Chapter 11-16

C H A P T E R 1 1
Vectors and the Geometry of Space
Section 11.1
Vectors in the Plane................................................................................2
Section 11.2
Space Coordinates and Vectors in Space ............................................13
Section 11.3
The Dot Product of Two Vectors.........................................................22
Section 11.4
The Cross Product of Two Vectors in Space ......................................30
Section 11.5
Lines and Planes in Space ....................................................................37
Section 11.6
Surfaces in Space..................................................................................50
Section 11.7
Cylindrical and Spherical Coordinates ................................................57
Review Exercises ..........................................................................................................68
Problem Solving ...........................................................................................................76
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 1
Vectors and the Geometry of Space
Section 11.1 Vectors in the Plane
5 1, 4 2
1. (a) v
4, 2
(b)
6 0, 2 3
6, 5
v
9 3, 5 10
6, 5
u
v
7. u
y
5
4
11 4 , 4 1
8. u
3
(4, 2)
2
25 0, 10 13
v
v
1
u
x
1
2
3
4
v
5 2, 5 0
9. (b) v
0, 6
y
(b)
(c) v
(a), (d)
1
y
−1
(5, 5)
4
−2
3
v
−3
v
2
−4
1
−5
(2, 0)
x
−1
(0, − 6)
−6
4 2, 3 3
3. (a) v
(3, 5)
5
3
2
3, 5
3i 5 j
x
−3 −2 −1
15, 3
5
3 3, 2 4
2. (a) v
15, 3
(b)
6, 0
10. (b) v
(c) v
y
−1
1
2
4
5
3 4, 6 6
1, 12
i 12 j
(a), (d)
4
3
y
(−1, 12)
2
(−6, 0)
−8
−6
v
v
−4
8
6
4
2
x
−2
−2
x
−4
−8 − 6 − 4 −2
1 2, 3 1
4. (a) v
(b)
(3, 6)
−4
−6
3, 2
11. (b) v
y
(c) v
3
(− 3, 2)
6 8 10
(4, −6)
6 8, 1 3
2, 4
2i 4 j
y
(a), (d)
2
2
6
v
4
1
(8, 3)
2
−3
2
−2
−1
v
x
x
−4 −2
5. u
5 3, 6 2
2, 4
v
3 1, 8 4
2, 4
u
v
6. u
1 4 , 8 0
5, 8
v
7 2, 7 1
5, 8
u
v
2
4
(6, − 1)
8
(−2, −4)
−6
© 2010 Brooks/Cole, Cengage Learning
Section 11.1
12. (b) v
(c) v
5 0, 1 4
5, 3
17. (a) 2 v
Vectors in the Plane
2 3, 5
5i 3j
3
6, 10
y
y
(a) and (d).
(6, 10)
10
8
(− 5, 3)
v
4
6
(3, 5)
2
4
v
2v
2
x
−6
−4
−2
x
2
(− 5, −1)
−2
−2
2
−2
4
6
8
10
(0, − 4)
(b) 3v
13. (b) v
(c) v
9, 15
y
6 6, 6 2
0, 4
3
−15 −12 −9 −6 −3
y
(0, 4)
−9
−15
v
(c)
2
(6, 2)
7v
2
4
21 , 35
2 2
y
x
2
6
( 212, 352 (
18
3 7, 1 1
15
10, 0
12
(3, 5)
9
10i
7
v
2
6
v
3
y
(a) and (d).
6
−12
(−9, − 15)
4
3
−3v −6
(6, 6)
6
(c) v
v
x
(a) and (d).
14. (b) v
(3, 5)
6
4j
x
−3
−3
3
3
6
9
12 15 18
2
1
(−10, 0)
v
(d)
x
−8 −6 −4 −2
(−3, −1)
2 4 6 8
y
(7, −1)
−2
4
3
(c) v
1
2
32 , 3 i (3, 5)
5
−3
15. (b) v
2, 10
3
2v
3
2
v
3
2
1, 53
4
3
(2, 103 (
v
1
x
5
j
3
−1
−1
1
2
3
4
5
y
(a) and (d)
( 12 , 3(
3
(− 1, 53( 2
v
( 32 , 43(
−2
16. (b) v
(c) v
x
−1
1
2
0.84 0.12, 1.25 0.60
0.72, 0.65
0.72i 0.65 j
(a) and (d).
y
(0.12, 0.60)
(0.84, 1.25)
1.25
1.00
0.75
(0.72, 0.65)
0.50
0.25
v
x
0.25 0.50 0.75 1.00 1.25
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
4
Vectors and the Geometry of Space
4 2, 3
18. (a) 4 v
8, 12
y
21.
y
u
(− 8, 12) 12
10
u−v
8
−v
6
4v
x
4
(−2, 3)
v
x
−8 − 6 − 4 −2
2
4
6
y
22.
(b) 12 v
1, 32
u + 2v
y
(−2, 3)
2v
3
2
u
v
x
x
−3 −2 −1
− 12 v
3
(1, − 32(
−2
2u
3
23. (a)
−3
2
3
(b) v u
(c) 0 v
0, 0
2, 5 4, 9
(c) 2u 5 v
y
8, 6
3
4, 9
2, 14
2 4, 9 5 2, 5
18, 7
(−2, 3)
3
2u
3
24. (a)
2
3
3, 8
2, 16
3
2
v
(b) v u
1
0v
x
−3
−2
−1
1
8, 25 3, 8
(c) 2u 5 v
11, 33
2 3, 8 5 8, 25
34, 109
−1
2i j
3
2
25. v
(d) 6u
12, 18
3i 3j
2
3, 32
y
y
1
(− 2, 3)
v
−6
x
−2
2
−6
6
10
(12, −18)
−18
3
u
3
u
2
−2
−14
19.
2
−1
− 6v
−10
x
14
−3
y
2i j i 2 j
26. v
3i j
3, 1
y
−u
2
w
x
v
1
x
1
2
3
u
20. Twice as long as given vector u.
−1
y
u
2u
x
© 2010 Brooks/Cole, Cengage Learning
Section 11.1
2i j 2 i 2 j
27. v
4i 3 j
4, 3
y
38. v
u
2w
5
5, 15
25 225
v
4
Vectors in the Plane
250
5, 15
v
v
5 10
5 10
10 3 10
,
unit vector
10
10
u + 2w
2
39. v
x
4
u
6
3 5
,
2 2
−2
2
§ 3·
§5·
¨ ¸ ¨ ¸
2
© ¹
© 2¹
v
5u 3w
28. v
5 2, 1 3 1, 2
7, 11
u
x
−4 −2
4
6
8
10
v
v
5u
−3w
34
2
§ 3· § 5·
¨ ¸, ¨ ¸
© 2¹ © 2¹
34
2
y
2
2
3
,
34
3 34 5 34
,
unit vector
34
34
v
−6
5
34
−8
−10
−12
29. u1 4
1
u1
3
3
u2
5
Q
3, 5
u2 2
30. u1 5
u2 3
Q
31.
4
u1
9
9
u2
6
9, 6
3
2
7
0
v
33.
v
42 32
34.
v
122 5
35.
v
6 2 5
36.
v
37. v
2
v
6.2
u
2
2
32
(b)
v
1 4
5
(c)
u v
(e)
3 12
v
u
v
v
3, 12
153
153
17 4 17
,
17
17
1
1
1, 1
2
1
1
1, 2
5
1
u v
u v
u v
u v
3
12
,
153
153
01
v
v
v
v
109
31 2 17 2
,
unit vector
50
50
0, 1
u
u
u
u
13
5 2
1, 2
2
(f )
2
1, 1 , v
50
5 2
11
3, 12
2
2
u
(d)
61
3.4
6.2, 3.4
v
v
u v
5
2
(a)
3
32.
10
6.2, 3.4
41. u
Terminal point
0 72
v
40. v
0, 1
1
unit vector
© 2010 Brooks/Cole, Cengage Learning
6
Chapter 11
42. u
Vectors and the Geometry of Space
3, 3
0, 1 , v
2, 4 , v
44. u
5, 5
(a)
u
01
1
(a)
u
4 16
(b)
v
99
3 2
(b)
v
25 25
(c)
u v
(c)
u v
3, 2
u v
9 4
u
u
(d)
u
u
1
1
v
v
(e)
1
3, 2
13
(b)
(c)
1
,v
2
1
4
u
u
u
u
1
1
v
v
2
1
1,
2
5
1
1
7, 1
5 2
u v
u v
1
2, 1
v
v
(f )
5, 4
5
4
41 | 6.403
v
u v
7, 5
u v
u+v
3
v
2
1
74 | 8.602
u
1
−1
x
2
3
4
5
6
7
u v d u v
85
2
74 d
46.
u
5 1
2, 3
13
41
3, 2
y
13 | 3.606
u
v
1, 2
u v
3
u
u+v
5 | 2.236
u v
2, 0
2
1
x
−3 −2 −1
−1
−2
1
2
3
v
−3
2
u+ v d u v
1
u v
u v
u v
u v
7
6
1
v
v
y
5 | 2.236
v
(e)
5 2
5, 5
5 2
u v
u v
v
13
49
9
4
u
u
1
2, 4
2 5
u
5
2
7
3,
2
u v
u
u
u
49
v
45.
2, 3
1
u v
(d)
(f )
1
u
49 1
v
v
u v
u v
1,
(e)
1
u v
u v
(f )
(a)
3, 3
3 2
v
v
43. u
(d)
0, 1
5 2
7, 1
u v
13
2 5
2
7
3,
2
85
1
2 d
47.
u
u
1
0, 3
3
§ u ·
6¨¨
¸¸
© u ¹
v
13 6 0, 1
5
0, 1
0, 6
0, 6
© 2010 Brooks/Cole, Cengage Learning
Section 11.1
1
1, 1
2
u
u
48.
§ u ·
4¨¨
¸¸
© u ¹
v
49.
58.
§ u ·
5¨¨
¸¸
© u ¹
v
7
5¬ªcos 0.5 º¼ i 5ª¬sin 0.5 º¼ j
u
5 cos 0.5 i 5 sin 0.5 j
5 cos 0.5 i 5 sin 0.5 j
v
2 2 1, 1
u v
10 cos 0.5 i
10 cos 0.5, 0
2 2, 2 2
1
1, 2
5
u
u
Vectors in the Plane
59. Answers will vary. Sample answer: A scalar is a real
number such as 2. A vector is represented by a directed
line segment. A vector has both magnitude and direction.
1
2
,
5
5
5, 2 5
S
3, 1 has direction
For example
1
2
5 ,
5
5
6
and a magnitude
of 2.
5, 2 5
60. See page 766:
(ku1, ku2)
u
u
50.
§ u ·
2¨¨
¸¸
© u ¹
v
1
(u1 + v1, u2 + v2)
3, 3
2 3
(u1, u2)
3
v1
52. v
5ª¬ cos 120q i sin 120q jº¼
5
5 3
i j
2
2
54. v
3i
v2
ku1
(b) Scalar. The price is a number.
62. (a) Scalar. The temperature is a number.
5 5 3
,
2 2
u1
u1
61. (a) Vector. The velocity has both magnitude and
direction.
3, 0
(b) Vector. The weight has magnitude and direction.
For Exercises 63–68,
au bw
a i 2j b i j
2 ª¬ cos 150q i sin 150q jº¼
3i j
63. v
4 ª¬ cos 3.5q i sin 3.5q jº¼
| 3.9925i 0.2442 j
2i j. So, a b
2, 2a b
simultaneously, you have a
64. v
3 j. So, a b
v
56.
§2 3 2·
3 2
j
¨¨
¸¸i 2
2
©
¹
u
4 cos 0q i 4 sin 0q j
u v
u
v
u v
i
65. v
3 2
3 2
i j
2
2
3 cos 45q i 3 sin 45q j
u v
v
57.
cos 0q i sin 0q j
2 cos 30q i 2 sin 30q j
5i 3j
5,
23 2 3 2
,
2
2
4i
3j
3
2 cos 4 i 2 sin 4 j
3i 3 j. So, a b
i j. So, a b
3. Solving
1.
0. Solving
1, b
2.
3, 2a b
2, b
1, 2a b
simultaneously, you have a
68. v
1. Solving
1.
1, b
3, 2a b
simultaneously, you have a
67. v
i 3i. So, a b
simultaneously, you have a
66. v
1, b
0, 2a b
simultaneously, you have a
u
a b i 2a b j.
3, 1
3.9925, 0.2442
55.
u2
(v1, v2)
v
1,
(u1, u2)
u2
u
3, 3
3ª¬ cos 0q i sin 0q jº¼
ku2
u
1
3
51. v
53. v
ku
u+v
i 7 j. So, a b
simultaneously, you have a
3. Solving
1.
1. Solving
2, b
3
1.
3
1, 2a b
2, b
7. Solving
3.
cos 2 i sin 2 j
2 cos 4 cos 2 i 2 sin 4 sin 2 j
2 cos 4 cos 2, 2 sin 4 sin 2
© 2010 Brooks/Cole, Cengage Learning
8
Chapter 11
69. f x
Vectors and the Geometry of Space
x2 , f c x
2 x, f c 3
y
6
10
(a) m
6. Let w
1
r
1, 6 .
37
w
37, then r
w
1, 6 , w
8
(a)
(b)
6
4
(b) m
70. f x
(a) m
(b) m
71. f x
16 . Let w
x 2 5, f c x
2. Let w
2 x, f c 1
3x 2
(a) m
3. Let w
(b) m
1
. Let w
3
2
(3, 9)
x
−2
2
4
w
w
3
2
1
2, 1 .
5
r
(1, 4)
1
−3
−1
x
1
−1
2
3
y
10, then
3, 1 , w
10
(b)
4
1.
1, 3 , w
8
(a)
1
1, 2 .
5
r
w
w
5, then r
3 at x
6
y
5, then r
2, 1 , w
x3 , f c x
1
6, 1 .
37
r
2
1, 2 , w
1
. Let w
2
w
w
37, then r
6, 1 , w
w
w
10, then
r
w
w
1
1, 3 .
10
2
(1, 1)
1
1
3, 1 .
10
r
(a)
(b)
x
1
72. f x
(a) m
(b) m
x3 , f c x
3x 2
12. Let w
2.
12 at x
1, 12 , w
1
. Let w
12
y
145, then
12, 1 , w
2
w
w
145, then
x
1
1, 12 .
145
r
w
w
r
−6
−4
−2
2
(a)
−4
1
12, 1 .
145
4
−6
(b)
−10
73. f x
fc x
(a) m
(b) m
74. f x
fc x
25 x 2
y
x
25 x 2
3
at x
4
4
(b)
(3, 4)
3
3
. Let w
4
4
. Let w
3
(a)
3.
4, 3 , w
3, 4 , w
1
r 4, 3 .
5
w
5, then
w
5, then
w
w
r
1
3, 4 .
5
2
1
−1
x
1
2
3
4
5
tan x
sec 2 x
y
2 at x
S
2.0
(a)
1.5
4
1.0
(a) m
(b) m
2. Let w
1
. Let w
2
1, 2 , w
2, 1 , w
5, then
w
w
5, then
r
w
w
1
1, 2 .
5
r
1
2, 1 .
5
(b)
0.5
−π
2
−π
4
π
4
π
2
x
−1.0
© 2010 Brooks/Cole, Cengage Learning
Section 11.1
75.
uv
(c)
uv u
2
2
i
j
2
2
6
2
2
,
2
2
(6, 5)
5
4
v
3
u v
3i 3 3 j
v
u v u
6, 5
y
2 3i 2 j
u
76.
2
1
x
−1
3 2 3 i 3 3 2 j
(d)
3 2 3, 3 3 2
1
2
3
4
5
6
6 2 52
v
61
F1
2, T F1
33q
77. (a)–(c) Programs will vary.
F2
3, T F2
125q
(d) Magnitude | 63.5
F3
2.5, T F3
R
F1 F2 F3 | 1.33
79.
Direction | 8.26q
TR
80.
F1
2, T F1
10q
F2
4, T F2
140q
F3
3, T F3
200q
110q
T F1 F2 F3 | 132.5q
F1 F2 F3 | 4.09
R
TR
81.
9
6i 5 j
(b) v
2j
v
9 3, 1 4
78. (a) v
2
2
i
j
2
2
u
Vectors in the Plane
T F1 F2 F3 | 163.0q
F1 F2
500 cos 30qi 500 sin 30q j 200 cos 45q i 200 sin 45q j
F1 F2
tan T
250 3 100 2
2
2
250 100 2
250 3 100 2 i 250 100 2 j
| 584.6 lb
250 100 2
Ÿ T | 10.7q
250 3 100 2
82. (a) 180 cos 30qi sin 30q j 275i | 430.88i 90 j
§ 90 ·
Direction: D | arctan ¨
¸ | 0.206 | 11.8q
© 430.88 ¹
430.882 902 | 440.18 newtons
Magnitude:
(b) M
D
(c)
(d)
275 180 cos T
2
180 sin T
2
ª 180 sin T
º
arctan «
»
275
180
cos
T
¬
¼
T
0q
30q
60q
90q
120q
150q
180q
M
455
440.2
396.9
328.7
241.9
149.3
95
D
0q
11.8q
23.1q
33.2q
40.1q
37.1q
0
500
50
M
0
α
180
0
0
180
0
(e) M decreases because the forces change from acting in the same direction to acting in the opposite direction as T increases
from 0q to 180q.
© 2010 Brooks/Cole, Cengage Learning
10
Chapter 11
Vectors and the Geometry of Space
83. F1 F2 F3
75 cos 30qi 75 sin 30q j 100 cos 45qi 100 sin 45q j 125 cos 120qi 125 sin 120q j
3 50 2 75
2
125
2
i 75
2
50 2 125
2
3 j
F1 F2 F3 | 228.5 lb
R
TR
T F1 F2 F3 | 71.3q
ª¬400 cos 30q i sin 30q j º¼ ª¬280 cos 45q i sin 45q j º¼ ª¬350 cos 135q i sin 135q j º¼
84. F1 F2 F3
ª200 3 140 2 175 2 º i ª200 140 2 175 2 º j
¬
¼
¬
¼
200 3 35 2
R
2
200 315 2
2
| 385.2483 newtons
§ 200 315 2 ·
arctan ¨¨
¸¸ | 0.6908 | 39.6q
© 200 3 35 2 ¹
TR
85. (a) The forces act along the same direction. T
(b) The forces cancel out each other. T
q.
180q.
(c) No, the magnitude of the resultant can not be greater than the sum.
10 cos T sin T
20, 0 , F2
86. F1
20 10 cos T , 10 sin T
F1 F2
(a)
400 400 cos T 100 cos 2T 100 sin 2 T
500 400 cos T
(b)
40
2␲
0
0
(c) The range is 10 d F1 F2 d 30.
The maximum is 30, which occur at T
The minimum is 10 at T
0 and T
2S .
S.
(d) The minimum of the resultant is 10.
87. 4, 1 , 6, 5 , 10, 3
y
y
8
8
6
6
4
(8, 4)
4
(1, 2)
(3, 1)
2
4
6
x
−4 −2
−2
8
−4
−4
88.
u
1u
3
8
(6, 5)
(1, 2)
7 1, 5 2
4
6
(8, 4)
(1, 2)
(10, 3)
2
(3, 1)
2
6
(8, 4)
4
2
2
(− 4, −1)
y
x
8
x
−2
−2
(3, 1)
4
6
8
10
−4
6, 3
2, 1
P1
1, 2 2, 1
P2
1, 2 2 2, 1
3, 3
5, 4
© 2010 Brooks/Cole, Cengage Learning
Section 11.1
JJJG
CB
JJJG
CA
89. u
v
u cos 30qi sin 30q j
91. Horizontal component
Vertical component
50°
u
C
3000
Horizontal components: u cos 30q v cos 130q
0
u
u cos 60qi sin 60q j
v
v cos 110qi sin 110q j
So, u sin 60q v sin 110q
u | 1958.1 pounds
100.
And u cos 60q v cos 110q
v | 2638.2 pounds
§1·
u ¨ ¸ v cos 110q
© 2¹
§ 24 ·
arctan ¨ ¸ | 0.8761 or 50.2q
© 20 ¹
T2
u
u cos T 1 i sin T 1 j
v
v cos T 2 i sin T 2 j
Vertical components: u sin T1 v sin T 2
Horizontal components: u cos T1 v cos T 2
0 or
0.
Multiplying the last equation by
§ 24 ·
arctan ¨
¸ S | 1.9656 or 112.6q
© 10 ¹
3 and adding to the
first equation gives
u sin 110q 3 cos 110q
100 Ÿ v | 65.27 lb.
§1·
Then, u ¨ ¸ 65.27 cos 110q
© 2¹
5000
0
u | 2169.4 and v | 3611.2.
0 gives
u | 44.65 lb.
(a) The tension in each rope: u
44.65 lb,
v
65.27 lb
Solving this system, you obtain
(b) Vertical components: u sin 60q | 38.67 lb,
y
v sin 110q | 61.33 lb
θ2
B
20°
u
v
C
100, or
§ 3·
u ¨¨
¸¸ v sin 110q
© 2 ¹
Solving this system, you obtain
A
v sin T
92. To lift the weight vertically, the sum of the vertical
components of u and v must be 100 and the sum of the
horizontal components must be 0.
x
30°
Vertical components: u sin 30q v sin 130q
90. T1
v cos T
1200 sin 6q | 125.43 ft sec
130° 30° B
v
11
1200 cos 6q | 1193.43 ft sec
v cos 130qi sin 130q j
y
A
Vectors in the Plane
x
θ1
v
30°
u
100 lb
93. u
v
900 cos 148q i sin 148q j
100 cos 45q i sin 45q j
u v
900 cos 148q 100 cos 45q i 900 sin 148q 100 sin 45q j
| 692.53i 547.64 j
§ 547.64 ·
¸ | 38.34q; 38.34q North of West
© 692.53 ¹
T | arctan ¨
u v |
692.53
2
547.64
2
| 882.9 km h
© 2010 Brooks/Cole, Cengage Learning
12
Chapter 11
94.
Vectors and the Geometry of Space
u
400i plane
v
50 cos 135qi sin 135q j
u v
25 2i 25 2 j wind
400 25 2 i 25 2 j | 364.64i 35.36 j
35.36
Ÿ T | 5.54q
364.64
tan T
Direction North of East: | N 84.46q E
Speed: | 336.35 mi h
95. True
102. Let the triangle have vertices at 0, 0 , a, 0 , and
96. True
b, c . Let u be the vector joining 0, 0 and b, c , as
97. True
indicated in the figure. Then v, the vector joining the
midpoints, is
98. False
v
a
b
c
§a b a·
¸i j
¨
2¹
2
© 2
b
c
i + j
2
2
1
1
bi cj
u.
2
2
0
99. False
ai b j
2 a
100. True
101.
u
cos 2 T sin 2 T
1,
v
sin 2 T cos 2 T
1
y
(b, c)
( a +2 b , 2c (
u
v
x
(0, 0)
( 2a , 0(
(a, 0)
103. Let u and v be the vectors that determine the
parallelogram, as indicated in the figure. The two
diagonals are u v and v u. So,
r
x u v ,s
4 v u . But,
r s
u
xu v y v u
So, x y
x
1 and x y
x y u x y v.
0. Solving you have
1.
2
y
s
u
r
v
u v v u
104. w
u ª¬ v cos T v i v sin T v jº¼ v ª¬ u cos Tu i u sin Tu jº¼
v ª¬ cos Tu cos T v i sin Tu sin T v jº¼
u
ª §T Tv ·
§ Tu T v ·
§ Tu T v ·
§ Tu T v · º
v «cos¨ u
¸ cos¨
¸i sin ¨
¸ cos¨
¸j
2 ¹
2 ¹
2 ¹
2 ¹ »¼
©
©
©
¬ ©
2 u
tan T w
§T
sin ¨ u
©
§T
cos¨ u
©
Tv ·
§ Tu
¸ cos¨
2 ¹
©
Tv ·
§ Tu
¸ cos¨
2 ¹
©
Tv ·
¸
2 ¹
Tv ·
¸
2 ¹
So, T w
Tu T v 2 and w bisects the angle between u and v.
§T Tv ·
tan ¨ u
¸
2 ¹
©
105. The set is a circle of radius 5, centered at the origin.
u
x, y
x2 y 2
5 Ÿ x2 y2
25
© 2010 Brooks/Cole, Cengage Learning
Section 11.2
v0t cos D and y
106. Let x
1 2
gt .
2
v0t sin D x tan D g 2
x sec 2 D
2v02
x tan D gx 2
1 tan 2 D
2v02
13
y
§
· 1 §
·
x
x
v0 sin D ¨
¸ g¨
¸
D
D
cos
2
cos
v
v
© 0
¹
© 0
¹
x
Ÿ y
v0 cos D
t
Space Coordinates and Vectors in Space
2
(x, y)
α
x
v02
gx 2
gx 2
v2
2 2 tan 2 D x tan D 0
2g
2v0
2v0
2g
§ v2 ·
v02
gx 2
gx 2 ª
v4 º
2 2 «tan 2 D 2 tan D ¨ 0 ¸ 20 2 »
2g
2v0
2v0 ¬
© gx ¹ g x ¼
gx 2 §
v2 ·
v02
gx 2
2 2 ¨ tan D 0 ¸
2g
2v0
2v0 ©
gx ¹
v02
gx 2
2 , then D can be chosen to hit the point x, y . To hit 0, y : Let D
2g
2v0
If y d
90q. Then
2
·
v2
v02
v2 § g
0 ¨ t 1¸ , and you need y d 0 .
2g
2g
2 g © v0
¹
1 2
gt
2
v0t y
2
The set H is given by 0 d x, 0 y and y d
Note: The parabola y
v02
gx 2
2
2g
2v0
v02
gx 2
2 is called the “parabola of safety.”
2g
2v0
Section 11.2 Space Coordinates and Vectors in Space
z
5.
1. A 2, 3, 4
B 1, 2, 2
(5, − 2, 2) − 3
2. A 2, 3, 1
4
B 3, 1, 4
3.
3
2
1
1
1 2
3
2
3
−2
−3
y
x
(5, − 2, − 2)
z
(2, 1, 3)
6.
6
5
4
3
z
8
6
(4, 0, 5)
(−1, 2, 1)
2
1
4
3
2
2 3
4
6
y
−2
x
x
−4
6
−6
(0, 4, − 5)
y
z
4.
8
7. x
3, y
4, z
8. x
7, y
2, z
5: 3, 4, 5
(3, −2, 5) 6
4
2
7, 2, 1
6
x
1:
6
( 32 , 4, −2(
y
9. y
10. x
z
0, y
0, x
3, z
12: 12, 0, 0
2: 0, 3, 2
© 2010 Brooks/Cole, Cengage Learning
14
Chapter 11
Vectors and the Geometry of Space
11. The z-coordinate is 0.
30. A 3, 4, 1 , B 0, 6, 2 , C 3, 5, 6
12. The x-coordinate is 0.
AB
9 41
13. The point is 6 units above the xy-plane.
AC
0 1 25
26
14. The point is 2 units in front of the xz-plane.
BC
9 1 16
26
15. The point is on the plane parallel to the yz -plane that
passes through x
3.
16. The point is on the plane parallel to the xy-plane that
passes through z
5 2.
Because AC
14
BC , the triangle is isosceles.
31. A 1, 0, 2 , B 1, 5, 2 , C 3, 1, 1
AB
0 25 16
17. The point is to the left of the xz-plane.
AC
419
18. The point is in front of the yz-plane.
BC
4 36 1
19. The point is on or between the planes y
y
3.
20. The point is in front of the plane x
3 and
Because AB
41
14
41
BC , the triangle is isosceles.
32. A 4, 1, 1 , B 2, 0, 4 , C 3, 5, 1
4.
21. The point x, y, z is 3 units below the xy-plane, and
below either quadrant I or III.
22. The point x, y, z is 4 units above the xy-plane, and
AB
419
AC
1 36 0
37
BC
1 25 9
35
14
Neither
above either quadrant II or IV.
23. The point could be above the xy-plane and so above
quadrants II or IV, or below the xy-plane, and so below
quadrants I or III.
33. The z-coordinate is changed by 5 units:
24. The point could be above the xy-plane, and so above
quadrants I and III, or below the xy-plane, and so below
quadrants II or IV.
34. The y-coordinate is changed by 3 units:
4 0
25. d
2
16 4 49
2 2
26. d
2
6 1
2
42
2
2
5 3
96
2 2
25 0 36
28. d
70
§ 5 2 9 3 7 3 ·
,
,
35. ¨
¸
2
2
2 ¹
©
2
2 2
§ 4 8 0 8 6 20 ·
36. ¨
,
,
¸
2
2
© 2
¹
2
4 6
2
2 4
2
5 2
AC
62 42 12
BC
4 2
BC
2
245
2
49 196
2
z 5
2
z 1
2
4
Radius: 5
2
2
x 4
49
15
y 2
38. Center: 4, 1, 1
2
62
22 62 32
2
x 0
63
6, 4, 7
Radius: 2
2
29. A 0, 0, 4 , B 2, 6, 7 , C 6, 4, 8
2
§3
·
¨ , 3, 5¸
©2
¹
37. Center: 0, 2, 5
61
4 49 9
AB
3, 7, 1 , 0, 9, 2 , 3, 8, 6
69
16 64 16
27. d
2
20
0, 0, 9 , 2, 6, 12 , 6, 4, 3
7
196
2
245
AB AC
2
2, 0, 0 0, 6, 0
2
39. Center:
14
Radius:
7 5
x 1
y 1
2
2
25
1, 3, 0
10
y 3
2
z 0
2
10
2
Right triangle
© 2010 Brooks/Cole, Cengage Learning
Section 11.2
Space Coordinates and Vectors in Space
15
40. Center: 3, 2, 4
r
3
tangent to yz -plane
2
x 3
y 2
2
z 4
2
9
x2 y2 z 2 2x 6 y 8z 1
41.
0
x 2 x 1 y 6 y 9 z 8 z 16
2
2
1 1 9 16
2
x 1
2
2
y 3
z 4
2
25
Center: 1, 3, 4
Radius: 5
x 2 y 2 z 2 9 x 2 y 10 z 19
42.
81 ·
§ 2
2
2
¨ x 9 x ¸ y 2 y 1 z 10 z 25
4¹
©
2
9·
2
§
¨x ¸ y 1 z 5
2
©
¹
2
0
19 81
1 25
4
109
4
§ 9
·
Center: ¨ , 1, 5 ¸
© 2
¹
109
2
Radius:
9 x 2 9 y 2 9 z 2 6 x 18 y 1
43.
x y z 2
x 2
2x
3
2
x 1,
3
Center:
1
3
1
9
2
2
2x
3
2y y 2y 1 z
2
y 1
2
z 0
0
1
9
0
2
19 2
1
9
1
1
1, 0
Radius: 1
44. 4 x 2 4 y 2 4 z 2 24 x 4 y 8 z 23
x 6x 9 y y 2
x 3
2
y 1
2
2
1
4
2
z 1
0
z 2z 1
2
2
23
4
9
1
4
1
16
Center: 3, 12 , 1
Radius: 4
45. x 2 y 2 z 2 d 36
Solid sphere of radius 6 centered at origin.
46. x 2 y 2 z 2 ! 4
Set of all points in space outside the ball of radius 2 centered at the origin.
© 2010 Brooks/Cole, Cengage Learning
16
Chapter 11
Vectors and the Geometry of Space
x 2 y 2 z 2 4 x 6 y 8 z 13
47.
x 2 4 x 4 y 2 6 y 9 z 2 8 z 16 4 9 16 13
x2
2
y 3
2
z 4
2
16
Interior of sphere of radius 4 centered at 2, 3, 4 .
x 2 y 2 z 2 ! 4 x 6 y 8 z 13
48.
x 2 4 x 4 y 2 6 y 9 z 2 8 z 16 ! 13 4 9 16
x 2
2
y 3
2
z 4
2
! 16
Set of all points in space outside the ball of radius 4 centered at 2, 3, 4 .
2 4, 4 2, 3 1
49. (a) v
2, 2, 2
53.
1, 1, 6
2i 2 j 2k
(b) v
4 3, 1 2, 6 0
1 1 36
z
(c)
Unit vector:
5
4
1, 1, 6
−3
2
54.
1
1
1
2
2
3
3
5, 12, 5
y
4
1 4, 7 5 , 3 2
4 0, 0 5, 3 1
5, 12, 5
55.
z
(c)
5 4 , 3 3, 0 1
1, 0, 1
Unit vector:
4
2
4
2
4
6
x
56. 2 1, 4 2 , 2 4
y
0 3, 3 3, 3 0
3, 0, 3
3i 3k
(b) v
1, 6, 6
Unit vector:
⟨−3, 0, 3⟩
5
1, 6, 6
3
−3
2
−2
1
2
3
3
5
y
4
6
6
,
73
73
4, 1, 1
z
(a), (d)
1
2
1
,
73
73
4i j k
(c) v
1
73
3 1 , 3 2, 4 3
57. (b) v
4
1, 6, 6
1 36 36
z
(c)
1
1
, 0,
2
2
2
2
51. (a) v
2
1, 0, 1
6
⟨4, − 5, 2 ⟩
1, 0, 1
11
8
194
5
12
5
,
,
194
194
194
194
4, 5, 2
4i 5 j 2h
6
6
38
5, 12, 5
25 144 25
Unit vector:
(b) v
1
1
,
,
38
38
38
x
50. (a) v
38
⟨− 2, 2, 2⟩
3
−2
1, 1, 6
4
x
(3, 3, 4)
(−1, 2, 3)
3
(0, 0, 0) 2
2 2, 3 3, 4 0
52. (a) v
(b) v
4
z
4
−2
(4, 1, 1) 2
4k
(c)
0, 0, 4
v
2
4
y
x
⟨ 0, 0, 4 ⟩
3
2
1
3
x
2
1
1
2
3
y
© 2010 Brooks/Cole, Cengage Learning
Section 11.2
4 2, 3 1 , 7 2
58. (b) v
6, 4, 9
Space Coordinates and Vectors in Space
62. (a) v
2, 2, 1
6i 4 j 9k
(c) v
z
4
z
(a), (d)
17
(− 4, 3, 7)
12
(− 6, 4, 9)
3
2
9
1
6
3
3
9
9
x
(2, −1, − 2)
z
59. q1 , q2 , q3 0, 6, 2
Q
y
4, 4, 2
(b) 2 v
y
⟩− 2, 2, −1⟩
3
x
8
3, 5, 6
6
4
⟩4, − 4, 2⟩
3, 1, 8
2
6
60. q1 , q2 , q3 0, 2,
1, 23 , 12
5
2
(c)
1, 43 , 3
Q
6
x
1, 1,
1v
2
y
1
2
z
61. (a) 2 v
2, 4, 4
2
z
⟩1, −1, 12 ⟩
5
1
4
3
⟨2, 4, 4⟩
2
−2
1
1
2
x
(d)
2
3
y
4
5v
2
y
5, 5,
5
2
z
x
(b) v
8
1, 2, 2
⟨5, −5, 25 ⟨
6
4
z
2
3
2
−3
−2
−3
⟨− 1, −2, −2⟩
1
2
2
3
63. z
u v
64. z
u v 2w
y
1, 2, 3 2, 2, 1
−3
1, 2, 3 2, 2, 1 8, 0, 8
(c)
3v
2
3 , 3, 3
2
65. z
z
2
−2
66. z
⟨ 32 , 3, 3⟩
y
3, 4, 20
−2
x
−3
(d) 0 v
67. 2z 3u
0, 0, 0
z
3
2
−3
1
2
3
x
−2
1
−2
−1
5u 3v 12 w
5, 10, 15 6, 6, 3 2, 0, 2
1
2
6, 12, 6
−3
−2
3
7, 0, 4
2u 4 v w
2, 4, 6 8, 8, 4 4, 0, 4
3
−3
1, 0, 4
y
3
−2
x
6
x
−2
−3
⟨0, 0, 0⟩
1
2
3
2 z1 3
4 Ÿ z1
7
2
2 z2 6
0 Ÿ z2
3
2 z3 9
4 Ÿ z3
z
y
2 z1 , z2 , z3 3 1, 2, 3
4, 0, 4
5
2
7 , 3, 5
2
2
−2
−3
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
18
Vectors and the Geometry of Space
68. 2u v w 3z
2 1, 2, 3 2, 2, 1 4, 0, 4 3 z1 , z2 , z3
0, 6, 9 3 z1 , 3z2 , 3 z3
0 3 z1
0 Ÿ z1
0, 0, 0
0
6 3 z2
0 Ÿ z2
2
9 3 z3
0 Ÿ z3
3
0, 2, 3
z
69. (a) and (b) are parallel because
6, 4, 10
2 3, 2, 5 and
2, 43 , 10
3
2
3
77. A 2, 9, 1 , B 3, 11, 4 , C 0, 10, 2 , D 1, 12, 5
JJJK
AB
1, 2, 3
JJJK
CD
1, 2, 3
JJJK
AC
2, 1, 1
JJJK
BD
2, 1, 1
JJJK
JJJK
JJJK
JJJK
CD and AC
BD, the given points
Because AB
form the vertices of a parallelogram.
3, 2, 5 .
70. (b) and (d) are parallel because
i 3i
4
71. z
4j
3
32 k
j 98 k
2 12 i 3 1i
2 2
2j
3
2j
3
34 k and
34 k .
3i 4 j 2k
(a) is parallel because 6i 8 j 4k
72. z
0, 0, 0
2 z.
7, 8, 3
(b) is parallel because z z
14, 16, 6 .
73. P 0, 2, 5 , Q 3, 4, 4 , R 2, 2, 1
JJJK
PQ
3, 6, 9
JJJK
PR
2, 4, 6
3 2, 4, 6
3, 6, 9
2
JJJK
JJJK
So, PQ and PR are parallel, the points are collinear.
74. P 4, 2, 7 , Q 2, 0, 3 , R 7, 3, 9
JJJK
PQ
6, 2, 4
JJJK
PR
3, 1, 2
3, 1, 2
12 6, 2, 4
JJJK
JJJK
So, PQ and PR are parallel. The points are collinear.
75. P 1, 2, 4 , Q 2, 5, 0 , R 0, 1, 5
JJJK
PQ
1, 3, 4
JJJK
PR
1, 1, 1
JJJK
JJJK
Because PQ and PR are not parallel, the points are not
collinear.
76. P 0, 0, 0 , Q 1, 3, 2 , R 2, 6, 4
JJJK
PQ
1, 3, 2
JJJK
PR
2, 6, 4
JJJK
JJJK
Because PQ and PR are not parallel, the points are not
collinear.
78. A 1, 1, 3 B 9, 1, 2 , C 11, 2, 9 , D 3, 4, 4
JJJK
8, 2, 5
AB
JJJK
8, 2, 5
DC
JJJK
2, 3, 7
AD
JJJK
BC
2, 3, 7
JJJK
JJJK
JJJK
JJJK
DC and AD
BC , the given points
Because AB
form the vertices of a parallelogram.
79. v
v
80. v
v
81. v
v
82. v
v
83. v
v
84. v
v
0, 0, 0
0
1, 0, 3
109
3j 5k
10
0, 3, 5
0 9 25
34
2i 5 j k
2, 5, 1
4 25 1
30
i 2 j 3k
1, 2, 3
1 49
14
4i 3j 7k
4, 3, 7
16 9 49
74
© 2010 Brooks/Cole, Cengage Learning
Section 11.2
2, 1, 2
85. v
(a)
v
v
(b) v
v
1
2, 1, 2
3
v
v
c
r 87
u
u
10
10
u
u
3
38
1
3, 2, 5
38
v
(b) v
1
1, 0, 0
8
3 2
1
1
,
2
2
1
1
1
,
,
3
3
3
3 2, 2, 1
2
3
3 2 2 1
, ,
2 3 3 3
7
97. v
2 ¬ªcos r30q j sin r30q k º¼
4, 6, 2
7
1, 1,
1
2
14 21 7
,
,
14
14 14
2 14
3j r k
3, r1
0,
z
2
3, 1⟩
y
−1
x
−2
⟨0,
3, − 1⟩
5 cos 45qi sin 45qk
98. v
5 2
i k or
2
5 cos 135qi sin 135qk
v
90. The terminal points of the vectors tu, u tv and
su tv are collinear.
⟨0,
−1
1
2
u v | 8.732
−2
1
−2
4, 7.5, 2
v | 9.019
3
3
3
,
,
3
3
3
96. v
u
u
10 10
,
2
2
3
3 u
2 u
1
1, 0, 0
8
u | 5.099
0,
95. v
89. (a)–(d) Programs will vary.
(e) u v
4
1, 1, 1
3
3
8
v
v
c 2 4c 2 9 c 2
0, 3, 3
10 0,
94. v
7
4
16
1
6, 0, 8
10
8, 0, 0
v
2
14c 2
93. v
4c 2 4 c 2 c 2
c i 2 j 3k
1
3, 2, 5
38
v
v
(b) r 73
10
9 4 25
v
49
c
14c
19
7
2
cu
92.
3, 2, 5
87. v
(a)
2
9c
1
6, 0, 8
10
v
v
(b) 88. v
36 0 64
v
(a)
9c
6, 0, 8
86. v
(a)
3
1
2, 1, 2
3
v
v
c 2i 2 j k
cv
91.
41 4
v
Space Coordinates and Vectors in Space
5 2
i k
2
z
5 2
(i + k)
2
su + tv
8
6
5 2
(− i + k)
2
4
2
u + tv
su
6
x
6
y
tv
u
v
99.
v
3, 6, 3
2v
3
2, 4, 2
4, 3, 0 2, 4, 2
2, 1, 2
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
20
Vectors and the Geometry of Space
100.
5, 6, 3
v
10 ,
3
2v
3
1, 2, 5 10 ,
3
4, 2
108.
r r0
x 1
2
y 1
2
z 1
2
x 1
2
y 1
2
z 1
2
4, 2
13 , 6, 3
3
2
4
This is a sphere of radius 2 and center 1, 1, 1 .
z
101. (a)
1
v
1
1
u
The tension vector T in each wire is
y
x
au bv
(b) w
ai a b j bk
0, a b
a
0, b
0
(c) ai a b j bk
a
1, a b
w
u v
2, b
(d) ai a b j bk
1, a b
2, b
T
8
h
T
1
182 h 2
8
i 2 j 3k
L2 182
8L
3
L2 182
Not possible
8.
8
0, 18, h and
h
So, T
i 2j k
24
3
c 0, 18, h where ch
T
0
So, a and b are both zero.
a
L2 182 .
109. (a) The height of the right triangle is h
JJJK
The vector PQ is given by
JJJK
PQ
0, 18, h .
182 L2 182
, L ! 18.
Q (0, 0, h)
102. A sphere of radius 4 centered at x1 , y1 , z1 .
v
x x2 y y1 , z z1
x x1
x x1
2
2
y y1
y y1
2
L
2
z z1
2
z z1
2
4
(0, 18, 0)
16
(b)
103. x0 is directed distance to yz-plane.
y0 is directed distance to xz-plane.
P
18
(0, 0, 0)
L
20
25
30
35
40
45
50
T
18.4
11.5
10
9.3
9.0
8.7
8.6
z0 is directed distance to xy-plane.
(c)
104. d
x x0
105.
x2 x1
2
2
y2 y1
y y0
2
2
z2 z1
z z0
2
30
L = 18
2
r2
T=8
0
106. Two nonzero vectors u and v are parallel if u
some scalar c.
107.
100
0
cv for
x 18 is a vertical asymptote and y
horizontal asymptote.
B
(d)
C
lim
L o18
lim
L of
A
JJJK JJJK
JJJK
AB BC
AC
JJJK JJJK JJJK
So, AB BC CA
8L
L2 182
8L
L2 182
(e) From the table, T
JJJK JJJK
AC CA
0
8 is a
f
lim
L of
8
1 18 L
10 implies L
2
8
30 inches.
a will be a vertical
110. As in Exercise 109(c), x
f.
asymptote. So, lim T
r0 o a © 2010 Brooks/Cole, Cengage Learning
Section 11.2
JJJK
113. AB
JJJK
AC
JJJK
AD
111. Let D be the angle between v and the coordinate axes.
cos D i cos D j cos D k
v
3 cos D
v
1
3
cos D
1
3
3
F
3
1, 1, 1
3
(
3
,
3
3
,
3
3
3
(
C3
0, 0, 500
60C2 45C3
0
65C3
0
C2 C3
500
104 , C
2
69
28 ,
23
and
112 . So:
69
F1 | 202.919 N
y
0.4
0.2
F1 F2 F3
C3 45, 65, 115
Solving this system yields C1
0.6
0.2
C2 60, 0, 115
45, 65, 115 , F3
115 C1 z
0.4
60, 0, 115 , F2
70C1
21
C1 0, 70, 115
0, 70, 115 , F1
So:
3
i jk
3
v
Space Coordinates and Vectors in Space
F2 | 157.909 N
0.4
0.6
F3 | 226.521N
x
550
112.
302,500
c2
c 75i 50 j 100k
18,125c 2
16.689655
c | 4.085
F | 4.085 75i 50 j 100k
| 306i 204 j 409k
114. Let A lie on the y-axis and the wall on the x-axis. Then A
JJJK
JJJK
AB
8, 10, 6 , AC
10, 10, 6 .
0, 10, 0 , B
10, 0, 6 and
8, 0, 6 , C
AB
10 2, AC
2 59
JJJK
JJJK
AB
AC
Thus, F1
420 JJJJK , F2
650 JJJJJK
AB
AC
F
F1 F2 | 237.6, 297.0, 178.2 423.1, 423.1, 253.9 | 185.5, 720.1, 432.1
F | 860.0 lb
115. d AP
2d BP
x2 y 1
2
z 1
2
x2 y2 z 2 2 y 2z 2
0
6 16
1
9
9
9
44
9
2
x 1
2
y 2
2
z2
4 x2 y 2 z 2 2 x 4 y 5
3 x 2 3 y 2 3 z 2 8 x 18 y 2 z 18
16 ·
2
1·
§ 2 8
§ 2
2
¨x x ¸ y 6y 9 ¨z z ¸
3
9¹
3
9¹
©
©
2
4·
§
¨x ¸ y 3
3¹
©
2
1·
§
¨z ¸
3¹
©
2
1·
2 11
§4
Sphere; center: ¨ , 3, ¸, radius:
3¹
3
©3
© 2010 Brooks/Cole, Cengage Learning
22
Chapter 11
Vectors and the Geometry of Space
Section 11.3 The Dot Product of Two Vectors
1. u
1, 5
3, 4 , v
(a) u ˜ v
3 1 4 5
(b) u ˜ u
33 44
2
(c)
u
(d)
u˜v v
32 42
2. u
17
25
25
17 1, 5
(e) u ˜ 2 v
17, 85
2u˜v
2 17
4 2 10 3
22
(b) u ˜ u
4 4 10 10
116
2
u
(d)
u˜v v
42 102
44, 66
2u˜v
3. u = 6, 4 , v
2 22
26
(b) u ˜ u
6 6 4 4
52
2
(d)
u˜v v
2u˜v
4, 8 , v
(a) u ˜ v
(b) u ˜ u
2
(c)
u
(d)
u˜v v
78, 52
2 26
9.
12
4 4 8 8
4
2
82
80
10.
84, 60
2u˜v
2 12
(b) u ˜ u
2 2 3 3 4 4
29
u
(d)
u˜v v
(e) u ˜ 2v
2 3
2 0, 6, 5
2u˜v
4
2u˜v
2
(b) u ˜ u
2 2 1 1 1 1
(c)
u
2
22 1
(d)
u˜v v
v
4
2
12
1
6
6
i k
2u˜v
2i j 2k , v
2
i 3 j 2k
(a) u ˜ v
2 1 1 3 2 2
5
(b) u ˜ u
2 2 1 1 2 2
9
22 12 2
2
(c)
u
(d)
u˜v v
u˜v
u v
12. u
cos T
9
5 i 3 j 2k
2u˜v
8 5 cos
5i 15 j 10k
2 5
10
S
20
3
cos T
40 25 cos
5S
6
500 3
2, 2
1, 1 , v
cos T
2
cos T
u˜v
u v
T
0, 12, 10
i k
2 1 1 0 1 1
29
22
2
(a) u ˜ v
11. u
2
(c)
i
2i j k , v
u˜v
0, 6, 5
2
u˜v v
24
2 0 3 6 4 5
2
(d)
80
(a) u ˜ v
2
u
u˜v
12 7, 5
2, 3, 4 , v
1
(c)
(e) u ˜ 2 v
52
7, 5
4 7 8 5
(e) u ˜ 2v
5. u
52
26 3, 2
(e) u ˜ 2 v
4. u
2
62 4
1
2
8. u
3, 2
6 3 4 2
u
(b) u ˜ u
(e) u ˜ 2 v
44
(a) u ˜ v
(c)
1
(e) u ˜ 2 v
116
22 2, 3
(e) u ˜ 2 v
(a) u ˜ v
7. u
(a) u ˜ v
(c)
i
34
2, 3
4, 10 , v
i, v
6. u
u˜v
u v
0
2 8
0
S
2
2, 1
3, 1 , v
u˜v
u v
T
5
10 5
1
2
S
4
© 2010 Brooks/Cole, Cengage Learning
Section 11.3
3i j, v
13. u
u˜v
u v
cos T
14.
2i 4 j
20. u
2
10 20
1
5 2
T
1 ·
§
arccos¨ ¸ | 98.1q
© 5 2¹
u
§S ·
§S ·
cos¨ ¸i sin ¨ ¸ j
6
© ¹
©6¹
v
cos T
2
i 2
ª 2
arccos «
1
¬ 4
T
1, 1, 1 , v
15. u
cos T
u˜v
u v
T
arccos
T =
v
2i 3 j
0
19. u
4, 0 , v
Neither
j 6k , v
i 2j k
2i 3j k , v
2i j k
2, 3, 1 , v
1, 1, 1
cos T , sin T , 1 ,
sin T , cos T , 0
2, 1, 0 and 0, 0, 0 : 1, 2, 0 ˜ 2, 1, 0
8
5 13
28. Consider the vector 3, 0, 0 joining 0, 0, 0 and
3, 0, 0 , and the vector 1, 2, 3 joining 0, 0, 0 and
1, 2, 3 : 3, 0, 0 ˜ 1, 2, 3
9
2 21
1, 1
4 z 0 Ÿ not orthogonal
3 0
The triangle has an obtuse angle, so it is an obtuse triangle.
i 2j k
9
14 6
0
The triangle has a right angle, so it is a right triangle.
8 13
65
u z cv Ÿ not parallel
u˜v
Ÿ parallel
perpendicular to the vector 2, 1, 0 joining
2
§ 3 21 ·
arccos¨¨
¸¸ | 10.9q
© 14 ¹
T
2i 4 j
27. The vector 1, 2, 0 joining 1, 2, 0 and 0, 0, 0 is
u˜v
u v
cos T
16 v
u z cv Ÿ not parallel
u˜v
0 Ÿ orthogonal
S
2i 3 j k , v
18. u
13 i 2 j , v
u z cv Ÿ not parallel
u˜v
8 z 0 Ÿ not orthogonal
Neither
26. u
§ 8 13 ·
arccos¨¨ ¸¸ | 116.3q
© 65 ¹
T
32
u z cv Ÿ not parallel
u˜v
0 Ÿ orthogonal
2
| 61.9q
3
u˜v
u v
cos T
u
25. u
2
3
3 2 2 3 0
u v
1,
2
u z cv Ÿ not parallel
u˜v
0 Ÿ orthogonal
3i 4 j, v = 2 j 3k
17. u
22. u
24. u
105q
2
3 6
u˜v
u v
cos T
º
3»
¼
3
2, 1, 1
3i 2 j k , v
16. u
2
j
2
23. u
2
1
4
4, 3 , v
u z cv Ÿ not parallel
u˜v
0 Ÿ orthogonal
3
1
i j
2
2
3§
2 · 1§ 2 ·
¨¨ ¸ ¨
¸
2 © 2 ¸¹ 2 ¨© 2 ¸¹
16
3
,
2
2, 18 , v
23
u z cv Ÿ not parallel
u˜v
0 Ÿ orthogonal
21. u
§ 3S ·
§ 3S ·
cos¨ ¸i sin ¨ ¸ j
© 4 ¹
© 4 ¹
u˜v
u v
The Dot Product of Two Vectors
3 21
14
29. A 2, 0, 1 , B 0, 1, 2 , C 12 , 32 , 0
JJJK
JJJK
AB
2, 1, 1
BA
2, 1, 1
JJJK
JJJK
5 3
5
AC
2 , 2 , 1
CA
, 32 , 1
2
JJJK
JJJK
1, 1, 2
BC
12 , 12 , 2
CB
2
2
JJJK JJJK
3
AB ˜ AC
5 2 1 ! 0
JJJK JJJK
BA ˜ BC
1 12 2 ! 0
JJJK JJJK
5
CA ˜ CB
43 2 ! 0
4
The triangle has three acute angles, so it is an acute
triangle.
© 2010 Brooks/Cole, Cengage Learning
24
Chapter 11
Vectors and the Geometry of Space
30. A 2, 7, 3 , B 1, 5, 8 , C 4, 6, 1
JJJK
JJJK
AB
3, 12, 5
BA
3, 12, 5
JJJK
JJJK
2, 13, 4
AC
2, 13, 4
CA
JJJK
JJJK
5, 1, 9
BC
5, 1, 9
CB
JJJK JJJK
AB ˜ AC
6 156 20 ! 0
JJJK JJJK
BA ˜ BC 15 12 45 ! 0
JJJK JJJK
CA ˜ CB 10 13 36 ! 0
32. u
i 2 j 2k , u
cos D
1
3
cos E
2
3
cos J
2
3
cos E
cos J
34. u
cos D
cos E
cos J
cos D
cos E
cos J
36. u
cos D
cos E
cos J
0, 6, 4 , u
cos D
0
cos E
3
13
2
13
4
9
4
9
1
52
cos 2 D cos 2 E cos 2 J
25
9
1
35 35 35
1
2 13
0
9
4
13 13
1
a 2 b2 c2
a, b, c , u
a
a 2 b2 c2
b
a 2 b2 c2
c
a 2 b2 c2
cos 2 D cos 2 E cos 2 J
35. u
33. u
cos J
1
9
35
cos 2 D cos 2 E cos 2 J
3
cos 2 D cos 2 E cos 2 J
u
5
35
3
35
1
35
cos D
The triangle has three acute angles, so it is an acute
triangle.
31. u
5, 3, 1
3, 2, 2
u
a2
b2
c2
2
2
2
2
2
2
a b c
a b c
a b2 c2
2
1
17
3
Ÿ D | 0.7560 or 43.3q
17
2
Ÿ E | 1.0644 or 61.0q
17
2
Ÿ y | 2.0772 or 119.0q
17
4, 3, 5
u
50
5 2
4
Ÿ D | 2.1721 or 124.4q
5 2
3
Ÿ E | 1.1326 or 64.9q
5 2
S
5
1
Ÿ J |
or 45q
4
5 2
2
© 2010 Brooks/Cole, Cengage Learning
Section 11.3
1, 5, 2
37. u
u
30
1
Ÿ D | 1.7544 or 100.5q
30
5
Ÿ E | 0.4205 or 24.1q
30
2
Ÿ J | 1.1970 or 68.6q
30
cos D
cos E
cos J
2, 6, 1
38. u
u
41
2
Ÿ D | 1.8885 or 108.2q
41
6
Ÿ E | 0.3567 or 20.4q
41
1
Ÿ J | 1.4140 or 81.0q
41
cos D
cos E
cos J
50
| 4.3193
F1
39. F1: C1
80
| 5.4183
F2
F2 : C2
JJJK
41. OA
| 4.3193 10, 5, 3 5.4183 12, 7, 5
108.2126, 59.5246, 14.1336
F | 124.310 lb
cos D |
108.2126
Ÿ D | 29.48q
F
cos E |
59.5246
Ÿ E | 61.39q
F
cos J |
14.1336
Ÿ J | 96.53q
F
0
0 Ÿ D
90q
0 102 102
10
cos J
2
0 102 102
1
J
Ÿ E
45q
2
2
cos E
42. F1
C1 0, 10, 10 .
200
F1
0, 100 2, 100 2
F2
C2 4, 6, 10
F3
C3 4, 6, 10
F
0, 0, w
F F1 F2 F3
4C2 4C3
43. u
40. F1: C1
F2 : C2
100
| 6.3246
F2
F
0 Ÿ C2
(a) w1
36.062
Ÿ E | 98.57q
F
cos J |
65.4655
Ÿ J | 74.31q
F
800 2
3
61 7 4
1, 4
12 42
34
1, 4
2, 8
17
44. u
u w1
9, 7 , v
(a) w1
6, 7 2, 8
projvu
§u ˜ v·
v
¨¨ v 2 ¸¸
©
¹
91 7 3
1 32
30
1, 3
10
(b) w 2
4, 1
1, 3
230.239, 36.062, 65.4655
cos E |
25 2
N
3
C3
§u ˜ v·
v
¨¨
2 ¸
¸
© v ¹
projvu
| 13.0931 20, 10, 5 6.3246 5, 15, 0
230.239
cos D |
Ÿ D | 162.02q
F
0 Ÿ C2
1, 4
F1 F2
F | 242.067 lb
C3
10C2 10C3 100 2
6, 7 , v
10 2 and
0
100 2 6C2 6C3
(b) w 2
300
| 13.0931
F1
C110 2 Ÿ C1
F1
W
25
0, 10, 10
cos D
F1 F2
F
The Dot Product of Two Vectors
u w1
9, 7
1, 3
3, 9
3, 9
6, 2
© 2010 Brooks/Cole, Cengage Learning
26
Chapter 11
45. u
2i 3 j
(a) w1
Vectors and the Geometry of Space
2, 3 , v
5i j
§u ˜ v·
v
¨¨
2 ¸
¸
© v ¹
2 5 31
projvu
52 1
13
5, 1
26
(b) w 2
u w1
2, 3 5, 1
50. u
v
i 4k
3i 2k
(a) w1
2i 3j
(a) w1
2, 3 , v
5 1
,
2 2
13 4 2
3, 0, 2
32 22
11
33
22
3, 0, 2
, 0,
13
13
13
1 5
,
2 2
3i 2 j
47. u
u w1
(a) w1
51. u ˜ v
48. u
u w1
2, 1, 1
8 2 2 1 0 1
2, 1, 1
22 1 1
18
2, 1, 1
6, 3, 3
6
(b) w 2
49. u
v
u w1
2i j 2k
3j 4k
(a) w1
projvu
8, 2, 0 6, 3, 3
2, 1, 3
u w1
cos D
v1
, cos E
v
u Ÿ u
v3
. D , E , and J
v
cv Ÿ u and v are parallel.
§u ˜ v·
0 Ÿ u˜v
(b) ¨¨
2 ¸
¸v
© v ¹
orthogonal.
57. Yes,
u˜v
v
v 2
v
v 2
§u ˜ v·
¨¨
2 ¸
¸v
© v ¹
8 6
2, ,
25 25
v2
, cos J
v
v1 , v2 , v3 are
55. See figure 11.29, page 787.
0, 3, 4
33 44
2,1, 2 0, ,
25 25
0. The
u˜v
.
u v
are the direction angles. See Figure 11.26.
u˜v
2, 1, 2
2 0 13 2 4
0, 3, 4
32 42
11
33 44
0, 3, 4
0, ,
25
25 25
(b) w 2
54. See page 786. Direction cosines of v
§u ˜ v·
56. (a) ¨
¨ v 2 ¸¸ v
©
¹
§u ˜ v·
v
¨¨
2 ¸
¸
© v ¹
projvu
u1v1 u2v2 u3v3
53. (a) and (b) are defined. (c) and (d) are not defined
because it is not possible to find the dot product of a
scalar and a vector or to add a scalar to a vector.
2, 1, 1
8, 2, 0 , v
(a) w1
0, 3, 3 2, 2, 2
u1 , u2 , u3 ˜ v1 , v2 , v3
angle T between u and v is given by cos T
2, 3
0 1 3 1 3 1
1, 1, 1
111
6
1, 1, 1
2, 2, 2
3
(b) w 2
20
30
, 0,
13
13
52. The vectors u and v are orthogonal if u ˜ v
§u ˜ v·
v
¨¨
2 ¸
¸
© v ¹
projvu
33
22
, 0,
13
13
1, 0, 4 1, 1, 1
0, 3, 3 , v
u w1
3, 2
2 3 3 2
3, 2
32 22
0 3, 2
0, 0
(b) w 2
§u ˜ v·
¨¨ v 2 ¸¸ v
©
¹
projvu
5 1
,
2 2
§u ˜ v·
v
¨¨
2 ¸
¸
© v ¹
projvu
3, 0, 2
5, 1
(b) w 2
46. u
1, 0, 4
(c) Obtuse,
2
v˜u
1
u
u
v
(b) Acute, 0 T S
v˜u
u
u 2
1
v
58. (a) Orthogonal, T
0 Ÿ u and v are
u
u 2
S
2
S
2
T S
© 2010 Brooks/Cole, Cengage Learning
Section 11.3
1.35, 2.65, 1.85
v
u˜v
cos 10qi sin 10q j
v
3240 1.35 1450 2.65 2235 1.85
F˜v
v
v 2
w1
$12,351.25
v
| 8335.1 cos 10qi sin 10q j
w1 | 8335.1 lb
3240, 1450, 2235
1.35, 2.65, 1.85
Increase prices by 4%: 1.04 v
48,000 j 8335.1 cos 10qi sin 10q j
New total amount: 1.04 u ˜ v
8208.5i 46,552.6 j
1.04 12,351.25
w 2 | 47,270.8 lb
JJJK
72. OA
61. (a)–(c) Programs will vary.
10, 5, 20 , v
0, 0, 1
JJJK
20
projv OA
0, 0, 1
0, 0, 20
12
JJJK
projv OA
20
u | 9.165
v | 5.745
T
90q
63. Programs will vary.
64.
F w1
(b) w 2
$12,845.30
62.
F˜v v
48,000 sin 10q v
This represents the total amount that the restaurant
earned on its three products.
60. u
73. F
21 63 42
, ,
26 26 13
v
§1
3 ·
85¨¨ i j¸¸
2
2
©
¹
10i
F˜v
W
65. Because u and v are parallel, projvu
u
66. Because u and v are perpendicular, projvu
74. F
0
v
50i
F˜v
W
u
1
3
i j. Want u ˜ v
4
2
v
12i 2 j and v
0.
12i 2 j are orthogonal to u.
75. F
9i 4 j. Want u ˜ v
v
4i 9 j and v 4i 9 j
v
2000i
W
F˜v
| 2900.2 km-N
JJJK
76. PQ
F
69. Answers will vary. Sample answer:
u
3, 1, 2 . Want u ˜ v
v
0, 2, 1 and v
0.
0, 2, 1 are orthogonal to u.
70. Answers will vary. Sample answer:
u
4, 3, 6 . Want u ˜ v
v
0, 6, 3 and v
0
0, 6, 3
1600 2000 cos 25q
| 2,900,184.9 Newton meters (Joules)
0.
are orthogonal to u.
1250 cos 20q | 1174.6 ft-lb
1600 cos 25q i sin 25q j
68. Answers will vary. Sample answer:
u
425 ft-lb
25 cos 20qi sin 20q j
67. Answers will vary. Sample answer:
are orthogonal to u.
27
48,000 j
71. (a) Gravitational Force F
3240, 1450, 2235
59. u
The Dot Product of Two Vectors
W
40i
100 cos 25qi
JJJK
4000 cos 25q | 3625.2 Joules
F ˜ PQ
77. False.
1, 1 , v
For example, let u
w
u˜w
1, 4 . Then u ˜ v
1 4
23
2, 3 and
5 and
5.
78. True
w˜ u v
w˜u w˜v
00
0 so, w and
u v are orthogonal.
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
28
79. Let s
Vectors and the Geometry of Space
length of a side.
v
s, s, s
v
s 3
cos D
D
x1 3 intersect at
1, 1 , 0, 0 and 1, 1 .
(b) y1c
cos E
E
x3 and y2
82. (a) The graphs y1
s
s 3
cos J
1
3
At 0, 0 , r 1, 0 is tangent to y1 and r 0, 1 is
§ 1 ·
arcos¨
¸ | 54.7q
© 3¹
J
1
.
3x 2 3
3 x 2 and y2c
tangent to y2 .
At 1, 1 , y1c
z
3 and y2c
1
.
3
1
1
1, 3 is tangent to y1 , r
3, 1 is tangent
10
10
to y2 .
r
s
v
y
At 1, 1 , y1c
s
3 and y2c
s
1
1
1, 3 is tangent to y1 , r
3, 1 is tangent
10
10
to y2 .
x
80.
1
.
3
r
s, s , s
v1
z
s 3
v1
y
s, s , 0
v2
s
v2
2
(1, 1)
1
(s, s, s)
T
6
3
arcos
−2
x 2 and y2
81. (a) The graphs y1
(s, s, 0)
x
6
| 35.26q
3
x1 3 intersect at
0, 0 and 1, 1 .
1
.
3x 2 3
2 x and y2c
(b) y1c
(c) At 0, 0 , the vectors are perpendicular 90q .
At 1, 1 ,
2 and y2c
1
.
3
(c) At 0, 0 , the vectors are perpendicular 90q .
1
1
1, 2 ˜
3, 1
5
10
1 1
45q
(b) y1c
At 1, 0 , y1c
5
50
1
2
tangent to y1 , r
2. r
1
1, 2 is
5
1
1, 2 is tangent to y2 .
5
2 and y2c
2. r
1
1, 2 is
5
1
1, 2 is tangent to y2 .
5
1
1
1, 2 ˜
1, 2
5
5
3
.
5
T | 0.9273 or 53.13q
(1, 1)
y = x 1/3
(0, 0) 1
x2 1
2 x.
2 and y2c
At 1, 0 , y1c
(c) At 1, 0 , cos T
y = x2
2
1 x 2 and y 2
2 x and y2c
y
−1
3
.
5
intersect at 1, 0 and 1, 0 .
tangent to y1 , r
At 1, 1 ,
1
6
10
T | 0.9273 or 53.13q
83. (a) The graphs of y1
1
1
1, 2 is tangent to y1 , r
3, 1 is tangent
5
10
to y2 .
T
1
1
1, 3 ˜
3, 1
10
10
1 1
By symmetry, the angle is the same at 1, 1 .
r
cos T
−1
−2
At 0, 0 , r 1, 0 is tangent to y1 and r 0, 1 is
At 1, 1 , y1c
2
1
cos T
tangent to y2 .
x
−1
(−1, −1)
v2
y = x3
(0, 0)
y
2 2
2 3
cos T
y = x 1/3
2
v1
x
By symmetry, the angle is the same at 1, 0 .
2
−1
© 2010 Brooks/Cole, Cengage Learning
Section 11.3
84. (a) To find the intersection points, rewrite the second
equation as y 1 x3. Substituting into the first
equation
y 1
2
x Ÿ x6
x Ÿ x
1, 0 , as indicated in the figure.
86. If u and v are the sides of the parallelogram, then the
diagonals are u v and u v, as indicated in the
figure.
œ u˜v
0
œ 2u ˜ v
2u ˜ v
œ u v ˜ u v
œ u v
y
y = x 3−1
x
1
u v ˜ u v
u v
2
2
œ The diagonals are equal in length.
1
−1
29
the parallelogram is a rectangle.
0, 1.
There are two points of intersection, 0, 1 and
(1, 0)
The Dot Product of Two Vectors
2
u+
u
−
v
u
(0, −1)
x = (y +1) 2
v
v
−2
z
87. (a)
(b) First equation:
y 1
2
(0, k, k)
x Ÿ 2 y 1 yc
At 1, 0 , yc
1 Ÿ yc
(k, 0, k)
1
2 y 1
1
.
2
k
Second equation: y
1, 0 , yc 3.
x 3 1 Ÿ yc
(k, k, 0)
x
3x 2 . At
y
k
k
k 2 k 2 02
(b) Length of each edge:
k
2
k
1
2
1
2, 1 unit tangent vectors to first curve,
5
1
r
1, 3 unit tangent vectors to second curve
10
(c) cos T
At 0, 1 , the unit tangent vectors to the first curve
JK
(d) r1
k, k, 0 k k k
, ,
2 2 2
k k k
, ,
2 2 2
JK
r2
0, 0, 0 k k k
, ,
2 2 2
k k k
, ,
2 2 2
r
are r 0, 1 , and the unit tangent vectors to the
k
§1·
arccos¨ ¸
©2¹
T
second curve are r 1, 0 .
2 k
(c) At 1, 0 ,
1
2, 1 ˜
5
cos T
T |
S
4
1
1, 3
10
5
50
1
.
2
cos T
or 45q
T
At 0, 1 the vectors are perpendicular, T
85. In a rhombus, u
90q.
v . The diagonals are u v and
u v.
u v ˜ u v
u v ˜u u v ˜v
u˜u v˜uu˜v v˜v
u 2 v
2
So, the diagonals are orthogonal.
u−v
0
88. u
k2
4
2
§k·
¨ ¸ ˜3
©2¹
109.5q
cos D , sin D , 0 , v
2
2
60q
1
3
cos E , sin E , 0
The angle between u and v is D E . Assuming that
D ! E . Also,
cos D E
u˜v
u v
cos D cos E sin D sin E
1 1
cos D cos E sin D sin E .
u
u+v
v
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
30
u v
89.
Vectors and the Geometry of Space
u v ˜ u v
2
u v
91.
u v ˜ u v
2
u v ˜u u v ˜v
u v ˜u u v ˜v
= u˜u v˜uu˜v v˜v
u˜u v˜uu˜v v˜v
u˜v
90.
u
2
u˜v u˜v v
u
2
v 2 2u ˜ v
v cos T
u
u˜v
v cos T
u
v cos T
u
v because cos T d 1.
d u
u 2 2u ˜ v v
2
d u 2 2 u
2
v v
2
d
u v
2
So, u v d u v .
92. Let w1
projvu, as indicated in the figure. Because
w1 is a scalar multiple of v, you can write
u
w1 w 2
cv w 2 .
Taking the dot product of both sides with v produces
u˜v
cv w 2 ˜ v
cv ˜ v w 2 ˜ v
c v 2, because w 2 and v are orthogonol.
So, u ˜ v
Ÿ c
u˜v
and
v 2
u˜v
v.
v 2
cv
projvu
w1
2
c v
u
w2
θ v
w1
Section 11.4 The Cross Product of Two Vectors in Space
1. j u i
i
j k
0
1 0
1 0
i
k
3. j u k
j k
0 1 0
0
0 0
i
1
z
z
1
1
k
j
j
1
1
i
x
−k
1
y
−1
i
2. i u j
x
i
1
j k
1 0
0
i
k
4. k u j
0 1 0
0
−i
j
j
1
1
1
y
−1
1 0
1
k
k
x
i
1
z
1
i
j k
0 0
z
1
y
−1
x
y
−1
© 2010 Brooks/Cole, Cengage Learning
Section 11.4
i
5. i u k
j k
1 0
0
0 0
1
The Cross Product of Two Vectors in Space
12, 3, 0 , v
11. u
j
i
uuv
z
−j
1
0
0
5
2 0 5 0 0 54
0 Ÿ v A uuv
y
1, 1, 2 , v
12. u
j k
0 0
1
1 0
0
i
j
uuv
j k
1 1 2
u˜ uuv
2i k
2, 0, 1
1 2 1 0 2 1
0 Ÿ u A uuv
1
v˜ uuv
k
0 2 1 0 0 1
j
x
0, 1, 0
0 1 0
z
1
0, 0, 54
12 0 3 0 0 54
v˜ uuv
1
i
54k
0 Ÿ u A uuv
−1
6. k u i
12 3
2
k
i
x
2, 5, 0
j k
u˜ uuv
1
−1
31
i
0 Ÿ v A uuv
1
y
2, 3, 1 , v
13. u
−1
i
i
7. (a) u u v
2 4
0
3 2
5
(b) v u u
uu v
(c) v u v
0
i
8. (a) u u v
uuv
j k
20i 10 j 16k
1
1 2
1
u˜ uuv
3 0
5
(b) v u u
uu v
(c) v u v
0
1 1 2 1 1 1
0 Ÿ v A uuv
15i 16 j 9k
10, 0, 6 , v
14. u
i
15i 16 j 9k
j k
7
3
17i 33j 10k
2
1 1 5
(b) v u u
uu v
(c) v u v
0
i
10. (a) u u v
10
0
5
(b) v u u
uu v
(c) v u v
0
18i 30 j 30k
18, 30, 30
10 18 0 30 6 30
0
Ÿ u A uuv
v˜ uuv
5 18 3 30 0 30
17i 33 j 10k
15. u
i j k, v
i
j
6
5 3 0
0
Ÿ v A uuv
uuv
k
3 2 2
1
uuv
5, 3, 0
j k
u˜ uuv
i
1, 1, 1
2 1 3 1 1 1
v˜ uuv
k
i j k
0 Ÿ u A uuv
2 3 2
9. (a) u u v
j k
2 3
20i 10 j 16k
j
1, 2, 1
8i 5 j 17k
1
8i 5 j 17k
2i j k
j
k
1 1
1
2i 3j k
2, 3, 1
2 1 1
u˜ uuv
1 2 1 3 1 1
0 Ÿ u A uuv
v˜ uuv
2 2 1 3 1 1
0 Ÿ v A uuv
v u u
vuu
uuv
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
32
Vectors and the Geometry of Space
i 6 j, v
16. u
2i j k
i
uuv
8, 6, 4
v
10, 12, 2
uuv
60, 24, 156
j k
1 6
2
6i j 13k
0
1
u˜ uuv
v˜ uuv
u
22.
1
1 6 6 1
uuv
uuv
0 Ÿ u A uuv
2 6 1 1 1 13
1
36 22
0 Ÿ v A uuv
60, 24, 156
5
2
13
,
,
3 22 3 22 3 22
z
17.
6
5
4
3
2
1
4
3
2
23.
3, 2, 5 , v
u
uu v
v
u
3.6, 1.4, 1.6
uu v
uu v
1
4
6
0.4, 0.8, 0.2
1.8
0.7
, ,
4.37
4.37
0.8
4.37
y
x
24.
0, 0,
u
z
18.
6
5
4
3
2
1
uu v
uu v
uu v
v
1
4
3
2
u
4
6
y
26.
z
4
2
21
,0
20
0, 1, 0
uu v
50, 40, 34
u u v | 72.498
6
5
4
3
2
1
3
27.
v
u
j
v
jk
1
u
4
6
i
y
uu v
x
j k
0 1 0
0 1
i
1
z
20.
6
5
4
3
2
1
uu v
A
4
3
2
u
4
6
i jk
v
jk
uu v
1 1
1
0 1
1
i
y
x
u
uuv
uuv
uuv
4, 3.5, 7 , v
2.5, 9, 3
2.94
,
11.8961
0.22
,
11.8961
1.79
11.8961
29.
j k
uu v
A
73.5, 5.5, 44.75
1
i
u
28.
v
1
21.
3
31
, 0,
2
5
25. Programs will vary.
x
19.
0,
7
,v
10
j k
u
3, 2, 1
v
1, 2, 3
i
uu v
j
uuv
2
k
3 2 1
1 2
A
j k
8, 10, 4
3
8, 10, 4
180
6 5
© 2010 Brooks/Cole, Cengage Learning
Section 11.4
30. u
2, 1, 0
v
1, 2, 0
j k
2 1 0
1
A
2
uuv
0, 0, 3
3
A
A
i
j k
1 2
3
5 4
JJJK JJJK
AB u AD
0
12, 15, 6
144 225 36
32. A 2, 3, 1 , B 6, 5, 1 , C 7, 2, 2 , D 3, 6, 4
JJJK
AB
4, 8, 2
JJJK
DC
4, 8, 2
JJJK
BC
1, 3, 3
JJJK
AD
1, 3, 3
JJJK
JJJK
JJJK
JJJK
DC and BC
AD, the figure ABCD is
Because AB
a parallelogram.
JJJK
JJJK
AB and AD are adjacent sides
JJJK JJJK
AB u AD
A
i
j
4
8 2
1 3
JJJK JJJK
AB u AD
A
1
2
i
1
2
i
JJJK JJJK
AB u AC
A
1
2
0
1
2
k
3 12
5
113, 2, 63
36. A 1, 2, 0 , B 2, 1, 0 , C 0, 0, 0
JJJK
JJJK
AB
3, 1, 0 , AC
1, 2, 0
i
A
j k
3 1 0
1 2
JJJK JJJK
AB u AC
1
2
5k
0
5
2
37. F
20k
JJJK
1 cos 40q j sin 40qk
PQ
2
i
JJJK
PQ u F
j
k
0 cos 40q 2 sin 40q 2
0
JJJK
PQ u F
10 cos 40qi
20
0
10 cos 40q | 7.66 ft-lb
z
PQ
18, 14, 20
40°
F
2 230
x
JJJK
PQ u F
6, 9, 2
36 81 4
y
38. F 2000 cos 30q j sin 30qk
JJJK
PQ 0.16k
j k
3 2
JJJK JJJK
AB u AC
j
1
ft
2
324 196 400
3
11
2 13 4
JJJK JJJK
1 16,742
AB u AC
2
3
1 0
6i 2 j 2k
35. A 2, 7, 3 , B 1, 5, 8 , C 4, 6, 1
JJJK
JJJK
AB
3, 12, 5 , AC
2, 13, 4
k
33. A 0, 0, 0 , B 1, 0, 3 , C 3, 2, 0
JJJK
JJJK
AB
1, 0, 3 , AC
3, 2, 0
JJJK JJJK
AB u AC
k
2 4 2
JJJK JJJK
AB u AC
9 5
j
3 5 4
JJJK JJJK
1 44
AB u AC
2
0
0, 0, 3
i
JJJK JJJK
AB u AC
31. A 0, 3, 2 , B 1, 5, 5 , C 6, 9, 5 , D 5, 7, 2
JJJK
AB
1, 2, 3
JJJK
1, 2, 3
DC
JJJK
5, 4, 0
BC
JJJK
5, 4, 0
AD
JJJK
JJJK
JJJK
JJJK
DC and BC
AD, the figure ABCD is
Because AB
a parallelogram.
JJJK
JJJK
AB and AD are adjacent sides
JJJK JJJK
AB u AD
33
34. A 2, 3, 4 , B 0, 1, 2 , C 1, 2, 0
JJJK
JJJK
AB
2, 4, 2 , AC
3, 5, 4
i
uuv
The Cross Product of Two Vectors in Space
1000 3j 1000k
i
j
k
0
0
0.16
z
0 1000 3 1000
11
2
JJJK
PQ u F
PQ
0.16 ft
160 3i
60°
F
160 3 ft-lb
y
x
© 2010 Brooks/Cole, Cengage Learning
34
Chapter 11
Vectors and the Geometry of Space
39. (a) Place the wrench in the xy-plane, as indicated in the figure.
JJJK
The angle from AB to F is 30q 180q T
210q T
JJJK
OA
18 inches 1.5 feet
JJJK
OA
3 3
3
i j
4
4
56 ª¬cos 210q T i sin 210q T jº¼
i
j
y
␪
18
1.5ª¬cos 30q i sin 30q jº¼
F
JJJK
OA u F
3 3
4
56 cos 210q T
in.
F
30⬚
x
O
k
100
3
4
56 sin 210q T
0
0
y = 84 sin θ
ª42 3 sin 210q T 42 cos 210q T ºk
0
¬
¼
ª42 3 sin 210q cos T cos 210q sin T 42 cos 210q cos T sin 210q sin T ºk
¬
¼
ª
º
§ 1
·
§
·
3
3
1
84 sin T k
sin T ¸¸ 42¨¨ cos T sin T ¸¸»k
«42 3 ¨¨ cos T 2
2
2
2
©
¹
©
¹¼»
¬«
0
JJJK
OA u F
JJJK
45q, OA u F
84
dT
dT
84 cos T
0 when T
This is reasonable. When T
40. (a) AC
15 inches
BC
JJJK
AB
12 inches
90q.
90q, the force is perpendicular to the wrench.
5
feet
4
1 foot
F
JJJK
AB u F
(c) When T
␪
B
12 in.
i
j
k
0
54
1
0 180 cos T
15 in.
C
A
180 sin T
225 sin T 180 cos T i
225 sin T 180 cos T
JJJK
30q, AB u F
§ 3·
§1·
225¨ ¸ 180¨¨
¸¸ | 268.38
© 2¹
© 2 ¹
225 sin T 180 cos T , T
For 0 T 141.34, T c T
(e)
42 2 | 59.40
5
jk
4
180 cos T j sin T k
JJJK
AB u F
(d) If T
2
2
84 sin T
(c) Let T
(b)
180
84 sin T , 0 d T d 180q
(b) When T
F
30⬚
B
␪ A
0 for 225 sin T
225 cos T 180 sin T
4
Ÿ T | 141.34q.
5
JJJK
5
Ÿ T | 51.34q. AB and F are perpendicular.
4
180 cos T Ÿ tan T
0 Ÿ tan T
400
0
180
0
From part (d), the zero is T | 141.34q, when the vectors are parallel.
© 2010 Brooks/Cole, Cengage Learning
Section 11.4
1 0 0
41. u ˜ v u w
0 1 0
1
v
3, 0, 0
w
1, 1, 5
0 4 0
1 1 1
u˜ vuw
1
2 1 0
3 0 0
u˜ vuw
V
2 0 1
0 3 0
4 15
60
1 1 5
0 0 1
43. u ˜ v u w
35
0, 4, 0
48. u
0 0 1
42. u ˜ v u w
The Cross Product of Two Vectors in Space
49. u u v
6
u˜v
0 0 1
60
0 Ÿ u and v are parallel.
0 Ÿ u and v are orthogonal.
So, u or v (or both) is the zero vector.
2 0 0
44. u ˜ v u w
1 1 1
50. (a) u ˜ v u w
0
vuw ˜u b
0 2 2
1 1 0
45. u ˜ v u w
0 1 1
2
(e) u ˜ w u v
1 0 1
V
u˜ vuw
So, a
1 3
1
0 6
6
u˜ vuw
47. u
3, 0, 0
v
0, 5, 1
w
2, 0, 5
51. u u v
72
72
v˜ wuu
wuu ˜v h
f
b
c
d
u u v ˜ w g
h and e
f
g
u1 , u2 , u3 ˜ v1 , v2 , v3
52. See Theorem 11.8, page 794.
53. The magnitude of the cross product will increase by a
factor of 4.
54. From the vectors for two sides of the triangle, and
compute their cross product.
0 5 1
x2 x1 , y2 y1 , z2 z1 u x3 x1 , y3 y1 , z3 z1
75
55. False. If the vectors are ordered pairs, then the cross
product does not exist.
2 0 5
V
ux w ˜ v d
u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k
3 0 0
u˜ vuw
v˜ wuu
w˜ vuu
4 0 4
V
uuv ˜w c
w˜ vuu
2
46. u ˜ v u w
w˜ uuv
u˜ vuw
75
56. False. In general, u u v
57. False. Let u
Then, u u v
1, 0, 0 , v
uuw
vuu
1, 0, 0 , w
1, 0, 0 .
0, but v z w.
58. True
59. u
u1 , u2 , u3 , v
uu v w
v1 , v2 , v3 , w
w1 , w2 , w3
i
j
k
u1
u2
u3
v1 w1 v2 w2
v3 w3
ª¬u2 v3 w3 u3 v2 w2 º¼ i ª¬u1 v3 w3 u3 v1 w1 º¼ j ª¬u1 v2 w2 u2 v1 w1 º¼ k
u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k u2 w3 u3w2 i u1w3 u3w1 j u1w2 u2 w1 k
uuv uuw
© 2010 Brooks/Cole, Cengage Learning
36
Chapter 11
Vectors and the Geometry of Space
u1 , u2 , u3 , v
60. u
i
cu u v
v1 , v2 , v3 , c is a scalar:
j
k
cu1 cu2
cu3
v1
v3
v2
cu2v3 cu3v2 i cu1v3 cu3v1 j cu1v2 cu2v1 k
c ª¬ u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k º¼
cuuv
61. u = u1 , u2 , u3
i
uuu
j
k
u1 u2
u3
u1 u2
u3
62. u ˜ v u w
uuv ˜w
u2u3 u3u2 i u1u3 u3u1 j u1u2 u2u1 k
u1
u2
u3
v1
v2
v3
w1
w2
w3
w˜ uuv
w1
w2
w3
u1
u2
u3
v1
v2
v3
0
w1 u2v3 v2u3 w2 u1v3 v1u3 w3 u1v2 v1u2
u1 v2 w3 w2v3 u2 v1w3 w1v3 u3 v1w2 w1v2
uuv
63.
u˜ vuw
u2v3 u3v2 i u1v3 u3v1 j u1v2 u2v1 k
uuv ˜u
u2v3 u3v2 u1 u3v1 u1v3 u2 u1v2 u2v1 u3
0
uuv ˜v
u2v3 u3v2 v1 u3v1 u1v3 v2 u1v2 u2v1 v3
0
So, u u v A u and u u v A v.
64. If u and v are scalar multiples of each other, u
uu v
cv u v
If u u v
u
65.
c vu v
v sin T
0, then u
cv for some scalar c.
0
0. Assume u z 0, v z 0. So, sin T
0, T
0, and u and v are parallel. So,
cv for some scalar c.
uuv
u
v sin T
If u and v are orthogonal, T
66. u
c0
a1 , b1 , c1 , v
vuw
a2 , b2 , c2 , w
i
j
k
a2
b2
c2
a3
b3
c3
i
uu vuw
uu vuw
S 2 and sin T
1. So, u u v
u
v .
a3 , b3 , c3
b2c3 b3c2 i a2c3 a3c2 j a2b3 a3b2 k
j
k
a1
b1
c1
b2c3 b3c2
a3c2 a2c3
a2b3 a3b2
ª¬b1 a2b3 a3b2 c1 a3c2 a2c3 º¼ i ª¬a1 a2b3 ª¬a1 a3c2 a2c3 b1 b2c3 b3c2 º¼k
ª¬a2 a1a3 b1b3 c1c3 a3 a1a2 b1b2 c1c2 º¼ i
ª¬c2 a1a3 b1b3 c1c3 c3 a1a2 b1b2 c1c2
a3b2 c1 b2c3 b3c2 º¼ j
ª¬b2 a1b3 b1b3 c1c3 b3 a1a2 b1b2 c1c2 º¼ j
º¼k
a1a3 b1b3 c1c3 a2 , b2 , c2 a1a2 b1b2 c1c2 a3 , b3 , c3
u˜w v u˜vw
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
Lines and Planes in Space
37
Section 11.5 Lines and Planes in Space
1 3t , y
1. x
2 t, z
2 5t
z
(a)
x
y
(b) When t
0, P
JJJK
PQ
9, 3, 15
1, 2, 2 . When t
10, 1, 17 .
3, Q
JJJK
The components of the vector and the coefficients of t are proportional because the line is parallel to PQ.
(c) y
0 when t
2. So, x
7 and z
12.
Point: 7, 0, 12
x
0 when t
13. So, y
7
3
z
0 when t
52 . So, x
15 and y
2. x
2
3t , y
1.
3
and z
Point: 0, 73 , 13
12 .
5
Point: 15 , 12
,0
5
1t
2, z
4.
z
(a)
x 3
2
(a)
z 2
7, 23, 0 : Substituting, you have
73
2
2
y
x
y 7
8
23 7
8
2
2
0 2
Yes, 7, 23, 0 lies on the line.
(b) When t
Q
JJJK
PQ
0, P
2, 2, 1 . When t
(b) 1, 1, 3 : Substituting, you have
2,
13
2
1
4, 2, 1 .
6, 0, 2
The components of the vector and the coefficients of
JJJK
t are proportional because the line is parallel to PQ.
(c) z
0 when t
1 and y
1. So, x
2.
Point: 1, 2, 0
x
0 when t
2 t , y
y
2 and z
1
3
(b)
0
t
2. Then y
z
4 2
32
2 t , you have
6 and
6. Yes, 0, 6, 6 lies on the line.
2, 3, 5 : For x
t
4t
3t , z
0, 6, 6 : For x
(a)
Yes, 1, 1, 3 lies on the line.
5. Point: 0, 0, 0
Direction vector: 3, 1, 5
2 . So,
3
Point: 0, 2, 13
3. x
1 7
3 2
8
1 1
4. Then y
not lie on the line.
2
34
2 t , you have
12 z 3. No, 2, 3, 5 does
Direction numbers: 3, 1, 5
3t , y
(a) Parametric: x
(b) Symmetric:
x
3
t, z
5t
z
5
y
6. Point: 0, 0, 0
Direction vector: v
5
2, , 1
2
Direction numbers: 4, 5, 2
(a) Parametric: x
(b) Symmetric:
x
4
4t , y
y
5
5t , z
2t
z
2
© 2010 Brooks/Cole, Cengage Learning
38
Chapter 11
Vectors and the Geometry of Space
7. Point: 2, 0, 3
13. Points: 7, 2, 6 , 3, 0, 6
Direction vector: 10, 2, 0
2, 4, 2
Direction vector: v
Direction numbers: 10, 2, 0
Direction numbers: 2, 4, 2
(a) Parametric: x
(b) Symmetric:
2 2t , y
x 2
2
3 2t
4t , z
8. Point: 3, 0, 2
0, 6, 3
Direction vector: v
3, y
y
(b) Symmetric:
2
Direction numbers: 2, 2, 5
3
(a) Parametric: x
9. Point: 1, 0, 1
(b) Symmetric:
3i 2 j k
(b) Symmetric:
1 3t , y
x 1
3
2t , z
25 5t
z 25
5
k
Direction numbers: 0, 0, 1
z 1
1
y
2
Parametric: x
2, y
4t
3, z
16. Point: 4, 5, 2
Directions numbers: 3, 2, 1
(a) Parametric: x
Direction vector: v
3 3t , y
x 3
3
5 2t , z
y 5
2
z 4
y 3
11
Parametric: x
2 9t
z 2
9
12. Points: 0, 4, 3 , 1, 2, 5
(b) Symmetric: x
y 4
2
2
3i 2 j k
2 3t , y
Direction vector: v
3 2t , z
4t
i 2 j k
Direction numbers: 1, 2, 1
Parametric: x
4 t , y
Direction vector: v
Direction numbers: 1, 2, 2
t, y
5 t, z
5 2t , z
2t
19. Point: 5, 3, 4
Direction vector: 1, 2, 2
(a) Parametric: x
4, y
18. Point 4, 5, 2
3 11t , z
x 5
17
Parametric: x
Direction numbers: 3, 2, 1
Direction numbers: 17, 11, 9
(a) Parametric:
x
5 17t , y
j
Direction numbers: 0, 1, 0
Direction vector: v
17
11
i j 3k
3
3
Direction vector: v
4t
17. Point: 2, 3, 4
§ 2 2 ·
11. Points: 5, 3, 2 , ¨ , , 1¸
© 3 3 ¹
(b) Symmetric:
2t , z
y
2
Direction vector: v
1t
10. Point: 3, 5, 4
(b) Symmetric:
x
2
2t , y
15. Point: 2, 3, 4
Direction numbers: 3, 2, 1
(a) Parametric: x
6
Direction vector: 10, 10, 25
2t
2t , z
z 2, x
Direction vector: v
2 2t , z
14. Points: 0, 0, 25 , 10, 10, 0
Direction numbers: 0, 2, 1
(a) Parametric: x
7 10t , y
(b) Symmetric: Not possible because the direction
number for z is 0. But, you could describe the
x 7
y 2
line as
,z
6.
10
2
z 3
2
y
4
(a) Parametric: x
4 2t , z
z 3
2
3 2t
2, 1, 3
Direction numbers: 2, 1, 3
Parametric: x
5 2t , y
3 t , z
4 3t
20. Point: 1, 4, 3
Direction vector: v
5i j
Direction numbers: 5, 1, 0
Parametric: x
1 5t , y
4 t, z
3
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
30. L1: v
21. Point: 2, 1, 2
Lines and Planes in Space
2, 1, 2
3, 2, 2 on line
Direction vector: 1, 1, 1
L2 : v
4, 2, 4
1, 1, 3 on line
Direction numbers: 1, 1, 1
L3: v
1, 12 , 1
2, 1, 3 on line
L4 : v
2, 4, 1
3, 1, 2 on line
2 t, y
Parametric: x
1 t, z
2t
L1 , L2 and L3 have same direction.
22. Point: 6, 0, 8
3, 2, 2 is not on L2 nor L3
Direction vector: 2, 2, 0
1, 1, 3 is not on L3
Direction numbers: 2, 2, 0
6 2t , y
Parametric: x
23. Let t
v
24. Let t
v
3, 1, 2
0: P
1, 2, 0
2t , z
8
other answers possible
any nonzero multiple of v is correct
0: P
0, 5, 4
4, 1, 3
other answers possible
any nonzero multiple of v is correct
25. Let each quantity equal 0:
P
7, 6, 2
v
4, 2, 1
other answers possible
any nonzero multiple of v is correct
P
3, 0, 3 other answers possible
v
5, 8, 6
L2 : v
any nonzero multiple of v is correct
3, 2, 4
6, 2, 5 on line
6, 4, 8
6, 2, 5 on line
L3: v
6, 4, 8
L4 : v
6, 4, 6
6, 2, 5 not online
not parallel to L1 , L2 , nor L3
L1 and L2 are identical. L1
28. L1: v
L2 : v
So, the three lines are parallel, not identical.
31. At the point of intersection, the coordinates for one line
equal the corresponding coordinates for the other line.
So,
(i) 4t 2
(iii) t 1
L2 and is parallel to L3 .
2 s 2, (ii) 3
s 1.
2 s 3, and
From (ii), you find that s
0 and consequently, from
(iii), t
0. Letting s
t
0, you see that equation (i)
is satisfied and so the two lines intersect. Substituting
zero for s or for t, you obtain the point 2, 3, 1 .
u
4i k
First line
v
2i 2 j k
Second line
u˜v
u v
cos T
26. Let each quantity equal 0:
27. L1: v
39
81
17 9
7
3 17
32. By equating like variables, you have
(i) 3t 1
(iii) 2t 4
3s 1, (ii) 4t 1
s 1.
t
1
2
2 s 4, and
t , and consequently from (ii),
From (i) you have s
and from (iii), t
3. The lines do not intersect.
33. Writing the equations of the lines in parametric form you
have
x
3t
y
2t
z
1 t
1 4s
y
2 s
z
3 3s.
2, 6, 2
3, 0, 1 on line
x
2, 1, 3
1, 1, 0 on line
For the coordinates to be equal, 3t
L3: v
2, 10, 4
1, 3, 1 on line
2t
L4 : v
2, 1, 3
5, 1, 8 on line
s
L2 and L4 are parallel, not identical, because 1, 1, 0 is
7 17
51
1 4s and
2 s. Solving this system yields t
11. When
7
17
7
and
using these values for s and t, the z
coordinates are not equal. The lines do not intersect.
not on L4 .
29. L1: v
4, 2, 3
L2 : v
2, 1, 5
L3: v
8, 4, 6
L4 : v
2, 1, 1.5
8, 5, 9 on line
8, 5, 9 on line
L1 and L3 are identical.
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
40
Vectors and the Geometry of Space
34. Writing the equations of the lines in parametric form you
have
x
2 3t
y
2 6t
z
3t
x
3 2s
y
5 s
z
2 4s.
By equating like variables, you have
2 4 s.
2 3t
3 2s, 2 6t
5 s, 3 t
1, s 1 and the point of intersection is
So, t
38. 2 x 3 y 4 z
P 0, 0, 1 , Q 2, 0, 0 , R 3, 2, 2
JJJK
JJJK
(a) PQ
2, 0, 1 , PR
3, 2, 3
JJJK JJJK
(b) PQ u PR
3, 6, 1
First line
v
2, 1, 4
Second line
2t 3
x
2 s 7
y
5t 2
y
s 8
z
t 1
z
2s 1
4
966
39. x 2 y 4 z 1
(a)
(b)
8
2
−
4
6
x
−8
8
y
(b)
(7, 8, −1)
x
5s 12
y
4t 10
y
3s 11
z
t
z
2s 4
x
0
−2
(3, 2, 2)
j
0, 1, 0
0
42. Point: 0, 1, 4
−3
Normal vector: n
3
k
0, 0, 1
0 x 0 0 y 1 1z 4
y
0
0
6
(a) P 0, 0, 1 , Q 0, 2, 0 , R 3, 4, 1
JJJK
JJJK
0, 2, 1 , PR
3, 4, 0
PQ
JJJK JJJK
(b) PQ u PR
0
y 3
z 4
37. 4 x 3 y 6 z
6 z 0
Point is not in plane
0 x 1 1 y 3 0 z 7
2
2
0
1, 5, 1 : 2 1 5 3 1 6
x = − 5s − 12
z y = 3s + 11
z = − 2s − 4
2
0
3, 6, 2 : 2 3 6 3 2 6
Normal vector: n
3
3
5, 2, 2 : 5 2 2 4 2 1
41. Point: 1, 3, 7
Point of intersection: 3, 2, 2
x = 2t − 1
y = − 4t + 10
z=t
0
Point is in plane
10
2t 1
36. x
7, 2, 1 : 7 2 2 4 1 1
40. 2 x y 3 z 6
(a)
10
0
Point is in plane
0
z
4
2, 3, 4
Point is in plane
4
6
2 0 1
variables in the equation. The cross product is
parallel to the normal vector.
2 966
483
Point of intersection: 7, 8, 1
2 and s
k
and R, they are the same to the coefficients of the
4
46 21
35. x
Note: t
j
The components of the cross product are
proportional for this choice of P, Q,
u
u˜v
u v
i
3 2 3
5, 4, 2 .
cos T
4
i
j
k
0 2 1
3
4
43. Point: 3, 2, 2
Normal vector: n
2 x 3 3 y 2 1z 2
2x 3y z
4, 3, 6
0
The components of the cross product are
proportional to the coefficients of the variables in
the equation. The cross product is parallel to the
normal vector.
2i 3 j k
0
10
44. Point: 0, 0, 0
Normal vector: n
3i 2k
3 x 0 0 y 0 2 z 0
0
3 x 2 z
0
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
45. Point: 1, 4, 0
2, 1, 2
Normal vector: v
2 x 1 1 y 4 2 z 0
Lines and Planes in Space
50. 1, 2, 3 , Normal vector: v
i, 1 x 1
0, x
51. 1, 2, 3 , Normal vector: v
k, 1 z 3
0, z
3
52. The plane passes through the three points
0
0, 0, 0 , 0, 1, 0 ,
46. Point: 3, 2, 2
3, 0, 1 .
The vector from 0, 0, 0 to 0, 1, 0 : u
j
4i j 3k
4x 3 y 2 3z 2
0
4 x y 3z
8
The vector from 0, 0, 0 to
Normal vector: u u v
3i k
3, 0, 1 : v
i
j k
0
1 0
47. Let u be the vector from 0, 0, 0 to
2, 0, 3 : u
1
0
2x y 2z 6
Normal vector: v
41
i 3k
3 0 1
2, 0, 3
x
3z
0
Let u be the vector from 0, 0, 0 to
3, 1, 5 : v
53. The direction vectors for the lines are
u
2i j k , v
3i 4 j k.
3, 1, 5
Normal vectors: u u v
i
j k
2
0
3
3, 19, 2
Normal vector: u u v
3 1 5
3 x 0 19 y 0 2 z 0
0
3 x 19 y 2 z
0
2, 1, 4
k
uu v
3
1
1
0
x y z
5
2
2i j k. Choose any
2i 2 j k
i
5, 10, 5
5 1, 2, 1
Normal vector: u u v
x 2y z 1
0
x 2 2 z 1
0
0
x 2z
0
2, 2, 1 : v
2i 2k
2i 4 j k
k
1 0 1
4i 3 j 4k
1
4 x 1 3 y 2 4 z 3
4x 3 y 4z
3i j 2k
Because v and n both lie in the plane P, the normal
vector to P is
Normal vector:
2 4
i 2k
Let n be a vector normal to the plane
2x 3y z
3: n
2i 3 j k
Let v be the vector from 1, 2, 3 to
j
k
55. Let v be the vector from 1, 1, 1 to
49. Let u be the vector from 1, 2, 3 to
i
j
2 1 1
2 2 1
1x 3 2 y 1 z 2
1, 2, 2 : v
5 i j k
point on the line, ª¬ 0, 4, 0 , for exampleº¼ , and let v be the
vector from 0, 4, 0 to the given point 2, 2, 1 :
2 1 4
u v
1
x 1 y 5 z 1
v
Normal vector:
i
j
1u
2
2
54. The direction of the line is u
Let u be the vector from 3, 1, 2 to 1, 2, 2 :
3, 2, 1 : u
k
Point of intersection of the lines: 1, 5, 1
1, 2, 3
v
j
3 4 1
48. Let u be the vector from 3, 1, 2 to 2, 1, 5 :
u
i
vun
i
j
k
3
1
2
7i j 11k
2 3 1
0
10
7 x 2 1 y 2 11 z 1
0
7 x y 11z
5
© 2010 Brooks/Cole, Cengage Learning
42
Chapter 11
Vectors and the Geometry of Space
56. Let v be the vector from 3, 2, 1 to
j 6k
3, 1, 5 : v
vun
j
Because u and v both lie in the plane P, the normal
vector to P is:
i
uuv
7
40i 36 j 6k
ª¬ y 2 º¼ ª¬ z 1 º¼
y z
2 20i 18 j 3k
7 j k
0
1
0
20 x 18 y 3z
27
k and let v be the vector from 4, 2, 1 to 3, 5, 7 : v
58. Let u
7 j 7 k
1 7 7
2
20 x 3 18 y 2 3 z 1
j k
1 0 0
k
0 1 6
6
i 7 j 7k
2, 5, 6 : v
Let n be the normal to the given plane:
n
6i 7 j 2k
Because v and n both lie in the plane P, the normal
vector to P is:
i
i and let v be the vector from 1, 2, 1 to
57. Let u
7i 3j 6k
Because u and v both lie in the plane P, the normal vector to P is:
uuv
i
j k
0
0 1
3i 7 j
3i 7 j
7 3 6
3x4 7 y 2
0
3x 7 y
59. xy-plane: Let z
Then 0
y
26
0.
z
4t Ÿ t
4 Ÿ x
2 3 4
1 24
7 and
−8
6
−6
4 −4
2
10. Intersection: 7, 10, 0
2
xz-plane: Let y
Then 0
z
2 3t Ÿ t
4 Then 0
z
z
0 Ÿ t
1 z
2 3
2 3t , y
32
32
Ÿ x
52 .
2 31
3 21
Then 2 3t
10
− 1 , 0, −
)0, − 12 , − 72 ) ) 3 3 )
12 and
1
2
1 t , z
3 2t
2 3
32
Intersection:
52 ,
52
and
52 ,
0
z
5
0, − 3 , 5
3
(5, 0, 5)
(
−4
x
5
−3
4
−2
3
)
6
4
−1
2
1
( − 52 , − 52 , 0 )
−1
2
−2
−3
y
0.
1Ÿ x
yz -plane: Let x
13 and
2
3
0.
xz-plane: Let y
Then t
1 2
72 . Intersection: 0, 12 , 72
1
2
xy -plane: Let z
y
Ÿ x
Ÿ y
1
2
60. Parametric equations: x
Then 3 2t
2
3
y
10
0.
1 2t Ÿ t
4 6 8
10
. Intersection: 13 , 0, 10
3
3
2
3
yz -plane: Let x
2
x
0.
(− 7, 10, 0)
5 and
5. Intersection: 5, 0, 5
0.
0 Ÿ t
3 2 23
23 Ÿ y
5
.
3
1 2
3
53 and
Intersection: 0, 53 ,
5
3
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
Lines and Planes in Space
43
61. Let x, y, z be equidistant from 2, 2, 0 and 0, 2, 2 .
x 2
2
2
y 2
2
z 0
2
x 0
x2 4x 4 y 2 4 y 4 z 2
y 2
2
z 2
2
x2 y2 4 y 4 z 2 4z 4
4x 8
4z 8
x z
0 Plane
62. Let x, y, z be equidistant from 1, 0, 2 and 2, 0, 1 .
x 1
2
y 0
2
z 2
2
2
x2
x2 2 x 1 y 2 z 2 4 z 4
y 0
2
z 1
2
x2 4x 4 y 2 z 2 2 z 1
2 x 4 z 5
4 x 2 z 5
2x 2z
0
x z
0 Plane
63. Let x, y, z be equidistant from 3, 1, 2 and 6, 2, 4 .
x3
2
y 1
2
z 2
2
x6
x2 6 x 9 y2 2 y 1 z 2 4z 4
2
y 2
2
2
z 4
x 2 12 x 36 y 2 4 y 4 z 2 8 z 16
6 x 2 y 4 z 14
12 x 4 y 8 z 56
18 x 6 y 4 z 42
0
9 x 3 y 2 z 21
0 Plane
64. Let x, y, z be equidistant from 5, 1, 3 and 2, 1, 6
x 5
2
y 1
2
z 3
2
x 2
x 2 10 x 25 y 2 2 y 1 z 2 6 z 9
n1 ˜ n 2
n1 n 2
0.
S 2 and the planes are orthogonal.
66. The normal vectors to the planes are
n1
3, 1, 4 , n 2
Because n 2
9, 3, 12 .
cos T
So, T
46
27
§ 2 138 ·
arccos¨¨
¸¸ | 83.5q.
© 207 ¹
cos T
4 138
414
n1 ˜ n 2
n1
38 2
n2
14
7 6
42
21
6
.
6
§ 6·
arccos¨¨
¸¸ | 65.9q.
© 6 ¹
69. The normal vectors to the planes are n1
n2
67. The normal vectors to the planes are
n1
i 3 j 6k , n 2
5i j k ,
536
68. The normal vectors to the planes are
n1
3i 2 j k , n 2
i 4 j 2k ,
So, T
3n1 , the planes are parallel, but not equal.
n1 ˜ n 2
n1 n 2
2
0 Plane
65. The normal vectors to the planes are
So, T
z 6
0
7 x 2 y 9z 3
1, 4, 7 , cos T
2
4 x 2 y 12 z 41
14 x 4 y 18 z 6
5, 3, 1 , n 2
y 1
x 2 4 x 4 y 2 2 y 1 z 2 12 z 36
10 x 2 y 6 z 35
n1
2
5, 25, 5 . Because n 2
1, 5, 1 and
5n1 , the planes are
parallel, but not equal.
2 138
.
207
70. The normal vectors to the planes are
n1
cos T
So, T
2, 0, 1 , n 2
n1 ˜ n 2
n1 n 2
S
2
4, 1, 8 ,
0
and the planes are orthogonal.
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
44
Vectors and the Geometry of Space
71. 4 x 2 y 6 z
12
77. x
5
z
z
3
6
4
(0, 0, 2)
5
(0, 6, 0)
4
6
(3, 0, 0)
x
6
72. 3x 6 y 2 z
(5, 0, 0)
y
78. z
6
8
z
z
3
y
5
x
8
(0, 0, 3)
(0, 1, 0)
3
2
2
(2, 0, 0)
x
y
3
5
73. 2 x y 3 z
4
79. 2 x y z
z
(0, − 4, 0)
(
6
z
3
0, 0, 4
3
y
5
x
(2
6
−4
2
4
2
4
6
y
x
−1
y
−6
(2, 0, 0)
3
x
Generated by Maple
74. 2 x y z
80. x 3z
4
3
z
4
z
(0, 0, 4)
2
(0, −4, 0)
1
2
−4
3
1
1
2
y
y
(2, 0, 0)
Generated by Mathematica
x
75. x z
2
x
1
3
1
81. 5 x 4 y 6 z 8
6
0
z
z
−2
8
(0, 0, 6)
(6, 0, 0)
1
−1
2
y
8
8
x
76. 2 x y
y
x
Generated by Maple
82. 2.1x 4.7 y z 3
8
0
z
z
3
8
2
1
(4, 0, 0)
(0, 8, 0)
8
x
8
y
1
x
2
y
Generated by Mathematica
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
83. P1: n
15, 6, 24
0, 1, 1 not on plane
P2 : n
5, 2, 8
0, 1, 1 on plane
P3: n
6, 4, 4
P4 : n
3, 2, 2
i
n 2 u n1
3, 5, 2
P3: n
8, 4, 12
P4 : n
4, 2, 6
P2 : n
0
14
x
4, 0, 0 not on plane
2
Substituting 2 for x in the second equation, you have
4 y 2 z
2 or z
2 y 1. Letting y 1, a
point of intersection is 2, 1, 1 .
3, 2, 5
1, 1, 1 on plane
6, 4, 10
1, 1, 1 not on plane
x
1 t, z
2, y
1 2t
6i 3 j k , n 2
92. (a) n1
n1 ˜ n 2
n1 n 2
cos T
1, 1, 1 on plane
i j 5k
4
46 27
60, 90, 30 or 2, 3, 1
9 on plane
0, 0, 10
P2 : n
6, 9, 3 or 2, 3, 1
P3: n
20, 30, 10 or 2, 3, 1
0, 0, 23 on plane
P4 : n
12, 18, 6 or 2, 3, 1
0, 0,
5
6
n1 u n 2
6 3
1
on plane
16, 31, 3 .
1
1 5
Find a point of intersection of the planes.
z
6x 3 y z 5 Ÿ 6x 3y 5
5 Ÿ 6 x 6 y 30 z
30
3 y 31z
35
x y 5z
P1 , P2 , and P3 are parallel.
x
c, 0, 0 , 0, c, 0 , and 0, 0, c .
y
0, z
1
z is a plane parallel to
c
x-axis and passing through the points 0, 0, 0 and
0 Ÿ y
0, 1, c .
4 Ÿ 4, 9, 2 .
9 31t , z
2 3t
1
t, y
2
Substituting t
t, z
1 2t
·
t ¸ 1 2t
¹
12, t
3
2
3 2 into the parametric equations for
the line you have the point of intersection 2, 3, 2 .
The line does not lie in the plane.
90. x cz
0
0 is the yz-plane.
If c 0, z
If c z 0, x cz
cos T
4 16t , y
3
2
§1
·
§ 3
2¨ t ¸ 2¨
©2
¹
© 2
x
0 is xy-plane.
If c z 0, cy z
91. (a) n 1
2 Ÿ x
93. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:
c
Each plane is parallel to the z-axis.
89. If c
9, z
Let y
87. Each plane passes through the points
88. x
2 138
207
T | 1.6845 | 96.52q
(b) The direction vector for the line is
i
j k
P4 and is parallel to P2 .
86. P1: n
14
7x
P1 and P4 are identical.
P1
7 j 2k .
2 1
x 4 y 2z
4, 0, 0 on plane
75, 50, 125
P4 : n
2
6x 4 y 2z
3, 2, 5
P3: n
k
Find a point of intersection of the planes.
P1 and P3 are parallel.
85. P1: n
j
1 4
3
2, 1, 3
P2 : n
45
(b) The direction vector for the line is
Planes P1 and P2 are parallel.
84. P1: n
Lines and Planes in Space
0 is a plane parallel to the y-axis.
3i 2 j k and n 2
n1 ˜ n 2
n1 n 2
i 4 j 2k
7
14
Ÿ T | 1.1503 | 65.91q
21
6
6
94. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:
x
1 4t , y
2 1 4t 3 2t
Substituting t
2t , z
5, t
3 6t
1
2
12 into the parametric equations for
the line you have the point of intersection 1, 1, 0 .
The line does not lie in the plane.
© 2010 Brooks/Cole, Cengage Learning
46
Chapter 11
Vectors and the Geometry of Space
95. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:
x
1 3t , y
1 2t , z
2 1 3t 3 1 2t
3t
10, 1
10, contradiction
So, the line does not intersect the plane.
96. Writing the equation of the line in parametric form and
substituting into the equation of the plane you have:
x
4 2t , y
1 3t , z
5 4 2t 3 1 3t
2 5t
17, t
97. Point: Q 0, 0, 0
Plane: 2 x 3 y z 12
Normal to plane: n
0
2, 3, 1
Point in plane: P 6, 0, 0
JJJK
6, 0, 0
Vector PQ
JJJK
PQ ˜ n
12
D
n
14
Normal to plane: n
0
5, 1, 1
2 2
1, 3, 4 and
101. The normal vectors to the planes are n1
n2
1, 3, 4 . Because n1
n 2 , the planes are
parallel. Choose a point in each plane.
P 10, 0,0 is a point in x 3 y 4 z
10.
Q 6, 0, 0 is a point in x 3 y 4 z
JJJK
PQ ˜ n1
JJJK
PQ
4, 0, 0 , D
n1
6.
4
26
4, 4, 9 . Because n1
3
n 2 , the planes are
parallel. Choose a point in each plane.
P 5, 0, 3 is a point in 4 x 4 y 9 z
7.
Q 0, 0, 2 is a point in 4 x 4 y 9 z
JJJK
PQ
5, 0, 1
JJJK
PQ ˜ n1
11
11 113
D
113
n1
113
18.
3, 6, 7 and
n2
6, 12, 14 . Because n 2
2n1 , the planes are
parallel. Choose a point in each plane.
P 0, 1, 1 is a point in 3 x 6 y 7 z
5
2, 1, 1
Point in plane: P 0, 0, 5
JJJK
2, 8, 1
Vector: PQ
JJJK
PQ ˜ n
11
11 6
D
6
n
6
2 26
13
4, 4, 9 and
103. The normal vectors to the planes are n1
99. Point: Q 2, 8, 4
Normal to plane: n
3, 4, 5
Point in plane: P 2, 0, 0
JJJK
Vector PQ: 1, 3, 1
JJJK
PQ ˜ n
20
D
n
50
n2
6 14
7
Point in plane: P 0, 9, 0
JJJK
Vector PQ
0, 9, 0
JJJK
PQ ˜ n
9
D
n
27
Plane: 2 x y z
Normal to plane: n
0
102. The normal vectors to the planes are n1
98. Point: Q 0, 0, 0
Plane: 5 x y z 9
Plane: 3x 4 y 5 z 6
0
Substituting t
0 into the parametric equations for the
line you have the point of intersection 4, 1, 2 .
The line does not lie in the plane.
100. Point: Q 1, 3, 1
1.
§ 25
·
Q¨ , 0, 0 ¸ is a point in 6 x 12 y 14 z
© 6
¹
JJJK
PQ
D
25
, 1, 1
6
JJJK
PQ ˜ n1
27 2
n1
94
27
2 94
25.
27 94
188
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
104. The normal vectors to the planes are n1
n2
2, 0, 4 . Because n1
2, 0, 4 and
n 2 , the planes are
4.
Q 5, 0, 0 is a point in 2 x 4 z
JJJK
PQ ˜ n1
JJJK
PQ
3, 0, 0 , D
n1
10.
6
20
3 5
5
4, 0, 1 is the direction vector for the line.
3, 2, 3
i
JJJK
PQ u u
106. u
j
JJJK
PQ u u
2, 9, 8
149
17
u
2533
17
u
103
5
0.
j k
2 1 2
JJJK
PQ u u
5
u
9
1, 2, 1 .
3, 6, 3 .
3v1 , the lines are parallel.
i
j
k
2
2
0
5
3
v2
396
54
22
3
110. The direction vector for L1 is v1
6, 9, 12 .
The direction vector for L2 is v 2
4, 6, 8 .
3v ,
2 2
Because v1
1, 1, 2 is the direction vector for the line.
6, 6, 18
3 6 3
JJJK
PQ u v 2
36 36 324
9 36 9
0, 2, 1
1 1 2
515
5
v 2 is the direction vector for L2 .
D
66
3
the lines are parallel.
0 in the parametric equations for the line .
Let Q 3, 2, 1 to be a point on L1 and P 1, 3, 0 a point
JJJK
on L2 . PQ
4, 5, 1 .
3, 1, 3
u
Q 2, 1, 3 is the given point, and P 1, 2, 0 is on the line
i
JJJK
PQ u u
D
206
10
Let Q 2, 3, 4 to be a point on L1 and P 0, 1, 4 a point
JJJK
on L2 . PQ
2, 0, 0 .
on the line let t
JJJK
PQ
1, 1, 2
let t
JJJK
PQ
14, 1, 3
u
Q 1, 2, 4 is the given point, and P 0, 3, 2 is a point
107. u
1
JJJK
PQ u u
JJJK
PQ u v 2
D
4
3
Because v 2
2, 1, 2 is the direction vector for the line.
i
1 2
0
The direction vector for L2 is v 2
k
3 2 3
JJJK
PQ u u
j k
109. The direction vector for L1 is v1
4 0 1
D
i
142 1 9
9 1
Q 1, 5, 2 is the given point, and P 2, 3, 1 is on the
line.
JJJK
PQ
0, 3, 1 is the direction vector for the line.
JJJK
PQ u u
D
105. u
47
Q 4, 1, 5 is the given point, and P 3, 1, 1 is on the line.
JJJK
PQ
1, 2, 4
parallel. Choose a point in each plane.
P 2, 0, 0 is a point in 2 x 4 z
108. u
Lines and Planes in Space
j
k
3 1
3
1
JJJK
PQ u u
u
j
k
4 5
1
4
1 2
98
6
i
JJJK
PQ u v 2
1, 9, 4
1 81 16
11 4
v 2 is the direction vector for L2 .
7
3
7 3
3
D
34, 36, 44
6 8
JJJK
PQ u v 2
v2
342 362 442
16 36 64
4388
116
1097
29
31813
29
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
48
Vectors and the Geometry of Space
111. The parametric equations of a line L parallel to
v
a, b, c, and passing through the point
115. (a) The planes are parallel if their normal vectors are
parallel:
a1 , b1 , c1
t a2 , b2 , c2 , t z 0
P x1 , y1 , z1 are
x
x1 at , y
y1 bt , z
The symmetric equations are
x x1
y y1
z z1
.
a
b
c
z1 ct.
(b) The planes are perpendicular if their normal vectors
are perpendicular:
a1 , b1 , c1 ˜ a2 , b2 , c2
0
116. Yes. If v1 and v 2 are the direction vectors for the lines
L1 and L2 , then v
v1 u v 2 is perpendicular to both
L1 and L2 .
112. The equation of the plane containing P x1 , y1 , z1 and
a, b, c is
having normal vector n
a x x1 b y y1 c z z1
0.
117. An equation for the plane is
x
y
z
1 Ÿ bcx acy abz
abc
a
b
c
For example, letting y
z
0, the x-intercept is
You need n and P to find the equation.
113. Simultaneously solve the two linear equations
representing the planes and substitute the values back
into one of the original equations. Then choose a value
for t and form the corresponding parametric equations
for the line of intersection.
a, 0, 0 .
118. (a) Matches (iii)
(b) Matches (i)
114. x
a: plane parallel to yz-plane containing a, 0, 0
(c) Matches (iv)
y
b: plane parallel to xz-plane containing 0, b, 0
(d) Matches (ii)
z
c: plane parallel to xy-plane containing 0, 0, c
119. Sphere
x 3
2
y 2
2
z 5
2
16
120. Parallel planes
4x 3y z
121. 0.92 x 1.03 y z
(a)
0.02 Ÿ z
10 r 4 n
10 r 4 26
0.02 0.92 x 1.03 y
Year
1999
2000
2001
2002
2003
2004
2005
x
1.4
1.4
1.4
1.6
1.6
1.7
1.7
y
7.3
7.1
7.0
7.0
6.9
6.9
6.9
z
6.2
6.1
5.9
5.8
5.6
5.5
5.6
Model z
6.25
6.05
5.94
5.76
5.66
5.56
5.56
The approximations are close to the actual values.
(b) According to the model, if x and z decrease, then so will y. (Answers will vary.)
122. On one side you have the points 0, 0, 0 , 6, 0, 0 , and 1, 1, 8 .
n1
i
j k
6
0
0
48 j 6k
1 1 8
On the adjacent side you have the points 0, 0, 0 , 0, 6, 0 , and 1, 1, 8 .
n2
i
j k
0
6
0
z
(− 1, − 1, 8)
48i 6k
6
1 1 8
cos T
n1 ˜ n 2
n1 n 2
4
36
2340
1
65
2
6
T
1
arccos
| 89.1q
65
x
(0, 0, 0) 4
(0, 6, 0)
6
y
(6, 0, 0)
© 2010 Brooks/Cole, Cengage Learning
Section 11.5
8 t , z1
6 t ; y1
123. L1: x1
(a) At t
49
3t
2 t , z2
1 t , y2
L2 : x2
Lines and Planes in Space
2t
0, the first insect is at P1 6, 8, 3 and the second insect is at P2 1, 2, 0 .
2
6 1
Distance
x1 x2
(b) Distance
82
2
2
30
y1 y2
2
2
70 | 8.37 inches
z1 z2
2
2
52 6 2t
(c) The distance is never zero.
(d) Using a graphing utility, the minimum distance is 5 inches when t
3t
2
5t 2 30t 70, 0 d t d 10
3 minutes.
15
0
15
0
124. First find the distance D from the point Q 3, 2, 4 to the plane. Let P 4, 0, 0 be on the plane.
n
D
2, 4, 3 is the normal to the plane.
JJJK
PQ ˜ n
7, 2, 4 ˜ 2, 4, 3
14 8 12
4 16 9
n
29
18
29
18 29
29
The equation of the sphere with center 3, 2, 4 and radius 18 29 29 is x 3
125. The direction vector v of the line is the normal to the
3, 1, 4 .
plane, v
The parametric equations of the line are
5 3t , y
x
4 t, z
3 4t.
7
26t
t
4
13
,4 4,
13
4
13
126. The normal to the plane, n
to the direction vector v
2
324
.
29
127. The direction vector of the line L through 1, 3, 1 and
3, 4, 2 is v
2, 1, 1 .
Substituting these equations into the equation of the
plane gives
1 2t 3 t 1 t
2
4t
3
t
34 .
4
13
77 , 48 ,
13 13
23
13
2, 1, 3 is perpendicular
2, 4, 0 of the line because
4, 1, 4 ˜ 2, 1, 3
1 2 34 , 3 34 , 1 19
14
19 14
14
3
4
12 , 94 , 14
128. The unknown line L is perpendicular to the normal vector
n
1, 1, 1 of the plane, and perpendicular to the direction
1, 1, 1 . So, the direction vector of L is
vector u
0.
So, the plane is parallel to the line. To find the distance
between them, let Q 2, 1, 4 be on the line and
JJJK
P 2, 0, 0 on the plane. PQ
4, 1, 4 .
JJJK
PQ ˜ n
D
n
4 19
z 4
Point of intersection:
3 4
2, 1, 3 ˜ 2, 4, 0
2
8
Point of intersection:
53
y 2
The parametric equations for L are
x 1 2t , y
3 t , z 1 t.
To find the point of intersection, solve for t in the
following equation:
3 5 3t 4 t 4 3 4t
2
i
v
j
k
1 1
1
2, 2, 0 .
1 1 1
The parametric equations for L are x
z
2.
1 2t , y
2t ,
129. True
130. False. They may be skew lines.
See Section Project
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
50
Vectors and the Geometry of Space
131. True
132. False. The lines x
t, y
0, z 1 and x
y
t , z 1 are both parallel to the plane z
the lines are not parallel.
0,
0, but
133. False. Planes 7 x y 11z 5 and 5 x 2 y 4 z 1
are both perpendicular to plane 2 x 3 y z
3, but
are not parallel.
134. True.
Section 11.6 Surfaces in Space
10. x 2 z 2
1. Ellipsoid
Matches graph (c)
2. Hyperboloid of two sheets
25
The y-coordinate is missing so you have a right circular
cylinder with rulings parallel to the y-axis. The generating
curve is a circle.
Matches graph (e)
z
6
3. Hyperboloid of one sheet
4
Matches graph (f )
8
x
4. Elliptic cone
8
y
Matches graph (b)
x2
11. y
5. Elliptic paraboloid
Matches graph (d)
6. Hyperbolic paraboloid
The z-coordinate is missing so you have a parabolic
cylinder with rulings parallel to the z-axis. The generating
curve is a parabola.
Matches graph (a)
7. y
z
4
5
Plane is parallel to the xz-plane.
z
4
2
−2
1
−1
1
4
12. y 2 z
2
−2
x
3
y
−3
1
2
2
x
3
3
3
4
5
−3
y
6
The x-coordinate is missing so you have a parabolic
cylinder with the rulings parallel to the x-axis. The
generating curve is a parabola.
z
8. z
2
Plane is parallel to the xy-plane.
z
2
3
−2
2
2
y
x
3
2
x
−1
2
3
−2
13. 4 x 2 y 2
y
−3
9. y 2 z 2
9
The x-coordinate is missing so you have a right circular
cylinder with rulings parallel to the x-axis. The generating
curve is a circle.
z
z
4
3
−3
7 6
x
4
x2
y2
1
1
4
The z-coordinate is missing so you have an elliptic
cylinder with rulings parallel to the z-axis. The generating
curve is an ellipse.
4
y
2
3
x
2
3
y
© 2010 Brooks/Cole, Cengage Learning
Section 11.6
14. y 2 z 2
Surfaces in Space
51
(b) From 0, 10, 0 :
16
z
y2
z2
1
16 16
The x-coordinate is missing so you have a hyperbolic
cylinder with rulings parallel to the x-axis. The generating
curve is a hyperbola.
y
z
6
(c) From 10, 10, 10 :
4
z
2
−4
2
6
x
4
3
y
6
−4
−6
3
15. z
3
y
x
sin y
The x-coordinate is missing so you have a cylindrical
surface with rulings parallel to the x-axis. The generating
curve is the sine curve.
19.
z
x2
y2
z2
1
4
1
1
Ellipsoid
2
z
2
1
x
y
1
4
xy-trace:
y
3
2
2
1 ellipse
3
xz-trace: x 2 z 2
4
x
2
ey
16. z
z
20.
x2
y2
z2
16
25 25
15
xy-trace:
10
5
3
2
x y
17. z
3
4
y
(b) You are viewing the paraboloid from above, but not
on the z-axis: 10, 10, 20
(c) You are viewing the paraboloid from the z-axis:
0, 0, 20
(d) You are viewing the paraboloid from the y-axis:
0, 20, 0
18. y z
2
x2
y2
16
25
4
3
2
1
ellipse
1
5
2
x
z
xz-trace:
16
25
1
ellipse
yz-trace: y 2 z 2
25
circle
21. 16 x 2 y 2 16 z 2
4
3
2
1
1 2
3 4
5
x
4 x2 4
4
2
y
4z2
4
1
Hyperboloid of one sheet
xy-trace: 4 x 2 y2
4
1 hyperbola
xz-trace: 4 x 2 z 2
yz-trace:
2
z
5
2
(a) You are viewing the paraboloid from the x-axis:
20, 0, 0
z
1 circle
y2
4z2
4
1 hyperbola
z
(a) From 10, 0, 0 :
3
2
−2
y
−3
2
3
x
−2
y
−2
1
2
1
2
2
x
1 ellipse
Ellipsoid
20
x
2
y
z
4
1
yz-trace:
The x-coordinate is missing so you have a cylindrical
surface with rulings parallel to the x-axis. The generating
curve is the exponential curve.
1 circle
2
3
y
−3
© 2010 Brooks/Cole, Cengage Learning
y
Chapter 11
52
Vectors and the Geometry of Space
22. 8 x 2 18 y 2 18 z 2
9 y2 9z 2 4x2
25. x 2 y z 2
2
1
0
Elliptic paraboloid
xy-trace: y
xy-trace: 9 y 4 x
xz-trace: x z
yz-trace: 9 y 9 z
2
2
1 hyperbola
2
2
1 hyperbola
−3
1
2
0,
1
2
3
3
x
26. z
1
y
2
y
−3
1: x 2 z 2
1
2
−2
4
−2
z2
yz-trace: y
y
z
2
point 0, 0, 0
1 circle
xz-trace: 9 z 2 4 x 2
x
3
2
Hyperboloid of one sheet
2
z
z
x2 4 y 2
4
2
Elliptic paraboloid
1
2
x
xy-trace: point 0, 0, 0
y
23. 4 x 2 y 2 z 2
1
xz-trace: z
x 2 parabola
yz-trace: z
4 y 2 parabola
3
1
2
1
x
Hyperboloid of two sheets
xy-trace: 4 x 2 y 2
1
hyperbola
yz-trace: none
xz-trace: 4 x 2 z 2
hyperbola
1
z
27. x 2 y 2 z
z
Hyperbolic paraboloid
xy-trace: y
rx
xz-trace: z
x2
yz-trace: z
y2
3
y
x
0
r1: z
3
2 3
3 2
x
y
1 x2
2
3
y
28. 3z
−3
y2
4
24. z 2 x 2 y2 x2
Hyperbolic paraboloid
xy-trace: y
rx
1
Hyperboloid of two sheets
xz-trace: z
1 x2
3
yz-trace: z
13 y 2
xy-trace: none
xz-trace: z 2 x 2
1 hyperbola
y2
yz-trace: z 4
1 hyperbola
2
r 10:
z
x2
y2
9
36
29. z 2
5
5
y
10
x
z
1
Elliptic cone
−3
xy-trace: point 0, 0, 0
1 ellipse
xz-trace: z
rx
yz-trace: z
r
When z
1
x
3
y
−1
y
3
r1, x 2 x
30. x 2
y
10
y2
9
x2 z
5
z
28
24
20
y2
9
1 ellipse
2 y2 2z2
z
5
Elliptic Cone
xy-trace: x
r 2y
xz-trace: x
r
5
x
2z
5
y
yz-trace: point: 0, 0, 0
© 2010 Brooks/Cole, Cengage Learning
Section 11.6
16 x 2 9 y 2 16 z 2 32 x 36 y 36
31.
16 x 2 x 1 9 y 4 y 4 16 z
2
2
16 x 1
2
9 y 2
x 1
1
2
2
0
2
2
1
16 z 2
y 2
16 9
53
z
36 16 36
2
Surfaces in Space
16
1
2
z2
1
2
x
1
4
−2
y
Ellipsoid with center 1, 2, 0 .
9 x 2 y 2 9 z 2 54 x 4 y 54 z 4
32.
0
9 x2 6x 9 y 2 4 y 4 9 z 2 6 z 9
z
81 81
10
5
9x 3
2
y 2
2
9 z 3
2
0
15 10
15
x
Elliptic cone with center 3, 2, 3 .
20
25
y
−15
33. z
2 cos x
§ 2·
¨ ¸
©z¹
37. x 2 y 2
z
3
r
y
3
2π
x
2
4
x2
z2
y
z
4
x 2 0.5 y 2
34. z
z
4
4
4
x
e z
38. x 2 y 2
−2
1
2
ln x 2 y 2
1
2
x
y
y
z
z
35. z
x 7.5 y
2
2
r
z
2
−4
4
−3
−3
−4
x 2 7.5 y 2
z
4
4
y
x
2
−2
−2
2
10 39. z
2
y
x
xy
z
12
36. 3.25 y
z
2
z
x2 z 2
3.25 y x
z
2
r 3.25 y x
−8
2
−4
8
8
y
x
2
1
−1
1
2
x
−1
2
3
y
−2
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
54
Vectors and the Geometry of Space
x
8 x2 y 2
40. z
44. z
4 x2
y
4 x2
z
x
0, y
0, z
4
5
2
2
2
4
4
x
y
4 3
x
41. 6 x 4 y 6 z
2
36
6z
2
4 y 2 6 x 2 36
3z 2
2 y 2 3 x 2 18
2
2
r
z
1
3
2 y 2 3 x 2 18
4
45. x 2 y 2
z
y
1
x z
2
0
z
4
z
3
6
2
4
−6
2
2
x
0
z
y
6 −2
2
3
x
−4
y
3
−6
4 x2 y2
46. z
42. 9 x 2 4 y 2 8 z 2
r
z
9 x2
8
1 y2
2
72
y
2z
9
z
0
z
z
3
20
−3
10
3
20
20
x
x
y
47. x 2 z 2
x z
2
z
2
z
2
x2 y 2
2
x2 y 2
1
43.
2
y
3
10
10
x2 y 2
2
48. x 2 z 2
x z
2
2
49. x 2 y 2
2
ª¬r y º¼ and z
r y
r2
r y
3 y; so,
r z
z
; so,
2
y ; so,
4 y.
2
ª¬r y º¼ and z
2
9y .
2
ª¬r z º¼ and y
z
x2 y2
3
z2
, 4x2 4 y 2
4
z 2.
2
50. y 2 z 2
−2
y2 z2
−2
2
x
1
2
2
ª¬r x º¼ and z
1
4
r x
1
2
4 x2 , x2 4 y 2 4z 2
4 x 2 ; so,
4.
y
© 2010 Brooks/Cole, Cengage Learning
Section 11.6
2
51. y 2 z 2
ª¬r x º¼ and y
y2 z2
§2·
2
2
¨ ¸ ,y z
© x¹
2
52. x y
x y
2
2
2
2
ª¬r z º¼ and y
2
; so,
x
r x
(a) When z
e ; so,
r z
2z
e .
53. x 2 y 2 2 z
x2 y2
0
2z
Major axis: 2 8
4 2
Minor axis: 2 4
4
c
2
2
a b ,c
2
2 z or x
cos 2 y
Equation of generating curve: x
cos y or z
56. The trace of a surface is the intersection of the surface
with a plane. You find a trace by setting one variable
2.
equal to a constant, such as x
0 or z
2
Major axis: 2 32
8 2
Minor axis: 2 16
8
c
2
32 16
4
0
218S
3
x2
4,
2
4 you have z
§1·
4¨ ¸ z 4
© 2¹
ª 4 x3
x º
»
2S «
3
4 ¼0
¬
4
x2
y2
2
4
62. z
x 2 is a cylinder.
2S ³ x 4 x x 2 dx
16, c
Foci: 0, r 4, 8
x 2 is a parabola.
4 4
x2
y2
, or
2
4
8 we have 8
(a) When y
57. See pages 814 and 815.
59. V
4, c
x2
y2
.
16
32
1
cos y
55. Let C be a curve in a plane and let L be a line not in a
parallel plane. The set of all lines parallel to L and
intersecting C is called a cylinder. C is called the
generating curve of the cylinder, and the parallel lines
are called rulings.
In three-space, z
2
2z
(b) When z
58. In the xz-plane, z
2
Foci: 0, r 2, 2
Equation of generating curve: y
54. x 2 z 2
x2
y2
, or
2
4
2 we have 2
x2
y2
4
8
1
z
55
x2
y2
2
4
61. z
4
.
x2
Surfaces in Space
x2.
9·
§
Focus: ¨ 0, 4, ¸
2¹
©
(b) When x
z
z
4
2 you have
2
2
y
,4 z 2
4
y 2.
Focus: 2, 0, 3
3
63. If x, y, z is on the surface, then
2
h ( x)
1
y 2
x
1
2
3
2
y2 4 y 4
4
p ( x)
x2 z 2
S
2S ³ y sin y dy
60. V
S
z
1.0
0.5
π
y
z2
x2 y 2 4 y 4 z 2
8y
2S 2
64. If x, y, z is on the surface, then
2
z2
x2 y 2 z 4
z2
x 2 y 2 z 2 8 z 16
x2
y2
2
8
8
Elliptic paraboloid shifted up 2 units. Traces parallel to
xy-plane are circles.
8z
π
2
2
Elliptic paraboloid
Traces parallel to xz-plane are circles.
0
2S >sin y y cos y@0
x2 y 2
x 2 y 2 16 Ÿ z
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
56
65.
Vectors and the Geometry of Space
x2
y2
z2
2
2
3963
3963
39502
67. z
1
y2
x2
2, z
2
b
a
bx ay
z
bx ay
4000
a 4b 2 ·
1§ 2
x a 2bx ¸
2¨
a ©
4 ¹
§
a 2b ·
¨x ¸
2 ¹
©
2
a
4000
y
4000
x
x2 y2
66. (a)
ª¬r z º¼
2
ª 2 z 1º
¬
¼
x2 y2 2z 2
2
y
2
0
Letting x
x
at , y
y
z
y2
x2
2
2
b
a
a 2b 4 ·
1§ 2
y ab 2 y ¸
2¨
b ©
4 ¹
2
§
ab 2 ·
¨y ¸
2 ¹
©
2
b
b§
a 2b · ab 2
r ¨x ¸
2 ¹
2
a©
at , you obtain the two intersecting lines
bt , z
0 and x
at ,
bt ab 2, z
2abt a 2b 2 .
4
68. Equating twice the first equation with the second equation:
3
2 x2 6 y 2 4z 2 4 y 8
2 x 2 6 y 2 4 z 2 3x 2
4y 8
−2
1
2
2
3 x 2
3x 4 y
y
3
6, a plane
x
2 ª
§1
·º
2S ³ x «3 ¨ x 2 1¸» dx
0
©2
¹¼
¬
(b) V
2 §
1 ·
2S ³ ¨ 2 x x3 ¸ dx
0
2 ¹
©
2
ª
x4 º
2S « x 2 »
8 ¼0
¬
4S | 12.6 cm3
69. True. A sphere is a special case of an ellipsoid (centered
at origin, for example)
x2
y2
z2
a2
b2
c2
having a
b
1
c.
70. False. For example, the surface x 2 z 2
y
formed by revolving the graph of x
y-axis, as the graph of z
3
2
1
x
1
2
e
y
e 2 y can be
e y about the
about the y-axis.
71. False. The trace x
2 of the ellipsoid
2
2
x
y
z2
1 is the point 2, 0, 0 .
4
9
3
72. False. Traces perpendicular to the axis are ellipses.
ª
§1
·º
2S ³ x «3 ¨ x 2 1¸» dx
12
2
©
¹¼
¬
2
(c) V
2S ³
1 3·
§
¨ 2 x x ¸ dx
2 ¹
©
2
12
73. The Klein bottle does not have both an “inside” and an
“outside.” It is formed by inserting the small open end
through the side of the bottle and making it contiguous
with the top of the bottle.
2
ª
x4 º
2S « x 2 »
8 ¼12
¬
4
31S
64
1
2
225S
| 11.04 cm3
64
y
3
2
1
x
3
© 2010 Brooks/Cole, Cengage Learning
Section 11.7
Cylindrical and Spherical Coordinates
57
Section 11.7 Cylindrical and Spherical Coordinates
1. 7, 0, 5 , cylindrical
x
r cos T
7 cos 0
7
y
r sin T
7 sin 0
0
z
5
x
7, 0, 5 , rectangular
y
z
2. 2, S , 4 , cylindrical
r cos T
2 cos S
2
y
r sin T
2 sin S
0
z
4
x
§ S ·
3. ¨ 3, , 1¸, cylindrical
© 4 ¹
x
3 cos
S
y
3 sin
S
z
1
4
4
3 2
2
3 2
2
§3 2 3 2 ·
,
, 1¸¸, rectangular
¨¨
2
© 2
¹
S ·
§
4. ¨ 6, , 2 ¸, cylindrical
4 ¹
©
y
z
1
4S
cos
2
3
1
4S
sin
2
3
8
1
4
3
4
§1
3 ·
¨¨ 4 , 4 , 8¸¸, rectangular
©
¹
7. 0, 5, 1 , rectangular
2, 0, 4 , rectangular
x
4S ·
§
6. ¨ 0.5,
, 8 ¸, cylindrical
3 ¹
©
§ S·
6 cos¨ ¸
© 4¹
§ S·
6 sin ¨ ¸
© 4¹
2
3 2
3 2
3 2, 3 2, 2 , rectangular
§ 7S ·
5. ¨ 4,
, 3¸, cylindrical
© 6 ¹
7S
4 cos
x
2 3
6
7S
2
y
4 sin
6
z
3
2 3, 2, 3 , rectangular
r
0
2
T
arctan
z
1
5
2
5
S
5
0
2
§ S ·
¨ 5, , 1¸, cylindrical
© 2 ¹
8. 2 2, 2 2, 4 , rectangular
r
2 2
2
T
arctan 1
z
4
2 2
2
4
S
4
S ·
§
¨ 4, , 4 ¸, cylindrical
4 ¹
©
9. 2, 2, 4 , rectangular
2 2 2
r
T
arctan 1
z
4
2
2 2
S
4
S
§
·
¨ 2 2, , 4 ¸, cylindrical
4
©
¹
10. 3, 3, 7 , rectangular
32 3
r
T
arctan 1
z
7
2
18
3 2
S
4
S ·
§
¨ 3 2, , 7 ¸, cylindrical
4 ¹
©
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
58
11. 1,
Vectors and the Geometry of Space
1 2
r
3
T
arctan 3
z
4
2
r
3
x2 y2
3
x2 y2
9
21.
3, 4 , rectangular
2
S
z
3
3
2
§ S ·
¨ 2, , 4 ¸, cylindrical
© 3 ¹
3
4 3
x
12 4
4
5S
6
§ 1 ·
arctan ¨ ¸
3¹
©
6
T
z
22. z
2
Same
z
3
S ·
§
¨ 4, , 6 ¸, cylindrical
6 ¹
©
13. z
4 is the equation in cylindrical coordinates.
1
2
(plane)
x
9, rectangular equation
r cos T
15. x y z
2
2
r2 z2
2
6
1
3
17, rectangular equation
17, cylindrical equation
x
x r 2 11, cylindrical equation
z
r cos T
sin T
r cos 2 T
18. x 2 y 2
r
1
2
x
2
y
−2
8 x, rectangular equation
r
8 cos T , cylindrical equation
y2
r sin T
2
r 2 sin 2 T z 2
10 z 2 , rectangular equation
10 z 2
x2 y 2
x2 y2 z2
4
z
2
z
2
0
z
10, cylindrical equation
4
20. x 2 y 2 z 2 3z
r z 3z
r
24.
8r cos T
2
−2
1
2
2
19.
2
2
sec T ˜ tan T , cylindrical equation
r
0
z
x , rectangular equation
r sin T
3y
3y
2
17. y
6
y
x
y
x
S
tan
x 2 y 2 11, rectangular equation
16. z
S
T
23.
9
9 sec T , cylindrical equation
r
y
3
3
14. x
y
−3
12. 2 3, 2, 6 , rectangular
r
4
0, rectangular equation
0, cylindrical equation
−2
−2
2
2
y
x
© 2010 Brooks/Cole, Cengage Learning
Section 11.7
r2 z2
5
x2 y2 z 2
5
25.
Cylindrical and Spherical Coordinates
29. 4, 0, 0 , rectangular
U
z
tan T
3
I
3
x
3
59
42 0 2 0 2
y
0 ŸT
x
4
0
S
arccos 0
2
S·
§
¨ 4, 0, ¸, spherical
2¹
©
y
−3
30. 4, 0, 0 , rectangular
26. z
r 2 cos 2 T
U
x2
tan T
z
z
1
3 2
2
y
x
0 2 02
4
0 Ÿ T
0
§z·
arccos ¨ ¸
arccos 0
©e¹
S·
§
¨ 4, 0, ¸, spherical
2¹
©
T
9
y
r
2 sin T
r2
2r sin T
27.
x2 y 2
2
x2 y 1
2
tan T
y
x
2S
3
T
0
2
I
1
2
1
arccos
1
2
U
2 3
2
42
4 2
3
S
1
2
4
tan T
y
x
2 10
S
1Ÿ T
4
2 cos T
r2
2r cos T
I
2x
2 ·
S
§
¨ 2 10, , arccos
¸, spherical
4
5¹
©
x y
2
2
r
28.
2
x y 2x
2
2
22 2 2 4 2
−2
x 1
2 3
y
2
−1
2
2
32. 2, 2, 4 2 , rectangular
−2
2
2
2S S ·
§
, ¸, spherical
¨ 4 2,
3 4¹
©
z
x
U
2y
x y 2y
2
S
31. 2, 2 3, 4 , rectangular
1 2
3 4
5 6
x
4
0
y2
1
2
arccos
5
3, 1, 2 3 , rectangular
33.
z
U
2
3 1 12
tan T
−2
2
T
y
3
y
x
S
6
x
I
4
1
3
arccos
3
2
S
6
§ S S·
¨ 4, , ¸, spherical
© 6 6¹
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
60
Vectors and the Geometry of Space
34. 1, 2, 1 , rectangular
U
1
tan T
y
x
I
§
¨
©
2
22 12
6
2 Ÿ T
arctan 2 S
§ 1 ·
arccos¨
¸
© 6¹
z
6, arctan 2 S , arccos
1 ·
¸, spherical
6¹
§ S S·
35. ¨ 4, , ¸, spherical
© 6 4¹
x
4 sin
y
z
4 sin
4 cos
6,
S
4
cos
S
4
sin
S
S
S
S
| 11.276
z
S
y
6 sin
S
z
6 cos
S
2
§ S·
12 sin 0 sin ¨ ¸
© 4¹
12 cos 0 12
0, 0, 12 , rectangular
y
9 sin S sin
z
9 cos S
S
4
S
4
9
5 2
2
cos S
6
sin S
0
0
2
6, 0, 0 , rectangular
41. y
2, rectangular equation
U sin I sin T
2
U
42. z
2 csc I csc T , spherical equation
6, rectangular equation
U cos I
U
6
6 sec I , spherical equation
43. x 2 y 2 z 2
49, rectangular equation
2
49
0
0
7, spherical equation
44. x 2 y 2 3 z 2
x2 y 2 z 2
U
0, rectangular equation
4z2
2
4 U 2 cos 2 I
1
4 cos 2 I
cos I
§ S ·
38. ¨ 9, , S ¸, spherical
© 4 ¹
9 sin S cos
U
§ S·
12 sin 0 cos¨ ¸
© 4¹
x
2
U
S ·
§
37. ¨12, , 0 ¸, spherical
4 ¹
©
y
6 sin
2
6
9
2.902, 2.902, 11.276 , rectangular
x
x
2 2
4
3S
4
§5 5 5 2 ·
¨¨ , , ¸, rectangular
2 ¸¹
©2 2
6
6
2, 2 2 , rectangular
12 cos
5 cos
5
2
5
2
S·
§
40. ¨ 6, S , ¸, spherical
2¹
©
§ 3S S ·
36. ¨12, , ¸, spherical
4 9¹
©
S
3S
x 12 sin cos
| 2.902
9
4
S
3S
y 12 sin sin
| 2.902
9
4
z
§ S 3S ·
39. ¨ 5, ,
¸, spherical
© 4 4 ¹
S
3S
x
5 sin
cos
4
4
3S
S
5 sin
sin
y
4
4
I
1
2
S
3
, cone spherical equation
0
0
0, 0, 9 , rectangular
© 2010 Brooks/Cole, Cengage Learning
Section 11.7
45. x 2 y 2
61
16, rectangular equation
U sin I sin 2 T U 2 sin 2 I cos 2 T
2
Cylindrical and Spherical Coordinates
2
U sin I sin T cos T
2
2
2
2
U 2 sin 2 I
U sin I
U
16
16
16
4
4 csc I , spherical equation
13, rectangular equation
46. x
U sin I cos T
13
U
13 csc I sec T , spherical equation
x2 y 2
47.
2 z 2 , rectangular equation
U 2 sin 2 I cos 2 T U 2 sin 2 I sin 2 T
2 U 2 cos 2 I
U 2 sin 2 I ª¬cos 2 T sin 2 T º¼
2 U 2 cos 2 I
U 2 sin 2 I
2 U 2 cos 2 T
sin I
cos 2 I
2
tan 2 I
2
2
tan I
48. x 2 y 2 z 2 9 z
0, rectangular equation
U 9 U cos I
2
S
51. I
6
0
U
49. U
r 2, spherical equation
5
x y2 z2
2
z
cos I
9 cos I , spherical equation
x y2 z2
2
z
3
2
25
z
x2 y 2 z 2
z2
x2 y2 z 2
3
4
6
5
3x 2 3 y 2 z 2
z
6 5
x
5
6
2
y
−2
−6
−1
x
3S
4
50. T
0, z t 0
2
−2
−1
1
1
−1
2
y
y
x
y
x
0
tan T
1
x y
z
3
−3
3
y
x
−3
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
62
Vectors and the Geometry of Space
S
52. I
U
56.
2
4
sin I cos T
z
cos I
U sin I cos T
x2 y 2 z 2
z
0
x
4
4
x2 y 2 z 2
z
4 csc I sec I
z
0
6
4
xy-plane
z
3
4
x
−3
−3
y
3
3
§ S ·
57. ¨ 4, , 0 ¸, cylindrical
© 4 ¹
−2
x
−3
53. U
U
4 cos I
x2 y2 z 2
x2 y 2 z 2 4 z
x y z 2
2
T
4z
x2 y2 z 2
2
I
0
2
4, z t 0
5
3
2
−2
1
2
2
U
54.
4
arccos 0
U
−3
3
y
2 sec I
U cos I
z
2
z
2
32 02
3
S
T
I
§0·
arccos¨ ¸
©9¹
4
S
2
§ S ·
59. ¨ 4, , 4 ¸, cylindrical
© 2 ¹
3
U
1
3
3
y
T
x
55.
S
S S·
§
¨ 3, , ¸, spherical
4 2¹
©
2
2
4
S
S ·
§
58. ¨ 3, , 0 ¸, cylindrical
4 ¹
©
4
3
42 02
§ S S·
¨ 4, , ¸, spherical
© 4 2¹
z
x
y
6
6
2
U
csc I
I
z
U sin I
1
2
x2 y 2
1
1
x2 y 2
1
−2
x
1
−1
1
2
4 2
S
2
§ 4 ·
arccos¨
¸
©4 2¹
S
4
S S·
§
¨ 4 2, , ¸, spherical
2 4¹
©
−2
2
42 4 2
y
−2
© 2010 Brooks/Cole, Cengage Learning
Section 11.7
§ 2S
·
, 2 ¸, cylindrical
60. ¨ 2,
© 3
¹
U
22 2
2
T
2S
3
I
§ 1 ·
arccos¨
¸
© 2¹
2 2
3S
4
S S ·
§
61. ¨ 4, , 6 ¸, cylindrical
6 6 ¹
©
42 6 2
2 13
S
I
arccos
6
3
13
I
4
2
4
2
4 2
S
3
1
arccos
2
S
4
S S·
§
¨ 4 2, , ¸, spherical
3 4¹
©
63. 12, S , 5 , cylindrical
U
122 52
T
S
I
arccos
13
5
13
§ S ·
64. ¨ 4, , 3¸, cylindrical
© 2 ¹
T
42 32
5
S
2
3
5
3·
§ S
¨ 5, , arccos ¸, spherical
5¹
© 2
I
arccos
2
10
S
6
S
10 cos
0
2
§ S ·
¨10, , 0 ¸, cylindrical
6 ¹
©
§ S S·
66. ¨ 4, , ¸, spherical
© 18 2 ¹
r
4 sin
S
2
4
S
18
4 cos
S
2
0
§ S ·
¨ 4, , 0 ¸, cylindrical
© 18 ¹
r
U sin I
T
S
z
U cos I
S
36 sin
2
36 cos
S
2
36
0
36, S , 0 , cylindrical
§ S S·
68. ¨18, , ¸, spherical
3 3¹
©
r
T
z
5·
§
¨13, S , arccos ¸, spherical
13 ¹
©
U
S
10 sin
S·
§
67. ¨ 36, S , ¸, spherical
2¹
©
S ·
§
62. ¨ 4, , 4 ¸, cylindrical
3 ¹
©
T
z
z
3 ·
S
§
¨ 2 13, , arccos
¸, spherical
6
13 ¹
©
U
r
T
T
63
§ S S·
65. ¨10, , ¸, spherical
6 2¹
©
T
2S 3S ·
§
,
¨ 2 2,
¸, spherical
3 4 ¹
©
U
Cylindrical and Spherical Coordinates
U sin I
18 sin
S
3
9
S
3
U cos I
18 cos
S
3
9 3
§ S
·
¨ 9, , 9 3 ¸, cylindrical
© 3
¹
S S·
§
69. ¨ 6, , ¸, spherical
6 3¹
©
r
6 sin
T
z
6 cos
S
3
3 3
S
6
S
3
3
S ·
§
¨ 3 3, , 3¸, cylindrical
6 ¹
©
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
64
Vectors and the Geometry of Space
5S ·
§
70. ¨ 5, , S ¸, spherical
6
©
¹
5 sin S
0
r
T
z
5S
6
5 cos S
§ S 3S ·
72. ¨ 7, ,
¸, spherical
© 4 4 ¹
r
T
5
5S
§
·
¨ 0, , 5¸, cylindrical
6
©
¹
z
8 sin
T
7S
6
z
8 cos
S
6
S
6
3S
4
7 2
2
3S
4
S
4
7 cos
7 2
2
§7 2 S 7 2 ·
, ,
¨¨
¸, cylindrical
2 ¸¹
© 2 4
§ 7S S ·
71. ¨ 8,
, ¸, spherical
© 6 6¹
r
7 sin
4
8 3
2
§ 7S
·
, 4 3 ¸, cylindrical
¨ 4,
© 6
¹
Rectangular
Cylindrical
Spherical
73. 4, 6, 3
7.211, 0.983, 3
7.810, 0.983, 1.177
74. 6, 2, 3
6.325, 0.322, 3
7.000, 0.322, 2.014
75. 4.698, 1.710, 8
§ S ·
¨ 5, , 8¸
© 9 ¹
9.434, 0.349, 0.559
76. 7.317, 6.816, 6
10, 0.75, 6
11.662, 0.750, 1.030
77. 7.071, 12.247, 14.142
14.142, 2.094, 14.142
2S S ·
§
, ¸
¨ 20,
3 4¹
©
78. 6.115, 1.561, 4.052
6.311, 0.25, 5.052
7.5, 0.25, 1
79. 3, 2, 2
3.606, 0.588, 2
4.123, 0.588, 1.064
80. 3 2, 3 2, 3
6, 0.785, 3
6.708, 0.785, 2.034
§5 4 3·
81. ¨ , , ¸
©2 3 2¹
2.833, 0.490, 1.5
3.206, 0.490, 2.058
82. 0, 5, 4
5, 1.571, 4
6.403, 1.571, 0.896
83. 3.536, 3.536, 5
§ 3S
·
¨ 5, , 5 ¸
© 4
¹
7.071, 2.356, 2.356
84. 1.732, 1, 3
11S ·
§
, 3¸
¨ 2,
6
©
¹
3.606, 2.618, 0.588
ª
§ 5S ·º
«Note: use the cylindrical coordinate ¨ 2, 6 , 3¸»
©
¹¼
¬
© 2010 Brooks/Cole, Cengage Learning
Section 11.7
Rectangular
Cylindrical
85. 2.804, 2.095, 6
3.5, 2.5, 6
Cylindrical and Spherical Coordinates
65
Spherical
6.946, 5.642, 0.528
ª¬Note: Use the cylindrical coordinates 3.5, 5.642, 6 º¼
86. 2.207, 7.949, 4
8.25, 1.3, 4
87. 1.837, 1.837, 1.5
2.598, 2.356, 1.5
§ 3S S ·
¨ 3, , ¸
© 4 3¹
88. 0, 0, 8
0, 0.524, 8
S ·
§
¨ 8, , S ¸
6 ¹
©
89. r
9.169, 1.3, 2.022
97. Rectangular to spherical: U 2
5
Cylinder
S
4
Plane
Sphere
U sin I cos T
y
U sin I sin T
z
U cos I
a
Cylinder with z-axis symmetry
T
b
Plane perpendicular to xy-plane
S
z
c
Plane parallel to xy-plane
4
(b) U
a
Sphere
T
b
Vertical half-plane
Matches graph (a)
I
c
Half-cone
z, x 2 y 2
z
99. x 2 y 2 z 2
Paraboloid
(a) r z
2
Matches graph (f )
94. U
4 sec I , z
U cos I
(b) U 2
4
2
Matches graph (b)
95. Rectangular to cylindrical: r
x y
2
2
y
x
tan T
z
2
z
Cylindrical to rectangular: x
r cos T
y
r sin T
z
z
c is a half-plane because of the restriction r t 0.
25
25 Ÿ U
5
z2
z 2 Ÿ 2r
(a) 4r 2
z
(b) 4 U sin I cos T U 2 sin 2 I sin 2 T
2
2
cos 2 I ,
tan 2 I
1
,
4
1
Ÿ I
2
tan I
U 2 cos 2 I
2
4 sin 2 I
101. x 2 y 2 z 2 2 z
(a) r z 2 z
2
2
25
100. 4 x 2 y 2
Plane
96. T
2
98. (a) r
Cone
93. r 2
·
¸
x y z ¸¹
z
2
Spherical to rectangular: x
5
Matches graph (c)
92. I
§
arccos¨
¨
©
I
Matches graph (e)
91. U
y
x
tan T
Matches graph (d)
90. T
x2 y 2 z 2
arctan
1
2
0
0 Ÿ r2 z 1
2
(b) U 2 U cos I
2
U U 2 cos I
U
2
1
0
0
2 cos I
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
66
102. x 2 y 2
(a) r 2
Vectors and the Geometry of Space
z
z
(b) U 2 sin 2 I
U cos I
U sin I
S
0 d z d r cos T
cos I
2
S
dT d
2
2
0 d r d 3
108. z
U
cos I
sin 2 I
U
csc I cot I
4
3
−4
103. x y
2
(a) r
2
4y
4r sin T , r
2
4 sin T
U sin 2 I
4 U sin I sin T
2
(b)
104. x 2 y 2
(a) r
0 d r d a
U
4 sin T
sin I
r d z d a
U
4 sin T csc I
z
−a
6
36
U 2 sin 2 I
36
2
2
U
105. x y
2
9
2
2
110. 0 d T d 2S
2 d r d 4
9
4
3
9
cos 2 T sin 2 T
(b) U 2 sin 2 I cos 2 T U 2 sin 2 I sin 2 T
−5
U 2 sin 2 I
9
cos 2 T sin 2 T
U2
9 csc 2 I
cos 2 T sin 2 T
x
5
111. 0 d T d 2S
S
0 d I d
6
0 d U d a sec I
z
(a) r sin T
4 Ÿ r
(b) U sin I sin T
107. 0 d T d
y
5
9
4
U
y
z
2
r2
106. y
a
z 2 d r 2 6r 8
(a) r cos T r sin T
2
6 csc I
−a
a
x
(b) U sin I cos T U 2 sin 2 I sin 2 T
2
2
a
36
36 Ÿ r
2
109. 0 d T d 2S
0
U sin I U sin I 4 sin T
y
4
4
x
4 csc T
a
30°
4,
4 csc I csc T
x
y
S
112. 0 d T d 2S
2
0 d r d 2
S
S
d I d
4
2
0 d U d1
0 d z d 4
z
5
z
2
3
2
1
2
3
x
2
3
y
−2
−2
2
2
y
x
© 2010 Brooks/Cole, Cengage Learning
Section 11.7
S
113. 0 d T d
67
117. Spherical
4 d U d 6
2
S
0 d I d
Cylindrical and Spherical Coordinates
z
2
0 d U d 2
8
z
−8
8
2
y
x
−8
y
2
2
118. Cylindrical
d r d 3
1
2
x
0 d T d 2S
114. 0 d T d S
S
0 d I d
9 r2 d z d
9 r2
z
2
1d U d 3
4
z
−4
3
4
2
−3
y
x
y
−2
4
−4
−1
1
2
3
119. Cylindrical coordinates:
x
r 2 z 2 d 9,
r d 3 cos T , 0 d T d S
115. Rectangular
0 d x d 10
z
0 d y d 10
3
0 d z d 10
−2
−3
z
2
10
4
3
3
y
x
10
10
y
x
120. Spherical coordinates:
U t 2
116. Cylindrical:
0.75 d r d 1.25
U d 3
S
0 d I d
4
0 d z d 8
z
z
3
8
2
−2
x
−2
2
2
−2
−1
2
y
1
1
2
y
x
121. False. r
z Ÿ x2 y 2
z 2 is a cone.
122. True. They both represent spheres of radius 2 centered at
the origin.
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
68
Vectors and the Geometry of Space
123. False. r , T , z
0, 0, 1 and r , T , z
represent the same point x, y, z
0, S , 1
0, 0, 1 .
sin T , r
125. z
1
y
r
sin T
z
y
1
y
The curve of intersection is the ellipse formed by the
intersection of the plane z
y and the cylinder r 1.
124. True except for the origin .
126. U
2 sec I Ÿ U cos I
U
2 Ÿ z
2 plane
4 sphere
The intersection of the plane and the sphere is a circle.
Review Exercises for Chapter 11
1. P
1, 2 , Q
JJJK
(a) u
PQ
JJJK
v
PR
4 1, 1 2
3, 1
5 1, 4 2
4, 2
4 2
2
v
(d) 2u v
2
20
or in the fourth quadrant x ! 0, y 0 . The
z-coordinate can be any number.
2, 1 , Q
5, 1 , R
JJJK
PQ
7, 0
(a) u
JJJK
v
4, 5
PR
(b) u
9. x 3
2 5
2 3, 1 4, 2
2. P
(c)
8. Looking towards the xy-plane from the positive z-axis.
The point is either in the second quadrant x 0, y ! 0
5, 4
3i j
(b) u
(c)
4, 1 , R
10, 0
2, 4
2
y 2
2
z 6
§ 15 ·
¨ ¸
©2¹
2
§0 4· §0 6· § 4 0·
10. Center: ¨
¸, ¨
¸, ¨
¸
© 2 ¹ © 2 ¹ © 2 ¹
2
2, 3, 2
Radius:
2
20
x2
2
2
30
y3
2
24
z2
2
2
494
17
17
7i
4 2 52
v
(d) 2u v
11. x 2 4 x 4 y 2 6 y 9 z 2
41
2 7, 0 4, 5
x2
18, 5
2
y 3
3
z2
4 4 9
9
Center: 2, 3, 0
3. v
v cos T i sin T j
Radius: 3
8 cos 60q i sin 60q j
§1
3 ·
j¸¸
8¨¨ i 2
2
©
¹
z
4i 4 3 j
4. v = v cos T i + v sin T j
1
1
cos 225qi sin 225q j
2
2
2
2
i j
4
4
4
3
2
4, 4 3
6
5
4
3
4 5
6
12. x 2 10 x 25 y 2 6 y 9
2
2
,
4
4
z2 4z 4
x5
5. z
0, y
6. x
z
4, x
0, y
5: 5, 4, 0
7: 0, 7, 0
y
x
2
y3
2
z2
2
34 25 9 4
4
Center: 5, 3, 2
Radius: 2
z
7. Looking down from the positive x-axis towards the
yz-plane, the point is either in the first quadrant
y ! 0, z ! 0 or in the third quadrant y 0, z 0 .
6
4
2
The x-coordinate can be any number.
2
y
4
6
8
x
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 11
2, 1, 3 , Q
0, 5, 1 , R
JJJK
(a) u
PQ
2, 6, 2
JJJK
v
PR
3, 6, 3
z
13. (a), (d)
20. P
(2, −1, 3) 3
2
1
4
5
3
1 2
3
−2
y
5
(b) u ˜ v
x
Because u ˜ v
4 2, 4 1 , 7 3
(b) v
2, 5, 10
2i 5 j 10k
(c) v
v
2
6
5
3
5
u
3 6, 3 2, 8 0
(c) v
3i 5 j 8k
1 3, 6 4, 9 1
5 3, 3 4, 6 1
Because 2w
·
j¸
¹
i 3j
3
T
4, 2, 10
2, 1, 5
8 5, 5 4, 5 7
3, 1, 2
w
11 5, 6 4, 3 7
6, 10, 4
2
4
arcos
24. u
v, the points lie in a straight line.
16. v
u˜v
u v
cos T
3, 5, 8
2
v
(6, 2, 0)
(b) v
w
5 2
>i j@
2
y
4
x
15. v
5 2
1
2
u˜v
1
16, 12, 24
2S
2S
§
2¨ cos
i sin
3
3
©
v
1
1
0, the vectors are orthogonal.
3S
3S ·
§
i sin
j¸
5¨ cos
4
4 ¹
©
23. u
8
7
6
5
54
4u, the vectors are parallel.
Because v
z
36
1, 4, 5
4, 3, 6 , v
22. u
14. (a), (d) (−3, −5, 8)
(3, −3, 8)
9 36 9
7, 2, 3 , v
21. u
(4, 4, −7)
(2, 5, −10)
5, 5, 0
2 3 6 6 2 3
(c) v ˜ v
−8
−9
−10
69
5 2 2 1
52
6
ª 3S 2S
15q «or,
˜
4
3
¬
6i 2 j 3k , v
u˜v
u v
cos T
3
2 4
S
12
6
º
or 15q»
¼
i 5 j
6 10
49
26
4
7 26
T | 83.6q
10, 5, 15 , v
25. u
Because v and w are not parallel, the points do not lie in
a straight line.
u
2, 1, 3
5 v Ÿ u is parallel to v and in the opposite
direction.
17. Unit vector:
18. 8
6, 3, 2
49
2, 3, 5
38
u
u
8
6, 3, 2
7
19. P
5, 0, 0 , Q
4, 4, 0 , R
JJJK
PQ
1, 4, 0
(a) u
JJJK
v = PR
3, 0, 6
(b) u ˜ v
(c) v ˜ v
2
,
38
48 24 16
, ,
7
7 7
2, 0, 6
1 3 4 0 0 6
9 36
3
,
38
5
38
T
S
26. u
1, 0, 3
v
2, 2, 1
u˜v
1
u
10
v
cos T
3
3
u˜v
u v
1
3 10
T | 83.9q
45
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
70
Vectors and the Geometry of Space
j kº
ª i
«
»
« 2 4 3»
«1 2 2»
¬
¼
27. There are many correct answers.
r 6, 5, 0 .
For example: v
JJJK
F ˜ PQ
28. W
JJJK
PQ cos T
F
75 8 cos 30q
300 3 ft-lb
3, 2, 1 , v
In Exercises 29–38, u
w
vuw
33. n
2, 4, 3 ,
n
5
n
n
1
1
2i j
2i j , unit vector or
5
5
34. u u v
j kº
ªi
«
»
3
2
1»
«
«2 4 3»
¬
¼
10i 11j 8k
vuu
j kº
ªi
«
»
«2 4 3»
«3 2
1»¼
¬
10i 11j 8k
1, 2, 2 .
29. u ˜ u
3 3 2 2 1 1
14
14
u˜v
u v
30. cos T
31. proju w
2
u
11
14 29
§
arccos¨
©
T
2
So, u u v
11
·
¸ | 56.9q
14 29 ¹
36. u u v w
u˜w
3, 2, 1 ˜ 2, 1, 0
3 4 2
5
uuv
10, 11, 8
102 112 8
2
4i 4 j 4k
uu v
j kº
ªi
«
»
3
2
1»
«
«2 4 3»
¬
¼
10i 11j 8k
uuw
j kº
ª i
«
»
« 3 2 1»
«1 2 2»
¬
¼
6i 7 j 4k
uuv uuw
37. Area parallelogram
4
3, 2, 1 u 1, 2, 1
j kº
ªi
«
»
3
2
1»
«
«1 2 1»
¬
¼
15 5 5
, ,
14 7 14
32. Work
vuu
u˜ vuw
35. V
§u ˜ w·
¨
¸u
¨ u 2 ¸
©
¹
5
3, 2, 1
14
15 10 5
, ,
14 14 14
2i j
4i 4 j 4k
uu v w
(See Exercises 34, 36)
285
38. Area triangle
1
vuw
2
1
2
2
2
1
2
5
See Exercise 33
2
39. F
c cos 20q j sin 20qk
JJJK
PQ
2k
i
JJJK
PQ u F = 0
200
c
F
F
z
j
k
0
2
PQ
2c cos 20qi
2 ft
0 c cos 20q c sin 20q
JJJK
PQ u F
2c cos 20q
F
y
100
cos 20q
x
100
cos 20q j sin 20qk
cos 20q
100 1 tan 2 20q
70°
100 j tan 20qk
100 sec 20q | 106.4 lb
© 2010 Brooks/Cole, Cengage Learning
4
Review Exercises for Chapter 11
ª2 1 0º
«
»
«0 2 1»
«0 1 2»
¬
¼
40. V
u˜ v˜w
41. v
9 3, 11 0, 6 2
10
4, x y 2 z
Solving simultaneously, you have z 1. Substituting
z 1 into the second equation, you have y
x 1.
Substituting for x in this equation you obtain two points
on the line of intersection, 0, 1, 1 , 1, 0, 1 . The
2 4t
x 3
6
y
11
z 2
4
1 t , z
(a) x
t, y
(b) x
y 1, z
(c)
x 1
(b) Symmetric equations:
9
y 4
6
1
1
4
3
3 2t
i j.
z
9, 6, 2
(a) Parametric equations:
x
1 9t , y
4 6t , z
43. v
3
direction vector of the line of intersection is v
8 1, 10 4, 5 3
42. v
25
6, 11, 4
(a) Parametric equations:
x
3 6t , y 11t , z
(b) Symmetric equations:
45. 3x 3 y 7 z
71
−4
−3
2
y
−4 −3 −2
z 3
2
2
1
3
4
2
3
−2
4
x
−3
−4
j
(a) x
2 t, z
1, y
i
3
uu v
46.
(b) None
(c)
j k
2 5
3
z
21i 11j 13k
1
1 4
Direction numbers: 21, 11, 13
4
(a) x
−4
2
−4
−2
−2
y 1
11
x
21
(b)
2
4
1 11t , z
21t , y
4 13t
z 4
13
4
y
x
(c)
−4
z
4
44.
Direction numbers: 1, 1, 1
1 t, y
(a) x
(b) x 1
2 t, z
y 2
(c)
−4
2
−4
−2
3t
2
4
z 3
−2
4
y
x
−4
z
6
5
4
3
47. P
JJJK
PQ
1
3, 4, 2 , Q
JJJK
0, 8, 1 , PR
3, 4, 1 , R
1, 1, 2
>4, 5, 4@
1
2
6
5
4
3
3
4 5
6
i
JJJK JJJK
PQ u PR
y
n
x
j
k
0 8
1
27i 4 j 32k
4 5 4
27 x 3 4 y 4 32 z 2
27 x 4 y 32 z
0
33
z
−4
−2
−2
2
4
−2
2
4
y
x
−4
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
72
Vectors and the Geometry of Space
3i j k
48. n
51. Q 1, 0, 2 point
3 x 2 1 y 3 1z 1
3x y z 8
2x 3 y 6z
0
0
A point P on the plane is 3, 0, 0 .
JJJK
PQ
2, 0, 2
z
3
2, 3, 6 normal to plane
JJJK
PQ ˜ n
8
n
7
n
2
1
−2
1
x
2
−1
D
(−2, 3, 1)
3
4
−2
6
y
52. Q 3, 2, 4 point
49. The two lines are parallel as they have the same direction
numbers, 2, 1, 1. Therefore, a vector parallel to the
2i + j k. A point on the first line
plane is v
is 1, 0, 1 and a point on the second line is 1, 1, 2 . The
vector u
2i j 3k connecting these two points is
also parallel to the plane. Therefore, a normal to the
plane is
vuu
i
j
k
2
1
1
x 2y
1
z
4
−4
−2
2
2
−2
y
5 2, 1 2, 3 1
3, 3, 2 be the direction
vector for the line through the two points. Let
n
2, 1, 1 be the normal vector to the plane. Then
i
vun
5, 3, 1 .
point in the second plane, Q 0, 0, 3 .
JJJK
PQ
0, 0, 5
JJJK
PQ ˜ n
5
5
D
n
35
35
35
7
j
k
3 3
2
1, 2, 1 direction vector
u
4
x
50. Let v
D
54. Q 5, 1, 3 point
−2
4
2, 5, 1 normal to plane
JJJK
PQ ˜ n
10
30
n
3
30
n
Choose a point in the first plane P 0, 0, 2 . Choose a
0
2
A point P on the plane is 5, 0, 0 .
JJJK
PQ
2, 2, 4
n
2 i 2 j .
Equation of the plane: x 1 2 y
−4
10
53. The normal vectors to the planes are the same,
2 1 3
2i 4 j
2x 5 y z
P 1, 3, 5 point on line
JJJK
PQ
6, 2, 2
i
JJJK
PQ u u
j
k
6 2 2
1 2
JJJK
PQ u u
5, 7, 3
D
2 1 1
is the normal to the unknown plane.
5 x 5 7 y 1 3 z 3
5 x 7 y 3z 27
0
0
264
6
u
55. x 2 y 3z
2, 8, 14
1
2 11
6
Plane
Intercepts: 6, 0, 0 , 0, 3, 0 , 0, 0, 2 ,
z
z
4
3
2
(5, 1, 3)
2
2
(2, −2, 1)
(0, 0, 2)
3
y
(0, 3, 0)
6
4
x
y
(6, 0, 0)
x
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 11
60. 16 x 2 16 y 2 9 z 2
z2
56. y
Because the x-coordinate is missing, you have a
cylindrical surface with rulings parallel to the x-axis.
The generating curve is a parabola in the yz-coordinate
plane.
xy-trace: point 0, 0, 0
xz-trace: z
r
4x
3
yz-trace: z
r
4y
3
2
1
3
4
0
Cone
z
2
73
4, x 2 y 2
z
y
x
9
z
4
−3
−3
1
z
2
57. y
2 3
3
x
y
Plane with rulings parallel to the x-axis.
z
2
61.
y
2
6
x
x2
y2
z2
16
9
1
y2
x2
z2
9
16
1
Hyperboloid of two sheets
58. y
xy-trace:
cos z
y2
x2
9
16
1
xz-trace: None
Because the x-coordinate is missing, you have a
cylindrical surface with rulings parallel to the x-axis.
The generating curve is y
cos z.
yz-trace:
y2
z2
9
1
z
z
4
2
−2
5
5
y
x
−2
59.
2
2
x
y
62.
x2
y2
z2
16
9
x2
y2
z2
25
4
100
1
Hyperboloid of one sheet
1
Ellipsoid
xy-trace:
x2
y2
16
9
xz-trace:
x2
z2
16
1
yz-trace:
y2
z2
9
1
1
xy-trace:
x2
y2
25
4
1
xz-trace:
x2
z2
25 100
1
yz-trace:
y2
z2
4
100
1
z
12
z
−5
2
−4
x
4
x
5
5
y
y
−2
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
74
63. x 2 z 2
Vectors and the Geometry of Space
69. 2 2, 2 2, 2 , rectangular
4.
Cylinder of radius 2 about y-axis
z
(a) r
2 2
T
arctan 1
2
y
−2
64. y z
2
2
(b) U
16.
2 2
3S
,I
4
T
Cylinder of radius 4 about x-axis
z
−2
2
y
r x
66. x 2 2 y 2 z 2
x2 z 2
3y
3 y 2 y 2 is a generating curve. Revolve
the curve about the y-axis.
x z2
ª¬r y º¼
x2 z 2
2y
arccos
2 5,
1
,
5
2
3 3 § 3 S 3 3·
,¨
, ,
¸, cylindrical
2 ¨© 2 2 2 ¸¹
2
3
,
2
§ 3·
§3 3·
§ 3·
¨¨
¸¸ ¨ ¸ ¨¨
¸¸
4
4
© ¹
©
¹
© 2 ¹
(b) U
arccos
2
3 § 30 S
,¨
, , arccos
10 ¨© 2 3
30
,T
2
S
3
2
§1 2x ·
¨
¸ , Cone
© 3 ¹
U
1002 502
50 5
S
T
I
§ 50 ·
arccos¨
¸
© 50 5 ¹
6
§ 1 ·
arccos¨
¸ | 63.4q or 1.107
© 5¹
S
S
§
·
§
·
¨ 50 5, , 63.4q ¸, sperical or ¨ 50 5, , 1.1071¸
6
6
©
¹
©
¹
5S
§
·
72. ¨ 81, , 27 3 ¸, cylindrical
6
©
¹
U
6561 2187
T
5S
6
I
§ 27 3 ·
arccos¨¨
¸¸
© 54 3 ¹
54 3
arccos
1
2
,
3 ·
¸, spherical
10 ¸¹
S
§
·
71. ¨100, , 50 ¸, cylindrical
6
©
¹
2y
1 revolved about the x-axis
ª¬r x º¼
S
3
,
2
z
I
1 2x
3
y2 z2
2
arctan 3
r 2y
2
z
2
2 5
2
2 y revolved about y-axis
68. 2 x 3z
arccos
2
T
3y 2 y2
3 y 2 y 2 (Trace in xy-plane)
Then x
z
2
2 2
2
x-axis.
67. z 2
2
2
§ 3·
§ 3·
¨¨
¸¸ ¨ ¸
© 4¹
© 4 ¹
x and revolve the curve about the
2
(a) r
Let x 2
2,
§ 3 3 3 3·
, ,
70. ¨¨
¸¸, rectangular
© 4 4 2 ¹
−2
65. Let y
3S
,z
4
4,
§
3S
5·
¨¨ 2 5, 4 , arccos 5 ¸¸, spherical
©
¹
2
x
2
2 2
§ 3S ·
¨ 4, , 2 ¸, cylindrical
© 4 ¹
2
x
2
S
3
5S S ·
§
¨ 54 3, , ¸, spherical
6 3¹
©
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 11
S 3S ·
§
73. ¨ 25, ,
¸, spherical
4 4 ¹
©
2
§
§ 3S · ·
¨ 25 sin ¨ ¸ ¸ Ÿ r
© 4 ¹¹
©
r2
78. z
4, cylindrical equation
z
4, rectangular equation
25 2
2
z
S
4
T
z
U cos I
4
25 cos
3S
4
25 2
2
−2
T
4
2
§ 2S ·
12 cos¨ ¸
© 3 ¹
75. x y
S
1
4
y
x
1
y
x, x t 0, rectangular coordinates, half-plane
6
z
3
4
2
tan
6 3
S
§
·
¨ 6 3, , 6 ¸, cylindrical
2
©
¹
2
, spherical coordinates
tan T
S
U cos I
z
S
79. T
2
r
y
x
S 2S ·
§
74. ¨12, ,
¸, spherical
2 3 ¹
©
§
§ 2S · ·
¨12 sin ¨ ¸ ¸ Ÿ r
© 3 ¹¹
©
2
2
§ 25 2 S 25 2 ·
, ,
¨¨
¸, cylindrical
4
2 ¸¹
© 2
2
75
2z
3
2
1
3
x
4
y
−3
(a) Cylindrical:
r 2 cos 2 T r 2 sin 2 T
2 z Ÿ r 2 cos 2T
2z
(b) Spherical:
U 2 sin 2 I cos 2 T U 2 sin 2 I sin 2 T
U sin 2 I cos 2 T 2 cos I
U
2
2
16
(a) Cylindrical: r 2 z 2
(b) Spherical: U
80. U
3 cos T , spherical coordinates
16
3·
§
x2 y 2 ¨ z ¸
2¹
©
4
x2 y2
0
2
2
§ 3·
¨ ¸ , rectangular coordinates, sphere
© 2¹
z
4
3
5r cos T
r2
x y2 z2
2
x 2 y 2 z 2 3z
5 cos T , cylindrical equation
77. r
3z
x2 y 2 z 2
0
2 sec 2T cos I csc2 I
76. x y z
2
2 U cos I
−2
5x
1
2
25
x2 5x y2
4
2
25
4
x
1
2
y
2
5·
§
2
¨x ¸ y
2¹
©
§5·
¨ ¸ , rectangular equation
© 2¹
z
3
x
3
2
y
© 2010 Brooks/Cole, Cengage Learning
Chapter 11
76
Vectors and the Geometry of Space
Problem Solving for Chapter 11
abc
1.
3. Label the figure as indicated.
0
bu abc
0
bua buc
0
aub
buc
buc
b c sin A
aub
a b sin C
From the figure, you see that
JJK
JJJK
JJK
1
1
SP
a b
RQ and SR
2
2
JJK JJJK
JJK JJJK
Because SP RQ and SR PQ,
1
1
a b
2
2
JJJK
PQ.
PSRQ is a parallelogram.
Q
Then,
buc
a b c
sin A
a
a
aub
a b c
1
2
sin C
.
c
R
a− 1b S
1
2
b
2
a+ 1b
2
4. Label the figure as indicated.
JJJK
PR
ab
JJJK
SQ
b a
a
c
P
ab ˜ ba
b
2
a
2
0, because
b
sin A
a
The other case,
x
³0
2. f x
(a)
b in a rhombus.
a
sin B
is similar.
b
S
t 4 1 dt
R
a
y
P
Q
b
4
2
−4
5. (a) u
x
−2
2
−2
−4
(b) f c x
fc 0
T
u
(c) r
0, 1, 1 is the direction vector of the line
determined by P1 and P2 .
JJJJK
P 1Q u u
D
u
4
2, 0, 1 u 0, 1, 1
x4 1
tan T
1
2
1, 2, 2
S
2
4
1
i j
2
3 2
2
z
2
2
,
2
2
6
5
4
3
2
2
,
2
2
(d) The line is y
3
2
P1
P2
1
x: x
t, y
t.
4
3
2
Q
2 3
4
y
x
(b) The shortest distance to the line segment
is P1Q
2, 0, 1
5.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 11
JJJK
JJJK
6. n PP 0 A n PP 0
9. (a) U
77
2 sin I
Torus
Figure is a square.
JJJG
So, PP 0
n and the points P form a circle of radius
z
2
n in the plane with center at P0 .
−3
n − PP0
n + PP0
n
3
n
x
P0
S³
7. (a) V
1
ª z2 º
«S »
¬ 2¼0
z dz
0
1
Note: (base)(altitude)
2
x
2
ca
At z
1
y
2
cb
2
10. (a) r
S abc.
cb
ªS abc º
«
»
¬ 2 ¼0
S abk
2
2
z2
r
0
x2
y2
d2
2 2
2
a
b
c
x2
y2
2
2
a
b
x2
a2 c2 d 2
c2
2³
c
0
c
2
c2
·§ b 2 c 2 d 2
¸¨
¸¨
c2
¹©
c
S ab
c2
2t 1; Q
4, 3, s
3, 1, 1 point on line
1, 2, s 1
i
j
1
2 s 1
2 1
k
4
7 s i 6 2 s j 5k
c2 d 2
D
2
d3º
2S ab ª 2
c d »
2 «
c ¬
3 ¼0
·
¸
¸
¹
1
t 1, z
2
2, 1, 4 direction vector for line
JJJK
PQ u u
1.
ª¬ u u v ˜ zº¼ w ª¬ u u v ˜ wº¼ z.
t 3, y
P
JJJK
PQ
c2 d 2
c2
c d dd
2
uuv u wuz
(a) u
y2
b2 c2 d 2
§ a2 c2 d 2
¨
¨
c2
©
S ab
4 3
Sr
3
12. x
d2
c2
x2 y 2
11. From Exercise 66, Section 11.4,
1
1
r 2 cos 2 T
Hyperbolic paraboloid
d ! 0,
(b) At height z
V
ª
x3 º
2S «r 2 x »
3 ¼0
¬
r
S
(b) z
1
(area of base)(height)
2
2 ³ S r 2 x 2 dx
2 cos T
Cylinder
2 k
1
S abk k
2
8. (a) V
2
−2
x
1
k
(c) V
1
3
³ 0 S abc ˜ dc
V
Area
−2
c
2
ca
y
1
S
2
c, figure is ellipse of area
S
3
2 cos I
Sphere
z
1
S1
2
z: slice at z
2
(b) U
1
S
2
−3
x2
y2
2
2
a
b
(b)
y
P
2
1
3
−2
JJJK
PQ u u
u
7 s
2
6 2 s
2
25
21
4
S abc
3
© 2010 Brooks/Cole, Cengage Learning
78
Chapter 11
Vectors and the Geometry of Space
(b)
10
−11
10
−4
The minimum is D | 2.2361 at s
1.
(c) Yes, there are slant asymptotes. Using s
1
21
D(s)
u cos 0 i sin 0 j
13. (a) u
5
21
5 x 2 10 x 110
x 2 2 x 22
5
21
x1
2
21 o r
5
x1
21
105
s 1 slant asymptotes.
21
r
y
x, you have
u i
j
Downward force w
T cos 90q T i sin 90q T j
T
T sin T i cos T j
0
u w T
u
sin T T
u i j T sin T i cos T j
cos T T
1
If T
30q, u
(b) From part (a), u
1 2 T and 1
tan T and T
3 2 T Ÿ T
2
| 1.1547 lb and u
3
1§ 2 ·
| 0.5774 lb
2 ¨© 3 ¸¹
sec T .
Domain: 0 d T d 90q
(c)
(d)
T
0q
10q
20q
30q
40q
50q
60q
T
1
1.0154
1.0642
1.1547
1.3054
1.5557
2
u
0
0.1763
0.3640
0.5774
0.8391
1.1918
1.7321
2.5
T
⎜⎜u ⎜⎜
0
60
0
(e) Both are increasing functions.
(f)
lim T
T o S 2
f and lim
T o S 2
u
f.
Yes. As T increases, both T and u increase.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 11
79
14. (a) The tension T is the same in each tow line.
T cos 20q cos 20 i T sin 20q sin 20q j
6000i
2T cos 20qi
6000
| 3192.5 lb
2 cos 20q
Ÿ T
2T cos T
(b) As in part (a), 6000i
3000
cos T
Ÿ T
Domain: 0 T 90q
(c)
(d)
T
10q
20q
30q
40q
50q
60q
T
3046.3
3192.5
3464.1
3916.2
4667.2
6000.0
10,000
0
90
0
(e) As T increases, there is less force applied in the direction of motion.
15. Let T
D E , the angle between u and v. Then
uuv
u v
sin D E
vuu
.
u v
For u
cos D , sin D , 0 and v
i
j
k
vuu
cos E
sin E
0
cos D
sin D
0
So, sin D E
vuu
cos E , sin E , 0 , u
v
1 and
sin D cos E cos D sin E k.
sin D cos E cos D sin E .
16. (a) Los Angeles: 4000, 118.24q, 55.95q
Rio de Janeiro: 4000, 43.23q, 112.90q
(b) Los Angeles: x
4000 sin 55.95q cos 118.24q
Rio de Janeiro: x
4000 sin 112.90q cos 43.23q
y
4000 sin 55.95q sin 118.24q
y
4000 sin 112.90q sin 43.23q
z
4000 cos 55.95q
z
4000 cos 112.90q
x, y, z | 1568.2, 2919.7, 2239.7
(c) cos T
u˜v
u v
x, y, z | 2684.7, 2523.8, 1556.5
1568.2 2684.7 2919.7 2523.8 2239.7 1556.5
4000 4000
| 0.02047
T | 91.17q or 1.59 radians
(d) s
rT
4000 1.59 | 6360 miles
© 2010 Brooks/Cole, Cengage Learning
80
Chapter 11
Vectors and the Geometry of Space
(e) For Boston and Honolulu:
a. Boston: 4000, 71.06q, 47.64q
Honolulu: 4000, 157.86q, 68.69q
b. Boston: x
4000 sin 47.64q cos 71.06q
Honolulu: x
4000 sin 68.69q cos 157.86q
y
4000 sin 47.64q sin 71.06q
y
4000 sin 68.69q sin 157.86q
z
4000 cos 47.64q
z
4000 cos 68.69q
959.4, 2795.7, 2695.1
959.4 3451.7 2795.7 1404.4 2695.1 1453.7
| 0.28329
4000 4000
u˜v
u v
c. cos T
3451.7, 1404.4, 1453.7
T | 73.54q or 1.28 radians
d. s
rT
| 5120 miles
17. From Theorem 11.13 and Theorem 11.7 (6) you have
JJJK
PQ ˜ n
D
n
w˜ uuv
uuv ˜w
u˜ vuw
uuv
uuv
uuv
z
D
a b c
2
2
d1 d 2
a b c
2
2
d1 d 2
2
a b c
2
2
2
.
plane
Introduce a coordinate system in the plane z
The new v-axis is the line z
2 y.
2 y, x
0.
Then the intersection of the cylinder and plane satisfies
the equation of an ellipse:
x2 y2
§ z·
x2 ¨ ¸
© 2¹
2
cylinder
2y
between this point and the second plane is
a d1 a b 0 c 0 d 2
1
The new u-axis is the original x-axis.
.
18. Assume one of a, b, c, is not zero, say a. Choose a point
in the first plane such as d1 a, 0, 0 . The distance
19. x 2 y 2
z2
4
x2 1
2
1
1
ellipse
z
3
2
(0, 1, 2)
y
2
−2
2
x
(0, −1, −2)
20. Essay.
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 2
Vector-Valued Functions
Section 12.1
Vector-Valued Functions .....................................................................82
Section 12.2
Differentiation and Integration of Vector-Valued Functions .............92
Section 12.3
Velocity and Acceleration..................................................................102
Section 12.4
Tangent Vectors and Normal Vectors ...............................................113
Section 12.5
Arc Length and Curvature..................................................................131
Review Exercises ........................................................................................................153
Problem Solving .........................................................................................................161
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 2
Vector-Valued Functions
Section 12.1 Vector-Valued Functions
1
t
i j 3t k
t 1
2
1. r t
3. r t
1
t 1
t
2
3t
Component functions: f t
g t
ht
ln ti et j tk
Component functions: f t
ln t
gt
e t
ht
t
Domain: 0, f
4. r t
Domain: f, 1 ‰ 1, f
sin ti 4 cos tj tk
Component functions: f t
4 t 2 i t 2 j 6tk
2. r t
4 t2
Component functions: f t
2
sin t
gt
4 cos t
ht
t
Domain: f, f
gt
t
ht
6t
Domain: > 2, 2@
Ft Gt
5. r t
cos ti sin tj t k cos ti sin tj
2 cos ti tk
Domain: >0, f
6. r t
Ft Gt
ln ti 5tj 3t 2k i 4tj 3t 2k
ln t 1 i 5t 4t j 3t 2 3t 2 k
ln t 1 i tj
Domain: 0, f
i
7. r t
Ft uGt
j
k
sin t cos t
cos 2 ti sin t cos tj sin 2 tk
0
sin t
cos t
i
j
k
t3
t
t
0
Domain: f, f
8. r t
Ft uGt
3
1
t 2
t 1
t
§ t3
·
t ·
§
3
3
t 3 t ¸k
¨ t t 2 ¸i t t 2 t t j ¨
t
t
1
1
©
¹
©
¹
Domain: f, 1 , 1, f
9. r t
1 t 2i
2
t 1j
(a) r 1
1i
2
(b) r 0
j
(c) r s 1
1
2
2
s 1 i s 11j
(d) r 2 't r 2
1
2
2
s 1 i sj
2
2 't i 2 't 1 j 2i j
2't 82
1
2
1
2
't
2
i 't j
1
't
2
2 2' t 1
2
't
2
i 1 ' t j 2i j
't 4 i ' t j
© 2010 Brooks/Cole, Cengage Learning
Section 12.1
10. r t
Vector-Valued Functions
83
cos ti 2 sin tj
(a) r 0
i
§S ·
(b) r¨ ¸
©4¹
2
i 2
(c) r T S
2j
cos T S i 2 sin T S j
§S
·
§S ·
(d) r¨ 't ¸ r¨ ¸
6
©
¹
©6¹
cos T i 2 sin T j
§ §S ·
S ·
§S
·
§S
·
cos¨ 't ¸i 2 sin¨ 't ¸ j ¨ cos¨ ¸i 2 sin j¸
6
6
6
6 ¹
©
¹
©
¹
© © ¹
1
ln ti j 3tk
t
1
(a) r 2
ln 2i j 6k
2
11. r t
(b) r 3 is not defined. ln 3 does not exist.
(c) r t 4
1
j3t 4k
t 4
1
ln 1 't i j 3 1 't k 0i j 3k
1 't
ln t 4 i (d) r 1 't r 1
t i t 3 2 j e t 4k
12. r t
(a) r 0
k
(b) r 4
2i 8 j e 1k
(c) r c 2
c 2i c 2
(d) r 9 't r 9
32
je
ª¬ c 2 4º¼
k
9 't i 9 't
9 't 3 i 32
j e >9 ' t 4@k 3i 27 j e 9 4 k
9 't
32
rt
t
2
3t
2
4t
t 9t 2 16t 2
27 j e
ª¬ 9 ' t 4º¼
2
rt
t 1 25t
sin 3ti cos 3tj tk
rt
sin 3t
2
cos 3t
x
t
2
1t
2
3ti tj 2tk , 0 d t d 1
3t , y
t, z
2t , 0 d t d 1, Parametric equation
Answers may vary
16. P 0, 2, 1 , Q 4, 7, 2
JJJG
v
PQ
4, 5, 3
rt
x
2 5t , z
3 12t ,
Answers may vary
18. P 1, 6, 8 , Q 3, 2, 5
JJJG
v
PQ
4, 4, 3
rt
1 4t i 6 4t j 8 3t k , 0 d t d 1
x 1 4t , y
6 4t , z
8 3t ,
0 d t d 1, Parametric equation
(Answers may vary)
19. r t ˜ u t
4ti 2 5t j 1 3t k , 0 d t d 1
4t , y
5 t, z
0 d t d 1, Parametric equation
2
15. P 0, 0, 0 , Q 3, 1, 2
JJJG
v
PQ
3, 1, 2
rt
2 t i 5 t j 3 12t k , 0 d t d 1
2 t , y
x
rt
14.
e 9 4 k
17. P 2, 5, 3 , Q 1, 4, 9
JJJG
v
PQ
1, 1, 12
t i 3tj 4tk
rt
13.
§ 't ·
ln 1 't i ¨
¸ j 3't k
© 1 't ¹
1 3t ,
0 d t d 1, Parametric equation
3t 1 t 2 3t 3 t 2 2t 3 4t 3
1 t3
4
8 4 t 3
5t 3 t 2 , a scalar.
No, the dot product is a scalar-valued function.
Answers may vary
© 2010 Brooks/Cole, Cengage Learning
84
Chapter 12
20. r t ˜ u t
Vector-Valued Functions
3 cos t 4 sin t 2 sin t 6 cos t t 2 t 2
t 3 2t 2 , a scalar.
No, the dot product is a scalar-valued function.
ti 2tj t 2k , 2 d t d 2
21. r t
x
t, y
x 2 . Matches (b)
So, z
cos S t , y
So, x 2 y 2
sin S t , z
t
x 2 . Matches (d)
2t
k , 0.1 d t d 5
3
2t
ln t , z
3
ti ln tj 24. r t
2
x
1. Matches (c)
e0.75t
t2, z
t, y
So, y
cos S t i sin S t j t 2k , 1 d t d 1
22. r t
x
x
t2
2t , z
ti t 2 j e 0.75t k , 2 d t d 2
23. r t
t, y
2x
3
So, z
and y
ln x. Matches (a)
25. (a) View from the negative x-axis: 20, 0, 0
(b) View from above the first octant: 10, 20, 10
(c) View from the z-axis: 0, 0, 20
(d) View from the positive x-axis: 20, 0, 0
ti tj 2k
26. r t
x
t, y
(a)
0, 0, 20
2 Ÿ x
t, z
y
(b) 10, 0, 0
(c)
5, 5, 5
z
z
y
1
2
3
3
−3
2
1
−3
−2
1
−2
−1
−1
2
2
1
2
3
3
y
1
x
y
t
Ÿ t
4
t 1
y
4x 1
27. x
2
4x
29. x
t3, y
y
x2 3
t2
y
7
6
5
4
3
2
y
4
3
2
1
1
2
3
4
5t Ÿ t
y
t
y
5 x
Domain: t t 0
1 2 3 4 5
−2
−3
−2
28. x
x
− 5 − 4 − 3 − 2 −1
x
−4 − 3 − 2 − 1
y
−3
x
3
3
−2
t 2 t, y
30. x
5 x
t2 t
y
y
9
8
7
6
5
4
3
2
1
5
4
3
2
1
x
− 5 −4 − 3 − 2 − 1
1 2 3 4 5
x
−1
−1
1
2
3
4
5
© 2010 Brooks/Cole, Cengage Learning
Section 12.1
cos T , y
31. x
37. x
y
2 cos t , y
2 sin t , z
1, Ellipse
x2
y2
4
4
z
t
2
1
x
−3 −2
z
t
2
3
1
Circular helix
−3
3
x
32.
x y
2
x
2 cos t
y
2 sin t
2
4, circle
38. x
y
t, y
y2 z2
1
3 cos t , z
3 sin t
3
3 sin t
3 cos t
2 tan T
y
9
y
4
8
x
12
y2
1, Hyperbola
4
9
4
4
x2
9
3
z
1
−1
3 sec T , y
3
Circular helix along cylinder y 2 z 2
x
−1
33. x
85
7
y2
9
x2 3 sin T
Vector-Valued Functions
9
6
3
39. x
x
−12 −9 −6
6
−3
2 sin t , y
et
2 cos t , z
9 12
x2 y2
−6
−9
z
−12
e
4
t
z
2 cos3 t , y
34. x
§ x·
¨ ¸
©2¹
23
2 sin 3 t
§ y·
¨ ¸
© 2¹
23
y
23
6
cos 2 t sin 2 t
1
x
23
−3
22 3
y
3
3
x
y
3
40. x
2
x
−3
−2
2
x
3
t2, y
2t , z
y2
,z
4
3
y
4
−2
−3
35. x
t 1
y
4t 2
z
2t 3
z
(0, 6, 5)
5
t
2
1
0
1
2
x
4
1
0
1
4
y
4
2
0
2
4
z
3
32
0
3
2
3
4
3
(2, − 2, 1)
Line passing through
the points: 0, 6, 5 , 1, 2, 3
(1, 2, 3)
z
1
3
4 5
3
y
6
3
2
x
36. x
t
y
2t 5
y
3t
−4 − 3
− 2 −1
z
( 25 , 0, 152 )
5
4
−6
2
−4
−2
5 , 0, 15
2
2
2
4
x
6
1
2
3
4
y
−2
6
(0, −5, 0)
1
−1
Line passing through
the points: 0, 5, 0 ,
3
t
2
−6
−3
x
2
4
y
−4
−6
© 2010 Brooks/Cole, Cengage Learning
y
Chapter 12
86
41. x
t2, z
t, y
2t3
3
z
1
3 2
t 2 i tj t k
2
2
43. r t
2 x3
3
x2 , z
y
Vector-Valued Functions
−3
Parabola
t
2
1
0
1
2
x
2
1
0
1
2
y
4
1
0
1
4
z
16
3
23
0
2
3
−2
−2
1
−1
2
3
−2
x
y
−3
−4
−5
16
3
3 2
1
t j t 2k
2
2
ti 44. r t
z
Parabola
z
1
−2
−1
)2, 4, )
16
3
6
1
2
3
x
2
−3
1
2
4
3
−2
y
−3
2
x
−3
5
−2
y
−4
42. x
cos t t sin t
y
sin t t cos t
z
t
Helix
z
z
2
4
−2
−1
1
2
x
x2 y2
1 t2
or x y z
2
z
§ 3
§1
1 ·
3·
sin ti ¨¨
cos t t ¸¸ j ¨¨ cos t ¸¸k
2
2
2
2
©
¹
©
¹
45. r t
)− 2, 4, − 163 )
−6
2
2
1 z2
2
3
2
3
x
1
4
2
y
3
t
y
Helix along a hyperboloid of one sheet
2 sin ti 2 cos tj 46. r t
2 sin tk
z
Ellipse
2
1
−1
−1
1
2
−1
1
y
x
−2
z
47.
z
(a)
2π
π
−2
−2
−2
−3
π
1
−2
−2
−2
−2
1
x
y
2π
4π
π
2
z
(c)
z
8π
−2
2
x
(b)
2π
2
2
2
2
x
y
The helix is translated 2
units back on the x-axis.
(d)
The height of the helix
increases at a faster rate.
(e)
z
x
y
2
y
The orientation of the
helix is reversed.
z
2
−2
π
x
2
−2
−6
π
−6
y
2π
6
x
The axis of the
helix is the x-axis.
6
y
The radius of the helix is
increased from 2 to 6.
© 2010 Brooks/Cole, Cengage Learning
Section 12.1
ti t 2 j 12 t 3h
48. r t
49. y
t 5
t , then y
ti t 5 j
rt
5
87
x 5
Let x
z
Vector-Valued Functions
4
3
50. 2 x 3 y 5
2
1
1
2
3
1 2
3 4
5
Let x
y
ti 1
3
x 2
2
rt
x
(a) u t
r t 2 j is a
1
3
t , then y
4
5
0
2t 5 .
2t 5 j
z
translation 2 units to the
left along the y-axis.
51. y
5
4
3
Let x
2
−2
1
1 2
3
2
2
ti t 2 j
rt
y
2
t 2 .
t , then y
3
4
5
52. y
x
(b) u t
t 2i tj 12 t 3k
Let x
z
4 t 2.
t , then y
ti 4 t 2 j
rt
5
has the roles of x and y
interchanged. The graph
is a reflection in the
plane x
y.
4 x2
4
3
53. x 2 y 2
2
−2
1
1 2
3
2
25
Let x
y
5 cos t , then y
5 sin t.
3
4
5 cos ti 5 sin tj
rt
5
x
(c) u t
r t 4k is
54.
z
an upward shift 4 units.
x 2
2
y2
Let x 2
5
4
2 cos t , y
2 sin t.
2 2 cos t i 2 sin tj
rt
3
4
2
1
1
1 2
3 4
5
2
3
y
55.
4
x2
y2
16
4
1
5
Let x
x
(d) u t
ti t 2 j 18 t 3k
rt
z
shrinks the z-value by a
factor of 4. The curve
rises more slowly.
5
4
56.
3
2
1
2
x2
y2
9
16
y
5
3
4
x
rt
r t reverses
z
the orientation.
5
57. r1 t
4
3
2
1
2
4
5
1 2
3
5
2
x
y
9
16
4 sin t
cos 2 t sin 2 t
1
3 cos ti 4 sin tj
ti t 2 j, 0 d t d 2 y
r2 t
2 t i 4 j, 0 d t d 2
r3 t
4 t j, 0 d t d 4
1
3
1
3 cos t and y
2
Then
5
2 tan t.
4 sec ti 2 tan tj
Let x
1
(e) u t
4 sec t , y
x2
y
(Other answers possible)
x
© 2010 Brooks/Cole, Cengage Learning
Chapter 12
88
Vector-Valued Functions
ti, 0 d t d 10 r1 0
58. r1 t
0, r1 10
10 cos ti sin tj ,
r2 t
S §
0 d t d
So, x y
2
§S ·
10i, r2 ¨ ¸
©4¹
¨ r2 0
4 ©
·
5 2i 5 2 j ¸
¹
x
0 d t d 1 r3 0
5 2i 5 2 j, r3 1
0
x
x2 y 2 , x y
Let x
t , then y
and z
x y
So, x
t, y
2
x
2
t
2
2t 2 .
ti tj 2t 2k
rt
z
2, −
(
2, 4 ) 5
(−
2, 4)
2,
6
2 sin t , z
4.
x
2
2
x2
4, z
2 cos t
4 sin t
t
0
S
S
S
6
4
2
x
0
1
y
2
z
0
rt
2
x
2
2t .
t , z
z
4 or
2 sin t , y
z
0
4
2 cos ti 2 sin tj 4k
61. x 2 y 2
Other answers possible
59. z
2
2 cos t , y
rt
5 2 1 t i 5 2 1 t j,
r3 t
x2 y2 , z
60. z
10i
3
1
3S
4
2
2
2
0
4
2
2
S
0
2
2
2
0
2 sin ti 2 cos tj 4 sin 2 tk
z
4
−3
1
2
2
y
3
3
x
−3
3
3
y
x
62. 4 x 2 4 y 2 z 2
16, x
16 4t 4 t 2 .
1
2
t 2 and y
t , then x
If z
z2
z
4
t
1.3
1.2
1
0
1
1.2
x
1.69
1.44
1
0
1
1.44
y
0.85
1.25
1.66
2
1.66
1.25
z
1.3
1.2
1
0
1
1.2
2
t 2i rt
4, x z
1 sin t , then z
1 sin t
2
y 1 sin t
2
2
2 x
2
2 cos 2 t , y
r 2 cos t
x
1 sin t , y
r 2 cos t
z
1 sin t
y2
y
2
16 4t 4 t 2 j tk
1
2
63. x 2 y 2 z 2
Let x
2
4
x
1 sin t and x 2 y 2 z 2
2 2 sin t y
2
2
4
1 sin t i 2 cos tj 1 sin t k and
rt
1 sin t i 2 cos tj 1 sin t k
S
t
x
0
1
2
y
0
r
z
2
3
2
2
S
6
0
S
S
6
2
z
3
−3
3
3
rt
4.
y
x
−3
6
2
1
3
2
r 2
r
1
1
2
2
6
2
0
0
© 2010 Brooks/Cole, Cengage Learning
y
Section 12.1
64. x 2 y 2 z 2
10, x y
Vector-Valued Functions
89
4
z
2 sin t , then y
Let x
S
S
2 sin t and z
S
S
6
2
2 1 sin 2 t
x
1
3
2
2
5
2
3
2
y
3
5
2
2
3
2
1
2
z
0
0
0
6
6
2
6
2
2
2 sin t i 2 sin t j rt
65. x 2 z 2
4, y 2 z 2
4
S
t
2
2 cos t.
y
4
4
x
2
2 cos tk
z
4
3
Subtracting, you have x 2 y 2
0 or y
So, in the first octant, if you let x
t i tj rt
66. x 2 y 2 z 2
Let x
t , then x
t, y
4 t2 .
t, z
16, xy
3
4
4 t2k
2
4
y
(2, 2, 0)
x
4 first octant
4
and x 2 y 2 z 2
t
t , then y
1
t
z
(0, 0, 2)
r x.
t2 16
z2
t2
z
16.
4
t 4 16t 2 16
84 3 d t d
8 4 3
4
y
4
x
8 4 3
t
1.5
2
2.5
3.0
3.5
8 4 3
x
1.0
1.5
2
2.5
3.0
3.5
3.9
y
3.9
2.7
2
1.6
1.3
1.1
1.0
z
0
2.6
2.8
2.7
2.3
1.6
0
ti rt
67. y 2 z 2
4
1
j
t 4 16t 2 16k
t
t
2t cos t
2
2t sin t
2
4t 2
4x2
e t cos t
68. x 2 y 2
z
7
6
5
e t sin t
2
e 2t
z2
z
16
160
12
120
8
80
4
x
2
40
4
8
40
120 80
12
16
x
y
40
80
120
y
69. lim ti cos tj sin tk
Si j
2
1 ·
§
70. lim ¨ 3ti 2
j k¸
t o2 ©
t 1
t ¹
6i t oS
2
1
j k
3
2
© 2010 Brooks/Cole, Cengage Learning
90
Chapter 12
Vector-Valued Functions
1 cos t º
ª
71. lim «t 2i 3tj k»
t o0¬
t
¼
because
1 cos t
lim
t o0
t
§
72. lim ¨
t o1 ©
ti lim
t o0
sin t
1
0. L'Hôpital's Rule
ln t
1 ·
j
k¸
t2 1
t 1 ¹
does not exist because lim
t o1
t 2 i t 3 j tk
81. r t
0
1
does not exist.
t 1
(a) s t
r t 3k
t 2i t 3 j t 3 k
(b) s t
r t 2i
t 2 2 i t 3 j tk
(c) s t
r t 5j
82. A vector-valued function r is continuous at t
a if the
limit of r t exists as t o a and lim r t
ra.
t oa
­ i j t t 2
is not continuous at
®
¯i j t 2
The function r t
sin t
ª
º
73. lim «et i j et k »
t o0 ¬
t
¼
i jk
t
because
0.
83. One possible answer is
sin t
lim
t o0
t
cos t
lim
t o0
1
1 L'Hôpital's Rule
1
t
ª
º
74. lim «e t i j 2
k
t of ¬
t
t 1 »¼
1.5 cos ti 1.5 sin tj rt
Note that r 2S
0
1
S
tk , 0 d t d 2S
1.5i 2k.
z
3
because
1
t of t
lim et
0, lim
t of
0, and lim
t of
t
2
t 1
2
0.
1
−2
−1
75. r t
t 2 i t 2 j tk
−2
−1
1
1
ti j
t
2
−1
x
2
y
Continuous on f, 0 , 0, f
84. (a) x
76. r t
ti t 1j
x 1
9
Continuous on >1, f
77. r t
ti arcsin tj t 1 k
(b) x
2e t , e t , ln t 1
Continuous on t 1 ! 0 or t ! 1: 1, f .
79. r t
e t , t 2 , tan t
Discontinuous at t
S
2
nS
8,
t,
3
y 2
25
1, z
3 cos t 1, z
2
z 2
25
2
y 2
25
2
y 2
25
4
4
5 sin t 2
2
1, x
4
5 sin t 2, z
4
2
1, z
3 cos 2t 1, y
x 1
9
5 sin t 2, z
2
3 cos t 1, y
x 1
9
(d) x
S
§ S
·
Continuous on ¨ nS , nS ¸
2
© 2
¹
80. r t
(c) x
2
4, y
y 1
9
Continuous on >1, 1@
78. r t
3 cos t 1, y
4
5 sin 2t 2, z
4
2
1, z
4
(a) and (d) represent the same graph
t
Continuous on >0, f
© 2010 Brooks/Cole, Cengage Learning
Section 12.1
x1 t i y1 t j z1 t k and u t
85. Let r t
Vector-Valued Functions
x2 t i y2 t j z2 t k. Then:
^
lim ¬ªr t u u t º¼
t oc
91
`
lim ª¬ y1 t z2 t y2 t z1 t º¼ i ª¬ x1 t z2 t x2 t z1 t º¼ j ª¬ x1 t y2 t x2 t y1 t º¼ k
t oc
ªlim y t lim z t lim y t lim z t º i ªlim x t lim z t lim x t lim z t º j
2
1
2
1
«¬t o c 1 t o c 2
»¼
«¬t o c 1 t o c 2
»¼
t oc
t oc
t oc
t oc
ª«lim x1 t lim y2 t lim x2 t lim y1 t º» k
t oc
t oc
t oc
¬t o c
¼
ªlim x t i lim y t j lim z t k º u ªlim x t i lim y t j lim z t k º
1
1
2
2
»¼ «¬t o c 2
»¼
t oc
t oc
t oc
t oc
¬«t o c 1
lim r t u lim u t
t oc
t oc
x1 t i y1 t j z1 t k and u t
86. Let r t
lim ¬ªr t ˜ u t ¼º
t oc
x2 t i y2 t j z2 t k. Then:
lim ª¬ x1 t x2 t y1 t y2 t z1 t z2 t º¼
t oc
lim x1 t lim x2 t lim y1 t lim y2 t lim z1 t lim z2 t
t oc
t oc
t oc
t oc
t oc
t oc
ªlim x t i lim y t j lim z t k º ˜ ªlim x t i lim y t j lim z t k º
1
1
2
2
»¼ ¬«t o c 2
»¼
t oc
t oc
t oc
t oc
¬«t o c 1
lim r t ˜ lim u t
t oc
t oc
x t i y t j z t k. Because r is
87. Let r t
continuous at t
c, then lim r t
t oc
rc.
are defined at c.
xt
r
2
xc
lim r
t oc
2
yt
yc
2
2
zt
zc
Equating components
t
2 s 3
t2
2
t
2
c
12s 2
2 s 3
rc
­ 1, if t t 0
®
¯1, if t 0
8s
4 s 2 12 s 9
8s
4 s 20 s 9
0
2s 9 2s 1
0
For s
1,
2
t
2
1
2
3
2.
For s
9,
2
t
2
9
2
3
6 and
f t i. Then r is not continuous at
0, whereas, r
2
2
88. Let
and r t
8s
3
So, r is continuous at c
f t
2s 3 i 8sj 12 s 2 k
us
x ci y c j z ck Ÿ x c,y c,z c
rc
ti t 2 j t 3k
90. r t
1 is continuous for all t.
t2
8
9
2
36 and t 3
12
9
2
54. Impossible.
The paths intersect at 2, 4, 8 , but at different times
t 2i 9t 20 j t 2k
89. r t
t
2 and s
3s 4 i s 2 j 5s 4 k.
us
9t 20
t
2
So, 3s 4
3s 4
s
. No collision.
91. No, not necessarily. See Exercise 90.
Equating components:
t2
1
2
92. Yes. See Exercise 89.
2
93. True
5s 4
5s 4 Ÿ s
94. False. The graph of x
y
z
t 3 represents a line.
4
9t 20
s2
16 Ÿ t
4.
The paths intersect at the same time t
16, 16, 16 . The particles collide.
95. True. See Exercises 89 and 90.
4 at the point
96. True. y 2 z 2
t 2 sin 2 t t 2 cos 2 t
t2
x
© 2010 Brooks/Cole, Cengage Learning
92
Chapter 12
Vector-Valued Functions
Section 12.2 Differentiation and Integration of Vector-Valued Functions
t 2i tj, t0
1. r t
t2, y t
xt
x
2
t
r′
(4, 2)
xt
2
x2 y2
r
4i 2 j
2
2ti j
−2
rc 2
4i j
−4
4
6
§S ·
r¨ ¸
©2¹
rc t
8
rc t0 is tangent to the curve at t 0 .
ti t 1 j, t0
2
xt
y
t, y t
§S ·
rc¨ ¸
©2¹
1
r′
1
i
rc 1
i 2j
−3 − 2 − 1
1
2
x
3
−3
2
2
r
1
1
j
2
1
2ti 2 j
t
1
4i j
4
rc 2
(4, (
1
2
x
y
2
7. r t
1 t i t 3 j, t0
(b) r 1
1
y
1t
t3
x 1
i 3t 2 j
rc 1
i 3j
2
1, ellipse
y
3i
3
2
3 cos ti 4 sin tj
1
−4
4 j
(3, 0)
r
−2 −1
1
−2
x
2
4
r′
−3
e t , e 2 t , t0
xt
et , y t
y
x2 , x ! 0
0
e 2t
et
2
y
3
r0
1, 1
2
rc t
e t , 2e 2 t
1
rc 0
1, 2
r′
(1, 1)
r
x
1
2
3
rc t0 is tangent to the curve at t 0 .
3
3
r
2
2i j
rc t
2
4 cos t
rc t0 is tangent to the curve at t 0 .
r′
rc t0 is tangent to the curve at t 0 .
4. (a) r t
S
x
4i rc t
i
3 sin t , y t
§S ·
rc¨ ¸
©2¹
y
1
y2
r2
xt
§S ·
r¨ ¸
©2¹
rc t
1
t
t2, y t
x
sin ti cos tj
2
1
t 2i j, t0
t
xt
x
1
j
§ x·
§ y·
¨ ¸ ¨ ¸
©3¹
© 4¹
−2
r c t0 is tangent to the curve at t 0 .
3. r t
r
3 sin ti 4 cos tj, t0
6. r t
(1, 0)
r
i 2tj
(0, 1)
1
3
2
rc t
r′
rc t0 is tangent to the curve at t 0 .
y
t2 1
x2 1
r1
sin t
x
rc t
2. r t
cos t , y t
y
2
4
y2
r2
S
cos ti sin tj, t0
5. r t
y
1
(2, 1)
r
x
−1
1
−1
2
3
(1, 0)
rc t0 is tangent to the curve at t 0 .
© 2010 Brooks/Cole, Cengage Learning
Section 12.2
e t , et , t0
8. r t
1
, yt
et
et
xt
0
r0
1, 1
rc t
e t , et
rc 0
1, 1
15. r t
y
et
r′
x
−1
1
2
rc t
−1
rc t
4, z
§ 3S ·
rc¨ ¸
© 2 ¹
2i k
)0, − 2, 32π )
19. r t
1
2
r2
2i 4 j x
i 4j
4
rc t
−4
3t 2 , 3 sin 3t , 3 cos 3t
21. r t
r
−4
−6
t sin t , t cos t , t
rc t
y
sin t t cos t , cos t t sin t , 1
r'
−6
22. r t
t 3i 3tj
rc t
t 3 , cos 3t , sin 3t
−2
2
3k
2
e t i 5et 5tet k
20. r t
2
−2
arcsin t , arccos t , 0
1
rc t
1 t2
3t 2i 3 j
23. r t
ti 1 t3 j
12. r t
1
i 3t 2 j
2 t
rc t
13. r t
2 cos t , 5 sin t
rc t
t sin t cos t , 2 cos t
1
1 t2
,0
1 2
t j
2
3t 2i tj
(b) rcc t
6ti j
3t 2 6t t
18t 3 t
t2 t i t2 t j
(a) rc t
2t 1 i 2t 1 j
(b) rcc t
2i 2 j
(c) rc t ˜ rcc t
25. r t
,
(a) rc t
24. r t
t cos t , 2 sin t
rc t
t 3i (c) rc t ˜ rcc t
2 sin t , 5 cos t
14. r t
t2 ·
2
¸j k
t
2 t¹
e t i 4 j 5tet k
rc t
y
z
3
2
i 2tj
rc 2
2
t 2
5t 3 2
2
i j k
2
t
t
r π
2
t j ln t 2k
§
2
i ¨ 2t
t
©
2π
x
rc t
4 ti t 2
rc t
−2
x2 , z
3a cos 2 t sin ti 3a sin 2 t cos tj
r′
3S
k
2
ti t 2 j 32 k , t0
11. r t
18. r t
z
t
2 sin ti 2 cos tj k
2 j y
3S
2
2 cos ti 2 sin tj tk , t0
§ 3S ·
r¨ ¸
© 2 ¹
10. r t
a cos3 ti a sin 3 tj k
17. r t
9. (a) and (b) r t
rc t
1
t2
i 16tj k
t
2
1
2 i 16 j tk
t
16. r t
r
rc t0 is tangent to the curve at t 0 .
x2 y 2
6i 14tj 3t 2k
(1, 1)
1
93
6ti 7t 2 j t 3k
rc t
2
1
, x ! 0
x
y
Differentiation and Integration of Vector-Valued Functions
2t 1 2 2t 1 2
8t
4 cos ti 4 sin tj
(a) rc t
4 sin ti 4 cos tj
(b) rcc t
4 cos ti 4 sin tj
(c) rc t ˜ rcc t
4 sin t 4 cos t 4 cos t 4 sin t
0
© 2010 Brooks/Cole, Cengage Learning
94
Chapter 12
26. r t
Vector-Valued Functions
8 cos ti 3 sin tj
(a) rc t
8 sin ti 3 cos tj
(b) rcc t
8 cos ti 3 sin tj
(c) rc t ˜ rcc t
8 sin t 8 cos t 3 cos t 3 sin t
1 2
1
t i tj t 3k
2
6
27. r t
(a) rc t
ti j (b) rcc t
i tk
(c) rc t ˜ rcc t
1 2
t k
2
t1 10 1 2
t t
2
t t3
2
ti 2t 3 j 3t 5 k
28. r t
(a) rc t
i 2 j 3k
(b) rcc t
0
(c) rc t ˜ rcc t
0
cos t t sin t , sin t t cos t , t
29. r t
(a) rc t
sin t sin t t cos t , cos t cos t t sin t , 1
(b) rcc t
cos t t sin t , sin t t cos t , 0
(c) rc t ˜ rcc t
t cos t , t sin t , 1
t cos t cos t t sin t t sin t sin t t cos t
t
e t , t 2 , tan t
30. r t
(a) rc t
e t , 2t , sec 2 t
(b) rcc t
e t , 2, 2 sec 2 t tan t
e 2t 4t 2 sec 4 t tan t
(c) rc t ˜ rcc t
31.
55 sin t cos t
rt
cos S t i sin S t j t 2k , t0
rc t
S sin S t i S cos S t j 2tk
2S
i 2
§ 1·
rc¨ ¸
© 4¹
§ 1·
rcc¨ ¸
© 4¹
rcc 1 4
rcc 1 4
2S
1
j k
2
2
2
rc 1 4
rc 1 4
§ 1·
rcc¨ ¸
© 4¹
1
4
2
2
§ 2S ·
§ 2S ·
§ 1·
¨¨ 2 ¸¸ ¨¨ 2 ¸¸ ¨© 2 ¸¹
©
¹
©
¹
§1·
rc¨ ¸
© 4¹
rcc t
1
2S i 4S 1
2
S2 1
4
4S 2 1
2
z
r″
⎜⎜r ″ ⎜⎜
2S j k
S 2 cos S t i S 2 sin S t j 2k
2S 2
i 2
2S 2
j 2k
2
2
§
¨¨ ©
§
2S 2 ·
¸ ¨¨
2 ¸¹
©
1
2 S4 4
x
y
2
2S 2 ·
¸ 2
2 ¸¹
2S 2i r′
⎜⎜r ′ ⎜⎜
2
S4 4
2S 2 j 4k
© 2010 Brooks/Cole, Cengage Learning
Section 12.2
32.
rc t
§1·
rc¨ ¸
© 4¹
3
1
i j e 1 4k
2
2
§1·
r c¨ ¸
© 4¹
rc 1 4
rc 1 4
9 1
e 1 2
4 4
3
10 4e 1 2
1
10 4e1 2
2
1
i 10 4e 1 2
§1·
2 j e t k , rcc¨ ¸
© 4¹
rcc t
§1·
rcc¨ ¸
© 4¹
2
4e
rc 0
0
1 2
j
e 1 4
4 e 1 2
rc t
39. r t
1
2 cos3 T i 3 sin 3 T j
6 cos 2 T sin T i 9 sin 2 T cos T j
1 cos T i sin T j
Smooth on
37. r T
rc T
1
t 1 i j t 2k
t
1
i 2 j 2tk z 0
t
r is smooth or all t z 0: f, 0 , 0, f
40. r t
et i e t j 3tk
et i e t j 3k z 0
r is smooth for all t: f, f
T sin T i 1 cos T j
rc 2 n 1 S
rc t
rc t
0
§ nS n 1 S ·
Smooth on ¨ ,
¸, n any integer.
2
© 2
¹
rc T
2.
r is not continuous when t
Smooth on f, 2 , 2, f
Smooth on f, 1 , 1, f
§ nS ·
rc¨ ¸
© 2 ¹
2t
2t 2
i j
3
8t
8 t3
16 4t 3
32t 2t 4
i
j
2
2
t3 8
t3 8
rc t z 0 for any value of t.
1
i 3tj
t 1
1
i 3j
2
t 1
Not continuous when t
36. r T
k
k
Smooth on f, 0 , 0, f
rc T
10 4e 1 2
38. r t
2ti 3t 2 j
35. r T
2e 1 4
2 j e 1 4k
t 2i t 3 j
rc t
rc t
j
4 e 1 2
§1·
rcc¨ ¸
©4¹
§1·
rcc¨ ¸
©4¹
34. r t
95
3
1
ti t 2 j e t k , t 0
2
4
3
1
3
1
§ ·
i 2tj e t k , r¨ ¸
i j e 1 4k
2
8
16
© 4¹
rt
33. r t
Differentiation and Integration of Vector-Valued Functions
0, n any integer
2n 1 S , 2 n 1 S
T 2 sin T i 1 2 cos T j
1 2 cos T i 2 sin T j
rc T z 0 for any value of T
Smooth on f, f
41. r t
rc t
ti 3tj tan tk
i 3 j sec 2 tk z 0
r is smooth for all t z
S
2
nS
2n 1
S.
2
S
§ S
·
Smooth on intervals of form ¨ nS , nS ¸,
2
2
©
¹
n is an integer.
42. r t
rc t
ti t2 1 j 1
i 2tj 1
tk
4
1
k z 0
4
2 t
r is smooth for all t ! 0: 0, f
© 2010 Brooks/Cole, Cengage Learning
96
Chapter 12
Vector-Valued Functions
ti 3tj t 2k , u t
43. r t
(a) rc t
i 3 j 2tk
(b) rcc t
2k
(c) r t ˜ u t
4ti t 2 j t 3k
4t 2 3t 3 t 5
Dt ¬ªr t ˜ u t ¼º
8t 9t 2 5t 4
(d) 3r t u t
ti 9t t 2 j 3t 2 t 3 k
Dt ª¬3r t u t º¼
(e) r t u u t
i 9 2t j 6t 3t 2 k
2t 4i t 4 4t 3 j t 3 12t 2 k
Dt ¬ªr t u u t º¼
8t 3i 12t 2 4t 3 j 3t 2 24t k
10t 2 t 4
(f ) r t
t 10 t 2
10 2t 2
Dt ª¬ r t º¼
10 t 2
ti 2 sin tj 2 cos tk
44. r t
1
i 2 sin tj 2 cos tk
t
ut
(a) rc t
i 2 cos tj 2 sin tk
(b) rcc t
2 sin tj 2 cos tk
(c) r t ˜ u t
1 4 sin 2 t 4 cos 2 t
Dt ¬ªr t ˜ u t ¼º
0, t z 0
1·
§
¨ 3t ¸i 4 sin tj 4 cos tk
t¹
©
(d) 3r t u t
1·
§
¨ 3 2 ¸i 4 cos tj 4sin tk
t ¹
©
Dt ª¬3r t u t º¼
j
kº
ªi
«
»
«t 2 sin t 2 cos t »
«1
»
« 2 sin t 2 cos t »
¬t
¼
(e) r t u u t
t2 4
1 2
t 4
2
Dt r t
45. r t
ti 2t 2 j t 3k , u t
(a) r t ˜ u t
1·
§1
·
§
2 cos t ¨ t ¸ j 2 sin t ¨ t ¸k
t¹
©t
¹
©
ª
ª
1·
1 ·º
§1
·
§ 1
·º
§
§
«2 sin t ¨ t t ¸ 2 cos t ¨ t 2 1¸» j «2 cos t ¨ t t ¸ 2 sin t ¨1 t 2 ¸» k
©
¹
©
¹
©
¹
©
¹¼
¬
¼
¬
Dt ª¬r t u t º¼
(f ) r t
5
1 2
2t
t
t 4
2
t 4k
t7
(i) Dt ¬ªr t ˜ u t ¼º
7t 6
(ii) Alternate Solution:
Dt ª¬r t ˜ u t º¼
r t ˜ uc t rc t ˜ u t
ti 2t 2 j t 3k ˜ 4t 3k i 4tj 3t 2k ˜ t 4k
4t 6 3t 6
7t 6
© 2010 Brooks/Cole, Cengage Learning
Section 12.2
(b) r t u u t
i
j
k
t
2
t3
2t
Differentiation and Integration of Vector-Valued Functions
2t 6i t 5 j
0 t4
0
(i) Dt ª¬r t u u t º¼
12t 5i 5t 4 j
r t u uc t u rc t u u t
(ii) Alternate Solution: Dt ¬ªr t u u t º¼
i
j
k
t
2
t 1 4t 3t 2
0
cos ti sin tj tk , u t
46. r t
(a) r t ˜ u t
97
2t
i
j
k
3
0 4t
3
0
0
t
12t 5i 5t 4 j
4
j tk
sin t t 2
(i) Dt ¬ªr t ˜ u t ¼º
cos t 2t
(ii) Alternate Solution:
r t ˜ uc t rc t ˜ u t
Dt ª¬r t ˜ u t º¼
cos ti sin tj tk ˜ k sin ti cos tj k ˜ j tk
i
(b) r t u u t
j
(i) Dt ª¬r t u u t º¼
2t cos t
k
cos t sin t
0
t cos t t
t sin t t i t cos t j cos tk
t
t
1
t cos t sin t 1 i cos t t sin t j sin tk
(ii) Alternate Solution:
r t u uc t rc t u u t
Dt ª¬r t u u t º¼
i
j
k
cos t sin t
0
47.
0
rt
3 sin ti 4 cos tj
rc t
3 cos ti 4 sin tj
r t ˜ rc t
ª
arccos «
«
«¬
T
j
1
t
0
1
7 sin t cos t
9 sin 2 t 16 cos 2 t
º
»
9 cos 2 t 16 sin 2 t »»
¼
7 sin t cos t
9 sin 2 t 16 cos 2 t
1.855 maximum at t
3.927
T
1.287 minimum at t
2.356
§ 3S ·
¨ ¸ and t
© 4 ¹
2
␲
9 cos 2 t 16 sin 2 t
T
S
sin t t cos t 1 i t sin t cos t j sin tk
7 sin t cos t
§ 5S ·
¨ ¸ and t
© 4 ¹
T
k
1
9 sin t cos t 16 cos t sin t
r t ˜ rc t
rt
rc t
cos T
i
t sin t cos t
1.571 for t
nS
,n
2
0.785
5.498
§S ·
−1
7
0
¨ ¸.
©4¹
§ 7S ·
¨ ¸.
© 4 ¹
0, 1, 2, 3, !
© 2010 Brooks/Cole, Cengage Learning
98
Chapter 12
Vector-Valued Functions
rt
t 2 i tj
rc t
2ti j
r t ˜ rc t
2t 3 t
48.
t 4 t 2 , rc t
rt
cos T
T
T
1.0
4t 2 1
2t 3 t
−0.5
t 4 t 2 4t 2 1
ª
º
2t 3 t
arccos «
»
4
2
2
4t 1 ¼
¬ t t
0.340 | 19.47q maximum at t
T z
S
2
49. rc t
2
ª¬3 t 't 2º¼ i ª1 t 't º j 3t 2 i 1 t 2 j
¬
¼
lim
't o 0
't
r t 't r t
lim
't o 0
't
3't i 2t 't 't
2
r t 't r t
't
3
ª
« t 't i t 't j 2 t 't
¬
lim
't o 0
't
lim
't o 0
't
t 't t
lim
lim
ª
« t 't lim «
't o 0
't
«
¬
ª
1
lim «
¬ t 't 't o 0 «
t
i
t
3
3
º
t
t
t j 2k »
'
i
»
't
»
¼
º
3
j 2k »
t 't t
»¼
1
2 t
i
3
j 2k
t2
r t 't r t
't
t 't , 0, 2 t 't
t 2 , 0, 2t
't
't o 0
't o 0
3
º ª
º
k » « t i j 2tk »
t
¼ ¬
¼
º
3't
j 2k »
»
t 't t 't
¼
i
2
r t 't r t
't
2
2t't 't , 0, 2't
lim
lim
0, sin t 't , 4 t 't
't o 0
't
0, 0 cos t , 4
³
2ti j k dt
54.
³
4t 3i 6tj 4 t k dt
55.
³ ¨© t i 56.
³ «¬ln ti 1
º
j k » dt
t
¼
lim 2t 't , 0, 2
't o 0
2t , 0, 2
0, sin t , 4t
't
lim 0,
't o 0
sin t cos 't 1
't
§ sin 't
cos t ¨
© 't
·
¸, 4
¹
0, cos t , 4
53.
·
j t 3 2k ¸ dt
¹
't
't o 0
0, sin t ˜ cos 't sin 't cos t sin t , 4't
't o 0
ª
3i 2tj
lim
lim
§1
lim 3i 2t 't j
't o 0
't o 0
ª
lim «
't o 0 « 't
¬
52. rc t
j
't
't o 0
51. rc t
§ 2·
¨¨
¸¸.
© 2 ¹
0.707
for any t.
lim
50. rc t
␲
0
t 2 i tj tk C
t 4i 3t 2 j 83 t 3 2k C
ln ti tj 2 52
t k C
5
t ln t t i ln tj tk C
(Integration by parts)
© 2010 Brooks/Cole, Cengage Learning
Section 12.2
57.
³ ª¬ 2t 1 i 4t
58.
³ ª¬e i sin tj cos tk º¼ dt
59.
³ «¬sec
2
60.
³ ª¬e
sin ti e t cos tjº¼ dt
3
j 3 t k º¼ dt
t
ti t 2 t i t 4 j 2t 3 2k C
1 º
j dt
1 t 2 »¼
tan ti arctan tj C
et
et
sin t cos t i cos t sin t j C
2
2
1
1
61.
³0
62.
³ 1 ti t
63.
³0
64.
³0
1
ªt 2 º
1
ª¬4t 2iº¼ « j» >tk @0
0
2
¬ ¼0
8ti tj k dt
1
1
S 2
S 4
3
j
3
99
et i cos tj sin tk C
t
ª
Differentiation and Integration of Vector-Valued Functions
1
jk
2
4i 1
1
ªt 2 º
ªt 4 º
ª3 4 3 º
« i» « j» « t k »
2
4
¼ 1
¬ ¼ 1 ¬ ¼ 1 ¬ 4
t k dt
S 2
>a sin ti@0
ª¬ a cos t i a sin t j k º¼ dt
0
S 2
>a cos tj@0
ª¬ sec t tan t i tan t j 2 sin t cos t k º¼ dt
S 2
>tk @0
ai aj S
2
S 4
ª¬sec ti ln sec t j sin 2 tk º¼
0
k
2 1 i ln
2 j 12 k
2
65.
66.
2
³0
2
ªt 2 º
t 2
t
« i» ª¬e jº¼ 0 ª¬ t 1 e k º¼ 0
2
¬ ¼0
ti et j tet k dt
ti t 2 j
3
³0
67. r t
r0
rt
t2 t4
3
³0 t
ti t 2 j dt
³
t 1 t 2 for t t 0
1 t 2 dt
4e 2t i 3et j dt
2i 3 j C
2i e 2 1 j e 2 1 k
ª1 1 t 2
¬« 3
32 3
2e 2t i 3et j C
2i Ÿ C
3 j
2e 2 t i 3 e t 1 j
68. r t
³
r0
C
rt
i 2 t 3 j 4t 3 2k
3t j 6 t k dt
2
t j 4t k C
3
32
i 2j
º
¼» 0
1
3
103 2 1
70. rcc t
4 sin tj 3 cos tk C1
rc 0
3k
³ 32 j dt
32tj C1
rc 0
C1
rc t
600 3i 600 32t j
rt
³ ª¬600
600 3i 600 j
3i 600 32t jº¼ dt
600 3ti 600t 16t 2 j C
r0
C
rt
600 3ti 600t 16t 2 j
0
3k C1 Ÿ C1
rt
4 cos tj 3 sin tk C2
r0
4 j C2
rt
4 cos tj 3 sin tk
71. r t
69. rc t
4 cos tj 3 sin tk
rc t
³
2
4 j Ÿ C2
te t i e1j k dt
0
0
1 2
e t i e t j tk C
2
r0
1
i jC
2
1
i jk Ÿ C
2
rt
1 t 2 ·
§
t
¨1 e ¸i e 2 j t 1 k
2
©
¹
i 2j k
§ 2 et 2 ·
¨
¸i e t 2 j t 1 k
¨
¸
2
©
¹
© 2010 Brooks/Cole, Cengage Learning
100
Chapter 12
ª
Vector-Valued Functions
1
³ «¬1 t 2 i 72. r t
1
1 º
j k » dt
2
t
t ¼
74. To find the integral of a vector-valued function, you
integrate each component function separately. The
constant of integration C is a constant vector.
1
arctan ti j ln tk C
t
S
r1
4
i jC
2i Ÿ C
75. At t
S·
§
¨ 2 ¸i j
4¹
©
and z directions simultaneously.
76. The graph of u t does not change position relative to
1·
S
ª
º
§
«2 4 arctan t » i ¨1 t ¸ j ln tk
¬
¼
©
¹
rt
t0 , the graph of u t is increasing in the x, y,
the xy-plane.
73. See “Definition of the Derivative of a Vector-Valued
Function” and Figure 12.8 on page 842.
x t i y t j z t k. Then
77. Let r t
cr t
cx t i cy t j cz t k and
Dt ª¬cr t º¼
cxc t i cyc t j czc t k
c ª¬ xc t i yc t j zc t k º¼
78. Let r t
x1 t i y1 t j z1 t k and u t
rt rut
Dt ª¬r t r u t º¼
x2 t i y2 t j z2 t k.
ª¬ x1 t r x2 t º¼ i ª¬ y1 t r y2 t º¼ j ª¬ z1 t r z2 t º¼ k
ª x1c t r x2c t º i ª y1c t r y2c t º j ª z1c t r z2c t º k
¬
¼
¬
¼
¬
¼
ª x c t i y1c t j z1c t k º r ª x2c t i y2c t j z2c t k º
¬ 1
¼ ¬
¼
79. Let r t
x t i y t j z t k , then w t r t
Dt ª¬w t r t º¼
ª¬w t xc t wc t x t º¼ i ª¬w t yc t wc t y t º¼ j ª¬w t zc t wc t z t º¼ k
x1 t i y1 t j z1 t k and u t
rt uut
D1 ª¬r t u u t º¼
rc t r uc t
w t x t i w t y t j w t z t k.
w t ª¬ xc t i yc t j zc t k º¼ wc t ª¬ x t i y t j z t k º¼
80. Let r t
crc t .
w t rc t wc t r t
x2 t i y2 t j z2 t k.
ª¬ y1 t z2 t z1 t y2 t º¼ i ª¬ x1 t z2 t z1 t x2 t º¼ j ª¬ x1 t y2 t y1 t x2 t º¼ k
ª y1 t z2c t y1c t z2 t z1 t y2c t z1c t y2 t º i ª x1 t z2c t x1c t z2 t z1 t x2c t z1c t x2 t º j
¬
¼
¬
¼
ª x1 t y2c t x1c t y2 t y1 t x2c t y1c t x2 t º k
¬
¼
^» y t z c t
^ª y c t z
¬
1
1
`
z1 t y2c t º i ª x1 t z2c t z1 t x2c t º j ª x1 t y2c t y1 t x2c t º k
¼
¬
¼
¬
¼
2
2
`
t z1c t y2 t º i ª x1c t z2 t z1c t x2 t º j ª x1c t y2 t y1c t x2 t º k
¼
¬
¼
¬
¼
r t u uc t rc t u u t
81. Let r t
x t i y t j z t k. Then r w t
Dt ª¬r w t º¼
x w t i y w t j z w t k and
xc w t wc t i yc w t wc t j zc w t wc t k Chain Rule
wc t ª¬ xc w t i yc w t j zc w t k º¼
82. Let r t
x t i y t j z t k. Then rc t
r t u rc t
Dt ª¬r t u rc t º¼
wc t rc w t .
xc t i yc t j zc t k.
ª¬ y t zc t z t yc t º¼ i ª¬ x t zc t z t xc t º¼ j ª¬ x t yc t y t xc t º¼ k
ª¬ y t zcc t yc t zc t z t ycc t zc t yc t º¼ i ª¬ x t zcc t xc t zc t z t xcc t zc t xc t º¼ j
ª¬ x t ycc t xc t yc t y t xcc t yc t xc t º¼k
ª¬ y t zcc t z t ycc t º¼ i ª¬ x t zcc t z t xcc t º¼ j ª¬ x t ycc t y t xcc t º¼ k
r t u rcc t
© 2010 Brooks/Cole, Cengage Learning
Section 12.2
x1 t i y1 t j z1 t k , u t
83. Let r t
r t ˜ ª¬u t u v t º¼
Differentiation and Integration of Vector-Valued Functions
x2 t i y2 t j z2 t k , and v t
101
x3 t i y3 t j z3 t k. Then:
x1 t ª¬ y2 t z3 t z2 t y3 t º¼ y1 t ª¬ x2 t z3 t z2 t x3 t º¼ z1 t ª¬ x2 t y3 t y2 t x3 t º¼
x1 t y2 t z3c t x1 t y2c t z3 t x1c t y2 t z3 t x1 t y3 t z2c t
Dt ª¬r t ˜ u t u v t º¼
x1 t y3c t z2 t x1c t y3 t z2 t y1 t x2 t z3c t y1 t x2c t z3 t y1c t x2 t z3 t
y1 t z2 t x3c t y1 t z2c t x3 t y1c t z2 t x3 t z1 t x2 t y3c t z1 t x2c t y3 t
z1c t x2 t y3 t z1 t y2 t x3c t z1 t y2c t x3 t z1c t y2 t x3 t
^x c t ª¬ y
1
`
t z3 t y3 t z2 t º¼ y1c t ª
¬ x2 t z3 t z2 t x3 t º¼ z1c t ª¬ x2 t y3 t y2 t x3 t º¼
2
^
^x t ª y
¬
`
t x c t º`
¼
x1 t ª y2c t z3 t y3 t z2c t º y1 t ª x2c t z3 t z2c t x3 t º z1 t ª x2c t y3 t y2c t x3 t º
¬
¼
¬
¼
¬
¼
1
2
t z3c t y3c t z2 t º y1 t ª x2 t z3c t z2 t x3c t º z1 t ª x2 t y3c t y2
¼
¬
¼
¬
3
rc t ˜ ª¬u t u v t º¼ r t ˜ ª¬uc t u v t º¼ r t ˜ ª¬u t u vc t º¼
x t i y t j z t k. If r t ˜ r t is constant, then:
84. Let r t
x2 t y2 t z 2 t
C
Dt >C @
Dt ª¬ x t y t z t º¼
2 x t xc t 2 y t yc t 2 z t zc t
0
2 ª¬ x t xc t y t yc t z t zc t º¼
0
2 ª¬r t ˜ rc t º¼
0.
2
So, r t ˜ rc t
2
2
0.
t sin t i 1 cos t j
85. r t
5
(a)
2 cos ti 3 sin tj
86. r t
(a) Ellipse
y
2
0
40
1
0
x
−3
The curve is a cycloid.
(b) rc t
rcc t
rc t
1 cos t i sin tj
−1
1
3
−2
sin ti cos tj
1 2 cos t cos 2 t sin 2 t
2 2 cos t
is 0, t
0.
Maximum of rc t
is 2, t
S .
sin 2 t cos 2 t
(b)
rc t
2 sin ti 3cos tj
rcc t
2 cos ti 3 sin tj
rc t
Minimum of rc t
rcc t
−1
4 sin 2 t 9 cos 2 t
Minimum of rc t
is 2, t
S 2.
Maximum of rc t
is 3, t
0.
1
Minimum and maximum of rc t
is 1.
© 2010 Brooks/Cole, Cengage Learning
102
Chapter 12
Vector-Valued Functions
et sin ti et cos tj
87. r t
rc t
et cos t et sin t i et cos t et sin t j
rcc t
et sin t et cos t et sin t et cos t i et cos t et sin t et sin t et cos t j
r t ˜ rcc t
2e 2t sin t cos t 2e 2t sin t cos t
2et cos ti 2et sin tj
0
So, r t is always perpendicular to rcc t .
ti 4 t 2 j
88. (a) r t
89. True
y
90. False. The definite integral is a vector, not a real number.
5
r(1)
cos t i sin t j k.
91. False. Let r t
3
rt
2
r(1.25)
1
r (1.25) − r (1)
d
ª rt º
¼
dt ¬
rc t
x
−3
−1
−1
3
r1
(b)
i 3j
r 1.25
1.25i 2.4375 j
r 1.25 r 1
0.25i 0.5625 j
(c)
2
0
sin t i cos t j
rc t
1
92. False.
rc t
i 2tj
rc 1
D ¬ªr t ˜ u t ¼º
i 2j
(See Theorem 2.2, part 4)
r 1.25 r 1
1.25 1
0.25i 0.5625 j
0.25
r t ˜ uc t rc t ˜ u t
i 2.25 j
This vector approximates rc 1 .
Section 12.3 Velocity and Acceleration
1. r t
3ti t 1 j
vt
rc t
3i j
at
rcc t
0
x
3t , y
y
x
1
3
v1
v
x
t 1,
4
−2
6
vt
2ti j
at
rcc t
2i
t, x
At 4, 2 , t
2.
v2
4i j
a2
2i
1.
6 t i tj
vt
rc t
i j
at
rcc t
0
x
6 t, y
(4, 2)
v
a
y2
x
2
4
6
−2
−4
0
ti t 2 4 j
4. r t
2. r t
4
2
t2, y
x
−4
3i j, a 1
y
rc t
2
(3, 0)
At 3, 0 , t
t 2 i tj
3. r t
y
6 x
t, y
vt
rc t
at
rcc t
x
i 2tj
t 4
t, y
At 1, 3 , t
y
2 j
4 x2
2
i 2 j, a 1
1, v 1
2 j
y
5
4
v
(1, 3)
3
(3, 3)
2
2
1
x
2
4
a
v
x
−3
−1
© 2010 Brooks/Cole, Cengage Learning
8
Section 12.3
t 2i t 3 j
5. r t
vt
at
x
2ti 3t 2 j
rcc t
2i 6tj
t2, y
t3, x
At 1, 1 , t
y2 3
1.
v1
2i 3 j
a1
2i 6 j
rc t
1 cos t , sin t
at
rcc t
sin t , cos t
t sin t , y
x
At S , 2 , t
v
(1, 1)
x
2 3 4 5 6 7 8
1 cos t cycloid
S.
y
4
vS
2, 0
aS
0, 1
2i
j
vt
rc t
3 t 2i
4
at
r cc t
3 ti
2
j
At 3, 2 , t
1
1 y3
4
3
4 x 1
3i j, a 2
2, v 2
vt
rc t
e t , et
at
rcc t
e t , et
x
3i
y
4
(3, 2)
0.
v0
1, 1
a0
1, 1
v
2
1
,y
et
et
At 1, 1 , t
6
a
2
4
et , y
11. r t
1
x
2
1
v
(1, 1)
a
i j
x
1
i j
2
ti 5tj 3tk
vt
i 5 j 3k
2 cos ti 2 sin tj
st
vt
vt
rc t
2 sin ti 2 cos tj
at
at
rcc t
2 cos ti 2 sin tj
x
2 cos t , y
2 sin t , x 2 y 2
−6
2,
2 ,t
§S ·
v¨ ¸
©4¹
§S ·
a¨ ¸
©4¹
S
4
12. r t
4
y
v
(
2)
2,
a
2i 2j
x
−3
4ti 4tj 2tk
4i 4 j 2k
st
vt
at
0
16 16 4
vt
3 cos ti 2 sin tj
st
vt
vt
3 sin ti 2 cos tj
at
2j k
at
3 cos ti 2 sin tj
At 3, 0 , t
2j
−3
14. r t
x2
y2
2 sin t ,
9
4
3 cos t , y
1 Ellipse
y
0.
6
3
13. r t
2i 35
0
vt
3
.
1 52 32
t2
k
2
i 2tj tk
x
y
6
−4
8. r t
x
x
− 6 −4 −2
At
2π
e t , et
10. r t
t Ÿ x
1, y
v
a
π
Ÿ y
7. r t
(π , 2)
2
1 i tj
1 t3
4
1 t3
4
a
−1
6. r t
x
vt
y
8
7
6
5
4
3
2
1
103
t sin t , 1 cos t
9. r t
rc t
Velocity and Acceleration
ti t 2 j 1 4t 2 t 2
1 5t 2
3ti tj 14 t 2k
vt
3i j 12 tk
st
vt
at
1k
2
9 1 14 t 2
10 14 t 2
3
v0
2j
a0
3i
1
v
a
−2
−1
−1
1
x
2
(3, 0)
−3
© 2010 Brooks/Cole, Cengage Learning
104
Chapter 12
15. r t
vt
st
at
Vector-Valued Functions
t i tj i j
9 t2k
t
9 t2
9
9 t2
32
vt
2ti j 3 t k
st
vt
at
2i 4t 2 1 9t
4, 3 sin t , 3 cos t
16 9 sin t 9 cos t
st
vt
at
2 cos t , 2 sin t , 2
vt
rc t
1,
rc 3
3 3
1, , 4 4
2
e
2t
2
cos t sin t
e
2t
1
ln t , , t 4
t
1 1
, , 4t 3
t t2
t
,
25 t 2
4
z
3
3
t
4
3
3
0.1 , 4 0.1
4
4
i j k, v 0
0, r 0
vt
³
i j k dt
ti tj tk C
v0
C
0, v t
23. a t
2et sin ti 2et cos tj et k
at
25 t 2 , t0
3.100, 3.925, 3.925
3
vt
1
3
0.1
4
4
2 1 t2
1
1
16t 6
t2 t4
1
t2
1 t 16t
2
10
0
t i t j tk , v t
t i jk
2
t
i jk C
2
rt
³
r0
C
r2
2i jk
24. a t
2i 3k , v 0
4 j, r 0
vt
³
2ti 3tk C
v0
C
rt
³
r0
C
r2
4i 8 j 6k
vt
st
t
25 t 2
3 t, y
x
et cos t et sin t i et sin t et cos t j et k
et
25 t 2 ,
t,
3 cos tj 3 sin tk
4 sin 2 t 4 cos 2 t 4t 2
cos t sin t
1
3
t
4 4
1 2t , z
22. (a) r t
5
et cos t , et sin t , et
e
1 t, y
(b) r 3 0.1 | 3 0.1, 4 2 sin t , 2 cos t , 2t
20. r t
2
2 cos t , 2 sin t , t 2
2t
3
4
1.100, 1.200, 0.325
4i 3 sin tj 3 cos tk
2
vt
at
1, 2,
4t , 3 cos t , 3 sin t
0, 3 cos t , 3 sin t
st
rc 1
4t 2 9t 1
3
k
2 t
at
vt
3t 2
4
1
(b) r 1 0.1 | 1 0.1, 1 2 0.1 ,
vt
19. r t
1, 2t ,
18 t 2
9 t2
t2
9 t2
st
18. r t
rc t
x
t 2i tj 2t 3 2k
vt
t3
, t0
4
k
16. r t
17. r t
t , t 2 ,
k
11
vt
21. (a) r t
ti tj tk dt
t2
i jk,
2
2i 2 j 2k
0, r t
2i 3k dt
4j Ÿ v t
2ti 4 j 3tk
2ti 4 j 3tk dt
0 Ÿ rt
0
t 2i 4tj 32 t 2k C
t 2i 4tj 32 t 2k
1 2
, , 12t 2
t 2 t3
© 2010 Brooks/Cole, Cengage Learning
Section 12.3
25. a t
tj tk , v 1
5 j, r 1
t2
t2
j k C
2
2
9
1
5j Ÿ C
j k
2
2
³
v1
1
1
j k C
2
2
vt
§ t2
§ t2
9·
1·
¨ ¸ j ¨ ¸k
2¹
2¹
©2
©2
rt
³ «¬¨© 2
r1
14
1
j k C
3
3
rt
r2
26. a t
ª§ t 2
§ t2
9·
1· º
¸ j ¨ ¸k » dt
2¹
2
2¹ ¼
©
0 Ÿ C
§ t3
§ t3
9 ·
1 ·
¨ t ¸ j ¨ t ¸k C
6
2
6
2 ¹
©
¹
©
14
1
j k
3
3
§ t3
§ t3
9
14 ·
1
1·
¨ t ¸ j ¨ t ¸k
6
2
3
6
2
3¹
©
¹
©
17
2
j k
3
3
32k
vt
³ 32k dt
v0
C
rt
³ ª¬3i 2 j r0
C
r2
6i j 60k
32tk C
3i 2 j k Ÿ v t
3i 2 j 1 32t k
1 32t k º¼ dt
5 j 2k Ÿ r t
3ti 2tj t 16t 2 k C
3ti 5 2t j 2 t 16t 2 k
cos ti sin tj, v 0
j k, r 0
vt
³
sin ti cos tj C
v0
jC
vt
sin ti cos tj k
rt
³
r0
i C
rt
cos ti sin tj tk
27. a t
r2
28. a t
cos ti sin tj dt
jk Ÿ C
sin ti cos tj k dt
i Ÿ C
i
k
cos ti sin tj tk C
0
cos 2 i sin 2 j 2k
et i 8k
29. r t
vt
³
v0
i C
vt
et 1 i 3 j 1 8t k
rt
105
0
vt
tj tk dt
Velocity and Acceleration
et i 8k dt
³ ª¬ e
t
et i 8tk C
2i 3 j k Ÿ C
88 cos 30q ti ª¬10 88 sin 30q t 16t 2 º¼ j
44 3ti 10 44t 16t 2 j
i 3j k
50
1 i 3j 1 8t k º¼ dt
et t i 3tj t 4t 2 k C
r0
i C
0 Ÿ C
rt
et t 1 i 3tj t 4t 2 k
0
300
0
i
© 2010 Brooks/Cole, Cengage Learning
106
Chapter 12
Vector-Valued Functions
900 cos 45q ti ª¬3 900 sin 45q t 16t 2 º¼ j
30. r t
The maximum height occurs when yc t
450 2t i 3 450 2t 16t 2 j
450 2 32t
0, which implies that t
225 2 16.
The maximum height reached by the projectile is
§ 225 2 ·
§ 225 2 ·
3 450 2 ¨¨
¸¸ 16¨¨
¸¸
© 16 ¹
© 16 ¹
y
2
50,649
8
6331.125 feet.
3 450 2t 16t 2
The range is determined by setting y t
0 which implies that
450 2 405,192
| 39.779 seconds
32
t
Range: x
§ 450 2 405,192 ·
450 2 ¨¨
¸¸ | 25,315.500 feet
32
©
¹
v0
v
§
·
ti ¨ 3 0 t 16t 2 ¸ j
2
2
©
¹
1 º
ª
v0 cos T ti «h v0 sin T t gt 2 » j
2 ¼
¬
31. r t
v0
t
2
300 when 3 v0
t 16t 2
2
3.
§ 300 2 ·
300 2 v0 § 300 2 ·
,
¨¨
¸¸ 16¨¨
¸¸
v0
v
2©
0
¹
© v0 ¹
t
v02
300 32 , v0
9600
2
40 6, v0
0, 300 3002 32
v02
0
40 6 | 97.98 ft sec
The maximum height is reached when the derivative of the vertical component is zero.
tv0
16t 2
2
yt
3
yc t
40 3 32t
40 3
32
t
3
rt
3 40 3t 16t 2
0
5 3
4
§5 3·
Maximum height: y¨¨
¸¸
© 4 ¹
32. 50 mi h
40 6
t 16t 2
2
§5 3·
§5 3·
3 40 3 ¨¨
¸¸ 16¨¨
¸¸
4
©
¹
© 4 ¹
2
78 feet
220
ft sec
3
ª
º
§ 220
·
§ 220
·
cos 15q ¸ti «5 ¨
sin 15q ¸t 16t 2 » j
¨
3
3
©
¹
©
¹
¬
¼
The ball is 90 feet from where it is thrown when x
220
cos 15qt
3
90 Ÿ t
27
| 1.2706 seconds.
22 cos 15q
2
The height of the ball at this time is y
33. x t
yt
y
t v0 cos T or t
·
§
·
27
27
§ 220
·§
5¨
sin 15q ¸¨
¸ 16¨
¸ | 3.286 feet.
q
q
3
22
cos
15
22
cos
15
©
¹©
¹
©
¹
x
v0 cos T
t v0 sin T 16t 2 h
§
·
x
x2
v0 sin T 16¨ 2
¸ h
2
T
v0 cos T
v
cos
© 0
¹
§ 16
·
tan T x ¨ 2 sec 2 T ¸ x 2 h
v
© 0
¹
© 2010 Brooks/Cole, Cengage Learning
Section 12.3
x 0.005 x 2
34. y
(a) y
(b)
16
sec 2 T
v02
16
2
v02
negative of coefficient of x
0.005 or v0
107
ti 0.004t 2 0.37t 6 j
35. r t
From Exercise 33 we know that tan T is the coefficient
of x. So, tan T
1, T
S 4 rad 45q. Also
Velocity and Acceleration
0.004 x 2 0.37 x 6
18
2
80 ft sec
0
120
0
40 2t i 40 2t 16t 2 j. Position function
rt
When 40 2t
vt
§3 2 ·
v¨¨
¸¸
© 4 ¹
0 Ÿ x
(d) From Exercise 33, tan T
3 2
4
16 sec T
v02
2
40 2i 40 2 32t j
45.8375
36. r t
8 2 5i 2 j . Direction
§3 2 ·
v¨¨
¸¸
© 4 ¹
8 2
25 4
0.3667 Ÿ T | 20.14q
16 sec 2 T
0.004
0.004 Ÿ v02
4000
cos 2 T
Ÿ v0 | 67.4 ft sec.
40 2i 40 2 24 2 j
Speed
0.008 x 0.37
and y 45.8375 | 14.56 ft
60,
60
40 2
t
(c) yc
140 cos 22q ti 2.5 140 sin 22q t 16t 2 j
When x
375, t | 2.889
and y | 20.47 feet.
8 58 ft sec
So, the ball clears
the 10-foot fence.
60
0
450
0
miles ·§
feet · §
sec ·
§
¨100
¸¨ 5280
¸ ¨ 3600
¸
hr ¹©
mile ¹ ©
hour ¹
©
37. 100 mi h
(a) r t
ª
º
§ 440
·
§ 440
·
cos T 0 ¸ti «3 ¨
sin T 0 ¸t 16t 2 » j
¨
3
3
©
¹
©
¹
¬
¼
Graphing these curves together with y
(b)
440
ft sec
3
10 shows that T 0
20q.
100
θ 0 = 20
θ 0 = 25
500
0
0
θ 0 = 10
θ 0 = 15
(c) You want
xt
§ 440
·
cos T ¸t t 400 and y t
¨
© 3
¹
§ 440
·
3¨
sin T ¸t 16t 2 t 10.
© 3
¹
30 11 cos T . Substituting this for t in y t yields:
From x t , the minimum angle occurs when t
2
§ 30 ·
§ 440
·§ 30 ·
3¨
sin T ¸¨
¸ 16¨
¸
© 3
¹© 11 cos T ¹
© 11 cos T ¹
14,400
400 tan T sec 2 T
121
14,400
1 tan 2 T 400 tan T 7
121
14,400 tan 2 T 48,400 tan T 15,247
tan T
T
10
7
0
0
48,400 r
48,4002 4 14,400 15,247
2 14,400
§ 48,400 1,464,332,800 ·
tan 1 ¨¨
¸¸ | 19.38q
28,800
©
¹
© 2010 Brooks/Cole, Cengage Learning
108
Chapter 12
7 feet, T
38. h
Vector-Valued Functions
35q, 30 yards
90 feet
v0 cos 35q ti ª¬7 v0 sin 35q t 16t 2 º¼ j
rt
(a) v0 cos 35qt
90 when 7 v0 sin 35q t 16t 2
4
t
§
·
§
·
90
90
7 v0 sin 35q ¨
¸ 16¨
¸
© v0 cos 35q ¹
© v0 cos 35q ¹
90
v0 cos 35q
2
90 tan 35q 3
v02
4
129,600
cos 2 35q
129,600
cos 2 35q 90 tan 35q 3
v02
v0 | 54.088 ft sec
(b) The maximum height occurs when yc t
v0 sin 35q 32t
0.
v0 sin 35
| 0.969 sec
32
t
At this time, the height is y 0.969 | 22.0 ft.
(c) x t
90 Ÿ v0 cos 35q t
90
90
| 2.0 sec
54.088 cos 35q
t
v cos T ti ª¬ v sin T t 16t 2 º¼ j
(a) You want to find the minimum initial speed v as a function of the angle T . Because the bale must be thrown to the position
16, 8 , you have
39. r t
16
8
t
v cos T t
v sin T t 16t 2 .
16 v cos T from the first equation. Substituting into the second equation and solving for v, you obtain:
8
1
§
·
1
512¨ 2
¸
2
© v cos T ¹
1
v2
v2
You minimize f T
fcT
fcT
§ 16 ·
§ 16 ·
v sin T ¨
¸ 16¨
¸
© v cos T ¹
© v cos T ¹
§ sin T ·
§
·
1
2¨
¸ 512¨ 2
¸
2
© cos T ¹
© v cos T ¹
§ sin T ·
2¨
¸ 1
© cos T ¹
§ sin T
· cos 2 T
1¸
¨2
© cos T
¹ 512
2
2 sin T cos T cos 2 T
512
512
2 sin T cos T cos 2 T
512
.
2 sin T cos T cos 2 T
§
·
2cos 2 T 2 sin 2 T 2 sin T cos T ¸
512¨
2
¨¨
¸¸
2 sin T cos T cos 2 T
©
¹
0 Ÿ 2 cos 2T sin 2T
0
tan 2T
2
T | 1.01722 | 58.28q
Substituting into the equation for v, v | 28.78 ft sec.
© 2010 Brooks/Cole, Cengage Learning
Section 12.3
(b) If T
109
45q,
v cos T t
16
v
2
t
2
v sin T t 16t 2
8
v
2
t 16t 2
2
512
From part (a), v 2
2
2 2 2 2
2 2
40. Place the origin directly below the plane. Then T
512
12
2
0, v0
v0 cos T ti 30,000 v0 sin T t 16t j
1024 Ÿ v
32 ft sec.
792 and
792ti 30,000 16t 2 j
2
rt
792i 32tj.
vt
At time of impact, 30,000 16t 2
34,294.6i
v 43.3
792i 1385.6 j
1596 ft sec
0 Ÿ t2
α
(0, 0)
1088 mi h
v0 cos T ti ª¬ v0 sin T t 16t 2 º¼ j
v0 sin T t 16t
2
0 when t
0 and t
42. From Exercise 41, the range is
v0 sin T
.
16
x
The range is
v0 cos T t
x
34,295
30,000
| 0.8748 Ÿ D | 0.7187 41.18q
34,294.6
tan D
41. r t
α
1875 Ÿ t | 43.3 seconds.
30,000
r 43.3
v 43.3
v0 cos T
v0 sin T
16
v02
sin 2T .
32
v02
sin 2T
32
v02
sin 24q
32
So, x
200
Ÿ v02
6400 sin 24q
Ÿ v0 | 125.4 ft sec
So,
2
x
Velocity and Acceleration
1200
sin 2T
32
43. (a) T
rt
10q, v0
1
Ÿ T | 1.91q.
15
3000 Ÿ sin 2T
66 ft sec
66 cos 10q ti ª¬0 66 sin 10q t 16t 2 º¼ j
5
r t | 65t i 11.46t 16t 2 j
Maximum height: 2.052 feet
0
Range: 46.557 feet
(b) T
rt
10q, v0
50
0
146 ft sec
146 cos 10q ti ª¬0 146 sin 10q t 16t 2 º¼ j
15
r t | 143.78t i 25.35t 16t j
2
Maximum height: 10.043 feet
Range: 227.828 feet
(c) T
rt
45q, v0
300
0
0
66 ft sec
40
66 cos 45q ti ª¬0 66 sin 45q t 16t º¼ j
2
r t | 46.67t i 46.67t 16t 2 j
Maximum height: 34.031 feet
0
200
0
Range: 136.125 feet
© 2010 Brooks/Cole, Cengage Learning
110
Chapter 12
(d) T
Vector-Valued Functions
45q, v0
146 ft sec
200
146 cos 45q ti ª¬0 146 sin 45q t 16t º¼ j
2
rt
r t | 103.24t i 103.24t 16t 2 j
Maximum height: 166.531 feet
Range: 666.125 feet
60q, v0
66 ft sec
(e) T
0
800
0
60
66 cos 60q ti ª¬0 66 sin 60q t 16t º¼ j
2
rt
r t | 33t i 57.16t 16t 2 j
Maximum height: 51.047 feet
Range: 117.888 feet
(f ) T
60q, v0
146 ft sec
140
0
0
300
146 cos 60q ti ª¬0 146 sin 60q t 16t º¼ j
2
rt
r t | 73t i 126.44t 16t 2 j
Maximum height: 249.797 feet
Range: 576.881 feet
44. (a) r t
0
t v0 cos T i tv0 sin T 16t 2 j
t v0 sin T 16t
Range: x
0 when t
v0 sin T
.
16
100 cos 30q ti ª¬1.5 100 sin 30q t 4.9t 2 º¼ j
§ v02 ·
¨ ¸ sin 2T
© 32 ¹
§ v sin T ·
v0 cos T ¨ 0
¸
© 32 ¹
§ v02 ·
¨ ¸ 2 cos 2T
© 32 ¹
2T
(b) y t
dy
dt
2
,T
S
4
The projectile hits the ground when
4.9t 2 100
1
2
t 1.5
0 Ÿ t | 10.234 seconds.
So the range is 100 cos 30q 10.234 | 886.3 meters.
The maximum height occurs when dy dt
0
100 sin 30
or
S
v0 cos T ti ª¬h v0 sin T t 4.9t 2 º¼ j
45. r t
The range will be maximum when
dx
dt
600
0
0.
9.8t Ÿ t | 5.102 sec
The maximum height is
y
rad.
2
| 129.1 meters.
tv0 sin T 16t 2
v0 sin T 32t
1.5 100 sin 30q 5.102 4.9 5.102
v0 sin T
.
32
0 when t
Maximum height:
§ v sin T ·
y¨ 0
¸
© 32 ¹
v02 sin 2 T
v 2 sin 2 T
16 0 2
32
32
Minimum height when sin T
46. r t
x
v02 sin 2 T
64
S
1, or T
2
.
v0 cos T ti ª¬h v0 sin T t 4.9t 2 º¼ j
v0 cos 8q ti ª¬ v0 sin 8q t 4.9t 2 º¼ j
50
50 when v0 cos 8q t
50 Ÿ t
. For this value of t , y
v0 cos 8q
§ 50 ·
§ 50 ·
v0 sin 8q ¨
¸ 4.9¨
¸
© v0 cos 8q ¹
© v0 cos 8q ¹
0:
2
50 tan 8q
0
4.9 2500
Ÿ v02
v02 cos 2 8q
4.9 50
| 1777.698 Ÿ v0 | 42.2 m sec
tan 8q cos 2 8q
© 2010 Brooks/Cole, Cengage Learning
Section 12.3
Velocity and Acceleration
111
b Z t sin Z t i b 1 cos Z t j
47. r t
vt
b Z Z cos Z t i bZ sin Z t j
at
bZ 2 sin Z t i bZ 2 cos Z t j
2bZ
vt
bZ 1 cos Z t i bZ sin Z t j
bZ 2 ª¬sin Z t i cos Z t jº¼
1 cos Z t
bZ 2
at
0 when Z t
0, 2S , 4S , !.
(a)
vt
(b)
v t is maximum when Z t
S , 3S , ! , then v t
2bZ.
b Z t sin Z t i b 1 cos Z t j
48. r t
bZ ª¬ 1 cos Z t i sin Z t jº¼
vt
bZ
vt
Speed
1 2 cos Z t cos 2 Z t sin 2 Z t
The speed has a maximum value of 2bZ when Z t
60 mi h
88 rad sec since b
88 ft sec
2 bZ
1 cos Z t .
S , 3S , !
1.
So, the maximum speed of a point on the tire is twice the speed of the car:
2 88 ft sec
bZ sin Z t i bZ cos Z t j
vt
49.
120 mi h
b 2Z sin Z t cos Z t b 2Z sin vt cos Z t
rt ˜vt
0
So, r t and v t are orthogonal.
50. (a) Speed
b 2Z 2 sin 2 Z t b 2Z 2 cos 2 Z t
v
(b)
b 2Z 2 ª¬sin 2 Z t cos 2 Z t º¼
bZ
10
−10
10
−10
The graphing utility draws the circle faster for greater values of Z.
51. a t
bZ 2 cos Z t i bZ 2 sin Z t j
bZ 2 ª¬cos Z t i sin Z t jº¼
Z 2r t
a t is a negative multiple of a unit vector from 0, 0 to cos Z t , sin Z t and so a t is directed toward the origin.
52.
at
bZ 2 cos Z t i sin vt j
53.
at
Z 2 b, b
1
F
Z
vt
vt
30 mi h
vt
2Z 2
8 10 ft sec
bZ 2
10
F
44 ft sec
44
rad sec
300
b
at
1
32
4 10 rad sec
bZ
54.
Z
2
m 32
m Z 2b
bZ 2
m bZ 2
3400
§ 44 ·
300 ¨
¸
32
© 300 ¹
2
2057
lb
3
Let n be normal to the road.
n cos T
3400
n sin T
2057
3
121
600
T | 11.4q
Dividing, tan T
© 2010 Brooks/Cole, Cengage Learning
112
Chapter 12
Vector-Valued Functions
55. To find the range, set
yt
0 then
§1 · 2
¨ g ¸t v0 sin T t h. By the Quadratic
©2 ¹
Formula, discount the negative value
0
v0 sin T t
2
v0 sin T
4 ª¬ 1 2 g º¼ h
2 ¬ª 1 2 g ¼º
v0 sin T § v sin T v0 cos T ¨ 0
¨
©
v02 sin 2 T 2 gh ·
¸
¸
g
¹
v0 cos T §
¨ v0 sin T g ¨
©
§
2 gh · ·
v02 ¨ sin 2 T 2 ¸ ¸
v0 ¹ ¸
©
¹
v02 cos T §
¨¨ sin T g
©
56. h
6 feet, v0
2 gh ·
sin T 2 ¸¸ feet.
v0 ¹
2
45 feet per second, T
m xt
rt
vt
x t i ª¬m x t bº¼ j
xc t i mxc t j
st
ª¬ xc t º¼ ª¬mxc t º¼
45 sin 42.5q 2
2
xcc t
0
xcc t i mxcc t j
at
rc t
(a) v t
3 4 sin 2 t cos 2 t
vc t
at
t
0
Speed
3
S
S
4
2
3
10
2
6
2S
3
S
3
13
2
3
y
(c)
8
6
4
2
−8
vt
xc t i yc t j zc t k Velocity vector
at
xcc t i ycc t j zcc t k Acceleration vector
xc t
vt
2
yc t
2
zc t
yc t
2
zc t º
¼
2
2 xc t xcc t 2 yc t ycc t 2 zc t zcc t
2 ª¬ xc t xcc t yc t ycc t zc t zcc t º¼
vt ˜at
2
4
6
8
−4
−6
−8
(d) The speed is increasing when the angle between
v and a is in the interval
ª S·
«0, 2 ¸.
¬
¹
2
C , C is a constant.
2
x
−4 −2
−2
x t i y t j z t k Position vector
Orthogonal
3 3 sin 2 t 1
6 cos ti 3 sin tj
(b)
At this time, x t | 69.02 feet.
dª
xc t
dt ¬
6 sin t i 3 cos t j
36 sin 2 t 9 cos 2 t
vt
32
Speed
0.
6 cos ti 3 sin tj
59. r t
| 2.08 seconds.
57. r t
C , C is a constant.
1 m2
42.5q. From
45 sin 2 42.5q 2 32 6
2
C
So, xc t
Exercise 55,
t
b, m and b are constants.
yt
v02 sin 2 T 2 gh
second.
g
At this time,
xt
xti yt j
58. r t
1
h v0 sin T t gt 2
2
The speed is decreasing when the angle is in the interval
§S º
¨ , S ».
©2 ¼
0
0
0
0
a cos Z t i b sin Z t j
60. r t
(a) rc t
Speed
(b) a t
vt
vt
vc t
aZ sin Z t i bZ cos Z t j
a 2Z 2 sin 2 Z t b 2Z 2 cos 2 Z t
aZ 2 cos Z t i bZ 2 sin Z t j
Z 2 a cos Z t i b sin Z t j
Z 2 r t
61. The velocity of an object involves both magnitude and
direction of motion, whereas speed involves only
magnitude.
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
Tangent Vectors and Normal Vectors
62. (a) The speed is increasing.
64. a t
(b) The speed is decreasing.
vt
63. (a) r1 t
r2 t
sin ti cos tj
³a t
cos ti
dt
xt i yt j ztk
v0
i
r1 2t
vt
cos ti sin tj
rt
³v t
Velocity: r2c t
2r1c 2t
Acceleration: r2cc t
r1 Z t , then:
(b) In general, if r3 t
Velocity: r3c t
4r1cc 2t
Z r1c Z t
Acceleration: r3cc t
Z r1cc Z t
2
113
sin tj C1
i C1 Ÿ C1
0
sin ti cos tj C2
dt
j C2 Ÿ C 2
r0
j
rt
sin ti cos tj
0
The path is a circle.
65. False. The acceleration is the derivative of the velocity.
66. True
67. True
68. False. For example, 6t r t
at
t 3i. Then v t
3t 2 i and
6t i. v t is not orthogonal to a t .
Section 12.4 Tangent Vectors and Normal Vectors
y
1.
5. r t
t 2i 2tj, t
1
rc t
2ti 2 j, rc t
T1
rc 1
rc 1
4t 2 4
1
i j
2
2
i 2
2 t2 1
2
j
2
x
2.
6. r t
t 3i 2t 2 j
rc t
3t 2i 4tj
y
rc t
T
T
N
T1
N
N
9t 4 16t 2
rc 1
rc 1
1
3i 4 j
9 16
3
4
i j
5
5
T
x
7.
y
3.
rt
4 cos ti 4 sin tj, t
rc t
4 sin ti 4 cos tj
rc t
§S ·
T¨ ¸
©4¹
x
4.
8.
y
T
N
N
N
x
§S ·
rc¨ ¸
©4¹
§S ·
rc¨ ¸
©4¹
2
i 2
6 cos ti 2 sin tj
rc t
6 sin ti 2 cos tj
§S ·
T¨ ¸
©3¹
4
16 sin 2 t 16 cos 2 t
rt
rc t
T
T
S
4
2
j
2
36 sin 2 t 4 cos 2 t
§S ·
rc¨ ¸
© 3¹
S·
§
rc¨ ¸
© 3¹
3 3i j
36 3 4 1 4
1
3 3i j
28
© 2010 Brooks/Cole, Cengage Learning
114
Chapter 12
Vector-Valued Functions
3ti ln tj, t
9. r t
e
14. r t
1
3i j
t
1
3i j
e
rc e
rc e
rc t
rc e
Te
rc t
3ei j
9e 2 1
e cos ti e j, t
t
| 0.9926i 0.1217 j
et cos t et sin t i et j
rc 0
i j
T0
rc 0
rc 0
i j
2
2
i 2
0, rc 0
rc 0
rc 0
i k , ª¬t
4
§
k , P¨1, 1,
3
©
t 2 i tj 12. r t
rc t
t, y
1, rc t
rc 1
rc 1
2i j
5
rc 0
T0
§
1 at ¨1, 1,
©
2, b
1, c
2t 1, y
t 1, z
4
4
3
3, y
3t , z
1
t
2, 4
4
at
§S ·
, rc¨ ¸
©4¹
2,
2,
2, 0 ,
º
2, 4 ».
¼
1
2,
2
2, 0
2, c
2 sin t , 2 cos t , 4 sin 2 t , P 1,
rc t
2 cos t , 2 sin t , 8 sin t cos t
When t
S
S
at 1,
6
§S ·
T¨ ¸
©6¹
6
§S ·
, rc¨ ¸
©6¹
2t 2,
3, 1
3, 1, 2 3 ,
º
3, 1 ».
¼
rc S 6
rc S 6
1
4
3, 1, 2 3
1, c
3, b
3t 1, y
Parametric equations: x
z
0
2, y
16. r t
T3
3, c
2,
2t ª¬t
3j k
10
Parametric equations: x
t 1, y
Parametric equations: x
z
4
2 3
t 3,
2 3t 1
2
t, t 2 , t 3
3
z
1, 2t , 2t 2
When t
3j k
1
3
t 1,
1, c
2, b
rc t
0 at P 3, 0, 0
0, b
S
rc S 4
rc S 4
17. r t
3 sin ti 3 cos tj k
Direction numbers: a
1, b
Direction numbers: a
ª
«t
¬
3 º.
¼
1 at 1, 1,
21
1
1, 1, 7
3
Direction numbers: a
0
3 cos ti 3 sin tj tk , P 3, 0, 0
rc 0
rc 0
4 ·º
¸.
3 ¹»¼
5
2i j
5
Parametric equations: x
t
4·
¸
3¹
ª
2i j «t
¬
rc 1
Direction numbers: a
rc t
t
2ti j
When t
13. r t
1
0, z
S
§S ·
T¨ ¸
©4¹
0 at 0, 0, 0 º¼.
0, c
1, b
Parametric equations: x
T1
ª
«t
¬
1
,t
3
2 sin t , 2 cos t , 0
When t
2
i k
2
Direction numbers: a
1, 1, 3
2 cos t , 2 sin t , 4 , P
rc t
2
j
2
i 2tj k
T0
1, rc 1
Direction numbers: a
15. r t
ti t 2 j tk , P 0, 0, 0
When t
t
4 t2
Parametric equations: x
1
z
t 3
3
0
rc t
rc t
1, 1, rc 1
rc 1
T1
t
11. r t
4 t 2 , P 1, 1,
When t
1
3i j
e
1
9 2
e
10. r t
t, t,
3, rc 3
1, 6, 18 ,
9
3 at 3, 9, 18 º¼.
rc 3
rc 3
6
18
15
12
9
6
3
3
−3
x
12
15
18
y
1
1, 6, 18
19
Direction numbers: a
1, b
6, c
Parametric equations: x
t 3, y
18
6t 9, z
18t 18
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
1
k
2
1
3 sin ti 4 cos tj k
2
3 cos ti 4 sin tj 18. r t
rc t
§S ·
, rc¨ ¸
2
©2¹
S
When t
ª
«t
¬
§S ·
T¨ ¸
©2¹
S ·º
§
at ¨ 0, 4, ¸».
2 ©
4 ¹¼
ti ln tj 1
1
i j
k ; rc 1
t
2 t
T1
rc t
rc t
1
6i k
37
1
t 4, z
i j
S
1 t, y
t, z
y
1
1
cos 2 s 2
2
uc 0
1, 0, 1
cos T
rc 0 ˜ uc 0
rc 0 uc 0
1
k
2
1
1 t
2
r 1.1 | 1.1i 0.1j 1.05k
i tj
Tt
rc t
rc t
Tc t
e t i 2 cos tj 2 sin tk , t0
rc t
e t i 2 sin tj 2 cos tk
r0
i 2 j, rc 0
T0
rc 0
rc 0
r t0 0.1
5
i 2k
5
u8
2, 16, 2
t
t 1
2
32
i Tc 2
Tc 2
1
t 1
2
32
j
2 5
5
i j
5
5
1
2i j
5
24. r t
ti 6
j, t
t
rc t
i 6
j
t2
2s
rc t
rc t
3
6 ·
§
¨ i 2 j¸
t
©
¹
1 36 t
1
4
6 ·
§
¨ i 2 j¸
t
¹
t 36 ©
t2
4
1
1, 2t , , rc 4
2
0.9, 2, 0.2
1
1, 8,
2
uc s
1
1
, 2, s 2 3 , uc 8
4
3
cos T
rc 4 ˜ uc 8
rc 4 uc 8
|
1 t2
N2
Tc t
72t
t 4 36
32
i Tc 2
144
96
i 32j
523 2
52
N2
Tc 2
Tc 2
So the curves intersect.
rc t
i tj
2
1
i 32j
53 2
5
Tt
| 1 0.1, 2, 2 0.1
2, 16, 2
2
Tc 2
r 0 0.1
21. r 4
2
0
i 2k , rc 0
Parametric equations:
xs
1 s, y s
2, z s
S
0 Ÿ T
1 2
t j, t
2
ti rc t
1.1, 0.1, 1.05
20. r t
1, 0, 1
sin s cos s cos s , sin s cos s cos s,
4
2
2
1
i j k
3
3
3
11 14
r t0 0.1
5
23. r t
i j 12k
Tangent line: x
uc s
4
1
rc t
1, sin t , cos t , rc 0
1
0, c
6t , y
t k , t0
0, 1, 0
rc t
2
5 4
6, b
Parametric equations: x
19. r t
3
2 §
1 ·
¨ 3i k ¸
2 ¹
37 ©
Direction numbers: a
u0
So the curves intersect.
4
x
rc S 2
rc S 2
0, 1, 0
5
1
S
22. r 0
115
z
1
k,
2
3i Tangent Vectors and Normal Vectors
12t 3
t 4 36
32
j
1
3i 2 j
13
1
1
, 2,
4 12
16.29167
Ÿ T | 1.2q
16.29513
© 2010 Brooks/Cole, Cengage Learning
116
Chapter 12
25. r t
rc t
Tt
Tc t
Tc 2
N2
27. r t
rc t
Tt
Tc t
Vector-Valued Functions
ln ti t 1 j, t
2
1
i j
t
1
1
t2
rc t
rc t
t
1t
2
32
i 1
1t
2
Tc t
Tc 0
N0
rc t
j
32
2
1
i 32j
32
5
5
Tc 2
Tc 2
2 5
5
i j
5
5
ti t 2 j ln tk , t
S
6
S
Tt
rc t
rc t
Tc t
cos ti sin tj, Tc t
sin ti cos tj
§S ·
Tc¨ ¸
©6¹
§S ·
N¨ ¸
©6¹
§S ·
Tc¨ ¸
©6¹
§S ·
Tc¨ ¸
©6¹
1
3
1
i j
2
2
3
1
i j
2
2
1
1
i 2tj k
t
1
i 2tj k
t
1
2
1 4t 2
t
rc t
rc t
1 4t 4
4t 4 t 2 1
32
i i 2 j 3k
14
Tt
S sin ti S cos tj
1 t2
N1
rc t
rc t
i tj
3
6
9
i 3 2 j 3 2k
63 2
6
6
rc t
S cos ti S sin tj, t
1
i j
t
Tc 1
28. r t
26. r t
ti 2t 2 j k
4t 4 t 2 1
2t 3 4t
4t 4 t 2 1
8t 3 t
j
32
4t 4 t 2 1
14
2 14
3 14
i j
k
14
14
14
0
29. r t
2i e t j e t k
2e
e
rc t
rc t
e e
t
2 t
e e
t
t
2
e e
t
t
2
i
1
1
j k
2
2
2
2
j
k
2
2
2
e e
t
t
2
j
2
e e
t
t
2
k
6 cos ti 6 sin tj k , t
rc t
6 sin ti 6 cos tj
Tt
rc t
rc t
Tc t
cos ti sin tj, Tc t
t
2i et j e t k
et e t
2 e t et
k
3
>i 2 j 3k@
63 2
2ti et j e t k , t
2t
32
§ 3S ·
N¨ ¸
© 4 ¹
30. r t
3S
4
sin ti cos tj
Tc 3S 4
Tc 3S 4
1
2
2
i j
2
2
cos 3ti 2 sin 3tj k , t
rc t
3 sin 3ti 6 cos 3tj
Tt
rc t
rc t
S
3 sin 3ti 6 cos 3tj
9 sin 2 3t 36 cos 2 3t
The normal vector is perpendicular to T t and points
toward the z-axis:
6 cos 3ti 3 sin 3tj
Nt
9 sin 2 3t 36 cos 2 3t
NS
6i
36
i
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
31. r t
4ti
vt
4i
at
0
35. r t
at
Tt
vt
vt
Tc t
0
Nt
Tc t
Tc t
4i
4
Tt
i
T1
4i 2 j
at
0
vt
vt
Tc t
0
Nt
Tc t
Tc t
2t
Nt
is undefined.
N1
a˜T
aN
a˜N
36. r t
8ti
at
8i
vt
vt
Tc t
0
Nt
Tc t
Tc t
8ti
8t
vt
2tj
at
2j
Tt
vt
vt
Tc t
0
Nt
Tc t
Tc t
2i
vt
1
4t 4
vt
2
1
i j
2
1
2ti 2 j
t 1
2
ti j
2
2
i j
2
2
t2 1
Tc t
Tc t
32
i t
t2 1
32
j
1
t2 1
1
t2 1
2tj
2t
j
2i 2 j
is undefined.
t jk
32
i t2j
1
2
t4 1
1
i
N1
34. r t
t 2i 2tj, t
2i, a 1
The path is a line and the speed is variable.
2t 3
2
at
Nt
i 2
2ti 2 j, v 1
T1
t 2i j
2
i j
2
vt
4t 2i
vt
1
i j
2
aT
t 1
2t
t4 1
t 1
Tt
Tt
t4 1
Tc t
Tc t
32
1
1
2i j
5
1
4
2
i j
2
4
The path is a line and the speed is constant.
33. r t
1 ·
§
¨ i 2 j¸
t
¹
t 1©
4
1
i j
2
117
i j,
t2
vt
vt
4ti 2tj
vt
Tt
1
1
ti j, v t
i 2 j, v 1
t
t
2
j, a 1
2j
t3
is undefined.
The path is a line and the speed is constant.
32. r t
Tangent Vectors and Normal Vectors
i tj
2
2
i j
2
2
aT
a˜T
2
aN
a˜N
2
j
is undefined.
The path is a line and the speed is variable.
© 2010 Brooks/Cole, Cengage Learning
118
Chapter 12
37. r t
Vector-Valued Functions
t t 3 i 2t 2 j, t
vt
1 3t 2 i 4tj, v 1
at
6ti 4 j, a 1
Tt
T1
2i 4 j
20
i 2 j
5
16t 3t 2 1
9t 4 10t 2 1
N1
2i j
5
40. r t
a˜N
e t i e t j tk , t
at
e i e j, a 0
T0
Tc t
Tc 0
vt
9t 4 10t 2 1
3t 4 i 2tj, v 0
6ti 2 j, a 0
Tt
vt
vt
T0
4i
16
8 5
5
2j
3t 4 i 2tj
2
9t 4 20t 2 16
i
4t 3t 2 4
9t 20t 16
4
j
32
Tc 1
32
j
163 2
N1
j
aT
a˜T
aN
a˜N
39. r t
4i
2
2 t
et i 4e 2t j, a 0
i 2j
5
N0
2i j
5
32
0
at
T0
9t 20t 2 16
2
e i 2e
vt
vt
32 18t 4
4
0
vt
Tt
i 1
j
2
et i e 2t j, t
t
32
j, v 0
e i 2e
t
i 2j
i 4j
2 t
j
4e 4t e 2t
aT
a˜T
1
18
5
7 5
5
aN
a˜N
1
2 4
5
6 5
5
i jk
i j
et i e t j k
e 2t e 2t 1
vt
i jk
3
e 2 t e 2t 2
e 4t e 2t 1
i 32
e 2 t 2e 2 t 1
e 4 t e 2t 1
j
32
et 1 e 4t
e 4t e 2t 1
32
k
3
3
i 32j
3
3
32
aT
2
2
i j
2
2
a˜T
0
aN
a˜N
N0
4
0
2
0
et i e t j k , v 0
t
4 36t
t 3 4t i t 2 1 j, t
Tc t
14 5
5
1
12 4
5
vt
Tt
5
i 2j
5
i 32
1
68
5
a˜T
t
at
9t 4 10t 2 1
64
32
i 32j
203 2
20
aN
vt
6i 4 j
Tc 1
aT
2i 4 j
1 3t 2 i 4tj
vt
vt
Tc t
38. r t
1
2
© 2010 Brooks/Cole, Cengage Learning
j
Section 12.4
et cos t i et sin t j
41. r t
42. r t
vt
et cos t sin t i et cos t sin t j
at
et 2 sin t i et 2 cos t j
S
At t
2
1
i j
2
v
v
, T
vt
2
i j .
2
Motion along r is counterclockwise. So,
N
1
i j
2
2
i j.
2
aT
a˜T
2eS
2
aN
a˜N
2eS
2
Tangent Vectors and Normal Vectors
119
a cos Z t i b sin Z t j
aZ sin Z t i bZ cos Z t j
v0
bZ j
at
aZ 2 cos Z t i bZ 2 sin Z t j
a0
aZ 2i
T0
v0
v0
j
Motion along r t is counterclockwise . So, N 0
aT
a˜T
0
aN
a˜N
aZ 2
i.
cos Z t0 Z t0 sin Z t0 i sin Z t0 Z t0 cos Z t0 j
43. r t0
Z 2t0 cos Z t0 i Z 2t0 sin Z t0 j
v t0
Z 2 ª¬ cos Z t0 Z t0 sin Z t0 i Z t0 cos Z t0 sin Z t0 jº¼
a t0
v
v
T t0
cos Z t0 i sin Z t0 j
Motion along r is counterclockwise. So
sin Z t0 i cos Z t0 j.
N t0
aT
a˜T
Z2
aN
a˜N
Z 2 Z t0
Z 3t0
Z t0 sin Z t0 i 1 cos Z t0 j
44. r t0
v t0
Z ª¬ 1 cos Z t0 i sin Z t0 jº¼
a t0
Z 2 ª¬ sin Z t0 i cos Z t0 jº¼
T
46. T t points in the direction that r is moving. N t points
in the direction that r is turning, toward the concave side
of the curve.
y
1 cos Z t0 i sin Z t0 j
v
v
a
T
2 1 cos Z t0
N
Motion along r is clockwise. So,
sin Z t0 i 1 cos Z t0 j
N
.
2 1 cos Z t0
aT
aN
45. r t
Z 2 sin Z t0
2 1 cos Z t0
a˜T
a˜N
Z2
2
1 cos Z t0
a cos Z ti a sin Z tj
vt
aZ sin Z ti aZ cos Z tj
at
aZ 2 cos Z ti aZ 2 sin Z tj
Tt
vt
vt
Nt
Tc t
Tc t
a
Z2
2
x
1 cos Z t0
aZ
47. Speed: v t
The speed is constant because aT
0.
48. If the angular velocity Z is halved,
aN
§Z ·
a¨ ¸
©2¹
2
aZ 2
.
4
aN is changed by a factor of
1
.
4
sin Z ti cos Z tj
cos Z ti sin Z tj
aT
a˜T
0
aN
a˜N
aZ 2
© 2010 Brooks/Cole, Cengage Learning
120
Chapter 12
Vector-Valued Functions
1
ti j, t0
2
t
1
x
t, y
Ÿ xy 1
t
1
rc t
i 2j
t
Tt
t 2i j
Nt
i t2j
t4 1
t 1
t i tj, t0
3
50. r t
N
1
)2, 12 )
rc t
T
1
rc t
rc t
x1 3
3t 2i j
9t 4 1
y
3i j
10
2
1
3 10
10
i j
10
10
(1, 1)
x
10
3 10
i j
10
10
4ti 4t 2 j, t0
51. r t
T
N
−1
N1
1
2
−1
1
4
rc t
2i 2tj
2
§1·
r¨ ¸
© 4¹
rc t
Tt
§1·
T¨ ¸
© 4¹
§ x·
4¨ ¸
© 4¹
2
x2
4
1
T
−2
1
i j
4
−1
1
−1
4i 8tj
i 1
1j
2
14
)1, )
x
i 2tj
2
N
−6
T
i tj
4 4t 2
1 t2
2i j
, perpendicular to T 2
5
2 cos ti 2 sin tj, t0
S
4
2 sin t Ÿ x y 2
2
rc t
2 sin ti 2cos tj
Tt
1
2 sin ti 2 cos tj
2
Nt
cos ti sin tj
§S ·
r¨ ¸
©4¹
2i §S ·
T¨ ¸
©4¹
2
i j
2
§S ·
N¨ ¸
©4¹
2
i j
2
4
sin ti cos tj
y
T
2j
1
(
2,
2)
N
x
−1
3 cos ti 2 sin tj, t0
cos t ,
y
2
1
−1
S
x2
y2
9
4
sin t Ÿ
1, Ellipse
3 sin ti 2 cos tj
rc S
2 j Ÿ T S
NS
i
y
3
j
1
N
T
4i 8tj
16 64t
2
1
4
6
−8
2i 2tj
2 cos t , y
4
(5, − 4)
−4
N2
rc t
N
x
2
2
i 2j
5
x
3
3
2
T2
54. r t
y
4t ,
4t 2
§ x 1·
¨
¸
© 2 ¹
5i 4 j
x
y 3 or y
2
−6 − 4 − 2
r2
53. r t
3
3t i j
T1
y
t 2
2
Tt
x
x
2
1
t Ÿ x
t3, y
y
2
17
i 4j
17
N2
x
3
17
4i j
17
T2
2t 1,
y
1
j
2
2i r2
x
Tt
4
y
2t 1 i t 2 j, t0
52. r t
49. r t
−2 −1
x
−1
1
2
−3
1 4t 2
2i j
5
2 5
5
i j
5
5
§1·
§1·
N¨ ¸ is perpendicular to T¨ ¸:
4
© ¹
© 4¹
§1·
N¨ ¸
© 4¹
i 2 j
5
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
55. r t
ti 2tj 3tk , t
vt
i 2 j 3k
at
0
1
57. r t
v
v
Tt
14
i 2 j 3k
14
T1
Tc
is undefined.
Tc
at
cos ti sin tj, a
4i 4 j 2k
at
0
aT
at ˜Tt
§S ·
a¨ ¸
©3¹
4ti 4tj 2tk
vt
vt
vt
§S ·
T¨ ¸
©3¹
3
1
sin 2 t cos 2 t 4
Tt
aN
aT , aN are not defined.
56. r t
sin ti cos tj 2k
121
S
cos ti sin tj 2tk , t
vt
vt
1
i 2 j 3k
14
Nt
Tangent Vectors and Normal Vectors
5
1
sin ti cos tj 2k
5
0
a 2 a 2T
10
1
·
1 §
3
1
i j 2k ¸¸
¨¨ 2
5© 2
¹
1
3
j
i 2
2
aTT an N
N
1
2i 2 j k
3
Tt
v
v
Nt
Tc
is undefined.
Tc
aT , aN are not defined.
58. r t
3ti tj t 2k , t
1
vt
3i j 2tk , v t
at
2k , a
Tt
vt
vt
a 1
2k , T 1
2
3i j 2tk
aT
a 1 ˜ T 1
aN
a 2 a 2T
a
N
2k
10 4t 2
10 4t 2
aTT aN N
1
3i j 2k
14
4
14
4
16
14
20
7
2 35
7
4 ª 1
º 2 35
N
3i j 2k » 7
14 «¬ 14
¼
7 ª
2
º
2k 3i j 2k »
7
2 35 «¬
¼
35 ª 6
2
10 º
i j k»
10 «¬ 7
7
7 ¼
© 2010 Brooks/Cole, Cengage Learning
122
Chapter 12
Vector-Valued Functions
ti t 2 j 59. r t
t2
k, t
2
et cos t et sin t i et sin t et cos t j et k
i 2j k
v0
i jk
2j k
at
2et cos ti 2et sin tj et k
a0
2i k
i 2tj tk
v1
at
1
v
v
i 2tj tk
1 5t 2
1 5t 2
32
Nt
Tc
Tc
N1
30
5i 2 j k
30
a˜T
5 6
6
aN
a˜N
30
6
vt
2i 2tj 4k ,
vt
20 4t 2
2 j, a
Tt
2i 2tj 4k
a2
2 j, T 2
aT
a2 ˜T2
N
a
2j
2
1
5 t2
i tj 2k
Nt
1
ª sin t cos t i cos t sin t jº¼
2¬
N0
2
2
i j
2
2
aT
a˜T
3
aN
a˜N
2
16
9
2
3
et i 2 j e t k , v t
at
et i e t k , a t
v0
i 2j k, a 0
T0
i 2j k
6
a˜T
i k
4
2
i 2 j 2k 9
3
e 2t 4 e 2t
e 2t e 2t
i k
0
a 2 aT2
Ÿ N
5
0
vt
a
4
et i 2tj e t k , t
aN
4
3
aTT aN N
1
>i j k@
3
aT
1
i 2 j 2k
3
a 2T
T0
62. r t
2
2 5 t2
aN
2
2 5 t2
at
a
5 1 5t 2
2t 1 i t 2 j 4tk , t
60. r t
1
ª cos t sin t i sin t cos t j k º¼
3¬
5ti 2 j k
5
1 5t 2
aT
v
v
Tt
6
i 2j k
6
5ti 2 j k
T1
et sin ti et cos tj et k
vt
vt
Tt
61. r t
1
2
aTT aN N
2N
1
i k
2
5N
3 ª
4
º
2 j i 2 j 2k »
9
2 5 «¬
¼
5
4 5
2 5
i j
k
15
3
15
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
S
4ti 3 cos tj 3 sin tk , t
63. r t
65. r t
2
4i 3 sin tj 3 cos tk
vt
§S ·
v¨ ¸
©2¹
4i 3 j
3 cos tj 3 sin tk
at
3k
Tt
v
v
§S ·
T¨ ¸
©2¹
at
6j k
2π
k
3
0
aN
a˜N
3
1
S
1
j k
3
aT
aN
cos ti sin tj
66. r t
aS
i
Tt
v
v
T S
3 10 §
1 ·
¨j k¸
10 ©
3 ¹
3 10 §
1 ·
¨ sin ti cos tj k ¸
10 ©
3 ¹
NS
i
3 10
cos ti sin tj
10
3 10
10
a˜T
aT
a˜N
0, aN
z
cos t i sin tj
a˜T
a˜N
2i 2k
at
2i
Tt
v
v
T1
1
i k
2
Nt
Tc
Tc
N1
4
1
i tk
2
t2 1
t2 1
32
t 1
32
2
i tk
t2 1
1
i k
2
aT
a˜T
aN
a˜N
2
2
T
x
3
2
1
y
ti k
2ti 2k
z
4
2
N
8
1
1
2
2
6
37
149
4t 2 4
1t
2
T
t 2i j 2tk , t
v1
2
4
x
37
5513
2ti 2k
z
4
N
74
149
vt
1
>37ti 6 j k@
1
>74i 6 j k@
37 149
1
74i 6 j k
5513
N2
10
3
6 j k@
37
1 37 2
x
1
sin ti cos tj k , v t
3
Tc
Tc
>37ti
37 1 37t 2
t
2 cos t i 1 sin t j k , t
3
Nt
32
y
4π
a˜T
at
Tc
Tc
1 37t 2
3
i 6tj tk
1
i 12 j 2k
149
1
N
cos tj sin tk
1
1 37t 2
z
T
aT
vS
i 12 j 2k
Nt
1
4i 3 j
5
§S ·
N¨ ¸
©2¹
vt
v2
T2
1
4i 3 sin tj 3 cos tk
5
Tc
Tc
64. r t
i 6tj tk
v
v
123
t2
k
2
ti 3t 2 j vt
Tt
§S ·
a¨ ¸
©2¹
Nt
Tangent Vectors and Normal Vectors
3
2
2
T
N
1
2
5
x
4
3
2
1
2
3
y
y
© 2010 Brooks/Cole, Cengage Learning
124
Chapter 12
Vector-Valued Functions
67. Let C be a smooth curve represented by r on an open
interval I. The unit tangent vector T t at t is defined as
68. The unit tangent vector points in the direction of motion.
69. (a) If aN
rc t
, rc t z 0.
rc t
Tt
(b) If aT
70. r t
The principal unit normal vector N t at t is defined as
Tc t
, Tc t z 0.
Tc t
Nt
0, then the motion is in a straight line.
The tangential and normal components of acceleration
are defined as a t
a TT t aN N t .
0, then the speed is constant.
3ti 4tj
vt
rc t
3i 4 j, v t
at
vc t
0
Tt
vt
vt
Tc t
0 Ÿ N t does not exist.
9 16
5
3
4
i j
5
5
The path is a line. The speed is constant (5).
S t sin S t, 1 cos S t
71. r t
The graph is a cycloid.
(a) r t
S t sin S t, 1 cos S t
vt
S S cos S t, S sin S t
at
S 2 sin S t, S 2 cos S t
Tt
vt
vt
Nt
Tc t
Tc t
aT
a˜T
aN
a˜N
1
1
2 1 cos S t
1
2 1 cos S t
S2
When t
1: aT
0, aN
(b) Speed:
3
: aT
2
s
ª¬S 2 sin 2 S t S 2 cos S t 1 cos S t º¼
2 1 cos S t
1
: aT
2
2S 2
, aN
2
2
When t
1
: aT
2
When t
1: aT
When t
3
: aT
2
2S
2
S 2 1 cos S t
2 1 cos S t
S2
2 1 cos S t
2
2
t=1
t = 23
x
2 1 cos S t
S 2 sin S t
2 1 cos S t
ds
dt
S 2 sin S t
2 1 cos S t
y
t = 21
2S
, aN
2
S
2S 2
2
S2
2
vt
sin S t, 1 cos S t
ª¬S 2 sin S t 1 cos S t S 2 cos S t sin S tº¼
1
When t
When t
1 cos S t, sin S t
2 1 cos S t
aT
2S 2
! 0 Ÿ the speed in increasing.
2
0 Ÿ the height is maximum.
2S 2
0 Ÿ the speed is decreasing.
2
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
vt
S sin S t S sin S t S 2t cos S t, S cos S t S cos S t S 2t sin S t
at
S 2 cos S t S 3t sin S t, S 2 sin S t S 3t cos S t
vt
Tt
aT
cos S t S 2 cos S t S 3t sin S t sin S t S 2 sin S t S 3t cos S t
a˜T
Tt
Nt
S 2 , aN
1, aT
S 3 . When t
t
k , t0
2
1
2 sin ti 2 cos tj k
2
2 cos ti 2 sin tj S
2j §S ·
T¨ ¸
©2¹
2 17 §
1 ·
¨ 2i k ¸
17 ©
2 ¹
§S ·
N¨ ¸
©2¹
j
4
rc t
Tt
Nt
r1
T1
N1
B1
2S 3 .
1 and t
2.
S
2
z
3
2
B
1
−2
−1
T
−2
N
( 0, 2, π2 )
1
2
17
4i k
17
§S ·
§S ·
T¨ ¸ u N¨ ¸
©2¹
©2¹
t3
k , t0
3
1
ti t 2 j y
x
k
i
§S ·
B¨ ¸
©2¹
S 2 , aN
2, aT
2 17 §
1 ·
¨ 2 sin ti 2 cos tj k ¸
17 ©
2 ¹
cos ti sin tj
§S ·
r¨ ¸
©2¹
S2
S 3t
S 2 ! 0 for all values of t, the speed is increasing when t
(b) Because aT
rc t
S 4 1 S 2t 2 S 4
a 2 aT 2
When t
S 2t cos S t, S 2t sin S t
cos S t, sin S t
vt
aN
73. r t
125
cos S t S t sin S t, sin S t S t cos S t
72. (a) r t
74. r t
Tangent Vectors and Normal Vectors
j
4 17
17
0
0
1
k
17
17
0
17
4 17
i k
17
17
17
i 4k
17
i 2tj t 2k
1
1 4t 2 t 4
i 2tj t 2k
1
ª 2t t 3 i 1 t 4 j t 2t 3 k º
¼
1 4t t 4 1 t 2 t 4 ¬
z
1
i j k
3
1
2
N
1
i 2j k
B
6
2
1
3i 3k
6 3
T1 u N1
2
i k
2
i
j
k
6
6
6
3
6
6
2
2
0
2
2
x
1
2
T
−1
2
( 1, 1, 31 )
3
3
3
i j
k
3
3
3
1
y
3
i jk
3
© 2010 Brooks/Cole, Cengage Learning
126
Chapter 12
Vector-Valued Functions
S
i sin tj cos tk , t0
75. r t
rc t
4
cos tj sin tk ,
rc t
1
§S ·
rc¨ ¸
©4¹
§S ·
T¨ ¸
©4¹
Tc t
2
2
j
k
2
2
sin tj cos tk ,
§S ·
N¨ ¸
©4¹
§S ·
B¨ ¸
©4¹
2
2
j
k
2
2
§S ·
§S ·
T¨ ¸ u N¨ ¸
©4¹
©4¹
i
j
0
2
2
2
2
0 2
2
2
2
2et i et cos tj et sin tk , t0
76. r t
k
i
0
rc t
2et i et cos t et sin t j et sin t et cos t k
rc 0
2i j k Ÿ T 0
rc t
2
1
2i j k
6
4e2t e 2t cos 2 t e 2t sin 2 t 2e2t cos t sin t e2t sin 2 t e 2t cos 2 t 2e 2t sin t cos t
4e2t 2e 2t cos 2 t sin 2 t
rc t
Tt
6e 2 t
6e t
rc t
rc t
1
ª¬2i cos t sin t j sin t cos t k º¼
6
Tc t
1
ª¬ sin t cos t j cos t sin t k º¼
6
Tc 0
1
> j k@ Ÿ N 0
6
B0
T0 uN0
2
j
2
i
j
k
2
6
1
6
1
6
0 2
2
2
2
2
k
2
3
3
3
i j
k
3
3
3
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
77. r t
rc t
3
4 cos ti 4 sin tj 2k ,
16 cos 2 t 16 sin 2 t 4
§S ·
rc¨ ¸
©3¹
2i 2 3j 2k
§S ·
T¨ ¸
©3¹
1
2i 2 3 j 2k
2 5
Tc t
1
2 5
20
5
i 5
5
5
§S ·
§S ·
T¨ ¸ u N¨ ¸
©3¹
© 3¹
3 cos 2ti 3 sin 2tj tk , t
rc t
6 sin 2ti 6 cos 2tj k
rc t
37
§S ·
rc¨ ¸
©4¹
6i k
§S ·
T¨ ¸
©4¹
Nt
rc t
rc t
j
k
15
5
5
5
1
2
0
3
2
78. r t
Tc t
15
5
j
k
5
5
5
i 5
3j k
3
1
i j
2
2
Tt
2 5
4 sin ti 4 cos tj
i
§S ·
B¨ ¸
©3¹
127
S
4 sin ti 4 cos tj 2tk , t0
rc t
§S ·
N¨ ¸
©3¹
Tangent Vectors and Normal Vectors
5
15
4 5
i j
k
10
10
10
5
i 10
3 j 4k
S
4
1
6 sin 2ti 6 cos 2tj k
37
1
12 cos 2ti 12 sin 2tj
37
1
6i k
37
Tc t
Tc t
cos 2ti sin 2tj
§S ·
N¨ ¸
©4¹
j
§S ·
B¨ ¸
©4¹
§S ·
§S ·
T¨ ¸ u N¨ ¸
4
© ¹
©4¹
i
6
37
0
j
0
1
k
1
37
0
1
i 37
6
k
37
© 2010 Brooks/Cole, Cengage Learning
128
Chapter 12
Vector-Valued Functions
79. From Theorem 12.3 you have:
rt
v0t cos T i h v0t sin T 16t 2 j
vt
v0 cos T i v0 sin T 32t j
at
32 j
v0 cos T i v0 sin T 32t j
Tt
v02 cos 2 T v0 sin T 32t
2
v0 sin T 32t i v0 cos T j
Nt
v02 cos 2 T v0 sin T 32t
32 v0 sin T 32t
aT
a˜T
aN
a˜N
v02
32v0 cos T
v02 cos 2 T v0 sin T 32t
At maximum height, aT
45q, v0
v0 cos T
v0 sin T 32t
0; vertical component of velocity
0 and aN
32.
2
2
150 ˜
75 2
2
32t
2
75 2 32t
32 75 2 32t
16 32t 75 2
11250 75 2 32t
2
256t 2 1200 2t 5625
2
256t 1200 2t 5625
32 75 2
1200 2
aN
11250 75 2 32t
At the maximum height, aT
81. (a) r t
2
150
150 ˜
aT
2
cos 2 T v0 sin T 32t
Maximum height when v0 sin T 32t
80. T
Motion is clockwise.
2
0 and aN
v0 cos T ti ª¬h v0 sin T t 2
32.
1 gt 2 º j
¼
2
120 cos 30q ti ª¬5 120 sin 30q t 16t 2 º¼ j
(b)
60 3ti ª¬5 60t 16t 2 º¼ j
70
0
400
0
Maximum height | 61.25 feet
range | 398.2 feet
(c) v t
Speed
at
60 3i 60 32t j
vt
3600 3 60 32t
2
8 16t 2 60t 225
32 j
© 2010 Brooks/Cole, Cengage Learning
Section 12.4
(d)
T
0.5
1.0
1.5
2.0
2.5
3.0
Speed
112.85
107.63
104.61
104.0
105.83
109.98
(e) From Exercise 79, using v 0
120 and T
Tangent Vectors and Normal Vectors
129
30q,
32 60 32t
aT
60 3
2
60 32t
2
32 60 3
aN
60 3
2
60 32t
2
40
aN
0
4
aT
−20
At t
1.875, aT
0 and the projectile is at its maximum height. When a T and a N have opposite signs, the speed is
decreasing.
v0 cos T ti ª¬h v0 sin T t 82. (a) r t
1 gt 2 º j
2
¼
220 cos 45q ti ª¬4 220 sin 45q t 16t 2 º¼ j
110 2ti ª¬4 110 2t 16t 2 º¼ j
(b)
450
0
−10
1800
Maximum height | 382.125 at t | 4.86
Range | 1516.4
110 2i ª¬110 2 32t º¼ j
(c) v t
vt
110 2
2
t
0.5
1.0
1.5
2.0
2.5
3.0
Speed
208.99
198.67
189.13
180.51
172.94
166.58
10 cos 10S t, 10 sin 10S t, 4 4t , 0 d t d
83. r t
(a)
110 2 32t
32 j
at
(d)
2
rc t
rc t
(b) aT
aT
1
20
100S sin 10S t , 100S cos 10S t , 4
100S
2
sin 2 10S t 100S
100S
2
16
0 and aN
2
cos 2 10S t 16
4 625S 2 1 | 314 mi h
1000S 2
0 because the speed is constant.
© 2010 Brooks/Cole, Cengage Learning
130
Chapter 12
84. 600 mi h
Vector-Valued Functions
880 ft sec
rt
880ti 16t 2 36,000 j
vt
880i 32tj
at
32 j
55i 2tj
16 4t 3025
a˜N
mv 2
r
4t 3025
2
4t 2 3025
0 and
2w. Then
aZ02
a˜N
a 2Z
2
m 2 2
rZ
r
mv 2
r
4aZ 2
GM
,v
r
GM
r
87. v
GM
r
9.56 u 104
| 4.82 mi sec
4000 115
88. v
GM
r
9.56 u 104
| 4.75 mi sec
4000 245
89. v
9.56 u 104
| 4.67 mi sec
4385
aZ 2 .
(a) Let Z0
m rZ 2
GMm 2
,v
r2
1760
From Exercise 45, we know a ˜ T
a˜N
m at
(b) By Newton’s Law:
a cos Z t i a sin Z t j
85. r t
v
rZ 2
(a) F
64t
aN
rZ
rZ 2 cos Z t i rZ cos Z t j
at
4t 2 3025
Motion along r is clockwise, therefore
2ti 55 j
Nt
4t 2 3025
a˜T
rZ 1
at
2
aT
rZ sin Z t i rZ cos Z t j
vt
vt
880i 32tj
Tt
r cos Z t i r sin Z t j
86. r t
or the centripetal acceleration is increased by a factor
of 4 when the velocity is doubled.
a 2. Then
(b) Let a0
§a· 2
¨ ¸Z
©2¹
a0Z 2
a˜N
§1· 2
¨ ¸ aZ
© 2¹
or the centripetal acceleration is halved when the
radius is halved.
90. Let x
distance from the satellite to the center of the earth x
v
4S 2 x 2
24
2
2
3600
x3
v |
2S x
t
2S x
24 3600
r 4000 . Then:
9.56 u 104
x
9.56 u 104
x
9.56 u 104 24
2
3600
4S 2
2
Ÿ x | 26,245 mi
2S 26,245
| 1.92 mi sec | 6871 mi h
24 3600
91. False. You could be turning.
92. True. All the motion is in the tangential direction.
cosh bt i sinh bt j, b ! 0
93. (a) r t
x
cosh bt , y
x y
2
(b) v t
at
2
sinh bt
cosh bt sinh 2 bt
2
1, hyperbola
b sinh bt i b cosh bt j
b 2 cosh bt i b 2 sinh bt j
b 2r t
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
cos Ii sin I j be the unit tangent vector.
94. Let T t
vua
sin Ii cos I j
dI
dt
M
v aT T u T v aN T u N
v aN T u N
vua
sin Ii cos I j
v aN T u N
97.
If dI dt ! 0, then the curve bends to the left and
M has the same direction as Tc. y
So,p M has the same
direction as
a
aTT aN N ˜ aTT aN N
aT 2 T
Because aN ! 0, we have aN
x
98. F
ma
T
M
again points to the
concave side of the curve.
N
x
dv
dt
x
at bt 2 ct 3
v
dx
dt
φ
a 2 aT 2 .
dv
Force
dt
a 2bt 3ct 2
dv
dt
2b 6ct
F2
4b 2 24bct 36c 2t 2
xt i yt j
4b 2 12c 2bt 3ct 2
b, m and b are constants.
yt
m xt
rt
x t i ª¬m x t
vt
xc t i mxc t j
4b 2 12c v a
bº¼ j
2
ª¬ xc t º¼ ª¬mxc t º¼
r i mj
vt
vt
1
y
Tc
Tc
So, Tc t
2
M
M has the opposite direction
as Tc. Thus,
Tt
2aT aN T ˜ N aN 2 N
a 2 aT 2
aN 2
If dI dt 0, then the curve bends to the right and
vt
2
aT 2 aN 2
φ
which is toward the
concave side of the curve.
95. r t
a˜a
2
T
Tc
,
Tc
v aN
vua
.
v
So, aN
and is rotated counterclockwise through an angle of
S 2 from T.
N
0, and T u N
v T u aTT aN N
dI
.
dt
cos ª¬I S 2 º¼ i sin ª¬I S 2 º¼ j
N
131
you have:
dT
dt
d T dI
dI dt
Tc t
M
aTT aN N, T u T
96. Using a
Then
Arc Length and Curvature
1 m2
F
2
f v
r 4b 2 12ac 12cv
The sign of the radical is the sign of 2b 6ct , which
cannot change.
xc t
1 m2
, constant
0.
Section 12.5 Arc Length and Curvature
3ti tj, >0, 3@
1. r t
dx
dt
s
3,
3
³0
dy
dt
1,
3
dz
dt
3 1
2
y
2
0
(0, 0)
x
dt
3
ª 10t º
¬
¼0
6
3 10
−3
9
(9, − 3)
−6
© 2010 Brooks/Cole, Cengage Learning
1,
132
Chapter 12
Vector-Valued Functions
y
ti t 2 j
2. r t
(4, 16)
16
dx
dt
s
dy
1,
dt
dz
2t ,
dt
0
12
8
4
³0
1 4t 2 dt
4
4
1ª
2t 1 4t 2 ln 2t 4¬
s
dy
dt
dz
dt
0
9t 4 4t 2 dt
³0
3t 2 ,
1
³0
2t ,
1,
dy
dt
6
³0
1
dx
dt
s
1ª 2
9t 4
27 ¬«
32 1
º
¼» 0
1
133 2 8 | 1.4397
27
4t 2 1 2t S 2
12a ³
6
º
4t 2 1»
¼0
1
t
2
40
35
30
25
20
15
10
5
1
ln
4
sin t cos t dt
(1, 0)
x
145 12 3 145
−1
3a sin 2 t cos t
1 2 3 4 5 6 7 8
a
2
3a ³
S 2
0
2 sin 2t dt
x
−a
>3a cos 2t@S0 2
a
−a
6a
a cos ti a sin tj
6. r t
(7, 36)
y
2
S 2
0
dy
dt
1
y
ª¬3a cos 2 t sin t º¼ ª¬3a sin 2 t cos t º¼ dt
0
4
x
(0, 0)
1 4t 2 dt
3a cos 2 t sin t ,
4³
3
9t 2 4 t dt
a cos3 ti a sin 3 tj
5. r t
2
(1, 1)
2t
ª1
« 4 ln
¬
1
1
t 1 i t 2 j, 0 d t d 6
4. r t
s
x
(0, 0)
y
12
1 1 2
9t 4 18t dt
18 ³ 0
dx
dt
65 º | 16.819
¼
t 3i t 2 j, >0, 1@
3. r t
dx
dt
1ª
8 65 ln 8 4¬
1 4t º
¼0
y
a
dx
dt
s
a sin t ,
2S
³0
7. (a) r t
dy
dt
a cos t
a 2 sin 2 t a 2 cos 2 t dt
2S
³0
a dt
2S
>at@0
a
x
1 º
ª
v0 cos T ti «h v0 sin T t gt 2 » j
2 ¼
¬
1
ª
º
100 cos 45q ti «3 100 sin 45q t 32 t 2 » j
2
¬
¼
(b) v t
2S a
50 2ti ª¬3 50 2t 16t 2 º¼ j
50 2i 50 2 32t j
50 2 32t
0 Ÿ t
25 2
16
§ 25 2 ·
§ 15 2 ·
Maximum height: 3 50 2 ¨¨
¸¸ 16¨¨ 16 ¸¸
16
©
¹
©
¹
2
81.125 ft
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
(c) 3 50 2t 16t 2
Arc Length and Curvature
133
0 Ÿ t | 4.4614
Range: 50 2 4.4614 | 315.5 feet
4.4614
³0
(d) s
2
50 2
2
50 2 32t
dt | 362.9 feet
1 º
ª
v0 cos T ti « v0 sin T t gt 2 » j
2 ¼
¬
8. (a) r t
1 2
gt
2
yt
v0 sin T t yc t
v0 sin T gt
Maximum height when sin T
Range: x t
S
1, or T
2
2v0 sin T
g
§ 2v sin T ·
v0 cos T ¨ 0
¸
g
©
¹
v0 2
sin 2 T
g
The range x t is a maximum for sin 2T
(c) xc t
S
1, or T
4
.
v0 cos T
yc t
xc t
.
0 Ÿ t
1 2
gt
2
v0 sin T t (b) y t
v0 sin T
.
g
0 when t
v0 sin T gt
2
yc t
2
v0 2 cos 2 T v0 sin T gt
2 v0 sin T g
³0
sT
Because v0
v0 2 cos 2 T v0 2 sin 2 T 2v0 2 g sin T t g 2t 2
v0 2 2v0 g sin T t g 2t 2
12
ª¬v0 2 2v0 g sin T t g 2t 2 º¼
dt
96 ft sec, you have
6 sin T
³0
sT
2
12
ª¬962 6144 sin T t 1024t 2 º¼ .
dt
Using a computer algebra system, s T is a maximum for T | 0.9855 | 56.5q.
ti 4tj 3tk , >0, 1@
9. r t
dx
dt
s
1,
1
³0
dy
dt
4,
dz
dt
z
4
3
3
2
ª 26t º
¬
¼0
1 16 9 dt
26
3
x
s
2
−3
(−1, 4, 3)
1
−1
−2
1
2
3
4
5
y
i t 2 j t 3k , >0, 2@
10. r t
dx
dt
−2
(0, 0, 0)
1
0,
dy
dt
2t ,
dz
dt
2
4t 2 9t 4 dt
2
4 9t 2 t dt
³0
³0
z
8
7
6
5
4
3
2
1
3t 2
1
4 9t 2
27
3 2º
2
»
¼0
1
403 2 43 2
27
1ª
80 10 8º¼
27 ¬
(1, 0, 0)
3
4
(1, 4, 8)
−4
5 6
y
x
© 2010 Brooks/Cole, Cengage Learning
Chapter 12
134
Vector-Valued Functions
ª 3S º
4t , cos t , sin t , «0, »
¬ 2¼
11. r t
dx
dt
dy
dt
4,
3S 2
³0
s
dz
dt
sin t ,
cos t
3S 2
³0
16 sin 2 t cos 2 t dt
3S
2
3S 2
ª 17t º
¬
¼
17 dt
0
17
z
(0, − 1, 0)
− 12
21
x
−9
9
6
−6
6
−3
−6
−9
(6 π , 0, − 1) − 12
18
15
12
dx
dt
2 cos t ,
S
³0
dx
dt
2S
³0
dx
dt
dz
dt
2 sin t
S
³0
29S
29 dt
dy
dt
a cos t ,
dy
dt
t cos t ,
(a, 0, 2π b)
dz
dt
b
S 2
³0
S 2
2
t cos t
2S
a2 b2
(0, 5 ␲, − 2)
2π b
x
(a, 0, 0)
y
z
2t
2
2
2t
2
(π2 , 1, π4 )
2
(1, 0, 0)
dt
1
2
2
2 S 2
5S 2
8
t º
5 »
2 ¼0
5t 2 dt
−4
−6
3
dz
dt
t sin t
4 6 8
10 12 14
πb
2S
t sin t ,
− 10
−8
−6
z
ª a2 b2 tº
¬
¼0
a 2 b 2 dt
−4
y
x
a 2 sin 2 t a 2 cos 2 t b 2 dt
³0
3
y
3
x
t 2i tj ln tk
15. r t
dx
dt
2t ,
dy
dt
1,
dz
dt
1
t
2
3
³1
2t
2
2
§1·
1 ¨ ¸ dt
©t ¹
3
³1
4t 4 t 2 1
dt
t2
3
³1
4t 4 t 2 1
dt | 8.37
t
sin S ti cos S tj t 3k
16. r t
s
5,
cos t t sin t , sin t t cos t , t 2
14. r t
dx
dt
dy
dt
a sin t ,
2S
s
z
12
10
8
6
4
(0, 0, 2)
4 cos 2 t 25 4 sin 2 t dt
³0
s
y
a cos ti a sin tj btk
13. r t
s
9
2 sin t , 5t , 2 cos t , >0, S @
12. r t
s
6
S cos S t ,
2
³0
dy
dt
S cos S t
S sin S t ,
2
dz
dt
S sin S t
2
3t 2
2
3t 2 dt
2
³0
S 2 9t 4 dt | 11.15
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
ti 4 t 2 j t 3k , 0 d t d 2
17. r t
(a) r 0
0, 4, 0 , r 2
2 4 8
2
distance
2
(a) r 0
2
84
(b) r 0
6i
6, 0, 0
2 j 2k
r2
2 21 | 9.165
(b) r 0
1, 3, 1
r 1.5
1.5, 1.75, 3.375
r2
44
2 11 | 6.633
0.5, 3.75, 0.125
r1
0, 2, 2
6 2 22 2 2
distance
0, 4, 0
r 0.5
135
§ St ·
§ St ·
6 cos¨ ¸i 2 sin ¨ ¸ j tk , 0 d t d 2
©4¹
©4¹
18. r t
2, 0, 8
Arc Length and Curvature
2, 0, 8
distance |
0.5
0.5
0.5
0.5
2
0.25
2
0.125
2
2
0.75
2
0.875
2
2
1.25
2
2.375
2
2
1.75
2
4.625
2
6, 0, 0
r 0.5
5.543, 0.765, 0.5
r 1.0
4.243, 1.414, 1.0
r 1.5
2.296, 1.848, 1.5
r 2.0
0, 2, 2
distance | 6.9698
(c) Increase the number of line segments.
(d) Using a graphing utility, you obtain
s
2
³0
rc t dt | 7.0105.
| 0.5728 1.2562 2.7300 4.9702
| 9.529
(c) Increase the number of line segments.
(d) Using a graphing utility, you obtain 9.57057.
19. r t
2 cos t , 2 sin t , t
t
s
5
(b)
2
³0
(a) s
2
t
2
³0
ª¬ xc u º¼ ª¬ yc u º¼ ª¬ zc u º¼ du
2 sin u
§ s ·
2 cos¨
¸, y
© 5¹
§ s ·
2 sin ¨
¸, z
© 5¹
§ s ·
§ s ·
2 cos¨
¸i 2 sin ¨
¸j © 5¹
© 5¹
r s
(c) When s
1
2
du
t
³0
5 du
t
ª 5u º
¬
¼0
5t
5: x
2 cos 1 | 1.081
y
2 sin 1 | 1.683
z
1
s
k
5
When s
2
§ 2
§
§ 2
§ s ··
sin
cos¨¨
¨
¨
¸ ¸ ¨¨
5
5
5
©
¹
©
¹
©
©
rc s
(a) s
2
s
5
1.081, 1.683, 1.000
20. r t
2 cos u
t
x
(d)
2
4
| 0.433
5
4
y
2 sin
| 1.953
5
4
z
| 1.789
5
0.433, 1.953, 1.789
4: x
2 cos
2
2
s ··
§ 1 ·
¸ ¸¸ ¨
¸
¸
5 ¹¹
© 5¹
4 1
5 5
1
3
4 sin t t cos t , 4 cos t t sin t , t 2
2
t
³0
t
³0
2
2
2
ª¬ xc u º¼ ª¬ yc u º¼ ª¬ zc u º¼ du
4u sin u
2
4u cos u
2
3u
2
du
t
³0
16u 9u 2 du
t
³ 0 5u du
5 2
t
2
© 2010 Brooks/Cole, Cengage Learning
136
Chapter 12
Vector-Valued Functions
2s
5
(b) t
x
§
4¨¨ sin
©
2s
5
2s
cos
5
2s ·
¸
5 ¸¹
y
§
4¨¨ cos
©
2s
5
2s
sin
5
2s ·
¸
5 ¸¹
z
3§
¨
2 ¨©
2s ·
¸
5 ¸¹
2
§
4¨¨ sin
©
r s
3s
5
2s
5
(c) When s
2s
cos
5
§
2s ·
¸¸i 4¨¨ cos
5 ¹
©
2s
5
2s ·
3s
¸¸ j k
5 ¹
5
2s
sin
5
5:
x
§
4¨ sin
¨
©
2 5
5
2 5
cos
5
2 5·
¸ | 1.030
5 ¸
¹
y
§
4¨ cos
¨
©
2 5
5
2 5
sin
5
2 5·
¸ | 5.408
5 ¸
¹
z
3 5
| 1.342
5
1.030, 5.408, 1.342
When s
4:
§
4¨¨ sin
©
x
y
z
8
5
8
cos
5
8·
¸ | 2.291
5 ¸¹
§
8
4¨¨ cos
5
©
12
2.4
5
8
sin
5
8·
¸ | 6.029
5 ¸¹
2.291, 6.029, 2.400
(d)
21. r s
rc s
Ts
Tc s
22. r s
2
§4
¨¨ sin
©5
rc s
§4
2s ·
¸ ¨¨ cos
5 ¸¹
©5
2
2s ·
§ 3·
¸ ¨ ¸
5 ¸¹
©5¹
§
§
2 ·
2 ·
s ¸¸i ¨¨1 s¸j
¨¨1 2 ¹
2 ¸¹
©
©
2
2
i j and rc s
2
2
rc s
rc s
2
16
9
25 25
23. r s
1 1
2 2
Ts
0 The curve is a line.
K
3 si j
rc s
i and rc s
Ts
rc s
Tc s
0 Ÿ K
s
k
5
rc s
Tc s
Tc s
§ s ·
§ s ·
2 cos¨
¸i 2 sin ¨
¸j © 5¹
© 5¹
1
rc s
0 Ÿ K
1
2
2
1
§ s ·
§ s ·
k
sin ¨
cos¨
¸i ¸j 5
5
5
© 5¹
© 5¹
2
2 § s ·
§ s ·
cos¨
i sin ¨
j
¸
5
5 © 5 ¸¹
© 5¹
Tc s
2
5
1
Tc s
0 The curve is a line.
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
24. r s
§
4¨¨ sin
©
2s
5
2s
cos
5
Ts
rc s
4
sin
5
Tc s
4
25
5
cos
2s
K
4
25
Tc s
2s
4
i cos
5
5
2s
4
i 5
25
5
2s
4ti 2tj
25. r t
vt
§
2s ·
¸i 4¨¨ cos
5 ¸¹
©
2s
5
2s
sin
5
Arc Length and Curvature
2s ·
3s
¸j k
5 ¸¹
5
2s
3
j k
5
5
5
sin
2s
2s
j
5
2 10 s
25s
ti 28. r t
4i 2 j
1 3
t j
9
vt
1
i t2j
3
v2
i 0
at
2
tj
3
(The curve is a line.)
a2
4
j
3
Tt
1
i t2j
3
t4
1
9
T2
3i 4 j
5
N2
4i 3j
5
1
2i j
5
Tt
Tc t
0
Tc t
K
26. r t
rc t
t 2i j
vt
2ti
Tt
i
Tc t
0
K
27. r t
Tc t
0
rc t
1
ti j
t
K
1
j
t2
vt
i v1
i j
rc t
at
2
j
t3
§S ·
rc¨ ¸
©2¹
a1
2j
Tt
Nt
N1
K
1
i j
2
a˜N
v 2
2
2
i t2j
45
25 9
36
125
1 cos 2 t
1
§S ·
0, sin t , a¨ ¸
©2¹
1
1, 0
§S ·
N¨ ¸
©2¹
0, 1
a˜N
v 2
1
1
0, 1
1, cos t
1 cos 2 t
§S ·
T¨ ¸
©2¹
K
5
3
9 t4
§S ·
1, 0 , rc¨ ¸
©2¹
Tt
12
16
9
3i t 2 j
1, cos t , rc t
t 2i j
t4 1
1
t , sin t
at
1
4
j, v 2
2
a˜N
v 2
29. r t
t4 1
137
1
© 2010 Brooks/Cole, Cengage Learning
138
Chapter 12
30. r t
5 cos t , 4 sin t , t
xt
K =
Vector-Valued Functions
5 cos t , y t
S
34. r t
3
rc t
2
yc º
¼
ª¬25 sin 2 t 16 cos 2 t º¼
32
º
¼
Tt
sin 2 S ti cos 2tj
Tc t
2S cos 2S ti 2S sin 2S tj
rc t
rc t
Tt
Tc t
K
160 91
8281
2S
8S
rc t
4 sin 2 S t cos 2 S t
a˜N
K
4 sin 2 S t cos 2 S t
a˜N
K
rc t
4 sin 2 S t cos 2 S t
32
aZ sin Z ti aZ cos Z tj
sin Z ti cos Z tj
Tc t
Z cos Z ti Z sin Z tj
Tc t
Z
rc t
aZ
1 cos Z t
at ˜Nt
vt 2
2
4a 1 cos Z t
Z 3t
at ˜Nt
v 2
37. r t
Z 3t
Z 4t 2
rc t
i 2tj tk
Tt
i 2tj tk
Tc t
K
1
Zt
t2
k
2
ti t 2 j 1 5t 2
5ti 2 j k
1 5t 2
a cos Z ti a sin Z tj
Tt
K
aZ 2
˜
2
32
2S
4 sin 2 S t cos 2 S t
32
From Exercise 43, Section 12.4, we have:
2
33. r t
2
cos Z t Z t sin Z t , sin Z t Z t cos Z t
36. r t
2S cos S ti 4S sin S tj
4 sin 2 S t cos 2 S t
Z a 2 sin 2 Z t b 2 cos 2 Z t
From Exercise 44, Section 12.4, we have:
4 sin 2 S t cos 2 S t
S
rc t
§ aZ 2 ·
¨
¸ 1 cos Z t
© 2¹
2a 2Z 2 1 cos Z t
2 sin S ti cos S tj
rc t
abZ
a 2 sin 2 Z t b 2 cos 2 Z t
a Z t sin Z t , a 1 cos Z t
1
4
2S sin S ti S cos S tj
Tc t
32
Tc t
ª¬a sin Z t b 2 cos 2 Z t º¼
2 cos S ti sin S tj
S
ª¬a 2 sin 2 Z t b 2 cos 2 Z t º¼
2
35. r t
8S sin 2S ti 8S cos 2S tj
32. r t
K
4 cos 2S ti 4 sin 2S tj
Tc t
ab 2Z cos Z t i a 2bZ sin Z t j
ab
20
ª25 3 4 16 1 4
¬
Tc t
32
32
rc t
K
a 2 sin 2 Z t b 2 cos 2 Z t
2 32
ª¬25 sin 2 t 16 cos 2 t º¼
20
31. r t
a sin Z t i b cos Z t j
Tt
5 sin t 4 sin t 4 cos t 5 cos t
§S ·
K¨ ¸
©3¹
aZ sin Z t i bZ cos Z t j
4 sin t
xcycc ycxcc
ª xc
¬
a cos Z t i b sin Z t j
32
Tc t
5
1 5t 2
rc t
1 5t
5
2
1 5t 2
32
1
a
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
1 2
t k
2
2t 2i tj 38. r t
rc t
4ti j tk
Tt
4ti j tk
39. r t
4ti 3 cos tj 3 sin tk
rc t
4i 3 sin tj 3 cos tk
Tt
1
>4i 3 sin tj 3 cos tk@
5
Tc t
1
>3 cos tj 3 sin tk@
5
1 17t 2
4i 17tj k
Tc t
1 17t 2
32
K
K
Tc t
289t 2 17
rc t
1 17t
1 17t 2
32
2
Arc Length and Curvature
12
Tc t
rc t
35
5
139
3
25
17
1 17t 2
32
e 2t i e 2t cos tj e 2t sin tk
40. r t
rc t
2e 2t i 2e 2t cos t e 2t sin t j 2e2t sin t e 2t cos t k
e 2t ª¬2i 2 cos t sin t j 2 sin t cos t k º¼
12
e 2t >9@
e 2t ª¬4 4 cos 2 t 4 cos t sin t sin 2 t 4 sin 2 t 4 sin t cos t cos 2 t º¼
rc t
Tt
rc t
rc t
Tc t
1
1
§ 2
·
§2
·
¨ sin t cos t ¸ j ¨ cos t sin t ¸k
3
3
© 3
¹
©3
¹
12
2
1
1
§2
·
§2
·
i ¨ cos t sin t ¸ j ¨ sin t cos t ¸k
3
3
3
3
3
©
¹
©
¹
12
ª§ 4
1
4
1
4
· §4
·º
2
2
2
2
«¨ 9 sin t 9 cos t 9 sin t cos t ¸ ¨ 9 cos t 9 sin t 9 cos t sin t ¸»
¹ ©
¹¼
©
Tc t
K
5
3
3e 2t
Tc t
rc t
x
3t , xc
y
2t 2 , yc
K
5
9e
3, xcc
2
4t , ycc
4
34 0
2 32
yc º
¼
ª9 4t 2 º
¬
¼
12
1, K
At t
9 16
x
e , xc
e , xcc
e
y
4t , yc
4, ycc
0
K
32
et i 4tj, P 1, 0 Ÿ t
42. r t
t
t
xcycc ycxcc
ª xc
¬
2
2 32
yc º
¼
1
43. r t
ti t 2 j t3
k , P 2, 4, 2 Ÿ t
4
rc t
i 2tj 3 2
t k
4
rc 2
i 4 j 3k , rc 2
rcc t
2j rcc 2
2 j 3k
0
xcycc ycxcc
ª xc
¬
5
3
2t
3ti 2t 2 j, P 3, 2 Ÿ t
41. r t
3e 2t
2
26
32
12
125
3
tk
2
i
0
rc 2 u rc 2
j k
6i 3 j 2k
1 4 3
0 2 3
t
04
1 16
32
4
173 2
rc 2 u rcc 2
49
rc u rcc
rc 3
7
263 2
K
7
7 26
676
© 2010 Brooks/Cole, Cengage Learning
140
Chapter 12
44. r t
Vector-Valued Functions
et cos ti et sin tj et k , P 1, 0, 1 Ÿ t
rc t
et cos t et sin t i et sin t et cos t j et k
rc t
cos 2 t 2 cos t sin t sin 2 t sin 2 t 2 sin t cos t cos 2 t 1
et
Tt
1
ª cos t sin t i sin t cos t j k º¼
3¬
Tc t
1
ª¬ sin t cos t i cos t sin t jº¼
3
rc 0
i j k Ÿ rc 0
45. y
Tc 0
2
rc 0
46. y
49. y
0, K
ycc
4
48. y
yc
ycc
K
1
K
0, K
0, and the radius of curvature is
ycc
4 cos 2 x
At x
2S , y
1
K
50. y
4
ª1 4 2 º
¬
¼
32
4
| 0.057
173 2
4
2x , x
x
1
4
, yc 1
x2
8
, ycc 1
x3
K
1
K
2
51. y
8
ycc
ª1 yc 2 º
¬
¼
yc
At x
173 2
| 17.523 radius of curvature
4
2
2 sin 2 x
0, and the radius of curvature is
1
yc
8
32
1 4
53 2
radius of curvature
8
32
8
53 2
2S
cos 2 x, x
yc
K
2 x 2 3, x
4x
1
K
2
3
mx b
yc
K
2
3
3x 2
Because ycc
undefined.
47. y
3
3
Because ycc
undefined.
3et
3
1
i j Ÿ Tc 0
3
Tc 0
K
0
ycc
At x
1, yc
4
ª¬1 02 º¼
0, ycc
4
4
32
1
4
e3 x , x
0
3e3 x , ycc
0, y
9e 3 x
1, yc
9
3, ycc
9
9
103 2
2 32
ª¬1 3 º¼
10 10
9
a2 x2 , x
0
x
a x2
2
a2
a2 x2
32
0: yc
0
ycc
1
a
K
1
K
1a
1 0
2
32
1
a
a radius of curvature
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
3
16 x 2
4
9 x
16 y
52. y
yc
At x
0
ycc
x3 , x
53. y
yc
3
16
32
8, yc
12, ycc
12
145
ycc
n n 1 xn 2
At x
1, yc
1, y
83
10
1
K
8
3
32
2
3
8
3·
§
Equation: x 2 ¨ y ¸
8
©
¹
57. y
1, n t 2
nx
8
3
K
x
1
, yc
x
1
, ycc
x2
2
9
64
n n 1
n n 1
2
x3
2 at 1, 2
32
1 02
1 2. Because the tangent line is
Radius of curvature
n, ycc
2
(b) The circles have different radii since the curvature is
1
different and r
.
K
32
n 1
yc
72
27
r
12
145 145
12
xn , x
0
K
6x
ª1 12 º
¬
¼
54. y
0: yc
3
§ 3·
Center: ¨ 0, ¸
© 8¹
2 32
1
K
K
2
12
K
x2 3
ycc
2
2, y
2
72 1 x 2
16
radius of curvature
3
3x 2 , ycc
At x
x 3
At x
3
16
1 0
1
K
24 x
2
ycc
3 16
K
2
yc
0: yc
141
4x2
x 3
56. (a) y
2
ª9 16 yc º
¬
¼
16 y
ycc
Arc Length and Curvature
horizontal at 1, 2 , the normal line is vertical. The center
of the circle is 1 2 unit above the point 1, 2 at 1, 5 2 .
2 32
ª¬1 n º¼
Circle: x 1
§S ·
55. (a) Point on circle: ¨ , 1¸
©2 ¹
2
5·
§
¨y ¸
2¹
©
2
1
4
4
§S ·
Center: ¨ , 0 ¸
©2 ¹
(1, 2)
−6
6
2
S·
§
Equation: ¨ x ¸ y 2
2¹
©
1
−4
(b) The circles have different radii because the curvature
1
is different and r
.
K
© 2010 Brooks/Cole, Cengage Learning
142
Chapter 12
58. y
ln x,
Vector-Valued Functions
x
1
yc
,
x
yc 1
1,
1
1
K
1 1
yc
1
K
2
32
2 2
The slope of the tangent line at 1, 0 is yc 1
x 1
2
1 x
0 y
2
2
2
x 1
x 1
2x2 4 x 2
8
2 x 2x 3
0
2 x 3 x 1
0
2
x
3
(1, 0)
−3
1 x
9
1
59. y
ex ,
y
yc
ex ,
ycc
yc 0
2.
3 and y
K
8
32
1
1
23 2
1
2 2
1
K
,r
2
x 1
2
x 1
2
x 1.
2
1
0 or x
2
0 and
3
(0, 1)
2
−1
2
x
π
B
A
x or y
x 1
−2π
The center of the circle is on the normal line 2 2 units
away from the point 0, 1 .
y
62.
4
0 x
2
1 y
2
x2 x2
x
3
2 2
B
8
A
−4
2
4
x
r2
Equation of circle: x 2
−3
−4
63. y
(0, 1)
−6
y 3
x 1
3
0
2
x
4
2
6
Because the circle is above
the curve, x
2 and y
3.
Center of circle: 2, 3
−2
2
8
2
3, yc
2 x 1 , ycc
2
K
1 ª¬2 x 1 º¼
x of
2
2
2 32
(a) K is maximum when x
(b) lim K
4
3
2 units
π
The slope of the normal line is 1.
Equation of normal line: y 1
1.
y
61.
2 2
The slope of the tangent line at 0, 1 is yc 0
2
2
4·
§
Equation of circle: x 2 ¨ y ¸
3¹
©
ex
1
1 12
2
1
K
Because the circle is above the curve, x
4
y
.
3
§ 4·
Center of circle: ¨ 0, ¸
−3
© 3¹
0
1, ycc 0
2
x
Center of circle: 3, 2
y 2
§1
·
¨ y¸
©3
¹
−5
Because the circle is below the curve, x
Equation of circle: x 3
2
1 x
3 or x
2
11
1
,r
2
32
The center of the circle is on the normal line
§ 1·
away from the point ¨1, ¸.
© 3¹
2 2
8
2
2
§ 1·
The slope of the tangent line at ¨1, ¸ is yc 1
© 3¹
The slope of the normal line is 1.
1
Equation of normal line: y x 1 or y
3
1.
The center of the circle is on the normal line 2 2 units
away from the point 1, 0 .
1 x
2
1, ycc 1
K
The slope of the normal line is 1.
Equation of normal line: y
1
x 2 , ycc
yc 1
1
, r
23 2
2 32
1 3
x , x
3
60. y
1
2
ycc
x
1
ycc 1
ª1 4 x 1 2 º
¬
¼
32
1 or at the vertex 1, 3 .
0
© 2010 Brooks/Cole, Cengage Learning
3
Section 12.5
64. y
3x 2 , ycc
x 3 , yc
6x
67. y
32
1 9x4
K
§ 1
,
(a) K is maximum at ¨ 4
© 45
(b) lim K
· § 1
,
¸, ¨ 4
3
45 ¹ © 45
1
4
1 ·
¸.
4
453 ¹
2 1 3
x , ycc
3
x 2 3 , yc
2
x 4 3
9
32
(b) lim K
0
1
, yc
x
x of
x1 3 9 x 2 3 4
K
1
, ycc
x2
32
68. y
K
2
. Assume x ! 0.
x3
ycc
2 x3
2 32
ª1 yc º
¬
¼
1 1 x4
dK
dx
x4 1
x4 1
K
1
.
2
ycc
ex
ycc
ex
2 32
ª1 yc º
¬
¼
1 e2 x
32
e x 1 2e 2 x
1 e2 x
52
0 Ÿ e2 x
1
Ÿ x
2
K has maximum curvature at x
52
(b) lim K
x of
1 §1·
ln¨ ¸
2 © 2¹
1
ln 2
2
1
ln 2.
2
0
0
x of
69. y
e x , yc
(a) 1 2e 2 x
32
(a) K has a maximum at x 1
and x
1 by symmetry .
(b) lim K
32
0
2 x3
32
6x2 1 x4
dK
dx
x 1
2
52
x2 1
x of
6
ª¬1 4 9 x 2 3 º¼
32
2 x 2 1
(b) lim K
2 9 x 4 3
x
ª1 1 x 2 º
¬
¼
(a) K has a maximum when x
(a) K o f as x o 0. No maximum
66. y
1
x2
1 x 2
dK
dx
0
x of
K
1
, ycc
x
ln x, yc
143
6x
K
65. y
Arc Length and Curvature
sinh x, yc
cosh x, ycc
sinh x
sinh x
ª1 cosh x 2 º
¬
¼
32
(a) By symmetry, consider x t 0, so sinh x
2 32
ª1 cosh x º
¬
¼
Kc
cosh x sinh x
sinh x:
3
2
1 cosh x
2 12
2 sinh x cosh x
2 3
ª1 cosh x º
¬
¼
Setting K c
0:
ª1 cosh x 2 º cosh x
¬
¼
1 cosh x
2
1 1 sinh 2 x
sinh 2 x
x
Also, x
sinh x
2
3 sinh x
3 cosh x
2
3 sinh 2 x
1
arcsinh 1 | 0.8814, Maximum
arcsinh 1
Maximum curvature at r arcsinh 1 , 1
(b) lim K
x of
0
© 2010 Brooks/Cole, Cengage Learning
144
Chapter 12
Vector-Valued Functions
cosh x, yc
70. y
sinh x, ycc
cosh x
K
cosh x
ª¬1 sinh xº¼
2
(b) lim K
ª¬cosh xº¼
2
1
cosh 2 x
32
0 when x
74. y
3 x 2 , ycc
cos x, ycc
sin x
6 x
b
³a
(b) Plane: K
6 x
ª¬1 9 x 4 º¼
32
Space: K
Curvature is 0 at x
3
x 1
72. y
0: 0, 1 .
3 x 1 , ycc
6 x 1
6 x 1
ycc
K
2 32
0 at x
4 32
ª1 y ' º
¬
¼
cos x, yc
ª1 9 x 1 º
¬
¼
sin x, ycc
1.
77. K
S
32
2
dt
b
³a
rc t
dt
Tc s
Tc t
rc t u rcc t
rc t
rc t
3
ycc
ª1 yc 2 º
¬
¼
32
0, so K
ycc .
4
Yes, for example, y
cos x
x has a curvature of 0 at its
relative minimum 0, 0 . The curvature is positive at any
cos x
ª1 yc 2 º
¬
¼
zc t
At the smooth relative extremum yc
ycc
K
dT
dS
2
76. The curve is a line.
Curvature is 0 at 1, 3 .
73. y
yc t
Answers will vary
2
3, yc
2
xc t
nS .
nS : nS , 0
Curvature is 0 for x
0, Maximum
sin x
0 for x
32
1 cos 2 x
75. (a) s
1 x 3 , yc
K
x
sin x, yc
K
0
x of
71. y
32
2 sinh x
cosh 3 x
(a) K c
cosh x
1 sin 2 x
32
0 for
other point on the curve.
KS .
2
§S
·
Curvature is 0 at ¨ KS , 0 ¸.
2
©
¹
ti t 2 j
78. r t
(a)
dx
dt
s
1,
dy
dt
2
³0
2t
1 2
2³0
1 4t 2 dt
1ª
4 17 ln 4 4¬
x 2 , yc
(b) Let y
At t
0, x
>1 0@
32
At t
K
2t
1 1ª
˜ 2t 1 4t 2 ln 2t 2 2 ¬«
2
1 4t 2 º Theorem 8.2
¼» 0
17 º | 4.647
¼
2 x, ycc
0, y
1 4t 2 2t dt u
0, yc
2
0, ycc
2, K
2
2
1, x
1, y
2
2 32
ª1 2 º
¬
¼
At t
2, x
2, y
K
>1 16@
2
32
1, yc
2, ycc
2
2
| 0.179
53 2
4, yc
4 ycc
2
2
| 0.0285
173 2
(c) As t changes from 0 to 2, the curvature decreases.
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
79. Endpoints of the major axis: r 2, 0
x2 4 y2
2 x 8 yyc
16 x 6 16 x 4 4 x 2 1
4
4 y 2 x2
16 y 2
16 y 3
12 y 4
2
1
4 y3
32
16 y 2 x 2
32
16
16 3x 2
32
2
1·
5
§
x2 ¨ y ¸
.
2
4
©
¹
To graph these circles, use
r 2 and
y2
x
, y2c
x 2
At P, y1c 0
a b 2 x , y1cc
2
x 2
ab and y2c 0
2
2a
4
, y2cc
x 2
2
0 2
3
1
.
2
2
82. y
Because the curves have a common tangent at P ,
1
1
y1c 0
y2c 0 or ab
. So, y1c 0
. Because the
2
2
curves have the same curvature at P, K1 0
K2 0 .
(a)
ª1 y 0
1
«¬
º
»¼
ª1 y 0
2
¬«
º
¼»
(b) V
ª1 1 2 2 º
¬
¼
(c) yc
32
1
1
r or a
r . In order that the curves
2
4
intersect at only one point, the parabola
y
must be concave downward. So,
1
1
4
and b
2.
a
y2
2a
4
2
1
x 2 x and y2
4
x
x 2
K
4
y
−2
x
2
§ 58 5 x8 5 ·
¸ dx
4
4 ¹
2 35
6 2 5
x , ycc
x
5
25
6
25 x 2 5
32
4 6 5º
ª
«¬1 25 x »¼
6
5
³ 0 2S x¨©
4 6 5º
ª
25 x 2 5 «1 x »
25
¬
¼
y2
P
−4
2
(rotated about y-axis)
32
So, 2a
y1
3
2
1 2
2 32
−2
x
ª1 1 2 2 º
¬
¼
y2cc 0
K2 0
3
z
2a
2 32
−3
4
2
K1 0
0. At 1, 0 , the
1 85
x ,0 d x d 5
4
4
y1cc 0
1
.
4
y
equations. So, the point P is origin.
ax b x , y1c
¹
1
1
1
5
r
x 2 and y
r
x2 .
2
4
2
4
(c) The curvature tends to be greatest near the extrema
of f, and K decreases as x o rf. f and K,
however, do not have the same critical numbers.
5
Critical numbers of f :
2
| r0.7071
x
0, r
2
−3
Critical numbers of K:
x
0, r0.7647, r0.4082
−2
x
x 2
You observe that 0, 0 is a solution point to both
y1
2 5 5. f 1
1, K
2
5
1
.
K
2
Using the graph of f, you see that the center of
2
§ 1·
curvature is ¨ 0, ¸. So,
f
© 2¹
0.
ax b x , y2
1
. Using the symmetry of
2
circle of curvature has radius
So, because 2 d x d 2, K is largest when x
smallest when x
For x
16
32
0. At 0, 0 , the circle
©
4 y x 2 y
ª1 x 4 y 2 º
¬
¼
16
32
the graph of f , you obtain x 2 §¨ y 1 ·¸
2
1 4 y 3
K
2. f 0
of curvature has radius
4 y 1 x 4 yc
16 y 2
ycc
0, K
(b) For x
x
4y
yc
80. y1
2 6x2 1
(a) K
0
145
x4 x2
81. f x
Endpoints of the minor axis: 0, r1
Arc Length and Curvature
4
y1
32
125S 53 5
| 114.6 cm3
9
6
25 x 2 5
1
5
0
0
(d) No, the curvature approaches f as x o 0. So, any
spherical object will hit the sides of the goblet before
touching the bottom 0, 0 .
−4
© 2010 Brooks/Cole, Cengage Learning
146
Chapter 12
Vector-Valued Functions
83. (a) Imagine dropping the circle x 2 y k
2
16
2
into the parabola y
x . The circle will drop to the point where the tangents
to the circle and parabola are equal.
x 2 and x 2 y k
y
2
16 Ÿ x 2 x 2 k
0 and yc
Taking derivatives, 2 x 2 y k yc
y k yc
So,
x
yk
2
16
2 x. So,
y
x
.
y k
x Ÿ yc
15
10
2x Ÿ x
2 x y k Ÿ 1
1
.
2
2 x2 k Ÿ x2 k
So,
−10
−5
x
5
10
2
§ 1·
x2 ¨ ¸
16 Ÿ x 2 15.75.
© 2¹
1
Finally, k
x2 16.25, and the center of the circle is 16.25 units from the vertex of
2
the parabola. Because the radius of the circle is 4, the circle is 12.25 units from the vertex.
x2 x2 k
2
y 2 or z
(b) In 2-space, the parabola z
x 2 has a curvature of K
the largest sphere that will touch the vertex has radius
84. s
c
K
y
1 3
x
3
yc
x2
ycc
2x
K
1
.
2
2x
1 x4
When x
32
1
2
c
1: K
s
1
4
30
At x
1K
2 at 0, 0 . The radius of
3
, K
2
s
4
2c Ÿ c
3
ª¬1 81 16 º¼
§ 3·
¨ ¸
© 2¹
2c
2
c
K
32
30
2
4
| 0.201
30
4
K
2
| 56.27 mi h
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
f x . Let D , E be the center of curvature. The radius of curvature is
85. P x0 , y0 point on curve y
yc
Arc Length and Curvature
f c x . Slope of normal line at x0 , y0 is
147
1
.
K
1
.
f c x0
1
x x0
c
f x0
Equation of normal line: y y0
D , E is on the normal line: f c x0 E y0
2
x0 , y0 lies on the circle: x0 D
y0 E
D x0
2
Equation 1
§1·
¨ ¸
©K¹
2
ª
« 1 f c x0
«
f cc x0
«
¬
2
2 3 2º
»
»
»
¼
Equation 2
Substituting Equation 1 into Equation 2:
2
ª¬ f c x0 E y0 º¼ y0 E
E y0
2
§1·
¨ ¸
©K¹
2
2
y
1 f c x0
ª1 f c x0 º
¬
¼
2
f cc x0
2 3
2
P ( x0, y0)
1
K
(␣ , ␤ )
2 2
E y0
2
ª1 f c x0 º
¬
¼
2
f cc x0
x
When f cc x0 ! 0, E y0 ! 0, and if f cc x0 0, then E y0 0.
So E y0
E
z
ex , f c x
1 fc 0
f cc 0
D, E
z
z
e x , 0, 1
2
x, ycc
2, 3
§ 1·
1, ¨1, ¸
© 2¹
2
2
1
§
·
¨1 2, 2 ¸
2
©
¹
x 2 , yc
2 x, ycc
1 fc 0
f cc 0
2
D, E
f cc x
0 2, 1 2
1 fc1
f cc 1
(c) y
y0 z
2
x2
, yc
2
D, E
2
x0 f c x0 z.
f x
(b) y
1 f c x0
f cc x0
y0 Similarly, D
86. (a) y
2
1 f c x0
f cc x0
§
¨ 1,
©
5·
¸
2¹
2, 0, 0
1
2
1·
§
¨ 0, 0 ¸
2¹
©
§ 1·
¨ 0, ¸
© 2¹
© 2010 Brooks/Cole, Cengage Learning
148
Chapter 12
87. r T
Vector-Valued Functions
r cos T i r sin T j
f T cos T i f T sin T j
xT
f T cos T
yT
f T sin T
xc T
f T sin T f c T cos T
yc T
f T cos T f c T sin T
xcc T
f T cos T f c T sin T f c T sin T f cc T cos T
f T cos T 2 f c T sin T f cc T cos T
ycc T
f T sin T f c T cos T f c T cos T f cc T sin T
f T sin T 2 f c T cos T f cc T sin T
ª xc
¬
2
2
yc º
¼
rc
cos T
r cc
sin T
2 rc
K
2
ª rc
¬
(b) r
T
rc
1
r cc
0
2 rc
K
ªf 2 T fc T
«¬
rr cc r 2
2
2
ª rc
¬
2 32
º
»¼
ªr 2 r c 2 º
¬
¼
2
32
r
2º
32
2 3
r2º
¼
32
2
3 1 sin T
8 1 sin T
ªcos 2 T 1 sin T º
¬
¼
¼
rr cc r 2
2
2 cos 2 T 1 sin T sin T 1 sin T
3
3
2 2 1 sin T
2 T2
1 T2
32
a sin T
(c) r
rc
a cos T
r
a sin T
2 rc
K
2
ª rc
¬
(d) r
eT
rc
eT
r cc
eT
2 rc
K
2
ª rc
¬
rr cc r 2
2
r
2º
r
2
2
2
2
¬ªa cos T a sin T ¼º
¼
rr cc r 2
2
2a 2 cos 2 T a 2 sin 2 T a 2 sin 2 T
32
2º
32
¼
2e 2T
2e
2T
3
2a 2
a3
2
,a ! 0
a
1
2eT
32
e aT , a ! 0
rc
ae aT
r cc
a 2e aT
2 rc
K
32
r 2 rr cc 2 r c
1 sin T
88. (a) r
89. r
f 2 T f T f cc T 2 f c T
xcycc ycxcc
K
2
2
ª rc
¬
rr cc r 2
2
r
2º
32
¼
2a 2e 2 aT a 2e 2 aT e 2 aT
2 2 aT
ª¬a e
e
2 aT 3 2
º¼
1
e
aT
a2 1
(a) As T o f, K o 0.
(b) As a o f, K o 0.
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
90. At the pole, r
2 rc
K
2
ª rc
¬
0.
rr cc r
2
r
2º
4 sin 2T
rc
8 cos 2T
2 rc
rc
32
¼
2
rc 0
At the pole: K
rc
2
2
rc
3
2
8
3t 2 , xcc t
1 2
t , yc t
2
yt
149
6t
t , ycc t
1
3t 2 1 t 6t
K
91. r
92. r
t 3 , xc t
94. x t
2
Arc Length and Curvature
ª 3t 2
«¬
2
32
2
t º
»¼
3t 2
1
4
t
3
3
9t 1
2
32
t 9t 1
2
32
K o 0 as t o rf
6 cos 3T
5
18 sin 3T
At the pole,
T
S
6
§S ·
, r c¨ ¸
©6¹
18,
−4
and
K
93. x
yc
ycc
2
2
18
rc S 6
f t,y
dy
dx
1
.
9
gt
dy
dt
dx
dt
d ª gc t
«
dt ¬« f c t
dx
dt
f c t g cc
95. x T
a T sin T
yT
a 1 cos T
xc T
a 1 cos T
yc T
a sin T
xcc T
a sin T
ycc T
a cos T
xc T ycc T yc T xcc T
K
ª xc T
¬
gc t
fc t
2
a 2 sin 2 T º
¼
32
cos T 1
1
a >2 2 cos T @3 2
1
1 cos T
1 cos t 0
a 2 2 >1 cos T @3 2
t g c t f cc t
2
f c t g cc t g c t f cc t
ª¬ f c t º¼
¬ª f c t ¼º
ycc
2 32
Minimum:
3
1
4a
T
ª § gc t · º
«1 ¨
¸ »
« © fc t ¹ »
¬
¼
f c t g cc t g c t f cc t
ª¬ f c t º¼
3
f c t g cc t g c t f cc t
2 3
ª¬ g c t º¼ ½°
¾
2
º¼
°¿
2
ª¬ f c t º¼ ª¬ g c t ¼º
2 32
S
K o f as T o 0
Maximum: none
2 32
ª1 yc º
¬
¼
1
§T ·
csc¨ ¸
4a
©2¹
1
2a 2 2 cos T
3
f c t g cc t g c t f cc t
­° ª f c t º ¬
¼
®
ª¬ f c t
°¯
32
a 2 1 cos T cos T a 2 sin 2 T
º
»
¼»
2
2
yc T º
¼
2
ªa 2 1 cos T
¬
ª¬ f c t º¼
fc t
K
4
0
96. (a) r t
3t 2i 3t t 3 j
vt
6ti 3 3t 2 j
ds
dt
vt
K
3 1 t2 ,
d 2s
dt 2
6t
2
2
3 1 t2
aT
d 2s
dt 2
aN
§ ds ·
K¨ ¸
© dt ¹
6t
2
2
31 t
2
2
˜ 9 1 t2
2
6
© 2010 Brooks/Cole, Cengage Learning
150
Chapter 12
(b) r t
Vector-Valued Functions
1 2
t k
2
ti t 2 j vt
i 2tj tk
ds
dt
vt
d 2s
dt 2
at
5t
98. F
5t 2 1
vt uat
aT
aN
j
k
1 2t
t
0
1
2
99. y
5
3
d 2s
dt 2
§ ds ·
mK ¨ ¸
© dt ¹
maN
j 2k
rc t u rcc t
rc t
2
2
5t 2 1
32
5t
cosh x
2
e x e x
2
yc
e x e x
2
sinh x
ycc
e x e x
2
cosh x
5t 1
3327.5 lb
2
§ 6400 lb ·§ 1 ·§ 35 5280 ft ·
¸
¨
¸¨
2 ¸¨
© 32 ft sec ¹© 250 ft ¹© 3600 sec ¹
94864
| 2108.1 lb
45
2j k
i
§ ds ·
mK ¨ ¸
© dt ¹
maN
§ 5500 lb ·§ 1 ·§ 30 5280 ft ·
¸
¨
¸¨
2 ¸¨
© 32 ft sec ¹© 100 ft ¹© 3600 sec ¹
5t 2 1
rc t u rcc t
K
97. F
2
§ ds ·
K¨ ¸
© dt ¹
2
K
5
5t 2 1
32
5t 2 1
cosh x
cosh x
2 32
ª1 sinh x º
¬
¼
2
cosh x
32
1
cosh 2 x
1
y2
5
5t 1
2
100. (a) K
(b) T t
dT
ds
Tc s
d T dt
˜
, by the Chain Rule
dt ds
d T dt
ds dt
Tc t
Tc t
vt
rc t
rc t
rc t
rc t
ds dt
rc t
ds
Tt
dt
rcc t
§ d 2s ·
ds
¨ 2 ¸T t Tc t
dt
© dt ¹
2
§ ds ·§ d 2 s ·
§ ds ·
¨ ¸¨ 2 ¸ ª¬T t u T t º¼ ¨ ¸ ª¬T t u Tc t º¼
dt
dt
© ¹©
© dt ¹
¹
rc t u rcc t
Because T t u T t
0 and
ds
dt
rc t , you have:
rc t u rcc t
rc t
2
ª¬T t u Tc t º¼
rc t u rcc t
rc t
2
T t u Tc t
So,
(c) K
rc t u rcc t
rc t
3
rc t u rcc t
rc t
3
rc t
2
Tt
Tc t
rc t
2
1 K rc t from (a )
K.
rc t u rcc t
rc t
rc t
2
vt uat
vt
rc t
2
at ˜Nt
rc t 2
© 2010 Brooks/Cole, Cengage Learning
Section 12.5
Arc Length and Curvature
151
101. False
102. False
1
radius
Curvature
103. True
104. True
§ ds ·
K¨ ¸
© dt ¹
aN
x t i y t j z t k. Then r
105. Let r
§ dr ·
r¨ ¸
© dt ¹
106. F
2
r
2
^
a
`
§d·
¨ ¸ r u rc
© dt ¹
1 2
xc t i yc t j zc t k. Then,
º
˜ 2 x t xc t 2 y t yc t 2 z t zc t »
¼
GmM
r
r3
GM
3 r
r
Because r is a constant multiple of a, they are parallel. Because a
d ªr º
dt «¬ r »¼
2
2
2
2 ª1
2
2
2
ª¬ x t º¼ ª¬ y t º¼ ª¬ z t º¼ « ª¬ x t º¼ ª¬ y t º¼ ª¬ z t º¼
2
¬
x t xc t y t yc t z t zc t
r ˜ rc.
ma Ÿ ma
107. Let r
2
ª¬ x t º¼ ª¬ y t º¼ ª¬ z t º¼ and r c
rc u rc r u rcc
00
rcc is parallel to r, r u rcc
0. Also,
0. So, r u rc is a constant vector which we will denote by L.
xi yj zk where x, y, and z are function of t, and r
r .
rrc r ª¬ r ˜ rc r º¼
rrc r dr dt
r2
r2
r 2rc r ˜ r c r
using Exercise 105
r3
x 2 y 2 z 2 xci ycj zck xxc yyc zzc xi yj zk
r3
1ª
xcy 2 xcz 2 xyyc xzzc i x 2 yc z 2 yc xxcy zzcy j x 2 zc y 2 zc xxcz yycz k ¼º
r3 ¬
i
j
k
1
1
c
c
c
c
c
yz y z xz x z xy xcy
^>r u rc@ u r`
r3
r3
x
y
z
108.
d ª rc
rº
uL »
dt «¬ GM
r¼
1
1
>rc u 0 rcc u L@ 3 ^>r u rc@ u r`
GM
r
º
1 ª
1
§ GMr ·
0 ¨ 3 ¸ u >r u rc@» 3 ^>r u rc@ u r`
GM «¬
© r ¹
¼ r
1
r
3 u >r u rc@ 3 ^>r u rc@ u r`
r
r
1
^>r u rc@ u r >r u rc@ u r` 0
r3
§ rc ·
§r·
So, ¨
¸ u L ¨ ¸ is a constant vector which we will denote by e.
© GM ¹
©r¹
© 2010 Brooks/Cole, Cengage Learning
152
Chapter 12
Vector-Valued Functions
109. From Exercise 106, you have concluded that planetary motion is planar. Assume that the planet moves in the
xy-plane with the sum at the origin. From Exercise 108, you have
§r
·
GM ¨ e ¸.
©r
¹
rc u L
Because rc u L and r are both perpendicular to L, so
is e. So, e lies in the xy-plane. Situate the coordinate system so that e lies along the positive x-axis
and T is the angle between e and r. Let e
e . Then r ˜ e
r e cos T
re cos T . Also,
L˜L
2
L
r u rc ˜ L
y
r ˜ rc u L
ª
r ·º
§
r ˜ «GM ¨ e ¸»
r ¹¼
©
¬
r ˜ rº
ª
GM «r ˜ e r »¼
¬
GM >re cos T r @.
So,
L 2 GM
1 e cos T
Planet
Sun
r
θ
x
e
r
and the planetary motion is a conic section. Because the planet returns to its initial position periodically,
the conic is an ellipse.
110.
r u rc
L
r cos T i sin T j
Let: r
rc
dT § dr
¨
dt © dt
r sin T i cos T j
Then: r u rc
dr dT ·
˜
¸
dT dt ¹
i
j
k
r cos T
r sin T
0
r sin T
dT
dt
r cos T
dT
dt
0
r2
dT
k and L
dt
r u rc
r2
dT
.
dt
1 E 2
r dT
2 ³D
111. A
So,
dA
dt
dA dT
dT dt
1 2 dT
r
2 dt
1
L
2
and r sweeps out area at a constant rate.
112. Let P denote the period. Then
P dA
³0
A
dt
dt
1
L P.
2
Also, the area of an ellipse is S ab where 2a and 2b are the lengths of the major and minor axes.
S ab
P
P2
1
L P
2
2S ab
L
4S 2 a 2 2
a c2
L 2
4S 2 a 2 2
a 1 e2
L 2
4S 2 a 4 § ed ·
¨ ¸
L 2 © a ¹
4S 2ed 3
a
L 2
4S 2 L
2
L
2
GM
a3
4S 2 3
a
GM
Ka 3
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 12
153
Review Exercises for Chapter 12
tan t i j t k
1. r (t )
(a) Domain: t z
S
(b) Continuous for all t z
S
2
(b) Continuous for all t ! 0
nS , n an integer
2t 1 i t 2 j t k
4. r t
(a) Domain: f, f
1
jk
t 4
ti 2. r (t )
(a) Domain: 0, f
nS , n an integer
2
In t i t j t k
3. r (t )
(b) Continuous for all t
(a) Domain: [0, 4) and 4, f
(b) Continuous except at t
2t 1 i t 2 j 5. r t
i (a) r 0
(b) r 2
4
t 2k
2k
3 i 4 j
2
(c) r c 1
2c 1 i c 1 j c 1k
2
(d) r 1 't r 1
ª¬2 1 't 1º¼ i 1 't j 2't i 't 2 2't j 3 't k 3 i j 3 't 3k
3 k
3i j
6. (a) r 0
§S ·
(b) r¨ ¸
©2¹
S
k
2
(c) r s S
3 cos s S i 1 sin s S j s S k
(d) r S 't r S
3 cos(S 't i 1 sin S 't j S 't k ) 3 i j S k
3 cos 't 3 i sin 't 't k
S cos t , S sin t
7. r t
S cos t , y
x
y
z
4
S sin t
2
2
S 2 , circle
x2 y2
i t j t2k
9. r t
1
x
−4
− 2 −1
1
2
4
x
1
y
t
z
t2 Ÿ z
1
y2
−2
2
−4
y
t 2 i 3t j t 3 k
10. r t
t 2, t 2 1
8. r t
1
x
z
x
t 2 Ÿ t
y
t2 1
x2
4
2
x 2 1, parabola
y
2
−4
4
2
x
3
2
2
4
y
−2
−4
1
x
−1
−1
1
2
3
4
5
−2
© 2010 Brooks/Cole, Cengage Learning
154
Chapter 12 Vector-Valued Functions
i sin t j k
11. r t
x
1, y
sin t , z
16. One possible answer is:
z
1
3
r1 t
4t i, 0 d t d 1
r2 t
4 cos t i 4 sin t j, 0 d t d
r3 t
4 t j, 0 d t d 4
2
1
−2
1
1
2
y
2
S
2
3
x
2 cos t i t j 2 sin t k
12. r t
x
17. The vector joining the points is 7, 4, 10 . One path is
2 cos t , y
x2 z 2
t, z
z
2 7t , 3 4t , 8 10t .
rt
3
2 sin t
18. The x- and y-components are 2 cos t and 2 sin t. At
4
2
x
3S
,
2
t
2π
y
the staircase has made
t i ln t j 13. r t
1 2
t k
2
3
of a revolution and is 2 meters
4
high. So, one answer is
z
2 cos t i 2 sin t j rt
3
2
x2 y 2 , x y
19. z
1
1
1
x
y
2
2
t, y
0, t
z
x
5
2t 2
t i t j + 2t 2 k
rt
3
t , z
4
t k.
3S
x
−3
2
1
ti 2
14. r t
tj 1
2
3
y
3
1 3
t k
4
x
20. x 2 z 2
z
x
6
t, y
4, x y
t, z
0, t
x
z
r 4 t2
5
4
rt
ti t j 4 t2 k
rt
ti t j 4 t2 k
3
2
−3
x
−2
1
1
−1
−1
2
3
1
2
−2
3
−3
y
y
21. lim t i t o 4
r1 t
3t i 4t j, 0 d t d 1
r2 t
3 i 4 t j, 0 d t d 4
r3 t
3 t i, 0 d t d 3
3t i t 1 j, u t
(a) rc t
3i j
(b) rcc t
0
(c) r t ˜ u t
4 tj+ k
§ sin2t
·
i e t j et k ¸
22. lim¨
t o 0©
t
¹
t i t2 j 3t 2 t 2 t 1
Dt ª¬r t ˜ u t º¼
3
5
15. One possible answer is:
23. r t
4
x
4i k
2 cos 2t ·
§
¨ lim
¸i j k
© t o0 1 ¹
2i j k
2 3
t k
3
t 3 2t 2
3t 2 4t
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 12
(d) u t 2r t
5t i t 2 2t 2 j Dt ª¬u t 2r t º¼
2 3
t k
3
5 i 2t 2 j 2t 2 k
10t 2 2t 1
rt
(e)
155
10t 1
Dt ª¬ r t º¼
10t 2 2t 1
2 4
t t 3 i 2t 4 j 3t 3 t 2 t k
3
§8 3
2·
3
2
Dt ª¬r t u u t º¼
¨ t 2t ¸ i 8t j 9t 2t 1 k
3
©
¹
(f ) r t u u t
sin t i cos t j t k , u t
24. r t
(a) rc t
cos t i sin t j k
(b) rcc t
sin t i cos t j
(c) r t ˜ u t
2
Dt ª¬r t ˜ u t º¼
0
§1
·
sin t i cos t j ¨ 2t ¸k
©t
¹
(d) u t 2r t
§ 1
·
cos t i sin t j ¨ 2 2 ¸k
© t
¹
Dt ª¬u t 2r t º¼
rt
(e)
1 t2
t
Dt ª¬ r t º¼
(f ) r t u u t
1
sin t i cos t j k
t
1 t2
§1
·
§1
·
¨ cos t t cos t ¸i ¨ sin t t sin t ¸ j
©t
¹
©t
¹
Dt ¬ªr t u u t º¼
1
1
§ 1
·
§1
·
¨ sin t 2 cos t t sin t cos t ¸i ¨ cos t 2 sin t t cos t sin t ¸ j
t
t
t
t
©
¹
©
¹
25. x t and y t are increasing functions at t
t0 , and z t is a decreasing function at t
t0.
26. The graph of u is parallel to the yz-plane.
27.
³
cos t i cos t j dt
28.
³
ln t i t ln t j k dt
29.
³
cos ti sin tj t k dt
30.
³
tj t 2k u i tj tk dt
sin t i t sin t cos t j C
t ln t t i ³
t2
1 2 ln t j t k C
4
1 t 2 dt
³ ª¬ t
2
1ª
t 1 t 2 ln t 2 ¬«
t 3 i t 2 j t k º¼ dt
2
31.
2
³ 2
3t i 2t 2 j t 3 k dt
ª 3t 2
2t 3
t4 º
j k»
« i 3
4 ¼ 2
¬ 2
1 t2 º C
¼»
§ t3
t4 ·
t3
t2
¨ ¸i j k C
4¹
3
2
©3
32
j
3
© 2010 Brooks/Cole, Cengage Learning
156
Chapter 12 Vector-Valued Functions
t j t sin t k dt
2
et 2 i 3t 2 j k dt
1
t 3 i arcsin t j t 2 k dt
³0
33.
³0
34.
ª2 3 2
« 3 t j sin t t cos t
¬
1
32.
³ 1
35. r t
³
2
ª¬2et 2 i t 3 j t k º¼
0
2t i et j e t k dt
rt
t 2 1 i + et 2 j e t 4 k
i 3j 5k Ÿ C
4t i t 3 j t k
rc t
vt
4 i 3t 2 j k
v
16 9t 4 1
Speed
at
t3
k C
3
C
rt
§ t3
·
ln sec t tan t i ln cos t j ¨ 3¸k
3
©
¹
17 9t 4
6t j
t i 5t j 2t 2 k
rt
38.
vt
1
i 5 j 4t k
2 t
Speed
v
1
25 16t 2
4t
at
rc t
3k
1
i 4k
4t 3 2
cos3 t , sin 3 t , 3t
rc t
3 cos 2 t sin t , 3 sin 2 t cos t , 3
9 cos 4 t sin 2 t 9 sin 4 t cos 2 t 9
vt
at
2
k
3
rt
37.
i 2 j 4k
r0
vt
t3 º
k»
3 ¼ 1
sec t i tan t j t 2 k dt
ln sec t tan t i ln cos t j 39. r t
1
1 t2 j t 2 i et j e t k C
jk C
³
2
j sin 1 cos 1 k
3
2e 2 i 8 j 2 k
ªt 4
« i t arcsin t ¬4
r0
36. r t
1
º
k»
¼0
vc t
3 cos 2 t sin 2 t cos 2 t sin 2 t 1
3 cos 2 t sin 2 t 1
6 cos t sin 2 t 3 cos 2 t cos t , 6 sin t cos 2 t 3 sin 2 t sin t , 0
3 cos t 2 sin 2 t cos 2 t , 3 sin t 2 cos 2 t sin 2 t , 0
t , tan t , et
40. r t
rc t
1, sec 2 t , et
vt
1 sec 4 t e 2t
vt
rcc t
0, 2 sec 2 t ˜ tan t , et
at
rc t
1
1
, 2t ,
t 3
2
rc 4
1, 8,
Because r 4
x
t, y
r t0 0.1
4
3 sinh t , cos h t , 2
rc 0
0, 1, 2 direction numbers
Because r 0
3, y
2
1
t.
2
r 4.1 | 0.1, 16.8, 2.05
2t.
r 0.1 | 3, 0.1, 0.2
v0 cos T t , v0 sin T t 1
direction numbers
2
16 8t , z
3, 0, 0 , the parametric equations are
t, z
r t0 0.1
43. r t
0, 16, 2 , the parametric equations are
0
rc t
x
1
ln t 3 , t 2 , t , t0
2
41. r t
3 cosh t , sinh t , 2t , t0
42. r t
1 2
gt
2
42 3 t , 42t 16t 2
42t
Range
16t 2 Ÿ t
0,
§ 21 ·
42 3 ¨ ¸
©8¹
21
8
441 3
| 190.96 ft
4
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 12
16t 2 6
44. y
16
6
Range
x
95
19
1 4t i 2 3t j
48. r t
4i 3 j
vt
v
ª¬ 20 cos 30q t º¼ i ª¬ 20 sin 30q t 4.9t º¼ j
2
at
5
0
1
4i 3 j
5
N t does not exist.
Tt
a˜T
0
Maximum height | 5.1 m; Range | 35.3 m
ª¬ 20 cos 45q t º¼ i ª¬ 20 sin 45q t 4.9t 2 º¼ j
0
a ˜ N does not exist.
49. r t
ti tj
vt
i 1
j
2 t
20
vt
at
0
0
a ˜ N does not exist
45
(b) r t
1
i 3 j
10
(The curve is a line)
20
0
10
0
a˜T
1
ª
º
ª¬ v0 cos T t º¼ i « v0 sin T t 9.8 t 2 » j
2
¬
¼
(a) r t
v
N t is not defined
v0 2
sin 2T
9.8
v0 | 38.06 m/sec
46. r t
i 3 j
Tt
9.8 95
sin 40q
v0 2
vt
at
8 6
| 6.532 ft/sec
3
v0
2 t i 3t j
47. r t
6
4
§ 6·
v0 ¨¨
¸¸ Ÿ v0
© 4 ¹
v0t Ÿ 4
x
45.
0 Ÿ t
157
45
4t 1
2 t
1
j
4t t
0
Maximum height | 10.2 m; Range | 40.8 m
(c) r t
Tt
ª¬ 20 cos 60q t º¼ i ª¬ 20 sin 60q t 4.9t 2 º¼ j
Nt
20
a˜T
0
45
0
a˜N
i 12 t j
4t 1 2 t
2 ti j
4t 1
i 2 tj
4t 1
1
4t t 4t 1
1
2t 4t 1
Maximum height | 15.3 m; Range | 35.3 m
(Note that 45q gives the longest range)
© 2010 Brooks/Cole, Cengage Learning
158
Chapter 12 Vector-Valued Functions
2
j
t 1
2t 1i 50. r t
2
2i vt
2
vt
t 1
2
t 1
4
t 1
2
4
at
t 1
3
vt
j
1 5t 2
v
1
at
2j k
Tt
i 2t j t k
1 5t 2
5t i 2 j k
j
Nt
2
t 1 i j
Tt
1 2
t k
2
i 2t j t k
ti t2 j 53. r t
4
t 1
1
5 1 5t 2
5t
a˜T
1 5t 2
2
i t 1 j
Nt
t 1
4
t 1
3
4t 1
a˜N =
t 1
3
vt
et i e t j
§t 1
¨
©
1
t 1
4
1 ·¸
¹
et i e t j
ei e j
e2t e 2t
e t i et j
Nt
rc t
t sin t cos t
2
t cos t sin t
2
t cos t 2 sin t i t sin t 2 cos t j
t2 1
t cos t sin t i t sin t cos t j
at ˜Tt
at ˜Nt
x
2
i 2t 4 1
S 3
3j S
3, S 3
k. Point: 1,
3
2 sin t i 2 cos t j k
3i j k
1
When t
t 1
rc t
2, x
2, y
t2, z
t, y
2t3
3
16 .
3
4, z
i 2t j 2t 2 k
Direction numbers when t
x
S 3t
3 t, z
3 t, y
t i t 2 j 23 t 3 k , x
56. r t
2
t2 1
2
t 2
t
rc S 3
t sin t cos t i t cos t sin t j
t
2
2t 4 1
4
t3
Direction numbers: 3, 1, 1
t2 1
Nt
2t 4 1
2 cos t i 2 sin t j tk , t
rc t
t sin t cos t i t cos t sin t j
speed
vt
vt
i j 2t 2k
rS 3
t cos t i t sin t j
Tt
Nt
2t 4 1
55. r t
e 2t e 2t
rcc t
t2i t2 j k
a˜N
e 2t e2t
2
at
Tt
a˜T
e 2t e 2t
a˜N
2
k
t3
2
e 2 t e 2 t
a˜T
2t 4 1
t2
t
t
Tt
vt
at
e 2t e 2t
vt
vt
4
4
1 5t 2
1
t 1 i tj k
t
1
i j 2k
t
54. r t
vt
et i e t j
vt
1
2
t 1
51. r t
52. r t
4
t 1
5
5 1 5t 2
4
a˜T
at
5
a˜N
1
t 2, y
4t 4, z
2, a
8t 1, b
4, c
8
16
3
t2 1
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 12
rc t
10
2S
³0
s
2t i 3t j, 0 d t d 5
rc t
10 sin t i 10 cos t j
rc t
58. Factor of 4
59. r t
10 cos t i 10 sin t j
62. r t
9.56 u 104
| 4.58 mi/sec
4000 550
GM
r
57. v
b
³a
rc t
20S
10 dt
2 i 3j
s
y
5
³0
dt
5
ª 13 t º
¬
¼0
4 9 dt
8
6
4
2
5 13
y
x
−8 −6 −4 −2
2
x
2 4 6 8 10 12 14
−4
−6
−8
−10
−12
−14
−16
3t i 2t j 4t k , 0 d t d 3
63. r t
rc t
(10, − 15)
t i 2t k , 0 d t d 3
rc t
3 i 2 j 4 k
b
³a
s
2
60. r t
b
³a
rc t
rc t
3
³0
2t i 2k
s
3
³0
dt
2 4 6 8
−4
−6
−8
(0, 0)
−4 −2
159
4t 2 4 dt
3
ªln
«¬
t2 1 t t
ln
10 3 3 10 | 11.3053
29 dt
t 2 1º
»¼ 0
3 29
(0, 0, 0)
2
2 4
6 8
x
z
9 4 16 dt
(− 9, 6, 12)
z
12
10
8
6
4
2
3
³0
dt
y
10
6
t i t 2 j 2t k , 0 d t d 2
64. r t
5
4
rc t
3
2
(0, 0, 0)
−2
(9, 0, 6)
1
1
b
³a
s
y
2
i 2t j 2 k , rc t
rc t
2
³0
dt
5 4t 2
5 4t 2 dt
2
21 3
4
5
5
4
ln 5 5
4
ln
105 4 5 | 6.2638
z
6
7
4
9
3
x
(2, 4, 4)
2
61. r t
10 cos3 t i 10 sin 3 t j
rc t
1
30 cos 2 t sin t i 30 sin 2 t cos t j
1
1
2
2
rc t
30 cos t sin t sin t cos t
4
2
4
2
30 cos t sin t
s
4³
S 2
0
S 2
30 cos t ˜ sin t dt
ª sin 2 t º
«120
»
2 ¼0
¬
60
rc t
b
³a
s
2
rc t
S 2
³0
dt
65 dt
65
S
65
2
z
π
2
2
)0, 8, π2 )
x
2
10
8
−10
S
8 sin t , 8 cos t , 1 , rc t
10
−2
y
4
8 cos t , 8 sin t , t , 0 d t d
65. r t
y
−10
3
x
x
6
4
4
(8, 0, 0)
6
8
y
© 2010 Brooks/Cole, Cengage Learning
160
Chapter 12 Vector-Valued Functions
rc t
s
S
2 sin t t cos t , 2 cos t t sin t , t , 0 d t d
66. r t
2t sin t , 2t cos t ,1 , rc t
b
³a
rc t
1
ln
4
S 2
³0
dt
2
4t 2 1
4t 2 1 dt
S2 1 S 1 2
1
§ 1 1·
t i t j t 3 k , P¨ , 1, ¸ Ÿ t
2
3
© 2 3¹
71. r t
rc t
t i j t 2 k , rc 1
rcc t
i 2t k , rcc 1
i
S
S 2 1 | 3.055
4
rc u rcc
1
i jk
i 2k
j k
2i j k
1 1 1
1 0 2
3t i 2t j
67. r t
Line
K
rc u rcc
rc 3
K
4 11
6
3 3
3
3
2
3
0
1
rc t
1 9t
t
1
9
t
i 3 j, rc t
t
1
t 3 2 i
2
rcc t
4 cos t i 3 sin t j t k , P 4, 0, S Ÿ t
72. r t
2 t i 3t j
68. r t
rc t
4 sin t i 3 cos t j k , rc S
rcc t
4 cos t i 3 sin t j, rcc S
i
rc u rcc
j
rc u rcc
j k
3 3 2
t k ; rc u rcc
2
3 0
t
1
t 3 2
2
0 0
rc t u rcc t
K
rc t
2t i 69. r t
3/2t 3 2
3
1 9t
3
2 1 9t
t3 2
rcc t
j 2k
i j k
2 t 2t
0 1 2
rc u rcc
rc 3
32
ycc
1
K
2 i t j 2t k , rc
K
32
1 2
t j t2 k
2
rc t
rc u rcc
yc
1 2
x 2
2
x
73. y
5t 2 4
At x
74. y
4 j 2 k , rc u rcc
20
5t 2 4
20
yc
4 5t 2
32
K
2t i 5 cos t j 5 sin t k
70. r t
rc t
2 i 5 sin t j 5 cos t k , rc t
rcc t
29
At x
j
k
2
5 sin t
5 cos t
0 5 cos t
5 sin t
r
91
1
and r
173 2
173 2
17 17.
1 x 2
e
4
1 x 2
e
4
32
1 x º
ª
«¬1 4 e »¼
ycc
ª1 yc 2 º
¬
¼
32
1 x2
e x 2
1
e x 2 , ycc
2
0, K
2
5
32
1
3 2
14
54
i
16 144
ª1 yc º
¬
¼
5 cos t j 5 sin t k
rc u rcc
0
2 32
2 5
32
0
ycc
4, K
4i
4 j 12 k
rc u rcc
rc 3
K
3
2t 3 2
3 j k
k
0 3 1
4
i
1
S
32
2
53 2
2
5 5
2 5
,
25
5 5
.
2
25i 10 sin t j 10 cos t k
rc u rcc
K
725
rc u rcc
rc 3
725
29
32
25 ˜ 29
29 29
5
29
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 12
75. y
ln x
1
yc
x
1
ycc 2
x
yc
ycc
5ax 4 3bx 2 c
20ax3 6bx
20ax 4 6bx
K
ª1 5ax 4 3bx 2 c 2 º
«¬
»¼
1 x2
ª1 yc 2 º
¬
¼
At x
ax5 bx3 cx
78. y
ycc
K
32
ª1 1 x 2 º
¬
¼
1
23 2
1, K
1
3 2
At x
2
and r
4
2 2
161
0 Ÿ 20a 6b
0
0 Ÿ 5a 3b c
0
1Ÿ a b c 1
k
yc
y1
1:
2 2.
3 2
Solving these 3 equations for a, b, c, you obtain
76. y
tan x
sec 2 x
ycc
2 sec 2 x tan x
ª1 yc 2 º
¬
¼
S
At x
4
3 2
3 2
ª¬1 sec 4 xº¼
4
53 2
4
5 5
symmetry, the same holds
y
(1, 1)
1
,K
15 . By
8
3 5 5 3 15
x x x
8
4
8
y
2 sec 2 x tan x
54 , c
b
1.
at x
ycc
K
3,
8
a
yc
4 5
and r
25
5 5
.
4
x
−1
77. The curvature changes abruptly from zero to a nonzero
constant at the points B and C.
1
(− 1, − 1)
−1
Problem Solving for Chapter 12
§ Su2 ·
¸ du , y t
2 ¹
t
³ 0 cos¨©
1. x t
§St ·
cos¨
¸, y c t
© 2 ¹
2
xc t
a
³0
(a) s
xc t
2
§St ·
sin ¨
¸
© 2 ¹
2
yc t dt
K
1
At t
(c) K
yc
a
³0
dt
a, K
Sa
S a.
S (length)
a
§St2 ·
S t cos¨
¸
© 2 ¹
2
§St2 ·
2§St ·
¸ S t sin ¨
¸
© 2 ¹
© 2 ¹
S t cos 2 ¨
a2 3
2 1 3
2
y 1 3 yc
x
3
3
2
§ St2 ·
S t sin ¨
¸, ycc t
© 2 ¹
(b) xcc t
2. x 2 3 y 2 3
§ Su2 ·
¸ du
2 ¹
t
³ 0 sin¨©
St
0
y1 3
Slope at P x, y .
x1 3
rt
cos3 t i sin 3 t j
rc t
3 cos 2 t sin t i 3 sin 2 t cos t j
rc t i
3 cos t sin t
Tt
rc t
rc t
Tc t
sin t i cos t j
cos t i sin t j
Q 0, 0, 0 origin
P
cos3 t , sin 3 t , 0 on curve.
i
JJJK
PQ u T
D
K
j
k
cos3 t sin 3 t 0
cos t sin t 0
JJJK
PQ u T
T
Tc t
rc t
cos3 t sin t sin 3 t cos t k
cos t sin t
1
3 cos t sin t
1
, is three times the
K
distance from the origin to the tangent line.
So, the radius of curvature,
© 2010 Brooks/Cole, Cengage Learning
162
Chapter 12 Vector-Valued Functions
5000 400t , 3200 16t 2
3. Bomb: r1 t
5. xc T
v0 cos T t , v0 sin T t 16t 2
Projectile: r2 t
1 cos T , yc T
xc T
2
2
yc T
5 v0 sin T 16 25
1600
v0 sin T
400.
xcc T
At t
10, bomb is at 5000 400 10
At t
5, projectile is at 5v0 cos T .
cos T 1
200.
2 Ÿ T | 63.43q.
4 sin
1600 Ÿ t
10
Projectile will travel 5 seconds:
5 v0 sin T 16 25
1600
v0 sin T
400.
2
t
and
2
§t·
§t·
16 cos 2 ¨ ¸ 16 sin 2 ¨ ¸
2
© ¹
© 2¹
4 sin
rc
sin T
st
³S
t
1 cos T
2 rc
Horizontal position:
10, bomb is at 5000 400 10
At t
5, projectile is at v0 cos T 5.
K
9000.
2
ª rc
¬
2
sin 2 T dT
Tº
ª
«¬4 cos 2 »¼
S
2
r2º
¼
8 sin
2
Ÿ T | 12.5q.
9
U
1800
| 1843.9 ft/sec
cos T
1
K
s 2 9U 2
7.
rt
d
dt
2
rt
2
t
2
3 T
2
4 sin
2
T
2
T
sin
3
2
4 sin 3 T
2
3 3 cos T
Combining,
v0
4 cos
2 sin 2 T 1 cos T cos T 1 cos T
8 sin 3
400
Ÿ tan T
1800
2 2 cos T dT
3 2
9000
1800.
v0 sin T
v0 cos T
t
³S
rr cc r 2
So,
5v0 cos T
v0 cos T
16.
t
T
t
³ S 2 sin 2 dT
At t
3
1 cos T
6. r
At 1600 feet: Bomb:
3200 16t
T·
§
¨ 2 sin ¸
2¹
©
2
s2 U 2
5000 400t , 3200 16t 2
2
cos T
T
1
K
So, U
v0 cos T t , v0 sin T t 16t 2
t
2
1
200
| 447.2 ft/sec
cos T
Projectile: r2 t
4 cos
T
8 sin 3
4. Bomb: r1 t
2
1 cos T cos T sin T sin T
Combining,
v0
Tº
ª
«4 cos 2 »
¬
¼S
sin T , ycc T
K
1000.
T
t
T
t
³ S 2 sin 2 dT
st
Horizontal position:
400
Ÿ tan T
200
4 sin 2
10 seconds.
Projectile will travel 5 seconds:
v0 sin T
v0 cos T
sin 2T
2 2 cos T
1600 Ÿ t
So, v0 cos T
2
1 cos T
At 1600 feet: Bomb:
3200 16t 2
sin T , 0 d T d S
3
4 sin
T
2
T
2
3
16 cos 2
T
2
16 sin 2
T
2
16
r t ˜r t
2
2 rt
d
rt
dt
r t ˜ rc t r c t ˜ r t Ÿ
d
rt
dt
r t ˜ rc t
rt
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 12
163
x i y j position vector
8. (a) r
r cos T i r sin T j
r
dT º
dT º
ª dr
ª dr
« dt cos T r sin T dt » i « dt sin T r cos T dt » j
¬
¼
¬
¼
dr
dt
d 2r
dt 2
a
2
ª d 2r
dr
dT
dr
dT
d 2T º
§ dT ·
sin T
sin T
r cos T ¨ ¸ r sin T 2 » i
« 2 cos T dt
dt
dt
dt
dt ¼»
© dt ¹
¬« dt
2
ª d 2r
dr
dT
dr
dT
d 2T º
§ dT ·
cos T
cos T
r sin T ¨ ¸ r cos T 2 »
« 2 sin T dt
dt
dt
dt
dt ¼»
© dt ¹
¬« dt
a ˜ cos T i sin T j
a ˜ ur
ar
2
ª d 2r
dr
dT
d 2T º
§ dT ·
r cos 2 T ¨ ¸ r cos T sin T 2 »
« 2 cos 2 T 2 sin T cos T
dt
dt
dt ¼»
© dt ¹
¬« dt
2
ª d 2r
dr
dT
d 2T º
§ dT ·
« 2 sin 2 T 2
sin T cos T
r sin 2 T ¨ ¸ r cos T sin T 2 »
dt
dt
dt ¼»
© dt ¹
¬« dt
a ˜ sin T i cos T j
aT
a ˜ uT
a
a ˜ u r u r a ˜ uT uT
2
d 2r
§ dT ·
r¨ ¸
2
dt
© dt ¹
2
dr dT
d 2T
r 2
dr dt
dt
2
ª d 2r
ª dr dT
d 2T º
§ dT · º
r 2 » uT
« 2 r ¨ ¸ » u r «2
dt ¼
© dt ¹ »¼
«¬ dt
¬ dt dt
§ St ·
§St ·
42,000 cos¨ ¸ i 42,000sin ¨ ¸ j
12
© ¹
© 12 ¹
(b) r
42,000,
r
dT
dt
dr
dt
S d 2T
,
12 dt 2
d 2r
dt 2
0,
0
0
2
So, a
§S ·
42,000¨ ¸ u r
© 12 ¹
Radial component: 875 2
S ur .
3
875 2
S
3
Angular component: 0
9. r t
4 cos t i 4 sin t j 3t k , t
S
rc t
2
4 sin t i 4 cos t j 3k , rc t
rcc t
4 cos t i 4 sin t j
T
Tc
N
B
At t
5
4
4
3
sin t i cos t j k
5
5
5
4
4
cos t i sin t j
5
5
cos ti sin tj
4
3
3
TuN
sin t i cos t j k
5
5
5
S
2
§S ·
, T¨ ¸
©2¹
§S ·
N¨ ¸
©2¹
§S ·
B¨ ¸
©2¹
4
3
i k
5
5
z
6π
B
T
N
3π
B
T
N
4
3
2
1
4
y
x
j
3
4
i k
5
5
© 2010 Brooks/Cole, Cengage Learning
164
Chapter 12 Vector-Valued Functions
S
cos t i sin t j k , t
10. r t
rc t
sin t i cos t j, rc t
11. (a)
2
y
ycc
−1
x
2
2
i
j
2
2
§S ·
N¨ ¸
©4¹
2
2
i
j
2
2
§S ·
B¨ ¸
©4¹
k
15 1 2
x
128
32
25 3 ·
§
x ¸
¨1 4096 ¹
©
−2
K
At the point 4,1 , K
1 constant length Ÿ
d
TuN
ds
dB
A B
ds
(a)
−3
T u Nc Tc u N
dB
dB
dB
So,
A B and
A T Ÿ
ds
ds
ds
−2
0
2
³0
(b) Length
2
³0
WN
T u N. Using Section 11.4, exercise 66,
TuN uT
(c)
dT
ds
Now, KN
Tc s
Tc s
Tc s
S 2 t 2 1 dt
K 0
K1
ª¬S 2 t 2 1º¼
32
2S
S S2 2
S2 1
32
| 1.04
K 2 | 0.51
(d)
5
dT
ds
0
d
BuT
B u Tc Bc u T
ds
B u KN W N u T
KT W B.
dt
S S 2 t2 2
T u T u N
Finally,
Nc s
K
N u T u N
ª¬ T ˜ N T T ˜ T Nº¼
N.
rc t
| 6.766 graphing utility
ª¬ N ˜ N T N ˜ T Nº¼
T
BuT
5
0
(e) lim K
t of
0
(f ) As t o f, the graph spirals outward and the
curvature decreases.
14. (a) Eliminate the parameter to see that the Ferris wheel has a radius of15 meters and is centered at 16 j. At t
the friend is located at r1 0
32
89
| 7.
120
3
for some scalar W .
TuN uN
1
K
2
T ˜ T u Nc T ˜ Tc u N
BuN
32
Ÿr
t cos S t , t sin S t , 0 d t d 2
13. r t
§
Tc ·
T u T ˜ Nc T ˜ ¨ Tc u
¸
¨
Tc ¸¹
©
(b) B
120
89
TuN
dB
ds
dB
T˜
ds
−1
2
B
yc
2
§S ·
, T¨ ¸
4 ©4¹
S
−2
1 −1
−2
1 52
x
32
5 32
x
64
15 1 2
x
128
12. y
1
T sin t i cos t j
Tc cos t i sin t j
N cos t i sin t j
B TuN k
At t
z
4
0,
j, which is the low point on the Ferris wheel.
(b) If a revolution takes ' t seconds, then
S t 't
St
10
10
and so ' t
2S
20 seconds. The Ferris wheel makes three revolutions per minute.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 12
(c) The initial velocity is r c2 t0
8.03 i 11.47 j. The speed is
165
8.032 11.47 2 | 14 m/sec. The angle of inclination
§ 11.47 ·
is arctan ¨
¸ | 0.96 radians or 55q.
© 8.03 ¹
(d) Although you may start with other values, t0
0 is a fine choice. The graph at
the right shows two points of intersection. At t
3.15 sec the friend is near the
20
vertex of the parabola, which the object reaches when
t t0
11.47
| 1.17 sec.
2 4.9
0
So, after the friend reaches the low point on the Ferris wheel, wait t0
2 sec
30
0
before throwing the object in order to allow it to be within reach.
(e) The approximate time is 3.15 seconds after starting to rise from the low point on
the Ferris wheel. The friend has a constant speed of r1c t
15 m/sec. The speed
of the object at that time is
rc2 3.15
8.032 ª¬11.47 9.8 3.15 2 º¼
2
| 8.03 m/sec.
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 3
Functions of Several Variables
Section 13.1
Introduction to Functions of Several Variables.................................167
Section 13.2
Limits and Continuity.........................................................................174
Section 13.3
Partial Derivatives ..............................................................................183
Section 13.4
Differentials ........................................................................................197
Section 13.5
Chain Rules for Functions of Several Variables ...............................203
Section 13.6
Directional Derivatives and Gradients ..............................................212
Section 13.7
Tangent Planes and Normal Lines.....................................................221
Section 13.8
Extrema of Functions of Two Variables ...........................................236
Section 13.9
Applications of Extrema of Functions of Two Variables.................246
Section 13.10 Lagrange Multipliers ..........................................................................256
Review Exercises ........................................................................................................269
Problem Solving .........................................................................................................281
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 3
Functions of Several Variables
Section 13.1 Introduction to Functions of Several Variables
1. No, it is not the graph of a function. For some values of x
0, 0 , there are 2 z-values.
and y for example, x, y
2. Yes, it is the graph of a function.
3. x z 3 y xy
2
2
10 xy 3 y 2
x z
10 xy 3 y 2
x2
Yes, z is a function of x and y.
z
4. xz 2 2 xy y 2
x2
y2
5.
z2
1
4
9
No, z is not a function of x and y. For example,
r1.
x, y
0, 0 corresponds to both z
6. z x ln y 8 yz
z 1 8y
z
0
x ln y
8y 1
3e 2
(d) f 5, y
5e y
(e) f x, 2
xe 2
ln x y
10. g x, y
4
(c) f 30, 5
30 5
150
(d) f 5, y
5y
(e) f x, 2
2x
(f ) f 5, t
5t
(b) g 0, 1
(c) g 0, e
ln 0 e
(d) g 1, 1
ln 1 1
e·
¸
2¹
404
(c) f 2, 3
4 4 36
(d) f 1, y
4 1 4 y2
3 4 y2
(e) f x, 0
4 x2 0
4 x2
© 2010 Brooks/Cole, Cengage Learning
12. f x, y, z
t 2
e
2
ln 3 ln e ln 2
ln 7
23
9
2
3
10
1
0
2 3
4
54
6
3
2
10
3
x y z
0
36
ln 2
xy
z
(d) h 5, 4, 6
(b) f 0, 1
0
1
ln 2 5
(c) h 2, 3, 4
4
ln 1
§ 3e ·
ln ¨ ¸
©2¹
1 ln 3 ln 2
ln e (b) h 1, 0, 1
(a) f 0, 0
0
ln 0 1
(a) h 2, 3, 9
4 x2 4 y 2
4 t2 4
ln 1 0
(a) g 1, 0
6
1 4
2
e
tet
11. h x, y, z
32
5
2e 1
(c) f 2, 1
(f ) g 2, 5
xy
(b) f 1, 4
(f ) f t , 1
(b) f 3, 2
§
(e) g ¨ e,
©
x ln y
Yes, z is a function of x and y.
8. f x, y
5e0
(f ) f t , t
4
No, z is not a function of x and y. For example,
r2.
x, y
1, 0 corresponds to both z
(a) f 3, 2
(a) f 5, 0
10
2
7. f x, y
xe y
9. f x, y
(a) f 0, 5, 4
(b) f 6, 8, 3
(c) f 4, 6, 2
(d) f 10, 4, 3
05 4
3
683
46 2
10 4 3
11
12
2 3
3
167
168
Chapter 13
13. f x, y
Functions of Several Variables
x sin y
§ S·
(a) f ¨ 2, ¸
© 4¹
(b) f 3, 1
2 sin
S
3 sin
y
§ S·
(d) f ¨ 4, ¸
© 2¹
4 sin
§ 3·
3¨¨
¸¸
© 2 ¹
S
3
S
3 3
2
S r 2h
14. V r , h
0 16 12
(b) g 4, 1
1 3 16 12
(c) g 4,
3
2
0
³x
90S
9
4
y
16. g x, y
S 32 10
(a) V 3, 10
y 2 3 y x 2 3x
(a) g 4, 0
(d) g 32 , 0
4
2
2t 3 dt
ª¬t 2 3t º¼
x
2
4
3 sin 1
S·
§
(c) f ¨ 3, ¸
3¹
©
y
³x
15. g x, y
9
2
y
º
ln t »
¼x
(b) V 5, 2
S 52 2
50S
(a) g 4, 1
ln
1
4
ln 4
(c) V 4, 8
S 42 8
128S
(b) g 6, 3
ln
3
6
ln 2
(d) V 6, 4
S 62 4
144S
(c) g 2, 5
ln
5
2
17. f x, y
f x 'x, y f x, y
'x
2 x 'x y 2 2 x y 2
(b)
f x , y 'y f x , y
'y
2 x y 'y 2 x y 2
'y
'x
2
2
(a)
3 x 'x
(b)
f x, y 'y f x, y
'y
3 x 2 2 y 'y 3x 2 2 y
2'x
'x
2, 'x z 0
2 y'y 'y 2
2 y 'y, 'y z 0
'y
2 y 3x 2 2 y
'x
'y
x2 y 2
6 x'x 3 'x
'x
2
6 x 3'x, 'x z 0
2'y
'y
2, 'y z 0
22. f x, y
y
x
Domain:
Range: z t 0
Range: all real numbers
e
xy
23. z
Domain: Entire xy-plane
Range: z ! 0
21. g x, y
Domain:
y
x
ln 14
1
2
Domain:
^ x, y : x is any real number, y is any real number`
20. f x, y
ln
3x 2 2 y
f x 'x, y f x, y
'x
19. f x, y
7
ln y ln x
2x y2
(a)
18. f x, y
ln
25
4
9
4
1
dt
t
§1 ·
(d) g ¨ , 7 ¸
©2 ¹
6
16 12
9
2
9
4
4
x
x ! 0`
x y
xy
Domain:
^ x, y :
x z 0 and y z 0`
Range: all real numbers
y
^ x, y :
^ x, y :
y t 0`
Range: all real numbers
© 2010 Brooks/Cole, Cengage Learning
Section 13.1
24. z
xy
x y
x z y`
(a) View from the positive x-axis: 20, 0, 0
(b) View where x is negative, y and z are positive:
15, 10, 20
Range: all real numbers
4 x2 y2
25. f x, y
(c) View from the first octant: 20, 15, 25
Domain: 4 x 2 y 2 t 0
y 2 d 4`
Range: 0 d z d 2
4 x 4y
2
x2 4 y 2 d 4
x2
y2
d1
4
1
­
½
x2
y2
d 1¾
® x, y :
4
1
¯
¿
Range: 0 d z d 2
Range: 2 d z d 2
(b) z
0 when x
y-axis.
^ x, y : 1 d
0 which represents points on the
(c) No. When x is positive, z is negative. When x is
negative, z is positive. The surface does not pass
through the first octant, the octant where y is
negative and x and z are positive, the octant where y
is positive and x and z are negative, and the octant
where x, y and z are all negative.
33. f x, y
arccos x y
Domain:
32. (a) Domain:
2
Domain: 4 x 4 y t 0
2
27. f x, y
20, 20, 0
^ x, y : x is any real number, y is any real number`
2
26. f x, y
x in the xy-plane:
(d) View from the line y
x2 y 2 d 4
^ x, y : x 2 z
4
Plane: z
5
4
3
x y d 1`
2
Range: 0 d z d S
3
2
1
1
2 3
4
y
5
§ y·
arcsin ¨ ¸
© x¹
28. f x, y
y
­
½
d 1¾
Domain: ® x, y : 1 d
x
¯
¿
Range: 29. f x, y
S
2
d z d
S
2
x
6 2x 3y
34. f x, y
z
6
Plane
Domain: entire xy-plane
Range: f z f
ln 4 x y
4
Domain: 4 x y ! 0
x y 4
^ x, y :
y x 4`
Range: all real numbers
30. f x, y
ln xy 6
3
2
3
4
y
x
y2
35. f x, y
Because the variable x is missing, the surface is a
cylinder with rulings parallel to the x-axis. The
generating curve is z
y 2 . The domain is the entire
xy-plane and the range is z t 0.
z
Domain: xy 6 ! 0
5
xy ! 6
^ x, y :
169
4 x
x2 y2 1
31. f x, y
^ x, y :
Domain:
Introduction to Functions of Several Variables
4
xy ! 6`
Range: all real numbers
1
4
2
3
y
x
© 2010 Brooks/Cole, Cengage Learning
170
Chapter 13
36. g x, y
Plane: z
Functions of Several Variables
1
y
2
1
y
2
z
1
144 16 x 2 9 y 2
12
42. f x, y
1
−2
z
Semi-ellipsoid
−1
1
2
y
2
4
Domain: set of all points
lying on or inside the ellipse
−1
x
37. z
x y
2
2
Paraboloid
Domain: entire xy-plane
Range: z d 0
38. z
1
2
§ x2 · § y 2 ·
¨ ¸¨ ¸
© 9 ¹ © 16 ¹
z
4
−2
4
−4
Range: 0 d z d 1
−2
2
x 2e xy
43. f x, y
y
2
2
x
z
x2 y 2
z
y
2
3
2
x
1
1
2
y
3
44. f x, y
x sin y
x
z
e x
z
4
8
Because the variable y is missing,
the surface is a cylinder with
rulings parallel to the y-axis. The
generating curve is z
e x .
The domain is the entire xy-plane
and the range is z ! 0.
−4
−4
6
4
4
x
y
2
−4
4
4
y
x
40. f x, y
x
1
Cone
Domain of f : entire xy-plane
Range: z t 0
39. f x, y
1
­xy, x t 0, y t 0
®
¯0, else where
e1 x
45. z
2 y2
Level curves:
z
25
e1 x
c
20
Domain of f : entire xy-plane
Range: z t 0
1 x2 y2
ln c
15
x y
2
10
5
2 y2
1 ln c
2
y
Circles centered at 0, 0
5
41. z
y 2 x2 1
x
Matches (c)
z
e1 x
46. z
Hyperbolic paraboloid
2 y2
Level curves:
Domain: entire xy-plane
c
Range: f z f
y
ln c
x y
2
x
2
e1 x
2 y2
1 x2 y 2
1 ln c
Hyperbolas centered at 0, 0
Matches (d)
© 2010 Brooks/Cole, Cengage Learning
y
Section 13.1
ln y x 2
47. z
re
c=2
c=1
c=0
The level curves are of the form
ln y x
c
y x
y
x re
2
171
y
9 x2 y 2
52. f x, y
Level curves:
c
Introduction to Functions of Several Variables
2
x y
2
2
c
1
9 x2 y 2
c
9 c , circles.
2
x y
2
−1
2
2
c=3
x
1
−1
0 is the point 0, 0 .
Parabolas
53. f x, y
Matches (b)
y
The level curves are
hyperbolas of the form
xy
c.
§ x 2y ·
cos¨
¸
4
©
¹
2
48. z
xy
1
−1
Level curves:
x2 2 y 2
4
4 cos 1 c
cos 1 c
x2 2 y 2
2
y
Matches (a)
x y
2
x
4 c=4
2
6 2x 3y
x
x2 y2
c
y
2
1·
§
2
¨x ¸ y
c¹
2
©
3
x
−2
c = 10
The level curves are ellipses of the form
x2 4 y 2
c
0 is the point 0, 0 .
c=0
c=2
c=4
c=6
x
x y2
2
2
c = −3
c=1
2
c=2
0
x
2
2
§1·
¨ ¸ .
© 2c ¹
c= 3
c = −2
2
c= 1
c = −1
2
So, the level curves are circles passing through the origin
and centered at r1 2c, 0 .
56. f x, y
ln x y
c
y
ln x y
ec
x y
y
x ec .
57. f x, y
c = −2
c = − 21
6
So, the level curves are parallel
lines of slope 1 passing through
the fourth quadrant.
x
y
c = −1
The level curves are of the form
x2 4 y 2
−2
c
c=0
c=8
c=0
c=1
c=2
c=3
c=4
2
2
c = −1
y
1
c = 14
c = 13
−2
The level curves are of the form
c=2
2
−1
c = 12
x
x2 y2
55. f x, y
2
−2
except x 2 4 y 2
x
−1
4
c.
−2
51. z
−2
So, the level curves are
hyperbolas.
y
The level curves are of the
c or
form 6 2 x 3 y
2x 3y
6 c. So, the
level curves are straight
lines with a slope of 23 .
1
xy
.
2
c, or ln c
c=4
c=3
c=2
2
The level curves are of the form
e xy
Level curves are parallel
lines of the form x y
e xy
54. f x, y
Ellipses
50. f x, y
1
−1
§ x2 2 y2 ·
cos¨
¸
4
©
¹
c
49. z
c=6
c=5
c=4
c=3
c=2
c=1
x
c = −1
c = −2
c = −3
c = −4
c = −5
c = −6
−4
c=1
c=0
c = 21
c = −1
−6
c=2
c = ± 23
x2 y 2 2
2
6
−2
−9
9
−6
© 2010 Brooks/Cole, Cengage Learning
x
172
Chapter 13
58. f x, y
Functions of Several Variables
4
xy
−6
65. The surface is sloped like a saddle. The graph is not
unique. Any vertical translation would have the same
level curves.
6
One possible function is
f x, y
−4
8
1 x2 y2
59. g x, y
4
xy .
66. The surface could be an ellipsoid centered at 0, 1, 0 .
−6
One possible function is
6
y 1
4
x2 f x, y
−4
1.
10
3 sin x y
60. h x, y
2
1
ª1 0.06 1 R º
1000 «
»
1 I
¬
¼
67. V I , R
−1
1
Inflation Rate
−1
Tax Rate
0
0.03
0.05
0
1790.85
1332.56
1099.43
0.28
1526.43
1135.80
937.09
0.35
1466.07
1090.90
900.04
61. The graph of a function of two variables is the set of all
points x, y, z for which z
f x, y and x, y is in
the domain of f. The graph can be interpreted as a
surface in space. Level curves are the scalar fields
f x, y
c, where c is a constant.
62. No, the following graphs are not hemispheres.
68. A r , t
5000e rt
x2 y 2
z
e
z
x2 y 2
Rate
5
10
15
20
x
y
0.02
5525.85
6107.01
6749.29
7459.12
0.03
5809.17
6749.29
7841.56
9110.59
0.04
6107.01
7459.12
9110.59
11,127.70
0.05
6420.13
8243.61
10,585.00
13,591.41
63. f x, y
Number of Year
The level curves are the lines c
x
or y
y
1
x.
c
These lines all pass through the origin.
xy, x t 0, y t 0
64. f x, y
(a)
69. f x, y, z
z
25
1
20
x y z, c
1
z
x y z , Plane
2
1
15
−2
10
5
−1
y
1
1
2
y
2
5
x
x
(b) g is a vertical translation of f three units downward.
(c) g is a reflection of f in the xy-plane.
(d) The graph of g is lower than the graph of f. If
z
f x, y is on the graph of f, then 12 z is on the
graph of g.
4x y 2z
c
4
4
4x y 2z
z
3
2
Plane
z
(e)
70. f x, y, z
3
x
25
2
1
4
y
20
15
10
5
y
5
x
© 2010 Brooks/Cole, Cengage Learning
Section 13.1
x2 y 2 z 2
71. f x, y, z
Introduction to Functions of Several Variables
z
c
9
9
x2 y2 z 2
5
78. V x, y
4
173
y
25 x y
2
2
c= 1
2
c= 1
3
c=1
25
4
−4
Sphere
4
x
y
4
x
−25
25
−4
−25
x 2
72. f x, y, z
z
1 y2
4
z
1
1
x2 1 y2
4
100 x 0.6 y 0.4
f x, y
79.
5
c
f 2 x, 2 y
z
100 2 x
100 2
Elliptic paraboloid
Vertex: 0, 0, 1
100 2
3
x
5
y
0.6
0.6 0.6
x
0.6
0.4
2y
2
2
0.4
0.4 0.6 0.4
x y
2 ª¬100 x y º¼
0.6 0.4
4x 4 y z
2
73. f x, y, z
2
2
0
0
4x2 4 y 2 z 2
ln C a ln x 1 a ln y
ln z
−2
1
2
Cx a y1 a
80. z
−2
Elliptic cone
2 f x, y
z
2
c
y 0.4
ln z ln y
y
2
ln C a ln x a ln y
x
ln
sin x z
74. f x, y, z
c
0
ln C a ln
x
y
z
1.20 xy 2 0.75 xz 2 0.75 yz
81. C
2
0
sin x z or z
z
y
base
sin x
4
x
front and back
2 ends
1.20 xy 1.50 xz yz
y
8
2
x
§ d 4·
¨
¸ L
© 4 ¹
75. N d , L
z
y
2
(a) N 22, 12
§ 22 4 ·
¨
¸ 12
© 4 ¹
243 board-feet
4
3
S r 2l S r 3
82. V
Sr2
3
3l 4r
2
(b) N 30, 12
76. w
(a)
(b)
(c)
(d)
77. T
§ 30 4 ·
¨
¸ 12
© 4 ¹
r
507 board-feet
l
1
, y x
x y
1
1
w 15, 9
h 10 min
15 9
6
1
1
w 15, 13
h
30 min
15 13
2
1
1
w 12, 7
h 12 min
12 7
5
1
1
w 5, 2
h
20 min
52
3
600 0.75 x 2 0.75 y 2
c = 600
c = 500
c = 400
The level curves are of the form
c
600 0.75 x 0.75 y
2
600 c
.
0.75
The level curves are circles
centered at the origin.
x2 y 2
83. PV
kT
(a) 26 2000
(b) P
kT
V
k 300 Ÿ k
520
3
520 § T ·
¨ ¸
3 ©V ¹
The level curves are of the form
y
30
2
c = 300
c = 200
c = 100
c=0
c
520 § T ·
¨ ¸, or V
3 ©V ¹
520
T.
3c
These are lines through the origin with slope
520
.
3c
x
− 30
30
− 30
© 2010 Brooks/Cole, Cengage Learning
174
Chapter 13
84. (a) z
Functions of Several Variables
0.026 x 0.316 y 5.04
f x, y
Year
2002
2003
2004
2005
2006
2007
z
35.2
39.5
43.6
49.4
53.2
61.6
Model
35.3
39.4
44.0
49.4
53.3
61.8
(b) y has the greater influence because its coefficient (0.316) is greater than x’s coefficient (0.026).
0.026 x 0.316 95 5.04
(c) f x, 95
0.026 x 35.06
This gives the shareholder’s equity in terms of net sales x, assuming total assets of y
85. (a) Highest pressure at C
95 billion .
89. False. Let
(b) Lowest pressure at A
f x, y
2 xy
(c) Highest wind velocity at B
f 1, 2
f 2, 1 , but 1 z 2.
86. Southwest
90. False. Let
87. (a) No; the level curves are uneven and sporadically
spaced.
f x, y
5.
5 z 2 2 f x, y .
Then, f 2 x, 2 y
(b) Use more colors.
88. (a) The different colors represent various amplitudes.
(b) No, the level curves are uneven and sporadically
spaced.
91. True
92. False. If there were a point x, y on the level curves
f x, y
C1 and f x, y
C2 , then C1
C2 .
Section 13.2 Limits and Continuity
1.
lim x
1
3.
x , y o 1, 0
f x, y
x, L
lim
x , y o 1, 3
G -neighborhood about 1, 3 such that
We need to show that for all H ! 0, there exist a
G -neighborhood about 1, 0 such that
x 1 H
x 1
x 1
x 1
So, choose G
2
2
whenever x, y z 1, 3 lies in the neighborhood.
y 0
d
2
x 1
From 0 G , it follows that
2
y 0
2
G.
H and the limit is verified.
y 3
x4 H
Take G
2
yb
2
x4
2
2
x4
2
y 1
2
G
y 3
d
2
x 1
G it follows that
2
y 3
2
G.
H and the limit is verified.
y b H
whenever 0 y 1
2
G.
H.
Then if 0 x4
2
y 3
f x, y L
G
xa
2
4. Let H ! 0 be given. We need to find G ! 0 such that
whenever
0
x 1
So, choose G
2. Let H ! 0 be given. We need to find G ! 0 such that
f x, y L
y 3 H
f x, y L
Whenever x, y z 1, 0 lies in the neighborhood.
From 0 3
y, L
We need to show that for all H ! 0, there exists a
1
f x, y L
3. f x, y
y
H , we have
xa
2
y b
2
G . Take
H.
Then if 0 y b
2
xa
2
y b
2
G
H , we have
H
y b H.
H
x 4 H.
© 2010 Brooks/Cole, Cengage Learning
Section 13.2
5.
6.
7.
8.
9.
lim
¬ª f x, y g x, y º¼
lim
ª 5 f x, y º
«
»
«¬ g x, y »¼
x, y o a, b
x, y o a, b
x, y o a, b
lim
ª f x, y g x, y º
«
»
f x, y
»¼
¬«
2x2 y
lim
x , y o 2,1
54
3
lim
x, y o a, b
lim
x, y o a, b
f x, y lim
x, y o a, b
81
9
lim
x, y o a, b
11.
lim
x , y o 1, 2
e
12
e
f x, y
0 40 1
x y
x2 1
lim
x , y o 2, 4
20.
e
2
x
y
lim
x , y o 0, 2
0
2
43
4
14.
15.
lim
x , y o 1, 2
2 4
22 1
6
5
21.
22.
0
0
§ x·
arccos¨ ¸
© y¹
1 xy
lim
x , y o 0,1
S
arccos 0
1
2
lim
x , y , z o 1, 3, 4
x y z
x
d S
y
13 4
2 2
lim
x , y , z o 2,1, 0
xe yz
2 e1 0
2
Continuous everywhere
23.
1 2
1 2
1
3
Continuous for all x z y.
xy
x2 y 2
25.
lim
arcsin 0
1
Continuous for x y z t 0
24.
x , y o 1,1
arcsin xy
1 xy
lim
x , y o 0,1
Continuous for all y z 0
x y
x y
7
4
Continuous for xy z 1, y z 0, 0 d
Continuous everywhere
13.
12
1
Continuous everywhere
12.
43
g x, y
19.
Continuous everywhere
xy
1
Continuous for xy z 1, xy d 1
x 4y 1
lim
x , y o 0, 0
175
20
3
Continuous everywhere
10.
43
g x, y
ª
ºª
º
f x, y »« lim g x, y »
«¬ x , ylim
o a, b
x
,
y
o
a
,
b
¼¬
¼
¬ª f x, y g x, y ¼º
x, y o a, b
f x, y ª
º
5« lim
f x, y »
¬ x, y o a, b
¼
lim g x, y
lim
x, y o a, b
lim
x, y o a , b
Limits and Continuity
1
2
Continuous except at 0, 0
lim
x , y o 1,1
lim
xy 1
1 xy
x , y o 1, 1
x2 y
1 xy 2
11
11
0
1
11
1
2
1
does not exist
x y
Because the denominator x y approaches 0 as
lim
x , y o 0, 0
x, y o 0, 0 .
16.
lim
x , y o 1,1
x
x y
1
11
2
2
26.
Continuous for x y ! 0
17.
lim
x , y o S 4, 2
y cos xy
2 cos
lim
x , y o 0, 0
1
does not exist because the denominator
x2 y2
xy approaches 0 as x, y o 0, 0 .
S
2
0
27.
lim
x , y o 2, 2
x2 y 2
x y
Continuous everywhere
lim
x y x y
x y
lim
x y
x , y o 2, 2
x , y o 2, 2
18.
lim
x , y o 2S , 4
sin
x
y
sin
2S
4
Continuous for all y z 0
1
28.
lim
x , y o 0, 0
x4 4 y 4
x2 2 y 2
lim
x , y o 0, 0
lim
x , y o 0, 0
4
x2 2 y2 x2 2 y2
x2 2 y 2
x2 2 y2
0
© 2010 Brooks/Cole, Cengage Learning
176
29.
Chapter 13
Functions of Several Variables
x y
x y
lim
x , y o 0, 0
34.
negative values of x and y.
lim
x , y o 2,1
x y 1
˜
x y 1
x y 1
x y 1
x y 1
lim
35. The limit does not exist because along the path
x
0, y
0, you have
x y 1
lim
x , y , z o 0, 0, 0
x y 1
lim
x , y o 2,1
xy yz xz
x2 y 2 z 2
lim
0, 0, z o 0, 0, 0
whereas along the path x
x y 1
x , y o 2,1
ln x 2 y 2 does not exist
because ln x 2 y 2 o f as x, y o 0, 0 .
does not exist because you can’t approach 0, 0 from
30.
lim
x , y o 0, 0
lim
2
x , y , z o 0, 0, 0
y
xy yz xz
x2 y2 z 2
0
z2
0
z , you have
lim
x , x , x o 0, 0, 0
x2 x2 x2
x2 x2 x2
1
31. The limit does not exist because along the line y
you have
lim
x , y o 0, 0
x y
x2 y
lim
x , 0 o 0, 0
x
x2
lim
x , 0 o 0, 0
0
36. The limit does not exist because along the path
y
z
0, you have
1
x
lim
x , y , z o 0, 0, 0
which does not exist.
xy yz 2 xz 2
x2 y2 z 2
0, x
However, along the path z
32. The limit does not exist because along the line
x
y you have
lim
x , y o 0, 0
x
x2 y2
lim
x , x o 0, 0
x
x2 x2
lim
lim
x , x o 0, 0
x , y , z o 0, 0, 0
x
.
0
xy yz 2 xz 2
x2 y2 z 2
33.
lim
x , y o 0, 0
37.
0
1 1
0
x2
0
y, you have
lim
x , x , 0 o 0, 0, 0
x2
x x2
2
1
2
Because the denominator is 0, the limit does not exist.
x2
2
x 1 y2 1
lim
x , 0, 0 o 0, 0, 0
0
lim
x , y o 0, 0
e xy
1
Continuous everywhere
38.
lim
x , y o 0, 0
ª
cos x 2 y 2
«1 x2 y 2
«¬
º
»
»¼
f
The limit does not exist.
Continuous except at 0, 0
39. f x, y
xy
x2 y2
Continuous except at 0, 0
Path: y
0
x, y
1, 0
f x, y
Path: y
0
0.5, 0
0
0.1, 0
0
0.01, 0
0
0.001, 0
0
x
x, y
1, 1
f x, y
1
2
0.5, 0.5
1
2
0.1, 0.1
1
2
0.01, 0.01
1
2
The limit does not exist because along the path y
along the path y
0.001, 0.001
1
2
0 the function equals 0, whereas
x the function equals 12 .
© 2010 Brooks/Cole, Cengage Learning
Section 13.2
40. f x, y
Limits and Continuity
177
y
x2 y2
Continuous except at 0, 0
Path: y
x
x, y
1, 1
f x, y
1
2
Path: y
0.5, 0.5
0.1, 0.1
1
0.01, 0.01
5
50
500
0
x, y
1, 0
f x, y
0
0.5, 0
0
0.1, 0
0.01, 0
0
0.001, 0
0
0
The limit does not exist because along the path y
to infinity.
41. f x, y
0.001, 0.001
0 the function equals 0, whereas along the path y
x the function tends
xy 2
x2 y4
Continuous except at 0, 0
Path: x
y2
x, y
1, 1
f x, y
12
Path: x
0.25, 0.5
12
0.01, 0.1
0.0001, 0.01
12
12
0.000001, 0.001
12
y2
1, 1
x, y
0.25, 0.5
1
2
f x, y
0.01, 0.1
1
2
1
2
0.0001, 0.01
0.000001, 0.001
1
2
1
2
y 2 the function equals 12 , whereas along the path x
The limit does not exist because along the path x
y 2 the function
equals 12 .
42. f x, y
2x y2
2x2 y
Continuous except at 0, 0
Path: y
0
x, y
1, 0
f x, y
Path: y
1
0.25, 0
4
0.01, 0
100
0.001, 0
1000
0.000001, 0
1,000,000
x
x, y
1, 1
f x, y
1
3
0.25, 0.25
1.17
0.01, 0.01
1.95
The limit does not exist because along the y
tends to 2.
0.001, 0.001
1.995
0.0001, 0.0001
2.0
0 the function tends to infinity, whereas along the line y
x the function
© 2010 Brooks/Cole, Cengage Learning
178
43.
Chapter 13
lim
x , y o 0, 0
Functions of Several Variables
x4 y4
x2 y 2
x2 y 2 x2 y 2
lim
x2 y 2
lim
x2 y2
x , y o 0, 0
x , y o 0, 0
0
So, f is continuous everywhere, whereas g is continuous everywhere except at 0, 0 . g has a removable discontinuity at 0, 0 .
44.
lim
x , y o 0, 0
4 x4 y4
2x2 y2
2x2 y2 2x2 y 2
lim
2x2 y2
x , y o 0, 0
lim
x , y o 0, 0
2x2 y 2
0
So, g is continuous everywhere, whereas f is continuous everywhere except 0, 0 . f has a removable discontinuity at 0, 0 .
45.
lim
x , y o 0, 0
So,
4x2 y2
x2 y 2
lim
x , y o 0, 0
51.
0
f x, y
g x, y
lim
x , y o 0, 0
5 xy
x2 2 y2
lim
x , y o 0, 0
Does not exist. Use the paths x
0.
0 and x
y.
z
f is continuous at 0, 0 , whereas g is not continuous at
2
0, 0 .
−3
−3
46.
lim
x , y o 0, 0
f x, y
lim
x , y o 0, 0
lim
x , y o 0, 0
§ x 2 2 xy 2 y 2 ·
¨
¸
x2 y 2
©
¹
§
2 xy 2 ·
¨1 2
¸
x y2 ¹
©
3
y
3
−2
x
1
(same limit for g)
So, f is not continuous at 0, 0 , whereas g is continuous
52.
at 0, 0 .
47.
lim
x , y o 0, 0
sin x sin y
0
6 xy
x2 y 2 1
lim
x , y o 0, 0
0
z
z
4
−4
2
y
−4
x
−2
2
48.
lim
x , y o 0, 0
1
1
sin cos
x
x
4
z
x
2
−2
4
y
−4
Does not exist
53.
4
6
x
lim
x , y o 0, 0
xy 2
2
x y2
lim
r o0
r cos T r 2 sin 2 T
r2
lim r cos T sin 2 T
6
r o0
y
0
2
49.
lim
x , y o 0, 0
x y
x4 2 y2
z
54.
2
Does not exist. Use the
x2.
paths x
0 and y
lim
x , y o 0, 0
2
−2
lim
x y
x2 y
x , y o 0, 0
r o0
r 3 cos3 T sin 3 T
r2
lim r cos3 T sin 3 T
0
y
2
50.
lim
r o0
x
2
x3 y 3
x2 y 2
55.
lim
x , y o 0, 0
x2 y2
x y2
2
lim
r o0
r 4 cos 2 T sin 2 T
r2
lim r 2 cos 2 T sin 2 T
2
z
r o0
0
18
Does not exist
4
x
4
y
© 2010 Brooks/Cole, Cengage Learning
Section 13.2
r cos T , y
56. x
lim
x2 y 2
lim
cos x 2 y 2
58.
lim
sin
59.
x , y o 0, 0
x , y o 0, 0
lim
x , y o 0, 0
60.
lim
x , y o 0, 0
61. x 2 y 2
lim
lim
x , y o 0, 0
lim cos r 2
cos 0
1
lim sin r
sin 0
0
r o0
r o0
x2 y2
sin
sin x 2 y 2
x y
2
lim
r o 0
x y2
2
sin r
r
sin r 2
r o0
r2
lim
2
1
lim
r o0
2r cos r 2
2r
lim cos r 2
r o0
1
r2
1 cos x 2 y 2
lim
x2 y 2
1 cos r 2
r2
x o0
0
r2
x 2 y 2 ln x 2 y 2
lim r 2 ln r 2
r o0
By L’Hôpital’s Rule, lim 2r 2 ln r
r o 0
lim
r o 0
2 ln r
1 r2
x y z
2
lim 2r 2 ln r
r o 0
lim
r o 0
2r
2 r 3
lim r 2
r o 0
2
2
For x 2 z y 2 , let z
Continuous except at 0, 0, 0
lim
z o0
z
2
x y2 4
Continuous for x 2 y 2 z 4.
65. f x, y, z
69. f t
sin z
e ey
2x 3y
t 2 , g x, y
f g x, y
f 2x 3 y
2x 3y
2
Continuous everywhere
f t
70.
xy sin z
67. For xy z 0, the function is clearly continuous.
For xy z 0, let z
sin z
z o0
z
1
x
Continuous everywhere
lim
sin z
z
x 2 y 2 . Then
implies that f is continuous for all x, y.
Continuous everywhere
66. f x, y, z
0
68. For x 2 z y 2 , the function is clearly continuous.
1
63. f x, y, z
64. f x, y, z
0
r
x , y o 0, 0
62. x 2 y 2
lim r cos 2 T sin 2 T
r o0
r
179
r 2 cos 2 T sin 2 T
r, x2 y2
r 2 cos 2 T sin 2 T
r o0
x2 y2
x2 y2
x2 y 2
lim
x2 y2
x , y o 0, 0
57.
r sin T ,
Limits and Continuity
g x, y
f g x, y
1
t
x2 y 2
f x2 y 2
xy. Then
1
x2 y 2
Continuous except at (0, 0)
1
implies that f is continuous for all x, y.
71. f t
1
, g x, y
t
f g x, y
2x 3y
f 2x 3y
Continuous for all y z
1
2x 3 y
2
x
3
© 2010 Brooks/Cole, Cengage Learning
180
Chapter 13
Functions of Several Variables
1
, g x, y
1t
72. f t
x2 y2
f x2 y 2
f g x, y
1
1 x2 y 2
Continuous for x 2 y 2 z 1
73. f x, y
x2 4 y
f x 'x, y f x, y
(a) lim
'x o 0
'x
(b) lim
'y o 0
74. f x, y
f x, y 'y f x, y
'y
ª x 'x
lim ¬
4 yº x 2 4 y
¼
'x
'x o 0
ª x 2 4 y 'y º¼ x 2 4 y
lim ¬
'y o 0
'y
'y o 0
ª x 'x
lim ¬
2
'x o 0
2
y 2 º x2 y 2
¼
'x
'x o 0
f x, y 'y f x, y
(b) lim
'y o 0
'y
ª x 2 y 'y 2 º x 2 y 2
¼
lim ¬
'y o 0
'y
f x, y 'y f x, y
(b) lim
'y o 0
'y
x 'x
x
y
y
lim
'x o 0
'x
x
x
y 'y y
lim
'y o 0
'y
lim
(b) By symmetry, lim
'y o 0
2 x'x 'x
'x
2
lim
2 y'y 'y
'y
2
lim
'x o 0
'y o 0
lim
'x o 0
1
y
xy xy x'y
y 'y y'y
lim
'y o 0
lim 4
4
'y o 0
lim 2 x 'x
2x
lim 2 y 'y
2y
'x o 0
'y o 0
1
y
lim
'y o 0
x'y
y 'y y'y
lim
'y o 0
x
y 'y y
x
y2
1
x y
2
x y x 'x y
x 'x y x y 'x
lim
'x o 0
'x
x 'x y x y 'x
f x, y 'y f x, y
'y
lim
'x o 0
1
x 'x y x y
1
x y
2
.
3x xy 2 y
(a) lim
f x 'x, y f x, y
'x
(b) lim
f x, y 'y f x, y
'y
'y o 0
'x
y
lim
'x o 0 'x
1
1
x 'x y
x y
lim
'x o 0
'x
'x o 0
'x o 0
4'y
'y
2x
1
x y
f x 'x, y f x, y
(a) lim
'x o 0
'x
77. f x, y
lim
lim 2 x 'x
'x o 0
x
y
f x 'x, y f x, y
(a) lim
'x o 0
'x
76. f x, y
2 x'x 'x
'x
x2 y 2
f x 'x, y f x, y
(a) lim
'x o 0
'x
75. f x, y
2
lim
3 x 'x x 'x y 2 y 3 x xy 2 y
'x
3'x y'x
lim
lim 3 y
3 y
'x o 0
'x o 0
'x
3x x y 'y 2 y 'y 3 x xy 2 y
lim
'y o 0
'y
lim
'x o 0
lim
'y o 0
x'y 2'y
'y
lim x 2
'y o 0
x2
© 2010 Brooks/Cole, Cengage Learning
Section 13.2
(a) lim
f x 'x, y f x, y
'x
(b) lim
f x, y 'y f x, y
'y
'y o 0
181
y y 1
78. f x, y
'x o 0
Limits and Continuity
y y 1 'x
lim
'x o 0
lim
y 'y
y y 1
32
y 'y
32
32
12
0
y 3 2 y1 2
'y
'y o 0
12
y 'y
y 'y
y
y1 2
lim
'y o 0
'y o 0
'y
'y
3 1 2 1 1 2
y y
LcHôpital's Rule
2
2
3y 1
2 y
lim
79. True. Assuming f x, 0 exists for x z 0.
83.
xy
.
x y2
80. False. Let f x, y
lim
x , y o 0, 0
x2 y 2
xy
(a) Along y
2
See Exercise 39.
lim
x , ax o 0, 0
ax:
x 2 ax
x ax
­°ln x y , x, y z 0, 0
.
®
0, y
0
x
°̄0,
2
81. False. Let f x, y
2
2
lim
x2 1 a2
ax 2
x o0
1 a2
,a z 0
a
If a
82. True
0, then y
0 and the limit does not exist.
(b) Along
y
2
x :
x2 x2
lim
x , x 2 o 0, 0
x x2
2
1 x2
x o0
x
lim
Limit does not exist.
(c) No, the limit does not exist. Different paths result in
different limits.
84. f x, y
(a) y
x2 y
x y2
4
ax: f x, ax
If a z 0,
(b) y
lim
x , x2
lim
x , ax o 0, 0
x 2 : f x, x 2
x4
2x4
x 2 ax
x ax
4
ax
x2 a2
2
ax
x2 a2
0.
x2 x2
x4 x2
2
x4
2x4
1
2
(c) No, the limit does not exist. f approaches different numbers along different paths.
85.
lim
x , y , z o 0, 0, 0
xyz
x2 y 2 z 2
lim
U o 0
U sin I cos T U sin I sin T U cos I
U2
lim U ª¬sin 2 I cos T sin T cos I º¼
U o 0
86.
lim
x , y , z o 0, 0, 0
ª
º
1
tan 1 « 2
2
2»
¬x y z ¼
ª1º
lim tan 1 « 2 »
¬U ¼
U o 0
0
S
2
© 2010 Brooks/Cole, Cengage Learning
182
Chapter 13
Functions of Several Variables
87. As x, y o 0, 1 , x 2 1 o 1 and x 2 y 1
So,
88.
lim
x , y o 0,1
lim
x , y o 0, 0
ª
x2 1 º
»
tan 1 «
2
2
«¬ x y 1 »¼
0 lim
Because
0 2
lim
xa
2
r 2 cos 2 T r 2 sin 2 T
r2
lim r 2 ª¬cos T sin T cos 2 T sin 2 T º¼
L1 , then for H 2 ! 0, there corresponds G 1 ! 0 such that f x, y L1 H 2 whenever
f x, y
2
G 1.
L2 , then for H 2 ! 0, there corresponds G 2 ! 0 such that g x, y L2 H 2 whenever
g x, y
y b
2
G 2.
Let G be the smaller of G1 and G 2 . By the triangle inequality, whenever
f x, y g x, y L1 L2
So,
lim
x, y o a, b
0
r o0
0.
y b
x, y o a, b
o 0.
.
r o0
x, y o a, b
xa
2
lim r cos T r sin T
f x, y
So, define f 0, 0
89. Because
S
2
x a
2
y b
2
G , we have
f x, y L1 g x, y L2 d f x, y L1 g x, y L2 ¬ª f x, y g x, y º¼
H
2
H
2
H.
L1 L2 .
90. Given that f x, y is continuous, then
lim
x, y o a, b
f x, y
f a, b 0, which means that for each H ! 0, there corresponds
a G ! 0 such that f x, y f a, b H whenever
0 xa
Let H
2
f a, b
y b
2
G.
2, then f x, y 0 for every point in the corresponding G neighborhood because
f x, y f a , b f a, b
2
Ÿ f a, b
2
f x, y f a , b f a, b
2
3
1
Ÿ f a, b f x, y f a, b 0.
2
2
91. See the definition on page 899. Show that the value of
lim
x , y o x0 , y 0
f x, y is not the same for two different paths to x0 , y0 .
92. See the definition on page 902.
93. (a) True
(b) False. The convergence along one path does not imply convergence along all paths.
(c) False. Let f x, y
ª x2
4«
«¬ x 2
2
2
2
2
y 3 º
» .
2
y 3 »¼
(d) True
94. (a) No. The existence of f 2, 3 has no bearing on the existence of the limit as x, y o 2, 3 .
(b) No, f 2, 3 can equal any number, or not even be defined.
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
Partial Derivatives
183
Section 13.3 Partial Derivatives
1. f x 4, 1 0
15. z
wz
wx
wz
wy
2. f y 1, 2 0
3. f y 4, 1 ! 0
4. f x 1, 1
0
16. z
5. No, y only occurs in the numerator.
6. Yes, x occurs in both the numerator and denominator.
7. Yes, x occurs in both the numerator and denominator.
8. No, y only occurs in the numerator.
2x 5 y 3
9. f x, y
f x x, y
2
f y x, y
5
x 2y 4
2
10. f x, y
f x x, y
2x
f y x, y
4 y
11. f x, y
x2 y3
f x x, y
2 xy 3
f y x, y
3x 2 y 2
12. f x, y
4 x3 y 2
2
f x x, y
12 x 2 y 2
f y x, y
8 x3 y 3
z
13.
wz
wx
wz
wy
14. z
wz
wx
wz
wy
x
y
x 2 4 xy 3 y 2
2x 4 y
4 x 6 y
y 3 2 xy 2 1
wz
wx
2 y 2
wz
wy
3 y 2 4 xy
17. z
wz
wx
wz
wy
18. z
e xy
ye xy
xe xy
ex
y
wz
wx
1 x
e
y
wz
wy
x x
e
y2
19. z
wz
wx
wz
wy
20. z
x
y2
x
4y
2 x 2e 2 y
ye y
x
ye yx
1
ye yx ª¬ yx 2 º¼
wz
wy
ey
wz
wx
wz
wy
22. z
x
y
2 xe 2 y
y2 y
e
x2
1
ln
x
2 y2
y
wz
wx
21. z
y
1
x 2e 2 y
y
2
e xy
x
x
y
1 y
ye
x
x
y·
§
e y x ¨1 ¸
x¹
©
x
ln x ln y
1
x
ln
1
y
xy
1
ln xy
2
wz
wx
1 y
2 xy
1
2x
wz
wy
1 x
2 xy
1
2y
© 2010 Brooks/Cole, Cengage Learning
184
Chapter 13
Functions of Several Variables
ln x 2 y 2
30. f x, y
wz
wx
2x
x y2
wf
wx
1
2 x y3
2
1 2
wz
wy
2y
x2 y 2
wf
wy
1
2 x y3
2
1 2
23. z
24. z
2
x y
x y
ln
ln x y ln x y
31. z
wz
wx
1
1
x y
x y
2 y
x y x y
wz
wy
1
1
x y
x y
2x
x y x y
2
25. z
x
3y
2y
x
wz
wx
2x 3 y2
2
2y
x
x3 3 y 3
x2 y
wz
wy
x2
6y
2 y2
x
12 y 3 x3
2 xy 2
26. f x, y
x y
2
x y
2
f y x, y
27. h x, y
wz
wx
wz
wy
wz
wx
wz
wy
x y
2
x xy 2 y
2
x y
2
e
y3 x2 y
2
2
34. z
x2 y 2
2
35.
x2 y 2
hx x, y
2 xe
h y x, y
2 ye
z
wz
wx
wz
wy
x2 y 2
x2 y2
3y2
2 2 x y3
y sin xy
x sin xy
cos x 2 y
2 cos x 2 y
2 sec 2 2 x y
sec 2 2 x y
sin 5 x cos 5 y
wz
wx
wz
wy
2
x3 xy 2
2
2
2
2x y3
tan 2 x y
33. z
x 2 y 2 y xy 2 x
f x x, y
3y2
1
sin x 2 y
32. z
xy
x2 y2
2
cos xy
wz
wx
wz
wy
2
2 x y3
5 cos 5 x cos 5 y
5 sin 5 x sin 5 y
e y sin xy
ye y cos xy
e y sin xy xe y cos x
e y x cos xy sin xy
28. g x, y
ln
x2 y2
1
ln x 2 y 2
2
g x x, y
1 2x
2 x2 y 2
x
x y2
g y x, y
1 2y
2 x2 y2
y
x2 y 2
36.
wz
wx
wz
wy
2
37.
29. f x, y
x2 y 2
f x x, y
1 2
x y2
2
1 2
f y x, y
1 2
x y2
2
1 2
2x
2y
z
x
x y
2
2
y
z
wz
wx
wz
wy
cos x 2 y 2
2 x sin x 2 y 2
2 y sin x 2 y 2
sinh 2 x 3 y
2 cosh 2 x 3 y
3 cosh 2 x 3 y
x2 y2
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
38.
z
wz
wx
wz
wy
39.
cosh xy 2
2 xy sinh xy 2
y
³x
f x, y
f x 'x, y f x, y
'x
3 x 'x 2 y 3 x 2 y
lim
'x o 0
'x
3'x
lim
3
'x o 0 'x
wf
wx
'x o 0
wf
wy
'y o 0
lim
t 2 1 dt
y
ªt 3
º
« t»
¬3
¼x
f x x, y
x2 1
f y x, y
y2 1
§ y3
· § x3
·
y¸ ¨
x¸
¨
© 3
¹ ©3
¹
1 x2
³x
y
2t 1 dt ³y
x
2t 1 dt
y
2t 1 dt ³x
y
2t 1 dt
³x
y
³x
f x x, y
2
f x x, y
2
>2t@xy
2 dt
x 2 2 xy y 2
42. f x, y
x y
lim
'x o 0
lim
'x o 0
lim
'x o 0
wf
wy
2'y
'y
2
x 'x
2
x 2 2 x y 'y y 'y
'y
2
lim 2 x 'x 2 y
'x o 0
x 2 2 xy y 2
2x y
lim 2 x 2 y 'y
2 y x
'y o 0
x y
43. f x, y
wf
wx
lim
f x 'x , y f x , y
'x
wf
wy
'y o 0
3 x 2 y 'y 3 x 2 y
'y
2
2 x 'x y y 2 x 2 2 xy y 2
'x o 0
'x
f x, y 'y f x, y
lim
'y o 0
'y
lim
lim
2 y 2x
'x o 0
lim
f x, y 'y f x, y
'y
'y o 0
wf
wx
lim
lim
'y o 0
>You could also use the Second Fundamental Theorem
of Calculus.@
40. f x, y
185
3x 2 y
41. f x, y
y 2 sinh xy 2
Partial Derivatives
lim
'y o 0
f x 'x, y f x, y
'x
x 'x y 'x
x 'x y 'x
x y
x y
x 'x y x 'x y f x, y 'y f x, y
'y
lim
'y o 0
lim
'y o 0
lim
'y o 0
x y
x y
lim
'x o 0
x y 'y 'y
x y 'y 'y
1
x y 'y 1
x 'x y x y
2
1
x y
x y
x y
x y 'y x y 'y x y
2
x y
x y
1
x y
© 2010 Brooks/Cole, Cengage Learning
186
Chapter 13
Functions of Several Variables
1
x y
44. f x, y
wf
wx
f x 'x, y f x, y
lim
'x o 0
'x
1
1
x 'x y
x y
lim
'x o 0
'x
wf
wy
f x, y 'y f x, y
lim
'y o 0
'y
1
1
x y '
x y
lim
'y o 0
'y
45. f x, y
f x x, y
e y sin x
f x x, y
e
x
f x x, y
cos y
e x cos y
§S S ·
f x x, y
f x x, y
0.
cos 2 x y
f y x, y
S·
§S
2 sin ¨ ¸
2
3¹
©
1.
§S
S·
sin ¨ ¸
3¹
©2
1
.
2
51. f x, y
f x x, y
sin xy
§ y·
¨ 2 ¸
© x ¹
1
§1·
¨ ¸
1 y 2 x2 © x ¹
2
y
x2 y2
x
x2 y2
1
4
arccos xy
y
1 x2 y 2
x
1 x2 y 2
xy
x y
y x y xy
x y
2
At 2, 2 : f x 2, 2
y cos xy
S
4
cos
S
2
0.
f y x, y
2 cos
S
2
At 2, 2 : f y 2, 2
0.
x y
y2
x y
2
2
1
4
x x y xy
x cos xy
§ S·
§ S·
At ¨ 2, ¸, f y ¨ 2, ¸
4
©
¹
© 4¹
x y
At 1, 1 , f y is undefined.
sin 2 x y
§S S ·
1
At 1, 1 , f x is undefined.
2 sin 2 x y
§ S·
§ S·
At ¨ 2, ¸, f x ¨ 2, ¸
© 4¹
© 4¹
f y x, y
50. f x, y
2
x y
1
4
1.
At ¨ , ¸, f y ¨ , ¸
©4 3¹
©4 3¹
48. f x, y
1
1 y 2 x2
At 2, 2 : f y 2, 2
e x sin y
1
y
x
0.
§S S ·
§S S ·
At ¨ , ¸, f x ¨ , ¸
4
3
©
¹
©4 3¹
f y x, y
arctan
At 2, 2 : f x 2, 2
f y x, y
At 0, 0 , f y 0, 0
47. f x, y
f x x, y
1.
e y sin x
At 0, 0 , f x 0, 0
f y x, y
1
x y 'y x y
e y cos x
At S , 0 , f y S , 0
46. f x, y
lim
'y o 0
49. f x, y
At S , 0 , f x S , 0
f y x, y
1
x 'x y x y
lim
'x o 0
x2
x y
2
1
4
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
2 xy
52. f x, y
4 x2 5 y 2
f x x, y
4x2 5 y 2
10
.
27
4x 5 y2
2
8
93 2
At 1, 1 , f y 1, 1
hx x, y
wz
wy
2
4
59.
2 y
At 2, 1 : hy 2, 1
2
49 x 2 y 2 , x
2, 2, 3, 6
60.
45 y 2
Intersecting curve: z
y
45 y 2
At 2, 3, 6 :
wz
wy
y=3
9x2 9
4
18 1
wz
wy
2x
h y x, y
y
3
45 9
1
2
61.
z
x=2
10
18
3
2
4
z
1, 1, 3, 0
40
9 y2
2 3
6
4
4
x
H x, y , z
sin x 2 y 3 z
H x x, y , z
cos x 2 y 3 z
H y x, y , z
2 cos x 2 y 3 z
H z x, y , z
3 cos x 2 y 3z
f x, y , z
3 x 2 y 5 xyz 10 yz 2
f x x, y , z
6 xy 5 yz
f y x, y , z
3 x 2 5 xz 10 z 2
f z x, y , z
5 xy 20 yz
w
x2 y2 z 2
ww
wx
x y2 z2
ww
wy
x y2 z2
ww
wz
x2 y2 z 2
x
2
y
2
z
8
x
8
y
62.
w
ww
wx
ww
wy
ww
wz
y
x
2 y
At 1, 3, 0 :
At 2, 1 : hx 2, 1
4
18 x
9x2 y 2 , x
58. z
x2 y 2
54. h x, y
4
x
3, 1, 3, 0
Intersecting curve: z
At 1, 1 : g y 1, 1
4
160
wz
At 1, 3, 0 :
wx
2 y
g y x, y
wz
wy
wz
wx
2
At 1, 1 : g x 1, 1
22
Intersecting curve: z
8
.
27
2 x
g x x, y
wz
wx
z
4 x2 y 2
53. g x, y
x2 4
9 x2 y2 , y
57. z
32
20
2x
At 2, 1, 8 :
8 x3
f y x, y
55. z
wz
wx
32
10
93 2
At 1, 1 , f x 1, 1
1, 2, 1, 8
Intersecting curve: z
10 y 3
187
z
x2 4 y2 , y
56. z
Partial Derivatives
7 xz
x y
7 xz x y
x y 7 z 7 xz
x y
2
1
7 yz
x y
2
7 xz
x y
2
7x
x y
© 2010 Brooks/Cole, Cengage Learning
y
188
Chapter 13
63. F x, y, z
68.
Fy x, y, z
y
x2 y 2 z 2
f x 3, 1, 1
Fz x, y, z
z
x2 y 2 z 2
f y x, y , z
1
G x, y , z
x
1 x2 y2 z 2
3
x yz
f x x, y , z
3 x 2 yz 2
f y x, y , z
f y 1, 1, 1
f z x, y , z
f z 1, 1, 1
3 2
x z
67.
f x x, y , z
z cos x y
2 xy 3 2 yz
f x 2, 1, 2
4 4
f y x, y , z
3x y 2 xz 3 z
f y 2, 1, 2
12 8 6
f x x, y , z
0
2 2
2
f x 1, 2, 1
f y x, y , z
2 xy 3 y
4 3
7
f y 1, 2, 1
f z x, y , z
f x x, y , z
1
yz
f z 1, 2, 1
f z x, y , z
f z 1, 1, 1
2
S
0
2
z cos x y
4 cos
S
0
2
sin x y
sin
S
1
2
3x 2 y 2 2 z 2
x
yz
f y 1, 1, 1
x y z
2 3
f x, y , z
f y x, y , z
xy
2
f x x, y , z
f x 1, 1, 1
4 cos
§ S
·
f z ¨ 0, , 4 ¸
© 2
¹
2 x3 yz
x y 2 xyz 3 yz
f z 2, 1, 2
z sin x y
f z x, y , z
1
2
1
3
f x, y , z
§ S
·
f y ¨ 0, , 4 ¸
© 2
¹
3
f x, y , z
f z x, y , z
3
9
f y x, y , z
70.
66.
x y z
§ S
·
f x ¨ 0, , 4 ¸
2
©
¹
2
f x, y , z
f x 1, 1, 1
69.
32
x y z
2
2
x 2 xz
93
2
32
3
x y z 0 xy
32
z
Gz x, y , z
2
x y z
y 2 yz
x y z
11
0
32
x y z x xy
f z 3, 1, 1
1 x2 y2 z 2
2
x y z
f z x, y , z
32
y
G y x, y , z
x y z y xy
f y 3, 1, 1
1 x2 y2 z 2
1 x2 y2 z 2
xy
x y z
f x, y , z
f x x, y , z
2
Gx x , y , z
65.
1
ln x 2 y 2 z 2
2
x2 y2 z 2
ln
x
x y2 z2
Fx x, y, z
64.
Functions of Several Variables
6x
3x
2 3x 2 y 2 2 z 2
3x 2 y 2 2 z 2
6
2 342
2y
3
5
2 3x y 2 z
2
3x y 2 2 z 2
2 3x 2 y 2 2 z 2
3x 2 y 2 2 z 2
2
2
2
5
3 5
5
y
2
2 5
5
4 z
2
5
2 z
2 5
5
1
x
y2 z
71. z
3 xy 2
w2 z
wywx
1
wz
wx
3y2 ,
w2 z
wx 2
0,
x
yz 2
wz
wy
6 xy,
w2 z
wy 2
6 x,
w2 z
wxwy
6y
6y
1
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
x2 3 y2
72. z
73.
2 x,
wz
wy
w2 z
6 y, 2
wy
wz
wx
w2 z
wx 2
w2 z
wywx
wz
wy
74.
z
wz
wx
w2 z
wx 2
0
w2 z
wywx
2
wz
wy
w2z
wy 2
2
2
w2 z
wxwy
2 x 6 y
2
w2 z
wxwy
w2 z
6,
wxwy
z
wz
wx
0
2x 2 y
w2 z
wxwy
w2z
wy 2
w2 z
wywx
x 2 xy 3 y
6
z
2,
2
w2z
wy 2
wz
wx
w2 z
wx 2
w2 z
wywx
wz
wy
75.
w2 z
wx 2
wz
wx
z
76.
So,
77.
x 4 3x 2 y 2 y 4
4 x 6 xy
2
12 x 6 y
2
3
2
12 xy
w2 z
wywx
x y
2
2
x
x2 y 2
w2 z
wx 2
x2 y2
w2 z
wywx
x2 y2
wz
wy
x y
y2
32
xy
32
y
2
w2z
wy 2
x2 y2
w2 z
wxwy
x2 y2
2
1
x y
2
1
x y
1
y x
1
x y
2
w2 z
.
wxwy
e x tan y
w2 z
wywx
e x sec 2 y
wz
wy
e x sec 2 y
wz
wx
w2 z
wx 2
w2 z
wywx
wz
wy
w2z
wy 2
w2 z
wxwy
2
1
x y
w2 z
wx 2
z
2
1
x y
e x tan y
w2 z
wxwy
78.
wz
wx
6 x 2 12 y 2
12 xy
1
x y
e x tan y
w2z
wy 2
189
ln x y
z
6 x 2 y 4 y 3
Partial Derivatives
2e x sec 2 y tan y
e x sec 2 y
2 xe y 3 ye x
2e y 3 ye x
3 ye x
2e y 3 ye x
2 xe y 3e x
2 xe y
2e y 3e x
x2
32
xy
32
© 2010 Brooks/Cole, Cengage Learning
190
Chapter 13
79. z
Functions of Several Variables
cos xy
wz
wx
x sin xy,
w2 z
wxwy
x y
85. f x x, y
y
x2 y2
y
y x
2
2
2
fx
2
x 2x
2
12 x 3 y 2
0 Ÿ y
4y
2
4x
y 2 4, you
Solving for x in the second equation, x
2
4 y.
x2 y2
0
f y x, y
x 2y 2
0
0 Ÿ y
x 2 2 2x 2
64 y Ÿ y
0 or y
4
31 3
Ÿ x
0 or x
1 § 16 ·
¨
¸
4 © 32 3 ¹
y2 x2
2
2x y 2
2
§ 4 4 ·
Points: 0, 0 , ¨ 2 3 , 1 3 ¸
3 ¹
©3
2 2x
0 Ÿ 3 x 6
0 Ÿ x
2,
2
87. f x x, y
2x y ex
2 xy y 2
f y x, y
x 2 y ex
2 xy y 2
2x y
Point: 2, 2
0 Ÿ y
x 2 2 x
2x y 5
0
x 2 y 1
0 Ÿ y
x 2 2x 5 1
0
0
2 x
0 Ÿ x
0 Ÿ y
0
Point: 0, 0
0
2x 5
0 Ÿ 3x 9
0 Ÿ x
3,
1
88. f x x, y
2x
x2 y 2 1
0 Ÿ x
0
f y x, y
2y
x2 y2 1
0 Ÿ y
0
Points: 0, 0
Point: 3, 1
fy
1
y2
0 Ÿ 3x 2
3 y 12 x
x
x2 y 2
81. f x x, y
83. f x x, y
0
x
1
0: 9 x 2 12 y
fy
3y4
2
x y
2x y 5
1
x
y2
2
2
f y x, y
0 and 9 x 2 12 y, f y x, y
obtain 3 y 2 4
x2 y2
x y
y4 Ÿ y
1
x
y2
1
and x
x2
2
2 xy
82. f x x, y
1
y
x2
2
x2 y2
1
§1·
¨ ¸
1 y 2 x2 © x ¹
2x y 2
0: fy
86. f x x, y
y 2y
2
x y
w2 z
wxwy
1
y , f y x, y
x2
2
2
2
0
Points: 1, 1
2
x y
w2z
wy 2
0, y
y
2
wz
wy
fx
0 Ÿ x
x 2 cos xy
2 xy
2
w z
wywx
2x
Point: 0, 0
w2 z
wy 2
1
§ y·
¨ ¸
1 y 2 x2 © x2 ¹
2
0
y
x
arctan
w2 z
wx 2
y
0 Ÿ y
x 2 2x
fx
wz
wx
y
2x y
xy cos xy sin xy
z
0
x 2 y
f y x, y
y 2 cos xy
yx cos xy sin xy
wz
wy
80.
w z
wx 2
y sin xy,
w2 z
wywx
2x y
84. f x x, y
2
2 x 4 y 4, f y x, y
0: 2 x 4 y
4x 2 y
4 x 2 y 16
4
16
Solving for x and y,
x
6 and y
4.
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
89.
z
x sec y
wz
wx
wz
wx
0
w2 z
wywx
w2 z
wx 2
sec y tan y
wz
wy
w z
wy 2
x sec y sec2 y tan 2 y
w2 z
wxwy
sec y tan y
w2 z
So,
wywx
w2z
wy 2
w2 z
wxwy
There are no points for which z x
wz
wx
z y , because
92.
25 x y
2
25 x y
2
2
32
w2 z
wx 2
32
w2 z
wywx
xy
25 x y
2
wz
wy
2
y
25 x y
2
w2z
wy 2
wz
wy
2
x 2 25
25 x y
2
w2 z
wxwy
2
w2z
wy 2
32
xy
25 x y
2
0 if x
y
2
wz
wy
z
wz
wx
2
y 2 25
w2 z
wywx
2
x2 x2 y 2
4 xy
2
x2 y2
2y
x2 y 2
2 y 2 x2
2
x2 y2
4 xy
2
x2 y2
There are no points for which z x
x
w2 z
wx 2
wz
wx
0
y 2 x2
x x2 y 2
x4 4x2 y 2 y4
w2 z
wxwy
sec y z 0.
25 x 2 y 2
90. z
wz
wy
w2 z
wxwy
32
93.
y x y xy
0.
y2
2
x y
x y
2
2 y2
x y
3
x y
2
2 y y 2 2 x y 1
x y
x x y xy
x y
2 xy
4
x y
x2
2
x y
2
2 x2
x y
3
x y
2
2x x2 2 x y
x y
f x, y , z
xyz
f x x, y , z
yz
f y x, y , z
xz
f yy x, y, z
0
f xy x, y, z
z
f yx x, y, z
z
f yyx x, y, z
0
f xyy x, y, z
0
f yxy x, y, z
0
So, f xyy
zy
xy
x y
4
There are no points for which z x
0
191
ln x ln x 2 y 2
1
2x
2
x
x y2
w2 z
wywx
x sec y tan y
2
wz
wx
§
·
x
ln ¨ 2
2¸
©x y ¹
91. z
sec y
w2 z
wx 2
Partial Derivatives
f yxy
f yyx
2 xy
x y
zy
3
0.
0.
© 2010 Brooks/Cole, Cengage Learning
3
192
94.
Chapter 13
f x, y , z
x 2 3xy 4 yz z 3
f x x, y , z
2x 3y
f y x, y , z
3 x 4 z
f yy x, y, z
0
f xy x, y, z
3
f yx x, y, z
3
f yyx x, y, z
0
f xyy x, y, z
0
f yxy x, y, z
0
So, f xyy
95.
f x, y , z
wz
wx
w2z
wx 2
wz
wy
x
f x x, y , z
e
f y x, y , z
ze x cos yz
f yy x, y, z
z 2e x sin yz
f xy x, y, z
ze x cos yz
f yx x, y, z
ze x cos yz
f yyx x, y, z
z 2e x sin yz
f yxy x, y, z
So, f xyy
f x, y , z
f x x, y , z
f y x, y , z
f yy x, y, z
f xy x, y, z
f yx x, y, z
f yyx x, y, z
f xyy x, y, z
f yxy x, y, z
2 x
sin yz
2 x
sin yz
z e
f yxy
2 z
§ e y e y ·
cos x¨
¸
2
©
¹
w2z
wx 2
§ e y e y ·
sin x¨
¸
2
©
¹
wz
wy
§ e y e y ·
sin x¨
¸
2
©
¹
w2 z
wy 2
§ e y e y ·
sin x¨
¸
2
©
¹
e x sin y
wz
wx
e x sin y
2
3
3
w2z
wx 2
wz
wy
w2 z
wy 2
4z
3
So,
12 z
x y
4
12 z
x y
4
0.
e x cos y
e x sin y
w2 z
w2z
2
2
wx
wy
4
100. z
§ e y e y ·
x¨
¸
2
©
¹
e x sin y
arctan
12 z
x y
§ e y e y ·
sin x¨
¸ sin
2
©
¹
z
99.
4z
x y
0.
wz
wx
2
4z
x y
00
§ e y e y ·
sin x¨
¸
2
©
¹
2 z
x y
0
w2 z w2 z
wx 2 wy 2
2z
x y
x y
5x
So,
f yyz .
x y
0
z
98.
sin yz
z e
5y
0.
sin yz
x
5 xy
w2 z
w2 z
2
2
wx
wy
So,
f yyx
e
z
97.
w2 z
wy 2
f yxy
f xyy x, y, z
96.
Functions of Several Variables
e x sin y e x sin y
0.
y
x
From Exercise 80, we have
w2 z
w2z
2
2
wx
wy
2 xy
x y
2
2
2
2 xy
x y2
2
2
0.
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
sin x ct
z
101.
wz
wt
c cos x ct
w2 z
wt 2
wz
wx
c 2 sin x ct
cos x ct
w2 z
wx 2
102.
wz
wt
w2 z
wt 2
wz
wx
cos 4 x 4ct
16c 2 cos 4 x 4ct
w2 z
wt 2
c 2 16 cos 4 x 4ct
z
ln x ct
wz
wt
c
x ct
w2 z
wt 2
c 2
wz
wx
w z
wx 2
2
w2 z
wt 2
z
wz
wt
w2 z
wt 2
wz
wx
w2 z
wx 2
w2 z
So, 2
wt
x ct
x
c
x
1 t
e cos
c
c
wz
wt
1 t
x
e sin
c2
c
§ w2 z ·
c 2 ¨ 2 ¸.
© wx ¹
107. Yes. The function f x, y
cos 3 x 2 y satisfies
both equations.
does not exist.
109.
2
w
ª f x x, y, z ¼º
wx ¬
0
There are no xs in the expression.
w
110.
ª f x, y, z ¼º
wx ¬
1
2
c 2
x ct
x
c
108. A function f x, y with the given partial derivatives
1
x ct
x ct
So,
§ w2z ·
c2 ¨ 2 ¸
© wx ¹
§ w2 z ·
c 2 ¨ 2 ¸.
© wx ¹
e t sin
w2 z
wx 2
4 sin 4 x 4ct
1 t
x
e cos
c2
c
e t sin
wz
wt
wz
wx
4c sin 4 x 4ct
16 cos 4 x 4ct
z
106.
w2 z
wx 2
103.
104.
§ w2z ·
c 2 ¨ 2 ¸.
© wx ¹
x
c
1
x
e t sin
c
c
wz
wt
So,
z
e t cos
w2 z
wx 2
sin x ct
w2 z
wt 2
So,
wz
wt
wz
wx
193
x
c
e t cos
z
105.
Partial Derivatives
2
§ w2 z ·
c2 ¨ 2 ¸
© wx ¹
y2 2
y 1 z
f x, y , then to find f x you consider y constant
111. If z
and differentiate with respect to x. Similarly, to find
f y , you consider x constant and differentiate with
respect to y.
sin wct sin wx
wc cos wct sin wx
y·
§
¨ sinh ¸
z¹
©
z
112.
z
(x0, y0, z 0 )
(x0, y0, z 0 )
w2c 2 sin wct sin wx
w sin wct cos wx
w2 sin wct sin wx
2§ w
z·
c ¨ 2 ¸.
© wx ¹
2
y
x
Plane: y = y0
y
x
Plane: x = x0
wf
denotes the slope of surface in the x-direction.
wx
wf
denotes the slope of the surface in the y-direction.
wy
© 2010 Brooks/Cole, Cengage Learning
194
Chapter 13
Functions of Several Variables
x y
113. The plane z
f x, y satisfies
wf
wf
! 0.
0 and
wy
wx
118. (a) C
z
wC
wx
4
4
y
4
x
x y
114. The plane z
wC
wy
y
175
x
z
6
16
16 4 205
x
119. IQ M , C
−6
115. In this case, the mixed partials are equal, f xy
sin x 2 y
cos x 2 y
f xx x, y
sin x 2 y
f xy x, y
2 sin x 2 y
f y x, y
2 cos x 2 y
f yy x, y
4 sin x 2 y
f yx x, y
2 sin x 2 y
So, f xy x, y
100
IQM
120. f x, y
200 x 0.7 y 0.3
wf
wx
140 x 0.3 y 0.3
(a)
f yx x, y .
At x, y
200 x1 200 x2 4 x12 8 x1 x2 4 x22
wr
(a)
200 8 x1 8 x2
wx1
wR
4, 12 ,
200 32 96
At x1 , x2
wx1
wR
200 8 x1 8 x2
(b)
wx2
wR
4, 12 , 2
At x1 , x2
72.
wx
M
C
100
, IQM 12, 10
10
C
100M
IQc
, IQc 12, 10
12
C2
When the chronological age is constant, IQ increases at a
rate of 10 points per mental age year.
When the mental age is constant, IQ decreases at a rate
of 12 points per chronological age year.
f yx .
See Theorem 13.3.
f x, y
237
(b) The fireplace-insert stove results in the cost
wC
wC
increasing at a faster rate because
!
.
wy
wx
y
f x x, y
183
x
205
y
8
116.
1
175
4
16
wC º
»
wy ¼ 80, 20
f x, y satisfies
wf
wf
! 0.
! 0 and
wy
wx
16
wC º
wx »¼ 80, 20
2
2
xy 175 x 205 y 1050
32
117. R
72.
(b)
§ 500 ·
140¨
¸
© 1000 ¹
wf
wx
60 x 0.7 y 0.7
wf
wx
0.3
1000, 500 ,
wf
wx
At x, y
§ y·
140¨ ¸
© x¹
0.3
§1·
140¨ ¸
© 2¹
§ x·
60¨ ¸
© y¹
0.3
| 113.72.
0.7
1000, 500 ,
§ 1000 ·
60¨
¸
© 500 ¹
0.7
60 2
0.7
| 97.47.
121. An increase in either price will cause a decrease in demand.
10
122.
V I, R
VI I , R
VI 0.03, 0.28
ª1 0.06 1 R º
1000 «
»
1 I
¬
¼
9
ª1 0.06 1 R º ª 1 0.06 1 R º
»
10,000 «
» «
2
1 I
»¼
1 I
¬
¼ «¬
11,027.20
9
VR I , R
VR 0.03, 0.28
ª1 0.06 1 R º ª 0.06 º
10,000 «
» «
»
1 I
¬
¼ ¬ 1 I¼
ª 1 0.06 1 R
10,000 «
11
«
1 I
¬
ª 1 0.06 1 R
600 «
10
«
1 I
¬
9
10
º
»
»
¼
º
»
»
¼
653.26
The rate of inflation has the greater negative influence.
© 2010 Brooks/Cole, Cengage Learning
Section 13.3
123. T
wT
wx
wT
wy
500 0.6 x 2 1.5 y 2
wT
1.2 x,
2, 3
2.4q m
wx
wT
3 y
2, 3
9q m
wy
0.885t 22.4h 1.20th 0.544
wA
(a)
0.885 1.20h
wt
wA
30q, 0.80
0.885 1.20 0.80
1.845
wt
wA
22.4 1.20t
wh
wA
30q, 0.80
22.4 1.20 30q
13.6
wh
(b) The humidity has a greater effect on A because its
coefficient 22.4 is larger than that of t.
PV
T
P
V
wT wP wV
˜
˜
wP wV wT
n
RT
xB
wT
PV
V
Ÿ
n
n
w
P
R
R
xB
xB
n
n
RT
RT
wP
xB
xB
Ÿ
wV
V
V2
n
n
RT
R
wV
xB
xB
Ÿ
wT
P
P
§
·§ n
·§ n ·
¨ V ¸¨ xB RT ¸¨ xB R ¸
¨ n ¸¨ ¸
2 ¸¨
¸¨ V ¸¨
¸¨ P ¸¸
R ¸¨
¨¨
© xB ¹©
¹©
¹
n
n
RT
RT
xB
xB
1
n
VP
RT
xB
195
5 x 2 xy 3 y 2
126. U
(a) U x
10 x y
(b) U y
x 6y
17 and U y 2, 3
(c) U x 2, 3
16. The person
should consume one more unit of y because the rate
of decrease of satisfaction is less for y.
124. A
125.
Partial Derivatives
(d)
z
1
−2
2
1
x
127. z
(a)
2
y
0.92 x 1.03 y 0.02
wz
wx
wz
wy
0.92
1.03
(b) As the consumption of flavored milk (x) increases,
the consumption of plain light and skim milk (z)
decreases. As the consumption of plain reduced-fat
milk (y) decreases, the consumption of plain light
and skim milk decreases.
1.2225 x 2 0.0096 y 2 71.381x 4.121y 354.65
wz
(a)
2.445 x 71.381
wx
w2 z
2.445
wx 2
wz
0.0192 y 4.121
wy
128. z
w2 z
wy 2
0.0192
§ w2 z
·
(b) Concave downward ¨ 2 0 ¸
© wx
¹
The rate of increase of Medicare expenses z is declining with respect to worker’s compensation expenses x .
§ w2 z
·
(c) Concave upward ¨ 2 ! 0 ¸
y
w
©
¹
The rate of increase of Medicare expenses z is increasing with respect to public assistance expenses ( y).
© 2010 Brooks/Cole, Cengage Learning
196
Chapter 13
Functions of Several Variables
129. False
x y 1.
Let z
130. True
131. True
132. True
­ xy x 2 y 2
° 2
, x, y z 0, 0
® x y2
°
x, y
0, 0
¯0,
133. f x, y
x 2 y 2 3x 2 y y 3 x3 y xy 3 2 x
(a) f x x, y
x2 y 2
y x4 4x2 y 2 y4
2
x 2 y 2 x3 3 xy 2 x3 y xy 3 2 y
f y x, y
x2 y 2
x x4 4 x2 y 2 y 4
2
(b) f x 0, 0
2
0 ª 'x º 0
¬
¼
lim
'x o 0
'x
0
f y 0, 0
f 0, 'y f 0, 0
lim
'y o 0
'y
2
0 ª 'y º 0
¬
¼
lim
'y o 0
'y
0
w § wf ·
¨ ¸
wy © wx ¹
w § wf ·
¨ ¸
wx © wy ¹
f yx 0, 0
lim
0, 0
0, 0
'y o 0
2
x2 y2
f 'x, 0 f 0, 0
lim
'x o 0
'x
(c) f xy 0, 0
2
x2 y2
f x 0, 'y f x 0, 0
'y
'y o 0
f y 'x, 0 f y 0, 0
lim
'x o 0
'x
4
'y 'y
lim
'y
2 2
lim
'x
2 2
1
4
'x 'x
'x o 0
lim 1
'y o 0
'y
lim 1
1
'x o 0
'x
(d) f yx or f xy or both are not continuous at 0, 0 .
y
³x
134. f x, y
1 t 3 dt
By the Second Fundamental Theorem of Calculus,
wf
wx
d y
1 t 3 dt
dx ³ x
wf
wy
d y
dy ³ x
x3 y 3
135. f x, y
d x
1 t 3 dt
dx ³ y
1 t 3 dt
(a) f x 0, 0
1 x3
1 y3 .
f y 0, 0
13
f 0 'x, 0 f 0, 0
'x
'x
lim
1
'x o 0 'x
f 0, 0 'y f 0, 0
lim
'y o 0
'y
'y
lim
1
'y o 0 'y
lim
'x o 0
(b) f x x, y and f y x, y fail to exist for
y
136. f x, y
x2 y2
23
For x, y z 0, 0 , f x x, y
For x, y
f x 0, 0
x, x z 0.
2 2
x y2
3
1 3
4x
2x
3 x y2
2
13
.
0, 0 , use the definition of partial derivative.
lim
'x o 0
f 0 'x f 0, 0
'x
43
lim
'x o 0
'x
'x
lim 'x
'x o 0
13
0
© 2010 Brooks/Cole, Cengage Learning
Section 13.4
Differentials
197
Section 13.4 Differentials
2 x2 y3
1. z
dz
z
2.
11. f x, y
4 xy dx 6 x y dy
3
2 2
(a) f 2, 1
'z
2x
x
dx 2 dy
y
y
(b) dz
12. f x, y
1
x2 y2
3. z
dz
x y
2
dx 2
x dx y dy
2
x y
2
dw
2
2y
2
x y
2
2
2
'z
(b) dz
x y
z 3y
13. f x, y
1
3x z
x y
dx dy 2
z 3y
z 3y
z 3y
2
dz
5
5.5125
f 2.1, 1.05 f 2, 1
(b) dz
cos y y sin x dx x sin y cos x dy
0.5125
2 x dx 2 y dy
0.5
16 x 2 y 2
11
f 2.1, 1.05
'z
0.05
x2 y 2
(a) f 2, 1
dz
0.05
2 0.1 3 0.05
2 2 0.1 2 1 0.05
x cos y y cos x
5. z
2 dx 3 dy
f 2.1, 1.05
dy
1.05
f 2.1, 1.05 f 2, 1
(a) f 2, 1
2x
2
4. w
1
f 2.1, 1.05
x2
y
2
dz
2x 3y
10.4875
f 2.1, 1.05 f 2.1
0.5125
2 x dx 2 y dy
2 2 0.1 2 1 0.05
0.5
cos y y sin x dx x sin y cos x dy
§1·
¨ ¸e
© 2¹
6. z
dz
14. f x, y
x2 y 2
§e
2 x¨
¨
©
x2 y2
e
x2 y2
e
2
(a) f 2, 1
x2 y2
·
¸ dx
¸
¹
§ e x2 y 2 e x2 y2 ·
¸ dy
2 y¨
¨
¸
2
©
¹
ex
2 y2
e x
y
x
2 y2
x dx y dy
0.5
f 2.1, 1.05
'z
(b) dz
15. f x, y
f 2.1, 1.05 f 2, 1
y
1
dx dy
x2
x
z
dz
8.
w
dw
9. w
dw
10. w
dw
e x sin y
'z
(b) dz
e y cos x z 2
16. f x, y
2 z 3 y cos x dx 2 z 3 sin x dy 6 z 2 y sin x dz
x yz sin yz
2
2
2 xyz dx x z z cos yz dy
2
2 2
2 x 2 yz y cos yz dz
0
1.05e 2.1 | 8.5745
f 2.1, 1.05 f 2, 1
1.1854
ye dx e dy
x
x
e 2 0.1 e 2 0.05 | 1.1084
e y sin x dx e y cos x dy 2 z dz
2 z 3 y sin x
1
1
0.1 0.05
4
2
e 2 | 7.3891
f 2.1, 1.05
e x sin y dx e x cos y dy
0
ye x
(a) f 2, 1
7.
0.5
(a) f 2, 1
x cos y
2 cos 1 | 1.0806
f 2.1, 1.05
2.1 cos 1.05 | 1.0449
'z
f 2.1, 1.05 f 2, 1
(b) dz
cos y dx x sin y dy
0.0357
cos 1 0.1 2 sin 1 0.05 | 0.0301
© 2010 Brooks/Cole, Cengage Learning
198
Chapter 13
Functions of Several Variables
x 2 y, x
17. Let z
2, y
9, dx
0.01, dy
0.02.
2 xy dx x dy
2
Then: dz
2
2.01 9.02 22 ˜ 9 | 2 2 9 0.01 22 0.02
x2 y2 , x
18. Let z
5, y
3, dx
0.44
0.05, dy
0.1.
Then:
x
dz
2
5.05
dz
2
3.1
1 x2
19. Let z
x2 y 2
dy
5
52 32 |
y2, x
3, y
52 32
6, dx
3
0.05 52 32
0.55
| 0.094
34
0.1
0.05. Then:
0.05, dy
2 1 x 2
2x
dx dy
2
y
y3
2
1 3.05
5.95
y
dx x2 y 2
2
2 1 32
23
1 32
| 2 0.05 0.05 | 0.012
2
6
6
63
sin x 2 y 2 , x
20. Let z
sin ª 1.05
¬
2
y
1, dx
0.05. Then: dz
0.05, dy
2 x cos x 2 y 2 dx 2 y cos x 2 y 2 dy
2
0.95 º sin 2 | 2 1 cos 12 12 0.05 2 1 cos 12 12 0.05
¼
21. See the definition on page 918.
A
25.
22. In general, the accuracy worsens as 'x and
'y increases.
0
lh
Δh
dA
l dh h dl
'A
1 dl h dh lh
h
h dl l dh dl dh
23. The tangent plane to the surface z
f x, y at the point
'A dA
P is a linear approximation of z.
dV
'z
dz
is the relative error.
|
z
z
and
V
26.
f x, y , then 'z | dz is the propagated error,
24. If z
ΔA
dA
dA
l
Δl
dl dh
S r 2h
π r 2dh
2S rh dr S r dh
2
ΔV − dV
Δh
2πrhdr
Δr
r
S r 2h
27. V
3
,r
4, h
8
dV
2S rh
Sr2
dr dh
3
3
'V
Sª
r 'r
2
3¬
Sr
3
2h dr r dh
Sª
h 'h r 2 h º
¼
3¬
4S
16 dr 4 dh
3
4 'r
2
8 'h 128º
¼
'h
dV
'V
'V dV
0.1
0.1
8.3776
8.5462
0.1686
0.1
0.1
5.0265
5.0255
0.0010
0.001
0.002
0.1005
0.1006
0.0001
0.0001
0.0002
0.0034
0.0034
0.0000
'r
dA
© 2010 Brooks/Cole, Cengage Learning
Section 13.4
S r r 2 h2 , r
28. S
dS
dr
S r 2 h2
dS
dh
S
12
6, h
199
16
1 2
S r 2 r 2 h2
S
2r 2 h 2
r 2 h2
rh
r h2
2
S
S
ª 2r 2 h 2 dr rh dhº
¼
r h ¬
dS
2
S 6, 16
2
292
>328 dr
96 dh@
322.101353
S r 'r
'S
Differentials
r 'r
2
h 'h
2
S 6 'r
'r
'h
dS
'S
'S dS
0.1
0.1
7.7951
7.8375
0.0424
0.1
0.1
4.2653
4.2562
0.0091
0.001
0.002
0.0956
0.0956
0.0000
0.0001
0.0002
0.0025
0.0025
0.0000
6 'r
2
16 'h
2
322.101353
0.92 x 1.03 y 0.02
29. z
(a) dz
0.92 dx 1.03 dy
(b) dz
0.92>r0.25@ 1.03>r0.25@
B0.23 r 0.2575
Maximum error: r0.4875
Relative error:
30.
dz
z
r0.4875
| 0.081
0.92 1.9 1.03 7.5 0.02
7.2, 2.5 , dx d 0.05, dy d 0.05
x, y
x
x 2 y 2 Ÿ dr
r
8.1%
x y
2
2
dx y
x y
2
2
dy
7.2
7.2 2.5
2
2
dx 2.5
7.2 2.52
2
dy
| 0.9447 dx 0.3280 dy
dr d 1.2727 0.05 | 0.064
§ y·
arctan ¨ ¸ Ÿ dT
© x¹
T
y
x
dx 2
dy | 0.0430 dx 0.1239 dy
x2 y2
x y2
Using the worst case scenario, dx
31.
V
dV
V
32.
A
dA
S r 2 h Ÿ dV
2
dr
dh
r
h
1 ab sin
2
1ª
2¬
0.05 and dy
0.05, you see that dT d 0.0083.
2S rh dr S r 2 dh
2 0.04 0.02
0.10
10%
C
b sin C da a sin C db ab cos C dC º¼
1 ª4
2¬
1 3 sin 45q r 1 12 cos 45q r0.02 º | r0.24 in.2
sin 45q r 16
16
¼
© 2010 Brooks/Cole, Cengage Learning
200
33.
Chapter 13
35.74 0.6215T 35.75v 0.16 0.4275Tv 0.16
C
wC
wT
wC
wv
0.6215 0.4275v 0.16
5.72v 0.84 0.0684Tv 0.84
wC
wC
0.16
dT dv
0.6215 0.4275 23
r1 5.72 23
wT
wv
r1.3275 r 1.1143 r2.4418 Maximum propagated error
dC
0.84
0.0684 8 23
0.84
r3
r2.4418
| r0.19 or 19%
12.6807
dC
C
v2
r
2v
v2
dv 2 dr
r
r
dv dr
2
2 0.03 0.02
v
r
a
34.
Functions of Several Variables
da
da
a
0.08
8%
Note: The maximum error will occur when dv and dr differ in signs.
1
bhl
2
35. (a) V
T ·§
T·
§
¨18 sin ¸¨18 cos ¸ 16 12
2 ¹©
2¹
©
V is maximum when sin T
1 or T
31,104 sin T in.3
18 sin T ft 3
b
2
S 2.
s2
sin T l
2
(b) V
dV
h
18
s2
s2
l cos T dT sin T dl
2
2
S·
S ·§ S · 182 § S ·§ 1 ·
§
§ 1 · 182
§
18¨ sin ¸ 16 12 ¨ ¸ 16 12 ¨ cos ¸¨ ¸ ¨ sin ¸¨ ¸
2¹
2
2 ¹© 90 ¹
2 ©
2 ¹© 2 ¹
©
© 2¹
©
θ
2
18
s sin T l ds 1809 in.3 | 1.047 ft 3
36. (a) Using the Law of Cosines:
a2
b 2 c 2 2bc cos A
3302 4202 2 330 420 cos 9q
a | 107.3 ft.
330 ft
b 2 4202 2b 420 cos T
(b) a
da
420 ft
1ª 2
S ·º
§
330 4202 840 330 ¨ cos ¸»
2 «¬
20
©
¹¼
1
1 2
| >11512.79@ >r1774.79@ | r8.27 ft
2
37. P
9°
1 2
1 2
ªb 4202 840b cos T ¼º ª¬ 2b 840 cos T db 840b sin T dT º¼
2¬
E 2 dE
,
E
R
3%
0.03,
dP
2E
E2
dE 2 dR
R
R
dP
P
ª 2E
º
E2
dE 2 dR» P
«
R
¬R
¼
Using the worst case scenario,
dR
R
4%
1 2
ª§
S ·
S ·§ S ·º
§
«¨ 2 330 840 cos 20 ¸ 6 840 330 ¨ sin 20 ¸¨ 180 ¸»
¹
©
¹©
¹¼
©
0.04
ª 2E
º
E2
dE 2 dR»
«
R
¬R
¼
dE
E
0.03 and
dR
R
E2 R
0.04:
2
1
dE dR
E
R
dP
d 2 0.03 0.04
P
0.10
10%.
© 2010 Brooks/Cole, Cengage Learning
Section 13.4
38.
1
R
1
1
R1
R2
R
R1R2
R1 R2
dR1
'R1
0.5
dR2
'R2
2
wR
wR
dR dR2
wR1
wR2
'R | dR
When R1
39.
L
dL
40.
R1 R2
2
15, we have 'R |
10 and R2
R12
'R1 R1 R2
152
10 15
2
0.5 2
'R2
102
10 15
2
2
0.00021 ln 100 0.75 r dL | 8.096 u 104 r 6.6 u 106 micro henrys
T
2S
L
g
dg
32.23 32.09
dL
2.48 2.50
When g
'z
0.14 ohm.
ª r1 100
r1 16 º
6
0.00021«
» | r6.6 u 10
100
2
¬
¼
ª dh dr º
0.00021«
»
r¼
¬h
L
z
201
§ 2h
·
0.00021¨ ln
0.75¸
© r
¹
0.14
0.02
wT
wT
dg dL
wg
wL
'T | dT
41.
R22
Differentials
32.09 and L
f x, y
S
g
2.50, 'T |
L
dg g
S
32.09
S
Lg
dL
2.5
0.14 32.09
S
2.5 32.09
0.02 | 0.0108 seconds.
x2 2x y
f x 'x, y 'y f x, y
2 x 'x 'x
2
2 ' x 'y
2
x 2 2 x 'x ' x
2 x 2 'x y ' y
x2 2 x y
2 x 2 'x ' y ' x ' x 0 ' y
f x x, y 'x f y x, y 'y H1'x H 2'y where H1
'x and H 2
0.
As 'x, 'y o 0, 0 , H1 o 0 and H 2 o 0.
42.
z
'z
f x, y
x2 y2
f x 'x, y 'y f x, y
x 2 2 x 'x ' x
2 x 'x 2 y 'y 'x 'x 'y 'y
2
y 2 2 y 'y 'y
2
x2 y 2
f x x, y 'x f y x, y 'y H1'x H 2'y where H1
'x and H 2
'y.
As 'x, 'y o 0, 0 , H1 o 0 and H 2 o 0.
43.
z
'z
f x, y
x2 y
f x 'x, y 'y f x, y
2 xy 'x y 'x
2
x 2 2 x ' x 'x
2
x 2 ' y 2 x 'x ' y ' x ' y
f x x, y 'x f y x, y 'y H1'x H 2'y where H1
2
y 'y x 2 y
2
2 xy 'x x 2'y y'x 'x ª2 x'x 'x º 'y
¬
¼
y 'x and H 2
2
2 x'x 'x .
As 'x, 'y o 0, 0 , H1 o 0 and H 2 o 0.
© 2010 Brooks/Cole, Cengage Learning
202
44.
Chapter 13
z
5 x 10 y y 3
f x, y
'z
Functions of Several Variables
f x 'x, y 'y f x, y
5 x 5'x 10 y 10'y y 3 3 y 2 'y 3 y 'y
5 'x 3 y 2 10 'y 0 'x 3 y 'y 'y
2
'y
2
3
5 x 10 y y 3
'y
f x x, y 'x f y x, y 'y H1'x H 2 'y where H1
2
0 and H 2
3 y 'y ' y .
As 'x, 'y o 0, 0 , H1 o 0 and H 2 o 0.
45. f x, y
­ 3x 2 y
,
° 4
2
®x y
°0,
¯
x, y z 0, 0
x, y
0, 0
0
f x 0, 0
lim
'x o 0
f 'x, 0 f 0, 0
'x
4
'x
lim
lim
'y o 0
f 0, 'y f 0, 0
'y
0
'x
'x o 0
0
f y 0, 0
0
'y
lim
2
0
0
'y
'y o 0
So, the partial derivatives exist at 0, 0 .
Along the line y
x:
x , y o 0, 0
x 2:
Along the curve y
f x, y
lim
lim
x , y o 0, 0
lim
x o0
3x3
x x2
4
3x 4
2x4
f x, y
lim
x o0
3x
x2 1
0
3
2
f is not continuous at 0, 0 . So, f is not differentiable at 0, 0 . See Theorem 12.5
46. f x, y
f x 0, 0
f y 0, 0
­ 5x2 y
,
° 3
3
®x y
°0,
¯
x, y z 0, 0
x, y
0, 0
lim
f 'x, 0 f 0, 0
'x
'x o 0
lim
f 0, 'y f 0, 0
'y
'y o 0
'x o 0
'y o 0
lim
00
'x
0
lim
00
'y
0
So, the partial derivatives exist at 0, 0 .
Along the line y
x:
Along the line x
0,
lim
f x, y
lim
f x, y
x , y o 0, 0
x , y o 0, 0
lim
x o0
5 x3
2 x3
5
.
2
0.
So, f is not continuous at 0, 0 . Therefore f is not differentiable at 0, 0 .
47. If f is differentiable at x0 , y0 , then the partial derivative
f x x0 , y0 exists, which implies that f x, y0 is
differentiable at x0 , y0 . If f x, y
f x, 0
at x
x2
0, f x, y
x 2 y 2 , then
x . Because x is not differentiable
x y is not differentiable.
2
48. f x, y
x2 y 2
1 4
(a) f 1, 2
1.05 2.12 | 2.3479
f 1.05, 2.1
(b) 'z
f 1.05, 2.1 f 1, 2
2
(c) dz
5 | 2.2361
2
x
x y
2
2
dx 0.1118
y
x y2
2
dy
1
2
0.5 0.1 | 0.1118
5
5
The results are the same.
© 2010 Brooks/Cole, Cengage Learning
Section 13.5
Chain Rules for Functions of Several Variables
203
Section 13.5 Chain Rules for Functions of Several Variables
1.
x2 y2
2t , y
3t
ww dx ww dy
2x 2 2 y 3
wx dt
wy dt
4x 6 y
8t 18t
26t
w
x
dw
dt
5. w
(a)
dw
dt
x y
w
2.
x
cos t , y
2
dw
dt
et
ww dx
ww dy
wx dt
wy dt
x
sin t x2 y2
dw
dt
x sin t yet
x2 y 2
3. w
6. w
y
x y
2
et
(a)
et e 2t
cos 2 t e 2t
(b) w
S t
et , y
7. w
ww dx ww dy
wx wt
wy dt
dw
dt
sin y et x cos y 1
sin S t et et cos S t
cos t 2 1 ,
t2
z
arccos t
sin t , z
et
ww dx ww dy
ww dz
wx dt
wy dt
wz dt
dw
dt
cos 2 t sin 2 t e 2t
2e 2 t
1 e 2t
2e 2 t
§
y cos z 1 x cos z 2t xy sin z ¨ ©
dw
dt
·
¸
1t ¹
1
2
t 2 t t t 2t t t 2
t 3 2t 3 t 3
(b) w
t4,
dw
dt
xy xz yz , x
dw
dt
w
dw
dt
§
1 t2 ¨
©
·
¸
1t ¹
1
2
4t 3
4t 3
t 1, y
ww dx
ww dy
ww dz
wx dt
wy dt
wz dt
t 2 1, z
t
y z x z 2t x y
t 2 1 t t 1 t 2t t 1 t 2 1
(b)
cos t , y
xy cos z
y
(a)
2t sin t 2 1
2 x sin t 2 y cos t 2 z et
(b) w
t
9. w
dw
dt
2t sin t 2 1
2 cos t sin t 2 sin t cos t 2e 2t
x
(a)
dw
dt
et sin t et cos t
§1·
§ 1 ·
¨ ¸ sin t ¨ ¸ cos t
© x¹
© y¹
1
tan t cot t
sin t cos t
dw
dt
1
sin x y 2t sin x y 0
x2 y 2 z 2 , x
ln
x
y
8. w
(a)
y
x
cos t
sin t
w
t2, y
2t sin x y
cos t sin t e 2t
e t
e t
cos x y , x
dw
dt
e 2t et et 2e 2t
et
x sin y
x
4.
2
ww dx
ww dy
wx dt
wy dt
y et x 2e 2t
(b) w
2
e 2t
et , y
xy, x
3 2t 2 1
t 1 t2 1 t 1 t t2 1 t
2t t 1 t 2 1 2t 1 3t 2 1
3 2t 2 1
© 2010 Brooks/Cole, Cengage Learning
204
Chapter 13
Functions of Several Variables
xy 2 x 2 z yz 2 , x
10. w
t2, y
2t , z
2
ww dx ww dy
ww dz
wx dt
wy dt
wz dt
dw
(a)
dt
y 2 2 xz 2t 2 xy z 2 2 x 2 2 yz 0
t 2 4t 2 t 4 2 2t 4
(b) w
dw
dt
4t 2 4t 2 2t 4t 3 4 2
24t 3 8
6t 4 8t
24t 3 8
x1 x2
2
y1 y2
1ª
10 cos 2t 7 cos t
2¬
2
2
6 sin 2t 4 sin t º
¼
f t
11. Distance
fc t
2
10 cos 2t 7 cos t
2
6 sin 2t 4 sin t
2
1 2
ªª¬2 10 cos 2t 7 cos t 20 sin 2t 7 sin t º¼ ª¬2 6 sin 2t 4 sin t 12 cos 2t 4 cos t º¼º
¬
¼
§S ·
f c¨ ¸
©2¹
1ª
10
2¬
2
42 º
¼
fc t
2
y2 y1
ª48 ¬
2
et
et , y
ww dx
dw dy
wx dt
wy dt
dw
dt
et e t et e t et e t et e t
d 2w
dt 2
e e
t
t
At t
4
11
2
11 29
| 2.04
29
2º
¼
2
48t 8 2 2 2 6
15. w
x2 y 2
x
s t, y
ww
ws
ww
wt
1.
s t
2x 1 2 y 1
2s t 2s t
2 x 1 2 y 1
1 and t
ww
ws
0,
17. w
dw
dt
ww dx ww dy
wx dt
wy dt
2
4
2t 2t
t
2
t 1
t 1
2x
x2
2t 2 1
y
y
t 1 4t
t
3
t 1
4
3t 4t
4
2
t 1
2
t 1 12t 3 12t 2 3t 4 4t 3 2 t 1
d 2w
dt 2
t 1
1:
d 2w
dt 2
4 24 7 4
16
4
68
16
2
3
es , y
0.
1 and t
2,
ww
ws
6e 2 s t
6e s e t e s
6 xy 0 3 y 2 3 x 2 et
3e 2t 3e2 s et
3e3t 3e 2 s t
ww
6 and
3e6 3.
wt
sin 2 x 3 y
s t
y
s t
2 cos 2 x 3 y 3 cos 2 x 3 y
5 cos 2 x 3 y
ww
wt
ww
wt
et
6 xy e s 3 y 2 3x 2 0
x
ww
ws
4 and
4t
y 3 3x 2 y
When s
1
4s
2s t 2s t
t 1
y
At t
2
x
t2
x
2 º ª48t 1 ¼
¬
16. w
2
x2
y
14. w
t
d 2w
dt 2
0,
3 44
When s
2
4
et e t
1 2
ww
ws
ww
wt
et e t
et e t
1
1
et e t
x y
x y
22
2 29
1
116
2
fc1
48 8 2 2 2 6
ln x y , x
13. w
ª
º
¬ª¬2 10 7 º¼ 2 4 12 ¼
x2 x1
f t
12. Distance
1 2
5 cos 5s t
2 cos 2 x 3 y 3 cos 2 x 3 y
cos 2 x 3 y
4.25
When s
0 and t
cos 5s t
S ww
2
,
ws
0 and
ww
wt
0.
© 2010 Brooks/Cole, Cengage Learning
Section 13.5
18. w
x2 y2
x
s cos t
y
s sin t
ww
ws
yz
r T, z
r T
, x T 2, y
x
ww
yz
z
y
z y
2r
(a)
0 1 1
wr
x2
x
x
x
T2
z
y
ww
yz
2T 1 1
x2
x
x
wT
r T r T
r T r T
2T 4
2
2 x cos t 2 y sin t
2s cos 2t
2 x s sin t 2 y s cos t
When s
4
,
2T r
2
S ww
3 and t
T
2 s 2 sin 2t
0 and
ws
ww
wt
18.
ww
wr
ww
wT
(a)
(b)
x 2 2 xy y 2 , x
2 x 2 y 1 2 x 2 y 1
ww
wT
2 x 2 y 1 2 x 2 y 1
(a)
(b)
2
4x 4 y
2r T r T r T
y
, x
x
r cos T , y
r2
T
T2
2
1
T2
2r 2
T3
2
4ª¬ r T r T º¼
4x y
8T
r 2 2rT T 2 2 r 2 T 2 r 2 2rT T 2
r sin T cos T
r cos T sin T
r2
r2
y
x
r sin T 2
r cos T
x2 y2
x y2
arctan
r sin T
r cos T
arctan tan T
4T 2
0
r sin T r sin T
r cos T r cos T
r2
r2
1
T
0
1
r cos T , y
25 5 x 2 5 y 2 , x
ww
wT
r T r T
r sin T
ww
wT
ww
wr
T3
8T
y
x
cos T 2
sin T
x2 y2
x y2
22. w
2r
0
ww
wr
w
yz
x
2r
2
T
0
arctan
ww
wr
ww
wT
(a)
r T
w
2
r T
ww
wr
ww
wr
ww
wT
21. w
r T, y
T
2
T3
(b) w
20. w
205
19. w
2 s cos 2 t 2s sin 2 t
ww
wt
Chain Rules for Functions of Several Variables
5 x
25 5 x 5 y
2
2
5 x
25 5 x 2 5 y 2
cos T r sin T
5 y
25 5 x 5 y
r sin T 2
2
sin T
5 y
25 5 x 2 5 y 2
5r cos 2 T 5r sin 2 T
r cos T
25 5 x 5 y
2
2
5 r
25 5r 2
5r 2 sin 2 T cos T 5r 2 sin T cos T
25 5 x 2 5 y 2
0
© 2010 Brooks/Cole, Cengage Learning
206
Chapter 13
25 5r 2
(b) w
ww
wr
23. w
ww
ws
Functions of Several Variables
5r
25 5r
2
;
ww
wT
s t, y
xyz , x
0
s t, z
st 2
yz 1 xz 1 xy t 2
s t st 2 s t st 2 s t s t t 2
ww
wt
yz 1 xz 1 xy 2 st
2 s 2t 2 s 2t 2 t 4
3s 2 t 2 t 4
s t st 2 s t st 2 s t s t 2 st
t 2 3s 2 t 2
2 st 3 2 s 3t 2 st 3
2 s 3t 4st 3
2st s 2 2t 2
24. w
ww
ws
x2 y 2 z 2 , x
t sin s, y
dy
dx
ww
ws
ze , x
s t s t
2t 4 s 2t 3
s t, z
29. ln
dy
dx
ª¬ s t st s t st t º¼
ª¬2s 2t t º¼
te s
2 t2
s t
s t s t
30.
s2, y
x cos yz , x
se
s2 t 2
t2, z
1 2t
2
x
y2 6
x2 y2
dy
dx
0
2 xy
x2 y 2
x2 y 2
y x
2
2
2
2y
2
2
cos st 2 2t 3 2 s s 2t 2 sin st 2 2t 3
2 xy 2 y x 2 y 2
cos yz 0 xz sin yz 2t xy sin yz 2
y 2 x2
2 xy 2 yx 4 4 x 2 y 3 2 y 5
6 s t 2 s t sin st 2t
3
2
27. x 2 xy y 2 x y
Fx x, y
Fy x, y
x x2 y2
y x2 y 2
0
y 2 x2
cos yz 2 s xz sin yz 0 xy sin yz 1
2 2
Fx x, y
Fy x, y
s 2t
4
x
1
x2 y2
y
1
x2 y2
Fx x, y
Fy x, y
2s 2 1
ª
¬ s t st s t st sº¼
ª¬2st 2 sº¼
x2 y2 x y
1
ln x 2 y 2 x y 4
2
2 s 2t s 2t sin st 2 2t 3 2 s 2t 2 sin st 2 2t 3
dy
dx
y sec xy tan xy sec 2 xy
st
yze xy 1 xze xy 1 e xy s
e
ww
wt
s t
y sec xy tan xy y sec2 xy
x sec xy tan xy x sec 2 xy
2st 4
yze xy 1 xze xy 1 e xy t
e s t
ww
ws
s t, y
xy
e s t
26. w
Fx x, y
Fy x, y
2 x sin s 2 y cos s 2 z 2 st
e
ww
wt
0
x sec xy tan xy sec 2 xy
2t sin 2 s 2t cos 2 s 4 s 2t 3
25. w
28. sec xy tan xy 5
2 x cos s 2 y t sin s 2 z t 2
2t 2 sin s cos s 2t 2 sin s cos s 2st 4
ww
wt
st 2
t cos s, z
31. F x, y, z
x2 y 2 z 2 1
3
0
2x y 1
x 2 y 1
y 2x 1
2y x 1
Fx
2 x, Fy
2 y, Fz
wz
wx
Fx
Fz
x
z
wz
wy
Fy
Fz
y
z
2z
© 2010 Brooks/Cole, Cengage Learning
y
x
Section 13.5
xz yz xy
32. F x, y, z
Fx
z y
Fy
z x
Fz
x y
wz
wx
F
x
Fz
y z
x y
wz
wy
Fy
Fz
x z
x y
Fx x, y, z
Fz x, y, z
ze xz y
xe xz
wz
wy
Fy x, y, z
Fz x, y, z
x
xe xz
2 x
2 y 2z
x
y z
wz
wy
F x, y , z
y
Fz x, y, z
2 z
2 y 2z
z
y z
wz
(i) 1 cos y z
wx
wz
1
cos y z
wx
0 implies
sec y z .
§
wz ·
(ii) ¨1 ¸ cos y z
y¹
w
©
wz
0 implies
wy
tan x y tan y z 1
1.
ln y
y2 2z
wz
(ii)
wy
Fy x, y, z
Fz x, y, z
x
2 yz
y
2
y 2z
Fx
y w
Fy
x z
Fz
y w
Fw
z x
ww
wx
Fx
Fw
y w
z x
y w
z x
ww
wy
Fy
Fw
x z
z x
x z
z x
ww
wz
Fz
Fw
y w
z x
y w
z x
40. x 2 y 2 z 2 5 yw 10 w2 2
sec 2 x y
Fy
sec 2 x y sec 2 y z
Fx
2 x, Fy
Fz
sec y z
ww
wx
Fx
Fw
2 x
5 y 20 w
ww
wy
Fy
Fw
5w 2 y
20 w 5 y
ww
wz
Fz
Fw
2z
5 y 20 w
wz
wx
wz
wy
Fx
Fz
Fy
Fz
sec x y
sec 2 y z
sec x y sec y z
sec 2 y z
2
2
§ sec 2 x y
·
¨¨ 2
1¸¸
y
z
sec
©
¹
36. F x, y, z
e sin y z z
x
Fx
e x sin y z
Fy
e x cos y z
Fz
e x cos y z 1
wz
wx
Fx
Fz
e sin y z
1 e x cos y x
wz
wy
Fy
Fz
e x cos y z
1 e x cos y z
x
2
41. F x, y, z , w
x 2 y2 z
y 3 2 yz
xy yz wz wx s
Fx
2
e xz
0
Fx x, y, z
Fz x, y , z
39. F x, y, z , w
0
1
e xz
wz
wx
0
F x, y , z
x
Fz x, y, z
35. F x, y, z
wz
wx
(i)
207
0
38. x ln y y 2 z z 2 8
wz
wx
34. x sin y z
e xz xy
37. F x, y, z
x 2 2 yz z 2 1
33. F x, y, z
Chain Rules for Functions of Several Variables
2 y 5w, Fz
F x, y , z , w
2 z , Fw
5 y 20 w
2x
5 y 20 w
cos xy sin yz wz 20
ww
wx
Fx
Fw
y sin xy
z
ww
wy
Fy
Fw
x sin xy z cos yz
z
ww
wz
Fz
Fw
y cos zy w
z
© 2010 Brooks/Cole, Cengage Learning
208
Chapter 13
42. F x, y, z , w
Functions of Several Variables
w
x y y z
1 2
ww
wx
Fx
Fw
1 x y
2
1
ww
wy
Fy
Fw
1
x y
2
ww
wz
Fz
Fw
2
1
x y
2
1 2
0
1
y z
2
1 2
2
1
x y
2
1
y z
1
y z
xy
43. (a) f x, y
x2 y2
§
t¨
¨
©
tx ty
f tx, ty
2
tx
ty
2
·
¸
x y ¹¸
xy
2
tf x, y
2
Degree: 1
§
y3
x¨
¨¨ x 2 y 2
©
(b) xf x x, y yf y x, y
32
·
§
x3
¸ y¨
¨¨ x 2 y 2
¸¸
¹
©
32
·
¸
¸¸
¹
xy
x y2
2
1 f x, y
x3 3xy 2 y 3
44. (a) f x, y
f tx, ty
tx
3
3 tx ty
Degree: 3
(b) xf x x, y yf y x, y
ex
45. (a) f x, y
ty
3
t 3 x3 3xy 2 y 3
t 3 f x, y
x 3x 2 3 y 2 y 6 xy 3 y 2
3x3 9 xy 2 3 y 3
3 f x, y
y
etx ty
f tx, ty
2
ex
y
f x, y
Degree: 0
§1
·
§ x
·
x¨ e x y ¸ y ¨ 2 e x y ¸
y
y
©
¹
©
¹
(b) xf x x, y yf y x, y
0
x2
46. (a) f x, y
x y2
2
tx
2
§
t¨
¨
©
2
tx
f tx, ty
ty
2
·
¸
x y ¸¹
x2
2
2
tf x, y
Degree: 1
ª 3
º
ª
x 2 xy 2 »
x2 y
x«
y«
3
2
« x2 y 2 »
« x2 y2
¬
¼
¬
(b) xf x x, y yf y x, y
47.
dw
dt
ww dx
ww dy
wx dt
wy dt
At t
2, x
dw
So,
dt
4, y
wf dg
wf dh
wx dt
wy dt
5 and f y 4, 3
3, f x 4, 3
5 1 7 6
47
º
»
x4 x2 y 2
¼
x2 y2
3 2»
48.
ww
ws
7.
ww
wt
32
x2 x2 y 2
x2 y 2
ww wx ww wy
wx ws
wy ws
wf wg
wf wh
wx ws
wy ws
ww wx ww wy
wx wt
wy wt
wf wg
wf wh
wx wt
wy wt
x2
32
x2 y2
f x, y
5 3 7 5
50
5 2 7 8
66
© 2010 Brooks/Cole, Cengage Learning
Section 13.5
dw
dt
ww dx ww dy
Page 925
wx dt
wy dt
ww
50.
ws
ww
wt
ww wx ww wy
wx ws
wy ws
ww wx ww wy
Page 925
wx wt
wy wt
49.
51.
wz
wx
f x x, y , z
f z x, y , z
wz
wy
f y x, y , z
f z x, y , z
et
yz 2t xz 2 xy e t
4t 2e t 2t 2e t 2t 3e t
(b) f
df
dt
t 2 2t e t
2t 2e t 3 t
2t 3e t
2t 3e t 6t 2e t
2t 2e t 3 t
The results are the same.
S r 2h
dV
dt
S
dS
dt
V
dV
dt
S
dS
dt
§
©
S ¨ 2rh
dr
dh ·
r2 ¸
dt
dt ¹
dr
dh º
ª
r »
2S « 2r h
dt
dt ¼
¬
1 2
Sr h
3
1 §
dr
dh ·
S ¨ 2rh r 2 ¸
3 ©
dt
dt ¹
ª§
dT
dt
A
dA
dt
S 12 ª¬2 36 6 12 4 º¼
2S ª¬ 24 36 6 12 4 º¼
1 ª
S 2 12 36 6 12
3 ¬
· dr
2r ¸
r h
¹ dt
r2
S «¨ r 2 h 2 ¬«©
ª§
S «¨ 122 362 ¬«©
©
T
dr
dh ·
r ¸
dt
dt ¹
4608S in.3 min
2
4 º
¼
624S in.2 min
1536S in.3 min
S r r 2 h 2 S r 2 Surface area includes base.
ª§
55. pV
§
©
S r ¨ 2h
2S r r h
S Ǭ12 10 56.
2t , z
209
wf dx wf dy
wf dz
wx dt
wy dt
wz dt
df
dt
(a)
t2, y
xyz , x
52. f x, y, z
f x x, y
f y x, y
dy
dx
53. V
54.
Chain Rules for Functions of Several Variables
2
2
dh º
»
r h dt ¼»
rh
2
·
2 12 ¸ 6 12 36
¹
144
2
2
º
12 ·
36
4 »
¸ 6 144 10 ¹
10
¼
2
º
4 »
12 36
¼»
36 12
2
2
648S
144S in.2 min
10
36S
20 9 10 in.2 min
5
mRT
1
pV
mR
1 ª dp
dV º
V
p »
mR «¬ dt
dt ¼
1
bh
2
T ·§
T·
§
¨ x sin ¸¨ x cos ¸
2 ¹©
2¹
©
wA dx wA dT
wx dt wT dt
x sin T
x2
sin T
2
b
2
dx x 2
dT
cos T
2
dt
dt
S ·§ 1 · 62 §
S ·§ S ·
§
6¨ sin ¸¨ ¸ ¨ cos ¸¨ ¸
4 ¹© 2 ¹
2©
4 ¹© 90 ¹
©
3 2 S 2 2
m hr | 2.566 m 2 h
2
10
x
h
θ
2
x
© 2010 Brooks/Cole, Cengage Learning
210
57.
Chapter 13
I
dI
dt
58.
V
dV
dt
Functions of Several Variables
1
m r12 r22
2
1 ª dr1
dr º
m 2r1
2r2 2 »
2 «¬ dt
dt ¼
S
m ª¬ 6 2 8 2 º¼
28m cm 2 sec
r 2 rR R 2 h
3
Sª
3 «¬
2r R h
dr
dR
dh º
r 2R h
r 2 rR R 2
dt
dt
dt »¼
Sª
ª2 15 25º¼ 10 4 ª¬15 2 25 º¼ 10 4 ª 15
¬
3 «¬¬
S
2
2
15 25 25 º 12 »º
¼
¼
19,500
3
6,500S cm3 min
S
dS
dt
S Rr
­ª
°
°«¬
¯
S ®«
Rr
R r
2
R r
­ª
°
°«¬
¯
S ®«
25 15
ª
«
«
¬
2
h2
h2 R r
h
R r
2
25 15
2
R r
R r
R r
2
R r
h2 R r
Rr
2
º
» dR
2 » dt
h ¼
½
dh °
¾
h 2 dt °¿
102 25 15
2
2
º
ª
» dr «
»
«
h 2 ¼ dt
¬
25 15
25 15
102 25 15
2
º
»4
»
102 ¼
25 15
25 15
2
º
» 4 25 15
»
102 ¼
ª
«
«
¬
10
25 15
2
º½
°
12 »¾
»°
102
¼¿
320 2S cm 2 min
59.
w
f x, y
x
y
u v
vu
ww
wu
ww
wv
ww ww
wu
wv
60.
w
ww
wx
ww
wy
ww ww
wx
wy
ww dx
ww dy
wx du
wy du
ww dx ww dy
wx dv
wy dv
ww ww
wx
wy
ww ww
wx
wy
0
x y sin y x
x y cos y x sin y x
x y cos y x sin y x
0
© 2010 Brooks/Cole, Cengage Learning
Section 13.5
61. w
r cos T , y
f x, y , x
ww
ww
cos T sin T
wx
wy
ww
wT
ww
ww
r sin T r cos T
wx
wy
ww
wr
ww
sin T
wT
ww
ww
r cos T
sin T
wr
wT
ww
r
wx
ww
wx
ww
r sin T
wr
ww
cos T
wT
ww
ww
r cos 2 T r sin T cos T
wx
wy
ww
ww
r sin 2 T r sin T cos T
wx
wy
ww
r cos 2 T r sin 2 T
wx
ww
ww
r cos T sin T
wr
wT
ww
ww sin T
cos T First Formula
wr
wT r
ww
ww
r sin T cos T r sin 2 T
wx
wy
ww
ww
r sin T cos T r cos 2 T
wx
wy
r cos T
r sin T
ww
ww
cos T
wr
wT
r
2
1 § ww ·
§ ww ·
(b) ¨ ¸ 2 ¨ ¸
r
r
w
© ¹
© wT ¹
ww
r sin 2 T r cos 2 T
wy
ww
wy
ww
ww
r sin T cos T
wr
wT
ww
wy
ww
ww cos T
sin T wr
wT r
2
Second Formula
2
2
2
arctan
y
,x
x
§ ww ·
ww ww
sin T cos T ¨ ¸ cos 2 T
wx wy
© wy ¹
r cos T , y
arctan tan T
T for ww
y
,
2
x y wy
x
ww
,
2
x y wr
0,
2
2
§ ww ·
§ ww ·
¨ ¸ ¨ ¸
© wx ¹
© wy ¹
2
§ ww ·
§ 1 ·§ ww ·
¨ ¸ ¨ 2 ¸¨ ¸
© wr ¹
© r ¹© wT ¹
2
§ ww ·
§ ww ·
So, ¨ ¸ ¨ ¸
© wx ¹
© wy ¹
2
2
2
y2
x y
2
2
0
1
1
r2
2
2
§ ww ·
§ ww ·
¨ ¸ ¨ ¸
x
w
© ¹
© wy ¹
2
r sin T
§ r sin T ·
arctan ¨
¸
© r cos T ¹
ww
wx
2
§ ww ·
ww ww
§ ww ·
§ ww ·
2
sin T cos T ¨ ¸ sin 2 T ¨ ¸ sin 2 T
¨ ¸ cos T 2
x
x
y
y
w
w
w
w
© ¹
© wx ¹
© ¹
2
62. w
211
r sin T
ww
wr
(a)
Chain Rules for Functions of Several Variables
2
2
S
2
ww
wT
x y
S
2
1
x2
2
T 2
2
1
x2 y 2
1
r2
1
r2
2
1 § ww ·
§ ww ·
¨ ¸ 2¨ ¸ .
r © wT ¹
© wr ¹
© 2010 Brooks/Cole, Cengage Learning
212
Chapter 13
wu
wv
and
wy
wy
wu
wx
63. Given
Functions of Several Variables
wv
, x
wx
wu
wr
wu
wu
cos T sin T
wx
wy
wv
wT
wv
wv
r sin T r cos T
wx
wy
So,
wu
wr
r sin T .
wv
wv
cos T sin T
wy
wx
ª wv
º
wv
r « cos T sin T »
y
x
w
w
¬
¼
1 wv
.
r wT
wv
wr
wv
wv
cos T sin T
wx
wy
wu
wT
wu
wu
r sin T r cos T
wx
wy
So,
r cos T and y
wv
wr
wu
wu
cos T sin T
wy
wx
ª wu
º
wu
r « cos T sin T »
wx
¬ wy
¼
1 wu
.
r wT
64. Note first that
wu
wx
wv
wy
wu
wy
wu
wr
x
y
cos T 2
sin T
x y2
x y2
wv
wT
y
x
r sin T 2
r cos T
x y2
x y2
So,
x
x2 y 2
wv
wx
y
.
x2 y2
r cos 2 T r sin 2 T
r2
2
2
wu
wr
r 2 sin 2 T r 2 cos 2 T
r2
1
1 wv
.
r wT
r sin T cos T r sin T cos T
r2
wv
wr
y
x
cos T 2
sin T
x2 y2
x y2
wu
wT
x
y
r sin T 2
r cos T
2
x y
x y2
So,
1
r
2
wv
wr
0
r 2 sin T cos T r 2 sin T cos T
r2
0
1 wu
.
r wT
Section 13.6 Directional Derivatives and Gradients
1. f x, y
v
’f x , y
3 x 4 xy 9 y
3
4
i j
5
5
3 4 y i 9 4x j
’f 1, 2
5i 5 j
v
v
3
4
i j
5
5
u
Du f 1, 2
2.
’f 1, 2 ˜ u
f x, y
’f x , y
3 x 2i 3 y 2 j
’f 4, 3
48i 27 j
u
Du f 4, 3
3 4
1
2
i j
2
x3 y 3 , v
v
v
2
i 2
2
j
2
’f 4, 3 ˜ u
24 2 27
2
2
21
2
2
© 2010 Brooks/Cole, Cengage Learning
Section 13.6
3. f x, y
v
xy
1
i 2
’f x , y
h x, y
8.
yi xj
’h
2 xe
Du f 0, 2
1
i 2 ye
’h 0, 0 ˜ u
x2 y 2
j
0
x2 y 2 z 2
9. f x, y, z
’f 0, 2 ˜ u
x2 y2
0
Du h 0, 0
1
3
i j
2
2
213
x2 y 2
i j
’h 0, 0
2 i
v
v
e
v
3j
’f 0, 2
u
Directional Derivatives and Gradients
3
i jk
3
2 xi 2 yj 2 zk
’f x , y , z
v
x
y
j
1
x
i 2j
y
y
f x, y
4.
v
’f x , y
’f 1, 1
’f 1, 1, 1
u
i j
v
v
u
j
v
’h
§ S·
’h¨1, ¸
© 2¹
ei
u
v
v
§ S·
Du h¨1, ¸
© 2¹
6. g x, y
v
11.
h x, y , z
’h
x
i 2
1 xy
2
’h 2, 1, 1
j
u
j
Du h 2, 1, 1
j
Du g 1, 0
g x, y
v
’g
’g 3, 4
u
Du g 3, 4
’g 1, 0 ˜ u
1
12.
x y
2
h x, y , z
v
x2 y2
’h x, y, z
3i 4 j
x
2
y
i x y2
2
3
4
i j
5
5
3
4
v
i j
5
5
v
’g 3, 4 ˜ u
’f 1, 2, 1 ˜ u
e
v
y
v
v
i 3k
1
2i j k
6
v
v
2
3
6
6
j
1 xy
y zi x z j y xk
Du f 1, 2, 1
i
arccos xy
’g 1, 0
7.
u
§ S·
’h¨1, ¸ ˜ u
© 2¹
’g x , y
u
’f 1, 2, 1
e x sin yi e x cos yj
j
’h 4, 1, 1
u
7
25
3
2i j k
’f x, y, z
i
v
2
3
’f 1, 1, 1 ˜ u
xy yz xz
10. f x, y, z
1
e x sin y
h x, y
5.
3
3
3
i j
k
3
3
3
v
v
Du f 1, 1, 1
’f 1, 1 ˜ u
Du f 1, 1
2i 2 j 2k
Du h 4, 1, 1
6
6
xyz
2, 1, 2
yzi xzj xyk
i 2 j 2k
v
v
2
1
2
i j k
3
3
3
’h 2, 1, 1 ˜ u
8
3
x arctan yz
1, 2, 1
arctan yzi xz
1 yz
2
j
xy
1 yz
2
k
S
i 2 j 2k
4
v
1
2
1
,
,
v
6
6
6
’h 4, 1, 1 ˜ u
S 8
S 8
4 6
24
6
© 2010 Brooks/Cole, Cengage Learning
214
Chapter 13
Functions of Several Variables
x2 y 2
13. f x, y
u
’f
1
1
i j
2
2
2 xi 2 yj
Du f
’f ˜ u
14. f x, y
y
x y
u
’f
x y
2
’f
’f ˜ u
Du f
i 2x y
u
’f
2
y
2
2 x y
x
x y
3y
2x y
2
2
2
’g
Du g
x
2x y
v
2
’g
3y x
1
3
i j
2
2
2 cos 2 x y i cos 2 x y j
cos 2 x y ’f
3
cos 2 x y
2
v
u
’h
v
v
Du f 1, 1
2i 6 j
3
4
i j
5
5
’f 1, 1 ˜ u
6 24
5
5
6
23.
8
5
1
i jk
x y z
i j k.
1
3i 3 j k
19
7
19
’h ˜ u
7 19
19
3x 5 y 2 1
21. f x, y
22.
4
4
5
5
3i 3 j k
v
v
Du h
3x 1
1
2
i j
5
5
At 1, 0, 0 , ’h
u
ey
2
4i 2 j 8k.
ln x y z
20. h x, y, z
3i 4 j
2 xi 6 yj, ’f 1, 1
ye z i xe z j xye z k
’g ˜ u
Du g
1
3
i j
2
2
e y i xe y j
1
3 y
xe
ey 2
2
2i 4 j
v
v
u
sin 2 x y
5
sin x y
5
0.
At 2, 4, 0 , ’g
x2 3 y 2
17. f x, y
1
2
sin x y sin x y
5
5
xye z
19. g x, y, z
xe y
u
v
At 0, S , Du f
j
§2 3·
¨¨
¸¸ cos 2 x y
2
©
¹
16. g x, y
1
2
i j
5
5
v
v
u
1
sin x y
5
1
f x, y
15.
i Sj
2
sin x y i sin x y j
Du f
3
1
i j
2
2
y
S
v
2
x 2
’f ˜ u
Du f
cos x y
18. f x, y
’f x, y
3i 10 yj
’f 2, 1
3i 10 j
g x, y
2 xe y
x
’g x, y
§ 2y y
¨ e
© x
’g 2, 0
2i 2 j
z
x
·
2e y x ¸i 2e y x j
¹
ln x 2 y
’z x, y
2x
1
i 2
j
x2 y
x y
’z 2, 3
4i j
© 2010 Brooks/Cole, Cengage Learning
Section 13.6
cos x 2 y 2
z
24.
’z x , y
32. f x, y
2 x sin x 2 y 2 i 2 y sin x 2 y 2 j
’z 3, 4
’f x , y
6 sin 25i 8 sin 25 j | 0.7941i 1.0588 j
’w x , y , z
6 xi 10 yj 4 zk
’w 1, 1, 2
6i 10 j 8k
’w x , y , z
33.
x sec2 y z k
’w 4, 3, 1
tan 2i 4 sec 2 j 4 sec 2k
2
2
JJJG
27. PQ
2
2
i j, u
i j
2
2
’g x , y
2 xi 2 yj, ’g 1, 2
’g ˜ u
JJJG
28. PQ
’f
6 xi 2 yj, ’f 1, 4
’f ˜ u
JJJG
29. PQ
’f
JJJG
30. PQ
’f
6i 8 j
4 5
2
5
’f ˜ u
’f x , y
y sin x y i
§ S·
’h¨ 0, ¸
© 3¹
§ 3 3S ·
3S
i ¨¨
¸¸ j
6
6
©
¹
§ S·
’h¨ 0, ¸
© 3¹
3S 2
9 6 3S 3S 2
36
36
6
2 5
5
36.
2
5
2 5
5
x 2 2 xy
2 x 2 y i 2 xj
’f 1, 0
2i 2 j
’f 1, 0
2 2
37.
ye x
’g x, y
ye x i e x j
’g 0, 5
5i j
’g 0, 5
2i
’f ˜ u
y cos x y
3 2S 2 2 3S 3
S
f x, y
17
35. g x, y
1
2
i S j, u
i j
2
5
5
2 cos 2 x cos yi sin 2 x sin yj
Du f
5
tan yi x sec 2 yj
§ S·
’h¨ 2, ¸
© 4¹
h x, y
1
4
x tan y
2
1
i j
5
5
i
’f S , 0
1
1
4 16
ª¬cos x y y sin x y º¼ j
e cos xi e sin xj
Du f
1
1
i j
2
4
i 4j
y
’f 0, 0
1
1 x
i j
2
y 1
y 1
§ S·
’h¨ 2, ¸
© 4¹
’h x , y
3 2
12
8
5
5
2i j, u
y
34.
2
1
i j
5
5
4i 2 j, u
Du f
31.
2 2 2
2i 4 j
h x, y
’h x, y
tan y z i x sec y z j
2
215
x y
y 1
’f 0, 1
x tan y z
w
26.
’f 0, 1
3x 2 5 y 2 2 z 2
w
25.
Du g
Directional Derivatives and Gradients
g x, y
26
ln 3 x 2 y 2
1
ln x 2 y 2
3
’g x, y
º
1 ª 2x
2y
i 2
j
3 «¬ x 2 y 2
x y 2 »¼
’g 1, 2
1§ 2
4 ·
¨ i j¸
3© 5
5 ¹
’g 1, 2
2 5
15
x2 y2 z 2
f x, y , z
1
’f x , y , z
x y2 z2
2
xi yj zk
1
i 4 j 2k
21
’f 1, 4, 2
’f 1, 4, 2
2
i 2j
15
1
© 2010 Brooks/Cole, Cengage Learning
216
Chapter 13
Functions of Several Variables
1
38. w
1 x y z
2
2
1
’w
1 x y z
2
’w 0, 0, 0
2
3
xi yj zk
’w
(b)
0
v
9 16
xe
’f x , y , z
e i xze j xye k
’f 2, 0, 4
i 8j
’f 2, 0, 4
65
3
9 16
u
3
4
i j
5
5
Du f
’f ˜ u
v
i 3j
v
10
u
1
i 10
y
x
and
3
2
§ 1·
§ 1·
¨ ¸ cos T ¨ ¸ sin T .
© 3¹
© 2¹
3
v
yz
(d)
For exercises 41–46, f x , y
(3, 2, 1)
45.
y
5
1 2
5 5
1
5
11 10
60
3
j
10
11
6 10
§1·
§1·
¨ ¸ i ¨ ¸ j
© 3¹
© 2¹
44. ’f
z
6
’f ˜ u
Du f
x
y
3
2
3
5
v 3i 4 j
yz
yz
5 2
12
5
1 2
5 5
’f ˜ u
(c)
yz
3
4
i j
5
5
Du f
33
f x, y , z
3
3i 4 j
i 4 j 4k
’w 2, 1, 1
41. f x, y
v
u
’w 2, 1, 1
§1· 1
§1· 1
¨ ¸
¨ ¸
© 3¹ 2 © 2 ¹ 2
’f ˜ u
Du f
y 2 z 2i 2 xyz 2 j 2 xy 2 zk
DT f x , y
§ 1 ·
¨
¸i j
© 2¹
u
xy 2 z 2
w
40.
2
0
’w 0, 0, 0
39.
43. (a)
2
1 1
9 4
’f
1
13
6
9
x
46.
42. (a) DS
4
f 3, 2
§1· 2 § 1 · 2
¨ ¸
¨ ¸
© 3¹ 2
© 2¹ 2
(b) D2S 3 f 3, 2
§ 1 ·§ 1 · § 1 · 3
¨ ¸¨ ¸ ¨ ¸
© 3 ¹© 2 ¹ © 2 ¹ 2
(c) D4S 3 f 3, 2
3·
§ 1 ·§ 1 · § 1 ·§
¨ ¸¨ ¸ ¨ ¸¨¨ ¸
© 3 ¹© 2 ¹ © 2 ¹© 2 ¸¹
5 2
12
23 3
12
1
1
i j
3
2
1
2i 3j
13
’f
’f
’f
So, u
13 3i 2 j and
1
’f ˜ u
Du f 3, 2
23 3
12
(d) DS 6 f 3, 2
§ 1 ·§ 3 · § 1 ·§ 1 ·
¨ ¸¨¨
¸ ¨ ¸¨ ¸
© 3 ¹© 2 ¸¹ © 2 ¹© 2 ¹
32 3
12
0. ’f is the direction of greatest
rate of change of f. So, in a direction orthogonal to
’f , the rate of change of f is 0.
47. (a) In the direction of the vector 4i j
(b) ’f
1
10
2x 3y i ’f 1, 2
1
10
4 i 1
10
1
10
3 x 2 y j
1j
52 i 1 j
10
Same direction as in part a
(c) ’f
2i
5
1 j, the
10
direction opposite that of the
gradient
© 2010 Brooks/Cole, Cengage Learning
Section 13.6
48. (a) In the direction of the vector i j
(b) ’f
1
1
1
y
i 2 2 x
2
xj
y
4
Directional Derivatives and Gradients
1
i 2
x
xj
Ÿ 4y
1
1
i j
2
2
’f 1, 2
8y
1 x2 y 2
50. (a) f x, y
y 2
2
2
1 x2 y2
y2 4 y 4 x2 1
4
Same direction as in part a
x2
3
Circle: center: 0, 2 , radius:
1
1
i j, the direction opposite that of the
2
2
(c) ’f
16 xy
(b) ’f
49. f x, y
x 2 y 2 , 4, 3, 7
’f
i 3
8 8x2 8 y2
1 x2 y 2
2
j
3
i
2
3, 2
z
(a)
2
1 x2 y 2
gradient
217
y
4
x
3
y
2
’f x, y ˜ u
(b) Du f x, y
2 x cos T 2 y sin T
1
8 cos T 6 sin T
Du f 4, 3
−2
x
−1
1
2
(c) The directional derivative of f is 0 in the direction r j.
Du f
(d)
12
z
8
6
4
π
−4
2π
θ
−6
−8
−12
6
x
y
Generated by Mathematica
−6
(c) Zeros: T | 2.21, 5.36
These are the angles T for which Du f 4, 3 equals
zero.
(d) g T
gc T
c
8 cos T 6 sin T
Du f 4, 3
’f 0, 0
64 36
10,
the maximum value of Du f 4, 3 , at T | 0.64.
x2 y 2
(f ) f x, y
’f 4, 3
7
8i 6 j is perpendicular to the level
curve at 4, 3 .
y
52. f x, y
c
−6 −4
6
−4
3, 4
25
’f 3, 4
6i 8 j
3, P
’f x , y
4
x2 y2
x2 y2
c
2
2i 3j
2 xi 2 yj
x
−2
2x 3 y
25, P
53. f x, y
2
6
’f x , y
6
4
2i 3 j
0
maximum 0.64 and minimum 3.79 .
2 4 i 2 3 j
0, 0
6 2x 3y
These are the angels for which Du f 4, 3 is a
’f 4, 3
6 2x 3y
6, P
’f x , y
8 sin T 6 cos T
Critical numbers: T | 0.64, 3.79
(e)
51. f x, y
xy
xy
1, 3
yi xj
3
’f 1, 3
3i j
−6
Generated by Mathematica
© 2010 Brooks/Cole, Cengage Learning
218
Chapter 13
1
,P
2
4
2
y2 x2
2
x2 y2
x
2
x y2
i x2 y 2
2
0
−4
57. f x, y
3x 2 2 y 2
(a) ’f
6 xi 4 yj
’f 1, 1
(b)
16i j
16i j
257
curve 3x 2 2 y 2
1
16i j is a unit vector normal to the level
257
curve 4 x y
2
y 1
’f 4, 1
tangent line
1
x
1
2
3
−2
−3
58. f x, y
9x2 4 y2
(a) ’f
18 xi 8 yj
’f 2, 1
5
10
15
(b)
i 2j
5
1
i 2 j is a unit vector normal to the level
5
curve x y 2
1360
4 85
1
9i 2 j is a unit vector normal to the level
85
i 2 yj
’f 4, 1
36i 8 j
’f 2, 1
x y2
(a) ’f x, y
1
2
2
−3 −2 −1
−10
56. f x, y
3
x
−5
x 1
3x
2
(d)
16 x 22 Tangent line
−15 −10 −5
1 at 1, 1 .
y
y
(d)
3
2
y
16 x 2
y
2 13
(c) The vector 2i 3 j is tangent to the level curve.
3
Slope
.
2
6 at 2, 10 .
(c) The vector i 16 j is tangent to the level curve.
16
Slope
16
1
y 10
36 16
1
3i 2 j is a unit vector normal to the level
13
8 xi j
’f 2, 10
6i 4 j
’f 1, 1
4x y
(a) ’f x, y
6
(4, −1)
−2
2
55. f x, y
2
j
1
j
2
’f 1, 1
x
−2
2 xy
1
2
x2 y2 2 x
(b)
y
1, 1
’f x , y
(b)
(d)
x
x2 y2
54. f x, y
c
Functions of Several Variables
3 at 4, 1 .
(c) The vector 2i j is tangent to the level curve.
1
Slope .
2
1
y 1 x 4
2
1
y
x 1
Tangent line
2
curve 9 x 2 4 y 2
40 at 2, 1 .
(c) The vector 2i 9 j is tangent to the level curve.
9
Slope
.
2
9
y 1
x2
2
9
y
x 10 Tangent line
2
y
(d)
4
2
x
−4
4
−2
(2, −1)
−4
© 2010 Brooks/Cole, Cengage Learning
Section 13.6
59. See the definition, page 934.
Directional Derivatives and Gradients
x
x2 y 2
65. T
60. Let f x, y be a function of two variables and
wf
.
wx
(a) If T
0q, then Du f
(b) If T
90q, then Du f
y2 x2
’T
cos T i sin T j a unit vector.
u
x y
2
’h
i 2 xy
x y2
2
2
j
1
7i 24 j
625
0.002 xi 0.008 yj
’h 500, 300
z
62.
2
5000 0.001x 2 0.004 y 2
66. h x, y
61. See the definition, pages 936 and 937.
2
7
24
i j
625
625
’T 3, 4
wf
.
wy
219
5’h
i 2.4 j or
5i 12 j
3
67.
3
x
18
00
1671
y
5
P
B
1994
A
00
63. The gradient vector is normal to the level curves. See
Theorem 13.12.
68. The wind speed is greatest at B.
9 x 2 y 2 and
64. f x, y
DT f x, y
18
2 x cos T 2 y sin T
69. T x, y
2 x cos T y sin T
dx
dt
9 x y
2
(a) f x, y
2
10
9
dy
dt
4 x
C1e 4t
xt
z
400 2 x 2 y 2 , P
x0
yt
C1
10
10e4t
xt
yt
10, 10
2 y
C2e 2t
y0
C2
10e 2t
(1, 2, 4)
x
3
3
y2
y
x
(b) DS
(c) DS
(d)
3
4
§ 2
2¨¨
© 2
f 1, 2
§1
2¨ ©2
f 1, 2
’f 1, 2
2i 4 j
’f 1, 2
4 16
(e) ’f 1, 2
·
3¸
¹
2
1 2 3
Therefore, u
4
20
2 5
1
i 2 j
5
1
dx
dt
100e 4t
y2 t
10 x
100 x 2 2 y 2 , P
70. T x, y
xt
2i 4 j
’f 1, 2
’f 1, 2
Du f 1, 2
·
2 ¸¸
¹
y2
10
dy
dt
2 x
C1e 2t
x0
xt
4e 2t
3x 2
16
e 4t
yt
C1
3
yt
y Ÿ u
4, 3
4 y
C2e 4t
y0
C2
3e 4t
3 2
x
16
5 2i j and
’f 1, 2 ˜ u
0.
© 2010 Brooks/Cole, Cengage Learning
220
Chapter 13
Functions of Several Variables
z
71. (a)
72. (a)
D
500
400
300
1
2
1
x
x
6
6
y
2
y
(b) ’T x, y
’T 3, 5
400e
x2 y 2
ª¬ x i 400e 7 ª¬3i 250 30 x 2 50 sin S y 2
(b) The graph of D
1 jº
2 ¼
would model the ocean floor.
1 jº
2 ¼
250 30 1 50 sin
(c) D 1, 0.5
There will be no change in directions perpendicular
to the gradient: r i 6 j
S
4
| 315.4 ft
wD
wD
60 x and
1, 0.5
60
wx
wx
wD
Sy
(e)
25S cos
and
wy
2
wD
S
| 55.5
1, 0.5
25S cos
wy
4
(d)
(c) The greatest increase is in the direction of the
gradient: 3i 12 j
(f ) ’D
§S y ·
60 xi 25S cos¨ ¸ j
© 2 ¹
’D 1, 0.5
60i 55.5 j
73. True
74. False
2 ! 1 when u
Du f x, y
S·
S·
§
§
¨ cos ¸i ¨ sin ¸ j.
4¹
4¹
©
©
75. True
76. True
z2
C. Then ’f x, y, z
e x cos yi e x sin yj zk.
2
78. We cannot use Theorem 13.9 because f is not a differentiable function of x and y. So, we use the definition of directional
derivatives.
e x cos y 77. Let f x, y, z
Du f x, y
Du f 0, 0
If f 0, 0
lim
t o0
f x t cos T , y t sin T f x, y
t
ª
§ t ·
§ t ·º
f «0 ¨
¸, 0 ¨
¸» f 0, 0
© 2¹
© 2 ¹¼
lim ¬
t o0
t
2, then Du f 0, 0
§
f ¨0 lim ©
t
2
t o0
,0 t
ª § t ·§ t · º
«4
»
1 ¨ 2 ¸¨ 2 ¸
lim « © 2 ¹© 2 ¹ »
t o0 t « § t ·
§t · »
« ¨ ¸¨ ¸ »
«¬ © 2 ¹ © 2 ¹ »¼
t ·
¸2
2¹
1 ª 2t 2 º
lim « 2 »
t o0 t t
¬ ¼
º
1 ª 2t 2
lim « 2 2»
t o0 t t
¬
¼
lim
t o0
2
which does not exist.
t
0
which implies that the directional derivative exists.
79. (a) f x, y
3
xy is the composition of two continuous functions, h x, y
xy and g z
z1 3 , and
therefore continuous by Theorem 13.2.
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
(b) f x 0, 0
f y 0, 0
f 0 'x, 0 f 0, 0
'x
lim
f 0, 0 'y f 0, 0
'y
'y o 0
S
cos T i sin T j, T z 0,
Let u
Du f 0, 0
lim
t o0
2
,S,
221
13
lim
'x o 0
Tangent Planes and Normal Lines
lim
0 ˜ 'x
0
'x
lim
0 ˜ 'y
0
'y
'x o 0
0
13
'y o 0
0
S
. Then
2
f 0 t cos T , 0 t sin T f 0, 0
t
3
lim
t o0
t 2 cos T sin T
t
3
lim
t o0
cos T sin T
, does not exist.
t1 3
z
(c)
3
−2
−1
2
y
2
x
Section 13.7 Tangent Planes and Normal Lines
1. F x, y, z
3 x 5 y 3z 15
0
3x 5 y 3z
2. F x, y , z
7. F x, y , z
15 Plane
x 2 y 2 z 2 25
’F
x2 y 2 z 2
4x2 9 y 2 4 z 2
4. F x, y , z
5. F x, y, z
’F
n
6. F x, y, z
’F
n
2
3 x 4 y 12 z
’F
’F
0
’F 3, 4, 5
0
n
9 16 144
x2 y2 z
F x, y , z
8.
13
’F
’F
3
4
12
i j k
13
13
13
x
x y
2
2
i y
x y2
2
jk
3
4
i jk
5
5
5 §3
4
·
¨ i j k¸
5
5 2©5
¹
1
3i 4 j 5k
5 2
2
3i 4 j 5k
10
x y z 4
9. F x, y, z
i jk
1
i jk
3
2 6
1
1
2
i j
k
6
6
6
’F x, y, z
0 Hyperbolic paraboloid
3i 4 j 12k , ’F
’F
’F
4 z 2 Elliptic cone
16 x 9 y 36 z
16 x 2 9 y 2 36 z
’F
’F
n
4 4 16
0
4x2 9 y 2
2
2i 2 j 4k
’F 1, 1, 2
25
Sphere, radius 5, centered at origin.
3. F x, y , z
2 xi 2 yj 2 zk
’F 1, 1, 2
0
x2 y2 z 2 6
3
i jk
3
’F
x3 z
3x 2i k
’F 2, 1, 8
12i k
’F 2, 1, 8
145
n
’F
’F
1
12i k
145
© 2010 Brooks/Cole, Cengage Learning
222
Chapter 13
F x, y , z
10.
Functions of Several Variables
x2 y4 z
’F x , y , z
2 xy 4i 4 x 2 y 3 j k
’F 1, 2, 16
32i 32 j k
’F
’F
n
’F x, y , z
F x, y , z
’F x , y , z
’F 2, 1, 2
’F
’F
n
12. F x, y, z
’F
1
32i 32 j k
2049
n
’F 1, 1, 1
n
’F
’F
F x, y , z
13.
’F x, y, z
’F 1, 4, 3
n
’F
’F
F x, y , z
14.
2 § 1
·
¨ i 3 3j k ¸
113 © 2
¹
1
i 6 3j 2k
113
113
i 6 3j 2k
113
4i 3 j 12k
’F x , y , z
2
cos x y i cos x y j k
§S S 3·
’F ¨ , , ¸
© 3 6 2¹
x y y z 2 xz 4
2 3
sin x y z 2
F x, y , z
16.
1
4i 3j 12k
13
3
n
4i 5 j 7k
3
3
i jk
2
2
·
2 § 3
3
i j k ¸¸
¨¨
2
10 © 2
¹
1
3i 3 j 2k
10
’F
’F
3 10
1
4i 5 j 7k
3 10
10
10
§ x ·
ln ¨
ln x ln y z
¸
© y z¹
1
1
1
i j
k
x
y z
y z
17. f x, y
i jk
1
i jk
3
ze x
2 y2
i 2 yze x
F x, y , z
x2 y2 3 z
Fx x, y, z
2x
’F 2, 2, 3
12i 12 j k
1
12i 12 j k
17
j ex
2 y2
Fy x, y, z
Fy 2, 1, 8
4
4x 2 y z
18.
2 y2
3 j 2k
2y
k
f x, y
F x, y , z
1
Fz x, y, z
Fz 2, 1, 8
2
4 x 2 2 y 1 1z 8
3
i jk
3
3
2 y2
3i x 2 y 2 3, 2, 1, 8
Fx 2, 1, 8
2 xze x
’F
’F
’F
’F
2 xi 3 j 3z 2k
’F x , y , z
n
1
i 3 3j k
2
x2 3 y z3 9
2 xy 3 2 z 3 i 3 x 2 y 2 2 yz j 6 xz 2 y 2 k
’F 1, 1, 1
sin yi x cos yj k
§ S ·
’F ¨ 6, , 7 ¸
© 6 ¹
2049
32i 32 j k
2049
11.
x sin y z 4
F x, y , z
15.
1
0
2
y
, 1, 2, 2
x
y
z
x
y
x2
Fx x, y, z
Fx 1, 2, 2
2
Fy x, y, z
1
x
Fy 1, 2, 2
1
2 x 1 y 2 z 2
0
2 x y z 2
0
2x y z
2
Fz x, y, z
1
Fz 1, 2, 2
1
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
x2 y2 z
F x, y , z
x
Fx x, y, z
x y
2
0
3x 4 y 5 z
0
G x, y , z
y x2
Gx x , y , z
1
Fz 3, 4, 5
1
4
5
0
y
x2 y 2
1 y2 x2
Gx 1, 0, 0
G y x, y , z
G y 1, 0, 0
0
1x
1 y2 x2
1
Gz x , y , z
1
Gz 1, 0, 0
1
x 2 y 2 , 1, 1, 2
g x, y
x2 y 2 z
Gx x , y , z
2x
G y x, y , z
2y
Gz x , y , z
1
Gx 1, 1, 2
2
G y 1, 1, 2
2
Gz 1, 1, 2
1
2 x 1 2 y 1 1z 2
2x 2 y z
22. z
x
x2 y2
0
G x, y , z
0
2
x 2 2 xy y 2 , 1, 2, 1
F x, y , z
x 2 2 xy y 2 z
Fx x, y, z
2x 2 y
Fx 1, 2, 1
2
Fy x, y, z
Fy 1, 2, 1
2 x 1 2 y 2 z 1
2 x 2 y z 1
2x 2 y z
23.
Fz x, y, z
y
arctan , 1, 0, 0
x
y
arctan z
x
g x, y
21.
x y2
2
Fy 3, 4, 5
3
4
x 3 y 4 z 5
5
5
3x3 4 y 4 5z 5
y z
y
Fy x, y, z
2
3
5
Fx 3, 4, 5
20.
223
x 2 y 2 , 3, 4, 5
f x, y
19.
Tangent Planes and Normal Lines
f x, y
2
2x
3
y, 3, 1, 1
F x, y , z
2
2x
3
y z
Fx x, y , z
23 ,
Fy x, y, z
2 x 2 y
Fz x, y, z
Fz 1, 2, 1
2
1
1
0
0
1
1,
23 x 3 y 1 z 1
0
23 x y z 2
0
2 x 3 y 3z
6
Fz x, y, z
1
© 2010 Brooks/Cole, Cengage Learning
224
Chapter 13
§ S ·
e x sin y 1 , ¨ 0, , 2 ¸
© 2 ¹
z
24.
Functions of Several Variables
F x, y , z
e x sin y 1 z
Fx x, y , z
e x sin y 1
§ S ·
Fx ¨ 0, , 2 ¸
© 2 ¹
2x z
§ S ·
Fy ¨ 0, , 2 ¸
© 2 ¹
2
ln
x 2 y 2 , 3, 4, ln 5
H x, y , z
ln
x2 y 2 z
x
x2 y2
H x x, y , z
§ S ·
Fz ¨ 0, , 2 ¸
© 2 ¹
0
1
1
1
ln x 2 y 2 z
2
H y x, y , z
3
25
H x 3, 4, ln 5
H y 3, 4, ln 5
3
4
x 3 y 4 z ln 5
25
25
3 x 3 4 y 4 25 z ln 5
h x, y
H x, y , z
4
25
H z x, y , z
1
H z 3, 4, ln 5
1
0
25 1 ln 5
§ S
2·
cos y, ¨¨ 5, ,
¸¸
4
2
©
¹
cos y z
H x x, y , z
H y x, y , z
0
§ S
2·
H x ¨¨ 5, ,
¸¸
4
2
©
¹
0
§ S
2·
H y ¨¨ 5, ,
¸¸
4
2
©
¹
2§
S· §
2·
¸
¨ y ¸ ¨¨ z 2 ©
4¹ ©
2 ¸¹
0
2
y z 2
2S
2
8
2
0
4 2 y 8z
27. x 2 4 y 2 z 2
F x, y , z
y
x2 y2
0
3x 4 y 25 z
26.
Fz x, y, z
2
h x, y
25.
e x cos y
Fy x, y, z
sin y
2
2
H z x, y , z
§ S
2·
H z ¨¨ 5, ,
¸¸
4
2
©
¹
1
1
2S 4
36, 2, 2, 4
x 2 4 y 2 z 2 36
Fx x, y, z
2x
Fy x, y, z
Fx 2, 2, 4
4
Fy 2, 2, 4
8y
16
4 x 2 16 y 2 8 z 4
0
x 2 4 y 2 2z 4
0
x 4 y 2z
Fz x, y, z
2z
Fz 2, 2, 4
8
18
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
28. x 2 2 z 2
F x, y , z
x2 y 2 2z 2
2x
Fy x, y, z
2 y
Fz x, y, z
4z
Fx 1, 3, 2
2
Fy 1, 3, 2
6
Fz 1, 3, 2
8
2 x 1 6 y 3 8 z 2
0
x 1 3 y 3 4 z 2
0
x 3 y 4z
0
29. xy 2 3 x z 2
8, 1, 3, 2
xy 2 3 x z 2 8
Fx x, y, z
y2 3
Fy x, y, z
2 xy
Fz x, y, z
2 z
Fx 1, 3, 2
12
Fy 1, 3, 2
6
Fz 1, 3, 2
4
12 x 1 6 y 3 4 z 2
30. x
225
y 2 , 1, 3, 2
Fx x, y, z
F x, y , z
Tangent Planes and Normal Lines
0
12 x 6 y 4 z
22
6x 3y 2z
11
y 2 z 3 , 4, 4, 2
F x, y , z
x 2 yz 3 y
Fx x, y, z
1
Fy x, y, z
2 z 3
Fz x, y, z
2 y
Fx 4, 4, 2
1
Fy 4, 4, 2
1
Fz 4, 4, 2
8
x 4 1 y 4 8 z 2
0
x y 8z
16
x y 8z
31. x y z
16
9, 3, 3, 3
F x, y , z
x y z 9
Fx x, y, z
1
Fy x, y, z
1
Fz x, y, z
1
Fx 3, 3, 3
1
Fy 3, 3, 3
1
Fz 3, 3, 3
1
x 3 y 3 z 3
0
x y z
9 same plane!
Direction numbers: 1, 1, 1
Line: x 3
y 3
z 3
32. x 2 y 2 z 2 9, 1, 2, 2
F x, y , z
x2 y2 z 2 9
Fx x, y, z
2x
Fy x, y, z
2y
Fz x, y, z
2z
Fx 1, 2, 2
2
Fy 1, 2, 2
4
Fz 1, 2, 2
4
Direction numbers: 1, 2, 2
Plane: x 1 2 y 2 2 z 2
Line:
x 1
1
y 2
2
0, x 2 y 2 z
9
z 2
2
© 2010 Brooks/Cole, Cengage Learning
226
Chapter 13
Functions of Several Variables
33. x 2 y 2 z
9, 1, 2, 4
F x, y , z
x2 y 2 z 9
Fx x, y, z
2x
Fy x, y, z
2y
Fz x, y, z
1
Fx 1, 2, 4
2
Fy 1, 2, 4
4
Fz 1, 2, 4
1
Direction numbers: 2, 4, 1
Plane: 2 x 1 4 y 2 z 4
Line:
x 1
2
y 2
4
0, 2 x 4 y z
14
z 4
1
16 x 2 y 2 , 2, 2, 8
34. z
16 x 2 y 2 z
F x, y , z
Fx x, y, z
2 x
Fy x, y, z
2 y
Fz x, y, z
1
Fx 2, 2, 8
4
Fy 2, 2, 8
4
Fz 2, 2, 8
1
4 x 2 4 y 2 z 8
0
4 x 4 y z
4x 4 y z
24
24
Direction numbers: 4, 4, 1
x 2
y 2
Line:
z 8
4
4
x 2 y 2 , 3, 2, 5
35. z
x2 y 2 z
F x, y , z
Fx x, y, z
2x
Fy x, y, z
2 y
Fz x, y, z
1
Fx 3, 2, 5
6
Fy 3, 2, 5
4
Fz 3, 2, 5
1
6x3 4 y 2 z 5
0
6x 4 y z
5
Direction numbers: 6, 4, 1
Line:
x 3
6
36. xy z
y 2
4
z 5
1
0, 2, 3, 6
F x, y , z
xy z
Fx x, y, z
y
Fx 2, 3, 6
Fy x, y, z
3
Fy 2, 3, 6
Direction numbers: 3, 2, 1
Plane: 3 x 2 2 y 3 z 6
Line:
37. xyz
x 2
3
y 3
2
x
2
Fz x, y, z
1
Fz 2, 3, 6
1
0, 3x 2 y z
z 6
1
10, 1, 2, 5
F x, y , z
xyz 10
Fx x, y, z
yz
Fy x, y, z
xz
Fz x, y, z
xy
Fx 1, 2, 5
10
Fy 1, 2, 5
5
Fz 1, 2, 5
2
Direction numbers: 10, 5, 2
Plane: 10 x 1 5 y 2 2 z 5
Line:
6
x 1
10
y 2
5
0, 10 x 5 y 2 z
30
z 5
2
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
Tangent Planes and Normal Lines
227
ye 2 xy , 0, 2, 2
38. z
F x, y , z
ye 2 xy z
Fx x, y, z
2 y 2e2 xy
Fy x, y, z
Fx 0, 2, 2
8
Fy 0, 2, 2
1 2 xy e 2 xy
1
8x0 y 2 z 2
0
8x y z
0
Fz x, y, z
1
Fz 0, 2, 2
1
Direction number: 8, 1, 1
Line:
x
8
y 2
1
z 2
1
y §
S·
arctan , ¨1, 1, ¸
x ©
4¹
y
F x, y , z
arctan z
x
y
Fx x, y, z
Fy x, y, z
x2 y2
39. z
1
S·
S·
§
§
Fx ¨1, 1, ¸
Fy ¨1, 1, ¸
4¹
2
4¹
©
©
Direction numbers: 1, 1, 2
S·
§
Plane: x 1 y 1 2¨ z ¸
4¹
©
Line:
x 1
1
40. y ln xz 2
F x, y , z
1
S·
§
Fz ¨1, 1, ¸
4¹
©
1
0, x y 2 z
S
2
2, e, 2, 1
y>ln x 2 ln z@ 2
y
x
2
e
Fx e, 2, 1
Fy x, y, z
Fy e, 2, 1
ln x 2 ln z
2 xi 2 yj
i
(a) ’F u ’G
’G 1, 1, 1
j
4
8
’G x, y, z
2i 2 j
2y
z
0
x 2 y 2 5 G x, y , z
’F x, y, z
Fz x, y, z
Fz e, 2, 1
1
2
x e y 2 4 z 1
e
2
x y 4z
e
2
Direction numbers: , 1, 4
e
x e
y 2
z 1
2e
1
4
’F 1, 1, 1
1
2
Fz x, y, z
z S 4
2
y 1
1
Fx x, y, z
41. F x, y, z
x
x2 y2
x z
i k
i k
k
2i 2 j 2k
2 2 0
1 0 1
2 i j k
Direction numbers: 1, 1, 1
Line: x 1
y 1
1
z 1
© 2010 Brooks/Cole, Cengage Learning
228
Chapter 13
(b) cos T
Functions of Several Variables
’F ˜ ’ G
’F ’G
2
2 2
1
2
2
Not orthogonal
42. F x, y, z
x2 y2 z
G x, y , z
4 y z
’F x, y, z
2 xi 2 yj k
’G x , y , z
j k
’F 2, 1, 5
4i 2 j k
’G 2, 1, 5
j k
i
(a) ’F u ’G
j
k
4 2 1
i 4 j 4k
0 1 1
Direction numbers: 1, 4, 4.
(b) cos T
43. F x, y, z
’F
’F ˜ ’G
’F ’G
x 2
1
y 1
4
3
21 2
3
42
x 2 z 2 25 G x, y, z
2 xi 2 zk
’F 3, 3, 4
’G
6i 8k
i
(a) ’F u ’G
42
; not orthogonal
14
y 2 z 2 25
2 yj 2 zk
’G 3, 3, 4
j
z 5
4
6 j 8k
k
48i 48 j 36k
6 0 8
12 4i 4 j 3k
0 6 8
Direction numbers: 4, 4, 3.
(b) cos T
’F ˜ ’ G
’F ’G
x3
4
64
10 10
y 3
4
16
; not orthogonal
25
x2 y2 z
44. F x, y, z
x
’F x , y , z
x y
2
2
G x, y , z
y
i x y2
2
i
(a) ’F u ’G
j
5i 2 j 3k
’G 3, 4, 5
5i 2 j 3k
2
22
k
2
34
26
i j
k
5
5
5
3 5 4 5 1
5
5 x 2 y 3z
j k ’G x, y, z
3
4
i jk
5
5
’F 3, 4, 5
z 4
3
3
Direction numbers: 1, 17, 13
x 3
1
(b) cos T
45. F x, y, z
’F x, y, z
’F 3, 1, 2
y 4
17
z 5
Tangent line
13
’F ˜ ’ G
’F ’G
85
2
38
8
Not orthogonal
5 76
x 2 y 2 z 2 14 G x, y, z
2 xi 2 yj 2 zk
6i 2 j 4k
’G x, y, z
’G 3, 1, 2
x y z
i jk
i jk
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
i
(a) ’F u ’G
j
k
6 2
4
2i 10 j 8k
Tangent Planes and Normal Lines
229
2>i 5 j 4k @
1 1
1
Direction numbers: 1, 5, 4
x 3
1
Line:
y 1
5
’F ˜ ’G
’F ’ G
(b) cos T
z 2
4
0 Ÿ orthogonal
x2 y2 z
46. F x, y, z
’F x, y, z
G x, y , z
2 xi 2 yj k ’G x, y, z
’F 1, 2, 5
2i 4 j k
i
(a) ’F u ’G
j
’G 1, 2, 5
i j 6k
25i 13 j 2k
6
Direction numbers: 25, 13, 2.
’F ˜ ’G
’F ’ G
x 1
25
y 2
13
z 5
2
0; orthogonal
3 x 2 2 y 2 z 15, 2, 2, 5
47. F x, y, z
i j 6k
k
2 4 1
1 1
(b) cos T
x y 6 z 33
x 2 y 2 5, 2, 1, 3
50. F x, y, z
’F x, y, z
6 xi 4 yj k
’F x, y, z
’F 2, 2, 5
12i 8 j k
’F 2, 1, 3
’F 2, 2, 5 ˜ k
cos T
1
209
cos T
1 ·
¸ | 86.03q
209 ¹
T
’F 2, 2, 5
§
arccos¨
©
T
2 xy z 3 , 2, 2, 2
’F
2 yi 2 xj 3 z 2k
’F 2, 2, 2
12
’F 2, 2, 2
176
§ 3 11 ·
arccos¨¨
¸¸ | 25.24q
© 11 ¹
T
x 2 y 2 z , 1, 2, 3
49. F x, y, z
’F x, y, z
’F 1, 2, 3
cos T
T
2 xi 2 yj k
2i 4 j k
’F 1, 2, 3 ˜ k
’F 1, 2, 3
arccos
’F 2, 1, 3 ˜ k
3 11
11
90q
arccos 0
3 x2 y2 6 y z
’F x, y, z
2 xi 2 y 6 j k
2 x
0
z
0, x
0, y
3
3 02 32 6 3
0, 3, 12
3x 2 2 y 2 3x 4 y z 5
’F x, y, z
6x 3 i 4 y 4 j k
6x 3
0, x
1
2
4y 4
0, y
1
2 1
2
1,
2
3
1
2
12
vertex of paraboloid
52. F x, y, z
z
1
21
0
’F 2, 1, 3
2 y 6
4i 4 j 12k
’F 2, 2, 2 ˜ k
cos T
4i 2 j
F x, y , z
51.
48. F x, y, z
2 xi 2 yj
2
3
1
2
4 1 5
31
4
1, 31
4
1
| 77.40q
21
© 2010 Brooks/Cole, Cengage Learning
230
Chapter 13
Functions of Several Variables
x 2 xy y 2 2 x 2 y z
53. F x, y, z
’F x, y, z
2x y 2 i x 2 y 2 j k
2x y 2
x 2 y 2
5 xy z
55. F x, y, z
’F x, y, z
0
0
5y
0
5x
0
5 yi 5 xj k
y
2x 2 Ÿ x 2 2x 2 2
x
y
3x 6 0 Ÿ x
2, z
4
Point: 0, 0, 0
2
Point: 2, 2, 4
y
z
’F x, y, z
8x 4 y 8 i 4x 4 y 5 j k
’F x , y , z
8x 4 y 8
4x 4 y 5
0
0
y
1
x2
x
1
y2
Adding, 12 x 3
0 Ÿ x
14 Ÿ y
32 , and
54
Point: 14 , 32 , 54
57. F x, y, z
1
1
z
x
y
§
1·
1 ·
§
¨ y 2 ¸i ¨ x 2 ¸ j k
x ¹
y
©
©
¹
x4 Ÿ x
1, y
1, z
3
Point: 1, 1, 3
x 2 2 y 2 3 z 2 3, 1, 1, 0
Fx x, y, z
2x
Fy x, y, z
4y
Fz x, y, z
6z
Fx 1, 1, 0
2
Fy 1, 1, 0
4
Fz 1, 1, 0
0
2 x 1 4 y 1 0 z 0
0
2 x 4 y
x 2 y
6
3
G x, y , z
xy 56. F x, y, z
4 x 2 4 xy 2 y 2 8 x 5 y 4 z
54. F x, y, z
z
0
x 2 y 2 z 2 6 x 10 y 14, 1, 1, 0
Gx x, y , z
2x 6
G y x, y , z
2 y 10
Gz x , y , z
2z
Gx 1, 1, 0
4
G y 1, 1, 0
8
Gz 1, 1, 0
0
4 x 1 8 y 1 0 z 0
4 x 8 y 12
x 2 y
0
0
3
The tangent planes are the same.
58. F x, y, z
Fx x, y, z
Fx 2, 3, 3
x 2 y 2 z 2 8 x 12 y 4 z 42, 2, 3, 3
2x 8
4
2 y 12
Fy x, y, z
Fy 2, 3, 3
6
4 x 2 6 y 3 2 z 3
G x, y , z
Fz 2, 3, 3
2z 4
2
0
4 x 6 y 2 z 20
2x 3 y z
Fz x, y, z
0
10
x 2 y 2 2 z 7, 2, 3, 3
Gx x, y , z
2x
G y x, y , z
2y
Gz x , y , z
2
Gx 2, 3, 3
4
G y 2, 3, 3
6
Gz x , y , z
2
4x 2 6 y 3 2z 3
4 x 6 y 2 z 20
2x 3y z
0
0
10
The tangent planes are the same.
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
2 xy 2 z , F 1, 1, 2
59. (a) F x, y, z
22
G x, y , z
231
0
8 x 5 y 8 z 13, G 1, 1, 2
2
Tangent Planes and Normal Lines
8 5 16 13
2
0
So, 1, 1, 2 lies on both surfaces.
(b) ’F
2 y 2i 4 xyj k , ’F 1, 1, 2
’G
2i 4 j k
16 xi 10 yj 8k , ’G 1, 1, 2
’F ˜ ’G
16i 10 j 8k
2 16 4 10 1 8
0
The tangent planes are perpendicular at 1, 1, 2 .
x 2 y 2 z 2 2 x 4 y 4 z 12
60. (a) F x, y, z
F 1, 2, 1
0
G x, y , z
4 x 2 y 2 16 z 2 24
G 1, 2, 1
0
So, 1, 2, 1 lies on both surfaces.
(b) ’F
2x 2 i 2 y 4 j 2z 4 k
’F 1, 2, 1
’G
4i 8 j 2k
8 xi 2 yj 32 zk
’G 1, 2, 1
’F ˜ ’G
8i 4 j 32k
32 32 64
0
The planes are perpendicular at 1, 2, 1 .
x2 4 y2 z 2 9
61. F x, y, z
’F
2 xi 8 yj 2 zk
This normal vector is parallel to the line with direction number 4, 8, 2.
4t Ÿ x
So, 2 x
8t Ÿ y
8y
2t Ÿ z
2z
x 4y z 9
2
2
2
2t
t
t
4t 2 4t 2 t 2 9
0 Ÿ t
r1
There are two points on the ellipse where the tangent plane is perpendicular to the line:
2, 1, 1 t 1
2, 1, 1
62. F x, y, z
’F
1
t
x2 4 y2 z 2 1
2 xi 8 yj 2 zk
i 4j k
The normal to the plane, n
must be parallel to ’F .
So,
2x
t Ÿ x
8y
4t Ÿ y
2 z
t Ÿ z
x2 4 y 2 z 2
t
2
t
2
t
2
t2
t2
t2 4
4
§1 1 1·
Two points: ¨ , , ¸
© 2 2 2¹
t
t2
1Ÿ t
r1.
§ 1 1 1·
1 and ¨ , , ¸ t
© 2 2 2¹
1
© 2010 Brooks/Cole, Cengage Learning
232
Chapter 13
Functions of Several Variables
63. Fx x0 , y0 , z0 x x0 Fy x0 , y0 , z0 y y0 Fz x0 , y0 , z0 z z0
0
Theorem 13.13
64. For a sphere, the common object is the center of the sphere. For a right circular cylinder, the common object is the axis of the
cylinder.
65. Answers will vary.
66. (a) x 2 y 2 z 2
0, 5, 13, 12
x2 y 2 z 2
F x, y , z
Fx x, y, z
2x
Fy x, y, z
2 y
Fz x, y, z
2z
Fx 5, 13, 12
10
Fy 5, 13, 12
26
Fz x, y, z
24
Direction numbers: 5, 13, 12
Plane: 5 x 5 13 y 13 12 z 12
0
5 x 13 y 12 z
0
(b) Line:
67. z
x 5
5
y 13
13
4 xy
, 2 d x d 2, 0 d y d 3
x2 1 y2 1
f x, y
4 xy
z
x2 1 y 2 1
(a) Let F x, y, z
§
·
§
·
4 y ¨ x2 1 2 x2 ¸
4x ¨ y2 1 2 y2 ¸
i 2
jk
2
2
y 1¨¨ x 2 1
x 1¨¨ y 2 1
¸¸
¸¸
©
¹
©
¹
’F x, y, z
’F 1, 1, 1
z 12
12
2
4 y 1 x2
y2 1 x2 1
2
i 4x 1 y2
x2 1 y 2 1
2
jk
k
Direction numbers: 0, 0, 1
Line: x
1, y
1, z
1t
Tangent plane: 0 x 1 0 y 1 1 z 1
4·
§
(b) ’F ¨ 1, 2, ¸
5¹
©
1, y
Line: x
Plane: 0 x 1 (c)
0i 2
4 3
2 5
2
4
t
5
6
4·
§
y 2 1¨ z ¸
25
5
©
¹
6 y 12 25 z 20
0
6 y 25 z 32
0
z
1
2
3
2
−1
0
z
1
x
1
6
jk
25
jk
6
t, z
25
0 Ÿ z
y
x
−2
2
−1
3
y
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
Tangent Planes and Normal Lines
233
sin y
, 3 d x d 3, 0 d y d 2S
x
sin y
Let F x, y, z
z
x
sin y
cos y
’F x , y , z
i jk
x2
x
1
§ S 1·
’F ¨ 2, , ¸
i k
2
2
4
©
¹
1
Direction numbers: , 0, 1 or 1, 0, 4
4
S
1
Line: x
2 t, y
,z
4t
2
2
68. (a) f x, y
S·
1·
§
§
Tangent plane: 1 x 2 0¨ y ¸ 4¨ z ¸
0 Ÿ x 4z 4 0
2¹
2¹
©
©
9
§ 2 3S 3 ·
i k
(b) ’F ¨ , , ¸
3
2
2
4
©
¹
9
Direction numbers: , 0, 1 or 9, 0, 4
4
2
3S
3
Line: x
9t , y
4t
,z
3
2
2
2·
3S ·
3·
§
§
§
Tangent plane: 9¨ x ¸ 0¨ y 0 Ÿ 9 x 4 z 12
¸ 4¨ z ¸
3¹
2 ¹
2¹
©
©
©
z
(c)
0
3
x
3
−3
2π
y
69. f x, y
6 x2 y2
, g x, y
4
z x2 (a) F x, y, z
’F x, y, z
’F 1, 2, 4
2x y
y2
6
4
1
yj k
2
2i j k
2 xi z 2x y
G x, y , z
’G x, y, z
2i j k
’G 1, 2, 4
2i j k
The cross product of these gradients is parallel to the curve of intersection.
i
j k
’F 1, 2, 4 u ’G 1, 2, 4
2
1
1
2i 4 j
2 1 1
Using direction numbers 1, 2, 0, you get x
’F ˜ ’G
’F ’G
cos T
4 1 1
6 6
1 t, y
2 2t , z
4.
4
Ÿ T | 48.2q
6
z
(b)
8
(1, 2, 4)
6
8
y
x
© 2010 Brooks/Cole, Cengage Learning
234
Chapter 13
Functions of Several Variables
16 x 2 y 2 2 x 4 y
70. (a) f x, y
2
2
g x, y
1 3x 2 y 2 6 x 4 y
f x, y
(b)
16 x 2 y 2 2 x 4 y
32 2 x 2 2 y 2 4 x 8 y
x 2 2 x 31
x 2 2 x 1 42
2
x 1
42
g x, y
3 y 2
y 2
2
3 y2 4 y 4
3 y 2
5
5
y
x
2
1. Then
14
2 r
14
14
2 j, ’g 1, 2 ’f 1, 2 14
1
2 j. The normals to f and g at this point are 2 j k and
2 j k , which are orthogonal.
Similarly, ’f 1, 2 1
g
42
y
1
5
f
3 y 2 12 y
To find points of intersection, let x
2
z
1
1 3x 2 y 2 6 x 4 y
2
1 3x 2 y 2 6 x 4 y
14
2 j and ’g 1, 2 14
1
2 j and the normals are
2 j k and
2 j k , which are also orthogonal.
(c) No, showing that the surfaces are orthogonal at 2 points does not imply that they are orthogonal at every point of
intersection.
71. F x, y, z
Fx x, y, z
Fy x, y, z
Fz x, y, z
Plane:
x2
y2
z2
2 2 1
2
a
b
c
2x
a2
2y
b2
2z
c2
2 x0
2y
2z
x x0 20 y y0 20 z z0
a2
b
c
x0 x
y y
z z
02 02
a2
b
c
72. F x, y, z
x02
y2
z2
02 02
2
a
b
c
0
1
x2
y2
z2
2 2 1
2
a
b
c
Fx x, y, z
2x
a2
Fy x, y, z
2y
b2
Fz x, y, z
2 z
c2
Fx x, y, z
2a 2 x
Fy x, y, z
2b 2 y
Fz x, y, z
2 z
Plane:
2a 2 x0 x x0 2b 2 y0 y y0 2 z0 z z0
a x0 x b y0 y z0 z
2
2x
2y
2z
Plane: 20 x x0 20 y y0 0 z z0
a
b
c2
x0 x
y y
z z
02 02
2
a
b
c
a2 x2 b2 y 2 z 2
73. F x, y, z
x02
y2
z2
02 02
2
a
b
c
0
2
a 2 x02
b 2 y02
z02
0
0
So, the plane passes through the origin.
1
© 2010 Brooks/Cole, Cengage Learning
Section 13.7
§ y·
xf ¨ ¸
© x¹
§ y·
xf ¨ ¸ z
© x¹
§ y·
§ y ·§ y ·
f ¨ ¸ xf c¨ ¸¨ 2 ¸
© x¹
© x ¹© x ¹
y ·§ 1 ·
§
§ y·
xf c¨ ¸¨ ¸
f c¨ ¸
© x ¹© x ¹
© x¹
1
z
74.
F x, y , z
Fx x, y, z
Fy x, y, z
Fx x, y, z
Tangent Planes and Normal Lines
235
§ y· y § y·
f ¨ ¸ f c¨ ¸
© x¹ x © x¹
Tangent plane at x0 , y0 , z0 :
ª § y0 · y0 § y0 ·º
§y ·
f c¨ ¸» x x0 f c¨ 0 ¸ y y0 z z0
«f¨ ¸ x
x
x
0
© 0 ¹¼
© x0 ¹
¬ © 0¹
0
ª § y0 · y0 § y0 ·º
§y ·
§y ·
§y ·
§y ·
§y ·
f c¨ ¸» x x0 f ¨ 0 ¸ y0 f c¨ 0 ¸ yf c¨ 0 ¸ y0 f c¨ 0 ¸ z x0 f ¨ 0 ¸
«f¨ ¸ © x0 ¹
© x0 ¹
© x0 ¹
© x0 ¹
© x0 ¹
¬ © x0 ¹ x0 © x0 ¹¼
0
ª § y0 · y0 § y0 ·º
§y ·
f c¨ ¸» x f c¨ 0 ¸ y z
«f ¨ ¸ x
x
x
0
0
0
© ¹¼
© x0 ¹
¬ © ¹
0
So, the plane passes through the origin x, y, z
75. f x, y
0, 0, 0 .
ex y
f x x, y
e x y , f y x, y
e x y
f xx x, y
e x y , f yy x, y
e x y , f xy x, y
e x y
(a) P1 x, y | f 0, 0 f x 0, 0 x f y 0, 0 y
(b) P2 x, y | f 0, 0 f x 0, 0 x f y 0, 0 y 1 x y
1 f
2 xx
0, 0 x 2 f xy 0, 0 xy 1 f
2 yy
0, 0 y 2
(c) If x
0, P2 0, y
1 y 1 y 2 . This
2
is the second-degree Taylor polynomial for e y .
If y
0, P2 x, 0
1 x 1 x 2 . This
2
is the second-degree Taylor polynomial for e x .
(d)
x
0
y
0
f x, y
1
P1 x, y
P2 x, y
1
1
(e)
0.1
0.9048
0.9000
z
P1
4
0.1
1.1052
1.1000
1.1050
0.2
0.5
0.7408
0.7000
0.7450
1
0.5
1.6487
1.5000
1.6250
76. f x, y
−2
0.9050
0.2
1 y2
2
f
P2
−2
0
1 x y 12 x 2 xy 2
x
1
−2
−4
2
y
cos x y
f x x, y
sin x y , f y x, y
sin x y
f xx x, y
cos x y , f yy x, y
cos x y , f xy x, y
(a) P1 x, y | f 0, 0 f x 0, 0 x f y 0, 0 y
1
(b) P2 x, y | f 0, 0 f x 0, 0 x f y 0, 0 y 1 f
2 xx
1
1 x2
2
xy cos x y
0, 0 x 2 f xy 0, 0 xy 1 f
2 yy
0, 0 y 2
1 y2
2
(c) If x
0, P2 0, y
1
1 y 2 . This
2
is the second-degree Taylor polynomial for cos y.
If y
0, P2 x, 0
1
1 x 2 . This
2
is the second-degree Taylor polynomial for cos x.
© 2010 Brooks/Cole, Cengage Learning
236
Chapter 13
(d)
x
Functions of Several Variables
y
f x, y
P1 x, y
P2 x, y
0
0
1
1
1
0
0.1
0.9950
1
0.9950
z
(e)
5
5
x
0.2
0.1
0.9553
1
0.9950
0.2
0.5
0.7648
1
0.7550
1
0.5
0.0707
1
0.1250
77. Given z
5
y
f x, y , then:
f x, y z
F x, y , z
’F x0 , y0 , z0
0
f x x0 , y0 i f y x0 , y0 j k
’F x0 , y0 , z0 ˜ k
cos T
’F x0 , y0 , z0
k
1
2
2
¬ª f x x0 , y0 ¼º ¬ª f y x0 , y0 ¼º 1
2
1
2
2
ª¬ f x x0 , y0 º¼ ¬ª f y x0 , y0 ¼º 1
78. Given w
F x, y, z where F is differentiable at
x0 , y0 , z0 and ’F x0 , y0 , z0 z 0,
the level surface of F at x0 , y0 , z0 is of the form F x, y, z
F x, y , z C
G x, y , z
Then ’G x0 , y0 , z0
C for some constant C. Let
0.
’F x0 , y0 , z0 where ’G x0 , y0 , z0 is normal to F x, y, z C
0 at x0 , y0 , z0 . So,
’F x0 , y0 z0 is normal to the level surface through x0 , y0 , z0 .
Section 13.8 Extrema of Functions of Two Variables
x 1
1. g x, y
2
y 3
2
t 0
Check: g x
2 x 1
0 Ÿ x
gy
2 y 3
0 Ÿ y
2, g yy
2, g xy
0, d
Check: g x
1
3
2 2 0
4 ! 0
At critical point 1, 3 , d ! 0 and g xx ! 0 Ÿ relative
minimum at 1, 3, 0 .
2
2
d 5
0 Ÿ x
3
y 2
Relative maximum: 3, 2, 5
Relative minimum: 1, 3, 0
g xx
s x3
2. g x, y
2 x 3
gy
2 y 2
g xx
2, g yy
d
0 Ÿ y
2, g xy
2 2 0
2
0
4 ! 0
At critical point 3, 2 , d ! 0 and g xx 0 Ÿ relative
maximum at 3, 2, 5 .
© 2010 Brooks/Cole, Cengage Learning
Section 13.8
x2 y2 1 t 1
3. f x, y
Extrema of Functions of Two Variables
Relative minimum: 0, 0, 1
0 Ÿ x
x y2 1
2
y
fy
0 Ÿ y
x2 y2 1
0
Check: f x
x2 y 2 1
fy
0
32
f xx
f yy
f xy
x2 1
f yy
x2 y2 1
25 x 2
x2 y2 1
32
At the critical point 0, 0 , f xx ! 0 and
f xx f yy f xy
2
! 0.
2
y2
2
y2
y
25 x 2
0 Ÿ x
2
0 Ÿ y
0
25 y 2
ª25 x 2
¬
32
xy
f xy
y2 d 5
x2
y2 1
f xx
2
Relative maximum: 2, 0, 5
x
Check: f x
25 x 2
4. f x, y
237
y2º
¼
2
25 x 2
ª25 x 2
¬
2
32
2
y2º
¼
32
y2º
¼
32
y x 2
ª25 x 2
¬
2
At the critical point 2, 0 , f xx 0
So, 0, 0, 1 is a relative minimum.
and f xx f yy f xy
2
! 0.
So, 2, 0, 5 is a relative maximum.
x2 y2 2 x 6 y 6
5. f x, y
x1
2
y 3
2
4 t 4
Relative minimum: 1, 3, 4
Check: f x
2x 2
0 Ÿ x
1
fy
2y 6
0 Ÿ y
3
f xx
2, f yy
2, f xy
0
At the critical point 1, 3 , f xx ! 0 and f xx f yy f xy
2
! 0. So, 1, 3, 4 is a relative minimum.
x 2 y 2 10 x 12 y 64
6. f x, y
x 2 10 x 25 y 2 12 y 36 25 36 64
x5
2
y 6
2
3 d 3
Relative maximum: 5, 6, 3
Check: f x
2 x 10
0 Ÿ x
5
fy
2 y 12
0 Ÿ y
6
f xx
2, f yy
2, f xy
0, d
2 2 0
4 ! 0
At critical point 5, 6 , d ! 0 and f xx 0 Ÿ relative maximum at 5, 6, 3 .
7. f x, y
3 x 2 2 y 2 6 x 4 y 16
fx
fy
6x 6
4y 4
0½
¾x
0¿
f xx
6, f yy
4, f xy
1, y
0, d
8. f x, y
1
64 0
24 ! 0.
At the critical point 1, 1 , d ! 0
and f xx ! 0 Ÿ 1, 1, 11 is a relative minimum.
3 x 2 2 y 2 3 x 4 y 5
fx
6 x 3
0 when x
fy
4 y 4
0 when y
f xx
6, f yy
4, f xy
At the critical point
and f xx f yy f xy
So,
1
,
2
2
1
,
2
1
.
2
1.
0
1 , f xx 0
! 0.
1, 31
is relative maximum.
4
© 2010 Brooks/Cole, Cengage Learning
238
Chapter 13
9. f x, y
Functions of Several Variables
x 2 5 y 2 10 x 10 y 28
fy
2 x 10
10 y 10
f xx
2, f yy
fx
0½
¾x
0¿
1
5, y
10, f xy
At the critical number 5, 1 , d ! 0
and f xx 0 Ÿ 5, 1, 2 is a relative maximum.
13
3x y
2
2
23
2y
hy
3 x2 y 2
2
½
0°
°°
¾x
0°
°
°¿
2x
hx
2 10 ! 0.
0, d
x2 y2
14. h x, y
23
0, y
0
Because h x, y t 2 for all x, y , 0, 0, 2 is a relative
minimum.
10. f x, y
2 x 2 2 xy y 2 2 x 3
4 x y
15. g x, y
fx
fy
4x 2 y 2
2x 2 y
0
f xx
4, f yy
0½ Solving simultaneously
¾
1 and y 1.
¿ yields x
2, f xy
2
Since f x, y t 2 for all x, y , the relative minima
! 0.
x 2 xy z
of f consist of all points x, y satisfying x y
1
2
y2 2x y
2x y 2
0½ Solving simultaneously
¾
4
3, y
x y 1 0 ¿ yields x
f xx
2, f yy
21 1
1, d
4 x
x2 y2 1
17. z
fx
fy
1, f xy
x y 2
16. f x, y
So, 1, 1, 4 is a relative minimum.
11. f x, y
all x, y , 0, 0, 4 is a relative maximum.
2
At the critical point 1, 1 , f xx ! 0
and f xx f yy f xy
0, 0 is the only critical point. Because g x, y d 4 for
1 ! 0.
z
4
Relative minimum: 1, 0, 2
−4
Relative maximum: 1, 0, 2
x
−4
and f xx ! 0 Ÿ 3, 4, 5 is a relative minimum.
12. f x, y
5 x 4 xy y 16 x 10
2
fx
fy
10 x 4 y 16
4x 2 y
0
f xx
10, f yy
2, f xy
0½ Solving simultaneously
¾ yields x 8 and y 16.
¿
and f xx f yy f xy
y 3 3 yx 2 3 y 2 3 x 2 1
18. f x, y
z
Relative maximum: 0, 0, 1
40
20
Saddle points:
3
0, 2, 3 , r
3, 1, 3
y
3
x
4
At the critical point 8, 16 , f xx 0
2
y
4
5
At the critical point 3, 4 , d ! 0
2
0.
x 2 4 y 2 e1 x
19. z
2 y2
z
6
Relative minimum: 0, 0, 0
! 0.
So, 8, 16, 74 is a relative maximum.
5
Relative maxima: 0, r1, 4
−4
−4
Saddle points: r1, 0, 1
x2 y 2
13. f x, y
fx
fy
x
x2 y 2
y
x2 y2
½
0°
°
¾x
0°
°
¿
x
4
4
e xy
20. z
Saddle point: 0, 0, 1
y
0
z
100
Because f x, y t 0 for all x, y and
f 0, 0
0, 0, 0, 0 is a relative minimum.
x
3
3
y
© 2010 Brooks/Cole, Cengage Learning
y
Section 13.8
80 x 80 y x 2 y 2
21. h x, y
hx
hy
80 2 x
80 2 y
0½
¾x
0¿
hxx
2, hyy
2, hxy
d
2 2 0
y
Extrema of Functions of Two Variables
23. g x, y
40
0,
4 ! 0
xy
gx
gy
y½
¾x
x¿
g xx
0, g yy
0 and y
0
0, g xy
1
2
At the critical point 0, 0 , g xx g yy g xy
At the critical point 40, 40 , d ! 0 and
239
0.
So, 0, 0, 0 is a saddle point.
hxx 0 Ÿ 40, 40, 3200 is a relative maximum.
x 2 3 xy y 2
24. h x, y
x y x y
2
22. g x, y
2
gx
gy
2x 1 0 ½x
¾
2 y 1 0¿ y
g xx
2, g yy
12
1 2
2, g xy
2 2 0
0, d
4 0
At the critical point 1 2, 1 2 , d 0
hxx
2, hyy
2, hxy
3
At the critical point 0, 0 , hxx hyy hxy
2
0.
x 2 xy y 2 3x y
25. f x, y
fx
2x y 3
fy
x 2 y 1
0
0
Solving simultaneously yields x
d
2x 3y
0 ½Solving simultaneously
¾
0¿ yields x
3x 2 y
0 and y
0.
So, 0, 0, 0 is a saddle point.
Ÿ 1 2, 1 2, 0 is a saddle point.
f xx
hx
hy
2, f yy
2, f xy
2 2 1
1, y
1.
1
2
5 0
At the critical point 1, 1 , d 0 Ÿ 1, 1, 1 is a saddle point.
2 xy 26. f x, y
1 4
x y2 1
2
fy
2 y 2 x3 ½° Solving by substitution yields 3 critical points:
¾
2 x 2 y 3 °¿ 0, 0 , 1, 1 , 1, 1
f xx
6 x 2 , f yy
fx
6 y 2 , f xy
At 0, 0 , f xx f yy f xy
At 1, 1 , f xx f yy f xy
2
2
At 1, 1 , f xx f yy f xy
fy
e x sin y
e
0 Ÿ 0, 0, 1 saddle point.
! 0 and f xx 0 Ÿ 1, 1, 2 relative maximum.
2
! 0 and f xx 0 Ÿ 1, 1, 2 relative maximum.
e x sin y
27. f x, y
fx
2
x
cos y
0½° Because e x ! 0 for all x and sin y and cos y are never
¾
0 °¿ both zero for a given value of y, there are no critical points.
© 2010 Brooks/Cole, Cengage Learning
Chapter 13
240
Functions of Several Variables
§1
2
2 · 1 x 2 y 2
¨ x y ¸e
©2
¹
28. f x, y
2 x 3 2 xy 2 3 x e1 x
fx
0½°
¾ Solving yields the critical points 0, 0
0 °¿
2 y2
1 x 2 y 2
fy
2 x2 y 2 y3 y e
f xx
4 x 4 4 x 2 y 2 12 x 2 2 y 2 3 e1 x
f yy
4 y 4 4 x 2 y 2 2 x 2 8 y 2 1 e1 x
f xy
4 x3 y 4 xy 3 2 xy e1 x
points r
2 y2
2 y2
2
0. So, 0, 0, e 2 is a saddle point. At the critical
2
2 2 , f xx 0 and f xx f yy f xy
2
6 2, 0 , f xx ! 0 and f xx f yy f xy
! 0. So, 0, r
! 0. So, r
4
x y
t 0. z
x2 y 2
29. z
e are relative maxima. At the critical
2 2,
6 2, 0, e e are relative minima.
f xx f yy f xy 2
35. d
y z 0.
0 if x
2· §
6 ·
, 0 ¸¸.
¸¸, ¨¨ r
2 ¹ © 2
¹
2 y2
At the critical point 0, 0 , f xx f yy f xy
points 0, r
§
, ¨¨ 0, r
©
2 8 f xy 2
16 f xy 2 ! 0
Ÿ f xy 2 16 Ÿ 4 f xy 4
Relative minimum at all points x, x , x z 0.
f xx f yy f xy 2 0 if f xx and f yy have opposite
36. d
z
60
signs. So, a, b, f a, b is a saddle point. For example,
40
consider f x, y
x 2 y 2 and a, b
0, 0 .
x3 y 3
37. f x, y
3
3
x
(a) f x
3x 2
fy
3y2
2
x2 y2
30. z
y
x2 y2
t 0. z
y 2 z 0.
0 if x 2
Relative minima at all points x, x and x, x , x z 0.
0 ½°
¾x
0°¿
y
0
Critical point: 0, 0
(b) f xx
6 x, f yy
6 y, f xy
At 0, 0 , f xx f yy f xy
z
2
0
0.
2
0, 0, 0 is a saddle point.
5
5
x
y
(c) Test fails at 0, 0 .
(d)
z
2
31. f xx f yy f xy
2
9 4 62
0
Insufficient information.
x
2
32. f xx 0 and f xx f yy f xy
2
3 8 2 ! 0
2
f has a relative maximum at x0 , y0
y
−2
1
2
−2
Saddle point
(0, 0, 0)
−2
33. f xx f yy f xy
2
9 6 10 0
2
f has a saddle point at x0 , y0 .
34. f xx ! 0 and f xx f yy f xy
2
25 8 102 ! 0
f has a relative minimum at x0 , y0
© 2010 Brooks/Cole, Cengage Learning
Section 13.8
38. f x, y
x 3 y 3 6 x 2 9 y 2 12 x 27 y 19
40. f x, y
0 °½Solving yields
¾
2 and y
3.
0°¿ x
(a) f x
(a) f x
3 x 2 12 x 12
fy
3 y 2 18 y 27
6 x 12, f yy
(b) f xx
Extrema of Functions of Two Variables
6 y 18, f xy
At 2, 3 , f xx f yy f xy
2
0
x 1
x 1
0.
4
y
2
y2
2
y2
x 1
2
32
2
32
x 1 y 2
ª x 1 2 y 2 2º
¬
¼
2
At 1, 2 , f xx f yy f xy
−2
½
0°
°°Solving yields
¾ x 1 and y 2.
0°
°
°¿
2
ª x 1 2 y 2 2º
¬
¼
f xy
−4
t 0
2
ª x 1 2 y 2 2º
¬
¼
f yy
(2, −3, 0)
x
2
y 2
(b) f xx
z
2
y2
(c) Test fails at 2, 3 .
(d)
y 2
x 1
fy
2, 3, 0 is a saddle point.
2
x 1
241
2
32
is undefined.
1, 2, 0 is an absolute minimum.
2
x 1
39. f x, y
y 4
2 x 1 y 4
(a) f x
2 x 1
fy
2
2
t 0
(c) Test fails at 1, 2 .
0½°critical points:
¾
0¿° 1, a and b, 4
2
y 4
z
4
2
(b) f xx
2 y 4
f yy
2 x 1
f xy
4 x 1 y 4
2
−6
−4
(1, −2, 0)
2
0.
Because f x, y t 0, there are absolute minima
at 1, a, 0 and b, 4, 0 .
(c) Test fails at 1, a and b, 4 .
x2 3 y 2 3 t 0
41. f x, y
2 ½ f and f are undefined
x
y
3x 1 3 °°
0 and y
0.
¾ at x
2 °
Critical
point:
0,
0
3 y 1 3 °¿
(a) f x
fy
z
6
2
, f yy
9x4 3
(b) f xx
4
6
x
Absolute
minimum
(b, −4, 0)
−2
−4
y
Absolute
minimum
(1, a, 0)
2
, f xy
9 y4 3
At 0, 0 , f xx f yy f xy
4
2
y
2
2
4
x
At both 1, a and b, 4 , f xx f yy f xy
(d)
(d)
2
0
is undefined.
0, 0, 0 is an absolute minimum.
(c) Test fails at 0, 0 .
(d)
z
6
6
x
4
2
2
4
6
y
Absolute
minimum (0, 0, 0)
© 2010 Brooks/Cole, Cengage Learning
242
Chapter 13
Functions of Several Variables
x2 y2
42. f x, y
23
½
4x
(a) f x
3x y
2
t 0
2
13°
° f x and f y are undefined at x
¾ Critical Point: 0, 0
4y
°
13
3 x 2 y 2 °¿
fy
(b) f xx
f yy
f xy
0, y
0.
4 x2 3 y 2
9 x2 y2
43
4 3x 2 y 2
9 x2 y2
43
8 xy
9 x y2
2
43
2
At 0, 0 , f xx f yy f xy
is undefined.
0, 0, 0 is an absolute minimum.
(c) Test fails at 0, 0 .
(d)
z
5
−4 −3
−2
x
2
4 3
1
2
3
y
4
(0, 0, 0)
2
x2 y 3
43. f x, y, z
fx
fy
2x
0
2 y 3
fz
2 z 1
z 1
2
t 0
½
°
0¾ Solving yields the critical point 0, 3, 1 .
0 °¿
Absolute minimum: 0 at 0, 3, 1
2
9 ª¬ x y 1 z 2 º¼ d 9
44. f x, y, z
The absolute maximum value of f is 9, and realized at all points where x y 1 z 2
0.
So, the critical points are of the form 0, a, b , c, 1, d , e, f , z
where a, b, c, d , e, f are real numbers.
x 2 4 xy 5, R
45. f x, y
fx
fy
2x 4 y
0
4 x
0½
¾x
¿
^ x, y : 1 d
x d 4, 0 d y d 2`
4
y
0 (not in region R)
Along y
0, 1 d x d 4: f
x 2 5, f 1, 0
Along y
2, 1 d x d 4: f
x 2 8 x 5, f c
f 1, 2
y
2, f 4, 2
3
2
6, f 4, 0
2x 8
21.
1
0
1, 0 d y d 2: f
4 y 6, f 1, 0
Along x
4, 0 d y d 2: f
21 16 y, f 4, 0
1
2
3
4
−1
11.
Along x
x
−1
6, f 1, 2
21, f 4, 2
2.
11.
So, the maximum is 4, 0, 21 and the minimum is 4, 2, 11 .
© 2010 Brooks/Cole, Cengage Learning
Section 13.8
46. f x, y
fx
fy
2x y
x
0
f 0, 0
^ x, y :
x 2 xy, R
0½
¾x
¿
y
Extrema of Functions of Two Variables
243
x d 2, y d 1`
y
0
2
0
Along y
Thus, f 2, 1
Along y
2, f 12 , 1
6, f
1
,
2
14 and f 2, 1
6.
x 2 x, f c
1
Along x
2, 1 d y d 1, f
Along x
2, 1 d y d 1, f
So, the maxima are f 2, 1
f x, y
2x 1
1, 2 d x d 2, f
Thus, f 2, 1
47. f x, y
x 2 x, f c
1, 2 d x d 2, f
4 2y Ÿ f c
6 and f 2, 1
1
.
2
2 z 0.
6 and the minima are f 12 , 1
14 and f
1
,
2
1
14 .
x 1, 0 d x d 1,
5 x 10
12 3 x 2 2 x 4
f x
1
2 z 0.
4 2y Ÿ f c
12 3 x 2 x 1
x
−1
2.
and the maximum is10, the minimum is 5. On the line y
f x, y
0 Ÿ x
12 3 x 2 y has no critical points. On the line y
f ( x)
12 .
−2
2x 1
14 , f 2, 1
0 Ÿ x
2 x 4, 1 d x d 2,
y
x 4
3
y=x+1
and the maximum is 6, the minimum is 5. On the line y
f x, y
12 3 x 2 12 x 1
f x
12 x 1, 0 d x d 2,
(1, 2)
2
2 x 10
y = −2x + 4
(0, 1)
1
(2, 0)
and the maximum is10, the minimum is 6.
1
2
x
3
y = − 21 x + 1
Absolute maximum: 10 at 0, 1
Absolute minimum: 5 at 1, 2
48. f x, y
2x y
fx
4 2x y
fy
2 2 x y
On the line y
f x, y
f x
2
0 Ÿ 2x
y
0 Ÿ 2x
y
x 1, 0 d x d 1,
2x x 1
2
x 1
2
and the maximum is 1, the minimum is 0. On the line y
f x, y
f x
2 x 12 x 1
2
5
x
2
1
f x
2 x 2 x 4
2
4x 4
y
3
2
(1, 2)
and the maximum is 16, the minimum is 0. On the line y
f x, y
12 x 1, 0 d x d 2,
2 x 4, 1 d x d 2,
2
y = 2x
1
2
(0, 1)
(2, 0)
x
1
and the maximum is16, the minimum is 0.
2
3
Absolute maximum: 16 at 2, 0
Absolute Minimum: 0 at 1, 2 and along the line y
2 x.
© 2010 Brooks/Cole, Cengage Learning
244
Chapter 13
49. f x, y
fx
fy
Functions of Several Variables
3x 2 2 y 2 4 y
6x
0 Ÿ x
0
4y 4 0 Ÿ y
2
4, 2 d x d 2,
On the line y
f x, y
½
¾ f 0, 1
1¿
f x
3 x 2 32 16
3x 2 16
and the maximum is 28, the minimum is16. On the curve y
f x, y
f x
3x 2 2 x 2
2
4x2
2 x4 x2
x 2 , 2 d x d 2,
y
(−2, 4)
x2 2x2 1
(2, 4)
3
and the maximum is 28, the minimum is 18 .
2
Absolute maximum: 28 at r 2, 4
1
Absolute minimum: 2 at 0, 1
50. f x, y
fx
fy
−2
½
¾ f 1, 1
1¿
2
y
1
2
1, 1 d x d 1,
On the line y
f x
On the curve y
f x, y
x
1
2 x 2 xy y 2
2 2y
0 Ÿ y 1
2 y 2x
0 Ÿ y
x Ÿ x
f x, y
−1
f x
(−1, 1)
2x 2x 1
(1, 1)
1.
x 2 , 1 d x d 1
x
−1
2x 2 x x2 x2
2
1
x 4 2 x3 2 x
11
and the maximum is1, the minimum is 16
.
Absolute maximum: 1 at 1, 1 and on y
0.6875 at 12 ,
11
Absolute minimum: 16
51. f x, y
fx
fy
0½
¾y
0¿
1, 2 d x d 2,
f
y
x
Along y
Along y
d 2, y d 1`
x
x2 2x2 x2
x 2 2 x 1, f c
1
4
2
f x, x
f
^ x, y : x
x 2 2 xy y 2 , R
2x 2 y
2x 2 y
1
−1
0
2x 2
1
−2
0 Ÿ x
1, f 2, 1
1, f 1, 1
0, f 2, 1
0 Ÿ x
1, f 2, 1
9, f 1, 1
0, f 2, 1
9.
1, 2 d x d 2,
x 2 2 x 1, f c
2x 2
Along x
2, 1 d y d 1, f
Along x
2, 1 d y d 1, f
So, the maxima are f 2, 1
4 4 y y2 , f c
2 y 4 z 0.
4 4 y y2, f c
9 and f 2, 1
1.
2 y 4 z 0.
9, and the minima are f x, x
0, 1 d x d 1.
© 2010 Brooks/Cole, Cengage Learning
Section 13.8
fx
fy
^ x, y : x 2 x 2 2 xy y 2 , R
52. f x, y
2x 2 y
2x 2 y
f x, x
0½
¾y
0¿
Extrema of Functions of Two Variables
245
y 2 d 8`
x
x2 2x2 x2
On the boundary x 2 y 2
0
8 x 2 and y
8, we have y 2
r
8 x 2 . Thus,
y
f
fc
x2 r 2 x
8 x2 8 x2
r ª 8 x2
«¬
Then, f c
1 2
2 x 2 2 8 x 2
f 2, 2
fx
fy
º
»¼
r
4 xy
,R
x 1 y2 1
4 1 x2 y
y2 1 x2 1
^ x, y :0
0 Ÿ x
2
16 4 x
8 x2
f 2, 2
16 and f 2, 2
2
4
2
2
.
−4
x
−2
2
4
−2
16 and f 2, 2
So, the maxima are f 2, 2
53. f x, y
12
8 x2
r 2.
4x 2 or x
0 implies 16
f 2, 2
8 r 2x
−4
0
16, and the minima are f x, x
0, x d 2.
d x d 1, 0 d y d 1`
1 or y
y
0
1
4 1 y2 x
x2 1 y 2 1
2
Ÿ x
0 or y
1
R
x
For x
0, y
0, also, and f 0, 0
For x
1, y
1, f 1, 1
0.
1
1.
The absolute maximum is 1
f 1, 1 .
The absolute minimum is 0
f 0, 0 . In fact, f 0, y
fx
^ x, y : x
4 xy
,R
x 1 y2 1
54. f x, y
2
4 1 x2 y
y 1 x 1
2
2
2
0 Ÿ x
f x, 0
0.
t 0, y t 0, x 2 y 2 d 1`
1 or y
y
0
1
fy
4 1 y2 x
x2 1 y2 1
2
0 Ÿ y
1 or x
0
R
x
0, also, and f 0, 0
For x
0, y
For x
1 and y
For x 2 y 2
1
0.
1, the point 1, 1 is outside R.
1, f x, y
Absolute maximum is
8
9
The absolute minimum is 0
f x,
1 x2
4x 1 x2
, and the maximum occurs at x
2 x2 x4
2
,y
2
2
.
2
§ 2
2·
f ¨¨
,
¸¸.
2
2
©
¹
f 0, 0 . In fact, f 0, y
f x, 0
0
© 2010 Brooks/Cole, Cengage Learning
Chapter 13
246
Functions of Several Variables
55. In this case, the point A will be a saddle point. The
xy.
function could be f x, y
x 2 y 2 , g x, y
60. f x, y
(a) f x
56. A and B are relative extrema. C and D are saddle points.
2x
0, f y
2 y
2x
0, g y
2y
gx
0 Ÿ 0, 0 is a critical
point.
75
60
(b) f xx
45
g xx
2
x
2
d
y
No extrema
2, f xy
2, f yy
0
2 2 0 0 Ÿ 0, 0 is a saddle point.
d
30
2, g yy
2, g xy
0
2 2 0 ! 0 Ÿ 0, 0 is a relative minimum.
61. False.
z
58.
0 Ÿ 0, 0 is a critical
point.
z
57.
x2 y 2
1 x y.
Let f x, y
4
0, 0, 1 is a relative maximum, but f x 0, 0 and
3
f y 0, 0 do not exist.
2
3
x2 y 2.
62. False. Consider f x, y
y
4
4
x
Then f x 0, 0
Extrema at all x, y
point.
63. False. Let f x, y
z
59.
f y 0, 0
7
6
6
x 2 y 2 (See Example 4 on page 958).
64. False.
x4 2x2 y2.
Let f x, y
x
0, but 0, 0, 0 is a saddle
3
Relative minima: r1, 0, 1
y
Saddle point: 0, 0, 0
−3
Saddle point
Section 13.9 Applications of Extrema of Functions of Two Variables
2. A point on the plane is given by
x, y , z
x, y, 3 x y . The square of
1. A point on the plane is given by
x, y , z
x, y, 3 x y . The square
the distance from 1, 2, 3 to this point is
of the distance from 0, 0, 0 to this point is
S
2
x2 y 2 3 x y .
Sx
2x 2 3 x y
Sy
2y 2 3 x y
From the equations S x
4x 2 y
2 x 4 y
S
0 and S y
0 we obtain
6
2
y 2
2
3 x y 3
x 1
2
y 2
2
y x .
Sx
2 x 1 2 y x
Sy
2 y 2 2 y x
0 and S y
From the equation S x
6.
Solving simultaneously, we have x
So, the distance is
x 1
12 1
2
12
1, z
1, y
3.
1.
4x 2 y
2 x 4 y
2
2
0 we obtain
2
4.
Solving simultaneously, we have
x
4 3, y
5 3, z 10 3.
So, the distance is
2
2
§4
·
§5
·
§5 4·
¨ 1¸ ¨ 2 ¸ ¨ ¸
3
3
©
¹
©
¹
©3 3¹
2
13
.
3
© 2010 Brooks/Cole, Cengage Learning
Section 13.9
3. A point on the surface is given by x, y, z
Applications of Extrema of Functions of Two Variables
247
1 2 x 2 y . The square of the distance
x, y ,
from 2, 2, 0 to a point on the surface is given by
2
S
x 2
Sx
2 x 2 2
Sy
2 y 2 2
2
y 2
From the equations S x
0 and S y
1 2
So, the distance is
1 2x 2 y 0
2
2
0, we obtain
1 2
2
4. A point on the surface is given by x, y , z
5
x 2
2x 2
2y 2
2
x, y ,
2
2
y 2
0½
¾ Ÿ x
0¿
1 2 x 2 y.
1, z
y
5.
7.
1 2 x 2 y . The square
of the distance from 0, 0, 2 to a point on the surface is given by
2
x2 y2 2 S
Sx
2x Sy
2y 22
1 2x 2 y .
1 2x 2 y
1 2x 2 y
22
1 2x 2 y
1 2x 2 y
0 and S y
From the equation S x
2x 4
2
1 2x 2 y
2y 4
2
1 2x 2 y
0, we obtain x
y.
4
2
0.
1 4x
Using a graphing utility, x
y | 0.3221.
So, 2 x ªx2 y 2 2 «¬
S
The minimum distance is
5. Let x, y, and z be the numbers. Because xyz
Z
27
.
xy
S
x y z
1
Sx
x2 y
xy
2
So, x
| 0.667.
7. Let x, y, and z be the numbers and let
27,
x 2 y 2 z 2 . Because
27
x2 y
1
0, S y
x y z
27
.
xy
x y 27
xy 2
0.
2
Sx
2 x 2 30 x y 1
Sy
2 y 2 30 x y 1
y
32, z
10, and z
32 y 2 2 xy 2 y 3
Py
64 xy 2 x 2 y 3 xy 2
y 2 32 2 x y
Ignoring the solution y
xyz
S
0
y 64 x 2 x 2 3 xy
0.
Sx
1, z
64 x 2 x 2 3x 32 2 x
0
4x x 8
0.
16, and z
8.
30
30.
10,
10.
2x 2
x3 y 2
y x2 y3
y
z
x2 y2 0, S y
1½°
4 2
¾x y
1°¿
x x3 y 2
So, x
1 xy.
x2 y 2 z 2
0 and substituting
0, we have
0 ½° 2 x y
¾
0°¿ x 2 y
8. Let x, y, and z be the numbers. Because
32 x y. So,
32 xy 2 x 2 y 2 xy 3
Px
8, y
2
Solving simultaneously yields x
xy 2 z
32 2 x into Py
30, we have
x y 30 x y
2
S
27½°
y
3
¾x
27 °¿
y
z
3.
P
So, x
12
S
6. Because x y z
y
2
1 2 x 2 y º»
¼
2y x2 y 4 Ÿ x
1
x2 y2
2
x2 y3
0
y
1.
© 2010 Brooks/Cole, Cengage Learning
248
Chapter 13
Functions of Several Variables
C
C
Cx
Cy
668.25
.
xy
xyz Ÿ z
9. The volume is 668.25
§ 668.25 668.25 ·
0.12¨
¸ 0.11 xy
y ¹
© x
0.06 2 yz 2 xz 0.11 xy
80.19 80.19
0.11 xy
x
y
80.19
0.11 y
0
x2
8.19
0.11x
0
y2
Solving simultaneously, x
y
z
y
x
9 and z
8.25.
80.19 80.19
Minimum cost:
0.11 xy
9
9
$26.73
10. Let x, y, and z be the length, width, and height, respectively. Then C0
C0 1.5 xy
.
2x y
1.5 xy 2 yz 2 xz and z
The volume is given by
C0 xy 1.5 x 2 y 2
V
xyz
2x y
Vx
Vy
y 2 2C0 3x 2 6 xy
4x y
2
x 2 2C0 3 y 2 6 xy
4x y
2
In solving the system Vx
Substituting y
x 2C0 9 x
2
.
0 and Vy
x into Vx
0, we note by the symmetry of the equations that y
0 yields
2
0, 2C0
16 x 2
11. Let a b c
x.
1
3
9 x2 , x
4S abc
3
k . Then V
2C0 , y
1
3
2C0 , and z
4
S ab k a b
3
1
4
2C0 .
4
S kab a 2b ab 2
3
4S
½
0° kb 2ab b 2
kb 2ab b 2
0
°
3
¾
4S
0°ka a 2 2ab
Vb
ka a 2 2ab
0
°¿
3
Solving this system simultaneously yields a
b and substitution yields b
Va
12. Consider the sphere given by x 2 y 2 z 2
k 3. So, the solution is a
r 2 and let a vertex of the rectangular box be x, y,
b
c
k 3.
r 2 x2 y2 .
Then the volume is given by
V
2 x 2 y 2 r 2 x2 y 2
8 xy
r 2 x2 y2
§
·
x
8¨ xy
y r 2 x2 y2 ¸
¨
¸
r 2 x2 y 2
©
¹
§
·
y
x r 2 x2 y 2 ¸
Vy
8¨ xy
2
2
2
¨
¸
r x y
©
¹
Solving the system
Vx
2 x2 y 2
r2
x2 2 y 2
r2
yields the solution x
y
z
r
8y
r x2 y 2
2
8x
r x2 y2
2
r 2 2 x2 y 2
0
r 2 x2 2 y 2
0.
3.
© 2010 Brooks/Cole, Cengage Learning
Section 13.9
Applications of Extrema of Functions of Two Variables
249
13. Let x, y, and z be the length, width, and height, respectively and let V0 be the given volume.
Then V0
xyz and z
V0 xy. The surface area is
§
V
V ·
2¨ xy 0 0 ¸
x
y¹
©
2 xy 2 yz 2 xz
S
Sx
V ·
§
2¨ y 02 ¸
x ¹
©
Sy
§
V ·
2¨ x 02 ¸
y ¹
©
½
0° x 2 y V0
°
¾
0° xy 2 V0
°
¿
0
0.
3
Solving simultaneously yields x
14.
3
V0 , y
V0 , and z
1
ª 30 2 x 30 2 x 2 x cos T º¼ x sin T
2¬
A
wA
wx
wA
wT
30 sin T 4 x sin T 2 x sin T cos T
3
V0 .
30 x sin T 2 x 2 sin T x 2 sin T cos T
0
30 cos T 2 x 2 cos T x 2 2 cos 2 T 1
0
2 x 15
.
x
From
wA
wx
0 we have 15 2 x x cos T
From
wA
wT
§ § 2 x 15 · 2
·
§ 2 x 15 ·
2 § 2 x 15 ·
2
0 we obtain 30 x¨
¸ 2x ¨
¸ x ¨¨ 2¨
¸ 1¸¸
x
x
x ¹
©
¹
©
¹
© ©
¹
0 Ÿ cos T
30 2 x 15 2 x 2 x 15 2 2 x 15
2
x2
0
3x 30 x
0
2
x
1
ŸT
2
Then cos T
10.
60q.
5 x12 8 x2 2 2 x1 x2 42 x1 102 x2
15. R x1 , x2
Rx1
10 x1 2 x2 42
Rx2
16 x2 2 x1 102
0, 5 x1 x2
0, x1 8 x2
Solving this system yields x1
Rx1x1
10
Rx1x2
2
Rx2 x2
16
3 and x2
Rx1x1 0 and Rx1x1 Rx2 x2 Rx1x2
2
So, revenue is maximized when x1
16. P x1 , x2
0
21
51
6.
! 0
3 and x2
6.
15 x1 x2 C1 C2
15 x1 15 x2 0.02 x12 4 x1 500 0.05 x2 2 4 x2 275
Px1
0.04 x1 11
0, x1
275
Px 2
0.10 x2 11
0, x2
110
Px1x1
0.04
Px1x 2
0
Px 2 x 2
0.10
Px1x1 0 and Px1x1Px 2 x 2 Px1x 2
So, profit is maximized when x1
2
0.02 x12 0.05 x2 2 11x1 11x2 775
! 0
275 and x2
110.
© 2010 Brooks/Cole, Cengage Learning
250
Chapter 13
Functions of Several Variables
2 pq 2 pr 2qr.
17. P p, q, r
p q r
1 p q.
1 implies that r
2 pq 2 p 1 p q 2q 1 p q
P p, q
2 pq 2 p 2 p 2 2 pq 2q 2 pq 2q 2
wP
wp
2q 2 4 p;
Solving
wP
wp
and so p
2 p 2 4 q
0 gives q 2 p
1
p 2q
1
1
§1 1·
and P¨ , ¸
3
© 3 3¹
q
§1·
§1·
§1·
§1·
§1·
2¨ ¸ 2¨ ¸ 2¨ ¸ 2¨ ¸ 2¨ ¸
©9¹
© 3¹
© 3¹
©9¹
©9¹
x ln x y ln y z ln z , x y z
H
18. (a)
wP
wq
wP
wq
Hx
1 ln x 1 ln 1 x y
0
Hy
1 ln y 1 ln 1 x y
0
ln 1 x y
ln x
So, ln 1 2 x
ln x Ÿ 1 2 x
13 ln
(b) H
1
3
1
3
ln
ln y Ÿ x
1
3
3k
Cx
§
3k ¨
©
Cy
§
2k ¨
¨¨
©
§
3k ¨
©
x 2 4 2k
yx
§
·
2k ¨
¸
¨¨
x2 4 ¹
©
yx
2
·
§ 1·
¸ 2k ¨ ¸
© 2¹
x 4¹
x
x
x 4
2
9x
x2
x
4 y x
y x
2
y
So, x
y
1
3
1.
3
z
ln 3
y x
2
1. The distance from R to S is10 y.
1 k 10 y
2
·
¸
¸
1 ¸¹
0
yx
yx
2
1
1
2
0
1
3
x2 4
x 4
1
2
2
2
2
2 y x
2
ln
1
3
0Ÿ
3x
2
2
yx
·
¸k
¸
1 ¸¹
2
ln
2
.
3
y.
x Ÿ x
yx
x
yx
1
3
6
9
x ln x y ln y 1 x y ln 1 x y
1
x 2 4. The distance from Q to R is
19. The distance from P to Q is
C
2 pq 2 p 2q 2 p 2 2q 2
y x
y x
2
2
1
1
3
1
1
3
2
2
| 0.707 km and y
2
1
2 3 3 2
6
2 3 3 2
| 1.284 km.
6
© 2010 Brooks/Cole, Cengage Learning
Section 13.9
20. S
d1 d 2 d3
00
2
y 2 4 y 2
dS
dy
1
2 y 2
4 y 2
The sum of the distance is minimized when y
21. (a) S x, y
02
2
y 2
2
0 2
2
y 0
x2 y 2 2 3
3
62 3
.
3
23
3
| 0.845.
3
2
x 2
x 2
2
y 2
y 2
2
2
2
x4
x4
2
y 2
2
2
From the graph we see that the surface has a minimum.
x
x 2
x 4
(b) S x x, y
2
2
2
x2 y 2
x 2 y 2
x4 y 2
y
S y x, y
x2 y 2
(c) ’S 1, 1
(d)
x2 , y2
2
y 2
d1 d 2 d3
x 0
tan T
2
251
2
2
0 when y
2
2
y 0
Applications of Extrema of Functions of Two Variables
y 2
x 2
2
S x 1, 1 i S y 1, 1 j
2
10 1
1
2
2
y 2
2
y 2
2
S
2
24
20
y 2
x4
1
§ 1
i ¨
2
© 2
2
y 2
2
4
2 ·
¸j
10 ¹
8
6
4
2
2
4
6
8
y
x
2
Ÿ T | 186.027q
5
1
x1 S x x1 , y1 t , y1 S y x1, y1 t
§
1
1 ··
§ 2
t, 1 ¨
¨1 ¸t ¸
2
10
2¹ ¹
©
©
§ 2 10
·
§
2 5
2·
2 2 ¸¸t ¨¨1 ¸¸t 2
2 ¨¨
5
5
5
©
¹
©
¹
§
1
1 ··
§ 2
S ¨1 t, 1 ¨
¸t ¸
2
2¹ ¹
© 10
©
§ 2 10
·
§
2 5
2·
2 2 ¸¸t ¨¨1 ¸¸t 2
10 ¨¨
5
5¹
© 5
¹
©
§ 2 10
·
§
2 5
2·
4 2 ¸¸t ¨¨1 ¸¸t 2
10 ¨¨
5
5¹
© 5
¹
©
Using a computer algebra system, we find that the minimum occurs when t | 1.344. So x2 , y2 | 0.05, 0.90 .
(e)
x3 , y3
x2 S x x2 , y2 t , y2 S y x2 , y2 t | 0.05 0.03t , 0.90 0.26t
S 0.05 0.03t , 0.90 0.26t
0.05 0.03t
2
0.90 0.26t
2
2
2.05 0.03t
3.95 0.03t
2
1.10 0.26t
2
1.10 0.26t
2
Using a computer algebra system, we find that the minimum occurs when t | 1.78. So x3 , y3 | 0.10, 0.44 .
x4 , y4
x3 S x x3 , y3 t , y3 S y x3 , y3 t | 0.10 0.09t , 0.44 0.01t
S 0.10 0.09t , 0.45 0.01t
0.10 0.09t
2
0.45 0.01t
3.90 0.09t
2
2
1.55 0.01t
2.10 0.09t
2
1.55 0.01t
2
2
Using a computer algebra system, we find that the minimum occurs when t | 0.44. So, x4 , y4 | 0.06, 0.44 .
Note: The minimum occurs at x, y
0.0555, 0.3992
(f ) ’S x, y points in the direction that S decreases most rapidly. You would use ’S x, y for maximization problems.
© 2010 Brooks/Cole, Cengage Learning
252
Chapter 13
Functions of Several Variables
x 4
22. (a) S
2
y2 x 1
2
y 6
2
x 12
2
y 2
2
z
30
The surface appears to have a minimum near x, y
x 4
(b) S x
2
x 4
y
2
y
Sy
2
x 4
y
2
x 1
2
x 1
y 6
2
x 1
y 6
x 12
2
y 6
x 12
y 2
2
x
y 2
2
2
x 12
2
y 2
4
2
−2
−4
2
4
6
2
8
y
0.258 i 0.03 j. Direction | 6.6q
1, 5 . Then ’S 1, 5
(c) Let x1 , y1
1, 5 .
(d) t | 0.94, x2 | 1.24, y2 | 5.03
(e) t | 3.56, x3 | 1.24, y3 | 5.06, t | 1.04, x4 | 1.23, y4 | 5.06
Note: Minimum occurs at x, y
1.2335, 5.0694
(f ) ’S x, y points in the direction that S decreases most rapidly.
23. Write the equation to be maximized or minimized as a function of two variables. Set the partial derivatives equal to zero (or
undefined) to obtain the critical points. Use the Second Partials Test to test for relative extrema using the critical points. Check
the boundary points, too.
24. See pages 964 and 965.
25. (a)
x
y
xy
x2
2
0
0
4
0
1
0
0
2
3
6
4
¦ xi
a
¦ yi
0
36 04
3 8 02
¦ xi yi
4
1ª
3 º
4 0»
3 «¬
4 ¼
3
,b
4
2
(b) S
26. (a)
8
4
,y
3
3
4
x
4
3
2
1
6
x
y
xy
x2
3
0
0
9
1
1
1
1
1
1
1
1
3
2
6
9
¦ yi
0
46 04
4 20 0
2
(b) S
2
¦ xi 2
§ 3 4
·
§4
·
§3 4
·
¨ 0 ¸ ¨ 1¸ ¨ 3¸
2
3
3
2
3
©
¹
©
¹
©
¹
¦ xi
a
6
2
¦ xi yi
4
3
,b
10
2
6
1ª
3 º
4
0
4 «¬
10 »¼
2
¦ xi 2
20
1, y
§1
·
§7
·
§ 13
·
§ 19
·
¨ 0 ¸ ¨ 1¸ ¨ 1¸ ¨ 2¸
© 10
¹
© 10
¹
© 10
¹
© 10
¹
2
3
x 1
10
1
5
© 2010 Brooks/Cole, Cengage Learning
Section 13.9
Applications of Extrema of Functions of Two Variables
253
27. (a)
x
y
xy
x2
0
4
0
0
1
3
3
1
1
1
1
1
2
0
0
4
¦ xi
¦ yi
4
44 48
4 6 42
a
44
(b) S
2
¦ xi yi
8
1
ª8 2 4 º¼
4¬
2, b
23
2
2 1
2
¦ xi 2
4
2 x 4
4, y
00
2
6
2
28. (a)
x
y
xy
x2
3
0
0
9
1
0
0
1
2
0
0
4
3
1
3
9
4
1
4
16
4
2
8
16
5
2
10
25
6
2
12
¦ xi
¦ yi
28
8 37 28 8
a
8 116 28
2
8
¦ xi yi
72
144
1
,b
2
2
2
37
¦ xi
1ª
1
º
8 28 »
8 «¬
2
¼
2
2
116
3
,y
4
1
3
x
2
4
2
2
2
§3
·
§ 1
·
§1
·
§3
·
§5
·
§5
·
§7
·
§9
·
¨ 0 ¸ ¨ 0 ¸ ¨ 0 ¸ ¨ 1¸ ¨ 1¸ ¨ 2 ¸ ¨ 2 ¸ ¨ 2 ¸
©4
¹
© 4
¹
©4
¹
©4
¹
©4
¹
©4
¹
©4
¹
©4
¹
(b) S
29. 0, 0 , 1, 1 , 3, 4 , 4, 2 , 5, 5
¦ xi
¦ xi yi
a
36
2
13,
46,
¦ yi
¦ xi 2
5 46 13 12
5 51 13
b
1ª
37
º
12 13 »
«
5¬
43
¼
y
37
7
x
43
43
¦ xi
¦ xi yi
51
7
43
37
43
9,
39,
¦ yi
¦ xi 2
3 39 9 9
a
3 35 9
2
1ª
3 º
9 9»
3 «¬
2 ¼
3
3
x
2
2
b
y
9,
35
36
24
3
2
9
6
3
2
7
7
y = 37
x + 43
43
7
3
2
30. 1, 0 , 3, 3 , 5, 6
12,
74
86
2
2
(5, 5)
(3, 4)
(4, 2)
−2
(1, 1)
(0, 0)
−1
10
6
−1
−1
© 2010 Brooks/Cole, Cengage Learning
254
Chapter 13
Functions of Several Variables
31. 0, 6 , 4, 3 , 5, 0 , 8, 4 , 10, 5
¦ xi
¦ xi yi
¦ yi
¦ xi 2
27,
70,
increases by about 2.09 million.
205
1ª
º
§ 175 ·
0 ¨
¸ 27 »
5 «¬
© 148 ¹
¼
b
n
350
296
2
5 205 27
(b) For each one-point increase in the percent x, y
0,
5 70 27 0
a
¦ yi
37. S a, b, c
175
148
945
148
8
wS
wa
¦2 xi 2
wS
wb
¦2 xi
wS
wc
2¦ yi axi 2 bxi c
n
n
18
−6
(10, − 5)
yi axi 2 bxi c
n
n
n
i 1
i 1
i 1
n
n
n
i 1
i 1
i 1
n
¦ xi
¦ xi yi
¦ yi
¦ xi 2
42
275
6 275 42 31
a
a ¦ xi b¦ xi cn
1§
29 ·
42 ¸
¨ 31 6©
53 ¹
b
i 1
or x
9
| 1.3365
29
425
x
53
318
y
33. (a) y
(b)
14
−1
¦ yi
i 1
108 2 y 2 z. The volume is given by
108 zy 2 zy 2 2 yz 2
xyz
Vy
108 z 4 yz 2 z 2
z 108 4 y 2 z
0
Vz
108 y 2 y 4 yz
y 108 2 y 4 z
0.
2
and 2 y 4 z
108
108, we obtain the solution
36 inches, y
18 inches, and z
18 inches.
39. 2, 0 , 1, 0 , 0, 1 , 1, 2 , 2, 5
80
50
(c) For each one-year increase in age, the pressure
changes by approximately1.6, the slope of
the line.
34. (a) Using a graphing utility, y
(b) When x
1.59, y
300 x 832.
300 1.59 832
355.
35. 1.0, 32 , 1.5, 41 , 2.0, 48 , 2.5, 53
7, ¦ yi
14, b
When x
i 1
V
x
1.6 x 84
0
a
i 1
Solving the system 4 y 2 z
−1
250
¦ xi
¦ xi yi
respectively. Then the sum of the length and girth is
108
given by x 2 y 2 z
400
425
318
i 1
n
38. Let x, y, and z be the length, width, and height,
31
29
| 0.5472
53
2
6 400 42
n
¦ xi 2 yi
n
2
6
0
i 1
n
32. 6, 4 , 1, 2 , 3, 3 , 8, 6 , 11, 8 , 13, 8 ; n
0
n
a ¦ xi 3 b ¦ xi 2 c ¦ xi
y = − 175
x + 945
148
148
0
i 1
a ¦ xi 4 b¦ xi 3 c ¦ xi 2
(4, 3)
(5, 0)
(8, − 4)
yi axi 2 bxi c
i 1
(0, 6)
−4
2
axi 2 bxi c
i 1
175
945
x
148
148
y
2.09 x 60.2
36. (a) y
19, y
1.6, y
174, ¦ xi yi
322, ¦ xi
¦ xi
¦ yi
¦ xi 2
¦ xi3
¦ xi 4
¦ xi yi
¦ xi 2 yi
8
8
3
,
7
(2, 5)
(−1, 0)
10
(1, 2)
−9
0
(0, 1)
(−2, 0)
6
−2
34
12
22
34a 10c
a
2
0
b
12, 10a 5c
22, 10b
6
,
5
c
26
,
35
y
3 2
x
7
65 x 8
26
35
13.5
14 x 19
41.4 bushels per acre.
© 2010 Brooks/Cole, Cengage Learning
Section 13.9
Applications of Extrema of Functions of Two Variables
40. 4, 5 , 2, 6 , 2, 6 4, 2
¦ xi
¦ yi
¦ xi 2
¦ xi3
¦ xi 4
¦ xi yi
¦ xi 2 yi
0, 0 , 2, 15 , 4, 30 ,
6, 50 , 8, 65 , 10, 70
0
8
(−2, 6)
(− 4, 5)
19
¦ xi
¦ yi
¦ xi 2
¦ xi3
¦ xi 4
¦ xi yi
¦ xi 2 yi
(2, 6)
40
(4, 2)
−9
9
0
−4
544
12
160
544a 40c
a
43. (a)
12, 40a 4c
160, 40b
245 , b
103 , c
41
,
6
y
245 x 2 19
3
x
10
230
220
1800
15,664
1670
13,500
1800a 220b 30c
220a 30b 6c
25 2
112
x
y
9
(b)
14
20
541
x
56
25
14
13,500
1670
230
| 0.22 x 2 9.66 x 1.79
120
(4, 12)
29
(3, 6)
99
−5
353
(2, 2)
14
−20
70
44. (a) y
254
(b) y
254
99a 29b 9c
70
29a 9b 4c
20
1, c
1, b
−1
7
(0, 0)
−2
353a 99b 29c
a
30
15,664a 1800b 220c
41
6
41. 0, 0 , 2, 2 , 3, 6 , 4, 12
¦ xi
¦ yi
¦ xi 2
¦ xi3
¦ xi 4
¦ xi yi
¦ xi 2 yi
255
(c)
0.075 x 5.3
0.002 x 2 0.12 x 5.1
7
x2 x
0, y
5
18
5.5
42. 0, 10 , 1, 9 , 2, 6 , 3, 0
¦ xi
¦ yi
¦ xi 2
¦ xi3
¦ xi 4
¦ xi yi
¦ xi 2 yi
11
(0, 10)
25
(2, 6)
(3, 0)
36
−9
−1
y
0.075 24 5.3
y
0.002 24
7.1 (billion)
2
0.12 24 5.1 | 6.8 (billion).
The quadratic model is less accurate because of the
negative x 2 coefficient.
21
33
33
36a 14b 6c
21
14a 6b 4c
25
54 ,
9
98
24 and
(1, 9)
14
98a 36b 14c
a
(d) For 2014, x
6
b
9
,
20
c
199
,
20
y
54 x 2 9
x
20
199
20
© 2010 Brooks/Cole, Cengage Learning
256
Chapter 13
Functions of Several Variables
45. (a) ln P
0.1499h 9.3018
(b) ln P
0.1499h 9.3018
e 0.1499 h 9.3018
P
(c)
46. (a)
10,957.7e 0.1499 h
14,000
1
y
ax b
y
1
0.0074 x 0.445
(b)
−2
0.0074 x 0.445
40
24
0
−2,000
60
0
(d) Same answers
(c) No. For x
1000 which seems inaccurate.
y
n
¦ axi
47. S a, b
70, y | 14, which is nonsense.
2
b yi
i 1
n
n
i 1
i 1
n
2a ¦ xi 2 2b¦ xi 2¦ xi yi
S a a, b
i 1
n
n
i 1
i 1
2a ¦ xi 2nb 2¦ yi
Sb a , b
n
S aa a, b
2¦ xi 2
Sbb a, b
2n
S ab a, b
2¦ xi
i 1
n
i 1
S aa a, b ! 0 as long as xi z 0 for all i. (Note: If xi
n
§ n ·
4n ¦ xi 2 4¨ ¦ xi ¸
i 1
©i 1 ¹
S aa Sbb Sab 2
d
2
0 for all i, then x
0 is the least squares regression line.)
2
2
n
ª n
§ n ·
§ n · º
4 «n¦ xi 2 ¨ ¦ xi ¸ » t 0 since n¦ xi 2 t ¨ ¦ xi ¸ .
«¬ i 1
i 1
©i 1 ¹
© i 1 ¹ »¼
As long as d z 0, the given values for a and b yield a minimum.
Section 13.10 Lagrange Multipliers
1. Maximize f x, y
2. Maximize f x, y
xy.
Constraint: x y
Constraint: 2 x y
10
O ’g
’f
yi x j
O½
y
y
¾x
O¿
x
x y 10 Ÿ x
f 5, 5
y
5
y
2O
x
O
2x y
2O i O j
4 Ÿ 4O
O
25
f 1, 2
y
4
O ’g
’f
Oi j
yi x j
xy.
4
1, x
1, y
2
2
12
10
Level curves
8
6
4
2
Constraint
x
2
4
6
8
10
12
© 2010 Brooks/Cole, Cengage Learning
Section 13.10
x2 y2.
3. Minimize f x, y
Constraint: x y
Constraint: 2 y x 2
4
y
¾x
2y
O¿
x y
4 Ÿ x
f 2, 2
2 xO i 2O j
2 xi 2 y j
O½
2x
y
2
8
0
O ’g
’f
Oi O j
2 xi 2 y j
257
x2 y 2.
6. Maximize f x, y
O’g
’f
Lagrange Multipliers
2x
2 xO Ÿ x
If x
0, then y
If O
1,
2O
2 y
0 or O
1
0 and f 0, 0
2 Ÿ y
0.
1 Ÿ x2
2 Ÿ x
2.
y
f
Constraint
4
2 1
2, 1
1, Maximum
2 x 2 xy y.
7. Maximize f x, y
x
−4
Constraint: 2 x y
4
O ’g
’f
Level curves
−4
Constraint: 2 x 4 y
5
O’g
’f
2O i 4O j
2 xi 2 y j
2O Ÿ x
4O Ÿ y
5 Ÿ 10O
2x
2y
2x 4 y
O
f
2x 1
O Ÿ x
2x y
100 Ÿ 4 x
f 25, 50
5
1,
2
1,
2
y
2x
O½ x
°
Constraint: x 2 y
2y
x 2y 5
2O
f 1, 2
0
3
1
x2 y
O
2, x
1, y
50
3 x y 10.
6
2 xyO i x 2O j
2 xyO Ÿ O
x 2O Ÿ O
3 ½
2 xy °° 2
¾3 x
1
°
x 2 °¿
§ 3x ·
6 Ÿ x2 ¨ ¸
©2¹
6
x3
4
x
3
0
5 Ÿ O
5
25, y
O ’g
3i j
O 2
¾ y
2O °¿
100
1
O i 2j
2 xi 2 y j
2x
°
¿
2600
O’g
’f
O 1½
°
O 1 ¾y
2
8. Minimize f x, y
x
x2 y2.
Constraint: x 2 y 5
2
2O Ÿ y
’f
5. Minimize f x, y
O
2 2y
x
O
2O
5
4
1,1
2
2O i O j
2 2 y i 2x 1 j
x2 y2.
4. Minimize f x, y
100
4, y
2
§
33 4 ·
f ¨¨ 3 4,
¸
2 ¸¹
©
3x
2
x z 0
2 xy Ÿ y
33 4
2
9 3 4 20
2
© 2010 Brooks/Cole, Cengage Learning
258
Chapter 13
Functions of Several Variables
6 x 2 y 2 is maximum when
9. Note: f x, y
Constraint: x y z
g x, y is maximum.
Maximize g x, y
6 x2 y 2.
Constraint: x y
2
2 x
O½
2 y
O¿
¾x
2 Ÿ x
f 1, 1
g 1, 1
y
2
x2 y 2.
Minimize g x, y
Constraint: 2 x 4 y
2y
2O ½
¾y
4O ¿
2x 4 y
1
2
15 Ÿ 10 x
§3 ·
g ¨ , 3¸
©2 ¹
Constraint: x y z
2y
O¾x
2z
°
y
O °¿
x y z
f 3, 3, 3
9 Ÿ x
xz
x y z
10 O 1
2
14 O
10
y
4, y
6
16 40 36 84 28
44
x 2 3 xy y 2 .
Case 1: On the circle x 2 y 2
z
2x 3y
3x 2 y
2 xO ½ 2
¾x
2 yO ¿
x2 y 2
1Ÿ x
1
y2
r
2
,y
2
§
2
2·
,r
Maxima: f ¨¨ r
¸¸
2
2
©
¹
3
§
2
2·
Minima: f ¨¨ r
,
¸
2 ¸¹
© 2
3
y
2
Constraint: x 2 y 2 d 1
9
r
2
2
5
2
1
2
Case 2: Inside the circle
xy Ÿ x
3 Ÿ x
1
14 O
15. Maximize or minimize f x, y
xyz.
Constraint: x y z
f 1, 1, 1
3 5
2
27
O½
°
O ¾ yz
O °¿
10 O
1
2
1
2
10
f 4, 6
3
z
12. Maximize f x, y, z
yz
xz
xy
10
x
x2 y2 z 2.
11. Minimize f x, y , z
O½
x 2 10 x y 2 14 y 28.
O½ x
¾
O¿ y
15
3
,y
2
1
3
z
O
x
2x
y
1
3
Constraint: x y
x y
15
2x
§3 ·
f ¨ , 3¸
©2 ¹
z
14. Minimize f x, y
2 x 10
2 y 14
1
1Ÿ x
1, 1, 1
3 3 3
f
1
is minimum.
2x
y
x y z
x 2 y 2 is minimum when g x, y
10. Note: f x, y
O½
°
O¾x
O °¿
2x
2y
2z
y
x y
x2 y2 z 2.
13. Minimize f x, y, z
y
z
z
1
fx
fy
2x 3 y
3x 2 y
f xx
2, f yy
0½
x
0¾¿
2, f xy
Saddle point: f 0, 0
y
0
3, f xx f yy f xy
2
d 0
0
By combining these two cases, we have a maximum
5 §
2
2·
of at ¨¨ r
,r
¸ and a minimum of
2 © 2
2 ¸¹
2
2·
1 §
at ¨ r
,B
¸.
2 ¸¹
2 ¨© 2
© 2010 Brooks/Cole, Cengage Learning
Section 13.10
e xy 4 .
16. Maximize or minimize f x, y
Constraints: x 2 z
Case 1: On the circle x 2 y 2
y 4e
x 4 e xy
4
x2 y2
1Ÿ x
r
y2
2
2
e1 8 | 1.1331
§
2
2·
Minima: f ¨¨ r
,r
¸¸
2
2
©
¹
e 1 8 | 0.8825
0½°
¾ Ÿ x
0°¿
4
xy 4
fy
x 4e
f xx
y 2 xy 4
e
, f yy
16
y
At 0, 0 , f xx f yy f xy
2
Saddle point: f 0, 0
1
x y z
O P½
xz
O P ¾ yz
O P °¿
°
6 Ÿ z
6 x
2
x y
12 Ÿ y
12 x
x
2
3
x·
9
§
2 12 x ¨ 3 ¸ Ÿ x
2¹
2
©
6, z
27 Ÿ x
72
x 0
19. f x, y
2
y 0
O½
x
O 2½
2y
O¿
y
O 2¿
x y
1
x
y
x2 y2
1
2x
¾
2
¾ Ÿ x
y
1
2
§ 1·
§ 1·
¨© ¸¹ ¨© ¸¹
2
2
2
2
2
20. Minimize the square of the distance f x, y
32
subject to the constraint 2 x 3 y
0
2x
Oi jk Pi jk
2y
xy Ÿ x
32½
¾2x 2 z
0¿
z
6
0
2
2O ½
¾y
3O ¿
x2 y 2
1.
3x
2
1 Ÿ x
2
,y
13
3
13
3·
§ 2
The point on the line is ¨ , ¸ and the desired
© 13 13¹
32 Ÿ x
y
f 8, 16, 8
2y z
x 2z
2x 3y
yz
x y z
2O
Minimum distance:
O’g P’h
x y z
2z
°
¾2 x
°
¿
xyz.
Constraints: x y z
xy
P
Constraint: x y
0.
§
2
2·
of e 1 8 at ¨¨ r
,r
¸.
2 ¸¹
© 2
yzi xz j xyk
2y
f 6, 6, 0
1·
§1
e xy ¨ xy ¸
16
4¹
©
Combining the two cases, we have a maximum
§
2
2·
,B
of e1 8 at ¨¨ r
¸¸ and a minimum
2
2
©
¹
’f
O P½
x
0
x 2 xy 4
e
, f xy
16
17. Maximize f x, y, z
2x
2x
Case 2: Inside the circle
y 4 e xy
O i 2k P i j
2 xi 2 y j 2 z k
§
2
2·
Maxima: f ¨¨ r
,B
¸
2 ¸¹
© 2
fx
12
O’g P’h
’f
2 xO½°
2
¾ Ÿ x
2 yO °¿
6
x y
1
259
x2 y2 z 2.
18. Minimize f x, y, z
Constraint: x 2 y 2 d 1
xy 4
Lagrange Multipliers
z
8
2
distance is d
§ 2·
§ 3·
¨© ¸¹ ¨© ¸¹
13
13
2
13
.
13
16
1024
© 2010 Brooks/Cole, Cengage Learning
260
Chapter 13
Functions of Several Variables
x2 y 2
21. f x, y
Constraint: x y
2 y 2
x y
§4 O·
¨
¸
2 © 2 ¹
4
O
6
x 1
22. f x, y
32 1 2
2
4O
x 4y
2
3 2
y2
1, y
2
x 3
y2
O
3
0
O
x2
y
3
2x2
2y
2x 3
2 2 x3
O 2 16O
6
17 O
4 x3 2 x 6
4
2 x 1 2x 2x 3
O
x
19
,y
17
1
Minimum distance:
8
17
x 12
25. f x, y
2
Minimum distance:
0
2
4
17
§ 19·
§ 8·
¨© ¸¹ ¨© ¸¹
17
17
2
5 17
17
2
1
2
2 y 1
0 Ÿ x
1, y
12
2
2
y 1
Constraint: y x 2 1
2x 11
3
4
2O x
2y
2O
2
11
2
Minimum distance:
2x 3
2
4 2O
§ 1·
52¨ ¸
© 2¹
3.
r 52
5 2, x
Constraint: y x 2
O 2
9 Ÿ distance
0, and f 0, 0
0, y
24. f x, y
3
½
O°x
¾
°y
¿
2 x 1
x
O
2
f r 5 2, 5 2
Constraint: x 4 y
2
2y 3
If x z 0, O
Minimum distance:
O 2
4O
2
1
3, y
2y
2x
y
0
2 xO
O 2
If x
O
x
Constraint: y x 2
4
2
x2 y 3
23. f x, y
4
½ x
°
¾
O ° y
¿
O
2x
2
1,
2
0
2 xO
O
y
x 1
O
2y 2
2
2x 1
2
2 x2 1 2
2 x 2x2
4 x3 2 x 1
2x2
4 x3
0
x | 0.3855
y | 1.1486
Minimum distance:
f 0.3855, 1.1486 | 0.188
© 2010 Brooks/Cole, Cengage Learning
Section 13.10
2
x 2 y 10 subject to the constraint x 4
26. Minimize the square of the distance f x, y
2 x 4 O °½ x
¾
2 yO
°¿ x 4
2x
2 y 10
Lagrange Multipliers
y 10
Ÿ y
y
2
y2
261
4.
5
x 10
2
§ 25
·
4 Ÿ x 2 8 x 16 ¨ x 2 50 x 100 ¸
4
© 4
¹
29 2
x 58 x 112
0
4
Using a graphing utility, we obtain x | 3.2572 and x | 4.7428 or by the Quadratic Formula,
x 4
2
y2
58 r
x
582 4 29 4 112
58 r 2 29
29 2
2 29 4
4 29
.
29
§
29 ·
4 ¨1 ¸ and y
29
©
¹
Using the smaller value, we have x
ª §
The point on the circle is «4 ¨1 ¬« ©
4r
10 29
| 1.8570.
29
29 · 10 29 º
,
» and the desired distance is d
29 ¸¹
29 ¼»
2
§
16¨1 ©
§ 10 29
·
29 ·
¨
10¸
29 ¸¹
29
©
¹
2
| 8.77.
The larger x-value does not yield a minimum.
27. Minimize the square of the distance
x 2
f x, y , z
2
y 1
2
29. Maximize f x, y, z
z 1
subject to the constraint x y z
2x 2
O½
2 y 1
O¾ y
°
x y z
x2 y2 z 2
1.
1Ÿ x 2x 1
x
1, y
z
0
4 2z
1 2
is d
0 1
2
01
2
2
0
4 Ÿ x
0 z
2
0
3 z 16 z 16
0
3z 4 z 4
0
2
4
3
z
3.
4.
0
4 2z
2
The point on the plane is 1, 0, 0 and the desired distance
2
0 and x 2 z
x2 y 2 z 2
x 2z
1
z subject to the constraints
2 xO P
2 yO Ÿ y
2 zO 2 P
0
0
1
x 1
z and y
O°¿
2z 1
2
or z
4
The maximum value of f occurs when z
28. Minimize the square of the distance
x 4
f x, y , z
2
y2 z2
x
2 x 4
x2 y 2
y
2y
x2 y 2
O
O
2z
x2 y 2 z
0, x
0.
x ½
O
z °
°
°°2 x 4
¾
2y
°
°
°
°¿
y
O
z
O
30. Maximize f x, y, z
x2 y 2 z
subject to the constraint
2, y
0, z
x2 y2 z 2
0
0
2 x
2 y
2
The point on the plane is 2, 0, 2 and the desired distance
is d
2 4
2
02 2 2
2 2.
4 at the point
of 4, 0, 4 .
1
2x y z
2y
36 and 2 x y z
2 xO 2P ½
°
2 yO P ¾ x
2 zO P °¿
x2 y2 z 2
2
z subject to the constraints
2.
2y
36
2 Ÿ z
y2 5 y 2
2x y 2
2
36
30 y 20 y 32
0
15 y 2 10 y 16
0
2
5y 2
5r
265
15
Choosing the positive value for y we have the point
y
§ 10 2 265 5 265 1 265 ·
,
,
¨¨
¸¸.
15
15
3
©
¹
© 2010 Brooks/Cole, Cengage Learning
262
Chapter 13
Functions of Several Variables
31. Optimization problems that have restrictions or
constraints on the values that can be used to produce the
optimal solution are called contrained optimization
problems.
32. See explanation at the bottom of page 971.
x2 y2 z 2.
33. Minimize f x, y, z
2x
O Ÿ x
2y
O Ÿ y
2z
O Ÿ z
x y z
3
O 2
O xyz½
1
O xz Ÿ y
O xyz¾ Ÿ x
1
O xy Ÿ z
y2z
O
2 xyz
O
xy 2
O
3O
2
3
x y z
O
2
xy 2
1, z
1
1 1
Minimum distance
x 1
34. Minimize f x, y, z
2z 3
x y z
y 2
2
2 O
4 O
6 O
2
2
2
6
O
2
3
5
,z
3
x y z
3
xy Ÿ 2 x
2
y
32
x
8
y
16
z
8
0.06 2 yz 2 xz 0.11 xy .
Constraint: g x, y, z
yzO
0.12 z 0.11x
xzO
0.12 y x
xyO
2
3
3
§ 0.24·
xz ¨
© x ¸¹
§ 11 ·
x 2 ¨ x¸
© 12 ¹
33·
§
f ¨ 9, 9, ¸
©
4¹
xyzO
x 2O Ÿ O
0.12 z 0.11x
xyz
668.25
668.25
0.12 xz 0.11yx
10
3
xyz
0.12 z 0.11 y
xyz
3
2
y z 0
z
2
37. Minimize f x, y, z
§ 1·
§ 1·
§ 1·
¨© ¸¹ ¨© ¸¹ ¨© ¸¹
3
3
3
32
2
0.12 2x
2
Minimum distance
3
xy 2 z.
xy Ÿ 2 x y
2
z 3 .
3
3O 4
z
32
x 2x x
3
2 O
2
4 O
O Ÿ y
2
6 O
O Ÿ z
2
2y 2
4
,y
3
1
2
O Ÿ x
2x 1
x
2
2
x y z
Constraint: g x, y, z
y
y2z Ÿ x
2 xyz
2
z
27
Constraint: g x, y, z
3
y
O xyz °¿
27 Ÿ x
3
3
1, y
°
36. Maximize P x, y, z
O 2
27
O yz Ÿ x
O
§ O·
¨ ¸ 2 © 2¹
2
x
xyz
1
x
O 2
O
Constraint: g x, y, z
xyz
x y z
Constraint: g x, y, z
x y z.
35. Minimize f x, y, z
0.12 yz 0.11xy Ÿ x
y
0.24
x
0.24 z Ÿ z
668.25 Ÿ x
y
0.11x
0.12
9, z
11x
12
33
4
$26.73
© 2010 Brooks/Cole, Cengage Learning
Section 13.10
38. Maximize f x, y, z
xyz
yz
1.5 yO 2 zO
xz
1.5 xO 2 zO
xy
2 xO 2 yO
1.5 xy 2 xz 2 yz
xyz
x>1.5 yO 2 zO @
y>1.5 xO 2 zO @
2 yzO
x
x
xz
y
2 xO 2 xO Ÿ O
§ x·
§ x·
1.5 x¨ ¸ 2 z ¨ ¸ Ÿ z
© 4¹
© 4¹
2C
,z
3
3
x
4
z
2
C Ÿ x
9
C Ÿ x2
2C
,
3
y
x
2C
4
4S abc
.
3
39. Maximize f x, y, z
Constraint: a b c
4S bc
3
4S ac
3
4S ab
3
(also by symmetry)
x 4.
§3 ·
§3 ·
1.5 x 2 2 x¨ x ¸ 2 x¨ x ¸
4
© ¹
©4 ¹
y
C
C
2 xzO
2
½
° 4S bc
¾
3
O°
°¿
O°
p q r
Constraint: g p, q, r
K , constant
4S ac
Ÿa
3
2 pq 2 pr 2qr.
41. Maximize P p, q, r
bŸa
b
c
K
3
2q 2r
O½
2 p 2r
O¾ p
2 p 2q
°
q
O°¿
p q r
O
P 13 , 13 , 13
3
2
9
r2
and let a vertex of the rectangular box be x, y, z .
Maximize V x, y, z
2x 2 y 2z
Constraint: g x, y, z
x2 y 2 z 2
8 yz
2 xO Ÿ 4 xyz
O x2 ½
8 xz
2 yO Ÿ 4 xyz
8 xy
2 zO Ÿ 4 xzy
O y ¾x
O z 2 °¿
x2 y2 z 2
2°
(a) ln x 1
ln y 1
8 xyz.
ln z 1
r2
x y z
(b) H 13 , 13 , 13
y
z
and
x ln x y ln y y ln z.
x y z
Constraint: g x, y, z
40. Consider the sphere given by x 2 y 2 z 2
1
3
2.
3
42. Maximize H x, y, z
a2
1
r
1Ÿ p
3p
So, ellipse is a sphere:
x2 y2 z 2
263
(volume).
1.5 xy 2 xz 2 yz
Constraint: g x, y, z
Lagrange Multipliers
1
O½
°
O¾x
y
O °¿
3x
z
1Ÿ x
3ª¬ 13 ln
1
3
º
¼
y
z
1
3
ln 3
r2
So, the rectangular box is a cube.
© 2010 Brooks/Cole, Cengage Learning
264
Chapter 13
Functions of Several Variables
43. Maximize V x, y, z
2x 2 y 2z
2x ½
O
a2 °
°
2 y ° x2
O
¾
b2 ° a 2
2z °
O°
c2 ¿
8 yz
8 xz
8 xy
y2
b2
8 xyz subject to the constraint
x2
y2
z2
2 2
2
a
b
c
1.
z2
c2
3x 2
3y2
x2
y2
z2
1Ÿ 2
1, 2
2 2
2
a
b
c
a
b
a
b
c
,y
,z
x
3
3
3
So, the dimensions of the box are
1,
3z 2
c2
1
2 3a 2 3b 2 3c
u
u
.
3
3
3
44. (a) Yes. Lagrange multipliers can be used.
xyz subject to the constraint x 2 y 2 z
(b) Maximize V x, y, z
O ½
°
2O ¾ y
yz
xz
z and x
2O °¿
xy
108.
2y
z
x 2 y 2z
108 Ÿ 6 y
x
108, y
36, y
18
z
x
y
18
Volume is maximum when the dimensions are 36 u 18 u 18 inches.
5 xy 3 2 xz 2 yz xy subject to the constraint xyz
45. Minimize C x, y, z
yzO ½
°
xzO ¾ x
xyO °¿
8 y 6z
8x 6 z
6x 6 y
480 Ÿ
xyz
4
3
y, 4 y
y3
3
360 u
3
O½
O¾x
°
O°¿
xy
x y z
y
360, z
360 u
4 3
3
4 3
3
360
360 feet.
z
x
y
xyz subject to the
constraint x y z
yz
3
y
46. (a) Maximize P x, y, z
xz
3z
480
x
Dimensions:
480.
S.
z
S Ÿ x
y
z
S
3
§ S ·§ S ·§ S ·
xyz d ¨ ¸¨ ¸¨ ¸, x, y, z ! 0
© 3 ¹© 3 ¹© 3 ¹
So,
S3
27
S
xyz d
3
x y z
.
xyz d
3
xyz d
3
3
© 2010 Brooks/Cole, Cengage Learning
Section 13.10
x1 x2 x3"x n subject to the constraint
(b) Maximize P
Lagrange Multipliers
265
n
¦ xi
S.
i 1
x2 x3"xn
O
x1x3"xn
O
½
°
°
°
O ¾ x1
x1x2"xn
°
#
°
O°¿
x1x2 x3"xn 1
n
¦ xi
S Ÿ x1
x2
x2
"
x3
"
x3
xn
S
n
xn
i 1
So,
§ S· § S· § S· § S·
x1x2 x3"xn d ¨ ¸ ¨ ¸ ¨ ¸ "¨ ¸ , xi t 0
© n¹ © n¹ © n¹ © n¹
§ S·
x1x2 x3"xn d ¨ ¸
© n¹
n
n
x1 x2 x3"xn d
S
n
n
x1 x2 x3"xn d
x1 x2 x3 " xn
.
n
47. Minimize A S , r
2S h 4S r
2S r
S r 2h
2S rh 2S r 2 subject to the constraint S r 2 h
2S rhO ½
¾h
Dimensions: r
3
V0
V0
and h
2S
2x
0
23
V0
2S
100 x 2 y 2 subject to the constraints x 2 y 2 z 2
48. Maximize T x, y, z
2y
2r
S r 2O ¿
V0 Ÿ 2S r 3
V0 .
50 and x z
0.
2 xO P ½
°
¾
2 zO P °¿
2 yO
If y z 0, then O
1 and P
So, x
z
T 0,
50, 0
If y
0 then x 2 z 2
§ 50
T¨
, 0,
© 2
0 and y
0.
50.
100 50
50 ·
2 ¸¹
0, z
150
2x2
100 50
4
50 and x
z
50 2.
112.5
So, the maximum temperature is150.
© 2010 Brooks/Cole, Cengage Learning
266
Chapter 13
Functions of Several Variables
Distance
, minimize T x, y
Rate
49. Using the formula Time
O°
d12 x 2
°
¾
O ° v1
°
¿
y
v2
d22 y 2
x y
y
d22 y 2
v2
d12 x 2
d22 y 2
d12 x 2
v1
x
Medium 1
P
x
y
and sin T 2
sin T1
v1
x
d12 x 2
a
Because sin T1
we have
d1
θ1
,
y
x
d2
θ2
d2 y
or
v2
2
y
2
Q
a
Medium 2
sin T 2
.
v2
§ Sl ·
2h l ¨ ¸ subject to
© 2¹
50. Case 1: Minimize P l , h
§ Sl ·
the constraint lh ¨
© 8 ¸¹
51. Maximize P x, y
S
2
Sl ·
§
¨© h ¸¹ O
4
2
lO Ÿ O
l
2h
A.
2
S
,1 l
2
§ Sl 2 ·
lh ¨
subject to
© 8 ¸¹
Case 2: Minimize A l , h
§ Sl ·
the constraint 2h l ¨ ¸
© 2¹
h
Sl
4
2h S
l
2
l
2O Ÿ O
h
l
or l
2
250,000.
§ y·
72O Ÿ ¨ ¸
© x¹
0.75
25 x 0.75 y 0.75
72O
25
§ x·
60O Ÿ ¨ ¸
© y¹
0.25
75 x 0.25 y 0.25
60O
75
§ y·
¨© ¸¹
x
0.75
§ y·
¨© ¸¹
x
0.25
y
x
y
P.
§ 18 ·
72 x 60¨ x¸
©5 ¹
S·
§
¨ A ¸O
2¹
©
100 x 0.25 y 0.75 subject to the
constraint 72 x 60 y
2
1
a.
½
x
v1
d22 y 2
subject to the constraint x y
v2
d12 x 2
v1
§ 72O · § 75 ·
¨©
¸¨
¸
25 ¹ © 60O ¹
18
5
18
x
5
288 x
250,000 Ÿ x
y
l
Sl
,h 2
4
l
Sl
2
4
15,625
18
3125
§ 15625
·
P¨
, 3125¸ | 226,869
© 18
¹
2h
h
l
© 2010 Brooks/Cole, Cengage Learning
Section 13.10
52. Maximize P x, y
100 x 0.4 y 0.6 subject to the
constraint 72 x 60 y
§ y·
72O Ÿ ¨ ¸
© x¹
0.6
§ x·
60O Ÿ ¨ ¸
© y¹
0.4
60 x 0.4 y 0.4
§ y·
¨© ¸¹
x
0.6
§ y·
¨© ¸¹
x
0.4
72O 1
˜
40 O
y
x
9
Ÿ y
5
§9 ·
72 x 60¨ x¸
©5 ¹
constraint 100 x
72O
40
60O
60
O
125,000
9
250,000 Ÿ x
§ 125,000
·
P¨
, 2500¸ | 496,399
© 9
¹
§ y·
60 x 0.4 y 0.4 O Ÿ ¨ ¸
© x¹
0.4
§ x·
40 x 0.6 y 0.6 O Ÿ ¨ ¸
© y¹
0.6
60
0.4
0.6
y
x
§4 ·
100 x 0.6 ¨ x ¸
©5 ¹
x
§ y·
¨© ¸¹
x
0.25
60
75O
72 75O
˜
25O 60
18
Ÿ y
5
0.75
18
x
5
3.6 x
50,000
x
500
| 191.3124
3.6 0.75
y
3.6 x | 688.7247
50,000.
72
60O
60
40O
3
2O
72 2O
˜
60O 3
4
Ÿ y
5
4
x
5
0.4
50,000
500
45
y
0.75
72
25O
C 191.3124, 688.7247 | 55,097.97
72 x 60 y subject to the constraint 100 x 0.6 y 0.4
72
§ y· § y·
¨ ¸ ¨ ¸
© x¹ © x¹
§ x·
75 x 0.25 y 0.25O Ÿ ¨ ¸
© y¹
0.25
60
100 x 0.25 3.6 x
2500
50,000.
0.75
y
x
y
54. Minimize C x, y
y
§ y·
25 x 0.75 y 0.75O Ÿ ¨ ¸
© x¹
9
x
5
180 x
0.25 0.75
72
§ y·
¨© ¸¹
x
267
72 x 60 y subject to the
53. Minimize C x, y
250,000.
40 x 0.6 y 0.6
Lagrange Multipliers
0.4
400
45
0.4
§ 500
400 ·
C¨
,
| $65,601.72
0.4
0.4 ¸
45 ¹
© 45
55. (a) Maximize g D , E , O
cos D cos E cos J subject to the constraint D E J
sin D cos E cos J
O½
cos D sin E cos J
O ¾tan D
cos D cos E sin J
D E J
§S S S·
g¨ , ˜ ¸
© 3 3 3¹
°
tan E
O °¿
S ŸD
E
J
tan J Ÿ D
E
S.
J
S
3
1
8
© 2010 Brooks/Cole, Cengage Learning
268
Chapter 13
Functions of Several Variables
(b) D E J
S Ÿ J
gD E
S D E
cos D cos E cos S D E
cos D cos E ª¬cos S cos D E
sin S sin D E º¼
cos D cos E cos D E
γ
3
2
α
56. Let r
3
3
radius of cylinder, and h
2S rh 2S r
S
h2 r 2
2S r h
3
2
r 2 h subject to g r , h
ªC 2 r 4 º
r2 «
»
¬ 2Cr ¼
5r 4
C
2
2C
5r 4
r2
C
5
C.
Cr
r5
2
2C
0
10r 3
C
C2 r4
2Cr
h
h2 r 2
C2 r4
2Cr
F r
C2
F cc r
rh r
r4
h
Fc r
constant surface area
r 2 h2 r 2
C 2 2Crh
f r, h
height of cone.
2
We maximize f r , h
C rh
height of cylinder
5S r h
volume
3
2
S r 2h V
β
C2 C2 5
2C C 2 5
14
45C
2 C2 5
14
2C
5r
2
5r 2
5r
2 5
r
5
So,
h
r
2 5
.
5
By the Second Derivative Test, this is a maximum.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
269
Review Exercises for Chapter 13
1. f x, y , z
x2 y z 2
y
x2 z 2 2
2
1 x2 y 2
4. f x, y
z
(a)
Elliptic paraboloid
1
z
−1
2
y
1
1
x
−2
2
f x 2, y is the graph of f shifted
(b) g x, y
y
3
x
−2
2 units back on the x-axis. The hemisphere has
equation
2
x 2
4 x2 y2 4 z 2
2. f x, y, z
y2 z2
4 f x, y is the graph of f reflected in
(c) g x, y
0
1, z t 0.
the xy-plane, and shifted 4 units upward.
Elliptic cone.
z
z
z
(d)
3
2
1
1
1
3 −1
x
−1
y
3
−2
−3
x2 y2
3. f x, y
(a)
−1
1
x
x
f 0, y
f x, 0
1
1 x2 y 2
z
z
ex
5. f x, y
5
y
1
1
2 y2
The level curves are of the form
4
ex
c
x2 y2.
ln c
−2
1
2
The level curves are circles centered at the origin.
y
2
2 y2
y
x
f x, y 2 is a vertical translation
(b) g x, y
2
c = 10
of f two units upward.
c=1
f x, y z is a horizontal translation
(c) g x, y
x
−2
2
of f two units to the right. The vertex moves
−2
from 0, 0, 0 to 0, 2, 0 .
Generated by Mathematica
(d)
z
z
5
5
4
4
z = f (1, y)
6. f x, y
The level curves are of the form
ln xy
c
ec
z = f (x, 1)
2
2
2
y
z
xy.
The level curves are hyperbolas
x
2
x
ln xy
y
f 1, y
z
1
2
f x, 1
c= 2
3
c= 2
c=1
c=0
−3
3
−2
© 2010 Brooks/Cole, Cengage Learning
y
Chapter 13
270
Functions of Several Variables
x2 y2
7. f x, y
11.
The level curves are of the form
x2 y2
1
x2
y2
.
c
c
12.
The level curves are hyperbolas.
lim
x , y o 1,1
c = 12
13.
x
−1
lim
x , y o 0, 0
4
1
xy
x2 y 2
y xe y
1 x2
−4
14.
x
x y
8. f x, y
lim
00
1 0
0
x , y o 0, 0
x2 y
x y2
4
For y
x2 ,
For y
0,
The level curves are of the form
c
x
x y
y
§ 1 c·
¨©
¸ x.
c ¹
x2 y
x y2
e x cos y
fx
e x cos y
fy
e x sin y
1
c=1
−3
xy
x y
16. f x, y
3
3
c= 2
c=2
fx
y x y xy
x y
−2
x y
x2 y 2
17.
z
wz
wx
3
−3
−3
18.
3
3
−3
y
1 x
z
y
z
wz
wx
wz
wy
2
y2
x y
2
x2
fy
e
0 for x z 0.
4
15. f x, y
c= 2
2
9. f x, y
x4
1
o .
x x4
2
4
Continuous to all x, y z 0, 0
c = − 12
3 c = −1
c=−2
c = −2
x2 y
x y2
4
The limit does not exist.
The level curves are passing through the origin with
1c
slope
.
c
10. g x, y
2
Continuous everywhere.
Generated by Mathematica
x
r x.
Continuous except when y
1
−4
1
2
Does not exist.
c = −12 c = −2
c=2
4
xy
x2 y2
Continuous except at 0, 0 .
c
y
lim
x , y o 1,1
2
e y e x
e x ,
wz
wy
e y
ln x 2 y 2 1
2x
x2 y 2 1
2y
x2 y 2 1
z
60
5
x
5
y
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
xy
x2 y2
19. g x, y
24. u x, t
y x 2 y 2 xy 2 x
gx
x2 y 2
xx y
2
gy
x2 y 2
2
x2 y2
2
2
25.
2
c sin akx cos kt
wu
wx
wu
wt
y y 2 x2
271
akc cos akx cos kt
kc sin akx sin kt
z
3
20.
w
x2 y 2 z 2
ww
wx
1 2
x y2 z2
2
1 2
x
2x
x y2 z2
2
−1
y
3
3
ww
wy
y
x y2 z2
ww
wz
x y z
x
2
z
2
21. f x, y , z
26. z
2
z arctan
2
wz
wx
y
x
yz
2
x y2
fy
z
§1·
¨ ¸
1 y 2 x2 © x ¹
fz
arctan
xz
x2 y 2
y
x
1
1 x2 y2 z 2
1 x2 y2 z 2
wf
wx
1
1 x2 y 2 z 2
2
x
1 x2 y 2 z 2
wf
wy
1 x2 y 2 z 2
wf
wz
1 x y2 z2
3 2
1 2
2x
wz
wy
wu
wx
wu
wt
2
ce n t sin nx
6x y
fy
x 6 y2
f xx
f yy
6
12 y
fx y
1
f yx
1
28. h x, y
32
hy
hxx
hyy
2
cne n t cos nx
2 n2 t
cn e
0.
4.
3 x 2 xy 2 y 3
fx
hx
32
x2
wz
. At 2, 0, 0 ,
1 y
wy
27. f x, y
32
z
23. u x, t
wz
wx
Slope in y-direction.
y
2
2 x ln y 1 . At 2, 0, 0 ,
Slope in x-direction.
z
§ y·
fx
¸
2
2 ¨
1 y x © x2 ¹
22. f x, y, z
x 2 ln y 1
hx y
sin nx
hyx
x
x y
y
x y
2
x
x y
2
2 y
x y
3
2x
x y
3
x y
2
2y x y
4
x y
x y
2
2y x y
x y
4
x y
x y
3
x y
x y
3
© 2010 Brooks/Cole, Cengage Learning
272
Chapter 13
Functions of Several Variables
x sin y y cos x
29. h x, y
wz
wx
w2 z
wx 2
wz
wy
sin y y sin x
x cos y cos x
hx
hy
hxx
hyy
y cos x
x sin y
hx y
cos y sin x
hyx
cos y sin x
w2 z
wy 2
cos x 2 y
30. g x, y
So,
gx
sin x 2 y
gy
2 sin x 2 y
g xx
cos x 2 y
g yy
4 cos x 2 y
gx y
2 cos x 2 y
g yx
2 cos x 2 y
2x
2
2 y
2
w2 z
w2 z
2
2
wx
wy
0.
x3 3 xy 2
32. z
wz
wx
w2 z
wx 2
wz
wy
w2 z
wy 2
So,
3x 2 3 y 2
6x
6 xy
6 x
w2 z
w2 z
2
2
wx
wy
0.
y
x2 y2
33. z
wz
wx
2 xy
x y2
2
w2 z
wx 2
2
ª
4 x 2
2 y «
« x2 y 2
¬
3
x2 y
w2 z
wy 2
x2 y2
w2 z
w2 z
2
2
wx
wy
2 2
2
º
»
1
x2 y
x2 y 2 2 y
wz
wy
So,
x2 y2
31. z
2 2»
2y
¼
3x 2 y 2
x2 y 2
3
x2 y2
x2 y 2
2
2 y 2 x 2 y 2 x 2 y 2 2 y
x y
2
2 4
2 y
3x 2 y 2
x2 y2
3
0.
34. z
e y sin x
wz
e y cos x
wx
w2 z
e y sin x
wx 2
wz
e y sin x
wy
w2 z
e y sin x
wy 2
So,
w2 z
w2z
2
2
wx
wy
e y sin x e y sin x
0.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
z
35.
x sin xy
wz
wz
dx dy
wx
wy
dz
273
xy cos xy sin xy dx x 2 cos xy dy
xy
36. z
x y2
2
wz
wz
dx dy
wx
wy
dz
ª
«
«
¬«
x y
2
2
x2 y 2 º
» dy
»
¼»
x 2 y 2 x xy y
x y
2
2
y3
x2 y 2
32
dx x3
x2 y2
32
dy
x2 y 2
z2
37.
ª
x2 y2 º
» dx «
»
«
«¬
¼»
x 2 y 2 y xy x
2 x dx 2 y dy
2 z dz
5 § 1 · 12 § 1 ·
¨ ¸ ¨ ¸
13 © 2 ¹ 13 © 2 ¹
x
y
dx dy
z
z
dz
Percentage error:
dz
z
17
| 0.654 cm
26
17 26
| 0.0503 | 5%
13
38. From the accompanying figure we observe
h
tan T
or h
x tan T
x
wh
wh
dh
dx dT
tan T dx x sec 2 T dT.
wx
wT
1
11S
S
r ,T
r
Letting x 100, dx
, and dT
.
2
60
180
h
θ
x
(Note that we express the measurement of the angle in radians.) The maximum error is approximately
dh
39.
40.
§ 11S · § 1·
§ 11S · § S ·
r
100 sec 2¨
r
| r0.3247 r 2.4814 | r2.81 feet.
tan ¨
© 60 ¸¹ ¨© 2¸¹
© 60 ¸¹ ¨© 180¸¹
V
1
3
S r 2h
dV
2
3
S rh dr 13S r 2 dh
A
dA
S 2 5 r 18 13S 2
2
r 18
r 56 S r 16 S
rS in.3
S r r 2 h2
§
¨© S
Sr 2
r 2 h2 S 2r 2 h 2
r h
2
41. w
2
3
2
ln x 2 y , x
(a) Chain Rule:
dw
dt
r h
2
2
·
¸¹ dr S rh
dr r h
2
2t , y
2
dh
S rh
r 2 h2
dh
S 8 25 § 1·
29
4t
¨© r ¸¹ 8
10S § 1·
¨r ¸
29 © 8¹
42. u
ww dx ww dy
wx dt
wy dt
r
43S
8 29
y 2 x, x
cos t , y
(a) Chain Rule:
du
dt
(b) Substitution: w
dw
dt
ln x 2 y
wu wx wu wy
wx wt
wy wt
1 sin t 2 y cos t
2x
1
2 2
1
x y
x y
2
8t 1
4t 2 4 t
sin t
sin t 2 sin t cos t
sin t 1 2 cos t
ln 4t 2 4 t
1
8t 1
2
4t 4 t
(b) Substitution: u
du
dt
sin 2 t cos t
2 sin t cos t sin t
sin t 1 2 cos t
© 2010 Brooks/Cole, Cengage Learning
274
Chapter 13
43. w
xy
,x
z
Functions of Several Variables
2r t , y
(a) Chain Rule:
ww
wr
2r t
rt , z
ww wx ww wy
ww wz
wx wr
wy wr
wz wr
y
x
xy
2 t 2 2
z
z
z
2r t t
2 2r t rt
2rt
2
2r t
2r t
2r t
4r 2t 4rt 2 t 3
2r t
ww
wt
2
ww wx ww wy
ww wz
wx wt
wy wt
wz wt
y
x
xy
1
1 r
z
z
z2
4r 2t rt 2 4r 3
2r t
(b) Substitution: w
ww
wr
ww
wt
44. u
2r t rt
xy
z
2r t
2
4r t 4rt 2 t 3
2r t
wu
wr
2r 2t rt 2
2r t
2
4r 2t rt 2 4r 3
2r t
x2 y 2 z 2 , x
(a) Chain Rule:
2
r cos t , y
2
r sin t , z
t
wu wx wu wy
wu wz
wx wr
wy wr
wz wr
2 x cos t 2 y sin t 2 z 0
2 r cos 2 t r sin 2 t
wu
wt
2r
wu wx wu wy
wu wz
wx wt
wy wt
wz wt
2 x r sin t 2 y r cos t 2 z
(b) Substitution: u r , t
wu
wr
wu
wt
r 2 cos 2 t r 2 sin 2 t t 2
wz
wz
2z
wx
wx
0
x 2y y
46.
2 xz
x 2y z
y 2z
xz 2 y sin z
0
wz
wz
z 2 y cos z
wx
wx
0
wz
wx
2 x y
y 2z
wz
wz
z 2z
wy
wy
wz
wy
r2 t2
2t
0
wz
wx
2t
2r
45. x 2 xy y 2 yz z 2
2x y y
2 r 2 sin t cos t r 2 sin t cos t 2t
0
2 xz
wz
wz
y cos z
sin z
wy
wy
wz
wy
z2
y cos z 2 xz
0
sin z
2 xz y cos z
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
47. f x, y
x2 y
’f
2 xy i x 2 j
’f 5, 5
u
50 i 25 j
’z
3
4
i j unit vector
5
5
’z 1, 1
Du f 5, 5
’f
’f 1, 4
1
v
5
u
’w 1, 2, 2
2i 4 j k
1
3v
u
Du w 1, 2, 2
50. w
1
3
’w 1, 2, 2 ˜ u
4
3
2
3
10 i 2 j
2
3
’z
4i 4 j
’z 2, 1
4 2
z
’z
§ S·
’z ¨ 0, ¸
© 4¹
1
x
−6 −4
12
3
4
−2
6
−4
−6
4 3
Tangent line
4 y sin x y, c
56. f x, y
(a) ’f x, y
§S ·
3, P¨ , 1¸
©2 ¹
4 y cos x i 4 sin x 1 j
§S ·
’f ¨ , 1¸
3j
©2 ¹
(b) Unit normal vector: j
e x cos y i e x sin y j
2
2
i j
2
2
Unit normal
vector
2
e x cos y
§ S·
’z ¨ 0, ¸
© 4¹
1
27 i 8 j
793
y
4
2 xy i x 2 j
’ z 2, 1
54 i 16 j
54 i 16 j
6
x2 y
z
65, P 3, 2
54i 16 j
(d)
’w 1, 0, 1 ˜ u
10
2
3
3
j
27
.
8
27
y z
x 3
8
27
65
Tangent line
y
x 8
8
1
i jk
3
Du w 1, 0, 1
52.
10 x 2 y i 2 x 6 yz j 3 y 2 k
’w 1, 0, 1
51.
4
3
2
(c) Slope
j 32 k
5 x 2 2 xy 3 y 2 z
’w
u
2
3i
x y
18 x i 8 y j
(b) Unit normal:
z i 2y j xk
x2
i 9 x2 4 y 2 , c
’f 3, 2
’w
j
4
(a) ’f x, y
y 2 xz
w
49.
2
x y
’z 2, 1
2 5
5
2
4j
55. f x, y
4 5
2 5
5
5
x2 y 2
1
,0
2
x 2 2 xy
’z 2, 1
’f 1, 4 ˜ u
Du f 1, 4
x2 y 2
x2
x y
’z
2 5
5
i j
5
5
x2 y 2
i 2
1
2
z
54.
2 xy
1
i
2
’z 1, 1
50
1 2
y x2
4
1
2x i + y j
2
2 i 2 j
f x, y
’f 5, 5 ˜ u
30 20
48.
y
x2 y2
z
53.
275
(c) Tangent line horizontal: y
(d)
1
y
2
2
,
2
2
3
2
(π2 , 1(
π
−2
π
2
π
x
© 2010 Brooks/Cole, Cengage Learning
276
57.
Chapter 13
Functions of Several Variables
x2 y z
F x, y , z
’F
2 xy i x j k
2
’F 2, 1, 4
y2 z 9
61. F x, y, z
0
x y
G x, y , z
4i 4 j k
’F
2y j k
So, the equation of the tangent plane is
’G
i j
4x 2 4 y 1 z 4
’F 2, 2, 5
4x 4 y z
0 or
4t 2, y
x
4t 1, z
’F u ’ G
t 4.
y 2 z 2 25
F x, y , z
58.
’F
6 j 8k
Tangent line:
0
2 3 j 4k
62.
So, the equation of the tangent plane is
3y 3 4z 4
0 or 3 y 4 z
y 3
3
2,
2x i 2 y j k
k
F x, y , z
x2 y2 z 2 9
’F
2x i 2 y j 2z k
’F 1, 2, 2
4i 2 j k
2i 4 j 4k
k
0
2 i 2j
1
0
3.
x 2 y 2 z 2 14
f x, y , z
’f x, y , z
2 i 2 j 2k
j
4 2 1
y 1
,z
2
’f 2, 1, 3
60.
0
So, the equation of the tangent line is
63.
4 t.
z 5
4
0
’F
4,
3, z
2, y
y 2
1
’G
x 2
1
and the equation of the normal line is
x
i j 4k
3 z
0
k
0 or z
1
G x, y , z
’F u ’G
0
So, the equation of the tangent plane is
z 4
4
i
2x 4 i 2 y 6 j k
’F 2, 3, 4
0
x 2
1
’F 2, 1, 3
x2 y 2 4x 6 y z 9
’F
k
x2 y 2 z
25,
z 4
.
4
F x, y , z
59.
j
F x, y , z
and the equation of the normal line is
x
i
1 1 0
2 y j 2z k
’F 2, 3, 4
4j k
8,
and the equation of the normal line is
0, 2, 2, 5
2x i 2 y j 2z k
4 i 2 j 6 k Normal vector to plane.
cos T
n˜k
n
T
36.7q
6
56
3 14
14
So, the equation of the tangent plane is
x 1 2 y 2 2z 2
x 2 y 2z
0 or
9,
and the equation of the normal line is
x 1
1
64. (a)
y 2
2
z 2
.
2
cos x sin y, f 0, 0
f x, y
fx
sin x, f x 0, 0
fy
cos y, f y 0, 0
P 1 x, y
0
1
1 y
(b) f xx
cos x, f xx 0, 0
1
f yy
sin y, f yy 0, 0
0
f xy
0, f xy 0, 0
P2 x, y
1
0
1 y 12 x 2
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
277
0, you obtain the 2nd degree Taylor polynomial for cos x.
(c) If y
(d)
x
y
0
0
0
f x, y
P1 x, y
P2 x, y
1.0
1.0
1.0
0.1
1.0998
1.1
1.1
0.2
0.1
1.0799
1.1
1.095
0.5
0.3
1.1731
1.3
1.175
1
0.5
1.0197
1.5
1.0
z
z
(e)
z
3
2
2
2
−2
1
1
y
2
y
1
−1
1
2
y
x
1
1
−1
−2
−1
−1
x
x
The accuracy lessens as the distance from 0, 0 increases.
2 x 2 6 xy 9 y 2 8 x 14
65. f x, y
fx
4x 6 y 8
fy
6 x 18 y
4 3 y 6 y
xy 67. f x, y
0
3 y
0, x
8 Ÿ y
4
3,
1
x2
1
x 2
y
0, x 2 y
1
0, xy 2
1
xy 2 or x
y and substitution yields the
y fx
4
x
1
1
x
y
fy
f xx
4
f yy
18
So, x 2 y
f xy
6
critical point 1, 1 .
f xx f yy f x y
2
4 18 6
2
36 ! 0.
So, 4, 43 , 2 is a relative minimum.
fx
fy
2x 3 y 5
3x 2 y
4x 9x
2
y3
f yy
0
At the critical point 1, 1 , f xx
Ÿ y
0
§ 3 ·
2 x 3¨ x¸
© 2 ¹
3
x
2
f xx f yy f x y
2
2 ! 0 and
3 ! 0.
So, 1, 1, 3 is a relative minimum.
5
z
10
x
f xx
fxy
x 2 3 xy y 2 5 x
66. f x, y
2
x3
1
f xx
2, f yy
2, y
20
3
2, f xy
3, d
49 0
Ÿ 2, 3 is a saddle point.
3 4
4
x
−20
−24
y
(1, 1, 3)
© 2010 Brooks/Cole, Cengage Learning
278
68.
Chapter 13
z
Functions of Several Variables
50 x y 0.1x3 20 x 150 0.05 y 3 20.6 y 125
zx
50 0.3 x 2 20
zy
50 0.15 y 20.6
r10
0, x
2
r14
0, y
Critical Points: 10, 14 , 10, 14 , 10, 14 , 10, 14
z xx
0.6 x, z yy
0.3 y, z x y
At 10, 14 , z xx z yy z x y
2
0
6 4.2 02 ! 0, z xx 0.
10, 14, 199.4 is a relative maximum.
At 10, 14 , z xx z yy z x y
2
6 4.2 02 0.
10, 14, 349.4 is a saddle point.
At 10, 14 , z xx z yy z xy
2
6 4.2 02 0.
10, 14, 200.6 is a saddle point.
At 10, 14 , z xx z yy xx y
2
6 4.2 02 ! 0, z xx 0.
10, 14, 749.4 is a relative minimum.
69. The level curves are hyperbolas. There is a critical point at 0, 0 , but there are no relative extrema. The gradient is normal to
the level curve at any given point x0 , y0 .
70. The level curves indicate that there is a relative extremum at A, the center of the ellipse in the second quadrant, and that there is
a saddle point at B, the origin.
R C1 C2
71. P x1 , x2
2
2
¬ª225 0.4 x1 x2 ¼º x1 x2 0.05 x1 15 x1 5400 0.03 x2 15 x2 6100
0.45 x12 0.43 x2 2 0.8 x1 x2 210 x1 210 x2 11,500
Px 1
0.9 x1 0.8 x2 210
0.9 x1 0.8 x2
Px 2
210
0.86 x2 0.8 x1 210
0.8 x1 0.86 x2
0
0
210
Solving this system yields x1 | 94 and x2 | 157.
Px 1 x1
0.9
Px 1 x 2
0.8
Px 2 x 2
0.86
Px 1 x 1 0
Px 1 x 1 Px 2 x 2 Px 1 x 2
2
! 0
So, profit is maximum when x1 | 94 and x2 | 157.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 13
0.25 x12 10 x1 0.15 x2 2 12 x2
72. Minimize C x1 , x2
subject to the constraint x1 x2
¾5 x1 3 x2
O¿
0.30 x2 12
x1 x2
4 y
20
1000 Ÿ 3 x1 3x2
x 2
500
5x y
6
10 x
3020
x1
377.5
x2
622.5
6
2000 Ÿ 5 x y
20
8 x1
C 377.5, 622.5
2000.
20O ½
¾5 x y
4O ¿
20 x 4 y
3000
5 x1 3 x2
4 x xy 2 y subject to the
constraint 20 x 4 y
1000.
O½
0.50 x1 10
73. Maximize f x, y
494
x
y
f 49.4, 253
104,997.50
279
49.4
253
13,201.8
74. Minimize the square of the distance:
x 2
f x, y , z
2
y 2
fx
2 x 2 2 x2 y2 2x
fy
2 y 2 2 x y 2y
2
2
2
x2 y 2 0 .
0 °½ x 2 2 x3 2 xy 2
¾
0°¿ y 2 2 y 3 2 x 2 y
2
y and hence: 4 x3 x 2
Clearly x
So, distance
2
0.6894 2
2
75. (a) y
2
2
2
ª2 0.6894 º | 4.3389.
¬
¼
0.004 x 2 0.07 x 19.4
(b) When x
76. (a) y
0
0. Using a computer algebra system, x | 0.6894.
0.6894 2
Distance | 2.08
0
80, y
0.004 80
2
0.07 80 19.4 | 50.6 Kg.
2.29t 2.0
20
0
9
0
(b)
Yes, the data appear linear.
(c) y
(d)
1.24 8.37 ln t
20
−1
3
−5
© 2010 Brooks/Cole, Cengage Learning
280
Chapter 13
Functions of Several Variables
xy yz xz subject to the
77. Optimize f x, y, z
constraint x y z
y z
O½
x z
O¾x
°
x y z
QR
y 2 1,
RS
z; x y z
z
y
1 1 1
, ,
3 3 3
z
3x
x 4
x 2 y subject to the
O½
2 xy
¾x
2O ¿
x2
x 2y
If x
2.
4 xy Ÿ x
2
0 or x
4y
2
0, y
1. If x
Maximum: f
4 1
,
3 3
Minimum: f 0, 1
4 y, then y
1
3,
x
Constraint:
2y
y2 1
3x
O
x2 4
2y
O
y2 1
1
O
9x2
x2 4 Ÿ x2
4 y2
y2 1 Ÿ y2
0
jk
2
| 0.707 km,
2
y
3
| 0.577 km,
3
10 O >i j k @
1
2
1
3
So, x
z
80. f x, y
i 4
3.
16
27
10
O ’g
’C
1
3
constraint x 2 y
y2 1 z
Constraint: x y z
1
3
2
78. Optimize f x, y
10
3 x2 4 2
C
1Ÿ x
Maximum: f
x 2 4,
1.
y
O°¿
x y
79. PQ
2
3
| 8.716 km.
2
3
ax by, x, y ! 0
x2
y2
64
36
1
4 x 3 y are lines of form y
(a) Level curves of f x, y
Using y
4
x C.
3
4
x 12.3, you obtain x | 7, y | 3, and f 7, 3
3
28 9
37.
8
−10
10
−8
Constraint is an ellipse.
(b) Level curves of f x, y
Using y
4 x 9 y are lines of form y
4
x C.
9
4
x 7, you obtain x | 4, y | 5.2, and f 4, 5.2
9
62.8.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 13
281
Problem Solving for Chapter 13
1. (a) The three sides have lengths 5, 6, and 5.
16
2
Thus, s
8 and A
(b) Let f a, b, c
area
83 2 3
2
12.
s s a s b s c,
subject to the constraint
a bc
Using Lagrange multipliers,
O
s s a s c
O
s s a s b
O.
Fx
yz , Fy
s a Ÿ a
to Area
1
O s s a s b
2. V
V
s b Ÿ a
Tangent plane
b
c which is an
3
x0 z0
3
So,
M
dM
dr
16
8S r S r
3
8 ·
§
r3¨ S ¸
©3 ¹
r3
Then, h
Base
3
y0 z0
4. (a) As x o rf, f x
x3 1
hence lim ª¬ f x g x º¼
x of
b and a
b
c.
(b) Let x0 , x03 1
13
13
o x and
lim ª f x g x º¼
x of ¬
0.
be a point on the graph of f .
The line through this point perpendicular
x x0 to g is y
4S r 2S rh
1000 Ÿ h
y
3
2
M
9
2
3
x0 y 0
b.
4 S r S r 2h
3
Material
3
1
base height
3
3
Using Lagrange multipliers,
1 O s s b s c
O s s a s c
0
z
s s a s b s c constant.
1
3 x0 y0 z0
1 § 1 3 3 ·§ 3 ·
¨
¸¨
¸
3 © 2 y0 z0 x0 z0 ¹© x0 y0 ¹
x
So, s a
(−
a b c, subject
(c) Let f a, b, c
xy
y0 z0 x x0 x0 z0 y y0 x0 y0 z z0
(b) V
Similarly, b
c and hence a
equilateral triangle.
2
xz , Fz
Tangent plane:
From the first 2 equations
s b
0
y0 z0 x x0 z0 y x0 y0 z
constant (perimeter).
s s b s c
xyz 1
3. (a) F x, y, z
3
x03 1.
This line intersects g at the point
1000 4 3 S r 3
Sr2
§1ª
¨ ¬ x0 ©2
§ 1000 4 3 S r 3 ·
4S r 2 2S r ¨¨
¸¸
Sr2
©
¹
2000
8
S r2
4S r 2 r
3
2000 16
8S r 2 S r
0
r
3
2000
r2
3
1
x03 1º, ª x0 ¼ 2¬
3
·
x03 1º ¸.
¼¹
The square of the distance between these two points
2
1
is h x0
x0 3 x03 1 .
2
h is a maximum for x0
3
1
. So, the point
2
1 ·
§ 1
on f farthest from g is ¨ 3 , 3 ¸.
2
2¹
©
4
2000
−6
750
S
6
13
Ÿ r
1000 4 3 S 750 S
Sr2
§6·
5¨ ¸ .
©S ¹
−4
0.
13
The tank is a sphere of radius r
§6·
5¨ ¸ .
©S ¹
© 2010 Brooks/Cole, Cengage Learning
282
Chapter 13
Functions of Several Variables
y
5. (a)
(−
2,
2(
c = −1
k 6 xy 6 xz 6 yz
c=1
V
x
3
−1
(
2, − 2 (
2, x y
O’g :
1
2O x
1
2O y
x y
2
2O x
2O y Ÿ x
2 x2
4 Ÿ x
2, f
2
2
2
4
1000
Ÿ H
xy
z
5y Hx
4
y
Hy
6000
x2
B
2 2, f 2,
2
So, x
y
0 Ÿ 5 yx 2
2 3 150 and z
2 2
2
y
z
10.
§
6000 6000 ·
k ¨ 5 xy ¸.
y
x ¹
©
y Ÿ x3
By symmetry, x
r 2, y
0, you obtain x
k 5 xy 6 xz 6 yz
7. H
2
1000
.
xy
§
1000 1000 ·
6k ¨ xy ¸.
y
x ¹
©
Setting H x
2 2.
x y.
Constraint: g x, y
’f
2
1000 Ÿ z
xyz
Then H
Maximum value of f is f
Maximize f x, y
k 5 xy xy 3 xz 3xz 3 yz 3 yz
H
6. Heat Loss
c=0
3
y3
6000
1200.
53
150.
3
x y
(b) f x, y
Constraint: x 2 y 2
0 Ÿ x, y
0, 0
Maximum and minimum values are 0.
Lagrange multipliers does not work:
1
1
2O x ½
¾x
2O y¿
y
Note that ’g 0, 0
` 8. (a) T x, y
0, a contradiction.
0.
2 x 2 y 2 y 10
10
1
4
1
4
2x2 y2 y 1·
§
2x2 ¨ y ¸
2¹
©
y 12
x2
18
14
(b) On x 2 y 2
Tc y
1, T x, y
2 y 1
4 x 0, Ty
§ 1·
T ¨ 0, ¸
© 2¹
39
minimum
4
1
2
1
4
x
−
1
2
−
1
2
1
2
2
ellipse
1
2 1 y 2 y 2 y 10
T y
0 Ÿ y
Inside: Tx
§
3 1·
T ¨¨ r
, ¸¸
2¹
© 2
2
y
1
,x
2
2y 1
r
12 y 2 y
3
.
2
§ 1·
0 Ÿ ¨ 0, ¸
© 2¹
49
maximum
4
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 13
wf
wx
9. (a)
x
Cax a 1 y1 a ,
wf
wf
y
wx
wy
wf
wy
283
C 1 a xa y a
Cax a y1 a C 1 a x a y1 a
ª¬Ca C 1 a º¼ x a y1 a
Cx a y1 a
(b) f tx, ty
C tx
r cos T , y
10. x
a
ty
f
1 a
r sin T , z
Ct a x at1 a y1 a
Cx a y1 a t
tf x, y
z
wu
wT
wu wx
wu wy
wu wz
wx wT
wy wT
wz wT
wu
wr
wu
wu
cos T sin T .
wx
wy
wu
wu
r cos T Similarly,
r sin T wx
wy
ª w 2u wx
w 2u wy
w 2u wz º
wu
r sin T « 2
» r cos T r cos T
x
T
x
y
T
x
z
T
w
w
w
w
w
w
w
w
wx
¬
¼
w 2u
wT 2
ª w 2u wx
w 2u wy
w 2u wz º
wu
2
«
» r sin T
y
x
T
y
T
y
z
T
wy
w
w
w
w
w
w
w
w
¬
¼
w 2u 2
w 2u
w 2u 2
wu
wu
r sin 2 T 2 r 2 cos 2 T 2
r sin T cos T r cos T r sin T
2
wx
wy
wxwy
wx
wy
Similarly,
w 2u
wr 2
w 2u
w 2u
w 2u
cos 2 T 2 sin 2 T 2
cos T sin T .
2
wx
wy
wxwy
Now observe that
ª w 2u
º 1 ª wu
º
w 2u
w 2u
wu
2
2
cos T sin T » « cos T sin T »
« 2 cos T 2 sin T 2
x
y
x
y
r
x
y
w
w
w
w
w
w
¬
¼
¬
¼
w 2u 1 wu
w 2u
1 w 2u
2
2
2
2
r wr
r wT
wr
wz
ª w 2u
º w 2u
1 wu
1 wu
w 2u
w 2u
sin T cos T cos T sin T » 2
« 2 sin 2 T 2 cos 2 T 2
r wx
r wy
wy
wxwy
¬ wx
¼ wz
w 2u
w 2u
w 2u
2 2.
2
wx
wy
wz
So, Laplace’s equation in cylindrical coordinates, is
64 cos 45q t
11. (a) x
(b) tan D
D
(c)
dD
dt
(d)
0.
32 2 t
64 sin 45q t 16t 2
y
w 2u
1 wu
1 w 2u
w 2u
2
2
2
2
wr
r wr
r wT
wz
32 2 t 16t 2
y
x 50
§ y ·
arctan ¨
¸
© x 50 ¹
§ 32 2 t 16t 2 ·
arctan ¨¨
¸¸
© 32 2 t 50 ¹
1
§ 32 2 t 16t 2 ·
1 ¨¨
¸¸
© 32 2 t 50 ¹
2
˜
64 8 2 t 2 25t 25 2
32 2 t 50
2
16 8 2 t 2 25t 25 2
64t 4 256 2 t 3 1024t 2 800 2 t 625
30
0
4
−5
No. The rate of change of D is greatest when the projectile is closest to the camera.
© 2010 Brooks/Cole, Cengage Learning
284
Chapter 13
(e)
dD
dt
Functions of Several Variables
0 when
8 2 t 2 25t 25 2
0
25 t
252 4 8 2 25 2
28 2
No, the projectile is at its maximum height when dy dt
x2 y 2
12. (a) d
(b)
2
32 2 t 16t 2
2
32 2 32t
2 | 1.41 seconds.
0 or t
4096t 2 1024 2 t 3 256t 4
16t
t 2 4 2 t 16
32 t 2 3 2 t 8
dd
dt
t 2 4 2 t 16
(c) When t
2:
32 12 6 2
dd
dt
(d)
32 2 t
| 0.98 second.
| 38.16 ft/sec
20 8 2
32 t 3 6 2 t 2 36t 32 12
d 2d
dt 2
t 2 4 2 t 16
when t
0 when t | 1.943 seconds. No. The projectile is at its maximum height
32
2.
13. (a) There is a minimum at 0, 0, 0 , maxima at 0, r1, 2 e and saddle point at r1, 0, 1 e :
x2 2 y2 e
fx
e
x2 y 2
x2 y 2
e
x2 y2
x2 y2
2
ª 2
º
¬ x 2 y 2 x 2 x ¼
x2 2 y 2 e
fy
2 x 2 x e
x2 y2
2 y 4 y e
e
x2 y2
ª¬2 x3 4 xy 2 2 xº¼
0 Ÿ x3 2 xy 2 x
0
x2 y2
ª x 2 2 y 2 2 y 4 yº
¬
¼
Solving the two equations x3 2 xy 2 x
e
x2 y2
ª¬4 y 3 2 x 2 y 4 yº¼
0 and 2 y 3 x 2 y 2 y
0 Ÿ 2 y3 x2 y 2 y
0
0, you obtain the following critical points:
0, r1 , r1, 0 , 0, 0 . Using the second derivative test, you obtain the results above.
z
1
2
2
y
x
(b) As in part (a), you obtain
fx
fy
e
e
x2 y 2
x2 y 2
ª2 x x 2 1 2 y 2 º
¬
¼
z
ª2 y 2 x 2 2 y 2 º
¬
¼
1
The critical numbers are 0, 0 , 0, r1 , r1, 0 .
These yield
r1, 0, 1 e minima
1
x
2
y
−1
0, r1, 2 e maxima
0, 0, 0 saddle
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 13
285
(c) In general, for D ! 0 you obtain
0, 0, 0 minimum
0, r1, E e maxima
r1, 0, D e saddle
For D 0, you obtain
r1, 0, D e minima
0, r1, E e maxima
0, 0, 0 saddle
14. Given that f is a differentiable function such that
’f x0 , y0
0 and f y x0 , y0
0, then f x x0, y0
Therefore, the tangent plane is z z0
z
z0
0.
r0.05, dT
dr
0 or
f x0 , y0 which is horizontal.
15. (a)
§ S·
¨ 5, ¸
© 18 ¹
16. r , T
6 cm
r0.05
x
r cos T
5 cos
y
r sin T
5 sin
1 cm
S
18
S
18
| 4.924
| 0.868
y
5
(b)
6 cm
4
1 cm
3
2
1
(c) The height has more effect since the shaded region
in (b) is larger than the shaded region in (a).
(d) A
hl Ÿ dA
π
θ = 18
(
(r, θ ) = 5,
π
18
)
5
x
1
2
3
4
5
(a) dx should be more effected by changes in r.
l dh h dl
If dl
0.01 and dh
0, then dA
1 0.01
0.01.
If dh
0.01 and dl
0, then dA
6 0.01
0.06.
cos T dr r sin T dT
dx
| 0.985 dr 0.868 dT
dx is more effected by changes in r because
0.985 ! 0.868.
(b) dy should be more effected by changes in T .
dy
sin T dr r cos T dT
| 0.174 dr 4.924 dT
dy is more effected by T because 4.924 ! 0.174.
17. Let g x, y
§ x·
yf ¨ ¸.
© y¹
g y x, y
§ x·
§ x ·§ x ·
f ¨ ¸ yf c ¨ ¸¨ 2 ¸
y
© ¹
© y ¹© y ¹
g x x, y
§ x ·§ 1 ·
yf c ¨ ¸¨ ¸
© y ¹© y ¹
§ x· x § x·
f ¨ ¸ f c¨ ¸
© y¹ y © y¹
§ x·
f c¨ ¸
© y¹
ª § x · x § x ·º
§
§x ·
§ x ··
Tangent plane at x0 , y0 , z0 is f c ¨ 0 ¸ x x0 « f ¨ 0 ¸ 0 f c ¨ 0 ¸» y y0 1¨ z y0 f ¨ 0 ¸ ¸
© y0 ¹
© y0 ¹ ¹
¬ © y0 ¹ y0 © y0 ¹¼
©
ª §x · x
§x ·
§ x ·º
Ÿ f c ¨ 0 ¸ x « f ¨ 0 ¸ 0 f c ¨ 0 ¸» y z
© y0 ¹
¬ © y0 ¹ y0 © y0 ¹¼
0
0.
This plane passes through the origin, the common point of intersection.
© 2010 Brooks/Cole, Cengage Learning
286
Chapter 13
18. x 2 y 2
x 1
Functions of Several Variables
2x
2
y2
y
1 Circle
x2
y2
2
1
Ellipse
2
a
b
The circle and ellipse intersect at x, y and x, y for a unique value of x.
2
x
−1
2
b 2
a x2
a2
b2
x2 2 a2 x2
a
y2
(x, y)
1
1
−1
Ellipse
(x, −y)
−2
Circle
2x
§
b2 · 2
2
0 Quadratic
¨1 2 ¸ x 2 x b
a ¹
©
For these to be a unique x-value, the discriminant must be 0.
§
b2 ·
4 4¨1 2 ¸b 2
a ¹
©
0
a 2 a 2b 2 b 4
0
We use lagrange multipliers to minimize the area f a, b
a ab b
2
g a, b
’f
2 2
S ab of the ellipse subject to the constraint
0.
O’g
S b, S a
O 2a 2ab 2 , 2a 2b 4b3
Sb
O 2a 2ab 2
Sa
O 2a 2b 4b3
O
4
Sb
2a 2ab 2
Sa
Ÿ 4b 4 2a 2b 2
4b3 2a 2b
Using the constraint, a 2 a 2b 2 b 4
0, a 2 a 2
2a 2 2a 2b 2 Ÿ 2b 4
a
a2
2
2
3
2
a
Ellipse:
19.
x2
y2
92
32
1
ª cos x t cos x t º¼
2¬
w 2u
wt 2
wu
wx
1
ª sin x t sin x t º¼
2¬
1
ªcos x t cos x t º¼
2¬
w 2u
wx 2
1
ª sin x t sin x t º¼
2¬
w 2u
wt 2
w 2u
.
wx 2
0
a
2
3
2
6
.
2
2, b
1
wu
wt
Then,
a
2
a2 Ÿ b2
20. u x, t
Let r
Then
wu
wt
w 2u
wt 2
1
ª f x ct f x ct
2¬
x ct and s
x ct.
1
u r, s
ª f r f s º¼.
2¬
wu wr wu ws
1 df
c
wr wt
ws wt
2 dr
1 d2 f
1 d2 f 2
2
c c
2
2 dr
2 ds 2
wu
wx
w 2u
wx 2
So,
wu wr wu ws
wr wx ws wx
1 d2 f 2 1
1 2 2dr 2
2
w 2u
wt 2
c2
º¼
1 df
c
2 ds
c2 ª d 2 f
d2 f º
« 2 »
2 ¬ dr
ds 2 ¼
1 df
1 df
1 1
2 dr
2 ds
d2 f 2
1 ªd 2 f
d2 f º
1
« 2 »
2
ds
2 ¬ dr
ds 2 ¼
w 2u
.
wx 2
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 4
Multiple Integration
Section 14.1
Iterated Integrals and Area in the Plane.............................................288
Section 14.2
Double Integrals and Volume ............................................................297
Section 14.3
Change of Variables: Polar Coordinates ...........................................308
Section 14.4
Center of Mass and Moments of Inertia ............................................316
Section 14.5
Surface Area .......................................................................................325
Section 14.6
Triple Integrals and Applications.......................................................332
Section 14.7
Triple Integrals in Cylindrical and Spherical Coordinates ...............344
Section 14.8
Change of Variables: Jacobians.........................................................350
Review Exercises ........................................................................................................358
Problem Solving .........................................................................................................367
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 4
Multiple Integration
Section 14.1 Iterated Integrals and Area in the Plane
1.
2.
x
³0
x
ª¬ xy y 2 º¼
0
x 2 y dy
x2
x2
ª1 y2 º
«
»
¬2 x ¼x
y
dy
x
³x
x2 x2
2y
³1
2 x2
3.
³ x3
7.
³ ey
8.
9.
x
³
1 y2
x y
2
1 y2
x3
³0
ye y
u
10.
x
y, du
S 2
³y
y
ª1
2 º
« 2 y ln x» y
¬
¼e
y ln x
dx
x
y
ª¬ x 2 y y 3 º¼ 3
x
x 2 3 y 2 dy
dy
2
x 2
x 1
2
cos y
³0
4.
³0
e y
x
S 2
³y
1
2
12.
³ 1 ³ 2
1
2
x y dy dx
x 2 y 2 dy dx
ª1
2« 1 y 2
¬3
1 y2
x
4
x 2 2 y 2 dx dy
2
3
x y 2 dx dy
³1 ³ 0
³ 1 ³ 1
1
³ 0 ª¬xy 1
ª
1
§
³ 1 «¬x
2
³1
1
3
S 2
cos3 x cos yº
¼y
2
1 2º
y
2 ¼0
1
³0
dx
y 2
y3 º
» dx
3 ¼ 2
16 ·
¸ dx
3¹
288
x3
cos y 1
ª
³ 1 «¬2 x
1 2º
»
¼
4 x2 x4
2
2 1 y2
1 2 y2
3
2
x 2 1 e x x 2e x
2
2
³1
1
2
ª§ 9
3
8
8º
2 x 2 » dx
3
3¼
ª 64
2º
« 3 8 y » dy
¬
¼
³ 1 «¬¨© 2
cos3 y cos y
ª¬ x 2 2 xº¼
0
1
§ 4 16 · § 4 16 ·
¨ ¸ ¨ ¸
3¹ © 3
3¹
©3
2
8 3º
ª 64
« 3 y 3y »
¬
¼1
8
§ 128 64 · § 64 8 ·
¸
¨
¸¨
3¹ © 3
3¹
© 3
8
3
· §1
·º
3 y 2 ¸ ¨ y 2 ¸» dy
¹ ©2
¹¼
2
4 2 y 2 dy
2
1
3
ª 4 x3 16 º
x»
«
3 ¼ 1
¬ 3
ª x3
2º
« 2 xy » dy
¬3
¼0
ª x2
2º
³ 1 «¬ 2 xy »¼ dy
1
³ 1
4 x2
x5 2 x 3 2 x 5 x9
y2 1 y2
2 x 2 dx
2
2
2
2
ª1 2 2 º
«2 x y »
¬
¼0
x 2 y dy
x
3
14.
y cos y
y 2 º, y ! 0
¼
2
4
2
y ln 2 y, y ! 0
1 cos 2 x sin x cos y dx
³ 1 ¨© 4 x
13.
32
2
cos y
3
x 4e x ª¬ x 2e y x º¼
0
dy
xe y
dy, v
ª cos x ¬
³0 ³0
1 y2
x3
x3
dy, dv
yª
ln y
2¬
> yx@0
y dx
4 x2
3
x ·¸ x 2 x3 x3
¹
x 1
y ªln 2 y ln 2 e y º¼
2 ¬
ª¬ xye y x º¼ x ³ e y
0
0
sin 3 x cos y dx
11.
§ x2
¨
©
x
ª1 3
2 º
« 3 x y x»
¬
¼
dx
> y ln x@12 y
y ln 2 y 0
1 § x4
x2 ·
¨
¸
2© x
x¹
5.
6.
y
dx
x
2 3º
ª
«4 y 3 y »
¬
¼ 1
16 · §
2·
§
¨8 ¸ ¨ 4 ¸
3¹ ©
3¹
©
18
© 2010 Brooks/Cole, Cengage Learning
Section 14.1
S 2
1
ln 4
ln 3
15.
³ 0 ³ 0 y cos x dy dx
16.
³0 ³0
S 2
³0
ln 4
ª¬e x y º¼ dx
0
ln 4
ª¬e x ln 3 e x º¼ dx
³0
e x y dy dx
³0
S
sin x
4
x
17.
³0 ³0
18.
³1 ³1
19.
³0 ³0
20.
³ 4 ³ 0
21.
1
x
x2
5
3y
³ 1 ³ 0
S
³0
4
³1
2 ye x dy dx
³ 0 ª¬ y
1
ln 4
ª¬e x ln 3 e x º¼
0
sin x
x
4
³1
dx
x
1
³0 x
64 x3 º dx
¼0
4
S
4
ª¬ xe x º¼
1
ª 1
¬« 2
1 x 2 dx
4
³ 4 x
³0 ³ y
2
³0
2
2 y y2
24.
³ 0 ³ 3 y2 6 y 3 y dx dy
25.
2
³0 ³0
3
y
4 y2
ª 9 2 39 º
« 2 y 16 y »
¬
¼ 1
27.
³0 ³0
28.
S 2
S 4
³0 ³
2 cos T
2
r dr dT
3 cos T
3
20 3 º
ª
«10 y 3 y » dy
¬
¼
xyº¼
0
2
1 y 2
r dr dT
2
³0
dx dy
ª
«
«¬
S 2
³0
2
8 y 2 4 y 3 dy
0
º
»
2
4 y »¼ 0
dy
y
2 cos T
ªr2 º
« »
¬ 2 ¼0
³0
S 4
32 4
º
¼» 4
0
2
9
128
32
2048
9
2
ªr 2 º
« »
¬2¼
3
³1
S 2
³0
dT
3 cos T
dT
3
16 3
2
· §
·º
y 4 y 3 ¸ ¨10 y y 3 2 y 3 ¸» dy
3
3
¹ ©
¹¼
2
80
3
20 y 1 y 2 º dy
¼
140
3
ª1 y ¬« 2
1 y3
6
1 2
2 3
1 y2
32 1
º
¼» 0
2
3
2
ª3 8 y 3 y 4 º
¬ 3
¼0
2
³ 0 2 dy >2 y@0
S 4
2
4 §S ·
¨ ¸ dy
y ©4¹
2 cos 2 T dT
³0
3·
§3
¨ cos 2T ¸ dT
4¹
©4
ª§
1629
16
4 y2
§ x ·º
« arctan ¨ ¸» dy
© y ¹¼ 0
¬y
S 4
³0
3³
2
2
1
2x
3 ª4
³1
1
4
4
e
e
39
§9
· § 9 39 ·
625 ¸ ¨ ¨ 25 ¸
16
©2
¹ © 2 16 ¹
³ 0 «¬¨© 20 y ª 2 5 y4 º
«5 y »
3 ¼0
¬
³ 0 ª¬ 12 1 y
dy
2 y y2
2
4
dx dy
x2 y 2
³1 ³ 0
ª
º
2 x3
2 y 2 x» dy
«10 x 3
¬
¼y
³ 0 >3xy@3 y2 6 y dy
2
4 y
26.
1
³ 0 ª¬ 12 x
x y dx dy
2
1
3
ª 2 64 x3
¬« 9
5
39 3 º
ª
³ 1 «¬9 y 4 y »¼ dy
5
³0
1 y 2
11
3y
2
1
S
sin 2 xº¼
0
5
10 2 x 2 2 y 2 dx dy
³0 ³0
º
¼» 0
1
2
6
ª
x3
1 2º
³ 1 «¬3x 3 4 xy »¼ dy
0
1 2·
§
2
¨ 3 x y ¸ dx dy
4 ¹
©
23.
32 1
1 x2
2
3
12 4 3 1
ª cos x ¬
4e 4 e 1
64 x3 dx
2
2y
2y
1
2
³ 0 >sin x sin x cos x@ dx
dx
x2
³ 4 ¬ª y
3 3º
ª
3
³ 1 «¬9 y 9 y 4 y »¼ dy
2
º
x»
¼0
eln 4 ln 3 eln 4 eln 3 1
xe x e x dx
1 x 2 º dx
¼0
5
22.
S 2
ª1
« 2 sin
¬
1
cos x dx
2
ª¬ y y cos x º¼ 0
ª¬ y 2e x º¼
1
64 x3 dy dx
S 2
³0
289
ln 3
1 cos x dy dx
1 x 2 dy dx
4
1
ª y2
º
« cos x» dx
2
¬
¼0
Iterated Integrals and Area in the Plane
3
³1
S
y
4
>S ln y@13
dy
S 2
1
ª
º
«T 2 sin 2T »
¬
¼0
§ 3 cos 2 T
3·
¸ dT
¨
2
2¹
©
S 4
3 º
ª3
« sin 2T 4T »
¬8
¼0
S 4
³0
S ln 3
S
2
3·
§3
¨ 1 cos 2T ¸ dT
2¹
©4
3 3S
8 16
© 2010 Brooks/Cole, Cengage Learning
290
29.
Chapter 14 Multiple Integration
S 2
sin T
³0 ³0
sin T
ª r2 º
«T »
¬ 2 ¼0
S 2
³0
T r dr dT
1 S
4³0
S 4
cos T
30.
³0 ³0
31.
³1 ³ 0
32.
33.
f
3
1x
f
³0 ³0
f
f
³1 ³1
S 4
³0
³0
³1
³0
f
3
³0
2
¬ª x arctan y¼º 0 dx
f
ª1
º
« y ln x» dy
¬
¼1
f
³1
32
ª cos 4 T º
«
»
4 ¼0
¬
cos3 sin T dT
ª 1º
« 2 x »
¬
¼1
1 f 1
dx
2 ³1 x2
S2
S 4
S 4
dT
f
3
f
1
dx dy
xy
cos T
ª¬r 3 sin T º¼
0
1x
x2
dy dx
1 y2
1 ªT 2
T
§1
·º
¨ cos 2T sin 2T ¸»
«
4¬ 2
2
©4
¹¼ 0
T T cos 2T dT
ª y2 º
« » dx
¬ 2 ¼0
f
³1
y dy dx
1
T sin 2 T dT
2
S 2
2
3r 2 sin T dr dT
S 2
³0
dT
0
1
2
1
8
4
º
1 ª§ 1 ·
Ǭ
1»
¸
4 «¬© 2 ¹
»¼
3
16
1
2
3
§S ·
x 2 ¨ ¸ dx
©2¹
ªS x 3 º
« ˜ »
¬ 2 3 ¼0
9S
2
ª1
1 º
« y f y 0 » dy
¬
¼
Diverges
34.
f
f
³0 ³0
xye
x2 y2
f
³0
dx dy
8
3
³ 0 > y@0 dx
3
8
³ 0 > x@0 dy
35. A
³ 0 ³ 0 dy dx
A
³ 0 ³ 0 dx dy
8
3
f
ª 1 x2 y2 º
« 2 ye
» dy
¬
¼0
8
³ 0 3 dx >3x@0
8
3
f
1 ye y 2
2
ª 1 e y 2 º
¬« 4
¼» 0
dy
1
4
y
24
8
>8 y@30
3
³ 0 8 dy
8
f
³0
24
6
4
2
x
2
2
3
³ 1 > y@1 dx
³1
3
2
³ 1 > x@1 dy
³ 1 dy
36. A
³ 1 ³ 1 dy dx
A
³ 1 ³ 1 dx dy
2
2
3
3
>2 x@12
2 dx
> y@13
3
2
2
2
4
6
8
y
3
2
1
x
1
4 x2
2
³0 ³0
37. A
4
³0 ³0
A
4 x
³ 0 > y@0
2
dy dx
4 y
2
2
dx
4 y
³ 0 > x@0 dy
4
dx dy
2
2
³0
4 x 2 dx
4
³0
4 y dy
ª
x3 º
«4 x »
3 ¼0
¬
³
4
0
4 y
3
16
3
12
1 dy
ª 2
« 3 4 y
¬
3 2º
4
»
¼0
2
8
3
16
3
y
y = 4 − x2
4
3
2
1
−1
x
1
2
3
© 2010 Brooks/Cole, Cengage Learning
Section 14.1
5
x 1
1
³2 ³0
38. A
12
x 1
³ 2 > y@0
5
dy dx
5
1
1 1 y 2
1
³ 0 ³ 2 dx dy ³ 1 2 ³ 2
A
1 1 y 2
³ 0 > x@2 dy ³ 1 2 > x@2
12
1
5
1
dx
x 1
5
³2
dx
ª2
¬
Iterated Integrals and Area in the Plane
y
5
x 1º¼
2
2
5
4
dx dy
3
12
³0
dy
§ 1
·
3 dy ³ ¨ 2 1¸ dy
12 y
©
¹
1
>3 y@0
4
³0 ³0
4
³0 ³0
A
2
x
2
y
2
dy dx
2
³ 0 > y@0
dx dy
8
3
4
2
x
2
4
³0
dx
44
x x−1
2
2
1
x
4
ª
8
«4 x x
3
¬
x x dx
1
y=
1
ª 1
º
« y»
y
¬
¼1 2
12
1
39. A
291
x2 º
»
2 ¼0
2
4
3
5
y
8
3
4
3
x )2
y = (2 −
2
Integration steps are similar to those above.
1
x
1
2x
4
³ 0 ³ x3 2 dy dx
40. A
42. A
³ 0 > y@x3 2 dx
4
4
³0
2x
y2 3
8
³0 ³ y 2
A
16 8
³0
dx dy
8
ª3 5 3
y2 º
« y »
4 ¼0
¬5
2
32
5
y·
§ 23
¨ y ¸ dy
2¹
©
−1
41. A
16
5
(4, 8)
1
9
3
y=x
x
9 y
3
§9
³1 ¨© y
2x 3
3
³0 ³0
dy dx 3
2x 3
3
2x
dx 3
³0
dx 5 x
5
³3 ³0
5 x
³ 3 > y@0
5
³3
5
3
³0
2
³0
5 y
2
3
ba
a2 x2
dy dx
³ 0 > y@0
a
ba
9
1 ln 9
2
a2 x2
2
1 cos 2T dT
y=5−x
dx
y
S 2
y = 23 x
8
S 2
b a
cos 2 T dT
a 2 x 2 dx
ab ³
0
a ³0
a sin T , dx
a cos T dT
ª ab §
1
·º
« 2 ¨T 2 sin 2T ¸»
¹¼ 0
¬ ©
4
b
y= b
a
S ab
1
3y ·
§
¨5 y ¸ dy
2¹
©
5y ·
¸ dy
2¹
y
3
> x@3 y 2 dy
§
5
a
³0 ³0
ab S
2 ³0
2
5 y
³ 0 ¨© 5 x
5
³ 0 ³ 3 y 2 dx dy
2
dx
5 x dx
1 2º
ª1 2 º
ª
« 3 x » «5 x 2 x »
¬
¼0 ¬
¼3
2
43. A
4
dy dx
6
·
y ¸ dy
¹
1
8
4
−2
1 2º
1 2º
ª
ª
«9 y 2 y » «9 ln y 2 y »
¬
¼0 ¬
¼1
x
1 2 3 4 5 6 7
y = 9x
(9, 1)
2
9 y
9 y dy (3, 3)
2
dy
1
y = x 3/2
4
³ 0 > x@ y dy ³1 > x@ y
³0
9
dx
x
6
dx dy
3
9
4
y
3
9
3
³ 0 x dx ³ 3
³ 0 ³ y dx dy ³ 1 ³ y
1
y = 2x
³ 0 > y@0
A
9 x
9
9 ln 9 ln 3
2
9
1 ln 9
2
A
6
5
4
3
2
1
dx
9
x
9
ª1 2 º
« 2 x » >9 ln x@3
¬
¼0
y
8
7
³ 0 > y@0 dx ³ 3 > y@0
3
16
5
3
32 16
5
9 x
9
dy dx
3
2 x x3 2 dx
4
ª 2 2 5 2º
«x 5 x »
¬
¼0
x
3
³ 0 ³ 0 dy dx ³ 3 ³ 0
2
a2 − x2
a
4
x
1
2
−1
2
5 2º
ª
«5 y 4 y »
¬
¼0
5
3
4
5
So, A
S ab.
b
a b b2 y 2
A
S ab
dx dy
³
³
0
0
4
4
So, A S ab. Integration steps are similar to those
above.
© 2010 Brooks/Cole, Cengage Learning
x
292
Chapter 14 Multiple Integration
44. A
y
2
4
2
³ 0 ³ y 2 dx dy ³ 2 ³ y 2 dx dy
y
dy 2
2
³0
4
§
y·
y
4
³0 ³0
f x, y dx dy, 0 d x d y, 0 d y d 4
4
y = 2x
4
³ 2 ¨© 2 2 ¸¹ dy
2
47.
y
4
³0 ³ x
3
f x, y dy dx
y
4
ª y2 º
ª
y2 º
« » «2 y »
4 ¼2
¬ 4 ¼0 ¬
1 4 3
2
1
y=x
2
3
x
1
2
3
4
2
2
A
2x
2
³0 ³ x
2
³0
dy dx
ª x2 º
« »
¬ 2 ¼0
2 x x dx
1
2
x
1
45. A
4 x2
1
³ 2 ³ x2
dy dx
4 x2
³ 2 > y@ x 2
1
dx
4 x x 2 dx
1
2 x x 2 dx
ª2 x ¬
y2
3
³0 ³
4 y
3
3
³0
46. A
2
³0 ³0
2
³0
4³
2³
4 x2
4
3
4
³0
3
−2
2
3
y = x2
4 y
x
−1
1
2
2
1
4 y
dx dy
x
1
> x@0 4 y dy
4
49.
4 y dy
3
32 3
º ª4 4 y
¼0 ¬ 3
2
³ 2 ³ 0
2
3
4 x2
f x, y dy dx, 0 d y d
2
32 4
º
¼3
³0 ³
9
2
4
4 y2
4 y2
y
y = 4 − x2
1
2
0
S 2
0
dx dy
y
3
4 x dx
S 2
4 x2 , 2 d x d 2
dy dx
2
−2
cos 2 T dT
x
−1
1
2
−1
1
1 cos 2T dT
ª2 T ¬
x
9
2
4 y dy 2 ³
ª1 y2 2 y ¬2
f x, y dy dx
3
1
y2
4
y=x+2
1 3º
x
3 ¼ 2
dy 2 ³
4 y
x2
1
dx dy 2³
y 2
³ 0 > x@
y d x d 2, 0 d y d 4
f x, y dx dy,
y
y
2
1 2
x
2
2
4
³0 ³0
(1, 3)
3
2
³ 2
4
³0 ³
3
2
1
³ 2
A
y=4−
48.
y
x2
2
1
2
x
S 2
sin 2T º
¼0
2 sin T , dx
1
S
2
50.
4 x2
2
³0 ³0
f x, y dy dx, 0 d y d 4 x 2 , 0 d x d 2
4 y
4
2 cos T dT ,
4 x2
³0 ³0
2 cos T
f x, y dx dy
y
A
2
³0 ³0
4³
S 2
0
4 y2
³0
dx dy
2³
cos 2 T dT
ª2 T ¬
y
2
1
2
S 2
sin 2T º
¼0
2 sin T , dy
4 y dy
2
S 2
0
4
3
1 cos 2T dT
2
1
S
2 cos T dT ,
x
−1
4 y2
2 cos T
1
2
3
4
−1
© 2010 Brooks/Cole, Cengage Learning
Section 14.1
51.
ln y
10
³1 ³ 0
f x, y dx dy, 0 d x d ln y, 1 d y d 10
ln 10
10
³ 0 ³ ex
Iterated Integrals and Area in the Plane
55.
1
2
2
³ 0 ³ 0 dy dx
1
³ 0 ³ 0 dx dy
293
2
y
f x, y dy dx
y
3
2
8
1
6
4
x
1
2
2
3
x
1
2
3
56.
52.
e x
2
³ 1 ³ 0
2
4
4
³ 1 ³ 2 dx dy
f x, y dy dx, 0 d y d e x , 1 d x d 2
2
³ 2 ³ 1 dy dx
2
y
4
e2
2
³ 0 ³ 1 f
x, y dx dy e
ln y
³ e2 ³ 1
f x, y dx dy
3
2
y
1
3
x
1
2
57.
x
−1
1
2
2
1 y2
1
³0 ³
1 y2
3
4
1
³ 1 ³ 0
dx dy
1 x2
S
dy dx
2
y
53.
1
1
³ 1 ³ x2
f x, y dy dx, x 2 d y d 1, 1 d x d 1
1
³0 ³
y
y
1
f x, y dx dy
x
−1
1
y
4
58.
3
2
³ 2 ³ 2
³ 2 ³ 2
4 x2
4 x2
4 y2
4 y2
2
dy dx
³ 2
dx dy
4S
4 x2 4S
4 x 2 dx
y
−2
x
−1
1
2
1
54.
S 2
cos x
³ S 2 ³ 0
1
f x, y dy dx, 0 d y d cos x, arccos y
³ 0 ³ arccos y
S
2
d xd
S
x
−1
2
1
−1
f x, y dx dy
y
2
59.
3
2
2
x
4 x
4
³ 0 ³ 0 dy dx ³ 2 ³ 0
dy dx
2
4 y
³0 ³ y
dx dy
4
y
1
2
−π
4
3
π
x
2
4
1
x
1
2
3
4
−1
© 2010 Brooks/Cole, Cengage Learning
294
60.
Chapter 14 Multiple Integration
x 2
4
³0 ³0
6 x
6
³4 ³0
dy dx x
dx 2
6
³4
6 x dx
y
4 2
6
6
5
2
6 y
2
4
³0
dy dx
³0 ³2y
2
³0
dx dy
ª
3y2 º
«6 y »
2 ¼0
¬
6 3 y dy
4
6
3
y= x
2
2
(4, 2) y = 6 − x
1
x
−1
−1
61.
2
1
2y
1
³ 0 ³ x 2 dy dx
³0 ³0
dx dy
1
1
3 y
³ 0 ³ y2
63.
y
1
2
3
4
1
³ 0 ³ x3
dx dy
5
x
6
dy dx
5
12
x= 3 y
y
2
x = y2
2
1
1
(1, 1)
x
1
2
x
1
62.
9
³0 ³
3
x
9
³0
dy dx
3
x dx
2 3 2º
ª
«3 x 3 x »
¬
¼0
3
y2
2
27 18
9
2
ª y3 º
« »
¬ 3 ¼0
dy
4
³0 ³
dx dy
4 x
4 x
32
3
dy dx
y
3
3
³0 y
dx dy
4 y2
³2 ³0
64.
9
³0 ³0
2
2
9
1
x
y
1
5
4
3
2
1
2
3
−1
x = 4 − y2
−2
x
y=
x
−1
1 2 3 4 5 6 7 8 9
−2
−3
−4
−5
65. The first integral arises using vertical representative rectangles. The second two integrals arise using horizontal representative
rectangles.
50 x 2
5
³0 ³ x
5
y
³0 ³0 x
y dx dy 2 2
50 y 2
5 2
³5 ³0
x 2 y 2 dy dx
³ 0 ª«¬13 x
x 2 y 2 dx dy
³ 0 13 y
5
15,625
18
5
5
2
50 x 2
dy 32
5 2
³5
15,625
S
18
1
3
13 x5 º dx
»¼
50 y 2
15,625
S
24
32
y=
50 − x 2
5
y 2 dy
(5, 5)
y=x
15,625
S
24
15,625
18
y
(0, 5 2 )
x
5
66. (a) A
(b) A
2
2y
³ 0 ³ y2
dx dy
2
³0
2
> x@ y2 dy
2
³0
2y
³ 0 ³ x 2 dy dx ³ 0 > y@x 2 dx
4
x
4
x
2 4
ª2 3 2
x º
« x »
4 ¼0
¬3
Integrals (a) and (b) are the same.
2 y y 2 dy
4
§
³ 0 ¨©
x 16
4
3
x·
¸ dx
2¹
4
3
ª 2
y3 º
«y »
3 ¼0
¬
4
8
3
4
3
y
4
x
2
x = 2y
y=
3
2
y= x
x = y2
(4, 2)
1
x
1
2
3
4
© 2010 Brooks/Cole, Cengage Learning
Section 14.1
67.
2
2
³0 ³ x x
y
2
³0 ³0
1 y 3 dy dx
x 1 y 3 dx dy
Iterated Integrals and Area in the Plane
70.
2
2
³0 ³ x e
y2
y
2
³0
ª
x2 º
3
« 1 y ˜ » dy
2 ¼0
¬
1 2
2³0
2
2
³ 0 ye
2
y2
y2 º
dx dy
y
dy
¼» 0
y2
dy
2
ª1 1 2
3
«2 ˜ 3 ˜ 3 1 y
¬
1
1
27 1
9
9
y
³ 0 ª¬«xe
1 y y dy
3
y
2
³0 ³0 e
dy dx
295
3 2º
ª 1 y2 º
« 2 e »
¬
¼0
2
»
¼0
1 4
1
e
e0
2
2
1§
1·
¨1 4 ¸ | 0.4908
e ¹
2©
26
9
3
y
2
(2, 2)
2
1
1
x
1
2
3
x
1
68.
4
³0 ³
2
x
3
dy dx
2 y3
y2
2
³0 ³0
2
³0
2
³0
2
3
dx dy
2 y3
y2
71.
ª 3x º
dy
«
3»
¬ 2 y ¼0
3y2
dy
2 y3
ln 10 ln 2
1
1
³ 0 ³ y sin x
2
1
1
³ 0 ª¬ y sin x
2
ªln 2 y 3 º
¬
¼0
1
³ 0 x sin x
ln 5
(4, 2)
2
y
1
2
3
1
2
³ 0 ³ 2 x 4e
y2
dx
1
x
69.
2
x
º dx
¼0
1
1
cos 1 1
2
2
1
1 cos 1 | 0.2298
2
3
1
2
dy dx
1
4
x
2
ª 1
2 º
« 2 cos x »
¬
¼0
y
y=
x
³ 0 ³ 0 sin x
dx dy
4
dy dx
2
y 2
2
³0 ³0
4e y dx dy
³ 0 ª¬«4 xe
y 2
y2 º
x
1
2
2
ªe y 2 º
«¬ »¼ 0
dy
¼» 0
e4 1
2
³ 0 2 ye
y2
dy
72.
2
4
³ 0 ³ y2
x sin x dx dy
4
³0 ³0
x
4
³ 0 ª¬ y
y
3
x sin x dy dx
x sin xº¼
0
x
dx
4
³ 0 x sin x dx
2
1
>sin x y
x
1
2
x cos x@0
4
sin 4 4 cos 4 | 1.858
4
3
3
(4, 2)
2
1
x
1
2
3
4
© 2010 Brooks/Cole, Cengage Learning
296
73.
74.
75.
Chapter 14 Multiple Integration
2
2x
1
2y
4
y
³ 0 ³ x2
x3 3 y 2 dy dx
³0 ³ y
2
³0 ³0
dx dy
x 1 y 1
ln 5
2
8
| 2.590
x1 3
³ 0 ³ x2 32
(b)
x1 3
4 2 y œ x2
x
sin 2 sin 3
| 0.408
2
3
sin x y dx dy
y3 œ y
77. (a) x
| 15.848
1664
105
32 y œ y
x 2 y xy 2 dy dx
(c) Both integrals equal
76.
a
ax
³0 ³0
a4
6
x 2 y 2 dy dx
x2
32
67,520
| 97.43.
693
y
4
x = y3
(8, 2)
2
x
2
−2
4 x2 œ x
78. (a) y
(b)
2
2
³0 ³
4 y2
6
8
x = 4 2y
4 y2
x2
œ x
4
4
y
4
16 4 y
xy
dx dy x y2 1
2
3
2
³2 ³0
xy
dx dy x y2 1
2
4
³3 ³0
16 4 y
xy
dx dy
x2 y 2 1
y
4
3
2
1
x
1
2
(c) Both orders of integration yield 1.11899.
2
4 x2
2
2
79.
³0 ³0
80.
³0 ³ x
81.
³0 ³0
82.
e xy dy dx | 20.5648
16 x3 y 3 dy dx | 6.8520
2S
1 cos T
S 2
1 sin T
³0 ³0
6r 2 cos T dr dT
15 T r dr dT
15S
2
45S
32
2
84. A region is vertically simple if it is bounded on the left
and right by vertical lines, and bounded on the top and
bottom by functions of x. A region is horizontally simple
if it is bounded on the top and bottom by horizontal lines,
and bounded on the left and right by functions of y.
85. The region is a rectangle.
86. The integrations might be easier. See Exercises 59–62.
135
| 30.7541
8
87. True
88. False, let f x, y
x.
83. An iterated integral is integration of a function of several
variables. Integrate with respect to one variable while
holding the other variables constant.
© 2010 Brooks/Cole, Cengage Learning
Section 14.2
Double Integrals and Volume
297
Section 14.2 Double Integrals and Volume
For Exercises 1–4, 'xi
1, 1
2 2
3 1
,
2 2
,
,
'yi
5 1
,
2 2
,
7, 1
2 2
2
3
4
1 and the midpoints of the squares are
1, 3
2 2
,
,
3 3
,
2 2
,
5 3
,
2 2
7, 3
2 2
,
.
y
4
3
2
1
x
1
1. f x, y
8
¦f
x y
xi , yi 'xi 'yi
1 2 3 4 23 4 5
24
i 1
2
4
2
³0 ³0
2. f x, y
f
¦f
4
³0
x y dy dx
1 2
x y dy dx
2
2
3. f x, y
8
¦f
2 x 2 dx
2
4
³0
ª x2 y2 º
«
» dx
¬ 4 ¼0
4
4
³0 x
2
x3 º
»
3 ¼0
dx
2
³0 ³0
8
¦f
i 1
4
2
³0
ª 2
y3 º
«x y » dx
3 ¼0
¬
4
4
³0 ³0
³0
8·
§ 2
¨ 2 x ¸ dx
3¹
©
1
x 1 y 1
4
4
4
4
4
4
4
4
9 15
21 27 15
25 35
45
dy dx
4
ª 2 x3
8x º
«
»
3 ¼0
¬ 3
160
3
7936
| 1.680
4725
2
4
ª 1
º
« x 1 ln y 1 » dx
¬
¼0
4
ln 3
dx
x 1
³0
³0
5.
4
52
1
x 1 y 1
xi , yi 'xi 'yi
³0 ³0
4
x 2 y 2 dy dx
4. f x, y
21
64
| 21.3
3
2 10
26 50 10 18 34 58
4
4
4
4
4
4
4
4
xi , yi 'xi 'yi
2
24
x2 y 2
i 1
4
4
2
¬ª x 2 x¼º 0
1
9
25
49
3
27
75 147
16 16 16
16 16 16
16
16
i 1
4
4
³0
1 2
x y
2
xi , yi 'xi 'yi
³0 ³0
ª
y2 º
« xy » dx
2 ¼0
¬
4
ª¬ln 3 ˜ ln x 1 º¼ 0
ln 3 ln 5 | 1.768
f x, y dy dx | 32 31 28 23 31 30 27 22 28 27 24 19 23 22 19 14
400
Using the corner of the ith square furthest from the origin, you obtain 272.
6.
2
2
³0 ³0
f x, y dy dx | 4 2 8 6
20
© 2010 Brooks/Cole, Cengage Learning
298
7.
Chapter 14 Multiple Integration
2
1
³0 ³0
2
³ 0 ª¬ y 2 xy 1 2 x 2 y dy dx
1
y 2 º¼ dx
0
2
³0
2 2 x dx
2
ª¬2 x x 2 º¼
0
8
y
3
2
1
x
1
8.
S
2
S 2
³0 ³0
3
sin 2 x cos 2 y dy dx
S 2
S
ª1
2
« 2 sin
¬
S
1
§S ·
sin 2 x¨ ¸ dx
2
©2¹
³0
³0
1
§
·º
x¨ y sin 2 y ¸» dx
2
©
¹¼ 0
S S
1 cos 2 x dx
8 ³0
S
ªS §
1
·º
« 8 ¨ x 2 sin 2 x ¸»
¹¼ 0
¬ ©
S2
8
y
3
2
1
x
1
9.
6
3
³0 ³ y 2
2
3
6
³0
x y dx dy
3
ª1 2
º
« 2 x xy» dy
¬
¼y 2
6
³0
5 2·
§9
¨ 3 y y ¸ dy
8 ¹
©2
6
3 2
5 3º
ª9
« y 2 y 24 y »
¬2
¼0
36
y
(3, 6)
6
4
2
x
2
4
6
y
10.
y
4
³0 ³ 1 2 y x
4
ª x3 y 2 º
dy
»
3 ¼12 y
³ 0 «¬
2 2
y dx dy
4
§ y7 2
y5 ·
¸ dy
3
24 ¹
³ 0 ©¨
4
ª 2 y9 2
y6 º
«
»
144 ¼ 0
¬ 27
1024 256
27
9
256
27
y
4
(2, 4)
3
2
1
x
1
11.
a
³ a ³ 2
a2 x2
a2 x2
3
4
x y dy dx
a
³ a
1 2º
ª
« xy 2 y »
¬
¼
a2 x2
dx
a2 x2
a
³ a
2 x a 2 x 2 dx
ª 2 2
2
« 3 a x
¬
3 2º
a
»
¼a
0
y
a
−a
a
x
−a
© 2010 Brooks/Cole, Cengage Learning
Section 14.2
12.
1
0
³ 0 ³ y 1 e
x y
1 y
1
³0 ³0
dx dy 1
³ 0 ª¬e
e x y dx dy
1
³0
x y 0
º¼
y 1
dy 1
³ 0 ª¬e
x y 1 y
º¼
0
Double Integrals and Volume
y
dy
2
1
ª¬ey 12 e 2 y 1 º¼
0
e e 2 y 1 dy
e e 1
1
2
y = −x + 1
y=x+1
x
−1
13.
5
3
³ 0 ³ 0 xy dx dy
3
5
³ 0 ³ 0 xy dy dx
3
³ 0 ª¬ 12 xy
2º
299
1
y
5
³
3
25
2 0
¼ 0 dx
5
3
ª 25 x 2 º
¬ 4 ¼0
x dx
225
4
4
3
2
1
x
1
14.
S 2
S
S
³ 0 ³ S sin x sin y dx dy
S 2
³ S ³ 0
2
4
3
5
y
sin x sin y dy dx
S
S 2
³ S >sin x cos y@0
5π
2
S
³ S sin x dx
dx
2π
0
3π
2
π
x
− 3π − π − π
2
2
15.
2
y
³1 ³1
y
dx dy x2 y2
4
2
³ 2 ³ y 2 x2
y
dx dy
y2
2
2x
³1 ³ x
π
2
3π
2
π
y
dy dx
x2 y 2
y
2x
1 2ª
ln x 2 y 2 º¼ dx
³
¬
1
x
2
2
1
ln 5 x 2 ln 2 x 2 dx
2 ³1
4
4 x
³0 ³0
4 y
4
³0 ³0
xe y dy dx
4
y 4 x
º¼
0
dx
4
³0
x=2
2
x=1
1
ª1 § 5 · º
« 2 ¨ ln 2 ¸ x»
¹ ¼1
¬ ©
xe y dx dy
y=x
1 5
ln
2 2
x
1
2
3
4
y
4
For the first integral, you obtain:
³ 0 ª¬xe
y = 2x
3
2
1 5 2
ln
dx
2 2 ³1
16.
4
3
xe 4 x x dx
2
4
ª 4 x
x2 º
1 x »
«e
2 ¼0
¬
17.
4
4 y
³ 3 ³ 4 y
2 y dx dy
4 x2
1
³ 0 ³ 4 x
1
³ 0 ª¬ y
2
5 8 e 4
1
e 4 13.
x
1
2 y dy dx
º¼
dx
4 x
1
2
2
4 x º dx
¼»
4
2
y = 4 − x2
3
(1, 3)
4
y=4−x
1
x
1
1
ª
º
x
«3x3 4x2 »
5
¬
¼0
5
4
2
³ ª¬16 8 x x 16 8 x x º¼ dx
0
2
3
y
4 x2
1
³ ª 4 x2
0 ¬
«
2
3
4
6
5
© 2010 Brooks/Cole, Cengage Learning
300
18.
Chapter 14 Multiple Integration
2
y
4
³ 0 ³ y2 1 4
³0 ³0
dx dy
x2
x
y
y
dy dx
1 x2
1 4 ª y2 º
«
»
2 ³ 0 ¬1 x 2 ¼ 0
4
x
4
ª1
2 º
« 4 ln 1 x »
¬
¼0
1 4 x
dx
2 ³ 0 1 x2
dx
3
1
ln 17
4
y=
x
2
1
x
1
19.
4
3x 4
³0 ³0
x dy dx 5
³4 ³0
25 x 2
25 y 2
3
³0 ³ 4y 3
x dy dx
3
³ 0 ª¬ 12 x
³
3
25
18 0
2
2
3
4
y
x dx dy
5
25 y 2
º¼
4y 3
25 − y 2
x=
4
dy
ª 25 9 y ¬ 18
9 y 2 dy
x= 4y
3
3
3
º
¼0
1 y3
3
(4, 3)
2
25
1
x
1
20.
4 y2
2
³0 ³
x 2 y 2 dy dx
1 3º
ª 2
³ 2 «¬x y 3 y »¼ 0
2
ª 2
³ 2 «¬x
ª x
2
« 4 x
¬ 4
³0 ³0
32
4
x=−
4 x2
3
x=
4 − y2
1
3 2º
» dx
¼
−2
4
³0
ª y2 º
« » dx
¬ 4 ¼0
4
³0
dx
4
x
−1
x·
1§
1ª
2
x 4 x2
¨ x 4 x 4 arcsin ¸ 2©
2 ¹ 12 ¬«
2
y
dy dx
2
4 − y2
dx
1
4 x2 4 x2
3
2
2
1
32
2
2
x ºº
4 x 2 24 arctan »»
2 ¼¼ 2
6x
y
2
³0 ³0
23. V
³0
4
4 x y dx dy
2
ª
º
x2
xy» dy
«4 x 2
¬
¼0
§
³ 0 ¨© 4 y 3
2
·
y
y 2 ¸ dy
2
¹
2
x
3
4
8
22. V
4
2
³0 ³0
y
2
ª 2
y3
y3 º
«2 y »
6
3 ¼0
¬
1
2
4S
y
2
y
1
5
y
4 x2
³2 ³0
4
4
3
x 2 y 2 dx dy
4 y2
2
21. V
2
2
1
y=x
x
8 8
6 3
1
4
2
6 2 y dy dx
4
³ 0 ª¬6 y 2
y 2 º¼
0
dx
4
³ 0 8 dx
32
2
³ 0 4 x dx
4 dy dx
2
2 x2 º
»¼ 0
8
y
y
4
x
2
³0 ³0
24. V
2
y=x
3
1
2
1
x
x
1
2
3
1
2
4
© 2010 Brooks/Cole, Cengage Learning
Section 14.2
6
³0 ³0
25. V
2 3 x 4
§ 12 2 x 3 y ·
¨
¸ dy dx
4
©
¹
6
³0
6
Double Integrals and Volume
y
2 3 x 4
ª1 §
3 2 ·º
« 4 ¨12 y 2 xy 2 y ¸»
¹¼ 0
¬ ©
§1
³ 0 ¨© 6 x
2
·
2 x 6 ¸ dx
¹
301
dx
5
y = − 23 x + 4
4
6
ª1 3
º
2
« x x 6 x»
¬18
¼0
3
12
2
1
x
2
1
3
4
5
6
−1
2 x
2
³0 ³0
26. V
³0
³0
2
1
2 x
2
2
2 x
ª
y2 º
«2 y xy »
2 ¼0
¬
2
2 x y dy dx
2
1
3º
x2 »
6
¼0
dx
y
dx
2
4
3
y=2−x
1
x
1
y
1
³0 ³0
27. V
1 xy dx dy
y
y
ª
x2 y º
«x » dy
2 ¼0
¬
1
³0
2
1
1§
y3 ·
³ 0 ©¨ y 2 ¹¸ dy
ª y2
y4 º
«
»
8 ¼0
¬2
3
8
1
y=x
x
1
y
2
³0 ³0
28. V
y
4 y 2 dx dy
2
2
2
³0
4y y
3
ª 2
y4 º
«2 y »
4 ¼0
¬
dy
4
1
y=x
x
1
2
f
f
f
1
29. V
³0 ³0
30. V
³0 ³0
31. V
4³
32. V
³0 ³0
33. V
³ 0 ³ 0 xy dy dx
f
f
e x y
2
³0
2
0
1
x
1
x
2
x 1
2
8 x2
y 1
dy dx
2
dy dx
f
³0
f
³0
ª
º
1
«
» dx
2
«¬ x 1 y 1 »¼ 0
f
ª2e x y 2 º dx
¬
¼0
1
2e x 2 dx
³0
f
1
x 1
f
ª¬4e x 2 º¼
0
2
dx
ª
1 º
«
»
x
1 »¼ 0
¬«
1
4
32 2 S
3
8 x 2 y 2 dy dx
1 x 2 dy dx
f
³0
f
1
3
³ 0 ª¬ 12 xy
2º
x
¼ 0 dx
1
1
x3
2 0
³
dx
1
ª1 x4 º
¬8 ¼ 0
1
8
y
1
y=x
x
1
© 2010 Brooks/Cole, Cengage Learning
302
Chapter 14 Multiple Integration
x
5
³ 0 ³ 0 x dy dx
34. V
³ 0 > xy@0 dx
5
5
³0 x
x
5
ª1 x3 º
¬3 ¼0
2
2
4
³0 ³0 x
35. V
2
³0
dx
2
dy dx
2
ª 4 x3 º
«
»
¬ 3 ¼0
125
3
y
2
4
³ 0 4x
ª¬ x 2 yº¼ dx
0
2
dx
32
3
y
5
y=x
4
4
3
3
2
2
1
1
x
1
2
4
3
5
x
−1
8³
36. V
r
0
³0
r 2 x2
1
r 2 x 2 y 2 dy dx
y
º
»
r 2 x2 ¼0
y
r 2 x 2 y 2 r 2 x 2 arcsin
§S · r
4¨ ¸ ³ r 2 x 2 dx
©2¹ 0
r
ª § 2
1 3 ·º
«2S ¨ r x 3 x ¸»
©
¹¼ 0
¬
r 2 x2
2³
0
x
³0
1 x 2 dy dx
1
2³ x 1 x 2 dx
0
r2 − x2
r
dx
4S r 3
3
x
r
37. Divide the solid into two equal parts.
V
3
y=
r ª
4³ « y
0
¬
1
2
y
x
y=x
2 ³ ª y 1 x 2 º dx
0 ¬
¼0
1
ª 2 1 x 2
¬« 3
32 1
º
¼» 0
1
2
3
x
1
4 x2
2
³0 ³0
38. V
2
³0
2
³0
4 x 2 dy dx
y
4
4 x 2 4 x 2 dx
16 8 x 2 x 4 dx
3
5 2
ª
8x
x º
»
«16 x 3
5 ¼0
¬
3
32 64 32
3
5
y = 4 − x2
2
256
15
1
x
1
2
³0 ³0
39. V
2
³0
4 x2
x y dy dx
2
³ 0 ª¬xy 12 y
2º
¼0
4 x2
dx
x 4 x 2 2 12 x 2 dx
ª 1 4 x 2
«¬ 3
32
2
³0
2
2 x 16 x3 º
»¼ 0
y
f
2
³0 ³0
40. V
16
3
S
2
2
3
1
dy dx
1 y2
4
f
³ 0 >arctan y@0
2
dx
2
ªS x º
«2»
¬ ¼0
dx
S
y
2
y=
2
4−x
2
1
1
x
1
2
x
1
2
© 2010 Brooks/Cole, Cengage Learning
Section 14.2
41. V
2
2³
0
2³
0
2
³0
³0
1 x 1 2
ª¬4 x 2 y 2 º¼ >4 2 x@ dy dx
1 x 1 2
Double Integrals and Volume
2 2 y 1 2
2
³0 ³
45. V
2 2 y 1 2
303
ª4 y x 2 2 y 2 º dx dy
¬
¼
y
2 x x 2 y 2 dy dx
x2
+ (y − 1) 2 = 1
2
3
y
(x − 1) 2 + y 2 = 1
1
1
−2
x
x
−1
1
2
−1
1
−1
46. V
2³
42. V
2
0
³0
1 x 1 2
3
4³
ª2 x x 2 y 2 º dy dx
¬
¼
9 x2
3
³ 3 ³ 0
y
9 x2
9 x2
³0
ª¬18 x 2 y 2 º¼ ª¬ x 2 y 2 º¼ dy dx
18 2 x 2 2 y 2 dy dx
y
(x − 1) 2 + y 2 = 1
x2 + y2 = 9
1
2
1
x
1
−2 −1
x
1
−1
2
−2
−1
4³
43. V
2
0
³0
4 x2
x 2 y 2 dy dx
y
47. z
V
x 2 + y2 = 4
1
0
9 x2
³0
9
9 y
2
0.5 x 1
³0 ³0
49. V
³0 ³0
50. V
³0 ³0
0
9 x 2 y 2 dy dx
9 y dx dy
81S
2
81
2
1
−1
S
5
³0 ³0
44. V
3
4³
48. V
x
−1
9 x2 y 2 , z
sin 2 x dx dy
y
16
4
y
2
dy dx | 1.2315
1 x2 y 2
ln 1 x y dx dy | 38.25
51. f is a continuous function such that
0 d f x, y d 1 over a region R of area 1. Let
5
4
f m, n
the minimum value of f over R and
3
f M, N
2
1
x
1
2
3
4
f m, n
5
Because
the maximum value of f over R. Then
³ R ³ dA d ³ R ³ f
³ R ³ dA
x, y dA d f M , N
³ R ³ dA.
1 and
0 d f m, n d f M , N d 1, you have
0 d f m, n 1 d
So, 0 d
³R ³ f
³R ³ f
x, y dA d f M , N 1 d 1.
x, y dA d 1.
© 2010 Brooks/Cole, Cengage Learning
304
52.
Chapter 14 Multiple Integration
z
x
y
z
a
b
c
1
c
x
y·
§
c¨1 ¸
a
b¹
©
z
³R ³ f
V
R
a ª
xy
y º
c³ « y »
0
2b ¼ 0
a
¬
x
y·
§
c¨1 ¸ dy dx
a
b¹
©
bª¬1 x a º¼
a
³0 ³0
x, y dA
2 b ¬ª1 x a ¼º
a
dx
b
y
x
2
a ª §
x · xb §
x · b2 §
x· º
1 ¸
1 ¸ » dx
c ³ «b¨1 ¸ ¨
¨
0
a¹
a©
a ¹ 2b ©
a ¹ »¼
¬« ©
a
2
3
ª ab §
x·
x 2b
x3b
ab §
x· º
2 1
c « ¨1 ¸ ¨
¸ »
2a
3a
6©
a¹
a ¹ »¼
«¬ 2 ©
0
53.
1
12
³0 ³ y 2 e
x2
2x
12
³0 ³0
dx dy
12
³0
2 xe
ª§ ab
ab ·º
ab · § ab
c «¨ ¸ ¨
¸
6 ¹»¼
3¹ © 2
© 2
2
y
e x dy dx
x2
y = 2x
ªe
¬«
dx
abc
6
12
x2
º
¼» 0
e 1 4 1
1
1 e 1 4 | 0.221
1
2
x
1
2
54.
ln 10
10
³ 0 ³ ex
1
dy dx
ln y
y
1
dx dy
ln y
ln y
10
³1 ³ 0
10
8
ln y
10
³1
10
³1
55.
2
³2 ³
4 x2
ª x º
«
» dy
¬ ln y ¼ 0
y = ex
4
> y@1
2
9
x
1
2
³ 2 ³
4 y 2 dy dx
4 x2
6
10
dy
4 y2
4 y2
2
4 y 2 dx dy
3
4
2
³ 2
2
2
³ 2
1
ª
2 y3 º
«8 y »
3 ¼ 2
¬
2 4 y 2 dy
5
ªx 4 y 2 º
¬
¼
y
4 y2
4 y2
x2 + y2 = 4
dy
1
16 · §
16 ·
§
¨16 ¸ ¨ 16 ¸
3¹ ©
3¹
©
3
³0
1
³ y 3 1 x 4 dx dy
1
1
3x
³0 ³0
1
1
dy dx
1 x4
3x
dx
x4
³0 1 3
2
57.
1
arccos y
³0 ³0
1
³0
y
3x
ª y º
«1 x 4 » dx
¬
¼0
3
y = 3x
(1, 3)
1
3
º
arctan x 2 »
2
¼0
2
1
3S
8
x
2
3
sin x 1 sin 2 x dx dy
S 2
cos x
³0 ³0
S 2
³0
§S ·
¨ ¸
©4¹
y
sin x 1 sin 2 x dy dx
1 sin 2 x
12
1
−1
64
3
56.
x
−1
sin x cos x dx
2
ª1 ˜
¬« 2
2
3
1 sin 2 x
32 S 2
º
¼» 0
y = cos x
1 ª2
3¬
2 1º¼
1
π
2
π
x
© 2010 Brooks/Cole, Cengage Learning
Section 14.2
58.
2
2
³ 0 ³ 1 2 x2
y cos y dx dy
2
2
³0
305
y
2y
2
³0 ³0
y cos y dy dx
Double Integrals and Volume
2y
(2, 2)
2
2 ³ y cos y dy
y cos y dy
0
2 >cos y y sin y@0
1
2 >cos 2 2 sin 2 1@
2
y = 12 x 2
x
1
59. Average
1 4
8³0
60. Average
1 5 3
2 xy dy dx
15 ³ 0 ³ 0
2
³0
1 5
9 x dx
15 ³ 0
61. Average
1 4
2 x dx
8³0
x dy dx
5
2
1 325 250
100 x 0.6 y 0.4 dx dy
1250 ³ 300 ³ 200
65. Average
250
1 325 ª
x1.6 º
100 y 0.4
«
» dy
³
1250 300 ¬
1.6 ¼ 200
15
2
128,844.1 325 0.4
y dy
1250 ³ 300
1 2 2 2
x y 2 dx dy
4³0 ³0
2
1 2 ª x3
2º
« xy » dy
4³0 ¬ 3
¼0
2
ª1 § 8
2 3 ·º
« 4 ¨ 3 y 3 y ¸»
¹¼ 0
¬ ©
62. Average
4
ª x2 º
« »
¬ 8 ¼0
3
1 5
ª xy 2 º dx
15 ³ 0 ¬ ¼ 0
1 ª9 x2 º
«
»
15 ¬ 2 ¼ 0
1
12
325
³0 ³0
1
8
3
0
4
³ ³0
20 4 x 2 y 2 dy dx
28 q
C
3
1 224
8 3
67. See the definition on page 994.
2³ >ln 2 x ln x@ dx
2 ³ ª¬ln x y º¼ 0 dx
0
2 ³ ln 2 dx
1 2
8 0
66. Average
1
x
1
ª y1.4 º
103.0753«
| 25,645.24
»
¬ 1.4 ¼ 300
1 2 §8
2·
¨ 2 y ¸ dy
4³0 ©3
¹
1
dy dx
x y
x
1
2
0
2> x ln 2@0
1
68. The value of
2 ln 2
³R ³ f
x, y dA would be kB.
69. (a) The total snowfall in the county R.
y
(b) The average snowfall in R.
y=x
1
(1, 1)
70. Part (b) is invalid. You cannot have the variable of
integration y as a limit of integration.
x
71. No, the maximum possible value is Area 6
6S .
72. The second is integrable. The first contains
63. Average
1 1 1 x y
e
dy dx
1 2³0 ³ x
1
1 º
ª
2 «e x 1 e 2 x »
2 ¼0
¬
e 2e 1
2
64. Average
1
S
1
S
S 2 ³0
S
³0
1
2³ e x 1 e 2 x dx
0
1
1º
ª
2 «e 2 e 2 e »
2
2¼
¬
e 1
1
S
1
S
S 2 ³0
sin y 2 dy which does not have an elementary
antiderivation.
73. f x, y t 0 for all x, y and
2
f
f
³ f ³ f f
x, y dA
sin x y dy dx
P 0 d x d 2, 1 d y d 2
1
>2 sin x@S0
2
2
³ 0 ³1
2
cos x S cos x dx
S2
2
5
S
2 cos x dx
5
³ 0 ³ 0 101 dy dx
³ 0 15 dx
ªcos x y º¼ 0 dx
S 2 ³0 ¬
S 2 ³0
³
1
10
³ 0 101 dx
1
dy dx
1.
5
0
© 2010 Brooks/Cole, Cengage Learning
306
Chapter 14 Multiple Integration
74. f x, y t 0 for all x, y and
f
f
³ f ³ f f
2
2
³0 ³0
x, y dA
1
2
³ 0 ³1
P 0 d x d 1, 1 d y d 2
x
dx
2
1 3x
³ 0 8 dx
1
xy dy dx
4
1
xy dy dx
4
2
³0
1
3
.
16
75. f x, y t 0 for all x, y and
f
f
³ f ³ f
f x, y dA
P 0 d x d 1, 4 d y d 6
6
1ª
y2 º
«9 y xy » dx
27 ¬
2 ¼3
3
6
1
9 x y dy dx
27
³0
1
6
1
9 x y dy dx
27
³ 0 27 4 x
³0 ³3
³0 ³4
3
1
2
3
³0
3
ª x x2 º
« »
¬ 2 18 ¼ 0
§1 1 ·
¨ x ¸ dx
©2 9 ¹
1
7
.
27
dx
76. f x, y t 0 for all x, y and
f
f
³ f ³ f f
x, y dA
P 0 d x d 1, x d y d 1
f
f
³0 ³0
1
1
³0 ³ x e
e x y dy dx
x y
f
³0
1
³ 0 ª¬e
dy dx
1
ª 1 2 x
º
e x 1 »
« 2 e
¬
¼0
V | 4 3 6 7 3 2 100
³0
x y 1
º¼
x
0
dx
1
³0
1 2
1
e e 1 2
2
77. Divide the base into six squares, and assume the height
at the center of each square is the height of the entire
square. So,
f
b
lim ªe x y º¼ dx
b of ¬
79.
e x dx
lim ªe x º¼
b of ¬
b
1
0
e 2 x e x 1 dx
2
1 1
e 1 | 0.1998.
2
1
2
³ 0 ³ 0 sin
x y dy dx
m
4, n
8
(a) 1.78435
2500 m3 .
(b) 1.7879
z
80.
7
(15, 5, 6)
(25, 5, 4)
(15, 15, 7)
(5, 5, 3)
2
4
³ 0 ³ 0 20e
x3 8
dy dx
m
10, n
x dx dy
m
4, n
8
m
6, n
4
20
(a) 129.2018
(5, 15, 2)
(b) 129.2756
20
y
30
x
81.
(25, 15, 3)
6
2
³4 ³0
y cos
(a) 11.0571
78. Sample Program for T1-82:
Program: DOUBLE
: Input A
: Input B
: Input M
: Input C
: Input D
: Input N
: 0 o V
(b) 11.0414
82.
4
2
³1 ³1
x 3 y 3 dx dy
(a) 13.956
(b) 13.9022
83. V | 125
Matches d.
z
(4, 0, 16) 16
(4, 4, 16)
: B A M o G
: D C N o H
: For (I, 1, M, 1)
: For (J, 1, N, 1)
: A 0.5G 2I 1 o X
: C 0.5H 2J 1 o Y
: V sin
X Y uGuH o V
(0, 4, 0)
(4, 0, 0)
5
5
x
y
(4, 4, 0)
84. V | 50
z
Matches a.
4
3
: End
: End
: Disp V
x
3
3
y
© 2010 Brooks/Cole, Cengage Learning
Section 14.2
85. False
8³
V
1
0
³0
2
1 x 2 y 2 dx dy
xy
e x e 2 x
x
ª 1 xy º
« x e »
¬
¼1
dy
307
So,
f
³0
86. True
1
³0 f
87. Average
2
³1 e
88.
1 y2
Double Integrals and Volume
³
1
0
t
³0 e
t2
x
1
³ 0 ³1 e
x dx
1
³
dt dx
1
1
0
1
³ xe
t2
f
2
2
f
³ 0 ³1 e
³1 ³ 0
dt dx
12 e 1
1
2
³1
1 e
>ln y@12
ln 2.
9 x 2 y 2 is a paraboloid opening downward
89. z
1
e xy dx dy
1
dy
y
2
³1
t
dx dy
ª e xy º
«
» dy
¬ y ¼0
2
0
xy
f
2
³ tet dt
dx dt
ª 1 et 2 º
«¬ 2 »¼ 0
t2
e x e 2 x
dx
x
with vertex 0, 0, 9 . The double integral is maximized if
z t 0. That is,
x
^ x, y : x 2
R
1
y 2 d 9`.
ª
«The maximum value is
¬
³R ³
81S º
.
2 »¼
9 x 2 y 2 dA
x 2 y 2 4 is a paraboloid opening upward with vertex
90. z
0, 0, 4 . The double integral is minimized if z d 0. That is,
^ x, y : x 2
R
y 2 d 4`.
[The minimum value is 8S .]
91.
2
³0
2
2
³0
y
Sx
³0 ³ x
tan 1 S x tan 1 x dx
ª x º
«1 y 2 » dy ¬
¼y S
2S
³2
1
dy dx
1 y2
2
ª x º
«1 y 2 » dy
¬
¼y S
2
1
y
³0 ³ y S 1 2
ª
y
³ 0 «¬1 y
2
y2
dx dy 2S
2
y S º
dy 1 y 2 »¼
2S
³2
3
³0 ³0
9 y2
9 x 2 y 2 dx dy
a
b
³0 ³0 e
^
max b 2 x 2 , a 2 y 2
dx dy
ª 2
y S º
«1 y 2 1 y 2 » dy
¬
¼
y
7
5
4
3
2
y=x
1
−2 −1
x
1
2
3
4
5
9 in the first octant.
` dy dx.
y
Divide the rectangle into two parts by the diagonal line ay
bx. On lower triangle,
b
b 2 x 2 t a 2 y 2 because y d x.
a
a bx a
b ay b
a bx 2 2
b ay 2 2
b2 x2
a2 y2
b x
a y
I
³ 0 ³ 0 e dy dx ³ 0 ³ 0 e dx dy ³ 0 a e dx ³ 0 b e dy
1 ª b2 x 2 º a
1 ª a2 y2 º b
e
e
«
»
»¼ 0
¼0
2ab ¬
2ab «¬
y = πx
6
9S
2
because this double integral represents the portion of the sphere x 2 y 2 z 2
1 4
9S
3
V
˜ S 3
8 3
2
93. Let I
y2
2S
2
ª1 §
1·
1
ª
1
2 º
2 º
« 2 ¨1 S ¸ ln 1 y » «2tan y 2S ln 1 y »
¼2
¹
¬ ©
¼0 ¬
1§
1·
1
1
1
ln 1 4S 2 2 tan 1 2 ln 5
¨1 ¸ ln 5 2 tan 2S 2©
2S
2S
S¹
1
1
ln 5 2 tan 1 2S 2 tan 1 2 ln 1 4S 2 | 0.8274
2
2S
92.
1
³2 ³ y S 1 2 2
1 ª b2 a 2
e
1 e a b 1º»
¼
2ab «¬
e
a 2b 2
(a, b)
b
ay = bx
a
x
1
ab
© 2010 Brooks/Cole, Cengage Learning
308
Chapter 14 Multiple Integration
94. Assume such a function exists.
1
1 O ³ u y u y x dy; O !
u x
x
1
³0 u x
D
1
1
1
³ 0 dx O ³ 0 ³ x u
dx
1
,0 d x d 1
2
y u y x dy dx
Change the order of integration.
1
³0 u x
D
dx
1 O³
Hold y fixed and let z
1
0
y
³0 u
y x, dz
dx.
1 O ³ u y ª³ u z dz º dy
0
¬« y
¼»
1
D
y
³0 u z
Let f y
0
1
y
1 O ³ u y ª³ u z dzº dy
»¼
0
¬« 0
dz. Then f c y
u y, f 0
1
1
1 O ³ f c y f y dy
D
1
y
1 O ³ u y ª³ u y x dxº dy
»¼
0
¬« 0
y u y x dx dy
0
ª f y 2º
»
1 O«
«¬ 2 »¼ 0
0, f 1
D.
1
2
2º
ª1
1 O« f 1 f 0 »
2
2
¬
¼
1
1 O D2
2
OD 2 2D 2 0.
For D to exist, the discriminant of this quadratic must be nonnegative.
4 8O t 0 Ÿ O d
b 2 4ac
But, O !
1
2
1
, a contradiction.
2
Section 14.3 Change of Variables: Polar Coordinates
1. Rectangular coordinates
2. Polar coordinates
3. Polar coordinates
4. Rectangular coordinates
5. R
^ r, T
: 0 d r d 8, 0 d T d S `
6. R
^ r, T
: 0 d r d 4 sin T , 0 d T d S `
7. R
^ r, T
: 0 d r d 3 3 sin T , 0 d T d 2S ` Cardioid
8. R
^ r, T
: 0 d r d 4 cos 3T , 0 d T d S `
9.
S
cos T
³0 ³0
π
2
r dr dT
2 cos T
S
ªr º
« »
¬ 2 ¼0
S
1
cos 2 T dT
2
1
1 cos 2T dT
4
³0
³0
S
³0
10.
dT
S
1ª
1
º
T sin 2T »
4 «¬
2
¼0
0
1
S
4
2
S
sin T
³0 ³0
r 2 dr dT
1 S 3
sin T dT
3³ 0
S
³0
sin T
ªr3 º
« »
¬ 3 ¼0
dT
1 S
1 cos 2 T sin T dT
3³ 0
S
1ª
cos3T º
«cos T »
3¬
3 ¼0
π
2
1 ª§
1· §
1 ·º
¨1 ¸ ¨ 1 ¸»
«
3 ©
3¹ ©
3 ¹¼
4
9
0
1
© 2010 Brooks/Cole, Cengage Learning
Section 14.3
11.
2S
6
³ 0 ³ 0 3r
2
2S
³0
sin T dr dT
2S
³0
Change of Variables: Polar Coordinates
6
ª¬r 3 sin T º¼ dT
0
13.
S 2
3
³0 ³2
S 2
³0
9 r 2 r dr dT
216 sin T dT
ª 1
2
« 3 9 r
¬
S 2
2S
>216 cos T @0
ª5 5 º
T»
«
¬ 3 ¼0
0
309
3 2º
3
» dT
¼2
5 5S
6
π
2
π
2
0
4
0
1
12.
S 4
4
³0 ³0 r
2
4
ªr3
º
« sin T cos T » dT
3
¬
¼0
S 4
³0
sin T cos dr dT
S 4
ª§ 64 · sin 2 T º
«¨ ¸
»
¬© 3 ¹ 2 ¼ 0
14.
S 2
3
³0 ³0
2
3
S 2
2
³0
re r dr dT
3
ª 1 r2 º
« 2 e » dT
¬
¼0
S 2
ª 1 9
º
« 2 e 1 T »
¬
¼0
16
3
S§
1·
¨1 9 ¸
e ¹
4©
π
2
π
2
0
1
2
3
4
0
1
15.
S 2 1 sin T
³0 ³0
ªT r 2 º
«
»
¬ 2 ¼0
S 2
1
T 1 sin T
2
³0
3
1 sin T
S 2
³0
T r dr dT
2
π
2
dT
2
dT
S 2
1 § 1
1 · 1 2 º
ª1 2
«8T sin T T cos T 2 T ¨ 2 cos T ˜ sin T 2T ¸ 8 sin T »
©
¹
¬
¼0
0
3 2 9
S 32
8
16.
S 2 1 cos T
³0 ³0
sin T r dr dT
S 2
³0
1
1 cos T
ª
r2 º
« sin T
»
2 ¼0
¬
dT
S 2
³0
sin T
1 cos T
2
2
dT
ª1
« 6 1 cos T
¬
3º
S 2
»
¼0
2
1
6
π
2
(x, y) = (0, 1)
0
1
17.
a
³0 ³0
a2 y2
y dx dy
S 2
a
³0 ³0 r
2
sin T dr dT
a3 S 2
sin T dT
3 ³0
S 2
ª a3
º
« cos T »
3
¬
¼0
a3
3
© 2010 Brooks/Cole, Cengage Learning
310
18.
19.
Chapter 14 Multiple Integration
a
a2 x2
2
4 x2
³0 ³0
³ 2 ³ 0
³0 ³
a
x x2
³0 ³0
S 2
S
³0
r 2 r dr dT
2
ªr4 º
« » dT
¬ 4 ¼0
S
³0
4 dT
2
1· 1
§
¨ x 2 x ¸ 4¹ 4
©
cos T
³ S
4S
cos T
ªr4 º
2 « 4 »
¬ ¼0
S 2
r 2 r dr dT
a3
3
1·
§
x x2 Ÿ y2 ¨ x ¸
2¹
©
1 §
1·
¨ x ¸ . So y
4 ©
2¹
³ S 2 ³ 0
x 2 y 2 dy dx
x x2
2
S 2
ª a3
º
« sin T »
3
¬
¼0
a3 S 2
cos T dT
3 ³0
cos T dr dT
2
S
x 2 y 2 dy dx
20. Note that x x 2
1
S 2
³0 ³0 r
x dy dx
22.
3
9 x2
2
8 y2
³0 ³0
³0 ³ y
2
23.
³0 ³0
24.
³0 ³0
4
32
x2 y 2
2 x x2
4 y y2
1
1 x 2
2
4 x2
³ 1 ³ 0
26.
³0 ³0
S 4
3
2
2
³0 ³0
x y dx dy
2
2
S 2
2 cos T
xy dy dx
³0 ³0
x 2 dx dy
³0 ³0
S 2
64³
25.
S 2
³0 ³0 r
dy dx
y = x − x2
r = cos θ
0
2
x
S
1
x 2 y 2 dy dx
³ 0 ³ 0 sin
2
dT
S 2
0
2 2
3
S 2
cos5 T sin dT
ª 4 cos 6 T º
«
»
6
¬
¼0
3
˜
S
4 2S
3
4
2
3
64 sin 4 T cos 2 dT
S 2
³ 0 ³ 0 cos r
x 2 y 2 dy dx 4³
S 2
º
64 ª 5
sin 3 T cos T
3
T sin T cos T »
«sin T cos T 6¬
4
8
¼0
sin 4 T sin 6 T dT
S 2
S 2
³0 ³0
S 4
ª 2 2 3 º
«
»
« 3 T»
«¬
»¼ 0
3
3
³0
r 3 cos 2 T dr dT
³0
27.
³0
r dr dT
2
cos x 2 y 2 dy dx
sin
2 2
S 4
0
2
243S
10
243 S 2
dT
5 ³0
dr dT
r 3 cos T sin T dr dT
4 sin T
S 2
4
1
.
4
π
2
dT
1 S2
1 S
cos 4 T dT
cos 4 T dT
4 ³ S 2
2³0
1§ 1 3 S ·
3S
Wallis's Formula
¨ ˜ ˜ ¸
2© 2 4 2 ¹
32
21.
2
2
³2 ³0
8 x2
2
1
ª1
2 º
« 2 sin r » dT
¬
¼0
S
³0
r dr dT
2
S
³0
S
sin 1 | 1.3218
2
1
sin 1 dT
2
S 2
2
³ 0 >sin r r cos r@0 dT
r r dr dT
2S
>Integration by parts@
S
sin 2 2 cos 2 | 2.7357
2
sin 2 2 cos 2 dT
S 4
2
³0 ³0
x 2 y 2 dy dx
S 4
³0
2
π
2
r 2 dr dT
4 2S
3
16 2
dT
3
0
1
28.
³0
5 2 2
x
5
³ 0 xy dy dx ³ 5
2 2
³0
25 x 2
xy dy dx
S 4
5
³0 ³0 r
S 4
³0
625
4
3
2
3
sin T cos T dr dT
sin T cos dT
S 4
ª 625 sin 2 T º
¬ 8
¼0
625
16
© 2010 Brooks/Cole, Cengage Learning
Section 14.3
29.
2
³0 ³0
4 x2
S 2
2
³0 ³0
x y dy dx
8 S 2
3 0
³
30.
S 2
5
³ S 2 ³ 0 e
r2 2
r dr dT
S 2
³ S 2 ª¬«e
S 2
³ S 2
r2 2 º
1
2
4 y2
³
5
25 2
2
³1
2 ³y
S 4
2
³ 0 ³1
S 4
16
3
y
5
4
3
2
1
dT
S 1 e 25 2
2
4 y2
x
−5 −4 −3 −2 −1
1 2 3 4
−2
−3
−4
−5
y
dx dy
x
arctan
π
2
(
1
,
2
1
2
(
T r dr dT
( 2, 2)
2 S 4
ª 3T º
«
»
¬ 4 ¼0
3
T dT
2
³0
3S
64
2
0
1
32.
3
³0 ³0
9 x2
S 2
3
9 r 2 r dr dT
S 2
3
9r r 3 dr dT
³0 ³0
9 x 2 y 2 dy dx
³0 ³0
33. V
S 2
1
³0 ³0
1 S 2
2 0
³
S 2
1
³0 r
1
4³
35. V
³0 ³0 r
36. V
³ R ³ ln x
2S
³0
5
3
1 S 2
8 0
³
sin 2T dr dT
4³
r 2 3 r dr dT
2
2
dr dT
y 2 dA
2S
³0
2³
S 2
0
4 cos T
³0
S 2
0
2S
2
³ 0 ³1
2S
0
2S
4
³ 0 ³1
81 S 2
dT
4 ³0
81S
8
1
4³
S 2
0
2³
ln r 2 r dr dT
2S
0
7S
2
7
dT
4
2
³ 1 r ln r dr dT
2
ªr2
º
« 1 2 ln r » dT
4
¬
¼1
2³
2S
³0
1
8
250S
3
16 r 2 r dr dT
16 r 2 r dr dT
S 2
ªr4
3r 2 º
« » dT
2 ¼0
¬4
S 2
0
128 S 2 ª
1 sin T 1 cos 2 T ¼º dT
3 ³0 ¬
38. V
³0
3
1 4º
ª9 2
« 2 r 4 r » dT
¬
¼0
ª 1 cos 2T º
¬ 16
¼0
sin 2T dT
125
dT
3
2³
37. V
S 2
2
r cos T r sin T r dr dT
34. V
0
311
cos T sin T r 2 dr dT
S 2
»¼ 0 dT
1e
y
dx dy x
arctan
1 y2
2
ª 8 sin T cos T º
¬3
¼0
cos T sin T dT
S 2
³0
S 2
³0 ³0
r cos T r sin T r dr dT
ª 1 e 25 2 T º
¬
¼ S
31.
Change of Variables: Polar Coordinates
ª 1
« 3
¬
ª 1
« 3
¬
2³
2S
0
3·
§
¨ ln 4 ¸ dT
4¹
©
4 cos T
3
º
16 r 2 »
¼0
dT
S 2
128 ª
cos3 T º
«T cos T »
3 ¬
3 ¼0
4
3
º
16 r 2 » dT
¼1
2S
³0
2 S
3³0
3·
§
4S ¨ ln 4 ¸
4¹
©
2
64 sin 3 T 64 dT
64
3S 4
9
5 15 dT
10 15S
© 2010 Brooks/Cole, Cengage Learning
312
Chapter 14 Multiple Integration
2S
4
³0 ³a
39. V
2S
³0
16 r 2 r dr dT
4
ª 1
« 3
¬
3
º
16 r 2 » dT
¼a
1
3
16 a 2
3
2S
One-half the volume of the hemisphere is 64S 3.
2S
16 a 2
3
16 a 2
64S
3
32
32
32
16 a 2
322 3
16 322 3
a2
4 4 23 2
a
40. x 2 y 2 z 2
V
8³
8³
S 2
0
S 2
0
a
³0
a2 Ÿ z
2 4 2 3 2 | 2.4332
a2 x2 y2
a 2 r 2 r dr dT
ª 1 2 2
2
« 2 ˜ 3 a r
¬
2S
a
8³
» dT
¼0
4
a2 r 2
8 times the volume in the first octant
3 2º
³ 0 ³ 0 25e
V
41. Total volume
16 8 3 2
r2 4
S 2
0
r dr dT
S 2
ª 8a 3 º
« T»
¬ 3 ¼0
a3
dT
3
2S
³0
4S a 3
3
4
ª50e r 2 4 º dT
»¼ 0
¬«
2S
³0
50 e 4 1 dT
1 e 4 100S | 308.40524
Let c be the radius of the hole that is removed.
c
c
2S
2S
2
1
ª50e r 2 4 º dT
V
25e r 4 r dr dT
³
³
³
«
»
0
0
0
¬
¼0
10
2S
³0
50 e c
2 4
100S 1 e c
1 dT Ÿ 30.84052
Ÿ ec
2 4
c2
4
c2
c
0.90183
0.10333
0.41331
0.6429
Ÿ diameter
42.
2 4
2c
1.2858
9
9
1
1
d z d
; d r d 1 cos 2 T
2
4 x2 y 2 9
4 x2 y 2 9 4
(a)
9
9
d z d
2
4r 36
4r 36
0.7
2
−1
1
−0.7
(b) Perimeter
r
dr
dT
³D
2
1
1 cos 2 T
2
1
1
1
cos 2 T
2
2
1
x
1
cos T sin T
Perimeter
(c) V
z
2
§ dr ·
r ¨ ¸ dT .
© dT ¹
E
2³
2S
0
2³
S
0
1
1 cos 2 T
4
1 2 1 cos 2 T
³1 4
2
y
cos 2 T sin 2 T dT | 5.21
9
r dr dT | 0.8000
4r 2 36
© 2010 Brooks/Cole, Cengage Learning
Section 14.3
S
6 cos T
43. A
³0 ³0
44. A
³ 0 ³ 2 r dr dT
45. A
³0 ³0
46. A
2S
4
2S
1 cos T
2S
S
³ 0 18 cos
r dr dT
2S
³0
6 dT
2 sin T
9³
T dT
S
0
S
1 cos 2T dT
48. A
8³
49. r
1
A
1 cos 2T ·
§
¨1 2 cos T ¸ dT
2
©
¹
1 2S
2³0
1 2S
2³0
1 2S
4 4 sin T sin 2 T dT
2³0
1 cos 2T ·
§
¨ 4 4 sin T ¸ dT
2
©
¹
r dr dT
A
S 4
2³
³0
S 3
2 cos T
³1
0
S 3
0
2
2 sin T
r dr dT
3 S 3
2 0
r dr dT
4³
2³
2S 3
0
2S 3
³0
>5T
S 3
³0
>3T
9 cos 2 2T dT
18³
³1
2 cos T
ªr 2 º
« »
¬ 2 ¼1
S 3
0
2³
dT
S 3
0
S 3
sin 2T º
ª1
2« T 2 ¼» 0
¬2
S 3
sin 6T º¼
0
S
S 4
9S
2
1
ª
º
18«T sin 4T »
4
¬
¼0
2³
2S 3
0
ªS
3º
2« »
4 ¼
¬6
3
3
2
2 2 cos T
ªr2 º
« »
¬ 2 ¼1
dT
π
2
³0
1
Ÿ T
2
1 cos T Ÿ cos T
S 3
S 3
r = 2 + 2 cos θ
1
ª¬3 8 cos T 2 1 cos 2T º¼ dT
r
2 3 cos T
ªr º
dT
« »
¬ 2 ¼1 cos T
8 cos 2 T 2 cos T 1 dT
2 sin 2T 2 sin T @0
2S 3
10S
3
4 3 3
2
0
r=1
r=1
ª¬3 8 cos T 4 cos T º¼ dT
2³
0
3
S
1º dT
¼
2S 3
r = 2 cos θ
1·
§
2
¨ 2 cos T ¸ dT
2¹
©
2S 4S
,
3 3
1
ŸT
2
2
3 cos T
1 cos 4T dT
1
6
3
r dr dT
³ 1 cos T r dr dT
S 4
0
3ª¬T 1 cos 6T dT
π
2
2³
2
0
9S
2
1
>8S 4 S 4@
2
S
1 Ÿ cos T
2 2 cos T
S 3
3³
0
3S
2
1ª
1§
1
·º
T 2 sin T ¨T sin 2T ¸»
2 «¬
2©
2
¹¼ 0
dT
4 sin 2 3T dT
S 4
8 sin T sin 2T @0
S 3
0
r
r dr dT
ª 2 2 cos T
¬
3 cos T
2³
³
1·
§
¨1 cos 2T ¸ dT
2¹
©
2 2 cos T
³0
A
3 cos 2T
0
2S 3
51. r
2 sin 3T
³0
2 cos T Ÿ T
2³
50. r
S 3
0
9S
2S
1 2S
2³0
2S
3³
ª §
1
·º
«9¨T 2 sin 2T ¸»
¹¼ 0
¬ ©
12S
1ª
1
1
º
4T 4 cos T T sin 2T »
2 «¬
2
4
¼0
47. A
313
1 2S
1 2 cos T cos 2 T dT
2³0
r dr dT
³0 ³0
2
Change of Variables: Polar Coordinates
S 3
³0
§S ·
3¨ ¸ ©3¹
3
10S
7 3
3
2
S
3
π
2
S 3
³0
ª9 cos 2 T 1 cos T 2 º dT
¬
¼
ª¬4 1 cos 2T 2 cos T 1º¼ dT
3 0
2
3
S
r = 3 cos θ
0
1
r = 1 + cos θ
© 2010 Brooks/Cole, Cengage Learning
314
Chapter 14 Multiple Integration
52. r
1 cos T
S 2
1
A
2
1 cos T
³ S 3 ³ 3 cos T
S 2
S 2
r dr dT 2 1 cos T
ªr º
³ S 3 «¬ 2 »¼
³S 3
dT 3 cos T
S
1 cos T
³S 2 ³ 0
³ S 2 «¬ 2 »¼
S
53. r
A
S 2
0
§1
S
³ S 2 ¨© 2 cos T
4 sin 3T
5S 18
S 18
4 sin 3T
³2
S 5S
1
Ÿ 3T
2
2 Ÿ sin 3T
3³
r dr dT
5S 18
S 18
6
,
A
2 2 cos T Ÿ cos T
2³
2³
S 2
0
S 2
0
S 2
³0
S 2
³0
S 2
³0
4 sin 3T
ªr 2 º
« »
¬ 2 ¼2
S
³S 2
1 cos T
2
dT
r = 1 + cos θ
1 cos 2T ·
¸ dT
4
¹
3
3 · § 3S
3S
§ 3S
· § S
·
1¸ ¨¨
1¸
¸¨
¨
¸
4
2
2
2
4
8
©
¹ ©
¹
¹ ©
S
8
³ 2 2 cos T r dr dT
π
2
3 5S 18 ª
4 sin 3T
2 ³ S 18 ¬
dT
r = 4 sin 3θ
4º dT
¼
2
5S 18
0
3ª
4
º
4T sin 6T »
2 «¬
3
¼S 18
0 Ÿ T
r
2
r = 2 − 2 cos θ
1
3
4
r=2
4
S 2 3
3
S
56. See Theorem 14.3.
2
57. r-simple regions have fixed bounds for T
π
2
T-simple regions have fixed bounds for r.
2
ªr2 º
dT
« »
¬ 2 ¼ 2 2 cos T
58. (a) Horizontal or polar representative elements
r=2
(b) Polar representative element
0
1
ª4 2 2 cos T 2 º dT
¬
¼
8 cos T 4 cos 2 T dT
8 cos T 2 1 cos 2T
>8 sin T
S 2
2T sin 2T @0
dT
8S
55. Let R be a region bounded by the graphs of
r
g1 T and r
g 2 T , and the lines T
T
2
dT 1
2
S 5S
,
18 18
Ÿ T
6
3 ª§ 10
4 § 3 · · § 2S
4 § 3 · ·º
¨¨
«¨ S ¨¨
¸¸ ¸¸ ¨¨
¸ ¸»
¨
2 «¬© 9
3© 2 ¹¹ © 9
3 © 2 ¸¹ ¸¹»¼
2
S
3
9 cos T
2
2
sin 2T º
ª3
« T sin T 4
8 »¼S
¬
3 5S 18
ª8 1 cos 6T 4º¼ dT
2 ³ S 18 ¬
54. r
1 cos T
2
.
4
3³
r = 3 cos θ
3
³S 3
dT
1 2 cos T 4 1 cos 2T
dT 2
S 2
π
2
S
0
ªr º
S
r
r dr dT
2 1 cos T
ª 3
º
« 2T sin T sin 2T »
¬
¼S
So, A
1
Ÿ T
2
3 cos T Ÿ cos T
(c) Vertical or polar
3
59. (a)
³ 3 ³ (b)
³0 ³0
2S
3
9 x2
9 x2
f x, y dy dx
f r cos T , r sin T r dr dT
(c) In general, the integral in part (b) is easier to
evaluate. The endpoints of the region of integration
are constants.
a and
b.
When using polar coordinates to evaluate a double
integral over R, R can be partitioned into small polar
sectors.
60. 0 d r d 4, 0 d T d 2S , x 2 y 2
r 2.
Answer (c)
61. You would need to insert a factor of r because of the
r dr dT nature of polar coordinate integrals. The plane
regions would be sectors of circles.
© 2010 Brooks/Cole, Cengage Learning
Section 14.3
Change of Variables: Polar Coordinates
62. (a) The volume of the subregion determined by the point 5, S 16, 7 is base u height
315
5 ˜ 10 ˜ S 8 7 .
Adding up the 20 volumes, ending with 45 ˜ 10 ˜ S 8 (12), you obtain
V | 10 ˜
S
ª5 7 9 9 5 15 8 10 11 8 25 10 14 15 11
8¬
35 12 15 18 16 45 9 10 14 12 º¼
5S
5S
>150 555 1250 2135 2025@ | >6115@ | 24,013.5 ft 3
4
4
63.
(b)
57 24,013.5
(c)
7.48 24,103.5 | 179,621 gallons
S 2
5
³S 4 ³ 0 r
1,368,769.5 pounds
T dr dT | 56.051
1 r 3 sin
ªNote: This integral equals
«¬
64.
S 4
4
³ 0 ³ 0 5e
rT
S 2
³S 4
1 r 3 dr .º»
¼
5
³0 r
T dT
sin
r dr dT | 87.130
base u height
65. Volume
base u height |
66. Volume
| 8S u 12 | 300
9S
4
u 3 | 21
Answer (a)
Answer (c)
z
z
6
16
4
2
y
4
2
6
4
x
4
4 6
x
y
67. False
Let f r , T
r 1 where R is the circular sector
0 d r d 6 and 0 d T d S . Then,
³R ³
r 1 dA ! 0 but r 1 ! 0 for all r.
68. True
f
(b) So, I
70. (a) Let u
(b) Let u
71.
7
³ 7 ³ f
³ f ³ f e
69. (a) I 2
49 x 2
49 x 2
x2 y2
2
4³
dA
S 2
0
f
³0
e r
2 2
r dr dT
4³
1
du
2
1
2
S 2
0
f
ªe r 2 2 º dT
«¬
»¼ 0
4³
S 2
0
dT
2S
2S .
2 x, then
2 x, then
4000e
f
³ f e
f
³ f e
x2
4 x 2
0.01 x2 y 2
dx
dx
dy dx
f
³ f e
f
u 2 2
³ f e
u 2
2S
7
1
du
2
³ 0 ³ 0 4000e
2S
S.
1
S.
2
0.01r 2
r dr dT
2S 200,000 e 0.49 1
2S
³0
7
ª200,000e 0.01r 2 º dT
«¬
»¼ 0
400,000S 1 e 0.49 | 486,788
© 2010 Brooks/Cole, Cengage Learning
316
72.
Chapter 14 Multiple Integration
f
f
³0 ³0
ke
x2 y2
S 2
f
³0 ³0
dy dx
2
ke r r dr dT
74. (a) 4 ³
S
³0
2
(b) 4 ³
0
(c) 2 ³
0
For f x, y to be a probability density function,
kS
4
2
³2
4 y2
f dx dy
f
ª k r2 º
« 2 e » dT
¬
¼0
2 k
kS
dT
2
4
S 2
³0
2
0
³0
S 2
4 x2 2
4 cos T
³0
y
f dy dx
fr dr dT
(x − 2) 2 + y 2 = 4
2
1
1
4
k
S
.
x
1
3
−1
73. (a)
(b)
³2 ³ y
2
³2
3
³2
4
3
S 3
f dx dy
3
³2
(c)
y
4
3x
−2
f dy dx
'T r22
'T r12
2
2
75. A
³x
4 csc T
³ S 4 ³ 2 csc T
3x
4
4
³4
f dy dx 3
³x
f dy dx
§ r r2 ·
'T ¨ 1
¸ r2 r1
© 2 ¹
r'r'T
fr dr dT
y
5
y=
y=x
3x
(4, 4)
4
,4
3
(
3
(2, 2)
2
,2
3
(
1
1
2
3
(
(
x
4
5
Section 14.4 Center of Mass and Moments of Inertia
2
1. m
2. m
3. m
4. m
2
2
3
9 x2
³ 0 ³ 0 xy dy dx
³0 ³0
S 2
1
³0 ³0
3
3
³0 ³3
2
ª xy 2 º
» dx
2 ¼0
³ 0 «¬
xy dy dx
3
³0
2
2
¬ª x ¼º 0
3
x 9 x2
9 x2
ª xy 2 º
«
»
¬ 2 ¼0
xy dy dx
3
³0
³0
dx
S 2
³0
r cos T r sin T r dr dT
9 x2
2
³ 0 2 x dx
3
ª y2 º
«x »
¬ 2 ¼3
1ª
2
«2 9 x
2¬
2
4
2
dx
1
ª
9 x2
« 1
« 4
3
¬
ª
r4 º
« cos T sin T
» dT
4 ¼0
¬
9 x2
dx
3
³0
x§
3
2 ¨©
3
32
x4 º
9 x2
»
2
4 ¼0
S 2
³0
9 x2
2
3 3
º
»
»
¼0
0
1
243
4
1
cos T sin T dT
4
·
9 ¸ dx
¹
1 ª 81 81
º
54»
2 «¬ 2
4
¼
243
4
S 2
ª 1 sin 2 T º
« ˜
»
2 ¼0
¬4
1
8
1 3ª
6 x 9 x 2 9 x x3 º dx
¼
2³0 ¬
297
8
© 2010 Brooks/Cole, Cengage Learning
Section 14.4
5. (a)
m
³ 0 ³ 0 kx dy dx
m
My
x
x, y
m
Mx
My
x
x, y
m
a
a
My
a
,y
2
m
§a a·
¨ , ¸
©2 2¹
a
y
a
y
a
y
³ 0 ³ 0 k dx dy
³ 0 ³ 0 kx dx dy
My
a
m
3
§ a 2a ·
¨ , ¸
©3 3 ¹
a
y
a
y
a
y
³ 0 ³ 0 kx dx dy
Mx
³ 0 ³ 0 kxy dx dy
My
³ 0 ³ 0 kx
x
x, y
ka 2
dx
2
ka 3
2
ka 3
2
Mx
m
x
My
m
ka 3b 2 6
ka 2b 2 4
ka 2b 2
4
ka 2b3
6
ka 3b 2
6
2a
,
3
y
Mx
m
ka 2b3 6
ka 2b 2 4
2
b
3
2
My
a
m
2
§ a 3a ·
¨ , ¸
©2 4 ¹
dx dy
1 2
ka
2
1 3
ka
3
1 3
ka
6
Mx
y
m
1 3
ka
6
1 4
ka
8
1
ka 4
12
Mx
y
m
2a
3
a
b
³ 0 ³ 0 kx
m
2
a
b
a
b
a
b
³0 ³0 k x
³0 ³0
My
³0 ³0 k x
2
a
dy dx
y dy dx
y 2 dy dx
k x 2 y y 3 dy dx
3
xy 2 dy dx
kab 2
a b2
3
kab 2
2a 2 3b 2
12
ka 2b 2
3a 2b 2
12
x
My
m
ka 2b 12 3a 2 2b 2
a 3a 2 2b 2
kab 3 a 2 b 2
4 a 2 b2
y
Mx
m
kab 2 12 2a 2 3b 2
b 2a 2 3b 2
kab 3 a 2 b 2
4 a 2 b2
§ a 3a 2 2b 2 b 2a 2 3b 2
¨
,
¨ 4 a 2 b2
4 a 2 b2
©
1 3
ka
3
a
y
1 4
2
Mx
³ 0 ³ 0 ky dx dy 4 ka
a
y
1 4
My
³ 0 ³ 0 kxy dx dy 8 ka
My
Mx
3a
x
y
,
m
m
8
§ 3a 3a ·
x, y
¨ , ¸
©8 4¹
m
2
317
§ 2a 2b ·
¨ , ¸
© 3 3¹
Mx
x, y
(b)
b
My
x, y
a
2
a
³ 0 ³ 0 kxy
center of square
2a
3
b
Mx
a
2
(b)
a
³ 0 ³ 0 kxy dy dx
m
a
My
m
a
³0
6. (a)
1 3
ka
2
a
a
1 4
2
³ 0 ³ 0 ky dy dx 3 ka
a
a
1 4
³ 0 ³ 0 kyx dy dx 4 ka
My
a
Mx
,
y
2
m
m
§ a 2a ·
¨ , ¸
©2 3 ¹
a
a
1 3
³ 0 ³ 0 kx dy dx 2 ka
a
a
1 4
³ 0 ³ 0 kxy dy dx 4 ka
a
a
1 3
2
³ 0 ³ 0 kx dy dx 3 ka
My
2a
Mx
,
y
3
m
m
§ 2a a ·
¨ , ¸
© 3 2¹
a
³ 0 ³ 0 ky dx dy
x, y
ka 2
³ 0 ³ 0 ky dy dx
Mx
x
(c)
a
My
Mx
7. (a)
a
³ 0 ³ 0 ky dy dx
x, y
(c)
a
Mx
x
(b)
a
³ 0 ³ 0 k dy dx
Center of Mass and Moments of Inertia
·
¸
¸
¹
y
³ 0 ³ 0 ky dx dy
3a
4
3a
4
© 2010 Brooks/Cole, Cengage Learning
318
Chapter 14 Multiple Integration
Mx
My
x
a y 2
k dx dy
Mx
My
x
y = 2x
(b) The x-coordinate changes by 5: x , y
a 5
y = − 2x + 2a
§a a·
¨ , ¸
© 2 3¹
x
x
y
x, y
§ 11a a ·
, ¸
¨
© 20 2 ¹
a·
§a
¨ 5, ¸
2¹
©2
2a ·
§a
¨ 5,
¸
2
3¹
©
a
10. The x-coordinate changes by c units horizontally and d
units vertically. This is not necessarily true for variable
densities. See Exercise 9.
Mx
My
x
y
x, y
(a, 0)
1
2
ka a 5 25
2
a 5 a
1 2
2
³ 5 ³ 0 kxy dy dx 4 ka a 5 25
a 5 a
1
3
2
³ 5 ³ 0 kx dy dx 3 ka a 5 125
3
2 ª a 5 125º
2 a 2 15a 75
My
¬
¼
2
m
3 a 10
3ª a 5 25º
¬
¼
Mx
a
m
2
§ 2 a 2 15a 75 a ·
¨
, ¸
¨
3 a 10
2¸
©
¹
My
m
a
2
(0, 0)
³ 5 ³ 0 kx dy dx
m
Mx
11.
( a2 , a)
a
kxy dx dy
9. (a) The x-coordinate changes by 5: x , y
(c)
y
1
ka 4
12
a
a y 2
1
2
5
³ 0 ³ y 2 kxy dx dy 24 ka
a
a y 2
11
2
5
³ 0 ³ y 2 kx y dx dy 240 ka
My
11a
Mx
a
,y
, x, y
20
2
m
m
a
³0 ³ y 2
m
(b)
a y 2
1 2
ka
2
a
a y 2
1 3
³ 0 ³ y 2 ky dx dy 6 ka
a
a y 2
1 3
³ 0 ³ y 2 kx dx dy 4 ka
My
a
Mx
a
,y
, x, y
m
m
2
3
a
³0 ³ y 2
m
8. (a)
1
³0 ³0
x
1
x
1
x
³0 ³0
³0 ³0
ky dy dx
ky 2 dy dx
kxy dy dx
My
2
3
m
Mx
8
m
15
§2 8 ·
¨ , ¸
© 3 15 ¹
1
k
4
2
k
15
1
k
6
12.
x2
2
x2
³0 ³0
My
³0 ³0
y
y= x
x, y
(23 , 158 (
2
Mx
y
(1, 1)
x2
³0 ³0
x
1
2
m
16
k
3
32
k
3
64
k
7
kxy dy dx
kxy 2 dy dx
kx 2 y dy dx
My
12
m
7
Mx
2
m
§ 12 ·
¨ , 2¸
©7 ¹
y
4
y = x2
(2, 4)
3
2
1
x
1
x
1
2
3
4
© 2010 Brooks/Cole, Cengage Learning
Section 14.4
13.
4 x
4
m
³1 ³ 0
Mx
³1 ³ 0
My
³1 ³ 0
kx3 dy dx
My
84k
30k
24k
30k
4 x
4
4 x
4
x
m
My
y
m
kx 2 dy dx
30k
kx 2 y dy dx
Center of Mass and Moments of Inertia
(b)
24k
84k
14
5
4
5
14.
x
m
Mx
y
2
3
4
1
1 1 x2
1
1 1 x2
³ 1 ³ 0
Mx
m
k dy dx
kS
2
(b)
§ 2 S·
¨ 0,
¸
4S ¹
©
x, y
y
2
y=
e2 1
,y
2 e2 1
1
e x
1
e x
1
e x
³0 ³0
Mx
³0 ³0
My
³0 ³0
kxy dy dx
x
My
m
1 3e 2
2 1 e 2
y
Mx
m
9 1 e 2
1
e x
1
e x
1
e x
³0 ³0
Mx
³0 ³0
My
³0 ³0
kxy 2 dy dx
x
My
m
1 4e 3
3 1 e 3
y
Mx
m
ex
1
ex
1
ex
Mx
³0 ³0
My
x
y
x, y
k dy dx
k e 1
ky dy dx
1
k e2 1
4
³0 ³0
kx dy dx
k
My
m
Mx
m
1
,
e 1
e2 1
4e 1
§ 1 e 1·
,
¨
¸
©e 1 4 ¹
e1
,
4
17.
ky 2 dy dx
ky 3 dy dx
2
4 x2
2
4 x2
m
³ 2 ³ 0
Mx
³ 2 ³ 0
1
1 e 3 k
9
1
1 e 4 k
16
1
1 4e 3 k
27
9 1 e 4
16 1 e 3
kx dy dx
kxy dy dx
x
0 by symmetry
y
Mx
m
x, y
·
¸
¸
¹
§ 1 4e 3 9 1 e 4
¨
,
¨ 3 1 e 3 16 1 e 3
©
x, y
1
ky 2 dy dx
m
1
³0 ³0
9 e2 1
1
1 e 2 k
4
1
1 e 3 k
9
1
1 3e 2 k
8
ky dy dx
§ 1 3e 2 4 1 e 3
¨
,
¨ 2 1 e 2 9 1 e 2
©
1
1 + x2
m
4 e3 1
Mx
m
4 1 e 3
x
−1
15. (a)
kxy dy dx
m
x, y
k
ky dy dx
2S
8
k
2
2S
2S ˜
kS
8
4S
ky 2 dy dx
§ e 2 1 4 e3 1 ·
¨
¸
,
¨ 2 e2 1 9 e2 1 ¸
©
¹
0 by symmetry
³ 1 ³ 0
e2 1
k
4
e3 1
k
9
e2 1
k
8
ky dy dx
My
m
x
1
ex
³0 ³0
2
1
1
My
4
y = 4x
ex
³0 ³0
y
3
1
Mx
x, y
16. (a)
ex
³0 ³0
x
§ 14 4 ·
¨ , ¸
© 5 5¹
x, y
1
m
319
·
¸
¸
¹
256
k
15
4096
k
105
y
16
7
y = 4 − x2
3
§ 16 ·
¨ 0, ¸
© 7¹
2
1
x
−2
−1
1
2
© 2010 Brooks/Cole, Cengage Learning
,
320
Chapter 14
Multiple Integration
3
9 y2
3
9 y2
3
9 y2
m
³ 3 ³ 0
Mx
³ 3 ³ 0
18.
kxy dx dy
³ 3 ³ 0
My
x, y
0
19.
by symmetry
23,328 k
35
kx 2 dx dy
My
36
,
m
7
§ 36 ·
¨ , 0¸
©7 ¹
x
648 k
5
kx dx dy
y
x
L
2
m
³0 ³0
L
sin S x 2
L
sin S x L
³0 ³0
Mx
Mx
m
y
0
by symmetry
k dy dx
2kL
ky dy dx
S
y
kL
4
y = sin π x
L
S
8
§L S ·
¨ , ¸
©2 8¹
x, y
x
L
2
L
y
20.
6
4
2
x
2
−2
4
6
y
³R ³
³R ³
My
m
Mx
m
S 4
S 4
a
³ 0 ³ 0 kr
kx dA
2
2
8
ka 3 2
˜
S a 2k
6
ka 2 3
6
§
4a 2 ¨ 4a 2 ,
¨ 3S
3S
©
a
a2 x2
a
a2 x2
Mx
³0 ³0
My
M x by symmetry
y
a
³ 0 ³ 0 kr
ky dA
³0 ³0
x, y
kxy dy dx
L2 k S 2 4
32S 2
LS 4
2
4S 2
16
9S
8
m
x
ky 2 dy dx
kL
8
2kL
9S
S a2k
x, y
22.
ky dy dx
§ L S 2 4 16 ·
¨
¸
,
¨ 4S 2
9S ¸
©
¹
x, y
x
cos S x L
My
m
Mx
m
y
My
L2
³0 ³0
x
Mx
cos S y L
³0 ³0
My
m
L2
Mx
−6
21.
cos S x L
³0 ³0
10
−4
L2
m
x = 9 − y2
2
˜
2
6
ka
cos T dr dT
3
2
π
2
6
4a 2
3S
8
S a 2k
y=x
r=a
4a 2 2
3S
a
0
2 ·
¸
¸
¹
k x 2 y 2 dy dx
k x 2 y 2 y dy dx
My
m
ka 3 2 sin T dr dT
8
ka 5
˜ 4
5
ka S
S 2
a
³ 0 ³ 0 kr
S 2
a
3
³ 0 ³ 0 kr
dr dT
4
ka 4S
8
sin T dr dT
ka 5
5
8a
5S
§ 8a 8a ·
¨ ,
¸
© 5S 5S ¹
© 2010 Brooks/Cole, Cengage Learning
Section 14.4
23.
My
x
2
e x
2
e x
kxy 2 dy dx
1 7e 6
k
27
³0 ³0
kx 2 y dy dx
1 13e 4
k
8
My
m
e 4 13
e4 5
8e 7
27 e 5e
6
k
x
e ln x k
³1 ³ 0 x
e ln x k
³1 ³ 0 x
My
k
m
1
Mx
k
6
m
1
§
·
¨ 2, ¸
© 2¹
e
ln x
Mx
My
x
y
x, y
27. m
Ix
Iy
x
y
S 6
1
r = 2 cos 3θ
81 3
| 1.12
40S
0
1
π
θ =−6
2
y
1
1
3
2S
1
1 cos T
³0 ³0
2S
kr dr dT
1 cos T
2
e 3
x
3S k
2
k 2S
cos T 1 3 cos T 3 cos 2 T cos3 T dT
3³0
3
1
5kS
2
ª
2
2
«cos T 2 1 cos T 3 cos T 1 sin T 4 1 cos 2T dT
4
¬
y 2 dy dx
³ 0 ³ 0 x dy dx
Ix
m
My
m
π
θ= 6
x , y | 1.12, 0
³0 ³0
kr 2 cos T dr dT
2
bh
3
bh 1
˜
3
bh
3
1
bh
˜
3
bh
r = 1 + cos θ
0
1
2
b
3
2
h
3
b
3
h
3
3
b
3
3
h
3
b
h hx b
b
h hx b
b
h hx b
28. m
³0 ³0
Ix
³0 ³0
Iy
³0 ³0
3
b3h
3
3
π
2
5
6
bh
m
π
2
y = ln x
My
5kS
2
˜
4
3kS
m
§5 ·
¨ , 0¸
©6 ¹
Iy
kr 2 cos T dr dT
x
2
kx dA
h
2 cos 3T
³ S 6 ³ 0
kS
3
kr dr dT
2
2
˜
k
2
˜
k
³R ³
b
kx dA
2 cos 3T
3
My
h
³R ³
y = e −x
·
¸
¸
¹
k
k dA
b
My
S 6
³ S 6 ³ 0
27 3
k | 1.17 k
40
1
x dy dx
³R ³
³0 ³0
k dA
x
k
6
m
x, y
³R ³
2
y dy dx
0 by symmetry
x
m
321
k
2
dy dx
y
k 2S
3³0
0 by symmetry
2
6
³1 ³ 0
m
y
y
§ e 4 13 8 e6 7
¨ 4
,
¨ e 5 27 e6 5e 2
©
x, y
25.
1 5e 4
k
8
kxy dy dx
Mx
m
y
26.
e x
³0 ³0
Mx
24.
2
³0 ³0
m
Center of Mass and Moments of Inertia
x
y
dy dx
y 2 dy dx
x 2 dy dx
bh
2
bh3
12
b3h
12
m
b3h 12
bh 2
b
6
6
b
6
Ix
m
bh3 12
bh 2
h
6
6
h
6
Iy
© 2010 Brooks/Cole, Cengage Learning
322
Chapter 14
29. m
Ix
Iy
I0
x
30. m
Multiple Integration
S a2
³R ³ y
2
³R ³ x
2
dA
dA
2S
2S
a
³0 ³0 r
cos 2 T dr dT
3
a 4S
4
a 4S
4
Ix
Iy
a 4S
2
I0
a
2
x
S
a
S
a
dA
³0 ³0 r
Iy
³R ³ x
2
dA
³0 ³0 r
I0
Ix Ix
x
y
3
sin 2 T dr dT
a 4S
8
3
cos 2 T dr dT
a 4S
8
a 4S
a 4S
8
8
2
a 4S
˜
S a2
8
Ix
m
2
S a4
16
S 2
a
3
3
sin 2 T dr dT
cos 2 T dr dT
S a4
S a4
16
8
Sa
Ix
m
y
a
³0 ³0 r
dA
Ix I y
S 2
³0 ³0 r
dA
4
16
˜
S a4
16
S a4
16
a
2
4
S a2
S ab
4³
a
³0
0
ba
a2 x2
4³
y 2 dy dx
b
b2 y 2
a
0
b3 2
a x2
3a 3
32
dx
I y Ix
x
Iy
m
1
a 3bS
˜
4
S ab
a
2
y
Ix
m
1
ab3S
˜
4
S ab
b
2
x 2 dx dy
a 3bS
ab3S
4
4
a
a
0
a
Ix
k³
0
Iy
k³
0
I0
Ix I y
a
m
3kab 4 2kb 2 a 3
12
m
ka 3b 2 6
kab 2 2
Ix
m
kab 4 4
kab 2 2
Iy
2
kab
2
b
kab 4
3
³ 0 y dy dx
4
b
ka 3b 2
2
x
yy
dy
dx
³0
6
b
³ 0 y dy dx
a 2 x 2 º dx
¼
a2 x2 x2
ab3S
4
x ºº
a 2 x 2 a 4 arcsin »»
a ¼¼ 0
34. U
ky
k³
4b3 a ª 2
a
3a 3 ³ 0 ¬
a 3bS
4
abS 2
a b2
4
I0
0
³0
ab
4³
y
³R ³ x
2
a
2
Iy
x
³R ³ y
a 4S
4
x · 1ª
4b3 ª a 2 §
2
2
2
2
2
¨ x a x a arcsin ¸ « x 2 x a
3«
a ¹ 8¬
3a ¬ 2 ©
m
4
2
2
33. U
S a2
S a2
³R ³ y
Ix
sin 2 T dr dT
a 4S
1
˜
4
S a2
Ix
m
Ix
32. m
3
a 4S
a 4S
4
4
Ix I y
y
a
³0 ³0 r
31. m
a2
3
b2
2
a
3
b
2
3
a
3
2
b
2
ky
2k ³
a
0
³0
a
a2 x2
a 2 x2
a
y dy dx
k³
y 3 dy dx
4ka 5
15
0
Ix
k³
a ³ 0
Iy
k³
a
I0
Ix I y
x
Iy
m
2ka 5 15
2ka 3 3
a2
5
y
Ix
m
4ka 5 15
2ka 3 3
2a 2
5
a
³0
a 2 x2
x 2 y dy dx
a 2 x 2 dx
2ka 3
3
2ka 5
15
2ka 5
5
a 5
5
a 10
5
© 2010 Brooks/Cole, Cengage Learning
Section 14.4
35. U
38. U
kx
4 x2
2
m
k³
Ix
k³
0
Iy
k³
0
I0
Ix I y
y
36. U
m
Ix
Iy
I0
Ix
Iy
I0
4 x2
³0
32k
3
16k
3
xy 2 dy dx
x3 dy dx
16k
x 2 y 2 dy dx
1
x
x 2 y 2 y 2 dy dx
1
x
x 2 y 2 x 2 dy dx
Ix
³ 0 ³ x2
Iy
³ 0 ³ x2
I0
Ix I y
2
3
2 3
3
x
Ix
m
32k 3
4k
8
3
4
6
2 6
3
Iy
m
y
Ix
m
k 48
k 24
k 60
k 24
39. U
k
24
k
60
k
48
2 xy dy dx
Ix
m
m
Ix
Iy
I0
2
2
10
5
4
³0 ³0
x
4
x
4
x
³0 ³0
³0 ³0
kxy dy dx
16k
kx3 y dy dx
512k
5
Iy
Ix
m
16k
3
˜
1
32k
b ³ 0
2 k ª³ x 2
«¬ b
2
³0 ³0
Iy
I0
m
b
Ix
592k
5
512k
3
˜
5
32k
b
m
kxy 3 dy dx
Ix I y
2k ³
40. U
b2 x2
xa
48
5
3
2
2
k x 6
2
dy dx
2k ³
4
³0
b
b
b
xa
2
b
b
dx
kx dy dx
m
k 20
˜
18 3k
30
9
Ix
m
3k 20
˜
56 3k
70
14
ky
512k
21
2
4x
k
32,768
2 ³ ³ 3 ky 3 dy dx
0
x
65
2
4x
2048k
2 ³ ³ 3 kx 2 y dy dx
x
0
45
321,536k
Ix I y
585
2³
b 2 x 2 dxº
»¼
4
2
x
2
0
Iy
4x
³ x3 ky dy dx
m
2048k
21
˜
45
512k
28
15
Ix
m
32,768k
21
˜
65
512k
8 1365
65
2 105
15
b 2 x 2 dx
x b 2 x 2 dx a 2 ³
2k x 6
351,945
891
351,945
891
x
Iy
y
b
395
891
3k
20
1
x
3k
2
³ 0 ³ x2 kxy dy dx 56
x
1
k
3
³ 0 ³ x2 kx dy dx 18
55k
Ix Iy
504
1
x
6
2
dy dx
b 2 x 2 dx 2a ³
4 15
5
316
2079
158 35
˜
2079 6
³ 0 ³ x2
y
32k
3
158
2079
158
2079
kx
x
kxy
4
x
³ 0 ³ x2
4
3
x
6
35
1
m
16k 3
4k
1
323
x2 y2
m
Iy
m
y
42. I
4k
k 1 3
x x5 dx
0 ³x
2³0
1 x
k 1 5
k ³ ³ 2 xy 3 dy dx
x x9 dx
0 x
4³0
1 x
k 1 5
k ³ ³ 2 x3 y dy dx
x x 7 dx
0 x
2³0
9k
3k
Ix I y
240
80
x
41. I
³0
2
k³
y
m
4 x2
2
x dy dx
kxy
x
37. U
³0
0
Iy
x
Center of Mass and Moments of Inertia
3º
ª 2k
«3 x 6 »
¬
¼0
ªS b 4
S a 2b 2 º
2k «
0
»
2 ¼
¬ 8
kS b 2 2
b 4a 2
4
416k
3
© 2010 Brooks/Cole, Cengage Learning
324
Chapter 14
x
4
43. I
³0 ³0
44. I
³ a ³ 0
a
Multiple Integration
kx x 6
a2 x2
2
dy dx
ky y a
2
ª y4
2ay 3
a2 y2 º
³ a k «¬ 4 3 2 »¼
0
ª1
a
4
24 7 2
72 5 2 º
ª2
k « x9 2 x x »
7
5
¬9
¼0
x x 2 12 x 36 dx
42,752k
315
dy dx
a
³ a k «¬ 4 a
4
4
³ 0 kx
a 2 x2
dx
2a 2 x 2 x 4 2a 2
a
3
a2 x2 x2
a2 x2 º
a2 2
a x 2 » dx
2
¼
ª1 §
2a 2 x 3
x 5 · 2a ª a 2 §
x·
2
2
2
k « ¨ a4 x ¸
« ¨ x a x a arcsin ¸
4
3
5
3
2
a
©
¹
¹
¬
¬ ©
1§
¨ x 2 x2 a2
8©
a
a 2 x 2 a 4 arcsin
x3 ·º
x ·º a 2 § 2
¨a x ¸»
¸» 2©
3 ¹¼ a
a ¹¼
ª1 §
2
1 · 2a § a 4S
a 4S · a 2 § 3 a 3 ·º
2k « ¨ a 5 a 5 a 5 ¸ ¨
¸
¨a ¸»
3
5 ¹
3© 4
16 ¹
2©
3 ¹¼
¬4©
a
³0 ³0
45. I
a2 x2
k a y y a
2
a
³0 ³0
dy dx
a2 x2
k a y
3
§ 7a5
a 5S ·
2k ¨
¸
8 ¹
© 15
a
³0
dy dx
§ 56 15S ·
ka 5 ¨
¸
60
©
¹
4º
ª k
« 4 a y »
¬
¼0
a2 x2
dx
a2 x2
k a 4
ª¬a 4a 3 y 6a 2 y 2 4ay 3 y 4 º¼
dx
³
0
0
4
k a
³ ªa 4 4a 3 a 2 x 2 6a 2 a 2 x 2 4a a 2 x 2 a 2 x 2 a 4 2a 2 x 2 x 4 a 4 º dx
¼
4 0¬
a
k
³ ª7 a 4 8a 2 x 2 x 4 8a 3 a 2 x 2 4ax 2 a 2 x 2 º dx
¼
4 0¬
a
8a 2 3
kª
x
x · a§
§
x «7 a 4 x 4a 3 ¨ x a 2 x 2 a 2 arcsin ¸ ¨ x 2 x 2 a 2
4¬
3
5
a ¹ 2©
©
8
1
1
17 ·
k§
·
§ 7S
a5k ¨
¨ 7 a 5 a 5 a 5 2a 5S a 5S ¸
¸
4©
3
5
4
15 ¹
¹
© 16
2
4 x2
³ 2 ³ 0
46. I
k y 2
2
dy dx
4 x2
3º
ªk
³ 2 «¬ 3 y 1 »¼ 0
2
k 2
16 12 x 2 6 x 4 x 6 dx
3 ³ 2
dx
2
planar lamina R.
The movements of mass with respect to the x- and y-axes
are
³ R ³ yU
x, y dA and M y
³ R ³ xU
x, y dA.
If m is the mass of the lamina, then the center of mass is
x, y
48. I x
Iy
§ My Mx ·
,
¨
¸.
© m m ¹
³R ³ y
2
U x, y dA, Moment of inertia about x-axis
³R ³ x U
2
k
ªk §
6 5 1 7 ·º
3
« 3 ¨16 x 4 x 5 x 7 x ¸»
¹¼ 2
¬ ©
47. Let U x, y be a continuous density function on the
Mx
2
³ 2 3 ¬ª 2 x
2
a 2 x 2 a 4 arcsin
x ·º
¸»
a ¹¼ 0
8¼º dx
2k §
192 128 ·
¨ 32 32 ¸
3©
5
7 ¹
1408k
105
49. See the definition on page 1017.
50. (a) U x, y
ky
y will increase.
(b) U x, y
k2 x
x , y will be the same.
(c) U x, y
kxy
Both x and y will increase.
(d) U x, y
k 4 x 4 y
Both x and y will decrease.
x, y dA, Moment of inertia about y -axis
© 2010 Brooks/Cole, Cengage Learning
Section 14.5
L
,A
2
y
51.
b
ya
ªªy L 2 º º
¬
¼ » dx
»
3
¬
¼0
a
,A
2
y
52.
b
³0 ³0
Iy
a
a b 12
ª¬L a 2 º¼ ab
2
2L
,A
3
y
53.
2³
Iy
b2
0
bL
,h
2
L
y 2 dy dx
LS a 2
2S
a
³0 ³0 r
3
sin 2 T dr dT
a 4S
4
a4
sin 2 T dT
4
a 4S 4
2S ³
a2
4L
³ x dA
R
§
·
¨ ³ R ³ x dA ¸
2S ¨
¸ ³ R ³ dA
¨ ³ R ³ dA ¸
©
¹
2S xA.
3 2L a
By our positioning, x
2
§
a 2 x2
L
³ R ³ 2S x dA
a 3b
12
L
3
³ 2 Lx b ¨© y a2 x2
325
55. Orient the xy-coordinate system so that L is along the yaxis and R is the first quadrant. Then the volume of the
solid is
a 3L 2a
3
ya
L
3
V
2
a·
§
¨ y ¸ dy dx
2¹
©
a
ya
a
2
L ³ a ³ ³0
L3b
12
L
L3b 12
L 2 bL
2
ab, h
a
2S
³ 0 ««
Iy
y hA
S a2 , h
0, A
Iy
2
3 L
b
y
54.
L·
§
¨ y ¸ dy dx
2¹
©
L
³0 ³0
Iy
L
2
bL, h
Surface Area
2L ·
¸ dy dx
3 ¹
r. So, V
2S rA.
y
L
3
2 b 2 ª§
2L · º
y dx
«
»
¨
¸
³
3 0 ¬«©
3 ¹ ¼»
2 Lx b
L
R
( x, y )
3
2 b2ª L
2L · º
§ 2 Lx
¨
«
» dx
¸
³
3 0 ¬« 27 © b
3 ¹ ¼»
b § 2 Lx
2 ª L3 x
2L ·
«
¨
¸
3 «¬ 27
8L © b
3 ¹
L3b 36
2L
2
Lb 6
3
ya
4 b2
º
»
»¼ 0
x
L3b
36
L
2
Section 14.5 Surface Area
2x 2 y
1. f x, y
fx
fy
2
4
4 x
³0 ³0
S
fx
2
1 fx
fy
2
1 4 4
3³
3 dy dx
4
0
4 x dx
3
2, f y
2
1 fx
3
3
fy
3
³0 ³0
S
2
14
3
³0 3
14 dy dx
14 dx
9 14
y
4
ª
x2 º
3«4 x »
2 ¼0
¬
15 2 x 3 y
2. f x, y
24
3
R
y
2
4
1
y=4−x
3
x
1
2
2
3
1
x
1
2
3
4
© 2010 Brooks/Cole, Cengage Learning
326
Chapter 14
7 2x 2 y
3. f x, y
fx
Multiple Integration
fy
2
2
1 fx
2S
fy
2
1 4 4
2S
2
³ 0 ³ 0 3 r dr dT
S
³0
6 dT
3
R
square with vertices 0, 0 , 3, 0 , 0, 3 , 3, 3
fx
0, f y
12S
3
2y
2
1 fx
x2 + y2 = 4
2
fy
3
³0 ³0
S
y
1 4 y2
3
³0 3
1 4 y 2 dx dy
3
4
x
6 37 ln 6 1 4 y 2 dy
3
ª 3 2 y 1 4 y 2 ln 2 y ¬« 4
1
−1
y2
6. f x, y
1 4 y2 º
¼» 0
37
1
y
−1
3
R
2
12 2 x 3 y
4. f x, y
fx
2
1 fx
2S
fy
3
³0 ³0
S
1
3
2, f y
x
2
1 4 9
2S
³0
14 r dr dT
1
14
9 14
dT
2
9 14 S
2
3 x3 2
7. f x, y
fx
y
3
3 12
x , fy
2
0
2
2
x2 + y2 = 9
1 fx
fy
1
2
1
1
−1
3
2
4 9x
dy dx
2
4
³0 ³0
S
x
−2 −1
ª4
« 27 4 9 x
¬
−2
3 2º
3
4 9x
2
9
x
4
4³
3
0
4 9x
dx
2
4
31 31 8
27
»
¼0
y
9 x
5. f x, y
2
2 x, f y
fx
2
1 fx
4
0
fy
3
2
2
1 4x2
1
2
2
³0 ³0
S
2³
1 4 x dy dx
2
2
0
1 4 x dx
2
x
1
2
3
2
ª1
2 « ln
¬4
1 4 x2 2 x ª1
2 « ln
¬4
17 4 2 17 1
ln 4 2
º
17 »
¼
17
x
º
1 4x2 »
2
¼0
2
8. f x, y
fx
2
2
S
y
2
2 y3 2
3
y1 2
0, f y
1 fx
4
fy
2 y
2
1 y
³0 ³0
1 y dx dy
ª2 1 y
¬
32
2 ˜ 33 2 2
5
2
5
1 y
2
³0
1 y 2 y dy
52 2
˜ 35 2 2 º
¼0
2
5
12
5
3 8
5
y
1
2
y=2−x
x
1
2
1
R
x
1
2
© 2010 Brooks/Cole, Cengage Learning
Section 14.5
9. f x, y
ln sec x
fx
2
1 fx
S 4
tan x
³0 ³0
S
S 4
³0
2
fy
1 tan 2 x
^ x, y : 0
R
0 d
d f x, y d 1`
x 2 y 2 d 1, x 2 y 2 d 1
x
fx
sec x
327
x2 y 2
11. f x, y
S
­
½
® x, y : 0 d x d , 0 d y d tan x¾
4
¯
¿
tan x, f y
0
R
Surface Area
y
, fy
x2 y2
x2 y 2
y
sec x dy dx
1 fx
2
sec x tan x dx
S 4
>sec x@0
y = tan x
S
1
2 1
R
2
2
fy
1 x2
1
³ 1 ³ 1
2S
1
³0 ³0
2 dy dx
1 x2
x2
y2
2
2
x y
x y2
2
2
2r dr dT
2S
y
π
4
10. f x, y
13 x 2 y 2
fx
2 y
2 x, f y
2
1 fx
2S
2
³0 ³0
S
2S
³0
2S
³0
S
x
1 4x2 4 y2
1 4r 2 r dr dT
12. f x, y
ª 1 1 4r 2
¬«12
1
12
32 2
º dT
¼» 0
R
y, f y
S
4
³4 ³
2S
x2 + y2 = 2
fy
16 x 2
16 x 2
4
³0 ³0
1
y 2 d 16`
x
2
1 fx
y
2
xy
^ x, y : x 2
fx
173 2 1 dT
17 17 1
6
x 2 + y2 = 1
1
1
2
fy
x
π
2
2
1 y 2 x2
1 y 2 x 2 dy dx
2S
17 17 1
3
1 r 2 r dr dT
y
x
−2
−1
1
x 2 + y 2 = 16
2
−1
2
−2
x
−2
2
−2
a2 x2 y2
13. f x, y
R
^ x, y : x 2 x
fx
a2 x2 y2
1 fx
S
y 2 d b 2 , 0 b a`
b
2
³ b ³ fy
b2 x2
b2 x2
2
y
, fy
1
y
a
a2 x2 y2
b
x2
y2
2
2
2
a x y
a x2 y2
2
a
a2 x2 y 2
dy dx
2S
b
³0 ³0
a
a2 r 2
x 2 + y 2 ≤ b2
a
−b
a x2 y2
2
r dr dT
2S a a b
a
−b
a 2 b2
© 2010 Brooks/Cole, Cengage Learning
x
328
Chapter 14
Multiple Integration
14. See Exercise 13.
a
³ a ³ S
2S
a
a2 x2
a x y
2
a
a
³0 ³0
16 x 2 y 2
16. z
a2 x2
a r
2
2
2
r dr dT
1 fy
dy dx
2
2S a 2
2
1 fx
S 2
4
2
1 4 x2 4 y2
1 4 x 2 y 2 dy dx
1 4r 2 r dr dT
S
24
65 65 1
y
14
6
3 2 x 12
8
³0 ³0
S
16 x 2
³0 ³0
2
fy
fy
4
³0 ³0
S
24 3 x 2 y
15. z
2
14 dy dx
y = 16 − x 2
48 14
4
y
2
16
x
12
2
4
6
8
4
x
4
8
16
25 x 2 y 2
17. z
2
1 fx
S
12
2³
fy
3
3 ³ y
2
1
9 x2
x2
y2
2
2
25 x y
25 x 2 y 2
5
9 x2
25 x y
2
dy dx
2
2³
2S
0
5
3
³0
25 r 2
5
25 x 2 y 2
r dr dT
20S
x 2 + y2 = 9
2
1
−2 −1
x
2
1
−1
−2
18. z
x2 y 2
2
2
1 fx
S
2S
fy
2
³0 ³0
2
1
5r dr dT
4S
4x2
4 y2
2
2
x y
x y2
2
5
5
y
x2 + y2 = 4
1
x
−1
1
−1
© 2010 Brooks/Cole, Cengage Learning
Section 14.5
2 y x2
19. f x, y
R
2
1 fx
fy
x
1
³0 ³0
S
2
5 4 x2
d f x, y d 16`
0 d x y d 16
2
fx
1
27 5 5
12
5 4 x 2 dy dx
^ x, y : 0
R
329
x2 y 2
22. f x, y
triangle with vertices 0, 0 , 1, 0 , 1, 1
Surface Area
2
2 x, f y
2
1 fx
2y
fy
2
1 4x2 4 y2
y
4
³4 ³
S
y=x
1
2S
16 x 2
4
³0 ³0
R
1 4 x 2 4 y 2 dy dx
16 x 2
65 65 1 S
1 4r 2 dr dT
6
y
x
x 2 + y 2 = 16
1
2
2x y
20. f x, y
R
2
triangle with vertices 0, 0 , 2, 0 , 2, 2
2
1 fx
fy
x
2
³0 ³0
S
x
−2
2
2
−2
5 4 y2
5 4 y 2 dy dx
5 § 8 21 37 ·
ln ¨
¸¸ 4 ¨©
5
¹
4 x2 y2
23. f x, y
21 5 5
4
12
R
^ x, y : 0
fx
2 x, f y
y
2
1 fx
d x d 1, 0 d y d 1`
2 y
fy
2
1 4x2 4 y2
3
1
R
1
1
2
9 x y
2
^ x, y : 0
0 d 9 x y 2 Ÿ x2 y 2 d 9
1 fx
S
2
3
³ 3 ³ 2S
3
³0 ³0
S
6
9 x2
9 x2
2
2
1
1 4x2 4 y2
1
³0 ³0
S
2 y
fy
d x d 1, 0 d y d 1`
x1 2 sin x, f y
1 fx
d f x, y `
2 x, fy
cos x
0
2
2
fx
2 32
x
3
^ x, y : 0
R
3
fx
R
1 4 x 2 4 y 2 dy dx | 1.8616
24. f x, y
x
21. f x, y
1
³0 ³0
S
y=x
2
fy
1
2
1
x sin x
25. Surface area ! 4 ˜ 6
x sin x
2
2
dy dx | 1.02185
24
Matches (e)
1 4 x 4 y dy dx
2
2
z
10
1 4r 2 r dr dT
37 37 1 | 117.3187
5
5
y
x
© 2010 Brooks/Cole, Cengage Learning
330
Chapter 14
Multiple Integration
26. Surface area | 9S
Matches (c)
^ x, y : 0
R
z
fx
3
2
y
3
2
1
1
³0 ³0
S
ex
^ x, y : 0
R
fx
ex , f y
2
1 fx
fx
3x2 3 y
0
fy
3x 3 y 2
1 e
1 e
³0
1 y 3 dy | 1.1114
d x d 1, 0 d y d 1`
2x
1
³0 ³0
S
1
³0
1 y 3 dx dy
square with vertices 1, 1 , 1, 1 , 1, 1 , 1, 1
2
1
1 y3
R
fy
1
2
fy
x3 3xy y 3
29. f x, y
27. f x, y
d x d 1, 0 d y d 1`
y3 2
0, f y
1 fx
3
x
2 y5 2
5
28. f x, y
1 e2 x
S
1
1
³ 1 ³ 1
3 y2 x
1 9 x2 y
2
9 y2 x
2
dy dx
dy dx
x 2 3 xy y 2
30. f x, y
| 2.0035
2x
3 x2 y ,
R
fx
^ x, y : 0
d x d 4, 0 d y d x`
2 x 3 y, f y
1 fx f y
3 x 2 y
2
3x 2 y
1 2x 3y
2
3x 2 y
1 13 x 2 y 2
S
e x sin y, f y
2
³2 ³
1 e 2 x sin 2 y e 2 x cos 2 y
4 x2
1 e 2 x dy dx
4 x2
R
S½
­
2
2
® x, y : x y d ¾
2¿
¯
fx
2 x sin x 2 y 2 , f y
2
1 fx
³
S 2
S 2
fx
³
^ x, y : 0
xy
ye , f y
1 fx
S
fy
2
S 2 x2
S 2 x2
2 y sin x 2 y 2
1 4 x 2 sin 2 x 2 y 2 4 y 2 sin 2 x 2 y 2
1 4 ª¬sin 2 x 2 y 2 º¼ x 2 y 2
1 4 x 2 y 2 sin 2 x 2 y 2 dy dx
e xy
33. f x, y
R
1 e 2 x
cos x 2 y 2
32. f x, y
S
1 13 x 2 y 2 dy dx
e x cos y
1 f x2 f y2
S
x
e x sin y
31. f x, y
fx
4
³0 ³0
4
2
10
³0 ³0
d x d 4, 0 d y d 10`
xe xy
fy
2
1 y 2e 2 xy x 2e 2 xy
1 e 2 xy x 2 y 2
1 e 2 xy x 2 y 2 dy dx
© 2010 Brooks/Cole, Cengage Learning
2
Section 14.5
^ x, y : 0
e
fx
x
fy
x
4
d x d 4, 0 d y d x`
e x cos y
sin y, f y
2
1 fx
³0 ³0
S
331
e x sin y
34. f x, y
R
Surface Area
2
1 e2 x sin 2 y e 2 x cos 2 y
1 e 2 x
1 e 2 x dy dx
38. (a) Yes. For example, let R be the square given by
35. See the definition on page 1021.
0 d x d 1, 0 d y d 1,
x 2 y 2 is a paraboloid opening upward.
36. f x, y
and S the square parallel to R given by
Using the figure below, you see that the surface areas
satisfy:
0 d x d 1, 0 d y d 1, z
b c a
1.
(b) Yes. Let R be the region in part (a) and S the surface
given by f x, y
xy.
y
x2 + y2 = 4
(c) No.
2
x
1 x2 ; f x
39. f x, y
12 x 2
1
S
x
1
2
16 ³
37. No, the surface area is the same.
z
f x, y
16 ³
f x, y k
z
and
³R ³
2
1 fx
1
0
1
0
1
x
³0
1 x2
x
1 x2
2
fy
, fy
0
dA
dy dx
dx
ª16 1 x 2
¬«
Sr2
k2 1
12 1
º
¼» 0
16
have the same partial derivatives.
40. f x, y
1 fx
S
³R ³
41. (a) V
x2 y 2
k
fy
2
1 fx
2
2
k 2 x2
k 2 y2
x2 y 2
x2 y2
1
fy
2
³R ³
dA
k 1³
k 2 1 dA
R
³ dA
A k2 1
y
³R ³ f
x, y
8³
³
625 x 2 y 2 dA where R is the region in the first quadrant
8³
S 2
R
0
24
³R ³
8³
R
³4
4³
625 r 2 r dr dT
1 fx
³
20
16
25
8
S
ª¬0 609 609 º¼ ˜
3
2
(b) A
k2 1
2
fy
25
625 x y
2
2
2
812S
dA
dA
S
lim ª200 625 r 2 º ˜
¼4 2
b o 25 ¬
b
2
« 3 625 r
¬
0
3 2º
25
» dT
¼4
12
R
8
4
609 cm3
8³
R
8³
S 2 ª2
³
S 2
0
1
25
³4
100S
x
4
8
12
16
20
24
x2
y2
dA
2
2
625 x y
625 x 2 y 2
25
625 r 2
r dr dT
609 cm 2
© 2010 Brooks/Cole, Cengage Learning
332
Chapter 14
42. (a) z
Multiple Integration
1 3
4 2 16
y y y 25
75
25
15
15 §
(b) V | 2 50
³0
(c) f x, y
fx
0, f y
50
2³
S
0
1 3
4 2 16
·
y y 25 ¸ dy
¨ y 25
15
© 75
¹
100 266.25
26,625 cubic feet
1 3
4 2 16
y y y 25
75
25
15
1
8
16
y2 y 25
25
15
15
³0
1 f y2 f x2 dy dx | 3087.58 sq ft
(d) Arc length | 30.8758
Surface area of roof | 2 50 30.8758
3087.58 sq ft
Section 14.6 Triple Integrals and Applications
1.
3
2
1
³0 ³0 ³0
3
3
³0
2.
1
1
1
³ 1 ³ 1 ³ 1 x
2
³ ³
y z dx dy dz
³ ³
x
xy
9
y 3
4
1
³0 ³0 ³0
4.
³0 ³0 ³0
5.
³ 1 ³ 0 ³ 0 2 ze
x dz dy dx
y2 9 x2
x
x2
³ 0 ³ 0 > xz@0
x
1
xy
y 3
9
³ ³0
1
2 0
z dz dx dy
³ 1 ³ 0 ª¬« 2 ze
4
dy dx dz
1
2 1 ª y 3 z 2 º1
¼ 1
9 1 ¬
³
1
x
³0 ³0
x2
x
yº dx dz
¼» 0
1
³1
4
e2
³ 0 «¬ 2 z yz z2 º
» dy
2 ¼0
18
4 1 z2
9 1
³
dz
1 ª x2 y 2 º
³
9
1
2 0
4
dz
x
1
³ 0 «¬ 2 »¼ dx
0
x 2 y dy dx
y 2 9 x 2 dx dy
ª ze x2 º dz
¬«
¼» 0
4
3
2
3 ª1
·
y z ¸ dz dy
¹
dy dz
y 2 z 2 dy dz
dy dx
2§ 1
3
³ 0 ³ 0 ¨© 2 ª¬3 y y 2 º¼
0
1 2 y 2 dy
2 1 1
3 1 1
1
1
º
xy xz » dz dy
¼0
1
1 1 1 ªx3 y 2 z 2 º
¼ 1
3 1 1 ¬
2 2 2
3.
ª x2
³ 0 ³ 0 «¬ 2
x y z dx dz dy
³0
y 3
ª¬ xy 2 3x3 º¼ dy
0
1
³ 1 ³ 0 2 zxe
x2
³1
2
ª
1 z º
«1 e
»
2 ¼1
¬
z 1 e1 dz
1
ª x5 º
« »
¬10 ¼ 0
x4
dx
2
³
8
27
9
2
18 0
y 3 dy
1
10
9
ª 1 y4 º
¬ 36 ¼ 0
729
4
dx dz
4
4
1
ª 4 z3 º
¬ 27 ¼ 1
15 §
1·
¨1 ¸
2©
e¹
e2
4
e2
4
S 2
1 xz
6.
³1 ³1 ³ 0
7.
³0 ³0 ³0
ln z dy dz dx
1 x
³1 ³1
x cos y dz dy dx
³0
S 2
y 2
3
9 y2
1y
³0 ³0 ³0
9.
³0 ³
9 y2
sin y dz dx dy
y2
³0
y dz dx dy
S 2
4
³0 ³0
1 x
ª¬ x cos y z º¼ 0
S 2
y 2
³0 ³0
ln z
dz dx
xz
sin y
dx dy
y
4
S 2
³0 ³0
dy dx
ª¬ x 1 x sin yº¼ 0 dx
S 2
e2
³1 ³1
ª¬ ln z yº¼ 0 dz dx
4
8.
4
1 xz
4
³1
ª ln z 2 º
«
» dx
«¬ 2 x »¼1
³1
2
dx
x
8
64
3
4
³0
x 1 x dx
1 S2
sin y dy
2³0
2 ln 4
x 1 x cos y dy dx
4
4
4
¬ª2 ln x ¼º1
ª x2
x3 º
« »
3 ¼0
¬2
S 2
ª 1
º
« 2 cos y»
¬
¼0
40
3
1
2
324
5
© 2010 Brooks/Cole, Cengage Learning
Section 14.6
2
10.
³0 ³0
11.
³0 ³0
2 x2
4 x2
2
4 y2
4
³1
³0 ³0
x 2 sin y
dz dy dx
z
2
3
2 2y 3
6 2 y 3z
³0 ³0
³0
ze x
2 y2
4 x2
2
³0 ³0
³ 0 ¬ªx
12.
2 x2
2
³ 2 x2 y2 y dz dy dx
2
4
6
6 x 2
5
5 x
3
2x
5 x y
³0 ³0 ³0
14. V
³0 ³0 ³0
15. V
³ 6 ³
16. V
³4 ³
17. z
1
2
9 x2
4
V
18. V
z
19. V
6 x2 y 2
16 x 2
³0
16 x2
4
4 2 x2
1
x2
y2
2
43
0 Ÿ z 8 z 10
0 Ÿ z
2
ª16 y ¬
8 Ÿ x2 y2
2z
16
ellipse
x
2
³ 0 dz dx dy
2 xy
20. V
³0 ³0 ³0
21. V
8³
dz dy dx
a2 x2
³0
4 y2
³ 2 ³0
³
0 ³0
dz dx dy
dz dy dx
1 2
2 2
a
6 x2 y 2
³0
6 y2
³ 2 2 dz dy dx
3 x 3y
2x2 3 y 2
2
6 y2
4 x2
3
4
³2 ³0
2
6
³ 6 ³
x2 3 y2
4 y2
2
dz dy dx
2
80 Ÿ z 2 2 z 80
³ 1 2 x2 y 2
4 2 x2
4 x2
2 y2
dz dy dx
80 x 2 y 2
16 x 2
2
ze x
ln 4 ª1 cos 4 x 2 º dx | 2.44167
¬
¼
x2 y2
2z z2
16 x 2
³ 2 ³
dz dy dx
16 x2 y 2
x y 2 Ÿ 2z
³4 ³
6 x2y 3
2
dz dy dx
³0
6 x2
x2 y2 z 2
³0
2
³0 x
dx
dz dy dx
6 x2
6
4 x2
1 § 6 x 2 y · x2 y 2
dy dx | 2.118
¨
¸ e
2©
3
¹
3 x 2
³0 ³0
13. V
16 2
15
4 y 2 x 2 y 2 y 3 dy dx
ln 4 cos y º¼
0
6
333
ª¬ x 2 sin y ln z º¼ dy dx
1
³0 ³0
dx dz dy
Triple Integrals and Applications
2
x dx dy
4 y2
2
2
2
³ 0 ³ 0 2 xy dy dx
a 2 x2 y 2
dz dy dx
8³
4³
³0
dy
³ 0 ª¬xy
a
0 ³0
a
0
ª
«y
¬«
16 8 y 2 y 4 dy
2 2
º¼
0
a2 x2
dx
2
³ 0 4 x dx
2
1 y5 º
5 ¼0
256
15
8
a 2 x 2 y 2 dy dx
a x y a x
2
8 3
y
3
2
§S · a
4¨ ¸ ³ a 2 x 2 dx
©2¹ 0
2
2
2
§
arcsin ¨
©
a
ª § 2
1 3 ·º
«2S ¨ a x 3 x ¸»
¹¼ 0
¬ ©
·º
¸»
2
2
a x ¹¼» 0
y
a 2 x2
dx
4 3
Sa
3
© 2010 Brooks/Cole, Cengage Learning
334
Chapter 14
4³
22. V
6
36 x 2
³0
0
Multiple Integration
36 x 2 y 2
³0
4³
dz dy dx
6
0
³0
36 x 2
36 x y
2
6ª
4³ «36 36 x 2 x 2
0
¬
4 x2
2
³0 ³0
23. V
4 x2
³0
2
2 x2
2
2 x2
9 x3
³0 ³0
24. V
³0
2
³0
2
3 2º
ª
§ x· 1
4 «9 x 36 x 2 324 arcsin ¨ ¸ x 36 x 2
©6¹ 6
¬
3 2º
4 x2
2
2
³0
dx
36 x 2 dx
6
4 162S
»
¼0
2
648S
256
15
z
9 x 3 dy dx
9 x 3 2 x 2 dx
−1
y
1
x
18 9 x 2 2 x 3 x 5 dx
1
4
18 2 6 2 2 3
2
12 2 3
3
4 y2
2
³ 0 ³ 0 ³ 2 y
3
2
³0 ³0
z
dy dz dx
z
28.
1
2
dz dy dx
1
x
y
−1
ª¬4 y 2 2 yº¼ dy dx
z
1
1 y2
³ 1 ³ 0
2
ª
y3
y2 º
«2 y » dx
3
2 ¼0
¬
4
1 z
³ y2
dx dz dy
29. Plane: 3x 6 y 4 z
3§
8
·
³ 0 ¨© 4 3 2 ¸¹ dx
3
3
ª10 º
« 3 x»
¬
¼0
1
³ 0 ³ 0 ³ 1
ª
x º
1 4
3
«18 x 3x x »
2
6 ¼0
¬
³0
36 x 2
» dx
¼
ª16 x 83 x3 15 x5 º
¬
¼0
16 8 x 2 x 4 dx
27.
2
3
0
ª
y3 º
2
«36 y x y »
3 ¼0
¬
1
36 x 2
3
dz dy dx
6
25. V
4³
dy dx
6
1
³0 ³0
³0
2
³0
dz dy dx
2
3
10
2
12 4 z 3
³0 ³0
1
3
y
12
12 4 z 3 x 6
³0
dy dx dz
z
x
3
26. The region in the xy-plane is:
x2
2
x
2
³ 0 ³ x2 ³ 0 dz dy dx
V
x2
³ 0 > xy@x2
2
2
³0
dx
x2
³ 0 ³ x2
x dy dx
x x 2 x3 dx
2
ª x3
x4 º
2
« x »
4 ¼0
¬3
8
4 4
3
2
3
y
4
8
3
x
30. Top plane: x y z
6
Side cylinder: x 2 y 2
9
z
y
y = x +2
3
³0 ³0
4
9 y2
6 x y
³0
6
dz dx dy
3
2
y = x2
3
1
x
1
2
3
4
3
6
y
6
x
© 2010 Brooks/Cole, Cengage Learning
Section 14.6
Triple Integrals and Applications
335
z
31. Top cylinder: y z
2
Side plane: x
1
x
³0 ³0 ³0
1 y2
2
z
1
y
32. Elliptic cone: 4x z
2
4
4
³0 ³ z ³0
1
y2 z2 2
2
y
2
5
4
dx dy dz
3
2
dz dy dx
1
3
1
x
1
2
x
1
5
y
33. Q
^ x, y , z : 0
³ ³³ xyz dV
Q
3
d x d 1, 0 d y d x, 0 d z d 3`
1
1
³ 0 ³ 0 ³ y xyz dx dy dz
3
y
x
1
³ 0 ³ 0 ³ 0 xyz dy dx dz
1
3
1
1
1
3
³ 0 ³ 0 ³ y xyz dx dz dy
³0 ³ y ³0
34. Q
^ x, y , z : 0 d
³ ³³ xyz dV
Q
2
xyz dz dx dy
1
3
x
1
x
3
³ 0 ³ 0 ³ 0 xyz dy dz dx
³0 ³0 ³0
R
x
1
9
16
xyz dz dy dx
x d 2, x 2 d y d 4, 0 d z d 2 x`
2 x
4
³ 0 ³ x2 ³ 0
xyz dz dy dx
z
4
2 x
y
4
³0 ³0 ³0
2
2 x
4
2
2 z
4
xyz dz dx dy
2
³ 0 ³ 0 ³ x2 xyz dy dz dx
2
2
³0 ³0
2 z 2
2
4
y
³0 ³0
^ x, y , z : x 2
4
4
³ 0 ³ 0 ³ x2 xyz dy dx dz
35. Q
y=x
1
y
(2, 4)
x
2 z
³0
y
xyz dx dy dz ³0 ³ 2 z 2 ³ 0
³0
y
xyz dx dz dy ³0 ³ 2 y ³ 0
2
4
4
2
2 z
xyz dx dy dz
104
21
dx dz dy
y 2 d 9, 0 d z d 4`
z
5
³³ ³ xyz dV
Q
4
3
³ 0 ³ 3 ³ 3
4
³ 3 ³ 0 ³ 3
4
³ 3 ³ 0 ³ 36. Q
^ x, y , z : 0 d
³ ³³ xyz dV
Q
9 x2
9 x2
9 y2
9 y2
9 x2
9 x2
4
3
xyz dy dx dz
³ 0 ³ 3 ³ xyz dx dz dy
³ 3 ³ xyz dy dz dx
³ 3 ³ 3
3
9 y2
9 y2
9 y2
9 y2
9 x2
9 x2
xyz dx dy dz
4
³ 0 xyz dz dx dy
4
³ 0 xyz dz dy dx
4
0
x d 1, y d 1 x 2 , 0 d z d 6`
1
1 x 2
1
6
1
6
6
³ 0 ³ 0 ³ 0 xyz dz dy dx
³0 ³0 ³0
1 y
1 x2
³0 ³0 ³0
xyz dx dz dy
xyz dy dz dx
³ 0 xyz dz dx dy
6
1
1 y
³0 ³0 ³0
6
1
4
y
6
6
1 x 2
³0 ³0 ³0
3
z
1 y
1
³0 ³0
3
x
xyz dx dy dz
xyz dy dx dz
3
2
1
2
1
2
y
x
© 2010 Brooks/Cole, Cengage Learning
y
336
Chapter 14 Multiple Integration
^ x, y , z : 0 d
37. Q
1 y 2
1
1 y
³0 ³0 ³0
y d 1, 0 d x d 1 y 2 , 0 d z d 1 y`
1
1 x
1
2 z z2
1
1
³0 ³0
dz dx dy
1 y
³0
³0
1 y 2
1 y
³0 ³0 ³0
^ x, y , z : 0 d
38. Q
1−x
1
³0
1 x
z=1−
1
1 x
1
³ 0 ³ 2 z z2 ³ 0
dy dx dz 1 z
³ 0 ³ 1
1
dz dy dx
1 z
³0 ³0
z
dy dz dx 1
1 1 x
³0 ³0
1 z
1
³0
1 y 2
³0 ³0 ³0
dx dz dy
y+z=1
dy dx dz
1 x
dy dz dx
1
5
12
dx dy dz
1
x
y
x = 1 − y2
x d 3, 0 d y d x, 0 d z d 9 x 2`
z
9
x
3
9 x2
³0 ³0 ³0
z = 9 − x2
9 x2
3
3
3
9 x2
³0 ³ y ³0
dz dy dx
9
³ 0 dy dz dx
9 z
9
z = 9 − y2
x
³0 ³0
³0 ³0
dz dx dy
9 z
³y
³0 ³0
9 z
3
9 y2
³0 ³0
dx dy dz
x
³ 0 dy dx dz
³y
9 z
dx dz dy
81
4
y=x
3
x
3
y
39.
40.
41.
4 2x 3
6
m
k³
0
M yz
k³
0
³0
4 2x 3
6
³0
x
M yz
m
m
k³
³0
0 ³0
M xz
k³
5
5 x
5 x
m
k³
4
1 5 15 3 x 3 y
1 5 15 3 x 3 y
4k ³
4
4
0
4
4 x
x dz dy dx
y 2 dz dy dx
4
4 x
k³
4
0
xz dz dy dx
2
k³
3
0
m
M xz
y
M xy
m
b
k³
0
k³
0
b
M xz
m
4
³0 x 4 x
4
0
4
³0 x
4x
2
a ¬ª1 y b ¼º
a ª¬1 y b º¼
³0
c ¬ª1 y b x a ¼º
³0
c ª¬1 y b x a º¼
³0
kab 2c 24
kabc 6
b
4
dz dx dy
y dz dx dy
kabc
6
kab 2c
24
b
b
b
b
b
b
M yz
k³
0
M xz
k³
xy
0 ³0 ³0
M xy
k³
0
b
b
b
³0 ³0 x
2
y dz dy dx
kb 6
6
³ 0 ³ 0 xyz dz dy dx
kb 6
8
m
kb 6 6
kb5 4
2b
3
M xz
m
kb6 6
kb5 4
2b
3
M xy
kb 6 8
kb5 4
b
2
M yz
m
a
b
c
b
c
b
c
b
c
k³
0
M xy
k³
0
M yz
k³
xz dz dy dx
0 ³0 ³0
M xz
k³
0
x
y
z
a
a
a
³ 0 ³ 0 z dz dy dx
³0 ³0 z
2
dz dy dx
³ 0 ³ 0 yz dz dy dx
m
ka 2bc 2 4
kabc 2 2
a
2
M xz
m
kab 2c 2 4
kabc 2 2
b
2
M xy
kabc3 3
kabc 2 2
2c
3
M yz
m
kb5
4
kb6
6
dz dy dx
2
m
1
³0
b
xy dz dy dx
0 ³0 ³0
dy dx
128k
3
b
k³
z
44.
2
b
m
x
dy dx
128k
3
2k ³ 16 x 8 x x dx
42.
12k
125
k
8
125
k
4
y dz dy dz
x dz dy dx
4
z
43.
y
4 x x 2 dx
³0 ³0
0
8k
2
³0 ³0
0
dz dy dx
3
2
³0
0 ³0
y
k³
2 y 2 x 3
³0
12k
8k
5
M xz
m
M xy
2 y 2 x 3
³0
kabc 2
2
kabc3
3
ka 2bc 2
4
kab 2c 2
4
© 2010 Brooks/Cole, Cengage Learning
Section 14.6
45. x will be greater than 2, whereas y and z will be
unchanged.
50.
46. z will be greater than 8 5, whereas x and y will be
Triple Integrals and Applications
m
M xz
k³
3 ³ 0
M xy
k³
3
x
49.
1
kS r 2 h
3
y
0 by symmetry
x
4k ³
M xy
r 2 x2
³0
y
h
³h
x2 y 2 r
z
z dz dy dx
r 2 x2
2kh 2 r
r 2 x 2 y 2 dy dx
2 ³0 ³0
r
32
kS r 2 h 2
4kh 2 r 2
r x2
dx
2 ³0
3r
4
M xy
kS r 2 h 2 4
3h
m
kS r 2 h 3
4
z
m
x
z
M xy
³0
3
3
³0
M yz
9 x2
9 x2
y
³0
y
³0
y
³0
18k
x dz dy dx
0
y dz dy dx
81S
k
8
z dz dy dx
81S
k
16
0
m
M xz
m
M xy
x, y, z
9 x2
y
³ 0 dz dy dx
9S
16
9S
32
m
§ 9S 9S ·
,
¨ 0,
¸
© 16 32 ¹
128kS
3
y
0 by symmetry
42 x 2 y 2
4k ³
2k ³
4
0
4
0
42 x 2
³0
42 x 2
³0
³0
42 x 2 y 2
z dz dy dx
42 x 2 y 2 dy dx
32
4k 4 2
4 x2
dx
3 ³0
1024k S 2
cos 4 T dT
let x
3 ³0
64S k
by Wallis's Formula
z
x, y, z
3
9 x2
3h ·
§
¨ 0, 0, ¸
4¹
©
x, y, z
51.
r
0
3
³0
k³
48. x , y and z will all be greater than their original values.
m
3
0
M yz
unchanged.
47. y will be greater than 0, whereas x and z will be
unchanged.
2k ³
337
M xy
m
64kS
3
˜
1
128kS
4ª
1 º
2k ³ «16 y x 2 y y 3 »
0
3 ¼0
¬
42 x 2
dx
4 sin T
3
2
3·
§
¨ 0, 0, ¸
2¹
©
© 2010 Brooks/Cole, Cengage Learning
338
Chapter 14 Multiple Integration
x
0
m
2k ³
0
M xz
2k ³
0
M xy
2k ³
0
52.
k³
x, y, z
20
m
k³
0
M yz
k³
0
y
z
20
20
k³
0
k³
0
20
5
k³
0
M yz
k³
0
M xz
k³
M xy
k³
x, y, z
1
³ 0 y2
y
dy dx
1
k³
2
ln 2 dx
0
y 1
k ln 4
z dz dy dx
1
2
kS
2
§1 S · 2
k ¨ ¸ ³ dx
8¹ 0
©4
2ª
º
y
1
arctan y» dx
k³ «
0 2( y 2 1)
2
¬
¼0
dy dx
§1 S ·
k¨ ¸
4¹
©2
2S
4S
3 5 x 12
3 5 x 12
³0
3 5 x 12
³0
3 5 x 12
³0
³0
5 12 y
³0
5 12 y
³0
5 12 y
³0
5 12 y
dz dy dx
200k
x dz dy dx
y
20
1000k
16
y dz dy dx
y = − 35 x + 12
12
1200k
8
z dz dy dx
250k
4
x
4
1000k
200k
1200k
200k
250k
200k
8
12
16
20
5
6
5
4
1
60 12 x 20 y
15
m
z
2
0
§S · 2
2k ¨ ¸ ³ dx
©4¹ 0
1
dy dx
1
5·
§
¨ 5, 6, ¸
4¹
©
54. f x, y
y
³0
0
1
³ 0 y2
2k ³
y dz dy dx
1
1
³0
M yz
m
M xz
m
M xy
m
x, y, z
x
2
1 y 2 1
1
³0 ³0
2
0
5
y
12
53. f x, y
x
2
1 y 2 1
1
³0 ³0
2k ³
dz dy dx
k ln 4
ln 4
S
kS
S
1
§
·
k ¨ ¸ kS
4¹
©2
§ ln 4 2 S ·
,
¨ 0,
¸
4S ¹
© S
z
M xy
2
1 y 2 1
1
³0 ³0
M xz
m
M xy
m
y
M xz
2
5
5
0
5
0
M yz
m
M xz
m
M xy
m
3 5 x3
³0
3 5 x3
³0
³0
³0
3 5 x3
³0
3 5 x3
³0
1 15 60 12 x 20 y
dz dy dx
1 15 60 12 x 20 y
1 15 60 12 x 20 y
³0
1 15 60 12 x 20 y
³0
25k 2
10k
15k 2
10k
10k
1
10k
5
4
3
4
x dz dy dx
y dz dy dx
z dz dy dx
10k
25k
2
15k
2
10k
y
5
4
y = 35 (5 − x)
3
2
1
x
1
2
3
4
5
§5 3 ·
¨ , , 1¸
©4 4 ¹
© 2010 Brooks/Cole, Cengage Learning
Section 14.6
55. (a) I x
k³
a
0
a
a
³0 ³0
a
a ª1
º
ka ³ « y 3 z 2 y» dz
0 3
¬
¼0
Ix
Iy
(b) I x
k³
a
a
a
³0 ³0
a 2
a 2
56. (a) I xy
k³
I xz
I yz
Ix
Iy
(b) I xy
k³
I xy
k³
I yz
I xz by symmetry
Ix
I xy I xz
a7k
30
Iy
I xy I yz
a7k
30
Iz
I yz I xz
7 ka 7
180
57. (a) I x
k³
z
a 2 ³a 2 ³a 2
a 2
a 2
2
a 2
a 2
a 2
a 2
³a 2 ³ a 2 z
2
y
a 2 ³a 2 ³ a 2
4
0
4
4
0
4 x
³0 ³0
4
2
x 2 y 2 dz dy dx
a 3k a 2 a 2 2
x y 2 dy dx
12 ³ a 2 ³ a 2
x 2 y 2 dz dy dx
ka ³
k³
y 2 z 2 dz dy dx
4 x
³0 ³0
x 2 z 2 dz dy dx
4ª
1
3º
4k ³ «4 x 2 x3 4 x » dx
0
3
¬
¼
Iz
k³
4
0
4
ka8
8
ka 5
6
4
k³
a
ka 5
12
dz dy dx
4 ª y3
y
3º
k ³ « 4 x 4 x » dx
0
3
3
¬
¼0
Iy
2ka 5
3
ª ka 4 § a 2 z 2
2 z 4 ·º
«
¨
¸»
4 ¹»¼ 0
«¬ 8 © 2
ka 4 a 2
a z 2 z 3 dz
8 ³0
ka 5
by symmetry
12
ka 5
ka 5
Iz
12
12
a 2
a
ª §1 3
1 3 ·º
«ka¨ 3 a z 3 az ¸»
©
¹¼ 0
¬
ka8
by symmetry
8
Iz
a 2
y 2 z 2 dy dz
ka 2 a a 3
y z yz 3 dy dz
2 ³0 ³0
y 2 z 2 xyz dx dy dz
a
Iy
a
³0
a §1
·
ka ³ ¨ a 3 az 2 ¸ dz
0 3
©
¹
ka 2 a ª y 4 z
y2 z3 º
«
» dz
³
2 0¬ 4
2 ¼0
Ix
a
0
339
2ka 5
by symmetry
3
Iz
0
ka ³
y 2 z 2 dx dy dz
Triple Integrals and Applications
4 x
³0 ³0
x 2 y 2 dz dy dx
4
º
4 ª§
y3 ·
k ³ «¨ x 2 y ¸ 4 x » dx
0
3¹
©
¼0
4
0
4
ª
³ 0 «¬ y
2
a 2
a 2
a 2 ³a 2
x 2 y 2 y 4 dy dx
4 x 1
3º
4 x » dy dx
3
¼
4 ª 64
4
3º
k ³ « 4 x 4 x » dx
0
3
3
¬
¼
k³
4
0
4
ª
³ 0 «¬x
2
4 x 4
4
0
4
³0
7 ka 7
360
ª 32
4 x
k «
¬ 3
4
2
1
4º
4 x »
3
¼0
256k
1
3º
4 x » dy dx
3
¼
1
1
4º
ª4
4k « x 3 x 4 4 x »
4
12
¬3
¼0
k³
a7k
72
512k
3
x 2 y 2 4 x dy dx
4§
64 ·
k ³ ¨ 4 x2 ¸ 4 x dx
0
3¹
©
256k
© 2010 Brooks/Cole, Cengage Learning
340
Chapter 14 Multiple Integration
(b) I x
k³
4
0
4 x
4
³0 ³0
4
4 ª y4
y2
3º
4x 4 x » dx
k³ «
0
4
6
¬
¼0
Iy
k³
4
0
4 x
4
³0 ³0
k³
4
0
4 x
4
³0 ³0
0
4
ª
³ 0 «¬ y
4x 3
1
3º
y 4 x » dy dx
3
¼
4ª
8
3º
k ³ «64 4 x 4 x » dx
0
3
¬
¼
k³
y x 2 z 2 dz dy dx
4
0
ª
4
³ 0 «¬x
2
1
1
4º
ª4
8k « x 3 x 4 4x »
4
12
¬3
¼0
k³
y x 2 y 2 dz dy dx
4
4
³0
0
ª
k «32 4 x
¬
4
2
2
4º
4 x »
3
¼0
2048k
3
1
3º
y 4 x » dy dx
3
¼
y4 x 4
4ª
1
3º
8k ³ «4 x 2 x3 4 x » dx
0
3
¬
¼
Iz
4
k³
y y 2 z 2 dz dy dx
1024k
3
x 2 y y 3 4 x dx
4
º
4 ª§ x 2 y 2
y4 ·
k ³ «¨
¸ 4 x » dx
0
4¹
© 2
¼0
8k ³
58. (a) I xy
k³
4
0
k³
4
8 x 2 64 4 x dx
0
4
ª §
4 3 1 4 ·º
2
«8k ¨ 32 x 4 x 3 x 4 x ¸»
¹¼ 0
¬ ©
32 8 x 4 x 2 x3 dx
4
4 y2
2
0 ³0 ³0
z 3 dz dy dx
k³
4
2
1
³0 4 4 0
4
y2
2
I xz
I yz
k³
4
4 y2
2
0
³0 ³0
4
2
0
k³
0
³0 ³0
4
2
k³
0
4 y2
2
³0
I xz I xy
(b) I xy
³0 ³0 ³0
k³
I xz
4
Ix
4 y2
2
4 y2
2
4
k³
4
0
2
³0 ³0
k³
4
0 ³0 ³0
2
4 y2
k³
4
2
4 y2
0 ³0 ³0
I xz I xy
4 y2
2
4 16,384
0
945
dx
dy dx
2
k 4 ª16 y 3
8 y5
y7 º
«
» dx
³
2 0¬ 3
5
7 ¼0
2
1
³0 2x
2
4 y2
2
65,536k
315
k 4 1024
dx
2 ³ 0 105
dy dx
2
k 4 ª 2§
8 y3
y 5 ·º
« x ¨16 y ¸» dx
³
0
2 ¬ ©
3
5 ¹¼ 0
8192k
, Iz
21
I yz I xy
2048k
105
I yz I xz
k 4 256 2
x dx
2 ³ 0 15
8192k
45
63,488k
315
4 z 2 dz dy dx k ³
4
0
2
4 y2
2
4 y2
2
4 y2
³0 ³0
z 3 dz dy dx
32,768k
65,536k
105
315
32,768k
315
y 2 4 z dz dy dx
4 y2
0
2
k³
z 2 4 z dz dy dx
³0 ³0
³0 ³0 ³0
k³
I yz
4
0
x 2 z dz dy dx
2048k
, Iy
9
4 y2
2
1
2
³0 2 y
0
1 2
x 16 8 y 2 y 4 dy dx
2
Ix
4
4
k³
1
16 y 2 8 y 4 y 6 dy dx
2
k³
4
³0
y 2 z dz dy dx
k 4 2
256 256 y 2 96 y 4 16 y 6 y 8 dy dx
4³0 ³0
dy dx
256 y 3
96 y 5 16 y 7
k 4ª
y9 º
«256 y » dx
³
0
4 ¬
3
5
7
9 ¼0
2048k
3
4
4 y 2 dz dy dx k ³
0
³0 ³0
y 2 z dz dy dx
1024k
2048k
15
105
1024k
21
x 2 z dz dy dx
4096k
8192k
9
45
11,264k
I xz I yz
35
4096k
15
x 2 4 z dz dy dx
4 x 2 dz dy dx k ³
48,128k
, Iy
315
4
0 ³0 ³0
I yz I xy
118,784k
, Iz
315
© 2010 Brooks/Cole, Cengage Learning
Section 14.6
k³
59. I xy
L2
a
L 2
³a ³
a2 x2
a2 x2
z 2 dz dx dy
k³
L2
L 2
2
a
³a 3 a
2
x2
S a 2 Lk , I xy
Because m
k³
I xz
L2
2k ³
k³
I yz
a
L 2
a
a2 x2
a
³ a ³
Iy
I xy I yz
Iz
I xz I yz
a2 x2
a2 x2
a 2
b2
c2
a 2
b2
c2
a 2
b2
³ c 2 ³ a 2 ³ b 2 y
I yz
³ c 2 ³ a 2 ³ b 2 x
Ix
I xy I xz
Iy
I xy I yz
Iz
I xz I yz
1 x
62.
³ 1 ³ 1
63. U
1 x2
2
a 2 x 2 dx dy
L2
2
a
L 2
³a x
kS a 2 ³
L2
L 2
dz dy dx
2
dz dy dx
xº
dy
a »¼ a
ka 4S L 2
dy
4 ³L 2
ka 4S L
4
ma 2
4
b3 c 2 a 2
dy dx
12 ³ c 2 ³ a 2
1 2
1
b abc
mb 2
12
12
c 2
a 2
1 2
ba 3 c 2
ba 3c
b³
y 2 dy dx
dx
a abc
³
c
c 2 ³ a 2
2
12
12
12
c2
1 2
1
abc3
c abc
mc 2
ab ³
x 2 dx
c 2
12
12
12
dz dy dx
2
1
mL2
12
a 2 x 2 dx dy
2
a
a 2 x 2 a 4 arcsin
2kS a 2 § L3 ·
¨ ¸
3 ©8¹
y 2 dy
1
ma 2
12
1
m a 2 b2
12
1
m b2 c2
12
1
m a2 c2
12
x2 y2
4 x2 y 2
1 x2
2k ³
ma 2
2
m
3a 2 L2
12
I xz
1
x 2 dz dx dy
ma 2
ma 2
4
4
2
mL
ma 2
12
4
c2
a
³ a y
x ·º
¸» dy
a ¹¼ a
m
3a 2 L2
12
³ c 2 ³ a 2 ³ b 2 z
1
L2
L 2
ma 2
mL2
4
12
60. I xy
³ 1 ³ 1 ³ 0
2k ³
a 2 x 2 a 2 arcsin
1ª
x 2x2 a2
L 2 8«
¬
I xy I xz
61.
y 2 dz dx dy
L2
Ix
a 4S Lk
4
a
ª y2 §
x
L 2 « 2 ¨
¬ ©
L2
x ºº
x 2 a 2 a 4 arcsin »» dy
a ¼¼ a
ma 2 4.
a2 x2
L2
L 2
2k ³
³ a ³
341
a 2 x 2 dx dy
x· 1
2 L 2 ª a2 §
k « ¨ x a 2 x 2 a 2 arcsin ¸ x 2 x 2 a 2
a¹ 8
3³L 2 ¬ 2 ©
a 4S ·
2k L 2 § a 4S
2¨
¸ dy
³
L
2
3
16 ¹
© 4
Triple Integrals and Applications
³0
x 2 y 2 z 2 dz dy dx
kx 2 x 2 y 2 dz dy dx
kz
2
(a) m
³ 2 ³ (b) x
y
z
(c) I z
4 x2
4 x2
4 x2 y2
³0
§
kz dz dy dx ¨
©
z
32kS ·
¸
3 ¹
4
z = 4 − x2 − y2
0 by symmetry
M xy
m
2
³2 ³
1 2
m³2 ³
4 x2
4 x2
4 x2
4 x2
4 x2 y2
³0
4 x2 y2
³0
kz 2 dz dy dx
§
x 2 y 2 kz dz dy dx ¨
©
2
32kS ·
¸
3 ¹
y
2
2
x
x2
+
y2
=4
© 2010 Brooks/Cole, Cengage Learning
Chapter 14 Multiple Integration
342
64. U
kxy
(a) m
5
³0 ³0
25 x 2
M yz
(b) x
25 x 2 y 2
1 5
m³0 ³0
m
y
³0
25 x 2
§
kxy dz dy dx ¨
©
³0
625 ·
k¸
3 ¹
25 x 2 y 2
§
x kxy dz dy dx ¨
©
25S ·
¸
32 ¹
25 x 2 y 2
§
z kxy dz dy dx ¨
©
25 ·
¸
16 ¹
z
x 2 + y 2 + z 2 = 25
5
x by symmetry
M xy
z
1 5
m³0 ³0
m
(c) I z
5
³0 ³0
25 x2
³0
25 x 2
³0
25 x 2 y 2
§
x 2 y 2 kxy dz dy dx ¨
©
5
y
5
62500 ·
k¸
21 ¹
65. See the definition, page 1027.
x
69. V
1 unit cube
See Theorem 14.4, page 1028.
Average value
66. Because the density increases as you move away from
the axis of symmetry, the moment of inertia will
increase.
1
1
1
1
³0 ³0
(b) The annular solid on the right has the greater
moment of inertia.
1
z 2 4 dx dy dz
z 2 4 dy dz
1
ª z3
º
« 4 z»
3
¬
¼0
(c) The solid on the left will reach the bottom first. The
solid on the right has a greater resistance to
rotational motion.
70. V
68. The region of integration is a cube:
1
4
3
1
³0
z 2 4 dz
13
3
64 cube with sides of length 4
Average value
z
1
f x, y , z dV
V ³ Q³³
1 4 4 4
xyz dx dy dz
64 ³ 0 ³ 0 ³ 0
1 4 4
8 yz dy dz
64 ³ 0 ³ 0
4
1 4
8 z dz
³ 0 z dz 8
8³0
1
2
1
f x, y , z dV
V ³ Q³³
³0 ³0 ³0
67. (a) The annular solid on the right has the greater density.
3
x 2 + y 2 = 25
y
x
Answer: (a)
71. V
1
base u height
3
1§ 1
4
·
¨ 2 2¸2
3© 2
3
¹
f x, y , z
z
2
(0, 0, 2)
x + y+ z = 2
x y z
Plane: x y z
Average value
2
1
f x, y , z dV
V ³ Q³³
(0, 2, 0)
(2, 0, 0)
x
2
3 2 2 x 2 x y
x y z dz dy dx
4³0 ³0 ³0
3 21
3
3
2
x 4 x 2 dx
2
4³0 6
4
2
2
y
3 2 2 x 1
2 x y x y 2 dy dx
4³0 ³0 2
© 2010 Brooks/Cole, Cengage Learning
Section 14.6
4
S
3
72. V
3
3
Triple Integrals and Applications
343
4 3S
1
f x, y, z dV
V ³ Q³³
Average value
1
4 3S ³ 3
3
³
3 x2
3 x2
³
3 x2 y2
3 x2 y2
x y dz dy dx
0, by symmetry
73. 1 2 x 2 y 2 3 z 2 t 0
z
2 x 2 y 2 3z 2 d 1
^ x, y , z : 2 x 2
Q
1
2
³ 1 2 ³ 1 2 x2
1 2 x2
Exact value:
y 2 3 z 2 d 1` ellipsoid
1 2 x2 y2
³
3
1 2 x2 y 2
3
1
1 2 x 2 y 2 3z 2 dz dy dx | 0.684
2x 2 + y 2 + 3z 2 = 1
1
1
y
x
4 6S
45
−1
74. 1 x 2 y 2 z 2 t 0
x2 y2 z 2 d 1
^ x, y , z : x 2
Q
1
³ 1 ³ 1 x2
³
1 x2
Exact value:
75.
14
15
1
`
y 2 z 2 d 1 sphere
1 x2 y2
1 x2 y2
1 x 2 y 2 z 2 dz dy dx | 1.6755
8S
15
3 a y2
4 x y2
³0 ³0
³a
dz dx dy
3 a y2
3 a y2
1ª
x2 º
³ 0 «¬ 4 y a x 2 »¼
0
2
So, 3a 2 22a 32
0
a
2,
y2
z2
2
9
b
dy
4 x y 2 a dx dy
1ª
2
2
³ 0 «« 4 y a 3 a y ¬
3 a y2 º
» dy
2
»¼
94 11a
1
a2
15
3
2
0
a 2 3a 16
76. x 2 1
³0 ³0
16
.
3
1
By symmetry, the volume in the first octant is
1
16S
8
2S
2S .
1
b 1 x2
³0 ³0
3 1 x 2 y 2 b2
³0
By trial and error, b
1 dz dy dz
4.
ª
x2
y2
z2
«Note: Volume at ellipsoid 2 2 2
a
b
c
¬
º
4
1 is S abc.»
3
¼
© 2010 Brooks/Cole, Cengage Learning
344
Chapter 14 Multiple Integration
1 xk .
77. Let yk
S
2n
S
x1 " xn
2n
S
n y1 y2 " yn
2
S
y1 " yn
2n
So,
1
1
1
0
0
³ 0 ³ 0 "³ 0 cos
I1
0
³ 1 ³ 1 "³ 1 sin
I1 I 2
S
½
x1 " xn ¾ dx1 dx2 " dxn
®
¯ 2n
¿
2­
S
½
y1 " yn ¾ dy1 dy2 " dyn
®
¯ 2n
¿
2­
1
1
1
³ 0 ³ 0 "³ 0 sin
S
½
x1 " xn ¾ dx1 dx2 " dxn
®
¯ 2n
¿
2­
I2
1
.
2
1 Ÿ I1
1
.
2
Finally, lim I1
nof
Section 14.7 Triple Integrals in Cylindrical and Spherical Coordinates
5
S 2
3
1.
³ 1 ³ 0 ³ 0 r cos T dr dT dz
2.
³0 ³0 ³0
S 4
6r
6
rz dz dr dT
S 4
S 4
3.
2 cos 2 T
³0 ³0
4 r2
³0
6
6r
ª rz 2 º
»
2 ¼0
S
2
r sin T dz dr dT
3
³0 ³0 ³0
5.
³0 ³0 ³0
6.
³0 ³0 ³0
2S
S 4
S 4
e U U 2 d U dT dI
cos I
S 4
S 2
2 cos 2 T
S 2
³0 ³0
S
³0 ³0
U 2 sin I d U dI dT
cos T
6
6
³0
S 2
S 4
U 2 sin I cos I d U dT dI
3
S 4
³0
dz
5
5
³ 1 92 dz
ª 9 zº
¬ 2 ¼ 1
9
2
5 1
§S ·
54¨ ¸
©4¹
1
108 dT
2
r 4 r sin T dr dT
S 2
³0
27S
2
2 cos 2 T
ª§ 2
º
r4 ·
«¨ 2r ¸sin T »
4
¹
¬«©
¼» 0
dT
S 2
ª¬8 cos 4 T 4 cos8 T º¼ sin T dT
2
S 4
³0
S 2
S1
³0 ³0 3
cos3 I sin I dI dT
1 S
3³ 0
1 S
3³ 0
S 4
4
³0
4
³0
S 4
27
12r 2 36r dr dT
2
ª 1 U3 º
« 3 e » dT dI
¬
¼0
1 2S
3³ 0
1
³0 ³0 2 r
dr dT
1 ªr4
3
2º
« 4r 18r » dT
2¬ 4
¼0
S 2
4.
S 2
5
³ 1 ª¬ 92 sin T º¼ 0
cos T dT dz
9
2
³ 0 ³ 0 «¬
³0
S 2
S 2
5
³ 1 ³ 0
ª 8 cos5 T
4 cos9 T º
«
»
9
¬
¼0
S2
1 e 8 dT dI
6
S 4
1 2S
ª¬cos 4 I º¼ dT
³
0
0
12
52
45
1 e 8
S
8
cos3 T sin I cos I dT dI
sin I cos I ª¬cos T 1 sin 2 T º¼ dT dI
S 4
ª
1 S4
sin 3 T º
sin I cos I «sin T »
³
0
3
3 ¼0
¬
5 2 S4
sin I cos I dI
36 ³ 0
4
S 2
z
7.
³0 ³0 ³0
8.
³0 ³0 ³0
S 2
S
re r dT dr dz
sin T
dI
S 4
ª 5 2 sin 2 I º
«
»
2 »¼ 0
«¬ 36
5 2
144
S e4 3
2 cos I U 2 d U dT dI
8
9
© 2010 Brooks/Cole, Cengage Learning
Section 14.7
S 2
e r
3
2
³0 ³0 ³0
9.
r dz dr dT
S 2
3
³ 0 ³ 0 re
r2
Triple Integrals in Cylindrical and Spherical Coordinates
3
S 2ª
1 r2 º
« 2 e » dT
¬
¼0
³0
dr dT
S 2
³0
1
1 e9 dT
2
S
4
345
1 e 9
z
3
2
1
3
1
y
2
3
x
2S
5 r2
5
³0 ³0 ³2
10.
r dz dr dT
2S
³0 ³0
5
2S
³0
5r r 3 dr dT
5
ª 5r 2
r4 º
» dT
«
4 ¼0
¬ 2
2S
³0
§ 25 25 ·
¨
¸ dT
4¹
© 2
25
˜ 2S
4
25S
2
z
5
−3
3
2S
S 2
4
³ 0 ³S 6 ³ 0 U
11.
y
3
−1
x
64 2S S 2
sin I dI dT
3 ³0 ³S 6
sin I d U dI dT
2
64 2S
>cos I @SS 62 dT
3 ³0
64 3S
3
32 3 2S
dT
3 ³0
z
4
2S
S
5
³0 ³0 ³0 U
12.
y
4
4
x
2
sin I d U dI dT
117 2S S
sin I dI dT
3 ³0 ³0
117 2S
S
>cos I @0 dT
3 ³0
468S
3
156S
z
r=5
7
r=2
7
13.
2S
2
2S
arctan 1 2
4
³ 0 ³ 0 ³ r2 r
2
³0 ³0
14.
cos T dz dr dT
4 sec I
³0
S 2
2
16 r 2
S 2
S 6
4
2S
a
³0 ³0 ³0
³0 ³0 ³0 U
15.
y
7
x
a
³0 ³0 ³a
S 4
2S
3
U 3 sin 2 I cos T d U dI dT r 2 dz dr dT
8S 2
2S
3
sin 2 I d U dI dT a2 r 2
2 a cos I
³ 0 ³ 0 ³ a sec I
0
r 2 cos T dz dr dT
S 2
2S
S 2
cot I csc I
³ 0 ³ arctan 1 2 ³ 0
U 3 sin 2 I cos T d U dI dT
0
3
S 2
2 csc I
³ 0 ³S 6 ³ 4
U 3 sin 2 I d U dI dT
8S 2
2S
3
3
0
U 3 sin 2 I cos T d U dT dI
0
© 2010 Brooks/Cole, Cengage Learning
346
16.
Chapter 14 Multiple Integration
S 2
3
9 r2
S 2
S 2
3
³0 ³0 ³0
³0 ³0 ³0 U
17. V
4³
S 2
0
a cos T
18. V
2
S 4
3
2
3
81S
8
sin I d U dI dT
3
³0
4 3 S
a
3 ³0
81S
8
r 2 z 2 r dz dr dT
a2 r 2
³0
2
2
2
S 2
a cos T
³0
0
r
a 2 r 2 dr dT
S 2
4 3ª
1
º
a T cos T sin 2 T 2 »
3 «¬
3
¼0
1 sin 3 T dT
ª S
4 «³
0
¬«
4³
r dz dr dT
r
³ 0 ³ 0 r dz dr dT
S 2
4
³0 ³2 2 ³0
16 r 2
4 3§ S
2·
a ¨ ¸
3 ©2
3¹
2a 3
3S 4
9
º
r dz dr dT »
¼»
z
7
Volume of lower hemisphere 4 Volume in the first octant
V
128S
ª S
4 «³
3
¬ 0
2
2
2
³0
ª8 2S
128S
4«
3
¬« 3
S 2
S 2ª
1
2
« 3 16 r
¬
³0
ª8 2S
128S
8 2S º
4«
»
3
3
3 ¼
¬
Ÿ x 12
cos T
S
cos T
2 r cos T
³ 0 ³ 0 ³ 2r2
³0 ³0
4 cos T
S
§2
cos 4 T ·
4
¨ cos T ¸ dT
2 ¹
©3
2S
128S
64 2S
3
3
2 r2
2S
12
14
2
1
4 1
ª 2 r º
«r » dT
2 ¼0
¬
0
2³
0
S
2
a cos T
³0
2S
2
2S
2
³0 ³0 ³r
³0
1
23. m
r
a2 r 2
y
r dz dr dT
a 2 r dr dT
r
4 r2
3 2º
a cos T
»
¼0
dT
2S
2
2S
2
4 r 2 r 2 dr dT
32
2S
³0
9 r cos T 2 r sin T
³0 ³0 ³0
2
2a 3
3S 4
9
r dz dr dT
ª 1
2
« 4 r
3
¬
³ 0 ³ 0 kr
S
³0
S
³0 ³0
dr dT
1
2S
2
a cos T
³0
2a 3 ª
cos3 T º
«T cos T »
3 ¬
3 ¼0
22. V
2
7
2a 3 S
1 sin 3 T dT
3 ³0
2S
2
2
S
2³
S ª 1
2 ³ « a 2 r 2
0
¬ 3
dT
r dz dr dT
³ 0 ³ 0 r 2 2r
³0
2
7
2
16
2
³ 0 ³ 0 ³ r2
2S
64S
2
3
21. V
y
x y Ÿ x y
2
1
2
S
1 S
cos 4 T dT
6³0
20. 2 x y
»
¼2
x
º
dT »
¼»
2r 2 cos T 2r 3 dr dT
3
2
4
r dz dr dT
ª 2r
r º
cos T »
«
3
2 ¼0
¬
³0
V
2
S
³0
º
r 16 r 2 dr dT »
¼
cos T for this circle.
In polar coordinates, use r
V
3 2º
x2 x y 2 Ÿ x2 x 1 4 y 2
S
2
2x2 2 y 2 Ÿ
19. In the xy-plane, 2 x
0
4
³0 ³2
r 2 dr dT r3 º
»
3 ¼0
2
8S
2
3
dT
2
kr r dz dr dT
9 r cos T 2r sin T dr dT
2
ª
º
r4
r4
cos T sin T » dT
k «3r 3 4
2
¬
¼0
2S
³ 0 k >24 4 cos T
8 sin T @ dT
2S
k >24T 4 sin T 8 cos T @0
k >48S 8 8@
48kS
© 2010 Brooks/Cole, Cengage Learning
Section 14.7
24.
S 2
12 e r
2
2
³0 ³0 ³0
S 2
2
³ 0 ³ 0 12ke
k r dz dr dT
S 2
ª6ke
¬«
³0
S 2
³0
Triple Integrals in Cylindrical and Spherical Coordinates
r2
r dr dT
2
r2
º
¼» 0
6ke 4 6k dT
28. U
kz
h
r0
h
y
m
4k ³
4³
V
S 2
0
h r 0 r r0
r0
³0 ³0
4h S
r0 ³ 0
2
r0
³0
r dz dr dT
r0 r r 2 dr dT
4h § r03
·§ S ·
¨ ¸¨ ¸
r0 ¨© 6 ¸¹© 2 ¹
4k ³
2kh
r02
S 2
0
2
h r0 r r0
r0
³0 ³0
S 2
27. U
³0 ³0
m
r02 r 2r0 r 2 r 3 dr dT
x2 y 2
x
y
m
4k ³
M xy
z
x, y, z
2h
5
h r0 r r0
r0
³0 ³0
r0
2
³0
r 3 dz dr dT
r0 r 3 r 4 dr dT
1
kS r04 h
10
30. I z
kr02 h 2S
12
4k ³
h
4
0
4k ³
S 2
0
h r0 r r0
r0
r0
r 2 dz dr dT
h r0 r r0
³0 ³0
1
kS r03 h 2
30
M xy
kS r03h 2 30
m
kS r03h 6
3 2
mr0 .
10
y 2 U x, y, z dV
S 2
0
4kh ³
kr
³0 ³0
2
§1
·
k ¨ S r02 h ¸
©3
¹
Q
0 by symmetry
S 2
1 § 3m · 4
¨
¸S r0 h
10 © S r02 h ¹
1
kS r04 h
10
³³³ x
kV
3m S r02 h. So,
from Exercise 25, we have k
zr dz dr dT
kr02 h 2S § 3 ·
¨
¸
12 © S r02 hk ¹
M xy
k
z 2 r dz dr dT
Because the mass of the core is m
Iz
r0
2kh 2 § r04 ·§ S ·
¨ ¸¨ ¸
r02 © 12 ¹© 2 ¹
z
S 2
0
4kh § r05 ·§ S ·
¨ ¸¨ ¸
r0 © 20 ¹© 2 ¹
1 2
S r0 hk from Exercise 25
3
M xy
4k ³
4kh S
r0 ³ 0
1 2
S r0 h
3
h r0 r r0
r0
³0 ³0
0 by symmetry
y
m
S 2
0
1
kS r02 h 2
12
zr dz dr dT
2h ·
§
¨ 0, 0, ¸
5¹
©
x, y, z
29. I z
h r0 r r0
r0
³0 ³0
1
kS r02 h3
30
M xy
kS r02 h3 30
m
kS r02 h 2 12
z
4h S 2 r03
dT
r0 ³ 0 6
26. x
S 2
0
4k ³
M xy
h
r0 r
r0
x2 y 2
0 by symmetry
x
3kS 1 e 4
25. z
347
1
kS r03h
6
r 2 z dz dr dT
31. m
Iz
S 2
0
r0 r0
³0
S 2 ªr5
4kh ³
0
4kh ³
0
«
¬5
S 2
4k ³
S 2
0
r0
r6 º
» dT
6r0 ¼ 0
1 5
r0 dT
30
b
h
³a ³0 r
S 2
0
b
4kh
³a r
3
3
4kh ³
1 5S
r0
30 2
Iz
r05 º
« » dT
6¼
¬5
1 5
r0 S kh
15
dz dr dT
dr dT
kh ³
S 2
0
b 4 a 4 dT
kS b 2 a 2 b 2 a 2 h
2
1
m a2 b2
2
32. m
S 2 ª r05
0
kS h b 2 a 2
kS b 4 a 4 h
h·
§
¨ 0, 0, ¸
5¹
©
r 4 dz dr dT
r 4
r dr dT
r0
k S b2h S a 2h
4kh ³
h
5
h r0 r r0
r0
³0 ³0
2
kS a 2 h
2k ³
S 2
0
2 a sin T
³0
h
³0 r
3
dz dr dT
3
kS a 4 h
2
3 2
ma
2
© 2010 Brooks/Cole, Cengage Learning
348
Chapter 14 Multiple Integration
33. V
2S
S 2
2S
S 2
3
³ 0 ³S 4 ³ 0 U
sin I d U dI dT
2
z
3
³ 0 ³ S 4 9 sin I dI dT
2S
S 2
S 4
³ 0 >9 cos I @
2S
³0
dT
§ 2·
9¨¨
¸¸ dT
© 2 ¹
34. x 2 y 2 z 2
2
§ 2·
18S ¨¨
¸¸
© 2 ¹
z
1·
§
x2 y 2 ¨ z ¸
2¹
©
2
1
4
S 4
cos I
³0 ³0 ³0
2S
S
4 sin I
35. V
³0 ³0 ³0
36. V
8³
S 4
S 2
−1
−1
1
1
cos I
2S
S 4
U 2 sin I d U dI dT
³0 ³0
U 2 sin I d U dI dT
16S 2
b
y
x
³0 ³a U
0
1
1
4
1·
§
Sphere with center ¨ 0, 0, ¸: U
2¹
©
2S
2
z
1·
§
x2 y 2 ¨ z 2 z ¸
4¹
©
V
9S
y
3
3
x
2
sin I d U dT dI
cos3 I
sin I dI dT
3
2S
³0
S 4
ª cos 4 I º
«
»
¬ 12 ¼ 0
S 4
37. m
8k ³
S 2
S 2
2ka 4 ³
0
S 2
kS a 4 ³
38. m
8k ³
S 2
0
S 2
S 2
S 2
0
3
a
S 2
³0
3
§
2 · 4S 3
b a3
¨¨1 ¸¸
2
©
¹ 3
2S
2
3
a
x
S 2
kS a 4
b
2
kS r 3
3
y
0 by symmetry
M xy
4k ³
S 2
0
S 2
r
³0 ³0 U
3
cos I sin I d U dT dI
1 4 S2 S2
kr
sin 2I dT dI
2 ³0 ³0
kr 4S S 2
sin 2I dI
4 ³0
sin I d U dT dI
2
sin 2 I dT dI
S 2
ª 1
º
4
« 8 kS r cos 2I »
¬
¼0
sin 2 I dI
S 2
kS a 4
y
2 b3 a3
39. m
ª¬kS a 4 cos I º¼
0
ª
1
·º
4§ 1
«kS a ¨ 2I 4 sin 2I ¸»
©
¹¼ 0
¬
8
a
b
x
sin I d U dT dI
sin I dI
S
b
sin I dT dI
³0 ³0 U
0
kS a 4 ³
S 2
³0
S 2
0
2ka 4 ³
a
³0 ³0 U
0
1§
1·
¨1 ¸ dT
12 ©
4¹
z
includes upper and lower cones
S 4 S 2
8 3
b a3 ³
³ 0 sin I dT dI
0
3
S 4
4S 3
b a 3 ³ sin I dI
0
3
ª 4S 3
º
3
« 3 b a cos I »
¬
¼0
2S
³0
dT
S
4
1 2 4
kS a
4
z
x, y, z
M xy
m
kS r 4 4
2kS r 3 3
1
kS r 4
4
3r
8
3r ·
§
¨ 0, 0, ¸
8¹
©
© 2010 Brooks/Cole, Cengage Learning
Section 14.7
40. x
y
0 by symmetry
42. I z
2
§2
·
k ¨ S R3 S r 3 ¸
3
©3
¹
m
Triple Integrals in Cylindrical and Spherical Coordinates
4k ³
M xy
S 2
0
2
kS R 3 r 3
3
S 2
R
³0 ³r
M xy
m
z
4
3R r
4
4
2kS R 3 r 3 3
4k ³
S 2
S 2
cos I
³0 ³0
S 4
S 2
T2
I2
U2
4
r2
y
r sin T
tan T
z
z
z
y
x
z
44. x
U sin I cos T
U2
y
U sin I sin T
tan T
z
U cos I
cos I
4
45.
T2
g T
h 2 r cos T , r sin T
³ T 1 ³ g12T ³ h1 r cos T , r sin T
x2 y2 z 2
y
x
z
x y2 z2
2
f r cos T , r sin T , z r dz dr dT
kS
192
f U sin I cos T , U sin I sin T , U cos I U 2 sin I d U dI dT
T
T 0 : plane parallel to z-axis
z
z0 : plane parallel to xy-plane
U0 : sphere of radius U0
(b) U
(b)
x2 y2
r0 : right circular cylinder about z-axis
47. (a) r
48. (a)
r cos T
3
1 8 ·º
ª2 § 1 6
« 5 kS ¨ 6 cos I 8 cos I ¸»
©
¹¼S
¬
³T 1 ³I 1 ³ U 1
43. x
8 R3 r 3
4 S2 S2
k
cos5 I sin 3 I dT dI
5 ³S 4 ³ 0
S 2
2
cos5 I 1 cos 2 I sin I dI
kS
5 ³S 4
46.
U 4 sin 3 I d U dT dI
S 2
U sin I d U dT dI
4
R
4kS 5
R r5
15
1
kS R 4 r 4
4
§
3 R4 r 4 ·
¨ 0, 0,
¸
¨
8 R 3 r 3 ¸¹
©
x, y, z
41. I z
kS R r
4
S 2
³0 ³r
ª 2kS 5
§
cos3 I ·º
R r 5 ¨ cos I «
¸»
3 ¹»¼ 0
«¬ 5
©
S 2 S 2
1
k R4 r 4 ³
³ 0 sin 2I dT dI
0
2
S 2
1
kS R 4 r 4 ³ sin 2I dI
0
4
S 2
S 2
0
S 2 S 2
4k 5
3
R r5 ³
³ 0 sin I dT dI
0
5
S 2
2kS 5
R r 5 ³ sin I 1 cos 2 I dI
0
5
U 3 cos I sin I d U dT dI
ª 1
º
4
4
« 8 kS R r cos 2I »
¬
¼0
4k ³
349
T
T 0 : plane parallel to z-axis
I
I0 : cone
S 2
a
a2 r 2
S 2
S 2
a
³0 ³0 ³0
³0 ³0 ³0 U
3
r 2 z 2 r dz dr dT
sin I d U dI dT
Integral (b) appears easier. The limits of integration are all constants.
z
a
a
a
y
x
© 2010 Brooks/Cole, Cengage Learning
350
Chapter 14 Multiple Integration
49. 16³
a
0
³0
a 2 x2
S 2
16 ³
0
S 2
16 ³
8³
50.
f
f
a2 x2 y 2
³0
a
³0 ³0
0
³0
S 2
a
0
³0
f
³ f ³ f ³ f
2S
S
2
S
f
2S
k of 0
lim
k of
S
16³
dw dz dy dx
a
0
a2 x2
³0
a r
2
2
z a r
2
2
4S ³
a 2 r 2 r dr dT
x2 y2 z 2
2
arcsin
S 2 ª a 2r 2
0
º
»
2
2
a r ¼0
z
«
¬ 2
r4 º
» dT
4 ¼0
a 4S ³
3 U2
³0 ³0 U e
S
2S
ª U 1 e
2 2S lim «
k of «
2
¬
2
k
3 U2
2
d 36 x 2 y 2
2
d 36r 2
r 2 z 2 8 d 6r
r 2 6r 9 z 2 1 d 0
dU
r 3
k
º
»
»
¼0
a 4S 2
2
dT
r2 z2 8
³0 U e
U2
S 2
0
In cylindrical coordinates,
sin I d U dI dT
dT
r dr dT
51. x 2 y 2 z 2 8
dx dy dz
2
³ 0 sin I dI ³ 0
a 2 x 2 y 2 z 2 dz dy dx
a2 r 2
a
U e U U 2 sin I d U dI dT
k
a 2 x2 y 2
³0
a 2 r 2 z 2 dz r dr dT
x2 y 2 z 2 e
³0 ³0 ³0
lim ³
a 2 x2 y 2 z 2
a2 r 2
1ª
«z
2¬
a
³0
ª1º
4S « »
¬2¼
2
z 2 d 1.
This is a torus: rotate x 3
2S
2
z2
1 about the
z-axis. By Pappus' Theorem,
V
2S 3 S
6S 2 .
Section 14.8 Change of Variables: Jacobians
1. x
y
4. x
1
u v
2
y
1
u v
2
wx wy
wy wx
wu wv wu wv
§ 1 ·§ 1 · § 1 ·§ 1 ·
¨ ¸¨ ¸ ¨ ¸¨ ¸
© 2 ¹© 2 ¹ © 2 ¹© 2 ¹
uv 2u
uv
wx wy
wy wx
wu wv
wu wv
1
2
5. x
u cos T v sin T
u sin T v cos T
2. x
au bv
y
y
cu dv
wx wy
wy wx
wu wv
wu wv
wx wy
wy wx
wu wv
wu wv
3. x
y
u a
u v2
y
v a
u v
wx wy
wy wx
wu wv
wu wv
7. x
eu sin v
y
eu cos v
wx wy
wy wx
wu wv
wu wv
8. x
y
ad cb
6. x
wx wy
wy wx
wu wv
wu wv
1 1 1 2v
1 2v
eu sin v eu sin v eu cos v eu cos v
v 2 u vu
2u
cos 2 T sin 2 T
1 1 0 0
1
1
e 2u
u
v
u v
wx wy
wy wx
wu wv
wu wv
§1·
§ u·
¨ ¸ 1 1 ¨ 2 ¸
©v¹
© v ¹
1
u
2
v
v
u v
v2
© 2010 Brooks/Cole, Cengage Learning
Section 14.8
3u 2v
9. x
y
3v
v
y
3
x, y
u, v
(0, 0)
(0, 0)
(3, 0)
(1, 0)
(2, 3)
(0, 1)
1
x 2 y 3
3
10. x
1
3
4u v
y
1
3
u v
u
x y
v
x 4y
x 2y
3
9
(0, 1)
(1, 0)
u
1
12. x
1
3
v u
y
1
3
2v u
u
y 2x
v
x y
v
x, y
1
(0, 0)
(0, 0)
(4, 1)
(3, 0)
−2
(2, 2)
0, 6
−4
(6, 3)
3, 6
−6
1
2
u v
y
1
2
u v
−1
(0, 0)
(3, 0)
1
−1
x, y
2
4
6
−3
−5
(0, −6)
(0, 4)
(2, 4)
(0, 1)
(2, 1)
4
u
5
u, v
(3, −6)
13 , 43
(2, 1)
1, 2
3 3
(0, 1)
4, 8
3 3
(0, 4)
2 , 10
3 3
(2, 4)
3
2
1
u
1
2
3
4
x y
x y
u
v
x, y
u, v
1, 1
2 2
(1, 0)
(0, 1)
1, 1
(1, 2)
3, 1
3, 3
2 2
13.
v
u, v
11. x
351
v
x 2v
3
u
Change of Variables: Jacobians
x 2y
x y
x 2y
x y
x 2y
x y
x 2y
x y
u
v
1
(1, 0)
u
−1
(1, −1)
(3, −1)
−2
(3, 0)
3y
0½
¾ Ÿ
y
4¿
3y
4½
¾
y
4 ¿
3y
4½
¾
y
1 ¿
3y
0½
¾
y
1¿
23
(3, 0)
2
4
4,
3
x
8
3
x
4
3
x
32
8,
3
5,
3
x − 2y = 0
x
−1
x
x+y=4
(43 , 83 (
2 1
(3, 3 ( (8, 4 (
3 3
1
1
1,
2
x − 2y = −4
3
(− 23 , 53 (
5
x4 2
³ 2 3 ³ 1 x
y
8
3y Ÿ y
x y ½ u v
¾
x 2 y¿ 2u v
3x Ÿ x
³³ 3 xy dA
R
v
2
3
x+y=1
2
3
1
3
3 xy dy dx u v
1
3
2u v
43
x4 2
³2 3 ³x 2
3 xy dy dx 83
4 x
³4 3 ³x 2
3 xy dy dx
32
27
164
27
296
27
164
9
© 2010 Brooks/Cole, Cengage Learning
352
14.
Chapter 14 Multiple Integration
³³ x 2
y
1
x 1
³ 0 ³ 1 x
sin 2 x y dA
R
f x dy dx 3 x
2
³ 1 ³ x 1
§ cos 2 1
5
15
17 ·
sin 1 cos 1 sin 2 1 ¸
¨¨ 16
3
16
16 ¸¹
©
13 13
13 13
sin 1 cos 1
sin 2 | 2.363
3
3
3
6
y
y=x+1
f x dy dx
2
8
ª 2
º
«cos 1 3 sin 1 cos 1 7 3»
¬
¼
y=x−1
y = −x + 3
1
x
1
2
y = −x + 1
15. x
y
1
u v
2
1
u v
2
v
(− 1, 1)
wx wy
wy wx
wu wv
wu wv
³³ 4 x
2
§ 1 ·§ 1 · § 1 ·§ 1 ·
¨ ¸¨ ¸ ¨ ¸¨ ¸
© 2 ¹© 2 ¹ © 2 ¹© 2 ¹
1
1
1
1
R
³ 1 ³ 1
16. x
y
1
u v,
u
2
1
u v, v
2
wx wy
wy wx
wu wv
wu wv
³R ³ 60 xy dA
ª1
1
2
1
2 º§ 1 ·
u v » ¨ ¸ dv du
4
¼© 2 ¹
u 2 v 2 dv du
§ 2 1·
³ 1 2¨© u 3 ¸¹du
2
1
x y
1 § 1 · § 1 ·§ 1 ·
¨ ¸ ¨ ¸¨ ¸
2 © 2 ¹ © 2 ¹© 2 ¹
1
3
1
3
§1
1
2
·§ 1
·§ 1 ·
u v ¸¨ ¸ dv du
³ 1 ³ 1 60¨© 2 u v ¸¨
¹© 2
¹© 2 ¹
15
3
u v
x, y
u, v
(0, 1)
1, 1
(2, 1)
(1, 3)
(−1, 1)
(1, 2)
1, 3
−2
(1, 0)
(1, 1)
1 0 1 1
3
4
³ 0 ³ 0 uv 1
y dA
3
³ 0 8u du
v
4
(−1, 3)
(1, 3)
2
(1, 1)
u
−1
1
1
15 §
³ 1 2 ¨© 2u
2
26 ·
¸ du
3¹
1
ª15 § 2 3 26 º
« 2 ¨ 3 u 3 u»
¬ ©
¼ 1
1
dv du
y
§ 2 26 ·
15¨ ¸
3¹
©3
1
u v
2
1
u v
2
wx wy
wy wx
wu wv
wu wv
120
v
u
1
−1
1
2
2
v=u−2
−2
36
³R ³ 4 x y e x y dA
2
0
v§ 1 ·
³ 0 ³ u 2 4ue ¨© 2 ¸¹ dv du
2
³ 0 2u 1 e
3
2
−1
18. x
u
wx wy
wy wx
wu wv
wu wv
8
3
v 2 u 2 dv du
ª 15 § v3
º
2 ·
³ 1 ¬« 2 ©¨ 3 u v ¹¸¼» du
1
³R ³ y x 1
ª § u3
u ·º
¸»
«2¨
3 ¹¼ 1
¬ ©3
(1, − 1)
v
1
y
(− 1, −1)
x y
³ 1 ³ 1 2
17. x
u
³ 1 ³ 1 4«¬ 4 u v
y 2 dA
(1, 1)
u2
du
2
ªu 2
º
2 « ueu 2 eu 2 »
¬2
¼0
2
1
u
1
2
3
4
2 1 e 2
© 2010 Brooks/Cole, Cengage Learning
Section 14.8
19.
³R ³ e
xy 2
dA
x
1
,y
x
2 x, y
4
y = 2x
3
y
,v
x
uv Ÿ u
v u, y
4
x
353
y = x1
y
x
,y
4
R: y
Change of Variables: Jacobians
y = 4x
xy
y = 41 x
2
R
1
wx
wu
wy
wu
w x, y
w u, v
wx
wv
wy
wv
12
1v
2 u3 2
1
2
1
u1 2v1 2
1 v1 2
2 u1 2
1
2
u1 2
v1 2
x
1§ 1
1·
¨ ¸
u¹
4© u
1
1
2u
2
3
4
Transformed Region:
y
y
y
y
³R
1
Ÿ yx
x
4
Ÿ ux
x
y
2x Ÿ
x
x
y
Ÿ
x
4
xy
³e
2
v
1Ÿ v
1
4 Ÿ v
4
3
S
2
2 Ÿ u
2
1
Ÿ u
4
1
4
2
4
³1 4 ³1
dA
u
1
§ 1 ·
e v 2 ¨ ¸ dv du
© 2u ¹
2
ª¬ e 2 e 1 2 ln u º¼
14
20. x
y
3
4
³
ª e v 2 º
du
14« u »
¬
¼1
2
³
2
14
1
v
³ R ³ y sin xy dA
³ 1 ³ 1 v sin u
4
4
1
dv du
v
21. u
x y
1,
v
x y
1
u
x y
1, v
x y
3
1·
§
e 2 e 1 2 ¨ ln 2 ln ¸
4¹
©
4
³ 1 3 sin u du
1
du
u
e1 2 e 2 ln 8 | 0.9798
>3 cos u@1
4
y=x+1
1
u v
2
1
v u
2
w x, y
w u, v
(1, 2)
y = −x + 3
(2, 1)
R
1
y=x−1
x
wx wy
wy wx
wu wv wu wv
³ R ³ 48 xy dA
3 cos 1 cos 4 | 3.5818
y
2
y
e 2 e 1 2
u
v
v
wx wy
wy wx
wu wv
wu wv
x
4
3
³1
1 § 1 · § 1 ·§ 1 ·
¨ ¸ ¨ ¸¨ ¸
2 © 2 ¹ © 2 ¹© 2 ¹
1
1
2
y = −x + 1
§1·
§1·
§1·
³ 1 48¨© 2 ¸¹ u v ¨© 2 ¸¹ v u ¨© 2 ¸¹ du dv
1
3§
2·
6 ³ ¨ 2v 2 ¸ dv
1
3¹
©
3
ª 2v3
2 º
v»
6«
3
3
¬
¼1
2
3
1
³ 1 ³ 1 6 v
2
u du dv
2 2º
ª
6 «18 2 »
3
3¼
¬
2
6³
3ª
1
1
u3 º
«uv » dv
3 ¼ 1
¬
2
96
© 2010 Brooks/Cole, Cengage Learning
354
Chapter 14 Multiple Integration
22. u
2y x
0, v
3x 2 y
0
u
2y x
8, v
3x 2 y
16
x
y
1
v u
4
1
v 3u
8
w x, y
x
−2
16
v
x y
0
u
x y
8,
v
x y
4
x
1
u v
2
y
1
u v
2
³ R³
8
³0 ³0 v u
2 y x dA
4,
w u, v
1
y= x
2
−1
1
8
x y
12
8
16
³0
x−y=0
x−y=4
x
8
4
³4 ³0
§1·
uev ¨ ¸ dv du
© 2¹
S,
v
x y
0
2S ,
v
x y
S
x
1
u v,
2
y
1
u v
2
4
6
8
ª1 2 4
º
« u e 1»
¬4
¼4
1 8
u e 4 1 du
2³4
3π
2
x−y=0
π
x− y=π
x+y=π
S
2S
³ 0 ³S
sin 2 x y dA
25. u
x 4y
0,
v
x y
0
u
x 4y
5,
v
x y
5
x
1
u 4v ,
5
y
1
u v
5
26. u
u
3x 2 y
x
1
u v,
4
wx wy
wy wx
wu wv
wu wv
³ R³
§1·
u 2 sin 2 v¨ ¸ du dv
© 2¹
0,
16,
v
ª 1 § u 3 · 1 cos 2v º
« ¨ ¸
» dy
2
¬2© 3 ¹
¼S
S
ª 7S 3 §
1
·º
«
¨ v sin 2v ¸»
12
2
©
¹
¬
¼0
7S 4
12
x−y=0
x + 4y = 5
x
−1
5
2y x
2y x
y
1
u 3v
8
8
3
4
−1
1
5
x−y=5
x + 4y = 0
−2
5 ª1 § 2 ·
§1·
uv ¨ ¸ du dv
©5¹
5
³0 ³0
dA
x
1
v
32
³0
3π
22S
2
5
³ 0 «¬5 ¨© 3 ¸¹u
y
0
º
v » dv
¼0
32
(2, 5)
3
(4, 2)
2
16
³0 ³0
100
9
3x + 2y = 16
(−2, 3)
1
8
5
ª2 5 § 2 · 3 2º
«
¨ ¸v »
¬ 3 © 3 ¹ ¼0
2y − x = 8
5
8
1 § 3 · 1§ 1 ·
¨ ¸ ¨ ¸
4 © 8 ¹ 8© 4 ¹
3x 2 y 2 y x
S
π
y
§ 1 ·§ 1 · § 1 ·§ 4 ·
¨ ¸¨ ¸ ¨ ¸¨ ¸
© 5 ¹© 5 ¹ © 5 ¹© 5 ¹
x y x 4 y dA
3x 2 y
x + y = 2π
π
2
1
2
wx wy
wy wx
wu wv
wu wv
12 e 4 1
y
π
2
³ R³
4096 2
3
2
x y
³ R³
·
2¸
¹
x+y=8
x+y=4
x y
2
162 § 32
¨
2 © 3
2 v dv
4
u
x y
32
3
6
24. u
w u, v
4
y
1
2
x y e x y dA
3
ª2 3 2º
« 3 vu » dv
¬
¼0
16
³0
du dv
2
(0, 0)
−1
2
w x, y
(4, 2)
R
2
3
y=− x
2
23. u
w x, y
3
y=− x+8
2
3
2
3x 2 y
(2, 5)
(−2, 3)
§ 1 ·§ 1 · § 3 ·§ 1 ·
¨ ¸¨ ¸ ¨ ¸¨ ¸
© 4 ¹© 8 ¹ © 8 ¹© 4 ¹
w u, v
³ R³
y
1
y= x+4 5
2
2y − x = 0
3x + 2 y = 0
−2
−1
§1·
u v3 2 ¨ ¸ du dv
©8¹
x
1
−1
2
3
4
(0, 0)
8
³0
8
16v3 2 dv
§ 2 · 5 2º
¨ ¸16v »
©5¹
¼0
4096
5
2
© 2010 Brooks/Cole, Cengage Learning
Section 14.8
x y, v
27. u
wx wy
wy wx
wu wv
wu wv
³ R³
1
u v,y
2
x y, x
355
1
u v
2
1
2
a
u
³ 0 ³ u
x y dA
Change of Variables: Jacobians
§1·
u ¨ ¸ dv du
© 2¹
y
a
³0
a
u u du
ª2 5 2º
« u »
¬5
¼0
2 52
a
5
v
v=u
a
a
x+y=a
u
2a
x
a
−a
v = −u
28. u
x
1,
v
xy
1
u
x
4,
v
xy
4
x
u,
y
u
v
y
x=1
4
3
xy = 4
2
x=4
wx wy
wy wx
wu wv
wu wv
1
u
1
x
1
2
3
4
xy = 1
³ R³ 1 xy
dA
x2 y2
4
4
v
§1·
³ 1 ³ 1 1 v 2 ¨© u ¸¹ dv du
2x y
29. u
4
º 1
» du
¼1 u
ª1
º
« 2 >ln 17 ln 2@ln u »
¬
¼1
v 1
3
u v
y
y
y= 1x+2
(6, 4)
4
1
3
2v u .
(3, 3)
y=x
R
2
y=x−2
(4, 2)
(1, 1)
y= 7x
x
2
2
8
1 § 17 ·
¨ ln ¸ ln 4
2© 2 ¹
3
u v
1
3
v x
Then y
4
2
6
u v Ÿ x
3x
ª1
30.
x y
v
4
³ 1 ¬« 2 ln 1 v
y= 1x+ 2
3
(2, 7)
6
3
y = −x + 9
6
4
(6, 3)
R
y= 1x
2
2
x
2
4
6
x, y
8
(0, 0)
(6, 3)
(9, 9)
(2, 7)
3, 9
6
2
Let u
y x and v
1
2
3 y x.
v 3u and y
x, y
1
2
1
2
vu.
v 3u ,
1
2
v u .
Note that:
x, y
12
u, v
(1, 1)
(0, 2)
(4, 2)
(–2, 2)
(6, 4)
(–2, 6)
(3, 3)
(0, 6)
(9, 9)
8
6
3y x
2, 3 y x
So, T u , v
v
(− 3, 9)
0,
y x
Then x
u, v
(0, 0)
y x
R
u
−4 − 2
2
4
6
8
10
One side is parallel to the u-axis.
© 2010 Brooks/Cole, Cengage Learning
356
Chapter 14
Multiple Integration
x2
y2
2
2
a
b
31.
2
au
a2
u 2 v2
(a)
bv
1
b2
2
au , y
2
bv
1, x
1
2
x
y
2
a2
b
u 2 v2
1
y
1
v
1
b
S
R
x
u
1
a
(b)
w x, y
w u, v
wx wy
wy wx
wu wv wu wv
³ S ³ ab dS
(c) A
ab S 1
2
ab
S ab
16 x 2 y 2
32. (a) f x, y
R:
a b 0 0
x2
y2
d1
16
9
V
³ R³ f
Let x
x, y dA
4u and y
³ R ³ 16 3v.
x 2 y 2 dA
1 u2
1
³ 1 ³ 2S
1 u2
1
³0 ³0
2S
12³
0
12³
0
2S
16 16u 2 9v 2 12 dv du
1
4
ª 2
4
2
2 º
«8r 4r cos T r sin T » dT
¬
¼0
12³
ª
§ 1 cos 2T · 9 § 1 cos 2T ·º
¸» dT
¸ ¨
«8 4¨
2
2
©
¹ 4©
¹¼
¬
2S
ªS
A cos «
¬« 2
r sin T .
16 16r 2 cos 2 T 9r 2 sin 2 T 12r dr dT
7
ª 39
º
12« T sin 2T »
8
16
¬
¼0
(b) f x, y
r cos T , v
Let u
ª 39S º
12«
»
¬ 4 ¼
2S
0
ª
2
2 º
«8 4 cos T sin T » dT
¬
¼
12³
2S
0
ª 39 7
º
cos 2T » dT
«
8
¬8
¼
117S
x2
y2 º
2»
2
a
b »¼
x2
y2
2 d1
2
a
b
Let x
au and y
R:
³R ³ f
x, y dA
r cos T , v
Let u
Aab ³
2S
0
bv.
1
³ 1 ³ 1 u2
1 u2
º
u 2 v 2 » ab dv du
¼
r sin T .
ªS º
³ 0 cos «¬ 2 r »¼ r dr dT
1
ªS
A cos «
¬2
1
ª 2r
4
§Sr ·
§ S r ·º
Aab « sin ¨ ¸ 2 cos¨ ¸» 2S
© 2 ¹ S
© 2 ¹¼ 0
¬S
ª§ 2
4 ·º
· §
2S Aab «¨ 0 ¸ ¨ 0 2 ¸»
S
S
¹ ©
¹¼
©
4 S 2 Aab
S
© 2010 Brooks/Cole, Cengage Learning
Section 14.8
w x, y
33. Jacobian
Change of Variables: Jacobians
357
wx wy
wy wx
wu wv
wu wv
w u, v
34. See Theorem 14.5.
35. x
u1 v,y
w x, y , z
w u , v, w
36. x
uv 1 w , z
uvw
1 v
u
0
v1 w
u 1 w
uv
vw
uw
uv
4u v, y
4v w, z
4 1
w x, y , z
w u , v, w
1 v ª¬u 2v 1 w u 2vwº¼ u ª¬uv 2 1 w uv 2 wº¼
u w
1
u v,y
2
37. x
0
0
4 1
1
0
17
w x, y , z
w u , v, w
1
1 v u 2v u uv 2
1
u v,z
2
12
12
0
12
1 2
0
2vw
2uw 2uv
2uv>1 4 1 4@
u v w, y
38. x
1
w x, y , z
w u , v, w
2uvw
uv
u v w
2uv, z
1 1
2v 2u 0
1
1
1
1 2u 1 2v 1 2v 2u
U sin I cos T , y
39. x
w x, y , z
w U, T , I
U sin I sin T , z
u 2v
4v
U cos I
sin I cos T
U sin I sin T
U cos I cos T
sin I sin T
U sin I cos T
U cos I sin T
cos I
0
U sin I
cos I ª¬ U sin I cos I sin T U 2 sin I cos I cos2 T º¼ U sin I ª¬U sin 2 I cos 2 T U sin 2 I sin 2 T º¼
cos I ª¬ U 2 sin I cos I sin 2 T cos 2 T º¼ U sin I ª¬U sin 2 I cos 2 T sin 2 T º¼
2
2
U 2 sin I cos 2 I U 2 sin 3 I
40. x
r cos T , y
w x, y , z
w r, T , z
r sin T , z
r sin T
sin T
r cos T
0
0
0
1
y Ÿ
U 2 sin I
z
cos T
x
,v
3
41. Let u
U 2 sin I cos 2 I sin 2 I
1ª¬r cos 2 T r sin 2 T º¼
w x, y
3 0
w u ,v
0 1
x
Ÿ v
2
3, y
r
3u
.
2
v
y
3
y=
2
x
2
v
y
1
3
v = 32 u
1
2
v = 3mu
B′
y = mx
A′
A
−3
−1
1
−2
−3
x
3
x2 + y2 = 1
9
B
u
−1
1
−1
x
−3
−1
1
−2
u2 + v2 = 1
−3
3
x2 + y2 = 1
9
u
−1
1
−1
u2 + v2 = 1
Region A is transformed to region Ac, and region B is transformed to region Bc.
2
2
3m Ÿ m
3
9
Note: You could also calculate the integrals directly.
Ac
Bc Ÿ
© 2010 Brooks/Cole, Cengage Learning
358
Chapter 14
Multiple Integration
Review Exercises for Chapter 14
1.
x2
³1
x2
ª¬ xy 1 ln y º¼1
x ln y dy
x3 1 ln x 2 x
2y
2y
2.
³y
3.
³0 ³0
ª x3
2º
« xy »
3
¬
¼y
x 2 y 2 dx
1 x
1
10 y 3
3
1
³ 0 ª¬3xy 3 x 2 y dy dx
1
³0
x x3 x3 ln x 2
1 x
y 2 º¼
0
y
dx
ª 43 x3 ¬
4 x 2 5 x 1 dx
5 x2
2
3
1
xº¼
0
y=x+1
29
6
2
1
x
1
4.
2x
2
³ 0 ³ x2
2
³ 0 ª¬x
x 2 2 y dy dx
2
³0
2
1 x4
2
(2, 4)
4
4x 2x 2x
ª 4 x3 ¬3
3
y
2x
y y 2 º¼ 2 dx
x
2
2
y = 2x
3
4
2 x5 º
5 ¼0
dx
2
3
2
88
15
y = x2
1
x
1
5.
9 x2
3
³0 ³0
3
³ 0 4x
4 x dy dx
2
3
4
y
9 x 2 dx
4
32 3
ª 4 9 x 2
«¬ 3
º
»¼ 0
y=
9 − x2
2
3
3
36
2
1
x
1
6.
2
3
³ 0 ³ 2
4 y2
4 y2
2³
dx dy
3
ª
«y
¬
y
4 y 2 dy
0
4
y=
(2, 2)
3
2
yº
4 y 2 4 arcsin »
2 ¼0
3
1
x
1
4S
3 3
2
3
(x − 2) 2 + y 2 = 4
7.
3
(3 x ) 3
³0 ³0
1
8.
2
x
3
6 2x
2
(6 y ) 2
³0 ³ y
3
³ 5 ³ A
25 x 2
25 x 2
2³
3
dx dy
dy dx
5 ³ 0
25 x2
dx dy
1
³0
dx dy
³ 0 ³ 0 dy dx ³ 2 ³ 0
A
9.
33y
³0 ³0
A
33y
1
³0 ³0
dy dx
3 3 y dy
1 2 (6
2 0
³
4
³ 5 ³ dy dx
(6 y ) 2
2
³0 ³ y
dy dx
3 y ) dy
25 y 2
25 y 2
2³
dx dy 3
5
ª3 y ¬
1
3 2º
y
2 ¼0
3
2
dx dy
ª1 6 y ¬2
4
3 2
y
2
3
³4 ³
25 y 2
2
º
¼0
3
dx dy 5
³4 ³
25 y 2
25 y 2
dx dy
3
25 x 2 dx
xº
ª
2
« x 25 x 25 arcsin 5 »
¬
¼ 5
25S
3
12 25 arcsin | 67.36
2
5
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 14
10.
4
6 x x2
1
0
³ 0 ³ x2 2 x dy dx
1 y
³ 1 ³ 1 1 y
6 x x2
4
4
³ 0 ³ x2 2 x dy dx
A
1
11. A
4³
0
A
4³
0
x 1 x2
³0
12
(1 ³
y2 1
2
³0 ³0
13. A
³ 2 ³ x3
14. A
³0 ³ y
x 1
5
1
8
1 y
³ 0 ³ 3
9 y
0
1 4 y2 ) 2
9 y
9 y
dx dy
64
3
1
3 2º
4
3
»
¼0
dx dy
1
2
5
³ 0 ³ 0 dy dx ³ 1 ³
dy dx 2 ³
2
1
x 1
³0
1
0
³ 3 ³ x
dx dy
4
2 3º
x
3 ¼0
ª 4
2
« 3 1 x
¬
1
3
9
³ 8 ³ 3
dx dy ª4 x 2 ¬
8 x 2 x 2 dx
4 ³ x 1 x 2 dx
dy dx
dx dy
2 y y2
3
³0
(1 1 4 y 2 ) 2
12. A
dy dx x 1
14
3
dy dx
y 3
2
³ 1 ³ y2 1 dx dy
dy dx
1 x
2
dy dx 1
1
³ 0 ³ 1
1 x
1 x
9
2
9
2
dy dx
15. Both integrations are over the common region R shown in the figure. Analytically,
1
2
³0 ³2y
2
2 y2
x y dx dy
4
3
4
3
y
2
y = 12 x
2
(2, 1)
x 2
³0 ³0
2
2
³2 ³0
x y dy dx 359
8 x2 2
x y dy dx
5
3
4
3
2 1
3
4
3
y = 12
1
4
3
2
8 − x2
R
x
1
3
2
−1
16. Both integrations are over the common region R shown in the figure. Analytically,
2
5 y
³ 0 ³3y 2 e
3
2x 3
³0 ³0
x y
85 e5
2
5
dx dy
e x y dy dx y
5
5 x
³3 ³0
5
4
e x y dy dx
3 5
e
5
e3 2
5
e5 e 3
8 5
e
5
2
5
3
(3, 2)
2
1
x
1
17. V
4
x2 4
³0 ³0
4
³ 0 ª¬x
4
³0
2
x 2 y 4 dy dx
1 x4
2
dx
4 x 2 8 dx
ª 1 x5 ¬10
4 x3
3
4
8 xº¼
0
3296
15
3
4
5
16
Average Value
x2 4
y 12 y 2 4 yº¼
0
19. Area R
2
1 2 2
16 x 2 y 2 dy dx
16 ³ 2 ³ 2
2
y3 º
1 2 ª
2
«16 y x y » dx
³
2
16
3 ¼ 2
¬
1 2 ª
16 º
64 4 x 2 » dx
16 ³ 2 «¬
3¼
2
18. V
3
x
³0 ³0
x y dy dx
3
³ 0 ª¬xy 3 3 2
x
2 0
³
3
ª 1 x3 º
¬ 2 ¼0
x
1 y2º
2 ¼0
dx
1ª
4 x3 16 º
x»
«64 x 16 ¬
3
3 ¼ 2
1ª
64 64 º
256 16 «¬
3
3 »¼
40
3
dx
27
2
© 2010 Brooks/Cole, Cengage Learning
360
Chapter 14
20. Area R
Multiple Integration
9
21. Area R
3
1 3ª 2
y3 º
«2 x y » dx
³
0
9 ¬
3 ¼0
1 3 3
2 x 2 y 2 dy dx
9³0 ³0
Average Value
35
f
f
³0 ³0
60 50
1
ª192 x
150 45 40 ¬
³ ³
5
3
³0
f
³0
f
1
>600 270 125@
15
25. Volume | base height
ª kxe x y y 1 º dx
¬
¼0
kxe
x
f
z
9
2
27
2
3
4
Matches (c)
k
2
y
4
(3, 3)
2
1.
1
(3, 0)
1
³ 0 ³ 0 xye
x y
2
13 qC
3
6
|
dx
ª¬ k x 1 e x º¼
0
P
1ª
125 x º
200 x 10 x3 15 «¬
3 »¼ 0
576 y x 2 5 y 2 2 xy 5000º¼ dx dy | 13,246.67
f
kxye x y dy dx
So, k
9
1 3ª
y3 º
40 y 6 x 2 y «
» dx
³
15 0 ¬
3 ¼0
1 3 5
40 6 x 2 y 2 dy dx
15 ³ 0 ³ 0
1 3ª
125 º
dx
200 30 x 2 15 ³ 0 «¬
3 »¼
23.
3
1
ª2 x3 9 x¼º
¬
0
9
15
Average temperature
22. Average
1 3
6 x 2 9 dx
9³0
4
x
dy dx | 0.070
26. Matches (c)
z
24.
1
1 ª kxy
x
³ 0 ³ 0 kxy dy dx
³0
0.25
0.5
0.25
0.25
h
x
³0 ³0
1
k
8
y
2
x
8.
27. True
8 xy dy dx
ª¬4 xy 2 º¼ dx
0
28. False,
x dx
29. True
0.25
30. True,
ª x2 º
« »
¬ 2 ¼0
31.
ª kx º
« »
¬ 8 ¼0
kx
dx
2
0.5
³0
³0
2
4 1
3
1, you have k
³0 ³0
P
3
2
k
8
Because
º
» dx
2 ¼0
³ 0 «¬
1
2 x
1
0.125
x 2 y 2 dy dx
4
³0 ³0
16 y 2
33. A
2³
34. A
4³
S
S 4
h sec T
³0 ³0
S 2
0
S 2
2 cos T
r dr dT
2 sin 2T
³0
r dr dT
4
³0 ³0
x 2 y 2 dx dy
0 ³0
1
³0 ³0 1 S
³0
r 3 dr dT
0
S 2 ªr 4 º
³0
S
4
« » dT
¬ 4 ¼0
ª
³ 0 «¬4 4 cos T 2
S 2
2
2
³1 ³1
z
1
dx dy x2 y2
S
h3
ª¬sec T tan T ln sec T tan T º¼ 0
6
2 cos T dT
2³
1
x dy dx
1
1
1
³0 ³0 1 x2
dx dy
S
4
r 2 dr dT
h3 S 4 3
sec T dT
3 ³0
32.
1
³ 0 ³ 0 x dy dx
2
2 sin 2T dT
8³
S 21
0
S 2
³0
64 dT
1 cos 2T º
» dT
2
¼
cos 4T
dT
2
4
h3 ª
2 ln
6¬
2 1º
¼
32S
S
1
sin 2T º
ª
«4T 4 sin T 2 T 4 »
¬
¼0
S 2
sin 4T º
ª
4 «T 4 »¼ 0
¬
9S
2
2S
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 14
4³
³1
0 ³0
36. V
8³
0
37. (a)
x2 y 2
S 2
R
³b
2
r2
r
1 z 2
S 2
h
35. V
2³
r dr dT dz
0 ³0
83 ³
R 2 r 2 r dr dT
2
S 2
h
S 2
0
ª R2 r 2
«¬
0
32 R
º dT
»¼ b
8
3
9 x2 y2
R 2 b2
39. (a)
9 r 2 cos 2 T r 2 sin 2 T
9 cos T sin T
2
2
2
9 cos 2T
4
3 cos 2T
r
−6
6
4³
(b) A
4³
(c) V
0
S 4
0
3 cos 2 T
³0
3 cos 2 T
³0
12 13
8 13
38. tan T
r dr dT
³0
4
9
³0
k³
0
Mx
k³
0
My
k³
0
(b)
3
Ÿ T | 0.9828
2
m
Mx
My
x
288
.
13
y
y
(8/ 13, 12/ 13)
4
3
S h3
1
3
4S
3
R 2 b2
k
4
16k
55
8k
45
2x
³ 2 x3 xy dy dx
2x
³ 2 x3 xy
2x
³ 2 x3 x
2
2
32
dy dx
y dy dx
y
32
45
64
55
y = 2x
2
1
§ 32 64 ·
¨ , ¸
© 45 55 ¹
y = 2x 3
x
1
2
17 k
30
1 2x
392k
k ³ ³ 3 y x 2 y 2 dy dx
0 2x
585
1 2x
156k
2
2
k ³ ³ 3 x x y dy dx
0 2x
385
My
936
m
1309
Mx
784
m
663
k³
x, y
x 2 + y 2 =16
1
dT
My
m
Mx
m
x, y
9 r r dr dT | 20.392
1
m
2
r cos T r sin T r dr dT
S 2
³0
y
The polar region is given by 0 d r d 4 and
0 d T d 0.9828. So,
arctan 3 2
32
x
−4
S 4
h
ª § 1 3 ·º
«S ¨ 3 z ¸»
¹¼ 0
¬ ©
h
S ³ z 2 dz
1 z 2 1 dT dz
361
y = 23 x
1
0
2x
³ 2 x3
x 2 y 2 dy dx
§ 936 784 ·
,
¨
¸
© 1309 663 ¹
2
1
θ
x
1
40.
8/ 13
m
k³
Mx
k³
4
kh L §
x
x2 ·
2
2 ¸ dx
¨
³
2 0©
L
L ¹
L
h 2 ª2 x L x 2 L2 º
«¬
»¼
dy dx
L
h 2 ª2 x L x 2 L2 º
»¼
¬«
y dy dx
0 ³0
0
³0
2
kh 2 L §
x
x2 ·
2 ¸ dx
¨2 ³
0
8
L
L ¹
©
kh 2 L ª
4 x 3x 2
2 x3
x4 º
4
2 3 4 » dx
«
³
8 0 ¬
L
L
L
L ¼
My
k³
L
0 ³0
h 2 ª2 x L x 2 L2 º
»¼
¬«
y
My
m
Mx
m
5khL2
12
˜
24
7 khL
17 kh 2 L
12
˜
80
7 khL
L
kh 2 ª
2 x2
x3
x4
x5 º
2 3 4»
«4 x 8 ¬
L
L
2L
5L ¼ 0
kh 2 17 L
˜
10
8
x dy dx
kh L §
x2
x3 ·
2 ¸ dx
¨ 2x ³
0
L
L ¹
2
©
x
7 khL
12
y
L
kh ª 2
x3
x4 º
2»
«x 2¬
3L 4 L ¼ 0
5L
14
51h
140
17 kh 2 L
80
kh 5 L2
˜
2 12
5khL2
24
h
(
2
x
y = h 2 − L − x2
2
L
(
x
L
© 2010 Brooks/Cole, Cengage Learning
362
Chapter 14
³R ³ y
41. I x
2
Multiple Integration
³R ³ x U
2
Iy
I0
Ix I y
m
³R ³ U
x
y
a
b
a
b
³ 0 ³ 0 kxy
U x, y dA
³ 0 ³ 0 kx
x, y dA
Iy
I0
m
x, y dA
Iy
1 4 kba 4
m
1 2 kba
2
Ix
m
1 6 kb3a 2
Ix
m
16,384k 315
128k 15
b 3
3
4
7
2
1 fx
³R ³
1 4 x 2 4 y 2 dA
S 2
0
5
³0
2
fy
dA
1 4r 2 r dr dT
32 5
2
ª 1 4r 2
«¬
º dT
»¼ 0
2
ª 101 3 2 1º dT
¬
¼
Sª
101 101 1º¼
6¬
2 7
7
128
21
8 42
21
16 x y 2
44. f x, y
^ x, y : 0
1, f y
2
1 fx
2
2
d x d 2, 0 d y d x`
2 y
fy
2
2 4 y2
³ 0 ª¬2
2
³0 ³ y
2 4 y 2 dx dy
ª1
«2 2 y
¬
2 4 y 2 2 ln 2 y 2 4 y2 y
2 4 y2
18
6 2 ln 4 3 2 9 2
ln
2
2 2 4 y 2 º dy
¼
1
2 4 y2
12
1
º ª
18 18 » «ln
12
¼ ¬«
ª1
« 2 4 18 2 ln 4 ¬
2
6
2 3 2º
2
»
¼0
2 2º
»
12 ¼»
5 2
ln 2 2 3
3
9 y2
45. f x, y
S
b2
3
2 y
³R ³
1 S
3³ 0
1 S
3³ 0
16,384
2
3
³ R ³ y U x, y dA ³ 0 ³ 0 ky dy dx 315 k
2 4 x2
512
2
2
³ R ³ x U x, y dA ³ 0 ³ 0 kx y dy dx 105 k
16,384k 512k 17,920
512
Ix I y
k
k
315
105
315
9
2 4 x2
128
³ R ³ U x, y dA ³ 0 ³ 0 ky dy dx 15 k
y
fx
a 2
2
2 x, f y
4³
4 x2
512k 105
128k 15
S
S
2
a2
2
1 2 kba 2
Iy
m
fx
fx
1
kba 4
4
dy dx
25 x 2 y 2
43. f x, y
ka b 2
2b 3a 2
12
a b
1
2
³ 0 ³ 0 kx dy dx 2 kba
x
R
1 3 2
kb a
6
dy dx
1 3 2
1
kb a kba 4
6
4
2
42. I x
3
2
0, f y
³R ³
3
2 y
1 fx
y
³0 ³y
2
fy
1 4 y 2 dx dy
2
dA
3
³0
y
ª 1 4 y 2 xº dy
¬
¼ y
3
³0
2 y 1 4 y 2 dy
12
43
1 4 y2
32 3
º
»¼ 0
1ª
6¬
37
32
1º
¼
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 14
4 x2 , f x
46. f x, y
363
y
2 x , f y
0
4
³R ³
S
0
2
³ 2 ³ x
1 4 x dA
2
2
2
³0 ³ x
1 4 x dy dx 2
1 4 x dy dx
2
3
(− 2, 2)
(2, 2)
These integrals are equal by symmetry.
2³
S
2
0
2
³x
³0
1 4 x dy dx
2 ª 12 ln
«¬
50
502 x 2
1
12
32 2
º
¼» 0
1 4x2
xy
x y·
§
¨ 20 ¸ dy dx
100
5 ¹
©
50
³0
2 ª 12 ln
¬
2
fy
1 50
100 ³ 0 ³ 0
S
−1
1
17 4 2 17 ª
x
x
2
2
502 x 2 «20 50 x 200
5
¬
2
17
12
17 1º
12 ¼
502 x 2 | 7.0717
502 x 2 º
» dy
10 ¼
50
32
250 x x3 º
3
» | 30,415.74 ft
30 ¼ 0
xy
100
20 1 fx
y=x
x
−2
ª §
x · 25 2
x4
1
2
2
x 502 x 2
«10¨ x 50 x 50 arcsin ¸ 50
4
800
15
©
¹
¬
(b) z
1
y = −x
ª2 1 4 x 2 x 1 4 x 2 º dx
¬
¼
1 4 x2 2 x x 1 4 x2 ³0 ³0
47. (a) V
2
2
2
1
502 x2
48. (a) Graph of f x, y
1 S
100 ³ 0
1002 x 2 y 2 dy dx
ª
x2 y2
25«1 e
¬
z
over region R
1002 x 2 y 2
100
y2
x2
2
100
1002
1000
2
50
³0
1002 r 2 r dr dT | 2081.53 ft 2
§ x 2 y 2 ·º
cos 2 ¨
¸»
© 1000 ¹¼
z
50
R
50
x
(b) Surface area
50
³R ³
y
1 f x x, y
2
f y x, y
2
dA
Using a symbolic computer program, you obtain surface area | 4540 sq. ft.
49.
3
³ 3 ³ 9 x2
9
³ x2 y2
9 x2
2S
3
2S
3
³0 ³0
2
4 x2
50.
³2 ³
51.
³0 ³0 ³0
a
b
4 x2
c
³0
x2 y2
2
a
a
52.
5
³0 ³0
25 x 2
³0
25 x 2 y 2
b
1 c3
3
1 bc 3
3
2S
³0
2
r2 2
r 3 dz dr dT
3
ª 3 r5 º
«3r » dT
5 ¼0
¬
1 2S
2³0
2
³0 r
5
162 2S
dT
5 ³0
324S
5
16 2S
dT
3 ³0
dr dT
32S
3
cy 2 cz 2 dy dz
13 b3c bcz 2 dz
1
dz dy dx
1 x y2 z2
2
dz dr dT
9r 2 r 4 dr dT
2S
³0 ³0
³0
2
³0 ³0 ³0
x 2 y 2 dz dy dx
x 2 y 2 z 2 dx dy dz
9
³ 0 ³ 0 ³ r2 r
x 2 y 2 dz dy dx
S 2
S 2
S 2
S 2
1 abc3
3
5
13 ab3c 13 a 3bc
1 abc
3
a 2 b2 c2
U2
³ 0 ³ 0 ³ 0 1 U 2 sin I d U dI dT
³ 0 ³ 0 >U
S 2
³0
arctan U @0 sin I dI dT
5
S 2
ª¬ 5 arctan 5 cos I º¼ 0
dT
S
2
5 arctan 5
© 2010 Brooks/Cole, Cengage Learning
364
Chapter 14
1
53.
³ 1 ³ 54.
³0 ³0
2
1 x2
56. V
³0
S 2
4³
55. V
1 x2 y2
³
1 x2
4 x2
Multiple Integration
1 x2 y2
4 x2 y2
2 cos T
³0
0
S 2 ª4
³
0
2³
0
2³
0
«3
¬
S 2
S 2
2 cos T
»
¼0
16 r 2
2 sin T
³0
³0
4³
r dz dr dT
3 2º
4 r2
dT
1
1 r 2
1 r 2
r 3 dz dr dT
S 2
0
32 S
3 ³0
r dz dr dT
2³
32 sin 2 T 4 sin 4 T dT
2
S 2
0
8³
2 cos T
³0
r
4 r 2 dr dT
1 sin 3 T dT
2 sin T
³0
S 2
0
S 2
r 16 r 2 dr dT
8 sin 2 T sin 4 T dT
S 2
m
4k ³
S 2
S 4
S 2
cos I
³0 ³0
M xy
4k ³
k³
z
x
S 4
S 2
cos I
³0 ³0
S 2
S 4 ³0
m
y
4
kS
24
cos5 I sin I dT dI
S 2
1
cos5 I sin I dI
kS
2 ³S 4
S 2
ª 1
6 º
« 12 kS cos I »
¬
¼S
4
kS
96
1
4
1·
§
¨ 0, 0, ¸
4¹
©
S 2
2k ³
0
M xz
2k ³
0
M xy
2k ³
0
S 2
S 2
a
cr sin T
³0 ³0
a
cr sin T
a
cr sin T
³0 ³0
³0 ³0
S 2
2
2
kca 3 ³ sin T dT
kca 3
0
3
3
S 2 a
S 2
1
1
r 2 sin T dz dr dT
kca 4 ³ sin 2 T dT
S kca 4
2kc ³ ³ r 3 sin 2 T dr dT
0
0
0
2
8
S 2 a
1 2 4 S2 2
1
kc a ³ sin T dT
S kc 2 a 4
rz dz dr dT
kc 2 ³ ³ r 3 sin 2 T dr dT
0
0
0
4
16
r dz dr dT
x
0
y
M xz
m
S kca 4 8
M xy
S kc 2 a 4 16
x, y, z
S 2
ª 2 §1
4 ·º
« 3 kS ¨ 4 cos I ¸»
©
¹¼S
¬
0 by symmetry
m
z
S 2
2
kS
cos3 I sin I dI
3 ³S 4
U 3 cos I sin I d U dT dI
kS 96
kS 24
M xy
x, y, z
58.
S 2
29S
2
U 2 sin I d U dT dI
4 S2 S2
k
cos3 I sin I dT dI
3 ³S 4 ³ 0
S 2
32 § S
2·
¨ ¸
3©2
3¹
32 ª
1
º
T cos T cos3 T »
3 «¬
3
¼0
ª
1
3§ 1
1
·º
8«4T 2 sin 2T sin 3 T cos T ¨ T sin 2T ¸»
4
4
2
4
©
¹¼ 0
¬
57.
8S
15
4
3
xyz dz dy dx
4 r2
³0
2S
³0 ³0 ³
x 2 y 2 dz dy dx
m
2kca 3 3
3
2kca 3
2kc ³
S 2
0
a
³0 r
2
sin T dr dT
3S a
16
3S ca
32
§ 3S a 3S ca ·
,
¨ 0,
¸
© 16 32 ¹
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 14
S 2
m
k³
0
M xy
k³
0
59.
x
S 2
y
a
S 2
a
³0 ³0
2
kS a 4
16
U cos I U 2 sin I d U dT dI
kS a 4 § 6 ·
¨
¸
16 © kS a 3 ¹
3a
8
§ 3a 3a 3a ·
¨ , , ¸
©8 8 8¹
500S
3
m
25 r 2
2S
3
³0 ³0 ³
x
2S
3
4
³0 ³0 ³
3 ª1
2S
³0 ³0
32
3
2S
81S 1
4 162S
5
³0 ³3 ³
2S
³0
3
3
2S
³0 ³0
25 r 2 4r dT dr
r
500S
125 º
ª 64
2S «
18 3
3 »¼
¬ 3
º
2r 2 »
¼0
zr dz dr dT 9 º
« 2 r 2 r » dr dT
¬
¼
M xy
m
z
25 r 2
500S
3
r dz dT dr
500S
ª 1
2S « 25 r 2
3
¬ 3
y
0 by symmetry
M xy
kS a 3
6
sin I d U dT dI
M xy
m
z
x, y, z
60.
S 2
³0 ³0 U
25 r 2
25 r 2
3
3ª
2S
2S
ª 81 º
« 8 T »
¬
¼0
162S
1
º
25 r 2 » r dr dT 0
2
¼
³ 0 ³ 0 «¬8 zr dz dr dT
ª1 4 9 2 º
« 8 r 4 r » dT
¬
¼0
500S
14S
3
3
81
S
4
1
8
1·
§
¨ 0, 0, ¸
8¹
©
x, y, z
S 2
61. I z
4k ³
62. I z
k³
63. z
f x, y
16 r 2
4
³3 ³0
0
S
0
0 d r d
2S
a
³0 ³0 U
2
4k ³
r 3 dz dr dT
S 2
0
4
³3
4kS a 6
9
sin 2 I U U 2 sin I d U dT dI
a2 x2 y 2
833S k
3
16r 3 r 5 dr dT
z
a2 r 2
a−h
2ah h 2
a
h
a
a
x
y
(a) Disc Method
S³
V
a
y
ª§
a
ª
S «a 2 y a 2 y 2 dy
y3 º
»
3 ¼a h
S Ǭ a3 ǩ
¬
ª 3 a
a
h3 º
S «a a3 a 2h a 2 h ah 2 »
3
3
3¼
¬
Equivalently, use spherical coordinates.
ah
¬
3
2S
(b) M xy
3
cos 1 a h a
³0 ³0
V
z
365
2S
cos 1 a h a
³0 ³0
M xy
V
a
³ a h sec I U
a
³ a h sec I
1 2
2
h S 2a h
4
1 2
h S 3a h
3
2
3
a h ·º
a3 · § 2
¸»
¸ ¨¨ a a h ¸
3¹
3
©
¹»¼
ª
h3 º
1 2
S «ah 2 »
S h >3a h@
3
3
¬
¼
2a
a2 − x2
y=
a
a−h
−a
x
a
sin I d U dI dT
U cos I U 2 sin I d U dI dT
3 2a h
4 3a h
1 2
h S 2a h
4
2
2
2
§
3 2a h ·
¸
Centroid: ¨ 0, 0,
¨
4 3a h ¸¹
©
© 2010 Brooks/Cole, Cengage Learning
366
Chapter 14
(c) If h
Multiple Integration
3a
a, z
2
3
a.
8
4 2a
3 ·
§
Centroid of hemisphere: ¨ 0, 0, a ¸
8
©
¹
(d) lim z
lim
ho0
ho0
cos 1 a h a
2S
a, I z
64. x 2 y 2 z2
a2
65.
³ ³Q ³
2S
4 3a h
S
a
12a
a
³0 ³0
(f ) If h
Iz
3 4a 2
U 2 sin 2 I
(e) x 2 y 2
Iz
2
3 2a h
³ a h sec I
a 3S
20a 2 15a 2 3a 2
30
4 5
a S.
15
1
a
³ a ³ x 2 y 2 dV
6 sin I
h3
20a 2 15ah 3h 2 S
30
U 2 sin 2 I U 2 sin I d U dI dT
1 z 2 a 2
1 z 2 a 2
³
1 y 2 z 2 a 2
1 y 2 z 2 a 2
x 2 y 2 dx dy dz
8
Sa
15
³0 ³0 ³0
U 2 sin I d U dI dT
Because U
6 sin I represents (in the yz-plane) a circle of radius 3 centered at 0, 3, 0 , the integral represents the volume of
the torus formed by revolving 0 T 2S this circle about the z-axis.
66.
S
2
1 r 2
³0 ³0 ³0
r dz dr dT
1 r 2 represents a paraboloid with vertex 0, 0, 1 , this integral represents the volume of the solid below the
Because z
4 x 2 in the xy-plane.
paraboloid and above the semi-circle y
67.
w x, y
w u, v
wx wy
wy wx
wu wv wu wv
68.
w x, y
w u, v
wx wy
wy wx
wu wv wu wv
2u 2v 2u 2v
69.
w x, y
w u, v
wx wy
wx wy
wu wv
wv wu
1§ 1 · 1§ 1 ·
¨ ¸ ¨ ¸
2© 2 ¹ 2© 2 ¹
x
1
u v,y
2
1 3 2 3
1
u v Ÿ u
2
9
x y, v
8uv
1
2
y
3
x y
2
Boundaries in xy -plane
Boundaries in uv-plane
x y
3
u
3
x y
5
u
5
x y
1
v
1
x y
1
v
1
1
³R ³ ln x y dA
5
1
§1
³ 3 ³ 1 ln¨© 2 u v
y = −x + 5
y=x+1
y=x−1
y = −x + 3
x
2
1
1
·§ 1 ·
u v ¸¨ ¸ dv du
2
¹© 2 ¹
5 ln 5 5 3 ln u 3
5
1
1
³ 3 ³ 1 2 ln u dv du
5
3
³ 3 ln u du
>u ln u
u@3
5
5 ln 5 3 ln 3 2 | 2.751
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 14
70.
w x, y
wx wy
wx wy
wu wv
wv wu
w u, v
x
v
Ÿ u
u
u, y
§1·
1¨ ¸ 0
©u¹
x, v
1
u
xy
Boundary in xy -plane
Boundary in uv-plane
x
1
u
1
x
5
u
5
y
5
4
xy
1
v
1
xy
5
v
5
x
³R ³ 1 x 2 y 2 dA
5
5
³1 ³1
367
2
§1·
du dv
2¨ ¸
1 u2 v u © u ¹
u
4 arctan v@1
5
y = 1x
x=1
3
x=5
5
5
³1 ³1
1
du dv
1 v2
5
³1
4
dv
1 v2
1
x
y = 1x
1
4
5
4 arctan 5 S
Problem Solving for Chapter 14
z
16³
1. V
16³
R
³
1 x 2 dA
S 4
0
1
³0
1 r 2 cos 2 T r dr dT
16
>sec T cos T tan T @S0
3
4
16 S 4 1 ª
1 cos 2 T
3 ³ 0 cos 2 T ¬«
82
32
1º dT
¼»
1
2 | 4.6863
y
1
1
R
y=x
x
1
d ax by Plane
c
a
b
, fy
c
c
2. z
fx
1 fx
S
3. (a)
2
³R ³
³ a2
fy
1
1
a2
b2
2
2
c
c
a 2 b2 c2
³R ³ dA
c
a2
b2
2 dA
2
c
c
a2 b2 c2
AR
c
du
u2
Then
(b) I1
³
³0
³0
1
u
arctan c. Let a 2
2 u2, u
v.
a
a
1
1
v
arctan
dv
C.
2
2 u 2 v2
2u
2 u2
2 2ª
«
¬
2 2
2
2 u2
§
¨ arctan
2u ©
2
2 sin T , du
4³
arctan
2
Let u
I1
2
S 6
0
u
º
» du
2
2 u ¼ u
v
u
2u
2
arctan
2 cos T dT , 2 u 2
§
1
arctan ¨¨
2 cos T
©
2 sin T ·
¸˜
2 cos T ¹¸
·
¸ du
2u ¹
u
2
2 2 sin 2 T
2 cos T dT
4³
S 6
0
³0
4
2 2
2u
2
u
arctan
2 u2
du
2 cos 2 T .
S 6
arctan tan T dT
4T 2 º
»
2 ¼0
§S ·
2¨ ¸
©6¹
2
S2
18
© 2010 Brooks/Cole, Cengage Learning
368
Chapter 14
(c) I 2
2
Multiple Integration
ª
«
2
¬
³
2
³
2 2
2 u2
§
1
arctan ¨¨
2 cos T
©
S 2
S 6
4³
S 2
S 6
³0
4
2
³
§ 1 sin T ·
arctan ¨
¸ dT
© cos T ¹
2 u2
2 2
§
arctan ¨¨
©
1 xy xy
1
³ 0 1 xy dx dy
S 2
S 6
1
2 u·
¸ du
2 u 2 ¸¹
"
1
ª1 xy xy
¬
f
¦ ³0
1
K 0
x y
y x
,v
2
2
2x
u
u v
Ÿ x
2
2y
u
u v
Ÿ y
2
§ 1 sin T ·
arctan ¨
¸ dT
© cos T ¹
2
§ § 1§S
···
arctan ¨ tan ¨ ¨ T ¸ ¸ ¸ dT
2
2
©
¹¹¹
© ©
2
³0 ³0
S 2
S 6
1 sin T
1 sin T 1 sin T
ª§ S 2
S 2 · § S 2 S 2 ·º
2 Ǭ
¸¨
¸»
8 ¹ © 12
72 ¹¼
© 4
6
1
4³
4³
2 cos T dT
1 sin T
1 sin T
1 cos S 2 T
S 2
1
1 xy
2 2 sin T ·
¸¸ ˜
2 cos T
¹
1 cos S 2 T
ªS
T2º
2« T »
2 ¼S
¬2
1
du
2
ª
§ u 2 ·
§ u 2 ·º
arctan ¨¨
arctan ¨¨
«
¸
¸» du
¸
2
2 ¸
2 u 2 ¬«
© 2u ¹
© 2 u ¹»¼
§ 1§S
··
(d) tan ¨ ¨ T ¸ ¸
¹¹
© 2© 2
(f)
2
2 sin T .
4³
(e) I 2
u º
»
2
2 u ¼u v
arctan
2
2
Let u
I2
2
4³
1§S
·
¨ T ¸ dT
2© 2
¹
S 2
S 6
2
1 sin T
cos 2 T
ª18 9 6 1 2 º
2«
S »
72
¬
¼
2³
1 sin T
cos T
S 2 §S
S 6
·
¨ T ¸ dT
©2
¹
S2
4 2
S
36
9
xy 1
"º dx dy
¼
2
f
yK
dy
K 1
¦
1
y K 1
K 0
K 1
1 f
³0 ³0 ¦
1
f
K 0
0
K
f
¦ ³0
dx dy
K 0
¦
2
xy
K 0
f
1
K 1
1
x K 1 y K
K 1
1
dy
0
1
¦ n2
2
n 1
(g) u
v
2
v
2
w x, y
1
2
1
2
w u, v
1
2
1
2
R
y
v
S
2
0, 0
l
1, 0
1 ·
§ 1
l ¨
,
¸
2¹
© 2
0, 1
1 ·
§ 1
l ¨
,
¸
2¹
© 2
1, 1
l
1
1
1
1
0, 0
( 12 , 12 )
1
(
S
R
1
2, 0)
u
2
3
4
−1
x
−2
( 12 , − 12 )
2, 0
³ 0 ³ 0 1 xy dx dy
³0
2 2
1
u
³ u
1
2
2
u
v
2
2
dv du ³
2
u 2 2 ³u
1
2
2
1
2
2
u
v
2
2
dv du
I1 I 2
S2
18
S2
S2
9
6
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 14
4. A:
5§
2S
r
³ 0 ³ 4 ¨© 16
r
r2 ·
¨
¸r dr dT
© 16 160 ¹
³0 ³9
9. From Exercise 69, Section 14.3,
f
523S
| 1.71 ft 3
960
10 §
2S
B
1333S
| 4.36 ft 3
960
r2 ·
¸ r dr dT
160 ¹
The distribution is not uniform. Less water in region of
greater area.
In one hour, the entire lawn receives
10 §
2S
r
r2 ·
¨
¸r dr dT
© 16 160 ¹
³0 ³0
5. Boundary in xy -plane
³ f e
So,
f
³0
f
2x
u
2
ev
v
3
³0
v
4
Let u
w x, y
w u, v
2§ v ·
¨ ¸
3© u ¹
2S
³ S ³ 1 w u, v
2
2S
1§ u ·
¨ ¸
3© v ¹
S 4
8 r2
2
³0
1
3
2
2x
6 x
³0 ³0 ³ x
7. V
1
,x
x
8S
4 2 5
3
f
0
v, u2
f
x y
°­ke
®
°̄0
f
1
1
n n
f
f
³0 ³0
x, y dA
ke x y
18
b
e x a dx
lim ª¬ a e x a º¼
0
bof
6
1
5
2
a
dx dy
f
³0
e y
a
dy
(0, 6, 0)
6
1
n
y
1
.
k
ka 2 or a
lim
nof
b
2S xe
b of ³ 0
lim
f
dy
f
³ f ³ f e
4³
f
0
f
³0
f
e
x2 y2
n 1
2
S.
dy dx
y
x2 y 2
2
0
1
2
lim ª«S e x º»
¼0
b of ¬
dx
dy dx
f
2
y = e− x
2
4 ³ e x dx ³ e y dy
1
n 1
b
x2
This same volume is given by
S
x n 1 n º
³ 0 «¬ n 1 y »¼ dy
0
a.
12. By the shell method,
V
1
1ª
y dx dy
elsewhere
So, assuming a, k ! 0, you obtain
z
1
³0 ³0 x
S
2
x t 0, y t 0
a
f
f
ª y n 1 º
«
»
2
«¬ n 1 »¼ 0
lim
§ S·
2¨¨
¸¸
© 4 ¹
2
0
³0
³0 n 1y
nof
2
These two integrals are equal to
dy dz dx
1
dv.
k ³ e x a dx ˜
(3, 3, 0)
y dx dy
ve v dv
u e u 2u du
sin I d U dI dT
3
³0 ³0 x
f
³0
v e v dv
v, 2u du
³0
ln 1 x dx
³ f ³ f f
x
8.
e v dv
³f
ln 1 x dx
11. f x, y
(0, 0, 0)
n n
dx
.
x
e v , dx
4
1
S
4
(See Problem Solving #9.)
1
3
(3, 3, 6)
1
S
2
f
8S
4 2 5
3
3
2
f
0
dA
S
2
e x dx
2 ³ u 2e u du
r dz dr dT
³ 0 ³ 0 ³ 2 sec I U
(b) V
23
w x, y
³0 ³0 ³2
6. (a) V
1
13
13
³ R ³ 1 dA
A
1
f
³0
1 f x2
ª 1 x2 º
« 2 xe » 2 ³ 0 e dx
¬
¼0
§1·
ln ¨ ¸, dv
© x¹
10. Let v
y
2§ u ·
¨ ¸
3© v ¹
2S
and
2
dx
1
˜
2
1
23
2 2
2
u
1§ v ·
¨ ¸
3© u ¹
e x
x 2e x dx
x
y
2S .
dx
125S
| 32.72 ft 3.
12
Boundary in uv-plane
1 2
x
3
1 2
x
4
x2 2
³0
y
y
369
4 ª³ e
«¬ 0
2
0
0
f
So,
f
³0
2
x2
e x dx
2
dxº .
»¼
x
b
S
2
.
© 2010 Brooks/Cole, Cengage Learning
Chapter 14
370
Multiple Integration
13. Essay
§ 'x ·
¨
¸ 'y
© cos T ¹
l ˜w
14. A
Δx
cos θ
sec T 'x 'y
P
Δy
Δy
θ
Δx
Area in xy-plane: 'x 'y
15. The greater the angle between the given plane and the xy-plane, the greater the surface area. So: z2 z1 z4 z3
16. Converting to polar coordinates,
f
f
³0 ³0
1
1 x y
2
2
2
dx dy
f
S 2
³0 ³0
f
³0
lim
t of
1
1 r2
r
1 r
t
S
2
2
2
§S ·
¨ ¸ dr
©2¹
³0 4 1 r
2
2
t
1 º
ªS
lim
˜
2»
t of «
¬ 4 1 r ¼0
1
1
³0 ³0
17.
1
1
³0 ³0
x y
x y
3
dx dy
3
dy dx
1
2
x y
x y
r dT dr
2r dr
S
4
1
2
The results are not the same. Fubini's Theorem is not valid because f is not continuous on the region 0 d x d 1, 0 d y d 1.
18. The volume of this spherical block can be determined as follows. One side is length 'U . Another side is U 'I . Finally, the
third side is given by the length of an arc of angle 'T in a circle of radius U sin I . Thus:
'V | 'U U 'I 'T U sin I
U 2 sin I 'U 'I 'T
z
ρi sin φi Δ θi
Δ ρi
ρi Δφi
y
x
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 5
Vector Analysis
Section 15.1
Vector Fields.......................................................................................372
Section 15.2
Line Integrals ......................................................................................383
Section 15.3
Conservative Vector Fields and Independence of Path ....................395
Section 15.4
Green’s Theorem ................................................................................402
Section 15.5
Parametric Surfaces............................................................................410
Section 15.6
Surface Integrals .................................................................................418
Section 15.7
Divergence Theorem ..........................................................................427
Section 15.8
Stokes’s Theorem ...............................................................................433
Review Exercises ........................................................................................................440
Problem Solving .........................................................................................................449
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R
Vector Analysis
1 5
Section 15.1 Vector Fields
1. All vectors are parallel to x-axis.
11. F x, y, z
z
3 yj
4
Matches (d)
F
3 y
c
2. All vectors are parallel to y-axis.
Matches (c)
2
x
3. Vectors are in rotational pattern.
Matches (e)
12. F x, y
4. All vectors point outward.
F
xi
x
y
4
y
3
c
2
Matches (b)
1
5. Vectors are parallel to x-axis for y
nS .
x
−3
−2
−1
1
2
3
Matches (a)
6. Vectors along x-axis have no x-component.
Matches (f )
i j
7. F x, y
F
y
y
16 x 2 y 2
x2
y2
2
c 16
c
2
F
4 xi yj
13. F x, y
2
c
1
2
x
−2
1
2
−2
−4
F
1
x
−4
8. F x, y
−1
y
2i
x2 y 2 i j
14. F x, y
F
1 x2 y 2
y
10
2
8
2
6
4
1
x
−2
−1
1
2
x
−2
2
−2
i jk
15. F x, y, z
yi xj
9. F x, y
F
5
F
z
y
3
y
4
y x2
2
4
−4
x
−5
5
4
x
−4
−5
10. F x, y
F
16. F x, y, z
yi 2 xj
F
y
y 2 4 x2
xi yj zk
x2 y 2 z 2
x2 y2 z 2
2
z
2
c
−2
−2
c2
2
x
−2
2
2
x
y
−2
−2
372
© 2010 Brooks/Cole, Cengage Learning
Section 15.1
f x, y
x2 2 y2
f x x, y
2x
2
f y x, y
4y
1
F x, y
2 xi 4 yj
1
2 xyi y 2 j
8
17. F x, y
21.
y
Note that ’f
x
−2
−1
1
2
Vector Fields
373
F.
−1
22.
−2
2 y 3x i 2 y 3x j
18. F x, y
y
6
f x, y
x2 f x x, y
2x
f y x, y
12 y
F x, y
2 xi g x, y
5 x 2 3 xy y 2
1 y2
4
1
yj
2
4
2
−6
−4
23.
x
−2
2
4
6
−6
xi yj zk
19. F x, y , z
x2 y 2 z 2
24.
g x x, y
10 x 3 y
g y x, y
3x 2 y
G x, y
10 x 3 y i 3 x 2 y j
g x, y
sin 3 x cos 4 y
z
2
1
1
1
2
3 cos 3x cos 4 y
g y x, y
4 sin 3 x sin 4 y
G x, y
3 cos 3x cos 4 yi 4 sin 3 x sin 4 yj
y
25.
2
x
xi yj zk
20. F x, y, z
g x x, y
z
2
f x, y , z
6 xyz
f x x, y , z
6 yz
f y x, y , z
6 xz
f z x, y , z
6 xy
F x, y , z
6 yzi 6 xzj 6 xyk
1
1
1
2
y
2
x
26. f x, y, z
fx
fy
fz
F x, y , z
x2 4 y 2 z 2
x
x 4 y2 z2
4y
2
x2 4 y 2 z 2
z
x2 4 y 2 z 2
x
x 4y z
2
2
2
i 4y
x 4y z
2
2
2
j
z
x 4 y2 z2
2
k
© 2010 Brooks/Cole, Cengage Learning
374
27.
28.
29.
30.
Chapter 15
g x, y , z
Vector Analysis
z ye x
2
2
g x x, y , z
2 xye x
g y x, y , z
ex
g z x, y , z
1
G x, y , z
2 xye x i e x j k
g x, y , z
y
z
xz
z
x
y
2
2
2
z
z
x2
y
g x x, y , z
g y x, y , z
xz
1
2
z
y
g z x, y , z
G x, y , z
§ z
§1
§ y
z·
xz ·
x·
1
¨ 2 ¸ i ¨ 2 ¸ j ¨ 2 ¸k
y¹
y ¹
x
y¹
© x
©z
© z
h x, y , z
xy ln x y
y
x
1
2
z
x
y
hx x, y, z
y ln x y xy
x y
h y x, y , z
x ln x y xy
x y
hz x, y, z
0
H x, y , z
ª xy
º
ª xy
º
y ln x y » i «
x ln x y » j
«
x
y
x
y
¬
¼
¬
¼
h x, y , z
hx x, y, z
h y x, y , z
hz x, y, z
H x, y , z
arcsin yz
xz
1 y2 z2
xy
1 y2 z2
arcsin yz i xz
1 y z
2 2
j
xy
1 y2z2
xy 2i x 2 yj
31. F x, y
M
x arcsin yz
xy 2 and N
2 xy
x 2 y have continuous first partial
M
M
sin y and N
x cos y have continuous first
partial derivatives.
wM
Ÿ F conservative
wy
1
yi xj
x2
32. F x, y
sin yi x cos yj
33. F x, y
derivatives.
wN
wx
k
y x 2 and N
y
1
i j
2
x
x
1 x have continuous first partial
wN
wx
cos y
1
yi xj
xy
34. F x, y
M
wM
Ÿ F is conservative.
wy
1 x and N
1
1
i j
x
y
1 y have continuous first partial
derivatives for all x z 0.
derivatives for all x, y z 0.
wN
wx
wN
wx
1
x2
wM
Ÿ F is conservative.
wy
0
wM
Ÿ F is conservative.
wy
© 2010 Brooks/Cole, Cengage Learning
Section 15.1
5 y 2 yi 3 xj
35. F x, y
M
5 y3, N
wN
wx
15 y 2
36. M
wM
Ÿ Conservative
wy
2 x 2 x
e
y2
wN
wx
2 y 2 x
y3
1
37. M
x
x y
2
2
32
wN
wx
x2 y2
wM
wy
z
xy 2
w
> y@
wy
f x x, y
40. F x, y
x y
2
32
32
x
1 xy
f x x, y
3x 2 y 2
f y x, y
2 x3 y
x Ÿ f x, y
xy k
f x, y
x y K
2y
2y
K
15 y 3i 5 xy 2 j
w
ª5 xy 2 º¼
wx ¬
45 y 2 z
5 y 2
1
yi 2 xj
y2
1
y2
2
y2
2y
x2
i 2j
x
y
2
x
w ª x2 º
« »
wx ¬ y 2 ¼
2x
y2
Not conservative
46. F x, y
x
y
i 2
j
x2 y2
x y2
º
wª
x
« 2
2»
wy ¬ x y ¼
º
wª
y
«
»
wx ¬ x 2 y 2 ¼
2 xy
x y2
2
2
2 xy
x y2
2
2
Conservative
Conservative
f x x, y
2y
Not conservative
w ª2 y º
wy «¬ x »¼
2 xyi x 2 j
2x
2y
1
2x
i 2j
y
y
45. F x, y
3 2
w 2
ªx º
wx ¬ ¼
ex
w ª 2x º
« »
wx ¬ y 2 ¼
6x2 y
2x
f x, y
w ª1º
« »
wy ¬ y ¼
6x2 y
w
>2 xy@
wy
x 2e x
wM
Ÿ Conservative
wy
Conservative
41. F x, y
f y x, y
44. F x, y
3x 2 y 2i 2 x3 yj
w
ª2 x3 yº¼
wx ¬
2 x3 ye x
2y
Not conservative
y , f y x, y
w
ª3 x 2 y 2 º¼
wy ¬
2 xye x
w
ª15 y 3 º¼
wy ¬
y
2
w
> x@ Ÿ Conservative
wx
1
f x x, y
43. F x, y
yi xj
39. F x, y
2y
2 x3 ye x
1
y
,N
1 xy
2 xy 1
2 xe x
2y
Conservative
y
Ÿ Not conservative
38. M
375
2 yi xj
2 xe x
w ª 2 x2 y º
xe »
¼
wx «¬
wM
Ÿ Conservative
wy
y
,N
x2 y2
wN
wx
e2 x
2y
2
wª
2 xye x y º»
¼
wy ¬«
15 xy 2
2 2x y
e ,N
y
xe x
42. F x, y
Vector Fields
2 xy, f y x, y
2
x , f x, y
x y K
2
f x x, y
x
x2 y 2
f y x, y
y
x2 y 2
f x, y
1
ln x 2 y 2 K
2
© 2010 Brooks/Cole, Cengage Learning
376
Chapter 15
47. F x, y
Vector Analysis
e x cos yi sin yj
w x
ªe cos yº¼
wy ¬
e x sin y
w
ª e x sin yº¼
wx ¬
Conservative
e sin y
e x cos y
f y x, y
e x sin y
f x, y
48. F x, y
2x
x2 y 2
º
»
2»
¼
ª
2y
w«
«
2
wx x y 2
¬
Conservative
º
»
2»
¼
f y x, y
f x, y
51. F x, y, z
curl F
curl F
w
wz
xyz
xyz
xyz
6 2i 3 2j 36k
4i j 3k
i 2
2y
x2 y 2
curl F
8 xy
x2 y 2
3
i
j
k
w
wx
w
wy
w
wz
x 2 z 2 xz
yz
z 2 x i 0 x 2 j 2 z 0 k
8 xy
x2 y2
x 2 zi 2 xz j yzk , 2, 1, 3
50. F x, y, z
j
2
3
z 2 x i x 2 j 2 zk
curl F 2, 1, 3
7i 4 j 6k
2x
x y2
2
2
2y
x y2
2
2
1
K
x2 y2
e x sin y i e x cos y j, 0, 0, 1
j
k
w
w
wx
wy
e x sin y e x cos y
w
wz
0
e x cos y e x cos y k
2e x cos y k
2k
e xyz i j k , 3, 2, 0
i
w
wx
j
w
wy
k
w
wz
e xyz
e xyz
e xyz
xz xy e xyz i yz xy e xyz j yz xz e xyz k
6i 6 j
curl F 3, 2, 0
53. F x, y, z
k
curl F 2, 1, 3
curl F 0, 0, 1
52. F x, y, z
j
w
wy
xz xy i yz xy j yz xz k
i
curl F
i
w
wx
e x cos y K
ª
2x
w«
«
2
wy x y 2
¬
f x x, y
curl F
x
f x x, y
xyz i xyz j xyz k , 2, 1, 3
49. F x, y, z
§ x·
arctan ¨ ¸i ln
© y¹
x2 y2 j k
i
w
wx
j
w
wy
k
w
wz
§ x·
arctan ¨ ¸
© y¹
1
ln x 2 y 2
2
1
ª
x y2
x
« 2
2
«¬ x y
1 x y
2
º
»k
»¼
2x
k
x2 y2
© 2010 Brooks/Cole, Cengage Learning
Section 15.1
i
j
k
w
wx
w
wy
w
wz
yz
y z
xz
x z
xy
x y
ª
1
x2 «
«¬ x y
2
ª x2
«
«¬ x y
2
º
ª y2
»i «
x z »¼
«¬ x y
x2
º
ª
1
»i y 2 «
«¬ x y
x z »¼
2
1
2
2
2
º
ª z2
»j «
y z »¼
«¬ x z
y2
2
º
ª
1
» j z2 «
«¬ y z
y z »¼
1
2
2
2
º
»k
y z »¼
z2
2
º
»k
x z »¼
1
2
sin x y i sin y z j sin z x k
55. F x, y, z
i
j
k
w
wx
w
wy
w
wz
sin x y
sin y z
sin z x
curl F
cos y z i cos z x j cos x y k
x2 y 2 z 2 i j k
56. F x, y, z
curl F
i
j
k
w
wx
w
wy
w
wz
x2 y 2 z 2
x2 y 2 z 2
xy 2 z 2i x 2 yz 2 j x 2 y 2 zk
57. F x, y, z
curl F
377
yz
xz
xy
i j
k
y z
x z
x y
54. F x, y, z
curl F
Vector Fields
y zi z x j x yk
x2 y 2 z 2
x2 y2 z 2
y 2 z 3i 2 xyz 3 j 3 xy 2 z 2k
58. F x, y, z
i
j
k
i
j
k
w
wx
w
wy
w
wz
w
wx
w
wy
w
wz
xy 2 z 2
x 2 yz 2
x2 y 2 z
0
Conservative
curl F
y 2 z3
0
2 xyz 3 3 xy 2 z 2
Conservative
2 2
f x x, y , z
xy z
f y x, y , z
x 2 yz 2
f z x, y , z
x2 y 2 z
f x, y , z
1 2 2 2
x y z K
2
xy 2 z 3 K
f x, y , z
sin zi sin xj sin yk
59. F x, y, z
curl F
i
j
k
w
wx
w
wy
w
wz
sin z sin x sin y
cos yi cos zj cos xk z 0
Not conservative
ye z i ze x j xe y k
60. F x, y, z
curl F
i
j
k
w
wx
w
wy
w
wz
ye z
ze x
xe y
xe y e x i e y ye z j ze x e z k z 0
Not conservative
© 2010 Brooks/Cole, Cengage Learning
378
Chapter 15
z
xz
x
i 2j k
y
y
y
61. F x, y, z
curl F
Vector Analysis
i
j
k
w
wx
w
wy
w
wz
z
y
xz
y2
§ x
§1
§ z
1·
x ·
z ·
¨ 2 2 ¸i ¨ ¸ j ¨ 2 2 ¸ k
y
y
y
y
y
y
©
¹
©
¹
©
¹
0
x
y
Conservative
f x x, y , z
z
y
f y x, y , z
f z x, y , z
x
y
f x, y , z
xz
K
y
xz
y2
x
y
i 2
j k
x y2
x y2
62. F x, y, z
63. F x, y
x 2i 2 y 2 j
2
i
j
k
w
wx
w
wy
w
wz
x
x y2
y
x y2
1
curl F
2
2
w 2
w
2 y2
x wx
wy
div F x, y
0
64. F x, y
xe x i ye y i
div F x, y
f y x, y , z
x
x2 y 2
y
x2 y 2
f z x, y , z
1
f x, y , z
³ x2
f x, y , z
f x, y , z
f x, y , z
66. F x, y, z
div F x, y, z
w
w
xe x ye y
wx
wy
xe x e x ye y e y
Conservative
f x x, y , z
2x 4 y
65. F x, y, z
div F x, y, z
sin xi cos yj z 2k
w
w
w
>sin x@ >cos y@ ª¬ z 2 º¼
wx
wy
wz
cos x sin y 2 z
x
dx
y2
1
ln x 2 y 2 g y, z K1
2
y
³ x 2 y 2 dy
1
ln x 2 y 2 h x, z K 2
2
³ dz z p x, y K3
1
ln x 2 y 2 z K
2
ln x 2 y 2 i xyj ln y 2 z 2 k
wª
w
wª
ln x 2 y 2 º¼ > xy@ ln y 2 z 2 º¼
wx ¬
wy
wz ¬
2x
2z
x 2
x2 y2
y z2
© 2010 Brooks/Cole, Cengage Learning
Section 15.1
xyzi xyj zk
67. F x, y, z
1 21
div F 2, 1, 1
73. See the definition on page 1064.
4
74. See the definition on page 1066.
75.
x 2 zi 2 xzj yzk
68. F x, y, z
div F x, y, z
2 xz y
div F 2, 1, 3
11
F x, y , z
i 3 xj 2 yk
G x, y , z
xi yj zk
FuG
e x sin yi e x cos yj z 2k
e sin y e sin y 2 z
div F 3, 0, 0
0
x
x
ln xyz i j k
1
1
1
3
2
div F 3, 2, 1
j
k
1
3x
2y
z
3xz 2 y 2 i z 2 xy j y 3 x 2 k
curl F u G
1
1
1
x
y
z
div F x, y, z
i
x y
div F x, y , z
70. F x, y, z
379
yz x 1
div F x, y, z
69. F x, y, z
Vector Fields
i
j
k
w
wx
w
wy
w
wz
3xz 2 y 2
z 2 xy y 3 x 2
1 1 i 6 x 3 x j 2 y 4 y k
11
6
9 xj 2 yk
71. See the definition, page 1058. Examples include velocity
fields, gravitational fields, and magnetic fields.
72. See the definition of Conservative Vector Field on page
1061. To test for a conservative vector field, see
Theorem 15.1 and 15.2.
xi zk
76. F x, y, z
x 2i yj z 2k
G x, y , z
i
j
x
0 z
x2
y
FuG
curl F u G
yzi xz 2 x 2 z j xyk
z2
i
j
k
w
wx
w
wy
w
wz
yz
xz 2 x 2 z
xy
x 2 xz x 2 i y y j z 2 2 xz z k
x x 2z 1 i z z 2x 1 k
xyzi yj zk
77. F x, y, z
curl F
k
i
j
k
w
wx
w
wy
w
wz
xyz
y
z
i
j
k
w
wx
w
wy
w
wz
0
xy
xz
curl curl F
xyj xzk
zj yk
© 2010 Brooks/Cole, Cengage Learning
380
Chapter 15
Vector Analysis
x 2 zi 2 xzj yzk
78. F x, y, z
curl F
i
j
k
w
wx
w
wy
w
wz
x z 2 xz
2
curl curl F
z 2 x i x 2 j 2 zk
i
j
k
w
wx
w
wy
w
wz
x
2
i 3xj 2 yk
G x, y , z
xi yj zk
i
j
k
1
3x
2y
x y
FuG
j
x
0 z
2
2
j 2 xk
curl F
xyzi yj zk
i
j
k
w
wx
w
wy
w
wz
xyz
y
z
x x
div curl F
curl F
3z 2 x
Mi Nj Pk and G
F G
0
i
j
k
w
wx
w
wy
w
wz
x 2 z 2 xz
yz
2 2
0
div curl F
83. Let F
xyj xzk
x 2 zi 2 xzj yzk
82. F x, y, z
z
z
yzi xz 2 x 2 z j xyk
0
81. F x, y, z
2 z
k
y
div F u G
3xz 2 y 2 i z 2 xy j y 3 x 2 k
div F u G
i
x
F x, y , z
FuG
x 2i yj z 2k
G x, y , z
yz
z 2x
79.
xi zk
80. F x, y, z
z 2 x i x 2 j 2 zk
Qi Rj Sk where M , N , P, Q, R, and S have continuous partial derivatives.
M Qi N R j P S k
curl F G
i
j
k
w
wx
w
wy
w
wz
M Q
N R P S
ªw
º
ªw
º
w
w
w
ªw
º
N R »i « P S M Q »j « N R M Q »k
« P S w
y
w
z
w
x
w
z
w
x
w
y
¬
¼
¬
¼
¬
¼
§ wP
§ wN
§ wS
§ wR
wN ·
wM ·
wM ·
wR ·
wQ ·
wQ ·
§ wP
§ wS
¨
¸i ¨
¸k ¨
¸i ¨
¸k
¸j ¨
¸j ¨
wz ¹
wz ¹
wy ¹
wz ¹
wz ¹
wy ¹
© wx
© wx
© wy
© wx
© wy
© wx
curl F curl G
84. Let f x, y, z be a scalar function whose second partial derivatives are continuous.
’f
wf
wf
wf
i j
k
wx
wy
wz
curl ’f
i
j
k
w
wx
w
wy
w
wz
wf
wx
wf
wy
wf
wz
§ w2 f
§ w2 f
§ w2 f
w2 f ·
w2 f ·
w2 f ·
¨
¸i ¨
¸j ¨
¸k
wzwy ¹
wzwx ¹
wywx ¹
© wywz
© wxwz
© wxwy
0
© 2010 Brooks/Cole, Cengage Learning
Section 15. 1
85. Let F
Mi Nj Pk and G
381
Ri Sj T k.
w
w
w
M R N S P T
wx
wy
wz
div F G
Vector Fields
wM
wR wN
wS
wP wT
wx
wx
wy
wy
wz
wz
ª wM
wN
wP º ª wR
wS
wT º
« wx wy wz » « wx wy wz »
¬
¼ ¬
¼
div F div G
86. Let F
Mi Nj Pk and G
FuG
div F u G
i
j
k
M
N
P
R
S
T
Ri Sj T k.
NT PS i MT PR j MS NR k
w
w
w
NT PS PR MT MS NR
wx
wy
wz
N
wT
wN
wS
wP
wR
wP
wT
wM
wS
wM
wR
wN
T
P
S
P
R
M
T
M
S
N
R
wx
wx
wx
wx
wy
wy
wy
wy
wz
wz
wz
wz
ª§ wP
§ wN
§ wS
wN ·
wP ·
wM · º ª § wT
wS ·
wT ·
wR ·º
§ wM
§ wR
Ǭ
¸R ¨
¸T » «M ¨
¸ N¨
¸»
¸S ¨
¸ P¨
wz ¹
wx ¹
wy ¹ ¼ ¬ © wy
wz ¹
wx ¹
wy ¹¼
© wz
© wz
© wx
© wx
© wy
curl F ˜ G F ˜ curl G
87. F
Mi Nj Pk
’ u ª¬’f ’ u F º¼
curl ’f ’ u F
curl ’f curl ’ u F
Exercise 83
curl ’ u F
Exercise 84
’u ’uF
88. Let F
’u fF
Mi Nj Pk.
i
j
k
w
wx
w
wy
w
wz
fM
fN
fP
§ wf
§ wf
wP wf
wN ·
wP wf
wM ·
wN
wf
wM ·
§ wf
N f
M f
M f
¨ P f
¸i ¨ P f
¸k
¸j ¨ N f
y
y
z
z
x
x
z
z
x
x
y
w
w
w
w
w
w
w
w
w
w
w
wy ¹
©
¹
©
¹
©
i
ª§ wP wN ·
§ wN
wM · º
wf
§ wP wM ·
f Ǭ
¸i ¨
¸k » ¸j ¨
y
z
x
z
x
y
w
w
w
w
w
w
wx
©
¹
¹
©
¹ ¼
©
M
89. Let F
div f F
Mi Nj Pk , then f F
j
k
wf
wy
wf
wz
N
P
f >’ u F@ ’f u F
f Mi f Nj f Pk.
w
w
w
fM fN fP
wx
wy
wz
f
wM
wf
wN
wf
wP
wf
M
f
N
f
P
wx
wx
wy
wy
wz
wz
§ wM
wN
wN · § wf
wf
wf ·
f¨
N P¸
¸¨ M w
x
w
y
w
z
w
x
w
y
wz ¹
©
¹ ©
f div F ’f ˜ F
© 2010 Brooks/Cole, Cengage Learning
382
Chapter 15
Vector Analysis
Mi Nj Pk.
90. Let F
§ wP
§ wN
wN ·
wM ·
wM ·
§ wP
¨
¸i ¨
¸k
¸j ¨
wz ¹
wz ¹
wy ¹
© wx
© wy
© wx
curl F
w ª wP
wN º
w ª wP
wM º
w ª wN
wM º
«
» «
»
wx ¬ wy
wz ¼ wy «¬ wx
wz »¼ wz ¬ wx
wy ¼
div curl F
w2P
w2 N
w2P
w2M
w2N
w2M
wxwy
wxwz
wywx
wywz
wzwx
wzwy
0
because the mixed partials are equal
In Exercises 91-93, F x , y , z = xi + yj + zk and f x , y , z = F x , y , z
1
ln x 2 y 2 z 2
2
x
y
z
i 2
j 2
k
x2 y2 z 2
x y2 z2
x y2 z2
ln f
91.
’ ln f
1
f
92.
xi yj zk
x2 y 2 z 2
F
f2
1
x y2 z2
2
§1·
’¨ ¸
©f¹
x
x y z
2
2
32
2
x2 y 2 z 2
93. f n
’f n
x2 + y2 + z2 .
=
n
x2 y 2 z 2
n
x2 y 2 z 2
y
i x y z
2
2
32
j
z
x y z
2
2
2
32
xi yj zk
k
x2 y 2 z 2
3
F
f3
n
n 1
x
x y z
2
n2
2
x y
2
2
in
x2 y 2 z 2
n 1
y
x y z
2
2
2
j n
x2 y 2 z 2
n 1
z
x y2 z2
2
k
nf n 2F
xi yj zk
xi yj
94. (a) F x, y
2
y
2
4
3
2
1
x
− 4 − 3 −2
2
3
4
2
3
−2
−3
−4
xi yj
(b) G x, y
y
x2 y 2
4
3
2
1
x
− 4 − 3 −2
4
−2
−3
−4
(c) All the vectors are unit vectors. Those of F point away from origin. Answers will vary.
95. True. F x y
16 x 2 y 4 o 0 as x, y o 0, 0 .
96. True. If x, y is on the positive y-axis, then
x
0 and y ! 0. So,
F x, y
F 0, y
97. False. Curl is defined on vector fields, not scalar fields.
98. False. See Example 7.
y 2 j.
© 2010 Brooks/Cole, Cengage Learning
Section 15.2
Line Integrals
383
Section 15.2 Line Integrals
°­ti tj,
®
°̄ 2 t i 1. r t
0 d t d1
2 t j,
x2
y2
16
9
1
cos 2 t sin 2 t
1
6.
1d t d 2
­ti t 2 j,
0 d t d 2
°
® 4 t i 4 j, 2 d t d 4
° 8 t j,
4 d t d 8
¯
2. r t
­ti,
°
°3i t 3 j,
®
° 9 t i 3 j,
° 12 t j,
¯
3. r t
0 d t d 3
x
y2
9
4 cos t
3 d t d 6
y
3 sin t
sin 2 t
6 d t d 9
x2 y 2
5.
2
0 d t d 2S
4ti 3tj, 0 d t d 1
7. r t
rc t
4i 3 j
1
³ C xy ds
³0
9
42 32 dt
4t 3t
1
³ 0 60t
2
x
y
9
9
cos 2 t sin 2 t
4 cos ti 3 sin tj
rt
9 d t d 12
­ti 4 tj,
0 d t d 5
5
°°
®5i 9 t j, 5 d t d 9
° 14 t i,
9 d t d 14
°̄
4. r t
x2
16
cos 2 t
2
1
ª¬20t 3 º¼
0
dt
20
1
rc t
x2
9
y2
9
3 cos t
3 sin t
cos 2 t
sin 2 t
x
y
ti 2 t j,
8. r t
1
i j
³C 3 x y ds
2
³0 3 t
3 2³
2t
2
0
12 1
2
dt
2t 2 dt
2
3 2 ª¬t 2 2t º¼
0
3 cos ti 3 sin tj
rt
0 d t d 2
0
0 d t d 2S
sin ti cos tj 2k ,
9. r t
rc t
³C
0 d t d
S
2
cos t i sin tj
x 2 y 2 z 2 ds
S 2
³0
sin 2 t cos 2 t 4
cos 2 t sin 2 t dt
S 2
³0
5dt
5S
2
12ti 5tj 84tk , 0 d t d 1
10. r t
rc t
12i 5 j 84k
1
³C
2 xyz ds
11. (a) r t
(b) rc t
³C
1
³0
2 12t 5t 84t
12
2
52 84
ti tj, 0 d t d 1
i j, rc t
x 2 y 2 ds
³0
dt
1
³0
10,080 t 3 85 dt
12. (a) r t
(b) rc t
2
1
2
t2 t2
2 dt
³C
ªt 4 º
856,800 « »
¬ 4 ¼0
ti 2tj,
0 d t d 2
i 2 j, rc t
x 2 y 2 ds
214,200
5
2
³0
t 2 4t 2
2
1
ªt 3 º
2 2« »
¬ 3 ¼0
2 2
3
ª
5t 3 º
« 5 »
3 ¼0
¬
5 dt
40 5
3
© 2010 Brooks/Cole, Cengage Learning
384
Chapter 15
13. (a) r t
(b)
³C
14. (a) r t
(b)
³C
15. (a) r t
(b) rc t
Vector Analysis
S
cos ti sin tj, 0 d t d
x 2 y 2 ds
S 2
³0
2
ª¬cos 2 t sin 2 t º¼
2 cos ti 2 sin tj, 0 d t d
x 2 y 2 ds
S 2
³0
16. (a) r t
(b) rc t
³C
17. (a) r t
(b)
2
cos t
2
S 2
³0
dt
S
dt
2
S
2
ª¬4 cos 2 t 4 sin 2 t º¼
2 sin t
2
2 cos t
2
S 2
³0
dt
4S
8 dt
ti , 0 d t d 1
i , rc t
1
1
³C
sin t
x 4
1
³0
y ds
ªt 2 º
« »
¬ 2 ¼0
t dt
1
2
tj, 1 d t d 9
j, rc t
x 4
y ds
1
9
³1
9
ª8 t 3 2 º
¬ 3 ¼1
4 t dt
8
3
27 1
208
3
­ti,
0 d t d1
°
® 2 t i t 1 j, 1 d t d 2
° 3 t j,
2 d t d 3
¯
³ C1
x 4
y ds
³ C2
x 4
y ds
1
2
1
³ 0 t dt
2
2
³1
ª
t2
8
3 2º
t 1 »
2 «2t 2
3
¬
¼1
ª 2 t 4 t 1º 1 1 dt
¬
¼
3
³ C3
x 4
y ds
³2 4
³C
x 4
y ds
1 19 2
8
2
6
3
3 t dt
ª 8
« 3 3 t
¬
3 2º
3
8
3
»
¼2
19 19 2
6
19 2
6
y
(0, 1)
19 1 2
C2
C3
6
C1
18. (a) r t
(b)
³ C1
³ C2
³ C3
0 d t d 2
­ti,
°
°2i t 2 j,
®
° 6 t i 2 j,
° 8 t j,
¯
x 4
x 4
x 4
³ C4
x 4
³C
x 4
y ds
y ds
y ds
y ds
y ds
(1, 0)
2 d t d 4
4 d t d 6
6 d t d 8
2
³ 0 t dt
4
³2
y
2 4 t 2 dt
6
³4
8
2
³6 4
6 t 4 2 dt
16 2
4
3
C3
2
28 2
(2, 2)
C4
1
8 t dt
2 4
16 2
3
16 2
16 2
28 2 3
3
C2
C1
8
56
3
x
1
2
2
© 2010 Brooks/Cole, Cengage Learning
x
Section 15.2
ti, 0 d t d 1, rc t
19. (a) C1 : 0, 0, 0 to 1, 0, 0 : r t
³ C1
t2º
»¼ 0
1
1
1
i tk , 0 d t d 1, rc t
C2 : 1, 0, 0 to 1, 0, 1 : r t
385
1
³ 0 2t dt
2 x y 2 z ds
i , rc t
Line Integrals
k , rc t
1
z
1
³ C2
1
³0
2 x y 2 z ds
ª
t2 º
«2t »
2 ¼0
¬
2 t dt
3
2
i tj k , 0 d t d 1, rc t
C3 : 1, 0, 1 to 1, 1, 1 : r t
1
(1, 0, 1)
j, rc t
C
1
(0, 0, 0)
(1, 0, 0)
³ C3
1
³0
2 x y 2 z ds
(b) Combining,
³C
1
³0 t
1
³0
2 x y 2 z ds
1
³0
2 x y 2 z ds
³C
2
t º
»
3 ¼0
dt
ª1 t
¬
2 x y 2 z ds
2
rc t
2 sin ti 2 cos tj k
4 sin 2 t 4 cos 2 t 1
³C U
4S
1tº
¼
2
6
³0
5ª
64S 3 º
«16S »
2 ¬
3 ¼
(0, 0, 0)
1
1
2
x
1
³0
2 dt
5
2
6
j k , rc t
2
2 dt
5
3
4S 2 3 | 795.7
2
6
.
z
rt
2 cos ti 2 sin tj tk , 0 d t d 4S
rc t
2 sin ti 2 cos tj k
4 sin 2 t 4 cos 2 t 1
³C U
4S
³0
t
23. r t
rc t
Mass
5
x, y, z ds
ªt 2
«
¬2
5 dt
4S
8S
1
ªt
t2 º
2« »
2 ¼0
¬3
3
t2 t
5 dt
5ª
t3 º
«4t »
2 ¬
3 ¼0
y
(0, 1, 0)
1
Mass
x, y, z ds
5 4S
4 t 2 dt
2 ³0
k , rc t
rc t
5
1
4 cos 2 t 4 sin 2 t t 2
2
(0, 1, 1)
C
22. U x, y, z
2 cos ti 2 sin tj tk , 0 d t d 4S
y
z
1
ª
t º
«t »
2 ¼0
¬
1 1
3 2
rt
rc t
1
1
3
1 2
x y2 z2
2
21. U x, y, z
Mass
j, rc t
2 1
1 t dt
(1, 1, 1)
x
1 t j 1 t k , 0 d t d 1, rc t
C3 : 0, 1, 1 to 0, 0, 0 : r t
(b) Combining,
3 1
1
4
3
23
.
6
j tk , 0 d t d 1, rc t
C2 : 0, 1, 0 to 0, 1, 1 : r t
³ C3
3
4
2
3
1
tj, 0 d t d 1, rc t
2 x y 2 z ds
³ C2
ª
t º
«t »
3 ¼0
¬
2 t 2 1 dt
2 x y 2 z ds
20. (a) C1 : 0, 0, 0 to 0, 1, 0 : r t
³ C1
3 1
4S
º
5»
¼0
8S 2
5
cos ti sin tj, 0 d t d S
sin ti cos tj,
³C U
S
³0
rc t
³C
x, y ds
1
x y ds
cos t sin t dt
>sin t
11
S
cos t @0
2
© 2010 Brooks/Cole, Cengage Learning
386
Chapter 15
Vector Analysis
24. r t
t 2i 2tj,
0 d t d1
rc t
2ti 2 j,
rc t
³C U
Mass
³ C 34 y ds
³
1
3
0 4
1
³ 0 3t t
rc t
1
2
t2 1
25. r t
4t 2 4 dt
2t
12
32 1
º
»¼ 0
dt
rc t
³C U
Mass
x, y, z ds
³1
S 2
³ C F ˜ dr
³0
12
3xi 4 yj
ti 4 t 2 j, 2 d t d 2
Ft
3ti 4 4 t 2 j
rc t
i t
4 t2
j
2
32 3
º
»
»
¼1
12
3 cos t sin t 4 sin t cos t dt
ª sin 2 t º
«
»
¬ 2 ¼0
C: r t
4t 2 5 dt
k 4t 2 5
³ C F ˜ dr
kª
41 41 27º¼
12 ¬
26.
sin ti cos tj
30. F x, y
³ C kz ds
kt
3 cos ti 4 sin tj
rc t
2 2 1
4t 2 5
3
Ft
S 2
t 2i 2tj tk , 1 d t d 3
2ti 2 j k ,
cos ti sin tj, 0 d t d S 2
C: r t
4t 2 4
x, y ds
3xi 4 yj
29. F x, y
31. F x, y, z
2
³ 2
ti t 2 j 2tk , 0 d t d 1
C: r t
2 cos ti 2 sin tj 3tk , 0 d t d 2S
Ft
t 3i 2t 2 j 2t 3k
rc t
2 sin ti 2 cos tj 3k
rc t
i 2tj 2k
49
³C U
Mass
1
13
x, y, z ds
³C
³C
k z ds
2S
³0
k 3t
13 dt
2S
§
3t 2 ·º
13 ¨ kt ¸»
2 ¹¼ 0
©
13 2S k 6S 2
F ˜ dr
32. F x, y, z
C: r t
Ft
rc t
³0
Ft
ti tj
rc t
i j
³ C F ˜ dr
1 2
t k, 0 d t d S
2
1
4 sin 2 ti 4 cos 2 tj t 4k
4
2 cos ti 2 sin tj tk
S
S
1
³0
t t dt
1
ª¬t 2 º¼
Ft
16 sin t cos ti 4 sin tj
rc t
4 sin ti 4 cos tj
F ˜ dr
1 5·
§
2
2
¨ 8 sin t cos t 8 cos t sin t t ¸ dt
4 ¹
©
ª8 3
t6 º
8
3
« sin t cos t »
3
24 ¼ 0
¬3
1
0
4 cos ti 4 sin tj, 0 d t d
C: r t
9
4
2 sin ti 2 cos tj xyi yj
28. F x, y
ª 9t 4 º
« »
¬ 4 ¼0
t 3 4t 3 4t 3 dt
x 2 i y 2 j z 2k
³ C F ˜ dr ³ 0
ti tj, 0 d t d 1
C: r t
³C
1
xi yj
27. F x, y
0
xyi xzj yzk
rt
rc t
ª t2 º
« »
¬ 2 ¼ 2
3t 4t dt
S 2
³0
S
33. F x, y, z
2
64 sin 2 t cos t 16 sin t cos t dt
S 2
ª 64 3
2 º
« 3 sin t 8 sin t »
¬
¼0
40
3
8 S6 8
3 24 3
S6
24
x 2 zi 6 yj yz 2k
rt
ti t 2 j ln tk , 1 d t d 3
Ft
t 2 ln ti 6t 2 j t 2 ln 2 tk
dr
16
3
1 ·
§
¨ i 2tj k ¸ dt
t ¹
©
³ C F ˜ dr
3
³1
ªt 2 ln t 12t 3 t ln t 2 º dt | 249.49
¬
¼
© 2010 Brooks/Cole, Cengage Learning
Section 15.2
xi yj zk
34. F x, y, z
rt
Ft
dr
t i tj e k , 0 d t d 2
ti tj e t k
2t 2 e 2t
³ C F ˜ dr
rt
2t 2 e 2t
cos3 t , y
cos3 ti sin 3 tj, 0 d t d
cos6 ti cos3 t sin 3 tj
3 cos8 t sin t 3 cos 4 t sin 5 t
3 cos 4 t sin t cos 4 t sin 4 t
3 cos 4 t sin t 2 cos 4 t 2 cos 2 t 1
ªt 2
6º
« t »
2
¬
¼0
66
³ C F ˜ dr
S 2
³0
ª¬6 cos8 t sin t 6 cos6 t sin t 3 cos 4 t sin t º¼ dt
S 2
xi yj
37. F x, y
On C1 , F t
ti , r c t
³ C1 F ˜ dr
On C2 , F t
43
105
yi xj
C: counterclockwise along the semicircle
1
2
1
³ 0 t dt
2
³1
4 x 2 from 2, 0 to 2, 0
y
i
2 t i t 1 j, rc t
³ C 2 F ˜ dr
38. F x, y
­ti
0 d t d1
°
® 2 t i t 1 j, 1 d t d 2
°3 t j
2 d t d 3
¯
C: r t
2
3 cos 4 t sin t ªcos 4 t 1 cos 2 t º
«¬
»¼
6 cos8 t sin t 6 cos6 t sin t 3 cos 4 t sin t
ª 2 cos9 t
6 cos 7 t
3 cos5 t º
«
»
7
5 ¼0
¬ 3
Work
t 6t 5 dt
S
Ft
Work
2
³ C F ˜ dr ³ 0
sin 3 t from 1, 0 to 0, 1
2
3 cos 2 t sin ti 3 sin 2 t cos tj
Work
ti 2t 3 j
2t e 2t dt | 6.91
rc t
F ˜ rc
Ft
x 2i xyj
36. F x, y
C: x
i 3t 2 j
Work
1
³0
rc t
2
i j et k dt
2
ti t 3 j, 0 d t d 2
C: r t
t
387
xi 2 yj
35. F x, y
x2 y 2 z 2
Line Integrals
i j
2 cos ti 2 sin tj,
rc t
2 sin ti 2 cos tj
Ft
2 sin ti 2 cos tj
F ˜ rc
4 sin 2 t 4 cos 2 t
Work
³ C F ˜ dr
4 ³ cos 2t dt
0
2
ª¬t 2 3t º¼
1
On C3 , F t
3 t j, rc t
4 cos 2t
S
ª¬ t 2 t 1 º¼ dt
>2 sin 2t@S0
46 13
0 d t d S
rt
0
0
xi yj 5 zk
39. F x, y, z
j
C: r t
2 cos ti 2 sin tj tk ,
0 d t d 2S
3
Work
³ C3 F ˜ dr
3
³2
t 3 dt
ªt 2
º
« 3t »
2
¬
¼2
§9
·
¨ 9¸ 2 6
©2
¹
Total work
1
1
0
2
2
0
1
2
rc t
2 sin ti 2 cos tj k
Ft
2 cos ti 2 sin tj 5tk
F ˜ rc
5t
Work
³ C F ˜ dr
2S
³0
5t dt
10S 2
© 2010 Brooks/Cole, Cengage Learning
388
Chapter 15
Vector Analysis
yzi xzj xyk
40. F x, y, z
45. F x, y
C : line from 0, 0, 0 to 5, 3, 2
rt
5ti 3tj 2tk ,
rc t
5i 3j 2k
Ft
6t 2i 10t 2 j 15t 2k
F ˜ rc
90t
Work
³ C F ˜ dr
(a)
0 d t d1
1
r1 t
2ti t 1 j,
r1c t
2i j
Ft
4t 2i 2t t 1 j
³ C1 F ˜ dr
2
³ 0 90t
x 2i xyj
2
dt
3
³1
1d t d 3
8t 2 2t t 1 dt
236
3
Both paths join 2, 0 and 6, 2 . The integrals are
30
negatives of each other because the orientations are
different.
41. Because the vector field determined by F points in the
general direction of the path C, F ˜ T ! 0 and work will
be positive.
(b)
42. Because the vector field determined by F points for the
most part in the opposite direction of the path C,
F ˜ T 0 and work will be negative.
r2 t
2 3 t i 2 t j, 0 d t d 2
r2c t
2i j
Ft
³ C2 F ˜ dr
2
43t i 23t 2t j
2
³0
ª8 3 t
¬
2
2 3 t 2 t º dt
¼
236
3
43. Because the vector field determined by F is
perpendicular to the path, work will be 0.
44. Because the vector field is perpendicular to the path,
work will be 0.
46. F x, y
x 2 yi xy 3 2 j
t 1 i t 2 j,
r1 t
(a)
r1c t
i 2tj
2
t 1 t 2i t 1 t 3 j
Ft
2
³ C1 F ˜ dr ³ 0 ª¬ t 1
(b)
0dt d 2
t 2t 4 t 1 º dt
¼
2 2
256
5
r2 t
1 2 cos t i 4 cos 2 tj, 0 d t d
r2c t
2 sin ti 8 cos t sin tj
1 2 cos t
Ft
S 2
³ C2 F ˜ dr
³0
2
S
2
4 cos 2 t i 1 2 cos t 8 cos3 t j
ª 1 2 cos t
¬
2
4 cos 2 t 2 sin t 8 cos t sin t 1 2 cos t 8 cos3 t dt
256
5
Both paths join 1, 0 and 3, 4 . The integrals are negatives of each other because the orientations are different.
3 yi xj
47. F x, y
yi xj
48. F x, y
C: r t
ti 2tj
C: r t
ti t 3 j
rc t
i 2j
rc t
i 3t 2 j
Ft
2ti tj
Ft
3t 3i tj
F ˜ rc
3t 3 3t 3
F ˜ rc
So,
2t 2t
³ C F ˜ dr
0.
0
So,
³ C F ˜ dr
0
0.
© 2010 Brooks/Cole, Cengage Learning
Section 15.2
y·
§
x3 2 x 2 i ¨ x ¸ j
2¹
©
49. F x, y
C: r t
ti t 2 j
rc t
i 2tj
Ft
§
t2 ·
t 2t i ¨ t ¸ j
2¹
©
F ˜ rc
§
t2 ·
t 3 2t 2 2t ¨ t ¸
2¹
©
So,
51. x
3
³ C F ˜ dr
52. x
x 3y
3 cos ti 3 sin tj
Ft
3 sin ti 3 cos tj
F ˜ rc
So,
10t , 0 d t d 1 Ÿ y
10
9 sin t cos t 9 sin t cos t
³ C F ˜ dr
0
0.
³0
10t ,
0 d t d1Ÿ y
y
, 0 d y d 10
5
5 x or x
§y
2·
¨ 3 y ¸ dy
5
©
¹
dy
2
2t , y
3 sin ti 3 cos tj
rc t
0.
2t , y
³C
C: r t
0
389
xi yj
50. F x, y
2
Line Integrals
10
ª y2
º
y3 »
«
10
¬
¼0
1010
5 x, 0 d x d 2
2
³C
53. x
y
³0
x 75 x 2 dx
10
³0
xy dx y dy
ª x2
º
25 x3 »
«
¬2
¼0
202
y
, 0 d y d 10, dx
5
10t , 0 d t d 1 Ÿ x
2t , y
³C
2
x 3 y 2 dx
§ y2
·
y ¸ dy
¨
25
©
¹
10
ª y3
y2 º
«
»
2 ¼0
¬ 75
190
or
3
5 dx, 0 d x d 2
5 x, dy
2
³C
54. x
2
³0
xy dx y dy
3 y x dx y 2 dy
2
³0
5 x,
dy
2
2
125 3 º
x ¼
3
0
28 ti , 0 d t d 5
55. r t
xt
t, y t
dx
dt , dy
190
3
5 dx, 0 d x d 2
3 5 x x dx 5 x 5 dx
ª 2
¬7 x ³C
ª 5 x3
25 x 2 º
«
»
2 ¼0
¬ 3
5 x 2 25 x dx
10t , 0 d t d 1 Ÿ y
2t , y
³C
1
dy
5
125
3
8
2
³0
14 x 125 x 2 dx
1084
3
56. r t
xt
0
dx
0
2 x y dx x 3 y dy
5
³ 0 2t dt
25
³C
tj, 0 d t d 2
0, y t
t
dy
dt
0,
2 x y dx x 3 y dy
y
2
³ 0 3t dt
2
ª3t2º
¬ 2 ¼0
y
3
2
2
1
1
x
1
2
3
4
5
−1
−2
x
−1
1
© 2010 Brooks/Cole, Cengage Learning
6
390
Chapter 15 Vector Analysis
0 d t d 3
°­ti,
®
3
t
3
,
3
i
j
d t d 6
°̄
57. r t
C1 : x t
t, y t
dx
dt , dy
³ C1
C2 : x t
dx
y
(3, 3)
3
0,
2
0
C2
3
³ 0 2t dt
2 x y dx x 3 y dy
1
9
C1
x
t 3
3, y t
0, dy
1
³C
6
³3
2 x y dx x 3 y dy
2 x y dx x 3 y dy
ª¬3 3 t 3 º¼ dt
9
45
2
ª 3t 2
º
6t »
«
2
¬
¼3
45
2
63
2
0 d t d 3
°­tj,
®
d t d 5
i
j
t
3
3
,
3
°̄
58. r t
3
dt
6
³ C2
2
y
x
C1 : x t
dx
C2 : x t
dx
59. x t
³C
t 3, y t
3
C1
3
³ 0 3t dt
−2
27
2
−3
3
C2
(2, − 3)
0
5
³ 3 ª¬2 t
2 x y dx x 3 y dy
2 x y dx x 3 y dy
t, y t
2
−1
2 x y dx x 3 y dy
dt , dy
³ C2
1
dt
0, dy
³ C1
³C
t
0, y t
10
27
2
1 t 2 , 0 d t d 1, dx
2 x y dx x 3 y dy
3 3º¼ dt
dt , dy
1
³ 0 ª¬ 2t
ªt 3
¬
2
5
3t º
¼3
10
47
2
2t dt
1 t 2 t 3 3t 2 2t º¼ dt
1
1
³0
60. x t
³C
6t 3 t 2 4t 1 dt
t 3 2 , 0 d t d 4, dx
t, y t
2 x y dx x 3 y dy
4
³0
4
t, y t
dx
dt , dy
³C
62. x t
dx
³C
3 12
t
2
12 t 3 2 2t dt
9 2
t
2
11
6
dt
ª 2t t 3 2 t 3t 3 2
¬
³0
61. x t
dt , dy
ª 3t 4
º
t3
2t 2 t »
«
3
¬ 2
¼0
3 12
t
2
º dt
¼
4
ª3t3 1t5 2 t2º
5
¬2
¼0
96 1
5
32 16
592
5
2t 2 , 0 d t d 2
4t dt
2 x y dx x 3 y dy
4 sin t , y t
4 cos t dt , dy
2
³0
2t 2t 2 dt t 6t 2 4t dt
3 cos t , 0 d t d
2
³0
2
ª6t 4 2 t 3 t 2 º
3
¬
¼0
24t 3 2t 2 2t dt
316
3
S
2
3 sin t dt
2 x y dx x 3 y dy
S 2
³0
S 2
³0
8 sin t 3 cos t 4 cos t dt 4 sin t 9 cos t 3 sin t dt
S 2
5 sin t cos t 12 cos 2 t 12 sin 2 t dt
ª5
º
2
« 2 sin t 12t »
¬
¼0
5
6S
2
© 2010 Brooks/Cole, Cengage Learning
Section 15.2
63. f x, y
h
65. f x, y
1 from 1, 0 to 0, 1
3ti 4tj, 0 d t d 1
rt
cos ti sin tj, 0 d t d
rc t
3i 4 j
rc t
sin ti cos tj
rc t
5
rc t
1
r
³C
³0
f x, y ds
64. f x, y
5h dt
³C
5h
y
³0
S 2
C: x y
2
ti tj, 0 d t d 4
rt
cos ti sin tj, 0 d t d
rc t
i j
rc t
sin ti cos tj
rc t
2
rc t
1
³0
67. f x, y
S
2
Lateral surface area:
Lateral surface area:
³C
1
2
1 from 1, 0 to 0, 1
rt
4
ª sin 2 t º
«
»
¬ 2 ¼0
cos t sin t dt
x y
2
C: line from 0, 0 to 4, 4
S 2
f x, y ds
66. f x, y
f x, y ds
S
2
Lateral surface area:
Lateral surface area:
1
391
xy
C : x2 y2
C: line from 0, 0 to 3, 4
Line Integrals
t
2 dt
³C f
8 2
S 2
x, y ds
³0
cos t sin t dt
>sin t
cos t @0
S 2
2
h
1 x 2 from 1, 0 to 0, 1
C: y
2
1 t i ª1 1 t º j, 0 d t d 1
¬
¼
i 2 1 t j
rt
rc t
rc t
1 41 t
2
Lateral surface area:
³C f
x, y ds
1
³0 h
1 41 t
2
hª
2 5 ln 2 4¬
68. f x, y
C: y
rt
rc t
rc t
dt
hª
«2 1 t
4¬
1
2
1 41 t
ln 2 1 t 1 41 t
2
º
»¼
0
5 º | 1.4789h
¼
y 1
1 x 2 from 1, 0 to 0, 1
2
1 t i ª1 1 t º j, 0 d t d 1
¬
¼
i 2 1 t j
1 41 t
2
Lateral surface area:
³C f
x, y ds
1
³ 0 ª¬2 2
1 t º 1 41 t
¼
ª
12 «2 1 t
¬
1 ª2
2¬
23
32
2
2³
1
0
1 41 t
ln 2 1 t 1 t ª2 4 1 t
¬
¬
5 ln 2 33
64
dt
2
dt 1
³0
1t
2
1 41 t
2
dt
1
1 41 t
1 ª2
64 «
5 2
ln 2 5º ¼
5
2
1 41 t
1º 1 4 1 t
¼
1 ª18
64 ¬
1 ª46
64 ¬
5 ln 2 2
º
»¼
0
1
2
ln 2 1 t 1 41 t
2
º
»¼
0
5º
¼
5 33 ln 2 5 º | 2.3515
¼
© 2010 Brooks/Cole, Cengage Learning
392
Chapter 15 Vector Analysis
69. f x, y
xy
1 x 2 from 1, 0 to 0, 1
C: y
You could parameterize the curve C as in Exercises 67 and 68. Alternatively, let x
1 cos 2 t
y
sin 2 t
rt
cos ti sin 2 tj, 0 d t d
rc t
sin ti 2 sin t cos tj
rc t
cos t , then:
S
2
sin 2 t 4 sin 2 t cos 2 t
sin t 1 4 cos 2 t
Lateral surface area:
³C f
S 2
³0
x, y ds
1 4 cos 2 t
sin 2 t and dv
Let u
³C f
ª 1
2
2
«12 sin t 1 4 cos t
¬
x, y ds
S 2
³0
cos t sin 2 t sin t 1 4 cos 2 t dt
ª 1
2
2
«12 sin t 1 4 cos t
¬
12
sin t cos t , then du
3 2º
S 2
»
¼0
32
1 S
6³0
2
sin 2 t ª 1 4 cos 2 t
«¬
sin t cos t º dt
»¼
2 sin t cos t dt and v
1 4 cos 2 t
1
1 4 cos 2 t
120
12
5 2º
S 2
»
¼0
32
1
1 4 cos 2 t
12
32
.
sin t cos t dt
1 ·
1
§ 1
5
¨ ¸
© 12 120 ¹ 120
1
25 5 11 | 0.3742
120
52
x2 y 2 4
70. f x, y
C : x2 y2
4
rt
2 cos ti 2 sin tj, 0 d t d 2S
rc t
2 sin ti 2 cos tj
rc t
2
Lateral surface area:
³C f
2S
³0
x, y ds
4 cos 2 t 4 sin 2 t 4 2 dt
2S
0
ª8 t ¬
1 cos 2t dt
1
2
2S
sin 2t º
¼0
16S
1 y2
71. (a) f x, y
rt
2 cos ti 2 sin tj, 0 d t d 2S
rc t
2 sin ti 2 cos tj
rc t
2
³C f
S
8³
2S
³0
x, y ds
1 4 sin 2 t 2 dt
2S
ª¬2t 4 t sin t cos t º¼ 0
12S | 37.70 cm 2
12S
| 7.54 cm3
5
(b) 0.2 12S
(c)
z
5
4
−3
3
3
y
x
© 2010 Brooks/Cole, Cengage Learning
Section 15.2
1
x
4
x3 2 , 0 d x d 40
C: y
rt
ti t 3 2 j, 0 d t d 40
rc t
i 3 12
t j
2
§9·
1 ¨ ¸t
©4¹
rc t
Lateral surface area:
³C f
40
³0
x, y ds
§9·
1 ¨ ¸t , then t
©4¹
Let u
40
1 ·
§
§9·
¨ 20 t ¸ 1 ¨ ¸t dt
4
©
¹
© 4¹
4 2
u 1 and dt
9
1 ·
§
§9·
¨ 20 t ¸ 1 ¨ ¸t dt
4
©
¹
© 4¹
³1
91
8
u du.
9
1 2
ª
º §8 ·
«20 9 u 1 » u ¨ 9 u ¸ du
¬
¼ © ¹
8 ªu5
179u 3 º
« »
81 ¬ 5
3 ¼1
rc t
rc t
a sin ti a cos tj,
³C y
2
2S
³0
U x, y ds
a3 ³
Iy
91
³C x U
2
2S
a3 ³
1 sin 2t
Ix
2S
0
u 4 179u 2 du
a cos ti a sin tj, 0 d t d 2S
³C y
2
U x, y ds
cos 2 t dt
2S
³0
a4 ³
Iy
³C x U
2
x, y ds
1 2 sin t cos t
a
a 2 sin 2 t sin t a 2 dt
2S
0
2S
³0
a4 ³
a 3S
b Ÿ sin t
rc t
a sin ti a cos tj,
a 3S
a 2 cos 2 t 1 a dt
b constant, 3 sin t
2
rc t
a
sin 2 t dt
91
3 cos ti 3 sin tj 1 sin 2 2t k , 0 d t d 2S
75. (a) Graph of: r t
For y
2S
³0
x, y ds
74. r t
a 2 sin 2 t 1 a dt
0
8
81 ³ 1
850,304 91 7184
| 6670.12
1215
a cos ti a sin tj, 0 d t d 2S
73. r t
Ix
393
20 72. f x, y
³0
Line Integrals
sin 3 t dt
0
a 2 cos 2 t sin t a 2 dt
2S
0
cos 2 t sin t dt
0
z
b
and
3
3
2
2
1
1 4 sin 2 t cos 2 t
1 4 sin 2 t 1 sin 2 t
1
4 2§
b2 ·
b ¨1 ¸.
9 ©
9¹
(b) Consider the portion of the surface in the first quadrant. The curve z
r1 t
4
y
x
1 sin 2 2t is over the curve
3 cos ti 3 sin tj, 0 d t d S 2. So, the total lateral surface area is
4 ³ f x, y ds
C
3
3
4
4³
S 2
0
1 sin 2 2t 3 dt
§ 3S ·
12¨ ¸
© 4 ¹
9S cm 2 .
2
(c) The cross sections parallel to the xz-plane are rectangles of height 1 4 y 3 1 y 2 9 and base 2 9 y 2 . So,
Volume
§
3
y2 §
y2 ··
2³ 2 9 y 2 ¨¨1 4 ¨1 ¸ ¸ dy
0
9©
9 ¹ ¸¹
©
27S
| 42.412 cm3 .
2
© 2010 Brooks/Cole, Cengage Learning
Chapter 15 Vector Analysis
394
ti t 2 j, 0 d t d 1
76. r t
rc t
x, y
i 2tj
³ C F ˜ dr
10
ª5 4 4 2 4 4 6 11º¼
34 ¬
|
16
3
3 sin ti 3 cos tj 77. r t
10
tk ,
2S
§1 1 ·
¨ , ¸
© 4 16 ¹
§1 1·
¨ , ¸
©2 4¹
§3 9 ·
¨ , ¸
© 4 16 ¹
1, 1
F x, y
5i
3.5i j
2i 2 j
1.5i 3 j
i 5j
rc t
i
i 0.5 j
i j
i 1.5 j
i 2j
F ˜ rc
5
4
4
6
11
0 d t d 2S
175k
F
10 ·
§
k ¸ dt
¨ 3 cos ti 3 sin tj 2S ¹
©
dr
³C
0, 0
2S
³0
F ˜ dr
1750
dt
2S
³ C F ˜ dr
78. W
2S
ª1750 º
« 2S t »
¬
¼0
³ C M dx M
15 4 x 2 y
N
15 xy
15 x c cx 2
dx
dx, dy
2cx dx
1750 ft ˜ lb
y
N dy
c
60 15 x 2 c cx 2
1
W
³ 1 ª¬60 15 x
Wc
16c 4
2
c cx 2 15 x c cx 2
0 Ÿ c
1
4
2 cx º dx
¼
120 4c 8c 2
parabola
−1
yields the minimum work, 119.5. Along the straight line path, y
79. See the definition of Line Integral, page 1070. See
Theorem 15.4.
82. (a) Work
y = c ) 1 − x 2)
x
1
0, the work is 120.
0
(b) Work is negative, because against force field.
(c) Work is positive, because with force field.
80. See the definition, page 1074.
83. False
81. The greater the height of the surface over the curve, the
greater the lateral surface area. So, z3 z1 z2 z4 .
³ C xy ds
y
4
2
1
³0 t
2
dt
84. False, the orientation of C does not affect the form.
3
³C f
2
1
x, y ds.
85. False, the orientations are different.
x
1
2
3
4
86. False. For example, see Exercise 32.
y x i xyj
87. F x, y
rt
kt 1 t i tj,
rc t
k 1 2t i j
Work
³ C F ˜ dr
1
y
1
kt 1 t i kt 2 1 t j¼º ˜ ª¬k 1 2t i jº¼ dt
1
kt 1 t k 1 2t kt 1 t º¼ dt
³ 0 ¬ª t
³ 0 ª¬ t
1
³0
k
0 d t d1
2k 2t 3 kt 3 kt 2 3k 2t 2 k 2t kt dt
12
2
2
1
k
12
(0, 1)
(0, 0)
−1
x
1
2
−1
© 2010 Brooks/Cole, Cengage Learning
Section 15.3
Conservative Vector Fields and Independence of Path
395
Section 15.3 Conservative Vector Fields and Independence of Path
x 2i xyj
1. F x, y
(a) r1 t
ti t 2 j, 0 d t d 1
r1c t
i 2tj
Ft
t 2i t 3 j
1
³ C F ˜ dr
(b) r2 T
³0
11
15
t 2 2t 4 dt
sin T i sin 2 T j, 0 d T d
r2c T
2
cos T i 2 sin T cos T j
Ft
sin 2 T i sin 3 T j
³C
S 2
³0
F ˜ dr
S 2
sin 2 T cos T 2 sin 4 T cos T dT
x 2 y 2 i xj
2. F x, y
ti t j, 0 d t d 4
r1c t
i 1
Ft
t t i tj
2 t
(a) r1 T
j
2
³ C F ˜ dr
4§
³0
1
2
¨t t 2
©
ªt 3
t2
t º
« »
3
2
3 ¼0
¬
11
15
yi xj
sec T i tan T j, 0 d T d
r1c T
sec T tan T i sec 2 T j
FT
tan T i sec T j
³ C F ˜ dr
·
t ¸ dt
¹
32 4
(b) r2 w
ª sin 3 T
2 sin 5 T º
«
»
5 ¼0
¬ 3
3. F x, y
(a) r1 t
2 wi j
Fw
w4 w2 i w2 j
F ˜ dr
2
³0
S 3
³0
S 3
³0
80
3
³
S
3
sec T tan 2 T sec3 T dT
ªsec T sec 2 T 1 sec3 T º dT
¬
¼
S 3
0
sec T dT
S 3
¬ªln sec T tan T ¼º 0
w2i wj, 0 d w d 2
r2c w
³C
S
ln 2 t 1i (b) r2 t
ª2 w w4 w2 w2 º dw
¬
¼
2
ª w6
w4
w3 º
«
»
2
3 ¼0
¬3
80
3
r2c t
3 | 1.317
t j, 0 d t d 3
1
1
i j
2 t 1
2 t
Ft
³ C F ˜ dr
ti t 1j
3ª
³0 «2
¬
1 3
³
2 0
1 3
³
2 0
1 3
2³0
t 1º
» dt
2 t ¼
t
t 1
1
dt
t t 1
1
t2 t 1 4 1 4
1
dt
dt
2
ª¬t 1 2 º¼ 1 4
ª 1 §
1·
« ln ¨ t ¸ 2¹
¬ 2 ©
3
º
t2 t »
¼0
1ª §7
·
§ 1 ·º
«ln ¨ 2 3 ¸ ln ¨ ¸»
2¬ © 2
¹
© 2 ¹¼
1
ln 7 4 3 | 1.317
2
© 2010 Brooks/Cole, Cengage Learning
396
Chapter 15 Vector Analysis
yi x 2 j
4. F x, y
2 t i 3 t j,
(a) r1 t
r1c t
i j
Ft
3t i 2t j
³C
0 d t d 3
2
3
³0
F ˜ dr
ª 3 t 2 t 2 º dt
¬
¼
r2c w
1
1
i j
w
w
Fw
3 ln w i 2 ln w j
ª
2 § 1 ·º
§1·
« 3 ln w ¨ w ¸ 2 ln w ¨ w ¸» dw
© ¹
© ¹¼
¬
e3
³1
e x sin yi e x cos yj
5. F x, y
wM
wy
e x cos y
wN
wx
Because
e x cos y
wM
, F is conservative.
wy
wM
wy
30 x 2 y
wN
Because
wx
69
2
30 x 2 y
wM
, F is conservative.
wy
wM
wy
1
y2
1
y2
9. F x, y , z
curl F
10. F x, y, z
2 ln w
3
Ft
2t 3i t 2 j
1
³ 0 4t
i 3t 2 j
Ft
2t 4i t 2 j
1
³ 0 5t
3
dt
4
dt
1
r1c t
i j
Ft
t 3 e3t t i te3t t j
2
³ C F ˜ dr
³ 0 ª¬« t
3
3
2
2
2
3 e3t t te3t t º» dt
¼
3t t 2
ªe3t t 2 º
¬«
¼» 0
2
curl F z 0, so F is not conservative.
1
ti t 3 j, 0 d t d 3
3
sin yz i xz cos yz j xy sin yz k
69
2
ye xy i xe xy j
³0 e
0 Ÿ F is conservative.
ti t 3 j, 0 d t d 1
r2c t
(a) r1 t
º
»
»¼1
ti t 2 j, 0 d t d 1
i 2tj
(b) r2 t
3
3 e
2 xyi x 2 j
r1c t
wN ·
¸
wz ¹
y zi 2 xyzj xy k
2
(a) r1 t
12. F x, y
xy
y ln zi x ln zj k
8. F x, y, z
z
curl F z 0 so F is not conservative.
x
x
z z
z
2
³ C F ˜ dr
wN
wM
, F is not conservative.
z
wx
wy
Because
ª 3 ln w
«
2
«¬
³ C F ˜ dr
1
x
i 2j
y
y
7. F x, y
§ wP
¨
© wy
3
2t º
»
3 »
¼0
11. F x, y
15 x 2 y 2i 10 x3 yj
6. F x, y
wN
wx
2
³ C F ˜ dr
wN
wx
3
2
2 ln w i 3 ln w j, 1 d w d e3
(b) r2 w
wN
wx
ª 3t
«
2
¬«
3 2t dt
e0 e0
0
(b) F x, y is conservative because
wM
wy
xye xy e xy
wN
.
wx
e xy K .
The potential function is f x, y
By Theorem 15.7,
³ C F ˜ dr
0.
© 2010 Brooks/Cole, Cengage Learning
Section 15.3
Conservative Vector Fields and Independence of Path
yi xj
13. F x, y
r1c t
i j
Ft
ti t j
³ C F ˜ dr
0 d t d1
(a) r1 t
r1c t
0
Ft
ti t 2 j, 0 d t d 1
(b) r2 t
r2c t
i 2tj
Ft
t 2 i tj
³ C F ˜ dr
1
2
13
dt
r3c t
r2c t
i 3t j
³ C F ˜ dr
i 1
³ 0 2t
3
dt
ln 3
1
t 3 j, 0 d t d 2
3
³ C F ˜ dr ³ 0 «¬9 t 1
12
3
ª¬ln t º¼1
1
j
3
2 ª1
t 3i tj
1
1
2
2
2
t 1 t 3 i t 1 t 3j
9
3
Ft
2
Ft
3
³1 t dt
t 1i (b) r2 t
³0 t
1
ti j, 1 d t d 3
t
1
i 2j
t
1
i 2tj
t
³ C F ˜ dr
ti t 3 j, 0 d t d 1
(c) r3 t
xy 2i 2 x 2 yj
14. F x, y
ti tj,
(a) r1 t
t3
2
2
2
º
t 1 t 3 » dt
9
¼
1 2 3
3t 7t 2 7t 3 dt
9³0
2
º
1 ª3t 4 7t 3 7t 2
3t»
«
9¬ 4
3
2
¼0
15.
³C y
2
397
44
27
dx 2 xy dy
Because wM wy
wN wx
y 2i 2 xyj is conservative. The potential function is f x, y
2 y , F x, y
xy 2 k . So, you
can use the Fundamental Theorem of Line Integrals.
4, 4
(a)
³C y
2
dx 2 xy dy
ª¬ x 2 yº¼
0, 0
(b)
³C y
2
dx 2 xy dy
ª¬ x 2 yº¼
1, 0
64
1, 0
(c) and (d) Because C is a closed curve,
16.
³ C 2x 3 y 1
Because wM wy
f x, y
0
³C y
2
dx 2 xy dy
0.
dx 3 x y 5 dy
wN wx
3, F x, y
2 x 3 y 1 i 3 x y 5 j is conservative. The potential function is
x 2 3 xy y 2 2 x 5 y k .
(a) and (d) Because C is a closed curve,
³C
2 x 3 y 1 dx 3x y 5 dy
0.
0,1
(b)
³C
2 x 3 y 1 dx 3 x y 5 dy
ª 2
º
y2
x 5 y»
« x 3 xy 2
¬
¼ 0, 1
2 x 3 y 1 dx 3 x y 5 dy
ª 2
º
y2
x 5 y»
« x 3 xy 2
¬
¼ 0,1
2, e2
(c)
³C
10
1
3 2e 2 e 4
2
© 2010 Brooks/Cole, Cengage Learning
398
17.
Chapter 15
³ C 2 xy dx Vector Analysis
x 2 y 2 dy
Because wM wy
wN wx
2 xyi x 2 y 2 j is conservative.
2 x, F x, y
x2 y The potential function is f x, y
y3
k.
3
0, 4
³C
2 xy dx x 2 y 2 dy
ª 2
y3 º
«x y »
3 ¼ 5, 0
¬
(b)
³C
2 xy dx x 2 y 2 dy
ª 2
y3 º
«x y »
3 ¼ 2, 0
¬
³C
x 2 y 2 dx 2 xy dy
(a)
64
3
0, 4
18.
Because wM wy
wN wx
64
3
x 2 y 2 i 2 xyj is conservative. The potential function is
2 y , F x, y
f x, y
x3 3 xy 2 k .
(a)
³C
x 2 y 2 dx 2 xy dy
ª x3
2º
« xy »
3
¬
¼ 0, 0
(b)
³C
x 2 y 2 dx 2 xy dy
ª x3
2º
« xy »
3
¬
¼ 2, 0
8, 4
896
3
0, 2
Because curl F
20. F x, y, z
0, F x, y, z is conservative. The
xyz k .
potential function is f x, y, z
³C
ti 2 j tk ,
F ˜ dr
(b) r2 t
8
3
yzi xzj xyk
19. F x, y, z
(a) r1 t
> xyz@ 0, 2, 0
4, 2, 4
Because curl F
³C
32
t 2 i t j t 2k ,
4, 2, 4
³ C F ˜ dr > xyz@ 0, 0, 0
x yz k .
cos ti sin tj t 2k , 0 d t d S
F ˜ dr
(b) r2 t
0 d t d 2
0, F x, y, z is conservative. The
potential function is f x, y, z
(a) r1 t
0 d t d 4
i zj yk
1, 0, S 2
yz@ 1, 0, 0
2
1 2t i S 2tk , 0 d t d 1
³ C F ˜ dr
32
>x
>x
1, 0, S 2
yz@ 1, 0, 0
2
2 y x i x2 z j 2 y 4z k
21. F x, y, z
F x, y, z is not conservative.
(a) r1 t
ti t 2 j k , 0 d t d 1
r1c t
i 2tj
Ft
2t 2 t i t 2 1 j 2t 2 4 k
³ C F ˜ dr
(b) r2 t
r2c t
Ft
1
³0
2
3
2t 3 2t 2 t dt
2
ti tj 2t 1 k ,
0 d t d1
i j 4 2t 1 k
2
2
3ti ªt 2 2t 1 º j ª2t 4 2t 1 º k
¬
¼
¬
¼
³ C F ˜ dr
1
³ 0 ª¬3t t
2
2t 1
2
3
8t 2t 1 16 2t 1 º dt
¼
1
1
2
3
2
³ 0 ª¬17t 5t 2t 1 16 2t 1 º¼ dt
3
ª17t 3
º
2t 1
5t 2
4
«
2 2t 1 »
3
2
6
¬«
¼» 0
17
6
© 2010 Brooks/Cole, Cengage Learning
Section 15.3
Conservative Vector Fields and Independence of Path
yi xj 3 xz 2k
22. F x, y, z
23. F x, y, z
F x, y, z is not conservative.
r1c t
sin ti cos tj k
Ft
sin ti cos tj 3t cos tk
S
³ 0 ª¬sin
S
³0
(a) r1 t
>t@0
2
t cos t 3t cos t º¼ dt
2
2
(b) r2 t
S
S
3ª¬t sin t º¼ 6 ³ t sin t dt
0
0
2
(a) r1 t
5S
r2c t
2i S k
Ft
t 4 cos t 2k
Ft
1 2t j 3S 2t 2 1 2t k
³ C F ˜ dr
³ 0 3S
t 1 2t dt
(b) r2 t
3S 3 ³ t 2 2t 3 dt
r2c t
3 2
1
0
1
ªt 3
t4 º
3S 3 « »
2 ¼0
¬3
Ft
S3
³C
3, 2
ª x y 2º
¬
¼ 1,1
26.
³ C ª¬2 x 27.
3S 2, S 2
³ C cos x sin y dx sin x cos y dy >sin x sin y@ 0, S
31.
³C
y i 2 x y jº¼ ˜ dr
y dx x dy
x2 y 2
x
2 3, 2
ª
§ x ·º
«arctan ¨ ¸»
© y ¹¼ 1,1
¬
x y
2
2
2
dx x y
S
S
4
12
2S , 0
2
2
dy
0
4i 4 j
16t 2 cos 4t k
1
³ 0 0 dt
3 yi 3 xj ˜ dr
0
>3xy@ 3,0, 80
72
25
1
0
1, 5
2y
2
3
ª¬e x sin yº¼
0, 0
sin y dx e x cos y dy
2x
S
52 0
2
³ 0 0 dt
4ti 4tj, 0 d t d 1
³ C F ˜ dr
2
25.
³C
t 2i t 2 j, 0 d t d 2
2ti 2tj
1
0
y sin zi x sin zj xy cos xk
r1c t
³ C F ˜ dr
30.
4, 0, 3
ª¬ xye z º¼
4, 0, 3
1 2t i S tk , 0 d t d 1
(b) r2 t
0
4 8t i 3k , 0 d t d 1
24. F x, y, z
S
³C e
4, 0, 3
ª¬ xye z º¼
4, 0, 3
³ C F ˜ dr
ª¬t 3t 2 sin t 6 sin t t cos t º¼
0
29.
4 cos ti 4 sin tj 3k , 0 d t d S
³ C F ˜ dr
ª¬1 3t 2 cos t º¼ dt
S
³C
xye z k .
f x, y , z
2
³ C F ˜ dr
28.
e z yi xj xyk
F x, y, z is conservative. The potential function is
cos ti sin tj tk , 0 d t d S
(a) r1 t
399
ª
º
1
« 2
2»
x
y
¬
¼ 7, 5
1
1
26
74
2
481
z 2 y dx 2 x z dy x y dz
xz 2 xy yz
F x, y, z is conservative and the potential function is f x, y, z
(a)
> xz
2 xy yz@ 0, 0, 0
(b)
> xz
2 xy yz@ 0, 0, 0 > xz 2 xy yz@ 0, 0,1
(c)
> xz
2 xy yz@ 0, 0, 0 > xz 2 xy yz@ 1, 0, 0 > xz 2 xy yz@ 1,1, 0
1,1,1
0, 0,1
1, 0, 0
20
2
1,1,1
1,1, 0
0 2
2
1,1,1
0 2 2 2
2
© 2010 Brooks/Cole, Cengage Learning
400
32.
Chapter 15
³ C zy dx Vector Analysis
xz dy xy dz
yzi xzj xyk is conservative and the potential function is f x, y, z
Note: Because F x, y, z
xyz k , the integral
is independent of path as illustrated below.
33.
(a)
> xyz@ 1,1,1
0, 0, 0
(b)
> xyz@ 0,0, 0,1
0, 0
> xyz@ 0, 0,1
(c)
> xyz@ 1,0,0,0,00
> xyz@ 1, 0, 0 > xyz@ 1,1, 0
1
1,1,1
01
1,1, 0
³ C sin x dx 1
1,1,1
>cos x
z dy y dz
0 01
S 2, 3, 4
yz@ 0, 0, 0
34. F x, y, z is conservative: f x, y, z
ª¬3 x 2 4 yz 10 z 2 º¼
0, 0, 0
5, 9
ª¬3x3 y 2 yº¼
0, 0
Work
11
27
9 x 2 y 2i 6 x3 y 1 j is conservative.
35. F x, y
12 1
3x 2 4 yz 10 z 2
3, 4, 0
³ C F ˜ dr
1
x2
y
36. F x, y is conservative. f x, y
30,366
3, 2
Work
ª x2 º
« »
¬ y ¼ 1,1
9
1
2
7
2
2 cos 2S ti 2 sin 2S tj
37. r t
rc t
4S sin 2S ti 4S cos 2S tj
at
8S 2 cos 2S ti 8S 2 sin 2S tj
Ft
ma t
W
³ C F ˜ dr
1
at
32
³C
S2
S2
4
cos 2S ti sin 2S tj
cos 2S ti sin 2S tj ˜ 4S sin 2S ti cos 2S tj dt
4
S 3 ³ 0 dt
C
0
a1i a2 j a3k
38. F x, y, z
Because F x, y, z is conservative, the work done in moving a particle along any path from P to Q is
>a1x a2 y
f x, y , z
39. F
a3 z@P
Q
q1 , q 2 , q 3
p1 , p 2 , p 3
a1 q1 p1 a2 q2 p2 a3 q3 p3
175 j
(a) r t
³C
ti 50 t j, 0 d t d 50
50
³0
F ˜ dr
(b) r t
dr
40. No. The force field is conservative.
i j dt
dr
JJJK
F ˜ PQ.
ti i ³ C F ˜ dr
1
25
42. A line integral is independent of path if
8750 ft ˜ lbs
175 dt
2
50 t j, 0 d t d 50
1
50
41. See Theorem 15.5.
³ C F ˜ dr does
not depend on the curve joining P and Q. See Theorem
15.6.
50 t j
50
³0
175
1
25
50 t dt
50
ª
t2 º
7 «50t »
2 ¼0
¬
8750 ft ˜ lbs
© 2010 Brooks/Cole, Cengage Learning
Section 15.3
a cos ti a sin tj, 0 d t d 2S , you have x 2 y 2
43. (a) For the circle r t
2S
³ C F ˜ dr
³0
Conservative Vector Fields and Independence of Path
a cos t ·
§ a sin t
i j¸ ˜ a sin ti a cos tj dt
¨
2
a2 ¹
© a
2S
³0
401
a 2 , and
sin 2 t cos 2 t dt
2S .
(b) For this curve, the answer is the same, 2S .
(c) For the opposite overtation, the answer is 2S .
(d) For the curve away from the origin, the answer is 0.
44. (a) The direct path along the line segment joining
4, 0 to 3, 4 requires less work than the path
46. Not conservative. The value of
negative if the path is below the x-axis.
(b) The closed curve given by the line segments joining
4, 0 , 4, 4 , 3, 4 , and 4, 0 satisfies
45. Conservative.
1, 0 to
1, 0 is positive if the path is above the x-axis, and
going from 4, 0 to 4, 4 and then to 3, 4 .
³ C F ˜ dr
³ C F ˜ dr from
47. False, it would be true if F were conservative.
z 0.
48. True
³ C F ˜ dr is independent of path.
49. True
50. False, the requirement is wM wy
wN wx.
51. Let
F
Mi Nj
Then
wM
wy
wf
wf
i j.
wy
wx
w § wf ·
¨ ¸
wy © wy ¹
w2 f
wy 2
and
wN
wx
w § wf ·
¨ ¸
wx © wx ¹
So, F is conservative. Therefore, by Theorem 15.7, you have
w2 f
w2 f
w2 f
Because
.
2
wx 2
wx
wy 2
§ wf
³ C ¨© wy dx ·
wf
dy ¸
wx ¹
³C
0 you have
M dx N dy
wM
wy
wN
.
wx
³ C F ˜ dr
0
for every closed curve in the plane.
52. Because the sum of the potential and kinetic energies remains constant from point to point, if the kinetic energy is decreasing
at a rate of 15 units per minute, then the potential energy is increasing at a rate of 15 units per minute.
y
x
i 2
j
2
x y
x y2
53. F x, y
(a) M
2
y
x2 y2
x2 y2 1 y 2 y
wM
wy
N
x2 y2
(b) r t
F
dr
x2 y2
x2 y 2
2
x
x2 y 2
x 2 y 2 1 x 2 x
wN
wx
So,
2
x2 y 2
wN
wx
2
x2 y 2
x2 y2
2
wM
.
wy
cos ti sin tj, 0 d t d S
sin ti cos tj
sin ti cos tj dt
³ C F ˜ dr
S
³0
sin 2 t cos 2 t dt
>t@S0
S
© 2010 Brooks/Cole, Cengage Learning
402
Chapter 15
Vector Analysis
cos ti sin tj, 0 d t d S
(c) r t
sin ti cos tj
F
sin ti cos tj dt
dr
³ C F ˜ dr
S
³0
S
>t@0
sin 2 t cos 2 t dt
S
cos ti sin tj, 0 d t d 2S
(d) r t
sin ti cos tj
F
sin ti cos tj dt
dr
³ C F ˜ dr
2S
³0
2S
>t@0
sin 2 t cos 2 t dt
2S
This does not contradict Theorem 15.7 because F is not continuous at 0, 0 in R enclosed by curve C.
§
x·
(e) ’¨ arctan ¸
y¹
©
1y
1 x y
2
i x y
2
2
1 x y
y
x
i 2
j
x2 y 2
x y2
j
F
Section 15.4 Green's Theorem
2
0 d t d1
°­ti t j,
®
d t d 2
2
t
2
t
,
1
i
j
°̄
1. r t
³ C y dx 2
x 2 dy
1
³ 0 ª¬t
2
³1 ª¬ 2 t
dt t 2 2t dt º¼ 4
2
dt 2 t
2
dt º
¼
2
1
³0
t 4 2t 3 dt 2
³1
22t
2
1
ª2 2 t 3 º
ªt 5
t4 º
«
»
«
»
2 ¼ 0 «¬
3
»¼1
¬5
dt
7
2
10 3
By Green's Theorem,
§ wN
³ R ³ ©¨ wx
y
wM ·
¸ dA
wy ¹
x
1
³ 0 ³ x2
2 x 2 y dy dx
1
³ 0 ª¬2 xy y 2 º¼
x
dx
x2
1
1
³0
°­ti tj,
®
°̄ 2 t i 2. r t
³ C y dx 2
x 2 dy
1
30
x 2 2 x 3 x 4 dx
ª x3
x4
x5 º
« »
2
5 ¼0
¬3
y=x
1
(1, 1)
C2
1
30
C1
y = x2
x
1
0 d t d1
2 t j, 1 d t d 2
1
³ 0 ª¬t
1
2
³ 0 2t
2
dt t 2 dt º¼ dt 2
ª
³ 1 «¬ t 2
2
ª
1
·º
2§
dt 2 t ¨ dt ¸»
© 2 2 t ¹¼
1
2t
2
3 2º
³ 1 «¬ 2 t
1
ªt 22
2t
ª 2t 3 º
« » «
3
2
5
¬ ¼ 0 «¬
2 1 1
1
3 2 5
30
» dt
¼
52 2
º
»
¼»1
y
y=
1
x
(1, 1)
C2
C1
y=x
By Green's Theorem,
§ wN
wM ·
³ R ³ ©¨ wx wy ¹¸ dA
x
1
1
³0 ³ x
x
2 x 2 y dy dx
1
³ 0 ª¬2 xy y 2 º¼
x
x
dx
1
1
³0
2 x3 2 x x 2 dx
ª4 5 2
x2
x3 º
»
« x 2
3 ¼0
¬5
1
30
© 2010 Brooks/Cole, Cengage Learning
Section 15.4
0 d t
­ti
°
d t
t
i
j
1
1
°
®
°3 t i j 2 d t
°4t j
3 d t
¯
3. r t
2
d1
d 3
d 4
³ 0 ª¬0 dt t
2
2
0 º¼ 2
3
³ 1 dt ³ 2 dt
2
³1 ª¬ t 1
11
0 1 dt º ¼
2
3
³ 2 ª¬1 dt
2
3t
1
1
³0 ³0
0 0 dt º
¼
2
y
2 x 2 y dy dx
1
1
2
³ 0 ª¬2 xy y º¼ 0 dx
³0
(1, 1)
C4
C2
x
1
ª¬ x 2 xº¼
0
2 x 1 dx
1
C1
0
1
0 d t d 3
­ti
°
3
i
t
3
j
3
d t d 7
°
®
° 10 t i 4 j 7 d t d 10
° 14 t j
10 d t d 14
¯
4. r t
3
³ C y dx x dy ³ 0 ª¬0 dt
2
2
t 2 0 º¼ 7
10
³ 3 9 dt ³ 7
7
³ 3 ª¬ t 3
16 dt
2
0 9 dt º ¼
10
³7
ª16 dt 10 t
¬
9 7 3 16 10 7
12
§ wN
³ R ³ ¨© wx
wM ·
¸ dA
wy ¹
3
4
³0 ³0
³ C xe
§ wN
³ R ³ ¨© wx
dx e x dy
3
ª¬4 x 16 xº¼
0
2
y
§ wN
³ R ³ ¨© wx
wM ·
¸ dA
wy ¹
dx e x dy
In Exercises 7–10,
14 t
2
0º
¼
C3
4
(3, 4)
C4
y 2 ¼º dx
0
C2
36 48
12
x
C1
4
2 sin t , 0 d t d 2S .
2S
³0
ª¬2 cos te 2 sin t 2 sin t e 2 cos t 2 cos t º¼ dt | 19.99
2
³2 ³ 4 x2
e x xe y dy dx
4 x2
2
wM ·
¸ dA
wy ¹
1
³0
3
xe x 3 x 2e x dx 1
x
³ 0 ³ x3
wN
wM
wx
wy
y x dx 2 x y dy
e x xe y dy dx
0
³1
ª
³ 2 ¬«2
6. C: boundary of the region lying between the graphs of y
³ C xe
³ 10 ª¬0 dt
4
2 cos t and y
y
3
³ 0 ¬ª2 xy 2 x 2 y dy dx
³ 0 >8 x 16@ dx
Let x
14
0º
¼
4
3
5. C : x 2 y 2
2
y
By Green's Theorem,
³C
³ 3 ª¬ 4 t
C3
1
7.
4
0º ¼
0
By Green's Theorem,
§ wN
wM ·
³ R ³ ¨© wx wy ¸¹dA
403
d 2
1
³ C y dx x dy
Green’s Theorem
4 x 2 e x xe
x and y
4 x2
xe 4 x2
º dx | 19.99
¼»
x3
xe x e x dx | 2.936 2.718 | 0.22
1
³0
3
xe x x3e x dx | 0.22
1.
3
³0 ³
3
x
y
dy
x2 2 x
³ 0 ¬ªx dx
x 2 2 x dx¼º
3
2 3
ª x
3x º
«
»
2 ¼0
¬ 3
3
9 27
2
9
2
(3, 3)
y=x
2
y = x 2 − 2x
1
x
1
2
3
4
−1
© 2010 Brooks/Cole, Cengage Learning
404
Chapter 15
Vector Analysis
8. Because C is an ellipse with a
³C
1, then R is an ellipse of area S ab
2 and b
³ R ³ 1 dA
y x dx 2 x y dy
Area of ellipse
2S . So, Green's Theorem yields
2S .
9. From the accompanying figure, we see that R is the shaded region. So, Green's Theorem yields
³C
³ R ³ 1 dA
y x dx 2 x y dy
6 10 2 2
Area of R
56.
y
(− 5, 3)
4
(5, 3)
2
(− 1, 1)
(1, 1)
x
(− 1, − 1)
2
(1, − 1)
−2
(− 5, − 3) −4
4
(5, − 3)
10. R is the shaded region of the accompanying figure.
³C
³ R ³ 1 dA
4
Area of shaded region
2
1
y x dx 2 x y dy
1S
2
11.
³ C 2 xy dx >25 9@
§ wN
³ R ³ ¨© wx
x y dy
1
y
1 x 2
−5 −4 −3 −2 −1
x
1 2 3 4 5
−2
−3
−4
−5
wM ·
¸ dA
wy ¹
³ 1 ³ 0
8S
1 x 2
³ 1 > y 2 xy@0
1
1 2 x dy dx
dx
1
³ 1 ª¬ 1 x
1
ª
x3
x4 º
x2 «x »
3
2 ¼ 1
¬
1
2
3
³ 1 ª¬1 x 2 x 2 x º¼ dx
2 x 1 x 2 º¼ dx
2
1 7
6 6
4
3
12. The given curves intersect at 0, 0 and 9, 3 . So, Green's Theorem yields
³C y
2
³R ³
dx xy dy
9
³0 ³0
13.
³C
x
x 2 y 2 dx 2 xy dy
14. In this case, let y
³C
y 2 y dA
§ wN
³R ³ ¨© wx
r sin T , x
x 2 y 2 dx 2 xy dy
§ wN
³ R ³ ¨© wx
wM
wy
2e x sin 2 y
wM ·
¸ dA
wy ¹
0.
³0
ª y2 º
«
»
¬ 2 ¼0
wM ·
¸ dA
wy ¹
2S
0
1 cos T
³0
4³
2S
0
x
dx
4
³ 4 ³ r cos T . Then dA
³ R ³ 4 y dA
4³
15. Because
9
y dy dx
³0
x
dx
2
16 x 2
16 x 2
9
ª x2 º
«
»
¬ 4 ¼0
81
.
4
4
³ 4 ¬ª2 y
2 y 2 y dy dx
2
¼º 16 x 2
16 x 2
dx
0
r dr dT and Green's Theorem yields
1 cos T
³0
r 2 sin T dr dT
wN
you have
wx
9
r sin T r dr dT
4 2S
sin T 1 cos T
3³0
16. Because
2S
3
wM
wy
ª 1 cos T 4 º
«
»
3
»¼ 0
¬«
dT
2x
x y2
2
0.
wN
,
wx
you have path independence and
§ wN
³ R ³ ¨© wx
wM ·
¸ dA
wy ¹
0.
© 2010 Brooks/Cole, Cengage Learning
Section 15.4
Green’s Theorem
405
17. By Green's Theorem,
³ C cos y dx ³R ³
xy x sin y dy
1§
³0 ³ x
y sin y sin y dA
x
³ 0 ¨© 2
1
x2 ·
¸ dx
2¹
x
1
ª x2
x3 º
« »
6 ¼0
¬4
³ 0 «¬ 2 »¼
y dy dx
1 1
4 6
x
1 ª y2 º
dx
x
1
12
18. By Green's Theorem,
³C
e x
2 2
y dx e y
2 2
³ R ³ 2 dA
x dy
2 ªS 6
¬
2 Area of R
S 2 3º
¼
2
60S .
19. By Green's Theorem,
³C
³R ³ 1 3
x 3 y dx x y dy
4>Area Large Circle Area Small Circle@
dA
4>9S S @
20. By Green's Theorem,
³C
y
³R
3 x 2 e y dx e y dy
2
2 y
³ 3x e dA
2
³1 ³ 2
³
1
2
2
2 y
2
16e 16e
2
2
³ 1 ³ 1
dy dx x
1
1
(2, 2)
3x 2e y dy dx
³ 1 ³ 2 3x e
2 y
2e e 7e e
2
2
dy dx
2e
(− 2, − 2)
1
e
(2, − 2)
(− , − )
2
(1, − 1)
1
2e 2 e .
xyi x y j
21. F x, y
C: x y
2
1
³ C xy dx 2S
³0
22. F x, y
1
2
2
(1, 1)
(− , )
³ 2 3x e
7 e e
Work
(− , )
3 x 2e y dy dx 2
2
32S
x y dy
³R ³ 1 x
1
ªr2
º
r2
cos T » dT
« 2
2
¬
¼0
2S
³0
dA
2S
1
³ 0 ³ 0 1 r cos T
1
1 cos T dT
2
r dr dT
2S
1
ª1
º
« 2T 2 sin T »
¬
¼0
S
ex 3 y i e y 6x j
C: r
2 cos T
Work
³C e
x
x3 2 3 y i 6 x 5
23. F x, y
³ R ³ 9 dA
3 y dx e y 6 x dy
2 cos T is a circle with a radius of one.
9S because r
y j
C : boundary of the triangle with vertices 0, 0 , 5, 0 , 0, 5
Work
24. F x, y
³C
x3 2 3 y dx 6 x 5
³ R ³ 9 dA
y dy
25. C: let x
A
1
2
225
2
5 5
3 x 2 y i 4 xy 2 j
C : boundary of the region bounded by the graphs of y
Work
9
³ C 3x
2
y dx 4 xy 2 dy
a cos t , y
1
x dy y dx
2³C
9
³0 ³0
x
x, y
4 y 2 1 dy dx
0, x
9
³0
9
4 x3 2
3
x1 2 dx
558
5
a sin t , 0 d t d 2S . By Theorem 15.9, you have
1 2S
ªa cos t a cos t a sin t a sin t º¼ dt
2³0 ¬
1 2S 2
a dt
2³0
2S
ª a2 º
« t»
¬ 2 ¼0
S a2.
© 2010 Brooks/Cole, Cengage Learning
406
Chapter 15
Vector Analysis
26. From the figure you see that
3
3
x, dy
dx, 0 d x d 2
2
2
x
1
4, dy
dx
2
2
0, dx
0.
C1 : y
C2 : y
C3 : x
y
C2
4
x
C3
2
C1
3x − 2y = 0
1
A
x
1 2§3
3 ·
1 0§ 1
1
·
¨ x x ¸ dx ³ 2 ¨ x 4 ¸ dx 0
2³0 © 2
2 ¹
2 © 2
2
2
¹
2
1 0
4 dx
2³ dx
4
0
2³2
x 2 1,
27. C1 : y
dy
5 x 3, dy
C2 : y
x
1
2 x dx
3
(4, 17)
1 4
1 1
x 2 x x 2 1 dx ³ x 5 5 x 3 dx
³
1
2
2 4
4
º
1ªx
1
1
« x» >3x@4
2¬ 3
2
¼1
4
y
15
3
2
5 dx
y = 5x − 3
So, by Theorem 15.9 you have
A
2y = 8
(2, 3)
3
1
1
>18@ >9@
2
2
10
C1
C2
5
y = x2 + 1
(1, 2)
9
.
2
x
1
2
3
4
28. Because the loop of the folium is formed on the interval 0 d t d f,
dx
3 1 2t 3
t3 1
2
3 2t t 4
dt and dy
t3 1
2
dt ,
you have
A
4
3
ª
§ 3t 2 · 3 1 2t
1 f «§ 3t · 3 2t t
¨ 3
¸
¨
¸
³
2
2
3
2 0 «© t 1 ¹ t 3 1
© t 1¹ t 3 1
¬
f
2 3
9 ft t 1
dt
2 ³ 0 t3 1 3
9 f t5 t 2
dt
2 ³ 0 t3 1 3
º
» dt
»
¼
2
3 f 2 3
3t t 1 dt
2³0
ª 3 º
«
»
«¬ 2 t 3 1 »¼
0
3
.
2
29. See Theorem 15.8, page 1093.
³
1
2 C
30. See Theorem 15.9: A
x dy y dx.
³ R ³ y dA. Let
31. For the moment about the x-axis, M x
³C
Mx
y2
dx
2
For the moment about the y-axis, M y
³C
My
x2
dy
2
1
x 2 dy and x
2³C
32. By Theorem 15.9 and the fact that x
A
1
2
³ x dy y dx
Mx
2A
1
y 2 dx and y
2³C
1
2
³
0 and M
y 2 2. By Green's Theorem,
1
y 2 dx.
2A³C
³R ³ x dA. Let
My
2A
N
N
x 2 2 and M
0. By Green's Theorem,
1
x 2 dy.
2A³C
r cos T , y
r sin T , you have
r cos T r cos T dT r sin T r sin T dT
³
1
r2
2 C
dT .
© 2010 Brooks/Cole, Cengage Learning
Section 15.4
2
2
³ 2
33. A
x
ª
x3 º
«4 x »
3 ¼ 2
¬
4 x 2 dx
Green’s Theorem
407
32
3
1
1
x 2 dy x 2 dy
³
C
1
2A
2 A ³ C2
2 x dx and for C2 , dy
For C1 , dy
To calculate y , note that y
1
2 32 3
y
2
³2
0. So, x
2
2
1
2 32 3
³2
y
ª 3 § x 4 ·º
« ¨ ¸»
«¬ 64 © 2 ¹»¼ 2
x 2 x dx
2
0.
y = 4 − x2
3
0 along C2 . So,
2
2
x5 º
3ª
8 x3
»
«16 x 64 ¬
3
5 ¼ 2
3
16 8 x 2 x 4 dx
64 ³ 2
2
2
4 x 2 dx
8
.
5
1
x
−2
§ 8·
¨ 0, ¸
© 5¹
x, y
34. Because A
Let x
S a2
1
, you have
2A
2
a sin t , 0 d t d S , then
area of semicircle
a cos t , y
x
1 S 2
a cos 2 t a cos t dt
S a2 ³ 0
S
S ³0
y
1 S 2
a sin 2 t a sin t dt
S a2 ³ 0
a
a
1
S a2
a
S
S ³0
cos3 t dt
. Note that y
0 and dy
S
S
S ³0
aª
cos3 t º
«cos t »
S¬
3 ¼0
sin 3 t dt
−1
C2 1
0 along the boundary y
aª
sin 3
«sin t S¬
3
1 sin 2 t cos t dt
C1
2
0.
S
tº
»
¼0
0
4a
.
3S
§ 4a ·
¨ 0,
¸
© 3S ¹
x, y
1
1
³0
35. Because A
have y
x x3 dx
x, dy
x
2 ³ x 2 dy
y
2 ³ y 2 dx
C
C
ª x2
x4 º
«
»
4 ¼0
¬2
1
1
, you have
4
2A
2. On C1 you have y
x3 , dy
3 x 2 dx and on C2 you
y
dx. So,
2³
C1
x 2 3 x 2 dx 2 ³
1
C2
x 2 dx
0
2 ³ x 6 dx 2 ³ x 2 dx
0
1
1
6 2
5 3
0
6³ x 4 dx 2³ x 2 dx
0
1
2 2
7
3
8
15
(1, 1)
1
C2
8
.
21
C1
x
1
§8 8·
¨ , ¸
© 15 21 ¹
x, y
1
2a c
2
0, dy
0
36. Because A
C1 : y
C2 : y
C3 : y
ac, you have
c
x a , dy
b a
c
x a , dy
b a
1
2A
1
,
2ac
y
(b, c)
2a
c
dx
b a
c
dx.
b a
C3
−a
So,
x
1
x 2 dy
2ac ³ C
1 ª a
0
2ac «¬³ a
y
1
y 2 dx
2ac ³ C
1 ª
«0 2ac «¬
x, y
§b
¨ ,
©3
b§
b
³a x
c
2
c
dx ba
·
³ a ¨© b a ¸¹
a
³b
2
2
x a dx x2
C2
C1
c
º
dx
b a »¼
1 ª
2abc º
0
2ac «¬
3 »¼
2
º
c ·
2
¨
¸ x a dx»
©b a¹
»¼
a §
³b
x
a
b
3
c2 b a º
1 ª c 2 b a
«
»
2ac ¬
3
3
¼
c
.
3
c·
¸
3¹
© 2010 Brooks/Cole, Cengage Learning
408
Chapter 15
Vector Analysis
37. A
1 2S 2
a 1 cos T
2³0
38. A
1 S 2
a cos 2 3T dT
2³0
2
a 2 S 1 cos 6T
dT
2 ³0
2
a 2 ª 3T
1
º
2 sin T sin 2T »
2 ¬« 2
4
¼0
S
a2 ª
sin 6T º
T «
4¬
6 »¼ 0
3S a 2
2
a2
3S
2
S a2
4
a cos 3T where 0 d T d S .
Note: In this case R is enclosed by r
2S
4S
. So,
dT d
3
3
39. In this case the inner loop has domain
A
2S
a 2 2S §
1 cos 2T ·
¨1 2 cos T ¸ dT
2 ³0 ©
2
2 ¹
dT
1 4S 3
1 4 cos T 4 cos 2 T dT
2 ³ 2S 3
1 4S 3
3 4 cos T 2 cos 2T dT
2 ³ 2S 3
sin T
, cos T
1 cos T
40. In this case, 0 d T d 2S and you let u
1 u2
, dT
1 u2
1
>3T 4 sin T sin 2T @42SS
2
3
3
3 3
.
2
S 2 du
.
1 u2
Now u Ÿ f as T Ÿ S and you have
A
9
§1· S
2¨ ¸ ³
© 2 ¹ 0 2 cos T
18³
f
0
2
9³
dT
2du
1 u2
f
0
1 u2
§1 u2 ·
4 4¨
2¸
©1 u ¹
1 u2
f
f
6 §S ·
6 ª u º
ª 6
arctan
«
¨ ¸
3© 2 ¹
3 ¬«1 3u 2 ¼» 0
¬ 3
41. (a)
³ C1 y
3
³ R ³ ª¬ 27 3x
dx 27 x x3 dy
2S
1
³0 ³0
(b) You want to find c such that
27c 2
3
c4
2
4
27c 3c 2 Ÿ c
f c
fc c
Maximum Value:
2S
3
³0 ³0
C
³0
1 u2
1 3u 2
2
du
2
f
3S
3S
0
3
3
º
3u »
¼0
2
f
0
12 § 1 · ª u
º
3 u» ¨ ¸«
3 © 2 ¹ ¬1 3u 2
¼0
ª 6
« 3 arctan
¬
f
13
23
du 18³
du
2
0
1 3u 2
1 3u 2
18³
2
f
º
3
³ 1 3u 2 du»
¼0
2 3S .
3 y 2 º¼ dA
27 3r 2 r dr dT
2S
³0
1
ª 27 r 2
3r 4 º
«
» dT
4 ¼0
¬ 2
2S
³0
51
dT
4
51
S
2
27 3r 2 r dr dT is a maximum:
3
27 3r 2 r dr dT
243S
2
­ti,
0 d t d 4
°
i
j
4
t
4
,
4 d t d 8
®
° 12 t i 12 t j, 8 d t d 12
¯
42. (a) r t
³ C y dx x dy
2
2
4
³ 0 ª¬0 dt t
0 64 By Green's Theorem,
³R
2
0 º¼ 128
3
8
³ 4 ª¬ t 4
0 16 dt º ¼
2
12
³8
ª 12 t
¬
2
dt 12 t
64
3
§ wN
wM ·
³ ¨© wx wy ¸¹ dA
dt º
¼
2
y
(4, 4)
4
x
³0 ³0
2 x 2 y dy dx
4
³0 x
2
dx
64
.
3
4
y=x
3
2
1
x
1
2
3
4
© 2010 Brooks/Cole, Cengage Learning
Section 15. 4
Green’s Theorem
cos ti sin tj, 0 d t d 2S
(b) r t
³ C y dx x dy
2
2
2S
³0
2S
³0
2S
³0
ª¬sin 2 t sin t dt cos 2 t cos t dt º¼
cos3 t sin 3 t dt
2S
ª
sin 3 t
cos3 t º
cos t «sin t »
3
3 ¼0
¬
ªcos t 1 sin 2 t sin t 1 cos 2 t º dt
¬
¼
By Green's Theorem,
1 x 2
1
³ 1 ³ 2S
1
2r cos T 2r sin T r dr dT
2 2S
cos T sin T dT
3³0
x 2 + y 2 =1
1
2 x 2 y dy dx
1 x 2
³0 ³0
³C
0
y
§ wN
wM ·
³ R ³ ¨© wx wy ¸¹ dA
43. I
409
2
0
3
x
−1
1
0.
−1
y dx x dy
x2 y 2
y
x
i 2
j.
x2 y 2
x y2
(a) Let F
wN
wx
F is conservative because
wM
wy
x2 y 2
x2 y 2
2
.
F is defined and has continuous first partials everywhere except at the origin. If C is a circle (a closed path) that does not
contain the origin, then
³ C F ˜ dr
³ C M dx N dy
§ wN
³ R ³ ¨© wx
wM ·
¸ dA
wy ¹
0.
a cos ti a sin tj, 0 d t d 2S be a circle C1 oriented clockwise inside C (see figure). Introduce line segments
(b) Let r
C2 and C3 as illustrated in Example 6 of this section in the text. For the region inside C and outside C1 , Green's Theorem
applies. Note that since C2 amd C3 have opposite orientations, the line integrals over them cancel. So,
C4
C1 C2 C C3 and
³ C4 F ˜ dr
³ C1 F ˜ dr ³ C F ˜ dr
0.
y
3
But,
³ C1 F ˜ dr
2S
³0
2S
³0
Finally,
³ C F ˜ dr
ª a sin t a sin t
a cos t a cos t
2
« 2
2
2
2
a
cos
t
a
sin
t
a
cos 2 t a 2 sin 2
¬
sin 2 t cos 2 t dt
³
C1
F ˜ dr
2S
>t@0
C
2
º
» dt
t¼
C1
C2
x
4
C3
2S .
−2
−3
2S .
Note: If C were oriented clockwise, then the answer would have been 2S .
44. (a) Let C be the line segment joining x1 , y1 and x2 , y2 .
y
dy
y2 y1
x x1 y1
x2 x1
y2 y1
dx
x2 x1
x2
³ C y dx x dy ³ x1
ª y2 y1
§ y y1 ·º
x x1 y1 x¨ 2
«
¸» dx
x
x
2
1
© x2 x1 ¹¼
¬
x2
ªª § y2 y1 ·
º º
«« x1 ¨
¸ y1 » x»
x
x
1¹
¼ ¼» x1
¬«¬ © 2
x2
³ x1
ª § y2 y1 ·
º
« x1 ¨
¸ y1 » dx
x
x
2
1
©
¹
¬
¼
ª § y2 y1 ·
º
« x1 ¨
¸ y1 » x2 x1
x
x
1¹
¬ © 2
¼
x1 y2 y1 y1 x2 x1
x1 y2 x2 y1
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
410
Vector Analysis
1
y dx x dy
2³C
(b) Let C be the boundary of the region A
1
1 1 dA
2³R ³
³ R ³ dA.
So,
1ª
y dx x dy 2 ¬«³ C1
³ R ³ dA
³ C2 y dx x dy " x dyº
¼»
³ Cn y dx where C1 is the line segment joining x1 , y1 and x2 , y2 , C2 is the line segment joining x2 , y2 and x3 , y3 , ", and
Cn is the line segment joining xn , yn and x1 , y1 . So,
1
ª x1 y2 x2 y1 x2 y3 x3 y2 " xn 1 yn xn yn 1 xn y1 x1 yn º¼.
2¬
³ R ³ dA
45. Pentagon: 0, 0 , 2, 0 , 3, 2 , 1, 4 , 1, 1
1ª
2¬
A
0 0 4 0 12 2 1 4 0 0 º¼
19
2
46. Hexagon: 0, 0 , 2, 0 , 3, 2 , 2, 4 , 0, 3 , 1, 1
1ª
2¬
A
47. Because
0 0 4 0 12 4 6 0 0 3 0 0 º¼
³ C F ˜ N ds
³ C fDN g ds
48.
³C
³ R ³ div F dA, then
³ C f ’g ˜ N ds
fDN g gDN f ds
³ R ³ div
³C f
50. F
x dx g y dy
f ’g dA
³ C fDN g ds ³ C gDN f
³R ³
49.
21
2
³R ³
ªw
y f ’ 2 g ’f ˜ ’g dA.
ds
f ’ 2 g ’f ˜ ’g dA ³ R ³ «¬ wx g
³R ³
f div ’g ’f ˜ ’g dA
º
w
f x » dA
wy
¼
³R ³
³R ³
g’ 2 f ’g ˜ ’f dA
³R ³
0 0 dA
f ’ 2 g g’ 2 f dA
0
Mi Nj
wN
wx
wM
wN
wM
Ÿ
wy
wx
wy
³ C F ˜ dr
³ C M dx N dy
0
§ wN
³ R ³ ¨© wx
wM ·
¸ dA
wy ¹
³R ³
0 dA
0
Section 15.5 Parametric Surfaces
ui vj uvk
1. r u , v
z
xy
Matches (e)
x y
2
2
2
z , cone
Matches (f )
2y
x z , plane
1
2
u v j vk
4. r u , v
4y
ui 14 v3 j vk
z 3 , cylinder
Matches (a)
2 cos v cos ui 2 cos v sin uj 2 sin vk
5. r u , v
x y z2
2
ui Matches (b)
u cos vi u sin vj uk
2. r u , v
3. r u , v
2
4 cos 2 v cos 2 u 4 cos 2 v sin 2 u 4 sin 2 v
4 cos2 v 4 sin 2 v
4, sphere
Matches (d)
4 cos ui 4 sin uj vk
6. r u , v
x y
2
2
4, circular cylinder
Matches (c)
© 2010 Brooks/Cole, Cengage Learning
Section 15.5
ui vj 7. r u , v
y 2z
v
k
2
Parametric Surfaces
2 cos v cos ui 4 cos v sin uj sin vk ,
12. r u , v
0 d u d 2S , 0 d v d 2S
0
2
2
z
5
4
3
2
x
y
z
4
16
1
Plane
411
1
−5
−4
−5
z
3
2
5
4
3
3 4 5
5
−3
−4
−5
x
−4
y
5
y
x
2 sinh u cos vi sinh u sin vj cosh uk ,
13. r u , v
2u cos vi 2u sin vj 12 u 2k
8. r u , v
x y
1 2
u ,
2
z
2
4u Ÿ z
2
1
8
2
x y
2
0 d u d 2, 0 d v d 2S
2
2
2
z
x
y
1
4
1
Paraboloid
z
2
9
1
6
z
9
4
4
4
0 d u d 1, 0 d v d 3S
y
x
4
4
−4
x
5
y
2
4
4
0 d u d S,
−3
0 d v d 2S
z
5
3 cos v cos ui 3 cos v sin uj 5 sin vk
9 cos 2 v cos 2 u 9 cos 2 v sin 2 u
x2 y 2 z 2
9
25
2
cos 2 v sin 2 v
4
3
9 cos 2 v
−2
1
−3
3
2
x
y
z
9
9
25
−2
2
1
−1
−3
2
y
3
x
1
Ellipsoid
cos3 u cos vi sin 3 u sin vj uk ,
16. r u , v
z
0 d u d
5
S
2
, 0 d v d 2S
z
2
x
4
3
3 4
y
−1
11. r u , v
2u cos vi 2u sin vj u k ,
4
z
1
y
z
0 d u d 1, 0 d v d 2S
x2 y2
x
−1
1
3
2
2
16
1
2
x
2
y
u sin u cos vi 1 cos u sin vj uk ,
15. r u , v
5
2
−4
−2
2
z
3
x2 y 2
y
9
8
Cylinder
10. r u , v
6
z
y
x
tan z
2 cos ui vj 2 sin uk
x2 z 2
x
3
2u cos vi 2u sin vj vk ,
14. r u , v
9. r u , v
6
x
y
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
412
Vector Analysis
For Exercises 17–20, r u, v
u cos vi u sin vj u 2 k , 0 d u d 2, 0 d v d 2S .
Eliminating the parameter yields z
x 2 y 2 , 0 d z d 4.
z
5
y
2
2
x
u cos vi u sin vj u 2k , 0 d u d 2, 0 d v d 2S
17. s u , v
z
x2 y 2
The paraboloid is reflected (inverted) through the xy-plane.
u cos vi u 2 j u sin vk , 0 d u d 2, 0 d v d 2S
18. s u , v
y
x2 z 2
The paraboloid opens along the y-axis instead of the z-axis.
u cos vi u sin vj u 2k , 0 d u d 3, 0 d v d 2S
19. s u , v
The height of the paraboloid is increased from 4 to 9.
4u cos vi 4u sin vj u 2k , 0 d u d 2, 0 d v d 2S
20. s u , v
z
x2 y 2
16
The paraboloid is "wider." The top is now the circle x 2 y 2
21. z
y
26. 4 x 2 y 2
ui vj vk
r u, v
22. z
6 x y
r u, v
27. z
ui vj 6 u v k
r u, v
23. y
64. It was x 2 y 2
4 x2 9z 2
xi r x, y
4 x 2 9 z 2 j zk
16
2 cos ui 4 sin uj vk
x2
r u, v
28.
4.
ui vj u 2k
x2
y2
z2
1
9
4
1
r u, v
3 cos v cos ui 2 cos v sin uj sin vk
or,
cos vi uj 13 u sin vk ,
r u, v
1u
2
u t 0,
0 d v d 2S
24. x
16 y 2 z 2
r y, z
16 y 2 z 2 i yj zk
29. z
4 inside x 2 y 2
r u, v
30. z
v cos ui v sin uj 4k , 0 d v d 3
x 2 y 2 inside x 2 y 2
r u, v
ui u t 0,
0 d v d 2S
25. x 2 y 2
r u, v
1
4
u cos vj u sin vk ,
25
5 cos ui 5 sin uj vk
9.
v cos ui v sin uj v 2k , 0 d v d 3
or,
r u, v
9.
31. Function: y
x
, 0 d x d 6
2
Axis of revolution: x-axis
u
u
cos v, z
sin v
2
2
0 d u d 6, 0 d v d 2S
x
u, y
© 2010 Brooks/Cole, Cengage Learning
Section 15.5
x, 0 d x d 4
32. Function: y
x
u, y
u cos v, z
0 d u d 4,
33. Function: x
u sin v
0 d v d 2S
sin u sin v, z
u
y 1, 0 d y d 2
u 2 1 cos v, y
0 d u d 2,
35. r u , v
ru u , v
N
j k
0 3
4
4 0
0
ru u , v
2 cosh vi 2 sinh vj uk
i jk
rv u , v
2u sinh vi 2u cosh vj
1.
At 4, 0, 2 , u
i j, rv 0, 1
i j k
i
j k
1
1 0
ru 2, 0
N
i j 2k
1
i 8i 8k
Tangent plane: x 4 z 2
u
j
k
2 uv
1.
j
i
4 j
Direction numbers: 1, 0, 1
x z
0
2
4ui vj vk ,
ru u , v
4i, rv u , v
i
j k
ru u rv
4
0
0
0 1
1
uvk , 1, 1, 1
1
k , rv 1, 1
2
ru 1, 1 u rv 1, 1
ru u rv
39. r u , v
1 and v
0.
0
v
i k , rv u , v
2 uv
At 1, 1, 1 , u
2 and v
2i 2k , rv 2, 0
0
(The original plane!)
ui vj 0
12
i j, rv u , v
0 and v
16 j 12k
2u cosh vi 2u sinh vj 12 u 2k ,
38. r u , v
x y 2z
N
i
§ S·
§ S·
ru ¨ 2, ¸ u rv ¨ 2, ¸
© 2¹
© 2¹
4 y 3z
0 d v d 2S
Tangent plane: x 1 y 1 2 z 1
ru 1, 1
4i
Tangent plane: 4 y 6 3 z 4
1 1
ru u , v
§ S·
3 j 4k , rv ¨ 2, ¸
© 2¹
u 2 1 sin v
u, z
ru 0, 1 u rv 0, 1
36. r u , v
0, 6, 4
S 2.
2 and v
u v i u v j vk , 1, 1, 1
At 1, 1, 1 , u
ru 0, 1
2u sin vi 3u cos vj
413
Direction numbers: 0, 4, 3
2
Axis of revolution: y-axis
x
rv u , v
N
0 d u d S , 0 d v d 2S
34. Function: z
2 cos vi 3 sin vj 2uk
§ S·
ru ¨ 2, ¸
© 2¹
sin z , 0 d z d S
sin u cos v, y
ru u , v
At 0, 6, 4 , u
Axis of revolution: z-axis
x
2u cos vi 3u sin vj u 2k ,
37. r u , v
Axis of revolution: x-axis
Parametric Surfaces
j
1
2
1
2
1 0
0 1
ru u rv
1
k
2
k
A
1
j k
4 j 4k
16 16
2
³0 ³0 4
0 d u d 2, 0 d v d 1
2 du dv
4 2
4 2 2 1
8 2
1
1
i jk
2
2
Direction numbers: 1, 1, 2
Tangent plane: x 1 y 1 2 z 1
x y 2z
0
0
© 2010 Brooks/Cole, Cengage Learning
414
Chapter 15
Vector Analysis
2u cos vi 2u sin vj u 2k , 0 d u d 2, 0 d v d 2S
40. r u , v
ru u , v
2 cos vi 2 sin vj 2u k
rv u , v
2 u sin vi 2 u cos vj
ru u rv
i
j
k
2 cos v
2 sin v
2u
2u sin v 2u cos v
ru u rv
A
0
16u cos v 16u sin 2 v 64u 2
4
2S
2
³0 ³0
2
4
2S
³0
4u u 2 4 du dv
a cos ui a sin uj vk ,
41. r u , v
ru u , v
a sin ui a cos uj
rv u , v
k
ru u rv
a sin u a cos u
i
j
0
ru u rv
A
4u 2 cos vi 4u 2 sin vj 8uk
b
ª4 2
«3 u 4
¬
4u u 2 4
3 2º
» dv
¼0
0
1
a du dv
2S ab
rv u , v
a sin u sin vi a sin u cos vj
ru u rv
a cos u cos v
i
j
0 d u d S , 0 d v d 2S
k
a cos u sin v a sin u
a sin u sin v a sin u cos v
a 2 sin 2 u cos vi a 2 sin 2 u sin vj a 2 sin u cos uk
0
2
a sin u
S
³0 ³0 a
2
sin u du dv
4S a 2
au cos vi au sin vj uk , 0 d u d b, 0 d v d 2S
43. r u , v
ru u , v
a cos vi a sin vj k
rv u , v
au sin vi au cos vj
ru u rv
i
j
k
a cos v
a sin v
1
au sin v au cos v
ru u rv
A
64S
2 2 1
3
a cos ui a sin uj
0
a cos u cos vi a cos u sin vj a sin uk
A
4
16 2 8 2S
3
k
ru u , v
2S
4
8 8 8 dv
3
0 d u d 2S , 0 d v d b
a sin u cos vi a sin u sin vj a cos uk ,
ru u rv
2S
³0
a
2S
³0 ³0
42. r u , v
2
2S
au cos vi au sin vj a 2 uk
0
au 1 a 2
b
³0 ³0 a
1 a 2 u du dv
S ab 2 1 a 2
© 2010 Brooks/Cole, Cengage Learning
Section 15.5
a b cos v cos ui a b cos v sin uj b sin vk , a ! b, 0 d u d 2S ,
44. r u , v
ru u , v
a b cos v sin ui a b cos v cos uj
rv u , v
b sin v cos ui b sin v sin uj b cos vk
ru u rv
i
j
k
a b cos v sin u
a b cos v cos u
0
b sin v cos u
b sin v sin u
b cos v
Parametric Surfaces
415
0 d v d 2S
b cos u cos v a b cos v i b sin u cos v a b cos v j b sin v a b cos v k
ru u rv
A
b a b cos v
2S
2S
³0 ³0
u cos vi 45. r u , v
u sin vj uk , 0 d u d 4, 0 d v d 2S
ru u , v
cos v
sin v
i jk
2 u
2 u
rv u , v
u sin vi ru u rv
cos v
2 u
u sin v
ru u rv
u u cos vj
i
A
2S
4
³0 ³0
4S 2 ab
b a b cos v du dv
u j
k
sin v
1
2 u
u cos v 0
u cos vi 1
du dv
4
S
6
17 17 1 | 36.177
ru u , v
cos u cos vi j cos u sin vk
rv u , v
sin u sin vi sin u cos vk
ru u rv
sin u cos vi cos u sin uj sin u sin vk
ru u rv
A
sin u
2S
S
³0 ³0
sin u
1
k
2
1
4
sin u cos vi uj sin u sin vk , 0 d u d S ,
46. r u , v
u sin vj 0 d v d 2S
1 cos 2 u
1 cos 2 u du dv
ª
S «2 2 ln
¬«
2 1º
»
2 1 ¼»
47. See the definition, page 1102.
50. (a) From 10, 10, 0
48. See the definition, page 1106.
(b) From 10, 10, 10
49. Function: z
x
(c) From 0, 10, 0
Axis of revolution: z-axis
x
u cos v, y
u sin v, z
r u, v
(d) From 10, 0, 0
u cos vi u sin vj uk
0 d v d 2S
u d 0,
51. r u , v
u
a sin 3 u cos3 vi a sin 3 u sin 3 vj a cos3 uk
0 d u d S,
0 d v d 2S
x
a sin u cos3 v Ÿ x 2 3
a 2 3 sin 2 u cos 2 v
y
a sin 3 u sin 3 v Ÿ y 2 3
a 2 3 sin 2 u sin 2 v
z
a cos3 u Ÿ z 2 3
3
x2 3 y 2 3 z 2 3
a 2 3 cos 2 u
a 2 3 ª¬sin 2 u cos 2 v sin 2 u sin 2 v cos 2 u º¼
a 2 3 ª¬sin 2 u cos 2 u º¼
a2 3
© 2010 Brooks/Cole, Cengage Learning
416
Chapter 15
Vector Analysis
u cos vi u sin vj vk
52. Graph of r u , v
0 d u d S,
0 d v d S from
(a) 10, 0, 0
(b)
0, 0, 10
(c) 10, 10, 10
z
z
−3
3
3
y
y
−3
3
3
3
−3
3
x
y
x
4 cos v cos ui 53. (a) r u , v
4 2 cos v cos ui (b) r u , v
4 cos v sin uj sin vk ,
4 2 cos v sin uj 2 sin vk ,
0 d u d 2S , 0 d v d 2S
0 d u d 2S , 0 d v d 2S
z
z
4
−6
4
−6
6
x
6
−4
y
8 cos v cos ui (c) r u , v
6
6
x
y
8 3 cos v cos ui (d) r u , v
8 cos v sin uj sin vk ,
8 3 cos v sin uj 3 sin vk ,
0 d u d 2S , 0 d v d 2S
0 d u d 2S , 0 d v d 2S
z
z
12
9
3
3
12
y
x
x
12
−9
y
−12
The radius of the generating circle that is revolved about the z-axis is b, and its center is a units from the axis of revolution.
54. r u , v
(a) If u
2u cos vi 2u sin vj vk , 0 d u d 1, 0 d v d 3S
Helix
1:
z
2 cos vi 2 sin vj vk
r 1, v
10
x2 y2
4
8
0 d z d 3S
4
2
−2
(b) If v
2S
:
3
§ 2S ·
r¨ u,
¸
© 3 ¹
y
z
3x
2S
3
−2
2
x
2
y
Line
z
ui 2S
3uj k
3
2
1
−2
−2
−1
−1
1
2
x
1
2
y
(c)If one parameter is held constant, the result is a curve in 3-space.
© 2010 Brooks/Cole, Cengage Learning
Section 15.5
20 sin u cos vi 20 sin u sin vj 20 cos uk , 0 d u d S 3,
55. r u , v
ru
20 cos u cos vi 20 cos u sin vj 20 sin uk
rv
20 sin u sin vi 20 sin u cos vj
i
ru u rv
j
Parametric Surfaces
417
0 d v d 2S
k
20 cos u sin v 20 sin u
20 cos u cos v
20 sin u sin v 20 sin u cos v
0
400 sin 2 u cos vi 400 sin 2 u sin vj 400 cos u sin u cos 2 v cos u sin u sin 2 v k
400 ª¬sin 2 u cos vi sin 2 u sin vj cos u sin uk º¼
ru u rv
S
400 sin 4 u cos 2 v sin 4 u sin 2 v cos 2 u sin 2 u
2S
³ S ³ dS
³0 ³0
56. x 2 y 2 z 2
Let x
S 3
400 sin 4 u cos 2 u sin 2 u
2S
2S
S 3
³ 0 >400 cos u@0 dv
400 sin u du dv
³0
400 sin 2 u
400 sin u
400S m 2
200 dv
1
u
ru u , v
cos vi sin vj rv u , v
u sin vi u cos vj.
At 1, 0, 0 , u
u 2 1. Then,
u sin v, and z
u cos v, y
1 and v
u 1
2
k
0. ru 1, 0 is undefined and rv 1, 0
j. The tangent plane at 1, 0, 0 is x
1.
u cos vi u sin vj 2vk , 0 d u d 3, 0 d v d 2S
57. r u , v
ru u , v
cos vi sin vj
rv u , v
u sin vi u cos vj 2k
z
i
ru u rv
j
cos v
k
sin v
ru u rv
A
2S
2 sin vi 2 cos vj uk
0
u sin v u cos v
2
2π
4 u2
3
³0 ³0
ª
§3 ¬«
©
S «3 13 4 ln ¨¨
4 u 2 du dv
−4
13 ·º
¸¸»
2
¹¼»
−2
2
4
4
y
x
ui f u cos vj f u sin vk , a d u d b, 0 d v d 2S
58. r u , v
ru u , v
i f c u cos vj f c u sin vk
rv u , v
f u sin vj f u cos vk
ru u rv
i
j
k
1
f c u cos v
f c u sin v
0 f u sin v
f u cos v
ru u rv
A
4π
2S
b
³0 ³a
f u f c u i f u cos vj f u sin vk
2
f u
1 ª¬ f c u º¼
f u
1 ª¬ f c u º¼ du dv
2
b
2S ³ f x
a
2
1 ª¬ f c x º¼ dx
because u
x
59. Answers will vary.
60. Answers will vary.
© 2010 Brooks/Cole, Cengage Learning
418
Chapter 15
Vector Analysis
Section 15.6 Surface Integrals
1. S : z
³S ³
2. S : z
³S ³
3. S : z
³S ³
4 x, 0 d x d 4,
4
15 2 x 3 y,
x 2 y z dS
2
1
³ 1 ³ 2S
³S ³
1 x2
1 x2
1 0 02 dy dx
x 2y 2
x1 2 ,
wz
wy
1
x§
2 3 2·
x ¸ 1 x dy dx
3
¹
2
0
5 § 1 ·ª 3 2
» ¨ ¸ x 1 x
¼ 0 12 © 3 ¹ ¬
2S
2
dy dx
2
32 1
1
º 5 x1 2 1 x dx
³
0
¼0
24
2 1
ln
4
x2 x 1
3
ln 4
2
2 1
1,
2
wz
wy
dy dx
1·
§
¨x ¸ 2¹
©
1
º
x2 x »
¼0
1
1º
ln »
4
2¼
61 2
5
ln 3 2 2 | 0.2536
288
192
1
3³
z
3ª
0
3
3 x
y2 º
«x »
¬ 2 ¼0
3
dx
2
2
2S 3
3
2
wz
wx
first octant ,
2
r cos T 2r sin T 2 r dr dT
x x 2 dx
2 15 2
5
1
ln
18
96
192 3 2 2
3 3
x3 x
2 ³0
1
1·
1
§
¨ x ¸ dx
2¹
4
©
2
5 ª3
18
48 «¬ 2
xy 1 1
2S
³0 ³0
5 1 32
1 x dx
» 12 ³ 0 x
¼0
2
5 § 1 · ª§
1·
¨ ¸ «¨ x ¸
18
24 © 2 ¹ ¬©
2¹
³0 ³0
128 14
1
3 2º
1
3 2º
2
5 1
18
24 ³ 0
³ S ³ xy dS
15 x y dy dx
14 dy dx
x 1 dx
2
5 2
5 1
3
18
24 ³ 0
3 x
4
³0
12 2
0
2 3 2·
x ¸ 1 x1 2
3
¹
ª1 5 2
« x 1 x
¬6
3
0
1 4 9 dy dx
2S
x§
2 ª1 5 2
x 1 x
3 «¬ 4
3 x y
2
0
4
2 ³ 3 dx
4 2 y dy dx
2
ª1
º
« 3 sin T 3 cos T T »
¬
¼0
1
³ 0 ³ 0 ¨© x 2 y ³ 0 ³ 0 ¨© x 2 y 5. S : z
3
³0
3, dS
14 ³
14 dy dx
2
ª1
º
« 3 cos T 3 sin T 1» dT
¬
¼
2 1 52
x
3³0
wz
wy
2,
4
0
0
2 32
wz
x , 0 d x d 1, 0 d y d x,
3
wx
x 2 y z dS
2³
02 dy dx
wz
wx
0 d y d 4,
wz
wy
0
2
1 1
x 2 y 15 2 x 3 y
wz
wx
³0
4. S : z
4
³0 ³0
wz
wy
1,
x 2y 4 x
0 d x d 2,
2, x 2 y 2 d 1,
x 2 y z dS
3
³0 ³0
x 2 y z dS
wz
wx
0 d y d 3,
3 ª x4
9 x2 º
3
« 2x »
2 ¬4
2 ¼0
3 ª 27 º
2 «¬ 4 »¼
27 3
8
3
3
y
y=3−x
x
© 2010 Brooks/Cole, Cengage Learning
Section 15.6
h, 0 d x d 2, 0 d y d
6. S : z
2
³ S ³ xy dS
wz
wy
2 x,
2
³ 0 ³ y xy
4
³ S ³ xy dS
4
1 wz
y,
2 wy
y2
x2
dy dx
4
4
1
2
cos x, 0 d x d
³R ³
³
6
7
6 0
S
2
x 2 2 xy
, 0 d y d
S 2
x2
³0 ³0
x 2 2 xy dS
U x, y , z
2
³0 ³0
11. S : 2 x 3 y 6 z
1
x
2
3904 160 5
15
3
1 4 x 2 4 y 2 dy dx | 11.47
x
2
x 2 2 xy
first octant Ÿ z
12
S 2
³0
1 sin 2 x dy dx
2 13 x x3
4
1 sin 2 x dx | 0.52
1y
2
y
5
x2 y2
x2 y2
ªx 2 4 «¬
1 13
2x
3
y = 4 − 23 x
4
1
3
2
12
2
6
4 23 x º dx
»¼
4 2x 3
³ ³0
7
6 0
dA
3
3
7 ª 4 x3
6«
¬3
x 2 y 2 dy dx
2
R
1
16 x 4
1
8
4 6
4 23 x º
»¼ 0
x
364
3
−1
1
2
3
4
5
6
a2 x2 y 2
12. S : z
U x, y , z
kz
³ S ³ kz dS
³R ³ k
13. S : r u , v
ru
2
391 17 1
240
1 4 x 2 dx dy
³ 0 ³ 0 xy
x 2 2 xy dS
10. S : z
m
2
1ª 2
x4 º
«2 x »
2¬
4 ¼0
10 x 2 y 2 , 0 d x d 2, 0 d y d 2
9. S : z
m
0
1 2
x 4 x 2 dx
2³0
xy dy dx
1
wz
xy, 0 d x d 4, 0 d y d 4,
2
wx
8. S : z
³S ³
wz
wy
0
2
³ S ³ xy dS
³S ³
4 x2
wz
wx
419
9 x 2 , 0 d x d 2, 0 d y d x,
7. S : z
wz
wx
³0 ³0
4 x2 ,
Surface Integrals
i,
a2 x2 y2
·
¸ dA
a 2 x 2 y 2 ¸¹
ui vj 2vk ,
0 d u d 1,
a
2
·
§
¸ ¨
2
2
2 ¸
¨
a x y ¹
©
x
§
a2 x2 y2 ¨
¨
©
³ R ³ ka dA
ka ³
R
z
2
·
¸ dA
2
2
2 ¸
a x y ¹
y
³ dA
ka 2S a 2
a
2ka 3S
a
a
x
0 d v d 2
j 2k
rv
i
ru u rv
³R ³ k
§
1¨
¨
©
j k
2 j k
1 0 0
0 1 2
ru u rv
5
2
³S ³
y 5 dS
2
1
³0 ³0
v 5
5 du dv
2
³0
v5
5 dv
ªv2
º
5 « 5v»
2
¬
¼0
12 5
© 2010 Brooks/Cole, Cengage Learning
y
420
Chapter 15
Vector Analysis
2 cos ui 2 sin uj vk , 0 d u d S 2, 0 d v d 1
14. r u , v
2 sin ui 2 cos uj,
ru
i
ru u rv
j
ru u rv
15. S : r u , v
1
³0 ³0
2 cos ui 2 sin uj vk ,
rv
j
0
0
ru u rv
0 d u d S 2, 0 d v d 1
1
4 cos u 4 sin u
16. S : r u , v
S 2
³0 ³0
2
2 cos u 2 sin u 2 du dv
12u cos vi 12u sin vj 16uk
x y dS
17. f x, y, z
4
³0 ³0
S 2
4³ >sin u cos u@0
1
0
1
4³ 2 dv
dv
0
8
0 d v d S
4u cos vi 4u sin vj 3uk , 0 d u d 4,
S
4
2 cos ui 2 sin uj
2
1
dv
k
0
2
x y dS
1 ª sin 2 u º
8³ «
»
0
¬ 2 ¼0
k
2 sin u 2 cos u
ru u rv
2
2 cos u 2 sin u 2 du dv
2 sin ui 2 cos uj,
ru u rv
³S ³
1
S 2
S 2
i
³S ³
2 cos ui 2 sin uj
0
0
4 cos 2 u 4 sin 2 v
³ xy dS
ru
k
k
2 sin u 2 cos u
0
³S
rv
20u
4u cos v 4u sin v 20u du dv
10,240
3
x2 y 2 z 2
S: z
x y,
x 2 y 2 d 1,
³S ³ f
x, y, z dS
³ 1 ³ 1
3³
1 x 2
1 x 2
1
1 ³ 2 3³
wz
wx
2S
0
wz
wy
1
ª x 2 y 2 x y 2 º 1 12 12 dy dx
¬
¼
1 x 2
1 x 2
3³
ª¬2 x 2 2 y 2 2 xyº¼ dy dx
1
ªr
º
r4
cos T sin T » dT
« 4
4
¬
¼0
2S
0
1
³0
2r 2 2r cos T r sin T r dr dT
3 2S
1 cos T sin T dT
2 ³0
2S
3ª
sin 2T º
«T »
2 ¬
2 ¼0
3S
xy
z
x 2 y 2 , 4 d x 2 y 2 d 16
18. f x, y, z
S: z
³S ³
f x, y, z dS
³S ³
2S
xy
2
x y2
4
³0 ³2 r
1 4 x 2 4 y 2 dy dx
1 4r 2 sin T cos T dr dT
2S
4
³0 ³2
2S
³0
r 2 sin T cos T
r2
ª1
2
«12 1 4r
¬
3 2º
1 4r 2 r dr dT
4
» sin T cos T dT
¼2
2S
ª 65 65 17 17 § sin 2 T ·º
«
¨
¸»
12
«¬
© 2 ¹»¼ 0
0
© 2010 Brooks/Cole, Cengage Learning
Section 15.6
19. f x, y, z
x2 y2 z 2
S: z
x2 y 2 , x2 y 2 d 4
³S ³ f
x, y, z dS
³ 2 ³ 2³
2
2 ³ ³
2
4 x2
2
x2 y 2 , x 1
4 x2
x, y, z dS
S : x2 y 2
2
·
¸ dy dx
x 2 y 2 ¸¹
y
2³
2S
0
2
³0
2³
r 2 dr dT
2S
0
2S
2
ªr3 º
« » dT
¬ 3 ¼0
32S
3
ª16 º
« 3 T»
¬
¼0
y2 d 1
³S ³
x y ³S ³
2 x2 y2
2
x y
2
2
2
x y
2
2
2
§
1 ¨
¨
©
2
2 x2 y 2
2
·
§
¸ ¨
2
2 ¸
¨
x y ¹
©
·
¸ dy dx
2
2 ¸
x y ¹
x
2³
dy dx
³
S
y
x 2 y 2 dy dx
2³
S
0
2 cos T
³0
r 2 dr dT
S
ª16 §
sin 3 T ·º
« ¨ sin T ¸»
3 ¹¼ 0
¬3©
16 S
1 sin 2 T cos T dT
3 ³0
0
9, 0 d x d 3, 0 d y d 3, 0 d z d 9
9 y 2 , 0 d y d 3, 0 d z d 9.
Project the solid onto the yz-plane; x
x, y, z dS
3
9
3
9
³ 0 ³ 0 ¬ª 9 ³0 ³0
324 ³
y
9 z
2
3
2
dz dy
9
§
z 3 ·º
«
¨ 9 z ¸» dy
2
3 ¹»¼
«¬ 9 y ©
0
³0
§S
·
972¨ 0 ¸
©2
¹
ª
§ y ·º
«972 arcsin ¨ 3 ¸»
© ¹¼ 0
¬
dy
2
3
3
9 y2
0
2
·
¸ 0
2 ¸
9 y ¹
y
3ª
dz dy
9 y2
3
3
§
y z ¼ 1¨
¨
©
2º
2
486S
x2 y 2 z 2
22. f x, y, z
S : x2 y2
9, 0 d x d 3, 0 d z d x
9 x2 .
Project the solid onto the xz-plane; y
f x, y, z dS
3
x
3
x
³0 ³0
³0 ³0
3
³0
Let u
2
·
§
¸ ¨
¨
x 2 y 2 ¸¹
©
x
x2 y 2 z 2
21. f x, y, z
³S ³
§
1¨
¨
©
2
2
x2 y2 x2 y 2
dy dx
x2 y 2
x 2 y 2 dy dx
16 S
cos3 T dT
3 ³0
³S ³ f
2
x2 y2
4 x2
4 x2
x y
2
421
x2 y2 z 2
20. f x, y, z
³S ³ f
x y 2
4 x2
2
2³
S: z
4 x2
2
Surface Integrals
x 2 , dv
§
2
2º
ª 2
¬x 9 x z ¼ 1 ¨
©
9 z2
3
9 x2
§
x3 ·
¨9x ¸ dx
3¹
9 x ©
x 9 x2
2
1 2
dx, then du
3
ª27 9 x 2 º ªª x 2
¬
¼ 0 «¬¬
2
dz dx
x
3ª
§
z 3 ·º
3
9z «
¨
¸» dx
2
3 ¹¼ 0
¬ 9 x ©
³0
dz dx
3
2
·
¸ 0
2
9 x ¹
x
3
³ 0 27 x 9 x
2 x dx, v
3
9 x2 º ¼0
2
1 2
dx 3
³0 x
3
9 x2
1 2
dx
9 x2 .
3
³0
º
2 x 9 x 2 dx»
¼
2
ª
2
«81 3 9 x
¬
3 2º
3
»
¼0
81 18
99
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
422
Vector Analysis
3 zi 4 j yk
23. F x, y, z
1 x y
S: z
y
first octant
x y z 1
G x, y , z
’G x , y , z
1
y = −x + 1
i j k
R
³S
³ F ˜ N dS
³R
1 x
1
³ F ˜ ’G dA
³0 ³0
3 z 4 y dy dx
1 1 x
³0³0
1 1 x
³0³0
x
1
ª¬3 1 x y 4 yº¼ dy dx
1
xi yj
S: z
3x 2 y z 6
G x, y , z
’G x , y , z
³ S ³ F ˜ N dS
3
3 x
2
2
³0 ³0
³0
1 x2 y 2 ,
3 x 2 y dy dx
ª¬3 xy y 2 º¼
0
dx
³0
2
º
9 ª x3
« 4 x»
4 ¬3
¼0
2 xi 2 yj k
³ R ³ F ˜ ’G dA
³R ³
2 x 2 2 y 2 z dA
³R ³
2 x 2 2 y 2 1 x 2 y 2 dA
2S
2S
³0
§ 9 ·§ 16 ·
¨ ¸¨
¸
© 4 ¹© 3 ¹
1
³0 ³0
9 2
x 4 dx
4
2
³0
2
³R ³ 1 2
ª §
3 · §
3 · º
«3 x¨ 3 x ¸ ¨ 3 x ¸ » dx
2 ¹ ©
2 ¹ ¼»
¬« ©
2
z t 0
x y z 1
2
³ S ³ F ˜ N dS
3
3 x
2
2
xi yj zk
’G x , y , z
³ R ³ F ˜ ’G dA
43
0
G x, y , z
3i 2 j k
dx
0
³ 2 2 x 2 dx
25. F x, y, z
6 3x 2 y, first octant
S: z
y 2 º¼
1
2
³ ª 1 x 3x 1 x 1 x º dx
0¬
¼
24. F x, y, z
1 x
1
³ 0 ª¬ y 3xy 1 3x 2 y dy dx
12
x 2 y 2 dA
r 2 1 r dr dT
1
ªr 4
r2 º
« » dT
2 ¼0
¬4
2S
³0
3
dT
4
3S
2
y
x2 + y2 ≤ 1
z
R
6
x
1
3
2
R 3
y
6
y = 3 − 3x
2
6
x
xi yj zk
26. F x, y, z
S : x y z2
2
z
2
36
36 x y
2
first octant
y
2
x2 + y2 = 62
6
5
G x, y , z
’G x , y , z
F ˜ ’G
³ S ³ F ˜ N dS
z 36 x 2 y 2
x
36 x y
2
x
2
36 x 2 y 2
2
³ R ³ F ˜ ’G dA
i 4
3
y
36 x y
2
y2
36 x 2 y 2
³R ³
2
j k
1
1
2
3
4
5
6
36 x 2 y 2
36
36 x y
x
36
z
2
R
2
2
dA
S 2
6
³0 ³0
36
36 r 2
r dr dT
improper
108S
© 2010 Brooks/Cole, Cengage Learning
Section 15.6
x2 y 2 , x2 y 2 d 4
y
x 2 + y2 ≤ 4
x2 y 2 z
G x, y , z
’G x , y , z
2 xi 2 yj k
³ S ³ F ˜ N dS
1
³ R ³ F ˜ ’G dA
³R ³
8 x 6 y 5 dA
2S
³ 0 ³ 0 >8r cos T
2S
³0
2S
³0
2
ª 8 r 3 cos T 2r 3 sin T 5 r 2 º dT
2 ¼0
¬ 3
ª 64 cos T 16 sin T 10º dT
¬ 3
¼
2S
y
a2 x2 y 2
z G x, y , z
a
a2 x2 y 2
x
F ˜ ’G
2
a2 x2 y 2
³ S ³ F ˜ N dS
x 2 + y2 ≤ a 2
a2 x2 y2
x
’G x, y, z
i 2
³ R ³ F ˜ ’G dA
³R ³
3 x 2 3 y 2 2a 2
2S
0
a2 x2 y2
r3
a
³0
a r
ª 2S ª
3«³ « r 2
0
¬
¬«
3³
2S
0
2
x
−a
3 x 3 y 2a
2
2 a2 x2 y 2
a2 x2 y2
a
j k
a2 x2 y2
y
−a
y
3³
2
dA
a2 r 2 2S
a
2S
0
2 2
a r2
3
2S
2 3
a dT 2a 2 ³ a dT
0
3
2
a2 x2 y2
³0 ³0
dr dT 2a 2 ³
2
3r 2 2a 2
a2 r 2
r dr dT
r
a
³0
a r2
2
dr dT
a
º
2S
2
2
2º
ª
» dT » 2a ³ 0 ¬ a r ¼ 0 dT
¼ 0 »¼
3 2º
a
0
x y i yj zk
29. F x, y, z
16 x 2 y 2 , z
0
z x y 16
2
G x, y , z
’G x , y , z
F ˜ ’G
20S
xi yj 2 zk
28. F x, y, z
S: z
x
1
−1
6r sin T 5@ r dr dT
2
R
−1
ª 64 sin T 16 cos T 10T º
¬ 3
¼0
S: z
423
4i 3 j 5k
27. F x, y, z
S: z
Surface Integrals
2
2 xi 2 yj k
2x x y 2 y2 z
³ S ³ F ˜ N dS
2 x 2 2 xy 2 y 2 16 x 2 y 2
x 2 y 2 2 xy 16
³ R ³ F ˜ ’G dA
2S
4
³0 ³0
2S
³0
r 2 2r 2 cos T sin T 16 r dr dT
4
ªr 4
º
r4
cos T sin T 8r 2 » dT
« 2
¬4
¼0
(The flux across the bottom z
2S
³ 0 >192 128 cos T sin T @ dT
2S
ª¬192 64 sin 2 T º¼
0
384S
0 is 0)
© 2010 Brooks/Cole, Cengage Learning
424
Chapter 15
Vector Analysis
4 xyi z 2 j yzk
30. F x, y, z
z
S: unit cube bounded by
1
x
0, x
1, y
0, y
1, z
0, z
1
S1 : The top of the cube
N
k, z
1
³ S1 ³ F ˜ N dS
1
³0 ³0 y 1
1
2
dy dx
y
1
1
1
x
S 2 : The bottom of the cube
N
k , z
0
1
³ S2 ³ F ˜ N dS
1
³0 ³0 y 0
dy dx
0
y dy dz
2
S3 : The front of the cube
N
i, x
1
1
³ S3 ³ F ˜ N dS
1
³0 ³0 4 1
S 4 : The back of the cube
N
i , x
0
1
³ S4 ³ F ˜ N dS
1
³0 ³0 4 0
y dy dx
0
S5 : The right side of the cube
N
j, y
1
1
³ S5 ³ F ˜ N dS
1
³0 ³0 z
2
dz dx
1
3
S6 : The left side of the cube
N
j, y
³ S6 ³ F ˜ N dS
0
1
1
³0 ³0 z
2
13
dz dx
So,
³ S ³ F ˜ N dS
31. E
S: z
1
2
0 2 0
1
3
1
3
5
.
2
yzi xzj xyk
1 x2 y 2
³ S ³ E ˜ N dS
³R ³ E ˜
g x x, y i g y x, y j k dA
§
yzi xzj xyk ˜ ¨
¨
©
³R ³
§
2 xyz
©
1 x y
³ R ³ ¨¨
2
2
x
1 x2 y 2
·
xy ¸ dA
¸
¹
i ³ R ³ 3xy dA
·
j k ¸ dA
¸
1 x2 y 2
¹
y
1
³ 1 ³ 1 x2
1 x2
3xy dy dx
0
© 2010 Brooks/Cole, Cengage Learning
Section 15.6
1 x2 y 2
S: z
³ S ³ E ˜ N dS
g x, y
³R ³ E ˜
g x x, y i g y x, y j k dA
§
xi yj 2 zk ˜ ¨
¨
©
³R ³
§
x2
©
1 x y
³ R ³ ¨¨
2
³R ³
2
x
1 x y
2
1 x y
2
2
·
j k ¸ dA
¸
1 x y
¹
y
i 2
2
2
2
1 x y
2
2
2 x2 y 2
³R ³
dA
2
·
2 z ¸ dA
¸
¹
y2
x y 2 1 x y2
2
1 x y
2
2S
1
³0 ³0
dA
2
2 r2
8S
3
r dr dT
1 r2
x2 y 2 , 0 d z d a
33. z
m
³S
Iz
³S ³ k
³ k dS
k³
r
m
Iz
2³
2³
³
·
§
¸ ¨
¨
x 2 y 2 ¸¹
©
x
³R ³ k
x2 y2
2
·
¸ dA
x 2 y 2 ¸¹
y
2k ³
2 dA
2S
0
k³
a
³0 r
³
R
2kS a 4
2
2ka 4
2S
4
dr dT
3
2 kS a 2
2 dA
a2
2
2kS a 2
a 2m
2
a2
a2 x2 y 2
³ k dS
S
S
³k
ª
2ka « r 2
¬
35. x 2 y 2
2k ³
R
³
2k ³
R
³
§
1¨
¨
©
a
a x y
2k ³
r a2 r 2
a2 r 2 1 2
2
·
§
¸ ¨
2
2
2 ¸
¨
a x y ¹
©
x
2
x 2 y 2 dS
r 2 , dv
Let u
R
dA
2
2
³
x2 y2
dr , du
2 2
a r2
3
2ka ³
a
» 2S
¼0
r
a
³0
a r
2
a
a x y
2
2
2
2
2ka ª
¬
dr dT
2ka ³
2S
a
a 2 r 2 º 2S
¼0
r3
a
³0
a r2
2 2
a 4S ka 2
3
2 2
a m
3
dA
0
2
4S ka 2
dr dT use integration by parts
a2 r 2 .
2r dr , v
3 2º
2S
0
2
·
¸ dA
2
2
2 ¸
a x y ¹
y
§2 ·
2ka¨ a 3 ¸ 2S
©3 ¹
a2 , 0 d z d h
U x, y , z
r
R
2
§
1 ¨
¨
©
x 2 y 2 dS
34. x 2 y 2 z 2
y
425
xi yj 2 zk
32. E
z
Surface Integrals
z
1
h
a2 x2
Project the solid onto the xz-plane.
Iz
4³
S³
4a 3 ³
x 2 y 2 1 dS
h
a x
2
h
a
0 ³0
§
2
2 º
ª 2
¬x a x ¼ 1 ¨
©
a
1
a
0 ³0
4³
2
dx dz
hª
xº
4a 3 ³ «arcsin » dz
0
a ¼0
¬
2
·
¸ 0
2
2
a x ¹
x
§S ·
4a 3 ¨ ¸ h
©2¹
2
a
dx dz
a
y
x
2S a 3h
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
426
36. z
Vector Analysis
x2 y 2 , 0 d z d h
z
Project the solid onto the xy-plane.
Iz
³S ³
2S
³0 ³0
h
1 4h
60
37. S : z
h
³ h ³
x 2 y 2 1 dS
r 2 1 4r 2 r dr dT
32
S
h x2
x2 y2
h x2
1 4 x 2 4 y 2 dy dx
ªh
2S « 1 4h
¬12
S
32
S ª
1 4h
60 ¬
ª¬10h 1 4h º¼ 60
1
1 4h
120
2S
» 120
¼
5 2º
y
6h 1 1º
¼
32
x
16 x 2 y 2 , z t 0
F x, y , z
0.5 zk
³ S ³ UF ˜ N dS
³ R ³ UF ˜
0.5 U ³
2S
0
³ R ³ 0.5U zk ˜
g x x, y i g y x, y j k dA
³ R ³ 0.5U z dA
4
³0
³ R ³ 0.5U 16 x
2
y 2 dA
0.5U ³
16 r 2 r dr dT
2 xi 2 yj k dA
2S
0
64 dT
64SU
16 x 2 y 2
38. S : z
F x, y , z
0.5 zk
³ S ³ UF ˜ N dS
³ R ³ UF ˜
g x x, y i g y x, y j k dA
ª
³ R ³ 0.5U zk ˜ ««
x
¬ 16 x y
³ R ³ 0.5 U z dA
0.5 U ³
2S
0
4
³0
2
³R ³ 0.5U
2
i n
x, y, z dS
lim
' o0
¦f
º
j k » dA
»¼
16 x y
y
2
2
16 x 2 y 2 dA
16 r 2 r dr dT
0.5 U ³
39. The surface integral of f over a surface S, where S is
g x, y , is defined as
given by z
³S ³ f
h
2S
0
64
dT
3
64SU
3
z
43. (a)
4
−6
−6
xi , yi , zi 'Si . (page 1112)
i 1
See Theorem 15.10, page 1112.
40. A surface is orientable if a unit normal vector N can be
defined at every nonboundary point of S in such a way
that the normal vectors vary continuously over the
surface S.
41. See the definition, page 1118.
x
6
6
−4
y
(b) If a normal vector at a point P on the surface is
moved around the Möbius strip once, it will point in
the opposite direction.
4 cos 2u i 4 sin 2u j
(c) r u , 0
This is circle.
z
See Theorem 15.11, page 1118.
4
42. Orientable
−2
2
2
x
y
−4
(d) (construction)
(e) You obtain a strip with a double twist and twice as
long as the original Möbius strip.
© 2010 Brooks/Cole, Cengage Learning
Section 15.7
44. (a) ru
rv
Divergence Theorem
427
i j 2uk
2vi j
ru u rv
i
j
k
1
1
2u
2v 1
2ui 4uvj 1 2v k
0
ru u rv is a normal vector to the surface.
u 2i u v 2 j u v k
(b) F u , v
F ˜ ru u rv
(c) x
3
y
1
z
4
2u 3 4uv u v 2 u v 1 2v
u v2 ½
°
u v ¾
°
u2
¿
1
4i 8 j 3k
ru u rv
(e)
v
2 not in domain
4i 3 j k
ru u rv 2, 1
F˜
2 u
ru u rv
at P.
ru u rv
(d) Calculate F ˜
F 3, 1, 4
u
2u 3 4u 2v 4uv3 v u 2v 2 2uv
89
ru u rv
ru u rv
1
16 24 3
89
³ S ³ F ˜ N dS
³R ³ F ˜
1
2
³ 1 ³ 0
37
89
37 89
89
ru u rv dA
2u 3 4u 2v 4uv 3 v u 2v 2 2uv du dv
1
§
³ 1 ¨©8v
3
4v 2 26v
·
6 ¸ dv
3
¹
44
3
Section 15.7 Divergence Theorem
1. Surface Integral: There are six surfaces to the cube, each with dS
z
0,
N
k ,
F˜N
z2,
³ S1 ³ 0 dA
z
a,
N
k,
F˜N
z2,
³ S2 ³ a
x
0,
N
i ,
F˜N
2 x ,
³ S3 ³ 0 dA
x
a,
N
i,
F˜N
2 x,
³ S4 ³ 2a dy dz
y
0,
N
j,
F˜N
2 y,
³ S5 ³ 0 dA
y
a,
N
j,
F˜N
2 y ,
³ S6 ³ 2a dA
So,
³ s ³ F ˜ N dS
a 4 2a 3 2a 3
Divergence Theorem: Because div F
³³³ div F dV
Q
a
a
a
³ 0 ³ 0 ³ 0 2 z dz dy dx
2
1 dA.
0
a
a
³0 ³0 a
dA
2
dx dy
a4
0
a
a
³ 0 ³ 0 2a dy dz
2a 3
0
a
a
³ 0 ³ 0 2a dz dx
2 a 3
a4.
2 z , the Divergence Theorem yields
a
a
³0 ³0 a
2
dy dx
a4.
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
428
Vector Analysis
2. Surface Integral: There are three surfaces to the cylinder.
Bottom: z
0, N
³ S1 ³ 0 dS
0
Top: z
³ S2 ³ h
k, F ˜ N
h, N
2
ru u rv
z2
4S h 2
2 cos ui 2 sin uj vk , 0 d u d 2S , 0 d v d h
2 sin ui 2 cos uj, rv
ru
z2
h 2 Area of circle
dS
Side: r u ,v
k
2 cos ui 2 sin uj
F ˜ ru u rv
8 cos 2 u 8 sin 2 u
2S
h
³ S3 ³ F ˜ N dS
So,
k , F ˜ N
³0 ³0
³ s ³ F ˜ N dS
8 cos 2 u 8 sin 2 u du dv
0 4S h 2 0
2S
h
2
³ 0 ³ 0 ³ 0 2 zr dz dr dT
Q
0
h
4S h 2 .
2 2 2z
Divergence Theorem: div F
³ ³ ³ 2 z dV
z
2z
4S h .
2
2
x
y
3. Surface Integral: There are four surfaces to this solid.
z
0, N
³ S1 ³ 0 dS
y
2 y z , dS
6 z
³0
³0 ³0
0, N
dx dz
z 2 6 z dz
36
y 2 x, dS
62y
³0
³0 ³0
x 2y z
6
z dx dz
dA
i , F ˜ N
3
³ S3 ³ y dS
So,
j, F ˜ N
6
³ S2 ³ z dS
³ S4 ³
z
0
0, N
x
k , F ˜ N
6, N
y dz dy
3
dz dy
6 y 2 y 2 dy
i 2j k
,F ˜ N
6
2 x 5 y 3z dz dy
³ s ³ F ˜ N dS
3
dA
62y
³0 ³0
0 36 9 45
Divergence Theorem: Because div F
9
2 x 5 y 3z
, dS
6
18 x 11y dx dy
6 dA
z
3
³0
90 90 y 20 y 2 dy
45
6
18.
1, you have
3
³³³ dV
Q
Volume of solid
1
Area of base u Height
3
1
9 6
3
y
6
18.
x
© 2010 Brooks/Cole, Cengage Learning
Section 15.7
Divergence Theorem
429
xyi zj x y k
4. F x, y, z
S: surface bounded by the planes y
4 x and the coordinate planes
4, z
Surface Integral: There are five surfaces to this solid.
z
k , F ˜ N
0, N
³ S1 ³ y
4
32
3
1
> xy x y@, dS
2
4
4
³0 ³0
xy x y dy dx
32 32
0 128
3
3
64 4 x
4
32
3
dx
0
³0 ³0 ³0
2 dA
128
64.
y , you have
Divergence Theorem: Because div F
Q
64
xy
1
> xy x y@ 2 dA
2
4
dx
2
2
i k
, F˜N
2
³³³ div F dV
4 x 8 dx
2
2
0
4 x
4
³0
z dz dx
4
³S ³ F ˜ N dS
4
0
z
³ 0 ³ 0 0 dS
4, N
4 x
4
³
z dz dx
i, F ˜ N
0, N
³ S4 ³ xy dS
So,
4 x
4
³0 ³0
³
x y dy dx
z
j, F ˜ N
³ S3 ³ z dS
³ S5 ³
4 x
4
³0 ³0
4, N
x z
4
j, F ˜ N
³ S2 ³ z dS
x
4
³0 ³0 x y dS
0, N
y
x y
y dz dy dx
64.
xzi yzj 2 z 2k
5. F x, y, z
Surface Integral: There are two surfaces.
Bottom: z
k , F ˜ N
0, N
³ S1 ³ F ˜ N dS
³ R³ 2z
2
2 z 2
³³ 0 dA
dA
0
z
Side: Outward unit normal is
N
F˜N
2 xi 2 yj k
1
4x2 4 y 2 1
−1
1
ª¬2 x z 2 y z 2 z º¼
2
4x 4 y 1
³ S2 ³ F ˜ N dS
2
2
³ S2 ³ ª¬2 x
2S
2
³ 0 ³ 0 ª¬«2r
1
2
³³³ div F dV
Q
2S
1
2S
1
³0 ³0 ³0
2
1
1
x
y
−1
y 2 z 2 z 2 º¼ dA
2
1 r 2 2 1 r 2 º r dr dT
¼»
z z 4z
Divergence Theorem: div F
1 r 2
2
2S
1
³0 ³0
2S
1
dT
2
S
ª3 3 1º
« 2 2 2 » dT
¬
¼
S
2r 2r 3 dr dT
³0
6z
6 z r dz dr dT
³0 ³0 3 1 r
2
2
r dr dT
2S
1
³0 ³0
3 6r 2 3r 4 r dr dT
2S
³0
© 2010 Brooks/Cole, Cengage Learning
430
Chapter 15
Vector Analysis
xy 2i yx 2 j ek
6. F x, y, z
x 2 y 2 and z
S: Surface bounded by z
4
Surface Integral: There are two surfaces.
Top: z
4, N
k, F ˜ N
e
³ S1 ³ F ˜ N dS
Side: z
³ S2
16S e
area circle e
³ F ˜ N dS
x
x2 y 2 , g x
g ( x, y )
ª
«
«¬
³ S2 ³
x2 y 2
x2 y 2
x2 y 2
x2 y2
º
e» dA
»¼
x2 y 2
y
, gy
x2 y2
³0 ³0
2S
³0
So,
7. Because div F
³³³ div F dV
Q
2S
4
4
³0 ³0 ³r
³³³ div F dV
Q
a
a
a
a
a
³0 ³0 ³0
Q
a
a
a
2a 2 a
2S
a
2S
S /2
2S
³0
256
dz
5
512S
5
a
ª¬a 2 x 2 2a 3 xº¼
0
2a 2 x 2a 3 dx
3a 4 .
a
a
1 a6
3
2S
S /2
a2 x2
2 xa 3
3
2a 3 xya dy dx
2a 3 34 a 5 .
2 U sin I cos T U sin I sin T U cos I U 2 sin I dI dT d U
1 U sin T cos T dT d U
2
a2 x2
a
³0 ³0
2 U 5 sin T cos T sin 3 I cos I dI dT d U
y z y
³0 ³0
Q
a
³0
dx
³0 ³0 ³0
Q
2S
³³³ 3 dV
³0
r5 º
4
«r » dz
5 ¼0
¬
2 xyz ,
a
³³³ 2 xyz dV
a
11. Because div F
3 xa 3
2
2 x 2 x 2 xyz
³a ³
Q
4
2S ª
4r 3 r 4 dr dz
2 xz 2 2 3 xy dz dy dx
2 xa 4
3
³0 ³0
³³³ div F dV
³0 ³0
2ax 2ay a 2 dy dx
³0 ³0 ³0
10. Because div F
4
2 x 2 y 2 z dz dy dx
³0 ³0 ³0
a
³³³ div F dV
2S
r 2 r dz dr dT
2 xz 2 2 3 xy, you have
³0
9. Because div F
§ 512
·
16e ¸ S
¨
© 5
¹
2 x 2 y 2 z , you have
³0 ³0
8. Because div F
ª 2048 sin 2 T cos 2 T
º
8e» dT
«
5
¬
¼
y2 x2
Divergence Theorem: div F
Q
¨
©
512
S.
5
³ S ³ F ˜ N dS
³³³ div F dV
·
cos 2 T r 2 sin 2 T
e ¸ r dr dT
r
¹
4 § 2r 2
2S
a ª§
³ 0 ««¨©
U 5 · sin 2 T º
¸
¬ 2 ¹
2S
» dU
¼» 0
2
0.
z , you have
³0
a2 x2 y2
ª a 2r
r3 º
» dr dT
«
2¼
¬ 2
z dz dy dx
2S
³0
2S
a
³0 ³0 ³0
a
a2 r 2
ª a 2r 2
r4 º
» dT
«
8 ¼0
¬ 4
zr dz dr dT
2S
³0
a4
dT
8
S a4
4
.
3, you have
3 Volume of Sphere
3ª¬ 43 S 33 º¼
108S .
© 2010 Brooks/Cole, Cengage Learning
Section 15.7
5
4 y2
2
³ 0 ³ 2 ³ Q
7
25 y 2
5
³³³ 2 y dV ³ 0 ³ 5 ³ 25 y 2
Q
2S
256 x 2
16
2S
³0
8
6
³³³
4 y
6
4
2
4 y2
0.
7 ª 4
³ 0 «¬ 3
25 y 2 dy dz
25 y 2
3/2 º
5
» dz
¼ 5
0.
4 y
³ 1/2
x2 y 2
x 2 y 2 e z dz dy dx
2S
100e8 dT
16
³0 ³0
r 2 e z r dz dr dT
262,104
S
5
8r 3 re8 12 r 4 re r /2 dr dT
200e8S .
4 x 2 dz dy dx
6
4
³ 0 ³ 0 4x
2
4 y dy dx
6
³ 0 32 x
2
dx
2304.
3e z , you have
3e z dz dy dx
6
4
³ 0 ³ 0 3ª¬e
4 y
1º¼ dy dx
6
³0 3 e
4
5 dx
18 e 4 5 .
y 4 x. Use spherical coordinates.
y 4 x dV
Q
4
S
2S
4
S
2S
4
S
³0 ³0 ³0
U sin I sin T U sin I cos T 4 U 2 sin I dT dI d U
³0 ³0 ³0
³ 0 ³ 0 8SU
18. div F
5
³ 0 ³ 2 0 dy dz
dy dz
4 x 2 , you have
ez ez ez
³0 ³0 ³0
dV
Q
17. div F
4
³0 ³0 ³0
16. Because div F
z
131,052
5
3x 2 x 2 0
15. Because div F
³³³ 3e
5
8
256 x 2
³ 0 ³ 0 ³ r /2
Q
7
³ 0 ³ 5 4 y
2 y dx dy dz
16
Q
dV
4 y2
2 y , you have
³0 ³
x 2 y 2 e z dV
2
³0
ª 2x2 º
³ 2 «¬ 2 »¼
2
y 2 x 2 e z , you have
14. Because div F
³³³ 4 x
5
xz dx dy dz
4 y2
1 2y 1
13. Because div F
³³³
431
xz , you have
12. Because div F
³³³ xz dV
Divergence Theorem
U 3 sin 2 I sin T U 3 sin 2 I cos T 4 U 2 sin I dT dI d U
2
4
³ 0 16SU
sin I dI d U
2
dU
1024S
3
2
³ S ³ F ˜ N dS
³³³ div F dV
³³³ 2 dV .
Q
Q
The surface S is the upper half of a hemisphere of radius 2. Because the volume is
³ S ³ F ˜ N dS
1 4S
2 3
23
16S /3, you have
32S
.
3
2 Volume
19. Using the Divergence Theorem, you have
³ S ³ curl F ˜ N dS
³³³ div
curl F dV
Q
curl F x, y, z
i
j
k
w
wx
w
wy
w
wz
4 xy z 2
div curl F
So,
³³³ div
6 yi 2 z 2 z j 4 x 4 x k
6 yi
2 x 2 6 yz 2 xz
0.
curl F dV
0.
Q
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
432
Vector Analysis
20. Using the Divergence Theorem, you have
³ S ³curl F ˜ N dS
³³³ div
curl F dV
Q
i
j
k
w
wx
w
wy
w
wz
xy cos z
yz sin x
xyz
curl F x, y, z
z y cos x z x sin z y cos x x sin z
Now, div curl F x, y, z
³ S ³ curl F ˜ N dS
xz y sin x i yz xy sin z j yz cos x x cos z k.
³³³ div
curl F dV
0. So,
0.
Q
21. See Theorem 15.12.
22. If div F x, y, z ! 0, then source.
If div F x, y, z 0, then sink.
If div F x, y, z
0, then incompressible.
23. (a) Using the triple integral to find volume, you need F so that
wM
wN
wP
div F
1.
wx
wy
wz
xi, F
So, you could have F
yj, or F
zk.
For dA
dy dz consider F
xi, x
f ( y, z ), then N
For dA
dz dx consider F
yj, y
f ( x, z ), then N
For dA
dx dy consider F
zk , z
Correspondingly, you then have V
(b) v
a
a
³ 0 ³ 0 x dy dz
Similarly,
a
a
³0 ³0
a
a
³ 0 ³ 0 a dy dz
a
f ( x, y ), then N
³ S ³ F ˜ N dS
a
³0 a
a
³ 0 ³ 0 z dx dy
y dz dx
2
dz
i f y j f zk
1 f y2 f z2
f xi j f zk
1 f x2 f z2
f xi f y j k
1 f x2 f y2
³ S ³ x dy dz
and dS
1 f y2 f z2 dy dz.
and dS
1 f x2 f z2 dz dx.
and dS
³ S ³ y dz dx
1 f x2 f y2 dx dy.
³ S ³ z dx dy.
a3
a3.
xi yj zk
24. F x, y, z
111
Divergence Theorem: div F
³³³ 3 dV
3 Volume of cube
3
3
Q
Surface Integral: There are six surfaces.
x
0: N
i,
F˜N
x,
³S1 ³ 0 dS
0
x
1:
i,
F˜N
x,
³S2 ³ 1 dS
1
y
0: N
j,
F˜N
y,
³S3 ³ 0 dS
0
y
1:
j,
F˜N
y,
³S4 ³ 1 dS
1
z
0: N
k , F ˜ N
z,
³S5 ³ 0 dS
0
z
1:
k,
z,
³S6 ³ 1 dS
1
So,
N
N
N
³ S ³ F ˜ N dS
F˜N
111
3.
© 2010 Brooks/Cole, Cengage Learning
Section 15.8
25. Using the Divergence Theorem, you have
³³³ div
433
curl F dV . Let
Q
Mi Nj Pk
F x, y , z
§ wP
§ wN
wN ·
wM ·
wM ·
§ wP
¨
¸i ¨
¸k
¸j ¨
w
y
w
z
w
x
w
z
w
x
wy ¹
©
¹
©
¹
©
curl F
w2P
w2 N
w2P
w 2M
w2N
w 2M
wxwy
wxwz
wywx
wywz
wzwx
wzwy
div curl F
³ S ³ curl F ˜ N dS
So,
³ S ³ curl F ˜ N dS
Stokes’s Theorem
³³³ 0 dV
0.
Q
a1i a2 j a3k , then div F
26. If F x, y, z
0.
0.
So,
³ S ³ F ˜ N dS
³³³ div F dV
Q
³ S ³ F ˜ N dS
³³³ div F dV
Q
30.
³ S ³ F ˜ N dS
³ S ³ f DN g dS
³S ³
3.
³³³ 3 dV
3V .
Q
xi yj zk , then div F
28. If F x, y, z
29.
0.
Q
xi yj zk , then div F
27. If F x, y, z
1
F
³³³ 0 dV
1
F
1
F
³³³ div F dV
Q
³ S ³ f ’g ˜ N dS
3.
Q
³³³ div
f ’g dV
Q
³ S ³ f DN g dS ³ S ³ gDN f
fDN g gDN f dS
³³³
3
F
³³³ 3 dV
Q
³³³
³³³
f div’g ’f ˜ ’g dV
Q
f ’ 2 g ’f ˜ ’g dV
Q
dS
f ’ 2 g ’f ˜ ’g dV Q
³³³ dV
³³³
³³³
g’ 2 f ’g ˜ ’f dV
Q
f ’ 2 g g’ 2 f dV
Q
Section 15.8 Stokes's Theorem
2 y z i e z j xyzk
1. F x, y, z
i
j
k
w
wx
w
wy
w
wz
2y z
ez
xyz
curl F
xz e z i yz 1 j 2k
curl F
curl F
4. F x, y, z
x 2 i y 2 j x 2k
2. F x, y, z
i
j
k
w
wx
w
wy
w
wz
x2
y2
x2
curl F
2xj
2 zi 4 x 2 j arctan xk
3. F x, y, z
i
j
k
w
wx
w
wy
w
wz
2z
4 x 2
arctan x
1 ·
§
¨2 ¸ j 8 xk
1 x2 ¹
©
x sin yi y cos xj yz 2k
i
j
k
w
wx
w
wy
w
wz
x sin y y cos x
yz 2
z 2i y sin x x cos y k
© 2010 Brooks/Cole, Cengage Learning
434
Chapter 15
Vector Analysis
ex
5. F x, y, z
curl F
e
2 y2
i ey
2 z2
j
k
w
wx
w
wy
w
wz
e
xz 2 ze
z x 2e
y2 z2
x
y2 z2
i yzj 2 ye
x2 y2
x2 y 2
³ C F ˜ dr ³ 0
k
2S
³0
k
j
k
w
wx
w
wy
w
wz
1 x2
y z, N
x z, P
cos t , y
’F
’F
2k
1
³C
2 area circle
18S
x y and C is the circle x 2 y 2
sin t dt , dy
cos t dt and
sin 2 t cos 2 t dt
2S .
2 Area of circle of radius 1
2S .
y
1, z
0, dz
³ C y dx ³ C y dx 0.
x dy
x dy
2S
³0
xi yj zk.
x2 y 2 z 2
2 x
2z
2k , you have
4
C
x 2 y 2 z 2 1.
2 xi 2 yj 2 zk
Now, because curl F
4
x
y z dx x z dy x y dz
1 x2 y 2 , zx
Because z 2
³ R ³ 2 dA
2
sin t , you have dx
2
2 y
1
Double Integral: Consider F x, y , z
Then N
2 x , g y
y2
º
»k
1 y 2 »¼
Letting x
18S
9 dt
z
ª
x
2 yi «
«¬ 1 x 2
³ C F ˜ dr
3 cos t dt
9
º
»k
1 y »¼
Line Integral:
3 sin t , dy
ª¬ 3sin t 3sin t 3cos t 3cos t º¼ dt
³ S ³ curl F ˜ N dS
ª x
2 yi «
«¬ 1 x 2
8. In this case, M
x dy
Double Integral:
g x, y
9 x2 y2 , g x
1 x 2 j y 2k
i
arcsin y
0
3 sin t dt , y
3 cos t , dx
curl F
curl F
0, dz
³ C y dx 2S
i yzj 2 ye
arcsin yi 6. F x, y, z
³ C F ˜ dr
xyz
y2 z2
z
9,
Line Integral:
i
x2 y2
7. C : x 2 y 2
j xyzk
x
, and z y
z
³S ³
y
, dS
z
1
x2
y2
2 dA
2
z
z
§1·
³ R ³ 2 z¨© z ¸¹ dA ³ R ³ 2 dA
curl F ˜ N dS
1
dA.
z
9. Line Integral:
From the figure you see that
C1 : z
0, dz
0
C2 : x
0, dx
0
C3 : y
0, dy
0
2
0
12
³ C F ˜ dr ³ C xyz dx y dy z dz ³ C1 y dy ³ C2 y dy zdz ³ C3 z dz ³ 0 y dy ³ 2 y dy ³ 0
³S ³
12 6 x 6 y
curl F ˜ N dS
0
³ 12 z dz
xyj xzk
Double Integral: curl F
Letting z
z dz g x, y , g x
2
2 x
³0 ³0
2
z
6
gy.
curl F ˜ >6i 6 j k @ dA
³R ³
³ 0 ª¬6 xy
12
³R ³
ª¬6 xy x 12 6 x 6 y º¼ dy dx
2
0
12 xy 6 x 2 yº¼
2 x
0
(0, 0, 12)
6 xy xz dA
2
2 x
³0 ³0
C3
C2
12 xy 12 x 6 x 2 dy dx
(2, 0, 0)
dx
0
x
4
(0, 2, 0)
C1
4
y
© 2010 Brooks/Cole, Cengage Learning
Section 15.8
y2 , x
0, dx
C3 : y
a, z
a 2 , dy
C4 : z
2
y ,x
0, dz
a, dx
a2
C3
2 y dy
dz
0
0, dz
435
z
10. Line Integral: From the figure you see that
C1 : y
0, z
0, dy
dz
0
C2 : z
Stokes's Theorem
C2
C1
2 y dy.
a
a
C4
x
y
So,
³ C F ˜ dr
³C z
³ C1 0 dx ³ C2 2 y
dx x 2 dy y 2 dz
2
0
³a 2y
3
0
³a a
dy 4
dx a
³0
2 yj k
1 4 y2
4
dx a
0
³ C4 ª¬a
2
dy 2 y 3 dyº¼
a5 a3
a3 1 a .
0, you have
1 4 y 2 dA.
and dS
Furthermore, curl F
2 yi 2 zj 2 xk. So,
³S ³
³R ³
curl F ˜ N dS
³ C3 a
ª¬a 4 xº¼ ª¬a 2 yº¼
0
a
a 2 2 y 3 dy
Double Integral: Because S is given by y 2 z
N
dy 3
a
³0
4 yz 2 x dA
a
a
³0 ³0
a
ª¬ a 4 x ax 2 º¼
0
a 4 2ax dx
a
a
³0 ³0
4 yz 2 x dA
a5 a3
4 y 3 2 x dydx
a3 1 a 2 .
11. These three points have equation:
x y z
2.
i jk
Normal vector: N
curl F
3i j 2k
³ S ³ curl F ˜ N dS
³R ³
6 dA
6 area of triangle in xy -plane
6 2
12. Let A
0, 0, 0 , B
UuV
UuV
N
2i 2 j
2 2
z
i
j
k
w
wx
w
wy
w
wz
z2
2x
y2
G x, y
JJJK
AB
JJJK
AC
i j k , and V
2k , and
2 dA. Because curl F
2x
k , you have
x2 y 2
³S ³
curl F ˜ N dS
³R ³ 0 dS
0.
2 yi 2 zj 2k
1 x 2 y 2 , Gx
³ S ³ curl F ˜ N dS
0, 0, 2 . Then U
i j
.
2
x y and dS
So, F x, y, z
13. curl F
1, 1, 1 , and C
12
³R ³
1
1
2 y
³ R ³ ª¬4 xy 4 y 1 x
2 yi 2 zj 2k ˜ 2 xi 2 yj k dA
³ 1 ³ ³ 1 4
2 x , G y
1 x 2
1 x2
2
y 2 2º¼ dA
ª¬4 xy 4 y 4 x 2 y 4 y 3 2º¼ dy dx
1 x 2 dx
1
2 ªarcsin x x 1 x 2 º
¬
¼ 1
2S
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
436
4 xzi yj 4 xyk , S : 9 x 2 y 2 , z d 0
14. F x, y, z
curl F
4 xi 4 x 4 y j
x2 y 2 z 9
G x, y , z
’G x , y , z
³S ³
Vector Analysis
2 xi 2 yj k
³ R ³ ª¬8 x
curl F ˜ N dS
3
³ 3 16 x
15. curl F
z
i
j
k
w
wx
w
wy
w
wz
z2
y
z
³ S ³ curl F ˜ N
³R ³
§
2 zj ˜ ¨
¨
©
x
4 x y
2
4 x2 y 2
j
k
w
wx
w
wy
w
wz
x2
z2
xyz
’G x , y , z
³S ³
x
2 yz
i
G x, y , z
z 32
9 x2
4 x2 y2
2
9 x2
dx
ª¬8 x 2 8 xy 8 y 2 º¼ dy dx
0
y
, Gy
4 x2 y 2
·
j k ¸ dA
¸
4 x y
¹
y
i ³R ³
dA
x 2i z 2 j xyzk , S : z
16. F x, y, z
16
3
9 x2
2 zj
³R ³
curl F
9 x2 2
4 x 2 y 2 , Gx
G x, y
3
³ 3 ³ 2 y 4 x 4 y º¼ dA
2
2
2
4 x2 y2
2y
4 x2 y 2
dA
2
³ 2 ³ 4 x2
4 x2
2 y dy dx
0
4 x2 y2
xz 2 z i yzj
4 x2 y2
x
4 x2 y 2
curl F ˜ N dS
i y
4 x2 y 2
ª z x 2 x
³ R ³ ««
¬
4 x2 y 2
³ R ³ ª¬ x x 2
ª 2
y3 º
³ 2 «¬ x y 2 xy 3 »¼
2
ª
2
ª 8
2
³ 2 «¬ 3 x
2
º
» dA
4 x 2 y 2 »¼
y2 z
2
³ 2 ³ y 2 º¼ dA
2
³ 2 «¬2 x
j k
4 x2
x 2 2 x y 2 dy dx
4 x2
4 x2
dx
4 x2
4 x2 4x 4 x2 2
4 x2
3
4 x2 4 x 4 x2 8
3
ª 8 § 1 ·ª
2
« 3 ¨ 8 ¸ « x 2 x 4
©
¹¬
¬
º
4 x 2 » dx
¼
º
4 x 2 » dx
¼
xº 4
4 x 2 16 arcsin » 4 x 2
2¼ 3
4
1
4
ª§ 1 ·
º
«¨ 3 ¸ 8S 3 2S 3 8S 3 2S »
¬© ¹
¼
32
2
8 § 1 ·ª
x ºº
2
¨ ¸ x 4 x 4 arcsin » »
3 © 2 ¹¬«
2 ¼ ¼ 2
0
© 2010 Brooks/Cole, Cengage Learning
Section 15.8
i
w
wx
j
w
wy
k
w
wz
1 2 ln x 2 y 2
arctan x y
1
curl F
’G x , y , z
2i 3 j k
³ R ³ x2
curl F ˜ N dS
S 2
4 sin T cos T
i
j
k
w
wx
w
wy
w
wz
yz
2 3y
x2 y2
z G x, y , z
16 x
x
’G x, y, z
16 x 2
2 yi y 2 x j zk
ª
³0 ³0
4
³0
16 over x 2 y 2
16
2 xy
16 x
16 x2
2
º
z » dA
¼
ª 2 xy
«
¬ 16 x 2
ª
³ R ³ «¬
2 xy
16 x
º
16 x 2 » dy dx
¼
ª x 16 x 2 16 x 2 º dx
¬
¼
º
16 x 2 » dA
¼
2
4
³0
ª 1
2
« 16 x
3
¬
ª
x
y2 «
2
16
x
¬
4
32
16 x x3 º
»
3 ¼0
º
16 x 2 y»
¼0
16 x 2
dx
64 · § 64 ·
§
¨ 64 ¸ ¨ ¸
3¹ © 3¹
©
64
3
xyj xzk
x 2 , Gx
G x, y
³ S ³ curl F ˜ N
³R ³
2 x, G y
0
a
³ R ³ xz dA
xyj xzk ˜ 2 xi k dA
a
³0 ³0 x x
2
dy dx
a5
4
xyzi yj zk
20. F x, y, z
i
j
k
w
wx
xyz
w
wy
y
w
wz
z
xyj xzk
S: the first octant portion of z
³S ³
8
3
i k
³ R ³ «¬
curl F ˜ N dS
curl F
³0
S 2
ª8 sin 3 T º
«
»
¬ 3 ¼0
8 sin 2 T cos T dT
2
4
z
2r sin T
r dr dT
r2
S 2
2 sin T dr dT
S : the first octant portion of x 2 z 2
19. curl F
2 sin 2T
³0 ³0
yzi 2 3 y j x 2 y 2 k
18. F x, y, z
³S ³
S 2
2y
dA
y2
³0 ³0
curl F
ª 2y º
« x 2 y 2 »k
¬
¼
2x 3 y z 9
G x, y , z
³S ³
ª
1y
y º
«
»k
2
2
2
x y2 »
«¬1 x y
¼
2 sin 2T in the first octant.
9 2 x 3 y over one petal of r
S: z
437
x
x 2 y 2 i arctan j k
y
ln
17. F x, y, z
Stokes's Theorem
curl F ˜ N dS
x 2 over x 2 y 2
³ R ³ xz dA
a
³0 x
3
³R ³ x
a 2 x 2 dx
3
dA
a 2 . You have N
a
³0 ³0
a 2 x2
ª 1 2 2
2
« 3 x a x
¬
2 xi k
1 4 x2
and dS
1 4 x 2 dA.
x3 dy dx
32
2 2
a x2
15
5 2º
a
»
¼0
2 5
a
15
© 2010 Brooks/Cole, Cengage Learning
438
Chapter 15
i j 2k
21. F x, y, z
curl F
Letting N
Vector Analysis
S: x y
2
i
j
k
curl F
w
wx
w
wy
w
wz
z
0
y
2
i
j
k
w
wx
w
wy
w
wz
1
1
2
zi yk
22. F x, y, z
0
³S ³
k , you have
curl F ˜ N dS
0.
1
Letting
N
k , curl F ˜ N
i j
0 and ³
S
³
curl F ˜ N dS
0.
23. See Theorem 15.13.
24. curl F measures the rotational tendency. See page 1135.
25. (a)
³ S ³ curl> f ’g @ ˜ N dS (Stokes's Theorem)
³ C f ’ g ˜ dr
f ’g
f
wg
wg
wg
i f
j f k
wx
wy
wz
i
j
k
w
wx
w
wy
w
wz
f wg wx
f wg wy
f wg wz
curl f ’g
ªª § w 2 g · § wf ·§ wg ·º ª § w 2 g · § wf ·§ wg ·ºº
«« f ¨
¸ ¨ ¸¨ ¸» « f ¨
¸ ¨ ¸¨ ¸»» i
¬«¬ © wywz ¹ © wy ¹© wz ¹¼ ¬ © wzwy ¹ © wz ¹© wy ¹¼¼»
ªª § w 2 g · § wf ·§ wg ·º ª § w 2 g · § wf ·§ wg ·ºº
«« f ¨
¸ ¨ ¸¨ ¸» « f ¨
¸ ¨ ¸¨ ¸»» j
¬«¬ © wxwz ¹ © wx ¹© wz ¹¼ ¬ © wzwx ¹ © wz ¹© wx ¹¼»¼
ªª § w 2 g · § wf ·§ wg ·º ª § w 2 g · § wf ·§ wg ·º º
«« f ¨
¸ ¨ ¸¨ ¸» « f ¨
¸ ¨ ¸¨ ¸» » k
«¬¬ © wxwy ¹ © wx ¹© wy ¹¼ ¬ © wywx ¹ © wy ¹© wx ¹¼ »¼
ª§ wf ·§ wg · § wf ·§ wg ·º
ª§ wf ·§ wg · § wf ·§ wg ·º
ª§ wf ·§ wg · § wf ·§ wg ·º
«¨ ¸¨ ¸ ¨ ¸¨ ¸» i «¨ ¸¨ ¸ ¨ ¸¨ ¸» j «¨ ¸¨ ¸ ¨ ¸¨ ¸»k
¬© wx ¹© wz ¹ © wz ¹© wx ¹¼
¬© wx ¹© wy ¹ © wy ¹© wx ¹¼
¬© wy ¹© wz ¹ © wz ¹© wy ¹¼
(b)
i
j
k
wf
wx
wf
wy
wf
wz
wg
wx
wg
wy
wg
wz
So,
³ C f ’ g ˜ dr
³C
f ’f ˜ dr
’ f u ’g
³ S ³ curl > f ’g @ ˜ N dS
³S ³
³C
f ’g g’f ˜ dr
u ’g @ ˜ N dS .
’f u ’f ˜ N dS using part a
0 because ’f u ’f
(c)
³ S ³ >’f
³C
0.
f ’g ˜ dr ³C
g ’ f ˜ dr
³S ³
’f u ’g ˜ N dS ³S ³
³S ³
’f u ’g ˜ N dS ³ S ³ ’f
’g u ’f ˜ N dS
u ’g ˜ N dS
using part a
0
© 2010 Brooks/Cole, Cengage Learning
Section 15.8
26. f x, y, z
xyz , g x, y, z
(a) ’g x, y, z
³ C ¬ª f
4 x2 y 2
z, S : z
xyzk
0 d t d 2S
2 cos ti 2 sin tj 0k ,
x, y, z ’g x, y, z º¼ ˜ dr
(b) ’f x, y, z
0
yzi xzj xyk
’g x , y , z
k
’f u ’ g
i
j
k
yz
xz
xy
0
0
1
x
N
4 x2 y 2
xzi yzj
y
i jk
4 x2 y 2
2
dS
§
1¨
¨
©
¬ f
³ S ³ ª’
x, y, z u ’g x, y, z º¼ ˜ N dS
·
§
¸ ¨
¨
4 x 2 y 2 ¸¹
©
x
2
·
¸ dA
4 x 2 y 2 ¸¹
y
ª
2 x2 y 2
4 x2 y 2
2S
³0 ³0
ai bj ck , then
i
because C u r
1
C u r ˜ dr
2³C
4 x2 y 2
4 x2 y2
¬
2
2
x2 z
³ S ³ ««
³S ³
27. Let C
439
k
f x, y , z ’g x , y , z
rt
Stokes's Theorem
dA
º
»
4 x 2 y 2 »¼
y2z
2
4 x2 y 2
dA
dA
2r 2 cos 2 T sin 2 T
4 r2
2ª
1
curl C u r ˜ N dS
2 ³S ³
2S
§1
·º
¨ sin 2T ¸» dr
¹¼ 0
4 r2 © 2
2r 3
³ 0 «¬
r dT dr
1
2 C ˜ N dS
2³S ³
0
³ S ³ C ˜ N dS
j k
a b
c
x
z
y
and curl C u r
bz cy i az cx j ay bx k
i
j
k
w
wx
w
wy
w
wz
2 ai bj ck
2C.
bz cy cx az ay bx
e y zi
28. (a) F x, y, z
z
From the figure you have
1
C1 : r1 t
ti ,
0 d t d 1, r1c
i
C2 : r2 t
i tj,
0 d t d 1, r2c
j
C3 : r3 t
1 t i j, 0 d t d 1, r3c
i
C4 : r4 t
1 t j,
0 d t d 1, r4c
j
³ C F ˜ dr
³ C1 F ˜ r1c dt ³ C2 F ˜ r2c dt ³ C3 F ˜ r3c dt ³ C4 F ˜ r4c dt
z
0, N
³ S ³ curl F ˜ N dS
(0, 1, 0)
1
x
1
C3
y
dt 0 ³ 0 e
C2
(1, 1, 0)
1
³0 e
0
1
1 0
dt 0
1 e
e y z j e y zk
Double Integral: curl F
G x, y
(0, 0, 0)
C4
C1
(1, 0, 0)
k
³ R ³ e
y
dA
1
1
³ 0 ³ 0 e
y
dx dy
1
³ 0 e
y
dy
1
ª¬e y º¼
0
1e
© 2010 Brooks/Cole, Cengage Learning
Chapter 15
440
Vector Analysis
z 2 i x 2 j y 2k
(b) F x, y, z
From the figure you have
2 cos ti 2 sin tj 4k ,
C: r t
0 d t d 2S .
2S
³ C F ˜ dr ³ C M dx N dy P dz ³ 0
x2 y2 ,
g x, y
³ S ³ curl F ˜ N dS
³0
³R ³
gx
2 x,
gy
2S
2
³0
4
2
ªª16 sin T ¬¬
C
y 2 2 xº¼ dA
³ 0 ³ 0 ª¬4 r sin T r cos T
2S
z
2y
2 yi 2 zj 2 xk ˜ 2 xi 2 yj k dA
³ R ³ ª¬4 xy 4 y x
16 º
3¼
4 r sin T r 2 2 r cos T º¼ r dr dT
cos T 128
5
sin T º dT
¼
x 4y z
2
x
S: z =
4 − x 2 − 4y 2
y
2
1
M , N , P exists, then
³S ³
0
2
2 cos t , sin t , 0 , 0 d t d 2S , be the boundary of S.
Let C : r t
If F
2
z
4, z t 0
2
−2
−2
0
29. Let S be the upper portion of the ellipsoid
2
ª¬32 sin t 8 cos3 t º¼ dt
2 yi 2 zj 2 xk
Double Integral: curl F
z
2S
ª16 2 sin t 4 cos 2 t 2 cos t 0º dt
¬
¼
curl F ˜ N dS
by i
³ C F ˜ dr
Stokes's Theorem
³ C G ˜ dr
by iii
sin t 2 cos t
,
, 0 ˜ 2 sin t , cos t , 0 dt
4
4
2S
³0
y
1
C
2
x
1 2S
2 sin 2 t 2 cos 2 t dt
4³0
S
So, there is no such F.
Review Exercises for Chapter 15
xi j 2k
1. F x, y, z
x 2 12 22
F
2 x 2 xy z 2
3. f x, y, z
x2 5
’f
F x, y , z
4 x y i xj 2 zk
z
x 2e yz
4. f x, y, z
3
2 xe yz i x 2 ze yz j x 2 ye yz k
F x, y , z
2
xe yz 2i xzj xyk
2
3
x
4
i 2 yj
2. F x, y
F
y
From M
N
1 4 y2
U
y
1 x 2
wU wx
wU wy
y x and
1 x, partial integration yields
y x h y and U
6. Because wM wy
x
2
4
wN wx, F is conservative.
2
suggests that U x, y
5
4
3
2
−1
−2
−3
−4
−5
5. Because wM wy
y x g x which
y x C.
1 y 2 z wN wx, F is not
conservative.
© 2010 Brooks/Cole, Cengage Learning
0
Review Exercises for Chapter 15
wM
wy
7. Because
wU
wx
From M
wN
wx
2 xy and
2 xy, F is conservative.
xy 2 x 2 and
N
wU
wy
U
1 2 2
x3
x y h y
2
3
x y y , partial integration yields
2
2
11. Because
wM
wy
1
y2 z
wN wM
,
wx wz
wN
wz
x
y2z2
wP
,
wy
1
yz 2
wP
,
wx
F is conservative. From
M
and
U
441
wU
wx
1
,
yz
wU
wy
N
x
,
y2z
wU
wz
P
x
yz 2
you obtain
1 2 2
y3
x y g x.
2
3
x3 3. So,
y 3 3 and g x
So, h y
1 2 2
x3
y3
x y C.
2
3
3
U x, y
8. Because wM wy
6 y 2 sin 2 x
conservative. From M
2 y 3 sin 2 x and
N
wU wy
U
y 3 cos 2 x h y and
U
y 3 1 cos 2 x g x which suggests that
3 y 2 1 cos 2 x , you obtain
y3 , g x
h y
wM
wy
U
x
h x, y Ÿ f x , y , z
yz
wM
wy
8 xy and
wN
wx
4 x,
(a) div F
wM
wN
z
, so F is
wy
wx
(a) div F
wM
wz
wN
wz
wP
2z
,
wx
wP
6y z
,
wy
wM
wz
wN
,
wx
y cos z z
wP
,
wx
x 2 i xy 2 j x 2 z k :
2 x 2 xy x 2
i
j
k
w
wx
w
wy
w
wz
x2
xy 2
x2 z
(b) curl F
2 xz j y 2k
y2 j z2 k :
14. Because F x, y, z
10. Because
4x
x
K.
yz
F is not conservative.
(b) curl F
not conservative.
wM
wy
wN
,
wx
sin z
C , and
y 1 cos 2 x C.
9. Because
x
g x, z ,
yz
U
13. Because F x, y, z
3
U x, y
x
f y, z ,
yz
12. Because
wN wx, F is
wU wx
U
2 y 2z
i
j
k
w
wx
w
wy
w
wz
0
y2
z2
0
F is not conservative.
15. Because F
(a) div F
(b) curl F
16. Because F
(a) div F
(b) curl F
cos y y cos x i sin x x sin y j xyzk :
y sin x x cos y xy
xzi yzj cos x sin y sin y cos x k
xzi yzj
3x y i y 2 z j z 3x k :
17. Because F
311
2i 3 j k
5
(a) div F
(b) curl F
arcsin xi xy 2 j yz 2k :
1
1 x2
2 xy 2 yz
z 2 i y 2k
© 2010 Brooks/Cole, Cengage Learning
442
Chapter 15
Vector Analysis
x 2 y i x sin 2 y j:
18. Because F
21. (a) Let x
2 x 2 sin y cos y
(a) div F
(b) curl F
³C
ln x 2 y 2 i ln x 2 y 2 j zk :
(a) div F
(b) Let x
(a) div F
cos t , y
³C
22. (a) Let x
4t , 0 d t d 1, then ds
5t , y
(b) C1 : x
1
³ 0 20t
t, y
2
4 4t , y
C3 : x
0, y
xy ds
1 sin t , y
cos t ,
³C
x 2 y 2 ds
y
dt
2t , 0 d t d 1, ds
2 t , 0 d t d 2, ds
4
2 5 dt
3
dt
dy
dt
4
³0
0 dt 1
³0
cos t
sin t , ds
2S
³0
cos t t sin t , y
25. (a) Let x
³ C 2x (b) Let x
³C
26. x
³C
3t , y
³0
ªt 2
t3 º
16 5 « »
3 ¼0
¬2
0 dt
C3
8 5
.
3
C2
(4, 0)
x
C1
3
4
1 cos t , 0 d t d 2S
ª 1 sin t
¬
2
sin t
2S
³0
ª cos t t sin t
¬
dt
2
dt
2S
³0
ª¬1 2 sin t sin 2 t 1 2 cos t cos 2 t º¼ dt
2S
2 cos t 2 sin t @0
dx
dt
t cos t ,
dy
dt
6S
t sin t
2
sin t t cos t º t 2 cos 2 t t 2 sin 2 t dt
¼
2S
³0
2S 2 1 2S 2
ª¬t 3 t º¼ dt
0 d t d 1.
1
³ 0 ª¬ 6t 3t 3 y dx x 2 y dy
2S
³0
S 2
³0
1
³0
27t 9t dt
1
18t 2 º¼
0
S
2
, dx
18
3 cos t dt , 0 d t d 2S .
ª¬ 6 cos t 3 sin t 3 sin t 3 cos t 6 sin t 3 cos t º¼ dt
sin t t sin t , 0 d t d
2 x y dx x 3 y dy
3t 6t 3 º¼ dt
3 sin t dt , dy
3 sin t , dx
2 x y dx x 2 y dy
cos t t sin t , y
>3t
sin t t cos t , 0 d t d 2S ,
3t ,
3 cos t , y
2
2
1 cos t º dt
¼
2
2S
x 2 y 2 ds
2
8t 8t 2 2 5 dt y = − 12 x + 2
(0, 2)
2
³ 0 >3 2 sin t 2 cos t@ dt
³C
2S
dt
20 41
3
41 dt
0, 0 d t d 4, ds
C2 : x
dx
dt
24. x
1.
41 dt.
1
23. x
2S
³0
x 2 y 2 ds
2
§
1
1 ·
z¨ 2 2 2 ¸
x
y
©
¹
z
z
2 2z
x2
y
³ C xy ds
³C
cos t
1
1
i j
y
x
(b) curl F
So,
2
125
3
0 d t d 2S ,
sin t ,
sin t
then ds
z
z
i j z 2k :
x
y
20. Because F
9t 2 16t 2 5 dt
1
2x 2 y
k
x2 y2
(b) curl F
1
³0
x 2 y 2 ds
5 dt.
ª
t3 º
«125 »
3 ¼0
¬
2x 2 y
1
x2 y 2
2x
2y
2
1
x2 y 2
x y2
0 d t d 1,
4t ,
9 16 dt
then ds
0
19. Because F
3t , y
t cos t dt , dy
2S
³0
9 dt
18S
cos t t cos t sin t dt
ªsin t cos t 5t 2 6t 2 cos 2 t t 1 sin 2 t 2t 3 º dt | 1.01
¬
¼
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 15
27.
³C
2 x y ds, r t
a cos3 ti a sin 3 tj, 0 d t d
xc t
3a ˜ cos 2 t sin t
yc t
3a ˜ sin 2 t cos t
³C
2 a ˜ cos3 t a ˜ sin 3 t
S
2
xc t
2
yc t
2
9a 2
5
dt
ti t 2 j t 3 2k , 0 d t d 4
28. r t
xc t
³C
S 2
³0
2 x y ds
1, yc t
C: y
3 12
t
2
2t , zc t
4
³0
x 2 y 2 z 2 ds
t 2 t4 t3
1 4t 2 9
t dt | 2080.59
4
3 sin x y
29. f x, y
12 x y
30. f x, y
2 x from 0, 0 to 2, 4
x 2 from 0, 0 to 2, 4
C: y
rt
ti 2tj, 0 d t d 2
rt
ti t 2 j, 0 d t d 2
rc t
i 2j
rc t
i 2tj
rc t
5
rc t
Lateral surface area:
³C f
³ 0 ª¬3 sin t
x, y ds
2
0
2t º¼ 5 dt
³C f
>3 sin 3t@ dt
1
ª
º
5 «3t cos 3t »
3
¬
¼0
1
1º
ª
5 «6 cos 6 »
3
3¼
¬
F
rt
t 2i t 2 j,
rc t
2ti 2tj
1
³ 0 ª¬t
1
0 d t d1
2
³ 0 6t
5
t 2 2t 2 t 2 t 2 2t º¼ dt
dt
1
t 6 º¼
0
1
4 cos t 3 sin t i 4 cos t 3 sin t j, 0 d t d 2S
2S
2S
³0
ª
7 sin 2 t º
«12t »
2 ¼0
¬
12 7 sin t cos t dt
24S
ª¬ 2 sin t i 2 cos t j k º¼ dt
2 cos t i 2 sin t j tk , 0 d t d 2S
³ C F ˜ dr
34. x
1 4t 2 dt | 41.532
ª¬ 4 sin t i 3 cos tjº¼ dt
³ C F ˜ dr
33. dr
12 t t 2
xyi 2 xyj
³ C F ˜ dr
5
19 cos 6 | 13.446
3
2
³0
x, y ds
31. F x, y
2
F
1 4t 2
Lateral surface area:
2
5³
32. dr
443
2S
³0
2 t, y
t dt
2S 2
2 t, z
dr
ª
«i j ¬
F
4 2t 4t t 2 , 0 d t d 2
º
k » dt
4t t ¼
2t
2
4t t 2 i 4t t 2 2 t j 0k
2
³ C F ˜ dr
2
³0
t 2 dt
ªt 2
º
« 2t »
2
¬
¼0
2
© 2010 Brooks/Cole, Cengage Learning
444
Chapter 15
Vector Analysis
y zi x z j x yk
35. F x, y, z
Curve of intersection: x
rt
ti tj 2t 2k ,
rc t
i j 4tk
2
³ 0 ª¬ t 2t
³ C F ˜ dr
36. Let x
2 sin t , y
S
³0
37. For y
x 2 , r1 t
For y
2 x, r2 t
³ C xy dx 0 d t d 2
³ 0 ª¬12t
2
2
2t º¼ dt
ª¬4t 3 t 2 º¼
0
36
4 sin 2 t , 0 d t d S .
2 cos t , z
³C
ª8 cos t ¬
8 sin t 16 sin 2 t cos t dt
16
3
S
sin 3 t º¼
0
16
ti t 2 j, 0 d t d 2
y
y = 2x
4
2 t i 4 2t j, 0 d t d 2
³ C1 xy dx x 2 y 2 dy
³ C F ˜ dr
x 2 y 2 dy 32
³ C2 xy dx 3
x 2 y 2 dy
4
3
(2, 4)
C2
y = x2
2
1
C1
x
1
2 x y dx 2 y x dy
2
3
4
2 cos t 2t sin t i 2 sin t 2t cos t j, 0 d t d S
rt
³ C F ˜ dr
4S 2 4S
xi y j is conservative.
ª1 x2 ¬2
Work
4, 8
2 y3 2 º
3
¼ 0, 0
1
2
10 sin ti 10 cos tj 40. r t
F
2
t 2t 2 2t 4t º¼ dt
2
100
3
39. F
2t 2
0i 4 j 2 sin t k
³ C F ˜ dr
38.
t2 t2
t, z
ª¬ 2 cos t i 2 sin t j 8 sin t cos t k º¼ dt
dr
F
t, y
16 2
3
8
3
83 2
2000 5280
tk
S 2
3 4 2
10 sin ti 10 cos tj 25
S
tk , 0 d t d
33S
2
20k
25 ·
§
k¸
¨10 cos ti 10 sin tj 33S ¹
©
S 2 500
250
³ C F ˜ dr ³ 0 33S dt 33 mi ˜ ton
dr
41.
³ C 2 xyz dx 42.
³ C y dx 43. (a)
³C y
2
x dy 1
dz
z
4, 4, 4
1
³ 0 ª¬ 1 t
dx 2 xy dy
1
³C y
2
So,
4
y 2i 2 xy j
³ C F ˜ dr
42
2
2
2
16 ln 4
3 2 1 3t 1 t º dt
¼
2t 1 2 3t 2 4t 1 º¼ dt
ª
³ 1 «¬t 1
dx 2 xy dy
(c) F x, y
6
ª¬ xy ln z º¼ 0, 0,1
³0 3 t
(b)
1, 3, 2
ª¬ x 2 yz º¼
0, 0, 0
x 2 z dy x 2 y dz
2t
t
1 º
» dt
2 t¼
’f where f x, y
11
2
4
³1
1
³0
t t dt
9t 2 14t 5 dt
4
ª¬t 2 º¼
1
1
ª¬3t 3 7t 2 5t º¼
0
15
15
xy 2 .
15.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 15
a T sin T , y
44. x
a 1 cos T , 0 d T d 2S
1
x dy y dx.
2³C
(a) A
445
y
Because these equations orient the curve backwards, you will use
1
y dx x dy
2³
1 2S 2
1 2S a
ªa 1 cos T 1 cos T a 2 T sin T sin T º¼ dT ³
0 0 dT
2³0 ¬
2 0
a 2 2S
ª1 2 cos T cos 2 T T sin T sin 2 T º¼ dT
2 ³0 ¬
a 2 2S
2 2 cos T T sin T dT
2 ³0
a2
6S
3S a 2 .
2
A
45.
1 2S 3
a 1 cos T
2A³0
1
y 2 dx
2A³C
1 § wN
1
³ C y dx 2 x dy
³ 0 ³ 0 ¨© wx
1
1
³0 ³0
46.
³ C xy dx 2 1 dy dx
2
2
³0 ³0
x 2 y 2 dy
³ C xy
³C y
dx x y dy
dx x 4 3 dy
2
48.
1
32
1 x 2 3
13
1 x2 3
32
S
3
³ C xy dx x 2 dy
, 0 d v d 2S
§ wN
³ R ³ ¨© wx
1
1
1
1
³ 1
0
4 y dy dx
0
wM ·
¸ dA
wy ¹
2 x x dy dx
³ 1 > xy@x2
1
dx
x x3 dx
1
2 y dy dx
1
³ 1 ª¬ 43 x
13
ª 8 x 2 3 1 x 2 3
¬« 7
sec u cos vi 1 2 tan u sin vj 2uk
0 d u d
49.
³ 1 ³ x 2
dx
a2 x2
a
4
4 x1 3
3
a2 x2
³ a 0 dx
ª x2
x4 º
«
»
4 ¼ 1
¬2
³ 1 ³ 1 x 2 3 3 2
1
a
³a ³
x 2 y 2 dx 2 xy dy
1
2 xy 2 xy dA
³ 1 83 x
51. r u , v
³C
5
a.
6
§ wN
wM ·
³ R ³ ¨© wx wy ¸¹ dA
2
³R ³
50.
1
a 3 5S
2 3 S a2
1 cos T dT
2 x x dy dx
2
47.
2
wM ·
¸ dy dx
wy ¹
³ 0 2 x dx
2
x
2π a
C2
S a. From Section 15.4,
(b) By symmetry, x
y
C1
52
0
32
1 x2 3
y y 2 º¼
3 2 dx
1 x 2 3
16
35
1 x2 3
52. r u , v
52 1
º
¼» 1
0
e u 4 cos vi e u 4 sin vj u
k
6
0 d u d 4, 0 d v d 2S
z
z
2
6
−4
2
4
x
2
−2
2
4
y
x
2
y
© 2010 Brooks/Cole, Cengage Learning
446
Chapter 15
Vector Analysis
z
53. (a)
z
(b)
z
(c)
3
3
3
−4
−4
−4
2
−4 −3
−4 −3
4
4
4
y
−2
x
3
2
−1
3
y
4
3
4
2
−2
x
−3
−4
3
y
4
−3
−3
−3
(d)
2
−2
x
−2
z
3
−4
4
−2
1
−4 −3
3
2
3
y
4
−2
x
−3
§ S·
The space curve is a circle: r¨ u , ¸
© 4¹
(e) ru
rv
3 2
3 2
cos ui sin uj 2
2
2
k
2
3 cos v sin u i 3 cos v cos u j
3 sin v cos u i 3 sin v sin u j cos v k
ru u rv
i
j
k
3 cos v sin u
3 cos v cos u
0
3 sin v cos u 3 sin v sin u cos v
3 cos 2 v cos u i 3 cos 2 v sin u j 9 cos v sin v sin 2 u 9 cos v sin v cos 2 u k
3 cos 2 v cos u i 3 cos 2 v sin u j 9 cos v sin v k
ru u rv
9 cos 4 v cos 2 u 9 cos 4 v sin 2 u 81 cos2 v sin 2 v
Using a Symbolic integration utility,
(f ) Similarly,
S 4
S 2
³0 ³0
ru u , v
i j
rv u , v
i j cos vk
i
j
k
1
1
0
2S
ru u rv du dv | 14.44.
ru u rv dv du | 4.27.
u v i u v j sin vk ,
54. S : r u , v
ru u rv
S 2
³S 4 ³ 0
9 cos 4 v 81 cos 2 v sin 2 v
0 d u d 2, 0 d v d S
cos vi cos vj 2k
1 1 cos v
ru u rv
³ S ³ z dS
2 cos 2 v 4
S
rv u , v
2 cos 2 v 4 du dv
ª
2« 6 ¬«
§
2 ln ¨¨
©
2 ·º
¸»
2 ¸¹¼»
6 6 u cos vi u sin vj u 1 2 u k , 0 d u d 2, 0 d v d 2S
55. S : r u , v
ru u , v
2
³ 0 ³ 0 sin v
z
2
cos vi sin vj 3 2u k
u sin vi u cos vj
3
ru u rv
i
j
k
cos v
sin v
3 2u
u sin v u cos v
ru u rv
³S ³
u
x y dS
2u 3
2S
x
y
3
−2
2u 3 u cos vi 2u 3 u sin vj uk
0
2
1
2
u cos v u sin v u
³0 ³0
−3
−3
2u 3
2
1 du dv
2
2S
³0 ³0
cos v sin v u 2
2u 3
2
1 dv du
© 2010 Brooks/Cole, Cengage Learning
0
Review Exercises for Chapter 15
56. (a) z
aa z
x2 y 2 , 0 d z d a2
a
z
0 Ÿ x2 y2
(b) S : g x, y
U x, y
m
k
R³
x2 y 2
x2 y2
k
³ S ³ e x, y , z
k³
2
a2
a2 a
z
447
³R ³ k
dS
x2 y2
2S
a 2 1³
0
1
a
³0 r
2
x2 y 2
1 g x2 g 2y dA
a2 x2
a2 y2
dA
2
2
x y
x y2
k³
2
dr dT
k
a 2 1³
2S
0
a3
dT
3
R
³
2
k
3
a2 1
y
a
a
x
x 2 y 2 dA
a 2 1 a 3S
x 2i xyj zk
57. F x, y, z
Q: solid region bounded by the coordinates planes and the plane 2 x 3 y 4 z
12
Surface Integral: There are four surfaces for this solid.
z
0, N
k ,
F˜N
z,
³ S1 ³ 0 dS
0
y
0,
N
j,
F˜N
xy,
³ S2 ³ 0 dS
0
x
0,
N
i,
F˜N
x2 ,
³ S3 ³ 0 dS
0
2i 3j 4k
, dS
29
§1· § 9 ·
1 ¨ ¸ ¨ ¸dA
© 4 ¹ © 16 ¹
2x 3y 4z
12, N
³ S4 ³ F ˜ N dS
1
2 x 2 3 xy 4 z dA
4 ³R ³
1 6 4 2x 3
2 x 2 3xy 12 2 x 3 y dy dx
4³0 ³0
29
dA
4
2
2
1 6 ª 2 § 12 2 x · 3 x § 12 2 x ·
§ 12 2 x ·
§ 12 2 x · 3 § 12 2 x · º
2
12
2
x
x
«
¨
¸
¨
¸
¨
¸
¨
¸
¨
¸ » dx
4 ³ 0 «¬ ©
3 ¹
2©
3 ¹
3 ¹
3 ¹ 2©
3 ¹ »¼
©
©
1 6
x3 x 2 24 x 36 dx
6³0
6
º
1 ª x4
x3
12 x 2 36 x»
«
6¬ 4
3
¼0
Divergence Theorem: Because div F
³³³ div F dV
Q
6
12 2 x 3
6
12 2 x 3
³0 ³0
³0 ³0
12 2 x 3 y 4
³0
66
2x x 1
3x 1, Divergence Theorem yields
3 x 1 dz dy dx
§ 12 2 x 3 y ·
3x 1 ¨
¸ dy dx
4
©
¹
12 2 x 3
1 6
3 º
ª
3 x 1 «12 y 2 xy y 2 »
4³0
2 ¼0
¬
dx
2
ª
1 6
§ 12 2 x · 3 § 12 2 x · º
3 x 1 «4 12 2 x 2 x¨
¨
» dx
¸
¸
³
4 0
3 ¹ 2©
3 ¹ »¼
©
¬«
1 62 3
3 x 35 x 2 96 x 36 dx
4³0 3
6
º
35 x3
1 ª3x 4
48 x 2 36 x»
«
3
6¬ 4
¼0
66.
© 2010 Brooks/Cole, Cengage Learning
448
Chapter 15
Vector Analysis
xi yj zk
58. F x, y, z
Q: solid region bounded by the coordinate planes and the plane 2 x 3 y 4 z
Surface Integral: There are four surfaces for this solid.
(0, 0, 3)
z
0, N
k , F ˜ N
z,
³ S1 ³ 0 dS
0
y
0,
N
j,
F˜N
y,
³ S2 ³ 0 dS
0
x
0,
N
i,
F˜N
x,
³ 0 dS
0
2x 3y 4z
³ S4
³ S3
2i 3 j 4k
, dS
29
12, N
1
2 x 3 y 4 z dy dx
4³R ³
³ N ˜ F dS
Triple Integral: Because div F
³³³ div F dV
³³³ 3 dV
Q
12
z
(0, 4, 0)
x
y
(6, 0, 0)
§1· § 9 ·
1 ¨ ¸ ¨ ¸ dA
© 4 ¹ © 16 ¹
29
dA
4
6
ª
x2 º
3«4 x »
3 ¼0
¬
6§
2x ·
3³ ¨ 4 ¸ dx
0
3¹
©
1 6 12 2 x 3
12 dy dx
4³0 ³0
36
3, the Divergence Theorem yields
3 Volume of solid
Q
ª1
º
3« Area of base Height »
¬3
¼
1
6 4 3
2
36.
cos y y cos x i sin x x sin y j xyzk
59. F x, y, z
y 2 over the square in the xy-plane with vertices 0, 0 , a, 0 , a, a , 0, a
S: portion of z
Line Integral: Using the line integral you have:
C1 : y
0,
dy
0
C2 : x
0,
dx
0,
C3 : y
a, dy
C4 : x
a, dx
³ C F ˜ dr
³C
y 2 , dz
z
0, z
0
2
2 y dy
a , dz
z
0,
y , dz
cos y y cos x dx sin x x sin y dy xyz dz
³ C1 dx ³ C2 0 ³ C3
a
2 y dy
2
0
³ 0 dx ³ a
³ C4
cos a a cos x dx cos a a cos x dx a
³0
sin a a sin y dy ay 3 2 y dy
sin a a sin y dy a
³ 0 2ay
4
dy
a
ª y5 º
a
0
a > x cos a a sin x@a > y sin a a cos y@0 «2a »
¬ 5 ¼0
2a 6
a a cos a a sin a a sin a a cos a a 5
Double Integral: Considering f x, y, z
N
So,
’f
’f
³S ³
2 yj k
1 4 y2
curl F ˜ N dS
a
a
³0 ³0 2y
2
1
C3
z y 2 , you have:
1 4 y 2 dA, and curl F
, dS
z
2a 6
5
z dy dx
a
a
³0 ³0 2y
4
C4
C1
xzi yzj.
dy dx
a
³0
y
x
2a 5
dx
5
C2
a
a
2a 6
.
5
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 15
449
x z i y z j x 2k
60. F x, y, z
S: first octant portion of the plane 3x y 2 z
12
Line Integral:
C1 : y
0, dy
0, z
C2 : x
0, dx
0,
z
C3 : z
0,
dz
0,
y
³ C F ˜ dr
³C
12 3x
, dz
2
12 y
, dz
2
12 3x, dy
3
dx
2
1
dy
2
3 dx
x z dx y z dy x 2 dz
ª
12 3 x
§ 3 ·º
x 2 ¨ ¸» dx 2
© 2 ¹¼
³ C1 «¬x 0§
12 § 3
5
·
x 6 ¸ dx ³ ¨ y 0
2
¹
©2
12 3 x y
Double Integral: G x, y, z
z
2
3
1
’G x , y , z
i jk
2
2
curl F
i 2x 1 j
3
³ 4 ¨© 2 x
2
³S ³
4
4
³0
³ C3 ª¬x 10 x 36 dx
12 3x 3 º¼ dx
8
z
(0, 0, 6)
x
(4, 0, 0)
y
(0, 12, 0)
12 3 x
xi yj zk , then div curl F
61. If curl F
·
6 ¸ dy ¹
³0 ³0
curl F ˜ N dS
12 y º
dy 2 ¼»
ª
³ C2 ¬« y 4
³0
x 1 dy dx
111
3x 2 15 x 12 dx
8
3, contradicting Theorem 15.3.
Problem Solving for Chapter 15
25
1. (a) ’T
N
x y2 z2
2
xi dA
1 x2
³ S ³ k’T ˜ N dS
Flux
yi zk @
1 x2 k
1
dS
> xi
32
25k ³
ª
x2
«
R³« 2
x y2 z2
¬
ª
12
1
x2
«
25k ³
32
1 2 ³ 0 « 2
2
2
1 x2
¬ x y z
12
25k ³
(b) r u , v
ru
1 y
2
32
sin u , 0, cos u , rv
1 x
2
12
dy dx
º
» dA
z
x2 y 2 z 2
3 2»
¼
º
1 x2
» dy dx
32
12
1 x 2 »¼
x2 y 2 z 2
25k ³
1
1
0
1 y
2
32
dy ³
1
12
1 2
1 x
2
12
dx
§ 2 ·§ S ·
25k ¨¨
¸¸¨ ¸
© 2 ¹© 3 ¹
25k
2S
6
0, 1, 0
cos u , 0, sin u
25
x y z
2
2
’T ˜ ru u rv
Flux
1
1
³0
1 x2
12
cos u , v, sin u
ru u rv
’T
1 2
12
32
1
32
> xi 25
v 1
2
2S 3
³ 0 ³S 3
2
32
25k
v 1
2
32
yj zk @
25
v 1
2
cos 2 u sin 2 u
du dv
25k
32
>cos ui vj sin uk@
25
v 1
2
32
2S
6
© 2010 Brooks/Cole, Cengage Learning
450
Chapter 15
Vector Analysis
1 x2 y2 ,
2. (a) z
wz
wx
x
1 x y
2
2
2
§ wz ·
§ wz ·
¨ ¸ ¨ ¸ 1 dA
© wx ¹
© wy ¹
dS
25
’T
x y2 z2
2
N
32
2
xi yj 1 x2 y 2 k
³S ³ k’T
Flux
k³
R
(b) r u , v
k³
˜ N dS
R
25
³
1 x y
2
25 xi yj zk
x
1 x2 y 2
³ 25 xi 25k ³
·
j k ¸ 1 x2 y 2
¸
1 x2 y 2
¹
y
i yj zk ˜ xi yj zk
2S
0
1
1
³0
1 r2
r dr dT
1
1 x2 y 2
dA
50S k
sin u cos v, sin u sin v, cos u
ru
cos u cos v, cos u sin v, sin u
rv
sin u sin v, sin u cos v, 0
ru u rv
ru u rv
sin 2 u cos v, sin 2 u sin v, sin u cos u sin 2 v sin u cos u cos 2 v
sin u
25k ³
Flux
3. r t
1 x2 y2
xi yj zk
dA
2
y
1
§
¨
¨
©
§ wz ·
§ wz ·
¨ ¸ ¨ ¸ 1
x
w
© ¹
© wy ¹
wz
wy
1 x2 y2
xi yj zk
wz
wz
i jk
wx
wy
2
,
2
2S
0
S 2
³0
sin u du dv
50S k
3 cos t , 3 sin t , 2t
rc t
3 sin t , 3 cos t , 2 , rc t
2S
Ix
³C
y 2 z 2 U ds
³0
Iy
³C
x 2 z 2 U ds
³0
Iz
³C
x 2 y 2 U ds
³0
2S
2S
13
9 sin 2 t 4t 2
13 dt
9 cos 2 t 4t 2
13 dt
9 cos 2 t 9 sin 2 t
1
13S 32S 2 27
3
1
13S 32S 2 27
3
13 dt
18S
13
2 2t 3 2
t2
, t,
2
3
4. r t
rc t
t , 1,
2t1 2 , rc t
U ds
1
t 1 dt
1t
t 1
1
1 § t4
Iy
³C
x 2 z 2 U ds
³ 0 ¨© 4
Ix
³C
y 2 z 2 U ds
³ 0 ¨© t
Iz
³C
x 2 y 2 U ds
³ 0 ¨© 4
1§
2
1 § t4
8 3·
t ¸ dt
9 ¹
49
180
8 3·
t ¸ dt
9 ¹
5
9
·
t 2 ¸ dt
¹
23
60
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 15
5.
w
1
f
ww
wx
ww
wy
ww
wz
x y z
³C y
n
2
2
32
x2 y2 z 2
32
z
x y z
2
2
52
2 y 2 x2 z 2
x2 y 2 z 2
52
2 z 2 x2 y2
x2 y 2 z 2
52
w2w w2w w2w
wx 2
wy 2
wz 2
’2w
32
2
x2 y 2 z 2
w2w
wz 2
y
x2 y2 z 2
2 x2 y 2 z 2
w2w
wy 2
x
0
1
is harmonic.
f
Therefore w
6.
w2w
wx 2
1
2
451
§ wN
³ R ³ ¨© wx
dx x n dy
wM ·
¸ dA
wy ¹
For the line integral, use the two paths
C1 : r1 x
xi, a d x d a
xi C2 : r2 x
³ C1 y
dx x n dy
n
³ C2 y
a 2 x 2 j, x
n
(a) For n
2a
a
³a
dx x dy
§ wN
wM ·
¸ dA
wy ¹
ª 2
2
«a x
¬
a
³ a ³ 0
a2 x2
³
1
2 C
n2
x
n
y = a2 − x2
º
» dx
2
2
a x ¼
x
C2
ª¬nx n 1 ny n 1 º¼ dy dx
−a
C1
x
a
1, 3, 5, 7, both integrals give 0.
(b) For n even, you obtain
4
16
n
2: a 3
n
4: a 5
3
15
(c) If n is odd then the integral equals 0.
7.
y
0
n
³ R ³ ©¨ wx
a
a to x
x dy y dx
³
2S
1
2 0
³
6: n
32 7
a
35
n
8: 256 9
a
315
ª¬a T sin T a sin T dT a 1 cos T a 1 cos T
2S
1 a2
2
0
ª¬T sin T sin 2 T 1 2 cos T cos 2 T º¼ dT
dT º¼
³
2S
1 a2
2
0
T sin T 2 cos T 2 dT
3 S a 2
So, the area is 3 S a 2 .
8.
³
1
2 C
x dy y dx
2³
S 2
0
ª 1 sin 2t cos t sin t cos 2t º dt
¬2
¼
2
2
3
So, the area is 43 .
9. (a) r t
rc t
W
(b) r t
rc t
W
tj, 0 d t d 1
j
1
³ C F ˜ dr
³0
ti j ˜ j dt
1
³ 0 dt
1
t t 2 i + t j, 0 d t d 1
1 2t i j
F ˜ dr
1§
³ 0 ¨© 2t t
1
³ 0 ª¬ 1 2t
2
i ª t t2
«¬
2
1º j¸· ˜ 1 2t i j dt
»¼ ¹
2t t 2 t 4 2t 3 t 2 1 º¼ dt
1
³0
t 4 4t 2 2t 1 dt
13
15
© 2010 Brooks/Cole, Cengage Learning
452
Chapter 15
Vector Analysis
c t t 2 i tj, 0 d t d 1
(c) r t
rc t
c 1 2t i j
F ˜ dr
c t t 2 t c 1 2t
c2 t t 2
2
1 1
c 2t 4 2c 2t 2 c 2t 2ct 2 ct 1
dW
dc
1
1
c 15
6
d 2W
dc 2
1
! 0 c
15
5
minimum.
2
3x 2 y 2i 2 x3 yj is conservative.
10. F x, y
f 2, 4 f 1, 1
8 16 1
S ab
12. Area
x3 y 2 potential function.
f x, y
Work
1 2
1
c c 1
30
6
5
0 Ÿ c
2
³ C F ˜ dr
W
rt
a cos ti b sin tj, 0 d t d 2S
rc t
a sin ti b cos tj
127
12 b sin ti 12 a cos tj
F
11. v u r
a1 , a2 , a3 u x, y , z
a2 z a3 y , a1 z a3 x, a1 y a2 x
curl v u r
2a1 , 2a2 , 2a3
W
2v
v u r dr
13. F x, y
³ S ³ curl v u r
x2 y2
wM
wy
ª 5
3mxy « x 2 y 2
¬ 2
x y
2
wN
wx
2
7 2
wN
wx
52
1
ab
2
2S
S ab
ª5 y 2 x 2 y 2 º
¬
¼
7 2
7 2
7 2
ª3 xyi 2 y 2 x 2 jº
¬
¼
5 2
º
2 y » x2 y 2
¼
ª 5
m 2 y 2 x 2 « x 2 y 2
¬ 2
mx x 2 y 2
x y
2
m 2 y 2 x2 x2 y2
52
mx x 2 y 2
So,
7 2
m 2 y 2 x2
N
m
2
3mxy x 2 y 2
52
3mx x 2 y 2
F ˜ dr
³ S ³ 2v ˜ N dS .
˜ N dS
M x, y i N x , y j
3mxy
M
2S
³0
1 ab
2
Same as area.
By Stokes's Theorem,
³C
ª 1 ab sin 2 t 1 ab cos 2 t º dt
2
¬2
¼
F ˜ dr
5 2
3mx
3mx x 2 4 y 2
x2 y2
72
5 2
º
2x » x2 y2
¼
5 2
2mx
ª 2 y 2 x 2 5 x 2 y 2 2 º
¬
¼
3 x 2 12 y 2
3mx x 2 4 y 2
x2 y 2
72
wM
and F is conservative.
wy
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 6
Additional Topics in Differential Equations
Section 16.1
Exact First-Order Equations...............................................................454
Section 16.2
Second-Order Homogeneous Linear Equations................................464
Section 16.3
Second-Order Nonhomogeneous Linear Equations .........................472
Section 16.4
Series Solutions of Differential Equations ........................................480
Review Exercises ........................................................................................................491
Problem Solving .........................................................................................................500
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R 1 6
Additional Topics in Differential Equations
Section 16.1 Exact First-Order Equations
1. 2 x xy 2 dx 3 x 2 y dy
wM
wy
wN
wx
wM
wy
wM
wy
2 xy
f x, y
0
x cos y
wN
cos y
wx
wM
wN
Not exact
z
wy
wx
wN
wx
wM
wy
e xy xye xy
3 x g c y
f y x, y
2 y 3x Ÿ g c y
Ÿ g y
3. x sin y dx x cos y dy
wM
wy
³ M x, y dx
³ 2 x 3 y dx
x 2 3 xy g y
x
4. ye xy dx xe xy dy
0
x 2 3 xy y 2 C1
x 2 3xy y 2
C
6. ye x dx e x dy
0
wM
wy
wN
Exact
wx
ex
f x, y
³N
ye x g c x
Ÿ g x
C1
f x, y
ye x C1
ye x
³e
x, y dy
f x x, y
0
2y
y 2 C1
f x, y
x
ye x g x
dy
ye x Ÿ g c x
e xy xye xy
wM
wy
wN
Exact
wx
f x, y
0
C
7. 3 y 2 10 xy 2 dx 6 xy 2 10 x 2 y
6 y 20 xy
³M
0
wN
Exact
wx
³ 3y
x, y dx
2
10 xy 2 dx
3xy 2 5 x 2 y 2 g y
f y x, y
6 xy 10 x 2 y g c y
Ÿ gc y
f x, y
2 Ÿ g y
6 xy 2 10 x 2 y
2 y C1
3 xy 2 5 x 2 y 2 2 y C1
3xy 2 5 x 2 y 2 2 y
454
0
wN
Exact
wx
3
wN
Exact
wx
wN
y
wx
wM
wN
Not exact
z
wy
wx
wM
wy
5. 2 x 3 y dx 2 y 3 x dy
2 xy
2. 1 xy dx y xy dy
wM
wy
0
C
© 2010 Brooks/Cole, Cengage Learning
Section 16.1
8. 2 cos 2 x y dx cos 2 x y dy
wM
wy
0
12. xe
wN
wx
2 sin 2 x y
wM
wy
2 sin 2 x y Exact
³ 2 cos
2 x y dx
f y x, y
0 Ÿ g y
9. 4 x 6 xy
3
2
dx 4 y 6 xy dy
3
e
0
12 xy
wN
wx
6 y
10. 2 y e
x2 y2
1 x2 y2
e
g y
2
gc y
ye
dx 2 xye
xy 2
4 xy 3 y e xy
wN
wx
2 xy y e
dy
wN
wx
0
x2 y2
C1
2
0
2 xy
x y
3
2 xy
x y
3
Not exact
2
14. ye y cos xy dx e y x cos xy sin xy dy
3
xy 2
Not exact
y
x
dx 2
dy
11. 2
x y2
x y2
y 2 x2
x2 y 2
f x, y
f y x, y
³M
2
0
x, y dx
§ x·
arctan ¨ ¸
© y¹
§ x·
arctan ¨ ¸ C1
© y¹
wN
wx
e y >cos xy xy sin xy y cos xy@
wN
Exact
wx
³ M x, y dx
y
³ ye cos xy dx
Ÿ gc y
0 Ÿ g y
C1
e y sin xy g y
e y sin xy xe y cos xy g c y
f y x, y
x
gc y
x2 y2
x
Ÿ gc y
x y2
e y cos xy ye y cos xy xye y sin xy
f x, y
§ x·
arctan ¨ ¸ g y
© y¹
0
wM
wy
wM
wy
wN
Exact
wx
2
f x, y
Ÿ gc y
C
wM
wy
wM
wy
wM
wy
x2 y 2
dx
§ y ·
§ x ·
13. ¨
¸ dx ¨
¸ dy
x
y
©
¹
© x y¹
Not exact
2 xy 2
wN
Exact
wx
x2 y2
2
wM
wy
0
1 x2 y 2
e
C1
2
f x, y
C
dy
455
x, y dx
Ÿ g y
C1
sin 2 x y C1
sin 2 x y
ye
x2 y 2
x2 y 2
³M
sin 2 x y g y
cos 2 x y Ÿ g c y
f x, y
2 xye
³ xe
cos 2 x y g c y
f y x, y
dx ye
f x, y
³ M x, y dx
f x, y
x2 y2
Exact First-Order Equations
f x, y
e y sin xy
0 Ÿ g y
C1
e y sin xy C1
C
C
© 2010 Brooks/Cole, Cengage Learning
0
456
Chapter 16
Additional Topics in Differential Equations
15. (a) and (c)
17. (a) and (c)
4
y
y
6
4
−6
4
6
−9
2
9
x
−4
−4
−4
x
−2
2
4
−6
−2
−4
−4
(b)
§1·
0, y¨ ¸
© 2¹
2 x tan y 5 dx x 2 sec 2 y dy
wM
wy
³M
4
(b)
wN
Ÿ Exact
wx
2 x sec 2 y
f x, y
S
x y
wM
wy
³ 2 x tan y 5
x, y dx
x
2
2
y
dx x y
2
³M
f x, y
32
x sec y
0 Ÿ g y
x 2 tan y 5 x
f x, y
§1 S ·
f¨ , ¸
©2 4¹
1 5
4 2
y
f y x, y
x y2
y
2
C
x2 y 2
11
4
Answer: x 2 tan y 5 x
16. (a) and (c)
f x, y
x2 y2
f 3, 4
32 42
1 2
2 x dx
gc y
Ÿ gc y
0Ÿ g y
C
25
5
5 or x 2 y 2
x2 y2
Solution:
y
dx
x2 y2 g y
C
C
11
4
x y2
2
1 2
x y2
2³
2
Ÿ gc y
x
³
x, y dx
x 2 sec2 y g c y
2
3
wN
Exact
wx
xy
2
0, y 4
dx
x 2 tan y 5 x g y
f y x, y
dy
x y2
2
C
25
4
18. (a) and (c)
4
−6
2
8
6
−12
12
x
−4
4
−4
−8
(b)
(b) 2 xy dx x 2 cos y dy
wM
wy
S
0, y 1
wN
Exact
wx
2x
f x, y
³ M x, y dx
f y x, y
x2 gc y
x 2 y sin y
C
f 1, S
C
S
Answer: x y sin y
2
wM
wy
f x, y
³ 2 xy dx
Ÿ gc y
cos y
Ÿ g y
sin y C1
S
2
1
f y x, y
³M
5
x, y dx
2
y dx
x gc y
Ÿ gc y
f x, y
0, y 4
wN
Exact
wx
³x
x2 y g y
x 2 cos y
x 2 y sin y C1
f x, y
x y dx y x dy
2
x3
xy g y
3
y2 x
y2 Ÿ g y
y3
C1
3
x3
y3
xy C1
3
3
x3
y3
xy
C
3
3
f 4, 5
43 C
Solution: x3 y 3 3 xy
129
© 2010 Brooks/Cole, Cengage Learning
C
Section 16.1
19.
y
dx ª¬ln x 1 2 yº¼ dy
x 1
wM
wy
f x, y
³M
f y x, y
ln x 1 g c y
³M
f x, y
³
2y Ÿ g y
y 2 C1
y ln x 1 y C1
y ln x 1 y 2
4: 4 ln 2 1 16
C Ÿ C
16
x3
xy 2 C1
3
x3
xy 2
3
y3
x
y
dx 2
dy
x2 y2
x y2
0
x3
xy 2
3
x2 y 2
³M
f x, y
2
1
ln x 2 y 2 g y
2
Ÿ gc y
ln x 2 y 2
16
0
x, y dx
³e
sin 3 y dx
Ÿ gc y
1 3x
e sin 3 y g y
3
0 Ÿ g y
1 3x
e sin 3 y C1
3
f x, y
3x
e sin 3 y
3x
2y 1
Ÿ g y
y 2 y C1
0
Solution: e sin 3 y
dx
x 2 y 3 x 2 y 2 y C1
3: 9 3
C
C
6
6
24. 2 xy 2 4 dx 2 x 2 y 6 dy
0
4 xy
f y x, y
wN
Exact
wx
³ M x, y dx
2
³ 2 xy 4 dx
2x2 y gc y
Ÿ g y
C1
f x, y
x2 y 2 4x g y
2x2 y 6 Ÿ gc y
6
6 y C1
x 2 y 2 4 x 6 y C1
x2 y2 4 x 6 y
y 1
C
S: C
Ÿ gc y
f x, y
e3 x cos 3 y g c y
f y x, y
2
2 y x2 1
Solution: x 2 y 3 x 2 y 2 y
wM
wy
wN
Exact
wx
³M
3x
x2 gc y
f y x, y
y0
21. e3 x sin 3 y dx e3 x cos 3 y dy
f x, y
0
³ 2 xy 9 x
x, y dx
x 2 y 3x 2 y 2 y
ln 16
3e3 x cos 3 y
36
wN
Exact
wx
³M
f x, y
C
Solution: x 2 y 2
y0
C1
C
ln x 2 y 2
C
12
x 2 y 3x3 g y
0 Ÿ g y
4: ln 16
C1
12
2x
f x, y
1
ln x 2 y 2 C1
2
f x, y
wM
wy
wM
wy
y
gc y
x2 y 2
f y x, y
y0
1: 9 3
23. 2 xy 9 x 2 dx 2 y x 2 1 dy
x, y dx
x
³ x 2 y 2 dx
0 Ÿ g y
C
Solution: x3 3 xy 2
wN
Exact
wx
x3
xy 2 g y
3
x 2 y 2 dx
f x, y
16
2 xy
x, y dx
2 xy g c y Ÿ g c y
Solution: y ln x 1 y 2
wM
wy
0
f y x, y
C
457
wN
Exact
wx
2y
2
f x, y
20.
wM
wy
y ln x 1 g y
x, y dx
Ÿ gc y
y2
22. x 2 y 2 dx 2 xy dy
0
wN
Exact
wx
1
x 1
Exact First-Order Equations
C
8: 1 64 4 48
13
C
Solution:
0
x2 y 2 4x 6 y
13
© 2010 Brooks/Cole, Cengage Learning
458
Chapter 16
Additional Topics in Differential Equations
25. y dx x 6 y 2 dy
wN wx wM wy
M
k
Integrating factor: e ³
Exact equation:
28. 5 x 2 y 2 dx 2 y dy
0
2
y
y dy
wM wy wN wx
N
k y
eln y
2
§ x
·
1
dx ¨ 2 6 ¸ dy
y
y
©
¹
gc y
6
g y
6 y C1
x
6y
y
0
Integrating factor: e ³
g y
C1
y
x x
2
g y
C1
2
x
h x dx
2
dx
h x
eln cos x
1
x2
0
gc y
0
g y
C1
x sin x cos x y sin x
30. 2 x 2 y 1 dx x3 dy
0
wM wy wN wx
N
1
x
dx
h x
eln 1 x
Integrating factor: e ³
h x dx
h x
eln x
2
5x gc y
0
g y
C1
y
g y
x
1
x2
0
gc y
0
g y
C1
x 2 y ln x
C
31. y 2 dx xy 1 dy
0
wN wx wM wy
M
k
Integrating factor: e ³
C
0
x 2 y ln x g y
f x, y
y·
1
§
Exact equation: ¨ 5 2 ¸ dx dy
x
x
©
¹
f x, y
1
x
1·
§
Exact equation: ¨ 2 xy ¸ dx x 2 dy
x¹
©
0
0
C
C
2
x
cos x
x sin x cos x y sin x g y
f x, y
eln x
C
Exact equation: x y cos x dx sin x dy
h x
0
0
tan x
hx
Integrating factor: e ³
wM wy wN wx
N
y
x
0
hx
Integrating factor: e ³
0
y
g y
x
27. 5 x 2 y dx x dy
5x gc y
29. x y dx tan x dy
y·
1
§
Exact equation: ¨ 2 x 2 ¸ dx dy
x
x
©
¹
0
e x
5 x 2e x 10 xe x 10e x y 2e x g y
f x, y
wM wy wN wx
N
wM wy wN wx
N
gc y
dx
y 2e x 5 x 2e x 10 xe x 10e x
26. 2 x 3 y dx x dy
x2 h x
Exact equation: 5 x 2 y 2 e x dx 2 ye x dy
C
f x, y
1
hx
Integrating factor: e ³
1
y2
x
g y
y
f x, y
0
y dy
1
y
k y
eln 1 y
§
1·
Exact equation: y dx ¨ x ¸ dy
y¹
©
1
y
0
xy g y
f x, y
1
y
gc y
g y
ln y C1
xy ln y
C
© 2010 Brooks/Cole, Cengage Learning
Section 16.1
32.
35. 4 x 2 y 2 y 2 dx 3x3 4 xy dy
x 2 2 x y dx 2 dy
0
wM wy wN wx
N
h x
1
2
hx
Integrating factor: e ³
g y
x2
4 x3 y 3 2 xy 4 dy 3 x 4 y 2 4 x 2 y 3 dy
dx 2e
x2
dy
0
2 x2 2 x 4 y ex 2 g y
0
0
g y
C1
C1
x y x2 y 4
4 3
x 2x 4 y e
33. 2 y dx x sin
x2
y dy
wN wx wM wy
M
gc y
sin
g y
2 cos
Exact equation:
k y
e
ln 1
3 x 2 y 3 5 x 4 y 2 dx 3 x3 y 2 2 x5 y dy
1
y
0
g y
C1
x y x5 y 2
3 3
C
37. y 5 x 2 y dx 2 xy 4 2 x3 dy
y
y C1
y
Exact equation:
§ y2
§ y
1 ·
x·
¨ 2 2 ¸ dx ¨ 2 2 3 ¸ dy
y ¹
y ¹
© x
© x
C
3
x
hx
Integrating factor: e ³
dx
0
h x
e
ln 1 x3
1
x3
y3
1
g y
2
2x2
x
1
g y
y C1
2
0
gc y
0
g y
C1
y2
x
2
x
y
C
38. y 3 dx xy 2 x 2 dy
0
Integrating factor: x 2 y 2
Exact equation:
C
0
y2
x
2 g y
x
y
f x, y
§ 2 y
§ 3y
·
1·
Exact equation: ¨ 3 3 ¸ dx ¨ 2 1¸ dy
x ¹
© x
© x
¹
3
gc y
0
Integrating factor: x 2 y 3
wM wy wN wx
N
1
y3
y
2x2
x2
gc y
yx g y
34. 2 y 3 1 dx 3 xy 2 x3 dy
f x, y
0
0
x 3 y 3 x5 y 2 g y
f x, y
y
y
y x cos
0
Integrating factor: x 2 y
0
§ x
sin y ·
y dy ¨
¸dy
¨ y
y ¸¹
©
Exact equation: 2
2
y dy
C
36. 3 y 2 5 x 2 y dx 3 xy 2 x3 dy
1
2y
k
Integrating factor: e ³
f x, y
C
0
x4 y3 x2 y 4 g y
f x, y
gc y
2
0
Integrating factor: xy 2
ex 2
Exact equation: x 2 x y e
f x, y
459
Exact equation:
dx
2
gc y
Exact First-Order Equations
f x, y
gc y
g y
y
1
x
y
§1
y
1 ·
dx ¨ 2 ¸ dy
x2
y ¹
©x
0
y
g y
x
1
2
y
1
C1
y
C
© 2010 Brooks/Cole, Cengage Learning
460
Chapter 16
Additional Topics in Differential Equations
39. y dx x dy
0
wM
wy
wN
wx
(a)
1 y
1
, 2 dx dy
2
x x
x
0,
(b)
1 1
x
, dx 2 dy
y2 y
y
0,
(c)
1 1
1
, dx dy
xy x
y
(d)
1
y
x
,
dx 2
dy
x2 y2 x2 y 2
x y2
wM
wy
0,
x2 y 2
x y
2
2
1
x2
wM
wy
1
y2
wM
wy
wN
wx
0
0,
wN
wx
2
40. axy 2 by dx bx 2 y ax dy
wM
wy
Exact equation:
wN
wx
2axy b,
0
wN
wx
2bxy a,
wM
wy
wM
only if a
wx
b.
Integrating factor: x m y n
ax m 1 y n 2 bx m y n 1 dx bx m 2 y n 1 ax m 1 y n dy
wM
wy
½
a n 2 x m 1 y n 1 b n 1 x m y n ° a n 2
°
¾
b m 2 x m 1 y n 1 a m 1 x m y n ° b n 1
°¿
wN
wx
2 b a ½° abn b 2 m
¾
a b °¿ abn a 2 m
an bm
bn am
2b a a b
2ab a 2 a 2 b 2
a b
2a b
a b
n
y
dy
dx
x y
2
y x
2
2
2
i x
y
y dy x dx
2b a
a b
a b
bn
41. F x, y
a m 1
aa b
2
m
§ 2b a ·
bn a¨ ¸
© a b¹
bm 2
2b b a
a b m
2
0
0
x
x y
2
2
j
b 2a b
a b
x y
2
dy
dx
x dy y dx
0
xy
C
C
2
i y
x y2
2
j
y
x
Family of hyperbolas
Family of circles
4
4
c=4
x
42. F x, y
c=9
−6
6
−6
6
c=1
−4
−4
© 2010 Brooks/Cole, Cengage Learning
Section 16.1
§
x ·
4 x 2 yi ¨ 2 xy 2 2 ¸ j
y ¹
©
43. F x, y
45.
3
wM
wy
8y
dy
2 y4 1
ln 2 y 4 1
§1·
ln ¨ 2 ¸ ln C
©x ¹
2
dx
x
2 x2 y 4 x2
3y2
C1
2
x 2 2 xy 3 y 2
C
g y
C
−3
Particular solution: x 2 2 xy 3 y 2
3
c=2
46.
−2
y
dy
dx
wM
wy
3
3
0
wN
wx
2x
x2 y g y
f x, y
§ 1 ·
ln C
ln ¨
2¸
©1 x ¹
C
1 x2
C, C
2 xy
x2 y2
2 xy dx x 2 y 2 dy
1 x 2 i 2 xyj
2 xy
dy
dx
1 x2
1
2x
dy
dx
y
1 x2
ln y
1, 4 4 3
Initial condition: y 2
c=6
44. F x, y
x2
xy g y
2
3y
gc y
2
c=4
0
wN
wx
1
f x, y
C
x2
2 y4 1
461
y x
3y x
x y dx 3 y x dy
y
1
2x
4 xy 3
dy
dx
dy
dx
Exact First-Order Equations
gc y
y2
g y
y3
C1
3
3x 2 y y 3
C
Initial condition: y 0
2, 8
C
4
Particular solution: 3x 2 y y 3
−6
8
6
−4
47. E x
20 x y
2 y 10 x
x dy
y dx
20 xy y 2 dx 10 x 2 2 xy dy
wM
wy
20 x 2 y
wN
wx
10 x 2 y xy 2 g y
f x, y
gc y
0
g y
C1
10 x y xy 2
2
0
K
500, 100 d x, K
Initial condition: C 100
10 x 2 y xy 2
xy 2 10 x 2 y 25,000,000
y
25,000,000
25,000,000
0 Quadratic Formula
10 x 2 100 x 4 4 x 25,000,000
2x
5 x2 x 4 1,000,000 x
x
© 2010 Brooks/Cole, Cengage Learning
462
48.
Chapter 16
Additional Topics in Differential Equations
dy
P x y
Q x
dx
dy P x y dx
Q x dx
ª¬P x y Q x º¼ dx dy
0
P x y Q x and N x, y
M x, y
1
d
d
ªP x y Q x º¼ >1@
dy ¬
dx
1
M y x, y N x x, y
N x, y
P x
e³
By Theorem 16.2, u x
P x dx
is an integrating factor.
50. (a) y 5 | 6.6980
49. (a) y 4 | 0.5231
16
2
−1
5
−3
6
−2
−1
(b)
xy
x y2
dy
dx
(b)
2
xy dx x 2 y 2 dy
1
ª N x M y ¼º
M¬
alone.
0
xy dx x y y
2
2
f x, y
³ xy
6 x y 2 dx 2 xy 3 y 2 dy
1
>2 x x@
xy
1y
Integrating factor: e ³
3
e
dy
ln y
y
³ 6x y 2 dx
3x 2 xy 2 y 3
f x, y
2
x y
g y
2
dx
f x, y
Initial condition: y 2
Particular solution:
1,
x2 y 2
y4
2
4
For x
(c)
9
4
C
Particular solution: 3x xy y
2
y4
C1
4
4 1
2 4
y 3 C1
1 Ÿ 1
Initial condition: y 0
C
3x 2 xy 2 g y
2 xy g c y Ÿ g y
fy
0
0
wN
Exact
wx
2 2
x2 y2
y4
2
4
(c)
2y
f x, y
dy
x2 y gc y Ÿ g y
For x
wM
wy
1
function of y
y
f y x, y
2 x2 y 2 y4
6x y2
y 3 y 2x
dy
dx
2
5, 75 5 y y 1
2
3
C
3
1
0Ÿ y
6.695.
16
C
−3
9
or
4
6
−2
9.
4, 32 y 2 y 4
9 Ÿ y4
0.528
2
−1
5
−1
© 2010 Brooks/Cole, Cengage Learning
Section 16.1
51. (a) y 4 | 0.408
463
52. (a) y s | 6.708
16
2
−1
5
−3
6
−2
−1
xy
x y2
dy
(b)
dx
1
ª N x M y ¼º
M¬
6 x y 2 dx 2 xy 3 y 2 dy
0
wM
wy
1
>2 x x@
xy
1
function of y alone.
y
³1 y dy
e
xy dx x y y dy
0
Integrating factor: e
2
2
f x, y
³ xy
3
ln y
f y x, y
x2 y gc y Ÿ g y
f x, y
x2 y2
y4
2
4
Initial condition: y 2
Particular solution:
For x
3x 2 xy 2 y 3
y 3 C1
C
1 Ÿ 1
Initial condition: y 0
C
Particular solution: 3x 2 xy 2 y 3
y4
C1
4
5, 75 5 y 2 y 3 1
For x
(c)
1
0 Ÿ y
6.695.
16
C
1,
4 1
2 4
x2 y2
y4
2
4
2 x2 y 2 y4
4, 32 y 2 y 4
3x 2 xy 2 g y
y 2 dx
2 xy g c y Ÿ g y
f x, y
x y
g y
2
dx
³ 6x fy
y
0
wN
Exact
wx
2y
f x, y
2 2
2
6x y2
y 3 y 2x
dy
dx
(b)
2
xy dx x 2 y 2 dy
(c)
Exact First-Order Equations
9
4
−3
C
The solution is less accurate. For #50, Euler’s
Method gives y 5 | 6.698, whereas in #52, you
9
or
4
9.
9 Ÿ y4
6
−2
obtain y 5 | 6.708. The errors are
6.695 6.698
6.695 6.708
0.528
0.003 and
0.013.
2
53. If M and N have continuous partial derivatives on an
open disc R, then M x, y dx N x, y dy
0 is exact
−1
5
if and only if
−1
wM
wy
The solution is less accurate. For #49, Euler’s
Method gives y 4 | 0.523, whereas in #51, you
54. See Theorem 16.2.
obtain y 4 | 0.408. The errors are
55. False
0.528 0.523
0.528 0.408
0.005 and
0.120.
wM
wy
2 x and
wN
wx
wN
.
wx
2 x
56. False
y dx x dy
0 is exact, but xy dx x 2 dy
0 is not
exact.
57. True
w
ª f x M º¼
wy ¬
wM
w
and
ª g y N º¼
wy
wx ¬
wN
wx
© 2010 Brooks/Cole, Cengage Learning
464
Chapter 16
Additional Topics in Differential Equations
58. True
61. M
w
ª f x º¼
wy ¬
w
ª g y º¼
wx ¬
0 and
xy 2 kx 2 y x3 , N
59. M
wM
wy
2 xy kx 2 ,
wN
wx
wM
wy
wN
Ÿ k
wx
3
ye 2 xy 2 x, N
60. M
wM
wy
e
wM
wy
wN
Ÿ k
wx
2 xy
2 xye
0
x3 x 2 y y 2
3x 2 2 xy
2 xy
wN
,
wx
ke
2 xy
2kxye
wM
wy
g c y sin x,
wN
wx
y2 f c x
wM
wy
wN
: g c y sin x
wx
f c x y2
y2 Ÿ g y
fc x
sin x Ÿ f x
g y ey, N
wM
wy
2 xy
y3
C1
3
gc y
62. M
kxe2 xy 2 y
y2 f x
g y sin x, N
xy
gc y e y g y e y ,
gc y e y g y e y
1
cos x C2
ª¬ g y e y º¼c
wN
wx
y
y
y
y2
C
2
g y ey
ª y2
º
e y «
C»
¬2
¼
g y
Section 16.2 Second-Order Homogeneous Linear Equations
1.
y
C1e 3 x C2 xe 3 x
yc
3C1e 3 x C2e 3 x 3C2 xe3 x
ycc
9C1e 3 x 6C2e3 x 9C2 xe 3 x
9C1e3 x 6C2e 3 x 9C2 xe 3 x 18C1e 3 x 6C2e 3 x 18C2 xe 3 x 9C1e 3 x 9C2 xe 3 x
ycc 6 yc 9 y
0
y approaches zero as x o f.
y
6
x
−4
−2
2.
2
4
6
y
C1e 2 x C2e 2 x
yc
2C1e 2 x 2C2e 2 x
ycc
4C1e 2 x 4C2e 2 x
ycc 4 y
4y 4y
3.
4y
0
The graphs are different combinations of the graphs of
e2 x and e2 x .
y
C1 cos 2 x C2 sin 2 x
yc
2C1 sin 2 x 2C2 cos 2 x
ycc
4C1 cos 2 x 4C2 sin 2 x
ycc 4 y
4 y 4 y
4 y
0
The graphs are basically the same shape, with left and
right shifts and varying ranges.
y
y
2
1
x
2
2
−1
x
−4
−2
2
4
−2
−2
© 2010 Brooks/Cole, Cengage Learning
Section 16.2
4. y
yc
C1e x cos 3 x C2e x sin 3 x
Second-Order Homogeneous Linear Equations
e x C1 cos 3 x C2 sin 3 x
e x C1 cos 3x C2 sin 3 x e x 3C1 sin 3x 3C2 cos 3 x
e x >C1 cos 3 x C2 sin 3x 3C1 sin 3 x 3C2 cos 3x@
ycc
y
>C1 cos 3x C2 sin 3x 3C1 sin 3x 3C2 cos 3x@
e >3C1 sin 3 x 3C2 cos 3 x 9C1 cos 3x 9C2 sin 3x@
2 yc 10 y
e x >C1 cos 3 x C2 sin 3x 3C1 sin 3 x 3C2 cos 3 x
e
x
x
ycc 465
1
x
2
3
4
3C1 sin 3 x 3C2 cos 3 x 9C1 cos 3 x 9C2 sin 3 x@
2e x >C1 cos 3 x C2 sin 3 x 3C1 sin 3 x 3C2 cos 3x@
10e x >C1 cos 3x C2 sin 3 x@
0
y approaches zero as x o f. The graphs are the same only reflected.
5. ycc yc
11. ycc 6 yc 9 y
0
Characteristic equation: m 2 m
Roots: m
y
y
Roots: m
0, 2
7. ycc yc 6 y
Roots: m
y
Characteristic equation: m m 6
y
3, 2
y
0
Characteristic equation: m 6m 5
Roots: m
y
y
0
Roots: m
1,
2
Roots: m
C1e 1 2 x C2e2 x
Roots: m
y
y
1, 3
4 4
C1e 1 4 x C2e 3 4 x
0
0
Characteristic equation: m 2 4
Roots: m
y
0
i, i
C1 cos x C2 sin x
16. ycc 4 y
0
Characteristic equation: 16m 2 16m 3
0
Characteristic equation: m 2 1
2
10. 16 ycc 16 yc 3 y
C1e 2 3 x C2 xe 2 3 x
15. ycc y
0
0
2, 2
3 3
Roots: m
C1e x C2e5 x
Characteristic equation: 2m 2 3m 2
0
Characteristic equation: 9m 2 12m 4
1, 5
9. 2 ycc 3 yc 2 y
y
0
0
C1e 1 4 x C2 xe 1 4 x
14. 9 ycc 12 yc 4 y
2
0
1, 1
4 4
Roots: m
C1e3 x C2e2 x
8. ycc 6 yc 5 y
C1e5 x C2 xe5 x
Characteristic equation: 16m 2 8m 1
0
0
5, 5
13. 16 ycc 8 yc y
0
2
0
Characteristic equation: m 2 10m 25
0
C1 C2e 2 x
Roots: m
C1e3 x C2 xe3 x
12. ycc 10 yc 25 y
0
0
3, 3
Roots: m
0, 1
Characteristic equation: m 2 2m
y
Characteristic equation: m 2 6m 9
0
C1 C2e x
6. ycc 2 yc
0
0
2i, 2i
C1 cos 2 x C2 sin 2 x
© 2010 Brooks/Cole, Cengage Learning
466
Chapter 16
17. ycc 9 y
Additional Topics in Differential Equations
25. y 4 y
0
Characteristic equation: m 9
2
Roots: m
C1e3 x C2e3 x
18. ycc 2 y
y
0
Characteristic equation: m 2 2
2,
Roots: m
y
C1e
Roots: m
2x
y
1
3i, 1 20. ycc 4 yc 21 y
Roots: m
3x
0
Characteristic equation: m 2 4m 21
2
17i, 2 0
Characteristic equation: m 2 3m 1
3
y
5 3
,
2
ª 3
C1e ¬
5 2º x
¼
22. 3 ycc 4 yc y
C2
y
ª 2 C1e ¬
y
5 2¼º x
7 2 ,
3
7 3º x
¼
ª 2 e¬
y
3
C2
y
7 3º x
¼
7i 2 ,
3
0
Characteristic equation: 2m 2 6m 7
Roots: m
y
0
7i
e
32x
3
2
5i 3 ,
0
5i
2
ª
§ 5 ·
§ 5 ·º
x ¸¸ C2 sin ¨¨
x ¸¸»
«C1 cos¨¨
2
©
¹
© 2 ¹¼»
¬«
0
0
1, 1, 1
C1e x C2 xe x C3 x 2e x
31. ycc 100 y
0
ª
§ 7 ·
§ 7 ·º
e 2 3 x «C1 cos¨¨
x ¸¸ C2 sin ¨¨
x ¸¸»
3
©
¹
© 3 ¹¼»
¬«
24. 2 ycc 6 yc 7 y
C1e x e x C2 cos 2 x C3 sin 2 x
Roots: m
7
0
1, 1 2i, 1 2i
Characteristic equation: m3 3m 2 3m 1
0
Characteristic equation: 9m 2 12m 11
2
0
30. yccc 3 ycc 3 yc y
0
23. 9 ycc 12 yc 11 y
Roots: m
1, 1, 1
Characteristic equation: m3 3m 2 7 m 5
2
2 3
0
C1e x C2 xe x C3e x
Roots: m
Characteristic equation: 3m 2 4m 1
Roots: m
0
29. yccc 3 ycc 7 yc 5 y
0
5
ª3 e¬
C1e C2e 2 x C3e3 x
Roots: m
17 x
0
1, 2, 3
Characteristic equation: m3 m 2 m 1
y
21. ycc 3 yc y
0
x
28. yccc ycc yc y
0
17i
17 x C2 sin
e 2 x C1 cos
Roots: m
C1 C2 x C3e x C4e x
Characteristic equation: m3 6m 2 11m 6
y
y
0
0, 0, 1, 1
27. yccc 6 ycc 11 yc 6 y
0
3i
3 x C2 sin
e x C1 cos
Roots: m
0
0
Characteristic equation: m 2 2m 4
Roots: m
1, 1, i, i
Characteristic equation: m 4 m 2
y
19. ycc 2 yc 4 y
0
C1e x C2e x C3 cos x C4 sin x
26. y 4 ycc
0
2
C2 2x
0
Characteristic equation: m 4 1
3, 3
Roots: m
y
0
0
y
C1 cos 10 x C2 sin 10 x
yc
10C1 sin 10 x 10C2 cos 10 x
(a) y 0
2: 2
C1
yc 0
0: 0
10C2 Ÿ C2
Particular solution: y
2 cos 10 x
(b) y 0
0: 0
C1
yc 0
2: 2
10C2 Ÿ C2
Particular solution: y
(c) y 0
yc 0
1: 1
3: 3
0
1
5
1
5
sin 10 x
C1
10C2 Ÿ C2
Particular solution: y
3
10
cos 10 x 3
10
sin 10 x
© 2010 Brooks/Cole, Cengage Learning
Section 16.2
32. y
C sin
yc
ycc
3t
3C sin
yc 0
Z
34. ycc 7 yc 12 y
3t
3C cos
ycc Z y
Second-Order Homogeneous Linear Equations
Roots: m
3C sin
3t Z sin
0 Ÿ Z
3C
5: 5
3t
3C Ÿ C
Solving simultaneously: C1
Particular solution: y
1, yc 0
0, y 0
4
Characteristic equation: m 2 m 30
y
C1 C2 , 4
1
, C2
11
Solving simultaneously: C1
36. ycc 2 yc 3 y
4 12
1 r
2
y
e x C1 cos 2 x C2 sin
yc
e x C1
2 sin
Initial conditions: y 0
2
C1
yc 0
1
C2
y
yc
1C
3 1
2
C2
C1 cos 4 x C2 sin 4 x
yc
4C1 sin 4 x 4C2 cos 4 x
Initial conditions: y 0
0
C1
yc 0
2
4C2 Ÿ C2
1
2
2 x e x C1 cos
2, yc 0
C2
3
sin
2
2x 2 x C2 sin
Particular solution: y
sin 4 x
2x
3
2
2 2 Ÿ C2
·
2x¸
¹
38. 4 ycc 4 yc y
1
0
0, y 0
3, yc 0
Roots: m
2, yc 0
1
1
Characteristic equation: 4m 2 4m 1
2m 1
13 C2 xe 1 3 x C2e 1 3 x
½°
¾ Ÿ C1
1°¿
1
2
0
2 C1
C1e 1 3 x C2 xe 1 3 x
Initial conditions: y 0
C1
y
1, 1
3 3
1C e 1 3 x
3 1
0
2i
Characteristic equation: 9m 2 6m 1
Roots: m
2
r4i
Particular solution: y
2 cos
§
e x ¨ 2 cos
©
0, y 0
4x
2x
2 x C2
Particular solution: y
6e
0, yc 0
0, y 0
6
9, C2
3x
3
1
Characteristic equation: m 2 2m 3
37. 9 ycc 6 yc y
6C1 5C2
10
11
2, yc 0
0, y 0
2 r
Roots: m
1 6x
e 10e 5 x
11
Particular solution: y
9e
3, 3C1 4C2
Characteristic equation: m 2 16
6C1e6 x 5C2e5 x
Initial conditions:
y0
1, yc 0
4, 1
Roots: m
35. ycc 16 y
0
6,5
C1e6 x C2e 5 x , yc
3C1e3 x 4C2e4 x
Initial conditions:
y0
3, yc 0
3, C1 C2
5 3
and
3
0
3, 4
C1e3 x C2e 4 x , yc
y
5 3
Roots: m
3
Characteristic equation: m 2 7 m 12
3t
33. ycc yc 30 y
3, yc 0
0, y 0
467
2
0
0
12 , 12
y
C1e 1 2 x C2 xe 1 2 x
yc
12 C1e 1 2 x 12 C2 xe 1 2 x C2e 1 2 x
Initial conditions:
2, C2
1
3
2e x 3 13 xe x 3
y0
3
C1
yc 0
1
12 C1 C2 Ÿ C2
Particular solution: y
3e x 2 5
2
5 xe x 2
2
© 2010 Brooks/Cole, Cengage Learning
468
Chapter 16
Additional Topics in Differential Equations
39. ycc 4 yc 3 y
0, y 0
1, y 1
Characteristic equation: m 2 4m 3
Roots: m
y
C1e x C2e3 x
y1
3: C1e C2e3
y
1
e3 3
, C2
e3 e
2, y S
0, y 0
yS
5: C2
Solution: y
41. ycc 9 y
y
2
5
2 cos
0, y 0
1x
2
5 sin
1x
2
3, y S
y
0
yS
5: C1
5
0, y 0
3, y 2
0
3 r 6i
C1e3 x cos 6 x C2e 3 x sin 6 x
y0
4: C1
yS
8: C1e 3S
0
Characteristic equation: 4m 20m 21
m
0
4
8
0
2k r
4k 2 4k
2
k 0
k2 1
(a) For k 1 and k ! 1, k 2 1 ! 0 and there are 2
distinct real roots.
3 7
,
2 2
C1e 3 2 x C2e 7 2 x
y0
3: C1 C2
y2
0: C1e 3 C2e 7
(b) For k
r1, k 2 1
0 and the roots are repeated.
(c) For 1 k 1, the roots are complex.
3
Solving simultaneously, C1
Solution: y
36 180
2
Characteristic equation: m 2 2km k
2
y
6 r
48. ycc 2kyc ky
No solution
Roots: m
8
47. Two functions y1 and y2 are linearly independent if the
0 is the
only solution to the equation C1 y1 C2 y2
trivial solution C1
C2
0.
3
42. 4 ycc 20 yc 21 y
4, y S
0, y 0
46. Answers will vary. See Theorem 16.4.
C1 cos 3 x C2 sin 3 x
3: C1
§ 1 2e7 2 · 7 2 x
2e 7 2 x ¨
¸ xe
e7 2
©
¹
45. No, it is not homogeneous because of the nonzero term
sin x.
r3i
y0
1 2e7 2
e7 2
1 Ÿ C 2
No solution
5
Characteristic equation: m 2 9m
Roots: m
1: C1e7 2 C2e 7 2
Roots: m
C1 cos 12 x C2 sin 12 x
2: C1
y1
2
Characteristic equation: m 2 6m 45
0
r 12 i
y0
2: C1
44. ycc 6 yc 45 y
5
Characteristic equation: 4m 2 1
Roots: m
y0
Solution: y
e3 3 x
3 e 3x
e 3
e
e3 e
e e
Solution: y
40. 4 ycc y
3e
e3 e
0
C1e 7 2 x C2 xe 7 2 x
3
Solving simultaneously, C1
1
2, y 1
7 7
,
2 2
Roots: m
1: C1 C2
0, y 0
Characteristic equation: 4m 2 28m 49
0
1, 3
y0
y
43. 4 ycc 28 yc 49 y
3
0 Ÿ C1 C2e 4
3
, C2
e 1
4
3 3 2 x
3e 4 7 2 x
e
4
e
e 1
e 1
0
49. By Hooke’s Law, F
3e 4
e 1
kx
F
x
k
4
Also, F
ma, and m
4
So, y
32
23
F
a
32
32
48.
1.
1
cos 4 3t
2
© 2010 Brooks/Cole, Cengage Learning
Section 16.2
50. By Hooke’s Law, F
kx
k
F
x
C1 cos
k m
32
32
k m t C2 sin
48
1.
1
2
yc t
4 3 C1 sin 4 3t 4 3C2 cos 4 3t
yc 0
1
1
Ÿ C2
2
8 3
1
1
cos 4 3t sin 4 3t
2
8 3
4 3 C2
C1 cos 4 3t C2 sin 4 3t
2
3
C1
yc t
4 3 C1 sin 4 3t 4 3 C2 cos 4 3t
yc 0
4 3 C2
1
Ÿ C2
2
1
8 3
3
24
2
3
sin 4 3t
cos 4 3t 3
24
53. By Hooke’s Law, 32
m
1
2
1
2
y0
yt
C1
yt
2
, yc 0
3
1
, yc 0
2
y0
k mt ,
469
C1 cos 4 3t C2 sin 4 3t
Initial conditions: y 0
4 3
Initial conditions: y 0
y
48.
2
cos 4 3t .
3
So, y
51. y
52. y
32
23
F
a
ma, and m
Also, F
Second-Order Homogeneous Linear Equations
w g
32 32
k 2 3 , so k
48. Moreover, because the weight w is given by mg, it follows that
1. Also, the damping force is given by 1 8 dy dt . So, the differential equation for the
oscillations of the weight is
§ d2y ·
m¨ 2 ¸
© dt ¹
§ d 2 y · 1 § dy ·
m¨ 2 ¸ ¨ ¸ 48 y
© dt ¹ 8 © dt ¹
1 § dy ·
¨ ¸ 48 y
8 © dt ¹
0.
In this case the characteristic equation is 8m 2 m 384
So, the general solution is y t
§
e t 16 ¨¨ C1 cos
©
Using the initial conditions, you have y 0
yc t
yc 0
ª§
e t 16 «¨¨ ¬«©
12,287
C ·
C1 2 ¸¸ sin
16
16 ¹
12,287
C
C2 1
16
16
0 Ÿ C2
0 with complex roots m
12,287t
C2 sin
16
C1
1 16 r
12,287 16 i.
12,287t ·
¸¸.
16
¹
1
2
12,287t §
¨¨
16
©
12,287
C ·
C2 1 ¸¸ cos
16
16 ¹
12,287t º
»
16
¼»
12,287
24,574
and the particular solution is
yt
e t 16 §
¨ cos
2 ¨©
12,287t
12,287
sin
16
12,287
12,287t ·
¸¸.
16
¹
© 2010 Brooks/Cole, Cengage Learning
470
Chapter 16
Additional Topics in Differential Equations
54. By Hooke’s Law, 32
k 2 3 , so k
§ d2y ·
m¨ 2 ¸
© dt ¹
48. Also, m
1. The damping force is given by 1 4 dy dt . So,
0.
The characteristic equation is 4m 2 m 192
§
e t 8 ¨¨ C1 cos
©
32 32
1 § dy ·
¨ ¸ 48 y
4 © dt ¹
§ d 2 y · 1 § dy ·
m¨ 2 ¸ ¨ ¸ 48 y
© dt ¹ 4 © dt ¹
yt
w g
3071t
C2 sin
8
1 8 r
0 with complex roots m
3071 8 i. So, the general solution is
3071t ·
¸.
8 ¸¹
Using the initial conditions, you have
1
2
y0
C1
yc t
ª§
e t 8 «¨¨ ¬«©
3071
C ·
C1 2 ¸¸ sin
8
8 ¹
3071
C
C2 1
8
8
yc 0
0 Ÿ C2
3071t §
¨¨
8
©
3071C2
C ·
1 ¸¸ cos
8
8¹
3071t º
»
8 ¼»
3071
6142
and the particular solution is
yt
e t 8 ª
«cos
2 ¬
55. ycc 9 y
3071t
3071
sin
8
3071
3071t º
».
8 ¼
57. ycc 2 yc 10 y
0
Undamped vibration
Period:
Damped vibration
2S
3
Matches (c)
58. ycc yc Matches (b)
56. ycc 25 y
0
37 y
4
0
Damped vibration
0
Matches (a)
Undamped vibration
Period:
2S
5
Matches (d)
59. Because m
a·
§
¨m ¸
2¹
©
a 2 is a double root of the characteristic equation, you have
2
m 2 am a2
4
0
and the differential equation is ycc ayc a 2 4 y
y
ycc ayc C1 C2 x e a 2 x
yc
Ca ·
§ C1a
C2 2 x ¸ e a 2 x
¨
2
2 ¹
©
ycc
§ C1a 2
C a2 ·
aC2 2 x ¸e a 2 x
¨
4 ¹
© 4
a2
y
4
0. The solution is
ª§ C a 2
C a2 · § C a2
C a2 · § C a2
C a 2 ·º
e a 2 x «¨ 1 C2 a 2 x ¸ ¨ 1 C2 a 2 x ¸ ¨ 1 2 x ¸»
4 ¹ © 2
2 ¹ © 4
4 ¹¼
© 4
0.
© 2010 Brooks/Cole, Cengage Learning
Section 16.2
Second-Order Homogeneous Linear Equations
471
D r E i are roots to the characteristic equation, you have
60. Because m
¬ªm D E i ºª
¼¬m D E i º¼
m 2 2D m D 2 E 2
and the differential equation is ycc 2D yc D 2 E 2 y
0
0. Note: i 2
1. The solution is
y
eD x C1 cos E x C2 sin E x
yc
eD x ª¬ C1D C2 E cos E x C2D C1E sin E x
ycc
eD x ª¬ C1D 2 C1E 2 2C2DE cos E x C2D 2 C2 E 2 2C1DE sin E xº¼
eD x ª¬ 2C1D 2 2C2DE cos E x 2C2D 2 2C1DE sin E xº¼
2D yc
eD x ª¬ C1D 2 C1E 2 cos E x C2D 2 C2 E 2 sin E xº¼
D2 E2 y
So, ycc 2D yc D 2 E 2 y
0.
61. False. The general solution is y
C1e3 x C2 xe3 x .
62. True
63. True
64. False. The solution y
x 2e x requires that m 1 is a triple root of the characteristic equation. Because the characteristic
equation is quadratic, m 1 can be at most a double root.
ebx , a z b
e ax , y2
65. y1
e ax
W y1 , y2
ae
66. y1
ebx
ax
be
e ax , y2
W y1 , y2
bx
b a e ax bx z 0 for any value of x.
e ax sin bx
W y1 , y2
W y1 , y2
dy
dx
d2y
dx 2
e ax axe ax
e 2 ax z 0 for any value of x.
be 2 ax z 0 for any value of x.
x2
x
x2
1
2x
69. x 2 ycc axyc by
(a)
ae ax
ae ax sin bx be ax cos bx aeax cos bx be ax sin bx
x, y2
Let x
xe ax
e ax cos bx
be 2 ax sin 2 bx be 2 ax cos 2 bx
68. y1
e ax
e ax cos bx, b z 0
eax sin bx, y2
67. y1
xe ax
x 2 z 0 for x z 0.
0, x ! 0
et .
dy dt
dx dt
e t
dy
dt
d dt ª¬e t dy dt º¼
et
x 2 ycc axyc by
ª d2y
dy º
e t «e t 2 e t »
dt ¼
¬ dt
ªd 2 y
dy º
e 2t « 2 »
dt ¼
¬ dt
0
ª § d2y
dy ·º
t § t dy ·
e 2t «e 2t ¨ 2 ¸» ae ¨ e
¸ by
dt
dt
© dt ¹
¹¼
¬ ©
d2y
dy
a 1
by
dt 2
dt
0
0
© 2010 Brooks/Cole, Cengage Learning
472
Chapter 16
Additional Topics in Differential Equations
(b) x 2 ycc 6 xyc 6 y
Let x
0
t
e . From part (a), you have:
2
d y
dy
5
6y
dt 2
dt
m 2 5m 6
0
0
m3 m 2
0
3, m2
m1
2
C1e3t C2e 2t
y
70. ycc Ay
C1e3 ln x C2e 2 ln x
C1e
ln 1 x3
C2 e
ln 1 x 2
C1
C
22 .
x3
x
0
If A ! 0, then y
y0
0 Ÿ C1
yS
0 Ÿ 0
Ax C2 sin
C1 cos
0 and y
C2 sin
AS Ÿ
C2 sin
Ax.
Ax
r1, r 2, r3, ! Ÿ A
A
1, 4, 9, 16, !
So, A is a perfect square integer.
If A 0, then y
C1e
y0
0 Ÿ C1 C2
yS
0 Ÿ C1e
If A
0, then ycc
AS
Ax
C2 e 0
C2 e AS
Ax
.
½°
¾ C1
0°¿
C2
0 Ÿ y
0 and the initial condition gives y
0
0.
Section 16.3 Second-Order Nonhomogeneous Linear Equations
1. y
2e 2 x 2 cos x
yc
4e 2 x 2 sin x
ycc
8e 2 x 2 cos x
ycc y
2. y
8e2 x 2 cos x 2e 2 x cos x
2 sin x 12 x sin x
y'
2 cos x 12 x cos x ycc
2 sin x 12 x sin x cos x
ycc y
3. y
1
2
sin x
2 sin x 12 x sin x cos x 2 sin x 1
2
sin x
cos x
3 sin x cos x ln sec x tan x
yc
3 cos x 1 sin x ln sec x tan x
ycc
3 sin x tan x cos x ln sec x tan x
ycc y
4. y
10e 2 x
3 sin x tan x cos x ln sec x tan x 3 sin x cos x ln sec x tan x
tan x
5 ln sin x cos x x sin x
yc
5 ln sin x sin x cos x cot x sin x x cos x
ycc
6 cos x cos x cos x ln sin x cos x csc 2 x 1 sin x cot x x
6 sin x sin x ln sin x cos x cot x x
5 cos x cos x ln sin x csc x cot x x sin x
ycc y
cos x 5 ln sin x
csc x cot x x sin x 5 ln sin x cos x x sin x
csc x cot x
© 2010 Brooks/Cole, Cengage Learning
Section 16.3
5. ycc 7 yc 12 y
3x 1
7. ycc 8 yc 16 y
ycc 7 yc 12 y
0
m 7 m 12
m3 m4
2
3x
C2 e
yh
C1 e
yp
A0 A1x
ycp
A1, yccp
7 A1 12 A0 A1 x
½°
¾ Ÿ A1
1°¿
161 14 x
7 A1 12 A0
Solution: y p
1,
4
4
ycc yc 6 y
0
m m6
m3 m 2
2
C1 e
yp
A, ycp
C2 e
yccp
A0
0
m 8m 16
m4
yh
C1 e4 x C2 xe 4 x
yp
Ae3 x , ycp
161
0 when m
9 A 24 A 16 A
23
4 A 2 A 3A
Solution: y p
9. ycc 2 yc 15 y
ycc 2 yc 15 y
m 2 2m 15
1
3x
e2 x
0
1 r
0 Ÿ m
Ae2 x , ycp
23
Solution: y p
e
ycc yc 3 y
4 Ÿ A
9 Ae3 x
e3 x
1Ÿ A
yp
6 A
4
9 Ae3 x 8 3 Ae3 x 16 Ae3 x
m2 m 3
3, 2
0 when m
yccp 8 ycp 16 y p
Solution: y p
2 x
2
3 Ae3 x , yccp
3x 1
0
yccp ycp 6 y p
ycc 8 yc 16 y
8. ycc yc 3 y
ycc yc 3 y
6. ycc yc 6 y
yh
3, 4
0
3
3x
0 when m
473
e3 x
2
4x
yccp 7 ycp 12 y p
12 A1
Second-Order Nonhomogeneous Linear Equations
11
2
2 Ae 2 x , yccp
4 Ae 2 x
4 Ae 2 x 2 Ae 2 x 3 Ae 2 x
1Ÿ A
e2 x
1
9
1 2x
e
9
sin x
0
m5 m3
yp
ycp
A sin x B cos x
yccp
A sin x B cos x
0 Ÿ m
5, 3
A cos x B sin x
yccp 2 ycp 15 y p
A sin x B cos x 2 A cos x B sin x 15 A sin x B cos x
A 2 B 15 A sin x B 2 A 15 B cos x
16 A 2 B
2 A 16 B
Solution: y p
sin x
sin x
1½
4
1
,B
¾ Ÿ A
0¿
65
130
4
1
sin x cos x
65
130
10. ycc 4 yc 5 y
ycc 4 yc 5 y
e x cos x
0
m 2 4m 5
0 Ÿ m
4 r 4
2
2 r i
yp
Ae x cos x Be x sin x
ycp
e x A cos x B sin x e x A sin x B cos x
yccp
e x B A sin x A B cos x e x B A cos x A B sin x
yccp 4 ycp 5 y p
e x A cos x B sin x
e x B A sin x A B cos x
e x 2 A sin x 2 B cos x
e x 2 A sin x 2 B cos x 4 e x B A sin x A B cos x 5 Ae x cos x Be x sin x
2 A 4 B A 5 B sin x 2 B 4 A B 5 A cos x
e x cos x
cos x
0½
6 A 9 B
3
2
,B
¾ Ÿ A
9 A 6B 1 ¿
39
39
3 x
2 x
Solution: y p
e cos x e sin x
39
39
© 2010 Brooks/Cole, Cengage Learning
474
Chapter 16
Additional Topics in Differential Equations
11. ycc 3 yc 2 y
14. ycc 9 y
2x
ycc 3 yc 2 y
ycc 9 y
0
m 3m 2
2
5e3 x
0 when m
0
m 9
1, 2.
2
0 when m
yh
C1e x C2e 2 x
yh
C1e 3 x C2e3 x
yp
A 0 A1 x
yp
Axe3 x
ycp
A1
ycp
Ae3 x 3x 1
yccp
0
yccp
Ae3 x 9 x 6
yccp 3 ycp 2 y p
2 A0 3 A1
2 A1
2 A0 3 A1 2 A1 x
1, A0
C1e x C2e 2 x x y
12. ycc 2 yc 3 y
3
2
m 2m 3
0 when m
C2 e
1, 3.
3x
yh
C1 e
yp
A0 A1 x a2 x 2
ycp
A1 2 A2 x
yccp
2 A2
3 A1 4 A2
½
°
¾ A0
1°¿
0
3 A0 2 A1 2 A2
C1 e x C2 e3 x y
13. ycc 2 yc
2e
1
3
m 2m
0 when m
yp
C1 C2 e
Ae
x
yccp 2 ycp
y
A0 A1e x
ycp
yccp
ycp
5
,
27
x 2 94 x 1
,
5
A1
5
27
4
,
9
A2
1
3
25 A0 16 A1e x
0, 2.
2e x or A
2 x
e
3
5 6e x
3
8
A1
C1 C2 x e5 x 83 e x 1
5
4 x ex
16 ycc 8 yc y
0
16m 8m 1
0 when m
yh
C1 C2 x e 1 4 x
yp
A0 A1 x A2e x
ycp
A1 A2e x
yccp
A2 e x
1 1
, .
4 4
A0 8 A1 A1x 9 A2e x
4 x 4e x
4
,
9
or A2
2
3
A1
4, A0
C1 C2 x e
y
yccp
3 Ae x
C1 C2 e
2 x
2 x
A1e x
16. 16 ycc 8 yc y
x2 1
5, 5.
5x
16 yccp 8 ycp y p
0
yh
yp
2
x
ycc 2 yc
2
0 when m
C2 xe
C1e
or A0
3 A2 x 2 3 A1 4 A2 x
1
0
yccp 10 ycp 25 y p
3 A0 2 A1 2 A2
3 A2
5x
yh
y
yccp 2 ycp 3 y p
5
6
5 6e x
m 2 10m 25
0
2
5e3 x or A
15. ycc 10 yc 25 y
ycc 10 yc 25 y
x2 1
ycc 2 yc 3 y
x
3
2
6 Ae3 x
C1e 3 x C2 56 x e3 x
y
0½°
¾ A1
°¿
2
yccp 9 y p
2x
3, 3.
17. ycc 9 y
ycc 9 y
m 9
2
14x
32
32 4 x 94 e x
sin 3 x
0
0 when m
3i, 3i.
yh
C1 cos 3x C2 sin 3 x
yp
A0 sin 3x A1 x sin 3 x A2 cos 3x A3 x cos 3x
yccp
9 A0 6 A3 sin 3 x 9 A1 x sin 3 x
6 A1 9 A2 cos 3x 9 A3 x cos 3 x
yccp 9 y p
6 A3 sin 3 x 6 A1 cos 3x
sin 3 x,
16
A1
0, A3
y
C1 16 x cos 3 x C2 sin 3 x
© 2010 Brooks/Cole, Cengage Learning
Section 16.3
yccc 3 yc 2 y
m3 3m 2
21. ycc yc
2e 2 x
18. yccc 3 yc 2 y
0
C1e C2 xe C3e
yh
yp
A0e
ycp
x
2 x
1, 1, 2.
0 when m
x
A1 xe
2 x
2 x
2 A1 xe
2 x
4 A1 xe2 x
yccp
4 A0 4 A1 e
yccc
p
8 A0 12 A1 e 2 x 8 A1 xe 2 x
9 A1e 2 x
yccc
p 3 ycp 2 y p
C1e C2 xe C3 x
y
19. ycc y
x
2
x
9
e
1, yc 0
x3 , y 0
ycc y
0
m 1
0 when m
2
2e 2 x or A1
2 x
A1 2 A2 x 3 A3 x 2
yccp
2 A2 6 A3 x
x3
1, A2
0, A1
6, A0
0
C1 cos x C2 sin x x3 6 x
yc
C1 sin x C2 cos x 3x 2 6
Initial conditions:
y 0
1, yc 0
0, 1
C2 6, C2
C1 , 0
6
cos x 6 sin x x3 6 x
Particular solution: y
1, yc 0
4, y 0
yccp
A cos x B sin x
yccp ycp
A B cos x A B sin x
A B
0½
¾A
2¿
1, B
C1 C2e x cos x sin x
yc
C2e x sin x cos x
0, yc 0
2i, 2i.
0 when m
C1 cos 2 x C2 sin 2 x
A0
yccp
0
y
C1 cos 2 x C2 sin 2 x 1
yc
2C1 sin 2 x 2C2 cos 2 x
C1
Particular solution: y
1 2e x cos x sin x
1, yc 0
2
0
m m2
1, 2.
0 when m
yh
yp
ycp
C1e x C2e 2 x
A cos 2 x B sin 2 x
yccp
4 A cos 2 x 4 B sin 2 x
2 A sin 2 x 2 B cos 2 x
yccp ycp 2 y p
6 A 2 B cos 2 x 2 A 6 B sin 2 x
3cos 2 x
6 A 2 B
3½
¾A
0¿
209 , B
y
C1e x C2e 2 x yc
C1e 2C2e
x
2 x
1, yc 0
y0
1
C1 2C2 C1 C2
C1 2C2
1, yc 0
0, 6
C2 1,
1
2, C1
cos 2 x 9
20
3
20
9
10
3
20
sin 2 x sin 2 x
3
10
cos 2 x
Initial conditions:
4 or A0
Initial conditions: y 0
C1 C2 1, 3
2
2
4 A0
3,
3 cos 2 x, y 0
ycc yc 2 y
2 sin x
1
y
2 A 6 B
6
0
yh
yp
yccp 4 y p
A sin x B cos x
22. ycc yc 2 y
y
m 4
A cos x B sin x
ycp
Particular solution: y
A3 x3 A2 x 2 A1 6 A3 x A0 2 A2
3
0, 1.
yp
0
ycp
2
C1 C2e
475
x
C2
A0 A1x A2 x 2 A3 x3
ycc 4 y
0 when m
i, i.
yp
20. ycc 4 y
m2 m
Initial conditions: y 0
C1 cos x C2 sin x
or A3
0
A B
0
yh
yccp y p
2
9
0, yc 0
2 sin x, y 0
ycc yc
yh
2 x
2 x
2 A0 A1 e
Second-Order Nonhomogeneous Linear Equations
6, 1
2C2 , C2
C1 1,
3
2, 1
C1 C2 9
,
20
3
10
11
½
20 °
C
17 ¾ 1
10 °
¿
1
,
5
C2
34
Particular solution:
y
1
20
4e x 15e 2 x 9 cos 2 x 3 sin 2 x
3 sin 2 x 1
© 2010 Brooks/Cole, Cengage Learning
476
Chapter 16
23. y ' 4 y
Additional Topics in Differential Equations
xe x xe 4 x , y 0
y ' 4 y
m4
25. ycc y
1
3
sec x
0
ycc y
0
0 when m
m 1
0 when m
2
4.
i, i.
yh
C1 cos x C2 sin x
yp
A0 A1x e x A2 x A3 x 2 e 4 x
yp
v1 cos x v2 sin x
ycp
A0 A1x e x A1e x
vc1 cos x vc2 sin x
yh
Ce 4 x
4 A2 x A3 x e
2
ycp 4 y p
A2 2 A3 x e
4x
3 A0 3 A1x e x A1e x A2e 4 x
2 A3 xe 4 x
19 ,
A0
A1
C y
13 ,
1 2
x
2
e
0, A3
1
9
1 1
,
3 3
4
9
Particular solution: y
24. yc 2 y
yc 2 y
m 2
§S ·
sin x, y¨ ¸
©2¹
0
v1
C 19 , C
12 x 2 e 4 x 1
9
4
9
vc2
2
5
ln cos x
sin x sec x
cos x sin x
1
sin x cos x
v2
³ dx
y
C1 ln cos x cos x C2 x sin x
x
2 x
yh
yp
Ce
A cos x B sin x
ycp
A sin x B cos x
ycp 2 y p
26. ycc y
A sin x B cos x 2 A cos x B sin x
2 B A sin x 2 A B cos x
2B A
y
³ tan x dx
cos x 0
1 3x e x
2.
0 when m
tan x
sin x cos x
1 3x e x
Initial conditions: y 0
sec x cos x
cos x sin x
vc1
12
sec x
sin x
0
xe x xe 4 x
A2
4x
0
vc1 sin x vc2 cos x
4x
1, 2 A B
yh y p
Ce2 x
2
1
,A 5
5
1
2
cos x sin x
5
5
0Ÿ B
§S ·
Initial conditions: y¨ ¸
©2¹
Particular solution: y
sin x
2 2
,
5 5
Ce S sec x tan x
ycc y
0
m 1
0 when m
2
i, i.
yh
C1 cos x C2 sin x
yp
v1 cos x v 2 sin x
vc1 cos x vc2 sin x
0
vc1 sin x vc2 cos x
2
,C
5
sin x
0
0
vc1
2
1
sin x cos x
5
5
sec x tan x
sec x tan x cos x
cos x sin x
tan 2 x
sin x cos x
v1
³ tan
2
³ sec
x dx
2
x 1 dx
tan x x
cos x
vc2
0
sin x sec x tan x
cos x sin x
tan x
sin x cos x
v2
³ tan x dx
y
yh y p
ln cos x
ln sec x
C1 cos x C2 sin x x tan x cos x
ln sec x sin x
C1 x tan x cos x C2 ln sec x sin x
© 2010 Brooks/Cole, Cengage Learning
Section 16.3
27. ycc 4 y
29. ycc 2 yc y
csc 2 x
ycc 4 y
0
m 4
2
2i, 2i.
0 when m
yh
C1 cos 2 x C2 sin 2 x
yp
v1 cos 2 x v 2 sin 2 x
vc1 cos 2 x vc2 sin 2 x
v1
³
y
yh
C1 C2 x e x
yp
v1 v 2 x e x
vc1
x ln x
v1
³ x ln x dx
e x ln x
ln x
v2
³ ln x dx
0
y
C1 C2 x e x 2 sin 2 x csc 2 x
cos 2 x
sin 2 x
1
cot 2 x
2
30. ycc 4 yc 4 y
1
ycc 4 yc 4 y
1
ln sin 2 x
4
³ 2 cot 2 x dx
1 ·
1
§
§
·
¨ C1 x ¸ cos 2 x ¨ C2 ln sin 2 x ¸ sin 2 x
2 ¹
4
©
©
¹
28. ycc 4 yc 4 y
m 2 4m 4
1, 1.
0
vc2
2 sin 2 x 2 cos 2 x
v2
0 when m
1
x
2
cos 2 x
vc2
m 2 2m 1
vc1e vc2 x 1 e x
2 sin 2 x 2 cos 2 x
1
dx
2
0
x
1
2
x2
x2
ln x 2
4
x ln x x
x 2e x
ln x 2 3
4
e2 x
x
0
0 when m
yh
C1 C2 x e 2 x
yp
v1 v 2 x e 2 x
2, 2.
2 2x
xe
ycc 4 yc 4 y
vc1e 2 x vc2 xe 2 x
0
m 2 4m 4
0 when m
0
v1e 2 x 2 vc2 2 x 1 e 2 x
2, 2.
e2 x
x
yh
C1 C2 x e 2 x
vc1
1
yp
v1 v 2 x e
v1
³ 1 dx
vc2
1
x
v2
³
y
C1 C2 x x x ln x e 2 x
2x
vc1e 2 x vc2 xe 2 x
0
vc1 2e2 x vc2 2 x 1 e2 x
x 2e 2 x
xe2 x
0
vc1
x 2e 2 x
2 x 1 e2 x
e2 x
xe 2 x
x 3e 4 x
e4 x
2e 2 x 2 x 1 e 2 x
v1
³
x3 dx
vc2
v2
y
0
³x
dx
1
3
x 2e 4 x
e4 x
x
3
1 4 · 2x
§
x ¸e
¨ C1 C2 x 12 ¹
©
1
dx
x
31. ycc yc 12 y
m 2 m 12
14 x 4
2e 2 x x 2 e 2 x
e4 x
2
x3
Let y p
e2 x
477
e x ln x
ycc 2 yc y
vc1e x vc2 xe x
csc 2 x
sin 2 x
csc 2 x 2 cos 2 x
cos 2 x
sin 2 x
vc1
0
0
vc1 2 sin 2 x vc2 2 cos 2 x
0
Second-Order Nonhomogeneous Linear Equations
F x
x2
ln x
0
m4 m3
0 Ÿ m
4, 3
Ax 2 Bx C. This is a generalized form of
x2.
32. ycc yc 12 y
m m 12
2
Because yh
yp
x
0
m4 m3
0 Ÿ m
4, 3
C1e 4 x C2e 3 x , let
Axe 4 x .
© 2010 Brooks/Cole, Cengage Learning
478
Chapter 16
Additional Topics in Differential Equations
33. Answers will vary. See the “Variation of Parameters”
box on page 1163.
34. (a) Because yccp
(b) y p
2
(c) y p
4
0 and 3 y p
35. qcc 10qc 25q
34
m 2 10m 25
12.
0, qc 0
6 sin 5t , q 0
37.
0
ycc 48 y
24
32
48
yh
C1 cos 8t C2 sin 8t
yp
A sin 4t B cos 4t
ycp
4 A cos 4t 4 B sin 4t
yccp
16 A sin 4t 16 B cos 4t
24
32
qp
A cos 5t B sin 5t
y
qcp
5 A sin 5t 5 B cos 5t
Initial conditions: y 0
qccp
25 A cos 5t 25B sin 5t
6 sin 5t , A
C1 C2 t e5t q
3
25
B
0, qc 0
C1
3
,
25
C2
38.
Particular solution: q
36. qcc 20qc 50q
m 20m 50
C1e
10 5
0,
e
5te
5t
cos 5t
10 sin 5t
2
qh
0, 5C1 C2
3
5
5t
C2 e
2
32
10 r 5
10 5
ycc 4 y
2
32
64 At 16 B sin 8t 16 A 64 Bt cos 8t
yccp 4 y p
2
,A
85
8
2
cos 5t sin 5t
85
85
Initial conditions:
q0
0, qc 0
10 5
C1
0, C1 C2 2 C1 10 5
87 2
, C2
170
8
85
2 C2 0,
2
17
4 sin 8t , A
ycc yc 4 y
8 7 2 10 5 2 t
8 7 2 10 5
e
e
170
170
8
2
cos 5t sin 5t
85
85
m 4
1
4
14
0, B
cos 8t 1
32
sin 8t 14 t cos 8t
1
,
4
yc 0
3
0.3
yp
A sin 8t B cos 8t
ycp
8 A cos 8t 8 B sin 8t
yccp
64 A sin 8t 64 B cos 8t
0
4
−0.05
8 B sin 8t 8 A cos 8t
4 sin 8t 8B
2
32
1
4
Initial conditions:
1
y0
, yc 0
3,
4
Ÿ B
321 , 8 A
C1 1
4
8C1 C2 Ÿ C2
1
32
0Ÿ A
Ÿ C1
9
,
32
34
2t
Particular solution: y
Ÿ C2
1
4
8, 8.
C1 C2 t e8t
3
1
32
4 sin 8t , y 0
0 when m
yccp ycp 4 y p
8C2 C1 , 0
1
4
2
32
0,
87 2
170
6.28
B sin 8t A cos 8t
yh
2
32
Particular solution:
q
2
32
1
m2
16
8
85
2 t
39.
0
C1 cos 8t C2 sin 8t 14 t cos 8t
y
10 sin 5t
r 8i.
−2
2
32
2 t
25 B 100 A sin 5t
10 5
0
yccp
2.
25 A 100 B cos 5t
C2 e
yc 0
1
,
4
At sin 8t Bt cos 8t
Particular solution: y
2 t
sin 8t sin 4t
1
2
yp
25 A cos 5t 25B sin 5t
10 5
cos 8t C1 cos 8t C2 sin 8t
qccp
C1e
C1 ,
12
yh
Initial conditions:
1
y0
, yc 0
0,
4
q
1
4
0 when m
A cos 5t B sin 5t
25 B 100 A
1
4
4
5 B cos 5t 5 A sin 5t
0½
¾B
10¿
0,
4 sin 8t , y 0
qcp
25 A 100 B
yc 0
8C2 4 Ÿ C2
qp
qccp 20qcp 50q p
1
,
4
2
m2
32
2
32
0 when m
2 t
1
2
3
25
3
25
0, A
0
cos 5t
0, C1 −2
C1 cos 8t C2 sin 8t sin 4t
Particular solution: y
Initial conditions:
q0
yh y p
0
253 ,
2␲
0
48 sin 4t , B
C1 C2 t e 5t
50 B cos 5t 50 A sin 5t
2
36 A sin 4t 36 B cos 4t
qh
qccp 10qcp 25q p
0
r 8i.
0 when m
yccp 48 y p
yc 0
1
,
4
48 sin 4t , y 0
24 2
m
32
24
32
5, 5.
0 when m
24
32
9
32
43 t e 8t 1
32
cos 8t
© 2010 Brooks/Cole, Cengage Learning
0
Section 16.3
40.
4
1
25
ycc yc y
32
2
2
1
, yc 0
2
0, y 0
Second-Order Nonhomogeneous Linear Equations
479
4
0.5
1 2 1
25
m m
8
2
2
0
m 2 4m 100
2 r 4
0 when m
y C1e 2t cos 4
6 t C2e 2t sin 4
0
1.57
6 i.
6t
−0.5
Initial conditions:
1
, yc 0
2
y0
C2
4,
3
4
6
1
2
C1 , 4
2C1 4
6 C2 ,
6
8
1 2t
e cos 4
2
Particular solution: y
6t 6 2t
e sin 4
8
6t
41. In Exercise 37,
yh
42. (a)
4
32
ª
1
1
5
§ 1 ·º
cos 8t sin 8t sin «8t S arctan ¨ ¸»
4
2
4
© 2 ¹¼
¬
5
1·
§
sin ¨ 8t S arctan ¸ |
4
2¹
©
ycc 43. x 2 ycc xyc y
25
2
y
0
C1 cos 10 x C2 sin 10 x
y
y0
1 1
:
2 2
yc 0
4: 4
y
1
2
y1
C1
cos 10 x 10C2 Ÿ C2
2
5
The motion is undamped.
and u1c
(b) If b ! 0, the motion is damped.
(c) If b ! 52 , the solution to the differential equation is
m x
m2 x
0 Ÿ u1c
4
x ln x
3
3
yh y p
4
ln x
x
2
4
3
ln x and u2
3
yp
uc2 ln x
4
ln x Ÿ u2c
x
4
ln x
x
. There would be
1
1
B sin ln x
x
x
A cos ln x
yccp
1
A cos ln x B sin ln x
x2
x 2 yccp xyc 4 y
2 ln x
2 x ln x
3
2
2
x ln x
3
C1x C2 x ln x 3
2
x ln x
3
3
1
A cos ln x B sin ln x
x
1§
1
1·
¨ A sin ln x B cos ln x ¸
x©
x
x¹
sin ln x
B A B 4 A sin ln x A B A 4 B cos ln x
1, 3B
1
1
B A sin ln x 2 A B cos ln x
x2
x
B A sin ln x A B cos ln x A cos ln x B sin ln x
4 A sin ln x B cos ln x
So, y p
x ln x
A sin ln x B cos ln x .
ycp
3A
u1
y
44. Let y p
x and y2
u1c u2c 1 ln x
sin 10 x
of the form y
C1e 1 C2e
no oscillations in this case.
4 x ln x
u1c x uc2 x ln x
25
5
sin 8t 2.6779 .
4
0 Ÿ A
sin ln x
1
3
1
sin ln x and y
3
yh y p
C1 sin ln x 2 C2 cos ln x 2 1
sin ln x .
3
© 2010 Brooks/Cole, Cengage Learning
480
Chapter 16
Additional Topics in Differential Equations
e 2 x cos e x
45. True. y p
ycp
e 2 x sin e x e x 2e 2 x cos e x
e x sin e x 2e 2 x cos e x
yccp
ªe x cos e x e x e x sin e x º ª2e 2 x sin e x e x 4e 2 x cos e x º
¬
¼ ¬
¼
So,
ª¬cos e x e x sin e x 2e x sin e x 4e 2 x cos e x º¼ 3ª¬e x sin e x 2e 2 x cos e x º¼ 2e 2 x cos e x
yccp 3 ycp 2 y p
ª¬e x 2e x 3e x º¼ sin e x ª¬1 4e 2 x 6e2 x 2e 2 x º¼ cos e x
cos e x .
46. True.
18 e 2 x , ycp
yp
yccp 6 ycp
14 e 2 x , yccp
12 e 2 x 6 14 e 2 x
47. ycc 2 yc y
0 Ÿ m
1, 1
C1e x C2 xe x , y p
x 2e x , particular solution
C1 C2 x e x x 2e x
General solution: f x
fc x
e2 x
2e x
m 2 2m 1
yh
12 e 2 x
C2 2 x C1 C2 x x 2 e x
C1 C2 x x 2 e x
x 2 C2 2 x C1 C2 e x
(a) No. If f x ! 0 for all x, then x 2 C2 x C1 ! 0 œ C2 2 4C1 0 for all x.
So, let C1
C2
x 2 3 x 2 e x and f c 32
1. Then f c x
14 0.
(b) Yes. If f c x ! 0 for all x, then
C2 2
2
4 C1 C2 0
Ÿ C2 2 4C1 4 0
C2 2 4C1 4
C2 2 4C1 0
Ÿ f x ! 0 for all x.
Section 16.4 Series Solutions of Differential Equations
f
1. yc y
0. Letting y
¦ an x n :
n 0
f
f
n 1
n 0
¦ nan x n 1 ¦ an x n
yc y
n 1 an 1
an 1
a1
a0 , a2
f
¦
y
n 0
f
¦
n 1 an 1 x n n 0
f
¦ an x n
0
n 0
an
an
n 1
a1
a0
, a3
2
2
a0 n
x
n!
a2
3
a0
, ! , an
1˜ 2˜3
a0
n!
a0e x
Check: By separation of variables, you have:
dy
³ y ³ dx
ln y
y
x C1
Ce x
© 2010 Brooks/Cole, Cengage Learning
Section 16.4
Series Solutions of Differential Equations
481
f
2. yc ky
¦ an xn :
0. Letting y
n 0
f
f
n 1
n 0
f
¦ nan xn 1 k ¦ an x n
yc ky
n 1 an 1
¦
n 1 an 1 x n n 0
f
¦ kan x n
0
n 0
kan
kan
n 1
an 1
a1
f
¦
y
n 0
k 2 a0
, a3
2
ka1
2
ka0 , a2
f
kn
a0 x n
n!
kx
n!
a0 ¦
n 0
ka2
3
k 3a0
, ! , an
1˜ 2 ˜3
kn
a0
n!
n
a0e kx
Check: By separation of variables, you have:
³
dy
y
³ k dx
kx C1
ln y
Ce kx
y
f
3. ycc 9 y
¦ an xn :
0. Letting y
n 0
f
¦ n n 1 an x n 2
ycc 9 y
n 2
n 2 n 1 an 2
n 2 n 1 an 2 x n n 0
a0
a1
a1
a2
9a0
2
a3
9a1
3˜2
a4
9a2
4˜3
a5
9a3
5˜4
92 a0
4 ˜ 3˜ 2 ˜1
#
9 a0
2n !
f
¦
n 0
92 a1
5 ˜ 4 ˜ 3˜ 2 ˜1
9n a1
2n 1 !
a2 n 1
9n a0 2 n
x 2n !
f
¦
n 0
9n a1
x 2 n 1
2n 1 !
f
a0 ¦
n 0
2n
2 n 1
3x
a f 3x
1¦
2n !
3 n 0 2n 1 !
C0e3 x C1e 3 x where C0 C1
Check: ycc 9 y
m 9
y
4. y
0
n 0
#
n
2
f
¦ 9an x n
9an
n 2 n 1
a0
y
n 0
f
¦
9an
an 2
a2 n
f
9 ¦ an x n
f
C0 ¦
n 0
3x
n!
n
f
C1 ¦
n 0
a0 and C0 C1
3 x
n!
n
a1
.
3
0 is a second-order homogeneous linear equation.
0 Ÿ m1
3 and m2
3
C1e3 x C2e3 x
C0e kx C1e kx . Follow the solution to Exercise 3 with 9 replaced by k 2 .
© 2010 Brooks/Cole, Cengage Learning
482
Chapter 16
Additional Topics in Differential Equations
f
5. ycc 4 y
¦ an x n :
0. Letting y
n 0
f
¦ n n 1 an x n 2
ycc 4 y
n 2
n 2 n 1 an 2
a0
a1
a1
a2
4a0
2
a3
4a1
3˜2
a4
4a2
4˜3
2
4 a0
4!
n 0
0
n 0
2
4 a1
5!
4a3
5˜4
a5
#
#
n
n
1 4
a0
2n !
a2 n
n
a2 n 1
n
1 4n a0 2 n
x 2n !
f
¦
n 0
Check: ycc 4 y
m 4
1 4n
a1
2n 1 !
n
1 4n a1 2 n 1
x
2n 1 !
f
¦
n 0
f
a0 ¦
1
n 0
n
2x
2n !
2n
n
2 n 1
a1 f 1 2 x
¦ 2n 1 !
2n 0
C0 cos 2 x C1 sin 2 x
0 is a second-order homogeneous linear equation.
0 Ÿ m
2
6. y
n 0
f
n 2 n 1 an 2 x n ¦ 4an x n
4an
n 2 n 1
a0
y
f
¦
4 an
an 2
y
f
4 ¦ an x n
r 2i
C1 cos 2 x C2 sin 2 x
C0 cos kx C1 sin kx. Follow the solution to Exercise 5 with 4 replaced by k 2 .
7. yc 3 xy
f
¦ an x n :
0. Letting y
n 0
f
n 1
a0
a2
a4
f
Ÿ a1
a0
3a
0
2
3 § 3a0 ·
¨
¸
4© 2 ¹
32
a0
23
a8
3§
33 a
¨¨ 3 0
8© 2 3 ˜ 2
0
0 and an 2
3 x 2 n
2n n!
un 1
un
lim
n 0
·
¸¸
¹
a1
0
a3
3a1
0
3
3 § 3a ·
¨ 1 ¸
5© 3 ¹
a5
33 a0
2 3˜2
3
34 a0
2 4˜3˜2
4
3an
n 2
0
a7
3 § 32 a1 ·
¨
¸
7© 3 ˜ 5 ¹
a9
3§
33 a1 ·
¨
¸
9© 3 ˜ 5 ˜ 7 ¹
0
0
n
a0 ¦
f
n of
n 0
n 0
3 § 32 ·
¨ 3 a0 ¸
6© 2
¹
lim
n 1
¦ 3an x n 1
n 2 an 2 x n 1
a6
y
f
¦ nan x n 1 ¦ 3an x n 1
yc 3 xy
¦
f
n 1
n of
3
x 2n 2
2n n!
˜
n 1
2
n 1 ! 3 n x 2 n
lim
n of
3x 2
2n 1
0
The interval of convergence for the solution is f, f .
© 2010 Brooks/Cole, Cengage Learning
Section 16.4
Series Solutions of Differential Equations
483
f
8. yc 2 xy
¦ an x n :
0. Letting y
n 0
f
n 1
f
n 1
n 0
¦ nan x n 1 ¦ 2an x n 1
yc 2 xy
¦
f
f
¦ 2an x n 1
n 2 an 2 x n 1
a0
0
a2
2a0
2
a4
2 § 2a0 ·
¨
¸
4© 2 ¹
a6
2 § 22 a0 ·
¨
¸
6 © 22 ˜ 2 ¹
a8
2 § a0 ·
¨ ¸
8 © 3! ¹
2an
n 2
a0
f
a0 ¦
n 0
lim
n of
0 and
n 0
an 2
y
Ÿ a1
0
22 a0
22 ˜ 2
a0
2
23 a0
2 ˜3˜2
a0
3!
3
a0
4!
a1
0
a3
2a1
3
a5
2 § 2a1 ·
¨
¸
5© 3 ¹
a7
2 § 22 a1 ·
¨
¸
7© 3 ˜ 5 ¹
a9
2 § 23 a1 ·
¨
¸
9© 3 ˜ 5 ˜ 7 ¹
0
0
0
0
x2n
n!
u n 1
un
lim
n of
x 2n 2
n!
˜
n 1 ! 2n
lim
n of
x2
n 1
0
The interval of convergence for the solution is f, f .
f
9. ycc xyc
¦ an x n :
0. Letting y
n 0
ycc xyc
f
¦ n n 1 an x n 2
n 2
f
¦ n n 1 an x n 2
n 2
f
¦
n 2 n 1 an 2 x n
n 0
f
¦ nan x n
n 0
f
¦ nan x n
nan
n 2 n 1
a0
a0
a1
a1
a2
0
a3
a1
3˜2
There are no even powered terms. a5
a7
f
lim
3a3
5˜4
5a5
7˜6
1 ˜ 3 ˜ 5 ˜ 7" 2n 1 x 2 n 1
2n 1 !
0
un 1
un
3a1
5!
5 ˜ 3a1
7!
f
a0 a1 ¦
n
n of
0
n 1
n 0
an 2
y
f
x ¦ nan x n 1
a0 a1 ¦
n 0
2 n! 2n 1
x 2n 3
˜
n 1 ! 2n 3
x 2 n 1
2n !x 2 n 1
2 n! 2n 1 !
n
n
lim
n of
2
n 1
f
a0 a1 ¦
n 0
2n 1 x
2 n 1 2n 3
x 2 n 1
2 n! 2n 1
n
2
lim
n of
0
Interval of convergence: f, f
© 2010 Brooks/Cole, Cengage Learning
484
Chapter 16
Additional Topics in Differential Equations
f
10. ycc xyc y
¦ an x n :
0. Letting y
n 0
ycc xyc y
f
¦ n n 1 an x n 2
n 2
f
¦
n 2 n 1 an 2 x n
n 0
a8
a6
8
a0
24 4!
y
a0 ¦
a6
n 0
lim
u n 1
un
lim
u n 1
un
n of
a1
a3
f
n of
an
n 2
a1
a0
8
a0
233!
a4
0
n 0
n 1 an x n
a0
a0
2
a2
4
a4
6
a2
n 1
f
¦ an x n
n 0
an 2
a0
f
¦
f
x ¦ nan x n 1 a0
22 2!
a5
a7
a9
a1
3
a3
5
a5
7
a1
3˜5
a1
3˜5˜7
a7
9
a1
3˜5˜7˜9
f
x 2n
x 2 n 1
a1 ¦
n
2 n!
n 0 1 ˜ 3 ˜ 5 ˜ 7" 2 n 1
lim
x2n 2
2n n!
˜ 2n
2
n 1! x
lim
1 ˜ 3 ˜ 5 ˜ 7" 2n 1
x2n 3
˜
1 ˜ 3 ˜ 5 ˜ 7" 2n 3
x 2 n 1
n of
n of
lim
n 1
n of
x2
2n 1
0
lim
n of
x2
2n 3
0
Because the interval of convergence for each series is f, f , the interval of convergence for the solution is f, f .
11.
x 2 4 ycc y
f
0. Letting y
¦ an x n :
n 0
x 2 4 ycc y
f
¦ n n 1 an x n
n 2
f
¦
f
4 ¦ n n 1 an x n 2 n 2
n 2 n 1 an x n n 0
an 2
¦4n2
n 0
n 1 an 2 x n
0
n 0
n 2 n 1 an
4n 2 n 1
a0
a0
a2
a0
42 1
a0
8
a4
3a2
44 3
a0
128
y
f
f
¦ an x n
a1
a5
a1
a3
a1
43 2
7 a3
45 4
7 a1
1920
a1
24
§
·
§
·
x2
x4
x3
7 x5
a0 ¨1 "¸ a1 ¨ x "¸
8
128
24 1920
©
¹
©
¹
© 2010 Brooks/Cole, Cengage Learning
Section 16.4
12. ycc x 2 y
Series Solutions of Differential Equations
485
f
0. Letting y
¦ an x n :
n 0
f
¦
f
n 2
f
n 0
0
¦ an x n 2
n 4 n 3 an 4 x n 2
n 2
f
¦ n n 1 an x n 2 ¦ an x n 2
ycc x 2 y
n 0
an
n 4 n3
an 4
Also:
y
a0 a1x a2 x 2 a3 x3 " an x n "
ycc
2a2 3 ˜ 2a3 x " n n 1 an x n 2 "
ycc x 2 y
2 a2
So, a2
a0
2a2 3 ˜ 2a3 x a0 4 ˜ 3a4 x 2 a1 5 ˜ 4a5 x3 "
0, 6a3
0, 12a4 a0
0 Ÿ a6
0 and a3
0, 20a5 a1
0
0, a10
0, and a11
0, a7
a0
a
0
4˜3
a
a0
4
8˜7
8˜7˜4˜3
a8
a0
12 ˜ 11
12 ˜ 11 ˜ 8 ˜ 7 ˜ 4 ˜ 3
a4
a8
a12
ycc x 2 y
0, y 0
0 and a4 n 3
0.
a1
5˜4
a
a1
5
9˜8
9˜8˜5˜4
a9
a1
13 ˜ 12
13 ˜ 12 ˜ 9 ˜ 8 ˜ 5 ˜ 4
2
1 2x y
yc 0
0
1 2 x yc 2 y
ycc 0
2
yccc
1 2 x ycc 4 yc
yccc 0
10
4
2
1 2 x yccc 6 ycc
y x
a13
ycc
#
a5
a9
yc
y
a1
0. Therefore, a4 n 2
§
·
x4
x8
x12
a0 ¨1 "¸
4 ˜ 3 8 ˜ 7 ˜ 4 ˜ 3 12 ˜ 11 ˜ 8 ˜ 7 ˜ 4 ˜ 3
©
¹
5
7
9
§
·
x
x
x
a1 ¨ x "¸
5 ˜ 4 9 ˜ 8 ˜ 5 ˜ 4 13 ˜ 12 ˜ 9 ˜ 8 ˜ 5 ˜ 4
©
¹
13. yc 2 x 1 y
4
a1
0
y
0
#
2
2 2 10 3
2 4
2 x x x x "
1!
2!
3!
4!
§1·
Using the first five terms of the series, y¨ ¸
© 2¹
Using Euler’s Method with 'x
i
xi
yi
0
0
2
1
0.1
2.2
2
0.2
2.376
3
0.3
2.51856
4
0.4
2.61930
5
0.5
2.67169
163
| 2.547.
64
0.1 you have yc
1 2 x y.
© 2010 Brooks/Cole, Cengage Learning
Chapter 16
486
14. yc 2 xy
Additional Topics in Differential Equations
0, y 0
15. Given a differential equation, assume that the solution
1
yc
2 xy
yc 0
0
ycc
2 xyc y
ycc 0
2
yccc
2 xycc 2 yc
yccc 0
0
y4
2 xyccc 3 ycc
y4 0
12
y
5
2 xy
4
4 yccc
y
2 xy 5 5 y 4
y6
5
0
derivatives into the differential equation. You should
then be able to determine the coefficients a0 , a1 , ! .
16. A recursion formula is a formula for determining
the next term of a sequence from one or more of the
120
preceding terms. See Example 1.
#
17. (a) From Exercise 9, the general solution is
2 2 12 4 120 6
x x x "
2!
4!
6!
1
1
1 x2 x4 x6 "
2
6
Using the first four terms of the series,
8
y1
| 2.667.
3
1
y x
Using Euler’s Method with 'x
i
0
xi
0.1 you have yc
0
f
a0 a1 ¦
y
n 0
0 Ÿ a0
y0
a1 ¦
yc
n 0
2 xy.
yc 0
n 0
1
(b) P3 x
1
0.1
1
2
0.2
1.02
3
0.3
1.0608
P5 x
0
f
a1 ¦
n 0
x2n
2n n!
a1
2
2¦
y
x 2 n 1
.
2 n! 2n 1
n
2n 1 x 2 n
22 n! 2n 1
f
f
yi
y and its
n 0
0
y6 0
#
f
¦ an x n . Then substitute
is of the form y
x 2 n 1
2 n! 2n 1
n
ª
x3 º
2 «x »
2 ˜ 3¼
¬
2x 2x x3
x5
2
3
4˜2˜5
x3
3
2x x3
x5
3
20
12
4
0.4
1.1244
5
0.5
1.2144
6
0.6
1.3358
7
0.7
1.4961
8
0.8
1.7056
9
0.9
1.9785
10
1.0
2.3346
−4
4
P3(x)
P5(x)
−12
(c) The solution is symmetric about the origin.
So, y 1 | 2.335.
18. (a) m 2 9
r 3i
C1 cos 3 x C2 sin 3 x
y
y0
yc
yc 0
yp
0 Ÿ m
2
C1
3C1 sin 3 x 3C2 cos 3x
6
3C2 Ÿ C2
2
2 cos 3 x 2 sin 3 x
© 2010 Brooks/Cole, Cengage Learning
Section 16.4
f
f
n 0
9 ¦ an x n
n 2
f
n 2
f
¦ n n 1 an x n 2
ycc 9 y
¦ n n 1 an x n 2
n 1
f
487
f
¦ an xn , yc ¦ nan x n 1, ycc
(b) Let y
Series Solutions of Differential Equations
0
n 0
f
¦
n 2 n 1 an 2 x n 9 ¦ an x n
n 0
0
n 0
n 2 n 1 an 2
9an
9
an
n 2 n 1
an 2
For n even,
9
a2
a0
2
For n odd,
9
a3
a1
3˜2
2
9
a0
4!
a4
9
a2
4˜3
a6
9
a0
6!
a5
9
a3
5˜4
a7
9
a1
7!
3
2
9
a1
5!
3
n
n
9
a0
2n !
and in general, a2 n
and in general, a2 n 1
9
a1
2n 1 !
So,
f
¦ an x n
y
f
f
n 0
n 0
¦ a2n x2n ¦ a2n 1x 2n 1
n 0
n
9
a0 x 2 n 2n !
f
¦
n 0
n
9
a1x 2 n 1
2n 1 !
f
¦
n 0
f
a0 ¦
n 0
1
n
3x
2n !
(c)
a1
2, and y
f
a1 ¦
n 0
ª
1 3x
2 «¦
2n !
«¬n 0
f
Applying the initial conditions, a0
2n
n
2 n 1
n
1 3 x
2n 1 !
2n
f
¦
n 0
n
2 n 1
º
1 3 x
».
2n 1 ! »
¼
y
P5
3
2
y
x
1
2
P3
19. ycc 2 xy
0, y 0
1, yc 0
20. ycc 2 xyc y
3
0, y 0
1, yc 0
2
ycc
2 xy
ycc 0
0
ycc
2 xyc y
ycc 0
1
yccc
2 xyc y
yccc 0
2
yccc
2 xycc yc
yccc 0
2
y
4
2 xycc 2 yc
y
5
2 xyccc 3 ycc
y
6
2 xy
4
4 yccc
2 xy 5 5 y 4
y7
#
y
4
y
5
y
6
12
0
0
0
0
16
120
y7 0
1
y
5
y
6
2 xyccc 3 ycc
#
y
2 xy
4
5 yccc
2 xy
5
7y
4
2 xy 6 9 y 5
y7
#
3
2
12 4 16 6 120 7
x x3 x x x
1!
3!
4!
6!
7!
§1·
Using the first six terms of the series, y¨ ¸ | 0.253.
© 4¹
y
y
4
y
4
0
3
y
5
0
10
y
6
0
21
y7 0
90
#
2
1
2
3
10
21
90 7
1 x x 2 x3 x 4 x5 x 6 x
1!
2!
3!
4!
5!
6!
7!
§1·
Using the first eight terms of the series, y¨ ¸ | 1.911.
© 2¹
© 2010 Brooks/Cole, Cengage Learning
488
Chapter 16
Additional Topics in Differential Equations
21. ycc x 2 yc cos x y
3, yc 0
0, y 0
2
ycc
x 2 yc cos x y
ycc 0
3
yccc
2 x 2 yc x 2 ycc sin x y cos x yc
yccc 0
2
ycc 0
1
yccc 0
1 1 2
y
2
3
2
x x 2 x3
1!
2!
3!
3
§1·
Using the first four terms of the series, y¨ ¸ | 3.846.
© 3¹
22. ycc e x yc sin x y
2, yc 0
0, y 0
ycc
e x yc sin x y,
yccc
e x yc e x ycc cos x y sin x yc
1
e x yc ycc cos x y sin x yc
y
2 2
1
1
2
x x 2 x3
1!
2!
3!
§1·
Using the first four terms of the series, y¨ ¸ | 1.823.
©5¹
ex , f c x
23. f x
e x , yc y
ycc y
f
¦ an x n , then:
Assume y
cos x, f c x
24. f x
0.
0.
¦ an x n , then:
Assume y
f
¦ nan x n 1
n 0
n 1
f
¦ an x n
n 1
f
¦
n 1 an 1 x
ycc
f
¦ nan x n 1
f
¦
n 0
an x
n
0,
a1
n
1,
a2
n 2
n 0
f
¦
an
,n t 0
n 1
a1
2
a2
3
a0
2
a0
23
a3
n
3,
a4
a3
4
a0
23 4
n
4,
a5
a4
5
a0
23 4 5
n 0
an 1
a0 ¦
n 0
a0
Ÿ an
n 1!
0
f
¦ an x n
n 0
an 2
an
,n t 0
n 1 n 2
a0
a0
a1
a1
a2
a
0
1 2
a3
a1
2 3
a4
a5
a3
4 5
a2
3 4
a0
4!
#
a2 n
a0
n!
1, you have the Maclaurin Series for f x
a0 ¦
n 0
xn
which converges on f, f . When
n!
n
1 a0
2n !
f
y
a1
5!
#
n
#
f
n 2 n 1 an 2 x n
a0
2,
a0
f
¦ n n 1 an x n 2 ¦ an x n 2
n
n
y
f
n 0
an 1
f
¦ n n 1 an x n 2
n 2
n 0
n
cos x,
f
n 0
yc
sin x, f cc x
a2 n 1
n
1 a1
2n 1 !
n
f
1 x 2 n
1 x 2 n 1
which
a1 ¦
2n !
2n 1 !
n 0
converges on f, f
ex.
When a0
1 and a1
Series for f x
0, you have the Maclaurin
cos x.
© 2010 Brooks/Cole, Cengage Learning
Section 16.4
25.
f x
arctan x
fc x
1
1 x2
2 x
f cc x
ycc
1 x 2 ycc 2 xyc
Series Solutions of Differential Equations
489
2
1 x2
2 x
yc
1 x2
0
f
¦ an xn , then:
Assume y
n 0
yc
f
¦ nan x n 1
n 1
ycc
f
¦ n n 1 an x n 2
n 2
1 x 2 ycc 2 xyc
f
¦ n n 1 an x n 2
n 2
f
¦
n 2 n 1 an 2 x n
n 0
f
f
f
n 2
n 0
n 0
¦ n n 1 an x n 2 ¦ n n 1 an x n ¦ 2nan x n
f
¦ n n 1 an x n n 0
0
f
¦ 2nan x n
n 0
f
¦ n n 1 an x n
n 0
n 2 n 1 an 2
an 2
n n 1 an
n
an , n t 0
n 2
n
0 Ÿ a2
0 Ÿ all the even-powered terms have a coefficient of 0.
n
1,
a3
n
3,
a5
n
5,
a7
n
7,
a9
1
a1
3
3
a3
5
5
a5
7
7
a7
9
1
a1
5
1
a1
7
1
a1
9
#
n
a2 n 1
f
y
a1 ¦
n 0
1 a1
2n 1
n
1 x 2 n 1
which converges on 1, 1 . When a1
2n 1
1, you have the Maclaurin Series for f x
arctan x.
© 2010 Brooks/Cole, Cengage Learning
490
Chapter 16
Additional Topics in Differential Equations
f x
26.
arcsin x
1
fc x
1 x2
x
f cc x
32
1 x2
1
ycc
1 x
1 x 2 ycc xyc
x
1 x2
˜
2
x
yc
1 x2
0
f
f
f
f
n 0
n 2
n 0
n 0
¦ an x n , then: ¦ ann n 1 x n 2 ¦ ann n 1 x n ¦ an nx n
Assume y
f
¦
n 2 n 1 an 2 x n
n 0
0 Ÿ a2
f
¦ n 2an x n
n 0
an 2
n
0
n2
an , n t 0
n 1 n 2
0 Ÿ all the even-powered terms have a coefficient of 0.
a1
a1
1
n
1,
a3
n
3,
a5
9
a3
4 5
9
a1
2 3 4 5
n
5,
a7
25
a5
6 7
9 25
a1
2 3 4 5 6 7
n
7,
a9
49
a7
8 9
9 25 49
a1
2 3 4 5 6 7 8 9
n
9,
a11
81
a9
10 11
2 3
a1
3
a1
2 4 5
3 5
a1
2 4 6 7
3 5 7
a1
2 4 6 8
9 25 49 81
a1
2 3 4 5 6 7 8 9 10 11
3 5 7 9
a1
2 4 6 8 10 11
#
2n !
a2 n 1
f
y
a1 ¦
n 0
2n !
n
2 n!
2
2n 1
2n n!
2
2n 1
a1
x 2 n 1 which converges on 1, 1 . When a1
1, you have the Maclaurin Series for f x
arcsin x.
f
27. ycc xy
0. Let y
¦ an x n .
n 0
f
¦ n n 1 an x n 2
ycc xy
n 2
2a2 f
¦ ª¬ n 3
n 0
So, a2
f
¦
n 1
n 0
n 2 an 3 an º¼ x n 1
0 and an 3
f
x ¦ an x n
an
for n
n3 n 2
n 3 n 2 an 3 x n 1 f
¦ an x n 1
0
n 0
0
0, 1, 2, !
The constants a0 and a1 are arbitrary.
a0
a3
a6
So, y
a0
a0
3˜2
a3
6˜5
a1
a1
a1
a4
4˜3
a0
a4
a1
a7
6˜5˜3˜2
7˜6
7˜6˜4˜3
a
a
a
a
a0 a1 x 0 x3 1 x 4 0 x 6 1 x 7 .
6
12
180
504
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 16
491
Review Exercises for Chapter 16
1.
y x3 xy 2 dx x dy
wM
wy
1 2 xy z
wN
wx
6. 3 x 2 5 xy 2 dx 2 y 3 5 xy 2 dy
0
wM
wy
1
Not exact
0
7.
wN
wx
1
3. 10 x 8 y 2 dx 8 x 5 y 2 dy
wM
wy
5 x 2 8 xy 2 x g y
8x 5 y 2
gc y
5y 2
g y
5 2
y 2 y C1
2
³
wM
wy
gc y
xy cos xy sin xy
wN
Exact
wx
³ y sin xy dx
cos xy g y
x sin xy g c y
cos xy y2
2
x sin xy y
2
y
C1
2
y Ÿ g y
C
y
9. (a)
0
0
4
wN
wx
6 y 2 1
f x, y
1
y2
Not exact
f y x, y
4. 2 x 2 y 3 y dx x 6 xy 2 dy
wM
wy
x
wN
z
y2
wx
f x, y
5
f x, y
5 x 2 8 xy 2 x y 2 2 y C1
2
5
5 x 2 8 xy 2 x y 2 2 y C
2
Exact:
0
8. y sin xy dx x sin xy y dy
10 x 8 y 2 dx
8x gc y
f y x, y
0
wN
wx
8
³
f x, y
x
x
dx 2 dy
y
y
wM
wy
Exact
Exact:
wN
5 y2
wx
10 xy z
Not exact
2. 5 x y dx 5 y x dy
wM
wy
0
x
2 x 2 y 3 y dx
−4
2
4
x 2 2 xy 3 xy g y
6 xy 2 x g c y
f y x, y
gc y
0
g y
C1
x 2 xy xy
3
(b)
x 2 xy xy C1
C
f x, y
3
5. x y 5 dx x 3 y z dy
wM
wy
2 x y dx 2 y x dy
wM
wy
2
f x, y
2
−4
x 6 xy 2
wN
Exact
wx
1
f x, y
³
f y x, y
x gc y
2
x
xy 5 x g y
2
x 3y 2
x y 5 dx
gc y
3 y 2
g y
3 2
y 2 y C1
2
³
x 2 xy g y
x gc y
gc y
2y
g y
y 2 C1
x 2 xy y 2
y2
2 x y dx
2y x
C
2: 44 4
4
C
Particular solution: x 2 xy y 2
(c)
x2
3
xy 5 x y 2 2 y C1
2
2
x 2 2 xy 10 x 3 y 2 4 y
wN
Exact
wx
1
f y x, y
0
0
C
0
4
4
−6
6
−4
© 2010 Brooks/Cole, Cengage Learning
492
Chapter 16
Additional Topics in Differential Equations
4
10. (a) and (c)
12. 3x 2 y 2 dx 2 x3 y 3 y 2 dy
−6
wM
wy
6
−4
6 xy y 3 dx 4 y 3 x 2 3xy 2 dy
(b)
wM
wy
6x 3 y
wN
Exact
wx
2
f x, y
³
f y x, y
3x 2 3 xy 2 g c y
gc y
6 xy y 3 dx
4y Ÿ g y
3x 2 y xy 3 2 y 2
y0
3 x 2 y xy 3 g y
4 y 3x 2 3 xy 2
C
11. 2 x y 3 dx x 3 y 1 dy
wM
wy
1
2
0
wN
wx
f x, y
³
2 x y 3 dx
x xy 3 x g y
2
f y x, y
x gc y
x 3y 1
gc y
g y
f x, y
3y 1
3
y 2 y C1
2
x 2 xy 3 x
2 x 2 2 xy 6 x 3 y 2 2 y
3 2
y y C1
2
C
Initial condition:
y2
0
³ 3x
f y x, y
2 x3 y g c y
y dx
gc y
3y2
g y
y 3 C1
x3 y 2 y 3
C Ÿ C
2: 4 8
Initial condition: y 1
wM wy wN wx
N
C
4
0
2 y 2 y
2 xy
hx
Integrating factor: e ³
dx
eln x
2
x
1
x2
2
§
y2 ·
2y
dy
Exact equation: ¨ 3 2 ¸ dx x ¹
x
©
§
y2 ·
¸ dx
x2 ¹
f x, y
³ ¨© 3 f y x, y
2y
gc y
x
gc y
3x 0 Ÿ g y
y2
x
y2
g y
x
C1
0
wN wx wM wy
M
4
2 x 2 x
2 xy
k y dy
eln y
2
§
2x
x2 ·
dx ¨1 2 ¸ dy
y
y ¹
©
³
2x
dx
y
f y x, y
x2
gc y
y2
1Ÿ g y
2
y
k y
1
y2
0
x2
g y
y
f x, y
x2
y
y
0
2y
x
14. 2 xy dx y 2 x 2 dy
gc y
3x h x
C
Exact equation:
2 x 2 xy 6 x 3 y 2 y
2 x3 y 3 y 2
C
4
Particular solution:
2
x3 y 2 g y
2 2
Integrating factor: e ³
8 0 12 0 0
2
f x, y
13. 3 x 2 y 2 dx 2 xy dy
C
1: 2
2
wN
Exact
wx
6x2 y
Particular solution: x3 y 2 y 3
2 y 2 C1
Particular solution: 3x 2 y xy 3 2 y 2
Exact:
0
0, y 1
1
x2
y2
y C1
C
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 16
15. dx 3x e 2 y dy
³
f x, y
f y x, y
3 xe
gc y
e y
g y
e y C1
xe
3y
e
y
y dy
k y
3
e3 y
dx 3 xe
3y
e
y
dy
0
xe3 y g y
e3 y dx
3y
m m2
30
1
k
Integrating factor: e ³
Exact equation: e
gc y
3 xe
e
3y
y
C1e 2 x C2e x
yc
2C1e 2 x C2e x
0
C1 C2
yc 0
3
2C1 C2
m 2 4m 5
wN wx wM wy
M
k
Integrating factor: e ³
y dy
y0
yc 0
cos y
³ cos
f y x, y
2 x cos y sin y g c y
2
x cos 2 y g y
y dx
2 r i
e2 x >2 cos x C2 sin x@
C2 2 2 Ÿ C2
3
2e 2 x cos x 3e 2 x sin x
−1
Ÿ g y
5
−1
21. ycc 2 yc 3 y
2 y sin y cos y cos 2 y
m 2m 3
2
y cos 2 y C1
x cos y y cos y
2
C
7
y
C1e 2 x C2e 2 x
yc
2C1e 2 x 2C2e 2 x
ycc
4C1e 2 x 4C2e 2 x
y1
4C1e 2 x 4C2e 2 x
4 C1e 2 x C2e 2 x
y
yc
C1 cos 2 x C2 sin 2 x
ycc
4C1 cos 2 x 4C2 sin 2 x
y2
−5
y3
−1
0
2
m 3 m 1
y
C1e 3 x C2e x
yc
3C1e3 x C2e x
y0
2
C1 C2
yc 0
0
3C1 C2
0
0 Ÿ m
C2
3
.
2
y
3 x
e
2
−6
6
0
4C1 Ÿ C1
1
2
and
12 e 3 x
0
m6
2
0, m
y
C1e 6 x C2 xe 6 x
yc
6C1e 6 x C2e6 x 6C2 xe 6 x
y0
2
yc 0
1
6, 6
2
C1
−1
6 2 C2 Ÿ C2
3
3, 1
8
Subtracting these equations, 2
5
m 2 12m 36
4C1 cos 2 x 4C2 sin 2 x
4 C1 cos 2 x C2 sin 2 x
0
22. ycc 12 yc 36 y
2C1 sin 2 x 2C2 cos 2 x
−3
16 20
2
3
f x, y
ycc 4 y
4 r
m
0
0
2 x sin y cos y 2 y sin y cos y cos 2 y
18.
7
y
cos 2 y dx ª¬2 x y sin y cos y cos 2 yº¼
ycc 4 y
1 and
e 2 x >2 sin x C2 cos x@ 2e 2 x >2cos x C2 sin x@
Exact equation:
17.
−8
3C1 Ÿ C1
0
C1 Ÿ y
2
yc
k y
12
C1e 2 x cos x C2e 2 x sin x
y
2 sin y sin y
cos y
tan y
8
e2 x e x
20. ycc 4 yc 5 y
0
2, 1
0, m
−12
y0
y
C
2
m2 m1
Adding these equations, 3
C2
1.
y
16. cos y dx ª¬2 x y sin y cos yº¼ dy
gc y
0
2
wN wx wM wy
M
3y
19. ycc yc 2 y
0
493
2
13
−1
y
−2
2e 6 x 13xe6 x
© 2010 Brooks/Cole, Cengage Learning
494
Chapter 16
Additional Topics in Differential Equations
23. ycc 2 yc 5 y
0
m 2 2m 5
27.
2 r
0 Ÿ m
4 20
2
1 r 2i
4
y1
e 1 C1 cos 2 C2 sin 2
4
y2
−1
5
e 2 C1 cos 4 C2 sin 4
0
−2
Solving this system, you obtain C1
C2 7.8161.
y
e
x
yp
A0 A1 x A2 x 2 A3 x3
y pc
A1 2 A2 x 3 A3 x 2
y pcc
2 A2 6 A3 x
A0 2 A2 A1 6 A3 x A2 x 2 A3 x3
x3 x
0, A1 5, A2
A0
0 Ÿ m
ri
28.
C1 cos x C2 sin x
2
§S ·
y¨ ¸
©2¹
C1
−4
C2
1
5
2 cos x sin x
−3
25. ycc is always positive according to the graph (concave
upwards), but yc is negative when x 0 (decreasing),
so ycc z yc.
m 1
2
0 when m
2 i,
2 x C2 sin
C1 cos
yp
Ae 2 x B 0 B1 x
y pc
2 Ae 2 x B1
y pcc
4 Ae 2 x
6 Ae 2 x 2 B 0 2 B1x
A
1
,
6
y
C1 cos
B0
0, B1
2 i.
2x
e2 x x
1
2
2 x C2 sin
2 x 16 e 2 x 2 cos x
i, i.
0 when m
yh
C1 cos x C2 sin x
yp
Ax cos x Bx sin x
y pc
Bx A cos x B Ax sin x
y pcc
2 B Ax cos x Bx 2 A sin x
y pcc y p
1
e2 x x
yh
y pcc 2 y p
26. ycc is positive for x ! 0 (concave upwards), but
12 yc 0 for x ! 0 (increasing). So, ycc z 12 yc.
ycc y
ycc 2 y
m2 2
3
0, A3
C1 cos x C2 sin x 5 x x3
y
y0
29.
0 when m i, i.
C1 cos x C2 sin x
0
m2 1
y
m 1
yh
y pcc y p
9.0496,
9.0496 cos 2 x 7.8161 sin 2 x
24. ycc y
y
x3 x
2
e x C1 cos 2 x C2 sin 2 x
y
ycc y
2 B cos x 2 A sin x
2 cos x
A
0, B
y
C1 cos x C2 x sin x
1
30. ycc 5 yc 4 y
m 2 5m 4
x 2 sin 2 x
0 when m
x
1, 4.
C2 e 4 x
yh
C1e
yp
A0 A1x A2 x 2 B 0 sin 2 x B1 cos 2 x
y pc
A1 2 A2 x 2 B 0 cos 2 x 2 B1 sin 2 x
y pcc
2 A2 4 B 0 sin 2 x 4 B1 cos 2 x
y pcc 5 y pc 4 y p
A0
y
4 A0 5 A1 2 A2 4 A1 10 A2 x 4 A2 x 2 10 B1 sin 2 x 10 B 0 cos 2 x
21
,
32
A1
85 , A2
C1e x C2e4 x 1
,
4
21
32
B0
85 x 0, B1
1
4
x 2 sin 2 x
101
x 2 101 cos 2 x
© 2010 Brooks/Cole, Cengage Learning
1
2
x
Review Exercises for Chapter 16
31. ycc 2 yc y
33. ycc yc 6 y
2 xe x
m 2 2m 1
0 when m
yh
C1 C2 x e x
yp
v1 v2 x e x
v1ce x v2c xe x
m2 m 6
1, 1.
v1ce x v2c x 1 e x
2 xe x
2 x 2
v1
³ 2x
2
23 x3
dx
yh
C1e
3x
yp
9 by inspection
v2
³ 2 x dx
yc 0
y
C1 C2 x x2
x3 e x
y
1
x 2e x
32. ycc 2 yc y
m 2 2m 1
1, 1.
x
C1 C2 x e
yp
v1 v 2 x e x
v1ce x v2c xe x
0
v1
³ x dx
v 2c
1
x2
1
³ x 2 dx
v2
y
1
1
ex x2
ln x
C1 C2 9 Ÿ C1 C2
0: 0
3C1 2C2 Ÿ C1
0, yc 0
ex , y 0
C1 cos 5 x C2 sin 5 x
yp
Ae x , y pc
yh y p
y0
0: 0
yc 0
0: 0
y
22
,
5
11
C2
33
5
2e3 x 3e 2 x 9
yh
y
1
x
11
5
2: 2
Ae x 25 Ae x
v1c e x v 2c x 1 e x
v1c
C1e3 x C2e 2 x 9
yh y p
34. ycc 25 y
0 when m
yh
C2e 2 x
Initial conditions:
y0
1
3
2
3, m2
2x
v2c
0
m1
y
v1c
0
0
m3 m 2
0
2, yc 0
54, y 0
495
y pcc
0
Ae x
1
26
1 x
C1 cos 5 x C2 sin 5 x e
26
e x Ÿ 26 A
1Ÿ A
1
1
Ÿ C1
26
26
1
1
Ÿ C2
5C2 26
130
C1 1
1
1 x
cos 5 x sin 5 x e
26
130
26
1
x
C1 C2 x ln x 1 e x
35. ycc 4 y
cos x
m2 4
0 Ÿ m
r 2i
yh
C1 cos 2 x C2 sin 2 x
yp
A cos x B sin x
y pc
A sin x B cos x
y pcc
A cos x B sin x
y pcc 4 y p
A cos x B sin x 4 A cos x B sin x
3 A cos x 3B sin x
yp
y
1
3
cos x Ÿ A
1
3
and B
cos x
0
cos x
yh y p
Initial conditions:
C1 cos 2 x C2 sin 2 x y0
6: 6
yc 0
Particular solution:
y
6: 6
17
3
C1 1
3
1
3
cos x
Ÿ C1
2C2 Ÿ C2
cos 2 x 3 sin 2 x 1
3
17
3
3
cos x
© 2010 Brooks/Cole, Cengage Learning
496
Chapter 16
Additional Topics in Differential Equations
ycc 3 yc
6x
m 3m
0 Ÿ m1
36.
2
3
0 and m2
3 x
yh
C1 C2e
yp
Ax3 Bx 2 Cx D
y pc
3 Ax 2 2 Bx C
y pcc
6 Ax 2 B
y pcc 3 y pc
6 Ax 2 B 3 3 Ax 2 2 Bx C
x2 yp
C1 C2e 3 x x 2 Initial conditions: y 0
3C2 2
3
43 e 3 x x 2 2
3
10 10
: 3
3
10
3
Particular solution: y
m2 m 2
1 xe x , y 0
1, yc 0
m 2 m 1
0 Ÿ m
2
3
x D
C3 C2e3 x x 2 Ÿ C2
x
1
3
43 and C3
A Bx Cx 2 e x
y pc
Bx Cx 2 e x B 2Cx e x
y pcc
B 2C B x Cx 2 e x 2C B 2Cx e x
C1e 2 x C2e x 1
yc 0
3
So, C2
Cx 2 B 4C x 2C 2 B e x
1 xe x Ÿ A
1
, 6C
2
1 and 2C 3B
0.
1
.
9
Initial conditions: y 0
83
Ÿ C1
18
Adding, 3C1
10
3
2C 2 B 4C B x Cx 2 e x B 2C B x Cx 2 e x 2 A Bx Cx 2 e x
1
and B
6
yh y p
x
B 2C B x Cx 2 e x
2 A 6Cx 2C 3B e x
y
2
3
2, 1
yp
23
3
C1e 2 x C2e x
So, C
1, and C
10 4e 3 x 3x 2 2 x
yh
y pcc y pc 2 y p
0, B
C3 C2
2: 2
yc 0
37. ycc yc 2 y
6 x, A
x D
2
3
yh y p
y
9 Ax 2 6 A 6 B x 2 B 3C
1 § 1
1 ·
¨ x x 2 ¸e x
2 © 9
6 ¹
1
3
Ÿ C1 C2
2
2
1
28
2C1 C2 Ÿ 2C1 C2
9
9
C1 C2 83
.
54
1
.
27
Particular solution: y
83 2 x
1 x 1 § 1 1 · x
e e ¨ x ¸ xe
54
27
2 ©9 6 ¹
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 16
38. yccc ycc
4 x2 , y 0
yccc ycc
0
m3 m 2
0 when m
1, yc 0
1, ycc 0
1
0, 0, 1.
yh
C1 C2 x C3e x
yp
A0 x 2 A1 x3 A2 x 4
y pc
2 A0 x 3 A1x 2 4 A2 x3
y pcc
2 A0 6 A1x 12 A2 x 2
y pccc
6 A1 24 A2 x
y pccc y pcc
497
2 A0 6 A1 6 A1 24 A2 x 12 A2 x 2
4 x 2 or A0
4, A1
43 , A2
13
C1 C2 x C3e x 4 x 2 43 x3 13 x 4
y
y'
C2 C3e x 8 x 4 x 2 ycc
C3e x 8 8 x 4 x 2
1, yc 0
Initial conditions: y 0
Particular solution: y
39. By Hooke’s Law, F
d y § 48 ·
¨ ¸y
dt 2
© 2¹
1, ycc 0
8 8 x 4 x 2 kx, k
F x
4
3
x3
C1 C3 , 1
1, 1
4
3
C2 C3 , 1
C3 8, C1
ma and m
F a
8, C2
8, C3
9
x3 13 x 4 9e x
64 4 3
48. Also, F
64 32
2. So,
2
y
0
6 t C2 sin 2
C2 cos 2
1
2
Because y 0
you have C1
40. From Exercise 39 you have k
6t .
1
2
and yc 0
48 and m
0 yields C2
0. So, y
1
2
cos 2
6t .
2. Also, the damping force is given by 1 8 dy dt .
§ d2y·
1 dy
2¨ 2 ¸
48 y
8 dt
© dt ¹
1
0
ycc yc 24 y
16
16 ycc yc 384 y
0
The characteristic equation 16m 2 m
m
So, y t
1
r
32
24,575 i
32
384
0 has complex roots
1
5 983
r
i.
32
32
ª
§ 5 983 ·
§ 5 983 ·º
e t 32 «C1 cos¨¨
t ¸¸ C2 sin ¨¨
t ¸¸».
© 32
¹
© 32
¹»¼
¬«
Initial conditions:
y0
1
Ÿ C1
2
yc 0
0 Ÿ
1
2
C
5 983
C2 1
32
32
Particular solution: y t
0 Ÿ C2
983
9830
ª1
§ 5 983 ·
§ 5 983 ·º
983
sin ¨¨
e t 32 « cos ¨¨
t ¸¸ t ¸¸»
9830
© 32
¹
© 32
¹»¼
¬« 2
© 2010 Brooks/Cole, Cengage Learning
498
Chapter 16
Additional Topics in Differential Equations
1
12S
24
cos 2t 2
sin 2t sin S t
2
4 S2
S 4
41. (a) (i) y
1
4
42. y p
cos x, y pc
y pcc 4 y pc 5 y p
12
14 sin x, y pcc
14 cos x
14 cos x 4 14 sin x 5 14 cos x
cos x sin x
0
10
False.
43. (a) y pcc
−12
1ª
1 6
2¬
(ii) y
2 t 3 sin 2
2 t cos 2
2t º
¼
A sin x and 3 y p
So, y pcc 3 y p
A sin x 3 A sin x
2 A sin x
60
(b) y p
0
14
(c) If y p
−60
y pcc
e ª
«199 cos
398 ¬
t 5
(iii) y
199 t
5
199 t º
»
5 ¼
199 sin
3 A sin x.
12 sin x
5
cos x
2
A cos x B sin x, then
A cos x B sin x, and solving for A and
B would be more difficult.
5, because yc
44. y
ycc
0 and 6 5
30
1
0
8
−1
1 2t
e cos 2t sin 2t
2
(iv) y
0.6
0
3
−0.2
(b) The object comes to rest more quickly. It may not
even oscillate, as in part (iv).
(c) It would oscillate more rapidly.
(d) Part (ii). The amplitude becomes increasingly large.
45.
x 4 yc y
f
¦ an x n :
0. Letting y
n 0
xyc 4 yc y
f
¦ nan x n
n 0
f
¦
f
4¦ nan x n 1 n 1
n 1 an x n
n 0
n 1 an
an 1
f
¦ 4nan x
n 1
f
¦ an x n
n 0
n 1
f
¦
n 1 an x n f
¦ 4 n 1 an 1x n
0
n 1
n 0
4 n 1 an 1
1
an
4
a0
a 0 , a1
y
a0 ¦
f
n 0
1
a0, a2
4
1
a1
4
1
a0, ! , an
42
1
a0
4n
xn
4n
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 16
499
f
46. ycc 3xyc 3 y
¦ an x n :
0. Letting y
n 0
f
¦ n n 1 an x n 2
ycc 3 xyc 3 y
n 2
f
¦
f
¦
n 2 n 1 an 2 x n
n 0
f
f
n 1
n 0
3 x ¦ nan x n 1 3¦ an x n
0
3 3n an x n
n 0
3 1 n an
n 2 n 1
an 2
a0
a0
a1
a1
a2
3
a0
2 ˜1
a3
0
There are no odd-powered terms for n ! 1.
a4
a6
3 § 3
·
a0 ¸
¨
4 ˜ 3© 2 ˜ 1 ¹
3 3 § 3 3 a0 ·
¨
¸
6 ˜ 5©
4! ¹
3 3 a0
4!
33 3 a 0
6!
a8
3 5 § 33 3 a 0 ·
¨
¸
8 ˜ 7 ¨© 6! ¸¹
a10
3 7 § 34 5 ˜ 3 a 0 ·
¨
¸¸
10 ˜ 9 ¨©
8!
¹
y
34 5 ˜ 3 a 0
8!
35 7 ˜ 5 ˜ 3 a 0
10!
f
1
3
a 0 x2 a 0 ¦
2
n 2
a0 47. ycc yc e x y
n 1 n
3 ª¬3 ˜ 5 ˜ 7" 2n 3 º¼ 2 n
x
2n !
2, yc 0
0, y 0
0
ycc
yc e x y
ycc 0
2
yccc
ycc e x y yc
yccc 0
2 2
y
4
yccc e y 2 yc ycc
y
5
y
x
e y 3 yc 3 ycc yccc
4
x
y | y 0 yc 0 x y
4
0
4
y
5
0
4 8
0
4
ycc 0 2
yccc 0 3
y4 0 4
y5 0 5
x x x x
2!
3!
4!
5!
2 x2 1 4
1 5
x x
6
30
§1·
Using the first four terms of the series, y¨ ¸ | 2.063.
© 4¹
48. ycc xy
1, yc 0
0, y 0
1
ycc
xy
ycc 0
0
yccc
xyc y
yccc 0
1
y
4
xycc yc yc
y
5
xycc 2 yc
xyccc ycc 2 ycc
y
6
y
7
xyccc 3 ycc
xy
4
yccc 3 yccc
xy
5
y
4
4y
y | y 0 yc 0 x 4
xy
4 yccc
4
xy
5
5y
4
ycc 0 2
y7 0 7
x "
x
2!
7!
y
4
0
2
y
5
0
0
y
6
0
4
y
7
0
10
1 x x3
x4
x6
x7
6
12 180 504
§1·
y¨ ¸ | 1.474
© 2¹
© 2010 Brooks/Cole, Cengage Learning
500
Chapter 16
Additional Topics in Differential Equations
Problem Solving for Chapter 16
1. 3 x 2 kxy 2 dx 5 x 2 y ky 2 dy
wM
wy
wN
wx
wM
wy
10 xy
wN
Ÿ k
wx
f x, y
³
f y x, y
5 x 2 y g c y
3 x 5 xy
2
wN
wx
2x y2
x
0
rE i
0 Ÿ m
Then C1 cos I
C2
, 0 d I 2S .
C1
C2 sin I .
C1
sin I
C2
. Then
cos I
C1 cos E x C2 sin E x
C sin I cos E x C cos I sin E x
0
y ·
¸ dx
x2 ¹
2
2 y
gc y
x
y
C sin E x . And if C2
0
y
S·
§
C sin ¨ E x ¸
2¹
©
y
y
g y
x
0 is
.
B1 B2 .
C1 C2
.
2
§ C1 C2 · r s x § C1 C2 · r s x
¨
¨
¸e
¸e
2
2
©
¹
©
¹
sx
sx
sx
ª §e e ·
§ e e sx ·º
e rx «C1 ¨
¸ C2 ¨
¸»
2
2
¹
©
¹¼
¬ ©
So y
2 y
Ÿ g y
x
B2e
rs x
C1 C2
and B2
2
Then B1
2
C1
e rx >C1 cosh sx C2 sinh sx@.
6. The roots of the characteristic equation m 2 am b
a 2 4b
0 and y
a
r E i, then y
2
(because cos E x and sin E x are bounded).
(ii) If the roots are complex, m
a (iii) If the roots are real and distinct, then y
For the first term, note that
0, then
C cos E x .
B1 B2 and C2
Let C1
0 Exact
2x B1e
rs x
C sin E x I .
0 and
5. The general solution to ycc ayc by
2
C
(i) If the roots are equal, then
0, then I
Note that if C1
0
ky
Ÿ k
x2
§
y2 ·
§ 2y ·
(b) ¨ 2 2 ¸ dx ¨ ¸ dy
x
© x ¹
©
¹
f y x, y
C1 C2 ax C1 C2 ax
e e
2
2
Let I be given by cot I
y
1
kxy dy
x2
y2 ·
ky
dy
2 ¸ dx x ¹
x
§
2
Let C
y 2 dx ³ ¨© 2 ra
C1 cos E x C2 sin E x
y
5 x 2 y 5 y 2
C
2. kx 2 y 2 dx kxy dy
f x, y
m E
2
C2
6 x3 15 x 2 y 2 10 y 3
2y
x2
4. ycc E 2 y
5 3
y C1
3
5y Ÿ g y
1
kx 2 x2
§
¨k
©
wM
wy
0 Exact
5
x3 x 2 y 2 g y
2
dx
2
5
5
x3 x 2 y 2 y 3
2
3
0 Ÿ m
§ e ax e ax ·
§ e ax e ax ·
C1 ¨
¸ C2 ¨
¸
2
2
©
¹
©
¹
C1 cosh ax C2 sinh ax
5
2
m a ma
B1e ax B2e ax
y
gc y
0, y ! 0
m2 a 2
2kxy
3 x 2 5 xy 2 dx 5 x 2 y 5 y 2 dy
(a)
3. ycc a 2 y
0
a 2 4b
C1e
a 1
a r
0 a, b ! 0 are m
a x
C1 C2 x e 2
a
o 0 as x o f.
a
x
a 2 4b
. You consider three cases:
2
x
C1e 2 cos E x C2e 2 sin E x o 0 as x o f
a 2 4b
x
2
a C2 e
4b
a. So y
a2
a 2 4b
x
2
. The second term clearly tends to 0 as x o f.
§ a a
4b ·
1
¨¨
¸¸
2 2
a2 ¹ x
C1e©
o 0 as x o f.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 16
7. ycc ay
0, y 0
0, ycc
(a) If a
and y L
y L
0Ÿ y
cx d . y 0
cL Ÿ c
0
8. ycc ay
0
0. So y
0
d
m2 a
0 is the
y
solution.
(b) If a 0, ycc ay
m2 a
y
a x
C1e
r
C2 e y0
0
C1 C2 Ÿ C1
y L
0
C1e
a L
C1e
a L
§e
2C1 ¨
¨
©
a L
C1e a L
e
2
9.
d 2T
g
T
dt 2
L
(a) T t
0,
(b)
ax
aL
nS
2
§ nS ·
¨ ¸ , n an integer.
© L ¹
·
¸¸
¹
0
C2
g
! 0
L
§
C1 sin ¨¨
©
§
Then C2 sin ¨¨
©
T t
aL
ax.
0 is the only solution.
§
g ·
t ¸¸ C2 cos¨¨
L ¹
©
§
Let I be given by tan ¨¨
©
Let A
a L
So
0
ai
a x C2 sin
C2 sin
0
a
a L Ÿ C1
2C1 sinh
So, y
C2
C2 e a L
C2 sin
y L
a x
r
y L
C1
0
y
a .
0 Ÿ m
C1 cos
y0
0 has characteristic equation
0 Ÿ m
0, a ! 0, y 0
501
g ·
I¸
L ¸¹
C2
§ g ·
cos ¨¨
I ¸¸
© L ¹
§
C1 sin ¨¨
©
g ·
I¸
L ¸¹
g ·
t¸
L ¸¹
C1 S
S
, I .
C2 2
2
§ g ·
C1 cos¨¨
I ¸¸.
© L ¹
C1
§ g ·
sin ¨¨
I ¸¸
© L ¹
§
g ·
t ¸¸ C2 cos¨¨
L ¹
©
g ·
t¸
L ¸¹
§
A sin ¨¨
©
g · §
I ¸ sin ¨
L ¸¹ ¨©
§
g ·
t ¸¸ A cos¨¨
L ¹
©
§
g ·
I ¸¸ cos¨¨
L ¹
©
g ·
t¸
L ¸¹
ª
A cos «
¬
º
g
t I »
L
¼
ª g
º
9.8, L
0.25
A cos «
t I », g
¬ L
¼
T 0
A cos ª¬ 39.2 I º¼
0.1
ª g
º
g
A
Tc t
sin «
t I »
L
¬ 4
¼
A 39.2 sin ª¬ 39.2 I º¼
Tc 0
0.5
5
Ÿ I | 0.1076 Ÿ A | 0.128.
Dividing, tan ª¬ 39.2 I º¼
39.2
T t
T t
0.128 cos ª¬ 39.2 t 0.108 º¼
2S
| 1 sec
39.2
(d) Maximum is 0.128.
(c) Period
(e) T t
0 at t | 0.359 sec, and at t | 0.860 sec.
(f ) T c 0.359 | 0.801, T c 0.860 | 0.801
© 2010 Brooks/Cole, Cengage Learning
502
Chapter 16
10. (a) Aycc
ycc
yc
y
y0
y2
Additional Topics in Differential Equations
12. ycc 2 yc 26 y
1 2
Wx , A ! 0
2
W 2
W
2W
x x
4x x2
A
2A
2A
W§ 2
x3 ·
¨ 2x ¸ C1
2 A©
3¹
W § 2 x3
x4 ·
¨
¸ C1 x C2
2 A© 3
12 ¹
2Wx 0 Ÿ C2
(a) O
1, Z
O Z
2
25 0, underdamped
1 5i
t
C1e cos 5t C2e t sin 5t
y
y0
C1
1
e t C1 cos 5t C2 sin 5t
yc t
0
e t 5C1 sin 5t 5C2 cos 5t
W § 16 16 ·
¸ 2C1
0 Ÿ
¨
2 A© 3
12 ¹
2W
2C1 Ÿ C1
A
yc 0
0
W
A
·
W § 2x
x
2x¸
¨
2 A© 3
12
¹
(b) Using a graphing utility, the maximum deflection is
at x | 1.1074, and the deflection is
W
W
1.43476 | 0.7174 .
A
2A
3
4
26,
2
1 5i, m2
(b) m1
1, yc 0
0, y 0
y
(c)
C1 5C2 Ÿ C2
4
e
t
1
cos 5t sin 5t
2
4
y
11. ycc 8 yc 16 y
(a) O
4, Z
(b) m1
m2
1, yc 0
0, y 0
4, O Z
2
2
0, critically damped
C1 C2 t e 4t ,
yc
4 C1 C2 t e 4t C2e 4t
y0
1
C1
yc 0
1
4 C2 Ÿ C2
1 5t e
(c)
5
−2
The solution oscillates.
13. ycc 20 yc 64 y
(a) O
10, Z
m1
(b)
10 6
10 6
16
16t
C1e C2e
2 C1 C2
yc t
4C1e 4t 16C2e 16t
yc 0
20
4C1 16C2
2 ½
¾C1
5¿
1, C2
1
e 4t e 16t
y
(c)
20
36 ! 0, overdamped
4, m2
4t
C1 4C2
4 t
2
8, O 2 Z 2
C1 C2
5
2, yc 0
0, y 0
y
y0
4
y
y
1
0
2
2
0
0
0
The solution tends to zero quickly.
2
0
The solution tends to zero quickly.
14. ycc 2 yc y
(a) O
1, Z
(b) m1
1, O 2 Z 2
y0
2
yc 0
1
y
1
0, critically damped
1
m2
C1 C2 t e t , yc
y
(c)
2, yc 0
0, y 0
C1 C2 t e t C2e t
C1
2 C2 Ÿ C2
1
2 t et
2
0
5
−2
The solution tends to zero quickly.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 16
15. Airy’s Equation: ycc xy
ycc xy y y
f
¦ an
Let y
0
ycc x 1 y y
f
¦ nan
n
x 1 , yc
n 0
f
¦ n n 1 an
n 2
f
¦
n 1
0
x 1
n 1
f
¦ n n 1 an
, ycc
n 1
ycc x 1 y y
n2
.
n 2
x 1
n2
x 1
n 1
f
¦ an
x 1
n 0
f
¦ an
n
f
¦ an
f
¦ ª¬ n 3
n 0
x 1
n
0
n 0
x 1
n 1
f
¦
n 1
n 0
2a2 a0
x 1
0
n 3 n 2 an 3 x 1
2a2 a0 503
n 2 an 3 an an 1º¼ x 1
an 1 x 1
n 1
n 1
0
0
1
a0 ; a0 , a1 arbitrary
2
an an 1
.
n3 n 2
0 Ÿ a2
In general, an 3
a3
a0 a1
6
a4
a1 a2
12
a5
a2 a3
20
§1 ·
a1 ¨ a0 ¸
2a1 a0
©2 ¹
12
24
1
a0 a1
a0 4a0 a1
2
6
20
120
§ a0 a1 · § 2a1 a0 ·
¨
¸¨
¸
5a0 6a1
© 6 ¹ © 24 ¹
a6
30
720
§ 2a1 a0 · § 4a0 a1 ·
¨
¸¨
¸
a4 a5
9a0 11a1
© 24 ¹ © 120 ¹
a7
42
42
5040
So, the first eight terms are
2a a0
4a a1
a
a a1
2
3
4
5
y
a0 a1 x 1 0 x 1 0
x 1 1
x 1 0
x 1
2
6
24
120
5a 6a1
9a 11a1
6
7
x 1 0
x 1 .
0
720
5040
a3 a4
30
16. (a) Tn 1 x
2 xTn x Tn 1 x
T0
1, T1
x
T2
2x x 1
T3
2x 2 x2 1 x
T4
2 x 4 x3 3x 2 x 2 1
2 x2 1
(b) 1 x 2 ycc xyc k 2 y
4 x3 3x
8x4 8x2 1
0
Substituting T0 , !, T4 into this equation shows that the polynomials
satisfy Chebyshev’s equation. For example, for T4 ,
1 x 2 ª¬96 x 2 16º¼ x ª¬32 x3 16 xº¼ 16ª¬8 x 4 8 x 2 1º¼
(c) T5
2 x 8 x 4 8 x 2 1 4 x3 3x
0
16 x5 20 x3 5 x
T6
2 x 16 x 5 20 x3 5 x 8 x 4 8 x 2 1
32 x 6 48 x 4 18 x 2 1
T7
2 x 32 x 6 48 x 4 18 x 2 1 16 x5 20 x3 5 x
64 x 7 112 x5 56 x3 7 x
© 2010 Brooks/Cole, Cengage Learning
504
Chapter 16
Additional Topics in Differential Equations
17. x 2 ycc xyc x 2 y
0 Bessell equation of order zero
f
f
¦ an x n , yc
(a) Let y
n 0
x 2 ycc xyc x 2 y
f
¦ nan x n 1, ycc
¦ n n 1 an xn 2 .
n 1
n 2
0
f
f
f
n 2
n 1
n 0
x 2 ¦ n n 1 an x n 2 x ¦ nan x n 1 x 2 ¦ an x n
f
f
n 2
f
n 1
f
¦ n n 1 an x n ¦ nan x n ¦ an x n 2
¦
n 2 n 1 an 2 x n 2 n 0
a1 x f
¦ ª¬ n 2
an
0 and an 2
2
n 2
All odd terms ai are 0.
a0
a2
22
a2
1
a4
a0 2
42
2 ˜ 42
a6
a4
62
y
a0 ¦
f
n 0
a0
n 0
f
¦
n 1
n 2 an 2 x n 2 f
¦ an x n 2
a0
2
2 1˜ 2
4
a0
6
2 3!
2
n
1 x 2 n
22 n n!
2
f
¦ an x n , yc
¦ nan x n 1, ycc
n 0
n 1
x ycc xyc x 1 y
2
2
1 ).
f
¦ n n 1 an x n 2
n 2
0
f
f
n 2
n 1
x 2 ¦ n n 1 an x n 2 x ¦ nan x n 1 x 2 1
f
¦ an x n
f
f
f
f
n 2
n 1
n 0
n 0
n 2 n 1 an 2 x n 2 n 0
a0 a1 a1 x f
¦
n 1
f
¦ ª¬ n 2
n 0
a0
f
y
n 2 an 2 x n 2 2a1 ¦
n 0
0
f
¦ an x n 2
n 0
f
¦
n 2
an 2 x n 2
n 1 an 2 n 2 an 2 an an 2 º¼ x n 2
0 and ª¬ n 2 n 1 n 2 1º¼ an 2
All even terms
a1
a3
2˜4
a3
a5
4˜6
a5
a7
6˜8
0
n 0
¦ n n 1 an x n ¦ nan x n ¦ an x n 2 ¦ an x n
f
0
.
1
22 ˜ 42 ˜ 6 2
f
¦
0
n 0
(b) This is the same function (assuming a0
18. (a) Let y
0
n 1 an 2 n 2 an 2 an º¼ x n 2
n 0
a1
0
an Ÿ ª¬n 2 4n 3º¼ an 2
0
0
an Ÿ an 2
an
n 1 n 3
ai are 0.
a1
23
a1
2a1
25 3!
25 ˜ 2! 3!
2a1
27 3! 4!
n
1 x 2 n 1
2 n 1
2
n! n 1 !
(b) This is the same function (assuming 2a1
1 ).
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 16
f
19. (a) Let y
¦ an x n , yc
n 0
ycc 2 xyc 8 y
¦ n n 1 an x n 2
n 2
f
n 0
an 2
f
f
n 1
n 0
a2
48
2x
0!
H1 x
2x
1!
1
¦
n 0
1
¦
n 0
2
H4 x
n 0
n 0
0
0
1
a2 Ÿ a2
3
2 4
a0
2
48
4a0 Ÿ a0
12
16 x 4 48 x 2 12
(b) H 0 x
H3 x
f
¦ 2nan x n ¦ 8an x n
n 1 an 2 2nan 8an º¼ x n
2 2
a2
43
16
H2 x
f
0
2n4
an
n 2 n 1
a4
H4 x
n 2
2 x ¦ nan x n 1 8 ¦ an x n
n 0
f
¦ n n 1 an xn 2 .
n 1
n 2 n 1 an 2 x n ¦ ª¬ n 2
f
¦ nan x n 1, ycc
0
f
¦
f
505
¦
n 0
0
1
1
2x
n
2 2n
1 2! 2 x
n! 2 2n !
n
3 2n
1 3! 2 x
n! 3 2n !
n
4 2n
1 4! 2 x
n! 4 2n !
2 2x
2!
3! 2 x
3!
4! 2 x
4!
2
2
1
3! 2 x
1
4! 2 x
2!
3
4
4 x2 2
1
8 x3 12 x
2
4!
2!
16 x 4 48 x 2 12
© 2010 Brooks/Cole, Cengage Learning
506
Chapter 16
Additional Topics in Differential Equations
20. (a) xycc 1 x yc ky
0
f
f
¦ an x n , yc
Let y
f
¦ nan x n 1, ycc
n 0
¦ n n 1 an xn 2 .
n 1
f
x ¦ n n 1 an x n 2 1 x
n 2
n 2
f
f
n 1
n 0
¦ nan x n 1 k ¦ an x n
f
f
f
f
n 2
n 1
n 1
n 0
0
¦ n n 1 an x n 1 ¦ nan x n 1 ¦ nan x n ¦ kan x n
f
¦
n 1 nan 1x n n 1
f
¦
f
¦ ª¬ n 1 nan 1 n 1
a1 ka0
Ÿ a1
n 1 a n 1 k n an
0 Ÿ an 1
Let a0
1.
For k
0, a1
a2
For k
1, a1
1, a2
For k
2, a1
2, a2
"
n 1
n 0
n 1 an 1 nan kan º¼ x n
0
2
f
¦ nan x n ¦ kan x n
n 1 an 1x n n 0
a1 ka0 f
"
nk
n 1
In general, for a given integer k t 0, ak 1
an
nk
1
2
2
k n 1 k n 2
0
¦
n 0
1
L1 x
¦
n 0
2
L2 x
¦
n 0
3
L3 x
¦
n 0
4
L4 x
¦
n 0
2
2
an
1 x.
an 2
1 2x "
ak 2
1 2
x .
2
0. Furthermore, in the given formula for
an . Finally, you can see that for k t n,
1 k 1 n n 2 k
˜
an 2
2
n2
n 1
an 1
1
"
k n 1 k n 2 "k k n !
n!
(b) L0 x
2
n2
n2 n 1
1
n 1
1 k n 1
n 1 k
an 1
n2
0
1.
1
Ÿ L2 x
2
Lk x , you can verify that an 1
2
0 Ÿ L1 x
1
a1
22
0
ka0
0 Ÿ L0 x
a3
0
k n!
n
k n 1 k n 2 " k 0
2
n 2 n 1 "22 ˜ 12
a0
n
a0
1 k!
k n ! n!
2
n
1 0! x n
0 n ! n!
2
1
2
1 x
2
1 2x x2
2
2
1 3x 3 2
x3
x 2
6
2
1 4 x 3x 2 n
1 1! x n
1 n ! n!
n
1 2! x n
2 n ! n!
n
1 3! x n
3 n ! n!
n
1 4! x n
4 n ! n!
2 3
1 4
x x
3
24
© 2010 Brooks/Cole, Cengage Learning