SOL Cálculo 2 de varias variables, 9na Edición - Ron Larson & Bruce H. Edwards (IN)
Descripción
SIGUENOS EN:
LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN DESCARGA DIRECTA VISITANOS PARA DESCARGARLOS GRATIS.
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C H A P T E R 1 0 Vectors and the Geometry of Space Section 10.1 Vectors in the Plane
. . . . . . . . . . . . . . . . . . . . 474
Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 479 Section 10.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 483 Section 10.4 The Cross Product of Two Vectors in Space . . . . . . . . 487 Section 10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 491 Section 10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 496 Section 10.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 499 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
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C H A P T E R 1 0 Vectors and the Geometry of Space Section 10.1
Vectors in the Plane
Solutions to Even-Numbered Exercises 2. (a) v 3 3, 2 4 0, 6 (b)
4. (a) v 1 2, 3 1 3, 2 (b)
y
−3 −2 −1
y
x 1
−1
2
3
3
(− 3, 2)
−2
2
v
−3
v
−4
1
−5
(0, − 6)
−6
−3
6. u 1 4, 8 0 5, 8
−2
x
−1
8. u 11 4, 4 1 15, 3
v 7 2, 7 1 5, 8
v 25 0, 10 13 15, 3
uv
uv
10. (b) v 3 2, 6 6 1, 12 (a) and (c).
12. (b) v 5 0, 1 4 5, 3
y
(a) and (c).
y
(1, 12)
12 10 8
(− 5, 3)
v
6 4 2
(3, 6) −6
1
2
3
4
5
6
−4
2
7
x
−2
(− 5, −1)
x
−1
4
v
2 −2
(0, − 4)
−4
(2, −6)
−6
14. (b) v 3 7, 1 1 10, 0 (a) and (c).
16. (b) v 0.84 0.12, 1.25 0.60 0.72, 0.65 (a) and (c).
y
y
3
1.25
2
1.00
1
(−10, 0)
(0.84, 1.25)
0.75
v
(0.72, 0.65)
x
−8 −6 −4 −2
(−3, −1)
(0.12, 0.60)
0.50
2 4 6 8
−2
(7, −1)
18. (a) 4v 4, 20
v
0.25
x
−3
0.25 0.50 0.75 1.00 1.25
1 (b) 2 v
12 , 52
y
y
(−1, 5)
(− 4, 20) 20
v 4v 1
(− 1, 5)
−4 −3 −2 −1
v −12 −8 −4
474
x 4
8
12
−2 −3
—CONTINUED—
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x
− 1v
3
2
( 12, − 25 (
4
Section 10.1
Vectors in the Plane
18. —CONTINUED— (d) 6v 6, 30
(c) 0v 0, 0
y
y
(− 1, 5) v
6
(− 1, 5)
−15 −10 −5
x 5
−10
10
15
−6v
−15
v
−20
0v
−25 x
−3 −2 −1
1
2
(6, − 30)
−30
3
20. Twice as long as given vector u.
22.
y
y
u + 2v u 2v 2u u x
x
2 2 16 24. (a) 3u 3 3, 8 2, 3
26. v 2i j i 2j
(b) v u 8, 25 3, 8 11, 33
3i j 3, 1
(c) 2u 5v 23, 8 58, 25 34, 109
y
2
w
v
1
x 1
2
3
u −1
28. v 5u 3w 52, 1 31, 2 7, 11
30. u1 3 4 u2 2 9
y 2 −4 −2
u1 7
x 4
6
8
10
u2 7
5u
−3w −6
Q 7, 7
v
−8 −10 −12
32. v 144 25 13
34. v 100 9 109
40. u 6.22 3.42 50 52
38. u 52 152 250 510 v
36. v 1 1 2
5, 15 1 3 u , unit vector u 10 10 510
v
1.24 0.68 6.2, 3.4 u , unit vector 52 2 2 u
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475
476
Chapter 10
Vectors and the Geometry of Space 44. u 2, 4, v 5, 5
42. u 0, 1, v 3, 3 (a) u 0 1 1
(a) u 4 16 25
(b) v 9 9 32
(b) v 25 25 52
(c)
u v 3, 2
(c)
u v 49 1 52
u v 9 4 13 u 0, 1 u
(d)
uu 1
1 v 3, 3 v 32
vv 1
1 uv 3, 2 u v 13
1 uv 7, 1 u v 52
(f)
uu vv 1
uu vv 1
u 3, 2
46.
1 v 5, 5 v 52
(e)
vv 1 (f)
1 u 2, 4 u 25
(d)
uu 1 (e)
u v 7, 1
1 u 1, 1 u 2
48.
u 13 3.606 4
v 1, 2
uu 22 1, 1 v 22, 22
v 5 2.236 u v 2, 0 u v 2 u v ≤ u v 1 u 0, 3 u 3
50. 3
52. v 5 cos 120i sin 120j 5 53 i j 2 2
u 0, 3 u v 0, 3
54. v cos 3.5i sin 3.5j
u 4i
56.
v i 3 j
0.9981i 0.0610j 0.9981, 0.0610
u v 5i 3 j 58.
u 5 cos0.5 i 5 sin0.5 j
60. See page 718:
5 cos0.5 i 5 sin0.5 j v 5 cos0.5 i 5 sin0.5 j u v 10 cos0.5 i
(ku1, ku2) (u1 + v1, u2 + v2) (u1, u2)
ku
u+v
ku2
u
u2
(u1, u2) u2
u (v1, v2) v2
v v1
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u1
u1 ku1
Section 10.1
Vectors in the Plane
477
62. See Theorem 10.1, page 719. For Exercises 64–68, au bw ai 2j bi j a bi 2a bj. 64. v 3j. Therefore, a b 0, 2a b 3. Solving simultaneously, we have a 1, b 1.
66. v 3i 3j. Therefore, a b 3, 2a b 3. Solving simultaneously, we have a 2, b 1.
68. v i 7j. Therefore, a b 1, 2a b 7. Solving simultaneously, we have a 2, b 3. 72. f x tan x
70. y x3, y 3x2 12 at x 2.
fx sec2 x 2 at x
(a) m 12. Let w 1, 12, then 1 w ± 1, 12. w 145
. 4
(a) m 2. Let w 1, 2, then w 1 1, 2. ± w 5
1 (b) m 12 . Let w 12, 1, then
1 w ± 12, 1. w 145
(b) m 12. Let w 2, 1, then w 1 ± 2, 1. w 5
74.
u 23 i 2j
76. magnitude 63.5 direction 8.26
u v 3i 33 j v u v u 3 23 i 33 2 j 78. F1 2, F1 10 F2 4, F2 140 F3 3, F3 200 R F1 F2 F3 4.09
R F1 F2 F3 163.0 80.
F1 F2 500 cos 30i 500 sin 30j 200 cos45 i 200 sin45 j 2503 1002 i 250 1002 j F1 F2 2503 1002 250 1002 584.6 lb 2
tan
2
250 1002 ⇒ 10.7 2503 1002
82. F1 F2 F3 400cos30i sin30j 280cos45i sin45j 350cos135i sin135j 2003 1402 1752i 200 1402 1752 j R
2003 3522 200 31522 385.2483 newtons 200 315 2
0.6908 39.6 200 3 35 2
R arctan
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478
Chapter 10
Vectors and the Geometry of Space
84. F1 20, 0, F2 10cos , sin (a) F1 F2 20 10 cos , 10 sin 400 400 cos 100 cos2 100 sin2 500 400 cos (b)
(c) The range is 10 ≤ F1 F2 ≤ 30.
40
The maximum is 30, which occur at 0 and 2. The minimum is 10 at . 2
0
(d) The minimum of the resultant is 10.
0
86.
u 7 1, 5 2 6, 3 1 u 2, 1 3 P1 1, 2 2, 1 3, 3 P2 1, 2 22, 1 5, 4
0.8761 or 50.2 24 20 24 arctan 1.9656 or 112.6 10
y
88. 1 arctan
2
θ2
A v C
u ucos 1 i sin 1 j
B u
θ1
x
v vcos 2 i sin 2 j Vertical components: u sin 1 v sin 2 5000 Horizontal components: u cos 1 v cos 2 0 Solving this system, you obtain u 2169.4 and v 3611.2. 90. 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. u u cos 60i sin 60j
20°
v
30° u
v v cos 110i sin 110j Thus, u sin 60 v sin 110 100, or
23 v sin 110 100.
u
And u cos 60 v cos 110 0 or u
12 v cos 110 0
Multiplying the last equation by 3 and adding to the first equation gives u sin 110 3 cos 110 100 ⇒ v 65.27 lb. Then, u
12 65.27 cos 110 0 gives
u 44.65 lb. (a) The tension in each rope: u 44.65 lb, v 65.27 lb. (b) Vertical components: u sin 60 38.67 lb. v sin 110 61.33 lb.
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100 lb
Section 10.2
Space Coordinates and Vectors in Space
u 400iplane
92.
v 50cos 135i sin 135j 252i 252j wind u v 400 252i 252j 364.64i 35.36j tan
35.36 ⇒ 5.54 364.64
Direction North of East: N 84.46 E Speed: 336.35 mph 94. u cos2 sin2 1, v sin2 cos 2 1 96. 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. Therefore, r xu v, s yv u. But, urs xu v yv u x yu x yv. 1 Therefore, x y 1 and x y 0. Solving we have x y 2 . s u r v
98. The set is a circle of radius 5, centered at the origin. u x, y x2 y2 5 ⇒ x2 y 2 25 100. True
102. False
104. True
ab0
Section 10.2 2.
Space Coordinates and Vectors in Space z
4.
z
8
8
(3, −2, 5)
(4, 0, 5)
x
x
y
( 32 , 4, −2( 6. A2, 3, 1 B3, 1, 4
y
(0, 4, − 5)
8. x 7, y 2, z 1:
10. x 0, y 3, z 2: 0, 3, 2
7, 2, 1
12. The x-coordinate is 0.
14. The point is 2 units in front of the xz-plane.
16. The point is on the plane z 3.
18. The point is behind the yz-plane.
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479
480
Chapter 10
Vectors and the Geometry of Space 22. The point x, y, z is 4 units above the xy-plane, and above either quadrant II or IV.
20. The point is in front of the plane x 4.
24. The point could be above the xy-plane, and thus above quadrants I or III, or below the xy-plane, and thus below quadrants II or IV. 26. d 2 22 5 32 2 22
28. d 4 22 5 22 6 32
16 64 16 96 46
4 49 9 62
30. A5, 3, 4, B7, 1, 3, C3, 5, 3
32. A5, 0, 0, B0, 2, 0, C0, 0, 3
AB 4 4 1 3 AC 4 4 1 3 BC 16 16 0 42 Since AB AC , the triangle is isosceles.
AB 25 4 0 29 AC 25 0 9 34 BC 0 4 9 13 Neither
34. The y-coordinate is changed by 3 units:
36.
5, 6, 4, 7, 4, 3, 3, 8, 3 38. Center: 4, 1, 1
40. Center: 3, 2, 4
Radius: 5
r3
x 42 y 12 z 12 25 x 2
y2
4 2 8, 0 2 8, 6 2 20 6, 4, 7
z2
(tangent to yz-plane)
8x 2y 2z 7 0
x 32 y 22 z 42 9
x2 y2 z2 9x 2y 10z 19 0
42.
x
2
9x
81 81 y2 2y 1 z2 10z 25 19 1 25 4 4
x 29 Center: Radius:
2
y 12 z 52
109 4
29, 1, 5 109
2 4x2 4y2 4z2 4x 32y 8z 33 0
44.
x2 y2 z2 x 8y 2z
x
2
x
1 33 1 y2 8y 16 z2 2z 1 16 1 4 4 4
x 21 y 4 2
2
Center:
33 0 4
z 12 9
12, 4, 1
Radius: 3
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Section 10.2
Space Coordinates and Vectors in Space
x2 y2 z2 < 4x 6y 8z 13
46.
x2 4x 4 y2 6y 9 z2 8z 16 < 4 9 16 13 x 22 y 32 z 42 < 16 Interior of sphere of radius 4 centered at 2, 3, 4. 48. (a) v 4 0 i 0 5 j 3 1k
50. (a) v 2 2 i 3 3 j 4 0k
4i 5j 2k 4, 5, 2 (b)
4k 0, 0, 4 (b)
z
z
8
4
〈 0, 0, 4 〉
〈4, − 5, 2〉
x
x
y
52. 1 4, 7 5, 3 2 5, 12, 5
54. 2 1, 4 2, 2 4 1, 6, 6
5, 12, 5 25 144 25 194 Unit vector:
5, 12, 5 12 5 5 , , 194 194 194 194
y
1, 6, 6 1 36 36 73
173, 673, 673
Unit vector:
56. (b) v 4 2 i 3 1 j 7 2k
z
6i 4j 9k 6, 4, 9
12
(− 4, 3, 7) (− 6, 4, 9)
(a) and (c).
x
y
(2, −1, − 2)
5 2 1 58. q1, q2, q3 0, 2, 2 1, 3, 2
Q 1, 3, 3 8
60. (a) v 2, 2, 1
(b) 2v 4, 4, 2
z
z
4
8
〈4, − 4, 2〉
〈− 2, 2, −1〉 x
(c)
1 2
x
y
v 1, 1, 12
(d)
5 2v
y
5, 5, 52
z
z
2
8
〈5, −5, 〈 5 2
〈1, −1, 12 〈 x
y
x
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y
481
482
Chapter 10
Vectors and the Geometry of Space
62. z u v 2w 1, 2, 3 2, 2, 1 8, 0, 8 7, 0, 4 1 64. z 5u 3v 2 w 5, 10, 15 6, 6, 3 2, 0, 2 3, 4, 20
66. 2u v w 3z 21, 2, 3 2, 2, 1 4, 0, 4 3z1, z2, z3 0, 0, 0 0, 6, 9 3z1, 3z2, 3z3 0, 0, 0 0 3z1 0 ⇒ z1 0 6 3z2 0 ⇒ z2 2 9 3z3 0 ⇒ z3 3 z 0, 2, 3 68. (b) and (d) are parallel since i 43 j 32 k 2 12 i 23 j 34 k and 34 i j 98 k 32 12 i 23 j 34 k. 72. P4, 2, 7, Q2, 0, 3, R7, 3, 9
70. z 7, 8, 3 (b) is parallel since zz 14, 16, 6.
\
PQ 6, 2, 4 \
PR 3, 1, 2 3, 1, 2 12 6, 2, 4 → → Therefore, PQ and PR are parallel. The points are collinear. 74. P0, 0, 0, Q1, 3, 2, R2, 6, 4
76. A1, 1, 3, B9, 1, 2, C11, 2, 9, D3, 4, 4
\
\
PQ 1, 3, 2
AB 8, 2, 5
\
\
PR 2, 6, 4 \
DC 8, 2, 5
\
\
AD 2, 3, 7
Since PQ and PR are not parallel, the points are not collinear.
\
BC 2, 3, 7 \
\
\
\
Since AB DC and AD BC , the given points form the vertices of a parallelogram. 78. v 1 0 9 10
80.
v 4, 3, 7
82.
v 16 9 49 74 u 6, 0, 8
84.
(a)
u 8
1 u 6, 0, 8 u 10
(b)
(a)
u 1 6, 0, 8 u 10
88. (a) u v 4, 7.5, 2
v 1 9 4 14
u 8, 0, 0
86.
u 36 0 64 10
v 1, 3, 2
u 1, 0, 0 u
(b)
90.
u 1, 0, 0 u
c u c, 2c, 3c
(b) u v 8.732
c u c 2 4c 2 9c 2 3
(c) u 5.099
14c 2 9
(d) v 9.014
c±
314 14
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Section 10.3 92. v 3
u 1 1 1 3 3 3 3 , , , , u 3 3 3 3 3 3
94. v 5
The Dot Product of Two Vectors
u 2 3 1 5 , , u 14 14 14
96. v 5cos 45i sin 45k
52 i k or 2
52 v 5cos 135i sin 135k i k 2
483
7 70 , 31470 , 1470 v 5, 6, 3
98.
103, 4, 2 1, 2, 5 103, 4, 2 133, 6, 3 2 3v
z
5 2 (i + k) 2
8
x
y
102. x x02 y y02 z z02 r 2
100. x0 is directed distance to yz-plane. y0 is directed distance to xz-plane. z0 is directed distance to xy-plane. 104. A sphere of radius 4 centered at x1, y1, z1. v x x2, y y1, z z1
106. As in Exercise 105(c), x a will be a vertical asymptote. Hence, lim T . r0 →a
x x1 y y1 z z1 4 2
2
2
x x12 y y12 z z12 16 sphere 550 c75i 50j 100k
108.
302,500
18,125c 2
c 2 16.689655 c 4.085 F 4.08575i 50j 100k
306i 204j 409k
110. Let A lie on the y-axis and the wall on the x-axis. Then A 0, 10, 0, B 8, 0, 6, C 10, 0, 6 and → → AB 8, 10, 6 , AC 10, 10, 6 . → → AB 102, AC 259 → → AB AC Thus, F1 420 → , F2 650 → 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
Section 10.3
The Dot Product of Two Vectors
2. u 4, 10 , v 2, 3
4. u i, v i
(a) u v 42 103 22
(a) u v 1
(b) u u 44 1010 116
(b) u u 1
(c)
u2
116
(d) u vv 222, 3 44, 66 (e) u
2v 2u v 222 44
(c) u2 1 (d) u v v i (e) u 2v 2u
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v 2
484
Chapter 10
Vectors and the Geometry of Space 8. u 3240, 1450, 2235
6. u 2i j 2k, v i 3j 2k (a) u
v 21 13 22 5
v 2.22, 1.85, 3.25
(b) u u 22 11 22 9 (c)
u2
Increase prices by 4%: 1.042.22, 1.85, 3.25 . New total amount: 1.04u v 1.0417,139.05
9
(d) u v v 5i 3j 2k 5i 15j 10k (e) u
10.
$17,824.61
2v 2u v 25 10
uv cos u v
12. u 3, 1 , v 2, 1
u v 4025 cos
cos
5 5003 6
u cos
14.
v cos cos
6 i sin 6 j
3
2
1 i j 2
uv u v
3
2
3
uv 32 23 0 0 u v u v
2
18. u 2i 3j k, v i 2j k cos
cos
22 21 22 42 1 2 arccos 1 3 105 4
4
16. u 3i 2j k, v 2i 3j
2 2 3 3 i sin j i j 4 4 2 2
uv 5 1 u v 105 2
uv u v
20. u 2, 18 , v
32, 61
u cv ⇒ not parallel
9 321 9 14 146 221
u v 0 ⇒ orthogonal
3 1421 10.9
arccos
1 22. u 3 i 2j, v 2i 4j
u
1 6
v ⇒ parallel
26. u cos , sin , 1 , v sin , cos , 0 u c v ⇒ not parallel u
v0
⇒ orthogonal
24. u 2i 3j k, v 2i j k u cv ⇒ not parallel u v 0 ⇒ orthogonal
28. u 5, 3, 1 cos cos cos
u 35
5 35
3 35
1 35
cos2 cos2 cos2
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25 9 1 1 35 35 35
Section 10.3
The Dot Product of Two Vectors
485
30. u a, b, c , u a2 b2 c 2 cos cos cos
a2
a b2 c2
a 2
b b2 c 2
a2
c b2 c2
cos2 cos2 cos2
32. u 4, 3, 5 cos cos cos
u 50 52
4 52 3 52 5 52
36. F1: C1 F2: C2
a2 b2 c2 1 a2 b2 c 2 a2 b2 c 2 a2 b2 c 2
1 2
34. u 2, 6, 1
⇒ 2.1721 or 124.4
cos
⇒ 1.1326 or 64.9
cos
⇒
or 45 4
cos
300
13.0931 F1
38.
230.239, 36.062, 65.4655
cos
65.4655 ⇒ 74.31 F
v1 s, s, s
cos
z
v1 (s, s, s)
40. F1 C10, 10, 10. F1 200 C1 102 ⇒ C1 102
6
3
v2 x
35.26
42. w2 u w1 9, 7 3, 9 6, 2
and F1 0, 1002, 1002 F2 C2 4, 6, 10 F2 C34, 6, 10 F 0, 0, w F F1 F2 F3 0 4C2 4C3 0 ⇒ C2 C3 1002 6C2 6C3 0 ⇒ C2 C3
y
s2 6 3 s3
arccos
F 242.067 lb
36.062 ⇒ 98.57 F
⇒ 1.4140 or 81.0
v2 s2
13.093120, 10, 5 6.32465, 15, 0
cos
1 41
⇒ 0.3567 or 20.4
v2 s, s, 0
F F1 F2
230.239 ⇒ 162.02 F
6 41
⇒ 1.8885 or 108.2
v1 s3
100
6.3246 F2
cos
2 41
u 41
252 N 3
44. w2 u w1 8, 2, 0 6, 3, 3 2, 1, 3
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(s, s, 0)
486
Chapter 10
Vectors and the Geometry of Space 48. u 1, 0, 4 , v 3, 0, 2
46. u 2, 3 , v 3, 2
uv vv 0v 0, 0
(a) w1
(a) w1
2
(b) w2 u w1 2, 3
11 33 22 3, 0, 2 , 0,
uv vv 13 13 13 2
(b) w2 u w1 1, 0, 4
50. The vectors u and v are orthogonal if u v 0.
3313, 0, 2213
20 30 , 0, 13 13
52. (a) and (b) are defined.
The angle between u and v is given by uv . u v
cos
54. See figure 10.29, page 739.
uv v v vu u u
56. Yes,
2
2
v
u
u vv2 v uu2 1 1 v u u v
58. (a) u 5, v 8.602, 91.33
60. (a)
(b) u 9.165, v 5.745, 90
6417, 1617
(b)
21 63 42 , , 26 26 13
62. Because u appears to be a multiple of v, the projection of u onto v is u. Analytically, projv u
uv 3, 2 6, 4 6, 4 v v2 6, 4 6, 4 26 6, 4 3, 2 u. 52
64. u 8i 3j. Want u v 0.
66. u 0, 3, 6 . Want u v 0.
v 3i 8j and v 3i 8j are orthogonal to u. \
68. OA 10, 5, 20 , v 0, 0, 1 \
projvOA
20 0, 0, 1 0, 0, 20 12
\
projvOA 20
v 0, 6, 3 and v 0, 6, 3 are orthogonal to u. 70. F 25cos 20i sin 20j v 50i W F v 1250 cos 20 1174.6 ft lb
\
72. PQ 4, 2, 10 \
V 2, 3, 6 \
W PQ
V 74 \
74. True w u v w u w v 000 ⇒ w and u v are orthogonal.
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Section 10.4
The Cross Product of Two Vectors in Space \
(d) r1 k, k, 0
z
76. (a)
(0, k, k) (k, 0, k)
k
\
r2 0, 0, 0
y
k
k
cos
(k, k, 0)
x
k 2
2k, 2k, 2k 2k, 2k, 2k
2k, 2k, 2k 2k, 2k, 2k
k2 4
2
3
1 3
(b) Length of each edge:
109.5
k2 k2 02 k2
(c) cos
k2 1 2 k2 k2
arccos
12 60
78. The curves y1 x2 and y2 x1 3 intersect at 0, 0 and at 1, 1. At 0, 0: 1, 0 is tangent to y1 and 0, 1 is tangent to y2. The angle between these vectors is 90.
2
At 1, 1: 1 5 1, 2 is tangent to y1 and 310 1, 1 3 1 10 3, 1 is tangent to y2. To find the angle between these vectors,
1
cos
80.
y
u
y1
y2
1 1 1 3 2 ⇒ 45. 5 10 2
(1, 1) x
(0, 0)
1
2
v u v cos
u v u v cos u v cos ≤ u v since cos
≤ 1.
82. Let w1 projv u, as indicated in the figure. Because w1 is a scalar multiple of v, you can write u w1 w2 cv w2. u v cv w2 v cv
u
w2
Taking the dot product of both sides with v produces
v w2 v
θ v
c v 2, since w2 and v are orthogonol.
w1
uv uv v. Thus, u v c v 2 ⇒ c and w1 projv u cv v 2 v 2
Section 10.4
The Cross Product of Two Vectors in Space
i 2. i j 1 0
j k 0 0 k 1 0
i 4. k j 0 0
z
k j
x
−1
1
−i
k j
j 1
1 y
x
1
1 −1
j k 0 1 j 0 0
z
1
k
i
i 6. k i 0 1
z
1
1
j k 0 1 i 1 0
y
http://librosysolucionarios.net
x
i
1 −1
y
487
488
Chapter 10
Vectors and the Geometry of Space
i 8. (a) u v 3 2
j k 0 5 15, 16, 9 3 2
i j k 10. (a) u v 3 2 2 8, 5, 17 1 5 1
(b) v u u v 15, 16, 9
(b) v u u v 8, 5, 17
(c) v v 0
(c) v v 0
12. u 1, 1, 2, v 0, 1, 0
i u v 1 0 u
14. u 10, 0, 6, v 7, 0, 0
j 1 1
k 2 2i k 2, 0, 1 0
i u v 10 7
u v 12 10 21
u
k 6 42j 0, 42, 0 0
u v 100 042 60
0 ⇒ uu v
0 ⇒ uu v
v u v 02 10 01
v u v 70 042 00
0 ⇒ vu v
16. u v
j 0 0
0 ⇒ vu v
i 1 2
j k 6 0 6i j 13k 1 1
u u v 16 61 0 ⇒ u u v v u v 26 11 113 0 ⇒ v u v 18.
20.
z 6 5 4 3 2 1
z 6 5 4 3 2 1
v 1
1 4
3
2
v
u
4 6
4
y
3
2
u
4 6
y
x
x
22. u 8, 6, 4 v 10, 12, 2 u v 60, 24, 156 u v 1 60, 24, 156 u v 3622
3522 , 3222 , 31322
24.
2 u k 3 1 v i 6k 2
1 u v 0, , 0 3
u v 0, 1, 0 u v
26. (a) u v 18, 12, 48
28. u i j k
u v 52.650
vjk
(b) u v 50, 40, 34 u v 72.498
i u v 1 0
j k 1 1 j k 1 1
A u v j k 2
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Section 10.4
The Cross Product of Two Vectors in Space
30. u 2, 1, 0 v 1, 2, 0 u v
i j 2 1 1 2
k 0 0, 0, 3 0
A u v 0, 0, 3 3 32. A2, 3, 1, B6, 5, 1, C3, 6, 4, D7, 2, 2 \
\
\
\
AB 4, 8, 2, AC 1, 3, 3, CD 4, 8, 2, BD 1, 3, 3 \
\
\
\
Since AB CD and AC BD , the figure is a parallelogram. \
\
AB and AC are adjacent sides and i j k AB AC 4 8 2 18, 14, 20. 1 3 3 \
\
\
\
Area AB AC 920 2230 34. A2, 3, 4, B0, 1, 2, C1, 2, 0 \
36. A1, 2, 0, B2, 1, 0, C0, 0, 0
\
\
\
AB 2, 4, 2, AC 3, 5, 4
AB 3, 1, 0, AC 1, 2, 0
i AB AC 2 3
i j AB AC 3 1 1 2
\
\
j k 4 2 6i 2j 2k 5 4
\
1 1 A AB AC 44 11 2 2 \
\
k 0 5k 0
1 5 A AB AC 2 2
\
\
38. F 2000cos 30 j sin 30 k 10003 j 1000k
\
z
\
PQ 0.16 k
i PQ F 0 0 \
j 0 10003
k 0.16 1603 i 1000
PQ F 1603 ft lb \
PQ 0.16 ft
60°
F
y
x
15 5 40. (a) B is 12 4 to the left of A, and one foot upwards:
AB 5 4 j k
\
(d) If T AB F ,
\
F 200cos j sin k
i j k (b) AB F 0 5 4 1 0 200 cos 200 sin \
250 sin 200 cos i
\
AB F 250 sin 200 cos
5 dT 2510 cos 8 sin 0 ⇒ tan d 4 ⇒ 51.34. The vectors are orthogonal. \
(e) The zero is 141.34, the angle making AB parallel to F. 400
2510 sin 8 cos 0
(c) For 30, \
12 8 23
AB F 25 10
−300
25 5 43 298.2.
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180
489
490
Chapter 10
Vectors and the Geometry of Space
1 42. u v w 2 0
1 1 0
1 0 4
46. u v w
1 0 1 1
44. u
v w
2 1 0
0 1 2
0 1 0 2
3 1 6 6 72 0 4
V u v w 72 48. u 1, 1, 0
50. See Theorem 10.8, page 746.
v 1, 0, 2 w 0, 1, 1 u
v w
1 1 0
1 0 1
0 2 3 1
V u v w 3
52. Form the vectors for two sides of the triangle, and compute their cross product: x2 x1, y2 y1, z 2 z1 x3 x1, y3 y1, z3 z1 54. False, let u 1, 0, 0, v 1, 0, 0, w 1, 0, 0. Then, u v u w 0, but v w. 56. u u1, u2, u3, v v1, v2, v3, w w1, w2, w3 u u1i u2 j u3 k v w v2w3 v3w2i v1w3 v3w1j v1w2 v2w1 k
u1 u2 u3 u v w u1v2w3 v3w2 u2v1w3 v3w1 u3v1w2 v2w1 v1 v2 v3 w1 w2 w3 58. u u1, u2, u3, v v1, v2, v3, c is a scalar.
i j c u v cu1 cu2 v1 v2
k cu3 v3
cu2v3 cu3v2i cu1v3 cu3v1j cu1v2 cu2v1k cu2v3 u3v2i u1v3 u3v1j u1v2 u2v1k cu v
u1 u2 60. u v w v1 v2 w1 w2
u3 v3 w3
w1 w2 u v w w u v u1 u2 v1 v2
w3 u3 v3
w1u 2v3 v2 u 3 w2u 1v3 v1u 3 w3u 1v2 v1u 2 u 1v2w3 w2v3 u 2v1w3 w1v3 u 3v1w2 w1v2 u v w
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Section 10.5
Lines and Planes in Space
491
62. If u and v are scalar multiples of each other, u cv for some scalar c. u v cv v c v v c 0 0 If u v 0, then u v sin 0. Assume u 0, v 0. Thus, sin 0, 0, and u and v are parallel. Therefore, u cv for some scalar c. 64. u a1, b1, c1 , v a2, b2, c2, w a3, b3, c3
i j v w a2 b2 a3 b3 u v w
k c2 b2c3 b3c2i a2c3 a3c2j a2b3 a3b2k c3
i j k a1 b1 c1 b2c3 b3c2 a3c2 a2c3 a2b3 a3b2
u v w b1a2b3 a3b2 c1a3c2 a2c3i a1a2b3 a3b2 c1b2c3 b3c2 j
a1a3c2 a2c3 b1b2c3 b3c2k a2a1a3 b1b3 c1c3 a3a1a2 b1b2 c1c2 i
b2a1a3 b1b3 c1c3 b3a1a2 b1b2 c1c2 j c2a1a3 b1b3 c1c3 c3a1a2 b1b2 c1c2k a1a3 b1b3 c1c3a2, b2, c2 a1a2 b1b2 c1c2a3, b3, c3 u wv u vw
Section 10.5
Lines and Planes in Space
2. x 2 3t, y 2, z 1 t (a)
(b) When t 0 we have P 2, 2, 1. When t 2 we have Q 4, 2, 1.
z
\
PQ 6, 0, 2 The components of the vector and the coefficients of t are proportional since the line is parallel to PQ . \
x
y
(c) z 0 when t 1. Thus, x 1 and y 2. Point: 1, 2, 0 2 x 0 when t . Point: 3
0, 2, 13
6. Point: 3, 0, 2
4. Point: 0, 0, 0
Direction vector: v 0, 6, 3
5 Direction vector: v 2, , 1 2
Direction numbers: 0, 2, 1
Direction numbers: 4, 5, 2
(a) Parametric: x 3, y 2t, z 2 t
(a) Parametric: x 4t, y 5t, z 2t
(b) Symmetric:
y z x (b) Symmetric: 4 5 2
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y z 2, x 3 2
492
Chapter 10
Vectors and the Geometry of Space 10. Points: 2, 0, 2, 1, 4, 3
8. Point: 3, 5, 4 Directions numbers: 3, 2, 1
Direction vector: 1, 4, 5
(a) Parametric: x 3 3t, y 5 2t, z 4 t
Direction numbers: 1, 4, 5
(b) Symmetric:
x3 y5 z4 3 2
(a) Parametric: x 2 t, y 4t, z 2 5t (b) Symmetric: x 2
12. Points: 0, 0, 25, 10, 10, 0
14. Point: 2, 3, 4 Direction vector: v 3i 2j k
Direction vector: 10, 10, 25
Direction numbers: 3, 2, 1
Direction numbers: 2, 2, 5
Parametric: x 2 3t, y 3 2t, z 4 t
(a) Parametric: x 2t, y 2t, z 25 5t (b) Symmetric:
y z 25 x 2 2 5
16. Points: 2, 0, 3, 4, 2, 2
18. L1: v 4, 2, 3
Direction vector: v 2i 2j k
L 2: v 2, 1, 5
Direction numbers: 2, 2, 1
L 3: v 8, 4, 6
Parametric: x 2 2t, y 2t, z 3 t Symmetric:
z2 y 4 5
1 1 2
8, 5, 9 on line
L 4: v 2, 1, 1.5
x2 y z3 2 2 1
(a) Not on line 1
8, 5, 9 on line
L1 and L 2 are identical.
(b) On line (c) Not on line
32 32 1
20. By equating like variables, we have (i) 3t 1 3s 1, (ii) 4t 1 2s 4, and (iii) 2t 4 s 1. 1 From (i) we have s t, and consequently from (ii), t 2 and from (iii), t 3. The lines do not intersect.
22. Writing the equations of the lines in parametric form we have x 2 3t
y 2 6t
z3t
x 3 2s
y 5 s
z 2 4s.
By equating like variables, we have 2 3t 3 2s, 2 6t 5 s, 3 t 2 4s. Thus, t 1, s 1 and the point of intersection is 5, 4, 2. u 3, 6, 1
(First line)
v 2, 1, 4
(Second line)
cos
u v u v
24. x 2t 1
4 4621
x 5s 12
y 4t 10
y 3s 11
zt
z 2s 4
4 966
2966 483
x = 2t − 1 y = − 4t + 10 z=t
z
x = − 5s − 12 y = 3s + 11 z = − 2s − 4
3 2
Point of intersection: 3, 2, 2 3 x
−2
−3
2 2
(3, 2, 2)
3
y
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Section 10.5
P 0, 0, 1, Q 2, 0, 0, R 3, 2, 2 \
(a) PQ 2, 0, 1, PR 3, 2, 3
i (b) PQ PR 2 3 \
\
493
28. Point: 1, 0, 3
26. 2x 3y 4z 4 \
Lines and Planes in Space
j k 0 1 2, 3, 4 2 3
n k 0, 0, 1 0x 1 0y 0 1z 3 0 z30
The components of the cross product are proportional (for this choice of P, Q, and R, they are the same) to the coefficients of the variables in the equation. The cross product is parallel to the normal vector. 32. Point: 3, 2, 2
30. Point: (0, 0, 0
Normal vector: v 4i j 3k
Normal vector: n 3i 2k 3x 0 0y 0 2z 0 0
4x 3 y 2 3z 2 0
3x 2z 0
4x y 3z 8
34. Let u be vector from 2, 3, 2 to 3, 4, 2: 1, 1, 4.
36. 1, 2, 3, Normal vector: v i, 1x 1 0, x 1
Let v be vector from 2, 3, 2 to 1, 1, 0: 1, 4, 2. Normal vector: u v
i j 1 1 1 4
k 4 18, 6, 3 2 36, 2, 1
6x 2 2 y 3 1z 2 0
6x 2y z 8 38. The plane passes through the three points 0, 0, 0, 0, 1, 0 3, 0, 1. The vector from 0, 0, 0 to 0, 1, 0: u j
The vector from 0, 0, 0 to 3, 0, 1: v 3 i k
i 0 Normal vector: u v 3 x 3 z 0
j 1 0
k 0 i 3 k 1
42. Let v be the vector from 3, 2, 1 to 3, 1, 5: v j 6k Let n be the normal to the given plane: n 6i 7j 2k Since v and n both lie in the plane P, the normal vector to P is:
i j k v n 0 1 6 40i 36j 6k 6 7 2 220i 18j 3k
40. The direction of the line is u 2i j k. Choose any point on the line, 0, 4, 0, for example, and let v be the vector from 0, 4, 0 to the given point 2, 2, 1: v 2i 2j k
i j Normal vector: u v 2 1 2 2
x 2 2z 1 0 x 2z 0
k 1 i 2k 1
44. Let u k and let v be the vector from 4, 2, 1 to 3, 5, 7: v 7i 3j 6k Since u and v both lie in the plane P, the normal vector to P is: uv
i 0 7
j 0 3
k 1 3i 7j 3i 7j 6
3x 4 7y 2 0 3x 7y 26
20x 3 18y 2 3z 1 0 20x 18y 3z 27
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494
Chapter 10
Vectors and the Geometry of Space
46. The normal vectors to the planes are n1 3, 1, 4, n2 9, 3, 12. Since n2 3n1, the planes are parallel, but not equal 48. The normal vectors to the planes are
50. The normal vectors to the planes are
n1 3i 2j k, n2 i 4j 2k, cos
n1 n2 3 8 2 n1 n2
Therefore, arccos
1421
n1 2, 0, 1, n2 4, 1, 8,
1 . 6
cos
16 65.9 .
Thus,
52. 3x 6y 2z 6
n1 n2 0. n1 n2
and the planes are orthogonal. 2
54. 2x y z 4
56. x 2y 4
z
z
z 4
4
3
2 −4 3
2
2
1
y
3
1
x
y y
x
x
58. z 8
3
4
3
60. x 3z 3 z
62. 2.1x 4.7y z 3 0 z
z
8 2
3
1 2 3
2
1 1
2
1
y
x
1
Generated by Mathematica 5 x
5
x
y
64. P1: n 60, 90, 30 or 2, 3, 1 P2: n 6, 9, 3 or 2, 3, 1 P3: n 20, 30, 10 or 2, 3, 1
2
y
Generated by Mathematica
0, 0, 109 on plane 0, 0, 23 on plane 0, 0, 56 on plane
P4: n 12, 18, 6 or 2, 3, 1 P1, P2, and P3 are parallel. 66. If c 0, z 0 is xy-plane. 1 z is a plane parallel to c x-axis and passing through the points 0, 0, 0 and 0, 1, c. If c 0, cy z 0 ⇒ y
68. The normals to the planes are n1 6, 3, 1. and n2 1, 1, 5.
The direction vector for the line is n1 n2
i j 6 3 1 1
k 1 16, 31, 3. 5
Now find a point of intersection of the planes. 6x 3y z 5 6x 3y z 5 ⇒ x y 5z 5 ⇒ 6x 6y 30z 30 3y 31z 35 Let y 9, z 2 ⇒ x 4 ⇒ 4, 9, 2. x 4 16t, y 9 31t, z 2 3t
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Section 10.5
70. Writing the equation of the line in parametric form and substituting into the equation of the plane we have:
72. Writing the equation of the line in parametric form and substituting into the equation of the plane we have: 54 2t 31 3t 17, t 0
1 2
Substituting t 12 into the parametric equations for the line we have the point of intersection 1, 1, 0. The line does not lie in the plane. 74. Point: Q0, 0, 0
Substituting t 0 into the parametric equations for the line we have the point of intersection 4, 1, 2. The line does not lie in the plane.
76. Point: Q3, 2, 1
Plane: 8x 4y z 8
Plane: x y 2z 4
Normal to plane: n 8, 4, 1
Normal to plane: n 1, 1, 2
Point in plane: P1, 0, 0
Point in plane: P4, 0, 0
\
\
Vector: PQ 1, 0, 0
Vector: PQ 1, 2, 1
PQ n 8 8 \
D
n
81
9
P 5,0, 3 is a point in 4x 4y 9z 7. Q 0, 0, 2 is a point in 4x 4y 9z 18. \
PQ 5, 0, 1
PQ n1
11 113
PQ 1, 1, 2
i PQ u 1 2 \
\
D
6
6
P 2, 0, 0 is a point in 2x 4z 4. Q 5, 0, 0 is a point in 2x 4z 10.
PQ n1 \
11113 113
P 0, 3, 2 is a point on the line let t 0.
1 6
80. The normal vectors to the planes are n1 2, 0, 4 and n2 2, 0, 4. Since n1 n2, the planes are parallel. Choose a point in each plane.
\
82. u 2, 1, 2 is the direction vector for the line. \
n
6
PQ 3, 0, 0, D
\
n1
PQ n 1 \
D
78. The normal vectors to the planes are n1 4, 4, 9 and n2 4, 4, 9. Since n1 n2, the planes are parallel. Choose a point in each plane.
D
495
x 4 2t, y 1 3t, z 2 5t
x 1 4t, y 2t, z 3 6t 21 4t 32t 5, t
Lines and Planes in Space
n1
6 20
35 5
84. The equation of the plane containing Px1, y1, z1 and having normal vector n a, b, c is ax x1 by y1 cz z1 0. You need n and P to find the equation.
j k 1 2 0, 2, 1 1 2
PQ u 5 5 9 u 3
86. x a: plane parallel to yz-plane containing a, 0, 0 y b: plane parallel to xz-plane containing 0, b, 0 z c: plane parallel to xy-plane containing 0, 0, c
88. (a) t v represents a line parallel to v. (b) u t v represents a line through the terminal point of u parallel to v. (c) su t v represent the plane containing u and v.
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496
Chapter 10
Vectors and the Geometry of Space
90. On one side we have the points 0, 0, 0, 6, 0, 0, and 1, 1, 8. n1
i j 6 0 1 1
k 0 48j 6k 8
On the adjacent side we have the points 0, 0, 0, 0, 6, 0, and 1, 1, 8. n2
i j 0 6 1 1
z
(− 1, − 1, 8)
k 0 48i 6k 8
n1 n2 cos n1 n2
6 4 2
36 1 2340 65
y
6
(6, 0, 0)
x
arccos
(0, 6, 0)
(0, 0, 0) 4
6
1 89.1 65
92. False. They may be skew lines. (See Section Project)
Section 10.6
Surfaces in Space
2. Hyperboloid of two sheets
4. Elliptic cone
Matches graph (e)
6. Hyperbolic paraboloid
Matches graph (b)
8. x 4
10. x2 z 2 25
z
Plane parallel to the yz-coordinate plane
Matches graph (a)
The y-coordinate is missing so we have a cylindrical surface with rulings parallel to the y-axis. The generating curve is a circle.
4
z 4 x
2
4
y
6 4
8 x 8
12. z 4 y 2
14. y 2 z 2 4
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola. z 8
y
16. z ey
y2 z2 1 4 4 The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a hyperbola.
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is the exponential curve. z
4
20
z 8
4
8
12
15
5
y
10
x
5 3
5 x
5
y
http://librosysolucionarios.net
x
1
2
3
4
y
Section 10.6
Surfaces in Space
497
18. y2 z2 4 (b) From 0, 10, 0:
(a) From 10, 0, 0:
(c) From 10, 10, 10:
z
z
z 3
y y 3
3
y
x
20.
x2 y2 z2 1 16 25 25
5 4 3 2 1
Ellipsoid x2 y2 1 ellipse 16 25
xy-trace:
x2 z2 1 ellipse 16 25
xz-trace:
y2 1 4 Hyperboloid of two sheets xy-trace: none
22. z 2 x 2
z
5
4
3
2
1
xz-trace: z 2 x2 1 hyperbola
1 2 3 4 5
y
yz-trace: z2
x
yz-trace: y2 z2 25 circle
24. z x2 4y2
z 5
5
5
x
y2 1 hyperbola 4
x2 y2 1 ellipse 9 36
z ± 10 :
26. 3z y 2 x 2
28. x2 2y 2 2z 2
Elliptic paraboloid
Hyperbolic paraboloid
Elliptic Cone
xy-trace: point 0, 0, 0
xy-trace: y ± x
xy-trace: x ± 2 y
xz-trace: z x2 parabola
1 xz-trace: z 3 x2
xz-trace: x ± 2 z
yz-trace: z 4y2 parabola
yz-trace: z
yz-trace: point: 0, 0, 0
13 y2
z
z
z
5
28 24 20
4
5 x 10 x
3
2
1
1
2
10
5
y
y
y
x
9x2 y2 9z2 54x 4y 54z 4 0
30.
z
4
9x2 6x 9 y2 4y 4 9z2 6z 9 81 4 81 9x 32 y 22 9z 32 4 2
x 32 y 22 z 32 1 49 4 49
1
x
2 3 4 5
y
Hyperboloid of one sheet with center 3, 2, 3. 32. z x2 0.5y 2
34. z 2 4y x 2 z ± 4y
z
36. x2 y 2 ez lnx2 y2 z
x2
z
z
4
−4
8 4
−8
4 −3
−3
−4
−2 4 1 2 x
8 1 2
y
x
4
−4 −8
x 3
y
http://librosysolucionarios.net
4 y
y
498
Chapter 10
38. z
Vectors and the Geometry of Space
x 8 x2 y2
z±
z
98 x
2
y 4 x2
1 2 y 9 2
x 0, y 0, z 0 z
2
20
5
10
2 4
4
2
z
4
x
42. z 4 x2
40. 9x 2 4y 2 8z 2 72
y 10 10 x
20
20 y x
44. z 4 x2 y2
4
y
46. x2 z 2 ry 2 and z ry 3y; therefore,
z
y 2z
4 3
3
x2 z 2 9y 2.
z0 −3 3
y
3 x
1 48. y2 z2 rx 2 and z rx 4 x2 ; therefore, 2 1 y 2 z 2 4 x2, x 2 4y 2 4z 2 4. 4 52. x 2 z 2 cos2 y
x2 y 2 e2z.
60. z
y sin y dy
0
2 sin y y cos y
0
56. About x-axis: y2 z2 r x 2
54. The trace of a surface is the intersection of the surface with a plane. You find a trace by setting one variable equal to a constant, such as x 0 or z 2.
Equation of generating curve: x cos y or z cos y
58. V 2
50. x2 y 2 rz 2 and y rz ez; therefore,
2 2
About y-axis: x2 z2 r y 2 About z-axis: x2 y2 r z 2
x2 y 2 2 4
(a) When y 4 we have z
z
Focus:
1.0
0, 4, 92
(b) When x 2 we have z2
0.5
π 2
π
y2 , 4z 2 y 2. 4
Focus: 2, 0, 3
y
62. If x, y, z is on the surface, then z2 x2 y2 z 42 z2 x2 y2 z2 8z 16 8z x2 y2 16 ⇒ z
x2 y2 2 8 8
Elliptic paraboloid shifted up 2 units. Traces parallel to xy-plane are circles.
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x2 1 4, 4 z 4 x2. 2 2
Section 10.7
Cylindrical and Spherical Coordinates
499
64. z 0.775x2 0.007y2 22.15x 0.54y 45.4 (a)
(b)
z
Year
1980
1985
1990
1995
1996
1997
z
37.5
72.2
111.5
185.2
200.1
214.6
200
Model
37.8
72.0
112.2
185.8
204.5
214.7
150
250
100
10 20
100 200
y
x
(c) For y constant, the traces parallel to the xz-plane are concave downward. That is, for fixed y (public assistance), the rate of increase of z (Medicare) is decreasing with respect to x (worker’s compensation).
(d) The traces parallel to the yz-plane (x constant) are concave upward. That is, for fixed x (worker’s compensation), the rate of increase of z (Medicare) is increasing with respect to y (public assistance).
66. Equating twice the first equation with the second equation, 2x2 6y2 4z2 4y 8 2x2 6y2 4z2 3x 2 4y 8 3x 2 3x 4y 6, a plane
Section 10.7 2.
Cylindrical and Spherical Coordinates
4, 2 , 2, cylindrical
4.
6, 4 , 2, cylindrical
x 4 cos
0 2
x 6 cos
y 4 sin
4 2
y 6 sin
4 32 32 4
z 2
0, 4, 2, rectangular
8. 22, 22, 4, rectangular r 22 2 22 2 4
arctan1
4
z4
4, 4 , 4, cylindrical 14. z x2 y2 2 rectangular equation z r2 2
6.
cylindrical equation
32, 32, 2 10. 23, 2, 6, rectangular
y sin
3 1 2
r 32 22 13
1 5 3 6
arctan
23 arctan 23
z 1
z1
3 0 2
12. 3, 2, 1, rectangular
r 12 4 4
x cos
z1 0, 1, 1, rectangular
z2
arctan
1, 32, 1, cylindrical
13, arctan 23, 1, cylindrical
4, , 1 , cylindrical 6
16. x2 y2 8x
rectangular equation
r 2 8r cos r 8 cos cylindrical equation
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500
Chapter 10
Vectors and the Geometry of Space 20. r
18. z 2
z 2
22. r 2 cos
Same x2 y2
z 3
x2 y 2
r 2 2r cos
z 2
x2 y2 2x x2 y 2 2x 0
z2 0 4
x 12 y 2 1
z
z
4 2
1
2
3
3
y
x
−2
−2 2
2
−2 y 2
x
3 x
26. 1, 1, 1, rectangular
24. z r 2 cos2 z
12 12 12 3
x2
4
arctan 1
z
9
arccos
3,
1 3 2 x
1 2 3 4 5 6
30. 4, 0, 0, rectangular
42 02 02 4
42 210 arctan 1 4 2 arccos 5 2 210, , arccos , spherical 4 5
2
22
32.
arccos 0
12, 34, 9 , spherical
2
4, , 2 , spherical
34.
9, 4 , , spherical
x 12 sin
3 cos 2.902 9 4
x 9 sin cos
0 4
y 12 sin
3 sin 2.902 9 4
y 9 sin sin
0 4
z 12 cos
11.276 9
z 9 cos 9 0, 0, 9, rectangular
2.902, 2.902, 11.276, rectangular
36.
1 , arccos , spherical 4 3
y
28. 2, 2, 42 , rectangular 22
1 3
6, , 2 , spherical x 6 sin cos 6 2 y 6 sin sin 0 2 z 6 cos 0 2 6, 0, 0, rectangular
38. (a) Programs will vary. (b) , , 5, 1, 0.5
x, y, z 1.295, 2.017, 4.388
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y
Section 10.7 40. x2 y2 3z2 0
42. x 10
rectangular equation
10 csc sec spherical equation
4 cos 2
14
44.
cos
1 2
3
rectangular equation
sin cos 10
x2 y2 z2 4z2 2
Cylindrical and Spherical Coordinates
2
cos2
(cone) spherical equation
3 4
46.
2
48. 2 sec
cos 2
z x2 y2 z2 z 0 x2 y2 z2 z0 cos
y tan x y 1 x xy0
z2 z 3
xy-plane
z
z
1
2
3
3
3 3
y
x
2 −3
−3
−3 y
3 x
3 −3
x
3
y
−2 −3
50. 4 csc sec
52.
4 sin cos
3, 4 , 0, cylindrical
54.
22 22 22
32 02 3
sin cos 4
x4 z
4
arccos
6
4
12 34
2 3 22, , , spherical 3 4
y
6
6
2 3
arccos
0 9 2
3, 4 , 2 , spherical
4
2, 23, 2, cylindrical
x
56.
4, 3 , 4, cylindrical
58.
4, 2 , 3, cylindrical
42 42 42
42 32 5
3
2
arccos
1 2
4
42, , , spherical 3 4
arccos
60.
4, 18 , 2 , spherical r 4 sin
3 5
3 5, , arccos , spherical 2 5
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4 2
18
z 4 cos
0 2
4, 18 , 0, cylindrical
501
502
Chapter 10
Vectors and the Geometry of Space
62.
18, 3 , 3 , spherical r sin 18 sin
64.
9 3
r 7 sin
5 6
z 5 cos 5
93 3
Spherical
68. 6, 2, 3
6.325, 0.322, 3
7.000, 0.322, 2.014
70. 7.317, 6.816, 6
10, 0.75, 6
11.662, 0.750, 1.030
72. 6.115, 1.561, 4.052
6.311, 0.25, 4.052
7.5, 0.25, 1
74. 32, 32, 3
6, 0.785, 3
6.708, 0.785, 2.034
76. 0, 5, 4
5, 1.571, 4
6.403, 1.571, 0.896
78. 1.732, 1, 3
2, 116, 3
3.606, 2.618, 0.588
Note: Use the cylindrical coordinate 2, 80. 2.207, 7.949, 4
82.
5 ,3 6
8.25, 1.3, 4
4
3 72 4 2
7 2 2, 4 , 7 2 2, cylindrical
Cylindrical
Rectangular
3 72 4 2
4
z 7 cos
0, 56, 5, cylindrical
9, , 93 , cylindrical 3
7, 4 , 34, spherical
66.
r 5 sin 0
3
z cos 18 cos
5, 56, , spherical
84.
9.169, 1.3, 2.022
4
Plane
Cone
Matches graph (e)
Matches graph (a)
86. 4 sec , z cos 4 Plane Matches graph (b) 90. a Sphere
88. r a Cylinder with z-axis symmetry
b Plane perpendicular to xy-plane
b Vertical half-plane
z c Plane parallel to xy-plane
c Half-cone
92. 4x2 y 2 z 2
94. x2 y 2 z
(a) 4r 2 z 2, 2r z
(a) r 2 z
(b) 42 sin2 cos2 2 sin2 sin2 2 cos2 ,
(b) 2 sin2 cos , sin2 cos ,
1 4 sin2 cos 2 , tan2 , 4
cos , csc cot sin2
1 1 tan , arctan 2 2 98. y 4
96. x2 y 2 16
(a) r sin 4, r 4 csc
(a) r 2 16, r 4 (b)
2
16, 16 0, sin 4 sin 4 0, 4 csc sin2
2
sin2
(b) sin sin 4, 4 csc csc
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Review Exercises for Chapter 10
100.
≤ ≤ 2 2
104. 0 ≤ ≤ 2
102. 0 ≤ ≤ 2 2 ≤ r ≤ 4
0 ≤ r ≤ 3
≤ ≤ 4 2
z 2 ≤ r 2 6r 8
0 ≤ z ≤ r cos
503
0 ≤ ≤ 1
z
z z
4 3
4 3
2
−5
−4
y
5 4
x
−2
−2
5
x
y
4
y
2
2 x
106. Cylindrical: 0.75 ≤ r ≤ 1.25, z 8 108. Cylindrical 1 2
110. 2 sec ⇒ cos 2 ⇒ z 2 plane
z
≤ r ≤ 3
4 sphere
4
0 ≤ ≤ 2
The intersection of the plane and the sphere is a circle.
9 r2 ≤ z ≤ 9 r2
−4 y
4 x −4
Review Exercises for Chapter 10 2. P 2, 1, Q 5, 1 R 2, 4 \
4. v v cos i v sin j
\
1 1 cos 225 i sin 225 j 2 2
(a) u PQ 7, 0 7i, v PR 4, 5 4i 5j
(b) v 42 52 41
2
4
i
2
4
j
(c) 2u v 14i 4i 5j 18i 5j 6. (a) The length of cable POQ is L.
y
\
OQ 9i yj
O
L 292 y2 ⇒
9
θ
\
Tension: T c OQ c81 y 2
P
250 250 81 y 2 ⇒ T y L24 81 Domain: L > 18 inches cy 250 ⇒ T
(c)
L
2
19
20
21
22
23
24
25
T
780.9
573.54
485.36
434.81
401.60
377.96
360.24
(d) The line T 400 intersects the curve at
1000
18 in.
250L L2 324
L
L 23.06 inches. 18
Q
500 lb
Also,
(b)
x
−9
L2 81 y 4
25 0
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(e) lim T 250 L→
The maximum tension is 250 pounds in each side of the cable since the total weight is 500 pounds.
504
Chapter 10
Vectors and the Geometry of Space
8. x z 0, y 7: 0, 7, 0
12. Center:
10. Looking towards the xy-plane from the positive z-axis. The point is either in the second quadrant x < 0, y > 0 or in the fourth quadrant x > 0, y < 0. The z-coordinate can be any number.
0 2 4, 0 2 6, 4 2 0 2, 3, 2
Radius: 2 02 3 02 2 42 4 9 4 17
x 22 y 32 z 22 17 z
14. x2 10x 25 y 2 6y 9 z 2 4z 4 34 25 9 4
x 52 y 32 z 22 4 Center: 5, 3, 2 Radius: 2
6 4 2
2
y
4 6 8 x
16. v 3 6, 3 2, 8 0 3, 5, 8 (3, −3, 8)
8 7 6 5 4 3 2 1
v
1 2
6
5
Since v and w are not parallel, the points do not lie in a straight line.
1
3
y
4
(6, 2, 0)
x
20. 8
18. v 8 5, 5 4, 5 7 3, 1, 2 w 11 5, 6 4, 3 7 6, 10, 4
z
6, 3, 2 8 48 24 16 6, 3, 2 , , 49 7 7 7 7
22. P 2, 1, 3, Q 0, 5, 1, R 5, 5, 0 \
(a) u PQ 2, 6, 2 2i 6j 2k, \
v PR 3, 6, 3 3i 6j 3k (b) u v 23 66 23 36 (c) v 24. u 4, 3, 6, v 16, 12, 24 Since v 4u, the vectors are parallel.
v 9 36 9 54
26. u 4, 1, 5, v 3, 2, 2 u
v0
⇒ is orthogonal to v.
2
28. u 1, 0, 3
30. W F PQ F PQ cos 758cos 30
v 2, 2, 1
3003 ft lb
\
u
v 1
u 10 v 3 cos
u v u v
1 310
83.9
http://librosysolucionarios.net
\
Review Exercises for Chapter 10
<
>
<
>
<
>
In Exercises 32–40, u 3, 2, 1 , v 2, 4, 3 , w 1, 2, 2 .
32. cos
u v u v
arccos
11
34. Work u
1429
w 3 4 2 5
141129 56.9
i 36. u v 3 2
j 2 4
k 1 10i 11j 8k 3
i v u 2 3
j 4 2
k 3 10i 11j 8k 1
Thus, u v v u.
i 38. u v w 3, 2, 1 1, 2, 1 3 1
i u v 3 2 u w
j 2 4
i 3 1
k 1 10i 11j 8k 3
j 2 2
j 2 2
k 1 4i 4j 4k 1
k 1 6i 7j 4k 2
u v u w 4i 4j 4k u v w 5 1 1 40. Area triangle v w 22 12 (See Exercise 35) 2 2 2
2 1 42. V u v w 0 2 0 1
46. u v
i 2 3
j 5 1
0 1 25 10 2
k 1 21i 11j 13k 4
(a) x 21t, y 1 11t, z 4 13t y1 z4 x 21 11 13
(a) x 1 t, y 2 t, z 3 t (b) x 1 y 2 z 3
Direction numbers: 21, 11, 13
(b)
44. Direction numbers: 1, 1, 1
48. P 3, 4, 2, Q 3, 4, 1, R 1, 1, 2 \
\
PQ 0, 8, 1, PR 4, 5, 4
i n PQ PR 0 4 \
\
j 8 5
k 1 27i 4j 32k 4
27x 3 4 y 4 32z 2 0 27x 4y 32z 33
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505
506
Chapter 10
Vectors and the Geometry of Space 52. Q5, 1, 3 point
50. The normal vectors to the planes are the same, n 5, 3, 1.
u 1, 2, 1 direction vector
Choose a point in the first plane, P 0, 0, 2. Choose a point in the second plane, Q 0, 0, 3.
P 1, 3, 5 point on line \
PQ 6, 2, 2
\
PQ 0, 0, 5
\
i j k PQ u 6 2 2 2, 8, 14 1 2 1 \
PQ n 5 35 5 D n 7 35 35
\
D
54. y z 2
PQ u 264 211 6 u 58. 16x 2 16y 2 9z 2 0
56. y cos z Since the x-coordinate is missing, we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is y cos z.
Since the x-coordinate is missing, we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola in the yz-coordinate plane.
Cone xy-trace: point 0,0, 0 xz-trace: z ±
z
z
z±
yz-trace:
4
2
4y 3
z 4, x 2 y 2 9
1 2
3
z 4
y −2
x
4
x
2
2
y −3
−3 x
60.
4x 3
x2 y2 z2 1 25 4 100
3
2 3
y
62. Let y r x 2x and revolve the curve about the x-axis.
z 12
Hyperboloid of one sheet xy-trace:
64.
−5
y2 x2 1 25 4
xz-trace:
z2 x2 1 25 100
yz-trace:
z2 y2 1 4 100
y
5
x
43, 34, 3 2 3 , rectangular
(a) r
43 34
(b)
43 34 3 2 3
2
2
2
2
3
2
, arctan3
2
30
2
,
33 , z , 3 2
23, 2 , 3 2 3 , cylindrical
3 , arccos , 3 10
30
2
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3 , , arccos , spherical 3 10
Review Exercises for Chapter 10
100.
≤ ≤ 2 2
104. 0 ≤ ≤ 2
102. 0 ≤ ≤ 2 2 ≤ r ≤ 4
0 ≤ r ≤ 3
≤ ≤ 4 2
z 2 ≤ r 2 6r 8
0 ≤ z ≤ r cos
503
0 ≤ ≤ 1
z
z z
4 3
4 3
2
−5
−4
y
5 4
x
−2
−2
5
x
y
4
y
2
2 x
106. Cylindrical: 0.75 ≤ r ≤ 1.25, z 8 108. Cylindrical 1 2
110. 2 sec ⇒ cos 2 ⇒ z 2 plane
z
≤ r ≤ 3
4 sphere
4
0 ≤ ≤ 2
The intersection of the plane and the sphere is a circle.
9 r2 ≤ z ≤ 9 r2
−4 y
4 x −4
Review Exercises for Chapter 10 2. P 2, 1, Q 5, 1 R 2, 4 \
4. v v cos i v sin j
\
1 1 cos 225 i sin 225 j 2 2
(a) u PQ 7, 0 7i, v PR 4, 5 4i 5j
(b) v 42 52 41
2
4
i
2
4
j
(c) 2u v 14i 4i 5j 18i 5j 6. (a) The length of cable POQ is L.
y
\
OQ 9i yj
O
L 292 y2 ⇒
9
θ
\
Tension: T c OQ c81 y 2
P
250 250 81 y 2 ⇒ T y L24 81 Domain: L > 18 inches cy 250 ⇒ T
(c)
L
2
19
20
21
22
23
24
25
T
780.9
573.54
485.36
434.81
401.60
377.96
360.24
(d) The line T 400 intersects the curve at
1000
18 in.
250L L2 324
L
L 23.06 inches. 18
Q
500 lb
Also,
(b)
x
−9
L2 81 y 4
25 0
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(e) lim T 250 L→
The maximum tension is 250 pounds in each side of the cable since the total weight is 500 pounds.
504
Chapter 10
Vectors and the Geometry of Space
8. x z 0, y 7: 0, 7, 0
12. Center:
10. Looking towards the xy-plane from the positive z-axis. The point is either in the second quadrant x < 0, y > 0 or in the fourth quadrant x > 0, y < 0. The z-coordinate can be any number.
0 2 4, 0 2 6, 4 2 0 2, 3, 2
Radius: 2 02 3 02 2 42 4 9 4 17
x 22 y 32 z 22 17 z
14. x2 10x 25 y 2 6y 9 z 2 4z 4 34 25 9 4
x 52 y 32 z 22 4 Center: 5, 3, 2 Radius: 2
6 4 2
2
y
4 6 8 x
16. v 3 6, 3 2, 8 0 3, 5, 8 (3, −3, 8)
8 7 6 5 4 3 2 1
v
1 2
6
5
Since v and w are not parallel, the points do not lie in a straight line.
1
3
y
4
(6, 2, 0)
x
20. 8
18. v 8 5, 5 4, 5 7 3, 1, 2 w 11 5, 6 4, 3 7 6, 10, 4
z
6, 3, 2 8 48 24 16 6, 3, 2 , , 49 7 7 7 7
22. P 2, 1, 3, Q 0, 5, 1, R 5, 5, 0 \
(a) u PQ 2, 6, 2 2i 6j 2k, \
v PR 3, 6, 3 3i 6j 3k (b) u v 23 66 23 36 (c) v 24. u 4, 3, 6, v 16, 12, 24 Since v 4u, the vectors are parallel.
v 9 36 9 54
26. u 4, 1, 5, v 3, 2, 2 u
v0
⇒ is orthogonal to v.
2
28. u 1, 0, 3
30. W F PQ F PQ cos 758cos 30
v 2, 2, 1
3003 ft lb
\
u
v 1
u 10 v 3 cos
u v u v
1 310
83.9
http://librosysolucionarios.net
\
Review Exercises for Chapter 10
<
>
<
>
<
>
In Exercises 32–40, u 3, 2, 1 , v 2, 4, 3 , w 1, 2, 2 .
32. cos
u v u v
arccos
11
34. Work u
1429
w 3 4 2 5
141129 56.9
i 36. u v 3 2
j 2 4
k 1 10i 11j 8k 3
i v u 2 3
j 4 2
k 3 10i 11j 8k 1
Thus, u v v u.
i 38. u v w 3, 2, 1 1, 2, 1 3 1
i u v 3 2 u w
j 2 4
i 3 1
k 1 10i 11j 8k 3
j 2 2
j 2 2
k 1 4i 4j 4k 1
k 1 6i 7j 4k 2
u v u w 4i 4j 4k u v w 5 1 1 40. Area triangle v w 22 12 (See Exercise 35) 2 2 2
2 1 42. V u v w 0 2 0 1
46. u v
i 2 3
j 5 1
0 1 25 10 2
k 1 21i 11j 13k 4
(a) x 21t, y 1 11t, z 4 13t y1 z4 x 21 11 13
(a) x 1 t, y 2 t, z 3 t (b) x 1 y 2 z 3
Direction numbers: 21, 11, 13
(b)
44. Direction numbers: 1, 1, 1
48. P 3, 4, 2, Q 3, 4, 1, R 1, 1, 2 \
\
PQ 0, 8, 1, PR 4, 5, 4
i n PQ PR 0 4 \
\
j 8 5
k 1 27i 4j 32k 4
27x 3 4 y 4 32z 2 0 27x 4y 32z 33
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505
506
Chapter 10
Vectors and the Geometry of Space 52. Q5, 1, 3 point
50. The normal vectors to the planes are the same, n 5, 3, 1.
u 1, 2, 1 direction vector
Choose a point in the first plane, P 0, 0, 2. Choose a point in the second plane, Q 0, 0, 3.
P 1, 3, 5 point on line \
PQ 6, 2, 2
\
PQ 0, 0, 5
\
i j k PQ u 6 2 2 2, 8, 14 1 2 1 \
PQ n 5 35 5 D n 7 35 35
\
D
54. y z 2
PQ u 264 211 6 u 58. 16x 2 16y 2 9z 2 0
56. y cos z Since the x-coordinate is missing, we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is y cos z.
Since the x-coordinate is missing, we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola in the yz-coordinate plane.
Cone xy-trace: point 0,0, 0 xz-trace: z ±
z
z
z±
yz-trace:
4
2
4y 3
z 4, x 2 y 2 9
1 2
3
z 4
y −2
x
4
x
2
2
y −3
−3 x
60.
4x 3
x2 y2 z2 1 25 4 100
3
2 3
y
62. Let y r x 2x and revolve the curve about the x-axis.
z 12
Hyperboloid of one sheet xy-trace:
64.
−5
y2 x2 1 25 4
xz-trace:
z2 x2 1 25 100
yz-trace:
z2 y2 1 4 100
y
5
x
43, 34, 3 2 3 , rectangular
(a) r
43 34
(b)
43 34 3 2 3
2
2
2
2
3
2
, arctan3
2
30
2
,
33 , z , 3 2
23, 2 , 3 2 3 , cylindrical
3 , arccos , 3 10
30
2
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3 , , arccos , spherical 3 10
Problem Solving for Chapter 10
66.
81, 56, 273, cylindrical
68.
6561 2187 543
12, 2 , 23, spherical
r 2 12 sin
5 6
2754 33 arccos 21 3
arccos
23
2
⇒ r 63
2
z cos 12 cos
543, 56, 3 , spherical
23 6
63, 2 , 6, cylindrical
70. x 2 y 2 z 2 16 (a) Cylindrical: r 2 z 2 16 (b) Spherical: 4
Problem Solving for Chapter 10
x
2. f x
t 4 1 dt
0 y
(a)
(b) f x x 4 1 f 0 1 tan
4 2 −4
x
−2
2
4
4
−2
u
−4
(c) ±
22, 22
1 2
i j
(d) The line is y x: x t, y t.
4. Label the figure as indicated.
S
R
\
PR a b a
\
SQ b a
a b b a b 2 a 2 0, because
a b in a rhombus. \
22, 22
P
Q
b
\
6. n PP0 n PP0
n + PP0
Figure is a square. → Thus, PP0 n and the points P form a circle of radius
n in the plane with center at P.
n − PP0
n
n
P0
P
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507
508
Chapter 10
Vectors and the Geometry of Space
r
r 2 x2 dx 2 r 2x
8. (a) V 2
0
x3 3
r 0
4 r 3 3
(b) At height z d > 0, x2 y2 d2 2 2 1 2 a b c x2 y2 d 2 c2 d 2 21 2 2 a b c c2 x2 y2 2 2 1. 2 d b c d 2 2 2 c c
a2
c2
Area
a c c d b c c d cab c 2
2
2
2
2
2
2
2
2
2
d 2
ab 2 2 2 c d dd 0 c c
V2
2ab 2 d3 cd 2 c 3
c 0
4 abc 3 10. (a) r 2 cos
(b) z r 2 cos 2 z2 x2 y2
Cylinder
Hyperbolic paraboloid
1 12. x t 3, y t 1, z 2t 1; Q 4, 3, s 2 (a) u 2, 1, 4 direction vector for line P 3, 1, 1 point on line \
PQ 1, 2, s 1 \
PQ u \
D (b)
i 1 2
j k 2 s 1 7 si 6 2sj 5k 1 4
PQ u 7 s2 6 2s2 25 21
u
(c) Yes, there are slant asymptotes. Using s x, we have
10
Ds −11
10
−4
The minimum is D 2.2361 at s 1.
y±
1 21 5 21
105
21
5x2 10x 110
5 21
x 12 21 → ±
215 x 1
s 1 slant asymptotes.
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x2 2x 22
Problem Solving for Chapter 10 14. (a) The tension T is the same in each tow line. 6000i Tcos 20 cos20i Tsin 20 sin20 j 2T cos 20 i ⇒ T
6000
3192.5 lbs 2 cos 20
(b) As in part (a), 6000i 2T cos ⇒ T
3000 cos
Domain: 0 < < 90 (c)
(d)
10
20
30
40
50
60
T
3046.3
3192.5
3464.1
3916.2
4667.2
6000.0
10,000
0
90 0
(e) As increases, there is less force applied in the direction of motion. 16. (a) Los Angeles: 4000, 118.24 , 55.95 Rio de Janeiro: 4000, 43.22 , 112.90 (b) Los Angeles: x 4000 sin 55.95 cos118.24 y 4000 sin 55.95 sin118.24 z 4000 cos 55.95
1568.2, 2919.7, 2239.7 Rio de Janeiro: x 4000 sin 112.90 cos43.22 y 4000 sin 112.90 sin43.22 z 4000 cos 112.90
2685.2, 2523.3, 1556.5 (c) cos
u v 1568.22685.2 2919.72523.3 2239.71556.5
u v
40004000
91.18 1.59 radians (d) s 40001.59 6366 miles —CONTINUED—
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509
510
Chapter 10
Vectors and the Geometry of Space
16. —CONTINUED— (e) For Boston and Honolulu: a. Boston: 4000, 71.06 , 47.64 Honolulu: 4000, 157.86 , 68.69 b. Boston: x 4000 sin 47.64 cos71.06 y 4000 sin 47.64 sin71.06 z 4000 cos 47.64
959.4, 2795.7, 2695.1 Honolulu: x 4000 sin 68.69 cos157.86 y 4000 sin 68.69 sin157.86 z 4000 cos 68.69
3451.7, 1404.4, 1453.7 (f) cos
u v 959.43451.7 2795.71404.4 2695.11453.7
u v
40004000
73.5 1.28 radians (g) s 40001.28 5120 miles 18. Assume one of a, b, c, is not zero, say a. Choose a point in the first plane such as d1a, 0, 0. The distance between this point and the second plane is D
20. Essay.
ad1a b0 c0 d2 a2 b2 c2
d1 d2
a2 b2 c2
d1 d2
a2 b2 c2
.
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C H A P T E R 1 0 Vectors and the Geometry of Space Section 10.1 Vectors in the Plane
. . . . . . . . . . . . . . . . . . . . 227
Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 232 Section 10.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 238 Section 10.4 The Cross Product of Two Vectors in Space . . . . . . . . 241 Section 10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 244 Section 10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 249 Section 10.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 252 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
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C H A P T E R 1 0 Vectors and the Geometry of Space Section 10.1
Vectors in the Plane
Solutions to Odd-Numbered Exercises 3. (a) v 4 3, 2 2 7, 0
1. (a) v 5 1, 3 1 4, 2 (b)
(b)
y
y
5
4
4
2
(− 7, 0)
3
(4, 2)
−8
2
v
−6
−4
−2
v
1
−4
x
1
x
−2
2
3
4
5
5. u 5 3, 6 2 2, 4
7. u 6 0, 2 3 6, 5
v 1 1, 8 4 2, 4
v 9 3, 5 10 6, 5
uv
uv
9. (b) v 5 1, 5 2 4, 3
11. (b) v 6 10, 1 2 4, 3
y
(a) and (c).
y
(a) and (c). (5, 5)
6 4
4
(10, 2)
2
(4, 3)
x
−4
(1, 2)
2
v
2
10
(6, −1)
(− 4, − 3)
v x
2
4
13. (b) v 6 6, 6 2 0, 4 (a) and (c).
15. (b) v
12 32 , 3 43 1, 53
(a) and (c).
y
y
( 12 , 3(
(6, 6)
3
6
4
2
(−1, 53 (
2
(0, 4)
( 32 , 43(
v
v (6, 2)
−2
x
−1
1
2
x 4
2
17. (a) 2v 4, 6
6
(b) 3v 6, 9 y
y
4 6
(4, 6)
v −8
2v
4
(2, 3)
−4
x
4
− 3v −4
(2, 3) 2
v
−8
x 2
4
6
(− 6, − 9)
—CONTINUED— 227
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228
Chapter 10
Vectors and the Geometry of Space
17. —CONTINUED— (c)
7, 21 2
7 2v
2 3
(d)
v
y
43 , 2
y
(7, 212 (
12
(2, 3)
3 8
v 2
7 v 2
( 43 , 2(
4
1
(2, 3) v
x 4
19.
2 v 3
8
x
12
1
21.
y
2
3
y
u −u u−v −v
x
x
23. (a)
2 3u
23 4, 9
83, 6
3 3 25. v 2 2i j 3i 2 j
3, 32
(b) v u 2, 5 4, 9 2, 14 (c) 2u 5v 24, 9 52, 5 18, 7
y
3
1
v = 2u x
2
3
u −1 3 u 2
−2 −3
27. v 2i j 2i 2j
29. u1 4 1
y
v = u + 2w
4i 3j 4, 3
u2 2 3
4
2w
u1 3
v
2
u2 5 Q 3, 5
x
u
4
6
−2
31. v 16 9 5
33. v 36 25 61
39. u
37. u 32 122 153
u 3 12 3, 12 v , 153 153 153 u
1717, 41717
v
35. v 0 16 4
32 52 2
2
34
2
u 5 3 2, 5 2 3 , u 34 2 34 34
unit vector
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33434, 53434 unit vector
Section 10.1
41. u 1, 1, v 1, 2
21 , v 2, 3
43. u 1,
(a) u 1 1 2 (a) u
(b) v 1 4 5 u v 0, 1
(c)
(c)
1 u 1, 1 u 2
(d)
1 v 1, 2 v 5
1 (e)
uv 0, 1 u v
uv u v
2
27
9 494
85
2
2 1 u 1, u 5 2
uu 1
v v
(f)
5
u v 3, u v
uu 1 (e)
1 41
(b) v 4 9 13
u v 0 1 1 (d)
Vectors in the Plane
1 v 2, 3 v 13
vv 1
1 (f)
7 2 uv 3, u v 85 2
uu vv 1 u 2, 1
45.
1 u 1, 1 u 2
47.
u 5 2.236 4
v 5, 4
uu 22 1, 1 v 22, 22
v 41 6.403 u v 7, 5 u v 74 8.602 u v ≤ u v u 1 3, 3 u 23
49. 2
51. v 3cos 0 i sin 0 j 3i 3, 0
uu 13 3, 3 v 1, 3
53. v 2cos 150i sin 150j 3i j 3, 1
55.
ui v uv
32 32 i j 2 2
2 23 2 i 3 2 2 j
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229
230 57.
Chapter 10
Vectors and the Geometry of Space
u 2cos 4i 2sin 4j
59. A scalar is a real number. A vector is represented by a directed line segment. A vector has both length and direction.
v cos 2i sin 2j u v 2 cos 4 cos 2 i 2 sin 4 sin 2j 61. To normalize v, you find a unit vector u in the direction of v: u
v . v
For Exercises 63–67, au bw ai 2j bi j a bi 2a bj. 63. v 2i j. Therefore, a b 2, 2a b 1. Solving simultaneously, we have a 1, b 1.
65. v 3i. Therefore, a b 3, 2a b 0. Solving simultaneously, we have a 1, b 2.
67. v i j. Therefore, a b 1, 2a b 1. Solving simultaneously, we have a 23 , b 13 . 71. f x 25 x2
69. y x 3, y 3x 2 3 at x 1. (a) m 3. Let w 1, 3, then
fx
w 1 1, 3. ± w 10
x 25 x2
3 at x 3. 4
3 (a) m 4 . Let w 4, 3, then
1 (b) m 3 . Let w 3, 1, then
1 w ± 4, 3. w 5
1 w ± 3, 1. w 10
4 (b) m 3. Let w 3, 4, then
1 w ± 3, 4 w 5
73.
u
2
2
i
2
2
j
75. Programs will vary.
u v 2 j v u v u
2
2
i
2
2
j
77. F1 2, F1 33 F2 3, F2 125 F3 2.5, F3 110 R F1 F2 F3 1.33
R F1 F2 F3 132.5 79. (a) 180cos 30i sin 30j 275i 430.88i 90j Direction: arctan
90 430.88 0.206 11.8
(b) M 275 180 cos 2 180 sin 2 sin 275180 180 cos
arctan
Magnitude: 430.882 902 440.18 newtons —CONTINUED—
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Section 10.1
Vectors in the Plane
231
79. —CONTINUED— (c)
(d)
0
30
60
90
120
150
180
M
455
440.2
396.9
328.7
241.9
149.3
95
0
11.8
23.1
33.2
40.1
37.1
0
500
(e) M decreases because the forces change from acting in the same direction to acting in the opposite direction as increases from 0 to 180.
50
M
α
0
180
0
180 0
0
81. F1 F2 F3 75 cos 30 i 75 sin 30j 100 cos 45i 100 sin 45 j 125 cos 120 i 125 sin 120 j
75 125 3 j i 502 7523 502 125 2 2 2
R F1 F2 F3 228.5 lb
R F1 F2 F3 71.3 83. (a) The forces act along the same direction. 0.
(b) The forces cancel out each other. 180.
(c) No, the magnitude of the resultant can not be greater than the sum. 85. 4, 1, 6, 5, 10, 3 y
y
6 4
4
(1, 2)
4
6
(6, 5) (1, 2)
x 8
−4
−4 −2 −2
4
6
(8, 4)
(1, 2)
(10, 3)
2
(3, 1) 2
6
(8, 4)
4
2
(3, 1) 2
8
6
(8, 4)
2
(− 4, −1)
y
8
8
x
−2 −2
8
−4
→ 87. u CB ucos 30 i sin 30 j → v CA vcos 130 i sin 130 j
−4
y
A
Vertical components: u sin 30 v sin 130 2000 Horizontal components: u cos 30 v cos 130 0
50°
130° 30° B
v
u C
x
30°
Solving this system, you obtain u 1305.5 and v 1758.8. 89. Horizontal component v cos 1200 cos 6 1193.43 ft sec Vertical component v sin 1200 sin 6 125.43 ft sec
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x
(3, 1)
4
6
8
10
232
Chapter 10
Vectors and the Geometry of Space
u 900 cos 148 i sin 148 j
91.
v 100 cos 45 i sin 45 j u v 900 cos 148 100 cos 45 i 900 sin 148 100 sin 45 j 692.53 i 547.64 j
arctan
547.64 692.53 38.34.
38.34 North of West.
u v 692.532 547.642 882.9 kmhr. 93. F1 F2 F3 0 3600j T2cos 35i sin 35 j T3cos 92i sin 92j 0 T2 cos 35 T3 cos 92 0 T2 cos 35 T3 sin 92 3600 T3 cos 92 T cos 92 ⇒ 3 sin 35 T3 sin 92 3600 and T30.97495 3600 ⇒ T3 3692.48 cos 35 cos 35 Finally, T2 157.32 T2
95. Let the triangle have vertices at 0, 0, a, 0, and b, c. Let u be the vector joining 0, 0 and b, c, as indicated in the figure. Then v, the vector joining the midpoints, is v
y
(b, c)
a 2 b a2 i 2c j
( a +2 b , 2c (
u v
b 1 c 1 i j bi cj u 2 2 2 2
x
(0, 0)
( 2a , 0(
(a, 0)
97. w uv vu u v cos v i v sin v j v u cos ui u sin u j u v cos u cos vi sin u sin v j
2u v cos
cos
tan w
v u v cos 2 2 u v tan u v 2 u v cos 2 2
u
sin
u v u v u v u v cos i sin cos j 2 2 2 2
Thus, w u v2 and w bisects the angle between u and v. 99. True
101. True
103. False
a i bj 2 a
Section 10.2 1.
Space Coordinates and Vectors in Space
(2, 1, 3)
z
3.
z 6 5 4 3
(5, − 2, 2) − 2 (−1, 2, 1) 4
1 4
3
2
3 2 1
2 3 4
1 3
2 −2 −3
x y
(5, − 2, − 2)
x
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1 2 3
y
Section 10.2 5. A2, 3, 4
Space Coordinates and Vectors in Space
7. x 3, y 4, z 5: 3, 4, 5
233
9. y z 0, x 10: 10, 0, 0
B1, 2, 2 11. The z-coordinate is 0.
13. The point is 6 units above the xy-plane.
15. The point is on the plane parallel to the yz-plane that passes through x 4.
17. The point is to the left of the xz-plane.
19. The point is on or between the planes y 3 and y 3.
21. The point x, y, z is 3 units below the xy-plane, and below either quadrant I or III.
23. The point could be above the xy-plane and thus above quadrants II or IV, or below the xy-plane, and thus below quadrants I or III. 25. d 5 02 2 02 6 02
27. d 6 12 2 22 2 42
25 4 36 65
25 0 36 61
29. A0, 0, 0, B2, 2, 1, C2, 4, 4
31. A1, 3, 2, B5, 1, 2, C1, 1, 2
AB 4 4 1 3 AC 4 16 16 6 BC 0 36 9 35 BC2 AB2 AC2
AB 16 4 16 6 AC 4 16 16 6 BC 36 4 0 210 Since AB AC, the triangle is isosceles.
Right triangle
35.
33. The z-coordinate is changed by 5 units:
0, 0, 5, 2, 2, 6, 2, 4, 9
2, 0, 0 0, 6, 0 1, 3, 0 2 Radius: 10
37. Center: 0, 2, 5
39. Center:
Radius: 2
x 02 y 22 z 52 4 x 2
y2
z2
5 22, 92 3, 7 2 3 32, 3, 5
x 12 y 32 z 02 10
4y 10z 25 0
x2 y 2 z2 2x 6y 0
x2 y2 z2 2x 6y 8z 1 0
41.
x2 2x 1 y2 6y 9 z2 8z 16 1 1 9 16 x 12 y 32 z 42 25 Center: 1, 3, 4 Radius: 5
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234
Chapter 10
Vectors and the Geometry of Space
43.
9x2 9y2 9z2 6x 18y 1 0
45. x2 y2 z2 ≤ 36 Solid ball of radius 6 centered at origin.
2 1 x 2 y2 z2 x 2y 0 3 9
x
2
2 1 1 1 x y2 2y 1 z2 1 3 9 9 9
x 31
2
y 12 z 02 1
13, 1, 0
Center:
Radius: 1 47. (a) v 2 4 i 4 2 j 3 1k
49. (a) v 0 3 i 3 3 j 3 0k
2i 2j 2k 2, 2, 2 (b)
3i 3k 3, 0, 3 (b)
z
z
− 3, 0, 3
5
5
4
4
−2, 2, 2
3
3
−3
2 −2
1
−2
1
1
1
2
−3
2
1 2
3
1
2 3
y
4
2
3
x
3
4
y
x
53. 5 4, 3 3, 0 1 1, 0, 1
51. 4 3, 1 2, 6 0 1, 1, 6 1, 1, 6 1 1 36 38
1, 0, 1 1 1 2
1 6 1 1, 1, 6 , , 38 38 38 38
Unit vector:
55. (b) v 3 1i 3 2j 4 3k 4i j k 4, 1, 1
12, 0, 12
Unit vector:
57. q1, q2, q3 0, 6, 2 3, 5, 6 Q 3, 1, 8
(a) and (c). z 5 4
(3, 3, 4) (−1, 2, 3)
3
(0, 0, 0) 2 −2
v
(4, 1, 1) 2
2 4
4
y
x
59. (a) 2v 2, 4, 4
(b) v 1, 2, 2 z
z 5
3
4
2
3
2, 4, 4
2
−3
1 2 3
1
1
2 3
2
4 x
y
x
−3
−2
− 1, −2, − 2
−2
−2
−2 −3
—CONTINUED—
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2
3
Section 10.2
Space Coordinates and Vectors in Space
235
59. —CONTINUED— (c)
3 2
v
32 , 3, 3
(d) 0v 0, 0, 0 z
z 3
3 2 −3
−2
−3
2 3 , 2
−2
0, 0, 0 1
2 y
3
−2
x
−2
1
2 3
−3
3, 3
−3
−2
1
−1
1
2
y
3
−2
x
−3
−3
61. z u v 1, 2, 3 2, 2, 1 1, 0, 4 63. z 2u 4v w 2, 4, 6 8, 8, 4 4, 0, 4 6, 12, 6 67. (a) and (b) are parallel since 6, 4, 10 23, 2, 5 2 and 2, 43 , 10 3 3 3, 2, 5.
65. 2z 3u 2z1, z2, z3 31, 2, 3 4, 0, 4 2z1 3 4 ⇒ z1
7 2
2z2 6 0 ⇒ z2 3 2z3 9 4 ⇒ z3 52 z
72 , 3, 52
69. z 3i 4j 2k
71. P0, 2, 5, Q3, 4, 4, R2, 2, 1 \
(a) is parallel since 6i 8j 4k 2z.
PQ 3, 6, 9 \
PR 2, 4, 6 3, 6, 9 32 2, 4, 6 \
\
Therefore, PQ and PR are parallel. The points are collinear. 73. P1, 2, 4, Q2, 5, 0, R0, 1, 5
75. A2, 9, 1, B3, 11, 4, C0, 10, 2, D1, 12, 5
\
\
PQ 1, 3, 4
AB 1, 2, 3
\
\
PR 1, 1, 1 \
CD 1, 2, 3
\
\
AC 2, 1, 1
Since PQ and PR are not parallel, the points are not collinear.
\
BD 2, 1, 1 \
\
\
\
Since AB CD and AC BD , the given points form the vertices of a parallelogram. 77. v 0
79.
v 1, 2, 3 v 1 4 9 14
u 2, 1, 2
83.
u 4 1 4 3 (a)
1 u 2, 1, 2 u 3
(b)
u 1 2, 1, 2 u 3
u 3, 2, 5
85.
u 9 4 25 38 (a)
1 u 3, 2, 5 u 38
(b)
u 1 3, 2, 5 u 38
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81.
v 0, 3, 5 v 0 9 25 34
87. Programs will vary.
236
Chapter 10
Vectors and the Geometry of Space
c v 2c, 2c, c
89.
91. v 10
c v 4c 2 4c 2 c 2 5
0,
9c2 25 c±
93. v
u 1 1 10 0, , u 2 2 ,
10
5 3
3 u 3 2 2 1 1 , , 1, 1, 2 u 2 3 3 3 2
95. v 2 cos± 30j sin± 30k
v 3, 6, 3
97.
3 j ± k 0, 3, ± 1
2 3v
2
2, 4, 2
4, 3, 0 2, 4, 2 2, 1, 2
z
−2
0,
1
−2
10
2 2
3, 1
−1 1 y
−1
2 x
0,
−2
3, −1
99. (a)
(b) w au bv ai a bj bk 0
z
a 0, a b 0, b 0 1
Thus, a and b are both zero. v
u
1
1 y
x
(c) ai a bj bk i 2j k
(d) a i a bj bk i 2j 3k
a 1, b 1
a 1, a b 2, b 3
wuv
Not possible
101. d x2 x12 y2 y12 z2 z12
103. Two nonzero vectors u and v are parallel if u cv for some scalar c.
105. (a) The height of the right triangle is h L2 182. The vector PQ is given by
Q (0, 0, h)
\
\
PQ 0, 18, h. L
The tension vector T in each wire is 24 T c0, 18, h where ch 8. 3 8 Hence, T 0, 18, h and h
(0, 18, 0) (0, 0, 0)
18
P
8 8L 8 182 L2 182 T T 182 h2 h L2 182 L2 182 (b)
L
20
25
30
35
40
45
50
T
18.4
11.5
10
9.3
9.0
8.7
8.6
—CONTINUED—
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Section 10.2
30
v cos i cos j cos k
L = 18
v 3 cos 1 cos T=8 0
3 1 3 3
100 0
v
x 18 is a vertical asymptote and y 8 is a horizontal asymptote.
3
3
i j k
z 0.6
8L (d) lim L→18 L2 182
0.4
(
3 , 3
3 , 3
8L 8 lim 8 L→ L2 182 L → 1 18L2
0.2
0.6 x
\
109. AB 0, 70, 115, F1 C10, 70, 115 \
AC 60, 0, 115, F2 C2 60, 0, 115 \
AD 45, 65, 115, F3 C3 45, 65, 115 F F1 F2 F3 0, 0, 500 Thus: 60C2 45C3
0
65C3
0
C2 C3 500
104 28 112 Solving this system yields C1 69 , C2 23, and C3 69 . Thus:
F1 202.919N F2 157.909N F3 226.521N 111. dAP 2dBP x2 y 12 z 12 2x 12 y 22 z 2
x2 y2 z2 2y 2z 2 4x2 y2 z2 2x 4y 5 0 3x2 3y2 3z2 8x 18y 2z 18 6
16 1 8 16 2 1 9 x2 x y2 6y 9 z2 z 9 9 3 9 3 9
4 44 x 9 3 Sphere; center:
( y
0.4
0.4
(e) From the table, T 10 implies L 30 inches.
115C1
3 3
0.2
lim
70C1
237
107. Let be the angle between v and the coordinate axes.
105. —CONTINUED— (c)
Space Coordinates and Vectors in Space
2
y 32 z
1 3
2
43, 3, 31 , radius: 2 311
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3
3
1, 1, 1
238
Chapter 10
Vectors and the Geometry of Space
Section 10.3
The Dot Product of Two Vectors
1. u 3, 4, v 2, 3
3. u 2, 3, 4, v 0, 6, 5
(a) u v 32 43 6
(a) u
(b) u u 33 44 25
(b) u u 22 33 44 29
(c)
u2
25
(c) u2 29
(d) u vv 62, 3 12, 18 (e) u
(e) u 2v 2u 7.
(a) u v 21 10 11 1
v 22 4
u 3240, 1450, 2235 v 2.22, 1.85, 3.25
(b) u u 22 11 11 6
u v $17,139.05
(c) u2 6
This gives the total amount that the person earned on his products.
(d) u v v v i k (e) u 2v 2u
9.
(d) u v v 20, 6, 5 0, 12, 10
2v 2u v 26 12
5. u 2i j k, v i k
v 20 36 45 2
v 2
uv cos u v
11. u 1, 1, v 2, 2
u v 85 cos
20 3
cos
13. u 3i j, v 2i 4j cos
uv 2 1 u v 1020 52
arccos
1
98.1 52
17. u 3i 4j, v 2j 3k cos
uv 8 813 u v 513 65
arccos
21. u 4, 3, v
813
116.3 65
12, 32
u cv ⇒ not parallel u v 0 ⇒ orthogonal
uv 0 0 u v 28
2
15. u 1, 1, 1, v 2, 1, 1 cos
2 uv 2 u v 36 3
arcos
2
3
61.9
19. u 4, 0, v 1, 1 u cv ⇒ not parallel u v 4 0 ⇒ not orthogonal Neither
23. u j 6k, v i 2j k u c v ⇒ not parallel u v 8 0 ⇒ not orthogonal Neither
25. u 2, 3, 1, v 1, 1, 1 u cv ⇒ not parallel u v 0 ⇒ orthogonal
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Section 10.3 27. u i 2j 2k, u 3 cos
29. u 0, 6, 4, u 52 213 cos 0
1 3
13
cos
2 cos 3
3
cos
2 cos 3
cos2
The Dot Product of Two Vectors
cos2
31. u 3, 2, 2
cos2
1 4 4 1 9 9 9
cos2 cos2 cos2 0
33. u 1, 5, 2
u 17
cos
3 ⇒ 0.7560 or 43.3 17
cos
cos
2 ⇒ 1.0644 or 61.0 17
cos
cos
2 ⇒ 2.0772 or 119.0 17
cos
35. F1: C1
50
4.3193 F1
F2: C2
80
5.4183 F2
2 13
1 30
5 30
u 30 ⇒ 1.7544 or 100.5 ⇒ 0.4205 or 24.1
2 ⇒ 1.1970 or 68.6 30
37. Let s length of a side. v s, s, s v s3
F F1 F2
cos cos cos
4.319310, 5, 3 5.418312, 7, 5 F 124.310 lb 108.2126 ⇒ 29.48 F
s v
14.1336 ⇒ 96.53 F
s x
\
41. w2 u w1 6, 7 2, 8 4, 1
0 02 102 102
cos cos 1 2
y
s
39. OA 0, 10, 10 cos
z
59.5246 ⇒ 61.39 cos
F cos
s 1 s3 3
13 54.7
arccos
108.2126, 59.5246, 14.1336
cos
9 4 1 13 13
02
0 ⇒ 90
10 102 102
⇒ 45
43. w2 u w1 0, 3, 3 2, 2, 2 2, 1, 1
45. u 2, 3, v 5, 1 (a) w1
13 5 1 5, 1 , uv v v 26 2 2 2
1 5 (b) w2 u w1 , 2 2
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239
240
Chapter 10
Vectors and the Geometry of Space 49. u v u1, u2, u3
47. u 2, 1, 2, v 0, 3, 4 (a) w1
v1, v2, v3 u1v1 u2v2 u3v3
uv v v 2
11 33 44 0, 3, 4 0, , 25 25 25
(b) w2 u w1 2,
51. (a) Orthogonal,
8 6 , 25 25
2
(b) Acute, 0 < <
2
53. See page 738. Direction cosines of v v1, v2, v3 are cos
v1 v v , cos 2 , cos 3 . v v v
(c) Obtuse,
uv v v u ⇒ u cv ⇒ u and v are parallel.
55. (a)
2
uv v v 0 ⇒ u v 0 ⇒ u and v
(b)
2
, , and are the direction angles. See Figure 10.26. 57. Programs will vary.
< < 2
are orthogonal. 59. Programs will vary.
61. Because u appears to be perpendicular to v, the projection of u onto v is 0. Analytically, projv u
uv 2, 3 6, 4 v 6, 4 06, 4 0. v2 6, 42
1 2 63. u i j. Want u v 0. 2 3
65. u 3, 1, 2. Want u v 0.
v 8i 6j and v 8i 6j are orthogonal to u.
(b) w2 F w1 48,000 j 8335.1cos 10 i sin 10 j
67. (a) Gravitational Force F 48,000 j v cos 10 i sin 10 j w1
v 0, 2, 1 and v 0, 2, 1 are orthogonal to u.
8208.5 i 46,552.6 j
Fv v F vv 48,000sin 10v v2
w2 47,270.8 lb
8335.1cos 10 i sin 10 j w1 8335.1 lb
12 i
69. F 85
3
2
\
71. PQ 4, 7, 5
j
v 1, 4, 8
v 10i
\
W PQ
W F v 425 ft lb
v 72
73. False. Let u 2, 4, v 1, 7 and w 5, 5. Then u v 2 28 30 and u w 10 20 30. 75. In a rhombus, u v. The diagonals are u v and u v.
u v u v u v u u v v uuvuuvvv
u−v u u+v v
u2 v2 0 Therefore, the diagonals are orthogonal.
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Section 10.4
The Cross Product of Two Vectors in Space
77. u cos , sin , 0, v cos , sin , 0 The angle between u and v is . Assuming that > . Also, cos
uv cos cos sin sin cos cos sin sin . u v 11
79. u v2 u v u v
81. u v2 u v u v
u v u u v v
u v u u v v
uuvuuvvv
uuvuuvvv
u2 u v u v v2
u2 2u v v2
u2 v2 2u v
≤ u 2 2u v v 2 from Exercise 66 ≤ u v2 Therefore, u v ≤ u v.
Section 10.4
The Cross Product of Two Vectors in Space
i 1. j i 0 1
i 3. j k 0 0
j k 1 0 k 0 0
j k 1 0 i 0 1
i 5. i k 1 0
z
z
z
1
1
1
k
x
1
i
−k
1
−1
x
y
i 7. (a) u v 2 3
−1
−j
k
j
j 1
j k 0 0 j 0 1
i
1
1 y
−1
j k 3 4 22, 16, 23 7 2
x
i j 3 9. (a) u v 7 1 1
i
1 −1
k 2 17, 33, 10 5
(b) v u u v 22, 16, 23
(b) v u u v 17, 33, 10
i (c) v v 3 3
(c) v v 0
j k 7 2 0 7 2
11. u 2, 3, 1, v 1, 2, 1 i j u v 2 3 1 2
k 1 i j k 1, 1, 1 1
u u v 21 31 11 0 ⇒ u u v v u v 11 21 11 0 ⇒ v u v
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y
241
242
Chapter 10
Vectors and the Geometry of Space
13. u 12, 3, 0, v 2, 5, 0
i j 12 3 2 5
uv u
15. u i j k, v 2i j k
i uv 1 2
k 0 54k 0, 0, 54 0
u v 120 30 054
u
u v 12 13 11 0 ⇒ uuv
0 ⇒ uuv
v u v 22 13 11
v u v 20 50 054
0 ⇒ vuv
0 ⇒ vuv 17.
19.
z 6 5 4 3 2 1
z 6 5 4 3 2 1
v
3
v
1
1 4
j k 1 1 2i 3j k 2, 3, 1 1 1
2
u 4
4
y
6
x
3
2
u
4
y
6
x
23. u 3i 2j 5k
21. u 4, 3.5, 7 v 1, 8, 4
u v 70, 23,
57 2
1 3 1 v i j k 2 4 10
140 46 57 uv , , u v 24,965 24,965 24,965
uv
71 11 5 , , 20 5 4
uv 20 71 11 5 , , u v 7602 20 5 4
71 7602
,
44
,
25
7602 7602
25. Programs will vary. 27. u j
29. u 3, 2, 1
vjk
v 1, 2, 3
i uv 0 0
j k 1 0 i 1 1
i uv 3 1
A u v i 1
A u v 8, 10, 4 180 6 5
31. A1, 1, 1,, B2, 3, 4, C6, 5, 2, D7, 7, 5 \
\
\
AB 1, 2, 3, AC 5, 4, 1, CD 1, 2, 3, BD 5, 4, 1 \
\
\
\
\
Since AB CD and AC BD , the figure is a parallelogram. AB and AC are adjacent sides and \
\
i AB AC 1 5 \
\
\
j k 2 1 8, 10, 4 2 3
j k 2 3 10i 14j 6k. 4 1
33. A0, 0, 0, B1, 2, 3, C3, 0, 0 \
\
AB 1, 2, 3, AC 3, 0, 0 \
\
AB AC
i 1 3
j 2 0
k 3 9j 6k 0
1 1 3 A AB AC 117 13 2 2 2 \
\
A AB AC 332 2 83
http://librosysolucionarios.net
\
Section 10.4 35. A2, 7, 3, B1, 5, 8, C4, 6, 1 \
The Cross Product of Two Vectors in Space
243
37. F 20k
\
AB 3, 12, 5, AC 2, 13, 4
i AB AC 3 2 \
\
1 PQ cos 40 j sin 40 k 2 i j k PQ F 0 cos 40 2 sin 40 2 10 cos 40 i 0 0 20 \
j k 12 5 113, 2, 63 13 4
1 1 Area AB AC 16,742 2 2 \
\
\
PQ F 10 cos 40 7.66 ft lb \
z
PQ 1 ft 2
40° F
y
x
3 OA k 2 \
39. (a)
z
F 60sin j cos k
100
OA
1.5 ft
F
θ
i j k OA F 0 0 3 2 90 sin i 0 60 sin 60 cos
\
\
OA F 90 sin
\
180 0
x
22 45
(b) When 45 : OA F 90 (c) Let T 90 sin .
0
y
2 63.64.
dT 90 cos 0 when 90 . d
This is what we expected. When 90 the pipe wrench is horizontal.
1 41. u v w 0 0
0 1 0
0 0 1 1
2 43. u v w 0 0
0 3 0
1 0 6 1
1 45. u v w 0 1
1 1 0
0 1 2 1
V u v w 2
47. u 3, 0, 0 v 0, 5, 1 w 2, 0, 5 u
v w
3 0 2
0 5 0
0 1 75 5
V u v w 75 49. u v u1, u2, u3
v1, v2, v3 u 2v3 u 3v2 i u 1v3 u 3v1j u 1v2 u 2v1k
51. The magnitude of the cross product will increase by a factor of 4.
53. If the vectors are ordered pairs, then the cross product does not exist. False.
55. True
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244
Chapter 10
Vectors and the Geometry of Space
57. u u1, u2, u3 , v v1, v2, v3 , w w1, w2, w3 u v w
i u1 v1 w1
j u2 v2 w2
k u3 v3 w3
u2v3 w3 u3v2 w2 i u1v3 w3 u3v1 w1 j u1v2 w2 u2v1 w1k u2v3 u3v2i u1v3 u3v1j u1v2 u2v1k u2w3 u3w2i
u1w3 u3w1j u1w2 u2w1k u v u w 59. u u1, u2, u3
i j u u u1 u 2 u1 u2
k u3 u2u3 u3u2 i u1u3 u3u1j u1u2 u2u1k 0 u3
u v u2v3 u3v2i u1v3 u3v1j u1v2 u2v1k
61.
u v u u2v3 u3v2u1 u3v1 u1v3u2 u1v2 u2v1u3 0 u v v u2v3 u3v2v1 u3v1 u1v3v2 u1v2 u2v1v3 0 Thus, u v u and u v v. 63. u v u v sin If u and v are orthogonal, 2 and sin 1. Therefore, u v u v .
Section 10.5
Lines and Planes in Space
1. x 1 3t, y 2 t, z 2 5t (a)
z
(b) When t 0 we have P 1, 2, 2. When t 3 we have Q 10, 1, 17. \
PQ 9, 3, 15 The components of the vector and the coefficients of t are proportional since the line is parallel to PQ . \
x y
(c) y 0 when t 2. Thus, x 7 and z 12. Point: 7, 0, 12
0, 73, 13 2 1 12 z 0 when t . Point: , , 0 5 5 5
1 x 0 when t . Point: 3
3. Point: (0, 0, 0
5. Point: 2, 0, 3
Direction vector: v 1, 2, 3
Direction vector: v 2, 4, 2
Direction numbers: 1, 2, 3
Direction numbers: 2, 4, 2
(a) Parametric: x t, y 2t, z 3t
(a) Parametric: x 2 2t, y 4t, z 3 2t
z y (b) Symmetric: x 2 3
(b) Symmetric:
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y z3 x2 2 4 2
Section 10.5
Lines and Planes in Space
9. Points: 5, 3, 2,
7. Point: 1, 0, 1 Direction vector: v 3i 2j k
23, 23, 1 11 17 i j 3k 3 3
Direction numbers: 3, 2, 1
Direction vector: v
(a) Parametric: x 1 3t, y 2t, z 1 t
Direction numbers: 17, 11, 9
(b) Symmetric:
y z1 x1 3 2 1
(a) Parametric: x 5 17t, y 3 11t, z 2 9t (b) Symmetric:
x5 y3 z2 17 11 9
13. Point: 2, 3, 4
11. Points: 2, 3, 0, 10, 8, 12 Direction vector: 8, 5, 12
Direction vector: v k
Direction numbers: 8, 5, 12
Direction numbers: 0, 0, 1
(a) Parametric: x 2 8t, y 3 5t, z 12t
Parametric: x 2, y 3, z 4 t
(b) Symmetric:
z x2 y3 8 5 12
15. Point: (2, 3, 1
17. Li: v 3, 2, 4
Direction vector: v 4i k Direction numbers: 4, 0, 1 Parametric: x 2 4t, y 3, z 1 t Symmetric:
x2 z1 ,y3 4 1
6, 2, 5 on line
L 2: v 6, 4, 8
6, 2, 5 on line
L 3: v 6, 4, 8
6, 2, 5 not on line
L 4: v 6, 4, 6
not parallel to L1, L 2, nor L 3
Hence, L1 and L 2 are identical.
(a) On line
L1 L 2 and L 3 are parallel.
(b) On line (c) Not on line y 3 (d) Not on line
6 4 2 21 1
19. At the point of intersection, the coordinates for one line equal the corresponding coordinates for the other line. Thus, (i) 4t 2 2s 2, (ii) 3 2s 3, and (iii) t 1 s 1. From (ii), we find that s 0 and consequently, from (iii), t 0. Letting s t 0, we see that equation (i) is satisfied and therefore the two lines intersect. Substituting zero for s or for t, we obtain the point (2, 3, 1. u 4i k
(First line)
v 2i 2j k
(Second line)
cos
u v u v
7 7 17 81 51
17 9 3 17
21. Writing the equations of the lines in parametric form we have x 3t
y2t
z 1 t
x 1 4s
y 2 s
z 3 3s.
17 11 For the coordinates to be equal, 3t 1 4s and 2 t 2 s. Solving this system yields t 7 and s 7 . When using these values for s and t, the z coordinates are not equal. The lines do not intersect.
23. x 2t 3
245
x 2s 7
y 5t 2
ys8
z t 1
z 2s 1
Point of intersection: 7, 8, 1
z 4
10
8
6
4
2 −2
4
−8
(7, 8, − 1)
x
6
8
10
y
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246
Chapter 10
Vectors and the Geometry of Space
25. 4x 3y 6z 6 (a) P 0, 0, 1, Q 0, 2, 0, R 3, 4, 1 \
\
PQ 0, 2, 1, PR 3, 4, 0
i j (b) PQ PR 0 2 3 4 \
\
k 1 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. 27. Point: 2, 1, 2
29. Point: 3, 2, 2
n i 1, 0, 0
Normal vector: n 2i 3j k
1x 2 0y 1 0z 2 0
2x 3 3y 2 1z 2 0
x20
2x 3y z 10
31. Point: 0, 0, 6
33. Let u be the vector from 0, 0, 0 to 1, 2, 3: u i 2j 3k
Normal vector: n i j 2k
Let v be the vector from 0, 0, 0 to 2, 3, 3: v 2i 3j 3k
1x 0 1y 0 2z 6 0 x y 2z 12 0
Normal vector: u v
x y 2z 12
i 1 2
j 2 3
k 3 3
3i 9j 7k 3x 0 9 y 0 7z 0 0 3x 9y 7z 0 35. Let u be the vector from 1, 2, 3 to 3, 2, 1: u 2i 2k Let v be the vector from 1, 2, 3 to 1, 2, 2: v 2i 4j k Normal vector:
12 u v
i 1 2
j k 0 1 4i 3j 4k 4 1
4x 1 3y 2 4z 3 0 4x 3y 4z 10
37. 1, 2, 3, Normal vector: v k, 1z 3 0, z 3
39. The direction vectors for the lines are u 2i j k, v 3i 4j k.
i Normal vector: u v 2 3
j k 1 1 5i j k 4 1
Point of intersection of the lines: 1, 5, 1
x 1 y 5 z 1 0 xyz5 41. Let v be the vector from 1, 1, 1 to 2, 2, 1: v 3i j 2k Let n be a vector normal to the plane 2x 3y z 3: n 2i 3j k
Since v and n both lie in the plane p, the normal vector to p is i j 1 vn 3 2 3
k 2 7i j 11k 1
7x 2 1y 2 11z 1 0 7x y 11z 5
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Section 10.5 43. Let u i and let v be the vector from 1, 2, 1 to 2, 5, 6: v i 7j 7k
247
45. The normal vectors to the planes are n1 5, 3, 1, n2 1, 4, 7, cos
Since u and v both lie in the plane P, the normal vector to P is:
i uv 1 1
Lines and Planes in Space
n1 n2 0. n1 n2
Thus, 2 and the planes are orthogonal.
j k 0 0 7j 7k 7j k 7 7
y 2 z 1 0 y z 1 49. The normal vectors to the planes are n1 1, 5, 1 and n2 5, 25, 5. Since n2 5n1, the planes are parallel, but not equal.
47. The normal vectors to the planes are n1 i 3j 6k, n2 5i j k,
n1 n2 5 3 6 4 138 .
cos
n1 n2
46 27
414
4 414138 83.5 .
Therefore, arccos
51. 4x 2y 6z 12
53. 2x y 3z 4
55. y z 5
z
z 6
3
4
2
z 6
−4 6
−1
4
x
6
y
y
6 6
x
3
y
x
57. x 5
59. 2x y z 6 z
z
z
3 6
4
2
2
−2 4
6 y
x
−6
5 x
61. 5x 4y 6z 8 0
5
2
y Generated by Maple
63. P1: n 3, 2, 5 P2: n 6, 4, 10
1, 1, 1 on plane 1, 1, 1 not on plane
x
c, 0, 0, 0, c, 0, and 0, 0, c.
1, 1, 1 on plane
P1 and P4 are identical. P1 P4 is parallel to P2.
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1
y
Generated by Maple
65. Each plane passes through the points
P3: n 3, 2, 5 P4: n 75, 50, 125
−1
248
Chapter 10
Vectors and the Geometry of Space
67. The normals to the planes are n1 3i 2j k and n2 i 4j 2k. The direction vector for the line is
i j k n2 n1 1 4 2 7 j 2k. 3 2 1 Now find a point of intersection of the planes. 6x 4y 2y 14
x 2
3 1 t, y t, z 1 2t 2 2
12 t 2 32 t 1 2t 12, t 23
Substituting t 3 2 into the parametric equations for the line we have the point of intersection 2, 3, 2. The line does not lie in the plane.
x 4y 2z 0 14
7x
69. Writing the equation of the line in parametric form and substituting into the equation of the plane we have:
x 2 Substituting 2 for x in the second equation, we have 4y 2z 2 or z 2y 1. Letting y 1, a point of intersection is 2, 1, 1. x 2, y 1 t, z 1 2t 71. Writing the equation of the line in parametric form and substituting into the equation of the plane we have: x 1 3t, y 1 2t, z 3 t 21 3t 31 2t 10, 1 10, contradiction Therefore, the line does not intersect the plane.
73. Point: Q0, 0, 0 Plane: 2x 3y z 12 0 Normal to plane: n 2, 3, 1 Point in plane: P6, 0, 0 \
Vector PQ 6, 0 0
PQ n 12 6 14 D \
n
75. Point: Q2, 8, 4 Normal to plane: n 2, 1, 1
P 10, 0, 0 is a point in x 3y 4z 10. Q 6, 0, 0 is a point in x 3y 4z 6.
Point in plane: P0, 0, 5
PQ n1 PQ 4, 0, 0, D
\
\
Vector: PQ 2, 8, 1 D
PQ n n
11
6
\
n1
11 6 6
79. The normal vectors to the planes are n1 3, 6, 7 and n2 6, 12, 14. Since n2 2n1, the planes are parallel. Choose a point in each plane. P 0, 1, 1 is a point in 3x 6y 7z 1. Q \
256, 0, 0 is a point in 6x 12y 14z 25.
PQ
256, 1, 1 \
n1
94
4 2 26 13
26
81. u 4, 0, 1 is the direction vector for the line. Q1, 5, 2 is the given point, and P2, 3, 1 is on the line. Hence, PQ 3, 2, 3 and \
i PQ u 3 4 \
\
D
PQ n1 27 2 D
7
77. The normal vectors to the planes are n1 1, 3, 4 and n2 1, 3, 4. Since n1 n2, the planes are parallel. Choose a point in each plane.
Plane: 2x y z 5
\
14
j k 2 3 2, 9, 8 0 1
PQ u 149 2533 u 17
17
27 27 94 2 94 188
83. The parametric equations of a line L parallel to v a, b, c, and passing through the point Px1, y1, z1 are x x1 at, y y1 bt, z z1 ct.
85. Solve the two linear equations representing the planes to find two points of intersection. Then find the line determined by the two points.
The symmetric equations are x x1 y y1 z z1 . a b c
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Section 10.6
Surfaces in Space
249
(b) Parallel planes
87. (a) Sphere
x 3 y 2 z 5 16 x2 y2 z2 6x 4y 10z 22 0 2
2
4x 3y z 10 ± 4n 10 ± 426
2
89. (a) z 28.7 1.83x 1.09y Year
1980
1985
1990
1994
1995
1996
1997
z (approx.)
16.16
14.23
9.81
8.60
8.42
8.27
8.23
(b) An increase in x or y will cause a decrease in z. In fact, any increase in two variables will cause a decrease in the third. z
(c) 30
(0, 0, 28.7)
(15.7, 0, 0)
(0, 26.3, 0)
30 x
30
y
91. True
Section 10.6
Surfaces in Space
1. Ellipsoid
3. Hyperboloid of one sheet
Matches graph (c)
5. Elliptic paraboloid
Matches graph (f)
7. z 3
9. y2 z2 9
z
Plane parallel to the xy-coordinate plane
Matches graph (d)
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a circle.
2
z
2 3 x
2
4 y
11. y x2
4
7 6
x
y
13. 4x2 y2 4
The z-coordinate is missing so we have a cylindrical surface with rulings parallel to the z-axis. The generating curve is a parabola. z 4
x2 y 2 1 1 4 The z-coordinate is missing so we have a cylindrical surface with rulings parallel to the z-axis. The generating curve is an ellipse. z 3
x
4
3
2 3 4
y
−3 2 3 x
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2
3
y
250
Chapter 10
Vectors and the Geometry of Space
15. z sin y
z 2
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is the sine curve.
1
3
y
3 4 x
17. x x2 y2 (a) You are viewing the paraboloid from the x-axis: 20, 0, 0 (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
19.
x2 y 2 z2 1 1 4 1 2
Ellipsoid 2
xz-trace: x 2 z 2 1 circle y2
4
4x 2
y2
x xy-trace: 1 ellipse 1 4
yz-trace:
21. 16x 2 y 2 16z 2 4
z
z2 1
1 ellipse
y2 4
2
x
y
−2
xy-trace: y
xy-trace: y ± x
point 0, 0, 0 yz-trace: y z 2 y 1:
z2
xy-trace: point 0, 0, 0 xz-trace: z ± x
yz-trace: z y 2
1
x
yz-trace: z
z
z ± 1: x2
3
±1
2
y
y2 1 4
z x
−3
3 2
2 3
y
2
1
1 3
4
−2
y
−2 2
2
x
−3
−2
29.
y2 4
Elliptic Cone
3
2
−3
27. z 2 x2
y ± 1: z 1 x 2
2
3
−2
x
y 2 4z 2 1 hyperbola 4
xz-trace: z x2
z
1
3
xz-trace: 4x 2 z 2 1 circle
Hyperbolic paraboloid
xz-trace: x2 z 2 0,
−3
2
y2 1 hyperbola 4
25. x2 y 2 z 0
x2
2 −2
xy-trace: 4x 2
Elliptic paraboloid
x2
4z 2 1
Hyperboloid on one sheet 2
yz-trace:
23. x2 y z 2 0
z 3
16x2 9y2 16z2 32x 36y 36 0
z
16x2 2x 1 9y2 4y 4 16z2 36 16 36
2 1
16x 12 9y 22 16z2 16
x 12 y 22 z2 1 1 169 1
1 2 2
x −2
Ellipsoid with center 1, 2, 0.
http://librosysolucionarios.net
4
y
y
3
y
Section 10.6 31. z 2 sin x
33. z 2 x 2 4y 2
35. x2 y2
z ± x2 4y2
z
Surfaces in Space
y±
z 3
251
2z
2
z4 x
2
2
5
z 4
−2
π x
3
−1
y
1 2 y
x
4 4
x
37. z 4 xy
39. 4x 2 y 2 4z 2 16
z
z±
5
y4 x 2
41. z 2x2 y 2 z2
4
2
2x2 y 2 2
z
x2 y2 1
8 5
4
3 3
6 4
5
x
4
y
−4
−6
z
−8 3
−2 8
6
4
x
−2 −4 −6
2 2 y −2
−8
−2 2
1
x
43. x2 y2 1 4
z0
3
y
2
45. x2 z 2 ry 2 and z r y ± 2y; therefore,
z
xz2
y
x2 z 2 4y.
2
3 x
2
3
y
z 47. x2 y 2 rz 2 and y rz ; therefore, 2 z2 x2 y 2 , 4x2 4y 2 z 2. 4
2 49. y 2 z 2 rx 2 and y r x ; therefore, x 2 2 2 4 2 2 2 y z , y z 2. x x
51. x 2 y 2 2z 0
53. 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.
x2
y2
2z
2
Equation of generating curve: y 2z or x 2z
z
4
55. See pages 765 and 766.
57. V 2
x4x x2 dx
4
0
2
4x3 x4 3 4
4 0
128 3
3 2
h ( x)
1
x 1
2
p ( x)
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3
4
252
Chapter 10
59. z
Vectors and the Geometry of Space
x2 y2 2 4
(a) When z 2 we have 2
x2 y2 x2 y2 , or 1 2 4 4 8
(b) When z 8 we have 8
x2 y 2 y2 x2 , or 1 . 2 16 32 4
Major axis: 28 42
Major axis: 232 82
Minor axis: 24 4
Minor axis: 216 8
c2
a2
b2,
c2
4, c 2
c 2 32 16 16, c 4
Foci: 0, ± 2, 2
Foci: 0, ± 4, 8
61. If x, y, z is on the surface, then
63.
y2 z2 x2 1 2 2 3963 3963 39422
y 22 x2 y 2)2 z2 y2
z
4y 4 x y 4y 4 z 2
2
2 4000
x2 z2 8y Elliptic paraboloid 4000
Traces parallel to xz-plane are circles.
y
4000 x
65. z
y2 x2 2 , z bx ay 2 b a bx ay
67. 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.
y2 x2 2 2 b a
a4b2 1 a2b4 1 2 x a2 bx 2 y2 ab2y a2 4 b 4
x a2b y ab2 2
a2
2
2 2
y±
b2
a2b ab2 b x a 2 2
Letting x at, you obtain the two intersecting lines x at, y bt, z 0 and x at, y bt ab2 z 2abt a2b2.
Section 10.7 1. 5, 0, 2, cylindrical x 5 cos 0 5 y 5 sin 0 0 z2
5, 0, 2, rectangular
Cylindrical and Spherical Coordinates 3.
2, 3 , 2, cylindrical
5.
4, 76, 3, cylindrical
x 2 cos
1 3
x 4 cos
7 23 6
y 2 sin
3 3
y 4 sin
7 2 6
z2
1, 3, 2, rectangular
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z3
23, 2, 3, rectangular
Section 10.7 9. 1, 3, 4, rectangular
7. 0, 5, 1, rectangular r 02 52 5
z1
2
r 22 22 22
arctan3
3
arctan1
1 3 2
4
z 4
z4
5, 2 , 1, cylindrical
22, 4, 4, cylindrical
2, 3 , 4, cylindrical
13. x2 y2 z2 10 rectangular equation
15. y x2
rectangular equation
r sin r cos
r z 10 cylindrical equation 2
11. 2, 2, 4, rectangular
2
r
5 0 2
arctan
Cylindrical and Spherical Coordinates
2
2
sin r cos2 r sec
19.
17. r 2 x2
y2
2 z
y 1 x 3 x 3 y
3 2 −3
x
3
r 2 2r sin
2
x2 y2 2y
1
x2 y2 2y 0
−2 x
2
cylindrical equation
21. r 2 sin
z
y tan 6 x
x2 y 2 4
2
6
tan
1
x2 y 12 1
2 y −2
z
x 3 y 0
2
2 3
−2
1
y −2
−3
1
2 x
−1 −2
23. r 2 z 2 4 x 2
y2
z2
25. 4, 0, 0, rectangular
z
4
42 02 02 4
2
arctan 0 0
1 −2 x
−2 2
1 −1
2
y
arccos 0
2
4, 0, 2 , spherical 27. 2, 23, 4, rectangular
22 23 2 42 42 arctan 3 arccos
1
4
2 3
2 2 42, , , spherical 3 4
29. 3, 1, 23 , rectangular
3 1 12 4 arctan
1 3
arccos
3
2
6
6
4, 6 , 6 , spherical
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2
y
253
254
Chapter 10
Vectors and the Geometry of Space
31.
4, 6 , 4 , spherical x 4 sin
cos 6 4 6
y 4 sin
sin 2 4 6
z 4 cos
22 4
33.
12, 4, 0, spherical x 12 sin 0 cos 0 4 y 12 sin 0 sin 0 4 z 12 cos 0 12
0, 0, 12, rectangular
6, 2, 22 , rectangular
5, 4 , 34, spherical
35.
x 5 sin
3 5 cos 4 4 2
y 5 sin
3 5 sin 4 4 2
z 5 cos
3 52 4 2
52, 52, 5 2 2 , rectangular
39. x2 y2 z2 36 rectangular equation
37. (a) Programs will vary. (b) x, y, z 3, 4, 2
2 36
spherical equation
, , 5.385, 0.927, 1.190 43. 2
41. x2 y2 9 rectangular equation
x 2
2 sin2 cos2 2 sin2 sin2 9
y2
z
z2
4
2
2 sin2 9
1 −2
sin 3
x
−2 2
1
2
−1
y
3 csc spherical equation
45.
6
47. 4 cos
z 2
z x2 y2 z2 3 z 2 x2 y2 z2 z2 3 2 4 x y2 z2
cos
x2 y2 z2
−2 −1 x
2
−1
4z x2 y2 z2
4 3
x2 y2 z2 4z 0
−2
−1
1
z 5
2
x2 y2 z 22 4
1 2
−2
y x
3
2
1
3x 2 3y 2 z 2 0 49. csc
51.
sin 1
4, 4 , 0, cylindrical 42 02 4
x2 y2 1
x2 y2 1
z
1 −2 x
−2 2
−1
1
2
4, 2 , 4, cylindrical 42 42 42
arccos 0
2
1
4
53.
2
4, , , spherical 4 2
y
−2
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2
4 4 2 4
arccos
42, 2 , 4 , spherical
2
3
−3
y
Section 10.7
55.
4, 6, 6, cylindrical
57. 12, , 5, cylindrical
42 62 213
2
13,
122
52
13
3
5 13
z 10 cos
3 , arccos , 6 13
36, , 2 , spherical r sin 36 sin
63.
36 2
6, 6 , 3 , spherical r 6 sin
z cos 36 cos
0 2
8, 76, 6 , spherical
65.
33 3
r 8 sin
6
z 6 cos
36, , 0, cylindrical
3 3
Spherical
7.810, 0.983, 1.177
69. 4.698, 1.710, 8
7.211, 0.983, 3 5, , 8 9
71. 7.071, 12.247, 14.142
14.142, 2.094, 14.142
20, 23, 4
73. 3, 2, 2
3.606, 0.588, 2
4.123, 0.588, 1.064
2.833, 0.490, 1.5
3.206, 0.490, 2.058
77. 3.536, 3.536, 5
5, 34, 5
7.071, 2.356, 2.356
79. 2.804, 2.095, 6
3.5, 2.5, 6
6.946, 5.642, 0.528
52, 43, 32
83 6 2
4, 76, 43, cylindrical
Cylindrical
67. 4, 6, 3
4 6
7 6
z 8 cos
33, 6 , 3, cylindrical
Rectangular
75.
0 2
10, 6 , 0, cylindrical
spherical
61.
6
13, , arccos 135 , spherical
13
10 2
r 10 sin
arccos
255
10, 6 , 2 , spherical
59.
6
arccos
Cylindrical and Spherical Coordinates
9.434, 0.349, 0.559
[Note: Use the cylindrical coordinates 3.5, 5.642, 6 83. 5
81. r 5
85. r 2 z, x 2 y 2 z
Cylinder
Sphere
Paraboloid
Matches graph (d)
Matches graph (c)
Matches graph (f)
87. Rectangular to cylindrical: r 2 x2 y2 tan
89. Rectangular to spherical: 2 x2 y2 z2
y x
tan
zz
arccos
Cylindrical to rectangular: x r cos y r sin
y x z x2 y2 z2
Spherical to rectangular: x sin cos y sin sin
zz
z cos
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256
Chapter 10
Vectors and the Geometry of Space
91. x2 y2 z2 16
93. x2 y2 z2 2z 0
(a) r 2 z 2 16
(a) r 2 z 2 2z 0, r 2 z 12 1
(b) 2 16, 4
(b) 2 2 cos 0, 2 cos 0,
2 cos 95. x2 y 2 4y
97. x2 y2 9
(a) r 2 4r sin , r 4 sin
(a) r 2 cos2 r 2 sin2 9,
(b) 2 sin2 4 sin sin ,
r2
sin sin 4 sin 0,
(b) 2 sin2 cos2 2 sin2 sin2 9,
4 sin , 4 sin csc sin
2 sin2 2
99. 0 ≤ ≤
9 cos2 sin2
2
9 , cos2 sin2
9 csc2 cos2 sin2
101. 0 ≤ ≤ 2
103. 0 ≤ ≤ 2
0 ≤ r ≤ a
0 ≤ r ≤ 2
r ≤ z ≤ a
0 ≤ z ≤ 4
6
0 ≤ ≤
0 ≤ ≤ a sec
z z
z a
5
a −a
3
30°
−a
2 1
a
x 2 3
2
3
a
y
x
y y
x
105. Rectangular
107. Spherical
z
0 ≤ x ≤ 10
z
4 ≤ ≤ 6
10
8
0 ≤ y ≤ 10 0 ≤ z ≤ 10
−8 8
10 10
y
x
y
x −8
109. z sin , r 1 z
y y y r 1
The curve of intersection is the ellipse formed by the intersection of the plane z y and the cylinder r 1.
Review Exercises for Chapter 10 1. P 1, 2, Q 4, 1, R 5, 4
3. v v cos i v sin j 8 cos 120 i 8 sin 120 j
\
(a) u PQ 3, 1 3i j,
4i 43j
\
v PR 4, 2 4i 2j (b) v 42 22 25 (c) 2u v 6, 2 4, 2 10, 0 10i
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Review Exercises for Chapter 10 5. 120 cos 100
arccos tan y
2 ft
56 y
2 2 ⇒ y y tan
120 lb 100 lb
θ
10 2 2 3.015 ft tan arccos5 6 11 5 11
7. z 0, y 4, x 5: 5, 4, 0
11. x 32 y 22 z 62
9. 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. The x-coordinate can be any number.
152
2
13. x2 4x 4 y 2 6y 9 z 2 4 4 9
15. v 4 2, 4 1, 7 3 2, 5, 10
x 22 y 32 z 2 9 Center: 2, 3, 0 Radius: 3
z
(2, − 1, 3) 3 2 1
z 5
4 3 2
6
5
4
4
3
−2
1 2 3 5
y
x
3
4 5 6
y
x
(4, 4, − 7)
17. v 1 3, 6 4, 9 1 4, 2, 10
19. Unit vector:
w 5 3, 3 4, 6 1 2, 1, 5
u 2 3 5 2, 3, 5 , , u 38 38 38 38
Since 2w v, the points lie in a straight line. 21. P 5, 0, 0, Q 4, 4, 0, R 2, 0, 6
23. u 7, 2, 3, v 1, 4, 5
\
(a) u PQ 1, 4, 0 i 4j,
Since u
v 0, the vectors are orthogonal.
\
v PR 3, 0, 6 3i 6k (b) u v 13 40 06 3 (c) v
v 9 36 45
3 3 52 i sin j
i j 4 4 2
2 2 i sin j i 3 j 3 3
25. u 5 cos v 2 cos u v
27. u 10, 5, 15, v 2, 1, 3 u 5v ⇒ u is parallel to v and in the opposite direction.
52 1 3 2
u 5 v 2 cos
257
u v 52 2 1 3 2 6 52
u v
arccos
2 6
4
4
15
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258
Chapter 10
Vectors and the Geometry of Space
29. There are many correct answers. For example: v ± 6, 5, 0.
<
>
<
>
<
>
In Exercises 31–39, u 3, 2, 1 , v 2, 4, 3 , w 1, 2, 2 . 31. u
u 33 22 11 14 14 u 2
uu w u
33. projuw
2
2
5 3, 2, 1 14
15 10 5 , , 14 14 14
15 5 5 , , 14 7 14
35. n v w n 5
i 2 1
j 4 2
k 3 2i j 2
v w 3, 2, 1 2, 1, 0 4 4
37. V u
1 n 2i j n 5 39. Area parallelogram u v 102 112 82 (See Exercises 36, 38) 285 41. F ccos 20 j sin 20 k
z
\
PQ 2k
i PQ F 0 0 \
\
PQ
j k 0 2 2c cos 20 i c cos 20 c sin 20
200 PQ F 2c cos 20
2 ft
70°
F
y
x
100 c cos 20 F
100 cos 20 j sin 20 k 100 j tan 20 k cos 20
F 1001 tan2 20 100 sec 20 106.4 lb 43. v j (a) x 1, y 2 t, z 3 (b) None
45. 3x 3y 7z 4, x y 2z 3 Solving simultaneously, we have z 1. Substituting z 1 into the second equation we have y x 1. Substituting for x in this equation we obtain two points on the line of intersection, 0, 1, 1, 1, 0, 1. The direction vector of the line of intersection is v i j. (a) x t, y 1 t, z 1 (b) x y 1, z 1
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Review Exercises for Chapter 10 49. Q 1, 0, 2
47. The two lines are parallel as they have the same direction numbers, 2, 1, 1. Therefore, a vector parallel to the plane is v 2i j k. A point on the first line 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
i v u 2 2
j 1 1
259
2x 3y 6z 6 A point P on the plane is 3, 0, 0. \
PQ 2, 0, 2 n 2, 3, 6
PQ n 8 D \
k 1 3
n
7
2i 4j 2i 2j. Equation of the plane: x 1 2y 0 x 2y 1 51. Q3, 2, 4 point
53. x 2y 3z 6 Plane
P5, 0, 0 point on plane
Intercepts: 6, 0, 0, 0, 3, 0, 0, 0, 2
n 2, 5, 1 normal to plane
z
\
PQ 2, 2, 4
PQ n \
D
n
3
(0, 0, 2)
30 10 30 3
3
(0, 3, 0)
6 x
1 55. y z 2
57.
(6, 0, 0)
x2 y2 z2 1 16 9 2
xy-trace:
z 2
2
y2 x2 1 16 9
xz-trace:
x2 z2 1 16
yz-trace:
y2 z2 1 9
y
6 x
y2 x2 z 2 1 16 9
z 2
Hyperboloid of two sheets x2 y2 xy-trace: 1 4 16
−2 5
5
y
x
xz-trace: None yz-trace:
z
Ellipsoid
Plane with rulings parallel to the x-axis
59.
y
y2 z2 1 9
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−4 4 x
5
−2
y
260
Chapter 10
Vectors and the Geometry of Space z
x2 y 2 rz 2
61. (a)
2z 1
4
2
3
x 2 y 2 2z 2 0 −2 1
2
2
3
y
x
2
(b) V 2
x 3
0
2
2
12 x
1
2
dx
y
3
1 2x x 3 dx 2
0
2 x 2
x4 8
2
2
1
0 x
4 12.6 cm3
2
(c) V 2
x 3
1 2 2
2
1 2
2
1
dx
3
1
2
3
3
x4 8
2
y
1 2x x 3 dx 2
2 x 2 4
12 x
1
2
1
2 1 2
x
31 225 11.04 cm 3 64 64
63. 22, 22, 2, rectangular (a) r
22 22 2 4, arctan1 3, z 2, 4, 3, 2 , cylindrical 4
(b) 22 22 22 25, 2
65.
2
67.
1002 502 505
5050 5 arccos 15 63.4
505, 6 , 63.4 , spherical
25, 34, arccos 55 , spherical
25, 4 , 34 , spherical r 2 25 sin
6
arccos
4
3 2 1 , arccos arccos , 4 5 25
100, 6 , 50 , cylindrical
2
34
2
⇒ r 25
2
2
2
4
z cos 25 cos
25
2
2 3 25 4 2
252 , , , cylindrical 4 2
69. x2 y 2 2z (a) Cylindrical: r 2 cos2 r 2 sin2 2z, r 2 cos 2 2z (b) Spherical: 2 sin2 cos2 2 sin2 sin2 2 cos , sin2 cos 2 2 cos 0, 2 sec 2 cos csc2
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256
Chapter 10
Vectors and the Geometry of Space
91. x2 y2 z2 16
93. x2 y2 z2 2z 0
(a) r 2 z 2 16
(a) r 2 z 2 2z 0, r 2 z 12 1
(b) 2 16, 4
(b) 2 2 cos 0, 2 cos 0,
2 cos 95. x2 y 2 4y
97. x2 y2 9
(a) r 2 4r sin , r 4 sin
(a) r 2 cos2 r 2 sin2 9,
(b) 2 sin2 4 sin sin ,
r2
sin sin 4 sin 0,
(b) 2 sin2 cos2 2 sin2 sin2 9,
4 sin , 4 sin csc sin
2 sin2 2
99. 0 ≤ ≤
9 cos2 sin2
2
9 , cos2 sin2
9 csc2 cos2 sin2
101. 0 ≤ ≤ 2
103. 0 ≤ ≤ 2
0 ≤ r ≤ a
0 ≤ r ≤ 2
r ≤ z ≤ a
0 ≤ z ≤ 4
6
0 ≤ ≤
0 ≤ ≤ a sec
z z
z a
5
a −a
3
30°
−a
2 1
a
x 2 3
2
3
a
y
x
y y
x
105. Rectangular
107. Spherical
z
0 ≤ x ≤ 10
z
4 ≤ ≤ 6
10
8
0 ≤ y ≤ 10 0 ≤ z ≤ 10
−8 8
10 10
y
x
y
x −8
109. z sin , r 1 z
y y y r 1
The curve of intersection is the ellipse formed by the intersection of the plane z y and the cylinder r 1.
Review Exercises for Chapter 10 1. P 1, 2, Q 4, 1, R 5, 4
3. v v cos i v sin j 8 cos 120 i 8 sin 120 j
\
(a) u PQ 3, 1 3i j,
4i 43j
\
v PR 4, 2 4i 2j (b) v 42 22 25 (c) 2u v 6, 2 4, 2 10, 0 10i
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Review Exercises for Chapter 10 5. 120 cos 100
arccos tan y
2 ft
56 y
2 2 ⇒ y y tan
120 lb 100 lb
θ
10 2 2 3.015 ft tan arccos5 6 11 5 11
7. z 0, y 4, x 5: 5, 4, 0
11. x 32 y 22 z 62
9. 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. The x-coordinate can be any number.
152
2
13. x2 4x 4 y 2 6y 9 z 2 4 4 9
15. v 4 2, 4 1, 7 3 2, 5, 10
x 22 y 32 z 2 9 Center: 2, 3, 0 Radius: 3
z
(2, − 1, 3) 3 2 1
z 5
4 3 2
6
5
4
4
3
−2
1 2 3 5
y
x
3
4 5 6
y
x
(4, 4, − 7)
17. v 1 3, 6 4, 9 1 4, 2, 10
19. Unit vector:
w 5 3, 3 4, 6 1 2, 1, 5
u 2 3 5 2, 3, 5 , , u 38 38 38 38
Since 2w v, the points lie in a straight line. 21. P 5, 0, 0, Q 4, 4, 0, R 2, 0, 6
23. u 7, 2, 3, v 1, 4, 5
\
(a) u PQ 1, 4, 0 i 4j,
Since u
v 0, the vectors are orthogonal.
\
v PR 3, 0, 6 3i 6k (b) u v 13 40 06 3 (c) v
v 9 36 45
3 3 52 i sin j
i j 4 4 2
2 2 i sin j i 3 j 3 3
25. u 5 cos v 2 cos u v
27. u 10, 5, 15, v 2, 1, 3 u 5v ⇒ u is parallel to v and in the opposite direction.
52 1 3 2
u 5 v 2 cos
257
u v 52 2 1 3 2 6 52
u v
arccos
2 6
4
4
15
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258
Chapter 10
Vectors and the Geometry of Space
29. There are many correct answers. For example: v ± 6, 5, 0.
<
>
<
>
<
>
In Exercises 31–39, u 3, 2, 1 , v 2, 4, 3 , w 1, 2, 2 . 31. u
u 33 22 11 14 14 u 2
uu w u
33. projuw
2
2
5 3, 2, 1 14
15 10 5 , , 14 14 14
15 5 5 , , 14 7 14
35. n v w n 5
i 2 1
j 4 2
k 3 2i j 2
v w 3, 2, 1 2, 1, 0 4 4
37. V u
1 n 2i j n 5 39. Area parallelogram u v 102 112 82 (See Exercises 36, 38) 285 41. F ccos 20 j sin 20 k
z
\
PQ 2k
i PQ F 0 0 \
\
PQ
j k 0 2 2c cos 20 i c cos 20 c sin 20
200 PQ F 2c cos 20
2 ft
70°
F
y
x
100 c cos 20 F
100 cos 20 j sin 20 k 100 j tan 20 k cos 20
F 1001 tan2 20 100 sec 20 106.4 lb 43. v j (a) x 1, y 2 t, z 3 (b) None
45. 3x 3y 7z 4, x y 2z 3 Solving simultaneously, we have z 1. Substituting z 1 into the second equation we have y x 1. Substituting for x in this equation we obtain two points on the line of intersection, 0, 1, 1, 1, 0, 1. The direction vector of the line of intersection is v i j. (a) x t, y 1 t, z 1 (b) x y 1, z 1
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Review Exercises for Chapter 10 49. Q 1, 0, 2
47. The two lines are parallel as they have the same direction numbers, 2, 1, 1. Therefore, a vector parallel to the plane is v 2i j k. A point on the first line 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
i v u 2 2
j 1 1
259
2x 3y 6z 6 A point P on the plane is 3, 0, 0. \
PQ 2, 0, 2 n 2, 3, 6
PQ n 8 D \
k 1 3
n
7
2i 4j 2i 2j. Equation of the plane: x 1 2y 0 x 2y 1 51. Q3, 2, 4 point
53. x 2y 3z 6 Plane
P5, 0, 0 point on plane
Intercepts: 6, 0, 0, 0, 3, 0, 0, 0, 2
n 2, 5, 1 normal to plane
z
\
PQ 2, 2, 4
PQ n \
D
n
3
(0, 0, 2)
30 10 30 3
3
(0, 3, 0)
6 x
1 55. y z 2
57.
(6, 0, 0)
x2 y2 z2 1 16 9 2
xy-trace:
z 2
2
y2 x2 1 16 9
xz-trace:
x2 z2 1 16
yz-trace:
y2 z2 1 9
y
6 x
y2 x2 z 2 1 16 9
z 2
Hyperboloid of two sheets x2 y2 xy-trace: 1 4 16
−2 5
5
y
x
xz-trace: None yz-trace:
z
Ellipsoid
Plane with rulings parallel to the x-axis
59.
y
y2 z2 1 9
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−4 4 x
5
−2
y
260
Chapter 10
Vectors and the Geometry of Space z
x2 y 2 rz 2
61. (a)
2z 1
4
2
3
x 2 y 2 2z 2 0 −2 1
2
2
3
y
x
2
(b) V 2
x 3
0
2
2
12 x
1
2
dx
y
3
1 2x x 3 dx 2
0
2 x 2
x4 8
2
2
1
0 x
4 12.6 cm3
2
(c) V 2
x 3
1 2 2
2
1 2
2
1
dx
3
1
2
3
3
x4 8
2
y
1 2x x 3 dx 2
2 x 2 4
12 x
1
2
1
2 1 2
x
31 225 11.04 cm 3 64 64
63. 22, 22, 2, rectangular (a) r
22 22 2 4, arctan1 3, z 2, 4, 3, 2 , cylindrical 4
(b) 22 22 22 25, 2
65.
2
67.
1002 502 505
5050 5 arccos 15 63.4
505, 6 , 63.4 , spherical
25, 34, arccos 55 , spherical
25, 4 , 34 , spherical r 2 25 sin
6
arccos
4
3 2 1 , arccos arccos , 4 5 25
100, 6 , 50 , cylindrical
2
34
2
⇒ r 25
2
2
2
4
z cos 25 cos
25
2
2 3 25 4 2
252 , , , cylindrical 4 2
69. x2 y 2 2z (a) Cylindrical: r 2 cos2 r 2 sin2 2z, r 2 cos 2 2z (b) Spherical: 2 sin2 cos2 2 sin2 sin2 2 cos , sin2 cos 2 2 cos 0, 2 sec 2 cos csc2
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Problem Solving for Chapter 10
261
Problem Solving for Chapter 10 abc0
1.
3. Label the figure as indicated.
b a b c 0
a
c
From the figure, you see that
b a b c 0
1 1 SP a b RQ and 2 2 \
b
a b b c
\
1 1 SR a b PQ . 2 2 \
b c b c sin A a b a b sin C
\
\
\
\
\
Since SP RQ and SR PQ , PSRQ is a parallelogram.
Then, Q
sin A b c a a b c
a
a b a b c 1 2
sin C . c The other case,
P
R
a− 1b S
1 2
b
2
z 6 5 4 3
\
2
sin A sin B is similar. a b
5. (a) u 0, 1, 1 direction vector of line determined by P1 and P2. D
a+ 1b
P1Q u u
P2
P1
2, 0, 1 0, 1, 1 2
1 4
3
2
Q
2 3 4
y
x
1, 2, 2 3 32 2 2 2
(b) The shortest distance to the line segment is P1Q 2, 0, 1 5.
1
7. (a) V
0
2
z2
2 dz
2 1 0
1 2
9. (a) 2 sin
1 1 1 Note: basealtitude 1 2 2 2 (b)
z
Torus
2 −3
x2 y2 z: (slice at z c) a2 b2
3
x2 y2 1 2 ca cb2
(b) 2 cos
At z c, figure is ellipse of area
V
0
abc dc
3
y
z
−3
−2 1
abc2 k 2
y
Sphere
cacb abc. k
3 −2
x
abk2 2 0
1 1 (c) V abkk baseheight 2 2 11. From Exercise 64, Section 10.4, u v w z u v z w u v w z.
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1
2 3 x
−2
2
262
Chapter 10
Vectors and the Geometry of Space
13. (a) u ucos 0 i sin 0 j u i
(d)
2.5
Downward force w j
T
T Tcos90 i sin90 j
u
Tsin i cos j
0
60 0
0 u w T u i j Tsin i cos j
(e) Both are increasing functions.
u sin T
(f)
lim
→ 2
T and
lim
→ 2
u .
1 cos T If 30 , u 12T and 1 32T ⇒ T
2 1.1547 lb 3
and u
1 2 0.5774 lb 2 3
(b) From part (a), u tan and T sec . Domain: 0 ≤ ≤ 90 (c)
0
10
20
30
40
50
60
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
15. Let , the angle between u and v. Then sin
u v v u . u v u v
For u cos , sin , 0 and v cos , sin , 0, u v 1 and
i j v u cos sin cos sin
k 0 sin cos cos sin k. 0
Thus, sin v u sin cos cos sin . 17. From Theorem 10.13 and Theorem 10.7 (6) we have PQ n D \
n
w u v u v w u v w. u v
u v
u v
19. a1, b1, c1, and a2, b2, c2 are two sets of direction numbers for the same line. The line is parallel to both u a1i b1j c1k and v a 2 i b2 j c2 k. Therefore, u and v are parallel, and there exists a scalar d such that u dv, a1i b1 j c1k da2i b2 j c2k, a1 a2d, b1 b2d, c1 c2d.
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C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . 268 Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . 273 Section 11.3 Velocity and Acceleration
. . . . . . . . . . . . . . . . . . . . 278
Section 11.4 Tangent Vectors and Normal Vectors . . . . . . . . . . . . . . . 283 Section 11.5 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . 289 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
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C H A P T E R 11 Vector-Valued Functions Section 11.1
Vector-Valued Functions
Solutions to Even-Numbered Exercises 2. rt 4 t 2 i t 2j 6t k
4. rt sin t i 4 cos tj t k
Component functions: f t 4 t
Component functions: f t sin t
2
gt t 2
gt 4 cos t
ht 6t
ht t
Domain: 2, 2
Domain: ,
6. rt Ft Gt ln t i 5t j 3t 2k i 4t j 3t 2k ln t 1i 5t 4tj 3t 2 3t 2k ln t 1i t j Domain: 0,
j
k
t 1 t1
t
i
3 8. rt Ft Gt t 3 t
t2
Domain: , 1, 1,
tt 2
10. rt cos t i 2 sin tj (a) r0 i (b) r
4
2
2
i 2 j
(c) r cos i 2 sin j cos i 2 sin j (d) r
6 t r6 cos6 t i 2 sin6 t j cos6 i 2 sin 6 j
12. rt t i t3 2j et 4k (a) r0 k (b) r4 2i 8j e1k (c) rc 2 c 2i c 23 2j ec2 4k (d) r9 t r9 9 t i 9 t3 2j e9t 4k 3i 27j e9 4k 9 t 3i 9 t3 2 27j e9t 4 e9 4k 14.
t t3 3 t j 3 t k i t 3t 2 t t t1 t1
rt t i 3t j 4t k rt t 2 3t2 4t2 t 9t2 16t2 t1 25t
268
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Section 11.1
Vector-Valued Functions
269
16. rt ut 3 cos t4 sin t 2 sin t6 cos t t 2t 2 t 3 2t 2, a scalar. The dot product is a scalar-valued function. 18. rt cos t i sin t j t 2k, 1 ≤ t ≤ 1
20. rt t i ln tj
2t k, 0.1 ≤ t ≤ 5 3
x cos t, y sin t, z t 2 x t, y ln t, z
Thus, x2 y 2 1. Matches (c)
2t 3
2 Thus, z 3 x and y ln x. Matches (a)
22. rt ti tj 2k x t, y t, z 2 ⇒ x y (a) 0, 0, 20
(c) 5, 5, 5
(b) 10, 0, 0
z
z y 1
2
3 3
−3
2
1 −3
−2
2
−2
1
−1
−1
2
1
2
3
3 y
1
x
2
26. x t 2 t, y t 2 t
24. x 1 t, y t y 1 x
x 2 cos t
28.
y 2 sin t
y
x2
5
Domain: t ≥ 0
y
−3
x
3
3
−2
y2 4
4 y
y
3 2
6 5
1
4 −1
3 2
1
x 1
−1
2
3
4
5
1
2
3
1 −1
x
−4 −3 −2 −1
x
−1
4
−2
30. x 2 cos3 t, y 2 sin3 t
2x
2 3
2y
2 3
cos2 t sin2 t 1
34. x 3 cos t, y 4 sin t, z
32. x t y 2t 5
y2 x2 1 9 16
y 3t Line passing through the points:
x2 3 y2 3 22 3
0, 5, 0,
y 3
−3
2
z 4
4
x
−2
Elliptic helix
6
(0, −5, 0)
t 2
z
( 25 , 0, 152 )
2
52, 0, 152
z
−6
3
−4
2 −2
−2 2
−3
4 x
6
−6
2 4
y
−4 −6
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4 x
4
y
t 2
270
Chapter 11
Vector-Valued Functions
3 36. x t2, y 2t, z t 2
38. x cos t t sin t y sin t t cos t
3 y2 x ,z y 4 4
zt
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
x2 y 2 1 t 2 1 z 2 or x2 y 2 z 2 1 zt Helix along a hyperboloid of one sheet z 4
z 3
3
−4 −3 −2 −1
2
2
3
y
4
x
2 1 1
2
−1
3
y
4
−2 −3
5 x
40. rt t i
3
2
1 t 2 j t 2k 2
42. rt 2 sin t i 2 cos tj 2 sin tk z
Ellipse
z
Parabola
2 1 1 −1 1 2 x
3
−2
−3
−1
−1 1
1
2
2 −2
3
1
−1
y
x y
−2
−3
1 44. rt t i t2 j t3h 2
(a) ut r t 2j is a translation 2 units to the left along the y-axis.
z z
1 (b) u t t2i t j t3k has the roles 2 of x and y interchanged. The graph is a reflection in the plane x y.
5 5
4
z
4
3 2 1 1
1 2 3 4 5
2 3
−2 y
3
5
2
4 3
1
1 2 3
2
2 y −2
3
4 4
5
1
1 2 3
2
5
x
y
3
x
4 5 x
(c) ut r t 4k is an upward shift 4 units. z
1 (d) u t ti t2 j t3k shrinks the 8 z-value by a factor of 4. The curve rises more slowly.
(e) ut r t reverses the orientation. z
z
5
5
4
1 2 3
4
3
5
3
2
4
2
1
3
1
2 1 2 3 4 5
1
1
y
2 3
1
4
4
2
5
3
5
5
y
4
x 5 x
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x
1 2 3
5
y
Section 11.1
Vector-Valued Functions
48. y 4 x2
46. 2x 3y 5 0
Let x t, then y 4 t 2.
1 Let x t, then y 2t 5. 3
rt ti 4 t 2 j
1 rt ti 2t 5j 3
50. x 22 y2 4
52.
Let x 2 2 cos t, y 2 sin t.
x2 y2 1 16 9 Let x 4 cos t, y 3 sin t.
rt 2 2 cos ti 2 sin t j
rt 4 cos ti 3 sin tj
54. One possible answer is rt 1.5 cos t i 1.5 sin tj
1 tk, 0 ≤ t ≤ 2
Note that r2 1.5i 2k. 56. r1t ti,
0 ≤ t ≤ 10 r10 0, r110 10i
r2t 10cos t i sin tj,
0 ≤ t ≤
r3t 521 ti 521 tj,
4
r 0 10i, r 4 52i 52 j 2
2
0 ≤ t ≤ 1 r30 52i 52 j, r31 0
(Other answers possible) 58. r1t ti t j,
0 ≤ t ≤ 1 y x
r2t 1 ti 1 tj,
0 ≤ t ≤ 1 y x
(Other answers possible) 60. z x2 y 2, z 4
z
Therefore, x2 y2 4 or
6
x 2 cos t, y 2 sin t, z 4. rt 2 cos ti 2sin tj 4k 2
2
y
x
62. 4x2 4y2 z2 16, x z2
z 4
1 If z t, then x t2 and y 16 4t4 t2. 2 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
2 4 x
1 rt t2i 16 4t4 t2j tk 2
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2
y
271
272
Chapter 11
Vector-Valued Functions
64. x2 y 2 z 2 10, x y 4 Let x 2 sin t, then y 2 sin t
2
t
x
1
y
3
z
z 21 sin2 t 2 cos t. z
0
6
2
3 2
2
5 2
3
2
5 2
2
3 2
1
2
6
6
0
and
2
2
4
y
4
4 x
6
0
2
2
rt 2 sin ti 2 sin tj 2 cos tk 66. x2 y 2 z 2 16, xy 4 (first octant)
z 4
Let x t, then y
4 t
x2 y 2 z2 t 2
and
16 z 2 16. t2
1 z t 4 16t 2 16 t
4
y
4
x
8 43 ≤ t ≤ 8 43 t
8 43
1.5
2
2.5
3.0
3.5
8 43
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
4 1 rt ti j t 4 16t 2 16k t t 68. x2 y2 et cos t2 et sin t2 e2t z2
70. lim et i t→0
sin t j et k i j k t
z
since
160 120
lim
80
t→0
40 40 120 80
40
x
72. lim t i t→1
80 120
(L’Hôpital’s Rule)
y
ln t 1 j 2t 2 k i j 2k t2 1 2
since lim
sin t cos t lim 1 t→0 1 t
1 t 74. lim et i j 2 k 0 t→ t t 1 since
ln t 1 t 1 lim . (L’Hôpital’s Rule) 1 t→1 2t 2
t→1 t 2
lim et 0, lim
t→
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t→
1 t 0, and lim 2 0. t→ t 1 t
Section 11.2
Differentiation and Integration of Vector-Valued Functions 78. rt 2et, et, lnt 1
76. rt t i t 1 j
Continuous on t 1 > 0 or t > 1: 1, .
Continuous on 1, 3 t
80. rt 8, t,
82. No. The graph is the same because rt ut 2. For example, if r0 is on the graph of r, then u2 is the same point.
Continuous on 0, 84. A vector-valued function r is continuous at t a if the limit of rt exists as t → a and lim rt ra. t→a
The function rt
ii jj
t ≥ 0 is not continuous at t 0. t < 0
86. Let rt x1ti y1tj z1tk and ut x2ti y2tj z2tk. Then: lim rt ut lim x1tx2t y1ty2t z1tz2t t→c
t→c
lim x1t lim x2t lim y1t lim y2t lim z1t lim z2t t→c
t→c
t→c
t→c
t→c
t→c
lim x1ti lim y1tj lim z1tk lim x2ti lim y2tj lim z2tk t→c
t→c
lim rt lim ut t→c
t→c
t→c
t→c
t→c
t→c
90. False. The graph of x y z t 3 represents a line.
88. Let f t
1,1,
if t ≥ 0 if t < 0
and rt f ti. Then r is not continuous at c 0, whereas, r 1 is continuous for all t.
Section 11.2
Differentiation and Integration of Vector-Valued Functions
2. rt ti t 3 j, t0 1 xt t, yt t 3
1 4. rt t 2i j, t0 2 t xt t 2, yt
y x3 r1 i j
x
rt i 3t 2j rt0 is tangent to the curve. y 4 3
r
x
−4 −3 −2
1 −2
2
3
y
4 2
r
−3 −4
1 y2
rt0 is tangent to the curve.
r' (1, 1)
1
1 t
1 r2 4i j 2 1 rt 2t i 2 j t 1 r2 4 i j 4
r1 i 3j
2
273
1
r' x 8 −1
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274
Chapter 11
Vector-Valued Functions
3 8. rt ti t 2 j k, t0 2 2
6. rt t i 4 t 2j (a)
y
y x 2, z
5
r(1)
rt i 2tj
3
r(1.25)
2
r (1.25) − r (1)
1
3 r2 2 i 4 j k 2
x −3
−1 −1
3
r2 i 4j
r1 i 3j
(b)
z
r1.25 1.25i 2.4375j
2 −2
r1.25 r1 0.25i 0.5625j
−4 −2
2
rt i 2tj
(c)
3 2
x
4
r
−6
−4
r 1 i 2j
y
r' −6
r1.25 r1 0.25i 0.5625j i 2.25j 1.25 1 0.25 This vector approximates r1. t2 1 10. rt i 16tj k t 2
12. rt 4 t i t2 t j ln t2 k rt
1 rt 2 i 16j tk t 14. rt sin t t cos t, cos t t sin t, t 2
rt t sin t, t cos t, 2t
t
i 2t t
t2 2 t
j 2t k
16. rt arcsin t, arccos t, 0
rt
18. rt t2 ti t2 tj
2
1 1 t , 1 1 t , 0 2
2
20. rt 8 cos ti 3 sin tj
(a) rt 2t 1i 2t 1 j
(a) rt 8 sin ti 3 cos tj
r t 2 i 2j
r t 8 cos t i 3 sin tj
(b) rt rt 2t 12 2t 12 8t
(b) rt rt 8 sin t8 cos t 3 cos t 3 sin t 55 sin t cos t
22. rt ti 2t 3j 3t 5k (a) rt i 2j 3k r t 0 (b) rt rt 0
24. rt et, t2, tant
(a) rt et, 2t, sec2 t
r t et, 2, 2 sec2 t tan t
(b) rt rt e2t 4t 2 sec4 t tan t
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Section 11.2 rt t i t 2j e0.75t k, t0
26.
Differentiation and Integration of Vector-Valued Functions
1 4
rt i 2tj 0.75e0.75t k r
0.1875 k
1 r 4
( 1) 1 r'' ( 4 ) r'' 4
−2
1 3 i j e316k 2 4
−2 −1 1 2
−1
1 2 y
x
1 12 2
2
−1
14 i 12 j 0.75e 2
z
( 1) 1 r' ( 4 ) r' 4
3 e316 4
2
275
−2
20 9e 5 9 e38 4 16 4
38
1 r14 4i 2j 3e316k r1 4 20 9e38 rt 2i r
9 0.75t e k 16
14 2 i 169 e
316k
14 2 169 e r
2
316
2
81 e 4 256
38
1024 81e38
16
r14 1 32j 9e316k r14 1024 81e38
28.
rt
1 i 3tj t1
rt
r sin i 1 cos j
30.
r 1 cos i sin j
1 i 3j t 12
r2n 1 0, n any integer Smooth on 2n 1 , 2n 1
Not continuous when t 1 Smooth on , 1, 1, 32. rt rt
2t 2t2 i j 3 8t 8 t3
34. rt et i et j 3tk rt et i et j 3k 0
16 4t3 32t 2t 4 i 3 j 3 2 t 8 t 82
1 36. rt t i t 2 1 j tk 4 rt
r is smooth for all t: ,
rt 0 for any value of t.
1 1 i 2tj k 0 4 2 t
r is smooth for all t > 0: 0,
r is not continuous when t 2. Smooth on , 2, 2, . 38. rt ti 2 sin tj 2 cos tk 1 ut i 2 sin tj 2 cos tk t (a) rt i 2 cos tj 2 sin tk
(b) rt 2 sin tj 2 cos tk
(c) rt ut 1 4 sin2 t 4 cos2 t 5
(d) 3rt ut 3t
Dtrt ut 0, t 0
1 i 4 sin tj 4 cos tk t
Dt3rt ut 3
—CONTINUED—
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1 i 4 cos tj 4 sin tk t2
276
Chapter 11
Vector-Valued Functions
38. —CONTINUED— (e) rt ut
i t 1t
j k 2 sin t 2 cos t 2 sin t 2 cos t
2 cos t
1t t j 2 sin t t 1t k
Dtrt ut 2 sin t
1t t 2 cos t t1 1 j 2
2 cos t t
1 1 2 sin t 1 2 t t
k
(f) rt t2 4 1 t Dtrt t 2 4122t t 2 4 2 rt t 2 i tj
40.
1.0
rt 2t i j rt rt 2t 3 t rt t 4 t 2, rt 4t 2 1 cos
2t 3
0
−0.5
t
t 4 t 2 4t 2 1
arccos
t 4
2t 3 t t 2 4t 2 1
22 .
0.340 19.47maximum at t 0.707
for any t. 2
42. rt lim
t→0
lim
rt t rt t
t t i t 3 t j 2t tk t i 3t j 2tk t
t→0
3 3 t t t t t t lim i j 2k t→0 t t
lim
t
lim
t→0
t→0
44.
t t t t
i
3t j 2k t ttt
1 3 i j 2k t tt t t t
3 1 i 2 j 2k t 2 t
8 4t 3 i 6tj 4 tk dt t 4 i 3t 2 j t 32 k C 3
46.
1 ln t i j k dt t ln t ti ln tj t k C t
(Integration by parts)
48.
et i sin tj cos tk dt et i cos tj sin tk C
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Section 11.2
50.
52.
1
3 t k dt t i t 3j
2
54.
ti etj tetk dt
0
t2 i
1
2
1
t4 j 4
1 1
34t k
1
43
1
2 i e j t 1e k t2
277
et et sin t cos ti cos t sin t j C 2 2
et sin t i et cos t j dt
1
Differentiation and Integration of Vector-Valued Functions
2
2
2
t
t
0
0
0
0
56. rt
2i e2 1j e2 1k
3t 2 j 6 tk dt t 3 j 4t 32 k C
r0 C i 2j rt i 2 t 3j 4t32k
58. rt 4 cos ti 3 sin tk rt 4 sin t i 3 cos tk C1 r0 3k 3k C1 ⇒ C1 0 rt 4 cos t i 3 sin t k C2 r0 4i C2 4j ⇒ C2 4j 4i rt 4 cos t 4i 4 j 3 sin tk
60. rt r1
1 1 1 1 i 2 j k dt arctan t i j ln t k C 1 t2 t t t
i j C 2i ⇒ C 2 ij 4 4
rt 2
1 arctan t i 1 j ln tk 4 t
62. To find the integral of a vector-valued function, you integrate each component function separately. The constant of integration C is a constant vector.
64. The graph of ut does not change position relative to the xy-plane.
66. Let rt x1ti y1tj z1tk and ut x2ti y2tj z2tk. rt ± ut x1t ± x2ti y1t ± y2t j z1t ± z2t k Dtrt ± ut x1t ± x2t i y1t ± y2t j z1t ± z2t k x1ti y1tj z1tk ± x2ti y2tj z2tk rt ± ut 68. Let rt x1ti y1tj z1tk and ut x2ti y2tj z2tk. rt ut y1tz2t z1ty2t i x1tz2t z1tx2t j x1ty2t y1tx2t k Dtrt ut y1tz2t y1tz2t z1ty2t z1ty2t i x1tz2t x1tz2t z1tx2t z1tx2t j
x1ty2t x1ty2t y1tx2t y1tx2t k y1tz2t z1ty2ti x1tz2t z1tx2t j x1ty2t y1tx2tk
y1tz2t z1ty2ti x1tz2t z1tx2t j x1ty2t y1tx2t k rt ut rt ut
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278
Chapter 11
Vector-Valued Functions
70. Let rt xti ytj ztk. Then rt xti ytj ztk. rt rt ytzt ztyti xtzt ztxt j xtyt ytxtk Dtrt rt ytzt ytzt ztyt ztyti xtzt xtzt ztxt ztxt j
xtyt xtyt ytxt ytxtk ytzt ztyti xtzt ztxt j xtyt ytxtk rt rt 72. Let rt xti ytj ztk. If rt rt is constant, then:
74. False Dtrt ut rt ut rt ut
x2t y 2t z 2t C
(See Theorem 11.2, part 4)
Dtx2t y 2t z2t DtC 2xtxt 2ytyt 2ztzt 0 2xtxt ytyt ztzt 0 2rt rt 0 Therefore, rt rt 0.
Section 11.3
Velocity and Acceleration
2. rt 6 t i t j
4. rt t2i t3j
y
y 8 7 6 5 4 3 2 1
vt rt 2ti 3t2j
vt rt i j at rt 0
at rt 2i 6tj
4
v
x 6 t, y t, y 6 x
(3, 3)
x
2
x y23
x t2, y t3
2
4
At 1, 1, t 1.
a
v
(1, 1) x
−1
v1 2i 3j
2 3 4 5 6 7
8
a1 2i 6j y
8. rt et, et
6. rt 3 cos ti 2 sin tj vt 3 sin ti 2 cos tj
vt rt et, et
at 3 cos ti 2 sin tj
at rt
et,
2
et 1
x 3 cos t, y 2 sin t,
x2 y2 1 Ellipse 9 4
At 3, 0, t 0.
At 1, 1, t 0. v0 1, 1 i j
3
a0 3i
1 −2
−1
−1
a0 1, 1 i j
v a 1
x 2
a
1 1 x et t, y et, y e x
y
v0 2j
v (1, 1)
(3, 0)
−3
10. rt 4t i 4tj 2t k vt 4i 4j 2k st vt 16 16 4 6 at 0
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x 1
2
Section 11.3 1 12. rt 3ti tj t2k 4
279
14. rt t 2 i tj 2t 32k vt 2ti j 3 t k
1 vt 3i j tk 2 st
Velocity and Acceleration
st 4t 2 1 9t 4t 2 9t 1
9 1 41t 10 41t 2
at 2i
2
3 k 2 t
1 at k 2 16. rt et cos t, et sin t, et
18. (a)
vt et cos t et sin ti et sin t et cos tj et k
rt t, 25 t 2, 25 t 2 , t0 3
rt 1,
st e2tcos t sin t2 e2tcos t sin t2 e2t
t
3 3 r3 1, , 4 4
et 3 at 2et sin ti 2et cos tj et k
,
t
25 t 2 25 t 2
3 x 3 t, y z 4 t 4
3 3 (b) r3 0.1 3 0.1, 4 0.1, 4 0.1 4 4 3.100, 3.925, 3.925 20. at 2i 3k vt
22. at cos t i sin tj, v0 j k, r0 i
2i 3k dt 2ti 3tk C
vt
v0 C 4j ⇒ vt 2ti 4j 3tk rt
cos t i sin t j dt sin t i cos t j C
v0 j C j k ⇒ C k
3 2ti 4j 3tk dt t2i 4tj t2k C 2
vt sin t i cos t j k rt
3 r0 C 0 ⇒ rt t2i 4tj t2k 2
sin t i cos tj k dt
cos ti sin tj tk C
r2 4i 8j 6k
r0 i C i ⇒ C 0 rt cos t i sin tj tk r2 cos 2i sin 2j 2k
24. (a) The speed is increasing. (b) The speed is decreasing. 26. rt 900 cos 45 t i 3 900 sin 45 t 16t 2 j 450 2t i 3 450 2t 16t 2j The maximum height occurs when yt 450 2 32t 0, which implies that t 225 2 16. The maximum height reached by the projectile is y 3 450 2
22516 2 1622516 2
2
50,649 6331.125 feet. 8
The range is determined by setting yt 3 450 2t 16t 2 0 which implies that t
450 2 405,192 39.779 seconds. 32
Range: x 450 2
450
2 405,192
32
25,315.500 feet
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280
Chapter 11
28. 50 mph rt
Vector-Valued Functions
220 ft/sec 3 220 cos 15 t i 5 sin 15 t 16t j 220 3 3 2
The ball is 90 feet from where it is thrown when x
27 220 cos 15 t 90 ⇒ t 1.2706 seconds. 3 22 cos 15
The height of the ball at this time is y5
27 27 sin 15 16 220 3 22 cos 15 22 cos 15
2
3.286 feet.
30. y x 0.005x 2 From Exercise 34 we know that tan is the coefficient of x. Therefore, tan 1, 4 rad 45 . Also 16 sec2 negative of coefficient of x 2 v02 16 2 0.005 or v0 80 ft/sec v02 rt 40 2ti 40 2t 16t 2j. Position function. When 40 2t 60, t
60 3 2 4 40 2
vt 40 2i 40 2 32tj v
3 4 2 40 2 i 40 2 24 2 j 8 25i 2j
Speed
direction
v 3 4 2 8
32. Wind: 8 mph
2 25 4 8 58 ftsec
176 ftsec 15
rt 140cos 22 t
176 i 2.5 140 sin 22 t 16t2j 15
50
0
450 0
When x 375, t 2.98 and y 16.7 feet. Thus, the ball clears the 10-foot fence.
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Section 11.3
Velocity and Acceleration
281
34. h 7 feet, 35 , 30 yards 90 feet rt v0 cos 35 ti 7 v0 sin 35 t 16t 2 j (a) v0 cos 35 t 90 when 7 v0 sin 35 t 16t 2 4 t 7 v0 sin 35
v
90 90 16 v0 cos 35 0 cos 35
2
90 v0 cos 35
4
90 tan 35 3 v02
129,600 v02 cos2 35 129,600 cos2 35 90 tan 35 3
v0 54.088 feet per second (c) xt 90 ⇒ v0 cos 35 t 90
(b) The maximum height occurs when yt v0 sin 35 32t 0. t
90 2.0 seconds 54.088 cos 35
t
v0 sin 35 0.969 second 32
At this time, the height is y0.969 22.0 feet. 36. Place the origin directly below the plane. Then 0, v0 792 and
α
rt v0 cos ti 30,000 v0 sin t 16t2j 792ti 30,000 16t2j
30,000
vt 792i 32tj.
α
At time of impact, 30,000
16t2
0 ⇒
t2
1875 ⇒ t 43.3 seconds.
(0, 0)
34,295
r43.3 34,294.6i v43.3 792i 1385.6j v43.3 1596 ft sec 1088 mph tan
30,000 0.8748 ⇒ 0.718741.18 34,294.6
38. From Exercise 37, the range is x
v02 sin 2 . 32
Hence, x 150
v02 4800 sin24 ⇒ v02 ⇒ v0 108.6 ftsec. 32 sin 24
40. (a) rt tv0 cos i tv0 sin 16t 2j tv0 sin 16t 0 when t Range:
x v0 cos
v0 sin
. 16
v 32sin v32 sin 2
0
The range will be maximum when
dx v2 0 2 cos 2 0 dt 32
2
0
(b) yt tv0 sin 16t 2 dy v sin
v0 sin 32t 0 when t 0 . dt 32 Maximum height: y
v 32sin v 0
sin2
v 2 sin2 v 2 sin2
16 0 2 0 32 32 64
Minimum height when sin 1, or
or 2
2
0
, rad. 2 4
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. 2
282
Chapter 11
Vector-Valued Functions
42. rt v0 cos t i h v0 sin t 4.9t 2 j v0 cos 8 t i v0 sin 8 t 4.9t 2 j x 50 when v0 cos 8 t 50 ⇒ t
v0 sin 8
v
50 . For this value of t, y 0: v0 cos 8
50 50 4.9 v0 cos 8 0 cos 8
2
0
50 tan 8
4.92500 4.950 ⇒ v02 1777.698 v02 cos2 8 tan 8 cos2 8 ⇒ v0 42.2 msec
44. rt bt sin t i b1 cos t j vt b1 cos t i sin tj Speed vt 2b 1 cos t and has a maximum value of 2b when t , 3 , . . . . 55 mph 80.67 ftsec 80.67 radsec since since b 1 Therefore, the maximum speed of a point on the tire is twice the speed of the car: 280.67 ftsec 110 mph 46. (a) Speed v b22 sin2t b22 cos2t
(b)
10
b22sin2t cos2t b −10
10
−10
The graphing utility draws the circle faster for greater values of . 48. at b2 costi sintj b2 50. vt 30 mph 44 ftsec
605
vt 44 radsec b 300
n
at b2
3000
44 3000 300 F mb 32 300 2
θ 2
605 lb
Let n be normal to the road. n cos 3000 n sin 605 Dividing the second equation by the first: tan
605 3000
arctan
605 3000 11.4 .
52. h 6 feet, v0 45 feet per second, 42.5 . From Exercise 47, t
45 sin 42.5 452 sin2 42.5 2326 2.08 seconds. 32
At this time, xt 69.02 feet.
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Section 11.4 54. rt xti ytj
Tangent Vectors and Normal Vectors
283
56. r1t xti ytj ztk
yt mxt b, m and b are constants.
r2t r12t
rt xti mxt b j
Velocity: r2t 2r12t
vt xti mxtj
Acceleration: r2t 4r12t
st xt2 mxt2 C, C is a constant.
In general, if r3t r1t, then: Velocity: r3t r1t
C Thus, xt 1 m2
Acceleration: r3t 2r1t
xt 0 at xti mxtj 0.
Section 11.4 2.
Tangent Vectors and Normal Vectors
rt t3i 2t2j
rt 6 cos ti 2 sin tj
4.
rt 3t2i 4tj
rt 6 sin ti 2 cos tj
rt 9t 16t 4
Tt T1
rt 36 sin2 t 4 cos2 t
2
rt 1 3t2i 4tj rt 9t4 16t2 1 9 16
Tt
3 4 3i 4j i j 5 5
T
4 6. rt t2i tj k 3
1 j 33i j 3 3633 43i 1 4 28
8. rt t, t, 4 t 2
rt 2ti j
rt 1, 1,
When t 1, rt r1 2i j T1
rt 6 sin ti 2 cos tj rt 36 sin2 t 4 cos2 t
4 t 1 at 1, 1, 3
.
When t 1, r1 1, 1,
r1 2i j 5 2i j 5 r1 5
T1
Direction numbers: a 2, b 1, c 0 4 Parametric equations: x 2t 1, y t 1, z 3
t 4 t 2
T
1 3
Parametric equations: x t 1, y t 1,
rt 2 cos t, 2 sin t, 8 sin t cos t
Direction numbers: a 1, b 1, c
10. rt 2 sin t, 2 cos t, 4 sin2 t
, r 3, 1, 23 , 6 6
, t 1 at 1, 1, 3 .
21 r1 1 1, 1, r1 7 3
z
When t
1 3
t 6 at 1, 3, 1 .
6 rr 6 6 41 3, 1, 23
Direction numbers: a 3, b 1, c 23 Parametric equations: x 3t 1, y t 3, z 23 t 1
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1 3
t 3
284
Chapter 11
Vector-Valued Functions
1 12. rt 3 cos ti 4 sin t j k 2
z 5
1 rt 3 sin t i 4 cos tj k 2 When t T
1 , r 3i k, 2 2 2
4 3
2
t 2 at 0, 4, 4 .
1 1
r 2 2 1 1 3i k 6i k 2 r 2 2 37 37
x
5 4
4
5
y
Direction numbers: a 6, b 0, c 1 Parametric equations: x 6t, y 4, z t
4
14. rt et i 2 cos tj 2 sin t k, t0 0 rt et i 2 sin t j 2 cos t k r0 i 2j, r0 i 2k, r0 5 T0
r0 i 2k r0 5
Parametric equations: xs 1 s, ys 2, zs 2s rt0 0.1 r0 0.1 1 0.1, 2, 20.1 0.9, 2, 0.2 16. r0 0, 1, 0 u0 0, 1, 0 Hence the curves intersect. rt 1, sin t, cos t, r0 1, 0, 1
us sin s cos s cos s, sin s cos s cos s,
1 1 cos 2s 2 2
u0 1, 0, 1 cos
18.
r0 u0 0 ⇒ r0 u0 2
6 rt ti j, t 3 t rt i
20.
rt 6 1 Tt i 2j rt 1 36 t4 t
t
t4 36
i t6 j 2
Tt
72t Tt 4 i 4 j t 363 2 t 363 2
rt sin t i 2 cos tj rt sin2 t 4 cos2 t
The unit normal vector is perpendicular to this vector and points toward the z-axis: Nt
12t3
N2
4
rt sin t i 2 cos t j
6 j t2
2
rt cos ti 2 sin tj k, t
2 cos ti sin tj sin2 t 4 cos2 t
T2 1 3i 2 j T2 13
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.
Section 11.4 22. rt 4t i 2tj
24. rt t 2 j k
vt 4i 2j
vt 2tj
at O
at 2j
Tt
1 vt 2i j vt 5
Tt
Tt O Nt
Tt is undefined. Tt
Nt
28. rt a costi b sintj
vt 2ti 2j, v1 2i 2j
vt a sinti b costj
at 2i, a1 2i
T1
v0 bj
vt 1 1 2ti 2j ti j t2 1 vt 4t2 4 1 2
i j
2
2
i
2
2
j
2
i
2
2
a0 a2i v0 j v0
Motion along rt is counterclockwise. Therefore, N0 i. aT a T 0
1 i tj t2 1 2
at a2 costi b2 sint j
T0
t 1 i 2 j Tt t2 13 2 t 13 2 Nt Tt 1 t2 1
N1
Tt is undefined. Tt
The path is a line and the speed is variable.
rt t2i 2tj, t 1
Tt
2t j vt j vt 2t
Tt O
The path is a line and the speed is constant. 26.
Tangent Vectors and Normal Vectors
aN a N a2
j
aT a T 2 aN a N 2 30. rt0 t0 sin t0i 1 cos t0j vt0 1 cos t0i sin t0j at0 2sin t0i cos t0j T
1 cos t0i sin t0 j v v 21 cos t0
Motion along r is clockwise. Therefore, N aT a T aN a N
sin t0i 1 cos t0j . 21 cos t0
2 sin t0 2 1 cos t0 21 cos t0 2 2 2
1 cos t0
32. Tt points in the direction that r is moving. Nt points in the direction that r is turning, toward the concave side of the curve.
y a
T N a
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x
285
286
Chapter 11
Vector-Valued Functions
34. If the angular velocity is halved, aN a
2
2
a2 . 4
aN is changed by a factor of 14 . rt 2 cos ti 2 sin t j, t0
36.
4
38. rt 4ti 4t j 2tk vt 4i 4j 2k
x 2 cos t, y 2 sin t ⇒ x 2 y 2 4
at 0
rt 2 sin ti 2 cos t j 1 Tt 2 sin ti 2 cos tj sin ti cos tj 2 Nt cos ti sin tj r
y
4 2i 2 j
T
4
2
N
4
2
2 2
i j
T 1
Tt
1 v 2i 2j k v 3
Nt
T is undefined. T
aT, aN are not defined.
( 2, 2 )
N x
−1
1 −1
i j
40. rt et sin t i et cos tj et k
rt ti 3t 2j
42.
vt et cos t et sin ti et sin t et cos tj et k
vt i 6tj tk
v0 i j k
v2 i 12j 2k
at 2et cos ti 2et sin tj et k
at 6 j k
a0 2i k v 1 Tt cos t sin ti sin t cos tj k v 3 1 T0 i j k 3
Tt T2
2
2
aT a
i
2
2
1 v i 6tj tk v 1 37t 2 1
149
i 12j 2k
1 37ti 6j k 1 37t 23 2 T Nt T 37 1 372
1 Nt sin t cos ti cos t sin tj 2 N0
t2 k 2
j
T 3
aN a N 2
N2
1 371 37t 2
1 37149
1 5513
74i 6j k
74i 6j k
aT a T aN a N
74 149
37 5513
z 4 2
N x
4
2
4
6
8
T
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37ti 6j k
y
37 149
Section 11.4
Tangent Vectors and Normal Vectors
287
46. If aT 0, then the speed is constant.
44. The unit tangent vector points in the direction of motion. 48. (a) rt cos t t sin t, sin t t cos t
vt sin t sin t 2t cos t, cos t cos t 2t sin t 2t cos t, 2t sin t at 2 cos t 3t sin t, 2 sin t 3t cos t Tt
vt cos t, sin t vt
aT a T cos t 2 cos t 3t sin t sin t 2 sin t 3t cos t 2 aN a2 aT2 41 2t 2 4 3t When t 1, aT 2, aN 3. When t 2, aT 2, aN 2 3. (b) Since aT 2 > 0 for all values of t, the speed is increasing when t 1 and t 2.
50.
rt ti t 2 j
t3 k, t0 1 3
z
1 2
rt i 2tj t 2 k
N
B
Tt
1 1 4t 2 t 4
i 2t j t 2 k
x
1 2
T −1 2
1 2t t 3i 1 t 4j t 2t 3k Nt 1 4t 2 t 41 t 2 t 4
( 1, 1, 31 )
1
y
1 r1 i j k 3 T1 N1
1 6
i 2j k
1 63
3i 3k
B1 T1 N1
2
2
i k
i
j
k
6
6
6
6
3
6
0
2
2
2
3
3
i
3
3
j
3
3
k
3
3
i j k
2
1 52. (a) rt v0 cos t i h v0 sin t gt2 j 2
(b)
60
100 cos 30 t i 5 100 sin 30 t 16t 2 j 503t i 5 50t 16t 2 j
−20
300 −10
Maximum height 44.0625 Range 279.0325 (c)
vt 503 i 50 32t j Speed vt 25003 50 32t2
464t 2
200t 625at 32j
(d) t Speed
0.5
1.0
1.5
2.0
2.5
3.0
93.04
88.45
86.63
87.73
91.65
98.06
—CONTINUED—
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288
Chapter 11
Vector-Valued Functions
52. —CONTINUED— (e)
Tt
253i 25 16t j 264t 2 200t 625
Nt
25 16ti 253 264t 2 200t 625
(f ) 0
1616t 25
aT a T
50
3
−50
64t2 200t 625
The speed is increasing when aT and aN have opposite signs.
4003 aN a N 64t2 200t 625 aTT aNN 32j 54. 600 mph 880 ft sec rt 880ti
16t2
56.
vt r sin ti r cos t j
36,000j
vt r1 r v
vt 880i 32tj
at r2 cos ti r2 sin t j
at 32j Tt
at r2
880i 32tj 55i 2tj 164t2 3025 4t2 3025
(a) F mat mr2
Motion along r is clockwise, therefore Nt
2ti 55j
mv2 GMm 2 GM , v , v r r2 r
GMr
1760 4t 3025
aN a N
2
9.56420010
m 2 2 mv 2 r r r
(b) By Newton’s Law:
4t2 3025
64t aT a T 4t2 3025
58. v
rt r cos ti r sin tj
4
4.77 mi sec
60. Let x distance from the satellite to the center of the earth x r 4000. Then: v
2 x 2 x t 243600
9.56 x 10
4
9.56 104 4 2x 2 2 2 24 3600 x x3 v
62.
9.56 1042423600 2 ⇒ x 26,245 mi 4 2 226,245 1.92 mi sec 6871 mph 243600 64. a 2 a a
rt xti ytj yt mxt b, m and b are constants.
aTT aNN aTT aNN
rt xti mxt b j
aT2T 2 2aTaNT
vt xti mxtj
aT2 aN2
vt xt2 mxt2 xt 1 m2 Tt
vt ± i mj , constant vt 1 m2
N aN2N 2
aN2 a2 aT2 Since aN > 0, we have aN a2 aT2.
Hence, Tt 0.
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Section 11.5
Section 11.5
y
dx dy dz 1, 0, 2t dt dt dt
12 8
1 4t 2 dt
4
0
1 2t1 4t 2 ln 2t 1 4t 4
(4, 16)
16
4
0 4
x
(0, 0)
1
2
3
4
1 865 ln 8 65 16.819 4
4. rt a cos ti a sin tj
y a
dx dy a sin t, a cos t dt dt s
289
Arc Length and Curvature
2. rt ti t 2k
s
Arc Length and Curvature
2
a2 sin2 t a2 cos2 t dt
a
x
0
2
2
a dt at
0
0
2a
2v sin 1 (b) yt v0 sin t gt2 0 ⇒ t 0 2 g
1 6. (a) rt v0 cos ti v0 sin t gt2 j 2 1 yt v0 sin t gt2 2 yt v0 sin gt 0 when t
Range: xt v0 cos v0 sin . g
Maximum height when sin 1, or
. 2
yt v0 sin gt xt2 yt2 v02 cos2 v0 sin gt2 v02 cos2 v02 sin2 2v02g sin t g2t2
v02 2v0 g sin t g 2t2
2v0 sin g
0
v
2
0
2v0 g sin t g 2t2
12 dt
Since v0 96 ftsec, we have s
6 sin
0
96
2
6144 sin t 1024t2
v2 2v0 sin 0 sin2 g g
The range xt is a maximum for sin 2 1, or
(c) xt v0 cos
s
12
.
dt
Using a computer algebra system, s is a maximum for 0.9855 56.5.
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. 4
290
Chapter 11
Vector-Valued Functions
8. rt 3t, 2 cos t, 2 sin t
10. rt cos t t sin t, sin t t cos t, t2
dy dz dx 3, 2 sin t, 2 cos t dt dt dt s
dy dz dx t cos t, t sin t, 2t dt dt dt
2
32 2 sin t2 2 cos t2 dt
s
0
2
2
2
13 dt 13t
0
0
13
2
2
3
3
2
1
(0, 2, 0) 2
2
3
4
5
2
5 2
0
8
(π2 , 1, π4 ) 2
(1, 0, 0) 1
y
2
2
3
3
y
3
5 x
x
12. rt sin ti cos tj t 3 k
14. rt 6 cos
dx dy dz cos t, sin t, 3t 2 dt dt dt
4t i 2 sin 4t j tk, 0 ≤ t ≤ 2
(a) r0 6i 6, 0, 0 r2 2j 2k 0, 2, 2
2
cos t2 sin t2 3t 22 dt
distance 62 22 22 44 211 6.633
0 2
t2 2
z
5 4 2
( 32π , 0, 2)
5t 2 dt 5
0
z
s
t cos t2 t sin t2 2t2 dt
0
(b)
2 9t 4 dt 11.15
r0 6, 0, 0 r0.5 5.543, 0.765, 0.5
0
r1.0 4.243, 1.414, 1.0 r1.5 2.296, 1.848, 1.5 r2.0 0, 2, 2 (c) Increase the number of line segments. (d) Using a graphing utility, you obtain
2
s
rt dt 7.0105.
0
16. rt 4sin t t cos t, 4cos t t sin t,
3 2 t 2
t
(a) s
xu2 yu2 zu2 du
0
t
0
t
4u sin u2 4u cos u2 3u2 du
0
16u 9u 2 du
t
0
5 5u du t 2 2
—CONTINUED—
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Section 11.5
Arc Length and Curvature
291
16. —CONTINUED—
2s5 2s 2s 2s x 4 sin cos 5 5 5 2s 2s 2s y 4 cos sin 5 5 5 3s 2s 3 z 2 5 5 3s 2s 2s 2s 2s 2s 2s rs 4 sin cos i 4 cos sin j k 5 5 5 5 5 5 5
(b) t
2
When s 4:
(c) When s 5:
2 5 5 2 5 5 cos2 5 5 6.956
x 4 sin
2 5 5 2 5 5 sin2 5 5 14.169
y 4 cos z
z
35 1.342 5
4 sin 5
2s 5
2
12 2.4 5
2.291, 6.029, 2.400
6.956, 14.169, 1.342 (d) rs
85 85 cos85 2.291 8 8 8 y 4 cos sin 6.029 5 5 5 x 4 sin
4 cos 5
2s 5
2
9 1 35 16 25 25 2
18. rs 3 s i j rs i
and
rs 1
Ts rs Ts 0 ⇒ K Ts 0
20.
(The curve is a line.)
2s5 2s5 cos2s5 i 4 cos2s5 2s5 sin2s5 j 3s5 k
rs 4 sin
2s5 i 54 cos2s5 j 53 k 4 5 2s 4 5 2s cos i sin j Ts 25 2s 5 25 2s 5 4 5 2 10s K Ts 25 2s 25s Ts rs
4 sin 5
22.
rt t 2j k vt 2tj Tt j Tt 0 K
Tt 0 rt
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292
Chapter 11
24.
rt t i t 2 j
Vector-Valued Functions
26.
vt i 2t j
rt 2 sin ti cos tj
v1 i 2j
rt 4 sin2 t cos2 t
at 2 j
Tt
a1 2j Tt
i 2tj 1 4t 2
Tt
K
28.
1 5
2i j
From Exercise 22, Section 11.4, we have:
a sin t i b cos tj a2 sin2 t b2 cos2 t
a
N
ab2 cos t i a2b sin t j a2 sin2 t b2 cos2 t32
K
ab
Tt a2 sin2 t b2cos2 t K rt
a2 sin2 t b2 cos2 t
32.
2 4 sin2 t cos2 t32
30. rt a t sin t, a1 cos t
rt a sin t i b cos t j
Tt
2 cos ti 4 sin tj 4 sin2 t cos2 t32
2 a N v 2 55
rt a cos t i b sin t j
Tt
2 sin t i cos tj 4 sin2 t cos2 t
2 Tt 4 sin2 t cos2 t K rt 4 sin2 t cos2 t
1 Nt 2ti j 1 4t 2 N1
rt 2 cos ti sin tj
a 2 2
at Nt vt 2
a
2 2
34.
rt 4i 4j 2k
2 4a1 cos t
1 rt 2t2 i tj t2 k 2
Tt
Tt 0
Tt
Tt 0 K rt
K
4ti j tk 1 17t2 4i 17tj k 1 17t232
Tt 289t2 17 1 17t212 rt 1 17t232
rt et cos ti et sin tj et k rt et sin t et cos ti et cos t et sin tj et k 1 sin t cos ti cos t sin tj k 3 1 Tt cos t sin ti sin t cos tj 3 Tt
K
rt 4ti j tk
1 Tt 2i 2j k 3
36.
1 cos t
2a2 21 cos t
ab a2 sin2 t b2 cos2 t32
rt 4t i 4tj 2tk
1 cos t
Tt 13 cos t sin t2 sin t cos t2 2 t rt 3et 3e
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17 1 17t232
Section 11.5
Arc Length and Curvature
293
38. y mx b Since y 0, K 0, and the radius of curvature is undefined. 3 42. y 16 x2 4
4 40. y 2x , x 1 x y 2
4 , y1 2 x2
y
9x 16y 9 16y2 16y
y
8 , y 1 8 x3
y
K
y 8 8 1 y232 1 432 532
At x 0:
532 1 K 8
y 0 y
(radius of curvature) K
3 16
1 16 K 3 4x2 3
44. (a) y
y ln x,
46.
x2
y
24x 32
y
721 x2 x2 33
Center:
y
y1 1, K
At x 0: y 0
(radius of curvature)
x1
1 y , x
x2
316 3 1 0232 16
1 x2
y 1 1
1
1 1232
1 1 , r 232 22 K 232
y
72 8 27 3
The slope of the tangent line at 1, 0 is y1 1.
K
8 83 1 0232 3
Equation of normal line: y x 1 x 1
r
1 3 K 8
The slope of the normal line is 1. The center of the circle is on the normal line 22 units away from the point 1, 0. 1 x2 0 y2 22
1 x2 x 12 8
0, 38
3 Equation: x2 y 8
2
2x2 4x 2 8
9 64
2x2 2x 3 0 2x 3x 1 0
(b) The circles have different radii since the curvature is different and r
1 . K
x 3 or x 1 Since the circle is below the curve, x 3 and y 2. Center of circle: 3, 2 Equation of circle: x 32 y 22 8 y
2
(1, 0) x
2
4
−2 −4
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6
294
Chapter 11
48.
1 y x3, 3
x1 y 1 2x
y x2, y1 1, K
Vector-Valued Functions
y 1 2
2 1 , 1 r 2 1 132 2 K
1 The slope of the tangent line at 1, 3 is y1 1.
The slope of the normal line is 1. 1 Equation of normal line: y 3 x 1 or y x
4 3
The center of the circle is on the normal line 2 units away from the point 1, 3 . 1
y
1 x2 13 y 2 2
3
1 x2 x 1 2 2
2
x 12 1
1
x
−2
x 0 or x 2
( 1, 31 ) 1
2
−1
4 Since the circle is above the curve, x 0 and y 3 . 4 Center of circle: 0, 3
4 Equation of circle: x2 y 3 2 2
52. y x3, y 3x2, y 6x
y
50. 4
K
3
B A −4
−2
x
2
4
145, 451 , 145 , 1
. 45 4
4
3
4
4
3
(b) lim K 0
−4
x→
1 1 54. y ln x, y , y 2 x x
6x 1 9x 432
(a) K is maximum at
−3
K
56. y cos x
x 1x2 2 1 1x232 x 132
2x2 1 dK 2 dx x 152
(a) K has a maximum when x
y sin x y cos x K
1 2
y cos x 0 for x K. 1 y232 1 sin2 x32 2
Curvature is 0 at
.
(b) lim K 0 x→
58. y cosh x
ex ex 2
60. See page 828.
y
ex ex sinh x 2
y
ex ex cosh x 2
K
cosh x cosh x 1 1 1 sinh x232 cosh2 x32 cosh2 x y 2 http://librosysolucionarios.net
2 K, 0 .
Section 11.5
Arc Length and Curvature
295
y
62. K
1 y232
At the smooth relative extremum y 0, so K y . Yes, for example, y x 4 has a curvature of 0 at its relative minimum 0, 0. The curvature is positive for any other point of the curvature. 64. y1 axb x, y2
x x2
y
y1 axb x, y2
x , x2
y1 ab 2x, 2 , x 22
y2
4
y2
We observe that 0, 0 is a solution point to both equations. Therefore, the point P is the origin.
2
y1 2a y2
y2
P −4
4 x 23
−2
x
2
4
y1 −4
At P, y10 ab and y20
2 1 . 0 22 2
1 1 Since the curves have a common tangent at P, y10 y20 or ab 2 . Therefore, y10 2 . Since the curves have the same curvature at P, K10 K20.
K10 K20
y1 0 2a 1 y10232 1 12232 y2 0 12 1 y20232 1 12232
Therefore, 2a ± 12 or a ± 14 . In order that the curves intersect at only one point, the parabola must be concave downward. Thus, a
1 4
and
1 2. 2a
b
1 y1 x2 x and 4
y2
x x2
1 66. y x 85, 0 ≤ x ≤ 5 4
5
(a)
(b) V
z
2x
0
4
2
2
2
4
y
4
x
2
14 x dx 85
5
x135 dx
0
(shells)
x185 2 185
5 0
5 185 143.25 cm3 5 36
(rotated about y-axis) 2 6 6 (c) y x35, y x25 5 25 25x25 6 25x25 6 K 4 65 32 4 1 x 25x 25 1 x 65 25 25
(d) No, the curvature approaches as x → 0. Hence, any spherical object will hit the sides of the goblet before touching the bottom 0, 0.
32
1
5
0 0
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296
Chapter 11
68. s
Vector-Valued Functions
c K
1 y x3 3 y x 2 y 2x K
2x 1 x 432
When x 1:
K
s
1 2
c
4 2c
12
4 2c ⇒ c 30
30 4 2
3 3 At x , K 0.201 2 [1 811632 s
32 cK 30K 2 56.27 mi hr. 4
70. r r cos i r sin j f cos i f sin j x f cos y f sin x f sin f cos y f cos f sin x f cos f sin f sin f cos f cos 2 f sin f cos y f sin f cos f cos f sin f sin 2 f cos f sin K
x2y yx2 32 f 2 f f 2 f2 r 2 rr 2r2
x y
f 2 f232
r 2 r 2 32
72. r
74. r e
r 1
r e
r 0
r e
K
2r 2 rr r 2 r r 2
2 32
2 2 1 232
76. At the pole, r 0. K
2r2 rr r 2 r2 r 232
2r2 2 r 3 r
K
2r 2 rr r 2 r 2
r 2 32
2e2 1 2e232 2 e
78. r 6 cos 3 r 18 sin 3 At the pole,
, r 18, 6 6
and K
2
2
r6 18
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1 . 9
Section 11.5 80. xt t3, xt 3t 2, x t 6t
Arc Length and Curvature
297
5
1 yt t 2, yt t, y t 1 2 K
3t 21 t6t
−4
4 0
3t 22 t232 3t 2 3 3 2 32 t 9t 2 132 t 9t 1
K → 0 as t → ± 1 (b) rt ti t2j t2 k 2
82. (a) rt 3t 2i 3t t 3j vt 6ti 3 3t 2j
vt i 2tj tk
ds d 2s vt 31 t 2, 2 6t dt dt K aT
ds vt 5t2 1 dt
2 31 t 22
d 2s 5t 5t2 1 dt2
d 2s 6t dt 2
ds aN K dt
2
at 2j k
2 31 t 22
91
t2 2
6
K
rt r t 2 rt3 5t 132
aT
d 2s 5t 5t2 1 dt2
5
aN K
ddsT dTdt dsdt , Tt Tt d Tdt rt dsdt vt
84. (a) K Ts
(b)
Tt
rt rt rt dsdt
rt
ds Tt dt
r t
ddt s Tt dsdt Tt
rt r t
i j k rt r t vt at 1 2t t j 2k 0 2 1
dsdt
2
5 5 5t2 1 5t2 1 5t2 132
by the Chain Rule
2
2
dsdt ddt s Tt Tt dsdt Tt Tt 2
2
2
Since Tt Tt 0 and
ds rt, we have: dt
rt r t rt2 Tt Tt rt r t rt2Tt Tt rt2Tt Tt rt 21K rt from (a) Therefore,
rt r t K. rt 3
rt r t (c) K rt 3
vt at rt r t rt vt at Nt rt 2 rt 2 rt 2
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298
Chapter 11
Vector-Valued Functions
86. F ma ⇒ ma
GmM r r3
a
GM r r3
Since r is a constant multiple of a, they are parallel. Since a r is parallel to r, r r 0. Also,
dtd r r r r r r 0 0 0. Thus, r r is a constant vector which we will denote by L.
88.
d r r 1 1 L r 0 r L 3r r r dt GM r GM r
1 GMr 1 0 r r 3 r r r GM r3 r
Thus,
1 r r r 3 r r r r3 r
1 r r r r r r 0 r3
r GM L rr is a constant vector which we will denote by e.
90. L r r
92. Let P denote the period. Then
Let: r rcos i sin j r rsin i cos j
Then: r r
r2
P
d dt
drdt ddr ddt
i
j
k
r cos r sin d dt
r sin r cos d dt
0 0
d d k and L r r r 2 . dt dt
A
0
dA 1 dt L P. dt 2
Also, the area of an ellipse is ab where 2a and 2b are the lengths of the major and minor axes. 1
ab L P 2 P P2
2 ab L 4 2a 2 2 4 2a 2 2 a c 2 a 1 e2 L 2 L 2
4 2a 4 ed 4 2ed 3 a 2 L a L 2
4 2 L 2GM 3 4 2 3 a a Ka 3 L 2 GM
Review Exercises for Chapter 11 2. rt t i
1 jk t4
(a) Domain: 0, 4 and 4, (b) Continuous except at t 4
4. rt 2t 1 i t 2j tk (a) Domain: , (b) Continuous for all t
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Review Exercises for Chapter 11
299
6. (a) r0 3i j (b) r
2 2 k
(c) rs 3 coss i 1 sins j s k (d) r t r 3 cos t i 1 sin t j tk 3i j k 3 cos t 3 i sin t t k 8. rt ti
t j t1
y
10. rt 2ti tj t 2k
4
x 2t, y t, z t 2, y 12 x, z y 2
3
t xt t, yt t1
z
2
5
1
x y x1
x −2 −1
2
1
3
4
−2
12. rt 2 cos ti tj 2 sin tk
t
0
1
1
2
x
0
2
2
4
y
0
1
1
2
z
0
1
1
4
3
x2 z 2 4
3
1 1 14. rt 2 ti tj 4 t 3k
z
x 2 cos t, y t, z 2 sin t
4 x
z 6
2
5
x
t
0
2
3
2
x
2
0
2
0
y
0
2
3
2
z
0
2
0
2
4 2π
3 y
2 −3
x
−2
1
1
2
−2
−1
−1 3
1
2
3
−3
y
18. The x- and y-components are 2 cos t and 2 sin t. At
16. One possible answer is: r1t 4ti,
0 ≤ t ≤ 1
r2t 4 cos ti 4 sin tj,
0 ≤ t ≤
r3t 4 t j,
0 ≤ t ≤ 4
t
2
3
, 2
the staircase has made 34 of a revolution and is 2 meters high. Thus, one answer is rt 2 cos ti 2 sin tj
20. x2 z 2 4, x y 0, t x
22. lim t→0
x t, y t, z ± 4 t 2
sint 2t i e
z
et k lim t→0
rt ti tj 4 t 2 k 3
x 5 y
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2 cos 2t ijk 1
2i j k
rt ti tj 4 t 2 k 4
t j
4 tk. 3
y
300
Chapter 11
Vector-Valued Functions
1 24. rt sin ti cos tj t k, ut sin ti cos t j k t (a) rt cos ti sin tj k
(b) rt sin t i cos t j
(c) rt ut 2
(d) ut 2rt sin ti cos tj
Dtrt ut 0
1t 2tk
Dtut 2rt cos t i sin tj
1 2 k t2
(e) rt 1 t 2 Dt rt (f) rt ut
t 1 t 2
1t cos t t cos t i 1t sin t t sin tj
1 1 1 1 cos t 2 sin t t cos t sin t j Dtrt ut sin t 2 cos t t sin t cos t i t t t t 26. The graph of u is parallel to the yz-plane.
28.
30.
t2 ln t i t ln tj k dt t ln t t i 1 2 ln tj tk C 4
tj t 2k i tj tk dt
32. rt
t2 t3i t 2j tk dt
t3 t4 i t3 j t2 k C 3
4
sec t i tan tj t 2k dt ln sec t tan t i ln cos t j
3
2
t3 kC 3
r0 C 3k
rt ln sec t tan t i ln cos t j
1
34.
0
1
36.
1
t j t sin t k dt
23 t
32 j
t3i arcsin tj t2k dt
t3 3k 3
sin t t cos tk
1 0
2 j sin 1 cos 1k 3
t4 i t arcsin t 1 t j t3 k 4
3
1
2
1
2 k 3 38.
rt t, tan t, et
40.
rt vt 1, sec2 t, et
rt 3 sinh t, cosh t, 2
vt 1 sec t e 4
r0 0, 1, 2 direction numbers
2t
rt at 0, 2 sec t tan t, e 2
rt 3 cosh t, sinh t, 2t, t 0 0
t
Since r0 3, 0, 0, the parametric equations are x 3, y t, z 2t. rt 0 0.1 r0.1 3, 0.1, 0.2
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Review Exercises for Chapter 11
42. Range 4
v02 16
v02 sin cos 16 6 213
3v 2 4 0 104 213
2 13 6
416 v02 ⇒ v0 11.776 ft sec 3 4
44. rt v0 cos ti v0 sin t 12 9.8t2 j (b) rt 20 cos 45ti 20 sin 45t 4.9t 2 j
(a) rt 20 cos 30ti 20 sin 30t 4.9t 2 j
20
20
0
0
45
45 0
0
Maximum height 10.2 m; Range 40.8 m
Maximum height 5.1 m; Range 35.3 m (c) rt 20 cos 60ti 20 sin 60t 4.9t 2 j 20
0
45 0
Maximum height 15.3 m; Range 35.3 m (Note that 45 gives the longest range)
46.
rt 1 4t i 2 3tj
rt 2t 1 i
48.
vt 4 i 3j
vt 2i
v 5 at 0
vt
1 Tt 4i 3j 5
2t 14 1 t 12 4 j t 13
Tt
t 12i j t 14 1
Nt
i t 12j t 14 1
a
T
4 t 13t 14 1
a
N
4t 12 t 13t 14 1
a T0
N
2 j t 12
at
Nt does not exist
a
2 j t1
does not exist
4 t 1t 14 1
http://librosysolucionarios.net
301
302
Chapter 11
Vector-Valued Functions
rt t cos ti t sin t j
50.
vt rt t sin t cos t i t cos t sin t j vt speed t sin t cos t2 t cos t sin t2 t2 1 at rt t cos t 2 sin t i t sin t 2 cos t j Tt
vt t sin t cos t i t cos t sin t j vt t2 1
Nt
t cos t sin t i t sin t cos t j t2 1 t
at Tt
t 2 1
t2 2 t 2 1
at Nt
rt t 1 i tj
52.
1 k t
54. rt t i t2j 23 t 3k, x t, y t 2, z 23 t 3 When t 2, x 2, y 4, z 16 3 .
1 vt i j 2 k t
Direction numbers when t 2, a 1, b 4, c 8
2t 4 1
vt
a
rt i 2tj 2t 2k
t2
x t 2, y 4t 4, z 8t 16 3
at
2 k t3
Tt
t 2i t 2 j k 2t 4 1
Nt
i j 2t 2k 22t 4 1 2 2t 4 1
T t3
4 t22t 4 1
a N
56. Factor of 4 58. rt t2i 2tk, 0 ≤ t ≤ 3 rt 2ti 2k
b
s
rt dt
a
z 6
5
3
4
4t2 4 dt
3 2
0
ln t2 1 t tt2 1
3
(0, 0, 0)
−2 1
(9, 0, 6)
0
ln10 3 310 11.3053
1
2
2 3 4
5 6 7 9 x
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y
Review Exercises for Chapter 11 60.
rt 10 cos ti 10 sin tj
62. rt ti t2j 2tk, 0 ≤ t ≤ 2
rt 10 sin ti 10 cos tj
rt i 2tj 2k, rt 5 4t2
rt 10
s
b
2
303
s
a
10 dt 20
5 4t2 dt
0
5 5 ln 5 ln105 45 6.2638 4 4
21
0
2
rt dt
y z 8 6 4 2
4 3 x
−8 −6 −4 −2
(2, 4, 4)
2
2 4 6 8
−4 −6 −8
1 1
1
2
2
3
4
y
x
64. rt 2sin t t cos t, 2cos t t sin t, t, 0 ≤ t ≤
2
66.
xt 2t sin t, 2t cos t, 1, rt 4t2 1
b
s
rt dt
a
2
4t2 1 dt
rt et cos t et sin t i et sin t et cos t k rt et cos t et sin t2 et sin t et cos t2 2et
0
17
rt et sin ti et cos tk, 0 ≤ t ≤
1 ln17 4 4.6468 4
s
rt dt
0
2
0
68.
rt 2ti 3tj rt
1 t
i 3j, rt
1t 9 1 t 9t
1 r t t 32i 2
j i 1 3 r r t 1 32 0 t 2 K
70.
k 0
3 3 t 32k; r r 32 2 2t
0
32t32 3 rt r t 3 rt 1 9t32t32 21 9t32
rt 2ti 5 cos tj 5 sin tk rt 2i 5 sin tj 5 cos tk, rt 29 rt 5 cos tj 5 sin tk
i r r 2 0
r r 725 K
j k 5 sin t 5 cos t 25i 10 sin tj 10 cos tk 5 cos t 5 sin t
725 25 29 r r 5 r 3 2932 29 29 29
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et dt 2et
0
2e 1
298
Chapter 11
Vector-Valued Functions
86. F ma ⇒ ma
GmM r r3
a
GM r r3
Since r is a constant multiple of a, they are parallel. Since a r is parallel to r, r r 0. Also,
dtd r r r r r r 0 0 0. Thus, r r is a constant vector which we will denote by L.
88.
d r r 1 1 L r 0 r L 3r r r dt GM r GM r
1 GMr 1 0 r r 3 r r r GM r3 r
Thus,
1 r r r 3 r r r r3 r
1 r r r r r r 0 r3
r GM L rr is a constant vector which we will denote by e.
90. L r r
92. Let P denote the period. Then
Let: r rcos i sin j r rsin i cos j
Then: r r
r2
P
d dt
drdt ddr ddt
i
j
k
r cos r sin d dt
r sin r cos d dt
0 0
d d k and L r r r 2 . dt dt
A
0
dA 1 dt L P. dt 2
Also, the area of an ellipse is ab where 2a and 2b are the lengths of the major and minor axes. 1
ab L P 2 P P2
2 ab L 4 2a 2 2 4 2a 2 2 a c 2 a 1 e2 L 2 L 2
4 2a 4 ed 4 2ed 3 a 2 L a L 2
4 2 L 2GM 3 4 2 3 a a Ka 3 L 2 GM
Review Exercises for Chapter 11 2. rt t i
1 jk t4
(a) Domain: 0, 4 and 4, (b) Continuous except at t 4
4. rt 2t 1 i t 2j tk (a) Domain: , (b) Continuous for all t
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Review Exercises for Chapter 11
299
6. (a) r0 3i j (b) r
2 2 k
(c) rs 3 coss i 1 sins j s k (d) r t r 3 cos t i 1 sin t j tk 3i j k 3 cos t 3 i sin t t k 8. rt ti
t j t1
y
10. rt 2ti tj t 2k
4
x 2t, y t, z t 2, y 12 x, z y 2
3
t xt t, yt t1
z
2
5
1
x y x1
x −2 −1
2
1
3
4
−2
12. rt 2 cos ti tj 2 sin tk
t
0
1
1
2
x
0
2
2
4
y
0
1
1
2
z
0
1
1
4
3
x2 z 2 4
3
1 1 14. rt 2 ti tj 4 t 3k
z
x 2 cos t, y t, z 2 sin t
4 x
z 6
2
5
x
t
0
2
3
2
x
2
0
2
0
y
0
2
3
2
z
0
2
0
2
4 2π
3 y
2 −3
x
−2
1
1
2
−2
−1
−1 3
1
2
3
−3
y
18. The x- and y-components are 2 cos t and 2 sin t. At
16. One possible answer is: r1t 4ti,
0 ≤ t ≤ 1
r2t 4 cos ti 4 sin tj,
0 ≤ t ≤
r3t 4 t j,
0 ≤ t ≤ 4
t
2
3
, 2
the staircase has made 34 of a revolution and is 2 meters high. Thus, one answer is rt 2 cos ti 2 sin tj
20. x2 z 2 4, x y 0, t x
22. lim t→0
x t, y t, z ± 4 t 2
sint 2t i e
z
et k lim t→0
rt ti tj 4 t 2 k 3
x 5 y
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2 cos 2t ijk 1
2i j k
rt ti tj 4 t 2 k 4
t j
4 tk. 3
y
300
Chapter 11
Vector-Valued Functions
1 24. rt sin ti cos tj t k, ut sin ti cos t j k t (a) rt cos ti sin tj k
(b) rt sin t i cos t j
(c) rt ut 2
(d) ut 2rt sin ti cos tj
Dtrt ut 0
1t 2tk
Dtut 2rt cos t i sin tj
1 2 k t2
(e) rt 1 t 2 Dt rt (f) rt ut
t 1 t 2
1t cos t t cos t i 1t sin t t sin tj
1 1 1 1 cos t 2 sin t t cos t sin t j Dtrt ut sin t 2 cos t t sin t cos t i t t t t 26. The graph of u is parallel to the yz-plane.
28.
30.
t2 ln t i t ln tj k dt t ln t t i 1 2 ln tj tk C 4
tj t 2k i tj tk dt
32. rt
t2 t3i t 2j tk dt
t3 t4 i t3 j t2 k C 3
4
sec t i tan tj t 2k dt ln sec t tan t i ln cos t j
3
2
t3 kC 3
r0 C 3k
rt ln sec t tan t i ln cos t j
1
34.
0
1
36.
1
t j t sin t k dt
23 t
32 j
t3i arcsin tj t2k dt
t3 3k 3
sin t t cos tk
1 0
2 j sin 1 cos 1k 3
t4 i t arcsin t 1 t j t3 k 4
3
1
2
1
2 k 3 38.
rt t, tan t, et
40.
rt vt 1, sec2 t, et
rt 3 sinh t, cosh t, 2
vt 1 sec t e 4
r0 0, 1, 2 direction numbers
2t
rt at 0, 2 sec t tan t, e 2
rt 3 cosh t, sinh t, 2t, t 0 0
t
Since r0 3, 0, 0, the parametric equations are x 3, y t, z 2t. rt 0 0.1 r0.1 3, 0.1, 0.2
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Review Exercises for Chapter 11
42. Range 4
v02 16
v02 sin cos 16 6 213
3v 2 4 0 104 213
2 13 6
416 v02 ⇒ v0 11.776 ft sec 3 4
44. rt v0 cos ti v0 sin t 12 9.8t2 j (b) rt 20 cos 45ti 20 sin 45t 4.9t 2 j
(a) rt 20 cos 30ti 20 sin 30t 4.9t 2 j
20
20
0
0
45
45 0
0
Maximum height 10.2 m; Range 40.8 m
Maximum height 5.1 m; Range 35.3 m (c) rt 20 cos 60ti 20 sin 60t 4.9t 2 j 20
0
45 0
Maximum height 15.3 m; Range 35.3 m (Note that 45 gives the longest range)
46.
rt 1 4t i 2 3tj
rt 2t 1 i
48.
vt 4 i 3j
vt 2i
v 5 at 0
vt
1 Tt 4i 3j 5
2t 14 1 t 12 4 j t 13
Tt
t 12i j t 14 1
Nt
i t 12j t 14 1
a
T
4 t 13t 14 1
a
N
4t 12 t 13t 14 1
a T0
N
2 j t 12
at
Nt does not exist
a
2 j t1
does not exist
4 t 1t 14 1
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301
302
Chapter 11
Vector-Valued Functions
rt t cos ti t sin t j
50.
vt rt t sin t cos t i t cos t sin t j vt speed t sin t cos t2 t cos t sin t2 t2 1 at rt t cos t 2 sin t i t sin t 2 cos t j Tt
vt t sin t cos t i t cos t sin t j vt t2 1
Nt
t cos t sin t i t sin t cos t j t2 1 t
at Tt
t 2 1
t2 2 t 2 1
at Nt
rt t 1 i tj
52.
1 k t
54. rt t i t2j 23 t 3k, x t, y t 2, z 23 t 3 When t 2, x 2, y 4, z 16 3 .
1 vt i j 2 k t
Direction numbers when t 2, a 1, b 4, c 8
2t 4 1
vt
a
rt i 2tj 2t 2k
t2
x t 2, y 4t 4, z 8t 16 3
at
2 k t3
Tt
t 2i t 2 j k 2t 4 1
Nt
i j 2t 2k 22t 4 1 2 2t 4 1
T t3
4 t22t 4 1
a N
56. Factor of 4 58. rt t2i 2tk, 0 ≤ t ≤ 3 rt 2ti 2k
b
s
rt dt
a
z 6
5
3
4
4t2 4 dt
3 2
0
ln t2 1 t tt2 1
3
(0, 0, 0)
−2 1
(9, 0, 6)
0
ln10 3 310 11.3053
1
2
2 3 4
5 6 7 9 x
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y
Review Exercises for Chapter 11 60.
rt 10 cos ti 10 sin tj
62. rt ti t2j 2tk, 0 ≤ t ≤ 2
rt 10 sin ti 10 cos tj
rt i 2tj 2k, rt 5 4t2
rt 10
s
b
2
303
s
a
10 dt 20
5 4t2 dt
0
5 5 ln 5 ln105 45 6.2638 4 4
21
0
2
rt dt
y z 8 6 4 2
4 3 x
−8 −6 −4 −2
(2, 4, 4)
2
2 4 6 8
−4 −6 −8
1 1
1
2
2
3
4
y
x
64. rt 2sin t t cos t, 2cos t t sin t, t, 0 ≤ t ≤
2
66.
xt 2t sin t, 2t cos t, 1, rt 4t2 1
b
s
rt dt
a
2
4t2 1 dt
rt et cos t et sin t i et sin t et cos t k rt et cos t et sin t2 et sin t et cos t2 2et
0
17
rt et sin ti et cos tk, 0 ≤ t ≤
1 ln17 4 4.6468 4
s
rt dt
0
2
0
68.
rt 2ti 3tj rt
1 t
i 3j, rt
1t 9 1 t 9t
1 r t t 32i 2
j i 1 3 r r t 1 32 0 t 2 K
70.
k 0
3 3 t 32k; r r 32 2 2t
0
32t32 3 rt r t 3 rt 1 9t32t32 21 9t32
rt 2ti 5 cos tj 5 sin tk rt 2i 5 sin tj 5 cos tk, rt 29 rt 5 cos tj 5 sin tk
i r r 2 0
r r 725 K
j k 5 sin t 5 cos t 25i 10 sin tj 10 cos tk 5 cos t 5 sin t
725 25 29 r r 5 r 3 2932 29 29 29
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et dt 2et
0
2e 1
304
Chapter 11
Vector-Valued Functions
72. y ex2
74. y tan x y sec2 x
1 1 y ex2, y ex2 2 4
K
y 2 32
1 y
At x 0, K
y 2 sec2 x tan x
1 x2 e 4 1 1 ex 4
K
32
2 2 25 55 14 ,r . 25 2 5432 532 55
x y 2 32 2 sec2 x 4tan32
1 y
At x
1 sec x]
4 45 4 55 , K 32 . and r 4 5 55 25 4
Problem Solving for Chapter 11 4. Bomb: r1t 5000 400t, 3200 16t2
x23 y23 a23
2.
Projectile: r2t v0 cos t, v0 sin t 16t2
2 13 2 13 y y 0 x 3 3
At 1600 feet: Bomb:
y13 y 13 Slope at Px, y. x
Projectile will travel 5 seconds:
rt cos3 ti sin3 tj rt 3 cos2 t sin ti 3 sin2 t cos tj
3200 16t2 1600 ⇒ t 10
5v0 sin 1625 1600 v0 sin 400.
rti 3 cos t sin t Tt
Horizontal position:
rt cos ti sin tj rt
At t 10, bomb is at 5000 40010 9000. At t 5, projectile is at v0 cos 5.
Tt sin ti cos tj Q0, 0, 0 origin P
cos3
\
t,
PQ T
sin3
Thus, t, 0 on curve.
i cos3 t cos t
5v0 cos 9000
j sin3 t sin t
v0 cos 1800.
k 0 0
Combining,
cos3 t sin t sin3 t cos tk
v0
\
PQ T D 3 cos t sin t T
K
400 2 v0 sin ⇒ tan ⇒ 12.5 . v0 cos 1800 9
1800 1843.9 ftsec cos
Tt 1 rt 3 cos t sin t
1 Thus, the radius of curvature, , is three times the K distance from the origin to the tangent line.
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Problem Solving for Chapter 11
6.
r 1 cos r sin
t
st
t
K
2 sin
d 4 cos 2 2
t
2 2 cos d
4 cos
t 2
r r
2 sin2 1 cos cos 1 cos 2 2
3 3 cos 8 sin3
3 4
2
2 3 sin3 4 sin 2 2 sin2
1 K
4 sin
2
3
s2 9 2 16 cos2
8. (a)
2r2 2 rr2 32r 2 8 sin3
t
1 cos 2 sin2 d
16 sin2 16 2 2
r xi yj position vector r r cos i r sin j dr dr d dr d cos r sin sin r cos i j dt dt dt dt dt
a
d 2r d 2r dr d dr d d sin r cos cos sin 2 dt dt2 dt dt dt dt dt
ddtr sin drdt cos ddt drdt cos ddt r sin ddt 2
2
2
2
r sin
d 2 i dt2
r cos
d 2 dt2
ar a ur a cos i sin j
dt
d 2r 2
cos2 2
ddtr sin 2
2
2
d 2r d r dt2 dt
dr d d sin cos r cos2 dt dt dt
2
dr d d sin cos r sin2 dt dt dt
r cos sin
2
dr d d 2 r 2 dr dt dt
urur a u u
ddtr rddt u 2drdt ddt r ddt u 2
2
2
2
r
2
d 2 dt2
r cos sin
2
a a u a sin i cos j 2 a a
2
—CONTINUED—
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d 2 dt2
305
306
Chapter 11
Vector-Valued Functions
10. rt cos ti sin tj k, t
8. —CONTINUED—
t t i 42,000 sin j (b) r 42,000 cos 12 12
r 42,000,
1 52 x 32
y
5 32 x 64
y
15 12 x 128
K
−1 2 −1 2
2
2
r
r
At t
875 2 3
−2
x
BTNk
15 12 x 128 25 3 x 1 4096
−2
N cos ti sin tj
u. 12 u 875 3
2 2 ,T i j 4 4 2 2
N
4 22i
B
4 k
2
2
j
32
At the point 4, 1, K
1 8932 120 ⇒ r 7. 32 89 K 120
14. (a) Eliminate the parameter to see that the Ferris wheel has a radius of 15 meters and is centered at 16j. At t 0, the friend is located at r10 j, which is the low point on the Ferris wheel. (b) If a revolution takes t seconds, then
t t t 2 10 10 and so t 20 seconds. The Ferris wheel makes three revolutions per minute. (c) The initial velocity is r2t 0 8.03i 11.47j. The speed is 8.032 11.472 14 msec. The angle of inclination is arctan11.47803 0.96 radians or 55 . (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 vertex of the parabola, which the object reaches when t t0
11.47 1.17 sec. 24.9
20
0
30 0
Thus, after the friend reaches the low point on the Ferris wheel, wait t0 2 sec 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 r1t 15 msec. The speed of the object at that time is r23.15 8.032 11.47 9.83.15 22 8.03 msec.
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−2
1 −1
T cos ti sin tj
Angular component: 0
12. y
2
T sin ti cos tj
d d 2 , 2 0 dt 12 dt
Radial component:
z
rt sin ti cos tj, rt 1
dr d 2r 0, 2 0 dt dt
Therefore, a 42000
4
y
C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . 39 Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . . 44 Section 11.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . 48 Section 11.4 Tangent Vectors and Normal Vectors Section 11.5 Arc Length and Curvature
. . . . . . . . . . . . . . . 54
. . . . . . . . . . . . . . . . . . . . .60
Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
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C H A P T E R 11 Vector-Valued Functions Section 11.1
Vector-Valued Functions
Solutions to Odd-Numbered Exercises 1 1. rt 5t i 4t j k t
3. rt ln t i et j t k Component functions: f t ln t
Component functions: f t 5t
gt et
gt 4t ht
ht t
1 t
Domain: 0,
Domain: , 0 0, 5. rt Ft Gt cos t i sin t j t k cos t i sin t j 2 cos t i t k Domain: 0,
i 7. rt Ft Gt sin t 0 Domain: ,
j cos t sin t
k 0 cos2 t i sin t cos t j sin2 t k cos t
1 9. rt 2 t 2 i t 1j 1 (a) r1 2 i
(b) r0 j 1 1 (c) rs 1 2 s 12i s 1 1j 2 s 12 i sj 1 (d) r2 t r2 2 2 t2i 2 t 1j 2i j
2 2t 12 t2i 1 tj 2i j 1 2 t 2 t2i tj
1 11. rt ln t i j 3t k t 1 (a) r2 ln 2i j 6k 2 (b) r3 is not defined.
ln3 does not exist.
(c) rt 4 lnt 4i
1 j 3t 4k t4
(d) r1 t r1 ln1 ti ln1 ti
1 j 31 tk 0i j 3k 1 t
1 1 t 1j 3tk 39
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40
13.
Chapter 11
Vector-Valued Functions 15. rt ut 3t 1t 2 14 t 38 4t 3
rt sin 3ti cos 3t j tk rt sin 3t2 cos 3t2 t 2 1 t 2
3t 3 t 2 2t 3 4t 3 5t 3 t 2, a scalar. The dot product is a scalar-valued function.
17. rt t i 2tj t 2k, 2 ≤ t ≤ 2
19. rt t i t 2j e0.75t k, 2 ≤ t ≤ 2
x t, y 2t, z t 2
x t, y t 2, z e0.75t
Thus, z x 2. Matches (b)
Thus, y x 2. Matches (d)
21. (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
yt1
y
x2 3
x2
x 1 3
y
27. x cos , y 3 sin
25. x t3, y t2
23. x 3t
y
y2 1 Ellipse 9 y
7 6 5 4 3 2
y
2 −5 −4 −3 −2 −1
x
4
2 1 x
x
− 3 −2
1 2 3 4 5
2
3
−2 −3
6
−2 −4
29. x 3 sec , y 2 tan
33. x 2 cos t, y 2 sin t, z t
31. x t 1 y 4t 2
x2 y2 1 Hyperbola 9 4 y
z 2t 3
x2 y 2 1 4 4
Line passing through the points:
zt
12
0, 6, 5, 1, 2, 3
9 6
Circular helix
z
z
3 −12 −9 −6
x
−3
6
(0, 6, 5)
5
9 12
7
4
−6
3
(2, − 2, 1)
−9
(1, 2, 3)
−12 1 3
3
4 5
6
y −3 3
x
3
y
x
2 37. x t, y t 2, z 3 t 3
35. x 2 sin t, y 2 cos t, z et x2 y 2 4
y x 2, z
z
z 6
2 3 3x
(2, 4, 163 )
4
6
z et
2
t
−3 3 x
3
y
2
1
0
1
2
x
2
1
0
1
2
y
4
1
0
1
4
z
16 3
23
0
2 3
16 3
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2 x
−2
5
y
−4 −6
(− 2, 4, − 163 )
Section 11.1 3 2 1 t k 39. rt t 2 i tj 2 2
Parabola
41. rt sin t i
z −3
−2
−2
1
−1
23 cos t 21 t j 12 cos t 23k
Helix
−3
z 2
2 3
2 −2
x
Vector-Valued Functions
−2
−1 1
x
y
−3
2
−4
3
−5
4 y
z
43.
(a)
2π
z
(b)
z 2π
π
8π
−2
−2
−3
2
1
x
y
π −2
−2 −2
−2
1
2 x
2π
4π
π −2
−2
2
2 2
2
x
y
The helix is translated 2 units back on the x-axis. (d)
y
x
The height of the helix increases at a faster rate. (e)
z
z
2 −2
π x
2 −2
π
−6
−6
y
2π 6 6
x
The axis of the helix is the x-axis.
y
The radius of the helix is increased from 2 to 6. 47. y x 22
45. y 4 x
Let x t, then y t 22.
Let x t, then y 4 t. rt ti 4 t j
rt ti t 22 j
49. x2 y 2 25
51.
Let x 5 cos t, then y 5 sin t.
x2 y2 1 16 4 Let x 4 sec t, y 2 tan t.
rt 5 cos ti 5 sin t j
rt 4 sec t i 2 tan tj
53. The parametric equations for the line are
z 8 7 6 5 4 3 2 1
x 2 2t, y 3 5t, z 8t. One possible answer is rt 2 2t i 3 5t j 8tk. 4 x
55. r1t t i,
z
(c)
3
2
(0, 8, 8)
1 4 5 6 7 8
(2, 3, 0)
y
0 ≤ t ≤ 4 r10 0, r14 4i
r2t 4 4ti 6tj,
0 ≤ t ≤ 1 r20 4i, r21 6j
r3t 6 t j,
0 ≤ t ≤ 6 r30 6j, r36 0
(Other answers possible)
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2
y
The orientation of the helix is reversed.
41
42
Chapter 11
Vector-Valued Functions 59. z x2 y 2, x y 0
0 ≤ t ≤ 2 y x2
57. r1t ti t 2j,
Let x t, then y x t and z x2 y 2 2t 2. Therefore,
r2t 2 ti, 0 ≤ t ≤ 2 r3t 4 tj, 0 ≤ t ≤ 4
x t, y t, z 2t 2.
(Other answers possible)
rt ti tj 2t2k z
(
2, −
(−
2, 4 ) 5
−3
1
2
2, 4)
2,
2
y
3
3 x
61. x2 y 2 4, z x2
z
x 2 sin t, y 2 cos t
4
z x2 4 sin2 t t
6
0
4
2
3
4
−3 3
x
0
1
2
2
2
y
2
3
2
0
2
2
z
0
1
2
4
2
0
y
3
x
0
rt 2 sin t i 2 cos tj 4 sin2 tk 63. x2 y 2 z 2 4, x z 2
z
Let x 1 sin t, then z 2 x 1 sin t and
x2
y2
z2
4.
3
1 sin t2 y 2 1 sin t2 2 2 sin2 t y 2 4 y 2 2 cos2 t,
−3
y ± 2 cos t
3
3
x 1 sin t, y ± 2 cos t
−3
z 1 sin t
2
t
x
0
y
0
z
2
6
1 2 ±
6
2
0
6
2
1
3 2
2
± 2
3 2
1
y
x
±
6
2 1 2
rt 1 sin t i 2 cos tj 1 sin t k and rt 1 sin t i 2 cos tj 1 sin t k
0 0
65. x2 z2 4, y 2 z2 4
z
Subtracting, we have x2 y 2 0 or y ± x.
3
(0, 0, 2)
Therefore, in the first octant, if we let x t, then x t, y t, z 4 t 2. rt ti tj 4 t2 k
4 x
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3
2 4
(2, 2, 0)
y
Section 11.1
67. y2 z2 2t cos t2 2t sin t2 4t2 4x2
69. lim t i t→2
Vector-Valued Functions
t2 4 1 1 j k 2i 2j k t 2 2t t 2
z
since
16
t2 4 2t lim 2. (L’Hôpital’s Rule) t→2 t 2 2t t→2 2t 2
12
lim
8 4
7
6
4
5
8
x
71. lim t 2 i 3t j t→0
12
16
y
1 cos t k 0 t
73. lim t→0
since lim t→0
1t i cos t j sin t k
does not exist since lim t→0
1 cos t sin t lim 0. t→0 1 t
1 does not exist. t
(L’Hôpital’s Rule)
1 75. rt t i j t
77. rt t i arcsin t j t 1k Continuous on 1, 1
Continuous on , 0, 0, 79. rt et, t 2, tan t
81. See the definition on page 786.
Discontinuous at t
Continuous on
n
2
n , n
2 2
83. rt t2i t 3j tk (a) st rt 2k t2i t 3j t 3k (b) st rt 2i t2 2i t 3j tk (c) st rt 5j t2i t 2j tk 85. Let rt x1t y1tj z1tk and ut x2ti y2tj z2tk. Then: lim rt ut lim y1tz2t y2tz1t i x1tz2t x2tz1t j x1ty2t x2ty1t k t→c
t→c
lim y1t lim z2t lim y2t lim z1t i lim x1t lim z2t lim x2t lim z1t j t→c
t→c
t→c
t→c
t→c
t→c
t→c
lim x1t lim y2t lim x2t lim y1t k t→c
t→c
t→c
t→c
t→c
lim x1ti lim y1tj lim z1tk lim x2ti lim y2tj lim z2tk t→c
t→c
t→c
t→c
t→c
lim rt lim ut t→c
t→c
87. Let rt xti ytj ztk. Since r is continuous at t c, then lim rt rc.
89. True
t→c
rc xci ycj zck ⇒ xc, yc, zc are defined at c. r xt2 yt2 zt2 lim r xc2 yc2 zc2 rc t→c
Therefore, r is continuous at c.
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t→c
43
44
Chapter 11
Vector-Valued Functions
Section 11.2
Differentiation and Integration of Vector-Valued Functions
1. rt t 2i t j, t0 2
3. rt cos ti sin t j, t0
y
4
xt t 2, yt t
(4, 2)
xt cos t, yt sin t
r′
2
x y2
r x
2
r2 4i 2j
4
6
r′
x
1
rt sin t i cos t j
rt0 is tangent to the curve.
r
2 i
rt0 is tangent to the curve.
7. rt 2 cos ti 2 sin t j t k, t0
5. rt t i t 2j (a)
y
rt 2 sin ti 2 cos tj k
6 16
1
4 16
r 2
2 16
r 4
1
r 4
1
r 2
2 16
4 16
6 16
32 2j 32 k
r
32 2i k
8 16
(0, − 2, 32π )
1 1 1 i j r 4 4 16 r r
r
1
x
(b)
3 2
x2 y 2 4, z t
8 16
z
2π
r′
12 21 i 41 j
r π −2
1 1 1 3 r i j 2 4 4 16
2 x
1 2
y
rt i 2tj
(c)
r
14 i 21 j
r12 r14 14i 316j 3 i j 12 14 14 4 This vector approximates r 14 . 9. rt 6ti 7t 2j t 3k rt 6i 14tj 3t 2k 13. rt et i 4 j rt e i t
11. rt a cos3 ti a sin3 tj k rt 3a cos2 t sin t i 3a sin2 t cos tj 15. rt t sin t, t cos t, t rt sin t t cos t, cos t t sin t, 1
1 17. rt t3i t2j 2 (a) rt 3t2i tj
(0, 1)
r
r j 2
−4
r2 4i j
y
x2 y2 1
8
−2
rt 2t i j
2
(b) rt rt 3t26t t 18t3 t
r t 6t i j
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Section 11.2
Differentiation and Integration of Vector-Valued Functions 1 1 21. rt t2i tj t3k 2 6
19. rt 4 cos ti 4 sin tj (a) rt 4 sin ti 4 cos tj
1 (a) rt ti j t2k 2
r t 4 cos t i 4 sin tj (b) rt rt 4 sin t4 cos t 4 cos t4 sin t
r t i tk
0
1 t3 (b) rt rt t1 10 t2t t 2 2
23. rt cos t t sin t, sin t t cos t, t (a) rt sin t sin t t cos t, cos t cos t t sin t, 1 t cos t, t sin t, 1 rt cos t t sin t, sin t t cos t, 0 (b) rt rt t cos tcos t t sin t t sin tsin t t cos t t 25.
rt cos ti sin tj t 2k, t0
1 4
z
r′′
rt sin ti cos tj 2tk
41
r
2
i
2
2
2
r′′
1 j k 2
x
r14 22 22 21
2
2
2
2
1 4
4 2
1
r′ r′
2
r14 1 2 i 2 j k r1 4 4 2 1 rt 2 cos ti 2 sin tj 2k
41
r
2 2
2
r 41
i
2 2
2 2
2
2
j 2k
2
2 2
2
2 2
2
4 4
1 r14 2 2i 2 2j 4k r14 2 4 4 27. rt t 2i t 3j
r 2 cos3 i 3 sin3 j
29.
rt 2t i 3t 2j r0 0 Smooth on , 0, 0,
r 6 cos2 sin i 9 sin2 cos j r
n2 0
Smooth on
31. r 2 sin i 1 2 cos j r 1 2 cos i 1 2 sin j r 0 for any value of Smooth on ,
n2, n 2 1, n any integer.
1 33. rt t 1i j t 2k t 1 rt i 2 j 2tk 0 t r is smooth for all t 0: , 0, 0,
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y
45
46
Chapter 11
Vector-Valued Functions
35. rt t i 3t j tan tk rt i 3j sec2 tk 0 r is smooth for all t
2n 1 n . 2 2
Smooth on intervals of form
n , n 2 2
37. rt ti 3tj t 2 k, ut 4ti t 2j t 3k (a) rt i 3j 2tk
(b) r t 2k
(c) rt ut 4t 2 3t 3 t 5
(d) 3rt ut t i 9t t 2j 3t 2 t 3k
Dt rt ut 8t 9t 2 5t 4
Dt 3rt ut i 9 2tj 6t 3t 2k
(e) rt ut 2t 4i t 4 4t 3j t 3 12t 2k
(f) rt 10t 2 t 4 t 10 t 2
Dt rt ut 8t 3i 12t 2 4t 3j 3t 2 24tk
Dt rt
10 2t 2
10 t 2
rt 3 sin t i 4 cos tj
39.
π
rt 3 cos ti 4 sin tj rt rt 9 sin t cos t 16 cos t sin t 7 sin t cos t cos
rt rt 7 sin t cos t rt rt 9 sin2 t 16 cos2 t 9 cos2 t 16 sin2 t
9 sin
arccos
2
1.855 maximum at t 3.927 1.287 minimum at t 2.356
7 sin t cos t t 16 cos2 t9 cos2 t 16 sin2 t
−1
7 0
54 and t 0.7854 .
34 and t 5.49874.
1.571 for t n , n 0, 1, 2, 3, . . . 2 2 41. rt lim
t→0
rt t rt t
3t t 2 i 1 t t2 j 3t 2i 1 t 2j t→0 t
lim
3t i 2tt t 2j t→0 t
lim
lim 3i 2t tj 3i 2tj t→0
43.
47.
49.
2t i j k dt t 2i tj t k C
45.
2t 1i 4t 3j 3 t k dt t 2 ti t 4j 2t 32k C
sec2 t i
1 2 i j t 32k dt ln ti tj t 52k C t 5
1 j dt tan t i arctan t j C 1 t2
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Section 11.2
1
51.
2 j t k
8t i t j k dt 4t 2i
0
1
t2
0
2
53.
1
1
0
0
0
55. rt
2
0
2
a cos t j
4e2t i 3et j dt 2e2t i 3et j C
0
2
tk
ai aj
0
57. rt
k 2
32j dt 32t j C1
r0 2i 3j C 2i ⇒ C 3j
r0 C1 600 3 i 600j
rt
rt 600 3 i 600 32tj
2e2ti
47
1 4i j k 2
a cos ti a sin tj k dt a sin t i
Differentiation and Integration of Vector-Valued Functions
3 1j et
rt
600 3 i 600 32tj dt
600 3 ti 600t 16t 2j C r0 C 0 rt 600 3 t i 600t 16t 2j
59. rt
1 2 2 tet i etj k dt et i etj t k C 2
1 1 r0 i j C i j k ⇒ C i 2j k 2 2 1 2 2 et i et 2j t 1k rt 1 et i et 2j t 1k 2 2
2
61. See “Definition of the Derivative of a Vector-Valued Function” and Figure 11.8 on page 794.
63. At t t0, the graph of ut is increasing in the x, y, and z directions simultaneously.
65. Let rt xti ytj ztk. Then crt cxti cytj cztk and Dt crt cxti cytj cztk c xti ytj ztk crt. 67. Let rt xti ytj ztk, then f trt f txti f tytj f tztk. Dt f trt f txt ftxt i f tyt ftyt j f tzt ftzt k f t xti ytj ztk ft xti ytj ztk f trt ftrt 69. Let rt xti ytj ztk. Then r f t x f ti y f tj z f tk and Dt r f t x f t fti y f t ftj z f t ftk
(Chain Rule)
ft x f ti y f tj z f tk ftr f t.
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48
Chapter 11
Vector-Valued Functions
71. Let rt x1ti y1tj z1tk, ut x2ti y2tj z2tk, and vt x3ti y3tj z3tk. Then: rt ut vt x1t y2tz3t z2ty3t y1tx2tz3t z2tx3t z1tx2ty3t y2tx3t Dtrt ut vt x1ty2tz 3t x1ty2tz 3t x1ty2tz 3t x1ty3tz 2t x1ty3tz2t x1ty3tz2t y1tx2tz 3t y1tx2tz 3t y1tx2tz3t y1tz2tx3t y1tz2tx3t y1tz2tx3t z1tx2ty3t z1tx2ty3t z1tx2ty3t z1ty2tx3t z1ty2tx3t z1ty2tx3t x1t y2tz3t y3tz2t y1tx2tz3t z2tx3t z1tx2ty3t y2tx3t
x1t y2tz3t y3tz2t y1tx2tz3t z2tx3t z1tx2ty3t y2tx3t x1t y2tz3t y3tz2t y1tx2tz3t z2tx3t z1tx2ty3t y2tx3t rt ut vt rt ut vt rt ut vt 73. False. Let rt cos t i sin tj k. rt 2 d rt 0 dt rt sin t i cos tj rt 1
Section 11.3
Velocity and Acceleration
1. rt 3t i t 1j
3. rt t2 i t j
y
vt rt 3i j
2
at rt 0
v
(3, 0)
x
x x 3t, y t 1, y 1 3 At 3, 0, t 1.
4 −2 −4
6
vt rt 2t i j
4
at rt 2i
2
(4, 2)
x t2, y t, x y 2 −2
v2 4i j
−4
a2 2i 7. rt t sin t, 1 cos t
vt rt 2 sin t i 2 cos tj
vt rt 1 cos t, sin t
at rt 2 cos ti 2 sin tj
at rt sin t, cos t
x 2 cos t, y 2 sin t, x2 y 2 4
x t sin t, y 1 cos t (cycloid)
At 2, 2 , t . 4
At , 2, t .
v 2 i 2j 4
a
2 i 2j 4
v 2, 0 2i a 0, 1 j y 4
y 2
3
v
(π , 2)
2)
2,
π
a x
−3
v
a
(
3
−3
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2π
v a x
2
At 4, 2, t 2.
v1 3i j, a1 0
5. rt 2 cos t i 2 sin t j
y
x
4
6
8
Section 11.3
vt i 2j 3k
vt i 2tj t k
st vt 1 4 9 14
st 1 4t 2 t 2 1 5t 2
at 0
at 2j k
13. rt t i tj 9 t 2 k vt i j st
17. (a)
t 9 t 2
15. rt 4t, 3 cos t, 3 sin t vt 4, 3 sin t, 3 cos t 4i 3 sin tj 3 cos tk
k
st 16 9 sin2 t 9 cos2 t 5
1 1 9 t t 189 tt 2
2
at
2
at 0, 3 cos t, 3 sin t 3 cos tj 3 sin tk
2
9 k 9 t 232
t3 ,t 1 4 0
rt 1, 2t,
3t 2 4
3 4
rt t, t 2,
r1 1, 2,
(b) r1 0.1 1 0.1, 1 20.1,
1 3 t 4 4
19. at i j k, v0 0, r0 0
21. at tj t k, v1 5j, r1 0
i j k dt t i tj t k C
vt
v0 C 0, vt t i tj tk, vt t i j k rt
ti tj tk dt
r0 C 0, rt
1 3 0.1 4 4
1.100, 1.200, 0.325
x 1 t, y 1 2t, z
vt
t2 k 2
11. rt t i t 2j
9. rt t i 2t 5j 3t k
Velocity and Acceleration
t2 t2 j kC 2 2
tj tk dt
1 9 1 1 v1 j k C 5j ⇒ C j k 2 2 2 2
t2 i j k C 2
vt
t2 i j k, 2
rt
r2 2i j k 2i 2j 2k
t2 29 j t2 21 k 2
2
t6 29 t j t6 21 t k C 3
3
r1
14 1 14 1 j kC0 ⇒ C j k 3 3 3 3
rt
t6 29 t 143 j t6 21 t 31 k
r2
17 2 j k 3 3
3
3
23. The velocity of an object involves both magnitude and direction of motion, whereas speed involves only magnitude. 25. rt 88 cos 30 ti 10 88 sin 30 t 16t 2 j
t2 9 t2 1 j k dt 2 2 2 2
50
44 3 t i 10 44t 16t 2j
0
300 0
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49
50
Chapter 11
Vector-Valued Functions
1 v v 27. rt v0 cos ti h v0 sin t gt 2 j 0 t i 3 0 t 16t 2 j 2 2 2 v0 v t 300 when 3 0 t 16t 2 3. 2 2 t
300 2 v0 300 2 300 2 , 16 v0 v0 v0 2
2
0, 300
300232 0 v02
v02 30032, v0 9600 40 6, v0 40 6 97.98 ftsec The maximum height is reached when the derivative of the vertical component is zero. yt 3
tv0 40 6 16t 2 3 t 16t 2 3 40 3t 16t 2 2 2
yt 40 3 32t 0 t
40 3 5 3 32 4
Maximum height: y
5 4 3 3 40 3 5 4 3 16 5 4 3
29. xt tv0 cos or t
2
78 feet
x v0 cos
yt tv0 sin 16t 2 h y
x2 16 x v0 sin 16 2 h tan x sec2 x2 h v0 cos v0 cos2 v02
31. rt ti 0.004t2 0.3667t 6j, or (a) y 0.004x2 0.3667x 6
(b)
18
0
120 0
(c) y 0.008x 0.3667 0 ⇒ x 45.8375 and
(d) From Exercise 29,
y45.8375 14.4 feet.
tan 0.3667 ⇒ 20.14
16 sec2 4000 16 sec2 0.004 ⇒ v02 v02 0.004 cos2 ⇒ v0 67.4 ftsec.
33. 100 mph 100 (a) rt (b)
miles hr
feet ftsec 5280 mile 3600 sechour 440 3
440 cos t i 3
sin t 16t j
440 3 3 0
0
Graphing these curves together with y 10 shows that 0 20 .
100
0
2
500 0
—CONTINUED—
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Section 11.3
Velocity and Acceleration
33. —CONTINUED— (c) We want xt
cos t ≥ 400
440 3
and
yt 3
sin t 16t
440 3
2
≥ 10.
From xt, the minimum angle occurs when t 3011 cos . Substituting this for t in yt yields: 3
30 30 16
sin
440 3 11 cos 11 cos 400 tan
2
10
14,400 sec2 7 121
14,400 1 tan2 400 tan 7 0 121 14,400 tan2 48,400 tan 15,247 0 tan
48,400 ± 48,4002 414,40015,247 214,400
tan1
1,464,332,800
48,400 28,800 19.38
35. rt v cos t i v sin t 16t 2 j (a) We want to find the minimum initial speed v as a function of the angle . Since the bale must be thrown to the position 16, 8, we have 16 v cos t 8 v sin t 16t 2. t 16v cos from the first equation. Substituting into the second equation and solving for v, we obtain: 8 v sin 12 512
16 16 16
v cos v cos
1 sin 512 2 cos v cos2
2
sin 1 2 1 v2 cos2 cos sin 1 cos2 2 sin cos cos2 2 1 2 v cos 512 512
512 2 sin cos cos2 512 . We minimize f 2 sin cos cos2 v2
f 512
2 cos2 2 sin2 2 sin cos 2 sin cos cos2 2
f 0 ⇒ 2 cos2 sin2 0 tan2 2
1.01722 58.28
Substituting into the equation for v, v 28.78 feet per second. (b) If 45 , 16 v cos t v
2
2
t
8 v sin t 16t2 v From part (a), v 2
2
2
t 16t2
512
2 22 22 22
2
512 1024 ⇒ v 32 ftsec. 12
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51
52
Chapter 11
Vector-Valued Functions
37. rt v0 cos t i v0 sin t 16t 2 j
v0 sin t 16t2 0 when t 0 and t
v0 sin . 16
The range is x v0 cos t v0 cos
v0 sin v02 sin 2 . 16 32
Hence, x
12002 1 ⇒ 1.91 . sin2 3000 ⇒ sin 2 32 15
39. (a) 10 , v0 66 ftsec
(b) 10 , v0 146 ftsec
rt 66 cos 10 ti 0 66 sin 10 t 16t j
rt 146 cos 10 ti 0 146 sin 10 t 16t 2 j
rt 65ti 11.46t 16t 2j
rt 143.78ti 25.35t 16t 2j
Maximum height: 2.052 feet
Maximum height: 10.043 feet
Range: 46.557 feet
Range: 227.828 feet
2
5
0
15
50
0
0
300 0
(c) 45 , v0 66 ftsec
(d) 45 , v0 146 ftsec
rt 66 cos 45 ti 0 66 sin 45 t
16t 2
j
rt 146 cos 45 ti 0 146 sin 45 t 16t 2 j
rt 46.67ti 46.67t 16t 2j
rt 103.24ti 103.24t 16t 2j
Maximum height: 34.031 feet
Maximum height: 166.531 feet
Range: 136.125 feet
Range: 666.125 feet
40
0
200
200
0
0
800 0
(e) 60 , v0 66 ftsec
(f ) 60 , v0 146 ftsec
rt 66 cos 60 ti 0 66 sin 60 t
16t 2
j
rt 146 cos 60 ti 0 146 sin 60 t 16t 2 j
rt 33ti 57.16t 16t 2j
rt 73ti 126.44t 16t 2j
Maximum height: 51.074 feet
Maximum height: 249.797 feet
Range: 117.888 feet
Range: 576.881 feet 300
60
0
140 0
0
600 0
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Section 11.3
Velocity and Acceleration
41. rt v0 cos t i h v0 sin t 4.9t 2 j 100 cos 30 ti 1.5 100 sin 30 t 4.9t 2 j 1 The projectile hits the ground when 4.9t2 100 2 t 1.5 0 ⇒ t 10.234 seconds.
The range is therefore 100 cos 30 10.234 886.3 meters. The maximum height occurs when dydt 0. 100 sin 30 9.8t ⇒ t 5.102 sec The maximum height is y 1.5 100 sin 30 5.102 4.95.1022 129.1 meters. 43. rt b t sin t i b1 cos t j vt b cos ti b sin t j b 1 cos ti b sin tj at b 2 sin ti b 2 cos t j b 2sin ti cos t j vt 2 b 1 cos t at b 2 (a) vt 0 when t 0, 2 , 4 , . . . .
45.
(b) vt is maximum when t , 3 , . . . , then vt 2b .
vt b sin t i b cos t j rt vt b2 sin t cos t b2 sin t cos t 0 Therefore, rt and vt are orthogonal.
47. at b 2 cos ti b 2 sin tj b 2cos ti sin tj 2rt at is a negative multiple of a unit vector from 0, 0 to cos t, sin t and thus at is directed toward the origin. 49. at 2b 1 m32 F m 2b
1 2 2 10 32
4 10 radsec vt b 8 10 ftsec 1 1 51. To find the range, set yt h v0 sin t 2 gt 2 0 then 0 2 gt 2 v0 sin t h. By the Quadratic Formula, (discount the negative value)
t
v0 sin v0 sin 2 412gh v0 sin v02 sin2 2gh . 212g g
At this time, xt v0 cos
v
0
v sin 2gh
v cos .
sin sin 2gh g v
sin v02 sin2 2gh v cos 0 v0 sin g g
v02
2
0
2
2
0
2
2
0
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53
54
Chapter 11
Vector-Valued Functions
53. rt xti ytj ztk Position vector vt xti ytj ztk Velocity vector at xti ytj ztk Acceleration vector Speed vt xt2 yt2 zt2 C, C is a constant. d xt2 yt2 zt2 0 dt 2xtxt 2ytyt 2ztzt 0 2xtxt ytyt ztzt 0 vt at 0 Orthogonal 55. rt 6 cos t i 3 sin tj (a) vt rt 6 sin t i 3 cos t j
(b)
vt 36 sin2 t 9 cos2 t 34 sin2 t cos2 t 33
sin2
t
0
4
2
2 3
Speed
3
3 10 2
6
3 13 2
3
t1
at vt 6 cos t i 3 sin t j (c)
(d) The speed is increasing when the angle between v and a is in the interval
6
−9
0, 2 .
9
−6
The speed is decreasing when the angle is in the interval
2 , . Section 11.4
Tangent Vectors and Normal Vectors
1. rt t2i 2tj
rt 4 cos ti 4 sin tj
3.
rt 2ti 2j, rt 4t2 4 2t2 1 Tt
rt 1 2ti 2j ti j rt 2 t2 1 t2 1
rt 4 sin ti 4 cos tj rt 16 sin2 t 16 cos2 t 4 Tt
2 2 1 i j i j T1 2 2 2
T
5. rt t i t 2j tk
rt sin ti cos tj rt
4 22 i
2
2
j
7. rt 2 cos t i 2 sin tj tk
rt i 2tj k
rt 2 sin t i 2 cos tj k
When t 0, r0 i k, t 0 at 0, 0, 0.
When t 0, r0 2j k, t 0 at 2, 0, 0.
T0
2 r0 i k r0 2
T0
5 r0 2j k r0 5
Direction numbers: a 1, b 0, c 1
Direction numbers: a 0, b 2, c 1
Parametric equations: x t, y 0, z t
Parametric equations: x 2, y 2t, z t
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Section 11.4
Tangent Vectors and Normal Vectors
9. rt 2 cos t, 2 sin t, 4 rt 2 sin t, 2 cos t, 0 When t T
2, 2, 0, , r 4 4
t 4 at 2, 2, 4 .
4 rr 4 4 21 2, 2, 0
Direction numbers: a 2, b 2, c 0 Parametric equations: x 2t 2, y 2t 2, z 4
2 11. rt t, t 2, t 3 3
z 18 15 12 9 6 3 3 6 9 −3
rt 1, 2t, 2t 2 When t 3, r3 1, 6, 18, t 3 at 3, 9, 18. T3
1 r3 1, 6, 18 r3 19
x
12
15 18
y
Direction numbers: a 1, b 6, c 18 Parametric equations: x t 3, y 6t 9, z 18t 18 13. rt ti ln tj t k, 1 1 k rt i j t 2t
15. r4 2, 16, 2
t0 1
u8 2, 16, 2
1
r1 i j 2 k
Hence the curves intersect.
rt i j 1 2k 2 2 1 i j k T1 rt 1 1 1 4 3 3 3
rt 1, 2t,
1 Tangent line: x 1 t, y t, z 1 t 2
us
rt 0 0.1 r1.1 1.1i 0.1j 1.05k cos
1.1, 0.1, 1.05
17.
1 rt ti t2j, t 2 2
Tt T2 N2
14, 2, 13s , u8 14, 2, 121 2 3
r4 u8 16.29167 ⇒ 1.2
r4 u8 16.29513
rt 6 cos ti 6 sin tj k, t
19.
3 4
rt 6 sin ti 6 cos tj
rt i tj Tt
1 1 , r4 1, 8, 2 2
rt i tj rt 1 t2
Tt
rt sin ti cos tj rt
Tt cos t i sin t j, Tt 1
t 1 i 2 j t 2 13 2 t 13 2 N
2 1 i 3 2 j 53 2 5
34
2
2
5 T2 25 1 2i j i j T2 5 5 5
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i
2
2
j
55
56
Chapter 11
Vector-Valued Functions 23. rt 4t 2 i
21. rt 4t i vt 4i
vt 8t i
at O
at 8i
Tt
4i vt i vt 4
Tt
Tt O
Tt O Nt
Tt is undefined. Tt
Nt
The path is a line and the speed is constant.
1 1 25. rt t i j, vt i 2 j, v1 i j, t t 2 at 3 j, a1 2j t Tt
vt 1 1 i 2j t 2i j vt t 4 1 t t 4 1
T1
2 1 i j i j 2 2
t2
1 2
1 t 4 1
i j
2
2
Tt is undefined. Tt
The path is a line and the speed is variable.
27. rt et cos ti et sin tj vt etcos t sin ti etcos t sin tj at et2 sin ti et2 cos tj At t
2 1 v , T i j i j. 2 v 2 2
Motion along r is counterclockwise. Therefore,
2t 3 2t i 4 j 4 3 2 Tt t 1 t 13 2 Nt Tt 2t t 4 1
N1
8ti vt i vt 8t
N
1 2
i j
aT a T 2e 2 aN a N 2e 2
i t 2j
i j
aT a T 2 aN a N 2 29. rt0 cos t0 t0 sin t0i sin t0 t0 cos t0j vt0 2t0 cos t0i 2t0 sin t0j at0 2cos t0 t0 sin t0i t0 cos t0 sin t0j Tt0
v cos t0i sin t0j v
Motion along r is counterclockwise. Therefore Nt0 sin t0i cos t0j. aT a T 2 aN a N 2 t0 3t0
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2
2
i j.
Section 11.4 31. rt a cos ti a sin tj
Tangent Vectors and Normal Vectors
57
33. Speed: vt a The speed is constant since aT 0.
vt a sin ti a cos tj at a 2 cos ti a 2 sin tj Tt
vt sin ti cos tj vt
Nt
Tt cos ti sin tj Tt
aT a T 0 aN a N a 2
35.
1 rt ti j, t0 2 t x t, y
Tt
t2i j t 4 1
Nt
i t 2j t 4 1
at 0
y
17
17 17
17
Tt
14 v 1 i 2j 3k i 2j 3k v 14 14
Nt
T is undefined. T
aT, aN are not defined.
3
2
N
1 r2 2i j 2
N2
vt i 2j 3k
1 ⇒ xy 1 t
1 j t2
rt i
T2
37. rt ti 2t j 3tk
1 1
2, 2
T 1
2
x 3
4i j i 4j
39. rt ti t 2j
t2 k 2
rt 4ti 3 cos tj 3 sin t k
41.
vt 4i 3 sin tj 3 cos tk
vt i 2tj tk v1 i 2j k
v
at 2j k
at 3 cos t j 3 sin t k
1 v Tt i 2tj tk v 1 5t 2 T1
6
6
N1
a
i 2j k
T Nt T 30
30
2 4i 3j
Tt
5t i 2j k 1 5t 23 2 5ti 2j k 5 51 5t 2 2 1 5t
5i 2j k
56 aT a T 6 aN a N
30
2 3k
T
v 1 4i 3 sin tj 3 cos tk v 5
2 514i 3j Nt
T cos tj sin tk T
z
T
N k 2
N 3
aT a T 0
2π 4π
aN a N 3
6 x
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3
y
58
Chapter 11
43. Tt
rt rt
Nt
Tt Tt
Vector-Valued Functions
If at aTTt aNNt, then aT is the tangential component of acceleration and aN is the normal component of acceleration. 45. If aN 0, then the motion is in a straight line. 47. rt t sin t, 1 cos t The graph is a cycloid. (a) rt t sin t, 1 cos t
y
vt cos t, sin t t = 21
at 2 sin t, 2 cos t Tt
vt 1 1 cos t, sin t vt 21 cos t
Nt
Tt 1 sin t, 1 cos t Tt 21 cos t
t=1
t = 23 x
aT a T
1 2 sin t 2 sin t1 cos t 2 cos t sin t 21 cos t 21 cos t
aN a N
1 21 cos t 221 cos t 2 sin2 t 2 cos t1 cos t 21 cos t 21 cos t 2
2 2 2 2 2 1 When t : aT , aN 2 2 2 2
When t 1: aT 0, aN 2 2 2 2 2 3 When t : aT , aN 2 2 2
(b) Speed: s vt 21 cos t
2 sin t ds aT dt 21 cos t 2 2 1 When t : aT > 0 ⇒ the speed in increasing. 2 2
When t 1: aT 0 ⇒ the height is maximum. 2 2 3 < 0 ⇒ the speed is decreasing. When t : aT 2 2
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Section 11.4
Tangent Vectors and Normal Vectors
t rt 2 cos ti 2 sin tj k, t0 2 2
49.
z
3
1 rt 2 sin ti 2 cos tj k 2
2
217 1 Tt 2 sin ti 2 cos t j k 17 2
−1
( 0, 2, π2 ) y
x
2 2j 4 k
T
2 2 1717 2i 21 k
N
2 j
T
−2
2
r
N 1
Nt cos ti sin tj
T B 2 2
B
1
−2
17
17
4i k
i j k 4 17 17 17 i 417 k 17i 4k N 0 2 17 17 17 17 17 0 1 0
51. From Theorem 11.3 we have: rt v0t cos i h v0t sin 16t2 j vt v0 cos i v0 sin 32tj at 32 j Tt Nt
v0 cos i v0 sin 32t j v02 cos2 v0 sin 32t2
v0 sin 32ti v0 cos j v02 cos2 v0 sin 32t2
aT a T aN a N
(Motion is clockwise.)
32v0 sin 32t v02 cos2 v0 sin 32t2 v0
2
cos2
32v0 cos v0 sin 32t2
Maximum height when v0 sin 32t 0; (vertical component of velocity) At maximum height, aT 0 and aN 32. 53. rt 10 cos 10 t, 10 sin 10 t, 4 4t, 0 ≤ t ≤ (a)
1 20
rt 100 sin10 t, 100 cos10 t, 4 rt 1002 sin210 t 1002 cos210 t 16 1002 16 4625 2 1 314 mi hr
(b) aT 0 and aN 1000 2 aT 0 because the speed is constant. 55. rt a cos t i a sin tj From Exercise 31, we know a
T 0 and a N a 2.
(a) Let 0 2 . Then
(b) Let a0 a 2. Then
a N a 0 a2 4a 2
2
2
or the centripetal acceleration is increased by a factor of 4 when the velocity is doubled.
a N a0 2
a2 12 a 2
2
or the centripetal acceleration is halved when the radius is halved.
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59
60
Chapter 11
57. v
Vector-Valued Functions
9.56410010
4
4.83 mi sec
59. v
9.56438510
4
4.67 misec
61. Let Tt cos i sin j be the unit tangent vector. Then Tt
d T d T d d d sin i cos j M . dt d dt dt dt
M sin i cos j cos 2 i sin 2 j and is rotated counterclockwise through an angle of 2 from T. If ddt > 0, then the curve bends to the left and M has the same direction as T. Thus, M has the same direction as
If ddt < 0, then the curve bends to the right and M has the opposite direction as T. Thus,
T , T
N
N
which is toward the concave side of the curve. y
T T
again points to the concave side of the curve. y
T T φ
φ
M
M
N
x
x
63. Using a aTT aNN, T T O, and T N 1, we have: v a vT aTT aNN vaTT T vaNT N vaNT N v a v aNT N v aN v a . v
Thus, aN
Section 11.5
Arc Length and Curvature
1. rt ti 3tj
3. rt a cos3 ti a sin3 tj
dx dy dz 1, 3, 0 dt dt dt
dx dy 3a cos2 t sin t, 3a sin2 t cos t dt dt
4
s
1 9 dt
0
s4
2
3a cos2 tsin t2 3a sin2 t cos t2 dt
0
4
10
2
12a
dt
0
sin t cos t dt
0
10t
4 0
4 10
3a
2
0
y
y
(4, 12) 12 a
8 x
−a
a
4 −a
(0, 0)
x
4
8
12
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2
2 sin 2t dt 3a cos 2t
0
6a
Section 11.5
Arc Length and Curvature
1 5. (a) rt v0 cos ti h v0 sin t gt2 j 2
1 100 cos 45 ti 3 100 sin 45 t 32t2 j 2 50 2ti 3 50 2t 16t2 j (b) vt 50 2i 50 2 32tj 50 2 32t 0 ⇒ t
25 2 16
Maximum height: 3 50 2
2516 2 16 2516 2 81.125 ft
2
(c) 3 50 2t 16t2 0 ⇒ t 4.4614 Range: 50 24.4614 315.5 feet
4.4614
(d) s
50 22 50 2 32t2dt 362.9 feet
0
7. rt 2ti 3tj tk
z
(4, − 6, 2)
dx dy dz 2 3, 1 dt dt dt
2
(0, 0, 0)
2
s
22 32 12 dt
2
x
2
14 dt 14 t
0
2 0
2 14
9. rt a cos ti a sin tj btk
11. rt t 2i tj ln tk
dx dy dz a sin t, a cos t, b dt dt dt s
y −2
0
4
2
dx dy dz 1 2t, 1, dt dt dt t s
0
2
2t2 12
1
2
a2 b2 dt a2 b2 t
0
0
3
2 a2 b2
1
3
z
(a, 0, 2π b)
3
a2 sin2 t a2 cos2 t b2 dt
4t 4 t 2 1
πb
x
(a, 0, 0)
y
13. rt t i 4 t 2j t 3k,
2
dt
4t 4 t 2 1 dt t2
1
2π b
1t
0 ≤ t ≤ 2
(a) r0 0, 4, 0, r2 2, 0, 8 distance 22 42 82 84 2 21 9.165 —CONTINUED—
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t
dt 8.37
61
62
Chapter 11
Vector-Valued Functions
13. —CONTINUED— (b)
r0 0, 4, 0 r0.5 0.5, 3.75, .125 r1 1, 3, 1 r1.5 1.5, 1.75, 3.375 r2 2, 0, 8 distance 0.52 .252 .1252 .52 .752 .8752 0.52 1.252 2.3752 0.52 1.752 4.6252
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. 15. rt 2 cos t, 2 sin t, t
t
(a) s
xu2 yu2 zu2 du
s
(b)
5
0
t
t
2 sin u2 2 cos u2 12 du
x 2 cos
0 t
t
5 du 5 u
0
(c) When s 5:
s5, y 2 sin s5, z s5
0
rs 2 cos
5t
s5i 2 sin s5j s5 k
x 2 cos 1 1.081 y 2 sin 1 1.683 z1
1.081, 1.683, 1.000 When s 4:
4
x 2 cos y 2 sin z
4 5
5
4 5
0.433 1.953
1.789
0.433, 1.953, 1.789 (d) rs
17.
25 sin s5 25 cos s5 15 45 51 1
rs 1 rs Ts
2
2
2
2
i
2
s i 1
2
2
j
2
2
and
2
19. rs 2 cos
s j
rs
12 21 1
rs rs rs
Ts 0 ⇒ K Ts 0
2
s5i 2 sin s5 j s5 k
Ts rs
2 5
sin
s5i 25 cos s5j 15 k
2 s 2 s i sin j Ts cos 5 5 5 5 (The curve is a line.)
K Ts
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2 5
Section 11.5
21.
rt 4t i 2tj
23.
vt 4i 2j
1 rt t i j t vt i
1 Tt 2i j 5
2 j t3
at
Tt 0 K rt
(The curve is a line.)
a1 2j Tt
t 2i j t 4 1
Nt
1 i t 2j t 4 112
N1 K
rt 4 cos2 ti 4 sin2 t j
27.
i j
a N 2 v 2 2
rt a cos ti a sin t j rt a sin ti a cos tj
Tt sin2 ti cos2 tj
Tt sin ti cos tj
K
Tt cos ti sin t j
Tt 2 1 rt 8 4
K
rt et cos t i et sin tj
Tt Tt K
t
1
K
From Exercise 21, Section 11.4, we have: a
N 3t K
1
3t at Nt 42 2 v
t
t
Tt 1 2 t e rt 2et 2 t2 k 2
rt i 2t j tk
Tt
t
cos t sin ti sin t cos t j
rt ti t 2j
Tt
t
1 sin t cos ti cos t sin tj 2 2
Tt
1 rt a a
31. rt cos t t sin t, sin t t cos t
rt e sin t e cos t i e cos t e sin tj t
33.
1 2
rt 8 sin2 ti 8 cos2 tj Tt 2 cos2 ti 2 sin2 t j
29.
1 j t2
v1 i j
Tt 0
25.
Arc Length and Curvature
i 2tj tk 1 5t 2 5t i 2j k 1 5t 232 Tt rt
35.
rt 4ti 3 cos tj 3 sin tk rt 4i 3 sin tj 3 cos tk 1 Tt 4i 3 sin tj 3 cos tk 5 1 Tt 3 cos tj 3 sin tk 5 K
3 Tt 35 rt 5 25
5 5 1 5t 2 1 5t 232 1 5t 2
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63
64
Chapter 11
Vector-Valued Functions
37. y 3x 2
39. y 2x2 3
Since y 0, K 0, and the radius of curvature is undefined.
y 4x y 4 K
4 4 0.057 1 4232 1732
1732 1 17.523 K 4
41. y a2 x2 x y a2 x2 y
43. (a) Point on circle:
2 , 0 Equation: x 2
y 0
r
1 1a K 1 0232 a 1 a K
y2 1
1 . K
(radius of curvature)
1 1 2 45. y x , y 1 2, y 3 x x x
Radius of curvature 12. Since the tangent line is horizontal at 1, 2, the normal line is vertical. The center of the circle is 12 unit above the point 1, 2 at 1, 52.
5 Circle: x 12 y 2
2
1 4
y ex,
x0
y ex,
y ex
y0 1,
y0 1
47.
2 2 1 0232
K
1 1 1 1 , r 2 2 1 1232 232 2 2 K
The slope of the tangent line at 0, 1 is y0 1. The slope of the normal line is 1. Equation of normal line: y 1 x or y x 1 The center of the circle is on the normal line 2 2 units away from the point 0, 1.
4
(1, 2) −6
2
(b) The circles have different radii since the curvature is different and
1 y a
K
2 , 1
Center:
2x2 a2 a2 x232
At x 0:
(radius of curvature)
0 x 2 1 y2 2 2
6
x2 x2 8
−4
x2 4 x ±2 Since the circle is above the curve, x 2 and y 3. Center of circle: 2, 3 Equation of circle: x 22 y 32 8 6
(0, 1)
−6 0
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3
Section 11.5
65
51. y x 12 3, y 2x 1, y 2
y
49.
Arc Length and Curvature
π
K x
π
B
2 2 1 2x 1232 1 4x 1232
(a) K is maximum when x 1 or at the vertex 1, 3.
A
(b) lim K 0 x→
−2π
2 2 53. y x23, y x13, y x43 3 9
55. y x 13 3
29x43 6 K 13 23 1 49x2332 x 9x 432 (a) K ⇒
as x ⇒ 0. No maximum
x→
y
y 6x 1 K
(b) lim K 0
57. K
y 3x 12
y 6x 1 0 at x 1. 1 y232 1 9x 1432
Curvature is 0 at 1, 3.
b
59. s
1 y
2 32
rt dt
61. The curve is a line.
a
The curvature is zero when y 0. 63. Endpoints of the major axis: ± 2, 0 Endpoints of the minor axis: 0, ± 1 x2 4y2 4 2x 8yy 0 y y K
x 4y
4y1 x4y 4y x2y 4y2 x2 1 3 16y2 16y2 16y3 4y
14y3
1 x4y232
16 16 16 16y2 x232 12y2 432 16 3x232
Therefore, since 2 ≤ x ≤ 2, K is largest when x ± 2 and smallest when x 0. 65. f x x4 x2 (a) K
2 6x2 1 16x 16x4 4x2 132 6
1 (b) For x 0, K 2. f 0 0. At 0, 0, the circle of curvature has radius 2 . Using the symmetry of the graph of f, you obtain
x2 y
1 2
2
1 . 4
For x 1, K 2 5 5. f 1 0. At 1, 0, the circle of curvature has radius 5
2
1 . K
Using the graph of f, you see that the center of curvature is 0, 12 . Thus, x2
1 y 2
2
5 . 4
2
f −3
3
To graph these circles, use y
1 ± 2
1 x2 and 4
y
1 ± 2
5 x2. 4
—CONTINUED—
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−2
66
Chapter 11
Vector-Valued Functions
65. —CONTINUED— (c) The curvature tends to be greatest near the extrema of f, and K decreases as x → ± . However, f and K do not have the same critical numbers. 2
Critical numbers of f: x 0, ±
2
± 0.7071
5
−3
3
Critical numbers of K: x 0, ± .7647, ± 0.4082
−2
67. (a) Imagine dropping the circle x2 y k2 16 into the parabola y x2. The circle will drop to the point where the tangents to the circle and parabola are equal.
y
and x2 y k2 16 ⇒ x2 x2 k2 16
y x2
Taking derivatives, 2x 2 y ky 0
y ky x ⇒ y
15
and y 2x. Hence,
10
x . yk
−10
−5
x 5
10
Thus, 1 x 2x ⇒ x 2x y k ⇒ 1 2x2 k ⇒ x2 k . yk 2 Thus,
21
x2 x2 k2 x2
2
16 ⇒ x2 15.75.
1 Finally, k x2 2 16.25, and the center of the circle is 16.25 units from the vertex of the parabola. Since the radius of the circle is 4, the circle is 12.25 units from the vertex.
(b) In 2-space, the parabola z y2 or z x2 has a curvature of K 2 at 0, 0. The radius of the largest sphere that will 1 touch the vertex has radius 1K 2 .
69. Given y f x: K R
y 1 y 232 1 K
The center of the circle is on the normal line at a distance of R from x, y. 1 Equation of normal line: y y0 x x0 y
x x y1x x
2
0
2
0
x x02 1
1 y232 y
Thus, x0, y0 x yz, y z. For y ex, y ex, y ex, z
1 y23 1 y2 y2
x x02
When x 0: x0 x yz 0 12 2
y 1 y y2 2
1 e2x ex ex. ex
y0 y z 1 2 3
2 2
y1 y2 x x0 yz y
Center of curvature: 2, 3 (See Exercise 47)
x0 x yz 1 y y0 x x yz z y y0 y z
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Section 11.5 71. r 1 sin
73. r a sin
r cos
r a cos
r sin K
r a sin
2r
rr r2 r 232 2
r2
K
2 cos2 1 sin sin 1 sin 2
cos2
1 sin
2 3
3 31 sin 81 sin 3 2 21 sin
75. r ea , a > 0 r
2r 2 rr r 2 r2 r 232
2a 2 cos2 a2 sin2 a2 sin2 a2 cos2 a2 sin2 3
2a2 2 ,a > 0 a3 a
77. r 4 sin 2
aea
r 8 cos 2
r a2ea
2r
rr r2 r 232 1 a 2 e a 1
K
Arc Length and Curvature
2
2a2e2a
r2
(a) As ⇒
, K ⇒ 0.
(b) As a ⇒
, K ⇒ 0.
a2e2a
e2a
a2e2a e2a 32
At the pole: K
2
r0
2 1 8 4
79. x f t
81. x a sin
y g t
x a1 cos
y a sin
x a sin
y a cos
dy dy dt gt y dx dx ft dt
K
d gt ftgt gtft dt ft ft2 y dx ft dt
K
y
ftgt gtft ft3
1 y232
ftgt gt f t ft 3 gt 2 32 1 ft
ftgt gtft ft3
x y2 y 2x32 x y
1 cos cos a2 sin2 a 21 cos 2 a 2 sin2 32
a2
1 cos 1 a 2 2 cos 32
1 1 cos a 2 21 cos 32
1 1 csc 2 2a 2 2 cos 4a
Minimum:
83. aN mK
1 4a
Maximum: none
2 3
ftgt gtf t ft2 gt 232
lb 1 305280 ft
dsdt 325500 ft sec 100 ft 3600 sec 2
2
2
1 cos ≥ 0
2
ft ftgt 2
y a1 cos
3327.5 lb
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K → as → 0
67
68
Chapter 11
Vector-Valued Functions
85. Let r xti ytj ztk. Then r r xt 2 yt 2 zt 2 and r xti ytj ztk. Then, r
drdt xt
2
yt 2 zt 2
12 xt
2
yt 2 zt 2 1 2 2xtxt 2ytyt 2ztzt
xtxt ytyt ztzt r r. 87. Let r xi yj zk where x, y, and z are functions of t, and r r. d r rr rdr dt rr rr r r r 2r r r r dt r r2 r2 r3
(using Exercise 77)
x2 y2 z2x i yj z k xx yy zz xi yj zk r3
1 xy 2 xz 2 xyy xzzi x 2y z 2y xxy zz yj x 2z y 2z xxz yyzk r3
i 1 yz yz 3 r x
j xz xz y
k 1 xy xy 3 r r r r z
89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have r L GM
rr e.
y
Planet Sun
Since r L and r are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate the coordinate system so that e lies along the positive x-axis and is the angle between e and r. Let e e. Then r e r e cos re cos . Also,
r
θ e
L2 L L r r L
r r L r GM e
r r
GM r e r r r GMre cos r
Thus, L 2 GM r 1 e cos and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse.
91. A
1 2
r d 2
Thus, dA dA d 1 2 d 1 r L dt d dt 2 dt 2 and r sweeps out area at a constant rate.
Review Exercises for Chapter 11 1. rt ti csc tk
3. rt ln ti tj tk
(a) Domain: t n , n an integer
(a) Domain: 0,
(b) Continuous except at t n , n an integer
(b) Continuous for all t > 0
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x
Review Exercises for Chapter 11 5. (a) r0 i 8 (b) r2 3i 4j 3 k 1 (c) rc 1 2c 1 1 i c 12j 3 c 13k
2c 1i c 12j 13 c 13k 1 1 (d) r1 t r1 21 t 1 i 1 t 2j 3 1 t 3k 3i j 3 k 1 2t i tt 2j 3t 3 3t 2 3t k
7. rt cos ti 2 sin2 tj xt cos t, yt 2
sin2
9. rt i tj t 2k x1
t
x 1, y sin t, z 1
yt
y 1 2
x2
11. rt i sin tj k
z t2 ⇒ z y2
y 21 x2
z
1 ≤ x ≤ 1
2
y 1
t
0
2
x
1
1
1
1
y
0
1
0
1
z
1
1
1
1
3 2
z
1 2
y 3
x
x
−1
1
2
1
1 −2
1
1
2
2 3
x
1 13. rt ti ln tj 2 t 2k
15. One possible answer is:
z 3 2 1
1
1
2
r1t 4ti 3tj,
0 ≤ t ≤ 1
r2t 4i 3 t j,
0 ≤ t ≤ 3
r3t 4 t i,
0 ≤ t ≤ 4
y
2 3
x
17. The vector joining the points is 7, 4, 10. One path is rt 2 7t, 3 4t, 8 10t.
19. z x2 y 2, x y 0, t x x t, y t, z 2t 2 rt ti tj 2t 2k z 5
−3 2
1
3
x
21. lim t 2 i 4 t 2 j k 4i k t→2
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2
3
y
y
69
70
Chapter 11
Vector-Valued Functions
2 23. rt 3ti t 1j, ut ti t 2j t 3k 3 (a) rt 3i j
(b) rt 0
(c) rt ut 3t 2 t 2t 1 t 3 2t 2
2 (d) ut 2rt 5ti t 2 2t 2 j t 3k 3
Dtrt ut 3t 2 4t
Dtut 2rt 5i 2t 2j 2t 2k 2 (f ) rt ut t 4 t 3 i 2t 4j 3t 3 t 2 tk 3
(e) rt 10t 2 2t 1 Dtrt
10t 1
Dtrt ut
10t 2 2t 1
25. xt and yt are increasing functions at t t0, and zt is a decreasing function at t t 0.
29.
cos t i sin tj tk dt
31. rt
1 t 2 dt
27.
83t
3
2t 2 i 8t 3j 9t 2 2t 1k
cos t i t cos tj dt sin t i t sin t cos t j C
1 t1 t 2 ln t 1 t 2 C 2
2
2ti etj et k dt t 2i etj et k C
33.
2
3t i 2t 2j t 3k dt
3t2 i 2t3 j t4 k 2
3
4
2 2
r0 j k C i 3j 5k ⇒ C i 2j 4k rt t2 1i et 2j et 4k
2
35.
et 2i 3t2j k dt 2et 2i t3j tk
0
2 0
2e 2i 8j 2k
37. rt cos3 t, sin3 t, 3t vt rt 3 cos2 t sin t, 3 sin2 t cos t, 3 vt 9 cos4 t sin2 t 9 sin4 t cos2 t 9 3cos2 t sin2 tcos2 t sin2 t 1 3cos2 t sin2 t 1 at vt 6 cos tsin2 t 3 cos2 t cos t, 6 sin t cos2 t 3 sin2 tsin t, 0 3 cos t2 sin2 t cos2 t, 3 sin t2 cos2 t sin2 t, 0
39.
1 rt lnt 3, t 2, t , t 0 4 2 rt
41. Range x
t 1 3, 2t, 21
r4 1, 8,
1 2
direction numbers
Since r4 0, 16, 2, the parametric equations are 1 x t, y 16 8t, z 2 2 t. rt 0 0.1 r4.1 0.1, 16.8, 2.05
43. Range x
v02 sin 2 80 ⇒ v0 9.8
9.8 34.9 m sec 80sin40
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v02 752 sin 2 sin 60 152 feet 32 32
32 j 3
Review Exercises for Chapter 11 rt 5ti
45.
rt ti tj
47.
vt 5i
vt i
vt 5 at 0
vt
4t 1
2t
Tt i at
Nt does not exist aT0 a
N
i 1 2t j 2t i j 4t 1 2t 4t 1
Nt
i 2t j 4t 1
a
T
1 4tt4t 1
a
N
does not exist
rt et i etj
1 j 4tt
Tt
(The curve is a line.)
49.
1 j 2t
1 2t4t 1
1 rt ti t 2j t 2k 2
51.
vt et i etj
vt i 2tj tk
vt e2t e2t
v 1 5t 2
at et i et j
at 2j k
et i et j Tt e2t e2t
Tt
Nt
et i etj e2t e2t
i 2tj tk 1 5t2
Nt
aT
e2t e2t e2t e2t
5t i 2j k 51 5t2
a
N
a
T
a
N
2 e2t e2t
5t 1 5t 2
5 51 5t 2
53. rt 2 cos ti 2 sin tj tk, x 2 cos t, y 2 sin t, z t 3 3 , x 2, y 2, z . 4 4
When t
rt 2 sin ti 2 cos tj k Direction numbers when t
3 , a 2, b 2, c 1 4
x 2t 2, y 2t 2, z t
55. v
9.56460010
4
4.56 mi sec
57. rt 2ti 3tj, 0 ≤ t ≤ 5
s
rt dt
a
5 0
−4 −2
5
0
13t
y 2
rt 2i 3j b
3 4
513
4 9 dt
(0, 0) x 2 4 6
8 10 12 14
−4 −6 −8 −10 −12 −14
(10, − 15)
−16
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5 1 5t 2
71
72
Chapter 11
Vector-Valued Functions
59.
rt 10 cos3 ti 10 sin3 tj
y
rt 30 cos2 t sin ti 30 sin2 t cos tj
10
rt 30cos4 t sin2 t sin4 t cos2 t
2
30 cos t sin t
s4
2
30 cos t
0
−10
sin t dt 120
sin2 t 2
2
x
−2
60
2
− 10
0
61. rt 3ti 2tj 4tk, 0 ≤ t ≤ 3
b
3
rt dt
a
12 10 8 6 4 2
3
9 4 16 dt
0
29 dt 329
0 2
2
s
rt dt
a
2
65 dt
0
1 rt i cos tj sin tk 2
65 2
s
z
(
0, 8,
π 2
)
rt dt
4 6 8
0
4 8 6
(8, 0, 0)
0
π 2
x
y
67. rt 3ti 2tj
69.
Line
5
2
1 cos2 t sin2 t dt 4 dt
0
25 t
0
5
2
1 rt 2ti t2j t2k 2 rt 2i tj 2tk, r 5t2 4
k0
rt j 2k
i r r 2 0 K
k 2t 4j 2k, r r 20 2
20 25 r r 2 r3 5t 43 2 4 5t23 2
1 1 y , y 2 x x
y x y 1
y 2 3 2
1 y
At x 4, K
j t 1
73. y ln x
1 71. y x2 2 2
K
y
1 65. rt ti sin tj cos tk, 0 ≤ t ≤ 2
rt < 8 sin t, 8 cos t, 1, rt 65 b
(0, 0, 0) 2 4 6 8 10
x
63. rt 8 cos t, 8 sin t, t, 0 ≤ t ≤
(−9, 6, 12)
z
rt 3i 2j 4k s
10
K 1 1 x23 2
y 2 3 2
1 y
At x 1, K
1 and r 173 2 1717. 173 2
75. The curvature changes abruptly from zero to a nonzero constant at the points B and C.
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1 x2 1 1 x2]3 2
2 1 1 and r 22. 23 2 22 4
68
Chapter 11
Vector-Valued Functions
85. Let r xti ytj ztk. Then r r xt 2 yt 2 zt 2 and r xti ytj ztk. Then, r
drdt xt
2
yt 2 zt 2
12 xt
2
yt 2 zt 2 1 2 2xtxt 2ytyt 2ztzt
xtxt ytyt ztzt r r. 87. Let r xi yj zk where x, y, and z are functions of t, and r r. d r rr rdr dt rr rr r r r 2r r r r dt r r2 r2 r3
(using Exercise 77)
x2 y2 z2x i yj z k xx yy zz xi yj zk r3
1 xy 2 xz 2 xyy xzzi x 2y z 2y xxy zz yj x 2z y 2z xxz yyzk r3
i 1 yz yz 3 r x
j xz xz y
k 1 xy xy 3 r r r r z
89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have r L GM
rr e.
y
Planet Sun
Since r L and r are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate the coordinate system so that e lies along the positive x-axis and is the angle between e and r. Let e e. Then r e r e cos re cos . Also,
r
θ e
L2 L L r r L
r r L r GM e
r r
GM r e r r r GMre cos r
Thus, L 2 GM r 1 e cos and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse.
91. A
1 2
r d 2
Thus, dA dA d 1 2 d 1 r L dt d dt 2 dt 2 and r sweeps out area at a constant rate.
Review Exercises for Chapter 11 1. rt ti csc tk
3. rt ln ti tj tk
(a) Domain: t n , n an integer
(a) Domain: 0,
(b) Continuous except at t n , n an integer
(b) Continuous for all t > 0
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x
Review Exercises for Chapter 11 5. (a) r0 i 8 (b) r2 3i 4j 3 k 1 (c) rc 1 2c 1 1 i c 12j 3 c 13k
2c 1i c 12j 13 c 13k 1 1 (d) r1 t r1 21 t 1 i 1 t 2j 3 1 t 3k 3i j 3 k 1 2t i tt 2j 3t 3 3t 2 3t k
7. rt cos ti 2 sin2 tj xt cos t, yt 2
sin2
9. rt i tj t 2k x1
t
x 1, y sin t, z 1
yt
y 1 2
x2
11. rt i sin tj k
z t2 ⇒ z y2
y 21 x2
z
1 ≤ x ≤ 1
2
y 1
t
0
2
x
1
1
1
1
y
0
1
0
1
z
1
1
1
1
3 2
z
1 2
y 3
x
x
−1
1
2
1
1 −2
1
1
2
2 3
x
1 13. rt ti ln tj 2 t 2k
15. One possible answer is:
z 3 2 1
1
1
2
r1t 4ti 3tj,
0 ≤ t ≤ 1
r2t 4i 3 t j,
0 ≤ t ≤ 3
r3t 4 t i,
0 ≤ t ≤ 4
y
2 3
x
17. The vector joining the points is 7, 4, 10. One path is rt 2 7t, 3 4t, 8 10t.
19. z x2 y 2, x y 0, t x x t, y t, z 2t 2 rt ti tj 2t 2k z 5
−3 2
1
3
x
21. lim t 2 i 4 t 2 j k 4i k t→2
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2
3
y
y
69
70
Chapter 11
Vector-Valued Functions
2 23. rt 3ti t 1j, ut ti t 2j t 3k 3 (a) rt 3i j
(b) rt 0
(c) rt ut 3t 2 t 2t 1 t 3 2t 2
2 (d) ut 2rt 5ti t 2 2t 2 j t 3k 3
Dtrt ut 3t 2 4t
Dtut 2rt 5i 2t 2j 2t 2k 2 (f ) rt ut t 4 t 3 i 2t 4j 3t 3 t 2 tk 3
(e) rt 10t 2 2t 1 Dtrt
10t 1
Dtrt ut
10t 2 2t 1
25. xt and yt are increasing functions at t t0, and zt is a decreasing function at t t 0.
29.
cos t i sin tj tk dt
31. rt
1 t 2 dt
27.
83t
3
2t 2 i 8t 3j 9t 2 2t 1k
cos t i t cos tj dt sin t i t sin t cos t j C
1 t1 t 2 ln t 1 t 2 C 2
2
2ti etj et k dt t 2i etj et k C
33.
2
3t i 2t 2j t 3k dt
3t2 i 2t3 j t4 k 2
3
4
2 2
r0 j k C i 3j 5k ⇒ C i 2j 4k rt t2 1i et 2j et 4k
2
35.
et 2i 3t2j k dt 2et 2i t3j tk
0
2 0
2e 2i 8j 2k
37. rt cos3 t, sin3 t, 3t vt rt 3 cos2 t sin t, 3 sin2 t cos t, 3 vt 9 cos4 t sin2 t 9 sin4 t cos2 t 9 3cos2 t sin2 tcos2 t sin2 t 1 3cos2 t sin2 t 1 at vt 6 cos tsin2 t 3 cos2 t cos t, 6 sin t cos2 t 3 sin2 tsin t, 0 3 cos t2 sin2 t cos2 t, 3 sin t2 cos2 t sin2 t, 0
39.
1 rt lnt 3, t 2, t , t 0 4 2 rt
41. Range x
t 1 3, 2t, 21
r4 1, 8,
1 2
direction numbers
Since r4 0, 16, 2, the parametric equations are 1 x t, y 16 8t, z 2 2 t. rt 0 0.1 r4.1 0.1, 16.8, 2.05
43. Range x
v02 sin 2 80 ⇒ v0 9.8
9.8 34.9 m sec 80sin40
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v02 752 sin 2 sin 60 152 feet 32 32
32 j 3
Review Exercises for Chapter 11 rt 5ti
45.
rt ti tj
47.
vt 5i
vt i
vt 5 at 0
vt
4t 1
2t
Tt i at
Nt does not exist aT0 a
N
i 1 2t j 2t i j 4t 1 2t 4t 1
Nt
i 2t j 4t 1
a
T
1 4tt4t 1
a
N
does not exist
rt et i etj
1 j 4tt
Tt
(The curve is a line.)
49.
1 j 2t
1 2t4t 1
1 rt ti t 2j t 2k 2
51.
vt et i etj
vt i 2tj tk
vt e2t e2t
v 1 5t 2
at et i et j
at 2j k
et i et j Tt e2t e2t
Tt
Nt
et i etj e2t e2t
i 2tj tk 1 5t2
Nt
aT
e2t e2t e2t e2t
5t i 2j k 51 5t2
a
N
a
T
a
N
2 e2t e2t
5t 1 5t 2
5 51 5t 2
53. rt 2 cos ti 2 sin tj tk, x 2 cos t, y 2 sin t, z t 3 3 , x 2, y 2, z . 4 4
When t
rt 2 sin ti 2 cos tj k Direction numbers when t
3 , a 2, b 2, c 1 4
x 2t 2, y 2t 2, z t
55. v
9.56460010
4
4.56 mi sec
57. rt 2ti 3tj, 0 ≤ t ≤ 5
s
rt dt
a
5 0
−4 −2
5
0
13t
y 2
rt 2i 3j b
3 4
513
4 9 dt
(0, 0) x 2 4 6
8 10 12 14
−4 −6 −8 −10 −12 −14
(10, − 15)
−16
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5 1 5t 2
71
72
Chapter 11
Vector-Valued Functions
59.
rt 10 cos3 ti 10 sin3 tj
y
rt 30 cos2 t sin ti 30 sin2 t cos tj
10
rt 30cos4 t sin2 t sin4 t cos2 t
2
30 cos t sin t
s4
2
30 cos t
0
−10
sin t dt 120
sin2 t 2
2
x
−2
60
2
− 10
0
61. rt 3ti 2tj 4tk, 0 ≤ t ≤ 3
b
3
rt dt
a
12 10 8 6 4 2
3
9 4 16 dt
0
29 dt 329
0 2
2
s
rt dt
a
2
65 dt
0
1 rt i cos tj sin tk 2
65 2
s
z
(
0, 8,
π 2
)
rt dt
4 6 8
0
4 8 6
(8, 0, 0)
0
π 2
x
y
67. rt 3ti 2tj
69.
Line
5
2
1 cos2 t sin2 t dt 4 dt
0
25 t
0
5
2
1 rt 2ti t2j t2k 2 rt 2i tj 2tk, r 5t2 4
k0
rt j 2k
i r r 2 0 K
k 2t 4j 2k, r r 20 2
20 25 r r 2 r3 5t 43 2 4 5t23 2
1 1 y , y 2 x x
y x y 1
y 2 3 2
1 y
At x 4, K
j t 1
73. y ln x
1 71. y x2 2 2
K
y
1 65. rt ti sin tj cos tk, 0 ≤ t ≤ 2
rt < 8 sin t, 8 cos t, 1, rt 65 b
(0, 0, 0) 2 4 6 8 10
x
63. rt 8 cos t, 8 sin t, t, 0 ≤ t ≤
(−9, 6, 12)
z
rt 3i 2j 4k s
10
K 1 1 x23 2
y 2 3 2
1 y
At x 1, K
1 and r 173 2 1717. 173 2
75. The curvature changes abruptly from zero to a nonzero constant at the points B and C.
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1 x2 1 1 x2]3 2
2 1 1 and r 22. 23 2 22 4
Problem Solving for Chapter 11
73
Problem Solving for Chapter 11
t
1. xt
cos
0
2 du, yt sin 2 du u2
3. Bomb: r1t 5000 400t, 3200 16t2
0
t2 t2 xt cos , yt sin 2 2
Projectile: r2t v0 cos t, v0 sin t 16t2
a
(a) s
u2
t
At 1600 feet: Bomb: 3200 16t2 1600 ⇒ t 10 seconds.
a
xt2 yt2 dt
0
dt a
Projectile will travel 5 seconds:
0
(b) x t t sin
K
t cos2
t2
t2
t2 t2 t sin2 2 2 1
2 , yt t cos 2
5v0 sin 1625 1600 v0 sin 400. Horizontal position:
t
At t 10, bomb is at 5000 40010 1000.
At t a, K a.
At t 5, projectile is at 5v0 cos .
(c) K a length
Thus, v0 cos 200. Combining, v0
v0 sin 400 ⇒ tan 2 ⇒ 63.4 . v0 cos 200
200 447.2 ft sec cos
5. x 1 cos , y sin , 0 ≤ ≤ 2 x2 y2 1 cos 2 sin2
2 2 cos
t
st
2 sin
d 4 cos 2 2
t
4 sin 2 2
4 cos
t 2
x sin , y cos K
1 1 cos cos sin sin 1 cos 1 3 2 sin 8 sin3 2 2
1
4 sin Thus,
2
1 t 4 sin and K 2
s2 2 16 cos2
2t 16 sin 2t 16. 2
7. r2t rt rt d d rt2 2rt rt dt dt rt rt rt rt ⇒
d rt rt rt dt rt
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74
Chapter 11
Vector-Valued Functions
9. rt 4 cos ti 4 sin tj 3tk, t
2
11. (a) B T N 1 constant length ⇒
rt 4 sin ti 4 cos tj 3k, rt 5
dB d T N T N T N ds ds
rt 4 cos ti 4 sin tj T
4 4 3 T sin ti cos tj k 5 5 5
dB T T N T T N ds
T T N T T
4 4 T cos ti sin tj 5 5
At t
3 3 4 sin ti cos tj k 5 5 5
(b) B T N. Using Exercise 10.3, number 64,
B N T N N N T N N
2 j
B T T N T T T N T
z 6π
B
Now, KN
T
N 4
3
dTds TTss Ts dTds .
Finally,
1
2
NT T TN
N.
T
N B
NT N TN
T
3 4 B i k 2 5 5
3π
Hence,
4 3 ,T i k 2 2 5 5 N
T 0 T
dB dB dB B and T ⇒ N ds ds ds for some scalar .
N cos ti sin tj BT N
dB B ds
y
4
x
d B T B T B T ds
Ns
B KN N T KT B. 13. rt t cos t, t sin t , 0 ≤ t ≤ 2 (a)
2
(b) Length
2
rt dt
0
2
−3
3
2t 2 1 dt 6.766
(graphing utility)
0
−2
(c)
K
2t 2 2 2t 2 1 3 2
(d)
5
K0 2 K1
2 2 1.04 2 13 2
5
0 0
K2 0.51 (e) lim K 0 t→
(f ) As t → , the graph spirals outward and the curvature decreases.
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C H A P T E R 1 2 Functions of Several Variables Section 12.1 Introduction to Functions of Several Variables . . . . . . . 308 Section 12.2 Limits and Continuity Section 12.3 Partial Derivatives
. . . . . . . . . . . . . . . . . . . 312
. . . . . . . . . . . . . . . . . . . . . 315
Section 12.4 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . 321 Section 12.5 Chain Rules for Functions of Several Variables
. . . . . .325
Section 12.6 Directional Derivatives and Gradients . . . . . . . . . . . 330 Section 12.7 Tangent Planes and Normal Lines . . . . . . . . . . . . . 334 Section 12.8 Extrema of Functions of Two Variables . . . . . . . . . . 340 Section 12.9 Applications of Extrema of Functions of Two Variables
. 345
Section 12.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 350 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
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C H A P T E R 1 2 Functions of Several Variables Section 12.1
Introduction to Functions of Several Variables
Solutions to Even-Numbered Exercises 4. z x ln y 8 0
2. xz 2 2xy y 2 4 No, z is not a function of x and y. For example, x, y 1, 0 corresponds to both z ± 2
z 8 x ln y Yes, z is a function of x and y.
8. gx, y ln x y
6. f x, y 4 x 2 4y 2
(c) ge, 0 lne 0 1 (d) g0, 1 ln0 1 0 (e) g2, 3 ln2 3 ln 1 0 (f) ge, e lne e ln 2e
(a) f 0, 0 4
(a) g2, 3 ln 2 3 ln 5
(b) f 0, 1 4 0 4 0
(b) g5, 6 ln 5 6 ln 11
(c) f 2, 3 4 4 36 36 (d) f 1, y 4 1 4y 2 3 4y 2 (e) f x, 0 4 x 2 0 4 x 2 (f) f t, 1 4 t 2 4 t 2
ln 2 ln e ln 2 1 12. Vr, h r 2h
10. f x, y, z x y z (a) f 0, 5, 4 0 5 4 3
(a) V3, 10 3 210 90
(b) f 6, 8, 3 6 8 3 11
(b) V5, 2 522 50
y
14. gx, y
x
1 dt t
1
(a) g4, 1
4
1 dt ln t t
1 4
3
ln 4
(b) g6, 3
6
16. f x, y 3xy y 2 (a)
f x x, y f x, y 3x x y y 2 3xy y 2 x x
(b)
3xy 3xy y 2 3xy y 2 3x y 3y, x 0 x x
f x, y y f x, y 3x y y y y2 3xy y 2 y y
3xy 3xy y 2 2yy y2 3xy y 2 y
y3x 2y y 3x 2y y, y 0 y
1 dt ln t t
308
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3 6
ln 3 ln 6 ln
12
Section 12.1 18. f x, y 4 x 2 4y 2
20. f x, y arccos
Domain: 4 x 2 4y 2 ≥ 0
4
x, y:
y2 1
y x
y ≤ 1 x
xy 6 > 0
Domain:
xy > 6
x, y: xy > 6
Range: 0 ≤ z ≤
≤ 1
309
22. f x, y lnxy 6
Domain: x, y: 1 ≤
x 2 4y 2 ≤ 4 x2
Introduction to Functions of Several Variables
Range: all real numbers
x2 y2 ≤ 1 4 1
Range: 0 ≤ z ≤ 2
24. z
xy xy
26. f x, y x 2 y 2
28. gx, y xy
Domain: x, y: x is any real number,
Domain: x, y: x y
y is any real number
Range: all real numbers
Domain: x, y: y ≥ 0 Range: all real numbers
Range: z ≥ 0
30. (a) Domain: x, y: x is any real number, y is any real number Range: 2 ≤ z ≤ 2 (b) z 0 when x 0 which represents points on the y-axis. (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.
32. f x, y 6 2x 3y Plane
1 Plane: z 2 x
Domain: entire xy-plane Range: < z <
1 36. z x 2 y 2 2 Cone
34. gx, y 12 x
Domain of f: entire xy-plane
z
Range: z ≥ 0
4 3 2
z 6
z −4
4 x
2 4
−2 −3 −4
y
3
2
1
1
2
y
3
x 4
2
3
3
4
y
x
x ≥ 0, y ≥ 0 elsewhere Domain of f : entire xy-plane
38. f x, y
0,
xy,
Range: z ≥ 0 z
1 144 16x 2 9y 2 40. f x, y 12
42. f x, y x sin y
Semi-ellipsoid
z
Domain: set of all points lying on or inside the ellipse x 29 y 216 1
4 −4
−4
Range: 0 ≤ z ≤ 1
25 20
z
5
y
5
x
x
4
−2
y −4
4
10
4
x
15
4
y
−4
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310
Chapter 12
Functions of Several Variables
44. f x, y xy, x ≥ 0, y ≥ 0 (a)
(b) g is a vertical translation of f 3 units downward
z 25
(c) g is a reflection of f in the xy-plane
20 15 10 5
y
5
x
(e)
(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.
z 25 20 15 10 5
y
5
46. z e1x
2 y 2
48. z cos
Level curves:
x
x 42y
2
50. f x, y 6 2x 3y The level curves are of the form 6 2x 3y c or 2x 3y 6 c. Thus, the level curves are straight lines with a slope of 23 .
Level curves: 1x 2 y 2
ce
ln c 1
x2
c cos
y2
x 2 y 2 1 ln c Hyperbolas centered at 0, 0
cos1 c
x
2
2y 2 4
x 2 2y 2 4
y 3
x 2 2y 2 4 cos1 c
Matches (d)
Ellipses
x
−2
Matches (a)
c = 10 c=8
54. f x, y e xy2
52. f x, y x2 2y2
The level curves are of the form
The level curves are ellipses of the form x2 2y2 c (except x2 2y2 0 is the point 0, 0).
e xy2 c, or ln c
y 3
Thus, the level curves are hyperbolas.
c=8 c=6 c=4 c=2 c=0
y
c=4 c=3 c=2
2
x
−3
3 1 −3
xy . 2
−2
x
−1 −1
c = 12
−2
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1
2
c = 14 c = 13
c=0 c=2 c=4 c=6
Section 12.1 56. f x, y lnx y
y
58. f x, y xy
c = − 21
c = −2
Introduction to Functions of Several Variables
The level curves are of the form
4
x
6
c lnx y
311
−6
6
c=0 −4
ec x y
c=1
c = 21
−4
c = −1 −6
y x ec
c=2
c = ± 23
Thus, the level curves are parallel lines of slope 1 passing through the fourth quadrant.
60. hx, y 3 sin x y
62. 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, for c, a constant.
1
−1
1
−1
64. f x, y
x y
66. The surface could be an ellipsoid centered at 0, 1, 0. One possible function is
The level curves are the lines f x, y x 2
1 x c or y x y c
y 1 2 1. 4
These lines all pass through the origin. 68. Ar, t 1000e rt Number of years Rate
5
10
15
0.08
$1491.82
$2225.54
$3320.12
$4953.03
0.10
$1648.72
$2718.28
$4481.69
$7389.06
0.12
$1822.12
$3320.12
$6049.65
$11,023.18
0.14
$2013.75
$4055.20
$8166.17
$16,444.65
70. f x, y, z 4x y 2z c4
20
72. f x, y, z x2 14 y2 z
c0
c1
4 4x y 2z
0 sin x z or z sin x
1 x2 14 y2 z
Plane
74. f x, y, z sin x z
z
Elliptic paraboloid
2
z
Vertex: 0, 0, 1
3
4
z
2
x
5 3 x
2
1 4
y
3 x
5
y
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8
y
312
Chapter 12
76. W x, y
Functions of Several Variables
1 , y < x xy
(a) W 15, 10 (c) W 12, 6
1 1 hr 12 min 15 10 5
(b) W 12, 9
1 1 hr 10 min 12 6 6
(d) W 4, 2
1 1 hr 20 min 12 9 3
1 1 hr 30 min 42 2
f x, y 100x 0.6 y 0.4
78.
f 2x, 2y 1002x0.62y0.4 10020.6x 0.620.4y 0.4 10020.620.4x 0.6y 0.4 2100x 0.6y 0.4 2f x, y 4 r 2 80. V r 2l r 3 3l 4r 3 3
r
l
82. (a)
Year
1995
1996
1997
1998
1999
2000
z
12.7
14.8
17.1
18.5
21.1
25.8
Model
13.09
14.79
16.45
18.47
21.38
25.78
(b) x has the greater influence because its coefficient 0.143 is larger than that of y0.024. (c) f x, 25 0.143x 0.02425 0.502 0.143x 1.102 This function gives the shareholder’s equity z in terms of net sales x and assumes constant assets of y 25. 84. Southwest
86. Latitude and land versus ocean location have the greatest effect on temperature.
88. True
90. True
Section 12.2
Limits and Continuity
2. Let > 0 be given. We need to find > 0 such that f x, y L x 4 < whenever 0 < x a y b x 4 y 1 < . Take . 2
2
2
2
Then if 0 < x 42 y 12 < , we have x 42 <
x 4 < . 4.
lim
x, y → a, b
4f x, y gx, y
4
lim
f x, y
lim
gx, y
x, y → a, b
x, y → a, b
45 20 3 3
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Section 12.2
6.
8.
f x, y gx, y f x, y
lim
5x y 1 0 0 1 1
x, y → a, b
lim
x, y → a, b
lim
f x, y lim
x, y → a, b
lim
x, y → a, b
gx, y
f x, y
10.
x, y → 0, 0
lim
x, y → 4, 2
y cosxy 2 cos
0 2
14.
lim
x, y, z → 2, 0, 1
x
lim
x, y → 1, 1
x y
1 1 1
2
2
lim
x, y → 1, 1
xy 1 x2 y2 2
Continuous except at 0, 0
Continuous everywhere
16.
53 2 5 5
Continuous for x y > 0
Continuous everywhere
12.
Limits and Continuity
18. f x, y
xe yz 2e0 2
Continuous everywhere
x2
x2 1 y 2 1
x2 0 0 x, y → 0, 0 x 1 y 2 1 0 10 1 lim
2
Continuous everywhere
1 cosxxy y 2
20.
lim
x, y → 0, 0
2
2
2
The limit does not exist. Continuous except at 0, 0 22. f x, y
y x2 y2
Continuous except at 0, 0 Path: y 0
x, y f x, y
Path: y x
1, 1
0.5, 0.5
0.1, 0.1
0.01, 0.01
0.001, 0.001
1 2
1
5
50
500
x, y
1, 0
0.5, 0
0.1, 0
0.01, 0
0.001, 0
f x, y
0
0
0
0
0
The limit does not exist because along the path y 0 the function equals 0, whereas along the path y x the function tends to infinity.
24. f x, y
2x y 2 2x 2 y
Continuous except at 0, 0 Path: y 0
Path: y x
x, y
1, 0
0.25, 0
0.01, 0
0.001, 0
0.000001, 0
f x, y
1
4
100
1000
1,000,000
x, y
1, 1
0.25, 0.25
0.01, 0.01
0.001, 0.001
0.0001, 0.0001
1 3
1.17
1.95
1.995
2.0
f x, y
The limit does not exist because along the line y 0 the function tends to infinity, whereas along the line y x the function tends to 2.
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313
314
26.
Chapter 12
Functions of Several Variables
4x2y2 0 x, y → 0, 0 x2 y2
28.
lim
Hence,
lim
x, y → 0, 0
f x, y
lim
x, y → 0, 0
lim
x, y → 0, 0
sin 1x cos 1x
z 2
Does not exist
gx, y 0.
f is continuous at 0, 0, whereas g is not continuous at 0, 0.
4 x
6 6 y
30.
lim
x, y → 0, 0
x2 y2 x2y
32. f x, y
Does not exist
2xy x2 y2 1
The limit equals 0.
z z 18 5
x
5
5
y
4 4
x
34.
36.
y
lim
xy 2 r cos r 2 sin2 lim lim r cos sin2 0 2 r→0 r→0 y r2
lim
x 2y 2 r 4 cos2 sin2 lim lim r 2 cos2 sin2 0 2 r→0 r→0 y r2
x, y → 0, 0 x 2
x, y → 0, 0 x 2
38. f x, y, z
z x2 y2 9
40. f x, y, z xy sin z Continuous everywhere
Continuous for x 2 y 2 9
42.
f t
1 t
44. f t
gx, y x 2 y 2
gx, y x 2 y 2
f gx, y f x 2 y 2
1 4t
f gx, y f x 2 y 2
1 x2 y2
Continuous for x 2 y 2 4
Continuous except at 0, 0 46. f x, y x 2 y 2 (a) lim
x→0
x x 2 y 2 x 2 y 2 f x x, y f x, y lim x→0 x x 2x x x2 lim 2x x 2x x→0 x→0 x
lim (b) lim
y→0
1 4 x2 y2
f x, y y f x, y x 2 y y2 x 2 y 2 lim y→0 y y lim
y→0
2y y y2 lim 2y y 2y y→0 y
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Section 12.3
Partial Derivatives
48. f x, y y y 1 (a) lim
y y 1 y y 1 f x x, y f x, y lim 0 x→0 x x
(b) lim
f x, y y f x, y y y32 y y12 y32 y12 lim y→0 y y
x→0
y→0
y y32 y32 y y12 y12 lim y→0 y→0 y y
lim
3 1 y12 y12 2 2
(L’Hôpital’s Rule)
3y 1 2y
50. See the definition on page 854.
52.
x2 y2 x, y → 0, 0 xy lim
(a) Along y ax:
lim
x, ax → 0, 0
lim
x→0
x2 ax2 xax
(b) Along y x2 :
x21 a2 1 a2 , a0 ax2 a
lim
x, x 2 → 0, 0
x2 x22 1 x2 lim x→0 xx2 x
limit does not exist
If a 0, then y 0 and the limit does not exist. (c) No, the limit does not exist. Different paths result in different limits. 54. Given that f x, y is continuous, then
lim
x, y → a, b
f x, y f a, b < 0, which means that for each > 0, there corresponds
a > 0 such that f x, y f a, b < whenever 0 < x a2 y b2 < .
Let f a, b 2, then f x, y < 0 for every point in the corresponding neighborhood since f a, b f a, b f x, y f a, b < 2 ⇒ 2 ⇒
56. False. Let f x, y
< f x, y f a, b <
f a, b 2
3 1 f a, b < f x, y < f a, b < 0. 2 2
xy . x2 y2
58. True
See Exercise 21.
Section 12.3 2. fy1, 2 < 0
Partial Derivatives 4. fx 1, 1 0
6. f x, y x 2 3y 2 7 fx x, y 2x fy x, y 6y
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8. z 2y2x z y2 x x z 4yx y
315
316
Chapter 12
Functions of Several Variables
10. z y3 4xy2 1
16.
12.
z xe xy
z 4y2 x
z x x e xy e xy e xy 1 x y y
z 3y2 8xy y
x x2 z xe xy 2 2 e xy y y y
18. f x, y
1 lnxy 2
1 y 1 z x 2 xy 2x z 1 x 1 y 2 xy 2y
xy x2 y2
z 1 2x 2x 2 x x2 y2 x y2
fx x, y
2y z y x2 y2
x 2 y 2 y xy2x y 3 x 2y 2 x 2 y 2 2 x y 2 2
fy x, y
x 2 y 2x xy2y x3 xy 2 2 2 2 2 x y x y 2 2
20. gx, y ln x 2 y 2 gx x, y
1 lnx 2 y 2 2
22.
f x, y 2x y3 f 1 1 2x y312 2 2x y3 x 2
1 2x x 2 x2 y2 x2 y2
3y2 1 f 2x y3123y2 y 2 22x y3
1 2y y gy x, y 2 x2 y2 x2 y2 24.
z lnx2 y2
z ln xy
14.
z sin 3x cos 3y
26.
z cosx 2 y 2
z 3 cos 3x cos 3y x
z 2x sinx 2 y 2 x
z 3 sin 3x sin 3y y
z 2y sinx 2 y 2 y
y
28. f x, y
x
y
2t 1 dt
x
2t 1 dt
x
y
2t 1 dt
y
y
x
2t 1 dt
2 dt 2t
x
y
x
2y 2x
fx x, y 2 fy x, y 2 30. f x, y x 2 2xy y 2 x y 2 f f x x, y f x, y lim x x→0 x lim
x→0
x x2 2x xy y 2 x 2 2xy y 2 lim 2x x 2y 2x y x→0 x
f f x, y y f x, y lim y y→0 y x2 2x y y y y 2 x 2 2xy y 2 lim 2x 2y y 2 y x y→0 y→0 y
lim
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Section 12.3
32. f x, y
Partial Derivatives
317
1 xy
1 1 f f x x, y f x, y x x y x y 1 1 lim lim lim x→0 x→0 x x yx y x x→0 x x x y2 1 1 f f x, y y f x, y x y y x y 1 1 lim lim lim y→0 y→0 x y yx y y y→0 y y x y2 34. hx, y x 2 y 2
36. z cos2x y z 2 sin2x y x
hxx, y 2x At 2, 1: hx 2, 1 4 hy x, y 2y
At
At 2, 1: hy 2, 1 2
z sin2x y1 sin2x y y At
40. f x, y
38. f x, y arccosxy fxx, y
y 1 x2y2
At 1, 1, fx is undefined. x fyx, y 1 x2y2 At 1, 1, fy is undefined.
4 , 3 , x z 2 sin
6 1
fxx, y
4 , 3 , y z sin
6 21
6xy
42. z x 2 4y 2, y 1, 2, 1, 8
4x2 5y2
Intersecting curve: z x 2 4
30y3 2 4x 5y232
At 1, 1, fx1, 1 fyx, y
z 2x x
30 10 27 9
At 2, 1, 8:
24x3 4x2 5y232
z 22 4 x
z 20
At 1, 1, fy1, 1
8 9
4 x
44. z 9x2 y 2, x 1, 1, 3, 0 Intersecting curve: z 9 y 2
z 23 6 y
z
46. fx x, y 9x 2 12y, fyx, y 12x 3y 2 fx fy 0: 9x2 12y 0 ⇒ 3x 2 4y Solving for x in the second equation, x y 24, you obtain 3 y 242 4y. 3y4 64y ⇒ y 0 or y
40
⇒ x 0 or x Points: 0, 0, 4
x
y
3y 2 12x 0 ⇒ y 2 4x
z 2y y At 1, 3, 0:
4
y
4
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34 , 34 23
13
4 313
1 16 4 323
318
Chapter 12
Functions of Several Variables
48. fx x, y
2x 0 ⇒ x0 x 2 y2 1
fy x, y
2y 0 ⇒ y0 x2 y2 1
50. (a) The graph is that of fx. (b) The graph is that of fy.
Points: 0, 0
52. w
3xz xy
54. Gx, y, z
1 1 x 2 y 2 z 2
w x y3z 3xz 3yz x x y2 x y2
Gxx, y, z
x 1 x 2 y 2 z 232
w 3xz y x y2
Gyx, y, z
y 1 x 2 y 2 z 232
3x w z xy
Gzx, y, z
z 1 x 2 y 2 z 232
56. f x, y, z 3x 2 y 5xyz 10yz 2
58.
fxx, y, z 6xy 5yz
z x 4 3x 2y 2 y 4
60.
z 1 x x y
z 4x 3 6xy 2 x
fy x, y, z 3x 2 5xz 10z 2
2z 1 x2 x y 2
2z 12x 2 6y 2 x2
fzx, y, z 5xy 20yz
z lnx y
1 2z y x x y 2
2z 12xy y x
z 1 1 y x y y x
z 6x 2 y 4y 3 y
2z 1 y2 x y 2
2z 6x 2 12y 2 y2
1 2z x y x y 2
2z 12xy x y
Therefore,
62.
z 2xey 3yex z 2ey 3yex x 2z 3yex x2 2z 2ey 3yex y x z 2xey 3ex y 2z 2xey y2 2z 2ey 3ex x y
64.
z sinx 2y
66.
2z 2z . y x x y
z 9 x 2 y 2
z cosx 2y x
z x x 9 x 2 y 2
2z sinx 2y x2
2z y2 9 2 x 9 x 2 y 232
2z 2 sinx 2y y x
2z xy y x 9 x 2 y 232
z 2 cosx 2y y
z y y 9 x 2 y 2
2z 4 sinx 2y y2
2z x2 9 2 y 9 x 2 y 232
2z 2 sinx 2y x y
xy 2z x y 9 x 2 y 232 Therefore,
2z 2z . y x x y
z z 0 if x y 0 x y
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Section 12.3
68.
z
xy xy
70.
Partial Derivatives
319
f x, y, z x 2 3xy 4yz z 3 fxx, y, z 2x 3y
z yx y xy y 2 x x y2 x y2
fy x, y, z 3x 4z fyyx, y, z 0
2z 2y 2 x2 x y3
fxyx, y, z 3
2z x y22y y 22x y1 2xy y x x y 4 x y3
fyxx, y, z 3 fyyxx, y, z 0
z xx y xy x2 2 y x y x y2
fxyyx, y, z 0 fyxyx, y, z 0
2z 2x 2 y2 x y3
Therefore, fxyy fyxy fyyx 0.
2z x y22x x 22x y 2xy x y x y4 x y3 There are no points for which zx zy 0. 72.
ey 2
2z x y2
z e y ey cos x x 2
fyx, y, z
2z x y2
2z e y ey sin x 2 x 2
fyyx, y, z
4z x y3
z e y ey sin x y 2
fxyx, y, z
4z x y3
2z e y ey sin x 2 y 2
fyxx, y, z
4z x y3
Therefore,
f x, y, z
2z xy
fxx, y, z
74.
z sin x
e
0.
12z fxyyx, y, z x y4
76. z arctan
2z 2z e y ey e y ey 2 sin x sin x 2 x y 2 2
12z fyyxx, y, z x y4
fyxyx, y, z
y
12z x y4
y x
78.
From Exercise 53, we have 2z 2xy 2xy 2z 2 2 0. 2 x y x y 22 x 2 y 22
z sinwct sinwx z wc coswct sinwx t 2z w 2c 2 sinwct sinwx t 2 z w sinwct coswx x 2z w 2 sinwct sinwx x 2 Therefore,
2z 2z c 2 2. 2 t x
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320
80.
Chapter 12
z et sin
Functions of Several Variables
x c
84. The plane z x y f x, y satisfies
82. If z f x, y, then to find fx you consider y constant and differentiate with respect to x. Similarly, to find fy, you consider x constant and differentiate with respect to y.
x z et sin t c 1 x z et cos x c c
f f < 0 and > 0. x y z
4
1 x 2z 2 et sin x 2 c c
2
4
2z z c 2 2. Therefore, t x
2 4
y
x
86. In this case, the mixed partials are equal, fxy fyx. See Theorem 12.3. 88. f x, y 200x0.7y0.3 (a)
f y 140x0.3y0.3 140 x x
At x, y 1000, 500, (b)
f 500 140 1000 x
f x 60x0.7y0.7 60 x y
At x, y 1000, 500,
0.3
0.3
140
12
0.3
113.72
0.7
f 1000 60 500 x
0.7
6020.7 97.47
1 R 1 0.10 1I
10
VI, R 1000
90.
1 R 1 0.10 1I
VII, R 10,000
9
1 0.101 R
1 0.101 R10 10,000 2 1 I 1 I 11
VI0.03, 0.28 14,478.99
1 R
1 0.101 R 1000 1 0.10 0.10 1I 1 I 1 I
VRI, R 10,000
9
9
10
VR0.03, 0.28 1391.17 The rate of inflation has the greater negative influence on the growth of the investment. (See Exercise 61 in Section 12.1.) 92. A 0.885t 22.4h 1.20th 0.544 (a)
94. U 5x 2 xy 3y 2 (a) Ux 10x y
A 0.885 1.20h t A 30 , 0.80 0.885 1.200.80 1.845 t A 22.4 1.20t h
(b) Uy x 6y (c) Ux2, 3 17 and Uy2, 3 16. The person should consume one more unit of y because the rate of decrease of satisfaction is less for y. (d)
A 30 , 0.80 22.4 1.2030 13.6 h
z 1 −2
2
x
(b) The humidity has a greater effect on A since its coefficient 22.4 is larger than that of t.
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1
2
y
Section 12.4 z 1.55x 22.15 x
96. (a)
(b) Concave downward
2z 1.55 x2
2z < 0 x2
y z > 0 2
The rate of increase of Medicare expenses z is increasing with respect to public assistance expenses y.
2z 0.014 y2 98. False
100. True
Let z x y 1.
y
102. f x, y
1 t 3 dt
x
By the Second Fundamental Theorem of Calculus, f d x dx d f y dy
y
1 t 3 dt
x
dz
1 t 3 dt 1 x3
y
1 t 3 dt 1 y 3.
x
Differentials
x2 y
4.
2x x2 dx 2 dy y y
6. z
12e
dz 2x
8.
x
y
Section 12.4 2. z
d dx
e
x 2 y 2
x 2 y 2
w dw
ex
ex 2
2 y 2
2 y 2
xy z 2y 1 z 2x xy dx dy dz z 2y z 2y2 z 2y2
dx 2y e
x 2 y 2
ex 2
w ey cos x z2
2 y 2
dy e
x 2 y 2
10.
dw ey sin x dx ey cos x dy 2z dz
ex
2 y 2
x dx y dy
w x 2 yz 2 sin yz dw 2xyz 2 dx x 2z 2 z cos yz dy
2x 2yz y cos yz dz 12. (a) f 1, 2 5 2.2361
14. (a) f 1, 2 e2 7.3891
f 1.05, 2.1 5.5125 2.3479
f 1.05, 2.1 1.05e2.1 8.5745
z 0.11180
z 1.1854
(b) dz
x x 2 y 2
dx
y x 2 y 2
dy
x dx y dy 0.05 20.1 0.11180 x 2 y 2 5
321
2
(c) Concave upward
The rate of increase of Medicare expenses z is declining with respect to worker’s compensation expenses x.
z 0.014y 0.54 y
Differentials
(b) dz ey dx xey dy e20.05 e20.1 1.1084
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322
Chapter 12
16. (a) f 1, 2
Functions of Several Variables
1 0.5 2
f 1.05, 2.1
1.05 0.5 2.1
z 0 (b) dz
x 1 dx 2 dy y y 1 1 0.05 0.1 0 2 4
18. Let z x 21 y3, x 2, y 9, dx 0.03, dy 0.1. Then: dz 2x1 y3 dx 3x 21 y2 dy
2.0321 8.93 221 93 221 930.03 3221 920.1 0 20. Let z sinx 2 y 2, x y 1, dx 0.05, dy 0.05. Then: dz 2x cosx 2 y 2 dx 2y cosx 2 y 2 dy sin 1.052 0.952 sin 2 21 cos12 120.05 21 cos12 120.05 0 24. If z f x, y, then z dz is the propagated error, z dz and is the relative error. z z
22. In general, the accuracy worsens as x and y increase.
26.
V r 2h dV 2 rh dr r 2 dh π r 2dh
∆V − dV
∆h 2πrhdr ∆r
r
28. S rr 2 h2 r 8, h 20 20
dS r 2 h21 2 r 2r 2 h21 2 dr
r 2 h2 r 2 2r 2 h2 2 2 1 2 r h r 2 h2
rh dS r r 2 h21 2h dh r 2 h2 2r 2 h2 rh dS dr dh r 2 h2 r 2 h2
r 2 h2
2r 2 h2 dr rh dh
S8, 20 541.3758
8
r
h
dS
S
0.1
0.1
10.0341
10.0768
0.0427
0.1
0.1
5.3671
5.3596
0.0075
0.12368
0.12368
0.683 105
0.00303
0.00303
0.286 107
0.001
0.002
0.0001
0.0002
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S dS
Section 12.4
30.
Differentials
C 1 0.0817 3.71 v1 2 0.25 T 91.4 v 2
0.1516 0.0204 T 91.4 v1 2
C 0.08173.71v 5.81 0.25v T dC Cv dv CT dT
0.02048 91.4± 3 0.08173.7123 5.81 0.2523± 1 0.1516 23 1 2
± 2.79 ± 1.46 ± 4.25 Maximum propagated error ± 4.25 dC ± 0.14 C 30.24
32. x, y 8.5, 3.2, dx ≤ 0.05, dy ≤ 0.05 r x 2 y 2 ⇒ dr
dr
x x 2 y 2
dx
8.5 8.52 3.22
y
dy
x 2 y 2
dx
3.2 8.52 3.22
dy 0.9359 dx 0.3523 dy
≤ 1.2880.05 0.064
y arctan x
y x2 ⇒ d y 1 x
2
1 x
dx 1
y x
2
dy
3.2 y x 8.5 dx 2 dy dx dy x2 y2 x y2 8.52 3.22 8.52 3.22
Using the worst case scenario, dx 0.05 and dy 0.05, you see that
d 34.
a da
≤ 0.00194 0.00515 0.0071.
v2 r v2 2v dv 2 dr r r
da dv dr 2 20.03 0.02 0.08 8% a v r Note: The maximum error will occur when dv and dr differ in signs. 36. (a) Using the Law of Cosines: a2 b2 c2 2bc cos A 3302 4202 2330420cos 9
330 ft 420 ft
a 107.3 ft. 9˚
(b) a b2 4202 2b420cos da
1 2 b 4202 840b cos 2
1 2
2b 840 cos db 840b sin d
1 3302 4202 840330 cos 2 20
2330 840 cos 20 6 840330sin 20 180 1 2
1 11512.79 1 2 ± 1774.79 ± 8.27 ft 2
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323
324
Chapter 12
38.
1 1 1 R R1 R2 R
Functions of Several Variables
R1R 2 R1 R2
dR1 R1 0.5 dR2 R2 2 R R22 R12 R dR dR R1 R2 R1 R2 2 R1 R22 R1 R22
R dR
When R1 10 and R2 15, we have R
40.
T 2
152 102 2 0.14 ohm. 2 0.5 10 15 10 152
Lg
dg g 32.24 32.09 0.15 dL L 2.48 2.5 0.02 T dT
T T dg dL g L g
Lg g
When g 32.09 and L 2.5, we have T
42.
Lg
32.09
L
2.5 0.15 32.09
2.532.09
0.02 0.0111 sec.
z f x, y x2 y2 z f x x, y y f x, y x2 2xx x2 y2 2yy y2 x2 y2 2xx 2yy xx yy fxx, y x fyx, y y 1x 2y where 1 x and 2 y. As x, y → 0, 0, 1 → 0 and 2 → 0.
44.
z f x, y 5x 10y y3 z f x x, y y f x, y 5x 5x 10y 10y y3 3y2y 3yy 2 y3 5x 10y y3 5x 3y2 10y 0x 3yy y2 y fxx, y x fyx, y y 1x 2y where 1 0 and 2 3yy y2. As x, y → 0, 0, 1 → 0 and 2 → 0.
46. f x, y
5x2y , y3 0 ,
x3
x, y 0, 0 x, y 0, 0
f x, 0 f 0, 0 00 lim 0 x→0 x→0 x x
(a) fx0, 0 lim
fy0, 0 lim
y→0
f 0, y f 0, 0 00 lim 0 y→0 y y
Thus, the partial derivatives exist at 0, 0.
(b) Along the line y x: Along the line x 0,
5x3 5 . x→0 2x3 2
lim
f x, y lim
lim
f x, y 0.
x, y → 0, 0 x, y → 0, 0
Thus, f is not continuous at 0, 0. Therefore f is not differentiable at 0, 0. (See Theorem 12.5)
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Section 12.5
Section 12.5 2.
Chain Rules for Functions of Several Variables
Chain Rules for Functions of Several Variables
w x2 y2
w ln
4.
x cos t, y et
y x
x cos t
dw y x sin t et 2 2 2 dt x y x y2
y sin t 1 1 dw sin t cos t dt x y
x sin t yet cos t sin t e2t x2 y2 cos2 t e2t
tan t cot t
1 sin t cos t
6. w cosx y, x t2, y 1 (a)
dw sinx y2t sinx y0 dt 2t sinx y 2t sint2 1
(b) w cost2 1,
dw 2t sint2 1 dt
8. w xy cos z xt y t2 z arccos t (a)
1 1t t 2t t 4t 1t
dw 1 y cos z1 x cos z2t xy sin z dt 1 t2 t2t tt2t tt2
(b) w t 4,
2
3
3
3
3
2
dw 4t3 dt
10. w xyz, x t2, y 2t, z et (a)
dw yz2t xz2 xyet dt 2tet 2t t2et 2 t22tet 2t2et2 1 t 2t2et3 t
(b)
w t22tet 2t3et dw 2t3et et6t2 2t2ett 3 dt
12. Distance f t x2 x12 y2 y12 48 3 2 48t1 2 2
48t8 22 26 ft 488 22 26 f1
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2
325
326
14.
Chapter 12
w
Functions of Several Variables
x2 , y
16.
w y3 3x2y x es
x t2,
y et
y t 1,
w 6xyes 3y2 3x20 6e2st s
t1 dw w dx w dy dt x dt y dt
2x x2 2t 2 1 y y
t4 2t 22t t 12 t1
t 14t3 t4 t 12
3t4 4t3 t 12
w 6xy0 3y2 3x2et t 3ete2t e2s When s 0 and t 1,
d 2w t 1212t3 12t2 3t4 4t32t 1 dt 2 t 14 At t 1:
18.
d 2w 424 74 68 4.25 dt 2 16 16
w sin2x 3y xst yst w 2 cos2x 3y 3 cos2x 3y s 5 cos2x 3y 5 cos5s t w 2 cos2x 3y 3 cos2x 3y t cos2x 3y cos5s t When s 0 and t
w w , 0 and 0. 2 s t
20. w 25 5x2 5y2, x r cos , y r sin (a)
w 5x 5y cos sin r 25 5x2 5y2 25 5x2 5y2
5r cos2 5r sin2 5r 25 5x2 5y2 25 5r2
w 5x 5y r sin r cos 2 2 25 5x 5y 25 5x2 5y2
5r2 sin2 cos 5r2 sin cos 0 25 5x2 5y2
(b) w 25 5r2 5r w w ; 0 r 25 5r2
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w w 6e and 3ee2 1. s t
Section 12.5
22. w (a)
yz , x 2, y r , z r x
w yz z y 2 2 1 1 x x x
r r r r 2 4 2
22 r2 2 2r2 3 3
w cos yz2s xz sin yz0 xy sin yz1 s cosst2 2t32s s2t2 sinst2 2t3 w cos yz0 xz sin yz2t xy sin yz2 t 2s2ts 2t sinst2 2t3 2s2t2 sinst2 2t3 6s2t2 2s3t sinst2 2t3
yz r r r2 21 2 x
w 2r 2 r w 2r2 3 26. w x2 y2 z2, x t sin s, y t cos s, z st2 w 2x cos s 2yt sin s 2zt2 s
28. cos x tan xy 5 0 F x, y sin x y sec2 xy dy x dx Fyx, y x sec2 xy
2t2 sin s cos s 2t2 sin s cos s 2st4 2st4 w 2x sin s 2y cos s 2z2st t 2t sin2 s 2t cos2 s 4s2t3 2t 4s2t3
30.
x y2 6 0 x2 y2
32. Fx, y, z xz yz xy Fx z y
Fxx, y dy dx Fyx, y
Fy z x Fz x y
y 2 x 2 x 2 y 22 2xy x 2 y 22 2y
2xy 2yx 2 y 22
2xy 2yx 4 4x2y 3 2y 5
y2
Fx yz z x Fz xy
x2
y2
Fy z xz y Fz xy
x2
34. Fx, y, z ex sin y z z Fx
ex
sin y z
Fy
ex
cos y z
Fz
ex
cos y z 1
Fx z ex sin y z x Fz 1 ex cos y x
327
24. w x cos yz, x s2, y t2, z s 2t
w yz z y z y 2r 2 0 1 1 2 r x x x x
(b) w
Chain Rules for Functions of Several Variables
36. x sin y z 0 z cos y z 0 implies (i) 1 x 1 z sec y z. x cos y z (ii)
1 yz cos y z 0 implies yz 1.
Fy ex cos y z z y Fz 1 ex cos y z
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328
Chapter 12
Functions of Several Variables 40. x2 y2 z2 5yw 10w2 2 Fx, y, z, w
38. x ln y y2z z2 8 0 (i)
Fxx, y, z ln y z 2 x Fzx, y, z y 2z
Fx 2x, Fy 2y 5w, Fz 2z, Fw 5y 20w w F 2x 2x x x Fw 5y 20w 5y 20w
x 2yz Fyx, y, z z x 2y2z y (ii) 2 3 y Fzx, y, z y 2z y 2yz
w Fy 5w 2y y Fw 20w 5y Fz w 2z z Fw 5y 20w
42. Fx, y, z, w w x y y z 0
44.
w Fx 1 x y1 2 1 x Fw 2 1 2x y
f x, y x3 3xy2 y3 f tx, ty tx3 3txty2 ty3 t3x3 3xy2 y3 t3f x, y
w Fy 1 1 x y1 2 y z1 2 y Fw 2 2
Degree: 3 x fxx, y y fyx, y x3x2 3y2 y6xy 3y2
1 1 2 xy 2 yz
3x3 9xy2 3y3 3f x, y
w Fz 1 z Fw 2y z
46.
f x, y f tx, ty
x2 y2
x2
tx2 x2 t tf x, y tx2 ty2 x2 y2
Degree: 1
xx
x fxx, y y fyx, y x
48.
3
2
2xy2 x2y y 2 y23 2 x y23 2
x4 x2y2 x2x2 y2 2 2 2 3 2 x y x y23 2
x2 f x, y x2 y2
w w x w y s x s y s
50.
w w x w y (Page 878) t x t y t
f x, y dy x dx fyx, y z f x, y, z x x fzx, y, z fyx, y, z z y fzx, y, z
52. (a)
V r 2h
dV dr dh dr dh 2rh r 2 r 2h r 122366 124 4608 in.3 min dt dt dt dt dt (b)
S 2 r r h
dh dr dS 2 2r h r 2 24 366 124 624 in.2 min dt dt dt
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Section 12.5
54. (a)
V
Chain Rules for Functions of Several Variables
2 r rR R 2h 3
dV dr dR dh 2r Rh r 2Rh r 2 rR R2 dt 3 dt dt dt
(b)
215 25104 15 225104 152 1525 25212 3
19,500 6,500 cm3 min 3
S R rR r2 h2 dS dt
R r2 h2 R r
R r
h
dh
R r2 h2 dt
R r
dr
R r2 h2 dt
R r2 h2 R r
R r
dR
R r2 h2 dt
25 152 102 25 15
4 25 15 4 25 15 25 15 10 25 15
25 152 102
25 152 102 25 15
2
2
10 12 25 152 102
3202 cm2 min 58. gt f xt, yt t n f x, y
56. pV mRT T
Let u xt, v yt, then
1 pV mR
dp 1 dV dT V p dt mR dt dt
gt
f u
du f dt v
dv f f x y dt u v
and gt ntn1f x, y. Now, let t 1 and we have u x, v y. Thus, f f x y nf x, y. x y
60.
w x y sin y x w x y cos y x sin y x x w x y cos y x sin y x y w w 0 x y
62.
y w arctan , x r cos , y r sin x arctan
sin arctantan for < < rr cos 2 2
w w w y x w 2 , 2 , 0, 1 x x y2 y x y2 r w w x y
2
1 w w r r
2
2
2
2
y2 x2 1 1 2 2 2 x y x y22 x2 y2 r 2 2
0
1 1 1 2 r2 r
w w w x y r 2
Therefore,
2
2
1 w 2 . r2
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329
330
Chapter 12
Functions of Several Variables
64. Note first that x u v x y x 2 y 2 v y u 2 . y x x y 2 x y r cos2 r sin2 1 u 2 cos 2 sin r x y2 x y2 r2 r y x r 2 sin2 r 2 cos2 v r sin 2 r cos 1 x2 y2 x y2 r2 Thus,
u 1 v . r r
y x r sin cos r sin cos v 2 cos 2 sin 0 r x y2 x y2 r2 x y r 2 sin cos r 2 sin cos u r sin 2 r cos 0 x2 y2 x y2 r2 Thus,
v 1 u . r r
Section 12.6 2.
Directional Derivatives and Gradients
f x, y x3 y3, v
2
2
i j
4.
x y
v j
f x, y 3x2i 3y2j f 4, 3 48i 27j u
f x, y
f x, y
2 2 v i j v 2 2
1 x i 2j y y
f 1, 1 i j
Du f 4, 3 f 4, 3 u 242
u
27 21 2 2 2 2
v j v
Du f 1, 1 f 1, 1 u 1 6.
gx, y arccos xy, v i 5j gx, y
y 1 xy2
i
8.
x 1 xy2
h 2xex
2 y2
i
2yex
Dug1, 0 g1, 0 u
5 26
Du h0, 0 h0, 0 u 0 526 26
f x, y, z x2 y2 z2
12.
hx, y, z xyz v 2, 1, 2
v i 2j 3k
h yz i xz j xyk
f 2x i 2yj 2zk
h2, 1, 1 i 2j 2k
f 1, 2, 1 2i 4j 2k 1 v 2 3 i j k v 14 14 14
6 Du f 1, 2, 1 f 1, 2, 1 u 14 7
u
2 1 2 v i j k v 3 3 3
Du h2, 1, 1 h2, 1, 1 u
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j
2 y2
h0, 0 0
v 5 1 u i j 26 v 26
u
2 y2
vij
j
g1, 0 j
10.
hx, y ex
8 3
Section 12.6
14. f x, y u
y xy 3
2
16. gx, y xey
i
3 1 u i j 2 2
1 j 2
g ey i xey j
y x f i j x y2 x y2 Du f f u
3 y 1 ey xe 3x 1 Dug ey 2 2 2
3y x 2x y2 2x y2
1 3y x 2x y2
18. f x, y cosx y v
20. gx, y, z xyez
i j 2
v 2i 4j g yez i xez j xyez k
f sinx yi sinx yj
At 2, 4, 0, g 4i 2j 8k.
v 1 2 u i j v 5 5 Du f
Directional Derivatives and Gradients
1 5
1 5
sinx y
sinx y
u 2 5
5
5
sinx y
v 2 1 i j v 5 5
Du g g u
4 5
sinx y
At 0, , Du f 0. 22.
gx, y 2xey x
gx, y
24.
2y y x e 2ey x i 2ey xj x
z lnx2 y zx, y
2x 1 i 2 j x2 y x y
z2, 3 4i j
g2, 0 2i 2j w x tan y z
26.
wx, y, z tany zi x sec2y zj x sec2y zk w4, 3, 1 tan 2i 4 sec2 2j 4 sec2 2k \
28.
PQ 2i 7j, u
2 53
i
7 53
j
f x, y 6xi 2yj, f 3, 1 18i 2j Du f f u
\
30.
PQ
36 14 50 5053 53 53 53 53
1 2 i j i j, u 2 5 5
f x, y 2 cos 2x cos yi sin 2x sin yj f 0, 0 2i Du f f u
2 25 5 5
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4 5
8 5
331
332
Chapter 12
Functions of Several Variables
hx, y y cosx y
32.
hx, y y sinx yi cosx y y sinx y j
3
h 0,
3
i
6
3 6 3 j
h0, 3 336 9 6
3 3 2
2
36
3 2 2 23 3 6
gx, y yex
2
34.
36.
gx, y 2xyex i ex j 2
w
2
w
g0, 5 j g0, 5 1
1 1 x2 y2 z2
1
x2
1 xi yj zk y2 z2 3
w0, 0, 0 0 w0, 0, 0 0
38.
w xy2z2 w y2z2i 2xyz2j 2xy2zk w2, 1, 1 i 4j 4k w2, 1, 1 33
For Exercises 40 – 46, f x, y 3
40. (a) D 4 f 3, 2
u
13 22 12 22 5122
(b) D2 3 f 3, 2
42. (a)
x y 1 1 and D f x, y cos sin . 3 2 3 2
13 21 12 23 2 123
12 i j
3
44. f
13 i 12 j
Du f f u (b)
13 12 12 12 5122
v 3i 4j v 9 16 5 4 3 u i j 5 5 Du f f u
46.
1 2 3 5 5 5
1 1 f i j 3 2 f 1 2i 3j f 13 Therefore, u 1 13 3i 2j and Du f 3, 2 f u 0. f is the direction of greatest rate of change of f. Hence, in a direction orthogonal to f, the rate of change of f is 0.
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Section 12.6
Directional Derivatives and Gradients
333
For Exercises 48 and 50, f x, y 9 x2 y2 and D f x, y 2x cos 2y sin 2 x cos y sin .
22 2
48. (a) D 4 f 1, 2 2 (b) D 3 f 1, 2 2
2
50. f 1, 2 2i 4j f 1, 2 1 i 2j f 1, 2 5
12 3 1 23
Therefore, u 1 5 2i j and
Du f 1, 2 f 1, 2 u 0.
52. (a) In the direction of the vector i j. 1 1 y 1 1 (b) f y i x j i x j 2 2x 2 2 4x f 1, 2
1 1 i j 2 2
(Same direction as in part (a).) (c) f 12 i 12 j, the direction opposite that of the gradient.
54. (a) f x, y
8y 2 1 x2 y2
(b) f
⇒ 4y 1 x2 y2
8 8x2 8y2 16xy i j 2 2 2 1 x y 1 x2 y22
f 3, 2
4 y2 4y 4 x2 1
3 i 2
y 4 3
y 22 x2 3
2
Circle: center: 0, 2, radius: 3
1
−2
(c) The directional derivative of f is 0 in the directions ± j.
(d)
z 6
−6
6
x
y
−6
56. f x, y 6 2x 3y c 6, P 0, 0 f x, y 2i 3j 6 2x 3y 6 0 2x 3y
58. f x, y xy c 3, P 1, 3 f x, y y i xj xy 3 f 1, 3 3i j
f 0, 0 2i 3j
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−1
x 1
2
334
Chapter 12
Functions of Several Variables
60. 3x2 2y2 1
62. xey y 5
y
f x, y 3x2 2y2
1
f x, y 6xi 4y j f 1, 1 6i 4j
x
−1
13
13
f x, y xey y
6
f x, y ey i xey 1j
4
f 5, 0 i 4j
1
1 f 1, 1 3i 2j
f 1, 1 13
y
2
1 f 5, 0 i 4j
f 5, 0 17
−1
3i 2j
64. hx, y 5000 0.001x2 0.004y2
17
17
x 2
4
6
i 4j
66. The directional derivative gives the slope of a surface at a point in an arbitrary direction u cos i sin j.
h 0.002x i 0.008y j h500, 300 i 2.4j or 5h 5i 12j 68. See the definition, page 887.
70. The gradient vector is normal to the level curves.
72. The wind speed is greatest at A.
See Theorem 12.12. 74. Tx, y 100 x2 2y2,
P 4, 3
dx 2x dt
dy 4y dt
xt C1e2t
yt C2e4t
4 x0 C1
3 y0 C2
xt
yt 3e4t
4e2t
3x 2 3 e4t y ⇒ u x 2 16 16
z
76. (a) 500
x
6
6
y
xi 12 j T 3, 5 400e73i 12 j
(b) T x, y 400ex
2 y
2
There will be no change in directions perpendicular to the gradient: ± i 6j (c) The greatest increase is in the direction of the gradi1 ent: 3i 2 j 78. False
80. True
Du f x, y 2 > 1 when
u cos
i sin j. 4 4
Section 12.7
Tangent Planes and Normal Lines
2. Fx, y, z x2 y2 z2 25 0 x2 y2 z2 25
4. Fx, y, z 16x2 9y2 144z 0 16x2 9y2 144z 0 Hyperbolic paraboloid
Sphere, radius 5, centered at origin.
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Section 12.7 6.
Fx, y, z x2 y2 z2 11
8.
Tangent Planes and Normal Lines
Fx, y, z x3 z
Fx, y, z 2xi 2yj 2zk
Fx, y, z 3x2i k
F3, 1, 1 6i 2j 2k
F2, 1, 8 12i k
n
F 1 6i 2j 2k
F 44
n
11 1 3i j k 3i j k 11 11
Fx, y, z x2 3y z3 9
10.
12.
Fx, y, z 2xi 3j 3z2k
145
12i k
145
Fx, y, z zex
2
y2
Fx, y, z 2xzex
2
F2, 1, 2 4i 3j 12k n
F 1 12i k
F 145
3
y2i
2yzex
2
y2j
F2, 2, 3 12i 12j k
F 1 4i 3j 12k
F 13
n
F 1 12i 12j k
F 17
Fx, y, z sinx y z 2
14.
Fx, y, z cosx yi cosx yj k F
3 , 6 , 23 n
2
i
3
2
jk
3 3 F 2 i jk
F 10 2 2
16.
3
1 10 10
10
3 i 3 j 2k 3 i 3 j 2k
y f x, y , 1, 2, 2 x Fx, y, z
y z x
Fxx, y, z
y x2
1 x
Fzx, y, z 1
Fy1, 2, 2 1
Fz1, 2, 2 1
Fyx, y, z
Fx1, 2, 2 2
2x 1 y 2 z 2 0 2x y z 2 0 2x y z 2
18.
y gx, y arctan , 1, 0, 0 x Gx, y, z arctan Gxx, y, z
y z x
yx2 y 1 y2x2 x2 y2
Gx1, 0, 0 0
Gyx, y, z
1x x 1 y2x2 x2 y2
Gy1, 0, 0 1
yz0
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Gzx, y, z 1 Gz1, 0, 0 1
ex
2
y2k
335
336 20.
Chapter 12
Functions of Several Variables
f x, y 2 23 x y, 3, 1, 1 Fx, y, z 2 23 x y z Fxx, y, z 23 , 23 x
Fyx, y, z 1,
Fzx, y, z 1
3 y 1 z 1 0 2 3 x y z 2 0
2x 3y 3z 6 22. z x2 2xy y2, 1, 2, 1 Fx, y, z x2 2xy y2 z Fxx, y, z 2x 2y
Fyx, y, z 2x 2y
Fzx, y, z 1
Fx1, 2, 1 2
Fy1, 2, 1 2
Fz1, 2, 1 1
2x 1 2 y 2 z 1 0 2x 2y z 1 0 2x 2y z 1
24.
hx, y cos y,
5, 4 , 22
Hx, y, z cos y z Hxx, y, z 0
Hyx, y, z sin y
0 H 5, 4 , 22 22 2 2 y z 0 2 4 2
2 Hx 5, , 4 2
y
Hzx, y, z 1
2 1 Hz 5, , 4 2
2
2
yz
2
8
2
2
0
4 2y 8z 2 4 26. x2 2z2 y2, 1, 3, 2 Fx, y, z x2 y2 2z2 Fxx, y, z 2x
Fyx, y, z 2y
Fx1, 3, 2 2
Fy1, 3,2 6
Fzx, y, z 4z Fz1, 3, 2 8
2x 1 6 y 3 8z 2 0
x 1 3 y 3 4z 2 0 x 3y 4z 0 28. x y2z 3, 4, 4, 2 Fx, y, z x 2yz 3y Fxx, y, z 1
Fyx, y, z 2z 3
Fzx, y, z 2y
Fx4, 4, 2 1
Fy4, 4, 2 1
Fz4, 4, 2 8
x 4 1y 4 8z 2 0 x y 8z 16 x y 8z 16
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Section 12.7
Tangent Planes and Normal Lines
337
30. x2 y2 z2 9, 1, 2, 2 Fx, y, z x2 y2 z2 9 Fxx, y, z 2x
Fyx, y, z 2y
Fzx, y, z 2z
Fx1, 2, 2 2
Fy1, 2, 2 4
Fz1, 2, 2 4
Direction numbers: 1, 2, 2 Plane: x 1 2 y 2 2z 2 0, x 2y 2z 9 Line:
x1 y2 z2 1 2 2
32. x2 y2 z2 0, 5, 13, 12 Fx, y, z x2 y2 z2 Fxx, y, z 2x
Fyx, y, z 2y
Fzx, y, z 2z
Fx5, 13, 12 10
Fy5, 13, 12 26
Fzx, y, z 24
Direction numbers: 5, 13, 12 Plane
Line:
5x 5 13 y 13 12z 12 0
x 5 y 13 z 12 5 13 12
5x 13y 12z 0 34. xyz 10, 1, 2, 5 Fx, y, z xyz 10 Fxx, y, z yz
Fyx, y, z xz
Fzx, y, z xy
Fx1, 2, 5 10
Fy1, 2, 5 5
Fz1, 2, 5 2
Direction numbers: 10, 5, 2 Plane: 10x 1 5 y 2 2z 5 0, 10x 5y 2z 30 Line:
x1 y2 z5 10 5 2
36. See the definition on page 897.
40.
Fx, y, z x2 y2 z Fx, y, z 2xi 2yj k F2, 1, 5 4i 2j k
i (a) F G 4 0
j 2 1
Gx, y, z 4 y z Gx, y, z j k G2, 1, 5 j k
k 1 i 4j 4k 1
Direction numbers: 1, 4, 4, (b) cos
38. 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.
x2 y1 z5 1 4 4
42 F G 3 3 ; not orthogonal
F G 21 2 42 14
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338 42.
Chapter 12
Functions of Several Variables
Fx, y, z x2 y2 z x
Fx, y, z
x2 y2
i
3 4 F3, 4, 5 i j k 5 5
i j k F G 35 45 1 5 2 3
Gx, y, z 5x 2y 3z 22 y x2 y2
jk
Gx, y, z 5i 2j 3k G3, 4, 5 5i 2j 3k
34 26 2 i j k 5 5 5
Direction numbers: 1, 17, 13 x3 y4 z5 Tangent line 1 17 13 cos
44.
F G
F G
85 2 38
Fx, y, z x2 y2 z Fx, y, z 2xi 2y j k F1, 2, 5 2i 4j k
i (a) F G 2 1
8 Not orthogonal 5 76 Gx, y, z x y 6z 33 Gx, y, z i j 6k G1, 2, 5 i j 6 k
k 1 25i 13j 2k 6 x1 y2 z5 Direction numbers: 25, 13, 2, 25 13 2
(b) cos
j 4 1
F G 0; orthogonal
F G
46. (a) f x, y 16 x2 y2 2x 4y gx, y
2
2
z 5
1 3x2 y2 6x 4y
f
g
f x, y gx, y
(b)
16 x2 y2 2x 4y
1 1 3x2 y2 6x 4y 2
5
5
y
x
32 2x2 2y2 4x 8y 1 3x2 y2 6x 4y x2 2x 31 3y2 12y
x2 2x 1 42 3 y2 4y 4 x 12 42 3 y 22 To find points of intersection, let x 1. Then 3 y 22 42
y 22 14 y 2 ± 14 f 1, 2 14 2 j, g 1, 2 14 1 2 j. The normals to f and g at this point are 2j k and 1 2 j k, which are orthogonal. Similarly, f 1, 2 14 2 j and g 1, 2 14 1 2 j and the normals are 2 j k and 1 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.
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Section 12.7 48. Fx, y, z 2xy z3, 2, 2, 2
Tangent Planes and Normal Lines
50. Fx, y, z x2 y2 5, 2, 1, 3
F 2yi 2xj 3z2k
Fx, y, z 2xi 2yj
F2, 2, 2 4i 4j 12k
F2, 1, 3 4i 2j
cos
F2, 2, 2 k 12 3 11
F2, 2, 2
176
cos
11
3 1111 25.24
52. Fx, y, z 3x2 2y2 3x 4y z 5
z
Fx, y, z 6x 3i 4y 4j k 6x 3 0, x
30 25
1 2
4y 4 0, y 1 z 3
1 2,
1 2 2
1,
−3
21 3
31 4
F2, 1, 3
arccos 0 90
arccos
F2, 1, 3 k 0
2
1 2
41 5
−2
−3
31 4
3
3
x
y
x2 y2 z2 2 21 2 a b c
56. Fx, y, z
54. T x, y, z 100 3x y z2, 2, 2, 5 dx 3 dt
dy 1 dt
dz 2z dt
xt 3t C1
yt t C2
zt C3e2t
x0 C1 2
y0 C2 2
z0 C3 5
x 3t 2
y t 2
z 5e2t
Fxx, y, z
2x a2
Fyx, y, z
2y b2
Fzx, y, z
2z c2
2y 2z 2x0 x x0 20 y y0 20 z z0 0 a2 b c
Plane:
x0x y0 y z0z x02 y02 z02 2 2 2 2 2 1 a2 b c a b c
58.
z xf Fx, y, z x f Fxx, y, z f
yx yx z
yx x f yx xy f yx xy f yx 2
Fyx, y, z x f
yx1x f yx
Fxx, y, z 1 Tangent plane at x0, y0, z0:
f x x f x x x f x y y z z 0 y0
y0
y0
y0
0
0
0
0
0
0
0
f x x f x x x f x y f x yf x y f x z x f x 0 y0
y0
y0
y0
0
0
0
0
y0
y0
y0
0
0
0
0
y0
0
0
0
0
f x x f x x f x y z 0 y0
y0
y0
y0
0
0
0
0
Therefore, the plane passes through the origin x, y, z 0, 0, 0.
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339
340
Chapter 12
Functions of Several Variables
60. f x, y cosx y fxx, y sinx y
fyx, y sinx y
fxxx, y cosx y,
fyyx, y cosx y,
fxyx, y cosx y
(a) P1x, y f 0, 0 fx0, 0x fy0, 0y 1 1 1 (b) P2x, y f 0, 0 fx0, 0x fy0,0y 2 fxx0, 0x2 fxy0, 0xy 2 fyy0, 0y2
1 12 x2 xy 12 y2 1 (c) If x 0, P20, y 1 2 y2. This is the second–degree Taylor polynomial for cos y. 1 If y 0, P2x, 0 1 2 x2. This is the second–degree Taylor polynomial for cos x.
(d)
x
y
f x, y
P1x, y
0
0
1
1
1
0
0.1
0.9950
1
0.9950
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
(e)
P2x, y
z 5
5 x
5
y
62. Given z f x, y, then: Fx, y, z f x, y z 0 Fx0, y0, z0 fxx0, y0i fyx0, y0j k cos
Section 12.8
Fx0, y0, z0 k Fx0, y0, z0 k
1
fxx0, y0 fyx0, y02 12 2
1 fxx0, y02 fyx0, y02 1
Extrema of Functions of Two Variables
2. gx, y 9 x 32 y 22 ≤ 9
z
(3, − 2, 9)
Relative maximum: 3, 2, 9
8
gx 2x 3 0 ⇒ x 3
6
gy 2 y 2 0 ⇒ y 2
4 2 1 x
y
6
4. f x, y 25 x 22 y2 ≤ 5
5
Relative maximum: 2, 0, 5 Check: fx fy fxx
x2 25 x 22 y2
y 25 x 22 y2
z
(2, 0 , 5)
0 ⇒ x2 x
5
5
y
0 ⇒ y0
25 y2 25 x 22 yx 2 ,f , f 2 2 3 2 yy 25 x 2 y 25 x 22 y23 2 xy 25 x 22 y23 2
At the critical point 2, 0, fxx < 0 and fxx fyy fxy2 > 0. Therefore, 2, 0, 5 is a relative maximum.
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Section 12.8
Extrema of Functions of Two Variables
6. f x, y x2 y2 4x 8y 11 x 22 y 42 9 ≤ 9
z
Relative maximum: 2, 4 , 9
8
(2, 4 , 9)
6
Check: fx 2x 4 0 ⇒ x 2 fy 2y 8 0 ⇒ y 4 6
fxx 2, fyy 2, fxy 0
4
2
x
8
y
At the critical point 2, 4, fxx < 0 and fxx fyy fxy2 > 0. Therefore, 2, 4, 9 is a relative maximum. 8. f x, y x2 5y2 10x 30y 62 fx 2x 10 0 x 5, y 3 fy 10y 30 0
fxx 2, fyy 10, fxy 0 2 > 0. At the critical point 5, 3, fxx < 0 and fxx fyy f xy
Therefore, 5, 3, 8 is a relative maximum. 10. f x, y x2 6xy 10y2 4y 4 fx 2x 6y 0
fy 6x 20y 4 0
Solving simultaneously yields x 6 and y 2.
fxx 2, fyy 20, fxy 6 At the critical point 6, 2, fxx > 0 and fxx fyy fxy2 > 0. Therefore, 6, 2, 0 is a relative minimum. 14. hx, y x2 y21 3 2
12. f x, y 3x2 2y2 3x 4y 5 fx 6x 3 0 when x 12 . fy 4y 4 0 when y 1. fxx 6, fyy 4, fxy 0 1 At the critical point 2 , 1, fxx < 0 and fxx fyy fxy2 > 0. Therefore, 12 , 1, 31 4 is a relative maximum.
2x 0 3x2 y22 3 2y hy 0 3x2 y22 3
hx
x 0, y 0
Since hx, y ≥ 2 for all x, y, 0, 0, 2 is a relative minimum.
18. f x, y y3 3yx2 3y2 3x2 1
16. f x, y x y 2
Relative maximum: 0, 0, 1
Since f x, y ≥ 2 for all x, y, the relative minima of f consist of all points x, y satisfying x y 0.
Saddle points: 0, 2, 3, ± 3, 1, 3 z 40 20 3 x
20. z exy
3
z
Saddle point: 0, 0, 1
100
x
3
3
y
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y
341
342
Chapter 12
Functions of Several Variables
22. gx, y 120x 120y xy x2 y2
gx 120 y 2x 0
Solving simultaneously yields x 40 and y 40.
gy 120 x 2y 0
gxx 2, gyy 2, gxy 1 At the critical point 40, 40, gxx < 0 and gxx gyy gxy2 > 0. Therefore, 40, 40, 4800 is a relative maximum. 24. gx, y xy gx y
x 0 and y 0
gy x
gxx 0, gyy 0, gxy 1 At the critical point 0, 0, gxx gyy gxy2 < 0. Therefore, 0, 0, 0 is a saddle point. 1 26. f x, y 2xy x4 y2 1 2 fx 2y 2x3 Solving by substitution yields 3 critical points: fy 2x 2y3 0, 0, 1, 1, 1, 1
fxx 6x2, fyy 6y2, fxy 2 At 0, 0, fxx fyy fxy2 < 0 ⇒ 0, 0, 1 saddle point. At 1, 1, fxx fyy fxy2 > 0 and fxx < 0 ⇒ 1, 1, 2 relative maximum. At 1, 1, fxx fyy fxy2 > 0 and fxx < 0 ⇒ 1, 1, 2 relative maximum.
28. f x, y
12 x
2
y2 e1x
2 y2
fx 2x3 2xy2 3xe1x
0
2 y2
fy 2x y 2y ye 2
3
1x2 y2
0
Solving yields the critical points 0, 0, 0, ±
fxx 4x 4 4x2y2 12x2 2y2 3e1x
2
2
,
±
6
2
,0 .
2 y2
fyy 4y4 4x2y2 2x2 8y2 1e1x
2 y2
fxy 4x3y 4xy3 2xye1x
2 y2
At the critical point 0, 0, fxx fyy fxy2 < 0. Therefore, 0, 0, e 2 is a saddle point. At the critical points 0, ± 2 2, fxx < 0 and fxx fyy fxy2 > 0. Therefore, 0, ± 2 2, e are relative maxima. At the critical points ± 6 2, 0, fxx > 0 and fxx fyy fxy2 > 0. Therefore, ± 6 2, 0, e e are relative minima. 30. z
x2 y22 ≥ 0. z 0 if x2 y2 0. x2 y2
z 2
Relative minima at all points x, x and x, x, x 0.
5 x
32. fxx < 0 and fxx fyy fxy2 38 22 > 0 f has a relative maximum at x0, y0.
5
y
34. fxx > 0 and fxx fyy fxy2 258 102 > 0 f has a relative minimum at x0, y0.
36. See Theorem 12.17.
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Section 12.8 Extrema at all x, y
z
38.
Extrema of Functions of Two Variables
40.
Relative maximum
z 4
4
343
(2, 1, 4)
3
2
3
4
y
3
4
4
y
4
x
x
42. A and B are relative extrema. C and D are saddle points.
44. d fxx fyy fxy2 < 0 if fxx and fyy have opposite signs. Hence, a, b, f a, b is a saddle point. For example, consider f x, y x2 y2 and a, b 0, 0.
46. f x, y x3 y3 6x2 9y2 12x 27y 19 fx 3x2 12x 12 0
Solving yields x 2 and y 3.
fy 3y 2 18y 27 0
fxx 6x 12, fyy 6y 18, fxy 0 At 2, 3, fxx fyy fxy2 0 and the test fails. 1, 2, 0 is a saddle point. 48. f x, y x 12 y 22 ≥ 0 fx fy fxx
x1 x 12 y 22
0
y2 0 x 12 y 22
Solving yields x 1 and y 2.
y 22 x 12 x 1 y 2 , f , f x 12 y 223 2 yy x 12 y 223 2 xy x 12 y 223 2
At 1, 2, fxx fyy fxy2 is undefined and the test fails. Absolute minimum: 1, 2, 0 50. f x, y x2 y22 3 ≥ 0 fx
4x 3x2 y21 3
fy
4y 3x2 y21 3
fxx
fx and fy are undefined at x 0, y 0. The critical point is 0, 0.
4x2 3y2 43x2 y2 8xy , f 2 2 4 3 , fyy 9x y 9x2 y24 3 xy 9x2 y24 3
At 0, 0, fxx fyy fxy2 is undefined and the test fails. Absolute minimum: 0, 0, 0 52. f x, y, z 4 x y 1z 22 ≤ 4 fx 2x y 12z 22 0 fy 2x2 y 1z 22 0 fz 2x y 12z 2 0
Solving yields the critical points 0, a, b, c, 1, d, e, f, 2. These points are all absolute maxima.
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344
Chapter 12
Functions of Several Variables
54. f x, y 2x y2
y
fx 42x y 0 ⇒ 2x y
3
fy 22x y 0 ⇒ 2x y
(1, 2) 2
On the line y x 1, 0 ≤ x ≤ 1, f x, y f x 2x x 12 x 12
(0, 1)
1 and the maximum is 1, the minimum is 0. On the line y 2 x 1, 0 ≤ x ≤ 2,
f x, y f x 2x
12
y = 2x
1
1
2
5 2
(2, 0) x
x 1 x 1 2
2
3
and the maximum is 16, the minimum is 0. On the line y 2x 4, 1 ≤ x ≤ 2, f x, y f x 2x 2x 42 4x 42 and the maximum is 16, the minimum is 0. Absolute maximum: 16 at 2, 0 Absolute minimum: 0 at 1, 2 and along the line y 2x. 56. f x, y 2x 2xy y2
y
fx 2 2y 0 ⇒ y 1
f 1, 1 1
2
fy 2y 2x 0 ⇒ y x ⇒ x 1
(−1, 1)
On the line y 1, 1 ≤ x ≤ 1,
(1, 1)
f x, y f x 2x 2x 1 1. On the curve y x2, 1 ≤ x ≤ 1
−1
x 1
f x, y f x 2x 2xx2 x22 x4 2x3 2x and the maximum is 1, the minimum is 11 16 . Absolute maximum: 1 at 1, 1 and on y 1
11 1 1 Absolute minimum: 16 0.6875 at 2 , 4
58. f x, y x2 2xy y2, R x, y: x ≤ 2, y ≤ 1
y
y x f 2x 2y 0 fx 2x 2y 0
2
y
f x, x x2 2x2 x2 0
x
−1
Along y 1, 2 ≤ x ≤ 2, f x2 2x 1, f 2x 2 0 ⇒ x 1, f 2, 1 1, f 1, 1 0, f 2, 1 9.
1
−2
Along y 1, 2 ≤ x ≤ 2, f x2 2x 1, f 2x 2 0 ⇒ x 1, f 2, 1 9, f 1, 1 0, f 2, 1 1. Along x 2, 1 ≤ y ≤ 1, f 4 4y y2, f 2y 4 0. Along x 2, 1 ≤ y ≤ 1, f 4 4y y2, f 2y 4 0. Thus, the maxima are f 2, 1 9 and f 2, 1 9, and the minima are f x, x 0, 1 ≤ x ≤ 1. 60. f x, y x2 4xy 5, R x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ x
xy0
y
fx 2x 4y 0
4
fy 4x 0
3
f 0, 0 5
2
Along y 0, 0 ≤ x ≤ 4, f x2 5 and f 4, 0 21.
1
Along x 4, 0 ≤ y ≤ 2, f 16 16y 5, f 16 0 and f 4, 2 11. Along y x, 0 ≤ x ≤ 4, f x2 4x3 2 5, f 2x 6x1 2 0 on 0, 4. Thus, the maximum is f 4, 0 21 and the minimum is f 4, 2 11.
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x 1
2
3
4
Section 12.9
62. f x, y
Applications of Extrema of Functions of Two Variables
345
4xy , R x, y: x ≥ 0, y ≥ 0, x2 y2 ≤ 1 x2 1 y2 1
fx
41 x2 y 0 ⇒ x 1 or y 0 y 1x2 12
fy
41 y2x 0 ⇒ y 1 or x 0 x 1 y2 12
2
2
For x 0, y 0, also, and f 0, 0 0. For x 1 and y 1, the point 1, 1 is outside R. For x2 y2 1, f x, y f x, 1 x2 Absolute maximum is
2 2 4x1 x2 , and the maximum occurs at x ,y . 2 x2 x4 2 2
2 2 8 f , . 9 2 2
The absolute minimum is 0 f 0, 0. In fact, f 0, y f x, 0 0 y
1
R x 1
64. False Let f x, y x4 2x2 y2. Relative minima: ± 1, 0, 1 Saddle point: 0, 0, 0
Section 12.9
Applications of Extrema of Functions of Two Variables
2. A point on the plane is given by x, y, 12 2x 3y. The square of the distance from 1, 2, 3 to a point on the plane is given by S x 12 y 22 9 2x 3y2
4. A point on the paraboloid is given by x, y, x2 y2. The square of the distance from 5, 0, 0 to a point on the paraboloid is given by S x 52 y2 x2 y22
Sx 2x 1 29 2x 3y2
Sx 2x 5 4xx2 y2 0
Sy 2 y 2 29 2x 3y3.
Sy 2y 4yx2 y2 0.
From the equations Sx 0 and Sy 0, we obtain the system
From the equations Sx 0 and Sy 0, we obtain the system
5x 6y 19
2x3 2xy2 x 5 0
6x 10y 29.
2y3 2x2y y 0.
16 31 43 Solving simultaneously, we have x 14 , y 14 , z 14 and the distance is
1614 1 3114 2 4314 3 2
2
2
1 . 14
Solving as in Exercise 3, we have x 1.235, y 0, z 1.525 and the distance is 1.235 52 1.5252 4.06.
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346
Chapter 12
Functions of Several Variables
6. Since x y z 32, z 32 x y. Therefore,
8. Let x, y, and z be the numbers and let S x2 y2 z2. Since x y z 1, we have
P xy2z 32xy2 x2y2 xy3
S x2 y2 1 x y2
Px 32y2 2xy2 y3 y232 2x y 0
Sx 2x 21 x y 0 2x y 1
Py 64xy 2x2y 3xy2 y64x 2x2 3xy 0.
Sy 2y 21 x y 0 x 2y 1.
Ignoring the solution y 0 and substituting y 32 2x into Py 0, we have
Solving simultaneously yields x 13 , y 13 , and z 13 .
64x 2x2 3x32 2x 0 4xx 8 0. Therefore, x 8, y 16, and z 8. C0 1.5xy . 10. Let x, y, and z be the length, width, and height, respectively. Then C0 1.5xy 2yz 2xz and z 2x y The volume is given by V xyz
C0 xy 1.5x2y2 2x y
Vx
y22C0 3x2 6xy 4x y2
Vy
x22C0 3y2 6xy . 4x y2
In solving the system Vx 0 and Vy 0, we note by the symmetry of the equations that y x. Substituting y x into Vx 0 yields x22C0 9x2 1 1 1 0, 2C0 9x2, x 2C0 , y 2C0 , and z 2C0 . 16x2 3 3 4 12. Consider the sphere given by x2 y2 z2 r 2 and let a vertex of the rectangular box be x, y, r 2 x2 y2 . Then the volume is given by V 2x2y 2r 2 x2 y2 8xyr 2 x2 y2
V 8xy
Vx 8 xy
y
x r 2 x2 y2
y r 2 x2 y2
x y
yr 2 x2 y2 xr 2
2
2
8y r 2 x2 y2
8x r 2 x2 y2
r 2 2x2 y2 0
r 2 x2 2y2 0.
Solving the system 2x2 y2 r 2 x2 2y2 r 2 yields the solution x y z r 3. 14. Let x, y, and z be the length, width, and height, respectively. Then the sum of the two perimeters of the two cross sections is given by
2x 2z 2y 2z 108 or x 54 y 2z. The volume is given by V xyz 54yz y2z 2yz2 Vy 54z 2yz 2z2 z54 2y 2z 0 Vz 54y y2 4yz y54 y 4z 0. Solving the system 2y 2z 54 and y 4z 54, we obtain the solution x 18 inches, y 18 inches, and z 9 inches.
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Section 12.9
16.
Applications of Extrema of Functions of Two Variables
347
1 A 30 2x 30 2x 2x cos x sin 2 30x sin 2x2 sin x2 sin cos A 30 sin 4x sin 2x sin cos 0 x A 30 cos 2x2 cos x22 cos2 1 0 From
A 2x 15 0 we have 15 2x x cos 0 ⇒ cos . x x
From
A 0 we obtain
30x
2x x 15 2x 2x x 15 x 22x x 15 2
2
2
1 0
302x 15 2x2x 15 22x 152 x2 0 3x2 30x 0 x 10 Then cos
1 ⇒ 60. 2
18. P p, q, r 2pq 2pr 2qr.
R 515p1 805p2 1.5p1 p2 1.5p12 p22
20.
p q r 1 implies that r 1 p q.
R p1 515 1.5p2 3p1 0
P p, q 2pq 2p1 p q 2q1 p q
R p2 805 1.5p1 p2 0
2pq 2p 2p2 2pq 2q 2pq 2q2 2pq 2p 2q
2p2
Solving this system yields p1 $2296.67, p2 $4250.
P P 2q 2 4p; 2p 2 4q p q Solving
3p1 1.5p2 515 1.5p1 p2 805
2q2
P P 0 gives p q
q 2p 1 p 2q 1 and hence p q P
1 and 3
13, 13 219 213 213 219 219
6 2 . 9 3
22. S d1 d2 d3 0 02 y 02 0 22 y 22 0 22 y 22 y 24 y 22 dS 2 y 2 23 6 23 1 0 when y 2 . dy 3 3 4 y 22 The sum of the distance is minimized when y
2 3 3 0.845. 3
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348
Chapter 12
Functions of Several Variables
24. (a) S x 42 y2 x 12 y 62 x 122 y 22
z 30
The surface appears to have a minimum near x, y 1, 5. (b) Sx Sy
x4 x 42 y2
y x 4 2
y2
x1
x 12 y 62
x 12 x 122 y 22
−2
y6 y2 2 2 x 1 y 6 x 122 y 22
x
2
4
−4
2 4 6 8
y
(c) Let x1, y1 1, 5. Then S1, 5 0.258i 0.03j Direction 6.6 (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 ) Sx, y points in the direction that S decreases most rapidly. 26. See the last paragraph on page 915 and Theorem 12.18.
28. (a)
x
y
xy
x2
3
0
0
9
1
1
1
1
1
1
1
1
3
2
6
9
x 0 y 4 x y 6 x i
i
i i
2
i
(b) S
46 04 3 3 1 , b 4 0 1, 420 02 10 4 10
y
3 x1 10
30. (a)
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
a (b) S
2
1 5
36
x 28 y 8 x y 37 x i
2
x
i
2
20
a
19 1 2 101 0 107 1 13 10 10
i i
i
2
116
72 1 1 1 3 1 3 837 288 , b 8 28 , y x 8116 282 144 2 8 2 4 2 4
34 0 41 0 14 0 34 1 54 1 54 2 74 2 94 2 2
2
2
2
2
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2
2
2
3 2
2
Section 12.9
Applications of Extrema of Functions of Two Variables 34. 6, 4, 1, 2, 3, 3, 8, 6, 11, 8, 13, 8; n 6
32. 1, 0, 3, 3, 5, 6
x 9, y 9, x y 39, x 35 i
x 42 y 31 x y 275 x 400
i
2
i i
i
339 99 36 3 a 335 92 24 2
3 3 9 1 9 9 b 3 2 6 2 3 3 y x 2 2 7
i
i
i i
i
2
a
6275 4231 29 0.5472 6400 422 53
b
1 29 425 31 42 1.3365 6 53 318
y
425 29 x 53 318
9
−1
6 −1
−1
14 −1
38. (a) y 1.8311x 47.1067
36. (a) 1.00, 450, 1.25, 375, 1.50, 330
x 3.75, y 1,155, x x y 1,413.75 i
i
(b) For each 1 point increase in the percent x, y increases by about 1.83 (slope of line).
4.8125,
2
i
i i
a
31,413.75 3.751,155 240 34.8125 3.752
b
1 1,155 2403.75 685 3
y 240x 685 (b) When x 1.40, y 2401.40 685 349.
40. Sa, b
n
ax b y i
2
i
i1
Saa, b 2a
n
x
2
i
i1
Sba, b 2a
2b
n
n
x 2x y i
i1
n
n
x 2nb 2 y i
i
i1
Saaa, b 2
i i
i1
i1
n
x
2
i
i1
Sbba, b 2n Saba, b 2
n
x
i
i1
Saaa, b > 0 as long as xi 0 for all i. (Note: If xi 0 for all i, then x 0 is the least squares regression line.) d SaaSbb Sab2 4n
n
x
i1
2
i
x 4n x x ≥ 0 since n x ≥ x .
4
2
n
n
i1
i
i1
2
n
2
i
n
i1
As long as d 0, the given values for a and b yield a minimum.
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i
i1
2
n
2
i
i
i1
349
350
Chapter 12
Functions of Several Variables
42. 4, 5, 2, 6, 2, 6, 4, 2
x 0 y 19 x 40 x 0 x 544 x y 12 x y 160
44. 0, 10, 1, 9, 2, 6, 3, 0
x 6 y 25 x 14 x 36 x 98 x y 21 x y 33
8
i
(−2, 6) (− 4, 5)
(4, 2) −9
9
4
4
i
i i
i i
2
i
i
b
3 10 ,
c
41 6 ,
y
5 2 24 x
3 10
x
i
98a 36b 14c 33
544a 40c 160, 40b 12, 40a 4c 19 a
9 −1
i
i
5 24 ,
(3, 0)
−9
3
i
2
(2, 6)
i
−4
3
(1, 9)
2
i
i
(0, 10)
i
i
2
11
i
(2, 6)
41 6
36a 14b 6c 21 14a 6b 4c 25 a 54 , b
46. (a) y 0.078x 2.96
48. (a)
y
7
(b)
−5
5 2 9 199 c 199 20 , y 4 x 20 x 20
1 ax b 0.0029x 0.1640 y
(b) y 0.0001429x2 0.07229x 2.9886 (c)
9 20 ,
1 0.0029x 0.1640
50
45 0 0
60 0
(d) For the linear model, x 50 gives y 6.86 billion. For the quadratic model, x 50 gives y 6.96 billion.
(c) No. For x 60, y 100. Note that there is a vertical asymptote at x 56.6.
As you extrapolate into the future, the quadratic model increases more rapidly.
Section 12.10
Lagrange Multipliers
2. Maximize f x, y xy.
4. Minimize f x, y x2 y 2.
Constraint: 2x y 4
Constraint: 2x 4y 5
f g
f g
y i xj 2 i j
2x i 2yj 2 i 4 j
y 2
2x 2 ⇒ x
x
2y 4 ⇒ y 2
2x y 4 ⇒ 4 4
2x 4y 5 ⇒ 10 5
12 , x 12 , y 1
1, x 1, y 2 f 1, 2 2
f 2 , 1 4 1
5
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Section 12.10 6. Maximize f x, y x2 y 2.
Lagrange Multipliers
8. Minimize f x, y 3x y 10.
Constraint: 2y x2 0
Constraint: x2y 6
f g
f g
2x i 2yj 2x i 2 j
3i j 2xy i x2 j
2x 2x ⇒ x 0 or 1
3 3 2xy ⇒ 2xy
If x 0, then y 0 and f 0, 0 0.
1 x2 ⇒ 12 x
If 1, 2y 2 2 ⇒ y 1 ⇒ x2 2 ⇒ x 2.
x2y 6 ⇒ x2
f 2, 1 2 1 1 Maximum.
3x 2 x 0
3x2 2xy ⇒ y
3x2 6 x3 4 3 4, y x
3 4, f
10. Note: f x, y x2 y 2 is minimum when gx, y is minimum.
2 y 1 x
y 2x
y 2x
xy 32 ⇒ 2x2 32 x 4, y 8
2x 4y 15 ⇒ 10 x 15
f 4, 8 16
3 x , y3 2 f
Constraint: xy 32
Constraint: 2x 4y 15 2y 4
3 4 3 4 20 3 9 2 2
12. Minimize f x, y 2x y.
Minimize gx, y x2 y 2. 2x 2
3 4 3 2
32, 3 g32, 3 3 2 5
14. Maximize or minimize f x, y exy 4. Constraint: x 2 y 2 ≤ 1 Case 1: On the circle x 2 y 2 1 y 4exy 4 2x
x2 y2 1 ⇒ x ±
2
2
Maxima: f ± Minima: f ±
2 2
,
,±
⇒ x y 2
x 4exy 4 2y
2
2
e
1.1331
e
0.8825
2
2
2
2
2
1 8
1 8
Case 2: Inside the circle fx y 4exy 4 0
⇒ xy0
fy x 4exy 4 0 fxx
y2 xy 4 x2 1 1 e , fyy e xy 4 , fxy exy xy 16 16 16 4
At 0, 0, fxx fyy fxy2 < 0. Saddle point: f 0, 0 1
Combining the two cases, we have a maximum of e1 8 at ±
2
2
,
2
2
and a minimum of e
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1 8
at ±
2
2
,±
2
2
.
351
352
Chapter 12
Functions of Several Variables
16. Maximize f x, y, z xyz.
18. Minimize x2 10x y2 14y 70 Constraint: x y 10
Constraint: x y z 6
2x 10 2y 14 xy8
yz xz x y z xy
xyz6 ⇒ xyz2
x 1 2 10 y 1 2 14
1 1 x y 10 14 2 2
f 2, 2, 2 8
12 8 ⇒ 4 Then x 3, y 5. f 3, 5 9 30 25 70 70 4 22. Maximize f x, y, z xyz.
20. Minimize f x, y, z x2 y 2 z 2.
Constraints: x 2 z 2 5
Constraints: x 2z 6
x 2y 0
x y 12 f g h
f g h
2x i 2yj 2z k i 2k i j
yz i xz j xyk 2x i 2z k i 2j yz 2x
2x 2y 2x 2y z 2z 2
x 2z 6 ⇒ z
xz 2 ⇒
6x x 3 2 2
xy 2z ⇒
x y 12 ⇒ y 12 x
2x 212 x 3 x 6, z 0 f 6, 6, 0 72
x 2
⇒ 92x 27 ⇒ x 6
xy 2
xy 2z
x2 z2 5
⇒ z 5 x 2
x 2y 0
⇒ y
yz 2x
x 2
xy2z xz2
x5 x 2 x3 x5 x 2 2 25 x 2 2 x5 x 2
x3 25 x 2
2x5 x 2 x 3 0 3x 3 10x x3x 2 10
103, y 21103, z 53 10 1 10 5 5 15 f , , 3 2 3 3 9 x 0 or x
Note: f 0, 0, 5 0 does not yield a maximum.
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Section 12.10
Lagrange Multipliers
353
24. Minimize the square of the distance f x, y x2 y 102 subject to the constraint x 42 y 2 4. 2x 2x 4 2 y 10 2y
x4 x
y 10
5 ⇒ y x 10 2
y
x 42 y 2 4 ⇒ x 2 8x 16
254 x
2
50x 100 4
29 2 x 58x 112 0 4 Using a graphing utility, we obtain x 3.2572 and x 4.7428 or, by the Quadratic Formula, x
58 ± 582 429 4112 58 ± 229 429 4 ± . 229 4 29 2 29
Using the smaller value, we have x 4 1
29
The point on the circle is 4 1 and the desired distance is d
29
29
29
and
y
1029 1.8570. 29
, 102929
161
29
29
102929 10
2
2
8.77.
The larger x-value does not yield a minimum. 28. Maximize f x, y, z z subject to the constraints x2 y 2 z2 0 and x 2z 4.
26. Minimize the square of the distance f x, y, z x 4 2
subject to the constraint
y2
x2
z2
0 2x
y 2 z 0.
x x 2x 4 z x2 y2 y y 2y z x2 y 2 2z
0 2y ⇒ y 0 1 2z 2
2x 4 2x 2y 2y
x2 y 2 z2 0 x 2z 4 ⇒ x 4 2z
4 2z2 02 z2 0
x2 y 2 z 0, x 2, y 0, z 2
3z2 16z 16 0
The point on the plane is 2, 0, 2 and the desired distance is
3z 4z 4 0
d 2 42 02 22 22.
4 3
z
or z 4
The maximum value of f occurs when z 4 at the point of 4, 0, 4. 30. See explanation at the bottom of page 922. 32. Maximize Vx, y, z xyz subject to the constraint 1.5x y 2xz 2yz C.
34. Minimize A , r 2 rh 2 r 2 subject to the constraint r 2h V0.
yz 1.5y 2z 3 xz 1.5x 2z x y and z x 4 xy 2x 2y
2 h 4 r 2 rh h 2r 2 r r 2
r 2h V0 ⇒ 2 r 3 V0
3 3 1.5xy 2xz 2yz C ⇒ 1.5x 2 x2 x2 C 2 2 x
2C
Dimensions: r
3
Volume is maximum when xy
2C
3
and
z
2C
4
.
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2V
3
0
and
2V
h2
3
0
354
Chapter 12
Functions of Several Variables
36. (a) Maximize Px, y, z xyz subject to the constraint
(b) Maximize P x1x2x3 . . . xn subject to the constraint n
x y z S. yz
x S. i
i1
x2x3 . . . xn x1x3 . . . xn x1x2 . . . xn
xz x y z xy S xyzS ⇒ xyz 3
x1x2x3 . . . xn1
Therefore,
n
S 3
xyz ≤
S 3
S , x, y, z > 0 3
x S ⇒ x i
1
i1
x1 x2 x3 . . . xn
S x2 x3 . . . xn n
Therefore,
S3 27
x1x2x3 . . . xn ≤
SnSnSn . . . Sn, x
3 xyz ≤
S 3
x1x2x3 . . . xn ≤
Sn
3 xyz ≤
xyz . 3
xyz ≤
38. Case 1: Minimize Pl, h 2h l 1
i
≥ 0
n
n x x x . . .x ≤ 1 2 3 n
S n
n x x x . . .x ≤ 1 2 3 n
x1 x2 x3 . . . xn . n
2l subject to the constraint lh 8l A. 2
l h 2 4
2
2h
2 l ⇒ , 1 l 2 l 2 l 2h
h
Case 2: Minimize Al, h lh h
l 2
8 subject to the constraint 2h l 2l P.
l
l
4 2
l
l l
l l 2 ⇒ , h 2 4 2 4 h
l or l 2h 2
40. Maximize T x, y, z 100 x2 y 2 subject to the constraints x2 y 2 z 2 50 and x z 0. 2x 2x 2y 2y 0 2z
42. Maximize Px, y 100x0.4y0.6 Constraint: 48x 36y 100,000.
If y 0, then 1 and 0, z 0.
40x0.6y0.6 48 ⇒
yx
0.6
60x0.4y0.4 36 ⇒
xy
0.4
0.6
T 0, 50, 0 100 50 150 If y 0, then x 2 z 2 2x 2 50 and x z 50 2.
50
2
, 0,
50
2
48 40
36 60
yx yx
Thus, x z 0 and y 50.
T
100 504 112.5
Therefore, the maximum temperature is 150.
0.4
48403660
y 2 ⇒ y 2x x 2500 5000 ,y 48x 36y2x 100,000 ⇒ x 3 3 P
5000 , $126,309.71. 2500 3 3
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Review Exercises for Chapter 12
355
44. Minimize Cx, y 48x 36y subject to the constraint 100x0.6y0.4 20,000. 48 60x0.4y0.4 ⇒
yx
0.4
36 40x0.6y0.6 ⇒
xy
0.6
48 60
36 40
yx yx 0.4
0.6
6048 4036
8 y 8 ⇒ y x x 9 9 100x0.6y0.4 20,000 ⇒ x0.6
89 x
0.4
200
x
200 209.65 890.4
y
8 200 186.35 9 890.4
Therefore, C209.65, 186.35 $16,771.94. 46. f x, y ax by, x, y > 0 Constraint:
x2 y2 1 64 36
(a) Level curves of f x, y 4x 3y are lines of form 4 y x C. 3
(b) Level curves of f x, y 4x 9y are lines of form 4 y x C. 9
4 Using y x 12.3, you obtain 3 x 7, y 3, and
4 Using y x 7, you obtain 9
f 7, 3 28 9 37.
x 4, y 5.2, and f 4, 5.2 62.8.
8
−10
10
−8
Constraint is an ellipse.
Review Exercises for Chapter 12 2. Yes, it is the graph of a function. 6. f x, y
4. f x, y ln xy The level curves are of the form c ln xy
1
2
3
c= 2
c=1 c=0
−3
e c xy. The level curves are hyperbolas.
c= 2
−2
x xy
3
The level curves are of the form c y
x xy
1 c c x.
1 c = −1 c = − 2 c = − 23 2
c = −2
1
c= 2
c=1 −3
3 3
−2
c= 2 c=2
The level curves are passing through the origin with slope 1c . c
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356
Chapter 12
Functions of Several Variables
1 x
8. gx, y y
10. f x, y, z 9x2 y2 9z2 0
z
12.
Elliptic cone
lim
x, y → 1, 1
xy x2 y 2
Does not exist
60
Continuous except when y ± x.
z 2
2 x
5
5
x
y xey 00 0 x, y → 0, 0 1 x2 10 lim
16. f x, y
Continuous everywhere
20.
y
y
2
14.
5
xy xy
fx
yx y xy y2 2 x y x y2
fy
x2 x y2 22. f x, y, z
w x2 y 2 z2 x w 1 2 x y 2 z2122x x 2 x2 y 2 z2
z lnx 2 y 2 1
18.
2x z x x 2 y 2 1 2y z y x2 y 2 1
1 1 x2 y 2 z2
1 fx 1 x2 y 2 z2322x 2
w y y x2 y 2 z2
w z z x2 y 2 z 2
x 1 x2 y 2 z232
fy
y 1 x2 y 2 z232
fz
z 1 x2 y 2 z232
26. z x2 ln y 1
24. ux, t csin akx cos kt u akccos akx cos kt x
z z 2x ln y 1. At 2, 0, 0, 0. x x
u kcsin akx sin kt t
Slope in x-direction. z x2 z . At 2, 0, 0, 4. y 1 y y Slope in y-direction.
28. hx, y
x xy
30. gx, y cosx 2y gx sinx 2y
hx
y x y2
hy
x x y2
gxx cosx 2y
hxx
2y x y3
gxy 2 cosx 2y
gy 2 sinx 2y gyy 4 cosx 2y gyx 2 cosx 2y
2x hyy x y3 hxy
x y2 2yx y xy x y4 x y3
hyx
x y2 2yx y xy x y4 x y3
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Review Exercises for Chapter 12 34. z e x sin y
32. z x3 3xy 2 z 3x2 3y 2 x
z e x sin y x
2z 6x x2
2z e x sin y x2
z 6xy y
z e x cos y y
2z 6x y 2 Therefore,
36. z
2z e x sin y y 2
2z 2z 0. x2 y 2
Therefore,
2z 2z 2 0. 2 x y
xy x2 y 2
dz
z z dx dy x y
x2 y 2 y xy x x2 y 2
x2 y 2
dx
x2 y 2x xy y x2 y 2
x2 y 2
dy x y y 3
2
2 32
dx
38. From the accompanying figure we observe tan
h or h x tan x
h
h h dx d tan dx x sec2 d. dh x
θ
1 11 Letting x 100, dx ± , , and d ± . 2 60 180
x
(Note that we express the measurement of the angle in radians.) The maximum error is approximately dh tan
40.
1160 ± 12 100 sec 1160 ± 180 ± 0.3247 ± 2.4814 ± 2.81 feet. 2
A r r 2 h2
dA r 2 h2
r2 rh dr dh r 2 h2 r 2 h2
2r 2 h2 rh 8 25 1 43 10 1 dr dh ± ± ± 8 r 2 h2 r 2 h2 29 29 8 8 29
42. u y 2 x, x cos t, y sin t Chain Rule:
du u x u y dt x t y t 1sin t 2ycos t sin t 2sin t cos t sin t1 2 cos t
Substitution: u sin2 t cos t du 2 sin t cos t sin t sin t1 2 cos t dt
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x3 dy x2 y 232
357
358
Chapter 12
44. w
Functions of Several Variables
xy , x 2r t, y rt, z 2r t z
Chain Rule:
xz2 y sin z 0
46.
w w x w y w z r x r y r z r
2xz
z2 z x y cos z 2xz
y x xy 2 t 2 2 z z z
2rt 2r tt 22r trt 2r t 2r t 2r t2
4r2t 4rt2 t 3 2r t2
z z z2 y cos z 0 x x
2xz
z z y cos z sin z 0 y y sin z z y 2xz y cos z
w w x w y w z t x t y t z t y x xy 1 r 2 1 z z z w
Substitution:
4r2t rt 2 4r 3 2r t2 xy 2r trt 2r2t rt 2 z 2r t 2r t
w 4r2t 4rt2 t 3 r 2r t2 w 4r2t rt2 4r 3 t 2r t2
48.
1 f x, y y 2 x 2 4
w 6x2 3xy 4y 2z
50.
w 12x 3yi 3x 8yzj 4y 2k
1 f 2xi yj 2
w1, 0, 1 12i 3j
f 1, 4 2i 2j
u
5 2 5 1 i j v u 5 5 5
Du f 1, 4 f 1, 4 u
z
52.
z
1 3
v
3
3
i
3
3
j
3
3
k
Duw1, 0, 1 w1, 0, 1 u 4 3 3 0 5 3
2 5 4 5 2 5 5 5 5
x2 xy
54.
z x2y z 2xy i x2j
x2 2xy x2 i j 2 x y x y2
z2, 1 4i 4 j z2, 1 4 2
z2, 1 4 j z2, 1 4 56. 4y sin x y2 3
58.
F 2yj 2z k
f x, y 4y sin x y2 f x, y 4y cos xi 4 sin x 2yj f
2 , 1 2j
Normal vector: j
Fx, y, z y 2 z 2 25 0 F2, 3, 4 6j 8k 23j 4k Therefore, the equation of the tangent plane is 3 y 3 4z 4 0 or
3y 4z 25,
and the equation of the normal line is x 2,
y3 z4 . 3 4
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Review Exercises for Chapter 12 60.
Fx, y, z x2 y 2 z 2 9 0
62.
Fx, y, z y 2 z 25 0 Gx, y, z x y 0
F 2xi 2y j 2z k F1, 2, 2 2i 4j 4k 2i 2j 2k
F 2y i k
Therefore, the equation of the tangent plane is
G i j F4, 4, 9 8i k
x 1 2 y 2 2z 2 0 or
x 2y 2z 9,
i F G 8 1
and the equation of the normal line is x1 y2 z2 . 1 2 2
j 0 1
k 1 i j 8k 0
Therefore, the equation of the tangent line is x4 y4 z9 . 1 1 8
64. (a) f x, y cos x sin y,
f 0, 0 1
(b) fxx cos x,
fx sin x,
fx0, 0 0
fyy sin y,
fyy0, 0 0
fy cos y,
fy0, 0 1
fxy 0,
fxy0, 0 0
P1x, y 1 y
1 P2x, y 1 y 2 x2
(c) If y 0, you obtain the 2nd degree Taylor polynomial for cos x.
(e)
fxx0, 0 1
z
(d)
f x, y
0
0
1.0
1.0
1.0
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 3
2 2 −2 −1 1
1
y
1 x
−1
−2
−1 y
1
P2x, y
y
z
2
P1x, y
x
2
1 −1
1
2
y
x
x
The accuracy lessens as the distance from 0, 0 increases. 66. f x, y 2x2 6xy 9y 2 8x 14 fx 4x 6y 8 0 fy 6x 18y 0, x 3y 43y 6y 8 ⇒ y 43 , x 4 fxx 4 fyy 18 fxy 6 fxx fyy f xy 2 418 62 36 > 0 Therefore, 4, 43 , 2 is a relative minimum.
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359
360
Chapter 12
Functions of Several Variables
68. z 50x y 0.1x3 20x 150 0.05y3 20.6y 125 zx 50 0.3x2 20 0, x ± 10 zy 50 0.15y 2 20.6 0, y ± 14 Critical Points: 10, 14, 10, 14, 10, 14, 10, 14 zxx 0.6x, zyy 0.3y, zxy 0 At 10, 14, zxx zyy zxy2 64.2 02 > 0, zxx < 0.
10, 14, 199.4 is a relative maximum. At 10, 14, zxx zyy zxy2 64.2 02 < 0.
10, 14, 349.4 is a saddle point. At 10, 14, zxx zyy zxy2 64.2 02 < 0.
10, 14, 200.6 is a saddle point. At 10, 14, zxx zyy xxy2 64.2 02 > 0, zxx < 0.
10, 14, 749.4 is a relative minimum. 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. 72. Minimize Cx1, x2 0.25x12 10x1 0.15x22 12x2 subject to the constraint x1 x2 1000. 0.50x1 10 0.30x2 12
5x 3x 20 1
2
x1 x2 1000 ⇒ 3x1 3x2 3000 5x1 3x2
20
3020
8x1
x1 377.5 x2 622.5 C377.5, 622.5 104,997.50 74. Minimize the square of the distance: f x, y, z x 22 y 22 x2 y2 02. fx 2x 2 2x2 y22x 0 x 2 2x3 2xy2 0
fy 2y 2 2x2 y22y 0 y 2 2y3 2x2y 0 Clearly x y and hence: 4x3 x 2 0. Using a computer algebra system, x 0.6894. Thus, distance2 0.6894 22 0.6894 22 20.689422 4.3389. distance 2.08 76. (a) 25, 28, 50, 38, 75, 54, 100, 75, 125, 102
x 375, y 297, x x 382,421,875, x y 26,900, x i
i
4
i
34,375,
2
i i
2
i i
yi 2,760,000,
x
382,421,875a 3,515,625b 34,375c 2,760,000 3,515,625a
34,375b
375c
26,900
34,375a
375b
5c
297
a 0.0045, b 0.0717, c 23.2914, y 0.0045x2 0.0717x 23.2914 (b) When x 80 km/hr, y 57.8 km.
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3
i
3,515,625
Review Exercises for Chapter 12
355
44. Minimize Cx, y 48x 36y subject to the constraint 100x0.6y0.4 20,000. 48 60x0.4y0.4 ⇒
yx
0.4
36 40x0.6y0.6 ⇒
xy
0.6
48 60
36 40
yx yx 0.4
0.6
6048 4036
8 y 8 ⇒ y x x 9 9 100x0.6y0.4 20,000 ⇒ x0.6
89 x
0.4
200
x
200 209.65 890.4
y
8 200 186.35 9 890.4
Therefore, C209.65, 186.35 $16,771.94. 46. f x, y ax by, x, y > 0 Constraint:
x2 y2 1 64 36
(a) Level curves of f x, y 4x 3y are lines of form 4 y x C. 3
(b) Level curves of f x, y 4x 9y are lines of form 4 y x C. 9
4 Using y x 12.3, you obtain 3 x 7, y 3, and
4 Using y x 7, you obtain 9
f 7, 3 28 9 37.
x 4, y 5.2, and f 4, 5.2 62.8.
8
−10
10
−8
Constraint is an ellipse.
Review Exercises for Chapter 12 2. Yes, it is the graph of a function. 6. f x, y
4. f x, y ln xy The level curves are of the form c ln xy
1
2
3
c= 2
c=1 c=0
−3
e c xy. The level curves are hyperbolas.
c= 2
−2
x xy
3
The level curves are of the form c y
x xy
1 c c x.
1 c = −1 c = − 2 c = − 23 2
c = −2
1
c= 2
c=1 −3
3 3
−2
c= 2 c=2
The level curves are passing through the origin with slope 1c . c
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356
Chapter 12
Functions of Several Variables
1 x
8. gx, y y
10. f x, y, z 9x2 y2 9z2 0
z
12.
Elliptic cone
lim
x, y → 1, 1
xy x2 y 2
Does not exist
60
Continuous except when y ± x.
z 2
2 x
5
5
x
y xey 00 0 x, y → 0, 0 1 x2 10 lim
16. f x, y
Continuous everywhere
20.
y
y
2
14.
5
xy xy
fx
yx y xy y2 2 x y x y2
fy
x2 x y2 22. f x, y, z
w x2 y 2 z2 x w 1 2 x y 2 z2122x x 2 x2 y 2 z2
z lnx 2 y 2 1
18.
2x z x x 2 y 2 1 2y z y x2 y 2 1
1 1 x2 y 2 z2
1 fx 1 x2 y 2 z2322x 2
w y y x2 y 2 z2
w z z x2 y 2 z 2
x 1 x2 y 2 z232
fy
y 1 x2 y 2 z232
fz
z 1 x2 y 2 z232
26. z x2 ln y 1
24. ux, t csin akx cos kt u akccos akx cos kt x
z z 2x ln y 1. At 2, 0, 0, 0. x x
u kcsin akx sin kt t
Slope in x-direction. z x2 z . At 2, 0, 0, 4. y 1 y y Slope in y-direction.
28. hx, y
x xy
30. gx, y cosx 2y gx sinx 2y
hx
y x y2
hy
x x y2
gxx cosx 2y
hxx
2y x y3
gxy 2 cosx 2y
gy 2 sinx 2y gyy 4 cosx 2y gyx 2 cosx 2y
2x hyy x y3 hxy
x y2 2yx y xy x y4 x y3
hyx
x y2 2yx y xy x y4 x y3
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Review Exercises for Chapter 12 34. z e x sin y
32. z x3 3xy 2 z 3x2 3y 2 x
z e x sin y x
2z 6x x2
2z e x sin y x2
z 6xy y
z e x cos y y
2z 6x y 2 Therefore,
36. z
2z e x sin y y 2
2z 2z 0. x2 y 2
Therefore,
2z 2z 2 0. 2 x y
xy x2 y 2
dz
z z dx dy x y
x2 y 2 y xy x x2 y 2
x2 y 2
dx
x2 y 2x xy y x2 y 2
x2 y 2
dy x y y 3
2
2 32
dx
38. From the accompanying figure we observe tan
h or h x tan x
h
h h dx d tan dx x sec2 d. dh x
θ
1 11 Letting x 100, dx ± , , and d ± . 2 60 180
x
(Note that we express the measurement of the angle in radians.) The maximum error is approximately dh tan
40.
1160 ± 12 100 sec 1160 ± 180 ± 0.3247 ± 2.4814 ± 2.81 feet. 2
A r r 2 h2
dA r 2 h2
r2 rh dr dh r 2 h2 r 2 h2
2r 2 h2 rh 8 25 1 43 10 1 dr dh ± ± ± 8 r 2 h2 r 2 h2 29 29 8 8 29
42. u y 2 x, x cos t, y sin t Chain Rule:
du u x u y dt x t y t 1sin t 2ycos t sin t 2sin t cos t sin t1 2 cos t
Substitution: u sin2 t cos t du 2 sin t cos t sin t sin t1 2 cos t dt
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x3 dy x2 y 232
357
358
Chapter 12
44. w
Functions of Several Variables
xy , x 2r t, y rt, z 2r t z
Chain Rule:
xz2 y sin z 0
46.
w w x w y w z r x r y r z r
2xz
z2 z x y cos z 2xz
y x xy 2 t 2 2 z z z
2rt 2r tt 22r trt 2r t 2r t 2r t2
4r2t 4rt2 t 3 2r t2
z z z2 y cos z 0 x x
2xz
z z y cos z sin z 0 y y sin z z y 2xz y cos z
w w x w y w z t x t y t z t y x xy 1 r 2 1 z z z w
Substitution:
4r2t rt 2 4r 3 2r t2 xy 2r trt 2r2t rt 2 z 2r t 2r t
w 4r2t 4rt2 t 3 r 2r t2 w 4r2t rt2 4r 3 t 2r t2
48.
1 f x, y y 2 x 2 4
w 6x2 3xy 4y 2z
50.
w 12x 3yi 3x 8yzj 4y 2k
1 f 2xi yj 2
w1, 0, 1 12i 3j
f 1, 4 2i 2j
u
5 2 5 1 i j v u 5 5 5
Du f 1, 4 f 1, 4 u
z
52.
z
1 3
v
3
3
i
3
3
j
3
3
k
Duw1, 0, 1 w1, 0, 1 u 4 3 3 0 5 3
2 5 4 5 2 5 5 5 5
x2 xy
54.
z x2y z 2xy i x2j
x2 2xy x2 i j 2 x y x y2
z2, 1 4i 4 j z2, 1 4 2
z2, 1 4 j z2, 1 4 56. 4y sin x y2 3
58.
F 2yj 2z k
f x, y 4y sin x y2 f x, y 4y cos xi 4 sin x 2yj f
2 , 1 2j
Normal vector: j
Fx, y, z y 2 z 2 25 0 F2, 3, 4 6j 8k 23j 4k Therefore, the equation of the tangent plane is 3 y 3 4z 4 0 or
3y 4z 25,
and the equation of the normal line is x 2,
y3 z4 . 3 4
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Review Exercises for Chapter 12 60.
Fx, y, z x2 y 2 z 2 9 0
62.
Fx, y, z y 2 z 25 0 Gx, y, z x y 0
F 2xi 2y j 2z k F1, 2, 2 2i 4j 4k 2i 2j 2k
F 2y i k
Therefore, the equation of the tangent plane is
G i j F4, 4, 9 8i k
x 1 2 y 2 2z 2 0 or
x 2y 2z 9,
i F G 8 1
and the equation of the normal line is x1 y2 z2 . 1 2 2
j 0 1
k 1 i j 8k 0
Therefore, the equation of the tangent line is x4 y4 z9 . 1 1 8
64. (a) f x, y cos x sin y,
f 0, 0 1
(b) fxx cos x,
fx sin x,
fx0, 0 0
fyy sin y,
fyy0, 0 0
fy cos y,
fy0, 0 1
fxy 0,
fxy0, 0 0
P1x, y 1 y
1 P2x, y 1 y 2 x2
(c) If y 0, you obtain the 2nd degree Taylor polynomial for cos x.
(e)
fxx0, 0 1
z
(d)
f x, y
0
0
1.0
1.0
1.0
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 3
2 2 −2 −1 1
1
y
1 x
−1
−2
−1 y
1
P2x, y
y
z
2
P1x, y
x
2
1 −1
1
2
y
x
x
The accuracy lessens as the distance from 0, 0 increases. 66. f x, y 2x2 6xy 9y 2 8x 14 fx 4x 6y 8 0 fy 6x 18y 0, x 3y 43y 6y 8 ⇒ y 43 , x 4 fxx 4 fyy 18 fxy 6 fxx fyy f xy 2 418 62 36 > 0 Therefore, 4, 43 , 2 is a relative minimum.
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359
360
Chapter 12
Functions of Several Variables
68. z 50x y 0.1x3 20x 150 0.05y3 20.6y 125 zx 50 0.3x2 20 0, x ± 10 zy 50 0.15y 2 20.6 0, y ± 14 Critical Points: 10, 14, 10, 14, 10, 14, 10, 14 zxx 0.6x, zyy 0.3y, zxy 0 At 10, 14, zxx zyy zxy2 64.2 02 > 0, zxx < 0.
10, 14, 199.4 is a relative maximum. At 10, 14, zxx zyy zxy2 64.2 02 < 0.
10, 14, 349.4 is a saddle point. At 10, 14, zxx zyy zxy2 64.2 02 < 0.
10, 14, 200.6 is a saddle point. At 10, 14, zxx zyy xxy2 64.2 02 > 0, zxx < 0.
10, 14, 749.4 is a relative minimum. 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. 72. Minimize Cx1, x2 0.25x12 10x1 0.15x22 12x2 subject to the constraint x1 x2 1000. 0.50x1 10 0.30x2 12
5x 3x 20 1
2
x1 x2 1000 ⇒ 3x1 3x2 3000 5x1 3x2
20
3020
8x1
x1 377.5 x2 622.5 C377.5, 622.5 104,997.50 74. Minimize the square of the distance: f x, y, z x 22 y 22 x2 y2 02. fx 2x 2 2x2 y22x 0 x 2 2x3 2xy2 0
fy 2y 2 2x2 y22y 0 y 2 2y3 2x2y 0 Clearly x y and hence: 4x3 x 2 0. Using a computer algebra system, x 0.6894. Thus, distance2 0.6894 22 0.6894 22 20.689422 4.3389. distance 2.08 76. (a) 25, 28, 50, 38, 75, 54, 100, 75, 125, 102
x 375, y 297, x x 382,421,875, x y 26,900, x i
i
4
i
34,375,
2
i i
2
i i
yi 2,760,000,
x
382,421,875a 3,515,625b 34,375c 2,760,000 3,515,625a
34,375b
375c
26,900
34,375a
375b
5c
297
a 0.0045, b 0.0717, c 23.2914, y 0.0045x2 0.0717x 23.2914 (b) When x 80 km/hr, y 57.8 km.
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3
i
3,515,625
Problem Solving for Chapter 12
78. Optimize f x, y x2y subject to the constraint x 2y 2. 2xy
x 4xy ⇒ x 0 or x 4y 2
x2 2 x 2y 2
1 4 If x 0, y 1. If x 4y, then y 3 , x 3 . 4 1 16 Maximum: f 3 , 3 27
Minimum: f 0, 1 0
Problem Solving for Chapter 12 4 2. V r 3 r 2h 3 Material M 4 r 2 2 rh V 1000 ⇒ h
1000 43 r 3 r2
Hence, M 4 r 2 2 r 4 r 2
1000 r43 r 3
2
2000 8 2 r r 3
dM 2000 16 8 r 2 r 0 dr r 3 8 r
16 2000 r 2 3 r
r3
83 2000 r3
Then, h
750 6 ⇒ r5
13
.
1000 43750 0. r2
The tank is a sphere of radius r 5
6
13
.
4. (a) As x → ± , f x x3 113 → x and hence lim f x gx lim f x gx 0.
x→
x→
(b) Let x0, x03 113 be a point on the graph of f.
4
−6
6
The line through this point perpendicular to g is y x x0
3 x 3 0
−4
1.
This line intersects g at the point
12 x
0
3 x 3 1 , 0
1 x 3 x03 1 . 2 0
The square of the distance between these two points is hx0
1 x 3 x03 1 2. 2 0
h is a maximum for x0
1 3 2
. Hence, the point on f farthest from g is
12, 12. 3
3
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361
362
Chapter 12
Functions of Several Variables
6. Heat Loss H k5xy xy 3xz 3xz 3yz 3yz k6xy 6xz 6yz V xyz 1000 ⇒ z
Then H 6k xy
1000 . xy
1000 1000 . y x
Setting Hx Hy 0, you obtain x y z 10. 8. (a) Tx, y 2x2 y2 y 10 10 1 1 4 4
2x2 y2 y
2x2 y
1 2
y
2
1 2
1 4
x −
1 2
−
1 2
1 2
x2 y 122 1 ellipse 18 14 (b) On x2 y2 1, Tx, y Ty 21 y2 y2 y 10 12 y2 y 3 1 T y 2y 1 0 ⇒ y , x ± . 2 2
Inside: Tx 4x 0, Ty 2y 1 0 ⇒
0, 21
12 394 minimum
T 0,
T ±
3
2
,
1 49 maximum 2 4
10. x r cos , y r sin , z z u u x u y u z
x
y
z
u u r sin r cos Similarly, x y
u u u cos sin . r x y 2u x 2u 2u y 2u z u cos
r sin r 2 2
x x y x z
x
u x u y u z u r sin
y x y y z y
r cos
2
2
2
2
2u 2 2 2u 2u 2 u u r sin 2 r 2 cos2 2 r sin cos r cos r sin
x2 y x y x y
Similarly,
2u 2u 2u 2 u 2 cos2 2 sin2 2 cos sin . r 2 x y x y
Now observe that 2u 1 u 2u 2u 2u 2u 1 u u 1 2u cos2 2 sin2 2 2 2 2 cos sin cos sin
r 2 r r r
z x2 y x y r x y
2u
x
2u 2u 2u 2 2. x2 y z
sin2
2u 2u 1 u 1 u 2u cos2 2 sin cos cos sin 2 y2 x y r x r y z
2
Thus, Laplaces equation in cylindrical coordinates, is
1 2u 2u 2u 1 u 2 2 2 0. 2 r r r r
z
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Problem Solving for Chapter 12 12. (a) d x2 y2 322t2 322t 16t22
(b)
dd 32 t2 32t 8 dt t2 42t 16
(d)
d 2d 32 t3 62t2 36t 3212 0 dt2 t2 42t 1632
4096t2 10242t3 256t4 16tt2 42t 16 (c) When t 2: dd 32 12 62
38.16 ftsec dt 20 82
when t 1.943 seconds. No. The projectile is at its maximum height when t 2.
14. Given that f is a differentiable function such that f x0, y0 0, then fxx0, y0 0 and fyx0, y0 0. Therefore, the tangent plane is z z0 0 or z z0 f x0, y0 which is horizontal.
16. r, 5,
18
y
dr ± 0.05, d ± 0.05 x r cos 5 cos
4.924 18
y r sin 5 sin
0.868 18
5 4 3 2 1
(
(r, θ ) = 5,
π 18
)
5 x 1
(a) dx should be more effected by changes in r. dx cos dr r sin d
2
3
4
5
(b) dy should be more effected by changes in . dy sin dr r cos d
0.985dr 0.868 d
0.174 dr 4.924 d
dx is more effected by changes in r because 0.985 > 0.868.
18.
π θ = 18
dy is more effected by because 4.924 > 0.174.
u 1 cosx t cosx t
t 2 2u 1 sinx t sinx t
t 2 2 u 1 cosx t cosx t
x 2 2u 1 sinx t sinx t
x 2 2 Then,
2u 2u 2. t 2 x
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363
C H A P T E R 1 2 Functions of Several Variables Section 12.1 Introduction to Functions of Several Variables . . . . . . . 76 Section 12.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . 80 Section 12.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . 83 Section 12.4 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . 88 Section 12.5 Chain Rules for Functions of Several Variables . . . . . . . 92 Section 12.6 Directional Derivatives and Gradients . . . . . . . . . . . . 98 Section 12.7 Tangent Planes and Normal Lines . . . . . . . . . . . . . 103 Section 12.8 Extrema of Functions of Two Variables . . . . . . . . . . 109 Section 12.9 Applications of Extrema of Functions of Two Variables
. 113
Section 12.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 119 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
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C H A P T E R 1 2 Functions of Several Variables Section 12.1
Introduction to Functions of Several Variables
Solutions to Odd-Numbered Exercises 1. x 2z yz xy 10
3.
zx 2 y 10 xy
x2 y2 z2 1 4 9 No, z is not a function of x and y. For example, x, y 0, 0 corresponds to both z ± 1.
10 xy z 2 x y Yes, z is a function of x and y.
5. f x, y
x y
(a) f 3, 2
3 2
(b) f 1, 4
(d) f 5, y
5 y
(e) f x, 2
(c) f 30, 5
x 2
(f) f 5, t
(a) f 5, 0 5e0 5 (b) f 3, 2 3e 2 2 e
30 6 5
5 t
xy z
9. hx, y, z
7. f x, y xe y
(c) f 2, 1 2e1
1 4
(a) h2, 3, 9
23 2 9 3
(b) h1, 0, 1
10 0 1
(d) f 5, y 5e y (e) f x, 2 xe 2 (f) f t, t tet
y
11. f x, y x sin y
13. gx, y
2t 3 dt
x
4 2 sin 4 2
4
(a) f 2,
(a) g0, 4
0
(b) f 3, 1 3 sin 1
4
(b) g1, 4
1
4
2t 3 dt t 2 3t
2t 3 dt t 2 3t
0 4 1
4 6
15. f x, y x 2 2y (a)
f x x, y f x, y x x2 2y x2 2y x x
(b)
x 2 2xx x2 2y x 2 2y x2x x 2x x, x 0 x x
f x, y y f x, y x 2 2 y y x 2 2y x 2 2y 2y x 2 2y 2y 2, y 0 y y y y
76
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Section 12.1 17. f x, y 4 x 2 y 2 Domain: 4
x2
19. f x, y arcsinx y
y ≥ 0 2 2
Domain: 4 x y > 0 xy < 4
Range: ≤ z ≤ 2 2
x, y: x 2 y 2 ≤ 4
x, y: y < x 4
Range: 0 ≤ z ≤ 2
23. z
Range: all real numbers
xy xy
25. f x, y e xy
27. gx, y
Domain: x, y: y 0
Domain: x, y: x 0 and y 0
1 xy
Domain: x, y: x 0 and y 0
Range: z > 0
Range: all real numbers
29. f x, y
77
21. f x, y ln4 x y
Domain: x, y: 1 ≤ x y ≤ 1
x y ≤ 4 2
Introduction to Functions of Several Variables
Range: all real numbers except zero
4x x2 y2 1
(a) View from the positive x-axis: 20, 0, 0
(b) View where x is negative, y and z are positive: 15, 10, 20
(c) View from the first octant: 20, 15, 25
(d) View from the line y x in the xy-plane: 20, 20, 0
31. f x, y 5
33. f x, y y 2
z
Plane: z 5
Since 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 ≥ 0.
4
2
2
z
4
4
y 5
x
4
1 4
2
3
y
x
35. z 4 x 2 y 2
37. f x, y ex
z 4
Paraboloid Domain: entire xy-plane Range: z ≤ 4 −3
2
3
3
z
Since the variable y is missing, the surface is a cylinder with rulings parallel to the y-axis. The generating curve is z ex. The domain is the entire xy-plane and the range is z > 0.
y
8 6 4 2
x 4 x
39. z y 2 x 2 1
41. f x, y x 2exy2
z
Hyperbolic paraboloid
z
Domain: entire xy-plane Range: < z <
y
x
y x
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4
y
78
Chapter 12
Functions of Several Variables
43. f x, y x 2 y 2 (a)
(c) g is a horizontal translation of f two units to the right. The vertex moves from 0, 0, 0 to 0, 2, 0.
z 5 4
(d) g is a reflection of f in the xy-plane followed by a vertical translation 4 units upward. (e)
−2 1
2
y
2
z
z
x
5
5
4
4
(b) g is a vertical translation of f two units upward z = f (x, 1)
z = f (1, y)
2 2
x
y
2
2
x
45. z e1x
2 y 2
47. z ln y x 2
Level curves:
y
49. z x y Level curves are parallel lines of the form x y c.
Level curves:
c
2 2 e1x y
c ln y
ln c 1 x 2 y 2
x2
y
± ec y x2
x 2 y 2 1 ln c
4
y x2 ± e c
Circles centered at 0, 0
Parabolas
Matches (c)
Matches (b)
2
−2
2
x 4 c=4
−2
c=2 c = −1
53. f x, y xy
51. f x, y 25 x 2 y 2
The level curves are hyperbolas of the form xy c.
The level curves are of the form c 25 x 2 y 2, x2
y2
25
c=0
y
c 2.
Thus, the level curves are circles of radius 5 or less, y centered at the origin. 6
1
c=5 c=4 c=3 c=2
−1
1 −1
2 −6
x
−2
2
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
6
−2
c=1 c=0
−6
55. f x, y
x x2 y2
57. f x, y x2 y2 2
The level curves are of the form x c 2 x y2 x x y2 0 c
c=− c=−
3 2
−9
y
1 2 2
c=2 x
2
2
y2
2c1
2
c = −2
c=
c = −1
1 c= 2
9
−6
c=1
2
x 2c1
6
3 2
Thus, the level curves are circles passing through the origin and centered at 12c, 0.
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Section 12.1 59. gx, y
8 1 x2 y2
Introduction to Functions of Several Variables
61. See Definition, page 838.
63. No, The following graphs are not hemispheres. z ex
4
−6
2 y 2
z x2 y2
6
−4
1 R 1 0.10 1I
Inflation Rate
f x, y x 2 y 2.
69. f x, y, z x 2y 3z
10
67. VI, R 1000
65. The surface is sloped like a saddle. The graph is not unique. Any vertical translation would have the same level curves. One possible function is
Tax Rate
0
0.03
0.05
0
2593.74
1929.99
1592.33
0.28
2004.23
1491.34
1230.42
0.35
1877.14
1396.77
1152.40
71. f x, y, z x 2 y 2 z 2
73. f x, y, z 4x 2 4y 2 z 2
c6
c9
c0
6 x 2y 3z
9 x2 y2 z2
0 4x 2 4y 2 z 2
Plane
Sphere
Elliptic cone
z
z
z 4
3
2
−2
−3
−2
−4
y
2
4
4
x 6
y
1
2
y
x
−4
x
75. Nd, L
d 4 4
(a) N22, 12
2
L
22 4 4 12 243 board-feet 2
(b) N30, 12
The level curves are of the form c 600
0.75x 2
x2 y 2
600 c . 0.75
30 4 4 12 507 board-feet 2
20.40xz 20.40yz
79. C 0.75xy
77. T 600 0.75x 2 0.75y 2
base front & back two ends
0.75y 2
0.75xy 0.80xz yz
The level curves are circles centered at the origin. c = 600 c = 500 c = 400
y 30
c = 300 c = 200 c = 100 c=0
x z y
x
−30
79
30
−30
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80
Chapter 12
Functions of Several Variables
81. PV kT, 202600 k300 (a) k
202600 520 300 3
(b) P
kT 520 T V 3 V
The level curves are of the form: c V
T 520 3 V 520 T 3c
Thus, the level curves are lines through the origin with slope
520 . 3c
83. (a) Highest pressure at C
85. (a) The boundaries between colors represent level curves
(b) Lowest pressure at A
(b) No, the colors represent intervals of different lengths, as indicated in the box
(c) Highest wind velocity at B
(c) You could use more colors, which means using smaller intervals 89. False. Let
87. False. Let f x, y 2xy
f x, y 5.
f 1, 2 f 2, 1, but 1 2
Section 12.2
Then, f 2x, 2y 5 22 f x, y.
Limits and Continuity
1. Let > 0 be given. We need to find > 0 such that f x, y L y b < whenever 0 < x a2 y b2 < . Take . Then if 0 < x a2 y b2 < , we have y b2 <
y b < . 3.
5.
7.
lim
f x, y gx, y
lim
f x, ygx, y
lim
x 3y 2 2 312 5
x, y → a, b
x, y → a, b
lim
x, y → a, b
lim
x, y → a, b
f x, y
f x, y
lim
x, y → a, b
lim
x, y → a, b
gx, y 5 3 2
gx, y 53 15
9.
x, y → 2, 1
lim
x, y → 0, 1
arcsinx y arcsin 0 0 1 xy
13.
Continuous for xy 1, y 0, x y ≤ 1 15.
lim
x, y, z → 1, 2, 5
x y z 8 22
Continuous for x y z ≥ 0
xy 24 3 xy 24
Continuous for x y
Continuous everywhere
11.
lim
x, y → 2, 4
lim
x, y → 1, 2
e xy e2
Continuous everywhere 17.
lim
x, y → 0, 0
e xy 1
Continuous everywhere
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1 e2
Section 12.2 19.
lim
x, y → 0, 0
Limits and Continuity
81
lnx2 y2 ln0
The limit does not exist. Continuous except at 0, 0
21. f x, y
xy x2 y2
Continuous except at 0, 0 Path: y 0
Path: y x
x, y
1, 0
0.5, 0
0.1, 0
0.01, 0
0.001, 0
f x, y
0
0
0
0
0
x, y
1, 1
0.5, 0.5
0.1, 0.1
0.01, 0.01
0.001, 0.001
1 2
1 2
1 2
1 2
1 2
f x, y
The limit does not exist because along the path y 0 the function equals 0, whereas along the path y x the function equals 12 .
23. f x, y
x2
xy 2 y4
Continuous except at 0, 0 Path: x y 2
Path: x y 2
x, y
1, 1
0.25, 0.5
0.01, 0.1
0.0001, 0.01
0.000001, 0.001
f x, y
12
12
12
12
12
x, y
1, 1
0.25, 0.5
0.01, 0.1
0.0001, 0.01
0.000001, 0.001
1 2
1 2
1 2
1 2
1 2
f x, y
The limit does not exist because along the path x y 2 the function equals 12 , whereas along the path x y 2 the function equals 12 .
25.
lim
x, y → 0, 0
f x, y
2xy2 y2 x2 y2
lim
x
lim
1 x 2xy y 1
2
x, y → 0, 0
27.
lim
x, y → 0, 0
z
2
x, y → 0, 0
2
sin x sin y 0
2
(same limit for g)
y
Thus, f is not continuous at 0, 0, whereas g is continuous at 0, 0.
29.
lim
x, y → 0, 0
x2y x 4 4y 2
x
31. f x, y
Does not exist
10xy 2x2 3y2
The limit does not exist. Use the paths x 0 and x y.
z
z
y
x
y x
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82
Chapter 12
Functions of Several Variables
33.
sinx 2 y 2 sin r 2 2r cos r 2 lim 2 lim lim cos r 2 1 2 2 r→0 r→0 r→0 x, y → 0, 0 x y r 2r
35.
x3 y 3 r 3 cos3 sin3 lim lim rcos3 sin3 0 2 2 r→0 r→0 x, y → 0, 0 x y r2
lim
lim
37. f x, y, z
1
39. f x, y, z
x 2 y 2 z 2
Continuous except at 0, 0, 0
41.
sin z ex ey
Continuous everywhere
f t t 2
f t
43.
gx, y 3x 2y
1 t
gx, y 3x 2y
f gx, y f 3x 2y
f gx, y f 3x 2y
3x 2y2 9x 2 12xy 4y 2
Continuous for y
Continuous everywhere
1 3x 2y
3x 2
45. f x, y x 2 4y (a) lim
x→0
f x x, y f x, y x x 2 4y x 2 4y lim
x→0
x
x 2x x x2 lim 2x x 2x
x→0
x→0
x
lim (b) lim
y→0
f x, y y f x, y x 2 4 y y x 2 4y lim
y→0
y
y lim
y→0
4 y lim 4 4
y→0
y
47. f x, y 2x xy 3y (a) lim
x→0
f x x, y f x, y 2x x x xy 3y 2x xy 3y lim
x→0
x
x lim
x→0
(b) lim
y→0
2 x xy lim 2 y 2 y
x→0
x
f x, y y f x, y 2x x y y 3 y y 2x xy 3y lim
y→0
y
y lim
y→0
x y 3 y lim x 3 x 3
y→0
y
51. No.
49. See the definition on page 851. Show that the value of
lim
x, y → x0 , y0
for two different paths to x0, y0.
f x, y is not the same
The existence of f 2, 3 has no bearing on the existence of the limit as x, y → 2, 3.
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Section 12.3 53. Since
lim
x, y → a, b
Partial Derivatives
f x, y L1, then for 2 > 0, there corresponds 1 > 0 such that f x, y L1 < 2 whenever
0 < x a2 y b2 < 1. Since
lim
x, y → a, b
gx, y L2, then for 2 > 0, there corresponds 2 > 0 such that gx, y L 2 < 2 whenever
0 < x a2 y b2 < 2. Let be the smaller of 1 and 2. By the triangle inequality, whenever x a2 y b2 < , we have
f x, y gx, y L 1 L 2 f x, y L 1 gx, y L 2 ≤ f x, y L 1 gx, y L 2 < 2 2 . Therefore,
lim
x, y → a, b
f x, y gx, y L 1 L 2.
55. True
57. False. Let f x, y
nx2 y2,
0,l
x, y 0, 0 x 0, y 0
See Exercise 19.
Section 12.3
Partial Derivatives
1. fx 4, 1 < 0
3. fy 4, 1 > 0
5. f x, y 2x 3y 5 fx x, y 2 fy x, y 3
7.
13.
z xy z y x
z 2x 5y x
z 2xe 2y x
x z y 2y
z 5x 6y y
z 2x 2e 2y y
z lnx 2 y 2
15.
z 2x x x 2 y 2
z
z ln
xx yy lnx y lnx y
z 1 1 2y 2 x x y x y x y2
2y z y x 2 y 2
17.
11. z x 2e 2y
9. z x2 5xy 3y2
z 1 1 2x y x y x y x 2 y 2
4y 2 x2 2y x
19. hx, y ex
z 2x 4y 2 x 3 4y 3 2 x 2y x x2y
2 y 2
hx x, y 2xex
2 y 2
hy x, y 2yex
2 y 2
z 8y x 3 16y 3 x2 2 y 2y x 2xy 2 21. f x, y x 2 y 2
23.
z tan2x y
fx x, y
1 2 x x y 212 2x 2 x 2 y 2
z 2 sec22x y x
fy x, y
1 2 y x y 212 2y 2 x 2 y 2
z sec 22x y y
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83
84
Chapter 12
25.
z e y sin xy
Functions of Several Variables
y
27. f x, y
t 2 1 dt
x
z ye y cos xy x
z ey sin xy xey cos xy y
t3 t 3
y
x
y3 y x3 x 3
3
fx x, y x 1 1 x 2 2
fy x, y y 2 1
ey x cos xy sin xy
[You could also use the Second Fundamental Theorem of Calculus.] 29. f x, y 2x 3y f x x, y f x, y 2x x 3y 2x 3y 2 x f lim lim lim 2 x→0 x→0 x x x→0 x x f x, y y f x, y 2x 3 y y 2x 3y 3 y f lim lim lim 3 y→0 y→0 y y y→0 y y 31. f x, y x y x x y x y f f x x, y f x, y lim lim x→0 x x→0 x x
lim
x→0
lim
x→0
x x y x y x x y x y xx x y x y 1 x x y x y
1 2x y
x y y x y f f x, y y f x, y lim lim y→0 y y→0 y y
lim
y→0
lim
y→0
33. gx, y 4 x 2 y 2
gy x, y 2y At 1, 1: gy 1, 1 2
1 2x y
z ex sin y y z At 0, 0: 0 y
y x
39. f x, y
1 y y 2 2 1 y 2x 2 x x y2 1 At 2, 2: fx 2, 2 4
1 1 x 2 1 y 2x 2 x x y2 1 At 2, 2: fy 2, 2 4 fy x, y
z ex cos y x z At 0, 0: 1 x
At 1, 1: gx 1, 1 2
fx x, y
1 x y y x y
35. z ex cos y
gxx, y 2x
37. f x, y arctan
x y y x y x y y x y yx y y x y
xy xy
yx y xy y 2 2 x y x y2 1 At 2, 2: fx 2, 2 4 fx x, y
xx y xy x2 2 x y x y2 1 At 2, 2: fy 2, 2 4 fy x, y
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Section 12.3
43. z 9x2 y 2, y 3, 1, 3, 0
41. z 49 x2 y2, x 2, 2, 3, 6 Intersecting curve: z 45 y
Partial Derivatives
Intersecting curve: z 9x 2 9 2
z 18x x
y z y 45 y 2
z 181 18 x
At 1, 3, 0:
z 3 1 At 2, 3, 6: y 45 9 2
y=3 z
z
x=2 10
160
4
3
2 4
y
x 8 x
8
y
45. fx x, y 2x 4y 4, fyx, y 4x 2y 16
47. fx x, y
1 1 y, fyx, y 2 x x2 y
fx fy 0: 2x 4y 4 fx fy 0:
4x 2y 16 Solving for x and y,
1 1 y 0 and 2 x 0 x2 y y
x 6 and y 4.
1 1 and x 2 x2 y
y y4 ⇒ y 1 x Points: 1, 1 49. (a) The graph is that of fy.
51.
(b) The graph is that of fx.
55. Hx, y, z sinx 2y 3z Hxx, y, z cosx 2y 3z Hy x, y, z 2 cosx 2y 3z Hzx, y, z 3 cosx 2y 3z
57.
w x 2 y 2 z 2
53. Fx, y, z ln x 2 y 2 z 2
w x x x 2 y 2 z 2
w y y x 2 y 2 z 2
Fx x, y, z
x x2 y2 z2
z w z x 2 y 2 z 2
Fy x, y, z
y x2 y2 z2
Fz x, y, z
z x2 y2 z2
z x 2 2xy 3y 2 z 2x 2y x
59.
1 lnx 2 y 2 z 2 2
z x 2 y 2 z x x x 2 y 2
2z 2 x2
2z y2 x2 x 2 y 2 32
2z 2 yx
xy 2z yx x 2 y 2 32
z 2x 6y y
z y y x 2 y 2
2z 6 y2
2z x2 2 2 y x y 2 32
2z 2 xy
xy 2z xy x 2 y 2 32
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85
86
61.
Chapter 12
Functions of Several Variables
z e x tan y
z arctan
63.
z e x tan y x
1 z y y 2 2 x 1 y 2x 2 x x y2
e x tan y x2 2z
2z 2xy x2 x 2 y 2 2
2z e x sec2 y yx
x2 y2 y2y 2z y2 x2 2 2 2 2 yx x y x y 22
z e x sec2 y y
1 1 z x 2 y 1 y 2x 2 x x y2
2e x sec2 y tan y y2 2z
2z 2xy y2 x 2 y 2 2
2z e x sec2 y xy
65.
x 2 y 2 x2x y 2 x2 2z 2 2 2 2 xy x y x y 22
z x sec y
z ln
67.
z sec y x
x
2
x ln x lnx 2 y 2 y2
z y2 x2 1 2x 2 x x x y 2 xx 2 y 2
2z 0 x2
2z x 4 4x 2y 2 y 4 2 x x 2x 2 y 2 2
2z sec y tan y yx
2z 4xy yx x2 y 2 2
z x sec y tan y y
z 2y 2 y x y2
2z x sec ysec2 y tan2 y y2
2z 2 y 2 x 2 2 y2 x y 22
2z sec y tan y xy Therefore,
y x
2z 4xy xy x 2 y 2 2
2z 2z . yx xy
There are no points for which zx zy 0.
There are no points for which zx 0 zy, because z sec y 0. x 69.
f x, y, z xyz
71.
f x, y, z ex sin yz
fxx, y, z yz
fxx, y, z ex sin yz
fy x, y, z xz
fy x, y, z zex cos yz
fyyx, y, z 0
fyyx, y, z z2ex sin yz
fxyx, y, z z
fxyx, y, z zex cos yz
fyxx, y, z z
fyxx, y, z zex cos yz
73.
z 5xy z 5y x 2z 0 x 2 z 5x y
fyyxx, y, z 0
fyyxx, y, z z2ex sin yz
fxyyx, y, z 0
fxyyx, y, z z2ex sin yz
2z 0 y 2
fyxyx, y, z 0
fyxyx, y, z z2ex sin yz
Therefore,
Therefore, fxyy fyxy fyyx 0.
Therefore, fxyy fyxy fyyx.
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2z 2z 0 0 0. x 2 y 2
Section 12.3 z e x sin y
75.
z sinx ct
77.
z e x sin y x
z c cosx ct t
2z e x sin y x 2
2z c 2 sinx ct t 2
z e x cos y y
z cosx ct x
2z e x sin y y 2
2z sinx ct x 2
Therefore,
2z 2z 2 e x sin y e x sin y 0. 2 x y
z et cos
79.
Therefore,
x c
Partial Derivatives
2z 2z c 2 2. t 2 x
81. See the definition on page 859.
x z et cos t c x 1 z et sin x c c 1 x 2z 2 et cos x 2 c c Therefore,
83.
2z z c 2 2. t x
z
z
(x0, y0, z 0 )
85. The plane z x y f x, y satisfies
(x0, y0, z 0 )
f f > 0 and > 0. x y z 6
y
x
y
x
y
Plane: x = x0
Plane: y = y0
8 x
f denotes the slope of the surface in the x-direction. x
−6
f denotes the slope of the surface in the y-direction. y 87. (a) C 32xy 175x 205y 1050
yx 175 1 C 16 175 183 x 4 x C 16 205 y y C 16 x
(b) The fireplace-insert stove results in the cost increasing at a faster rate because C C > . y x
80, 20
C y
80, 20
164 205 237
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87
88
Chapter 12
Functions of Several Variables 91.
89. An increase in either price will cause a decrease in demand.
T 500 0.6x 2 1.5y 2 T T 1.2x, 2, 3 2.4m x x T T 3y 2, 3 9m y y
PV mRT
93.
T P
P
95. (a)
T
V PV T ⇒ mR P mR
P
mRT P mRT 2 ⇒ V V V
V
mRT V mR ⇒ P T P
V
V T mR V
97. f x, y
mRT V2
z 1.83 x z 1.09 x
(b) As the consumption of skim milk x increases, the consumption of whole milk z decreases. Similarly, as the consumption of reduced-fat milk y increases, the consumption of whole milk z decreases.
mRP
mRT mRT 1 VP mRT
xyx 2 y 2 , x2 y2 0,
x, y 0, 0 x, y 0, 0
(a) fxx, y
x 2 y 23x 2y y 3 x3 y xy 32x y x 4 4x 2y 2 y 4 x 2 y 22 x 2 y 22
fyx, y
x 2 y 2x3 3xy 2 x3y xy 32y x x 4 4x 2 y 2 y 4 x 2 y 22 x 2 y 22
(b) fx0, 0 lim
x→0
f x, 0 f 0, 0 0x 2 0 lim 0 x→0 x x
0y 2 0 f 0, y f 0, 0 lim 0 y→0 y→0 y y
fy0, 0 lim
(c) fxy0, 0
f y x
0, 0
fyx0, 0
f x y
0, 0
lim
y→0
lim
fx0, y fx0, 0 y y4 lim lim 1 1 y→0 y22y y→0 y fyx, 0 fy0, 0 x
x→0
lim
x→0
x x4 lim 1 1 x22x x→0
(d) fyx or fxy or both are not continuous at 0, 0. 99. True
Section 12.4 1.
101. True
Differentials
z 3x 2y 3
3. z
dz 6xy 3 dx 9x 2 y 2 dy dz
1 x2 y2 2x 2y dx 2 dy x 2 y 2 2 x y 2 2 2 x dx y dy x 2 y 22
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Section 12.4
Differentials
89
5. z x cos y y cos x dz cos y y sin x dx x sin y cos x dy cos y y sin x dx x sin y cos x dy 7. z e x sin y
w 2z 3 y sin x
9.
dz e x sin y dx e x cos y dy
dw 2z 3 y cos x dx 2z3 sin x dy 6z 2 y sin x dz
11. (a) f 1, 2 4
13. (a) f 1, 2 sin 2
f 1.05, 2.1 3.4875
f 1.05, 2.1 1.05 sin 2.1
z f 1.05, 2.1 f 1, 2 0.5125
z f 1.05, 2.1 f 1, 2 0.00293
(b) dz 2x dx 2y dy
(b) dz sin y dx x cos y dy
20.05 40.1 0.5
sin 20.05 cos 20.1 0.00385
15. (a) f 1, 2 5 f 1.05, 2.1 5.25 z 0.25 (b) dz 3 dx 4 dy 30.05 40.1 0.25
17. Let z x 2 y 2, x 5, y 3, dx 0.05, dy 0.1. Then: dz 5.052 3.12 52 32
5 52 32
0.05
x x 2 y 2
3 52 32
0.1
19. Let z 1 x 2y 2, x 3, y 6, dx 0.05, dy 0.05. Then: dz
dx 0.55
34
y x 2 y 2
dy
0.094
21 x 2 2x dx dy y2 y3
1 3.052 1 32 23 21 32 2 0.05 0.05 0.012 2 2 5.95 6 6 63 23. The tangent plane to the surface z f x, y at the point P is a linear approximation of z.
21. See the definition on page 869.
25.
A lh dA l dh h dl ∆h
∆A
dA
h
dA
dA l
∆l
r 2h 3
r
h
r3
0.1
0.1
4.7124
4.8391
0.1267
h6
0.1
0.1
2.8274
2.8264
0.0010
0.0565
0.0566
0.0001
0.0019
0.0019
0.0000
27. V
dV
r2
r 2 rh dr dh 2h dr r dh 3 3 3
0.001
0.002
0.0001
0.0002
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dV
V
V dV
90
Chapter 12
Functions of Several Variables
29. (a) dz 1.83 dx 1.09 dy z z dx dy x y
(b) dz
1.83± 0.25 1.09± 0.25 ± 0.73 Maximum propagated error: ± 0.73 Relative error:
31.
± 0.73 ± 0.73 dz ± 0.1166 11.67% z 1.837.2 1.098.5 28.7 6.259
V r 2h dV 2 rh dr r 2 dh dr dh dV 2 V r h 20.04 0.02 0.10 10%
1 33. A 2 ab sin C
dA 12 b sin C da a sin C db ab cos C dC 1 12 4sin 45 ± 16 3sin 45 ± 161 12cos 45± 0.02 ± 0.24 in.2
35. (a) V
1 bhl 2
18 sin
b 2
2
18 cos 21612
18
31,104 sin in.3
h
18
θ 2
18 sin ft3 V is maximum when sin 1 or 2. (b)
V
s2 sin l 2
dV ssin l ds
18 sin
s2 s2 l cos d sin dl 2 2
1 182
1612 1612 cos 2 2 2 2
90 182 sin 2 12 2
1809 in3 1.047 ft3
37.
P dP
E2 R E2 2E dE 2 dR R R
dE dR dP 2 20.02 0.03 0.07 7% P E R
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Section 12.4
39. L 0.00021 ln
h
dL 0.00021
dh
2h 0.75 r
dr ± 1100 ± 116 0.00021 ± 6.6 106 r 100 2
L 0.00021ln 100 0.75 8.096 104 ± dL 8.096 104 ± 6.6 106 micro–henrys 41.
z f x, y x2 2x y z f x x, y y f x, y x2 2xx x2 2x 2x y y x2 2x y 2xx x2 2x y 2x 2 x y xx 0y fxx, y x fyx, y y 1x 2y where 1 x and 2 0. As x, y → 0, 0, 1 → 0 and 2 → 0.
43.
z f x, y x2y z f x x, y y f x, y x2 2xx x2 y y x2y 2xyx yx2 x2y 2xxy x2 y 2xyx x2y yx x 2xx x2 y fxx, y x fyx, y y 1x 2y where 1 yx and 2 2xx x2. As x, y → 0, 0, 1 → 0 and 2 → 0.
45. f x, y
3x2y , x, y 0, 0 y2 x, y 0, 0 0,
x4
0 0 f x, 0 f 0, 0 x4 (a) fx0, 0 lim lim 0 x→0 x→0 x x 0 0 f 0, y f 0, 0 y2 fy0, 0 lim lim 0 y→0 y→0 y y Thus, the partial derivatives exist at 0, 0. (b) Along the line y x:
lim
x, y → 0, 0
Along the curve y x2:
lim
f x, y lim
x, y → 0, 0
x →0 x 4
f x, y
3x3 3x lim 0 x2 x →0 x2 1
3x 4 3 2x 4 2
f is not continuous at 0, 0. Therefore, f is not differentiable at 0, 0. (See Theroem 12.5) 47. Essay. For example, we can use the equation F ma: dF
F F dm da a dm m da. m a
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Differentials
91
92
Chapter 12
Functions of Several Variables
Section 12.5 1.
Chain Rules for Functions of Several Variables
w x2 y2
w x sec y
3.
x et
x et
y et
yt dw sec yet x sec y tan y1 dt
dw 2xet 2yet 2e2t e2t dt
et sec t1 tan t et sec t sec t tan t 5. w xy, x 2 sin t, y cos t (a)
dw 2y cos t xsin t 2y cos t x sin t dt 2cos2 t sin2 t 2 cos 2t
(b) w 2 sin t cos t sin 2t,
dw 2 cos 2t dt
7. w x2 y2 z2 x et cos t y et sin t z et (a)
dw 2xet sin t et cos t 2yet cos t et sin t 2zet 4e2t dt
(b) w 2e2t,
dw 4e2t dt
9. w xy xz yz, x t 1, y t 2 1, z t (a)
dw y z x z2t x y dt t 2 1 t t 1 12t t 1 t 2 1 32t 2 1
(b)
w t 1t 2 1 t 1t t 2 1t dw 2tt 1 t 2 1 2t 1 3t 2 1 32t 2 1 dt
11. Distance f t x1 x22 y1 y2 2 10 cos 2t 7 cos t2 6 sin 2t 4 sin t2 1 ft 10 cos 2t 7 cos t2 6 sin 2t 4 sin t212 2
210 cos 2t 7 cos t20 sin 2t 7 sin t 26 sin 2t 4 sin t12 cos 2t 4 cos t f
2 21 10
2
42122107 2412
22 1129 1 1161244 2.04 2 229 20
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Section 12.5
Chain Rules for Functions of Several Variables
13. w arctan2xy, x cos t, y sin t, t 0 dw w dx w dy dt x dt y dt
2y 2x sin t cos t 1 4x2y2 1 4x2y2
2 cos t 2 sin t sin t cos t 1 4 cos2 t sin2 t 1 4 cos2 t sin2 t
2 cos2 t 2 sin2 t 1 4 cos2 t sin2 t
d 2w 1 4 cos2 t sin2 t8 cos t sin t 2 cos2 t 2 sin2 t8 cos3 t sin t 8 sin3 t cos t dt 2 1 4 cos2 t sin2 t2 At t 0,
15.
8 cos t sin t1 2 sin4 t 2 cos 4 t 1 4 cos2 t sin2 t2
d 2w 0. dt 2
w x2 y2
17. w x2 y2
xst
x s cos t
yst
y s sin t
w 2x 2y 2x y 4s s
w 2x cos t 2y sin t s
w 2x 2y1 2x y 4t t
2s cos2 t 2s sin2 t 2s cos 2t w 2xs sin t 2ys cos t 2s2 sin 2t t
When s 2 and t 1, w w 8 and 4. s t
When s 3 and t
19. w x2 2xy y2, x r , y r (a)
w 2x 2y1 2x 2y1 0 r w 2x 2y1 2x 2y1 4x 4y 4x y 4r r 8
(b)
w r 2 2r r r 2 r 2 2r 2 2r 2 2 r 2 2r 2 4 2 w 0 r w 8
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w w , 0 and 18. 4 s t
93
94
Chapter 12
Functions of Several Variables
y 21. w arctan , x r cos , y r sin x (a)
y x r sin cos r cos sin w 2 cos 2 sin 0 r x y2 x y2 r2 r2 w y x r sin r sin r cos r cos 2 r sin 2 r cos 1 x y2 x y2 r2 r2
(b)
w arctan
r sin arctantan r cos
w 0 r w 1 25. w ze xy, x s t, y s t, z st
23. w xyz, x s t, y s t, z st2
w zx z e xy1 2 e xy1 e xyt s y y
w yz1 xz1 xyt2 s s tst2 s tst2 s ts tt2 2s2t2 s2t2 t4 3s2t2 t4 t23s2 t2
s st t ss ttst t
estst
2
sts t sst tst 2
estst
w yz1 xz1 xy2st t s tst2 s tst2 s ts t2st 2st3 2s3t 2st3 2s3t 4st3 2sts2 2t2
estst
estst estst
31. Fx, y, z x2 y2 z2 25
st sts t s st s t2
sts t sts t ss t2 s t2
ss2 t2 s t2
29. ln x2 y2 xy 4
Fxx, y 2x 3y 2 dy dx Fyx, y 3x 2y 1 3y 2x 2 2y 3x 1
w zx z e xy1 2 e xy1 e xys t y y
ts t2
ts2 4st t2 s t2
estst
27. x2 3xy y2 2x y 5 0
2
2
1 lnx2 y2 xy 4 0 2 x y Fxx, y dy x2 y2 x x2y y3 dx Fyx, y y y xy2 x3 x 2 2 x y 33. Fx, y, z tanx y tan y z 1
Fx 2x
Fx sec2x y
Fy 2y
Fy sec2x y sec2 y z
Fz 2z
Fz sec2 y z
Fx x z x Fz z
Fx z sec2x y 2 x Fz sec y z
Fy z y y Fz z
Fy sec2x y sec2 y z z y Fz sec2 y z
sec x y 1
sec y z
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2
2
Section 12.5
37. exz xy 0
35. x 2 2yz z2 1 0 (i) 2x 2y (ii) 2y
Chain Rules for Functions of Several Variables
Fxx, y, z z zexz y x Fzx, y, z xexz
z z x z 2z 0 implies . x x x yz
Fyx, y, z x 1 z xz xz exz y Fzx, y, z xe e
z z z z 2z 2z 0 implies . y y y yz
41. Fx, y, z, w cos xy sin yz wz 20
39. Fx, y, z, w xyz xzw yzw w2 5 Fx yz zw
w Fx y sin xy x Fw z
Fy xz zw
w Fy x sin xy z cos yz y Fw z
Fz xy xw yw Fw xz yz 2w
w Fz y cos zy w z Fw z
Fx zy w w x Fw xz yz 2w Fy zx w w y Fw xz yz 2w Fz xy xw yw w z Fw xz yz 2w
43. f x, y f tx, ty
xy x2 y2
txty
t
tx2 ty2
xy x2 y2
tf x, y
Degree: 1 x fxx, y y fyx, y x
45.
x
y3 x3 y 2 y232 x y232
2
xy x2 y2
1 f x, y
f x, y exy
47.
f tx, ty etxty exy f x, y
dw w dx w dy (Page 876) dt x dt y dt
Degree: 0 x fxx, y y fyx, y x
1y e y yx e 0 xy
xy
2
49. w f x, y is the explicit form of a function of two variables, as in z x2 y2. The implicit form is of the form F x, y, z 0, as in z x2 y2 0.
51.
1 A bh x sin 2 2
x cos 2 x2 sin 2
b 2
dx x2 d dA x sin cos dt dt 2 dt
6 sin 4
1 62 cos 2 2 4
32 2 2 m hr 90 2 10
x
h
θ 2
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x
95
96
Chapter 12
Functions of Several Variables
1 V r 2h 3
53. (a)
dr 1 dV 1 dh 212366 1224 1536 in.3min 2rh r 2 dt 3 dt dt 3 S rr 2 h2 r 2 (Surface area includes base.)
(b)
144 3612 12 36 212 6 4 12 36 12 36 12 36 12 10 6 144 4 10
10
dS dt
r 2 h2
2
55.
I
2
r2 rh dh dr 2r dt r 2 h2 r 2 h2 dt 2
2
2
2
36 648 144 in.2min 20 910 in.2min 5 10
1 mr12 r22 2
dr1 dr2 dI 1 m62 82 28m cm2sec m 2r1 2r2 dt 2 dt dt
tan
2 x
tan
4 x
57. (a)
8 6 4
tan tan 4 1 tan tan x
θ φ
x
tan 2x 4 1 2xtan x x tan 2 4
8 tan x
x2 tan 2x 8 tan 0 (b) Fx, x2 8tan 2x 0 Fx d 2x tan 2 2 cos2 2x sin cos 2 2 dx F sec x 8 x2 8 (c)
d 1 0 ⇒ 2 cos2 2x sin cos ⇒ cos x sin ⇒ tan dx x Thus, x2
59.
1x 2x 8 1x 0 ⇒ 8x x ⇒ x 22 ft.
w f x, y xuv yvu w dy w w w w dx u x du y du x y w w w w dx w dy v x dv y dv x y w w 0 u v
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Section 12.5 61.
Chain Rules for Functions of Several Variables
w f x, y, x r cos , y r sin w w w cos sin r x y w w w r sin r cos x y (a)
r cos
r cos
w w w r cos2 r sin cos r x y
sin
w w w r sin2 r sin cos x x
w w w sin r cos2 r sin2 r x r
w w w r cos sin x r w w w sin cos x r r
r sin cos r sin
w w w r sin cos r sin2 r x y w w w r sin cos r cos2 x y
w w w cos r sin2 r cos2 r y r
w w w r sin cos y r w cos w w sin y r r
(b)
w r
2
1 w r 2
cos w x
2
2
2
2
63. Given
2
w w w sin cos x y y
w w w sin cos x y y
cos 2
2
2
sin2
w x
w
w x
y 2
u v u v and , x r cos and y r sin . x y y x
u u u v v cos sin cos sin r x y y x v v v v v r sin r cos r cos sin x y y x
Therefore,
u 1 v . r r
v v v u u cos sin cos sin r x y y x u u u u u r sin r cos r cos sin x y y x
Therefore,
1 u v . r r
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2
2
sin2
97
98
Chapter 12
Functions of Several Variables
Section 12.6 1.
Directional Derivatives and Gradients
f x, y 3x 4xy 5y v
3.
vij
1 i 3 j 2
f x, y yi xj
f x, y 3 4yi 4x 5j
f 2, 3 3i 2j
f 1, 2 5i j
u
3 1 v i j u v 2 2
gx, y x2 y2
7.
g3, 4
v i h e x sin yi e x cos yj
y x i j 2 2 2 x y x y2
2 ei
h 1,
3 4 i j 5 5
u
3 4 v i j u v 5 5
2 h1, 2 u e
f x, y, z xy yz xz
11.
hx, y, z x arctan yz v 1, 2, 1
v 2i j k f x, y, z y zi x zj x yk f 1, 1, 1 2i 2j 2k u
v i v
Duh 1,
7 Du g3, 4 g3, 4 u 25 9.
52 2
hx, y e x sin y
v 3i 4j g
2 2 v i j v 2 2
Du f 2, 3 f 2, 3 u
1 5 3 2
Du f 1, 2 f 1, 2 u
5.
f x, y xy
6 6 6 v i j k v 3 6 6
Du f 1, 1, 1 f 1, 1, 1 u
hx, y, z arctan yz i h4, 1, 1
26 3
u
xz xy j k 1 yz2 1 yz2
i 2j 2k 4
v 2 1 1 , , v 6 6 6
Du h4, 1, 1 h4, 1, 1 u
1 2
i
8 86 24 46
15. f x, y sin2x y
13. f x, y x2 y2 u
1 2
u
j
f 2 cos2x y i cos2x y j
f 2x i 2y j Du f f u
3 1 i j 2 2
2 2
x
2 2
y 2 x y
Du f f u cos2x y
2 2 3 cos2x y
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3
2
cos2x y
Section 12.6
v 3i 3j k
v 2i 2j f 2x i 8yj
h
v 1 1 i j v 2 2
Du f
2 2
x
8 2
y 2x 4y
u
\
w 3x2y 5yz z2
27.
2
\
5
i
Du g g u
1 5
j
31.
1 5
i
2 5
j
2 8 10 25 5 5 5
hx, y x tan y hx, y tan yi x sec2 yj
f x, y ex cos yi ex sin yj
4 i 4j
f 0, 0 i
h 2,
Du f f u
37.
PQ 2i 4j, u
gx, y 2xi 2yj, g1, 2 2i 4j
w1, 1, 2 6i 13j 9k
2 5
3 x2 y2 gx, y ln
25 5
g1, 2
1 2 4 2 i j i 2j 3 5 5 15
35.
f 1, 4, 2
1 x2 y2 z2
1 21
f 1, 4, 2 1
25 15
17
f x, y, z x2 y2 z2 f x, y, z
1 2x 2y i 2 j 3 x2 y2 x y2
h2, 4
1 lnx2 y2 3
gx, y
g1, 2
719 19
z3, 4 6 sin 25i 8 sin 25j 0.7941i 1.0588j
wx, y, z 6xyi 3x2 5zj 2z 5yk
33.
zx, y 2x sinx2 y2i 2y sinx2 y2j
f 2, 1 3i 10j
PQ 2i j, u
7 19
z cosx2 y2
23.
f x, y 3i 10yj
29.
v 1 3i 3j k v 19
Du h h u
f x, y 3x 5y2 10
25.
1 i j k xyz
At 1, 0, 0, h i j k.
At P 3, 1, Du f 72.
21.
99
19. hx, y, z lnx y z
17. f x, y x2 4y2
u
Directional Derivatives and Gradients
f x, y, z xeyz f x, y, z eyz i xzeyz j xyeyz k f 2, 0, 4 i 8j f 2, 0, 4 65
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xi yj zk
i 4j 2k
100
Chapter 12
Functions of Several Variables
For Exercises 39–45, f x, y 3
39. f x, y 3
x y 1 1 and D f x, y cos sin . 3 2 3 2
x y 3 2
41. (a) D43 f 3, 2
z
(3, 2, 1)
3
6
y
13 21 12 23
2 33 12
(b) D6 f 3, 2
9 x
v 3i 4j
43. (a)
(b)
v 9 16 5
13 23 12 21
3 23 12
v i 3j v 10
3 4 u i j 5 5 1 2 1 Du f f u 5 5 5
u
1 10
i
Du f f u
3 10
j
11 1110 60 6 10
1 1 1 45. f 9 4 613
For Exercises 47 and 49, f x, y 9 x2 y2 and D f x, y 2x cos 2y sin 2x cos y sin . 47. f x, y 9 x2 y2
49.
f 1, 2 2i 4j f 1, 2 4 16 20 25
z 9
(1, 2, 4)
3
3
y
x
51. (a) In the direction of the vector 4i j. 1 1 (b) f 10 2x 3yi 10 3x 2yj 1 1 1 f 1, 2 10 4i 10 1j 25 i 10 j
(Same direction as in part (a).) 2 1 (c) f 5 i 10 j, the direction opposite that of the gradient.
53. f x, y x2 y2, 4, 3, 7 z
(a)
x y
—CONTINUED–
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Section 12.6
Directional Derivatives and Gradients
101
53. —CONTINUED— (b) Du f x, y f x, y u 2x cos 2y sin Du f 4, 3 8 cos 6 sin y 12 8 4
π
−4
2π
x
−8 −12
Generated by Mathematica
(c) Zeros: 2.21, 5.36 These are the angles for which Du f 4, 3 equals zero. (d) g Du f 4, 3 8 cos 6 sin g 8 sin 6 cos Critical numbers: 0.64, 3.79 These are the angles for which Du f 4, 3 is a maximum 0.64 and minimum 3.79. (e) f 4, 3 24i 23j 64 36 10, the maximum value of Du f 4, 3, at 0.64. (f )
f x, y x2 y2 7
y
f 4, 3 8i 6j is perpendicular to the level curve at 4, 3.
6 4 2
x
−6 −4
2
−2
4
6
−4 −6
Generated by Mathematica
57. f x, y
55. f x, y x2 y2 c 25, P 3, 4
x x2 y2
f x, y 2xi 2yj
1 c , P 1, 1 2
x2 y2 25
f x, y
f 3, 4 6i 8j
y2 x2 2xy i 2 j x2 y22 x y22
1 x x2 y2 2 x2 y2 2x 0 1 f 1, 1 j 2
59. 4x2 y 6
61. 9x2 4y2 40
y
f x, y 4x2 y
12
f x, y 8xi j
8
y
f x, y 9x2 4y2
4
f x, y 18xi 8yj
2
f 2, 10 16i j
f 2, 1 36i 8j
f 2, 10 1 16i j f 2, 10 257
f 2, 1 1 9i 2j f 2, 1 85
257
257
16i j
x
4
4
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85
85
9i 2j
x
−4
4 −2 −4
102
Chapter 12
Functions of Several Variables
x x2 y2
63. T
T
65. See the definition, page 885.
y2 x2 2xy i 2 j x2 y22 x y22
T3, 4
7 24 1 i j 7i 24j 625 625 625
67. Let f x, y be a function of two variables and u cos i sin j a unit vector.
z
69. 3
f . x
(a) If 0 , then Du f
(b) If 90 , then Du f
f . y
3 x
P
73. T x, y 400 2x2 y2,
71.
18
00
1671
B
5
y
P 10, 10
dx 4x dt
dy 2y dt
xt C1e4t
yt C2e2t
10 x0 C1
10 y0 C2
xt
yt 10e2t
1994
A 18
00
x
10e4t
y2 10
y2t 100e4t
y2 10x 75. (a)
(b) The graph of D 250 30x2 50 sin y2 would model the ocean floor.
D 400 300
1 2 x
1 2 y
(c) D1, 0.5 250 301 50 sin (e)
315.4 ft 4
D D y and 1, 0.5 25 cos 55.5 25 cos y 2 y 4
(d)
D D 60x and 1, 0.5 60 x x
(f ) D 60x i 25 cos
2yj
D1, 0.5 60i 55.5j 77. True
81. Let f x, y, z ex cos y
79. True z2 C. Then f x, y, z ex cos yi ex sin yj zk. 2
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Section 12.7
Section 12.7
Tangent Planes and Normal Lines
Tangent Planes and Normal Lines
1. Fx, y, z 3x 5y 3z 15 0
3. Fx, y, z 4x2 9y2 4z2 0 4x2 9y2 4z2 Elliptic cone
3x 5y 3z 15 Plane 5. Fx, y, z x y z 4
7.
F i j k n
Fx, y, z
F 1 i j k F 3
3
3
Fx, y, z x2 y2 z
i j k
n
F 5 3 4 i jk F 52 5 5
Fx, y, z x2y4 z
11.
Fx, y, z 2xy4 i 4x2y3j k F 1 32i 32j k F 2049
2049
2049
n
32i 32j k
1 F 6, , 7 i 33 j k 6 2
F 2 1 i 33 j k F 113 2
15.
1 113 113
113
i 63 j 2k i 63 j 2k
f x, y 25 x2 y2, 3, 1, 15 Fx, y, z 25 x2 y2 z Fxx, y, z 2x Fx3, 1, 15 6
3i 4j 5k
y x z ln x ln y z
F 1 i j k F 3
Fx, y, z sin yi x cos yj k
n
10
3i 4j 5k
F1, 4, 3 i j k
Fx, y, z x sin y z 4
2
Fx, y, z ln
13.
1 52
1 1 1 Fx, y, z i j k x yz yz
F1, 2, 16 32i 32j k n
x y i jk 2 2 x y2 y
3 4 F3, 4, 5 i j k 5 5
9.
x2
Fyx, y, z 2y Fy3, 1, 15 2
Fzx, y, z 1 Fz3, 1, 15 1
6x 3 2 y 1 z 15 0 0 6x 2y z 35 6x 2y z 35
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3
3
i j k
103
104 17.
Chapter 12
Functions of Several Variables
f x, y x2 y2, 3, 4, 5 Fx, y, z x2 y2 z Fxx, y, z Fx3, 4, 5
x
Fyx, y, z
x2 y2
3 5
Fy3, 4, 5
y
Fzx, y, z 1
x2 y2
4 5
Fz3, 4, 5 1
3 4 x 3 y 4 z 5 0 5 5 3x 3 4 y 4 5z 5 0 3x 4y 5z 0 19.
gx, y x2 y2, 5, 4, 9 Gx, y, z x2 y2 z Gxx, y, z 2x
Gyx, y, z 2y
Gzx, y, z 1
Gx5, 4, 9 10
Gy5, 4, 9 8
Gz5, 4, 9 1
10x 5 8y 4 z 9 0 10x 8y z 9 z exsin y 1,
21.
0, 2 , 2
Fx, y, z exsin y 1 z Fxx, y, z exsin y 1
Fyx, y, z ex cos y
Fx 0, , 2 2 2
Fy 0, , 2 0 2
Fzx, y, z 1
Fz 0, , 2 1 2
2x z 2 23.
hx, y ln x2 y2, 3, 4, ln 5 Hx, y, z ln x2 y2 z Hxx, y, z Hx3, 4, ln 5
x x2 y2 3 25
1 lnx2 y2 z 2 Hyx, y, z
Hy3, 4, ln 5
y x2 y2 4 25
Hzx, y, z 1 Hz3, 4, ln 5 1
3 4 x 3 y 4 z ln 5 0 25 25 3x 3 4 y 4 25z ln 5 0 3x 4y 25z 251 ln 5 25. x2 4y2 z2 36, 2, 2, 4 Fx, y, z x2 4y2 z2 36 Fxx, y, z 2x Fx2, 2, 4 4
Fyx, y, z 8y Fy2, 2, 4 16
Fzx, y, z 2z Fz2, 2, 4 8
4x 2 16 y 2 8z 4 0
x 2 4 y 2 2z 4 0 x 4y 2z 18
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Section 12.7
Tangent Planes and Normal Lines
27. xy2 3x z2 4, 2, 1, 2 Fx, y, z xy2 3x z2 4 Fxx, y, z y2 3
Fyx, y, z 2xy
Fx2, 1, 2 4
Fzx, y, z 2z
Fy2, 1, 2 4
FZ2, 1, 2 4
4x 2 4 y 1 4z 2 0 xyz1 29. x2 y2 z 9, 1, 2, 4 Fx, y, z x2 y2 z 9 Fxx, y, z 2x
Fyx, y, z 2y
Fzx, y, z 1
Fx1, 2, 4 2
Fy1, 2, 4 4
Fz1, 2, 4 1
Direction numbers: 2, 4, 1 Plane: 2x 1 4 y 2 z 4 0, 2x 4y z 14 Line:
x1 y2 z4 2 4 1
31. xy z 0, 2, 3, 6 Fx, y, z xy z Fxx, y, z y Fx2, 3, 6 3
Fyx, y, z x Fy2, 3, 6 2
Fzx, y, z 1 Fz2, 3, 6 1
Direction numbers: 3, 2, 1 Plane: 3x 2 2 y 3 z 6 0, 3x 2y z 6 Line:
x2 y3 z6 3 2 1
y 33. z arctan , x
1, 1, 4
Fx, y, z arctan Fxx, y, z
Fx 1, 1,
y z x
y x2 y2
1 4 2
Fyx, y, z
Fy 1, 1,
x x2 y2
1 4 2
Fzx, y, z 1
Fz 1, 1,
1 4
Direction numbers: 1, 1, 2
Plane: x 1 y 1 2 z Line:
0, x y 2z 4 2
x 1 y 1 z 4 1 1 2
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105
106
Chapter 12
35. z f x, y
Functions of Several Variables
4xy , 2 ≤ x ≤ z, 0 ≤ y ≤ 3 x2 1 y2 1
(a) Let Fx, y, z Fx, y, z
y2
4xy z x2 1 y2 1 4y x2 1 2x2 4x y2 1 2y2 i 2 jk 2 2 1 x 1 x 1 y2 12
y2
4y1 x2 4x1 y2 i 2 jk 2 2 1x 1 x 1 y2 12
F1, 1, 1 k. Direction numbers: 0, 0, 1. Line: x 1, y 1, z 1 t Tangent plane: 0x 1 0 y 1 1z 1 0 ⇒ z 1
(b) F 1, 2,
4 43 6 jk 0i jk 5 252 25
Line: x 1, y 2
6 4 t, z t 25 5
6 4 y 2 1 z 0 25 5
Plane: 0x 1
6y 12 25z 20 0 6y 25z 32 0 (c)
z 1
1 2
3
2 x
(d) At 1, 1, 1, the tangent plane is parallel to the xy-plane, implying that the surface is level there. At 1, 2, 45 , the function does not change in the x-direction.
z
y
−1
x
−2 2
−1
3 y
37. Fxx0, y0, z0x x0 Fyx0, y0, z0y y0 F2x0, y0, z0z z0 0 (Theorem 12.13) 39.
Fx, y, z x2 y2 5
Gx, y, z x z
Fx, y, z 2x i 2y j
Gx, y, z i k
F2, 1, 2 4i 2j
i (a) F G 4 1
j 2 0
k 0 2i 4j 2k 2i 2j k 1
Direction numbers: 1, 2, 1, (b) cos
41.
G2, 1, 2 i k
x2 y1 z2 1 2 1
10 F G 4 2 ; not orthogonal F G 202 10 5
Fx, y, z x2 z2 25 F 2x i 2z k F3, 3, 4 6i 8k
Gx, y, z y2 z2 25 G 2yj 2zk G3, 3, 4 6j 8k
—CONTINUED—
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Section 12.7
Tangent Planes and Normal Lines
41. —CONTINUED— i (a) F G 6 0
j 0 6
k 8 48i 48j 36k 124i 4j 3k 8
Direction numbers: 4, 4, 3, (b) cos
43.
x3 y3 z4 4 4 3
64 F G 16 ; not orthogonal F G 1010 25
Fx, y, z x2 y2 z2 6
Gx, y, z x y z
Fx, y, z 2x i 2yj 2zk
Gx, y, z i j k
F2, 1, 1 4i 2j 2k
G2, 1, 1 i j k
i (a) F G 4 1
j 2 1
k 2 6j 6k 6 j k 1
Direction numbers: 0, 1, 1, x 2,
45. f x, y 6 x2
(b) cos
F G 0; orthogonal F G
y1 z1 1 1
y2 , gx, y 2x y 4
(a) Fx, y, z z x2 Fx, y, z 2x i
y2 6 4
Gx, y, z z 2x y
1 yj k 2
Gx, y, z 2i j k G1, 2, 4 2i j k
F1, 2, 4 2i j k
The cross product of these gradients is parallel to the curve of intersection. F1, 2, 4 G1, 2, 4
i 2 2
j 1 1
k 1 2i 4j 1
Using direction numbers 1, 2, 0, you get x 1 t, y 2 2t, z 4. cos (b)
4 1 1 4 F G ⇒ 48.2
F G 6 6 6
z 8
(1, 2, 4)
6 8
y
x
47. Fx, y, z 3x2 2y2 z 15, 2, 2, 5
49. Fx, y, z x2 y2 z, 1, 2, 3
Fx, y, z 6xi 4yj k
Fx, y, z 2xi 2yj k
F2, 2, 5 12i 8j k
F1, 2, 3 2i 4j k
cos
F2, 2, 5 k F2, 2, 5
arccos
1 209
1
86.03
209
cos
F1, 2, 3 k F1, 2, 3
arccos
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1 21
1
77.40
21
107
108 51.
Chapter 12
Functions of Several Variables
Fx, y, z 3 x2 y2 6y z
53. Tx, y, z 400 2x2 y2 4z2, 4, 3, 10
Fx, y, z 2xi 2y 6j k 2x 0, x 0
z
2y 6 0, y 3 8
z 3 02 32 63 12
0, 3, 12 (vertex of paraboloid) 8
dx 4kx dt
dy 2ky dt
dz 8kz dt
xt C1e4kt
yt C2e2kt
zt C3e8kt
x0 C1 4
y0 C2 3
z0 C3 10
x 4e4kt
y 3e2kt
z 10e8kt
6
x 8 y
55. Fx, y, z
x2 y2 z2 2 21 2 a b c
57. Fx, y, z a2x2 b2y2 z2 Fxx, y, z 2a2x
2x Fxx, y, z 2 a Fyx, y, z
Fyx, y, z 2b2y Fzx, y, z 2z
2y b2
Plane: 2a2x0x x0 2b2y0 y y0 2z0z z0 0
2z Fzx, y, z 2 c Plane:
a2x0x b2y0y z0z a2x02 b2y02 z02 0 Hence, the plane passes through the origin.
2y 2z 2x0 x x0 20 y y0 20 z z0 0 a2 b c x0x y0 y z0z x02 y02 z02 2 2 2 2 2 1 a2 b c a b c
59. f x, y exy fxx, y exy,
fyx, y exy
fxxx, y exy,
fyyx, y exy,
fxyx, y exy
(a) P1x, y f 0, 0 fx0, 0x fy0, 0y 1 x y 1 1 (b) P2x, y f 0, 0 fx0, 0x fy0,0y 2 fxx0, 0x2 fxy0, 0xy 2 fyy0, 0y2
1 x y 12 x2 xy 12 y2 1 (c) If x 0, P20, y 1 y 2 y2. This is the second–degree Taylor polynomial for ey. 1 If y 0, P2x, 0 1 x 2 x2. This is the second–degree Taylor polynomial for ex.
(d)
x
y
f x, y
P1x, y
P2x, y
0
0
1
1
1
0
0
0.9048
0.9000
0.9050
(e)
f z
P2 P1
0.2
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
4
−2
2 x
−2 1 −2
2
−4
y
61. Given w Fx, y, z where F is differentiable at
x0, y0, z0 and Fx0, y0, z0 0, the level surface of F at x0, y0, z0 is of the form Fx, y, z C for some constant C. Let Gx, y, z Fx, y, z C 0. Then Gx0, y0, z0 Fx0, y0, z0 where Gx0, y0, z0 is normal to Fx0, y0, z0 C 0. Therefore, Fx0, y0z0 is normal to Fx0, y0, z0 C.
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Section 12.8
Section 12.8
Extrema of Functions of Two Variables
109
Extrema of Functions of Two Variables
1. gx, y x 12 y 32 ≥ 0
z
Relative minimum: 1, 3, 0
5
gx 2x 1 0 ⇒ x 1 gy 2 y 3 0 ⇒ y 3
1 3
2
1
x
(1, 3, 0)
4
y
3. f x, y x2 y2 1 ≥ 1
z 5
Relative minimum: 0, 0, 1 Check: fx
x x2 y2 1
0 ⇒ x0 −3
y fy 0 ⇒ y0 x2 y2 1 fxx
3
(0, 0, 1)
2 2
x
3
y
y2 1 x2 1 xy , f , f x2 y2 132 yy x2 y2 132 xy x2 y2 132
At the critical point 0, 0, fxx > 0 and fxx fyy fxy2 > 0. Therefore, 0, 0, 1 is a relative minimum. 5. f x, y x2 y2 2x 6y 6 x 12 y 32 4 ≥ 4
z 2 1
Relative minimum: 1, 3, 4
2 1
Check: fx 2x 2 0 ⇒ x 1
x
−1 −2 −3 −4
fy 2y 6 0 ⇒ y 3
1 7
y
(−1, 3, − 4)
fxx 2, fyy 2, fxy 0
At the critical point 1, 3, fxx > 0 and fxx fyy fxy2 > 0. Therefore, 1, 3, 4 is a relative minimum. 7. f x, y 2x2 2xy y2 2x 3
fx 4x 2y 2 0 fy 2x 2y 0
Solving simultaneously yields x 1 and y 1.
fxx 4, fyy 2, fxy 2 At the critical point 1, 1, fxx > 0 and fxx fyy fxy2 > 0. Therefore, 1, 1, 4 is a relative minimum. 9. f x, y 5x2 4xy y2 16x 10
fx 10x 4y 16 0 fy 4x 2y 0
Solving simultaneously yields x 8 and y 16.
fxx 10, fyy 2, fxy 4 At the critical point 8, 16, fxx < 0 and fxx fyy fxy2 > 0. Therefore, 8, 16, 74 is a relative maximum. 11. f x, y 2x2 3y2 4x 12y 13
13. f x, y 2x2 y2 3
fx 4x 4 4x 1 0 when x 1. fy 6y 12 6 y 2 0 when y 2.
fx
2x x2 y2
2y
0
x 0, y 0
fxx 4, fyy 6, fxy 0
fy
At the critical point 1, 2, fxx > 0 and fxx fyy fxy2 > 0. Therefore, 1, 2, 1 is a relative minimum.
Since f x, y ≥ 3 for all x, y, 0, 0, 3 is relative minimum.
x2 y2
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0
110
Chapter 12
Functions of Several Variables
15. gx, y 4 x y
0, 0 is the only critical point. Since gx, y ≤ 4 for all x, y, 0, 0, 4 is relative maximum. 4x x2 y2 1
17. z
19. z x2 4y2e1x
2 y2
Relative minimum: 0, 0, 0
Relative minimum: 1, 0, 2
Relative maxima: 0, ± 1, 4
Relative maximum: 1, 0, 2
Saddle points: ± 1, 0, 1
z
z 4 6 5 −4 y
4 x
−4
5
−4
−4 4
x
4
y
21. hx, y x2 y2 2x 4y 4 hx 2x 2 2x 1 0 when x 1. hy 2y 4 2 y 2 0 when y 2. hxx 2, hyy 2, hxy 0 At the critical point 1, 2, hxx hyy hxy2 < 0. Therefore, 1, 2, 1 is a saddle point. 23. hx, y x2 3xy y2 hx 2x 3y 0
hy 3x 2y 0
Solving simultaneously yields x 0 and y 0.
hxx 2, hyy 2, hxy 3 At the critical point 0, 0, hxx hyy hxy2 < 0. Therefore, 0, 0, 0 is a saddle point. 25. f x, y x3 3xy y3 fx 3x2 y 0
fy 3x y2 0
Solving by substitution yields two critical points 0, 0 and 1, 1.
fxx 6x, fyy 6y, fxy 3 At the critical point 0, 0, fxx fyy fxy2 < 0. Therefore, 0, 0, 0 is a saddle point. At the critical point 1, 1, fxx 6 > 0 and fxx fyy fxy2 > 0. Therefore, 1, 1, 1 is a relative minimum. 27. f x, y ex sin y
fx ex sin y 0 fy ex cos y 0
29. z
Since ex > 0 for all x and sin y and cos y are never both zero for a given value of y, there are no critical points.
x y4 ≥ 0. z 0 if x y 0. x2 y2
z 60
Relative minimum at all points x, x, x 0.
40
3 x
3
y
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Section 12.8 31. fxx fyy fxy2 94 62 0
Extrema of Functions of Two Variables
111
33. fxx fyy fxy2 96 102 < 0 f has a saddle point at x0, y0.
Insufficient information.
35. (a) The function f defined on a region R containing x0, y0 has a relative minimum at x0, y0 if f x, y ≥ f x0, y0 for all x, y in R. (b) The function f defined on a region R containing x0, y0 has a relative maximum at x0, y0 if f x, y ≤ f x0, y0 for all x, y in R. (c) A saddle point is a critical point which is not a relative extremum. (d) See definition page 906. z
37.
No extrema
z
39.
Saddle point
41. The point A will be a saddle point. The function could be
7 6
75 60
f x, y x2 y2.
45 30
2
x
x 2
6
3
y
−3
y
43. d fxx fyy fxy2 28 fxy2 16 fxy2 > 0 ⇒ fxy < 16 ⇒ 4 < fxy < 4 2
45. f x, y x3 y3 fx 3x2 0 fy 3y2 0
Solving yields x y 0
fxx 6x, fyy 6y, fxy 0 At 0, 0, fxx fyy fxy2 0 and the test fails. 0, 0, 0 is a saddle point. 47. f x, y x 12 y 42 ≥ 0 fx 2x 1 y 42 0 fy 2x 12y 4 0
Solving yields the critical points 1, a and b, 4.
fxx 2 y 42, fyy 2x 12, fxy 4x 1 y 4 At both 1, a and b, 4, fxx fyy fxy2 0 and the test fails. Absolute minima: 1, a, 0 and b, 4, 0 49. f x, y x23 y23 ≥ 0 fx fy
2 3 x 3
2 3 y 3
fxx
fx and fy are undefined at x 0, y 0. The critical point is 0, 0.
2 2 ,f ,f 0 3 3 yy 3 y xy 9x 9y
At 0, 0, fxx fyy fxy2 is undefined and the test fails. Absolute minimum: 0 at 0, 0 51. f x, y, z x2 y 32 z 12 ≥ 0 fx 2x 0
fy 2 y 3 0 fz 2z 1 0
Solving yields the critical point 0, 3, 1.
Absolute minimum: 0 at 0, 3, 1
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112
Chapter 12
Functions of Several Variables
53. f x, y 12 3x 2y has no critical points. On the line y x 1, 0 ≤ x ≤ 1,
y
f x, y f x 12 3x 2x 1 5x 10
3
and the maximum is 10, the minimum is 5. On the line y 2x 4, 1 ≤ x ≤ 2,
y=x+1 (1, 2)
2
f x, y f x 12 3x 22x 4 x 4 and the maximum is 6, the minimum is 5. On the line y 12 x 1, 0 ≤ x ≤ 2,
y = −2x + 4
(0, 1)
1
f x, y f x 12 3x 2 12 x 1 2x 10
(2, 0) 1
and the maximum is 10, the minimum is 6.
2
y=−
Absolute maximum: 10 at 0, 1
1x 2
x
3
+1
Absolute minimum: 5 at 1, 2 55. f x, y 3x2 2y2 4y
y
(−2, 4)
⇒ x0
fx 6x 0
(2, 4)
f 0, 1 2
fy 4y 4 0 ⇒ y 1
3
On the line y 4, 2 ≤ x ≤ 2,
2
f x, y f x 3x 32 16 3x 16 2
2
1
and the maximum is 28, the minimum is 16. On the curve y x2, 2 ≤ x ≤ 2, f x, y f x
3x2
2 x2 2
4x2
2x4
x2
x2
2x2
−2
1
x
−1
1
2
1
and the maximum is 28, the minimum is 8 . Absolute maximum: 28 at ± 2, 4 Absolute minimum: 2 at 0, 1
57. f x, y x2 xy, R x, y: x ≤ 2, y ≤ 1
xy0
fx 2x y 0 fy x 0
y 2
f 0, 0 0
x
−1
1 Along y 1, 2 ≤ x ≤ 2, f x2 x, f 2x 1 0 ⇒ x 2 .
1
1 1 Thus, f 2, 1 2, f 2 , 1 4 and f 2, 1 6.
−2
1 Along y 1, 2 ≤ x ≤ 2, f x2 x, f 2x 1 0 ⇒ x 2 . 1 1 Thus, f 2, 1 6, f 2 , 1 4 , f 2, 1 2.
Along x 2, 1 ≤ y ≤ 1, f 4 2y ⇒ f 2 0. Along x 2, 1 ≤ y ≤ 1, f 4 2y ⇒ f 2 0.
1 1 1 1 Thus, the maxima are f 2, 1 6 and f 2, 1 6 and the minima are f 2 , 1 4 and f 2 , 1 4 .
59. f x, y x2 2xy y2, R x, y: x2 y2 ≤ 8
y
y x
fx 2x 2y 0
4
fy 2x 2y 0 f x, x
x2
2x2
x2
2
0
On the boundary x2 y2 8, we have y2 8 x2 and y ± 8 x2. Thus,
−4
2 −2
f x2 ± 2x8 x2 8 x2 8 ± 2x8 x2 f ± 8 x2122x 28 x212 ±
x
−2
−4
16 4x2 . 8 x2
Then, f 0 implies 16 4x2 or x ± 2. f 2, 2 f 2, 2 16 and
f 2, 2 f 2, 2 0
Thus, the maxima are f 2, 2 16 and f 2, 2 16, and the minima are f x, x 0, x ≤ 2.
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4
Section 12.9
61. f x, y
Applications of Extrema of Functions of Two Variables
4xy , R x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 x2 1 y2 1
fx
41 x2y 0 ⇒ x 1 or y 0 1x2 1
fy
41 y2x ⇒ x 0 or y 1 2 x 1 y2 12
113
y
1
y2
R x 1
For x 0, y 0, also, and f 0, 0 0. For x 1, y 1, f 1, 1 1. The absolute maximum is 1 f 1, 1. The absolute minimum is 0 f 0, 0. In fact, f 0, y f x, 0 0 63. False
Let f x, y 1 x y .
0, 0, 1 is a relative maximum, but fx0, 0 and fy0, 0 do not exist.
Section 12.9
Applications of Extrema of Functions of Two Variables
1. A point on the plane is given by x, y, 12 2x 3y. The square of the distance from the origin to this point is S x2 y2 12 2x 3y2
3. A point on the paraboloid is given by x, y, x2 y2. The square of the distance from 5, 5, 0 to a point on the paraboloid is given by S x 52 y 52 x2 y22
Sx 2x 212 2x 3y2
Sx 2x 5 4xx2 y2 0
Sy 2y 212 2x 3y3 From the equations Sx 0 and Sy 0, we obtain the system 5x 6y 24
Sy 2 y 5 4yx2 y2 0. From the equations Sx 0 and Sy 0, we obtain the system 2x3 2xy2 x 5 0
3x 5y 18.
2y3 2x2y y 5 0
12 18 Solving simultaneously, we have x 7 , y 7 54 6 z 12 24 7 7 7 . Therefore, the distance from
the origin to
12 18 6 7 , 7 ,7
is
127 187 67 2
2
2
Multiply the first equation by y and the second equation by x, and subtract to obtain x y. Then, we have x 1, y 1, z 2 and the distance is
614 . 7
1 52 1 52 2 02 6.
5. Let x, y and z be the numbers. Since x y z 30, z 30 x y. P xyz 30xy x2y xy2 Px 30y 2xy y2 y30 2x y 0 2x y 30
Py 30x x2 2xy x30 x 2y 0 x 2y 30 Solving simultaneously yields x 10, y 10, and z 10. 7. Let x, y, and z be the numbers and let S x2 y2 z2. Since x y z 30, we have S x2 y2 30 x y2 Sx 2x 230 x y1 0 2x y 30
Sy 2y 230 x y1 0 x 2y 30. Solving simultaneously yields x 10, y 10, and z 10.
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114
Chapter 12
Functions of Several Variables
9. Let x, y, and z be the length, width, and height, respectively. Then the sum of the length and girth is given by x 2y 2z 108 or x 108 2y 2z. The volume is given by V xyz 108zy 2zy2 2yz2 Vy 108z 4yz 2z2 z108 4y 2z 0 Vz 108y 2y2 4yz y108 2y 4z 0. Solving the system 4y 2z 108 and 2y 4z 108, we obtain the solution x 36 inches, y 18 inches, and z 18 inches. 11. Let a b c k. Then V
13. Let x, y, and z be the length, width, and height, respectively and let V0 be the given volume.
4 abc 4 abk a b 3 3
Then V0 xyz and z V0 xy. The surface area is
4 kab a2b ab2 3
S 2xy 2yz 2xz 2 xy
V0 0 x2y V0 0 x2
V0 0 xy2 V0 0. y2
Va
4 kb 2ab b2 0 kb 2ab b2 0 3
Sx 2 y
Vb
4 ka a2 2ab 0 ka a2 2ab 0. 3
Sy 2 x
V0 V0 x y
3 V ,y 3 V , Solving simultaneously yields x 0 0 3 V . and z 0
Solving this system simultaneously yields a b and substitution yields b k 3. Therefore, the solution is a b c k 3.
15. The distance from P to Q is x2 4. The distance from Q to R is y x2 1 . The distance from R to S is 10 y. C 3kx2 4 2k y x2 1 k10 y
yx yx 1 k0 ⇒ 2 y x 1 y x 1 1 x 2k 0 3k 2 x 4 C 2k
Cx 3k y
y x x 2k 0 y x2 1 4
x2
2
2
2
x x2 4
1 3
3x x2 4 9x2 x2 4 x2 x
1 2 2
2
2 y x y x2 1 4 y x2 y x2 1
y x2 y Therefore, x
2
2
1 3 1 3
0.707 km and y
1 2
23 32 6
23 32 1.284 kms. 6
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Section 12.9
Applications of Extrema of Functions of Two Variables
115
17. Let h be the height of the trough and r the length of the slanted sides. We observe that the area of a trapezoidal cross section is given by
w 2r 2w 2r 2x w 2r xh
Ah
where x r cos and h r sin . Substituting these expressions for x and h, we have Ar, w 2r r cos r sin wr sin 2r 2 sin r 2 sin cos Now Arr, w sin 4r sin 2r sin cos sin w 4r 2r cos 0 ⇒ w r 4 2 cos Ar, wr cos 2r 2 cos r 2 cos 2 0. Substituting the expression for w from Arr, 0 into the equation Ar, 0, we have r 24 2 cos cos 2r 2 cos r 22 cos2 1 0 1 r 22 cos 1 0 or cos . 2 Therefore, the first partial derivatives are zero when 3 and r w 3. (Ignore the solution r 0.) Thus, the trapezoid of maximum area occurs when each edge of width w 3 is turned up 60 from the horizontal. 19. Rx1, x2 5x12 8x22 2x1x2 42x1 102x2 Rx1 10x1 2x2 42 0, 5x1 x2 21 Rx2 16x2 2x1 102 0, x1 8x2 51 Solving this system yields x1 3 and x2 6. Rx1x1 10 Rx1x2 2 Rx2x2 16 Rx1x1 < 0 and Rx1x1 Rx2x2 Rx1x22 > 0 Thus, revenue is maximized when x1 3 and x2 6. 21. Px1, x2 15x1 x2 C1 C2 15x1 15x2 0.02x12 4x1 500 0.05x22 4x2 275 0.02x12 0.05x22 11x1 11x2 775 Px1 0.04x1 11 0, x1 275 Px2 0.10x2 11 0, x2 110 Px1x1 0.04 Px1x2 0 Px2x2 0.10 Px1x1 < 0 and Px1x1 Px2x2 Px1x22 > 0 Therefore, profit is maximized when x1 275 and x2 110.
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116
Chapter 12
Functions of Several Variables
23. (a) Sx, y d1 d2 d3 x 02 y 02 x 22 y 22 x 42 y 22 x2 y2 x 22 y 22 x 42 y 22 From the graph we see that the surface has a minimum. (b) Sxx, y Syx, y
S 24
x2 x x4 x2 y2 x 22 y 22 x 42 y 22 y x2 y2
y2 x 22 y 22
20
y2 x 42 y 22
4
8
1 2 1 i j (c) S1, 1 Sx1, 1i Sy1, 1j 2 2 10 tan
2 10 1 2 1 1 2
2 5
1 2
t, 1
2 10
2
2
4
6
8
y
x
⇒ 186.027
(d) x2, y2 x1 Sxx1, y1t, y1 Syx1, y1t 1
S 1
6
4
1 2
t, 1
2 10
1 2
t
2 t 2 2 510 2 2 t 1 2 5 5 25 t
1
2
10 2 510 2 2 t 1 2 5 5 25 t
10 2 510 4 2 t 1 2 5 5 25 t
2
2
Using a computer algebra system, we find that the minimum occurs when t 1.344. Thus, x2, y2 0.05, 0.90. (e) x3, y3 x2 Sxx2, y2t, y2 Syx2, y2t 0.05 0.03t, 0.90 0.26t S0.05 0.03t, 0.90 0.26t 0.05 0.03t2 0.90 0.26t2 2.05 0.03t2 1.10 0.26t2 3.95 0.03t2 1.10 0.26t2 Using a computer algebra system, we find that the minimum occurs when t 1.78. Thus x3, y3 0.10, 0.44.
x4, y4 x3 Sxx3, y3t, y3 Syx3, y3t 0.10 0.09t, 0.44 0.01t S0.10 0.09t, 0.45 0.01t 0.10 0.09t2 0.45 0.01t2 2.10 0.09t2 1.55 0.01t2 3.90 0.09t2 1.55 0.01t2 Using a computer algebra system, we find that the minimum occurs when t 0.44. Thus, x4, y4 0.06, 0.44. Note: The minimum occurs at x, y 0.0555, 0.3992 (f) Sx, y points in the direction that S decreases most rapidly. You would use Sx, y for maximization problems. 25. 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.
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Section 12.9
27. (a)
y
xy
x2
4
0
4
0
0
0
0
1
3
3
1
6
4
1
1
1
1
2
0
0
4
y
xy
x2
2
0
0
0
1
2
3
x 0 y 4 xy 6 x i
29. (a)
i i
2
i
8
x 4 y 8 xy 4 x i
36 04 3 1 3 4 a , b 4 0 , 38 02 4 3 4 3
a
3 4 y x 4 3
2
4 1 3
2
3 4 3 2 3
x 13, x y 46,
2
(b) S 4 42 2 32 2 12 0 02 2
a
546 1312 74 37 551 132 86 43
b y
y 0, x 205
i
i
i
2
i i
i
a
175 570 270 350 5205 272 296 148
37 7 1 12 13 5 43 43
b
175 945 1 0 27 5 148 148
37 7 x 43 43
y
175 945 x 148 148
8
7 y = 37 x + 43 43
(0, 6)
(5, 5)
(4, 3)
(3, 4)
−4
18
(5, 0) (8, − 4)
(4, 2) (1, 1) (0, 0)
10
−6
−1
(10, − 5)
y = − 175 x + 945 148 148
37. 1.0, 32, 1.5, 41, 2.0, 48, 2.5, 53
35. (a) y 1.7236x 79.7334
x 7, y 174, x y 322, x
240
i
i
i i
a 14, b 19, y 14x 19
When x 1.6, y 41.4 bushels per acre. 0 100
6
44 48 1 2, b 8 24 4, 46 42 4
x 27, x y 70,
i
2
i i
7
2
i
33. 0, 6, 4, 3, 5, 0, 8, 4, 10, 5
y 12, x 51
i
(b)
i i
1 6
31. 0, 0, 1, 1, 3, 4, 4, 2, 5, 5
−2
i
y 2x 4
3 4 (b) S 0 2 3
117
x
x
i
Applications of Extrema of Functions of Two Variables
100
(c) For each one-year increase in age, the pressure changes by 1.7236 (slope of line).
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2
i
13.5
118
Chapter 12
Functions of Several Variables
n
y ax
39. Sa, b, c
i
bxi c2
2
i
41. 2, 0, 1, 0, 0, 1, 1, 2, 2, 5
i1
n S 2xi2 yi axi2 bxi c 0 a i1
n S 2xi yi axi2 bxi c 0 b i1
n S 2 yi axi2 bxi c 0 c i1
n
a
x
b
4
i
n
x
c
3
i
n
x
i
i1
i1
i1
n
n
n
a
x
3
i
b
i1
x
2
i
c
i1 n
x
a
2
i
x
i
yi
2
i
i i
i1
−2
i
45. 0, 0, 2, 15, 4, 30, 6, 50, 8, 65, 10, 70
x 30, y 230, x 220, x 1,800, x 15,664, x y 1,670, x y 13,500
i
i
i
i
2
2
i
i
3
3
i
i
4
4
i
i
i i
i i
i
2
i
353a 99b 29c 254
i
15,664a 1,800b 220c 13,500
99a 29b 9c 70
1,800a
29a 9b 4c 20
220b 30c 1,670
220a
a 1, b 1, c 0, y x2 x
y
25 2 112 x
30b
541 56
x
6c 25 14
14 120
(4, 12) (3, 6) (2, 2)
−5
(0, 0)
7
−1
−2
14 −20
47. (a) ln P 0.1499h 9.3018 (b) ln P 0.1499h 9.3018 P e0.1499h9.3018 10,957.7e0.1499h (c)
(0, 1)
3 2 6 26 a 37 , b 65 , c 26 35 , y 7 x 5 x 35
i
x 9 y 20 x 29 x 99 x 353 x y 70 x y 254 i
6
(−2, 0)
34a 10c 22, 10b 12, 10a 5c 8
n
x cn y
43. 0, 0, 2, 2, 3, 6, 4, 12
2
−9
i i
i1
i1
2
i
4
x x y
i
(1, 2)
i
2
n
n
(2, 5)
(−1, 0)
i
3
n
i
8
i
i
i1
i1
b
i1
2
x 0 y 8 x 10 x 0 x 34 x y 12 x y 22
14,000
−2
24
−2,000
(d) Same answers.
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230
0.22x 2 9.66x 1.79
Section 12.10
Section 12.10
Lagrange Multipliers
119
Lagrange Multipliers
1. Maximize f x, y xy.
3. Minimize f x, y x2 y 2.
y
Constraint: x y 10
10
f g
2x i 2yj i j
6 4
y
xy x
4
2
2x
constraint x 2
4
6
8
10
x y 10 ⇒ x y 5
12
constraint
f g
level curves
8
y i xj i j
y
Constraint: x y 4
12
2y
x −4
4
x y
−4
level curves
xy4 ⇒ xy2
f 5, 5 25
f 2, 2 8
5. Minimize f x, y x2 y 2.
7. Maximize f x, y 2x 2xy y.
Constraint: x 2y 6
Constraint: 2x y 100
f g
f g
2x i 2yj i 2 j
2 2y i 2x 1j 2 i j
⇒ x
2x
2
2 2y 2 ⇒ y 1 2x 1
2y 2 ⇒ y
1 y 2x 2
2x y 100 ⇒ 4x 100
3 ⇒ 6 2
x 2y 6
⇒ x
x 25, y 50
4, x 2, y 4
f 25, 50 2600
f 2, 4 12 9. Note: f x, y 6 x2 y 2 is maximum when gx, y is maximum. Maximize gx, y 6 x2 y 2.
2y
xe xy 2y
xy
xy2 f 2, 2
f 1, 1 g1, 1 2 13. Maximize or minimize f x, y x 2 3xy y 2. Constraint:
y2
xy
x2 y 2 8 ⇒ 2x2 8
xy2 ⇒ xy1
x2
Constraint: x2 y 2 8 ye xy 2x
Constraint: x y 2 2x
11. Maximize f x, y e xy.
e4
Case 2: Inside the circle
≤ 1
Case 1: On the circle
x2
fx 2x 3y 0
y2
2x 3y 2x 3x 2y 2y
2
Maxima: f ± Minima: f ±
2 2
,±
,
fxx 2, fyy 2, fxy 3, fxx fyy fxy2 ≤ 0
x2 y 2
x2 y2 1 ⇒ x ± 2
xy0
fy 3x 2y 0
1
2
2
2
2 2
2
Saddle point: f 0, 0 0
,y±
5 2
2
2
5
By combining these two cases, we have a maximum of 2 at
±
2
2
, ±
2
2
1
and a minimum of 2 at
12
±
2
2
,
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2
2
.
120
Chapter 12
Functions of Several Variables
15. Minimize f x, y, z x2 y 2 z 2.
17. Minimize f x, y, z x2 y 2 z 2.
Constraint: x y z 6
Constraint: x y z 1
2x 2y x y z 2z
2x 2y x y z 2z
xyz6 ⇒ xyz2
xyz1 ⇒ xyz
f 2, 2, 2 12
f 3 , 3 , 3 3 1 1 1
19. Maximize f x, y, z xyz.
1 3
1
21. Maximize f x, y, z xy yz.
Constraints: x y z 32
Constraints: x 2y 6
xyz0
x 3z 0
f g h
f g h
yz i xz j xyk i j k i j k
y i x z j yk i 2j i 3k
yz xz yz xy ⇒ x z xy
y 8 3 x z 2 y ⇒ xz y 4 3 y 3
x y z 32
2x 2z 32 ⇒ x z 8 xyz0 y 16
x 2y 6 ⇒ y 3 x 3z 0 ⇒ z
f 8, 16, 8 1024 x
x 2
x 3
x 8 x 3 3 3 2
3 x 3, y , z 1 2
3 f 3, , 1 6 2 23. Minimize the square of the distance f x, y x2 y 2 subject to the constraint 2x 3y 1. 2x 2 2y 3
y 3x2
2 3 The point on the line is 13 , 13 and the desired distance is
2 13
2
f x, y, z x 22 y 12 z 12 subject to the constraint x y z 1.
3 2 2x 3y 1 ⇒ x , y 13 13
d
25. Minimize the square of the distance
3 13
2
13
13
2x 2 2 y 1 y z and y x 1 2z 1
x y z 1 ⇒ x 2x 1 1 x 1, y z 0
.
The point on the plane is 1, 0, 0 and the desired distance is d 1 22 0 12 0 12 3.
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Section 12.10 27. Maximize f x, y, z z subject to the constraints x2 y 2 z2 36 and 2x y z 2. 0 2x 2 0 2y x 2y 1 2z
Lagrange Multipliers
121
29. Optimization problems that have restrictions or contstraints on the values that can be used to produce the optimal solution are called contrained optimization problems.
x2 y 2 z 2 36 2x y z 2 ⇒ z 2x y 2 5y 2
2y2 y 2 5y 22 36 30y 2 20y 32 0 15y 2 10y 16 0 y
5 ± 265 15
Choosing the positive value for y we have the point 265 5 265 1 265
10 152
,
,
15
3
.
31. Maximize Vx, y, z xyz subject to the constraint x 2y 2z 108.
33. Minimize Cx, y, z 5xy 32xz 2yz xy subject to the constraint x yz 480.
yz xz 2 y z and x 2y xy 2
8y 6z yz 8x 6z xz x y, 4y 3z 6x 6y xy
x 2y 2z 108 ⇒ 6y 108, y 18
xyz 480 ⇒ 43 y 3 480
x 36, y z 18 Volume is maximum when the dimensions are 36 18 18 inches
z
3 360, z 4 3 360 x y 3 3 360 3 360 4 3 360 Dimensions: feet 3
z x
y
y
35. Maximize Vx, y, z 2x2y2z 8xyz subject to the constraint 8yz
2x a2
8xz
2y b2
8xy
2z c2
x
x2 y2 z2 2 2 2 1. a b c
x2 y2 z2 2 2 2 a b c
x2 y2 z2 3x2 3y 2 3z 2 1 ⇒ 2 1, 2 1, 2 1 a2 b 2 c 2 a b c x
a 3
,y
b 3
,z
c 3
Therefore, the dimensions of the box are
23a 23b 23c . 3 3 3
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122
Chapter 12
Functions of Several Variables
37. Using the formula Time
d12 x2 d22 y 2 Distance , minimize Tx, y subject to the constraint x y a. Rate v1 v2
x v1d22 x 2 x y y v1d12 x2 v2d22 y 2 v2d22 y 2
Medium 1
P
d1
xya
θ1
Since sin 1 xd1 v1 2
x d12 x2
x2
and sin 2
yd2 v2 2
y2
y d22 y
, we have 2
θ2
39. Maximize P p, q, r 2pq 2pr 2qr.
subject to the constraint 48x 36y 100,000.
P g
⇒ 3 4 p q r 41 ⇒
4 3
25x0.75y0.75 48 ⇒
yx
0.75
75x0.25y0.25 36 ⇒
xy
0.25
pqr1
0.75
⇒ p
48 25
36 75
yx yx
pqr1 q r 23
Q
41. Maximize Px, y 100x 0.25y 0.75
Constraint: p q r 1 2q 2r 2p 2r 2p 2q
d2
a Medium 2
sin 1 sin 2 . v1 v2
or
y
x
1 3,
q 13 , r 13
0.25
48253675
y 4 x
P 13 , 13 , 13 2 13 13 2 13 13 2 13 13 23 .
y 4x 48x 36y 100,000 ⇒ 192x 100,000 x Therefore, P
3125 6250 6 , 3
43. Minimize Cx, y 48x 36y subject to the constraint 100x0.25y0.75 20,000. 48 25x0.75y0.75 ⇒
yx
0.75
36 75x0.25y0.25 ⇒
xy
0.25
48 25
36 75
yx yx 0.75
0.25
25487536
y 4 ⇒ y 4x x 100x0.25y0.75 20,000 ⇒ x0.254x0.75 200 x
200 200 502 40.75 22
y 4x 2002 Therefore, C 502, 2002 $13,576.45.
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147,314.
3125 6250 ,y 6 3
Review Exercises for Chapter 12
123
45. (a) Maximize g, , cos cos cos subject to the constraint . sin cos cos cos sin cos tan tan tan ⇒ cos cos sin ⇒ 3
g
3 , 3 . 3 81
γ 3
(b) ⇒
2
g cos cos cos α
cos cos cos cos sin sin
3
3
β
cos cos cos
Review Exercises for Chapter 12 1. No, it is not the graph of a function. 3. f x, y ex
5. f x, y x2 y 2
2 y 2
The level curves are of the form c
The level curves are of the form c x2 y 2
2 2 ex y
ln c x2 y 2.
1
The level curves are circles centered at the origin.
x2 y 2 . c c
The level curves are hyperbolas.
y
y 2
c = 10 4
c = −12 c = − 2 c=2
c=1 c = 12
x
−2
1
2
x
−4
1
−1
4
−2
Generated by Mathematica
−4
Generated by Mathematica
7. f x, y ex
2
9. f x, y, z x2 y z2 1
y 2
z
y x2 z2 1
z 3
3
Elliptic paraboloid
−3 −3
2 x 3
3 x
11.
y
y
−3
xy 1 x, y → 1, 1 x y 2 2 lim
3
2
Continuous except at 0, 0.
13.
lim
x, y → 0, 0
4x2y x4 y2
For y x 2, For y 0,
4x 4 4x 2y 4 2, for x 0 4 2 x y x x4 4x 2y 0, for x 0 x4 y 2
Thus, the limit does not exist. Continuous except at 0, 0.
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124
Chapter 12
Functions of Several Variables
15. f x, y ex cos y
z ey yex x
fx ex cos y fy e sin y x
19. gx, y
z xey yex
17.
z xey ex y
xy x2 y 2
21. f x, y, z z arctan
y x
gx
yx2 y 2 xy2x y y2 x2 2 x2 y 22 x y 22
fx
z y yz 2 1 y2 x2 x2 x y2
gy
xx2 y 2 x2 y 22
fy
z xz 1 2 1 y 2 x2 x x y2
fz arctan
23. ux, t cen t sinnx 2
y x
z
25.
u 2 cnen t cosnx x
3
u 2 cn2en t sinnx t
−1 3
y
3 x
27. f x, y 3x2 xy 2y 3
29. hx, y x sin y y cos x
fx 6x y
hx sin y y sin x
fy x
hy x cos y cos x
6y 2
z x2 y2
31.
z 2x x 2z 2 x2
fxx 6
hxx y cos x
fyy 12y
hyy x sin y
fxy 1
hxy cos y sin x
fyx 1
hyx cos y sin x
z 2y y 2z 2 y 2 Therefore,
33.
z
y x2 y 2
35. z x sin
z 2xy x x2 y 22
dz
y x
y y y z z y dx dy sin cos dx cos dy x y x x x x
2z 4x2 1 3x2 y 2 2y 2 2y 2 x2 x y 23 x2 y 22 x y 23
z x2 y2 x2 y 2 2y 2 y x2 y 22 x y 22
x2 y 222y 2x2 y 2x2 y 22y 2z 2 y x2 y 24 2y Therefore,
2z 2z 2 0. 2 x y
3x2 y 2 x2 y 23
2z 2z 2 0. 2 x y
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Review Exercises for Chapter 12 37.
z2 x2 y 2
39.
V 13 r 2h dV 23 rh dr 13 r 2 dh 23 25± 18 13 22± 18
2z dx 2x dx 2y dy dz
y 5 1 12 1 17 x dx dy 0.654 cm z z 13 2 13 2 26
Percentage error:
dz 17 26 0.0503 5% z 13
41. w lnx2 y2, x 2t 3, y 4 t Chain Rule:
dw w dx w dy dt x dt y dt
2x 2y 2 2 1 x2 y2 x y2
22t 32 24 t 2t 32 4 t2 2t 32 4 t2
10t 4 5t2 4t 25
Substitution: w lnx2 y2 ln2t 32 4 t2 10t 4 dw 22t 32 24 t 2 dt 2t 32 4 t2 5t 4t 25 43. u x2 y 2 z2, x r cos t, y r sin t, z t Chain Rule:
u u x u y u z r x r y r z r 2x cos t 2y sin t 2z0 2r cos2 t r sin2 t 2r u u x u y u z t x t y t z t 2xr sin t 2yr cos t 2z 2r 2 sin t cos t r 2 sin t cos t 2t 2t
Substitution: ur, t r 2 cos2 t r 2 sin2 t t 2 r 2 t 2 u 2r r u 2t t x2y 2yz xz z2 0
45. 2xy 2y
z z z x z 2z 0 x x x 2xy z 2xy z z x 2y x 2z x 2y 2z
x2 2y
125
z z z 2z x 2z 0 y y y z x2 2z x2 2z y 2y x 2z x 2y 2z
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5 1 ± 6 ± 6 ± in.3
126
Chapter 12
47.
f x, y x2y
Functions of Several Variables
w z i 2yj xk
f 2xyi x 2j
w1, 2, 2 2i 4 j k
f 2, 1 4 i 4j u
1
2
v
2
2
i
2
2
2 1 2 1 u v i j k 3 3 3 3
j
Duw1, 2, 2 w1, 2, 2 u
Du f 2, 1 f 2, 1 u 2 2 2 2 0 z
51.
y x2 y 2
4 4 2 2 3 3 3 3
z ex cos y
53.
z ex cos yi ex sin y j
2xy x2 y 2 z 2 i 2 j x y 22 x y 22
4 22 i
z 0,
1 1 z1, 1 i , 0 2 2 z1, 1
w y 2 xz
49.
2
2
j
2
2
,
z0, 4 1
1 2
55. 9x2 4y2 65
57.
f x, y 9x2 4y2
Fx, y, z x2y z 0 F 2xy i x 2j k
f x, y 18xi 8yj
F2, 1, 4 4i 4j k
f 3, 2 54i 16j
Therefore, the equation of the tangent plane is
Unit normal:
4x 2 4 y 1 z 4 0 or
54i 16j 1 27i 8j
54i 16j 793
4x 4y z 8, and the equation of the normal line is x2 y1 z4 . 4 4 1
59.
Fx, y, z x2 y2 4x 6y z 9 0 F 2x 4i 2y 6j k
61.
Fx, y, z x2 y 2 z 0 Gx, y, z 3 z 0
F2, 3, 4 k
F 2x i 2yj k
Therefore, the equation of the tangent plane is
G k
z 4 0 or
F2, 1, 3 4i 2j k
z 4,
and the equation of the normal line is
i F G 4 0
x 2, y 3, z 4 t.
j 2 0
k 1 2i 2j 1
Therefore, the equation of the tangent line is x2 y1 , z 3. 1 2 63.
f x, y, z x2 y2 z2 14 f x, y, z 2xi 2yj 2zk f 2, 1, 3 4i 2j 6k Normal vector to plane. cos
n k n
6
56
3 14 14
36.7
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2
2
Review Exercises for Chapter 12 65. f x, y x3 3xy y 2
127
z
fx 3x2 3y 3x2 y 0
30
fy 3x 2y 0
x
fxx 6x
2 −30
y
fyy 2 fxy 3 From fx 0, we have y x2. Substituting this into fy 0, we have 3x 2x2 x2x 3 0. Thus, x 0 or 32 . At the critical point 0, 0, fxx fyy fxy2 < 0. Therefore, 0, 0, 0 is a saddle point.
3 9 3 9 27 At the critical point 2 , 4 , fxx fyy fxy2 > 0 and fxx > 0. Therefore, 2 , 4 , 16 is a relative minimum.
67. f x, y xy
1 1 x y
fx y
1 0, x2y 1 x2
fy x
1 0, xy 2 1 y2
x2y
fxx
2 x3
Thus,
xy2
z
20
3 4
or x y and substitution yields the critical point 1, 1.
4 x
−20 −24
y
(1, 1, 3)
fxy 1 fyy
2 y3
At the critical point 1, 1, fxx 2 > 0 and fxx fyy fxy2 3 > 0. Thus, 1, 1, 3 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 at x0, y0. 71. Px1, x2 R C1 C2 225 0.4x1 x2x1 x2 0.05x12 15x1 5400 0.03x22 15x2 6100 0.45x12 0.43x22 0.8x1x2 210x1 210x2 11,500 Px1 0.9x1 0.8x2 210 0 0.9x1 0.8x2 210 Px2 0.86x2 0.8x1 210 0 0.8x1 0.86x2 210 Solving this system yields x1 94 and x2 157. Px1x1 0.9 Px1x2 0.8 Px2x2 0.86 Px1x1 < 0 Px1x1 Px2x2 Px1x22 > 0 Therefore, profit is maximum when x1 94 and x2 157.
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128
Chapter 12
Functions of Several Variables 75. (a) y 2.29t 2.34
73. Maximize f x, y 4x xy 2y subject to the constraint 20x 4y 2000. 4 y 20 x 2 4
30
5x y 6
20x 4y 2000 ⇒
5x y 500
−2
5x y 6 (b)
494
10x
11 −5 20
x 49.4 y 253
−1
3
f 49.4, 253 13,201.8
−5
Yes, the data appears more linear. (c) y 8.37 ln t 1.54 (d)
25
−1
10 −5
The logarithmic model is a better fit. 77. Optimize f x, y, z xy yz xz subject to the constraint x y z 1. yz xz xyz xy
x y z 1 ⇒ x y z 13 Maximum: f 13 , 13 , 13 13 79. PQ x2 4, QR y2 1, RS z; x y z 10 C 3 x2 4 2 y2 1 2 Constraint: x y z 10 C g 3x
x2 4
i
2y
y2 1
j k i j k
3x x2 4 2y y2 1 1 9x2 x2 4 ⇒ x2
1 2
4y2 y2 1 ⇒ y2
1 3
Hence, x
2
2
,y
3
3
, z 10
2
2
3
3
8.716 m.
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Review Exercises for Chapter 12
123
45. (a) Maximize g, , cos cos cos subject to the constraint . sin cos cos cos sin cos tan tan tan ⇒ cos cos sin ⇒ 3
g
3 , 3 . 3 81
γ 3
(b) ⇒
2
g cos cos cos α
cos cos cos cos sin sin
3
3
β
cos cos cos
Review Exercises for Chapter 12 1. No, it is not the graph of a function. 3. f x, y ex
5. f x, y x2 y 2
2 y 2
The level curves are of the form c
The level curves are of the form c x2 y 2
2 2 ex y
ln c x2 y 2.
1
The level curves are circles centered at the origin.
x2 y 2 . c c
The level curves are hyperbolas.
y
y 2
c = 10 4
c = −12 c = − 2 c=2
c=1 c = 12
x
−2
1
2
x
−4
1
−1
4
−2
Generated by Mathematica
−4
Generated by Mathematica
7. f x, y ex
2
9. f x, y, z x2 y z2 1
y 2
z
y x2 z2 1
z 3
3
Elliptic paraboloid
−3 −3
2 x 3
3 x
11.
y
y
−3
xy 1 x, y → 1, 1 x y 2 2 lim
3
2
Continuous except at 0, 0.
13.
lim
x, y → 0, 0
4x2y x4 y2
For y x 2, For y 0,
4x 4 4x 2y 4 2, for x 0 4 2 x y x x4 4x 2y 0, for x 0 x4 y 2
Thus, the limit does not exist. Continuous except at 0, 0.
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124
Chapter 12
Functions of Several Variables
15. f x, y ex cos y
z ey yex x
fx ex cos y fy e sin y x
19. gx, y
z xey yex
17.
z xey ex y
xy x2 y 2
21. f x, y, z z arctan
y x
gx
yx2 y 2 xy2x y y2 x2 2 x2 y 22 x y 22
fx
z y yz 2 1 y2 x2 x2 x y2
gy
xx2 y 2 x2 y 22
fy
z xz 1 2 1 y 2 x2 x x y2
fz arctan
23. ux, t cen t sinnx 2
y x
z
25.
u 2 cnen t cosnx x
3
u 2 cn2en t sinnx t
−1 3
y
3 x
27. f x, y 3x2 xy 2y 3
29. hx, y x sin y y cos x
fx 6x y
hx sin y y sin x
fy x
hy x cos y cos x
6y 2
z x2 y2
31.
z 2x x 2z 2 x2
fxx 6
hxx y cos x
fyy 12y
hyy x sin y
fxy 1
hxy cos y sin x
fyx 1
hyx cos y sin x
z 2y y 2z 2 y 2 Therefore,
33.
z
y x2 y 2
35. z x sin
z 2xy x x2 y 22
dz
y x
y y y z z y dx dy sin cos dx cos dy x y x x x x
2z 4x2 1 3x2 y 2 2y 2 2y 2 x2 x y 23 x2 y 22 x y 23
z x2 y2 x2 y 2 2y 2 y x2 y 22 x y 22
x2 y 222y 2x2 y 2x2 y 22y 2z 2 y x2 y 24 2y Therefore,
2z 2z 2 0. 2 x y
3x2 y 2 x2 y 23
2z 2z 2 0. 2 x y
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Review Exercises for Chapter 12 37.
z2 x2 y 2
39.
V 13 r 2h dV 23 rh dr 13 r 2 dh 23 25± 18 13 22± 18
2z dx 2x dx 2y dy dz
y 5 1 12 1 17 x dx dy 0.654 cm z z 13 2 13 2 26
Percentage error:
dz 17 26 0.0503 5% z 13
41. w lnx2 y2, x 2t 3, y 4 t Chain Rule:
dw w dx w dy dt x dt y dt
2x 2y 2 2 1 x2 y2 x y2
22t 32 24 t 2t 32 4 t2 2t 32 4 t2
10t 4 5t2 4t 25
Substitution: w lnx2 y2 ln2t 32 4 t2 10t 4 dw 22t 32 24 t 2 dt 2t 32 4 t2 5t 4t 25 43. u x2 y 2 z2, x r cos t, y r sin t, z t Chain Rule:
u u x u y u z r x r y r z r 2x cos t 2y sin t 2z0 2r cos2 t r sin2 t 2r u u x u y u z t x t y t z t 2xr sin t 2yr cos t 2z 2r 2 sin t cos t r 2 sin t cos t 2t 2t
Substitution: ur, t r 2 cos2 t r 2 sin2 t t 2 r 2 t 2 u 2r r u 2t t x2y 2yz xz z2 0
45. 2xy 2y
z z z x z 2z 0 x x x 2xy z 2xy z z x 2y x 2z x 2y 2z
x2 2y
125
z z z 2z x 2z 0 y y y z x2 2z x2 2z y 2y x 2z x 2y 2z
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5 1 ± 6 ± 6 ± in.3
126
Chapter 12
47.
f x, y x2y
Functions of Several Variables
w z i 2yj xk
f 2xyi x 2j
w1, 2, 2 2i 4 j k
f 2, 1 4 i 4j u
1
2
v
2
2
i
2
2
2 1 2 1 u v i j k 3 3 3 3
j
Duw1, 2, 2 w1, 2, 2 u
Du f 2, 1 f 2, 1 u 2 2 2 2 0 z
51.
y x2 y 2
4 4 2 2 3 3 3 3
z ex cos y
53.
z ex cos yi ex sin y j
2xy x2 y 2 z 2 i 2 j x y 22 x y 22
4 22 i
z 0,
1 1 z1, 1 i , 0 2 2 z1, 1
w y 2 xz
49.
2
2
j
2
2
,
z0, 4 1
1 2
55. 9x2 4y2 65
57.
f x, y 9x2 4y2
Fx, y, z x2y z 0 F 2xy i x 2j k
f x, y 18xi 8yj
F2, 1, 4 4i 4j k
f 3, 2 54i 16j
Therefore, the equation of the tangent plane is
Unit normal:
4x 2 4 y 1 z 4 0 or
54i 16j 1 27i 8j
54i 16j 793
4x 4y z 8, and the equation of the normal line is x2 y1 z4 . 4 4 1
59.
Fx, y, z x2 y2 4x 6y z 9 0 F 2x 4i 2y 6j k
61.
Fx, y, z x2 y 2 z 0 Gx, y, z 3 z 0
F2, 3, 4 k
F 2x i 2yj k
Therefore, the equation of the tangent plane is
G k
z 4 0 or
F2, 1, 3 4i 2j k
z 4,
and the equation of the normal line is
i F G 4 0
x 2, y 3, z 4 t.
j 2 0
k 1 2i 2j 1
Therefore, the equation of the tangent line is x2 y1 , z 3. 1 2 63.
f x, y, z x2 y2 z2 14 f x, y, z 2xi 2yj 2zk f 2, 1, 3 4i 2j 6k Normal vector to plane. cos
n k n
6
56
3 14 14
36.7
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2
2
Review Exercises for Chapter 12 65. f x, y x3 3xy y 2
127
z
fx 3x2 3y 3x2 y 0
30
fy 3x 2y 0
x
fxx 6x
2 −30
y
fyy 2 fxy 3 From fx 0, we have y x2. Substituting this into fy 0, we have 3x 2x2 x2x 3 0. Thus, x 0 or 32 . At the critical point 0, 0, fxx fyy fxy2 < 0. Therefore, 0, 0, 0 is a saddle point.
3 9 3 9 27 At the critical point 2 , 4 , fxx fyy fxy2 > 0 and fxx > 0. Therefore, 2 , 4 , 16 is a relative minimum.
67. f x, y xy
1 1 x y
fx y
1 0, x2y 1 x2
fy x
1 0, xy 2 1 y2
x2y
fxx
2 x3
Thus,
xy2
z
20
3 4
or x y and substitution yields the critical point 1, 1.
4 x
−20 −24
y
(1, 1, 3)
fxy 1 fyy
2 y3
At the critical point 1, 1, fxx 2 > 0 and fxx fyy fxy2 3 > 0. Thus, 1, 1, 3 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 at x0, y0. 71. Px1, x2 R C1 C2 225 0.4x1 x2x1 x2 0.05x12 15x1 5400 0.03x22 15x2 6100 0.45x12 0.43x22 0.8x1x2 210x1 210x2 11,500 Px1 0.9x1 0.8x2 210 0 0.9x1 0.8x2 210 Px2 0.86x2 0.8x1 210 0 0.8x1 0.86x2 210 Solving this system yields x1 94 and x2 157. Px1x1 0.9 Px1x2 0.8 Px2x2 0.86 Px1x1 < 0 Px1x1 Px2x2 Px1x22 > 0 Therefore, profit is maximum when x1 94 and x2 157.
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128
Chapter 12
Functions of Several Variables 75. (a) y 2.29t 2.34
73. Maximize f x, y 4x xy 2y subject to the constraint 20x 4y 2000. 4 y 20 x 2 4
30
5x y 6
20x 4y 2000 ⇒
5x y 500
−2
5x y 6 (b)
494
10x
11 −5 20
x 49.4 y 253
−1
3
f 49.4, 253 13,201.8
−5
Yes, the data appears more linear. (c) y 8.37 ln t 1.54 (d)
25
−1
10 −5
The logarithmic model is a better fit. 77. Optimize f x, y, z xy yz xz subject to the constraint x y z 1. yz xz xyz xy
x y z 1 ⇒ x y z 13 Maximum: f 13 , 13 , 13 13 79. PQ x2 4, QR y2 1, RS z; x y z 10 C 3 x2 4 2 y2 1 2 Constraint: x y z 10 C g 3x
x2 4
i
2y
y2 1
j k i j k
3x x2 4 2y y2 1 1 9x2 x2 4 ⇒ x2
1 2
4y2 y2 1 ⇒ y2
1 3
Hence, x
2
2
,y
3
3
, z 10
2
2
3
3
8.716 m.
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Problem Solving for Chapter 12
129
Problem Solving for Chapter 12 1. (a) The three sides have lengths 5, 6, and 5. Thus, s 16 2 8 and A 8323 12 (b) Let f a, b, c area2 ss as bs c, subject to the constraint a b c constant (perimeter). Using Lagrange multipliers, ss bs c ss as c ss bs b From the first 2 equations s b s a ⇒ a b. Similarly, b c and hence a b c which is an equilateral triangle. (c) Let f a, b, c a b c, subject to Area2 ss as bs c constant. Using Lagrange multipliers, 1 ss bs c 1 ss as c 1 ss as b Hence, s a s b ⇒ a b and a b c. 1 (b) V baseheight 3
3. (a) Fx, y, z xyz 1 0 Fx yz, Fy xz, Fz xy
1 1 3 3 3 2 y0 z0 x0 z0
Tangent plane:
z
3
9 3 x0 y0 2
3 x 0 y0
3 x 0 y0 Tangent plane
y0 z 0x x0 x0 z 0y y0 x0 y0z z0 0
3 x0 z0
y0 z 0 x x0 z 0 y x0 y0 z 3x0 y0 z 0 3
3 3 x
3 y0 z0
5. We cannot use Theorem 12.9 since f is not a differentiable function of x and y. Hence, we use the definition of directional derivatives. f x t cos , y t sin f x, y t
Du f x, y lim
t →0
f 0 Du f 0, 0 lim
t
t
,0
2
t →0
2
f 0, 0
t
1 lim t→0 t
t 2 t 2
4
t2 t2 2 2
lim
t →0
1 2t2 2 lim which does not exist. t →0 t t t2
If f 0, 0 2, then
f 0 Du f 0, 0 lim
t →0
t 2
,0 t
t 2
2 lim 1 t →0
tt
2t2 2
2 0
which implies that the directional derivative exists.
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Base
y
130
Chapter 12
Functions of Several Variables
7. H k5xy 6xz 6yz z
9. (a)
1000 6000 6000 ⇒ H k 5xy . xy y x
Hx 5y
f f Caxa1y1a, C1 ax aya x y x
f f y Cax ay1a C1 ax ay1a x y
6000 0 ⇒ 5yx2 6000 x2
Ca C1 a x ay1a Cx ay1a f
By symmetry, x y ⇒ x3 y3 1200. Thus, x y
3 150 2
a 1a Ct a x at1a y1a (b) f tx, ty Ctx ty
5 3 150. and z 3
Cx ay1at t f x, y
11. (a) x 64cos 45 t 322t y 64sin 45 t 16t2 322t 16t2 y (b) tan x 50
arctan (c)
d dt
(d)
30
x y 50 arctan32322t2t16t50
1 322t 16t2 1 322t 50
0
2
2
64 82t2 25t 252 16 82t2 25t 252 2 4 64t 2562t3 1024t2 8002t 625 322t 50
4
−5
No. The rate of change of is greatest when the projectile is closest to the camera. (e)
d 0 when dt 82t2 25t 252 0 t
25 252 4 82 252 2 82
0.98 second.
No, the projectile is at its maximum height when dy dt 322 32t 0 or t 2 1.41 seconds. 13. (a) There is a minimum at 0, 0, 0, maxima at 0, ± 1, 2 e and saddle point at ± 1, 0, 1 e:
z 1
fx x2 2y2ex
2
y2
ex
y2
2
2y
2 y2
x2
ex
2 y2
2
2
2
y
2
2x3 4xy2 2x 0 ⇒ x3 2xy2 x 0
fy x2 2y2ex
ex
2
x2 2y22x 2x
ex
2
2x 2xex y
y2
4yex
x
2 y2
2y22y 4y
4y3 2x2y 4y 0 ⇒ 2y3 x2y 2y 0
y2
Solving the two equations x3 2xy2 x 0 and 2y3 x2y 2y 0, you obtain the following critical points: 0, ± 1, ± 1, 0, 0, 0. Using the second derivative test, you obtain the results above. —CONTINUED—
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Problem Solving for Chapter 12
131
13. —CONTINUED— (c) In general, for > 0 you obtain
(b) As in part (a), you obtain fx e fy
0, 0, 0 minimum
2xx2 1 2y2
2y2 x2 2y2
x2 y2
2 2 ex y
0, ± 1, e maxima
The critical numbers are 0, 0, 0, ± 1, ± 1, 0. These yield
± 1, 0, e saddle For < 0, you obtain
± 1, 0, 1 e minima 0, ± 1, 2 e maxima
± 1, 0, e minima 0, ± 1, e maxima
0, 0, 0 saddle
0, 0, 0 saddle
z 1
1 x
2
y
−1
15. (a)
(b)
6 cm
6 cm
1 cm
1 cm
(c) The height has more effect since the shaded region in (b) is larger than the shaded region in (a).
(d) A hl ⇒ dA l dh h dl If dl 0.01 and dh 0, then dA 10.01 0.01. If dh 0.01 and dl 0, then dA 60.01 0.06.
17. Essay
19. ux, t
1 f x ct f x ct
2
1 Let r x ct and s x ct. Then ur, s f r f s . 2 u u r u s 1 df 1 df c c t r t s t 2 dr 2 ds 2u 1 d 2f 1 d 2f 2 c2 d 2f d 2f c2 c t2 2 dr2 2 ds2 2 dr2 ds2
u u r 1 df u s 1 df 1 1 x r x s x 2 dr 2 ds 2u 1 d 2f 2 1 d 2f 2 1 d 2f d 2f 1 1 x2 2 dr2 2 ds2 2 dr2 ds2
Thus,
2u 2u c2 2 . 2 t x
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C H A P T E R 13 Multiple Integration Section 13.1 Iterated Integrals and Area in the Plane
. . . . . . . . . . . . . 365
Section 13.2 Double Integrals and Volume . . . . . . . . . . . . . . . . . . . 369 Section 13.3 Change of Variables: Polar Coordinates . . . . . . . . . . . . . 375 Section 13.4 Center of Mass and Moments of Inertia . . . . . . . . . . . . . 379 Section 13.5 Surface Area
. . . . . . . . . . . . . . . . . . . . . . . . . . . 385
Section 13.6 Triple Integrals and Applications . . . . . . . . . . . . . . . . . 388 Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates . . . . 393 Section 13.8 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . 397 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405
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C H A P T E R 13 Multiple Integration Section 13.1
Iterated Integrals and Area in the Plane
Solutions to Even-Numbered Exercises
2
x
2.
x
y 1 y2 dy x 2 x
x
6.
3
x
1y2
2
13 x
x 2 y 2 dx
sin3 x cos y dx
y
2
2
1y2
1 cos2 x sin x cos y dx 2
cos x 31 cos x cos y
1
2
1
1 2
x2 y2 dy dx
4
16 4x3 16 dx x 3 3 3
4
64 x3 dy dx
4 0
4x2
2
x
y3 3
1
1
dx
2
x2y
1
1
4
2
2
16.
0
4
10 2x2 2y2 dx dy
y
2
0
0
2yy2
3y2 6y
2
20.
0
2
0
0
cos
r dr d
2
0
2
2 cos
4
0
0
4
0
y
d
3
sin
4 4
2
2 2048 2 0 1283 2 9 9
20y
0
16 3 2 y 4y3 10y y3 2y3 3 3
3 y
8y2 4y3 dy 3
2
cos
8
2 cos2 d
0
r
0
0
dy
dy 3
2
3r 2 sin dr d
2y
2
3y 6y
r2
43 163 34 163 8
14 3 7y4 y4 y 2y3 dy 5y2 3 6 2
10y
2yy2
3xy
0
2 cos
4
22.
2
3y dx dy
1
2 64 x33 2 9
2x3 2y2x 3
10x
0
2
1
x2
64 x3 x2 dx
2y
0
18.
8 8 dx 2x2 3 3
2x2
dx
y 64 x3
4
1
1 cos3 y cos y 3
cos y
3
y
14.
2 1 y 2 1 2y 2 3
y 21 y 21 2
2 3 2
y
12.
y cos y
0
x2x3 x33 x5 2 x3 2 x5 x9
13 1 y
1y2
cos y
y dx yx
0
3
y 2x
3
4.
x2 x x
x3
cos y
1 x 4 x2 x x 2 1 2 x x 2
1y2
10.
x
x2 3y2 dy x2y y3
x
8.
x2
3
y4
2
20
0
2
dy
56 140 8 3 3
16
0
1 sin 2 2
2
0
2
d
0
cos3 sin d
cos4 4
4
0
1 4
2 1
4
1
3 16 365
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366
Chapter 13
3
24.
0
x2 dy dx 1 y2
0
26.
0
Multiple Integration
xyex
2
y2
3
0
3
28. A
3
1
1
3
A
3
y
3
dx dy
1
2 ye
1
dy
1
dy
0
1 y2 1 2 ye dy ey 2 4
4
1
2 dy 2y
3
0
1 4
y
4
3
2 dx 2x
1
9 2
0
0
3
3
x
1
x2 y2
2 dx 2 x3
3 3
x2
0
3
dx
1
1
3
1
3
dy dx
3
0
0
dx
0
dx dy
x 2 arctan y
3
2
1
1
1
x
1
1 x1
5
30. A
2
5
x
1
3 dy
0
1 2
1 2
3y
0
1
1
0
2
3
2
y=
34. A
4 − x2
1
2
cos2 d
x 1
2
x 2 sin ,
4
1 sin 2 2
2
dx
x3 2
2x x3 2 dx
0
2
y 2 sin ,
1 sin 2 2
dx dy
y 2
y2 3
0
1 cos 2 d
0
2
0
2 3
8
4 y2 dy
2
4
y
0
0
cos2 d 2
8
A
2
dx dy
16 2 16 32 5 5
dx 2 cos d, 4 x 2 2 cos
0
2
y
0
4y 2
0
2x
4
2 x 2 x5 2 5
1 cos 2 d
6 5 4 3 2 1
0
2
2
8 7
4
0
A
dy dx
x 3 2
0
0
2
y
2x
0
2
0
4
y
dy dx
4 x2 dx
4
5
4
1 2
0
2
x−1
x 1
2
1
y=
1 1 dy y2
4x 2
2
3
dy
2
1 y y
2
2
5
2
1 2
1 2
32. A
x
2
0
5
4
1 1 y2
1
dy
3
y
dx dy
1 2 2
1 2
2
1 dx 2 x 1 x 1
1 1 y2
1
dx dy
2
5
dx
0
2
5
0
1 x1
y
0
1 2
A
5
dy dx
2
5y 3
5 3
y2 4
y dy 2
8 0
3 16 32 16 5 5
0
dy 2 cos d, 4 y 2 2 cos
http://librosysolucionarios.net
−1
(4, 8) y = 2x
y = x 3/2 x 1 2 3 4 5 6 7
8
Section 13.1
3
36. A
x
0
0
3
x
9 x
9 ln x
9
0
3
3
dx
0
3
3
9
x dx
0
3
A
0
y
dy
y
x
2
y dy 2
2 4
0
2
9 y dy
1
0
x
2x x dx
0
x2
2 2
2
0
y
y = 2x
4
dy
3 2
9 y dy y
1
1 9y y 2 2
2
dy dx
y
3
0
dx dy
9 y
1
2
2x
0
y
3
1
4
2
A
9 y
1
x
0
dx dy
y 2
2y y4
y2 4
9 9ln 9 ln 3 2
3
9
2
2
y dy 2
2
9
1
y 2
367
1 4 3 2
dx dy
0
4
dx dy
2
9 dx x
9 1 ln 9 2 1
y
0
y
0
1 x2 2
2
38. A
dy dx
0
9
dx
9 x
3
y
0
9
dy dx
Iterated Integrals and Area in the Plane
1
y=x x
3
1 9 ln y y 2 2
1
9 1 ln 9 2 1
2
3
4
y 6
y=x 4
(3, 3)
y = 9x
2
(9, 1) x 2
4
6
8
−2
4
40.
0
2
2
y
f x, y dx dy, y ≤ x ≤ 2, 0 ≤ y ≤ 4
0
0
2
4x
f x, y dy dx, 0 ≤ y ≤ 4 x2, 0 ≤ x ≤ 2
0
y
x2
2
42.
f x, y dy dx
4
4
y = x2
0
0
y
4y
f x, y dx dy
4
0
3
3 2 2
1
1
x
−1
2
44.
2
3
x
f x, y dy dx, 0 ≤ y ≤ ex, 1 ≤ x ≤ 2
2
e
2
1
0
e
f x, y dx dy
e
2
2
46.
f x, y dx dy
cos x
2 0
ln y
1
3
4
4
e
1 0
2
−1
x 1
1
f x, y dy dx, 0 ≤ y ≤ cos x,
1
0
≤ x ≤ 2 2 y
arccos y
arccos y
f x, y dx dy
2 3 2
y
3 1 2
2 −π 4
−1
x 1
2
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π 4
x
368
Chapter 13
2
48.
4
1
Multiple Integration 4
dx dy
2
2
2
2
dy dx 2
50.
4x2
2 4x2
1
2
y
2
dy dx
2
4 x2 4 x2 dx 4
4y 2
dx dy 4
2 4y 2
4
y
3 2
1
1 x 1
2
3
x
−1
4
1 −1
4
52.
x 2
0
6
dy dx
0
6x
4
0
2
0
6y
0
4
dy dx
2
dx dy
2y
0
x dx 2
y
6
6 x dx 4 2 6
6
4
5
3y2 6 3y dy 6y 2
2
4
6
3
0
2
y= x 2
(4, 2) y = 6 − x
1 −1 −1
9
54.
3
x
0
3
0
2
y
0
9
dy dx
3
dx dy
y2 dy
3
y3 3
9
2
5 4 3 2 1
0
9
4y 2
56.
dx dy
2 0
4
0
y=
y
4x
32 dy dx 3 4x
2
x 1
2
3
−1
x = 4 − y2
−2
58. The first integral arises using vertical representative rectangles. The second integral arises using horizontal representative rectangles.
2x
x2
0
2
x sin y dy dx
y
x cos2x x cosx 2 dx
0
4
0
y
y 2
4
x sin y dx dy
0
(2, 4) 4 3
1 1 1 cos4 sin4 4 2 4
5
6
2
1 1 y sin y y 2 sin y dy 2 8
1
1 1 1 cos4 sin4 4 2 4
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x 1
2
x
1 2 3 4 5 6 7 8 9
1
2
4
x
−1 −2 −3 −4 −5
2
3
y
27 18 9
0
0
0
2 3 x dx 3x x3 2 3
x 1
Section 13.2
2
60.
0
2
2
ey dy dx 2
x
0
ey dx dy 2
0
y
xey
2
2
0
4
y2
2
0
0
1
0
x
y x sin x
2y
sin x y dx dy
y
2 0
4
dx
0
0
64.
1 1 1 1 e4 e0 1 4 0.4908 2 2 2 e
x sin x dy dx
0
4
1 2 2 yey dy ey 2
x
4
x sin x dx dy
dy
0
0
62.
x sin x dx sin x x cos x
0
sin 4 4 cos 4 1.858
0
a
sin 2 sin 3 0.408 2 3
4
66.
0
ax
x 2 y 2 dy dx
0
a4 6
68. (a) y 4 x 2 ⇔ x 4 y 2 y4
2
(b)
x2 ⇔ x 16 4y 4
2
4y 2
0
xy dx dy x2 y 2 1
y 4
3
2
2
xy dx dy x2 y 2 1
0
3
164y
4
3
0
2
xy dx dy x2 y 2 1
1
(c) Both orders of integration yield 1.11899.
2
70.
0
369
y
2
Double Integrals and Volume
x 1
2
2
16 x3 y3 dy dx 6.8520
72.
x
0
1sin
15r dr d
0
2
45 2 135 30.7541 32 8
74. 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. 76. The integrations might be easier. See Exercise 59-62. 78. False, let f x, y x.
Section 13.2
Double Integrals and Volume
For Exercises 2 and 4, xi yi 1 and the midpoints of the squares are
12, 21 , 32, 12 , 52, 12 , 72, 12 , 12, 32 , 32, 32 , 52, 32 , 72, 32 .
y 4 3 2
1 x 1
1 2. f x, y x2y 2
1
9
1 2 x y dy dx 2
25
49
3
27
75
f x , y x y 16 16 16 16 16 16 16 i
i
i
i
i1
4
0
2
0
4
0
x2y 2 4
2 0
4
dx
0
x2 dx
x3 3
4 0
147 21 16
64 21.3 3
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2
3
4
370
Chapter 13
Multiple Integration
1 x 1 y 1
4. f x, y 8
4
4
4
4
4
4
4
4
7936
f x , y x y 9 15 21 27 15 25 35 45 4725 1.680 i
i
i
i
i1
4
0
2
0
4
1
x 1 y 1
dy dx
0
4
0
2
6.
0
2
sin2 x cos 2 y dy dx
0
0
0
y
4
0
4
x2y2 dx dy
1 2y
0
4
1
0
4 0
ln 3ln 5 1.768
0
y1
0
3
2
1
1 cos 2x dx
x
1
2
3
0
2 8
1 2y 5
3
y
y
dy
4
3
1
y dy 24
y6 144
(2, 4)
2
4
x
0
1
2
3
4
1024 256 256 27 9 27
1
exy dx dy
dx
0
7 2
9 2
y
2
8 x 21 sin 2x
y
2y27
1 1 2 sin x y sin 2y 2 2
1 2 sin x dx 2 2
8
x3y2 3
0
12.
ln 3 dx ln 3 lnx 1 x1
f x, y dy dx 4 2 8 6 20
10.
dx
0
0
0
2
2
8.
1 ln y 1 x1
0
1y
1
e xy dx dy
0
0
1
0
exy
dy
y1
0
y
1y
e xy
dy
0
1
e e2y1 dy
2
y=x+1
y = −x + 1
0
1 ey e2y1 2
1 0
1 e e1 2
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−1
x 1
Section 13.2
2
14.
sin x sin y dx dy
0
4
16.
sin x sin y dy dx
0
2
0
4
xey dy dx
0
0
dx
4x
4
0
4x
4y
xey dx dy
0
For the first integral, we obtain:
2
sin x cos y
Double Integrals and Volume
xey
x dx
0
0
4x
xe 4 xdx
0
sin x dx
0 y
e 4x1 x
x2 2
5 8
e4
13.
e4
y
5π 2
4
2π 3π 2
3
π
2 1
x − 3π − π − π 2 2
π 2
π
3π 2
x 1
2
18.
0
4
y 2 dx dy y2 1 x
0
0
0
4
1 2
0
4
1 2
0
y2 1 x2
4
4 3
x
y=
dx
0
2
x dx 1 x2
1
x
x 1
14 ln1 x
4
2
0
2
3
4
1 ln17 4 y
x 2 y 2 dx dy
4y 2
3
y
y dy dx 1 x2
4y 2
2
20.
x
4
2
2
4x2
1 y3 3
x2
2 0 2
x 2y
2 2
2
4
y2
x=−
dy dx
4 − y2
3
x=
4 − y2
1
4x 2
dx
−2
0
x
−1
1
2
1 x 2 4 x 2 4 x 23 2 dx 3
x 1 x 1 x 4 x23 2 x 4 x2 4 arcsin x4 x23 2 6x 4 x2 24 arctan 4 2 2 12 2
4
22.
0
2
4
6 2y dy dx
0
0
6y y 2
0
2
2
dx
24.
0
4x dx 2x2
2
y=x 4 1
3 x
2
1
1 x 1
2
3
4
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2
2 0
0
y
8 dx 32
0
y
2
4 dy dx
0
4
x
8
2 2
4
4 0
371
372
Chapter 13
2
26.
0
Multiple Integration
2x
2 x y dy dx
0
2
y2 2
2y xy
0
2
0
2
2x
dx
28.
0
y
0
0
2
y4 4
2y 2
4 3
0
4y y3 dy
0
1 2 x2 dx 2
1 x 23 6
2
4 y 2 dx dy
2 0
4 y
y 2
2
y=2−x 1
y=x
1
x 1
x 1
30.
0
0
exy 2 dy dx
0
1
32.
x
1 x2 dy dx
0
34. V
2e
xy 2
0
dx
xy
0
dx
3
dx
0
2
125 3 0
1 x 2
1
3
4
5
r 2x 2
r
0
r 2 x 2 y 2 dy dx
y
0
y=
r
y r 2 x 2 y 2 r 2 x 2 arcsin
4
0
2
y
r 2x 2
r 2 x2
r2 − x2
r
dx
0
r
r 2 x 2 dx
r
0
1 2 r 2x x3 3
x
r 0
4 r3 3
2
0
2
4x
40. V
y
0
0
4
4
4
x2
dx
x2
y=4−
x2
16 8x2 x4 dx
1
0
5 2
x x 16x 8 3 5
0
1 dy dx 1 y2
x 1
2
3
4
y
2
dx
0
dx 2
x 2
arctan y
2
0
3
0
0
2
2
32
2
3
0
2
4 x2 dy dx
2
4
y=x
4
x2
5
36. V 8
0
5
38. V
5
x
0
2ex 2 dx 4ex 2
1 3
x dy dx
5
4
y
0
1 x3 3
0
x
0
0
5
2
2
2 0
1
x 1
64 32 256 3 5 15
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2
Section 13.2
5
42. V
0
0
0
0
1
0
x 2
1
3
5
4
5 2
4 y
ln1 x y dx dy 38.25
0
x y z 1 a b c
V
81 2
3
5
zc 1
z
x y a b
a
b1 x a
a
f x, y dA
0
R
xy y2 y a 2b
c
0
a
c
ab x 1 2 a
2
ab
c
ln 10
0
10
e
x
1 dy dx ln y
10
1
10
1
10
ln y
0
dx
2
y
a
0
dx
x 2b ab x x3b 2 1 2a 3a 6 a
ab ab ab 3 2 6
2
3 a 0
abc6
2
52.
0
2
2
y cos y dy dx
1 2x
2
0
2y
y cos y dx dy
0
2
ln y
dy
y cos y 2y dy
0
0
y
2
10
dy y
1
1 dx dy ln y
x ln y
a x
b1 x a
0
R
x xb x b2 x 1 1 a a a 2b a
b 1
c
x y dy dx a b
c 1
0
a
50.
9 y dx dy
2
2 y
0
48.
9y
373
4
dy 2
16
46. V
9
44. V
5
0
5
y
sin2 x dx dy
Double Integrals and Volume
1
2
9
2
(2, 2)
y cos y dy
0
2 cos y y sin y
y 10
1
2
y = 12 x 2
0
x
2cos 2 2 sin 2 1
8
1
2
6
y = ex
4 2
x 1
2
3
1 54. Average 8
5
4
4
0
2
0
1 xy dy dx 8
4
0
x2 2x dx 8
4 0
2
56. Average
1 1 2
1
0
1
1
e xy dy dx 2
x
1 2 e x1 e2x 2 e2 2e 1 e 12
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e x1 e 2x dx
0
1 0
1 1 2 e2 e2 e 2 2
374
Chapter 13
Multiple Integration
58. The second is integrable. The first contains
sin y2 dy
60. (a) The total snowfall in the county R. (b) The average snowfall in R.
which does not have an elementary antiderivation.
62. Average
1 150
60
45
50
192x 576y x2 5y 2 2xy 5000 dx dy 13,246.67
40
64. f x, y ≥ 0 for all x, y and
2
f x, y dA
0
2
0
1
P0 ≤ x ≤ 1, 1 ≤ y ≤ 2
0
1 xy dy dx 4
2
1
2
0 1
1 xy dy dx 4
0
x dx 1 2 3 3x dx . 8 16
66. f x, y ≥ 0 for all x, y and
f x, y dA
0
lim
0
1
P0 ≤ x ≤ 1, x ≤ y ≤ 1
exy dy dx
0
b→
e
xy
b 0
1
x
1 e2x ex1 2
ex dx lim
b→
0
1
exy dy dx
0
dx
0
1 0
exy
1 x
: Input A : Input B
b 0
1
e2x ex1 dx
0
1 1 1 e2 e1 e1 12 0.1998. 2 2 2
70.
0
Program: DOUBLE
x
1
dx
2
68. Sample Program for TI-82:
e
4
20ex 8 dy dx 3
0
(a) 129.2018 (b) 129.2756
: Input M : Input C : Input D : Input N :0 → V : B A M → G : D C N → H : For I, 1, M, 1 : For J, 1, N, 1 : A 0.5G2I 1 → x : C 0.5H2J 1 → y
: V sin x y G H → V : End : End : Disp V
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m 10, n 20
Section 13.3
4
72.
1
Change of Variables: Polar Coordinates
2
x3 y3 dx dy
74. V 50
m 6, n 4
1
Matches a.
(a) 13.956
z
(b) 13.9022 4 3
y
3
3
x
76. True
2
78.
1
1 exy dy exy x
2 1
Thus,
ex
0
e2x dx x
ex e2x x
2
2
exy dx dy
1
0
1
exy dx dy
0
2
1
2
1
Section 13.3
exy y
dy
0
2
1 dy ln y y
1
ln 2.
Change of Variables: Polar Coordinates
2. Polar coordinates
4. Rectangular coordinates
6. R r, : 0 ≤ r ≤ 4 sin , 0 ≤ ≤
8. R r, : 0 ≤ r ≤ r cos 3, 0 ≤ ≤
4
10.
0
4
r 2 sin cos dr d
0
4
0
r3 sin cos
4
3
0
64 sin2 2
3
16 3
2
d
12.
0
3
rer dr d 2
0
2
0
4
0 1
2
3
4
r 2
3 0
1 e9 1 2
0
1 1 9 4 e
π 2
π 2
21e
0 1
2
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3
d 2
0
375
376
Chapter 13
2
14.
0
1cos
Multiple Integration
2
sin r dr d
0
0
2
a
16.
0
0
x dy dx
2
4
x 2 y 2 dx dy
0
4yy 2
x 2 dx dy
2
0
0
0
a3 3
4 sin
2
52 2
0
x
0
22 d 3 3
22 3 3
2
sin4 sin6 d
24.
2
r dr d
e
0
25x 2
52 2
xy dy dx
3
0
0
a3 3
4
3
42 4 3
64 sin4 cos2 d
4
4
2
0
5
r3 sin cos dr d
0
625 sin cos d 4 4
sin 625 8
625 16
2
0
y
2
e
5
r2 2
2
2
d
5 4 3 2 1
0
1 e25 2 d
2
22 3
0
0
1 e25 2
26.
64 sin3 cos 3 sin cos sin5 cos 6 4 8
0
r2 2
0
r3 cos2 dr d
5
2
r 2 dr d
2
a3 sin 3
cos d
0
5
xy dy dx
1
0
0
22.
0
1 6
2
0
64
(x, y) = (0, 1)
0
4
π 2
d
22
0
y
2
0
4
0
1 1 cos 3 6
0
8y 2
r 2 cos dr d
20.
1cos
a
0
0
2
18.
a2 x 2
r2 2
sin 1 cos 2 d 2
0
sin
9x 2
9 x 2 y 2 dy dx
0
2
2
0
2
2
0
3
−5 −4 −3 −2 −1
1 e25 2
9 r 2 r dr d
0 3
0
x 1 2 3 4
−2 −3 −4 −5
9r r 3 dr d
2
0
9 2 1 4 r r 2 4
3 0
d
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81 4
2
0
d
81 8
2
Section 13.3
2
28. V 4
1
0
2
r 2 3 r dr d 4
0
0
4
4
0
7
r 4 3r 2 4 2
1
d
30.
0
Change of Variables: Polar Coordinates
lnx2 y 2 dA
R
2
2
7 d 4
2
2
2
2
0
2
2
r ln r dr d
1
0
4 74
ln r 2 r dr d
1
0
2
r2 1 2 ln r 4
ln 4
0
4 ln 4
32. V
2
0
4
16 r 2 r dr d
1
2
0
4
1 16 r 2 3 3
1
d
2
3 4
2 1
3 d 4
515 d 1015
0
34. x 2 y 2 z 2 a2 ⇒ z a2 x 2 y 2 a2 r 2
2
V8
a
0
2
8
a2 r 2 r dr d
0
2
8
0
36.
(8 times the volume in the first octant)
0 a
3 a2 r 2 3 20 d
1 2
2
a3 8a3 d 3 3
2
0
4a3 3
9 9 ≤ z ≤ ; 4x2 y2 9 4x2 y2 9 (a)
1 1 ≤ r ≤ 1 cos2 4 2
9 9 ≤ z ≤ 2 4r 2 36 4r 36
(b) Perimeter
r2
ddr
0.7
−1
2
1
−0.7
d.
1 1 1 r 1 cos2 cos2 2 2 2
z
dr cos sin d Perimeter 2
1
0
(c) V 2
2
0
38. A
2
0
2
40.
0
1 21cos2
1 4
4
r dr d
2
2sin
2
1 1 cos2 2 cos2 sin2 d 5.21 4
1 1
x
y
9 r dr d 0.8000 4r 2 36
6 d 12
0
r dr d
0
1 2
2
0
2 sin 2 d
1 2
2
4 4 sin sin2 d
0
1 1 1 4 4 cos sin 2 2 2 4
2
0
1 2
2
4 4 sin
0
1 9 8 4 4 2 2
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1 cos 2 d 2
d
377
378
Chapter 13
4
3 cos 2
42. 8
0
Multiple Integration
r dr d 4
4
9 cos2 2 d 18
0
0
4
1 cos 4 d 18
0
44. See Theorem 13.3.
1 sin 4 4
4
0
9 2
46. (a) Horizontal or polar representative elements (b) Polar representative element (c) Vertical or polar
48. (a) The volume of the subregion determined by the point 5, 16, 7 is base height 5 10 volumes, ending with 45 10 8 12 , you obtain V 10 57 9 9 5 158 10 11 8 2510 14 15 11 8 3512 15 18 16 459 10 14 12 5 5 150 555 1250 2135 2025 6115 24013.5 ft3 4 4
8 7 . Adding up the 20
(b) 56 24013.5 1,344,759 pounds (c) 7.48 24103.5 179,621 gallons
4
50.
4
0
5er r dr d 87.130
52. Volume base height
0
9 4 3 21
z
6 4
Answer (a)
2 y 4 2 4 x
54. True
56. (a) Let u 2x, then
(b) Let u 2x, then
58.
0
kex
2 y2
2
ex dx
e4x dx
dy dx
2
2
0
0
eu 2
1
2
eu
2
2
2
0
0
1 2
2 .
1 1 du . 2 2
ker r dr d 2
du
k 2 er 2
0
d
2
0
k k d 2 4
For f x, y to be a probability density function, k 1 4 4 .
k
2
60. (a) 4
0
f dy dx
0
2
0
(x − 2) 2 + y 2 = 4
2
1
4 x2 2
(b) 4
(c) 2
f dx dy
2
2
0
y
2 4y2
4 cos
x 1
3
−1
f r dr d
−2
0
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Section 13.4
Section 13.4
3
2. m
0
2
9x
3
xy dy dx
0
0
3
0
0
xy2 2
1 9 x23 4 3
x
0
3
0
a
0
b
kxy dy dx
0
a
0
kxy2 dy dx
0
0
3 0
3 9x2
dx
3
x 3 9 x2 9 dx 2 3
1 2
y2 2
0
6x 9 x2 9x x3 dx
1 9x2 x 4 29 x232 2 2 4
3 0
297 1 81 81 54 2 2 4 8
(b) m
0
a
ka2b3 6
Mx
ka3b2 6
My
b
kx 2y dy dx
0
a
ka2b2 4
b
a
My
dx
0
3
xy dy dx
3
Mx
9x2
243 1 0 93 12 4
3 9x2
6. (a) m
x9 x22 dx 2
3
379
Center of Mass and Moments of Inertia
4. m
Center of Mass and Moments of Inertia
0
a
0
b
kx 2 y 2 dy dx
0
kab 2 a b2 3
b
kx 2y y3 dy dx
kab2 2 2a 3b2 12
kx3 xy 2 dy dx
ka2b 2 3a 2b2 12
0 b
0
x
My 2 2 2 a m ka b 4 3
x
My ka2b123a2 2b2 a 3a2 2b2 m kab3a2 b2 4 a2 b2
y
Mx ka2b36 2 2 2 b m ka b 4 3
y
Mx kab2122a2 3b2 b 2a2 3b2 m kab3a2 b2 4 a2 b2
8. (a) m
ka3b26
a2k 2
y
a
Mx
0
ax
ky dy dx
0
ka3 6
a
y=a−x
My Mx by symmetry xy
Mx ka36 a 2 m ka 2 3
a
x
—CONTINUED—
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380
Chapter 13
Multiple Integration
8. —CONTINUED—
a
(b) m
ax
0
x 2 y 2 dy dx
0
a
x 2y
0
a
My
a
ax
dx
0
0
1 a4 ax 2 x3 a x3 dx 3 6
ax
0
x3 xy 2 dy dx
0
a
y3 3
1 1 1 ax3 x 4 a3x a2x2 ax3 x 4 dx 3 3 3
0
a
a3x 3a2x2 6ax3 4x4 dx
0
a5 15
a515
x
My 2a 4 m a 6 5
y
2a by symmetry 5
10. The x-coordinate changes by h units horizontally and k units vertically. This is not true for variable densities.
a2 x 2
a
12. (a) m
0
0
ka2 4
0
kx dy dx
Mx
a 0
ka3 3
0
2
Mx
0
2
My
0
3
2
kx dy dx
kx 4 dx
0
0
32k 5
2
0
32k 3
5
My 32k m 3
y
Mx 16k 5 5 m 32k 2
5
32k 3
8
kx 2 dy dx 30k
0 4x
kx 2 y dy dx 24k
0
4
My
1
x
ka5 5
4x
kx3 dy dx 84k
0
x
My 84k 14 m 30k 5
y
My 24k 4 m 30k 5
y 4
y = 4x
3 2
1 x 1
2
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3
4
8a
ka4 5
4x
4
Mx
1
kx2 dy dx
kr 4 sin dr d
My ka5 m 5
1
3
x
a
0
4
16. m
3
0
kx 2 y 2 y dy dx
0
xy
x
kxy dy dx 16k
ka4 8
My Mx by symmetry
4a y by symmetry 3
x
kr3 dr d
0
0
My ka33 4a x m ka24 3
a
a2 x 2
0
k a2 x 232 3
2
2
a
x a2 x 2 dx
kx 2 y 2 dy dx
0
0
0
0
14. m
a2 x 2
0
a
k
a
(b) m
a2 x 2
a
My
k dy dx
Section 13.4
Center of Mass and Moments of Inertia
L2
18.
x 0 by symmetry
ky 2 dy dx
3 0
23,328k 35
L2
Mx
0
9x 2
3
Mx
0
9x 2
3
m
20. m
ky3
3 0
139,968k dy dx 35
M 139,968k y x m 35
cos xL
cos xL
0
35 6 23,328k
ky dy dx
kL 8
cos xL
kx dy dx
L 2 2k 2 2
0
My L 2 2k x m 2 2
y
y 12
Mx kL m 8
kL
kL
8
y = 9 − x2 t 6 3
−6
1
y = cos π x L
x
−3
3
6
x L 2
22. m
k x 2 y 2 dA
y
Mx ka4 2 2 m 8
e
Mx
1
e
My
1
4
0
My ka4 2 m 8
1
4
k x 2 y 2 dA
x
24. m
kr 2 dr d
ln x
0 ln x
0 ln x
0
12
ka3
ka3 12
a
0
R
e
a
0
k x 2 y 2 y dA
R
My
4
0
R
Mx
kr3 sin d
0
a
kr 3 cos d
0
π 2
ka4 2 2 8
y=x r=a
ka4 2 8
a
3 2a 2 12
ka3
3 2 2 a 2 y
k k dy dx x 2 k k y dy dx x 6
3
2
y = ln x 1
k x dy dx k x
x
My k m 1
k2
y
Mx k m 6
k3
1
2
e 3
x
2 2
kL
0
L2
My
k dy dx
0
1
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0
L 2 2
381
382 26.
Chapter 13
Multiple Integration
y 0 by symmetry m
k dA
R
My
2
0
2
x
0
2
k 3
0 1
kr 2 cos dr d
0
3 1 cos 1 cos2 3 cos 1 sin2 1 cos 22 d 2 4
0
5k 4 My 5k m 4
2
5
dy dx
bh 2
3k 6
h hxb
0 h hxb
b
0
h hxb
0
30. m
y2 dy dx
bh3 12
Ix
x2 dy dx
b3h 12
Iy
0
b
Iy
1cos
cos 1 3 cos 3 cos2 cos3 d
2
k 3
0
Ix
0
I bh 12 y m bh2 Iy m
x
3
a2 2
y 2 dA
R
x 2 dA
I0 Ix Ix
6 h h 6 6
xy
a4 8
r3 sin2 dr d
a4 8
r3 cos2 dr d
a4 8
a
0
0
R
6 b3h12 b b bh2 6 6
x
a
0
0
a4 a4 8 4
mI a8 2a 4
x
2
32. m ab
ba a2 x 2
a
Ix 4
0
y 2 dy dx
0
a
b3 2 4b3 a x 232 dx 3 3 3a 3a
4
0
a
0
a2 a2 x 2 x 2 a2 x 2 dx
x 1 x 4b3 a2 x a2 x 2 a2 arcsin x2x 2 a2 a2 x 2 a4 arcsin 3a3 2 a 8 a
ab b2 y 2
b
Iy 4
0
x 2 dx dy
0
I0 Iy Ix
a3b 4
a3b ab3 ab 2 a b2 4 4 4
m a 4b 1ab 2a I ab 1 b y ab 2 m 4
x
r = 1 + cos θ
3k kr dr d 2
0
b
28. m
1cos
0
kx dA
R
π 2
Iy
x
3
3
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a 0
ab3 4
a 2
Section 13.4 34. ky
36. kxy
m 2k
0
1
0
a2 x 2
a
Ix k
a 0
a2 x 2
a
Iy k
a 0
I0 Ix Iy
1
2ka3 3
y3 dy dx
Ix k
1
Iy k
2
3
x
0
x
1
Iy
x 2 y 2 y 2 dy dx
x2
0
0
x 2 y 2x 2 dy dx
x2
I0 Ix Iy
0
2
158 2079
Ix 2
0
2
158 2079
Iy 2
0
316 2079
4
2
0
2
0
a2 x 2
a 0
k
a a
k
a
k
y4 2ay3 a2y 2 4 3 2
32,768k 65
4x
kx 2 y dy dx
x3
2048k 45
321,536k 585
2kx 6 dx 2
2k3 x 6 3
4 0
416k 3
a2 x2
dx
0
1 4 2a a2 a 2a2x 2 x 4 a2 a2 x 2 x 2 a2 x 2 a2 x 2 dx 4 3 2 2 3
4
x5 2a a2 x x a2 x 2 a2 arcsin 5 3 2 a
1 x x2x 2 a2 a2 x 2 a4 arcsin 8 a
14 a
2k
ky3 dy dx
x3
ky y a2 dy dx
14 a x 2a3x
4x
x
0
a
512k 21
Iy
x
4
kx 6 dy dx
a
44. I
k 48
0
21 28 2 105 512k 15 15 m 2048k 45 I 32,768k 21 8 1365 y 512k 65 m 65
Iy
ky dy dx
x3
I0 Ix Iy
x
42. I
x5 x7 dx
1
9k 3k 240 80
4x
m2
158 35 395 6 891 m 2079 I 395 y x m 891 x
k 60
2
6 35
x 2 y 2 dy dx
x
x5 x9 dx
0
40. ky
x2
1
Ix
k 2
1
38. x 2 y 2 1
k 4
5
k 24
2
3
m
x3 x5 dx
0
x
5
x
x3y dy dx
1
Iy
x
5
x
1 m k48 k24 2 I k60 2 y m k24 5
2ka5 5
Iy
xy3 dy dx
I0 Ix Iy
a 15 a m 2ka 2ka 3 5 5 I 2a 4ka 15 2a y m 2ka 3 5 10 x
x
x2
0
2ka5 15
x 2y dy dx
k 2
x2
0
4ka5 15
xy dy dx
x2
0
0
a2 x 2 dx
x
mk
y dy dx
a
k
a2 x 2
a
Center of Mass and Moments of Inertia
a2 a x x3 2
3
a
2
a
2 1 2a a4 a4 a2 3 a3 a5 a5 a 3 5 3 4 16 2 3
2k 7a15 a8 ka 56 6015 5
5
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5
383
384
Chapter 13
4x 2
2
46. I
Multiple Integration
2 0
2
ky 22 dy dx
2
k y 13 3
2
k 3
2k 192 128 1408k 32 32 3 5 7 105
2
16 12x 2 6x 4 x6 dx
dx
0
3k 16x 4x
2
4x2
3
k 2 x 2 8 dx 3 2
6 5 1 x x7 5 7
2 2
48. x, y k 2 x .
50. x, y k4 x4 y. Both x and y will decrease
x, y will be the same.
52. Ix
y2 x, y dA Moment of inertia about x-axis.
R
Iy
x2 x, y dA Moment of inertia about y-axis.
R
54. Orient the xy-coordinate system so that L is along the y-axis and R is in the first quadrant. Then the volume of the solid is V
2 x dA
y
R
2
x dA
R
L R ( x, y )
x dA
2
R
dA
dA
x
R
R
2 x A. By our positioning, x r. Therefore, V 2 rA.
a a 56. y , A ab, h L 2 2
b
Iy
0
a
a
0
58. y 0, A a2, h L
y
a 2
2
dy dx
3
ab 12
a3b12 a3L 2a a ya 2 L a2 ab 32L a
Iy
2
a
r3 sin2 dr d
0
2
0
y 2 dy dx
a a2 x 2
0
a2 x 2
a4 2 sin d 4
a4 4
ya
a2 a44 2 La 4L
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Section 13.5
Section 13.5
Surface Area 4. f x, y 10 2x 3y
2. f x, y 15 2x 3y
R x, y: x 2 y 2 ≤ 9
fx 2, fy 3
fx 2, fy 3
1 fx 2 fy 2 14
3
S
3
0
Surface Area
1 fx 2 fy 2 14
3
14 dy dx
0
314 dx 914
9x 2
3
0
S
y
3
R
2
3
14 r dr d 914
0
0
2
14 dy dx
3 9x 2
y
y = 9 − x2
1 2
x
1
2
R
3
1
−2 −1
x
1
−1 −2
y = − 9 − x2
6. f x, y y 2
y
R square with vertices 0, 0, 3, 0, 0, 3, 3, 3
3
fx 0, fy 2y
2
R
1 fx 2 fy 2 1 4y 2
3
S
0
3
1
3
1 4y 2 dx dy
0
31 4y 2 dy
x
1
34 2y1 4y
2
3
3
10. f x, y 9 x 2 y 2 fx 2x, fy 2y
fx 0, fy y1 2 1 fx2 fy2 1 y
S
0
2y
1 fx 2 fy 2 1 4x 2 4y 2
2
1 y dx dy
0
1 y 2 y dy
S
3 2
2 33 2
2 5
2 1 y5 2 5
2
35 2 2 5
2
2
2
2
1 1 4r2 3 2 12
0
0
2
0
1 4r2 r dr d
0
0
0
21 y
12 8 3 5 5 R
y
1
x
−1 2
1
1 −1
y=2−x R
x 1
d 2 0
1 173 2 1 d 1717 1 12 6 y
3
0 4 637 ln 6 37
ln 2y 1 4y 2
2 8. f x, y 2 y 3 2 3
2
2
0
2
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385
386
Chapter 13
Multiple Integration
12. f x, y xy
y
R x, y:
x2
y2
x 2 + y 2 = 16
≤ 16 2
fx y, fy x 1 fx 2 fy 2 1 y 2 x 2
S
4 16x 2 2
1
y2
x2
4
1 r 2 r dr d
0
0
dy dx
2 1717 1 3
14. See Exercise 13.
a2 x 2
a
S
a
a dy dx 2 x2 y 2 2 2 a a x
2
0
a
0
a a2 r 2
1 fy 2 fy2 1 4x 2 4y 2
16x
4
0
1 fx2 fy2
2
1 4x 2 y 2 dy dx
0
2
0
r dr d 2a2
18. z 2x2 y2
16. z 16 x 2 y 2
S
2 −2
16x 2
4
x
−2
4
1 4r 2 r dr d
0
S
2
2
2
2
4y2 5 x2 y2
2
5r dr d 45
0
0
6565 1 24
1 x 4x y
y
x2 + y2 = 4
y 1 6
16 − x 2
y=
x
−1
4
1 −1
2
x
2
4
6
20. f x, y 2x y 2
22. f x, y x 2 y 2
R triangle with vertices 0, 0, 2, 0, 2, 2
R x, y: 0 ≤ f x, y ≤ 16
1 fx fy 5 4y
0 ≤ x 2 y 2 ≤ 16
2
2
S
2
x
0
5 4y 2 dy dx
0
2
1 2121 55 12
fx 2x, fy 2y 1 fx 2 fy 2 1 4x 2 4y 2
y
16x 2
4
S
3
y=x 2
R
1
4 16x 2 2
1 4x 2 4y 2 dy dx
4
1 4r 2 dr d
0
0
y
x 2 + y 2 = 16
x
1
2
3
2
x
−2
2 −2
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6565 1 6
Section 13.5
R x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 fx x
Matches (c)
sin x, fy 0
z
1 fx 2 fy 2 1 x sin x
1
S
0
1
2
3 2
1 x sin x dy dx 1.02185
0
2
3
3
x
28. f x, y 25 y5 2
y
30. f x, y x 2 3xy y 2
R x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
R x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ x
fx 0, fy y3 2
fx 2x 3y, fy 3x 2y 3x 2y
1 fx fy 1 2
1
S
0
2
1 fx fy 2 1 2x 3y2 3x 2y2
y3
1 13x 2 y 2
1
1 y3 dx dy
4
0
S
1
387
26. Surface area 9
24. f x, y 23 x3 2 cos x
1 2
Surface Area
1 y dy 1.1114
0
3
x
1 13x 2 y 2 dy dx
0
0
32. f x, y cosx 2 y 2
2
R x, y: x 2 y 2 ≤
fx 2x sinx 2 y 2, fy 2y sinx 2 y 2 1 fx 2 fy 2 1 4x 2 sin2x 2 y 2 4y 2sin2x 2 y 2 1 4 sin2x 2 y 2x 2 y 2
S
2
( 2)x 2
2
( 2)x 2
1 4x2 y 2 sin2x 2 y 2 dy dx
34. f x, y ex sin y
36. (a) Yes. For example, let R be the square given by
R x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ x
0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
fx ex sin y, fy ex cos y
and S the square parallel to R given by
1 fx 2 fy 2 1 e2x sin2 y e2x cos2 y
1 e
2x
4
S
0
0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z 1. (b) Yes. Let R be the region in part (a) and S the surface given by f x, y xy.
x
1 e2x dy dx
(c) No.
0
38. f x, y kx 2 y 2 1 fx 2 fy 2
S
R
1 x kx y 2 2
2
1 fx 2 fy 2 dA
2
R
k2y 2 k2 1 x y2 2
k2 1 dA k2 1
dA Ak2 1 r2k2 1
R
http://librosysolucionarios.net
388
Chapter 13
Multiple Integration
15
1 3 4 16 y y 2 y 25 75 25 15
40. (a) z
(b) V 250
0
1 3 4 16 y y 2 y 25 dy 75 25 15
100266.25 26,625 cubic feet 1 4 16 (c) f x, y y 3 y 2 y 25 75 25 15 fx 0, fy
50
(d) Arc length 30.8758 Surface area of roof 25030.8758 3087.58 sq ft
1 2 8 16 y y 25 25 15
15
S2
0
1 fy2 fx2 dy dx 3087.58 sq ft
0
42. False. The surface area will remain the same for any vertical translation.
Section 13.6
1
2.
1
1
1 1 1
0
0
2 3
z dz dx dy
1 2
0
e2
4
1
1 xz
0
1
1
9
2
8.
y 2
0
0
0
0
e2
2x 2y 2
0
3
0
2(2y 3)
1 ln z2 x 2
0
y dz dy dx
1 xz
ln zy
1
4y 2
2
0
e2 1
y 2
0
0
0
0
1
3 2
1
2 18
dz
9
4 9
1
1
z2 dz
1
1
274 z
1
3
1
2
0
9
4
0
4 1
2 ln 4
0
0
3(x 2)
0
2
1 sin y dy cos y 2
4y 2x 2y 2y3 dy dx
(6x) 2
729 4
ln z dz dx xz
2 dx 2 ln x x
361 y
y3 dy
0
e2
4
dz dx
0
y z
1
sin y 1 dx dy y 2
6
14.
1
dy
1
6
dx dz dy
0
2x
0
3
y 3
4
2x 2
62y3z 2y 2
dx
2
0
zex
2 9
y 2 9x 2 dx dy
xy 2 3x3
0
2x 2
0
12.
sin y dz dx dy
2
10.
1 y
dy dz
0
9
1
1
y 3
0
4
1
y 2z2 dy dz
1
x3y 2z2
1
1 1
4
ln z dy dz dx
1
1 1
1 2
6.
1
1 3
y 29x 2
y 3
x 2y 2z2 dx dy dz
9
4.
Triple Integrals and Applications
0
162 15
(6x2y) 3
zex
2y 2
dz dy dx
0
1 6 x 2y 2 3
2
ex
2 2
y
dy dx 2.118
2
9x
dz dy dx
0
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1 2
8 27
Section 13.6
Triple Integrals and Applications
1 16. z x2 y2 ⇒ 2z x2 y2 2 x2 y2 z2 2z z2 80 ⇒ z2 2z 80 0 ⇒ z 8z 10 0 ⇒ z 8 ⇒ x2 y2 2z 16
16x
4
16x
1
18.
2
4
1
0
80x2 y2
2
dz dy dx
1 2x y 2
xy
0
1
dz dy dx
0
0
2
36x y
36x2
6
2
0
1
xy dy dx
0
0
0
6
4
0
0
1 4
36x2
6
6
36 x2 y2dy dx 4
dz dy dx 4
0
1
x x2 dx 2 4
2
20. 4
0
1
0
0
2
0
2x
0
9x 2
2
dz dy dx
0
0
4162 648
9 x 22 x dx
2 0
50 3
26. Elliptic cone: 4x 2 z2 y 2
z 6
Side cylinder: x2 y2 9 9y2
dx
0
0
24. Top plane: x y z 6 3
6
2 3 2
9 2 1 18 9x 2x 2 x3 dx 18x x 2 x3 x 4 2 3 4
0
36x2
2
9 x 2 dy dx
0
2
6x 61x36 x
0
2x
0
0
y3 3
1 3636 x2 x236 x2 36 x23 2 dx 3
4 9x36 x2 324 arcsin
22.
36y x2y
z 5
6xy
dz dx dy
4
0
3 2 3
3
y
6
1
6
3
2
1
x
x
5
4
0
4
z
y
y 2z2 2
dx dy dz
0
28. Q x, y, z: 0 ≤ x ≤ 2, x 2 ≤ y ≤ 4, 0 ≤ z ≤ 2 x
2
xyz dV
0
2
2
2
0
4
0
2
2x
xyz dz dx dy
2
0
2x
0
x
4
x2
2z
0
0
4
x2
(2z)2
0
2 y
0
y
0
2
xyz dx dy dz
0
0 y
4
4
xyz dy dx dz
0
y
xyz dy dz dx
0
4
x2 0
4
z
2x
xyz dz dy dx
0
Q
4
4
xyz dx dz dy
0
4
2z
xyz dx dy dz
(2z)2 0 2
2 y
2z
0
dx dz dy
104 21
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(2, 4)
y
389
390
Chapter 13
Multiple Integration
30. Q x, y, z: 0 ≤ x ≤ 1, y ≤ 1 x2, 0 ≤ z ≤ 6
1
xyz dV
0
Q
0
0
0
0
0
1
0
xyz dz dy dx
6
xyz dz dx dy
0
1y2
xyz dx dz dy
0
1y2
xyz dx dy dz
5x
mk
0
5
0
2
1x2
5x
0
0
xyz dy dz dx
1x2
xyz dy dx dz
0
1 5153x3y
y dz dy dz
1 5153x3y
125 k 8
y 2 dz dy dx
0
125 k 4
a
b
0
a
0
b
c
a
0
b
a
0
b
0
z2 dz dy dx
kabc3 3
xz dz dy dx
ka2bc2 4
yz dz dy dx
kab2c2 4
0 c
0 c
Mxz k
0
0
b
0
0
0
Mxz kab2c2 4 b m kabc2 2 2
z
Mxy kabc3 3 2c m kabc2 2 3
38. z will be greater than 8 5, whereas x and y will be unchanged.
40. x, y and z will all be greater than their original values.
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dz dx dy
kabc 6
c[1(y b)(x a)]
0
kab2c 24
b Mxz m kabc 6 4
Myz ka2bc2 4 a x m kabc2 2 2 y
c1(y b)(x a)
0
a[1(y b)]
Mxz k
0
a1(y b)
mk
kabc2 2
c
Myz k
0
z dz dy dx
0
Mxy k
0
b
34.
y
mk
y
2
0
M y xz 2 m
36.
1
x
0
Mxz k
1
0
5
32.
6
0
6
1
0
1
6
0
6
1y 2
0
1
6
6
0
0
1
1x2
z
y dz dx dy
kab2c 24
Section 13.6
4x 2
2
42.
m 2k
0
y
dz dy dx
0
0
2
4 x 2 dx
k
0
2
y
4x 2
2
0
y
4x 2
2
0
x dz dy dx 0 y dz dy dx 2k
0
0
Mxy 2k
16k 3
0
0
Mxz 2k
44.
4x 2
2
Myz k
y
z dz dy dx k
0
0
x
Myz 0 0 m 16k 3
y
Mxz 2k 3 m 16k 3 8
z
Mxy k 3 m 16k 3 16
x0
2
1 (y 21)
1
m 2k
0
0
2
0
2
0
2
k
0
1
0
1
0
0
2
y dy dx k y2 1
ln 2 dx k ln 4
0
z dz dy dx
1 dy dx k y 2 12
2
0
Mxz k ln 4 ln 4 m k
z
Mxy 2 1 k k m 2 4 4
1 y arctan y 2 y 2 1 2
1 0
dx k
mk
0
5
Myz k
0
5
Mxz k
0
5
Mxy k
0
(3 5)x3
0 (3 5)x3
0 (3 5)x3
0 (3 5)x3
0
1 4 8
2
dx k
0
1 60 12x 20y 15 5
dx k
0
0
y
46. f x, y
2
2
1 (y 21)
1
Mxy 2k
0
0
y dz dy dx 2k
0
1 dy dx 2k y2 1 4
1
0
1 (y 21)
1
2
dz dy dx 2k
0
Mxz 2k
0
Triple Integrals and Applications
y 5
(1 15)(6012x20y)
dz dy dx 10k
0
y = 35 (5 - x)
3
(1 15)(6012x20y)
x dz dy dx
0
25k 2
2 1 x
(1 15)(6012x20y)
0
4
y dz dy dx
15k 2
1
2
3
4
(1 15)(6012x20y)
z dz dy dx 10k
0
x
Myz 25k 2 5 m 10k 4
y
Mxz 15k 2 3 m 10k 4
z
Mxy 10k 1 m 10k
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5
12 4
391
392
Chapter 13
Multiple Integration
a 2
48. (a) Ixy k
a 2
a 2
a 2 a 2 a 2
ka5 by symmetry 12
Ixz Iyz
Ix Iy Iz
ka5 ka5 ka5 12 12 6
a 2
(b) Ixy k
a 2
a 2
a 2
a3k 12
z2x 2 y 2 dz dy dx
a 2 a 2 a 2 a 2
Ixz k
ka5 12
z2 dz dy dx
a 2
a 2 a 2 a2
a 2
x 2 y 2 dy dx
a 2 a 2
a 2
y 2x 2 y 2 dz dy dx ka
a 2
a7k 72
a 2
a 2 a 2
7ka7 360
x 2y 2 y 4 dy dx
Iyz Ixz by symmetry Ix Ixy Ixz
a7k 30
Iy Ixy Iyz
a7k 30
Iz Iyz Ixz
7ka7 180
4
2
50. (a) Ixy k
0
k 4 k 4
0
4
4y 2
4
2
1 4 y 24 dy dx 4
z3 dz dy dx k
0
0
0
2
256 256y 2 96y4 16y6 y8 dy dx
0
0
4
256y
0
4
2
4y 2
Ixz k
0
0
4
2
k
0
0
4
2
0
4
2
k
0
0
4
y 2z dz dy dx k
0
0
2
0
4y 2
0
4
x 2z dz dy dx k
0
2
0
2 0
4
dx k
0
16,384 65,536k dx 945 315
4
0
16y3 8y5 y7 3 5 7
2 0
dx
k 2
4
0
1024 2048k dx 105 105
1 2 x 4 y 22 dy dx 2
1 2 k x 16 8y 2 y4 dy dx 2 2
Ix Ixz Ixy
1 2 y 4 y 22 dy dx 2
1 k 16y 2 8y4 y6 dy dx 2 2
Iyz k
0
256y3 96y5 16y7 y9 3 5 7 9
4
0
x 2 16y
8y3 y5 3 5
2 0
dx
k 2
2048k 8192k 63,488k , Iy Iyz Ixy , Iz Iyz Ixz 9 21 315
—CONTINUED—
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4
0
256 2 8192k x dx 15 45
Section 13.7
Triple Integrals in Cylindrical and Spherical Coordinates
50. —CONTINUED—
4
(b) Ixy
4y 2
2
0
0
z24 z dz dy dx
0
4
Ixz
0
0
4
2
4y 2
0
0
4
0
4
4
a2
Iyz
0
0
b2
z2 dz dy dx
b2
c2
a2
c2 a2
dy dx
Iz Ixz Iyz
1 ma2 c2 12
1 2 1 b abc mb2 12 12
a2
c2 a2
x 2 dz dy dx ab
1 mb2 c2 12
1 1x 2 0
4096k 8192k 4096k 9 45 15
0
y 2 dy dx
c2
Iy Ixy Iyz
4x 2y 2
0
c2
y 2 dz dy dx b
1 ma2 b2 12
x 2z dz dy dx
4y 2
2
b3 12
b2
c2 a2 b2
1x 2
1024k 2048k 1024k 15 105 21
48,128k 118,784k 11,264k , Iy Iyz Ixy , Iz Ixz Iyz 315 315 35
Ix Ixy Ixz
1
54.
a2
4
4x 2 dz dy dx k
c2 a2 b2 c2
4y 2
c2 a2 b2 c2
Ixz
y 2z dz dy dx
0
x 24 z dz dy dx
a2
0
0
2
0
4y 2
2
0
Ix Ixz Ixy
c2
32,768k 65,536k 32,768k 105 315 315
4y 2
2
0
4
0
k
52. Ixy
z3 dz dy dx
0
4y 2 dz dy dx k
Iyz k
0
0
y 24 z dz dy dx
4y 2
2
0
4y 2
2
0
k
0
4
4z2 dz dy dx k
0
0
4y 2
2
k
c2
x 2 dx
ba3 12
c2
c2
dx
ba3c 1 1 a2abc ma2 12 12 12
abc3 1 1 c2abc mc2 12 12 12
kx 2x 2 y 2 dz dy dx
56. 6
58. Because the density increases as you move away from the axis of symmetry, the moment of intertia will increase.
Section 13.7
4
2.
0
2
0
Triple Integrals in Cylindrical and Spherical Coordinates
2r
rz dz dr d
0
2
0
0
4
0
4.
1 2
2
0
e 2 d d d 3
2
0
4
0
rz2 2
2r
dr d
0
2
1 2
4r 4r 2 r 3dr d
0
2
0
0
2
1 3 e 3
0
d d
4
0
2
0
0
2r
2
4r 3 r 4 3 4
2
0
d
2 3
4
d
0
1 2 1 e8 d d 1 e8 3 6
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6
393
394
Chapter 13
4
6.
0
4
cos
0
Multiple Integration 1 3
2 sin cos d d d
0
1 3
1 3
2
0
0
0
3
0
3r2
0
r dz dr d
2
4
0
sin cos cos 1 sin2 d d
0
4
sin cos sin
0
sin3 3
4
sin cos d
0
r 3 r 2dr d
4
d
0
52 sin2 36 2
4
52 144
0
z
0
2
4
3r 2 r 4 2 4
0
4
3
0
0
cos3 sin cos d d
0
8 9
2 cos 2 d d d
2
10.
sin
4
0
52 36
8.
4
2
3
d
0
9 9 d 4 2
0
2
2
3
y
3
x
2
12.
0
0
5
2 sin d d d
2
117 3
117 3
2
14. (a)
0
2
(b)
0
0
16r2
0
0
1
0
2
0
sin d d
2
r 2 dz dr d
cos
0
2
2
0
7
3 sin2 d d d
4
y
8 2 23 3
2
8
(b)
0
2
r
r dz dr d
0
0
2 csc
6
rr 2 z2 dz dr d
0
0
r=2
d 7
1r2
22
x
0
2
8 2 23 3
3 sin2 d d d
r=5
7
0
4
2 18. V 43 4 3
z
0
468 3
0
6
2
16. (a)
2
16r2
4
22 0
0
r dz dr d
2
1
0
3 sin d d d
0
z 7
(Volume of lower hemisphere) 4(Volume in the first octant) V
128 4 3
2
0
22
0
128 82 4 3 3
r 2 dr d
2
0
2
0
r16 r 2 dr d
22
3 16 r 1
4
2 32
22
128 82 82 4 3 3 3
128 642 64 2 2 3 3 3
4
d
x
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7
7
y
8
Section 13.7
20. V
2
2
0
0
2
r dz dr d
22.
0
r
2
0
k r dz dr d
0
0
r4 r 2 r 2 dr d
1 r3 4 r 232 3 3
2
0
d
0
2
2
2kh 2 r0
26.
2
2kh r02
r0 r 2r0 2
r2
0
h2
kr0 2 12
r0 12
2
Mxy kr02h2 3 h z m 12 r02hk 4
28. Iz
4kh
0
2
0
2
2
4kh
0
4kh
0
h(r0 r)r0
2
2
r0
0
z
0
h(r0 r)r0
zr dz dr d
z2r dz dr d
0
1 k r02h3 30 Mxy k r02h330 2h m k r02h212 5
2
0
2a sin
0
3 ka4h 2 3 ma2 2
1 5 r d 30 0
1 5 r 30 0 2
1 5 r kh 15 0
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h
0
r 3 dz dr d
2
2
0
6ke4 6k d
0
Mxy 4k
r05 r05 d 5 6
0
2
r0 r 4 r dr d r0
r 6 r0 r5 d 5 6r0 0
0
4kh
r 4 dz dr d
0
r0
r
6ke
1 k r02h2 12
Iz 2k
0
4kh
dr d
r3
hr0 rr0
r0
r0
0
30. m ka2h
2
0
2
x2 y2x, y, z dV
Q
4k
z r dz dr d
r0
4
2
kz
m 4k
hr0 rr0
r dr d
0
x y 0 by symmetry
0
0
2
3k 1 e4
r0
0
r
12ke
0
1 m r02hk from Exercise 23 3 Mxy 4k
2
0
x y 0 by symmetry
2
2
12e
8 2 2 3
2
0
2
0
24.
2
r
2
0
4r2
Triple Integrals in Cylindrical and Spherical Coordinates
395
396
Chapter 13
Multiple Integration
4
32. V 8
2
0
b
2 sin d d d
0
4
8 b3 a3 3
0
4 3 b a3 3
4
3
1
b
a a
x
2
2
0
3
3
2 b3 a3
0
0
2
2
3 sin2 d d d
36.
ka4
x y 0 by symmetry mk
d d
sin2
23 R
sin2
3
Mxy 4k
d
2
0
1 1 sin 2 2 4
2 2 r 3 kR3 r3 3 3
0
2
0
y
a
2ka4
ka4
b
4
3 b a 23 2
0
sin d
2
ka4
b
a3cos
2 4
0
z
sin d d
0
34. m 8k
2
0
43 b
(includes upper and lower cones)
a
2
0
1 kR4 r 4 2
2
0
R
3 cos sin d d d
r
2
0
1 kR4 r 4 4
1 2 4 k a 4 4
2
sin 2 d d
0
2
sin 2 d
0
1 kR4 r 4 cos 2 8
2
0
1 kR4 r 4 4
Mxy kR4 r44 3R4 r4 z 3 3 m 2kR r 3 8R3 r3
2
38. Iz 4k
0
0
4k 5 R r5 5
R
4 sin3 d d d
2
0
2k 5 R r5 5
2
sin 1 cos2 d
4k 5 R r5 15 2
2
1
2 x2 y2 z2
y sin sin
tan
z cos
cos
y x z x2 y2 z2
0
1
sin3 d d
cos3 2k 5 R r5 cos 5 3
1
2
0
40. x sin cos
r
2
42.
2
2
0
f sin cos , sin sin , cos 2 sin d d d
44. (a) You are integrating over a cylindrical wedge.
(b) You are integrating over a spherical block.
46. The volume of this spherical block can be determined as follows. One side is length . Another side is . Finally, the third side is given by the length of an arc of angle in a circle of radius sin . Thus:
z
ρi sin φi ∆ θi ∆ ρi
V sin 2 sin
ρi ∆φi
y x
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Section 13.8
Section 13.8
Change of Variables: Jacobians
Change of Variables: Jacobians
2. x au bv
4. x uv 2u
y cu dv
y uv
x y y x ad cb u v u v
x y y x v 2u vu 2u u v u v
6. x u a
8. x
yva
1 10. x 4u v 3 1 y u v 3
y x 1 u 1 u uv x y 1 1 2 2 u v u v v v v v v2
x, y
u, v
0, 0
0, 0
1 −1
−1
3, 0
uxy
2, 2
0, 6
−4
v x 4y
6, 3
3, 6
−6
uxy
1 y u v, 2
vxy
x y y x 1 1 1 u v u v 2 2 2
12 21
v
4, 1
1 12. x u v, 2
u v
yuv
x y y x 11 00 1 u v u v
(0, 0) 1
(3, 0) 2
4
u 5
6
−2 −3 −5
(0, −6)
(3, −6)
x, y
u, v
0, 1
1, 1
2, 1
1, 3
1, 2
1, 3
1, 0
1, 1
60xy dA
R
1
1 u v 2
60
1 1 1
3
3
1 1 1
1 1
1
15
21u v12 dv du
15 2 v u2 dv du 2
15 v 3 u2v 2 3
15 26 2u2 du 2 3
152 23u
3
397
3 1
du
v
(−1, 3)
(1, 3) 2
(−1, 1) −2
(1, 1) u
−1
1 −1
26 u 3
1 1
23 263 120
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2
398
Chapter 13
Multiple Integration v
1 14. x u v 2
u 1
1 y u v 2
−1
y x 1 x y u v u v 2
v=u−2
−2
2
4x yexy dA
0
R
0
12 dv du
4uev
u2
2
u2 ue
2u1 eu2 du 2
0
16. x
2
2
u2
eu2
2
21 e2
0
u v
yv y x 1 x y u v u v v
4
y sin xy dA
1
R
4
vsin u
1
18. u x y ,
1 dv du v
4
3 sin u du 3 cos u
1
vxy0
u x y 2,
vxy
1 x u v, 2
1 y u v 2
1 x, y u, v 2
4 1
y
3π 2
π
x−y=0
x + y = 2π
π 2
x− y=π
x+y=π
x y2 sin2 x y dA
R
π 2
2
u2 sin2 v
0
0
π
12 du dv
1 u3 1 cos 2v 2 3 2
u 3x 2y 16,
v 2y x 8
1 x u v, 4
1 y u 3v 8
2
dv
712 v 21 sin 2v 3
0
y
8
0
R
−2
u v3 2
3
(4, 2) 2y − x = 0
−1
18 du dv
16v 3 2 dv
0
(2, 5)
2
8
2y − x = 8
3x + 2 y = 0
16
0
7 4 12
3x + 2y = 16
(−2, 3)
3x 2y2y x3 2 dA
5
x y y x 1 3 1 1 1 u v u v 4 8 8 4 8
x
3π 2
v 2y x 0
20. u 3x 2y 0,
3cos 1 cos 4 3.5818
2516v
8
5 2
0
4096
2 5
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x −1
1
(0, 0)
2
3
4
Section 13.8 22. u x 1,
v xy 1
u x 4,
v xy 4
x u,
y
y x 1 x y u v u v u
R
Change of Variables: Jacobians
y
x=1 4 3
v u
xy = 4 2
x=4 1 x
xy dA 1 x 2y 2
4
1
1
4
1
4
1
2
3
4
xy = 1
v 1 dv du 1 v2 u
4
1 ln1 v 2 2
1
4
1 1 du ln 17 ln 2 ln u u 2
1
1 17 ln ln 4 2 2
24. (a) f x, y 16 x 2 y 2 R:
y2 x2 ≤ 1 16 9
V
f x, y dA
R
Let x 4u and y 3v.
1u2
1
16 x 2 y 2 dA
R
1 1u2 2
16 16u2 9v2 12dv du
1
16 16r 2 cos2 9r 2 sin2 12r dr d
0
0
2
12
0
2
12
9 8r2 4r 4 cos2 r 4 sin2 4 84
0
2
398 167 sin 2
R:
1 0
0
2
d 12
1 cos 2 9 1 cos 2 2 4 2
12 (b) f x, y A cos
Let u r cos , v r sin .
0
2
2
2
394 117
12
x2 y 2 2 ≤ 1 a2 b
Let x au and y bv.
1
f x, y dA
R
1u2
2 u
2
A cos
1 1u
2
v 2 ab dv du
Let u r cos , v r sin .
2
Aab
0
1
0
2 rr dr d Aab 2r sin2r 4 cos2r
1
cos
2
0
2
2 0 0 4 4 2Aab
2 Aab
2
d 12
2 ax yb 2
8 4 cos
2
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2
0
9 2 sin d 4
398 87 cos 2 d
399
400
Chapter 13
Multiple Integration 28. x 4u v, y 4v w, z u w
26. See Theorem 13.5.
4 x, y, z 0 u, v, w 1 30. x r cos , y r sin , z z
cos r sin x, y, z sin r cos r, , z 0 0
1 4 0
0 1 17 1
0 0 1 r cos2 r sin2 r 1
Review Exercises for Chapter 13
2y
2.
x3 xy 2
y
y
2
4.
0
x2
dx dy 2
0
2 4y2
3
dy dx
0
2
0
0
2
y
3
2
0
4 3
0
1 1 1y
x2 2x
1 1y
8
0
3 9y
4
1
dx dy
0
5
dy dx
0
1
2
x1
1 1x
0
dx dy
2
3 x
dy dx
1
dy dx
4 0
dx dy
64 3
14 3
1 1x
1 1x
0
3 9y
3 9y
8
2 8x 2x2 dx 4x2 x3 3 0
dy dx
9
dx dy
dy dx
9 2
16. Both integrations are over the common region R shown in the figure. Analytically,
2
0
3
0
5y
e xy dx dy
2 8 5 e 5 5
exy dy dx
3y 2 2x 3
0
88 15
3
2
1 1y
2yy 2
0
0
1 2 x4 x5 2 5
dx dy
dy dx
0
3
14. A
3
3
y
126y 23 y
6xx2
y 21
2
0
6 3y dy
y 2
43 x
6y 2
2
0
0
4
12. A
0
dy dx
0
0
4x 2 2x 3 2x 4 dx
dx dy
dy dx
A
2
dx
y
6xx2
x2 2x
x2
2 4 y 2 dy y4 y 4 arcsin
62x
2
1 2
2x
6y 2
2
0
3
0
x
4
x 2y y 2
2 4y2
A
10.
10y3 3
0
2
2
x 2 2y dy dx
0
8.
2x
3
6.
2y
3
x 2 y 2 dx
5
3
5x
0
y
5 4 3
3 2 8 2 exy dy dx e5 e 3 e 5 e 3 e5 5 5 5 5
(3, 2)
2 1 x 1
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2
3
4
5
Review Exercises for Chapter 13
3
18. V
x
0
3
x y dy dx
20. Matches (c)
z
0 3
1 xy y 2 2
0
x
dx
2
0
1 y
3
3 2
2
x2
dx 2
0
x
3
1 x3 2
1
22.
0
27 2 0
x
1
kxy dy dx
0
0
1
0
kxy 2 2
1
24. False,
dx
0
0
1
2
x dy dx
0
1
2
x dy dx
1
kx 3 dx 2
kx8
4 1
x
0
k 8
Since k 8 1, we have k 8.
0.5
P
0.25
0
1
26. True,
0
1
0
4
28.
0
8xy dy dx 0.03125
0
1 dx dy < 1 x2 y2
16y2
x 2 y 2 dx dy
0
2
0
8 3
1
1
0
0
2
0
30. V 8
4
0
1 dx dy 1 x2 4
r 3 dr d
2
0
r4
4 4 0
d
2
64 d 32
0
R
R 2 r 2r dr d
32. tan
b
2
R 2 r 23 2
0
8 R2 b23 2 3
2
R b
d
The polar region is given by 0 ≤ r ≤ 4 and 0 ≤ ≤ 0.9828. Hence,
arctan3 2
d
4
r cos r sin r dr d
0
0
0
4 R 2 b23 2 3
1213 3 ⇒ 0.9828 2 813
y
(8/ 13, 12/ 13) 4 3
x 2 + y 2 =16 y = 23 x
2 1
θ x 1
8/ 13
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4
288 13
401
402
Chapter 13
h 2 2 x L x2 L2
L
34. m k
0
0
2
0
L
kh2 8
4
0
y
x x2 2 2 dx L L
L 0
kh2 8
17L 17kh2L 10 80
h 2 2 x L x2 L2
x dy dx
0
L
kh 2
2x
0
L 0
kh 2
5L2 5khL2 12 24
12 5L 7khL 14 12
51h
7khL 140
2
y2 x, y dA
0
x, y dA
0
R
2
4x
ky3 dy dx
0
2
x2
x2 x3 kh 2 x3 x4 2 dx x 2 L L 2 3L 4L
Mx 17kh2L m 80
16,384 k 315
2
4x
kx2y dy dx
0
512 k 105
16,384k 512k 17,920 512 I0 Ix Iy k k 315 105 315 9
2
x, y dA
0
R
2
4x
ky dy dx
0
128 k 15
4 512k 105 128k 15 7 I 16,384k 315 128 y m 128k 15 21 Iy m
x
x
38. f x, y 16 x y2 R x, y: 0 ≤ x ≤ 2, 0 ≤ y ≤ x fx 1, fy 2y 1 fx2 fy2 2 4y2
2
S
0
( x
4x 3x2 2x3 x4 2 3 4 dx L L L L
R
m
(
2 x y = h 2 − L − x2 2 L
L
My 5khL2 x m 24
Iy
0
x x 7khL dx L L2 12 h
kh2 2x 2 x3 x4 x5 4x 2 3 4 8 L L 2L 5L
0
36. Ix
2
2
y dy dx
L
My k
y
L
0
2
L
kh 2
h 2 2 x L x2 L2
Mx k
kh 8
dy dx
0
L
Multiple Integration
2
2
2 4y2 dx dy
y
0
22 4y2 y2 4y2 dy
122y
121 2 4y
2 4y2 2 ln 2y 2 4y2
2
2 3 2
0
2418 2 ln 4 18 12 1818 ln2 1
1
62 ln 4 32
22 12
2 92 52 ln2 ln 22 3 2 6 3
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Review Exercises for Chapter 13
(b) Surface area
40. (a) Graph of
403
1 fxx, y2 fyx, y2 dA
R
f x, y z
25 1 ex
2
y x 1000 2
y 2 1000
cos2
Using a symbolic computer program, you obtain surface area 4,540 sq. ft.
2
over region R z 50
R
50 x
4x2
2
42.
50
0
2
x 2 y 2 dz dy dx
0
25x2
25x2 y2
0
1
0
x2
r 2 2
2
0
0
5
(x2 y2) 2
2 4x2
44.
y
0
1 r3 dz dr d 2
2
1 dz dy dx y2 z2
0
2
0
2
0
5
0
2
2
0
0
arctan
5 0
2
0
4x 2
4x 2 y 2
0
xyz dz dy dx
0
2
48. V 2
2 sin
0
0
0
2
0
Mxz 2k
2
0
Mxy 2k
2
0
a
0
a
0
a
0
cr sin
32 3
0
d
5 arctan 5 2
r16 r 2 dr d
0
8 sin2 sin4 d
r dz dr d 2kc
2
0
0
0
2
0
0
r 2 sin dr d 2
0
rz dz dr d kc2
2
a
r 2 sin dz dr d 2kc
0 cr sin
2
5 arctan 5cos
1 3 3 1 1 sin cos sin 2 4 4 2 4
0 cr sin
d
0
0
8 4 2 sin 2
2
2
32 sin2 4 sin4 d 8
m 2k
2 sin
0
2
2
50.
2
r dz dr d 2
0
4 3
16r 2
16 3
sin d d
0
46.
r 5 dr d
2 sin d d d 1 2
0
2
2
2 kca3 3
2
sin d
0
a
0
1 r3 sin2 dr d kca4 2
a
r 3 sin2 dr d
0
29 2
1 2 4 kc a 4
2
2
Mxz kca4 8 3a m 2kca3 3 16
z
Mxy kc2a4 16 3ca m 2kca3 3 32
sin2 d
0
sin2 d
0
x0 y
2 kca3 3
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1 kca4 8
1 kc2a4 16
404
52.
Chapter 13
500 3
m
0
2
Mxy
3
0
0
2
0
0
by symmetry
60.
2
0
2
r25 r2 4rd dr
500 64 125 500 14 2 18 162 3 3 3 3 3
25r 2
1 4 9 2 r r 8 4
0
0
25r 2
5
3
0
1 3 9 r r dr d 2 2
3
0
zr dz dr d
2
zr dz dr d
2
0
0
3
d
0
81 8
3
2
1 8 25 r2 r dr d 0 2
0
81 4
0
a
2 sin2 2 sin d d d
0
4ka6 9
y2
z2 21 a
58.
0
a
1z2 a2
a
1z2 a2
1y 2z2 a2
1r2
2
0
r dz dr d
0
Since z 1 r 2 represents a paraboloid with vertex 0, 0, 1, this integral represents the volume of the solid below the paraboloid and above the semi-circle y 4 x2 in the xy-plane.
x2 y2 dV
Q
25r 2
54. Iz k
4
2
3
Mxy 81 1 1 m 4 162 8
z
Iz
0
500 3
r dz d dr
4
3
0
56. x
25r2
xy0
2
2
3
500 1 2 25 r23 2 2r 2 3 3
Multiple Integration
x 2 y 2 dx dy dz
1y 2z2 a2
8 a 15
x, y x y y x u, v u v u v 2u2v 2u2v 8uv y
x, y x y x y 1 1 1 0 62. u, v u v v u u u
5 4
y = 1x
x=1
v x u, y ⇒ u x, v xy u
3 2
x=5
Boundary in xy-plane
Boundary in uv-plane
x1
u1
x5
u5
xy 1
v1
xy 5
R
x dA 1 x2y2
5
1
5
1
x 1
v5
u 1 du dv 1 u2v u2 u
4 arctan v
5 1
1
5
1
5
1
1 du dv 1 v2
5
1
4 dv 1 v2
4 arctan 5
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y = 1x
4
5
400
Chapter 13
Multiple Integration 28. x 4u v, y 4v w, z u w
26. See Theorem 13.5.
4 x, y, z 0 u, v, w 1 30. x r cos , y r sin , z z
cos r sin x, y, z sin r cos r, , z 0 0
1 4 0
0 1 17 1
0 0 1 r cos2 r sin2 r 1
Review Exercises for Chapter 13
2y
2.
x3 xy 2
y
y
2
4.
0
x2
dx dy 2
0
2 4y2
3
dy dx
0
2
0
0
2
y
3
2
0
4 3
0
1 1 1y
x2 2x
1 1y
8
0
3 9y
4
1
dx dy
0
5
dy dx
0
1
2
x1
1 1x
0
dx dy
2
3 x
dy dx
1
dy dx
4 0
dx dy
64 3
14 3
1 1x
1 1x
0
3 9y
3 9y
8
2 8x 2x2 dx 4x2 x3 3 0
dy dx
9
dx dy
dy dx
9 2
16. Both integrations are over the common region R shown in the figure. Analytically,
2
0
3
0
5y
e xy dx dy
2 8 5 e 5 5
exy dy dx
3y 2 2x 3
0
88 15
3
2
1 1y
2yy 2
0
0
1 2 x4 x5 2 5
dx dy
dy dx
0
3
14. A
3
3
y
126y 23 y
6xx2
y 21
2
0
6 3y dy
y 2
43 x
6y 2
2
0
0
4
12. A
0
dy dx
0
0
4x 2 2x 3 2x 4 dx
dx dy
dy dx
A
2
dx
y
6xx2
x2 2x
x2
2 4 y 2 dy y4 y 4 arcsin
62x
2
1 2
2x
6y 2
2
0
3
0
x
4
x 2y y 2
2 4y2
A
10.
10y3 3
0
2
2
x 2 2y dy dx
0
8.
2x
3
6.
2y
3
x 2 y 2 dx
5
3
5x
0
y
5 4 3
3 2 8 2 exy dy dx e5 e 3 e 5 e 3 e5 5 5 5 5
(3, 2)
2 1 x 1
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2
3
4
5
Review Exercises for Chapter 13
3
18. V
x
0
3
x y dy dx
20. Matches (c)
z
0 3
1 xy y 2 2
0
x
dx
2
0
1 y
3
3 2
2
x2
dx 2
0
x
3
1 x3 2
1
22.
0
27 2 0
x
1
kxy dy dx
0
0
1
0
kxy 2 2
1
24. False,
dx
0
0
1
2
x dy dx
0
1
2
x dy dx
1
kx 3 dx 2
kx8
4 1
x
0
k 8
Since k 8 1, we have k 8.
0.5
P
0.25
0
1
26. True,
0
1
0
4
28.
0
8xy dy dx 0.03125
0
1 dx dy < 1 x2 y2
16y2
x 2 y 2 dx dy
0
2
0
8 3
1
1
0
0
2
0
30. V 8
4
0
1 dx dy 1 x2 4
r 3 dr d
2
0
r4
4 4 0
d
2
64 d 32
0
R
R 2 r 2r dr d
32. tan
b
2
R 2 r 23 2
0
8 R2 b23 2 3
2
R b
d
The polar region is given by 0 ≤ r ≤ 4 and 0 ≤ ≤ 0.9828. Hence,
arctan3 2
d
4
r cos r sin r dr d
0
0
0
4 R 2 b23 2 3
1213 3 ⇒ 0.9828 2 813
y
(8/ 13, 12/ 13) 4 3
x 2 + y 2 =16 y = 23 x
2 1
θ x 1
8/ 13
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4
288 13
401
402
Chapter 13
h 2 2 x L x2 L2
L
34. m k
0
0
2
0
L
kh2 8
4
0
y
x x2 2 2 dx L L
L 0
kh2 8
17L 17kh2L 10 80
h 2 2 x L x2 L2
x dy dx
0
L
kh 2
2x
0
L 0
kh 2
5L2 5khL2 12 24
12 5L 7khL 14 12
51h
7khL 140
2
y2 x, y dA
0
x, y dA
0
R
2
4x
ky3 dy dx
0
2
x2
x2 x3 kh 2 x3 x4 2 dx x 2 L L 2 3L 4L
Mx 17kh2L m 80
16,384 k 315
2
4x
kx2y dy dx
0
512 k 105
16,384k 512k 17,920 512 I0 Ix Iy k k 315 105 315 9
2
x, y dA
0
R
2
4x
ky dy dx
0
128 k 15
4 512k 105 128k 15 7 I 16,384k 315 128 y m 128k 15 21 Iy m
x
x
38. f x, y 16 x y2 R x, y: 0 ≤ x ≤ 2, 0 ≤ y ≤ x fx 1, fy 2y 1 fx2 fy2 2 4y2
2
S
0
( x
4x 3x2 2x3 x4 2 3 4 dx L L L L
R
m
(
2 x y = h 2 − L − x2 2 L
L
My 5khL2 x m 24
Iy
0
x x 7khL dx L L2 12 h
kh2 2x 2 x3 x4 x5 4x 2 3 4 8 L L 2L 5L
0
36. Ix
2
2
y dy dx
L
My k
y
L
0
2
L
kh 2
h 2 2 x L x2 L2
Mx k
kh 8
dy dx
0
L
Multiple Integration
2
2
2 4y2 dx dy
y
0
22 4y2 y2 4y2 dy
122y
121 2 4y
2 4y2 2 ln 2y 2 4y2
2
2 3 2
0
2418 2 ln 4 18 12 1818 ln2 1
1
62 ln 4 32
22 12
2 92 52 ln2 ln 22 3 2 6 3
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Review Exercises for Chapter 13
(b) Surface area
40. (a) Graph of
403
1 fxx, y2 fyx, y2 dA
R
f x, y z
25 1 ex
2
y x 1000 2
y 2 1000
cos2
Using a symbolic computer program, you obtain surface area 4,540 sq. ft.
2
over region R z 50
R
50 x
4x2
2
42.
50
0
2
x 2 y 2 dz dy dx
0
25x2
25x2 y2
0
1
0
x2
r 2 2
2
0
0
5
(x2 y2) 2
2 4x2
44.
y
0
1 r3 dz dr d 2
2
1 dz dy dx y2 z2
0
2
0
2
0
5
0
2
2
0
0
arctan
5 0
2
0
4x 2
4x 2 y 2
0
xyz dz dy dx
0
2
48. V 2
2 sin
0
0
0
2
0
Mxz 2k
2
0
Mxy 2k
2
0
a
0
a
0
a
0
cr sin
32 3
0
d
5 arctan 5 2
r16 r 2 dr d
0
8 sin2 sin4 d
r dz dr d 2kc
2
0
0
0
2
0
0
r 2 sin dr d 2
0
rz dz dr d kc2
2
a
r 2 sin dz dr d 2kc
0 cr sin
2
5 arctan 5cos
1 3 3 1 1 sin cos sin 2 4 4 2 4
0 cr sin
d
0
0
8 4 2 sin 2
2
2
32 sin2 4 sin4 d 8
m 2k
2 sin
0
2
2
50.
2
r dz dr d 2
0
4 3
16r 2
16 3
sin d d
0
46.
r 5 dr d
2 sin d d d 1 2
0
2
2
2 kca3 3
2
sin d
0
a
0
1 r3 sin2 dr d kca4 2
a
r 3 sin2 dr d
0
29 2
1 2 4 kc a 4
2
2
Mxz kca4 8 3a m 2kca3 3 16
z
Mxy kc2a4 16 3ca m 2kca3 3 32
sin2 d
0
sin2 d
0
x0 y
2 kca3 3
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1 kca4 8
1 kc2a4 16
404
52.
Chapter 13
500 3
m
0
2
Mxy
3
0
0
2
0
0
by symmetry
60.
2
0
2
r25 r2 4rd dr
500 64 125 500 14 2 18 162 3 3 3 3 3
25r 2
1 4 9 2 r r 8 4
0
0
25r 2
5
3
0
1 3 9 r r dr d 2 2
3
0
zr dz dr d
2
zr dz dr d
2
0
0
3
d
0
81 8
3
2
1 8 25 r2 r dr d 0 2
0
81 4
0
a
2 sin2 2 sin d d d
0
4ka6 9
y2
z2 21 a
58.
0
a
1z2 a2
a
1z2 a2
1y 2z2 a2
1r2
2
0
r dz dr d
0
Since z 1 r 2 represents a paraboloid with vertex 0, 0, 1, this integral represents the volume of the solid below the paraboloid and above the semi-circle y 4 x2 in the xy-plane.
x2 y2 dV
Q
25r 2
54. Iz k
4
2
3
Mxy 81 1 1 m 4 162 8
z
Iz
0
500 3
r dz d dr
4
3
0
56. x
25r2
xy0
2
2
3
500 1 2 25 r23 2 2r 2 3 3
Multiple Integration
x 2 y 2 dx dy dz
1y 2z2 a2
8 a 15
x, y x y y x u, v u v u v 2u2v 2u2v 8uv y
x, y x y x y 1 1 1 0 62. u, v u v v u u u
5 4
y = 1x
x=1
v x u, y ⇒ u x, v xy u
3 2
x=5
Boundary in xy-plane
Boundary in uv-plane
x1
u1
x5
u5
xy 1
v1
xy 5
R
x dA 1 x2y2
5
1
5
1
x 1
v5
u 1 du dv 1 u2v u2 u
4 arctan v
5 1
1
5
1
5
1
1 du dv 1 v2
5
1
4 dv 1 v2
4 arctan 5
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y = 1x
4
5
Problem Solving for Chapter 13
405
Problem Solving for Chapter 13
4. A:
S
R
0
2
2
a b c 2
2
2
c a2
6. (a) V
b2 c2
10
9
523 r r2 r dr d 1.71 ft3 16 160 960
The distribution is not uniform. Less water in region of greater area. In one hour, the entire lawn receives
2
0
dA
10
0
125 r r2 r dr d 32.72 ft3. 16 160 12
R
b2 c2 AR c
2
0
(b) V
a2 b2 2 dA c2 c
1
2
B
1 ac
r2 r 1333 4.36 ft3 r dr d 16 160 960
5
4
0
a b fx , fy c c 1 fx2 fy2
2
1 2. z d ax by Plane c
2
0
2
0
8r 2
r dz dr d
2
4
0
8 42 5 3
22
2 sec
2 sin d d d
8 42 5 3
8. Volume 5 6 5 5 4 84 m3
10. Let v ln
1x , dv dxx.
1 ev , x ev, dx ev dv x
1
0
ln1 x dx
0
v ev dv
v ev dv
0
Let u v, u2 v, 2u du dv.
1
0
12. Essay
ln1 x dx
0
u eu 2u du 2 2
0
u2eu du 2 2
4
2
(PS #9)
14. The greater the angle between the given plane and the xyplane, the greater the surface area. Hence: z2 < z1 < z4 < z3
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C H A P T E R 13 Multiple Integration Section 13.1 Iterated Integrals and Area in the Plane
. . . . . . . . . . . . . 133
Section 13.2 Double Integrals and Volume . . . . . . . . . . . . . . . . . . . 137 Section 13.3 Change of Variables: Polar Coordinates . . . . . . . . . . . . . 143 Section 13.4 Center of Mass and Moments of Inertia . . . . . . . . . . . . . 146 Section 13.5 Surface Area
. . . . . . . . . . . . . . . . . . . . . . . . . . . 153
Section 13.6 Triple Integrals and Applications . . . . . . . . . . . . . . . . . 157 Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates . . . . 162 Section 13.8 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . 166 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
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C H A P T E R 13 Multiple Integration Section 13.1
Iterated Integrals and Area in the Plane
Solutions to Odd-Numbered Exercises
x
1.
0
4x 2
5.
x 2y dy
y
e
y
12 x y
4x 2
2 2
y
y ln x 1 dx y ln2 x x 2
x3
9.
0
y x
ye
0
3.
y dx y ln x x
1
2y
y ln 2y 0 y ln 2y
1
4x 2 x 4 2
1 y y ln2y ln2ey ln y2 y 2 2 2
ey
y x
x3
x3
dy xye
2y
3 x2 2
0
0
7.
x
1 2x y dy 2xy y 2 2
x
0
2
0
x3
ey x dy x 4 ex x 2ey x
x 21 ex x 2ex 2
0
2
u y, du dy, dv ey x dy, v xey x
x y dy dx
1 x2 dy dx
x 2 2y 2 1 dx dy
1
11.
0
1
13.
0
2
15.
1
2
1
0
0
x
4
2
0
1
1y 2
1
x y dx dy
0
0
1
2
0
4y2
0
2
21.
0
sin
2 dx dy 4 y 2
r dr d
2
0
0
1 4
2
0
dx
1 3
4
dy
0
1y 2
1
2 3 2
0
0
3
0
21 23 1 x
1 3 x 2xy 2 x 3
1 2 x xy 2
1
x1 x2 dx
64 4 8y 2 4 dy 3 3
2
19 6y 2 dy
1
3 19y 2y 4
3
0
r2 2
2x 4 y 2
sin
0
2
20 3
2 3
1
dy
0
2
1
x
1 1 1 1 2 1 y 2 y1 y 2 dy y y3 1 y 23 2 2 2 6 2 3
0
19.
2x 2 dx x 2 2x
0
0
0
1
0
0
2
1
dx
y1 x 2
17.
2
1
0
1
1 2 y 2
xy
d
4y2
2
dy
0
2 dy 2y
2
0
1 0
4
0
0
2
1 sin2 d 2
cos 2 d
1 2 1 cos 2 sin 2 4 2 4 2
2
0
2 1 32 8
133
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134
Chapter 13
y dy dx
1 dx dy xy
1 x
23.
0
1
1
y2
2 1 x
1
1 2
dx
0
1
25.
Multiple Integration
y ln x 1
1
dy
1
1
1 1 dx x2 2x
1
0
1
1 1 2 2
y 1y 0 dy 1
Diverges
8
27. A
0
8
dy dx
3
dx dy
8
x
0
dx
24
8
0 6
3
dy
8
3 dx 3x
0
0
0
y
8
0
0
8
0
3
y
0
3
A
3
8 dy 8y
3
24
4
0
0
2 x 2
29. A
4x2
2
0
2
dy dx
y
dx
8
y = 4 − x2
4
0
3
2
6
y
4x2
0
0
4
4
x2
dx
2
0
x3 3
4x
0
x
4y
4
dy
0
0
dy dx
0
dx
−2
−1
4
0
y2
4y
3
2
0
3
0
0
2 y 1
1 2
9 2
4y
3
0
3
y = (2 −
0
4
dy 2
dx dy
x
x )2
2
4y
dy
1
0
3
x 1
4
3 0
2
4 y dy
3
3 4 y
2 2y 4 y3 2 3
4 4x x dx
4
0
8 3
4
dx dy
y 2 4 y dy 2
2
2 y 2
y
4y
4
dx
0
Integration steps are similar to those above.
dx dy 2
y2
x
2 x 2
8 x2 4x xx 3 2 4
y
0
0
x 1
4
dy dx
1
2
0
2
1 1 2x x 2 x3 2 3 3
2 16 8 3 3
2 x x 2 dx
2
0
y=x+2
4 x x 2 dx
2
4
2
x2
2
2 x
(1, 3)
3
4x 2
y
4
33.
1
A
3
y
y = 4 − x2
1
2
2 4 y1 21 dy 4 y3 2 3
2 x2 1
4
4 y dy
4x2
1
1
dx dy
0
0
31. A
x
−1
0
4
0
1
16 3
4y
4
A
2
4
4
3 2
3
9 2
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3
4
8 3
Section 13.1
3
2x 3
5
0
0
3
3
3
y
0
0
3
dx 0
3
3
1
5
1
2
2
0
2
A
3
3y 2
2
3y dy 2
0
0
2
cos2 d
0
ab 2
2
1 cos 2 d
0
2
2 21 sin 2 ab
0
ab 4
a bb2 y2
b
0
dx dy
0
ab 4
y
5y 5 dy 5y y 2 2 4
5
a2 x 2 dx ab
dx
0
Therefore, A ab. Integration steps are similar to those above.
5y
2
a
b a
b aa2 x2
y
0
0
A 4
dy
2
0
a
dy dx
Therefore, A ab.
5y
0
dx dy
x
5
3y 2
2
5y
0
b aa2 x2
a
135
x a sin , dx a cos d
5 x dx
3 x 5x 2 x
5
2x dx 3
0
5x
dx
y
dy dx
0
5
2x 3
A 37. 4
5x
dy dx
35. A
Iterated Integrals and Area in the Plane
2 0
5
y = ba
b
a2 − x2
y x
a
4 3
y = 23 x
2
y=5−x
1 x 1
2
3
4
5
−1
4
39.
0
2
y
f x, y dx dy, 0 ≤ x ≤ y, 0 ≤ y ≤ 4
0
4
0
41.
4x2
2 0
f x, ydy dx, 0 ≤ y ≤ 4 x2, 2 ≤ x ≤ 2
4y 2
2
4
f x, y dy dx
dx dy
4y 2
0
x
y
y 3
3 1
2 1
−2
10
43.
1
2
3
1
f x, ydx dy, 0 ≤ x ≤ ln y, 1 ≤ y ≤ 10
ln 10
0
2
4
ln y
0
1 −1
x 1
x
−1
45.
1
1 x2
10
e
x
f x, y dy dx, x 2 ≤ y ≤ 1, 1 ≤ x ≤ 1
1
f x, ydy dx
y
0
y
y
f x, y dx dy
y 4
8 3
6 2
4 2 x 1
2
3
−2
−1
x 1
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2
136
Chapter 13
1
47.
0
2
Multiple Integration
2
dy dx
0
1
0
1
dx dy 2
49.
0
1y2
1y2
0
y
1
dx dy
1
1x2
dy dx
0
2
y
3
1
2 x
−1
1
1
x
1
2
51.
0
2
x
3
4
dy dx
0
4x
2
2
dy dx
0
0
4y
2
dx dy 4
53.
y
0
1
1
dy dx
x 2
0
2y
dx dy 1
0
y
y
3
2 2
1
1 x 1
2
3
4
x
−1
1
1
55.
0
3 y
x
1
dx dy
y2
0
x= 3 y
5 dy dx 12
x3
2
y
x = y2
2
1
(1, 1) x
1
2
57. The first integral arises using vertical representative rectangles. The second two integrals arise using horizontal representative rectangles.
5
0
50x 2
0
x
5
0
y
x 2y 2 dx dy
0
52
5
50y 2
y=
0
0
1 2 1 x 50 x 23 2 x5 dx 3 3
15625 24 5
x 2y 2 dx dy
y
5
x 2y 2 dy dx
1 5 y dy 3
52
5
15625 24
50 − x 2
(0, 5 2 ) 5
1 15625 15625 15625 50 y23 2 y2 dy 3 18 18 18
(5, 5) y=x x 5
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Section 13.2
2
59.
0
2
x
1
0
1
y
0
2
0
x
2
y sin x 2
0
1 0
0
dx
0
1 1 1 cos 1 1 1 cos 1 0.2298 2 2 2
4
65.
0
y
2 dx dy ln 52 2.590 x 1y 1
0
y
x2 32
4
x = y3
(8, 2)
2
x
2
1 1 26 27 1 9 9 9
x
1664 15.848 105
x 2y xy 2 dy dx
x 2 −2
(c) Both integrals equal 67520693 97.43
69.
dy
0
13
x232
0
y
2
1
sinx 2 dy dx
x 42y ⇔ x2 32y ⇔ y (b)
3 32
67. (a) x y3 ⇔ y x13
8
x2 2
12 13 23 1 y
1 x sin x 2 dx cos x 2 2
x3 3y 2 dy dx
0
0
0
2x
1 y3
0
x
0
1
0
1 y3 y 2 dy
2
x1 y3 dx dy
2
1 2
1
sin x 2 dx dy
63.
y
0
61.
2
x1 y3 dy dx
Double Integrals and Volume
4
6
8
x = 4 2y
2
4x2
exy dy dx 20.5648
71.
0
1cos
0
6r 2 cos dr d
0
15 2
73. An iterated integral is a double integral of a function of two variables. First integrate with respect to one variable while holding the other variable constant. Then integrate with respect to the second variable. 75. The region is a rectangle.
Section 13.2
77. True
Double Integrals and Volume
For Exercise 1–3, xi yi 1 and the midpoints of the squares are
12, 21 , 32, 12 , 52, 12 , 72, 12 , 12, 32 , 32, 32 , 52, 32 , 72, 32 .
y 4 3 2
1 x 1
2
1. f x, y x y 8
f x , y x y 1 2 3 4 2 3 4 5 24 i
i
i
i
i1
4
0
2
0
4
x y dy dx
0
xy
y2 2
2 0
4
dx
0
2x 2 dx x 2 2x
4 0
24
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3
4
137
138
Chapter 13
Multiple Integration
3. f x, y x 2 y 2 8
2
f x , y x y 4 i
i
i
i
i1
4
0
4
5.
0
10 26 50 10 18 34 58 52 4 4 4 4 4 4 4
2
4
x 2 y 2 dy dx
0
y3 3
x2y
0
2 0
4
dx
2x 2
0
8 2x3 8x dx 3 3 3
4 0
160 3
4
f x, ydy dx 32 31 28 23 31 30 27 22 28 27 24 19 23 22 19 14
0
400 Using the corner of the ith square furthest from the origin, you obtain 272.
2
7.
0
1
y
2
1 2x 2y dy dx
0
y 2xy y
1
2
dx
0
0
3
2
2
2 2x dx
0
1
2x x 2
2 x
1
0
2
3
8
6
9.
0
3
6
x y dx dy
y2
0
6
0
1 2 x xy 2
3
dy
y
y2
(3, 6) 6
9 5 3y y 2 dy 2 8
5 9 3 y y 2 y3 2 2 24
4
2
6 0
x 2
36
a2 x2
a
11.
x y dy dx
a a2 x2
a
a
1 xy y 2 2
a2 x2
a2 x2
4
6
y
dx
a
a
a
2xa2 x 2 dx
2 a2 x232 3
5
13.
0
3
3
xy dx dy
0
0
25 2
x y dy dx
y 5
1 2 xy 2
5
dx
0
4 3 2
3
x dx
1
0
x 1
−a
0
0
0
a
a
5
3
a
−a
254 x
3
2
0
2
3
4
5
225 4
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x
Section 13.2
2
15.
0
y
4
y 2 2 dx dy y2 x y
2
y 2 2 dx dy y2 x y
2
2
0
3
4y
4
17.
4y
dx
2
x
1
2
ln
5x 2
1
0
x 1
0
dx
12 ln 2x
2
5
0
ln
5 2 y
2y ln x dy dx
4
ln x y2
(1, 3)
3
4x2
dx
2
4x
1
ln x4 x22 4 x2 dx
x
4
0
3x4
5
x dy dx
0
4
1
25x 2
0
0
4
0
2
0
y dy dx 2
4
0
y2 4
25y2
3
dy
4y3
25 18
x= 4y 3
9y y 25 18 3 1
3
3
0
x 2
1
25
y
dx
4
0
3
dx 4
2
0
1 x 1
2
23.
0
y
2
4 x ydx dy
0
4x
0
2
4y
0
2y2
x2 xy 2
y3 y3 6 3
8 8 8 4 6 3
2
2
3
4
y
dy
y
0
y2 y2 dy 2
(4, 3)
1
9 y 2 dy
0
2
25 − y 2
x=
2
3
4
5 4
1 2 x 2
0
x dx dy
4y3
3
4
y
25y 2
3
x dy dx
21.
3
26 25
3
0
0
19.
2
2
1 5 ln 2 2
1
y=x
dx
ln
2x 2
4x
0
x=2
3
2x
lnx 2 y 2
0
y = 2x
4
4x2
1
2y ln x dx dy
x
y
y dy dx x2 y 2
2
1 2
1 2
2x
Double Integrals and Volume
2
1
y=x
0
x 1
2
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3
4
5
4
139
140
Chapter 13
23x4
6
25.
0
0
Multiple Integration
6
12 2x 3y dy dx 4
0
6
0
3 1 12y 2xy y 2 4 2
dx
5
0
3
y = − 23 x + 4
4
1 2 x 2x 6 dx 6
181 x
y
23x4
3 2 1
6
x 2 6x
x
0
1
−1
2
3
4
12
1
27.
y
0
1
1 xy dx dy
0
0
1
0
y
x 2y x 2
y
dy
0 1
y3 y dy 2
y=x x
y2 y4 2 8
29.
0
0
1
0
3 8
1 dy dx x 12 y 12
0
x 11 y 1 2
0
dx
0
1 1 dx x 12 x 1
0
4x2
2
31. 4
1
0
4 x 2 y 2 dy dx 8
0
1
33. V
0
1
0
x
2
35. V
xy dy dx
0
0
x
1 2 xy 2
18 x
1
0
0
1 2
1
x 2 dy dx
0
2
x3 dx
0
dx
32 3
4 0
4x3
3 2 0
y
2
x 2y
0
1 8
4
dx
4
0
y
4
1
y=x 3 2 1
x 1
x
−1
37. Divide the solid into two equal parts.
1
0
y=x
1 x 2 dy dx
0
1
2
y
x
V2
1
x
y1 x 2
0
1
dx
0
x 1
1
2
x1 x 2 dx
0
2 1 x 232 3
1 0
2 3
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2
3
4x 2 dx
1
5
6
Section 13.2
0
0
0
xy
0
2
4x 2
0
0
1 1 4 x 232 2x x3 3 6
2 0
2
4
x 2 y 2 dy dx
1 x 24 x 2 4 x 232 dx, 3
4
dx
16 cos2
0
x 2 sin
32 cos4 d 3
32 3
4 3 16
4 16
16 3
8
y
y 2
x 2 + y2 = 4
4 − x2
y=
1
1 x
−1
1 −1
x 1
2
4x2
2
43. V 4
0
0.5x1
2
4 x 2 y 2 dy dx 8
45. V
0
0
0
2 dy dx 1.2315 1 x2 y 2
47. f is a continuous function such that 0 ≤ f x, y ≤ 1 over a region R of area 1. Let f m, n the minimum value of f over R and f M, N the maximum value of f over R. Then
f m, n
dA ≤
R
Since
f x, y dA ≤ f M, N
R
dA.
R
dA 1 and 0 ≤ f m, n ≤ f M, N ≤ 1, we have 0 ≤ f m, n1 ≤
R
Therefore, 0 ≤
f x, y dA ≤ f M, N1 ≤ 1.
R
f x, y dA ≤ 1.
R
1
49.
12
0
12
ex 2 dx dy
y2
0
2x
1
ex 2 dy dx
0
51.
arccos y
0
sin x1 sin2 x dx dy
0
12
2xex 2 dx
0
12
0
2
e14
cos x
sin x1 sin2 x dy dx
0
2
1 sin2 x12 sin x cos x dx
0
e14 1 1
0
ex 2
0.221
12 23 1 sin
2
y
y
y = 2x 1
2
1 2
1
y = cos x
x 1 2
1
141
0
2
1 2 y 2
1 x4 x 2 2 x 2 dx 2
0
4x 2
2
41. V 4
x y dy dx
2
4x 2
2
39. V
Double Integrals and Volume
π 2
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π
x
x32
2
0
1 22 1 3
142
Chapter 13 1 8
53. Average
Multiple Integration
4
0
2
x dy dx
0
1 8
4
2x dx
0
8 x2
4 0
1 4
55. Average
2
57. See the definition on page 946.
1 4
2
0
2
x 2 y 2 dx dy
0
2
0
x3 xy 2 3
2 0
14 83 y 3 y 2
2
3
0
59. The value of
dy
1 4
2
0
8 3
f x, y dA would be kB.
R
1 1250
61. Average
1 1250
325
300
250
100x 0.6y 0.4 dx dy
200
325
100y 0.4
300
x1.6 1.6
63. f x, y ≥ 0 for all x, y and
250 200
5
f x, y dA
0
0
2
P0 ≤ x ≤ 2, 1 ≤ y ≤ 2
2
0
2
1
dy
325
128,844.1 1250
1 dy dx 10 1 dy dx 10
5
0 2
0
y 0.4 dy 103.0753
300
y1.4
1.4 325 300
25,645.24
1 dx 1 5 1 1 dx . 10 5
65. f x, y ≥ 0 for all x, y and
3
f x, y dA
0
3
0
1
P0 ≤ x ≤ 1, 4 ≤ y ≤ 6
0
6
3
1 9 x y dy dx 27
1 y2 9y xy 27 2 6
4
6 3
3
dx
1 9 x y dy dx 27
0
1
0
1 1 x x2 x dx 2 9 2 18
3 0
1
2 7 4 x dx . 27 27
67. Divide the base into six squares, and assume the height at the center of each square is the height of the entire square. Thus,
z
V 4 3 6 7 3 2100 2500m3.
7
(15, 15, 7) (5, 5, 3)
(15, 5, 6)
(5, 15, 2)
(25, 5, 4)
20 y
30
(25, 15, 3)
x
1
69.
0
2
6
sin x y dy dx
0
m 4, n 8
71.
4
2
y cos x dx dy
0
(a) 1.78435
(a) 11.0571
(b) 1.7879
(b) 11.0414
http://librosysolucionarios.net
m 4, n 8
8 2y 2 dy 3
Section 13.3
73. V 125
75. False
z
(4, 0, 16) 16
Matches d.
Change of Variables: Polar Coordinates
1
(4, 4, 16)
V8
0
5
(4, 4, 0)
1
0
1
f x dx
x
0
1
0
1
e t dt dx 2
1
0
t
t
1 2
et dt dx
x 1
1
e t dx dt 2
0
2
te t dt
0
1 2 et 2
Section 13.3
1 x 2 y 2 dx dy
0
y
5
x
1y 2
(0, 4, 0)
(4, 0, 0)
77. Average
143
1 0
x
1 1 e 1 1 e 2 2
1
Change of Variables: Polar Coordinates
1. Rectangular coordinates
3. Polar coordinates
5. R r, : 0 ≤ r ≤ 8, 0 ≤ ≤
7. R r, : 0 ≤ r ≤ 3 3 sin , 0 ≤ ≤ 2 Cardioid
2
9.
0
6
3r 2 sin dr d
0
2
0
2
r 3 sin
6 0
π 2
d
216 sin d
0
0
4
216 cos
2
11.
0
3
9 r 2 r dr d
2
2
0
0
31 9 r
3
2 3 2
2
0
2
π 2
d
2
5 3 5
0
55 6
0 1
2
13.
0
1sin
r dr d
2
2 1sin 0
0
0
2r
2
0
8
3 2 9 32 8
2
3
d
π 2
1 1 sin 2 d 2
1
2
1 1 sin cos cos 2 2
2
sin 2 8 sin2 0 1
1
0 1
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2
144
Chapter 13
y dx dy
x 2 y 23 2 dy dx
a2 y 2
a
15.
0
0
a
r 2 sin dr d
0
2xx 2
xy dy dx
2
0
2
2
x 2 y 2 dy dx
0
2
r 4 dr d
243 5
2
r3 cos sin dr d 4
d
2
8x 2
cos5 sin d
4
x 2 y 2 dy dx
4x 2
0
x y dy dx
2
2
0
0
4y 2
1 2
25.
1y 2
0
8 3
r cos r sin r dr d
0
2
0
y arctan dx dy x
2
1 2
4
0
2
29. V
1
r3 sin 2 dr d
0
2
5
r 2 dr d
0
0
2
0
2
2
cos sin r2 dr d
0
2
83 sin cos
0
16 3
π 2
y arctan dx dy x
(
1 , 2
1 2
2
( ( 2, 2)
r dr d
3 3 2 d 2 4
4
0
3 2 64
0 1
2
r cos r sin r dr d
0
31. V 2
0
2
1 2
0 3
1
0
y
2
1
4
0
27. V
4y 2
1
2 3
162 d 3
0
cos sin d
0
π 2
r 2 dr d
42 3
2
0
4
0
2
a3 3
4 cos6 6
22
0
0
243 10
0
0
2
0
0
23.
3
0
3
2
a3 cos
sin d
2 cos
22
x
a3 3
0
0
0
0
0
0
21.
2
9x 2
2
19.
0
0
3
17.
Multiple Integration
128 3
4 cos
0
2
0
1 8
2
sin 2 d
0
1 cos 2 16
2
0
1 8
250 3
16 r2 r dr d 2
2
0
1 sin 1 cos2 d
4 cos
3 16 r 1
2 3
0
128 cos3 cos 3 3
d
2
0
2 3
2
64 sin3 64 d
0
64 3 4 9
http://librosysolucionarios.net
Section 13.3
33. V
2
4
16 r 2 r dr d
a
0
2
3 16 r 1
2 3
4 a
0
d
Change of Variables: Polar Coordinates
1 16 a2 32 3
One-half the volume of the hemisphere is 64 3. 2 64 16 a23 2 3 3
16 a23 2 32 16 a2 322 3 3 a2 16 322 3 16 8 2
a 35. Total Volume V
3 3 4 4 2 2 24 2 2 2.4332
2
4
50e r2 4
4 0
0
2
0
0
2
25er 4 r dr d
2
d
50e4 1 d
0
1 e4 100 308.40524 Let c be the radius of the hole that is removed. 1 V 10
2
c
25er
r dr d
4
2
0
0
2
2
0
c
50e
r 2 4
0
d
50ec 4 1 d ⇒ 30.84052 1001 ec 2
4
2
0
⇒ ec
4
2
0.90183
c2 0.10333 4
c2 0.41331 c 0.6429 ⇒ diameter 2c 1.2858
37. A
6 cos
0
0
2
39.
0
r dr d
1cos
r dr d
0
3
0
2 sin 3
0
1 2 1 2
r dr d
18 cos d 9 2
0
41. 3
1 cos 2 d 9
0
2
1 sin 2 2
0
9
1 2 cos cos2 d
0 2
2
2 1 2 cos 1 cos
d 21 2 sin 21 21 sin 2 2
0
0
3 2
3
0
4 sin2 3 d 3
3
1 cos 6 d 3
0
43. Let R be a region bounded by the graphs of r g1 and r g2, and the lines a and b.
1 sin 6 6
3
0
3 2
45. r-simple regions have fixed bounds for .
-simple regions have fixed bounds for r.
When using polar coordinates to evaluate a double integral over R, R can be partitioned into small polar sectors.
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145
146
Chapter 13
Multiple Integration
47. You would need to insert a factor of r because of the r dr d nature of polar coordinate integrals. The plane regions would be sectors of circles.
2
49.
5
4
r1 r 3 sin dr d 56.051
0
Note: This integral equals
2
5
sin d
4
r1 r3 dr
0
51. Volume base height
53. False
z 16
8 12 300
Let f r, r 1 where R is the circular sector 0 ≤ r ≤ 6 and 0 ≤ ≤ . Then,
Answer (c)
r 1 dA > 0
but
R
6
4
4 6
x
55. (a) I 2
e x
2 dA
2 y2
2
4
0
y
er
22
r dr d 4
2
0
0
e r 22
0
2
d 4
d 2
0
(b) Therefore, I 2.
49x 2
7
57.
4000e0.01 x
2 y2
7 49x 2
2
dy dx
7
4000e0.01r r dr d 2
0
0
2
0
200,000e
7
0.01r 2
0
d
2 200,000 e0.49 1 400,000 1 e0.49 486,788
4
59. (a)
3x
2
23
4
61. A
43
4 csc
2 csc
3x
x
f dy dx
(2, 2) 2 ,2 3
(
1
1
2
0
0
2
0
4
xy dy dx
0
Center of Mass and Moments of Inertia
3
xy2 2
3 0
4
dx
0
2
r cos r sin r dr d
0
9 9x2 x dx 2 4
2
0
4
36
0
2
cos sin
0
r 3 dr d
2
4 cos sin d
0
4
sin2 2
2
(4, 4) 4 ,4 3
(
3
43 x
f r dr d
y=x
3x
4
4
f dy dx
y=
r r2
r22 r12 1
r2 r1 r r 2 2 2
4
3. m
f dy dx
2
Section 13.4 1. m
5
2
3
(c)
f dx dy
y3
2
(b)
y
y
2
0
http://librosysolucionarios.net
3
(
( x
4
5
r 1 0 for all r.
Section 13.4
a
5. (a)
m
b
0
ky dy dx
0
a
My
k dy dx kab
m
(b)
0
0
a
b
0
a
b
0
Mx
Center of Mass and Moments of Inertia
kx dy dx
0
Mx
0
My
0
b
ky 2 dy dx
kab3 3
kxy dy dx
ka2b2 4
b
0
x
My b2 a m kab 2
My ka2b24 a x m kab22 2
y
Mx kab22 b m kab 2
y
ka2
a2, b2
x, y
a
(c) m
x, y
b
0
a
kx dy dx k
0
a
0
b
Mx
0
ka2b2 4
kx 2 dy dx
ka3b 3
b
My
0
0
Mx kab33 2 b m kab22 3
a2, 23b
1 xb dx ka2b 2
kxy dy dx
0
a
My ka3b3 2 2 a x m ka b2 3 y
Mx ka2b24 b 2 m ka b2 2
x, y
7. (a)
kab2 2
0
a
ka2b 2
ky dy dx
0
a
kab2 2
b
23 a, b2
k m bh 2
y
y=
b x by symmetry 2
b2
Mx
0
2hxb
h
b
ky dy dx
0
2hx b y=−
2h xb b
ky dy dx
b2 0
x b
kbh2 kbh2 kbh2 12 12 6 y
x, y
2h (x − b) b
Mx kbh26 h m kbh2 3
b2, h3
—CONTINUED—
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147
148
Chapter 13
Multiple Integration
7. —CONTINUED—
b2
(b) m
0
2hxb
0
b2
Mx
0
0
0 2hxb
0
0
2hxb
kxy dy dx
kb2h2 12
2h xb b
b
kx dy dx
0
kx dy dx
b2 0
1 2 1 1 kb h kb2h kb2h 12 6 4
b2
0
2hxb
2h xb b
b
kxy dy dx
0
kxy dy dx
b2 0
1 2 2 5 1 kh b kh2b2 kh2b2 32 96 12
b2
0
kbh3 12
kb2 2
b2
My
ky 2 dy dx
b2 0
Mx kbh312 h m kbh26 2
2h xb b
b
kxy dy dx
y
Mx
2hxb
b
2
kx dy dx
0
2h xb b
kx2 dy dx
b2 0
11 7 1 3 kb h kb3h kb3h 32 96 48
x
My 7kb3h48 7 b m kb2h4 12
y
Mx kh2b212 h m kb2h4 3
9. (a) The x-coordinate changes by 5: x, y
a2 5, 2b
a 2b 5, (b) The x-coordinate changes by 5: x, y 2 3
a5
(c) m
5
a5
Mx
5
a5
My
5
x
kbh2 6
b2 0
My h 12 b m kbh26 2
ky dy dx
2h xb b
b
ky 2 dy dx
x
(c) m
2h xb b
b2 0
2hxb
b2
My
b
ky dy dx
b
0
1 25 2 kx dy dx k a 5 b kb 2 2
b
0
1 25 2 2 2 kxy dy dx k a 5 b kb 4 4
b
kx 2 dy dx
0
1 125 3 k a 5 b kb 3 3
My 2 a 15a 75 m 3 a 10
M b y x m 2
11. (a)
x 0 by symmetry m
a2k 2
yk dy dx
Mx 2a3k m 3
a2k 3
a
Mx y
a 0
a
(b) m Mx
a2 x 2
a 0 a
My
a2 x 2
a 0 a
2
a2 x 2
a2 x 2
a 0
2
a4k
16 3 24
k a y y 2 dy dx
a5k
15 32 120
kx a y y dy dx 0
My 0 m
y
Mx a 15 32 m 5 16 3
http://librosysolucionarios.net
4a
k a y y dy dx
x
2a3k 3
Section 13.4
x
4
13. m
0
32k 3
kxy dy dx
0
Mx
0
0
1 1x 2
kx
Mx
dy dx 32k
2y
0
x
My 32k m 1
y
Mx 256k m 21
k dy dx
1 0 1 1x 2
1
x
My
m
256k kxy 2 dy dx 21
0
4
x 0 by symmetry 1
x
4
15.
Center of Mass and Moments of Inertia
1 0
k 2
k ky dy dx 2 8
M 2 k 2 y x 2 m 8 k 4
3
32k 3
y
3 8 32k 7
2
1 1 + x2
y=
y 3
y=
x
2
x
−1
1
1 x 1
2
3
4
−1
17. y 0 by symmetry
16y 2
4
m
4 0
x
kx dx dy
8192k 15
4 0
L by symmetry 2
sin xL
L
m
0
16y 2
4
My
19. x
kx 2 dx dy
My 524,288k m 105
524,288k 105
15
8192k
sin xL
L
Mx
0
64 7
y
y
ky dy dx
0
ky 2 dy dx
0
Mx 4kL m 9
4
16
kL 9
y
x = 16 − y 2
8
2
4 x 4
y = sin π x L
1
8
−4
x
−8
21. m Mx
a2k 8
ky dA
π 2
4
0
R
My
L
L 2
kx dA
R
a
kr 2 sin dr d
0
4
0
ka3 2 2 6
a
kr 2 cos dr d
0
x
My ka32 m 6
y
Mx ka3 2 2 m 6
8
a2k
y=x
ka32 6
4a2 3 8
a2k
4a 2 2 3
http://librosysolucionarios.net
kL 4
r=a
a
0
4kL 9
149
150
Chapter 13
ex
2
23. m
0
0
2
k
1 e4 4
y 0 by symmetry
k ky 2 dy dx 1 e6 9
m
k 1 5e4 8
My
x
e
0
25.
x
e
0
My
ky dy dx
0
2
Mx
Multiple Integration
kxy dy dx
0
2 cos 3
6 0
kr dr d
k 3
kx dA
R
My k e4 5 m 8e4
k e4 1 2 e4 1 0.46
y
Mx k e6 1 m 9e6
k e4 1 9 e6 e2 0.45
e4 5
4 e6 1
4e4
6
k dA
R
x
4e4
6
2 cos 3
6 0
kr 2 cos dr d 1.17k
My 3 1.17k 1.12 m k
x π 2
y
π θ= 6
2
r = 2 cos 3θ y = e −x
1
0 1
π θ =−6
x 1
2
29. m a2
27. m bh
b
Ix
0
h
0
b
Iy
0
h
3 I bh y m 3 Iy m
b3h 3
x
Ix
a2 4
y2
dA
y 2 dA
R
Iy
3 1 h bh 3 1
2
3
2
x 2 dA
b b 3 3
I0 Ix Iy
3 h h 3 3
xy
4
a4 4
r3 cos2 dr d
a4 4
0
2
a
0
0
a4
r3 sin2 dr d
a
0
R
b2
bh
2
0
x2
dA
2
0
R
I0 Ix Iy xy
a 4 a 4 4 2
mI a4 1a 4
x
2
a 2
33. ky
R
Iy
Ix
b3h 3
x 2dy dx
0
x
31. m
bh3 3
y 2 dy dx
r3
sin2
0
a4 dr d 16
a
r3
cos2
0
a4 dr d 16
mI 16a 4a 4
2
b
mk
0
a
0
a 2
kab2 2
b
kab4 4
y3 dy dx
0 b
Iy k
0
y dy dx
0
Ix k
a
a4 a4 a4 16 16 8 x
a
a
x 2y ydy dx
0
I0 Ix Iy
ka3b2 6
3kab4 2kb2a3 12
m kakabb 26 a3 a3 33 a I b kab 4 b 2 y b m kab 2 2 2 2 x
Iy
3 2
x
4 2
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2
2
2
Section 13.4 35. kx
37. kxy
2
4x
2
mk
0
0
4x 2
Ix k
0
xy 2 dy dx
0 4x 2
2
Iy k
0
x3 dy dx
0
32k 3
Ix
0
Iy
0
2 23 3 3
8 mI 32k3 4k 3
6
Iy
x
kxy3 dy dx 16k
x
kx3 y dy dx
0
I0 Ix Iy
4 m 16k3 4k 3
4
x
26 3
32k 3
kxy dy dx
0
4
16k 3
x
0
4
I0 Ix Iy 16k
y
4
m
x dy dx 4k
0
2
x
Center of Mass and Moments of Inertia
512k 5
592k 5
x
3 48 4 15 32k 5 5 m 512k 5
y
mI 16k1 32k3 32
Iy
x
39. kx
x
1
m
0
x
0
kxy 2 dy dx
2
x
x
1
Iy
3k 20
x
1
Ix
kx dy dx
2
0
x
kx3 dy dx 2
I0 Ix Iy
3k 56
k 18
55k 504
x
m 18k 203k
30
y
mI 563k 203k
70
Iy
x
b
b
x a 2 dy dx 2k
0
b
2k
2k
b
0
b
x a 2b2 x2 dx
b
x 2b2 x 2 dx 2a
b
b
xb2 x2 dx a2
b
b2 x 2 dx
b4 a2b2 kb2 2 0
b 4a2 8 2 4
4
43. I
14
b2 x 2
b
41. I 2k
9
x
0
4
kx x 6 2 dy dx
0
9 x
kxx x 2 12x 36 dx k
2
92
24 72 72 52 x x 7 5
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4 0
42,752k 315
6
2
151
152
Chapter 13
a 2x 2
a
45. I
0
Multiple Integration
0
0
a
k 4
a
a
0
3
0
a2 x2
k 4 a y 4
a2 x 2
dx
0
dx
0
a4 4a3a2 x2 6a2 a2 x2 4a a2 x2 a2 x2 a4 2a2x2 x4 a4 dx
0
k 4
a
k a y dy dx
0
a4 4a3y 6a2 y2 4ay3 y4
0
k 4
a2 x 2
a
k a y y a 2 dy dx
7a4 8a 2x 2 x 4 8a3a2 x 2 4ax 2a 2 x 2 dx
k 8a2 3 x5 x a x x 4a3 xa2 x 2 a2 arcsin 7a4x x 2x 2 a2 a2 x 2 a4 arcsin 4 3 5 a 2 a
k 8 1 1 7 17 7a5 a5 a5 2a5 a5 a5k 4 3 5 4 16 15
a 0
49. x, y kxy.
47. x, y ky. y will increase
Both x and y will increase 51. Let x, y be a continuous density function on the planar lamina R. The movements of mass with respect to the x- and y-axes are
y x, y dA and My
Mx
R
x x, y dA.
R
If m is the mass of the lamina, then the center of mass is
x, y
m , m . My Mx
53. See the definition on page 968
L L 55. y , A bL, h 2 2
b
Iy
0
b
0
ya y
L
0
y
L 2
57. y
2L bL L .A ,h 3 2 3
b2
2
y L2 3 3
L
Iy 2
dy dx
0
L 0
dx
L3b 12
Iy L L3b12 L hA 2
L2 bL 3
ya
2 3
2Lxb
y 2L3
b2
y
0
b2
2 3
0
2L 3
dx
2L xb
L 2Lx 2L 27 b 3
b 2Lx 2L 2 L3x 3 27 8L b 3 L3b36
2L L 2 3 L b6 2
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dy dx
3 L
2
dx 3
4 b2 0
L3b 36
Section 13.5
Section 13.5
Surface Area
1. f x, y 2x 2y
3. f x, y 8 2x 2y
R triangle with vertices 0, 0, 2, 0, 0, 2
R x, y: x 2 y 2 ≤ 4
fx 2, fy 2
fx 2, fy 2
1 fx fy 3 2
2
S
0
1 fx 2 fy 2 3
2
2x
2
3 dy dx 3
0
x2 2
2 0
4x 2
2
2 x dx
S
2 4x 2
0
3 2x
6
2
2
3r dr d 12
0
0
y = 4 − x2
R
1
x
−1
y = −x + 2
1 −1
R
1
3 dy dx
y
y
2
Surface Area
y = − 4 − x2 x 1
2
5. f x, y 9 x 2
y
R square with vertices, 0, 0, 3, 0, 0, 3, 3, 3
3
fx 2x, fy 0
2
R
1 fx 2 fy 2 1 4x 2
3
S
0
3
1
3
1 4x 2 dy dx
0
3 1 4x 2 dx
x
1
34 2x 1 4x
3
3
y
R rectangle with vertices 0, 0, 0, 4, 3, 4, 3, 0
4
3 fx x1 2, fy 0 2
3
3
0
4
4 9x
2
0
4 9x
3
dx
dy dx
27 4 9x
3
3 2
0
1 94 x
0
4
R
2
1 fx 2 fy 2
4
4 9x
2
1
2
x 1
2
3
4
4 31 31 8 27
9. f x, y ln sec x
R x, y: 0 ≤ x ≤
y
, 0 ≤ y ≤ tan x 4
2
y = tan x
fx tan x, fy 0
1
R
1 fx 2 fy 2 1 tan2 x sec x
S
4
0
3
0 4 6 37 ln 6 37
ln 2x 1 4x 2
2
7. f x, y 2 x3 2
S
2
0
tan x
0
sec x dy dx
4
0
4
sec x tan x dx sec x
0
2 1
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π 4
π 2
x
153
154
Chapter 13
Multiple Integration
11. f x, y x 2 y 2
y
R x, y: 0 ≤ f x, y ≤ 1
x 2 + y2 = 1
1
0 ≤ x 2 y 2 ≤ 1, x 2 y 2 ≤ 1 x
1
x y fx , fy x 2 y 2 x 2 y 2
1 x
1 fx 2 fy 2
1x 2
1
S
1 1x 2
2 dy dx
2
x2 y2 2 2 2 y x y2
2
0
1
2 r dr d 2
0
13. f x, y a2 x 2 y 2
y
R x, y: x 2 y 2 ≤ b2, b < a
a
x y fx , fy a2 x 2 y 2 a2 x 2 y 2
b x
b
S
2
2
2
b b2 x 2
a2
a dy dx x2 y 2
2
0
b
0
a a2 r 2
15. z 24 3x 2y 16
3 2x12
12
8
0
b
x
a
−b
r dr d 2a a a2 b2
y
1 fx 2 fy 2 14
S
−b
x2 y2 a 1 2 a x 2 y 2 a2 x 2 y 2 a2 x 2 y 2
1 fx fy 2
x 2 + y 2 ≤ b2
b
14 dy dx 48 14
8
0 4 x 4
8
12
16
17. z 25 x 2 y 2 1 fx 2 fy 2
9x
3
S2
1 25 x
x2 2
y2
y2 25 x 2 y 2
5
2
2
3
0
0
5 25 r 2
y
R triangle with vertices 0, 0, 1, 0, 1, 1
1
S
0
x
0
5 4x 2 dy dx
y=x
1
1 27 5 5 12
x
−1 −2
19. f x, y 2y x 2 1 fx 2 fy 2 5 4x 2
1
−2 −1
r dr d 20
R
x 1
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x 2 + y2 = 9
2
25 x 2 y 2
5 dy dx 2 y 2 2 25 x 9x
3
2
y
1
2
Section 13.5 21. f x, y 4 x 2 y 2
Surface Area
23. f x, y 4 x 2 y 2
R x, y: 0 ≤ f x, y
R x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
0 ≤ 4 x 2 y 2, x 2 y 2 ≤ 4
fx 2x, fy 2y
fx 2x, fy 2y
1 fx 2 fy 2 1 4x 2 4y 2
1 fx 2 fy 2 1 4x 2 4y 2
4x 2
2
S
2 4x 2 2
1 4x 2 4y 2 dy dx
2
1 4r 2 r dr d
1
0
1 4x2 4y2 dy dx 1.8616
0
17 17 1
0
0
1
S
6
y
x 2 + y2 = 4
1
x
−1
1 −1
25. Surface area > 4 6 24.
27. f x, y ex R x, y: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
Matches (e)
fx ex, fy 0
z
1 fx 2 fy 2 1 e2x
10
1
S
1
0
1 e2x dy dx
0
1
5
5
y
1 e2x 2.0035
0
x
29. f x, y x3 3xy y3
31. f x, y ex sin y
R square with vertices 1, 1, 1, 1, 1, 1, 1, 1
fx ex sin y, fy ex cos y
fx 3x 2 3y 3x 2 y, fy 3x 3y 2 3 y 2 x
1 fx2 fy2 1 e2x sin2 y e2x cos2 y
S
1
1
1
1
1 9x 2 y2 9 y 2 x2 dy dx
S
4x2
2
4x2
33. f x, y exy R x, y: 0 ≤ x ≤ 4, 0 ≤ y ≤ 10 fx yexy, fy xexy 1 fx 2 fy 2 1 y 2e2xy x 2e2xy 1 e2xy x 2 y 2
4
S
0
1 e2x
2
10
1 e2xyx 2 y 2 dy dx
0
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1 e2x dy dx
155
156
Chapter 13
Multiple Integration 37. f x, y 1 x 2; fx
35. See the definition on page 972.
S
x 12 x 2
, fy 0
1 fx2 fy2 dA
R
1
x
16
0
0
1 1 x 2
1
16
0
502 x 2
50
39. (a) V
0
20 50
2
0
dx 161 x 21 2
1 0
xy xy dy dx 20 100 5
0
50
x
1 x 2
dy dx
x2
x x 502 x 2 502 x 2 502 x 2 dy 200 5 10
10 x 50 x 2 502 arcsin
1 x 25 x4 x3 502 x 23 2 250x x2 50 4 800 15 30
50 0
30,415.74 ft3 xy 100
(b) z 20
1 fx 2 fy 2
S
41. (a) V
1 100 1 100
502 x 2
50
0
1 100y 2
2
1002 x 2 y 2 x2 2 100 100
1002 x 2 y 2 dy dx
0
2
0
50
1002 r 2 r dr d 2081.53 ft2
0
y
f x, y
24
R
8
20
625
x2
y2
dA
where R is the region in the first quadrant
R
8
2
4
625 r 2 r dr d
x
4
2
0
R 8
25
0
4
4
25
2 625 r 23 2 3
4
d
8 0 609 609 2 3
812 609 cm3 (b) A
16 12
1 fx 2 fy 2 dA 8
R
8
R
R
25 dA 8 625 x 2 y 2
lim 200 625 r 2 b→25
1
b 4
2
0
25
4
x2 y2 dA 625 x 2 y 2 625 x 2 y 2 25
625 r 2
r dr d
100 609 cm2 2
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8
12
16
20
24
16
Section 13.6
Section 13.6
3
1.
2
0
Triple Integrals and Applications
1
0
3
x y z dx dy dx
0
2
0
2
0
1
x
0
xy
0
1
x dz dy dx
0
4
1
1
0
0
1
1
2
0
1
zex
2
0
1
2
4
0
0
1x
2zex y
4
7.
1
4
0
x cos y dz dy dx
0
2
4
0
4x
0
0
4x 2
0
0
4
1
0
13.
0
0
dz
2 0
0
a
a2 x 2
a2 x 2y 2
19. 8
0
0
4x 2
2
0
0
z2 2
dy dx
2
2
4
0
0
4
1
15 1 1 2 e
x1 xcos y dy dx
0
4
dx
x1 xdx
x2 x3
3 4
2
8
0
0
64 40 3 3
128 15
x sin y ln z
4
2
dy dx
1
0
x 2 ln4cos y
4x2
0
2
dx
0
x 2 ln 4 1 cos 4 x 2 dx 2.44167
3
9x2
3
9x2
2
2
9x y
dz dy dx
0
4y 2
x dx dy
2
2
18
1 10
x3 dy dx 2
1
z1 e1 dz 1 e1
1x
2 0
1 2
0
0
15.
2
dz dx dy
0
3
2
1
dz dy dx
x
dz 3z z2
2zxex dx dz
0
4y 2
2
1
4xy
2
17.
1
x4 x5 dx 2 10
4
0
4x
4
2
4
0
4x
0
x cos yz
4x 2
x 2 sin y dz dy dx z
dx
dx dz
x1 xsin y
2
x dz dy dx
0
2
2
11.
x2
4x 2
2
0
1
x
0
0
4
x
x 2y 2 2
0
9.
1 1 y y 2 yz 2 2
dy dx
2
3
0
0
2zex dy dx dz
x 2y dy dx
x
dy dx
0
1 y z dy dz 2
x
0
5.
xz
0
1
0
1
xy
x
0
1 2 x xy xz 2
0
3
3.
Triple Integrals and Applications
2
4 y 22 dy
0
a
dz dy dx 8
0
0
a2 x 2
a y x
256 15
a2 x2
y a2 x 2 y 2 a2 x 2 arcsin
2
a2 x 2 y 2 dy dx
2
0
4
0
0
a
4
2
8 1 16 8y 2 y 4 dy 16y y3 y5 3 5
a
0
1 a2 x 2 dx 2 a2x x3 3
a 0
4 a3 3
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2
0
dx
157
158
Chapter 13
4x 2
2
21.
0
0
Multiple Integration
4x 2
2
dz dy dx
2
4 x 22 dx
0
0
0
8 1 16 8x 2 x 4 dx 16x x3 x5 3 5
23. Plane: 3x 6y 4z 12
2
0
256 15
25. Top cylinder: y 2 z2 1 Side plane: x y
z 3
z
1 2
y
3
4 x
1
x
3
0
(124z) 3
0
1
y
(124z3x) 6
dy dx dz
1
0
0
x
0
1y 2
dz dy dx
0
27. Q x, y, z: 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 3
3
xyz dV
0
Q
1
0
1
3
xyz dx dy dz
y
0
0
1
0
1
0
1
0
1
1
0
y
x
xyz dy dx dz
3
0
3
0
1
y
x
0
1
R
xyz dx dz dy
x
y
1
x
xyz dy dz dx
0 3
xyz dz dx dy
0 3
0
xyz dz dy dx
9 16
29. Q x, y, z: x2 y2 ≤ 9, 0 ≤ z ≤ 4
4
xyz dV
0
Q
4
0
3
3
xyz dy dx dz
9y2
xyz dx dy dz
3 9y2 4
4
9y2
3
3
x
xyz dx dz dy
9y2
9y2
4
3 0 3
5
9x2
4
xyz dz dx dy
3 9y2 0 3
z
3 9x2
3 0 3
3
y=x
1
0
9x2
xyz dy dz dx
9x2
9x2
4
3 9x2 0
xyz dz dy dx 0
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4
y
Section 13.6
dz dy dx
0
x dz dy dx
6
31.
4(2x 3)
mk
0
4(2x 3)
Myz k
0
2(y 2)(x 3)
0
8k
6
Triple Integrals and Applications
0
2(y 2)(x 3)
0
12k x
Myz 12k 3 m 8k 2
4
33.
4
4x
0
0
4
4k
0
4
0
0
Mxy k
0
4
2k
0
b
x4 x dy dx
b
0
0
b
xz dz dy dx k
0
0
4
0
4 x2 x dy dx 2
128k 16x 8x 2 x3 dx 3
b
0
b
b
0
b
0
z
3kh2 r2 4kh2 3r 2
r 2x 2
0
kb6 8
b
0
Mxy kb6 8 b 5 m kb 4 2
h
z dz dy dx
h x 2y 2 r
0
r
xyz dz dy dx
0
z
x y 0 by symmetry
kb6 6
Mxz kb6 6 2b 5 m kb 4 3
1 m kr 2h 3
0
xy 2 dz dy dx
b
y
39. y will be greater than 0, whereas x and z will be unchanged.
r
kb6 6
Myz kb6 6 2b 5 x m kb 4 3
37. x will be greater than 2, whereas y and z will be unchanged.
Mxy 4k
x 2y dz dy dx
0
Mxy k
0
r 2x 2
r 2 x 2 y 2 dy dx
0
r
r 2 x 23 2 dx
0
kr 2h2 4 Mxy kr 2h2 4 3h m kr 2h 3 4
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kb5 4
b
Mxz k
0
xy dz dy dx
0
Myz k
Mxy 1 z m
41.
b
mk
b
4
0
35.
0
4x
4
128k 4x x 2 dx 3 4
0
4
x dz dy dx k
mk
159
160
43.
Chapter 13
m
Multiple Integration
128k 3
x y 0 by symmetry z 42 x 2 y 2
42 x 2
4
Mxy 4k
0
0
2k
0
2
1024k 3
4
0
42 x2
0
dx
4k 3
4
4 x 23 2 dx 2
0
let x 4 sin
cos4 d
by Wallis’s Formula
0
0
Myz k
0
0
Mxz k
0
0
0
x dz dy dx 1000k
4 x 8
4
(5 12)y
12
20
16
y dz dy dx 1200k
(5 12)y
z dz dy dx 250k
0
x
Myz 1000k 5 m 200k
y
Mxz 1200k 6 m 200k
z
Mxy 250k 5 m 200k 4
a
a
0
a
a
0
0
Ix Iy Iz
a
a
a 0
a
dz ka
0
0
2ka5 by symmetry 3
y 2 z2xyz dx dy dz
0
0
0
ka2 2
a
y 4z y2z3 4 2
Ix Iy Iz
a 0
dz
ka4 8
y 2 z2 dy dz
0
1 1 3 1 2 a az dz ka a3z az3 3 3 3
a
(b) Ix k
0
a
y 2 z2 dx dy dz ka
1 3 y z2y 3
ka
a
47. (a) Ix k
0
y = − 35 x + 12
8
0
(3 5)x12
20
16 12
0
0
Mxy k
dz dy dx 200k
(5 12)y
(3 5)x12
20
20
(5 12)y
0
(3 5)x12
20
3
y
(3 5)x12
mk
3
128k 2
5 y 12 20
1 16y x 2y y3 3
0
Mxy 64k m 1
45. f x, y
42 x 2 y 2 dy dx 2k
0
64k z
z dz dy dx
0
42 x 2
4
42 x 2y 2
ka2 2
a
0
0
2ka 3
a
y3z yz3 dy dz
0
a
0
5
a
a2z 2z3 dz
ka8 a2z 4
2 2
2z4 4
a 0
ka8 by symmetry 8
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ka8 8
Section 13.6
4
4
0
0
4
4
0
k
0
0
4
0
4
4
0
4
0
4
0
y3 4 x 3
4 0
4
4
0
4
0
4
0
0
4
0
4x
4
0
0
4 0
a
a2 x 2
2k 3
L 2
k
L 2
L 2
2
L 2
4 0
1024k 3
x 2y y34 x dx
0
4
8x 2 644 x dx
dx k
0
L 2 a a2 x 2
4
yx 2 y 2 dz dy dx k
2 3
0
1 x 2y4 x y4 x3 dy dx 3
4 1 32 8x 4x 2 x3 dx 8k 32x 4x 2 x3 x4 3 4 0
L 2
8 644 x 4 x3 dx 3
4
4
0
2048k 3 4
x 2y 2 y 4 4 x 2 4
0
51. Ixy k
0
0
4
k
8k
4
yx 2 z2 dz dy dx k
Iz k
0
4
64 4 x dx 256k 3
1 4 1 1 4x 2 x3 4 x3 dx 8k x3 x4 4 x4 3 3 4 12
8k
4
0
512k 3
1 y34 x y4 x3 dy dx 3
0
dx k
4x
Iy k
0
4
y y 2 z2 dz dy dx k
2 k 324 x2 4 x4 3 4
4x 2
0
0
0
0
dx k
y4 y2 4 x 4 x3 4 6
4
4
x 2 y 24 x dy dx
0
4x
(b) Ix k
0
4
x 2 y 2 dz dy dx k
x 2y
4
1 x 24 x 4 x3 dy dx 3
4x
4
k
0
0
k
0
4
0
Iz k
4
4
x 2 z2 dz dy dx k
1 4 1 1 4x2 x3 4 x3 dx 4k x3 x 4 4 x4 3 3 4 12
0
4
64 4 4 x 4 x3 dx 3 3
256k
0
0
4k
0
4
4x
Iy k
0
0
4
dx k
0
1 3 y 24 x 4 x dy dx 3
0
4
32 1 4 x2 4 x4 3 3
4
4
y 2 z2 dz dy dx k
y3 y 4 x 4 x3 3 3
k
4x
49. (a) Ix k
Triple Integrals and Applications
L 2
z2 dz dx dy k
a
L 2 a
4 0
2048k 3
2 2 a x2 a2 x 2 dx dy 3
x 1 x a2 x a2 x 2 a2 arcsin x2x 2 a2 x 2 a2 a4 arcsin 2 a 8 a a4 a4 a4Lk dy 4 16 4
Since m a2Lk, Ixy ma2 4. —CONTINUED—
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a a
dy
161
162
Chapter 13
Multiple Integration
51. —CONTINUED—
L2
Ixz k
a2 x 2
a
L2 a a2 x 2 L2
2k
L2
a
a2 x 2
L2 a a2 x 2
a
L2 a
y2 x xa2 x 2 a2 arcsin 2 a
L2
L2
Iyz k
y 2 dz dx dy 2k
a a
L2
L2 a
2
Iy Ixy Iyz
ma2 ma2 ma2 4 4 2
Iz Ixz Iyz
mL2 ma2 m 3a2 L2 12 4 12
1
L2
a a
dy
ka4 4
L2
dy
L2
ka4L ma2 4 4
2
ma mL m 3a2 L2 4 12 12
2ka2 L3 1 mL2 3 8 12
y 2 dy
x 2a2 x 2 dx dy
1 x x2x 2 a2a2 x 2 a4 arcsin a L2 8
Ix Ixy Ixz
1
dy ka2
a
L2
2k
53.
L2
x 2 dz dx dy 2k
y 2a2 x 2 dx dy
1x
x 2 y 2x 2 y 2 z2 dz dy dx
1 1 0
55. See the definition, page 978. See Theorem 13.4, page 979.
57. (a) The annular solid on the right has the greater density. (b) The annular solid on the right has the greater movement of inertia. (c) The solid on the left will reach the bottom first. The solid on the right has a greater resistance to rotational motion.
Section 13.7
4
1.
0
2
0
Triple Integrals in Cylindrical and Spherical Coordinates
2
4
r cos dr d dz
0
0
4
0
2
3.
2 cos2
0
0
2
0
2 cos r2
2
d dz
0
2
4
2 cos d dz
0
0
4r2
r sin dz dr d
2
0
0
2 cos2
2 sin
2
r4 r 2sin dr d
2
4
0
0
0
z
0
2 sin d d d
0
4
7.
cos
1 3
2
0
0
2
8 cos4 4 cos8 sin d
2
0
4
0
cos3 sin d d
2dz 8
0
0
0
5.
4
dz
1 12
2r
2
2 cos2
r4 sin 4
8 cos5 4 cos9 5 9
2
cos4
0
2
rer d dr dz e4 3
0
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4
0
d
d
0
2
0
8
52 45
Section 13.7
2
9.
0
3
r
2
e
0
r dz dr d
2
3
0
0
2
2
11.
2
2
2 sin d d d
0
2
2
4 sec
0
0
2
0
2
2
6
0
2
sin d d
cos
0
2
323 3
z 4
2
6
d
d
4
4
x
y
0
643 3
2
a cos
2
2
cot csc
arctan12 0
3 sin2 cos d d d 0
3 sin2 cos d d d 0
2
0
0
0
a cos
ra2 r 2 dr d
0
a2 r2
r dz dr d
2
0
4 2 2a 3 a3 3 4 3 2 3 9
ra2 r 2 dr d
a cos
0
2a3 3 4 9
2
kr 29 r cos 2r sin dr d
0
2
k 3r 3
0
2
krr dz dr d
0
r4 r4 cos sin 4 2
0
k 24 4 cos 8 sin d
0
0
2a3 cos3 cos 3 3
d
1 sin3 d
2
9r cos 2r sin
2
0
0
1 a2 r 232 3
0
0
0
2
21. m
0
a cos
0
2
4 1 1 sin3 d a3 cos sin2 2 3 3
a cos
0
r dz dr d 4
0
a2 r2
0
4 a3 3
2a3 3
r 2 cos dz dr d 0
a sec
0
3 sin2 cos d d d
x
0
2a cos
0
2
64 3
y
3
a
4
2
64 3
3
1 2
a a2 r2
a
0
19. V 2
1
0
2
17. V 4
2
r2
arctan12
0
(b)
0
r 2 cos dz dr d 0
2
15. (a)
d
4
0
0
(b)
3
3
1 e9 4
13. (a)
z
1 1 e9d 2
4
6
0
2
1 2 er 2
0
rer dr d
0
0
Triple Integrals in Cylindrical and Spherical Coordinates
k 24 4 sin 8 cos k 48 8 8 48k
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2
0
2 0
d
163
164
Chapter 13
23. z h
h h x2 y2 r0 r r0 r0
2
V4
0
Multiple Integration
4h r0
0
2
r0
r0r r 2 dr d
1 r02h 2 3
2
0
0
r3 dz dr d
r0
0
r0r3 r 4 dr d
2
0
2
b
r3 dz dr d
0
2
b
r3 dr d
0
a
2
kh
h
a
4kh
1 k r04h 10
b4 a4 d
0
Since the mass of the core is m kV k3 r02h from Exercise 23, we have k 3m r02h. Thus,
kb4 a4h 2
1 Iz k r04h 10
kb2 a2b2 a2h 2
1
1 3m r04h 10 r02h
3 mr 2 10 0
2
0
1 ma2 b2 2
31. V
0
r 2 z dz dr d
0
Mxy k r03h230 h m k r03h6 5
Iz 4k
0
29. m kb2h a2h khb2 a2
0
4kh r05 r0 20
h(r0 r)r0
r0
1 k r03 h2 30
0
2
4kh r0
h(r0 r)r0
r0
2
0
z
r 2 dz dr d
0
Mxy 4k
27. Iz 4k
0
1 k r03h 6
r0 d 6
4h r03 r0 6
h(r0 r)r0
r0
0
3
0
2
m 4k
0
2
4h r0
r dz dr d
0
kx 2 y 2 kr x y 0 by symmetry
hr0 rr0
r0
0
25.
4 sin
2 sin d d d 16 2
33. m 8k
0
2
0
2ka4
2
0
0
2
2
0
ka4
a
3 sin d d d sin d d
0
2
sin d
0
ka4
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2
ka4cos
0
Section 13.7
35.
Triple Integrals in Cylindrical and Spherical Coordinates
2 m kr 3 3
37. Iz 4k
x y 0 by symmetry
2
Mxy 4k
2
0
1 kr 4 2
kr 4 4
0
2
2
0
3 cos sin d d d sin 2 d d
1 k r 4 cos 2 8 z
2
2
cos5 sin3 d d
0
2
4
cos5 1 cos2 sin d
k 192
2
1
6
1 cos8 8
41.
1
g2
g1
h2r cos , r sin
h1r cos , r sin
zz (b) 0: sphere of radius 0
43. (a) r r0: right circular cylinder about z-axis
0: plane parallel to z-axis
0: plane parallel to z-axis
z z0: plane parallel to xy-plane
0: cone
0
4
f r cos , r sin , zr dz dr d
y x
tan
zz
2
x2 y 2 r 2
y r sin
a
2
Mxy kr 44 3r m 2kr33 8
39. x r cos
45. 16
4
4 sin3 d d d
0
5 k 6 cos
1 k r 4 4
0
0
cos
sin 2 d
0
2
2
2 k 5
0
2
4
4 k 5
r
0
2
a2 x 2
0
a2 x 2y 2
0
a2 x2 y 2z2
0
dw dz dy dx
a2 x 2
a
16
0
0
a
0
0
2
16
0
2
0
4
a
0
2
0
a4
2
0
a2 r 2
a2 r 2 z2 dzr dr d
0
a
0
8
a2 x 2 y 2 z2 dz dy dx
0
2
16
a2 x 2y 2
1 z za2 r 2 z2 a2 r 2 arcsin 2 a2 r 2
2 a r 2r dr d 2 a2r 2 r 4 2 4
d
a 0
d
a4 2 2
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a2 r2
0
r dr d
165
166
Chapter 13
Multiple Integration
Section 13.8
Change of Variables: Jacobians
1 1. x u v 2
3. x u v 2 yuv
1 y u v 2
x y y x 11 12v 1 2v u v u v
y x 1 x y u v u v 2
12 1212
1 2
5. x u cos v sin y u sin v cos x y y x cos2 sin2 1 u v u v 7. x eu sin v y eu cos v x y y x eu sin veu sin v eu cos veu cos v e2u u v u v 9. x 3u 2v y 3v v
y 3
u
x 2v x 2 y3 3 3
x, y
u, v
0, 0
0, 0
3, 0
1, 0
2, 3
0, 1
v
(0, 1)
1
(1, 0) u
1
2y x 3 9
1 11. x u v 2 1 y u v 2 y x 1 x y u v u v 2
21 1212 21
1
4x 2 y 2 dA
1
1 1
R
1
12 dv du
14 u v
1 u v2 4
2
4
1
1 1
1
u2 v 2 dv du
1
2 u2
u3 u 1 du 2 3 3 3
13. x u v
R
1
8 3
4
y x x y 10 11 1 u v u v
3
yx y dA
1
v
yu
0
4
0
3
2
3
uv1 dv du
0
8u du 36
1 u 1
2
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3
4
Section 13.8
15.
xy2
e
Change of Variables: Jacobians
dA
y = x1
y
R
4
1 4 x R: y , y 2x, y , y 4 x x
y = 2x 3
y = 4x
x u x, y y u, v u
1 v12 x v 2 u32 y 1 v12 v 2 u12
R 1 x
1 1 2 u12v12 1 1 1 1 4 u u 2u 1 u12 2 v12
Transformed Region: y
y = 41 x
2
y x vu, y uv ⇒ u , v xy x
1
2
3
4
3
4
v
1 ⇒ yx 1 ⇒ v 1 x
3
S 2
4 y ⇒ yx 4 ⇒ v 4 x y 2x ⇒ y
u
y 2 ⇒ u2 x
1
y 1 1 x ⇒ ⇒ u 4 x 4 4
2
exy2 dA
4
14 1
R
2u1 dv du e u 2
ev2
17. u x y 4,
vxy0
u x y 8,
vxy4
1 x u v 2
1 y u v 2
2
du
1
14
e2 e12ln u
1 e2 e12 du u 14
2
e2 e12 ln 2 ln
14
1 e12 e2ln 8 0.9798 4
y
6
x−y=0
x+y=8
4
2
1 x, y u, v 2
v2 4
x−y=4
x+y=4
x
8
x yexy dA
4
R
1 2
2
4
uev
0
4
6
12 dv du
8
ue4 1 du
4
19. u x 4y 0,
vxy0
u x 4y 5,
vxy5
1 x u 4v, 5
1 y u v 5
14 u e 2
4
1
8 4
12e4 1
y
x−y=0 2
x + 4y = 5
y x 1 x y u v u v 5
1
51 1545 15
5
x yx 4y dA
5
5
uv
0
R
0
0
x
−1
3 −1
x + 4y = 0 −2
1 du dv 5
1 2 32 u v 5 3
5 0
dv
2 3 5 23v
5
32
0
100 9
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4
x−y=5
167
168
Chapter 13
Multiple Integration
1 1 21. u x y, v x y, x u v, y u v 2 2 x y y x 1 u v u v 2
a
x y dA
u
u
0
R
12 dv du
a
u
u u du
0
y
5 u 2
a
52
0
2 a52 5
y
v=u a a
x+y=a x 2a
a
23.
x −a
v = −u
x2 y 2 2 1, x au, y bv a2 b
(a)
x2 y 2 21 a2 b
u2 v 2 1 v
au2 bv2 2 1 a2 b
y
1
u2 v 2 1
b
S R
u x
a
(b)
x, y x y y x u, v u v u v ab 00 ab
(c) A
ab dS
S
ab12 ab 25. Jacobian
x y y x x, y u, v u v u v
27. x u1 v, y uv1 w, z uvw
1v x, y, z v1 w u, v, w vw
u 0 u1 w uv uw uv
1 v u2v1 w u2vw u uv21 w uv2w 1 vu2v uuv 2 u2v
29. x sin cos , y sin sin , z cos
sin cos x, y, z sin sin , , cos
sin sin sin cos 0
cos cos cos sin sin
cos
sin cos sin sin cos cos sin
sin2 cos2 sin2 sin2 2
2
2
2
cos
2 sin cos sin2 cos2 sin
sin2 cos2 sin2 2 sin cos2 2 sin3
2 sin cos2 sin2 2 sin
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1
Review Exercises for Chapter 13
169
Review Exercises for Chapter 13
x2
1.
1
1
3.
0
4x dy dx
dy dx
3x3
1
A2
5 0
25y 2
4
5
1
2
5
1 0
3 2
3
5
4 25y 2
dx dy
1 0
29 6
25y 2
dx dy
25y 2
4
25 x2 dx x25 x2 25 arcsin
1 0
x 5
3 5
25 3 12 25 arcsin 67.36 2 5
4 3
dx dy
x1
2
0
5 x2 x 2
2
5
36
3
4 x1 x2 dx 1 x232 3
0
1 14y 2
1 14y 2
0
3
dy dx 4
A4
dy dx 2
0
12
4
dx dy
25y 2
x1x 2
11. A 4
0
3 3 3y dy 3y y 2 2 0
25x 2
3
13. A
1
dy dx
3
dx dy
0
5 25x 2
0
43x
0
dx dy
25x 2
4x 2 5x 1 dx
33y
0
1
0
1
dx
33y
0
1x
4 4x9 x 2 dx 9 x 232 3
1
0
3
3xy y 2
3
0
A
9.
9x2
0
0
x31 ln x 2 x x x3 x3 ln x 2
0
3
1
1
3x 2y dy dx
0
0
7.
x2
1x
3
5.
x ln y dy xy1 ln y
x1
2
dy dx 2
1
x3
dy dx
0
2
y3
1
y 2 1
dx dy
9 2 y
15. Both integrations are over the common region R shown in the figure. Analytically,
22y 2
1
0
2y
2
0
0
4
0
4
0
(2, 1)
x 24
x2 y 4 dy dx
2
x y dy dx
x2 4
dx
2
4 4 5 4 1 2 2 3 3 3 3 3
19. Volume baseheight 27 9 3 2 2 Matches (c)
z 6 4 2 y
4
(3, 3)
2
(3, 0)
4 x
1 5 4 3 x x 8x 10 3
3
2
0
1 4 x 4x2 8 dx 2 4 0
3296 15
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8 − x2 x
1 −1
0
1 x2y y 2 4y 2
y=1
1
8x 2
0
2
4
x y dy dx
y = 12 x
2
4 4 2 3 3
22
x2
0
17. V
x y dx dy
170
Chapter 13
21.
0
Multiple Integration
kxyexy dy dx
0
kxexy y 1
0
0
dx
0
kxex dx kx 1ex
0
k
Therefore, k 1.
1
P
1
0
xyexy dy dx 0.070
0
23. True
h
27.
25. True
0
x
x 2 y 2 dy dx
0
4
h sec
0
2
h
29. V 4
0
h3 3
4
sec3 d
0
1z 2
0
r dr d dz
1
1 z2 1 d dz
2
0
0
4
0
h3
2 ln 2 1 6
31. (a) x2 y22 9x2 y2 r 2 9cos2 sin2 9 cos 2 r 3cos 2
h
z2
h3 sec tan ln sec tan 6
r 22 9r 2 cos2 r 2 sin2
2
h
r2 dr d
0
dz
4
0
13 z
h
3
0
h3 3
−6
6
−4
4
(b) A 4
0
(c) V 4
1
2x
0
2x
1
3
2x
1
3
xy2 dy dx
2x
My k
0
k 4
2x
Mx k
0
xy dy dx
4
2x
3
x2y dy dx
1
1
16k 55
Mx k
8k 45
My k
0
1
0
3cos 2
2x
3
x2 y2dy dx
17k 30
2x
2x
3
yx2 y2dy dx
2x
2x
3
xx2 y2dy dx
x
My 32 m 45
x
My 936 m 1309
y
Mx 64 m 55
y
Mx 784 m 663
y
y = 2x 2
1
y = 2x 3 x 1
9 r 2 r dr d 20.392
0
2x
(b) m k
0
r dr d 9
0
0
33. (a) m k
3cos 2
2
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392k 585 156k 385
Review Exercises for Chapter 13
a
y2 x, ydA
35. Ix
0
R
0
a
x2 x, ydA
Iy
b
37. S
0
a
x, ydA
16x2
4
0
4
1 4x2 4y2 dy dx
0
2
4
0
1 4r 2 r dr d
0
b
0
R
1 fx2 fy2 dA
4
1 kx3 dy dx kba4 4
1 1 ka2b 2 I0 Ix Iy kb3a2 kba 4 2b 3a2 6 4 12 m
R
b
0
R
1 kxy2 dy dx kb3a2 6
0
1 kx dy dx kba2 2
x
kba a a 2 m 14 12kba 2 2
y
kb a b b 3 mI 16 12kba 3 3
Iy
4
13 65
32
2
1
d
162 5
0
6565 1 6
2
2
3 2
x
2
2
39. f x, y 9 y2 fx 0, fy 2y S
1 fx2 fy2 dA
R
3
y
y
0
1 4y2 dx dy
3
1 4y2 x
0
y
dy
y
3
21 4y2 dy
0
9x2
3
41.
9
3
12 1 4y232 43
x 2 y 2 dz dy dx
3 9x2 x2 y2
0
2
a
0
b
0
c
a
x 2 y 2 z2 dx dy dz
0
0
a
0
1
45.
1x 2
b
0
r
2
r2 dz dr d
3
9r 2 r 4 dr d
0
2
3r 3
0
r5 5
3 0
2
0
d
324 5
1 3 c cy 2 cz2 dy dz 3
1 3 1 3 1 1 1 1 bc b c bcz 2 dz abc 3 ab 3c a 3bc abca 2 b 2 c 2 3 3 3 3 3 3
2
1x 2y 2
2 2 2 1 1x 1x y
9
0
2
0
3
0
43.
1 3732 1 6
x2 y 2 dz dy dx
0
1
0
1r 2
1r 2
r 3 dz dr d
8 15
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171
172
Chapter 13
2
47. V 4
2 cos
0
r4 r 2 dr d
0
2
2 cos
4 4 r 232 3
0
49.
r dz dr d
0
2 cos
0
4r 2
0
2
4
Multiple Integration
0
2
32 3
1 sin3 d
0
1 32 cos cos3 3 3
2
m 4k
2
0
32 2 3 2 3
2
2 cos3 sin d d k 3
0
2
2
4
cos
0
2
2
4
2 1 cos3 sin d k cos4 3 4
2
4
k 24
3 cos sin d d d
0
2
1 cos5 sin d d k 2
0
2 sin d d d
0
4
4
cos
0
2
Mxy 4k k
2
4
4 k 3
z
d
2
4
cos5 sin d
1 k cos6 12
2
4
k 96
Mxy k96 1 m k24 4
x y 0 by symmetry
51.
mk
2
0
Mxy k
2
a
0
2
0
2 sin d d d
0
2
53. Iz 4k
2
0
a
0
ka3 6
cos 2 sin d d d
0
ka4 16
4k
4
3
2
0
16r2
r3 dz dr d
0
4
16r 3 r 5 dr d
3
833k 3
Mxy ka4 3a 6 xyz m 16 ka3 8
55. z f x, y a2 x2 y2
z
a−h
a2 r 2
a
h
0 ≤ r ≤ 2ah h2 a
x
a
y
(a) Disc Method
a
V
a2 y2dy
y
ah
y3 3
a2y
a3
a
ah
a
3
a3 a h3 a2a h 3 3
2a
a
a3 a3 h3 h3 1 a3 a2h a2h ah2 ah2 h 2 3a h 3 3 3 3 3
a−h −a
Equivalently, use spherical coordinates V
2
0
1
cos
0
aha
a
ahsec
2 sin d d d
—CONTINUED—
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y = a2 − x2
x a
Review Exercises for Chapter 13 55. —CONTINUED—
0
cos1aha
2
(b) Mxy
a
ahsec
0
cos 2 sin d d d
1 h2 2a h2 4 1 2 h 2a h2 3 2a h2 4 z 1 2 4 3a h V h 3a h 3 Mxy
0, 0, 342a3a hh 2
centroid:
3a2 3 a 42a 8
(c) If h a, z
centroid of hemisphere: (d) lim z lim h →0
h →0
32a h2 34a2 a 43a h 12a
(e) x2 y2 2 sin2
2
Iz
0
0
0
a
ah sec
0
(f ) If h a, Iz
2
cos1aha
2 sin2 2 sin d d d
h3 20a2 15ah 3h2 30
57.
0, 0, 38 a
6 sin
a3 4 20a2 15a2 3a2 a5 30 15
2 sin d d d
x, y
x y
y x
u, v u v u v
59.
0
13 23 9
Since 6 sin 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 < < 2 this circle about the z-axis.
61.
x, y
x y x y 1 1 1 1 1
u, v u v v u 2 2 2 2 2
y
Boundaries in xy-plane
Boundaries in uv-plane
xy3
u3
xy5
u5
x y 1
v 1
xy1
ln x ydA
R
5
3
1
2
1
ln
y=x−1
y = −x + 3
x 1
v1
1
y = −x + 5
y=x+1
3
1 1 x u v, y u v ⇒ u x y, v x y 2 2
1 1 1 u v u v 2 2 2
5
dv du
3
1
1 ln u dv du 1 2
5 ln 5 5 3 ln u 3 5 ln 5 3 ln 3 2 2.751
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5
3
2
3
ln u du u ln u u
5 3
173
174
Chapter 13
Multiple Integration
Problem Solving for Chapter 13
1. (a) V 16
z
1 x2 dA
R
4
16
1
1
0
0
16 3
4
0
1 r 2 cos2 r dr d
1
1 cos2 32 1 d cos2
16 sec cos tan 3
1
1
4
R y=x
x
0
82 2 4.6863 (b) Programs will vary.
3. (a)
1 u du arctan c. Let a2 2 u2, u v. a2 u2 a a
Then
1 1 v dv arctan C. 2 u2 2 u2 2 u2 v2
22
(b) I1
0
u
2 v arctan 2 2 u 2 u2
22
2
2 u2
0
22
4
2 u2
0
arctan arctan
u
arctan
2 u2
u
du
u
u 2 u2
du
du
2 u2
Let u 2 sin , du 2 cos d, 2 u2 2 2 sin2 2 cos2 .
I1 4
6
0
4
1
2 cos
arctan
6
arctantan d
0
2
(c) I2
22
2
2 22 2 u 2
4 2 2
6
arctan
2 cos d
2
6
0
v 2 arctan 2 u2 2 u2
2
2 sin 2 cos
2
2 18
u 2
du
u 2
u 2 u 2 arctan 2 u2 2 u2
du
2 u 4 arctan du 2 u2 2 u2
22
Let u 2 sin .
I2 4
4
2
6
2 2 sin 1 arctan 2 cos 2 cos
2
arctan
6
2 cos d
1 cossin d
—CONTINUED—
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y
Problem Solving for Chapter 13 3. —CONTINUED— (d) tan
cos2 1 sin 12 2
11 cos 2 1 sin
1 sin1 sin1 sin 1 cossin 2
2
(e) I2 4
arctan
6
4
2
6
2 (f )
1 sin d 4 cos
2
1 d 2 2 2
2
2
2 2
2
6
6
2
4
2
1
0
1
0
2
12 2
d
arctan tan
6
2 2 2 8 12 72
1 dx dy 1 xy
1
0
1
xy < 1
1
1 xy xy2 . . . dx dy
0
1
xyK dx dy
0
0 K0
1
K0
0
K0
1
0
y K1 yK dy 2 K1 K0 K 1
uv uv
2x 2
2y 2
R
1, 0 ↔ 0, 1 ↔ 1, 1 ↔
0
0
⇒ y
y
uv 2
R
uv 2
x
12 1 12
S
v 2
0, 0 ↔ 0, 0
1
1
1 1 2 2 K 1 K0 n1 n
⇒ x
x, y 12 u, v 12
1
0
(
1 , 2
1 2
)
1
(
S
12, 12 1 1 , 2 2
xK1y K 1 dy K1 0
xy yx ,v 2 2
1 sin cos
18 9 6 1 2 4 2 2 72 36 9
(g) u
2 d
1 1 xy xy2 . . . 1 xy
2
2
1
2, 0) u
2
3
4
−1 −2
(
1 − 1 , 2 2
)
2, 0
1 dx dy 1 xy
22
0
u
1 dv du 2 v2 u u 1 2 2
I1 I2
2
u 2
22 u 2
1 dv du u2 v 2 1 2 2
2 2 2 18 9 6
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175
176
Chapter 13
Multiple Integration
5. Boundary in xy-plane
Boundary in uv-plane
y x
u1
y 2x
u2
1 y x2 3
v3
z
7. 6
(3, 3, 6)
5 4
(0, 0, 0) 3
2
x
1 y x2 4
v4
1 x, y 3 u, v 2 3 A
uv uv
23
13
1 dA
R
2 3 1 3
uv uv
23
13
(0, 6, 0)
(3, 3, 0)
6
3
1 3
V
2x
0
0
y
6x
dy dz dx 18
x
x, y 1 1 dA u, v 3
S
9. From Exercise 55, Section 13.3,
ex 2 dx 2 2
Thus,
ex 2 dx 2
0
2
0
1 2
ex dx 2
0
2 1 2 2 4
x ≥ 0, y ≥ 0 elsewhere
kexya 0
f x, y dA
ex dx
0
1 2 2 x2ex dx xex 2
11. f x, y
and
2
0
2
0
∆x cos θ
kexya dx dy
cos x y sec x y P
0
k
13. A l w
xa
e
0
dx
eya dy
0
∆y ∆y
θ ∆x
These two integrals are equal to
exa dx lim
0
ae xa
b→
b 0
a.
Area in xy-plane: x y
Hence, assuming a, k > 0, you obtain 1 ka2
or
a
1 k
.
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C H A P T E R 14 Vector Analysis Section 14.1 Vector Fields
. . . . . . . . . . . . . . . . . . . . . . . . . . . 407
Section 14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 Section 14.3 Conservative Vector Fields and Independence of Path . . . . . . 419 Section 14.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 423 Section 14.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 427 Section 14.6 Surface Integrals
. . . . . . . . . . . . . . . . . . . . . . . . . 431
Section 14.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 436 Section 14.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 439 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448
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C H A P T E R Vector Analysis Section 14.1
14
Vector Fields
Solutions to Even-Numbered Exercises 2. All vectors are parallel to x-axis. Matches (d)
4. Vectors are in rotational pattern.
6. Vectors along x-axis have no x-component.
Matches (e)
Matches (f) 10. Fx, y x i yj
8. Fx, y 2i
F x2 y2
F 2
y
y
5 4 3 2 1
1
−2
x
−1
1
2
−5 −4
x
−2 −1
1 2
4 5
−2 −3 −4 −5
−2
12. Fx, y x i
14. Fx, y x2 y2 i j
F 1 x2 y22
F x c
y y
10
3
8
2
6
1 −3
−2
4
x
−1
1
2
3
x
−2
16. Fx, y, z x i y j z k
2
18. Fx, y 2y 3x i 2y 3x j
F x y z c 2
2
2
y 6
x2 y2 z2 c2
4
z
2 −6
2
−4
x
−2
2
4
6
−2
−2
−6 2 2
x
y −2
407
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408
Chapter 14
Vector Analysis
20. Fx, y, z xi yj z k
22. f x, y sin 3x cos 4y fxx, y 3 cos 3x cos 4y
z
2
fyx, y 4 sin 3x sin 4y
1
Fx, y 3 cos 3x cos 4yi 4 sin 3x sin 4yj 1
1
2
y
2 x
y z xz z x y z z fxx, y, z 2 x y 1 xz fyx, y, z 2 z y y 1 x fzx, y, z 2 z x y z z 1 xz y 1 x Fx, y, z 2 i 2 j 2 k x y z y z x y
24. f x, y, z
28. Fx, y
26. gx, y, z x arcsin yz gxx, y, z arcsin yz gyx, y, z gzx, y, z
1 y 1 yi xj 2 i j 2 x x x
xz 1 y2z2
xy 1 y2z2
Gx, y, z arcsin yz i
30. Fx, y
xz 1 y2z2
j
xy 1 y2z2
1 1 1 y i x j i j xy x y
M y x2 and N 1 x have continuous first partial derivatives for all x 0.
M 1 x and N 1 y have continuous first partial derivatives for all x, y 0.
M 1 N 2 ⇒ F is conservative. x x y
N M 0 ⇒ F is conservative. x y
32. M
x x2 y2
,N
y
34. M
x2 y2
N M xy 2 ⇒ Conservative x x y23 2 y
y 1 x2y2
,N
x 1 x2y2
N M 1 1 x 1 x2y23 2 y 1 x2y23 2 ⇒ Not conservative
36. Fx, y
1 y i 2 x j y2 1 2x i 2j y y
1 1 2 y y y
2x 2 2 2 x y y
Not conservative
38. Fx, y 3x2 y2 i 2x3yj 3x2 y2 6x2 y y 2x3y 6x2y x Conservative fxx, y 3x2 y2 fyx, y 2x 3 y f x, y x3y2 K
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40. Fx, y
2y x2 i 2j x y
2y 2 y x x
x2 2x 2 2 x y y
Not conservative
k
Section 14.1
42. Fx, y
2x 2y i 2 j x2 y22 x y22
44. Fx, y, z x2z i 2xz j yz k, 2, 1, 3 k i j curl F x y z x2z 2xz yz
8xy 2x 2 y x2 y22 x y23
8xy 2y 2 x x 2 y 2 2 x y 23
z 2x i 0 x2 j 2z 0 k
Conservative
z 2x i x2 j 2zk
2x fxx, y 2 x y22 fyx, y
curl F 2, 1, 3 7i 4j 6k
2y x2 y22
f x, y
Vector Fields
1 K x2 y2
46. Fx, y, z exyz i j k, 3, 2, 0
curl F
i x
j y
k z
exyz exyz exyz
curl F 3, 2, 0 6i 6j
yz xz xy i j k yz xz xy
48. Fx, y, z
j k y z xz xy xz xy
i x curl F yz yz
xz xyexyz i yz xyexyzj yz xzexyz k
x x y 2
2
x 1 y
x2
2
x2 y2 z 2 y 2 z 2 i j k 2 2 2 2 x z x y y z x z y z2
1 1 1 1 1 i y2 j z2 k x z2 x y2 y z2 y z2 x z2
50. Fx, y, z x2 y2 z2i j k i j k curl F x y z x2 y2 z2 x2 y2 z2 x2 y2 z2 52. Fx, y, z e z y i x j k
y z i z x j x y k x2 y2 z2
54. Fx, y, z y 2z 3 i 2xyz 3 j 3xy2z 2 k
i j k curl F x y z xe z i yezj 0 yez xez ez
i j k curl F x 0 y z 2z 3 3 2z2 y 2xyz 3xy
Not conservative
Conservative
f x, y, z xy 2z3 K
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409
410
Chapter 14
Vector Analysis
56. Fx, y, z
x y i 2 jk x2 y2 x y2
i j x y curl F y x x2 y2 x2 y2
58.
Fx, y xe x i ye y j div Fx, y
k z 0
xe x ye y x y
xe x e x ye y e y e xx 1 e y y 1
1
Conservative
fx x, y, z
x x2 y2
fy x, y, z
y x2 y2
fz x, y, z 1 f x, y, z f x, y, z
60.
x dx x2 y2
1 lnx 2 y 2 g y, z K1 2
y dy x2 y2
1 lnx 2 y 2 hx, z K2 2
f x, y, z
f x, y, z
1 lnx 2 y 2 z K 2
dz z px, y K3
Fx, y, z lnx 2 y 2 i xy j ln y 2 z 2k div Fx, y, z
62.
2x 2z lnx 2 y 2 x y ln y 2 z2 2 x 2 x y z x y2 y z2
Fx, y, z x 2z i 2xz j yz k
64.
div Fx, y, z 2 xz y div F2, 1, 3 11
66. See the definition of Conservative Vector Field on page 1011. To test for a conservative vector field, see Theorem 14.1 and 14.2.
Fx, y, z lnxyzi j k 1 1 1 div Fx, y, z x y z 1 1 11 div F3, 2, 1 1 3 2 6
68. See the definition on page 1016.
70. Fx, y, z x i z k Gx, y, z x 2 i yj z 2 k
i j k F G x 0 z yz i xz 2 x 2 z j xyk x2 y z2
i j curlF G x y yz xz 2 x 2z
k z xy
x 2xz x i y y j z 2 2xz z k 2
xx 2z 1i zz 2x 1k
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Section 14.1
72. Fx, y, z x2z i 2xz j yz k i j curl F x y x2z 2xz
74. Fx, y, z x i z k Gx, y, z x 2 i y j z 2 k
k z 2x i x 2 j 2zk z yz
i curlcurl F x z 2x
Vector Fields
i FG x x2
j k j 2xk y z x2 2z
j 0 y
k z yzi xz 2 x 2zj xyk z2
divF G 0
76. Fx, y, z x 2 z i 2xz j yz k
i j k 2 curl F x y z z 2x i x j 2z k x2z 2xz yz divcurl F 2 2 0
78. Let f x, y, z be a scalar function whose second partial derivatives are continuous. f
f f f i j k x y z
j
i
k
curlf x
y
z
f x
f y
f z
f f f f f f i j k0 yz zy xz zx xy yx 2
2
2
2
2
2
80. Let F M i N j P k and G R i Sj T k.
i j k F G M N P NT PS i MT PR j MS NRk R S T divF G
NT PS PR MT MS NR x y z
N
M T N S P R P T S M R N T P S P R M T M S N R x x x x y y y y z z z z P
N
M
P
N
M
T
S
R
T
S
R
y z R z x S x y T M y z N z x P x y
curl F G F curl G
82. Let F M i Nj Pk.
i j f F x y fM fN
k z fP
f N f P f M f N f M Nf i Pf Mf j Nf Mf k yf P f P y z z x x z z x x y y
f
i j k f f f P N P M N M i j k x y z f F f F y z x z x y M N P
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411
412
Chapter 14
Vector Analysis
84. Let F M i Nj Pk. curl F
P M N M N i j k P y z x z x y
divcurl F
P N P M N M x y z y x z z x y
2P 2N 2P 2M 2N 2M 0 xy xz yx yz zx zy
(since the mixed partials are equal)
In Exercises 86 and 88, Fx, y, z x i yj zk and f x, y, z Fx, y, z x 2 y 2 z 2. 1 1 f x 2 y 2 z 2
86.
1f x w
88.
2
x y z x i y j zk F i 2 j 2 k 3 3 2 2 2 y 2 z 23 2 x y 2 z 23 2 x y 2 z 23 2 f x y z 2w 2x 2 y 2 z 2 2 x2 x y 2 z 25 2
1 1 f x 2 y 2 z 2
w x 2 x x y 2 z 23 2
2w 2y 2 x 2 z 2 2 y2 x y 2 z 25 2
w y 2 y x y 2 z 23 2
2z 2 x 2 y 2 2w 2 z 2 x y 2 z 2 5 2
w z 2 z x y 2 z 23 2
2w
2w 2w 2w 2 2 0 x 2 y z
Therefore w
Section 14.2
Line Integrals
t i 45 t j, 0 ≤ t ≤ 5 4. rt 5 i 9 t j, 5 ≤ t ≤ 9 14 t i, 9 ≤ t ≤ 14
x2 y2 1 16 9
2.
1 is harmonic. f
cos2 t sin2 t 1 cos2 t
x2 16
sin2 t
y2 9
x 4 cos t y 3 sin t rt 4 cos t i 3 sin t j 0 ≤ t ≤ 2
t i t 2 j, 0 ≤ t ≤ 2 6. rt 4 t i 4 j, 2 ≤ t ≤ 4 8 t j, 4 ≤ t ≤ 8 8. rt t i 2 t j, 0 ≤ t ≤ 2; rt i j
C
2
4xy ds
0
2
4t 2 t 1 1 dt 4 2
0
2t t 2 dt 4 2 t 2
t3 3
2 0
4 2 4
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8 16 2 3 3
Section 14.2
Line Integrals
10. rt 12t i 5t j 3 k, 0 ≤ t ≤ 2; rt 12i 5j
2
8xyz ds
0
C
2
812t5t3 122 52 02 dt
18,720t 2 dt 18,720
0
12. rt t j, 1 ≤ t ≤ 10
x 2 y 2 ds
C
10
8 6
10
t2
dt
4 2
10
1 3 t 3
1
333
14. rt 2 cos t i 2 sin t j, 0 ≤ t ≤
x 2 y 2 ds
C
−4
x
−2
2
4
2
y
2
2
4 cos2 t 4 sin2 t 2 sin t2 2 cos t2 d t 1
0
49,920
0
10
0 t 2 0 1 dt
1
3 2
y
1
t3
2
8 dt 4
x
0
1
16. rt t i 3t j, 0 ≤ t ≤ 3
C
x 4 y ds
3
0
y
t 4 3t 1 9 dt
10
6
0 2 4 6
t i, 2i t 2 j, 18. rt 6 t i 2 j, 8 t j,
C1
C2
C3
C4
x 4 y ds x 4 y ds
x 4 y ds x 4 y ds
C
2
(3, 9)
9
6
t 2 8 3 3 2 t 10 2 3
27 144
≤ ≤ ≤ ≤
t t t t
≤ ≤ ≤ ≤
3 3
0
57 10 2
−3
x 3
6
y
2 4 6 8
2
6
4
C3
2
(2, 2)
C4 1
C2
t dt 2
0
4
2
C1
2 4 t 2 ds 4
16 2 3
6 t 4 2 ds 2 8 2
8
4 8 t ds
6
x 4 y ds 2 4
16 2 3
16 2 16 2 56 2 8 2 8 2 3 3 3
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x 1
2
413
414
Chapter 14
Vector Analysis 22. Fx, y xyi yj
20. x, y, z z rt 3 cos t i 3 sin t j 2 t k, 0 ≤ t ≤ 4
C: rt 4 cos t i 4 sin t j, 0 ≤ t ≤
rt 3 sin t i 3 cos t j 2 k
Ft 16 sin t cos t i 4 sin t j rt 4 sin t i 4 cos t j
rt 3 sin t 3 cos t 2 13 2
Mass
2
x, y, z ds
C
4
2
2t 13 dt 16 2 13
0
C
F dr
2
24. Fx, y 3x i 4y j
64 sin2 t cos t 16 sin t cos t dt
0
2
64 3 sin t 8 sin2 t 3
0
rt 2 cos t i 2 sin t j t k
2
t2 F dr 3t 4t dt 2 2 C
2 2
0
C
F dr
0
1 8 sin2 t cos t 8 cos2 t sin t t 5 dt 4
83 sin
3
t
8 t6 cos3 t 3 24
0
8 6 8 6 16 3 24 3 24 3 xi yj zk x 2 y 2 z 2
rt t i t j e t k, 0 ≤ t ≤ 2 Ft
t i t j et k 2t 2 e 2t
d r i j e t k dt
C
F dr
2
1 2t e 2t dt 6.91 e 2t
2t 2
0
30. Fx, y x 2 i x y j C: x cos3 t, y sin3 t from 1, 0 to 0, 1 rt cos3 t i sin3 t j, 0 ≤ t ≤
2
rt 3 cos 2 t sin t i 3 sin2 t cos t j Ft cos 6 t i cos3 t sin3 t j F r 3 cos8 t sin t 3 cos4 t sin5 t 3 cos4 t sin t cos4 t sin4 t 3 cos4 t sin t cos4 t 1 cos2 t2 3 cos4 t sin t 2 cos4 t 2 cos2 t 1 6 cos8 t sin t 6 cos6 t sin t 3 cos4 t sin t Work
C
F dr
40 3
1 C: rt 2 sin t i 2 cos t j t 2 k, 0 ≤ t ≤ 2 1 Ft 4 sin2 t i 4 cos2 tj t 4 k 4
Ft 3t i 4 4 t 2 j t j rt i 4 t 2
28. Fx, y, z
26. Fx, y, z x 2 i y 2 j z 2 k
C: rt t i 4 t 2 j, 2 ≤ t ≤ 2
2
2
6 cos8 t sin t 6 cos6 t sin t 3 cos4 t sin t dt
0
2 cos3
9
t
6 cos7 t 3 cos5 t 7 5
2
0
43 105
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Section 14.2 32. Fx, y y i x j
C: line from 0, 0, 0 to 5, 3, 2 rt 5i 3j 2k
rt 2 sin t i 2 cos t j
Ft 6t 2 i 10t 2 j 15t 2 k
Ft 2 sin t i 2 cos tj
F
C
F d r 4
cos 2t dt 2 sin 2t
0
0
F dr
0, 0
14 , 161
12 , 14
34 , 169
1, 1
F dr
Fx, y
5i
3.5i j
2i 2j
1.5i 3j
i 5j
C
rt
i
i 0.5j
ij
i 1.5j
i 2j
5
4
4
6
11
C
x, y
rt i 2 t j
Work
0
36. rt t i t 2 j, 0 ≤ t ≤ 1 10 5 44 24 46 11 34 16 3
r 90t 2
r 4 sin2 t 4 cos2 t 4 cos 2t
0 ≤ t ≤ 1
rt 5t i 3t j 2t k,
0 ≤ t ≤
rt 2 cos t i 2 sin t j,
Work
F
r
1
90t 2 dt 30
0
38. Fx, y x2 y i xy3 2 j r1t t 1 i t 2 j, 0 ≤ t ≤ 2
(a)
r1t i 2 t j Ft t 12 t 2 i t 1t 3 j
C1
F dr
2
t 12 t 2 2t 4t 1 dt
0
256 3
r2t 1 2 cos t i 4 cos2 t j, 0 ≤ t ≤
(b)
2
r2t 2 sin t i 8 cos t sin t j
Ft 1 2 cos t24 cos2 t i 1 2 cos t8 cos3 t j
C2
F dr
2
1 2 cos t 4 cos 2
0
2
t2 sin t 8 cos t sin t1 2 cos t8 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. 40. Fx, y 3y i x j C: rt t i
42. Fx, y x i yj
t3j
C: rt 3 sin t i 3 cos t j
rt i 3t 2 j
rt 3 cos t i 3 sin t j
Ft 3t 3 i t j
Ft 3 sin t i 3 cos t j
F r Thus,
C
3t 3
F r 9 sin t cos t 9 sin t cos t 0
0
3t 3
F d r 0.
Thus,
C
F dr 0.
44. x 2t, y 10t, 0 ≤ t ≤ 1 ⇒ y 5x, 0 ≤ x ≤ 2
C
x 3y 2 dx
0
415
34. Fx, y, z yz i xz j xyk
C: counterclockwise along the semicircle y 4 x 2 from 2, 0 to 2, 0
F
Line Integrals
2
x 75x 2 dx
2 25x x2
2
3
0
202
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416
Chapter 14
Vector Analysis
46. x 2t, y 10t, 0 ≤ t ≤ 1 ⇒ y 5x, dy 5 dx, 0 ≤ x ≤ 2
3y x dx y 2 dy
C
2
35x x dx 5x25 dx
0
2
14x 125x2 dx
0
125 3 x 3
7x2
2
28
0
125 1084 8 3 3 y
48. rt t j, 0 ≤ t ≤ 2 xt 0, yt t
2
dx 0, dy dt
1
2
2x y dx x 3y dy
0
C
50. rt
2
3 3t dt t 2 2
0
6
1
0 ≤ t ≤ 3 t j, t 3 i 3j, 3 ≤ t ≤ 5
y
x 1
xt 0, yt t
C1:
x
−1
2
3
−1
dx 0, dy dt
C1
−2
3
2x y dx x 3y dy
3t dt
0
C1
27 2
−3
C2
(2, − 3)
xt t 3, yt 3
C2:
dx dt, dy 0
5
2x y dx x 3y dy
3
C2
2t 3 3 dt t 32 3t
5 3
10
27 47 10 2 2
2x y dx x 3y dy
C
3 52. xt t, yt t 3 2, 0 ≤ t ≤ 4, dx dt, dy t 1 2 dt 2
2x y dx x 3y dy
C
4
4
54. xt 4 sin t, yt 3 cos t, 0 ≤ t ≤ dx 4 cos t dt, dy 3 sin t dt
2x y dx x 3y dy
32 t dt 1 2
9 2 1 3 2 3 1 t t 2t dt t 3 t 5 2 t 2 2 2 2 5
0
2t t 3 2 t 3t 3 2
0
2
0
1 592 96 32 16 5 5
8 sin t 3 cos t4 cos t dt 4 sin t 9 cos t3 sin t dt
2
5 sin t cos t 12 cos 2 t 12 sin2 t dt
0
4
2
0
C
2
52 sin t 12t 2
0
5 6 2
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Section 14.2 56. f x, y y
58. f x, y x y
C: line from 0, 0 to 4, 4
C: x 2 y 2 1 from 1, 0 to 0, 1
rt t i t j, 0 ≤ t ≤ 4
rt cos t i sin t j, 0 ≤ t ≤
rt i j
rt 1
Lateral surface area:
4
f x, y ds
0
C
2
rt sin t i cos t j
rt 2
Line Integrals
Lateral surface area:
t 2 dt 8 2
f x, y ds
cos t sin t dt
2
0
C
sin t cos t
2
0
2
60. f x, y y 1 C: y 1 x 2 from 1, 0 to 0, 1 rt 1 t i 1 1 t2 j, 0 ≤ t ≤ 1 rt i 21 t j
rt 1 41 t2 Lateral surface area:
1
f x, y ds
2 1 t 2 1 41 t2 dt
0
C
1
0
1
1 41 t 2 dt
2
1 t2 1 41 t2 dt
0
0
1 21 t 1 41 t2 ln 21 t 1 41 t2 2
1
0
1 21 t241 t2 1 1 4 1 t2 ln 21 t 1 4 1 t2 64
1 1 2 5 ln 2 5 64 18 5 ln 2 5 2
33 1 23 5 ln 2 5 46 5 33 ln 2 5 2.3515 32 64 64
62. f x, y x 2 y 2 4 C: x 2 y 2 4 rt 2 cos t i 2 sin t j, 0 ≤ t ≤ 2 rt 2 sin t i 2 cos t j
rt 2 Lateral surface area:
C
f x, y ds
2
0
2π
4 cos2 t 4 sin2 t 42 dt 8
0
1 cos 2t dt 8 t
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1 sin 2t 2
2
0
16
1
417
418
Chapter 14
Vector Analysis
1 64. f x, y 20 x 4 C: y x 3 2, 0 ≤ x ≤ 40 rt t i t 3 2 j, 0 ≤ t ≤ 40 3 rt i t 1 2 j 2
1 94 t
rt
Lateral surface area:
0
C
Let u 1 t, then t 9 4
40
0
1 20 t 4
40
f x, y ds
1
4 2 9 u
1 20 t 4
1 94t dt
1 and dt 89 u du.
9 t dt 4
91
1
1 8 8 20 u2 1 u u du 9 9 81
8 u5 179u3 81 5 3
66. f x, y y
91
1
91
u4 179u 2 du
1
850,304 91 7184 6670.12 1215
z 4
C: y x 2 from 0, 0 to 2, 4
3
S8
2 1
Matches c. 4
3
2 4
x
68. W
C
F dr
y
(2, 4, 0)
M dx N dy
y
C c
M 154 x y 60 15x c cx 2
2
2
y = c ) 1 − x 2)
N 15xy 15xc cx 2 dx dx, dy 2cx dx
1
W
1
60 15x 2c cx 2 15xc cx 22 cx dx
−1
x 1
120 4c 8c 2 (parabola) w 16c 4 0 ⇒ c 14 yields the minimum work, 119.5. Along the straight line path, y 0, the work is 120. 72. (a) Work 0
70. See the definition, page 1024.
(b) Work is negative, since against force field. (c) Work is positive, since with force field. 74. False, the orientation of C does not affect the form
76. False. For example, see Exercise 32.
f x, y ds.
C
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Section 14.3
Section 14.3
Conservative Vector Fields and Independence of Path
Conservative Vector Fields and Independence of Path
2. F x, y x 2 y 2 i x j (b) r2 w w 2 i w j, 0 ≤ w ≤ 2
(a) r1 t t i t j, 0 ≤ t ≤ 4 r1 t i F t
t2
C
1 2t
r2 w 2w i j
j
F w w 4 w 2 i w 2 j
t i t j
F dr
4
0
1 t 2 t t dt 2
32 4
t3 t2 t 3 2 3
0
F dr
C
80 3
2
2w w 4 w 2 w 2 dw
0
w3
6
w4 w3 2 3
2 0
80 3
4. F x, y yi x 2 j (a) r1 t 2 t i 3 t j, 0 ≤ t ≤ 3 r1 t i j F t 3 t i 2 t 2 j
C
F dr
3
3 t 2 2 t 3 2 3
3
69 2
w1 2 ln w w1 dw 3 2ln w
3 t 2 t 2 dt
0
0
(b) r2 w 2 ln w i 3 ln w j, 1 ≤ w ≤ e3 r2 w
1 1 i j w w
F w 3 ln w i 2 ln w 2 j
3
C
F dr
e
2
3 ln w
2
1
Since
e3 1
8. F x, y, z y ln z i x ln z j
6. F x, y 15x 2 y 2 i 10x 3 y j N 30x 2 y x
2 ln w 3 3
M 30x 2 y y
69 2
xy k z
curl F 0 so F is not conservative. x x N P y z z z
N M , F is conservative. x y
10. F x, y, z sin yz i xz cos yz j xy sin yz k curl F 0, so F is not conservative. 12. F x, y ye xy i xe xy j (a) r1 t t i t 3 j, 0 ≤ t ≤ 3
(b) F x, y is conservative since
r1 t i j F t t 3 e
3tt 2
C
F dr
3
i te
3tt 2
3tt 2
t 3 e
M N xye xy e xy. y x
j
dt
3tt 2
te
The potential function is f x, y e xy k.
0 3
0
e 3tt 3 2t dt
e 3tt
2
3
2
e0 e0 0
0
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419
420
Chapter 14
Vector Analysis
14. F x, y xy 2 i 2x 2 y j (a) r1 t t i
1 j, t
r1 t i
1 j t2
F t
C
1 (b) r2 t t 1 i t 3 j, 3
1 ≤ t ≤ 3
r2 t i
1 i 2tj t
F dr
3
1
1 j 3
1 2 F t t 1 t 3 2 i t 1 2 t 3 j 9 3
1 dt t
3
ln t
1
C
2
F dr
ln 3
0
1 2 t 1 t 3 2 t 1 2 t 3 dt 9 9
2
1 9
3t 3 7t 2 7t 3 dt
0
1 3t 4 7t 3 7t 2 3t 9 4 3 2
16.
0 ≤ t ≤ 2
2 0
44 27
2x 3y 1 dx 3x y 5 dy
C
Since My Nx 3, F x, y 2x 3y 1 i 3x y 5 j is conservative. The potential function is f x, y x 2 3xy y 22 x 5y k. (a) and (d) Since C is a closed curve, (b)
2x 3y 1 dx 3x y 5 dy 0.
C
C
(c)
18.
y2 x 5y 2
y2 x 5y 2
0, 1
1 3 2e2 e4
2
Since curl F 0, F x, y, z is conservative. The potential function is f x, y, z x yz k.
Since My Nx 2y, F x, y x 2 y 2 i 2xy j
(a) r1 t cos t i sin t j t 2 k, 0 ≤ t ≤
is conservative. The potential function is f x, y x 33 xy 2 k.
2, e2
10
20. F x, y, z i z j yk
x 2 y 2 dx 2xy dy
C
(a)
0, 1
2x 3y 1 dx 3x y 5 dy x 2 3xy
C
0, 1
2x 3y 1 dx 3x y 5 dy x 2 3xy
C
x 2 y 2 dx 2xy dy
C
8, 4
x3 xy 2 3
0, 0
1, 0, 2
1, 0, 0
0, 2
C
8 3 2, 0
1, 0, 2
F dr x yz
1, 0, 0
22. F x, y, z y i x j 3xz 2 k F x, y, z is not conservative. (a) r1 t cos t i sin t j t k, 0 ≤ t ≤ r1 t sin t i cos t j k F t sin t i cos t j 3t 2 cos t k
C
F dr
0
t
sin2 t cos 2 t 3t 2 cos t dt
0
3 t 2 sin t
0
6
0
2
(b) r2 t 1 2t i 2 t k, 0 ≤ t ≤ 1
896 3
x3 xy 2 (b) x 2 y 2 dx 2xy dy 3 C
F dr x yz
1 3t 2 cos t dt
0
—CONTINUED—
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t sin t dt t 3t 2 sin t 6 sin t t cos t
0
5
2
Section 14.3
Conservative Vector Fields and Independence of Path
22. —CONTINUED— (b) r2 t 1 2t i t k, 0 ≤ t ≤ 1 r2 t 2 i k F t 1 2t j 3 2t 2 1 2t k
C
1
F dr
1
3 3 t 2 1 2t dt 3 3
0
t 2 2t 3 dt 3 3
0
24. F x, y, z y sin z i x sin z j xy cos x k
26.
C
(a) r1 t t 2 i t 2 j, 0 ≤ t ≤ 2
t3 t2 3
4 1 0
3 2
2 x y i 2 x y j dr x y 2
4, 3
3, 2
49
r1 t 2 t i 2 t j F t t 4 cos t 2 k
C
F dr
2
0 dt 0
0
(b) r2 t 4t i 4t j, 0 ≤ t ≤ 1 r2 t 4 i 4 j F t 16t 2 cos 4t k
C
28.
30.
32.
F dr
1
0 dt 0
0
y dx x dy x arctan x2 y2 y C
C
x2
23, 2
1, 1
3 4 12
2x 2y 1 dx 2 dy 2 y 2 2 x y 2 2 x y2
1, 5
7, 5
1 1 12 26 74 481
zy dx xz dy xy dz
C
Note: Since F x, y, z yz i xz j xyk is conservative and the potential function is f x, y, z xyz k, the integral is independent of path as illustrated below. 1, 1, 1
(a)
xyz
(b)
xyz
(c)
xyz
0, 0, 0
0, 0, 1
0, 0, 0
1, 0, 0
34.
0, 0, 0
1 1, 1, 1
xyz
0, 0, 1
1, 1, 0
xyz
1, 0, 0
011 1, 1, 1
xyz
1, 1, 0
0011
6x dx 4z dy 4y 20z dz 3x 2 4yz 10z 2
C
36. F x, y Work
4, 3, 1
0, 0, 0
46
2x x2 i 2 j is conservative. y y 2 1, 4
xy
3, 2
1 9 17 4 2 4
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421
422
Chapter 14
Vector Analysis
38. F x, y, z a 1 i a 2 j a 3 k Since F x, y, z is conservative, the work done in moving a particle along any path from P to Q is
f x, y, z a1x a 2 y a 3 z
Q q1, q2, q3
P p1, p2, p3
a1 q1 p1 a 2 q2 p2 a 3 q3 p3 F PQ . \
40. F 150j 1 (b) r t t i 50 50 t 2 j
(a) r t t i 50 t j, 0 ≤ t ≤ 50 dr i j dt
F dr
C
50
150 dt 7500 ft
0
lbs
C
50
F dr 6
50 t dt 7500 ft lbs
0
y x i 2 j x2 y2 x y2
42. F x, y (a) M
1 d r i 25 50 t j dt
y x2 y2
(b) r t cos t i sin t j, 0 ≤ t ≤
x2 y 2 M x 2 y 2 1 y 2y
2 y x 2 y 2 2 x y 2 2 N
x2
x y2
F sin t i cos t j dr sin t i cos t j dt
C
F dr
sin2 t cos2 t dt t
0
0
x2 y2 N x 2 y 2 1 x 2x
2 2 2 2 x x y
x y 2 2 Thus,
N M . x y
(c) r t cos t i sin t j, 0 ≤ t ≤
(d) r t cos t i sin t j, 0 ≤ t ≤ 2
F sin t i cos t j
F sin t i cos t j
dr sin t i cos t j dt
dr sin t i cos t j dt
C
F dr
sin2 t cos2 t dt
0
C
t
0
F dr
2
sin2 t cos2 t dt
0 2
t
0
2
This does not contradict Theorem 14.7 since F is not continuous at 0, 0 in R enclosed by curve C.
(e) arctan
x 1y xy 2 i j y 1 xy 2 1 xy 2
x y i 2 jF x2 y2 x y2
44. A line integral is independent of path if
C
F dr does not depend on the curve joining P and Q. See Theorem 14.6
46. No, the amount of fuel required depends on the flight path. Fuel consumption is dependent on wind speed and direction. The vector field is not conservative. 48. True
50. False, the requirement is My Nx.
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Section 14.4
Section 14.4
Green’s Theorem y
0 ≤ t ≤ 4 4 ≤ t ≤ 8 8 ≤ t ≤ 12
t i, 2. rt 4 i t 4 j, 12 t i 12 t j,
4
y 2 dx x 2 dy
(4, 4) 4
2
0 dt t 2 0
t 420 16 dt
1
4
x 1
12
12 t2dt 12 t2dt 0 64
8
By Green’s Theorem,
y=x
3
8
0
C
R
N M dA x y
4
0
x
4
2x 2y dy dx
0
y 2 dx x 2 dy
C
2
2
3
128 64 3 3
x 2 dx
0
64 . 3
4. rt cos t i sin t j, 0 ≤ t ≤ 2
Green’s Theorem
y
sin2 t sin t dt cos2 tcos t dt
1
x 2 + y 2 =1
0
2
cos3 t sin3 t dt
x
−1
1
0
2
−1
cos t1 sin2 t sin t1 cos2 t dt
0
sin3 t cos3 t cos t 3 3
sin t By Green’s Theorem,
R
N M dA x y
1
0
0
1x 2
1 1x2 2
0
2
1
2x 2y dy dx
2r cos 2r sin r dr d
0
2 3
2
0
2 cos sin d 0 0. 3
6. C: boundary of the region lying between the graphs of y x and y x 3
xe y dx e x dy
C
R
1
xe x 3x 2e x dx 3
0
N M dA x y
In Exercises 8 and 10,
1
0
0
xe x e x dx 2.936 2.718 0.22
1
x
e x xe y dy dx
x3
1
xe x x3 e x dx 0.22 3
0
N M 1. x y
8. Since C is an ellipse with a 2 and b 1, then R is an ellipse of area ab 2. Thus, Green’s Theorem yields
y x dx 2x y dy
C
1 dA Area of ellipse 2.
R
y
10. R is the shaded region of the accompanying figure.
C
y x dx 2x y dy
1 dA
R
Area of shaded region 1 25 9 8 2
4 2 1 −5 −4 −3 −2 −1
x 1 2 3 4 5
−2 −3 −4 −5
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4
423
424
Chapter 14
Vector Analysis
12. The given curves intersect at 0, 0 and 9, 3. Thus, Green’s Theorem yields
y 2 dx xy dy
C
y 2y dA
R
9
0
x
9
y dy dx
0
0
y 2 2
x
0
9
dx
0
x x 2 dx 2 4
9 0
81 4
14. In this case, let y r sin , x r cos . Then d A r dr d and Green’s Theorem yields
x 2 y 2 dx 2xy dy
C
2
0
0
1cos
1cos
r sin r dr d
0
r 2 sin dr d
0
2
4 3
0
1 cos 4 3
sin 1 cos 3 d 2
0
0.
N M we have 2ex sin 2y y x
R
N M dA 0. x y
18. By Green’s Theorem,
ex 2 y dx ey 2 x dy 2
2
C
2 dA 2Area of R 2 62 23 60.
R
20. By Green’s Theorem,
2
R
4
16. Since
4y dA 4
y
3x 2 e y dx e y dy
C
(−1 , 1 )
3x 2e y dA
R
2
1
2
1
2
3x 2e y dy dx
1
2
(2, 2)
2
1 1
2
3x 2e y dy dx
x
3x 2e y dy dx
1
2
(1, 1)
(−2 , 2 )
1
1 2
(− 2, − 2) (−1 , −1 )
3x 2e y dy dx
(2, − 2) (1, − 1)
7e 2 e2 2e 2 e 7e 2 e2 2e1 e2 16e 2 16e2 2e 2e1. 22. Fx, y e x 3y i ey 6x j C: r 2 cos Work
ex 3y dx ey 6x dy
C
9 dA 9 since r 2 cos is a circle with a radius of one.
R
24. Fx, y 3x 2 y i 4xy 2 j C: boundary of the region bounded by the graphs of y x, y 0, x 4 Work
C
4
3x 2 y dx 4xy 2 dy
0
x
0
4
4y 2 1 dy dx
4 32 3x
x12 dx
0
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176 15
Section 14.4 26. From the figure we see that
y
C2
4
3 3 C1: y x, dy dx, 0 ≤ x ≤ 2 2 2 1 x C2: y 4, dy dx 2 2
x + 2y = 8 (2, 3)
3
C3 2
C1 3x − 2y = 0
1
C3: x 0, dx 0. 1 2
A
1 2
2
0 0
3 3 1 x x dx 2 2 2
0
2
x 1
4
2
4 dx 2
2
we have
dx 4
0
1 2
t
0
9 2
0
3
,
32t t 4 3t 3t 2 31 2t 3 3 dt 3 2 1 t 1 t 1 t 3 12
t5 t 2 9 dt t 3 13 2
30. See Theorem 14.9: A
t 2t 3
0
3 1 dt t 3 13 2
Mx
C
x2 1 dy 2 C 2
2t 3 1
3t 2 t 3 12 dt
3
0
Mx 1 2A 2A
y 2 dx and y
C
3 . 2
My 1 2A 2A
x 2 dy and x
y 2 dx.
C
x dA. Let N x 22 and M 0. By Green’s Theorem,
R
C
x 2 dy.
C
(b) By Theorem 14.9 and the fact that x r cos , y r sin , we have 1 2
0
y dA. Let N 0 and M y 22. By Green’s Theorem,
R
For the moment about the y-axis, My My
y2 1 dx 2 2
1 x dy y dx. 2 C
32. (a) For the moment about the x-axis, Mx
A
3
31 2t3 32t t 4 dt and dy 3 dt, 3 2 t 1 t 12
dx
2
x 1 1 x 4 dx 0 2 2 2
28. Since the loop of the folium is formed on the interval 0 ≤ t ≤
A
Green’s Theorem
x dy y dx
1 2
r cos r cos d r sin r sin d
1 2
r 2 d.
C
1 a2 1 . Note that y 0 and dy 0 along the boundary y 0. , we have 2 2A a 2 Let x a cos t, y a sin t, 0 ≤ t ≤ , then
34. Since A area of semicircle
x y
1 a2 1 a2
a 2 cos 2 t a cos t dt
0
a 2 sin2 t a sin t dt
0
x, y 0,
a
4a 3
a
cos 3 t dt
0
0
sin3 t dt
a
1 sin2 t cos t dt
0
a cos 3 t cos t 3
0
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a sin3 t sin t 3
4a . 3
0
0
425
426
Chapter 14
Vector Analysis y
1 1 1 36. Since A 2ac ac, we have , 2 2A 2ac
(b, c) 2a
C1: y 0, dy 0
C3
C2
c c x a, d y C2: y dx ba ba c c x a, dy dx. ba ba
C3: y Thus,
1 2ac
x
1 2ac
y
x 2 dy
C
y 2 dx
C
x, y
38. A
1 2
1 2ac
a
b
a
1 0 2ac
0
b
a
x2
a
−a
c dx ba
a
x2
b
c 2 x a2 d x ba
x
C1
a
1 2abc b c 0 dx ba 2ac 3 3
a
b c a x a dx 2
2
b
1 c 2b a c 2b a c . 2ac 3 3 3
b3, 3c
a 2 cos 2 3 d
0
a2 2
1 cos 6 a2 sin 6 d 2 4 6
0
0
a 2 4
Note: In this case R is enclosed by r a cos 3 where 0 ≤ ≤ . 40. In this case, 0 ≤ ≤ 2 and we let u
sin 1 u2 2 du , cos , d . 1 cos 1 u2 1 u2
Now u ⇒ as ⇒ and we have
1 A2 2
18
0
0
9 d 9 2 cos 2
13 du 18 1 3u 2
0
6
3
2 63 1 u3u 2
0
0
2du 1 u2 18 1 u2 1 u 22 44 1 u2 1 u 22
3 arctan 3 u 6
23 6 du arctan 3 u 1 3u 22
3 0
3
3
0
0
0
1 u2 du 1 3u 22
1 u3u
12 1
3 2
3
3
2
3
1
2 3.
42. (a) Let C be the line segment joining x1, y1 and x2, y2. y dy
y2 y1 x x1 y`1 x2 x1 y2 y1 dx x2 x1
y dx x dy
C
x2
x1
y2 y1 y1 x 2 x1
x x 1
y y1 y2 y1 x x1 y1 x 2 x2 x1 x2 x1
x2 x1
x2
1
x1
y2 y1 y1 x 2 x1 2 x1
x
x1
dx x x
x1 y2 y1 y1x2 x1 x1 y2 x2 y1 —CONTINUED—
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y2 y1 y1 d x 2 x1
3u 2
du
0
Section 14.5 42. —CONTINUED— (b) Let C be the boundary of the region A Therefore,
dA
R
1 2
y dx x dy
C
1 2
y dx x dy y dx x dy . . .
1 2
C1
C2
1 1 dA
R
Parametric Surfaces
427
d A.
R
y dx x dy
Cn
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. Thus,
R
1 dA x1y2 x2 y1 x2 y3 x3 y2 . . . xn1 yn xn yn1 xn y1 x1yn . 2
44. Hexagon: 0, 0, 2, 0, 3, 2, 2, 4, 0, 3, 1, 1 A 12 0 0 4 0 12 4 6 0 0 3 0 0 21 2 46. Since
C
F N ds
f D N g ds
C
div F dA, then
R
f g
C
N ds
div f g dA
R
48.
f x dx g y dy
C
R
R
Section 14.5
f div g f g dA
g y f x dA x y
R
f 2g f g dA.
0 0 dA 0
R
Parametric Surfaces
2. ru, v u cos v i u sin v j uk
4. ru, v 4 cos ui 4 sin uj vk
x2 y2 z2
x 2 y 2 16
Matches d.
Matches a.
1 6. ru, v 2u cos v i 2u sin v j u 2 k 2 1 1 z u2, x 2 y 2 4u2 ⇒ z x2 y2 2 8 Paraboloid
8. ru, v 3 cos v cos ui 3 cos v sin u j 5 sin vk x 2 y 2 9 cos2 v cos2 u 9 cos2 v sin2 u 9 cos2 v x2 y2 z2 cos2 v sin2 v 1 9 25 x2 y2 z2 1 9 9 25
z 4
Ellipsoid z 4
4
y
5
x
x
4
3
3 4
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y
428
Chapter 14
Vector Analysis
For Exercises 10 and 12,
z
r u, v u cos vi u sin vj u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2.
5
Eliminating the parameter yields z x 2 y 2, 0 ≤ z ≤ 4. y
2
2 x
10. su, v u cos v i u 2 j u sin vk, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 y x2 z2 The paraboloid opens along the y-axis instead of the z-axis. 12. su, v 4u cos v i 4u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 z
x2 y2 16
The paraboloid is “wider.” The top is now the circle x 2 y 2 64. It was x 2 y 2 4. 14. ru, v 2 cos v cos ui 4 cos v sin u j sin vk, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2
16. ru, v 2u cos v i 2u sin v j v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 3 y z arctan x
z
x2 y2 z2 1 4 16 1
5 4 3
−5
−4
z
−5
8
4 5
3
4
−4 5
−3 −4 −5
x
0 ≤ u ≤
, 0 ≤ v ≤ 2 2
2 x
18. ru, v cos3 u cos v i sin3 u sin v j uk,
−4
−2
y 4
2 4
20. z 6 x y ru, v u i v j 6 u vk
z
2
−1
x
−1 1
1 y
22. 4x 2 y 2 16
24.
ru, v 2 cos ui 4 sin uj vk
x2 y2 z2 1 9 4 1 ru, v 3 cos v cos ui 2 cos v sin uj sin vk
28. Function: y x3 2, 0 ≤ x ≤ 4
26. z x 2 y 2 inside x 2 y 2 9. ru, v v cos u i v sin uj v 2 k, 0 ≤ v ≤ 3
Axis of revolution: x-axis x u, y u3 2 cos v, z u3 2 sin v 0 ≤ u ≤ 4, 0 ≤ v ≤ 2
30. Function: z 4 y 2, 0 ≤ y ≤ 2 Axis of revolution: y-axis x 4 u 2 cos v, y u, z 4 u 2 sin v 0 ≤ u ≤ 2, 0 ≤ v ≤ 2
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y
Section 14.5
1 34. ru, v 2u cosh v i 2u sinh v j u2 k, 2 ruu, v 2 cosh v i 2 sinh v j u k
32. ru, v u i v j uv k, 1, 1, 1 ruu, v i
Parametric Surfaces
v u k, rvu, v j k 2uv 2uv
rvu, v 2u sinh v i 2u cosh v j
At 1, 1, 1, u 1 and v 1.
At 4, 0, 2, u 2 and v 0.
1 1 ru1, 1 i k, rv1, 1 j k 2 2
ru2, 0 2i 2k, rv2, 0 4j
N ru rv 8i 8k
i j k N ru1, 1 rv1, 1 1 0 12 12 i 12 j k 1 0 1 2
Direction numbers: 1, 0, 1 Tangent plane: x 4 z 2 0 x z 2
Direction numbers: 1, 1, 2
Tangent plane: x 1 y 1 2z 1 0 x y 2z 0 36. ru, v 4u cos v i 4u sin v j u2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 ruu, v 4 cos v i 4 sin v j 2u k rvu, v 4u sin v i 4u cos v j ru rv
ru rv A
2
64u 4
256u2
8uu 2 4
2
8uu2 4 du dv
0
0
i j k 4 cos v 4 sin v 2u 8u 2 cos v i 8u 2 sin v j 16u k 4u sin v 4u cos v 0
2
0
128 1282 64 dv 22 1 3 3 3
38. ru, v a sin u cos v i a sin u sin v j a cos uk, 0 ≤ u ≤ π, 0 ≤ v ≤ 2 ruu, v a cos u cos v i a cos u sin v j a sin uk rvu, v a sin u sin v i a sin u cos v j
i j k ru rv a cos u cos v a cos u sin v a sin u a 2 sin2 u cos vi a 2 sin2 u sin v j a 2 sin u cos uk a sin u sin v a sin u cos v 0 ru rv A
2
0
a2
sin u
a2 sin u du dv 4a 2
0
40. ru, v a b cos v cos u i a b cos v sin u j b sin v k, a > b, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 ruu, v a b cos v sin u i a b cos v cos u j rvu, v b sin v cos u i b sin v sin u j b cos v k
i ru rv a b cos v sin u b sin v cos u
j a b cos v cos u b sin v sin u
k 0 b cos v
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 rv ba b cos v A
2
0
2
ba b cos v du dv 4 2ab
0
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429
430
Chapter 14
Vector Analysis
42. ru, v sin u cos v i u j sin u sin vk, 0 ≤ u ≤ , 0 ≤ v ≤ 2 ruu, v cos u cos v i j cos u sin vk rvu, v sin u sin v i sin u cos vk ru rv sin u cos v i cos u sin u j sin u sin vk ru rv sin u 1 cos2 u A
2
0
2 1
sin u 1 cos2 u du dv 22 ln
0
2 1
44. See the definition, page 1055. 46. Graph of ru, v u cos v i u sin v j vk 0 ≤ u ≤ , 0 ≤ v ≤ 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
x
3 x
48. ru, v 2u cos v i 2u sin v j vk, 0 ≤ u ≤ 1, 0 ≤ v ≤ 3 (b) If v
(a) If u 1: r1, v 2 cos v i 2 sin v j vk
r u,
x2 y2 4
z
0 ≤ z ≤ 3
2 2 u i 3 u j k 3 3
y 3 x
10 8
z
Helix 4
z
2 3
2
2
−2
−1
−1 1
−2 2
1
−2
Line
2 −2 x
2 : 3
2
1 2
x
y
(c) If one parameter is held constant, the result is a curve in 3-space. 50. x 2 y 2 z 2 1 Let x u cos v, y u sin v, and z u 2 1 . Then, ruu, v cos v i sin v j
u u 2 1
k
rvu, v u sin v i u cos v j. At 1, 0, 0, u 1 and v 0. ru1, 0 is undefined and rv1, 0 j. The tangent plane at 1, 0, 0 is x 1.
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y
y
Section 14.6
Surface Integrals
52. ru, v ui f u cos v j f u sin vk, a ≤ u ≤ b, 0 ≤ v ≤ 2 ruu, v i fu cos v j fu sin vk rvu, v f u sin v j f u cos vk
k i j ru rv 1 fu cos v fu sin v 0 f u sin v f u cos v ru rv f u1 fu2 A
2
0
f u fu i f u cos v j f u sin vk
b
f u1 fu2 du dv
a b
2
f x1 f x2 dx
since u x
a
Section 14.6
Surface Integrals z z 2, 3, dS 1 4 9 dy dx 14 dy dx x y
2. S: z 15 2x 3y, 0 ≤ x ≤ 2, 0 ≤ y ≤ 4,
2
x 2y z dS
4
0
S
x 2y 15 2x 3y14 dy dx
0
2
4
14
0
15 x y dy dx 12814
0
2 z z x 1 2, 0 4. S: z x 3 2, 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 3 x y
1
x 2y z dS
x
0
S
0
1
x
0
2 x 2y x 3 2 1 x 1 22 02 dy dx 3
0
2 x 2y x 3 2 1 x dy dx 3
1
2 3
x 5 2x 1 dx
0
1
2 1 5 2 x 1 x 3 2 3 4
6x 1
2
3 2
18 2
18 2
18 2
18
0
1 x 3 2
5 2
1 0
52 5 18 24 5 24
1
x3 21 x dx
0
x
5 1 12 3
3 2
1 x3 2
1 0
5 24
1
x1 21 x dx
0
1
x x2 dx
0
1
0
5 12
x
1 2
2
x 21x
1 dx 4
1
5 1 24 2
5 3 1 3 1 1 2 ln 2 ln 48 2 4 2 4 2
152 5 1 612 5 ln ln 3 22 0.2536 96 192 3 22 288 192
2
x
1 1 ln x x2 x 4 2
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0
431
432
Chapter 14
Vector Analysis z z 0 x y
6. S: z h, 0 ≤ x ≤ 2, 0 ≤ y ≤ 4 x 2,
4x 2
2
dx dS
0
S
xy dy dx
0
1 2
2
x4 x 2 dx
0
1 x4 2x 2 2 4
2 0
2
1 z 1 z 1 8. S: z xy, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, y, x 2 x 2 y 2
4
xy dS
0
S
160 1 y4 x4 dy dx 3904 15 3
4
2
5
2
xy
0
x , 0 ≤ y ≤ 2 2
10. S: z cos x, 0 ≤ x ≤
2
x 2 2xy dS
0
S
x 2
x 2 2xy1 sin2 x dy dx
0
2
0
x3 1 sin2 x dx 0.52 4
12. S: z a 2 x 2 y 2
x, y, z kz m
kz dS
S
z
a
1
ka2 x 2 y 2
R
ka 2 x 2 y 2
R
ka dA ka
R
x
a
y x2 y2
2
a 2 x 2 y 2
2
2
dA a
a dA a 2 x 2 y 2
x
dA ka2a2 2ka3
R
14. S: ru, v 2 cos u i 2 sin u j vk, 0 ≤ u ≤ 0 ≤ v ≤ 2
, 2
ru rv 2 cos ui 2 sin u j 2
2
2
x y dS
0
S
2 cos u 2 sin u2 du dv 16
0
16. S: ru, v 4u cos v i 4u sin v j 3uk, 0 ≤ u ≤ 4, 0 ≤ v ≤ ru rv 12u cos v i 12u sin v j 16uk 20u
x y dS
S
18. f x, y, z
4
4u cos v 4u sin v20u du dv
0
0
10,240 3
xy z
S: z x 2 y 2, 4 ≤ x 2 y 2 ≤ 16
f x, y, z dS
S
xy 1 4x 2 4y 2 dy dx x2 y2
S
2
r1 4r 2 sin cos dr d
2
0
4
2
0
2
0
65 1717 sin2
65
12
2
2
0
4
2
r 2 sin cos
1 4r 2 r dr d
r2
1 1 4r 23 2 12
0
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4 2
sin cos d
a
y
Section 14.6
Surface Integrals
20. f x, y, z x 2 y 2 z 2 S: z x 2 y 2, x 12 y 2 ≤ 1
f x, y, z dS
S
x 2 y 2 x 2 y 2
S
2
S
2
cos3 d
0
3 sin 16
x y y 2
2
2
2
dy dx
2
x 2 y 2 dy dx 2
16 3
x y2
2
2
S
x 2
2xx yy dy dx
2x 2 y 2
2
1
16 3
sin3
3
0
2 cos
r 2 dr d
0
1 sin2 cos d
0
0
0
22. f x, y, z x 2 y 2 z 2 S: x 2 y 2 9, 0 ≤ x ≤ 3, 0 ≤ z ≤ x Project the solid onto the xz-plane; y 9 x 2.
3
f x, y, z dS
0
S
x
0
3
0
3
0
1
x 2 9 x 2 z 2
x
9 z 2
0
3 dz dx 9 x 2
3 x3 9x dx 2 3 9 x
3
0
x 9 x 2
2
02 dz dx
z3 3 9z 3 9 x 2
3
x
dx
0
3
27x9 x 21 2 dx
0
x 39 x 21 2 dx
0
Let u x 2, dv x9 x 21 2 dx, then du 2x dx, v 9 x 2.
x 9 x
279 x 2
3
3
2
2
0
0
3
2 81 9 x 23 2 3
0
3
2x9 x 2 dx
0
81 18 99
24. Fx, y, z x i y j
y
S: 2x 3y z 6 (first octant)
3
Gx, y, z 2x 3y z 6
2
Gx, y, z 2 i 3 j k
S
F N dS
R
F G dA
3
0
3
0
2x 3 2
y = − 23 x + 2
1
R
2x 3y dy dx
x
1
0
4 3 2 x 2 4x x 2 3 2 3
4 3 2 x 3 2x 2 x 2 9 4 3
dx 2
3 3 0
12
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2
3
433
434
Chapter 14
Vector Analysis
26. Fx, y, z x i y j z k
y
S: x 2 y 2 z 2 36 (first octant)
5
z 36 x 2 y 2
4 3
Gx, y, z z 36 x 2 y 2 Gx, y, z F G
S
x2 + y2 = 62
6
R
2 1
x y i jk 2 2 36 x y 36 x 2 y 2
x 2
1
3
4
5
6
x2 y2 36 z 2 2 36 x y 36 x 2 y 2 36 x 2 y 2
F N dS
R
F G dA
36
36 x 2 y 2
R
2
6
0
36 36 r 2
0
dA
r dr d
(improper)
108 28. Fx, y, z x i y j 2z k
y
S: z a 2 x 2 y 2
x 2 + y2 ≤ a 2
a
Gx, y, z z a 2 x 2 y 2 −a
x y Gx, y, z i jk a 2 x 2 y 2 a 2 x 2 y 2
a
−a
x2 y2 3x 2 3y 2 2a 2 F G 2a 2 x 2 y 2 2 2 2 2 2 2 a x y a x y a 2 x 2 y 2
S
F N dS
R
F G dA
3x 2 3y 2 2a 2 dA a 2 x 2 y 2
R
2
a
3r 2 2a 2 r dr d
a 2 r 2
0
0
2
3
a
0
0
2
3
2
0
2
0
a
0
2 r 2a 2 r 2 a 2 r 23 2 3
0
3
r3 dr d 2a 2 a 2 r 2
2 3 a d 2a 2 3
2
r a 2 r 2 a
0
dr d
d 2a 2
2
0
a d 0
0
30. Fx, y, z x y i y j z k S: z 1 x 2 y 2, z 0 Gx, y, z z x 2 y 2 1 Gx, y, z 2x i 2y j k F G 2xx y 2y y 1 x 2 y 2 xx 2 2xy y 2 1
S
F N dS
R
F G dA
x 2 2xy y 2 1 dA
R
2
0
2
0
1
r 2 2r 2 cos sin 1r dr d
0
3 1 3 sin2
sin cos d 4 2 4 4
2
The flux across the bottom z 0 is zero.
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0
3 2
a
a
2
r2
0
d
x
Section 14.6 34. Orientable
32. 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. 36. E yz i xz j x y k S: z 1 x 2 y 2
N dS
E
S
E
1 xx
yz i xz j x yk
R
gxx, y i gyx, y j k dA
R
2xyz xy dA 1 x 2 y 2
R
y2
2
y
i
1 x 2 y 2
1x 2
1
3xy dA
R
j k dA
1 1x 2
3xy dy dx 0
38. x 2 y 2 z 2 a 2 z ± a 2 x 2 y 2
m2
k dS 2k
S
1
R
a
2
x x2 y2
a 2
a dA 2ka a 2 x 2 y 2
2k
R
2
a
0
0
2
y x2 y2
r a 2 r 2
2
dA
dr d
a
2ka a 2 r 2 2 4ka 2
0
kx 2 y 2 dS
Iz 2
S
x 2 y 2
2k
R
a dA 2ka x2 y2
a 2
2
0
a
0
r3 dr d (use integration by parts) r2
a 2
a
2 2ka r 2a 2 r 2 a 2 r 23 2 2 3 0 2ka
23a 2 32 a 4ka 32 a m 3
2
2
2
Let u r 2, dv ra 2 r 21 2 dr, du 2r dr, v a 2 r 2. 40. z x 2 y 2, 0 ≤ z ≤ h
z
Project the solid onto the xy-plane. Iz
h
x 2 y 21 dS
S
h
2
2
y
x y 1 4x 2 4y 2 dy dx 2
h hx 2
0
hx 2
2
x
h
r 21 4r 2 r dr d
0
12h 1 4h
3 2
1 2 1 4h5 2 120 120
1 4h3 2 1 4h3 26h 1 1 10h 1 4h 60 60 60
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Surface Integrals
435
436
Chapter 14
Vector Analysis
42. S: z 16 x 2 y 2 Fx, y, z 0.5zk
F N dS
S
F gxx, yi gyx, yj k dA
R
0.5z k
R
x 16 x 2 y 2
0.5 z dA
R
0.5
2
0
Section 14.7
i
y 16 x 2 y 2
j k dA
0.516 x 2 y 2 d
R
4
16 r 2r dr d 0.5
2
0
0
64 64 d 3 3
Divergence Theorem
2. Surface Integral: There are three surfaces to the cylinder. Bottom: z 0, N k, F N z 2
0 dS 0
S1
Top: z h, N k, F N z 2
h 2 dS h 2 Area of circle) 4h 2
S2
Side: ru, v 2 cos ui 2 sin u j v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ h ru 2 sin ui 2 cos u j, rv k ru rv 2 cos ui 2 sin uj F ru rv 8 cos2 u 8 sin2 u
S3
Therefore,
2
h
F N dS
0
8 cos2 u 8 sin2 u du dv 0
0
F N dS 0 4h2 0 4h2.
S
Divergence Theorem: div F 2 2 2z 2z
2z dV
2
0
Q
2
0
h
2zr dz dr d 4h 2.
0
z
h
x
2
2
y
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Section 14.7 4. Fx, y, z xyi z j x y k S: surface bounded by the planes y 4, z 4 x and the coordinate planes Surface Integral: There are five surfaces to this solid.
N x y
z 0, N k, F
4
x y dS
4
0
S1
0
4
z dS
4x
0
S2
0
4
z dS
0
S3
4
z dz dx
0
y 4, N j, F N z
4x 8 dx 64
0
y 0, N j, F N z
4
x y dy dx
4x
32 4 x2 dx 2 3
4
z dz dx
0
32 4 x2 dx 2 3
0
x 0, N i, F N xy
x z 4, N
S5
4
0
S4
4
xy dS
0 dS 0
0
i k 1 , FN xy x y , dS 2 dA 2 2
1 xy x y 2 dA 2
Therefore,
F N dS 64 S
4
0
32 3
4
xy x y dy dx 128
0
32 3 0 128 64.
Divergence Theorem: Since div F y, we have
4
div F dV
4
0
Q
4x
0
y dz dy dx 64.
0
6. Since div F 2xz 2 2 3xy we have
a
div F dV
0
Q
a
0
a
a
0
a
2xz2 2 3xy dz dy dx
0
0
a
0
2 3 xa 2a 3xya dy dx 3
2 4 3 xa 2a2 xa3 dx 3 2
3 1 a6 2a3 a5. 3 4 8. Since div F y z y z, we have
a 2x 2
a
div F dV
Q
a 2x 2y 2
a a 2x 2 0 2
a
0
0
z dz dy dx
a 2r r 3 dr d 2 2
2
0
2
0
a
0
a 2r 2 r 4 4 8
a 2r 2
0
a
0
d
zr dz dr d
2 4 a
0
8
d
10. Since div F xz, we have
xz dV
Q
4
0
3
9y 2
3 9y
4
xz dx dy dz 2
0
3
z 0 dy dz 0. 2 3
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a 4 . 4
Divergence Theorem
437
438
Chapter 14
Vector Analysis
12. Since div F y 2 x 2 ez, we have
16
x 2 y 2 ez dV
2
16
2
0
8
r 2
0
Q
16. div F 2
S
6
3ez dV
F N dS
4
4y
0
6
3e z dz dy dx
0
2
0
16
0
1 8r 3 re8 r 4 re r 2 dr d 2
131,052 262,104 100e8 d 200e8 5 5
14. Since div F e z e z e z 3e z, we have
x 2 y 2 ez dz dy dx
1 2x 2 y 2
r 2 ezr dz dr d
0
0
8
256x2
0
Q
256x2
4
0
div F dV
Q
6
3e 4y 1 dy dx
0
3e 4 5 dx 18e 4 5.
0
2 dV.
Q
The surface S is the upper half of a hemisphere of radius 2. Since the volume is 12 43 23 16 3, you have
32 . 3
F N dS 2Volume
S
18. Using the Divergence Theorem, we have
S
curl F N dS
div curlF dV
Q
j y yz sin x
i curl Fx, y, z x xy cos z
k z xyz
xz y sin x i yz xy sin z j yz cos x x cos z k.
Now, div curl Fx, y, z z y cos x z x sin z y cos x x sin z 0. Therefore,
S
curl F N dS
div curl F dV 0.
Q
a
20. If div Fx, y, z > 0, then source.
22. v
0
If div Fx, y, z < 0, then sink.
a
0
a
Similarly,
If div Fx, y, z 0, then incompressible.
a
x dy dz
0
0
a
0
a
a
y dz dx
0
a
a dy dz
0
a 2 dz a 3
0
a
z dx dy a 3.
0
24. If Fx, y, z a1 i a 2 j a 3 k, then div F 0. Therefore,
S
F N dS
div F dV
Q
0 dV 0.
Q
26. If Fx, y, z x i yj z k, then div F 3. 1 F
28.
S
F N dS
1 F
div F dV
1 F
Q
f D N g gD N f dS
S
S
3 dV
Q
f D N g dS
dV
Q
gD N f dS
S
f 2g f g dV
Q
3 F
g 2f g f dV
Q
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f 2g g 2f dV
Q
Section 14.8
Section 14.8
Stokes’s Theorem
2. F x, y, z x 2 i y 2 j x 2 k i curl F x
j y
2
2
x
Stokes’s Theorem
k z x2
y
4. F x, y, z x sin y i y cos x j yz 2 k
2x j
curl F
i x
j y
k z
x sin y
y cos x
yz 2
z 2 i y sin x x cos yk
6. F x, y, z arcsin y i 1 x 2 j y 2 k
curl F
i x
j y
k z
arcsin y 1 x2
2y i
1x x
2y i
1 x x
y2
2
2
1
1 y 2
1 1 y 2
k k
8. In this case C is the circle x 2 y 2 4, z 0, dz 0.
Line Integral:
C
F dr
y dx x dy
C
Let x 2 cos t, y 2 sin t, then dx 2 sin t dt, dy 2 cos t dt, and
y dx x dy
C
Double Integral: Fx, y, z z x 2 y 2 4, N curl F 2k, therefore
curl F NdS
2
2 dA
R
4x 2
2
2
4
4x 2
2
2 dy dx 2
2
24 x 2 dx
x 2
2
2
1
C3
C2: z y 2, x 0, dx 0, dz 2y dy C3: y a, z a2, dy dz 0
F dr
1
C4
x
y
z 2 dx x 2 dy y 2 dz
C
0
C1
2y3 dy
C2
a
C2
C1 1
C4: z y 2, x a, dx 0, dz 2y dy.
C
8.
z
C1: y 0, z 0, dy dz 0
4 dt 8.
0
10. Line Integral: From the accompanying figure we see that for
Hence,
2
F 2xi 2yj k , dS 1 4x 2 4y 2 dA F 1 4x 2 4y 2
4 x 2 dx 2 x4 x 2 4 arcsin
2
a
2y3 dy
0
C4
0
a4 dx
0
a2 dy y22y dy
a4 dx
C3
0
a 2 dy
a
a
a y
2y3 dy a4 x
a
0
2
0
—CONTINUED—
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a
a5 a 3 a 3a2 1.
439
440
Chapter 14
Vector Analysis
10. —CONTINUED— Double Integral: Since Fx, y, z y 2 z, we have N
2yj k 1 4y 2
and dS 1 4y 2 dA.
Furthermore, curl F 2y i 2z j 2 x k. Therefore,
S
curl F N dS
4yz 2x dA
a
0
R
a
a
4y2 2x dy dx
0
a 4 2ax dx a4 x ax 2
0
\
a 0
a 3 a2 1.
\
12. Let A 0, 0, 0, B 1, 1, 1, and C 0, 0, 2. Then U AB i j k, and V AC 2k, and N
2i 2 j i j U V . U V 2 22
Hence, Fx, y, z x y and dS 2 dA. Since curl F
2x k, we have x2 y2
S
curl F N dS
0 dS 0.
R
14. Fx, y, z 4xz i y j 4xyk, S: 9 x 2 y 2, z ≤ 0 curl F 4x i 4x 4y j Gx, y, z x 2 y 2 z 9 Gx, y, z 2x i 2y j k
S
curl F N dS
8x 2 2y 4x 4y dA
R
3
9x 2
3
9x 2
8x 2 8xy 8y 2 dy dx
3
16x29 x2
3
16 9 x23 2 dx 0 3
16. Fx, y, z x 2 i z 2 j xyz k, S: z 4 x 2 y 2 i curl F x
j y
k z
x2
z2
xyz
xz 2z i yzj
Gx, y, z z 4 x 2 y 2 Gx, y, z
S
x 4 x 2 y 2
curl F N dS
y
i
4 x 2 y 2
R
jk
zx 2x y 2z dA 2 2 4 x y 4 x 2 y 2
R
2
2 2
2 2
2
x 2 y 2xy
y3 3
4x 2
2 4x 2
x 2 2x y 2 dy dx
4x2
4x2
dx
2 2x 24 x 2 4x4 x 2 4 x 24 x 2 dx 3
8 8 x 24 x 2 4x4 x 2 4 x 2 dx 3 3
3 8 x2x
2
xx 2 y 2 dA
8 1
4 x 2 16 arcsin
24
3 8 3 2 3 8 3 2 0 1
4
1
x4 x
x 8 1 4 4 x 2 3 2 2 3 3 2
4
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2
4 arcsin
x 2
2 2
Section 14.8
18. Fx, y, z yz i 2 3y j x 2 y 2 k j y
i curl F x
k z
2 3y x 2 y 2
yz
2y i y 2x j z k
S: the first octant portion of x z 2 16 over x 2 y 2 16 2
Gx, y, z z 16 x 2 Gx, y, z
x 16 x 2
ik
curl F N dS
S
R
R
16 x 2
4
0
0
4
0
z dA
16 x 2 dA
162xy x
2
16x 2
dx 0
x 16 x 2 16 x 2 dx
64 64 64 3 3 3
1 x3 16 x 2 3 2 16x 3 3 64
16 x 2 dy dx
x y 2 16 x 2 y 16 x 2
0
2xy
16x 2
4
2xy 16 x 2
4 0
20. Fx, y, z xyz i y j z k
curl F
i x
j y
k z
xyz
y
z
xyj xzk
S: the first octant portion of z x 2 over x 2 y 2 a 2. We have N
S
2x i k and dS 1 4x 2 dA. 1 4x 2
curl F N dS
xz dA
R
a
0
a 2 x2
x 3 dA
R
x 3 dy dx
0
a
x 3a 2 x 2 dx
0
1 2 x 2a 2 x 2 3 2 a 2 x 2 5 2 3 15
a 0
2 5 a 15
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Stokes’s Theorem
441
442
Chapter 14
Vector Analysis
22. Fx, y, z z i yk
S: x 2 y 2 1 i curl F x z
j y 0
k z y
ij
Letting N k, curl F N 0 and
S
curl F N dS 0.
24. curl F measures the rotational tendency. See page 1084. 26. f x, y, z xyz, gx, y, z z, S: z 4 x 2 y 2 (a) gx, y, z k f x, y, z gx, y, z xyzk rt 2 cos t i 2 sin t j 0k, 0 ≤ t ≤ 2
C
f x, y, z gx, y, z dr 0
(b) f x, y, z yz i xz j xyk gx, y, z k
i f g yz 0
j xz 0
x
N
4 x 2 y 2
dS
1
S
i
k xy xz i yz j 1 y
4 x 2 y 2
x 4 x 2 y 2
jk
4 yx y 2
f x, y, z gx, y, z N dS
2 4 x 2 y 2
dA
S
x 2z y 2z 2 dA 2 2 4 x y 4 x 2 y 2 4 x 2 y 2
2
0
2
dA
2
S
2
2
0
2x 2 y 2 dA 4 x 2 y 2 2
0
2r 2cos 2 sin 2 r d dr 4 r 2 2r 3
1 sin 2
4 r 2 2
2 0
dr 0
Review Exercises for Chapter 14 2. Fx, y i 2yj
4. f x, y, z x 2eyz Fx, y, z 2xeyz i x 2ze yz j x 2 ye y z k
y 5 4 3 2
xeyz2i xz j x y k x
−1 −2 −3 −4 −5
2
4
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Review Exercises for Chapter 14
443
6. Since M y 1 x 2 N x, F is conservative. From M U x y x 2 and N U y 1 x, partial integration yields U y x h y and U y x gx which suggests that Ux, y y x C. 8. Since M y 6y 2 sin 2x N x, F is conservative. From M U x 2y3 sin 2x and N U y 3y 21 cos 2x, we obtain U y3 cos 2x h y and U y31 cos 2x gx which suggests that h y y3, gx C, and Ux, y y31 cos 2x C. 10. Since
12. Since
N M 4x , y x
M N sin z , y x
P M 2z , z x
M P y cos z , z x
F is not conservative.
P N 6y z y F is not conservative. 16. Since F 3x y i y 2z j z 3x k:
14. Since F xy 2 j z x 2 k; (a) div F 2xy
(a) div F 3 1 1 5
x2
(b) curl F 2xz j y
(b) curl F 2i 3j k
2k
z z 20. Since F i j z 2 k: x y
18. Since F x 2 y i x sin2 y j: (a) div F 2x 2 sin y cos y
(a) div F
(b) curl F 0
z 1 1 z 2z z 2 2 2 x2 y 2 x y
1 1 (b) curl F i j y x 22. (a) Let x 5t, y 4t, 0 ≤ t ≤ 1, then ds 41 dt.
1
xy ds
20t 2 41 dt
0
C
y
2041 3
4 3
(b) C1: x t, y 0, 0 ≤ t ≤ 4, ds dt
2
4
xy ds
0
C
0
C
2
t2 t3
2
3 1 0
4
0 dt
85 . 3
dy dx 1 cos t, sin t dt dt
t sin t1 cos t2 sin t2 dt
0
3
0
24. x t sin t, y 1 cos t, 0 ≤ t ≤ 2, x ds
x
C1
2
8t 8t 2 25 dt
165
1
0 dt
C2 (4, 0)
C3: x 0, y 2 t, 0 ≤ t ≤ 2, ds dt
y = − 12 x + 2
C3
C2: x 4 4t, y 2t, 0 ≤ t ≤ 1, ds 25 dt
Therefore,
(0, 2)
2
t sin t2 2 cos t dt
0
2
2
0
2
2
2
t1 cos t sin t1 cos t dt 2 32 1 cos t3 2 0 t1 cos t dt
0
8
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2
2
0
t1 cos t dt
444
Chapter 14
Vector Analysis
26. x cos t t sin t, y sin t t sin t, 0 ≤ t ≤
2x y dx x 3y dy
2
, dx t cos t dt, dy cos t t cos t sin t dt 2
sin t cos t5t 2 6t 2 cos2 tt 1 sin2 t2t 3 dt 1.01
0
C
28. r t t i t 2 j t 3 2 k, 0 ≤ t ≤ 4 3 x t 1, y t 2t, z t t1 2 2
1 4t
4
x 2 y 2 z2 ds
t 2 t 4 t3
0
C
2
9 t dt 2080.59 4
30. f x, y 12 x y C: y x 2 from 0, 0 to 2, 4 rt t i t 2 j, 0 ≤ t ≤ 2 r t i 2t j r t 1 4t 2 Lateral surface area:
2
f x, y ds
12 t t 21 4t 2 dt 41.532
0
C
32. dr 4 sin t i 3 cos t j dt F 4 cos t 3 sin t i 4 cos t 3 sin t j, 0 ≤ t ≤ 2
C
F dr
2
12 7 sin t cos t dt 12t
0
7 sin2 t 2
2
0
24
34. x 2 t, y 2 t, z 4t t 2, 0 ≤ t ≤ 2 dr
i j 24tt t k dt 2
F 4 2t 4t t 2 i 4t t 2 2 t j 0k
C
F dr
2
t 2 dt
0
t2 2t
2
2 0
2
36. Let x 2 sin t, y 2 cos t, z 4 sin2 t, 0 ≤ t ≤ . dr 2 cos t i 2 sin t j 8 sin t cos t k dt F 0 i 4 j 2 sin t k
F dr
F dr
2x y dx 2y x dy
C
38.
C
8 sin t 16 sin2 t cos t dt 8 cos t
0
16 3 sin t 3
0
16
C
rt 2 cos t 2t sin t i 2 sin t 2t cos t j, 0 ≤ t ≤
C
F dr 4 2 4
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Review Exercises for Chapter 14
40. rt 10 sin t i 10 cos t j 10 sin t i 10 cos t j
445
2000 5280 t k, 0 ≤ t ≤ 2 2 25 tk 33
F 20k
dr 10 cos t i 10 sin t j
F dr
y dx x dy
C
42.
2
500 250 dt mi ton 33 33
0
C
25 k 33
4, 4, 4)
0, 0, 1 16 ln 4
1 dz xy ln z z
44. x a sin , y a1 cos , 0 ≤ ≤ 2 1 (a) A 2
y
x dy y dx.
C
Since these equations orient the curve backwards, we will use A
1 2
1 2
C1
y dx x dy
2
C2
a21 cos 1 cos a2 sin sin d
0
a2 2 a2 2
2
1 2
2 a
2π a
0 0 d
0
1 2 cos cos2 sin sin2 d
0 2
2 2 cos sin d
0
a2 6 3a2. 2
(b) By symmetry, x a. From Section 14.4, y
46.
1 2A
y 2 dx
C
1 2A
2
0
2
xy dx x 2 y 2 dy
0
C
a31 cos 21 cos d
2
2x x dy dx
48.
0
5 1 a35 a 23a2 6
C
2
y 2 dx x 4 3 dy
C
1
1x 2 33 2
1
(1x 2 3 3 2
1
4 1 3 x yy 2 3
1 1
1
43 x
1 3
a2 x 2
a a2 x 2 a
2x dx 4
0
50.
a
x 2 y 2 dx 2xy dy
2y dy dx
1x2 33 2
1x 2 33 2
dx
8 1 3 x 1 x2 33 2 dx 3
8 16 x 2 3 1 x2 35 2 1 x2 35 2 7 35
1 1
0
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a
0 dx 0
4y dy dx
x
446
Chapter 14
Vector Analysis
u 52. r u, v eu 4 cos v i eu 4 sin v j k 6 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 z
2
2
2 x
y
54. S: ru, v u v i u v j sin v k,
0 ≤ u ≤ 2, 0 ≤ v ≤
ru u, v i j ru u, v i j cos v k
i j k ru rv 1 1 0 1 1 cos v
cos v i cos v j 2 k
ru rv 2 cos2 v 4
z dS
S
0
2
sin v2 cos2 v 4 du dv 2 6 2 ln
0
66 22
56. (a) z aa x2 y2, 0 ≤ z ≤ a2
z
z 0 ⇒ x2 y2 a2
a
2
(b) S: gx, y z a2 ax2 y2
x, y kx2 y2 m
ex, y, z dS
x
S
a
a
y
kx2 y2 1 gx2 gy2 dA
R
k
1 x ax y 2 2
x2 y2
2
R
k
R
2
a2y2 dA x y2 2
a2 1 x2 y2 dA
ka2 1
2
0
ka2 1
2
0
a
r2 dr d
0
a3 d 3
2 ka2 1 a3 3
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Review Exercises for Chapter 14 58. Fx, y, z x i y j z k Q: solid region bounded by the coordinate planes and the plane 2x 3y 4z 12
z
(0, 0, 3)
Surface Integral: There are four surfaces for this solid. z0 y 0,
2x 3y 4z 12, N
S4
N F dS
1 4
1 4
(0, 4, 0)
0 dS 0
x
(6, 0, 0)
S2
F N x,
N i,
0 dS 0
S1
F N y,
N j,
x 0,
F N z,
N k,
0 dS 0
S3
2i 3j 4k , dS 29
1 14 169 dA
29
dA
4
2x 3y 4z dy dx
R
6
(122x) 3
0
6
12 dy dx 3
0
4
0
2x x2 dx 3 4x 3 3
6 0
36
Triple Integral: Since div F 3, the Divergence Theorem yields.
div F dV
Q
13 Area of baseHeight 21 643 36.
3 dV 3Volume of solid 3
Q
60. Fx, y, z x z i y z j x 2 k
z
(0, 0, 6)
S: first octant portion of the plane 3x y 2z 12 Line Integral:
C
C1: y 0,
dy 0,
z
12 3x , 2
3 dz dx 2
C2: x 0,
dx 0,
z
12 y , 2
1 dz dy 2
C3: z 0,
dz 0,
y 12 3x,
F dr
(4, 0, 0) (0, 12, 0)
y
dy 3 dx
x z dx y z dy x 2 dz
C
x
C1 0
x
4
dx y 12 2 y dy
12 3x 3 x2 2 2
3 5 x 2 x 6 dx 2 2
Double Integral: Gx, y, z
12
0
C2
32 y 6 dy
x 12 3x3 dx
C3
4
10x 36 dx 8
0
12 3x y z 2
3 1 Gx, y, z i j k 2 2 curl F i 2x 1 j
S
curl F N dS
4
0
123x
0
4
x 1 dy dx
3x 2 15x 12 dx 8
0
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y
447
442
Chapter 14
Vector Analysis
22. Fx, y, z z i yk
S: x 2 y 2 1 i curl F x z
j y 0
k z y
ij
Letting N k, curl F N 0 and
S
curl F N dS 0.
24. curl F measures the rotational tendency. See page 1084. 26. f x, y, z xyz, gx, y, z z, S: z 4 x 2 y 2 (a) gx, y, z k f x, y, z gx, y, z xyzk rt 2 cos t i 2 sin t j 0k, 0 ≤ t ≤ 2
C
f x, y, z gx, y, z dr 0
(b) f x, y, z yz i xz j xyk gx, y, z k
i f g yz 0
j xz 0
x
N
4 x 2 y 2
dS
1
S
i
k xy xz i yz j 1 y
4 x 2 y 2
x 4 x 2 y 2
jk
4 yx y 2
f x, y, z gx, y, z N dS
2 4 x 2 y 2
dA
S
x 2z y 2z 2 dA 2 2 4 x y 4 x 2 y 2 4 x 2 y 2
2
0
2
dA
2
S
2
2
0
2x 2 y 2 dA 4 x 2 y 2 2
0
2r 2cos 2 sin 2 r d dr 4 r 2 2r 3
1 sin 2
4 r 2 2
2 0
dr 0
Review Exercises for Chapter 14 2. Fx, y i 2yj
4. f x, y, z x 2eyz Fx, y, z 2xeyz i x 2ze yz j x 2 ye y z k
y 5 4 3 2
xeyz2i xz j x y k x
−1 −2 −3 −4 −5
2
4
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Review Exercises for Chapter 14
443
6. Since M y 1 x 2 N x, F is conservative. From M U x y x 2 and N U y 1 x, partial integration yields U y x h y and U y x gx which suggests that Ux, y y x C. 8. Since M y 6y 2 sin 2x N x, F is conservative. From M U x 2y3 sin 2x and N U y 3y 21 cos 2x, we obtain U y3 cos 2x h y and U y31 cos 2x gx which suggests that h y y3, gx C, and Ux, y y31 cos 2x C. 10. Since
12. Since
N M 4x , y x
M N sin z , y x
P M 2z , z x
M P y cos z , z x
F is not conservative.
P N 6y z y F is not conservative. 16. Since F 3x y i y 2z j z 3x k:
14. Since F xy 2 j z x 2 k; (a) div F 2xy
(a) div F 3 1 1 5
x2
(b) curl F 2xz j y
(b) curl F 2i 3j k
2k
z z 20. Since F i j z 2 k: x y
18. Since F x 2 y i x sin2 y j: (a) div F 2x 2 sin y cos y
(a) div F
(b) curl F 0
z 1 1 z 2z z 2 2 2 x2 y 2 x y
1 1 (b) curl F i j y x 22. (a) Let x 5t, y 4t, 0 ≤ t ≤ 1, then ds 41 dt.
1
xy ds
20t 2 41 dt
0
C
y
2041 3
4 3
(b) C1: x t, y 0, 0 ≤ t ≤ 4, ds dt
2
4
xy ds
0
C
0
C
2
t2 t3
2
3 1 0
4
0 dt
85 . 3
dy dx 1 cos t, sin t dt dt
t sin t1 cos t2 sin t2 dt
0
3
0
24. x t sin t, y 1 cos t, 0 ≤ t ≤ 2, x ds
x
C1
2
8t 8t 2 25 dt
165
1
0 dt
C2 (4, 0)
C3: x 0, y 2 t, 0 ≤ t ≤ 2, ds dt
y = − 12 x + 2
C3
C2: x 4 4t, y 2t, 0 ≤ t ≤ 1, ds 25 dt
Therefore,
(0, 2)
2
t sin t2 2 cos t dt
0
2
2
0
2
2
2
t1 cos t sin t1 cos t dt 2 32 1 cos t3 2 0 t1 cos t dt
0
8
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2
2
0
t1 cos t dt
444
Chapter 14
Vector Analysis
26. x cos t t sin t, y sin t t sin t, 0 ≤ t ≤
2x y dx x 3y dy
2
, dx t cos t dt, dy cos t t cos t sin t dt 2
sin t cos t5t 2 6t 2 cos2 tt 1 sin2 t2t 3 dt 1.01
0
C
28. r t t i t 2 j t 3 2 k, 0 ≤ t ≤ 4 3 x t 1, y t 2t, z t t1 2 2
1 4t
4
x 2 y 2 z2 ds
t 2 t 4 t3
0
C
2
9 t dt 2080.59 4
30. f x, y 12 x y C: y x 2 from 0, 0 to 2, 4 rt t i t 2 j, 0 ≤ t ≤ 2 r t i 2t j r t 1 4t 2 Lateral surface area:
2
f x, y ds
12 t t 21 4t 2 dt 41.532
0
C
32. dr 4 sin t i 3 cos t j dt F 4 cos t 3 sin t i 4 cos t 3 sin t j, 0 ≤ t ≤ 2
C
F dr
2
12 7 sin t cos t dt 12t
0
7 sin2 t 2
2
0
24
34. x 2 t, y 2 t, z 4t t 2, 0 ≤ t ≤ 2 dr
i j 24tt t k dt 2
F 4 2t 4t t 2 i 4t t 2 2 t j 0k
C
F dr
2
t 2 dt
0
t2 2t
2
2 0
2
36. Let x 2 sin t, y 2 cos t, z 4 sin2 t, 0 ≤ t ≤ . dr 2 cos t i 2 sin t j 8 sin t cos t k dt F 0 i 4 j 2 sin t k
F dr
F dr
2x y dx 2y x dy
C
38.
C
8 sin t 16 sin2 t cos t dt 8 cos t
0
16 3 sin t 3
0
16
C
rt 2 cos t 2t sin t i 2 sin t 2t cos t j, 0 ≤ t ≤
C
F dr 4 2 4
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Review Exercises for Chapter 14
40. rt 10 sin t i 10 cos t j 10 sin t i 10 cos t j
445
2000 5280 t k, 0 ≤ t ≤ 2 2 25 tk 33
F 20k
dr 10 cos t i 10 sin t j
F dr
y dx x dy
C
42.
2
500 250 dt mi ton 33 33
0
C
25 k 33
4, 4, 4)
0, 0, 1 16 ln 4
1 dz xy ln z z
44. x a sin , y a1 cos , 0 ≤ ≤ 2 1 (a) A 2
y
x dy y dx.
C
Since these equations orient the curve backwards, we will use A
1 2
1 2
C1
y dx x dy
2
C2
a21 cos 1 cos a2 sin sin d
0
a2 2 a2 2
2
1 2
2 a
2π a
0 0 d
0
1 2 cos cos2 sin sin2 d
0 2
2 2 cos sin d
0
a2 6 3a2. 2
(b) By symmetry, x a. From Section 14.4, y
46.
1 2A
y 2 dx
C
1 2A
2
0
2
xy dx x 2 y 2 dy
0
C
a31 cos 21 cos d
2
2x x dy dx
48.
0
5 1 a35 a 23a2 6
C
2
y 2 dx x 4 3 dy
C
1
1x 2 33 2
1
(1x 2 3 3 2
1
4 1 3 x yy 2 3
1 1
1
43 x
1 3
a2 x 2
a a2 x 2 a
2x dx 4
0
50.
a
x 2 y 2 dx 2xy dy
2y dy dx
1x2 33 2
1x 2 33 2
dx
8 1 3 x 1 x2 33 2 dx 3
8 16 x 2 3 1 x2 35 2 1 x2 35 2 7 35
1 1
0
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a
0 dx 0
4y dy dx
x
446
Chapter 14
Vector Analysis
u 52. r u, v eu 4 cos v i eu 4 sin v j k 6 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 z
2
2
2 x
y
54. S: ru, v u v i u v j sin v k,
0 ≤ u ≤ 2, 0 ≤ v ≤
ru u, v i j ru u, v i j cos v k
i j k ru rv 1 1 0 1 1 cos v
cos v i cos v j 2 k
ru rv 2 cos2 v 4
z dS
S
0
2
sin v2 cos2 v 4 du dv 2 6 2 ln
0
66 22
56. (a) z aa x2 y2, 0 ≤ z ≤ a2
z
z 0 ⇒ x2 y2 a2
a
2
(b) S: gx, y z a2 ax2 y2
x, y kx2 y2 m
ex, y, z dS
x
S
a
a
y
kx2 y2 1 gx2 gy2 dA
R
k
1 x ax y 2 2
x2 y2
2
R
k
R
2
a2y2 dA x y2 2
a2 1 x2 y2 dA
ka2 1
2
0
ka2 1
2
0
a
r2 dr d
0
a3 d 3
2 ka2 1 a3 3
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Review Exercises for Chapter 14 58. Fx, y, z x i y j z k Q: solid region bounded by the coordinate planes and the plane 2x 3y 4z 12
z
(0, 0, 3)
Surface Integral: There are four surfaces for this solid. z0 y 0,
2x 3y 4z 12, N
S4
N F dS
1 4
1 4
(0, 4, 0)
0 dS 0
x
(6, 0, 0)
S2
F N x,
N i,
0 dS 0
S1
F N y,
N j,
x 0,
F N z,
N k,
0 dS 0
S3
2i 3j 4k , dS 29
1 14 169 dA
29
dA
4
2x 3y 4z dy dx
R
6
(122x) 3
0
6
12 dy dx 3
0
4
0
2x x2 dx 3 4x 3 3
6 0
36
Triple Integral: Since div F 3, the Divergence Theorem yields.
div F dV
Q
13 Area of baseHeight 21 643 36.
3 dV 3Volume of solid 3
Q
60. Fx, y, z x z i y z j x 2 k
z
(0, 0, 6)
S: first octant portion of the plane 3x y 2z 12 Line Integral:
C
C1: y 0,
dy 0,
z
12 3x , 2
3 dz dx 2
C2: x 0,
dx 0,
z
12 y , 2
1 dz dy 2
C3: z 0,
dz 0,
y 12 3x,
F dr
(4, 0, 0) (0, 12, 0)
y
dy 3 dx
x z dx y z dy x 2 dz
C
x
C1 0
x
4
dx y 12 2 y dy
12 3x 3 x2 2 2
3 5 x 2 x 6 dx 2 2
Double Integral: Gx, y, z
12
0
C2
32 y 6 dy
x 12 3x3 dx
C3
4
10x 36 dx 8
0
12 3x y z 2
3 1 Gx, y, z i j k 2 2 curl F i 2x 1 j
S
curl F N dS
4
0
123x
0
4
x 1 dy dx
3x 2 15x 12 dx 8
0
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y
447
448
Chapter 14
Vector Analysis
Problem Solving for Chapter 14
z
z x y ,
x 1 x2 y2 y 1 x2 y2
2. (a) z 1 x2 y2, dS
x z
y z
T
25 x i y i z k 25x i yj z k x2 y2 z232
2
N
1 dA
1 1 x2 y2
z
z i jk
x
y
x z
y z 2
2
x
2
i
1 x2 y2
1 y 1 x2 y2
j k 1 x2 y2
x i y j 1 x2 y2 k x i y j z k Flux
k T
N dS
S
25x i y j z k x i y j z k
k
R
25
k
1 x2 y2
R
25k
2
0
1
0
1
1 r 2
1 1 x2 y2
dA
dA r dr d 50k
(b) ru, v 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 rv sin2 u cos v, sin2 u sin v, sin u cos u sin2 v sin u cos u cos2 v ru rv sin u
Flux 25k
2
0
2
sin u du dv 50k
0
t2 , t, 2 32t 2
4. rt
32
r t t, 1, 2t12, r t t 1
ds Iy
1 t 1 dt 1 1t 1
C
Ix
0
y2 z2 ds
0
1
x2 y2 ds
C
0
t4 8 3 49 t dt 4 9 180
1
C
Iz
x2 z2 ds
8 5 t 2 t 3 dt 9 9
t4 23 t 2 dt 4 60
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Problem Solving for Chapter 14
6.
1 x dy y dx 2 2 C
2
0
12 sin 2t cos t sin t cos 2t dt 2 23
Hence, the area is 43. 8. F x, y 3x2 y 2 i 2x 3 y j is conservative. f x, y x3y2 potential function. Work f 2, 4 f 1, 1 816 1 127 10. Area ab rt a cos t i b sin t j, 0 ≤ t ≤ 2 r t a sin t i b cos t j 1 1 F b sin t i a cos t j 2 2 F
dr 2 ab sin2 t
W
1
2
0
1 1 ab cos2 t dt ab 2 2
1 F dr ab2 ab 2
Same as area.
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449
C H A P T E R 14 Vector Analysis Section 14.1 Vector Fields
. . . . . . . . . . . . . . . . . . . . . . . . . . . 178
Section 14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Section 14.3 Conservative Vector Fields and Independence of Path . . . . . . 190 Section 14.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 193 Section 14.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 198 Section 14.6 Surface Integrals
. . . . . . . . . . . . . . . . . . . . . . . . . 202
Section 14.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 208 Section 14.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 211 Review Exercises
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
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C H A P T E R Vector Analysis Section 14.1
14
Vector Fields
Solutions to Odd-Numbered Exercises 1. All vectors are parallel to y-axis. Matches (c)
3. All vectors point outward. Matches (b)
5. Vectors are parallel to x-axis for y n . Matches (a)
9. Fx, y x i yj
7. Fx, y i j
11. Fx, y, z 3yj
F x2 y2 c
F 2
x2
y
y2
F 3 y c
c2
z 4
y
5 1 x
−4
x
x
−5
3
2
y
4
5
−4 −5
13. Fx, y 4x i yj F x2 c 216
16x2
y2
15. Fx, y, z i j k c
17.
F 3
2
y 1 c2
2 z
1 y
4
−2
y
1
1
2
−2 4
x
−1
x
−1 −1
4 −4
2
−2
y
x
−4
2
−2
21. f x, y 5x2 3xy 10y2
z
19.
1
23. f x, y, z z ye x
2
2
fxx, y 10x 3y
fxx, y, z 2x ye x
1
fyx, y 3x 20y
fyx, y, z e x
1
2
y
Fx, y 10x 3y i 3x 20y j
2
2
fz 1 Fx, y, z 2 x ye x i e x j k 2
2 x
178
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2
Section 14.1
Vector Fields
179
25. gx, y, z xy lnx y xy xy xy gyx, y, z x lnx y xy gxx, y, z y lnx y
gzx, y, z 0 Gx, y, z
x xy y y lnx y i x xy y x lnx y j
27. Fx, y 12xyi 6x2 y j
29. Fx, y sin yi x cos yj
M 12xy and N 6x2 y have continuous first partial derivatives.
M sin y and N x cos y have continuous first partial derivatives.
N M ⇒ F is conservative. 12x x y
N M cos y ⇒ F is conservative. x y 2 2x 33. M e2xy, N 2 e 2xy y y
31. M 15y3, N 5xy2 M N 5y2 45y2 ⇒ Not conservative x y
35. Fx, y 2xy i x2 j
N 2 y 2x 2xy M e ⇒ Conservative x y3 y 39. Fx, y
37. Fx, y xe x y2yi x j 2
x y i 2 j x2 y2 x y2
2xy 2x y
2 2 2
2 xye x y 2xe x y 2x 3 ye x y y
2xy x 2 y x2 y2 x y22
2
x 2x x
2 x2y 2 2
x e 2xe x y 2x 3 ye x y x
2xy y 2 x x 2 y 2 x y 22
Conservative
Conservative
Conservative
fxx, y 2xye x
fxx, y 2xy
fyx, y x 2e x
fyx, y x 2
2y
2y
f x, y e x y K
fxx, y
x x2 y2
fyx, y
y x2 y2
f x, y
1 lnx2 y2 K 2
2
f x, y x 2 y K
41. Fx, y e xcos y i sin y j
43. Fx, y, z x yz i y j z k, 1, 2, 1
x
e cos y e x sin y y
i
x
e sin y e x sin y x
45. Fx, y, z e x sin y i e x cos y j, 0, 0, 3 i j k x curl F x y z 2e cos yk e x sin y ex cos y 0 curl F 0, 0, 3 2k
k
curl F 1, 2, 1 2j k
Not conservative
j
curl F xyj xzk x y z xyz y z
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180
Chapter 14
Vector Analysis
47. Fx, y, z arctan
curl F
xy i lnx
i
j
x
y
2
k
z
1 lnx 2 y 2 1 2
x arctan y
y2 j k
x
2
2x x xy2 k 2 k 2 y 1 xy2 x y2
49. Fx, y, z sinx y i sin y z j sinz xk i j k cos y z i cosz x j cosx y k curl F x y z sinx y sin y z sinz x
51. Fx, y, z sin y i x cos y j k i j curl F x y sin y x cos y
53. Fx, y, z e z y i xj xyk i j k curl F x y z 0 yez xe z xye z
k z 2 cos yk 0 1
Not conservative
Conservative
fx x, y, z ye z fy x, y, z xe z
fz x, y, z xye z f x, y, z xye z K
55. Fx, y, z
1 x i 2 j 2z 1k y y
i j curl F x y 1 x y y2
k z
57.
Fx, y 6x 2 i xy 2 j div Fx, y
0
6x 2 x y 2 x y
12x 2xy
2z 1
Conservative
fx x, y, z
1 y
fy x, y, z
x y2
fz x, y, z 2z 1 f x, y, z f x, y, z f x, y, z
1 x dx g y, z K1 y y
x x dy hx, z K2 y2 y
2z 1 dz
z 2 z px, y K3 f x, y, z
x z2 z K y
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Section 14.1 59.
Fx, y, z sin x i cos y j z 2 k div Fx, y, z
61.
Vector Fields
sin x cos y z2 cos x sin y 2z x y z
Fx, y, z xyz i y j z k
63.
Fx, y, z e x sin y i e x cos y j
div Fx, y, z yz 1 1 yz 2
div Fx, y, z e x sin y e x sin y
div F1, 2, 1 4
div F0, 0, 3 0
65. See the definition, page 1008. Examples include velocity fields, gravitational fields and magnetic fields.
67. See the definition on page 1014.
69. Fx, y, z i 2x j 3y k Gx, y, z x i y j zk
i j k F G 1 2x 3y 2xz 3y 2 i z 3x y j y 2x 2k x y z
i j curl F G x y 2xz 3y 2 3xy z
k 1 1 i 4x 2x j 3y 6yk 6xj 3yk z 2 y 2x
71. Fx, y, z xyz i y j z k i curl F x xyz
j y y
Gx, y, z xi y j zk
k x y j xz k z z
i curlcurl F x 0
73. Fx, y, z i 2x j 3yk
i j k F G 1 2x 3y x y z
j k z j yk y z xy xz
2xz 3y 2 i z 3x y j y 2x 2k divF G 2z 3x
75. Fx, y, z xyz i y j zk i curl F x xyz
j y y
k z xyj xzk z
divcurl F x x 0
77. Let F M i N j Pk and G Q i R j S k where M, N, P, Q, R, and S have continuous partial derivatives.
F G M Q i N R j P Sk i curlF G x MQ
j y NR
k z PS
P S N R i P S M Q j N R M Q k y z x z x y
N P M N M S R S Q R Q i j k i j k P y z x z x y y z x z x y
curl F curl G
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181
182
Chapter 14
Vector Analysis
79. Let F M i Nj P k and G R i Sj T k. divF G
M R N S P T M R N S P T x y z x x y y z z
M
N
P
R
S
div F div G 81. F M i N j Pk f F curl f F curl f curl F
(Exercise 77)
curl F
(Exercise 78)
F 83. Let F M i Nj Pk, then f F f M i f Nj f Pk. div f F
M f N f P f f M f N fP f M f N f P x y z x x y y z z f
N N f f f M N P M x y z x y z
f div F f F In Exercises 85 and 87, Fx, y, z xi yj zk and f x, y, z Fx, y, z x 2 y 2 z 2.
85.
1 lnx 2 y 2 z 2 2 x y z xi yj z k F i 2 j 2 k 2 2 ln f 2 x y2 z2 x y2 z2 x y2 z2 x y2 z2 f ln f
87. f n x 2 y 2 z 2
n
f n nx 2 y 2 z 2
n1
x
n x 2 y 2 z 2 n1
i nx 2 y 2 z 2
n1
x 2 y 2 z 2
z x 2 y 2 z 2
T
x y z x y z
y x 2 y 2 z 2
j
k
nx 2 y 2 z 2 n2x i yj z k n f n2 F 89. The winds are stronger over Phoenix. Although the winds over both cities are northeasterly, they are more towards the east over Atlanta.
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Section 14.2
Section 14.2
Line Integrals
t i, 3i t 3 j, 3. rt 9 t i 3j, 12 t j,
x2 y2 9
1.
Line Integrals
x2 y2 1 9 9
0 3 6 9
≤ ≤ ≤ ≤
t t t t
≤ ≤ ≤ ≤
3 6 9 12
cos2 t sin2 t 1 cos2 t
x2 9
sin2 t
y2 9
x 3 cos t y 3 sin t rt 3 cos t i 3 sin t j 0 ≤ t ≤ 2
5. r t
t2i tit j, 2 tj,
0 ≤ t ≤ 1 1 ≤ t ≤ 2
7. r t 4 t i 3tj, 0 ≤ t ≤ 2; rt 4 i 3j
C
2
x y ds
0
5t dt
0
9. rt sin t i cos t j 8t k, 0 ≤ t ≤
2
4t 3t42 32 dt
2
x 2 y 2 z 2 ds
2 2 0
; rt cos t i sin t j 8k 2
sin2 t cos2 t 64t 2cos t2 sin t2 64 dt
2
651 64t 2 dt 65 t
0
11. rt t i, 0 ≤ t ≤ 3
x 2 y 2 ds
C
64t 3 3
3
t 2 02 1 0 dt
2 83 3
65
6
1
3
t 2 dt
x 1
3 t 1
3
3
0
2
2
3
−1
9
13. rt cos t i sin t j, 0 ≤ t ≤
x 2 y 2 ds
0
65
2
0
2
y
0
10
0
C
5t2
2
y
1
cos 2 t sin2 t sin t2 cos t2 dt
0
C
2
0
dt
2
x 1
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3 16 2
183
184
Chapter 14
Vector Analysis
15. rt t i t j, 0 ≤ t ≤ 1
C
1
x 4y ds
0
y
t 4t 1 1 dt
2 3 t t2
2
8
1
3 2
0
(1, 1)
1
192 6
x 1
t i, 0 ≤ t ≤ 1 17. rt 2 t i t 1 j, 1 ≤ t ≤ 2 3 t j, 2 ≤ t ≤ 3
C1
C2
x 4y ds
x 4y ds
1
0
C3
C
x 4y ds
1
t2 8 t 13 2 2 3
2 1
3
2
C2
C3
2 t 4t 1 1 1 dt
2 2t
(0, 1)
1 2
t dt
2
y
C1
(1, 0)
x
192 6
8 43 t dt 3 t3 2 3
3 2
8 3
1 192 8 19 192 191 2 x 4y ds 2 6 3 6 6
1 19. x, y, z x 2 y 2 z 2 2 rt 3 cos t i 3 sin tj 2t k, 0 ≤ t ≤ 4 rt 3 sin t i 3 cos t j 2k rt 3 sin t2 3 cos t2 22 13 Mass
x, y, z ds
C
4
1 3 cos t2 3 sin t2 2t2 13 dt 2
0
13
2
4
9 4t 2 dt
0
13
2
4
9t 4t3 3
0
213 27 64 2 4973.8 3 23. Fx, y 3x i 4yi
21. Fx, y xyi yj C: rt 4t i tj, 0 ≤ t ≤ 1
C: rt 2 cos t i 2 sin t j, 0 ≤ t ≤
Ft 4t 2 i t j
Ft 6 cos t i 8 sin t j
rt 4i j
C
F dr
2
rt 2 sin t i 2 cos t j
1
16t 2 t dt
C
0
16 3 1 2 t t 3 2
1 0
F dr
2
35 6
12 sin t cos t 16 sin t cos t dt
0
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2
2 sin2 t
0
2
Section 14.2 25. Fx, y, z x 2 yi x zj x yzk
Line Integrals
27. Fx, y, z x 2 z i 6yj yz 2 k
C: rt t i t 2 j 2k, 0 ≤ t ≤ 1
rt t i t 2 j ln t k, 1 ≤ t ≤ 3
Ft t 4 i t 2j 2 t 3 k
Ft t 2 ln t i 6t 2 j t 2 ln 2 tk
r t i 2 t j
1 d r i 2 t j k dt t
C
F dr
1
2t t 2 dt t4
0
t5 2t3 5
3
2t 2
1
0
185
17 15
C
F dr
3
t 2 ln t 12t 3 t ln t2 dt
1
249.49
29. Fx, y x i 2y j C: y x3 from 0, 0 to 2, 8 rt t i t 3 j, 0 ≤ t ≤ 2 rt i 3t 2 j Ft t i 2t 3 j F r t 6t 5 Work
C
F dr
2
0
1 t 6t 5 dt t 2 t 6 2
2 0
66
31. Fx, y 2xi yj C: counterclockwise around the triangle whose vertices are 0, 0, 1, 0, 1, 1
t i, 0 ≤ t ≤ 1 1 ≤ t ≤ 2 rt i t 1 j, 3 t i 3 t j, 2 ≤ t ≤ 3 On C1:
Ft 2t i, rt i Work
F
C1
On C2:
1
2t dt 1
0
Ft 2 i t 1 j, rt j Work
C2
On C3:
dr
F dr
2
t 1 dt
1
Ft 23 t i 3 t j, rt i j Work
Total work
C
C3
F dr
F dr 1
3
2 3 t 3 t dt
2
35. rt 3 sin t i 3 cos t j
C: rt 2 cos t i 2 sin t j t k, 0 ≤ t ≤ 2 rt 2 sin t i 2 cos t j k
C
F dr
2
5t dt 10 2
10 t k, 0 ≤ t ≤ 2 2
F 150k
dr 3 cos t i 3 sin t j
Ft 2 cos t i 2 sin t j 5tk F r 5t
3 2
1 3 0 2 2
33. Fx, y, z x i yj 5zk
Work
1 2
C
F dr
0
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2
0
10 k dt 2
2
1500 1500 dt t 2 2
0
1500 ft lb
186
Chapter 14
Vector Analysis
37. Fx, y x2 i xyj r1t 2t i t 1 j, 1 ≤ t ≤ 3
(a)
r2t 23 t i 2 t j, 0 ≤ t ≤ 2
(b)
r1t 2 i j
r2t 2i j
Ft 4t i 2tt 1 j 2
C1
F dr
3
8t 2 2tt 1 dt
1
236 3
Ft 43 t2 i 23 t2 t j F dr
C2
Both paths join 2, 0 and 6, 2. The integrals are negatives of each other because the orientations are different.
236 3
rt i 2 t j
Ft 2 t i t j
Ft t 3 2t 2 i t
F r 2t 2t 0 F dr 0.
C
F dr 0.
y 43. x 2t, y 10t, 0 ≤ t ≤ 1 ⇒ y 5x or x , 0 ≤ y ≤ 10 5
10
x 3y 2 d y
0
C
5y 3y dy 10y y
10
2
2
3
0
1010
y 1 45. x 2t, y 10t, 0 ≤ t ≤ 1 ⇒ x , 0 ≤ y ≤ 10, dx dy 5 5
10
xy dx y dy
0
C
y2 y3 y2 y dy 25 75 2
10 0
190 OR 3
y 5x, d y 5 dx, 0 ≤ x ≤ 2
xy dx y dy
2
5x 2 25x dx
0
C
5x3
3
25x 2 2
2 0
190 3 y
47. rt t i, 0 ≤ t ≤ 5 3
xt t, yt 0 dx dt,
C
dy 0
2x y dx x 3y d y
2 1
x
5
0
2t dt 25
1
2
3
4
−1 −2
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5
t2 j 2
F r t 3 2t 2 2t t Thus,
y j 2
C: rt t i t 2 j
rt i 2 j
C
83 t2 23 t2 t dt
0
41. Fx, y x 3 2x 2 i x
C: rt t i 2 t j
2
39. Fx, y yi x j
Thus,
t2 0 2
Section 14.2
t3ii, t 3j,
49. rt
Line Integrals
0 ≤ t ≤ 3 3 ≤ t ≤ 6
xt t, yt 0,
C1:
dx dt, dy 0
2t dt 9
3 3t 3 dt
3
2x y dx x 3y dy
0
C1
C2: xt 3, yt t 3 dx 0, dy dt
2x y dx x 3y dy
6
3
C2
2x y dx x 3y dy 9
C
2
3t 2
6t
6
3
45 2
45 63 2 2
y
(3, 3) 3
2
C2 1
C1 x 1
2
3
51. xt t, yt 1 t2, 0 ≤ t ≤ 1, dx dt, dy 2t dt
2x y dx x 3y dy
C
1
2 t 1 t 2 t 3 3t 22t dt
0
1
6t 3 t 2 4t 1 dt
0
3t2
4
t3 2t 2 t 3
1
0
11 6
53. xt t, yt 2t 2, 0 ≤ t ≤ 2 dx dt, dy 4t dt
2x y dx x 3y dy
C
2
2t 2t 2 dt t 6t 2 4t dt
0
2
0
55. f x, y h
2 24t 3 2t 2 2t dt 6t 4 t 3 t 2 3
r 3t i 4t j, 0 ≤ t ≤ 1
316 3
rt cos t i sin t j, 0 ≤ t ≤ rt sin t i cos t j
rt 5
rt 1
Lateral surface area:
C
0
C: x 2 y 2 1 from 1, 0 to 0, 1
rt 3i 4j
1
f x, y ds
2
57. f x, y xy
C: line from 0, 0 to 3, 4
0
5h dt 5h
Lateral surface area:
f x, y ds
2
cos t sin t dt
0
C
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2
sin2 t 2
0
1 2
2
187
188
Chapter 14
Vector Analysis
59. f x, y h C: y 1 x 2 from 1, 0 to 0, 1 rt 1 t i 1 1 t2 j, 0 ≤ t ≤ 1 rt i 21 tj rt 1 41 t 2 Lateral surface area:
1
f x, y ds
h1 4 1 t2 dt
0
C
h 21 t1 41 t2 ln 21 t 1 41 t2 4
0 1
h 25 ln 2 5 1.4789h 4
61. f x, y xy C: y 1 x 2 from 1, 0 to 0, 1 You could parameterize the curve C as in Exercises 59 and 60. Alternatively, let x cos t, then: y 1 cos2 t sin2 t rt cos t i sin2 t j, 0 ≤ t ≤ 2 rt sin t i 2 sin t cos t j rt sin2 t 4 sin2 t cos2 t sin t1 4 cos2 t Lateral surface area:
f x, y ds
2
cos t sin2 t sin t1 4 cos2 t dt
0
C
2
sin2 t 1 4 cos2 t1 2 sin t cos t dt
0
1 Let u sin2 t and dv 1 4 cos2 t1 2 sin t cos t, then du 2 sin t cos t dt and v 12 1 4 cos2 t3 2.
C
2
2
1 1 sin2 t 1 4 cos2 t 3 2 1 4 cos2 t5 2 12 120
1 1 1 1 55 2 255 11 0.3742 12 120 120 120
1 sin2 t1 4 cos2 t3 2 12
f x, y ds
0
1 6
1 4 cos2 t3 2 sin t cos t dt
0
2
0
63. (a) f x, y 1 y 2
(c)
z
rt 2 cos t i 2 sin t j, 0 ≤ t ≤ 2
5 4
rt 2 sin t i 2 cos t j rt 2 S
f x, y ds
C
2
−3
1 4 sin2 t2 dt
0
(b) 0.212
2
2t 4t sin t cos t
0
3 x
12 37.70 cm2
12 7.54 cm3 5
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3
y
Section 14.2 65. S 25
Line Integrals
z 60
Matches b
50 40 30 20 10 3
3 x
y
67. (a) Graph of: rt 3 cos t i 3 sin t j 1 sin2 2t k 0 ≤ t ≤ 2 z
3
2 1
3
3 4
4
y
x
(b) Consider the portion of the surface in the first quadrant. The curve z 1 sin2 2t is over the curve r1t 3 cos t i 3 sin t j, 0 ≤ t ≤ 2. Hence, the total lateral surface area is 4
f x, y ds 4
2
1 sin2 2t3 d t 12
0
C
34 9 sq. cm
(c) The cross sections parallel to the xz-plane are rectangles of height 1 4 y 321 y 2 9 and base 29 y 2. Hence,
3
Volume 2
29 y 2 1 4
0
y2 y2 1 9 9
dy 42.412 cm
3
69. See the definition of Line Integral, page 1020. See Theorem 14.4.
71. The greater the height of the surface over the curve, the greater the lateral surface area. Hence, z 3 < z1 < z 2 < z 4 .
y 4 3 2
1 x 1
73. False
C
2
3
75. False, the orientations are different.
1
xy ds 2
4
t 2 dt
0
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189
190
Chapter 14
Vector Analysis
Section 14.3
Conservative Vector Fields and Independence of Path
1. Fx, y x 2 i xy j (b) r2 sin i sin2 j, 0 ≤ ≤
(a) r1t t i t 2 j, 0 ≤ t ≤ 1 r1t i 2t j
r2 cos i 2 sin cos j
Ft t 2 i t 3 j
C
F dr
2
Ft sin2 i sin3 j
1
t 2 2t 4 dt
0
11 15
C
2
F dr
sin2 cos 2 sin4 cos d
0
2
sin3 2 sin5 3
5
0
11 15
3. F x, y y i x j (a) r1 sec i tan j, 0 ≤ ≤
3
r1 sec tan i sec2 j F tan i sec j
C
F dr
3
0
sec tan2 sec3 d
3
sec d ln sec tan
F dr
3
t
2 t 1
0
1 2
3
0
t 1
2 t
N e x cos y x
dt 12
3
ln 2 3 1.317
0
1 1 dt 2 t t 1
1 1 dt ln 2 t 12 2 14
M e x cos y y
N M , F is conservative. x y
t 21 t
2
3
0
7. Fx, y
1 t 2 t 14 14
t
12 ln 7 4 3 1.317
7 1 1 ln 2 3 ln 2 2 2
5. Fx, y e x sin yi ex cos yj
Since
3
0
1 1 i j 2 t 1 2 t
Ft t i t 1 j
C
sec sec2 1 sec3 d
0 ≤ t ≤ 3
(b) r2t t 1 i t j,
3
0
0
r2t
0
1 x i 2j y y
N 1 2 x y Since
3
1 M 2 y y
N M
, F is not conservative. x y
9. Fx, y, z y 2 z i 2xyz j xy 2 k curl F 0 ⇒ F is conservative.
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dt
Section 14.3
Conservative Vector Fields and Independence of Path
191
11. Fx, y 2xyi x 2 j (a) r1t t i t 2 j, 0 ≤ t ≤ 1
(b) r2t t i t 3 j, 0 ≤ t ≤ 1
r1t i 2t j
r2 t i 3t 2 j
Ft 2t 3 i t 2 j
Ft 2t 4 i t 2 j
C
F dr
1
4t 3 dt 1
0
C
F dr
1
5t 4 dt 1
0
13. Fx, y yi xj (a) r1t t i t j,
0 ≤ t ≤ 1
(c) r3t t i t 3 j,
r2 t i 2t j
r3t i
Ft t i t j
Ft t 2 i t j
Ft t 3 i t j
C
(b) r2t t i t 2 j,
r1t i j
15.
0 ≤ t ≤ 1
F dr 0
C
F dr
1
t 2 dt
0
1 3
C
F dr
0 ≤ t ≤ 1
3t 2 j
1
2t 3 dt
0
1 2
y 2 dx 2 xy dy
C
Since M y N x 2y, Fx, y y 2 i 2xy j is conservative. The potential function is f x, y xy 2 k. Therefore, we can use the Fundamental Theorem of Line Integrals. (a)
4, 4
y 2 dx 2xy dy x 2 y
C
0, 0
64
(c) and (d) Since C is a closed curve,
(b)
C
1, 0
y 2 dx 2xy dy x 2 y
1, 0
0
y 2 dx 2xy dy 0.
C
17.
2x y dx x 2 y 2 dy
19. Fx, y, z yz i xz j xyk
C
Since curl F 0, Fx, y, z is conservative. The potential function is f x, y, z xyz k.
Since M y N x 2x, Fx, y 2xyi x 2 y 2 j is conservative. The potential function is f x, y x2y (a)
(b)
y3 3
2x y dx x 2 y 2 dy x 2 y
C
0, 4
y3 3
2x y dx x 2 y 2 dy x 2 y
C
(a) r1t t i 2j tk, 0 ≤ t ≤ 4
y3 k. 3 5, 0 0, 4
2, 0
64 3
64 3
C
C
(a) r1t t i t 2 j k, 0 ≤ t ≤ 1 r1t i 2t j Ft 2t 2 t i t 2 1 j 2t 2 4k
C
1
0
2t 3 2t 2 t dt
2 3
—CONTINUED—
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4, 2, 4
F dr xyz
Fx, y, z is not conservative.
F dr
0, 2, 0
32
(b) r 2t t 2 i t j t 2 k, 0 ≤ t ≤ 2
21. Fx, y, z 2y x i x2 z j 2y 4zk
4, 2, 4
F d r xyz
(0, 0, 0
32
192
Chapter 14
Vector Analysis
21. —CONTINUED— (b) r2t t i t j 2t 12 k, 0 ≤ t ≤ 1 r2t i j 4 2t 1k Ft 3t i t 2 2t 12 j 2t 4 2t 1 2 k
C
1
F dr
3t t 2 2t 12 8t 2t 1 162t 13 dt
0 1
17t3
17t 2 5t 2t 12 162t 13 dt
0
23. Fx, y, z e z y i x j x y k
25.
C
Fx, y, z is conservative. The potential function is f x, y, z xye z k.
3
5t 2 2t 1 3 22t 14 2 6
3, 8
yi x j dr xy
0, 0
24
(a) r1t 4 cos t i 4 sin tj 3k, 0 ≤ t ≤
C
4, 0, 3
F dr xye z
4, 0, 3
0
(b) r2t 4 8t i 3k, 0 ≤ t ≤ 1
C
27.
4, 0, 3
F dr xye z
4, 0, 3
0
cos x sin y dx sin x cos y dy sin x sin y
e x sin y dx e x cos y dy e x sin y
y 2z dx x 3z dy 2x 3y dz
C
29.
C
31.
32, 2
2, 0
0, 0
0,
1
0
C
Fx, y, z is conservative and the potential function is f x, y, z xy 3yz 2xz. 1, 1, 1
(a)
xy 3yz 2xz
(b)
xy 3yz 2xz
(c)
xy 3yz 2xz
0, 0, 0 0, 0, 1
0, 0, 0
1, 0, 0
33.
0, 0, 0
000
xy 3yz 2xz
0, 0, 1
1, 1, 0
1, 0, 0
2, 3, 4
sin x dx z dy y dz cos x yz
C
1, 1, 1
xy 3yz 2xz
0, 0, 0
000
1, 1, 1
xy 3yz 2xz
1, 1, 0
12 1 11
35. Fx, y 9x 2 y 2 i 6x 3 y 1 j is conservative.
5, 9
Work 3x 3 y 2 y
0, 0
30,366
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0 1 1 0
1 0
17 6
Section 14.4
Green’s Theorem
193
37. rt 2 cos 2 t i 2 sin 2 t j rt 4 sin 2 t i 4 cos 2 t j at 8 2 cos 2 t i 8 2 sin 2 t j Ft m a t W
C
1 2 at cos 2 t i sin 2 t j 32 4
F dr
C
2 cos 2 t i sin 2 t j 4sin 2 t i cos 2 t j dt 3 4
0 dt 0
C
39. Since the sum of the potential and kinetic energies remains constant from point to point, if the kinetic energy is decreasing at a rate of 10 units per minute, then the potential energy is increasing at a rate of 10 units per minute. 41. No. The force field is conservative.
43. See Theorem 14.5, page 1033.
45. (a) The direct path along the line segment joining 4, 0 to 3, 4 requires less work than the path going from 4, 0 to 4, 4 and then to 3, 4. (b) The closed curve given by the line segments joining 4, 0, 4, 4, 3, 4, and 4, 0 satisfies
C
47. False, it would be true if F were conservative.
F dr 0.
49. True
51. Let F Mi Nj Then
f f i j. y x
M 2f f 2 y y y y
and
f 2f N 2. Since x x x x
2f 2f M N 0 we have . x 2 y 2 y x Thus, F is conservative. Therefore, by Theorem 14.7, we have
C
f f dx dy y x
M dx N dy
C
F dr 0
C
for every closed curve in the plane.
Section 14.4
Green’s Theorem
0 ≤ t ≤ 4 4 ≤ t ≤ 8 8 ≤ t ≤ 12 12 ≤ t ≤ 16
t i, 4 i t 4 j, 1. r t 12 t i 4j, 16 t j,
C
4
y dx x dy 2
2
y
(4, 4) 4 3
2
8
0 dt t 0 2
0
t 4 0 16 dt 2
4
12
16dt 12 t20
R
1
16 t20 0dt
12
0 64 64 0 0
x
16
8
By Green’s Theorem,
1
N M dA x y
4
0
4
0
4
2x 2y dy dx
8x 16 dx 0.
0
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2
3
4
194
Chapter 14
3. rt
Vector Analysis
4
y 2 dx x 2 dy
0
C
4
0
By Green’s Theorem,
R
0 ≤ t ≤ 4 4 ≤ t ≤ 8
t i t 2 4 j, 8 t i 8 t j,
N M dA x y
y
8
t4 t dt t 2 dt 16 2
t4 t3 dt 16 2
4
3
8 t2dt 8 t2dt
C2
2
4
8
28 t2 dt
4
224 128 32 5 3 15
C1
1
x 1
x
4
2x 2y dy dx
x 2 4
0
(4, 4)
4
3
2
4
x3 32 x4 dx . 2 16 15
x2
0
5. C: x 2 y 2 4 Let x 2 cos t and y 2 sin t, 0 ≤ t ≤ 2.
xey dx e x dy
C
R
2
2 cos te2 sin t 2 sin t e2 cos t 2 cos t dt 19.99
0
N M dA x y
In Exercises 7 and 9,
7.
4x 2
2
2
2 4x 2
ex xey dy dx
2
2 4 x
2
2
2
N M 1. x y
2
y x dx 2x y dy
0
C
y
x
dy dx
(2, 2)
x2 x
2
y=x
2
2x x 2 dx
1
0
y = x2 − x
4 3
x 1
2
9. From the accompanying figure, we see that R is the shaded region. Thus, Green’s Theorem yields
y x dx 2x y dy
y
(− 5, 3)
1 dA
(− 1, 1)
Area of R
(− 1, − 1)
C
(5, 3) (1, 1) x
11. Since the curves y 0 and y 4 x 2 intersect at 2, 0 and 2, 0, Green’s Theorem yields 2xy dx x y dy
2
1 2x d A
R
4x2
2 0
1 2x dy dx
2
2
4x2
y 2xy
0
dx
2
2
−2
(− 5, − 3) −4
56.
C
4 2
R
610 22
e x xe 4x xe 4x dx 19.99
2
4 8x x 2 2x3 dx
4x 4x 2
x3 x4 3 2
2 2
8 8 32 16 . 3 3 3
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(1, − 1)
4
(5, − 3)
Section 14.4
Green’s Theorem
13. Since R is the interior of the circle x 2 y 2 a 2, Green’s Theorem yields
x 2 y 2 dx 2xy dy
2y 2y dA
C
R
a2 x 2
a
15. Since
a
a a2 x 2
4y dy dx 4
a
0 dx 0.
2x N M 2 , y x y2 x
we have path independence and
R
N M dA 0. x y
17. By Green’s Theorem,
sin x cos y dx xy cos x sin y dy
C
1
0
19. By Green’s Theorem,
y sin x sin y sin x sin y dA
R
xy dx x y dy
C
x
y dy dx
x
1
x x 2 dx
0
1 x 2 x3 2 2 3
1 0
1 . 12
1 x dA
R
1 2
2
3
1 r cos r dr d
1
0
2
0
4 263 cos d 8.
21. Fx, y xy i x y j C: x 2 y 2 4 Work
xy dx x y dy
C
1 x dA
R
2
0
2
1 r cos r dr d
0
2
0
2 38 cos d 4
23. Fx, y x 3 2 3y i 6x 5 y j C: boundary of the triangle with vertices 0, 0, 5, 0, 0, 5 Work
C
x 3 2 3y dx 6x 5 y dy
R
9 dA 9 12 55 225 2
25. C: let x a cos t, y a sin t, 0 ≤ t ≤ 2. By Theorem 14.9, we have A
1 2
C
x dy y dx
1 2
2
0
a cos t a cos t a sin t a sin t dt
1 2
2
0
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a 2 dt
2
a2 t 2
0
a 2.
195
196
Chapter 14
Vector Analysis
27. From the accompanying figure we see that
y
C1: y 2x 1, dy 2 dx C2: y 4
(1, 3) 2
dy 2x dx.
x 2,
−6
Thus, by Theorem 14.9, we have
1
1 2
A
3 1
1 2
3 1
1 2
1 dx
3
1 dx
1 2 1 2
3
3
1 2
x2 2x 1 dx
4
(− 3, − 5)
x2x 4 x 2 dx
−4 −6
1
x 2 4 dx
1 1
3
x
−4
x 2 4 dx
1
1 2
3
3 x 2 dx
29. See Theorem 14.8, page 1042.
1 x3 3x 2 3
1 3
32 . 3
31. Answers will vary. F1x, y yi xj F2x, y x2 i y2 j F3x, y 2xyi x2 j
33. A
x
1 2A
2
4 x 2 dx 4x
2
1 2A
x 2 dy
C1
x3 3
2 2
y
32 3
y = 4 − x2 3
x 2 dy 2
C2
For C1, dy 2x dx and for C2, dy 0. Thus, 1 232 3
x
2
x 22x dx
2
1
2
643 2 x4
C1
2
0.
−2
x
−1
C2 1
2
To calculate y, note that y 0 along C2. Thus, 1 232 3
y
2
4 x 22 dx
2
3 64
2
2
16 8x 2 x 4 dx
3 8x 3 x 5 16x 64 3 5
2 2
8 . 5
85
x, y 0,
1
35. Since A
x x 3 dx
0
y x, dy dx. Thus, x2
x 2 dy 2
C
0
y 2
0
1 1 , we have 2. On C1 we have y x 3, dy 3x 2 dx and on C2 we have 4 2A
x2 dx
y
6 2 8 5 3 15
(1, 1)
1
C2 C1
y 2 dx
1
x
0
x 6 dx 2
0
x, y
x 2 dx
1
C
2
4 1
C2
0
x 4 dx 2
2
x 23x 2 dx 2
C1
1
6
x2 x4
1
1
8 2 2 x 2 dx . 7 3 21
158 , 218
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Section 14.4
37. A
1 2
2
197
a21 cos 2 d
0
2
a2 2
1 2 cos 21 cos22 d a2 32 2 sin 41 sin 2 2
0
39. In this case the inner loop has domain
41. I
Green’s Theorem
A
1 2
1 2
4 3
2 3 4 3
2 3
2 0
a2 3a2 3 2 2
4 2 ≤ ≤ . Thus, 3 3
1 4 cos 4 cos2 d
3 4 cos 2 cos 2 d
1 3 4 sin sin 2
2
4 3
2 3
3 3 . 2
y dx x dy x2 y2 C
(a) Let F
x y i 2 j. x2 y2 x y2 x2 y2 N M 2 . x y x y 2 2
F is conservative since
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
C
F dr
M dx N dy
C
R
N M dA 0. x y
(b) Let r a cos t i a sin t j, 0 ≤ t ≤ 2 be a circle C1 oriented clockwise inside C (see figure). Introduce line segments 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 and C3 have opposite orientations, the line integrals over them cancel. Thus, C4 C1 C2 C C3 and
F dr
C4
But,
F dr
C1
Finally,
C
F dr
C1
2
2
C
F dr 0.
t a cos ta cos t dt aacossinttaa sin sin t a cos t a sin t 2
0
2
2
2
2
0
F dr
2
sin2 t cos2 t dt t
0
2
2
2
2.
F dr 2.
C1
Note: If C were orientated clockwise, then the answer would have been 2. y 3
C
2
C1
C2 x 4
C3 −2 −3
43. Pentagon: 0, 0, 2, 0, 3, 2, 1, 4, 1, 1 A 12 0 0 4 0 12 2 1 4 0 0 19 2
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198
45.
Chapter 14
y n dx x n dy
C
Vector Analysis
R
N M dA x y
y
2a
y = a2 − x2
For the line integral, use the two paths C1: r1 x x i, a ≤ x ≤ a
C2
C2: r2 x x i a 2 x 2 j, x a to x a
−a
x
C1
a
y n dx x n dy 0
C1
y n dx x n dy
a
a
C2
R
N M dA x y
a
2
a
a2 x 2
x 2 n 2 x n
a 0
x a 2 x 2
dx
nx n1 nyn1 dy dx
(a) For n 1, 3, 5, 7, both integrals give 0. (b) For n even, you obtain n 2 : 43 a 3
5 n 4 : 16 15 a
7 n 6 : 32 35 a
256
n 8 : 315 a 9
(c) If n is odd and 0 < a < 1, then the integral equals 0.
47.
f DN g gD N f ds
C
f D N g ds
C
R
gDN f ds
C
f 2g f g dA
g2 f g f dA
R
f 2g g 2 f dA
R
49. F M i N j N M N M 0 ⇒ 0. x y x y
C
F dr
M dx N dy
C
Section 14.5
R
N M dA x y
0 dA 0
R
Parametric Surfaces
1. r u, v u i vj uvk
3. r u, v 2 cos v cos ui 2 cos v sin uj 2 sin vk
z xy
x2 y2 z2 4
Matches c.
Matches b.
v 5. r u, v ui vj k 2
7. r u, v 2 cos ui vj 2 sin uk x2 z2 4
y 2z 0
Cylinder
Plane
z z 3 3 2 −4 3 4 5 5 x
y x
5
5 −3
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y
Section 14.5 For Exercises 9 and 11,
Parametric Surfaces
199
z
r u, v u cos vi u sin vj u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2.
5
Eliminating the parameter yields z x 2 y 2, 0 ≤ z ≤ 4. 2
2
y
x
9. s u, v u cos v i u sin v j u 2 k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2 z x 2 y 2
The paraboloid is reflected (inverted) through the xy-plane. 11. s u, v u cos v i u sin v j u 2 k, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 The height of the paraboloid is increased from 4 to 9. 15. r u, v 2 sinh u cos vi sinh u sin vj cosh uk,
13. r u, v 2u cos v i 2u sin v j u 4 k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2 z
x2
0 ≤ u ≤ 2, 0 ≤ v ≤ 2
z
16
y2 2
z2
3
1
2
x2 4
y2 1
z
1
9 6
1 9
6
3
6
9
x 2 2
x
y
17. r u, v u sin u cos v i 1 cos u sin v j uk, 0 ≤ u ≤ , 0 ≤ v ≤ 2
z 5 4 3
−2 −3 3
−2 2
−1
1
2
−3
3
y
x
21. x 2 y 2 16
19. z y r u, v ui vj vk 23. z x 2
25. z 4 inside x 2 y 2 9.
r u, v ui vj u 2 k x 27. Function: y , 0 ≤ x ≤ 6 2 Axis of revolution: x-axis x u, y
r u, v 4 cos ui 4 sin uj vk
u u cos v, z sin v 2 2
r u, v v cos u i v sin u j 4k, 0 ≤ v ≤ 3
29. Function: x sin z, 0 ≤ z ≤ Axis of revolution: z-axis x sin u cos v, y sin u sin v, z u 0 ≤ u ≤ , 0 ≤ v ≤ 2
0 ≤ u ≤ 6, 0 ≤ v ≤ 2
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y
200
Chapter 14
Vector Analysis
31. r u, v u v i u v j vk, 1, 1, 1
33. r u, v 2u cos v i 3u sin v j u 2 k, 0, 6, 4
ru u, v i j, rv u, v i j k
ru u, v 2 cos v i 3 sin v j 2uk
At 1, 1, 1 , u 0 and v 1.
rv u, v 2u sin v i 3u cos v j
ru 0, 1 i j, rv 0, 1 i j k
i N ru 0, 1 rv 0, 1 1 1
j 1 1
k 0 i j 2k 1
Tangent plane: x 1 y 1 2 z 1 0
At 0, 6, 4 , u 2 and v 2.
2 3j 4k, r 2, 2 4i
ru 2,
2 r 2, 2
N ru 2,
x y 2z 0 (The original plane!)
v
i 0 4
v
j 3 0
k 4 16j 12k 0
Direction numbers: 0, 4, 3 Tangent plane: 4 y 6 3 z 4 0 4y 3z 12 35. r u, v 2ui
v v j k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 1 2 2
1 1 ru u, v 2i, rv u, v j k 2 2
i ru rv 2 0
j 0 12
k 0 j k 1 2
ru rv 2
1
A
0
2
2 du dv 22
0
37. r u, v a cos ui a sin uj vk, 0 ≤ u ≤ 2, 0 ≤ v ≤ b ru u, v a sin ui a cos uj rv u, v k
i j k ru rv a sin u a cos u 0 a cos ui a sin uj 0 0 1 ru rv a
b
A
0
2
a du dv 2ab
0
39. r u, v au cos v i au sin v j uk, 0 ≤ u ≤ b, 0 ≤ v ≤ 2 ru u, v a cos vi a sin vj k rv u, v au sin v i au cos v j ru rv
ru rv au1 a 2 A
2
0
i j k a cos v a sin v 1 au cos v i au sin v j a 2uk au sin v au cos v 0
b
a1 a 2 u du dv ab21 a 2
0
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Section 14.5
Parametric Surfaces
41. r u, v u cos v i u sin v j uk, 0 ≤ u ≤ 4, 0 ≤ v ≤ 2 ru u, v
cos v sin v i jk 2u 2u
rv u, v u sin v i u cos v j
ru rv
i cos v 2u
k
j sin v 2u
u sin v u cos v
0
u 41 u 41 du dv 6 17
ru rv A
2
0
1 u cos vi u sin vj 1 k 2
4
17 1 36.177
0
45. (a) From 10, 10, 0
43. See the definition, page 1051.
(b) From 10, 10, 10
(c) From 0, 10, 0
(d) From 10, 0, 0
47. (a) r u, v 4 cos v cos ui
(b) r u, v 4 2 cos v cos ui
4 cos v sin u j sin vk,
4 2 cos v sin uj 2 sin vk,
0 ≤ u ≤ 2, 0 ≤ v ≤ 2
0 ≤ u ≤ 2, 0 ≤ v ≤ 2
z
z
4
4
−6
−6
6 x
(c)
6
−4
y
r u, v 8 cos v cos u i
6
6
x
y
(d) r u, v 8 3 cos v cos ui
8 cos v sin uj sin vk,
8 3 cos v sin uj 3 sin vk,
0 ≤ u ≤ 2, 0 ≤ v ≤ 2
0 ≤ u ≤ 2, 0 ≤ v ≤ 2
z
z 12
9
3 3 y
x
12 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.
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201
202
Chapter 14
Vector Analysis
49. r u, v 20 sin u cos v i 20 sin u sin v j 20 cos uk 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 ru 20 cos u cos v i 20 cos u sin v j 20 sin uk rv 20 sin u sin v i 20 sin u cos v j
i j k ru rv 20 cos u cos v 20 cos u sin v 20 sin u 20 sin u sin v 20 sin u cos v 0
400 sin2 u cos v i 400 sin2 u sin vj 400 cos u sin u cos2 v cos u sin u sin2 v k 400 sin2 u cos v i sin2 u sin vj cos u sin uk ru rv 400sin4 u cos2 v sin4 u sin2 v cos2 u sin2 u 400sin4 u cos2 u sin2 u 400sin2 u 400 sin u
S
dS
S
2
400 sin u du dv
0
0
3
2
2
0
3
400 cos u
dv
0
200 dv 400 m2
0
51. r u, v u cos v i u sin vj 2vk, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2 ru u, v cos v i sin vj rv u, v u sin v i u cos vj 2k ru rv
i j cos v sin v u sin v u cos v
k 0 2 sin vi 2 cos vj uk 2
ru rv 4 u2 A
2
3
4 u2 du dv 313 4 ln
0
0
3 2 13
z 4π
2π
−4
−2
2
4
4
y
x
53. Essay
Section 14.6
Surface Integrals
1. S: z 4 x, 0 ≤ x ≤ 4, 0 ≤ y ≤ 4,
S
4
x 2y z dS
z z 1, 0 x y
4
x 2y 4 x 1 1 2 0 2 dy dx
0
0
4
4
2
0
4 2y dy dx 0
0
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Section 14.6 z z 0 x y
3. S: z 10, x 2 y 2 ≤ 1,
1x 2
1
x 2y z dS
2
2
1
0
2 0
5. S: z 6 x 2y, (first octant) xy dS
0
y 5 4
xy1 1 2 dy dx 2
2
3 x2
1
dx
x
0 −1
6
6
2
y = 3 − 21 x
3 2
xy 2 2
0
1
2
3
4
5
6
1 x 9 3x x 2 dx 4
0
6 9x 2
2
z z 1, 2 x y
0
6
10
3 x2
6
1 2 cos sin 5 d 3 3
13 sin 32 cos 5
S
r cos 2r sin 10 r dr d
0
0
x 2y 10 1 0 2 0 2 dy dx
1 1x 2
S
6
Surface Integrals
2
x3
x4 16
6 0
276 2
7. S: z 9 x 2, 0 ≤ x ≤ 2, 0 ≤ y ≤ x, z z 2x, 0 x y
2
xy dS
0
S
2
xy1 4x 2 dx dy
y
39117 1 240
9. S: z 10 x 2 y 2, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
2
x 2 2xy dS
0
S
2
x 2 2xy 1 4x 2 4y 2 dy dx 11.47
0
1 1 11. S: 2x 3y 6z 12 (first octant) ⇒ z 2 x y 3 2 x, y, z x 2 y 2 m
7 6
7 6
0
6
0
5
2
1 2
y = 4 − 23 x
4
dA 1 1 3
x 2 y 2
R
6
y
2
3 2
R
1
4 2x3
x y dy dx 2
x
2
−1
0
2 1 2 x2 4 x 4 x 3 3 3
dx 67 43 x 3
3
1 1 2 x4 4 x 6 8 3
4 6 0
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364 3
1
2
3
4
5
6
203
204
Chapter 14
Vector Analysis
v 13. S: r u, v ui vj k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2 2 ru rv
21 j k
2
y 5 dS
2
1
0
S
5
v 5
5
0
2
du dv 65
, 0 ≤ v ≤ 2 2
15. S: r u, v 2 cos ui 2 sin uj vk, 0 ≤ u ≤ ru rv 2 cos ui 2 sin uj 2
2
xy dS
0
S
2
8 cos u sin u du dv 8
0
17. f x, y, z x 2 y 2 z 2 S: z x 2, x 2 y 2 ≤ 1
f x, y, z dS
S
1x 2
1
1 1x 2 2
2
0
1
0
1
0
2
2
r 2 r 2 cos2 4r cos 4 r dr d
0
2
2
r 2 r cos 2 2 r dr d
0
2
2
x 2 y 2 x 2 2 1 1 2 0 2 dy dx
0
d
r4 r4 4r 3 cos2 cos 2r 2 4 4 3
1 0
9 1 1 cos 2 4 cos d 4 4 2 3
2
94 81 21 sin 2 34 sin
2
0
184 4 19 4 2
2
19. f x, y, z x 2 y 2 z 2 S: z x 2 y 2, x 2 y 2 ≤ 4
f x, y, z dS
S
4x 2
2
2 4x 2
2 2 2
x 2 y 2 x 2 y 2
4x 2
2
2 4x 2
2
4x 2
2 4x 2 2
0
2
2
0
x y
x 2
2
2
1
x 2
x y2
y2 x2 y2 dy dx x2 y2
x 2 y 2 dy dx
r 2 dr d 2
2
0
r3 3
2 0
d
2
163
0
32 3
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x y y 2
2
2
2
dy dx
Section 14.6
Surface Integrals
21. f x, y, z x 2 y 2 z 2 S: x 2 y 2 9, 0 ≤ x ≤ 3, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9 Project the solid onto the yz-plane; x 9 y 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9.
3
f x, y, z dS
0
S
9 y 2 y 2 z 2
0
3
1
9
9
9 z 2
0
0
3
324
0
3
3 dz dy 9 y 2
0
y 9 y 2
2
0 2 dz dy
3 z3 9z 3 9 y 2
3
3 y d y 972 arcsin 3 9 y 2
0
972
9
dy
0
2 0 486
23. F x, y, z 3z i 4 j yk
y
S: x y z 1 (first octant) 1
G x, y, z x y z 1
y = −x + 1
G x, y, z i j k
S
F N dS
R
F G dA
0
1x
3 1 x y 4 y dy dx
0
1
1x
0
1 3x 2y dy dx
0
1
1
3z 4 y dy dx
0
0
x
1x
1
R
1
y 3xy y 2
0
1x
dx
0
1
1 x 3x 1 x 1 x 2 dx
0
1
2 2x2 dx
0
4 3
25. F x, y, z x i yj z k
y
S: z 9 x 2 y 2, 0 ≤ z
4
G x, y, z x 2 y 2 z 9
2
G x, y, z 2 x i 2y j k
S
F N dS
−4
R
F G dA
z dA
2y 2
R
2x 2 2y 2 9 x 2 y 2 dA
R
x 2 y 2 9 dA
R
2
0
2
0
3
r 2 9 r dr d
0
r 4 9r 2 4 2
3 0
d
243 2
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R x
−2
2 −2
2x 2
x2 + y2 ≤ 9
−4
4
205
206
Chapter 14
Vector Analysis
27. F x, y, z 4i 3j 5k S: z x y , 2
2
x2
y
x 2 + y2 ≤ 4
y ≤ 4 2
G x, y, z x y 2 z
1
2
G x, y, z 2 x i 2y j k
S
F N dS
x
−1
F G dA
R
R
8x 6y 5 dA
2
2
8r cos 6r sin 5 r dr d
0
2
0
−1
R
0
1
2
8 5 r 3 cos 2r 3 sin r 2 3 2
0
2
d
0
64 cos 16 sin 10 d 3
64 sin 16 cos 10 3
2
0
20
29. F x, y, z 4xyi z2 j yzk
z
S: unit cube bounded by x 0, x 1, y 0, y 1, z 0, z 1 1
S1: The top of the cube N k, z 1
S1
F N dS
1
0
1
y 1 dy dx
0
1 2
F N dS
S2
S3: The front of the cube N i, x 1
1
1
0
y 0 dy dx 0
0
S3
S4: The back of the cube N i, x 0
S4
F N dS
F N dS
S6
F N dS
1
0
N j, y 1
1
4 0 y dy dx 0
0
S5
F N dS
1
0
1
z 2 dz dx
0
1 3
Therefore,
S
F N dS
1 1 1 5 020 . 2 3 3 2
31. The surface integral of f over a surface S, where S is given by z g x, y , is defined as
S
1
0
1
4 1 y dy dz 2
0
S5: The right side of the cube
S6: The left side of the cube N j, y 0
y
x
S2: The bottom of the cube N k, z 0
1
1
f x, y, z dS lim
n
f x , y , z S . (page 1061)
→0 i1
i
i
i
i
See Theorem 14.10, page 1061. 33. See the definition, page 1067. See Theorem 14.11, page 1067.
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1
0
1
0
z 2 dz dx
1 3
Section 14.6 35. (a) 4 −6
−6
x
207
(c) r u, 0 4 cos 2u i 4 sin 2u j
(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.
z
Surface Integrals
This is a circle. z
6 6
−4
4
y
−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.
37. z x 2 y 2, 0 ≤ z ≤ a m
k dS k
S
Iz
R
k x 2 y 2 dS
S
2
0
2
2
2
2
2 dA 2 ka 2
dA k
R
k x 2 y 2 2 dA
a
2ka 4
r 3 dr d
4
0
2ka 4
x y y 2
x x 2 y 2
R
2k
1
2
2 a2
2ka 2 a 2m 2
39. x 2 y 2 a 2, 0 ≤ z ≤ h
z
x, y, z 1
h
y ± a 2 x 2 Project the solid onto the xz-plane.
x 2 y 2 1 dS
Iz 4
a
h
1
a
x 2 a 2 x 2
4
0
0
h
a
4a 3
0
0
y
x a 2 x 2
2
0 2 dx dz
1 dx dz x2
a 2
h
4a 3
a
x
S
arcsin
0
x a
a
0
dz 4a 3
2 h 2a h 3
41. S: z 16 x 2 y 2, z ≥ 0 F x, y, z 0.5zk
S
F N dS
R
F gx x, y i gy x, y j k dA 0.5 z dA
R
0.5
2
0
R
0.5 z k 2x i 2y j k dA
0.5 16 x 2 y 2 dA
R
4
0
16 r 2 r dr d 0.5
2
64 d 64
0
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208
Chapter 14
Vector Analysis
Section 14.7
Divergence Theorem
1. Surface Integral: There are six surfaces to the cube, each with dS 1 dA. z 0,
N k,
z a,
N j,
y a, Therefore,
S2
F N dS a
4
s
a
0
2a dy dz
S4
a 2 dx dy a 4
0
a
a
0
2a dy dz 2a3
0
0 dA 0
S5
F N 2y,
N j,
a
0 dA 0
N 2y,
F
S3
F N 2x,
N i,
y 0,
a 2 dA
F N 2x,
N i,
x a,
0 dA 0
S1
F N z 2,
N k,
x 0,
N z 2,
F
2a dA
S6
a
a
0
2a dz dx 2a 3
0
2a 3 2a 3 a 4.
Divergence Theorem: Since div F 2z, the Divergence Theorem yields
a
div F dV
a
0
Q
0
a
a
2z dz dy dx
0
0
a
a 2 dy dx a 4.
0
3. Surface Integral: There are four surfaces to this solid.
z
N z
z 0, N k, F
6
0 dS 0
S1
y 0, N j, F N 2y z, dS dA dx dz
6
z dS
0
S2
6z
z dx dz
0
3
y
6
6
z 2 6z dz 36
x
0
x 0, N i, F N y 2x, dS dA dz dy
3
y dS
0
S3
Therefore,
6y 2y 2 dy 9
0
i 2j k 2x 5y 3z , FN , dS 6 dA 6 6
3
2x 5y 3z dz dy
S4
3
y dz dy
0
x 2y z 6, N
62y
0
62y
3
18 x 11y dx dy
0
90 90y 20y 2 dy 45
0
F N dS 0 36 9 45 18. s
Divergence Theorem: Since div F 1, we have
1 1 dV Volume of solid Area of base Height 96 18. 3 3
Q
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Section 14.7
Divergence Theorem
5. Since div F 2x 2y 2z, we have
a
div F dV
0
Q
a
a
0
a
2x 2y 2z dz dy dx
0
0
a
0
div F dV
Q
0
Q
2
a
0
0
2
2
0
0
3a 4.
2 sin cos sin sin cos 2 sin d d d
0
0
2 5sin cos sin3 cos d d d
0
2
2
0
a
a
2xyz dV
a
2a 2x 2a 3 dx a 2x 2 2a 3x
0
7. Since div F 2x 2x 2xyz 2xyz
a
2ax 2ay a 2 dy dx
1 5 sin cos d d 2
a
0
5 sin2 2 2
2
0
d 0.
9. Since div F 3, we have
43 2 32.
3 dV 3Volume of sphere 3
3
Q
11. Since div F 1 2y 1 2y, we have
Q
4
2y dV
9y 2
3
3 9y2
0
6
4 x 2 dV
0
Q
4
0
4y
3
6
4x 2 dz dy dx
0
0
3
0
13. Since div F 3x 2 x 2 0 4x 2, we have
4
2y dx dy dz
4
4y9 y 2 dy dz
4
0
4 2 3 2 9 y 3
6
4x 24 y dy dx
0
32x 2 dx 2304.
0
15. Fx, y, z xyi 4yj xz k div F y 4 x
S
F N dS
div F dV
Q
3
0
3
0
3
0
3
0
0
0
2
sin sin sin cos 42 sin d d d
0 2
3 sin2 sin 3 sin2 cos 42 sin d d d
0
3 sin2 cos 3 sin2 sin 42 sin
0
82 sin d d
82 cos
3
162 d
0
2
0 d d
0
0
y x 4 dV
Q
3
0
d
163 3
3 0
144.
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3 3
dz 0.
209
210
Chapter 14
Vector Analysis
17. Using the Divergence Theorem, we have
S
curl F N dS
div curlF dV
Q
i curl Fx, y, z x 4xy z 2
j y 2x 2 6yz
k z 2xz
div curl F 0. Therefore,
6y i 2z 2z j 4x 4x k 6yi
div curl F dV 0.
Q
19. See Theorem 14.12, page 1073. 21. Using the triple integral to find volume, we need F so that div F
M N P 1. x y z
Hence, we could have F x i, F yj, or F z k. 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
i fy j fz k 1 fy2 fz2
fx i j fz k 1 fx2 fz2
fx i fy j k
For dA dx dy consider F z k, z f x, y, then N Correspondingly, we then have V
S
F N dS
23. Using the Divergence Theorem, we have
x dy dz
S
curl F N dS
S
Fx, y, z M i Nj Pk curl F div curl F Therefore,
S
1 fx2 fy2
and dS 1 fy2 fz2 dy dz. and dS 1 fx2 fz2 dz dx. and dS 1 fx2 fy2 dx dy.
y dz dx
S
z dx dy.
S
div curl F dV. Let
Q
N P M N M i j k P y z x z x y 2P 2N 2P 2M 2N 2M 0. x y x z y x y z z x z y
curl F N dS
0 dV 0.
Q
25. If Fx, y, z x i yj z k, then div F 3.
S
27.
F N dS
div F dV
Q
f D N g dS
S
f g
S
3 dV 3V.
Q
N dS
div f g dV
Q
f div g f g dV
Q
f 2g f g dV
Q
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Section 14.8
Section 14.8
Stokes’s Theorem
Stokes’s Theorem
1. F x, y, z 2y z i xyz j e z k
curl F
i x
j y
k z
2y z
xyz
ez
xyi j yz 2k
3. F x, y, z 2z i 4x 2 j arctan xk j y
k z
4x2
arctan x
i curl F x 2z
5. F x, y, z e x
curl F
2 y 2
i ey
i x
ex
2
y 2
j y
ey
2
2 z 2
2
1 j 8xk 1 x2
j xyzk
k z
z 2
xyz
xz 2ze
y 2 z 2
i yz j 2ye x
zx 2e y
2 z 2
i yz j 2ye x
2 y 2
2 y 2
k
k
7. In this case, M y z, N x z, P x y and C is the circle x 2 y 2 1, z 0, dz 0. Line Integral:
C
F dr
y dx x dy
C
Letting x cos t, y sin t, we have dx sin t dt, dy cos t dt and
y dx x dy
C
2
sin 2 t cos 2 t dt 2.
0
Double Integral: Consider F x, y, z x 2 y 2 z 2 1. Then N
F 2x i 2yj 2zk x i yj zk. F 2x 2 y 2 z 2
Since z 2 1 x 2 y 2, z x
2x x y , and z y , dS 2z z z
Now, since curl F 2k, we have
S
curl F N dS
R
2z
1 xz
2 2
y2 1 dA dA. z2 z
1z dA 2 dA 2Area of circle of radius 1 2. R
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211
212
Chapter 14
Vector Analysis
9. Line Integral: From the accompanying figure we see that for
z
C1: z 0, dz 0
6 4
C2: x 0, dx 0
C2
C3
2
C3 : y 0, dy 0. Hence,
C
F dr
C
y dy
3
0
y dy
0
4
y dy z dz
C2
C3
6
y dy
3
(0, 3, 0)
2
xyz dx y dy z dz
C1
(0, 0, 6)
x
(4, 0, 0)
C1
y
z dz
0
z dz
0
z dz 0.
6
Double Integral: curl F xyj xzk Considering Fx, y, z 3x 4y 2z 12, then N
3i 4j 2k F and dS 29 dA. F 29
Thus,
S
curl F N dS
4xy 2xz dy dx
R
4
(3x12) 4
0
0
4
4xy 2x6 2y 23 x dy dx
(123x) 4
0
8xy 3x 2 12x dy dx
0
4
0 dx 0.
0
\
\
11. Let A 0, 0, 0, B 1, 1, 1 and C 0, 2, 0. Then U AB i j k and V AC 2j. Thus, N
U V 2i 2k i k . U V 2 22
Surface S has direction numbers 1, 0, 1, with equation z x 0 and dS 2 dA. Since curl F 3i j 2k, we have
S
curl F N dS
R
1 2
2 dA
dA Area of triangle with a 1, b 2 1.
R
13. F x, y, z z 2 i x 2 j y 2 k, S: z 4 x 2 y 2, 0 ≤ z i curl F x
j y
k z 2yi 2z j 2xk
z2
x2
y2
Gx, y, z x 2 y 2 z 4 Gx, y, z 2x i 2yj k
S
curl F N dS
4xy 4yz 2x dA
R
2
4x 2
2 2
4x 2
2
4x 2
4x 2
4xy 4y 4 x 2 y 2 2x dy dx
4xy 16y 4x 2y 4y 3 2x dy dx
2
2
4x4 x 2 dx 0
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Section 14.8
Stokes’s Theorem
15. Fx, y, z z 2 i yj xzk, S: z 4 x 2 y 2
i curl F x
j y
k z
z2
y
xz
zj
Gx, y, z z 4 x 2 y 2 Gx, y, z
x y i jk 2 2 4 x y 4 x 2 y 2
curl F F dS
S
yz dA 4 x 2 y 2
R
x 17. Fx, y, z lnx 2 y 2 i arctan j k y
curl F
j y
k z
arctan x y
1
i x 1 2 ln
x2
y2
R
y4 x 2 y 2 dA 4 x 2 y 2
1 1 y x y x 2
2
2
2
4x 2
2 4x 2
y dy dx 0
y 2y k 2 k y2 x y2
S: z 9 2x 3y over one petal of r 2 sin 2 in the first octant. Gx, y, z 2x 3y z 9 Gx, y, z 2i 3j k
S
curl F N dS
R
2
0
2
0
2
2y dA x2 y2 2 sin 2
0
2r sin
r dr d
r2
4 sin cos
2 sin dr d
0
8 sin 2 cos d
0
19. From Exercise 10, we have N
S
curl F N dS
R
2x i k
a
0
a
j y
k z
1
1
2
Letting N k, we have
0
8 3
0
0
ax 3 dx
ax4
4 a 0
a5 . 4
23. See Theorem 14.13, page 1081.
0
S
2
a
x 3 dy dx
21. Fx, y, z i j 2k i curl F x
3
and dS 1 4x 2 dA. Since curl F xyj xzk, we have
1 4x 2
xz dA
8 sin3
curl F N dS 0.
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213
214
Chapter 14
25. (a)
f g
C
Vector Analysis
dr
S
g g g if jf k x y z
f g f
curl f g
curl f g N dS (Stoke’s Theorem)
i x
j y
f g x f g y f g z 2g
(b)
C
(c)
f
g
2g
f
g
2g
f
g
2g
f
g
f xy x y f yx y x k
f g f g f g f g i j k yf gz zf g y x z z x x y y x
f f dr
j
k
f y g y
f z g z
dr
f
S
S
f g gf dr
f g
C
curl f g N dS
f g dr f f
S
C
since
C r dr
i C r a x and
curlC r
1 2
j b y
S
k c z
i x
g N dS.
gf dr
g N dS
g N dS
S
S
27. Let C ai bj ck, then
C
f
S
f N dS (using part a.)
S
g
1 2
f
2g
0 since f f 0.
C
2g
f xz x z f zx z x j
f g
C
g
f x g x
f
f yz y z f zy z y i
i
Therefore,
k z
curl C r N dS
1 2
g f N dS (using part a.) f g N dS 0
S
2C
N dS
C
S
bz cy i az cx j ay bx k
j y
k z
bz cy cx az ay bx
2ai b j ck 2C.
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N dS
Review Exercises for Chapter 14
Review Exercises for Chapter 14 1. Fx, y, z x i j 2k
3. f x, y, z 8x 2 xy z 2 Fx, y, z 16x y i x j 2z k
z 3 2
3 4 x
4
y
5. Since My 1y 2 Nx, F is not conservative. 7. Since My 12xy Nx, F is conservative. From M Ux 6xy 2 3x 2 and N Uy 6x 2 y 3y 2 7, partial integration yields U 3x 2 y 2 x3 h y and U 3x 2 y 2 y3 7y gx which suggests h y y3 7y, gx x3, and Ux, y 3x 2 y 2 x3 y3 7y C. 9. Since N M 4x , y x P M 1 . z x F is not conservative. 11. Since M 1 N 2 , y y z x
M 1 P 2 , z yz x
P x N 22 , z y z y
F is conservative. From U x U 1 U x , N 2 , P 2 x yz y y z z yz
M we obtain U
x f y, z, yz
U
x x x gx, z, U hx, y ⇒ f x, y, z K yz yz yz
13. Since F x 2 i y 2 j z2 k: (a) div F 2x 2y 2z (b) curl F
N P M N M i j k 0i 0j 0k 0 P y z x z x y
15. Since F cos y y cos x i sin x x sin y j x yz k: (a) div F y sin x x cos y xy (b) curl F xz i yz j cos x sin y sin y cos xk xz i yz j
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215
216
Chapter 14
Vector Analysis
17. Since F arcsin x i xy 2 j yz2 k: 1 2xy 2yz (a) div F 1 x 2
19. Since F lnx 2 y 2i lnx 2 y 2j z k: (a) div F
(b) curl F z 2 i y 2 k
2y 2x 1 x2 y 2 x2 y 2 2x 2y 1 x2 y 2 2x 2y k x2 y 2
(b) curl F
21. (a) Let x t, y t, 1 ≤ t ≤ 2, then ds 2 dt.
2
x 2 y 2 ds
1
C
2t 22 dt 22
t3
3
2 1
62
(b) Let x 4 cos t, y 4 sin t, 0 ≤ t ≤ 2, then ds 4 dt.
2
x 2 y 2 ds
C
164 dt 128
0
23. x cos t t sin t, y sin t t cos t, 0 ≤ t ≤ 2,
x 2 y 2 ds
C
2
dx dy t cos t, t sin t dt dt
cos t t sin t2 sin t t cos t2 t 2 cos2 t t 2 sin2 t dt
0
2
t 3 t dt
0
2 21 2 2 25. (a) Let x 2t, y 3t, 0 ≤ t ≤ 1
1
2x y dx x 3y dy
0
C
1
7t2 7t3 dt
35t dt
0
35 2
(b) x 3 cos t, y 3 sin t, dx 3 sin t dt, dy 3 cos t dt, 0 ≤ t ≤ 2
2x y dx x 3y dy
C
27.
2
9 9 sin t cos t dt 18
0
2x y ds, r t a cos3 t i a sin3 t j, 0 ≤ t ≤
C
x t 3a
2
cos2 t sin t
y t 3a sin2 t cos t
2x y ds
2
0
C
2a cos3 t a sin3 tx t2 y t2 dt
9 a2 5
29. f x, y 5 sinx y C: y 3x from 0, 0 to 2, 6 rt t i 3t j, 0 ≤ t ≤ 2 r t i 3j r t 10 Lateral surface area:
C2
2
f x, y ds
0
2
5 sint 3t 10 dt 10
0
5 sin 4t dt
10
4
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41 cos 8 32.528
Review Exercises for Chapter 14 33. dr 2 sin t i 2 cos t j k dt
31. d r 2t i 3t 2 j dt
F 2 cos t i 2 sin t j t k, 0 ≤ t ≤ 2
F t 5 i t 4 j, 0 ≤ t ≤ 1
C
F dr
1
5t 6
0
5 dt 7
C
F dr
2
t dt 2 2
0
35. Let x t, y t, z 2t 2, 2 ≤ t ≤ 2, dr i j 4t k dt. F t 2t 2 i 2t 2 t j 2t k
C
F dr
2
2
4t 2 dt
4t3
3 2 2
64 3
37. For y x 2, r1t t i t 2 j, 0 ≤ t ≤ 2
y
y = 2x
For y 2x, r2t 2 t i 4 2t j, 0 ≤ t ≤ 2
xy dx x 2 y 2 dy
C
xy dx x 2 y 2 dy
C1
4
3
xy dx x 2 y 2 dy
y = x2
2
C2 1
4 100 32 3 3
(2, 4) C2
C1 x 1
2
3
4
39. F x i y j is conservative. Work
41.
2 x 1
2
4, 8
2 y 32 3
0, 0
2 32 8 1 16 8 3 42 2 3 3
C
43. (a)
1, 4, 3
2xyz dx x 2z dy x 2 y dz x 2 yz
y 2 dx 2xy dy
C
1
0, 0, 0
12
1 t23 21 3t1 t dt
0
1
3t 2 2t 1 23t 2 4t 1 dt
0
1
9t 2 14t 5 dt
0
3t 3 7t 2 5t (b)
y 2 dx 2xy dy
C
4
1
4
1
15
0
t 1 2tt
1 2t
dt
t t dt
1
t2
4 1
15
(c) Fx, y y 2 i 2xy j f where f x, y xy 2. Hence,
C
45.
C
F dr 42 11 15 2
2
y dx 2x dy
0
2
2
0
2
2 1 dy dx
0
2 dx 4
47.
xy 2 dx x 2y dy
C
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R
2xy 2xy dA 0
217
218
49.
Chapter 14
Vector Analysis
1
xy dx x 2 dy
0
C
x
x
1
x dy dx 2
x 2 x3 dx
0
1 12
z
51. ru, v sec u cos v i 1 2 tan u sin v j 2u k 0 ≤ u ≤
6
, 0 ≤ v ≤ 2 3 −4 2 4 x
53. (a)
2
y
4
−2
(b)
z
z 3
3
−4
−4 −4
−4 −3 4
4
y
4
−2
x
3
(d)
z
3
4
y
4
y
3
2
−2
−4 −3 4
2
z
3
3
2 −2
x
−1 −3
−3
(c)
2
−2
x
−3
−4
−4 −4 −3
3
4
y
4
1
−2
3
2 −2
x
3
−3
−3
The space curve is a circle:
4 3 2 2 cos u i 3 2 2 sin u j
r u,
(e) ru 3 cos v sin u i 3 cos v cos u j rv 3 sin v cos u i 3 sin v sin u j cos v k
i j k 3 cos v cos u 0 ru rv 3 cos v sin u 3 sin v cos u 3 sin v sin u cos v
3 cos2 v cos u i 3 cos2 v sin u j 9 cos v sin v sin2 u 9 cos v sin v cos2 uk 3 cos2 v cos u i 3 cos2 v sin u j 9 cos v sin vk ru rv 9 cos4 v cos2 u 9 cos4 v sin2 u 81 cos2 v sin2 v 9 cos4 v 81 cos2 v sin2 v Using a Symbolic integration utility,
2
4
2
ru rv du dv 14.44
0
(f) Similarly,
4
0
2
ru rv dv du 4.27
0
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2
2
k
Review Exercises for Chapter 14 0 ≤ u ≤ 2, 0 ≤ v ≤ 2
55. S: ru, v u cos v i u sin vj u 12 u k,
z
ru u, v cos v i sin vj 3 2u k rv u, v u sin v i u cos vj
i ru ru cos v u sin v
2
x y dS
S
2
x
3 −2
2
u cos v u sin v u2u 32 1 du dv
0
0
2
−3
3
j k sin v 3 2u 2u 3u cos v i 2u 3u sin v j u k u cos v 0
ru rv u2u 3 1
2 −3
0
2
cos v sin vu22u 32 1 dv du 0
0
57. Fx, y, z x2 i xyj z k Q: solid region bounded by the coordinates planes and the plane 2x 3y 4z 12 Surface Integral: There are four surfaces for this solid. z0 y 0, x 0,
S4
F N dS
1 4
1 4
1 4 1 6
0 dS 0
S2
F N x 2,
N i,
0 dS 0
S1
F N xy,
N j,
0 dS 0
S3
2i 3j 4k , dS 29
2x 3y 4z 12, N
F N z,
N k,
1 14 169 dA
29
dA
4
2x 2 3xy 4z dA
R
6
0
4(2x3)
2x 2 3xy 12 2x 3y dy dx
0
6
2x 2
0
12 3 2x 3x2 12 3 2x
6
x3 x 2 24x 36 dx
0
2
12
12 3 2x 2x12 3 2x 23 12 3 2x dx 2
1 x 4 x3 12x 2 36x 6 4 3
6 0
66
Divergence Theorem: Since div F 2x x 1 3x 1, Divergence Theorem yields
6
div F dV
0
Q
0
6
1 4
1 4
1 4
(122x3y)4
3x 1 dz dy dx
0
(122x)3
0
(122x)3
3x 1
0
6
0
6
0
6
0
12 2x4 3y dy dx (122x3
12 3 2x 2312 3 2x dx
3 3x 1 12y 2xy y 2 2
3x 1 412 2x 2x
dx
0 2
2 3 1 3x 4 35x3 3x 35x 2 96x 36 dx 48x 2 36x 3 6 4 3
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6 0
66.
y
219
220
Chapter 14
Vector Analysis
59. Fx, y, z cos y y cos x i sin x x sin y j xyz k S: portion of z y 2 over the square in the xy-plane with vertices 0, 0, a, 0, a, a, 0, a Line Integral: Using the line integral we have: dy 0
C1: y 0,
C
z 1
C2: x 0, dx 0,
z y 2,
dz 2y dy
C3: y a, dy 0,
z a2,
dz 0
C4: x a, dx 0,
z y 2,
dz 2y dy
F dr
C4 C1 a
cos y y cos x dx sin x x sin y dy xyz dz dx
C3
0
a
dx
sin a a sin y dy ay32y dy
C4
a
cos a a cos x dx
a
0
cos a a cos x dx
0
C2
0
a
sin a a sin y dy
2ay 4 dy
0
y sin a a cos y 2a 5
a x cos a a sin x
0
a
a
0
a a cos a a sin a a sin a a cos a a
y5
a 0
2a6 2a6 5 5
Double Integral: Considering f x, y, z z y 2, we have: f 2yj k , dS 1 4y 2 dA, and curl F xz i yz j. f 1 4y 2
N Hence,
S
curl F N dS
a
0
a
a
2y 2z dy dx
0
0
a
a
2y 4 dy dx
0
0
2a5 2a6 dx . 5 5
Problem Solving for Chapter 14 1. (a) T
25 xi yi zk x2 y2 z23 2
N xi 1 x2 k dS
1 1 x2
Flux
dy dx
kT
N dS
S
R
1 2
25k
1
1 2 0 1
25k
0
25k
1
1 2 0 1 2
25k
z x2 dA x2 y2 z23 21 x21 2 x2 y2 z23 2
25k
x2
x2
y2
x2 1 2
x2
y2
1 x2 dy dx z23 21 x21 2
1 dy dx 1 y23 21 x21 2
1 dy 1 y23 2
22 3 25k
1
z2 3 2
1 2
1 dx 1 x21 2 1 2
2
6
—CONTINUED—
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C2 a y
x
C
C1
C3
Problem Solving for Chapter 14 1. —CONTINUED— (b) ru, v cos u, v, sin u ru sin u, 0, cos u, rv 0, 1, 0 ru rv cos u, 0, sin u T
25 xi yj zk x2 y2 z23 2 25 cos ui vj sin uk v 2 13 2 25
25
ru rv v2 13 2 cos2 u sin2 u v2 13 2
T
1
Flux
0
2 3
3
2 25k du dv 25k v2 13 2 6
3. rt 3 cos t, 3 sin t, 2t r t 3 sin t, 3 cos t, 2, r t 13 Ix
y2 z2 ds
C
Iy
x2 z2 ds
2
0
x2 y2 ds
C
5.
2
0
C
Iz
2
1 9 sin2 t 4t213 dt 13 322 27 3
1 9 cos2 t 4t213 dt 13 322 27 3
9 cos2 t 9 sin2 t13 dt 1813
0
1 1 x dy y dx 2 C 2
2
0
a sin a sin d a1 cos a1 cos d
2
1 a2 2
sin sin2 1 2 cos cos2 d
0 2
1 a2 2
sin 2 cos 2 d
0
3a2 Hence, the area is 3a2. 7. (a) rt tj, 0 ≤ t ≤ 1 r t j W
1
F dr
0
C
ti j j dt
1
dt 1
0
(b) rt t t 2 i tj, 0 ≤ t ≤ 1 r t 1 2t i j W F dr
1
2t t 2 i t t 22 1 j 1 2ti j dt
0 1
1 2t2t t 2 t 4 2t 3 t 2 1 dt
0
1
0
t 4 4t2 2t 1 dt
13 15
—CONTINUED—
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221
222
Chapter 14
Vector Analysis
7. —CONTINUED— (c) rt ct t 2 i t j, 0 ≤ t ≤ 1 r t c1 2t i j F dr ct t 2 tc1 2t c2t t 22 11 c 2 t 4 2c 2 t 2 c 2 t 2ct 2 ct 1
W
F dr
C
1 2 1 c c1 30 6
dW 5 1 1 c 0 ⇒ c dc 15 6 2 5 d 2W 1 > 0 c minimum. dc 2 15 2 9. v r a1, a2, a3 x, y, z a 2 z a 3 y, a1z a 3 x, a 1 y a 2 x curlv r 2a 1, 2a 2, 2a 3 2v By Stoke’s Theorem,
curlv r N dS
v r dr
C
S
2v
S
N dS.
11. Fx, y Mx, y i Nx, y j M
m 3xy i 2y 2 x 2 j x 2 y 25 2
3mxy 3mxyx 2 y 25 2 x 2 y 2 5 2
M 5 3mxy x2 y27 22y x2 y25 23mx y 2
3mxx 2 y 27 2 5y 2 x 2 y 2 N
3mxx 2 4y 2 x 2 y 27 2
m2y 2 x 2 m2y 2 x 2x 2 y 25 2 x 2 y 25 2
N 5 m2y 2 x 2 x 2 y 27 22x x 2 y 25 22mx x 2
mxx 2 y 27 2 2y 2 x 25 x 2 y 22 mxx 2 y 27 2 3x 2 12y 2 Therefore,
3mxx 2 4y 2 x 2 y 27 2
N M and F is conservative. x y
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