ 21 36(x 2) 25(x 2) 8(x 2) (x 2)
24. f(x) œ 3x& x% 2x$ x# 2 Ê f w (x) œ 15x% 4x$ 6x# 2x, f ww (x) œ 60x$ 12x# 12x 2,
f www (x) œ 180x# 24x 12, f Ð4Ñ (x) œ 360x 24, f Ð5Ñ (x) œ 360, f ÐnÑ (x) œ 0 if n 6; f(1) œ 7,
f w (1) œ 23, f ww (1) œ 82, f www (1) œ 216, f Ð4Ñ (1) œ 384, f Ð5Ñ (1) œ 360, f ÐnÑ (1) œ 0 if n 6
216
384
360
#
$
%
&
Ê 3x& x% 2x$ x# 2 œ 7 23(x 1) 82
2! (x 1) 3! (x 1) 4! (x 1) 5! (x 1)
œ 7 23(x 1) 41(x 1)# 36(x 1)$ 16(x 1)% 3(x 1)&
25. f(x) œ x# Ê f w (x) œ 2x$ , f ww (x) œ 3! x% , f www (x) œ 4! x& Ê f ÐnÑ (x) œ (1)n (n 1)! xn2 ;
f(1) œ 1, f w (1) œ 2, f ww (1) œ 3!, f www (1) œ 4!, f ÐnÑ (1) œ (1)n (n 1)! Ê x"#
_
œ 1 2(x 1) 3(x 1)# 4(x 1)$ á œ ! (1)n (n 1)(x 1)n
nœ0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
742
Chapter 11 Infinite Sequences and Series
26. f(x) œ
Ê f w (x) œ (1 x)# , f ww (x) œ 2(1 x)$ , f www (x) œ 3! (1 x)% Ê f ÐnÑ (x) œ n! (1 x)n1 ;
x
1x
f(0) œ 0, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3! Ê
x
1x
_
œ x x# x$ á œ ! xn1
nœ0
27. f(x) œ ex Ê f w (x) œ ex , f ww (x) œ ex Ê f ÐnÑ (x) œ ex ; f(2) œ e# , f w (2) œ e# , á f ÐnÑ (2) œ e#
Ê ex œ e# e# (x 2)
e#
#
(x 2)#
e$
3!
_
(x 2)$ á œ !
nœ0
e#
n!
(x 2)n
28. f(x) œ 2x Ê f w (x) œ 2x ln 2, f ww (x) œ 2x (ln 2)# , f www (x) œ 2x (ln 2)$ Ê f ÐnÑ (x) œ 2x (ln 2)n ; f(1) œ 2, f w (1) œ 2 ln 2,
f ww (1) œ 2(ln 2)# , f www (1) œ 2(ln 2)$ , á , f ÐnÑ (1) œ 2(ln 2)n
Ê 2x œ 2 (2 ln 2)(x 1)
_
29. If ex œ !
nœ0
f ÐnÑ (a)
n!
2(ln 2)#
#
(x 1)#
2(ln 2)$
3!
_
(x 1)$ á œ !
nœ0
2(ln 2)n (x1)n
n!
(x a)n and f(x) œ ex , we have f ÐnÑ (a) œ ea f or all n œ 0, 1, 2, 3, á
!
Ê ex œ ea ’ (x 0!a)
(x a)"
1!
(x a)#
2!
á “ œ ea ’1 (x a)
(x a)#
2!
á “ at x œ a
30. f(x) œ ex Ê f ÐnÑ (x) œ ex for all n Ê f ÐnÑ (1) œ e for all n œ 0, 1, 2, á
Ê ex œ e e(x 1)
e
#!
(x 1)#
e
3!
(x 1)$ á œ e ’1 (x 1)
f ww (a)
f www (a)
#
$
w
# (x a) 3! (x a) á Ê f (x)
www
œ f w (a) f ww (a)(x a) f 3!(a) 3(x a)# á Ê f ww (x) œ f ww (a) f www (a)(x
Ðn2Ñ
Ê f ÐnÑ (x) œ f ÐnÑ (a) f Ðn1Ñ (a)(x a) f # (a) (x a)# á
w
w
Ðn Ñ
Ðn Ñ
(x 1)#
2!
31. f(x) œ f(a) f w (a)(x a)
Ê f(a) œ f(a) 0, f (a) œ f (a) 0, á , f
(a) œ f
a)
(x 1)$
3!
f Ð4Ñ (a)
4!
á“
4 † 3(x a)# á
(a) 0
32. E(x) œ f(x) b! b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n
Ê 0 œ E(a) œ f(a) b! Ê b! œ f(a); from condition (b),
lim
xÄa
Ê
Ê
Ê
f(x) f(a) b" (x a) b# (x a)# b$ (x a)$ á bn (x a)n
(x a)n
œ0
w
a)# á nbn (x a)n1
lim f (x) b" 2b# (x a) n(x3b$ (xa)
œ0
n1
xÄa
ww
f
(x)
2b
3!
b
(x
a)
n(n
á
")bn (x a)n2
#
$
b" œ f w (a) Ê xlim
œ
n(n 1)(x a)n2
Äa
f www (x) 3! b$ á n(n 1)(n 2)bn (x a)n3
" ww
b# œ # f (a) Ê xlim
œ0
n(n 1)(n #)(x a)n3
Äa
0
ÐnÑ
f (x) n! bn
f www (a) Ê xlim
œ 0 Ê bn œ n!" f ÐnÑ (a); therefore,
n!
Äa
ww
ÐnÑ
g(x) œ f(a) f w (a)(x a) f 2!(a) (x a)# á f n!(a) (x a)n œ Pn (x)
œ b$ œ
"
3!
33. f(x) œ ln (cos x) Ê f w (x) œ tan x and f ww (x) œ sec# x; f(0) œ 0, f w (0) œ 0, f ww (0) œ 1
#
Ê L(x) œ 0 and Q(x) œ x2
34. f(x) œ esin x Ê f w (x) œ (cos x)esin x and f ww (x) œ ( sin x)esin x (cos x)# esin x ; f(0) œ 1, f w (0) œ 1,
f ww (0) œ 1 Ê L(x) œ 1 x and Q(x) œ 1 x
35. f(x) œ a1 x# b
w
ww
"Î#
Ê f w (x) œ x a1 x# b
$Î#
x#
#
and f ww (x) œ a1 x# b
f (0) œ 0, f (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1
$Î#
3x# a1 x# b
&Î#
; f(0) œ 1,
#
x
#
36. f(x) œ cosh x Ê f w (x) œ sinh x and f ww (x) œ cosh x; f(0) œ 1, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x#
#
Section 11.9 Convergence of Taylor Series; Error Estimates
37. f(x) œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x; f(0) œ 0, f w (0) œ 1, f ww (0) œ 0 Ê L(x) œ x and Q(x) œ x
38. f(x) œ tan x Ê f w (x) œ sec# x and f ww (x) œ 2 sec# x tan x; f(0) œ 0, f w (0) œ 1, f ww œ 0 Ê L(x) œ x and Q(x) œ x
11.9 CONVERGENCE OF TAYLOR SERIES; ERROR ESTIMATES
_
1. ex œ 1 x
x#
#!
á œ !
2. ex œ 1 x
x#
#!
á œ !
_
œ!
nœ0
xn
n!
Ê e5x œ 1 (5x)
(5x)#
#!
á œ 1 5x
xn
n!
Ê exÎ2 œ 1 ˆ #x ‰
ˆ #x ‰#
#!
á œ1
_
(1)n x2n1
(#n1)!
Ê 5 sin (x) œ 5 ’(x)
(1)n x2n1
(#n1)!
Ê sin
nœ0
_
nœ0
x#
2# #!
x
#
5# x#
#!
x$
2$ 3!
_
5$ x$
3!
á œ!
nœ0
(1)n 5n xn
n!
á
(1)n xn
2n n!
3. sin x œ x
_
x$
3!
x&
5!
á œ!
x&
5!
á œ!
nœ0
(x)$
3!
(x)&
5!
á“
nb1 2nb1
x
œ ! 5((1)
#n1)!
nœ0
4. sin x œ x
_
x$
3!
_
nœ0
1x
#
1x
#
œ
ˆ 1#x ‰$
3!
ˆ 1#x ‰&
5!
ˆ 1#x ‰(
7!
á
n 2nb1 2nb1
1
x
œ ! (21)
2nb1 (#n1)!
nœ0
2n
_
nœ0
_
6. cos x œ !
(1)n x2n
(2n)!
x$
2†2!
x'
2# †4!
nœ0
œ1
_
nœ0
_
8. sin x œ !
nœ0
_
9. cos x œ !
nœ0
œ
x%
4!
xn
n!
7. ex œ !
Ê cos Èx 1 œ !
Ê
_
n œ0
(1)n x2nb1
(2n1)!
(1)n x2n
(2n)!
_
11. cos x œ !
nœ0
x&
5!
(1)n x2n
(2n)!
xn
n!
_
xnb1
n!
œ!
nœ0
_
Ê x# sin x œ x# Œ !
nœ0
Ê
(1)n x2nb1
(2n1)!
x)
8!
œ
$
cos ŒŠ x# ‹
x"!
10!
x#
#
1 cos x œ
"Î#
$
(1)n ŒŠ x# ‹
_
œ!
ax 1 b#
4!
a x 1 b$
6!
_
_
(1)n x3n
2n (2n)!
œ!
n œ0
x#
#
x$
#!
œ x x#
(1)n x2nb1
(#n1)!
_
1!
nœ0
_
œ!
nœ0
x%
3!
x&
4!
(1)n x2nb3
(2n1)!
(1)n x2n
(#n)!
œ
x#
#
á
œ x$
11
x&
3!
x#
2
x(
5!
x*
7!
á
x%
4!
_
(1)n x2n1
(2n1)!
x'
6!
x)
8!
n 2n
nœ2
á
2n
(#n)!
nœ0
x 1
#!
œ1
x
á œ ! ((1)
#n)!
Ê sin x x
x(
7!
"Î#
nœ0
(1)n ax1bn
(2n)!
á
Ê xex œ x Œ !
_
x$
3!
x*
2$ †6!
nœ0
$Î#
cos Š xÈ ‹
2
_
œ!
(2n)!
nœ0
x'
6!
10. sin x œ !
œ Šx
_ (1)n ’ ax1b"Î# “
(1)n x2n
(2n)!
5. cos x œ !
x*
9!
x""
11!
x$
3!
_
œ Œ!
nœ0
á‹ x
_
Ê x cos 1x œ x !
nœ0
(1)n x2nb1
(#n1)!
x$
3!
(1)n (1x)2n
(#n)!
œ
x&
5!
_
œ!
nœ0
x
x(
7!
x*
9!
x$
3!
(")n 12n x2nb1
(#n)!
x""
11!
á œ!
œx
nœ2
1 # x$
2!
1 % x&
4!
1 ' x(
6!
á
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x"!
10!
á
743
744
Chapter 11 Infinite Sequences and Series
_
nœ0
13. cos# x œ
œ1
"
#
_
(1)n x2n
(2n)!
12. cos x œ !
(2x)#
2†2!
Ê x# cos ax# b œ x# !
nœ0
"
#
(2x)%
2†4!
(2x)'
2†6!
"
#
"
#
2x ‰
14. sin# x œ ˆ 1cos
œ
#
_
œ!
nœ1
15.
x#
12x
(1)nb1 (2x)2n
#†(2n)!
_
n
2n
nœ0
(2x))
2†8!
"
#
_
nœ1
"
#
_
œ!
nœ0
"# ’1
á œ1!
cos 2x œ
2n
"# Š1
(2x)#
2!
(1)n (2x)2n
2†(2n)!
(2x)#
#!
(")n x4n2
(#n)!
(2x)%
4!
œ x#
(2x)'
6!
_
œ1!
nœ1
(2x)%
4!
(2x)'
6!
x'
2!
(2x))
8!
x"!
4!
x"%
6!
á
á“
(1)n 22n1 x2n
(2n)!
(2x)#
2†2!
á‹ œ
(2x)%
2†4!
(2x)'
2†6!
á
(1)n 22n1 x2n
(2n)!
nœ1
_
_
nœ0
nœ0
œ x# ˆ 1"2x ‰ œ x# ! (2x)n œ ! 2n xn2 œ x# 2x$ 2# x% 2$ x& á
_
nœ1
"
1 x
œ!
16. x ln (1 2x) œ x !
17.
_
œ
"
#
! (1) (2x) œ
(2n)!
cos 2x
#
(1)n ax# b
(#n)!
(1)nc1 (2x)n
n
_
œ!
nœ1
(1)nc1 2n xn1
n
_
œ ! xn œ 1 x x# x$ á Ê
_
nœ0
d
dx
œ 2x#
ˆ 1" x ‰ œ
2# x$
#
"
(1x)#
2$ x%
4
2% x&
5
á
_
œ 1 2x 3x# á œ ! nxn1
nœ1
œ ! (n 1)xn
n œ0
18.
2
a1 x b $
œ
d#
dx#
ˆ 1" x ‰ œ
d
dx
_
Š (1"x)# ‹ œ
d
dx
_
a1 2x 3x# á b œ 2 6x 12x# á œ ! n(n 1)xn2
nœ2
œ ! (n 2)(n 1)x
n
nœ0
19. By the Alternating Series Estimation Theorem, the error is less than
&
Ê kxk& 600 ‚ 10% Ê kxk È6 ‚ 10# ¸ 0.56968
20. If cos x œ 1
x#
#
kx k &
5!
Ê kxk& a5!b a5 ‚ 10% b
%
and kxk 0.5, then the error is less than ¹ (.5)
24 ¹ œ 0.0026, by Alternating Series Estimation Theorem;
since the next term in the series is positive, the approximation 1
x#
#
is too small, by the Alternating Series Estimation
Theorem
21. If sin x œ x and kxk 10$ , then the error is less than
a10c$ b
3!
$
¸ 1.67 ‚ 1010 , by Alternating Series Estimation Theorem;
$
The Alternating Series Estimation Theorem says R# (x) has the same sign as x3! . Moreover, x sin x
Ê 0 sin x x œ R# (x) Ê x 0 Ê 10$ x 0.
22. È1 x œ 1
x
#
x#
8
x$
16
#
á . By the Alternating Series Estimation Theorem the kerrork ¹ 8x ¹
œ 1.25 ‚ 10&
c $
3Ð0Þ1Ñ (0.1)$
3!
c $
(0.1)$
3!
23. kR# (x)k œ ¹ e3!x ¹
24. kR# (x)k œ ¹ e3!x ¹
1.87 ‚ 104 , where c is between 0 and x
œ 1.67 ‚ 10% , where c is between 0 and x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
(0.01)#
8
Section 11.9 Convergence of Taylor Series; Error Estimates
c &¸
e e
25. kR% (x)k ¸ cosh
5! x œ ¹ #
c
cc x&
5! ¹
"
1.65 1.65
#
†
(0.5)&
5!
&
œ (1.13) (0.5)
5! ¸ 0.000294
c #
26. If we approximate eh with 1 h and 0 Ÿ h Ÿ 0.01, then kerrork ¹ e #h ¹ Ÿ
e0Þ01 h†h
#
Ÿ Še
0Þ01
(0.0")
‹h
#
œ 0.00505h 0.006h œ (0.6%)h, where c is between 0 and h.
27. kR" k œ ¹ (1"c)#
x#
#! ¹
28. tan" x œ x
x$
3
x#
#
x&
5
œ ¸ #x ¸ kxk .01 kxk œ (1%) kxk Ê ¸ #x ¸ .01 Ê 0 kxk .02
x(
7
á Ê
1
4
œ tan" 1 œ 1
"
3
"
5
x%
5!
x'
7!
"
7
á ; kerrork
"
#n 1
.01
Ê 2n 1 100 Ê n 49
x$
3!
29. (a) sin x œ x
x&
5!
series representing
L s# œ
sin x
x
graph of y œ 1
á Ê
sin x
x
x#
3!
œ1
#
x
6
‹ 0. Therefore 1
#
x
6
sin x
x
x#
6 ; if
sin x
x 1
á , s" œ 1 and s# œ 1
then by the Alternating Series Estimation Theorem, L s" œ
sin x
x ,
Š1
(b) The graph of y œ
x(
7!
L is the sum of the
0 and
1
sin x
x , x Á 0, is bounded below by the
x#
6 and above by the graph of y œ 1 as
derived in part (a).
30. (a) cos x œ 1
x#
#!
x%
4!
x'
6!
á Ê 1 cos x œ
if L is the sum of the series representing
L s" œ
(b)
1 cos x
x#
"
#
0 and
1 cos x
x#
1 cos x
x#
Š "#
x
The graph of y œ 1 xcos
is bounded below by
#
"
x#
the graph of y œ # 24 and above by the graph
y œ "# as indicated in part (a).
x#
#!
x%
4!
x'
6!
x)
8!
á Ê
1 cos x
x#
œ
"
#
1
4
33. tan" x when x œ
; the sum is cos ˆ 14 ‰ œ
1
3
"
È2
x%
6!
, then by the Alternating Series Estimation Theorem
x#
4! ‹
0. Therefore
"
#
x#
#4
1 cos x
x#
"
#
of
31. sin x when x œ 0.1; the sum is sin (0.1) ¸ 0.099833417
32. cos x when x œ
x#
4!
¸ 0.707106781
; the sum is tan" ˆ 13 ‰ ¸ 0.808448
34. ln (1 x) when x œ 1; the sum is ln (1 1) ¸ 1.421080
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
.
x'
8!
á ;
745
746
Chapter 11 Infinite Sequences and Series
35. ex sin x œ 0 x x# x$ ˆ 3!"
"
3
#
$
œxx x
"
30
"
90
&
x
'
"
6
$
%
œ1x x x
2x ‰
37. sin# x œ ˆ 1 cos
œ
#
Ê
d
dx
asin# xb œ
œ 2x
(2x)$
3!
"
#
#
Š 2x
2!
d
dx
(2x)&
5!
"
#
"
30
(2x)(
7!
&
"‰
#!
œ1
2$ x%
4!
2& x'
6!
"‰
3!
x& ˆ 5!"
" "
#! 3!
"‰
4!
x' ˆ 5!"
" "
3! 3!
"‰
5!
á
x$ ˆ #"!
"‰
3!
x% ˆ 4!"
" "
#! 2!
"‰
4!
x& ˆ 4!"
" "
2! 3!
"‰
5!
á
x á
"
#
cos 2x œ
2$ x%
4!
2& x'
6!
"# Š1
(2x)#
2!
á ‹ œ 2x
(2x)%
4!
(2x)$
3!
(2x)'
6!
(2x)&
5!
2x#
#!
á‹ œ
(2x)(
7!
2$ x%
4!
2& x'
6!
á
á Ê 2 sin x cos x
á œ sin 2x, which checks
38. cos# x œ cos 2x sin# x œ Š1
2x#
#!
x% ˆ 3!"
x á
36. ex cos x œ 1 x x# ˆ #"!
"
3
"‰
#!
(2x)#
#!
(2x)%
4!
(2x)'
6!
"
3
á œ 1 x# x%
2
45
(2x))
8!
x'
#
á ‹ Š 2x
#!
"
315
2$ x%
4!
2& x'
6!
2( x)
8!
á‹
x) á
39. A special case of Taylor's Theorem is f(b) œ f(a) f w (c)(b a), where c is between a and b Ê
f(b) f(a) œ f w (c)(b a), the Mean Value Theorem.
40. If f(x) is twice differentiable and at x œ a there is a point of inflection, then f ww (a) œ 0. Therefore,
L(x) œ Q(x) œ f(a) f w (a)(x a).
41. (a) f ww Ÿ 0, f w (a) œ 0 and x œ a interior to the interval I Ê f(x) f(a) œ
Ê f(x) Ÿ f(a) throughout I Ê f has a local maximum at x œ a
(b) similar reasoning gives f(x) f(a) œ
local minimum at x œ a
f ww (c# )
#
(x a)#
f ww (c# )
#
(x a)# Ÿ 0 throughout I
0 throughout I Ê f(x)
f(a) throughout I Ê f has a
42. f(x) œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f Ð3Ñ (x) œ 6(1 x)%
Ê f Ð4Ñ (x) œ 24(1 x)& ; therefore
"
1 x
¸ 1 x x# x$ . kxk 0.1 Ê
&
%
Ð4Ñ
10
11
"
1 x
10
9
‰
Ê ¹ (1"x)& ¹ ˆ 10
9
&
%
‰ Ê the error e$ Ÿ ¹ max f 4! (x) x ¹ (0.1)% ˆ 10
‰ œ 0.00016935 0.00017, since ¹ f
Ê ¹ (1x x)& ¹ x% ˆ 10
9
9
&
Ð4Ñ
(x)
4! ¹
œ ¹ (1"x)& ¹ .
43. (a) f(x) œ (1 x)k Ê f w (x) œ k(1 x)k1 Ê f ww (x) œ k(k 1)(1 x)k2 ; f(0) œ 1, f w (0) œ k, and f ww (0) œ k(k 1)
Ê Q(x) œ 1 kx k(k # ") x#
"
(b) kR# (x)k œ ¸ 3†3!2†" x$ ¸ 100
Ê kx$ k
"
100
Ê 0x
"
100"Î$
or 0 x .21544
44. (a) Let P œ x 1 Ê kxk œ kP 1k .5 ‚ 10n since P approximates 1 accurate to n decimals. Then,
P sin P œ (1 x) sin (1 x) œ (1 x) sin x œ 1 (x sin x) Ê k(P sin P) 1k
œ ksin x xk Ÿ
kx k $
3!
0.125
3!
‚ 103n .5 ‚ 103n Ê P sin P gives an approximation to 1 correct to 3n
decimals.
_
_
nœ0
nœk
45. If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n 1)(n 2)â(n k 1)an xnk and f ÐkÑ (0) œ k! ak
Ê ak œ
f ÐkÑ (0)
k!
for k a nonnegative integer. Therefore, the coefficients of f(x) are identical with the
corresponding coefficients in the Maclaurin series of f(x) and the statement follows.
46. Note: f even Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w odd;
f odd Ê f(x) œ f(x) Ê f w (x) œ f w (x) Ê f w (x) œ f w (x) Ê f w even;
also, f odd Ê f(0) œ f(0) Ê 2f(0) œ 0 Ê f(0) œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.9 Convergence of Taylor Series; Error Estimates
(a) If f(x) is even, then any odd-order derivative is odd and equal to 0 at x œ 0. Therefore,
a" œ a$ œ a& œ á œ 0; that is, the Maclaurin series for f contains only even powers.
(b) If f(x) is odd, then any even-order derivative is odd and equal to 0 at x œ 0. Therefore,
a! œ a# œ a% œ á œ 0; that is, the Maclaurin series for f contains only odd powers.
47. (a) Suppose f(x) is a continuous periodic function with period p. Let x! be an arbitrary real number. Then f
assumes a minimum m" and a maximum m# in the interval [x! ß x! p]; i.e., m" Ÿ f(x) Ÿ m# for all x in
[x! ß x! p]. Since f is periodic it has exactly the same values on all other intervals [x! pß x! 2p],
[x! 2pß x! 3p], á , and [x! pß x! ], [x! 2pß x! p], á , and so forth. That is, for all real numbers
_ x _ we have m" Ÿ f(x) Ÿ m# . Now choose M œ max ekm" k , km# kf . Then
M Ÿ km" k Ÿ m" Ÿ f(x) Ÿ m# Ÿ km# k Ÿ M Ê kf(x)k Ÿ M for all x.
(b) The dominate term in the nth order Taylor polynomial generated by cos x about x œ a is
sin (a)
n!
causing the graph of Pn (x) to move away from cos x.
48. (b) tan" x œ x
"
3
œ
x#
5
x$
3
x&
5
á ; from the Alternating Series
x tan" x
x$
"
3
x tan" x
x$
á Ê
x tan" x
x$
Estimation Theorem,
Ê
x#
5
x#
Š "3
5
"
3
0
‹0 Ê
; therefore, the lim
xÄ0
"
3
x tan" x
x$
x tan" x
x$
œ
"
3
49. (a) ei1 œ cos (1) i sin (1) œ 1 i(0) œ 1
(b) ei1Î4 œ cos ˆ 14 ‰ i sin ˆ 14 ‰ œ
"
È2
i
È2
œ Š È" ‹ (1 i)
2
(c) ei1Î2 œ cos ˆ 1# ‰ i sin ˆ 1# ‰ œ 0 i(1) œ i
50. ei) œ cos ) i sin ) Ê ei) œ ei()) œ cos ()) i sin ()) œ cos ) i sin );
ei) eci)
;
#
ei) eci)
œ #i
ei) ei) œ cos ) i sin ) cos ) i sin ) œ 2 cos ) Ê cos ) œ
ei) ei) œ cos ) i sin ) (cos ) i sin )) œ 2i sin ) Ê sin )
51. ex œ 1 x
x#
#!
ei) œ 1 i)
Ê
œ
ei) eci)
œ
#
#
%
1 )#! )4!
ei) eci)
#i
œ)
œ
)$
3!
(x a)n or
(x a)n . In both cases, as kxk increases the absolute value of these dominate terms tends to _,
cos (a)
n!
x$
3!
(i))#
2!
)&
5!
Š1 i)
)'
6!
Š1 i)
(i))#
#!
(i))$
3!
(i))%
4!
œ 1 i)
(i))#
2!
á œ 1 i)
á‹ Š1 i)
(i))#
#!
(i))#
#!
(i))$
3!
(i))$
3!
(i))$
3!
(i))%
4!
(i))%
4!
á and
(i))%
4!
á
á‹
#
á œ cos );
(i))#
#!
)(
7!
x%
i)
4! á Ê e
(i))$
(i))%
3! 4!
$
%
#
$
%
))
))
))
))
(i3!
(i4!
á‹ Š1 i) (i#)!) (i3!
(i4!
á‹
#i
á œ sin )
52. ei) œ cos ) i sin ) Ê ei) œ eiÐ)Ñ œ cos ()) i sin ()) œ cos ) i sin )
(a) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2 cos ) Ê cos ) œ
ei) eci)
#
œ cosh i)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
747
748
Chapter 11 Infinite Sequences and Series
(b) ei) ei) œ (cos ) i sin )) (cos ) i sin )) œ 2i sin ) Ê i sin ) œ
53. ex sin x œ Š1 x
x#
#!
x$
3!
x%
4!
á ‹ Šx
x$
3!
x&
5!
x(
7!
œ (1)x (1)x# ˆ 6" #" ‰ x$ ˆ 6" 6" ‰ x% ˆ 1#"0
ei) eci)
2
œ sinh i)
á‹
"
1#
" ‰ &
#4 x
x
á œ x x# 3" x$
"
30
x& á ;
ex † eix œ eÐ1iÑx œ ex (cos x i sin x) œ ex cos x i aex sin xb Ê e sin x is the series of the imaginary part
_
of eÐ1iÑx which we calculate next; eÐ1iÑx œ !
nœ0
œ 1 x ix
Ð1iÑx
of e
is x
(xix)n
n!
œ 1 (x ix)
(x ix)#
#!
(x ix)$
3!
(x ix)%
4!
á
"
"
"
"
"
#
$
$
%
&
&
'
#! a2ix b 3! a2ix 2x b 4! a4x b 5! a4x 4ix b 6! a8ix b á Ê the imaginary
2 #
2 $
4 &
8 '
" $
" &
" '
#
#! x 3! x 5! x 6! x á œ x x 3 x 30 x 90 x á in agreement with our
x
part
product calculation. The series for e sin x converges for all values of x.
54.
d
dx
ˆeÐaibÑ ‰ œ
d
dx
ceax (cos bx i sin bx)d œ aeax (cos bx i sin bx) eax (b sin bx bi cos bx)
œ aeax (cos bx i sin bx) bieax (cos bx i sin bx) œ aeÐaibÑx ibeÐaibÑx œ (a ib)eÐaibÑx
55. (a) ei)" ei)# œ (cos )" i sin )" )(cos )# i sin )# ) œ (cos )" cos )# sin )" sin )# ) i(sin )" cos )# sin )# cos )" )
œ cos()" )# ) i sin()" )# ) œ eiÐ)" )# Ñ
"
"
) i sin ) ‰
(b) ei) œ cos()) i sin()) œ cos ) i sin ) œ (cos ) i sin )) ˆ cos
cos ) i sin ) œ cos ) i sin ) œ ei)
56.
a bi ÐabiÑx
C" iC# œ ˆ aa# bib# ‰ eax (cos bx i sin bx) C" iC#
a# b# e
ax
œ a# e b# (a cos bx ia sin bx ib cos bx b sin bx) C" iC#
ax
œ a# e b# [(a cos bx b sin bx) (a sin bx b cos bx)i] C" iC#
ax
ax
œ e (a cosa#bxb#b sin bx) C" ie (a sina#bxb#b cos bx) iC# ;
ÐabiÑx
ax ibx
ax
ax
ax
e
'e
œe e
ÐabiÑx
dx œ
œ e (cos bx i sin bx) œ e cos bx ie sin bx, so that given
a bi
a# b#
eÐabiÑx C" iC# we conclude that ' eax cos bx dx œ
and ' eax sin bx dx œ
e (a sin bx b cos bx)
a# b#
ax
eax (a cos bx b sin bx)
a# b#
C"
C#
57-62. Example CAS commands:
Maple:
f := x -> 1/sqrt(1+x);
x0 := -3/4;
x1 := 3/4;
# Step 1:
plot( f(x), x=x0..x1, title="Step 1: #57 (Section 11.9)" );
# Step 2:
P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x );
P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x );
P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x );
# Step 3:
D2f := D(D(f));
D3f := D(D(D(f)));
D4f := D(D(D(D(f))));
plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 11.9)" );
c1 := x0;
M1 := abs( D2f(c1) );
c2 := x0;
M2 := abs( D3f(c2) );
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.9 Convergence of Taylor Series; Error Estimates
749
c3 := x0;
M3 := abs( D4f(c3) );
# Step 4:
R1 := unapply( abs(M1/2!*(x-0)^2), x );
R2 := unapply( abs(M2/3!*(x-0)^3), x );
R3 := unapply( abs(M3/4!*(x-0)^4), x );
plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #57 (Section 11.9)" );
# Step 5:
E1 := unapply( abs(f(x)-P1(x)), x );
E2 := unapply( abs(f(x)-P2(x)), x );
E3 := unapply( abs(f(x)-P3(x)), x );
plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green],
linestyle=[1,1,1,3,3,3], title="Step 5: #57 (Section 11.9)" );
# Step 6:
TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 );
L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 );
# (a)
R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 );
L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 );
R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 );
L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 );
R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 );
plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2],
color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#57(a) (Section 11.9)" );
abs(`f(x)`-`P`[1](x) ) <= evalf( E1(x0) );
# (b)
abs(`f(x)`-`P`[2](x) ) <= evalf( E2(x0) );
abs(`f(x)`-`P`[3](x) ) <= evalf( E3(x0) );
Mathematica: (assigned function and values for a, b, c, and n may vary)
Clear[x, f, c]
f[x_]= (1 x)3/2
{a, b}= {1/2, 2};
pf=Plot[ f[x], {x, a, b}];
poly1[x_]=Series[f[x], {x,0,1}]//Normal
poly2[x_]=Series[f[x], {x,0,2}]//Normal
poly3[x_]=Series[f[x], {x,0,3}]//Normal
Plot[{f[x], poly1[x], poly2[x], poly3[x]}, {x, a, b},
PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,1,0], RGBColor[0,0,1], RGBColor[0,.5,.5]}];
The above defines the approximations. The following analyzes the derivatives to determine their maximum values.
f''[c]
Plot[f''[x], {x, a, b}];
f'''[c]
Plot[f'''[x], {x, a, b}];
f''''[c]
Plot[f''''[x], {x, a, b}];
Noting the upper bound for each of the above derivatives occurs at x = a, the upper bounds m1, m2, and m3 can be defined
and bounds for remainders viewed as functions of x.
m1=f''[a]
m2=-f'''[a]
m3=f''''[a]
r1[x_]=m1 x2 /2!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
750
Chapter 11 Infinite Sequences and Series
Plot[r1[x], {x, a, b}];
r2[x_]=m2 x3 /3!
Plot[r2[x], {x, a, b}];
r3[x_]=m3 x4 /4!
Plot[r3[x], {x, a, b}];
A three dimensional look at the error functions, allowing both c and x to vary can also be viewed. Recall that c must be a
value between 0 and x, so some points on the surfaces where c is not in that interval are meaningless.
Plot3D[f''[c] x2 /2!, {x, a, b}, {c, a, b}, PlotRange Ä All]
Plot3D[f'''[c] x3 /3!, {x, a, b}, {c, a, b}, PlotRange Ä All]
Plot3D[f''''[c] x4 /4!, {x, a, b}, {c, a, b}, PlotRange Ä All]
11.10 APPLICATIONS OF POWER SERIES
1. (1 x)"Î# œ 1 "# x
ˆ "# ‰ ˆ "# ‰ x#
2. (1 x)"Î$ œ 1 "3 x
ˆ "3 ‰ ˆ 23 ‰ x#
8. a1 x# b
"Î$
œ 1 3" x#
"Î#
9. ˆ1 1x ‰ œ 1 "# ˆ 1x ‰
$
(4)(3)x#
#!
12. a1 x# b œ 1 3x#
ˆ 3" ‰ ˆ 32 ‰ ˆ 53 ‰ x$
á œ 1 3" x 9" x#
5
81
x$ á
ˆ "# ‰ ˆ "# ‰ (2x)#
#!
#
#!
(2)(3) ˆ x# ‰
#
ˆ "3 ‰ ˆ 43 ‰ ax# b#
(3)(2) ax# b
#!
#
(3)(2)(2x)#
#!
%
14. ˆ1 #x ‰ œ 1 4 ˆ x# ‰
(4)(3) ˆ x# ‰
#
#!
3!
3!
$
á œ 1 "3 x# 29 x%
á œ1
"
#x
á œ1
1
8x#
"
16x$
2
3x
4
9x#
x* á
5
16
14
81
x' á
á
40
81x$
dy
dx
á
œ 1 4x 6x# 4x$ x%
œ 1 3x# 3x% x'
œ 1 6x 12x# 8x$
(4)(3)(2) ˆ x# ‰
3!
$
(4)(3)(2)(1) ˆ x# ‰
4!
%
œ 1 2x 32 x# "# x$
15. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
x$ á
á œ 1 x 12 x# 12 x$ á
á œ 1 "# x$ 38 x'
3!
(3)(2)(1)(2x)$
3!
5
16
á œ 1 x 34 x# "# x$ á
ˆ 3" ‰ ˆ 32 ‰ ˆ 35 ‰ ˆ x2 ‰$
(4)(3)(2)x%
4!
(3)(2)(1) ax# b
3!
13. (1 2x)$ œ 1 3(2x)
$
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ ax$ b$
á œ 1 "# x 38 x#
á œ 1 x 34 x# "# x$
3!
3!
#!
$
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 1x ‰$
ˆ "3 ‰ ˆ 23 ‰ ˆ 2x ‰#
(4)(3)(2)x$
3!
3!
ˆ 3" ‰ ˆ 43 ‰ ˆ 73 ‰ ax# b$
#!
Š "# ‹ Š "# ‹ Š 3# ‹ (2x)$
(2)(3)(4) ˆ x# ‰
#!
#!
3!
3!
#!
ˆ "# ‰ ˆ "# ‰ ˆ 1x ‰#
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (x)$
(2)(3)(4) ˆ x# ‰
ˆ "# ‰ ˆ 3# ‰ ax$ b#
"Î$
10. ˆ1 2x ‰ œ 1 "3 ˆ 2x ‰
11. (1 x)% œ 1 4x
x$ á
3!
(2)(3) ˆ x# ‰
œ 1 "# x$
"
16
#!
#
6. ˆ1 x# ‰ œ 1 # ˆ x# ‰
"Î#
á œ 1 "# x 8" x#
ˆ "# ‰ ˆ 3# ‰ (x)#
4. (1 2x)"Î# œ 1 "# (2x)
#
5. ˆ1 x# ‰ œ 1 # ˆ x# ‰
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ x$
3!
#!
3. (1 x)"Î# œ 1 "# (x)
7. a1 x$ b
#!
œ a" 2a# x á nan xn1 á
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
16
x%
Section 11.10 Applications of Power Series
Ê
dy
dx
y œ (a" a! ) (2a# a" )x (3a$ a# )x# á (nan an1 )xn1 á œ !
Ê a" a! œ 0, 2a# a" œ 0, 3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 1 when x œ 0 we have
a! œ 1. Therefore a" œ 1, a# œ
Ê y œ 1 x "# x#
"
3!
a "
2 †1
œ
x$ á
"
#
, a$ œ
(1)n
n!
a #
3
œ 3"†# , á , an œ
_
(")n xn
n!
xn á œ !
nœ0
an1
n
œ
(1)n
n!
œ ex
16. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
Ê
dy
dx
dy
dx
œ a" 2a# x á nan xn1 á
2y œ (a" 2a! ) (2a# 2a" )x (3a$ 2a# )x# á (nan 2an1 )xn1 á œ !
Ê a" 2a! œ 0, 2a# 2a" œ 0, 3a$ 2a# œ 0 and in general nan 2an1 œ 0. Since y œ 1 when x œ 0 we have
a! œ 1. Therefore a" œ 2a! œ 2(1) œ 2, a# œ
nc1
an œ ˆ 2n ‰ an1 œ ˆ 2n ‰ Š n2 1 ‹ an2 œ
œ 1 (2x)
(2x)#
2!
(2x)$
3!
á
2n
n!
(2x)n
n!
2
#
a" œ
2
#
(2) œ
Ê y œ 1 2x
_
(2x)n
n!
á œ !
nœ0
2#
#
2#
#
, a$ œ
x#
2$
3!
2
3
a# œ
2
3
x$ á
#
Š 2# ‹ œ
2n
n!
2$
3 †#
,á ,
xn á
œ e2x
17. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
Ê
dy
dx
dy
dx
œ a" 2a# x á nan xn1 á
y œ (a" a! ) (2a# a" )x (3a$ a# )x# á (nan an1 )xn1 á œ 1
Ê a" a! œ 1, 2a# a" œ 0, 3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 0 when x œ 0 we have
a! œ 0. Therefore a" œ 1, a# œ a#" œ "# , a$ œ a3# œ 3"†# , a% œ a4$ œ 4†3"†# , á , an œ ann1 œ n!"
Ê y œ 0 1x "# x#
œ ˆ1 1x "# x#
"
3 †#
"
3 †#
x$
x$
"
4†3†2
"
4†3†#
x% á
x% á
"
n!
"
n!
xn á
_
xn á ‰ 1 œ !
nœ0
xn
n!
1 œ ex 1
18. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
Ê
dy
dx
dy
dx
œ a" 2a# x á nan xn1 á
y œ (a" a! ) (2a# a" )x (3a$ a# )x# á (nan an1 )xn1 á œ 1
Ê a" a! œ 1, 2a# a" œ 0, 3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 2 when x œ 0 we have
a! œ 2. Therefore a" œ 1 a! œ 1, a# œ
Ê y œ 2 x "# x#
_
œ1!
nœ0
(1)n xn
n!
"
3 †#
x$ á
(")n
n!
a "
# †1
œ
"
#
, a$ œ
a #
3
œ 3"†# , á , an œ
xn á œ 1 Š1 x "# x#
"
3 †#
an1
n
œ
x$ á
(")n
n!
(")n
n!
xn á ‹
œ 1 ex
19. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
Ê
dy
dx
dy
dx
œ a" 2a# x á nan xn1 á
y œ (a" a! ) (2a# a" )x (3a$ a# )x# á (nan an1 )xn1 á œ x
Ê a" a! œ 0, 2a# a" œ 1, 3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 0 when x œ 0 we have
a! œ 0. Therefore a" œ 0, a# œ 1 # a" œ "# , a$ œ a3# œ 3"†# , a% œ a4$ œ 4†3"†# , á , an œ ann1 œ n!"
Ê y œ 0 0x "# x#
œ ˆ1 1x "# x#
"
3 †#
"
3 †#
x$
x$
"
4†3†#
"
4†3†#
x% á
x% á
"
n!
"
n!
xn á
_
xn á ‰ 1 x œ !
nœ0
xn
n!
1 x œ ex x 1
20. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
Ê
dy
dx
dy
dx
œ a" 2a# x á nan xn1 á
y œ (a" a! ) (2a# a" )x (3a$ a# )x# á (nan an1 )xn1 á œ 2x
Ê a" a! œ 0, 2a# a" œ 2, 3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 1 when x œ 0 we have
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
751
752
Chapter 11 Infinite Sequences and Series
a! œ 1. Therefore a" œ 1, a# œ
Ê y œ 1 1x "# x#
œ Š1 1x "# x#
"
3 †#
"
3 †#
2 a "
#
œ
x$ á
x$ á
(")n
n!
a #
"
# , a$ œ 3 œ
n
(")
n
n! x á
an1
n
3"†# , á , an œ
_
xn á ‹ 2 2x œ !
nœ0
(1)n xn
n!
(")n
n!
œ
2 2x œ ex 2x 2
21. yw xy œ a" (2a# a! )x (3a$ a" )x á (nan an2 )xn1 á œ 0 Ê a" œ 0, 2a# a! œ 0, 3a$ a" œ 0,
4a% a# œ 0 and in general nan an2 œ 0. Since y œ 1 when x œ 0, we have a! œ 1. Therefore a# œ a#! œ "# ,
a$ œ
a#
4
œ
Ê y œ 1 "# x#
"
#†4
a"
3
œ 0, a% œ
"
#†4
, a& œ
x%
"
#†4†6
a$
5
œ 0, á , a2n œ
"
#†4†6â2n
"
#†4†6â2n
x' á
and a2n1 œ 0
_
x2n á œ !
nœ0
_
x2n
2n n!
œ!
nœ0
n
#
Š x# ‹
n!
#
œ ex Î2
22. yw x# y œ a" 2a# x (3a$ a! )x# (4a% a" )x$ á (nan an3 )xn1 á œ 0 Ê a" œ 0, a# œ 0,
3a$ a! œ 0, 4a% a" œ 0 and in general nan an3 œ 0. Since y œ 1 when x œ 0, we have a! œ 1. Therefore
a$ œ a3! œ "3 , a% œ a4" œ 0, a& œ a5# œ 0, a' œ a6$ œ 3"†6 , á , a3n œ 3†6†9"â3n , a3n1 œ 0 and a3n2 œ 0
"
3
"
3 †6
$
Ê yœ1 x
'
x
"
3†6†9
*
x á
"
3†6†9â3n
_
x á œ !
3n
nœ0
n
x3n
3n n!
_ Š x$ ‹
3
œ!
nœ0
n!
$
œ ex Î3
23. (1 x)yw y œ (a" a! ) (2a# a" a" )x (3a$ 2a# a# )x# (4a% 3a$ a$ )x$ á
(nan (n 1)an1 an1 )xn1 á œ 0 Ê a" a! œ 0, 2a# 2a" œ 0, 3a$ 3a# œ 0 and in
general (nan nan1 ) œ 0. Since y œ 2 when x œ 0, we have a! œ 2. Therefore
_
a" œ 2, a# œ 2, á , an œ 2 Ê y œ 2 2x 2x# á œ ! 2xn œ
nœ0
2
1x
24. a1 x# b yw 2xy œ a" (2a# 2a! )x (3a$ 2a" a" )x# (4a% 2a# 2a# )x$ á (nan nan2 )xn1 á
œ 0 Ê a" œ 0, 2a# 2a! œ 0, 3a$ 3a" œ 0, 4a% 4a# œ 0 and in general nan nan2 œ 0. Since y œ 3 when
x œ 0, we have a! œ 3. Therefore a# œ 3, a$ œ 0, a% œ 3, á , a2n1 œ 0, a2n œ (1)n 3
_
_
nœ0
nœ0
n
Ê y œ 3 3x# 3x% á œ ! 3(1)n x2n œ ! 3 ax# b œ
3
1 x#
25. y œ a! a" x a# x# á an xn á Ê yww œ 2a# 3 † 2a$ x á n(n 1)an xn2 á Ê yww y
œ (2a# a! ) (3 † 2a$ a" )x (4 † 3a% a# )x# á (n(n 1)an an2 )xn2 á œ 0 Ê 2a# a! œ 0,
3 † 2a$ a" œ 0, 4 † 3a% a# œ 0 and in general n(n 1)an an2 œ 0. Since yw œ 1 and y œ 0 when x œ 0,
"
we have a! œ 0 and a" œ 1. Therefore a# œ 0, a$ œ 3"†# , a% œ 0, a& œ 5†4"†3†# , á , a2n1 œ (#n
1)! and
a2n œ 0 Ê y œ x
"
3!
x$
"
5!
_
x& á œ !
nœ0
x2nb1
(2n 1)!
œ sinh x
26. y œ a! a" x a# x# á an xn á Ê yww œ 2a# 3 † 2a$ x á n(n 1)an xn2 á Ê yww y
œ (2a# a! ) (3 † 2a$ a" )x (4 † 3a% a# )x# á (n(n 1)an an2 )xn2 á œ 0 Ê 2a# a! œ 0,
3 † 2a$ a" œ 0, 4 † 3a% a# œ 0 and in general n(n 1)an an2 œ 0. Since yw œ 0 and y œ 1 when x œ 0,
we have a! œ 1 and a" œ 0. Therefore a# œ "# , a$ œ 0, a% œ
Ê y œ 1 "2 x#
"
4!
_
x% á œ !
nœ0
(1)n x2n
(2n)!
"
4 † 3 †#
, a& œ 0, á , a2n1 œ 0 and a2n œ
(")n
(#n)!
œ cos x
27. y œ a! a" x a# x# á an xn á Ê yww œ 2a# 3 † 2a$ x á n(n 1)an xn2 á Ê yww y
œ (2a# a! ) (3 † 2a$ a" )x (4 † 3a% a# )x# á (n(n 1)an an2 )xn2 á œ x Ê 2a# a! œ 0,
3 † 2a$ a" œ 1, 4 † 3a% a# œ 0 and in general n(n 1)an an2 œ 0. Since yw œ 1 and y œ 2 when x œ 0,
we have a! œ 2 and a" œ 1. Therefore a# œ ", a$ œ 0, a% œ
"
4†3
, a& œ 0, á , a2n œ 2 †
(1)nb1
(#n)!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
and
Section 11.10 Applications of Power Series
a2n1 œ 0 Ê y œ 2 x x# 2 †
x%
4!
_
(1)n1 x2n
(2n)!
á œ2x2!
nœ1
753
œ x cos 2x
28. y œ a! a" x a# x# á an xn á Ê yww œ 2a# 3 † 2a$ x á n(n 1)an xn2 á Ê yww y
œ (2a# a! ) (3 † 2a$ a" )x (4 † 3a% a# )x# á (n(n 1)an an2 )xn2 á œ x Ê 2a# a! œ 0,
3 † 2a$ a" œ 1, 4 † 3a% a# œ 0 and in general n(n 1)an an2 œ 0. Since yw œ 2 and y œ 1 when x œ 0,
"
"
"
1
3
we have a! œ 1 and a" œ 2. Therefore a# œ "
# , a$ œ # , a% œ 2†3†4 , a& œ 5†4†# œ 5! , á , a2n œ (#n)!
and a2n1 œ
Ê y œ 1 2x #" x#
3
(2n1)!
3
3!
_
x$ á œ 1 2x !
x2n
(2n)!
nœ1
_
!
nœ1
3x2nb1
(2n1)!
29. y œ a! a" ax 2b a# ax 2b# á an ax 2bn á
Ê yww œ 2a# 3 † 2a$ ax 2b á n(n 1)an ax 2bn2 á Ê yww y
œ (2a# a! ) (3 † 2a$ a" )(x 2) (4 † 3a% a# )(x 2)# á (n(n 1)an an2 )(x 2)n2 á œ x
œ ax 2b 2 Ê 2a# a! œ 2, 3 † 2a$ a" œ 1, and n(n 1)an an2 œ 0 for n 3. Since y œ 0 when x œ 2,
2
we have a! œ 0, and since yw œ 2 when x œ 2, we have a" œ 2. Therefore a# œ 1, a$ œ "# , a% œ 41†3 a1b œ 4†
3 †2 †1 ,
2
3
a& œ 51†4 ˆ #" ‰ œ 5†4†33†2†1 , . . . , a2n œ a2n
b! , and a2n1 œ (2n1)! . Since a" œ 2, we have a" ax 2b œ a2bax 2b and
a2bax 2b œ a3 1bax 2b œ a3bax 2b a1bax 2b œ x 2 3ax 2b.
Ê y œ x 2 3ax 2b
2b5 . . .
Êyœx2
_
Ê y œ x 2!
nœ0
2
2! ax
2
3
4
2
3
2
3
2! ax 2b 3! ax 2b 4! ax 2b 5! ax
2b2 4!2 ax 2b4 . . . 3ax 2bx 3!3 ax 2b3
_
(x2)2n
(2n)!
3!
nœ0
3
5! ax
2b5 . . .
(x2)2nb1
(2n1)!
30. yww x# y œ 2a# 6a$ x (4 † 3a% a! )x# á (n(n 1)an an4 )xn2 á œ 0 Ê 2a# œ 0, 6a$ œ 0,
4 † 3a% a! œ 0, 5 † 4a& a" œ 0, and in general n(n 1)an an4 œ 0. Since yw œ b and y œ a when x œ 0,
we have a! œ a, a" œ b, a# œ 0, a$ œ 0, a% œ 3a†4 , a& œ 4b†5 , a' œ 0, a( œ 0, a) œ 3†4a†7†8 , a* œ 4†5b†8†9
Ê y œ a bx
a
3†4
x%
b
4†5
x&
a
3†4†7†8
x)
b
4†5†8†9
x* á
31. yww x# y œ 2a# 6a$ x (4 † 3a% a! )x# á (n(n 1)an an4 )xn2 á œ x Ê 2a# œ 0, 6a$ œ 1,
4 † 3a% a! œ 0, 5 † 4a& a" œ 0, and in general n(n 1)an an4 œ 0. Since yw œ b and y œ a when x œ 0,
we have a! œ a and a" œ b. Therefore a# œ 0, a$ œ #"†3 , a% œ 3a†4 , a& œ 4b†5 , a' œ 0, a( œ #†3"
†6†7
Ê y œ a bx
1
2†3
x$
a
3†4
x%
b
4†5
x&
1
2†3†6†7
ax)
3†4†7†8
x(
bx*
4†5†8†9
á
32. yww 2yw y œ (2a# 2a" a! Ñ (2 † 3a$ 4a# a" )x (3 † 4a% 2 † 3a$ a# )x# á
((n 1)nan 2(n 1)an" an2 )xn2 á œ 0 Ê 2a# 2a" a! œ 0, 2 † 3a$ 4a# a" œ 0,
3 † 4a% 2 † 3a$ a# œ 0 and in general (n 1)nan 2(n 1)an" an2 œ 0. Since yw œ 1 and y œ 0 when
1
when x œ 0, we have a! œ 0 and a" œ 1. Therefore a# œ 1, a$ œ "# , a% œ 16 , a& œ 24
and an œ (n"1)!
Ê y œ x x# "# x$ 6" x%
'00 2 sin x# dx œ '00 2 Šx# x3! x5!
Þ
33.
kE k Ÿ
Þ
(
(.2)
7†3!
Þ
x
œ ’x
'
"!
_
x& á œ !
nœ1
xn
(n1)!
$
á ‹ dx œ ’ x3
_
œ!
x(
7†3!
nœ0
_
xnb1
n!
á“
œx!
nœ0
!Þ#
!
$
xn
n!
¸ ’ x3 “
œ xex
!Þ#
!
¸ 0.00267 with error
¸ 0.0000003
'00 2 ec x " dx œ '00 2
Þ
34.
"
#4
x#
4
x$
18
"
x
Š1 x
á“
!Þ#
!
x#
#!
x$
3!
x%
4!
á 1‹ dx œ '0 Š1
¸ 0.19044 with error kEk Ÿ
0Þ2
(0.2)%
96
x
#
x#
6
x$
24
á ‹ dx
¸ 0.00002
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
754
Chapter 11 Infinite Sequences and Series
'00 1 È "
1 x%
dx œ '0 Š1
kE k Ÿ
(0.1)&
10
œ 0.000001
'00 25
$È
0Þ1
Þ
35.
Þ
36.
kE k Ÿ
1 x# dx œ '0
0Þ25
&
(0.25)
45
Þ
#
x#
3
x%
9
á“
x$
9
á ‹ dx œ ’x
!Þ"
¸ [x]!Þ"
! ¸ 0.1 with error
!
x&
45
!Þ#&
á“
¸ ’x
!
!Þ#&
x$
9 “!
¸ 0.25174 with error
%
'
$
&
(
$
&
!
(0.1)7
7†7!
¸ 2.8 ‚ 10
Þ
%
(0.1)9
216
¸ 0.0996676643, kEk Ÿ
39. a1 x% b
Ê
"Î#
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰
4!
'0
0Þ1
Š "# ‹
œ (1)"Î#
Š1
%
x
#
)
x
8
x
16
x&
10
x(
42
á“
!Þ"
!
¸ ’x
x$
3
x&
10
¸ 4.6 ‚ 10
ˆ "# ‰ ˆ "# ‰
#!
%
"#
$
12
"'
5x
128
x%
#
#
(1)$Î# ax% b
x)
8
á ‹ dx ¸ ’x
&
(1)(Î# ax% b á œ 1
)
'
(1)"Î# ax% b
1
!
12
'00 1 exp ax# b dx œ '00 1 Š1 x# x2! x3! x4! á ‹ dx œ ’x x3
Þ
40.
Š1
x&
10
á ‹ dx œ ’x
'00 1 sinx x dx œ '00 1 Š1 x3! x5! x7! á ‹ dx œ ’x 3x†3! 5x†5! 7x†7! á “ !Þ" ¸ ’x 3x†3! 5x†5! “ !Þ"
¸ 0.0999444611, kEk Ÿ
38.
3x)
8
¸ 0.0000217
Þ
37.
x%
2
x"#
16
!Þ"
ˆ "# ‰ ˆ "# ‰ ˆ 3# ‰
3!
5x"'
128
$
á
(0.1)9
72
¸ 0.100001, kEk Ÿ
x
10 “ !
(1)&Î# ax% b
¸ 1.39 ‚ 1011
"
x
'01 ˆ 1 xcos x ‰ dx œ '01 Š "# x4! x6! x8! 10!
á ‹ dx ¸ ’ x# 3x†4! 5x†6! 7x†8! 9†x10! “
#
%
'
)
$
&
(
*
#
¸ 0.4863853764, kEk Ÿ
!
¸ 1.9 ‚ 1010
1
11†12!
41.
'01 cos t# dt œ '01 Š1 t# 4!t t6! á ‹ dt œ ’t 10t 9t†4! 13t †6! á “ "
42.
'01 cos Èt dt œ '01 Š1 #t 4!t 6!t 8!t á ‹ dt œ ’t t4 3t†4! 4t†6! 5t†8! á “ "
%
)
"#
&
*
"$
Ê kerrork
!
#
Ê kerrork
x
Ê kerrork
t'
3!
"
15†7!
¸
x(
7†2!
x
"
33†34
Ê kerrork
(b) kerrork
t"%
7!
$
á ‹ dt œ ’ t3
t'
2!
x*
9†3!
t$
3
t)
3!
x""
11†4!
t&
5
t(
7
t"!
4!
t"#
5!
t(
7†3!
t""
11†5!
t
2
t#
3
t$
4
$
Ê kerrork
"
13†5!
%
&
t"&
15†7!
á“
x
¸
!
x$
3
x(
7†3!
x""
11†5!
#
#
x
#
%
x
3†4
á ‹ dt œ ’t
F(x) ¸ x
x#
##
t(
7†2!
t*
9†3!
t""
11†4!
t"$
13†5!
á“
x
!
¸ 0.00064
á ‹ dt œ ’ t2
(0.5)'
6# ¸ .00043
"
32# ¸ .00097 when
t&
5
á ‹ dt œ ’ t3
¸ .00089 when F(x) ¸
46. (a) F(x) œ '0 Š1
x
$
¸ 0.000013
45. (a) F(x) œ '0 Št
(b) kerrork
t"!
5!
x
x&
5
#
¸ 0.000004960
44. F(x) œ '0 Št# t%
x$
3
%
¸ .00011
!
"
5†8!
43. F(x) œ '0 Št#
$
"
13†6!
t%
1#
'
x
t'
30
x
5†6
á“ ¸
t#
2†2
t$
3†3
x$
3#
x%
4#
)
!
x
7†8
x#
#
x%
1#
Ê kerrork
á (1)"&
t%
4†4
t&
5†5
x
á (1)$"
¸ .00052
$#
x
31†32
á“ ¸ x
!
(0.5)'
30
x#
##
x$
3#
x$"
31#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x%
4#
x&
5#
!Þ"
x(
42 “ !
Section 11.10 Applications of Power Series
47.
"
x#
aex (1 x)b œ
"
x
"
x
’Š1 x
aex ex b œ
2x#
"
t%
3!
2x%
"
)&
Š )
)$
6
)Ä0
)
7!
)%
x$
3!
á ‹ 1 x‹ œ
"
#
x
3!
x#
4!
á Ê lim
x%
4!
á ‹ Š1 x
x#
#!
x$
3!
x%
4!
á ‹“ œ
e e x
x
œ x lim
Š2
Ä_
x
Š1
ex (1 x)
x#
xÄ0
t#
#
t%
4!
2x#
3!
2x%
5!
t'
6!
á ‹“ œ 4!"
)&
5!
á‹ œ
2x'
t#
6!
Š2x
2x$
3!
2x&
5!
2x(
7!
á‹
y%
7
á‹
x&
5
á‹
á‹ œ 2
7!
"
x
#
t%
8!
á Ê lim
" cos t Š t# ‹
t%
tÄ0
"
á ‹ œ 24
"
)&
)$
6
Š)
á‹ œ
9!
"
y$
x$
3!
"
#
t#
#
’1
t%
8!
ay tan" yb œ
"
5!
)#
7!
$
)%
9!
á Ê lim
sin ) ) Š )6 ‹
)&
)Ä0
"
1 #0
y$
3
’y Š y
)$
3!
)
y&
5
"
3
á ‹“ œ
y#
5
y%
7
y tan" y
y$
á Ê lim
yÄ0
œ lim Š 3"
yÄ0
y#
5
"
3
œ
52.
"
t%
sin )‹ œ
#
xÄ0
#
t#
6!
x#
#!
x#
#
á Ê lim
7!
Š1 cos t t# ‹ œ
œ lim Š 5!"
"
y$
2x'
5!
tÄ0
51.
á‹ œ
œ lim Š 4!"
50.
x#
4!
x
3!
œ2
49.
ŠŠ 1 x
œ lim Š "#
xÄ0
48.
"
x#
755
tanc" y sin y
y$ cos y
Ê lim
yÄ0
œ
Œy
y$
3
y&
5
á Œy
y$
tanc" y sin y
y$ cos y
Œ
œ lim
"
6
"
6x%
y&
5!
á
23y#
5!
"
x#
á
"
#x %
55.
56.
x# 4
ln (x 1)
œ
œ
x%
#
1 Š1
x#
#!
x'
3
á
x%
4!
á‹
œ
Œ1
Š #"!
(x 2)(x 2)
’(x 2)
(x c 2)#
#
x 2
œ lim
(x c 2)$
3
#
x Ä 2 ’1 x c# 2 (x 32) á“
"
3!(x 1)$
á“
x#
#
x#
4!
œ
x%
3
"
n10n
"
10)
x#
#
x$
3
"
6x'
23y&
5!
y$
cos y
á
á ‰ œ 1
x%
4
"
5!(x 1)&
"
3!(x 1)#
á
œ
"
6
Œ
23y#
5!
á
cos y
"
#x #
’1
á‹ œ 1
"
5!(x 1)%
#
"
6x%
x# Še1Îx 1‹
á Ê x lim
Ä_
x2
(x c 2)#
3
x2
#
x#
#
á“
"
3!(x 1)#
"
5!(x 1)%
á
á‹ œ 1
a1 x # b
lim ln1
cos x
xÄ0
Ê
á‹
Œ1
œ lim
Ê lim
Š #"!
xÄ0
x#
#
x#
4!
x%
3
á
á‹
œ 2! œ 2
x# 4
x Ä 2 ln (x 1)
œ4
x‰
57. ln ˆ 11
x œ ln (1 x) ln (1 x) œ Šx
58. ln (1 x) œ x
y$
6
œ "6
Ê x lim
(x 1) sin ˆ x " 1 ‰ œ x lim
Š1
Ä_
Ä_
#
Œx
Œ
á ‰ œ 1
54. (x 1) sin ˆ x " 1 ‰ œ (x 1) Š x " 1
ln a1 x# b
1 cos x
œ
cos y
yÄ0
#
"
#x#
cos y
53. x# Š1 e1Îx ‹ œ x# ˆ1 1
ˆ1
œ x lim
Ä_
y$
3!
á
Ê n10n 10) when n
(1)n1 xn
n
x$
3
x%
4
á ‹ Šx
x#
#
n1 n
á Ê kerrork œ ¹ (")n
x
¹œ
x$
3
"
n10n
x%
4
á ‹ œ 2 Šx
when x œ 0.1;
8 Ê 7 terms
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x$
3
756
Chapter 11 Infinite Sequences and Series
59. tan" x œ x
"
#n 1
"
10$
x$
3
x&
5
Ê n
60. tan" x œ x
x$
3
x*
9
á
(")n1 x2n1
2n1
n1 2n1
á Ê kerrork œ ¹ (1)2nx1
x(
7
x*
9
á
(1)n1 x2n1
2n1
2n1
(1)n
2n1
nœ1
(1)nc1
2n1
nœ1
2n 1
x2n1 ¹
á and n lim
†
¹x
Ä _ 2n 1
_
_
"
#n 1
when x œ 1;
st
Ê tan" x converges for kxk 1; when x œ 1 we have !
we have !
¹œ
œ 500.5 Ê the first term not used is the 501 Ê we must use 500 terms
1001
#
x&
5
x(
7
¸ 2n 1 ¸ œ x#
œ x# n lim
Ä _ #n 1
which is a convergent series; when x œ 1
which is a convergent series Ê the series representing tan" x diverges for kxk 1
(1)n1 x2n1
x$
x&
x(
x*
á and when the series representing 48
3 5 7 9 á
2n 1
"
'
error less than 3 † 10 , then the series representing the sum
" ‰
" ‰
" ‰
48 tan" ˆ 18
32 tan" ˆ 57
20 tan" ˆ #39
also has an error of magnitude less than 10' ;
" ‰
tan" ˆ 18
has an
61. tan" x œ x
thus
2nc1
kerrork œ 48
"
Š 18
‹
#n 1
"
3†10'
Ê n
4 using a calculator Ê 4 terms
62. ln (sec x) œ '0 tan t dt œ '0 Št
x
"Î#
x
t$
3
x#
3x%
# 8
2nb3
lim ¹ 1†3†5â(2n 1)(2n 1)x
†
n Ä _ 2†4†6â(2n)(2n 2)(2n 3)
63. (a) a1 x# b
¸1
2t&
15
á ‹ dt ¸
x#
#
x%
12
$
5x'
"
x ¸ x x6
16 Ê sin
2†4†6â(2n)(2n ")
1†3†5â(2n 1)x2nb1 ¹ 1 Ê
x'
45
3x&
40
á
5x(
112
; Using the Ratio Test:
(2n 1)(2n1)
x# n lim
¹
¹1
Ä _ (2n 2)(2n 3)
Ê kxk 1 Ê the radius of convergence is 1. See Exercise 69.
d
dx
(b)
acos" xb œ a1 x# b
64. (a) a1 t# b
œ1
"Î#
t#
#
65.
"
1x
5x(
112
Ê cos" x œ
¸ (1)"Î# ˆ "# ‰ (1)$Î# at# b
3t%
2# †2!
(b) sinh" ˆ 4" ‰ ¸
term,
"Î#
"
4
3†5t'
2$ †3!
"
384
1
#
sin" x ¸
x
, evaluated at t œ
"
4
Šx
ˆ "# ‰ ˆ #3 ‰ (1)&Î# at# b#
#!
Ê sinh" x ¸ '0 Š1
3
40,960
1
#
t#
#
3t%
8
5t'
16 ‹
x$
6
3x&
40
5x(
112 ‹
¸
1
#
x
x$
6
3x&
40
ˆ #" ‰ ˆ #3 ‰ ˆ #5 ‰ (1)(Î# at# b$
3!
dt œ x
x$
6
3x&
40
5x(
112
œ 0.24746908; the error is less than the absolute value of the first unused
since the series is alternating Ê kerrork
œ 1 "(x) œ 1 x x# x$ á Ê
d
dx
ˆ 11x ‰ œ
"
1 x#
œ
d
dx
5 ˆ "4 ‰
112
(
¸ 2.725 ‚ 10'
a1 x x# x$ á b
œ 1 2x 3x# 4x$ á
66.
"
1 x#
œ 1 x# x% x' á Ê
d
dx
ˆ 1 " x# ‰ œ
2x
a1 x # b #
œ
d
dx
a1 x# x% x' á b œ 2x 4x$ 6x& á
8â(2n 2)†(2n)
67. Wallis' formula gives the approximation 1 ¸ 4 ’ 3†23††45††45††67††67†â
(2n 1)†(2n 1) “ to produce the table
n
10
20
30
80
90
93
94
95
100
µ1
3.221088998
3.181104886
3.167880758
3.151425420
3.150331383
3.150049112
3.149959030
3.149870848
3.149456425
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
5x(
112
Section 11.11 Fourier Series
At n œ 1929 we obtain the first approximation accurate to 3 decimals: 3.141999845. At n œ 30,000 we still do
not obtain accuracy to 4 decimals: 3.141617732, so the convergence to 1 is very slow. Here is a Maple CAS
procedure to produce these approximations:
pie :=
proc(n)
local i,j;
a(2) := evalf(8/9);
for i from 3 to n do a(i) := evalf(2*(2*i2)*i/(2*i1)^2*a(i1)) od;
[[j,4*a(j)] $ (j = n5 .. n)]
end
68. ln 1 œ 0; ln 2 œ ln
¸ ln 2
&
Š "5 ‹
3
Š 3" ‹
¸ 2 "3
1 Š "3 ‹
$
2 "5
$
1 Š "3 ‹
3
&
Š 3" ‹
5
(
Š 3" ‹
7
¸ 0.69314; ln 3 œ ln 2 ln ˆ # ‰ œ ln 2 ln
3
1 Š 5" ‹
1 Š 5" ‹
(
Š "5 ‹
5
Š "5 ‹
7
¸ 1.09861; ln 4 œ 2 ln 2 ¸ 1.38628; ln 5 œ ln 4 ln ˆ 4 ‰ œ ln 4 ln
5
¸ 1.60943; ln 6 œ ln 2 ln 3 ¸ 1.79175; ln 7 œ ln 6 ln ˆ 76 ‰ œ ln 6 ln
"
1 Š 13
‹
1 Š 9" ‹
1 Š 9" ‹
¸ 1.94591; ln 8 œ 3 ln 2
"
1 Š 13
‹
¸ 2.07944; ln 9 œ 2 ln 3 ¸ 2.19722; ln 10 œ ln 2 ln 5 ¸ 2.30258
69. a1 x# b
"Î#
œ a1 ax# bb
"Î#
ˆ "# ‰ ˆ 3# ‰ ˆ 5# ‰ (1)(Î# ax# b$
3!
á œ1
Ê sin" x œ '0 a1 t b
x
œ (1)"Î# ˆ "# ‰ (1)$Î# ax# b
#
x
#
dt œ '0 Œ1 !
_
x
# "Î#
nœ1
%
1†3x
2# †#!
1†3†5x
2$ †3!
'
ˆ "# ‰ ˆ 3# ‰ (1)&Î# ax# b#
#!
_
1†3†5â(2n1)x2n
#n †n!
á œ1!
nœ1
1†3†5â(2n 1)x2n
#n †n!
_
1†3†5â(2n 1)x2nb1
#†4â(2n)(2n 1)
dt œ x !
nœ1
,
where kxk 1
_
70. ctan" td x œ
_
œ 'x ˆ t"#
tan" x œ 'x
"
t%
Ê tan" x œ
œ
_
1
#
1
#
"
t'
"
x
"t
lim
b Ä _
"
t)
"
3t$
_
t#
bÄ_
_
Š 1# ‹
œ 'x – t " — dt œ 'x
1Š ‹
á ‰ dt œ lim
"
3x$
dt
1 t#
"t
"
3t$
"
5t&
"
t#
"
7t(
ˆ1
"
t#
"
t%
"
t'
"
x
"
3x$
b
á ‘x œ
"
" x
td c_ œ tan" x 1#
5x& á , x 1; ctan
"
"
"
"
"
"
‘x
5t& 7t( á b œ x 3x$ 5x& 7x( á
œ
á ‰ dt
"
5x&
'_ 1 dt t
"
7x(
á
x
#
Ê tan" x œ 1#
"
x
"
3x$
x 1
71. (a) tan atan" (n 1) tan" (n 1)b œ
N
tan atanc" (n 1)b tan atan" (n 1)b
1 tan atan" (n 1)b tan atan" (n 1)b
œ
(n 1) (n 1)
1 (n 1)(n 1)
œ
2
n#
N
(b) ! tan" ˆ n2# ‰ œ ! ctan" (n 1) tan" (n 1)d œ atan" 2 tan" 0b atan" 3 tan" 1b
nœ1
nœ1
"
atan
_
4 tan
(c) ! tan" ˆ n2# ‰ œ
nœ1
"
2b á atan" (N 1) tan" (N 1)b œ tan" (N 1) tan" N
lim tan" (N 1) tan" N 14 ‘ œ
nÄ_
1
#
1
#
1
4
œ
1
4
31
4
11.11 FOURIER SERIES
1. a0 œ
1
21
'021 1 dx œ 1, ak œ 11 '021 cos kx dx œ 11 sinkkx ‘201 œ 0, bk œ 11 '021 sin kx dx œ 11 coskkx ‘201 œ 0.
Thus, the Fourier series for faxb is 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
5x&
á ,
757
758
Chapter 11 Infinite Sequences and Series
2. a0 œ
' 1 1 dx '121 1 dx • œ 0, ak œ 11 ”'01 cos kx dx '121 cos kx dx • œ 11 ” sinkkx ¹1 sinkkx ¹21 • œ 0,
1
21 ” 0
bk œ 11 ”'0 sin kx dx '1 sin kx dx • œ 11 ” cosk kx ¹
21
1
0
4
, k odd
2 cos k1b œ œ k1
.
0, k even
Thus, the Fourier series for faxb is 14 sin x
œ
1
0
1
21
cos kx
k ¹1
•œ
1
acos
k1 c
k1 1b acos 21k cos 1kb d
1
k 1 a2
3. a0 œ
sin 3x
3
sin 5x
5
. . . ‘.
' 1 x dx '121 ax 21b dx • œ 211 "# 12 "# a412 12 b 212 ‘ œ 0.
1
21 ” 0
Note,
'121 ax 21bcos kx dx œ '01 u cos ku du (Let u œ 21 x). So ak œ 11 ”'01 x cos kx dx '121 ax 21b cos kx dx • œ 0.
Note, '1 ax 21bsin kx dx œ '0 u sin ku du (Let u œ 21 x). So bk œ 11 ”'0 x sin kx dx '1 ax 21b sin kx dx •
21
œ
2
1
1
1
21
'01 x sin kx dx œ 12 xk cos kx k1 sin kx ‘01 œ 2k cos k1 œ 2k a1bk1 .
2
_
Thus, the Fourier series for faxb is ! a1bk1 2 sink kx .
k œ1
4. a0 œ
1
21
'021 faxb dx œ 211 '01 x2 dx œ 16 12 ,
2
œ 11 ’ Š xk
'021 faxb cos kx dx œ 11 '01 x2 cos kx dx
1
1
21
k2 ‹sin kx k# x cos kx “ œ k# cos k1 œ a1bk ˆ k# ‰, bk œ 11 '0 faxb sin kx dx œ 11 '0 x2 sin kx dx œ
œ 11 ’ Š k23
3
2
x
k
2
‹cos kx
#
k2 x
0
1
ak œ
1
1
2
sin kx “ œ 11 ’ Š k23
0
2
1
k
2
k
‹a1b
#
k3
k
k
“ œ 11 ’ Ša1b 1‹ k#3 “ 1k a1b
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 11.11 Fourier Series
œœ
14k3 1k , k odd
.
1k , k even
Thus, the Fourier series for faxb is 16 12 2 cos x Š 1
5. a0 œ
bk œ
1
21
1
1
'021 ex dx œ 211 ae21 1b,
'0
21
ex sin kx dx
Thus, the Fourier series
6. a0 œ
1
21
ak œ
1
1
4
1 ‹sin
x "# cos 2x
1
2
sin 2x 29 cos 3x Š 91271 4 ‹sin 3x . . .
2
'021 ex cos kx dx œ 11 1 e k acos kx k sin kxb ‘201 œ 1ea11k1 b ,
x
2
2
2
21
kˆ1 e21 ‰
œ 11 1e k2 asin kx k cos kxb ‘0 œ 1a1 k2 b .
_
21
kx
k sin kx ‰
for faxb is 211 ae21 1b e 1 1 ! ˆ cos
1 k2 1 k2 .
k œ1
x
x
2
1
k
1
1a1k2 b e a1b
œ
1 ex
asin
1 1 k2
1‘ œ
a1 e 1 b
1 a1 k 2 b ,
e1 1
1 a1 k 2 b ,
1
kxb ‘0
kx k cos
œ
k odd
k even
. bk œ
k 1
k
1a1 k2 b e a1b
Thus, the Fourier series for faxb is
7. a0 œ
2
'021 faxb dx œ 211 '01 ex dx œ e121 1 , ak œ 11 '021 faxb cos kx dx œ 11 '01 ex cos kx dx œ 11 1 e k acos kx k sin kxb ‘01
œ
e1 1
21
759
a1 e 1 b
21 cos
1
21
'0
21
x
faxb dx œ
a1 e 1 b
21 sin
1
21
'0
21
x
e1 1
51 cos
2x
cos x dx œ 0, ak œ
1
1
1
1
'021 faxb sin kx dx œ 11 '01 ex sin kx dx
1‘ œ
2 a1 e 1 b
sin
51
'0
21
k a1 e 1 b
1 a1 k 2 b ,
1 e1
1 a1 k 2 b ,
2x
k odd
k even
a1 e 1 b
101 cos
Ú
cos x cos kx dx œ Û
Ü
3x
.
3 a1 e1 b
101 sin
1 sinak 1bx
1 ’ 2 ak 1 b
3x . . .
sinak 1bx
2 ak 1 b
1 "
1 #x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
“ ,
0
1
"
sin
2x ‘0 ,
4
kÁ1
kœ1
760
Chapter 11 Infinite Sequences and Series
!,
œœ"
#,
bk œ
1
1
kÁ1
.
kœ1
'0
21
Ú
Ý 1 ’ cosak 1bx
2 ak 1 b
1
cos x sin kx dx œ Û
Ý
Ü
Thus, the Fourier series for faxb is
8. a0 œ
1
21
"
# cos
x!
k even
1
cosak 1bx
2 ak 1 b
0
411 cos
2k
1ak2 1b sin
œ 11 ”'0 2 cos kx dx '1 x cos kx dx • œ 11 cosk2kx
bk œ
œ
1
1
21
1
2x¹ , k œ 1
!,
œ
k odd
.
k even
2k
1ak2 1b ,
0
kx.
'021 faxb dx œ 211 ”'01 2 dx '121 x dx • œ 1 43 1,
1
kÁ1
“ ,
ak œ
x sin kx
k
1
1
'021 faxb cos kx dx
‘21 œ
1
1 a1bk
1 k2
œœ
12k2 , k odd
.
0, k even
'021 faxb sin kx dx œ 11 ”'01 2 sin kx dx '121 x sin kx dx • œ 11 ” 2k cos kx¹1 ˆ x cosk kx sink kx ‰¹21 •
1ˆ4
k 1
2
0
3‰,
1
k,
1
k odd
.
k even
Thus, the Fourier series for faxb is 1 34 1 12 cos x ˆ 14 3‰sin x "# sin 2x
2
91 cos
3x 13 ˆ 14 3‰sin 3x . . . .
9.
'021 cos px dx œ 1p sin px¹21 œ 0 if p Á 0.
10.
'021 sin px dx œ 1p cos px¹21 œ 1p c 1 1 d œ 0 if p Á 0.
11.
'021 cos px cos qx dx œ '021 "# c cos ap qbx cos ap qbx ddx œ "# p 1 q sin ap qbx p 1 q sin ap qbx ‘201 œ 0 if p Á q.
0
0
If p œ q then '0 cos px cos qx dx œ '0 cos2 px dx œ '0
21
12.
21
21
"
# a1
cos 2pxb dx œ "# Šx
1
2p sin
2px‹¹
21
0
œ 1.
'021 sin px sin qx dx œ '021 "# c cos ap qbx cos ap qbx ddx œ "# p 1 q sin ap qbx p 1 q sin ap qbx ‘201 œ 0 if p Á q.
If p œ q then '0 sin px sin qx dx œ '0 sin2 px dx œ '0
21
21
21
"
# a1
cos 2pxb dx œ "# Šx
1
2p sin
2px‹¹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
21
0
œ 1.
Chapter 11 Practice Exercises
13.
761
'021 sin px cos qx dx œ '021 "# c sin ap qbx sin ap qbx ddx œ "# p 1 q cos ap qbx p 1 q cos ap qbx ‘201
21
21
21
œ "# a1 1b p 1 q a1 1b p 1 q ‘ œ 0. If p œ q then '0 sin px cos qx dx œ '0 sin px cos px dx œ '0 "# sin 2px dx
œ 411 cos 2px¹
21
0
œ 411 a1 1b œ 0.
14. Yes. Note that if f is continuous at c, then the expression
facb b facc b
2
œ facb since fac b œ limb faxb œ facb and
xÄc
fac b œ limc faxb œ facb. Now since the sum of two piecewise continuous functions on Ò0, 21Ó is also continuous on Ò0, 21Ó,
x Äc
the function f g satisfies the hypothesis of Theorem 24, and so its Fourier series converges to
af gbacb b af gbacc b
2
for 0 c 21. Let sf axb denote the Fourier series for faxb. Then for any c in the interval a0, 21b
b
c
sfg acb œ af gbac b af gbac b œ " ’ lim+ af gbaxb limc af gbaxb “ œ " ’ limb faxb limb gaxb limc faxb limc gaxb “
#
2
"
# c afac b
œ
x Äc
x Äc
#
x Äc
x Äc
x Äc
x Äc
gac bb afac b gac bb d œ sf acb sg acb, since f and g satisfy the hypothesis of Theorem 24.
15. (a) faxb is piecewise continuous on Ò0, 21Ó and f w axb œ 1 for all x Á 1 Ê f w axb is piecewise continuous on Ò0, 21Ó. Then
by Theorem 24, the Fourier series for faxb converges to faxb for all x Á 1 and converges to "# afa1 b fa1 bb
œ "# a1 1b œ 0 at x œ 1.
_
(b) The Fourier series for faxb is ! a1bk1 2 sink kx . If we differentiate this series term by term we get the series
k œ1
_
! a1b
k 1
k œ1
2 cos kx, which diverges by the nth term test for divergence for any x since lim a1bk1 2 cos kx Á 0.
kÄ_
16. Since the Fourier series in discontinuous at x œ 1, by Theorem 24, the Fourier series will converge to
at x œ 1 we have
fa1b b fa1c b
2
œ
1 2
61
Ê
0 12
2
œ 16 12 2 cos 1 Š 1
Ê
0 12
2
œ 16 12 2
12
2
12
6
_
œ 2!
n œ1
1
n2
Ê
12
3
"
#
2
9
2
2
Š 1 1 4 ‹sin
2 cos x
4
1 ‹sin
1 "# cos 21
. . . œ 16 12 2ˆ1
_
œ 2!
n œ1
1
n2
Ê
12
6
_
œ!
n œ1
1
4
1
2
x
"
# cos
2x
1
2
sin 2x
2
9 cos
3x
facb b facc b
.
2
2
Š 91271 4 ‹sin
sin 21 29 cos 31 Š 91271 4 ‹sin 31 . . .
2
1
9
_
. . . ‰ œ 16 12 2! n12 Ê
n œ1
12
2
œ
12
6
_
2! n12
n œ1
1
n2 .
CHAPTER 11 PRACTICE EXERCISES
1. converges to 1, since n lim
a œ n lim
Š1
Ä_ n
Ä_
2. converges to 0, since 0 Ÿ an Ÿ
2
Èn
(1)n
n ‹
œ1
, n lim
0 œ 0, n lim
Ä_
Ä_
2
Èn
œ 0 using the Sandwich Theorem for Sequences
ˆ 1 2n2 ‰ œ lim ˆ #"n 1‰ œ 1
3. converges to 1, since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
n
4. converges to 1, since n lim
a œ n lim
c1 (0.9)n d œ 1 0 œ 1
Ä_ n
Ä_
5. diverges, since ˜sin
n1 ™
#
œ e0ß 1ß 0ß 1ß 0ß 1ß á f
6. converges to 0, since {sin n1} œ {0ß 0ß 0ß á }
7. converges to 0, since n lim
a œ n lim
Ä_ n
Ä_
ln n#
n
œ 2 n lim
Ä_
Š "n ‹
1
œ0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Thus,
3x . . . .
762
Chapter 11 Infinite Sequences and Series
ln (2n")
n
8. converges to 0, since n lim
a œ n lim
Ä_ n
Ä_
Š 2n 2b 1 ‹
œ n lim
Ä_
ˆ n nln n ‰ œ lim
9. converges to 1, since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
10. converges to 0, since n lim
a œ
Ä_ n
$
lim ln a2nn 1b
nÄ_
œ0
1
1Š "n ‹
œ1
1
Š
œ n lim
Ä_
6n#
‹
2n$ 1
1
ˆ n n 5 ‰n œ lim Š1
11. converges to ec5 , since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
ˆ1 n" ‰cn œ lim
, since n lim
a œ n lim
Ä_ n
Ä_
nÄ_
"
e
12. converges to
œ n lim
Ä_
(5)
n ‹
"
ˆ1 "n ‰n
n
œ
12n
6n#
œ n lim
Ä_
œ ec5 by Theorem 5
"
e
by Theorem 5
ˆ 3 ‰1În œ lim
13. converges to 3, since n lim
a œ n lim
Ä_ n
Ä_ n
nÄ_
3
n1În
œ
3
1
œ 3 by Theorem 5
ˆ 3 ‰1În œ lim
14. converges to 1, since n lim
a œ n lim
Ä_ n
Ä_ n
nÄ_
31În
n1În
œ
1
1
œ 1 by Theorem 5
n
15. converges to ln 2, since n lim
a œ n lim
n a21În 1b œ n lim
Ä_ n
Ä_
Ä_
21În 1
Š "n ‹
œ0
2
n
œ n lim
Ä_
–
Š21În ln 2‹
n#
—
œ n lim
21În ln 2
Ä_
Š "
‹
n#
œ 2! † ln 2 œ ln 2
2
n
È
16. converges to 1, since n lim
a œ n lim
2n 1 œ n lim
exp Š ln (2nn 1) ‹ œ n lim
exp Œ 2n1b 1 œ e! œ 1
Ä_ n
Ä_
Ä_
Ä_
17. diverges, since n lim
a œ n lim
Ä_ n
Ä_
(n 1)!
n!
œ n lim
(n 1) œ _
Ä_
18. converges to 0, since n lim
a œ n lim
Ä_ n
Ä_
(4)n
n!
Š "# ‹
Š "# ‹
19.
"
(2n 3)(2n 1)
œ
#n 3
Š "# ‹
Ê sn œ –
2n 1
Š "# ‹
"
Ê n lim
s œ n lim
Ä_ n
Ä _ –6
20.
2
n(n 1)
œ
2
n
2
n1
ˆ1
œ n lim
Ä_
21.
22.
9
(3n 1)(3n 2)
œ 3# 3n3#
œ
_
_
nœ0
nœ0
5
—–
Š "# ‹
5
Š "# ‹
7
Š "‹
#
— á – #n 3
Š "# ‹
2n 1 —
œ
Š "# ‹
3
3n 2
3
"
6
Ê sn œ ˆ #3 35 ‰ ˆ 35 38 ‰ ˆ 38
ˆ3
Ê n lim
s œ n lim
Ä_ n
Ä_ #
"
en
Š "# ‹
2 ‰
n1
œ #2
2
n1
Ê n lim
s
Ä_ n
œ 1
8
2
2
(4n 3)(4n 1) œ 4n 3 4n 1
œ 29 4n21 Ê n lim
s
Ä_ n
23. ! en œ !
œ
Ê sn œ ˆ #2 23 ‰ ˆ 32 42 ‰ á ˆ n2
2 ‰
n1
3
3n 1
2n 1 —
3
œ 0 by Theorem 5
3 ‰
3n 2
œ
á ˆ 3n 3 1
3 ‰
3n 2
3
#
Ê sn œ ˆ 92
ˆ
œ n lim
Ä_
3 ‰
11
2 ‰
2 ‰
ˆ 2
13 13 17
2
2 ‰
2
9 4n1 œ 9
, a convergent geometric series with r œ
"
e
ˆ 172
2 ‰
21
á ˆ 4n2 3
and a œ 1 Ê the sum is
"
1 Š "e ‹
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2 ‰
4n 1
e
e1
Š "# ‹
2n 1
Chapter 11 Practice Exercises
_
24. ! (1)n
3
4n
nœ1
ˆ 34 ‰
1ˆ c4" ‰
_
‰n a convergent geometric series with r œ "4 and a œ
œ ! ˆ 43 ‰ ˆ "
4
_
nœ1
Ê the sum is
œ 35
25. diverges, a p-series with p œ
26. !
3
4
nœ0
5
n
_
œ 5 !
nœ1
27. Since f(x) œ
_
"
x"Î#
nœ1
diverges since it is a nonzero multiple of the divergent harmonic series
Ê f w (x) œ #x"$Î# 0 Ê f(x) is decreasing Ê an1 an , and
_
(")n
Èn
series !
"
n,
"
#
converges by the Alternating Series Test. Since !
nœ1
"
Èn
lim a œ lim
nÄ_ n nÄ_
1
Èn
œ 0, the
diverges, the given series converges
conditionally.
28. converges absolutely by the Direct Comparison Test since
"
#n$
"
n$
for n
1, which is the nth term of a
convergent p-series
29. The given series does not converge absolutely by the Direct Comparison Test since
the nth term of
decreasing Ê
"
ln (n 1)
"
n1
, which is
"
a divergent series. Since f(x) œ ln (x" 1) Ê f w (x) œ (ln (x 1))
# (x 1) 0 Ê f(x) is
"
an1 an , and n lim
a œ n lim
œ 0, the given series converges conditionally
Ä_ n
Ä _ ln (n 1)
by the
Alternating Series Test.
30.
'2_ x(ln" x)
#
dx œ lim
bÄ_
'2b
"
x(ln x)#
b
dx œ lim c(ln x)" d 2 œ lim ˆ ln"b
bÄ_
bÄ_
" ‰
ln 2
œ
"
ln #
Ê the series
converges absolutely by the Integral Test
31. converges absolutely by the Direct Comparison Test since
ln n
n$
n
n$
œ
"
n#
, the nth term of a convergent p-series
n
n
32. diverges by the Direct Comparison Test for en n Ê ln ˆen ‰ ln n Ê nn ln n Ê ln nn ln (ln n)
Ê n ln n ln (ln n) Ê
33. n lim
Ä_
Š
ln n
ln (ln n)
"
n
È n# 1 ‹
Š n"# ‹
34. Since f(x) œ
œ Én lim
Ä_
3x#
x$ 1
n#
n# 1
Ê f w (x) œ
"
n
, the nth term of the divergent harmonic series
œ È1 œ 1 Ê converges absolutely by the Limit Comparison Test
3x a2 x$ b
ax $ 1 b#
0 when x
2 Ê an1 an for n
2 and n lim
Ä_
3n#
n$ 1
œ 0, the
series converges by the Alternating Series Test. The series does not converge absolutely: By the Limit
#
Comparison Test, n lim
Ä_
Š n$3n 1 ‹
ˆ n" ‰
œ n lim
Ä_
3n$
n$ 1
œ 3. Therefore the convergence is conditional.
35. converges absolutely by the Ratio Test since n lim
’ n 2 †
Ä _ (n 1)!
36. diverges since n lim
a œ n lim
Ä_ n
Ä_
(")n an# 1b
2n# n 1
n!
n1“
œ n lim
Ä_
n2
(n 1)#
œ01
does not exist
nb1
37. converges absolutely by the Ratio Test since n lim
†
’ 3
Ä _ (n 1)!
n!
3n “
œ n lim
Ä_
3
n1
œ01
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
763
764
Chapter 11 Infinite Sequences and Series
n 2 3
n
È
É
38. converges absolutely by the Root Test since n lim
an œ n lim
nn œ n lim
Ä_
Ä_
Ä_
n n
39. converges absolutely by the Limit Comparison Test since n lim
Ä_
40. converges absolutely by the Limit Comparison Test since n lim
Ä_
nb1
41. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 4) †
Ä_
Ä _ (n 1)3nb1
n3n
(x 4)n ¹
1 Ê
"
Š $Î#
‹
n
Š Èn(n "1)(n 2) ‹
6
n
œ01
Š n"# ‹
Š
n # an # 1 b
n%
" ‹ œ Én lim
Ä_
nÈ n# 1
kx 4 k
lim
3
nÄ_
n(n 1)(n 2)
n$
œ Én lim
Ä_
kx 4 k
3
ˆ n n 1 ‰ 1 Ê
_
(1)n 3n
n3n
Ê kx 4k 3 Ê 3 x 4 3 Ê 7 x 1; at x œ 7 we have !
nœ1
_
alternating harmonic series, which converges conditionally; at x œ 1 we have !
nœ1
3n
n3n
œ1
œ1
1
_
œ ! ("n ) , the
n
nœ1
_
œ!
nœ1
"
n
, the divergent
harmonic series
(a) the radius is 3; the interval of convergence is 7 Ÿ x 1
(b) the interval of absolute convergence is 7 x 1
(c) the series converges conditionally at x œ 7
42. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x1) † (2n1)! ¹ 1 Ê (x 1)# n lim
Ä_
Ä _ (2n1)! (x1)2nc2
Ä_
all x
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
2n
nb1
43. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (3x(n1)1)# †
Ä_
Ä_
n#
(3x 1)n ¹
1 Ê k3x 1k n lim
Ä_
Ê 1 3x 1 1 Ê 0 3x 2 Ê 0 x
_
"
n#
œ !
nœ1
_
have !
nœ1
"
(#n)(2n1)
2
3
n#
(n 1)#
_
œ 0 1, which holds for
1 Ê k3x 1k 1
nc1
_
2nc1
; at x œ 0 we have ! (1) n# (1) œ ! ("n)#
n
nœ1
nœ1
, a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x œ
_
(1)n1 (1)n
n#
(a) the radius is
n1
œ ! ("n)#
2
3
we
, which converges absolutely
nœ1
"
3
; the interval of convergence is 0 Ÿ x Ÿ
(b) the interval of absolute convergence is 0 Ÿ x Ÿ
2
3
2
3
(c) there are no values for which the series converges conditionally
44. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ n2 †
Ä_
Ä _ 2n 3
Ê
_
!
nœ1
k2x 1k
#
n1
2n 1
†
†
2n 1
n1
†
2n
(2x 1)n ¹
1 Ê
k2x 1k
lim
2
nÄ_
n2
¸ 2n
3 †
2n " ¸
n1
(1) 1 Ê k2x 1k 2 Ê 2 2x 1 2 Ê 3 2x 1 Ê 3# x
(2)n
#n
_
œ!
n1 ‰
lim ˆ 2n
1 œ
nÄ_
(2x 1)nb1
2nb1
nœ1
"
#
(")n (n1)
2n 1
which diverges by the nth-Term Test for Divergence since
Á 0; at x œ
"
#
_
n1
we have ! 2n
1 †
nœ1
2n
#n
_
n"
œ ! 2n
1 , which diverges by the nthnœ1
Term Test
(a) the radius is 1; the interval of convergence is 3# x
(b) the interval of absolute convergence is 3# x
"
#
"
#
(c) there are no values for which the series converges conditionally
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
#
1
; at x œ 3# we have
Chapter 11 Practice Exercises
nb1
45. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ (n 1)nb1
Ê
kx k
e
¸ˆ n ‰n ˆ n " 1 ‰¸ 1 Ê
1 Ê kxk n lim
Ä _ n1
nn
xn ¹
kx k
e n lim
Ä_
ˆ n " 1 ‰ 1
† 0 1, which holds for all x
(a) the radius is _; the series converges for all x
(b) the series converges absolutely for all x
(c) there are no values for which the series converges conditionally
nb1
46. n lim
†
¹ uunbn 1 ¹ 1 Ê n lim
¹ x
Ä_
Ä _ Èn 1
_
Èn
xn ¹
n
1 Ê kxk n lim
1 Ê kxk 1; when x œ 1 we have
Ä _ Én1
_
! (È1) , which converges by the Alternating Series Test; when x œ 1 we have !
n
n
nœ1
nœ1
"
Èn
, a divergent
p-series
(a) the radius is 1; the interval of convergence is 1 Ÿ x 1
(b) the interval of absolute convergence is 1 x 1
(c) the series converges conditionally at x œ 1
2nb1
47. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (n 32)x
nb1
Ä_
Ä_
_
_
nœ1
nœ1
the series ! nÈ31 and !
n1
È3
†
3n
(n 1)x2n1 ¹
1 Ê
x#
3 n lim
Ä_
2‰
È
È3;
ˆ nn
1 1 Ê 3x
, obtained with x œ „ È3, both diverge
(a) the radius is È3; the interval of convergence is È3 x È3
(b) the interval of absolute convergence is È3 x È3
(c) there are no values for which the series converges conditionally
2nb3
48. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ (x 2n1)x
3
Ä_
Ä_
†
2n 1
(x 1)2nb1 ¹
ˆ 2n 1 ‰ 1 Ê (x 1)# (1) 1
1 Ê (x 1)# n lim
Ä _ 2n 3
_
Ê (x 1)# 1 Ê kx 1k 1 Ê 1 x 1 1 Ê 0 x 2; at x œ 0 we have ! (1)#n(1)1
n
2nb1
nœ1
_
œ!
nœ1
_
(1)3nb1
2n 1
that !
nœ1
"
2n 1
_
œ!
nœ1
(1)nc1
2n 1
which converges conditionally by the Alternating Series Test and the fact
_
diverges; at x œ 2 we have !
nœ1
(1)n (1)2nb1
2n 1
_
œ!
nœ1
(1)n
2n 1
, which also converges
conditionally
(a) the radius is 1; the interval of convergence is 0 Ÿ x Ÿ 2
(b) the interval of absolute convergence is 0 x 2
(c) the series converges conditionally at x œ 0 and x œ 2
nb1
(n 1)x
49. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹ cschcsch
(n)xn
Ä_
Ä_
c"
2n1
Ê kxk n lim
¹ e1 ee 2n2 ¹ 1 Ê
Ä_
kx k
e
¹ 1 Ê kxk n lim
Ä_ »
Š enb1 c2ecnc1 ‹
ˆ en c2ecn ‰
»1
_
1 Ê e x e; the series ! a „ ebn csch n, obtained with x œ „ e,
nœ1
both diverge since n lim
a „ e)n csch n Á 0
Ä_
(a) the radius is e; the interval of convergence is e x e
(b) the interval of absolute convergence is e x e
(c) there are no values for which the series converges conditionally
50. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹x
Ä_
Ä_
nb1
coth (n 1)
xn coth (n) ¹
c2nc2
1 Ê kxk n lim
†
¹1e
Ä _ 1 ec2nc2
1 ec2n
1 ec2n ¹
1 Ê kxk 1
_
Ê 1 x 1; the series ! a „ 1bn coth n, obtained with x œ „ 1, both diverge since n lim
a „ 1bn coth n Á 0
Ä_
nœ1
(a) the radius is 1; the interval of convergence is 1 x 1
(b) the interval of absolute convergence is 1 x 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
765
766
Chapter 11 Infinite Sequences and Series
(c) there are no values for which the series converges conditionally
51. The given series has the form 1 x x# x$ á (x)n á œ
52. The given series has the form x
ln ˆ 53 ‰ ¸ 0.510825624
x#
#
x$
3
á (1)n1
53. The given series has the form x
x$
3!
x&
5!
á (1)n
x2n1
(2n 1)!
x#
2!
x%
4!
á (1)n
x2n
(2n)!
xn
n
"
1x
, where x œ
"
4
; the sum is
á œ ln (1 x), where x œ
2
3
"
1 ˆ "4 ‰
œ
4
5
œ
"
#
; the sum is
á œ sin x, where x œ 1; the sum is
sin 1 œ 0
54. The given series has the form 1
55. The given series has the form 1 x
56. The given series has the form x
tan" Š È"3 ‹ œ
57. Consider
"
1 2x
x$
3
x#
2!
x&
5
x#
3!
x2n1
(2n 1)
á (1)n
á œ tan" x, where x œ
as the sum of a convergent geometric series with a œ 1 and r œ 2x Ê
_
nœ0
nœ0
œ 1 (2x) (2x)# (2x)$ á œ ! (2x)n œ ! 2n xn where k2xk 1 Ê kxk
58. Consider
"
1 x$
"
È3
; the sum is
1
6
_
"
1 2x
"
#
as the sum of a convergent geometric series with a œ 1 and r œ x$ Ê
"
1 x$
œ
"
1 ax$ b
_
#
$
œ 1 ax$ b ax$ b ax$ b á œ ! (1)n x3n where kx$ k 1 Ê kx$ k 1 Ê kxk 1
nœ0
_
59. sin x œ !
nœ0
_
60. sin x œ !
nœ0
_
61. cos x œ !
nœ0
_
62. cos x œ !
nœ0
_
63. ex œ !
nœ0
_
64. ex œ !
nœ0
_
(1)n x2nb1
(2n 1)!
Ê sin 1x œ !
nœ0
(1)n x2nb1
(2n 1)!
_
nœ0
Ê sin
_ (1)n Š 2x ‹
3
œ!
2x
3
_
_
nœ0
(1)n ˆx&Î# ‰
(2n)!
2n
_
(1)n x2n
(2n)!
Ê cos ˆx&Î# ‰ œ !
(1)n x2n
(2n)!
Ê cos È5x œ cos ˆ(5x)"Î# ‰ œ !
nœ0
_ ˆ 1 x ‰n
xn
n!
#
Ê ex œ !
nœ0
_
3
8
œ
9
32
#
n!
a x # b
n!
nœ0
Ê f www (x) œ 3x$ a3 x# b
f (1) œ
nœ0
nœ0
Ê eÐ1xÎ2Ñ œ !
3
32
œ!
_
xn
n!
"Î#
&Î#
_
œ!
n œ0
n
_
œ!
nœ0
(1)n 22nb1 x2nb1
32nb1 (#n 1)!
œ!
(2n 1)!
nœ0
(1)n 12nb1 x2nb1
(#n 1)!
œ!
2nb1
65. f(x) œ È3 x# œ a3 x# b
www
(1)n (1x)2nb1
(2n 1)!
(1)n x5n
(#n)!
(1)n ˆ(5x)"Î# ‰
(2n)!
2n
_
œ!
nœ0
(1)n 5n xn
(#n)!
1 n xn
#n n!
(1)n x2n
n!
Ê f w (x) œ x a3 x# b
"Î#
Ê f ww (x) œ x# a3 x# b
$Î#
$Î#
a3 x# b
3x a3 x# b
; f(1) œ 2, f w (1) œ "# , f ww (1) œ "8
#
$
Ê È3 x# œ 2 (x 1) 3(x$ 1) 9(x& 1) á
2†1!
1
3
á œ ex , where x œ ln 2; the sum is eln Ð2Ñ œ 2
xn
n!
á
á œ cos x, where x œ 13 ; the sum is cos
2 †2!
"
#
2 †3!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
"Î#
3
8
,
Chapter 11 Practice Exercises
66. f(x) œ
"
1x
œ (1 x)" Ê f w (x) œ (1 x)# Ê f ww (x) œ 2(1 x)$ Ê f www (x) œ 6(1 x)% ; f(2) œ 1, f w (2) œ 1,
"
1 x
f ww (2) œ 2, f www (2) œ 6 Ê
"
x1 œ
œ 4"# ,
67. f(x) œ
w
f (3)
68. f(x) œ
"
x
f www (a) œ
œ 1 (x 2) (x 2)# (x 2)$ á
(x 1)" Ê f w (x) œ (x 1)# Ê f ww (x) œ 2(x 1)$ Ê f www (x) œ 6(x 1)% ; f(3) œ
ww
f (3) œ
www
, f (2) œ
2
4$
6
4%
Ê
"
x 1
œ
"
4
"
4#
(x 3)
"
4$
#
(x 3)
œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f www (x) œ 6x% ; f(a) œ
6
a%
Ê
"
x
"
a
œ
"
a#
(x a)
"
a$
(x a)#
"
a%
"
4%
"
a
"
4
,
2
a$
,
$
(x 3) á
, f w (a) œ a"# , f ww (a) œ
(x a)$ á
69. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
œ a" 2a# x á nan xn1 á Ê
dy
dx
y
œ aa" a! b a2a# a" bx (3a$ a# )x# á (nan an1 )xn1 á œ 0 Ê a" a! œ 0, 2a# a" œ 0,
3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 1 when x œ 0 we have a! œ 1. Therefore a" œ 1,
a# œ
a "
2 †1
a #
3
œ "# , a$ œ
a $
4
œ
"
3†2
"
3 †#
x$ á
Ê y œ 1 x "# x#
, a% œ
œ 4†3"†# , á , an œ
(1)n1
n!
an1
n
_
xn á œ !
nœ0
œ
(1)n xn
n!
1 (1)n
n (n1)!
œ
(1)n1
n!
œ ex
70. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
œ a" 2a# x á nan xn1 á Ê
dy
dx
y
#
œ aa" a! b a2a# a" bx (3a$ a# )x á (nan an1 )xn1 á œ 0 Ê a" a! œ 0, 2a# a" œ 0,
3a$ a# œ 0 and in general nan an1 œ 0. Since y œ 3 when x œ 0 we have a! œ 3. Therefore a" œ 3,
a# œ a2" œ #3 , a$ œ a3# œ 3†32 , an œ ann1 œ n!3 Ê y œ 3 3x 23†1 x# 33†# x$ á n!3 xn á
œ 3 Š1 x
x#
#!
x$
3!
á
xn
n!
_
á ‹ œ 3 !
nœ0
xn
n!
œ 3ex
71. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
œ a" 2a# x á nan xn1 á Ê
dy
dx
2y
#
œ aa" 2a! b a2a# 2a" bx (3a$ 2a# )x á (nan 2an1 )xn1 á œ 0. Since y œ 3 when x œ 0 we
#
have a! œ 3. Therefore a" œ 2a! œ 2(3) œ 3(2), a# œ #2 a" œ #2 (2 † 3) œ 3 Š 2# ‹ , a$ œ 23 a#
n1 n1
$
2
2
œ 23 ’3 Š 2# ‹“ œ 3 Š 32†# ‹ , á , an œ ˆ n2 ‰ an1 œ ˆ n2 ‰ Š3 Š ((n1) 1)!
‹‹ œ 3 Š (1)
n! ‹
#
#
$
n n
(1) 2
$
Ê y œ 3 3(2x) 3 (2)# x# 3 (2)
xn á
3 †# x á 3
n!
œ 3 ’1 (2x)
(2x)#
2!
(2x)$
3!
á
n n
(1)n (2x)n
n!
_
á “ œ 3!
nœ0
(1)n (2x)n
n!
œ 3e2x
72. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
œ a" 2a# x á nan xn1 á Ê
dy
dx
y
œ aa" a! b a2a# a" bx (3a$ a# )x# á (nan an1 )xn1 á œ 1 Ê a" a! œ 1, 2a# a" œ 0,
3a$ a# œ 0 and in general nan an1 œ 0 for n 1. Since y œ 0 when x œ 0 we have a! œ 0. Therefore
a" œ 1 a! œ 1, a# œ 2†a1" œ "# , a$ œ 3a# œ 3"†2 , a% œ 4a$ œ 4†3"†# , á , an
œ
an1
n
œ ˆ n1 ‰ (n(1)1)! œ
n
œ 1 ’1 x "# x#
"
3 †#
(1)n1
n!
Ê y œ 0 x "# x#
x$ á
(1)n
n!
"
3†#
x$ á
_
xn á “ 1 œ !
nœ0
(1)n xn
n!
(1)n1
n!
xn á
1 œ 1 ex
73. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
767
œ a" 2a# x á nan xn1 á Ê
dy
dx
y
#
œ aa" a! b a2a# a" bx (3a$ a# )x á (nan an1 )xn1 á œ 3x Ê a" a! œ 0, 2a# a" œ 3,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
768
Chapter 11 Infinite Sequences and Series
3a$ a# œ 0 and in general nan an1 œ 0 for n 2. Since y œ 1 when x œ 0 we have a! œ 1. Therefore
a" œ 1, a# œ 3 2 a" œ #2 , a$ œ a3# œ 32†2 , a% œ a4$ œ 4†32†# , á , an œ ann1 œ n!2
Ê y œ 1 x ˆ 2# ‰ x# 33†# x$ 4†23†# x% á n!2 xn á
œ 2 ˆ1 x "# x#
"
3 †#
x$
"
4†3†#
_
"
n!
x% á
xn á ‰ 3 3x œ 2 !
nœ0
xn
n!
3 3x œ 2ex 3x 3
74. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
œ a" 2a# x á nan xn1 á Ê
dy
dx
y
dy
dx
#
œ aa" a! b a2a# a" bx (3a$ a# )x á (nan an1 )xn1 á œ x Ê a" a! œ 0, 2a# a" œ 1,
3a$ a# œ 0 and in general nan an1 œ 0 for n 2. Since y œ 0 when x œ 0 we have a! œ 0. Therefore
a" œ 0, a# œ
" a"
2
"
#
œ
Ê y œ 0 0x "# x#
_
œ!
nœ0
(1)n xn
n!
a #
3
, a$ œ
"
3 †#
œ 3"†2 , á , an œ
x$ á
(1)n
n!
an1
n
œ
(1)n
n!
xn á œ Š1 x #" x#
"
3 †#
x$ á
(1)n
n!
xn á ‹ 1 x
1 x œ ex x 1
75. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
œ a" 2a# x á nan xn1 á Ê
y
dy
dx
#
œ aa" a! b a2a# a" bx (3a$ a# )x á (nan an1 )xn1 á œ x Ê a" a! œ 0, 2a# a" œ 1,
3a$ a# œ 0 and in general nan an1 œ 0 for n 2. Since y œ 1 when x œ 0 we have a! œ 1. Therefore
a" œ 1, a# œ 1 2 a" œ #2 , a$ œ a3# œ 32†2 , a% œ a4$ œ 4†32†# , á , an œ ann1 œ n!2
Ê y œ 1 x ˆ #2 ‰ x# 32†# x$ 4†22†# x% á n!2 xn á
œ 2 ˆ1 x "# x#
"
3 †#
x$
"
4†3†#
x% á
_
"
n!
xn
n!
xn á ‰ 1 x œ 2 !
nœ0
1 x œ 2ex x 1
76. Assume the solution has the form y œ a! a" x a# x# á an1 xn1 an xn á
Ê
dy
dx
œ a" 2a# x á nan xn1 á Ê
y
dy
dx
#
œ aa" a! b a2a# a" bx (3a$ a# )x á (nan an1 )xn1 á œ x Ê a" a! œ 0, 2a# a" œ 1,
3a$ a# œ 0 and in general nan an1 œ 0 for n 2. Since y œ 2 when x œ 0 we have a! œ 2. Therefore
a" œ 2, a# œ 1 2 a" œ 1# , a$ œ a3# œ 31†2 , a% œ a4$ œ 4†31†# , á , an œ ann1 œ n!1
Ê y œ 2 2x 1# x#
œ ˆ1 x "# x#
77.
"
4†3†#
x% á
1
4 †3 † #
x% á
'
"
#
"
#% † 4
"
n!
"
#( †7†2!
"
2"! †10†3!
"
2"$ †13†4!
*
&
'11Î2
¸
"
#
x""
11†3!
tanc" x
x
"
9†2$
x"(
17†5!
x#$
23†7!
dx œ '1 Š1
1Î2
"
5# †#&
¸ 0.4872223583
"
7# †#(
x#
3
"
"&
x#*
29†9!
x%
5
9# †2*
_
xn
n!
nœ0
"#
á ‹ dx œ ’x
"
2"' †16†5!
x#"
7!
xn á
x#(
9!
x%
4
1 x œ ex x 1
x(
7†2!
x"!
10†3!
x"!
3!
x"'
5!
x##
7!
x&
25
x(
49
x*
81
21# †##"
x"$
13†4!
á“
"Î#
!
¸ 0.484917143
á ‹ dx œ '0 Šx%
1
x#)
9!
á ‹ dx
"
á “ ¸ 0.185330149
!
1
n!
xn á ‰ 1 x œ !
*
'01 x sin ax$ b dx œ '01 x Šx$ x3! x5!
œ ’ x5
79.
x$
x$
'01Î2 exp ax$ b dx œ '01Î2 Š1 x$ x2! x3! x4!
¸
78.
"
3 †#
1
3 †#
x'
7
"
11# †2""
x)
9
x"!
11
"
13# †2"$
á ‹ dx œ ’x
"
15# †2"&
"
17# †#"(
x$
9
"
19# †#"*
x""
121
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
á“
"Î#
!
Chapter 11 Practice Exercises
80.
'01Î64
dx œ '0
1Î64
tanc" x
Èx
œ 23 x$Î#
2
#1
x(Î#
"
Èx
2
55
x$
3
Šx
x""Î#
2
105
$
7 sin x
2x
x Ä 0 e 1
81. lim
82.
x Ä 0 Š2x
œ lim
5!
#
á ‹ dx œ '0
1Î64
"Î'%
&
2$ x $
3!
œ ˆ 3†28$
#
á‹
x Ä 0 Š2
2# x
#!
á‹
83.
œ lim
tÄ0
"‰
t#
#
"
2 Š 4!
t6! á‹
#
Š1 2t4! á‹
Š sinh h ‹ cos h
84. lim
œ
#
h#
Π#!
$
œ lim
h%
5!
#
lim t # 2 2 cos t
t Ä 0 2t (1 cos t)
" cos# z
z Ä 0 ln (1 z) sin z
85. lim
2
105†8"&
á ‰ ¸ 0.0013020379
œ
7
#
$
&
$
&
2 Š )3! )5! á‹
œ lim
$
&
Š )3! )5! á‹
)Ä0
h%
4!
%
t# 2 2 Œ1 t# t4! á
œ lim
t Ä 0 2t# Š1 1
t#
#
t%
4!
á‹
%
t'
6!
2 Πt4!
œ lim
tÄ0
Št%
2t'
4!
á
á‹
"
1#
œ
Œ1
h#
3!
%
#
%
h5! á Œ1 h#! h4! á
h#
h'
6!
h'
7!
á
h#
hÄ0
2
55†8""
œ2
hÄ0
h#
3!
á‹
) Š) )3! )5! á‹
œ lim
h#
hÄ0
2
21†8(
%
2$ x #
3!
#
lim ˆ "
t Ä 0 # 2 cos t
ˆx"Î# 3" x&Î# 5" x*Î# 7" x"$Î# á ‰ dx
7 Š1 x3! x5! á‹
œ lim
$
)Ä0
Š 3!" )5! á‹
)Ä0
2# x #
2!
x(
7
Š1 ) )#! )3! á‹ Š1 ) )#! )3! á‹ 2)
œ lim
)#
x"&Î# á ‘ !
#
)
c)
lim e )e sin)2)
)Ä0
x&
5
7 Šx x3! x5! á‹
œ lim
2 Š 3!"
œ lim
hÄ0
1 Š1 z#
œ lim
#
z%
3
Š #"!
"
3!
h#
5!
h#
4!
h%
6!
h%
7!
á‹ œ
"
3
%
á‹
$
$
&
z Ä 0 Šz z# z3 á‹ Šz z3! z5! á‹
Šz# z3 á‹
œ lim
#
$
%
z Ä 0 Š z# 2z3 z4 á‹
#
Š1 z3 á‹
œ lim
#
z
z Ä 0 Š "# 2z
3 4 á‹
y#
cos
y
cosh y
yÄ0
86. lim
y#
œ lim
y Ä 0 Œ1
"
œ lim
œ 2
y Ä 0 Œ1 2y% á
6!
y#
#
y%
4!
y'
6!
á Œ1
xÄ0
Ê
r
x#
3
x#
88. (a) csc x ¸
r
x#
s‰ œ lim –
xÄ0
œ 0 and s
"
x
x
6
y%
4!
y'
6!
á
y#
œ lim
y Ä 0 Œ
2y#
#
'
2y6! á
œ 1
$
87. lim ˆ sinx$3x
y#
#!
9
#
&
(3x)
Š3x (3x)
6 120 á‹
x$
œ 0 Ê r œ 3 and s œ
Ê csc x ¸
(b) The approximation sin x ¸
6 x#
6x
6x
6 x#
Ê sin x ¸
r
x#
s— œ lim Š x3#
xÄ0
9
#
81x#
40
á
9
#
6x
6 x#
is better than
sin x ¸ x.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
r
x#
s‹ œ 0
769
770
Chapter 11 Infinite Sequences and Series
_
"
2n
89. (a) ! ˆsin
nœ1
" ‰
#n 1
sin
_
á œ ! (1)n sin
nœ2
"
n
lim sin
nÄ_
_
"
2n
nœ1
Ê tan
"
n
sin 3" ‰ ˆsin
; f(x) œ sin
_
œ 0 Ê ! (1)n sin
nœ2
" ¸
42
(b) kerrork ¸sin
90. (a) ! ˆtan
"
#
œ ˆsin
" ‰
#n 1
tan
Test
(b) kerrork ¸tan
Ê f w (x) œ
sin 5" ‰ ˆsin
cos Š "x ‹
x#
"
6
sin 7" ‰ á ˆsin
0 if x
2 Ê sin
"
n 1
" ¸
42
"
n
_
"
n
œ ! (1)n tan
nœ2
, and n lim
tan
Ä_
(see Exercise 89); f(x) œ tan
_
"
n
œ 0 Ê ! (1)n tan
nœ2
"
n
"
x
Ê f w (x) œ
sec# ˆ x" ‰
x#
0
2
3
nb1
kœ2
, and
¸ 0.02382 and the sum is an underestimate because the remainder is positive
(2n1)(2n3)(x1)
¸ 2n 3 ¸ 1 Ê kxk
92. n lim
† 34††59††714ââ(2n(5n1)x1)n ¹ 1 Ê kxk n lim
¹ 3†5†47†â
9†14â(5n1)(5n4)
Ä_
Ä _ 5n 4
Ê the radius of convergence is 52
"‰
k#
" ‰
#n 1
converges by the Alternating Series
nb1
n
"
n
sin
converges by the Alternating Series Test
(3n 1)(3n 2)x
â(2n)
¸ 3n 2 ¸ 1 Ê kxk
91. n lim
† #†52†8†4â†6(3n
¹ 2†5†8#â
†4†6â(2n)(2n 2)
1)xn ¹ 1 Ê kxk n lim
Ä_
Ä _ 2n 2
Ê the radius of convergence is 23
93. ! ln ˆ1
sin
"
#n
¸ 0.02381 and the sum is an underestimate because the remainder is positive
tan
"
n1
"
n
"
x
"
4
n
n
kœ2
kœ2
5
2
œ ! ln ˆ1 k" ‰ ln ˆ1 k" ‰‘ œ ! cln (k 1) ln k ln (k 1) ln kd
œ cln 3 ln 2 ln 1 ln 2d cln 4 ln 3 ln 2 ln 3d cln 5 ln 4 ln 3 ln 4d cln 6 ln 5 ln 4 ln 5d
á cln (n 1) ln n ln (n 1) ln nd œ cln 1 ln 2d cln (n 1) ln nd
after cancellation
n
"‰
k#
Ê ! ln ˆ1
kœ2
n
94. !
kœ2
"
k# 1 -
ˆ n " 1
_
Ê !
kœ2
"
#
œ
n
!ˆ
kœ2
" ‰‘
n1
"
k # 1
_
1 ‰
œ ln ˆ n2n
Ê ! ln ˆ1
"‰
k#
kœ2
"
k1
œ
"
#
" ‰
k1
ˆ 1"
"
#
"ˆ3
œ n lim
Ä_ # 2
œ
1
n
"
n
"
#
1‰
œ n lim
ln ˆ n 2n
œ ln
Ä_
" ‰
n1
1 ‰
n1
œ
œ
"
#
ˆ #3
"
n
" ‰
n1
œ
"
#
2(n 1) 2n
’ 3n(n 1)2n(n
“œ
1)
1†4†7â(3n 2)
(3n)!
nœ1
_
dy
dx
_
œ!
nœ1
_
1†4†7â(3n 2)
(3n 1)!
x3nc1
1†4†7â(3n5)
(3n3)!
x3nc2
(3n ")
(3n 1)(3n 2)(3n 3)
1†4†7â(3n 2)
(3n 2)!
x3nc2 œ x !
œ x Œ1 !
1†4†7â(3n 2)
(3n)!
x3n œ xy 0 Ê a œ 1 and b œ 0
x#
1x
œ x# x# (x) x# (x)# x# (x)$ á œ x# x$ x% x& á œ ! (1)n xn which
Ê
d# y
dx#
œ!
nœ1
_
nœ1
96. (a)
x3n Ê
3n# n 2
4n(n 1)
3
4
3nb3
_
is the sum
ˆ 11 3" ‰ ˆ #" 4" ‰ ˆ 3" 5" ‰ ˆ 4" 6" ‰ á ˆ n " # n" ‰
1)x
$
95. (a) n lim
† 1†4†7â(3n)!
¹ 1†4†7â(3n(3n2)(3n
3)!
(3n 2)x3n ¹ 1 Ê kx k n lim
Ä_
Ä_
œ kx$ k † 0 1 Ê the radius of convergence is _
(b) y œ 1 !
"
#
œ
x#
1 (x)
nœ2
_
nœ2
converges absolutely for kxk 1
_
_
nœ2
nœ2
(b) x œ 1 Ê ! (1)n xn œ ! (1)n which diverges
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 11 Practice Exercises
_
771
_
97. Yes, the series ! an bn converges as we now show. Since ! an converges it follows that an Ä 0 Ê an 1
nœ1
nœ1
_
_
nœ1
nœ1
for n some index N Ê an bn bn for n N Ê ! an bn converges by the Direct Comparison Test with ! bn
_
98. No, the series ! an bn might diverge (as it would if an and bn both equaled n) or it might converge (as it
nœ1
would if an and bn both equaled "n ).
_
_
nœ1
kœ1
!(xk1 xk ) œ lim (xn1 x" ) œ lim (xn1 ) x" Ê both the series and
99. ! (xn1 xn ) œ n lim
Ä_
nÄ_
nÄ_
sequence must either converge or diverge.
Š 1 banan ‹
100. It converges by the Limit Comparison Test since n lim
Ä_
an
œ n lim
Ä_
_
"
1 a n
œ 1 because ! an converges
nœ1
and so an Ä 0.
101. Newton's method gives xn1 œ xn
Lœ
_
102. !
nœ1
an
n
œ a"
103. an œ
"
ln n
"
ln #
104. (a) T œ
xn
"
40
, and if the sequence {xn } has the limit L, then
w
a#
#
"
11
á
"
#
ˆ "# ‰
#
a$
3
a%
4
"
3
"
#
á
a$
á , and
a%
"
ln #
"
ln 4
_
nœ2
#
Š0 2 ˆ "# ‰ e"Î# e‹ œ
x#
#
1
"
8
"
6
"
7
8" ‰ a)
(a# a% a) a"' á ) which is a divergent series
á ‰ which diverges so that 1 !
(b) x# ex œ x# Š1 x
39
40
a" ˆ "# ‰ a# ˆ 3" 4" ‰ a% ˆ 5"
á
" ‰
16 a"'
2 Ê a#
for n
ˆ1
39
40
ww
L
"
10
œ
Ê L œ 1 and Öxn × converges since ¹ faÒfxbafxbaÓx2 b ¹ œ
39
40
ˆ 9"
œ
"
40
axn 1b%!
40 axn 1b$*
"
n ln n
"
ln 8
á œ
"
ln #
"
# ln 2
"
3 ln 2
á
diverges by the Integral Test.
e"Î# 4" e ¸ 0.885660616
á ‹ œ x# x$
x%
#
á Ê
'01 Šx# x$ x# ‹ dx œ ’ x3
%
$
x%
4
"
x&
10 “ !
œ
41
60
(c) If the second derivative is positive, the curve is concave upward and the polygonal line segments used in
the trapezoidal rule lie above the curve. The trapezoidal approximation is therefore greater than
the actual area under the graph.
(d) All terms in the Maclaurin series are positive. If we truncate the series, we are omitting positive terms
and hence the estimate is too small.
(e)
'01 x# ex dx œ cx# ex 2xex 2ex d "! œ e 2e 2e 2 œ e 2 ¸ 0.7182818285
105. a0 œ
bk œ
1
21
1
1
'021 faxb dx œ 211 '121 1 dx œ "# ß ak œ 11 '021 faxb cos kx dx œ 11 '121 cos kx dx œ 0.
'021 faxb sin kx dx œ 11 '121 sin kx dx œ cos1kkx ¹21 œ 11k Š1 a1bk ‹ œ œ 1k ,
2
1
Thus, the Fourier series of faxb is
"
#
!
k odd
2
1k sin
k odd
.
0, k even
kx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 0.68333
772
Chapter 11 Infinite Sequences and Series
106. a0 œ
œ
' 1 x dx '121 1 dx • œ "# 4" 1, ak œ 11 ”'01 x cos kx dx '121 cos kx dx • œ 11 cosk kx x sink kx ‘01
1
21 ” 0
k
1
1k2 Ša1b
2
1‹ œ œ
12k2 ,
0,
k odd
.
k even
bk œ 11 ”'0 x sin kx dx '1 sin kx dx • œ 11 sink2kx
1
1ˆ
œk
21
‘1
0
21
cos kx
1k ¹1
œ
a1bkb1
k
1
1 k Š1
a1bk ‹
1 12 ‰, k odd
.
1k , k even
Thus, the Fourier series of faxb is
107. a0 œ
x cos kx
k
"
#
4" 1 12 cos x ˆ1 12 ‰sin x
"
#
sin 2x
2
91 cos
3x 13 ˆ1 12 ‰ sin 3x . . .
' 1 a1 xb dx '121 ax 21b dx • œ 211 ’'01 a1 xb dx '01 a1 ub du “ œ 0 where we used the
1
21 ” 0
substitution u œ x 1 in the second integral. We haveak œ 11 ”'0 a1 xb cos kx dx '1 ax 21b cos kx dx •. Using
1
21
the substitution u œ x 1 in the second integral gives '1 ax 21b cos kx dx œ '0 a1 ub cosaku k1b du
Ú
'01 a1 ub cos ku du, k odd
œÛ 1
.
Ü '0 a1 ub cos ku du, k even
21
#
Thus, ak œ 1
'01 a1 xb cos kx dx,
0,
Now, since k is odd, letting v œ 1 x Ê
Exercise 106). So, ak œ œ
4
1 k2 ,
0,
1
k odd
.
k even
#
1
'01 a1 xb cos kx dx œ 1# '01 v cos kv dv œ 1# ˆ k2 ‰ œ 14k , k odd. (See
2
k odd
.
k even
#
Using similar techniques we see that bk œ 1
'01 a1 ub sin ku du,
0,
2
, k odd
k odd
.
œœk
0, k even
k even
Thus, the Fourier series of faxb is ! ˆ 14k2 cos kx 2k sin kx‰.
k odd
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
Chapter 11 Additional and Advanced Exercises
'021 ksin xk dx œ 11 '01 sin x dx œ 12 . We have ak œ 11 '021 ksin xk cos kx dx
1
#1
œ 11 ’'0 sin x cos kx dx '1 sin x cos kx dx “. Using techniques similar to those used in Exercise 107, we find
108. a0 œ
1
21
ak œ
bk œ
1
1
2
1
'01 sin x cos kx dx,
0,
k odd
k even
œ œ
0,
k odd
.
k even
4
ak 2 1 b 1 ,
'021 ksin xk sin kx dx œ 11 ’'01 sin x sin kx dx '1#1 sin x sin kx dx “ œ 2 ' 1
1
0
0,
sin x sin kx dx,
k odd
k even
for all k.
2
1
Thus, the Fourier series of faxb is
! Š ak241b1 cos kx‹.
k even
CHAPTER 11 ADDITIONAL AND ADVANCED EXERCISES
1. converges since
"
(3n #)Ð2n1ÑÎ2
"
(3n 2)$Î#
"
Š $Î#
‹
n
lim
nÄ_
Š
ˆ 3n n 2 ‰
œ n lim
"
Ä_
‹
(3n 2)$Î#
2. converges by the Integral Test:
$
1
œ Š 24
1$
192 ‹
œ
$Î#
_
and !
nœ1
"
(3n 2)$Î#
converges by the Limit Comparison Test:
œ 3$Î#
'1_ atan" xb# x dx1 œ
#
" xb$
lim ’ atan 3
bÄ_
" bb$
b
tan
“ œ lim ’ a 3
"
bÄ_
1$
192 “
71 $
192
c2n
e
3. diverges by the nth-Term Test since n lim
a œ n lim
(1)n tanh n œ lim (1)n Š 11
(1)n
ec2n ‹ œ n lim
Ä_ n
Ä_
Ä_
bÄ_
does not exist
4. converges by the Direct Comparison Test: n! nn Ê ln (n!) n ln (n) Ê
Ê logn (n!) n Ê
logn (n!)
n$
"
n#
12
(3)(5)(4)#
n
, which is the nth-term of a convergent p-series
5. converges by the Direct Comparison Test: a" œ 1 œ
œ
ln (n!)
ln (n)
, a% œ ˆ 53††64 ‰ ˆ 42††53 ‰ ˆ 31††42 ‰ œ
"2
(4)(6)(5)#
12
(1)(3)(2)#
, a# œ
_
,á Ê 1!
nœ1
1 †2
3 †4
œ
12
(2)(4)(3)#
12
(n 1)(n 3)(n 2)#
, a$ œ ˆ 42††53 ‰ ˆ 31††42 ‰
represents the
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ0
773
774
Chapter 11 Infinite Sequences and Series
12
(n 1)(n 3)(n 2)#
given series and
12
n%
, which is the nth-term of a convergent p-series
anb1
an
6. converges by the Ratio Test: n lim
Ä_
œ n lim
Ä_
n
(n 1)(n 1)
œ01
"
1L
7. diverges by the nth-Term Test since if an Ä L as n Ä _, then L œ
Ê L# L 1 œ 0 Ê L œ
1 „ È 5
#
Á0
_
_
"
32nb1
8. Split the given series into !
nœ1
and !
nœ1
2n
32n
; the first subseries is a convergent geometric series and the
n
n n
È
2È
n 2n
É
second converges by the Root Test: n lim
32n œ n lim
Ä_
Ä_
1
3
9. f(x) œ cos x with a œ
"
#
cos x œ
È3
#
9
È3
ww
# ,f
1 ‰$
á
3
Ê f ˆ 13 ‰ œ 0.5, f w ˆ 13 ‰ œ
ˆx 13 ‰ 4" ˆx 13 ‰#
È3
1#
ˆx
œ
" †1
9
"
9
œ
1
ˆ 13 ‰ œ 0.5, f www ˆ 13 ‰ œ
È3
#
, f Ð4Ñ ˆ 13 ‰ œ 0.5;
10. f(x) œ sin x with a œ 21 Ê f(21) œ 0, f w (21) œ 1, f ww (21) œ 0, f www (21) œ 1, f Ð4Ñ (21) œ 0, f Ð5Ñ (21) œ 1,
f Ð6Ñ (21) œ 0, f Ð7Ñ (21) œ 1; sin x œ (x 21)
x#
#!
11. ex œ 1 x
x$
3!
(x 21)$
3!
(x 21)&
5!
(x 21)(
7!
á
á with a œ 0
12. f(x) œ ln x with a œ 1 Ê f(1) œ 0, f w (1) œ 1, f ww (1) œ 1,f www (1) œ 2, f Ð4Ñ (1) œ 6;
ln x œ (x 1)
(x 1)#
#
(x 1)$
3
(x 1)%
4
á
13. f(x) œ cos x with a œ 221 Ê f(221) œ 1, f w (221) œ 0, f ww (221) œ 1, f www (221) œ 0, f Ð4Ñ (221) œ ",
f Ð5Ñ (221) œ 0, f Ð6Ñ (221) œ 1; cos x œ 1 "# (x 221)# 4!" (x 221)% 6!" (x 221)' á
14. f(x) œ tan" x with a œ 1 Ê f(1) œ
1
4
tan" x œ
(x 1)
2
(x 1)#
4
1
4
(x 1)$
12
, f w (1) œ
"
#
, f ww (1) œ "# , f www (1) œ
ˆ ba ‰n ln ˆ ba ‰
a n
nÄ_ ˆ b ‰ 1
16. 1
2
10
3
10#
_
œ1!
nœ0
œ1
200
999
n1
17. sn œ !
kœ0
7
10$
'kkb1
2
10%
_
2
103n1
30
999
!
nœ0
dx
1 x#
0†ln ˆ ba ‰
01
œ ln b
7
999
3
7
10'
_
103n2
œ
3
10&
!
999237
999
Ê sn œ '0
1
_
á œ1!
nœ1
7
œ
n
lnˆˆ ba ‰ 1‰
n
nÄ_
Ê n lim
c œ ln b lim
Ä_ n
œ1
2 ‰
ˆ 10
$
1ˆ " ‰
10
2
_
103n2
!
nœ1
Š 103# ‹
$
1ˆ " ‰
10
3
_
103n1
!
nœ1
Š 107$ ‹
"‰
1 ˆ 10
7
103n
$
412
333
dx
1 x#
'1
2
dx
1 x#
Ê n lim
s œ n lim
atan" n tan" 0b œ
Ä_ n
Ä_
nb1
1În
œ ln b since 0 a b. Thus, n lim
c œ eln b œ b.
Ä_ n
103n3
nœ0
;
á
n
15. Yes, the sequence converges: cn œ aan bn b1În Ê cn œ b ˆˆ ba ‰ 1‰
œ ln b lim
"
#
á 'nc1 1 dxx# Ê sn œ '0
n
n
dx
1 x#
1
#
#
1)
18. n lim
† (n 1)(2x
¹ uunbn 1 ¹ œ n lim
¹ (n 1)x
¹ œ n lim
¹ x † (n 1) ¹ œ ¸ 2x x 1 ¸ 1
nxn
Ä_
Ä _ (n 2)(2x 1)n1
Ä _ 2x 1 n(n 2)
Ê kxk k2x 1k ; if x 0, kxk k2x 1k Ê x 2x 1 Ê x 1; if "# x 0, kxk k2x 1k
n
Ê x 2x 1 Ê 3x 1 Ê x "3 ; if x #" , kxk k2x 1k Ê x 2x 1 Ê x 1. Therefore,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 11 Additional and Advanced Exercises
the series converges absolutely for x 1 and x "3 .
19. (a) Each Anb1 fits into the corresponding upper triangular region, whose vertices are:
(nß f(n) f(n1)), (n1ß f(n1)) and (nß f(n)) along the line whose slope is f(n 1) f(n).
(b) If Ak œ
n1
! Ak œ
kœ1
œ
kb1
'k
f(k 1) f(k)
#
œ
n 1
f(1) f(2)
#
'1 f(x) dx '2 f(x) dx á 'nc1 f(x) dx
2
3
! f(k) '1 f(x) dx Ê ! Ak œ ! f(k)
n1
Ê ! An
f(x) dx, then
f(1) f(2) f(2) f(3) f(3) á f(n 1) f(n)
#
f(1) f(n)
#
_
f(1) f(2)
2
All the An 's fit into the first upper triangular region whose area is
n
kœ2
n1
n
kœ1
kœ1
n
f(1) f(n)
#
'1n f(x) dx f(1) # f(2) , from
n1
part (a). The sequence œ ! Ak is bounded above and increasing, so it converges and the limit in
kœ1
question must exist.
! f(k) ' f(x) dx " afa1b fanbb•, which exists by part (b). Since f is positive and
(c) Let L œ n lim
#
Ä_ ”
1
kœ1
n
n
! f(k) ' f(x) dx• œ L " afa1b Mb.
0 exists. Thus n lim
#
Ä_ ”
1
n
decreasing lim fanb œ M
nÄ_
n
kœ1
20. The number of triangles removed at stage n is 3n1 ; the side length at stage n is
at stage n is
È3
4
(a)
È3
4
b# 3
b
2nc1
; the area of a triangle
#
ˆ 2nbc1 ‰ .
È3
4
#
Š b2# ‹ 3#
È3
4
(b) a geometric series with sum
#
Š b2% ‹ 3$
Š
È 3 b# ‹
4
1 ˆ 34 ‰
È3
4
#
Š b2' ‹ á œ
È3
4
_
b# !
nœ0
3n
22n
œ
È3
4
_
n
b# ! ˆ 43 ‰
nœ0
œ È3b#
(c) No; for instance,the three vertices of the original triangle are not removed. However the total area removed
is È3b# which equals the area of the original triangle. Thus the set of points not removed has area 0.
21. (a) No, the limit does not appear to depend on the value of the constant a
(b) Yes, the limit depends on the value of b
(c) s œ Š1
œ n lim
Ä_
cos ˆ na ‰ n
n ‹
a
n
lim Š1
nÄ_
Ê ln s œ
sin ˆ na ‰ cos ˆ na ‰
1
cos ˆ na ‰
n
cos ˆ na ‰ n
bn ‹
œ
ln Œ1
cos ˆ na ‰
n
ˆ "n ‰
01
10
Ê n lim
ln s œ
Ä_
na sin ˆ na ‰ cos ˆ na ‰
n#
"
Š # ‹
n
"
cos ˆ na ‰ Œ
n
œ e1Îb
1În
_
"
#
1
œ 1 Ê n lim
s œ e" ¸ 0.3678794412; similarly,
Ä_
a n ‰n
22. ! an converges Ê n lim
a œ 0; n lim
’ˆ 1 sin
“
#
Ä_ n
Ä_
nœ1
œ
an ‰
ˆ 1sin
œ n lim
œ
#
Ä_
Ä_
1sin Šnlim an ‹
#
œ
1sin 0
#
Ê the series converges by the nth-Root Test
nb1 nb1
23. n lim
¹ uunbn 1 ¹ 1 Ê n lim
¹b x †
Ä_
Ä _ ln (n 1)
ln n
bn xn ¹
1 Ê kbxk 1 Ê "b x
"
b
œ5 Ê bœ „
"
5
24. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln x and
ex have infinitely many nonzero terms in their Taylor expansions.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
775
776
Chapter 11 Infinite Sequences and Series
sin (ax) sin xx
x$
xÄ0
xÄ0
26.
lim cos#axx# b
xÄ0
"
3!
(b)
28.
un
unb1
œ
(n 1)#
n#
un
unb1
œ
n1
n
œ
Ê Cœ
3
#
"
3!
œ
1
n
2
n
á‹ x
a# x#
#
á b
œ 1 Ê lim Š "2x#b
xÄ0
_
"
n#
Ê C œ 2 1 and !
nœ1
_
"
n
nœ1
œ1
Š 64 ‹
n
5n#
4n# 4n 1
1 and kf(n)k œ
a% x%
4!
Ê C œ 1 Ÿ 1 and !
0
n#
4n# 2n
4n# 4n 1
á “ is finite if a 2 œ 0 Ê a œ 2;
#x#
"
n#
x$
3!
œ 76
xÄ0
œ1
œ1
"
#
5! ‹ x
Œ1
œ 1 Ê lim
un
unb1
2n(2n 1)
(2n 1)#
&
Š a5!
œ 23!
Ê b œ 1 and a œ „ 2
27. (a)
á‹ Šx
x$
$
sin 2x sin x x
x$
lim
xÄ0
a$ x$
3!
xÄ0
a$
3!
œ lim ’ a x# 2
Šax
œ lim
25. lim
œ
Š4
5
4
n
_
_
_
nœ1
nœ1
nœ1
a# x#
48
á ‹ œ 1
converges
diverges
œ1
5
4n# 4n 1
a#
4
Š 3# ‹
n
5n#
– Š4n# c 4n b 1‹ —
n#
after long division
_
"
‹
n#
Ÿ 5 Ê ! un converges by Raabe's Test
nœ1
29. (a) ! an œ L Ê an# Ÿ an ! an œ an L Ê ! an# converges by the Direct Comparison Test
an
Š1c
an ‹
(b) converges by the Limit Comparison Test: n lim
Ä_
an
œ n lim
Ä_
"
1 an
_
œ 1 since ! an converges and
nœ1
therefore x lim
a œ0
Ä_ n
30. If 0 an 1 then kln (1 an )k œ ln (1 an ) œ an
a#n
#
a$n
3
á an an# an$ á œ
an
1 an
,
a positive term of a convergent series, by the Limit Comparison Test and Exercise 29b
_
31. (1 x)" œ 1 ! xn where kxk 1 Ê
nœ1
#
"
(1x)#
$
4 œ 1 2 ˆ "# ‰ 3 ˆ "# ‰ 4 ˆ "# ‰ á n ˆ "# ‰
_
32. (a) ! xn1 œ
nœ1
_
Ê !
nœ1
_
(b) x œ !
nœ1
x#
1 x
n(n 1)
xn
n(n ")
xn
_
Ê ! (n 1)xn œ
nœ1
œ
2
x
$
Š1 "x ‹
Ê xœ
œ
2x#
(x 1)$
2x#
(x 1)$
2xx#
(1x)#
œ
n 1
d
dx
_
(1 x)" œ ! nxn1 and when x œ
nœ1
"
#
we have
á
_
Ê ! n(n 1)xn1 œ
nœ1
2
(1x)$
_
Ê ! n(n 1)xn œ
nœ1
2x
(1x)$
, kxk 1
Ê x$ 3x# x 1 œ 0 Ê x œ 1 Š1
È57 "Î$
9 ‹
Š1
È57 "Î$
9 ‹
¸ 2.769292, using a CAS or calculator
33. The sequence {xn } converges to
1
#
from below so %n œ
Estimation Theorem %n1 ¸ 3!" (%n )$ with
overestimate Ê 0 %n1 "6 (%n )$ .
_
kerrork
"
5!
1
#
xn 0 for each n. By the Alternating Series
&
(%n ) , and since the remainder is negative this is an
34. Yes, the series ! ln (1 an ) converges by the Direct Comparison Test: 1 an 1 an
nœ1
a#n
#!
Ê 1 an ean Ê ln (1 an ) an
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
a$n
3!
á
Chapter 11 Additional and Advanced Exercises
"
(1x)#
35. (a)
œ
d
dx
ˆ 1" x ‰ œ
d
dx
_
a1 x x# x$ á b œ 1 2x 3x# 4x$ á œ ! nxn1
nœ1
_
! n ˆ 5 ‰ n 1
6
(b) from part (a) we have
nœ1
_
ˆ "6 ‰ œ ˆ "6 ‰ ’
(c) from part (a) we have ! npn1 q œ
q
(1 p)#
nœ1
_
_
ˆ "# ‰
36. (a) ! pk œ ! 2k œ
kœ1
1ˆ "# ‰
kœ1
œ
q
q#
2
"
“
1 ˆ 56 ‰
œ
œ6
"
q
_
_
kœ1
kœ1
œ 1 and E(x) œ ! kpk œ ! k2k œ
"
#
_
! k21k œ ˆ " ‰
#
kœ1
"
1ˆ "# ‰‘#
œ2
by Exercise 35(a)
_
_
kœ1
kœ1
(b) ! pk œ !
œ ˆ "6 ‰
"
1ˆ 56 ‰‘#
_
_
(c) ! pk œ !
kœ1
kœ1
_
œ!
kœ1
5kc1
6k
"
k1
_
5
_
_
kœ1
kœ1
! ˆ 5 ‰k œ ˆ " ‰ ’ ˆ 6 ‰5 “ œ 1 and E(x) œ ! kpk œ ! k
6
5
1ˆ ‰
"
5
œ
kœ1
6
5kc1
6k
"
6
œ
_
"
k(k1)
œ ! ˆ k"
kœ1
" ‰
k1
œ
lim ˆ1
kÄ_
" ‰
k1
_
_
kœ1
kœ1
œ 1 and E(x) œ ! kpk œ ! k Š k(k " 1) ‹
, a divergent series so that E(x) does not exist
C! ekt! ˆ1 enkt! ‰
1ekt!
Ê Rœ
lim R œ
nÄ_ n
¸ 0.58195028;
C! ekt!
1 ekt!
"
e1 a1 en b
e"! b
Ê R" œ e" ¸ 0.36787944 and R"! œ e 1a1e"
1 e"
R œ e" 1 ¸ 0.58197671; R R"! ¸ 0.00002643 Ê R RR"! 0.0001
Þ1n
eÞ1 ˆ1 eÞ1n ‰ R
Rn œ
, # œ #" ˆ eÞ1 " 1 ‰ ¸ 4.7541659; Rn R# Ê 1eÞ1 e 1 ˆ #" ‰ ˆ eÞ1 " 1 ‰
1 eÞ1
n
n
Ê 1 enÎ10 "# Ê enÎ10 "# Ê 10
ln ˆ "# ‰ Ê 10
ln ˆ "# ‰ Ê n 6.93
(b) Rn œ
38. (a) R œ
(b) t! œ
C!
ekt! 1
"
0.05
kœ1
œ6
37. (a) Rn œ C! ekt! C! e2kt! á C! enkt! œ
(c)
_
! k ˆ 5 ‰ k 1
6
Ê Rekt! œ R C! œ CH Ê ekt! œ
CH
CL
Ê t! œ
"
k
C!
ekt! 1
œ
Ê nœ7
ln Š CCHL ‹
ln e œ 20 hrs
(c) Give an initial dose that produces a concentration of 2 mg/ml followed every t! œ
"
0.0#
2 ‰
ln ˆ 0.5
¸ 69.31 hrs
by a dose that raises the concentration by 1.5 mg/ml
"
0.1 ‰
‰
(d) t! œ 0.2
ln ˆ 0.03
œ 5 ln ˆ 10
3 ¸ 6 hrs
_
39. The convergence of ! kan k implies that
nœ1
ka n k
1 ka n k
Ê
lim ka k œ 0. Let N 0 be such that kan k
nÄ_ n
2 kan k for all n N. Now kln a1 an bk œ ¹an
kan k kan k # kan k $ kan k % á œ
ka n k
1 ka n k
a#n
#
a$n
3
a%n
4
"
#
Ê 1 kan k
#
$
á ¹ Ÿ kan k ¹ a#n ¹ ¹ a3n ¹ ¹ a4n ¹ á
_
2 kan k . Therefore ! ln (1 an ) converges by the Direct
nœ1
_
Comparison Test since ! kan k converges.
nœ1
_
40. !
nœ3
"
n ln n(ln (ln n))p
lim
bÄ_
converges if p 1 and diverges otherwise by the Integral Test: when p œ 1 we have
'3b x ln x(lndx(ln x)) œ
cpb1
œ lim ’ (ln (ln1 x))p
bÄ_
b
lim cln (ln (ln x))d b$ œ _; when p Á 1 we have lim
bÄ_
“ œ
$
(ln (ln 3))cpb1
1p
bÄ_
,
_,
%
"
#
'3b x ln x(lndx(ln x))
if p 1
if p 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
p
777
778
Chapter 11 Infinite Sequences and Series
41. (a) s2n1 œ
c"
1
c#
#
c$
3
c2n1
2n1
á
œ
t"
1
t# t"
2
"
œ t" ˆ1 "# ‰ t# ˆ "# 3" ‰ á t2n ˆ 2n
_
(b) ecn f œ e(1)n f Ê !
nœ1
(1)n
n
t$ t#
3
" ‰
#n 1
t2n1 t2n
2n1
á
t2n1
2n1
2n
œ!
kœ1
tk
k(k1)
t2n1
2n1
.
"
3
"
5
"
6
converges
(c) ecn f œ e1ß 1ß 1ß 1ß 1ß 1ß 1ß 1ß 1ß á f Ê the series 1
"
#
"
4
"
7
á converges
42. (a) a1 t t# t$ á (1)n tn b (1 t) œ 1 t t# t$ á (1)n tn t t# t$ t% á (1)n tn1
œ 1 (1)n tn1 Ê 1 t t# t$ á (1)n tn
'0
x
(b)
dt œ '0 ’1 t t# á (1)n tn
x
"
1t
œ ’t
t#
2
t$
3
á
œx
x#
#
x$
3
á
(1)n tn1
n1 “!
x
(c) x 0 and Rn1 œ (1)n1
'0
x
tn1
1t
x
x
ktk nb1
1 kx k
dx œ
kxk nb2
a1 kxkb (n 2)
dt Ê kRn1 k œ '0
tnb1
1t
tnb1
1 t
(1)n1 tn1
n1
dt
dt Ÿ '0 tn1 dt œ
x
dt Ê kRn1 k œ ¹'0
kxk nb2
a1kxkb (n2)
"
1t
dt Ê ln k1 xk
x
since k1 tk
(e) From part (d) we have kRn1 k Ÿ
kxk nb2
lim
n Ä _ a1 kxkb (n 2)
(1)n1 tn1
n1
œ
dt Ê cln k1 tkd !x
x
(d) 1 x 0 and Rnb1 œ (1)nb1 '0
Ÿ '0
(1)nb1 tnb1
“
1t
Rn1 , where Rn1 œ '0
(1)n xn1
n1
x
'0
(1)n tnb1
1t
x
tnb1
1t
xn2
n#
dt¹ Ÿ '0 ¹ 1t t ¹ dt
x
nb1
1 kxk
Ê the given series converges since
œ 0 Ê kRn1 k Ä 0 when kxk 1. If x œ 1, by part (c) kRn1 k Ÿ
kxknb2
n2
Thus the given series converges to lna1 xb for 1 x Ÿ 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
1
n2
Ä 0.
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
1. The line through the point (#ß $ß !) parallel to the z-axis
2. The line through the point (1ß 0ß !) parallel to the y-axis
3. The x-axis
4. The line through the point (1ß !ß !) parallel to the z-axis
5. The circle x# y# œ 4 in the xy-plane
6. The circle x# y# œ 4 in the plane z = 2
7. The circle x# z# œ 4 in the xz-plane
8. The circle y# z# œ 1 in the yz-plane
9. The circle y# z# œ 1 in the yz-plane
10. The circle x# z# œ 9 in the plane y œ 4
11. The circle x# y# œ 16 in the xy-plane
12. The circle x# z# œ 3 in the xz-plane
13. (a) The first quadrant of the xy-plane
(b) The fourth quadrant of the xy-plane
14. (a) The slab bounded by the planes x œ 0 and x œ 1
(b) The square column bounded by the planes x œ 0, x œ 1, y œ 0, y œ 1
(c) The unit cube in the first octant having one vertex at the origin
15. (a) The solid ball of radius 1 centered at the origin
(b) The exterior of the sphere of radius 1 centered at the origin
16. (a) The circumference and interior of the circle x# y# œ 1 in the xy-plane
(b) The circumference and interior of the circle x# y# œ 1 in the plane z œ 3
(c) A solid cylindrical column of radius 1 whose axis is the z-axis
17. (a) The closed upper hemisphere of radius 1 centered at the origin
(b) The solid upper hemisphere of radius 1 centered at the origin
18. (a) The line y œ x in the xy-plane
(b) The plane y œ x consisting of all points of the form (xß xß z)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
780
Chapter 12 Vectors and the Geometry of Space
19. (a) x œ 3
(b) y œ 1
(c) z œ 2
20. (a) x œ 3
(b) y œ 1
(c) z œ 2
21. (a) z œ 1
(b) x œ 3
(c) y œ 1
22. (a) x# y# œ 4, z œ 0
(b) y# z# œ 4, x œ 0
(c) x# z# œ 4, y œ 0
23. (a) x# (y 2)# œ 4, z œ 0
(b) (y 2)# z# œ 4, x œ 0
(c) x# z# œ 4, y œ 2
24. (a) (x 3)# (y 4)# œ 1, z œ 1
(c) (x 3)# (z 1)# œ 1, y œ 4
(b) (y 4)# (z 1)# œ 1, x œ 3
25. (a) y œ 3, z œ 1
(b) x œ 1, z œ 1
(c) x œ 1, y œ 3
26. Èx# y# z# œ Èx# (y 2)# z# Ê x# y# z# œ x# (y 2)# z# Ê y# œ y# 4y 4 Ê y œ 1
27. x# y# z# œ 25, z œ 3 Ê x# y# œ 16 in the plane z œ 3
28. x# y# (z 1)# œ 4 and x# y# (z 1)# œ 4 Ê x# y# (z 1)# œ x# y# (z 1)# Ê z œ 0, x# y# œ 3
29. 0 Ÿ z Ÿ 1
30. 0 Ÿ x Ÿ 2, 0 Ÿ y Ÿ 2, 0 Ÿ z Ÿ 2
31. z Ÿ 0
32. z œ È1 x# y#
33. (a) (x 1)# (y 1)# (z 1)# 1
(b) (x 1)# (y 1)# (z 1)# 1
34. 1 Ÿ x# y# z# Ÿ 4
35. kP" P# k œ Éa3 1b# a3 1b# a0 1b# œ È9 œ 3
36. kP" P# k œ Éa2 1b# a5 1b# a0 5b# œ È50 œ 5È2
37. kP" P# k œ Éa4 1b# a2 4b# a7 5b# œ È49 œ 7
38. kP" P# k œ Éa2 3b# a3 4b# a4 5b# œ È3
39. kP" P# k œ Éa2 0b# a2 0b# a2 0b# œ È3 † 4 œ 2È3
40. kP" P# k œ Éa0 5b# a0 3b# a0 2b# œ È38
41. center (2ß 0ß 2), radius 2È2
42. center ˆ "# ß "# ß "# ‰ , radius
43. center ŠÈ2ß È2ß È2‹ , radius È2
44. center ˆ!ß "3 ß 3" ‰ , radius
È21
#
È29
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.2 Vectors
45. (x 1)# (y 2)# (D 3)# œ 14
46. x# (y 1)# (z 5)# œ 4
47. (x 2)# y# z# œ 3
48. x# (y 7)# z# œ 49
781
49. x# y# z# 4x 4z œ 0 Ê ax# 4x 4b y# az# 4z 4b œ 4 4
#
Ê (x 2)# (y 0)# (z 2)# œ ŠÈ8‹ Ê the center is at (2ß 0ß 2) and the radius is È8
50. x# y# z# 6y 8z œ 0 Ê x# ay# 6y 9b az# 8z 16b œ 9 16 Ê (x 0)# (y 3)# (z 4)# œ 5#
Ê the center is at (0ß 3ß 4) and the radius is 5
51. 2x# 2y# 2z# x y z œ 9 Ê x# "# x y# "# y z# "# z œ
Ê ˆx# "# x
" ‰
16
ˆy# "# y
" ‰
16
ˆz# "# z
Ê the center is at ˆ "4 ß 4" ß 4" ‰ and the radius is
" ‰
16
œ
9
#
3
16
9
#
È
5È 3
4
52. 3x# 3y# 3z# 2y 2z œ 9 Ê x# y# 23 y z# 23 z œ 3 Ê x# ˆy# 23 y "9 ‰ ˆz# 23 z "9 ‰ œ 3
#
#
Ê (x 0)# ˆy "3 ‰ ˆz 3" ‰ œ Š
È29 #
3 ‹
Ê the center is at ˆ0ß 3" ß 3" ‰ and the radius is
È29
3
53. (a) the distance between (xß yß z) and (xß 0ß 0) is Èy# z#
(b) the distance between (xß yß z) and (0ß yß 0) is Èx# z#
(c) the distance between (xß yß z) and (0ß 0ß z) is Èx# y#
54. (a) the distance between (xß yß z) and (xß yß 0) is z
(b) the distance between (xß yß z) and (0ß yß z) is x
(c) the distance between (xß yß z) and (xß 0ß z) is y
55. kABk œ Éa1 a1bb# a1 2b# a3 1b# œ È4 9 4 œ È17
kBCk œ Éa3 1b# a4 a1bb# a5 3b# œ È4 25 4 œ È33
kCAk œ Éa1 3b# a2 4b# a1 5b# œ È16 4 16 œ È36 œ 6
Thus the perimeter of triangle ABC is È17 È33 6.
56. kPAk œ Éa2 3b# a1 1b# a3 2b# œ È1 4 1 œ È6
kPBk œ Éa4 3b# a3 1b# a1 2b# œ È1 4 1 œ È6
Thus P is equidistant from A and B.
12.2 VECTORS
1. (a)
3a3b, 3a2b¡ œ 9, 6¡
(b) É92 a6b2 œ È117 œ 3È13
3. (a)
3 a2b, 2 5¡ œ 1, 3¡
(b) È12 32 œ È10
#
#
#
#
Ê ˆx 4" ‰ ˆy 4" ‰ ˆz 4" ‰ œ Š 5 4 3 ‹
2. (a)
2a2b, 2a5b¡ œ 4, 10¡
(b) É42 a10b2 œ È116 œ 2È29
4. (a)
3 a2b, 2 5¡ œ 5, 7¡
(b) É52 a7b2 œ È74
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
9
782
Chapter 12 Vectors and the Geometry of Space
5. (a) 2u œ 2a3b, 2a2b¡ œ 6, 4¡
3v œ 3a2b, 3a5b¡ œ 6, 15¡
2u 3v œ 6 a 4b, 4 15¡ œ 12, 19¡
6. (a) 2u œ 2a3b, 2a2b¡ œ 6, 4¡
5v œ 5a2b, 5a5b¡ œ 10, 25¡
2u 5v œ 6 a10b, 4 25¡ œ 16, 29¡
(b) É122 a19b2 œ È505
7. (a)
(b) Éa16b2 292 œ È1097
3
5u
œ ¢ 53 a3b, 53 a2b£ œ ¢ 59 , 56 £
4
5v
œ ¢ 45 a2b, 45 a5b£ œ ¢ 85 , 4£
3
5u
45 v œ ¢ 95 ˆ 85 ‰, 65 4£ œ ¢ 15 ,
2
‰2 œ
(b) Ɉ 15 ‰ ˆ 14
5
5
5
5
8. (a) 13
u œ ¢ 13
a3b, 13
a2b£ œ ¢ 15
13 ,
12
13 v
5
13
u
14
5 £
È"97
5
9.
2 1, 1 3¡ œ 1, 4¡
11.
0 2, 0 3¡ œ 2, 3¡
œ ¢ 12
13 a2b,
12
13 v
12
13 a5b£
ˆ 24 ‰
œ ¢ 15
13 13 ,
70 ‰2
(b) Éa3b2 ˆ 13
œ
10. ¢ 2a#4b 0,
œ ¢ 24
13 ,
"3
#
10
13
10
13 £
60
13 £
60
13 £
œ ¢3,
È6421
13
0£ œ 1, 1¡
Ä
Ä
Ä
Ä
12. AB œ 2 1, 0 a1b¡ œ 1, 1¡, CD œ 2 a1b, 2 3¡ œ 1, 1¡, AB CD œ 0, 0¡
13. ¢cos
21
3 ,
sin
21
3 £
œ ¢ "# ,
È3
2 £
14. ¢cos ˆ 341 ‰, sin ˆ 341 ‰£ œ ¢ È"2 , È"2 £
15. This is the unit vector which makes an angle of 120‰ 90‰ œ 210‰ with the positive x-axis;
È3
2 ,
"# £
cos 135‰ , sin 135‰ ¡ œ ¢ È" ,
"
È2 £
cos 210‰ , sin 210‰ ¡ œ ¢
16.
2
Ä
17. P" P# œ a2 5bi a9 7bj a2 a1bbk œ 3i 2j k
Ä
18. P" P# œ a3 1bi a0 2bj a5 0bk œ 4i 2j 5k
Ä
19. AB œ a10 a7bbi a8 a8bbj a1 1bk œ 3i 16j
Ä
20. AB œ a1 1bi a4 0bj a5 3bk œ 2i 4j 2k
21. 5u v œ 5 1, 1, 1¡ 2, 0, 3¡ œ 5, 5, 5¡ 2, 0, 3¡ œ 5 2, 5 0, 5 3¡ œ 3, 5, 8¡ œ 3i 5j 8k
22. 2u 3v œ 2 1, 0, 2¡ 3 1, 1, 1¡ œ 2, 0, 4¡ 3, 3, 3¡ œ 5, 3, 1¡ œ 5i 3j k
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
70
13 £
Section 12.2 Vectors
23. The vector v is horizontal and 1 in. long. The vectors u and w are
the horizontal. All vectors must be drawn to scale.
(a)
(b)
(c)
(d)
"1
16
783
in. long. w is vertical and u makes a 45‰ angle with
24. The angle between the vectors is 120‰ and vector u is horizontal. They are all 1 in. long. Draw to scale.
(a)
(b)
(c)
(d)
25. length œ k2i j 2kk œ È2# 1# (2)# œ 3, the direction is
26. length œ k9i 2j 6kk œ È81 4 36 œ 11, the direction is
9
2
6 ‰
œ 11 ˆ 11
i 11
j 11
k
2
3
9
11
i "3 j 32 k Ê 2i j 2k œ 3 ˆ 32 i 3" j 32 k‰
i
2
11
j
6
11
k Ê 9i 2 j 6k
27. length œ k5kk œ È25 œ 5, the direction is k Ê 5k œ 5(k)
9
28. length œ ¸ 35 i 45 k¸ œ É 25
29. length œ ¹ È16 i
Ê
1
È6
i
1
È6
1
È6
j
j
"
È6
"
È6
16
25
œ 1, the direction is
3
5
#
i 45 k Ê
3
5
k¹ œ Ê3 Š È"6 ‹ œ É #" , the direction is
k œ É #" Š È13 i
1
È3
j
"
È3
i 45 k œ 1 ˆ 35 i 45 k‰
1
È3
i
1
È3
j
"
È3
k
k‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
784
Chapter 12 Vectors and the Geometry of Space
30. length œ ¹ È13 i
Ê
1
È3
i
1
È3
1
È3
j
j
"
È3
"
È3
#
k¹ œ Ê3 Š È"3 ‹ œ 1, the direction is
k œ 1 Š È13 i
1
È3
j
31. (a) 2i
(b) È3k
32. (a) 7j
(b) 3 5 2 i
È
33. kvk œ È12# 5# œ È169 œ 13;
34. kvk œ É 4"
"
4
"
4
œ
È3 v
# ; kv k
œ
œ
v
kv k
"
13
"
È3
4È 2
5
vœ
1
È3
i
1
È3
j
3
35. (a) 3i 4j 5k œ 5È2 Š 5È
i
2
4
5È 2
j
"
È2
"
13
1
È3
i
1
È3
j
"
È3
k
k‹
k
(c)
3
10
(c)
1
4
j 25 k
(d) 6i 2j 3k
i 13 j k
(12i 5k) Ê the desired vector is
1
È3
a
È2
(d)
7
13
i
a
È3
j
a
È6
k
(12i 5k)
k Ê the desired vector is 3 Š È"3 i
"
È3
j
"
È3
k‹
œ È3i È3j È3k
(b) the midpoint is
k‹ Ê the direction is
i
4
5È 2
j
"
È2
k
ˆ "# ß 3ß 5# ‰
36. (a) 3i 6j 2k œ 7 ˆ 37 i 67 j 27 k‰ Ê the direction is
(b) the midpoint is ˆ 5# ß 1ß 6‰
37. (a) i j k œ È3 Š È13 i
(b) the midpoint is
3
5È 2
1
È3
j
1
È3
3
7
i 67 j 27 k
k‹ Ê the direction is È13 i
1
È3
j
1
È3
k
ˆ 5# ß 7# ß 9# ‰
38. (a) 2i 2j 2k œ 2È3 Š È13 i
1
È3
j
1
È3
k‹ Ê the direction is
1
È3
i
1
È3
j
1
È3
k
(b) the midpoint is ("ß "ß 1)
Ä
39. AB œ (5 a)i (1 b)j (3 c)k œ i 4j 2k Ê 5 a œ 1, 1 b œ 4, and 3 c œ 2 Ê a œ 4, b œ 3, and
c œ 5 Ê A is the point (4ß 3ß 5)
Ä
40. AB œ (a 2)i (b 3)j (c 6)k œ 7i 3j 8k Ê a 2 œ 7, b 3 œ 3, and c 6 œ 8 Ê a œ 9, b œ 0,
and c œ 14 Ê B is the point (9ß 0ß 14)
41. 2i j œ a(i j) b(i j) œ (a b)i (a b)j Ê a b œ 2 and a b œ 1 Ê 2a œ 3 Ê a œ
bœa"œ
"
#
3
#
and
42. i 2j œ a(2i 3j) b(i j) œ (2a b)i (3a b)j Ê 2a b œ 1 and 3a b œ 2 Ê a œ 3 and
b œ 1 #a œ 7 Ê u" œ a(2i 3j) œ 6i 9j and u# œ b(i j) œ 7i 7j
43. If kxk is the magnitude of the x-component, then cos 30° œ
kx k
kFk
Ê kxk œ kFk cos 30° œ (10) Š
È3
# ‹
œ 5È3 lb
Ê Fx œ 5È3 i;
if kyk is the magnitude of the y-component, then sin 30° œ
ky k
kFk
Ê kyk œ kFk sin 30° œ (10) ˆ 1# ‰ œ 5 lb Ê Fy œ 5 j.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.2 Vectors
44. If kxk is the magnitude of the x-component, then cos 45° œ
kx k
kFk
Ê kxk œ kFk cos 45° œ (12) Š
È2
# ‹
œ 6È2 lb
Ê Fx œ 6È2 i (the negative sign is indicated by the diagram)
if kyk is the magnitude of the y-component, then sin 45° œ
ky k
kFk
Ê kyk œ kFk sin 45° œ (12) Š
È2
# ‹
œ 6È2 lb
Ê Fy œ 6È2 j (the negative sign is indicated by the diagram)
45. 25‰ west of north is 90‰ 25‰ œ 115‰ north of east. 800 cos 155‰ , sin 115‰ ¡ ¸ 338.095, 725.046¡
46. 10‰ east of south is 270‰ 10‰ œ 280‰ ”north” of east. 600 cos 280‰ , sin 280‰ ¡ ¸ 104.189,590.885¡
Ä
È
47. (a) The tree is located at the tip of the vector OP œ (5 cos 60°)i (5 sin 60°)j œ 5# i 5 # 3 j Ê P œ Š 5# ß
Ä
Ä
(b) The telephone pole is located at the point Q, which is the tip of the vector OP PQ
œ Š 52 i
5È 3
#
j‹ (10 cos 315°)i (10 sin 315°)j œ Š 5#
Ê Q œ Š 5 10
#
48. Let t œ
q
pq
and s œ
È2
ß
10È2
# ‹i
È
Š5 # 3
5È 3
# ‹
10È2
# ‹j
5È3 10È2
‹
#
p
pq.
Choose T on OP1 so that TQ is
parallel to OP2 , so that ˜TP1 Q is similar to ˜OP1 P2 . Then
Ä
Ä
kOTk
kOP1 k œ t Ê OT œ t OP1 so that T œ at x1 , t y1 , t z1 b.
Ä
Ä
TQk
Also, kkOP
œ s Ê TQ œ s OP2 œ s x2 , y2 , z2 ¡.
2k
Letting Q œ ax, y, zb, we have that
Ä
TQ œ x t x1 , y t y1 , z t z1 ¡ œ s x2 , y2 , z2 ¡
Thus x œ t x1 s x2 , y œ t y1 s y2 , z œ t z1 s z2 .
(Note that if Q is the midpoint, then qp œ 1 and t œ s œ
so that x œ
"
#
x1 "# x2 œ
x1 x2
2 ,
yœ
y1 y2
2 ,
zœ
z1 z2
2
"
#
so that this result agress with the midpoint formula.)
Ä
49. (a) the midpoint of AB is M ˆ 5# ß 5# ß 0‰ and CM œ ˆ 5# 1‰ i ˆ 5# 1‰ j (! 3)k œ
Ä
(b) the desired vector is ˆ 23 ‰ CM œ 23 ˆ 3# i 3# j 3k‰ œ i j 2k
3
#
i 3# j 3k
(c) the vector whose sum is the vector from the origin to C and the result of part (b) will terminate
at the center of mass Ê the terminal point of (i j 3k) (i j 2k) œ 2i 2j k is the point
(2ß 2ß 1), which is the location of the center of mass
Ä
50. The midpoint of AB is M ˆ 3# ß 0ß 5# ‰ and ˆ 32 ‰ CM œ
ˆ 3# 1‰ i (0 2)j ˆ 5# 1‰ k‘ œ 32 ˆ 5# i 2j 7# k‰
Ä
œ 53 i 43 j 73 k. The terminal point of ˆ 53 i 43 j 73 k‰ OC œ ˆ 53 i 43 j 73 k‰ (i 2j k)
œ 23 i 23 j 43 k is the point ˆ 23 ß 23 ß 43 ‰ which is the location of the intersection of the medians.
2
3
51. Without loss of generality we identify the vertices of the quadrilateral such that A(0ß 0ß 0), B(xb ß 0ß 0),
C(xc ß yc ß 0) and D(xd ß yd ß zd ) Ê the midpoint of AB is MAB ˆ x#b ß 0ß 0‰ , the midpoint of BC is
MBC ˆ xb # xc ß y#c ß 0‰ , the midpoint of CD is MCD ˆ xc # xd ß yc # yd ß z#d ‰ and the midpoint of AD is
xb
MAD ˆ x#d ß y#d ß z#d ‰ Ê the midpoint of MAB MCD is Œ #
xb xc
of MAD MBC œ Œ #
#
x
b #d
ß yc 4 yd ,
b
xc xd
#
#
ß yc 4 yd ,
zd
4
which is the same as the midpoint
zd
4 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
785
786
Chapter 12 Vectors and the Geometry of Space
52. Let V" , V# , V$ , á , Vn be the vertices of a regular n-sided polygon and vi denote the vector from the center to
n
Vi for i œ 1, 2, 3, á , n. If S œ ! vi and the polygon is rotated through an angle of
i(21)
n
iœ1
where
i œ 1, 2, 3, á , n, then S would remain the same. Since the vector S does not change with these rotations we conclude
that S œ 0.
53. Without loss of generality we can coordinatize the vertices of the triangle such that A(0ß 0), B(bß 0) and
Ä
C(xc ß yc ) Ê a is located at ˆ b # xc ß y#c ‰ , b is at ˆ x#c ß y#c ‰ and c is at ˆ b# ß 0‰ . Therefore, Aa œ ˆ b# x#c ‰ i ˆ y#c ‰ j ,
Ä
Ä
Ä
Ä
Ä
Bb œ ˆ x#c b‰ i ˆ y#c ‰ j , and Cc œ ˆ b# xc ‰ i (yc ) j Ê Aa Bb Cc œ 0 .
54. Let u be any unit vector in the plane. If u is positioned so that its initial point is at the origin and terminal point is at ax, yb,
then u makes an angle ) with i, measured in the counter-clockwise direction. Since kuk œ 1, we have that x œ cos ) and
y œ sin ). Thus u œ cos ) i sin ) j. Since u was assumed to be any unit vector in the plane, this holds for every unit
vector in the plane.
12.3 THE DOT PRODUCT
)
NOTE: In Exercises 1-8 below we calculate projv u as the vector Š kuk kcos
vk ‹ v , so the scalar multiplier of v is
the number in column 5 divided by the number in column 2.
v†u
kvk
kuk
cos )
kuk cos )
projv u
1. 25
5
5
1
5
2i 4j È 5k
2.
3
1
13
3
13
3
3 ˆ 35 i 45 k‰
3.
25
15
5
"
3
5
3
"
9
4.
13
15
3
13
45
13
15
13
225
5.
2
È34
È3
2
È3 È34
2
È34
"
17
6. È3 È2
È2
3
È3 È2
3È 2
È3 È2
È2
È3 È2
#
( i j )
7. 10 È17
È26
È21
10 È17
È546
10 È17
È26
10 È17
È26
(5i j)
È30
6
È30
6
"
5
"
È30
"
"
"
5 ¢ È# , È3 £
8.
"
6
(1)(2) (0)(1)
9. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È2#(2)(1)
1 # 0# È 1# 2# (
10. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È
11. ) œ cos" Š kuuk†kvvk ‹ œ cos"
Î
1)#
(5j 3k)
œ cos" Š È910È25 ‹ œ cos" ˆ 23 ‰ ¸ 0.84 rad
ŠÈ3‹ ŠÈ3‹ (7)(1) (0)(2)
Ï ÊŠÈ3‹
(2i 10j 11k)
‹ œ cos" Š È54È6 ‹ œ cos" Š È430 ‹ ¸ 0.75 rad
(2)(3) (2)(0) (1)(4)
‹
2# (2)# 1# È3# 0# 4#
#
(10i 11j 2k)
#
Ñ
(7)# 0# ÊŠÈ3‹ (1)# (2)# Ò
7
œ cos" Š È352È
‹
8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.3 The Dot Product
787
1
œ cos" Š È26
‹ ¸ 1.77 rad
12. ) œ cos" Š kuuk†kvvk ‹ œ cos"
Î
Ñ
(1)(1) ŠÈ2‹ (1) ŠÈ2‹ (1)
#
Ï Ê(1)# ŠÈ2‹
#
ŠÈ2‹ È(1)# (1)# (1)# Ò
1
œ cos" Š È
‹
5 È3
1
œ cos" Š È15
‹ ¸ 1.83 rad
Ä
Ä
Ä
Ä
Ä
Ä
13. AB œ 3, 1¡, BC œ 1, 3¡, and AC œ 2, 2¡. BA œ 3, 1¡, CB œ 1, 3¡, CA œ 2, 2¡.
Ä
Ä
Ä
Ä
Ä
Ä
¹ AB ¹ œ ¹ BA ¹ œ È10, ¹ BC ¹ œ ¹ CB ¹ œ È10, ¹ AC ¹ œ ¹ CA ¹ œ 2È2,
Ä Ä
AB†AC
Ä Ä œ cos"
Angle at A œ cos" Œ ¹AB
¹ ¹AC¹
3 a2 b 1 a 2 b
ŠÈ10‹Š2È2‹
œ cos" Š È15 ‹ ¸ 63.435‰
Ä Ä
BC†BA
a1ba3b a3ba1b
" 3
‰
Ä Ä œ cos"
Angle at B œ cos" Œ ¹BC
œ cos ˆ 5 ‰ ¸ 53.130 , and
¹ ¹BA¹
È
È
Š
Ä Ä
CB†CA
Ä Ä œ cos"
Angle at C œ cos" Œ ¹CB
¹ ¹CA¹
10‹Š
10‹
1 a 2 b 3 a 2 b
ŠÈ10‹Š2È2‹
œ cos" Š È15 ‹ ¸ 63.435‰
Ä
Ä Ä
Ä
14. AC œ 2, 4¡ and BD œ 4, 2¡. AC † BD œ 2a4b 4a2b œ 0, so the angle measures are all 90‰ .
15. (a) cos ! œ
i†v
kik kvk
œ
a
kvk
, cos " œ
j †v
kjk kvk
œ
b
k vk
, cos # œ
#
#
k†v
kkk kvk
#
œ
cos# ! cos# " cos# # œ Š kvak ‹ Š kbvk ‹ Š kvck ‹ œ
(b) kvk œ 1 Ê cos ! œ
a
kv k
œ a, cos " œ
b
kv k
c
kvk
and
a # b # c#
k v k k vk
œ b and cos # œ
c
kvk
œ
kv k k v k
kvk kvk
œ1
œ c are the direction cosines of v
16. u œ 10i 2k is parallel to the pipe in the north direction and v œ 10j k is parallel to the pipe in the east
direction. The angle between the two pipes is ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È1042È101 ‹ ¸ 1.55 rad ¸ 88.88°.
17. u œ ˆ vv††uv v‰ ˆu
v†u
v†v
v‰ œ
3
#
(i j) (3j 4k) 3# (i j)‘ œ ˆ 3# i 3# j‰ ˆ 3# i 3# j 4k‰ , where
v‰ œ
"
#
v ˆu "# v‰ œ
v † u œ 3 and v † v œ 2
18. u œ ˆ vv††uv v‰ ˆu
v†u
v†v
1
#
(i j) (j k) 1# (i j)‘ œ ˆ 1# i 1# j‰ ˆ 1# i 1# j k‰ ,
where v † u œ 1 and v † v œ 2
28
ˆ 14
19. u œ ˆ vv††uv v‰ ˆu vv††uv v‰ œ 14
3 (i 2j k) (8i 4j 12k) 3 i 3 j
28
14 ‰
16
22 ‰
ˆ 10
œ ˆ 14
3 i 3 j 3 k 3 i 3 j 3 k , where v † u œ 28 and v † v œ 6
20. u œ ˆ vv††uv v‰ ˆu
v†u
v†v
v‰ œ
"
1
14
3
k‰‘
(A) (i j k) ˆ 1" ‰ A‘ œ (i) (j k), where v † u œ 1 and v † v œ 1; yes
21. The sum of two vectors of equal length is always orthogonal to their difference, as we can see from the equation
( v" v# ) † ( v " v # ) œ v " † v " v # † v " v " † v # v # † v # œ k v " k # k v # k # œ 0
Ä Ä
22. CA † CB œ (v (u)) † (v u) œ v † v v † u u † v u † u œ kvk # kuk # œ 0 because kuk œ kvk since both equal
Ä
Ä
the radius of the circle. Therefore, CA and CB are orthogonal.
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788
Chapter 12 Vectors and the Geometry of Space
23. Let u and v be the sides of a rhombus Ê the diagonals are d" œ u v and d# œ u v
Ê d" † d# œ (u v) † (u v) œ u † u u † v v † u v † v œ kvk # kuk # œ 0 because kuk œ kvk , since a rhombus
has equal sides.
24. Suppose the diagonals of a rectangle are perpendicular, and let u and v be the sides of a rectangle Ê the diagonals are
d" œ u v and d# œ u v. Since the diagonals are perpendicular we have d" † d# œ 0
Í (u v) † (u v) œ u † u u † v v † u v † v œ 0 Í kvk # kuk # œ 0 Í akvk kukb akvk kukb œ 0
Í akvk kukb œ 0 which is not possible, or akvk kukb œ 0 which is equivalent to kvk œ kuk Ê the rectangle is a square.
25. Clearly the diagonals of a rectangle are equal in length. What is not as obvious is the statement that equal
diagonals happen only in a rectangle. We show this is true by letting the adjacent sides of a parallelogram be
the vectors (v" i v# j) and (u" i u# j). The equal diagonals of the parallelogram are
d" œ (v" i v# j) (u" i u# j) and d# œ (v" i v# j) (u" i u# j). Hence kd" k œ kd# k œ k(v" i v# j) (u" i u# j)k
œ k(v" i v# j) (u" i u# j)k Ê k(v" u" )i (v# u# )jk œ k(v" u" )i (v# u# )jk
Ê È(v" u" )# (v# u# )# œ È(v" u" )# (v# u# )# Ê v#1 2v" u" u1# v## 2v# u# u##
œ v#1 2v" u" u1# v## 2v# u# u## Ê 2(v" u" v# u# ) œ 2(v" u" v# u# ) Ê v" u" v# u# œ 0
Ê (v" i v# j) † (u" i u# j) œ 0 Ê the vectors (v" i v# j) and (u" i u# j) are perpendicular and the parallelogram
must be a rectangle.
26. If kuk œ kvk and u v is the indicated diagonal, then (u v) † u œ u † u v † u œ kuk # v † u œ u † v kvk #
œ u † v v † v œ (u v) † v Ê the angle cos" Š k(uuvvk )k†uuk ‹ between the diagonal and u and the angle
cos" Š k(uuvvk )k†vvk ‹ between the diagonal and v are equal because the inverse cosine function is one-to-one.
Therefore, the diagonal bisects the angle between u and v.
27. horizontal component: 1200 cosa8‰ b ¸ 1188 ft/s; vertical component: 1200 sina8‰ b ¸ 167 ft/s
28. kwkcosa33‰ 15‰ b œ 2.5 lb, so kwk œ
2.5 lb
cos 18‰ .
Then w œ
2.5 lb
cos 18‰
cos 33‰ , sin 33‰ ¡ ¸ 2.205, 1.432¡
29. (a) Since kcos )k Ÿ 1, we have ku † vk œ kuk kvk kcos )k Ÿ kuk kvk (1) œ kuk kvk .
(b) We have equality precisely when kcos )k œ 1 or when one or both of u and v is 0. In the case of nonzero
vectors, we have equality when ) œ 0 or 1, i.e., when the vectors are parallel.
30. (xi yj) † v œ kxi yjk kvk cos ) Ÿ 0 when
1
#
Ÿ ) Ÿ 1. This
means (xß y) has to be a point whose position vector makes
an angle with v that is a right angle or bigger.
31. v † u" œ (au" bu# ) † u" œ au" † u" bu# † u" œ a ku" k # b(u# † u" ) œ a(1)# b(0) œ a
32. No, v" need not equal v# . For example, i j Á i 2j but i † (i j) œ i † i i † j œ 1 0 œ 1 and
i † (i 2j) œ i † i 2i † j œ 1 2 † 0 œ 1.
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Section 12.3 The Dot Product
789
33. P(x" ß y" ) œ P ˆx" , bc ba x" ‰ and Q(x# ß y# ) œ Q ˆx# ß bc ba x# ‰ are any two points P and Q on the line with b Á 0
Ä
Ä
Ê PQ œ (x# x" )i ba (x" x# )j Ê PQ † v œ (x# x" )i ba (x" x# )j‘ † (ai bj) œ a(x# x" ) b ˆ ba ‰ (x" x# )
Ä
œ 0 Ê v is perpendicular to PQ for b Á 0. If b œ 0, then v œ ai is perpendicular to the vertical line ax œ c.
Alternatively, the slope of v is ba and the slope of the line ax by œ c is ba , so the slopes are negative
reciprocals Ê the vector v and the line are perpendicular.
34. The slope of v is
b
a
and the slope of bx ay œ c is ba , provided that a Á 0. If a œ 0, then v œ bj is parallel to
the vertical line bx œ c. In either case, the vector v is parallel to the line ax by œ c.
35. v œ i 2j is perpendicular to the line x 2y œ c;
P(2ß 1) on the line Ê 2 2 œ c Ê x 2y œ 4
36. v œ 2i j is perpendicular to the line 2x y œ c;
P(1ß 2) on the line Ê (2)(1) 2 œ c
Ê 2x y œ 0
37. v œ 2i j is perpendicular to the line 2x y œ c;
P(2ß 7) on the line Ê (2)(2) 7 œ c
Ê 2x y œ 3
38. v œ 2i 3j is perpendicular to the line 2x 3y œ c;
P(11ß 10) on the line Ê (2)(11) (3)(10) œ c
Ê 2x 3y œ 8
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Chapter 12 Vectors and the Geometry of Space
39. v œ i j is parallel to the line x y œ c;
P(2ß 1) on the line Ê a2b 1 œ c Ê x y œ 1
or x y œ 1.
40. v œ 2i 3j is parallel to the line 3x 2y œ c;
P(0ß 2) on the line Ê 0 2(2) œ c Ê 3x 2y œ 4
41. v œ i 2j is parallel to the line 2x y œ c;
P(1ß 2) on the line Ê 2(1) 2 œ c Ê 2x y œ 0
or 2x y œ 0.
42. v œ 3i 2j is parallel to the line 2x 3y œ c;
P(1ß 3) on the line Ê (2)(1) (3)(3) œ c
Ê 2x 3y œ 11 or 2x 3y œ 11
Ä
Ä
43. P(0ß 0), Q(1ß 1) and F œ 5j Ê PQ œ i j and W œ F † PQ œ (5j) † (i j) œ 5 N † m œ 5 J
44. W œ kFk (distance) cos ) œ (602,148 N)(605 km)(cos 0) œ 364,299,540 N † km œ (364,299,540)(1000) N † m
œ 3.6429954 ‚ 10"" J
Ä
45. W œ kFk ¹ PQ ¹ cos ) œ (200)(20)(cos 30°) œ 2000È3 œ 3464.10 N † m œ 3464.10 J
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.3 The Dot Product
Ä
46. W œ kFk ¹ PQ ¹ cos ) œ (1000)(5280)(cos 60°) œ 2,640,000 ft † lb
In Exercises 47-52 we use the fact that n œ ai bj is normal to the line ax by œ c.
1
47. n" œ 3i j and n# œ 2i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È610È
‹ œ cos" Š È"2 ‹ œ
5
1
4
3 1
" ˆ
48. n" œ È3i j and n# œ È3i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
"2 ‰ œ
È ‹ œ cos
4
49. n" œ È3i j and n# œ i È3j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š
21
3
4
È3 È3
È4 È4 ‹
œ cos" Š
È3
2 ‹
1
6
œ
50. n" œ i È3j and n# œ Š1 È3‹ i Š1 È3‹ j Ê ) œ cos" Š knn""k†knn## k ‹
œ cos"
1 È3 È3 3
È 1 3 É 1 2È 3 3 1 2È 3 3
4
œ cos" Š 2È
‹ œ cos" Š È"2 ‹ œ
8
1
4
4
7
51. n" œ 3i 4j and n# œ i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È3 È
‹ œ cos" Š 5È
‹ ¸ 0.14 rad
25 2
2
10
"
52. n" œ 12i 5j and n# œ 2i 2j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È24169
Š
È8 ‹ œ cos
14
‹
26È2
¸ 1.18 rad
53. The angle between the corresponding normals is equal to the angle between the corresponding tangents. The
points of intersection are Š
y
3
4
œ f w Š
È3
# ‹ Šx
È3 3
# ß 4‹
Š
f(x) œ ˆ 3# ‰ x# is y
3
4
È3
# ‹‹
œ f w Š
and Š
È3 3
# ß 4‹ .
At Š
Ê y œ È 3 Šx
È3
# ‹ Šx
Š
È3
# ‹‹
È3 3
# ß 4‹
È3
# ‹
y œ È 3 Šx
È3
# ‹
3
4
œ È3x
9
4
21
3 ,
the angle is
1
3
3
4
Ê y œ È3x 34 , and the tangent line for
Ê y œ È 3 Šx
normals are n" œ È3i j and n# œ È3i j. The angle at Š
œ cos" Š È43È14 ‹ œ cos" ˆ "# ‰ œ
the tangent line for f(x) œ x# is
and
21
3 .
È3 3
# ß 4‹
At Š
and the tangent line for f(x) œ
È3
# ‹
3
4
œ È3x 94 . The corresponding
is ) œ cos" Š knn""k†knn## k ‹
È3 3
# ß 4‹
3
#
the tangent line for f(x) œ x# is
x# is y œ È3 Šx
È3
# ‹
œ È3x 34 . The corresponding normals are n" œ È3i j and n# œ È3i j. The angle at Š
) œ cos" Š knn""k†knn## k ‹ œ cos" Š È3È1 ‹ œ cos" ˆ "# ‰ œ
4 4
54. The points of intersection are Š!ß
tangent line at Š!ß
curve x œ y#
3
4
È3
# ‹
is y
È3
#
È3
# ‹
È3
# ‹.
"
#y
œ
the angle is
The curve x œ
œ È"3 (x 0) Ê n" œ
dy
dx
has derivative
and Š!ß
21
3 ,
"
È3
3
4
1
3
and
3
4
È3 3
# ß 4‹
is
21
3 .
y# has derivative
dy
dx
œ #"y Ê the
i j is normal to the curve at that point. The
Ê the tangent line at Š!ß
È3
# ‹
is y
È3
#
œ
"
È3
(x 0)
Ê n# œ È"3 i j is normal to the curve. The angle between the curves is ) œ cos" Š knn""k†knn## k ‹
œ cos"
"3 1
É 3"
1
É "3
"
œ cos
1
ˆ2‰
Š ˆ 34 ‰ ‹ œ cos" ˆ "# ‰ œ
3
1
3
and
21
3 .
Because of symmetry the angles between
the curves at the two points of intersection are the same.
$
55. The curves intersect when y œ x$ œ ay# b œ y' Ê y œ 0 or y œ 1. The points of intersection are (!ß !) and
("ß "). Note that y 0 since y œ y' . At (!ß 0) the tangent line for y œ x$ is y œ 0 and the tangent line for
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
791
792
Chapter 12 Vectors and the Geometry of Space
y œ Èx is x œ 0. Therefore, the angle of intersection at (0ß 0) is 1# . At (1ß 1) the tangent line for y œ x$ is
y œ 3x 2 and the tangent line for y œ Èx is y œ "# x "# . The corresponding normal vectors are
n" œ 3i j and n# œ "# i j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È" ‹ œ 14 , the angle is
2
1
4
and
31
4 .
56. The points of intersection for the curves y œ x# and y œ $Èx are (!ß 0) and (1ß 1). At (!ß 0) the tangent
line for y œ x# is y œ 0 and the tangent line for y œ $Èx is x œ 0. Therefore, the angle of intersection at (!ß 0)
is 1# . At (1ß 1) the tangent line for y œ x# is y œ 2x 1 and the tangent line for y œ $Èx is y œ
The corresponding normal vectors are n" œ 2i j and n# œ
œ cos" È
2
3
5
"
É 9"
"
œ cos
1
ˆ5‰
"
Œ È5 3È10 œ cos
3
"
3
i j Ê ) œ cos" Š knn""k†knn## k ‹
Š È"2 ‹ œ 14 , the angle is
1
4
and
31
4 .
12.4 THE CROSS PRODUCT
â
âi
â
1. u ‚ v œ â 2
â
â"
â
j
k â
â
2 " â œ 3 ˆ 23 i "3 j 32 k‰ Ê length œ 3 and the direction is 32 i 3" j 32 k;
â
0 " â
v ‚ u œ (u ‚ v) œ 3 ˆ 23 i "3 j 32 k‰ Ê length œ 3 and the direction is 32 i 3" j 32 k
â
â i
â
2. u ‚ v œ â 2
â
â "
â
j kâ
â
3 0 â œ 5(k) Ê length œ 5 and the direction is k
â
1 0â
v ‚ u œ (u ‚ v) œ 5(k) Ê length œ 5 and the direction is k
â
â i
â
3. u ‚ v œ â 2
â
â "
â
j
k â
â
2 4 â œ 0 Ê length œ 0 and has no direction
â
1 2 â
v ‚ u œ (u ‚ v) œ 0 Ê length œ 0 and has no direction
â
âi
â
4. u ‚ v œ â 1
â
â0
j
1
0
â
k â
â
1 â œ 0 Ê length œ 0 and has no direction
â
0 â
v ‚ u œ (u ‚ v) œ 0 Ê length œ 0 and has no direction
â
âi
â
5. u ‚ v œ â 2
â
â0
j
0
3
â
kâ
â
0 â œ 6(k) Ê length œ 6 and the direction is k
â
0â
v ‚ u œ (u ‚ v) œ 6(k) Ê length œ 6 and the direction is k
â
âi
â
6. u ‚ v œ (i ‚ j) ‚ (j ‚ k) œ k ‚ i œ â 0
â
â1
â
j kâ
â
0 1 â œ j Ê length œ 1 and the direction is j
â
0 0â
v ‚ u œ (u ‚ v) œ j Ê length œ 1 and the direction is j
â
â
j
k â
â i
â
â
7. u ‚ v œ â 8 2 4 â œ 6i 12k Ê length œ 6È5 and the direction is È"5 i È25 k
â
â
2
1 â
â 2
v ‚ u œ (u ‚ v) œ (6i 12k) Ê length œ 6È5 and the direction is " i 2 k
È5
È5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
3
x 23 .
Section 12.4 The Cross Product
âi
â
â
8. u ‚ v œ â 3#
â
â1
k ââ
â
1 â œ 2i 2j 2k Ê length œ 2È3 and the direction is È"3 i È13 j È"3 k
â
2â
v ‚ u œ (u ‚ v) œ (2i 2j 2k) Ê length œ 2È3 and the direction is " i 1 j " k
j
"#
1
È3
â
âi
â
9. u ‚ v œ â 1
â
â0
j
0
1
â
kâ
â
0⠜ k
â
0â
â
âi
â
10. u ‚ v œ â 1
â
â0
â
âi
â
11. u ‚ v œ â 1
â
â0
j
0
1
â
k â
â
1 ⠜ i j k
â
1 â
â
j
âi
â
12. u ‚ v œ â 2 1
â
â1 2
â
âi
â
13. u ‚ v œ â 1
â
â1
j
1
1
â
kâ
â
0 ⠜ 2k
â
0â
â
âi j
â
14. u ‚ v œ â 0 1
â
â1 0
â
â
j k â
â i
Ä
Ä
â
â
15. (a) PQ ‚ PR œ â 1 1 3 â œ 8i 4j 4k Ê Area œ
â
â
â 1 3 1 â
(b) u œ „
Ä Ä
PQ‚PR
Ä Ä
¹PQ‚PR¹
œ „
"
È6
j
0
1
"
#
Ä
Ä
¹ PQ ‚ PR ¹ œ
È3
È3
â
k â
â
1 ⠜ i k
â
0 â
â
kâ
â
0 ⠜ 5k
â
0â
â
kâ
â
2 ⠜ 2j k
â
0â
"
#
È64 16 16 œ 2È6
(2i j k)
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793
794
Chapter 12 Vectors and the Geometry of Space
â
â
j kâ
âi
Ä
Ä
â
â
16. (a) PQ ‚ PR œ â 1 0 2 â œ 4i 4j 2k Ê Area œ
â
â
â 2 2 0 â
(b) u œ „
Ä Ä
PQ‚PR
Ä Ä
¹PQ‚PR¹
Ä Ä
PQ‚PR
Ä Ä
¹PQ‚PR¹
œ „
"
È2
( i j ) œ „
"
È2
"
#
Ä
Ä
¹ PQ ‚ PR ¹ œ
Ä Ä
PQ‚PR
Ä Ä
¹PQ‚PR¹
"
#
È16 16 4 œ 3
œ „
"
È14
"
#
È1 1 œ
È2
#
(i j )
â
â
j
k â
âi
Ä
Ä
â
â
18. (a) PQ ‚ PR œ â 2 1 1 â œ 2i 3j k Ê Area œ
â
â
â 1 0 2 â
(b) u œ „
Ä
Ä
¹ PQ ‚ PR ¹ œ
œ „ 3" (2i 2j k)
â
â
â i j kâ
Ä
Ä
â
â
17. (a) PQ ‚ PR œ â 1 1 1 â œ i j Ê Area œ
â
â
â1 1 0â
(b) u œ „
"
#
"
#
Ä
Ä
¹ PQ ‚ PR ¹ œ
"
#
È4 9 1 œ
È14
#
(2i 3j k)
â
â
â a" a# a$ â
â
â
19. If u œ a" i a# j a$ k, v œ b" i b# j b$ k, and w œ c" i c# j c$ k, then u † (v ‚ w) œ â b" b# b$ â ,
â
â
â c" c# c$ â
â
â
â
â
â b" b # b $ â
â c" c# c$ â
â
â
â
â
v † (w ‚ u) œ â c" c# c$ â and w † (u ‚ v) œ â a" a# a$ â which all have the same value, since the
â
â
â
â
â a" a# a$ â
â b" b# b $ â
interchanging of two pair of rows in a determinant does not change its value Ê the volume is
â
â
â2 0 0â
â
â
k(u ‚ v) † wk œ abs â 0 2 0 â œ 8
â
â
â0 0 2â
â
â 1
â
20. k(u ‚ v) † wk œ abs â 2
â
â 1
â
1 1 â
â
1 2 ⠜ 4 (for details about verification, see Exercise 19)
â
2 1 â
â
â2 1
â
21. k(u ‚ v) † wk œ abs â 2 1
â
â1 0
â
0â
â
1 â œ k7k œ 7 (for details about verification, see Exercise 19)
â
2â
â
â
â 1 1 2 â
â
â
22. k(u ‚ v) † wk œ abs â 1 0 1 â œ 8 (for details about verification, see Exercise 19)
â
â
â 2 4 2 â
23. (a) u † v œ 6, u † w œ 81, v † w œ 18 Ê none
â
â
â
â
â
â
j
k â
j
k â
j k â
âi
â i
â i
â
â
â
â
â
â
1 1 â œ 0, v ‚ w œ â 0
1 5 â Á 0
(b) u ‚ v œ â 5 1 1 â Á 0, u ‚ w œ â 5
â
â
â
â
â
â
â 0 1 5 â
â 15 3 3 â
â 15 3 3 â
Ê u and w are parallel
24. (a) u † v œ 0, u ‚ w œ 0, u † r œ 31, v † w œ 0, v † r œ 0, w † r œ 0 Ê u ¼ v, u ¼ w, v ¼ w, v ¼ r
and w ¼ r
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.4 The Cross Product
795
â
â
â
â
â
â
j
k â
j k â
â i
âi j k â
â i
â
â
â
â
â
â
2 1 ⠜ 0
(b) u ‚ v œ â 1 2 1 â Á 0, u ‚ w œ â 1 2 1 â Á 0, u ‚ r œ â 1
â
â
â
â
â 1 1 1 â
â 1 1 1 â
â1 0 1 â
â #
# â
â
â
â
â
â
â
j
kâ
j
kâ
j kâ
â i
â i
â i
â
â
â
â
â
â
1
1 â Á 0, w ‚ r œ â 1
0
1âÁ0
v ‚ w œ â 1 1 1 â Á 0, v ‚ r œ â 1
â
â
â 1 1 1 â
â 1 1 1 â
â 1 0 1â
â #
â #
# â
# â
Ê u and r are parallel
Ä
Ä
25. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (60°) œ
2
3
Ä
Ä
26. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (135°) œ
† 30 †
2
3
È3
#
† 30 †
ft † lb œ 10È3 ft † lb
È2
#
ft † lb œ 10È2 ft † lb
27. (a) true, kuk œ Èa#1 a## a3# œ Èu † u
(b) not always true, u † u œ kuk #
â
â
â
â
j kâ
j kâ
â i
â i
â
â
â
â
(c) true, u ‚ 0 œ â a" a# a$ â œ 0i 0j 0k œ 0 and 0 ‚ u œ â 0 0 0 â œ 0i 0j 0k œ 0
â
â
â
â
â0 0 0â
â a" a# a$ â
â
â
j
k â
â i
â
â
a#
a$ â œ (a# a$ a# a$ )i (a" a$ a" a$ )j (a" a# a" a# )k œ 0
(d) true, u ‚ (u) œ â a"
â
â
â a" a# a$ â
(e)
(f)
(g)
(h)
not always true, i ‚ j œ k Á k œ j ‚ i for example
true, distributive property of the cross product
true, (u ‚ v) † v œ u † (v ‚ v) œ u † 0 œ 0
true, the volume of a parallelpiped with u, v, and w along the three edges is (u ‚ v) † w œ (v ‚ w) † u œ u † (v ‚ w),
since the dot product is commutative.
28. (a) true, u † v œ a" b" a# b# a$ b$ œ b" a" b# a# b$ a$ œ v † u
â
â
â
â
j kâ
j kâ
â i
â i
â
â
â
â
(b) true, u ‚ v œ â a" a# a$ â œ â b" b# b$ â œ (v ‚ u)
â
â
â
â
â b" b# b$ â
â a" a# a$ â
â
â
â
â
j
k â
j kâ
â i
â i
â
â
â
â
(c) true, (u) ‚ v œ â a" a# a$ â œ â a" a# a$ â œ (u ‚ v)
â
â
â
â
b#
b$ â
â b"
â b" b # b $ â
(d) true, (cu) † v œ (ca" )b" (ca# )b# (ca$ )b$ œ a" (cb" ) a# (cb# ) a$ (cb$ ) œ u † (cv) œ c(a" b" a# b# a$ b$ )
œ c(u † v)
â
â
â
â
â
â
j kâ
j
k â
j
k â
â i
â i
â i
â
â
â
â
â
â
a#
a$ â œ u ‚ (cv)
(e) true, c(u ‚ v) œ c â a" a# a$ â œ â ca" ca# ca$ â œ (cu) ‚ v œ â a"
â
â
â
â
â
â
â b" b# b$ â
â b" b# b$ â
â cb" cb# cb$ â
#
(f) true, u † u œ a#1 a## a3# œ ˆÈa#1 a## a3# ‰ œ kuk #
(g) true, (u ‚ u) † u œ 0 † u œ 0
(h) true, u ‚ v ¼ u and u ‚ v ¼ v Ê (u ‚ v) † u œ v † (u ‚ v) œ 0
29. (a) projv u œ Š kvukk†vvk ‹ v
(b) „ (u ‚ v)
(c) „ a(u ‚ v) ‚ wb
(d) k(u ‚ v) † wk
30. (a) (u ‚ v) ‚ (u ‚ w)
(b) (u v) ‚ (u v) œ (u v) ‚ u (u v) ‚ v œ u ‚ u v ‚ u u ‚ v v ‚ v
œ 0 v ‚ u u ‚ v 0 œ 2(v ‚ u), or simply u ‚ v
(c) kuk kvvk
(d) ku ‚ wk
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
796
Chapter 12 Vectors and the Geometry of Space
31. (a) yes, u ‚ v and w are both vectors
(c) yes, u and u ‚ w are both vectors
(b) no, u is a vector but v † w is a scalar
(d) no, u is a vector but v † w is a scalar
32. (u ‚ v) ‚ w is perpendicular to u ‚ v, and u ‚ v is perpendicular to both u and v Ê (u ‚ v) ‚ w is
parallel to a vector in the plane of u and v which means it lies in the plane determined by u and v.
The situation is degenerate if u and v are parallel so u ‚ v œ 0 and the vectors do not determine a plane.
Similar reasoning shows that u ‚ (v ‚ w) lies in the plane of v and w provided v and w are nonparallel.
33. No, v need not equal w. For example, i j Á i j, but i ‚ (i j) œ i ‚ i i ‚ j œ 0 k œ k and
i ‚ ( i j ) œ i ‚ i i ‚ j œ 0 k œ k.
34. Yes. If u ‚ v œ u ‚ w and u † v œ u † w, then u ‚ (v w) œ 0 and u † (v w) œ 0. Suppose now that v Á w.
Then u ‚ (v w) œ 0 implies that v w œ ku for some real number k Á 0. This in turn implies that
u † (v w) œ u † (ku) œ k kuk # œ 0, which implies that u œ 0. Since u Á 0, it cannot be true that v Á w, so
v œ w.
â
j
â i
Ä
Ä
Ä
Ä
â
35. AB œ i j and AD œ i j Ê AB ‚ AD œ â 1 1
â
â 1 1
â
âi
Ä
Ä
Ä
Ä
â
36. AB œ 7i 3j and AD œ 2i 5j Ê AB ‚ AD œ â 7
â
â2
â
âi
Ä
Ä
Ä
Ä
â
37. AB œ 3i 2j and AD œ 5i j Ê AB ‚ AD œ â 3
â
â5
â
kâ
Ä
Ä
â
0 â œ 2k Ê area œ ¹ AB ‚ AD ¹ œ 2
â
0â
â
j kâ
Ä
Ä
â
3 0 â œ 29k Ê area œ ¹ AB ‚ AD ¹ œ 29
â
5 0â
â
kâ
Ä
Ä
â
0 â œ 13k Ê area œ ¹ AB ‚ AD ¹ œ 13
â
0â
j
2
1
â
âi
Ä
Ä
Ä
Ä
â
38. AB œ 7i 4j and AD œ 2i 5j Ê AB ‚ AD œ â 7
â
â2
j
4
5
â
â i
Ä
Ä
Ä
Ä
â
39. AB œ 2i 3j and AC œ 3i j Ê AB ‚ AC œ â 2
â
â 3
â
âi
Ä
Ä
Ä
Ä
â
40. AB œ 4i 4j and AC œ 3i 2j Ê AB ‚ AC œ â 4
â
â3
j
3
1
â
kâ
Ä
Ä
â
0 â œ 43k Ê area œ ¹ AB ‚ AD ¹ œ 43
â
0â
â
kâ
â
0 â œ 11k Ê area œ
â
0â
â
j kâ
â
4 0 â œ 4k Ê area œ
â
2 0â
"
#
"
#
Ä
Ä
¹ AB ‚ AC ¹ œ
11
#
Ä
Ä
¹ AB ‚ AC ¹ œ 2
â
â i
Ä
Ä
Ä
Ä
â
41. AB œ 6i 5j and AC œ 11i 5j Ê AB ‚ AC œ â 6
â
â 11
j
5
5
â
kâ
â
0 â œ 25k Ê area œ
â
0â
"
#
Ä
Ä
¹ AB ‚ AC ¹ œ
â
â i
Ä
Ä
Ä
Ä
â
42. AB œ 16i 5j and AC œ 4i 4j Ê AB ‚ AC œ â 16
â
â 4
j
5
4
â
kâ
â
0 â œ 84k Ê area œ
â
0â
"
#
Ä
Ä
¹ AB ‚ AC ¹ œ 42
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
25
#
Section 12.5 Lines and Planes in Space
â
â i
â
43. If A œ a" i a# j and B œ b" i b# j, then A ‚ B œ â a"
â
â b"
"
#
kA ‚ Bk œ „
"
#
a"
ºb
"
j
a#
b#
â
kâ
a
â
0â œ º "
b"
â
0â
a#
k and the triangle's area is
b# º
a#
. The applicable sign is () if the acute angle from A to B runs counterclockwise
b# º
in the xy-plane, and () if it runs clockwise, because the area must be a nonnegative number.
Ä
Ä
44. If A œ a" i a# j, B œ b" i b# j, and C œ c" i c# j, then the area of the triangle is "# ¹ AB ‚ AC ¹ . Now,
â
â
i
j
kâ
â
Ä
Ä
Ä
Ä
b a"
b# a#
â
â
k Ê "# ¹ AB ‚ AC ¹
AB ‚ AC œ â b" a" b# a# 0 â œ º "
c" a"
c# a# º
â
â
â c" a" c# a# 0 â
k(b" a" )(c# a# ) (c" a" )(b# a# )k œ "# ka" (b# c# ) a# (c" b" ) (b" c# c" b# )k
â
â
â a" a# 1 â
â
" â
œ „ # â b" b# 1 â . The applicable sign ensures the area formula gives a nonnegative number.
â
â
â c" c# 1 â
œ
"
#
12.5 LINES AND PLANES IN SPACE
1. The direction i j k and P(3ß 4ß 1) Ê x œ 3 t, y œ 4 t, z œ 1 t
Ä
2. The direction PQ œ 2i 2j 2k and P(1ß 2ß 1) Ê x œ 1 2t, y œ 2 2t, z œ 1 2t
Ä
3. The direction PQ œ 5i 5j 5k and P(2ß 0ß 3) Ê x œ 2 5t, y œ 5t, z œ 3 5t
Ä
4. The direction PQ œ j k and P(1ß 2ß 0) Ê x œ 1, y œ 2 t, z œ t
5. The direction 2j k and P(!ß !ß !) Ê x œ 0, y œ 2t, z œ t
6. The direction 2i j 3k and P(3ß 2ß 1) Ê x œ 3 2t, y œ 2 t, z œ 1 3t
7. The direction k and P(1ß 1ß 1) Ê x œ 1, y œ 1, z œ 1 t
8. The direction 3i 7j 5k and P(2ß 4ß 5) Ê x œ 2 3t, y œ 4 7t, z œ 5 5t
9. The direction i 2j 2k and P(0ß 7ß 0) Ê x œ t, y œ 7 2t, z œ 2t
â
â
â i j kâ
â
â
10. The direction is A ‚ B œ â 1 2 3 â œ 2i 4j 2k and P(#ß $ß 0) Ê x œ 2 2t, y œ 3 4t, z œ 2t
â
â
â3 4 5â
11. The direction i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0
12. The direction k and P(0ß 0ß 0) Ê x œ 0, y œ 0, z œ t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
797
798
Chapter 12 Vectors and the Geometry of Space
Ä
13. The direction PQ œ i j 3# k and P(0ß 0ß 0) Ê x œ t,
y œ t, z œ
3
#
t, where 0 Ÿ t Ÿ 1
Ä
14. The direction PQ œ i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0,
where 0 Ÿ t Ÿ 1
Ä
15. The direction PQ œ j and P(1ß 1ß 0) Ê x œ 1, y œ 1 t,
z œ 0, where 1 Ÿ t Ÿ 0
Ä
16. The direction PQ œ k and P(1ß 1ß 0) Ê x œ 1, y œ 1, z œ t,
where 0 Ÿ t Ÿ 1
Ä
17. The direction PQ œ 2j and P(0ß 1ß 1) Ê x œ 0,
y œ 1 2t, z œ 1, where 0 Ÿ t Ÿ 1
Ä
18. The direction PQ œ $i 2j and P(0ß 2ß 0) Ê x œ 3t,
y œ 2 2t, z œ 0, where 0 Ÿ t Ÿ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.5 Lines and Planes in Space
799
Ä
19. The direction PQ œ 2i 2j 2k and P(2ß 0ß 2)
Ê x œ 2 2t, y œ 2t, z œ 2 2t, where 0 Ÿ t Ÿ 1
Ä
20. The direction PQ œ i 3j k and P(1ß 0ß 1)
Ê x œ 1 t, y œ 3t, z œ 1 t, where 0 Ÿ t Ÿ 1
21. 3(x 0) (2)(y 2) (1)(z 1) œ 0 Ê 3x 2y z œ 3
22. 3(x 1) (1)(y 1) (1)(z 3) œ 0 Ê 3x y z œ 5
â
â
j kâ
â i
Ä
Ä
Ä
Ä
â
â
23. PQ œ i j 3k, PS œ i 3j 2k Ê PQ ‚ PS œ â 1 1 3 â œ 7i 5j 4k is normal to the plane
â
â
â 1 3 2 â
Ê 7(x 2) (5)(y 0) (4)(z 2) œ 0 Ê 7x 5y 4z œ 6
â
â i
Ä
Ä
Ä
Ä
â
24. PQ œ i j 2k, PS œ 3i 2j 3k Ê PQ ‚ PS œ â 1
â
â 3
j
1
2
â
kâ
â
2 ⠜ i 3j k is normal to the plane
â
3â
Ê (1)(x 1) (3)(y 5) (1)(z 7) œ 0 Ê x 3y z œ 9
25. n œ i 3j 4k, P(2ß 4ß 5) œ (1)(x 2) (3)(y 4) (4)(z 5) œ 0 Ê x 3y 4z œ 34
26. n œ i 2j k, P(1ß 2ß 1) œ (1)(x 1) (2)(y 2) (1)(z 1) œ 0 Ê x 2y z œ 6
x œ 2t 1 œ s 2
2t s œ 1
4t 2s œ 2
Ê œ
Ê œ
Ê t œ 0 and s œ 1; then z œ 4t 3 œ 4s 1
27. œ
y œ 3t 2 œ 2s 4
3t 2s œ 2
3t 2s œ 2
Ê 4(0) 3 œ (4)(1) 1 is satisfied Ê the lines do intersect when t œ 0 and s œ 1 Ê the point of
intersection is x œ 1, y œ 2, and z œ 3 or P(1ß 2ß 3). A vector normal to the plane determined by these lines is
â
â
âi j k â
â
â
n" ‚ n# œ â 2 3 4 â œ 20i 12j k, where n" and n# are directions of the lines Ê the plane
â
â
â 1 2 4 â
containing the lines is represented by(20)(x 1) (12)(y 2) (1)(z 3) œ 0 Ê 20x 12y z œ 7.
xœ t
œ 2s 2
t 2s œ 2
Ê œ
Ê s œ 1 and t œ 0; then z œ t 1 œ 5s 6 Ê 0 1 œ 5(1) 6
28. œ
y œ t 2 œ s 3
t s œ 1
is satisfied Ê the lines do intersect when s œ 1 and t œ 0 Ê the point of intersection is x œ 0, y œ 2 and z œ 1
â
â
j kâ
âi
â
â
or P(0ß 2ß 1). A vector normal to the plane determined by these lines is n" ‚ n# œ â 1 " 1 â
â
â
â2 1 5â
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
800
Chapter 12 Vectors and the Geometry of Space
œ 6i 3j 3k, where n" and n# are directions of the lines Ê the plane containing the lines is represented by
(6)(x 0) (3)(y 2) (3)(z 1) œ 0 Ê 6x 3y 3z œ 3.
29. The cross product of i j k and 4i 2j 2k has the same direction as the normal to the plane
â
â
j k â
â i
â
â
Ê n œ â 1 " 1 â œ 6j 6k. Select a point on either line, such as P(1ß 2ß 1). Since the lines are given
â
â
â 4 2 2 â
to intersect, the desired plane is 0(x 1) 6(y 2) 6(z 1) œ 0 Ê 6y 6z œ 18 Ê y z œ 3.
30. The cross product of i 3j k and i j k has the same direction as the normal to the plane
â
â
j
k â
âi
â
â
n œ â 1 3 1 â œ 2i 2j 4k. Select a point on either line, such as P(0ß 3ß 2). Since the lines are
â
â
1 â
â1 1
given to intersect, the desired plane is (2)(x 0) (2)(y 3) (4)(z 2) œ 0 Ê 2x 2y 4z œ 14
Ê x y 2z œ 7.
â
âi j
â
31. n" ‚ n# œ â 2 1
â
â1 2
â
k â
â
1 ⠜ 3i 3j 3k is a vector in the direction of the line of intersection of the planes
â
1 â
Ê 3(x 2) (3)(y 1) 3(z 1) œ 0 Ê 3x 3y 3z œ 0 Ê x y z œ 0 is the desired plane containing
P! (2ß 1ß 1)
â
â
j
k â
âi
Ä
â
â
32. A vector normal to the desired plane is P" P# ‚ n œ â 2 0 2 â œ 2i 12j 2k; choosing P" (1ß 2ß 3) as a
â
â
â 4 1 2 â
point on the plane Ê (2)(x 1) (12)(y 2) (2)(z 3) œ 0 Ê 2x 12y 2z œ 32 Ê x 6y z œ 16
is the desired plane
â
âi
Ä
â
33. S(0ß 0ß 12), P(0ß 0ß 0) and v œ 4i 2j 2k Ê PS ‚ v œ â 0
â
â4
Ê dœ
Ä
¹PS‚v¹
kv k
œ
24È1 4
È16 4 4
œ
24È5
È24
â
kâ
â
12 â œ 24i 48j œ 24(i 2j)
â
2 â
j
0
2
œ È5 † 24 œ 2È30 is the distance from S to the line
â
j
â i
Ä
â
34. S(0ß 0ß 0), P(5ß 5ß 3) and v œ 3i 4j 5k Ê PS ‚ v œ â 5 5
â
4
â 3
Ê dœ
Ä
¹PS‚v¹
kv k
œ
È169 256 25
È9 16 25
œ
È450
È50
â
k â
â
3 ⠜ 13i 16j 5k
â
5 â
œ È9 œ 3 is the distance from S to the line
Ä
35. S(2ß 1ß 3), P(2ß 1ß 3) and v œ 2i 6j Ê PS ‚ v œ 0 Ê d œ
Ä
¹PS‚v¹
kv k
œ
0
È40
œ 0 is the distance from S to the line
(i.e., the point S lies on the line)
â
âi
Ä
â
ß
ß
ß
ß
œ
Ê
‚
œ
36. S(2 1 1), P(0 1 0) and v 2i 2j 2k
PS v â 2
â
â2
Ê dœ
Ä
¹PS‚v¹
kv k
œ
È4 36 16
È4 4 4
œ
È56
È12
j
0
2
â
k â
â
1 ⠜ 2 i 6 j 4 k
â
2 â
œ É 14
3 is the distance from S to the line
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 12.5 Lines and Planes in Space
â
j
â i
Ä
â
37. S(3ß 1ß 4), P(4ß 3ß 5) and v œ i 2j 3k Ê PS ‚ v œ â 1 4
â
â 1 2
Ê dœ
Ä
¹PS‚v¹
kv k
œ
È900 36 36
È1 4 9
œ
È972
È14
œ
È486
È7
œ
È81†6
È7
œ
9È42
7
â
â i
Ä
â
38. S(1ß 4ß 3), P(10ß 3ß 0) and v œ 4i 4k Ê PS ‚ v œ â 11
â
â 4
Ê dœ
Ä
¹PS‚v¹
kv k
œ
28È1 4 1
4È 1 1
â
kâ
â
9 ⠜ 30i 6j 6k
â
3â
is the distance from S to the line
j
7
0
â
kâ
â
3 â œ 28i 56j 28k œ 28(i 2j k)
â
4â
œ 7È3 is the distance from S to the line
Ä
39. S(2ß 3ß 4), x 2y 2z œ 13 and P(13ß 0ß 0) is on the plane Ê PS œ 11i 3j 4k and n œ i 2j 2k
Ä
9
Ê d œ ¹ PS † knnk ¹ œ ¹ È111 4648 ¹ œ ¹ È
¹œ3
9
Ä
40. S(0ß 0ß 0), 3x 2y 6z œ 6 and P(2ß 0ß 0) is on the plane Ê PS œ 2i and n œ 3i 2j 6k
Ä
Ê d œ ¹ PS † knnk ¹ œ ¹ È9 46 36 ¹ œ È649 œ 67
Ä
41. S(0ß 1ß 1), 4y 3z œ 12 and P(0ß 3ß 0) is on the plane Ê PS œ 4j k and n œ 4j 3k
Ä
Ê d œ ¹ PS † knnk ¹ œ ¹ È161639 ¹ œ 19
5
Ä
42. S(2ß 2ß 3), 2x y 2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2j 3k and n œ 2i j 2k
Ä
Ê d œ ¹ PS † knnk ¹ œ ¹ È4216 4 ¹ œ 83
Ä
43. S(0ß 1ß 0), 2x y 2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2i j and n œ 2i j 2k
Ä
4 1 0
Ê d œ ¹ PS † knnk ¹ œ ¹ È
¹ œ 53
414
Ä
44. S(1ß 0ß 1), 4x y z œ 4 and P(1ß 0ß 0) is on the plane Ê PS œ 2i k and n œ 4i j k
Ä
È
Ê d œ ¹ PS † knnk ¹ œ ¹ È16811 1 ¹ œ È918 œ 3 # 2
Ä
45. The point P(1ß 0ß 0) is on the first plane and S(10ß 0ß 0) is a point on the second plane Ê PS œ 9i, and
Ä
n œ i 2j 6k is normal to the first plane Ê the distance from S to the first plane is d œ ¹ PS † knnk ¹
œ ¹ È1 94 36 ¹ œ
9
È41
, which is also the distance between the planes.
46. The line is parallel to the plane since v † n œ ˆi j "# k‰ † (i 2j 6k) œ 1 2 3 œ 0. Also the point
Ä
S(1ß 0ß 0) when t œ 1 lies on the line, and the point P(10ß 0ß 0) lies on the plane Ê PS œ 9i. The distance
Ä
from S to the plane is d œ ¹ PS † knnk ¹ œ ¹ È1 49 36 ¹ œ È9 , which is also the distance from the line to the
41
plane.
47. n" œ i j and n# œ 2i j 2k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È22È1 9 ‹ œ cos" Š È"2 ‹ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
4
801
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Chapter 12 Vectors and the Geometry of Space
523
48. n" œ 5i j k and n# œ i 2j 3k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
‹ œ cos" (0) œ
27 È14
1
#
442
49. n" œ 2i 2j 2k and n# œ 2i 2j k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
‹ œ cos" Š 3"
È3 ‹ ¸ 1.76 rad
12 È9
50. n" œ i j k and n# œ k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È 1È ‹ ¸ 0.96 rad
3 1
5
51. n" œ 2i 2j k and n# œ i 2j k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š 2È94È61 ‹ œ cos" Š 3È
‹ ¸ 0.82 rad
6
18
26 ‰
52. n" œ 4j 3k and n# œ 3i 2j 6k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È825È
¸ 0.73 rad
‹ œ cos" ˆ 35
49
53. 2x y 3z œ 6 Ê 2(1 t) (3t) 3(1 t) œ 6 Ê 2t 5 œ 6 Ê t œ "# Ê x œ 3# , y œ 3# and z œ
Ê ˆ 3# ß 3# ß "# ‰ is the point
"
#
54. 6x 3y 4z œ 12 Ê 6(2) 3(3 2t) 4(2 2t) œ 12 Ê 14t 29 œ 12 Ê t œ 41
14 Ê x œ 2, y œ 3
41
20
27
and z œ 2 7 Ê ˆ2ß 7 ß 7 ‰ is the point
55. x y z œ 2 Ê (1 2t) (1 5t) (3t) œ 2 Ê 10t 2 œ 2 Ê t œ 0 Ê x œ 1, y œ 1 and z œ 0
Ê (1ß 1ß 0) is the point
56. 2x 3z œ 7 Ê 2(1 3t) 3(5t) œ 7 Ê 9t 2 œ 7 Ê t œ 1 Ê x œ 1 3, y œ 2 and z œ 5
Ê (4ß 2ß 5) is the point
â
â
â i j kâ
â
â
57. n" œ i j k and n# œ i j Ê n" ‚ n# œ â 1 1 1 â œ i j, the direction of the desired line; (1ß 1ß 1)
â
â
â1 1 0â
is on both planes Ê the desired line is x œ 1 t, y œ 1 t, z œ 1
â
âi
â
58. n" œ 3i 6j 2k and n# œ 2i j 2k Ê n" ‚ n# œ â 3
â
â2
â
j
k â
â
6 2 ⠜ 14i 2j 15k, the direction of the
â
1 2 â
desired line; (1ß 0ß 0) is on both planes Ê the desired line is x œ 1 14t, y œ 2t, z œ 15t
â
âi
â
59. n" œ i 2j 4k and n# œ i j 2k Ê n" ‚ n# œ â 1
â
â1
j
2
1
â
k â
â
4 ⠜ 6j 3k, the direction of the
â
2 â
desired line; (4ß 3ß 1) is on both planes Ê the desired line is x œ 4, y œ 3 6t, z œ 1 3t
â
â
j
k â
âi
â
â
60. n" œ 5i 2j and n# œ 4j 5k Ê n" ‚ n# œ â 5 2 0 â œ 10i 25j 20k, the direction of the
â
â
â 0 4 5 â
desired line; (1ß 3ß 1) is on both planes Ê the desired line is x œ 1 10t, y œ 3 25t, z œ 1 20t
2t 4s œ 2
2t 4s œ 2
Ê œ
61. L1 & L2: x œ 3 2t œ 1 4s and y œ 1 4t œ 1 2s Ê œ
4t 2s œ 2
2t s œ 1
Ê 3s œ 3 Ê s œ 1 and t œ 1 Ê on L1, z œ 1 and on L2, z œ 1 Ê L1 and L2 intersect at (5ß 3ß 1).
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41
7 ,
Section 12.5 Lines and Planes in Space
L2 & L3: The direction of L2 is
"
3
"
6
(4i 2j 4k) œ
"
3
803
(2i j 2k) which is the same as the direction
(2i j 2k) of L3; hence L2 and L3 are parallel.
2t 2r œ 0
t rœ0
Ê œ
Ê 3t œ 3
L1 & L3: x œ 3 2t œ 3 2r and y œ 1 4t œ 2 r Ê œ
4t r œ 3
4t r œ 3
Ê t œ 1 and r œ 1 Ê on L1, z œ 2 while on L3, z œ 0 Ê L1 and L2 do not intersect. The direction of L1
is È"21 (2i 4j k) while the direction of L3 is 3" (2i j 2k) and neither is a multiple of the other; hence
L1 and L3 are skew.
2t s œ 1
Ê 5s œ 3 Ê s œ 35 and t œ
62. L1 & L2: x œ 1 2t œ 2 s and y œ 1 t œ 3s Ê œ
t 3s œ 1
zœ
12
5
while on L2, z œ 1
while the direction of L2 is
3
5
œ
"
È11
2
5
Ê L1 and L2 do not intersect. The direction of L1 is
"
È14
4
5
Ê on L1,
(2i j 3k)
(i 3j k) and neither is a multiple of the other; hence, L1 and L2 are
skew.
s 2r œ 3
L2 & L3: x œ 2 s œ 5 2r and y œ 3s œ 1 r Ê œ
Ê 5s œ 5 Ê s œ 1 and r œ 2 Ê on L2,
3s r œ 1
z œ 2 and on L3, z œ 2 Ê L2 and L3 intersect at (1ß 3ß 2).
L1 & L3: L1 and L3 have the same direction È"14 (2i j 3k); hence L1 and L3 are parallel.
63. x œ 2 2t, y œ 4 t, z œ 7 3t; x œ 2 t, y œ 2 "# t, z œ 1 3# t
64. 1(x 4) 2(y 1) 1(z 5) œ 0 Ê x 4 2y 2 z 5 œ 0 Ê x 2y z œ 7;
È2 (x 3) 2È2 (y 2) È2 (z 0) œ 0 Ê È2x 2È2y È2z œ 7È2
65. x œ 0 Ê t œ "# , y œ "# , z œ 3# Ê ˆ!ß "# ß 3# ‰ ; y œ 0 Ê t œ 1, x œ 1, z œ 3 Ê (1ß 0ß 3); z œ 0
Ê t œ 0, x œ 1, y œ 1 Ê (1ß 1ß 0)
66. The line contains (0ß 0ß 3) and ŠÈ3ß 1ß 3‹ because the projection of the line onto the xy-plane contains the origin
and intersects the positive x-axis at a 30° angle. The direction of the line is È3i j 0k Ê the line in question
is x œ È3t, y œ t, z œ 3.
67. With substitution of the line into the plane we have 2(1 2t) (2 5t) (3t) œ 8 Ê 2 4t 2 5t 3t œ 8
Ê 4t 4 œ 8 Ê t œ 1 Ê the point (1ß 7ß 3) is contained in both the line and plane, so they are not parallel.
68. The planes are parallel when either vector A" i B" j C" k or A# i B# j C# k is a multiple of the other or
when k(A" i B" j C" k) ‚ (A# i B# j C# kk œ 0. The planes are perpendicular when their normals are
perpendicular, or(A" i B" j C" k) † (A# i B# j C# k) œ 0.
69. There are many possible answers. One is found as follows: eliminate t to get t œ x 1 œ 2 y œ
Ê x 1 œ 2 y and 2 y œ
z3
#
z3
#
Ê x y œ 3 and 2y z œ 7 are two such planes.
70. Since the plane passes through the origin, its general equation is of the form Ax By Cz œ 0. Since it meets
the plane M at a right angle, their normal vectors are perpendicular Ê 2A 3B C œ 0. One choice satisfying
this equation is A œ 1, B œ 1 and C œ 1 Ê x y z œ 0. Any plane Ax By Cz œ 0 with 2A 3B C œ 0
will pass through the origin and be perpendicular to M.
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804
Chapter 12 Vectors and the Geometry of Space
71. The points (aß 0ß 0), (0ß bß 0) and (0ß 0ß c) are the x, y, and z intercepts of the plane. Since a, b, and c are all
nonzero, the plane must intersect all three coordinate axes and cannot pass through the origin. Thus,
y
x
z
a b c œ 1 describes all planes except those through the origin or parallel to a coordinate axis.
72. Yes. If v" and v# are nonzero vectors parallel to the lines, then v" ‚ v# Á 0 is perpendicular to the lines.
Ä
Ä
73. (a) EP œ cEP" Ê x! i yj zk œ c c(x" x! )i y" j z" kd Ê x! œ c(x" x! ), y œ cy" and z œ cz" ,
where c is a positive real number
x !
(b) At x" œ 0 Ê c œ 1 Ê y œ y" and z œ z" ; at x" œ x! Ê x! œ 0, y œ 0, z œ 0; x lim
c œ x lim
Ä_
Ä _ x" x!
œ x lim
Ä_
!
"
1
!
!
œ 1 Ê c Ä 1 so that y Ä y" and z Ä z"
74. The plane which contains the triangular plane is x y z œ 2. The line containing the endpoints of the line
segment is x œ 1 t, y œ 2t, z œ 2t. The plane and the line intersect at ˆ 23 ß 23 ß 23 ‰ . The visible section of the line
#
#
#
segment is Ɉ "3 ‰ ˆ 23 ‰ ˆ 23 ‰ œ 1 unit in length. The length of the line segment is È1# 2# 2# œ 3 Ê
the line segment is hidden from view.
12.6 CYLINDERS AND QUADRIC SURFACES
1. d, ellipsoid
2. i, hyperboloid
3. a, cylinder
4. g, cone
5. l, hyperbolic paraboloid
6. e, paraboloid
7. b, cylinder
8. j, hyperboloid
9. k, hyperbolic paraboloid
10. f, paraboloid
11. h, cone
12. c, ellipsoid
13. x# y# œ 4
14. x# z# œ 4
15. z œ y# 1
16. x œ y#
17. x# 4z# œ 16
18. 4x# y# œ 36
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3
of
Section 12.6 Cylinders and Quadric Surfaces
19. z# y# œ 1
20. yz œ 1
21. 9x# y# z# œ 9
22. 4x# 4y# z# œ 16
23. 4x# 9y# 4z# œ 36
24. 9x# 4y# 36z# œ 36
25. x# 4y# œ z
26. z œ x# 9y#
27. z œ 8 x# y#
28. z œ 18 x# 9y#
29. x œ 4 4y# z#
30. y œ 1 x# z#
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805
806
Chapter 12 Vectors and the Geometry of Space
31. x# y# œ z#
32. y# z# œ x#
33. 4x# 9z# œ 9y#
34. 9x# 4y# œ 36z#
35. x# y# z# œ 1
36. y# z# x# œ 1
y#
4
37.
y#
4
z#
9
40.
y#
4
x#
4
z# œ 1
x#
4
œ1
z#
9
œ1
39. z# x# y# œ 1
41. x# y#
z#
4
œ1
42.
38.
x#
4
x#
4
y#
z#
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ1
Section 12.6 Cylinders and Quadric Surfaces
43. y# x# œ z
44. x# œ y# œ z
45. x# y# z# œ 4
46. 4x# 4y# œ z#
47. z œ 1 y# x#
48. y# z# œ 4
49. y œ ax# z# b
50. z# 4x# 4y# œ 4
51. 16x# 4y# œ 1
52. z œ x# y# 1
53. x# y# z# œ 4
54. x œ 4 y#
55. x# z# œ y
56. z#
x#
4
57. x# z# œ 1
y# œ 1
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807
808
Chapter 12 Vectors and the Geometry of Space
58. 4x# 4y# z# œ 4
59. 16y# 9z# œ 4x#
60. z œ x# y# 1
61. 9x# 4y# z# œ 36
62. 4x# 9z# œ y#
63. x# y# 16z# œ 16
64. z# 4y# œ 9
65. z œ ax# y# b
66. y# x# z# œ 1
67. x# 4y# œ 1
68. z œ 4x# y# 4
69. 4y# z# 4x# œ 4
70. z œ 1 x#
71. x# y# œ z
72.
x#
4
y# z# œ 1
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Section 12.6 Cylinders and Quadric Surfaces
74. 36x# 9y# 4z# œ 36
73. yz œ 1
75. 9x# 16y# œ 4z#
76. 4z# x# y# œ 4
77. (a) If x#
œ 1Š
y#
4
z#
9
œ 1 and z œ c, then x#
È 9 c#
È
#
‹ Š 2 93 c ‹
3
œ
x#
a#
(c)
Ê
x#
#
Š 9 9 c ‹
4 ˆ9 c# ‰
“
9
2 1 a9 z # b
9
œ 1 Ê A œ ab1
, where 3 Ÿ z Ÿ 3. Thus V œ 2 '0
3
!
y#
b#
z#
c#
x#
œ1 Ê
–
Ê V œ 2 '0
then V œ
41 r
3
$
1ab
c#
a# Šc# z# ‹
c#
y#
—
ac# z# b dz œ
21
9
a9 z# b dz
–
21ab
c#
b# Šc# z# ‹
c#
’c# z
œ 1 Ê A œ 1 Ša
È c# z#
È #
#
‹ Š b cc z ‹
c
—
c
z$
3 “!
œ
21ab
c#
ˆ 23 c$ ‰ œ
41abc
3
. Note that if r œ a œ b œ c,
, which is the volume of a sphere.
78. The ellipsoid has the form
of the barrel. Thus,
#
r
R#
x#
R#
#
V œ 1 'ch y# dz. Now,
h
y
R#
y#
R#
z#
c#
#
œ 1. To determine c# we note that the point (0ß rß h) lies on the surface
œ1 Ê c œ
h
c#
#
#
z
c#
h# R #
R# r#
. We calculate the volume by the disk method:
œ 1 Ê y# œ R# Š1
z#
c# ‹
œ R # ’1
Ê V œ 1 'ch ’R# Š R h# r ‹ z# “ dz œ 1 ’R# z "3 Š R h# r ‹ z$ “
h
4
3
y#
’
$
c
œ
'03 a9 z# b dz œ 491 ’9z z3 “ $ œ 491 (27 9) œ 81
41
9
9 c #
9
œ
2 1 a9 c # b
9
(b) From part (a), each slice has the area
œ
y#
4
#
#
#
#
h
ch
z# aR# r# b
“
h# R #
#
#
œ R# Š R h# r ‹ z#
#
œ 21 R# h "3 aR# r# b h‘ œ 21 Š 2R3 h
1R# h 23 1r# h, the volume of the barrel. If r œ R, then V œ 21R# h which is the volume of a cylinder of
radius R and height 2h. If r œ 0 and h œ R, then V œ
4
3
1R$ which is the volume of a sphere.
79. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z,
gives the ellipse
x#
#
Š zac ‹
y#
#
Š zbc ‹
œ 1. The area of this ellipse is 1 ˆaÈ cz ‰ ˆbÈ cz ‰ œ
the volume is given by V œ '0
h
Aœ
r# h
3 ‹
1abh
c ,
1abz
c
#
h
dz œ ’ 1abz
2c “ œ
as determined previously. Thus, V œ
!
1abh#
c
1abh#
c .
œ
"
#
1abz
c
x#
a#
y#
b#
œ
(see Exercise 77a). Hence
Now the area of the elliptic base when z œ h is
ˆ 1abh
‰h œ
c
"
#
z
c
(base)(altitude), as claimed.
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809
810
Chapter 12 Vectors and the Geometry of Space
80. (a) For each fixed value of z, the hyperboloid
x
–
#
a# Šc# z# ‹
c#
y
—
–
#
b# Šc# z# ‹
c#
h
h
1abh
3
Š2 1
(c) Am œ A ˆ h# ‰ œ
œ
œ
h#
c# ‹
1ab
c#
h
1ab
#
6 1ab c# a4c
1abh
#
#
3c# a3c h b œ
81. y œ y" Ê
z
c
œ
y#1
b#
œ
1ab
#
c # Šc
x#
a#
y#
b#
z#
c#
œ 1 results in a cross-sectional ellipse
œ 1. The area of the cross-sectional ellipse (see Exercise 77a) is
ac# z# b dz œ
1ab
c#
(b) A! œ A(0) œ 1ab and Ah œ A(h) œ
œ
—
A(z) œ 1 Š ac Èc# z# ‹ Š bc Èc# z# ‹ œ
V œ '0 A(z) dz œ '0
x#
a#
1abh
3
#
h
4
Š2
‹œ
h# b
1ab
4c#
1ab
c#
1ab
c#
ac# z# b . The volume of the solid by the method of slices is
1ab
c#
#
c# z "3 z$ ‘ h œ
!
c# h#
c# ‹
œ
h
3
21ab
a4c# h# b Ê
ac# h# b‘ œ
1abh
6c#
h
6
1ab
c#
ac# h# b‘ œ
1abh
3c#
h
3
1abh
3c#
a3c# h# b
a3c# h# b
(2A! Ah )
(A! 4Am Ah )
ac# 4c# h# c# h# b œ
1abh
6c#
a6c# 2h# b
V from part (a)
, a parabola in the plane y œ y" Ê vertex when
#
#
ˆc# h 3" h$ ‰ œ
ac h# b , from part (a) Ê V œ
#
a
#
1
Ê Vertex Š0ß y" ß cy
b# ‹ ; writing the parabola as x œ c z
1
Ê Focus Š0ß y" ß cy
b#
1ab
c#
a# y#1
b#
dz
dx
œ 0 or c
dz
dx
œ 2x
a# œ 0 Ê x œ 0
#
#
a
we see that 4p œ ac Ê p œ 4c
a#
4c ‹
82. The curve has the general form Ax# By# Dxy Gx Hy K œ 0 which is the same form as Eq. (1) in
Section 10.3 for a conic section (including the degenerate cases) in the xy-plane.
83. No, it is not mere coincidence. A plane parallel to one of the coordinate planes will set one of the variables
x, y, or z equal to a constant in the general equation Ax# By# Cz# Dxy Eyz Fxz Gx Hy Jz K
œ 0 for a quadric surface. The resulting equation then has the general form for a conic in that parallel plane.
For example, setting y œ y" results in the equation Ax# Cz# Dw x Ew z Fxz Gx Jz Kw œ 0 where
Dw œ Dy" , Ew œ Ey" , and Kw œ K By#1 Hy" , which is the general form of a conic section in the plane y œ y"
by Section 10.3.
84. The trace will be a conic section. To see why, solve the plane's equation Ax By Cz œ 0 for one of the
variables in terms of the other two and substitute into the equation Ax# By# Cz# á K œ 0. The result
will be a second degree equation in the remaining two variables. By Section 10.3, this equation will represent a
conic section. (See also the discussion in Exercises 82 and 83.)
85. z œ y#
86. z œ 1 y#
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Section 12.6 Cylinders and Quadric Surfaces
811
87. z œ x# y#
88. z œ x# 2y#
(a)
(c)
(b)
(d)
89-94. Example CAS commands:
Maple:
with( plots );
eq := x^2/9 + y^2/36 = 1 - z^2/25;
implicitplot3d( eq, x=-3..3, y=-6..6, z=-5..5, scaling=constrained,
shading=zhue, axes=boxed, title="#89 (Section 12.6)" );
Mathematica: (functions and domains may vary):
In the following chapter, you will consider contours or level curves for surfaces in three dimensions. For the purposes of
plotting the functions of two variables expressed implicitly in this section, we will call upon the function ContourPlot3D.
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812
Chapter 12 Vectors and the Geometry of Space
To insert the stated function, write all terms on the same side of the equal sign and the default contour equating that
expression to zero will be plotted.
This built-in function requires the loading of a special graphics package.
< with( plots );
r := t -> [sin(t)-t*cos(t),cos(t)+t*sin(t),t^2];
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
831
832
Chapter 13 Vector-Valued Functions and Motion in Space
t0 := 3*Pi/2;
lo := 0;
hi := 6*Pi;
P1 := spacecurve( r(t), t=lo..hi, axes=boxed, thickness=3 ):
display( P1, title="#58(a) (Section 13.1)" );
Dr := unapply( diff(r(t),t), t );
# (b)
Dr(t0);
# (c)
q1 := expand( r(t0) + Dr(t0)*(t-t0) );
T := unapply( q1, t );
P2 := spacecurve( T(t), t=lo..hi, axes=boxed, thickness=3, color=black ):
display( [P1,P2], title="#58(d) (Section 13.1)" );
62-63. Example CAS commands:
Maple:
a := 'a'; b := 'b';
r := (a,b,t) -> [cos(a*t),sin(a*t),b*t];
Dr := unapply( diff(r(a,b,t),t), (a,b,t) );
t0 := 3*Pi/2;
q1 := expand( r(a,b,t0) + Dr(a,b,t0)*(t-t0) );
T := unapply( q1, (a,b,t) );
lo := 0;
hi := 4*Pi;
P := NULL:
for a in [ 1, 2, 4, 6 ] do
P1 := spacecurve( r(a,1,t), t=lo..hi, thickness=3 ):
P2 := spacecurve( T(a,1,t), t=lo..hi, thickness=3, color=black ):
P := P, display( [P1,P2], axes=boxed, title=sprintf("#62 (Section 13.1)\n a=%a",a) );
end do:
display( [P], insequence=true );
58-63. Example CAS commands:
Mathematica: (assigned functions, parameters, and intervals will vary)
The x-y-z components for the curve are entered as a list of functions of t. The unit vectors i, j, k are not inserted.
If a graph is too small, highlight it and drag out a corner or side to make it larger.
Only the components of r[t] and values for t0, tmin, and tmax require alteration for each problem.
Clear[r, v, t, x, y, z]
r[t_]={ Sin[t] t Cos[t], Cos[t] t Sin[t], t2}
t0= 31 / 2; tmin= 0; tmax= 61;
ParametricPlot3D[Evaluate[r[t]], {t, tmin, tmax}, AxesLabel Ä {x, y, z}];
v[t_]= r'[t]
tanline[t_]= v[t0] t r[t0]
ParametricPlot3D[Evaluate[{r[t], tanline[t]}], {t, tmin, tmax}, AxesLabel Ä {x, y, z}];
For 62 and 63, the curve can be defined as a function of t, a, and b. Leave a space between a and t and b and t.
Clear[r, v, t, x, y, z, a, b]
r[t_,a_,b_]:={Cos[a t], Sin[a t], b t}
t0= 31 / 2; tmin= 0; tmax= 41;
v[t_,a_,b_]= D[r[t, a, b], t]
tanline[t_,a_,b_]=v[t0, a, b] t r[t0, a, b]
pa1=ParametricPlot3D[Evaluate[{r[t, 1, 1], tanline[t, 1, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.2 Modeling Projectile Motion
833
pa2=ParametricPlot3D[Evaluate[{r[t, 2, 1], tanline[t, 2, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
pa4=ParametricPlot3D[Evaluate[{r[t, 4, 1], tanline[t, 4, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
pa6=ParametricPlot3D[Evaluate[{r[t, 6, 1], tanline[t, 6, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];
Show[GraphicsArray[{pa1, pa2, pa4, pa6}]]
13.2 MODELING PROJECTILE MOTION
m‰
1. x œ (v! cos !)t Ê (21 km)ˆ 1000
œ (840 m/s)(cos 60°)t Ê t œ
1 km
v#!
g
2. R œ
21,000 m
(840 m/s)(cos 60°)
œ 50 seconds
v#
!
sin 2! and maximum R occurs when ! œ 45° Ê 24.5 km œ Š 9.8 m/s
# ‹ (sin 90°)
Ê v! œ È(9.8)(24,500) m# /s# œ 490 m/s
(b)
(c)
v!#
g
(500 m/s)#
9.8 m/s# (sin 90°) ¸ 25,510.2 m
5000 m
x œ (v! cos !)t Ê 5000 m œ (500 m/s)(cos 45°)t Ê t œ (500 m/s)(cos
45°) ¸ 14.14 s; thus,
"
"
#
y œ (v! sin !)t # gt Ê y ¸ (500 m/s)(sin 45°)(14.14 s) # a9.8 m/s# b (14.14 s)# ¸ 4020 m
!)#
45°))#
ymax œ (v! sin
œ ((5002 m/s)(sin
¸ 6378 m
2g
a9.8 m/s# b
3. (a) t œ
2v! sin !
g
œ
2(500 m/s)(sin 45°)
9.8 m/s#
¸ 72.2 seconds; R œ
sin 2! œ
4. y œ y! (v! sin !)t "# gt# Ê y œ 32 ft (32 ft/sec)(sin 30°)t "# a32 ft/sec# b t# Ê y œ 32 16t 16t# ;
the ball hits the ground when y œ 0 Ê 0 œ 32 16t 16t# Ê t œ 1 or t œ 2 Ê t œ 2 sec since t 0; thus,
x œ (v! cos !) t Ê x œ (32 ft/sec)(cos 30°)t œ 32 Š
È3
# ‹ (2)
¸ 55.4 ft
5. x œ x! (v! cos !)t œ 0 (44 cos 45°)t œ 22È2t and y œ y! (v! sin !)t "# gt# œ 6.5 (44 sin 45°)t 16t#
œ 6.5 22È2t 16t# ; the shot lands when y œ 0 Ê t œ
22È2 „ È968 416
3#
¸ 2.135 sec since t 0; thus
x œ 22È2t ¸ Š22È2‹ (2.135) ¸ 66.43 ft
6. x œ 0 (44 cos 40°)t ¸ 33.706t and y œ 6.5 (44 sin 40°)t 16t# ¸ 6.5 28.283t 16t# ; y œ 0
Ê t¸
28.283 È(28.283)# 416
3#
¸ 1.9735 sec since t 0; thus x ¸ (33.706)(1.9735) ¸ 66.52 ft Ê the
difference in distances is about 66.52 66.43 œ 0.09 ft or about 1 inch
7. (a) R œ
v#!
g
(b) 6m ¸
v#
#
# #
!
sin 2! Ê 10 m œ Š 9.8 m/s
Ê v! ¸ 9.9 m/s;
# ‹ (sin 90°) Ê v! œ 98 m s
(9.9 m/s)#
9.8 m/s#
(sin 2!) Ê sin 2! ¸ 0.59999 Ê 2! ¸ 36.87° or 143.12° Ê ! ¸ 18.4° or 71.6°
8. v! œ 5 ‚ 10' m/s and x œ 40 cm œ 0.4 m; thus x œ (v! cos !)t Ê 0.4m œ a5 ‚ 10' m/sb (cos 0°)t
Ê t œ 0.08 ‚ 10' s œ 8 ‚ 10) s; also, y œ y! (v! sin !)t "# gt#
#
Ê y œ a5 ‚ 10' m/sb (sin 0°) a8 ‚ 10) sb "# a9.8 m/s# b a8 ‚ 10) sb œ 3.136 ‚ 10"% m or
3.136 ‚ 10"# cm. Therefore, it drops 3.136 ‚ 10"# cm.
9. R œ
10. v! œ
v#!
g
v#
#
#
#
!
sin 2! Ê 3(248.8) ft œ Š 32 ft/sec
Ê v! ¸ 278.02 ft/sec ¸ 190 mph
# ‹ (sin 18°) Ê v! ¸ 77,292.84 ft /sec
80È10
3
ft/sec and R œ 200 ft Ê 200 œ
È
#
Š 80 3 10 ‹
2! ¸ 115.8° Ê ! ¸ 57.9°; If ! ¸ 32.1‰ , ymax
(sin 2!) Ê sin 2! œ 0.9 Ê 2! ¸ 64.2° Ê ! ¸ 32.1°; or
#
È
’Š 80 3 10 ‹ (sin 32.1°)“
œ
¸ 31.4 ft. If ! ¸ 57.9‰ , ymax ¸ 79.7 ft 75 ft. In
2(32)
32
order to reach the cushion, the angle of elevation will need to be about 32.1°. At this angle, the circus performer will go
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
834
Chapter 13 Vector-Valued Functions and Motion in Space
31.4 ft into the air at maximum height and will not strike the 75 ft high ceiling.
11. x œ (v! cos !)t Ê 135 ft œ (90 ft/sec)(cos 30°)t Ê t ¸ 1.732 sec; y œ (v! sin !)t "# gt#
Ê y ¸ (90 ft/sec)(sin 30°)(1.732 sec) "# a32 ft/sec# b (1.732 sec)# Ê y ¸ 29.94 ft Ê the golf ball will clip
the leaves at the top
12. v! œ 116 ft/sec, ! œ 45°, and x œ (v! cos !)t
Ê 369 œ (116 cos 45°)t Ê t ¸ 4.50 sec;
also y œ (v! sin !)t "# gt#
Ê y œ (116 sin 45°)(4.50) "# (32)(4.50)#
¸ 45.11 ft. It will take the ball 4.50 sec to travel
369 ft. At that time the ball will be 45.11 ft in
the air and will hit the green past the pin.
13. We do part b first.
"
315
#
(b) x œ (v! cos !)t Ê 315 ft œ (v! cos 20°)t Ê v! œ t cos
20° ; also y œ (v! sin !)t # gt
315 ‰
"
#
#
#
#
Ê 34 ft œ ˆ t cos
20° (t sin 20°) # (32)t Ê 34 œ 315 tan 20° 16t Ê t ¸ 5.04 sec Ê t ¸ 2.25 sec
(a) v! œ
315
(2.25)(cos 20°)
v#!
g
¸ 149 ft/sec
14. R œ
v#!
g
sin 2! œ
15. R œ
v#!
g
sin 2! Ê 16,000 m œ
(2 sin ! cos !) œ
v#!
g
(400 m/s)#
9.8 m/s#
[2 cos (90° !) sin (90° !)] œ
v#!
g
[sin 2(90° !)]
sin 2! Ê sin 2! œ 0.98 Ê 2! ¸ 78.5° or 2! ¸ 101.5° Ê ! ¸ 39.3°
or 50.7°
16. (a) R œ
(2v! )#
g
sin 2! œ
4v#!
g
v#
sin 2! œ 4 Š g! sin !‹ or 4 times the original range.
(b) Now, let the initial range be R œ
Ê
(pv! )#
g
v#!
g
sin 2!. Then we want the factor p so that pv! will double the range
v#!
sin 2! œ 2 Š g sin 2!‹ Ê p# œ 2 Ê p œ È2 or about 141%. The same percentage will approximately
double the height:
apv0 sin !b2
2g
œ
2av0 sin !b2
2g
Ê p# œ 2 Ê p œ È2.
17. x œ x! (v! cos !)t œ 0 (v! cos 40°)t ¸ 0.766 v! t and y œ y! (v! sin !)t "# gt# œ 6.5 (v! sin 40°)t 16t#
¸ 6.5 0.643 v! t 16t# ; now the shot went 73.833 ft Ê 73.833 œ 0.766 v! t Ê t ¸
#
when y œ 0 Ê 0 œ 6.5 (0.643)(96.383) 16 Š 96.383
v! ‹ Ê 0 ¸ 68.474
148,635
v#!
96.383
v!
sec; the shot lands
Ê v! ¸ É 148,635
68.474
¸ 46.6 ft/sec, the shot's initial speed
18. ymax œ
(v! sin !)#
2g
Ê
3
4
ymax œ
3(v! sin !)#
and
8g
# #
y œ (v! sin !)t "# gt# Ê
3(v! sin !)#
8g
œ (v! sin !)t "# gt#
Ê 3(v! sin !)# œ (8gv! sin !)t 4g t Ê 4g# t# (8gv! sin !)t 3(v! sin !)# œ 0 Ê 2gt 3v! sin ! œ 0 or
sin !
!
2gt v! sin ! œ 0 Ê t œ 3v!2gsin ! or t œ v! 2g
. Since the time it takes to reach ymax is tmax œ v! sin
,
g
then the time it takes the projectile to reach
3
4
of ymax is the shorter time t œ
v! sin !
2g
or half the time it takes
to reach the maximum height.
19.
dr
dt
œ ' (gj) dt œ gtj C" and
dr
dt
(0) œ (v! cos !)i (v! sin !)j Ê g(0)j C" œ (v! cos !)i (v! sin !)j
Ê C" œ (v! cos !)i (v! sin !)j Ê ddtr œ (v! cos !)i (v! sin ! gt)j ; r œ ' [(v! cos !)i (v! sin ! gt)j] dt
œ (v! t cos !)i ˆv! t sin ! "# gt# ‰ j C# and r(0) œ x! i y! j Ê [v! (0) cos !]i v! (0) sin ! "# g(0)# ‘ j C#
œ x! i y! j Ê C# œ x! i y! j Ê r œ (x! v! t cos !)i ˆy! v! t sin ! "# gt# ‰ j Ê x œ x! v! t cos ! and
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.2 Modeling Projectile Motion
y œ y! v! t sin ! "# gt#
20. From Example 3(b) in the text, v! sin ! œ È(68)(64) Ê v! sin 56.5° ¸ 65.97 Ê v! ¸ 79 ft/sec
21. The horizontal distance from Rebollo to the center of the cauldron is 90 ft Ê the horizontal distance to the
nearest rim is x œ 90 "# (12) œ 84 Ê 84 œ x! (v! cos !)t ¸ 0 Š v!90g
sin ! ‹ t Ê 84 œ
"
#
(90)(32)
È(68)(64)
t
#
Ê t œ 1.92 sec. The vertical distance at this time is y œ y! (v! sin !)t gt
¸ 6 È(68)(64) (1.92) 16(1.92)# ¸ 73.7 ft Ê the arrow clears the rim by 3.7 ft
22. The projectile rises straight up and then falls straight down, returning to the firing point.
23. Flight time œ 1 sec and the measure of the angle of elevation is about 64° (using a protractor) so that
tœ
Rœ
2v! sin !
g
v#!
g
Ê 1œ
Ê v! ¸ 17.80 ft/sec. Then ymax œ
2v! sin 64°
32
sin 2! Ê R œ
#
(17.80)
32
(17.80 sin 64°)#
2(32)
¸ 4.00 ft and
sin 128° ¸ 7.80 ft Ê the engine traveled about 7.80 ft in 1 sec Ê the engine
velocity was about 7.80 ft/sec
24. When marble A is located R units downrange, we have x œ (v! cos !)t Ê R œ (v! cos !)t Ê t œ
R
v! cos !
#
. At
"
R
R
that time the height of marble A is y œ y! (v! sin !)t "# gt# œ (v! sin !) Š v! cos
! ‹ # g Š v! cos ! ‹
#
R
Ê y œ R tan ! "# g Š v# cos
# ! ‹ . The height of marble B at the same time t œ
!
R
v! cos !
seconds is
#
R
h œ R tan ! "# gt# œ R tan ! "# g Š v# cos
# ! ‹ . Since the heights are the same, the marbles collide regardless
!
of the initial velocity v! .
25. (a) At the time t when the projectile hits the line OR we
have tan " œ yx ; x œ [v! cos (! " )]t and
y œ [v! sin (! " )]t "# gt# 0 since R is
below level ground. Therefore let
kyk œ "# gt# [v! sin (! " )]t 0
"# gt# (v! sin (! " ))t‘
" gt v sin (! " )‘
œ # v! cos! (! ")
[v! cos (! " )]t
v! cos (! " ) tan " œ "# gt v! sin (! " )
t œ 2v! sin (! ") 2vg ! cos (! ") tan " , which is the time
so that tan " œ
Ê
Ê
when the projectile hits the downhill slope. Therefore,
x œ [v! cos (! " )] ’ 2v! sin (! ") 2vg ! cos (! ") tan " “ œ
maximized, then OR is maximized:
dx
d!
œ
2v#!
g
2v#!
g
ccos# (! " ) tan " sin (! " ) cos (! " )d . If x is
[ sin 2(! " ) tan " cos 2(! " )] œ 0
Ê sin 2(! " ) tan " cos 2(! " ) œ 0 Ê tan " œ cot 2(! " ) Ê 2(! " ) œ 90° "
Ê ! " œ "# (90° " ) Ê ! œ "# (90° " ) œ "# of nAOR.
(b) At the time t when the projectile hits OR we have
tan " œ yx ; x œ [v! cos (! " )]t and
y œ [v! sin (! " )]t "# gt#
v! sin (! " ) "# gt‘
v! cos (! " )
v! cos (! " ) tan " œ v! sin (! " ) "# gt
t œ 2v! sin (! ") 2vg ! cos (! ") tan " , which is the
Ê tan " œ
Ê
Ê
cv! sin (! " )d t "# gt#
[v! cos (! " )]t
œ
time
when the projectile hits the uphill slope. Therefore,
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
835
836
Chapter 13 Vector-Valued Functions and Motion in Space
x œ [v! cos (! " )] ’ 2v! sin (! ") 2vg ! cos (! ") tan " “ œ
maximized, then OR is maximized: ddx! œ
2v#!
g
2v#!
g
csin (! " ) cos (! " ) cos# (! " ) tan " d . If x is
[cos 2(! " ) sin 2(! " ) tan " ] œ 0
Ê cos 2(! " ) sin 2(! " ) tan " œ 0 Ê cot 2(! " ) tan " œ ! Ê cot 2(! " ) œ tan "
œ tan (" ) Ê 2(! " ) œ 90° (" ) œ 90° " Ê ! œ "# (90° " ) œ "# of nAOR. Therefore v! would bisect
nAOR for maximum range uphill.
26. (a) ratb œ axatbbi ayatbbj; where xatb œ a145 cos 23‰ 14bt and yatb œ 2.5 a145 sin 23‰ bt 16t2 .
(b) ymax œ
av0 sin !b2
2g
2.5 œ
a145sin 23‰ b2
64
v0 sin !
g
2.5 ¸ 52.655 feet, which is reached at t œ
‰
œ
145sin 23‰
32
¸ 1.771 seconds.
(c) For the time, solve y œ 2.5 a145 sin 23 bt 16t œ 0 for t, using the quadratic formula
tœ
145 sin 23‰ Éa145 sin 23‰ b2 160
32
2
¸ 3.585 sec. Then the range at t ¸ 3.585 is about x œ a145 cos 23‰ 14ba3.585b
¸ 428.311 feet.
(d) For the time, solve y œ 2.5 a145 sin 23‰ bt 16t2 œ 20 for t, using the quadratic formula
tœ
145 sin 23‰ Éa145 sin 23‰ b2 1120
32
‰
¸ 0.342 and 3.199 seconds. At those times the ball is about
xa0.342b œ a145 cos 23 14ba0.342b ¸ 40.860 feet from home plate and xa3.199b œ a145 cos 23‰ 14ba3.199b
¸ 382.195 feet from home plate.
(e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate.
27. (a) (Assuming that "x" is zero at the point of impact:)
ratb œ axatbbi ayatbbj; where xatb œ a35 cos 27‰ bt and yatb œ 4 a35 sin 27‰ bt 16t2 .
(b) ymax œ
av0 sin !b2
2g
4 œ
a35sin 27‰ b2
64
4 ¸ 7.945 feet, which is reached at t œ
‰
v0 sin !
g
œ
35sin 27‰
32
¸ 0.497 seconds.
(c) For the time, solve y œ 4 a35 sin 27 bt 16t œ 0 for t, using the quadratic formula
tœ
35 sin 27‰ Éa35 sin 27‰ b2 256
32
2
¸ 1.201 sec. Then the range is about xa1.201b œ a35 cos 27‰ ba1.201b
¸ 37.453 feet.
(d) For the time, solve y œ 4 a35 sin 27‰ bt 16t2 œ 7 for t, using the quadratic formula
tœ
35 sin 27‰ Éa35 sin 27‰ b2 192
32
‰
¸ 0.254 and 0.740 seconds. At those times the ball is about
xa0.254b œ a35 cos 27 ba0.254b ¸ 7.921 feet and xa0.740b œ a35 cos 27‰ ba0.740b ¸ 23.077 feet the impact point,
or about 37.453 7.921 ¸ 29.532 feet and 37.453 23.077 ¸ 14.376 feet from the landing spot.
(e) Yes. It changes things because the ball won't clear the net (ymax ¸ 7.945).
28. The maximum height is y œ
(v! sin !)#
#g
and this occurs for x œ
v#!
#g
sin 2! œ
v#! sin ! cos !
g
. These equations describe
parametrically the points on a curve in the xy-plane associated with the maximum heights on the parabolic trajectories in
terms of the parameter (launch angle) !. Eliminating the parameter !, we have x# œ
œ
v%! sin# !
g#
v%! sin% !
g#
Ê x# 4 Šy
29.
d2 r
dt2
œ
v#!
4g ‹
#
v#!
g
œ
(2y) (2y)# Ê x# 4y# Š
v!%
4g#
, where x
2v#!
g ‹y
v%! sin# ! cos# !
g#
v#
œ 0 Ê x# 4 ’y# Š 2g! ‹ y
œ
ˆv%! sin# !‰ a1 sin# !b
g#
v%!
16g# “
œ
v%!
4g#
0.
k ddtr œ gj Ê Patb œ k and Qatb œ gj Ê ' Patb dt œ kt Ê vatb œ e' Patb dt œ ekt Ê
dr
dt
œ
1
vatb
' vatb Qatb dt
œ gekt ' ekt j dt œ gekt ek j C1 ‘ œ gk j Cekt , where C œ gC1 ; apply the initial condition:
kt
dr
dt ¹tœ0
œ av0 cos !bi av0 sin !bj œ gk j C Ê C œ av0 cos !bi ˆ gk v0 sin !‰j
Ê
œ ˆv0 ekt cos !‰i ˆ gk ekt ˆ gk v0 sin !‰‰j, r œ ' c ˆv0 ekt cos !‰i ˆ gk ekt ˆ gk v0 sin !‰‰j ddt
dr
dt
gt
v
œ ˆ k0 ekt cos !‰i Š k
eckt ˆ g
k
k
v0 sin !‰‹j C2 ; apply the initial condition:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.3 Arc Length and the Unit Tangent Vector T
v0 sin ! ‰
j C2 Ê C2 œ ˆ vk0 cos !‰i ˆ kg2
k
ˆ vk0 ˆ1 ekt ‰sin ! kg2 ˆ1 kt ekt ‰‰j
ra0b œ 0 œ ˆ vk0 cos !‰i ˆ kg2
Ê ratb œ ˆ vk0 ˆ1 ekt ‰cos !‰i
837
v0 sin ! ‰
j
k
152 ‰
30. (a) ratb œ axatbbi ayatbbj; where xatb œ ˆ 0.12
a1 e0.12t bacos 20‰ b and
152 ‰
32 ‰
0.12t
yatb œ 3 ˆ 0.12
a1 e0.12t basin 20‰ b ˆ 0.12
b
2 a1 0.12t e
(b) Solve graphically using a calculator or CAS: At t ¸ 1.484 seconds the ball reaches a maximum height of about 40.435
feet.
(c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.126 seconds. The range is
152 ‰ˆ
about xa3.126b œ ˆ 0.12
1 e0.12a3.126b ‰acos 20‰ b ¸ 372.311 feet.
(d) Use a graphing calculator or CAS to find that y œ 30 for t ¸ 0.689 and 2.305 seconds, at which times the ball is about
xa0.689b ¸ 94.454 feet and xa2.305b ¸ 287.621 feet from home plate.
(e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet above the
ground when it passes over the fence.
1 ‰
31. (a) ratb œ axatbbi ayatbbj; where xatb œ ˆ 0.08
a1 e0.08t ba152 cos 20‰ 17.6b and
152 ‰
32 ‰
0.08t
yatb œ 3 ˆ 0.08
a1 e0.08t basin 20‰ b ˆ 0.08
b
2 a1 0.08t e
(b) Solve graphically using a calculator or CAS: At t ¸ 1.527 seconds the ball reaches a maximum height of about 41.893
feet.
(c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.181 seconds. The range is
1 ‰ˆ
about xa3.181b œ ˆ 0.08
1 e0.08a3.181b ‰a152 cos 20‰ 17.6b ¸ 351.734 feet.
(d) Use a graphing calculator or CAS to find that y œ 35 for t ¸ 0.877 and 2.190 seconds, at which times the ball is about
xa0.877b ¸ 106.028 feet and xa2.190b ¸ 251.530 feet from home plate.
(e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) to find that
1 ‰ˆ
y œ 20 at t ¸ 0.376 and 2.716 seconds. Then define xawb œ ˆ 0.08
1 e0.08a2.716b ‰a152 cos 20‰ wb, and solve
xawb œ 380 to find w ¸ 12.846 ft/sec.
13.3 ARC LENGTH AND THE UNIT TANGENT VECTOR T
1. r œ (2 cos t)i (2 sin t)j È5tk Ê v œ (2 sin t)i (2 cos t)j È5k
#
Ê kvk œ Ê(2 sin t)# (2 cos t)# ŠÈ5‹ œ È4 sin# t 4 cos# t 5 œ 3; T œ
œ ˆ 23 sin t‰ i ˆ 23 cos t‰ j
È5
3
1
v
kv k
1
k and Length œ '0 kvk dt œ '0 3 dt œ c3td 1! œ 31
2. r œ (6 sin 2t)i (6 cos 2t)j 5tk Ê v œ (12 cos 2t)i (12 sin 2t)j 5k
Ê kvk œ È(12 cos 2t)# (12 sin 2t)# 5# œ È144 cos# 2t 144 sin# 2t 25 œ 13; T œ
‰
ˆ 12
‰
œ ˆ 12
13 cos 2t i 13 sin 2t j
5
13
1
1
v
kv k
k and Length œ '0 kvk dt œ '0 13 dt œ c13td 1! œ 131
#
3. r œ ti 23 t$Î# k Ê v œ i t"Î# k Ê kvk œ É1# at"Î# b œ È1 t ; T œ
)
and Length œ '0 È1 t dt œ 23 (1 t)$Î# ‘ ! œ
8
v
kv k
œ
"
È1 t
i
Èt
È1 t
52
3
4. r œ (2 t)i (t 1)j tk Ê v œ i j k Ê kvk œ È1# (1)# 1# œ È3 ; T œ
v
kv k
œ
$
and Length œ '0 È3 dt œ ’È3t“ œ 3È3
3
k
!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
È3
i
1
È3
j
"
È3
k
838
Chapter 13 Vector-Valued Functions and Motion in Space
5. r œ acos$ tb j asin$ tb k Ê v œ a3 cos# t sin tb j a3 sin# t cos tb k Ê kvk
œ Éa3 cos# t sin tb# a3 sin# t cos tb# œ Èa9 cos# t sin# tb acos# t sin# tb œ 3 kcos t sin tk ;
Tœ
v
kv k
œ
3 cos# t sin t
3 kcos t sin tk
1Î2
Length œ '0
j
3 sin# t cos t
3 kcos t sin tk
k œ ( cos t)j (sin t)k , if 0 Ÿ t Ÿ
1Î2
1Î2
3 kcos t sin tk dt œ '0 3 cos t sin t dt œ '0
3
#
1
#
, and
1Î#
sin 2t dt œ 34 cos 2t‘ !
œ
3
#
6. r œ 6t$ i 2t$ j 3t$ k Ê v œ 18t# i 6t# j 9t# k Ê kvk œ Éa18t# b# a6t# b# a9t# b# œ È441t% œ 21t# ;
Tœ
v
kv k
œ
"8t#
21t#
i
6t#
21t#
j
7. r œ (t cos t)i (t sin t)j
9t#
21t#
kœ
6
7
2È2 $Î#
k
3 t
i 27 j 37 k and Length œ '1 21t# dt œ c7t$ d " œ 49
2
#
Ê v œ (cos t t sin t)i (sin t t cos t)j ŠÈ2 t"Î# ‹ k
#
Ê kvk œ Ê(cos t t sin t)# (sin t t cos t)# ŠÈ2 t‹ œ È1 t# 2t œ È(t 1)# œ kt 1k œ t 1, if t
Tœ
v
kv k
œ ˆ cos tt t1sin t ‰ i ˆ sin ttt1cos t ‰ j Š
È2 t"Î#
t1 ‹ k
1
1
and Length œ '0 (t 1) dt œ ’ t2 t“ œ
#
!
1#
2
0;
1
8. r œ (t sin t cos t)i (t cos t sin t)j Ê v œ (sin t t cos t sin t)i (cos t t sin t cos t)j
œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t sin t)# œ Èt# œ ktk œ t if È2 Ÿ t Ÿ 2; T œ v
kv k
t‰
t‰
œ ˆ t cos
i ˆ t sin
j œ (cos t)i (sin t)j and Length œ 'È2 t dt œ ’ t2 “
t
t
2
#
#
È#
œ1
9. Let P(t! ) denote the point. Then v œ (5 cos t)i (5 sin t)j 12k and 261 œ '0 È25 cos# t 25 sin# t 144 dt
t!
œ '0 13 dt œ 13t! Ê t! œ 21, and the point is P(21) œ (5 sin 21ß 5 cos 21ß 241) œ (0ß 5ß 241)
t!
10. Let P(t! ) denote the point. Then v œ (12 cos t)i (12 sin t)j 5k and
131 œ '0 È144 cos# t 144 sin# t 25 dt œ '0 13 dt œ 13t! Ê t! œ 1, and the point is
t!
t!
P(1) œ (12 sin (1)ß 12 cos (1)ß 51) œ (0ß 12ß 51)
11. r œ (4 cos t)i (4 sin t)j 3tk Ê v œ (4 sin t)i (4 cos t)j 3k Ê kvk œ È(4 sin t)# (4 cos t)# 3#
œ È25 œ 5 Ê s(t) œ '0 5 d7 œ 5t Ê Length œ s ˆ 1# ‰ œ
t
51
#
12. r œ (cos t t sin t)i (sin t t cos t)j Ê v œ (sin t sin t t cos t)i (cos t cos t t sin t)j
t
œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t cos t)# œ œ Èt# œ t, since 1# Ÿ t Ÿ 1 Ê s(t) œ '0 7 d7 œ
Ê Length œ s(1) s ˆ 1# ‰ œ
1#
#
ˆ 1# ‰#
#
œ
t#
#
31 #
8
13. r œ aet cos tb i aet sin tb j et k Ê v œ aet cos t et sin tb i aet sin t et cos tb j et k
Ê kvk œ Éaet cos t et sin tb# aet sin t et cos tb# aet b# œ œ È3e2t œ È3 et Ê s(t) œ '0 È3 e7 d7
t
œ È3 et È3 Ê Length œ s(0) s( ln 4) œ 0 ŠÈ3 e ln 4 È3‹ œ
3È 3
4
14. r œ (1 2t)i (1 3t)j (6 6t)k Ê v œ 2i 3j 6k Ê kvk œ È2# 3# (6)# œ 7 Ê s(t) œ '0 7 d7 œ 7t
t
Ê Length œ s(0) s(1) œ 0 (7) œ 7
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.3 Arc Length and the Unit Tangent Vector T
#
839
#
15. r œ ŠÈ2t‹ i ŠÈ2t‹ j a1 t# b k Ê v œ È2i È2j 2tk Ê kvk œ ÊŠÈ2‹ ŠÈ2‹ (2t)# œ È4 4t#
œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’2 Š 2t È1 t#
1
"
#
"
ln Št È1 t# ‹‹“ œ È2 ln Š1 È2‹
!
16. Let the helix make one complete turn from t œ 0 to t œ 21.
Note that the radius of the cylinder is 1 Ê the
circumference of the base is 21. When t œ 21, the point P is
(cos 21ß sin 21ß 21) œ (1ß 0ß 21) Ê the cylinder is 21 units
high. Cut the cylinder along PQ and flatten. The resulting
rectangle has a width equal to the circumference of the
cylinder œ 21 and a height equal to 21, the height of the
cylinder. Therefore, the rectangle is a square and the portion
of the helix from t œ 0 to t œ 21 is its diagonal.
17. (a) r œ (cos t)i (sin t)j (" cos t)k, 0 Ÿ t Ÿ 21 Ê x œ cos t, y œ sin t, z œ 1 cos t Ê x# y#
œ cos# t sin# t œ 1, a right circular cylinder with the z-axis as the axis and radius œ 1. Therefore
P(cos tß sin tß 1 cos t) lies on the cylinder x# y# œ 1; t œ 0 Ê P(1ß 0ß 0) is on the curve; t œ 1# Ê Q(!ß 1ß 1)
Ä
Ä
is on the curve; t œ 1 Ê R(1ß 0ß 2) is on the curve. Then PQ œ i j k and PR œ 2i 2k
j k×
Ô i
Ä
Ä
Ê PQ ‚ PR œ 1 " " œ 2i 2k is a vector normal to the plane of P, Q, and R. Then the
Õ 2 0 2 Ø
plane containing P, Q, and R has an equation 2x 2z œ 2(1) 2(0) or x z œ 1. Any point on the curve
will satisfy this equation since x z œ cos t (1 cos t) œ 1. Therefore, any point on the curve lies on the
intersection of the cylinder x# y# œ 1 and the plane x z œ 1 Ê the curve is an ellipse.
(b) v œ ( sin t)i (cos t)j (sin t)k Ê kvk œ Èsin# t cos# t sin# t œ È1 sin# t Ê T œ kvvk
œ
( sin t)i (cos t)j (sin t)k
È1 sin# t
Ê T(0) œ j , T ˆ 1# ‰ œ
ik
È2
, T(1) œ j , T ˆ 3#1 ‰ œ
ik
È2
(c) a œ ( cos t)i (sin t)j (cos t)k ; n œ i k is
normal to the plane x z œ 1 Ê n † a œ cos t cos t
œ 0 Ê a is orthogonal to n Ê a is parallel to the
plane; a(0) œ i k , a ˆ 1# ‰ œ j , a a1b œ i k ,
‰œj
a ˆ 31
#
21
(d) kvk œ È1 sin# t (See part (b) Ê L œ '0 È1 sin# t dt
(e) L ¸ 7.64 (by Mathematica)
18. (a) r œ (cos 4t)i (sin 4t)j 4tk Ê v œ (4 sin 4t)i (4 cos 4t)j 4k Ê kvk œ È(4 sin 4t)# (4 cos 4t)# 4#
1Î2
œ È32 œ 4È2 Ê Length œ '0 4È2 dt œ ’4È2 t“
1Î#
!
œ 21 È 2
(b) r œ ˆcos #t ‰ i ˆsin #t ‰ j #t k Ê v œ ˆ "# sin #t ‰ i ˆ "# cos #t ‰ j "# k
#
#
#
Ê kvk œ Ɉ "# sin #t ‰ ˆ "# cos #t ‰ ˆ "# ‰ œ É 4"
"
4
œ
È2
#
41
Ê Length œ '0
È2
#
dt œ ’
È2
2
t“
%1
!
œ 21È2
(c) r œ (cos t)i (sin t)j tk Ê v œ ( sin t)i (cos t)j k Ê kvk œ È( sin t)# ( cos t)# (1)# œ È1 1
œ È2 Ê Length œ 'c21 È2 dt œ ’È2 t“
0
!
#1
œ 21 È 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
840
Chapter 13 Vector-Valued Functions and Motion in Space
19. nPQB œ nQOB œ t and PQ œ arc (AQ) œ t since
PQ œ length of the unwound string œ length of arc (AQ);
thus x œ OB BC œ OB DP œ cos t t sin t, and
y œ PC œ QB QD œ sin t t cos t
20. r œ acos t t sin tbi asin t t cos tbj Ê v œ asin t t cos t sin tbi acos t atasin tb cos tbbj
œ at cos tbi at sin tbj Ê kvk œ Éat cos tb2 at sin tb2 œ Èt2 œ ktk œ t, t
0ÊTœ
v
kv k
œ
t cos t
t i
t sin t
t j
œ cos t i sin t j
13.4 CURVATURE AND THE UNIT NORMAL VECTOR N
sin t ‰
È1# ( tan t)# œ Èsec# t œ ksec tk œ sec t, since
1. r œ ti ln (cos t)j Ê v œ i ˆ cos
t j œ i (tan t)j Ê kvk œ
1
1
v
"
tan
t
# t # Ê T œ kvk œ ˆ sec t ‰ i ˆ sec t ‰ j œ (cos t)i (sin t)j ; ddtT œ ( sin t)i (cos t)j
Ê ¸ ddtT ¸ œ È( sin t)# ( cos t)# œ 1 Ê N œ
,œ
1
kv k
†
¸ ddtT ¸
"
sec t
œ
ˆ ddtT ‰
¸ ddtT ¸
œ ( sin t)i (cos t)j ;
† 1 œ cos t.
t tan t ‰
2. r œ ln (sec t)i tj Ê v œ ˆ secsec
i j œ (tan t)i j Ê kvk œ È( tan t)# 1# œ Èsec# t œ ksec tk œ sec t,
t
1
1
v
tan
since # t # Ê T œ kvk œ ˆ sec tt ‰ i ˆ sec1 t ‰ j œ (sin t)i (cos t)j ; ddtT œ (cos t)i (sin t)j
Ê ¸ ddtT ¸ œ È(cos t)# ( sin t)# œ 1 Ê N œ
,œ
1
kv k
† ¸ ddtT ¸ œ
"
sec t
ˆ ddtT ‰
¸ ddtT ¸
œ (cos t)i (sin t)j ;
† 1 œ cos t.
3. r œ (2t 3)i a5 t# b j Ê v œ 2i 2tj Ê kvk œ È2# (2t)# œ 2È1 t# Ê T œ kvvk œ È 2 # i
2 1t
Í
#
#
Í
Í
t
"
t
"
dT ¸
¸
œ È " # i È t # j ; ddtT œ
i
j
Ê
œ
$
$
$
ŠÈ1 t# ‹$
dt
1 t
1 t
ŠÈ1 t# ‹
ŠÈ1 t# ‹
Ì ŠÈ1 t# ‹
œ É a1 "t# b# œ
,œ
1
kv k
"
1t#
† ¸ ddtT ¸ œ
Ê Nœ
"
#È1 t#
†
ˆ ddtT ‰
¸ ddtT ¸
"
1 t#
œ
œ
t
È1 t#
i
"
È1 t#
2t
2È1 t#
j;
"
# a1 t# b3/2
4. r œ (cos t t sin t)i (sin t t cos t)j Ê v œ (t cos t)i (t sin t)j Ê kvk œ È( t cos t)# (t sin t)# œ Èt# œ ktk
œ t, since t 0 Ê T œ
v
kv k
œ
(t cos t)i(t sin t)j
t
œ (cos t)i (sin t)j ;
Ê ¸ ddtT ¸ œ È( sin t)# (cos t)# œ 1 Ê N œ
5. (a) ,axb œ
1
kvaxbk
ˆ ddtT ‰
¸ ddtT ¸
dT
dt
œ ( sin t)i (cos t)j
œ ( sin t)i (cos t)j ; , œ
1
kv k
† ¹ dTdtaxb ¹. Now, v œ i f w axbj Ê kvaxbk œ É1 c f w axb d2 Ê T œ
1Î2
œ Š1 c f w axb d2 ‹
1Î2
i f w axbŠ1 c f w axb d2 ‹
j. Thus
dT
dt axb
Í
2
Í
2
f ax b f a x b
cf
Í
dTaxb
f ax b
3
2
Ê ¹ dt ¹ œ ” Š1 c f axb d2 ‹ •
3 2 œ Ë
2
Ì
Š 1 c f ax b d ‹
w
ww
ww
ww
œ
w
ww
Î
w
2
Š1 c f a x b d ‹
w
axb d2 Š1 c f axb d2 ‹
w
2
Š 1 c f ax b d ‹
w
3
"
t
"
t
†1œ
v
kv k
f ax b f a x b
Î
w
† ¸ ddtT ¸ œ
3Î2
œ
i
f ax b
ww
3Î2
2
Š1 c f axb d ‹
w
kf axbk
ww
2
¹ 1 c f ax b d ¹
w
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
j
j
Section 13.4 Curvature and the Unit Normal Vector N
Thus ,axb œ
1
a1 Òf axbÓ2 b1
(b) y œ ln (cos x) Ê
œ
"
sec x
†
2
Î
w
kf axbk
k1 Òf axbÓ2 k
ww
w
œ
kf axbk
ww
3Î2
2
Š1 c f axb d ‹
w
œ ˆ cos" x ‰ ( sin x) œ tan x Ê
dy
dx
œ cos x, since 1# x
d# y
dx#
œ sec# x Ê , œ
k sec# xk
c1 ( tan x)# d$Î#
œ
sec# x
ksec$ xk
1
#
(c) Note that f ww (x) œ 0 at an inflection point.
Þ
Þ
Þ
Þ
6. (a) r œ f(t)i g(t)j œ xi yj Ê v œ xi yj Ê kvk œ Èx# y# Ê T œ
dT
dt
œ
Þ Þ ÞÞ Þ ÞÞ
Þ Þ ÞÞ Þ ÞÞ
yay x x yb
xax y y x b
dT
Þ # Þ # 3/2 i Þ # Þ # 3/2 j Ê ¸ dt ¸
ax y b
ax y b
Þ ÞÞ Þ ÞÞ
ky x x yk
1
1
dT
Þ
Þ ; , œ kvk † ¸ dt ¸ œ È Þ # Þ #
kx# y# k
x y
œ
Þ Þ ÞÞ Þ ÞÞ
yay x x yb 2
Þ
Þ 3/2 “ ’
ax# y# b
Þ ÞÞ Þ ÞÞ
Þ ÞÞ Þ ÞÞ
k y x x yk
k y x x yk
Þ # Þ # œ Þ # Þ # 3/2 .
kx y k
ax y b
œ Ê’
†
v
kv k
Þ Þ ÞÞ Þ ÞÞ
xax y y xb 2
Þ
Þ 3/2 “
ax# y# b
œ
Þ
Þ
y
x
ÈxÞ # yÞ # i ÈxÞ # yÞ # j
œÊ
Þ
Þ Þ ÞÞ Þ ÞÞ
ay# x# bay x x yb2
Þ
Þ 3
ax# y# b
Þ
ÞÞ
Þ
(b) r(t) œ ti ln (sin t)j , 0 t 1 Ê x œ t and y œ ln (sin t) Ê x œ 1, x œ 0; y œ
Ê ,œ
k csc# t 0k
a1 cot# t)b$Î#
"
œ
csc# t
csc$ t
cos t
sin t
ÞÞ
œ cot t, y œ csc# t
œ sin t
Þ
t
"
(sinh t)i ln (cosh t)j Ê x œ tan" (sinh t) and y œ ln (cosh t) Ê x œ 1 cosh
sinh# t œ cosh t
$
#
ÞÞ
Þ
ÞÞ
ksech t sech t tanh tk
sinh t
#
œ sech t, x œ sech t tanh t; y œ cosh
œ ksech tk
asech# t tanh# tb
t œ tanh t, y œ sech t Ê , œ
(c) r(t) œ tan
œ sech t
7. (a) r(t) œ f(t)i g(t)j Ê v œ f w (t)i gw (t)j is tangent to the curve at the point (f(t)ß g(t));
n † v œ c gw (t)i f w (t)jd † cf w (t)i gw (t)jd œ gw (t)f w (t) f w (t)gw (t) œ 0; n † v œ (n † v) œ 0; thus,
n and n are both normal to the curve at the point
(b) r(t) œ ti e2t j Ê v œ i 2e2t j Ê n œ 2e2t i j points toward the concave side of the curve; N œ
knk œ È4e4t 1 Ê N œ
2e
È1 4e4t
2t
(c) r(t) œ È4 t# i tj Ê v œ
Nœ
n
knk
and knk œ É1
t#
4 t#
i
t
È4 t#
œ
"
È1 4e4t
and
j
i j Ê n œ i
Ê Nœ
2
È4 t#
n
knk
"
#
t
È4 t#
j points toward the concave side of the curve;
ŠÈ4 t# i tj‹
8. (a) r(t) œ ti "3 t$ j Ê v œ i t# j Ê n œ t# i j points toward the concave side of the curve when t 0 and
n œ t# i j points toward the concave side when t 0 Ê N œ
Nœ
"
È1 t%
2ktk
;
1 t%
at# i jb for t 0 and
at# i jb for t 0
(b) From part (a), kvk œ È1 t% Ê T œ
œ
"
È1 t%
Nœ
ˆ ddtT ‰
¸ ddtT ¸
œ
2t$
1 t%
2ktk Š a1 t% b$Î# i
N does not exist at t œ 0,
dt ¸
œ ¸ ddtT † ds
œ 0 at t œ 0
"
È1 t% i
t#
È1 t% j
dT
dt
t$
i
ktkÈ1 t%
œ
2t$
$Î# i
a1 t% b
2t
$Î# j
a1 t% b
6
2
Ê ¸ ddtT ¸ œ É a4t1 t%4tb$
t
j; t Á 0
ktkÈ1 t%
where the curve has a point of inflection; ddtT ¸ tœ0 œ 0 so the curvature , œ ¸ ddsT ¸
Ê N œ ," ddsT is undefined. Since x œ t and y œ 3" t$ Ê y œ 3" x$ , the curve is
2t
$Î# j‹
a1 t% b
œ
Ê
the
cubic power curve which is concave down for x œ t 0 and concave up for x œ t 0.
9. r œ (3 sin t)i (3 cos t)j 4tk Ê v œ (3 cos t)i (3 sin t)j 4k Ê kvk œ È(3 cos t)# (3 sin t)# 4#
œ È25 œ 5 Ê T œ kvvk œ ˆ 35 cos t‰ i ˆ 35 sin t‰ j 45 k Ê ddtT œ ˆ 35 sin t‰ i ˆ 35 cos t‰ j
#
#
Ê ¸ ddtT ¸ œ Ɉ 35 sin t‰ ˆ 35 cos t‰ œ
3
5
Ê Nœ
ˆ ddtT ‰
¸ ddtT ¸
œ ( sin t)i (cos t)j ; , œ
1
5
†
3
5
œ
3
25
10. r œ (cos t t sin t)i (sin t t cos t)j 3k Ê v œ (t cos t)i (t sin t)j Ê kvk œ È(t cos t)# (t sin t)# œ Èt#
œ ktk œ t, if t 0 Ê T œ
v
kv k
œ (cos t)i (sin t)j , t 0 Ê
Ê ¸ ddtT ¸ œ È( sin t)# (cos t)# œ 1 Ê N œ
ˆ ddtT ‰
¸ ddtT ¸
dT
dt
œ ( sin t)i (cos t)j
œ ( sin t)i (cos t)j ; , œ
"
t
†1œ
"
t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
841
842
Chapter 13 Vector-Valued Functions and Motion in Space
11. r œ aet cos tb i aet sin tb j 2k Ê v œ aet cos t et sin tb i aet sin t et cos tb j Ê
kvk œ Éaet cos t et sin tb# aet sin t et cos tb# œ È2e2t œ et È2 ;
Tœ
v
kv k
t sin t
cos t
œ Š cos È
‹ i Š sin tÈ
‹j Ê
2
2
#
dT
dt
t sin t
œ Š sinÈt 2 cos t ‹ i Š cos È
‹j
2
#
t sin t
Ê ¸ ddtT ¸ œ ÊŠ sinÈt 2 cos t ‹ Š cos È
‹ œ1 Ê Nœ
2
,œ
1
kv k
† ¸ ddtT ¸ œ
†1œ
1
et È2
ˆ ddtT ‰
¸ ddtT ¸
œ Š cosÈt 2 sin t ‹ i Š sinÈt 2 cos t ‹ j ;
1
et È2
12. r œ (6 sin 2t)i (6 cos 2t)j 5tk Ê v œ (12 cos 2t)i (12 sin 2t)j 5k
Ê kvk œ È(12 cos 2t)# (12 sin 2t)# 5# œ È169 œ 13 Ê T œ v
kv k
‰
ˆ 12
‰
œ ˆ 12
13 cos 2t i 13 sin 2t j
5
13
k Ê
‰
ˆ 24
‰
œ ˆ 24
13 sin 2t i 13 cos 2t j
dT
dt
‰# ˆ 24
‰# œ
Ê ¸ ddtT ¸ œ Ɉ 24
13 sin 2t
13 cos 2t
,œ
1
kv k
† ¸ ddtT ¸ œ
$
#
1
13
†
24
13
œ
24
13
ˆ ddtT ‰
¸ ddtT ¸
Ê Nœ
œ ( sin 2t)i (cos 2t)j ;
24
169 .
13. r œ Š t3 ‹ i Š t# ‹ j , t 0 Ê v œ t# i tj Ê kvk œ Èt% t# œ tÈt# 1, since t 0 Ê T œ
œ
t
Èt# t
i
1
Èt# 1
#
t
œ É at1#
œ
1 b$
j Ê
"
t# 1
œ
dT
dt
Ê Nœ
i
1
at# 1b$Î#
ˆ ddtT ‰
¸ ddtT ¸
14. r œ acos$ tb i asin$ tb j , 0 t
œ
1
#
1
Èt# 1
t
at# 1b$Î#
i
j Ê ¸ ddtT ¸ œ ÊŠ
t
Èt# 1
j; , œ
1
kv k
"
‹
at# 1b$Î#
† ¸ ddtT ¸ œ
#
Š
1
tÈt# 1
†
t
‹
at# 1b$Î#
1
t# 1
œ
v
kv k
#
"
.
t at# 1b$Î#
Ê v œ a3 cos# t sin tb i a3 sin# t cos tb j
Ê kvk œ Éa3 cos# t sin tb# a3 sin# t cos tb# œ È9 cos% t sin# t 9 sin% t cos# t œ 3 cos t sin t, since 0 t
Ê Tœ
v
kv k
œ ( cos t)i (sin t)j Ê
œ (sin t)i (cos t)j; , œ
1
kv k
† ¸ ddtT ¸ œ
dT
dt
œ (sin t)i (cos t)j Ê ¸ ddtT ¸ œ Èsin# t cos# t œ 1 Ê N œ
1
3 cos t sin t
†1œ
v
kv k
œ ˆsech at ‰ i ˆtanh at ‰ j Ê
Ê ¸ ddtT ¸ œ É a"# sech# ˆ at ‰ tanh# ˆ at ‰
,œ
1
kv k
† ¸ ddtT ¸ œ
1
cosh
t
a
†
"
a
sech ˆ at ‰ œ
"
a
"
a#
dT
dt
ˆ ddtT ‰
¸ ddtT ¸
1
3 cos t sin t .
15. r œ ti ˆa cosh at ‰ j , a 0 Ê v œ i ˆsinh at ‰ j Ê kvk œ É1 sinh# ˆ at ‰ œ Écosh# ˆ at ‰ œ cosh
Ê Tœ
1
#
œ ˆ "a sech
sech% ˆ at ‰ œ
"
a
t
a
t
a
tanh at ‰ i ˆ "a sech# at ‰ j
sech ˆ at ‰ Ê N œ
ˆ ddtT ‰
¸ ddtT ¸
œ ˆ tanh at ‰ i ˆsech at ‰ j ;
sech# ˆ at ‰.
16. r œ (cosh t)i (sinh t)j tk Ê v œ (sinh t)i (cosh t)j k Ê kvk œ Èsinh# t ( cosh t)# 1 œ È2 cosh t
Ê Tœ
v
kv k
œ Š È"2 tanh t‹ i
Ê ¸ ddtT ¸ œ É "# sech% t
,œ
1
kv k
† ¸ ddtT ¸ œ
1
È2 cosh t
†
"
#
"
È2
j Š È"2 sech t‹ k Ê
sech# t tanh# t œ
"
È2
sech t œ
"
#
"
È2
dT
dt
œ Š È"2 sech# t‹ i Š È"2 sech t tanh t‹ k
sech t Ê N œ
&Î#
œ (sech t)i (tanh t)k ;
sech# t.
17. y œ ax# Ê yw œ 2ax Ê yww œ 2a; from Exercise 5(a), ,(x) œ
Ê ,w (x) œ 3# k2ak a1 4a# x# b
ˆ ddtT ‰
¸ ddtT ¸
k2ak
a1 4a# x# b$Î#
œ k2ak a1 4a# x# b
$Î#
a8a# xb ; thus, ,w (x) œ 0 Ê x œ 0. Now, ,w (x) 0 for x 0 and ,w (x) 0 for
x 0 so that ,(x) has an absolute maximum at x œ 0 which is the vertex of the parabola. Since x œ 0 is the
only critical point for ,(x), the curvature has no minimum value.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.4 Curvature and the Unit Normal Vector N
18. r œ (a cos t)i (b sin t)j
â
i
j
â
â
œ â a sin t b cos t
â
â a cos t b sin t
Ê v œ (a sin t)i (b cos t)j Ê a œ (a cos t)i (b sin t)j Ê v ‚ a
â
kâ
â
0 â œ abk Ê kv ‚ ak œ kabk œ ab, since a b 0; , (t) œ kvkv‚k$ak
â
0â
œ ab aa# sin# t b# cos# tb
$Î#
; ,w (t) œ #3 (ab) aa# sin# t b# cos# tb
œ 3# (ab) aa# b# b (sin 2t) aa# sin# t b# cos# tb
points on the major axis, or t œ
1
#
0t
and for 1 t
31
# ;
1 31
#, #
w
&Î#
&Î#
a2a# sin t cos t 2b# sin t cos tb
; thus, ,w (t) œ 0 Ê sin 2t œ 0 Ê t œ 0, 1 identifying
identifying points on the minor axis. Furthermore, ,w (t) 0 for
, (t) 0 for
1
#
t 1 and
31
#
t 21. Therefore, the points associated
with t œ 0 and t œ 1 on the major axis give absolute maximum curvature and the points associated with t œ
and t œ
19. , œ
31
#
1
#
on the minor axis give absolute minimum curvature.
d,
da
Ê
a
a# b#
a b and
843
d,
da
a # b #
aa # b # b #
œ
;
d,
da
œ 0 Ê a# b# œ 0 Ê a œ „ b Ê a œ b since a, b
0 if a b Ê , is at a maximum for a œ b and , (b) œ
b
b# b#
œ
"
2b
0. Now,
d,
da
0 if
is the maximum value of , .
20. (a) From Example 5, the curvature of the helix r(t) œ (a cos t)i (a sin t)j btk, a, b 0 is , œ a# a b# ; also
kvk œ Èa# b# . For the helix r(t) œ (3 cos t)i (3 sin t)j tk, 0 Ÿ t Ÿ 41, a œ 3 and b œ 1 Ê , œ 3# 3 1# œ
and kvk œ È10 Ê K œ '0
41
3
10
È10 dt œ ’
3
È10
t“
%1
!
3
10
121
È10
œ
(b) y œ x# Ê x œ t and y œ t# , _ t _ Ê r(t) œ ti t# j Ê v œ i 2tj Ê kvk œ È1 4t# ;
Tœ
,œ
1
È1 4t# i
1
È1 4t#
œaÄ
lim
_
†
2t
dT
È1 4t# j; dt
2
1 4t#
'a0
œ
2
14t#
œ
4t
i
a1 4t# b3/2
2
3.
ŠÈ1 4t# ‹
dt lim
bÄ_
œaÄ
lim
a tan" 2ab lim
_
_
Then K œ 'c_
'0b 1 24t
bÄ_
2
j;
a1 4t# b3/2
#
¸ ddtT ¸ œ É
2
$
ŠÈ1 4t# ‹
16t2 4
a1 4t# b3
œ
2
1 4t# .
ŠÈ1 4t# ‹ dt œ '_ 124t# dt
!
dt œ a Ä
lim
ctan" 2td a lim
_
a tan" 2bb œ
1
#
1
#
Thus
_
bÄ_
œ1
ctan" 2td 0
b
21. r œ ti (sin t)j Ê v œ i (cos t)j Ê kvk œ È1# (cos t)# œ È1 cos# t Ê ¸v ˆ 1# ‰¸ œ É1 cos# ˆ 1# ‰ œ 1; T œ
œ
i cos t j
È1 cos2 t
Ê 3œ
"
1
Ê
dT
dt
œ
sin t cos t
i
a1 cos2 tb3/2
sin t
j
a1 cos2 tb3/2
Ê ¸ ddtT ¸ œ
ksin tk
1 cos2 t ;
¸ ddtT ¸
tœ 12
œ
¸sin 12 ¸
1 cos2 ˆ 12 ‰
œ
1
1
œ 1. Thus ,ˆ 12 ‰ œ
1
1
†1œ1
#
œ 1 and the center is ˆ 1# ß 0‰ Ê ˆx 1# ‰ y# œ 1
2
22. r œ (2 ln t)i ˆt "t ‰ j Ê v œ ˆ 2t ‰ i ˆ1 t"# ‰ j Ê kvk œ É t42 ˆ1 t12 ‰ œ
dT
dt
œ
2ˆt2 1‰
i
at2 1b2
œ
"
#
Ê 3œ
"
,
at2
4t
j
1 b2
2
Ê ¸ ddtT ¸ œ Ê 4at
1b2 16t2
at2 1b4
œ
2
t2 1 .
Thus , œ
1
kv k
t2 1
t2
† ¸ ddtT ¸ œ
ÊTœ
t2
t2 1
†
2
t2 1
v
kv k
œ
œ
2t2
at2 1b2
2t
t2 1 i
t2 1
t2 1 j;
Ê ,a1b œ
2
22
œ 2. The circle of curvature is tangent to the curve at P(0ß 2) Ê circle has same tangent as the curve
Ê v(1) œ 2i is tangent to the circle Ê the center lies on the y-axis. If t Á 1 (t 0), then (t 1)# 0
#
Ê t# 2t 1 0 Ê t# 1 2t Ê t t 1 2 since t 0 Ê t "t 2 Ê ˆt "t ‰ 2 Ê y 2 on both
sides of (0ß 2) Ê the curve is concave down Ê center of circle of curvature is (0ß 4) Ê x# (y 4)# œ 4
is an equation of the circle of curvature
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
v
kv k
844
Chapter 13 Vector-Valued Functions and Motion in Space
23. y œ x# Ê f w (x) œ 2x and f ww (x) œ 2
Ê ,œ
24. y œ
x%
4
k2 k
a1 (2x)# b$Î#
œ
2
a1 4x# b$Î#
Ê f w (x) œ x$ and f ww (x) œ 3x#
Ê ,œ
k3x# k
$Î#
#
Š1 ax$ b ‹
œ
3x#
a1 x' b$Î#
25. y œ sin x Ê f w (x) œ cos x and f ww (x) œ sin x
Ê ,œ
k sin xk
a1 cos# xb$Î#
œ
ksin xk
a1 cos# xb$Î#
26. y œ ex Ê f w (x) œ ex and f ww (x) œ ex
Ê ,œ
ke x k
# $Î#
Š1 aex b ‹
œ
ex
ˆ1 e2x ‰$Î#
27-34. Example CAS commands:
Maple:
with( plots );
r := t -> [3*cos(t),5*sin(t)];
lo := 0;
hi := 2*Pi;
t0 := Pi/4;
P1 := plot( [r(t)[], t=lo..hi] ):
display( P1, scaling=constrained, title="#27(a) (Section 13.4)" );
CURVATURE := (x,y,t) ->simplify(abs(diff(x,t)*diff(y,t,t)-diff(y,t)*diff(x,t,t))/(diff(x,t)^2+diff(y,t)^2)^(3/2));
kappa := eval(CURVATURE(r(t)[],t),t=t0);
UnitNormal := (x,y,t) ->expand( [-diff(y,t),diff(x,t)]/sqrt(diff(x,t)^2+diff(y,t)^2) );
N := eval( UnitNormal(r(t)[],t), t=t0 );
C := expand( r(t0) + N/kappa );
OscCircle := (x-C[1])^2+(y-C[2])^2 = 1/kappa^2;
evalf( OscCircle );
P2 := implicitplot( (x-C[1])^2+(y-C[2])^2 = 1/kappa^2, x=-7..4, y=-4..6, color=blue ):
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.5 Torsion and the Unit Binormal Vector B
845
display( [P1,P2], scaling=constrained, title="#27(e) (Section 13.4)" );
Mathematica: (assigned functions and parameters may vary)
In Mathematica, the dot product can be applied either with a period "." or with the word, "Dot".
Similarly, the cross product can be applied either with a very small "x" (in the palette next to the arrow) or with the word,
"Cross". However, the Cross command assumes the vectors are in three dimensions
For the purposes of applying the cross product command, we will define the position vector r as a three dimensional vector
with zero for its z-component. For graphing, we will use only the first two components.
Clear[r, t, x, y]
r[t_]={3 Cos[t], 5 Sin[t] }
t0= 1 /4; tmin= 0; tmax= 21;
r2[t_]= {r[t][[1]], r[t][[2]]}
pp=ParametricPlot[r2[t], {t, tmin, tmax}];
mag[v_]=Sqrt[v.v]
vel[t_]= r'[t]
speed[t_]=mag[vel[t]]
acc[t_]= vel'[t]
curv[t_]= mag[Cross[vel[t],acc[t]]]/speed[t]3 //Simplify
unittan[t_]= vel[t]/speed[t]//Simplify
unitnorm[t_]= unittan'[t] / mag[unittan'[t]]
ctr= r[t0] + (1 / curv[t0]) unitnorm[t0] //Simplify
{a,b}= {ctr[[1]], ctr[[2]]}
To plot the osculating circle, load a graphics package and then plot it, and show it together with the original curve.
<;
t0 := sqrt(3);
rr := eval( r, t=t0 );
v := map( diff, r, t );
vv := eval( v, t=t0 );
a := map( diff, v, t );
aa := eval( a, t=t0 );
s := simplify(Norm( v, 2 )) assuming t::real;
ss := eval( s, t=t0 );
T := v/s;
TT := vv/ss ;
q1 := map( diff, simplify(T), t ):
NN := simplify(eval( q1/Norm(q1,2), t=t0 ));
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
850
Chapter 13 Vector-Valued Functions and Motion in Space
BB := CrossProduct( TT, NN );
kappa := Norm(CrossProduct(vv,aa),2)/ss^3;
tau := simplify( Determinant(< vv, aa, eval(map(diff,a,t),t=t0) >)/Norm(CrossProduct(vv,aa),2)^3 );
a_t := eval( diff( s, t ), t=t0 );
a_n := evalf[4]( kappa*ss^2 );
Mathematica: (assigned functions and value for t0 will vary)
Clear[t, v, a, t]
mag[vector_]:=Sqrt[vector.vector]
Print["The position vector is ", r[t_]={t Cos[t], t Sin[t], t}]
Print["The velocity vector is ", v[t_]= r'[t]]
Print["The acceleration vector is ", a[t_]= v'[t]]
Print["The speed is ", speed[t_]= mag[v[t]]//Simplify]
Print["The unit tangent vector is ", utan[t_]= v[t]/speed[t] //Simplify]
Print["The curvature is ", curv[t_]= mag[Cross[v[t],a[t]]] / speed[t]3 //Simplify]
Print["The torsion is ", torsion[t_]= Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]]2 //Simplify]
Print["The unit normal vector is ", unorm[t_]= utan'[t] / mag[utan'[t]] //Simplify]
Print["The unit binormal vector is ", ubinorm[t_]= Cross[utan[t],unorm[t]] //Simplify]
Print["The tangential component of the acceleration is ", at[t_]=a[t].utan[t] //Simplify]
Print["The normal component of the acceleration is ", an[t_]=a[t].unorm[t] //Simplify]
You can evaluate any of these functions at a specified value of t.
t0= Sqrt[3]
{utan[t0], unorm[t0], ubinorm[t0]}
N[{utan[t0], unorm[t0], ubinorm[t0]}]
{curv[t0], torsion[t0]}
N[{curv[t0], torsion[t0]}]
{at[t0], an[t0]}
N[{at[t0], an[t0]}]
To verify that the tangential and normal components of the acceleration agree with the formulas in the book:
at[t]== speed'[t] //Simplify
an[t]==curv [t] speed[t]2 //Simplify
13.6 PLANETARY MOTION AND SATELLITES
1.
T#
a$
œ
41 #
GM
Ê T# œ
41 #
GM
a$ Ê T# œ
41 #
a6.6726‚10"" Nm# kg# b a5.975‚10#% kgb
(6,808,000 m)$
¸ 3.125 ‚ 10( sec# Ê T ¸ È3125 ‚ 10% sec# ¸ 55.90 ‚ 10# sec ¸ 93.2 min
2. e œ 0.0167 and perihelion distance œ 149,577,000 km and e œ
Ê 0.0167 œ
(149,577,000,000 m)v#!
a6.6726‚10"" Nm# kg# b a1.99‚10$! kgb
r! v#!
GM
1
1 Ê v#! ¸ 9.03 ‚ 10) m# /sec#
Ê v! ¸ È9.03 ‚ 10) m# /sec# ¸ 3.00 ‚ 10% m/sec
T#
41 #
GM #
$
a$ œ GM Ê a œ 41# T
c""
#
#
#%
$
a6.6726‚10
Nm kg b ˆ5.975‚10 kg‰
Ê a$ œ
(5535 sec)# œ 3.094 ‚ 10#! m$ Ê a ¸ È3.094
41 #
œ 6.764 ‚ 10' m ¸ 6764 km. Note that 6764 km ¸ "# a12,757 km 183 km 589 kmb.
3. 92.25 min œ 5535 sec and
‚ 10#! m$
#
4. T œ 1639 min œ 98,340 sec and mass of Mars œ 6.418 ‚ 10#$ kg Ê a$ œ GM
41# T
""
#
#
#$
#
$
‚10 kgb (98,340 sec)
œ a6.6726‚10 Nm kg b 4a6.418
¸ 1.049 ‚ 10## m$ Ê a ¸ È1.049 ‚ 10## m$
1#
œ 2.19 ‚ 10( m œ 21,900 km
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 13.6 Planetary Motion and Satellites
5. 2a œ diameter of Mars perigee height apogee height œ D 1499 km 35,800 km
Ê 2(21,900) km œ D 37,299 km Ê D œ 6501 km
6. a œ 22,030 km œ 2.203 ‚ 10( m and T# œ
Ê T# œ
#
41
a6.6720‚10"" Nm# kg# b a6.418‚10#$ kgb
41 #
GM
a$
(2.203 ‚ 10( m)$ ¸ 9.856 ‚ 10* sec#
Ê T ¸ È9.856 ‚ 10) sec# ¸ 9.928 ‚ 10% sec ¸ 1655 min
7. (a) Period of the satellite œ rotational period of the Earth Ê period of the satellite œ 1436.1 min
œ 86,166 sec; a$ œ
a6.6726‚10"" Nm# kg# b ˆ5.975‚10#% kg‰ (86,166 sec)#
GMT#
41 #
$
Ê a$ œ
41 #
$
##
È
#"
$
¸ 7.4980 ‚ 10 m Ê a ¸ 74.980 ‚ 10 m ¸ 4.2168 ‚ 10( m œ 42,168 km
(b) The radius of the Earth is approximately 6379 km Ê the height of the orbit is 42,168 6379 œ 35,789 km
(c) Symcom 3, GOES 4, and Intelsat 5
GMT#
41 #
a6.6726‚10c"" Nm# kg# b a6.418‚10#$ kgb (88,644 sec)#
œ
41 #
(
8. T œ 1477.4 min œ 88,644 sec Ê a$ œ
œ
$
8.524 ‚ 10#" m$ Ê a ¸ È
8.524 ‚ 10#" m$
¸ 2.043 ‚ 10 m œ 20,430 km
9. Period of the Moon œ 2.36055 ‚ 10' sec Ê a$ œ
œ
a6.6726‚10c"" Nm# kg# b ˆ5.975‚10#% kg‰ (2.36055‚10' sec)#
41 #
)
GMT#
41 #
$
¸ 5.627 ‚ 10#& m$ Ê a ¸ È
5.627 ‚ 10#& m$
¸ 3.832 ‚ 10 m œ 383,200 km from the center of the Earth.
10. r œ
Ê v# œ
GM
v#
11. Solar System:
T#
a$
œ
É a6.6726‚10
Ê kvk œ É GM
r œ
"" Nm# kg# b a5.975‚10#% kgb
41 #
a6.6726‚10"" Nm# kg# b a1.99‚10$! kgb
¸ 2.97 ‚ 10"* sec# /m$ ;
#
r
¸ 1.9967 ‚ 10( r"Î# m/sec
Earth:
#
T
a$
œ
41
a6.6726‚10"" Nm# kg# b a5.975‚10#% kgb
¸ 9.902 ‚ 10"% sec# /m$ ;
Moon:
T#
a$
œ
41 #
a6.6726‚10"" Nm# kg# b a7.354‚10## kgb
¸ 8.045 ‚ 10"# sec# /m$ ;
r! v#!
GM
12. e œ
GM
r
1 Ê v#! œ
GM(e 1)
r!
Ê v! œ É GM(er! 1) ;
Circle: e œ 0 Ê v! œ É GM
r!
2GM
Ellipse: 0 e 1 Ê É GM
r! v! É r!
Parabola: e œ 1 Ê v! œ É 2GM
r!
Hyperbola: e 1 Ê v! É 2GM
r!
13. r œ
Ê v# œ
GM
v#
14. ?A œ
"
#
GM
r
Ê v œ É GM
r which is constant since G, M, and r (the radius of orbit) are constant
kr(t ?t) ‚ r(t)k Ê
œ
"
#
¹ r(t ??t)t r(t) ‚ r(t)
œ
"
#
¸ ddtr ‚ r(t)¸ œ
"
#
?A
?t
œ
"
#
?t)
‚ r(t)¹ œ
¹ r(t
?t
"
#
r(t) r(t)
‚ r(t)¹
¹ r(t ?t) ?
t
r(t) ‚ r(t)¹ œ #" ¹ r(t ??t)t r(t) ‚ r(t)¹ Ê
¸r(t) ‚ ddtr ¸ œ "# kr ‚ rÞ k
"
?t
dA
dt
œ lim
"
?t Ä 0 #
¹ r(t ??t)t r(t) ‚ r(t)¹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
851
852
Chapter 13 Vector-Valued Functions and Motion in Space
#
# %
r v#
# %
#
! !
15. T œ Š 2r!1va! ‹ È1 e# Ê T# œ Š 4r1# va# ‹ a1 e# b œ Š 4r1# va# ‹ ”1 Š GM
1‹ • (from Equation 32)
! !
! !
r# v %
# %
r v#
# %
! !
œ Š 4r1# va# ‹ ’ G!# M!# 2 Š GM
‹“ œ Š 4r1# va# ‹ ’
!
!
# %
œ a41 a
!
2GM r v#
2 ‰
b Š 2r! GM! ! ‹ ˆ GM
!
2GMr! v#! r#! v%!
“
G# M#
œ
ˆ41# a% ‰ a2GM r! v#! b
r! G# M#
" ‰ˆ 2 ‰
#
œ a41 a b ˆ 2a
GM (from Equation 35) Ê T œ
# %
4 1 # a$
GM
Ê
T#
a$
œ
41 #
GM
16. Let rAB (t) denote the vector from planet A to planet B at time t. Then rAB (t) œ rB (t) rA (t)
œ [3 cos (1t) 2 cos (21t)]i [3 sin (1t) 2 sin (21t)]j
œ c3 cos (1t) 2 acos# (1t) sin# (1t)bd i [3 sin (1t) 4 sin (1t) cos (1t)]j
œ c3 cos (1t) 4 cos# (1t) 2d i [(3 4 cos (1t)) sin (1t)]j Ê parametric equations for the path are
x(t) œ 2 [3 4 cos (1t)] cos (1t) and y(t) œ [3 4 cos (1t)] sin (1t)
17. The graph of the path of planet B is the limacon
¸
at the right.
18. (i)
(ii)
(iii)
(iv)
(v)
Perihelion is the time t such that kr(t)k is a minimum.
Aphelion is the time t such that kr(t)k is a maximum.
Equinox is the time t such that r(t) † w œ 0 .
Summer solstice is the time t such that the angle between r(t) and w is a maximum.
Winter solstice is the time t such that the angle between r(t) and w is a minimum.
CHAPTER 13 PRACTICE EXERCISES
1. r(t) œ (4 cos t)i ŠÈ2 sin t‹ j Ê x œ 4 cos t
and y œ È2 sin t Ê
x#
16
y#
#
œ 1;
v œ (4 sin t)i ŠÈ2 cos t‹ j and
a œ (4 cos t)i ŠÈ2 sin t‹ j ; r(0) œ 4 i , v(0) œ È2j ,
a(0) œ 4i ; r ˆ 14 ‰ œ 2È2i j , v ˆ 14 ‰ œ 2È2i j ,
a ˆ 1 ‰ œ 2È2i j ; kvk œ È16 sin# t 2 cos# t
4
Ê aT œ
d
dt
kvk œ
at t œ 14 : aT œ
14 sin t cos t
È16 sin# t2 cos# t
7
È 8 1
œ
7
3
; at t œ 0: aT œ 0, aN œ Ékak# 0 œ 4, , œ
, aN œ É9
49
9
œ
4È 2
3
,,œ
aN
kv k #
œ
aN
kv k #
œ
4
2
œ 2;
4È 2
27
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 13 Practice Exercises
2. r(t) œ ŠÈ3 sec t‹ i ŠÈ3 tan t‹ j Ê x œ È3 sec t and y œ È3 tan t Ê
x#
3
y#
3
853
œ sec# t tan# t œ 1;
Ê x# y# œ 3; v œ ŠÈ3 sec t tan t‹ i ŠÈ3 sec# t‹ j
and
a œ ŠÈ3 sec t tan# t È3 sec$ t‹ i Š2È3 sec# t tan t‹ j ;
r(0) œ È3i , v(0) œ È3j , a(0) œ È3i ;
kvk œ È3 sec# t tan# t 3 sec% t
Ê aT œ
d
dt
kvk œ
6 sec# t tan$ t 18 sec% t tan t
2È3 sec# t tan# t 3 sec% t
;
at t œ 0: aT œ 0, aN œ Ékak# 0 œ È3,
,œ
3. r œ
œ
È3
3
œ
"
È1 t#
i
t
È1 t#
aN
kv k #
"
È3
j Ê v œ t a1 t# b
$Î#
i a 1 t# b
#
#
Ê kvk œ Ê’t a1 t# b$Î# “ ’a1 t# b$Î# “ œ
d kv k
dt
œ0 Ê
2t
a1 t# b#
œ 0 Ê t œ 0. For t 0,
"
1 t#
2t
a1 t# b#
$Î#
j
. We want to maximize kvk :
0; for t 0,
2t
a1 t# b#
d kv k
dt
œ
2t
a1 t# b#
and
0 Ê kvk max occurs when
t œ 0 Ê kvk max œ 1
4. r œ aet cos tb i aet sin tb j Ê v œ aet cos t et sin tb i aet sin t et cos tb j
Ê a œ aet cos t et sin t et sin t et cos tb i aet sin t et cos t et cos t et sin tb j
œ a2et sin tb i a2et cos tb j . Let ) be the angle between r and a . Then ) œ cos" Š krrk†kaak ‹
œ cos"
2e2t sin t cos t2e2t sin t cos t
Éaet cos tb# aet sin tb# Éa2et sin tb# a2et cos tb#
â
âi
â
5. v œ 3i 4j and a œ 5i 15j Ê v ‚ a œ â 3
â
â5
Ê ,œ
6. , œ
kv ‚ a k
kv k $
kyww k
$Î#
1 ayw b# ‘
œ ex a1 e2x b
d,
dx
œ
25
5$
œ
$Î#
&Î#
œ 0 Ê a1 2e2x b œ 0 Ê e2x œ
maximum at the point Š ln È2ß
x# y# œ 1, 2x
dx
dt
dx
dt
i
2y
dy
dt
dy
dt
d,
dx
Ê
3e3x a1 e2x b
7. r œ xi yj Ê v œ
for all t
â
kâ
â
0 â œ 25k Ê kv ‚ ak œ 25; kvk œ È3# 4# œ 5
â
0â
j
4
"5
"
5
œ ex a1 e2x b
$Î#
1
#
œ cos" Š 2e02t ‹ œ cos" 0 œ
"
#
œ ex a1 e2x b
œ ex a1 e2x b
$Î#
ex ’ #3 a1 e2x b
&Î#
&Î#
a2e2x b“
&Î#
ca1 e2x b 3e2x d œ ex a1 e2x b
a1 2e2x b ;
Ê 2x œ ln 2 Ê x œ "# ln 2 œ ln È2 Ê y œ È"2 ; therefore , is at a
"
È2 ‹
j and v † i œ y Ê
œ0 Ê
dy
dt
œ
x dx
y dt
dx
dt
œ y. Since the particle moves around the unit circle
Ê
dy
dt
œ yx (y) œ x. Since
dx
dt
œ y and
dy
dt
œ x, we have
v œ yi xj Ê at (1ß 0), v œ j and the motion is clockwise.
dy
dy
" # dx
# dx
dt œ 3x dt Ê dt œ 3 x dt . If r œ xi yj , where x and y are differentiable functions of
dy
dy
dx
" # dx
"
#
then v œ dx
dt i dt j. Hence v † i œ 4 Ê dt œ 4 and v † j œ dt œ 3 x dt œ 3 (3) (4) œ 12 at (3ß 3). Also,
#
#
#
#
#
‰# ˆ 3" x# ‰ ddt#x . Hence a † i œ 2 Ê ddt#x œ 2 and
a œ ddt#x i ddt#y j and ddt#y œ ˆ 23 x‰ ˆ dx
dt
#
a † j œ ddt#y œ 23 (3)(4)# "3 (3)# (2) œ 26 at the point (xß y) œ (3ß 3).
8. 9y œ x$ Ê 9
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
t,
854
9.
Chapter 13 Vector-Valued Functions and Motion in Space
dr
dt
orthogonal to r Ê 0 œ
†rœ
dr
dt
" dr
# dt
† r "# r †
œ
dr
dt
#
" d
# dt
#
(r † r) Ê r † r œ K, a constant. If r œ xi yj , where
x and y are differentiable functions of t, then r † r œ x y Ê x# y# œ K, which is the equation of a circle
centered at the origin.
10. (b) v œ (1 1 cos 1t)i (1 sin 1t)j
Ê a œ a1# sin 1tb i a1# cos 1tb j ;
v(0) œ 0 and a(0) œ 1# j ;
v(1) œ 21i and a(1) œ 1# j ;
v(2) œ 0 and a(2) œ 1# j ;
v(3) œ 21i and a(3) œ 1# j
(c) Forward speed at the topmost point is kv(1)k œ kv(3)k œ 21 ft/sec; since the circle makes
"
#
revolution per
second, the center moves 1 ft parallel to the x-axis each second Ê the forward speed of C is 1 ft/sec.
11. y œ y! (v! sin !)t "# gt# Ê y œ 6.5 (44 ft/sec)(sin 45°)(3 sec) "# a32 ft/sec# b (3 sec)# œ 6.5 66È2 144
¸ 44.16 ft Ê the shot put is on the ground. Now, y œ 0 Ê 6.5 22È2t 16t# œ 0 Ê t ¸ 2.13 sec (the
positive root) Ê x ¸ (44 ft/sec)(cos 45°)(2.13 sec) ¸ 66.27 ft or about 66 ft, 3 in. from the stopboard
12. ymax œ y!
(v! sin !)#
#g
œ 7 ft
[(80 ft/sec)(sin 45°)]#
(2) a32 ft/sec# b
¸ 57 ft
(v! sin !)t "# gt#
(v sin !) " gt
y
œ ! v! cos ! #
x œ
(v! cos !)t
2v! sin ! 2v! cos ! tan 9
, which is the time when
g
13. x œ (v! cos !)t and y œ (v! sin !)t "# gt# Ê tan 9 œ
Ê v! cos ! tan 9 œ v! sin ! "# gt Ê t œ
the golf ball
hits the upward slope. At this time
x œ (v! cos !) Š 2v! sin ! 2vg ! cos ! tan 9 ‹
œ Š 2g ‹ av#! sin ! cos ! v#! cos# ! tan 9b . Now
OR œ
x
cos 9
Ê OR œ Š g2 ‹ Š
v#! sin ! cos ! v#! cos# ! tan 9
‹
cos 9
œŠ
2v#! cos !
sin !
‹ Š cos
g
9
œŠ
2v#! cos !
9 cos ! sin 9
‹ Š sin ! cos cos
‹
#9
g
œŠ
2v#! cos !
g cos# 9 ‹ [sin (!
cos ! tan 9
cos 9 ‹
9)]. The distance OR is maximized
when x is maximized:
dx
d!
œŠ
2v#!
g ‹(cos
Ê cot 2! œ tan (9) Ê 2! œ
14. R œ
v#!
g
1
#
2! sin 2! tan 9) œ 0 Ê (cos 2! sin 2! tan 9) œ 0 Ê cot 2! tan 9 œ 0
9 Ê !œ
9
#
1
4
#
ft) a32 ft/sec b
sin 2! Ê v! œ É sinRg2! ; for 4325 yards: 4325 yards œ 12,975 ft Ê v! œ É (12,975(sin
90°)
#
ft) a32 ft/sec b
¸ 644 ft/sec; for 4752 yards: 4752 yards œ 14,256 ft Ê v! œ É (14,256(sin
¸ 675 ft/sec
90°)
15. (a) R œ
v#!
g
v#
#
#
#
!
sin 2! Ê 109.5 ft œ Š 32 ft/sec
Ê v! œ È3504 ft# /sec#
# ‹ (sin 90°) Ê v! œ 3504 ft /sec
¸ 59.19 ft/sec
(b) x œ (v! cos !)t and y œ 4 (v! sin !)t "# gt# ; when the cork hits the ground, x œ 177.75 ft and y œ 0
Ê 177.75 œ Šv!
Ê v! œ
(177.75)È2
t
"
È2 ‹ t
œ
and 0 œ 4 Šv!
4(177.75)È2
È181.75
"
È2 ‹ t
16t# Ê 16t# œ 4 177.75 Ê t œ
È181.75
4
¸ 74.58 ft/sec
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 13 Practice Exercises
855
5
16. (a) x œ v! (cos 40°)t and y œ 6.5 v! (sin 40°)t "# gt# œ 6.5 v! (sin 40°)t 16t# ; x œ 262 12
ft and y œ 0 ft
5
Ê 262 12
œ v! (cos 40°)t or v! œ
262.4167
#
#
and 0 œ 6.5 ’ (cos
40°)t “ (sin 40°)t 16t Ê t œ 14.1684
262.4167
(cos 40°)t
Ê t ¸ 3.764 sec. Therefore, 262.4167 ¸ v! (cos 40°)(3.764 sec) Ê v! ¸
#
(v! sin !)
2g
(b) ymax œ y!
a(91)(sin 40°)b
(2)(32)
¸ 6.5
2
262.4167
(cos 40°)(3.764 sec)
Ê v! ¸ 91 ft/sec
¸ 60 ft
#
#
17. x# œ av!# cos# !b t# and ˆy "# gt# ‰ œ av!# sin# !b t# Ê x# ˆy "# gt# ‰ œ v!# t#
Þ ÞÞ Þ ÞÞ
ÞÞ# ÞÞ# ÞÞ# ÞÞ# ÞÞ# axÞ ÞÞx yÞ ÞÞyb#
x x y y
Ê
x y s œ x y xÞ # yÞ #
Þ
Þ
#
#
Èx y
ÞÞ# ÞÞ# Þ # Þ #
Þ # ÞÞ#
Þ ÞÞ Þ ÞÞ Þ # ÞÞ#
Þ ÞÞ Þ ÞÞ
Þ ÞÞ
Þ ÞÞ
Þ ÞÞ Þ ÞÞ
ax y b ax y b ax x 2x x y y y y b
ax y y x b#
x# y# y# x# 2x x y y
œ
œ
œ
Þ# Þ#
Þ# Þ#
Þ
Þ
x y
x y
x# y#
Þ ÞÞ Þ ÞÞ
Þ # Þ # $Î#
Þ
Þ
#
#
kx y y x k
ax y b
ÞÞ
ÞÞ
ÞÞ
x y
Ê È x# y# s # œ È Þ # Þ # Ê ÈÞÞ# ÞÞ# ÞÞ# œ kxÞ ÞÞy yÞ ÞÞxk œ ," œ 3
x y
x y s
ÞÞ
18. s œ
d
dt
ÈxÞ # yÞ # œ
19. r(t) œ ’'0 cos ˆ "# 1)# ‰ d)“ i ’'0 sin ˆ "# 1)# ‰ d)“ j Ê v(t) œ cos Š 1#t ‹ i sin Š 1#t
â
i
j
â
â
#
1t#
#
#
â
sin Š 1#t ‹
a(t) œ 1t sin Š 1#t ‹ i 1t cos Š 1#t ‹ j Ê v ‚ a œ â cos Š # ‹
â
â 1t sin Š 1t# ‹ 1t cos Š 1t# ‹
â
#
#
t
t
œ 1 tk Ê , œ
20. s œ a) Ê ) œ
kv ‚ a k
kv k $
s
a
#
œ 1t; kv(t)k œ
Ê 9œ
s
a
1
#
ds
dt
Ê
#
‹ j Ê kvk œ 1;
k ââ
â
0 ââ
â
0 ââ
œ 1 Ê s œ t C; r(0) œ 0 Ê s(0) œ 0 Ê C œ 0 Ê , œ 1s
d9
ds
œ
"
a
Ê , œ ¸ "a ¸ œ
"
a
since a 0
21. r œ (2 cos t)i (2 sin t)j t# k Ê v œ (2 sin t)i (2 cos t)j 2tk Ê kvk œ È(2 sin t)# (2 cos t)# (2t)#
œ 2È1 t# Ê Length œ '0 2È1 t# dt œ ’tÈ1 t# ln ¹t È1 t# ¹“
1Î4
1Î%
!
œ
1
4
É1
1#
16
ln Š 14 É1
22. r œ (3 cos t)i (3 sin t)j 2t$Î# k Ê v œ (3 sin t)i (3 cos t)j 3t"Î# k
1#
16 ‹
$
#
Ê kvk œ É(3 sin t)# (3 cos t)# a3t"Î# b œ È9 9t œ 3È1 t Ê Length œ '0 3È1 t dt œ 2(1 t)$Î# ‘ !
3
œ 14
23. r œ
4
9
(1 t)$Î# i 49 (1 t)$Î# j 3" tk Ê v œ
#
2
3
(1 t)"Î# i 32 (1 t)"Î# j 3" k
#
#
Ê kvk œ É 23 (1 t)"Î# ‘ 23 (1 t)"Î# ‘ ˆ 3" ‰ œ 1 Ê T œ
i 23 j 3" k ;
"
3
(1 t)"Î# i
â
â
â
Ê N(0) œ È" i È" j ; B(0) œ T(0) ‚ N(0) œ ââ
2
2
â
â
Ê T(0) œ
2
3
dT
dt
œ
2
3
(1 t)"Î# i 23 (1 t)"Î# j 3" k
"
3
(1 t)"Î# j Ê ddtT (0) œ 3" i 3" j Ê ¸ ddtT (0)¸ œ
â
i
j
kâ
2
" â
"
"
4
â
23
3
3 â œ È i È j È k;
3 2
3 2
3 2
"
"
0 ââ
È2
È2
a œ "3 (1 t)"Î# i 3" (1 t)"Î# j Ê a(0) œ 3" i 3" j and v(0) œ 32 i 32 j 3" k Ê v(0) ‚ a(0)
â i
j
k ââ
È
â
Š 32 ‹
È2
È2
â 2 2 "â
kv ‚a k
"
"
4
i
j
k
k
v
a
k
ω 3
œ
Ê
‚
œ
Ê
,
(0)
œ
œ
â
3
3
9
9
9
3
1$ œ 3 ;
kv k $
â "
â
"
â 3
0â
3
Þ
Þ
a œ "6 (1 t)$Î# i "6 (1 t)$Î# j Ê a(0) œ 6" i 6" j Ê 7 (0) œ
â 2
â 3
â
â "
â 3
â "
â 6
23
"
3
"
6
kv ‚a k #
â
â
â
0 ââ
0 ââ
"
3
œ
2 ‰
ˆ "3 ‰ ˆ 18
Š
È2 ‹# œ
3
t œ 0 Ê ˆ 49 ß 49 ß 0‰ is the point on the curve
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
6
;
È2
3
856
Chapter 13 Vector-Valued Functions and Motion in Space
24. r œ aet sin 2tb i aet cos 2tb j 2et k Ê v œ aet sin 2t 2et cos 2tb i aet cos 2t 2et sin 2tb j 2et k
Ê kvk œ Éaet sin 2t 2et cos 2tb# aet cos 2t 2et sin 2tb# a2et b# œ 3et Ê T œ
œ ˆ "3 sin 2t
dT
dt
2
3
cos 2t‰ i ˆ 3" cos 2t
œ ˆ 23 cos 2t
Ê N(0) œ
2
3
sin 2t‰ j 23 k Ê T(0) œ
sin 2t‰ i ˆ 23 sin 2t
4
3
4
3
cos 2t‰ j Ê
dT
dt
È
Š2 3 5‹
i 3" j 32 k ;
i 43 j Ê ¸ ddtT (0)¸ œ 23 È5
â
j
kâ
1
2 â
4
2
5
â
3
3 â œ È i È j È k;
3 5
3 5
3 5
2
â
È5 0 â
(0) œ
â
â i
â 2
2
"
œ È5 i È5 j ; B(0) œ T(0) ‚ N(0) œ ââ 3
â "
â È5
ˆ 23 i 43 j‰
2
3
2
3
v
kv k
a œ a4et cos 2t 3et sin 2tb i a3et cos 2t 4et sin 2tb j 2et k Ê a(0) œ 4i 3j 2k and v(0) œ 2i j 2k
â
â
j kâ
âi
â
â
Ê v(0) ‚ a(0) œ â 2 " 2 â œ 8i 4j 10k Ê kv ‚ ak œ È64 16 100 œ 6È5 and kv(0)k œ 3
â
â
â 4 3 2 â
Ê ,(0) œ
6È 5
3$
œ
2È 5
9
t
â
â 2
â
â 4
â
â 2
1
3
11
;
Þ
t
a œ a4e cos 2t 8e sin 2t 3et sin 2t 6et cos 2tb i a3et cos 2t 6et sin 2t 4et sin 2t 8et cos 2tb j 2et k
Þ
œ a2et cos 2t 11et sin 2tb i a11 et cos 2t 2et sin 2tb j 2et k Ê a(0) œ 2i 11j 2k
Ê 7 (0) œ
kv ‚ a k #
â
2â
â
2â
â
2â
œ
80
180
œ 94 ; t œ 0 Ê (!ß "ß 2) is on the curve
25. r œ ti "# e2t j Ê v œ i e2t j Ê kvk œ È1 e4t Ê T œ
dT
dt
œ
2 e
ˆ1 e4t ‰$Î#
4t
i
2t
2e
ˆ1 e4t ‰$Î#
j Ê
dT
dt
(ln 2) œ
32
17È17
i
8
17È17
"
È1 e4t
i
e2t
È1 e4t
j Ê T (ln 2) œ
j Ê N (ln 2) œ È417 i
"
È17
"
È17
i
4
È17
j;
j;
â i
j
k ââ
â
â "
4
0 ââ œ k ; a œ 2e2t j Ê a(ln 2) œ 8j and v(ln 2) œ i 4j
È17
B (ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ È17
â
"
â 4
â
â È17 È17 0 â
â
â
â i j kâ
â
â
Þ
8
Ê v(ln 2) ‚ a(ln 2) œ â " 4 0 â œ 8k Ê kv ‚ ak œ 8 and kv(ln 2)k œ È17 Ê ,(ln 2) œ 17È
; a œ 4e2t j
17
â
â
â0 8 0â
Þ
Ê a(ln 2) œ 16j Ê 7 (ln 2) œ
â
â1
â
â0
â
â0
4
8
16
kv ‚a k #
â
0â
â
0â
â
0â
œ 0; t œ ln 2 Ê (ln 2ß 2ß 0) is on the curve
26. r œ (3 cosh 2t)i (3 sinh 2t)j 6tk Ê v œ (6 sinh 2t)i (6 cosh 2t)j 6k
Ê kvk œ È36 sinh# 2t 36 cosh# 2t 36 œ 6È2 cosh 2t Ê T œ kvvk œ Š È" tanh 2t‹ i
2
Ê T(ln 2) œ
#
15
17È2
i
"
È2
j
8
17È2
k;
8 ‰
8 ‰ ˆ 15 ‰
œ Š È22 ‹ ˆ 17
i Š È22 ‹ˆ 17
17 k œ
dT
dt
œ
Š È22
sech 2t‹ i
Š È22
sech 2t tanh 2t‹ k Ê
j Š È" sech 2t‹ k
dT
dt
(ln 2)
2
#
#
È
8 2
128
240
k Ê ¸ ddtT (ln 2)¸ œ ÊŠ 289
È2 ‹ Š 289È2 ‹ œ 17
â
â
j
k â
â i
â
â
"
15
8
"
8
15
8
Ê N(ln 2) œ 17
i 17
k ; B(ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ 17È2 È2 17È2 ââ œ 1715
È2 i È2 j 17È2 k ;
â 8
â
0 15
â 17
17 â
17 ‰
15 ‰
51
ˆ
ˆ
a œ (12 cosh 2t)i (12 sinh 2t)j Ê a(ln 2) œ 12 8 i 12 8 j œ # i 45
# j and
â i
j k ââ
â
â
45
51
45
51
‰
ˆ 17 ‰
6 ââ
v(ln 2) œ 6 ˆ 15
4
8 i 6 8 j 6k œ 4 i 4 j 6k Ê v(ln 2) ‚ a(ln 2) œ ââ 4
â
45
â 51
0â
2
#
128
289È2
i
#
"
È2
240
289È2
œ 135i 153j 72k Ê kv ‚ ak œ 153È2 and kv(ln 2)k œ
51
4
È2 Ê ,(ln 2) œ
153È2
$
Š 51 È2‹
œ
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
32
867
;
Chapter 13 Practice Exercises
Þ
Þ
a œ (24 sinh 2t)i (24 cosh 2t)j Ê a(ln 2) œ 45i 51j Ê 7 (ln 2) œ
â 45
â 4
â 5"
â 2
â
â 45
51
4
45
2
51
kv ‚a k #
â
6â
â
0â
â
0â
œ
32
867
857
;
45
‰
t œ ln 2 Ê ˆ 51
8 ß 8 ß 6 ln 2 is on the curve
27. r œ a2 3t 3t# b i a4t 4t# b j (6 cos t)k Ê v œ (3 6t)i (4 8t)j (6 sin t)k
Ê kvk œ È(3 6t)# (4 8t)# (6 sin t)# œ È25 100t 100t# 36 sin# t
"Î#
"
#
a25 100t 100t# 36 sin# tb
(100 200t 72 sin t cos t) Ê aT (0) œ ddtkvk (0) œ 10;
a œ 6i 8j (6 cos t)k Ê kak œ È6# 8# (6 cos t)# œ È100 36 cos# t Ê ka(0)k œ È136
Ê
d kv k
dt
œ
Ê aN œ Ékak# a#T œ È136 10# œ È36 œ 6 Ê a(0) œ 10T 6N
28. r œ (2 t)i at 2t# b j a1 t# b k Ê v œ i (1 4t)j 2tk Ê kvk œ È1# (1 4t)# (2t)#
"Î#
œ È2 8t 20t# Ê ddtkvk œ "# a2 8t 20t# b
(8 40t) Ê aT œ ddtkvk (0) œ 2È2; a œ 4j 2k
#
Ê kak œ È4# 2# œ È20 Ê aN œ Ékak# aT# œ Ê20 Š2È2‹ œ È12 œ 2È3 Ê a(0) œ 2È2T 2È3N
29. r œ (sin t)i ŠÈ2 cos t‹ j (sin t)k Ê v œ (cos t)i ŠÈ2 sin t‹ j (cos t)k
#
Ê kvk œ Ê(cos t)# ŠÈ2 sin t‹ (cos t)# œ È2 Ê T œ
v
kv k
œ Š È"2 cos t‹ i (sin t)j Š È"2 cos t‹ k ;
#
#
œ Š È"2 sin t‹ i (cos t)j Š È"2 sin t‹ k Ê ¸ ddtT ¸ œ ÊŠ È"2 sin t‹ ( cos t)# Š È"2 sin t‹ œ 1
â
â
i
j
k
â
â
â " cos t sin t
â
"
ˆ ddtT ‰
cos
t
"
"
â
â
È2
Ê N œ ¸ dT ¸ œ Š È2 sin t‹ i (cos t)j Š È2 sin t‹ k ; B œ T ‚ N œ â È2
â
dt
â " sin t cos t " sin t â
â È2
â
È2
â i
â
j
k
â
â
â
â
œ È"2 i È"2 k ; a œ ( sin t)i ŠÈ2 cos t‹ j (sin t)k Ê v ‚ a œ â cos t È2 sin t cos t â
â
â
â sin t È2 cos t sin t â
Þ
œ È2 i È2 k Ê kv ‚ ak œ È4 œ 2 Ê , œ kv‚$ak œ 2 $ œ " ; a œ ( cos t)i ŠÈ2 sin t‹ j (cos t)k
dT
dt
kv k
Ê 7œ
â
â cos t
â
â sin t
â
â
â cos t
â
cos t ââ
È2 sin t
È2 cos t sin t ââ
È2 sin t cos t ââ
kv ‚a k #
œ
ŠÈ2‹
È2
(cos t) ŠÈ2‹ ŠÈ2 sin t‹ (0) (cos t) ŠÈ2‹
4
œ0
30. r œ i (5 cos t)j (3 sin t)k Ê v œ (5 sin t)j (3 cos t)k Ê a œ (5 cos t)j (3 sin t)k
Ê v † a œ 25 sin t cos t 9 sin t cos t œ 16 sin t cos t; v † a œ 0 Ê 16 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0
Ê t œ 0, 1# or 1
31. r œ 2i ˆ4 sin #t ‰ j ˆ3 1t ‰ k Ê 0 œ r † (i j) œ 2(1) ˆ4 sin #t ‰ (1) Ê 0 œ 2 4 sin
Ê tœ
1
3
t
#
Ê sin
t
#
œ
"
#
Ê
t
#
œ
(for the first time)
32. r(t) œ ti t# j t$ k Ê v œ i 2tj 3t# k Ê kvk œ È1 4t# 9t% Ê kv(1)k œ È14
Ê T(1) œ È"14 i È214 j È314 k , which is normal to the normal plane
Ê
"
È14
(x 1)
2
È14
(y 1)
3
È14
(z 1) œ 0 or x 2y 3z œ 6 is an equation of the normal plane. Next we
calculate N(1) which is normal to the rectifying plane. Now, a œ 2j 6tk Ê a(1) œ 2j 6k Ê v(1) ‚ a(1)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
6
858
Chapter 13 Vector-Valued Functions and Motion in Space
â
âi
â
œ â"
â
â0
œ
"
#
œ
22
È14
â
j kâ
â
2 3 â œ 6i 6j 2k Ê kv(1) ‚ a(1)k œ È76 Ê ,(1) œ
â
2 6â
a1 4t# 9t% b
"Î#
2j3k
Š iÈ
‹
14
a8t 36t$ b¹
È19
7È14
#
œ
tœ1
22
È14
, so a œ
ŠÈ14‹ N Ê N œ
È14
2È19
d# s
dt#
È76
$
È
Š 14‹
œ
È19
7È14
;
ds
dt
œ kv(t)k Ê
d# s
dt# ¹ tœ1
#
‰ N Ê 2j 6k
T , ˆ ds
dt
8
9 ‰
11
8
9
ˆ 11
7 i 7 j 7 k Ê 7 (x 1) 7 (y 1) 7 (z 1)
œ 0 or 11x 8y 9z œ 10 is an equation of the rectifying plane. Finally, B(1) œ T(1) ‚ N(1)
â
â
j kâ
â i
È14
â
â
2 3 â œ È" (3i 3j k) Ê 3(x 1) 3(y 1) (z 1) œ 0 or 3x 3y z
œ Š 2È19 ‹ Š È" ‹ ˆ 7" ‰ â "
19
14
â
â
â 11 8 9 â
œ 1 is an equation of the osculating plane.
" ‰
33. r œ et i (sin t)j ln (1 t)k Ê v œ et i (cos t)j ˆ 1
t k Ê v(0) œ i j k ; r(0) œ i Ê (1ß 0ß 0) is on the line
Ê x œ 1 t, y œ t, and z œ t are parametric equations of the line
34. r œ ŠÈ2 cos t‹ i ŠÈ2 sin t‹ j tk Ê v œ ŠÈ2 sin t‹ i ŠÈ2 cos t‹ j k Ê v ˆ 14 ‰
œ ŠÈ2 sin 14 ‹ i ŠÈ2 cos 14 ‹ j k œ i j k is a vector tangent to the helix when t œ
is parallel to v ˆ 14 ‰ ; also r ˆ 14 ‰ œ ŠÈ2 cos 14 ‹ i ŠÈ2 sin 14 ‹ j
Ê x œ 1 t, y œ 1 t, and z œ
35. (a) ?SOT ¸ ?TOD Ê
Ê y! œ
6380#
6817
DO
OT
œ
1
4
1
4
1
4
Ê the tangent line
k Ê the point ˆ1ß 1ß 14 ‰ is on the line
t are parametric equations of the line
OT
SO
Ê
y!
6380
œ
6380
6380437
Ê y! ¸ 5971 km;
(b) VA œ '5971 21x Ê1 Š dx
dy ‹ dy
#
6380
6380
œ 21'5971 È6380# y# Š È6380
# y# ‹ dy
6817
œ 21 '5971 6380 dy œ 21 c6380yd ')"(
&*("
6817
œ 16,395,469 km# ¸ 1.639 ‚ 10( km# ;
(c) percentage visible ¸
16,395,469 km#
41(6380 km)#
¸ 3.21%
CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES
1. (a) The velocity of the boat at (xß y) relative to land is the sum of the velocity due to the rower and the
"
velocity of the river, or v œ 250
(y 50)# 10‘ i 20j . Now, dy
dt œ 20 Ê y œ 20t c; y(0) œ 100
"
Ê c œ 100 Ê y œ 20t 100 Ê v œ 250
(20t 50)# 10‘ i 20j œ ˆ 85 t# 8t‰ i 20j
8 $
Ê r(t) œ ˆ 15
t 4t# ‰ i 20tj C" ; r(0) œ 0i 100j Ê 100j œ C" Ê r(t)
8 $
œ ˆ 15
t 4t# ‰ i (100 20t) j
(b) The boat reaches the shore when y œ 0 Ê 0 œ 20t 100 from part (a) Ê t œ 5
8
100
‰
Ê r(5) œ ˆ 15
† 125 4 † 25‰ i (100 20 † 5)j œ ˆ 200
3 100 i œ 3 i ; the distance downstream is
therefore
100
3
m
2. (a) Let ai bj be the velocity of the boat. The velocity of the boat relative to an observer on the bank of the
river is v œ ai ’b
3x(20 x)
“j.
100
x œ at Ê v œ ai ’b
The distance x of the boat as it crosses the river is related to time by
3at(20 at)
“j
100
œ ai Šb
3a# t# 60at
‹j
100
Ê r(t) œ ati Šbt
a# t$
100
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
30at#
100 ‹ j
C;
Chapter 13 Additional and Advanced Exercises
a# t$ 30at#
‹j.
100
r(0) œ 0i 0j Ê C œ 0 Ê r(t) œ ati Šbt
Ê 20 œ at Ê t œ
œ
2000b 8000 12,000
100a
#
ˆ 20
‰$
a
The boat reaches the shore when x œ 20
‰
30a ˆ 20
a
100
#
$
#
(20) 30(20)
œ 20b
a
100a
#
#
È
È
Ê b œ 2; the speed of the boat is 20 œ kvk œ a b œ Èa# 4 Ê a# œ 16
20
a
‰
and y œ 0 Ê 0 œ b ˆ 20
a
a
Ê a œ 4; thus, v œ 4i 2j is the velocity of the boat
a# t$ 30at#
‹j
100
(b) r(t) œ ati Šbt
$
(c) x œ 4t and y œ 2t
œ
œ
16t$
100
œ 4ti Š2t
120t#
100 ‹ j
by part (a), where 0 Ÿ t Ÿ 5
#
16t
120t
100 100
4 $
6 #
2
#
25 t 5 t 2t œ 25 t a2t 15t 25b
2
25 t(2t 5)(t 5), which is the graph of
the cubic displayed here
3. (a) r()) œ (a cos ))i (a sin ))j b)k Ê
œ Èa# b#
(b)
d)
dt
d)
dt
Ê
)
œ É a#2gb
b# Ê
d)
dt
d)
È)
dr
dt
œ [(a sin ))i (a cos ))j bk]
)
É a#2gb
œ É a#2gz
b# œ
b# Ê
gbt#
2 aa # b # b
; z œ b) Ê z œ
œ [(a sin ))i (a cos ))j bk]
d)
dt
; kvk œ È2gz œ ¸ ddtr ¸
œ É a#41gbb# œ 2É a#1gbb#
gb# t#
2 aa # b # b
œ [(a sin ))i (a cos ))j bk] Š a# gbt
b# ‹ , from part (b)
i (a cos ))j bk
Ê v(t) œ ’ (a sin ))È
“ Š Èagbt
‹œ
# b#
a# b#
d# r
dt#
d) ¸
dt )œ#1
d)
dt
"Î#
œ É a#2gb
œ É a#2gb
b# dt Ê 2)
b# t C; t œ 0 Ê ) œ 0 Ê C œ 0
Ê 2)"Î# œ É a#2gb
b# t Ê ) œ
(c) v(t) œ
dr
dt
gbt
È a# b #
T;
#
œ [(a cos ))i (a sin ))j] ˆ ddt) ‰ [(a sin ))i (a cos ))j bk]
#
d# )
dt#
gb
œ Š a# gbt
b# ‹ [(a cos ))i (a sin ))j] [(a sin ))i (a cos ))j bk] Š a# b# ‹
#
i (a cos ))j bk
œ ’ (a sin ))È
“ Š Èa#gb b# ‹ a Š a# gbt
b# ‹ [( cos ))i (sin ))j]
a# b#
œ
gb
È a# b#
#
T a Š a# gbt
b# ‹ N (there is no component in the direction of B).
4. (a) r()) œ (a) cos ))i (a) sin ))j b)k Ê
dr
dt
# "Î#
kvk œ È2gz œ ¸ ddtr ¸ œ aa# a# )# b b
(b) s œ '0 kvk dt œ '0 aa# a# )# b# b
t
t
œ '0 aÉ a
)
#
b#
a#
(1 e)r!
1 e cos )
Ê
dr
d)
ˆ ddt) ‰ Ê
d)
dt
œ
œ
c#
#
)
ln ¹u Èc# u# ¹“ œ
(1 e)r! (e sin ))
(1 e cos ))#
!
;
dr
d)
œ0 Ê
Ê sin ) œ 0 Ê ) œ 0 or 1. Note that
dr
d)
a
#
;
È a# a# ) # b#
dt œ '0 aa# a# )# b# b
u# du œ a '0 Èc# u# du, where c œ
d)
dt
È2gb)
t
)
Ê s œ a ’ u# Èc# u#
5. r œ
"Î# d)
dt
œ [(a cos ) a) sin ))i (a sin ) a) cos ))j bk]
"Î#
d) œ '0 aa# a# u# b# b
)
"Î#
du
È a# b#
ka k
Š)Èc# )# c# ln ¹) Èc# )# ¹ c# ln c‹
(1 e)r! (e sin ))
(1 e cos ))#
œ 0 Ê (1 e)r! (e sin )) œ 0
0 when sin ) 0 and
dr
d)
0 when sin ) 0. Since sin ) 0 on
1 ) 0 and sin ) 0 on 0 ) 1, r is a minimum when ) œ 0 and r(0) œ
(1 e)r!
1 e cos 0
œ r!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
859
860
Chapter 13 Vector-Valued Functions and Motion in Space
6. (a) f(x) œ x 1
"
#
sin x œ 0 Ê f(0) œ 1 and f(2) œ 2 1
"
#
"
#
sin 2
since ksin 2k Ÿ 1; since f is continuous
on [0ß 2], the Intermediate Value Theorem implies there is a root between 0 and 2
(b) Root ¸ 1.4987011335179
7. (a) v œ
dx
dt
i
v†iœ
Ê
j and v œ
dr
dt
œ
dr
dt
d)
ˆ dr ‰
ˆ d) ‰
dt u) œ dt [(cos ))i (sin ))j] r dt [( sin ))i (cos ))j] Ê
dy
dx
dr
d)
dr
d)
dt œ dt cos ) r dt sin ); v † j œ dt and v † j œ dt sin ) r dt cos )
ur r
cos ) r ddt) sin ) Ê
dr
dt
dy
dt
dy
dt
v†i œ
dx
dt
and
sin ) r ddt) cos )
dy
(b) ur œ (cos ))i (sin ))j Ê v † ur œ dx
dt cos ) dt sin )
d)
d)
‰
ˆ dr
‰
œ ˆ dr
dt cos ) r dt sin ) (cos ) ) dt sin ) r dt cos ) (sin ) ) by part (a),
Ê v † ur œ
dr
dt
; therefore,
œ
dr
dt
dx
dt
cos )
dy
dt
sin );
u) œ (sin ))i (cos ))j Ê v † u) œ sin ) dy
dt cos )
dr
d)
dr
d)
ˆ
‰
ˆ
œ dt cos ) r dt sin ) ( sin )) dt sin ) r dt cos )‰ (cos )) by part (a) Ê v † u) œ r ddt) ;
dx
dt
therefore, r ddt) œ dx
dt sin )
8. r œ f()) Ê
dr
dt
œ f w () )
d)
dt
Ê
d# r
dt#
dy
dt
cos )
#
œ f ww ()) ˆ ddt) ‰ f w ())
d# )
dt#
;vœ
dr
dt
ur r
d)
dt
u)
"Î#
"Î#
#
d) ‰
r sin ) ddt) ‰ i ˆsin ) dr
r# ˆ ddt) ‰ “ œ ’af w b# f # “
dt r cos ) dt j Ê kvk œ
Þ ÞÞ Þ ÞÞ
d)
dr
kv ‚ ak œ kx y y xk , where x œ r cos ) and y œ r sin ). Then dx
dt œ (r sin )) dt (cos )) dt
œ ˆcos )
‰#
’ˆ dr
dt
dr
dt
#
#
d) dr
d)
ˆ d) ‰# (r sin )) ddt#) (cos )) ddt#r ; dy
dt dt (r cos )) dt
dt œ (r cos )) dt (sin
#
#
ˆ d) ‰# (r cos )) ddt#) (sin )) ddt#r . Then kv ‚ ak
Ê
œ (2 cos )) ddt) dr
dt (r sin )) dt
#
$
$
d) d# r
d) ˆ dr ‰#
œ (after much algebra) r# ˆ ddt) ‰ r ddt#) dr
œ ˆ ddt) ‰ Šf 2 f † f ww 2af w b2 ‹
dt r dt dt# 2 dt dt
Ê
d# x
dt#
d# y
dt#
Ê ,œ
œ (2 sin ))
kv ‚a k
kv k
œ
dr
dt
dr
dt
f 2 f†f ww 2af w b2
$Î#
af w b# f # ‘
9. (a) Let r œ 2 t and ) œ 3t Ê
vœ
))
ˆ ddt) ‰ ;
dr
dt
œ 1 and
d)
dt
d# r
dt#
œ3 Ê
#
ur r ddt) u) Ê v(1) œ ur 3u) ; a œ ’ ddt#r
œ
d# )
dt#
#
r ˆ ddt) ‰ “ ur
œ 0. The halfway point is (1ß 3) Ê t œ 1;
#
’r ddt#) 2 dr
dt
d)
dt “ u)
Ê a(1) œ 9ur 6u)
(b) It takes the beetle 2 min to crawl to the origin Ê the rod has revolved 6 radians
#
#
Ê L œ '0 É[f())]# cf w ())d# d) œ '0 Ɉ2 3) ‰ ˆ "3 ‰ d) œ '0 É4
6
6
œ '0 É 37 129 ) ) d) œ
6
#
œ È37
"
6
dL
dt
4)
3
)#
9
"
9
'06 È() 6)# 1 d) œ "3 ’ ()#6) È() 6)# 1 "# ln ¸) 6 È() 6)# 1¸“ '
!
dL
dt
œ ˆ ddtr ‚ mv‰ Šr ‚ m
d# r
dt# ‹
Ê
dL
dt
œ (v ‚ mv) (r ‚ ma) œ r ‚ ma ; F œ ma Ê krck$ r
œ r ‚ ma œ r ‚ Š krck$ r‹ œ krck$ (r ‚ r) œ 0 Ê L œ constant vector
â
â i
â
11. (a) ur ‚ u) œ â cos )
â
â sin )
j
sin )
cos )
â
kâ
â
0 â œ k Ê a right-handed frame of unit vectors
â
0â
œ ( sin ))i (cos ))j œ u) and ddu)) œ ( cos ))i (sin ))j œ ur
Þ
ÞÞ
Þ Þ
Þ
ÞÞ
ÞÞ
ÞÞ
Þ
Þ
ÞÞ
(c) From Eq. (7), v œ rur r)u) zk Ê a œ v œ a r ur r ur b ˆr )u) r) u) r) u) ‰ z k
Þ#
ÞÞ
ÞÞ
ÞÞ
ÞÞ
œ Š r r) ‹ ur ˆr) 2r )‰ u) z k
(b)
d)
ln ŠÈ37 6‹ ¸ 6.5 in.
10. L(t) œ r(t) ‚ mv(t) Ê
œ ma Ê
"
3
6
dur
d)
12. (a) x œ r cos ) Ê dx œ cos ) dr r sin ) d); y œ r sin ) Ê dy œ sin ) dr r cos ) d); thus
dx# œ cos# ) dr# 2r sin ) cos ) dr d) r# sin# ) d)# and
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 13 Additional and Advanced Exercises
dy# œ sin# ) dr# 2r sin ) cos ) dr d) r# cos# ) d)# Ê dx# dy# dz# œ dr# r# d)# dz#
(b)
(c) r œ e) Ê dr œ e) d)
Ê L œ '0
ln 8
Èdr# r# d)# dz#
œ '0 Èe#) e#) e#) d)
ln 8
œ '0 È3e) d) œ ’È3 e) “
ln 8
ln 8
0
œ 8È3 È3 œ 7È3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
861
862
Chapter 13 Vector-Valued Functions and Motion in Space
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
CHAPTER 14 PARTIAL DERIVATIVES
14.1 FUNCTIONS OF SEVERAL VARIABLES
1. (a)
(b)
(c)
(d)
(e)
(f)
Domain: all points in the xy-plane
Range: all real numbers
level curves are straight lines y x œ c parallel to the line y œ x
no boundary points
both open and closed
unbounded
2. (a)
(b)
(c)
(d)
Domain: set of all (xß y) so that y x 0 Ê y x
Range: z 0
level curves are straight lines of the form y x œ c where c
boundary is Èy x œ 0 Ê y œ x, a straight line
0
(e) closed
(f) unbounded
3. (a) Domain: all points in the xy-plane
(b) Range: z 0
(c) level curves: for f(xß y) œ 0, the origin; for f(xß y) œ c 0, ellipses with center (!ß 0) and major and minor
axes along the x- and y-axes, respectively
(d) no boundary points
(e) both open and closed
(f) unbounded
4. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves: for f(xß y) œ 0, the union of the lines y œ „ x; for f(xß y) œ c Á 0, hyperbolas centered at
(0ß 0) with foci on the x-axis if c 0 and on the y-axis if c 0
(d) no boundary points
(e) both open and closed
(f) unbounded
5. (a) Domain: all points in the xy-plane
(b) Range: all real numbers
(c) level curves are hyperbolas with the x- and y-axes as asymptotes when f(xß y) Á 0, and the x- and y-axes
when f(xß y) œ 0
(d) no boundary points
(e) both open and closed
(f) unbounded
6. (a) Domain: all (xß y) Á (0ß y)
(b) Range: all real numbers
(c) level curves: for f(xß y) œ 0, the x-axis minus the origin; for f(xß y) œ c Á 0, the parabolas y œ cx# minus the
origin
(d) boundary is the line x œ 0
864
Chapter 14 Partial Derivatives
(e) open
(f) unbounded
7. (a) Domain: all (xß y) satisfying x# y# 16
(b) Range: z "4
(c)
(d)
(e)
(f)
level curves are circles centered at the origin with radii r 4
boundary is the circle x# y# œ 16
open
bounded
8. (a)
(b)
(c)
(d)
(e)
(f)
Domain: all (xß y) satisfying x# y# Ÿ 9
Range: 0 Ÿ z Ÿ 3
level curves are circles centered at the origin with radii r Ÿ 3
boundary is the circle x# y# œ 9
closed
bounded
9. (a)
(b)
(c)
(d)
(e)
(f)
Domain: (xß y) Á (0ß 0)
Range: all real numbers
level curves are circles with center (!ß 0) and radii r 0
boundary is the single point (0ß 0)
open
unbounded
10. (a)
(b)
(c)
(d)
(e)
(f)
Domain: all points in the xy-plane
Range: 0 z Ÿ 1
level curves are the origin itself and the circles with center (0ß 0) and radii r 0
no boundary points
both open and closed
unbounded
11. (a) Domain: all (xß y) satisfying 1 Ÿ y x Ÿ 1
(b) Range: 1# Ÿ z Ÿ 1#
(c)
(d)
(e)
(f)
level curves are straight lines of the form y x œ c where 1 Ÿ c Ÿ 1
boundary is the two straight lines y œ 1 x and y œ 1 x
closed
unbounded
12. (a) Domain: all (xß y), B Á 0
(b) Range: 1# z 1#
(c)
(d)
(e)
(f)
level curves are the straight lines of the form y œ cx, c any real number and x Á 0
boundary is the line x œ 0
open
unbounded
13. f
14. e
15. a
16. c
17. d
18. b
Section 14.1 Functions of Several Variables
19. (a)
(b)
20. (a)
(b)
21. (a)
(b)
22. (a)
(b)
865
866
Chapter 14 Partial Derivatives
23. (a)
(b)
24. (a)
(b)
25. (a)
(b)
Section 14.1 Functions of Several Variables
26. (a)
(b)
27. (a)
(b)
28. (a)
(b)
#
#
867
29. f(xß y) œ 16 x# y# and Š2È2ß È2‹ Ê z œ 16 Š2È2‹ ŠÈ2‹ œ 6 Ê 6 œ 16 x# y# Ê x# y# œ 10
30. f(xß y) œ Èx# 1 and (1ß 0) Ê D œ È1# 1 œ 0 Ê x# 1 œ 0 Ê x œ 1 or x œ 1
31. f(xß y) œ 'x
y
1
1 t#
dt at ŠÈ2ß È2‹ Ê z œ tan" y tan" x; at ŠÈ2ß È2‹ Ê z œ tan" È2 tan" ŠÈ2‹
œ 2 tan" È2 Ê tan" y tan" x œ 2 tan" È2
_
n
32. f(xß y) œ ! Š xy ‹ at (1ß 2) Ê z œ
œ
n 0
Ê y œ 2x
"
1 Š xy ‹
œ
y
yx
; at (1ß 2) Ê z œ
2
#1
œ2 Ê 2œ
y
yx
Ê 2y 2x œ y
868
Chapter 14 Partial Derivatives
33.
34.
35.
36.
37.
38.
39.
40.
41. f(xß yß z) œ Èx y ln z at (3ß 1ß 1) Ê w œ Èx y ln z; at (3ß 1ß 1) Ê w œ È3 (1) ln 1 œ 2
Ê Èx y ln z œ 2
Section 14.1 Functions of Several Variables
869
42. f(xß yß z) œ ln ax# y z# b at ("ß #ß ") Ê w œ ln ax# y z# b ; at ("ß #ß ") Ê w œ ln (1 2 1) œ ln 4
Ê ln 4 œ ln ax# y z# b Ê x# y z# œ 4
_
(x b y)n
n! zn
43. g(xß yß z) œ !
nœ0
_
at (ln 2ß ln 4ß 3) Ê w œ !
œ eÐln 8ÑÎ3 œ eln 2 œ 2 Ê 2 œ eÐxyÑÎz Ê
44. g(xß yß z) œ 'x
y
d)
È1 )#
'È2
z
dt
tÈt# 1
xy
z
nœ0
(x b y)n
n! zn
œ eÐxyÑÎz ; at (ln 2ß ln 4ß 3) Ê w œ eÐln 2ln 4ÑÎ3
œ ln 2
at ˆ0ß "# ß 2‰ Ê w œ csin" )d x csec" td È2
y
z
œ sin" y sin" x sec" z sec" ŠÈ2‹ Ê w œ sin" y sin" x sec" z
Ê w œ sin"
"
#
sin" 0 sec" 2
1
4
œ
1
4
Ê
1
#
1
4
; at ˆ0ß "# ß 2‰
œ sin" y sin" x sec" z
45. f(xß yß z) œ xyz and x œ 20 t, y œ t, z œ 20 Ê w œ (20 t)(t)(20) along the line Ê w œ 400t 20t#
Ê
dw
dt
œ 400 40t;
dw
dt
œ 0 Ê 400 40t œ 0 Ê t œ 10 and
d# w
dt#
œ 40 for all t Ê yes, maximum at t œ 10
Ê x œ 20 10 œ 10, y œ 10, z œ 20 Ê maximum of f along the line is f(10ß 10ß 20) œ (10)(10)(20) œ 2000
46. f(xß yß z) œ xy z and x œ t 1, y œ t 2, z œ t 7 Ê w œ (t 1)(t 2) (t 7) œ t# 4t 5 along the line
Ê
dw
dt
œ 2t 4;
dw
dt
œ 0 Ê 2t 4 œ 0 Ê t œ 2 and
d# w
dt#
œ 2 for all t Ê yes, minimum at t œ 2 Ê x œ 2 1 œ 1,
y œ 2 2 œ 0, and z œ 2 7 œ 9 Ê minimum of f along the line is f(1ß 0ß 9) œ (1)(0) 9 œ 9
‰
47. w œ 4 ˆ Th
d
"Î#
"Î#
km)
œ 4 ’ (290 5K)(16.8
“
K/km
¸ 124.86 km Ê must be
"
#
(124.86) ¸ 63 km south of Nantucket
48. The graph of f(x" ß x# ß x$ ß x% ) is a set in a five-dimensional space. It is the set of points
(x" ß x# ß x$ ß x% ß f(x" ß x# ß x$ ß x% )) for (x" ß x# ß x$ ß x% ) in the domain of f. The graph of f(x" ß x# ß x$ ß á ß xn ) is a set
in an (n 1)-dimensional space. It is the set of points (x" ß x# ß x$ ß á ß xn ß f(x" ß x# ß x$ ß á ß xn )) for
(x" ß x# ß x$ ß á ß xn ) in the domain of f.
49-52. Example CAS commands:
Maple:
with( plots );
f := (x,y) -> x*sin(y/2) + y*sin(2*x);
xdomain := x=0..5*Pi;
ydomain := y=0..5*Pi;
x0,y0 := 3*Pi,3*Pi;
plot3d( f(x,y), xdomain, ydomain, axes=boxed, style=patch, shading=zhue, title="#49(a) (Section 14.1)" );
plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, orientation=[-90,0],
title="#49(b) (Section 14.1)" );
# (b)
L := evalf( f(x0,y0) );
# (c)
plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, contours=[L],
orientation=[-90,0], title="#49(c) (Section 14.1)" );
53-56. Example CAS commands:
Maple:
eq := 4*ln(x^2+y^2+z^2)=1;
implicitplot3d( eq, x=-2..2, y=-2..2, z=-2..2, grid=[30,30,30], axes=boxed, title="#53 (Section 14.1)" );
870
Chapter 14 Partial Derivatives
57-60. Example CAS commands:
Maple:
x := (u,v) -> u*cos(v);
y := (u,v) ->u*sin(v);
z := (u,v) -> u;
plot3d( [x(u,v),y(u,v),z(u,v)], u=0..2, v=0..2*Pi, axes=boxed, style=patchcontour, contours=[($0..4)/2], shading=zhue,
title="#57 (Section 14.1)" );
49-60. Example CAS commands:
Mathematica: (assigned functions and bounds will vary)
For 49 - 52, the command ContourPlot draws 2-dimensional contours that are z-level curves of surfaces z = f(x,y).
Clear[x, y, f]
f[x_, y_]:= x Sin[y/2] y Sin[2x]
xmin= 0; xmax= 51; ymin= 0; ymax= 51; {x0, y0}={31, 31};
cp= ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False];
cp0= ContourPlot[[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Contours Ä {f[x0,y0]}, ContourShading Ä False,
PlotStyle Ä {RGBColor[1,0,0]}];
Show[cp, cp0]
For 53 - 56, the command ContourPlot3D will be used and requires loading a package. Write the function f[x, y, z] so that
when it is equated to zero, it represents the level surface given.
For 53, the problem associated with Log[0] can be avoided by rewriting the function as x2 + y2 +z2 - e1/4
< x^2+y^3-3*x*y;
x0,x1 := -5,5;
y0,y1 := -5,5;
plot3d( f(x,y), x=x0..x1, y=y0..y1, axes=boxed, shading=zhue, title="#65(a) (Section 14.7)" );
plot3d( f(x,y), x=x0..x1, y=y0..y1, grid=[40,40], axes=boxed, shading=zhue, style=patchcontour, title="#65(b)
(Section 14.7)" );
fx := D[1](f);
# (c)
fy := D[2](f);
crit_pts := solve( {fx(x,y)=0,fy(x,y)=0}, {x,y} );
fxx := D[1](fx);
# (d)
fxy := D[2](fx);
fyy := D[2](fy);
discr := unapply( fxx(x,y)*fyy(x,y)-fxy(x,y)^2, (x,y) );
for CP in {crit_pts} do
# (e)
eval( [x,y,fxx(x,y),discr(x,y)], CP );
Section 14.8 Lagrange Multipliers
909
end do;
# (0,0) is a saddle point
# ( 9/4, 3/2) is a local minimum
Mathematica: (assigned functions and bounds will vary)
Clear[x,y,f]
f[x_,y_]:= x2 y3 3x y
xmin= 5; xmax= 5; ymin= 5; ymax= 5;
Plot3D[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, AxesLabel Ä {x, y, z}]
ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False, Contours Ä 40]
fx= D[f[x,y], x];
fy= D[f[x,y], y];
critical=Solve[{fx==0, fy==0},{x, y}]
fxx= D[fx, x];
fxy= D[fx, y];
fyy= D[fy, y];
discriminant= fxx fyy fxy2
{{x, y}, f[x, y], discriminant, fxx} /.critical
14.8 LAGRANGE MULTIPLIERS
1.
™ f œ yi xj and ™ g œ 2xi 4yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 4yj) Ê y œ 2x- and x œ 4yÊ x œ 8x-# Ê - œ „
È2
4
or x œ 0.
CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the ellipse so x Á 0.
CASE 2: x Á 0 Ê - œ „
È2
4
Therefore f takes on its extreme values at Š „
are „
2.
È2
#
#
Ê x œ „ È2y Ê Š „ È2y‹ 2y# œ 1 Ê y œ „ "# .
È2 "
2 ß #‹
and Š „
È2
"
2 ß #‹ .
The extreme values of f on the ellipse
.
™ f œ yi xj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 2yj) Ê y œ 2x- and x œ 2yÊ x œ 4x-# Ê x œ 0 or - œ „ 12 .
CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the circle x# y# 10 œ 0 so x Á 0.
CASE 2: x Á 0 Ê - œ „ 12 Ê y œ 2x ˆ „ "# ‰ œ „ x Ê x# a „ xb# 10 œ 0 Ê x œ „ È5 Ê y œ „ È5.
Therefore f takes on its extreme values at Š „ È5ß È5‹ and Š „ È5ß È5‹ . The extreme values of f on the
circle are 5 and 5.
3.
™ f œ 2xi 2yj and ™ g œ i 3j so that ™ f œ - ™ g Ê 2xi 2yj œ -(i 3j) Ê x œ -# and y œ 3#Ê ˆ -# ‰ 3 ˆ 3#- ‰ œ 10 Ê - œ 2 Ê x œ 1 and y œ 3 Ê f takes on its extreme value at (1ß 3) on the line.
The extreme value is f("ß $) œ 49 1 9 œ 39.
4.
™ f œ 2xyi x# j and ™ g œ i j so that ™ f œ - ™ g Ê 2xyi x# j œ -(i j) Ê 2xy œ - and x# œ Ê 2xy œ x# Ê x œ 0 or 2y œ x.
CASE 1: If x œ 0, then x y œ 3 Ê y œ 3.
CASE 2: If x Á 0, then 2y œ x so that x y œ 3 Ê 2y y œ 3 Ê y œ 1 Ê x œ 2.
Therefore f takes on its extreme values at (!ß 3) and (#ß "). The extreme values of f are f(0ß 3) œ 0 and
f(#ß 1) œ 4.
910
Chapter 14 Partial Derivatives
5. We optimize f(xß y) œ x# y# , the square of the distance to the origin, subject to the constraint
g(xß y) œ xy# 54 œ 0. Thus ™ f œ 2xi 2yj and ™ g œ y# i 2xyj so that ™ f œ - ™ g Ê 2xi 2yj
œ - ay# i 2xyjb Ê 2x œ -y# and 2y œ 2-xy.
CASE 1: If y œ 0, then x œ 0. But (0ß 0) does not satisfy the constraint xy# œ 54 so y Á 0.
CASE 2: If y Á 0, then 2 œ 2-x Ê x œ -" Ê 2 ˆ -" ‰ œ -y# Ê y# œ -2# . Then xy# œ 54 Ê ˆ -" ‰ ˆ -2# ‰ œ 54
Ê -$ œ " Ê - œ " Ê x œ 3 and y# œ 18 Ê x œ 3 and y œ „ 3È2.
27
3
Therefore Š$ß „ 3È2‹ are the points on the curve xy# œ 54 nearest the origin (since xy# œ 54 has points
increasingly far away as y gets close to 0, no points are farthest away).
6. We optimize f(xß y) œ x# y# , the square of the distance to the origin subject to the constraint g(xß y)
œ x# y 2 œ 0. Thus ™ f œ 2xi 2yj and ™ g œ 2xyi x# j so that ™ f œ - ™ g Ê 2x œ 2xy- and 2y œ x# ˆ 2y ‰ Ê x# œ 2y#
Ê - œ 2y
x# , since x œ 0 Ê y œ 0 (but g(0ß 0) Á 0). Thus x Á 0 and 2x œ 2xy x#
Ê a2y# b y 2 œ 0 Ê y œ 1 (since y 0) Ê x œ „ È2 . Therefore Š „ È2ß 1‹ are the points on the curve
x# y œ 2 nearest the origin (since x# y œ 2 has points increasingly far away as x gets close to 0, no points are
farthest away).
7. (a) ™ f œ i j and ™ g œ yi xj so that ™ f œ - ™ g Ê i j œ -(yi xj) Ê 1 œ -y and 1 œ -x Ê y œ
xœ
"
-
Ê
"
-#
"
4
œ 16 Ê - œ „ . Use - œ
"
4
"
-
and
since x 0 and y 0. Then x œ 4 and y œ 4 Ê the minimum
value is 8 at the point (4ß 4). Now, xy œ 16, x 0, y 0 is a branch of a hyperbola in the first quadrant
with the x-and y-axes as asymptotes. The equations x y œ c give a family of parallel lines with m œ 1.
As these lines move away from the origin, the number c increases. Thus the minimum value of c occurs
where x y œ c is tangent to the hyperbola's branch.
(b) ™ f œ yi xj and ™ g œ i j so that ™ f œ - ™ g Ê yi xj œ -(i j) Ê y œ - œ x y y œ 16 Ê y œ 8
Ê x œ 8 Ê f()ß )) œ 64 is the maximum value. The equations xy œ c (x 0 and y 0 or x 0 and y 0
to get a maximum value) give a family of hyperbolas in the first and third quadrants with the x- and yaxes as asymptotes. The maximum value of c occurs where the hyperbola xy œ c is tangent to the line
x y œ 16.
8. Let f(xß y) œ x# y# be the square of the distance from the origin. Then ™ f œ 2xi 2yj and
™ g œ (2x y)i (2y x)j so that ™ f œ - ™ g Ê 2x œ -(2x y) and 2y œ -(2y x) Ê
2y
2yx
œ-
Ê 2x œ Š 2y2yx ‹ (2x y) Ê x(2y x) œ y(2x y) Ê x# œ y# Ê y œ „ x.
CASE 1: y œ x Ê x# x(x) x# 1 œ 0 Ê x œ „
"
È3
and y œ x.
CASE 2: y œ x Ê x# x(x) (x)# 1 œ 0 Ê x œ „ 1 and y œ x. Thus f Š È"3 ß È"3 ‹ œ
2
3
œ f Š È"3 ß È"3 ‹ and f(1ß 1) œ 2 œ f(1ß 1).
Therefore the points (1ß 1) and (1ß 1) are the farthest away; Š È"3 ß È"3 ‹ and Š È"3 ß È"3 ‹ are the closest
points to the origin.
9. V œ 1r# h Ê 161 œ 1r# h Ê 16 œ r# h Ê g(rß h) œ r# h 16; S œ 21rh 21r# Ê ™ S œ (21h 41r)i 21rj and
™ g œ 2rhi r# j so that ™ S œ - ™ g Ê (21rh 41r)i 21rj œ - a2rhi r# jb Ê 21rh 41r œ 2rh- and
21r œ -r# Ê r œ 0 or - œ 2r1 . But r œ 0 gives no physical can, so r Á 0 Ê - œ 2r1 Ê 21h 41r
œ 2rh ˆ 2r1 ‰ Ê 2r œ h Ê 16 œ r# (2r) Ê r œ 2 Ê h œ 4; thus r œ 2 cm and h œ 4 cm give the only extreme
surface area of 241 cm# . Since r œ 4 cm and h œ 1 cm Ê V œ 161 cm$ and S œ 401 cm# , which is a larger
surface area, then 241 cm# must be the minimum surface area.
Section 14.8 Lagrange Multipliers
911
10. For a cylinder of radius r and height h we want to maximize the surface area S œ 21rh subject to the constraint
#
g(rß h) œ r# ˆ h# ‰ a# œ 0. Thus ™ S œ 21hi 21rj and ™ g œ 2ri h# j so that ™ S œ - ™ g Ê 21h œ 2-r and
21r œ
-h
#
Ê
1h
r
4r#
4
œ - and 21r œ ˆ 1rh ‰ ˆ h# ‰ Ê 4r# œ h# Ê h œ 2r Ê r#
œ a# Ê 2r# œ a# Ê r œ
a
È2
Ê h œ aÈ2 Ê S œ 21 Š Èa2 ‹ ŠaÈ2‹ œ 21a# .
#
#
x
11. A œ (2x)(2y) œ 4xy subject to g(xß y) œ 16
y9 1 œ 0; ™ A œ 4yi 4xj and ™ g œ 8x i 2y
9 j so that ™ A
2y
2y
32y
x
x
‰ ˆ 32y
‰
œ - ™ g Ê 4yi 4xj œ - ˆ 8 i 9 j‰ Ê 4y œ ˆ 8 ‰ - and 4x œ ˆ 9 ‰ - Ê - œ x and 4x œ ˆ 2y
9
x
Ê y œ „ 34 x Ê
Then y œ
3
4
x#
16
Š2È2‹ œ
ˆ „43 x‰#
œ 1 Ê x#
9
3È 2
# , so the length is
x#
a#
12. P œ 4x 4y subject to g(xß y) œ
y#
b#
and height œ 2y œ
2b#
È a# b#
2x œ 4È2 and the width is 2y œ 3È2.
1 œ 0; ™ P œ 4i 4j and ™ g œ
‰
ˆ 2y ‰
Ê 4 œ ˆ 2x
a# - and 4 œ b# - Ê - œ
œ 1 Ê aa# b# b x# œ a% Ê x œ
œ 8 Ê x œ „ 2È2 . We use x œ 2È2 since x represents distance.
2a#
x
a#
È a# b#
#
2x
a#
i
#
b
‰ 2a
and 4 œ ˆ 2y
b # Š x ‹ Ê y œ Š a# ‹ x Ê
#
, since x 0 Ê y œ Š ba# ‹ x œ
Ê perimeter is P œ 4x 4y œ
4a# 4b#
È a# b#
b#
È a# b#
2y
b#
x#
a#
j so that ™ P œ - ™ g
#
#
Š ba# ‹ x#
b#
œ1 Ê
Ê width œ 2x œ
x#
a#
b# x#
a%
2a#
È a# b #
œ 4Èa# b#
13. ™ f œ 2xi 2yj and ™ g œ (2x 2)i (2y 4)j so that ™ f œ - ™ g œ 2xi 2yj œ -[(2x 2)i (2y 4)j]
2#
#
Ê 2x œ -(2x 2) and 2y œ -(2y 4) Ê x œ -
1 and y œ -1 , - Á 1 Ê y œ 2x Ê x 2x (2x) 4(2x)
œ 0 Ê x œ 0 and y œ 0, or x œ 2 and y œ 4. Therefore f(0ß 0) œ 0 is the minimum value and f(2ß 4) œ 20 is the
maximum value. (Note that - œ 1 gives 2x œ 2x 2 or ! œ 2, which is impossible.)
14. ™ f œ 3i j and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 3 œ 2-x and 1 œ 2-y Ê - œ
#
Ê y œ x3 Ê x# ˆ x3 ‰ œ 4 Ê 10x# œ 36 Ê x œ „
yœ
2
È10 .
Therefore f Š È610 ß È210 ‹ œ
20
È10
6
È10
Ê xœ
6
È10
3
2x
3 ‰
and 1 œ 2 ˆ 2x
y
and y œ È210 , or x œ È610 and
6 œ 2È10 6 ¸ 12.325 is the maximum value, and
f Š È610 ß È210 ‹ œ 2È10 6 ¸ 0.325 is the minimum value.
15. ™ T œ (8x 4y)i (4x 2y)j and g(xß y) œ x# y# 25 œ 0 Ê ™ g œ 2xi 2yj so that ™ T œ - ™ g
Ê (8x 4y)i (4x 2y)j œ -(2xi 2yj) Ê 8x 4y œ 2-x and 4x 2y œ 2-y Ê y œ -2x1 , - Á 1
Ê 8x 4 ˆ -2x1 ‰ œ 2-x Ê x œ 0, or - œ 0, or - œ 5.
CASE 1: x œ 0 Ê y œ 0; but (0ß 0) is not on x# y# œ 25 so x Á 0.
CASE 2: - œ 0 Ê y œ 2x Ê x# (2x)# œ 25 Ê x œ „ È5 and y œ 2x.
CASE 3: - œ 5 Ê y œ
and y œ È5 .
2x
4
#
œ #x Ê x# ˆ #x ‰ œ 25 Ê x œ „ 2È5 Ê x œ 2È5 and y œ È5, or x œ 2È5
Therefore T ŠÈ5ß 2È5‹ œ 0° œ T ŠÈ5ß 2È5‹ is the minimum value and T Š2È5ß È5‹ œ 125°
œ T Š2È5ß È5‹ is the maximum value. (Note: - œ 1 Ê x œ 0 from the equation 4x 2y œ 2-y; but we
found x Á 0 in CASE 1.)
16. The surface area is given by S œ 41r# 21rh subject to the constraint V(rß h) œ
4
3
1r$ 1r# h œ 8000. Thus
™ S œ (81r 21h)i 21rj and ™ V œ a41r# 21rhb i 1r# j so that ™ S œ - ™ V œ (81r 21h)i 21rj
œ - ca41r# 21rhb i 1r# jd Ê 81r 21h œ - a41r# 21rhb and 21r œ -1r# Ê r œ 0 or 2 œ r-. But r Á 0
so 2 œ r- Ê - œ 2r Ê 4r h œ 2r a2r# rhb Ê h œ 0 Ê the tank is a sphere (there is no cylindrical part) and
4
3
1r$ œ 8000 Ê r œ 10 ˆ 16 ‰
"Î$
.
912
Chapter 14 Partial Derivatives
17. Let f(xß yß z) œ (x 1)# (y 1)# (z 1)# be the square of the distance from (1ß 1ß 1). Then
™ f œ 2(x 1)i 2(y 1)j 2(z 1)k and ™ g œ i 2j 3k so that ™ f œ - ™ g
Ê 2(x 1)i 2(y 1)j 2(z 1)k œ -(i 2j 3k) Ê 2(x 1) œ -, 2(y 1) œ 2-, 2(z 1) œ 3Ê 2(y 1) œ 2[2(x 1)] and 2(z 1) œ 3[2(x 1)] Ê x œ y # 1 Ê z 2 œ 3 ˆ y # 1 ‰ or z œ 3y # 1 ; thus
y1
ˆ 3y # 1 ‰ 13 œ 0 Ê y œ 2 Ê x œ 3# and z œ 5# . Therefore the point ˆ 3# ß 2ß 5# ‰ is closest (since no
# 2y 3
point on the plane is farthest from the point (1ß 1ß 1)).
18. Let f(xß yß z) œ (x 1)# (y 1)# (z 1)# be the square of the distance from (1ß 1ß 1). Then
™ f œ 2(x 1)i 2(y 1)j 2(z 1)k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê x 1 œ -x, y 1 œ -y
#
‰# ˆ 1 " - ‰# œ 4
and z 1 œ -z Ê x œ 1 " - , y œ 1 " - , and z œ 1" - for - Á 1 Ê ˆ 1 " - ‰ ˆ 1"
Ê
"
"-
œ „
2
È3
Ê xœ
2
È3
, y œ È23 , z œ
2
È3
or x œ È23 , y œ
2
È3
, z œ È23 . The largest value of f
occurs where x 0, y 0, and z 0 or at the point Š È23 ß È23 ß È23 ‹ on the sphere.
19. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(2xi 2yj 2zk) Ê 2x œ 2x-, 2y œ 2y-,
and 2z œ 2z- Ê x œ 0 or - œ 1.
CASE 1: - œ 1 Ê 2y œ 2y Ê y œ 0; 2z œ 2z Ê z œ 0 Ê x# 1 œ 0 Ê x œ „ 1 and y œ z œ 0.
CASE 2: x œ 0 Ê y# z# œ 1, which has no solution.
Therefore the points on the unit circle x# y# œ 1, are the points on the surface x# y# z# œ 1 closest to the originÞ
The minimum distance is 1.
20. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ yi xj k so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(yi xj k) Ê 2x œ -y, 2y œ -x, and 2z œ Ê xœ
-y
#
Ê 2y œ - Š -#y ‹ Ê y œ 0 or - œ „ 2.
CASE 1: y œ 0 Ê x œ 0 Ê z 1 œ 0 Ê z œ 1.
CASE 2: - œ 2 Ê x œ y and z œ 1 Ê x# (1) 1 œ 0 Ê x# 2 œ 0, so no solution.
CASE 3: - œ 2 Ê x œ y and z œ 1 Ê (y)y 1 1 œ 0 Ê y œ 0, again.
Therefore (0ß 0ß 1) is the point on the surface closest to the origin since this point gives the only extreme value
and there is no maximum distance from the surface to the origin.
21. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ yi xj 2zk so that ™ f œ - ™ g Ê 2xi 2yj 2zk œ -(yi xj 2zk) Ê 2x œ y-, 2y œ x-, and
2z œ 2z- Ê - œ 1 or z œ 0.
CASE 1: - œ 1 Ê 2x œ y and 2y œ x Ê y œ 0 and x œ 0 Ê z# 4 œ 0 Ê z œ „ 2 and x œ y œ 0.
CASE 2: z œ 0 Ê xy 4 œ 0 Ê y œ 4x . Then 2x œ
4
x
- Ê -œ
x#
#
#
, and 8x œ x- Ê 8x œ x Š x# ‹
Ê x% œ 16 Ê x œ „ 2. Thus, x œ 2 and y œ 2, or x = 2 and y œ 2.
Therefore we get four points: (#ß 2ß 0), (2ß 2ß 0), (0ß 0ß 2) and (!ß 0ß 2). But the points (!ß 0ß 2) and (!ß !ß 2)
are closest to the origin since they are 2 units away and the others are 2È2 units away.
22. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk and
™ g œ yzi xzj xyk so that ™ f œ - ™ g Ê 2x œ -yz, 2y œ -xz, and 2z œ -xy Ê 2x# œ -xyz and 2y# œ -yxz
Ê x# œ y# Ê y œ „ x Ê z œ „ x Ê x a „ xb a „ xb œ 1 Ê x œ „ 1 Ê the points are (1ß 1ß 1), ("ß 1ß 1),
("ß "ß "), and (1ß 1, 1).
23. ™ f œ i 2j 5k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i 2j 5k œ -(2xi 2yj 2zk) Ê 1 œ 2x-,
2 œ 2y-, and 5 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #5- œ 5x Ê x# (2x)# (5x)# œ 30 Ê x œ „ 1.
Section 14.8 Lagrange Multipliers
913
Thus, x œ 1, y œ 2, z œ 5 or x œ 1, y œ 2, z œ 5. Therefore f(1ß 2ß 5) œ 30 is the maximum value and
f(1ß 2ß 5) œ 30 is the minimum value.
24. ™ f œ i 2j 3k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i 2j 3k œ -(2xi 2yj 2zk) Ê 1 œ 2x-,
2 œ 2y-, and 3 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #3- œ 3x Ê x# (2x)# (3x)# œ 25 Ê x œ „ È514 .
Thus, x œ
5
È14
,yœ
10
È14
,zœ
15
È14
or x œ È514 , y œ È1014 , z œ È1514 . Therefore f Š È514 ß È1014 ß È1514 ‹
œ 5È14 is the maximum value and f Š È514 ß È1014 , È1514 ‹ œ 5È14 is the minimum value.
25. f(xß yß z) œ x# y# z# and g(xß yß z) œ x y z 9 œ 0 Ê ™ f œ 2xi 2yj 2zk and ™ g œ i j k so that
™ f œ - ™ g Ê 2xi 2yj 2zk œ -(i j k) Ê 2x œ -, 2y œ -, and 2z œ - Ê x œ y œ z Ê x x x 9 œ 0
Ê x œ 3, y œ 3, and z œ 3.
26. f(xß yß z) œ xyz and g(xß yß z) œ x y z# 16 œ 0 Ê ™ f œ yzi xzj xyk and ™ g œ i j 2zk so that
™ f œ - ™ g Ê yzi xzj xyk œ -(i j 2zk) Ê yz œ -, xz œ -, and xy œ 2z- Ê yz œ xz Ê z œ 0 or y œ x.
But z 0 so that y œ x Ê x# œ 2z- and xz œ -. Then x# œ 2z(xz) Ê x œ 0 or x œ 2z# . But x 0 so that
32
x œ 2z# Ê y œ 2z# Ê 2z# 2z# z# œ 16 Ê z œ „ È45 . We use z œ È45 since z 0. Then x œ 32
5 and y œ 5
32
4
which yields f Š 32
5 ß 5 ß È5 ‹ œ
4096
25È5
.
27. V œ 6xyz and g(xß yß z) œ x# y# z# 1 œ 0 Ê ™ V œ 6yzi 6xzj 6xyk and ™ g œ 2xi 2yj 2zk so that
™ V œ - ™ g Ê 3yz œ -x, 3xz œ -y, and 3xy œ -z Ê 3xyz œ -x# and 3xyz œ -y# Ê y œ „ x Ê z œ „ x
Ê x# x# x# œ 1 Ê x œ È"3 since x 0 Ê the dimensions of the box are È23 by È23 by È23 for maximum
volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.)
28. V œ xyz with xß yß z all positive and
x
a
y
b
z
c
œ 1; thus V œ xyz and g(xß yß z) œ bcx acy abz abc œ 0
Ê ™ V œ yzi xzj xyk and ™ g œ bci acj abk so that ™ V œ - ™ g Ê yz œ -bc, xz œ -ac, and xy œ -ab
Ê xyz œ -bcx, xyz œ -acy, and xyz œ -abz Ê - Á 0. Also, -bcx œ -acy œ -abz Ê bx œ ay, cy œ bz, and
a
cx œ az Ê y œ ba x and z œ ca x. Then xa by cz œ 1 Ê xa "b ˆ ba x‰ "c ˆ ca x‰ œ 1 Ê 3x
a œ 1 Ê xœ 3
Ê y œ ˆ ba ‰ ˆ 3a ‰ œ b3 and z œ ˆ ca ‰ ˆ 3a ‰ œ 3c Ê V œ xyz œ ˆ 3a ‰ ˆ 3b ‰ ˆ 3c ‰ œ abc
27 is the maximum volume. (Note that
there is no minimum volume since the box could be made arbitrarily thin.)
29. ™ T œ 16xi 4zj (4y 16)k and ™ g œ 8xi 2yj 8zk so that ™ T œ - ™ g Ê 16xi 4zj (4y 16)k
œ -(8xi 2yj 8zk) Ê 16x œ 8x-, 4z œ 2y-, and 4y 16 œ 8z- Ê - œ 2 or x œ 0.
CASE 1: - œ 2 Ê 4z œ 2y(2) Ê z œ y. Then 4z 16 œ 16z Ê z œ 43 Ê y œ 43 . Then
#
#
4x# ˆ 43 ‰ 4 ˆ 43 ‰ œ 16 Ê x œ „ 43 .
CASE 2: x œ 0 Ê - œ
2z
y
#
#
#
#
#
Ê 4y 16 œ 8z Š 2z
y ‹ Ê y 4y œ 4z Ê 4(0) y ay 4yb 16 œ 0
Ê y# 2y 8 œ 0 Ê (y 4)(y 2) œ 0 Ê y œ 4 or y œ 2. Now y œ 4 Ê 4z# œ 4# 4(4)
Ê z œ 0 and y œ 2 Ê 4z# œ (2)# 4(2) Ê z œ „ È3.
°
°
The temperatures are T ˆ „ 43 ß 43 ß 43 ‰ œ 642 23 , T(0ß 4ß 0) œ 600°, T Š0ß 2ß È3‹ œ Š600 24È3‹ , and
°
T Š0ß 2ß È3‹ œ Š600 24È3‹ ¸ 641.6°. Therefore ˆ „ 43 ß 43 ß 43 ‰ are the hottest points on the space probe.
30. ™ T œ 400yz# i 400xz# j 800xyzk and ™ g œ 2xi 2yj 2zk so that ™ T œ - ™ g
Ê 400yz# i 400xz# j 800xyzk œ -(2xi 2yj 2zk) Ê 400yz# œ 2x-, 400xz# œ 2y-, and 800xyz œ 2z-.
Solving this system yields the points a!ß „ 1ß 0b , a „ 1ß 0ß 0b , and Š „ "# ß „ "# ß „
È2
# ‹.
The corresponding
914
Chapter 14 Partial Derivatives
temperatures are T a!ß „ 1ß 0b œ 0, T a „ 1ß 0ß 0b œ 0, and T Š „ "# ß „ "# ß „
maximum temperature at Š "# ß "# ß „
Š "# ß "# ß „
È2
# ‹
and Š "# ß "# ß „
È2
# ‹
and Š "# ß "# ß „
È2
# ‹;
È2
# ‹
œ „ 50. Therefore 50 is the
50 is the minimum temperature at
È2
# ‹.
31. ™ U œ (y 2)i xj and ™ g œ 2i j so that ™ U œ - ™ g Ê (y 2)i xj œ -(2i j) Ê y # œ 2- and
x œ - Ê y 2 œ 2x Ê y œ 2x 2 Ê 2x (2x 2) œ 30 Ê x œ 8 and y œ 14. Therefore U(8ß 14) œ $128
is the maximum value of U under the constraint.
32. ™ M œ (6 z)i 2yj xk and ™ g œ 2xi 2yj 2zk so that ™ M œ - ™ g Ê (6 z)i 2yj xk
œ -(2xi 2yj 2zk) Ê 6 z œ 2x-, 2y œ 2y-, x œ 2z- Ê - œ 1 or y œ 0.
CASE 1: - œ 1 Ê 6 z œ 2x and x œ 2z Ê 6 z œ 2(2z) Ê z œ 2 and x œ 4. Then
(4)# y# 2# 36 œ 0 Ê y œ „ 4.
x
x ‰
CASE 2: y œ 0, 6 z œ 2x-, and x œ 2z- Ê - œ 2z
Ê 6 z œ 2x ˆ 2z
Ê 6z z# œ x#
Ê a6z z# b 0# z# œ 36 Ê z œ 6 or z œ 3. Now z œ 6 Ê x# œ 0 Ê x œ 0; z œ 3
Ê x# œ 27 Ê x œ „ 3È3.
Therefore we have the points Š „ 3È3ß 0ß 3‹ , (0ß 0ß 6), and a4ß „ 4ß 2b . Then M Š3È3ß 0ß 3‹
œ 27È3 60 ¸ 106.8, M Š3È3ß 0ß 3‹ œ 60 27È3 ¸ 13.2, M(0ß 0ß 6) œ 60, and M(4ß 4ß 2) œ 12
œ M(4ß 4ß 2). Therefore, the weakest field is at a4ß „ 4ß 2b .
33. Let g" (xß yß z) œ 2x y œ 0 and g# (xß yß z) œ y z œ 0 Ê ™ g" œ 2i j , ™ g# œ j k , and ™ f œ 2xi 2j 2zk
so that ™ f œ - ™ g" . ™ g# Ê 2xi 2j 2zk œ -(2i j) .(j k) Ê 2xi 2j 2zk œ 2-i (. -)j .k
Ê 2x œ 2-, 2 œ . -, and 2z œ . Ê x œ -. Then 2 œ 2z x Ê x œ 2z 2 so that 2x y œ 0
Ê 2(2z 2) y œ 0 Ê 4z 4 y œ 0. This equation coupled with y z œ 0 implies z œ 43 and y œ 43 .
Then x œ
œ
4
3
2
3
#
so that ˆ 23 ß 43 ß 43 ‰ is the point that gives the maximum value f ˆ 23 ß 43 ß 43 ‰ œ ˆ 23 ‰ 2 ˆ 43 ‰ ˆ 43 ‰
#
.
34. Let g" (xß yß z) œ x 2y 3z 6 œ 0 and g# (xß yß z) œ x 3y 9z 9 œ 0 Ê ™ g" œ i 2j 3k ,
™ g# œ i 3j 9k , and ™ f œ 2xi 2yj 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2yj 2zk
œ -(i 2j 3k) .(i 3j 9k) Ê 2x œ - ., 2y œ 2- 3., and 2z œ 3- 9.. Then 0 œ x 2y 3z 6
‰
œ "# (- .) (2- 3.) ˆ 9# - 27
# . 6 Ê 7- 17. œ 6; 0 œ x 3y 9z 9
"
9
27
81
Ê # (- .) ˆ3- # .‰ ˆ # - # .‰ 9 Ê 34- 91. œ 18. Solving these two equations for - and . gives
-.
2- 3.
3- 9.
78
81
9
- œ 240
œ 123
œ 59
. The minimum value is
59 and . œ 59 Ê x œ # œ 59 , y œ
#
59 , and z œ
#
21,771
81
123
9
369
f ˆ 59 ß 59 ß 59 ‰ œ 59# œ 59 . (Note that there is no maximum value of f subject to the constraints because
at least one of the variables x, y, or z can be made arbitrary and assume a value as large as we please.)
35. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject
to the constraints g" (xß yß z) œ y 2z 12 œ 0 and g# (xß yß z) œ x y 6 œ 0. Thus ™ f œ 2xi 2yj 2zk ,
™ g" œ j 2k , and ™ g# œ i j so that ™ f œ - ™ g" . ™ g# Ê 2x œ ., 2y œ - ., and 2z œ 2-. Then
0 œ y 2z 12 œ ˆ -# .# ‰ 2- 12 Ê 5# - "# . œ 12 Ê 5- . œ 24; 0 œ x y 6 œ .# ˆ -# .# ‰ 6
Ê
yœ
"
# - . œ 6 Ê - #. œ 12. Solving these two equations for - and . gives - œ 4 and . œ
-.
# œ 4, and z œ - œ 4. The point (2ß 4ß 4) on the line of intersection is closest to the origin.
maximum distance from the origin since points on the line can be arbitrarily far away.)
36. The maximum value is f ˆ 23 ß 43 ß 43 ‰ œ
4
3
from Exercise 33 above.
4 Ê xœ
.
#
(There is no
œ 2,
Section 14.8 Lagrange Multipliers
915
37. Let g" (xß yß z) œ z 1 œ 0 and g# (xß yß z) œ x# y# z# 10 œ 0 Ê ™ g" œ k , ™ g# œ 2xi 2yj 2zk , and
™ f œ 2xyzi x# zj x# yk so that ™ f œ - ™ g" . ™ g# Ê 2xyzi x# zj x# yk œ -(k) .(2xi 2yj 2zk)
Ê 2xyz œ 2x., x# z œ 2y., and x# y œ 2z. - Ê xyz œ x. Ê x œ 0 or yz œ . Ê . œ y since z œ 1.
CASE 1: x œ 0 and z œ 1 Ê y# 9 œ 0 (from g# ) Ê y œ „ 3 yielding the points a0ß „ 3ß 1b.
CASE 2: . œ y Ê x# z œ 2y# Ê x# œ 2y# (since z œ 1) Ê 2y# y# 1 10 œ 0 (from g# ) Ê 3y# 9 œ 0
#
Ê y œ „ È3 Ê x# œ 2 Š „ È3‹ Ê x œ „ È6 yielding the points Š „ È6ß „ È3ß "‹ .
Now f a!ß „ 3ß 1b œ 1 and f Š „ È6ß „ È3ß "‹ œ 6 Š „ È3‹ 1 œ 1 „ 6È3. Therefore the maximum of f is
1 6È3 at Š „ È6ß È3ß 1‹, and the minimum of f is 1 6È3 at Š „ È6ß È3ß "‹ .
38. (a) Let g" (xß yß z) œ x y z 40 œ 0 and g# (xß yß z) œ x y z œ 0 Ê ™ g" œ i j k , ™ g# œ i j k , and
™ w œ yzi xzj xyk so that ™ w œ - ™ g" . ™ g# Ê yzi xzj xyk œ -(i j k) .(i j k)
Ê yz œ - ., xz œ - ., and xy œ - . Ê yz œ xz Ê z œ 0 or y œ x.
CASE 1: z œ 0 Ê x y œ 40 and x y œ 0 Ê no solution.
CASE 2: x œ y Ê 2x z 40 œ 0 and 2x z œ 0 Ê z œ 20 Ê x œ 10 and y œ 10 Ê w œ (10)(10)(20)
œ 2000
â
â
âi j k â
â
â
" â œ 2i 2j is parallel to the line of intersection Ê the line is x œ 2t 10,
(b) n œ â " "
â
â
â " " " â
y œ 2t 10, z œ 20. Since z œ 20, we see that w œ xyz œ (2t 10)(2t 10)(20) œ a4t# 100b (20)
which has its maximum when t œ 0 Ê x œ 10, y œ 10, and z œ 20.
39. Let g" (Bß yß z) œ y x œ 0 and g# (xß yß z) œ x# y# z# 4 œ 0. Then ™ f œ yi xj 2zk , ™ g" œ i j , and
™ g# œ 2xi 2yj 2zk so that ™ f œ - ™ g" . ™ g# Ê yi xj 2zk œ -(i j) .(2xi 2yj 2zk)
Ê y œ - 2x., x œ - 2y., and 2z œ 2z. Ê z œ 0 or . œ 1.
CASE 1: z œ 0 Ê x# y# 4 œ 0 Ê 2x# 4 œ 0 (since x œ y) Ê x œ „ È2 and y œ „ È2 yielding the points
Š „ È2ß „ È2ß !‹ .
CASE 2: . œ 1 Ê y œ - 2x and x œ - 2y Ê x y œ 2(x y) Ê 2x œ 2(2x) since x œ y Ê x œ 0 Ê y œ 0
Ê z# 4 œ 0 Ê z œ „ 2 yielding the points a!ß !ß „ 2b .
Now, f a!ß !ß „ 2b œ 4 and f Š „ È2ß „ È2ß !‹ œ 2. Therefore the maximum value of f is 4 at a!ß !ß „ 2b and the
minimum value of f is 2 at Š „ È2ß „ È2ß !‹ .
40. Let f(xß yß z) œ x# y# z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject
to the constraints g" (xß yß z) œ 2y 4z 5 œ 0 and g# (xß yß z) œ 4x# 4y# z# œ 0. Thus ™ f œ 2xi 2yj 2zk ,
™ g" œ 2j 4k , and ™ g# œ 8xi 8yj 2zk so that ™ f œ - ™ g" . ™ g# Ê 2xi 2yj 2zk
œ -(2j 4k) .(8xi 8yj 2zk) Ê 2x œ 8x., 2y œ 2- 8y., and 2z œ 4- 2z. Ê x œ 0 or . œ "4 .
CASE 1: x œ 0 Ê 4(0)# 4y# z# œ 0 Ê z œ „ 2y Ê 2y 4(2y) 5 œ 0 Ê y œ
Ê y œ 56 yielding the points ˆ!ß "# ß "‰ and ˆ!ß 56 ß 53 ‰ .
CASE 2: . œ
#
"
4
"
#
, or 2y 4(2y) 5 œ 0
Ê y œ - y Ê - œ 0 Ê 2z œ 4(0) 2z ˆ 4" ‰ Ê z œ 0 Ê 2y 4(0) œ 5 Ê y œ
#
#
4 ˆ #5 ‰
(0) œ 4x
Ê no solution.
"
5
"
Then f ˆ!ß # ß 1‰ œ 4 and f ˆ!ß 56 ß 53 ‰ œ 25 ˆ 36
"9 ‰ œ
125
36
5
#
and
Ê the point ˆ!ß #" ß 1‰ is closest to the origin.
41. ™ f œ i j and ™ g œ yi xj so that ™ f œ - ™ g Ê i j œ -(yi xj) Ê 1 œ y- and 1 œ x- Ê y œ x
Ê y# œ 16 Ê y œ „ 4 Ê (4ß 4) and (%ß 4) are candidates for the location of extreme values. But as x Ä _,
y Ä _ and f(xß y) Ä _; as x Ä _, y Ä 0 and f(xß y) Ä _. Therefore no maximum or minimum value
exists subject to the constraint.
916
Chapter 14 Partial Derivatives
4
42. Let f(Aß Bß C) œ ! (Axk Byk C zk )# œ C# (B C 1)# (A B C 1)# (A C 1)# . We want
kœ1
to minimize f. Then fA (Aß Bß C) œ 4A 2B 4C, fB (Aß Bß C) œ 2A 4B 4C 4, and
fC (Aß Bß C) œ 4A 4B 8C 2. Set each partial derivative equal to 0 and solve the system to get A œ "# ,
B œ 3# , and C œ "4 or the critical point of f is ˆ "# ß 3# ß 4" ‰ .
43. (a) Maximize f(aß bß c) œ a# b# c# subject to a# b# c# œ r# . Thus ™ f œ 2ab# c# i 2a# bc# j 2a# b# ck and
™ g œ 2ai 2bj 2ck so that ™ f œ - ™ g Ê 2ab# c# œ 2a-, 2a# bc# œ 2b-, and 2a# b# c œ 2cÊ 2a# b# c# œ 2a# - œ 2b# - œ 2c# - Ê - œ 0 or a# œ b# œ c# .
CASE 1: - œ 0 Ê a# b# c# œ 0.
#
$
CASE 2: a# œ b# œ c# Ê f(aß bß c) œ a# a# a# and 3a# œ r# Ê f(aß bß c) œ Š r3 ‹ is the maximum value.
(b) The point ŠÈaß Èbß Èc‹ is on the sphere if a b c œ r# . Moreover, by part (a), abc œ f ŠÈaß Èbß Èc‹
#
$
Ÿ Š r3 ‹ Ê (abc)"Î$ Ÿ
r#
3
œ
abc
3
, as claimed.
n
44. Let f(x" ß x# ß á ß xn ) œ ! ai xi œ a" x" a# x# á an xn and g(x" ß x# ß á ß xn ) œ x#" x## á x#n 1. Then we
iœ1
want ™ f œ - ™ g Ê a" œ -(2x" ), a# œ -(2x# ), á , an œ -(2xn ), - Á 0 Ê xi œ
n
n
iœ1
iœ1
"Î#
Ê 4-# œ ! ai# Ê 2- œ Œ! ai#
n
n
iœ1
iœ1
ai
2-
Ê f(x" ß x# ß á ß xn ) œ ! ai xi œ ! ai ˆ #a-i ‰ œ
Ê
"
#-
a#"
4- #
a##
4- #
a#n
4- #
"Î#
á
n
n
iœ1
iœ1
! a#i œ Œ! a#i
the maximum value.
45-50. Example CAS commands:
Maple:
f := (x,y,z) -> x*y+y*z;
g1 := (x,y,z) -> x^2+y^2-2;
g2 := (x,y,z) -> x^2+z^2-2;
h := unapply( f(x,y,z)-lambda[1]*g1(x,y,z)-lambda[2]*g2(x,y,z), (x,y,z,lambda[1],lambda[2]) );
hx := diff( h(x,y,z,lambda[1],lambda[2]), x );
hy := diff( h(x,y,z,lambda[1],lambda[2]), y );
hz := diff( h(x,y,z,lambda[1],lambda[2]), z );
hl1 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[1] );
hl2 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[2] );
sys := { hx=0, hy=0, hz=0, hl1=0, hl2=0 };
q1 := solve( sys, {x,y,z,lambda[1],lambda[2]} );
q2 := map(allvalues,{q1});
for p in q2 do
eval( [x,y,z,f(x,y,z)], p );
``=evalf(eval( [x,y,z,f(x,y,z)], p ));
end do;
Mathematica: (assigned functions will vary)
Clear[x, y, z, lambda1, lambda2]
f[x_,y_,z_]:= x y y z
g1[x_,y_,z_]:= x2 y2 2
g2[x_,y_,z_]:= x2 z2 2
h = f[x, y, z] lambda1 g1[x, y, z] lambda2 g2[x, y, z];
hx= D[h, x]; hy= D[h, y]; hz= D[h,z]; hL1=D[h, lambda1]; hL2= D[h, lambda2];
critical=Solve[{hx==0, hy==0, hz==0, hL1==0, hL2==0, g1[x,y,z]==0, g2[x,y,z]==0},
# (a)
#(b)
# (c)
# (d)
is
œ1
Section 14.9 Partial Derivatives with Constrained Variables
{x, y, z, lambda1, lambda2}]//N
{{x, y, z}, f[x, y, z]}/.critical
14.9 PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES
1. w œ x# y# z# and z œ x# y# :
Î x œ x(yß z) Ñ
y
yœy
(a) Œ Ä
Ä w Ê Š ``wy ‹ œ
z
z
Ï zœz Ò
œ 2x `` xy 2y Ê 0 œ 2x `` xy 2y Ê
`x
`y
`y
`z
œ
"
#y
`x
`z
œ
1
2x
`w `x
`x `z
`z
`y
œ 0 and
œ 2x `` yx 2y `` yy
`w `y
`y `z
`w `z `x
`z `z ; `z
œ 0 and
`z
`z
œ 2x `` xz 2y `` yz
`z
`z
œ 2x `` xz 2y `` yz
"
Ê ˆ ``wz ‰x œ (2x)(0) (2y) Š 2y
‹ (2z)(1) œ 1 2z
`w `x
`x `z
`w `y
`y `z
`w `z `y
`z `z ; `z
œ 0 and
Ê ˆ ``wz ‰y œ (2x) ˆ #"x ‰ (2y)(0) (2z)(1) œ 1 2z
2. w œ x# y z sin t and x y œ t:
Î xœx Ñ
ÎxÑ
Ð yœy Ó
y
Ä Ð
Ä w Ê Š ``wy ‹ œ
(a)
Ó
zœz
xz
ÏzÒ
Ït œ x yÒ
ß
`t
`y
`w `z `z
`z `y ; `y
z
Î x œ x(yß z) Ñ
y
yœy
Ä w Ê ˆ ``wz ‰y œ
(c) Œ Ä
z
Ï zœz Ò
Ê 1 œ 2x `` xz Ê
`w `y
`y `y
œ yx Ê Š ``wy ‹ œ (2x) ˆ xy ‰ (2y)(1) (2z)(0) œ 2y 2y œ 0
Î xœx Ñ
x
y œ y(xß z)
Ä w Ê ˆ ``wz ‰x œ
(b) Œ Ä
z
Ï zœz Ò
Ê 1 œ 2y `` yz Ê
`w `x
`x `y
`w `x
`x `y
`w `y
`y `y
`w `z
`z `y
`w `t `x
`t `y ; `y
œ 0,
`z
`y
œ 0, and
œ 1 Ê Š ``wy ‹ œ (2x)(0) (1)(1) (1)(0) (cos t)(1) œ 1 cos t œ 1 cos (x y)
xßt
Îx œ t yÑ
ÎyÑ
Ð yœy Ó
z
Ä Ð
Ä w Ê Š ``wy ‹ œ
(b)
Ó
zœz
zt
ÏtÒ
Ï tœt Ò
ß
Ê
`x
`y
œ
`t
`y
`y
`y
`w `x
`x `y
`w `y
`y `y
`w `z
`z `y
`w `t `z
`t `y ; `y
œ 0 and
`t
`y
œ0
œ 1 Ê Š ``wy ‹ œ (2x)(1) (1)(1) (1)(0) (cos t)(0) œ 1 2at yb œ 1 2y 2t
zßt
Î xœx Ñ
ÎxÑ
Ð yœy Ó
y
Ä Ð
Ä w Ê ˆ ``wz ‰x y œ ``wx `` xz ``wy
(c)
Ó
œ
z
z
ÏzÒ
Ït œ x yÒ
Ê ˆ ``wz ‰ œ (2x)(0) (1)(0) (1)(1) (cos t)(0) œ 1
`y
`z
`w `z
`z `z
`w `t `x
`t `z ; `z
œ 0 and
`y
`z
œ0
`w `y
`y `z
`w `z
`z `z
`w `t `y
`t `z ; `z
œ 0 and
`t
`z
œ0
`w `z
`z `t
`w `t `x
`t `t ; `t
œ 0 and
`z
`t
œ0
ß
xßy
Îx œ t yÑ
ÎyÑ
Ð yœy Ó
z
Ä Ð
Ä w Ê ˆ ``wz ‰y t œ
(d)
Ó
zœz
ÏtÒ
Ï tœt Ò
ß
`w `x
`x `z
Ê ˆ ``wz ‰y t œ (2x)(0) (1)(0) (1)(1) (cos t)(0) œ 1
ß
Î xœx Ñ
ÎxÑ
Ð y œ t xÓ
z
Ä Ð
Ä w Ê ˆ ``wt ‰x z œ
(e)
Ó
zœz
ÏtÒ
Ï tœt Ò
ß
`w `x
`x `t
`w `y
`y `t
Ê ˆ ``wt ‰x z œ (2x)(0) (1)(1) (1)(0) (cos t)(1) œ 1 cos t
ß
917
918
Chapter 14 Partial Derivatives
Îx œ t yÑ
ÎyÑ
Ð yœy Ó
z
Ä Ð
Ä w Ê ˆ ``wt ‰y z œ
(f)
Ó
z
œ
z
ÏtÒ
Ï tœt Ò
ß
`w `x
`x `t
`w `y
`y `t
`w `z
`z `t
`w `t `y
`t `t ; `t
œ 0 and
`z
`t
œ0
Ê ˆ ``wt ‰y z œ (2x)(1) (1)(0) (1)(0) (cos t)(1) œ cos t 2x œ cos t 2(t y)
ß
3. U œ f(Pß Vß T) and PV œ nRT
Î PœP Ñ
P
VœV
Ä U Ê ˆ ``UP ‰V œ
(a) Œ Ä
V
Ï T œ PV Ò
nR
V ‰
œ ``UP ˆ ``UT ‰ ˆ nR
`U `P
`P `P
nRT
ÎP œ V Ñ
V
Ä U Ê ˆ ``UT ‰V œ
(b) Œ Ä
VœV
T
Ï TœT Ò
‰ `U
œ ˆ ``UP ‰ ˆ nR
V `T
`U `P
`P `T
4. w œ x# y# z# and y sin z z sin x œ 0
Î xœx Ñ
x
yœy
Ä w Ê ˆ ``wx ‰y œ
(a) Œ Ä
y
Ï z œ z(xß y) Ò
(y cos z)
Ê
`z
`x
(sin x)
ˆ ``wx ‰
yk (0ß1ß1)
`z
`x
z cos x œ 0 Ê
`z
`x
`x
`z
`U `V
`V `T
`U `T
`T `P
`w `y
`y `x
z cos x
y cos z sin x .
#
œ
`U `T
`T `T
`U
`P
‰ ˆ `U ‰
œ ˆ ``UP ‰ ˆ nR
V ` V (0)
`w `z `y
`z `x ; `x
V ‰
‰
ˆ ` U ‰ ˆ nR
ˆ `` U
V (0) ` T
`U
`T
œ 0 and
`z
`x
œ
1
1
œ1
œ (2x)
`x
`z
(2y)(0) (2z)(1)
At (0ß 1ß 1),
œ (2x)(1) (2y)(0) (2z)(1)k Ð0ß1ß1Ñ œ 21
Î x œ x(yß z) Ñ
y
yœy
Ä w Ê ˆ ``wz ‰y œ
(b) Œ Ä
z
Ï zœz Ò
œ (2x)
`w `x
`x `x
œ
`U `V
`V `P
`y
`z y
x) `` xz œ
2z. Now (sin z)
Ê y cos z sin x (z cos
`w `x
`x `z
`w `y
`y `z
cos z sin x (z cos x)
0 Ê
`x
`z
œ
y cos z sin x
.
z cos x
`w `z
`z `z
`x
`z
`y
`z œ 0
(!ß "ß 1), `` xz œ (11)(1)0
œ 0 and
At
œ
"
1
Ê ˆ ``wz ‰Ck (!,"ß1Ñ œ 2(0) ˆ 1" ‰ 21 œ 21
5. w œ x# y# yz z$ and x# y# z# œ 6
Î xœx Ñ
x
yœy
Ä w Ê Š ``wy ‹ œ
(a) Œ Ä
y
x
Ï z œ z(xß y) Ò
œ a2xy# b (0) a2x# y zb (1) ay 3z# b
`x
`y
œ 0 Ê 2y (2z)
`z
`y
œ0 Ê
#
`z
`y
`z
`y
`z
`y
`w `y
`y `y
`w `z
`z `y
œ 2x# y z ay 3z# b
`z
`y .
Now (2x)
œ yz . At (wß xß yß z) œ (4ß 2ß 1ß 1),
`z
`y
`x
`y
2y (2z)
`x
`y
`w `x
`x `y
a2x# y zb (1) ay 3z# b (0) œ a2x# yb
œ 0 Ê (2x)
`x
`y
2y œ 0 Ê
`x
`y
`w `y
`y `y
`x
`y
œv
`u
`y
u Š uv
`u
`y
u
`u
`y ‹
`v
`y ;
`w `z
`z `y
2x# y z. Now (2x)
œ yx . At (wß xß yß z) œ (4ß 2ß 1ß 1),
œ Šv
#
x œ u# v# and
u
v
#
‹
`u
`y
Ê
`u
`y
`x
`y
œ
œ 0 Ê 0 œ 2u
v
v# u# .
`u
`y
œ 0 and
x (4ß2ß1ßc1)
`x
`y
`x
`y
2y (2z)
œ "2 Ê Š ``wy ‹ ¹
z
œ (2)(2)(1)# ˆ "# ‰ (2)(2)# (1) (1) œ 5
6. y œ uv Ê 1 œ v
`z
`y
œ "1 œ 1 Ê Š ``wy ‹ ¹
#
œ c(2)(2) (1) (1)d c1 3(1) d (1) œ 5
Î x œ x(yß z) Ñ
y
yœy
Ä w Ê Š ``wy ‹ œ
(b) Œ Ä
z
z
Ï zœz Ò
œ a2xy# b
`w `x
`x `y
2v
`v
`y
At (uß v) œ ŠÈ2ß 1‹ ,
Ê
`u
`y
`v
`y
œ
œ ˆ uv ‰
"
#
1# ŠÈ2‹
`u
`y
`z
`y
(4ß2ß1ßc1)
Ê 1
œ 1
œ 0 and
Section 14.10 Taylor's Formula for Two Variables
Ê Š `` uy ‹ œ 1
x
r
x œ r cos )
7. Œ Ä Œ
Ê ˆ ``xr ‰) œ cos ); x# y# œ r# Ê 2x 2y
y œ r sin )
)
Ê ``xr œ xr Ê ˆ ``xr ‰ œ È #x #
`y
`x
`r
`x
œ 2r
and
`y
`x
8. If x, y, and z are independent, then ˆ ``wx ‰y z œ
ß
`w `x
`x `x
`w `y
`y `x
`w `z
`z `x
`w `t
`t `x
œ (2x)(1) (2y)(0) (4)(0) (1) ˆ ``xt ‰ œ 2x ``xt . Thus x 2z t œ 25 Ê 1 0
Ê ˆ ``wx ‰ œ 2x 1. On the other hand, if x, y, and t are independent, then ˆ ``wx ‰
yßz
œ
Ê 1
`r
`x
x y
y
`w `x
`x `x
œ 0 Ê 2x œ 2r
`t
`x
œ0 Ê
`t
`x
œ 1
yßt
``wy `` xy ``wz `` xz ``wt ``xt œ (2x)(1) (2y)(0) 4 `` xz (1)(0) œ
2 `` xz 0 œ 0 Ê `` xz œ "# Ê ˆ ``wx ‰yßt œ 2x 4 ˆ "# ‰ œ 2x 2.
9. If x is a differentiable function of y and z, then f(xß yß z) œ 0 Ê
`f `x
`x `x
`z
`x .
2x 4
`f `y
`y `x
`f `z
`z `x
Thus, x 2z t œ 25
œ0 Ê
`f
`x
`f `y
`y `x
œ0
Ê Š `` xy ‹ œ `` f/f/`` yz . Similarly, if y is a differentiable function of x and z, Š `` yz ‹ œ `` f/f/`` xz and if z is a
z
x
differentiable function of x and y, ˆ `` xz ‰y œ `` f/f/`` xy . Then Š `` xy ‹ Š `` yz ‹ ˆ `` xz ‰y
z
œ Š
` f/` y ˆ
` f/` z ‰
` f/` x
` f/` z ‹ ` f/` x Š ` f/` y ‹
10. z œ z f(u) and u œ xy Ê
œ x ˆ1 y
df ‰
du
y ˆx
df ‰
du
`z
`x
x
œ 1.
œ1
df ` u
du ` x
œ1y
df
du ;
also
`z
`y
œ0
df ` u
du ` y
œx
df
du
so that x
`z
`x
y
œ 0 and
`x
`y
œ0 Ê
`g
`y
œx
11. If x and y are independent, then g(xß yß z) œ 0 Ê
`g `x
`x `y
`g `y
`y `y
`g `z
`z `y
`y
Ê Š `` yz ‹ œ `` g/
g/` z , as claimed.
x
12. Let x and y be independent. Then f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and
Ê `` xf `` xx `` yf `` xy `` zf `` xz ``wf ``wx œ `` xf `` zf `` xz ``wf ``wx
`g `x
`g `y
`g `z
`g `w
`g
`g `z
`g `w
`x `x `y `x `z `x `w `x œ `x `z `x `w `x œ 0
`f
`z
`g
`z
`z
`x
`z
`x
`f
`w
`g
`w
`w
`x
`w
`x
œ `` xf
œ
`g
`x
Ê ˆ `` xz ‰y œ
c ``xf
» c `g
`x
`f
`z
» `g
`z
`f
`w
`g »
`w
`f
`w
`g »
`w
œ
`f
`y
`f `z
`z `y
`z
`z `y
`g
`f `z
`z `y
`f
`w
`g
`w
`w
`y
`w
`y
`f `w
`w `y
œ 0 and (similarly)
œ `` yf
œ
`g
`y
Ê Š ``wy ‹ œ
x
`f
`z
» `g
`z
`f
`z
» `g
`z
c ``yf
c `` gy »
`f
`w
`g »
`w
œ
œ0
imply
`g
`g `f
`w `x `w
`g `f
`f `g
`z `w
`z `w
`x
`y œ 0
`g
`g `z
`y `z `y
`y
`x
œ 0 and
``xf
Likewise, f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and
œ
`z
`y
œ
`f
`x
`f
`z
`g
`w
`g
`w
``wf
``wf
`g
`x
`g
`z
`f `x
`f `y
`f `z
`x `y `y `y `z `y
`g `w
` w ` y œ 0 imply
Ê
`g
`g `f
`y `z `y
`g `f
`f `g
`z `w `z `w
`` fz
œ
`f
`z
`f
`z
`g
`y
`g
`w
`g
`z
`f `g
`w `z
``yf
, as claimed.
`f `w
`w `y
, as claimed.
14.10 TAYLOR'S FORMULA FOR TWO VARIABLES
1. f(xß y) œ xey Ê fx œ ey , fy œ xey , fxx œ 0, fxy œ ey , fyy œ xey
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 1 y † 0 "# ax# † 0 2xy † 1 y# † 0b œ x xy quadratic approximation;
fxxx œ 0, fxxy œ 0, fxyy œ ey , fyyy œ xey
`g `z
`z `y
œ0
919
920
Chapter 14 Partial Derivatives
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (!ß !) 3x# yfxxy (0ß 0) 3xy# fxyy (!ß !) y$ fyyy (0ß 0)d
œ x xy "6 ax$ † 0 3x# y † 0 3xy# † 1 y$ † 0b œ x xy #" xy# , cubic approximation
2. f(xß y) œ ex cos y Ê fx œ ex cos y, fy œ ex sin y, fxx œ ex cos y, fxy œ ex sin y, fyy œ ex cos y
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (!ß 0) "# cx# fxx (!ß !) 2xyfxy (!ß !) y# fyy (0ß 0)d
œ 1 x † 1 y † 0 "# cx# † 1 2xy † 0 y# † (1)d œ 1 x "# ax# y# b , quadratic approximation;
fxxx œ ex cos y, fxxy œ ex sin y, fxyy œ ex cos y, fyyy œ ex sin y
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ 1 x "# ax# y# b 6" cx$ † 1 3x# y † 0 3xy# † (1) y$ † 0d
œ 1 x "# ax# y# b 6" ax$ 3xy# b , cubic approximation
3. f(xß y) œ y sin x Ê fx œ y cos x, fy œ sin x, fxx œ y sin x, fxy œ cos x, fyy œ 0
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (!ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 0 "# ax# † 0 2xy † 1 y# † 0b œ xy, quadratic approximation;
fxxx œ y cos x, fxxy œ sin x, fxyy œ 0, fyyy œ 0
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ xy "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ xy, cubic approximation
4. f(xß y) œ sin x cos y Ê fx œ cos x cos y, fy œ sin x sin y, fxx œ sin x cos y, fxy œ cos x sin y,
fyy œ sin x cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 1 y † 0 "# ax# † 0 2xy † 0 y# † 0b œ x, quadratic approximation;
fxxx œ cos x cos y, fxxy œ sin x sin y, fxyy œ cos x cos y, fyyy œ sin x sin y
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ x "6 cx$ † (1) 3x# y † 0 3xy# † (1) y$ † 0d œ x 6" ax$ 3xy# b, cubic approximation
5. f(xß y) œ ex ln (1 y) Ê fx œ ex ln (1 y), fy œ
ex
1y
, fxx œ ex ln (1 y), fxy œ
ex
1y
x
, fyy œ (1 e y)#
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 1 "# cx# † 0 2xy † 1 y# † (1)d œ y "# a2xy y# b , quadratic approximation;
fxxx œ ex ln (1 y), fxxy œ
Ê f(xß y) ¸ quadratic
ex
ex
2ex
1 y , fxyy œ (1 y)# , fyyy œ (1 y)$
"
$
#
#
6 cx fxxx (0ß 0) 3x yfxxy (!ß 0) 3xy fxyy (0ß 0)
$
#
#
$
y$ fyyy (0ß 0)d
œ y "2 a2xy y# b 6" cx † 0 3x y † 1 3xy † (1) y † 2d
œ y "# a2xy y# b 6" a3x# y 3xy# 2y$ b , cubic approximation
4
2
(2x y 1)# , fxy œ (2x y 1)# ,
"
#
#
fyy œ (2x "
y 1)# Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) # cx fxx (0ß 0) 2xyfxy (0ß 0) y fyy (0ß 0)d
œ 0 x † 2 y † 1 "# cx# † (4) 2xy † (2) y# † (1)d œ 2x y "# a4x# 4xy y# b
œ (2x y) "# (2x y)# , quadratic approximation;
fxxx œ (2x 16y 1)$ , fxxy œ (2x 8y 1)$ , fxyy œ (2x 4y 1)$ , fyyy œ (2x 2y 1)$
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ (2x y) "# (2x y)# 6" ax$ † 16 3x# y † 8 3xy# † 4 y$ † 2b
œ (2x y) "# (2x y)# 3" a8x$ 12x# y 6xy# y# b
œ (2x y) "# (2x y)# 3" (2x y)$ , cubic approximation
6. f(xß y) œ ln (2x y 1) Ê fx œ
2
2x y 1
, fy œ
"
#x y 1
, fxx œ
7. f(xß y) œ sin ax# y# b Ê fx œ 2x cos ax# y# b , fy œ 2y cos ax# y# b , fxx œ 2 cos ax# y# b 4x# sin ax# y# b ,
fxy œ 4xy sin ax# y# b , fyy œ 2 cos ax# y# b 4y# sin ax# y# b
Section 14.10 Taylor's Formula for Two Variables
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 0 "# ax# † 2 2xy † 0 y# † 2b œ x# y# , quadratic approximation;
fxxx œ 12x sin ax# y# b 8x$ cos ax# y# b , fxxy œ 4y sin ax# y# b 8x# y cos ax# y# b ,
fxyy œ 4x sin ax# y# b 8xy# cos ax# y# b , fyyy œ 12y sin ax# y# b 8y$ cos ax# y# b
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ x# y# "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ x# y# , cubic approximation
8. f(xß y) œ cos ax# y# b Ê fx œ 2x sin ax# y# b , fy œ 2y sin ax# y# b ,
fxx œ 2 sin ax# y# b 4x# cos ax# y# b , fxy œ 4xy cos ax# y# b , fyy œ 2 sin ax# y# b 4y# cos ax# y# b
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 0 y † 0 "# cx# † 0 2xy † 0 y# † 0d œ 1, quadratic approximation;
fxxx œ 12x cos ax# y# b 8x$ sin ax# y# b , fxxy œ 4y cos ax# y# b 8x# y sin ax# y# b ,
fxyy œ 4x cos ax# y# b 8xy# sin ax# y# b , fyyy œ 12y cos ax# y# b 8y$ sin ax# y# b
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0) y$ fyyy (0ß 0)d
œ 1 "6 ax$ † 0 3x# y † 0 3xy# † 0 y$ † 0b œ 1, cubic approximation
9. f(xß y) œ
"
1xy
Ê fx œ
"
(1 x y)#
œ fy , fxx œ
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0)
2
(1 x y)$
"
#
# cx fxx (0ß 0)
œ fxy œ fyy
2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 1 y † 1 "# ax# † 2 2xy † 2 y# † 2b œ 1 (x y) ax# 2xy y# b
œ 1 (x y) (x y)# , quadratic approximation; fxxx œ
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0)
6
œ fxxy œ fxyy œ fyyy
(1 x y)%
#
3xy fxyy (0ß 0) y$ fyyy (0ß 0)d
$
œ 1 (x y) (x y)# "6 ax$ † 6 3x# y † 6 3xy# † 6 y † 6b
œ 1 (x y) (x y)# ax$ 3x# y 3xy# y$ b œ 1 (x y) (x y)# (x y)$ , cubic approximation
10. f(xß y) œ
fxy œ
"
1 x y xy
1
(" x y xy)#
Ê fx œ
, fyy œ
1y
(1 x y xy)#
, fy œ
1x
(" x y xy)#
, fxx œ
2(1 y)#
(1 x y xy)$
,
#
2(" x)
(1 x y xy)$
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 1 y † 1 "# ax# † 2 2xy † 1 y# † 2b œ 1 x y x# xy y# , quadratic approximation;
fxxx œ
6(1 y)$
(1 x y xy)%
, fxxy œ
[4(1 x y xy) 6(1 y)(1 x)](1 y)
(1 x y xy)%
,
$
x)
[4(1 x y xy) 6(1 x)(1 y)](1 x)
, fyyy œ (1 6(1
(1 x y xy)%
x y xy)%
Ê f(xß y) ¸ quadratic "6 cx$ fxxx (0ß 0) 3x# yfxxy (!ß 0) 3xy# fxyy (0ß 0)
œ 1 x y x# xy y# "6 ax$ † 6 3x# y † 2 3xy# † 2 y$ † 6b
#
#
$
#
#
$
fxyy œ
y$ fyyy (0ß 0)d
œ 1 x y x xy y x x y xy y , cubic approximation
11. f(xß y) œ cos x cos y Ê fx œ sin x cos y, fy œ cos x sin y, fxx œ cos x cos y, fxy œ sin x sin y,
fyy œ cos x cos y Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 1 x † 0 y † 0 "# cx# † (1) 2xy † 0 y# † (1)d œ 1
x#
#
y#
#
, quadratic approximation. Since all partial
derivatives of f are products of sines and cosines, the absolute value of these derivatives is less than or equal
to 1 Ê E(xß y) Ÿ "6 c(0.1)$ 3(0.1)$ 3(0.1)$ 0.1)$ d Ÿ 0.00134.
12. f(xß y) œ ex sin y Ê fx œ ex sin y, fy œ ex cos y, fxx œ ex sin y, fxy œ ex cos y, fyy œ ex sin y
Ê f(xß y) ¸ f(0ß 0) xfx (0ß 0) yfy (0ß 0) "# cx# fxx (0ß 0) 2xyfxy (0ß 0) y# fyy (0ß 0)d
œ 0 x † 0 y † 1 "# ax# † 0 2xy † 1 y# † 0b œ y xy , quadratic approximation. Now, fxxx œ ex sin y,
fxxy œ ex cos y, fxyy œ ex sin y, and fyyy œ ex cos y. Since kxk Ÿ 0.1, kex sin yk Ÿ ke0Þ1 sin 0.1k ¸ 0.11 and
kex cos yk Ÿ ke0Þ1 cos 0.1k ¸ 1.11. Therefore,
921
922
Chapter 14 Partial Derivatives
E(xß y) Ÿ
"
6
c(0.11)(0.1)$ 3(1.11)(0.1)$ 3(0.11)(0.1)$ (1.11)(0.1)$ d Ÿ 0.000814.
CHAPTER 14 PRACTICE EXERCISES
1. Domain: All points in the xy-plane
Range: z 0
Level curves are ellipses with major axis along the y-axis
and minor axis along the x-axis.
2. Domain: All points in the xy-plane
Range: 0 z _
Level curves are the straight lines x y œ ln z with
slope 1, and z 0.
3. Domain: All (xß y) such that x Á 0 and y Á 0
Range: z Á 0
Level curves are hyperbolas with the x- and y-axes
as asymptotes.
4. Domain: All (xß y) so that x# y
Range: z 0
0
Level curves are the parabolas y œ x# c, c
0.
5. Domain: All points (xß yß z) in space
Range: All real numbers
Level surfaces are paraboloids of revolution with
the z-axis as axis.
Chapter 14 Practice Exercises
6. Domain: All points (xß yß z) in space
Range: Nonnegative real numbers
Level surfaces are ellipsoids with center (0ß 0ß 0).
7. Domain: All (xß yß z) such that (xß yß z) Á (0ß !ß 0)
Range: Positive real numbers
Level surfaces are spheres with center (0ß 0ß 0) and
radius r 0.
8. Domain: All points (xß yß z) in space
Range: (0ß 1]
Level surfaces are spheres with center (0ß 0ß 0) and
radius r 0.
9.
lim
Ðxß yÑ Ä Ð1ß ln 2Ñ
ey cos x œ eln 2 cos 1 œ (2)(1) œ 2
2y
10.
lim
Ðxß yÑ Ä Ð0ß 0Ñ x cos y
11.
lim
#
#
Ðx ß y Ñ Ä Ð 1 ß 1 Ñ x y
xÁ „y
12.
13.
14.
xy
œ
œ
20
0 cos 0
œ2
xy
lim
Ðxß yÑ Ä Ð1ß 1Ñ (x y)(x y)
xÁ „y
œ
1
lim
Ðxß yÑ Ä Ð1ß 1Ñ x y
(xy 1) ax# y# xy 1b
xy 1
lim
x$ y$ 1
xy 1
lim
ln kx y zk œ ln k1 (1) ek œ ln e œ 1
Ðx ß y Ñ Ä Ð 1 ß 1 Ñ
P Ä Ð1 ß 1 ß e Ñ
lim
P Ä Ð1 ß 1 ß 1 Ñ
œ
lim
Ð x ß y Ñ Ä Ð 1 ß 1Ñ
œ
œ
lim
Ðxß yÑ Ä Ð1ß 1Ñ
"
11
œ
"
#
ax# y# xy 1b œ 1# † 1# 1 † 1 1 œ 3
tan" (x y z) œ tan" (1 (1) (1)) œ tan" (1) œ 14
15. Let y œ kx# , k Á 1. Then
y
lim
#
Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x y
#
yÁx
œ
kx#
lim
#
#
axß kx# b Ä Ð0ß 0Ñ x kx
œ
k
1 k#
which gives different limits for
œ
1 k#
k
which gives different limits for
different values of k Ê the limit does not exist.
16. Let y œ kx, k Á 0. Then
lim
Ðxß yÑ Ä Ð0ß 0Ñ
xy Á 0
x# y#
xy
œ
lim
(xß kxÑ Ä Ð0ß 0Ñ
x# (kx)#
x(kx)
923
924
Chapter 14 Partial Derivatives
different values of k Ê the limit does not exist.
17. Let y œ kx. Then
x# y#
œ
lim
#
#
Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x y
x# k# x#
x # k# x#
1 k#
1 k#
œ
which gives different limits for different values
of k Ê the limit does not exist so f(0ß 0) cannot be defined in a way that makes f continuous at the origin.
sin (x y)
Ðx ß y Ñ Ä Ð 0 ß 0 Ñ k x y k
18. Along the x-axis, y œ 0 and
œ lim
lim
sin x
kx k
xÄ0
œœ
1, x 0
, so the limit fails to exist
", x 0
Ê f is not continuous at (0ß 0).
19.
`g
`r
œ cos ) sin ),
20.
`f
`x
œ
"
#
Š x# 2x
y# ‹
`f
`y
œ
"
#
Š x# 2y
y# ‹
21.
`f
` R"
œ R"# ,
"
`f
` R#
`g
`)
œ r sin ) r cos )
y
‹
x#
y #
1 ˆx‰
Š
Š 1x ‹
y #
1 ˆx‰
œ R"# ,
œ
x
x# y#
y
x# y#
œ
xy
x# y#
œ
y
x# y#
x
x# y#
œ
xy
x# y#
`f
` R$
#
,
œ R"#
$
22. hx (xß yß z) œ 21 cos (21x y 3z), hy (xß yß z) œ cos (21x y 3z), hz (xß yß z) œ 3 cos (21x y 3z)
23.
`P
`n
œ
RT
V
,
`P
`R
œ
nT
V
`P
`T
,
œ
nR
V
,
`P
`V
œ nRT
V#
24. fr (rß jß Tß w) œ 2r"# j É 1Tw , fj (rß jß Tß w) œ #r"j# É
25.
œ
"
4rj
É T1"w œ
`g
`x
œ
"
y
`g
`y
,
"
4rjT
œ1
, fT (rß jß Tß w) œ ˆ #"rj ‰ Š È"1w ‹ Š 2È" T ‹
T
1w
É 1Tw , fw (rß jß Tß w) œ ˆ #"rj ‰ É T1 ˆ "# w$Î# ‰ œ 4r"jw É 1Tw
x
y#
Ê
` #g
` x#
œ 0,
` #g
` y#
œ
2x
y$
,
` #g
` y` x
œ
` #g
` x` y
œ y"#
26. gx (xß y) œ ex y cos x, gy (xß y) œ sin x Ê gxx (xß y) œ ex y sin x, gyy (xß y) œ 0, gxy (xß y) œ gyx (xß y) œ cos x
27.
`f
`x
œ 1 y 15x#
2x
x# 1
,
`f
`y
œx Ê
` #f
` x#
œ 30x
22x#
ax # 1 b #
,
` #f
` y#
œ 0,
` #f
` y` x
œ
` #f
` x` y
œ1
28. fx (xß y) œ 3y, fy (xß y) œ 2y 3x sin y 7ey Ê fxx (xß y) œ 0, fyy (xß y) œ 2 cos y 7ey , fxy (xß y) œ fyx (xß y)
œ 3
29.
`w
`x
Ê
Ê
30.
`w
`x
Ê
Ê
31.
`w
`x
Ê
œ y cos (xy 1),
`w
`y
œ x cos (xy 1),
dx
dt
œ et ,
dy
dt
dw
t
ˆ " ‰
dt œ [y cos (xy 1)]e [x cos (xy 1)] t1 ;
dw ¸
ˆ " ‰
dt tœ0 œ 0 † 1 [1 † (1)] 01 œ 1
œ ey ,
`w
`y
œ xey sin z,
dw
y "Î#
axey
dt œ e t
dw ¸
dt tœ1 œ 1 † 1 (2 † 1
œ 2 cos (2x y),
`w
`r
`w
`y
`w
`z
œ y cos z sin z,
sin zb ˆ1
"‰
t
dx
dt
œ
"
t1
t œ 0 Ê x œ 1 and y œ 0
œ t"Î# ,
dy
dt
œ 1 "t ,
dz
dt
œ1
(y cos z sin z)1; t œ 1 Ê x œ 2, y œ 0, and z œ 1
0)(2) (0 0)1 œ 5
œ cos (2x y),
`x
`r
œ 1,
`x
`s
œ cos s,
`y
`r
œ s,
`y
`s
œr
œ [2 cos (2x y)](1) [ cos (2x y)](s); r œ 1 and s œ 0 Ê x œ 1 and y œ 0
Chapter 14 Practice Exercises
`w ¸
` r Ð1ß0Ñ
`w ¸
` s Ð1ß0Ñ
Ê
Ê
32.
33.
`w
`u
`w
`v
œ
`f
`x
œ y z,
œ
Ê
`w
`x
œ
`w
`s
œ [2 cos (2x y)](cos s) [ cos (2x y)](r)
œ (2 cos 21)(cos 0) (cos 21)(1) œ 2 1
`x
`u
`x
`v
œ ˆ 1 x x#
" ‰
u
x# 1 a2e cos vb ; u œ v œ 0 Ê
" ‰
`w ¸
u
x# 1 a2e sin vb Ê ` v Ð0ß0Ñ œ
`f
`y
`f
`z
œ ˆ 1 x x#
œ x z,
œ y x,
dx
dt
df
dt œ (y z)(sin t) (x z)(cos
df ¸
dt tœ1 œ (sin 1 cos 2)(sin 1)
Ê
34.
dw
dx
dw
dx
œ (2 cos 21) (cos 21)(0) œ 2;
dw ` s
ds ` x
œ (5)
dw
ds
and
`w
`y
œ
dw ` s
ds ` y
œ sin t,
œ ˆ 52 "5 ‰ (2) œ
1 y cos xy
2y x cos xy
dz
dt
œ 2 sin 2t
(cos 1 cos 2)(cos 1) 2(sin 1 cos 1)(sin 2)
œ (1)
dw
ds
œ
`w
`x
Ê
dw
ds
5
`w
`y
œ5
œ
dy
dx ¹ Ð0ß1Ñ
1"
2
dw
ds
5
dw
ds
dy
dx ¹ Ð0ßln 2Ñ
1 y cos xy
œ FFxy œ 2y
x cos xy
dy
dx
xby
2y e
œ FFxy œ 2x
exby
ß
1‰
4
#
œ
i
j Ê f increases most rapidly in the direction u œ
È2
#
i
u œ È12 i
1
È2
1
È2
i
1
È2
"
È2
œ
È2
#
and decreases most
38. ™ f œ 2xec2y i 2x# ec2y j Ê ™ f k Ð1ß0Ñ œ #i #j Ê k ™ f k œ È2# (2)# œ 2È2; u œ
Ê f increases most rapidly in the direction u œ
#
œ "# i "# j Ê k ™ f k œ Ɉ "# ‰ ˆ "# ‰ œ
È2
# j
È
È
È
È
rapidly in the direction u œ #2 i #2 j ; (Du f)P! œ k ™ f k œ #2 and (Dcu f)P! œ #2 ;
7
u" œ kvvk œ È33i # 4j4# œ 35 i 45 j Ê (Du" f)P! œ ™ f † u" œ ˆ "# ‰ ˆ 35 ‰ ˆ "# ‰ ˆ 45 ‰ œ 10
™f
k™f k
È2
#
dy
dx
2
œ 2 ln0 2
2 œ (ln 2 1)
37. ™ f œ ( sin x cos y)i (cos x sin y)j Ê ™ f k ˆ 14
È2
#
œ0
œ 1
36. F(xß y) œ 2xy exy 2 Ê Fx œ 2y exy and Fy œ 2x exy Ê
uœ
;
t) 2(y x)(sin 2t); t œ 1 Ê x œ cos 1, y œ sin 1, and z œ cos 2
Ê at (xß y) œ (!ß 1) we have
Ê at (xß y) œ (!ß ln 2) we have
2
5
ˆ 52 "5 ‰ (0) œ 0
œ cos t,
dy
dt
`w ¸
` u Ð0ß0Ñ
xœ2 Ê
35. F(xß y) œ 1 x y# sin xy Ê Fx œ 1 y cos xy and Fy œ 2y x cos xy Ê
œ
™f
k™f k
œ
1
È2
i
1
È2
j
j and decreases most rapidly in the direction
j ; (Du f)P! œ k ™ f k œ 2È2 and (Dcu f)P! œ 2È2 ; u" œ
v
kv k
œ
ij
È 1# 1#
œ
1
È2
i
1
È2
j
Ê (Du" f)P! œ ™ f † u" œ (2) Š È" ‹ (2) Š È" ‹ œ 0
2
2
2
3
6
39. ™ f œ Š 2x 3y
6z ‹ i Š 2x 3y 6z ‹ j Š 2x 3y 6z ‹ k Ê ™ f k Ð1ß1ß1Ñ œ 2i 3j 6k ;
uœ
™f
k™f k
œ
2i 3j 6k
È 2# 3# 6#
œ
2
7
i 37 j 67 k Ê f increases most rapidly in the direction u œ
2
7
i 37 j 67 k and
decreases most rapidly in the direction u œ 27 i 37 j 67 k ; (Du f)P! œ k ™ f k œ 7, (Du f)P! œ 7;
u" œ
v
kv k
œ
925
2
7
i 37 j 67 k Ê (Du" f)P! œ (Du f)P! œ 7
40. ™ f œ (2x 3y)i (3x 2)j (1 2z)k Ê ™ f k Ð0ß0ß0Ñ œ 2j k ; u œ
rapidly in the direction u œ
2
È5
j
"
È5
™f
k™f k
œ
2
È5
j
"
È5
k Ê f increases most
k and decreases most rapidly in the direction u œ È25 j
(Du f)P! œ k ™ f k œ È5 and (Du f)P! œ È5 ; u" œ
v
kv k
œ
ijk
È 1# 1# 1#
Ê (Du" f)P! œ ™ f † u" œ (0) Š È"3 ‹ (2) Š È"3 ‹ (1) Š È"3 ‹ œ
3
È3
œ
"
È3
œ È3
i
"
È3
j
"
È3
k
"
È5
k;
;
926
Chapter 14 Partial Derivatives
41. r œ (cos 3t)i (sin 3t)j 3tk Ê v(t) œ (3 sin 3t)i (3 cos 3t)j 3k Ê v ˆ 13 ‰ œ 3j 3k
Ê u œ È"2 j
"
È2
k ; f(xß yß z) œ xyz Ê ™ f œ yzi xzj xyk ; t œ
Ê ™ f k Ð1ß0ß1Ñ œ 1j Ê ™ f † u œ (1j) † Š È"2 j
"
È2
k‹ œ
1
3
yields the point on the helix (1ß 0ß 1)
1
È2
42. f(xß yß z) œ xyz Ê ™ f œ yzi xzj xyk ; at (1ß 1ß 1) we get ™ f œ i j k Ê the maximum value of
Du f k
œ k ™ f k œ È3
Ð1ß1ß1Ñ
43. (a) Let ™ f œ ai bj at (1ß 2). The direction toward (2ß 2) is determined by v" œ (2 1)i (2 2)j œ i œ u
so that ™ f † u œ 2 Ê a œ 2. The direction toward (1ß 1) is determined by v# œ (1 1)i (1 2)j œ j œ u
so that ™ f † u œ 2 Ê b œ 2 Ê b œ 2. Therefore ™ f œ 2i 2j ; fx a1, 2b œ fy a1, 2b œ 2.
(b) The direction toward (4ß 6) is determined by v$ œ (4 1)i (6 2)j œ 3i 4j Ê u œ 35 i 45 j
Ê ™f†uœ
14
5
.
44. (a) True
(b) False
(c) True
(d) True
45. ™ f œ 2xi j 2zk Ê
™ f k Ð0ß1ß1Ñ œ j 2k ,
™ f k Ð0ß0ß0Ñ œ j ,
™ f k Ð0ß1ß1Ñ œ j 2k
46. ™ f œ 2yj 2zk Ê
™ f k Ð2ß2ß0Ñ œ 4j ,
™ f k Ð2ß2ß0Ñ œ 4j ,
™ f k Ð2ß0ß2Ñ œ 4k ,
™ f k Ð2ß0ß2Ñ œ 4k
47. ™ f œ 2xi j 5k Ê ™ f k Ð2ß1ß1Ñ œ 4i j 5k Ê Tangent Plane: 4(x 2) (y 1) 5(z 1) œ 0
Ê 4x y 5z œ 4; Normal Line: x œ 2 4t, y œ 1 t, z œ 1 5t
48. ™ f œ 2xi 2yj k Ê ™ f k Ð1ß1ß2Ñ œ 2i 2j k Ê Tangent Plane: 2(x 1) 2(y 1) (z 2) œ 0
Ê 2x 2y z 6 œ 0; Normal Line: x œ 1 2t, y œ 1 2t, z œ 2 t
49.
`z
`x
œ
2x
x# y#
Ê
`z ¸
` x Ð0ß1ß0Ñ
œ 0 and
`z
`y
œ
2y
x# y#
2(y 1) (z 0) œ 0 or 2y z 2 œ 0
Ê
`z
` y ¹ Ð0ß1ß0Ñ
œ 2; thus the tangent plane is
Chapter 14 Practice Exercises
50.
`z
`x
œ 2x ax# y# b
#
`z ¸
` x ˆ1ß1ß 12 ‰
Ê
œ #" and
`z
`y
œ 2y ax# y# b
#
Ê
`z
` y ¹ ˆ1ß1ß 1 ‰
2
927
œ #" ; thus the tangent
plane is "# (x 1) "# (y 1) ˆz "# ‰ œ 0 or x y 2z 3 œ 0
51. ™ f œ ( cos x)i j Ê ™ f k Ð1ß1Ñ œ i j Ê the tangent
line is (x 1) (y 1) œ 0 Ê x y œ 1 1; the
normal line is y 1 œ 1(x 1) Ê y œ x 1 1
52. ™ f œ xi yj Ê ™ f k Ð1ß2Ñ œ i 2j Ê the tangent
line is (x 1) 2(y 2) œ 0 Ê y œ
"
#
x 3# ; the normal
line is y 2 œ 2(x 1) Ê y œ 2x 4
53. Let f(xß yß z) œ x# 2y 2z 4 and g(xß yß z) œ y 1. Then ™ f œ 2xi 2j 2kk a1 1 12 b œ 2i 2j 2k
â
â
â i j kâ
â
â
and ™ g œ j Ê ™ f ‚ ™ g œ â 2 2 2 â œ 2i 2k Ê the line is x œ 1 2t, y œ 1, z œ "# 2t
â
â
â0 " 0â
ß ß
54. Let f(xß yß z) œ x y# z 2 and g(xß yß z) œ y 1. Then ™ f œ i 2yj kk a 12 1 12 b œ i 2j k and
â
â
â i j kâ
â
â
™ g œ j Ê ™ f ‚ ™ g œ â 1 2 1 â œ i k Ê the line is x œ "# t, y œ 1, z œ "# t
â
â
â0 " 0â
ß ß
55. f ˆ 14 ß 14 ‰ œ
"
#
, fx ˆ 14 ß 14 ‰ œ cos x cos yk Ð1Î4ß1Î4Ñ œ
Ê L(xß y) œ
"
#
"# ˆx 14 ‰ "# ˆy 14 ‰ œ
"
#
"
#
"
#
, fy ˆ 14 ß 14 ‰ œ sin x sin yk Ð1Î4ß1Î4Ñ œ "#
x "# y; fxx (xß y) œ sin x cos y, fyy (xß y) œ sin x cos y, and
fxy (xß y) œ cos x sin y. Thus an upper bound for E depends on the bound M used for kfxx k , kfxy k , and kfyy k .
With M œ
È2
#
we have kE(xß y)k Ÿ
with M œ 1, kE(xß y)k Ÿ
"
#
"
#
Š
È2
ˆ¸
# ‹ x
#
14 ¸ ¸y 14 ¸‰ Ÿ
#
(1) ˆ¸x 14 ¸ ¸y 14 ¸‰ œ
"
#
È2
4
(0.2)# Ÿ 0.0142;
(0.2)# œ 0.02.
56. f(1ß 1) œ 0, fx (1ß 1) œ yk Ð1ß1Ñ œ 1, fy (1ß 1) œ x 6yk Ð1ß1Ñ œ 5 Ê L(xß y) œ (x 1) 5(y 1) œ x 5y 4;
fxx (xß y) œ 0, fyy (xß y) œ 6, and fxy (xß y) œ 1 Ê maximum of kfxx k , kfyy k , and kfxy k is 6 Ê M œ 6
Ê kE(xß y)k Ÿ
"
#
(6) akx 1k ky 1kb# œ
"
#
(6)(0.1 0.2)# œ 0.27
57. f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ y 3zk Ð1ß0ß0Ñ œ 0, fy (1ß 0ß 0) œ x 2zk Ð1ß0ß0Ñ œ 1, fz (1ß 0ß 0) œ 2y 3xk Ð1ß0ß0Ñ œ 3
Ê L(xß yß z) œ 0(x 1) (y 0) 3(z 0) œ y 3z; f(1ß 1ß 0) œ 1, fx (1ß 1ß 0) œ 1, fy (1ß 1ß 0) œ 1, fz ("ß "ß !) œ 1
Ê L(xß yß z) œ 1 (x 1) (y 1) 1(z 0) œ x y z 1
58. f ˆ0ß !ß 14 ‰ œ 1, fx ˆ!ß 0ß 14 ‰ œ È2 sin x sin (y z)¹
ˆ0ß0ß 1 ‰
œ 0, fy ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y z)¹
4
fz ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y z)¹
ˆ0ß0ß 1 ‰
È2
#
œ 1 Ê L(xß yß z) œ 1 1(y 0) 1 ˆz 14 ‰ œ 1 y z
Ê L(xß yß z) œ
È2
È2
È2
ˆ1 1 ‰
ˆ1 1 ‰
# , fy 4 ß 4 ß 0 œ # , fz 4 ß 4 ß 0 œ #
È
È
È
È
È
#2 ˆy 14 ‰ #2 (z 0) œ #2 #2 x #2
, fx ˆ 14 ß 14 ß 0‰ œ
È2
#
È2
#
ˆx 14 ‰
œ 1,
4
4
f ˆ 14 ß 14 ß 0‰ œ
ˆ0ß0ß 1 ‰
y
È2
#
z
1
4
;
928
Chapter 14 Partial Derivatives
59. V œ 1r# h Ê dV œ 21rh dr 1r# dh Ê dVk Ð1Þ5ß5280Ñ œ 21(1.5)(5280) dr 1(1.5)# dh œ 15,8401 dr 2.251 dh.
You should be more careful with the diameter since it has a greater effect on dV.
60. df œ (2x y) dx (x 2y) dy Ê df k Ð1ß2Ñ œ 3 dy Ê f is more sensitive to changes in y; in fact, near the point
(1ß 2) a change in x does not change f.
61. dI œ
"
R
dV
V
R#
"
100
dR Ê dI¸ Ð24ß100Ñ œ
dV
24
100#
dR Ê dI¸ dVœ1ßdRœ20 œ 0.01 (480)(.0001) œ 0.038,
" ‰
20 ‰
or increases by 0.038 amps; % change in V œ (100) ˆ 24
¸ 4.17%; % change in R œ ˆ 100
(100) œ 20%;
Iœ
24
100
œ 0.24 Ê estimated % change in I œ
dI
I
‚ 100 œ
0.038
0.24
‚ 100 ¸ 15.83% Ê more sensitive to voltage change.
62. A œ 1ab Ê dA œ 1b da 1a db Ê dAk Ð10ß16Ñ œ 161 da 101 db; da œ „ 0.1 and db œ „ 0.1
¸ ¸ 2.61
¸
Ê dA œ „ 261(0.1) œ „ 2.61 and A œ 1(10)(16) œ 1601 Ê ¸ dA
A ‚ 100 œ 1601 ‚ 100 ¸ 1.625%
63. (a) y œ uv Ê dy œ v du u dv; percentage change in u Ÿ 2% Ê kduk Ÿ 0.02, and percentage change in v Ÿ 3%
Ê kdvk Ÿ 0.03;
dy
y
Ÿ 2% 3% œ 5%
(b) z œ u v Ê dzz œ
œ
v du u dv
uv
du dv
uv
œ
œ
du
uv
Ê ¸ dzz ‚ 100¸ Ÿ ¸ du
u ‚ 100
64. C œ
Ê
dv
v
du
u
dv
uv
Þ
Þ
Þ
Þ
Þ
Þ
Ÿ
¸ du
Ê ¹ dy
y ‚ 100¹ œ u ‚ 100
du
u
dv
v
dv
v
¸ ¸ dv
¸
‚ 100¸ Ÿ ¸ du
u ‚ 100 v ‚ 100
(since u 0, v 0)
‚ 100¸ œ ¹ dy
y ‚ 100¹
(0.425)(7)
7
71.84w0 425 h0 725 Ê Cw œ 71.84w1 425 h0 725
2.975
5.075
dC œ 71.84w
1 425 h0 725 dw 71.84w0 425 h1 725
Þ
dv
v
Þ
and Ch œ
(0.725)(7)
71.84w0 425 h1 725
Þ
Þ
dh; thus when w œ 70 and h œ 180 we have
dCk Ð70ß180Ñ ¸ (0.00000225) dw (0.00000149) dh Ê 1 kg error in weight has more effect
65. fx (xß y) œ 2x y 2 œ 0 and fy (xß y) œ x 2y 2 œ 0 Ê x œ 2 and y œ 2 Ê (2ß 2) is the critical point;
#
fxx (2ß 2) œ 2, fyy (#ß 2) œ 2, fxy (#ß 2) œ 1 Ê fxx fyy fxy
œ 3 0 and fxx 0 Ê local minimum value
of f(#ß 2) œ 8
66. fx (xß y) œ 10x 4y 4 œ 0 and fy (xß y) œ 4x 4y 4 œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1) is the critical point;
#
fxx (0ß 1) œ 10, fyy (0ß 1) œ 4, fxy (0ß 1) œ 4 Ê fxx fyy fxy
œ 56 0 Ê saddle point with f(0ß 1) œ 2
67. fx (xß y) œ 6x# 3y œ 0 and fy (xß y) œ 3x 6y# œ 0 Ê y œ 2x# and 3x 6 a4x% b œ 0 Ê x a1 8x$ b œ 0
Ê x œ 0 and y œ 0, or x œ "# and y œ "# Ê the critical points are (0ß 0) and ˆ "# ß "# ‰ . For (!ß !):
#
œ 9 0 Ê saddle point with
fxx (!ß !) œ 12xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 12yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 Ê fxx fyy fxy
#
f(0ß 0) œ 0. For ˆ "# ß "# ‰: fxx œ 6, fyy œ 6, fxy œ 3 Ê fxx fyy fxy
œ 27 0 and fxx 0 Ê local maximum
"
"‰
"
ˆ
value of f # ß # œ 4
68. fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3y# 3x œ 0 Ê y œ x# and x% x œ 0 Ê x ax$ 1b œ 0 Ê the critical
points are (0ß 0) and (1ß 1) . For (!ß !): fxx (!ß !) œ 6xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 6yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3
#
Ê fxx fyy fxy
œ 9 0 Ê saddle point with f(0ß 0) œ 15. For (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3
#
Ê fxx fyy fxy
œ 27 0 and fxx 0 Ê local minimum value of f(1ß 1) œ 14
69. fx (xß y) œ 3x# 6x œ 0 and fy (xß y) œ 3y# 6y œ 0 Ê x(x 2) œ 0 and y(y 2) œ 0 Ê x œ 0 or x œ 2 and
y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (2ß 0), and (2ß 2) . For (!ß !): fxx (!ß !) œ 6x 6k Ð0ß0Ñ
#
œ 6, fyy (!ß !) œ 6y 6k Ð0ß0Ñ œ 6, fxy (!ß 0) œ 0 Ê fxx fyy fxy
œ 36 0 Ê saddle point with f(0ß 0) œ 0. For
#
(0ß 2): fxx (!ß 2) œ 6, fyy (0ß #) œ 6, fxy (!ß 2) œ 0 Ê fxx fyy fxy
œ 36 0 and fxx 0 Ê local minimum value of
Chapter 14 Practice Exercises
929
#
f(!ß 2) œ 4. For (#ß 0): fxx (2ß 0) œ 6, fyy (#ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy fxy
œ 36 0 and fxx 0
Ê local maximum value of f(2ß 0) œ 4. For (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0
#
Ê fxx fyy fxy
œ 36 0 Ê saddle point with f(2ß 2) œ 0.
70. fx (xß y) œ 4x$ 16x œ 0 Ê 4x ax# 4b œ 0 Ê x œ 0, 2, 2; fy (xß y) œ 6y 6 œ 0 Ê y œ 1. Therefore the critical
points are (0ß 1), (2ß 1), and (2ß 1). For (!ß 1): fxx (!ß 1) œ 12x# 16k Ð0ß1Ñ œ 16, fyy (!ß 1) œ 6, fxy (!ß 1) œ 0
#
Ê fxx fyy fxy
œ 96 0 Ê saddle point with f(0ß 1) œ 3. For (2ß 1): fxx (2ß 1) œ 32, fyy (2ß 1) œ 6,
#
fxy (2ß 1) œ 0 Ê fxx fyy fxy
œ 192 0 and fxx 0 Ê local minimum value of f(2ß 1) œ 19. For (#ß 1):
#
fxx (2ß 1) œ 32, fyy (#ß 1) œ 6, fxy (2ß 1) œ 0 Ê fxx fyy fxy
œ 192 0 and fxx 0 Ê local minimum value of
f(2ß 1) œ 19.
71. (i)
On OA, f(xß y) œ f(0ß y) œ y# 3y for 0 Ÿ y Ÿ 4
Ê f w (!ß y) œ 2y 3 œ 0 Ê y œ 3# . But ˆ!ß 3# ‰
is not in the region.
Endpoints: f(0ß 0) œ 0 and f(0ß 4) œ 28.
(ii) On AB, f(xß y) œ f(xß x 4) œ x# 10x 28
for 0 Ÿ x Ÿ 4 Ê f w (xß x 4) œ 2x 10 œ 0
Ê x œ 5, y œ 1. But (5ß 1) is not in the region.
Endpoints: f(4ß 0) œ 4 and f(!ß 4) œ 28.
(iii) On OB, f(xß y) œ f(xß 0) œ x# 3x for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x 3 Ê x œ
critical point with f ˆ 3# ß !‰ œ 94 .
3
#
and y œ 0 Ê ˆ 3# ß 0‰ is a
Endpoints: f(0ß 0) œ 0 and f(%ß 0) œ 4.
(iv) For the interior of the triangular region, fx (xß y) œ 2x y 3 œ 0 and fy (xß y) œ x 2y 3 œ 0 Ê x œ 3
and y œ 3. But (3ß 3) is not in the region. Therefore the absolute maximum is 28 at (0ß 4) and the
absolute minimum is 94 at ˆ 3# ß !‰ .
On OA, f(xß y) œ f(0ß y) œ y# 4y 1 for
0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2y 4 œ 0 Ê y œ 2 and
x œ 0. But (0ß 2) is not in the interior of OA.
Endpoints: f(0ß 0) œ 1 and f(0ß 2) œ 5.
(ii) On AB, f(xß y) œ f(xß 2) œ x# 2x 5 for 0 Ÿ x Ÿ 4
Ê f w (xß 2) œ 2x 2 œ 0 Ê x œ 1 and y œ 2
Ê (1ß 2) is an interior critical point of AB with
f(1ß 2) œ 4. Endpoints: f(4ß 2) œ 13 and f(!ß 2) œ 5.
(iii) On BC, f(xß y) œ f(4ß y) œ y# 4y 9 for 0 Ÿ y Ÿ 2 Ê f w (4ß y) œ 2y 4 œ 0 Ê y œ # and x œ 4. But
(4ß 2) is not in the interior of BC. Endpoints: f(4ß 0) œ 9 and f(%ß 2) œ 13.
(iv) On OC, f(xß y) œ f(xß 0) œ x# 2x 1 for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x 2 œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0)
is an interior critical point of OC with f(1ß 0) œ 0. Endpoints: f(0ß 0) œ 1 and f(4ß 0) œ 9.
(v) For the interior of the rectangular region, fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and
y œ 2. But (1ß 2) is not in the interior of the region. Therefore the absolute maximum is 13 at (4ß 2)
and the absolute minimum is 0 at (1ß 0).
72. (i)
930
73. (i)
Chapter 14 Partial Derivatives
On AB, f(xß y) œ f(2ß y) œ y# y 4 for
2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y 1 Ê y œ "# and
x œ 2 Ê ˆ2ß "# ‰ is an interior critical point in AB
with f ˆ2ß "# ‰ œ 17
4 . Endpoints: f(2ß 2) œ 2 and
f(2ß 2) œ 2.
On BC, f(xß y) œ f(xß 2) œ 2 for 2 Ÿ x Ÿ 2
Ê f w (xß 2) œ 0 Ê no critical points in the interior of
BC. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 2.
(iii) On CD, f(xß y) œ f(2ß y) œ y# 5y 4 for
2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y 5 œ 0 Ê y œ 5# and x œ 2. But ˆ#ß 5# ‰ is not in the region.
(ii)
Endpoints: f(2ß 2) œ 18 and f(2ß 2) œ 2.
(iv) On AD, f(xß y) œ f(xß 2) œ 4x 10 for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 4 Ê no critical points in the interior
of AD. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 18.
(v) For the interior of the square, fx (xß y) œ y 2 œ 0 and fy (xß y) œ 2y x 3 œ 0 Ê y œ 2 and x œ 1
Ê (1ß 2) is an interior critical point of the square with f(1ß 2) œ 2. Therefore the absolute maximum
"‰
ˆ
is 18 at (2ß 2) and the absolute minimum is 17
4 at #ß # .
On OA, f(xß y) œ f(0ß y) œ 2y y# for 0 Ÿ y Ÿ 2
Ê f w (!ß y) œ 2 2y œ 0 Ê y œ 1 and x œ 0 Ê
(!ß 1) is an interior critical point of OA with
f(0ß 1) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0.
(ii) On AB, f(xß y) œ f(xß 2) œ 2x x# for 0 Ÿ x Ÿ 2
Ê f w (xß 2) œ 2 2x œ 0 Ê x œ 1 and y œ 2
Ê (1ß 2) is an interior critical point of AB with
f(1ß 2) œ 1. Endpoints: f(0ß 2) œ 0 and f(2ß 2) œ 0.
(iii) On BC, f(xß y) œ f(2ß y) œ 2y y# for 0 Ÿ y Ÿ 2
Ê f w (2ß y) œ 2 2y œ 0 Ê y œ 1 and x œ 2
Ê (2ß 1) is an interior critical point of BC with f(2ß 1) œ 1. Endpoints: f(2ß 0) œ 0 and f(2ß 2) œ 0.
(iv) On OC, f(xß y) œ f(xß 0) œ 2x x# for 0 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2 2x œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0)
is an interior critical point of OC with f(1ß 0) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0.
(v) For the interior of the rectangular region, fx (xß y) œ 2 2x œ 0 and fy (xß y) œ 2 2y œ 0 Ê x œ 1 and
y œ 1 Ê (1ß 1) is an interior critical point of the square with f(1ß 1) œ 2. Therefore the absolute maximum
is 2 at (1ß 1) and the absolute minimum is 0 at the four corners (0ß 0), (0ß 2), (2ß 2), and (2ß 0).
74. (i)
On AB, f(xß y) œ f(xß x 2) œ 2x 4 for
2 Ÿ x Ÿ 2 Ê f w (xß x 2) œ 2 œ 0 Ê no critical
points in the interior of AB. Endpoints: f(2ß 0) œ 8
and f(2ß 4) œ 0.
(ii) On BC, f(xß y) œ f(2ß y) œ y# 4y for 0 Ÿ y Ÿ 4
Ê f w (2ß y) œ 2y 4 œ 0 Ê y œ 2 and x œ 2
Ê (2ß 2) is an interior critical point of BC with
f(2ß 2) œ 4. Endpoints: f(2ß 0) œ 0 and f(2ß 4) œ 0.
(iii) On AC, f(xß y) œ f(xß 0) œ x# 2x for 2 Ÿ x Ÿ 2
Ê f w (xß 0) œ 2x 2 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of AC with f(1ß 0) œ 1.
Endpoints: f(2ß 0) œ 8 and f(2ß 0) œ 0.
(iv) For the interior of the triangular region, fx (xß y) œ 2x 2 œ 0 and fy (xß y) œ 2y 4 œ 0 Ê x œ 1 and
y œ 2 Ê (1ß 2) is an interior critical point of the region with f(1ß 2) œ 3. Therefore the absolute maximum
is 8 at (2ß 0) and the absolute minimum is 1 at (1ß 0).
75. (i)
Chapter 14 Practice Exercises
76. (i)
(ii)
931
On AB, f(xß y) œ f(xß x) œ 4x# 2x% 16 for
2 Ÿ x Ÿ 2 Ê f w (xß x) œ 8x 8x$ œ 0 Ê x œ 0
and y œ 0, or x œ 1 and y œ 1, or x œ 1 and y œ 1
Ê (0ß 0), ("ß 1), (1ß 1) are all interior points of AB
with f(0ß 0) œ 16, f(1ß 1) œ 18, and f(1ß 1) œ 18.
Endpoints: f(2ß 2) œ 0 and f(2ß 2) œ 0.
On BC, f(xß y) œ f(2ß y) œ 8y y% for 2 Ÿ y Ÿ 2
Ê f w (2ß y) œ 8 4y$ œ 0 Ê y œ $È2 and x œ 2
Ê Š2ß $È2‹ is an interior critical point of BC with
f Š2ß $È2‹ œ 6 $È2. Endpoints: f(2ß 2) œ 32 and f(2ß 2) œ 0.
(iii) On AC, f(xß y) œ f(xß 2) œ 8x x% for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 8 4x$ œ 0 Ê x œ $È2 and y œ 2
Ê Š $È2ß 2‹ is an interior critical point of AC with f Š $È2ß 2‹ œ 6 $È#. Endpoints:
f(2ß 2) œ 0 and f(2ß 2) œ 32.
(iv) For the interior of the triangular region, fx (xß y) œ 4y 4x$ œ 0 and fy (xß y) œ 4x 4y$ œ 0 Ê x œ 0 and
y œ 0, or x œ 1 and y œ 1 or x œ 1 and y œ 1. But neither of the points (0ß 0) and (1ß 1), or (1ß 1) are interior
to the region. Therefore the absolute maximum is 18 at (1ß 1) and (1ß 1), and the absolute minimum is 32 at
(2ß 2).
On AB, f(xß y) œ f(1ß y) œ y$ 3y# 2 for
1 Ÿ y Ÿ 1 Ê f w (1ß y) œ 3y# 6y œ 0 Ê y œ 0
and x œ 1, or y œ 2 and x œ 1 Ê (1ß 0) is an
interior critical point of AB with f(1ß 0) œ 2; (1ß 2)
is outside the boundary. Endpoints: f(1ß 1) œ 2
and f(1ß 1) œ 0.
(ii) On BC, f(xß y) œ f(xß 1) œ x$ 3x# 2 for
1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 6x œ 0 Ê x œ 0
and y œ 1, or x œ 2 and y œ 1 Ê (0ß 1) is an
interior critical point of BC with f(!ß 1) œ 2; (2ß 1) is outside the boundary. Endpoints: f("ß 1) œ 0 and
f("ß 1) œ 2.
(iii) On CD, f(xß y) œ f("ß y) œ y$ 3y# 4 for 1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y# 6y œ 0 Ê y œ 0 and x œ 1, or
y œ 2 and x œ 1 Ê ("ß 0) is an interior critical point of CD with f("ß 0) œ 4; (1ß 2) is outside the boundary.
Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0.
(iv) On AD, f(xß y) œ f(xß 1) œ x$ 3x# 4 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 6x œ 0 Ê x œ 0 and y œ 1,
or x œ 2 and y œ 1 Ê (0ß 1) is an interior point of AD with f(0ß 1) œ 4; (#ß 1) is outside the
boundary. Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0.
(v) For the interior of the square, fx (xß y) œ 3x# 6x œ 0 and fy (xß y) œ 3y# 6y œ 0 Ê x œ 0 or x œ 2, and
y œ 0 or y œ 2 Ê (0ß 0) is an interior critical point of the square region with f(!ß 0) œ 0; the points (0ß 2),
(2ß 0), and (2ß 2) are outside the region. Therefore the absolute maximum is 4 at (1ß 0) and the
absolute minimum is 4 at (0ß 1).
77. (i)
932
Chapter 14 Partial Derivatives
On AB, f(xß y) œ f(1ß y) œ y$ 3y for 1 Ÿ y Ÿ 1
Ê f w (1ß y) œ 3y# 3 œ 0 Ê y œ „ 1 and x œ 1
yielding the corner points (1ß 1) and (1ß 1) with
f(1ß 1) œ 2 and f(1ß 1) œ 2.
(ii) On BC, f(xß y) œ f(xß 1) œ x$ 3x 2 for
1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 3 œ 0 Ê no
solution. Endpoints: f("ß 1) œ 2 and f("ß 1) œ 6.
(iii) On CD, f(xß y) œ f("ß y) œ y$ 3y 2 for
1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y# 3 œ 0 Ê no
solution. Endpoints: f(1ß 1) œ 6 and f("ß 1) œ 2.
(iv) On AD, f(xß y) œ f(xß 1) œ x$ 3x for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x# 3 œ 0 Ê x œ „ 1 and y œ 1
yielding the corner points (1ß 1) and (1ß 1) with f(1ß 1) œ 2 and f(1ß 1) œ 2
(v) For the interior of the square, fx (xß y) œ 3x# 3y œ 0 and fy (xß y) œ 3y# 3x œ 0 Ê y œ x# and
x% x œ 0 Ê x œ 0 or x œ 1 Ê y œ 0 or y œ 1 Ê (!ß 0) is an interior critical point of the square
region with f(0ß 0) œ 1; (1ß 1) is on the boundary. Therefore the absolute maximum is 6 at ("ß 1) and
the absolute minimum is 2 at (1ß 1) and (1ß 1).
78. (i)
79. ™ f œ 3x# i 2yj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê 3x# i 2yj œ -(2xi 2yj) Ê 3x# œ 2x- and
2y œ 2y- Ê - œ 1 or y œ 0.
CASE 1: - œ 1 Ê 3x# œ 2x Ê x œ 0 or x œ 23 ; x œ 0 Ê y œ „ 1 yielding the points (0ß 1) and (!ß 1); x œ
Ê yœ „
È5
3
yielding the points Š 23 ß
È5
3 ‹
and Š 23 ß
2
3
È5
3 ‹.
CASE 2: y œ 0 Ê x# 1 œ 0 Ê x œ „ 1 yielding the points (1ß 0) and (1ß 0).
Evaluations give f a!ß „ 1b œ 1, f Š 23 ß „
È5
3 ‹
œ
23
27
, f("ß 0) œ 1, and f("ß 0) œ 1. Therefore the absolute
maximum is 1 at a!ß „ 1b and (1ß 0), and the absolute minimum is 1 at ("ß !).
80. ™ f œ yi xj and ™ g œ 2xi 2yj so that ™ f œ - ™ g Ê yi xj œ -(2xi 2yj) Ê y œ 2-x and
xy œ 2-y Ê x œ 2-(2-x) œ 4-# x Ê x œ 0 or 4-# œ 1.
CASE 1: x œ 0 Ê y œ 0 but (0ß 0) does not lie on the circle, so no solution.
CASE 2: 4-# œ 1 Ê - œ "# or - œ "# . For - œ "# , y œ x Ê 1 œ x# y# œ 2x# Ê x œ C œ „ È" yielding the
2
points
Š È"2
ß
"
È2 ‹
and Š
"
È2
,
"
È2 ‹ .
"
#
#
#
#
For - œ , y œ x Ê 1 œ x y œ 2x Ê x œ „
"
È2
and
y œ x yielding the points Š È"2 ß È"2 ‹ and Š È"2 , È"2 ‹ .
Evaluations give the absolute maximum value f Š È"2 ß È"2 ‹ œ f Š È"2 ß È"2 ‹ œ
"
#
and the absolute minimum
value f Š È"2 ß È"2 ‹ œ f Š È"2 ß È"2 ‹ œ #" .
81. (i) f(xß y) œ x# 3y# 2y on x# y# œ 1 Ê ™ f œ 2xi (6y 2)j and ™ g œ 2xi 2yj so that ™ f œ - ™ g
Ê 2xi (6y 2)j œ -(2xi 2yj) Ê 2x œ 2x- and 6y 2 œ 2y- Ê - œ 1 or x œ 0.
CASE 1: - œ 1 Ê 6y 2 œ 2y Ê y œ "# and x œ „
È3
#
yielding the points Š „
È3
#
ß "# ‹ .
CASE 2: x œ 0 Ê y# œ 1 Ê y œ „ 1 yielding the points a!ß „ 1b .
Evaluations give f Š „
È3
#
ß "# ‹ œ
"
#
, f(0ß 1) œ 5, and f(!ß 1) œ 1. Therefore
"
#
and 5 are the extreme
values on the boundary of the disk.
(ii) For the interior of the disk, fx (xß y) œ 2x œ 0 and fy (xß y) œ 6y 2 œ 0 Ê x œ 0 and y œ "3
Ê ˆ!ß 13 ‰ is an interior critical point with f ˆ!ß "3 ‰ œ 3" . Therefore the absolute maximum of f on the
disk is 5 at (0ß 1) and the absolute minimum of f on the disk is "3 at ˆ!ß 3" ‰ .
Chapter 14 Practice Exercises
933
82. (i) f(xß y) œ x# y# 3x xy on x# y# œ 9 Ê ™ f œ (2x 3 y)i (2y x)j and ™ g œ 2xi 2yj so that
™ f œ - ™ g Ê (2x 3 y)i (2y x)j œ -(2xi 2yj) Ê 2x 3 y œ 2x- and 2y x œ 2yÊ 2x(" -) y œ 3 and x 2y(1 -) œ 0 Ê 1 - œ
x
2y
x
and (2x) Š 2y
‹ y œ 3 Ê x# y# œ 3y
Ê x# œ y# 3y. Thus, 9 œ x# y# œ y# 3y y# Ê 2y# 3y 9 œ 0 Ê (2y 3)(y 3) œ 0
Ê y œ 3, 3# . For y œ 3, x# y# œ 9 Ê x œ 0 yielding the point (0ß 3). For y œ 3# , x# y# œ 9
Ê x#
9
4
œ 9 Ê x# œ
Ê xœ „
27
4
È
¸ 20.691, and f Š 3 # 3 , 3# ‹ œ 9
27È3
4
3È 3
#
È
. Evaluations give f(0ß 3) œ 9, f Š 3 # 3 ß 3# ‹ œ 9
27È3
4
¸ 2.691.
(ii) For the interior of the disk, fx (xß y) œ 2x 3 y œ 0 and fy (xß y) œ 2y x œ 0 Ê x œ 2 and y œ 1
Ê (2ß 1) is an interior critical point of the disk with f(2ß 1) œ 3. Therefore, the absolute maximum of f on
the disk is 9
27È3
4
È
at Š 3 # 3 ß 3# ‹ and the absolute minimum of f on the disk is 3 at (2ß 1).
83. ™ f œ i j k and ™ g œ 2xi 2yj 2zk so that ™ f œ - ™ g Ê i j k œ -(2xi 2yj 2zk) Ê 1 œ 2x-,
1 œ 2y-, 1 œ 2z- Ê x œ y œ z œ -" . Thus x# y# z# œ 1 Ê 3x# œ 1 Ê x œ „ È"3 yielding the points
Š È"3 ß È"3 ,
"
È3 ‹
and Š È"3 ,
f Š È"3 ß È"3 ß È"3 ‹ œ
3
È3
"
È3
, È"3 ‹ . Evaluations give the absolute maximum value of
œ È3 and the absolute minimum value of f Š È"3 ß È"3 ß È"3 ‹ œ È3.
84. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin and g(xß yß z) œ z# xy 4. Then
™ f œ 2xi 2yj 2zk and ™ g œ yi xj 2zk so that ™ f œ - ™ g Ê 2x œ -y, 2y œ -x, and 2z œ 2-z
Ê z œ 0 or - œ 1.
CASE 1: z œ 0 Ê xy œ 4 Ê x œ 4y and y œ 4x Ê 2 Š 4y ‹ œ -y and 2 ˆ 4x ‰ œ -x Ê
8
-
œ y# and
8
-
œ x#
Ê y# œ x# Ê y œ „ x. But y œ x Ê x# œ 4 leads to no solution, so y œ x Ê x# œ 4 Ê x œ „ 2
yielding the points (2ß 2ß 0) and (2ß 2ß 0).
CASE 2: - œ 1 Ê 2x œ y and 2y œ x Ê 2y œ ˆ y# ‰ Ê 4y œ y Ê y œ 0 Ê x œ 0 Ê z# 4 œ 0 Ê z œ „ 2
yielding the points (0ß 0ß 2) and (!ß 0ß 2).
Evaluations give f(2ß 2ß 0) œ f(2ß 2ß 0) œ 8 and f(!ß !ß 2) œ f(!ß !ß 2) œ 4. Thus the points (!ß !ß 2) and
(!ß !ß 2) on the surface are closest to the origin.
85. The cost is f(xß yß z) œ 2axy 2bxz 2cyz subject to the constraint xyz œ V. Then ™ f œ - ™ g
Ê 2ay 2bz œ -yz, 2ax 2cz œ -xz, and 2bx 2cy œ -xy Ê 2axy 2bxz œ -xyz, 2axy 2cyz œ -xyz, and
2bxz 2cyz œ -xyz Ê 2axy 2bxz œ 2axy 2cyz Ê y œ ˆ bc ‰ x. Also 2axy 2bxz œ 2bxz 2cyz Ê z œ ˆ ac ‰ x.
Then x ˆ bc x‰ ˆ ac x‰ œ V Ê x$ œ
#
Height œ z œ ˆ ac ‰ Š cabV ‹
"Î$
#
c# V
ab
œ Š abcV ‹
#
Ê width œ x œ Š cabV ‹
"Î$
"Î$
#
, Depth œ y œ ˆ bc ‰ Š cabV ‹
"Î$
#
œ Š bacV ‹
"Î$
, and
.
86. The volume of the pyramid in the first octant formed by the plane is V(aß bß c) œ
"
3
ˆ #" ab‰ c œ
"
6
abc. The point
(2ß 1ß 2) on the plane Ê "b 2c œ 1. We want to minimize V subject to the constraint 2bc ac 2ab œ abc.
ac
ab
Thus, ™ V œ bc
6 i 6 j 6 k and ™ g œ (c 2b bc)i (2c 2a ac)j (2b a ab)k so that ™ V œ ac
ab
abc
Ê bc
6 œ -(c 2b bc), 6 œ -(2c 2a ac), and 6 œ -(2b a ab) Ê 6 œ -(ac 2ab abc),
abc
abc
6 œ -(2bc 2ab abc), and 6 œ -(2bc ac abc) Ê -ac œ 2-bc and 2-ab œ 2-bc. Now - Á 0 since
2
a
a Á 0, b Á 0, and c Á 0 Ê ac œ 2bc and ab œ bc Ê a œ 2b œ c. Substituting into the constraint equation gives
y
2
2
2
x
z
a a a œ 1 Ê a œ 6 Ê b œ 3 and c œ 6. Therefore the desired plane is 6 3 6 œ 1 or x 2y z œ 6.
87. ™ f œ (y z)i xj xk , ™ g œ 2xi 2yj , and ™ h œ zi xk so that ™ f œ - ™ g . ™ h
Ê (y z)i xj xk œ -(2xi 2yj) .(zi xk) Ê y z œ 2-x .z, x œ 2-y, x œ .x Ê x œ 0
™g
934
Chapter 14 Partial Derivatives
or . œ 1.
CASE 1: x œ 0 which is impossible since xz œ 1.
CASE 2: . œ 1 Ê y z œ 2-x z Ê y œ 2-x and x œ 2-y Ê y œ (2-)(2-y) Ê y œ 0 or
4-# œ 1. If y œ 0, then x# œ 1 Ê x œ „ 1 so with xz œ 1 we obtain the points (1ß 0ß 1)
and (1ß 0ß 1). If 4-# œ 1, then - œ „ "# . For - œ "# , y œ x so x# y# œ 1 Ê x# œ
Ê xœ „
"
È2
"
#
with xz œ 1 Ê z œ „ È2, and we obtain the points Š È"2 ß È"2 ß È2‹ and
Š È"2 ß È"2 ß È2‹ . For - œ
"
#
, y œ x Ê x# œ
"
#
Ê xœ „
"
È2
with xz œ 1 Ê z œ „ È2,
and we obtain the points Š È"2 ß È"2 , È2‹ and Š È"2 ß È"2 ß È2‹ .
Evaluations give f(1ß 0ß 1) œ 1, f(1ß 0ß 1) œ 1, f Š È"2 ß È"2 ß È2‹ œ
f Š È"2 ß È"2 ß È2‹ œ
3
#
, and f Š È"2 ß È"2 ß È2‹ œ
3
#
"
#
, f Š È"2 ß È"2 , È2‹ œ
. Therefore the absolute maximum is
Š È"2 ß È"2 ß È2‹ and Š È"2 ß È"2 ß È2‹ , and the absolute minimum is
"
#
3
#
"
#
,
at
at Š È"2 ß È"2 ß È2‹ and
Š È"2 ß È"2 ß È2‹ .
88. Let f(xß yß z) œ x# y# z# be the square of the distance to the origin. Then ™ f œ 2xi 2yj 2zk ,
™ g œ i j k , and ™ h œ 4xi 4yj 2zk so that ™ f œ - ™ g . ™ h Ê 2x œ - 4x., 2y œ - 4y.,
and 2z œ - 2z. Ê - œ 2x(1 2.) œ 2y(1 2.) œ 2z(1 2.) Ê x œ y or . œ "# .
CASE 1: x œ y Ê z# œ 4x# Ê z œ „ 2x so that x y z œ 1 Ê x x 2x œ 1 or x x 2x œ 1
(impossible) Ê x œ "4 Ê y œ 4" and z œ #" yielding the point ˆ 4" ß 4" ß #" ‰ .
CASE 2: . œ
"
#
Ê - œ 0 Ê 0 œ 2z(1 1) Ê z œ 0 so that 2x# 2y# œ 0 Ê x œ y œ 0. But the origin
(!ß 0ß 0) fails to satisfy the first constraint x y z œ 1.
Therefore, the point ˆ "4 ß 4" ß #" ‰ on the curve of intersection is closest to the origin.
89. (a) y, z are independent with w œ x# eyz and z œ x# y# Ê
œ a2xeyz b
`x
`y
`w
`y
`w `x
`w `y
`w `z
`x `y `y `y `z `y
œ 2x `` xy 2y Ê `` xy œ yx
œ
azx# eyz b (1) ayx# eyz b (0); z œ x# y# Ê 0
; therefore,
Š ``wy ‹ œ a2xeyz b ˆ yx ‰ zx# eyz œ a2y zx# b eyz
z
(b) z, x are independent with w œ x# eyz and z œ x# y# Ê
œ a2xeyz b (0) azx# eyz b
`y
`z
`w
`z
œ
`w `x
`x `z
ayx# eyz b (1); z œ x# y# Ê 1 œ 0 2y
1
ˆ ``wz ‰ œ azx# eyz b Š 2y
‹ yx# eyz œ x# eyz Šy
x
`w `y
`w `z
`y `z `z `z
`y
`y
"
` z Ê ` z œ #y
; therefore,
z
2y ‹
(c) z, y are independent with w œ x# eyz and z œ x# y# Ê
`w
`z
œ
œ a2xeyz b `` xz azx# eyz b (0) ayx# eyz b (1); z œ x# y# Ê 1
1 ‰
ˆ ``wz ‰ œ a2xeyz b ˆ 2x
yx# eyz œ a1 x# yb eyz
`w `x
`w `y
`w `z
`x `z `y `z `z `z
œ 2x `` xz 0 Ê `` xz œ #"x
; therefore,
y
90. (a) T, P are independent with U œ f(Pß Vß T) and PV œ nRT Ê ``UT œ ``UP `` TP
‰ ˆ ``VT ‰ ˆ ``UT ‰ (1); PV œ nRT Ê P ``VT œ nR Ê ``VT œ
œ ˆ ``UP ‰ (0) ˆ `` U
V
U ‰ ˆ nR ‰
ˆ ``UT ‰ œ ˆ `` V
P
P
`U `V
`U `T
`V `T `T `T
nR
P ; therefore,
`U
`T
`U `P
`U `V
(b) V, T are independent with U œ f(Pß Vß T) and PV œ nRT Ê `` U
V œ `P `V `V `V
U‰
œ ˆ ``UP ‰ ˆ ``VP ‰ ˆ `` V
(1) ˆ ``UT ‰ (0); PV œ nRT Ê V ``VP P œ (nR) ˆ ``VT ‰ œ 0 Ê
ˆ `` U
‰
V T
œ
ˆ ``UP ‰ ˆ VP ‰
`U
`V
91. Note that x œ r cos ) and y œ r sin ) Ê r œ Èx# y# and ) œ tan" ˆ yx ‰ . Thus,
`w
`x
œ
`w `r
`r `x
`w `)
`) `x
œ ˆ ``wr ‰ Š Èx#x y# ‹ ˆ ``w) ‰ Š x#yy# ‹ œ (cos ))
`w
`r
ˆ sinr ) ‰
`w
`)
;
`U `T
`T `V
`P
P
`V œ V
; therefore,
Chapter 14 Practice Exercises
`w
`y
œ
`w `r
`r `y
`u
`x
92. zx œ fu
93.
`u
`y
`v
`x
fv
`w
`x
`u
`x œ a
" `w
" `w
a `x œ b `y
œ
and
`w
`z
œ
2
rs
œ
Ê
Ê
œ
2
x# y# 2z
and
`w
`s
œ
`w
dw
` x œ du
b ``wx œ a
"
(r s)#
`w `x
`x `s
`w `y
`y `s
Solving this system yields
Ê ae cos vb
`u
`y
`u
`x
ae sin vb
`u
`y
`v
`y
`v
`y
fv
dw ` u
du ` y
œ
2(r s)
2 ar# 2rs s# b
œ
"
rs
œ
`w `x
`x `r
dw
du
`w `z
`z `s
and
œ
`w `y
`y `r
"
rs
rs
(r s)#
,
œb
`w
`y
`w `z
`z `r
ˆ cosr ) ‰
dw
du
œ
œ
Ê
2y
x# y# 2z
"
rs
`v
u
` x œ 1; e sin v
`v
u
sin v.
` x œ e
œ
œ
dw
du
2(r s)
#(r s)#
rs
(r s)#
and
" `w
b `y
œ
rs
(r s)#
œ
dw
du
,
’ (r " s)# “ (2s) œ
2r 2s
(r s)#
2
rs
`u
`x
aeu cos vb
`v
`x
œ 0.
Similarly, e cos v x œ 0
u
œ 0 and e sin v y œ 0 Ê aeu sin vb
`v
`y
`w
`)
y œ 0 Ê aeu sin vb
u
œ eu sin v and
" `w
a `x
’ (r " s)# “ (2r) œ
aeu sin vb
cos v and
`w
`r
œ bfu bfv
œ
`u
`x
u
œe
u
second system yields
`u
`y
`w
`y
œa
`w
`r
Ê
95. eu cos v x œ 0 Ê aeu cos vb
u
`u
`x
`w
`y
2(r s)
(r s)# (r s)# 4rs -
œ
2x
x# y# 2z
œ ˆ ``wr ‰ Š Èx#y y# ‹ ˆ ``w) ‰ Š x# x y# ‹ œ (sin ))
œ afu afv , and zy œ fu
œ b and
Ê
94.
`w `)
` ) )y
935
`u
`y
œ eu cos v. Therefore Š `` ux i
aeu cos vb
`u
`y
`v
`y
j‹ † Š `` vx i
œ 1. Solving this
`v
`y
j‹
œ caeu cos vb i aeu sin vb jd † caeu sin vb i aeu cos vb jd œ 0 Ê the vectors are orthogonal Ê the angle
between the vectors is the constant 1# .
96.
`g
`)
œ
Ê
`f `x
`x `)
` #g
` )#
`f `y
`y `)
œ (r sin ))
#
œ (r sin )) Š `` xf#
œ (r sin )) Š `` x)
`y
`) ‹
`x
`)
`f
`x
(r cos ))
` #f ` y
` y` x ` ) ‹
`f
`y
`f
`x
(r cos ))
(r cos )) (r cos )) Š `` x)
#
(r cos )) Š ``x`fy
`y
`) ‹
`x
`)
` #f ` y
` y# ` ) ‹
(r sin ))
`f
`y
(r sin ))
œ (r sin ) r cos ))(r sin ) r cos )) (r cos ) r sin )) œ (2)(2) (0 2) œ 4 2 œ 2 at
(rß )) œ ˆ2ß 1# ‰ .
97. (y z)# (z x)# œ 16 Ê ™ f œ 2(z x)i 2(y z)j 2(y 2z x)k ; if the normal line is parallel to the
yz-plane, then x is constant Ê `` xf œ 0 Ê 2(z x) œ 0 Ê z œ x Ê (y z)# (z z)# œ 16 Ê y z œ „ 4.
Let x œ t Ê z œ t Ê y œ t „ 4. Therefore the points are (tß t „ 4ß t), t a real number.
98. Let f(xß yß z) œ xy yz zx x z# œ 0. If the tangent plane is to be parallel to the xy-plane, then ™ f is
perpendicular to the xy-plane Ê ™ f † i œ 0 and ™ f † j œ 0. Now ™ f œ (y z 1)i (x z)j (y x 2z)k
so that ™ f † i œ y z 1 œ 0 Ê y z œ 1 Ê y œ 1 z, and ™ f † j œ x z œ 0 Ê x œ z. Then
z(1 z) (" z)z z(z) (z) z# œ 0 Ê z 2z# œ 0 Ê z œ "# or z œ 0. Now z œ "# Ê x œ "# and y œ
Ê ˆ "# ß "# ß "# ‰ is one desired point; z œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1ß 0) is a second desired point.
99. ™ f œ -(xi yj zk) Ê
`f
`x
œ -x Ê f(xß yß z) œ
"
#
-x# g(yß z) for some function g Ê -y œ
`f
`y
œ
`g
`y
-y# h(z) for some function h Ê -z œ `` zf œ `` gz œ hw (z) Ê h(z) œ #" -z# C for some arbitrary
constant C Ê g(yß z) œ "# -y# ˆ "# -z# C‰ Ê f(xß yß z) œ "# -x# "# -y# "# -z# C Ê f(0ß 0ß a) œ "# -a# C
Ê g(yß z) œ
"
#
and f(0ß 0ß a) œ
‰
100. ˆ df
ds u (0 0 0)
ß
ß ß
"
#
-(a)# C Ê f(0ß 0ß a) œ f(0ß 0ß a) for any constant a, as claimed.
œ lim
sÄ0
œ lim
sÄ0
f(0 su" ß 0 su# ß 0 su$ )f(0ß 0ß 0)
s
És# u#" s# u## s# u#$ 0
s
,s0
,s0
"
#
936
Chapter 14 Partial Derivatives
sÉu#" u## u#$
œ lim
œ lim kuk œ 1;
s
sÄ0
sÄ0
y
however, ™ f œ Èx# xy# z# i Èx# y# z# j Èx# zy# z# k fails to exist at the origin (0ß 0ß 0)
101. Let f(xß yß z) œ xy z 2 Ê ™ f œ yi xj k . At (1ß 1ß 1), we have ™ f œ i j k Ê the normal line is
x œ 1 t, y œ 1 t, z œ 1 t, so at t œ 1 Ê x œ 0, y œ 0, z œ 0 and the normal line passes through the
origin.
102. (b) f(xß yß z) œ x# y# z# œ 4
Ê ™ f œ 2xi 2yj 2zk Ê at (2ß 3ß 3)
the gradient is ™ f œ 4i 6j 6k which is
normal to the surface
(c) Tangent plane: 4x 6y 6z œ 8 or
2x 3y 3z œ 4
Normal line: x œ 2 4t, y œ 3 6t, z œ 3 6t
CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES
fx (0ß h) fx (0ß 0)
h
1. By definition, fxy (!ß 0) œ lim
hÄ0
so we need to calculate the first partial derivatives in the
numerator. For (xß y) Á (0ß 0) we calculate fx (xß y) by applying the differentiation rules to the formula for
fy (xß y) œ
hÄ0
`w
`x
#
x xy
x# y#
00
h
œ lim
2.
$
x# y y$
x# y#
ax# y# b (2x) ax# y# b (2x)
ax # y # b #
4x# y$
Ê fx (0ß h)
a x # y # b#
f(0ß0)
For (xß y) œ (0ß 0) we apply the definition: fx (!ß 0) œ lim f(hß 0)
œ lim 0 h 0 œ
h
hÄ0
hÄ0
fy (hß 0) fy (!ß 0)
h 0
fxy (0ß 0) œ lim
œ 1. Similarly, fyx (0ß 0) œ lim
, so for (xß y) Á
h
h
hÄ0
hÄ0
f(xß y): fx (xß y) œ
(xy)
$ #
4x y
ax # y # b #
Ê fy (hß 0) œ
$
x# y y$
x# y#
œ
$
œ hh# œ h.
0. Then by definition
(0ß 0) we have
œ h; for (xß y) œ (0ß 0) we obtain fy (0ß 0) œ lim
h
h#
h0
h
œ 0. Then by definition fyx (0ß 0) œ lim
hÄ0
œ 1 ex cos y Ê w œ x ex cos y g(y);
`w
`y
hÄ0
f(0ß h) f(!ß 0)
h
œ 1. Note that fxy (0ß 0) Á fyx (0ß 0) in this case.
œ ex sin y gw (y) œ 2y ex sin y Ê gw (y) œ 2y
Ê g(y) œ y# C; w œ ln 2 when x œ ln 2 and y œ 0 Ê ln 2 œ ln 2 eln 2 cos 0 0# C Ê 0 œ 2 C
Ê C œ 2. Thus, w œ x ex cos y g(y) œ x ex cos y y# 2.
3. Substitution of u u(x) and v œ v(x) in g(uß v) gives g(u(x)ß v(x)) which is a function of the independent
variable x. Then, g(uß v) œ 'u f(t) dt Ê
v
œ Š ``u
#
#
fzz œ Š ddr#f ‹ ˆ ``zr ‰
` #r
` y#
#
œ
` g du
` u dx
'uv f(t) dt‹ dudx Š ``v 'uv f(t) dt‹ dvdx
` g dv
` v dx
œ Š ``u
#
df ` # r
dr ` x#
. Similarly, fyy œ Š ddr#f ‹ Š ``yr ‹
Ê
` #r
` x#
y # z#
3
ˆÈ x # y # z # ‰
v
` '
dv
du
dv
dv
du
'vu f(t) dt‹ du
dx Š ` v u f(t) dt‹ dx œ f(u(x)) dx f(v(x)) dx œ f(v(x)) dx f(u(x)) dx
4. Applying the chain rules, fx œ
Ê
dg
dx
œ
df ` # r
dr ` z#
x # z#
3
ˆÈ x # y # z # ‰
#
df ` r
dr ` x
#
Ê fxx œ Š ddr#f ‹ ˆ ``xr ‰
. Moreover,
; and
`r
`z
œ
‰
Ê Š ddr#f ‹ Š x# xy# z# ‹ ˆ df
dr Œ ˆÈ
`r
`x
œ
x
È x # y # z#
z
È x # y # z#
y # z#
x # y # z# ‰
Ê
` #r
` z#
d# f
3
œ
œ
#
x# y#
3
ˆÈ x # y # z# ‰
y#
;
`r
`y
œ
#
y
È x # y # z#
. Next, fxx fyy fzz œ 0
x # z#
Š dr# ‹ Š x# y# z# ‹ ˆ dr ‰ Œ ˆÈx# y# z# ‰3
df
df ` # r
dr ` y#
and
Chapter 14 Practice Exercises
#
Ê
df
dr
d
dr
x# y#
#
‰
Š ddr#f ‹ Š x# zy# z# ‹ ˆ df
dr Œ ˆÈ
x # y # z# ‰
(f w ) œ ˆ 2r ‰ f w , where f w œ
œ Cr# Ê f(r) œ Cr b œ
Ê
df
dr
3
df
f
d# f
dr#
œ0 Ê
Š Èx# 2y# z# ‹
df
dr
d# f
dr#
œ0 Ê
2 df
r dr
937
œ0
œ 2 rdr Ê ln f w œ 2 ln r ln C Ê f w œ Cr# , or
w
w
b for some constants a and b (setting a œ C)
a
r
5. (a) Let u œ tx, v œ ty, and w œ f(uß v) œ f(u(tß x)ß v(tß y)) œ f(txß ty) œ tn f(xß y), where t, x, and y are
independent variables. Then ntnc1 f(xß y) œ ``wt œ ``wu ``ut ``wv ``vt œ x ``wu y ``wv . Now,
`w
`w `u
`w `v
`w
`w
ˆ `w ‰
ˆ `w ‰
ˆ " ‰ ˆ ``wx ‰ . Likewise,
` x œ ` u ` x ` v ` x œ ` u (t) ` v (0) œ t ` u Ê ` u œ t
`w
`y
œ
`w `u
`u `y
`w `v
`v `y
ntnc1 f(xß y) œ x
`w
`x
Ê
œ
`f
`x
`w
`u
`w
`v
y
`w
`y
and
œ ˆ ``wu ‰ (0) ˆ ``wv ‰ (t) Ê
œ
n(n 1)t
`f
`x
Ê nf(xß y) œ x
Also from part (a),
œ
`
`y
œ t#
ˆt
`w ‰
`v
` #w
` v` u
œt
` w `u
` u# ` t
#
` w
` x#
` #w
` x#
`w
`v .
` w `v
` v` u ` t
y
x
ˆ ``wx ‰ œ
t
` #w ` v
` v# ` y
œ
` #w
` u#
`
`x
` #w
` y#
y
ˆt
`w ‰
`u
` #w
` v#
œ t#
, ˆ t"# ‰
`f
`x
y
`f
`y
, as claimed.
Differentiating with respect to t again we obtain
#
`
`x
œ
` #w ` u
` u` v ` y
Ê ˆ t"# ‰
`w
`u
#
f(xß y) œ x
œ ˆ "t ‰ Š ``wy ‹ . Therefore,
œ ˆ xt ‰ ˆ ``wx ‰ ˆ yt ‰ Š ``wy ‹. When t œ 1, u œ x, v œ y, and w œ f(xß y)
(b) From part (a), ntnc1 f(xß y) œ x
nc2
`w
`v
œ
` #w ` u
` u` v ` t
œt
, and
` #w
` v#
` #w ` v
` v# ` t
y
#
` w `u
` u# ` x
t
` #w
` y` x
`
`y
œ
, and ˆ t"# ‰
#
` w `v
` v` u ` x
ˆ ``wx ‰ œ
` #w
` y` x
œ
` #w
` u#
œ x#
2xy
#
#
œ t#
` w
` u#
,
`
`y
`w ‰
`u
œt
ˆt
` #w
` u` v
` w
` y#
œ
y#
`
`y
` #w
` v#
.
Š ``wy ‹
` #w ` u
` u# ` y
t
` #w ` v
` v` u ` y
` #w
` v` u
‰ Š ``y`wx ‹ Š yt# ‹ Š `` yw# ‹ for t Á 0. When t œ 1, w œ f(xß y) and
Ê n(n 1)tnc2 f(xß y) œ Š xt# ‹ Š `` xw# ‹ ˆ 2xy
t#
#
#
#
#
#
#
#
#
we have n(n 1)f(xß y) œ x# Š `` xf# ‹ 2xy Š ``x`fy ‹ y# Š `` yf# ‹ as claimed.
6. (a) lim
rÄ0
sin 6r
6r
œ lim
tÄ0
(b) fr (0ß 0) œ lim
hÄ0
œ lim
hÄ0
(c) f) (rß )) œ lim
hÄ0
sin t
t
œ 1, where t œ 6r
f(0 hß 0) f(0ß 0)
h
36 sin 6h
12
œ lim
hÄ0
œ0
f(rß ) h) f(rß ))
h
ˆ sin6h6h ‰ 1
h
œ lim
hÄ0
sin 6h 6h
6h#
6 cos 6h 6
12h
i
y
È x # y # z#
hÄ0
(applying l'Hopital's
rule twice)
s
œ lim
hÄ0
ˆ sin6r6r ‰ ˆ sin6r6r ‰
h
œ lim
0
hÄ0 h
7. (a) r œ xi yj zk Ê r œ krk œ Èx# y# z# and ™ r œ
œ
œ lim
œ0
x
È x # y # z#
j
z
È x # y # z#
k
r
r
(b) rn œ ˆÈx# y# z# ‰
n
Ê ™ arn b œ nx ax# y# z# b
œ nrnc2 r
(c) Let n œ 2 in part (b). Then
"
#
ÐnÎ2Ñ1
i ny ax# y# z# b
ÐnÎ2Ñ1
j nz ax# y# z# b
™ ar# b œ r Ê ™ ˆ "# r# ‰ œ r Ê
r#
#
œ
"
#
ÐnÎ2Ñ1
k
ax# y# z# b is the function.
(d) dr œ dxi dyj dzk Ê r † dr œ x dx y dy z dz, and dr œ rx dx ry dy rz dz œ
x
r
dx
y
r
dy
z
r
dz
Ê r dr œ x dx y dy z dz œ r † dr
(e) A œ ai bj ck Ê A † r œ ax by cz Ê ™ (A † r) œ ai bj ck œ A
8. f(g(t)ß h(t)) œ c Ê 0 œ
df
dt
œ
` f dx
` x dt
` f dy
` y dt
œ Š `` xf i
`f
`y
j‹ † Š dx
dt i
dy
dt
j‹ , where
dx
dt
i
dy
dt
j is the tangent vector
Ê ™ f is orthogonal to the tangent vector
9. f(xß yß z) œ xz# yz cos xy 1 Ê ™ f œ az# y sin xyb i (z x sin xy)j (2xz y)k Ê ™ f(0ß 0ß 1) œ i j
Ê the tangent plane is x y œ 0; r œ (ln t)i (t ln t)j tk Ê rw œ ˆ "t ‰ i (ln t 1)j k ; x œ y œ 0, z œ 1
Ê t œ 1 Ê rw (1) œ i j k . Since (i j k) † (i j) œ rw (1) † ™ f œ 0, r is parallel to the plane, and
r(1) œ 0i 0j k Ê r is contained in the plane.
938
Chapter 14 Partial Derivatives
10. Let f(xß yß z) œ x$ y$ z$ xyz Ê ™ f œ a3x# yzb i a3y# xzb j a3z# xyb k Ê ™ f(0ß 1ß 1) œ i 3j 3k
$
Ê the tangent plane is x 3y 3z œ 0; r œ Š t4 2‹ i ˆ 4t 3‰ j (cos (t 2)) k
#
Ê rw œ Š 3t4 ‹ i ˆ t4# ‰ j (sin (t 2)) k ; x œ 0, y œ 1, z œ 1 Ê t œ 2 Ê rw (2) œ 3i j . Since
rw (2) † ™ f œ 0 Ê r is parallel to the plane, and r(2) œ i k Ê r is contained in the plane.
11.
`z
`x
œ 3x# 9y œ 0 and
`z
`y
œ 3y# 9x œ 0 Ê y œ
"
3
#
x# and 3 ˆ "3 x# ‰ 9x œ 0 Ê
"
3
x% 9x œ 0
Ê x ax$ 27b œ 0 Ê x œ 0 or x œ 3. Now x œ 0 Ê y œ 0 or (!ß 0) and x œ 3 Ê y œ 3 or (3ß 3). Next
` #z
` x#
œ 6x,
` #z
` y#
œ 6y, and
and for (3ß 3),
` #z ` #z
` x# ` y#
` #z
` x` y
#
` #z ` #z
` x# ` y#
œ 9. For (!ß 0),
#
Š ``x`zy ‹ œ 243 0 and
` #z
` x#
#
#
Š ``x`zy ‹ œ 81 Ê no extremum (a saddle point),
œ 18 0 Ê a local minimum.
12. f(xß y) œ 6xyeÐ2x3yÑ Ê fx (xß y) œ 6y(1 2x)eÐ2x3yÑ œ 0 and fy (xß y) œ 6x(1 3y)eÐ2x3yÑ œ 0 Ê x œ 0 and
y œ 0, or x œ "# and y œ 3" . The value f(0ß 0) œ 0 is on the boundary, and f ˆ #" ß 3" ‰ œ e"2 . On the positive y-axis,
f(0ß y) œ 0, and on the positive x-axis, f(xß 0) œ 0. As x Ä _ or y Ä _ we see that f(xß y) Ä 0. Thus the
absolute maximum of f in the closed first quadrant is e"2 at the point ˆ #" ß 3" ‰ .
13. Let f(xß yß z) œ
P! (x! ß y! ß y! ) is
y#
x#
a# b#
!‰
ˆ 2x
a# x
z#
c# 1
!‰
ˆ 2y
b# y
Ê ™fœ
ˆ 2zc#! ‰ z
2y
2x
a# i b# j
#
2y#!
!
œ 2x
a# b#
#
2z
c# k Ê an equation of the plane tangent
2z#!
ˆ x! ‰
ˆ y! ‰
ˆ z! ‰
c# œ 2 or a# x b# y c# z œ 1.
#
at the point
#
The intercepts of the plane are Š xa! ß 0ß 0‹ , Š0ß by! ß 0‹ and Š!ß !ß cz! ‹ . The volume of the tetrahedron formed
#
#
#
by the plane and the coordinate planes is V œ ˆ "3 ‰ ˆ #" ‰ Š xa! ‹ Š by! ‹ Š cz! ‹ Ê we need to maximize
V(xß yß z) œ
’
(abc)#
6
(abc)#
"
6 “ Š x# yz ‹
y#
x#
z#
a# b# c# œ 1.
(abc)#
"
2z
6 “ Š xyz# ‹ œ c# -.
(xyz)" subject to the constraint f(xß yß z) œ
œ
2x
a#
-, ’
(abc)#
"
6 “ Š xy# z ‹
œ
2y
b#
-, and ’
Thus,
Multiply the first equation
by a# yz, the second by b# xz, and the third by c# xy. Then equate the first and second Ê a# y# œ b# x#
Ê y œ ba x, x 0; equate the first and third Ê a# z# œ c# x# Ê z œ ca x, x 0; substitute into f(xß yß z) œ 0
Ê xœ
a
È3
Ê yœ
b
È3
Ê zœ
c
È3
Ê Vœ
È3
#
abc.
14. 2(x u) œ -, 2(y v) œ -, 2(x u) œ ., and 2(y v) œ 2.v Ê x u œ v y, x u œ .# , and
y v œ .v Ê x u œ .v œ .# Ê v œ
"
#
or . œ 0.
CASE 1: . œ 0 Ê x œ u, y œ v, and - œ 0; then y œ x 1 Ê v œ u 1 and v# œ u Ê v œ v# 1
1 „ È1 4
Ê no real solution.
#
"
"
"
#
v œ # and u œ v Ê u œ 4 ; x 4 œ #" y and y œ x 1 Ê x 4" œ
#
#
Ê x œ "8 Ê y œ 78 . Then f ˆ 8" ß 87 ß 4" ß #" ‰ œ ˆ 8" 4" ‰ ˆ 87 #" ‰
is 38 È2. (Notice that f has no maximum value.)
Ê v# v 1 œ 0 Ê v œ
CASE 2:
15. Let (x! ß y! ) be any point in R. We must show
lim
Ðh ß k Ñ Ä Ð 0 ß 0 Ñ
lim
Ðx ß y Ñ Ä Ð x ! ß y ! Ñ
x
"
#
#
Ê 2x œ 4"
œ 2 ˆ 83 ‰ Ê the minimum distance
f(xß y) œ f(x! ß y! ) or, equivalently that
kf(x! hß y! k) f(x! ß y! )k œ 0. Consider f(x! hß y! k) f(x! ß y! )
œ [f(x! hß y! k) f(x! ß y! k)] [f(x! ß y! k) f(x! ß y! )]. Let F(x) œ f(xß y! k) and apply the Mean Value
Theorem: there exists 0 with x! 0 x! h such that Fw (0 )h œ F(x! h) F(x! ) Ê hfx (0ß y! k)
œ f(x! hß y! k) f(x! ß y! k). Similarly, k fy (x! ß () œ f(x! ß y! k) f(x! ß y! ) for some ( with
y! ( y! k. Then kf(x! hß y! k) f(x! ß y! )k Ÿ khfx (0ß y! k)k kkfy (x! ß ()k . If M, N are positive real
numbers such that kfx k Ÿ M and kfy k Ÿ N for all (xß y) in the xy-plane, then kf(x! hß y! k) f(x! ß y! )k
Ÿ M khk N kkk . As (hß k) Ä 0, kf(x! hß y! k) f(x! ß y! )k Ä 0 Ê
lim
kf(x! hß y! k) f(x! ß y! )k
Ðh ß k Ñ Ä Ð 0 ß 0 Ñ
Chapter 14 Practice Exercises
939
œ 0 Ê f is continuous at (x! ß y! ).
16. At extreme values, ™ f and v œ
dr
dt
df
dt
are orthogonal because
œ ™f†
dr
dt
œ 0 by the First Derivative Theorem for
Local Extreme Values.
17.
`f
`x
œ 0 Ê f(xß y) œ h(y) is a function of y only. Also,
Moreover,
`f
`y
œ
`g
`x
`g
`y
œ
`f
`x
œ 0 Ê g(xß y) œ k(x) is a function of x only.
Ê hw (y) œ kw (x) for all x and y. This can happen only if hw (y) œ kw (x) œ c is a constant.
Integration gives h(y) œ cy c" and k(x) œ cx c# , where c" and c# are constants. Therefore f(xß y) œ cy c"
and g(xß y) œ cx c# . Then f(1ß 2) œ g(1ß 2) œ 5 Ê 5 œ 2c c" œ c c# , and f(0ß 0) œ 4 Ê c" œ 4 Ê c œ
Ê c# œ
9
#
. Thus, f(xß y) œ
"
#
y 4 and g(xß y) œ
"
#
"
#
x 9# .
18. Let g(xß y) œ Du f(xß y) œ fx (xß y)a fy (xß y)b. Then Du g(xß y) œ gx (xß y)a gy (xß y)b
œ fxx (xß y)a# fyx (xß y)ab fxy (xß y)ba fyy (xß y)b# œ fxx (xß y)a# 2fxy (xß y)ab fyy (xß y)b# .
19. Since the particle is heat-seeking, at each point (xß y) it moves in the direction of maximal temperature
increase, that is in the direction of ™ T(xß y) œ aec2y sin xb i a2ec2y cos xb j . Since ™ T(xß y) is parallel to
2ec2y cos x
ec2y sin x œ
È
œ 2 ln #2
the particle's velocity vector, it is tangent to the path y œ f(x) of the particle Ê f w (x) œ
2 cot x.
Integration gives f(x) œ 2 ln ksin xk C and f ˆ 14 ‰ œ 0 Ê 0 œ 2 ln ¸sin 14 ¸ C Ê C
œ ln Š È22 ‹
#
œ ln 2. Therefore, the path of the particle is the graph of y œ 2 ln ksin xk ln 2.
20. The line of travel is x œ t, y œ t, z œ 30 5t, and the bullet hits the surface z œ 2x# 3y# when
30 5t œ 2t# 3t# Ê t# t 6 œ 0 Ê (t 3)(t 2) œ 0 Ê t œ 2 (since t 0). Thus the bullet hits the
surface at the point (2ß 2ß 20). Now, the vector 4xi 6yj k is normal to the surface at any (xß yß z), so that
n œ 8i 12j k is normal to the surface at (2ß 2ß 20). If v œ i j 5k , then the velocity of the particle
†25 ‰
after the ricochet is w œ v 2 projn v œ v Š 2knvk†#n ‹ n œ v ˆ 2209
n œ (i j 5k) ˆ 400
209 i
œ 191
209 i
391
209
j
995
209
600
209
j
50
209
k‰
k.
21. (a) k is a vector normal to z œ 10 x# y# at the point (!ß 0ß 10). So directions tangential to S at (!ß 0ß 10) will
be unit vectors u œ ai bj . Also, ™ T(xß yß z) œ (2xy 4) i ax# 2yz 14b j ay# 1b k
Ê ™ T(!ß 0ß 10) œ 4i 14j k . We seek the unit vector u œ ai bj such that Du T(0ß 0ß 10)
œ (4i 14j k) † (ai bj) œ (4i 14j) † (ai bj) is a maximum. The maximum will occur when ai bj
has the same direction as 4i 14j , or u œ È"53 (2i 7j).
(b) A vector normal to S at (1ß 1ß 8) is n œ 2i 2j k . Now, ™ T(1ß 1ß 8) œ 6i 31j 2k and we seek the unit
vector u such that Du T(1ß 1ß 8) œ ™ T † u has its largest value. Now write ™ T œ v w , where v is parallel
to ™ T and w is orthogonal to ™ T. Then Du T œ ™ T † u œ (v w) † u œ v † u w † u œ w † u. Thus
Du T(1ß 1ß 8) is a maximum when u has the same direction as w . Now, w œ ™ T Š ™knTk#†n ‹ n
62 2 ‰
œ (6i 31j 2k) ˆ 124
(2i 2j k) œ ˆ6
41
œ 98
9 i
127
9
j
58
9
k Ê uœ
w
kwk
152 ‰
i
9
ˆ31
152 ‰
j
9
ˆ2
76 ‰
9 k
"
œ È29,097
(98i 127j 58k).
22. Suppose the surface (boundary) of the mineral deposit is the graph of z œ f(xß y) (where the z-axis points up
into the air). Then `` xf i `` yf j k is an outer normal to the mineral deposit at (xß y) and `` xf i `` yf j points in
the direction of steepest ascent of the mineral deposit. This is in the direction of the vector
`f
`x
i
`f
`y
j at (0ß 0)
(the location of the 1st borehole) that the geologists should drill their fourth borehole. To approximate this
vector we use the fact that (0ß 0ß 1000), (0ß 100ß 950), and (100ß !ß 1025) lie on the graph of z œ f(xß y).
The plane containing these three points is a good approximation to the tangent plane to z œ f(xß y) at the point
940
Chapter 14 Partial Derivatives
â
â
j
k â
â i
â
â
(0ß 0ß 0). A normal to this plane is â 0 "00 50 â œ 2500i 5000j 10,000k, or i 2j 4k. So at
â
â
â "00 0 25 â
(!ß 0) the vector
`f
`x
in the direction of
i
"
È5
`f
`y
j is approximately i 2j . Thus the geologists should drill their fourth borehole
(i 2j) from the first borehole.
23. w œ ert sin 1x Ê wt œ rert sin 1x and wx œ 1ert cos 1x Ê wxx œ 1# ert sin 1x; wxx œ
positive constant determined by the material of the rod Ê 1# ert sin 1x œ
"
c#
"
c#
wt , where c# is the
arert sin 1xb
# #
Ê ar c# 1# b ert sin 1x œ 0 Ê r œ c# 1# Ê w œ ec 1 t sin 1x
24. w œ ert sin kx Ê wt œ rert sin kx and wx œ kert cos kx Ê wxx œ k# ert sin kx; wxx œ
Ê k# ert sin kx œ
"
c#
"
c#
wt
# #
arert sin kxb Ê ar c# k# b ert sin kx œ 0 Ê r œ c# k# Ê w œ ec k t sin kx.
# #
Now, w(Lß t) œ 0 Ê ec k t sin kL œ 0 Ê kL œ n1 for n an integer Ê k œ
# # #
#
As t Ä _, w Ä 0 since ¸sin ˆ nL1 x‰¸ Ÿ 1 and ec n 1 tÎL Ä 0.
n1
L
# # #
#
Ê w œ ec n 1 tÎL sin ˆ nL1 x‰ .
CHAPTER 15 MULTIPLE INTEGRALS
15.1 DOUBLE INTEGRALS
1.
'03 '02 a4 y# b dy dx œ '03 ’4y y3 “ # dx œ 163 '03 dx œ 16
2.
'03 'c02 a(x# y 2xyb dy dx œ '03 ’ x 2y
xy# “
œ '0 a4x 2x# b dx œ ’2x#
œ0
$
!
# #
3
3.
$
2x$
3 “!
'c01 'c11 (x y 1) dx dy œ 'c01 ’ x2
#
4.
5.
#
yx x“
œ 'c1 (2y 2) dy œ cy# 2yd " œ 1
0
!
dx
"
"
dy
!
'121 '01 (sin x cos y) dx dy œ '121 c( cos x) (cos y)xd 1! dy
21
œ '1 (1 cos y 2) dy œ c1 sin y 2yd #1
1 œ 21
'01 '0x (x sin y) dy dx œ '01 c x cos yd !x dx
1
1
œ '0 (x x cos x) dx œ ’ x2 (cos x x sin x)“
#
œ
6.
1#
#
!
2
'01 '0sin x y dy dx œ '01 ’ y2 “ sin x dx œ '01 "# sin# x dx
#
œ
"
4
!
'0 (1 cos 2x) dx œ "4 x 2" sin 2x‘ !1 œ 14
1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
942
7.
Chapter 15 Multiple Integrals
'1ln 8 '0ln yexby dx dy œ '1ln 8 cexbyd !ln y dy œ '1ln 8 ayey eyb dy
œ c(y 1)ey ey d 1ln 8 œ 8(ln 8 1) 8 e
œ 8 ln 8 16 e
'12 'yy
#
8.
dx dy œ '1 ay# yb dy œ ’ y3
2
$
œ ˆ 83 2‰ ˆ "3 #" ‰ œ
7
3
œ
3
#
5
6
'01 '0y 3y$ exy dx dy œ '01 c3y# exy d 0y
#
9.
#
y#
# “"
#
dy
œ '0 Š3y# ey 3y# ‹ dy œ ’ey y$ “ œ e 2
1
$
"
$
!
10.
Èx
'14 '0
œ
3
#
3
#
eyÎÈx dy dx œ
'14 32 Èx eyÎÈx ‘ 0Èx dx
%
(e 1) '1 Èx dx œ 32 (e 1) ˆ 23 ‰ x$Î# ‘ " œ 7(e 1)
4
11.
'12 'x2x
x
y
dy dx œ '1 cx ln yd x2x dx œ (ln 2) '1 x dx œ
12.
'12 '12
1
xy
13.
'01 '01cx ax# y# b dy dx œ '01 ’x# y y3 “ "x dx œ '01 ’x# (1 x) (13x) “ dx œ '01 ’x# x$ (13x) “ dx
2
dy dx œ '1
2
2
"
x
3
#
ln 2
(ln 2 ln 1) dx œ (ln 2) '1 dx œ (ln 2)#
2
$
$
$
0
$
œ ’ x3
x%
4
"
(1x)%
“
1#
!
œ ˆ 3"
"
4
0‰ ˆ0 0
" ‰
1#
œ
"
6
14.
'01 '01 y cos xy dx dy œ '01 csin xyd 1! dy œ '01 sin 1y dy œ 1" cos 1y‘ "! œ 1" (1 1) œ 12
15.
'01 '01cu ˆv Èu‰ dv du œ '01 ’ v2
#
œ '0 Š "# u
1
u#
#
vÈ u “
"u
0
u"Î# u$Î# ‹ du œ ’ 2u
du œ '0 ’ 12u# u Èu(1 u)“ du
1
u#
#
u$
6
#
"
32 u$Î# 52 u&Î# “ œ
!
"
#
"
#
"
6
2
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
5
œ "#
2
5
"
œ 10
Section 15.1 Double Integrals
16.
'12 '0ln t es ln t ds dt œ '12 ces ln td 0ln t dt œ '12 (t ln t ln t) dt œ ’ t2
#
œ (2 ln 2 1 2 ln 2 2) ˆ "4 1‰ œ
17.
"
4
ln t
t#
4
t ln t t“
#
"
'c02 'vcv 2 dp dv œ 2'c02 cpd vv dv œ 2'c02 2v dv
œ 2 cv# d c2 œ 8
0
18.
È1cs
'01 '0
#
È1cs
8t dt ds œ '0 c4t# d 0
1
œ '0 4 a1 s# b ds œ 4 ’s
1
19.
#
ds
"
s$
3 “!
œ
8
3
'c11ÎÎ33 '0sec t 3 cos t du dt œ '11ÎÎ33 c(3 cos t)ud 0sec t
1Î3
œ 'c1Î3 3 dt œ 21
20.
'03 '142u 4v 2u dv du œ '03 2uv4 ‘ 142u du
3
$
œ '0 (3 #u) du œ c3u u# d ! œ !
#
21.
'24 '0Ð4y)Î2 dx dy
22.
'02 '0x2 dy dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
943
944
Chapter 15 Multiple Integrals
23.
'01 'xx dy dx
24.
'01 '1cy1cydx dy
25.
'1e 'ln1ydx dy
26.
'12 '0ln x dy dx
27.
'09 '0
28.
'04 '0
#
È
1
2
È9cy
È4cx
16x dx dy
y dy dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.1 Double Integrals
È1cx
29.
'c11 '0
30.
'c22 '0
31.
'01 'x1 siny y dy dx œ '01 '0y siny y dx dy œ '01 sin y dy œ 2
32.
#
È4cy
3y dy dx
#
6x dx dy
'02 'x2 2y# sin xy dy dx œ '02 '0y2y# sin xy dx dy
2
2
œ '0 c2y cos xyd 0y dy œ '0 a2y cos y# 2yb dy
#
œ c sin y# y# d ! œ 4 sin 4
33.
'01 'y1 x# exy dx dy œ '01 '0x x# exy dy dx œ '01 cxexyd 0x dx
œ '0 axex xb dx œ ’ "2 ex
1
È4cy
'02 '04cx 4xey dy dx œ '04 '0
#
34.
#
#
2y
œ '0 ’ #x(4ey) “
4
35.
'02
# 2y
Èln 3 Èln 3
'y/2
Èln 3
œ '0
È4cy
0
dy œ '0
4 2y
e
Èln 3
ex dx dy œ '0
#
#
#
#
Èln 3
2xex dx œ cex d 0
"
x#
# “!
xe2y
4 y
œ
e 2
#
dx dy
2y
%
dy œ ’ e4 “ œ
!
'02x ex
#
e) "
4
dy dx
œ eln 3 1 œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
945
946
36.
Chapter 15 Multiple Integrals
'03 'È1xÎ3 ey
$
dy dx œ '0
1
'03y
#
$
ey dx dy
œ '0 3y# ey dy œ cey d ! œ e 1
1
37.
$
$
"
'01Î16 'y1Î2 cos a161x& b dx dy œ '01Î2 '0x
%
"Î%
cos a161x& b dy dx
161x b
œ '0 x% cos a161x& b dx œ ’ sin a80
“
1
1Î2
38.
'08 'È2x
œ '0
39.
"
y % 1
$
2
&
$
dy dx œ '0
2
dy œ
y
y % 1
"
4
"Î#
!
œ
"
801
'0y y "1 dx dy
$
%
#
cln ay% 1bd ! œ
ln 17
4
' ' ay 2x# b dA
R
xb1
œ 'c1 'cxc1 ay 2x# b dy dx '0
0
1
'x1cc1x ay 2x# b dy dx
x"
1x
œ 'c1 "2 y# 2x# y‘ x1 dx '0 2" y# 2x# y‘ x1 dx
0
1
œ 'c1 "# (x 1)# 2x# (x 1) "# (x 1)# 2x# (x 1)‘dx
0
'0 "# (1 x)# 2x# (1 x) "# (x 1)# 2x# (x 1)‘ dx
1
œ 4 'c1 ax$ x# b dx 4 '0 ax$ x# b dx
0
1
%
œ 4 ’ x4
40.
0
x$
3 “ c1
"
x$
3 “!
%
4 ’ x4
%
œ 4 ’ (41)
(1)$
3 “
3
4 ˆ 4" 3" ‰ œ 8 ˆ 12
4 ‰
12
8
œ 12
œ 23
2x
' ' xy dA œ ' ' xy dy dx ' ' xy dy dx
0
x
2Î3 x
2Î3
R
2x
2Î3
1
2x
2x
œ '0 "2 xy# ‘ x dx '2Î3 2" xy# ‘ x dx
1
œ '0 ˆ2x$ "# x$ ‰ dx '2Î3 "# x(2 x)# "# x$ ‘ dx
2Î3
1
œ '0
2Î3
3
#
x$ dx '2Î3 a2x x# b dx
1
"
2Î3
2‰
8 ‰‘
‰ ˆ
4 ˆ 2 ‰ ˆ 27
œ 38 x% ‘ 0 x# 23 x$ ‘ #Î$ œ ˆ 38 ‰ ˆ 16
œ
81 1 3 9 3
41. V œ '0
1
'x2cx ax# y# b dy dx œ '01 ’x# y y3 “ 2cx dx œ '01 ’2x# 7x3
œ ˆ 23
$
x
7
12
2cx#
42. V œ 'c2 'x
1
œ ˆ 23
"
5
" ‰
12
ˆ0 0
16 ‰
12
œ
x# dy dx œ 'c2 cx# yd x
4" ‰ ˆ 16
3
32
5
16 ‰
4
(2x)$
3 “
27
81
ˆ 36
81
16 ‰
81
$
dx œ ’ 2x3
7x%
12
4
3
2cx#
1
$
6
81
dx œ 'c2 a2x# x% x$ b dx œ 23 x$ 15 x& 14 x% ‘ #
40
œ ˆ 60
1
12
60
"
15 ‰
60
ˆ 320
60
384
60
240 ‰
60
œ
189
60
œ
63
20
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
13
81
"
(2x)%
12 “ !
Section 15.1 Double Integrals
4cx#
43. V œ 'c4 '3x
1
4cx
(x 4) dy dx œ 'c4 cxy 4yd 3x
dx œ 'c4 cx a4 x# b 4 a4 x# b 3x# 12xd dx
1
1
#
œ 'c4 ax$ 7x# 8x 16b dx œ 14 x% 73 x$ 4x# 16x‘ % œ ˆ "4
1
œ
"
"
4
157
3
44. V œ '0
2
œ
‰
12‰ ˆ 64
3 64
625
12
È4cx
'0
7
3
#
È4cx
(3 y) dy dx œ '0 ’3y
2
œ ’ 32 xÈ4 x# 6 sin" ˆ x# ‰ 2x
y#
2 “0
#
x$
6 “!
dx œ '0 ’3È4 x# Š 4#x ‹“ dx
#
2
œ 6 ˆ 1# ‰ 4
#
8
6
œ 31
16
6
œ
9 1 8
3
45. V œ '0
'03 a4 y# b dx dy œ '02 c4x y# xd $! dy œ '02 a12 3y# b dy œ c12y y$ d #! œ 24 8 œ 16
46. V œ '0
'04cx
2
2
#
2
œ 8x 43 x$
47. V œ '0
2
4cx#
a4 x# yb dy dx œ '0 ’a4 x# b y
"
10
#
x& ‘ ! œ 16
32
3
œ
32
10
y#
2 “!
48032096
30
œ
#
(2x)%
4 “!
xb1
0
a4 x# b dx œ '0 Š8 4x#
2
#
x%
#‹
1
1Îx
2
50. V œ 4 '0
1Î3
'x1cc1x (3 3x) dy dx œ 6 'c01 a1 x# b dx 6 '01 (1 x)# dx œ 4 2 œ 6
2
2
2
$
0
1Î$
'1_ 'ec1 x"y dy dx œ '1_ ’ lnx y “ "
52.
'c11 'c1/1/È11ccxx
$
È
ˆ1 x" ‰‘ œ 2 '1 ˆ1 x" ‰ dx
$
c7 ln ksec x tan xk sec x tan xd !
x
"
x
'0sec x a1 y# b dy dx œ 4 '01Î3 ’y y3 “ sec x dx œ 4 '01Î3 Šsec x sec3 x ‹ dx
51.
$
ec x
bÄ_
1
1/ ˆ1cx# ‰
1Î#
c1/ a1c
_ _
'c_
'_ ax 1b"ay 1b -dx dy œ 2 '0_ Š y 21 ‹ Š
#
’7 ln Š2 È3‹ 2È3“
2
3
_
(2y 1) dy dx œ 'c1 cy# ydº
#
#
œ
dx œ '1 ˆ x$x ‰ dx œ lim
œ 4 lim c csin" b 0d œ 21
bÄ1
#
#
x# b1Î#
lim
bÄ_
dx œ 'c1 È 2
"x ‘ b œ lim
1
1
1 x #
bÄ_
ˆ "b 1‰ œ 1
dx œ 4 lim c csin" xd !
bÄ1
b
tan" b tan" 0‹ dy œ 21 lim
bÄ_
'0b y "1 dy
#
œ 21 Š lim tan" b tan" 0‹ œ (21) ˆ 1# ‰ œ 1#
bÄ_
54.
'0_ '0_ xecÐx2yÑ dx dy œ '0_ e2y
_
œ '0 ec2y dy œ
55.
"
# b lim
Ä_
lim
bÄ_
_
cxex ex d b0 dy œ '0 e2y lim
aec2b 1b œ
bÄ_
abeb eb 1b dy
"
#
' ' f(xß y) dA ¸ "4 f ˆ #" ß 0‰ 8" f(0ß 0) 8" f ˆ 4" ß 0‰ 4" f ˆ #" ß 0‰ 4" f ˆ #" ß #" ‰ 8" f ˆ!ß #" ‰ 8" f ˆ 4" ß #" ‰
R
œ
"
4
dx
128
15
49. V œ '1 'c1Îx (x 1) dy dx œ '1 cxy yd 1Î1xÎx dx œ '1 1
œ 2 cx ln xd #" œ 2(1 ln 2)
53.
"
#
œ 20
48. V œ 'c1 'cxc1 (3 3x) dy dx '0
2
3
2
2
x
'02cx a12 3y# b dy dx œ '02 c12y y$ d #
dx œ '0 c24 12x (2 x)$ d dx
!
œ ’24x 6x#
œ
dx œ '0
ˆ #"
"
#
0‰ 8" ˆ0
"
4
"
#
34 ‰ œ
3
16
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
947
948
56.
Chapter 15 Multiple Integrals
' ' f(xß y) dA ¸ "4 f ˆ 74 ß 94 ‰ f ˆ 94 ß 94 ‰ f ˆ 54 ß 114 ‰ f ˆ 74 ß 114 ‰ f ˆ 94 ß 114 ‰ f ˆ 114 ß 114 ‰ f ˆ 45 ß 134 ‰ f ˆ 47 ß 134 ‰
R
œ
"
16
‰
ˆ 11 13 ‰
ˆ 7 15 ‰
ˆ 9 15 ‰‘
f ˆ 94 ß 13
4 f 4 ß 4 f 4 ß 4 f 4 ß 4
(25 27 27 29 31 33 31 33 35 37 37 39) œ
57. The ray ) œ
1
6
È3 È4cx
È
'xÎÈ3 È4x# dy dx œ '0 3 ’a4 x# b Èx3 È4 x# “ dx œ ”4x
R
58.
#
'2_ '02 ax xb "(y1)
#
œ 6 ’ lim
1
0
_
dx œ 6 '2
cln (x 1) ln xd 2b œ 6 lim
lim
bÄ_
bÄ_
dx
x(x1)
[ln (b 1) ln b ln 1 ln 2]
$
x
7
12
$
" ‰
1#
(2x)$
3 “
dx œ
ˆ0 0
$
’ 2x3
16 ‰
12
œ
ˆ1 1" ‰ y
1y# dy
0
1
1
ˆ 21 ‰ ln 5
"
'
21
ˆ2 1y ‰
1 y #
2
"
2 tan
"
21 2 tan
21
2
7x%
12
"
(2x)%
12 “ !
4
3
'02 atan" 1x tan" xb dx œ '02 'x1x 1"y
œ 2 tan
3 ‰
x # x
'x2cx ax# y# b dy dx œ '01 ’x# y y3 “ 2cx dx
œ ˆ 23
œ'
2
"Î$
dy dx œ '0
2
#
'yyÎ1
"
1 y #
dx dy '2
21
# #
"
‰
dy œ ˆ 12"
y
1 cln a1 y bd ! 2 tan
"
21
"
21
ln a1 41# b 2 tan" 2
#
ln a1 41 b
"
#1
"
#1
'y2Î1 1"y
#
dx dy
21
ln a1 y# b‘ 2
ln 5
ln 5
#
61. To maximize the integral, we want the domain to include all points where the integrand is positive and to
exclude all points where the integrand is negative. These criteria are met by the points (xß y) such that
4 x# 2y# 0 or x# 2y# Ÿ 4, which is the ellipse x# 2y# œ 4 together with its interior.
62. To minimize the integral, we want the domain to include all points where the integrand is negative and to
exclude all points where the integrand is positive. These criteria are met by the points (xß y) such that
x# y# 9 Ÿ 0 or x# y# Ÿ 9, which is the closed disk of radius 3 centered at the origin.
63. No, it is not possible By Fubini's theorem, the two orders of integration must give the same result.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
x
È3
.
È3
$Î#
a4 x # b
3È 3
ln ˆ1 "b ‰ ln 2“ œ 6 ln 2
1
œ '0 ’2x# 7x3
2
_
1)
' ˆ x#3x
dy dx œ '2 ’ 3(y
ax# xb “ dx œ 2
'2b ˆ x" 1 x" ‰ dx œ 6
bÄ_
59. V œ '0
_
#Î$
bÄ_
œ
x$
3
20È3
9
œ 6 lim
60.
œ 24
meets the circle x# y# œ 4 at the point ŠÈ3ß 1‹ Ê the ray is represented by the line y œ
Thus, ' ' f(xß y) dA œ '0
œ
384
16
•
0
Section 15.1 Double Integrals
64. One way would be to partition R into two triangles with the
line y œ 1. The integral of f over R could then be written
as a sum of integrals that could be evaluated by integrating
first with respect to x and then with respect to y:
' ' f(xß y) dA
R
œ '0
1
'22ccÐ2yyÎ2Ñ f(xß y) dx dy '12 'y2cÐ1 yÎ2Ñ f(xß y) dx dy.
Partitioning R with the line x œ 1 would let us write the
integral of f over R as a sum of iterated integrals with
order dy dx.
65.
'bb 'bb ex y
#
#
dx dy œ 'b 'b ey ex dx dy œ 'b ey Œ'b ex dx dy œ Œ'b ex dx Œ'b ey dy
b
b
#
#
b
#
#
b
#
b
#
#
b
#
#
#
#
#
œ Œ'cb ecx dx œ Œ2 '0 ecx dx œ 4 Œ'0 ecx dx ; taking limits as b Ä _ gives the stated result.
b
66.
'01 '03 (yx1)
dy dx œ '0
3
#
œ
b
#Î$
"
3 b lim
Ä 1c
'0
b
dy
(y1)#Î$
'01 (yx1)
dx dy œ '0
3
#
#Î$
"
3
b
'b
3
lim
b Ä 1b
dy
(y1)#Î$
œ
"
(y1)#Î$
lim
b Ä 1c
$
"
’ x3 “ dy œ
!
"
3
'03 (ydy1)
#Î$
(y 1)"Î$ ‘ b lim (y 1)"Î$ ‘ 3
0
b
b Ä 1b
œ ’ lim c (b 1)"Î$ (1)"Î$ “ ’ lim b (b 1)"Î$ (2)"Î$ “ œ (0 1) Š0 $È2‹ œ 1 $È2
bÄ1
bÄ1
67-70. Example CAS commands:
Maple:
f := (x,y) -> 1/x/y;
q1 := Int( Int( f(x,y), y=1..x ), x=1..3 );
evalf( q1 );
value( q1 );
evalf( value(q1) );
71-76. Example CAS commands:
Maple:
f := (x,y) -> exp(x^2);
c,d := 0,1;
g1 := y ->2*y;
g2 := y -> 4;
q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d );
value( q5 );
plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0],
scaling=constrained, title="#71 (Section 15.1)" );
r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 );
value( r5);
value( q5-r5 );
67-76. Example CAS commands:
Mathematica: (functions and bounds will vary)
You can integrate using the built-in integral signs or with the command Integrate. In the Integrate command, the
integration begins with the variable on the right. (In this case, y going from 1 to x).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
949
950
Chapter 15 Multiple Integrals
Clear[x, y, f]
f[x_, y_]:= 1 / (x y)
Integrate[f[x, y], {x, 1, 3}, {y, 1, x}]
To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done
with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to
use the double equal sign for the equations of the bounding curves.
Clear[x, y, f]
< y/(x^2+y^2);
a,b := 0,1;
f1 := x -> x;
f2 := x -> 1;
plot3d( f(x,y), y=f1(x)..f2(x), x=a..b, axes=boxed, style=patchnogrid, shading=zhue, orientation=[0,180], title="#43(a)
(Section 15.3)" );
# (a)
q1 := eval( x=a, [x=r*cos(theta),y=r*sin(theta)] );
# (b)
q2 := eval( x=b, [x=r*cos(theta),y=r*sin(theta)] );
q3 := eval( y=f1(x), [x=r*cos(theta),y=r*sin(theta)] );
q4 := eval( y=f2(x), [x=r*cos(theta),y=r*sin(theta)] );
theta1 := solve( q3, theta );
theta2 := solve( q1, theta );
r1 := 0;
r2 := solve( q4, r );
plot3d(0,r=r1..r2, theta=theta1..theta2, axes=boxed, style=patchnogrid, shading=zhue, orientation=[-90,0],
title="#43(c) (Section 15.3)" );
fP := simplify(eval( f(x,y), [x=r*cos(theta),y=r*sin(theta)] ));
# (d)
q5 := Int( Int( fP*r, r=r1..r2 ), theta=theta1..theta2 );
value( q5 );
Mathematica: (functions and bounds will vary)
For 43 and 44, begin by drawing the region of integration with the FilledPlot command.
Clear[x, y, r, t]
< x^2*y^2*z;
q1 := Int( Int( Int( F(x,y,z), y=-sqrt(1-x^2)..sqrt(1-x^2) ), x=-1..1 ), z=0..1 );
value( q1 );
Mathematica: (functions and bounds will vary)
Due to the nature of the bounds, cylindrical coordinates are appropriate, although Mathematica can do it as is also.
Clear[f, x, y, z];
f:= x2 y2 z
Integrate[f, {x,1,1}, {y,Sqrt[1 x2 ], Sqrt[1 x2 ]}, {z, 0, 1}]
N[%]
topolar={x Ä r Cos[t], y Ä r Sin[t]};
fp= f/.topolar //Simplify
Integrate[r fp, {t, 0, 21}, {r, 0, 1},{z, 0, 1}]
N[%]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
968
Chapter 15 Multiple Integrals
15.5 MASSES AND MOMENTS IN THREE DIMENSIONS
cÎ2
bÎ2
aÎ2
cÎ 2
cÎ2
$
#
c#
12
; likewise Ry œ É a
bÎ2
cÎ2
bÎ2
1. Ix œ 'cÎ2 'bÎ2 'aÎ2 ay# z# b dx dy dz œ a 'cÎ2 'bÎ2 ay# z# b dy dz œ a 'cÎ2 ’ y3 yz# “
dz
bÎ2
b
b
œ a 'cÎ2 Š 12
bz# ‹ dz œ ab ’ 12
z
Rx œ É b
#
œ '3 '2 ’ 8y3
3
4
#
4
2y$
3
$
Ð42yÑÎ3
Iz œ 'c3 'c2 'c4Î3
3
4
3. Ix œ '0 '0
a
œ
M
3
b
and Rz œ É a
#
b#
12
c$
12 ‹
œ
abc
1#
8(2 y)$
81
x# (4 2y)
3
64
81 “
4x#
3
4
Ð42yÑÎ3
3
3
dy dx œ 'c3 ˆ12x#
3
64
81 “
ax# y# b dz dy dx œ '3 '2 ax# y# b ˆ 83
3
4
M
3
aa# c# b and Iz œ
M
3
2y ‰
3
dy dx œ 12 '3 ax# 2b dx œ 360
3
$
Myz œ '0
x dz dy dx œ '0
dx œ
abc ab# c# b
3
aa# b# b , by symmetry
1cx
x "# ‹ dx œ
"
6
;
'0 x(1 x y) dy dx œ "# '0 ax$ 2x# xb dx œ 24"
1
1cx
1cxcy
Ê x œ y œ z œ "4 , by symmetry; Ix œ '0 '0 '0
ay# z# b dz dy dx
1
1cx
1
"
"
œ '0 '0 ’y# xy# y$ (1 x3 y) “ dy dx œ "6 '0 (1 x)% dx œ 30
Ê Iy œ Ix œ 30
, by symmetry
1
'0 '0
c$ b
3 ‹
#
1cxcy
ax# z# b dz dy dx
dx œ 280;
'01cx '01cxcy dz dy dx œ '01 '01cx (1 x y) dy dx œ '01 Š x#
1cx
Ð42yÑÎ3
4
32 ‰
3
$
1
ab# c# b ;
ay# z# b dz dy dx
'0c ay# z# b dz dy dx œ '0a '0b Šcy# c3 ‹ dy dx œ '0a Š cb3
4. (a) M œ '0
M
12
' ' '
dy dx œ '3 104
3 dx œ 208; Iy œ 3 2 4Î3
ab# c# b where M œ abc; Iy œ
ab# c# b œ
, by symmetry
3
œ 'c3 'c2 ’ (4 812y)
3
c#
12
#
œ ab Š b12c
is the top of the wedge Ê Ix œ '3 '2 '4Î3
4 2y
3
2. The plane z œ
#
cÎ2
z$
3 “ cÎ2
$
1
1
$
È5
5
Ix
(b) Rx œ É M
œ É 5" œ
¸ 0.4472; the distance from the centroid to the x-axis is É0#
"
16
"
16
œ É 8" œ
È2
4
¸ 0.3536
5. M œ 4 '0
1
œ 2 '0
'01 '4y4 dz dy dx œ 4 '01 '01 a4 4y# b dy dx œ 16 '01 23 dx œ 323 ; Mxy œ 4 '01 '01 '4y4
#
#
z dz dy dx
12
'0 a16 16y% b dy dx œ 128
'0 dx œ 128
5
5 Ê z œ 5 , and x œ y œ 0, by symmetry;
1
1
4
1
1
64y
7904
%
' 1 1976
‰
Ix œ 4 '0 '0 '4y ay# z# b dz dy dx œ 4 '0 '0 ’ˆ4y# 64
3 Š4y 3 ‹“ dy dx œ 4 0 105 dx œ 105 ;
1
1
1
'
#
Iy œ 4 '0
1
œ
4832
63
'01 '4y4 ax# z# b dz dy dx œ 4 '01 '01 ’ˆ4x# 643 ‰ Š4x# y# 64y3 ‹“ dy dx œ 4 '01 ˆ 38 x# 128
‰ dx
7
'
#
; Iz œ 4 '0
1
œ 16 '0 Š 2x3
1
#
'01 '4y4 ax# y# b dz dy dx œ 16 '01 '01 ax# x# y# y# y% b dy dx
2
15 ‹
#
dx œ
ŠÈ4x# ‹Î2
256
45
2x
6. (a) M œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
x dz dy dx œ '2 'ŠÈ4x#‹Î2 x(2 x) dy dx œ '2 x(2 x) ŠÈ4 x# ‹ dx œ 21;
ŠÈ4x# ‹Î2
2x
y dz dy dx œ '2 'ŠÈ4x#‹Î2 y(2 x) dy dx
œ
"
#
4 x #
4 “
'c22 (2 x) ’ 44x
#
2
2
ŠÈ4x# ‹Î2
2
2x
Mxz œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
2
ŠÈ4x# ‹Î2
Myz œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
ŠÈ4x# ‹Î2
dz dy dx œ '2 'ŠÈ4x#‹Î2 (2 x) dy dx œ '2 (2 x) ŠÈ4 x# ‹ dx œ 41;
2
ŠÈ4x# ‹Î2
dx œ 0 Ê x œ "# and y œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.5 Masses and Moments in Three Dimensions
ŠÈ4x# ‹Î2
2x
(b) Mxy œ 'c2 'ŠcÈ4cx#‹Î2 '0
2
œ 51 Ê z œ
7. (a) M œ 4 '0
2
È4cx
Mxy œ '0
#
'22 'ŠŠÈ44xx ‹‹ÎÎ22
È
#
#
"
#
(2 x)# dy dx œ
'22 (2 x)# ŠÈ4 x# ‹ dx
'x4y dz dy dx œ 4 '01Î2 '02 'r 4 r dz dr d) œ 4 '01Î2 '02 a4r r$ b dr d) œ 4 '01Î2 4 d) œ 81;
#
#
#
'0 'r zr dz dr d) œ '0 '0
2
"
#
5
4
'0
21
z dz dy dx œ
21
4
2
#
by symmetry
(b) M œ 81 Ê 41 œ '0
21
r
#
a16 r% b dr d) œ
È
Èc
'0 c 'r c r dz dr d) œ '021 '0
#
32
3
'021 d) œ 6431
acr r$ b dr d) œ '0
21
Ê zœ
8
3
c#
4
c# 1
#
d) œ
, and x œ y œ 0,
Ê c# œ 8 Ê c œ 2È2,
since c 0
8. M œ 8; Mxy œ 'c1 '3 'c1 z dz dy dx œ 'c1 '3 ’ z2 “ dy dx œ 0; Myz œ 'c1 '3 'c1 x dz dy dx
"
1
5
1
1
5
"
#
1
5
1
œ 2 'c1 '3 x dy dx œ 4 'c1 x dx œ 0; Mxz œ 'c1 '3 'c1 y dz dy dx œ 2 'c1 '3 y dy dx œ 16 'c1 dx œ 32
1
5
1
1
5
1
Ê x œ 0, y œ 4, z œ 0; Ix œ ' ' ' ay# z# b dz dy dx œ ' ' ˆ2y# 23 ‰ dy dx œ 23 ' 100 dx œ 400
3 ;
1
5
1
1
c1
3
5
1
1
c1
c1
5
1
c1
3
1
5
1
1
5
1
Iy œ 'c1 '3 'c1 ax# z# b dz dy dx œ 'c1 '3 ˆ2x# 23 ‰ dy dx œ 43 'c1 a3x# 1b dx œ 16
3 ;
1
5
1
1
5
1
400
‰
Iz œ 'c1 '3 'c1 ax# y# b dz dy dx œ 2 'c1 '3 ax# y# b dy dx œ 2 'c1 ˆ2x# 98
3 dx œ 3
Ê Rx œ Rz œ É 50
3
and Ry œ É 23
Ð2yÑÎ2
9. The plane y 2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1
2
œ 'c2 'c2 ’ (y 6)#(4 y)
2
Mœ
"
#
4
#
(2 y)$
24
4
4
$
49
3 ‹
dt œ 1386;
IL
(3)(6)(4) œ 36 Ê RL œ É M
œ É 77
#
2
"
#
c(y 6)# z# d dz dy dx
#
"3 “ dy dx; let t œ 2 y Ê IL œ 4 'c2 Š 13t
24 5t 16t
Ð2yÑÎ2
10. The plane y 2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1
œ
4
c(x 4)# y# d dz dy dx
'c22 'c42 ax# 8x 16 y# b (4 y) dy dx œ 'c22 a9x# 72x 162b dx œ 696; M œ "# (3)(6)(4) œ 36
IL
Ê RL œ É M
œ É 58
3
11. M œ 8; IL œ '0
4
'02 '01 cz# (y 2)# d dz dy dx œ '04 '02 ˆy# 4y 133 ‰ dy dx œ 103 '04 dx œ 403
IL
Ê RL œ É M
œ É 53
12. M œ 8; IL œ '0
4
'02 '01 c(x 4)# y# d dz dy dx œ '04 '02 c(x 4)# y# d dy dx œ '04 2(x 4)# 83 ‘ dx œ 160
3
IL
Ê RL œ É M
œ É 20
3
'02cx '02cxcy 2x dz dy dx œ '02 '02cx a4x 2x# 2xyb dy dx œ '02 ax$ 4x# 4xb dx œ 43
2
2cx
2cxcy
2
2cx
2
8
8
Mxy œ '0 '0 '0
2xz dz dy dx œ '0 '0 x(2 x y)# dy dx œ '0 x(23 x) dx œ 15
; Mxz œ 15
by
2
2cx
2cxcy
2
2cx
2
#
symmetry; Myz œ '0 '0 '0
2x# dz dy dx œ '0 '0 2x# (2 x y) dy dx œ '0 a2x x# b dx œ 16
15
13. (a) M œ '0
2
(b)
$
Ê xœ
4
5
969
, and y œ z œ
2
5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
970
Chapter 15 Multiple Integrals
14. (a) M œ '0
2
È
'0 x '04cx
(b) Myz œ '0
2
Ê xœ
È
256È2k
231
œ
k
4
kxy dz dy dx œ k'0
'0 x '04cx
2
#
2
2
È
'0 x '04cx
40È2
77
Ê yœ
Èx
'0
kx# y dz dy dx œ k '0
; Mxz œ '0
5
4
œ
#
#
xy a4 x# b dy dx œ
Èx
'0
x# y a4 x# b dy dx œ
kxy# dz dy dx œ k'0
2
È
'0 x '04cx
; Mxy œ '0
2
'02 a16x# 8x% x' b dx œ 256k
105
#
Ê zœ
'02 a4x# x% b dx œ 32k
15
k
#
Èx
'0
k
#
'02 a4x$ x& b dx œ 8k3
xy# a4 x# b dy dx œ
kxyz dz dy dx œ '0
2
Èx
'0
k
3
'02 ˆ4x&Î# x*Î# ‰ dx
#
xy a4 x# b dy dx
8
7
15. (a) M œ '0
'01 '01 (x y z 1) dz dy dx œ '01 '01 ˆx y 3# ‰ dy dx œ '01 (x 2) dx œ 5#
1
1
1
1
1
1
4
‰
Mxy œ '0 '0 '0 z(x y z 1) dz dy dx œ "# '0 '0 ˆx y 53 ‰ dy dx œ "# '0 ˆx 13
6 dx œ 3
1
(b)
Ê Mxy œ Myz œ Mxz œ
(c) Iz œ '0
1
4
3
, by symmetry Ê x œ y œ z œ
8
15
'01 '01 ax# y# b (x y z 1) dz dy dx œ '01 '01 ax# y# b ˆx y 3# ‰ dy dx
œ '0 ˆx$ 2x# "3 x 34 ‰ dx œ
1
Ê Ix œ Iy œ Iz œ
11
6
11
6
, by symmetry
Iz
(d) Rx œ Ry œ Rz œ É M
œ É 11
15
16. The plane y 2z œ 2 is the top of the wedge.
Ð2yÑÎ2
(a) M œ 'c1 'c2 'c1
1
4
4
Ð2yÑÎ2
Mxz œ 'c1 'c2 'c1
1
4
Ð2yÑÎ2
Mxy œ 'c1 'c2 'c1
1
4
Ð2yÑÎ2
(c) Ix œ 'c1 'c2 'c1
1
4
Ð2yÑÎ2
Iy œ 'c1 'c2 'c1
1
4
Ð2yÑÎ2
Iz œ 'c1 'c2 'c1
1
4
1
Ð2yÑÎ2
(b) Myz œ 'c1 'c2 'c1
1
(x 1) dz dy dx œ 'c1 'c2 (x 1) ˆ2 y# ‰ dy dx œ 18
4
x(x 1) dz dy dx œ 'c1 'c2 x(x 1) ˆ2 y# ‰ dy dx œ 6;
1
4
y(x 1) dz dy dx œ 'c1 'c2 y(x 1) ˆ2 y# ‰ dy dx œ 0;
1
z(x 1) dz dy dx œ
"
#
4
'c11 'c42 (x 1) Š y4
#
y‹ dy dx œ 0 Ê x œ
(x 1) ay# z# b dz dy dx œ 'c1 'c2 (x 1) ’2y#
1
4
(x 1) ax# z# b dz dy dx œ 'c1 'c2 (x 1) ’2x#
1
4
$
1
3
, and y œ z œ 0
$
3" ˆ" 2y ‰ “ dy dx œ 45;
"
3
y
#
"
3
x# y
#
$
3" ˆ" 2y ‰ “ dy dx œ 15;
(x 1) ax# y# b dz dy dx œ 'c1 'c2 (x 1) ˆ2 y# ‰ ax# y# b dy dx œ 42
1
4
I
Ix
Iz
(d) Rx œ É M
œ É 5# ß Ry œ É My œ É 56 ß and Rz œ É M
œ É 37
17. M œ '0
1
Èz
'zc1c1z '0
(2y 5) dy dx dz œ '0
1
'zc1c1z ˆz 5Èz‰ dx dz œ '01 2 ˆz 5Èz‰ (1 z) dz
"
$Î#
œ 2 '0 ˆ5z"Î# z 5z$Î# z# ‰ dz œ 2 10
"# z# 2z&Î# 3" z$ ‘ ! œ 2 ˆ 93 #3 ‰ œ 3
3 z
1
È4cx
16c2 ˆx# by# ‰
18. M œ 'c2 'cÈ4cx# '2 ax#by#b
2
œ 4 '0
21
#
È4cx
Èx# y# dz dy dx œ ' ' È # Èx# y# c16 4 ax# y# bd dy dx
c2 c 4cx
2
'02 r a4 r# b r dr d) œ 4 '021 ’ 4r3
$
#
r5 “ d) œ 4 '0
&
#
21
!
64
15
d) œ
5121
15
19. (a) Let ?Vi be the volume of the ith piece, and let (xi ß yi ß zi ) be a point in the ith piece. Then the work done
by gravity in moving the ith piece to the xy-plane is approximately Wi œ mi gzi œ (xi yi zi 1)g ?Vi zi
Ê the total work done is the triple integral W œ '0
'01 '01 (x y z 1)gz dz dy dx
1
1
1
1
1
"
"
œ g '0 '0 "# xz# "# yz# 3" z$ #" z# ‘ ! dy dx œ g '0 '0 ˆ #" x #" y 56 ‰ dy dx œ g '0 2" xy 4" y# 56 y‘ ! dx
1
œ g '0 ˆ "# x
1
13 ‰
12
#
dx œ g ’ x4
13
12
"
16 ‰
x“ œ g ˆ 12
œ
!
4
3
g
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.5 Masses and Moments in Three Dimensions
971
8
8
8 ‰
(b) From Exercise 15 the center of mass is ˆ 15
ß 15
ß 15
and the mass of the liquid is 5# Ê the work done by
8 ‰
gravity in moving the center of mass to the xy-plane is W œ mgd œ ˆ 52 ‰ (g) ˆ 15
œ 34 g, which is the same as
the work done in part (a).
20. (a) From Exercise 19(a) we see that the work done is W œ g '0
2
œ kg '0
2
œ
kg
4
Èx "
'0
#
#
xy a4 x# b dy dx œ
8 &
" (‘ #
$
16
3 x 5 x 7 x ! œ
kg
4
'0
2
È
'0 x '04cx
#
x# a4 x# b dx œ
kg
4
#
kxyz dz dy dx
'0 a16x# 8x% x' b dx
2
256k†g
105
È
(b) From Exercise 14 the center of mass is Š 54 ß 4077 2 ß 78 ‹ and the mass of the liquid is
gravity in moving the center of mass to the xy-plane is W œ mgd œ
21. (a) x œ
Myz
M
ˆ 32k
‰ ˆ8‰
15 (g) 7
32k
15
œ
Ê the work done by
256k†g
105
' ' ' x$ (xß yß z) dx dy dz œ 0 Ê Myz œ 0
œ0 Ê
R
(b) IL œ ' ' ' kv hik# dm œ ' ' ' k(x h) i yjk# dm œ ' ' ' ax# 2xh h# y# b dm
D
D
D
œ ' ' ' ax# y# b dm 2h ' ' ' x dm h# ' ' ' dm œ Ix 0 h# m œ Ic m h# m
Þ
D
D
22. IL œ Ic m mh# œ
Þ
Þ
2
5
ma# ma# œ
7
5
ma#
#
23. (a) (xß yß z) œ ˆ #a ß #b ß #c ‰ Ê Iz œ Ic m abc ŠÉ a4
Þ
œ
abc aa# b# b
3
abc aa# b# b
4
#
(b) IL œ IcÞmÞ abc ŒÉ a4
œ
abc aa# 7b# b
3
3
4
Þ
b#
4‹
#
Ê IcÞmÞ œ Iz
# b #
abc aa# b# b
; RcÞmÞ œ É IcMÞmÞ œ É a 12
1#
#
#
#
#
#
ˆ b# 2b‰# œ abc aa12 b b abc aa 4 9b b
abc aa# b# b
4
œ
IL
; RL œ É M
œ Éa
Ð42yÑÎ3
24. M œ 'c3 'c2 'c4Î3
Þ
D
#
abc a4a# 28b# b
1#
7b#
3
dz dy dx œ '3 '2 23 (4 y) dy dx œ '3 23 ’4y
3
œ
4
3
#
%
y#
2 “ #
dx œ 12 '3 dx œ 72;
3
x œ y œ z œ 0 from Exercise 2 Ê Ix œ IcÞmÞ 72 ŠÈ0# 0# ‹ œ IcÞmÞ Ê IL œ IcÞmÞ 72 ŠÉ16
16
9 ‹
#
‰ œ 1488
œ 208 72 ˆ 160
9
25. MyzB"B# œ ' ' ' x dV" ' ' ' x dV# œ MÐyzÑ" MÐyzÑ# Ê x œ MÐyzÑ" MÐyzÑ# m" m# ; similarly,
B"
B#
y œ MÐxzÑ" MÐxzÑ# m" m# and z œ MÐxyÑ" MÐxyÑ# m" m# Ê c œ xi yj zk
œ m " m ˆMÐyzÑ" MÐyzÑ# ‰ i ˆMÐxzÑ" MÐxzÑ# ‰ j ˆMÐxyÑ" MÐxyÑ# ‰ k‘
"
œ
œ
#
"
m" m#
"
m" m#
cam" x" m# x# b i am" y" m# y# b j am" z" m# z# b kd
cm" ax" i y" j z" kb m# ax# i y# j z# kbd œ
i 13j 13
13
13
2 k
Ê x œ 13
7
14 , y œ 7 , z œ 14
2i 7j 12 k
" ‰
k‰ 12 ˆi 11
Ê x œ 1, y œ 27 , z œ 41
# j # k 12 12 œ
2
13i 74j 5k
" ‰
11
" ‰
37
5
ˆ
Ê x œ 13
# k 12 i # j # k 2 12 œ
14
14 , y œ 7 , z œ 14
25i 92j 7k
" ‰
k‰ 2 ˆ "# i 4j "# k‰ 12 ˆi 11
Ê x œ 25
# j # k 12 2 12 œ
26
26 ,
26. (a) c œ 12 ˆi 32 j k‰ 2 ˆ "# i 4j "# k‰12 2 œ
(b) c œ 12 ˆi 32 j
(c) c œ 2 ˆ "# i 4j
(d) c œ 12 ˆi 32 j
m" c" m# c#
m" m#
13
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
yœ
46
13
,zœ
7
26
972
Chapter 15 Multiple Integrals
#
$
#
Š 1a3 h ‹ Š h4 k‹ Š 213a ‹ Šc 3a
8 k‹
27. (a) c œ
Š a 31 ‹ Š h
œ
m" m#
# 3a#
4
k‹
m" m#
1 a# h
3
, where m" œ
and m# œ
21 a$
3
; if
h# 3a#
4
œ 0, or h œ aÈ3, then the centroid is on the common base
(b) See the solution to Exercise 55, Section 15.2, to see that h œ aÈ2.
#
28. c œ
#
Š s 3h ‹ Š 4h k‹ s$ ˆc #s k‰
s
Š 12
‹ ah# c 6s# b k‘
œ
m" m#
m" m#
, where m" œ
s# h
3
and m# œ s$ ; if h# 6s# œ 0,
or h œ È6s, then the centroid is in the base of the pyramid. The corresponding result in 15.2, Exercise 56, is h œ È3s.
15.6 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES
1.
'021 '01 'r
È2cr
#
dz r dr d) œ '0
21
œ '0 Š 2 3 23 ‹ d) œ
21
2.
È
'021 '03 'r Î318r
4.
'03 ’r a18 r# b"Î# r3 “ dr d) œ '021 ’ "3 a18 r# b$Î# 12r “ $ d)
$
%
!
2
1
$
’ 12)1#
21
21
)&
201% “ !
È
'01 '0 Î '3È44rr
) 1
dz r dr d) œ '0
#
1
%
ˆ
# ‰"Î#
)
) Î1
!
1
'0 Î
) 1
œ 4 '0 Š 21)#
1
#
21
4
3‹
1
)
1
d) œ
'02 Š 4)1
1
3
#
#
#
"
#
c9 a4 r# b a4 r# bd r dr d) œ 4 '0
)%
41 % ‹
1
d)
'0 Î a4r r$ b dr d)
) 1
371
15
d) œ
'01 ’r a2 r# b"Î# r# “ dr d) œ 3 '021 ’ a2 r# b"Î# r3 “ " d)
$
!
d) œ 1 Š6È2 8‹
6.
'021 '01 'c11ÎÎ22 ar# sin# ) z# b dz r dr d) œ '021 '01 ˆr$ sin# ) 12r ‰ dr d) œ '021 Š sin4 ) 24" ‹ d) œ 13
7.
'021 '03 '0zÎ3
8.
'c11 '021 '01bcos
9.
'01 '0 z '021 ar# cos# ) z# b r d) dr dz œ '01 '0
#
r$ dr dz d) œ '0
21
)
21
z
3
'03 324
dz d) œ '0 20
d) œ 3101
%
4r dr d) dz œ 'c1 '0 2(1 cos ))# d) dz œ 'c1 61 d) œ 121
1
21
È
1
È
1
Èz
œ '0 ’ 14r 1r# z# “
10.
4) %
161% ‹
171
5
3 dz r dr d) œ 3 '0
œ 3 '0 ŠÈ2
21
'0 Î2 a3r 24r$ b dr d) œ '02 32 r# 6r% ‘ !Î2
1
#
'021 '01 'r 2cr
œ
z dz r dr d) œ '0
#
œ 4 '0 ’2r# r4 “
5.
3
91 Š8È2 7‹
)
3
#
!
21
'021 '0 Î2 '0324r
œ
$
41 ŠÈ2 "‹
dz r dr d) œ '0
#
#
œ
3.
$Î#
'01 ’r a2 r# b"Î# r# “ dr d) œ '021 ’ "3 a2 r# b$Î# r3 “ " d)
%
Èz
!
1
#
$
È
#
$Î#
r# sin 2)
4
dz œ '0 Š 14z 1z$ ‹ dz œ ’ 112z
'02 'rc24cr '021 (r sin ) 1) r d) dz dr œ '02 'rc24cr
œ 21 ’ "3 a4 r# b
#
’ r2)
r$
3
#
z# )“
"
1 z%
4 “!
#1
!
r dr dz œ '0
1
Èz
'0
a1r$ 21rz# b dr dz
1
3
œ
21r dz dr œ 21'0 ’r a4 r# b
2
"Î#
r# 2r“ dr
#
r# “ œ 21 83 4 3" (4)$Î# ‘ œ 81
!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates
È4cr
'021 '01 '0
11. (a)
È3
#
dz r dr d)
È4cz
'021 '0 '01 r dr dz d) '021 'È23 '0
(b)
È4cr
'01 '0
(c)
#
#
r dr dz d)
'021 r d) dz dr
'021 '01 'r 2cr dz r dr d)
#
12. (a)
È2cz
(b)
'021 '01 '0z r dr dz d) '021 '12 '0
(c)
'0 'r '0
2cr#
1
21
r dr dz d)
r d) dz dr
13.
'c11ÎÎ22 '0cos '03r
14.
'c11ÎÎ22 '01 '0r cos
15.
'01 '02 sin '04cr sin
17.
'c1ÎÎ22 '11cos '04
19.
'0 Î4 '0sec '02r sin
21.
'01 '01 '02 sin 9 3# sin 9 d3 d9 d) œ 83 '01 '01 sin% 9 d9 d) œ 83 '01 Š’ sin 94cos 9 “ 1 34 '01 sin# 9 d9‹ d)
)
)
#
Î
r$ dz dr d) œ '1Î2 '0 r% cos ) dr d) œ
)
1
f(rß )ß z) dz r dr d)
1 2
1
)
"
5
'1ÎÎ22 cos ) d) œ 25
1
f(rß )ß z) dz r dr d)
16.
' ÎÎ22 '03 cos '05r cos
f(rß )ß z) dz r dr d)
18.
'1ÎÎ22 'cos2 cos '03r sin
20.
' ÎÎ42 '0csc '02r sin
)
)
1
)
f(rß )ß z) dz r dr d)
1
)
1
1
)
23.
24.
)
)
1
)
1
)
f(rß )ß z) dz r dr d)
f(rß )ß z) dz r dr d)
f(rß )ß z) dz r dr d)
$
!
1
1
1
1
1
œ 2 '0 '0 sin# 9 d9 d) œ '0 ) sin#2) ‘ ! d) œ '0 1 d) œ 1#
22.
)
'021 '01Î4 '02 (3 cos 9) 3# sin 9 d3 d9 d) œ '021 '01Î4 4 cos 9 sin 9 d9 d) œ '021 c2 sin# 9d !1Î% d) œ '021 d) œ 21
'021 '01 '0Ð1cos 9ÑÎ2 3# sin 9 d3 d9 d) œ 24" '021 '01 (1 cos 9)$ sin 9 d9 d) œ 96" '021 c(1 cos 9)% d !1 d)
21
21
" '
16 '
%
œ 96
a
2
0
b
d
)
œ
d) œ 6" (21) œ 13
96
0
0
'031Î2 '01 '01 53$ sin$ 9 d3 d9
œ
5
6
d) œ
5
4
'031Î2 '01 sin$ 9 d9 d) œ 54 '031Î2 Š’ sin 93cos 9 “ 1 23 '01 sin 9 d9‹ d)
'031Î2 c cos 9d 1! d) œ 53 '031Î2 d) œ 5#1
#
!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
973
974
25.
Chapter 15 Multiple Integrals
'021 '01Î3 'sec2 9 33# sin 9 d3 d9
d) œ '0
21
œ '0 (4 2) ˆ8 "# ‰‘ d) œ
21
26.
27.
5
#
'01Î3 a8 sec$ 9b sin 9 d9 d) œ '021 8 cos 9 "2 sec# 9‘ !1Î$ d)
'021 d) œ 51
'021 '01Î4 '0sec 9 3$ sin 9 cos 9 d3 d9 d) œ "4 '021 '01Î4 tan 9 sec# 9 d9 d) œ "4 '021 2" tan# 9‘ !1Î% d)
21
œ "8 '0 d) œ 14
2
0
'02 'c01 '11ÎÎ42 3$ sin 29 d9 d) d3 œ '02 '01 3$ cos229 ‘ 11Î#
d) d3 œ '0 '1 3#
Î%
$
d) d3 œ '0
2 $
3 1
#
d3
#
%
œ ’ 138 “ œ 21
!
28.
'11ÎÎ63 'csc2 csc9 9 '021 3# sin 9 d) d3 d9 œ 21 '11ÎÎ63 'csc2 csc9 9 3# sin 9 d3 d9 œ 231 '11ÎÎ63 c3$ sin 9d csc2 csc9 9 d9
œ
29.
141
3
281
3È 3
'01 '01 '01Î4 123 sin$ 9 d9 d) d3 œ '01 '01 Œ123 ’ sin 39 cos 9 “ 1Î% 83 '01Î4 sin 9 d9 d) d3
#
!
œ '0
1
œ
30.
'11ÎÎ63 csc# 9 d9 œ
'0 Š
1
23
È2
83 ccos
1Î%
9d ! ‹
d) d3 œ '0
1
'0 Š83
1
103
È2 ‹
d) d3 œ 1'0 Š83
1
103
È2 ‹
d3 œ 1 ’43#
"
53#
È2 “
!
Š4È2 5‹ 1
È2
'11ÎÎ62 '11ÎÎ22 'csc2 9 53% sin$ 9 d3 d) d9 œ '11ÎÎ62 '11ÎÎ22 a32 csc& 9b sin$ 9 d) d9 œ '11ÎÎ62 '11ÎÎ22 a32 sin$ 9 csc# 9b d) d9
œ 1 '1Î6 a32 sin$ 9 csc# 9b d9 œ 1 ’ 32 sin 39 cos 9 “
1Î2
œ
#
È
1 Š 3224 3 ‹
641
3
ccos
1Î#
9d 1Î'
1 ŠÈ 3 ‹ œ
È3
3
1
1Î#
1Î'
641
3
'11ÎÎ62 sin 9 d9 1 ccot 9d 11Î#
Î'
È
ˆ 6431 ‰ Š #3 ‹
œ
331È3
3
œ 111È3
31. (a) x# y# œ 1 Ê 3# sin# 9 œ 1, and 3 sin 9 œ 1 Ê 3 œ csc 9; thus
'021 '01Î6 '02 3# sin 9 d3 d9 d) '021 '11ÎÎ62 '0csc 9 3# sin 9 d3 d9 d)
'021 '12 '1sinÎ6
(b)
3# sin 9 d9 d3 d) '0
21
'02 '01Î6 3# sin 9 d9 d3 d)
'021 '01Î4 '0sec 9 3# sin 9 d3 d9 d)
'021 '01 '01Î4 3# sin 9 d9 d3 d)
32. (a)
(b)
'0
21
33. V œ '0
21
œ
" Ð1Î3Ñ
"
3
È
'1 2 'cos1Î"4 Ð"Î3Ñ 3# sin 9 d9 d3 d)
'01Î2 'cos2 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a8 cos$ 9b sin 9 d9 d)
31 ‰
'021 ’8 cos 9 cos4 9 “ 1Î# d) œ 3" '021 ˆ8 4" ‰ d) œ ˆ 12
(21) œ 3161
%
!
'01Î2 '11cos 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a3 cos 9 3 cos# 9 cos$ 9b sin 9 d9 d)
21
21
1Î#
111
' 21
ˆ 11 ‰
œ "3 '0 3# cos# 9 cos$ 9 14 cos% 9‘ ! d) œ 3" '0 ˆ 32 1 4" ‰ d) œ 11
12 0 d) œ 12 (21) œ 6
34. V œ '0
21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates
35. V œ '0
21
œ
"
12
(2)
36. V œ '0
21
œ
"
12
%
'01Î2 '01cos 9
'0
21
21
d) œ
1Î#
9)
'021 '01Î2 (1 cos 9)$ sin 9 d9 d) œ 3" '021 ’ (" cos
d)
“
4
%
!
1
6
3# sin 9 d3 d9 d) œ
"
6
"
3
(21) œ
8
3
'021 '11ÎÎ42 cos$ 9 sin 9 d9 d) œ 83 '021 ’ cos4 9 “ 1Î# d)
%
1Î%
1
3
21
4 '
81
'11ÎÎ32 '02 3# sin 9 d3 d9 d) œ 83 '021 '11ÎÎ32 sin 9 d9 d) œ 83 '021 c cos 9d 11Î#
Î$ d) œ 3 0 d) œ 3
1Î2
(c) 8 '0
2
(c)
'0
21
3# sin 9 d3 d9 d) œ
(21) œ
'11ÎÎ42 '02 cos 9
39. (a) 8'0
40. (a)
"
12
d) œ
" ‰
ˆ 83 ‰ ˆ 16
38. V œ '0
!
'021 d) œ 43 (21) œ 831
%
21
37. V œ '0
œ
1
9)
'01 '01ccos 9 3# sin 9 d3 d9 d) œ 3" '021 '01 (1 cos 9)$ sin 9 d9 d) œ 3" '021 ’ (" cos
“ d)
4
'01Î2 '02 3# sin 9 d3 d9 d)
È4cx
'0
#
È4cx cy
'0
È
'01Î2 '03Î 2 'r
#
È9r
#
(b) 8'0
1Î2
È4cr
'02 '0
21
dz r dr d)
#
dz dy dx
dz r dr d)
(b)
'01Î2 '01Î4 '03 3# sin 9 d3 d9 d)
'01Î2 '01Î4 '03 3# sin 9 d3 d9 d) œ 9 '01Î2 '01Î4 sin 9 d9 d) œ 9 '01Î2 Š È"
41. (a) V œ '0
#
'01Î3 'sec2 9 3# sin 9 d3 d9 d)
(b) V œ '0
21
È3 È3cx È4cx cy
(c) V œ 'cÈ3 'cÈ3cx '1
dz dy dx
È3
21
"Î#
#
#
È
91 Š2 È2‹
1‹ d) œ
2
È4cr
'0 3 '1
#
4
dz r dr d)
#
#
(d)
V œ '0 '0
œ
#
’r a4 r b
# $Î#
'021 d) œ 531
5
6
È
r#
#•
È$
!
d) œ '0 Š "3
21
3
#
4$Î#
3 ‹
d)
'01 '0 1cr r# dz r dr d)
1Î2
21
1
Iz œ '0 '0 '0 a3# sin# 9b a3# sin 9b d3 d9 d), since r# œ x# y# œ 3# sin# 9 cos# ) 3# sin# 9 sin# )
42. (a) Iz œ '0
21
(b)
21
r“ dr d) œ '0 ” a4 3r b
#
œ 3# sin# 9
(c) Iz œ '0
21
œ
2
15
1Î2
1Î2
44. V œ 4'0
œ 4 '0
!
41
15
'01 'r 414r
1Î2
dz r dr d) œ 4 '0
#
%
d) œ
1Î2
1Î2
#
(21) œ
43. V œ 4 '0
œ 4 '0
'01Î2 5" sin$ 9 d9 d) œ 5" '021 Œ’ sin 93cos 9 “ 1Î# 23 '01Î2 sin 9 d9 d) œ 152 '021 c cos 9d !1Î# d)
81
3
'01 '1Èr1r
ˆ "#
'01 a5r 4r$ r& b dr d) œ 4 '01Î2 ˆ 5# 1 "6 ‰ d)
"
3
1Î2
#
dz r dr d) œ 4 '0
"‰
3
d) œ 2'0 d) œ
1Î2
'01 Šr r# rÈ1r# ‹ dr d) œ 4 '01Î2 ’ r2
2 ˆ 1# ‰
#
r$
3
"3 a1 r# b
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
$Î# "
“ d)
!
975
976
Chapter 15 Multiple Integrals
45. V œ '31Î2 '0
21
3 cos )
'0cr sin
#1
œ 94 cos% )‘ $1Î# œ
c3 cos )
46. V œ 2 '1Î2 '0
1
œ 18 Œ’ cos
47. V œ '0
1Î2
#
dz r dr d) œ '31Î2 '0
)
21
0œ
9
4
1
È1r
'0sin '0
#
)
r# dr d) œ
2
3
'1Î2 27 cos$ ) d)
1
'11Î2 cos ) d) œ 12 csin )d 11Î# œ 12
Î
dz r dr d) œ '0
1 2
'0sin
)
Î
rÈ1r# dr d) œ '0 ’ "3 a1 r# b
1 2
$Î# sin )
“
!
d)
#
!
œ csin
2
9
48. V œ '0
1Î2
1Î#
)d !
1
6
'0cos '03
)
œ
È1r
1Î#
œ
Î
1 2
1Î2
23 ccos )d !
4 3 1
18
dz r dr d) œ '0
#
œ '0 ’ a1 cos# )b
1
#
2
3
21
'01Î2 ’a1 sin# )b$Î# 1“ d) œ 3" '01Î2 acos$ ) 1b d) œ 3" Œ’ cos )3 sin ) “ 1Î# 32 '01Î2 cos ) d) 3) ‘ !1Î#
œ 3"
œ
r# sin ) dr d) œ '31Î2 a9 cos$ )b (sin )) d)
9
4
'0r dz r dr d) œ 2 '1Î2 '0c3 cos
1
) sin )
“
3
1Î#
)
3 cos )
$Î#
1
#
'0cos
)
Î
3rÈ1r# dr d) œ '0 ’ a1 r# b
1 2
1“ d) œ '0 a1 sin$ )b d) œ ’)
1Î2
2
3
œ
'12Î13Î3 '0a 3# sin 9 d3 d9 d) œ '021 '12Î13Î3
50. V œ '0
'01Î2 '0a 3# sin 9 d3 d9 d) œ a3 '01Î6 '01Î2
51. V œ '0
'01Î3 'sec2 9
1Î6
21
“
2
3
!
d)
'01Î2 sin ) d)
31 4
6
49. V œ '0
21
1Î#
sin# ) cos )
“
3
!
$Î# cos )
$
a$
3
a$
3
sin 9 d9 d) œ
sin 9 d9 d) œ
a$
3
'021 c cos 9d #11Î$Î$ d) œ a3 '021 ˆ "# "# ‰ d) œ 213a
$
'01Î6 d) œ a181
$
3# sin 9 d3 d9 d)
'021 '01Î3 a8 sin 9 tan 9 sec# 9b d9 d)
21
1Î$
œ 3" '0 8 cos 9 2" tan# 9‘ ! d)
21
21
œ "3 '0 4 #" (3) 8‘ d) œ 3" '0 5# d) œ 56 (21) œ 531
œ
"
3
52. V œ 4 '0
1Î2
'01Î4 'sec2 sec9 9 3# sin 9 d3 d9 d)
œ
4
3
'01Î2 '01Î4 a8 sec$ 9 sec$ 9b sin 9 d9 d)
'01Î2 '01Î4 sec$ 9 sin 9 d9 d) œ 283 '01Î2 '01Î4 tan 9 sec# 9 d9 d) œ 283 '01Î2 2" tan# 9‘ !1Î% d)
1Î2
'
œ 14
d) œ 731
3
0
œ
28
3
53. V œ 4 '0
'01 '0r
54. V œ 4 '0
'01 'r r 1 dz r dr d) œ 4 '01Î2 '01 r dr d) œ 2 '01Î2 d) œ 1
55. V œ 8 '0
'1 2 '0r
56. V œ 8 '0
'1 2 '0
1Î2
1Î2
1Î2
œ
8
3
'0
dz r dr d) œ 4 '0
1Î2
'01 r$ dr d) œ '01Î2 d) œ 1#
#
#
1Î2
1Î2
#
È
1Î2
È2r
È
d) œ
dz r dr d) œ 8 '0
#
È2
'1
1Î2
dz r dr d) œ 8 '0
r# dr d) œ 8 Š 2
È2
'1
È2"
‹
3
'01Î2 d) œ 41 Š2 3 2"‹
È
1Î2
rÈ2 r# dr d) œ 8 '0 ’ "3 a2 r# b
$Î#
“
È#
1
d)
41
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
$
Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates
57. V œ '0
'02 '04cr sin
58. V œ '0
'02 '04cr cos cr sin
21
21
)
dz r dr d) œ '0
21
)
)
'02 a4r r# sin )b dr d) œ 8 '02 ˆ1 sin3 ) ‰ d) œ 161
1
dz r dr d) œ '0
21
'02 c4r r# (cos ) sin ))d dr d) œ 83 '02
1
(3 cos ) sin )) d) œ 161
59. The paraboloids intersect when 4x# 4y# œ 5 x# y# Ê x# y# œ 1 and z œ 4
Ê V œ 4 '0
1Î2
'01 '4r5r
#
#
1Î2
dz r dr d) œ 4 '0
'01 a5r 5r$ b dr d) œ 20 '01Î2 ’ r2
#
1Î2
r4 “ d) œ 5'0
%
"
!
60. The paraboloid intersects the xy-plane when 9 x# y# œ 0 Ê x# y# œ 9 Ê
V œ 4 '0
1Î2
'13 '09r
#
dz r dr d) œ 4 '0
1Î2
'13 a9r r$ b dr d) œ 4 '01Î2 ’ 9r2
dz r dr d) œ 8 '0
'01 r a4 r# b"Î# dr d) œ 8 '021 ’ "3 a4 r# b$Î# “ " d)
#
%
œ 64 '0 d) œ 321
1Î2
61. V œ 8 '0
21
È4cr
'01 '0
'0
21
œ 83
#
21
ˆ3$Î# 8‰ d) œ
1Î2
r4 “ d) œ 4 '0 ˆ 81
4
$
"
d) œ
17 ‰
4
51
#
d)
!
41 Š8 3È3‹
3
62. The sphere and paraboloid intersect when x# y# z# œ 2 and z œ x# y# Ê z# z 2 œ 0
Ê (z 2)(z 1) œ 0 Ê z œ 1 or z œ 2 Ê z œ 1 since z 0. Thus, x# y# œ 1 and the volume is
given by the triple integral V œ 4 '0
1Î2
œ 4 '0 ’ "3 a2 r# b
1Î2
63. average œ
"
64. average œ
œ
3
21
ˆ 431 ‰
È2r
#
#
1Î2
dz r dr d) œ 4 '0
r4 “ d) œ 4 '0 Š 2 3 2
$Î#
%
"
1Î2
È
!
'021 '01 'c11
"
#1
'01 'r
r# dz dr d) œ
È
'021 '01 'cÈ11ccrr
#
#
"
#1
7
12 ‹
d) œ
'01 ’r a2 r# b"Î# r$ “ dr d)
1 Š8È2 7‹
6
'021 '01 2r# dr d) œ 3"1 '021 d) œ 23
r# dz dr d) œ
3
41
'021 '01 2r# È1 r# dr d)
'021 ’ 8" sin" r 8" rÈ1 r# a1 2r# b“ " d) œ 1631 '021 ˆ 1# 0‰ d) œ 323 '021 d) œ ˆ 323 ‰ (21) œ 3161
!
65. average œ
"
ˆ 431 ‰
'021 '01 '01 3$ sin 9 d3 d9 d) œ 1631 '021 '01 sin 9 d9 d) œ 831 '021 d) œ 43
66. average œ
"
ˆ 231 ‰
'021 '01Î2 '01 3$ cos 9 sin 9 d3 d9 d) œ 831 '021 '01Î2
œ
3
161
cos 9 sin 9 d9 d) œ
3
81
'021 ’ sin2 9 “ 1Î# d)
'021 d) œ ˆ 1631 ‰ (21) œ 38
#
!
67. M œ 4 '0
'01 '0r dz r dr d) œ 4 '01Î2 '01 r# dr d) œ 43 '01Î2 d) œ 231 ; Mxy œ '021 '01 '0r z dz r dr d)
21
1
21
M
œ "# '0 '0 r$ dr d) œ 18 '0 d) œ 14 Ê z œ M œ ˆ 14 ‰ ˆ 231 ‰ œ 38 , and x œ y œ 0, by symmetry
1Î2
xy
68. M œ '0
'02 '0r dz r dr d) œ '01Î2 '02 r# dr d) œ 83 '01Î2 d) œ 431 ; Myz œ '01Î2 '02 '0r x dz r dr d)
1Î2
1Î2
1Î2
1Î2
2
2
r
2
œ '0 '0 r$ cos ) dr d) œ 4 '0 cos ) d) œ 4; Mxz œ '0 '0 '0 y dz r dr d) œ '0 '0 r$ sin ) dr d)
1Î2
1Î2
1Î2
1Î2
2
r
2
M
œ 4 '0 sin ) d) œ 4; Mxy œ '0 '0 '0 z dz r dr d) œ "# '0 '0 r$ dr d) œ 2 '0 d) œ 1 Ê x œ M
1Î2
yz
yœ
Mxz
M
œ
3
1
, and z œ
Mxy
M
œ
3
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
3
1
,
977
978
Chapter 15 Multiple Integrals
; Mxy œ '0
21
81
3
69. M œ
œ 4 '0 ’ sin2 9 “
21
#
'11ÎÎ32 '02 z3# sin 9 d3 d9 d) œ '021 '11ÎÎ32 '02 3$ cos 9 sin 9 d3 d9 d) œ 4 '021 '11ÎÎ32
d) œ 4 '0 ˆ "# 38 ‰ d) œ
1Î#
21
1Î$
"
#
'021 d) œ 1
Ê zœ
Mxy
M
œ (1) ˆ 831 ‰ œ
3
8
cos 9 sin 9 d9 d)
, and x œ y œ 0,
by symmetry
70. M œ '0
'01Î4 '0a 3# sin 9 d3 d9 d) œ a3 '021 '01Î4 sin 9 d9 d) œ a3 '021 2 #È2 d) œ 1a Š23 2‹ ;
1Î4
1Î4
21
a
21
21
a '
Mxy œ '0 '0 '0 3$ sin 9 cos 9 d3 d9 d) œ a4 '0 '0 sin 9 cos 9 d9 d) œ 16
d) œ 18a
0
21
$
$
$
%
Mxy
M
Ê zœ
%
È2
‰ 2
œ Š 18a ‹ – $ 3 È — œ ˆ 3a
8 Š #
1a Š2 2‹
%
‹œ
3 Š2È2‹ a
16
%
, and x œ y œ 0, by symmetry
È
71. M œ '0
È
È
'04 '0 r dz r dr d) œ '021 '04 r$Î# dr d) œ 645 '021 d) œ 1285 1 ; Mxy œ '021 '04 '0 r z dz r dr d)
21
4
'021 d) œ 6431 Ê z œ MM œ 65 , and x œ y œ 0, by symmetry
œ "# '0 '0 r# dr d) œ 32
3
21
xy
1Î3
È1r
1Î3
1Î3
72. M œ 'c1Î3 '0 'È1r# dz r dr d) œ '1Î3 '0 2rÈ1 r# dr d) œ '1Î3 ’ 23 a1 r# b
1
#
1
$Î# "
“ d)
!
È1r
1Î3
1
œ 23 'c1Î3 d) œ ˆ 23 ‰ ˆ 231 ‰ œ 491 ; Myz œ '1Î3 '0 'È1r r# cos ) dz dr d) œ 2 '1Î3 '0 r# È1 r# cos ) dr d)
1Î3
1Î3
#
1
#
1Î3
œ 2 'c1Î3 ’ 18 sin" r "8 rÈ1 r# a1 2r# b“ cos ) d) œ
"
!
Myz
M
Ê xœ
œ
9È 3
32
1
8
'11ÎÎ33 cos ) d) œ 18 csin )d 1Î13Î3 œ ˆ 18 ‰ Š2 † È#3 ‹ œ 1È8 3
, and y œ z œ 0, by symmetry
73. Iz œ '0
'12 '04 ax# y# b dz r dr d) œ 4 '021 '12 r$ dr d) œ '021 15 d) œ 301; M œ '021 '12 '04 dz r dr d)
21
2
21
I
œ '0 '1 4r dr d) œ '0 6 d) œ 121 Ê Rz œ É M
œ É #5
21
z
74. (a) Iz œ '0
'01 'c11 r$ dz dr d) œ 2 '021 '01 r$ dr d) œ "# '021 d) œ 1
21
1
1
21
1
21
Ix œ '0 '0 'c1 ar# sin# ) z# b dz r dr d) œ 2 '0 '0 ˆ2r$ sin# ) 2r3 ‰ dr d) œ '0 Š sin# ) "3 ‹ d)
21
(b)
#
œ 4)
sin 2)
8
#1
3) ‘ ! œ
1
#
21
3
œ
71
6
75. We orient the cone with its vertex at the origin and axis along the z-axis Ê 9 œ
which is through the vertex and parallel to the base of the cone Ê Ix œ '0
21
œ '0
%
76. Iz œ '0
21
œ 2 '0
'0a
21
77. Iz œ '0
21
21
. We use the the x-axis
1
'01 Šr$ sin# ) r% sin# ) 3r r3 ‹ dr d) œ '021 Š sin20 ) 10" ‹ d) œ 40) sin802) 10) ‘ #!1 œ #10 15 œ 14
21
œ '0
1
4
'0 'r ar# sin# ) z# b dz r dr d)
1
2
15
Èa cr
#
'
#
cÈa# cr#
r$ dz dr d) œ '0
21
a& d) œ
81 a
15
#
'0a 2r$ Èa# r# dr d) œ 2 '021 ’Š r5
#
ˆh‰
a
%
a
r&
5a “ !
2a#
#
15 ‹ aa
r# b
$Î# a
“ d)
&
'0a ' h r ax# y# b dz r dr d) œ '021 '0a
h ’ r4
d) œ '0 h Š a4
21
%
a&
5a ‹
h
'
hr
a
d) œ
' h r$ dz dr d) œ '021 '0a
hr
a
ha%
20
'021 d) œ 110ha
Šhr$
hr%
a ‹
dr d)
%
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
!
Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates
78. (a) M œ '0
'01 '0r z dz r dr d) œ '021 '01 "# r& dr d) œ 12" '021 d) œ 16 ; Mxy œ '021 '01 '0r
21
1
21
" '
1
œ "3 '0 '0 r( dr d) œ 24
d) œ 12
Ê z œ #" , and x œ y œ 0, by symmetry;
0
21
Iz œ '0
21
#
'01 '0r
(b) M œ '0
21
#
'01 '0r
zr$ dz dr d) œ
'021 '01
"
#
r# dz dr d) œ '0
21
#
r( dr d) œ
'021 d) œ 18
"
16
#
z# dz r dr d)
È3
#
Iz
Ê Rz œ É M
œ
'01 r% dr d) œ "5 '021 d) œ 215 ; Mxy œ '021 '01 '0r
#
zr# dz dr d)
'021 '01 r' dr d) œ 14" '021 d) œ 17 Ê z œ 145 , and x œ y œ 0, by symmetry; Iz œ '021 '01 '0r
21
1
21
I
œ '0 '0 r' dr d) œ "7 '0 d) œ 271 Ê Rz œ É M
œ É 57
"
2
œ
979
#
r% dz dr d)
z
79. (a) M œ '0
'01 'r 1 z dz r dr d) œ "# '021 '01 ar r$ b dr d) œ 8" '021 d) œ 14 ; Mxy œ '021 '01 'r 1 z# dz r dr d)
21
1
21
21
1
1
" '
œ 3" '0 '0 ar r% b dr d) œ 10
d) œ 15 Ê z œ 45 , and x œ y œ 0, by symmetry; Iz œ '0 '0 'r zr$ dz dr d)
0
21
1
21
I
1
" '
œ "# '0 '0 ar$ r& b dr d) œ 24
d) œ 12
Ê Rz œ É M
œ É 3"
0
21
z
(b) M œ '0
'01 'r 1 z# dz r dr d) œ 15 from part (a); Mxy œ '021 '01 'r 1 z$ dz r dr d) œ 4" '021 '01 ar r& b dr d)
21
21
1
1
21
1
" '
1
5
" ' '
# $
'
'
'
œ 12
d
)
œ
Ê
z
œ
,
and
x
œ
y
œ
0,
by
symmetry;
I
z
r
dz
dr
d
)
œ
ar$ r' b dr d)
z œ
6
6
3
0
0
0
r
0
0
21
I
5
" '
1
œ 28
d) œ 14
Ê Rz œ É M
œ É 14
0
21
z
80. (a) M œ '0
'01 '0a 3% sin 9 d3 d9 d) œ a5 '021 '01 sin 9 d9 d) œ 2a5 '021 d) œ 415a ;
1
1
1
21
a
21
21
Iz œ '0 '0 '0 3' sin$ 9 d3 d9 d) œ a7 '0 '0 a1 cos# 9b sin 9 d9 d) œ a7 '0 ’ cos 9 cos3 9 “ d)
21
&
&
&
(
'0
21
4a(
#1
œ
$
(
!
8a( 1
21
d) œ
Ê Rz œ
œ
Iz
ÉM
(b) M œ '0
É 10
21
a
'01 '0a 3$ sin# 9 d3 d9 d) œ a4 '021 '01 (1cos# 29) d9 d) œ 18a '021 d) œ 1 4a ;
1
1
21
a
21
Iz œ '0 '0 '0 3& sin% 9 d3 d9 d) œ a6 '0 '0 sin% 9 d9 d)
1
1
21
21
21
1
œ a6 '0 Š’ sin 49 cos 9 “ 34 '0 sin# 9 d9‹ d) œ a8 '0 9# sin429 ‘ ! d) œ 116a '0 d)
21
%
%
# %
'
$
'
'
'
!
a' 1 #
8
œ
81. M œ '0
21
Ê Rz œ
'0a '0
h
a
Èa cr
#
#
Iz
ÉM
œ
a
È2
dz r dr d) œ '0
21
'0a
rÈa# r# dr d) œ
h
a
Èa cr
'021 ’ 3" aa# r# b$Î# “ a d)
h
a
!
'0 3 d) œ
'0 '0 aa# r r$ b dr d)
; Mxy œ '0 '0 '0
z dz r dr d) œ
21
h '
œ 2a
Š a# a4 ‹ d) œ a h4 1 Ê z œ Š 1a4h ‹ ˆ 2ha3 1 ‰ œ 38 h, and x œ y œ 0, by symmetry
0
œ
21 $
a
h
a
21
2ha# 1
3
#
%
%
h
a
a
#
# #
#
21
h#
2a#
a
# #
#
#
82. Let the base radius of the cone be a and the height h, and place the cone's axis of symmetry along the z-axis
with the vertex at the origin. Then M œ
œ
h#
#
'021 ’ r2
#
a
r%
4a# “ !
d) œ
h#
#
'021 Š a#
#
1a# h
3
and Mxy œ '0
a#
4‹
d) œ
21
h # a#
8
'0a ' h r z dz r dr d) œ "# '021 '0a Šh# r ha
21
Mxy œ '0
'0a '0h (z 1) dz r dr d) œ '021 '0a
'021 d) œ h a4 1
# #
21
'0a '0h
Ê z œ ’ 1a
#
az# zb dz r dr d) œ '0
21
a2h$ 3h# b
“ ’ 1a# ah2# 2hb “
6
œ
#
Š h# h‹ r dr d) œ
'0a
2h# 3h
3h 6
$
Š h3
h#
# ‹r
r$ ‹ dr d)
a
Ê zœ
Mxy
M
# #
œ Š h a4 1 ‹ ˆ 1a3# h ‰ œ
x œ y œ 0, by symmetry Ê the centroid is one fourth of the way from the base to the vertex
83. M œ '0
#
#
ˆh‰
a# ah# 2hb
4
dr d) œ
'021 d) œ 1a ah# 2hb ;
a# a2h$ 3h# b
12
#
#
'021 d) œ 1a a2h6 3h b
#
$
, and x œ y œ 0, by symmetry;
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
#
3
4
h, and
980
Chapter 15 Multiple Integrals
Iz œ '0
21
'0a '0h
(z 1)r$ dz dr d) œ Š h
Iz
Rz œ É M
œ É 1a
%
ah# 2hb
4
†
2
1a# ah# 2hb
#
2h
# ‹
œ
'021 '0a r$ dr d) œ Š h # 2h ‹ Š a4 ‹'021 d) œ 1a ah4 2hb ;
#
%
%
#
a
È2
84. The mass of the plant's atmosphere to an altitude h above the surface of the planet is the triple integral
M(h) œ '0
'01 'Rh .! ecÐ3RÑ 3# sin 9 d3 d9 d) œ 'Rh '021 '01 .! ecÐ3RÑ 3# sin 9 d9 d) d3
h
21
h
21
h
1
œ 'R '0 .! ecÐ3RÑ 3# ( cos 9)‘ ! d) d3 œ 2 'R '0 .! ecR ec3 3# d) d3 œ 41.! ecR 'R ec3 3# d3
21
c3
#
œ 41.! ecR ’ 3 ec
œ 41.! ecR Š
h ech
#
c
23ec3
c#
2hech
c#
h
2ec3
c$ “ R
2ech
c$
(by parts)
R# ecR
c
2RecR
c#
2ecR
c$ ‹ .
The mass of the planet's atmosphere is therefore M œ lim
hÄ_
#
M(h) œ 41.! Š Rc
2R
c#
2
c$ ‹ .
85. The density distribution function is linear so it has the form $ (3) œ k3 C, where 3 is the distance from the
center of the planet. Now, $ (R) œ 0 Ê kR C œ 0, and $ (3) œ k3 kR. It remains to determine the constant
k: M œ '0
21
'01 '0R (k3 kR) 3# sin 9 d3 d9 d) œ '021 '01 ’k 34
%
1
21
œ '0 '0 k Š R4
%
R%
3 ‹
Ê $ (3) œ 13M
3
R%
$
R
kR 33 “ sin 9 d9 d)
!
21
21
sin 9 d9 d) œ '0 1k# R% c cos 9d 1! d) œ '0 6k R% d) œ k13R
3M
1R%
‰R œ
R . At the center of the planet 3 œ 0 Ê $ (0) œ ˆ 13M
R%
3M
1R$
%
Ê k œ 13M
R%
.
86. x2 y2 œ a2 Ê a3 sin 9 cos )b2 a3 sin 9 sin )b2 œ a2 Ê a32 sin2 9bacos2 ) sin2 )b œ a2 Ê 32 sin2 9 œ a2
Ê 3 sin 9 œ a or 3 sin 9 œ a Ê 3 sin 9 œ a or 3 œ a csc 9, since 0 Ÿ 9 Ÿ 1 and 3 0.
87. (a) A plane perpendicular to the x-axis has the form x œ a in rectangular coordinates Ê r cos ) œ a Ê r œ
Ê r œ a sec ), in cylindrical coordinates.
(b) A plane perpendicular to the y-axis has the form y œ b in rectangular coordinates Ê r sin ) œ b Ê r œ
Ê r œ b csc ), in cylindrical coordinates.
88. ax by œ c Ê aar cos )b bar sin )b œ c Ê raa cos ) b sin )b œ c Ê r œ
a
cos )
b
sin )
c
a cos ) b sin ) .
89. The equation r œ fazb implies that the point ar, ), zb
œ afazb, ), zb will lie on the surface for all ). In particular
afazb, ) 1, zb lies on the surface whenever afazb, ), zb does
Ê the surface is symmetric with respect to the z-axis.
90. The equation 3 œ fa9b implies that the point a3, 9, )b œ afa9b, 9, )b lies on the surface for all ). In particular, if
afa9b, 9, )b lies on the surface, then afa9b, 9, ) 1b lies on the surface, so the surface is symmetric wiith respect to the
z-axis.
15.7 SUBSTITUTIONS IN MULTIPLE INTEGRALS
1. (a) x y œ u and 2x y œ v Ê 3x œ u v and y œ x u Ê x œ
` (xßy)
` (ußv)
œ»
"
3
23
"
3
"
3
»œ
"
9
2
9
œ
"
3
(u v) and y œ
"
3
(2u v);
"
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates
(b) The line segment y œ x from (!ß 0) to (1ß 1) is x y œ 0
Ê u œ 0; the line segment y œ 2x from (0ß 0) to
(1ß 2) is 2x y œ 0 Ê v œ 0; the line segment x œ 1
from (1ß 1) to ("ß 2) is (x y) (2x y) œ 3
Ê u v œ 3. The transformed region is sketched at the
right.
2. (a) x 2y œ u and x y œ v Ê 3y œ u v and x œ v y Ê y œ
` (xßy)
` (ußv)
œ»
"
3
1
3
2
3
"3
"
»œ9
2
9
"
3
(u v) and x œ
"
3
(u 2v);
œ 3"
(b) The triangular region in the xy-plane has vertices (0ß 0),
(2ß 0), and ˆ 23 ß 23 ‰ . The line segment y œ x from (0ß 0)
to ˆ 23 ß 23 ‰ is x y œ 0 Ê v œ 0; the line segment
y œ 0 from (0ß 0) to (#ß 0) Ê u œ v; the line segment
x 2y œ 2 from ˆ 23 ß 23 ‰ to (2ß 0) Ê u œ 2. The
transformed region is sketched at the right.
3. (a) 3x 2y œ u and x 4y œ v Ê 5x œ 2u v and y œ
` (xßy)
` (ußv)
œ»
2
5
1
10
15
3
10
»œ
6
50
1
50
œ
"
#
(u 3x) Ê x œ
"
5
(2u v) and y œ
"
10
"
10
(b) The x-axis y œ 0 Ê u œ 3v; the y-axis x œ 0
Ê v œ 2u; the line x y œ 1
"
Ê "5 (2u v) 10
(3v u) œ 1
Ê 2(2u v) (3v u) œ 10 Ê 3u v œ 10. The
transformed region is sketched at the right.
4. (a) 2x 3y œ u and x y œ v Ê x œ u 3v and y œ v x Ê x œ u 3v and y œ u 2v;
" 3
` (xßy)
` (ußv) œ º 1 2 º œ 2 3 œ 1
(b) The line x œ 3 Ê u 3v œ 3 or u 3v œ 3;
x œ 0 Ê u 3v œ 0; y œ x Ê v œ 0; y œ x 1
Ê v œ 1. The transformed region is the parallelogram
sketched at the right.
5.
'04 'yÐÎy2Î2Ñ1 ˆx y# ‰ dx dy œ '04 ’ x2
#
œ
"
#
y
'0 (y 1 y) dy œ '0 dy œ
4
"
#
4
xy 2
# “y
"
dy œ
2
"
#
"
#
'04 ’ˆ y# 1‰# ˆ y# ‰# ˆ y# 1‰ y ˆ y# ‰ y“ dy
(4) œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
(3v u);
981
982
6.
Chapter 15 Multiple Integrals
' ' a2x# xy y# b dx dy œ ' ' (x y)(2x y) dx dy
R
ßy)
œ ' ' uv ¹ `` (x
(ußv) ¹ du dv œ
G
R
' ' uv du dv;
"
3
G
We find the boundaries of G from the boundaries of R,
shown in the accompanying figure:
xy-equations for
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
the boundary of R
y œ 2x 4
"
3
(2u v) œ (u v) 4
vœ4
y œ 2x 7
"
3
(2u v) œ 23 (u v) 7
vœ7
yœx2
"
3
uœ2
yœx1
"
3
Ê
7.
2
3
(2u v) œ
1
3
(u v) 2
(2u v) œ
1
3
(u v) 1
u œ 1
' ' uv du dv œ 3" ' ' uv dv du œ 3" ' u ’ v2# “ du œ
c1 4
c1
2
"
3
7
(
2
11
#
%
G
'c21 u du œ ˆ 11# ‰ ’ u2 “ #
#
"
‰
œ ˆ 11
4 (4 1) œ
33
4
' ' a3x# 14xy 8y# b dx dy
R
œ ' ' (3x 2y)(x 4y) dx dy
R
ßy)
œ ' ' uv ¹ `` (x
(ußv) ¹ du dv œ
G
"
10
' ' uv du dv;
G
We find the boundaries of G from the boundaries of R,
shown in the accompanying figure:
xy-equations for
the boundary of R
Simplified
for the boundary of G
uv-equations
3
#
yœ x1
"
10
(3v u) œ
(2u v) 1
uœ2
y œ #3 x 3
"
10
3
(3v u) œ 10
(2u v) 3
uœ6
"
4
yœ x
"
10
(3v u) œ
vœ0
y œ "4 x 1
"
10
1
(3v u) œ 20
(2u v) 1
Ê
8.
Corresponding uv-equations
"
10
' ' uv du dv œ
"
10
G
3
10
1
20
(2u v)
vœ4
'26 '04 uv dv du œ 10" '26 u ’ v2 “ % du œ 45 '26 u du œ ˆ 45 ‰ ’ u2 “ ' œ ˆ 54 ‰ (18 2) œ 645
#
#
!
#
' ' 2(x y) dx dy œ ' ' 2v ¹ `` (x(ußßy)v) ¹ du dv œ ' ' 2v du dv; the region G is sketched in Exercise 4
R
G
G
3c3v
"
Ê ' ' 2v du dv œ '0 'c3v 2v du dv œ '0 2v(3 3v 3v) dv œ '0 6v dv œ c3v# d ! œ 3
1
1
1
G
9. x œ
u
v
v" uv#
œ v" u v" u œ 2u
v ;
v
u º
Ê v œ 1, and y œ 4x Ê v œ 2; xy œ 1 Ê u œ 1, and xy œ 9 Ê u œ 3; thus
and y œ uv Ê
y œ x Ê uv œ
u
v
y
x
œ v# and xy œ u# ;
` (xßy)
` (ußv)
œ J(uß v) œ º
' ' ŠÉ yx Èxy‹ dx dy œ ' ' (v u) ˆ 2uv ‰ dv du œ ' ' Š2u 2uv # ‹ dv du œ ' c2uv 2u# ln vd #" du
1
1
1
1
1
3
2
3
2
3
R
œ '1 a2u 2u# ln 2b du œ u# 23 u# ln 2‘ " œ 8 23 (26)(ln 2) œ 8
3
$
52
3
(ln 2)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates
` (xßy)
` (ußv)
10. (a)
œ J(uß v) œ º
" 0
œ u, and
v uº
the region G is sketched at the right
(b) x œ 1 Ê u œ 1, and x œ 2 Ê u œ 2; y œ 1 Ê uv œ 1 Ê v œ "u , and y œ 2 Ê uv œ 2 Ê v œ
'1 '1
2
œ
2
3
#
y
x
dy dx œ '1
2
'1Îu ˆ uvu ‰ u dv du œ '1 '1Îu uv dv du œ '1
2Îu
1Îu
du œ '1 u ˆ u2#
" ‰
2u#
; thus,
du
1
R
` (xßy)
` (ußv)
u ’ v2 “
2
2
21
12.
2Îu
#
#
I! œ ' ' ax# y# b dA œ '0
œ
2
'12 u ˆ u" ‰ du œ 3# cln ud #" œ 3# ln 2; '12 '12 yx dy dx œ '12 ’ x1 † y2 “ 2 dx œ 3# '12 dxx œ 3# cln xd #" œ 3# ln 2
11. x œ ar cos ) and y œ ar sin ) Ê
ab
4
2Îu
2
2
u
` (xßy)
` (rß))
œ J(rß )) œ º
ar sin )
œ abr cos# ) abr sin# ) œ abr;
br cos ) º
'01 r# aa# cos# ) b# sin# )b kJ(rß ))k dr d) œ '021 '01 abr$ aa# cos# ) b# sin# )b dr d)
'021 aa# cos# ) b# sin# )b d) œ ab4 ’ a2) a
#
œ J(uß v) œ º
a cos )
b sin )
#
sin 2)
4
b# )
2
21
b# sin 2)
“
4
!
œ
ab1 aa# b# b
4
È1cu#
1
a 0
œ ab; A œ ' ' dy dx œ ' ' ab du dv œ 'c1 'cÈ1cu# ab dv du
º
0 b
R
G
œ 2ab 'c1 È1u# du œ 2ab ’ u2 È1 u#
1
"
#
sin" u“
"
"
œ ab csin" 1 sin" (1)d œ ab 1# ˆ 1# ‰‘ œ ab1
13. The region of integration R in the xy-plane is
sketched in the figure at the right. The
boundaries of the image G are obtained as
follows, with G sketched at the right:
xy-equations for
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
the boundary of R
xœy
"
3
(u 2v) œ
x œ 2 2y
"
3
(u 2v) œ 2 23 (u v)
yœ0
0œ
Also, from Exercise 2,
` (xßy)
` (ußv)
1
3
1
3
(u v)
(u v)
œ J(uß v) œ "3 Ê
vœ0
uœ2
vœu
'02Î3 'y22y (x 2y) eÐyxÑ dx dy œ '02 '0u uev ¸ 3" ¸ dv du
œ
"
3
'02 u cecv d !u du œ 3" '02 u a1 ecu b du œ 3" ’u au ecu b u#
œ
"
3
a3ec2 1b ¸ 0.4687
#
#
ecu “ œ
!
"
3
c2 a2 ec2 b 2 ec2 1d
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
983
984
Chapter 15 Multiple Integrals
14. x œ u
` (xßy)
` (ußv)
v
#
and y œ v Ê 2x y œ (2u v) v œ 2u and
" "#
v
º œ 1; next, u œ x #
0 "
and v œ y, so the boundaries of the region of
œ J(uß v) œ º
œx
y
#
integration R in the xy-plane are transformed to the
boundaries of G:
xy-equations for
Corresponding uv-equations
Simplified
for the boundary of G
uv-equations
œ
uœ0
the boundary of R
xœ
xœ
u
y
#
y
#
2
u
yœ0
œ
2
uœ2
vœ0
vœ2
2
"
4
œ
v
#
v
#
vœ0
yœ2
Ê '0
v
#
v
#
ÐyÎ2Ñ2
'yÎ2
ae
16
vœ2
y$ (2x y) eÐ2xyÑ dx dy œ '0
2
#
% #
1b ’ v4 “
!
'0
2
v$ (2u) e4u du dv œ '0 v$ ’ 4" e4u “ dv œ
2
#
#
#
!
"
4
'02 v$ ae16 1b dv
œe 1
16
15. (a) x œ u cos v and y œ u sin v Ê
` (xßy)
` (ußv)
ϼ
cos v u sin v
œ u cos# v u sin# v œ u
sin v
u cos v º
(b) x œ u sin v and y œ u cos v Ê
` (xßy)
` (ußv)
ϼ
sin v
u cos v
œ u sin# v u cos# v œ u
cos v u sin v º
â
â
â cos v u sin v 0 â
â
â
z)
u cos v 0 â œ u cos# v u sin# v œ u
œ â sin v
16. (a) x œ u cos v, y œ u sin v, z œ w Ê ``(u(xßßvyßßw)
â
â
0
"â
â 0
â
â
â2 0 0â
â
â
z)
(b) x œ 2u 1, y œ 3v 4, z œ "# (w 4) Ê ``(u(xßßvyßßw)
œ â 0 3 0 â œ (2)(3) ˆ #" ‰ œ 3
â
â
â 0 0 "# â
â
â sin 9 cos )
â
17. â sin 9 sin )
â
â cos 9
œ (cos 9) º
3 cos 9 cos )
3 cos 9 sin )
3 sin 9
3 cos 9 cos )
3 cos 9 sin )
â
3 sin 9 sin ) â
â
3 sin 9 cos ) â
â
0
â
3 sin 9 sin )
sin 9 cos )
(3 sin 9) º
3 sin 9 cos ) º
sin 9 sin )
3 sin 9 sin )
3 sin 9 cos ) º
œ a3# cos 9b asin 9 cos 9 cos# ) sin 9 cos 9 sin# )b a3# sin 9b asin# 9 cos# ) sin# 9 sin# )b
œ 3# sin 9 cos# 9 3# sin$ 9 œ a3# sin 9b acos# 9 sin# 9b œ 3# sin 9
18. Let u œ g(x) Ê J(x) œ
du
dx
œ gw (x) Ê 'a f(u) du œ 'g(a) f(g(x))gw (x) dx in accordance with Theorem 6 in
b
g(b)
Section 5.6. Note that gw (x) represents the Jacobian of the transformation u œ g(x) or x œ g" (u).
19.
'03 '04 'y1ÎÐ2 yÎ2Ñ ˆ 2x # y 3z ‰ dx dy dz œ '03 '04 ’ x2
#
œ '0 ’ (y 4 1)
3
#
y#
4
%
yz
3 “!
dz œ '0 ˆ 94
3
4z
3
xy
#
"ÐyÎ2Ñ
xz
3 “ yÎ2
"4 ‰ dz œ '0 ˆ2
3
dy dz œ '0
4z ‰
3
3
'04 "# (y 1) y# 3z ‘ dy dz
dz œ ’2z
$
2z#
3 “!
œ 12
â
â
âa 0 0â
#
#
#
â
â
20. J(uß vß w) œ â 0 b 0 â œ abc; the transformation takes the ellipsoid region xa# yb# zc# Ÿ 1 in xyz-space
â
â
â0 0 câ
into the spherical region u# v# w# Ÿ 1 in uvw-space ˆwhich has volume V œ 43 1‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 15 Practice Exercises
Ê V œ ' ' ' dx dy dz œ ' ' ' abc du dv dw œ
R
G
985
41abc
3
â
â
âa 0 0â
â
â
21. J(uß vß w) œ â 0 b 0 â œ abc; for R and G as in Exercise 19, ' ' ' kxyzk dx dy dz
â
â
R
â0 0 câ
œ ' ' ' a# b# c# uvw dw dv du œ 8a# b# c#
G
œ
4a# b# c#
3
'01Î2 '01Î2
'01Î2 '01Î2 '01 (3 sin 9 cos ))(3 sin 9 sin ))(3 cos 9) a3# sin 9b d3 d9 d)
a # b # c#
3
sin ) cos ) sin$ 9 cos 9 d9 d) œ
'01Î2 sin ) cos ) d) œ a b6 c
# # #
â 1
â
â
22. u œ x, v œ xy, and w œ 3z Ê x œ u, y œ vu , and z œ "3 w Ê J(uß vß w) œ â uv#
â
â 0
0
"
u
0
0 ââ
0 ââ œ
" â
â
3
"
3u
;
' ' ' ax# y 3xyzb dx dy dz œ ' ' ' u# ˆ vu ‰ 3u ˆ vu ‰ ˆ w3 ‰‘ kJ(uß vß w)k du dv dw œ "3 ' ' ' ˆv vw
‰ du dv dw
u
0
0
1
3
R
2
2
G
œ
"
3
'0 '0 (v vw ln 2) dv dw œ 3" '03 (1 w ln 2) ’ v2 “ # dw œ 32 '03 (1 w ln 2) dw œ 32 ’w w2
œ
2
3
ˆ3
3
2
#
#
!
9
#
ln 2“
$
!
ln 2‰ œ 2 3 ln 2 œ 2 ln 8
23. The first moment about the xy-coordinate plane for the semi-ellipsoid,
x#
a#
y#
b#
z#
c#
œ 1 using the
transformation in Exercise 21 is, Mxy œ ' ' ' z dz dy dx œ ' ' ' cw kJ(uß vß w)k du dv dw
D
œ abc#
G
' ' ' w du dv dw œ aabc# b † aMxy of the hemisphere x# y# z# œ 1, z
G
the mass of the semi-ellipsoid is
2abc1
3
#
3 ‰
Ê z œ Š abc4 1 ‹ ˆ 2abc
1 œ
3
8
0b œ
abc# 1
4
;
c
24. A solid of revolutions is symmetric about the axis of revolution, therefore, the height of the solid is solely a function of r.
That is, y œ faxb œ farb. Using cylindrical coordinates with x œ r cos ), y œ y and z œ r sin ), we have
V œ ' ' ' r dy d) dr œ 'a
G
b
'021 '0farb
r dy d) dr œ 'a
b
'021 c r y df0arb d) dr œ 'ab '021 r farb d) dr œ 'ab c r)farb d021 dr
'a 21rfarbdr. In the last integral, r is a dummy or stand-in variable and as such it can be replaced by any variable name.
b
Choosing x instead of r we have V œ 'a 21xfaxbdx, which is the same result obtained using the shell method.
b
CHAPTER 15 PRACTICE EXERCISES
1.
'110 '01Îyyexy dx dy œ '110 cexy d !"Îy dy
10
œ '1 (e 1) dy œ 9e 9
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
986
Chapter 15 Multiple Integrals
'01 '0x eyÎx dy dx œ '01 x eyÎx ‘ !x
$
2.
$
dx
œ '0 Šxex x‹ dx œ ’ "2 ex
1
3.
#
È
"
x#
# “!
#
'03Î2 'È994t4t
3Î2
t ds dt œ '0
#
#
È9 4t
œ
#
ctsd È # dt
9 4t
œ '0 2tÈ9 4t# dt œ ’ "6 a9 4t# b
3Î2
"
6
œ ˆ0$Î# 9$Î# ‰ œ
4.
27
6
œ
#
Èy
"
#
“
!
dy
'01 y ˆ4 4Èy y y‰ dy
œ '0 ˆ2y 2y$Î# ‰ dy œ ’y#
1
'c02 '2x4 cb x4
#
5.
$Î# $Î#
9
#
'01 'È2cy Èy xy dx dy œ '01 y ’ x2 “ 2cÈy
œ
e2
#
"
4y&Î#
5 “!
"
5
œ
dy dx œ 'c2 ax# 2xb dx
0
$
œ ’ x3 x# “
!
#
œ ˆ 83 4‰ œ
4
3
'04 'c(Èy c4 c4)/2y dx dy œ '04 ˆ y c2 4 È4 y‰ dy
4
2
œ ’ y2 2y 23 a4 yb3/2 “ œ 4 8
0
œ 4
6.
œ
16
3
2
3
† 43/2
4
3
'01 'yÈy Èx dx dy œ '01 23 x$Î# ‘ yÈy dy
œ
œ
2
3
2
3
'01 ˆy$Î% y$Î# ‰ dy œ 23 47 y(Î% 25 y&Î# ‘ "!
ˆ 47 25 ‰ œ
4
35
'01 'xx Èx dy dx œ '01 x1/2 ax x2 b dx œ '01 ˆx3/2 x5/2 ‰ dx
2
1
œ 25 x5/2 27 x7/2 ‘0 œ
7.
È9 x
'c33 '0Ð1Î2Ñ
#
2
5
œ
2
7
y dy dx œ 'c3 ’ y2 “
3
#
œ 'c3 "8 a9 x# b dx œ ’ 9x
8
3
œ ˆ 27
8
È
27 ‰
24
'03Î2 'È994y4y
#
#
4
35
ˆ 27
8
27 ‰
24
3Î2
È
Ð1Î2Ñ 9 x#
!
dx
$
x$
24 “ $
œ
27
6
œ
9
#
y dx dy œ '0 2yÈ94y# dy
œ "4 † 23 a94y# b3/2 º
3/2
œ
0
"
6
† 93/2 œ
27
6
œ
9
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 15 Practice Exercises
'02 '04 x 2x dy dx œ '02 c2xyd 04 x dx
2
2
œ '0 a2xa4 x2 bb dx œ '0 a8x 2x3 b dx
2
8.
2
2
x4
2 “!
œ ’4x2
È4 c y
'04 '0
œ 16
œ8
16
2
È4 c y
2x dx dy œ '0 cx2 d 0
4
œ '0 a4 yb dy œ ’ 4y
4
dy
4
y2
2
“ œ 16
16
2
0
œ8
9.
'01 '2y2 4 cos ax# b dx dy œ '02 '0x/2 4 cos ax# b dy dx œ '02 2x cos ax# b dx œ csin ax# bd #! œ sin 4
10.
'02 'y1Î2 ex
11.
'08 'È2x
y%
12.
'01 'È1y
21 sin a1x# b
x#
$
$
#
dx dy œ '0
1
"
1
'02x ex
dy dx œ '0
2
'0y
$
1
"
y% 1
dx dy œ '0
1
4 c x#
dy dx œ '0 2xex dx œ cex d ! œ e 1
#
#
dx dy œ
'02 y 4y 1 dy œ ln417
$
"
4
%
'0x 21 sinx a1x b dy dx œ '01 21x sin a1x# b dx œ c cos a1x# bd "! œ (1) (1) œ 2
$
#
#
13. A œ 'c2 '2x b 4 dy dx œ 'c2 ax# 2xb dx œ
0
15. V œ '0
1
"
#
0
14. A œ '1
4
4
3
'x2 c x ax# y# b dy dx œ '01 ’x# y y3 “ 2cx dx œ '01 ’2x# (23x)
$
$
x
"
12
œ ˆ 23
7 ‰
12
6 c x#
16. V œ 'c3 'x
2
"
6 c x#
dx œ 'c3 a6x# x% x$ b dx œ
2
xy dy dx œ '0 ’ xy2 “ dx œ '0
1
#
"
1
!
È1 c x
ˆ1‰
4
'01
4
7x$
3 “
$
dx œ ’ 2x3
(2x)%
12
37
6
"
7x%
12 “ !
'01 '0
#
xy dy dx œ
4
1
'01 ’ xy2 “
#
x
2
dx œ
È1 c x
#
!
125
4
"
4
dx œ
2
1
'01 ax x$ b dx œ #"1
'22Îx dy dx œ '12 ˆ2 2x ‰ dx œ 2 ln 4; My œ '12 '22Îx x dy dx œ '12 x ˆ2 2x ‰ dx œ 1;
Mx œ '1
2
'22Îx
y dy dx œ '1 ˆ2
2
2y c y#
2‰
x#
dx œ 1 Ê x œ y œ
20. M œ '0 'c2y dx dy œ '0 a4y y# b dy œ
4
4
2y c y
32
3
#
4
4
# #
4
"
# ln 4
2y c y#
; Mx œ '0 'c2y y dx dy œ '0 a4y# y$ b dy œ ’ 4y3
y
My œ '0 'c2y x dx dy œ '0 ’ a2y#y b 2y# “ dy œ ’ 10
2
dx dy œ '1 ˆÈy 2 y‰ dy œ
4
3
2
18. average value œ
2
œ
x# dy dx œ 'c3 cx# yd x
1
19. M œ '1
2%
12
17. average value œ '0
21. Io œ '0
'2Ècyy
&
%
y%
2 “!
4
$
œ 128
5 Ê xœ
My
M
œ 12
5 and y œ
%
y%
4 “!
Mx
M
œ
œ2
'2x4 ax# y# b (3) dy dx œ 3 '02 Š4x# 643 14x3 ‹ dx œ 104
$
'c22 'c11 ax# y# b dy dx œ 'c22 ˆ2x# 23 ‰ dx œ 403
a
b
a
'b'a
Ix œ 'ca 'cb y# dy dx œ 'ca 2b3 dx œ 4ab
3 ; Iy œ cb ca
22. (a) Io œ
(b)
$
œ
4ab
3
$
$
4a b
3
œ
#
#
4ab ab a b
3
$
x# dx dy œ 'cb
b
2a$
3
dy œ
4a$ b
3
Ê Io œ Ix Iy
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
64
3
;
987
988
Chapter 15 Multiple Integrals
23. M œ $ '0
3
'02xÎ3 dy dx œ $ '03 2x3 dx œ 3$ ; Ix œ $ '03 '02xÎ3 y# dy dx œ 818$ '03 x$ dx œ ˆ 818$ ‰ Š 34 ‹ œ 2$
%
24. M œ '0
Ê Rx œ É 32
"3
'xx (x 1) dy dx œ '01 ax x$ b dx œ "4 ; Mx œ '01 'xx y(x 1) dy dx œ #" '01 ax$ x& x# x% b dx œ 120
;
1
x
1
1
x
2
8
13
My œ '0 'x x(x 1) dy dx œ '0 ax# x% b dx œ 15
Ê x œ 15
and y œ 30
; Ix œ '0 'x y# (x 1) dy dx
1
I
17
' 1' x #
'1 $ &
œ "3 '0 ax% x( x$ x' b dx œ 280
Ê Rx œ É M
œ É 17
70 ; Iy œ 0 x x (x 1) dy dx œ 0 ax x b dx
1
#
#
#
#
x
#
œ
Iy
ÉM
Ê Ry œ
1
12
œ É 13
25. M œ 'c1 'c1 ˆx# y# "3 ‰ dy dx œ 'c1 ˆ2x# 43 ‰ dx œ 4; Mx œ 'c1 'c1 y ˆx# y# "3 ‰ dy dx œ 'c1 0 dx œ 0;
1
1
1
My œ ' ' x ˆx# y# 3" ‰ dy dx œ ' ˆ2x$ 43 x‰ dx œ 0
1
1
1
c1 c1
1
1
1
c1
26. Place the ?ABC with its vertices at A(0ß 0), B(bß 0) and C(aß h). The line through the points A and C is
yœ
h
a
x; the line through the points C and B is y œ
œ b$ '0 ˆ1 yh ‰ dy œ
h
È
27.
'11 'È11xx
28.
'c11 'cÈ11ccyy
È
"
#
h
21
2
a1 x # y # b
(x b). Thus, M œ '0
h
'ayÐaÎh bÑyÎh b $ dx dy
'ayÐaÎh bÑyÎh b y# $ dx dy œ b$ '0h Šy# yh ‹ dy œ $1bh#
$
'01
2r
a1 r # b#
ln ax# y# 1b dx dy œ '0
21
#
#
œ
; Ix œ '0
dy dx œ '0
#
#
$ bh
#
h
ab
"
dr d) œ '0 1 " r# ‘ ! d) œ
21
'01 r ln ar# 1b dr d) œ '021 '12
"
#
"
#
$
Ix
; Rx œ É M
œ
h
È6
'021 d) œ 1
ln u du d) œ
"
#
'021 cu ln u ud #" d)
'021 (2 ln 2 1) d) œ [ln (4) 1] 1
1Î3
29. M œ 'c1Î3 '0 r dr d) œ
3
9
#
'11ÎÎ33 d) œ 31; My œ '11ÎÎ33 '03 r# cos ) dr d) œ 9 '11ÎÎ33 cos ) d) œ 9È3
Ê xœ
3È 3
1
and y œ 0 by symmetry
30. M œ '0
1Î2
yœ
13
31
'13 r dr d) œ 4 '01Î2 d) œ 21; My œ '01Î2 '13
26
3
'01Î2 cos ) d) œ 263
Ê xœ
by symmetry
31. (a) M œ 2 '0
1Î2
'11bcos
)
1Î2
1Î2
1cos )
My œ 'c1Î2 '1
1 cos 2) ‰
#
1Î2
Ê xœ
d) œ
81
4
;
(r cos )) r dr d)
œ 'c1Î2 Šcos# ) cos$ )
32 151
24
(b)
r dr d)
œ '0 ˆ2 cos )
œ
r# cos ) dr d) œ
151 32
61 48
cos% )
3 ‹
d)
, and
y œ 0 by symmetry
32. (a) M œ 'c! '0 r dr d) œ 'c!
d) œ a# !; My œ 'c! '0 (r cos )) r dr d) œ 'c!
!
2a sin !
Ê x œ 2a 3sin
œ0
! , and y œ 0 by symmetry; lim c x œ lim c
3!
!
a
!
!
a#
#
!Ä1
!
a
a$ cos )
3
d) œ
2a$ sin !
3
!Ä1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
13
31
, and
,
Chapter 15 Practice Exercises
(b) x œ
and y œ 0
2a
51
Ècos 2)
33. ax# y# b ax# y# b œ 0 Ê r% r# cos 2) œ 0 Ê r# œ cos 2) so the integral is 'c1Î4 '0
1Î4
#
1Î4
œ 'c1Î4 ’ 2 a1 " r# b “
œ
"
#
1Î4
"
y # b#
œ '0 ’ "#
1Î3
œ
"
#
dx dy œ '0
1Î3
"
# a1 sec# )b “
’ È"2 tan"
''
"
y # b#
a1 x #
R
œ '0
35.
1Î4
"
" '
"
'11ÎÎ44 ˆ1 1 cos
‰
ˆ
‰
2) d) œ # 1Î4 1 # cos ) d)
#
#
a1 x #
R
(b)
!
d) œ "#
dr d)
'c1Î4 Š1 sec# ) ‹ d) œ "# ) tan2 ) ‘ 1Î14Î4 œ 14 2
''
34. (a)
Ècos 2)
r
a1 r # b #
1Î2
lim
bÄ_
u
È2 “
È$
!
d) œ
È2
4
œ
dx dy œ '0
1Î2
’ "#
'0sec
)
dr d) œ '0
1Î3
r
a1 r# b#
'01Î3 1 secsec) ) d); ”
#
"
#
#
"
# a1 b# b “d)
dr d) œ '0
œ
'0
1Î2
"
#
Î
1 2
r
a1 r# b#
d) œ
sec )
!
d)
u œ tan )
Ä
du œ sec# ) d) •
tan" É 3#
'0_
’ 2 a1 " r# b “
"
#
È3
'0
du
2 u #
b
lim
bÄ_
’ 2 a1 " r# b “ d)
0
1
4
'01 '01 '01 cos (x y z) dx dy dz œ '01 '01 [sin (z y 1) sin (z y)] dy dz
1
œ '0 [ cos (z 21) cos (z 1) cos z cos (z 1)] dz œ 0
36.
'lnln67 '0ln 2 'lnln45 eÐxyzÑ dz dy dx œ 'lnln67 '0ln 2 eÐxyÑ dy dx œ 'lnln67 ex dx œ 1
37.
'01 '0x '0xby (2x y z) dz dy dx œ '01 '0x Š 3x#
38.
'1e '1x '0z 2yz dy dz dx œ '1e '1x "z dz dx œ '1e ln x dx œ cx ln x xd 1e œ 1
#
#
#
3y#
# ‹
dy dx œ '0 Š 3x#
1
%
x'
#‹
dx œ
8
35
$
39. V œ 2 '0
1Î2
40. V œ 4 '0
2
'0cos y '02x dz dx dy œ 2 '01Î2 '0cos y
È4cx
'0
œ ’x a4 x# b
41. average œ
"
3
#
'04cx
$Î#
#
dz dy dx œ 4 '0
2
È4cx
'0
#
1Î2
2x dx dy œ 2 '0 cos# y dy œ 2 ’ y2
a4 x# b dy dx œ 4 '0 a4 x# b
2
$Î#
1Î2
sin 2y
4 “!
œ
1
#
dx
#
6xÈ4 x# 24 sin" x2 “ œ 24 sin" 1 œ 121
!
'01 '03 '01
30xzÈx# y dz dy dx œ
"
3
'01 '03 15xÈx# y dy dx œ 3" '03 '01 15xÈx# y dx dy
œ
"
3
'03 ’5 ax# yb$Î# “ " dy œ "3 '03 5(1 y)$Î# 5y$Î# ‘ dy œ 3" 2(1 y)&Î# 2y&Î# ‘ $! œ 3" 2(4)&Î# 2(3)&Î# 2‘
œ
"
3
2 ˆ31 3&Î# ‰‘
!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
989
990
Chapter 15 Multiple Integrals
42. average œ
43. (a)
'021 '01 '0a
3
4 1 a$
È
È
È
3$ sin 9 d3 d9 d) œ
'021 '01 sin 9 d9 d) œ 83a1 '021 d) œ 3a4
3a
161
'cÈ22 'cÈ22ccyy 'Èx4bcyx cy 3 dz dx dy
#
#
(b)
'0 '0 '0
(c)
'021 '0 2 'r
21
#
#
1Î4
2
È
È4cr
#
#
33# sin 9 d3 d9 d)
3 dz r dr d) œ 3 '0
21
#
È2
'0
’r a4 r# b
"Î#
r# “ dr d) œ 3 '0 ’ "3 a4 r# b
21
$Î#
$
r3 “
œ '0 ˆ2$Î# 2$Î# 4$Î# ‰ d) œ Š8 4È2‹'0 d) œ 21 Š8 4È2‹
21
21
'c11ÎÎ22 '01 'rr
1Î2
#
44. (a)
'c11ÎÎ22 '01 'rr
#
45. (a)
(b)
46. (a)
(c)
(d)
47.
#
!
d)
21(r cos ))(r sin ))# dz r dr d) œ '1Î2 '0 'r# 21r$ cos ) sin# ) dz r dr d)
1Î2
#
(b)
È#
21r$ cos ) sin# ) dz r dr d) œ 84 '0
1
r#
'01 r' sin# ) cos ) dr d) œ 12'01Î2 sin# ) cos ) d) œ 4
'021 '01Î4 '0sec 9 3# sin 9 d3 d9 d)
'021 '01Î4 '0sec 9 3# sin 9 d3 d9 d) œ 3" '021 '01Î4 (sec 9)(sec 9 tan 9) d9 d) œ 3" '021 2" tan# 9‘ !1Î4 d) œ 6" '021 d) œ 13
È
È
1Î2
1
r
'01 '0 1cx '0 x y (6 4y) dz dy dx
(b) '0 '0 '0 (6 4r sin )) dz r dr d)
'01Î2 '11ÎÎ42 '0csc 9 (6 43 sin 9 sin )) a3# sin 9b d3 d9 d)
#
#
#
'01Î2 '01 '0r (6 4r sin )) dz r dr d) œ '01Î2 '01 a6r# 4r$ sin )b dr d) œ '01Î2 c2r$ r% sin )d "! d)
1Î2
1Î#
œ '0 (2 sin )) d) œ c2) cos )d ! œ 1 1
È
È3 È3cx
È4cx cy
'01 'È13ccxx '1
#
#
z# yx dz dy dx '1
#
#
'0
#
È4cx cy
'1
#
#
z# yx dz dy dx
48. (a) Bounded on the top and bottom by the sphere x# y# z# œ 4, on the right by the right circular
cylinder (x 1)# y# œ 1, on the left by the plane y œ 0
(b)
È
'01Î2 '02 cos 'cÈ44ccrr dz r dr d)
)
#
#
49. (a) V œ '0
21
È8cr
'02 '2
#
dz r dr d) œ '0
21
'02 ŠrÈ8 r# 2r‹ dr d) œ '021 ’ "3 a8 r# b$Î# r# “ # d)
!
œ '0 "3 (4)$Î# 4 3" (8)$Î# ‘ d) œ '0
21
21
(b) V œ '0
21
œ
œ
50. Iz œ '0
32
5
Š2 3 2È8‹ d) œ
'0 '2 sec 9 3# sin 9 d3 d9 d) œ 83 '0 '0
1Î4
21
1Î4
4
3
Š4È2 5‹ '0 d) œ
Š2È2 sin 9 sec$ 9 sin 9‹ d9 d)
8
3
'021 '01Î4 Š2È2 sin 9 tan 9 sec# 9‹ d9 d) œ 83 '021 ’2È2 cos 9 "# tan# 9“ 1Î% d)
8
3
'0
21
œ
È8
4
3
21
!
21
Š 2
"
#
2È2‹ d) œ
8
3
'0
21
Š 5 #4
È2
‹ d) œ
81 Š4È2 5‹
3
'01Î3 '02 (3 sin 9)# a3# sin 9b d3 d9 d) œ '021 '01Î3 '02 3% sin$ 9 d3 d9 d)
'021 '01Î3 asin 9 cos# 9 sin 9b d9 d) œ 325 '021 ’ cos 9 cos3 9 “ 1Î$ d) œ 831
$
!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
81 Š4È2 5‹
3
Chapter 15 Additional and Advanced Exercises
51. With the centers of the spheres at the origin, Iz œ '0
21
'01 'ab $ (3 sin 9)# a3# sin 9b d3 d9 d)
'021 '01 sin$ 9 d9 d) œ $ ab 5 a b '021 '01 asin 9 cos# 9 sin 9b d9 d)
1
21
21
œ $ ab 5 a b '0 ’ cos 9 cos3 9 “ d) œ 4$ ab15 a b '0 d) œ 81$ ab15 a b
$ ab & a & b
5
œ
&
&
&
&
&
$
&
&
&
!
'01 '01ccos (3 sin 9)# a3# sin 9b d3 d9 d) œ '02 '0 '01ccos 3% sin$ 9 d3 d9 d)
1
1
21
21
œ "5 '0 '0 (1 cos 9)& sin$ 9 d9 d) œ '0 '0 (1 cos 9)' (1 cos 9) sin 9 d9 d);
52. Iz œ '0
21
)
1
u œ 1 cos 9
” du œ sin 9 d9 • Ä
"
5
œ
'021 2 56†2
$
&
d) œ
32
35
"
5
1
)
'021 '02 u' (2 u) du d) œ 5" '021 ’ 2u7
(
#
u)
8 “!
d) œ
"
5
'021 ˆ 7" 8" ‰ 2) d)
'021 d) œ 64351
53. x œ u y and y œ v Ê x œ u v and y œ v
" "
Ê J(uß v) œ º
œ 1; the boundary of the
0 "º
image G is obtained from the boundary of R as
follows:
xy-equations for
Corresponding uv-equations
Simplified
the boundary of R
for the boundary of G
uv-equations
yœx
yœ0
Ê
vœuv
_
uœ0
vœ0
_ _
'0 '0 esx f(x yß y) dy dx œ '0 '0
x
vœ0
esÐuvÑ f(uß v) du dv
$s "t
!$ "#
54. If s œ !x " y and t œ # x $ y where (!$ "# )# œ ac b# , then x œ
and J(sß t) œ
œ
"
Èac b#
"
(!$ "# )#
$
º #
'021 '0_ rer
#
"
œ
! º
dr d) œ
"
!$ "#
"
#Èac b#
Ê
_ _
'c_
'_ e as t b È
'021 d) œ È
#
1
ac b#
#
"
ac b#
,yœ
,
ds dt
1
Èac b#
. Therefore,
# s !t
!$ "#
œ 1 Ê ac b# œ 1# .
CHAPTER 15 ADDITIONAL AND ADVANCED EXERCISES
6cx#
1. (a) V œ 'c3 'x
2
6cx#
(c) V œ 'c3 'x
2
6cx#
(b) V œ 'c3 'x
2
x# dy dx
6cx#
x# dy dx œ 'c3 'x
2
a6x# x% x$ b dx œ ’2x$
x&
5
'0x
#
dz dy dx
#
x%
4 “ $
œ
125
4
2. Place the sphere's center at the origin with the surface of the water at z œ 3. Then
9 œ 25 x# y# Ê x# y# œ 16 is the projection of the volume of water onto the xy-plane
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
991
992
Chapter 15 Multiple Integrals
Ê V œ '0
21
'04 'ccÈ325cr
dz r dr d) œ '0
21
#
'04 ŠrÈ25 r# 3r‹ dr d) œ '021 ’ "3 a25 r# b$Î# 3# r# “ % d)
21
21
œ '0 "3 (9)$Î# 24 3" (25)$Î# ‘ d) œ '0
3. Using cylindrical coordinates, V œ '0
21
œ '0 ˆ1
21
4. V œ 4 '0
1Î2
"
3
cos )
'01 'r
È2r
#
#
œ 4 '0 Š 3"
1Î2
"
4
"
3
'01 '02crÐcos
sin )‰ d) œ )
1Î2
dz r dr d) œ 4 '0
2È 2
3 ‹
d) œ Š 8
!
26
3
"
3
)
521
3
d) œ
sin )Ñ
sin )
dz r dr d) œ '0
21
'01 a2r r# cos ) r# sin )b dr d)
#1
"
3
cos )‘ ! œ 21
'01 ŠrÈ2 r# r$ ‹ dr d) œ 4'01Î2 ’ "3 a2 r# b$Î# r4 “ " d)
%
!
'
1Î2
È 2 7
‹ 0
3
1 Š8È2 7‹
d) œ
6
5. The surfaces intersect when 3 x# y# œ 2x# 2y# Ê x# y# œ 1. Thus the volume is
V œ 4 '0
1
È1 c x
'0
6. V œ 8 '0
1Î2
œ
64
3
#
'2x3cx2ycy
#
#
1Î2
dz dy dx œ 4 '0
#
#
'01 '2r3r
#
#
1Î2
dz r dr d) œ 4 '0
'01 a3r 3r$ b dr d) œ 3'01Î2 d) œ 31#
'01Î2 '02 sin 9 3# sin 9 d3 d9 d) œ 643 '01Î2 '01Î2 sin% 9 d9 d)
'01Î2 ” sin 94cos 9 ¹1Î# 43 '01Î2 sin# 9 d9• d) œ 16 '01Î2 92 sin429 ‘ 1! Î# d) œ 41 '01Î2 d) œ 21#
$
!
7. (a) The radius of the hole is 1, and the
radius of the sphere is 2.
(b) V œ 2 '0
21
8. V œ '0
1
È4cz
È
'0 3 '1
È9cr
'03 sin '0
)
#
r dr dz d) œ '0
21
#
dz r dr d) œ '0
'03 sin
1
œ '0 ’ "3 a9 9 sin# )b
1
$Î#
)
È3
'0
a3 z# b dz d) œ 2È3 '0 d) œ 4È31
21
rÈ9 r# dr d) œ '0 ’ "3 a9 r# b
1
3" (9)$Î# “ d) œ 9'0 ’1 a1 sin# )b
1
œ '0 a1 cos ) sin# ) cos )b d) œ 9 ’) sin )
1
9. The surfaces intersect when x# y# œ
coordinates is V œ 4 '0
1Î2
œ
"
#
'0
1Î2
10. V œ '0
1Î2
œ
15
4
'0
d) œ
Î
ˆr# 1‰ 2
#
“
!
d)
“ d) œ 9'0 a1 cos$ )b d)
1
œ 91
Ê x# y# œ 1. Thus the volume in cylindrical
1Î2
dz r dr d) œ 4 '0
'01 Š #r r# ‹ dr d) œ 4'01Î2 ’ r4
$
#
%
"
r8 “ d)
!
1
4
'12 '0r sin
1Î2
'0 'r
1
x# y# 1
#
1
sin$ )
3 “!
$Î#
$Î# 3 sin )
#
) cos )
dz r dr d) œ '0
sin ) cos ) d) œ
1Î2
15
4
'12 r$ sin ) cos ) dr d) œ '0 Î2 ’ r4 “ # sin ) cos ) d)
1Î#
#
’ sin2 ) “
!
1
%
"
œ
15
8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 15 Additional and Advanced Exercises
11.
'0_ ec
ax
ecbx
x
_
dx œ '0
'ab exy dy dx œ 'ab '0_ exy dx dy œ 'ab Š
œ 'a lim ’ e y “ dy œ 'a lim Š "y
cxy
b
t
tÄ_
b
tÄ_
!
ecyt
y ‹
dy œ 'a
b
"
y
lim
tÄ_
'0t exy dx‹ dy
dy œ cln yd ab œ ln ˆ ba ‰
12. (a) The region of integration is sketched at the right
Ê '0
a sin "
œ '0
"
È
'y cota "c y
#
#
ln ax# y# b dx dy
'0a r ln ar# b dr d);
u œ r#
” du œ 2r dr • Ä
"
#
'0" '0a ln u du d)
#
'0" [u ln u u] !a d)
"
œ "# '0 ’2a# ln a a# lim
"
#
œ
(b)
13.
#
'0
a cos "
'0
t ln t“ d) œ
a#
#
ln ax y b dy dx 'a cos " '0
#
tÄ0
(tan ")x
#
Èa cx
a
#
'0" (2 ln a 1) d) œ a# " ˆln a "# ‰
#
ln ax# y# b dy dx
'0x '0u emÐxtÑ f(t) dt du œ '0x 't x emÐxtÑ f(t) du dt œ '0x (x t)emÐxtÑ f(t) dt; also
'0x '0v '0u emÐxtÑ f(t) dt du dv œ '0x 't x 't v emÐxtÑ f(t) du dv dt œ '0x 't x (v t)emÐxtÑ f(t) dv dt
x
x
x
œ '0 "2 (v t)# emÐxtÑ f(t)‘ t dt œ '0 (x # t) emÐxtÑ f(t) dt
#
14.
'01 f(x) Š'0x g(xy)f(y) dy‹ dx œ '01 '0x
œ '0
1
g(xy)f(x)f(y) dy dx
'y1 g(xy)f(x)f(y) dx dy œ '01 f(y) Œ'y1 g(xy)f(x) dx dy;
'01 '01 g akxykb f(x)f(y) dx dy œ '01 '0x g(xy)f(x)f(y) dy dx '01 'x1 g(yx)f(x)f(y) dy dx
1
1
1
1
œ '0 'y g(xy)f(x)f(y) dx dy '0 'x g(yx)f(x)f(y) dy dx
œ '0
1
'y1 g(xy)f(x)f(y) dx dy '01 'y1 g(xy)f(y)f(x) dx dy
ðóóóóóóóóóóóóñóóóóóóóóóóóóò
simply interchange x and y
variable names
œ 2'0
1
'y1 g(xy)f(x)f(y) dx dy, and the statement now follows.
15. Io (a) œ '0 '0
a
œ
a#
4
"
1#
xÎa#
ax# y# b dy dx œ '0 ’x# y
a
a# ; Iwo (a) œ
"
#
xÎa#
y$
3 “!
a 6" a$ œ 0 Ê a% œ
dx œ '0 Š xa#
"
3
a
$
Ê a œ %É 3" œ
x$
3a' ‹
"
%
È
3
%
x
dx œ ’ 4a
#
a
x%
12a' “ !
. Since Iwwo (a) œ
"
#
#" a% 0, the
value of a does provide a minimum for the polar moment of inertia Io (a).
16. Io œ '0
2
'2x4 ax# y# b (3) dy dx œ 3 '02 Š4x# 14x3
$
64
3 ‹
dx œ 104
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
993
994
Chapter 15 Multiple Integrals
17. M œ 'c) 'b sec ) r dr d) œ
)
'c Š a#
)
a
#
)
b#
#
sec# )‹ d)
œ a# ) b# tan ) œ a# cos" ˆ ba ‰ b# Š
È a# b#
‹
b
œ a# cos" ˆ ba ‰ bÈa# b# ; Io œ 'c) 'b sec ) r$ dr d)
)
a
'c aa% b% sec% )b d)
œ "4 'c ca% b% a1 tan# )b asec# )bd d)
)
"
4
œ
)
)
)
)
b% tan$ )
“
3
)
œ
"
4
œ
%
$
a% )
b% tan )
)
b tan
#
#
6
" %
" $È #
" ˆ b ‰
a
# a cos
a # b
œ
’a% ) b% tan )
2 ˆy# Î2‰
b# 6" b$ aa# b# b
18. M œ 'c2 '1cay#Î4b dx dy œ '2 Š1
2
œ 'c2 ’ x2 “
2
œ
19.
3
16
#
2c ˆy# Î2‰
1 ay# Î4b
ˆ32
64
3
2
y#
4‹
2
3 ‰ ˆ 32†8 ‰
œ ˆ 16
œ
15
32 ‰
5
#
y$
12 “ #
dy œ ’y
3
dy œ 'c2 32
a4 y# b dy œ
48
15
$Î#
œ
8
3
2
'c22 a16 8y# y% b dy œ 163 ’16y 8y3
3
32
Ê xœ
My
M
‰ ˆ 38 ‰ œ
œ ˆ 48
15
6
5
# #
# #
# #
"
ab
# #
#
y&
5 “!
# #
# #
# #
!
1‹
"
#ab
# #
Šea b 1‹
# #
Šea b 1‹
ßy)
'yy 'xx ``F(x
' y ` F(xßy) x
x ` y dx dy œ y ’ ` y “
"
!
# #
!
œ
$
, and y œ 0 by symmetry
'0a '0b emax ab x ßa y b dy dx œ '0a '0bxÎa eb x dy dx '0b '0ayÎb ea y dx dy
a
b
a
b
"
"
œ '0 ˆ ba x‰ eb x dx '0 ˆ ba y‰ ea y dy œ ’ 2ab
eb x “ ’ 2ba
ea y “ œ #"ab Šeb a
# #
20.
2 ˆy# Î2‰
; My œ '2 '1ay#Î4b x dx dy
"
#
"
!
!
" ßy)
dy œ 'y ’ ` F(x
`y
y"
"
!
x!
` F(x! ßy)
`y “
dx œ cF(x" ß y) F(x! ß y)d yy!"
œ F(x" ß y" ) F(x! ß y" ) F(x" ß y! ) F(x! ß y! )
21. (a) (i)
(ii)
(iii)
(iv)
Fubini's Theorem
Treating G(y) as a constant
Algebraic rearrangement
The definite integral is a constant number
(b)
'0ln 2 '01Î2 ex cos y dy dx œ Œ'0ln 2 ex dx Œ'01Î2 cos y dy œ aeln 2 e0 b ˆsin 1# sin 0‰ œ (1)(1) œ 1
(c)
'12 'c11 yx
#
dx dy œ Œ'1
2
"
y#
dy Œ'c1 x dx œ ’ y" “ ’ x2 “
#
1
#
"
"
"
œ ˆ #" 1‰ ˆ #" #" ‰ œ 0
22. (a) ™ f œ xi yj Ê Du f œ u" x u# y; the area of the region of integration is
Ê average œ 2'0
1
#
œ 2 ’u" Š x2
(b) average œ
"
area
_ _
23. (a) I# œ '0
œ "#
'0
'01Î2
_
x$
3‹
1cx
'0
(u" x u# y) dy dx œ 2 '0 u" x(1 x) "# u# (1 x)# ‘ dx
ˆ "# u# ‰
"
(1x)$
3 “!
' ' (u" x u# y) dA œ
R
1Î2
e ˆx y ‰ dx dy œ '0
#
lim
bÄ_
"
#
1
#
_
(b) > ˆ "# ‰ œ '0 t"Î# et dt œ '0 ay# b
u"
area
"
3
(u" u# )
M
u# ' '
' ' x dA area
y dA œ u" Š My ‹ u# ˆ MMx ‰ œ u" x u# y
R
R
'0_ aer b r dr d) œ '01Î2 ”
aecb 1b d) œ
#
œ 2 ˆ 6" u" 6" u# ‰ œ
#
"
#
'01Î2 d) œ 14
"Î# y#
e
Ê Iœ
_
lim
bÄ_
'0b rer
#
dr• d)
È1
#
(2y) dy œ 2 '0 ey dy œ 2 Š
#
È1
# ‹
œ È1, where y œ Èt
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Chapter 15 Additional and Advanced Exercises
24. Q œ '0
21
'0R kr# (1 sin )) dr d) œ kR3 '021 (1 sin )) d) œ kR3
$
25. For a height h in the bowl the volume of water is V œ
œ
È
È
Èh
'cÈhh 'cÈhhccxx ah x# y# b dy dx œ '021 '0
#
$
c) cos )d #!1 œ
È
È
'cÈhh 'cÈhhccxx 'xhby
#
#
#
#
dz dy dx
ah r# b r dr d) œ '0 ’ hr2 r4 “
21
#
21kR$
3
#
%
d) œ '0
Èh
21
!
h#
4
d) œ
h# 1
#
.
Since the top of the bowl has area 101, then we calibrate the bowl by comparing it to a right circular cylinder
h# 1
#
whose cross sectional area is 101 from z œ 0 to z œ 10. If such a cylinder contains
to a depth w then we have 101w œ
rain, w œ 3 and h œ È60.
h# 1
#
h#
20
Ê wœ
cubic inches of water
. So for 1 inch of rain, w œ 1 and h œ È20; for 3 inches of
26. (a) An equation for the satellite dish in standard position
is z œ "# x# "# y# . Since the axis is tilted 30°, a unit
vector v œ 0i aj bk normal to the plane of the
È3
#
È
#3
water level satisfies b œ v † k œ cos ˆ 16 ‰ œ
Ê a œ È1 b# œ "# Ê v œ "# j
Ê "# (y 1)
Ê zœ
"
È3
È3
#
y Š #"
k
ˆz "# ‰ œ 0
"
È3 ‹
is an equation of the plane of the water level. Therefore
the volume of water is V œ ' '
R
x# y# 23 y 1
and " œ
2
3
2
È3
1
2
3
#
1
3
#
1
2
#
Ê Vœ ' '
c
Š 23 yb1c È23 cy# ‹
œ y; y œ 1 Ê
dz
dy
!
y# and
dz
dy
"Î#
Š 23 yb1c È23 cy# ‹
"
3
"
#
1
1
2
œ 0. When x œ 0, then y œ ! or y œ " , where ! œ
Ê 49 4 Š È2 1‹
(b) x œ 0 Ê z œ
' Èx byby c È dz dy dx, where R is the interior of the ellipse
' È bb cÈ
1
3
"Î#
1
2
y
x#
1
2
1
2
2
3
Ê 49 4 Š È2 1‹
3
#
1
3
1 dz dx dy
y#
œ 1 Ê the tangent line has slope 1 or a 45° slant
Ê at 45° and thereafter, the dish will not hold water.
27. The cylinder is given by x# y# œ 1 from z œ 1 to _ Ê
œ '0
21
_
'0 '1
1
z
ar# z# b&Î#
'0 '0 '
21
dz r dr d) œ a lim
Ä_
1
' ' ' z ar# z# b&Î# dV
D
a
rz
1 ar# z# b&Î#
dz dr d)
œ a lim
Ä_
'021 '01 ’ˆ "3 ‰
œ a lim
Ä_
'021 ’ "3 ar# a# b"Î# 3" ar# 1b"Î# “ " d) œ a lim
' 21 ’ 3" a1 a# b"Î# 3" ˆ2"Î# ‰ 3" aa# b"Î# 3" “ d)
Ä_ 0
a
r
“
ar# z# b$Î# 1
dr d) œ a lim
Ä_
'021 '01 ’ˆ 3" ‰
r
ar# a# b$Î#
ˆ 3" ‰
r
“
ar# 1b$Î#
dr d)
!
œ a lim
21 ’ "3 a1 a# b
Ä_
"Î#
3" Š
È2
# ‹
3" ˆ "a ‰ 3" “ œ 21 ’ 3" ˆ 3" ‰
È2
# “.
28. Let's see?
The length of the "unit" line segment is: L œ 2'0 dx œ 2.
1
The area of the unit circle is: A œ 4'0
1
È1 c x
'0
The volume of the unit sphere is: V œ 8'0
1
2
dy dx œ 1.
È1 c x
'0
2
È1 c x c y
'0
2
2
dz dy dx œ 43 1.
Therefore, the hypervolume of the unit 4-sphere should be:
Vhyper œ 16'0
1
È1cx
'0
2
È1cx cy
'0
2
2
È1cx cy cz
'0
2
2
2
dw dz dy dx.
Mathematica is able to handle this integral, but we'll use the brute force approach.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
995
996
Chapter 15 Multiple Integrals
Vhyper œ 16'0
1
È1cx
œ 16'0
'0
œ 16'0
'0
œ 16'0
'0
1
1
1
È1cx
È1cx
È1cx
'0
2
2
2
2
È1cx cy
'0
È1cx cy
'0
2
2
È1cx cy cz
'0
2
2
2
dw dz dy dx œ 16'0
1
2
z2
È 1 x2 y2 É 1
1 c x2 c y2
dz dy dx œ –
È1cx
'0
È1cx cy
'0
2
2
È 1 x 2 y2 z2
œ cos )
1
È1cx
'0
2
a1 x2 y2 b'1/2 sin2 ) d) dy dx
0
3/2
x2 y2 b dy dx œ 41'0 ŠÈ1 x2 x2 È1 x2 3" a1 x2 b ‹ dx
1
1 x3
$
‘ dx œ 83 1' a1 x2 b3/2 dx œ ”
0
1
0
x œ cos )
œ 83 1'1/2 sin4 ) d)
•
dx œ sin ) d)
2 ) ‰2
œ 83 1'1/2 ˆ 1 cos
d) œ 23 1'1/2 a1 2 cos 2) cos2 2)bd) œ 23 1'1/2 ˆ 3# 2 cos 2)
2
0
dz dy dx
—
dz œ È1 x2 y2 sin ) d)
0
1
4 a1
2
z
È 1 x2 y2
a1 x2 y2 b'1/2 È1 cos2 ) sin ) d) dy dx œ 16'0
œ 41'0 È1 x2 a1 x2 b
1
2
0
0
cos 4) ‰
d)
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ
12
2
CHAPTER 16 INTEGRATION IN VECTOR FIELDS
16.1 LINE INTEGRALS
1. r œ ti (" t)j Ê x œ t and y œ 1 t Ê y œ 1 x Ê (c)
2. r œ i j tk Ê x œ 1, y œ 1, and z œ t Ê (e)
3. r œ (2 cos t)i (2 sin t)j Ê x œ 2 cos t and y œ 2 sin t Ê x# y# œ 4 Ê (g)
4. r œ ti Ê x œ t, y œ 0, and z œ 0 Ê (a)
5. r œ ti tj tk Ê x œ t, y œ t, and z œ t Ê (d)
6. r œ tj (2 2t)k Ê y œ t and z œ 2 2t Ê z œ 2 2y Ê (b)
7. r œ at# 1b j 2tk Ê y œ t# 1 and z œ 2t Ê y œ
z#
4
1 Ê (f)
8. r œ (2 cos t)i (2 sin t)k Ê x œ 2 cos t and z œ 2 sin t Ê x# z# œ 4 Ê (h)
9. r(t) œ ti (1 t)j , 0 Ÿ t Ÿ 1 Ê
dr
dt
œ i j Ê ¸ ddtr ¸ œ È2 j ; x œ t and y œ 1 t Ê x y œ t (" t) œ 1
Ê 'C f(xß yß z) ds œ '0 f(tß 1 tß 0) ¸ ddtr ¸ dt œ '0 (1) ŠÈ2‹ dt œ ’È2 t“ œ È2
1
"
1
!
10. r(t) œ ti (1 t)j k , 0 Ÿ t Ÿ 1 Ê
œ t (1 t) 1 2 œ 2t 2 Ê
dr
dt
œ i j Ê ¸ ddtr ¸ œ È2; x œ t, y œ 1 t, and z œ 1 Ê x y z 2
'C f(xß yß z) ds œ '01 (2t 2) È2 dt œ È2 ct# 2td "! œ È2
11. r(t) œ 2ti tj (2 2t)k , 0 Ÿ t Ÿ 1 Ê
dr
dt
œ 2i j 2k Ê ¸ ddtr ¸ œ È4 1 4 œ 3; xy y z
œ (2t)t t (2 2t) Ê 'C f(xß yß z) ds œ '0 a2t# t 2b 3 dt œ 3 23 t$ "# t# 2t‘ ! œ 3 ˆ 23
1
12. r(t) œ (4 cos t)i (4 sin t)j 3tk , 21 Ÿ t Ÿ 21 Ê
"
dr
dt
1
œ c20td ##
1 œ 801
dr
dt
œ È1 9 4 œ È14 ; x y z œ (1 t) (2 3t) (3 2t) œ 6 6t Ê
œ i 3 j 2k
'C f(xß yß z) ds
œ '0 (6 6t) È14 dt œ 6È14 ’t t2 “ œ Š6È14‹ ˆ "# ‰ œ 3È14
1
#
"
!
14. r(t) œ ti tj tk , 1 Ÿ t Ÿ _ Ê
_
dr
dt
13
#
'C f(xß yß z) ds œ 'c2211 (4)(5) dt
13. r(t) œ (i 2j 3k) t(i 3j 2k) œ (1 t)i (2 3t)j (3 2t)k , 0 Ÿ t Ÿ 1 Ê
Ê
2‰ œ
œ (4 sin t)i (4 cos t)j 3k
Ê ¸ ddtr ¸ œ È16 sin# t 16 cos# t 9 œ 5; Èx# y# œ È16 cos# t 16 sin# t œ 4 Ê
¸ ddtr ¸
"
#
œ i j k Ê ¸ ddtr ¸ œ È3 ;
È3
x # y # z#
_
Ê 'C f(xß yß z) ds œ '1 Š 3t#3 ‹ È3 dt œ 1t ‘ " œ lim ˆ b" 1‰ œ 1
œ
È3
t# t# t#
œ
È3
3t#
È
bÄ_
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
998
Chapter 16 Integration in Vector Fields
15. C" : r(t) œ ti t# j , 0 Ÿ t Ÿ 1 Ê
œ i 2tj Ê ¸ ddtr ¸ œ È1 4t# ; x Èy z# œ t Èt# 0 œ t ktk œ 2t
dr
dt
$Î#
0 Ê 'C f(xß yß z) ds œ '0 2tÈ1 4t# dt œ ’ "6 a" 4t# b “ œ
"
1
since t
!
"
C# : r(t) œ i j tk, 0 Ÿ t Ÿ 1 Ê
dr
dt
1
"
#
"
5
6
#
16. C" : r(t) œ tk , 0 Ÿ t Ÿ 1 Ê
(5)$Î#
"
6
œ
"
6
Š5È5 1‹ ;
œ k Ê ¸ ddtr ¸ œ 1; x Èy z# œ 1 È1 t# œ 2 t#
Ê 'C f(xß yß z) ds œ '0 a2 t# b (1) dt œ 2t "3 t$ ‘ ! œ 2
œ 'C f(xß yß z) ds 'C f(xß yß z) ds œ
"
6
È5
"
3
œ
5
3
; therefore 'C f(xß yß z) ds
3
#
œ k Ê ¸ ddtr ¸ œ 1; x Èy z# œ 0 È0 t# œ t#
dr
dt
Ê 'C f(xß yß z) ds œ '0 at# b (1) dt œ ’ t3 “ œ "3 ;
1
"
$
!
"
C# : r(t) œ tj k, 0 Ÿ t Ÿ 1 Ê
œ j Ê ¸ ddtr ¸ œ 1; x Èy z# œ 0 Èt 1 œ Èt 1
dr
dt
"
Ê 'C f(xß yß z) ds œ '0 ˆÈt 1‰ (1) dt œ 23 t$Î# t‘ ! œ
1
#
C$ : r(t) œ ti j k , 0 Ÿ t Ÿ 1 Ê
Ê 'C f(xß yß z) ds œ '0 (t)(1) dt œ ’ t2 “ œ
"
#
œ "6
Ê
"
#
!
$
17. r(t) œ ti tj tk , 0 a Ÿ t Ÿ b Ê
1 œ "3 ;
œ i Ê ¸ ddtr ¸ œ 1; x Èy z# œ t È1 1 œ t
dr
dt
1
2
3
dr
dt
Ê
'C f(xß yß z) ds œ 'C
œ i j k Ê ¸ ddtr ¸ œ È3 ;
"
f ds 'C f ds 'C f ds œ "3 ˆ 3" ‰
xyz
x # y # z#
#
œ
'C f(xß yß z) ds œ 'ab ˆ 1t ‰ È3 dt œ ’È3 ln ktk “ b œ È3 ln ˆ ba ‰ , since 0 a Ÿ b
$
ttt
t# t# t#
œ
"
#
1
t
a
18. r(t) œ (a cos t) j (a sin t) k , 0 Ÿ t Ÿ 21 Ê
Èx# z# œ È0 a# sin# t œ œ
1
dr
dt
œ (a sin t) j (a cos t) k Ê ¸ ddtr ¸ œ Èa# sin# t a# cos# t œ kak ;
1
21
kak sin t, 0 Ÿ t Ÿ 1
Ê 'C f(xß yß z) ds œ '0 kak# sin t dt '1 kak# sin t dt
kak sin t, 1 Ÿ t Ÿ 21
#1
œ ca# cos td ! ca# cos td 1 œ ca# (1) a# d ca# a# (1)d œ 4a#
19. r(x) œ xi yj œ xi
x#
#
j, 0 Ÿ x Ÿ 2 Ê
œ '0 (2x)È1 x# dx œ ’ 23 a1 x# b
2
#
dr ¸
œ i xj Ê ¸ dx
œ È1 x# ; f(xß y) œ f Šxß x# ‹ œ
dr
dx
$Î# #
“ œ
!
2
3
ˆ5$Î# 1‰ œ
10È5 2
3
20. r(t) œ a1 tbi 1# a1 tb2 j, 0 Ÿ t Ÿ 1 Ê ¸ ddtr ¸ œ É1 a1 tb# ; f(xß y) œ f Ša1 tbß 1# a1 tb2 ‹ œ
Ê
'C f ds œ '01 a1 tb
œ0
1
4 a1 tb
#
É1 a1 tb
ˆ "#
" ‰
#0
œ
4
œ 2x Ê 'C f ds
x$
#
Š x# ‹
É1 a1 tb# dt œ ' Ša1 tb 14 a1 tb4 ‹ dt œ ’ "# a1 tb2
0
1
a1 tb 14 a1 tb4
É1 a1 tb#
1
20 a1
tb5 “
"
!
11
#0
21. r(t) œ (2 cos t) i (2 sin t) j , 0 Ÿ t Ÿ
1
#
Ê
dr
dt
œ (2 sin t) i (2 cos t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 cos tß 2 sin t)
œ 2 cos t 2 sin t Ê 'C f ds œ '0 (2 cos t 2 sin t)(2) dt œ c4 sin t 4 cos td !
1Î2
22. r(t) œ (2 sin t) i (2 cos t) j , 0 Ÿ t Ÿ
œ 4 sin# t 2 cos t Ê
1Î#
1
4
Ê
dr
dt
œ 4 (4) œ 8
œ (2 cos t) i (2 sin t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 sin tß 2 cos t)
'C f ds œ '01Î4 a4 sin# t 2 cos t b (2) dt œ c4t 2 sin 2t 4 sin td 01Î%
œ 1 2Š1 È2‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.1 Line Integrals
23. r(t) œ at# 1b j 2tk , 0 Ÿ t Ÿ 1 Ê
dr
dt
999
œ 2tj 2k Ê ¸ ddtr ¸ œ 2Èt# 1; M œ 'C $ (xß yß z) ds œ '0 $ (t) Š2Èt# 1‹ dt
1
3/2
œ '0 ˆ 3# t‰ Š2Èt# 1‹ dt œ ’at# 1b “ œ 2$Î# 1 œ 2È2 1
"
1
!
24. r(t) œ at# 1b j 2tk , 1 Ÿ t Ÿ 1 Ê ddtr œ 2tj 2k
Ê ¸ dr ¸ œ 2Èt# 1; M œ ' $ (xß yß z) ds
dt
C
œ 'c1 ˆ15Èat# 1b 2‰ Š2Èt# 1‹ dt
1
œ 'c1 30 at# 1b dt œ ’30 Š t3 t‹“
1
$
"
"
œ 60 ˆ "3 1‰ œ 80;
Mxz œ 'C y$ (xß yß z) ds œ 'c1 at# 1b c30 at# 1bd dt
1
œ 'c1 30 at% 1b dt œ ’30 Š t5 t‹“
1
&
œ 48 Ê y œ
Mxz
M
"
"
œ 60 ˆ "5 1‰
3
'
'
œ 48
80 œ 5 ; Myz œ C x$ (xß yß z) ds œ C 0 $ ds œ 0 Ê x œ 0; z œ 0 by symmetry (since $ is
independent of z) Ê (xß yß z) œ ˆ!ß 35 ß 0‰
25. r(t) œ È2t i È2t j a4 t# b k , 0 Ÿ t Ÿ 1 Ê
dr
dt
œ È2i È2j 2tk Ê ¸ ddtr ¸ œ È2 2 4t# œ 2È1 t# ;
(a) M œ 'C $ ds œ '0 (3t) Š2È1 t# ‹ dt œ ’2 a1 t# b
1
(b)
$Î# "
“ œ 2 ˆ2$Î# 1‰ œ 4È2 2
!
"
1
M œ 'C $ ds œ '0 (1) Š2È1 t# ‹ dt œ ’tÈ1 t# ln Št È1 t# ‹“ œ ’È2 ln Š1 È2‹“ (0 ln 1)
!
œ È2 ln Š1 È2‹
26. r(t) œ ti 2tj 23 t$Î# k , 0 Ÿ t Ÿ 2 Ê
dr
dt
œ i 2j t"Î# k Ê ¸ ddtr ¸ œ È1 4 t œ È5 t;
#
M œ 'C $ ds œ '0 ˆ3È5 t‰ ˆÈ5 t‰ dt œ '0 3(5 t) dt œ 32 (5 t)# ‘ ! œ
2
2
Myz œ 'C x$ ds œ '0 t[3(5 t)] dt œ '0 a15t 3t# b dt œ "25 t#
2
2
#
t$ ‘ !
3
#
a7# 5# b œ
2
œ '0 ˆ10t$Î# 2t&Î# ‰ dt œ 4t&Î#
2
œ
38
36
œ
19
18
,yœ
Mxz
M
œ
76
36
œ
19
9
4 (Î# ‘ #
7 t
!
, and z œ
2
œ 4(2)&Î# 47 (2)(Î# œ 16È2
Mxy
M
œ
144È2
7†36
27. Let x œ a cos t and y œ a sin t, 0 Ÿ t Ÿ 21. Then
dx
dt
œ
4
7
(24) œ 36;
œ 30 8 œ 38;
Mxz œ 'C y$ ds œ '0 2t[3(5 t)] dt œ 2 '0 a15t 3t# b dt œ 76; Mxy œ 'C z$ ds œ '0
2
3
#
32
7
È2 œ
dz
dt
œ0
2 $Î#
[3(5
3 t
144
7
t)] dt
È2 Ê x œ
Myz
M
È2
œ a sin t,
dy
dt
œ a cos t,
‰ Š dy
ˆ dz ‰ dt œ a dt; Iz œ ' ax# y# b $ ds œ ' aa# sin# t a# cos# tb a$ dt
Ê Êˆ dx
dt
dt ‹ dt
C
0
#
#
21
#
Iz
œ '0 a$ $ dt œ 21$ a$ ; M œ 'C $ (xß yß z) ds œ '0 $ a dt œ 21$ a Ê Rz œ É M
œ É 2211aa$$ œ a.
21
21
28. r(t) œ tj (2 2t)k , 0 Ÿ t Ÿ 1 Ê
dr
dt
$
œ j 2k Ê ¸ ddtr ¸ œ È5; M œ 'C $ ds œ '0 $ È5 dt œ $ È5;
1
"
Ix œ 'C ay# z# b $ ds œ '0 ct# (2 2t)# d $ È5 dt œ '0 a5t# 8t 4b $ È5 dt œ $ È5 53 t$ 4t# 4t‘ ! œ
1
1
"
Iy œ 'C ax# z# b $ ds œ '0 c0# (2 2t)# d $ È5 dt œ '0 a4t# 8t 4b $ È5 dt œ $ È5 43 t$ 4t# 4t‘ ! œ
1
1
Iz œ 'C ax# y# b $ ds œ '0 a0# t# b $ È5 dt œ $ È5 ’ t3 “ œ
1
$
"
!
Iz
and Rz œ É M
œ
"
3
I
5
3
4
3
$ È5 ;
$ È5 ;
Ix
$ È5 Ê Rx œ É M
œ É 35 , Ry œ É My œ É 34 œ
"
È3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
2
È3
,
1000 Chapter 16 Integration in Vector Fields
29. r(t) œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ 21 Ê
œ ( sin t)i (cos t)j k Ê ¸ ddtr ¸ œ Èsin# t cos# t 1 œ È2;
dr
dt
(a) M œ 'C $ ds œ '0 $ È2 dt œ 21$ È2; Iz œ 'C ax# y# b $ ds œ '0 acos# t sin# tb $ È2 dt œ 21$ È2
21
21
Iz
Ê Rz œ É M
œ1
(b) M œ 'C $ (xß yß z) ds œ '0 $ È2 dt œ 41$ È2 and Iz œ 'C ax# y# b $ ds œ '0 $ È2 dt œ 41$ È2
41
41
Iz
Ê Rz œ É M
œ1
30. r(t) œ (t cos t)i (t sin t)j
2È2 $Î#
k,
3 t
0ŸtŸ1 Ê
dr
dt
œ (cos t t sin t)i (sin t t cos t)j È2t k
"
Ê ¸ ddtr ¸ œ È(t 1)# œ t 1 for 0 Ÿ t Ÿ 1; M œ 'C $ ds œ '0 (t 1) dt œ "2 (t 1)# ‘ ! œ
1
Mxy œ 'C z$ ds œ '0 Š 2 3 2 t$Î# ‹ (t 1) dt œ
È
1
œ
2È 2
3
ˆ 72 52 ‰ œ
2È 2
3
ˆ 24
‰
35 œ
16È2
35
Ê zœ
2È 2
3
'0 ˆt&Î# t$Î# ‰ dt œ 2È3 2
Mxy
M
œ Š 1635 2 ‹ ˆ 23 ‰ œ
32È2
105
œ '0 at# cos# t t# sin# tb (t 1) dt œ '0 at$ t# b dt œ ’ t4 t3 “ œ
1
1
%
"
$
!
"
4
a2# 1# b œ
3
#
;
27 t(Î# 25 t&Î# ‘ "
!
1
È
"
#
; Iz œ 'C ax# y# b $ ds
"
3
œ
7
12
Iz
7
Ê Rz œ É M
œ É 18
31. $ (xß yß z) œ 2 z and r(t) œ (cos t)j (sin t)k , 0 Ÿ t Ÿ 1 Ê M œ 21 2 as found in Example 4 of the text;
Ix
also ¸ ddtr ¸ œ 1; Ix œ 'C ay# z# b $ ds œ '0 acos# t sin# tb (2 sin t) dt œ '0 (2 sin t) dt œ 21 2 Ê Rx œ É M
1
1
œ1
32. r(t) œ ti
2È2 $Î#
j
3 t
t#
#
k, 0 Ÿ t Ÿ 2 Ê
dr
dt
œ i È2 t"Î# j tk Ê ¸ ddtr ¸ œ È1 2t t# œ È(1 t)# œ 1 t for
0 Ÿ t Ÿ 2; M œ 'C $ ds œ '0 ˆ t"1 ‰ (1 t) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t"1 ‰ (1 t) dt œ ’ t2 “ œ 2;
2
Mxz œ 'C y$ ds œ '
2È2 $Î#
3 t
0
yœ
Mxz
M
œ
16
15
2
, and z œ
Mxy
M
œ
2
dt œ
#
3
#
È
’ 4152 t&Î# “
!
œ
$
œ '0 ˆt# 89 t$ ‰ dt œ ’ t3 92 t% “ œ
#
!
Iz
Rz œ É M
œ
2
3
; Mxy œ 'C z$ ds œ '0
2 #
t
2
2
$
32
15
#
dt œ
#
$ #
’ t6 “
!
; Ix œ 'C ay# z# b $ ds œ '0 ˆ 89 t$ 4" t% ‰ dt œ ’ 29 t%
Iy œ 'C ax# z# b $ ds œ '0 ˆt# "4 t% ‰ dt œ ’ t3
2
2
8
3
32
9
œ
56
9
#
t&
20 “ !
œ
8
3
32
20
œ
64
15
Ix
Ê Rx œ É M
œ
2
3
!
œ
#
t&
20 “ !
%
3
œ
Ê xœ
Myz
M
œ
32
9
; Iz œ 'C ax# y# b $ ds
Iy
32
ÉM
É 29
œ É 15
, and
5 , Ry œ
È7
33-36. Example CAS commands:
Maple:
f := (x,y,z) -> sqrt(1+30*x^2+10*y);
g := t -> t;
h := t -> t^2;
k := t -> 3*t^2;
a,b := 0,2;
ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2):
'ds' = ds(t)*'dt';
F := f(g,h,k):
'F(t)' = F(t);
Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b );
`` = value(rhs(%));
Mathematica: (functions and domains may vary)
Clear[x, y, z, r, t, f]
f[x_,y_,z_]:= Sqrt[1 30x2 10y]
#
# (a)
# (b)
# (c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
32
20
œ 1,
232
45
;
Section 16.2 Vector Fields, Work, Circulation, and Flux 1001
{a,b}= {0, 2};
x[t_]:= t
y[t_]:= t2
z[t_]:= 3t2
r[t_]:= {x[t], y[t], z[t]}
v[t_]:= D[r[t], t]
mag[vector_]:=Sqrt[vector.vector]
Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}]
N[%]
16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX
1. f(xß yß z) œ ax# y# z# b
`f
`y
"Î#
# $Î#
œ y a x# y # z b
and
similarly,
œ
y
x # y # z#
and
`f
`z
"
#
2. f(xß yß z) œ ln Èx# y# z# œ
`f
`y
`f
`x
Ê
`f
`z
$Î#
# $Î#
œ z a x# y # z b
`f
`x
ln ax# y# z# b Ê
œ
3. g(xß yß z) œ ez ln ax# y# b Ê
œ "# ax# y# z# b
Ê ™fœ
z
x # y # z#
`g
`x
`g
`y
œ x# 2x
y# ,
(2x) œ x ax# y# z# b
Ê ™fœ
œ
"
#
$Î#
; similarly,
x i y j z k
ax# y# z# b$Î#
Š x# y"# z# ‹ (2x) œ
x
x # y # z#
;
x i y j zk
x# y# z#
œ x# 2y
y# and
`g
`z
œ ez
z
Ê ™ g œ Š x#2xy# ‹ i Š x# 2y
y# ‹ j e k
`g
`x
4. g(xß yß z) œ xy yz xz Ê
œ y z,
`g
`y
`g
`z
œ x z, and
œ y x Ê ™ g œ (y z)i (B z)j (x y)k
5. kFk inversely proportional to the square of the distance from (xß y) to the origin Ê È(M(xß y))# (N(xß y))#
œ
k
x# y#
, k 0; F points toward the origin Ê F is in the direction of n œ
Ê F œ an , for some constant a 0. Then M(xß y) œ
Ê È(M(xß y))# (N(xß y))# œ a Ê a œ
k
x# y#
ax
È x# y#
Ê Fœ
x
È x# y#
and N(xß y) œ
kx
ax# y# b$Î#
i
i
y
È x# y#
j
ay
È x# y#
ky
ax# y# b$Î#
j , for any constant k 0
6. Given x# y# œ a# b# , let x œ Èa# b# cos t and y œ Èa# b# sin t. Then
r œ ŠÈa# b# cos t‹ i ŠÈa# b# sin t‹ j traces the circle in a clockwise direction as t goes from 0 to 21
Ê v œ ŠÈa# b# sin t‹ i ŠÈa# b# cos t‹ j is tangent to the circle in a clockwise direction. Thus, let
F œ v Ê F œ yi xj and F(0ß 0) œ 0 .
7. Substitute the parametric representations for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F , and calculate the work W œ 'C F †
(a) F œ 3ti 2tj 4tk and
dr
dt
(b) F œ 3t# i 2tj 4t% k and
œ
7
3
2œ
dr
dt
.
œijk Ê F†
dr
dt
dr
dt
œ 9t Ê W œ '0 9t dt œ
1
œ i 2tj 4t$ k Ê F †
dr
dt
œ 7t# 16t( Ê W œ '0 a7t# 16t( b dt œ 73 t$ 2t) ‘ !
1
13
3
(c) r" œ ti tj and r# œ i j tk ; F" œ 3ti 2tj and
F# œ 3i 2j 4tk and
d r#
dt
œ k Ê F# †
d r#
dt
d r"
dt
9
#
œ i j Ê F" †
d r"
dt
"
œ 5t Ê W" œ '0 5t dt œ
1
œ 4t Ê W# œ '0 4t dt œ 2 Ê W œ W" W# œ
1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
9
#
5
#
;
1002 Chapter 16 Integration in Vector Fields
8. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate the work W œ 'C F †
dr
dt
.
" ‰
(a) F œ ˆ t#
1 j and
dr
dt
œijk Ê F†
" ‰
(b) F œ ˆ t#
1 j and
dr
dt
œ i 2tj 4t$ k Ê F †
dr
dt
Ê F# †
d r#
dt
œ 0 Ê W œ '0
"
t# 1
dt œ
1
œ
2t
t# 1 Ê W
and ddtr" œ i j
dr
dt
" ‰
(c) r" œ ti tj and r# œ i j tk ; F" œ ˆ t#
1 j
1
Ê W œ '0
"
t# 1
œ
"
t# 1
œ '0
"
dt œ ctan" td ! œ
1
Ê
2t
t# 1
F" † ddtr"
1
4
"
dt œ cln at# 1bd ! œ ln 2
œ
"
t# 1
; F# œ
"
#
j and
œk
d r#
dt
1
4
9. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate the work W œ 'C F †
(a) F œ Èti 2tj Ètk and
(b) F œ t# i 2tj tk and
dr
dt
dr
dt
dr
dt
.
œijk Ê F†
"
œ 2Èt 2t Ê W œ '0 ˆ2Èt 2t‰ dt œ 43 t$Î# t# ‘ ! œ
1
dr
dt
œ i 2tj 4t$ k Ê F †
dr
dt
œ 4t% 3t# Ê W œ '0 a4t% 3t# b dt œ 45 t& t$ ‘ ! œ "5
1
(c) r" œ ti tj and r# œ i j tk ; F" œ 2tj Èt k and
œ 1; F# œ Èti 2j k and
d r#
dt
œ k Ê F# †
"
3
œ i j Ê F" †
d r"
dt
d r"
dt
"
œ 2t Ê W" œ '0 2t dt
1
œ 1 Ê W# œ '0 dt œ 1 Ê W œ W" W# œ 0
1
d r#
dt
10. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate the work W œ 'C F †
dr
dt
.
œ 3t# Ê W œ '0 3t# dt œ 1
1
(a) F œ t# i t# j t# k and
dr
dt
œijk Ê F†
(b) F œ t$ i t' j t& k and
dr
dt
œ i 2tj 4t$ k Ê F †
%
œ ’ t4
t)
4
"
94 t* “ œ
!
dr
dt
œ t$ 2t( 4t) Ê W œ '0 at$ 2t( 4t) b dt
1
17
18
(c) r" œ ti tj and r# œ i j tk ; F" œ t# i and
F# œ i tj tk and
dr
dt
d r#
dt
œ k Ê F# †
d r#
dt
œ i j Ê F" †
d r"
dt
œ t Ê W# œ '0 t dt œ
1
d r"
dt
"
#
œ t# Ê W" œ '0 t# dt œ
1
Ê W œ W" W# œ
"
3
;
5
6
11. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate the work W œ 'C F †
dr
dt
.
œ 3t# 1 Ê W œ '0 a3t# 1b dt œ ct$ td ! œ 2
(a) F œ a3t# 3tb i 3tj k and
†
(b) F œ a3t# 3tb i 3t% j k
Ê F†
dr
dt
1
"
dr
&
$
#
dt œ 6t 4t 3t 3t
1
"
Ê W œ 0 a6t& 4t$ 3t# 3tb dt œ t' t% t$ 3# t# ‘ ! œ 3#
r" œ ti tj and r# œ i j tk ; F" œ a3t# 3tb i k and ddtr" œ i j Ê F" † ddtr" œ 3t#
1
"
Ê W" œ 0 a3t# 3tb dt œ t$ 32 t# ‘ ! œ "# ; F# œ 3tj k and ddtr# œ k Ê F# † ddtr#
Ê W œ W" W# œ 12
'
(c)
dr
dt œ i j k Ê F
and ddtr œ i 2tj 4t$ k
'
3t
œ 1 Ê W# œ '0 dt œ 1
12. Substitute the parametric representation for r(t) œ x(t)i y(t)j z(t)k representing each path into the vector
field F, and calculate the work W œ 'C F †
(a) F œ 2ti 2tj 2tk and
dr
dt
dr
dt
.
œijk Ê F†
(b) F œ at# t% b i at% tb j at t# b k and
dr
dt
dr
dt
œ 6t Ê W œ '0 6t dt œ c3t# d ! œ 3
œ i 2tj 4t$ k Ê F †
Ê W œ '0 a6t& 5t% 3t# b dt œ ct' t& t$ d ! œ 3
1
1
"
dr
dt
œ 6t& 5t% 3t#
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1
Section 16.2 Vector Fields, Work, Circulation, and Flux 1003
(c) r" œ ti tj and r# œ i j tk ; F" œ ti tj 2tk and
F# œ (1 t)i (t 1)j 2k and
d r#
dt
œ k Ê F# †
dr"
dt
œ i j Ê F" †
d r"
dt
Ê F†
dr
dt
œ 2t$ Ê work œ '0 2t$ dt œ
1
œ 2 Ê W# œ '0 2 dt œ 2 Ê W œ W" W# œ 3
1
d r#
dt
13. r œ ti t# j tk , 0 Ÿ t Ÿ 1, and F œ xyi yj yzk Ê F œ t$ i t# j t$ k and
1
œ 2t Ê W" œ '0 2t dt œ ";
œ i 2tj k
dr
dt
"
#
14. r œ (cos t)i (sin t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 2yi 3xj (x y)k
Ê F œ (2 sin t)i (3 cos t)j (cos t sin t)k and
œ 3 cos# t 2sin2 t
œ 32 t
3
4
"
6
sin 2t t
cos t
"
6
sin 2t
2
"
6
dr
dt
œ ( sin t)i (cos t)j 6" k Ê F †
sin t Ê work œ '0 ˆ3 cos# t 2 sin2 t
sin t
"
6
#1
t‘ !
cos
21
"
6
cos t
"
6
dr
dt
sin t‰ dt
œ1
15. r œ (sin t)i (cos t)j tk , 0 Ÿ t Ÿ 21, and F œ zi xj yk Ê F œ ti (sin t)j (cos t)k and
dr
dt
œ (cos t)i (sin t)j k Ê F †
œ cos t t sin t
t
2
sin 2t
4
dr
dt
œ t cos t sin# t cos t Ê work œ '0 at cos t sin# t cos tb dt
21
#1
sin t‘ ! œ 1
16. r œ (sin t)i (cos t)j 6t k , 0 Ÿ t Ÿ 21, and F œ 6zi y# j 12xk Ê F œ ti acos# tbj (12 sin t)k and
dr
dt
œ (cos t)i (sin t)j "6 k Ê F †
dr
dt
œ t cos t sin t cos# t 2 sin t
Ê work œ '0 at cos t sin t cos# t 2 sin tb dt œ cos t t sin t
21
1
3
#1
cos$ t 2 cos t‘ ! œ 0
17. x œ t and y œ x# œ t# Ê r œ ti t# j , 1 Ÿ t Ÿ 2, and F œ xyi (x y)j Ê F œ t$ i at t# b j and
dr
dt
œ i 2tj Ê F †
dr
dt
œ t$ a2t# 2t$ b œ 3t$ 2t# Ê 'C xy dx (x y) dy œ 'C F †
#
œ 34 t% 23 t$ ‘ " œ ˆ12
16 ‰
3
ˆ 34 23 ‰ œ
45
4
18
3
œ
dr
dt
dt œ 'c" a3t$ 2t# b dt
#
69
4
18. Along (0ß 0) to (1ß 0): r œ ti , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ ti tj and
dr
dt
œi Ê F†
dr
dt
œ t;
Along (1ß 0) to (0ß 1): r œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (1 2t)i j and
dr
dr
dt œ i j Ê F † dt œ 2t;
Along (0ß 1) to (0ß 0): r œ (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i (x y)j Ê F œ (t 1)i (1 t)j and
dr
dt
œ j Ê F †
dr
dt
œ t 1 Ê 'C (x y) dx (x y) dy œ '0 t dt '0 2t dt '0 (t 1) dt œ '0 (4t 1) dt
1
1
1
1
"
œ c2t# td ! œ 2 1 œ 1
19. r œ xi yj œ y# i yj , 2
Ê
c1
'C F † T ds œ '2
F†
1, and F œ x# i yj œ y% i yj Ê
y
dr
dy
dr
dt
œ 2yi j and F †
dr
dy
œ 2y& y
"
4‰
dy œ '2 a2y& yb dy œ 3" y' #" y# ‘ # œ ˆ 3" #" ‰ ˆ 64
3 # œ
20. r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ
Ê F†
c1
dr
dy
1
#
, and F œ yi xj Ê F œ (sin t)i (cos t)j and
œ sin# t cos# t œ 1 Ê
'C F † dr œ '0
1Î2
dr
dt
3
#
63
3
œ 39
#
œ ( sin t)i (cos t)j
(1) dt œ 1#
21. r œ (i j) t(i 2j) œ (1 t)i (1 2t)j , 0 Ÿ t Ÿ 1, and F œ xyi (y x)j Ê F œ a1 3t 2t# b i tj and
dr
dt
œ i 2j Ê F †
dr
dt
œ 1 5t 2t# Ê work œ 'C F †
dr
dt
dt œ '0 a1 5t 2t# b dt œ t 52 t# 23 t$ ‘ ! œ
1
22. r œ (2 cos t)i (2 sin t)j , 0 Ÿ t Ÿ 21, and F œ ™ f œ 2(x y)i 2(x y)j
Ê F œ 4(cos t sin t)i 4(cos t sin t)j and ddtr œ (2 sin t)i (2 cos t)j Ê F †
"
dr
dt
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
25
6
1004 Chapter 16 Integration in Vector Fields
œ 8 asin t cos t sin# tb 8 acos# t cos t sin tb œ 8 acos# t sin# tb œ 8 cos 2t Ê work œ 'C ™ f † dr
œ 'C F †
dt œ '0 8 cos 2t dt œ c4 sin 2td #!1 œ 0
21
dr
dt
23. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 21, F" œ xi yj , and F# œ yi xj Ê
F" œ (cos t)i (sin t)j , and F# œ ( sin t)i (cos t)j Ê F" †
dr
dt
dr
dt
œ ( sin t)i (cos t)j ,
œ 0 and F# †
dr
dt
œ sin# t cos# t œ 1
Ê Circ" œ '0 0 dt œ 0 and Circ# œ '0 dt œ 21; n œ (cos t)i (sin t)j Ê F" † n œ cos# t sin# t œ 1 and
21
21
F# † n œ 0 Ê Flux" œ '0 dt œ 21 and Flux# œ '0 0 dt œ 0
21
21
(b) r œ (cos t)i (4 sin t)j , 0 Ÿ t Ÿ 21 Ê
F# œ (4 sin t)i (cos t)j Ê F" †
dr
dt
dr
dt
œ ( sin t)i (4 cos t)j , F" œ (cos t)i (4 sin t)j , and
œ 15 sin t cos t and F# †
dr
dt
œ 4 Ê Circ" œ '0 15 sin t cos t dt
21
#1
œ "25 sin# t‘ ! œ 0 and Circ# œ '0 4 dt œ 81; n œ Š È417 cos t‹ i Š È"17 sin t‹ j Ê F" † n
21
œ
4
È17
cos# t
4
È17
sin# t and F# † n œ È1517 sin t cos t Ê Flux" œ '0 (F" † n) kvk dt œ '0 Š È417 ‹ È17 dt
21
21
# ‘ #1
œ 81 and Flux# œ '0 (F# † n) kvk dt œ '0 Š È1517 sin t cos t‹ È17 dt œ 15
2 sin t ! œ 0
21
21
24. r œ (a cos t)i (a sin t)j , 0 Ÿ t Ÿ 21, F" œ 2xi 3yj , and F# œ 2xi (x y)j Ê
œ (a sin t)i (a cos t)j ,
dr
dt
F" œ (2a cos t)i (3a sin t)j , and F# œ (2a cos t)i (a cos t a sin t)j Ê n kvk œ (a cos t)i (a sin t)j ,
F" † n kvk œ 2a# cos# t 3a# sin# t, and F# † n kvk œ 2a# cos# t a# sin t cos t a# sin# t
Ê Flux" œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t
21
sin 2t ‘ #1
4
!
3a# 2t
Flux# œ '0 a2a# cos# t a# sin t cos t a# sin# tb dt œ 2a# 2t
21
25. F" œ (a cos t)i (a sin t)j ,
d r"
dt
œ (a sin t)i (a cos t)j Ê F" †
sin 2t ‘ #1
4
!
d r"
dt
sin 2t ‘ #1
4
!
œ 1a# , and
a#
#
#1
csin# td ! a# 2t
sin 2t ‘ #1
4
!
œ 1a#
œ 0 Ê Circ" œ 0; M" œ a cos t,
N" œ a sin t, dx œ a sin t dt, dy œ a cos t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa# cos# t a# sin# tb dt
œ '0 a# dt œ a# 1;
1
1
F # œ ti ,
d r#
dt
œ i Ê F# †
d r#
dt
œ t Ê Circ# œ 'ca t dt œ 0; M# œ t, N# œ 0, dx œ dt, dy œ 0 Ê Flux#
a
œ 'C M# dy N# dx œ 'ca 0 dt œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ a# 1
a
26. F" œ aa# cos# tb i aa# sin# tb j ,
d r"
dt
œ (a sin t)i (a cos t)j Ê F" †
d r"
dt
œ a$ sin t cos# t a$ cos t sin# t
Ê Circ" œ '0 aa$ sin t cos# t a$ cos t sin# tb dt œ 2a3 ; M" œ a# cos# t, N" œ a# sin# t, dy œ a cos t dt,
1
$
dx œ a sin t dt Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos$ t a$ sin$ tb dt œ
1
F # œ t# i ,
d r#
dt
œ i Ê F# †
d r#
dt
œ t# Ê Circ# œ 'ca t# dt œ
a
2a$
3
4
3
a$ ;
; M# œ t# , N# œ 0, dy œ 0, dx œ dt
Ê Flux# œ 'C M# dy N# dx œ 0; therefore, Circ œ Circ" Circ# œ 0 and Flux œ Flux" Flux# œ
27. F" œ (a sin t)i (a cos t)j ,
d r"
dt
œ (a sin t)i (a cos t)j Ê F" †
d r"
dt
4
3
a$
œ a# sin# t a# cos# t œ a#
Ê Circ" œ '0 a# dt œ a# 1 ; M" œ a sin t, N" œ a cos t, dx œ a sin t dt, dy œ a cos t dt
1
Ê Flux" œ 'C M" dy N" dx œ '0 aa# sin t cos t a# sin t cos tb dt œ 0; F# œ tj ,
1
dr#
dt
œ i Ê F# †
d r#
dt
œ0
Ê Circ# œ 0; M# œ 0, N# œ t, dx œ dt, dy œ 0 Ê Flux# œ 'C M# dy N# dx œ 'ca t dt œ 0; therefore,
a
Circ œ Circ" Circ# œ a# 1 and Flux œ Flux" Flux# œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.2 Vector Fields, Work, Circulation, and Flux 1005
28. F" œ aa# sin# tb i aa# cos# tb j ,
d r"
dt
œ (a sin t)i (a cos t)j Ê F" †
Ê Circ" œ '0 aa$ sin$ t a$ cos$ tb dt œ
1
4
3
d r"
dt
œ a$ sin$ t a$ cos$ t
a$ ; M" œ a# sin# t, N" œ a# cos# t, dy œ a cos t dt, dx œ a sin t dt
Ê Flux" œ 'C M" dy N" dx œ '0 aa$ cos t sin# t a$ sin t cos# tb dt œ
1
2
3
a$ ; F# œ t# j ,
d r#
dt
œ i Ê F# †
d r#
dt
œ0
Ê Circ# œ 0; M# œ 0, N# œ t# , dy œ 0, dx œ dt Ê Flux# œ 'C M# dy N# dx œ 'ca t# dt œ 23 a$ ; therefore,
a
Circ œ Circ" Circ# œ
4
3
a$ and Flux œ Flux" Flux# œ 0
29. (a) r œ (cos t)i (sin t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
F œ (cos t sin t)i acos# t sin# tb j Ê F †
dr
dt
(b)
sin 2t
1
‘1
4 sin t ! œ #
r œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr œ 2i and F œ (1 2t)i (1 2t)# j Ê
1
"
F † ddtr œ 4t 2 Ê C F † T ds œ 0 (4t 2) dt œ c2t# 2td ! œ 0
r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê ddtr" œ i j and F œ (1 2t)i a1 2t
'
(c)
œ (sin t)i (cos t)j and
œ sin t cos t sin# t cos t Ê 'C F † T ds
œ '0 a sin t cos t sin# t cos tb dt œ "2 sin# t
1
dr
dt
Ê F†
t
#
'
œ (2t 1) a1 2t 2t# b œ 2t# Ê Flow" œ 'C F †
d r"
dt
"
0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
œ i a2t# 2t 1b j Ê F †
"
œ t# 23 t$ ‘ ! œ
"
3
dr#
dt
dr#
dt
dr"
dt
œ '0 2t# dt œ
1
2
3
2t# b j
; r# œ ti (t 1)j ,
œ i j and F œ i at# t# 2t 1b j
œ 1 a2t# 2t 1b œ 2t 2t# Ê Flow# œ 'C F †
dr#
dt
#
Ê Flow œ Flow" Flow# œ
2
3
"
3
œ '0 a2t 2t# b dt
1
œ1
30. From (1ß 0) to (0ß 1): r" œ (1 t)i tj , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
d r"
dt
œ i j ,
F œ i a1 2t 2t# b j , and n" kv" k œ i j Ê F † n" kv" k œ 2t 2t# Ê Flux" œ '0 a2t 2t# b dt
1
"
œ t# 23 t$ ‘ ! œ
"
3
;
From (0ß 1) to (1ß 0): r# œ ti (1 t)j , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
d r#
dt
œ i j ,
#
F œ (1 2t)i a1 2t 2t b j , and n# kv# k œ i j Ê F † n# kv# k œ (2t 1) a1 2t 2t# b œ 2 4t 2t#
Ê Flux# œ '0 a2 4t 2t# b dt œ 2t 2t# 23 t$ ‘ ! œ 23 ;
1
"
From (1ß 0) to (1ß 0): r$ œ (1 2t)i , 0 Ÿ t Ÿ 1, and F œ (x y)i ax# y# b j Ê
#
d r$
dt
œ 2i ,
#
F œ (1 2t)i a1 4t 4t b j , and n$ kv$ k œ 2j Ê F † n$ kv$ k œ 2 a1 4t 4t b
Ê Flux$ œ 2 '0 a1 4t 4t# b dt œ 2 t 2t# 43 t$ ‘ ! œ
1
31. F œ Èx#y y# i
"
x
È x# y#
2
3
Ê Flux œ Flux" Flux# Flux$ œ
j on x# y# œ 4;
at (2ß 0), F œ j ; at (0ß 2), F œ i ; at (2ß 0),
È
F œ j ; at (!ß 2), F œ i ; at ŠÈ2ß È2‹ , F œ #3 i "# j ;
at ŠÈ2ß È2‹ , F œ
Fœ
È3
#
È3
#
i "# j ; at ŠÈ2ß È2‹ ,
i "# j ; at ŠÈ2ß È2‹ , F œ
È3
#
i "# j
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
3
2
3
2
3
œ
"
3
1006 Chapter 16 Integration in Vector Fields
32. F œ xi yj on x# y# œ 1; at (1ß 0), F œ i ;
at (1ß 0), F œ i ; at (0ß 1), F œ j ; at (0ß 1),
F œ j ; at Š "# ß
È3
# ‹,
Fœ
"
#
at Š "# ß
È3
# ‹,
F œ "# i
at Š "# ß
È3
# ‹,
Fœ
at Š "# ß
È3
# ‹,
"
#
i
i
È3
#
È3
#
È3
#
j;
j;
j;
F œ "# i
È3
#
j.
33. (a) G œ P(xß y)i Q(xß y)j is to have a magnitude Èa# b# and to be tangent to x# y# œ a# b# in a
counterclockwise direction. Thus x# y# œ a# b# Ê 2x 2yyw œ 0 Ê yw œ xy is the slope of the tangent
line at any point on the circle Ê yw œ ba at (aß b). Let v œ bi aj Ê kvk œ Èa# b# , with v in a
counterclockwise direction and tangent to the circle. Then let P(xß y) œ y and Q(xß y) œ x
Ê G œ yi xj Ê for (aß b) on x# y# œ a# b# we have G œ bi aj and kGk œ Èa# b# .
(b) G œ ˆÈx# y# ‰ F œ ŠÈa# b# ‹ F .
34. (a) From Exercise 33, part a, yi xj is a vector tangent to the circle and pointing in a counterclockwise
direction Ê yi xj is a vector tangent to the circle pointing in a clockwise direction Ê G œ Èyxi #xjy#
is a unit vector tangent to the circle and pointing in a clockwise direction.
(b) G œ F
35. The slope of the line through (xß y) and the origin is
pointing away from the origin Ê F œ
xi yj
È x# y#
y
x
Ê v œ xi yj is a vector parallel to that line and
is the unit vector pointing toward the origin.
36. (a) From Exercise 35, Èxxi #yjy# is a unit vector through (xß y) pointing toward the origin and we want
kFk to have magnitude Èx# y# Ê F œ Èx# y# Š Èxxi #yjy# ‹ œ xi yj .
(b) We want kFk œ
C
È x# y#
37. F œ 4t$ i 8t# j 2k and
38. F œ 12t# j 9t# k and
dr
dt
where C Á 0 is a constant Ê F œ
dr
dt
œ i 2tj Ê F †
œ 3j 4k Ê F †
39. F œ (cos t sin t)i (cos t)k and
dr
dt
dr
dt
dr
dt
40. F œ (2 sin t)i (2 cos t)j 2k and
œ 12t$ Ê Flow œ '0 12t$ dt œ c3t% d ! œ 48
2
#
1
œ ( sin t)i (cos t)k Ê F †
dr
dt
yj
Š Èxxi #yjy# ‹ œ C Š xx#i
y# ‹.
œ 72t# Ê Flow œ '0 72t# dt œ c24t$ d ! œ 24
Ê Flow œ '0 ( sin t cos t 1) dt œ "2 cos# t
1
C
È x# y#
1
t‘ !
dr
dt
"
œ sin t cos t 1
œ ˆ "# 1‰ ˆ "# 0‰ œ 1
œ (2 sin t)i (2 cos t)j 2k Ê F †
dr
dt
œ 4 sin# t 4 cos# t 4 œ 0
Ê Flow œ 0
41. C" : r œ (cos t)i (sin t)j tk , 0 Ÿ t Ÿ
Ê F†
dr
dt
1
#
Ê F œ (2 cos t)i 2tj (2 sin t)k and
dr
dt
œ ( sin t)i (cos t)j k
œ 2 cos t sin t 2t cos t 2 sin t œ sin 2t 2t cos t 2 sin t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.2 Vector Fields, Work, Circulation, and Flux 1007
Ê Flow" œ '0
1Î2
C# : r œ j
1
#
1Î#
( sin 2t 2t cos t 2 sin t) dt œ "2 cos 2t 2t sin t 2 cos t 2 cos t‘ !
(1 t)k , 0 Ÿ t Ÿ 1 Ê F œ 1(1 t)j 2k and
dr
dt
Ê Flow# œ '0 1 dt œ c1td "! œ 1;
1
#
œ k Ê F†
dr
dt
œ 1 1 ;
œ 1
1
C$ : r œ ti (1 t)j , 0 Ÿ t Ÿ 1 Ê F œ 2ti 2(1 t)k and
dr
dt
œij Ê F†
Ê Flow$ œ '0 2t dt œ ct# d ! œ 1 Ê Circulation œ (1 1) 1 1 œ 0
1
42. F †
œx
dr
dt
dx
dt
y
dr
dt
œ 2t
"
dy
dt
z
` f dx
` x dt
œ
dz
dt
` f dy
` y dt
by the chain rule Ê Circulation œ 'C F †
dr
dt
` f dz
` z dt
dt œ 'a
, where f(xß yß z) œ
b
d
dt afaratbbb
"
#
ax# y# x# b Ê F †
dr
dt
œ
d
dt afaratbbb
dt œ farabbb faraabb. Since C is an entire ellipse,
rabb œ raab, thus the Circulation œ 0.
43. Let x œ t be the parameter Ê y œ x# œ t# and z œ x œ t Ê r œ ti t# j tk , 0 Ÿ t Ÿ 1 from (0ß 0ß 0) to (1ß 1ß 1)
Ê
œ
dr
dt
œ i 2tj k and F œ xyi yj yzk œ t$ i t# j t$ k Ê F †
œ t$ 2t$ t$ œ 2t$ Ê Flow œ '0 2t$ dt
1
"
#
44. (a) F œ ™ axy# z$ b Ê F †
œ 'a
(b)
dr
dt
b
d
dt afaratbbb
dr
dt
œ
` f dx
` x dt
` f dy
` y dt
` z dz
` z dt
œ
df
dt
, where f(xß yß z) œ xy# z$ Ê )C F †
dr
dt
dt
dt œ farabbb faraabb œ 0 since C is an entire ellipse.
Ð2ß1ß1Ñ
'C F † ddtr œ 'Ð1ß1ß1Ñ
Ð#ß"ß"Ñ
axy# z$ b dt œ cxy# z$ d Ð"ß"ß"Ñ œ (2)(1)# (1)$ (1)(1)# (1)$ œ 2 1 œ 3
d
dt
45. Yes. The work and area have the same numerical value because work œ 'C F † dr œ 'C yi † dr
œ 'b [f(t)i] † i
a
df
dt
j‘ dt
[On the path, y equals f(t)]
œ 'a f(t) dt œ Area under the curve
b
46. r œ xi yj œ xi f(x)j Ê
from the origin Ê F †
Ê
'C
dr
dx
œ
F † T ds œ 'C F †
dr
dx
dr
dx
[because f(t) 0]
œ i f w (x)j ; F œ
kx
È x# y#
k†y†f (x)
È x# y#
dx œ 'a k
b
w
d
dx
k
È x# y#
œ
(xi yj) has constant magnitude k and points away
kx k†f(x)†f (x)
Èx# [f(x)]#
w
œk
d
dx
Èx# [f(x)]# , by the chain rule
Èx# [f(x)]# dx œ k Èx# [f(x)]# ‘ b
a
œ k ˆÈb# [f(b)]# Èa# [f(a)]# ‰ , as claimed.
47-52. Example CAS commands:
Maple:
with( LinearAlgebra );#47
F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >;
r := t -> < 2*cos(t) | sin(t) >;
a,b := 0,2*Pi;
dr := map(diff,r(t),t);
F(r(t));
q1 := simplify( F(r(t)) . dr ) assuming t::real;
q2 := Int( q1, t=a..b );
value( q2 );
Mathematica: (functions and bounds will vary):
Exercises 47 and 48 use vectors in 2 dimensions
Clear[x, y, t, f, r, v]
f[x_, y_]:= {x y6 , 3x (x y5 2)}
# (a)
# (b)
# (c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1008 Chapter 16 Integration in Vector Fields
{a, b}={0, 21};
x[t_]:= 2 Cos[t]
y[t_]:= Sin[t]
r[t_]:={x[t], y[t]}
v[t_]:= r'[t]
integrand= f[x[t], y[t]] . v[t] //Simplify
Integrate[integrand,{t, a, b}]
N[%]
If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises
49 - 52 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied.
Clear[x, y, z, t, f, r, v]
f[x_, y_, z_]:= {y y z Cos[x y z], x2 x z Cos[x y z], z x y Cos[x y z]}
{a, b}={0, 21};
x[t_]:= 2 Cos[t]
y[t_]:= 3 Sin[t]
z[t_]:= 1
r[t_]:={x[t], y[t], z[t]}
v[t_]:= r'[t]
integrand= f[x[t], y[t],z[t]] . v[t] //Simplify
NIntegrate[integrand,{t, a, b}]
16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS
1.
`P
`y
œxœ
2.
`P
`y
œ x cos z œ
3.
`P
`y
œ 1 Á 1 œ
5.
`N
`x
œ0Á1œ
6.
`P
`y
œ0œ
7.
`f
`x
œ 2x Ê f(xß yß z) œ x# g(yß z) Ê
Ê
8.
`f
`x
`f
`z
`N
`z
`N
`z
,
,
`M
`z
`N
`z
œyœ
,
`N
`z
`M
`y
`M
`z
`M
`z
`P
`x
,
`N
`x
`M
`y
œzœ
œ y cos z œ
`P
`x
,
`f
`x
`N
`x
œ0œ
`P
`x
,
`N
`x
`M
`y
œ ex sin y œ
`f
`y
œ
`f
`z
`M
`y
Ê Not Conservative
`g
`y
œ 3y Ê g(yß z) œ
h(z) Ê
`f
`z
œ 2xe
y2z
w
œx
`f
`y œ
y2z
3y#
#
h(z)
2z# C
`g
`y
œxz Ê
œ z Ê g(yß z) œ zy h(z)
w
xey2z
`f
`y
`g
`y
œ xey2z Ê
`g
`y
œ 0 Ê f(xß yß z)
w
Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xey2z C
h (z) œ 2xe
`f
`z
`g
`y
h(z) Ê f(xß yß z) œ x#
w
œ y sin z Ê f(xß yß z) œ xy sin z g(yß z) Ê
Ê f(xß yß z) œ xy sin z h(z) Ê
`f
`y
3y
#
#
3y#
#
œ x y h (z) œ x y Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
œ ey2z Ê f(xß yß z) œ xey2z g(yß z) Ê
œ xy sin z C
œ 1 Á 1 œ
Ê Conservative
œ hw (z) œ 4z Ê h(z) œ 2z# C Ê f(xß yß z) œ x#
y2z
10.
Ê Conservative
4.
œ y z Ê f(xß yß z) œ (y z)x g(yß z) Ê
œ xe
`M
`y
Ê Not Conservative
œ (y z)x zy C
`f
`x
œ sin z œ
Ê Not Conservative
Ê f(xß yß z) œ (y z)x zy h(z) Ê
9.
`N
`x
Ê Conservative
œ x sin z
w
`g
`y
œ x sin z Ê
`g
`y
œ 0 Ê g(yß z) œ h(z)
w
œ xy cos z h (z) œ xy cos z Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1009
11.
`f
`z
œ
Ê f(xß yß z) œ
z
y # z#
"
#
œ (x ln x x) tan (x y) h(y) Ê f(xß yß z) œ
`f
`y
Ê
12.
œ
"
#
`f
`x
œ
œ
y
y # z#
#
#
"
#
`g
`x œ
#
œ
ln x sec# (x y) Ê g(xß y)
ln ay# z b (x ln x x) tan (x y) h(y)
sec# (x y) hw (y) œ sec# (x y)
y
y # z#
Ê hw (y) œ 0 Ê h(y) œ C Ê f(xß yß z)
ln ay z b (x ln x x) tan (x y) C
y
1 x# y#
`g
`y
œ
z
È 1 y # z#
Ê
`f
`z
œ
y
È 1 y # z#
hw (z) œ
Ê f(xß yß z) œ tan
"
(xy) sin
`f
`z
"
z
y
È 1 y # z#
`g
`y
x
1 x# y#
œ
"
z
Ê hw (z) œ
`P
`y
`N
`z
`M
`P `N
`M
`z œ 0 œ `x , `x œ 0 œ `y
`g
`f
#
` y œ ` y œ 2y Ê g(yß z) œ y
œ0œ
œ 2x Ê f(xß yß z) œ x# g(yß z) Ê
`P
`y
14. Let F(xß yß z) œ yzi xzj xyk Ê
œ yz Ê f(xß yß z) œ xyz g(yß z) Ê
`f
`z
œ xyz h(z) Ê
Ð3ß5ß0Ñ
'Ð1ß1ß2Ñ
Ê
`N
`z
œxœ
Ê M dx N dy P dz is
,
,
`f
`y
`M
`z
œyœ
`g
`y
œ xz
`P
`x
,
`N
`x
`M
`y
œzœ
`g
`y
œ xz Ê
h(z) Ê f(xß yß z) œ x# y# œ h(z)
'Ð0Ð2ß0ß3ß0ßÑ 6Ñ 2x dx 2y dy 2z dz
œ f(2ß 3ß 6) f(!ß !ß !) œ 2# 3# (6)# œ 49
exact;
z
È 1 y # z#
Ê h(z) œ ln kzk C
œ hw (z) œ 2z Ê h(z) œ z# C Ê f(xß yß z) œ x# y# z# C Ê
`f
`x
x
1 x# y#
(yz) ln kzk C
13. Let F(xß yß z) œ 2xi 2yj 2zk Ê
`f
`x
œ
Ê g(yß z) œ sin" (yz) h(z) Ê f(xß yß z) œ tan" (xy) sin" (yz) h(z)
"
exact;
`f
`y
Ê f(xß yß z) œ tan" (xy) g(yß z) Ê
Ê
Ê
`f
`x
ln ay# z# b g(xß y) Ê
Ê M dx N dy P dz is
œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z)
œ xy hw (z) œ xy Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xyz C
yz dx xz dy xy dz œ f(3ß 5ß 0) f(1ß 1ß 2) œ 0 2 œ 2
15. Let F(xß yß z) œ 2xyi ax# z# b j 2yzk Ê
Ê M dx N dy P dz is exact;
`f
`x
`P
`y
`N
`z
œ 2z œ
,
`M
`z
œ0œ
`P
`x
œ 2xy Ê f(xß yß z) œ x# y g(yß z) Ê
Ê g(yß z) œ yz# h(z) Ê f(xß yß z) œ x# y yz# h(z) Ê
`f
`z
,
`N
`x
`f
`y
w
œ 2x œ
œ x#
`g
`y
`M
`y
`g
`y
œ x# z# Ê
œ z#
œ 2yz h (z) œ 2yz Ê hw (z) œ 0 Ê h(z) œ C
Ê f(xß yß z) œ x# y yz# C Ê 'Ð0ß0ß0Ñ 2xy dx ax# z# b dy 2yz dz œ f("ß #ß $) f(!ß !ß !) œ 2 2(3)# œ 16
Ð1ß2ß3Ñ
16. Let F(xß yß z) œ 2xi y# j ˆ 1 4 z# ‰ k Ê
Ê M dx N dy P dz is exact;
Ê f(xß yß z) œ x#
œ x#
y$
3
œ ˆ9
27
3
y$
3
h(z) Ê
`f
`x
`P
`y
œ0œ
`N
`z
`f
`z
,
`N
`x
œ0œ
`f
`y
œ
`M
`y
`g
`y
$
œ y# Ê g(yß z) œ y3 h(z)
4
1 z#
dz œ f(3ß 3ß 1) f(!ß !ß !)
(! ! 0) œ 1
17. Let F(xß yß z) œ (sin y cos x)i (cos y sin x)j k Ê
Ê M dx N dy P dz is exact;
œ cos y sin x Ê
`P
`x
œ0œ
œ hw (z) œ 1 4 z# Ê h(z) œ 4 tan" z C Ê f(xß yß z)
Ð3ß3ß1Ñ
4†
`M
`z
œ 2x Ê f(xß yß z) œ x# g(yß z) Ê
4 tan" z C Ê 'Ð0ß0ß0Ñ 2x dx y# dy
1‰
4
,
`g
`y
`f
`x
`P
`y
œ0œ
`N
`z
,
`M
`z
œ0œ
`P
`x
,
`N
`x
œ cos y cos x œ
œ sin y cos x Ê f(xß yß z) œ sin y sin x g(yß z) Ê
œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ sin y sin x h(z) Ê
Ê f(xß yß z) œ sin y sin x z C Ê
' 100101
Ð ß ß Ñ
Ð ß ß Ñ
`f
`z
`f
`y
`M
`y
œ cos y sin x
`g
`y
œ hw (z) œ 1 Ê h(z) œ z C
sin y cos x dx cos y sin x dy dz œ f(0ß 1ß 1) f(1ß !ß !)
œ (0 1) (0 0) œ 1
18. Let F(xß yß z) œ (2 cos y)i Š "y 2x sin y‹ j ˆ "z ‰ k Ê
Ê M dx N dy P dz is exact;
œ
"
y
2x sin y Ê
`g
`y
œ
"
y
`f
`x
`P
`y
œ0œ
`N
`z
,
`M
`z
œ0œ
`P
`x
œ 2 cos y Ê f(xß yß z) œ 2x cos y g(yß z) Ê
,
`f
`y
`N
`x
œ 2 sin y œ
œ 2x sin y
Ê g(yß z) œ ln kyk h(z) Ê f(xß yß z) œ 2x cos y ln kyk h(z) Ê
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
`f
`z
`M
`y
`g
`y
œ hw (z) œ
"
z
1010 Chapter 16 Integration in Vector Fields
Ê h(z) œ ln kzk C Ê f(xß yß z) œ 2x cos y ln kyk ln kzk C
Ê 'Ð0ß2ß1Ñ
Ð1ß1Î2ß2Ñ
2 cos y dx Š "y 2x sin y‹ dy
1
#
œ ˆ2 † 0 ln
"
z
dz œ f ˆ1ß 1# ß 2‰ f(!ß #ß ")
ln 2‰ (0 † cos 2 ln 2 ln 1) œ ln
#
19. Let F(xß yß z) œ 3x# i Š zy ‹ j (2z ln y)k Ê
Ê M dx N dy P dz is exact;
`f
`x
`P
`y
œ
2z
y
1
#
`N
`z
œ
`M
`z
,
œ0œ
`P
`x
`f
`y
œ 3x# Ê f(xß yß z) œ x$ g(yß z) Ê
Ê f(xß yß z) œ x$ z# ln y h(z) Ê
œ x$ z# ln y C Ê 'Ð1ß1ß1Ñ 3x# dx
Ð1ß2ß3Ñ
`N
`x
,
œ0œ
œ
`g
`y
œ
`M
`y
z#
y
Ê g(yß z) œ z# ln y h(z)
`f
`z
œ 2z ln y hw (z) œ 2z ln y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)
z#
y
dy 2z ln y dz œ f(1ß 2ß 3) f("ß "ß ")
œ (1 9 ln 2 C) (1 0 C) œ 9 ln 2
#
`P
`y
20. Let F(xß yß z) œ (2x ln y yz)i Š xy xz‹ j (xy)k Ê
Ê M dx N dy P dz is exact;
x#
y
œ
`g
`y
xz Ê
`f
`x
`N
`z
œ x œ
,
`M
`z
œ y œ
`P
`x
,
`N
`x
œ 2x ln y yz Ê f(xß yß z) œ x# ln y xyz g(yß z) Ê
`f
`z
œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# ln y xyz h(z) Ê
œ
2x
y
`f
`y
œ
zœ
x#
y
`M
`y
xz
`g
`y
œ xy hw (z) œ xy Ê hw (z) œ 0
Ê h(z) œ C Ê f(xß yß z) œ x# ln y xyz C Ê 'Ð1ß2ß1Ñ (2x ln y yz) dx Š xy xz‹ dy xy dz
Ð2ß1ß1Ñ
#
œ f(2ß 1ß 1) f("ß 2ß 1) œ (4 ln 1 2 C) (ln 2 2 C) œ ln 2
21. Let F(xß yß z) œ Š "y ‹ i Š 1z
x
y# ‹ j
Ê M dx N dy P dz is exact;
Ê
`g
`y
œ
"
z
Ê g(yß z) œ
Ê f(xß yß z) œ
x
y
y
z
ˆ zy# ‰ k Ê
`f
`x
"
y
œ
C Ê 'Ð1ß1ß1Ñ
Ð2ß2ß2Ñ
"
y
œ z"# œ
Ê f(xß yß z) œ
h(z) Ê f(xß yß z) œ
y
z
`P
`y
x
y
dx Š 1z
y
z
x
y# ‹
x
y
`N
`z
`M
`z
,
`f
`y
zy#
g(yß z) Ê
`f
`z
h(z) Ê
dy
y
z#
`P
`x
œ0œ
œ
,
`N
`x
œ y1# œ
œ yx#
`g
`y
œ
"
z
`M
`y
x
y#
hw (z) œ zy# Ê hw (z) œ 0 Ê h(z) œ C
dz œ f(2ß 2ß 2) f("ß 1ß 1) œ ˆ 2#
2
#
C‰ ˆ 1"
œ0
22. Let F(xß yß z) œ
Ê
`f
`x
`P
`y
2xi 2yj 2zk
x # y # z#
œ 4yz
œ
3%
`N
`z
,
`M
`z
Šand let 3# œ x# y# z# Ê
œ
2x
x# y# z# Ê f(xß yß z) œ
Ê `` gy œ 0 Ê g(yß z) œ h(z)
w
œ x# 2z
y# z# Ê h (z) œ 0 Ê
Ð2ß2ß2Ñ
Ê 'Ð1ß1ß1Ñ
`P
`x
œ 4xz
œ
3%
2x dx 2y dy 2z dz
x # y # z#
,
`N
`x
œ 4xy
œ
3%
`3
`x
`M
`y
œ
x
3
`3
`y
,
œ
y
3
`3
`z
,
œ 3z ‹
Ê M dx N dy P dz is exact;
`f
`y
ln ax# y# z# b g(yß z) Ê
œ
2y
x # y # z#
Ê f(xß yß z) œ ln ax# y# z# b h(z) Ê
`g
`y
`f
`z
œ
œ
2y
x # y # z#
hw (z)
2z
x # y # z#
h(z) œ C Ê f(xß yß z) œ ln ax# y# z# b C
œ f(2ß 2ß 2) f("ß 1ß 1) œ ln 12 ln 3 œ ln 4
23. r œ (i j k) t(i 2j 2k) œ (1 t)i (1 2t)j (1 2t)k, 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 2 dt, dz œ 2 dt
Ð2ß3ß1Ñ
Ê 'Ð1ß1ß1Ñ y dx x dy 4 dz œ '0 (2t 1) dt (t 1)(2 dt) 4(2) dt œ '0 (4t 5) dt œ c2t# 5td ! œ 3
1
1
24. r œ t(3j 4k), 0 Ÿ t Ÿ 1 Ê dx œ 0, dy œ 3 dt, dz œ 4 dt Ê
' 000304
Ð ß ß Ñ
Ð ß ß Ñ
"
#
x# dx yz dy Š y# ‹ dz
œ '0 a12t# b (3 dt) Š 9t# ‹ (4 dt) œ '0 54t# dt œ c18t# d ! œ 18
1
25.
`P
`y
œ0œ
1
#
`N
`z
,
`M
`z
œ 2z œ
`P
`x
,
`N
`x
,
`M
`z
"
œ0œ
`M
`y
Ê M dx N dy P dz is exact Ê F is conservative
Ê path independence
26.
`P
`y
œ ˆÈ
yz
x # y # z# ‰
$
œ
`N
`z
œ ˆÈ
xz
$
x # y # z# ‰
œ
`P
`x
,
`N
`x
œ ˆÈ
xy
x # y # z# ‰
$
œ
`M
`y
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
"
1
C‰
Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1011
Ê M dx N dy P dz is exact Ê F is conservative Ê path independence
27.
`P
`y
`f
`x
œ0œ
œ
2x
y
`N
`z
,
œ0œ
Ê f(xß y) œ
Ê f(xß y) œ
28.
`M
`z
x#
y
"
y
`N
`z
,
`M
`z
#
x
y
`P
`x
`N
`x
,
œ 2x
y# œ
`P
`x
`f
`x
œ e ln y Ê f(xß yß z) œ e ln y g(yß z) Ê
œ
ex
y
"
y#
Ê gw (y) œ
Ê g(y) œ "y C
1
y ‹
œ cos z œ
x
`N
`x
#
`P
`y
,
1 x#
y#
œ yx# gw (y) œ
C Ê F œ ™ Šx
œ0œ
Ê F is conservative Ê there exists an f so that F œ ™ f;
#
`f
`y
g(y) Ê
`M
`y
œ
`M
`y
x
Ê F is conservative Ê there exists an f so that F œ ™ f;
`f
`y
ex
y
œ
œ y sin z h(z) Ê f(xß yß z) œ e ln y y sin z h(z) Ê
x
`g
ex
`y œ y
`f
`z œ y
x
`g
`y
sin z Ê
œ sin z Ê g(yß z)
w
cos z h (z) œ y cos z Ê hw (z) œ 0
Ê h(z) œ C Ê f(xß yß z) œ ex ln y y sin z C Ê F œ ™ ae ln y y sin zb
29.
`P
`y
œ0œ
`N
`z
`f
`x
œ x# y Ê f(xß yß z) œ
,
`M
`z
Ê f(xß yß z) œ
œ
"
3
x$ xy
œ0œ
" $
3 x xy
" $
z
3 y ze
(a) work œ 'A F †
B
dr
dt
`P
`x
,
"
3
"
3
`N
`x
`M
`y
œ1œ
Ê F is conservative Ê there exists an f so that F œ ™ f;
`f
`y
x$ xy g(yß z) Ê
y$ h(z) Ê
œx
`g
`y
œ y# x Ê
`f
`z
`g
`y
z
œ y# Ê g(yß z) œ
"
3
y$ h(z)
œ hw (z) œ zez Ê h(z) œ zez e C Ê f(xß yß z)
ez C Ê F œ ™ ˆ 3" x$ xy 3" y$ zez ez ‰
dt œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ ˆ 3" 0 0 e e‰ ˆ 3" 0 0 1‰
B
Ð"ß!ß"Ñ
œ1
(b) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1
B
Ð"ß!ß"Ñ
(c) work œ 'A F † dr œ "3 x$ xy 3" y$ zez ez ‘ Ð"ß!ß!Ñ œ 1
B
Ð"ß!ß"Ñ
Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 0) to (1ß 0ß 1).
B
30.
`P
`y
œ xeyz xyzeyz cos y œ
that F œ ™ f;
`f
`x
`N
`z
,
`M
`z
œ yeyz œ
`P
`x
`N
`x
,
œ zeyz œ
œ eyz Ê f(xß yß z) œ xeyz g(yß z) Ê
`f
`y
`M
`y
Ê F is conservative Ê there exists an f so
œ xzeyz
Ê g(yß z) œ z sin y h(z) Ê f(xß yß z) œ xe z sin y h(z) Ê
yz
`g
`y
`f
`z
œ xzeyz z cos y Ê
`g
`y
œ z cos y
w
œ xye sin y h (z) œ xyeyz sin y
yz
Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xeyz z sin y C Ê F œ ™ axeyz z sin yb
(a) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ
B
Ð"ß1Î#ß!Ñ
(b) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ
B
Ð"ß1Î#ß!Ñ
(c) work œ 'A F † dr œ cxeyz z sin yd Ð"ß!ß"Ñ
B
Ð"ß1Î#ß!Ñ
œ (1 0) (1 0) œ 0
œ0
œ0
Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 1) to ˆ1ß 1# ß 0‰ .
B
31. (a) F œ ™ ax$ y# b Ê F œ 3x# y# i 2x$ yj ; let C" be the path from (1ß 1) to (0ß 0) Ê x œ t 1 and
y œ t 1, 0 Ÿ t Ÿ 1 Ê F œ 3(t 1)# (t 1)# i 2(t 1)$ (t 1)j œ 3(t 1)% i 2(t 1)% j
and r" œ (t 1)i (t 1)j Ê dr" œ dt i dt j Ê
'C
"
F † dr" œ '0 c3(t 1)% 2(t 1)% d dt
1
œ '0 5(t 1)% dt œ c(t 1)& d ! œ 1; let C# be the path from (0ß 0) to (1ß 1) Ê x œ t and y œ t,
1
"
0 Ÿ t Ÿ 1 Ê F œ 3t% i 2t% j and r# œ ti tj Ê dr# œ dt i dt j Ê 'C F † dr# œ '0 a3t% 2t% b dt
1
1
œ '0 5t% dt œ 1
Ê 'C F † dr œ 'C F † dr" 'C
"
#
#
F † dr# œ 2
Ð1ß1Ñ
(b) Since f(xß y) œ x$ y# is a potential function for F, 'Ð1ß1Ñ F † dr œ f(1ß 1) f(1ß 1) œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1012 Chapter 16 Integration in Vector Fields
32.
`P
`y
`f
`x
œ0œ
`N
`z
`M
`z
,
œ0œ
`P
`x
,
`N
`x
œ 2x sin y œ
œ 2x cos y Ê f(xß yß z) œ x# cos y g(yß z) Ê
Ê f(xß yß z) œ x# cos y h(z) Ê
(a)
(b)
(c)
(d)
`M
`y
`f
`z
Ê F is conservative Ê there exists an f so that F œ ™ f;
`f
`y
œ x# sin y
`g
`y
œ x# sin y Ê
`g
`y
œ 0 Ê g(yß z) œ h(z)
œ hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x# cos y C Ê F œ ™ ax# cos yb
'C 2x cos y dx x# sin y dy œ cx# cos yd Ð!ß"Ñ
Ð"ß!Ñ œ 0 1 œ 1
'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ
Ð"ß1Ñ œ 1 (1) œ 2
'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ
Ð"ß!Ñ œ 1 1 œ 0
'C 2x cos y dx x# sin y dy œ cx# cos yd Ð"ß!Ñ
Ð"ß!Ñ œ 1 1 œ 0
33. (a) If the differential form is exact, then
all x, and
`N
`x
œ
`M
`y
`P
`y
`N
`z
œ
Ê 2ay œ cy for all y Ê 2a œ c,
`M
`z
œ
`P
`x
Ê 2cx œ 2cx for
Ê by œ 2ay for all y Ê b œ 2a and c œ 2a
(b) F œ ™ f Ê the differential form with a œ 1 in part (a) is exact Ê b œ 2 and c œ 2
34. F œ ™ f Ê g(xß yß z) œ 'Ð0ß0ß0Ñ F † dr œ 'Ð0ß0ß0Ñ ™ f † dr œ f(xß yß z) f(0ß 0ß 0) Ê
ÐxßyßzÑ
`g
`z
œ
`f
`z
ÐxßyßzÑ
`g
`x
œ
`f
`x
0,
`g
`y
œ
`f
`y
0, and
0 Ê ™ g œ ™ f œ F, as claimed
35. The path will not matter; the work along any path will be the same because the field is conservative.
36. The field is not conservative, for otherwise the work would be the same along C" and C# .
37. Let the coordinates of points A and B be axA , yA , zA b and axB , yB , zB b, respectively. The force F œ ai bj ck is
conservative because all the partial derivatives of M, N, and P are zero. Therefore, the potential function is
fax, y, zb œ ax by cz C, and the work done by the force in moving a particle along any path from A to B is
faBb faAb œ f axB , yB , zB b faxA , yA , zA b œ aaxB byB czB Cb aaxA byA czA Cb
Ä
œ aaxB xA b bayB yA b cazB zA b œ F † BA
38. (a) Let GmM œ C Ê F œ C ’
`P
`y
Ê
3yzC
ax# y# z# b&Î#
`f
`x
œ
œ
xC
ax# y# z# b$Î#
yC
Ê `` gy œ
ax# y# z# b$Î#
some f;
œ
œ
`N
`z
,
x
ax# y# z# b$Î#
`M
`z
œ
i
y
ax# y# z# b$Î#
3xzC
ax# y# z# b&Î#
Ê f(xß yß z) œ
œ
,
C
ax# y# z# b"Î#
0 Ê g(yß z) œ h(z) Ê
Ê h(z) œ C" Ê f(xß yß z) œ
`P
`x
C
ax# y# z# b"Î#
`f
`z
œ
j
`N
`x
œ
z
ax# y# z# b$Î#
3xyC
ax# y# z# b&Î#
g(yß z) Ê
zC
ax# y# z# b$Î#
`f
`y
k“
`M
`y
œ
œ
Ê F œ ™ f for
yC
ax# y# z# b$Î#
hw (z) œ
C" . Let C" œ 0 Ê f(xß yß z) œ
zC
ax# y# z# b$Î#
GmM
ax# y# z# b"Î#
`g
`y
is a potential
function for F.
(b) If s is the distance of (xß yß z) from the origin, then s œ Èx# y# z# . The work done by the gravitational field
F is work œ 'P F † dr œ ’ Èx#GmM
“
y # z#
P#
T#
"
T"
œ
GmM
s#
GmM
s"
œ GmM Š s"#
"
s" ‹ ,
as claimed.
16.4 GREEN'S THEOREM IN THE PLANE
1. M œ y œ a sin t, N œ x œ a cos t, dx œ a sin t dt, dy œ a cos t dt Ê
`N
`y
`M
`x
œ 0,
`M
`y
œ 1,
œ 0;
Equation (11):
)C M dy N dx œ '021 [(a sin t)(a cos t) (a cos t)(a sin t)] dt œ '021 0 dt œ 0;
' ' Š ``Mx ``Ny ‹ dx dy œ ' ' 0 dx dy œ 0, Flux
R
`N
`x
R
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
œ 1, and
Section 16.4 Green's Theorem in the Plane 1013
Equation (12):
)C M dx N dy œ '021 [(a sin t)(a sin t) (a cos t)(a cos t)] dt œ '021 a# dt œ 21a# ;
Èa c x
' ' Š ``Nx ``My ‹ dx dy œ ' '
ca cc
a
R
#
œ 2a
ˆ 1#
1‰
#
#
#
2 dy dx œ 'ca 4Èa# x# dx œ 4 ’ x2 Èa# x#
a
sin" xa “
a
ca
#
œ 2a 1, Circulation
2. M œ y œ a sin t, N œ 0, dx œ a sin t dt, dy œ a cos t dt Ê
Equation (11):
a#
#
)C M dy N dx œ '0
21
`M
`x
`M
`y
œ 0,
`N
`x
œ 1,
œ 0, and
`N
`y
œ 0;
a# sin t cos t dt œ a# "2 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Flux
#1
R
21
#1
Equation (12): )C M dx N dy œ '0 aa# sin# tb dt œ a# 2t sin4 2t ‘ ! œ 1a# ; ' ' Š ``Nx ``My ‹ dx dy
œ ' ' 1 dx dy œ '0
21
R
'0
a
r dr d) œ '0
21
R
a#
#
#
d) œ 1a , Circulation
3. M œ 2x œ 2a cos t, N œ 3y œ 3a sin t, dx œ a sin t dt, dy œ a cos t dt Ê
`N
`y
`M
`x
œ 2,
`M
`y
œ 0,
`N
`x
œ 0, and
œ 3;
Equation (11):
)C M dy N dx œ '021 [(2a cos t)(a cos t) (3a sin t)(a sin t)] dt
œ '0 a2a# cos# t 3a# sin# tb dt œ 2a# 2t
21
sin 2t ‘ #1
4
!
3a# 2t
sin 2t ‘ #1
4
!
œ 21a# 31a# œ 1a# ;
' ' Š ``Mx ``Ny ‹ œ ' ' 1 dx dy œ ' ' r dr d) œ ' a## d) œ 1a# , Flux
0
0
0
21
R
a
21
R
Equation (12):
)C M dx N dy œ '021 [(2a cos t)(a sin t) (3a sin t)(a cos t)] dt
#1
œ '0 a2a# sin t cos t 3a# sin t cos tb dt œ 5a# 12 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Circulation
21
R
4. M œ x# y œ a$ cos# t, N œ xy# œ a$ cos t sin# t, dx œ a sin t dt, dy œ a cos t dt
Ê ``Mx œ 2xy, ``My œ x2 , ``Nx œ y# , and ``Ny œ 2xy;
Equation (11):
)C M dy N dx œ '021 aa% cos$ t sin t a% cos t sin$ tb œ ’ a4
%
cos% t
a%
4
sin% t“
' ' Š ``Mx ``Ny ‹ dx dy œ ' ' (2xy 2xy) dx dy œ 0, Flux
R
#1
!
œ 0;
R
)C M dx N dy œ '021 aa% cos# t sin# t a% cos# t sin# tb dt œ '021 a2a% cos# t sin# tb dt
21
41
%1
œ '0 "# a% sin# 2t dt œ a4 '0 sin# u du œ a4 u2 sin42u ‘ ! œ 1#a ; ' ' Š ``Nx ``My ‹ dx dy œ ' ' ay# x# b dx dy
Equation (12):
%
%
21
a
21
œ '0 '0 r# † r dr d) œ '0 a4
%
5. M œ x y, N œ y x Ê
`M
`x
%
R
d) œ
1 a%
#
, Circulation
œ 1,
`M
`y
œ 1,
`N
`x
œ 1,
`N
`y
Circ œ ' ' [1 (1)] dx dy œ 0
R
œ 1 Ê Flux œ ' ' 2 dx dy œ '0
1
R
'01 2 dx dy œ 2;
R
6. M œ x# 4y, N œ x y# Ê
`M
`x
œ 2x,
`M
`y
œ 4,
`N
`x
œ 1,
`N
`y
œ 2y Ê Flux œ ' ' (2x 2y) dx dy
R
1
1
1
1
"
"
œ '0 '0 (2x 2y) dx dy œ '0 cx# 2xyd ! dy œ '0 (1 2y) dy œ cy y# d ! œ 2; Circ œ ' ' (1 4) dx dy
œ '0
1
'0 3 dx dy œ 3
R
1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1014 Chapter 16 Integration in Vector Fields
`M
`x
7. M œ y# x# , N œ x# y# Ê
`M
`y
œ 2x,
`N
`x
œ 2y,
œ 2x,
œ 2y Ê Flux œ ' ' (2x 2y) dx dy
`N
`y
R
3
x
3
$
œ '0 '0 (2x 2y) dy dx œ '0 a2x# x# b dx œ "3 x$ ‘ ! œ 9; Circ œ ' ' (2x 2y) dx dy
R
3
x
3
œ '0 '0 (2x 2y) dy dx œ '0 x# dx œ 9
`M
`x
8. M œ x y, N œ ax# y# b Ê
œ '0
1
`M
`y
œ 1,
'0 (1 2y) dy dx œ '0 ax x b dx œ
x
1
œ 2x,
œ 2y Ê Flux œ ' ' (1 2y) dx dy
`N
`y
; Circ œ ' ' (2x 1) dx dy œ '0
1
"
6
#
`N
`x
œ 1,
R
œ '0 a2x# xb dx œ 76
R
'0 (2x 1) dy dx
x
1
`M
`x
9. M œ x ex sin y, N œ x ex cos y Ê
Ècos 2)
1Î4
Ê Flux œ ' ' dx dy œ 'c1Î4 '0
R
œ 1 ex sin y,
Î
`M
`y
œ ex cos y,
1 4
1Î%
Ècos 2)
1Î4
R
R
y
x
, N œ ln ax# y# b Ê
Ê Flux œ ' ' Š x#yy#
R
Circ œ ' ' Š x# 2x
y#
x
x# y# ‹
R
`M
`x
11. M œ xy, N œ y# Ê
œ '0 Š 3x#
1
#
3x%
# ‹
2y
x# y# ‹
dx œ
`M
`x
dx dy œ '0
1
dx dy œ '0
1
`M
`y
œ y,
y
x# y#
œ
'1
`M
`y
œ
œ 0,
`M
`y
œ 0,
Ê Flux œ ' ' (x sin y) dx dy œ '0
1Î2
R
œ
2x
x# y#
,
`N
`y
œ
"
#
2y
x# y#
1
`N
`y
œ 2y Ê Flux œ ' ' (y 2y) dy dx œ '0
1
R
`M
`x
`N
`x
Î
1 4
#
1
12. M œ sin y, N œ x cos y Ê
,
;
r dr d) œ '1Î4 ˆ "# cos 2)‰ d) œ
'12 ˆ r sinr ) ‰ r dr d) œ '01 sin ) d) œ 2;
; Circ œ ' ' x dy dx œ '0
"
5
x
x# y#
"
#
œ ex sin y
)‰
ˆ r cos
r dr d) œ '0 cos ) d) œ 0
r#
2
`N
`x
œ x,
,
`N
`y
œ 1 ex cos y,
r dr d) œ '1Î4 ˆ "# cos 2)‰ d) œ 4" sin 2)‘ 1Î% œ
Circ œ ' ' a1 ex cos y ex cos yb dx dy œ ' ' dx dy œ 'c1Î4 '0
10. M œ tan"
`N
`x
œ cos y,
'0
1Î2
R
'xx 3y dy dx
#
'xx x dy dx œ '01 ax# x$ b dx œ 1"#
#
`N
`x
`N
`y
œ cos y,
œ x sin y
(x sin y) dx dy œ '0 Š 18 sin y‹ dy œ 18 ;
1Î2
#
#
1Î#
Circ œ ' ' [cos y ( cos y)] dx dy œ '0 '0 2 cos y dx dy œ '0 1 cos y dy œ c1 sin yd ! œ 1
1Î2
1Î2
1Î2
R
13. M œ 3xy
x
1 y#
`M
`x
, N œ ex tan " y Ê
Ê Flux œ ' ' Š3y
R
"
1 y#
œ 3y
R
$
ex
y
Ê
3 c x#
`M
`y
`N
`y
œ
"
1 y#
21
œ '0 a$ (1 cos ))$ (sin )) d) œ ’ a4 (1 cos ))% “
14. M œ y ex ln y, N œ
,
dx dy œ ' ' 3y dx dy œ '0
"
1 y# ‹
21
"
1 y#
œ1
ex
y
,
`N
`x
œ
ex
y
#1
!
'0aÐ1 cos Ñ
)
(3r sin )) r dr d)
œ 4a$ a4a$ b œ 0
Ê Circ œ ' ' ’ ey Š1
x
R
ex
y ‹“
œ 'c1 'x% b 1 dy dx œ 'c1 ca3 x b ax 1bd dx œ 'c1 ax x 2b dx œ
1
15. M œ 2xy$ , N œ 4x# y# Ê
œ '0
1
'0
x$
2xy dy dx œ '
#
1
#
`M
`y
œ 6xy# ,
`N
`x
"!
2
33
1
2
0 3
x
dx œ
%
1
%
#
dx dy œ ' ' (1) dx dy
R
44
15
œ 8xy# Ê work œ )C 2xy$ dx 4x# y# dy œ ' ' a8xy# 6xy# b dx dy
R
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.4 Green's Theorem in the Plane 1015
`M
`y
16. M œ 4x 2y, N œ 2x 4y Ê
œ 2,
`N
`x
œ 2 Ê work œ )C (4x 2y) dx (2x 4y) dy
œ ' ' [2 (2)] dx dy œ 4 ' ' dx dy œ 4(Area of the circle) œ 4(1 † 4) œ 161
R
R
`M
`y
17. M œ y# , N œ x# Ê
œ '0
1
1cx
'0
œ 2y,
œ 2x Ê )C y# dx x# dy œ ' ' (2x 2y) dy dx
`N
`x
R
(2x 2y) dy dx œ '0 a3x 4x 1b dx œ cx 2x# xd ! œ 1 2 1 œ 0
1
`M
`y
18. M œ 3y, N œ 2x Ê
œ 3,
#
`N
`x
œ 2 Ê )C 3y dx 2x dy œ ' ' (2 3) dx dy œ '0
`M
`y
œ 6,
1
R
œ '0 sin x dx œ 2
1
19. M œ 6y x, N œ y 2x Ê
"
$
`N
`x
'0sin x 1 dy dx
œ 2 Ê )C (6y x) dx (y 2x) dy œ ' ' (2 6) dy dx
R
œ 4(Area of the circle) œ 161
20. M œ 2x y# , N œ 2xy 3y Ê
`M
`y
œ 2y,
`N
`x
œ 2y Ê
)C a2x y# b dx (2xy 3y) dy œ ' ' (2y 2y) dx dy œ 0
R
21. M œ x œ a cos t, N œ y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt Ê Area œ
œ
'0
21
"
#
aa# cos# t a# sin# tb dt œ
"
#
'0
21
'021 aab cos# t ab sin# tb dt œ "# '021 ab dt œ 1ab
"
#
)C
"
#
)C x dy y dx
x dy y dx
a# dt œ 1a#
22. M œ x œ a cos t, N œ y œ b sin t Ê dx œ a sin t dt, dy œ b cos t dt Ê Area œ
œ
"
#
"
#
23. M œ x œ a cos$ t, N œ y œ sin$ t Ê dx œ 3 cos# t sin t dt, dy œ 3 sin# t cos t dt Ê Area œ
'0
21
œ
"
#
œ
3
16
a3 sin# t cos# tb acos# t sin# tb dt œ
u2
sin 2u ‘ %1
4
!
"
#
'0
21
a3 sin# t cos# tb dt œ
3
8
œ
3
8
1
24. M œ x œ t# , N œ y œ
t$
3
t Ê dx œ 2t dt, dy œ at# 1b dt Ê Area œ
È
È
'0
21
"
#
"
#
'cÈ33 ’t# at# 1b Š t3 t‹ (2t)“ dt œ "# 'cÈ33 ˆ 3" t% t# ‰ dt œ 12 151 t&
œ
8
5
È3
25. (a) M œ f(x), N œ g(y) Ê
`M
`y
œ 0,
`N
`x
R
(b) M œ ky, N œ hx Ê
`M
`y
œ k,
`N
`x
È$
31 t$ ‘ È$ œ
œ 0 Ê )C f(x) dx g(y) dy œ ' ' Š ``Nx
R
œ ' ' 0 dx dy œ 0
œ h Ê )C ky dx hx dy œ ' ' Š ``Nx
œ ' ' (h k) dx dy œ (h k)(Area of the region)
3
16
'0
sin# u du
)C x dy y dx
œ
$
sin# 2t dt œ
)C x dy y dx
41
R
`M
`y ‹
`M
`y ‹
"
15
Š9È3 15È3‹
dx dy
dx dy
R
26. M œ xy# , N œ x# y 2x Ê
`M
`y
œ 2xy,
`N
`x
œ 2xy 2 Ê )C xy# dx ax# y 2xb dy œ ' ' Š ``Nx
œ ' ' (2xy 2 2xy) dx dy œ 2 ' ' dx dy œ 2 times the area of the square
R
R
R
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
`M
`y ‹
dx dy
1016 Chapter 16 Integration in Vector Fields
27. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's
Theorem, with M œ 4x$ y and N œ x% , )C 4x$ y dx x% dy œ ' ' ’ ``x ax% b
`
`y
R
œ ' ' ðóóñóóò
a4x$ 4x$ b dx dy œ 0.
a4x$ yb“ dx dy
R
0
28. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with
`
`
$
$
M œ x$ and N œ y$ , )C y$ dy x$ dx œ ' ' ”ðñò
` x ay b ï
` y ax b • dx dy œ 0.
R
0
`M
`x
29. Let M œ x and N œ 0 Ê
œ 1 and
`N
`y
œ0 Ê
0
)C M dy N dx œ ' ' Š ``Mx ``Ny ‹ dx dy
œ ' ' (1 0) dx dy Ê Area of R œ ' ' dx dy œ )C x dy; similarly, M œ y and N œ 0 Ê
R
`N
`x
R
œ 0 Ê )C M dx N dy œ ' ' Š ``Nx
R
œ ' ' dx dy œ Area of R
Ê )C x dy
R
`M
`y ‹
`M
`y
œ 1 and
dy dx Ê )C y dx œ ' ' (0 1) dy dx Ê )C y dx
R
R
30.
'ab f(x) dx œ Area of R œ )C y dx, from Exercise 29
31. Let $ (xß y) œ 1 Ê x œ
My
M
' ' x $ (xßy) dA
œ 'R '
$ (xßy) dA
' ' x dA
œ 'R '
R
dA
' ' x dA
œ
Ê Ax œ ' ' x dA œ ' ' (x 0) dx dy
R
A
R
R
R
œ )C x#
dy, Ax œ ' ' x dA œ ' ' (0 x) dx dy œ ) xy dx, and Ax œ ' ' x dA œ ' ' ˆ 23 x 3" x‰ dx dy
œ)
#
#
"
C 3
R
R
"
3
"
#
x dy xy dx Ê
C
)C x
dy œ )C xy dx œ
#
"
3
)C x
dy xy dx œ Ax
32. If $ (xß y) œ 1, then Iy œ ' ' x# $ (xß y) dA œ ' ' x# dA œ ' ' ax# 0b dy dx œ
R
R
R
R
#
R
"
3
)C
x$ dy,
' ' x# dA œ ' ' a0 x# b dy dx œ ) x# y dx, and ' ' x# dA œ ' ' ˆ 34 x# "4 x# ‰ dy dx
C
R
R
œ)
"
C 4
33. M œ
`f
`y
"
4
$
#
x dy x y dx œ
, N œ `` xf Ê
`M
`y
"
4
œ
)C x
` #f
` y#
,
R
$
#
dy x y dx Ê
`N
`x
"
3
œ `` xf# Ê )C
#
)C x
`f
`y
R
$
dy œ )C x# y dx œ
dx
`f
`x
"
4
)C
dy œ ' ' Š `` xf#
#
R
x$ dy x# y dx œ Iy
` #f
` y# ‹
dx dy œ 0 for such
curves C
34. M œ
"
4
x# y 3" y$ , N œ x Ê
the ellipse
"
4
`M
`y
œ
1
4
x# y# ,
`N
`x
œ 1 Ê Curl œ
`N
`x
`M
`y
œ 1 ˆ "4 x# y# ‰ 0 in the interior of
x# y# œ 1 Ê work œ 'C F † dr œ ' ' ˆ1 4" x# y# ‰ dx dy will be maximized on the region
R œ {(xß y) | curl F}
R
0 or over the region enclosed by 1 œ
2y
35. (a) ™ f œ Š x# 2x
y# ‹ i Š x# y# ‹ j Ê M œ
2x
x# y#
,Nœ
"
4
x# y#
2y
x# y#
; since M, N are discontinuous at (0ß 0), we
compute 'C ™ f † n ds directly since Green's Theorem does not apply. Let x œ a cos t, y œ a sin t Ê dx œ a sin t dt,
dy œ a cos t dt, M œ
2
a
cos t, N œ
2
a
sin t, 0 Ÿ t Ÿ 21, so 'C ™ f † n ds œ 'C M dy N dx
œ '0 ˆ 2a cos t‰aa cos tb ˆ 2a sin t‰aa sin tb ‘dt œ '0 2acos2 t sin2 tbdt œ 41. Note that this holds for any
21
21
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
Section 16.4 Green's Theorem in the Plane 1017
a 0, so 'C ™ f † n ds œ 41 for any circle C centered at a0, 0b traversed counterclockwise and 'C ™ f † n ds œ 41
if C is traversed clockwise.
(b) If K does not enclose the point (0ß 0) we may apply Green's Theorem: 'C ™ f † n ds œ 'C M dy N dx
œ ' ' Š ``Mx
R
`N
`y ‹
dx dy œ ' ' Š ax2 y2 b2
2 ˆy 2 x 2 ‰
R
2 ˆx 2 y 2 ‰
‹
ax 2 y 2 b 2
dx dy œ ' ' 0 dx dy œ 0. If K does enclose the point
R
(0ß 0) we proceed as in Example 6:
Choose a small enough so that the circle C centered at (0ß 0) of radius a lies entirely within K. Green's Theorem
applies to the region R that lies between K and C. Thus, as before, 0 œ ' ' Š ``Mx
R
`N
`y ‹
dx dy
œ 'K M dy N dx 'C M dy N dx where K is traversed counterclockwise and C is traversed clockwise.
Hence by part (a) 0 œ ’ ' M dy N dx “ 41 Ê 41 œ
K
'K ™ f † n ds œ œ 0
'K M dy N dx
œ 'K ™ f † n ds. We have shown:
if (0ß 0) lies inside K
if (0ß 0) lies outside K
41
36. Assume a particle has a closed trajectory in R and let C" be the path Ê C" encloses a simply connected region
R" Ê C" is a simple closed curve. Then the flux over R" is )C F † n ds œ 0, since the velocity vectors F are
"
tangent to C" . But 0 œ )C F † n ds œ )C M dy N dx œ ' ' Š ``Mx
"
"
R"
`N
`y ‹
dx dy Ê Mx Ny œ 0, which is a
contradiction. Therefore, C" cannot be a closed trajectory.
37.
'gg yy
#Ð Ñ
"Ð Ñ
`N
`x
'cd 'gg yy ˆ ``Nx dx‰ dy œ 'cd [N(g# (y)ß y) N(g" (y)ß y)] dy
#Ð Ñ
dx dy œ N(g# (y)ß y) N(g" (y)ß y) Ê
"Ð Ñ
œ 'c N(g# (y)ß y) dy 'c N(g" (y)ß y) dy œ 'c N(g# (y)ß y) dy 'd N(g" (y)ß y) dy œ 'C N dy 'C N dy
d
38.
d
c
#
"
œ )C dy
Ê
'ab 'cd
dy dx œ 'a [M(xß d) M(xß c)] dx œ 'a M(xß d) dx 'a M(xß c) dx œ 'C M dx 'C M dx.
`M
`y
)C N dy œ ' '
d
R
`N
`x
dx dy
b
b
b
3
Because x is constant along C# and C% , 'C M dx œ 'C M dx œ 0
Ê
Š'C
"
M dx 'C
#
M dx 'C
$
#
M dx 'C
%
%
M dx‹ œ )C M dx Ê 'a
b
'cd
`M
`y
"
dy dx œ )C M dx.
39. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field F œ Mi Nj
can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero,
and whose i and j components are independent of z. For such a field to be conservative, we must have
`N
`M
`N
`M
` x œ ` y by the component test in Section 16.3 Ê curl F œ ` x ` y œ 0.
40. Green's theorem tells us that the circulation of a conservative two-dimensional field around any simple closed
curve in the xy-plane is zero. The reasoning: For a conservative field F œ Mi Nj , we have ``Nx œ ``My
(component test for conservative fields, Section 16.3, Eq. (2)), so curl F œ
`N
`x
`M
`y
œ 0. By Green's theorem,
the counterclockwise circulation around a simple closed plane curve C must equal the integral of curl F over the
region R enclosed by C. Since curl F œ 0, the latter integral is zero and, therefore, so is the circulation.
The circulation )C F † T ds is the same as the work )C F † dr done by F around C, so our observation that
circulation of a conservative two-dimensional field is zero agrees with the fact that the work done by a
conservative field around a closed curve is always 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
1018 Chapter 16 Integration in Vector Fields
41-44. Example CAS commands:
Maple:
with( plots );#41
M := (x,y) -> 2*x-y;
N := (x,y) -> x+3*y;
C := x^2 + 4*y^2 = 4;
implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#41(a) (Section 16.4)" );
curlF_k := D[1](N) - D[2](M):
# (b)
'curlF_k' = curlF_k(x,y);
top,bot := solve( C, y );
# (c)
left,right := -2, 2;
q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right );
value( q1 );
Mathematica: (functions and bounds will vary)
The ImplicitPlot command will be useful for 41 and 42, but is not needed for 43 and 44. In 44, the equation of the line
from (0, 4) to (2, 0) must be determined first.
Clear[x, y, f]
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