SOL Cálculo de Varias Variables - Dennis G. Zill, Warren Wright - 4ta Edición

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Chapter 10
Conics and Polar Coordinates
10.1
Conic Sections
y
1. vertex: (0, 0)
focus: (1, 0)
directrix: x = −1
axis: y = 0
x
y
2. vertex: (0, 0)
focus: (7/8, 0)
directrix: x = − 78
axis: y = 0
x
y
3. vertex: (0, 0)
focus: (0, −4)
directrix: y = 4
axis: x = 0
x
668
669
10.1. CONIC SECTIONS
4. vertex: (0, 0)
1
focus: 0, 40
1
directrix: y = − 40
axis: x = 0
y
x
5. vertex: (0, 1)
focus: (4, 1)
directrix: x = −4
axis: y = 1
y
x
6. vertex: (−2, −3)
focus: (−4, −3)
directrix: x = 0
axis: y = −3
y
x
7. vertex: (−5, −1)
focus: (−5, 2)
directrix: y = 0
axis: x = −5
y
x
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670
CHAPTER 10. CONICS AND POLAR COORDINATES
8. vertex: (2, 0) focus: 2, − 14
directrix: y = 14
axis: x = 2
y
9. (y + 6)2 = 4(x + 5)
vertex: (−5, −6)
focus: (−4, −6)
directrix: x = −6
axis: y = −6
y
10. (x + 3)2 = −(y + 2)
vertex: (−3, −2)
focus: (−3, −9/4)
directrix: y = −7/4
axis: x = −3
2
11. x + 52 = 14 (y − 1)
vertex: (−5/2, −1)
focus: (−5/2, −15/16)
directrix: y = −17/16
axis: x = − − 5/2
x
x
y
x
y
x
671
10.1. CONIC SECTIONS
12. (x − 1)2 = 4(y − 4)
vertex: (1, 4)
focus: (1, 5)
directrix: y = 3
axis: x = 1
y
x
13. (y − 4)2 = −2(x + 3)
vertex: (3, 4)
focus: (5/2, 4)
directrix: x = 7/2
axis: y = 4
y
x
14. (y − 2)2 = 4 x + 41
vertex: (−1/4, 2)
focus: (3/4, 2)
directrix: x = −5/4
axis: y = 2
y
x
15. x2 = 28
16. y 2 = −16x
17. y 2 = 10x
18. x2 = −40y
19. The parabola is of the form (y − k)2 = 4p(x − h) with (h, k) = (−2, −7) and p = 3. Thus the
equation is (y + 7)2 = 12(x + 2).
20. The parabola is of the form (x−h)2 = 4p(y −5) with (h, k) = (2, 0) and p = 3, so the equation
of the parabola is (x − 2)2 = 12y.
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CHAPTER 10. CONICS AND POLAR COORDINATES
21. The parabola is of the form x2 = 4py with (−2)2 = 4p(8). Thus p =
x2 = 12 y.
22. The parabola is of the form y 2 = 4px with
1
y 2 = 16
x.
1 2
4
= 4p(1) so p =
1
64 .
1
8
and the equation is
Thus the equation is
23. To find the x-intercept set y = 0. Solving 42 = 4(x + 1) gives x = 3. The x-intercept is (3, 0).
To find the y-intercept set x = 0. Solving (y + 4)2 = 4 gives y = −4 ± 2. The y-intercepts are
(0, −2) and (0, −6).
√
2
24. To find
√set y = 0. Solving (x − 1) = 2 gives x = 1 ± 2. The x-intercepts are
√ the x-intercept
(1 + 2, 0) and (1 − 2, 0). To find the y-intercept set x = 0. Solving 1 = −2(y − 1) gives
y = 12 . The y-intercept is (0, 1/2).
y
25. center: (0,√
0)
foci: (0, ± 15)
vertices: (0, ±4)
endpoints of the
√ minor axis: (±1, 0)
15
eccentricity:
4
x
26. center: (0, 0)
foci: (±4, 0)
vertices: (±5, 0)
endpoints of the minor axis: (0, ±3)
4
eccentricity:
5
y
x2
y2
+
= 1 center: (0, 0)
16
9√
foci: (± 7, 0)
vertices: (±4, 0)
endpoints of the
√ minor axis: (0, ±3)
7
eccentricity:
4
y
27.
x
x
673
10.1. CONIC SECTIONS
28.
x2
y2
+
= 1 center: (0, 0)
2
4 √
foci: (0, ± 2)
vertices: (0, ±2)
√
endpoints of the
√ minor axis: (± 2, 0)
2
eccentricity:
2
29. center: (1,√
3)
foci: (1 ± 13, 3)
vertices: (−6, 3), (8, 3)
endpoints of the
√ minor axis: (1, −3), (1, 9)
13
eccentricity:
7
30. center: (−1, 2)√
foci: (−1, 2 ± 11)
vertices: (−1, −4), (−1, 8)
endpoints of the
√ minor axis: (−6, 2), (4, 2)
11
eccentricity:
6
y
x
y
x
y
x
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CHAPTER 10. CONICS AND POLAR COORDINATES
31. center: (−5, −2)√
foci: (−5, −2 ± 15)
vertices: (−5, −6), (−5, 2)
endpoints of the
√ minor axis: (−6, −2), (−4, −2)
15
eccentricity:
4
32. center: (3, −4)√
foci: (3, −4 ± 17)
vertices: (3, −13), (3, 5)
endpoints of the
√ minor axis: (−5, −4), (11, −4)
17
eccentricity:
9
2
y + 12
=1
33. x +
4
center: (0, −1/2)√
foci: (0, −1/2 ± 3)
vertices: (0, −5/2), (0, 3/2)
endpoints of the
√ minor axis: (−1, −1/2), (1, −1/2)
3
eccentricity:
2
2
y
x
y
x
y
x
675
10.1. CONIC SECTIONS
34.
35.
36.
(x + 2)2
(y − 4)2
+
=1
2
72
center: (−2, 4)√
foci: (−2, 4 ± 70)√
vertices: (−2, 4 ± 6 2)
√
endpoints of the
√ minor axis: (−2 ± 2, 4)
35
eccentricity:
6
(x − 7)2
(y + 1)2
+
=1
9
25
center: (2, −1)
foci: (2, −5), (2, 3)
vertices: (2, −6), (2, 4)
endpoints of the minor axis: (−1, −1), (5, −1)
4
eccentricity:
5
(y − 1)2
(x + 1)2
+
=1
5
9
center: (−1, 1)
foci: (−1, −1), (−1, 3)
vertices: (−1, −2), (−1, 4)
√
endpoints of the minor axis: (−1 ± 5, 1)
2
eccentricity:
3
y
x
y
x
y
x
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676
37.
CHAPTER 10. CONICS AND POLAR COORDINATES
x2
(y + 3)2
+
=1
9
3
center: (0,
−3)
√
foci: (± 6, −3)
vertices: (±3, −3)
√
3)
endpoints of the
minor
axis:
(0,
−3
±
√
6
eccentricity:
3
(y − 1/2)2
=1
3
center: (1, 1/2)√
foci: (1, 1/2 ± 2)√
vertices: (1, 1/2 ± 3)
endpoints of the
√ minor axis: (0, 1/2), (2, 1/2)
2
eccentricity:
3
38. (x − 1)2 +
y
x
y
x
39. The center is (0, 0) with the x-axis as the major axis. a = 5 and c = 3, so b = 4. Thus the
x2
y2
equation is
+
= 1.
25 16
√
40. The center is (0, 0) with the x-axis as the major axis. a = 9 and c = 2, so b = 77. Thus the
x2
y2
equation is
+
= 1.
81 77
41. The center is (1, −3) with the x-axis as the major axis. a = 4 and b = 2. Thus the equation
(x − 1)2
(y + 3)2
is
+
= 1.
16
4
42. The center is (1, −2) with the y-axis as the major axis. a = 4 and b = 3. Thus the equation
(x − 1)2
(y + 2)2
is
+
= 1.
9
16
√
√
43. The center is (0, 0) with the x-axis as the major axis. c = 2 and b = 3, so a = 11. Thus
x2
y2
the equation is
+
= 1.
11
9
677
10.1. CONIC SECTIONS
44. The center is (0, 0) with the y-axis as the major axis. c =
x2
y2
the equation is
+
= 1.
59 64
√
5 and a = 8, so b =
√
59. Thus
√
45. The center is (0, 0) with the y-axis as the major axis. c = 3 thus 9 = a2 − b2 and a = 9 + b2 .
y2
x2
= 1. The ellipse passes through the point
Thus the equation is of the form 2 +
b
9 + b2
√
√
√
(−1)2 (2 2)2
(−1, 2 2), thus
+
= 1. Solving this for b, we obtain b = 3. Thus a2 = 12 and
2
2
b
9+b
y2
x2
+
= 1.
the equation is
3
12
46. The center is (0, 0) with the x-axis as the major axis and a = 5. The equation is of the form
√
x2 y 2
5
19
+ 2 = 1. The ellipse passes through the point ( 5, 4) so
+ 2 = 1. Solving for b2 , we
25 b
25 b
x2
y2
2
obtain b = 20. Thus the equation of the ellipse is
+
= 1.
25 20
47. The y-axis as the major axis with c = 3 and a = 4. Thus b =
(x − 1)2
(y − 3)2
ellipse is
+
=1
7
16
√
7 and the equation of the
48. The center is (15/2, 4) with the x-axis as the major axis. a = 1/2 and c = 7/2, thus b =
(x − 15/2)2
(y − 4)2
= 1.
Thus the equation is
+
(11/2)2
18
√
18.
y
49. center: (0,
√ 0)
foci: (± 41, 0)
vertices: (±4, 0)
asymptotes: y = ± 54 x
√
41
eccentricity:
4
x
y
50. center: (0,
√ 0)
foci: (± 8, 0)
vertices: (±2, 0)
asymptotes: y√= ±x
eccentricity: 2
x
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678
51.
52.
CHAPTER 10. CONICS AND POLAR COORDINATES
y2
x2
−
= 1 center: (0, 0)
20
4
√
foci: (0, ±2 6)√
vertices: (0, ±2 5)√
asymptotes: yr= ± 5x
6
eccentricity:
5
y
x
y2
x2
−
= 1 center: (0, 0)
9
16
foci: (0, ±5)
vertices: (0, ±3)
asymptotes: y = ± 34 x
5
eccentricity:
3
53. center: (5,√
−1)
foci: (5 ± 53, −1)
vertices: (3, −1), (7, −1)
asymptotes: y = −1 ± 72 (x − 5)
√
53
eccentricity:
2
y
x
y
x
679
10.1. CONIC SECTIONS
54. center: (−2,√
−4)
foci: (−2 ± 35,√−4)
vertices: (−2 ± 10, −4)
5x + 10
asymptotes: y = −4 ± √
10
r
7
eccentricity:
2
55. center: (0, 4)√
foci: (0, 4 ± 37)
vertices: (0, −2), (0, 10)
asymptotes: y√= 4 ± 6x
37
eccentricity:
6
y
x
y
x
56. center: (−3, 1/4)√
foci: (−3, 1/4 ± 13)
vertices: (−3, 9/4), (−3, −7/4)
1
asymptotes: y = ± 23 (x + 3)
√ 2
13
eccentricity:
2
y
x
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680
57.
58.
59.
CHAPTER 10. CONICS AND POLAR COORDINATES
(x − 3)2
(y − 1)2
−
= 1 center: (3, 1)
5
√ 25
foci: (3 ± 30,√1)
vertices: (3 ± 5, 1) √
asymptotes: y√= 1 ± 5(x − 3)
eccentricity: 6
y
x
(x + 1)2
(y − 1/2)2
−
= 1 center: (−1, 1/2)
10
√ 50
foci: (−1 ± 60,√1/2)
vertices: (−1 ± 10, 1/2)
√
asymptotes: y√= 1/2 ± 5(x + 1)
eccentricity: 6
(y − 1)2
(x − 2)2
−
= 1 center: (2, 1)
6
√ 5
foci: (2 ± 11,√1)
vertices: (2 ± 6, 1) r
5
asymptotes: y = 1 ±
(x − 2)
6
r
11
eccentricity:
6
y
x
y
x
681
10.1. CONIC SECTIONS
60.
(x − 8)2
(y − 3)2
−
= 1 center: (8, 3)
25
√ 16
foci: (8 ± 41, 3)
vertices: (13, 3), (3, 3)
asymptotes: y√= 3 ± 4/5(x − 8)
41
eccentricity:
5
(x − 1)2
= 1 center: (1, 3)
√1/4
foci: (1, 3 ± 5/2)
vertices: (1, 2), (1, 4)
asymptotes: y√= 3 ± 2(x − 1)
5
eccentricity:
2
61. (y − 3)2 −
y
x
y
x
62.
(y + 5)2
(x + 1)2
−
= 1 center: (−1, −5)
18
4√
foci: (−1, −5 ± 22)√
vertices: (−1, −5 ± 18)
√
asymptotes: y = −5 ± 218 (x + 1)
√
11
eccentricity:
3
y
63. The center is (0, 0) with the y-axis as the transverse axis. c = 4 and a = 2, thus b =
y2
x2
The equation is
−
=1
4
12
64. The center is (0, 0) with the y-axis as the transverse axis. c = 3 and a = 3/2, thus b =
x
√
12.
√
3 3
.
2
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682
CHAPTER 10. CONICS AND POLAR COORDINATES
The equation is
x2
y2
−
=1
9/4 27/4
65. The center is (1, −3) with the y-axis as the transverse axis. c = 3 and a = 2, thus b =
(y + 3)2
(x − 1)2
The equation is
−
=1
4
5
√
5.
66. The center is (2, 2) with the y-axis as the transverse axis. c = 5 and a = 32, thus b = 4. The
(x − 2)2
(y − 2)2
−
= 1.
equation is
9
16
67. The center is (−1, 3) with the y-axis as the transverse axis. a = 1 and the equation is of the
√
(y − 3)2
(x + 1)2
= 1. The hyperbola passes through the point (−5, 3 + 5) thus
form
−
2
1
b
√
(−5 + 1)2
(x + 1)2
2
2
2
(3 + 5 − 3) −
=
1.
Thus
b
=
4
and
the
equation
is
(y
−
3)
−
= 1.
b2
4
68. The center is (3, −5) with the y-axis as the transverse axis. a = 3 and the equation is of
(x + 5)2
(y − 3)2
−
= 1. The hyperbola passes through the point (1, −1) thus
the form
9
b2
2
2
(−4)
(6)
324
(y − 3)2
(x + 5)2
− 2 = 1. Thus b2 =
and the equation is
−
= 1.
9
b
7
9
324/7
69. The center is (2, 4) with the y-axis as the transverse axis and a = 1. After solving the
x+6
x
asymptote given in the problem for y, we obtain y =
= + 3. The equation of the
2
2
(x − 2)2
2
hyperbola is of the form (y − 4) −
= 1. The asymptote equations for this hyperbola
b2
x−2
−x + 2
x
2
are y − 4 =
and y − 4 =
(these are also equivalent to y = + 4 −
and
b
b
b
b 2
x
). Letting b equal 2 or -2 will yield one asymptote with the equation
y = − + 4+
b
b
x
(x − 2)2
y = + 3. In either case, the equation of the hyperbola is (y − 4)2 −
= 1.
2
4
√
√
c
70. The y-axis is the conjuate axis. The center is (−5, 7) with b = 3 and = 10. Thus c = 10a.
a
(y + 7)2
2
2
2
2
2
2
Since c = b +a then 10a = 9+a and a = 1. Thus the equation is (x+5)2 −
= 1.
9
71. We place the coordinate axes so that the origin is at the vertex of the parabola. The point
(2, 2) lies on the parabola. Thus the equation is x2 = 2y with p = 1/2. The focus of this
parabola occurs at the point (0, 1/2). Thus the light source is 6 inches from the vertex.
72. We place the coordinate axes so that the origin is at the vertex of the parabola. The point
(10, 4) lies on the parabola. Thus the equation is x2 = 25y with p = 25/4. The focus is located
at (0, 25/4). The eyepiece should be located 6.25 ft from the vertex.
73. We place the coordinate axes so that the origin is at the vertex of the parabola. The parabola
is of the form x2 = 4py and contains the point (20, 1). Thus the equation of the parabola is
x2 = 400y. The towers are located at x = 175 and x = −175. Hence the height of the towers
683
10.1. CONIC SECTIONS
is found by solving (175)2 = 400y. Solving this equation yields y = 76.5625. Therefore the
towers are 76.5625 ft above the road.
74. We place the coordinate axes so that the origin is at the vertex of the parabola. The parabola
is of the form x2 = 4py and contains the point (125, 75). Thus the equation of the parabola
is x2 = 625
12 . We need to find the y-value of the point on the parabola when we are 50 ft from
the tower or when x = 75f t. Hence this y-value is found by solving the equation 752 = 625
3 y
which yield the solution y = 27 ft. The height of the cable above the roadway at a point 50
ft from one of the towers is 27 ft.
75. We place the coordinate axes so that the origin is at end of the pipe with the parabola in
Quadrants 3 and 4. The equation is of the form x2 = 4py and the point (4, −2) lies on
the parabola. Therefore the equation is x2 = −8y. The water hits the ground at y = −20.
The point on the parabola with y-value -20 is found by solving x2 = −8(−20). This point
is x = 12.65. Thus the water hits the ground 12.65 m from the point on the ground directly
beneath the end of the pipe.
76. We place the coordinate axes with the x-axis along the ground and the y-axis to be through
the dart √
thrower. Thus the dart was released at the point (0, 5) and hits the ground√at the
point (10 10, 0). The parabola is of the form x2 = 4p(y−5) and contains the point (10 10, 0).
Therefore the equation of the parabola is x2 = −200y. To find the height of the dart 10 ft from
the thrower, we need to find the y-value of the point on the parabola corresponding to the
x-value of 10. Hence, we need to solve the equation 102 = −200y which yields y = −1/2f t.
So 10 feet from the thrower the dart will be 0.5 ft below the thrower or it will be 4.5 ft from
the ground.
77. Taking the center of the ellipse to be at the origin, we have
a = 3.6 × 107 and b = 3.52 × 107 . Since c2 = a2 − b2 ,
c2 = 12.96 × 1014 − 12.3904 × 1014 = 0.5696 × 1014 and
c ≈ 0.75 × 107 . The perihelion or least distance is a − c ≈
2.85 × 107 miles or 28.5 million miles. And the aphelion or
greatest distance is a + c ≈ 4.35 × 107 miles or 43.5 million
miles.
y
3.52x107
b
a
78. Using a = 3.6×107 and c = 0.75×107 , we compute the eccentricity e =
79. From a = 1.67 × 109 and 4.25 × 108 we obtain
c
a-c
0.75 × 107
≈ 0.20833.
3.6 × 107
c2 = a2 − b2 = 2.7889 × 1018 − 18.0625 × 106 = 2.78.89 × 1016 − 18.0625 × 1016
= 260.8275 × 1016 = 2.608275 × 1018 .
Then c ≈ 1.615 × 109 and the eccentricity is
1.615 × 109
c
≈
≈ 0.967.
a
1.67 × 109
3.6x107
x
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684
CHAPTER 10. CONICS AND POLAR COORDINATES
80. We place the coordinate axes so that the origin is
at the center of the ellipse. The length of the major axis is 200+2(4000)+1000 = 9200 so that a =
4600. Therefore c = 4600 − (200 + 4000) √
= 400.
Then b2 = a2 − c2 = 46002 − 4002 = (1000 21)2 ,
x2
y2
√
=1
and the equation is
+
2
4600
(1000 12)2
P
satellite
200 mi
1000 mi
center
of
earth
center of
elliptical
orbit
81. We place the coordinate axes so that the origin is at the point
across the base. Thus
s midway
2
x
2
a = 5 and b = 15 so the equation of the doorway is y = 15 1 −
. The height of the
25
s
32
= 12 ft.
doorway at a point on the base 4 ft from the center is y = 152 1 −
25
82. The base of the room is an ellipse with a = 20 and b = 16. We place the coordinate axes
so that the origin is at the point in the center of the room and let the major axes be in the
x-direction. c2 = a2 − b2 , so c2 = 202 − 162 , which gives c = 12. Thus the foci occur at
the points (−12, 0) and (12, 0). Therefore the listening and whispering posts occur along the
center line on the longer part of the base 12 ft in each direction from the center.
83.
84.
center
vertices
foci
ellipse
(0, 1)
(−2, 1),√(2, 1), (0, −2),
√ (0, 4)
(0, 1 − 5), (0, 1 + 5)
center
vertices
foci
ellipse
(1, 4)
(−2, 4),
√(4, 4), (1, 3),√(1, 5)
(1 − 2 2, 4), (1 + 2 2, 4)
shifted ellipse
(4, 1)
(2.1), (6.1),
√ (4. − 2),√(4.4)
(4, 1 − 5), (4, 1 + 5)
shifted ellipse
(−4, 7)
(−7, 7), √
(−1, 7), (−4, 6),√(−4.8)
(−4 − 2 2, 7), (−4 + 2 2, 7)
y2
x2
−
=1
144 25
(b) Conjugate hyperbolas have the same asymptotes and do not intersect.
85. (a)
(y + 5/2)2
(x − 3/2)2
−
= 1, The
5
5
equations of the asymptotes are y = 7/2−x and x = 1+x. These lines are perpendicular,
thus the hyperbola is a rectangular hyperbola.
86. (a) The equation of the hyperbola can be written as
(b) The hyperbola in Problem 50 is a rectangular hyperbola.
685
10.2. PARAMETRIC EQUATIONS
√
87. Since a = 4 and b = 20, we have c2 = a2 + b2 = 16 + 20 = 36 and hence c = 6. Thus
the foci occur at F1 = (−6, 0) and F2 = (6, 0). The line joining (−6, −5) and F2 is given by
5
5
y=
x − . The ray of light travels southwest along this line.
12
2
√
√
88. (a) The distance from (0, b) to (a, 0) is a2 + b2 . Thus R = a2 + b2 = r.
√
√
2
2
2
2
(b) From A = a +
√ r and B = R − b = a + b + r − b = a + b + (A − a) − b =
2
2
A − (a + b) + a + b we see that
p
p
p
p
p
A − B = a + b − a2 + b2 = (a + b)2 − a2 + b2 = a2 + 2ab + b2 − a2 + b2 > 0.
Thus, A > B.
10.2
1.
2.
Parametric Equations
t
x
y
-3
-5
6
-2
-3
2
t
x
y
0
1
0
π
√6
3
2
-1
-1
0
π
√4
2
2
1/4
1/2
0
1
0
1
3
2
π
3
1
2
3/4
3.
2
5
6
3
7
12
π
2
0
1
5π
6
√
- 23
1/4
7π
√4
2
2
1/2
4.
5.
y
y
y
x
x
x
6.
7.
y
8.
y
y
x
x
x
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686
CHAPTER 10. CONICS AND POLAR COORDINATES
10.
9.
y
y
x
x
11. y = (t2 )2 + 3t2 − 1 = x2 + 3x − 1;
12. − 21 y = t3 + t;
x = − 12 y + 4;
y = x2 + 3x − 1,
2x + y = 4
13. x = cos 2t = cos2 t − sin2 t = 1 − 2 sin2 t = 1 − 2y 2 ;
14. ln x = t;
y = ln(ln x),
x≥0
y = 1 − 2y 2 ,
x > 1. Alternatively, ey = t;
−1 ≤ y ≤ 1
y
x = ee , x > 1
15. t = x1/3 ; y = 3 ln x1/3 ; y = ln x, x > 0
16. x2 tan2 x, y 2 = sec2 t; x2 + 1 = tan2 t + 1 = sec2 t = y 2 ; y 2 − x2 = 1. y ≥ 1
17.
y
y
x
y=x
x
x = sin t
y = sin t
687
10.2. PARAMETRIC EQUATIONS
18.
y
y
x
x
√
x=− t
y=t
t≥0
y = x2
19.
y
y
x
x
y=
x2
−1
4
x = 2t
y = t2 − 1
−1 ≤ t ≤ 2
20.
y
y
x
x
y = −x2
x = at
y = −e2t
t≥0
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688
CHAPTER 10. CONICS AND POLAR COORDINATES
21.
y
y
x
x
x2 − y 2 = 1
x = cosh t
y = sinh t
22.
y
y
x
x
y = 2x − 2
x = x = t2 − 1
y = 2t2 − 4
23.
y
y
a
a
a
x
a
x = a cos t
y = a sin t
a>0
0≤t≤π
x = a sin t
y = a cos t
a>0
0≤t≤π
x
689
10.2. PARAMETRIC EQUATIONS
24.
y
y
a
x
b
-b
-a
x
x = a cos t
y = b sin t
a>b>0
π ≤ t ≤ 2π
x = a sin t
y = b cos t
a>b>0
π ≤ t ≤ 2π
25.
y
y
a
a
a
a
x
x
x = a cos t
y = a sin t
a>0
−π/2 ≤ t ≤ π/2
x = a cos 2t
y = a sin 2t
a>0
−π/2 ≤ t ≤ π/2
26.
y
y
a
a
a
x = a cos t/2
y = a sin t/2
a>0
0 ≤ t ≤ π/2
x
a
x = a cos(−t/2)
y = a sin(−t/2)
a>0
−π ≤ t ≤ 0
x
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CHAPTER 10. CONICS AND POLAR COORDINATES
28.
27.
y
y
x
x
29. This is the same as x = 1/y or xy = 1. The graphs are the same.
30. Since x = t1/2 ≥ 0 for all t, x can never be -1. But (−1, −1) is on the xy = 1, so the graphs
are not the same.
31. Since | cos t| ≥ 1, x can never be 2. But (2, 1/2) is on xy = 1, so the graphs are not the same.
32. Since t2 + 1 ≥ 1 for all t, x can never be -1. But (−1, −1) is on xy = 1, so the graphs are not
the same.
33. Since e−2t > 0 for all t, x can never be -1. But (−1, −1) is on xy = 1, so the graphs are not
the same.
34. This is the same as x = 1/y or xy = 1. The graphs are the same.
35. From sin φ = Ly we have t = L sin φ. Since (x, y) is on the circle x2 + y 2 = r2 ,
p
p
± r2 − y 2 = ± r2 − l2 sin2 φ.
x =
36. From the figure in the text, we see that x = r cos 3θ + R cos θ and y = r sin 3θ + r sin θ. (The
actual curve generated by these equations will have the general appearance of the curve in
Figure 10.2.12 in the text only when R > 3r.)
37. From the figure we see that β = θ − π/2 and α = β =
θ − π/2. The length of the line segment from R to P is
equal to the arc of the circle subtended by θ; that is aθ. Now,
x = aθ sin α = aθ sin(θ − π/2) = −aθ cos θ, t = a cos β =
a cos(θ − π/2) = a sin θ, b = a sin β = a sin(θ − π/2) =
−a cos θ, and c = aθ cos α = aθ cos(θ − π/2) = aθ sin θ.
Thus
x = c − b = aθ sin θ − (−a cos θ) = a(cos θ + θ sin θ)
y = s + t = −aθ cos θ = a(sin θ − θ cos θ).
P
aθ
s
α
y
x
R
b
a β
t
θ
x
c
691
10.2. PARAMETRIC EQUATIONS
38. The hypotenuse of the triangle OAB is b − a and
OB = (b − a) cos θ. The actue angle at A in the
right triangle with hypotenuse AP is φ − θ. Thus,
BC = a cos(φ − θ) and x = OB + BC = (b −
a) cos θ + a cos(φ − θ). Similarly, y = AB − AD =
(b − a) sin θ − a sin(φ − θ). Now, the arc on the
smaller circle subtended by the angle φ has length
aφ and the arc on the larger circle subended by the
angle θ has length bθ. From the definition of the
hypocycloid, aφ = bθ. Then φ = bθ
a and φ − θ =
bθ
−
θ.
Thus,
the
parametric
equations
of the
a
hypocycloid are x = (b−a) cos θ+a cos[(b−a)/a]θ,
y = (b − a) sin θ − a sin[(b − a)/a]θ.
a
0
θ
A
θ
a
φ
D
B C
P
b
39. (a) When b = 4a, the equations become x = 3a cos θ + a cos 3θ, y = 3a sin θ − a sin 3θ.
Using the identities cos 3θ = 4 cos3 θ − 3 cos θ and sin 3θ = 3 sin θ − 4 sin3 θ, the parametric equations of the hypocycloid of four cusps become x = 4a cos3 θ = b cos3 θ,
y = 4a sin3 θ = b sin3 θ.
(b)
y
x
(c) Writing x2/3 = b2/3 cos2 θ, y 2/3 = b2/3 sin2 θ we obtain x2/3 + y 2/3 = b2/3 .
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692
CHAPTER 10. CONICS AND POLAR COORDINATES
40. The hypotenuse of triangle OAB is a + b and
OB = (a + b) cos θ. The acute
angle at A in tri
angle ADP is φ − 12 − θ = (φ + θ) − π2 . Thus,
BC = DP = a sin(φ + θ − π/2) = −a cos(φ + θ)
and x = OB + BC = (a + b) cos θ − a cos(φ + θ).
Similarly,
π
y = AB − AD = (a + b) sin θ − a cos φ + θ −
2
= (a + b) sin θ − a sin(φ + θ).
Now, the arc on the smaller circle subtended by
the angle φ has length aφ and the arc on the large
circle subtended by the angle θ has length bθ.
From the definition of the epicycloid, aφ = bθ.
a+b
bθ
Then φ = bθ
θ.
a and φ + θ = a + θ =
a
Thus, the parametric equations
epicycloid
of the
a+b
θ, y = (a +
are x = (a + b) cos θ − a cos
a
a+b
b) sin θ − a sin
θ.
a
a
b
φ
A
a
D
P
θ
O
b B
C
41. (a) When b = 3a, the equations become x = 4a cos θ − a cos 4θ, y = 4a sin θ − a sin 4θ.
(b)
y
x
42. (a) The Q be the point (0, 2a) at the top of the circle and let (x, y) be the coordinates of
point P. Then the measure of angle ∠OBQ is equal to θ. The gives
tan θ =
2a
x
or x =
2a
.
tan θ
Also, the measure of angle ∠AQO is equal to θ. Letting r represent the length of the
segment AO, we have
r
sin θ =
or r = 2a sin θ.
2a
Since y = r sin θ, we have y = (2a sin θ) sin θ = 2a sin2 θ.
693
10.2. PARAMETRIC EQUATIONS
(b) Rewrite x as x =
2a cos θ
. Then
sin θ
x2 y =
4a2 cos2 θ
(2a sin2 θ) = 8a3 cos2 θ
sin2 θ
and 4a2 y = 8a3 sin2 θ. This gives
x2 y + 4a2 y = 8a3 cos2 θ + 8a3 sin2 θ
y(x2 + 4a2 ) = 8a3
y=
43.
8a3
x2 + 4a2
44.
45.
y
y
x
x
46.
y
47.
y
x
48.
y
y
x
x
x
x − x1
. Plugging this into the equation
x2 − x1
x − x1
y = y1 + (y2 − y1 )
x2 − x1
y2 − y1
= y1 +
(x − x1 )
x2 − x1
49. Using the equation from x to solve for t, we have t =
for y yields
which is the equation of a line joining (x1 , y1 ) and (x2 , y2 ). When 0 ≤ t ≤ 1, we get the line
segment with endpoints (x1 , y1 ) and (x2 , y2 ).
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CHAPTER 10. CONICS AND POLAR COORDINATES
50. (a) x = −2 + (4 − (−2))t = −2 + 6t
y = 5 + (8 − 5) = 5 + 3t
(b) y = 21 x + 6
(c) x = −2 + 6t, y = 5 + 3t; 0 ≤ t ≤ 1
51. If the launch point is designated as the origin, then the equations describing the skier’s motion
from launch until landing are given by
x = 75t and y = −16t2
where t = 0 at the moment of launch. At the moment of impact, we have
tan 33◦ =
Thus, t =
10.3
1.
2.
3.
4.
5.
6.
7.
16
|y|
=
t.
|x|
15
75
tan 33◦ ≈ 3.044, and therefore x ≈ 228.3 ft, y ≈ −148.25 ft.
16
Calculus and Parametric Equations
dx
= 3t2 − 2t;
dt
dx
4
= − 2;
dt
t
dy
= 6t2 − 1;
dt
dx
t
=√
;
2
dt
t +1
dy
2t + 5
= 2
;
dx
3t − 2t
dy
= 2t + 5;
dt
dy
= 4t3 ;
dt
dy
dx
=
t=−1
dy
6t2 − 1
6t4 − t2
=
;
=−
2
dx
−4/t
4
dy
4t
= √
dx
t/ t2 + 1
3
3
5
dy
dx
p
= 4t2 t2 + 1;
92
= −23
4
t=2
=−
dy
dx
√
t= 3
√
= 4(3) 4 = 24
dy
dy
4e−4t
dx
= 2e2t ;
= −4e−4t ;
= − 2t = −2e−6t
dt
dt
dx
2e
−6
dy
1
= −2e−6 ln 2 = −2eln 2 = −2(2−6 ) = −
dx t=ln 2
32
dx
dy
= 2 cos θ(− sin θ);
cos θ;
dθ
dθ
dy
1
= −1
=−
dx θ=π/6
2(1/2)
dy
cos θ
1
=
=−
dx
−2 sin θ cos θ
2 sin θ
dx
dy
dy
2 sin θ
sin θ
= 2 − 2 cos θ;
= 2 sin θ;
=
=
dθ
dθ
dx
2 − 2 cos θ
1 − cos θ
√
√
√
dy
2/2
2
√
√ = 2+1
=
=
dx θ=π/4
1 − 2/2
2− 2
dy
12t
4t
= 2
= 2
.
dx
3t + 3
t +1
At t = −1 we observe x = −4, y = 7, and m = dy/dx = −2. The tangent line is y = −2x − 1.
695
10.3. CALCULUS AND PARAMETRIC EQUATIONS
8.
9.
10.
11.
12.
13.
dy
2t + 1/t
1
=
=t+ .
dx
2
2t
3
At t = 1 we observe x = 6, y = 1, and m = dy/dx = 3/2. The tangent line is y = x − 8.
2
2t
4
4
dy
=
. At (2, 4), t = −2 and m = dy/dx = 4/3. The tangent line is y = x + .
dt
(2t + 1)
3
3
√
√
dy
1
(4t3 − 2t)
= 1 − 2 . At (0, 6), t = 3 or − 3 and m = 5/6. The tangent line is
=
3
dx
4t
2t
5
y = x + 6.
6
−2 sin t
dy
=
. When y = 1, cos t = 1/2 and t = π/3 or 5π/3. For t = π/3, x = 4 sin(2π/3) =
dx
8 cos 2t
√
√
√
dy
−2 sin(π/3)
3
4( 3/2) = 2 3, and m =
=
=
.
dx
8 cos(2π/3)
4
dy
3t2
3t
3t
=
= . The slope of the tangent line is -3. Solving
= −3, we obtain t = −2. At
dx
2t
2
2
t = −2 we observe x = 4 and y = −7. The point on the graph is (4, 7).
dx
dy
dy
2t − 4
= 2;
= 2t − 4;
=
=t−2
dt
dt
dx
2
We want t − 2 = 3. Then t = 5 and the point of tangency is (5, 8). The equation of the tangent
line is y − 8 = 3(x − 5) or y = 3x − 7.
14. For θ = π/2 we observe x = −2/π and y = 1 − 1 = 0. For θ = −π/2 we observe x = −2/π
and y = −1 + 1 = 0. Thus, the curve intersects itself when θ = π/2 and θ = −π/2.
dy
2 dy
cos θ − 2/π
2
dx
= − sin θ;
= cos θ − ;
=
= csc θ − cot θ
dθ
dθ
π dx
− sin θ
π
dy
2
When θ = π/2, the slope of the tangent line is
=
and its equation is y − 0 =
dx θ=π/2
π
dy
2
2
4
2
2
=−
x+
or y = x+ 2 . When θ = −π/2, the slope of the tangent line is
π
π
π
π
dx θ=−π/2
π
2
2
2
4
and its equation is y − 0 = −
x+
or y = − x − 2 .
π
π
π
π
15.
dx
dy
dy
2t
= 3t2 − 1;
= 2t;
= 2
. The tangent line is
dt
dt
dt
3t − 1
2
horizontal when
√ 2t = 0 or t = 0, and vertical when 3t −1 =
0 or t = ±1/ 3. Thus,√there is a horizontal
√ tangent at (0, 0)
and vertical at (−2/3 3, 1/3) and (2/3 3, 1/3).
y
2
1
x
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696
16.
17.
18.
CHAPTER 10. CONICS AND POLAR COORDINATES
dx
3
dy
dy
2t − 2
16t − 16
. The
= t2 ;
= 2y − 2;
=
=
2
dt
8
dt
dx
3t /8
3t2
tangent line is horizontal when 16t − 16 = 0 or t = 1, and
vertical when 3t2 = 0 or t = 0. Thus, there is a horizontal
tangent at (9/8, −1) and a vertical tangent at (1, 0).
dx
dy
dy
−3 sin 3t
= cos t;
= −3 sin 3t;
=
. The
dt
dt
dx
cos t
tangent line is horizontal when sin 3t = 0 or t =
0, π/3, 2π/3, π, 4π/3, 5π/3, 2π and vertical when
cos t = 0 or t = π/2,
there are horizontal
√ 3π/2. Thus,
√
√ tangents
√ at (0, 1), ( 3/2, −1), ( 3/2, 1), (0, −1), (− 3/2, 1),
( 3/2, 1) and vertical tangents at (1, 0) and (−1, 0).
18t2
dy
=
= 3t;
dx
6t
20.
dy
cos t
=
= − cot t;
dx
− sin t
22.
1
dy
dy
dx
= 1;
= 3t2 − 6t;
= 3t2 − 6t. The tangent line is
dt
dt
dx
horizontal when 3t2 − 6t = 3t(t − 2) = 0 or t = 0, 2. Thus,
there are horizontal tangents at (−1, 0) and (1, −4). There
are no vertical tangents.
19.
21.
y
d2 y
3
1
=
= ;
2
dx
6t
2t
1
x
y
1
x
1
y
1
1
x
d3 y
−1/2t2
1
=
=− 3
3
dx
6t
12t
d2 y
csc2 t
=
= − csc3 t;
dx2
− sin t
d3 y
3 csc3 t cot t
=
= −3 csc4 t cot t
dx3
− sin t
dy
2e2t + 3e3t
d2 y
−6e3t − 12e4t
3t
4t
=
=
−2e
−
3e
;
=
= 6e4t + 12e5t
dx
−e−t
dx2
−e−t
d3 y
24e4t + 60e5t
=
= −24e5t − 60e6t
3
dx
−e−t
t−1
dy
=
;
dx
t+1
d2 y
2
2/(t + 1)2
=
=
;
2
dx
t+1
(t + 1)3
6
d3 y
−6/(t + 1)4
=−
=
3
dx
t+1
(t + 1)5
d2 y
dy 0 /dt
(32 − 16t)/3t3
256 − 128t
128 2 − t
=
=
=
=
. Then
dx2
dx/dt
3t2 /8
9t5
9
t5
d2 y/dx2 is 0 when t = 2 and undefined when t = 0. The graph is concave downward on
(−∞, 0) and (2, ∞) and concave upward on (0, 2).
23. Using Problem 16,
t
y 00
-
0
und
+
2
0
-
697
10.3. CALCULUS AND PARAMETRIC EQUATIONS
24.
dy 0 /dt
dy
dy
d2 y
6t + 6
dx
=
= 2;
= 6t2 + 12t + 4;
= 3t2 + 6t + 2;
=
= 3t + 3
2
dt
dt
dx
dx
dx/dt
2
Solving 3t + 3 = 0 we obtain t = −1. Since d2 y/dx2 < 0 for t < −1 and d2 y/dx2 > 0 for
t > −1, the graph has a point of inflection when t = −1 or at (3, 0).
25. x (t) = 5t ,
0
y (t) = 12t ;
0
2
s=
2
Z
0
26. x0 (t) =√t2 , y 0 (t) = t
Z 3p
Z
s=
t4 + t2 dt =
Z
0
1
t
4
=
1
27. x0 (t)Z= et cos t + et sin t;
π
s=
0
=
Z
3
0
1 1/2
1
u du = u3/2
2
3
4
√
π
p
2
p
25t4
+
144t4 dt
= 13
Z
2
t2 dt =
0
t2 + 1dt
13 3
t
3
2
=
0
104
3
u = t2 + 1, du = 2tdt
1
7
(8 − 1) =
3
3
y 0 (t) = −et sin t + et cos t
[e2t (cos2 t + 2 sin t cos t + sin2 t) + e2t (sin2 t − 2 sin t cos t + cos2 t)]1/2 dt
et (2)1/2 dt =
0
28. x0 (θ) = a(1 − cos θ),
Z
q
√
2et
π
0
=
√
2(eπ − 1)
y 0 (θ) = a sin θ. Using symmetry,
Z
p
1 − 2 cos θ + cos2 θ + sin2 θdθ
0
0
√
√ Z π√
√ Z π√
1 + cos θ
√
= 2 2a
1 − cos θdθ = 2 2a
1 − cos θ
dθ
1 + cos θ
0
0
√
Z
Z
π
π
√
√
1 − cos2 θ
sin θ
√
√
dθ u = 1 + cos θ, du = − sin θdθ
= 2 2a
dθ = 2 2a
2
1 + cos θ
1 + cos θ
0
0
Z 2
√ Z 0 −1/2
√ Z 0 −du
√
√ = 2 2a
= 2 2a
u
du = 2 2a lim+
u1/2 du
u
b→0
2
2
b
√ √
√
√
√
2
= 2 2a lim 2 u b = 2 2a lim (2 2 − 2 b) = 8a.
s=2
π
a2 (1 − cos θ)2 + a2 sin2 θdθ = 2a
b→0+
π
b→0+
29. x0 (θ) = −3b cos2 θ sin θ; y 0 (θ) = 3b sin2 θ cos θ
Z π/2
Z
s=
(9b2 cos4 θ sin2 θ + 9b2 sin4 θ cos2 θ)1/2 dθ = 3|b|
0
= 3|b|
Z
0
π/2
sin θ cos θ
0
π/2
1
3
sin 2θdθ = − |b| cos 2θ
2
4
30. x0 (θ) = −4a sin θ + 4a sin 4θ;
π/2
0
3
3
= − |b|(−1 − 1) = |b|
4
2
y 0 (θ) = 4a cos θ − 4a cos 4θ
p
cos2 θ + sin2 θdθ
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698
CHAPTER 10. CONICS AND POLAR COORDINATES
s=
Z
2π/3
0
= 4|a|
Z
[(−4a sin θ + 4a sin 4θ)2 + (4a cos θ − 4a cos 4θ)2 ]1/2 dθ
2π/3
0
= 4|a|
Z
2π/3
0
Z
√
= 4 2|a|
(sin2 θ − 2 sin θ sin 4θ + sin2 4θ + cos2 θ − 2 cos θ cos 4θ + cos2 4θ)1/2 dθ
(2 − 2 sin θ sin 4θ − 2 cos θ cos 4θ)1/2 dθ
2π/3
0
Z
√
= 4 2|a|
[1 − (cos 4θ cos θ + sin 4θ sin θ)]1/2 dθ
2π/3
0
Z
√
= 4 2|a|
2π/3
0
p
q
Z
√
1 − cos(4θ − θ)dθ = 4 2|a|
2 sin2 3θ/2dθ = 8|a|
3θ
2
= 8|a| − cos
3
2
Z
0
√
0
2π/3
sin
0
2π/3
2π/3
1 − cos 3θdθ
3θ
dθ
2
16
32
|a|(cos π − cos 0) =
|a|
3
3
=−
31. (a) Setting x = 0 we have t2 −√
4t − 2 = 0 which implies t = 2 ±
y ≈ −0.6551. When t = 2 + 6, y ≈ 1390.66.
√
6. When t = 2 −
√
6,
(b) Using Newton’s Method to solve t5 −4t3 −1 = 0 we obtain t ≈ −1.96687, −0.654175, 2.02968
with corresponding x values 9.73606, 1.04465, −5.99912.
32. If y = F (x) and x = f (t), then g(y) = y = F (f (t)). Thus,
A=
=
Z
x2
x1
b
Z
F (x)dx x = f (t), dx = f 0 (t)dt
F (f (t))f (t)dt =
0
a
Z
b
g(t)f 0 (t)dt.
a
33. From Example 7 in Section 10.2, f (θ) = a(θ − sin θ) and g(θ) = a(1 − cos θ) for 0 ≤ θ ≤ 2π.
Then f 0 (θ) = a(1 − cos θ), and using symmetry,
A=2
Z
π
a2 (1 − cos θ)2 dθ = 2a2
Z
π
(1 − 2 cos θ + cos2 θ)dθ
0
0
Z π
1 1
3
1
2
= 2a
(1 − 2θ + + cos 2θ)dθ = 2a2
θ − 2 sin θ + sin 2θ
2 2
2
4
0
3
= 2a2
π = 3(πa2 ).
2
Thus, the area under an arch of the cycloid is three times the area of the circle.
π
0
699
10.4. POLAR COORDINATE SYSTEM
10.4
Polar Coordinate System
1.
2.
π
(3,π)
Ο
polar
axis
Ο
polar
axis
−π/2
(2,−π/2)
3.
4.
Ο
π/6
polar
axis
(−1,π/6)
π/2
Ο
polar
axis
(−1/2,π/2)
5.
6.
(−4,−π/6)
7π/4
Ο
polar
axis
Ο
−π/6
7.
(a)
2, − 5π
4
8.
(a)
5, − 3π
2
9.
10.
(a)
(a)
4, −
3, −
5π
3
7π
4
polar
axis
(b)
(b)
(b)
(b)
2, 11π
4
5, 5π
2
4, 7π
3
3, 9π
4
(2/3,7π/4)
(c)
(c)
(c)
(c)
−2, 7π
4
−5, 3π
2
−4, 4π
3
−3, 5π
4
(d)
(d)
(d)
(d)
−2, − π4
−5, − π2
−4, − 2π
3
−3, − 3π
4
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CHAPTER 10. CONICS AND POLAR COORDINATES
11.
(a)
1, − 11π
6
12.
(a)
3, − 5π
6
(b)
(b)
1, 13π
6
3, 19π
6
(c)
(c)
−1, 7π
6
−3, π6
(d)
(d)
−1, − 5π
6
−3, − 11π
6
13. With r = 1/2 and θ = 2π/3 we have x = 1/2 !
cos 2π/3 = 1/2(−1/2) = −1/4, y = 1/2 sin 2π/3 =
√
√
√
1
3
1/2( 3/2) = 3/4. The point is − ,
in rectangular coordinates.
4 4
√
√
14. With r = −1 and θ = 7π/4 we have x = −1 cos 7π/4 !
= −1( 2/2) = − 2/2, y =
√ √
√
√
2
2
−1 sin 7π/4 = −1(− 2/2) = 2/2. The point is −
,
in rectangular coordinates.
2
2
15. With √
r = −6 and√θ = −π/3 we have x =√−6cos(−π/3) = −6(1/2) = −3, y = −6 sin(−π/3) =
−6(− 3/2) = 3 3. The point is −3, 3 3 in rectangular coordinates.
√
√
√ √
√
16. With r = 2 and θ = 11π/6 we have x = 2 cos 11π/6 != 2( 3/2) = 6/2, y =
√
√
√
√
√
6
2
2 sin 11π/6 = 2(−1/2) = − 2/2. The point is
,−
in rectangular coordinates.
2
2
√
√
17. With
√ r = 4 and√θ = 5π/4 we have x√= 4 cos
√5π/4
= 4(− 2/2) = −2 2, y = 4 sin 5π/4 =
4(− 2/2) = −2 2. The point is −2 2, −2 2 in rectangular coordinates.
18. With r = −5 and θ = π/2 we have x = −5 cos π/2 = 0, y = −5 sin π/2 = −5. The point is
(0, −5) in rectangular coordinates.
19. With x√= −2 and
have r2 = 8 and tan θ = 1.
y = −2 we √
3π
(a) 2 2, − 4
(b) −2 2, π4
20. With x = 0 and y = −4 we have
r2 = 16 and tan θ undefined.
(a) 4, − π2
(b) −4, π2
√
√
21. With x = 1 and y = − 3 wehave r2 = 4 and tan θ = − 3.
(a) 2, − π3
(b) −2, 3π
3
√
√
√
22. With x√= 6 and y = 2 we
have r2 = 8 and tan θ = 1/ 3.
√
(a) 2 2, π6
(b) −2 2, − 5π
6
23. With x = 7 and y = 0 we have r2 = 49 and tan θ = 0.
(a) (7, 0)
(b) (−7, π) or (−7, −π)
24. With√x = 1 and y = 2 √
we have r2 = 5 and tan√
θ = 2.
√
−1
(a)
5, tan 2 or
5, 1.1071
(b) − 5, −π + tan−1 2 or − 5, −2.0344
701
10.4. POLAR COORDINATE SYSTEM
26.
25.
y
y
x
x
2
2
polar
axis
4
27.
4
polar
axis
28.
y
y
x
x
polar
axis
polar
axis
29.
30.
y
y
x
1
polar
axis
In Problems 31-40, we use x = r cos θ and y = r sin θ.
31. r sin θ = 5;
r = 5 csc θ
32. r cos θ + 1 = 0;
r = − sec θ
2
4
x
polar
axis
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CHAPTER 10. CONICS AND POLAR COORDINATES
33. r sin θ = 7r cos θ;
tan θ = 7;
34. 3r cos θ + 8r sin θ + 6 = 0;
θ = tan−1 7
r(3 cos θ + 8 sin θ) = −6;
r=−
6
(3 cos θ + 8 sin θ)
35. r2 sin2 θ = −4r cos θ; r2 (1 − cos2 θ) + 4r cos θ = 0; r2 − (r2 cos2 θ − 4r cos θ + 4) = 0; r2 −
(r cos θ − 2)2 = 0; [r − (r cos θ − 2)][r + (r cos θ) − 2] = 0
−2
2
Solving for r, we obtain r =
or r =
. Since replacement of (r, θ) by
(1 − cos θ)
(1 + cos θ)
(−r, θ + π) in the first equation gives the second equation, we take the polar equation to be
2
.
r=
1 + cos θ
36. r2 cos2 θ−12r sin θ−36 = 0; r2 (1−sin2 θ)−12r sin θ−36 = 0; r2 −(r2 sin2 θ+12r sin θ+36) = 0;
r2 − (r sin θ + 6)2 = 0; [r − (r sin θ + 6)][r + (r sin θ + 6)] = 0
−6
6
or r =
. Since replacement of (r, θ) by
Solving for r, we obtain r =
(1 − sin θ)
(1 + sin θ)
(−r, θ + π) in the second equation gives the first equation, we take the polar equation to be
6
r=
.
(1 − sin θ)
37. r2 = 36. Since r = −6 has the same graph as r = 6, we take the equation to be r = 6.
38.
(r cos θ)2 − (r sin θ)2 = 1
r2 cos2 θ − r2 sin2 θ = 1
r2 cos2 θ − sin2 θ = 1
r2 1 − 2 sin2 θ = 1
√
39. r2 + r cos θ = r2 = ±r; r(r + cos θ ∓ 1) = 0. Solving for r, we obtain r = 0 or r = ±1 − cos θ.
Since replacement of (r, θ) in r = −1 − cos θ by (−r, θ + π) gives r = 1 − cos θ, and since θ = 0
gives r = 0, we take the polar equation to be r = 1 − cos θ.
40. r3 cos3 θ + r3 sin3 θ − r2 sin θ cos θ = 0; r2 [r(cos3 θ + sin3 θ) − 21 sin 2θ] = 0
sin 2θ
Solving for r gives r = 0 or r =
. Since π = 0 gives r = 0 in the second
2(cos3 θ + sin3 θ)
sin 2θ
equation, we take the polar equation to be r =
.
2(cos3 θ + sin3 θ)
In Problems 41-52, we use r2 = x2 + y 2 , r cos θ = x, r sin θ = y, and tan θ = y/x.
41. r cos θ = 2;
x=2
42. x = −4
43. r = 12 sin θ cos θ;
r3 = 12r sin θr cos θ;
44. 2(x2 + y 2 )1/2 = y/x;
45. r2 = 8 sin θ cos θ;
(x2 + y 2 )3/2 = 12xy;
4x2 (x2 + y 2 ) = y 2
r4 = 8r sin θr cos θ;
(x2 + y 2 )2 = 8xy
(x2 + y 2 )3 = 144x2 y 2
703
10.4. POLAR COORDINATE SYSTEM
46. r2 (cos2 θ − r2 sin2 θ) = 16;
47. r2 + 5r sin θ = 0;
r2 cos2 θ − r2 sin2 θ = 16;
x2 − y 2 = 16
x2 + y 2 + 5y = 0
48. r2 = 2r+r cos θ; x2 +y 2 = 2(x2 +y 2 )1/2 +x; x2 +y 2 −x = 2(x2 +y 2 )1/2 ; (x2 +y 2 −x)2 = 4(x2 +y 2 )
49. r + 3r cos θ = 2; (x2 + y 2 )1/2 + 3x = 2; (x2 + y 2 )1/2 = 2 − 3x; x2 + y 2 = 4 − 12x + 9x2 ;
8x2 − y 2 − 12x + 4 = 0
p
p
50. 4r − r sin θ = 10; 4 x2 + y 2 − y = 10; 4 x2 + y 2 = y + 10; 16(x2 + y 2 ) = y 2 + 20y + 100;
16x2 + 15y 2 − 20y − 100 = 0
51. 3r cos θ + 8r sin θ = 5; 3x + 8y = 5
52. r cos θ = 3 cos θ + 3; r2 cos θ = 3r cos θ + 3r; r(r cos θ − 3) = 3r cos θ; (x2 + y 2 )1/2 (x − 3) = 3x;
(x2 + y 2 )(x − 3)2 = 9x2
p
p
53.
(x2 − x1 )2 + (y2 − y1 )2 = (r2 cos θ2 − r1 cos θ1 )2 + (r2 sin θ2 − r1 sin θ1 )2
q
= r22 cos2 θ2 − 2r2 r1 cos θ2 cos θ1 + r12 cos2 θ1 + r22 sin2 θ2 − 2r2 r2 sin θ2 sin θ1 + r12 sin2 θ1
q
= r22 + r12 − 2r1 r2 (cos θ2 cos θ1 + sin θ2 sin θ1 )
q
= r22 + r12 − 2r1 r2 cos(θ2 − θ1 )
54. Consider the general linear function y = ax + b. Transforming this function into polar coordinates, we have r sin θ = ar cos θ + b. To find a line passing through (r1 , θ1 ) and (r2 , θ2 ), we
would need to solve the following system for a and b:
r1 sin θ1 = ar1 cos θ1 + br2 sin θ2 = ar2 cos θ2 + b
To find the line passing through 3, 3π
and 1, π4 , we solve the system
4
√
√
2
3 2
= −3a
+b
2
2
√
√
2
2
=a
+b
2
2
for a and b. This yields a = − 21 , b =
√
3 2
4 .
Thus, the equation of the line is
√
1
3 2
r sin θ = − r cos θ +
2
4
or
√
3 42
r=
sin θ + 21 cos θ
√
3 2
The y-intercept occurs when r cos θ = 0 and hence r sin θ =
. Together, these yield θ = π2
4
!
√
√
3 2
3 2 π
and r =
. The y-intercept is thus
,
in polar coordinates. The x-intercept
4
4 2
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CHAPTER 10. CONICS AND POLAR COORDINATES
√
√
3 2
3 2
occurs when r sin θ = 0 and hence r cos θ =
. Together, these yield θ = 0 and r =
.
2
2
!
√
3 2
,0 .
The x-intercept is thus
2
55. Solutions of f (θ) = 0 are θ values at which the graph of r = f (θ) passes through the origin.
10.5
Graphs of Polar Equations
1.
2.
polar
axis
3.
1
circle
polar
axis
circle
4.
line through origin
5.
6.
polar
axis
polar
axis
7.
polar
axis
spiral
line through origin
spiral
8.
9.
polar
axis
cardioid
polar
axis
polar
axis
polar
axis
cardioid
cardioid
705
10.5. GRAPHS OF POLAR EQUATIONS
11.
10.
12.
polar
axis
cardioid
limacon with an interior loop
limacon with an interior loop
13.
14.
15.
polar
axis
polar
axis
dimpled limacon
16.
polar
axis
dimpled limacon
17.
convex limacon
18.
polar
axis
polar
axis
polar
axis
rose curve
convex limacon
19.
rose curve
20.
21.
polar
axis
polar
axis
rose curve
polar
axis
polar
axis
rose curve
polar
axis
rose curve
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CHAPTER 10. CONICS AND POLAR COORDINATES
23.
22.
24.
polar
axis
polar
axis
polar
axis
rose curve
circle
circle with center on x-axis
25.
26.
27.
circle with center on y-axis
28.
polar
axis
polar
axis
polar
axis
circle with center on y-axis
lemniscate
30.
29.
polar
axis
polar
axis
polar
axis
lemniscate
lemniscate
31.
lemniscate
32.
π
10
polar
axis
1
polar
axis
707
10.5. GRAPHS OF POLAR EQUATIONS
33. r = 2.5
34. r = −4 cos θ
35. r = 4 − 3 cos θ
36. r = 2 + 3 sin θ
37. r = 2 cos 4θ
38. r = 5 cos 2θ
39. Solving 4 sin θ = 2 we have sin θ = 1/2 and θ = π/6 and
5π/6. The points of intersection are (2, π/6) and (2, 5π/6).
polar
axis
1
40. Writing sin 2θ = 2 sin θ cos θ and setting sin θ = 2 sin θ cos θ,
we obtain sin θ(2 cos θ − 1) = 0. This gives θ = 0, π, π/3,
and 5π/3. For θ = 0 and θ = π we obtain √the pole (0, 0).
The
√ other two points of intersection are ( 3/2, π/3) and
(− 3/2, 5π/3).
41. Setting 1 − cos θ = 1 + cos θ, we obtain 2 cos θ = 0. This
gives θ = ±π/2. Two points of intersection are (1, π/2) and
(1, −π/2). From the figure we see that the pole ((0, 0) on
r = 1 − cos θ and (0, π) on r = 1 + cos θ) is also a point of
intersection.
42. Setting 3−3 cos θ−3 cos θ, we obtain 3 = 6 cos θ or 12 = cos θ.
This yields θ = ± π3 . Two points of intersection are 23 , π3
and 23 , − π3 . From the figure we see that the pole ((0, 0)
on r = 3 − 3 cos θ and (0, π/2) on r = 3 cos θ) is also a point
of intersection.
1
1
polar
axis
polar
axis
polar
axis
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CHAPTER 10. CONICS AND POLAR COORDINATES
43. Setting 6 sin 2θ = 3, we obtain sin 2θ = 1/2. Then 2θ = π/6,
5π/6, 13π/6, and 17π/6. This gives the points of intersection
(3, π/12), (3, 5π/12), (3, 13π/12), and (3, 17π/12). Writing
the second equation in the form −r = 6 sin 2(θ + π), we
obtain r = −6 sin 2θ. Setting −6 sin 2θ = 3, we obtain
sin 2θ = −1/2. Then 2θ = −π/6, −5π/6, −13π/6, and
−17π/6. This gives the points of intersection (3, −π/12),
(3, −5π/12), (3, −13π/12), and (3, −17π/12).
44. Using cos 2θ = 2 cos2 θ − 1, we have 2 cos2 θ − 1 = 1 + cos θ
or 2 cos2 θ −√cos θ − 2 = 0.√From the quadratic formula,
1 + 17
1 ± 17
. Since
> 1, we solve only cos θ =
cos θ =
4
4
√
1 − 17
. This gives θ ≈ 2.4667 and θ ≈ 3.8165. For both of
4
these values r ≈ 0.219. Writing −r = cos 2(θ + π), we have
r = − cos 2θ = −2 cos2 θ + 1. Solving this equation with r =
1 + cos θ, we have −2 cos2 θ + 1 = 1 + cos θ or cos θ(2 cos θ +
1) = 0. For cos θ = 0 we obtain θ = π/2 and θ = 3π/2. For
both of these values r = 1. From cos θ = −1/2 we obtain θ =
2π/3 and θ = 4π/3. For both of these values r = 1/2. Thus,
(0.219, 2.47), (0.219, 3.82), (1, π/2), (1, 3π/2), (1/2, 2π/3),
and (1/2, 4π/3) are points of intersection. From the graph
we see that the pole ((0, π/4) on r = cos 2θ and (0, π) on
r = 1 + cos θ) is also a point of intersection.
polar
axis
polar
axis
1
45. Setting 4 sin θ cos2 θ = sin θ we obtain sin θ(4 cos2 θ −1) = 0.
This gives θ = 0, θ = √
π/3, and θ = 2π/3.
√ The points of
intersection are (0, 0), ( 3/2, π/3), and ( 3/2, 2π/3).
1
polar
axis
709
10.5. GRAPHS OF POLAR EQUATIONS
46. From the figure we see that the graphs intersect at the pole
(which occurs for θ = π on the cardioid and θ = π/2 on the
lemniscate) and at (2, 0). Setting (1 + cos θ)2 = 4 cos θ we
have
cos2 θ − 2 cos θ + 1 = 0 =⇒ (cos θ − 1) = 0 =⇒ cos θ = 1
polar
axis
1
which yields only the point (2,0). The points of intersection
in the second and third quadrants occur when π/2 < θ <
3π/2 on the cardioid and when −π/2 < θ < π/2 and r < 0
on the lemniscate. If (r2 , θ2 ) represents a point in the second
or third quadrants on the cardioid, we have r2 = 1 + cos θ2 .
If (r1 , θ1 ) is a point in the second or third quadrants on the lemniscate, we have r12 = 4 cos θ.
At points of intersection then, we have r2 = 1 + cos θ, r12 = 4 cos θ, r2 = −r1 , and θ2 = θ1 + π.
Substituting the last two equations into the first equation, we obtain −r1 = 1 + cos(θ1 + π) =
1 − cos θ1 or r12 = 1 − 2 cos θ1 + cos2 θ1 . Combining with r12 = 4 cos θ1 we have
√
√
6 ± 36 − 4
2
2
= 3 ± 2 2.
1 − 2 cos θ1 + cos θ1 = 4 cos θ1 =⇒ cos θ1 − 6 cos θ1 + 1 = 0 =⇒ cos θ1 =
2
√
Since cos θ1 ≤ 1, cos θ1 = 3−2 2 and θ1 ≈ 1.40 or θ1 ≈ −1.40. In either case r1 ≈ −0.83. The point
of intersection in the second quadrant occurs when θ ≈ −1.40 and r ≈ −0.83 on the lemniscate
and when θ ≈ −1.40 + π ≈ 1.74 and r ≈ 0.83 on the cardioid. In the third quadrant the point of
intersection occurs when θ ≈ 1.40 and r ≈ −0.83 on the lemniscate and when θ ≈ 1.40 + π ≈ 4.54
and r ≈ 0.83 on the cardioid.
47.
(a)
(b)
polar
axis
polar
axis
48.
(a)
(b)
polar
axis
49. (d)
(c)
50. (c)
(d)
polar
axis
(c)
polar
axis
(d)
polar
axis
polar
axis
51. (b)
polar
axis
52. (a)
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CHAPTER 10. CONICS AND POLAR COORDINATES
53.
polar
axis
54. For a = 0, the graph is a circle.
For a = 14 , 12 , and 34 , the graph is a limacon with an interior loop.
For a = 1, the graph is a cardiod.
For a = 54 , 32 , and 74 , the graph is a dimpled limacon
For a = 2, 94 , 52 , 11
4 , and 3, the graph is a convex limacon.
As a → ∞, the graphs more closely approximate a circle.
56.
55.
(r,θ)
(r,θ)
θ
θ
(r, θ+π)
(−r, π−θ)
Symmetric with respect to the
origin.
Symmetric with respect to the
x-axis.
57.
58.
(r,θ)
θ
(−r, −θ)
(r,θ)
θ
(−r,θ+2π)
Symmetric with respect to the
origin.
Symmetric with respect to the
y-axis .
10.6. CALCULUS IN POLAR COORDINATES
711
59. Symmetric with respect to the x-axis.
60. The graph is symmetric with respect to the y-axis.
61. (a) The graphs are identical.
(b) The graphs are identical.
62. From the statement in the text preceding Problem 33 in Section 4.1 we have that the component of acceleration in the direction of the ramp is −g sin θ, where g is the acceleration due
to gravity and −π ≤ θ ≤ 0. Thus the distance traveled in time t along the ramp at angle θ
is r = − 21 gt2 sin θ. But this is the equation of a circle of radius gt2 /4 centered at (0, −gt2 /4),
whose topmost point is (0, 0) which is taken at the point of release.
10.6
1.
Calculus in Polar Coordinates
dy
= θ cos θ + sin θ
dθ
dx
= −θ sin θ + cos θ
dθ
θ cos θ + sin θ
2
dy
=
= − at θ =
dx
−θ sin θ + cos θ
π
π
2.
2. At θ = 3,
dy
cos θ sin θ
cos 3 sin 3
=
− 2 =
−
dθ
θ
θ
3
9
− sin θ cos θ
dx
sin 3 cos 3
=
− 2 =−
−
dθ
θ
θ
3
9
3 cos 3 − sin 3
dy
=
dx
−3 sin 3 − cos 3
3. At θ = π3 ,
dy
= (4 − 2 sin θ) cos θ + (−2 cos θ) sin θ
dθ
√ ! √
3
1
= 4 − 4 sin θ cos θ = 4 − 4
=4− 3
2
2
dx
= −(4 − 2 sin θ) sin θ + (−2 cos θ) cos θ
dθ
3
1
2
2
= −4 + 2 sin θ − 2 cos θ = −4 + 2
−2
4
4
3 1
= −4 + − = −3
√2 2
dy
4− 3
=
dx
−3
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CHAPTER 10. CONICS AND POLAR COORDINATES
3π
4. At θ =
,
4
dy
= (2 − cos θ) cos θ + (sin θ) sin θ
dθ
1
1
= 1 − cos2 θ + sin2 θ = 1 −
+
=1
2
2
dx
= −(1 − cos θ) sin θ + (sin θ) cos θ
θ
= − sin θ + 2 cos θ sin θ
√
√ ! √ !
2
2
2
=−
+2
2
2
2
√
2
=1−
2
dy
1
√
=
dx
1− 2
2
5. At θ = π/6,
dy
= sin θ cos θ + cos θ sin θ
dθ
√ ! √
3
3
1
= 2 cos θ sin θ = 2
=
2
2
2
1
3
dx
1
= − sin2 θ + cos2 θ = −
+
=
dθ
4
4
2
√
dy
= 3
dx
6. At θ = π/4,
dy
= (10 cos θ) cos θ + (−10 sin θ) sin θ
dθ
1
1
2
2
= 10 cos θ = 1 − sin θ = 10
− 10
2
2
=0
dx
= −(10 cos θ) sin θ + (−10 sin θ) cos θ
dθ
√ ! √ !
2
2
= −10 cos θ sin θ = −20
2
2
= −10
dy
0
=
=0
dx
−10
713
10.6. CALCULUS IN POLAR COORDINATES
7.
dy
= (2 + 2 cos θ) cos θ + (−2 sin θ) sin θ
dθ
= 2(1 + cos2 θ − sin2 θ)
dx
= −(2 + 2 cos θ) sin θ + (−2 sin θ) cos θ
dθ
= −2 − 4 cos θ sin θ
dy
2(1 + cos2 θ − sin2 θ)
1 + cos2 θ − sin2 θ
=
=
dx
2(−2 − 2 cos θ sin θ)
−1 − 2 cos θ sin θ
If the tangent line is horizontal, we must have
1 + cos2 θ − sin2 θ = 0
which requires sin θ = ±1 and thus θ = π2 or θ = 3π
2 . Hence, the polar coordinates of points
on the graph with horizontal tangents are (2, π/2) and (2, 3π/2). If the tangent line is vertical,
we must have
1
−1 − 2 cos θ sin θ = 0 or cos θ sin θ = −
2
3π
7π
which occurs at θ = 4 or θ = 4 . Hence the polar coordinates of points on the graph with
√
√
vertical tangents are (2 − 3, 3π/4) and (2 + 3, 7π/4).
8.
dy
= (1 − sin θ) cos θ + (− cos θ) sin θ
dθ
= 1 − 2 sin θ cos θ
dx
= −(1 − sin θ) sin θ + (− cos θ) cos θ
dθ
= − sin θ + sin2 θ − cos2 θ
dy
1 − 2 sin θ cos θ
=
dx
− sin θ + sin2 θ − cos2 θ
If the tangent line is horizontal, we must have
1 − 2 sin θ cos θ = 0
which occurs at θ =
or
sin θ cos θ =
1
2
and
= 5π
4 . Hence,
θ√
the√polar coordinates of points on the graph with
2 π
horizontal tangents are 1 − 2 , 4 and 1 + 22 , 5π
4 . If the tangent line is vertical, we must
have
− sin θ + sin2 θ − cos2 θ = 0
π
4
− sin θ + sin2 θ − (1 − sin2 θ) = 0
2 sin2 θ − sin θ − 1 = 0
9.
which gives sin θ = − 12 or sin θ = 1. This occurs at θ = π2 , θ = 7π
6 , and θ =
the polar
coordinates
of
points
on
the
graph
with
vertical
tangents
are 0, π2 ,
3 11π
2, 6 .
dy
= (4 cos 3θ)(cos θ) + (−12 sin 3θ)(sin θ)
dθ
dx
= −(4 cos 3θ)(sin θ) + (−12 sin 3θ)(cos θ)
dθ
11π
6 . Hence,
3 7π
2 , 6 , and
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714
CHAPTER 10. CONICS AND POLAR COORDINATES
The points on the graph correspond to θ =
π
and θ = 2π
3 . At θ = 3 , we have
√ 3
1
(−4)
+
(0)
2
2
dy
√ =
dx
−(−4) 23 + (0) 21
√
−2
3
= √ =−
3
2 3
√
and the rectangular coordinates
of the point are (−2, −2 3). Hence, the equation of the
√
√
3
tangent line is y = −2 3 −
(x + 2). At θ = 2π
3 , we have
3
√ 4 − 12 + (0) 23
dy
√ =
dx
−(4) 23 + (0) − 21
√
2
3
= √ =
3
2 3
√
and the rectangular
√coordinates of the point are (−2, 2 3). Hence, the equation of the tangent
√
3
line is y = 2 3 +
(x + 2).
3
10.
dy
= (1 + 2 cos θ) cos θ + (−2 sin θ) sin θ
dθ
= 1 + 2 cos2 θ − 2 sin2 θ
dx
= −(1 + 2 cos θ) sin θ + (−2 sin θ) cos θ
dθ
= − sin θ − 4 cos θ sin θ
At θ = π3 , we have
1+2
π
3
1 2
2
dy
=
√
dx
− 23 − 4
√ 2
− 2 23
√3 = 0.
1
2
2
√
Also, r = 2 so the rectangular
coordinates of the point are (1, 3). Hence, the equation of
√
the tangent line is y = 3.
5π
At θ =
, we have
3
√ 2
2
1 + 2 12 − 2 − 23
dy
= √
√3 = 0.
3
1
dx
−
4
− 2
2
2
√
Also, r = 2 so the rectangular
coordinates of the point are (1, − 3). Hence, the equation of
√
the tangent line is y = − 3.
dr
11. r = 0 when sin θ = 0 which occurs at θ = 0 and θ = π.
= −2 cos θ 6= 0 at either θ = 0 or
dθ
θ = π. Therefore, θ = 0 and θ = π define tangent lines to the graph at the origin.
715
10.6. CALCULUS IN POLAR COORDINATES
dr
= −3 sin θ 6= 0 at either θ =
dθ
define tangent lines to the graph at the origin.
12. r = 0 when cos θ = 0 which occurs at θ =
and θ =
3π
2 .
π
2
and θ = 3π
2
√
1
2
5π
7π dr √
13. r = 0 when sin θ = − √ = −
which occurs at θ =
and θ =
.
= 2 cos θ 6= 0 at
2
4
4 dθ
2
5π
7π
5π
7π
θ=
or θ =
. Therefore, θ =
and θ =
define tangent lines to the graph at the
4
4
4
4
origin.
or θ =
3π
2 .
Therefore, θ =
π
2
π
2
1
dr
.
which occurs at θ = π6 and θ = 5π
= −2 cos θ 6= 0 at θ =
6
2
dθ
5π
π
5π
θ = 6 . Therefore, θ = 6 and θ = 6 define tangent lines to the graph at the origin.
14. r = 0 when sin θ =
π
6
or
π
5π 7π 9π 11π 13π 15π 17π
19π
15. r = 0 when cos 5θ = 0 which occurs at θ = 10
, 3π
10 , 10 , 10 , 10 , 10 , 10 , 10 , 10 , and 10 .
dr
nπ
= −5 sin 5θ 6= 0 at any of these θ values. Therefore, θ =
defines a tangent line to the
dθ
10
graph at the origin for n = 1, 3, 5, . . . , 19.
16. r = 0 when sin 2θ = 0 which occurs at θ = 0,
π
2,
dr
= 4 cos 2θ 6= 0 at any of these
dθ
define tangent lines to the graph at the
π, and
θ values. Therefore, θ = 0, θ = π2 , θ = π and θ =
origin.
π
Z
1 π
1
=π
17. A =
4 sin2 θdθ = θ − sin 2θ
2 0
2
0
3π
2
3π
2 .
1
18. A =
1
2
Z
π
100 cos2 θdθ =
0
25θ +
25
sin 2θ
2
polar
axis
π
= 25π
0
polar
axis
1
19. A =
2
Z
0
2π
(4 + 4 cos θ) dθ = 8
2
Z
2π
(1 + 2 cos θ + cos2 θ)dθ
0
2π
= (8θ + 16 sin θ + 4θ + 2 sin 2θ)|0 = 24π
polar
axis
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716
CHAPTER 10. CONICS AND POLAR COORDINATES
20. A =
=
21. A =
=
1
2
Z
1
2
Z
2π
0
(1 − sin θ)2 dθ =
1
2
Z
2π
0
1
1
1
θ + cos θ + θ − sin 2θ
2
4
8
2π
(3 + 2 sin θ)2 dθ =
0
1
2
Z
(1 − 2 sin θ + sin2 θ)dθ
2π
=
0
2π
3
π
2
polar
axis
(9 + 12 sin θ + 4 sin2 θ)dθ
0
9
1
θ − 6 cos θ + θ − sin 2θ
2
2
2π
= 11π
0
polar
axis
22. A =
=
23. A =
1
2
1
2
9
=
4
24. A =
=
1
2
1
6
1
=
6
Z
2π
(2 + cos θ)2 dθ =
0
1
2
Z
2π
0
1
1
2θ + 2 sin θ + θ + sin 2θ
4
8
Z
2π
Z
4π
sin udu =
2
0
Z
π
cos2 3θdθ
2π
=
0
9
π
2
polar
axis
u = 2θ du = 2dθ
9 sin2 2θdθ
0
(4 + 4 cos θ + cos2 θ)dθ
9
9
u−
sin 2u
8
16
4π
=
0
9
π
2
polar
axis
u = 3θ, du = 3dθ
0
Z
3π
cos2 udu
0
1
1
u + sin 2u
2
4
3π
0
π
=
4
polar
axis
717
10.6. CALCULUS IN POLAR COORDINATES
1
25. A =
2
Z
3π/2
2 3
θ
3
4θ2 dθ =
0
3π/2
=
0
29 3
π
4
polar
axis
26. A =
1
2
Z
π
π/2
π
θ
dθ = −
π
π2
2θ
π/2
=−
π
π
−π =
2
2
polar
axis
27. A =
1
2
Z
π
e2θ dθ =
0
π
1 2θ
e
4
=
0
1 2θ 1
e −
4
4
polar
axis
1
28. A =
2
Z
2
100e−2θ dθ
1
= −25e−2θ
2
1
= −25(e−4 − e−2 ) = 25e−2 − 25e−4
polar
axis
29. A =
=
30. A =
1
2
Z
1
2
Z
=−
π/4
tan2 θdθ =
0
1
1
tan θ − θ
2
2
25
2
π/4
Z
0
=
0
π/4
(sec2 θ − 1)dθ
1 π
−
2
8
25
cot θ
2
!
√
√
3 √
25 3
− 3 =
3
3
π/3
π/6
1
2
25 csc2 θdθ = −
−0=
4−π
8
polar
axis
π/3
π/6
polar
axis
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718
CHAPTER 10. CONICS AND POLAR COORDINATES
Z
1
31. A =
4π/3(1 + 2 cos θ)2 dθ
2 2π/3
Z
1
4π/3(1 + 4 cos θ + 4 cos2 θ)dθ
=
2 2π/3
4π/3
1
1
1
θ + 4 sin θ + 4
θ + sin 2θ
=
2
2
4
2π/3
√
2π − 3 3
=
2
Z
Z
1 2π/3
1 4π/3
32. A =
(1 + 2 cos θ)2 dθ −
(1 + 2 cos θ)2 dθ
2 −2π/3
2 2π/3
Z
Z
1 4π/3
1 2π/3
(1 + 4 cos θ + 4 cos2 θ)dθ −
(1 + 4 cos θ + 4 cos2 θ)dθ
=
2 −2π/3
2 2π/3
2π/3
1
1
1
1
1
1
=
−
θ + 4 sin θ + 4
θ + sin 2θ
θ + 4 sin θ + 4
θ + sin 2θ
2
2
4
2
2
4
−2π/3
√
√
1
1
= (3 3 + 4π) − (2π − 3 3)
2√
2
=3 3+π
33. Solving 2 cos 3θ = 1 in the first quadrant, we obtain cos 3θ =
1/2, 3θ = π/3, and θ = π/9. Using symmetry,
"Z
#
π/9
A=6
0
=4
Z
0
π/3
(4 cos2 3θ − 1)dθ
u = 3θ, du = 3dθ
π/3
(cos2 u − 1)du = (2u + sin 2u − u)|0
=
√
4π/3
2π/3
polar
axis
π
3
+
.
3
2
34. The circles intersect at θ = π/4. Using symmetry,
" Z
# π/4
π 1
1 π/4 2
1
1
π−2
sin θdθ =
= − =
A=2
θ − sin 2θ
.
2 0
2
4
8
4
8
0
polar
axis
10.6. CALCULUS IN POLAR COORDINATES
35. Solving 5 sin θ = 3 − sin θ in the first quadrant, we obtain
sin θ = 1/2 and θ = π/6. Using symmetry,
" Z
#
1 π/2
2
2
A=2
(25 sin θ − (3 − sin θ) )dθ
2 π/6
Z π/2
=
(24 sin2 θ + 6 sin θ − 9)dθ
719
polar
axis
π/6
π/2
= (12θ − 6 sin 2θ − 6 cos θ − 9θ)|π/6
√ √
√
3π π
=
−
− 3 3 − 3 3 = π + 6 3.
2
2
36. From Problem 35, the point of intersection in the first quadrant is at θ = π/6. Using symmetry,
" Z
#
Z π/3
1 π/6
1
A=2
25 sin2 θdθ +
(3 − sin θ)2 dθ
2 0
2 π/6
Z π/6
Z π/3
2
= 25
sin θdθ +
(9 − 6 sin θ + sin2 θ)dθ
=
=
0
polar
axis
π/6
π/2
1
25
1
25
+ 9θ + 6 sin θ + θ − sin 2θ
θ−
sin 2θ
2
4
2
4
0
π/6
√ !
√ !
√
√
25π 25 3
3
19π
19π
21π
−
+
−
+3 3−
=
− 6 3.
12
8
4
12
8
4
π/6
37. Solving 4 − 4 cos θ = 6 in the second quadrant, we obtain
cos θ = −1/2 and θ = 2π/3. Using symmetry,
" Z
#
1 π
2
A=2
(4 − 4 cos θ) − 36 dθ
2 2π/3
Z π
=
(16 cos2 θ − 32 cos θ − 20)dθ
2π/3
π
= (8θ + 4 sin 2θ − 32 sin θ − 20θ)|2π/3
√
√
√
= −12π − (8π − 2 3 − 16 3) = 18 3 − 4π.
polar
axis
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720
CHAPTER 10. CONICS AND POLAR COORDINATES
38. From Problem 37, the point of intersection in the second
quadrant is at θ = 2π/3. Using symmetry,
" Z
#
Z
1 2π/3
1 π
2
A=2
(4 − 4 cos θ) dθ +
36dθ
2 0
2 2π/3
Z 2π/3
2π
2
(1 − 2 cos θ + cos θ)dθ + 36 π −
= 16
3
0
polar
axis
2π/3
= (16θ − 32 sin θ + 8θ + 4 sin 2θ)|0
+ 12π
√
√
√
= 16π − 16 3 − 2 3 + 12π = 28π − 18 3.
39.
40.
R 2π √
R 2π
dr
= 0 so we have L = 0
32 dθ = 0 3dθ = 6π
dθ
dr
= −6 sin θ so we have
dθ Z π p
36 cos2 θ + 36 sin2 θdθ
L=
0
Z π
=
6dθ = 6π
0
dr
1
41.
= eθ/2 ;
dθ
2
dr
dθ
2
5
1
+ r = eθ + eθ = eθ ;
4
4
2
s=
√
√
5 R 4 θ/2
e dθ = 5eθ/2
0
2
2
dr
dr
−θ
42.
= −2e ;
+ r2 = 4e−2θ + 4e−2θ = 8e−2θ ;
dθ √
dθ
√
√
√
Rπ
π
s = 2 2 0 e−θ dθ = −2 2e−θ 0 = −2 2(e−π−1 ) = 2 2(1 − e−π )
43.
dr
= 3 sin θ so we have
dθ
4
0
=
√
4(e2 − 1)
721
10.6. CALCULUS IN POLAR COORDINATES
L=
Z
2π
0
=
Z
2π
0
=
Z
2π
0
=
√
18
p
(3 − 3 cos θ)2 + (3 sin θ)2 dθ
p
9 − 18 cos θ + 9 cos2 θ + 9 sin2 θdθ
√
Z
18 − 18 cos θdθ
2π
√
0
1 − cos θdθ
θ
1
θ
= (1 + cos θ) −→ 1 − cos θ = 2 sin2
2
2
2
s
√ Z 2π
θ
= 18
2 sin2
dθ
2
0
Z 2π
θ
θ
=6
sin
dθ u = , du = 2dθ
2
2
0
Z π
= 12
sin udu
cos2
0
π
= 12 (− cos u)|0
= 24
44.
45.
dr
= sin2 θ3 cos θ3 so we have
dθ
Z πs
θ
θ
θ
4
6
2
+ sin
cos
dθ
sin
L=
3
3
3
0
Z πs
θ
4
=
sin
dθ
3
0
π/3
Z π3
1
1
2
=3
sin udu = 3
u − sin 2u
2
4
0
0
√
4π − 3 3
=
8
=
Z
0
π
θ
sin
dθ
3
2
u=
θ
1
, du = dθ
3
2
(a) The lemniscate r2 = 9 cos 2θ is only defined for − π4 ≤
5π
θ ≤ π4 and 3π
4 ≤θ ≤ 4 .
1
(b) A =
2
Z
1
9 cos 2θdθ +
2
−π/4
π/4
u = 2θ, du = 2dθ
Z
5π/4
9 cos 2θdθ
3π/4
9
9
π/2
5π/2
(sin u)|−π/2 + (sin u)|3π/2
4
4
9 9
= + =9
2 2
=
polar
axis
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722
CHAPTER 10. CONICS AND POLAR COORDINATES
46. Example 8 uses the fact that
q
cos2
= cos
on the interval 0 ≤ θ ≤ π. This same
equality does not hold on the interval π < θ < 2π since cos θ2 is negative on this interval.
√
In Problem 43, the equality sin2 θ2 = sin θ2 is used which holds for all θ in the interval
0 ≤ θ ≤ 2π.
θ
2
θ
2
1
47. To obtain the area, we can compute the area of half of one
of the petals and then use symmetry, multiplying by 16.
area of a half-petal =
1
2
1
=
4
Z
1
π/8
sin2 2θdθ
polar
axis
u = 2θ, du = 2dθ
0
Z
π/4
sin2 udu
0
1 1
1
u − sin 2u
4 2
4
π−2
=
32
π−2
π−2
Total area = 16 ·
=
32
2
π/4
=
48. Each of the areas will equal
1 + cos θ.
polar
axis
3π
2
0
since the graphs are simply rotations of the graph of r =
polar
axis
polar
axis
49. No; Let A1 denote the area of the graph of r = 2(1 + cos θ) and let A2 denote the area of the
graph ofZ r = 1 + cos θ. Then Z
1 2π
1 2π
2
2
A=
4(1 + cos θ) dθ = 4
(1 + cos θ) dθ
2 0
2 0
= 4A1
polar
axis
723
10.7. CONIC SECTIONS IN POLAR COORDINATES
polar
axis
polar
axis
50. The equations of the circles are r = 1, r = 2 sin θ, and
r = 2 cos θ. We split the area into three regions to get
Z
Z
Z
1 π/3
1 π
1 π/6
4 sin2 θdθ +
1dθ +
4 cos2 θdθ
A=
2 0
2 π/6
2 π/3
π/6
1
1
1
1
1
π/3
+ (θ)|π/6 + 2
=2
θ − sin 2θ
θ − sin 2θ
2
4
2
2
4
0
√
√
π
3
π
2π
3
= −
+
+
−
6
4
12
3
4
π
polar
axis
π/3
51. From L = mr3 dθ = dt we obtain r2 dθ = Ldt/m. Then
1
A=
2
10.7
Z
θ2
θ1
1
r dθ =
2
2
Z
a
b
L
L
dt =
(b − a).
m
2m
Conic Sections in Polar Coordinates
1. Identifying e = 1, the graph is a parabola.
polar
axis
2. Writing r =
is an ellipse.
1
, we identify e = 1/2. The graph
1 − (1/2) sin θ
polar
axis
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724
CHAPTER 10. CONICS AND POLAR COORDINATES
3. Writing r =
is an ellipse.
15/4
, we identify e = 1/4. The graph
1 − (1/4) cos θ
polar
axis
4. Writing r =
parabola.
5/2
, we identify e = 1. The graph is a
1 − sin θ
polar
axis
5. Identifying e = 2, the graph is a hyperbola.
polar
axis
6. Writing r =
is an ellipse.
2
, we identify e = 1/3. The graph
1 + (1/3) sin θ
polar
axis
7. Writing r =
hyperbola.
6
, we identify e = 2. The graph is a
1 + 2 cos θ
polar
axis
725
10.7. CONIC SECTIONS IN POLAR COORDINATES
8. Writing r =
parabola.
6
, we identify e = 1. The graph is a
1 − cos θ
polar
axis
9. Writing r =
is an ellipse.
2
, we identify e = 4/5. The graph
1 + (4/5) sin θ
polar
axis
1
, we identify e = 5/2. The graph
1 + (5/2) cos θ
is an hyperbola.
10. Writing r =
polar
axis
11. From r =
6
, we have e = 2. Converting to a rectangular equation, we get
1 + 2 sin θ
6
1 + 2 sin θ
r + 2r sin θ = 6
r=
p
x2
r = 6 − 2r sin θ
+ y 2 = 6 − 2y
x2 + y 2 = 36 − 24y + 4y 2
(y − 4)2
x2
−
=1
4
12
with a = 2 and b =
√
12 so c2 = a2 + b2 = 4 + 12 = 16 so c = 4. Thus e =
c
4
= = 2.
a
2
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726
CHAPTER 10. CONICS AND POLAR COORDINATES
12. From r =
get
10
5
, we have e = 23 . Converting to a rectangular equation, we
=
2 − 3 cos θ
1 − 32 cos θ
10
2 − 3 cos θ
2r = 10 + 3r cos θ
r=
2
p
x2 + y 2 = 10 + 3x
4(x2 + y 2 ) = 100 + 60x + 9x2
with a = 4 and b =
13. From r =
get
√
(x + 6)2
y2
−
=1
16
20
20 so c2 = a2 + b2 = 16 + 20. Thus c = 6 and e =
c
a
=
6
4
= 23 .
12
4
, we have e = 32 . Converting to a rectangular equation, we
=
2
3 − 2 cos θ
1 − 3 cos θ
12
3 − 2 cos θ
3r = 12 + 2r cos θ
r=
p
3 x2 + y 2 = 12 + 2x
2
x − 24
y2
5
+ 144 = 1
1296
25
with a =
36
5
and b =
12
√
5
5
so c = a − b =
2
2
2
1296
25
− 144
5 =
576
25 .
Thus c =
24
5
so e =
c
a
=
25/4
36/5
= 23 .
√
√
2 3
2
1
3
14. From r = √
=
, we have e = √ =
. Converting to a rectangular
1
3
1 + √3 sin θ
3 + sin θ
3
equation, we get
√
2 3
r= √
3 + sin θ
√
√
3r = 2 3 − r sin θ
√
√ p
2
3 x + y2 = 2 3 − y
√
x2
(y + 3)
+
=1
6
9
√
√
√
c
3
2
2
2
with a = 3 and b = 6 so c = a − b = 9 − 6 = 3. Hence c = 3 and e = =
.
a
3
15. Since e = 1, the conic is a parabola. The directrix is 3 units to the right of the focus and
3
perpendicular to the x-axis. Therefore, r =
.
1 + cos θ
16. Since e = 32 , the conic is a hyperbola. The directix is 2 units above the focus and parallel to
3
the x-axis. Therefore, r =
.
1 + 32 sin θ
727
10.7. CONIC SECTIONS IN POLAR COORDINATES
17. Since e = 23 , the conic is an ellipse. The directrix is 2 units below the focus and parallel to
the x-axis. Therefore, r =
18. Since e =
1−
4
3
.
2
3 sin θ
1
2,
the conic is an ellipse. The directrix is 4 units to the right of the focus and
2
.
perpendicular to the x-axis. Therefore, r =
1
1 + 2 cos θ
19. Since e = 2, the conic is a hyperbola. The directrix is 6 units to the right of the focus and
12
perpendicular to the x-axis. Therefore, r =
.
1 + 2 cos θ
20. Since e = 1, the conic is a parabola. The driectrix is 2 units below the focus and parallel to
2
.
the x-axis. Therefore, r =
1 − sin θ
21. r =
22. r =
3
1 + cos θ +
1+
3
2
2π
3
3
sin θ − π6
23. Since the vertex is
units below the focus, the directrix must be 3 units below the focus and
3
parallel to the x-axis. Therefore, r =
.
1 − sin θ
3
2
24. Since the vertex is 2 units to the left of the focus, the directrix must be 4 units tot he left of
4
.
the focus and perpendicular to the x-axis. Therefore, r =
1 − cos θ
25. Since the vertex is
units to the left of the focus, the directrix must be 1 units to the left of
1
.
the focus and perpendicular to the x-axis. Therefore, r =
1 − cos θ
1
2
26. Since the vertex is 2 units to the right of the focus, the directrix must be 4 units to the right
4
of the focus and perpendicular to the x-axis. Therefore, r =
.
1 + cos θ
27. Since the vertex is
1
4
28. Since the vertex is
3
2
units below the focus, the directrix must be
1/2
.
parallel to the x−axis. Therefore, r =
1 − sin θ
1
2
units below the focus and
units above the focus, the directrix must be 3 units above the focus and
3
parallel to the x-axis. Therefore, r =
.
1 + sin θ
4
rotated counterclockwise by π/4. The original parabola
1 + cos θ
had its focus at the origin and its directrix at x = 4. The original vertex therefore had polar
coordinates (2, 0). After rotation, the vertex is located at (2, π/4).
29. This is the parabola r =
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728
CHAPTER 10. CONICS AND POLAR COORDINATES
5
5/3
=
rotated counterclockwise by π/3. The
3 + 2 cos θ
1 + 2/3 cos θ
original ellipse had vertices at θ = 0 and θ = π. The polar coordinates of the vertices were
(1, 0) and (5, π). After rotation, the vertices are located at (1, π/3) and (5, 4π/3).
30. This is the ellipse r =
10
5
rotated clockwise by π/6. The original ellipse
=
1
2 − sin θ
2 − 2 sin θ
had vertices at θ = π/2 and θ = 3π/2. The polar coorinates of the vertices were (10, π/2) and
(10.3, 3π/2). After rotation, the vertices are located at (10.π/3) and (10/3, 4π/3).
31. This is the ellipse r =
6
rotated clockwise by π/3. The original hyperbola had
1 + 2 sin θ
vertices at θ = π/2 and θ = 3π/2. The polar coordinates of the vertices were (2, π/2) and
(−6, 3π/2). After rotation, the vertices are located at (2, π/6) and (−6, 7π/6).
32. This is the hyperbola r =
33. Identifying ra = 12, 000 and e = 0.2, we have from (7) in the text 0.2 =
for rp , we obtain rp = 8, 000 km.
12, 000 − rp
. Solving
12, 000 + rp
0.2p
. When θ = 0, r = 12, 000 so 12, 000 =
(1 − 0.2 cos θ)
0.2p
9, 600
= p4 . Thus, p = 48, 000 and the equation of the orbit is r =
.
(1 − 0.2)
(1 − 0.2 cos θ)
34. The equation of the orbit is r =
35. The equation of the orbit is r =
ep
. From (7) in the text,
(1 − e cos θ)
1.5 × 108 − 1.47 × 108
5
=
≈ 1.67 × 10−2 .
8
8
1.52 × 10 + 1.47 × 10
299
ep
When θ = 0, r = ra = 1.52 × 108 =
. Thus ep ≈ 1.52 × 108 − 2.52 × 106 ≈
(1 − 1.67 × 10−2 )
(1.49 × 108 )
1.49 × 108 and the equation of the orbit is r =
.
(1 − 1.67 × 10−2 cos θ)
e=
0.97p
. The length of the major axis is the sum
(1 − 0.97 cos θ)
0.97p
0.97p
0.97p
of ra = r(0) =
and rp = r(π) =
=
. That is
0.03
1 + 0.97
1.97
1
1
ra + rp = 0.97p
+
= 3.34 × 109 .
0.03 1.97
36. (a) The equation of the orbit is r =
Solving for p we obtain p = 1.02 × 108 . The equation of the orbit is
r=
0.97(1.02 × 108 )
.
(1 − 0.97 cos θ)
(b) From part (a)
ra = r(0) =
9.87 × 107
9.87 × 107
=
≈ 3.29 × 109 miles
1 − 0.97
0.03
729
10.7. CONIC SECTIONS IN POLAR COORDINATES
and
rp = r(π) =
9.87 × 107
9.87 × 107
=
≈ 5.01 × 107 miles .
1 + 0.97
1.97
38.
37.
polar
axis
polar
axis
39.
40.
polar
axis
polar
axis
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730
CHAPTER 10. CONICS AND POLAR COORDINATES
42.
41.
polar
axis
polar
axis
43. Continuing the development in the paragraph following (3) in the text, we see that r =
e(d + r cos θ) yields (1 − e2 )x2 − 2e2 dx + y 2 = e2 d2 which in turn gives
2e2 d
y2
x2 −
x+
2
1−e
1 − e2
2
2e2 d
e2 d
y2
x2 −
x+
+
2
2
1−e
1−e
1 − e2
2
y2
e2 d
+
x−
1 − e2
1 − e2
2
e2 d
x − 1−e
2
y2
h
i +h
i
1
(1−e2 )2
1
1−e2
=
=
=
e2 d2
1 − e2
e2 d2
+
1 − e2
e2 d
1 − e2
2
e2 d2 (1 − e2 )
e4 d2
1
+
=
2
2
2
2
(1 − e )
(1 − e )
(1 − e2 )2
= 1,
When 0 < e < 1, both the denominators are positive and the denominator of the fraction
involving the x term is smaller. Therefore, this is in the standard form for an ellipse with
center and foci on the x-axis. When e > 1, the first denominator is positive while the second
is negative. Therefore, this is in the standard form for a hyperbola with center and foci on
the x-axis.
ed
= ra .
1−e
ed
At θ = π, r = rp so
= rp .
1−e
44. At θ = 0, r = ra so
Solving the second equation for d, we have d =
(1 + e)rp
. Plugging this value for d into the
e
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CHAPTER 10 IN REVIEW
first equation, we have
e
(1+e)rp
e
= ra
ra
(1 + e)rp
= ra
1−e
rp + erp = ra − era
r(ra + rp ) = ra − rp
ra − rp
e=
ra + rp
Chapter 10 in Review
A. True/False
1. True
2. True
3. True
4. False; there are no y-intercepts since −y 2 /b2 = 1 has no real solution.
5. True
6. True
7. False; (−r, θ) and (r, θ + π) are the same point.
8. False; since x = t2 , x ≥ 0 in the parametric form, but (−1, 2) is on the graph of y = x2 + 1.
9. True; solving x = t2 + t − 12 = (t − 3)(t + 4) = 0 we obtain t = 3 and t = 4. Since
33 − 7(3) = 27 − 21 = 6, the graph intersects the y-axis at (0, 6).
10. True
11. True
12. False; since 6 is even the graph has 12 petals.
13. False; the same point can be expressed as (−4, π/2), which does satisfy the equation.
14. True
15. True
16. True; since e = 1/15 is close to 0.
17. True
18. True; since r = −5 sec θ is equivalent to r cos θ or x = −5.
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732
CHAPTER 10. CONICS AND POLAR COORDINATES
19. False; if r < 0, the point (r, θ) is in the same quadrant as the terminal side of θ + π.
20. True
21. True
22. True
23. True
24. False; this integral will compute the area inside the inner loop of the limacon twice.
25. False
26. False; r = cos θ and r = sin θ intersect at the pole which is (0, π/2) for r = cos θ and (0, 0) for
r = sin θ.
B. Fill in the Blanks
1. 4p = 1/2, p = 1/8. The focus is (0, 1/8).
2. c2 = 4 + 12 = 16. The foci are (±4, 0).
3. The center is (0, 2).
4. The asymptotes are 25y 2 − 4x2 = 0 or y = ±2x/5.
5. 4p = 8, p = 2. The directrix is y = −3 − 2 = −5.
6. a2 = 36, b2 = 16. The vertices are (−1 ± 6, −7) and (−1, −7 ± 4) or (−7, −7), (5, −7),
(−1, −11), and (−1, −3).
7. Completing the square, y + 10 = (x + 2)2 . The vertex is (−2, −10).
8. b2 = 9. The length of the conjugate axis is 2 · 3 = 6.
9. a2 = 4. The endpoints of the transverse axis are the vertices (4 ± 2, −1) or (2, −1) and (6, −1).
10. The major axis is on the line x = 3.
11. Completing the square, we have 25(x2 − 8x + 16) + (y 2 + 6y + 9) = −384 + 409 = 25 or
(x − 4)2 + (y + 3)2 /25 = 1. The center of the ellipse is at (4, −3).
12. Setting y = 0 and solving, we have (x + 1)2 + 64 = 100, (x + 1)2 = 36, or x = ±6 − 1. The
x-intercepts are -7 and 5.
√
13. Setting
x = 0 and solving, we have y 2 − 4 = 1, y 2 = 5, or y = ± 5. The y-intercepts are
√
± 5.
14. Using implicit differentiation, 2yy 0 − y 0 + 3 = 0 or y 0 =
tangent line is
15. line
3
(−1)
= −3.
3
. At (1, 1) the slope of the
1 − 2y
733
CHAPTER 10 IN REVIEW
16. x = 0 at t = ±1, so the y-intercepts occurs at (0, 3) and (0, −1).
17. circle
18. convex limacon
dr
19. r = 0 at θ = 0, π/3, 2π/3, π, 4π/3, and 5π/3.
= 3 cos 3θ 6= 0 at any of the θ values
dθ
nπ
defines a tangnet to the graph at the origin for
mentioned. Thus, the polar equation θ =
3
n = 0, . . . , 5.
20. From r =
1
2
,
5
2 sin θ
1
=
2 + 5 sin θ
1+
we have e = 52 .
21. The focus is the origin. The directrix is 10 units below the origin and parallel to the x-axis.
Therefore, the vertex is 5 units below the origin at (0, −5).
12
6
=
, we see that the conic is an ellipse with a directrix 12 units
2 + cos θ
1 + 21 cos θ
to the right of the focus at the origin and perpendicular to the x-axis. The two vertices must
therefore occur at θ = 0 and θ = π. Thus polar coordinates of the vertices are (4, 0) and
(12, π). The foci must therefore be at the origin and at (8, pi). The center is at (4, π).
22. From r =
C. Exercises
dy
sin t
1.
=
. At t = π/2,
dx
1 − cos t
equation is
"
1
1
y − = −√ x −
2
3
√
√
dy
( 3/2)
=
= 3. The slope of the normal line is
dx
(1/2)
√ !#
3
π
−
3
2
2. x0 (t) = 1 − cos t, y 0 (t) = sin t
Z 2π q
Z
s=
(1 − cos t)2 + sin2 tdt =
0
√ Z
= 2
0
2π
0
2π
√
1 − cos tdt
p
−1
√
3
and its
√
√
3
3π
1
π
or y = − √ x + √ = −
x+
.
3
9
3
3 3
1 − 2 cos t + cos2 t + sin2 tdt
by symmetry
√
√
√ Z π√
√ Z π√
√ Z π 1 − cos2 t
1 + cos t
√
=2 2
1 − cos tdt = 2 2
1 − cos t √
dt = 2 2
dt
1 + cos t
1 + cos t
0
0
0
Z
π
√
sin t
√
=2 2
dt u = 1 + cos t, du = − sin tdt
1
+ cos t
0
Z
Z 2
0
√ √
√
√
√
√
√
−du
2
√ = 2 2 lim
=2 2
u−1/2 du = 2 2 lim 2 u b = 2 2 lim (2 2 − 2 b) = 8
u
b→0+ b
b→0+
b→0+
2
(3t2 − 18t)
3t
= 3t
2 −9, t 6= 0. Solving 2 −9 = −6
2t
we obtain t = 2. Since x(2) = 8 and y(2) = −26, the point on the graph is (8, −26).
3. The slope of the line 6x+y = 8 is -6 and
dy
dx
=
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734
4.
CHAPTER 10. CONICS AND POLAR COORDINATES
dy
dt
2
= 2t
= 1t . The slope of the tangent line is 1t and the point on the curve is (t2 + 1, 2t). The
equation of a tangent line is then y − 2t = 1t [x − (t2 + 1)]. Since we want the tangent line to
pass through (1, 5), we have 5 − 2t = 1t (1 − t2 − 1) = −t. Solving for t we obtain t = 5. The
point on the curve is (x(5), y(5)) = (26, 10). Also, the tangent is vertical if t = 0. At t = 0,
the point on the graph is (1, 0). The tangent line at this point will also pass through (1, 5).
5. (a) Since 4x2 (1 − x2 ) = y 2 ≥ 0, 1 − x2 ≥ 0, x2 ≤ 1, and |x| ≤ 1.
(b) Letting x = sin t, we have y 2 = 4 sin2 t(1 − sin2 t) = 4 sin2 t cos2 t = sin2 2t. Parametric
equations for the curve are x = sin t, y = sin 2t, for 0 ≤ t ≤ 2π.
(c)
cos 2t)
= (2 cos
is horizontal when cos 2t = 0 or t = π/4, 3π/4, 5π/4, 7π/4.
t . The tangent line√
√
√
√
The points on the graph are ( 2/2, 1), ( 2/2, −1), (− 2/2, 1) and (− 2/2, −1).
dy
dx
(d)
y
1
x
6. The area inside the limacon is
Z
Z
1 2π
1 2π
(3 + cos θ)2 dθ =
(9 + 6 cos θ + cos2 θ)dθ
2 0
2 0
9
1
1
=
θ + 3 sin θ + θ + sin 2θ
2
4
8
2π
0
19π
=
.
2
polar
axis
The circle r = 4 cos θ has radius 2, so its area is 4π. Thus,
19π
the area outside the circle and inside the limacon is
−
2
11π
4π =
.
2
7. Solving 3 sin θ = 1 + sin θ in the first quadrant, we have
sin θ = 1/2 and θ = π/6. Using symmetry,
polar
axis
735
CHAPTER 10 IN REVIEW
#
" Z
Z
1 π/2
1 π/6
2
2
9 sin θdθ +
(1 + sin θ) dθ
A=2
2 0
2 π/6
π/6 9
1
9
1
=
+ θ − 2 cos θ + θ − sin 2θ
θ − sin 2θ
2
4
2
4
0
√ ! "
√ !#
3π 9 3
3π π √
5π
3
=
+
=
−
− 3−
.
4
8
4
4
8
4
π/2
π/6
1 R π/2
50 cos2 θdθ, we
2 0
√
5 2
see that the area corresponds to a semicircle of radius
.
2
8. Writing A =
R π/2
0
25(1 − sin2 θ)dθ =
9. From x = 2 sin 2θ cos θ, y = 2 sin 2θ sin θ, we find
2 sin 2θ cos θ + 4 cos 2θ sin θ
dy
=
dx
−2 sin 2θ sin θ + 4 cos 2θ cos θ
dy
dx
and
polar
axis
θ=π/4
√ !
√ !
2
2
2(1)
+ 4(0)
2
2
=
√ !
√ ! = −1.
2
2
−2(1)
+ 4(0)
2
2
√
√
√
(a) At θ = π/4,
√ x = 2 and y =
√ 2. The Cartesian equation of the tangent line is y − 2 =
−1(x − 2) or y = −x + 2 2.
√
(b) Using x √
= r cos θ and y = r sin θ in (a), we obtain r sin θ = −r cos θ = 2 2 or
2 2
r=
.
sin θ + cos θ
10. By writing r =
1
(1 −
1
2
sin θ)
we see that e = 1/2 and the graph is an ellipse with major axis
along the y-axis. The vertices on this axis have polar
coordinates (2, π/2) and (2/3, 3π/2). The corresponding
rectangular coordinates are (0, 2) and (0, −2/3). Thus, the
center of the ellipse is at (0, 2/3). Since once focus is at
the origin, c = 2/3. From a√= 4/3 and c = 2/3 we find
2 3
4
12
b2 = 16
the
9 − 9 = 9 and b = 3 . Thus,
vertices on the
minor axis are at
√
− 2 3 3 , 23
and
√
2 3 2
3 , 3
y
v
1
c
.
x
v
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736
CHAPTER 10. CONICS AND POLAR COORDINATES
11. Multiplying both sides of the equation by r, we have r2 = r cos θ + r sin θ. The corresponding
Cartesian equation is x2 + y 2 = x + y.
12. Multipying both sides by cos θ, we have
r cos θ = 1 − 5 cos2 θ
5x2
x+ y 2
5x2
x−1=− + 2
x y
5x2
x2 + y 2 = −
x−1
5x2
y 2 = −x2 −
x−1
3
−x
−
4x2
y2 =
x−1
x=1−
13. 2(r cos θ)(r sin θ) = 5
5
r2 =
2 cos θ sin θ
14. Using x2 + y 2 = r2 and x = r cos θ, we have
(r2 − 2r cos θ)2 = 9r2 =⇒ [r(r − 2 cos θ)]2 = 9r2 =⇒ r2 (r − 2 cos θ)2 = 9r2
=⇒ r − 2 cos θ = ±3 =⇒ r = ±3 + 2 cos θ.
Since replacement of (r, θ) by (−r, θ + π) in r = −3 + 2 cos θ gives r = 3 + 2 cos θ, we take the
polar equation to be r = 3 + 2 cos θ.
15. By writing the equation as r cos θ = −1 we see that the line is x = −2. The curve is then
a parabola with axis along the x-axis, directrix x = −1, and focus at the origin. Since the
1
.
vertex is at (−1/2, 0), p = 1 and the equation is r =
(1 − cos θ)
16. Since the transverse axis lie along the y − axis and e = 2, the form of the equation is
2p
2p
r =
we see that 2p = 4 and the equation is
. From 4/3 = r(3π/2) = (1+2)
(1 − 2 sin θ)
4
r=
.
(1 − 2 sin θ)
17. r = 3 sin(10θ)
18. r = 2.8 cos(7θ)
y 2 x2
y 2 x2
− 2 = 1. The asymptotes for the hyperbola are
−
=0
100 b
100 b2
or by = ±10x. Since the given asymptotes are 3y = ±5x, we have the proportion 3b = 10
5 .
y2
x2
Thus, b = 6 and the equation of the hyperbola is
−
= 1.
100 36
19. The form of the equation is
737
CHAPTER 10 IN REVIEW
sin θ
=1
1 + cos θ
cos θ
=0
x = r cos θ =
1 + cos θ
cos θ + cos2 θ + sin2 θ
cos θ(1 + cos θ) − sin θ(sin θ)
dy
=
=
dθ
(1 + cos θ)2
(1 + cos θ)2
1 + cos θ
=1
=
(1 + cos θ)2
dx
− sin θ(1 + cos θ) − cos θ(− sin θ)
=
dθ
(1 + cos θ)2
− sin θ − sin θ cos θ + sin θ cos θ
=
(1 + cos θ)2
− sin θ
=
= −1
(1 + cos θ)2
dy
1
=
= −1
dx
−1
The tangent line therefore has slope -1 and passes through the point (0, 1). Its equations is
y = 1 − (x − 1) or y = 1 − x.
20. At θ = π/2,
y = r sin θ =
21. Substituting y = tx into x3 + y 3 = 3axy we obtain
x3 + y 3 x = 3atx2 =⇒ (1 + t3 )x = 3at =⇒ x =
and y = tx =
22.
3at
,
1 + t3
3at2
.
(1 + t3 )
dx
3a(1 − 2t3 )
dy
3at(2 − t3 )
=
;
=
dt
(1 + t3 )2
dt
(1 + t3 )2
dy
is not zero at these values, the graph
Solving dt = 0 we obtain t = 0 and t = 21/3 . Since dx
√ dt
√
has horizontal tangent lines at (0, 0) and ( 3 2a, 3 4a).
23. (a) From x = r cos θ and y = r sin θ we obtain r3 (cos3 θ + sin3 θ) = 3ar2 cos θ sin θ or r =
(3a cos θ sin θ)
.
(cos3 θ + sin3 θ)
(b) The loop is formed from θ = 0 to θ = π/2, so the area is
Z
Z
1 π/2 9a2 cos2 θ sin2 θ
9a2 π/2
cos2 θ sin2 θ
A=
dθ
=
dθ
θ
2 0
2 0
(1 + tan3 θ)2 cos6 θ
(cos3 θ + sin )
Z
Z
9a2 π/2
sin2 θ
9a2 π/2
tan2 θ
=
dθ
=
dθ
2 0
2 0
(1 + tan3 θ)2 cos4 θ
(1 + tan3 θ) cos2 θ
Z
9a2 π/2 tan2 θ sec3 θ
=
dθ u = tan θ, du = sec2 θdθ
2 0
(1 + tan3 θ)θ
∞
Z
9a2 ∞
u2
1
1
1
9a2
3a2
3a2
=
−
lim
.
du
=
=
−
−
1
=
2 0 (1 + u3 )2
2
3 1 + u3
2 t→∞ 1 + t3
2
0
www.elsolucionario.org
738
CHAPTER 10. CONICS AND POLAR COORDINATES
3at2
3at
and
y(t)
=
. Then, assuming a > 0, we have
(1 + t3 )
(1 + t3 )
lim − y(t) = −∞. lim + x(t) = −∞, and lim + y(t) = ∞. Finally, using
24. From Problem 21, x(t) =
lim x(t) = ∞,
t→−1−
L’Hôpital’s Rule,
t→−1
t→−1
t→−1
3a + 6at
3at + 3at2 h
= lim
= −a
t→−1
t→−1
1 + t3
3t2
lim [x(t) + y(t)] = lim
t→−1
and x + y = −a or x + y + a = 0 is an asymptote.
25. Using symmetry
Z π/2 2
Z π/2 1
θ
1
2θ
A=2
2 sin
dθ = 4
1 − cos
dθ
2
3
2
3
0
0
!
√
√
π/2
3
2θ
π 3 3
3 3
=2
= 2 θ − sin
−
=π−
.
2
3 0
2
2 2
2
26. The circle centered at (1, 0) has polar equation r = 2 cos θ. Solving 1 = 2 cos θ, we obtain
θ = π/3. Using symmetry,
" Z
#
"
√ !#
√ π
1 π/2
π
π
3
π/2
A=4
(1 − 4 cos2 θ)dθ = 2(θ − 2θ − sin 2θ)|π/3 = 2 − − − −
= 3− .
2 π/3
2
3
2
3
27. (a) r = 2 cos θ −
π
4
(b) Note that r = 2 cos θ defines a circle of radius 1 centered at (1, 0). A rotation of π/4
√ √ √ 2
puts the center at 22 , 22 . The new rectangular equation is therefore x − 22 +
√ 2
y − 22 = 1
28. (a) r =
1
1 + cos(θ + π/6)
(b) Using the sum of angles identity for cosine, we have r =
1
√
r=
3
1
1 +√ 2 cos θ − 2 sinθ
r 1 + 23 cos θ − 12 sin θ = 1
√
r + 23 r cos θ − 12 r sin θ = 1
√
p
x2 + y 2 + 23 x − 12 y = 1
√
p
x2 + y 2 = 1 − 23 x + 12 y
√
√
2
x2 + y 2 = 34 x2 − 3xy
− 3x + y4 + y + 1
2
√
√
3xy
1 2
+ 3x + 34 y 2 − y = 1
4x +
2
1
1 + cos θ cos π/6 − sin θ sin π/6
739
CHAPTER 10 IN REVIEW
29. Taking the center of the ellipse to be at the origin, we have a = 5 × 108 and b = 3 × 108 , since
c2 = a2 − b2 , c2 = 1.6 × 1017 and c = 4 × 108 . The minimum distance is a − c = 108 m and
the maximum distance is a + c = 9 × 108 m.
30. To find the width, we need to first find the maximum y-value for points on the portion of the
petal satisfying 0 ≤ θ ≤ π/4. We know
y = r sin θ = cos 2θ sin 2θ and therefore
= (cos2 θ − sin2 θ) sin θ
dy
= (2 cos θ(− sin θ) − 2 sin θ(cos θ)) sin θ + (cos2 θ − sin2 θ) cos θ Thus,
dθ
= −5 sin2 θ cos θ + cos3 θ
dy
dθ
= 0 when cos θ =
= cos θ(cos2 θ − 5 sin2 θ)
0 or when cos2 θ − 5 sin2 θ = 0. Since cos θ 6= 0 for 0 ≤ θ ≤ π/4, we need to find where
cos2 θ − 5 sin2 θ = 0. This occurs when
cos2 θ = 5 sin2 θ
θ = tan
−1
1
√
5
sin2 θ
1
=
5
cos2 θ
1
= tan2 θ
5
1
√ = tan θ
5
≈ 0.4205
1
yields a
A cursory examination of the graph tells us that the critical point θ = tan−1 √
5
√
6
maximum for y on the interval 0 ≤ θ ≤ pi/4, and this maximum is ymax =
. The width is
9
√
2 6
.
therefore w = 2ymax =
9
www.elsolucionario.org
Multivariable Calculus
Complete Solutions Manual
Brian Fulton
Melanie Fulton
Fourth Edition
Contents
11 Vectors and 3-Space
11.1 Vectors in 2-Space . .
11.2 3-Space and Vectors .
11.3 Dot Product . . . . . .
11.4 Cross Product . . . . .
11.5 Lines in 3-Space . . .
11.6 Planes . . . . . . . . .
11.7 Cylinders and Spheres
11.8 Quadric Surfaces . . .
Chapter 11 in Review . . .
A. True/False . . . . .
B. Fill in the Blanks .
C. Exercises . . . . . .
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2
2
6
12
17
24
28
35
38
42
42
43
45
12 Vector-Valued Functions
12.1 Vector Functions . . . . . . .
12.2 Calculus of Vector Functions
12.3 Motion on a Curve . . . . . .
12.4 Curvature and Acceleration .
Chapter 12 in Review . . . . . . .
A. True/False . . . . . . . . .
B. Fill in the Blanks . . . . .
C. Exercises . . . . . . . . . .
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48
48
55
62
69
73
73
73
74
13 Partial Derivatives
13.1 Functions of Several Variables . . . .
13.2 Limits and Continuity . . . . . . . .
13.3 Partial Derivatives . . . . . . . . . .
13.4 Linearization and Differentials . . . .
13.5 Chain Rule . . . . . . . . . . . . . .
13.6 Directional Derivative . . . . . . . .
13.7 Tangent Planes and Normal Lines .
13.8 Extrema of Multivariable Functions
13.9 Method of Least Squares . . . . . . .
13.10Lagrange Multipliers . . . . . . . . .
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77
77
82
85
92
99
107
112
118
125
127
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ii
CONTENTS
Chapter 13 in Review . .
A. True/False . . . .
B. Fill in the Blanks
C. Exercises . . . . .
1
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132
132
133
134
14 Multiple Integrals
14.1 The Double Integral . . . . . . . . . . . . . .
14.2 Iterated Integrals . . . . . . . . . . . . . . . .
14.3 Evaluation of Double Integrals . . . . . . . .
14.4 Center of Mass and Moments . . . . . . . . .
14.5 Double Integrals in Polar Coordinates . . . .
14.6 Surface Area . . . . . . . . . . . . . . . . . .
14.7 The Triple Integral . . . . . . . . . . . . . . .
14.8 Triple Integrals in Other Coordinate Systems
14.9 Change of Variables in Multiple Integrals . .
Chapter 14 in Review . . . . . . . . . . . . . . . .
A. True/False . . . . . . . . . . . . . . . . . .
B. Fill in the Blanks . . . . . . . . . . . . . .
C. Exercises . . . . . . . . . . . . . . . . . . .
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139
139
141
148
161
169
180
185
193
200
208
208
209
209
15 Vector Integral Calculus
15.1 Line Integrals . . . . . . . . . .
15.2 Line Integrals of Vector Fields .
15.3 Independence of the Path . . .
15.4 Green’s Theorem . . . . . . . .
15.5 Parametric Surfaces and Area .
15.6 Surface Integrals . . . . . . . .
15.7 Curl and Divergence . . . . . .
15.8 Stokes’ Theorem . . . . . . . .
15.9 Divergence Theorem . . . . . .
Chapter 15 in Review . . . . . . . .
A. True/False . . . . . . . . . .
B. Fill in the Blanks . . . . . .
C. Exercises . . . . . . . . . . .
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218
218
225
232
239
245
252
263
268
274
280
280
281
282
16 Higher-Order Differential Equations
16.1 Exact First-Order Equations . . . .
16.2 Homogeneous Linear Equations . . .
16.3 Nonhomogeneous Linear Equations .
16.4 Mathematical Models . . . . . . . .
16.5 Power Series Solutions . . . . . . . .
Chapter 16 in Review . . . . . . . . . . .
A. True/False . . . . . . . . . . . . .
B. Fill in the Blanks . . . . . . . . .
C. Exercises . . . . . . . . . . . . . .
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288
288
291
295
303
307
316
316
316
317
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www.elsolucionario.org
Chapter 11
Vectors and 3-Space
11.1
Vectors in 2-Space
1.
(a)
6i + 12j
2.
(a)
h3, 3i
3.
(a)
h12, 0i
4.
(a)
1
2i
5.
(a)
−9i + 6j
6.
(a)
h3, 9i
7.
(a)
−6i + 27i
8.
(a)
h21, 30i
9.
(a)
h4, −12i − h−2, 2i = h6, −14i
10.
(a)
(4i + 4j) − (6i − 4j) = −2i + 8j
11.
(a)
(4i − 4j) − (−6i + 8j) = 10i − 12j
12.
(a)
h8, 0i − h0, −6i = h8, 6i
13.
(a)
h16, 40i − h−4, −12i = h20, 52i
14.
(a)
h8, 12i − h10, 6i = h−2, 6i
− 12 j
(b)
i + 8j
h3, 4i
(b)
(c)
(c)
h4, −5i
(b)
2
3i
(b)
(b)
+ 32 j
(b)
(b)
(d)
h4, 5i
(d)
− 13 i − j
(c)
−3i + 9j
0
h6, 18i
(c)
h8, 12i
−4i + 18j
(c)
h6, 8i
5
(e)
√
41
(d)
(d)
(d)
3
√
41
√
3 10
(e)
(e)
(e)
10/3
√
√
2 85
10
(−3i − 3j) − (−15i − 10j) = 18i − 17j
(b)
(−3i + 3j) − (−15i + 20j) = 12i − 17j
h−6, 0i − h0, −15i = h−6, 15i
h−12, −30i − h−10, −30i = h−2, 0i
h−6, −9i − h25, 15i = h−31, −24i
2
34
√
6 10
(e)
√
4 13
√
(e)
√
4 10
0
5
√
(e)
√
2 2/3
(d)
(d)
(b)
(b)
(b)
(e)
h−3, 9i − h−5, 5i = h2, 4i
(b)
(b)
65
−3i − 5j
(c)
(c)
√
(d)
h−1, −2i
(c)
h−4, −12i
(b)
3i
11.1. VECTORS IN 2-SPACE
3
16.
15.
P2
5
-5
P1P2
5
P1
P1P2
P1
-5
5
−−−→
P1 P2 = h2, 5i
P2
−−−→
P1 P2 = h6, −4i
17.
18.
5
P2
5
P1
P1
P1P2
P2
P1P2
5
5
−−−→
P1 P2 = h2, 2i
−−−→
P1 P2 = h2, −3i
−−−→ −−→ −−→
19. Since P1 P2 = OP2 − OP1 ,
terminal point is (1, 18)
−−−→ −−→ −−→
20. Since P1 P2 = OP2 − OP1 ,
point is (9, 8)
−−→ −−−→ −−→
OP2 = P1 P2 + OP1 = (4i + 8j) + (−3i + 10j) = i + 18j, and the
−−→ −−→ −−−→
OP1 = OP2 + P1 P2 = h4, 7i − h−5, −1i = h9, 8i, and the initial
21. a(= −a), b = (− 41 a), c(= 52 a), e(= 2a), and f (= − 21 a) are parallel to a.
22. We want −3b = a, so c = −3(9) = −27
23. h6, 15i
24. h5, 2i
25. |a| =
26. |a| =
√
√
√
4 + 4 = 2 2; (a)
9 + 16 = 5; (a)
27. |a| = 5; (a)
28. |a| =
√
u=
1
1
1
1
1
u = √ h2, 2i = √ , √
;
(b) −u = − √ , − √
2 2
2
2
2
2
1
−3 4
3 4
u = h−3, 4i =
,
;
(b) −u =
,−
5
5 5
5 5
1
h0, −5i = h0, −1i;
5
1 + 3 = 2; (a)
√
1
u = h1, − 3i =
2
(b)
*
−u = h0, 1i
√ +
1
3
,−
;
2
2
*
(b)
−u =
√ +
1
3
− ,
2 2
4
CHAPTER 11. VECTORS AND 3-SPACE
29. |a + b| = |h5, 12i| =
30. |a + b| = |h−5, 4i| =
31. |a| =
32. |a| =
√
9 + 49 =
q
1
4
+
1
4
=
√
√
25 + 144 = 13;
√
25 + 16 =
58;
√1
2
√
3b-a
35.
41;
1
13 h5, 12i
u=
b=3
( 1/1√2
( 12 i
=
√1 h−5, 4i
41
b = 2 ( √158 3i + 7j) =
58;
3
33. − 4 a = h−3 − 15/2i
√
u=
−
√6 i
58
1
2 j)
+
5 12
13 , 13
D
E
= − √541 , √441
√14 j
58
√
√
3 2
3 2
=
i−
j
2
2
34. 5(a + b) = 5h0, 1i = h0, 5i
36.
b = a+(b+c)
3b
b+c
a
a
b
c
37. x = −(a + b) = −a − b
38. x = 2(a − b) = 2a − 2b
39.
40.
c
-c
b
a
b = (−c) − a;
(b + c) + a = 0;
a+b+c=0
d
e
b
a
From Problem 39, e + c + d = 0.
but b = e − a and e = a + b, so
(a + b) + c + d = 0.
41. From 2i + 3j = k1 b + k2 c = k1 (i + j) + k2 (i − j) = (k1 + k2 )i + (k1 − k2 )j we obtain the
system of equations k1 + k2 = 2, k1 − k2 = 3. Solving, we find k1 = 25 and k2 = − 12 . Then
a = 52 b − 12 c.
42. From 2i + 3j = k1 b + k2 c = k1 (−2i + 4j) + k2 (5i + 7j) = (−2k1 + 5k2 )i + (4k1 + 7k2 )j we
1
obtain the system of equations −2k1 + 5k2 = 2, 4k1 + 7k2 = 3. Solving, we find k1 = 34
and
7
k2 = 17 .
43. From y 0 = 12 x we see that the slope of the tangent line at (2, 2) is 1. A vector with slope 1 is
√
i + j. A unit vector is (i + j)/|i + j| = (i + j)/ 2 = √12 i + √12 j. Another unit vector tangent
to the curve is − √12 i − √12 j.
44. From y 0 = −2x + 3 we see that the slope of the tangent
√ line at (0, 0) is 3. A vector with slope
3 is i + 3j. A unit vector is (i + 3j)/|i + 3j| = (i + 3j)/ 10 = √110 i + √110 j. Another unit vector
tangent to the curve is − √110 i − √110 j.
www.elsolucionario.org
11.1. VECTORS IN 2-SPACE
45.
5
P2
(a) Since the shortest distance between two point is a straight line,
|a + b| ≤ |a| + |b|.
a
P1
b
c
P3
(b) When P2 lies on the line segment between P1 and P3 , |a + b| = |a| + |b|.
46. Since y = 2a(L2 + y 2 )3/2 is an odd function on [−a, a], Fy = 0. Now using the fact that
L/(L2 + y 2 )3/2 is an even function, we have
Z
a
−a
L
L dy
=
2
2
3/2
a
2a(L + y )
Z
0
a
(L2
dy
+ y 2 )3/2
y = L tan θ, dy = L sec2 θ dθ
Z tan−1 a/L
L sec2 θ dθ
1
sec2 θ dθ
=
La 0
sec3 θ
L3 (1 + tan2 θ)3/2
0
tan−1 a/L
Z tan−1 a/L
1
1
=
cos θ dθ =
sin θ
La 0
La
0
1
a
1
√
=
= √
La L2 + a2
L L2 + a2 .
L
=
a
Z
tan−1 a/L
√
√
Then Fx = qQ/4π0 L L2 + a2 and F = (qQ/4π0 L L2 + a2 )i.
47. (a) Since Ff = −Fg , |Fg | = |Ff | = µ|Fn | and tan θ = |Fg |/|Fn | = µ|Fn |/|Fn | = µ
(b) θ = arctan 0.6 ≈ 31◦
48. Since w + F1 + F2 = 0,
−200j + |F1 | cos 20◦ i + |F2 | sin 20◦ j − |F2 | cos 15◦ i + |F2 | sin 15◦ j = 0
or
(|F1 | cos 20◦ − |F2 | cos 15◦ )i + (|F1 | sin 20◦ − |F2 | sin 15◦ − 200)j = 0.
Thus, |F1 | cos 20◦ − |F2 | cos 15◦ = 0; |F1 | sin 20◦ + |F2 | sin 15◦ = 0. Solving this system for
|F1 | and |F2 |, we obtain
|F1 | =
200 cos 15◦
200 cos 15◦
200 cos 15◦
=
=
≈ 336.8 lb
sin 15◦ cos 20◦ + cos 15◦ sin20◦
sin(15◦ + 20◦ )
sin 35◦
and
|F2 | =
sin 15◦
200 cos 20◦
200 cos 20◦
=
≈ 327.7 lb.
◦
◦
◦
cos 20 + cos 15 sin20
sin 35◦
√
√
√
√
√
49. Since Fq
2 = 200(i + j)/ 2 = 100 2i + 100 2j, F3 ) = F2 F1 = (100 2 − 200)i + 100 2j and
p
√
√
√
|F3 | = (100 2 − 200)2 + (100 2)2 = 200 2 − 2 ≈ 153 lb.
6
CHAPTER 11. VECTORS AND 3-SPACE
−→
50. We have OA = 150 cos 20◦ i + 150 sin 20◦ j,
240 cos 190◦ i + 240 sin 190◦ j. Then
−−→
AB = 200 cos 113◦ i + 200 sin 113◦ j,
−−→
BC =
r = (150 cos 20◦ + 200 cos 133◦ + 240 cos 190◦ )i + (150 sin 20◦ + 200 sin 113◦ + 240 sin 190◦ j and
≈ −173.55i + 193.73j
|r| ≈ 260.09 miles.
51.
P2
P2
M
N
P1
P1
−−→
Place one corner of the parallelogram at the origin, and let two adjacent sides be OP1 and
−−→
OP2 . Let M be the midpoint of the diagonal connecting P1 and P2 and N be the midpoint
−−→
−−→ −−→
−−→ −−→
of the other diagonal. By Problem 37, OM = 21 (OP1 + OP2 ). Since OP1 + OP2 is the main
−−→
−−→ −−→
−−→ −−→
diagonal of the parallelogram and N is its midpoint, ON = 12 (OP1 + OP2 . Thus, OM = ON
and the diagonals bisect each other.
−−→ −−→ −→
−−→ −−→ −−→ −→
B
52. By Problem 39, AB + BC + CA = 0 and AD + DE + ED + CA =
−−→ −−→
−→
0. From the first equation AB + BC = −CA. Since D and E
−−→
−−→
−−→
−−→
−−→ −−→
D
E
are midpoint, AD = 21 AB and EC = 12 BC. Then, 12 AB + DE +
−
−
→
−
→
1
2 BC + CA = 0 and
C
A
−→ 1 −→
−−→
−→ 1 −−→ −−→
1 −→
AB + BC = −CA −
DE = −CA −
−CA = − CA.
2
2
2
Thus, the line segment joining the midpoints D and E is parallel to the side AC and half its length.
11.2
3-Space and Vectors
1.- 6.
(0,0,4)
(1,1,5)
(5,-4,3)
(3,4,0)
(6,-2,0)
(6,0,0)
11.2. 3-SPACE AND VECTORS
7
7. A plane is perpendicular to the z-axis, 5 units above the xy-plane.
8. A plane perpendicular to the x-axis, 1 unit in front of the yz-plane.
9. A line perpendicular to the xy-plane at (2, 3, 0).
10. A single point located at (4, −1, 7).
11. (2, 0, 0), (2, 5, 0), (2, 0, 8), (2, 5, 8), (0, 5, 0), (0, 5, 8), (0, 0, 8), (0, 0, 0)
12.
z
(-1,3,7)
(-1,6,7)
(3,3,7)
(3,6,7)
(-1,3,4)
(3,3,4)
(-1,6,4)
(3,6,4)
y
x
13. (a) xy-plane: (−2, 5, 0), xz-plane: (−2, 0, 4), yz-plane: (0, 5, 4);
(b) (−2, 5, −2)
(c) Since the shortest distance between a point and a plane is a perpendicular line, the point
in the plane x = 3 is (3, 5, 4).
14. We find planes that are parallel to coordinate planes.
(a) z = −5;
(b) x = 1 and y = −1;
(c)
z=2
15. The union of the planes x = 0, y = 0, and z = 0.
16. The origin (0, 0, 0).
17. The point (−1, 2, −3).
18. The union of the planes x = 2 and z = 8.
19. The union of the planes z = 5 and z = −5.
20. The line through the points (1, 1, 1), (−1, −1, −1), and the origin.
p
√
21. d = (3 − 6)2 + (−1 − 4)2 + (2 − 8)2 = 70
www.elsolucionario.org
8
CHAPTER 11. VECTORS AND 3-SPACE
p
√
(−1 − 0)2 + (−3 − 4)2 + (5 − 3)2 = 3 6
p
(a) 7;
(b) d = (−3)2 + (−4)2 = 5
p
(a) 2;
(b) d = (−6)2 + 22 + (−3)2 = 7
p
√
d(P1 , P2 ) = p(32 + 62 + (−6)2 = 9; d(P1 , P3 ) = 22 + 12 + 22 = 3
√
d(P2 , P3 ) = ((2 − 3)2 + (1 − 6)2 + (2 − (−6))2 = 90. The triangle is a right triangle.
q
p
√
√
√
d(P1 , P2 ) = (12 + 22 + 42 = 21; d(P1 , P3 ) = 32 + 22 + (2 2)2 = 21
q
p
√
√
d(P2 , P3 ) = ((3 − 1)2 + (2 − 2)2 + (2 2 − 4)2 = 28 − 16 2.
The triangle is an isosceles triangle.
p
√
d(P1 , P2 ) = p((4 − 1)2 + (1 − 2)2 + (3 − 3)2 = 10
√
d(P1 , P3 ) = p(4 − 1)2 + (6 − 2)2 + (4 − 3)2 = 26
√
d(P2 , P3 ) = ((4 − 4)2 + (6 − 1)2 + (4 − 3)2 = 26; The triangle is an isosceles triangle.
p
d(P1 , P2 ) = p((1 − 1)2 + (1 − 1)2 + (1 − (−1))2 = 2
d(P1 , P3 ) = p(0 − 1)2 + (−1 − 1)2 + (1 − (−1))2 = 3
√
d(P2 , P3 ) = ((0 − 1)2 + (−1 − 1)2 + (1 − 1)2 = 5; The triangle is a right triangle.
p
√
d(P1 , P2 ) = p((−2 − 1)2 + (−2 − 2)2 + (−3 − 0)2 = 34
√
d(P1 , P3 ) = p(7 − 1)2 + (10 − 2)2 + (6 − 0)2 = 2 34
√
d(P2 , P3 ) = ((7 − (−2))2 + (10 − (−2))2 + (6 − (−3))2 = 3 34
Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear.
p
√
d(P1 , P2 ) = p((0 − 1)2 + (3 − 2)2 + (2 − (−1))2 = 11
√
d(P1 , P3 ) = p(1 − 1)2 + (1 − 2)2 + ((−3) − (−1))2 = 5
√
d(P2 , P3 ) = ((1 − 0)2 + (1 − 3)2 + ((−3) − 2)2 = 30
Since adding any two of the above distances will not result in the third, the points cannot be
collinear.
p
√
d(P1 , P2 ) = p((−4) − 1)2 + ((−3) − 0)2 + (5 − 4)2 = 35
√
d(P1 , P3 ) = p((−7) − 1)2 + ((−4) − 0)2 + (8 − 4)2 = 96
√
d(P2 , P3 ) = ((−7) − (−4))2 + ((−4) − (−3))2 + (8 − 5)2 = 19
Since adding any two of the above distances will not result in the third, the points cannot be
collinear.
p
√
d(P1 , P2 ) = p(1 − 2)2 + (4 − 3)2 + (4 − 2)2 = 6
√
d(P1 , P3 ) = p(5 − 2)2 + (0 − 3)2 + (−4 − 2)2 = 3 6
√
d(P2 , P3 ) = (5 − 1)2 + (0 − 4)2 + (−4 − 4)2 = 4 6
Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear.
p
√
(2 − x)2 + (1 − 2)2 + (1 − 3)2 = 21 −→ x2 − 4x + 9 = 21 −→ x2 − 4x + 4 = 16 −→
(x − 2)2 = 16 −→ x = 2 + −4 or x = 6, −2
p
(0 − x)2 + (3 − x)2 + (5 − 1)2 = 5 −→ 2x2 − 6x + 25 = 25 −→ x2 − 3x = 0 −→ x = 0, 3
22. d =
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
11.2. 3-SPACE AND VECTORS
9
1 + 7 3 + (−2) 1/2 + 5/2
,
,
= (4, 1/2, 3/2)
35.
2
2
2
0 + 4 5 + 1 −8 + (−6)
36.
,
,
= (2, 3, −7)
2
2
2
37. (x1 + 2)/2 = −1, x1 = −4, (y1 + 3)/2 = −4, y1 = −11;
The coordinates of P1 are (−4, −11, 10).
(z1 + 6)/2 = 8, z1 = 10
38. (−3 + (−5))/2 = x3 = −4, (4 + 8)/2 = y3 = 6; (1 + 3)/2 = z3 = 2.
The coordinates of P3 are (−4, 6, 2).
−3 + (−4) 4 + 6 1 + 2
,
,
= (−7/2, 5, 3/2)
(a)
2
2
2
−4 + (−5) 6 + 8 2 + 3
(b)
,
,
= (−9/2, 7, 5/2)
2
2
2
−−−→
39. P1 P2 = h−3, −6, 1i
−−−→
41. P1 P2 = h2, 1, 1i
−−−→
40. P1 P2 = h8, −5/2, 8i
−−−→
42. P1 P2 = h−3, −3, 7i
43.
44.
z
z
y
y
x
x
45.
46.
z
z
y
y
x
x
10
CHAPTER 11. VECTORS AND 3-SPACE
47. Since the k component is zero, while the i and j components are nonzero, the vector lies in
the xy-plane.
48. Since the j components is the only nonzero component, the vector lies on the y-axis.
49. Since the vector is a scalar multiple of k, the vector lies on the z-axis.
50. Since the i component is zero while the j and k components are nonzero, the vector lies in
the yz-plane.
51. a + (b + c) = h2, 4, 12i
52. 2a − (b − c) = h2, −6, 4i − h−3, −5, −8i = h5, −1, 12i
53. b + 2(a − 3c) = h−1, 1, 1i + 2h−5, −21, −25i = h−11, −41, −49i
54. 4(a + 2c) − 6b = 4h5, 9, 20i − h−6, 6, 6i = h26, 30, 74i
√
√
55. |a + c| = |h3, 3, 11i| = 9 + 9 + 121 = 139
√
√
√
56. |c||2b| = ( 4 + 36 + 81)(2)( 1 + 1 + 1) = 22 3
a
b
1
+5
=
|b| = 1 + 5 = 6
|a|
|b|
|b|
√
√
58. |b|a + |a|b = 1 + 1 + 1h1, −3, 2i + 1 + 9 + 4h−1, 1, 1i
√ √
√ √ √
√
= h 3, −3 3, 2 3i + h− 14, 14, 14i
√
√
√
√
√
√
= h 3 − 14, −3 3 + 14, 2 3 + 14i
57.
59. |a| =
60. |a| =
√
√
u=−
100 + 25 + 100 = 15;
1+9+4=
√
14;
1
h10, −5, 10i = h−2/3, 1/3, −2/3i
15
1
1
3
2
u = √ (i − 3j + 2k) = √ i − √ j + √ k
14
14
14
14
61. b = 4a = 4i − 4j + 4k
62. |a| =
√
36 + 9 + 4 = 7;
b=−
63.
z
a
1 (a+b)
2
b
y
x
1
2
1
3
3 1
h−6, 3, −2i =
,− ,
7
7 14 7
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11.2. 3-SPACE AND VECTORS
11
64. Following the hint, we first complete the following:
s
2 2 2
y1 + y2
z1 + z2
x1 + x2
− x1 +
− y1 +
− z1
2
2
2
s
2 2 2
y1 − y2
z1 − z2
x1 − x2
+
+
=
2
2
2
p
1
=
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
2
s
2 2 2
y1 + y2
z1 + z2
x1 + x2
d(M, P2 ) =
+ y2 −
+ z2 −
x2 −
2
2
2
s
2 2 2
x1 − x2
y1 − y2
z1 − z2
=
+
+
2
2
2
p
1
=
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
2
p
d(P1 , P2 ) = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
From the above, we see that d(P1 , P2 ) = d(P1 , M ) + d(M, P2 ).
Therefore, M is collinear with P1 and P2 and lies between them. Since d(P1 , M ) = d(M, P2 ),
we also have that M is equidistant from P1 and P2 . Therefore, M is the midpoint of the line
segment joining P1 and P2 .
√
3 1 √
65. xP = 1, yP = cos 30◦ + sin 30◦ =
+ ( 3 + 1),
2
√2
3
1
1 √
◦
◦
xP = − sin 30 + cos 30 = − +
= ( 3 − 1),
2
2√ 2
√
√
1 √
1 √
2 1 √
2
1 √
◦
◦
xR = cos 45 − ( 3 − 1) sin 45 =
− ( 3 − 1)
= (3 2 − 6), yr = ( 3 + 1),
2
2
4
2
√2
√2
√
1 √
1 √
2 1 √
2
◦
◦
+ ( 3 − 1)
= ( 2 + 6),
zr = sin 45 + ( 3 − 1) cos 45 =
2
2
2
2
4
√
√
√ 1 1 √
1 √
1 √
1 √
3
◦
◦
xS = (3 2 − 6) cos 60 + ( 3 + 1) sin 60 = (3 2 − 6) + ( 3 + 1)
4
2
4
2 2
2
√
√
1 √
= (3 2 − 6 + 6 + 2 3),
8
√
√
√
1 √
1 √
1 √
3 1 √
1
◦
◦
yS = − (3 2 − 6) sin 60 + ( 3 + 1) cos 60 = − (3 2 − 6)
+ ( 3 + 1)
4
2
4
2
2
2
√
√
√
1
= (−3 6 + 3 2 + 2 3 + 2),
8
√
1 √
zs = ( 2 + 6)
4
d(P1 , M ) =
Thus, xS ≈ 1.4072,
66. (a) MP =
1
0
0
yS ≈ 0.2948,
0
cos α
− sin α
0
sin α
cos α
,
zS ≈ 0.9659.
MR =
cos β
0
sin β
0 − sin β
1
0
0 cos β
12
CHAPTER 11. VECTORS AND 3-SPACE



 

x
xP
xS
(b) Mγ MR MP  y  = Mγ MR  yP  =  yS 
z
zP
zS
  


 

 
1 √0
0
1
1
1
1
0
0
1
√

3
1 
1  =  21 (√3 + 1) 
sin 30◦   1  =  0
(c) MP  1  =  0 cos 30◦
2
√2 
◦
◦
1
3
1
1
0 − sin 30 cos 30
1
0 − 12
2 ( 3 − 1)
2
√
√ 



 


2
1
1
1
0 − 22
cos 45◦ 0 − sin 45◦
2
√
√

  1 ( 3 + 1)  = 
0   21 (√3 + 1) 
0
1
0
MR MP  1  = 
 √0 1
2 √
√
◦
◦
1
1
2
2
1
sin 45
0 cos 45
0
2 ( 3 − 1)
2 ( 3 − 1)
2
2
√ 
 1 √
6)
4 (3 √2 −
=  12 √
( 3 +√1) 
1
6)
4( 2 +
√ 

 
 1 √
◦
6)
1
cos 60
sin 60◦
0
4 (3 √2 −
Mγ MR MP  1  =  sin sin 60◦ cos sin 60◦ 0   12 √
( 3 +√1) 
1
1
0
0
1
6)
4( 2 +


√
√
√
√
√
√



1
1
3
1
6)
6 + 6 +√2 3)
0
4 (3 √2 −
8 (3 √2 − √
2
 2√3

1
1
= −
( 3 +√1)  =  18 (−3 6 +√3 2 √
+ 2 3 + 2) 
0   2√
2
2
1
1
6)
6)
0
0 1
4( 2 +
4( 2 +
11.3
Dot Product
1. a · b = 2(−1) + (−3)2 + 4(5) = 12
2. b · c = (−1)3 + 2(6) + 5(−1) = 4
3. a · c = 2(3) + (−3)6 + 4(−1) = −16
4. a · (b + c) = 2(2) + (−3)8 + 4(4) = −4
5. a · (4b) = 2(−4) + (−3)8 + 4(20) = 48
6. b · (a − c) = (−1)(−1) + 2(−9) = 5(5) = 8
7. a · a = 22 + (−3)2 + 42 = 29
8. (2b) · (3c) = (−2)9 + 4(18) + 10(−3) = 24
9. a · (a + b + c) = 2(4) + (−3)5 + 4(8) = 25
10. (2a) · (a − 2b) = 4(4) + (−6)(−7) + 8(−6) = 10
2(−1) + (−3)2 + 4(5)
12
a·b
b=
h−1, 2, 5i =
h−1, 2, 5i = h−2/5, 4/5, 2i
11.
b·b
(−1)2 + 22 + 52
30
12. (c · b)a = [3(−1) + 6(2) + (−1)5]h2, −3, 4i = 4h2, −3, 4i = h8, −12, 16i
√
13. a · b = 10(5) cos(π/4) = 25 2
11.3. DOT PRODUCT
13
√
14. a · b = 6(12) cos(π/6) = 36 3
15. a · b = |a||b| cos θ = (2)(3) cos(2π/3) = 6(−1/2) = −3
√ √
16. a · b = |a||b| cos(θ) = (4)(1) cos(5π.6) = 4 − 23 = −2 3
√
√
5a · b = 3(2) + (−1)2 = 4; |a| = 10, |b| = 2 2
4
1
1
√ = √ −→ θ = arccos √ ≈ 1.11 rad ≈ 63.45◦
cos θ = √
( 10)(2 2
5
5
√
√
18. 5a · b = 2(−3) + 1(−4) = −10; |a| = 5, |b| = 5
√
−10
2
cos θ = √
= − √ −→ θ = arccos(−2/ 5) ≈ 2.68 rad ≈ 153.43◦
( 5)5
5
√
√
√
19. 5a · b = 2(−1) + 4(−1) + 0(4) = −6; |a| = 2 5, |b| = 3 2
√
−6
1
√ = − √ −→ θ = arccos(−1/ 10) ≈ 1.89 rad ≈ 108.43◦
cos θ = √
(2 5)(3 2
10
√
√
√
20. 5a · b = 12 (2) + 12 (−4) + 32 (6) = 8; |a| = 11/2, |b| = 2 14
√
8
8
√
cos θ = √
=√
−→ θ = arccos(8/ 154) ≈ 0.87 rad ≈ 49.86◦
( 11/2)(2 14)
154
17.
√
21. a and f, b and e, c and d
22. (a) a × b = 2 · 3 + (−c)2 + 3(4) = 0 −→ c = 9
(b) a · b = c(−3) + 12 (4) + c2 = c2 − 3c + 2 = (c − 2)(c − 1) = 0 −→ c = 1, 2
23. Solving the system of equations 3x1 + y1 − 1 = 0, −3x1 + 2y1 + 2 = 0 gives x1 = 4/9 and
y1 = −1/3. Thus, v = h4/9, −1/3, 1i.
24. If a and b represent adjacent sides of the rhombus, then |a| = |b|, then diagonals of the
rhombus are a + b and a − b, and
(a + b) · (a − b) = a · a − a · b + b · a − b · b = |a|2 − |b|2 = 0.
Thus, the diagonals are perpendicular.
25. Since
c·a=
b−
a·b
a·b
a·b 2
a
·a=a·a−
(a · a) = b · a −
|a| = b · a − a · b = 0,
2
2
|a|
|a|
|a|2
the vectors c and a are orthogonal.
√
√
26. a · b = 1(1) + c(1) = c + 1; |a| = 1 + c2 , |b| = 2
√
1
c+1
√ =⇒ 1 + c2 = c + 1 −→ 1 + c2 = c2 + 2c + 1 =⇒ c = 0
cos 45◦ = √ = √
2
2
1+c 2
√
√
√
√
27. |a| = 14; cos α − 1/ 14, α ≈ 74.50◦ ; cos β = 2/ 14, β ≈ 57.69◦ ; cos γ = 3/ 14,
γ = 36.70◦
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14
CHAPTER 11. VECTORS AND 3-SPACE
28. |a| = 9; cos α = 2/3, α ≈ 48.19◦ ; cos β = 2/3, β ≈ 48.19◦ ; cos γ = −1/3, γ = 109.47◦
√
29. |a| = 2; cos α = 1/2, α ≈ 60◦ ; cos β = 0, β ≈ 90◦ ; cos γ = − 3/2, γ = 150◦
√
√
√
√
30. |a| = 78; cos α = 5/ 78, α ≈ 55.52◦ ; cos β = 7/ 78, β ≈ 37.57◦ ; cos γ = 2/ 78,
γ = 76.91◦
−−→
−−→
31. Let θ be the angle between AD and AB and a be the length of an edge of an edge of the
−−→
−−→
cube. Then AD = ai + aj + ak, AB = ai and
−−→ −−→
AD · AB
a2
1
cos θ −−→ −−→ = √ √ = √
2
2
3
3a a
|AD||AB|
−−→
−→
so θ ≈ 0.955317 radian or 54.7356◦ . Letting φ be the angle between AD and AC and noting
−→
that AC = ai + aj we have
r
−−→ −→
a2 + a2
2
AD · AC
=
cos θ −−→ −→ = √ √
2 2a2
3
3a
|AD||AC|
so φ ≈ 0.61548 radian or 35.2644◦
√
√
√
32. a = h5, 7, 4i;√ |a| = 3 10; cos α = 5/3 10, α = 58.19◦ ; cos β = 7/3 10, β = 42.45◦ ;
cos γ = 4/3 10, γ ≈ 65.06◦
33. compb a = a · b/|b| = h1, −1, 3i · h2, 6, 3i/7 = 5/7
√
√
34. compa b = b · a/|a| = h2, 6, 3i · h1, −1, 3i/ 11 = 5/ 11
√
√
35. b − a = h1, 7, 0i; compa (b − a) = (b − a) · a/|a| = h1, 7, 0i · h1, −1, 3i/ 11 = −6/ 11
36. a + b = h3, 5, 6i; 2b = h4, 12, 6i; comp2b (a + b) · 2a/|2b| = h3, 5, 6i · h4, 12, 6i/14 = 54/7
√
√
−−→
−−→
−−→ −−→
−
→ a = a · OP /|OP | = (4i + 6j) · (3i + 10j)/ 109 =
37. OP = 3i + 10j; |OP | = 109; comp−
OP
√
72/ 109
√
√
−−→
−−→
−−→ −−→
−
→ a = a · OP /|OP | = h2, 1, −1i · h1, −1, 1i/ 3 = 0
38. OP = h1, −1, 1i; |OP | = 3; comp−
OP
39. (a) compb a = a · b/|b| = (−5i + 5j) · (−3i + 4j)/5 = 7
projb a = (compb a)b/|b| = 7(−3i + 4j)/5 = − 21
5 i+
28
5 j
= − 45 i
28
(b) projb⊥ a = a − projb a = (−5i + 5j) − (− 21
− 35 j
5 + 5 j)
√
√
40. (a) compb a = a · b/|b| = (4i + 2j)√· (−3i + j)/ √
10 = − 10
projb a = (compb a)b/|b| = − 10(−3i + j)/ 10 = 3i − j
(b) projb⊥ a = a − projb a = (4i + 2j) − (3i − j) = i + 3j
41. (a) compb a = a · b/|b| = (−i − 2j + 7k) · (6i − 3j − 2k)/7 = −2
6
4
projb a = (compb a)b/|b| = −2(6i − 3j − 2k)/7 = − 12
7 i + 7j + 7k
6
4
5
(b) projb⊥ a = a − projb a = (−i − 2j + 7k) − (− 12
7 i − 7 j + 7 k) = 7 i −
20
7 j
+
45
7 k
11.3. DOT PRODUCT
15
42. (a) compb a = a · b/|b| = h1, 1, 1i · h−2, 2, −1i/3 = −1/3
projb a = (compb a)b/|b| = − 13 h−2, 2, −1i/3 = h2/9, −2/9, 1/9i
(b) projb⊥ a = a − projb a = h1, 1, 1i − h2/9, −2/9, 1/9i = h7/9, 11/9, 8/9i
43. a + b = 3i + 4j; |a + b| = 5; compa+b a = a · (a + b)/|a + b| = (4i + 3j) · (3i + 4j)/5 =
72
96
24/5; proj(a+b) a = (comp(a+b) a)(a + b)/|a + b| = 24
5 (3i + 4j)/5 = 25 i + 25 j
√
√
√
44. a − b = 5i + 2j; |a − b| = 29; comp(a−b) b = b · (a − b)/|a − b| = (−i + j)/ 29 = −3/ 29
√
3
15
6
proj(a−b) b = (comp(a−b) )(a − b)/|a − b| = − √ (5i + 2j)/ 29 = − i − j
29
29
29
15
6
14
35
proj(a−b)⊥ b = b − proj(a−b) b = (−i + j) − − i − j = − i + j
29
29
29
29
45. We identify F = 29, θ = 60◦ and |d| = 100. Then W = |F||d| cos θ = 20(100)( 12 ) = 1000 ft-lb.
46. W = F · d = |F||d| cos θ Since the force is acting at a 45◦ angle to the direction of motion,
we have θ = 45◦ . Therefore,
√
W = (3000)(400) cos 45◦ = 600, 000 2 ft-lb.
47. We identify d = −i + 3j + 8k. Then W = F · d = h4, 3, 5i · h−1, 3, 8i = 45 N-m.
48. (a) Since w and d are orthogonal, W = w · d = 0.
√
(b) We identify θ = 0◦ . Then W = |F||d| cos θ = 30( 42 + 32 ) = 150 N-m.
78
49. Using d = 6i + 2j and F = 3 53 i + 45 j , W = F · d = 95 , 12
5 · h6, 2i = 5 ft-lb.
50. Let a and b be vectors from the center of the carbon atom to the center of two distinct
hydrogen atoms. The distance between two hydrogen atoms is then
p
√
|b − a| = (b − a) · (b − a) = b · b − 2a · b + a · a
p
= |b|2 + |a|2 − 2|a||b| cos θ
p
= (1.1)2 + (1.1)2 − 2(1.1)(1.1) cos 109.5◦
p
= 1.21 + 1.21 − 2.42(−0.333807) ≈ 1.80 angstroms.
51. If a and b are orthogonal, then a · b = 0 and
a 1 b1
a2 b2
a3 b3
+
+
a b
a b
a b
1
1
=
(a1 b1 + a2 b2 a3 b3 ) =
(a · b) = 0.
|a||b|
|a||b|
cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 =
52. We want cos α = cos β = cos γ or a1 = a2 = a3 . Letting a1 = a2 = a3 = 1 we obtain the
1
1
1
vector i + j + k. A unit vector in the same direction is √ i + √ j + √ k.
3
3
3
16
CHAPTER 11. VECTORS AND 3-SPACE
53. For the following, let a = ha1 , a2 , a3 i, b = hb1 , b2 , b3 i, and let k be any scalar.
Proof of (i): If a = 0 = h0, 0, 0i, then a · b = (0)b1 + (0)b2 + (0)b3 = 0. Similarly, if b = 0,
then a · b = a1 (0) + a2 (0) + a3 (0) = 0
Proof of (ii): Using the Commutative Property of real numbers, we have
a · b = a1 b1 + a2 b2 + a3 b3
= b1 a1 + b2 a2 + b3 a3
=b·a
Proof of (iv): a · (kb) = ha1 , a2 , a3 i · hkb2 , kb2 , kb3 i
= a1 kb1 + a2 kb2 + a3 kb3
= k(a1 b1 + a2 b2 + a3 b3 )
= k(a · b)
= hla1 , ka2 , ka3 i · hb1 , b2 , b3 i
= ka1 b1 + ka2 b2 + ka3 b3
= k(a1 b1 + a2 b2 + a3 b3 )
= k(a · b)
Therefore, a · (kb) = (ka) · b = k(a · b)
Proof of (v): Since x2 ≥ 0 for any real number x, we have a · a = a21 + a22 + a23 ≥ 0.
54. Using the fact that | cos θ| < 1, we have |a · b| = |a||b|| cos θ| ≤ |a||b|.
55. |a + b|2 = (a + b) · (a + b) = a · a + 2a · b + b · b = |a|2 + 2a · b + |b|2
≤ |a|2 + 2|a · b| + |b|2
since x ≤ |x|
≤ |a|2 + 2|a||b| + |b|2 = (|a| + |b|)2 by Problem 54
Thus, since |a + b| and |a| + |b| are positive, |a + b| ≤ |a| + |b|.
56. Let P1 (x1 , y1 ) and P2 (x2 , y2 ) be distinct points on the line ax + by = −c. Then
−−−→
n·P1 P2 = ha, bi·hx2 −x1 , y2 −y1 i = ax2 −ax1 = by2 −by1 (ax2 +by2 )−(ax1 +by1 ) = −c−(−c) = 0,
and the vectors are perpendicular. Thus, n is perpendicular to the line.
−−−→
57. Let θ be the angle between n and P2 P1 . Then
−−−→
−−−→
|ha, bi · hx1 − x2 , y1 − y2 i|
|ax1 − ax2 + by1 = by2 |
|n · P2 P1 |
√
√
=
=
d = ||P1 P2 | cos θ| =
2
2
|n|
a +b
a2 + b2
|ax1 + by1 − (ax2 + by2 )|
|ax1 + by1 − (−c)|
|ax1 + by1 + c|
√
√
√
=
=
=
.
2
2
2
2
a +b
a +b
a 2 + b2
58. (a) Since N = hx, yi is a unit normal, T = h−y, xi is a unit tangent at P (x, y). Now we
www.elsolucionario.org
11.4. CROSS PRODUCT
17
compute
−−→
N · P O = hx, yi · hc − x, d − yi = cx + dy − (x2 + y 2 ) = cx + dy − 1
−−→
T · OP = h−y, xi · hc − x, d − yi = dx − cy
−→
N · P S = hx, yi · ha − x, b − yi = ax + by − (x2 + y 2 ) = ax + by − 1
−→
−T · P S = hy, −xi · ha − x, b − yi = ay − bx.
Now, since |N| = |T| = 1,
−−→
−→
−−→
−→
N · PO
N · PS
T · PO
−T · P S
=
cos
θ
=
,
and
=
cos
φ
=
−−→
−→
−−→
−→ ,
|P O|
|P S|
|P O|
|P S|
we have
−−→
−→
N · PO
N · PS
cx + dy − 1
ax + by − 1
=
−−→
−→ or dx − cy = ay − bx .
TP O
−TP S
(b) With a = 2, b = 0, c = 0, and d = 3 the equation in (a) becomes (3y − 1)/3x =
2
2
(2x − 1)/2y or
− 2y. Substituting
y 2 = 1 − x2 we obtain 6x2 − 3x =
√ 6x − 3x = 6y
√
2
2
2
6(1 − x ) − 2 1 − x or 12x − 3x − 6 = −2 1 − x2 . Squaring both sides, we obtain
144x4 − 72x3 + 36x + 32 = 0.
(c) Newton’s method gives us the roots −0.6742, −0.48302, ; 0.76379, and 0.89343. Since S
and O are on the√positive x- and y-axes, respectively,√we can ignore the negative roots.
Computing y = 1 − 0.763792 ≈ 0.645465 and y = 1 − 0.893432 ≈ 0.449202 we see
that only P (x, y) = (0.76379, 0.645465) satisfies (3y − 1)/3x = (2x − 1)/2y.
11.4
Cross Product
1. a × b =
i
1
0
j
−1
3
k
0
5
=
−1
3
2. a × b =
i
2
4
j
1
0
k
0
−1
=
1
0
3. a × b =
i
1
2
j
−3
0
k
1
4
=
−3
0
4. a × b =
i
1
−5
j
1
2
k
1
3
=
1
2
5. a × b =
i
2
−1
j
−1
3
k
2
−1
=
1
0
0
5
i−
2
4
0
−1
1
i−
4
1
2
1
4
0
i−
5
0
−1
1
3
−1
3
i−
2
−1
1
−5
i−
2
4
j+
j+
−1
3
1
0
j+
1
2
k = −5i − 5j + 3k
1
k = −i + 2j − 4k
0
−3
0
1
j+
3
1
−5
2
−1
2
j+
−1
k = h−12, −2, 6i
1
k = h1, −8, 7i
2
2
−1
−1
3
k = −5i + 5k
18
CHAPTER 11. VECTORS AND 3-SPACE
6. a × b =
i
4
2
7. a × b =
i
1/2
4
8. a × b =
i
0
2
9. a × b =
i
2
−3
10. a × b =
i
8
1
j
1
3
k
−5
−1
j
k
0 1/2
6
0
j
5
−3
j
2
−3
j
1
−2
k
0
4
=
k
−6
10
0
6
5
−3
=
k
−4
6
−5
−1
1
3
=
=
=
4
2
−5
−1
j+
4
2
i−
1/2
4
1/2
0
j+
i−
1/2
0
0
2
0
i−
4
2 −4
−3 6
1
−2
−6
10
i−
i j
2 7
1 1
is perpendicular
13. a × b =
k
−2 4
4 =
−1 0
0
both a and b.
8
1
0
1
2
1
−4
−1
i−
−1
4
−4
6
−2
−3
1
2
0
k = h−3, 2, 3i
6
2
−3
2
−3
8
1
1
−2
k = h−2, −86, −17i
j+
0
1
2
1
4
j+
0
i
j
k
−2 1
5 1
5 −2 1 =
i−
j+
0 −7
2 −7
2 0 −7
a · (a × b) = h5, −2, 1i · h14, 37, 4i = 70 − 74 + 4 = 0
b · (a × b) = h2, 0, −7i · h14, 37, 4i = 28 + 0 − 28 = 0
15. a × b =
1/2
4
j+
−4
1
j+
j+
k = 14i − 6j + 10k
5
k = h20, 0, −10i
−3
−6
j+
10
i−
i−
k
7 −4
−4 =
i−
1 −1
−1
to both a and b.
i
j
−1 −2
4 −1
is perpendicular to
14. a × b =
1
2
0
2
j+
2
−3
i−
−−−→
−−−→
11. P1 P2 = (−2, 2, −4); P1 P3 = (−3, 1, 1)
i
j k
−−−→ −−−→
2 −4
P1 P2 × P1 P3 = −2 2 −4 =
1 1
−3 1 1
−−−→
−−−→
12. P1 P2 = (0, 1, 1); P1 P3 = (1, 2, 2)
i j k
−−−→ −−−→
1
P1 P2 × P1 P3 = 0 1 1 =
2
1 2 2
0
4
1
3
−2
−3
k = h0, 0, 0i
2
k = 6i + 14j + 4k
1
1
k=j−k
2
7
1
k = −3i − 2j − 5k
−1
4
−2
−1
k = h4, 16, 9i
5
2
−2
0
k = h14, 37, 4i
11.4. CROSS PRODUCT
19
i
j
k
1/2 −1/4
−1/4 −4
1/2 −1/4 −4 =
k
i+
2
−2
−2
6
2
−2
6
19
1
= − i − 11j − k
2
2
1
1
19
1
19 11
a · (a × b) =
i − j − 4k) · (− i − 11j − k = − +
+2=0
2
4
2
2
4
4
19
1
b · (a × b) = (2i − 2j + 6k) · − i − 11j − k = −19 + 22 − 3 = 0
2
2
16. a × b =
17. (a) b × c =
i
2
3
j
1
1
a × (b × c) =
k
1
1
i
1
0
=
j
−1
1
1
1
1
1
k
2
−1
−1
1
=
1
1
j+
2
3
1
1
2
−1
i−
1
0
2
−1
2
3
i−
k=j−k
j+
1
0
−1
1
k = −i + j + k
(b) a · c = (i − j + 2k) · (3i + j + k) = 4; (a · c)b = 4(2i + j + k) = 8i + 4j + 4k
a · b = (i − j + 2k) · (2i + j + k) = 3; (a · b)c = 3(3i + j + k) = 9i + 3j + 3k
a × (b × c) = (a · c)b − (a · b)c = (8i + 4j + 4k) − (9i + 3j + 3k) = −i + j + k
18. (a) b × c =
i
1
−1
a × (b × c) =
j k
2 −1
5 8
i
3
21
j
0
−7
=
2
5
k
−4
7
−1
8
=
i−
0
−7
1
−1
−4
7
−1
8
i−
j+
1
−1
3 −4
21 7
2
k = 21i − 7j + 7k
5
j+
3
21
0
−7
k
= −28i − 105j − 21k
(b) a · c = (3i − 4k) · (−i + 5j + 8k) = −35; (a · c)b = −35(i + 2j − k) = −35i − 70j + 35k
a · b = (3i − 4k) · (i + 2j − k) = 7; (a · b)c = 7(−i + 5j + 8k) = −7i + 35j + 56k
a×(b×c) = (a·c)b−(a·b)c = (−35i−70j+35k)−(−7i+35j+56k) = −28i−105j−21k
19. (2i) × j = 2(i × j) = 2k
20. i × (−3k) = −3(i × k) = −3(−j) = 3j
21. k × (2i − j) = k × (2i) + k × (−j) = 2(k × i) − (k × j) = 2j − (−i) = i + 2j
22. i × (j × k) = i × i = 0
23. [(2k) × (3j)] × (4j) = [2 · 3(k × j) × (4j)] = 6(−i) × 4j = (−6)(4)(i × j) = −24k
24. (2i − j + 5k) × i = (2i × i) + (−j × i) + (5k × i) = 2(i × i) + (i × j) + 5(k × i) = 5j + k
25. (i + j) × (i + 5k) = [(i + j) × i] + [(i + j) × 5k] = (i × i) + (j × i) + (i × 5k) + (j × 5k)
= −k + 5(−j) + 5i = 5i − 5j − k
26. i × k − 2(j × i) = −j − 2(−k) = −j + 2k
www.elsolucionario.org
20
CHAPTER 11. VECTORS AND 3-SPACE
27. k · (j × k) = k · i = 0
28. i · [j × (−k)] = i · [−(j × k)] = i · (−i) = −(i · i) = −1
√
29. |4j − 5(i × j)| = |4j − 5k| = 41
30. (i × j) · (3j × i) = k · (−3k) = −3(k · k) = −3
31. i × (i × j) = i × k = −j
32. (i × j) × i = k × i = j
33. (i × i) × j = 0 × j = 0
34. (i · i)(i × j) = 1(k) = k
35. 2j · [i × (j − 3k)] = 2j · [(i × j) + (i × (−3k)] = 2j · [k + 3(k × i)] = 2j · (k + 3j) = 2j · k + 2j · 3j
= 2(j · k) + 6(j · j) = 2(0) + 6(1) = 6
36. (i × k) × (j × i) = (−j) × (−k) = (−1)(−1)(j × k) = j × k = i
37. a × (3b) = 3(a × b) = 3(4i − 3j + 6k) = 12i − 9j + 18k
38. b × a = −a × b = −(a × b) = −4i + 3j − 6k
39. (−a) × b = −(a × b) = −4i + 3j − 6k
p
√
40. |a × b| = 42 + (−3)2 + 62 = 61
41. (a × b) × c =
i
4
2
j
−3
4
k
6
−1
=
−3 6
4 −1
i−
4
2
6
−1
j+
4
2
−3
4
k = −21i + 16j + 22k
42. (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10
43. a · (b × c) = (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10
44. (4a) · (b × c) = (4a × b) · c = 16(2) + (−12)4 + 24(−1) = −40
−−→
45. (a) Let A = (1, 3, 0), B = (2, 0, 0), C = (0, 0, 4), and D = (1, −3, 4). Then AB = i − 3j,
−→
−−→
−−→
−−→
−−→
AC = −i − 3j + 4k, CD = i − 3j, and BD = −i − 3j + 4k. Since AB = CD and
−→ −−→
AC = BD, the quadrilateral is a parallelogram.
(b) Computing
k
0 = −12i − 4j − 6k
4
√
we find that the area is | − 12i − 4j − 6k| = 144 + 16 + 36 = 14.
−−→ −→
AB × AC =
i
1
−1
j
−3
−3
46. (a) Let A = (3, 4, 1), B = (−1, 4, 2), C = (2, 0, 2), and D = (−2, 0, 3). Then
−−→
−→
−−→
−−→
AB = −4i + k, AC = −i − 4j + k, CD = −4i + k, and BD = −i − 4j + k. Since
−−→ −−→
−→ −−→
AB = CD and AC = BD, the quadrilateral is a parallelogram.
11.4. CROSS PRODUCT
21
(b) Computing
i
−4
−1
j k
0 1 = 4i + 3j + 16k
−4 1
√
√
we find that the area is |4i + 3j + 16k| = 16 + 9 + 256 = 281 ≈ 16.76.
−−→ −→
AB × AC =
−−−→
−−−→
47. P1 P2 = j; P2 P3 = −j + k
i
j k
−−−→ −−−→
P 1 P2 × P2 P 3 = 0 1 0
0 −1 1
A = 21 |i| = 12 sq. unit
=
1
−1
0
i−
1
−−−→
−−−→
48. P1 P2 = j + 2k; P2 P3 = 2i + j − 2k
i j k
−−−→ −−−→
1 2
P 1 P2 × P2 P 3 = 0 1 2 =
1 −2
2 1 −2
A = 21 | − 4i + 4j − 2k| = 3 sq. units
i−
−−−→
−−−→
49. P1 P2 = −3j − k; P2 P3 = −2i − k
i
j
k
−−−→ −−−→
0 −3 −1 =
P1 P2 × P2 P3 =
−2 0 −1
A = 21 |3i + 2j − 6k| = 72 sq. units
−1
−1
−−−→
−−−→
50. P1 P2 = −i + 3k; P2 P3 = 2i + 4j − k
i
j k
−−−→ −−−→
0
P1 P2 × P2 P3 = −1 0 3 =
4
2 4 −1
A = 12 | − 12i + 5j − 4k| =
√
185
2
−3
0
3
−1
i−
0
0
1
−1
0
0
0
1
0
2
2
−2
j+
0
2
i−
0
−2
−1
−1
j+
−1
2
j+
3
−1
j+
1
1
k=i
k = −4i + 4j − 2k
0
−2
−1
2
−3
0
k = 3i+2j−6k
0
k = −12i + 5j − 4k
4
sq. units
i
j k
−1 4
−1 0
4 0
−1 4 0 =
k = 8i + 2j − 10k
j+
i−
2 2
2 2
2 2
2 2 2
v = |a · (b × c)| = |(i + j) · (8i + 2j − 10k)| = |8 + 2 + 0| = 10 cu. units
51. b × c =
i j k
4 1
1 1
1 4
1 4 1 =
i−
j+
k = 19i − 4j − 3k
1 5
1 5
1 1
1 1 5
v = |a · (b × c)| = |(3i + j + k) · (19i − 4j − 3k)| = |57 − 4 − 3| = 50 cu. units
52. b × c =
j k
6 −6
−2 −6
−2 6
6 −6 =
i− 5
j+ 5
k = 21i − 14j − 21k
1
1
3
3
1
5
2
2
2
2
3
2
2
a · (b × c) = (4i + 6j) · (21i − 14j − 21k) = 84 − 84 + 0 = 0. The vectors are coplanar.
53. b × c =
i
−2
22
CHAPTER 11. VECTORS AND 3-SPACE
i
j k
1 1
−2 1
−2 1
−2 1 1
= 3
j+
k = − 27 i − 4j − 3k
i−
3
0
−2
0
−2
3
2
2
0
−2
2
7
7
a · (b × c) = (i + 2j − 4k) · (− 2 i − 4j − 3k) = − 2 − 8 + 12 = 0. The vectors are not coplanar.
54. b × c =
−−−→
−−−→
55. The four points will be coplanar if the three vectors P1 P2 = h3, −1, −1i, P2 P3 = h−3, −5, 13i,
−−−→
and P3 P4 = h−8, 7, −6i are coplanar.
i
j
k
−−−→ −−−→
−3 −5
−3 13
−5 13
k = h−61, −122, −61i
j+
i−
P2 P3 ×P3 P4 = −3 −5 13 =
−8 7
−8 −6
7 −6
−8 7 −6
−−−→ −−−→ −−−→
P1 P2 · (P2 P3 × P3 P4 ) = h3, −1, −1i · h−61, −122, −61i = −183 + 122 + 61 = 0
The four points are coplanar.
−−−→
−−−→
56. The four points will be coplanar if the three vectors P1 P2 = h−3, 3, −1i, P2 P3 = h1, 2, −6i,
−−−→
and P3 P4 = h4, −6, 5i are coplanar.
i
j
k
−−−→ −−−→
1 2
1 −6
2 −6
k = h−26, −29, −14i
j+
i−
P2 P3 × P3 P4 = 1 2 −6 =
4 −6
4 5
−6 5
4 −6 5
−−−→ −−−→ −−−→
P1 P2 · (P2 P3 × P3 P4 ) = h−3, 3, −1i · h−26, −29, −14i = 78 − 87 + 14 = 5
The four points are not coplanar.
57. (a) Since θ = 90◦ , |a × b| = |a||b|| sin 90◦ | = 6.4(5) = 32.
(b) The direction of a × b is into the fourth quadrant of the xy-plane or to the left of the
plane determined by a and b as shown in Figure 11.4.9 in the text. It makes an angle
of 30◦ with the positive x-axis.
√
√
(c) We identify n = ( 3i − j)/2. Then a × b = 32n = 16 3i − 16j.
√
√ √
58. Using Definition 11.4, a × b = 27(8) sin 120◦ n = 24 3( 3/2)n = 36n. By the right-hand
rule, n = j or n = −j. Thus, a × b = 36j or −36j.
59. b × c =
i
4
7
j
5
8
k
6
3
=
5
8
6
3
i−
4
7
6
3
j+
4
7
5
8
k = h−30, 30, −3i
a×b=
i
1
4
j
2
5
k
3
6
=
1
5
3
6
i−
1
4
3
6
j+
1
4
2
5
k = h−3, 6, −3i
3
−3
i−
1
−33
a×(b×c) =
i
1
−33
j
k
2
3
30 −3
=
2
30
i
j k
6 −3
−3 6 −3 =
8 3
7 8 3
Therefore, a × (b × c) 6= (a × b) × c
(a × b) × c =
i−
−3
7
3
1
j+
−3
−30
−3
3
j+
−3
7
2
k = h−96, −93, 96i
30
6
k = h42, −12, −66i
8
www.elsolucionario.org
11.4. CROSS PRODUCT
23
60. For the following, let a = ha1 , a2 , a3 i, b = hb1 , b2 , b3 i, and c = hc1 , c2 , c3 i be arbitrary vectors.
Proof of (iv):
(a + b) × c = −c × (ab )
by property (ii)
= (−c × a) + (−c × b)
= (a × c) + (b × c)
by property (iii)
by property (ii)
Proof of (v): Let k be any scalar. Then
i
a1
kb1
a × (kb) =
j
a2
kb2
k
a3
kb3
= (a2 kb3 − a3 kb2 )i − (a1 kb3 − a3 kb1 )j + (a1 kb2 − a3 kb1 )k
= k(a2 b3 − a3 b2 )i − k(a1 b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k
i
ka1
b1
(ka) × b =
j
ka2
b2
k
ka3
b3
= (ka2 b3 − ka3 b2 )i − (ka1 b3 − ka3 b1 )j + (ka1 b2 − ka2 b1 )k
= k(a2 b3 − a3 b2 )i − k(a − 1b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k
Using Definition 11.4.1, we have
k(a × b) = k[(a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k]
= k(a2 b3 − a3 b2 )i − k(a1 b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k
Therefore, a × (kb) = (ka) × b = k(a × b).
Proof of (vii): From Equation 11.4.7, we have
a · (a × b) =
a1
a1
b1
a2
a2
b2
a3
a3
b3
However, using Property (ii) of determinants in Appendix I in the text, we see that a·(a×b =
0.)
Proof of (viii): From Equation 11.4.7, we have
b · (a × b) =
b1
a1
b1
b2
a2
b2
b3
a3
b3
However, using Property (ii) of determinants in Appendix I in the text, we see that b × (a ×
b) = 0.
61. Using equation 9 in the text,
a · (b × c) =
a1
b1
c1
a2
b2
c2
a3
b3
c3
and (a × b) · c = c · (a × b) =
c1
a1
b1
c2
a2
b2
c3
a3
b3
.
24
CHAPTER 11. VECTORS AND 3-SPACE
The second determinant can be obtained from the first by an interchange of the second and
third rows followed by an interchange of the new first and second rows. Using Property (iii)
of determinates in Appendix I in the text, we see that a · (b × c) = (a × b) · c.
62. b × c = (b2 c3 − b3 c2 )i − (b1 c3 − b3 c1 )j + (b1 c2 − b2 c1 )k
a × (b × c) = [a2 (b1 c2 − b2 c1 ) + a3 (b1 c3 − b3 c1 )]i − [a1 (b1 c2 − b2 c1 ) − a3 (b2 c3 − b3 c2 )]j
+ [−a1 (b1 c3 − b3 c1 ) − a2 (b2 c3 − b3 c2 )]k
= (a2 b1 c2 − a2 b2 c1 + a3 b3 c1 )i − (a1 b1 c2 − a2 b2 c1 − a3 b2 c3 + a3 b3 c2 )j
− (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k
(a × c)b − (a · b)c = (a1 c1 + a2 c2 + a3 c3 )(b1 i + b2 j + b3 k) − (a1 b1 + a2 b2 + a3 b3 )(c1 i + c2 j + c3 k)
= (a2 b1 c2 − a2 b2 c1 + a3 b1 c3 − a3 b3 c1 )i − (a1 b1 c2 − a2 b2 c1 − a3 b2 c3 + a3 b3 c2 )j
− (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k
63. a × (b × c) + b × (c × a) + c × (a × b)
= (a · c)b − (a · b)c − (b · c)a + (c · b)a − (c · a)b
= [(a · c)b − (c · a)b] + [(b · a)c − (a · b)c] + [(c · b)a − (b · c)a] = 0
64. If either a, b or c is the zero vector, the result is trivial. Therefore, assume all three are
nonzero. If b is a scalar multiple of c, then b × c = 0 and a · (b × c) = a · 0 = 0
If b is not a scalar multiple of c, then b × c is orthogonal to the plane containing b and c.
This implies b × c is orthogonal to a since a lies in the same plane as b and c. Hence, by
Theorem 11.3.3, we have a · (b × c) = 0.
65. (a) We first note that a × b = k, b × c = 21 (i − k), c × a =
1
1
1
2 , b · (c × a) = 2 , and c · (a × b) = 2 . Then
A=
1
2 (i
− k)
1
2
= i − k, B =
1
2 (j
− k)
1
2
= j − k,
1
2 (j
− k), a · (b × c) =
and C =
k
1
2
= 2k.
(b) We need to compute A · (B × C). Using the formula from Problem 62 we have
(c × a) × (a × b)
([c × a) · b]a − [(c × a) · a]b
=
[b · (c × a)][c × (a × b)]
[b · (c × a)][c × (a × b)]
a
=
since (c × a) · a = 0.
c · (a × b
B×C=
Then
A · (B × C) =
b×c
a
1
·
=
a · (b × c) c · (a × b)
c · (a × b)
and the volume of the unit cell of the reciprocal lattice is the reciprocal of the volume
of the unit cell of the original lattice.
11.5
Lines in 3-Space
1. hx, y, zi = h4, 6, −7i + th3, 12 , − 32 i
11.5. LINES IN 3-SPACE
25
2. hx, y, zi = h1, 8, −2i + th−7, −7, 0i
3. hx, y, zi = h0, 0, 0i + th5, 9, 4i
4. hx, y, zi = h0, −3, 10i + th12, −5, −6i
−−→
−−→
The equation of a line through P1 and P2 is 3-space with r1 = OP1 and r2 = OP2 can be
expressed as r = r1 + t(ka) or r = r2 + t(ka) where a = r2 − r1 and k is any non-zero scalar.
Thus, the form of the equation of a line is not unique. (See the alternative solution to Problem
5.)
5. a = h1 − 3, 2 − 5, 1 − (−2)i = h−2, −3, 3i;
hx, y, zi = h1, 2, 1i + th−2, −3, 3i
Alternate Solution: a = h−31, 5 − 2, −2 − 1i = h2, 3, −3i;
6. a = h0 − (−2), 4 − 6, 5 − 3i = h2, −2, 2i;
hx, y, zi = h3, 5, −2i + th2, 3, −3i
hx, y, zi = h0, 4, 5i + th−2, −2, 2i
7. a = h1/2 − (−3/2), −1/2 − 5/2, 1 − (−1/2)i = h2, −3, 3/2i;
hx, y, zi = h1/2, −1/2, 1i + th2, −3, 3/2i
8. a = h10 − 5, 2 − (−3), 10 − 5i = h5, 5, −15i;
hx, y, zi = h10, 2, −10i + th5, 5, −15i
9. a = h1 − (−4), 1 − 1, −1 − (−1)i = h5, 0, 0i;
hx, y, zi = h1, 1, −1i + th5, 0, 0i
10. a = h3 − 5/2, 2 − 1, 1 − (−2)i = h1/2, 1, 3i;
hx, y, zi = h3, 2, 1i + th1/2, 1, 3i
11. a = h2 − 6, 3 − (−1), 5 − 8i = h−4, 4, −3i; x = 2 − 4t, y = 3 + 4t, z = 5 − 3t
12. a = h2 − 0, 0 − 4, 0 − 9i = h2, −4, −9i; x = 2 + 2t, y = −4t, z = −9t
13. a = h1 − 3, 0 − (−2), 0 − (−7)i = h−2, 2, 7i; x = 1 − 2t, y = 2t, z = 7t
14. a = h0 − (−2), 0 − 4, 5 − 0i = h2, −4, 5i; x = 2t, y = −4t, z = 5 + 5t
15. a = h4−(−)6, 1/2−(−1/4), 1/3−1/6i = h10, 3/4, 1/6i; x = 4+10t, y =
1 3
1 1
+ t, z = + t
2 4
3 6
16. a = h−3 − 4, 7 − (−8), 9 − 1(−1)i = h−7, 15, 10i; x = −3 − 7t, y = 7 + 15t, z = 9 + 10t
17. a1 = 10 − 1 = 9, a2 = 14 − 4 = 10, a3 = −2 − (−9) = 7;
x − 10
y − 14
z+2
=
=
9
10
7
18. a1 = 1 − 2/3 = 1/3, a2 = 3 − 0 = 3, a3 = 1/4 − (1/4) = 1/2;
19. a1 = −7 − 4 = −11, a2 = 2 − 2 = 0, a3 = 5 − 1 = 4;
20. a1 = 1 − (−5) = 6, a2 = 1 − (−2), a3 = 2 − (−4);
x−1
y−3
z − 1/4
=
=
1/3
3
1/2
x−7
z−5
=
, y=2
−11
4
x−1
y−1
z−2
=
=
6
3
6
21. a1 = 5 − 5 = 0, a2 = 10 − 1 = 9, a3 = −2 − (−14) = 12; x = 5,
y − 10
z+2
=
9
12
www.elsolucionario.org
26
CHAPTER 11. VECTORS AND 3-SPACE
22. a1 = 5/6 − 1/3 = 1/2, a2 = −1/4 − 3/8 = 5/8, a3 = 1/5 − 1/10 = 1/10;
x − 5/6
y + 1/4
z − 1/5
=
=
1/2
−5/8
1/10
23. Writing the given line in the form x/2 = (y − 1)/(−3) = (z − 5)/6, we see that a direction
vector is h2, −3, 6i. Parametric equations for the lines are x = 6+2t, y = 4−3t, z = −2+6t.
24. A direction vector is h5, 1/3, −2i. Symmetric equations for the line are
(x − 4)/5
= (y + 11)(1/3) = (z + 7)/(−2).
25. A direction vector parallel to both the xy- and xy-planes is i = h1, 0, 0i. Parametric equations
for the line are x = 2 + t, y = −2, z = 15.
26. (a) Since the unit vector j = h0, 1, 0i lies along the y-axis, we have x = 1, y = 2 + t, z = 8.
(b) Since the unit vector k = h0, 0, 1i is perpendicular to the xy-plane, we have z = 1,
y = 2, z = 8 + t.
27. Both lines go through the points (0, 0, 0) and (6, 6, 6). Since two points determine a line, the
lines are the same.
28. The direction vector of line L1 is v1 = h3, 6, −9i. The direction vector of line
L2 is v2 = h−1, −2, 3i. Since v1 = −3v2 , lines L1 and L2 are parallel. Hence, if we can find
a point that lies on both lines, then they must be parallel. Letting t = 0 for L1 and t = 3 for
L2 , we see that the point (2, −5, 4) lies on both lines. Therefore L1 and L2 are the same.
−7 − 3
= −5.
29. (a) Equating the x components, we have x = 3 + 2t, = −7, which gives t =
2
We can check our work by plugging this value of t into the y and z components to get
y = 4 − (−5) = 9 and z = −1 + 6(−5) = −31
(b) Equating the x components, we have x = 5 − x = −7 which gives s = 5 + 7 = 12.
We can check our work by plugging this value of s into the y and z components to get
1
y = 3 + (12) = 9 and z = 5 − 3(12) = −31
2
30. a and f are parallel since h9, −12, 6i = −3h−3, 4, 2i. c and d are orthogonal since
h2, −3, 4i · h1, 4, 5/2i = 0.
31. In the xy-plane, z = 9+3t = 0 and t = −3. Then x = 4−2(−3) = 10 and y = 1+2(−3) = −5.
The point is (10, −5, 0). In the xz-plane, y = 1+2t = 0 and t = −1/2. Then x = 4−2(−1/2) =
5 and z = 9 + 3(−1/2) = 15/2. The point is (5, 0, 15/2). In the yz-plane, x = 4 − 2t = 0 and
t = 2. Then y = 1 + 2(2) = 5 and z = 9 + 3(2) = 15. The point is (0, 5, 15).
32. The parametric equations for the line are x = 1+2t, y = −2+3t, z = 4+2t. In the xy-plane,
z = 4 + 2t = 0 and t = −2. Then x = 1 + 2(−2) = −3 and y = −2 + 3(−2) = −8. The point
is (−3, −8, 0). In the xz-plane, y = −2 + 3t = 0 and t = 2/3. Then x = 1 + 2(2/3) = 7/3
and z = 4 + 2(2/3) = 16/3. The point is (7/3, 0, 16/3). In the yz-plane, x = 1 + 2t = 0 and
t = −1/2. Then y = −2 + 3(−1/2) = −7/2 and z = 4 + 2(−1/2) = 3. The point is (0, −7/2, 3).
11.5. LINES IN 3-SPACE
27
33. Solving the system 4 + t = 6 + 2s, 5 + t = 11 + 4s, −1 + 2t = −3 + s, or t − 2s = 2, t − 4s =
6, 2t − s = −2 yields s = −2 and t = −2 in all three equations. Thus, the lines intersect at
the point x = 4 + (−2) = 2, y = 5 + (−2) = 3, z = −1 + 2(−2) = −5, or (2, 3, −5).
34. Solving the system 1 + t = 2 − s, 2 − t = 1 + s, 3t = 6s, or t + s = 1, t + s = 1, t − 2s = 0
yields s = 1/3 and t = 2/3 in all three equations. Thus, the lines intersect at the point
x = 1 + 2/3 = 5/3, y = 2 − 2/3 = 4/3, z = 3(2/3) = 2, or (5/3, 4/3, 2).
35. The system of equations 2 − t = 4 + s, 3 + t = 1 + s, 1 + t = 1 − s, or t + s = −2, t − s =
−2, t + s = 0 has no solution since −2 6= 0. Thus, the lines do not intersect.
36. Solving the system 3 + t = 2 + 2s, 2 + t = −2 + 3s, 8 + 2t = −2 + 8s, or t + 2s = 1, t − 3s =
−4, 2t − 8s = −10 yields s = 1 and t = −1 in all three equations. Thus, the lines intersect
at the point x = 3 − (−1) = 4, y = 2 + (−1) = 1, z = 8 + 2(−1) = 6, or (4, 1, 6).
37. Using the first two points, we determine the line x = 4 + 6t, y = 3 + 12t, z = −5 − 6t.
Letting t = −5/6 we see that (−1, −7, 0) is on the line. Thus, the points lie on the same line.
38. Using the first two points, we determine the line x = −1 − 12t, y = 6 + 4t,
z = 6 − 8t.
6 5 when
Setting x = −2 in the first equation, we obtain t = 1/4. Since z = 6 − 8 41 = 4 =
t = 1/4, the points do not lie on the same line.
39. A direction vector for the line is h6−2, −1−5, 3−9i = h4, −6, −6i. Thus, parametric equations
for the line segment are x = 2 + 4t, y = 5 − 6t, z = 9 − 6t, where 0 ≤ t ≤ 1.
40. The midpoint of the first line segment, obtained by letting t = 3/2, is (4, 1/2, −1/2). The
midpoint of the second line segment, obtained by letting t = 0, is (−2, 6, 5). A direction vector
for the line segment connecting the midpoints is h−2−4, 6−1/2, 5−(−1/2)i = h−6, 11/2, 11/2i.
Thus, parametric equations for the line segment are x = 4 − 6t, y = 1/2 + (11/2)t,
z = −1/2 + (11/2)t, where 0 ≤ t ≤ 1.
41. a = h−1, 2, −2i,
θ = arccos
b = h2, 3, −6i,
a · b = 16,
|a| = 3,
|b| = 7;
cos θ =
a·b
16
=
;
|a||b|
3·7
16
≈ 40.37◦
21
√
√
42. a = h2, 7, −1i, b = h−2, 1, 4i, a · b = −1, |a| = 3 6, |b| = 21; a·b
1
1
1
cos θ =
=− √ √
= − √ ; θ = arccos − √ √
≈ 91.70◦
|a||b|
(3 6)( 21)
9 14
(3 6)( 21)
43. A direction vector perpendicular to the given lines will be h1, 1, 1i × h−1, 1, −5i = h−6, 3, 3i.
Equations of the lines are x = 4 − 6t, y = 1 + 3t, z = 6 + 3t.
44. The direction vectors of the given lines are h3, 2, 4i and h6, 4, 8i = 2h3, 2, 4i. These are parallel,
so we need a third vector parallel to the plane containing the lines which is not parallel to
them. The point (1, −1, 0) is on the first line and (−4, 6, 10) is on the second line. A third
vector is then h1, −1, 0i − h−4, 6, 10i = h5, −7, −10i. Now a direction vector perpendicular to
the plane is h3, 2, 4i × h5, −, 7−, 10i = h8, 50, −31i. Equations of the line through (1, −1, 0)
and perpendicular to the plane are x = 1 + 8t, y = −1 + 50t, z = −31t.
28
CHAPTER 11. VECTORS AND 3-SPACE
45. In the system −3 + t = 4 + s, 7 + 3t = 8 − 2s, 5 + 2t = 10 − 4s, or t − s = 7, 3t + 2s =
1, 2t + 4s = 5, the first and second equations have solution t = 3 and s = −4. Substituting
into the third equation, we find 2(3) = 4(−4) = 6 − 16 = −10 6= 5. The direction vectors of
the lines are h1, 3, 2i and h1, −2, −4i, so the lines are not parallel. Thus, the lines are skew.
46. In the system 6 + 2t = 7 + 8s, 6t = 4 − 4s, −8 + 10t = 3 − 24s, or 2t − 8s = 1, 6t + 4s =
4, 10t + 24s = 11, the second and third equations have solution t = 1/2 and s = 1/4.
Substituting into the first equation, we find 2(1/2) − 8(1/4) = −1 6= 1. The direction vectors
of the lines are h2, 6, 10i and h8, −4, , −24i, so the lines are not parallel. Thus, the lines are
skew.
−−−→ −−−→ −−−→ −−−→
47. The vector (P1 P2 × P3 P4 )/|P1 P2 × P3 P4 | is a unit vector perpendicular to the two planes. To
find the shortest distance between the planes we compute the absolute value of the component
−−−→
of P1 P3 on this unit vector. Then
−−−→ −−−→ −−−→
−−−→ −−−→
−−−→ P1 P2 × P3 P4
|P1 P3 · P1 P2 × P3 P4 |
d = P1 P3 · −−−→ −−−→ =
−−−→ −−−→
| P1 P2 × P3 P4 |
| P1 P2 × P3 P4 |
−−−→
48. We take P1 = (−3, 7, 5), P2 = (−2, 10, 7), P3 + (4, 8, 10), and P4 = (5, 6, 6). Then P1 P3 =
−−−→
−−−→
−−−→ −−−→
h7, 1, 5i, P1 P2 = h1, 3, 2i, P3 P4 = h1, −2, −4i, and P1 P2 × P3 P4 = h−8, 6, −5i. The distance between the lines is then
d=
11.6
√
|h7, 1, 5i · h−8, 6, −5i|
75
= √ = 3 5.
|h−8, 6, −5i|
5 5
Planes
1. 2(x − 5) − 3(y − 1) + 4(z − 3) = 0;
2x − 3y + 4z = 19
2. 4(x − 1) − 2(y − 2) + 0(z − 5) = 0;
4x − 2y = 0
3. −5(x − 6) + 0(y − 10) + 3(z + 7) = 0;
−5x + 3z = −51
4. 6x − y + 3z = 0
5. 6(x − 1/2) + 8(y − 3/4) − 4(z − 1/2) = 0;
6. −(x + 1) + (y − 1) − (z − 0) = 0;
6x + 8y − 4z = 11
−x + y − z = 2
7. From the points (3, 5, 2) and (2, 3, 1) we obtain the vector u = i + 2j + k. From the points
(2, 3, 1) and (−1, −1, 4) we obtain the vector v = 3i + 4j − 3k. From the points (−1, −1, 4)
and (x, y, z) we obtain the vector w = (x + 1)i + (y + 1)j + (z − 4)k. Then, a normal vector is
u×v =
i
1
3
j
2
4
k
1
−3
= −10i + 6j − 2k
A vector equation of the plane is −10(x + 1) + 6(y + 1) − 2(z − 4) = 0 or 5x − 3y + z = 2.
www.elsolucionario.org
11.6. PLANES
29
8. From the points (0, 1, 0) and (0, 1, 1) we obtain the vector u = k. From the points (0, 1, 1)
and (1, 3, −1) we obtain the vector v = i + 2j − 2k. From the points (1, 3, −1) and (x, y, z) we
obtain the vector w = (x − 1)i + (y − 3)j + (z + 1)k. Then, a normal vector is
u×v =
i
0
1
j
0
2
k
1
−2
= −2i + j
A vector equation of the plane is −2(x − 1) + (y − 3) + 0(z + 1) = 0 or −2x + y = 1.
9. From the points (0, 0, 0) and (1, 1, 1) we obtain the vector u = i + j + k. From the points
(1, 1, 1) and (3, 2, −1) we obtain the vector v = 2i + j − 2k. From the points (3, 2, −1) and
(x, y, z) we obtain the vector w = (x − 3)i + (y − 2)j + (z + 1)k. Then, a normal vector is
i
1
2
u×v =
j
1
1
k
1
−2
= −3i + 4j − k
A vector equation of the plane is −3(x − 3) + 4(y − 2) − (z + 1) = 0 or −3x + 4y − z = 0.
10. The three points are not collinear and all satisfy x = 0, which is the equation of the plane.
11. From the points (1, 2, −1) and (4, 3, 1) we obtain the vector u = 3i + j + 2k. From the points
(4, 3, 1) and (7, 4, 3) we obtain the vector v = 3i + j + 2k. From the points (7, 4, 3) and (x, y, z)
we obtain the vector w = (x − 7)i + (y − 4)j + (z − 3)k. Since u × v = 0, the points are
collinear.
12. From the points (2, 1, 2) and (4, 1, 0) we obtain the vector u = 2i−2k. From the points (4, 1, 0)
and (5, 0, −5) we obtain the vector v = i − j − 5k. From the points (5, 0, −5) and (x, y, z) we
obtain the vector w = (x − 5)i + yj + (z + 5)k. Then, a normal vector is
u×v =
i
2
1
j
0
−1
k
−2
−5
= −2i + 8j − 2k
A vector equation of the plane is −2(x − 5) + 8y − 2(z + 5) = 0 or x − 4y = z = 0.
13. A normal vector to x + y − 4z = 1 is h1, 1, −4i. The equation of the parallel plane is (x − 2) +
(y − 3) − 4(z + 5) = 0 or x + y − 4z = 25.
14. A normal vector to 5x − y + z = 6 is h5, −1, 1i. The equation of the parallel plane is 5(x −
0) − (y − 0) + (z − 0) = 0 or 5x − y + z = 0.
15. A normal vector to the xy-plane is h0, 0, 1i. The equation of the parallel plane is z − 12 = 0
or z = 12.
16. A normal vector is h0, 1, 0i. The equation of the plane is y + 5 = 0 or y = −5.
30
CHAPTER 11. VECTORS AND 3-SPACE
17. Direction vectors of the lines are h3, −1, 1i and h4, 2, 1i. A normal vector to the plane is
h3, −1, 1i × h4, 2, 1i = h−3, 1, 10i. A point on the first line, and thus in the plane, is (1, 1, 2).
The equation of the plane is −3(x − 1) + (y − 1) + 10(z − 2) = 0 or −3x + y + 10z = 18.
18. Direction vectors of the lines are h2, −1, 6i and h1, 1, −3i. A normal vector to the plane is
h2, −1, 6i×h1, 1, −3i = h−3, 12, 3i. A point on the first line, and thus in the plane, is (1, −1, 5).
The equation of the plane is −3(x − 1) + 12(y + 1) + 3(z − 5) = 0 or −x + 4y + z = 0.
19. A direction vector for the two lines is h1, 2, 1i. Points on the lines are (1, 1, 3) and (3, 0, −2).
Thus, another vector parallel to the plane is h1 − 3, 1 − 0, 3 + 2i = h−2, 1, 5i. A normal vector
to the plane is h1, 2, 1i × h−2, 1, 5i = h9, −7, 5i. Using the point (3, 0, −2) in the plane, the
equation of the plane is 9(x − 3) − 7(y − 0) + 5(z + 2) = 0 or 9x − 7y + 5z = 17.
20. A direction vector for the line is h3, 2, −1i. Letting t = 0, we see that the origin is on the line
and hence in the plane. Thus, another vector parallel to the plane is h4 − 0, 0 − 0, −6 − 0i =
h4, 0, −6i. A normal vector to the plane is h3, 2, −2i × h4, 0, −6i = h−12, 10, −8i. The equation
of the plane is −12(x − 0) + 10(y − 0) − 8(z − 0) = 0 or 6x − 5y + 4z = 0.
21. A direction vector for the line, and hence a normal vector for the plane, is h−3, 1, −1/2i. The
equation of the plane is −3(x − 2) + (y − 4) − 21 (z − 8) = 0 or −3x + y − 21 z = −6.
22. A normal vector to the plane is h2 − 1, 6 − 0, −3 + 2i = h1, 6, −1i. The equation of the plane
is (x − 1) + 6(y − 1) − (z − 1) = 0 or x + 6y − z = 6.
23. Normal vectors to the plane are (a) h2, −1, 3i, (b) h1, 2, 2i, (c) h1, 1, −3/2i, (d) h−5, 2, 4i,
(e) h−8, −8, 12i, (f ) h−2, 1, −2i. Parallel planes are (c) and (e), and (a) and (f ). Perpendicular planes are (a) and (d), (b) and (c), (b) and (e), and (d) and (f ).
24. A normal vector to the plane is h−7, 2, 3i. This is the direction vector for the line and the
equations of the line are x − 4 − 7t, y = 1 + 2t, z = 7 + 3t.
25. A direction vector of the line is h−6, 9, 3i, and the normal vectors of the plane are (a) h4, 1, 2i,
(b) h2, −3, 1i, (c) h10, −15, −5i, (d) h−4, 6, 2i. Vectors (c) and (d) are multiples of the
direction vector and hence the corresponding planes are perpendicular to the line.
26. A direction vector of the line is h−2, 4, 1i, and the normal vectors of the plane are (a) h1, −1, 3i,
(b) h6, −3, 0i, (c) h1, −2, 5i, (d) h−2, 1, −2i. Since the dot product of each normal vector
with the direction vector is non-zero, none of the planes are parallel to the line.
27. Letting z = t in both equations and solving 5x − 4y = 8 + 9t,
x = 2 + t, y = 21 − t, z = t.
x + 4y = 4 − 3t, we obtain
28. Letting y = t in both equations and solving x − z = 2 − 2t, 3x + 2z = 1 + t, we obtain
x = 1 − 53 t, y = t, z = −1 + 57 t or, letting t = 5s, x = 1 − 3s, y = 5s, z = −1 + 7s.
29. Letting z = t in both equations and solving 4x − 2y = 1 + t,
x = 12 − 12 t, y = 12 − 32 t, z = t.
x + y = 1 − 2t, we obtain
30. Letting z = t and using y = 0 in the first equation, we obtain x = − 21 t, y = 0, z = t.
11.6. PLANES
31
31. Substituting the parametric equations into the equation of the plane, we obtain 2(1+2t)23(2−
t) + 2(−3t) = −7 or t = −3. Letting t = −3 in the equation of the line, we obtain the point
of intersection (−5, 5, 9).
32. Substituting the parametric equations into the equation of the plane, we obtain (3 − 2t) +
(1 + 6t) − 4(2 − 21 ) = 12 or 2t = 0. Letting t = 0 in the equation of the line, we obtain the
point of intersection (3, 1, 2).
33. Substituting the parametric equations into the equation of the plane, we obtain 1+2−(1+t) =
8 or t = −6. Letting t = −6 in the equation of the line, we obtain the point of intersection
(1, 2, −5).
34. Substituting the parametric equations into the equation of the plane, we obtain 4 + t − 3(2 +
t) + 2(1 + 5t) = 0 or t = 0. Letting t = 0 in the equation of the line, we obtain the point of
intersection (4, 2, 1).
In Problems 35 and 26, the cross product of the normal vectors to the two planes will be
a vector parallel to both planes, and hence a direction vector for a line parallel to the two
planes.
35. Normal vectors are h1, 1, −4i and h2, −1, 1i. A direction vector is
h1, 1, −4i × h2, −1, 1i = h−3, −9, −3i = −3h1, 3, 1i.
Equations of the line are x = 5 + t,
y = 6 + 3t,
z = −12 + t.
36. Normal vectors are h2, 0, 1i and h−1, 3, 1i. A direction vector is
h2, 0, 1i × h−1, 3, 1i = h−3, −3, 6i = −3h1, 1, −2i.
Equations of the line are x = −3 + t, y = 5 + t, z = −1 − 2t.
In Problems 37 and 38, the cross product of the direction vector of the line with the normal
vector of the given plane will be a normal vector to the desired plane.
37. A direction vector of the line is h3, −1, 5i and a normal vector to the given plane is h1, 1, 1i. A
normal vector to the desired plane is h3, −1, 5i × h1, 1, 1i = h−6, 2, 4i. A point on the line, and
hence in the plane is h4, 0, 1i. The equation of the plane is −6(x − 4) + 2(y − 0) + 4(z − 1) = 0
or 3x − y − 2x = 10.
38. A direction vector of the line is h3, 5, 2i and a normal vector to the given plane is h2, −4, −1i.
A normal vector to the desired plane is h−3, 5, 2i × h2, −4, −1i = h3, 1, 2i. A point on the line,
and hence in the plane is h2, −2, 8i. The equation of the plane is 3(x−2)+(y +2)+2(z −8) = 0
or 3x + y + 2x = 20.
www.elsolucionario.org
32
CHAPTER 11. VECTORS AND 3-SPACE
39.
40.
z
z
10
6
y
x
6
2
5
y
x
41.
42.
z
z
2
4
y
6
4
y
x
x
-6
43.
44.
z
z
6
4
2
y
1
x
2
y
x
45. (a) A direction vector for the line is a = −2i + j − k and a normal vector for the plane is
n = i + j − k. Since a · n = −2 + 1 + 1 = 0, the line is perpendicular to n and thus parallel
11.6. PLANES
33
to the plane. Since (0, 0, 0) is on the line and (0, 0, −1) is in the plane, the line is above
the plane.
(b) A normal vector for the plane is n − 3i − 4j + 2k. Since a · n = 6 − 4 − 2 = 0, the line is
parallel to the plane. Since (0, 0, 0) is one the line and (0, 0, 4) is in the plane, the line is
below the plane.
−−−→
46. The distance D will be the absolute value of compn P0 P1 . Thus, using ax1 + by1 + cz1 = −d,
−−−→ n
hx2 − x1 , y2 − y1 , z2 − z1 i · ha, b, ci
√
=
D = P 0 P1 ·
|n|
a2 + b2 + c2
|ax2 + by2 + cz2 − (zx1 + by1 + cz1 )|
|ax2 + by2 + cz2 + d|
√
√
=
=
.
a2 + b2 + c2
a2 + b2 + c2
47. Using, Problem 46, D =
3
|1(2) − 3(1) + 1(4) − 6|
√
=√ .
1+9+1
11
48. (a) The normal vectors are n1 = i − 2j + 3k and n2 = −4i + 8j − 12k = −4n. Since n1 and
n2 are parallel, the planes are parallel.
(b) To find the distance between the planes we choose (0, 0, 1) on the first plane. Then,
using Problem 46, the distance between the planes is
D=
19
| − 4(0) + +8(0) − 12(1) − 7|
p
=√
≈ 1.27.
2
2
2
224
(−4) + 8 + (−12)
49. Normal vectors are h1, −3, 2i and h−1, 1, 1i. Then
cos θ =
−2
2
h1, −3, 2i · h−1, 1, 1i
= √ √ = −√
|h1, −3, 2i||h−1, 1, 1i|
14 3
42
√
and θ = arccos(−2/ 42) ≈ 107.98◦
50. Normal vectors are h, 2, 6, 3i and h4, −2, 4i. Then
cos θ =
h2, 6, 3i · h4, −2, 4i
8
4
=
=
|h2, 6, 3i||h4, −2, 4i|
7(6)
21
and θ = arccos(−4/21) ≈ 79.02◦
51. Let the bottoms of the table legs be represented by points in 3-space. The rocking of a fourlegged table occurs when these four points are not coplanar. Hence, not all four legs can rest
on the plane of the floor simultaneously.
However, a three-legged table cannot have this problem. Given any three points in space, a
plane can be found passing through them. Therefore, the bottoms of the legs in a three-legged
table are coplanar. This implies that they will all rest on the plane of the floor, even if the
legs are of uneven lengths.
34
CHAPTER 11. VECTORS AND 3-SPACE
52. Let n1 = h1, −1, 2i which is normal to the plane x − y + 2z = 1. Let n2 = h1, 1, 1i which is
normal to the plane x + y + z = 3. Since L is perpendicular to both n1 and n2 , L must be
parallel to n1 × n2 .
n1 × n2 =
i
1
1
j
−1
1
k
2
1
− 3i + j + 2k
Therefore, L is parallel to v = h−3, 1, 2i.
To completely determine L, we need a point which L passes through. Hence we need a point
(x, y, z) which satisfies the equations of both planes. Since it satisfies both equations, it must
satisfy their sum:
x − y + 2z = 1
x+y+z =3
2x + 3z = 4
So the coordinates of the point satisfy 2x + 3z = 4. Since v as a nonzero k-component, the
line L passes through every possible z-value. This implies the existence of a point on the
intersection of the two planes with a z-value of zero. Letting z = 0, we must have x = 2 since
2x + 3z = 4 for every point on L. Plugging x = 2 and z = 0 into the equation of the first
place, we get y = 1. Therefore (2, 1, 0) lies on the line L.
Using this point and the parallel vector v, the parametric equations of L are
x = 2 − 3t; y = 1 + t; z = 2t
To show that this answer is equivalent to that found in Example 8, first note that both lines
1
pass through (2, 1, 0). Also, the parallel vector used in Example 8 is h−3/2, 1/2, 1i = v.
2
Therefore, the two solutions are the same.
53. (a) The plane should pass through the midpoint of the
−2, 3) and
line segment joining (1, 3 3
1 + 2 −2 + 5 3 − 1
,
,
=
, ,1 .
(2, 5, −2). This is given in Problem 11.2.64 as M =
2
2
2
2 2
The vector joining (1, −2, 3) and (2, 5, −1) should be perpendicular to the plane. This
vector is n = h1, 7, −4i. Using the point (3/2, 3/2, 1) and the normal vector n, the equation of the plane is given by z + 7y − 4z = 8.
(b) The distance from the plane to either of the two points is equal to half the length of the
1p 2
1√
line segment joining the two points. This is given by
1 + 72 + (−4)2 =
66
2
2
www.elsolucionario.org
11.7. CYLINDERS AND SPHERES
11.7
35
Cylinders and Spheres
1.
2.
z
3.
z
z
y
y
y
2
x
x
x
4.
5.
6.
z
z
z
5
3
y
y
y
x
x
x
7.
8.
9.
z
z
z
y
y
1
y
x
x
x
36
CHAPTER 11. VECTORS AND 3-SPACE
10.
11.
12.
z
z
z
6
y
3
y
x
y
1
x
x
13.
14.
15.
z
z
z
y
y
y
x
x
x
16.
17.
z
18.
z
z
3
y
y
x
y
x
x
center: (0, 0, 3)
radius: 4
11.7. CYLINDERS AND SPHERES
37
19.
20.
z
z
y
x
y
x
center: (−3, −4, 5)
radius: 2
center: (1, 1, 1)
radius: 1
21. (x2 + 8x + 16) + (y 2 − 6x + 9) + (z 2 − 4z + 4) = 7 + 16 + 9 + 4
(x + 4)2 + (y − 3)2 + (z − 2)2 = 36; center: (−4, 3, 2); radius: 6
22. 4(x2 + x + 1/4) + 4y 2 + 4(z 2 − 3z + 9/4) = −9 + 1 + 9
(x + 1/2)2 + y 2 + (z − 3/2)2 = 1/4; center: (−1/2, 0, 3/2);
23. x2 + y 2 + (z 2 − 16z = 64) = 64;
2
center: (0, 0, 8);
2
2
24. (x − x + 1/4) + (y + y + 1/4)√+ z = 1/4 + 1/4;
center: (1/2, −1/2, 0); radius: 2/2
radius: 1/2
radius: 8
(x − 1/2)2 + (y + 1/2)2 + z 2 = 1/2
25. (x + 1)2 + (y − 4)2 + (z − 6)2 = 3
26. x2 + (y − 3)2 + z 2 = 25/16
27. (x − 1)2 + (y − 1)2 + (z − 4)2 = 16
28. (x − 5)2 + (y − 2)2 + (z − 2)2 = 52
29. There are two solutions: one sphere is inside the given sphere and the other is outside.
x2 + (y − 8)2 + z 2 = 4 or x2 + (y − 4)2 + z 2 = 4.
p
30. (2t)2 + (3t)2 + (6t)2 = 21; t = 3; a = 2t = 6; b = 3t = 9; c = 6t = 18
(x − 6)2 + (y − 9)2 + (z − 18)2 = 25
p
√
31. The center is at (1, 4, 2) and the radius is (1 − 0)2 + (4 + 4)2 (2 − 7)2 = 3 10. The equation
is (x − 1)2 + (y − 4)2 + (z − 2)2 = 90.
p
√
32. The radius is (−3 − 0)2 + (1 − 0)2 + (2 − 0)2 = 14. The equation is (x + 3)2 + (y − 1)2 +
(z − 2)2 = 14.
33. The upper half of the sphere x2 + y 2 + (z − 1)2 = 4; a hemisphere
34. A circle√on the sphere x2 + y 2 + (z − 1)2 = 4; the circle is parallel to the xy-plane and has
radius 3.
35. All points on and outside the unit sphere centered at the origin
www.elsolucionario.org
38
CHAPTER 11. VECTORS AND 3-SPACE
36. All points inside the sphere of radius 1 centered at (1, 2, 3), except the center
37. x2 + y 2 + z 2 = 1 represents a sphere of radius 1 and x2 + y 2 + z 2 = 9 represents a sphere
of radius 3. Therefore 1 ≤ x2 + y 2 + z 2 ≤ 9 represents the set of points lying between these
two spheres. Thus, the geometric object is a hollowed out ball with outer radius 3 and inner
radius 1.
38. This set of points is identical to the found in Problem 11.7.37, with the added restriction
z ≥ 0. This restriction will remove points with negative z-coordinates, leaving only the upper
half of the hollowed out ball.
11.8
Quadric Surfaces
2. elliptical cone
1. paraboloid
z
z
y
x
y
x
3. x2 /4 + y 2 + z 2 /9 = 1; ellipsoid
4. −x2 /4 − y 2 /4 + z 2 /4 = 1
hyperboloid of two sheets
z
z
y
x
x
y
11.8. QUADRIC SURFACES
39
6. x2 /25 + y 2 /25 + z 2 /100 = 1
ellipsoid
5. x2 /4 − y 2 /144 + z 2 /16 = 1
hyperboloid of one sheet
z
z
10
4
y
5
x
y
5
x
8. y 2 /9 − x2 /16 = z
hyperbolic paraboloid
7. elliptical cone
z
z
y
y
x
x
10. x2 + y 2 = −9z
paraboloid
9. hyperbolic paraboloid
z
z
y
y
x
x
40
CHAPTER 11. VECTORS AND 3-SPACE
12. −z 2 /9 + y 2 + z 2 /9 = 1
hyperboloid of one sheet
11. x2 /4 − y 2 /4 − z 2 /4 = 1
hyperboloid of two sheets
z
z
3
y
y
1
x
x
z2
=x
1/4
paraboloid
13. y 2 +
14. hyperboloid of one sheet
z
z
y
y
x
x
15.
16.
z
z
4
x
3
y
x
y
www.elsolucionario.org
11.8. QUADRIC SURFACES
17.
41
18.
z
z
10
y
y
x
x
p
19. The equation can be written as x2 + (± y 2 + z 2 )2 = 1. The surface is generated by revolving
the circles x2 + y 2 = 1 or x2 + z 2 = 1 about the x-axis. [Alternatively, the surface is generated
by revolving the circles x2 + y 2 = 1 or y 2 + z 2 = 1 about the y-axis, or the circles x2 + z 2 = 1
or y 2 + z 2 = 1 about the z-axis.]
p
20. The equation can be written as −9x2 + (±2 y 2 + z 2 02 = 36. The surface is generated by
revolving the hyperbolas = 9x2 + 4y 2 = 36 or −9x2 + 4z 2 = 36 about the x-axis.
√
2
2 2
21. The equation can be written as y = e± x +z ) . The surface is generated by revolving the
2
2
curves y = ex or y = ez about the y-axis.
p
22. The equation can be written as (± x2 + y 2 )2 = sin2 z. The surface is generated by revolving
the curves x2 = sin2 z or y 2 = sin2 z about the z-axis.
p
√
23. Replacing x by ± x2 + y 2 we have y = ±2 x2 + z 2 or y 2 = 4x2 + 4z 2 .
√
√
24. Replacing z by x2 + z 2 we have y = ( x2 + z 2 )1/2 or y 4 = x2 + z 2 ; y ≥ 0.
p
p
25. Replacing z by ± y 2 + z 2 we have ± y 2 + z 2 = 9 − x2 or y 2 + z 2 = (9 − x2 )2 , x ≥ 0.
26. Replacing y by
p
p
x2 + y 2 we have z = 1 + ( x2 + y 2 )2 or z = 1 + x2 + y 2 .
27. Replacing z by
p
p
y 2 + z 2 we have x2 − (± y 2 + z 2 )2 = 4 or x2 − y 2 − z 2 = 4.
p
p
28. Replacing x by ± x2 + y 2 we have 3(± x2 + y 2 )2 + 4z 2 = 12 or 3x2 + 3y 2 + 4z 2 = 12.
29. Replacing y by
p
x2 + y 2 we have z = ln
p
x2 + y 2 .
p
p
30. Replacing y by ± y 2 + z 2 we have x(± y 2 + z 2 ) = 1 or x2 (y 2 + z 2 ) = 1.
31. The surface is Problem 11 is a surface of revolution about the x-axis. The surface in Problem
2 is a surface of revolution about the y-axis. The surface is Problems 1, 4, 6, 10, and 14 are
surfaces of revolution about the z
42
CHAPTER 11. VECTORS AND 3-SPACE
32.
z
π
1
y
x
33. The first equation is the lower nappe of the cone (z + 2)2 = x2 + y 2 whose axis of revolution
is the z-axis and whose vertex is at (0, 0, −2).
34. The first equation is the right-hand of the cone (y − 1)2 = x2 + z 2 whose axis of revolution is
the y-axis and whose vertex is at (0, 1, 0).
2
2
2
2
35. (a) Writing the equation of the ellipse in√the form
√ x /(c − z)a + y /(c − z)b = 1 we see
that the area of a cross-section is πa c − zb c − z = πab(c − z).
Rc
1
c
(b) V = 0 πab(c − x)dz = πab − 21 (c − z)2 0 = πabc2
2
36. (a) Using the formula for the area of an ellipse given in Problem 35(a) we see that a horizontal
cross-sectional area of the ellipsoid is πab(1 − z 2 /c2 ). Then
Z
c
V =2
0
z
πab 1 − 2 dz = 2πab
c
(b) When a = b = c the volume is
4
3
3 πa ,
z3
z− 2
3c
c
=
0
4
πabc.
3
which is the formula for the volume of a sphere.
37. Expressing the line in the form (x − 2)/4 = (y + 2)/(−6) = (z − 6)/3 we see that parametric
equations for the line are x = 2 + 4t, y = −2 = 6t, z = 6 + 3t. Writing the equation of
the ellipse as 36x2 + 9y 2 + 4z 2 = 324 and substituting, we obtain 36(2 + 4t)2 + 9(−2 − 6t)2 +
4(6 + 3t)2 = 936t2 + 936t + 324 or 936t(t + 1) = 0. When t = 0 we obtain the point (2, −2, 6),
and when t = −1 we obtain the point (−2, 4, 3).
Chapter 11 in Review
A. True/False
1. True
2. False; the points must be non-collinear.
3. False; since a normal to the plane is h2, 3, −4i which is not a multiple of the direction vector
h5, −2, 2i of the line.
CHAPTER 11 IN REVIEW
4. True
7. True
43
5. True
8. True
6. True
9. True
10. True; since a × b and c × d are both normal to the plane and hence parallel (unless a × b = 0
or c × d.)
11. True. The normal vector of the first plane is h1, 2, −1i while the normal vector of the second
plane is h−2, −4, 2i. Since the second vector is a scalar multiple of the first, the planes are
parallel.
12. False. Look at Figure 11.5.3 in the text.
13. True. This is a parabolic cylinder similar to that shown in Figure 11.7.6.
14. True. In the yz-plane, we have x = 0. Therefore, the equation of the surface becomes
1 or y 2 + z 2 = 2.
y2
2
2
+ z2 =
15. False. Find the equation of the plane containing the first three points, P1 (0, 1, 2), P2 (1, −1, 1),
−−−→
−−−→
and P3 (3, 2, 6). This plane must contain the vectors P1 P2 = h1, −2, 1i and P1 P3 = h3, 1, 4i.
i
j
k
−−−→ −−−→
Define n = P1 P2 × P1 P3 = 1 −2 −1 = h−7, −7, −7i. Then n must be normal to the
3 1
4
plane. Using n and the point P1 , the equation of the plane becomes −7x − 7y + 7z = 7 or
z +y −z = −1. The fourth point P4 (2, 1, 2) does not lie on the plane since (2)+(1)−(2) 6= −1.
16. True
17. False. The trace in the yz-plane is described by the equation 9y 2 + z 2 = 1 which represents
an ellipse.
18. True. This ellipsoid results from revolving the graph of the ellipse x2 + 9y 2 = 1 about the
y-axis.
19. True. |a × b| = |a||b|| sin θ| = |a||b| since θ = 90◦
20. False. Let a = i,
b = j, and c = k. Then a · b = a · c = 0 but b 6= c.
B. Fill in the Blanks
1. 9i + 2j + 2k
2. orthogonal
3. −5(k × j) = −5(−i) = 5i
4. i · (i × j) = i × k = 0
p
(−12)2 + 42 + 62 = 14
5.
6. k × (i + 2j − 5k) = k × i + 2(k × j) − 5(k × k)
= j − 2i − 5(0)
= h−2, 1, 0i
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44
CHAPTER 11. VECTORS AND 3-SPACE
7.
2
4
−5
3
= 2(3) − (−5)(4) = 6 + 20 = 26
8. (−1 − 20)i − (−2 − 0)j + (8 − 0)k = −21i + 2j + 8k
9. −6i + j − 7k
10. The smallest component is the j-component with magnitude 3. Therefore, the sphere cannot
have a radius larger than 3 or its interior will intersect the xz-plane. Thus, we need a sphere
with radius 3 and center (4, 3, 7). The equation is
(x − 4)2 + (y − 3)2 + (z − 7)2 = 9
11. Writing the line in parametric form, we have x = 1 + t, y = −2 + 3t z = −1 + 2t.
Substituting into the equation of the plane yields (1 + t) + 2(−2 + 3t) − (−1 + 2t) = 13 or
t = 3. Thus, the point of intersection is x = 1 + 3, y = −2 + 3(3) = 7, z = −1 + 2(3) = 5,
or (4, 7, 5).
12. |a| =
p
√
42 + 32 + (−5)2 = 5 2;
13. x2 − 2 = 3,
x2 = 5;
1
4
3
1
u = − √ (4i + 3j − 5k) = − √ i − √ j + √ k
5 2
5 2
5 2
2
y2 − 1 = 5,
y2 = 6;
z2 − 7 = −4,
z2 = 3;
P2 = (5, 6, 3)
14. (5, 1/2, 5/2)
√
15. (7.2)(10) cos 135◦ = −36 2
16. 2b = h−2, 4, 2i;
4c = h0, −8, 8i;
a · (2b + 4c) = h3, 1, 0i · h−2, −4, 10i = −10
17. 12, −8, 6
18. cos θ =
19. A =
a·b
1
= √ √ = 1/2;
|a||b|
2 2
θ = 60◦
√
1
|5i − 4j − 7k| = 3 10/2
2
20. (x + 5)2 + (y − 7)2 + (z + 9)2 = 6
21. | − 5 − (−3)| = 2
22. parallel: −2c = 5,
c = −5/2; orthogonal: 1(−2) + 3(−6) + c(5) = 0,
c=4
23. The equation can be transformed into something more recognizable by completing the square:
x2 + 2y 2 + 2z 2 − 4y − 12z = 0
=⇒ x2 + 2(y 2 − 2y) + 2(z 2 − 6z) = 0
=⇒ x2 + 2(y 2 − 2y + 1) + 2(z 2 − 6z = 9) = 20
=⇒ x2 + 2(y − 1)2 + 2(z − 3)2 = 20
This is the equation of an ellipsoid centered at (0, 1, 3).
24. Letting z = 1, the trace is described by the equation y = x2 − 1, which is a parabola.
CHAPTER 11 IN REVIEW
45
C. Exercises
i
j k
1 0
1 0
1 1 0 =
+
i−
1 1
−2 1
1 −2 1
A unit vector perpendicular to both a and b is
1. a × b =
1
1
1
−2
k = i − j + 3k
a×b
1
1
3
1
(i − j − 3k = √ i − √ j − √ k.
=√
|a × b|
1+1+9
11
11
11
s r
√
2 2
2
1
1
1
3
3
2. The magnitude of a is given by |a| =
+
+ −
=
=
. Letting
2
2
2
4
2
α, β, and γ represent the angles between a and i, j, and k respectively, we have cos α =
1
1
−1
1
1
1
√2 = √ , cos β = √2 = √ , and cos γ = √ 2 = − √ . From this we are able to
3
3
3
3
3
3
2
2
2
1
compute: α = cos−1 √
≈ 0.95532
3
1
β = cos−1 √
≈ 0.95532
3 1
γ = cos−1 − √
≈ 2.18628
3
3. compb a = a · b/|b| = h1, 2, −2i · h4, 3, 0i/5 = 2
4. compa b = b · a/|a| = h4, 3, 0i · h1, 2, −2i/3 = 10/3
proja b = (compa ba/|a| = (10/3)h1, 2, −2i/3 = h10/9, 20/9, −20/9i
√
5. First we compute 2a = h2, 4, −4i, |b| = 16 + 9 = 5, and 2a · b = 20. So projb 2a =
2a · b
20
16 12
b=
h4, 3, 0i = h , , 0i.
2
|b|
25
5 5
6. compb (a − b) = (a − b) · b/|b| = h−3, −1, −2i · h4, 3, 0i/5 = −3
projb (a − b) = (compb (a − b))b/|b| = −3h4, 3, 0i/5 = h−12/5, −9/5, 0i
projb⊥ (a−b) = (a−b)−projb (a−b) = h−3, −1, −2i−h−12/5, −9/5, 0i = h−3/5, 4/5, −10/5i
7.
x2
y2
+
= 1;
16
4
8.
x2
1
+ z 2 = − y;
2
4
9. −
10.
elliptical cylinder
paraboloid
x2
y2
z2
−
+
= 1;
9
9/4
9
x2
y2
(z − 5)2
+
+
= 1;
25 25
25
11. x2 − y 2 = 9z;
12. plane
hyperboloid of two sheets
sphere
hyperbolic paraboloid
46
CHAPTER 11. VECTORS AND 3-SPACE
√
√
13. Replacing x by ± x2 + z 2 we have (± x2 +p
z 2 )2 − y 2 = 1 or x2 + z 2 −p
y 2 = 1, which is a
2
2
2
hyperboloid of one sheet. Replacing y by (± y + z ) we have x − (± y 2 + z 2 )2 = 1 or
x2 − y 2 − z 2 = 1, which is a hyperboloid of two sheets.
14. The surface is generated by revolving y = 1+x, x ≥ 0, about the y-axis or
√ y = 1+z, z ≥ 0
about the z-axis. The restrictions on x and z are required since y = 1 + x2 + z 2 ≥ 1.
15. Let a = ha, b, ci and r = hx, y, zi. Then
(a) (r − a) · r = hx − a, y − b, z − ci · hx, y, zi = x2 − ax + y 2 − by + z 2 − ac = 0 implies
2 a 2
b
c 2
a2 + b2 + c2
x−
+ y−
+ z−
=
. The surface is a sphere.
2
x
2
4
(b) (r − a) · a = hx − a, y − b, z − ci · ha, b, ci = a(x − a) + b(y − b) + c(z − c) = 0
The surface is a plane.
16. h4, 2, −2i − h2, 4, −3i = h2, −2, 1i; h2, 4, −3i − h6, 7, −5i = h−4, −3, 2i;
h2, −2, 1i · h−4, −3, 2i = 0. The points are the vertices of a right triangle.
17. A direction vector of the given line is h4, −2, 6i. A parallel line containing (7, 3, −5) is
(x − 7)/4 = (y − 3)/(−2) = (z + 5)/6.
18. A normal to the plane is h8, 3, −4i. The line with this direction vector and through (5, −9, 3)
is x = 5 + 8t, y = −9 + 3t, z = 3 − 4t.
19. The direction vectors are h−2, 3, 1i and h2, 1, 1i. Since h−2, 3, 1i · h2, 1, 1i = 0, the lines are
orthogonal. Solving 1 − 2t = x = 1 + 2s, 3t = y = −4 + s, we obtain t = −1 and s = 1. The
point (3, −3, 0) obtained by letting t = −1 and s = 1 is common to the two lines, so they do
intersect.
20. Vectors in the plane are h2, 3, 1i and h1, 0, 2i. A normal vector is h2, 3, 1i×h1, 0, 2i = h6, −3, −3i =
3h2, −1, −1i. An equation of the plane is 2x − y − z = 0.
21. The lines are parallel with direction vector h1, 4, −2i. Since (0, 0, 0) is on the first line and
(1, 1, 3) is on the second line, the vector h1, 1, 3i is in the plane. A normal vector to the plane
is thus h1, 4, −2i × h1, 1, 3i = h14, −5, −3i. An equation of the plane is 14x − 5y − 3z = 0.
22. Letting z = t in the equations of the plane and solving −x + y = 4 + 8t, 3x − y = −2t,
we obtain x = 2 + 3t, y = 6 + 11t, z = t. Thus, a normal to the plane is h3, 11, 1i and an
equation of the plane is 3(x − 1) + 11(y − 7) + (z + 1) = 0 or 3x + 11y + z = 79.
23. A normal vector is (i − 2j) × (2i + 3k) = −6i − 3j + 4k. Thus, an equation of the plane is
−6(z − 1) − 3(y + 1) + 4(z − 2) = 0 or 6x + 3y − 4z = −5.
24. The points at the ends of the diameter, obtained from t = −1 and t = 0 are (2, 4, 2) and
(4, 7, 8). The center of the
√ sphere is the midpoint of the line segment or (3, 11/2, 5). The
diameter of the sphere is 22 + 32 + 62 = 7. The equation is (x−3)2 +(y −11/2)2 +(z −5)2 =
49/4.
www.elsolucionario.org
CHAPTER 11 IN REVIEW
47
25. We compute (a × b) · c. First a × b = −3i + 3j − 3k. Then (a × b) · c = −3(4) + 3(5) − 3(1) = 0,
and the vectors are coplanar.
1
|c|. Since
2
2
2
2
a · b = 0 we have |b − a| = (b − a) · (b − a) = a · a + 2a · b + b · b = |a| + |b| , we have
1p 2
1
1
|d| = 21 |a + b| =
|a| + |b|2 = |b − a| = |c|.
2
2
2
26. Let d be the vector from the right angle to M . We want to show that |d| =
27. (a) We have v = vj and B = Bi. Then F = q(v × B = q(vj × Bi) = q(−vBk) = −qvBk.
(b) We first note that L = mr × v and r × v = 0. Then
r × L = r × (mr × v) = m[r × (r × v)] = m[(r · v)r − (r · r)v] = −m|r|2 v,
and so v = −
1
1
(r × L) =
(L × r).
2
m|r|
m|r|2
√
√
10
a
= √ (i + j) = 5 2i + 5 2j; d = h7, 4, 0i − h4, 1, 0i = 3i + 3j
|a|
2√
√
√
W = F · d = 15 2 + 15 2 = 30 2 N-m
√
√
√
√
29. F = 5 2i + 5 2j + 50i = (5 2 + 50)i + 5 2j; d = 3i + 3j
28. F = 10
√
√
√
W = 15 2 + 150 + 15 2 = 30 2 + 150N-m ≈ 192.4 N-m
30. Let |F1 | = F1 and |F2 | = F2 . Then F1 = F1 [(cos 45◦ )i + (sin 45◦ )j]
and F2 = F2 [(cos 120◦ )i +
√ !
1
1
3
1
(sin 120◦ )j] or F1 = F1 √ i + √
j . Since w + F1 + F2 = 0,
and F2 = F2 − i +
2
2
2
2
F1
and
1
1
√ i + √ + F2
2
2
√ !
1
3
− i+
j = 50j,
2
2
1
1
√ F1 − F2 i +
2
2
!
√
3
1
√ F1 +
F2 j = 50j
2
2
√
1
1
1
3
√ F1 − F2 = 0, √ F1 +
F2 = 50.
2
2
2
2
√
√
√
Solving, we obtain F1 = 25( 6 − 2) ≈ 25.9lb and F2 = 50( 3 − 1) ≈ 36.6lb.
www.elsolucionario.org
Chapter 12
Vector-Valued Functions
12.1
Vector Functions
2
1. Since the square root function
S is only defined for nonnegative values, we must have t − 9 ≥ 0.
So the domain is (−∞, −3) [3, ∞).
2. Since the natural logarithm is only defined for positive values, we must have 1 − t2 > 0. So
the domain is (−1, 1).
3. Since the inverse sine function is only defined for values between -1 and 1, the domain is
[−1, 1].
4. The vector function is defined for all real numbers.
5. r(t) = sin πti + cos πtj − cos2 πtk
6. r(t) = cos2 πti + 2 sin2 πtj + t2 k
7. r(t) = e−t i + e2t j + e3t k
8. r(t) = −16t2 i + 50tj + 10k
9. x = t2 ,
y = sin t,
z = cos t
10. r(t) = t sin t(i + k) = t sin ti + 0j + t sin tk so x = t sin t,
11. x = ln t,
y = 1 + t,
12. x = 5 sin t sin 3t,
z = t3
y = 5 cos 3t,
z = 5 cos t sin 3t
48
y = 0,
z = t sin t
12.1. VECTOR FUNCTIONS
13.
49
14.
z
15.
z
z
y
y
4
y
x
x
16.
x
17.
z
18.
y
y
2
y
x
2
x
x
19.
20.
z
z
y
y
x
x
21.
z
y
x
Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper
loop shown intersects the xz-plane at about (286751, 0, 286751).
50
CHAPTER 12. VECTOR-VALUED FUNCTIONS
22.
23.
z
z
y
10
10
x
10
y
x
24.
z
y
x
25. r(t) = h4, 0i + h0 − 4, 3 − 0it = (4 − 4t)i + 3tj, 0 ≤ t ≤ 1
y
x
26. r(t) = h0, 0, 0i + h1 − 0, 1 − 0, 1 − 0it = ti + tj + tk, 0 ≤ t ≤ 1
www.elsolucionario.org
12.1. VECTOR FUNCTIONS
27. x = t, y = t, z = t2 + t2 = 2t2 ; r(t) = ti + tj + 2t2 k
z
y
x
√
√
√
28. x = t, y = 2t, z = ± t2 + 4t2 + 1 = ± 5t2 − 1; r(t) = ti + 2tj ± 5t2 − 1k
z
y
x
29. x = 3 cos t, z = 9 − 9 cos2 t = 0 sin2 t; y = 3 sin t; r(t) = 3 cos ti + 3 sin tj + 9 sin2 tk
z
y
x
30. x = sin t, z = 1, y = cos t; r(t) = sin ti + cos tj + k
51
52
CHAPTER 12. VECTOR-VALUED FUNCTIONS
z
y
x
31. x = t, y = t, z = 1 − 2t; r(t) = ti + tj + (1 − 2t)k
z
y
x
32. x = 11,
y = t,
z = 3 + 2t;
r(t) = i + tj + (3 + 2t)k
z
y
x
33. (b); Notice that the y and z values consistently increase while the x values oscillate rapidly
between -1 and 1. The only vector fucntion that describes this behavior is (b).
34. (c); The trace of the graph on the xy−plane would look like a circle, while the z value oscillates
between 0 and 1. The only vector function that describes this behavior is (c).
35. (d); Notice that the z value is contant. The only vector function that satisfies this constraint
is (d).
36. (a); Notice that the x values consistently increase while the trace of the graph on the yz-plane
would look like a circle. The only vector function that describes this behavior is (a).
12.1. VECTOR FUNCTIONS
53
37. Letting x = at cos t, y = bt sin t, and z = ct, we have
c2 t 2
z2
= 2 = t2 = t2 cos2 t + t2 sin2 t
2
c
c
a2 t2 cos2 t b2 t2 sin2
+
=
a2
b2
x2
y2
= 2 + 2
a
b
38.
z
y
x
39. Letting x = aekt cos t, y = bekt sin t, and z = cekt , we have
c2 ekt
z2
=
= e2kt = e2kt cos2 t + e2kt sin2 t
c2
c2
a2 e2kt cos2 t b2 e2kt sin2 t
=
+
a2
b2
2
2
x
y
+ 2
a2
b
40.
z
y
x
41. x2 + y 2 + z 2 = a2 sin2 kt cos2 t + a2 sin2 kt sin2 t + a2 cos2 kt
= a2 sin2 kt + a2 cos2 kt
= a2
www.elsolucionario.org
54
42.
CHAPTER 12. VECTOR-VALUED FUNCTIONS
k=1
k=2
k=3
z
z
z
y
y
y
x
x
x
k=4
k = 10
k = 20
z
z
z
y
y
y
x
x
43. (a)
z
y
x
(b) r1 (t) = ti + tj + (4 − t2 )k
r2 (t) = ti − tj + (4 − t2 )k
(c)
z
y
x
44. C lies on the surface of the sphere of radius a.
x
12.2. CALCULUS OF VECTOR FUNCTIONS
55
45.
46.
k = 0.1
k = 0.2
k = 0.3
z
z
z
y
y
y
x
x
47.
x
k=2
k=4
z
z
y
y
x
48.
k=
x
1
10
k=1
z
z
y
x
12.2
y
x
Calculus of Vector Functions
1. lim [t3 i + t4 j + t5 k] = 23 i + 24 j + 25 k = 8i + 16j + 32k
t→2
56
CHAPTER 12. VECTOR-VALUED FUNCTIONS
2. r(t) =
sin 2t
ln t
i + (t − 2)5 k +
k. Using L’Hôpital’s Rule,
t
1/t
lim+ r(t) =
t→0
1/t
2 cos 2t
k
= 2i − 32j
i + (t − 2)5 j +
1
−1/t2
3. Using
Rule, we have
L’Hôpital’s
t2 − 1 5t − 1 2et−1 − 2
2t 5t − 1 2et−1
lim
,
,
= lim =
,
,
i = h2, 2, 2
t→1
t→1
t−1 t+1
t−1
1 t+1
1
π
4. Since lim tan−1 t = , we have
t→∞
2
e2t
1
e−1
1
−1
−1
lim
,
,
tan
t
,
,
tan
t
=
lim
t→∞ 2e2t + t 2e−t + 5
t→∞ 2 + te−2t 2 + 5et
1
π
=
, 0,
2
2
The last equality follows from using L’Hôpital’s Rule to get
lim te−2t = lim
t→∞
t→∞
1
t
= lim
=0
t→∞ 2e2t
e2t
5. lim [−4r1 (t) + 3r2 (t)] = −4(i − 2j + k) + 3(2i + 5j + 7k) = 2i + 23j + 17k
t→α
6. lim r1 (t) · r2 (t) = (i − 2j + k) = (i − 2j + k) · (2i + 5j + 7k) = −1
t→α
7. Notice that the k component ln(t − 1) is not defined at t = 1. Therefore, r(t) is not continuous
at t = 1.
8. Notice that sin πt, tan πt, and cos πt are each continuous at t = 1 since the sine, cosine, and
tangent function are continuous on their domains. Therefore, since each of the component
functions are continuous at t = 1, we know that r(t) is continuous at t = 1.
9. r0 (t) = 3i + 8tj + (10t − 1)k
so r0 (1) = 3i + 8j + 9k = h3, 8, 9i
r(1.1) − r(1)
h3(1.1) − 1, 4(1.1)2 , 5(1.1)2 − (1.1)i − h3(1) − 1, 4(1)2 , 5(1)2 − (1)i
while
=
0.1
0.1
h2.3, 4.84, 4.95i − h2, 4, 4i
=
0.1
h0.3, 0.84, 0.95i
= h3, 8.4, 9.5i
=
0.1
−5
i + (6t + 1)j − 3(1 − t)2 k
(1 + 5t)2
−5
so r0 (0) =
i + j + 3k = h−5, 1, −3i
1
10. r0 (t) =
www.elsolucionario.org
12.2. CALCULUS OF VECTOR FUNCTIONS
57
1
, 3(0.05)2 + (0.05), (1 + 0.05)3
r(0.05) − r(0)
1 + 5(0.05)
while
=
0.05
0.05
h0.8, 0.0575, 0.857375i − h1, 0, 1i
=
0.05
h−0.2, 0.0575, −0.142625i
=
0.05
= h−4, 1.15, −2.8525i
11. r0 (t) =
1
1
i − 2 j;
t
t
r00 (t) = −
12. r0 (t) = h−t sin t, 1 − sin ti;
1
k;
1 + t2
−
1
, 3(0)2 + (0), (1 − 0)3
1 + 5(0)
1
2
i + 3j
t2
t
r00 (t) = h−t cos t − sin t, − cos ti
13. r0 (t) = h2te2t + e2t , 3t2 , 8t − 1i;
14. r0 (t) = 2ti + 3t2 j +
r00 (t) = h4te2t + 4e2t , 6t, 8i
r00 (t) = 2i + 6tj −
15. r0 (t) = −2 sin ti +√6 cos tj
r0 (π/6) = −i + 3 3j
2t
k
(1 + t2 )2
16. r0 (t) = 3t2 i + 2tj
r0 (−1) = 3i − 2j
y
y
x
x
8t
k
(1 + t2 )2
0
r (−1) = j − 2k
17. r0 (t) = j −
18. r0 (t) = −3 sin√
ti + 3 cos√tj + 2k
−3
2
3 2
r0 (π/4) =
i+
j + 2k
2
2
z
z
y
x
y
x
58
CHAPTER 12. VECTOR-VALUED FUNCTIONS
1
1
8
19. r(t) = ti + j + t3 k; r(2) = 2i + 2j + k; r0 (t) = i + tj + t2 k; r0 (2) = i + 2j + 4k
2
3
3
Using the point (2, 2, 8/3) and the direction vector r0 (2), we have x = 2 + t, y = 2 + 2t, z =
8/3 + 4t.
20. r(t) = (t3 −t)i+
6t
j+(2t+1)2 k;
t+1
r(1) = 3j+9k;
r0 (t) = (3t2 −1)i+
6
j+(8t+4)k;
(t + 1)2
3
r0 (1) = 2i + j + 12k. Using the point (0, 3, 9) and the direction vector r0 (1), we have x =
2
2t, y = 3 + 23 , z = 9 + 12t.
p
√
2 =
6
21. r0 (t) = het + tet , 2t + 2, 3t2 − 1i so r0 (0) = h1, 2, −1i and |r0 (0)| = 12 + 22 + (−1)
r0 (0)
1
2 −1
h1, 2, −1i
√
The unit tangent vector at t = 0 is given by 0
= √ ,√ ,√
=
|r (0)|
6
6
6
6
To find the parametric equations of the tangent line at t = 0, we first compute r(0) =
1
2 −1
h0, 0, 0i. The tangent line is then given in vector form as p(t) = h0, 0, 0i + t √ , √ , √
=
6
6
6
1
2 −1
1
2
−1
√ t, √ t, √ t or in parametric form as x = √ t, y = √ t, z = √ t.
6
6
6
6
6
6
p
√
2 + (2)2 + (1)2 =
22. r0 (t) = h3 cos 3t, 2 sec2 2t, 1i so r0 (π) = h−3, 2, 1i and |r0 (π)| = (−3)
14.
−3
2
1
r0 (π)
h−3, 2, 1i
= √ ,√ ,√
The unit tangent vector at t = π is given by 0
= √
|r (π)|
14
14
14
14
To find the parametric equations of the tangent line at t = π, we first compute r0 (π) = h1, 0, πi.
The tangent line is then given in vector
form as
−3
2
1
p(t) = h1, 0, πi + t √ , √ , √
14
14
14
−3
2
1
= 1 − √ t, √ t, π + √ t
14
14
14
2
1
−3
or in parametric form as x = 1 − √ t, y = √ t, z = π + √ t
14
14
14
* √
+
1
3 π
23. r(π/3) =
,
,
2 2 3
r0 (t) = h− *
sin t, cos t, 1i+
√
3 1
r0 (π/3) = −
, ,1
2 2
so the tangent
* √ line+is given
* by
+
√
1
3 π
3 1
p(t) =
,
,
+t −
, ,1
2 2 3
2 2
*
+
√
√
1
3
3 1 π
=
−
t,
+ t, + t
2
2
2
2 3
24. r(0) = h6, 1, 1i
r0 (t) = h−3e−t/2 , 2e2t , 3e3t i
12.2. CALCULUS OF VECTOR FUNCTIONS
59
r0 (0) = h−3, 2, 3i So the tangent line is given by
r(t) = h6, 1, 1i + th−3, 2, 2i
= h6 − 3t, 1 + 2t, 1 + 3ti
d
[r(t) × r0 (t)] = r(t) × r00 (t) + r0 (t) × r0 (t) = r(t) × r00 (t)
dt
d
d
[r(t) · (tr(t))] = r(t) · (tr(t))+ = r(t) · (tr0 (t) + r(t)) + r0 (t) · (tr(t))
26.
dt
dt
= r(t) · (tr0 (t)) + r(t) · r(t) + r0 (t) · (tr(t)) = 2t(r(t) · r0 (t)) + r(t) · r(t)
25.
27.
d
d
[r(t) · (r0 (t) × r00 (t))] = r(t) · (r0 (t) × r00 (t)) + r0 (t) · (r0 (t) × r00 (t))
dt
dt
= r(t) · (r0 (t) × r000 (t) + r00 (t) × r00 (t)) + r0 (t) · (r0 (t) × r00 (t))
= r(t) · (r0 (t) × r000 (t))
28.
d
d
[r1 (t) × (r2 (t) × r3 (t))] = r1 (t) × (r2 (t) × r3 (t)) + r0 (t) × (r2 (t) × r3 (t))
dt
dt
= r1 (t) × (r2 (t) × r03 (t) + r02 (t) × r3 (t) + r01 (t) × (r2 (t) × r3 (t))
= r1 (t) × (r2 (t) × r03 (t)) + r1 (t) × (r02 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t))
d
1
[r1 (2t) + r2 ( 1t )] = 2r0 (2t) − 2 r02 ( 1t )
dt
t
d 3 2
30.
[t r(t )] = t3 (2t)r0 (t2 ) + 3t2 r(t2 ) = 2t4 r0 (t2 ) + 3t2 r(t2 )
dt
Z 2
Z 2
Z 2
Z 2
2
1
2
3
31.
i + t3
r(t)dt =
tdt i +
3t dt j +
4t dt k = t2
2 −1
−1
−1
−1
−1
Z 4
Z 4
Z 4
Z 4
√
√
2t + 1dt i +
32.
− tdt j +
sin πtdt k
r(t)dt =
29.
0
0
0
4
4
0
4
2
−1
j + t4
2
−1
k=
3
i + 9j + 15k
2
26
1
2
1
16
(2t + 1)3/2 i − t3/2 j − cos πt k =
i− j
3
3
π
3
3
0
0
0
Z
Z
Z
Z
2
33.
r(t)dt =
tet dt i +
−e−2t dt j +
tet dt k
1 t2
1 2
1 −2t
1
t
t
+ c2 j + e + d3 k = et (t − 1)i + e−2t j + et k + c,
= [te − e + c1 ]i + e
2
2
2
2
where c = c1 i + c2 j + c3 k.
Z
Z
Z
Z
1
t
t2
34.
r(t)dt =
dt i +
dt j +
dt k
1 + t2
1 + t2
1 + t2
Z 1
1
−1
2
k
= [tan t + c1 ]i +
ln(1 + t ) + c2 j +
1−
2
1 + t2
1
= [tan−1 t + c1 ]i +
ln(1 + t2 ) + c2 j + [t − tan−1 t + c3 ]k
2
1
= tan−1 ti + ln(1 + t2 )j + (t − tan−1 t)k + c,
2
=
www.elsolucionario.org
60
CHAPTER 12. VECTOR-VALUED FUNCTIONS
where c = c1 i + c2 j + c3 k.
R
R
R
R
6dt i + 6tdt j + 3t2 dt k = [6t + c1 ]i + [3t2 + c2 ]j + [t3 + c3 ]k
35. r(t) = r0 (t)dt =
Since r(0) = i + 2j + k = c1 i + c2 j + c3 k, c1 − 1, c2 = −2, and c3 = 1. Thus,
r(t) = (6t + 1)i + (3t2 − 2)j + (t3 + 1)k
R
R
R
t sin t2 dt i + − cos 2tdt j = − 21 cos t2 + c1 i + − 21 sin 2t + c2 j
36. r(t) = r0 (t)dt =
Since r(0) = 32 = (− 12 + c1 )i + c2 j, c1 = 2, and c2 = 0. Thus,
r(t) =
1
1
2
− cos t + 2 i − sin 2tj.
2
2
R
R
R
R
37. r0 (t) = r00 (t)dt =
12tdt i +
−3t−1/2 dt j +
2dt k = [6t2 + c1 ]i + [−6t1/2 + c2 ]j +
[2t + c3 ]k
Since r0 (1) = j = (6 + c1 )i + (−6 + c2 )j + (2 + c3 )k, c1 = −6, c2 = 7, and c3 = −2. Thus,
r0 (t) = (6t2 − 6)i + (−6t1/2 + 7)j + (2t − 2)k.
Z
r(t) =
0
Z
r (t)dt =
Z
Z
1/2
(6t − 6)dt i +
(−6t + 7)dt j +
(2t − 2)dt k
2
= [2t3 − 6t + c4 ]i + [−4t3/2 + 7t + c5 ]j + [t2 − 2t + c6 ]k.
Since
r(1) = 2i − k = (−4 + c4 )i + (3 + c5 )j + (−1 + c6 )k,
c4 = 6, c5 = −3, and c6 = 0. Thus,
r(t) = (2t3 − 6t + 6)i + (−4t3/2 + 7t − 3)j + (t2 − 2t)k.
38. r0 (t) =
Z
r00 (t)dt =
Z
Z
Z
sec2 tdt i +
cos tdt j +
− sin tdt k
= [tan t + c1 ]i + [sin t + c2 ]j + [cos t + c3 ]k
Since r0 (0) = i + j + k = c1 i + c2 j + c3 k, c1 = 1, c2 = 1, and c3 = 0. Thus,
r0 (t) = (tan t + 1)i + (sin t + 1)j + cos tk.
Z
r(t) =
r0 (t)dt =
Z
Z
Z
(tan t + 1)dt i +
(sin t + 1)dt j +
cos tdt k.
= [ln | sec t| + c4 ]i + [− cos t + t + c5 ]j + [sin t + c6 ]k
Since r(0) = −j + 5k = (−1 + c5 )j + (c6 )k, c4 = 0, c5 = 0, and c6 = 5. Thus,
r(t) = (ln | sec t| + t)i + (− cos t + t)j + (sin t + 5)k.
p
√
39. r0 (t) = −a sin ti + a cos tj + ck; |r0 (t)| = (−a sin t)2 + (a cos t)2 + c2 = a2 + c2
√
√
R 2π √
2π
a2 + c2 dt = a2 + c2 t 0 = 2π a2 + c2
s= 0
12.2. CALCULUS OF VECTOR FUNCTIONS
61
40. r0 (t) = i +
p(cos t − t sin t)j + (sin t + t cos t)k
√
|r0 (t)| = 12 + (cos t − t sin t)2 + (sin t + t cos t)2 = 2 + t2
√
√
√
√
√
Rπ√
π
s = 0 2 + t2 dt = 2t 2 + t2 + ln |t + 2 + t2 | 0 = π2 2 + π 2 + ln(π + 2 + π 2 ) − ln 2
t
t
t
41. r0 (t) = (−2e
+ et sin 2t)j + et k
p sin 2t + e cos 2t)i + (2e cos 2t √
√
2
0
2t
2
2t
2t
|r (t)| = 5e cos 2t + 5e sin 2t + e = 6e2t = 6et
3π
√
√
R 3π √ t
= 6(e3π − 1)
s= 0
6e dt = 6et
0
q
√
√
√
42. r0 (t) = 3i + 2 3tj + 2t2 k; |r0 (t)| = 32 + (2 3t)2 + (2t2 )2 = 9 + 12t2 + 4t4 = 3 + 2t2
1
R1
= 3 + 23 = 11
s = 0 (3 + 2t2 )dt = 3t + 23 t3
3
0
Rt
s
43. From r0 (t) = h9 cos t, −9 sin ti, we find |r0 (t)| = 9. Therefore, s = 0 9du = 9t so that t = . By
9 E
D
D
s
s
sE
s
. Note that r0 (s) = sin , cos
substituting for t in r(t), we obtain r(s) = 9 sin , 9 cos
9
9
9
9
r
s
s
so that r0 (s) = sin2 + cos
= 1.
9
9
√
Rt
44. From r(t) = h−5 sin t, 12, 5 cos ti, we find |r0 (t)| = 169 = 13. Therefore, s = 0 13du = 13t
s
s 12
5
s
. By substituting for t in r(t), we obtain r(s) = h5 cos 13
, 13 s, 13
cos 13
i.
so that t =
13
5
s 12 5
s
0
Note that r (t) = − 13 sin 13 , 13 , 13 cos 13 so that
r
25 144
25
25
s
0
sin
+ 169 +
cos2 13
|r (s)| =
=1
169
13
169
√
√
Rt√
45. From r0 (t) = h2, −3, 4i, we find |r0 (t)| = 29. Therefore,
s = 0 29du = 20t so that
2
3
4
s
t = √ . By substituting for t in r(t), we obtain r(s) = 1 + √ s, 5 − √ s, 2 + √ s .
29
29
29
29
r
D
E
9
16
4
2
3
4
Note that r0 (s) = √29 , − √29 , √29 so that r0 (s) =
+
+
= 1.
29 29 29
t
t
t
t
46. From r0 (t)
p= he cos t − e sin t, e sin t + e cos t, 0i we find
√
√
|r0 (t)| = e2t cost −2e2t cos t sin t + e2t sin2 t + e2t sin2 t + 2e2t sin t cos t + e2t cos2 t = 2e2t = et 2.
√
Rt √
Therefore, s = 0 eu 2du = 2(et − 1) so that t = ln √s2 + 1 . By substituting for t in r(t),
we obtain
E
D
√s + 1 cos(ln √s + 1
√s + 1 sin ln √s + 1
r(s) =
,
, 1 Note that
2 2 D 2 2
E
r0 (s) = √12 cos ln √s2 + 1 − √12 sin ln √s2 + 1 , √12 sin ln √s2 + 1 + √12 cos ln √s2 + 1 , 0
so that v
u
u 1 cos2 ln √s + 1 − cos ln √s + 1 sin ln √s + 1 + 1 sin2 ln √s + 1
u 2
2
2
2
2
2
|r0 (s)| = t
+ 12 sin2 ln √s2 + 1 + sin ln √s2 + 1 cos ln √s2 + 1 + 21 cos2 ln √s2 + 1
s
s
s
= cos2 ln √ + 1
+ sin2 ln √ + 1
=1
2
2
62
CHAPTER 12. VECTOR-VALUED FUNCTIONS
d
d
d
(r · r) = |r|2 = c2 = 0 and
dt0
dt
dt
Thus, r is perpendicular to r.
47. Since
d
dt (r
· r) = r · r0 + r0 · r = 2r · r0 , we have r · r0 = 0.
48. Let v = ai + bj and r(t) = x(t)i + y(t)j. Then
Z
b
Z
v · r(t)dt =
a
b
Z
b
Z
Z
b
r(t)dt.
a
a
a
a
b
y(t)dt = v ·
x(t)dt + b
[ax(t) + by(t)]dt = a
Rt
49. From r(t) = r0 + tv, we get r0 (t) = v so that |r0 (t)| = |v|. Therfore s = 0 |r0 (t)|du =
Rt
s
|v|du = |v|t which gives t =
. Substituting for t in r(t), we have r0 (s) = r0 +
0
|v|
v
|v|
s
0
0
v . Note that r (s) = |v| so that |r (s)| = |v| = 1.
|v|v = r0 + s
|v|
p
s
3 −4
,
50. (a) |h3, −4i| = 32 + (−4)2 = 5 so r(s) = h1, 2i + h3, −4i = h1, 2i + s
5
5 5
√
√
(b) r(t)
+ th1, 2, −1i and |h1, 2, −1i| = 1 + 4 + 1 = 6 so r(s) = h1, 1, 10i +
= h1, 1, 10i 2 −1
1
s √ ,√ ,√
6
6
6
12.3
Motion on a Curve
1. v(t) = 2ti + t3 j; v(1) = 2i + j; |v(1)| =
a(t) = 2i + 3t2 j; a(1) = 2i + 3j
√
4+1=
√
5;
y
a
v
x
√
√
2
j; v(1) = 2i − 2j; |v(1)| = 4 + 4 = 2 2;
3
t
6
a(t) = 2i + 4 j; a(1) = 2i + 6j
t
2. v(t) = 2ti −
y
a
v
x
www.elsolucionario.org
12.3. MOTION ON A CURVE
63
3. v(t) = −2 sinh 2ti + 2 cosh 2tj; v(1) = 2j; |v(0)| = 2;
a(t) = −4 cosh 2ti + +4 sinh 2tj; a(0) = −4i
y
v
a
x
√
1
4. v(t) = −2 sin ti + cos tj; v(π/3) = − 3i + j;
2
p
√
1/4
=
13/2;
a(t)
=
−2
cos
ti − sin tj;
|v(π/3)| = 3 +
√
3
a(π/3) = −i −
j
2
y
v
a
x
5. v(t) = (2t − 2)i + k; v(2) = 2j + k; |v(2)| =
a(t) = 2j; a(2) = 2j
√
4+1=
√
5;
z
v
a
y
x
6. v(t) = i + j; v(2) = i + j + 12k; |v(2)| =
a(t) = 6tk; a(2) = 12k
√
1 + 1 + 144 =
√
146;
z
a
v
y
x
64
CHAPTER 12. VECTOR-VALUED FUNCTIONS
7. v(t) = i + 2tj + 3t2 k;
√
√
mathbf v(1) = i + 2j + 3k; |v(1)| = 1 + 1 + 9 = 14;
a(t) = 2j + 6tk; a(1) = 2j + 6k
z
a
v
y
x
8. v(t) = i + 3t2 j + k;
√
√
v(1) = i + 3j + k; |v(1)| = 1 + 9 + 1 = 11;
a(t) = 6tj; a(1) = 6j
z
v
a
y
x
9. The particle passes through the xy-plane when z(t) = t2 −5t = 0 or t = 0, 5 which gives us the
points (0, 0, 0) and (25, 115, 0). v(t) = 2ti + (3t2 − 2)j + (2t − 5)k; v(0) = −2j − 5k, v(5) =
10i + 73j + 5k; a(t) = 2i + 6tj + 2k; a(0) = 2i + 30j + 2k
10. If a(t) = 0, then v(t) = c1 and r(t) = c1 t + c2 . The graph of this equation is a straight line.
√
11. Initially we are given s0 = 0 and v0 = (480 cos 30◦ )i + (480 cos 30◦ )j = 240 3i + 240j. Using
a(t) = −32j we find
Z
v(t) =
a(t)dt = −32tj + c
√
240 3i + 240j = v(0) = c
√
√
v(t) = −32tj + 240 3i + 240j = 240 3i + (240 − 32t)j
Z
√
r(t) = v(t)dt = 240 3ti + (240t − 16t2 )j + b
0 = r(0) = b.
√
√
(a) The shell’s trajectory is given by r(t) = 240 3ti + (240t − 16t2 )j or x = 240 3t, y =
240 − 16t2 .
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum
altitude is y(15/2) = 900 ft.
(c) Solving y(t) = 240t − 16t2 = 16t(15 − t) = 0, we see that the
√ shell is at ground level
when t = 0 and t = 15. The range of the shell is s(15) = 3600 3 ≈ 6235 ft.
12.3. MOTION ON A CURVE
65
(d) From (c), impact is when t = 15. The speed at impact is
p
√
|v(15)| = |240 3i + (240 − 32 · 15)j| = 2402 · 3 + (−240)2 = 480 ft/s.
√
12. Initially we are given s0 = 1600j and v0 = (480 cos 30◦ )i + (480 sin 30◦ )j = 240 3i + 240j.
Using a(t) = −32j we find
Z
v(t) = a(t)dt = −32tj + c
√
240 3i + 240j = v(0) = c
√
√
v(t) = −32tj + 240 3i + 240j = 240 3i + (240 − 32t)j
Z
√
r(t) = v(t)dt = 240 3ti + (240t − 16t2 )j + b
1600j = r(0) = b.
√
√
(a) The shell’s trajectory is given by r(t) = 240 3ti+(240t−16t2 +1600)j or s = 240 3t, y =
240t − 16t2 + 1600.
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum
altitude is y(15/2) = 2400 ft.
(c) Solving y(t) = −16t2 + 240t + 1600 = −16(t − 20)(t + 5) = 0, we
√ see that the shell hits
the ground when t = 20. The range of the shell is x(20) = 4800 3 ≈ 8314 ft.
(d) From (c), impact is when t = 20. The speed at impact is
p
√
√
|v(20)| = |240 3i + (240 − 32 · 20)j| = 2402 · 3 + (−400)2 = 160 13 ≈ 577 ft/s.
13. We are given s0 = 81j and v0 = 4i. Using a(t) = −32j, we have
Z
v(t) =
a(t)dt = −32tj + c
4i = v(0) = c
v(t) = 4i − 32tj
Z
r(t) =
v(t)dt = 4ti − 16t2 j + b
81j = r(0) = b
r(t) = 4ti + (81 − 16t2 )j.
Solving y(t) = 81 − 16t2 = 0, we see that the car hits the water when t = 9/4. Then
p
√
|v(9/4)| = |4i − 32(9/4)j| = 42 + 722 = 20 13 ≈ 72.11ft/s.
www.elsolucionario.org
66
CHAPTER 12. VECTOR-VALUED FUNCTIONS
14. Let θ be the angle of elevation. Then v(0) = 98 cos θi + 98 sin θj. Using a(t) = −9.8j, we have
Z
v(t) = a(t)dt = −9.8tj + c
98 cos θi + 98 sin θj = v(0) = c
v(t) = 98 cos θi + (98 sin θ − 9.8t)j
r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j + b.
Since r(0) = 0, b = 0 and r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j. Setting y(t) = 98t sin θ −
4.9t2 = t(98 sin θ − 4.9t) = 0, we see that the projectile hits the ground when t = 20 sin θ.
Thus, using x(t) = 98t cos θ, 490 = s(t) = 98(20 sin θ) cos θ or sin 2θ = 0.5. Then 2θ = 30◦ or
150◦ . The angles of elevation are 15◦ and 75◦ .
√
√
s 2
s 2
i+
j. Using a(t) =
15. Let s be the initial speed. Then v(0) = s cos 45◦ i + s sin 45◦ j =
2
2
−32j, we have
Z
v(t) = a(t)dt = −32j + c
√
√
s 2
s 2
i+
j = v(0) = c
2
2
!
√
√
s 2
s 2
i+
− 32t j
v(t) =
2
2
!
√
√
s 2
s 2
r(t) =
ti +
t − 16t2 j + b.
2
2
Since r(0) = 0, b = 0 and
√
s 2
r(t) =
ti +
2
!
√
s 2
2
t − 16t j.
2
√
√
2
Setting
√ y(t) = s 2t/2 − 16t = t(2 √2/2 − 16t) = 0 we see that the ball hits the ground when
t√
= 2s/32. Thus, using x(t) = s 2t/2 and the fact that 100 yd = 300 ft, 300 = x(t) =
√
s 2 √
s2
( 2s/32) =
and s = 9600 ≈ 97.98 ft/s.
2
32
16. Let s be the initial speed and θ the initial angle. Then v() = s cos θi + s sin θj. Using a(t) =
−32j, we have
Z
v(t) = a(t)dt = −32tj + c
s cos θi + s sin θj = v(0) = c
v(t) = s cos θi + (s sin θ − 32t)j
r(t) = st cos θi + (st sin θ − 16t2 )j + b.
12.3. MOTION ON A CURVE
67
Since r(0) = 0, b = 0 and r(t) = st cos θi + (st sin θ − 16t2 )j. Setting y(t) = st sin θ −
16t2 = t(s sin θ − 16t) =, we see that the ball hits the ground when t = (s sin θ)/16. Using
x(t) = st cos θi, we see that the range of the ball is
s sin θ
s2 sin θ cos θ
s2 sin 2θ
x
=
=
.
16
16
32
√ 2
◦
2
◦
◦
2
◦
For
√ θ2 = 30 , the range is s sin 60 /32 = ◦3s /64 and for θ = 60 the range is s sin 120 /32 =
3s /64. In general, when the angle is 90 − θ then range is
[s2 sin 2(90◦ − θ)]/32 = s2 [sin(180◦ − 2θ)]/32 = s2 (sin 2θ)/32.
Thus, for angles θ and 90◦ − θ, the range is the same.
17. r0 (t) = v(t) = −r0 ω sin ωti + r0 ω cos ωtj; v = |v(t)| =
00
2
2
ω = v/r0 ; a(t)
q = r (t) = −r0 ω cos ωti − r0 ω sin ωtj
q
r02 ω 2 sin2 ωt + r02 ω 2 cos2 ωt = r0 ω
r02 ω 4 cos2 ωt + r02 ω 4 sin2 ωt = r0 ω 2 = r0 (v/r0 )2 = v 2 /r0 .
p
√
18. (a) v(t) = −b sin ti + b cos tj + ck; |v(t)| = b2 sin2 t + b2 cos2 t + c2 = b2 + c2
√
Rt
Rt√
ds √ 2
(b) s = 0 |v(t)|du = 0 b2 + c2 du = t b2 + c2 ;
= b + c2
dt
p
d2 s
= 0; a(t) = −b cos ti−b sin tj; |a(t)| = b2 cos2 t + b2 sin2 t = |b|. Thus, d2 s/dt2 =
(c)
6
2
dt
|a(t)|.
a = |a(t)| =
19. Let the initial speed of the projectile be s and let the target be
at (x0 , y0 ). Then vp (0) = s cos θi + s sin θj and vt (0) = 0. Using
a(t) = −32j, we have
R
vp (t) = a dt = −32tj + c
s cos θi + s sin θj = vp (0) = c
vp (t) = s cos θi + (s sin θ − 32t)j
rp (t) = st cos θi + (st sin θ − 16t2 )j + b.
y
(x0,y0)
x0 tan θ
θ
x0
x
Since rp (0) = 0, b = 0 and rp (t) = st cos θi + (st sin θ − 16t2 )j. Also, vt (t) = −32tj + c and since
vt (0) = 0, c = 0 and vt (t) = −32tj. Then rt (t) = −16t2 tj + b. Since rt (0) = x0 i + y0 j, bx0 i + y0 j
and rt (t) = x0 i+(y0 −16t2 )j. Now, the horizontal component of rp (t) will be x0 when t = x0 /s cos θ
at which time the vertical component of rp (t) will be
(sx0 /s cos θ) sin θ − 16(x0 /s cos θ)2 = x0 tan θ − 16(x0 /s cos θ)2 = y−) − 16(x0 /s cos θ)2 .
Thus, rp (x0 /s cos θ) = rt (x0 /s cos θ) and the projectile will strike the target as it falls.
20. The initial angle is θ = 0, the initial height is 1024 ft, and the initial speed is s = 180(5280)/3600 =
264 ft/s. Then x(t) = 264t and y(t) = −16t2 + 1024. Solving y(t) = 0 we see that the pack
hits the ground at t = 8 seconds. The horizontal distance tranvelled is x(8) = 2112 feet. From
the figure in the text, tan α = 1024/2112 = 16/33 and α ≈ 0.45 radian or 25.87◦ .
21. By Problem 17, a = v 2 /v0 = 15302 /(4000 · 5280) ≈ 0.1108. We are given mg = 192, so
m = 192/32 and we = 1192 − (192/32)(0.1108) ≈ 191.33 lb.
68
CHAPTER 12. VECTOR-VALUED FUNCTIONS
22. By problem 17, the centripetal acceleration is v 2 /r0 . Then
<mv2/r0, 32 m>
the horizontal force is mv 2 /r0 . The vertical force is 32m.
The resultant force is U = (mv 2 /r0 )i + 32mj. From the
figure, we see that tan φ = (mv 2 /r0 )/32m = v 2 /32r0 . Using
r0 = 60 and v = 44 we obtain tan φ = 442 /32(60) ≈ 1.0083
and φ ≈ 45.24◦ .
< 0, 32m>
φ
<mv2/r0, 0>
23. Solving x(t) = (v0 cos θ)t for t and substituting into y(t) − 21 gt2 + (v0 sin θ)t + s0 we obtain
1
y=− g
2
x
v0 cos θ
2
+ (v0 sin θ)
x
g
x2 + (tan θ)x + s0 ,
+ s) = − 2
v0 cos θ
2v0 cos2 θ
which is the equation of a parabola.
24. Since the projectile is launched from ground level, s0 = 0. To find the maximum height
we maximize y(t) = − 21 gt2 + (v0 sin θ)t. Solving y 0 (t) = −gt + v0 sin θ = 0, we see that
t = (v0 /g) sin θ is a critical point. Since y 00 (t) = −g ≤ 0,
H=y
v0 sin θ
g
=
1 v02 sin2 θ
v0 sin θ
v02 sin2 θ
g
+
v
sin
θ
=
0
2
g2
g
2g
is the maximum height. To find the range we solve y(t) = − 21 gt2 + (v0 sin θ)t = t(v0 sin θ −
1
2 gt) = 0. The positive solution to this equation is t = (2v0 sin θ)/g. The range is thus
x(t) = (v0 cos θ)
2v0 sin θ
v 2 sin 2θ
= 0
.
g
g
25. Letting r(t) = x(t)i + y(t)j + z(t)k, the equation dr/dt = v is equivalent to dx/dt =
6t2 x, dy/dt = −4ty 2 , dz/dt = 2t(z + 1). Separating the variables and integrating, we
obtain x/x = 6t2 dt, dy/y 2 = −4tdt, dz/(z + 1) = 2tdt, and ln x = 2t3 + c1 , −1/y =
2t2 + c2 , ln(z + 1) + t2 + c3 . Thus,
3
r(t) = k1 e2t i +
2
1
j + (k3 et − 1)k.
2t2 + k2
26. We require the fact that dr/dt = v. Then
d
dp dr
dL
= (r × p = r
+
× p = τ + v × p = τ + v × mv = τ + m(v × v) = τ + 0 = τ.
dt
dt
dt
dt
27. (a) Since F is directed along r we have F = cr for some constant c. Then
τ = r × F = r × (cr) = c(r × r) = 0.
(b) If τ = 0 then dL/dt = 0 and L is constant.
www.elsolucionario.org
12.4. CURVATURE AND ACCELERATION
69
28. (a) Since the cannon is pointing directly to the left, tha parmetric equations describing the
path of the cannon ball are given by
1
x(t) = v0 t, y(t) = − gt2 + s0
2
r
2s0
The cannon ball will touch the groun when y = 0, which occurs at t =
. At that
g
r
r
2s0
2s0
time, x is given by x =
= −v0
. Notice that this x value will be farther
g
g
to the left with increasing values of v0 . Therefore, the cannon ball travels farther with
more gunpowder.
r
2s0
(b) As shown in part (a), the cannon ball will touch the groun when t =
. This value
g
of t is independent of v0 . This occurs because v0 has no vertical component.
(c) If the cannon ball is dropped, we have v0 = 0. Therefore, the parametric equations
describing the cannon ball motion are given by
1
x(t) = 0, y(t) = − gt2 + s0 .
2
r
2s0
. Therefore the cannon ball touches the ground at the
As before, y = 0 when t =
g
same time regardless of whether it is fired or dropped.
12.4
Curvature and Acceleration
1. r0 (t) = −t sin ti + t cos tj + 2tk;
sin t
cos t
2
T=− √ i+ √ j+ √ k
5
5
5
|r0 (t)| =
p
√
t2 sin2 t + t2 cos2 t + 4t2 = 5t;
√
2. r0 (t) = et (− sin t + cos t)i + et (cos t + sin t)i + 2et k,
√
|r0 (t)| = [et (sin2 t−2 sin t cos t+cos2 t)+e2t (cos2√t+2 sin t cos t+sin2 t)+2e2t ]1/2 = 4e2t = 2et ;
1
1
2
T(t) = (− sin t + cos t)i + (cos t + sin t)j +
k
2
2
2
p
3. √
We assume a > 0. r0 (t) = −a sin ti + a cos tj + ck; |r0 (t)| = a2 sin2 t + a2 cos2 t + c2 =
a2 + c2 ;
a sin t
a cos t
c
dT
a cos t
a sin t
T(t) − √
i+ √
j+ √
k;
= −√
i− √
j,
2 + c2
2 + c2
2 + c2
2 + c2
dt
a
a
a
a
a2 + c2
s
a2 cos2 t a2 sin2 t
a
+ 2
=√
; N = − cos ti − sin tj;
2
a2 + c2
a + c2
a + c2
i
j
k
a cos t
c
a sin t
c sin t
c cos t
a
√
√
B = T×N = −√ 2
=√
i− √
+√
k;
2
2
2
2
a +c
a +c
a + c2
a2 + c2
a2 + c2
a2 + c2
− cos t
− sin t
0
√
|dT/dt|
a/ a2 + c2
a
κ=
= √
= 2
r0 (t)
a + c2
a 2 + c2
dT
=
dt
70
CHAPTER 12. VECTOR-VALUED FUNCTIONS
√
√
|r0 ; (1)| = 3;
1
T(t) = (1 + t2 + t4 )−1/2 (i + tj + t2 k), T(1) √ (i + j + k);
3
dT
1
t
2
4 −3/2
3
2
= − (1 + t + t )
(2t + 4t )i + [(1 + t + t)−1/2 − (1 + t2 + t)−3/2 (2t + 4t3 )]j
dt
2
2
2
2
4 −1/2 t
2
4 −3/2
3
[2t(1 + t + t )
(1 + t + t )
(2t + 4t )]k;
2
r
√
d
1
1 1
2
1
d
1
T(1) = − √ i + √ k,
T(1) =
+ = √ ; N(1) = − √ (i − k)k,
dt
dt
3 3
3
3
3
2
i√
j√
k√
1
B(1) = 1/ √3 1/ 3 1/√3 = √ (i − 2j + k);
6
−1/ 2
0
1/ 2
√ √
√
d
2/ 3
2
κ=
=
T(1) = |r0 (1)| = √
dt
3
3
√
1
5. From Example 1 in the text, a normal to the osculating plane is B(π/4) = 26
(3i − 3j + 2 2k).
√ √
The point on√the curve
An equation
√
√ when t = π/4 is ( 2, 2, 3π/4).
√
√ of the√plane is 3(x −
2) − 3(y − 2) + 2 2(z − 3π/4(= 0, 3x − 3y + 2 2z = 3π/2, or 3 2x − 3 2y + 4z = 3π.
4. r0 (t) = i + tj + t2 k;
|r0 (t)| =
1 + t2 + t4 ,
6. From Problem 4, a normal to the osculating plane is B(1) = √16 (i − 2j + k). The point on the
curve when t = 1 is (1, 1/2, 1/3). An equaiton of the plane is (x−1)−2(y −1/2)+(z −1/3) = 0
or x − 2y + z = 1/3.
√
7. v(t) = j + 2tk, |v(t)| = 1 + 4t2 ; a(t) = 2k; v · a = 4t, v × a = 2i, |v × a| = 2;
4t
2
aT = √
, aN = √
2
1 + 4t
1 + 4t2
8. v(t) = −3
p
psin ti + 2 cos tj + k,
√ p
|v(t)| = 9 sin2 t + 4 cos2 t = 1 = 5 sin2 t + 4 sin2 t + 4 cos2 t + 1 = 5 sin2 +1;
a(t) = −3 cos ti − 2 sin tj; v · a = 9 sin t cos
cos t,
qt − 4 sin t cos t = 5 sin t√
√
2
v × a = 2 sin ti − 3 cos tj + 6k, |v × a| = 4 sin +(cos2 t + 36 = 5 cos2 t + 8;
s
√
5 sin t cos t
cos2 t + 8
aT p
, aN =
sin2 t + 1
sin2 t + 1
√
9. v(t) = 2ti + 2tj + 4tk, |v(t)| = 2 6t, t > 0; a(t) = 2i + 2j + 4k; v · a = 24t, v × a = 0;
√
24t
aT = √ = 2 6, aN = 0, t > 0
2 6t
√
10. v(t) = 2ti − 3t2 j = 4t3 k, |v(t)| = t 4 + 9t2 + 16t4 , t >); a(t) = 2i −√
6tj + 12t2 k;
3
5
4
3
2
2
v · a = 4t + 18t + 48t ; v × a =√−12t i − 16t j − 6t k, |v × a| = 2t 36t4 + 64t2 + 9;
4 + 18t2 + 48t4
2t 36t4 + 64t2 + 9
aT = √
, aN = √
t>0
4 + 9t2 + 16t4
4 + 9t2 + 16t4
√
11. v(t) = 2i + 2tj, |v(t)| = 2 1 + t2 ; a(t) = 2j; v × a = 4k, |v × a| = 4;
2t
2
aT = √
, aN = √
2
1+t
1 + t2
12.4. CURVATURE AND ACCELERATION
71
√
t
2t
1 − t2
1
1 + t2
i+
j, |v(t)| =
; a(t) = −
i+
j;
12. v(t) =
2
2
2
2
2
1+t
1+t
1+t
(1 + t )
(1 + t2 )2
2t
t − t3
1
1
v·a=−
+
; v×a=
k, |v × a| =
;
(1 + t2 )3
(1 + t2 )3
(1 + t2 )2
(1 + t2 )2
2 3
2 2
t/(1 + t )
t
1
a/(1 + t )
aT = − √
=−
=
, aN = √
2
2
2
2
2
(1 + t )3/2
(1 + t2 )3/2
1 + t )/(1 + t
1 + t /(1 + t )
13. v(t) = −5 sin ti + 5 cos tj, |v(t)| = 5; a(t) = −5 cos ti − 5 sin tj;
v × a = 25k, |v × a| = 25; aT = 0, aN = 5
v · a = 0,
p
0
14. v(t) = sinh ti + cosh tj, |v(t)| = sinh t2 + cosh2 t a(t) = cosh ti + sinh tj
v · a = 2 sinh t cosh t; v × a = (sinh2 t − cosh2 t)k = −k, |v × a| = 1;
1
2 sinh t cosh t
, aN = p
aT = p
2
2
2
sinh + cosh
sinh + cosh2
√
15. v(t) = et (i + j + k), √|v(t)| = 3e−t ; a(t) = e−t (i + j + k); v · a = −3e−2t ; v × a = 0,
|v × a| = 0; aT = − 3e−t , aN = 0
√
16. v(t) = i + 2j + 4k, |v(t)| = 21; a(t) = 0; v · a = 0, v × a = 0, |v × a| = 0; aT =
0, aN = 0
p
17. v(t) = −a sin ti + b cos tj + ck, |v(t)| = a2p
sin2 t + b2 cos2 +c2 ; a(t) = −a cos ti − b sin tj;
v × a = bc sin tip− ac cos tj + abk, |v × a| = b2 c2 sin2 t + a2 c2 cos2 t + a2 b2
b2 c2 sin2 t + a2 c2 cos2 t + a2 b2
|v × a|
=
κ=
|v|3
(a2 sin2 t + b2 cos2 t + c2 )3/2
p
18. (a) v(t) = −a sin ti + b cos tj, |v(t)| = a2 sin2 t + b2 cos2 t; a(t) = −a cos ti − b sin tj;
ab
v × a = abk; |v × a| = ab; κ = 2 2
(a sin t + b2 cos2 t)3/2
(b) When a = b, |v(t)| = a, |v × a| = a2 , and κ = a2 /a3 = 1/a.
19. The equation of a line is v(t) = b + tc, when b and c are constant vectors.
v(t) = c, |v(t)| = |c|; a(t) = 0; v × a = 0; κ = |v × a|/|v|3 = 0
20. v(t) = a(1 − cos t)i + a sin tj; v(π) = 2ai, |v(π)| = 2a; a(t) = a sin ti + a cos tj,
i
j k
|v × a|
2a2
1
a(π) = −aj; |v × a| = 2a 0 0 = −2a2 k; |v × a| = 2a2 ; κ =
=
=
|v|3
8a3
4a
0 −a 0
p
21. v(t) = f 0 (t)i + g 0 (t)j, |v(t)| = [f 0 (t)]2 + [g 0 (t)]2 ; a(t) = f 00 (t)i + g 00 (t)j;
v × a = [f 0 (t)g 00 (t) − g 0 (t)f 00 (t)]k, |v × a| = |f 0 (t)g 00 (t) − g 0 (t)f 00 (t)|;
|v × a|
|f 0 (t)g 00 (t) − g 0 (t)f 00 (t)|
κ=
=
|v|3
([f ”(t)]2 + [g 0 (t)]2 )3/2
22. For y = F (x), r = xi + F (x)j. We identify f (x) = x and g(x) = F (x) in Problem 21. Then
f 0 (x) = 1, f 00 (x) = 0, g 0 (x) = F 0 (x), g 00 (x) = F 00 (x), and κ = |F 00 (x)|/(1 + [F 0 (x)]2 )3/2 .
www.elsolucionario.org
72
CHAPTER 12. VECTOR-VALUED FUNCTIONS
23. F (x) = x2 ,
F 00 (x) = 2,
F (0) = 0,
F (1) = 1;
F 00 (0) = 2,
F 00 (1) = 2;
F 0 (x) = 2x,
F 0 (0) = 0, F 0 (1) = 2;
2
1
κ(0) =
= 2; ρ(0) = ;
2
3/2
2
(1 + 0 )
2
2
= √ ≈ 0.18;
2
3/2
(1√+ 2 )
5 5
√
5 5
≈ 5.59; Since 2 > 2/5 5, the curve is ”sharper” at (0, 0).
ρ(1) =
2
κ(1) =
24. F (x) = x3 ,
F (−1) = −1,
F 0 (1/2) = 3/4;
F (1/2) = 1/8;
F 00 (x) = 6x, F 00 (−1) = −6,
F 0 (x) = 3x2 ,
F 0 (−1) = 3,
F 00 (1/2) = 3;
κ(−1) =
3
√ ≈ 0.19;
5 10
√
5 10
ρ(−1) =
≈ 5.27;
3
3
3
125
=
≈ 1.54; ρ( 21 ) =
≈ 0.65
κ( 12 ) =
125/64
192
[1 + (3/4)2 ]3/2
Since 1.54 > 0.19, the curve is ”sharper” at (1/2, 1/8).
|−6|
(1+32 )3/2
=
6
√
10 10
=
|F 00 (x)|
.
|1 + (F 0 (x))2 |3/2
2
Now, F 0 (x)2x, F 00 (x) = 2, and (F 0 (x))2 = 4x2 so that κ =
.
(1 + 4x2 )3/2
As x → ±∞, the denominator grows without bound. Therefore, κ(x) → 0 as x → ±∞.
25. Letting F (x) = x2 , we can use Problem 22 to get κ(x) =
y
x
26. (a)
√
3t(2t2 + 1) t4 + 4t2 + 1
2t(t2 + 2)
√
−
;
(t4 + t2 + 1)5/2
(t4 + t2 + 1)3/2 t4 + 4t2 + 1
critical numbers occur at t = −.271469, t = 0, and t = .271469.
(b) κ0 (t) =
(c) Maximum of 1.017182 occurs at t = −.271469 and t = .271469.
27. Since (c, F (c)) is an inflection point and F 00 exists on an interval containg c, we must have
F 00 (c) = 0. Therefore, using the formula from Problem 22, we see that the curvature is zero.
28. We use the fact that T · N = 0 and T · T = N · N = 1. Then
|a(t)|2 = a · a = (an N + at T) · (an N + at T) = a2N N · N + 2an at N · T + a2T T · T = a2N + a2T .
CHAPTER 12 IN REVIEW
73
Chapter 12 in Review
A. True/False
1. True; |v(t)| =
√
2
2. True; the curvature of a circle of radius a is κ = a1 .
3. True
4. False; consider r(t) = t2 i. In this case, v(t) = 2ti and a(t) = 2i. Since v · a = 4t, the velocity
and acceleration vectors are not orthogonal for t 6= 0.
5. True
6. False; see Problem 20c in Section 14.2
7. True
8. True
9. False; consider r1 (t) = r2 (t) = i.
10. True,
d
dr dr
dr
d
|r(t)|2 = (r · r) = r ·
+
· r = 2r · .
dt
dt
dt
dt
dt
B. Fill in the Blanks
1. y = 4
2. 0
3. r0 (t) = h1, 2t, t2 i so r0 (1) = h1, 2, 1i
4. r00 (t) = h0, 2, 2ti so r00 (1) = h0, 2, 2i
5. r0 (1) × r00 (1) =
i
1
0
j
2
2
j
1
2
= h2, −2, 2i so r0 (1) × r00 (1) =
√
12.
√
√
r0 (1) × r00 (1)
12
2
√
=
.
=
|r0 (1)|3
6
6 6
r0 (1)
h1, 2, 1i
1
2
1
6. T(1) = 0
= √
= √ ,√ ,√
|r (1)|
6
6
6
6
0
2
r (t)
h1, 2t, t i
1
2t
t2
7. T(t) = 0
=√
= √
,√
,√
|r (t)|
1 + 4t2 + t4
1 + 4t2 + t4
1 + 4t2 + t4 1 + 4t2 + t4
2
4
−2(t + 2)
−2(t − 1)
2t(2t2 + 1)
.
So T0 (t) =
, 4
, 4
4
2
3/2
2
3/2
(t + 4t
(t + 4t + 1)
(t + 4t2 + 1)3/2
+ 1)
q
−6
1
6
−1
1
1
1
0
√
√
√ .
This gives T0 (1) =
,
0,
=
,
0,
and
|T
(1)|
=
6 + 6 =
3/2
63/2
6
6
3
D
E 6
−1
√
√1
,
0,
0
T (1)
−1
1
6
6
Therefore N(1) = 0
=
= h √ , 0, √ i.
1
√
|T (1)|
( 3)
2
2
Since r0 (1)| =
√
6, we have κ(1) =
74
CHAPTER 12. VECTOR-VALUED FUNCTIONS
8. B(1) = T(1) × N(1) =
i
j
k
√1
6
−1
√
2
√2
6
√1
6
√1
2
0
=
9. A normal to the normal plane is T(1) =
D
1 −1 1
√ ,√ ,√
3
3
3
√1 , √2 , √1
6
6
6
E
so we can use n = h1, 2, 1i as a vector
1
3 i,
normal to the plane. Since r(1) = h1, 1,
the point (1, 1, 13 ) lies on the normal plane at t = 1.
Thus an equation of the normal plane is (x − 1) + 2(y − 1) + (z − 13 ) = 0 or x + 2y + z = 1)
3
or 3x + 6y + 3z = 10
D
E
−1 √1
10. A normal to the osculating plane is B(1) = √13 , √
,
. So we can use n = h1, −1, 1i as a
3
3
normal vector. Using the point (1, 1, 31 ), an equation of the osculating plane is (z − 1) − (y −
1) + (z − 13 ) = 0 or x − y + z = 31 or 3x − 3y + 3z = 1.
C. Exercises
√
Rπ√
cos2 t + sin2 +1dt = 0 2dt = 2π
√
√
√
√
Rt√
2. r0 (t) = 5i + j + 7k; s(t) = 0 25 + 1 + 49du = 5 3t; s(3) = 15 3. Solving 5 3t = 80 3,
√
we see that the distance traveled will be 80 3 when t = 16 or at the point (80, 17, 112).
1. r0 (t) = cos ti + sin tj + k;
3. r(3) = −27i + 8j + k;
s=
Rπp
0
r0 (t) = −6ti = √
is x = −27 − 18t, y = 8 + t, z = 1 + t.
2
+ k;
t+1
r0 (2) = −18i + j + k. The tangent line
5.
4.
z
z
y
x
y
x
6.
d
d
d
[r1 (t) × r2 (t)] = r1 (t) × r2 (t) + r1 (t) × r2 (t)
dt
dt
dt
= (t2 i + 2tj + t3 k) × (−i + 2tj + 2tk) + (2ti + 2j + 2t2 k) × [−ti + t2 j + (t2 + 1)k]
= (4t2 − 2t4 )i − 3t3 j + (2t3 + 2t)k + (2t2 + 2 − 3t4 )i − (5t3 + 2t)j + (2t3 + 2t)k
= (2 + 6t2 − 5t4 )i − (8t3 + 2t)j + (4t3 + 4t)k
d
d
[r1 (t) × r2 (t)] = [(2t3 + 2t − t5 )i − (2t4 + t2 )j + (t4 + 2t2 )k]
dt
dt
= (2 + 6t2 − 5t4 )i − (8t3 + 2t)j + (4t3 + 4t)k
www.elsolucionario.org
CHAPTER 12 IN REVIEW
7.
75
d
d
d
[r1 (t) · r2 (t)] = r1 (t) · r2 (t) + r1 (t) · r2 (t)
dt
dt
dt
= (cos ti − sin tj + 4t3 k) · (2ti + sin tj + 2e2t k)
(− sin ti − cos tj + 12t2 k) · (t2 i + sin tj + e2t k)
= (2t cos t − sin t cos t + 8t3 e2t − t2 sin t − sin t cos t + 12t2 e2t
= 2t cos t − t2 sin t − 2 sin t cos t + 8t3 e2t + 12t2 e2t
d
d
[r1 (t) · r2 (t)] = [t2 cos t − sin2 t + 4t3 e2t ] = −t2 sin t + 2t cos t − 2 sin t cos t + 8t3 e2t + 12t2 e2t
dt
dt
8.
d
d
[r1 (t) · (r2 (t) × r3 (t))] = r1 (t) · [r2 (t) × r3 (t)] + r0 (t) · [r2 (t) × r3 (t)]
dt
dt
= r1 (t) · [(r2 (t) × r03 (t)) + (r02 (t) × r3 (t))] + r01 (t) · (r2 (t) × r3 (t))
= r1 (t) · (r2 (t) × r03 (t)) + r1 (t) · r02 (t) × r3 (t)) = r01 (t) · (r2 (t) × r3 (t))
9. We are given F = ma = 2j; v(0) = i + j + k. and r(0) = i + j. Then
Z
Z
2
2
v(t) = a(t)dt =
jdt = tj + c
m
m
i = j + k = v(0) = c
2
v(t) = i +
t+1 j+k
m
1 2
t + t j + tk + b
r(t) = ti +
m
i + j = r(0) = b
1 2
r(t) = (t + 1)i +
t + t + 1 j + tk
m
The parametric equations are x = t, y =
1 2
t + t + 1, z = t.
m
10.
y
x
v
a
v(t) = i − 3t2 j, v(1)
√ a(t) = −6tj, a(1) = −6j
√ = i − 3j;
|v(1)| = |i − 3j| = 1 + 9 = 10
76
CHAPTER 12. VECTOR-VALUED FUNCTIONS
11. v(t) = 6i + j + 2tk; a(t) = 2k. To find when the particle passes through the plane, we solve
−6t + t + t2 = −4 or t2 − 5t + 4 = 0. This gives t = 1 and t = 4. v(1) = 6i + j + 2k, a(1) = 2k;
v(4) = 6i + j + 8k, a(4) = 2k
12. We are given r(0) = i + 2j + 3k.
Z
r(t) =
Z
v(t)dt =
(−10ti + (3t2 − 4t)j + k)dt = −5t2 i + (t3 − 2t2 )j + tk + c
i + 2j + 3k = r(0) = c
r(t) = (1 − 5t2 )i + (t3 − 2t2 + 2)j + (t + 3)k
r(t) = −19i + 2j + 5k
√
√
√
R √
13. v(t) = a(t)dt = ( 2 sin ti + 2 cos tj)dt = − 2 cos
√ ti + 2 sin
√tj + c;
−i + j + √
k = v(π/4)
=
−i
+
j
+
c,
c
=
k;
v(t)
=
−
2
cos
ti
+
2 sin tj + k;
√
r(t) = − 2 sin
ti−
2
cos
tj+tk+b;
i+2j+(π/4)k
=
r(π/4)
=
−i−j+(π/4)k+b,
b = 2i+3j;
√
√
r(t) = (2 − 2 2 sin t)i + (3 − 2 cos t)j + tk; r(3π/4) = i + 4j + (3π/4)k
√
3
3
14. v(t) = ti + t2 j − tk; |v| = t t2 + 2, t > 0; a(t) = i + 2tj − k; v · a = t + 2t
√ + t = 2t√+ 2t ;
3
2
2
√
2t + 2t
2 + 2t
t 2
2t
v × a = t2 bi + t2 k, |v × at2 2; aT = √
= √
, aN = √
= √
;
2
2
2
2
t t +2
t +2
t t +2
t +2
√
√
t2 2
2
=
κ= 3 2
3/2
2
t (t + 2)
t(t + 2)3/2
R
0
15. r0 (t) = sinh
= sinh 1i + cosh 1j + k;
p ti + cosh tj + k, r (1)√
√
√
2
2
0
|r (t)| = sinh t + cosh t + 1 = 2 cosh2 t = 2 cosh t; |r0 (1)| = 2 cosh 1;
1
1
1
1
T = √ tanh ti + √ j + √ sech tk, T(1) = √ (tanh 1i + j + sech 1k);
2
2
2
2
dT
1
d
1
1
1
2
= √ sech ti − √ sech t tanh tk;
T(1) = √ sech2 1i − √ sech 1 tanh 1k,
dt
dt
2
2
2
2
sech 1 p
d
1
2
2
T(1) == √
sech 1 + tanh +1 = √ sech 1; N(1) = sech 1i − tanh 1k;
dt
2
2
1
1
1
B(1) = T(1) × N(1) = − √ tanh 1i + √ (tanh2 1 + sech2 1)j − √ sech 1k
2
2
2
1
= √ (− tanh 1i + j − sech 1k)
2
√
(sech 1)/ 2
1
d
T(1) /|r0 (1)| = √
= sech2 1
κ=
dt
2
2 cosh 1
16. The parametric equations describing the path of the ball are
√
x(t) = 66 cos(30◦ )t = 33 3ty(t) = −16t2 + 66 sin(30◦ )t + 148 = −16t2 + 33t + 148
The ball touches the ground when y(t) = 0 or −16t2 + 33t + 148 = 0. This occurs when
t ≈ 4.243. The ball therefore strikes the ground √
at x(4.243) = 242.52 ft.
The velocity of the √
ball at time t is v(t) = h33 3, −32t + 33i. The impact velocity is given
by v(4.243) = h33 3, −32(4.243) + 33i ≈ h57.158, −102.776i. The impact speed is then
|v(4.243)| ≈ 117.6 ft/s.
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Chapter 13
Partial Derivatives
13.1
Functions of Several Variables
1. {(x, y)|(x, y) 6= (0, 0)}
2. {(x, y)|x 6= x ± 3y}
3. {(t, Y )|y 6= x2 }
4. {(x, y)|y ≥ −4}
5. {(s, t)|s, t any real numbers}
S
6. {(u, b)|(u, v(6= (0, 0)} {(u, v)|u2 + v 2 6= 1}
7. {(r, s)| |s| ≥ 1}
8. {(θ, φ) | tan θ tan φ 6= 1}
kπ, k an integer}
T
{(θ, φ)
θ 6= π/2 + kπ, k an integer}
T
{(θ, φ) | φ 6= π/2 +
9. (u, v, w)|u2 + v 2 + q 2 ≥ 16
10. {(x, y, z)|x2 + y 2 + < 25 and z 6= 5}
√ √
11. (c); The domain of f (x, y) = x+ y − x is {(x, y)|x ≥ 0, y−x ≥ 0} = {(x, y)|x ≥ 0, y ≥ x}
√
12. (e); The domain of f (x, y) = xy is {(x, y)|xy ≥ 0} = {(x, y)|x ≥ 0, y ≥ 0 or x ≤ 0, y ≤ 0}
13. (b); The domain of f (x, y) = ln(x − y 2 ) is (x, y)|x − y 2 > 0 = (x, y)|x > y 2
p
x2 + y 2 − 1 14. (h); The domain of f (x, y) =
is (x, y)|x2 + y 2 − 1 ≥ 0, y 6= x =
y−x
(x, y)|x2 + y 2 ≥ 1, y 6= x
n
o n
o
q
15. (d); The domain of f (x, y) = xy − 1 is (x, y)| xy − 1 ≥ 0 = (x, y)| xy ≥ 1
77
78
CHAPTER 13. PARTIAL DERIVATIVES
16. (g); The domain of f (x, y) =
x4 + y 4
is {(x, y)|xy 6= 0} = {(x, y)|x 6= 0, y 6= 0}
xy
17. (f ); The domain of f (x, y) = sin−1 (xy) is {(x, y)||xy| ≤ 1}
p
18. (a); The domain of f (x, y) = y − x2 is {(x, y)|y − x2 ≥ 0} = ∗x, y)y ≥ x2
19. {(x, y)|x ≥ 0 and y ≥ 0}
y
x
20. {(x, y)| x2 ≤ 1 and y 2 ≥ 4}
\
{(x, y)| x2 ≥ 1 and y 2 ≤ 4}
[
{(x, y)| |x| ≤ 1 and |y| ≥ 2} {(x, y)| |x| ≥ 1 and |y| ≤ 2}
y
2
1
x
21. {(x, y)|y − x ≥ 0}
y
22. {(x, y)|xy ≥ −1}
y
x
23. {z | z ≥ 10}
25. {w ||; −1 ≤ w ≤ 1}
R4
27. f (2, 3) = 2 (2t − 1)dt = (t2 − t)|42 = 12 − 2 = 10
R1
f (−1, 1) = −1 (2t − 1)dt = (t2 − t)|1−1 = 0 − 2 = −2
x
24. all real numbers
26. {x | w < 7}
13.1. FUNCTIONS OF SEVERAL VARIABLES
f (5, −5) = ln
28. f (3, 0) = ln 9/9 = ln 1 = 0;
29. f (−1, 1, −1) = (−2)2 = 4;
79
25
1
= ln = − ln 2
25 + 25
2
f (2, 3, −2) = 22 = 4
√ √ √
30. f ( 3, 3, 6) = 1/3 + 1/2 + 1/6 = 1;
f (1/4, 1/5, 1/3) = 16 + 25 + 9 = 50
31. A plane through the origin perpendicular to the xz-plane
32. A parabolic cylinder perpendicular to the yz-plane
33. The upper half of a cone lying above the xy-plane with axis along the positive z-axis
34. The upper half of a hyperboloid of two sheets with axis lying along the positive z-axis
35. The upper half of an ellipsoid
36. A hemisphere lying below the yy-plane
37. y = − 21 x + C
38. x = y 2 − c
y
y
x
x
39. x2 − y 2 = 1 + c2
40. 4x2 + 9y 2 = 36 − c2 , −6 ≤ c ≤ 6
y
y
x
x
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80
CHAPTER 13. PARTIAL DERIVATIVES
41. y = x2 + ln c, c > 0
42. y = x + tan c, −π/2 < c < π/2
y
y
x
x
43. x2 /9 + z 2 /4 = c; elliptical cylinder
44. Setting f (x, y, z) equal to a constant√c, we have (x − 1)2 + (y − 2)2 + (z − 2)2 = c which
is the equation of a sphere of radius c centered at (1, 2, 3). Therefore, the level curves are
concentric spheres centered at (1, 2, 3).
45. x2 + 3y 2 + 6z 2 = c; ellipsoid
46. 4y − 2z + 1 = c; plane
47.
c=0
c<0
z
x
c>0
z
z
y
x
y
x
y
48. Setting x = −4, y = 2, and z = −3 in x2 /16 + y 2 /4 + z 2 /9 = c we obtain c = 3. The equation
2
2
2
of the
the x-intercepts are
√ surface is x /16 + y /4 + z /9 =√3. Setting y = z = 0 we find √
±4 3. Similarly, the y-intercepts are ±2 3 and the z-intercepts are ±3 3.
13.1. FUNCTIONS OF SEVERAL VARIABLES
81
49.
P
v
50. From V = s2 h we obtain h = V /s2 .
51. C(r, h) = πr2 (1.8) + πr2 (1) + 2πh(2.3) = 2.8πr2 + 4.6πrh
250 − xy
. Thus,
x+y
52. Let the height of the box be h. Then 2xy + 2xh + 2yh = 500 and h =
250xy + x2 y 2
.
x+y
2
53. V + πr2 g + 13 πr2 32 h = 11
9 πr h
V = xyh =
54. From the figure, we see that t = x tan θ = x
z
p
y2 − z2
xz
=p
y2 − z2
!
x
θ
t
y
θ
My2-z2
z
55. X = 2(156)(50) = 15, 600 sq cm
√
√
56. h(20, −6.67) + (10 20 − 20 + 10.5)(33 + 6.67) = (20 5 − 9.5)(39.67) ≈ 1397 kcal/m2 h
57. (a) The distance the water falls in time t is s(t) = 12 gt2 + vt where vis the velocity of the
water at the top level (t = 0). The velocity of the water at time t is v(t) = gt + v. If t1
is the time it takes a cross-section of water to fall from the top level to the bottom level,
then V = gt1 + v and t1 = (V − v)/g. The distance traveled in time t1 is
2
1
1
V −v
V −v
h = gt21 + vt1 = g
+v
2
2
g
g
Simplifying the equation we obtain 2gh = V 2 − v 2 . Now the rates at the top and bottom
levels are Z = vπr2 and Q = V πr2 (recall that the flow rate is constant). Solving for
2
2
2 2
2 2
v and V and substituting into 2gh
√ = V − v we obtain 2gh = (Q/πr ) − (Q/πR ) .
πr2 R2 2gh
Solving for Q we find Q = √
.
R4 − r 4
(b) When r = 0.2 cm, R = 1 cm, and h = 10, Q ≈ 7.61 cm3 /s.
82
CHAPTER 13. PARTIAL DERIVATIVES
13.2
1.
Limits and Continuity
(x2 + y 2 ) = 25 + 1 = 26
lim
(x,y)→(5,−1)
2.
x2 − y
4−1
=
=3
2−1
(x,y)→(2,1) x − y
lim
5x2 + y 2
5x2
=
lim
= 5.
(x,y)→(0,0) x2 + y 2
(x,y)→(0,0) x2
y2
5x2 + y 2
=
lim
= 1. The limit does not exist.
On x = 0,
lim
2
2
(x,y)→(0,0) y 2
(x,y)→(0,0) x + y
3. On y = 0,
lim
4.
4x2 + y 2
4+4
1
=
=
4
4
16 + 16
4
(x,y)→(1,2) 16x + y
5.
4−1−1
4 − x2 − y 2
=
=1
x2 + y 2
1+1
(x,y)→(1,1)
lim
lim
−y
2x2 − y
=
lim
= ∞.
(x,y→(0,0) 2y 2
(x,y→(0,0) x2 + 2y 2
2x2 − y
2x2
On y = 0,
lim
=
lim
= 2.. The limit does not exist.
(x,y→(0,0) x2 + 2y 2
(x,y→(0,0) x2
6. On x = 0,
lim
x2 y
x3
x
=
lim
=
lim
= 0.
2
4
+y
(x,y→(0,0) x + x2
(x,y→(0,0) x2 + 1
1
x4
x2 y
=
lim
= . The limit does not exist.
On y = x2 ,
lim
4
2
4
4
2
(x,y→(0,0) x + x
(x,y→(0,0) x + y
7. On y = x,
lim
(x,y→(0,0) x4
6xy 2
6x3
6x
=
lim
=
lim
= 0.
(x,y→(0,0) x2 + y 4
(x,y→(0,0) x2 + x4
(x,y→(0,0) 1 + x2
2
4
x y
6y
On x = y 2 ,
lim
=
lim
= 3. The limit does not exist.
4
2
4
(x,y→(0,0) x + y
(x,y→(0,0) y + y 4
8. On y = x,
9.
lim
lim
x3 y 2 (x + y)3 = 1(4)(27) = 108
(x,y)→(1,2)
10.
lim
(x,y)→(2,3)
11.
6
xy
6
=
=−
x2 − y 2
4−9
5
exy
1
= =1
1
(x,y)→(0,0) x + y + 1
lim
sin xy
sin mx2
=
lim
2
2
(x,y)→(0,0) x + y
(x,y)→(0,0) (1 + m2 )x2
m sin mx2
m
=
lim
=
.
1 + m2
(x,y)→(0,0) 1 + m2 mx2
The limit does not exist.
12. On y = mx,
lim
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13.2. LIMITS AND CONTINUITY
13.
lim
(x,y)→(2,2)
14.
83
xy
4
1
=
=
x3 + y 2
8+4
3
√
cos(3x + y) = cos(3π + π/4) = cos 13π/4 = − 2/2
lim
(x,y)→(π,π/4)
15.
x2 − 3y + 1
1
=−
3
(x,y)→(0,0) x + 5y − 3
lim
16. On y = mx,
x2 m2 x2
m2
x2 y 2
=
lim
=
.
4
4
4
4
+ 5y
1 + 5m4
(x,y)→(0,0) x + 5m x
lim
(x,y)→(0,0) x4
The limit does not exist.
17.
lim
xy 2
(x,y)→(4,3)
18.
19.
x + 2y
4+6
= 4(9)
= 360
x−y
4−3
0
x2 y
=
=0
1+0
(x,y)→(1,0) x+ y 3
lim
xy − x − y + 1
(x − 1)(y − 1)
=
lim
2
+ y − 2x − 2y + 2 (x,y)→(1,1) (x − 1)2 + m2 (x − 1)2
On y − x = m(x − 1),
(x − 1)(y − 1)
(x − 1)m(x − 1)
m
lim
=
lim
=
.
2
2
2
2
2
2
1 + m2
(x,y)→(1,1) (x − 1) + m (x − 1)
(x,y)→(1,1) (x − 1) + m (x − 1)
The limit does not exist.
lim
(x,y)→(1,1) x2
20. On x = 0,
lim
(x,y)→(0,3)
21.
−3y
xy − 3y
=
lim
. The limit does not exist.
x2 + y 2 − 6y + 9 (x,y)→(0,3) (y − 3)2
x3 y + xy 3 − 3x2 − 3y 2
xy(x2 + y 2 ) − 3(x2 + y 2
=
lim
x2 + y 2
x2 + y 2
(x,y)→(0,0)
(x,y)→(0,0)
lim
=
lim
(xy − 3) = −3
(x,y)→(0,0)
22.
23.
4
y 3 + 2x3
8 − 16
=
=
−2 − 40
21
(x,y)→(−2,2) x + 5xy 2
lim
lim
ln(2x2 − y 2 ) = ln(2 − 1) = 0
(x,y)→(1,1)
24.
sin−1 (x/y)
sin−1 (1/2)
π/6
1
=
=
=
−1
(x − y
cos−1 (−1)
π
6
(x,y)→(1,2) cos
In Problems 25-30 let x = r cos θ and y = r sin θ. Then x2 + y 2 = r2 and (x, y) → (0, 0) if
lim
and only if r → 0. We also use the facts that | cos θ| ≤ 1 and | sin θ| ≤ 1 for all θ.
25.
(x2 − y 2 )2
(r2 cos2 θ − r2 sin2 θ)2
r4 (cos2 θ − sin2 θ)2
=
lim
=
lim
r→0
r→0
r2
r2
(x,y)→(0,0) x2 + y 2
lim
= lim r2 cos2 2θ = 0
r→0
84
26.
CHAPTER 13. PARTIAL DERIVATIVES
sin 3r2
sin(3x2 + 3y 2 )
=
lim
Use L’Hôpital’s Rule
r→0
x2 + y 2
r2
(x,y)→(0,0)
6r cos 3r2
= lim
= lim 3 cos 3r2 = 3
r→0
r→0
2r
lim
27.
6r2 cos θ sin θ
6xy
p
√
= lim 3|r| sin 2θ = 0
= lim
r→0
r→0
(x,y)→(0,0)
r2
x2 + y 2
28.
x2 − y 2
r2 cos2 θ − r2 sin2 θ
p
√
= lim |r| cos 2θ = 0
= lim
r→0
r→0
(x,y)→(0,0)
r2
x2 + y 2
29.
x3
r3 cos3 θ
=
lim
= lim r cos3 θ = 0
r→0
r→0
r2
(x,y)→(0,0) x2 + y 2
30.
x3 + y 3
r3 cos3 θ + r3 sin3 θ
= lim
= lim r(cos3 θ + sin3 θ) = 0
2
2
r→0
r→0
r2
(x,y)→(0,0) x + y
lim
lim
lim
lim
31. {(x, y) | x ≥ 0 and y ≥ −x}
32. {(x, y) | x 6= 0 and y 6= 0}
33. {(x, y) | y 6= 0 and x/y 6= π/2 + kπ, k and integer}
34. {(x, y) | x and y are real}
35. (a) For x2 + y 2 < 1, f (x, y) = 0 is continuous
(b) For x ≥ 0, f (x, y) is not continuous since it is discontinuous at (2, 0).
(c) For y > x, f (x, y) is not continuous since it is discontinuous at (2, 3).
36. (a) For y ≥ 3, f (x, y) is not continuous since it is not defined at (0, 3).
(b) For |x| + |y| < 1, f (x.y) is discontinuous since it is not defined at (0, 0).
(c) For (x − 2)2 + y 2 < 1, f (x, y) is discontinuous since it is not defined at (2, 0).
37. Since
lim
(x,y)→(0,0)
f (x, y) =
6x2 y 3
6r5 cos2 θ sin3 θ
=
lim
= lim 6r cos2 θ sin3 θ = 0 = f (0, 0)
r→0
r→0
r4
(x,y)→(0,0) (x2 + y 2 )2
lim
the function is continuous at (0, 0).
38. Since f (x, 0) = 0 for all x and f (0, y) = 0 for all y, f (x, 0) and f (0, y) are continuous at x = 0
and y = 0, respectively. On y = x,
lim
f (x, y) =
(x,y)→0,0)
so f (x, y) is not continuous at (0, 0).
x2
1
= ,
4
(x,y)→(0,0) 2x2 + 2x2
lim
13.3. PARTIAL DERIVATIVES
85
39. Choose > 0. Using x = r cos θ and y = r sin θ we have
3t cos θr2 sin2 θ 3
3xy 2
=
r cos θ sin2 θ.
2x2 + 2y 2
2r2
2
p
Let δ = 2
x2 + y 2 < δ, we have
3 . Now, whenever r =
3
3xy 2
3
3
3 2
2
| = |r cos θ sin θ| ≤ |r| < δ =
| 2
= .
2x + 2y 2
2
2
2
2 3
Thus
lim
3xy 2
= 0.
+ y2
(x,y)→(0,0) x2
40. Choose > 0. Using x = r cos θ and y = r sin θ we have
x2 y 2
r2 cos2 θr2 sin2 θ
=
= r2 cos2 θ sin2 θ.
x2 + y 2
r2
Now, whenever r =
lim
p
x2 + y 2 <
√
(for δ =
√
),
x2 y 2
= r2 cos2 θ sin2 θ ≤ r2 ≤ . Thus,
+ y2
x2
x2 y 2
= 0.
+ y2
(x,y)→(0,0) x2
41. Where y 6= x, we have
f (x, y) =
x3 − y 3
(x − y)(x2 + xy + y 2 )
=
= x2 + xy + y 2 .
x−y
x−y
When y = x, we have
x2 + xy + y 2 = x2 + x2 + x2 = 3x2 = f (x, y).
Therefore, f (x, y) = x2 + xy + y 2 throughout the entire plane. Since x2 + xy + y 2 is a
polynomial, f must be continuous throughout the plane and thus has no discontinuities.
p
42. Choose > 0. Then for δ = , whenever 0 < (x − a)2 + (y − b)2 < δ, we have
p
|f (x, y) − b| = |y − b| ≤ (x − a)2 + (y − b)2 < δ = .
Thus,
lim
y = b.
(x,y)→(a,b)
13.3
1.
Partial Derivatives
∂z
7(x = 4x) + 8y 2 − 7x − 8y 2
74x
= lim
= lim
=7
4→0 4x
∂x 4→0
4x
∂z
7x + 8(y + 4y)2 − 7x − 8y 2
16y 4 y + 8(4y)2
= lim
= lim
4y→0
∂y 4y→0
4y
4y
= lim (16y + 8 4 y) = 16y
4y→0
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86
CHAPTER 13. PARTIAL DERIVATIVES
2.
3.
∂z
(x + 4x)y − xy
y4x
= lim
= lim
= y;
4x→0
4x→0
∂x
4x
4x
∂z
x(y + 4y) − xy
x4y
= lim
= lim
=x
4y→0
4y→0
∂y
4y
4y
3(x + 4x)2 y + 4x + 4x)y 2 − 3x2 y − 4xy 2
∂z
= lim
∂x 4x→0
4x
3x2 y + 6x(4x)y + 3(4x)2 y + 4xy 2 + 4(4x)y 2 − 3x2 y − 4xy 2
= lim
4x→0
4x
6x(4x)y + 3(4x)2 y + 4(4x)y 2
= lim
= lim (6xy + 3(4x)y + 4y 2 ) = 6xy + 4y 2
4x→0
4x→0
4x
3x2 (y + 4y) + 4x(y + 4y)2 − 3x2 y − 4xy 2
∂z
= lim
∂y 4y→0
4y
3x2 y + 3x2 4 y + 4xy 2 + 8xy 4 y + 4x(4y)2 − 3x2 y − 4xy 2
= lim
4y→0
4y
3x2 4 y + 8xy 4 y + 4x(4y)2
= lim
= lim (3x2 + 8xy + 4x 4 y) = 3x2 + 8xy
4y→0
4y→0
4y
x+4
x
−
∂z
x2 + x 4 x + xy + (4x)y − x2 − x 4 x − xy
x+4+y x+y
4.
= lim
= lim
4x→0
∂x 4x→0
4x
(x + 4x + y)(x + y) 4 x
y
(4x)y
=
= lim
4x→0 (x + 4x + y)(x + y) 4 x
(x + y)2
x
x
−
x2 + xy − x2 − xy − x 4 y
∂z
x + y + 4y x + y
= lim
= lim
4y→0 (x + y + 4y)(x + y) 4 y
∂y 4y→0
4y
−x 4 y
x
= lim
=−
4y→0 (x + y + 4y)(x + y) 4 y
(x + y)2
5. zx = 2x − y 2 ; zy = −2xy + 20y 4
6. zx = −3x2 + 12xy 3 ; zy = 18x2 y 2 + 10y
7. zx = 20x3 y 3 − 2xy 6 + 30x4 ; zy = 15x4 y 2 − 6x2 y 5 − 4
8. zx = 3x2 y 2 sec2 (x3 y 2 ); zy = 2x3 sec2 (x3 y 2 )
√
2
24y x
9. zx = √
;
z
=
−
y
(3y 2 + 1)2
x(3y 2 + 1)
10. zx = 12x2 − 10x + 8; zy = 0
11. zx = −(x3 − y 2 )−2 (3x2 ) = −3x2 (x3 − y 2 )−2 ; zy = −(x3 − y 2 )−2 (−2y) = 2y(x3 − y 2 )−2
12. zx = 6(−x4 + 7y 2 + 3y)5 (−4x) = −24x( − x4 + 7y 2 − 3y)5 ; zy = 6(−x4 + 7y 2 + 3y)5 (14y + 3)
13. zx = 2(cos 5x)(− sin 5x)(5) = −10 sin 5x cos 5x; zy = 2(sin 5y)(cos 5y)(5) = 10 sin 5y cos 5y
13.3. PARTIAL DERIVATIVES
14. zx = (2x tan−1 y 2 )ex
3
2
tan−1 y 2
87
; zy =
3
2x2 y x2 tan−1 y2
e
1 + y4
3
15. fx = x(3x2 yex y + ex y ; fy = x4 ex y
θ
θ
1
θ
θ
θ
θ
2
2
16. fθ = φ cos
; fφ = φ cos
− 2 + 2φ sin = −θ cos + 2φ sin
φ
φ
φ
φ
φ
φ
φ
7y
(x + 2y)3 − (3x − y)
=
;
(x + 2y)2
(x + 2y)2
(x2 − y 2 )2 y − xy 2(x2 − y 2 )2x
18. fx =
=
(x2 −y 2 )4
(x2 − y 2 )x − xy 2(x2 − y 2 )(−2y)
fy =
(x2 − y 2 )4
17. fx =
fy =
(x + 2y)(−1) − (3x − y)(2)
−7x
=
(x + 2y)2
(x + 2y)2
−3x2 y − y 3
;
(x2 − y 2 )3
3xy 2 + x3
= 2
(x − y 2 )3
8u
15v 2
; gv = 2
3
− 5v
4u + 5v 3
√
√
1
s
r
1
20. hr = √ + 2 ; hx = − 2 − √
r
s
2s r
2s r
y y2
y
√
√
y
1 y/z
21. wx = √ ; wy = 2 x − y
e
=2 x−
+ 1 ey/z ; wz = −yey/z − 2 = 2 ey/z
z
z
z
z
x
1
xy
22. wx = xy
+ (ln xz)y = y + y ln xz; wy = x ln xz; wz =
x
z
19. gu =
4u2
23. Fu = 2uw2 − v 3 − vwt2 sin(ut2 ); Fv = −3uv 2 + w cos(ut2 );
Fx = 3(2x2 t)3 (4xt) = 16xt(2x2 t)3 = 128x7 t4 ; Ft = −2uvwt sin(ut2 ) + 64x8 t3
4 5
24. Gp = 2pq 3 e2r s
4 5
Gq = 3p2 q 2 e2r s
4 5
4 5
Gr = p2 q 3 (8r3 s5 )e2r s = 8p2 q 3 r3 s5 e2r s
4 5
4 5
Gs = p2 q 3 (10r4 s4 )e2r s = 10p1 q 3 r4 s4 e2r s
25. zy = 16x3 y 3 , zy (1, −1) = −16
26. zx = 12x2 y 4 , zx (1, −1) = 12
18x2
(x + y)18x − 18xy
=
, fy (−1, 4) = 2. An equation of the tangent line is given
(x + y)2
(x + y)2
by x = −1 and z + 24 = 2(y − 4). Parametric equations of the line are x = −1, y = 4 + t, z =
−24 + 2t.
27. fy =
(x + y)18y − 18xy
18y 2
=
, fx (−1, 4) = 32. An equation of the tangent line is given
(x + y)2
(x + y)2
z + 24
by y = 4 and z + 24 = 32(x + 1). Symmetric equations of the line are x + 1 =
, y = 4.
32
28. fx =
88
CHAPTER 13. PARTIAL DERIVATIVES
−x
29. zx = p
, zx (2, 2) = −2
9 − x− y 2
√
√ √
3
, zy ( 2, 3) = −
30. zy = p
2
2
2
9−x −y
−y
31.
∂z
∂2z
= y 2 exy
= yexy ;
∂x
∂x2
32.
∂3z
∂z
∂2z
4 −4
=
6x
y
;
= −24x4 y −5
= −2x4 y −3 ;
∂y
∂y 2
∂y 3
33. fx = 10xy 2 − 2y 3 ; fxy = 20xy − 6y 2
34. f (p, q) = ln(p + q) − 2 ln q, fq =
2
1
1
− , fqp = −
p+q
q
(p + q)2
35. wt = 3u2 v 3 t2 , wtu 6uv 3 t2 ; wtuv 18uv 2 t2
36. wv = −
u2 sin(u2 v)
u4 cos(u2 x)
3u4 cos(u2 v)
; wvv −
; wvvt =
3
3
t
t
t4
2
2
2
2
2
37. Fr = 2rer cos θ; Frθ − 2rer sin θ; Frθr − 2r(2rer ) sin θ − 2er sin θ = −2er (2r2 + 1) sin θ
2s
4s
(s − t) − (s + t)(−1)
=
; Htt =
;
2
2
(s − t)
(s − t)
(s − t)3
4
2
−8s − 4t
(s − t) − 4x(3)(s − t)
=
=
6
(x − t)
(s − t)4
38. Ht =
Htts
39.
40.
∂2z
∂z
∂2z
∂z
= −5x4 y 2 +8xy;
= −60x3 y 2 +8y;
= 6x5 −20x3 y 3 +4y 2 ;
= −60x3 y 2 +8y
∂y
∂x∂y
∂x
∂y∂x
∂z
2x
(1 + 4x2 y 2 )2 − 2x(8xy 2 )
2y
∂z
2 − 8x2 y 2
∂z
=
=
=
;
=
;
2
2
2
2
2
2
2
2
∂y
1 + 4x y
∂x∂y
(1 + 4x y )
(1 + 4x y )
∂x
1 + 4x2 y 2
2 2
2
2 2
(1 + 4x y )2 − 2y(8x y)
2 − 8x y
∂z
=
=
∂y∂x
(1 + 4x2 y 2 )2
(1 + 4x2 y 2 )2
41. wu = 3u2 v 4 − 8uv 2 t3 , wuv = 12u2 v 3 − 16uvt3 , wuvt = −48uvt2 ; wt = −12u2 v 2 t2 + v 2 ,
wtv = −24u2 vt2 + 2v, wtvu = −48uvt2 ; wv = 4u3 v 3 − 8u2 vt3 + 2vt, wvu = 12u2 v 3 − 16uvt3 ,
wvut = −48uvt2
42. Fη = 6η 2 (η 3 +ξ 2 +τ ) = 6η 5 +6η 2 ξ 2 +6η 2 τ, Fηξ = 12η 2 ξ, Fηξη = 24ηξ; Fξ = 4ξ(η 3 +ξ 3 +τ ) =
4η 3 ξ + 4xi3 + 4xiτ, Fξη = 12η 2 ξ, Fξηη 24ηξ; Fηη = 30η 4 + 12ηξ 2 , Fηηξ = 24ηξ
43. 2x + 2zzx = 0, zx = −x/z; 2y + 2zzy = 0, zy = −y/z
2x
;
2z − y 2
2yz
2zzy = y 2 zy + 2yz =⇒ (2z − y 2 )zy = 2yz =⇒ zy =
2z − y 2
44. 2zzx = 2x + y 2 zx =⇒ (2z − y 2 )zx = 2x =⇒ zx =
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13.3. PARTIAL DERIVATIVES
89
vz − 2uv 3
;
2z − uv
2 2
uz − 3u v
2zzv + 3u2 v 2 − uvzv − uz = 0 =⇒ (2z − uv)zv = uz − 3u2 v 2 =⇒ zv =
2z − uv
45. 2zzu + 2uv 3 − uvzu − vz = 0 =⇒ (2z − uv)zu = vz − 2uv 3 =⇒ zu =
46. sez zs + ez − test + 12s2 t = zs =⇒ (sez − 1)zs = tes − ez − 12s2 t =⇒ zs =
sez zt − sest + 4s3 = zt =⇒ (sez − 1)zt = sest − 4s3 =⇒ zt =
sest − 4s3
sez − 1
test − ez − 12s2 t
;
sez − 1
47. ax = y sin θ, Ay = x sin θ, Aθ = xy cos θ
48. Vh = (π/3)(r2 + rR + R2 ), Vr = (π/3)h(2r + R), VR = (π/3)h(r + 2R)
49.
50.
51.
52.
53.
∂2u
∂u
= −4π 2 (cosh 2πy + sinh 2πy) sin 2πx;
= 2π(cosh 2πy + sinh 2πy) cos 2πx;
∂x
∂x2
∂u
∂2u
= (2π sinh 2πy + 2π cosh 2πy) sin 2πx;
= (4π 2 cosh 2πy + 4π 2 sinh 2πy) sin 2πx;
∂y
∂y 2
∂2u ∂2u
+ 2 = −4π 2 (cosh 2πy + sinh 2πy) sin 2πx + 4π 2 (cosh 2πy + sinh 2πy) sin 2πx = 0
∂x2
∂y
nπ ∂ 2 u
nπ ∂u
nπ
n2 π 2 −(nπx/L)
= − e−(nπx/L sin
=
e
sin
;
y;
∂x
L
L
∂x2
L2
L 2
2 2
nπ −(nπx/L)
nπ
∂ u
n π
nπ
∂u
=
e
cos
y;
= − 2 e−(nπx/L) sin
y;
∂y
L
L
∂y 2
L
L
nπ n2 π 2
nπ n2 π 2 −(nπx/L)
∂2u ∂2u
+ 2 =
e
sin
−
e−(nπx/L) sin
=0
2
2
2
∂x
∂y
L
L
L
L
∂z
2x
∂2z
(x2 + y 2 )2 − 2x(2x)
2y 2 − 2x2 ∂z
2y
= 2
,
=
=
;
= 2
,
∂x
x + y 2 ∂x2
(x2 + y 2 )2
(x2 + y 2 ) ∂y
x + y2
2
2
2
2
2
2
2
2
2
2
∂ z
(x + y )2 − 2y(2y)
2x − 2y
∂ z
∂ z
2y − 2x + 2x − 2y 2
=
=
;
+
=
=0
∂y 2
(x2 + y 2 )2
(x2 + y 2 )2 ∂x2
∂y 2
(x2 + y 2 )2
∂z
2
2
2
2
2
2
= 2yex −y sin 2xy + 2xex −y cos 2xy = 2ex −y (x cos 2xy − y sin 2xy),
∂x
∂2z
2
2
2
2
= 2ex −y (−2xy sin 2xy + cos 2xy − 2y 2 cos 2xy) + 4xex −y (x cos 2xy − y sin 2xy);
∂x2
∂z
− 2
− 2
− 2
= −2xex y sin 2xy − 2yex y cos 2xy = −2ex y (x sin 2xy + y cos 2xy),
∂y
∂2z
− 2
− 2
= −2ex y (2x2 cos 2xy − 2xy sin 2xy + cos 2xy) + 4yex y (x sin 2xy + y cos 2xy);
∂y 2
− 2
∂2z
∂2z
+ 2 =2ex y (−2xy sin 2xy + cos 2xy − 2y 2 cos 2xy + 2x2 cos 2xy − 2xy sin 2xy
2
∂x
∂y
− 2x2 cos 2xy + 2xy sin 2xy − cos 2xy + 2xy sin 2xy − 2y 2 cos 2xy) = 0
∂u
x
=− 2
;
∂x
(x + y 2 + z 2 )3/2
∂2u
2x2 − y 2 − z 2
=
;
∂x2
(x2 + y 2 + z 2 )5/2
∂u
y
∂u
z
=− 2
;
=− 2
;
∂y
(x + y 2 + z 2 )3/2 ∂z
(x + y 2 + z 2 )3/2
∂2u
−x2 + 2y 2 − z 2
∂2u
−x2 − y 2 + 2z 2
=
;
=
;
∂y 2
(x2 + y 2 + z 2 )5/2 ∂z 2
(x2 + y 2 + z 2 )5/2
90
CHAPTER 13. PARTIAL DERIVATIVES
2x2 − y 2 − z 2 − x2 + 2y 2 − z 2 − x2 − y 2 + 2z 2
∂2u ∂2u ∂2u
+
+
=
=0
∂x2
∂y 2
∂z 2
(x2 + y 2 + z 2 )5/2
54.
√
√
∂u √ 2
∂2u
2
2
2
2
m2 +n2 x
=
(m
+
n
)e
= m + n2 e m +n x cos my sin nz;
cos my sin nz;
∂x
∂x2
2
√
√
∂u
∂ u
2
2
2
2
= −m2 e m +n x cos my sin nz;
= −me m +n x sin my sin nz;
∂y
∂y 2
√
√
∂u
∂2u
2
2
2
2
= −n2 e m +n x cos my sin nz;
= ne m +n x cos my cos nz;
2
∂z
∂z
√
√
∂2u ∂2u ∂2u
2
2
2
2
+ 2 + 2 = (m2 + n2 )e m +n x cos my sin nz − m2 e m +n x cos my sin nz
2
∂x
∂y
∂z
− n2 e
55.
56.
57.
√
m2 +n2 x
cos my sin nz = 0
∂u
∂u
∂2u
∂2u
=
−
cos
at
sin
x;
= −a1 cos at sin x;
= cos at cos x,
=
−a
sin
at
sin
x,
∂x
∂x2
∂t
∂t2
∂2u
∂2u
a2 2 = a2 (− cos at sin x) = 2
∂x
∂t
∂2u
∂u
= − sin(x + at) + cos(x − at),
= − cos(x + at) − sin(x − at);
∂x
∂x2
∂u
∂2u
∂2u
= −a sin(x + at) − a cos(x − at),
= −a2 cos(x + at) − a2 sin(x − at); a2 2 =
2
∂t
∂t
∂x
∂2u
2
2
−a cos(x + at) − a sin(x − at) = 2
∂t
2x
∂2C
4x2 −1/2 −x2 /kt
2
∂C
/
2
= − t−/12 e−x kt ,
=
t
e
− t−1/2 e−x /kt ;
2
2
2
∂x
kt
∂x
k t
kt
2
∂C
t−3/2 −x2 /kt k ∂ 2 C
x2 −1/2 −x2 /kt t−1/2 −x2 /kt
∂C
−1/2 x
−x2 /kt
=t
e
−
e
;
= 2t
e
−
e
=
2
2
∂t
kt
2
4 ∂x
kt
2t
∂t
58. (a) Pv = −k(T /V 2 )
(b) P V = kt, P VT = k, VT = k/P
(c) P V = kT, V = kTp , Tp = V /k
∂u
−gx/z, 0 ≤ x ≤ at
59. (a)
=
−gt,
x > at
∂t
For x > at, the motion is that of a freely falling body.
∂u
(b) For x > at,
= 0. For x > at, the string is horizontal.
∂x
60.
∂S
= 0.0790975w0.425 h−0.275 ; Sh (60, 36) + 0.0790975(60)0.425 (36)−0.275 ≈ 0.1682
∂h
The approximate increase in skin-area as h increases from 36 to 37 inches is 0.1682 sq ft.
∂2z
fx (x + ∆x, y) − fx (x, y)
= lim
2
∆x→0
∂x
∆x
∂2z
fy (x, y + ∆y) − fx (x, y)
(b)
= lim
∆y→0
∂y 2
∆y
61. (a)
13.3. PARTIAL DERIVATIVES
91
∂2z
fy (x + ∆x, y) − fy (x, y)
= lim
∂x∂y ∆x→0
∆x
(c)
62. Integrating zx = 2xy 3 + 2y + 1/x with respect to x, we obtain z = x2 y 3 + 2xy + ln x + φ(y).
Then 3x2 y 2 + 2x + 1 = zy = 3x2 y 2 + 2x + φ0 (y). Since φ0 (y) = 1, φ(y) = y + C, and
z = x2 y 3 + 2xy + ln x + y + C.
63. Consider the mixed partials:
∂
∂2z
=
∂y∂x
∂y
∂z
∂x
= 2y
and
∂2z
∂
=
∂x∂y
∂x
∂z
∂y
= 2x.
∂z ∂z
∂2z
∂2z
,
,
, and
are all continuous on an open set, we should have
∂x ∂y ∂y∂x
∂x∂y
∂2z
∂2z
=
on that set. But the mixed partials are equal only on the line y = x, which
∂y∂x
∂x∂y
contains no open set in the plane. Therefore, such a function cannot exist.
Since
64. (a) There are 10 different third-order partial derivatives: Fxxx , Fxxy , Fxxz , Fxyy , Fxyz , Fxzz ,
Fyyy , Fyyz , Fyzz , Fzzz
(b) Since the mixed partials are equal, the order in which differentiation occurs is irrelevant.
The nth order partial derivatives are given by
∂nz
∂nz
∂nz
∂nz
∂nz
,
,
,
.
.
.
,
,
.
∂xn ∂xn−1 ∂y ∂xn−2 ∂y 2
∂x∂y n−2 ∂y n
Hence, there are n + 1 different nth order partial derivatives.
65. (a) There slopes of the surface in the x and y directions are zero everywhere. This implies
that the surface must have constant height everywhere. Therefore f must have the form
f (x, y) = c.
(b) Since the mixed partials are both zero, we have
∂ ∂z
∂z
= 0 and df rac∂∂y
=0
∂x ∂y
∂x
∂z
∂z
is a function of y alone and
is a function of x alone. Therefore, z
∂y
∂x
has no term that depends on both x and y. Hence z is of the form z = g(x) + h(y) + c
where g and h are twice continuously differentiable functions of a single variable.
which implies
66. The level curves suggest that the surface height is decreasing as we move slightly to the right
∂z
of the point, and increasing as we move slightly up from the point. This implies
< 0 and
∂x
∂z
> 0.
∂y
67.
∂z
∂x
∂z
∂y
f (0 + ∆x, 0) − f (0, 0)
0/2(∆x)2
= lim
= 0;
∆x→0
∆x→0
∆x
∆x
(0,0)
f (0, 0 + ∆y) − f (0, 0)
0/2(∆y)2
= lim
= lim
=0
∆y→0
∆y→0
∆y
∆y
(0,0)
= lim
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92
CHAPTER 13. PARTIAL DERIVATIVES
68. (a)
(b)
13.4
1.
2.
3.
4.
5.
y 5 − 4x2 y 3 − x4 ∂z
∂z
;
=
∂x
(x2 + y 2 )2
∂x
= y;
(0,y)
∂z
−x5 + 4x3 y 2 + xy 4 ∂z
;
=
∂y
x2 + y 2 )2
∂y
= −x
(x,0)
∂2z
∂2z
∂2z
∂2z
= 1;
= −1 =⇒
6=
∂y∂x
∂x∂y
∂y∂x
∂x∂y
Linearization and Differentials
∂f
∂f
= 4y 2 − 6x2 y so
(1, 1) = −2
∂x
∂x
∂f
∂f
= 8zy − 2x3 so
(1, 1) = 6
∂y
∂y
f (1, 1) = 2 The linearization is L(x, y) = 2 − 2(x − 1) + 6(y − 1) = −2x + 6y − 2
∂f
∂f
3x2 y
so
= p
(2, 2) = 3
3
∂x
∂x
2 x y
∂f
∂f
x3
so
= p
(2, 2) = 1
3
∂y
∂y
2 x y
f (2, 2) = 4 The linearization is L(x, y) = 4 + 3(x − 2) + (y − 2) = 3x + y − 4
p
∂f
x2
∂f
353
= x2 + y 2 + p
so
(8, 15) =
2
2
∂x
∂x
17
x +y
∂f
xy
∂f
120
=p
so
(8, 15) =
∂y
∂y
17
x2 + y 2
120
f (8, 15) = 136 The linearization is L(x, y) = 136+ 353
17 (x−8)+ 17 (y −15) =
∂f π 3π −3
∂f
= 3 cos x cos y so
,
=
∂x
∂x 4 4
2
∂f
∂f π 3π −3
= 3 sin x sin y so
=
,
∂y
∂y 4 4
2
−3
π 3π
f 4, 4 =
The linearization is L(x, y) =
2
3
2 (π − 1)
−3
2
−
3
2
x−
π
4
−
3
2
y−
3π
4
353
120
17 x+ 17 y −136
=
−3
2 x
− 32 y +
2x
∂f
∂f
= 2
so
(−1, 1) = −1
∂x
x + y3
∂x
∂f
3y 2
∂f
3
= 2
so
(−1, 1) =
∂y
x + y3
∂y
2
3
f (−1, 1) = ln(2) The linearization is L(x, y) = ln(2) − (x + 1) + 32 (y − 1) = −x + y − 52 + ln(2)
2
∂f
∂f
2π
6.
= 3e−2y cos 3x so
0, π3 = 3e− 3
∂x
∂x
∂f
∂f
= −2e−2y sin 3x so
0, π3 = 0
∂y
∂y
2π
2π
f 0, π3 = 0 The linearization is L(x, y) = 3e− 3 (x − 0) = 3xe− 3
√
√
7. Note that we are trying to approximate f (102, 80) where f (x, y) = x + 4 y. Since (102, 80)
is reasonably close to (100, 81), we can use the linearization of f at (100, 81) to approximate
13.4. LINEARIZATION AND DIFFERENTIALS
93
the value at (102, 80). To do this, we compute
∂f
∂f
∂f
1
1
∂f
1
1
1
,
= √ ,
(100, 81) =
,
=
(100, 81) =
=
, and
3/2
∂x
∂x
20
∂y
∂y
4(27)
108
2 x
4y
f (100, 81) = 13
1
1
(x − 100) + 108
(y − 81). For the approximation, we have
The linearization is L(x, y) = 13 + 20
1
1
1
1
− 108
= 7069
L9102, 80) = 13 + 20 (102 − 100) + 108 (80 − 81) = 13 + 10
540 ≈ 13.0907
√
x
8. We are trying to approximate f (36, 63) where f (x, y) = √ . Use the linearization of f at the
y
point (36, 64). To do this, compute
√
∂f
∂f
1
1 ∂f
∂f
3
3
− x
= √ √ ,
(36, 64) =
,
= 2y
(35, 64) = −
, and f (36, 64) = .
3/2 ;
∂x
96 ∂y
∂y
512
4
2 x y ∂x
3
1
3
The linearization if L(x, y) = + (x − 36) −
(y − 64). For the approximation, we have
4 96
512
3
1
3
387
L(36, 63) = + (36 − 36) −
(63 − 64) =
≈ .7559.
4 96
512
512
9. First, linearize f at (2, 2). To do this, compute
∂f
∂f
∂f
∂f
= 2(x2 + y 2 )2x,
(2, 2) = 64,
= 2(x2 + y 2 )2y,
(2, 2) = 64, and f (2, 2, ) = 64.
∂x
∂x
∂y
∂y
The linearizationis L(x, y) = 64 + 64)x − 2) + 64)y − 2). For the approximation, we have
L(1.95, 2.01) = 64 + 64(−0.05) + 64(0.01) ≈ 61.44.
10. First, linearize f at 12 , 3 . To do this, compute
∂f 1 ∂f 1 ∂f
∂f
π
= −πy sin(πxy),
= −πx sin(πxy),
, 3 = 3π,
, 3 = , and f 21 , 3 =
2
2
∂x
∂x
∂y
∂y
2
0.
π
1
+ (y − 3). For the approximation, we have
The linearization is L(x, y) = 3π x −
2
2
π
L(0.52, 2.96) = 3π(0.02) + (−0.04) ≈ 0.1257.
2
11. dz = 2x sin 4ydx + 4x2 cos 4ydy
h
i
2
2
2
2
2
2
2
2
2
2
12. dz = x(2xex −y ) + ex −y dx − 2yxex −y dy = (2x2 + 1)ex −y dx − 2xyex −y dy
13. dz = p
2x
2x2 − 4y 3
dx − p
6y 2
2x2 − 4y 3
dy
14. dz = 45x2 y(5x3 y + 4y 5 )2 dx + (15x3 + 60y 4 )(5x3 y + 4y 5 )2 dy
15. df =
(s + 3t)2 − (2s − t)
(s + t)(−1) − (2s − t)3
7t
7s
ds +
dt =
−
dt
(s + 3t)2
(s + 3t)2
(s + 3t)2
(s + 3t)2
16. dg = 2r cos 3θdr − 3r2 sin 3θdθ
17. dw = 2xy 4 z −5 dx + 4x2 y 3 z −5 dy − 5x2 y 4 z −6 dz
2
2
2
18. dw = −2xe−z sin(x2 + y 4 )dx − 4y 3 e−z sin(x2 + y 4 )dy − 2ze−z cos(x2 + y 4 )dz
19. dF = 3r2 dr − 2s−3 ds − 2t−1/2 dt
94
CHAPTER 13. PARTIAL DERIVATIVES
20. dG = sin φ cos θdρ − ρ sin φ sin θdθ + ρ cos φ cos θdφ
21. w = ln u + ln v − ln s − ln t; dw =
22. dw = √
du dv ds dt
+
−
−
u
v
s
t
u
v
st2
s2 t
du − √
dv + √
ds + √
dt
u2 + s2 t2 − v 2
u2 + s2 t2 − v 2
u2 + s2 t2 − v 2
u2 + s2 t2 − v 2
23. ∆z = z(2.2, 3.9) − z(2, 4) = (6.6 + 15.6 + 8) − (6 + 16 + 8) = 0.1; dz = 3dx + 4dy
When x = 2, y = 4, dx = 0.2, and dy = −0.1, dz = 3(0.2) + 4(−0.1) = 0.2
24. ∆z = z(0.2, −0.1) − z(0, 0) = 2(0.2)2 (−0.1) + 5(−0.1) − 0 = −0.508; dz = 4xydx + (2x+ 5)dy
When x = y = 0, dx = 0.2, and dy = −0.1, dz = 5(−0.1) = −0.5.
25. ∆z = z(3.1, 0.8) − z(3, 1) = (3.1 + 0.8)2 − (3 + 1)2 = 15.21 − 16 = −0.79;
dz = 2(x + y)dx + 2(x + y)dy. When x = 3, y = 1, dx = 0.1, and dy = −0.2,
2(3 + 1)(0.1) + 2(3 + 1)(0.2) = 0.8 − 1.6 = −0.8
dz =
26. ∆z = z(0.9, 1.1) − z(1, 1) = [(0.9)2 + (0.9)2 (1.1)2 + 2] − [1 + 1 + 2] = 3.7901 − 4 = −0.2099;
dz = (2x + 2xy 2 )dx + 2xy dy. When x = y = 1, dx = −0.1, and dy = 0.1, dz = 4(−0.1) +
2(0.1) = −0.2.
27. ∆z = 5(x + ∆x)2 + 3(y + ∆y) − (x + ∆x)(y + ∆y) − (5x2 + 3y − xy)
= 10x∆x + 5(∆x)2 + 3∆y − x∆y − y∆x − ∆x∆y
= (10x − y)∆x + (3 − z)∆y + (5∆x)∆x − (∆x)∆y
1 = 5∆x, 2 = −∆x
28. ∆z = 10(y + ∆y)2 + 3(x + ∆x) − (x + ∆x) − (10y 2 + 3x − x2 )
= 20y∆y + 10(∆y)2 + 3∆x − 2x∆x − (∆x)2
= (3 − 2x)∆x + 20y∆y − (∆x)∆x + (10∆y)∆y
1 = −∆x, 2 = 10∆y
29. ∆x = (x + ∆x)2 (y + ∆y)2 − x2 y 2 = [x2 + 2x∆x + (∆x)2 ][y 2 + 2y∆y + (∆y)2 ] − x2 y 2
= 2x2 y∆y + x2 (∆y)2 + 2xy 2 ∆x + 4xy(∆x)∆y + 2x(∆x)(∆y)2 + y 2 (∆x)2 + 2y(∆x)2 ∆y + (∆x)2 (∆y)2
= 2xy 2 ∆x + 2x2 y∆y + [4xy∆y + 2x(∆y)2 + y 2 x]∆x + [x2 ∆y + 2y(∆x)2 + (∆x)2 ∆y]∆y
1 = 4xy∆y + 2x(∆y)2 + y 2 x, 2 = x2 ∆y + 2y(∆x)2 + (∆x)2 ∆y (Several other choices of 1
and 2 are possible.)
30. ∆z = (x + ∆x)3 − (y + ∆y)3 − (x3 − y 3 ) = 3x2 ∆x + 3x(∆x)2 + (∆x)3 − 3y 2 ∆y − 3y(∆y)2 − (∆y)3
= 3x2 ∆x − 3y 2 ∆y + [3x∆x + (∆x)2 ] − [3y∆y + (∆y)2 ]∆y
1 = 3x∆x + (∆x)2 , 2 = −3y∆y − (∆y)2
31. R =
R1 R2 R3
; ∆R1 = ±0.009R1 , ∆R! = ±0.009R2 , ∆R3 = ±0.0009R3
R2 R3 + R1 R3 + R1 R2
www.elsolucionario.org
13.4. LINEARIZATION AND DIFFERENTIALS
95
R12 R32
R22 R32
+
(±0.009R
)
(±0.009R2 )
1
(R2 R3 + R1 R3 + R1 R2 )2
(R2 R3 + R1 R3 + R1 R2 )2
R12 R22
+
(±0.009R3 )
(R2 R3 + R1 R3 + R1 R2 )2
R2 R3 + R1 R3 + R1 R2
= 0.009R
= 0.009R
R2 R3 + R1 R3 + R1 R2
The maximum percentage error is approximately 0.9%.
|∆R| ≈ |dR| ≤
32. We are given ∆T = ±0.006T and ∆V = ±0.008V. Then
|∆P | ≈ |dP | =
kT
k
kT
kT
(±0.006T ) − 2 (±0.008V ) ≤
(0.006) +
(0.008) = P (0.014).
V
V
V
V
Thus, the approximate maximum percentage error in P is 1.4%.
(2r2 + R2 − R(2R)
−R(4r)
2r2 − R2
4rR
dR
+
mg
dr
=
mg
dR − mg
dr
2
2
2
2
2
2
+
2
2
(2r + R )
2r + R )
(2r R )
(2r+ R2 )2
When R = 4, r = 0.8, dR = 0.1, and dr = 0.1,
33. dT = mg
2(0.8)2 − 42
4(0.8)4
∆T ≈ dT = mg
(0.1) −
(0.1)
[2(0.8)2 + 42 ]2
[2(0.8)2 + 42 ]2
−1.472 − 1.28
= mg
≈ −0.009 mg.
298.598
The tension decreases.
34. V = πr2 h, dV = 2πrhdr + πr2 dh. When r = 5, h = 10, dr = 0.3, and dh = 0.5,
∆V ≈ dV = 2π(5)(1−)(0.3) + π(52 )(0.5) = 42.5π cm3 .
Since V (5, 10) = 250π cm3 ,
V (5.3, 10.5) = V (5, 10) + ∆V ≈ V (5, 10) + dV = 250π + 42.5π = 292.5π cm3 .
35. V = lwh, dV = whdl + lhdq + lwdh. With dl = ±0.02l, dw = ±0.05w, and dh = ±0.08h,
|∆V | ≈ |dV | = |wh(±0.02l) + lh(±0.05w) + lw(±0.08h)| ≤ lwh(0.02 + 0.5 + 0.8) = 0.15V.
The approximate percentage increase in volume is 15%.
36. S = 2lw + 2lh + 2wh, dS = (2w + 2h)dl + (2l + 2h)dw + (2l + 2w)dh
With l = 3, w = 1, h = 2, sl = 0.06, dw = 0.05, and dh = 0.16,
∆S ≈ dS = (2 + 4)(0.06) + (6 + 4)(0.05) + (6 + 2)(0.16) = 2.14 ft2 .
Since S(3, 1, 2) = 22 ft2 , the new surface area is approximately S(3, 1, 2) + dS = 24.14 ft2 .
96
CHAPTER 13. PARTIAL DERIVATIVES
37. dS = 0.1091(0.425)w−0.575 h0.725 dw + 0.1091(0.725)w0.425 h−0.275 dh
With dw = ±0.03w and dh = ±0.05h,
|∆S| ≈ |dS| = 0.1091|0.425w−0.575 h0.725 (±0.03w) + 0.725w0.425 h−0.275 (±0.05h)|
≤ 0.1091[0.425w0.425 h0.725 (0.03)] + 0.1091[0.725w0.425 h0.725 (0.05)]
= 0.1091w0.425 h0.725 (0.013 + 0.036) = 0.049S.
The approximate maximum percentage error is 4.9%.
"
38. Z = R + 100L −
2
1
1000c
2 #1/2
;
"
2 #−1/2 1
1
1 2
r + 100L −
2RdR + 2 100L −
(100)dL
dZ =
2
1000c
1000c
1
1
2 100L −
dC
1000c
1000c2
X
2
2 −1/2
= (R + X )
RdR + 1000XdL +
dC
1000c2
With R = 4000, L + 0.4, C = 10−5 , dR = 25, d := 0.05, and dC = 1.1 × 10−5 − 10−5 =
10−6 , we have X = 300 and
300
2
2 −1/2
−6
dZ = (400 + 300 )
400(25+ 100(300)(0.05) +
10
1000(10)−10
1
=
(10000 + 15000 + 3000) = 56 ohms.
500
The new impedance is approximately Z(400, 0.4, 10−5 ) + dZ = 500 + 56 = 556 ohms.
39. (a) If a function w = f (x, y, z) is differentiable at a point (x0 , y0 , z0 ), then the function
L(x, y, z) = f (x0 , y0 , z0 )+fx (x0 , y0 , z0 )(x−x0 )+fy (x0 , y0 , z0 )(y−y0 )+fz (x0 , y0 , z0 )(z−z0 )
is a linearization of f at (x0 , y0 , z0 ).
p
(b) Let f (x, y, z) = x2 + y 2 + z 2 . Then we wish to approximate f (9.1, 11.75, 19.98). To do
this, linearize f at (9, 12, 20). Compute
x
∂f
9
∂f
=p
,
(9, 12, 20) =
2
2
2
∂x
∂x
25
x +y +z
∂f
y
∂f
12
=p
(9, 12, 20) =
,
2
2
2
∂y
∂y
25
x +y +z
∂f
z
∂f
4
=p
,
(9, 12, 20) = and f (9, 12, 20) = 25
∂z
5
x2 + y 2 + z 2 ∂z
13.4. LINEARIZATION AND DIFFERENTIALS
97
9
12
4
(x − 9) +
(y − 12) + (z − 20). For the
25
25
5
12
4
9
approximation, we have L(9.1, 11.75, 19.98) = 25 + (0.1) + (−0.25) + (−0.02) =
25
25
5
24.9
The linearization is L(x, y, z) = 25 +
40. According to Theorem 13.4.3, if f were differentiable at (0, 0), then f would have to be
continuous at (0, 0). However, as shown in Problem 38 in Exercises 13.2, f is not continuous
at (0, 0). Therefore, f cannot be differentiable at (0, 0).
41. (a) The graph of z = f (x, y) is an inverted cone with vertex at the origin. Since the graph
comes to a sharp ”point” at the origin, there is no possible increment formula for ∆z
that will work in every direction there.
(b) We show that the partial derivative fx does not exist at (0, 0). If h > 0,
√
√
h2 + 02 − 02 + 02
f (0 + h) − f (0, 0)
=
h
h
√
2
h
|h|
=
=
=1
h
h
f (0 + h) − f (0, 0)
= −1.
h
f (0 + h) − f (0, 0)
Therefore, lim
does not exist. But this means fx does not exist at
h→0
h
(0, 0) and thus f is not differentiable at (0, 0).
But if h < 0, then
www.elsolucionario.org
98
CHAPTER 13. PARTIAL DERIVATIVES
42.
Δz
z
z
x
y
x
y
Δy
Δx
∆V
y
x
z
z
x
y
dV
43.
∆V − dV
(a) From the figure we see that α = π − (θ + φ). Then
xh = L cos θ + l cos α = L cos θ + l cos(π − θ − φ)
= L cos θ − l cos(θ + φ)
and yh = L sin θ − l sin α = L sin θ − l sin(π − θ − φ)
= L sin θ − l sin(θ + φ).
L
θ
θ
φ
l
α
(x0,y0)
13.5. CHAIN RULE
99
(b) Using l sin(θ + φ) = ye − yh and l cos(θ + φ) = xe − xh , we have
dxh = (−L sin θ + l sin(θ + φ))dθ + l sin(θ + φ)dφ = −yh dθ + (ye − yh )dφ
dyh = (L cos θ − l cos(θ + φ))dθ − l cos(θ + φ)dφ = xh − (xe − xh )dφ.
(c) One position has the lower arm reaching straight up, with the elbow on the x-axis, so that
θ = 0 and φ = 270◦ . The other position has the lower arm reaching straight across, with
the elbow on the y-axis, so that θ = φ = 90◦ . In both cases, (xh , yh ) = (L, L). In the first
case, (xe , ye ) = (L, 0), and in the second case (xe , ye ) = (0, L). In general, the approximate
maximum error in xh is
|dxh | = | − hh dθ + (ye yh )dφ| ≤ L|dθ| + |ye − L||dφ| = (L + |ye − L|)
π
.
180
Thus, in the first case the approximate maximum error is 2πL/180, while in the second case
it is only πL/180.
44. (a) The horizontal and vertical components of velocity are v cos θ and v sin θ, respectively.
The projectile strikes the cliff wall at time t = D/v cos θ. At this time its height is
2
1
D
1
1 D2
H = tv sin θ − gt2 = D tan θ − g
= D tan θ − g 2 sec2 θ.
2
2
v cos θ
2 v
∂H
D2
d2
∂H
2
2
2
dv +
dθ = g 3 sec θdv + D sec θ − g 2 sec θ tan θ dθ
(b) dH =
∂v
∂θ
v
v
(c) When D = 100, g = 32, v = 100, and θ = 45◦ , we have H = 68 ft.
(d) Taking |dv|≤ 1 and
|dθ| ≤ π/180
we find
1002
1002
π
|dH| ≤ 32
sec2
sec2 4 (1) + 100 sec2 π4 − 32
3
100
1002
3.01 ft.
π
4
tan π4
π
180
=
34π
16
+
≈
25
45
(e) We have
dH =
∂H
∂H
∂H
+
+
∂v
∂θ
∂D
=g
D2
D2
D
2
2
2
2
sec
θdv
+
D
sec
θ
−
g
sec
θ
tan
θ
+
tan
θ
−
g
sec
θ
dD.
v3
v2
v2
With |dD| ≤ 2, we obtain dH ≤
13.5
1.
2.
16 34π 18
34 34π
+
+
=
+
≈ 3.73 ft.
25
45
25
25
45
Chain Rule
∂z dx ∂z dy
2x
dz
2y
(2t) + 2
(−2t−3 )
=
+
= 2
dt
∂x dt
∂y dt
x + y2
x + y2
4xt − 4yt−3
=
x2 + y 2
dz
∂z dx ∂z dy
=
+
dt
∂x dt
∂y dt
= (3x2 y − y 4 )(5e5t ) + (x3 − 4xy 3 ) (5 sec(t) tan(t))
100
CHAPTER 13. PARTIAL DERIVATIVES
dz
∂z dx ∂z dy
=
+
= −3 sin(3x + 4y)(2) − 4 sin(3x + 4y)(−1)
dt
∂x dt
∂y dt
5π
and y = −5π
At t = π, x =
4
2 15π
dz
15π
= −6 sin
so
− 5π + 4 sin
− 5π = −6 + 4 = −2
dt t=π
2
2
dz
∂z dx ∂z dy
−8
xy
4.
+ xexy (3)
=
+
= ye
dt
∂x dt
∂y dt
(2t + 1)2
At t = 0, x = 4 and y = 5
dz
so
= −40e20 + 12e20 = −28e20
dt t=0
2
r
1
r
1
2r
4r
dp
2u
√
− 3 −
− √
=
(2u)−
+
5.
=
du
2s + t
(2s + t)2
u
(2s + t)2 2 u
2s + t u3 (2s + t)2 2 u(2s + t)2
3.
y2
2xy
3xy 2
y 2 sin s 2xy cos s 3xy 2 sec2 s
dr
= 3 (− sin s) + 3 (cos s) − 4 (sec2 s) = −
+
−
ds
z
z
z
z3
z3
z4
2
2
∂z ∂x ∂z ∂y
+
= y 2 exy (3u2 ) + 2xyexy (1)
7. zu =
∂x ∂u ∂y ∂u
6.
2
2
= 3u2 y 2 exy + 2xyexy
2
2
∂z ∂x ∂z ∂y
+
= y 2 exy (0) + 2xyexy (−2v)
zv =
∂x ∂v
∂y ∂v
= −4vxyexy
2
∂z ∂x ∂z ∂y
+
= 2x cos 4y(2uv 3 ) − 4x2 sin 4y(3u2 )
∂x ∂u ∂y ∂u
= 4uv 3 x cos 4y − 12u2 x2 sin 4y
∂z ∂x ∂z ∂y
+
= 2x cos 4y(3u2 v 2 ) − 4x2 sin 4y(3v 2 )
zv =
∂x ∂v
∂y ∂v
= 6u2 v 2 x cos 4y − 12v 2 x2 sin 4y
8. zu =
9. zu = 4(4u3 ) − 10y[2(2u − v)(2)] = 16u3 − 40(2u − v)y
zv = 4(−24v 2 ) − 10y[2(2u − v)(−1)] = −96v 2 + 20(2u − v)y
2
2y
1
2xv 2
−2x
v
2y
10. zu =
+ 2
+
− 2 =
2
2
2
(x + y) v (x + y)
u
v(x + y)
u (x + y)2
u
2y
−2x
2yu
4xv
2v
zv =
− 2 +
=− 2
−
2
(x + y)
v
(x + y)2 u
v (x + y)2
u(x + y)2
3
3 2
(u + v 2 )1/2 (2u)(−e−t sin θ) + (u2 + v 2 )1/2 (2v)(−e−t cos θ)
2
2
= −3u(u2 + v 2 )1/2 e−t sin θ − 3v(u2 + v 2 )1/2 e−t cos θ
3
3
wθ = (u2 + v 2 )1/2 (2u)e−t cos θ + (u2 + v 2 )1/2 (2v)(−e−t sin θ)
2
2
= 3u(u2 + v 2 )1/2 e−t cos θ − 3v(u2 + v 2 )1/2 e−t sin θ
11. wt =
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13.5. CHAIN RULE
101
√
√
u/2 uv
rs2 u
v/2 uv
rv
(2r) +
(2rs2 ) = √
+√
12. wr =
1 + uv
1 + uv
uv(1 + uv)
uv(1 + uv)
√
√
v/2 uv
u/2 uv
−sv
r2 su
ws =
(−2s) +
(2r2 s) = √
+√
1 + uv
1 + uv
uv(1 + uv)
uv(1 + uv)
2
2
13. Ru = s2 t4 (ev ) + 2rst4 (−2uve−u ) + 4rs2 t3 (2uv 2 eu
2
2
2 2
v
2
2
) = s2 t4 ev − 4uvrst4 e−u + 8uv 2 rs2 t3 eu
2 2
2
2
16. sφ = 2pe3θ + 2q[− sin(φ + θ)] − 2rθ2 + 4(2) = 2pe3θ − 2q sin(φ + θ) − 2rθ2 + 8
sθ = 2p(3e3θ ) + 2q[− sin(φ + θ)] − 2r(2φθ) + 4(8) = 6pφe3θ − 2q sin(φ + θ) − 4rφθ + 32
17. (a) 3x2 − 2x2 (2yy 0 ) − 4xy 2 + y 0 = 0 =⇒ (1 − 4x2 y)y 0 = 4xy 2 − 3x2 =⇒ y 0 =
(b) fx = 3x2 − 4xy 2 , fy = −4x2 y + 1; y 0 = −
4xy 2 − 3x2
1 − 4x2 y
3x2 − 4xy 2
4xy 2 − 3x2
=
−4x2 y + 1
1 − 4x2 y
18. (a) 1 + 4yy 0 = ey y 0 =⇒ 1 = (ey − 4y)y 0 =⇒ y 0 =
ey
1
− 4y
1
1
= y
4y − ey
e − 4y
y cos xy
19. (a) y 0 = (cos xy)(xy 0 + y) =⇒ (1 − x cos xy)y 0 = y cos xy =⇒ y 0 =
1 − x cos xy
(b) f (x, y) = y − sin xy; fx = −y cos xy, fy = 1 − x cos xy;
−y cos xy
y cos xy
y0 = −
=
1 − x cos xy
1 − x cos xy
(b) f (x, y) = x + 2y 2 − ey ; fx = 1, fy = 4y − ey ; y 0 = −
2
(x + y)−1/3 (1 + y 0 ) = xy 0 + y =⇒ 2(x + y)−1/3 + 2(x + y)−1/3 y 0 = 3xy 0 + 3y
3
3y − 2(x + y)−1/3
=⇒ 2(x + y)−1/3 − 3x y 0 = 3y − 2(x + y)−1/3 =⇒ y 0 =
2(x + y)−1/3 − 3x
2
2
(b) f (x, y) = (x + y)2/3 − xy; fx = (x + y)−1/3 − y, fy = (x + y)−1/3 − x;
3
3
2
−1/3
−1/3
(x
+
y)
−
y
3y
−
2(x
+
y)
3
y0 = − 2
=
−1/3 − x
2(x + y)−1/3 − 3x
3 (x + y)
20. (a)
v
2 2
Rv = s2 t4 (2uvev ) + 2rst4 (e−u ) + 4rs2 t3 (2u2 veu v ) = 2s2 t4 uvev + 2rst4 e−u + 8rs2 t3 u2 veu
1 1
1
1
t2
t
1
t2
1/t
√
√
+
+ 2+
+
=
14. Qx =
2
2
2
2
2
P
q t
r 1 + (x/t)
qt
r(t + x2 )
1−x
p 1−x
1
2t sin−1 x
x
1
1
2x
−x/t2
2x
=
Qt = (2t sin−1 x) +
− 3 +
− 3−
2
P
q
t
r 1 + (x/t)
p
qt
r(t2 + x2 )
2x
u
2y
cosh rs
xu
y cosh rs
15. wt = p
+ p
=p
+ p
2
2
2
2
2
2
u
2 x + y rs + tu 2 x + y
x + y (rs + tu) u x2 + y 2
s
st sinh rs
yst sinh rs
2x
2y
xs
− p
wr = p
+ p
=p
2
2
2
2
2
2
rs
+
tu
u
2 x +y
2 x +y
x + y (rs + tu) u x2 + y 2
2x
yt cosh rs
t
2y
−t cosh rs
xt
wu = p
− p
+ p
=p
2
2
2
2
2
2
2
2
u
2 x + y rs + tu 2 x + y
x + y (rs + tu) u x2 + y 2
2 2
v
102
CHAPTER 13. PARTIAL DERIVATIVES
21. Fx = 2x, Fy = 2y, Fz = −2z;
∂z
2x
x ∂z
2y
y
=−
= ;
=−
=
∂x
−2z
z ∂y
−2z
z
z 1/3
2
2
∂z
(2/3)x−1/3
2 −1/3
=
−
;
x
, Fy = y −1/3 , Fz = z −1/3 ;
=−
3
3
3
∂x
(2/3)z −1/3
x1/3
∂z
z 1/3
(2/3)y −1/3
= − 1/3
=−
−1/3
∂y
(2/3)z
y
22. Fx =
23. F (x, y, z) = xy 2 z 3 + x2 − y 2 − 5z 2 , Fx = y 2 z 3 + 2x, Fy = 2xyz 3 − 2y, Fz = 3xy 2 z 2 − 10z
∂z
∂z
y 2 z 3 + 2x
y 2 z 3 + 2x
2xyz 3 − 2y
2xyz 3 − 2y
;
=−
=
=
−
=
∂x
3xy 2 z 2 − 10z
10z − 3xy 2 z 2 ∂y
3xy 2 z 2 − 10z
10z − 3xy 2 z 2
1
1
1 ∂z
−1/x
z
24. F (x, y, z) = z − ln(xyz); Fx = − , Fy = − , Fz = 1 − ;
=−
=
;
x
y
z ∂x
1 − 1/z
xz − x
∂z
−1/y
z
=−
=
∂y
1 − 1/z
yz − y
25. Let y = x + at and z = x − at. Then u(x, t) = F (y) + G(z) and
dF ∂y
dG ∂z
dF
dG
∂u
=
+
=
+
;
∂x
dy ∂x
dz ∂x
dy
dz
dF ∂y dG ∂z
dF
dG
∂u
=
+
=a
−a
;
∂t
dy ∂t
dz ∂t
dy
dz
Thus, a2
∂2u
d2 F ∂y
d2 G ∂z
d2 F
d2 G
=
+
=
+
;
∂x2
dy 2 ∂x
dz 2 ∂x
dy 2
dz 2
∂2u
d2 F ∂y
d2 G ∂z
d2 F
d2 G
=a 2
−a 2
= a2 2 + a2 2 .
2
∂xt
dy ∂t
dz ∂t
dy
dz
2
2
∂2u
∂2u
2d F
2d G
=
a
+
a
=
.
∂x2
dy 2
dz 2
∂t2
26. Solving η = x + at and ξ = x − at for x and t, we obtain x = (η + ξ)/2 and t = (η − ξ)/2a.
Then
∂u ∂x ∂u ∂t
1 ∂u
1 ∂u
∂u
=
+
=
−
∂ξ
∂x ∂ξ
∂t ∂ξ
2 ∂x 2a ∂t
and
Setting
∂2u
1 ∂ 2 u ∂x
1 ∂ 2 u ∂t
1 ∂2u
1 ∂2u
=
−
=
−
∂η∂ξ
2 ∂x2 ∂η
2a ∂t2 ∂η
4 ∂x2
4a2 ∂t2
1 ∂2u
1 ∂2u
∂2u
∂2u
∂2u
= 0, we have
− 2 2 = 0 or a2 2 = 2 .
2
∂η∂ξ
4 ∂x
4a ∂t
∂x
∂t
27. With x = r cos θ and y = r sin θ
∂u ∂x ∂u ∂y
∂u
∂u
∂u
=
+
=
cos θ +
sin θ
∂r
∂x ∂r
∂y ∂r
∂x
∂y
∂2u
∂ 2 u ∂x
∂ 2 u ∂y
∂2u
∂2u
2
=
cos
θ
+
sin
θ
=
cos
θ
+
sin2 θ
∂r2
∂x2 ∂r
∂y 2 ∂r
∂x2
∂y 2
∂u
∂u ∂x ∂u ∂y
∂u
∂u
∂ 2 u ∂y
=
+
=
(r sin θ) +
(−r sin θ) + 2
(r cos θ)
∂θ
∂x ∂θ
∂y ∂θ
∂y
∂y
∂y ∂θ
∂u
∂2u
∂u
∂2u
= −r
cos θ + r2 2 sin2 θ − r
sin θ − r2 2 cos2 θ.
∂x
∂x
∂y
∂y
13.5. CHAIN RULE
103
∂2u ∂2u
+ 2 = 0, we have
∂x2
∂y
∂ 2 u 1 ∂u
∂2u
∂2u
1 ∂u
1 ∂2u
∂u
2
2
+
cos θ + 2 sin θ +
+ 2 2 =
cos θ +
sin θ
∂r2
r ∂r
r ∂θ
∂x2
∂y
r ∂x
∂y
∂u
∂u
1
∂2u
∂2u
+ 2 −r
+ r2 2 sin2 θ − r
sin θ + r2 2 cos2 θ
r
∂x
∂x
∂y
∂y
2
∂2u
∂
u
∂u 1
1
2
2
2
2
=
(cos
θ
+
sin
θ)
+
(cos
θ
+
sin
θ)
+
cos
θ
−
cos
θ
∂x2
∂y 2
∂x r
r
∂u 1
1
+
sin θ − sin θ
∂y r
r
∂2u ∂2u
+ 2 =0
=
∂x2
∂y
Using
28.
∂z
dz ∂u ∂z
dz ∂u
=
,
=
∂x
du ∂x ∂y
du ∂y
29. Letting u = y/x in Problem 28, we have
∂z
dz ∂u
dz ∂u
dz y ∂z
dz
+y
=x
+y
=x
x
− 2 +y
∂x
∂y
du ∂x
du ∂y
du
x
du
1
dz −y
y
=
+
= 0.
x
du
x
x
30. We first compute
∂u
∂u ∂r
∂u
x
p
=
=
∂x
∂r ∂x
∂r x2 + y 2
x
∂ 2 u x2
∂u
y2
∂2u
∂u
y2
∂2u
+p
+ 2 2
=
=
2
2
2
2
3/2
2
2
3/2
∂x
∂r (x + y )
∂r (x + y )
∂r x + y 2
x2 + y 2 ∂r
∂u
x2
∂ 2 u y2
∂2u
=
.
∂y 2
∂r (x2 + y 2 )3/2 ∂r2 x2 + y 2
Then
∂u x2 + y 2
∂ 2 u 1 ∂u
∂2u ∂2u
∂ 2 u x2 + y 2
+ 2 =
=
+
.
+ 2 2
2
2
2
2
3/2
∂x
∂y
∂r (x + y )
∂r x + y
∂r2
r ∂r
31. We first compute
∂u
∂
= B erf
∂x
∂x
x
√
4kt
∂
=B
∂x
2
π
Z
√
x/ 4kt
!
e
0
−v 2
dv
2
1 −x2 /4kt
= B√ √
e
π 4kt
2
2
1 x −x2 /4kt
x
∂2u
√
√
=
B
−
e
= −B √ √ e−x /4kt
3
∂x2
2kt
π 4kt
2k π kt
!
Z x/√4kt
∂
x
∂u
∂ 2
2
x
1 −3/2
−v 2
= B erf √
=B
e dv = B √ − √
− t
e−x/4kt
∂t
∂t
∂t π 0
2
π
4kt
2 k
2
x
= −B √ √ e−x /4kt
3
2 π kt
Then k
∂2u
∂u
=
.
∂x2
∂t
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104
CHAPTER 13. PARTIAL DERIVATIVES
dI
∂I dE
∂I dR
1
E
=
+
= (2) − 2 (−1),
dt
∂E dt
∂R dt
R
R
1
dI
2
60
3/5
8
=
and when E = 60 and R = 50,
=
+
+
=
amp/min.
dt
50 502
25
25
125
32. We are given dE/dt = 2 and dR/dt = −1. Then
33. Since the height of the triangle is x sin θ, the area is given by A = 12 xy sin θ. Then
dA
∂A dx ∂A dy ∂A dθ
1
dx 1
dy 1
dθ
=
+
+
= y sin θ
+ x sin θ
+ xy cos θ .
dt
∂x dt
∂y dt
∂θ dt
2
dt
2
dt
2
dt
When x = 10, y = 8, θ = π/6, dz/dt = 0.3, dy/dt = 0.5, and dθ/dt = 0.1,
1
1
1
1
(0.3) + (10)
(0.5) + (10)(8)
2
2
2
2
√
√
= 0.6 + 1.25 + 2 3 = 1.85 + 2 2 ≈ 5.31 cm2 /s.
dA
1
= (8)
dt
2
34.
35.
√ !
3
(0.1)
2
dP
(V − 0.0427)(0.08)dT /dt
0.08T (dV /dt)
3.6 dV
=
−
+ 3
dt
(V − 0.0427)2
(V − 0.0427)2
V dt
0.08
3.6
dT
0.08T
dV
=
+
−
V − 0.0427 dt
V3
(V − 0.0427)2 dt
dS
dw
dh
= 0.1091 0.425w−0.575 h0.725
+ 0.725w0.425 h−0.275
dt
dt
dt
When w = 25, h = 29, dw/dt = 4.2, and dh/dt = 2,
dS
= 0.1091[0.425(25)−0.575 (29(0.725 (4.2) + 0.725(25)0.425 (29)−0.275 (2)] ≈ 0.5976in2 /yr.
dt
36.
dw
∂w dx ∂w dy ∂w dz
xdx/dt + ydy/dt + zdz/dt
−4x sin t + 4y cos t + 5z
p
=
+
+
=
=p
2
2
2
dt
∂x dt
∂y dt
∂z dt
x +y +z
16 cos2 t + 16 sin2 t + 25t2
−16 sin t cos t + 16 sin t cos t + 25t
25t
√
=
=√
2
16 + 25t
16 + 25t2
dw
125π
125π/2
=√
≈ 4.9743
=p
dt t=5π/2
64 + 625π 2
16 + 625π 2 /4
37. Since dT /dT = 1 and
0 = FT =
∂P
= 0,
∂T
∂F ∂P
∂F ∂V
∂F ∂T
∂V
∂F/∂T
1
+
+
=⇒
=−
=−
.
∂P ∂T
∂V ∂T
∂T ∂T
∂T
∂F/∂V
∂T /∂V
38. (a) From the law of sines,
r
500
500 sin φ
=
so r =
.
sin φ
sin(π − θ − φ)
sin(θ + φ)
(b) r = 500 sin 75◦ / sin 137◦ ≈ 708 yds
13.5. CHAIN RULE
105
(c) Using the chain rule, we obtain
dr
∂r dθ
∂r dφ
=
+
dt
∂θ dt
∂φ dt
500 sin φ cos(θ + φ) dθ
sin(θ + φ) cos φ − sin φ cos(θ + φ) dφ
=−
+ 500
dt
dt
sin2 (θ + φ)
sin2 (θ + φ)
500 sin φ cos(θ + φ) dθ
=−
dt
sin2 (θ + φ)
[sin θ cos φ + cos θ sin φ] cos φ − sin φ[cos θ cos φ − sin θ sin φ] dφ
+ 500
dt
sin2 (θ + φ)
500 sin φ cos(θ + φ) dθ
500 sin θ dφ
=−
+
.
2
dt
sin (θ + φ)
sin2 (θ + φ) dt
When dθ/dt = 5◦ = 5π/180 and dφ/dt = −10π/180, we have
500 sin 75◦ cos 137◦ 5π
500 sin 62◦
10π
dr
=−
+
−
≈ −99.4yd/min.
dt
180
180
sin2 137◦
sin2 137◦
The distance from C to A is decreasing.
39. (a) Using f = π, l = 6, V = 100, and c = 330, 000 we obtain f ≈ 380.04 cycles per second.
!
r
c
A −1/2
A
1
∂f
c
A 1
=
− 2 =−
=−
f and
(b) We first note that
∂V
4π lV
lv
4π lV V
2V
∂f
1
= − f.
∂l
2l
df
∂f dV
∂f dl
1 dV
1 dl
f 1 dV
1 dl
Then
=
+
=−
f
− f
=−
+
.
dt
∂V dt
∂l dt
2V dt
2l dt
2 V dt
l dt
Using dV /dt = −10, dl/dt = 1, V = 100, and l = 6 we find
df
f
1
1
f 1
1
=−
(−10) + (1) = −
−
< 0.
dt
2 100
6
2 6 10
The frequency is decreasing.
40. (a)
w
w
x
x
w
y
w
z
y
z
y
x
z
x
x
x
106
CHAPTER 13. PARTIAL DERIVATIVES
dw
∂w ∂w y
∂w dz
=
+
+
dx
∂x
∂y x
∂z dx
(b) Using the formula from Part (a), we have
dw
= (y 2 + 1) + (2xy − 2z)
dx
1
+ (−2y)(ex )
x
41.
z
z
u
z
v
u
u
t1 u
t2
t1
t2
z
w
w
v
v
t1
u
u t4
t3
t3
t4
t1
v
t2
t2
v
v t4
t3
t3
w
t1 w
t2
t4
t1
w
w t4
t3
t2
t3
∂z
∂z ∂u
∂z ∂v
∂z ∂w
=
+
+
∂t2
∂u ∂t2
∂v ∂t2
∂w ∂t2
∂z
∂z ∂u
∂z ∂v
∂z ∂w
=
+
+
∂t4
∂u ∂t4
∂v ∂t4
∂w ∂t4
42. Since w = F (x, y, z, u) = 0, ∂w/∂x = 0. Also dx/dx = 1, ∂y/∂x = 0, and ∂z/∂x = 0. Then
dx
∂u
∂z
∂u
∂w
= Fx (x, y, z, u)
+ Fu (x, y, z, u)
+ FZ (x, y, z, u)
+ Fu (x, y, z, u)
∂x
dx
∂x
∂x
∂x
implies ∂u/∂x = −Fx (x, y, z, u)/Fu (x, y, z, u). Similarly, ∂u/∂y = −FY (x, y, z, u)/Fu (x, y, z, u)
and ∂u/∂z = −FZ (x, y, z, u)/Fu (x, y, z, u).
43. Letting F (x, y, z, u) = −xyz + x2 yu + 2xy 3 u − u4 − 8 we find FZ = −yz + 2xyu + 2y 3 u,
Fy = −xz + x2 u + 6xy 2 u, Fz = −xy, and Fu = x2 y + 2xy 3 − 4u3 . Then
∂u
−yz + 2xyu + 2y 3 u
=− 2
,
∂x
x y + 2xy 3 − 4u3
∂u
−xz + x2 y + 6xy 2 u
=− 2
,
∂y
x y + 2xy 3 − 4u3
∂u
xy
= 2
.
∂z
x y + 2xy 3 − 4u3
44. (a) Let u = λx and v = λy. Then f (u, v) = λn f (x, y), and differentiating both sides with
respect to λ, we have
∂f ∂u ∂f ∂v
+
= nλn−1 f (x, y) or xfu (u, v) + yfu (u, v) = nλn−1 f (x, y).
∂u ∂λ
∂v ∂λ
Letting λ = 1, we have u = x and y = v, so xfx (x, y) + yfy (x, y) = nf (x, y).
(b) f (λx, λy) = 4(λx)2 (λy 3 ) − 3(λx)(λy)4 + (λx)5 = λ5 f (x, y)
t4
www.elsolucionario.org
13.6. DIRECTIONAL DERIVATIVE
107
(c) xfx + yfy = x(8xy 3 − 3y 4 + 5x4 ) + y(12x2 y 2 − 12xy 3 )
= 8x2 y 3 − 3xy 4 + 5x5 + 12x2 y 3 − 12xy 4 = 20x2 − 15xy 4 + 5x5
= 5(4x2 y 3 − 3xy 4 + x5 ) = 5f (x, y)
y
y
λy
= λ0 f
, we see that z = f
(d) By observing that f
=f
λx
x
x
of degree zero.
13.6
y
x
is homogeneous
Directional Derivative
1. ∇f = (2x − 3x2 y 2 )i + (4y 3 − 2x3 y)j
2
2
2. ∇f = 4xye−2x y i + (1 + 2x2 e−2x y )j
3. ∇F =
y2
2xy
3xy 2
i
+
j
−
k
z3
z3
z4
4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j − xy 2 sin yzk
5. ∇f = 2xi − 8yj; ∇f (2, 4) = 4i − 32j
x3 − 4y 3
27
3x2
5
6. ∇f = p
i+ p
j; ∇f (3, 2) = √ i − √ j
3
4
3
4
38
2
38
2 x y−y
2 x y−y
7. ∇F = 2xz 2 sin 4yi + 4x2 z 2 cos 4yj + 2x2 z sin 4yk
√
√
4π
4π
4π
i + 16 cos
j + 8 sin
k = 2 3i − 8j − 4 3k
∇F (−2, π/3, 1) = −4 sin
3
3
3
2x
2y
2z
4
3
1
i+ 2
j+ 2
k; ∇F (−4, 3, 5) = − i + j + k
x2 + y 2 + z 2
x + y2 + z2
x + y2 + z2
25
25
25
√
√
f (x + h 3/2, y + h/2) − f (x, y)
(x + h 3/2)2 + (y + h/2)2 )2x − y 2
9. Du f (x, y) = lim
= lim
h→0
h→0
h
h
√
2
2
√
√
h 3x + 3h /4 = hy + h /4
= lim
= lim ( 3x + 3h/4 + y + h/4) = 3x + y
h→0
h→0
h
√
√
f (x + h 2/2, y + h 2/2) − f (x, y)
10. Du f (x, y) = lim
h→0
h
√
√
3x + 34 2/2 − (y + h 2/2)2 − 3x + y 2
= lim
h→0
h
√
√
√
√
√
√
3h 2/2 − h 2yh2 /2
= lim
= lim (3 2/2 − 2y − h/2) = 3 2/2 − 2y
h→0
h→0
h
√
√
3 1
15 3
2 6
3 5
11. u =
i j; ∇f = 15x y i + 30x y j; ∇f (−1, 1) = 15i − 30j; Du f (−1, 1) =
− 15 =
2 2
2
√
15
( 3 − 2)
2
8. ∇F =
108
CHAPTER 13. PARTIAL DERIVATIVES
√
√
2 2
12. u =
i
j; ∇f = (4 + y 2 )i(2xy − 5)j; ∇f (3, −1) = 5i − 11j;
2 2
√
√
√
5 2 11 2
−
= −3 2
Du f (3, −1) =
2
2
√
√
x
1
3 10
−y
1
10
13. u =
i+ 2
j; ∇f (2, −2) = i + j;
i−
j; ∇f = 2
2
2
10
10
y
x +y
4
4
√
√ x +√
10 3 10
10
Du f (2, −2) =
−
=−
40
40
20
x2
6
8
y2
i
+
j; ∇f (2, −1) = i + 4j;
i + j; ∇f =
10
10
(x + y)2
(x + y)2
3 16
19
Du f (2, −1) = +
=
5
5
5
√
15. u = (2i + j)/ 5; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f (3, 2) = 28i + 42j;
2(28)
42
98
Du f (3, 2) = √ + √ = √
5
5
5
√
√
16. u = −i; ∇F = 2x tan yi + x2 sec2 yj; ∇f (1/2, π/3) = 3i + j; Du f (1/2, π/3) = − 3
14. u =
1
1
17. u = √ j + √ k; ∇f = 2xy 2 (2z + 1)2 i2x2 y(2z + 1)2 j + 4x2 y 2 (2z + 1)k; ∇f (1, −1, 1) =
2
2
√
18
12
6
18i − 18j + 12k; Du f (1, −1, 1) = − √ + √ = − √ = −3 2
2
2
2
2
1
2x
1
2y
2y 2 − 2x2
18. u = √ i − √ j + √ k; ∇f = 2 i − 2 j +
k; ∇f (2, 4, −1) = 4i − 8j − 24k;
z
z
z3
6
6
6
√
4
16
24
Du f (2, 4, −1) = √ − √ − √ = −6 6
6
6
6
x2 + 4z
y2
i+ p
j+ p
k;
x2 y + 2y 2 z
2 x2 y + 2y 2 z
x2 y + 2y 2 z
∇f (−2, 2, 1) = −i + j + k; Du f (−2, 2, 1) = −1
19. u = −k; ∇f = p
xy
√
2
1
2
20. u = −(4i−4j+2k)/ 36 = − i+ j− k; ∇f = 2i−2yj+2zk; ∇f (4, −4, 2) = 2i+8j+4k;
3
3
3
4 16 4
8
Du f (4, −4, 2) = − +
− =
3
3
3
3
√
21. u = (−4i − j/ 17; ∇f = 2(x − y)i − 2(x − y)j; ∇f (4, 2) = 4i − 4j;
16
4
12
Du f (4, 2) = − √ + √ = − √
17
17
17
√
22. u = (−2i + 5j/ 29; ∇f = (3x2 − 5y)i − (5x − 2y)j; ∇f (1, 1) = −2i − 3j;
4
15
11
Du f (1, 1) = √ − √ = − √
29
29
29
√
√
2
2x
2x
23. ∇f = 2e sin yi + e cos yj; ∇f (0, π/4) = 2i +
j
2
p
√
√
√
√ 2
1/2
The maximum Du is ( 2) + ( 2/2)2
= 5/2 in the direction 2i + ( 2/2)j.
13.6. DIRECTIONAL DERIVATIVE
109
24. ∇f = (xyex−y + yex−y i + (−xyex−y + xex−y j; ∇f (5, 5) = 30i − 20j
√
1/2
The maximum Du is 302 + (−20)2
= 10 13 in the direction 30i − 20j.
25. ∇f = (2x + 4z)i + 2z 2 j + (4x + 4yz)k; ∇f (1, 2, −1) = −2i + 2j − 4k
√
1/2
The maximum Du is (−2)2 + (2)2 + (−4)2
= 2 6 in the direction −2i + 2j − 4k.
26. ∇f = yzi + xzj + xyk; ∇f (3, 1, −5) = −5i − 15j + 3k
1/2 √
The maximum Du is (−5)2 + (−15)2 + (3)2
= 259 in the direction −5i − 15j + 3k.
2 2
2
2 2
2
27. ∇f =
p2x secp(x + y )ip+ 2y sec2 (x + y )j;
p
∇f ( π/6, π/6) = 2 p
π/6 sec (π/3)(i + j) = 8p π/6(i + j)
The minimum Du is −8 π/6(12 + 12 )1/2 = −8 π/3 in the direction −(i + j).
28. ∇f = 3x2 i − 3y 2 j; ∇f (2, −2) = 12i − 12j = 12(i − j)
√
1/2
The minimum Du is −12 12 + (−1)2
= −12 2 in the direction −(i − j) = −i + j.
√ y
√
√
ze
x
3
2
y
√
29. ∇f =
i + xze j + √ k; ∇f (16, 0, 9) = i + 12j + k. The minimum Du is
8
3
2 x
2 z
√
3
2
2
2
2 1/2
− (3/8) + 12 + (2/3)
= − 83281/24 in the direction − i − 12j − k.
8
3
1
1
1
i + j − k; ∇f (1/2, 1/6, 1/3) = 2i + 6j − 3k
x
y
z
1/2
The minimum Du is − 22 + 62 (−3)2
= −7 in the direction −2i − 6j + 3k.
30. ∇f =
31. Using implicit differentiation on 2x2 + y 2 = 9 we find y 0 =√−2x/y. At (2, 1) the slope of
the tangent line is −2(2)/1 = −4. √Thus, u√= ±(i − 4j)/
√ 17. Now, ∇f = i + 2yj and
∇f (3, 4) = i + 8j. Thus, Du = ±(1/ 17 − 32 17) = ±31/ 17.
2x + y − 1
x + 2y
3x + 3y − 1
√
√
32. ∇f = (2x + y − 1)i + (x + 2y)j; Du f (x, y) =
+ √
=
Solving
2
2
2
√
(3x + 3y − 1)/ 2 = 0 we see that Du is 0 for all points on the line 3x + 3y = 1.
33. (a) Vectors perpendicular to 4i + 3j are ±(3i − 4j). Take u = ±
3
4
i− j .
5
5
√
4
3
(b) u = (4i + 3j)/ 16 + 9 = i + j
5
5
4
3
(c) u = − i − j
5
5
34. D−u f (a, b) = ∇f (a, b) · (−u) = −∇f (a, b) · u = −Du f (a, b) = −6
35. (a) ∇f = (3x2 − 6xy 2 )i + (−6x2 y + 3y 2 )j
3(3x2 − 6xy 2 ) −6x2 y + 3y 2
9x2 − 18xy 2 − 6x2 y + 3y 2
√
√
√
Du f (x, y) =
+
=
10
10
10
www.elsolucionario.org
110
CHAPTER 13. PARTIAL DERIVATIVES
3
3
(b) F (x, y) = √ (3x2 − 3xy 2 − 2x2 y + y 2 ); ∇F = √ [(6x − 6y 2 − 4xy)i + (−12xy −
10
10
2x2 + 2y)j] 3
3
3
1
2
√
√
(6x − 6y − 4xy) + √
(−12xy − 2x2 + 2y)
Du F (x, y) = √
10
10
10
10
9
3
1
= (3x − 3y 2 − 2xy) + (−6xy − x2 + y) = (27x − 27y 2 − 36xy − 3x2 + 3y)
5
5
5
12
5
α − β = 7 and Dv f (a, b) =
36. Let ∇f (a, b) = αi + βj. Then Du f (a, b) = ∇f (a, b) · u =
13
13
5
12
∇f (a, b) · v =
α − β = 3. Solving for α and β, we obtain α = 13 and β = −13/6. Thus,
13
13
∇f (a, b) = 13i − (13/6)j.
37.
38. ∇f = h2x, −5yi, |∇f | =
p
10x2 + 25y 2 = 10, 4x2 + 25y 2 = 100,
x2
y2
+
=1
25
4
y
x
39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, the
maximum decrease in temperature) is in the direction −∇T (4, 3) = −16i − 4j.
40. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of
a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be −∇T (x, y), we
have x0 (t)i + y 0 (t)j = −∇T (x, y) = 4xi + 2yj. Separating variables in dx/dt = 4x, we obtain
dx/x = 4dt, ln x = 4t + c1 , and x = C1 e4t . Separating variables in dy/dt = 2y, we obtain
dy/y = 2dt, ln y = 2t + c2 , and y = C2 e2t . Since x(0) = 4 and y(0) = 2, we have x = 4e4t
and y = 2e2t . The equation of the path is 4e4t i + 2e2t j for t ≥ 0, or eliminating the parameter,
x = y 2 , y ≥ 0.
41. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of
a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be ∇T (x, y), we
have x0 (t)i + y 0 (t)j = ∇T (x, y) = −4xi − 2yj. Separating variables in dx/dt = −4x, we obtain
dx/x = −4dt, ln x = −4t + c1 , and x = C1 e−4t . Separating variables in dy/dt = −2y, we
obtain dy/y = −2dt, ln y = −2t + c2 , and y = C2 e−2t . Since x(0) = 3 and y(0) = 4, we have
x = 3e−4t and y = 4e−2t . The equation of the path is 3e−4t i + 4e−2t j, or eliminating the
parameter, 16x = 3y 2 , y ≥ 0.
13.6. DIRECTIONAL DERIVATIVE
111
42. Substituing x = 0, y = 0, z = 1, and T = 500 into t =
and T (x, y, z) =
k
we see that k = 500
x2 + y 2 + z 2
500
.
x2 + y 2 + z 2
1
1
2
2
h1, −2, −2i = i − j − k
3
3
3
3
1000y
1000z
1000x
i− 2
j− 2
k
∇T = − 2
2
2
2
2
2
2
(x + y + z )
(x + y + z )
(x + y 2 + z 2 )2
500
750
750
∇T (2, 3, 3) = −
i−
j−
k
121
121
121
1
500
2
750
2
750
2500
Du T (2, 3, 3) =
−
−
−
−
−
=
3
121
3
121
3
121
363
(a) u =
(b) The direction of maximum increase is
∇T (2, 3, 3) = −
500
750
750
252
i−
j−
k=
(−2i − 3j − 3k).
121
121
121
121
(c) The maximum rate of change of T is |∇T (2, 3, 3)| =
√
250 √
250 22
4+9+9=
.
121
121
Gmx
Gmy
Gm
i+ 2
j= 2
(xi + yj)
2
3/2
2
3/2
+y )
(x + y )
(x + y 2 )3/2
The maximum and minimum values of Du U (x, y) are obtained when u is in the directions
∇U and
−∇U , respectively. Thus, at a point (x,y), not (0,0), the directions of maximum and minimum
increase in U are xi+yj and −xi−yj, respectively. A vector at (x, y) in the direction ±(xi+yj)
lies on a line through the origin.
43. ∇U =
(x2
44. Since ∇f = fx (x, y)i + fy (x, y)j, we have ∂f /∂x = 3x2 + y 3 + yexy . Integrating, we obtain
f (x, y) = x3 + xy 3 + exy + g(y). Then fy = 3xy 2 + xexy + g 0 (y) = −2y 2 + 3xy 2 + xexy . Thus,
g 0 (y) = −2y 2 , g(y) = − 32 y 3 + c, and f (x, y) = x3 + xy 3 + exy − 23 + C.
45. ∇(cf ) =
∂
∂
(cf )i +
j = cfx i + cfy j = c(fx i + fy j) = c∇f
∂x
∂y
46. ∇(f + g) = (fx + gx )i + (fy + gy )j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g
47. ∇(f g) = (f gx + fx g)i + (f gy + fy g)j = f (gx i + gy j) + g(fx i + fy j) = f ∇g + g∇f
48. ∇(f /g) = (gfx − f gx )/g 2 i + (gfy − f gy )/g 2 j = g(fx i + fy j)/g 2 − f (gx i + gy j)/g 2
= g∇f /g 2 − f ∇g/g 2 = (g∇f − f ∇g)/g 2
p
∂r
x
x
∂r
y
y
x2 + y 2 so
=p
= and
=p
=
2
2
2
2
∂x
r
∂y
r
x +y
x +y
Dx yE 1
r
This gives ∇r =
,
= hx, yi =
r r
r
r
49. r(x, y) =
112
50.
CHAPTER 13. PARTIAL DERIVATIVES
∂ (f (r))
df ∂r
∂ (f (r))
df ∂r
∂ (f (r)) ∂ (f (r))
=
and
=
so that ∇f (r) = h
,
i
∂x
dr ∂x
∂y
dr ∂y
∂x
∂y
df ∂r df ∂r
df ∂r ∂r
=h
,
i=
h ,
i = f 0 (r)∇r = f 0 (r)r/r
dr ∂x dr ∂y
dr ∂x ∂y
51. Let u = u1 i + u2 j and v = v1 i + v2 j.
Dv f = (fx i+ fy j) · v = v1 fx + v2 fy
∂
∂
Du Dv f =
(v1 fx + v2 fy )i +
(v1 fx + v2 fy )j · u = [(v1 fxx + v2 fyz )i + (v1 fxy + v2 fyy )j] · u
∂x
∂y
= u1 v1 fxx + u1 v2 fyx + u2 v1 fxy + u2 v2 fyy
D − uf = (fx i + fy j) · u = u1 fx + u2 fy
∂
∂
(u1 fx + u2 fy )i +
(u1 fx + u2 fy )j · v = [(u1 fxx + u2 fyx )i + (u1 fxy + u2 fyy )j] · v
Dv Du f =
∂x
∂y
= u1 v1 fxx + u2 v1 fyx + u1 v2 fxy + u2 v2 fyy
Since the second partial derivatives are continuous, fxy = fyx and Du Dv f = Dv Du f. [Note
that this result is a generalization fxy = fyx since Di Dj f = fyx and Dj Di f = fxy ]
i
∂
∂x
f1
k
∂
52. ∇ × F =
∂z
f3
∂f3
∂f2
∂f3
∂f1
∂f2
∂f1
=
−
i−
−
j+
−
k
∂y
∂z
∂x
∂z
∂x
∂y
13.7
j
∂
∂y
f2
Tangent Planes and Normal Lines
1. Since f (6, 1) = 4, the level curve is x − 2y = 4. ∇f = i − 2j;
∇f (6, 1) = i − 2j
y
x
2. Since f (1, 3) = 5, the level curve is y+2x = 5x or y = 3x, x 6= 0.
y
1
∇f = − 2 i + j; ∇f (1, 3) = −3i + j
x
x
y
x
www.elsolucionario.org
13.7. TANGENT PLANES AND NORMAL LINES
113
3. Since f (2, 5) = 1, the level curve is y = x2 + 1. ∇f = −2xi + j;
∇f (2, 5) = −4i + j
y
x
4. Since f (−1, 3) = 10, the level curve is x2 + y 2 = 10.
∇f = 2xi + 2yj; ∇f (−1, 3) = −2i + 6j
y
x
5. Since f (−2, −3) = 2, the level curve is x2 /4 + y 2 /0 = 2
x
2y
x2 /8 + y 2 /18 = 1. ∇f = i + j; ∇f (−2, −3) = −i −
2
9
y
or
2
j
3
x
6. Since f (2, 2) = 2, the level curve is y 2 = 2x, x 6= 0.
2y
y2
∇f = − 2 i + j; ∇f (2, 2) = −i + 2j
x
x
y
x
114
CHAPTER 13. PARTIAL DERIVATIVES
7. Since f (1, 1) = −1, the level curve is (x − 1)2 − y 2 = −1 or
y 2 − (x − 1)2 = 1. ∇f = 2(x − 1)i − 2yj; ∇f (1, 1) = −2j
y
x
8. Since f (π/6, 3/2) = 1, the level curve is y − 1 = sin x or
−(y − 1) cos x
1
y = 1 + sin x, sin x 6= 0. ∇f =
i+
j;
2
sin x
sin x
√
∇f (π/6, 3/2) = − 3i + 2j
y
x
9. Since f (3, 1, 1) = 2, the level curve is y + z = 2 ∇f = j + k;
∇f (3, 1, 1) = j + k
z
2
2
y
x
10. Since f (1, 1, 3) = −1, the level curve is x2 + y 2 − z = −1 or
z = 1 + x2 + y 2 . ∇f = 2xi + 2yj − k; ∇f (1, 1, 3) = 2i + 2j − k
z
y
x
11. Since F (3, 4, 0) = 5, the level curve is x2 + y 2 + z 2 = 25.
x
y
z
∇F = p
i+ p
j+ p
k;
2
2
2
2
2
2
2
x +y +z
x +y +z
x + y2 + z2
3
4
∇F (3, 4, 0) = i + j
4
5
z
5
5
y
x
13.7. TANGENT PLANES AND NORMAL LINES
115
12. Since F (0, −1, 1) = 0, the level curve is x2 − y 2 + z = 0 or z = y 2 − x2 .
∇F = 2xi − 2yj + k; ∇F (0, −1, 1) = 2i + k
z
y
x
13. F (x, y, z) = x2 + y 2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c 4i + j + 21 k or 2x =
4c, 2y = c, −1 = c/2. From the third equation c = −2. Thus, x = −4 and y = −1. Since
z = x2 + y 2 = 16 + 1 = 17, the point on the surface is (−4, −1, −17).
14. F (x, y, z) = x3 + y 3 + z; ∇F = 3x2 i + 2yj + k. We want ∇F = c(27i + 8j + k) or 3x2 =
27c, 2y = 8c, 1 = c. From c = 1 we obtain x = ±3 and y = 4. Since z = 15 − x3 − y 2 =
15 − (±3)3 − 16 = −1 ∓ 27, the points on the surface are (3, 4, −28) and (−3, 4, 26).
15. F (x, y, z0 = x2 + y 2 + z 2 ; ∇F = 2xi + 2yj + 2zk. ∇F (−2, 2, 1) = −4i + 4j + 2k. The equation
of the tangent plane is −4(x + 2) + 4(y − 2) + 2(z − 1) = 0 or −2x + 2y + z = 9.
16. F (x, y, z) = 5x2 − y 2 + 4z 2 ; ∇F = 10xi − 2yj + 8zk; ∇F (2, 4, 1) = 20i − 8j + 8k.The equation
of the tangent plane is 20(x − 2) − 8(y − 4) + 8(z − 1) = 0 or 5x − 2y + 2z = 4.
17. F (x, y, z) = x2 − y 2 − 3z 2 ; ∇F = 2xi − 2yj − 6zk; ∇F (6, 2, 3) = 12i − 4j − 18k. The equation
of the tangent plane is 12(x − 6) − 4(y − 2) − 18(z − 3) = 0 or 6x − 2y − 9z = 5.
18. F (x, y, z) = xy + yz + zx; ∇F = (y + z)i + (x + z)j + (y + x)k; ∇F (1, −3, −5) = −8i − 4j − 2k.
The equation of the tangent plane is −8(x − 1) − 4(y + 3) − 2(z + 5) = 0 or 4x + 2y + z = −7.
19. F (x, y, z) = x2 + y 2 + z; ∇F = 2xi + 2yj + k; ∇F (3, −4, 0) = 6i − 8j + k. The equation of
the tangent plane is 6(x − 3) − 8(y + 4) + z = 0 or 6x − 8y + z = 50.
20. F (x, y, z) = xz; ∇F = zi + xk; ∇F (2, 0, 3) = 3i + 2k. The equation of the tangent plane is
3(x − 2) + 2(z − 3) = 0 or 3x + 2z = 12.
√
21. F (x, y,√z) = cos(2x+y)−z; ∇F = −2 sin(2x+y)i−sin(2x+y)j−k;
∇F (π/2, π/4, −1 2) =
√
√
√ 2
π
2
π
1
2i+
j−k. The equation of the tangent plane is 2 x −
+
y−
− z+√
=
2
2
2
4
2
√
π π √
1
5π
0, 2 x −
+ y−
− 2 z+√
= 0, or 2x + y − 2z =
+ 1.
2
4
4
2
22. F (x, y, z) = x2 y 3 + 6z; ∇F = 2xy 3 i + 3x2 y 2 j + 6k; ∇F (2, 1, 1) = 4i + 12j + 6k. The equation
of the tangent plane is 4(x − 2) + 12(y − 1) + 6(z − 1) = 0 or 2x + 6y + 3z = 13.
√
√
√
2x
2y
23. F (x, y, z) = ln(x2 + y 2 ) − z; ∇F = 2
i+ 2
j − k; ∇F (1/ 2, 1/ 2, 0) = 2i +
2
2
x +y
x +y
√
√
√
1
1
√
√
2j − k. The equation of the tangent plane is 2 x −
+ 2 y−
− (z − 0) =
2
2
√
√
√
1
1
+2 y− √
− 2z = 0, or 2x + 2y − 2z = 2 2.
0, 2 x − √
2
2
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116
CHAPTER 13. PARTIAL DERIVATIVES
−2y
24. F (x,
sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) =
√ y, z) = 8e
16 3i − 8j − k. The equation of the tangent plane is
√
√
√
2 3π
− 4.
16 3(x − π/24) − 8(y − 0) − (z − 4) = 0 or 16 3x − 8y − z =
3
25. The gradient of F (x, y, z) = x2 + y 2 + z 2 is ∇F = 2xi + 2yj + 2zk, so the normal vector to
the surface at (x0 , y0 , z0 ) is 2x0 i + 2y0 j + 2z0 k. A normal vector to the plane 2x + 4y + 6z = 1
is 2i + 4j + 6k. Since we want the tangent plane to be parallel to the given plane, we find c
so that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c, y0 = √
2c, z0 = 3c. Now, (x0 , y0 , z0 ) is on
2
2
2
2
2. Thus, the points on the surface
the surface,
so
c
+
(2c)
+
(3c)
=
14c
=
7
and
c
=
±1/
√
√ √
√
√
√
are ( 2/2, 2, 3 2/2) and − 2/2, − 2, −3 2/2).
26. The gradient of F (x, y, z) = x2 − 2y 2 − 3z 2 is ∇F (x, y, z) = 2xi − 4yj − 6zk, so a normal vector
to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i − 4y0 j − 6z0 k. A normal vector to the
plane 8x+4y+6z = 5 is 8i+4j+6k. Since we want the tangent plane to be parallel to the given
plane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c or x0 = 4c, y0 = −c, √z0 = −c.
2
2
2
Now (x0 , y0 , z0 ) is on the surface,
2(−c)2 − 3(−c)
√so (4c)
√ −√
√ =
√ 11c
√ = 33 and c = ± 3. Thus,
the points on the surface are (4 3, − 3, − 3) and (−4 3, 3, 3).
27. The gradient of F (x, y, z) = x2 +4x+y 2 +z 2 −2z is ∇F = (2x+4)i+2yj+(2z−2)k, so a normal
to the surface at (x0 , y0 , z0 ) is (2x0 + 4)i + 2y0 j + (2z0 − 2)k. A horizontal plane has normal ck
for c 6= 0. Thus, we want 2x0 + 4 = 0, 2y0 = 0, 2z0 − 2 = c or x0 = −2, y0 = 0, z0 = c + 1.
Since (x0 , y0 , z0 ) is on the surface, (−2)2 + 4(−2) + (c + 1)2 − 2(c + 1) = c2 − 5 = 11 and
c = ±4. The points on the surface are (−2, 0, 5) and (−2, 0, −3).
28. The gradient of F (x, y, z) = x2 + 3y 2 + 4z 2 − 2xy is ∇F = (2x − 2y)i + (6y − 2x)j + 8zk, so
a normal to the surface at (x0 , y0 , z0 ) is 2(x0 − y0 )i + 2(3y0 − x0 )j + 8z0 k.
(a) A normal to the xz plane is cj for c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) =
c, 8z0 = 0 or x0 = y0 , 3y0 −x0 = c/2, z0 = 0. Solving the first two equations, we obtain
2
x0 = y0 = c/4. Since (x0 , y0√
, z0 ) is on the surface, (c/4)2 +3(c/4)2 +4(0)√
−2(c/4)(c/4)
=
√
2
2c /16
=
16
and
c
=
±16/
2.
Thus,
the
points
on
the
surface
are
(4/
2,
4/
2,
0)
and
√
√
(−4 2, −4 2, 0).
(b) A normal to the yz-plane is ci for c 6= 0. Thus, we want 2(x0 − y0 ) = c, 2(3y0 − x0 ) =
0, 8z0 = 0 or x0 − y0 = c/2, x0 = 3y0 , z0 = 0. Solving the first two equations, we
obtain x0 = 3c/4 and y0 = c/4. Since (x0 , y0 , z0 )√is on the surface, (3c/4)2 + 3(c/4)2 +
4(0)2√− 2(3c/4)(c/4)
= 6c2 /16
√
√ = 16 and
√ c = ±16 6. Thus, the points on the surface are
(12/ 6, 4/ 6, 0) and (−12/ 6, −4/ 6, 0).
(c) A normal to the xy-plane is ckfor c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) =
0, 8z0 = c or x0 = y0 , 3y0 −x0 = 0, z0 = c/8. Solving the first two equations, we obtain
x0 = y0 = 0. Since (x0 , y0 , z0 ) is on the surface, 02 +3(0)2 +4(c/8)2 −2(0)(0) = c2 /16 = 16
and c = ±16. Thus, the points on the surface are (0, 0, 2) and (0, 0, −2).
29. If (x0 , y0 , z0 ) is on x2 /a2 + y 2 /b2 + z 2 /c2 = 1, then x20 /a2 + y02 /b2 + z02 /c2 = 1 and x0 , y0 , z0 )
is on the plane xx0 /a2 + yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is
13.7. TANGENT PLANES AND NORMAL LINES
117
∇F (x0 , y0 , z0 ) = (2x − 0/a2 )i + (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i +
(y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane,
the normal vectors are parallel and the plane is tangent to the surface.
30. If (x0 , y0 , z0 ) is on x2 /a2 − y 2 /b2 + z 2 /c2 = 1, then x20 /b2 − y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 )
is on the plane xx0 /a2 − yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is
∇F (x0 , y0 , z0 ) = (2x0 /a2 )i − (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i −
(y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane,
the normal vectors are parallel, and the plane is tangent to the surface.
31. F (x, y, z) = x2 + 2y 2 + z 2 ; ∇F = 2xi + 4yj + 2zk; ∇F (1, −1, 1) = 2i − 4j + 2k. Parametric
equations of the line are x = 1 + 2t, y = −1 − 4t, z = 1 + 2t.
32. F (x, y, z) = 2x2 − 4y 2 − z; ∇F = 4xi − 8yj − k; ∇F (3, −2, 2) = 12i + 16j − k. Parametric
equations of the line are x = 3 + 12t; y = −2 + 16t, z = 2 − t.
33. F (x, y, z) = 4x2 + 9y 2 − z; ∇F = 8xi + 18yj − k; ∇F (1/2, 1/3, 3) = 4i + 6j − k. Symmetric
y − 1/3
z−3
x − 1/2
=
=
.
equations of the line are
4
6
−1
34. F (x, y, z) = x2 + y 2 − z 2 ; ∇F = 2xi + 2yj − 2zk; ∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric
x−3
y−4
z−5
equations of the line are
=
=
.
6
8
−10
35. Let F (x, y, z) = x2 + y 2 − z 2 . Then ∇F = 2xi + 2yj − 2zk and a normal to the surface
at (x0 , y0 , z0 ) is x0 i + y0 j − z0 k. An equation of the tangent plane at (x0 , y0 , z0 ) is x0 (x −
x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0 or x0 x + y0 y − z0 z = x20 + y02 − z02 . Since (x0 , y0 , z0 ) is on
the surface, z02 = x20 + y02 and x20 + y02 − z02 = 0. Thus, the equation of the tangent plane is
x0 x + y0 y − z0 z = 0, which passes through the origin.
1
1
1
√ √
x+ y + z. Then ∇F = √ i+ √ j+ √ k and a normal to the surface
2 x 2 y
2 z
1
1
1
at (x0 , y0 , z0 ) is √ i + √ j + √ k. An equation of the tangent plane at (x0 , y0 , z0 ) is
2 x0
2 y0
2 z0
1
1
1
1
1
1
√
√
√
√ (x−x0 )+ √ (y−y0 )+ √ (z−z0 ) = 0 or √ x+ √ y+ √ z = x0 + y0 + z0 =
2 x0
2 y0
2 z0
x0
y0
z0
√
√ √
√ √
√
√ √
√ √
√
√a. The
√ sum of the intercepts is x0 a + y0 a + z0 a = ( x0 + y0 + z0 ) a =
a · a = a.
36. Let F (x, y, z) =
√
37. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i + 2y0 j + 2z0 k. Parametric
equations of the normal line are x = x0 + 2x0 t, y = y0 + 2y0 t, z = z0 + 2z0 t. Letting
t = −1/2, we see that the normal line passes through the origin.
38. The normal lines to F (x, y, z) = 0 and G(x, y, z) = 0 are Fx i+Fy j+Fz k and Gx i+Gy j+Gz k,
respectively. These vectors are orthogonal if and only if their dot product is 0. Thus, the
surfaces are orthogonal at P if and only if Fx Gx + Fy Gy + Fz Gz = 0.
39. We have F (x, y, z) = x2 + y 2 + z 2 and G(x, y, z) = x2 + y 2 − z 2 .
∇F = h2x, 2y, 2zi =
6 0 except at the origin
118
CHAPTER 13. PARTIAL DERIVATIVES
∇G = h2x, 2y, −2zi =
6 0 except at the origin
Therefore, the gradient vectors are nonzero at each of the intersection points. Now
Fx Gx + Fy Gy + Fx Gz = (2x)(2x) + (2y)(2y) + (2z)(−2z)
= 4x2 + 4y 2 − 4z 2
= 4(x2 + y 2 + z 2 ) = 4(0) = 0
The second to last equality follows from the fact that the intersection points lie on both
surfaces and hence satisfy the second equation x2 + y 2 − z 2 = 0.
40. Let F (x, y, z) = x2 − y 2 + z 2 − 4 and G(x, y, z) = 1/xy 2 − z. Then
Fx Gx + Fy Gy + Fz Gz = (2x)(−1/x2 y 2 ) + (−2y)(−2/xy 3 ) + (2z)(−1)
= −2/xy 2 + 4/xy 2 − 2z = 2(1/xy 2 − z).
For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 2(0)
and the surfaces are orthogonal at points of intersection.
13.8
Extrema of Multivariable Functions
1. fx = 2x; fxx = 2; fxy = 0; fy = 2y; fyy = 2; D = 4. Solving fx = 0 and fy = 0, we
obtain the critical point (0, 0). Since D(0, 0) = 4 > 0 and fxx (0, 0) = 2 > 0, f (0, 0) = 5 is a
relative minimum.
2. fx = 8x; fxx = 8; fxy = 0; fy = 16y; fyy = 16; D = 128. Solving fx = 0 and fy = 0, we
obtain the critical point (0, 0). Since D(0, 0) = 128 > 0 and fxx (0, 0) = 8 > 0, f (0, 0) = 0 is
a relative minimum.
3. fx = −2x + 8; fxx = −2; f xy = 0; fy = −2y + 6; fyy = −2; D = 4. Solving fx = 0 and
fy = 0 we obtain the critical point (4, 3). Since D(4, 3) = 4 > 0 and
fxx (4, 3) = −2 < 0, f (4, 3, ) = 25 is a relative maximum.
4. fx = 6x − 6; f xx = 6; f xy = 0; fy = 4y + 8; fyy = 4; D = 24. Solving fx = 0 and
fy = 0, we obtain the critical point (1, −2). Since D(1, −2) = 24 > 0 and
fxx (1, −2) = 6 > 0, f (1, −2) = −11 is a relative minimum.
5. fx = 10x + 20; fxx = 10; fxy = 0; fy = 10y − 10; fyy = 10; D = 100. Solving
fx = 0 and fy = 0, we obtain the critical point (−2, 1). Since D(−2, 1) = 100 > 0 and
fxx (−2, 1) = 10 > 0, f (−2, 1) = 15 is a relative minimum.
6. fx = −8x − 8; fxx = −8; fxy = 0; fy = −4y + 12; fyy = −4; D = 32. Solving
fx = 0 and fy = 0, we obtain the critical point (−1, 3). Since D(−1, 3) = 32 > 0 and
fxx (−1, 3) = −8 < 0, f (−1, 3) = 27 is a relative maximum.
7. fx = 12x2 − 12; fxx = 24x; fxy = 0; fy = 3y 2 − 3; fyy = 6y; D = 144xy. Solving
fx = 0 and fy = 0, we obtain the critical points (−1, −1), (−1, 1), (1, −1), and (1, 1).
Since D(−1, 1) = −144 < 0 and D(1, −1) = −144 < 0, these points do not give relative
extrema. Since D(−1, −1) = 144 > 0 and fxx (−1, −1) = −24 < 0, f (−1, −1) = 10 is a
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13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS
119
relative maximum. Since D(1, 1) = 144 > 0 and fxx (1, 1) = 24 > 0,
relative minimum.
f (1, 1) = −10 is a
8. fx = −3x2 + 27; fxx = −6x; fxy = 0; fy = 6y 2 − 24; fyy = 12y; D = −72xy.
Solving fx = 0, fy = 0, we obtain the critical points (−3, −2), (−3, 2), (3, −2), and (3, 2).
Since D(−3, −2) = −432 < 0 and D(3, 2) = −432 < 0, these points do no give relative
extrema. Since D(−3, 2) = 432 > 0 and fxx (−3, 2) = 18 > 0, f (−3, 2) = 432 > 0 and
fxx (3, −2) = −18 < 0, f (3, −2) = 89 is a relative maximum.
9. fx = 4x − 2y − 10; fxx = 4; fxy = −2; fy = 8y − 2x − 2; fyy = 8; D = 32 − (−2)2 = 28.
Setting fx = 0 and fy = 0, we obtain 4x − 2y = 10 and 8y − 2x = 2 or 2x − y = 5
and 4y − x = 1. Solving, we obtain the critical point (3, 1). Since D(3, 1) = 28 > 0 and
fxx (3, 1) = 4 > 0, f (3, 1) = −14 is a relative minimum.
10. fx = 10x + 5y − 10; fxx = 10; fxy = 5; fy = 10y + 5x − 5; fyy = 10; D = 100 − (5)2 = 75.
Setting fx = 0 and fy = 0, we obtain 10x + 5y = 10 and 10y + 5x = 5 or 2x + y = 2
and 2y + x = 1. Solving, we obtain the critical point (1, 0). Since D(1, 0) = 75 > 0 and
fxx (1, 0) = 10 > 0, f (1, 0) = 13 is a relative minimum.
11. fx = 2t − 8; fxx = 0; fxy = 2; fy = 2x − 5; fyy = 0;
D(x, y) = −4 < 0 for all (x, y), there are no relative extrema.
D = 0 − 22 = −4. Since
12. fx = 2y + 6; fxx = 0; fxy = 2; fy = 2x + 10; fyy = 0;
D(x, y) = −4 < 0 for all (x, y), there are no relative extrema.
D = 0 − 22 = −4. Since
13. fx = −6x2 + 6y; fxx = −12x; fxy = 6; fy = −6y 2 + 6x; fyy = −12y; D = 144xy − 36.
Setting fx = 0 and fy = 0, we obtain −6x2 + 6y = 0 and −6y 2 + 6x = 0 or y = x2 and
x = y 2 . Substituting x = y 2 into y = x2 , we obtain y = y 4 or y(y 3 − 1) = 0. Thus, y = 0 and
y = 1. The critical points are (0, 0) and (1, 1). Since D(0, 0) = −36 < 0, (0, 0) does not give
a relative extremum. Since D(1, 1) = 108 > 0 and fxx (1, 1) = −12 < 0, f (1, 1) = 12 is a
relative maximum.
14. fx = 3x2 − 6y; fxx = 6x; fx y = −6; fy = 3y 2 − 6x; fyy = 6y; D = 36xy − 36.
Setting fx = 0 and fy = 0, we obtain 3x2 − 6y = 0 and 3y 2 − 6x = 0 or x2 = 2y and
y 2 = 2x. Substituting y = x2 /2 into y 2 = 2x we obtain x4 = 8x or x(x3 − 8) = 0. Thus,
x = 0 and x = 2. The critical points (0, 0) and (2, 2). Since D(0, 0) = −36 < 0, f (0, 0) is
not an extremum. Since D(2, 2) = 108 > 0 and fxx (2, 2) = 12 > 0, f (2, 2) = 19 is a relative
minimum.
15. fx = y + 2/x2 ; fxx = −4/x3 ; fxy = 1; fy = x + 4/y 2 ; fyy = −8/y 3 ; D = 32/x3 Y 3 − 1.
Setting fx = 0 and fy = 0 we obtain y + 2/x2 = 0 and x + 4/y 2 = 0. Substituting y = −2/x2
into x + 4/y 2 = 0 we obtain x + x4 = x(1 + x3 ) = 0. Since x = 0 is not in the domain of f,
the only critical point is (−1, −2). Since D(−1, −2) = 3 > 0 and
fxx (−1, −2) = 4 > 0, f (−1, −2) = 14 is a relative minimum.
16. fx = −6xy − 3y 2 + 36y; fxx = −6y; fxy = −6x − 6y + 36 = 6(6 − x − y); fy =
−3x2 − 6xy + 36x; fyy = −6x; D = 36xy − 36(6 − x − y)2 . Setting fx = 0 and fy = 0
we obtain −6xy − 3y 2 + 36y = 0 and −3x2 − 6xy + 36x = 0 or −3y(2x + y − 12) = 0
and −3x(x + 2y − 12) = 0. Letting y = 0, the first equation is satisfied and the second
120
CHAPTER 13. PARTIAL DERIVATIVES
equation becomes −3x(x − 12) = 0. Thus, (0, 0) and (12, 0) are critical points. Similarly,
letting x = 0 we obtain the critical point (0, 12). Finally solving 2x + y = 12 and x + 2y = 12
we obtain the critical point (4, 4). Since D(0, 0) = −362 < 0, D(0, 12) = −362 < 0, and
D(12, 0) = −362 < 0, none of these points give relative extrema. Since D(4, 4) = 432 > 0 and
fxx (4, 4) = −24 < 0, f (4, 4) = 192 is a relative maximum.
17. fx = (xex + ex ) sin y; fxx = (xex + 2ex ) sin y; fxy = (xex + ex ) cos y; fy = xex cos y;
fyy = −xex sin y; D = −xe2x (x + 2) sin2 y − e2x (x + 1)2 cos2 y.
Setting fx (x, y) = 0
and fy (x, y) = 0 we obtain (xex + ex ) sin y = 0 and xex cos y = 0. Since ex > 0 for all
x, we have (x + 1) sin y = 0 and x cos y = 0. When x = −1, we must have cos y = 0
or y = π/2 + kπ, k an integer. When x = 0, we must have sin y = 0 or y = kπ, k
an integer. Thus, the critical points are (0, kπ) and (−1, π/2 + kπ), k an integer. Since
D(0, kπ) = 0−cos2 kπ < 0, (0, kπ) does not give a relative extrema. Now, D(−1, π/2+kπ) =
e−2 sin2 (π/2+kπ)−0 > 0 and fxx (−1, π/2+kπ) = e−1 sin(π/2+kπ). Since fxx (−1, π/2+kπ)
is positive for k even and negative for k odd, f (−1, π/2 + kπ) = −e−1 are relative minima for
k even, and f (−1, π/2 + kπ) = e−1 are relative maxima for k odd.
2
2
2
2
18. fx = (2x + 4)ey −3y+x +4x ; fxx = [(2x + 4)2 + 2]ey −3y+x +4x ;
2
2
2
2
fxy = (2x + 4)(2y − 3)ey −3y+x +4x ; fy = (2y − 3)ey −3y+x +4x ;
2
2
2
2
fyy = [(2y − 3)2 + 2]ey −3y+x +4x ; D = [(2x + 4)2 + 2][(2y − 3)2 + 2] · e2(y −3y+x +4x) − [(2x +
2
2
4)(2y − 3)]2 e2(y −3y+x +4x) . Setting fx = 0 and fy = 0 and using the fact that an exponential
function is always positive, we obtain 2x + 4 = 0 and 2y − 3 = 0. Thus, a critical point
is (−2, 3/2). Since D(−2, 3/2) = 4e2(9/4−9/2+4−8) > 0 and fxx (−2, 3/2) = 2e9/4−9/2+4−8 >
0, f (−2, 3/2) = e9/4−9/2+4−8 = e−25/4 is a relative minimum.
19. fx = cos x; fxx = − sin x; fxy = 0; fy = cos y; fyy = − sin y; D = sin x sin y. Solving
fx = 0 and fy = 0, we obtain the critical points (π/2 + mπ, π/2 + nπ) for m and n integers.
For m even and n odd or m odd and n even, D < 0 and no relative extrema result. For m
and n both even, D > 0 and fxx < 0 and f (π/2 + mπ, π/2 + nπ) = 2 are relative maxima.
For m and n both odd, D > 0 and f xx > 0 and f (π/2 + mπ, π/2 + nπ) = −2 are relative
minima.
20. fx = y cos xy; fxx = −y 2 sin xy; fxy = −xy sin xy + cos xy; fy = x cos xy;
fyy = −x2 sin xy; D = x2 y 2 sin2 xy − (−xy sin xy + cos xy)2 = 2xy sin xy cos xy − cos2 xy.
Setting fx = 0 and fy = 0 we see that (0, 0) is a critical point. Also, solving cos xy = 0 we
obtain xy = π/2 + kπ or y = π(1 + 2k)/2x for k an integer. Since D(0, 0) = −1 < 0, (0, 0)
does not give a relative extrema. For any of the critical points (x, π(1 + 2k)/2x), D = 0
and no conclusion can be drawn from the second partials test. Since −1 ≤ sin xy ≤ 1 for all
(x, y)f (x, π(1 + 2k)/2x) = −1 for k odd are relative minima and f (x, π(1 + 2k)/2x) = 1 for
k even are relative maxima.
21. Let the numbers be x, y, and 21 − x − y. We want to maximize P (x, y) = xy(21 − x − y) =
21xy − x2 y − xy 2 . Now Px = 21y − 2xy − y 2 ; Pxx = −2y; Pxy = 21 − 2x − 2y;
Py = 21x − x2 − 2xy; Pyy = −2x; D = 4xy − (21 − 2x − 2y)2 . Setting Px = 0 and Py = 0,
we obtain y(21 − 2x − y) = 0 and x(21 − x − 2y) = 0. Letting x = 0 and y = 0, we obtain the
critical points (0, 0), (0, 21), and (21, 0). Each of these results in P = 0 which is clearly not
a maximum. Solving 21 − 2x − y = 0 and 21 − x − 2y = 0, we obtain the critical point (7, 7).
13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS
121
Since D(7, 7) = 147 > 0 and Pxx (7, 7) = −14 < 0, P (7, 7) = 343 is a maximum. The three
numbers are 7,7, and 7.
22. Let the sides of the base of the box by x and y. Then, since the volume of the box is 1, its
height is 1/xy and S = 2xy + 2x(1/xy) + 2y(1/xy) = 2xy + 2/y + 2/x, x > 0, y > 0. Now
Sx = 2y − 2/x2 ; Sxx = 4/x3 ; Sxy = 2; Sy = 2x − 2/y 2 ; Syy = 4/y 3 ; D = 16/x3 y 3 − 4.
Setting Sx = 0 and Sy = 0 we obtain y = 1/x2 and x = 1/y 2 . The critical point is (1, 1).
Since D(1, 1) = 12 > 0 and Sxx (1, 1) = 4 > 0, S(1, 1) = 6 is a minimum. The box is 1 foot
on each side.
23. Let (x, y, 1 − x − 2y) be a point on the plane x + 2y + z = 1. We want to minimize f (x, y) =
x2 + y 2 + (1 − x − 2y)2 . Now fx = 2x − 2(1 − x − 2y); fxx = 4; fxy = 4;
fy = 2y − 4(1 − x − 2y); fyy = 10; D = 40 − 42 = 24. Setting fx = 0 and fy = 0 we obtain
2x−2(1−x−2y) = 0 and 2y−4(1−x−2y) = 0 or 2x+2y = 1 and 2x+5y = 2. Thus, (1/6, 1/3)
is a critical point. Since D = 24 > 0 and fxx = 4 > 0 for all (x, y), f (1/6, 1/3) = 1/6 is a
minimum. Thus, the point on the plane closest to the origin is (1/6, 1/3, 1/6).
24. Let (x, y, 1 − x − y) be a point on the plane x + y + z = 1. We want to minimize the square
of the distance between the point and the plane. This is given by
f (x, y) = (x − 2)2 + (y − 3)2 + (−x − y)2 = 2x2 + 2y 2 − 4x − 6y + 2xy + 13.
fx = 4x − 4 + 2y; fxx = 4; fxy = 2; fy = 4y − 6 + 2x; fyy = 4; D = 16 − 22 = 12.
Setting fx = 0 and fy = 0 we obtain 4x − 4 + 2y = 0 and 4y − 6 + 2x = 0 or 2x + y = 2 and
x + 2y = 3. Thus, (1/3, 4/3) is a critical point. Since D = 12 > 0 and fxx = 4 > 0 for all
(x, y), fp
(1/3, 4/3) =√25/3 is a minimum. Thus, the least distance between the point and the
plane is 25/3 = 5/ 3.
25. Let (x, y, 8/xy) be a point on the surface. We want to minimize the square of the distance
to the origin or f (x, y) = x2 + y 2 + 64/x2 y 2 . Now fx = 2x − 128/x3 y 2 ; fxx = 2 + 384/x4 y 2 ;
fxy = 256/x3 y 3 ; fy = 2y − 128/x2 y 3 ; fyy = 2 + 384/x2 y 4 ; D = (2 + 384/x4 y 2 )(2 +
384/x2 y 4 ) − (256)2 /x6 y 6 . Setting fx = 0 and fy = 0 we obtain 2x − 128/x3 y 2 = 0 and
2y − 128/x2 y 3 = 0 or x4 y 2 = 64 and x2 y 4 = 64; x 6= 0, y 6= 0. This gives x4 y 2 =
x2 y 4 , x2 y 2 (x2 − y 2 ) = 0 or x2 = y 2 . Thus, x6 = 64 and x = ±2. Similarly, y = ±2 and
the critical points are (−2, −2), (−2, 2), (2, −2), and (2, 2). Since D(±2, ±2) = 48 > 0 and
fxx (±2, ±2) = 8 > 0, f (±2, ±2) = 12 are minima. The points closest to√the origin
√ are
(−2, −2, 2), (−2, 2, −2), (2, −2, −2), and (2, 2, 2). The minimum distance is 12 = 2 3.
26. We will minimize the square of the distance between the lines. This is given by
f (s, t) = [(3 + 2s) − t]2 + [(6 + 2s) − (4 − 2t)]2 + [(8 − 2s) − (1 + t)]2
= (2s − t + 3)2 + (2s + 2t + 2)2 + (−2s − t − 7)2 = 12s2 + 6t2 + 8st − 8s + 12t + 62.
fs = 24s+8t−8; fss 24; fst = 8; ft = 12t+8s−12; ftt = 12; D = 24(12)−64 = 224. Solving
24s+8t−8 = 0 and 12t+8s−12 = 0 we obtain the critical point (0, 1). Since D(0, 1) = 224 > 0
and fss (0, 1) = 24 > 0, we see that f (0, 1) is a minimum. The corresponding
points
on√the
p
√
lines are (3, 6, 8) on L2 and (1, 2, 2) on L1 . The minimum distance is f (0, 1) = 56 = 2 14.
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122
CHAPTER 13. PARTIAL DERIVATIVES
27. We will maximize the square of the volume of the box in the first octant,
V (x, y) = x2 Y 2 z 2 = x2 y 2 (c2 − c2 x2 /a2 − c2 y 2 /b2 ).
Vx = 2c2 xy 2 −4c2 x3 y 2 /a2 −2c2 xy 4 /b2 ; Vxx = 2c2 y 2 −12c2 x2 y 2 /a2 −2c2 y 4 /b2 ; Vxy = 4c2 xy−
8c2 x3 y/a2 − 8c2 xy 3 /b2 ; Vy = 2c2 x2 y − 2c2 x4 /a2 − 4c2 x2 y 3 /b2 ; Vyy = 2c2 x2 − 2c2 x4 /a2 −
2
12c2 x2 y 2 /b2 ; D = Vxx Vyy − Vxy
. Setting Vx = 0 and Vy = 0 we obtain xy 2 − 2x3 y 2 /a2 −
4 2
2
4
2
2 3 2
2 2
2 2
2 2
xy /b = 0, x y−x y/a −2x y /b = 0, or, assuming x
√ y > 0, 2b x +a y = a b .
√> 0 and
2
2
2
2
Solving, we obtain x = a /3 and y = b /3. Thus, (a/ 3, b/ 3) is a critical point. Since
√
√
14
4
20 2 2 4
14
D(a/ 3, b/ 3) = (− b2 c2 )(− a2 c2 ) − (− abc2 )2 =
a b c >0
9
9
9
9
and vxx = −
√
√
14 2 2
b c < 0, V (a/ 3, b/ 3) = a2 b2 c2 /27. The maximum volume is
9
q
√
√
√
8 V (a/ 3, b/ 3) = 8 3abc/9.
28. Let a + b + c = k. Then c = k − a − b and we want to maximize
V (a, b) = 4πab(k − a − b)/3 = 4π(kab − a2 b − ab2 )/3.
4π
8π
4π
4π
(kb − 2ab − b2 ); Vaa = − ; Vab =
(k − 2a − 2b); Vb =
(ka − a2 − 2ab);
3
3
3
3
8π
64x2
16x2
Vbb = − a; D =
ab −
(k − 2a − 2b)2 . Setting Va = 0 and Vb = 0 we obtain
3
9
9
kb − 2ab − b2 = 0 or ka − a2 − 2ab = 0, a 6= 0, b 6= 0, or 2a + b = k and a + 2b = k. Solving,
we get a = b = k/3. Since D(k/3, k/3) = 16π 2 k 2 /27 > 0 and Vaa (k/3, k/3) = −8πk/9 < 0,
the volume is maximized when a = b = k/3. Since c = k − a − b = k/3, a = b = c and the
ellipsoid is a sphere.
Va =
29. The perimeter is given by P = 2x + 2y + 2x sec θ and the
area is a = 2xy + x2 tan θ. Solving P for 2y and substituting in A, we obtain A = P x − 2x2 (1 + sec θ) + x2 tan θ. Now
Ax (x, θ) = P −4x(1+sec θ)2x tan θ; Axx (xθ) = −4(1+sec θ)+2 tan θ;
Axθ (x, θ) = −4x sec θ tan θ + 2x sec2 θ; Aθ (xθ) = x2 sec θ(sec θ −
2 tan θ); Aθθ (x, θ) = 2x2 sec θ(tan θ − 2 sec2 θ + 1). We assume that
x > 0 and 0 ≤ θ ≤ π/2.
x tanθ
x secθ
θ
x
x
y
Setting Ax = 0 and Aθ = 0, we obtain P −4x(1+sec θ)+2x tan θ = 0 and x2 sec θ(sec θ−2 tan θ) = 0.
We note from the second equation and the fact that sec θ 6= 0 for all θ that sec θ − 2√
tan θ = 0.
Solving for θ, we obtain θ = 30◦ and solving Ax = 0 for x, we obtain x) = P/(4 + 2 3). Since
√
√
√
√
D(x0 , 30◦ ) = (−2 3 + 2)(4x20 ( √3 − 5)/3 3) − 02 > 9 and Axx = 2 − 2 3 < 0, A(x0 , 30◦ ) √
is a
maximum. Letting x = P/(4+2 3) and θ = 30√◦ in P = 2x+2y+2x
sec
θ,
we
obtain
P
=
2y+P
?
3.
√
√
Thus, the area is maximized for x = P/(4 + 2 3), y = P ( 3 − 1)/2 3, and θ = 30◦
13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS
(x sin θ)(24 −
1
2x + x cos θ) = 24x sin θ − 2x sin θ + x2 sin 2θ.
2
Now Ax = 24 sin θ − 4x sin θ + x sin 2θ; Axx =
−4 sin θ + sin 2θ; Axθ = 24 cos θ − 4x cos θ + 2x cos 2θ;
Aθ = 24x cos θ − 2x2 cos θ + x2 cos 2θ; Aθθ =
−24x sin θ + 2x2 sin θ − 2x2 sin 2θ.
30. We want to maximize A(x, θ)
123
=
2
x
θ
24-2x
x sinθ
θ
x cosθ
We assume 0 < x < 12 and =< θ < π/2. Setting Ax = 0 and Aθ = 0 we obtain 24 sin θ − 4x sin θ +
2x sin θ cos θ = 0 and 24x cos θ − 2x2 cos θ + x2 (2 cos2 θ − 1) = 0 or 12 − 2x + x cos θ = 0 and
2x cos2 θ − 2x cos θ + 2 cos θ − x = 0. Solving the first equation for cos θ and substituting into the
second equation, we obtain 2x(2 − 12/x)2 − 2x(12 − 12/x) + 24(2
− x = 0. Simplifying, we
√ − 12/x)√
find x = 8 and√cos θ = 1/2 or θ = 60◦ .√Since D(8, 60◦ ) = (−3 3/2)(−96 3) − (−12)2 = 288 > 0
and Axx = −3 3/2 < 0, A(8, 60◦ = 48 3) square inches is the maximum area.
2
2
31. fx = − x−1/3 , fy = − y −1/3 . Since fx = 0 and fy = 0 have no solutions, f (x, y) has no
3
3
critical points and Theorem 13.8.2 does not apply. However, for all (x, y),
f (0, 0) = 16 ≥ 16 − (x1/3 )2 − (y 1/3 )2 = f (x, y), and f (0, 0) = 16 is an absolute maximum.
32. fx = −4x3 y 2 ; fxx = −12x2 y 2 ; fxy = −8x3 y; fy = −2x4 y; fyy = −2x4 ;
D = 24x6 y 2 − 64x6 y 2 = −40x6 y 2 . Setting fx = 0 and fy = 0 we see that (0, y) and (x, 0)
are critical points for any x and y. Since, for any critical point, D = 0, Theorem 13.8.2 does
not apply. However, for all (x, y), f (0, 0) = 1 ≥ 1 − (x2 y)2 = f (x, y), and f (0, 0) = 1 is an
absolute maximum.
33. fx = 10x; fxx = 10; fxy = 0; fy = 4y 3 ; fyy = 12y 2 ; D = 120y 2 . Solving fx = 0 and
fy = 0 we obtain the critical point (0, 0). Since D(0, 0) = 0, Theorem 13.8.2 does not apply.
However, for any (x, y), f (0, 0) = −8 ≤ 5x2 +y 4 −8 = f (x, y) and f (0, 0) = −8 is an absolute
minimum.
x
y2
xy
y
34. fx = p
; fxx = 2
;
; fxy = − 2
; fy = p
(x + y 2 )3/2
(x + y 2 )3/2
x2 + y 2
x2 + y 2
x2
−xy
x2 y 2
fyy =
−( 2
; D =
)2 = 0. Since D = 0 for all (x, y),
2
2
3
2
2
3/2
(x + y )
(x + y )
(x + y 2 )3/2
p
Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 0 ≤ x2 + y 2 = f (x, y), so
f (0, 0) = 0 is an absolute minimum.
In Problems 35-38 we parameterize the boundary of R by letting x = cos t and y = sin t;
0 ≤ t ≤ 2π. Then, for (x(t), y(t)) on the boundary, we maximize or minimize F (t) =
f (cos t, sin t) on [0, 2π].
√
35. fx = 1; fy = 3. There
√ are no critical points on the interior
√ of R. On the boundary we
consider F (t) = cos t + 3 sin t. Solving F 0 (t) = − sin t + 3 cos t = 0, we obtain critical
points at t = π/3
2, and F (4π/3) = −2, we
√ and t = 4π/3. Comparing F (0) = 1, F (π/3) = √
see that f (1/2, 3/2) = 2 is an absolute maximum and f (−1/2, − 3/2) = −2 is an absolute
minimum.
124
CHAPTER 13. PARTIAL DERIVATIVES
36. fx = y; fy = x. Solving fx = 0 and fy = 0 we obtain the critical point (0, 0) with corre1
sponding function value f (0, 0) = 0. On the boundary we consider F (t) = cos t sin t = sin 2t.
2
Solving F 0 (t) = cos 2t = 0, we obtain critical points at π/4, 3π/4, 5π/4, and 7π/4. Comparing f (0, 0, ) = 0, F (0) = 0, F√
(π/4) √= 1/2, F (3π/4)
=√−1/2, F (5π/4) = 1/2,
√
2/2,
2/2)
=
f
(−
2/2,
−
2/2) = 1/2 are absolute
and F (7π/4) = −1/2,
we
see
that
f
(
√
√
√
√
maxima and f (− 2/2, 2/2) = f ( 2/2, − 2/2) = −1/2 are absolute minima.
37. fx = 2x + y; fy = x + 2y. Solving fx = 0 and fy = 0 we obtain the critical point
(0, 0) with corresponding function value f (0, 0) = 0. On the boundary we consider F (t) =
1
cos2 t + cos t sin t + sin2 t = 1 + sin 2t. Solving F 0 (t) = cos 2t = 0 we obtain critical points at
2
π/4, 3π/4, 5π/4, and 7π/4. Comparing f (0, 0) = 0, F (0) = 1, F (π/4) = 3/2;
√
√
F (3π/4)
=√
1/2, F (5π/4) = 3/2, and F (7π/4) = 1/2, we see that f ( 2/2, w/2) =
√
f (− 2/2, − 2/2) = 3/2 are absolute maxima and f (0, 0) = 0 is an absolute minimum.
38. fx = −2x; fy = −6y + 4. Solving fx = 0 and fy = 0, we obtain the critical point (0, 2/3),
which is inside R, with corresponding function value f (0, 2/3) = 5. On the boundary we
consider F (t) = − cos2 t − 3 sin2 t + 4 sin t + 1. Solving F 0 (t) = 2 cos t sin t − 6 sin t cos t +
4 cos t = 4 cos t − 4 sin t cos t = 0, we obtain critical points at π/2 and 3π/2. Comparing
f (0, 2/3) = 5, F (0) = 0, F (π/2) = 2, and F (3π/2) = −6, we see that f (0, −1) = −6 is an
absolute minimum and f (0, 2/3) = 5 is an absolute maximum.
39. fx = 4; fy = −6. There are no critical points over the region R, so absolute extrema must
occur on the boundary. We parameterize the boundary by x = 2 cos t and y = sin t for 0 ≤ t ≤
2π. Considering F (t) = 8 cos t − 6 sin t we obtain F 0 (t) = −8 sin t − 6 cos t. Solving F 0 (t) = 0
we find tan t = −3/4. Using 1 + tan2 t = sec2 t we see that sec2 t = 25/16 and cos t = −4/5, t
is in the second quadrant and sin t = 3/5. The corresponding points on the boundary of R are
(8/5, −3/5) and (−8/5, 3/5). Comparing f (0) = F (2π) = f (2, 0) = 8, f (8/5, −3/5) = 10,
and f (−8/5, 3/5) = −10 we see that the absolute minimum is f (−8/5, 3/5) = −10 and the
absolute maximum is f (8/5, −3/5) = 10.
40. fx = y − 2; fy = x − 1. Solving fx = 0 and fy = 0 we obtain the critical point (1, 2) in the
region. On x = 0, F (y) = f (0, y) = −y + 6, which has no critical points for 0 ≤ y ≤ 8. The
endpoints of the interval are (0, 0) and (0, 8). On y = 0, G(x) = f (x, 0) = −2x + 6, which
has no critical points for 0 ≤ x ≤ 4. The endpoints of the interval are (0, 0) and (4, 0). On
y = −2x + 8, H(x) = f (x, −2x + 8) = x(−2x + 8) = x(−2x + 8) − 2x − (−2x + 8) + 6 =
−2x2 + 8x − 2. Solving H 0 (x) = −4x + 8 = 0 we obtain x = 2. The corresponding point on
the triangle is (2, 4). Comparing f (0, 0) = 6, f (0, 8) = −2; f (4, 0) = −2; f (2, 4) = 6, and
f (1, 2) = 4 we see that absolute maxima are f (0, 0) = f (2, 4) = 6 and absolute minima are
f (0, 8) = f (4, 0) = −2.
41. (a) fx = y cos xy; fy = x cos xy. Setting fx = 0 and fy = 0 we obtain y cos xy = 0 and
x cos xy = 0. If y = 0 from the first equation, then necessarily x = 0 from the second
equation. Thus, (0, 0) is a critical point. For x 6= 0 and y 6= 0 we have cos xy = 0 or
xy = π/2. Thus, all points (x, π/2x) for 0 ≤ x ≤ π are also critical points.
(b) Since 0 ≤ sin xy ≤ 1 for 0 ≤ x ≤ π and 0 ≤ y ≤ 1, f (x, y) = sin xy has absolute
minima at any points for which sin xy = 0 and absolute maxima at any points for which
www.elsolucionario.org
13.9. METHOD OF LEAST SQUARES
125
sin xy = 1. Thus, f (x, y) has absolute minima when xy = 0 or xy = π, that is, at the
points (0, y), (x, 0), and (π, 1) which are in the region. Absolute maxima occur when
xy = π/2 or along the curve y = π/2x in the region
(c)
z
1
y
x
42. We want to maximize P (x, y) = R(x, y) − C(x, y) = 108x − 8x2 + 192y − 6y 2 − 4xy − 20. Now
Px = 108 − 16x − 4y; Pxx = −16; Pxy = −4; Py = 192 − 4x; Pyy = −12; D = 192 − 16 =
176. Setting Px = 0 and Py = 0 we obtain 108 − 16x − 4y = 0 and 192 − 12y − 4x = 0
or 4x + y = 27 and x + 3y = 48. Solving, we see that (3, 15) is a critical point. Since
D(3, 15) = 175 > 0 and Pxx (3, 15) = −16 < 0, P (3, 15) = 1582 is the maximum profit
43. Since the volume of the box is 60, the height is 60/xy. Then
C(x, y) = 10xy + 20xy + 2[2x60/xy + 2y60/xy] = 30xy + 240/y + 240/x.
Cx = 30y − 240/x2 ; Cxx = 480/x3 , Cxy = 30; Cy = 30x − 240/y 2 ; cyy = 480/y 3 ; D =
4802 /x3 y 3 −900. Setting Cx = 0 and Cy = 0 we obtain 30y −240/x2 = 0 and 30x−240/y 2 = 0
or y = 8/x2 and x = 8/y 2 . Substituting the first equation into the second, we have x = x4 /8
or x(x3 − 8) = 0. Thus, (2, 2) is a critical point. Since D(2, 2) = 2700 > 0 and
Cxx (2, 2) = 60 > 0, C(2, 2) is a minimum. Thus, the cost is minimized when the base of the
box is 2 feet square and the height is 15 feet.
13.9
1.
4
X
Method of Least Squares
xi = 14,
i=1
4
X
yi = 8,
i=1
4
X
xi yi = 30,
i=1
4
X
x2i = 54, m =
4(30) − 14(8)
= 0.4,
4(54) − (14)2
x2i = 14, m =
4(34) − 6(14)
= 2.6,
4(14) − (6)2
i=1
54(8) − 30(14)
b=
= 0.6, y = 0.4x + 0.6
4(54) − (14)2
2.
4
X
i=1
xi = 6,
4
X
i=1
yi = 14,
4
X
i=1
xi yi = 34,
4
X
i=1
14(14) − 34(6)
b=
= −0.4, y = 2.6x − 0.4
4(14) − (6)2
126
3.
CHAPTER 13. PARTIAL DERIVATIVES
5
X
5
X
xi = 15,
i=1
yi = 15,
5
X
i=1
5
X
xi yi = 56,
i=1
x2i = 55, m =
5(56) − 15(15)
= 1.1,
5(55) − (15)2
x2i = 54, m =
5(55) − 14(14)
≈ 1.06757,
5(54) − (14)2
i=1
55(15) − 56(15)
b=
= −0.3, y = 1.1x − 0.3
5(55) − (15)2
4.
5
X
4
X
xi = 14,
i=1
yi = 14,
5
X
i=5
5
X
xi yi = 55,
i=1
i=1
54(14) − 55(14)
≈ −0.189189, y ≈ 1.06757x − 0.189189
b=
5(54) − (14)2
5.
7
X
7
X
xi = 21,
i=1
yi = 42,
7
X
i=1
xi yi = 164,
i=1
7
X
x2i = 91, m =
i=1
7(164) − 21(42)
≈ 1.35714,
7(91) − (21)2
91(42) − 164(21)
≈ 1.92857, y ≈ 1.35714x + 1.92857
b=
7(91) − (21)2
6.
7
X
7
X
xi = 28,
i=1
yi = 17.2,
7
X
i=1
xi yi = 80.2,
i=1
7
X
x2i = 140, m =
i=1
7(80.2) − 28(17.2)
≈ 0.407143,
7(140) − (28)2
140(17.2) − 80.2(28)
≈ 0.828571, y ≈ 0.407143x + 0.828571
b=
7(140) − (28)2
7.
6
X
Ti = 420,
6
X
vi = 1055,
Ti vi = 68, 000,
i=1
i=1
i=1
6
X
6
X
Ti2 = 36, 400, m =
i=1
6(68, 000) − 420(1055)
≈
6(36, 400) − (420)2
36, 400(1055) − 68, 000(420)
≈ 234.333, v ≈ −0.835714T + 234.333.
−0.835714, b =
6(36, 400) − (420)2
8.
6
X
i=1
Ti = 3150,
6
X
Ri = 29.57,
6
X
i=1
i=1
Ti Ri = 17, 878,
6
X
Ti2 = 1, 697, 500, m =
i=1
6(17, 878) − 3150(29.57)
≈
6(1, 697, 500) − (3150)2
1, 697, 500(29.57) − 17, 878(3150)
0.05, b =
≈ −23.32, R ≈ 0.05T − 23.32 − 23.32. When
6(1, 697, 500) − (3150)2
T = 700, R ≈ 14.34.
9. (a) least-squares line: y = 0.5966x + 4.3665
least-squares quadratic: y = −0.0232x2 + 0.5618x + 4.5942
least-squares cubic: y = 0.00079x3 − 0.0212x2 + 0.5498x + 4.5840
(b)
least-squares line
least-squares quadratic
least-squares cubic
y
y
x
y
x
x
13.10. LAGRANGE MULTIPLIERS
127
10. The least-squares line is given by y = 2.0533x − 3837.115. Plugging 2020 in for x, we predict
that the population will be 310.551 million.
13.10
Lagrange Multipliers
1. f has constrained extrema where
the level lines intersect the circle.
2. f has constrained extrema where
the level curve intersect the line.
y
y
2
1
x
1
x
3. fx = 1; fy = 3; gx = 2x; gy = 2y. We need to solve 1 = 2λx, 3 = 2λy, x2 + y 2 − 1 = 0.
Dividing the second equation by the first, we obtain √
3 = y/x or y =√3x. Substituting
√ into
2
2
the third equation,
we
have
x
+
9x
=
1
or
x
=
±1/
10.
For
x
=
1/
10,
y
=
3/
√
√
√
√
√ 10 and
for x = −1/ 10, y = −3/ 10.
A
constrained
maximum
is
f
(1/
10,
3/
10)
+
10 and a
√
√
√
constrained minimum is f (−1/ 10, −3/ 10) = − 10.
4. fx = y; fy = x; gx = 1/2; gy = 1. We need to solve y = λ/2, x = λ, x/2 + y − 1 = 0.
From the first two equations y = x/2. Substituting into the second equation, we have x = 1.
Thus, f (1, 1/2) = 1/2 is a constrained extremum. Since (0, 1) satisfies the constraint and
f (0, 1) = 0 < 1/2, f (1, 1/2) = 1/2 is a constrained maximum.
5. fx = y; fy = x; gx = 2x; gy = 2y. We need to solve y = 2λx, x = 2λy, x2 + y 2 − 2 = 0.
Substituting the second equation into the first, we obtain y = 4λ2 y or y(4λ2 − 1) = 0. If
y = 0, then from the second equation x = 0. Since g(0, 0) = −2 6= 0, (0, 0) does not
satisfy the constraint. Thus, λ = ±1/2 and y = ±x. Substituting into the third equation,
we have 2x2 = 2 or x = ±1. Solutions of the system are x = 1, y = 1, λ = 1/2, x =
−1, y = −1; λ = 1/2, x = 1, y = −1, λ = −1/2, and x = −1, y = 1, λ = −1/2.
Thus, f (1, 1) = f (−1, −1) = 1 are constrained maxima and f (1, −1) = f (−1, 1) = −1 are
constrained minima.
6. fx = 2x; fy = 2y; gx = 2; gy = 1. We need to solve 2x = 2λ, 2y = λ, 2x + y − 5 = 0.
Substituting the second equation into the first, we find x = 2y. Substituting into the third
equation, we have 4y + y − 5 = 0 or y = 1. A constrained extremum is f (2, 1) = 5. Since (0, 5)
satisfies the constraint and f (0, 5) = 25 > 5, f (2, 1) = 5 is a constrained minimum.
7. fx = 6x; fy = 6y; gx = 1; gy = −1. We need to solve 6x = λ, 6y = −λ; x − y − 1 = 0.
From the first two equations, we obtain x + y = 0. Solving this with the third equation, we
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128
CHAPTER 13. PARTIAL DERIVATIVES
obtain x = 1/2, y = −1/2. Thus, f (1/2, −1/2) = 13/2 is a constrained extremum. Since
(1, 0) satisfies the constraint and f (1, 0) = 8 > 13/2, f (1/2 − 1/2) = 13/2 is a constrained
minimum.
8. fx = 8x; fy = 4y; gx = 8x; gy = 2y. We need to solve 8x = 8λx, 4y = 2λy, 4x2 + y 2 − 4 =
0 or x(λ − 1) = 0, y(λ − 2) = 0, 4x2 + y 2 = 4. If x = 0, then from the third equation
y = ±2. If y = 0, then from the third equation x = ±1. The cases λ = 1 and λ = 2 lead also
to y = 0 and x = 0, respectively. Thus, f (0, 2) = f (0, −2) = 18 are constrained maxima and
f (1, 0) = f (−1, 0) = 14 are constrained minima.
9. fx = 2x; fy = 2y; gx = 4x3 ; gy = 4y 3 . We need to solve 2x = 4λx3 , 2y = 4λy 3 , x4 +
y 4 − 1 = 0 or x(2λx2 − 1) = 0, y(2λy 2 − 1) = 0, x4 + y 4 = 1. If x = 0, the from the
2
third equation y = ±1. If y = 0, then x = ±1. From 2λx
= 1 = 2λy 2 we
have x2 = y 2 .
√
√
4
4
Substituting into the third equation, we obtain
x = ±1/ 2 and y = ±1/ 2. Solutions of
√
4
the system are (0, ±1), (±1, 0),
and
(±1/
2,
±1/[4]2).
Thus, f (0, ±1) = f (±1, 0) = 1 are
√
√
√
constrained minima and f (±1/ 4 2, ±1/ 4 2) = 2 are constrained maxima.
10. fx = 16x−8y; fy = 4y −8x; gx = 2x; gy = 2y. We need to solve 16x−8y = 2λx, 4y −8x =
2λy, x2 + y 2 − 10 = 0 or 8 − 47/x = λ, x2 + y 2 = 10. From the first two equations,
we obtain 6 − 4y/x = −4x/y, 6(y/x) − 4(y/x)2 = −4, and 2(y/x)2 − 3(y/x) − 2 = 0.
Factoring, we have (2y/x + 1)(y/x − 2) = 0. Then y = −x/2 and y = 2xs.
√ Substituting
y = −x/2 into the third equation, we have x2 + x2 /4 = 10 and x =√±2 2. Substituting
2
2
y = 2x into the
we have
of the
√ third
√ equation,
√ √
√ x √+ 4x = 10√ and x√ = ± 2. Solutions
√
√
system
(− 2,
√ are
√ (2 2, − 2), (−2 2, 2), ( 2, 2 2), and √
√−2 2). Thus,
√ f (2√ 2, − 2) =
f (−2 2, 2) = 100 are constrained maxima and f ( 2, 2 2) = f (− 2, −2 2) = 0 are
constrained minima.
√
√
√
11. fx = 3x2 y; fy = x3 ; gx = 1/2 x; gy = 1/2 y. We need to solve 3x2 y = λ/2 x, x3 =
√
√
√
√
√
λ/2 y,
x + y − 1 = 0 or 6x5/2 y = λ, 2x3 y 1/2 = λ,
x + y = 1. From the first two
√
√
equations, we obtain 3x5/2 y = x3 y 1/2 and 3 y = x. Substituting into the third equation,
√
√
√
we have 3 y + y = 4 y = 1. Then, y = 1/16 and x = 9/16. Since (1/4, 1/4) satisfies the
constraint and f (1/4, 1/4) = 1/256, f (9/16, 1/16) + 729/65, 536 is a constrained maximum.
We also consider x = 0, which requires y = 1; and y = 0, which requires x = √
1. Since x ≥ 0
√
and y ≥ 0, f (0, 1) = f (1, 0) = 0 ≤ x3 y = f (x, y) for all (x, y) which satisfy x + y = 1.
Thus, f (0, 1) = 0 and f (1, 0) = 0 are constrained minima.
12. fx = y 2 ; fy = 2xy; gx = 2x; gy = 2y. We need to solve y 2 = 2λx, 2xy = 2λy, x2 +
2
2
y 2 − 27 = 0 or y 2 = 2λx, y(x
√ − λ) = 0, x + y = 27. When y = 0 in the third equation,
2
we obtain x = 27 or x = ±3 3, and λ = 0. When x = λ in the second equation, we obtain
2
2
2
y 2 = 2x2 from the first
√ equation and
√ x + 2x = 3x = 27 from the√third equation.√This gives
x = ±3 and y = ±3 2. Since f (±3√ 3, 0) = 0, we√see that f (−3, 3 2) = f (−3, −3 2) = −54
are constrained minima and f (3, 3 2) = f (3, −3 2) = 54 are constrained maxima.
13. Fx = 1; Fy = 2; Fz = 1; gx = 2x; gy = 2y; gz = 2z. We need to solve 1 = 2λx, 2 =
2λy; 1 = 2λz, x2 + y 2 + z 2 − 30 = 0. From the first and second equations, we obtain
y = 2x. From the first and third equations, we obtain z √
= x. Substituting
into √
the fourth
√
2
2
2
2
equation,
we
have
x
+
4x
+
x
=
6x
=
30.
Thus,
x
=
±
5,
y
=
±2
5,
z
=
±
5.√Then,
√ √ √
√
√
√
√
F ( 5, 2 5, 5) = 6 5 is a constrained maximum and F (− 5, −2 5, − 5) = −6 5 is a
constrained minimum.
13.10. LAGRANGE MULTIPLIERS
129
14. Fx = 2x; Fy = 2y; Fx = 2z; gx = 1; fy = 2; gz = 3. We need to solve 2x =
λ, 2y = 2λ, 2z = 3λ, x + 2y + 3z − 4 = 0. From the first and second equations, y = 2x.
From the first and third equations, z = 3x. Substituting into the fourth equation, we have
x+4x+9x = 14x = 4. Thus, x = 2/7, y = 4/7, andz = 6/7. Then, F (2/7, 4/7, 6/7) = 56/49
is a constrained extremum. Since (4, 0, 0) satisfies the constraint and F (4, 0, 0) = 16 >
56/49, F (2/7, 4/7, 6/7) = 56/49 is a constrained minimum.
15. Fx = yz; Fy = xz; Fz = xy; gx = 2x; gy = y/2; gz = 2z/9. We need to solve
yz = 2λx, xz = λy/2, xy = 2λz/9 or xyz/2 = λx2 , xyz/2 = λy 2 /4, xyz/2 = λz 2 /9 along
with x2 + y 2 /4 + z 2 /9 − 1 = 0 or x2 + Y 2 /4 + z 2 /9 = 1 for x > 0, y > 0, z > 0. From
the first three equations and the fact that λ 6= 0, x2 = y 2 /4 = z 2 /9. Substituting into the
2
2
2
2
2
2
2
third equation,
we
√ obtain
√ x +√x + x = 3x = 1, so x = 1/3, y =√4/3, and z = 3.
√
Since
23/6, 1, 1)
Thus, F ( 3/3, 2 3/3,√ 3) = 2 3/3 is√a constrained
√ extremum.
√
√ ( √
√ satisfies
the constraint and F ( 23/6, 1, 1) = 23/6 < 2 3/3, F ( 3/3, 2 3/3, 3) = 2 3/3 is a
constrained maximum.
16. Fx = yz; Fy = xz; Fz = xy; gx = 3x2 ; gy = 3y 2 ; gz = 3z 2 . We need to solve
yz + 3λx2 , xz = 3λy 2 , xy = 3λz 2 or xyz = 3λx3 , syz = 3λy 3 , xyz
= 3λz 3 along
√
√ with
3
3
3
3
3
3
3
x + y + z√ − 24 = 0 or x + y + z = 24. Taking λ = 0 we see that ( 24, 0, 0), (0, 3 24, 0),
and (0, 0, 3 24) satisfy the system. If λ 6= 0, the the first three equations imply x3 = y 3 = z 3 .
Substituting into the fourth√
equation,√we obtain
x3 +x3 +x3 = 3x√3 = 24 or x = 2. Then
√
√ (2, 2, 2)
3
3
3
satisfies the system. Since 24 = 2 3, F (2 3, 0, 0) = F (0, 2 3 3, 0) = F (0, 0, 2 3 3) = 5 is a
constrained minimum and F (2, 2, 2) = 13 is a constrained maximum.
17. Fx = 3x2 ; Fy = 3y 2 ; Fz = 3z 2 ; gx = 1; gy = 1; gz = 1. We need to solve 3x2 = λ, 3y 2 =
λ, 3z 2 = λ, x + y + z − 1 = 0 for x > 0, y > 0, z > 0, and hence λ > 0. From the first three
equations x2 = y 2 = z 2 , and since x, y, and z are positive, x = y = z. Then, from the fourth
equation, x = y = z = 1/3 and F (1/3, 1/3, 1/3) = 1/9 is a constrained extremum. Since
(1/2, 1/4, 1/4) satisfies the constraint and F (1/2, 1/4, 1/4) = 5/32 > 1/9, F (1/3, 1/3, 1/3) =
1/9 is a constrained minimum.
18. Fx = 8xy 2 z 2 ; Fy = 8x2 yz 2 ; Fz = 8x2 Y 2 z; gx = 2x; gy = 2y; gz = 2z. We need to solve
8xy 2 z 2 = 2λx, 8x2 yz 2 = 2λy, 8x2 y 2 z = 2λz or 4x2 y 2 z 2 = λx2 , 4x2 y 2 z 2 = λy 2 , 4x2 y 2 z 2 =
λz 2 along with x2 + y 2 + z 2 − 9 = 0 or x2 + y 2 + z 2 = 9 for x > 0, y > 0, z > 0, and
hence lambda > 0. From the first three equations, we see x2 = y 2 = z 2 . Substituting into
2
2
2
2
the third equation,
we obtain
√
√ √ x √+ x + x = 3x = 9. Thus, since x, y, and z are positive,
x = y = z = 3 and F ( 3, 3, 3) = 108 is √
a constrained
extremum. Since (1, 2, 2) satisfies
√ √
the constraint and F (1, 2, 2) = 64 < 108, F ( 3, 3, 3) = 108 is a constrained maximum.
19. Fx = 2x; Fy = 2y; Fz = 2z; gx = 2; gy = 1; gz = 1; hx = −1; hy = 2; hz = −3. We
need to solve 2x = 2λ−µ, 2y = λ+2µ, 2z = λ−3µ subject to 2x+y+z = 1, −x+2y−3z = 4.
Solving the first three equations for x, y, and z, respectively, and substituting into the
constraint equations, we obtain 2λ−µ+λ/2−3µ/2 = 1, −λ+µ/2 +λ+ 2µ−3λ/2 +9µ/2 = 4
or 6λ − 3µ = 2, −3λ + 14µ = 8. From this, we obtain λ = 52/75 and µ = 54/75. Then
x = 1/3, y = 16/15, and z = −11/15. Thus, F (1/3, 16/15, −11/15) = 134/75 is a constrained
minimum.
20. Fx = 2x; Fy = 2y; Fz = 2z; gx = 4; gy = 0; gz = 1; hx = 2x; hy = 2y; hz = −2z. We
need to solve 2x = 4λ + 2xµ, 2y = 2yµ, 2z = λ − 2zµ subject to 4x + z = 7, z 2 = x2 + y 2 .
130
CHAPTER 13. PARTIAL DERIVATIVES
Consider the second equation. If y = 0, then the constraint equations become 4x + z = 7 and
z 2 = x2 . The solutions of these equations are x = z = 7/5 and x = −z = 7/3. In either case,
the first and third equations can be solved for λ and µ. Thus, (7/5, 0, 7/5) and (7/3, 0, −7/3)
are candidates for constrained extrema. Now, if y 6= 0, then from the second equations µ = 0.
In this case, 2x = 4λ and 2z = λ or x = 4z. Then the first constraint equation becomes
16z + z = 17z = 7, so z = 7/17. Then, x = 28/17 and y 2 = z 2 − x2 < 0. Hence, the system
has no solution when y 6= 0. Thus, F (7/5, 0, 7/5) = 98/25 is a constrained minimum and
F (7/3, 0, −7/3) = 98/9 is a constrained maximum.
21. We want p
to maximize A(x, yx y/2 subject to P (x, y) =
Mx2+y2
x + y + x2p+ y 2 − 4 = 0. Ax =p y/2; Ay = x/2;
y
; Py = 1 + y/ x2 + y 2 .pWe need
Px = 1 + x/ x2 + y 2p
to solve p
y/2 = λ + λx/ x2 + y 2 , x/2 = λ + λy/ x2 + y 2 ,
x
x + y + x2 + y 2 − 4 = 0 for x > 0, y > 0, and hence
p
x2 + y 2 or
λ > 0. Subtracting the second
equation
from
the
first,
we
have
(y
−
x)/2
=
λ(x
−
y)/
p
p
2
2
2
2
(y − x) = (y − x)(−2λ/ x + y ). Since −2λ/ x + y is negative,
1, and hence
√ it cannot equal
√
2 = (2 +
y − x = 0 or y √
= x. Substituting in the third equation gives
2x
+
2x
2)x
√
√
√ = 4. Thus,
x = y = 4/(2 + 2) and this maximum area is A(4/(2 + 2), 4/(2 + 2)) = 4/(3 + 2 2).
22. Let the base of the box have dimensions x and y and let the height be z. We want to
maximize V (x, y, z) = xyz subject to S(x, y, z) = xy + 2yz + 2xz − 75 = 0. Now Vx =
yz; Vy = xz; Vz = xy; Sx = y + 2z; Sy = x + 2z; Sz = 2y + 2x. We need to
solve yz = λ(y + 2z), xz = λ(x + 2z), xy = λ(2y + 2x), xy + 2yz + 2xz − 75 = 0 or
xyz = λ(xy + 2xz), xyz = λ(xy + 2yz), xyz = λ(2yz + 2xz), xy + 2yz + 2xz = 75,
for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we have
xy + 2xz = xy + 2yz, which gives xz = yz or x = y; and xy + 2yz = 2yz + 2xz, which
gives xy = 2xz or y = 2z. Substituting x = y = 2z into the fourth equation, we obtain
4z 2 + 4z 2 + 4z 2 = 12z 2 = 75. Thus, z = 5/2cm and x = y = 5cm. When the box is closed,
S(x, y, z) = 2(xy + yz + xz) − 75, Sx = 2(y + z), Sy = 2(x + z), Sz = 2(x + y), and we need
to solve xyz = 2λ(xy + xz), xyz = 2λ(xy + yz), xyz = 2λ(yz + xz), 2(xy + yz + xz) = 75
for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we have
xy+xz = xy+yz, which gives xz = yz or x = y; and xy+yz = yz +xz, which gives xy = xz or
2
2
2
2
y = z. Substituting x √
= y = z into the fourth equation, we obtain
√ 2(x + x + x ) = 6x = 75.
Thus, x = y = z = 5/ 2. The box is a cube with each side 5/ 2cm.
p
9 + y 2 − 81π.
23. We want to maximize V (x, y) = 9πx + 3πy subject
to
S(x,
y)
=
9π
+
6πx
+
3π
p
2
Now Vp
6π; Sy = 3πy/ 9 + y . We need to solve 9π = 6πλ, 3π =
x = 9π; Vy = 3π; Sx = p
3πλy/ 9 + y 2 , 9π + 6πx + 3 9 + y 2 − 81π = 0 for x > 0 and y >
√ 0. From the first
equation, λ = 3/2. Usingpλ = 3/2 in the second equation
gives
y
=
6/
5. From the third
√
equation, we √
have 6x + 3 9 +√36/5 = 72 or x = 12 − 9/2 5. The volume is maximum when
x = 12 − 9/2 5m and y = 6/ 5m.
1
2
1
24. Ux = x−2/3 y 2/3 ; Uy = x1/3 y −1/3 ; gx = 1; gy = 6. We need to solve x−2/3 y 2/3 =
3
3
3
2 1/3
1
λ,
x y−1/3 = 6λ, x + 6y − 18 = 0 or y = 3λx2/3 y 1/3 ,
x = 3λx2/3 y 1/3 , x +
3
3
6y = 18. From the first two equations, y = x/3. Substituting into the third equation, we
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13.10. LAGRANGE MULTIPLIERS
131
have x + 2x = 3x = 18. Thus, x = 6 and y = 2. Since (12, 1) satisfies the constraint and
U (12, 1) = 121/3 , U (6, 2) = 61/3 22/3 = 241/3 is a constrained maximum.
25. We want to maximize z(x, y) = P − x − y subject to z 2 /xy 3 = k or (P − x − y)2 − kxy 3 = 0.
Now zx = −1; zy = −1; gx = −2(P − x − y) − ky 3 ; gy = −2(P − x − y) − 3kxy 2 . We need
to solve −1 = −2λ(P − x − y) − λky 3 , −1 = −2λ(P − x − y) − 3λkxy 2 , (P − x − y)2 = kxy 3
for x > 0, y > 0, and z > 0. From the first two equations,
√ we have y = 3x. Substituting into
the third equation, we obtain (P − 4x)2 = 27kx4 or 27kx2 = P − 4x. (Since z > 0, z =
P − x − y = P − 4x > 0.) Using the quadratic formula and the fact that x > 0, we find
p
√
√
−2 + 4 + P 27k
16 + 4P 27k
√
=
x=
.
27k
2 27k
p
√
√
Then the maximum value of z is P − 4x = P + 4(2 − 4 + P 27k/ 27k.
−4 +
p
26. (a) See part (b).
(b) Maximizing 1/(x21 · · · x2n ) is equivalent to minimizing the denominator F (x1 , . . . xn ) =
x21 + · · · + x2n . The constraint is still x1 + · · · xn = 1, which we can write as g(x1 , . . . xn ) =
x1 + · · · + xn − 1 = 0. Since ∂F/∂xi = 2xi and ∂g/∂xi = 1, we can get the equations
2x1 = λ,, 2x2 = λ, , . . . , 2xn = λ, x1 + · · · xn = 1. The solution is x1 = . . . = 1/2 ( and
λ = 1/2n).
4
27. f (x, y) is the square of the distance from a point on the graph of x√
+ y 4 =√1 to the origin.
4
The points (0, ±1) and (±1, 0) are closest to the origin, while (±1/ 2, ±1/ 4 2) are farthest
from the origin.
28. F (x, y, z) is the square of the distance from a point on the plane x + 2y + 3z = 4 to the origin.
The point (2/7, 4/7, 6/7) is closest to the origin.
29. F is the square of the distance of points on the intersection of the planes 2x + y + z = 1 and
−x + 2y − 3z = 4 from the origin. The point (1/3, 16/15, −11/15) is closest to the origin.
30. F is the square of the distance of points on the intersection of the plane 4x + z = 7 and
the circular cone z 2 = x2 = y 2 . The point (7/5, 0, 7/5) is closest to the origin and the point
(7/3, 0, −7/3) is farthest from the origin.
31. We want to minimize f (x, ) = x2 + y 2 subject to xy 2 = 1. Now fx = 2x; fy = 2y; gx =
y 2 ; gy = 2xy. We need to solve 2x = λy 2 , 2y = 2λxy, xy 2 − 1 = 0 or 2xy = λy 3 , 2xy =
2λx2 y, xy 2 = 1 for x > 0, y > 0, and hence λ > 0. From the first two equations, we have
y 3 = 2x2 y or y 2 = 2x2 . Substituting into the third equation gives 2x3 = 1 or x = 2−1/3 . Again,
from the third equation we have y = 1/(2−1/3 )1/2 = 21/6 . Thus, the point closest to the origin
is (2−1/3 , 21/6 ). Since the surface is F (x, y, z) = xy 2 − 1 = 0, ∇F = y 2 i + 2xyj is normal to
the surface at (x, y, z). Thus, a normal to the surface at (2−1/3 , 21/6 , 0) is ∇F (2−1/3 , 21/6 , 0)
is ∇F (2−1/3 , 21/6, 0) = 21/3 i + 2(2−1/3 )(21/6 )j = 21/3 i + 25/6 j = 22/3 (2−1/3 i + 21/6 j). Since
∇F (2−1/3 , 21/6 , 0) is a multiple of the vector from the origin to P (2−1/3 , 21/6 , 0), this vector
is perpendicular to the surface.
132
CHAPTER 13. PARTIAL DERIVATIVES
1 −2/3 1/3 1/3
1
1
x
y z ; Fy = x1/3 y −2/3 z 1/3 ; Fz = x1/3 y 1/3 z −2/3 ; gx = 1; gy =
3
3
3
1 1/3 −2/3 1/3
1 1/3 1/3 −2/3
1
x y
z
= λ,
x y z
=
1; gz = 1. We need to solve x−2/3 y 1/3 z 1/3 = λ,
3
3
3
1/3 1/3
1/3 1/3 1/3
1/3 1/3 1/3
λ, x + y + z − k = 0 or x y Z1/3 = 3λx, x y z
= 3λy, x y z
=
λz, x + y + z = k. From the first three equations, x = y = z. Substituting into the fourth
equation, x + x + x = 3x = k. Thus, x = y = z = k/3 and F (k/3, k/3, k/3) = k/3 is a
constrained maximum.
32. Fx =
33. For any x + y + z = k, by Problem 32,
√
3
xyz ≤
k
x+y+z
=
.
3
3
34. Distance from the xz-plane is measured by |y|. Alternatively, we will find the extreme values
of F (x, y, z) = y 2 subject to g(x, y, z) = x2 + z 2 − 1 = 0 and h(x, y, z) = x + y + 2z − 4 = 0.
Now Fx = 0; Fy = 2y; Fz = 0; gx = 2x; gy = 0; gz = 2z; hx = 1; hy = 1; hz = 2.
We need to solve 0 = 2λx + µ, 2y = µ, 0 = 2λz + 2µ, x2 + z 2 = 1, x + y + 2z = 4.
By inspection we see that if λ = 0, then µ = 0 and y = 0. Similarly, if µ = 0, then λ = 0
and y = 0. Substituting y = 0 into the fourth and fifth equations, we obtain the system
x2 + z 2 = 1, x + 2z = 4, which is inconsistent. Thus, µ 6= 0 and λ 6= 0. Now, solving the
first three equations for x, y, and z and substituting into the fourth and fifth equations, we
obtain the system
µ2
µ
µ 2µ
µ2
+
= 1, −
+ −
= 4 or 5µ2 = 4λ2 , (λ − 5)µ = 8λ.
4λ2
λ2
2λ
2
λ
8λ 2
Solving the second equation for µ and substituting into the first, we obtain 5(
) = 4λ2
λ
−5
√
or 80 = (λ − 5)2 . Thus,
√ λ = 5 ± 4 5. From µ = 8λ/(λ − 5) we find that corresponding
2
values of µ are 8 ± 2 5. Since
√ 2y = µ we see that the objective
√ function F (x, y, z) = y is
minimized when µ = 8 − 2 5 and maximized
when µ = 8 + 2 5. Corresponding values of
√
√
µ
8±2 5
1
√ = ∓ √ ≈ |mp0.45, y − µ/2 = 4 ± 5 ≈ 6.24 and
x, y, and z are x = −
=−
2λ
10 ± 8 5
5
√
√
√
√
1.76, z = −µ/λ = 2x = ∓2/ 5 ≈ ∓0.89. The closest point is (−1/ 5, 4 − 5, −2 5) or
about (−4.5, 6.24, −0.89).
Chapter 13 in Review
A. True/False
1. False; see Example 3 in Section 13.2 in the text.
2. False; (0,4.1) is in the domain of g but not in the domain of f .
3. True
4. True
5. False; consider z = y 2 .
CHAPTER 13 IN REVIEW
133
6. False; consider f (x, y) = xy at (0, 0).
7. False; ∇f is perpendicular to the level curve f (x, y) = c.
8. True
9. True
10. False; at a saddle point fx = fy = 0, but there is no extremum.
B. Fill in the Blanks
1.
3x2 + xy 2 − 3xy − 2y 3
3+1−3−2
1
=
=−
2
2
5x − y
5−1
4
(x,y)→(1,1)
lim
2. where x − y + 1 = 0
3. 3x2 + y 2 = 3(2)2 + (−4)2 = 28
4.
∂
∂T ∂p ∂T ∂q
T (p, q) =
+
= Tp gξ + Tq hξ
∂ξ
∂p ∂ξ
∂q ∂ξ
5.
∂F dr
∂F ds
d
F (r, s) =
+
= Fr g 0 (w) + Fs h0 (w)
dw
∂r dw
∂s dw
6. dg = gs ∆s + gt ∆t =
4s
2
∆s − 3 ∆t
t2
t
7. fyyzx
8.
∂3f
∂y 2 ∂x
∂f
∂ R y
9. Using the Fundamental Theorem of Calculus, we have
(x, y) =
F (t)dt = F (y)
∂y
∂y x
i
i
∂ hR x
∂f
∂ R y
∂ h Rx
− y F (t)dt = −
F (t)dt = −F (x)
(x, y) =
F (t)dt =
x
y
∂x
∂x
∂x
∂x
10. ∇F (x0 , y0 , z0 ) = i + j + k
∂2
∂
fx (x, y)g(y)h(z) =
[fx (x, y)g 0 (y)h(z) + fxy (x, y)g(y)h(z)]
∂z∂y
∂z
= fx (x, y)g 0 (y)h0 (z) + fxy (x, y)g(y)h0 (z)
11. Fx,y,z =
12. The distinct fourth-order partial derivatives are fxxxx , fxxxy , fxxyy , fxyyy , and fyyyy .
www.elsolucionario.org
134
CHAPTER 13. PARTIAL DERIVATIVES
C. Exercises
3
1. zy = −x3 ye−x
2. zu = −
3. fr =
y
+ e−x
3
y
v sin uv
= −v tan uv
cos uv
3
3 2 3
r (r + θ2 )−1/2 ; frθ = − r2 θ(r3 + θ2 )−3/2
2
2
4.
∂f
∂2f
= 2(2 + y 2 )2
= 2(2x + xy 2 )(2 + y 2 ) = 2(2 + y 2 )2 x;
∂x
∂x2
5.
∂z
∂2z
= 3x2 y 2 sinh x2 y 3 ;
= 9x4 y 4 cosh x2 y 3 + 6x2 y sinh x2 y 3
∂y
∂y 2
6.
∂z
∂2z
∂3z
2
2
2
2
2
= −4y(ex + e−y );
= −8xyex ; 2
= −16x2 yex − 8yex
∂y
∂x∂y
∂ x∂y
7. Fs = 3s2 t5 v −4 ; Fst = 15s2 t4 v −4 ; Fstv = −60s2 t4 v −5
8.
xy x y
∂2w
x
x
1
∂3w
2x
∂4w
2
∂w
= 2 + + ;
=− 2 − 2 + ; 2
= 3;
= 3
2
∂z
z
y
x ∂y∂z
z
y
x ∂ y∂z
y
∂x∂ y∂z
y
9. ∇f = −
10. ∇F =
y
1
1
1
y
x
1
1
i+
j=− 2
i+ 2
j; ∇f (1, −1) = i + j
2
2
2
2
2
2
2
x 1 + y /x
x 1 + y /x
x +y
x +y
2
2
2x
9y 2
4(x2 − 3y 3 )
i− 4 j−
k; ∇F (1, 2, 1) − 2i − 36j + 92k
4
z
z
z5
6
1
2
11. ∇f = (2xy − y 2 )i + (x2 − 2xy)j; u = √ i + √ j = √ (i + 3j);
40
40
10
1
1
2
2
2
Du f = √ (2xy − y + 3x − 6xy) = √ (3x − 4xy − y 2 )
10
10
2x
2y
2z
2
1
2
i+ 2
j+ 2
k; u = − i + j + k;
2
2
2
2
2
2
+y +z
x +y +z
x +y +z
3
3
3
−4x + 2y + 4z
Du F =
3(x2 + y 2 + z 2 )
12. ∇F =
x2
CHAPTER 13 IN REVIEW
13. {(x, y)|(x + y)2 ≤
{(x, y)| |x + y| ≤ 1}
135
1}
=
14. {(x, y)|y > x, y 6= x + 1}
y
y
x
x
15. ∆z = 2(x + ∆x)(y + ∆y) − (y + ∆y)2 − (2xy − y 2 ) = 2x∆y + 2y∆x + 2∆x∆y − 2y∆y − (∆y)2
16. ∆z = (x + ∆x)2 − 4(y + ∆y)2 + 7(x + ∆x) − 9(y + ∆y) + 10 − (x2 − 4y 2 + 7x − 9y + 10)
= 2∆x + (∆x)2 − 8y∆y − 4(∆y)2 + 7∆x − 9∆y
4x + 3y − (x − 2y)4
11y
(4x + 3y)(−2) − (x − 2y)3
−11x
=
; zy =
=
;
(4x + 3y)2
(4x + 3y)2
(4x + 3y)2
(4x + 3y)2
11y
11x
dz =
dx −
dy
(4x + 3y)2
(4x + 3y)2
17. zx =
18. Ax = 2y + 2z; Ay = 2x + 2z; Az = 2y + 2x; dA = 2(y + z)dx + 2(x + z)dy + 2(x + y)dz
p
√
√
√
19. zy = 4y/ x2 + 4y 2 , zy (− 5, 1) = 4/3, z(− 5, 1) = 3. The line is given by x = − 5 and
√
z−3
y−1
4
=
.
z − 3 = (y − 1). Symmetric equations of the line are x = − 5,
3
4
3
−−→
1
1
20. The direction vector is P Q = 2i + 2j. ∇z = (y + 2x)i + xj. u = √ i + √ j; Du = ∇z · u =
2
2
√
√
√
√
(y + 2x + x)/ 2 = (y + 3x)/ 2; Du (2, 3) = 9/ 2. The slope of the tangent line is 9/ 2.
21. fx = 2xy 4 , fy = 4x2 y 3 .
(a) u = i, Du (1, 1) = fx (1, 1) = 2
√
√
√
(b) u = (i − j/ 2, Du (1, 1) = (2 − 4)/ 2 = −2/ 2
(c) u = j, Du (1, 1) = fy (1, 1) = 4
22. (a)
dw
∂w dx ∂w dxy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
x
y
z
=p
6 cos 2t + p
(−8 sin 2t) + p
15t2
2
2
2
2
2
2
2
x +y +z
x +y +z
x + y2 + z2
(6x cos 2t − 8y sin 2t + 15zt2 )
p
=
x2 + y 2 + z 2
136
CHAPTER 13. PARTIAL DERIVATIVES
(b)
dw
∂w dx ∂w dxy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
6
8r
2r
2t
y
z
x
sin
15t2 r3
cos + p
+p
=p
2
2
2
2
2
2
2
2
2
2
r
r
t
t
x +y +z
x +y +z
x +y +z
2r
6x
2t 8yr
cos + 2 sin
+ 15zt2 r3
r
r
t
t
p
=
x2 + y 2 + z 2
√
π
1
23. F (x, y, z) = sin xy − z; ∇F = y cos xyi + x cos xyj − k; ∇F (1/2, 2π/3, 3/2) = i + j − k.
3
4
√
π
1
1
2π
3
The equation of the tangent plane is (x− )+ (y − )−(z −
) = 0 or 4πx+3y −12z =
3
2
4
3
2
√
4π − 6 3.
24. We want to find a normal to the surface that is parallel to k. ∇F = (y −2)i+(x−2y)j+2zk.
We need y − 2 = 0 and x − 2y = 0. The tangent√plane is parallel√to z = 2 when y = 2 and
x = 4. In this case z 2 = 5. The points are (4, 2, 5) and (4, 2, − 5).
25. ∇F = 2xi + 2yj; The equation of the tangent plane is 6(x − 3) + 8(y − 4) = 0
or 3x + 4y = 25.
26. We want to minimize
1
3
Du f = u · ∇f = √ (i + j) · [(3x2 + 3y − 6x)i + (3x + 3y 2 )j] = √ (x2 + y − 2x + x + y 2 )
2
2
or equivalently, we want to minimize F (x, y) = x2 − x + y 2 + y. Now Fx = 2x − 1; Fxx =
2; Fxy = 0;
Fy = 2y + 1; Fyy = 2; D = 4. Solving Fx = 0 and Fy = 0 we obtain x = 1/2 and y = −1/2.
Since D = 4 > 0 and Fxx = 2 > 0, F, and hence Du f, has a minimum at (1/2, −1/2).
27. We want to maximize v(x, y, z) = xyz subject to x + 2y + z = 6. Now Vx = yz; Vy = xz;
VZ = xy; gx = 1; gy = 2; gz = 1. We need to solve yz = λ, xz = 2λ, xy = λ or xyz = λx,
xyz = 2λy, xyz = λz along with x + 2y + z − 6 = 0 or x + 2y + z = 6. From the first three
equations, we have x = 2y = z. Substituting into the fourth equation gives x + x + x = 3x = 6
or x = 2. Then y = 1 and z = 2 and V (2, 1, 2) = 4 is the maximum volume.
c2
Dθ2
G
c2
(b) dM = (θ2 dD + 2Dθdθ)
G
dM
c2 θ2
2Dθ
θ2
2Dθ
dD
dθ
(c) We have
=
dD +
dθ =
dD +
dθ =
+ 2 , so
M
G M
M
Dθ2
Dθ2
D
θ
28. (a) M =
dθ
dM
dD
dD
dθ
+2
=
≤
+2
≤ 0.10 + 2(0.02) = 0.14 = 14%.
M
D
θ
D
θ
www.elsolucionario.org
CHAPTER 13 IN REVIEW
137
√
29. We√are given v =
14 5ry −1/2 , dr = −1, dy = 1, r = 20, and y = 25. Now, dv =
√
14 5y −1/2 dr − 7 5ry −3/2 dy and the approximate change in volume is
√
√
√
∆v ≈ 14 5(25)−1/2 (−1) − 7 5(20)(25)−3/2 (1) = −98 5/25 ≈ −8.77cm/s.
30. ∆f = 2xi + 2yj, ∆f (3, 4) = 6i = 8j
√
√
− 2j)/ 5;
(a) ∆f (1, −2)2i −√4j; u =√(2i − 4j) √20 = (i √
Du f (3, 4) = 6 5 − 16 5 = −10 5 = −2 5
√
(b) v = (6i + 8j)/ 100 = (3i + 4j)/5; Dv f (3, 4) = 18/5 + 32/5 = 10
R2 − r 2
. Then, after a straightforward but lengthy comR2 − 2rR cos(θ − φ) + r2
putaiton, we find
31. Let g(r, θ) =
gr =
grr =
gθθ =
(2r2 R + 2R3 ) cos(θ − φ) − 4rR2
(R2 − 2rR cos(θ − φ) + r2 )2 ,
8R4 cos2 (θ − φ) + (−12rR3 − 4r3 R) cos(θ − φ) − 4R4 + 12r2 R2
,
[R2 − 2rR cos(θ − φ) + r2 ]3
(4r4 R2 − 4r2 R4 ) cos2 (θ − φ) + (2r5 R − 2rR5 ) cos(θ − φ) − 8r4 R2 + 8r2 R4
(R2 − 2rR cos θ + r2 )3
and r2 grr + rgr + gθθ . Then
Z
Z π
Z π
r2 π
r
1
2
r Urr + rUr + Uθθ =
g(r, θ)f (φ)dφ +
g(r, θ)f (φ)dφ +
g(r, θ)f (φ)dφ
2π π
2π π
2π π
Z π
1
=
(r2 grr + rgr + gθθ )dφ = 0.
2π π
αz
βAxα y β
βy
αAxα y β
=
; fy = Aβxα y β−1 =
=
;
x
x
y
y
xαzx − αz
xα(αz/x) − αz
α2 z − αz
α(α − 1)
fxx =
=
=
=
;
x2
x2
x2
x2
2
yβzy − βz
yβ(βz − βz)
β z − βz
β(β − 1)z
fyy =
=
=
=
;
y2
y2
y2
y2
α(βz)/y
αβz
fxy = fyz =
=
x
xy
32. fx = Aαxα−1 y β =
33. Since D = 4(6) − 52 = −1 < 0, f (a, b) is not a relative extremum.
34. Since D = 2(7) − 02 = 14 > 0 and fxx = 2 > 9, f (a, b) is a relative minimum.
35. Since D = (−5)(−9) − 62 = 9 > 0 and fxx = −5 < 0, f (a, b) is a relative maximum.
36. Since D = (−2)(−8) − 42 = 0, no determination is possible.
138
CHAPTER 13. PARTIAL DERIVATIVES
37. Since x = L cos θ and y = L sin θ,
1
1
1
A = xy = L2 sin θ cos θ = l2 sin 2θ.
2
2
4
L
y
θ
x
38. Substituting x = h cot φ into tan θ =
h=
tan θ
.
1 − tan θ cos φ
h
and solving, we obtain
1+x
h
φ
θ
1
x
39. A = xy − (y − 2z)(x − 2z) − z 2 = 2(x + y)z − 5z 2
40. We are given V (x, y, z) = xyz, x = 30, y = 40, z = 25, and dx = dy = −1 and dz = −1/2.
Then dV = yzdx + dzdy + dydz, so the approximate volume of plastic is
|dV | = 40(25)(1) + 30(25)(1) + 30(40)(1/2) = 2350cm3
.
p
p
41. V = (2x)(2y)z = 4xy 4 − x2 − y 2 = 16xy − 4xy x2 + y 2
42. C(x, y, z) = 1.5(2xy + 2xz + 2yz + xz + 5yz) =
3
(2xy + 3xz + 7yz)
2
www.elsolucionario.org
Chapter 14
Multiple Integrals
14.1
The Double Integral
1. With f (x, y) = x + 3y + 1 and ∆Ak = 1,
Z Z
(x + 3y + 1)dA ≈ f (1/2, 1/2) + f (3/2, 1/2) + f (5/2, 1/2) + f (1/2, 3/2)
R
+ f (3/2, 3/2) + f (5/2, 3/2) + f (1/2, 5/2) + f (3/2, 5/2)
= 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 52.
2. With f (x, y) = 2x + 4y and ∆Ak = 1/4,
Z Z
(2x + 4y)dA ≈
R
1
[2(3/2) + 4(1/2) + 2(2) + 4(1/2) + 2(5/2) + 4(1/2) + 2(1/2) + 4(1)
4
+ 2(3/2) + 4(1) + 2(1) + 4(1) + 2(1/2) + 4(3/2) + 2(1) + 4(3/2) + 2(3/2)
+ 4(3/2) + 2(1) + 4(2) + 2(1/2) + 4(2) + 2(1/2) + 4(5/2)]
1
= (3 + 2 + 4 + 2 + 5 + 2 + 4 + 4 + 3 + 4 + 2 + 4 + 1 + 6 + 2 + 6 + 3
4
93
+ 6 + 2 + 8 + 1 + 8 + 1 + 10) =
.
4
3. (a) With f (x, y) = x + y, and ∆Ak = 1,
Z Z
(x + y) dA ≈ (−3/2 + 1/2) + (−1/2 + 1/2) + (1/2 + 1/2) + (3/2 + 1/2)
R
+ (−3/2 + 3/2) + (−1/2 + 3/2) + (1/2 + 3/2) + (3/2 + 3/2)
1
16
= (−2 + 0 + 2 + 4 + 0 + 2 + 4 + 6) =
= 8.
2
2
(b) With f (x, y) + y + 4 and ∆Ak = 1,
139
140
CHAPTER 14. MULTIPLE INTEGRALS
Z Z
(x + y) dA ≈ (−2 + 1) + (−1 + 1) + (0 + 1) + (1 + 1) + (−2 + 2) + (−1 + 2) + (0 + 2) + (1 + 2)
R
= 8.
4. With
Z Z f (x, y) = xy and ∆Ak = 1/4,
1
xydA ≈ [0(1/2) + (1/2)(1/2) + (−1/2)(1) + (0)(1) + (1/2)(1)
4
R
+ (1)(1) + (−1/2)(3/2) + (0)(3/2) + (1/2)(3/2)
y
4
+ (1)(3/2) + (−1/2)(2) + (0)(2) + (1/2)(2) + (1)(2)
1
+ (−1)(5/2) + (−1/2)(5/2) + (0)(5/2) + (1/2)(5/2)
x
+ (1)(5/2) + (3/2)(5/2) + (−1)(3) + (−1/2)(3) + (0)(3)
+ (1/2)(3) + (1)(3) + (3/2)(3) + (−1)(7/2) + (−1/2)(7/2)
+ (0)(7/2) + (1/2)(7/2) + (1)(7/2) + (3/2)(7/2)] = 73/16
5.
RR
6.
RR
7.
RR
8.
RR
R
R
R
R
10dA = 10
RR
10dA = 10
RR
10dA = 10
RR
10dA = 10
RR
R
dA = 10(6) = 60
R
dA = 10(12) = 120
1
2
dA = 10 π(2) = 10π
R
4
1
dA = 10 (5)
R
2
5
125
=
2
2
9. No, since x + 5y is negative at (3, −1) which is in R.
10. Yes, since x2 + y 2 is nonnegative on R.
11.
RR
12.
RR
13.
RR
14.
RR
15.
RR
R
10dA = 10
RR
dA = 10(8) = 80
R
−5xdA = −5
R
(2x + 4y)dA = 2
R
(2x + 4y)dA =
R
(3x + 7y + 1)dA = 3
Z Z
R
R
RR
R
xdA + 4
xdA −
RR
R
RR
R
RR
R
RR
RR
R
Z Z
ydA +
R1
f (x, y)dA +
RR
R1
R
RR
R2
f (x, y)dA +
dA = 3(3) + 7(7) + 8 = 66
Z Z
y dA − 4
R
RR
RR
Z Z
2
(2 + y) dA =
f (x, y)dA =
f (x, y)dA = −5.
R2
R
ydA = 2(3) + 4(7) = 34
ydA = 3 − 7 = −4
xdA + 7
R
f (x, y)dA =
18. Since
RR
RR
2
y dA −
R
RR
xdA = −5(3) = −15
R
Z Z
2
16.
17.
R
RR
Z Z
dA − 4
R
y 2 dA
ydA −
R
= −4(8) − 4(7) = −60
R
f (x, y)dA = 4 + 14 = 18
RR
R2
f (x, y)dA, 25 = 30 +
RR
R2
f (x, y)dA and
14.2. ITERATED INTEGRALS
14.2
1.
R
Iterated Integrals
dy = y + c1 (x)
2. RBy holding y fixed,
(1 − 2y)dy = x − 2yx + c2 (y)
3. By
2
3
Z holding y fixed,
x √
x
√
2
(6x y − 3x y)dx = 6
y−3
y + c2 (y)
3
2
3 √
= 2x3 y − x2 y + c2 (y)
2
4. By
2
Z holding x fixed,
y 3/2
y
√
(6x2 y − 3x y)dy = 6x2
− 3x
+ c1 (x)
2
(3/2)
= 3x2 y 2 − 2xy 3/2 + c1 (x)
5. By holding x fixed,
R
1
ln |y + 1|
dy =
+ c1 (x)
x(y + 1)
x
6. By
2
Z holding x fixed,
x
− 5xy 4 + c2 (y)
(1 + 10x − 5y 4 )dx = x + 10
2
= x + 5x2 − 5xy 4 + c2 (y)
7. By
Z holding y fixed,
sin 4x
(12y cos 4x − 3 sin y)dx = 12y
− 3x sin y + c2 (y)
4
= 3y sin 4x − 3x sin y + c2 (y)
8. By
Z holding x fixed,
tan 3xy
+ c1 (x)
sec2 3xydy =
3x
9. By
Z holding y fixed, p
y
√
dy = y 2x + 3y + c2 (y)
2x + 3y
10. By
Z holding x fixed, 1 (2x + 5y)7
(2x + 5y)6 dy =
+ c1 (x)
5
7
(2x + 5y)7
=
+ c1 (x)
35
Z 3
3
11.
(6xy − 5ey ) dx = (3x2 y − 5xey ) −1 = (27y − 15ey ) − (3y + 5ey ) = 24y − 20ey
−1
141
www.elsolucionario.org
142
12.
CHAPTER 14. MULTIPLE INTEGRALS
Z
2
tan xy dy =
1
13.
Z
3x
x3 exy = x2 exy
1
Z
14.
15.
1
2
= x2 (e3x − ex )
y
2x
x2
xy
x
dy = ln(x2 + y 2 )
2
+y
2
x
e2y/x dy =
16.
x3
Z
1
ln | sec 2x − sec x|
x
=
(8x3 y − 4xy 2 ) dx = (2x4 y − 2x2 y 2 )
0
Z
3x
1
2
y3
√
Z
1
ln | sec xy|
x
x 2y/x
e
2
x
x3
=
2x
=
0
= (2y 13 − 2y 8 ) − (2y 3 − 2y 3 ) = 2y 13 − 2y 8
x
x
[ln(x2 + 4x2 ) − ln x2 ] = ln 5
2
2
3
2
x 2y/x
x
(e
− e2x /x ) = (e2 − e2x )
2
2
sec y
(2x + cos y) dx = (x2 + x cos y)
17.
y3
√
y
tan y
sec y
tan y
= sec2 y + sec y cos y − tan2 y − tan y cos y
= sec2 y + 1 − tan2 y − sin y = 2 − sin y
Z
18.
1
√
y ln x dx
Integration by parts
y
= y(x ln x −
Z
1
x)|√
y
π/2
cos x sin3 ydy = cos x cos y
19.
√
√
√
√
= y(0 − 1) − y( y ln y − y) = −y − y y
x
− sin2 y 2
−
3
3
π/2
= 0 − cos2 x
x
− sin2 x 2
−
3
3
1
ln y − 1
2
cos2 (1 − cos2 x) 2 cos2 x
cos2 x sin3 x 2 cos2 x
+
=
+
=
3
3
3
3
cos2 x cos4 x 2 cos2 x
1
=
−
+
= cos2 x − cos4 x
3
3
3
3
1
Z 1
1
sin xy cos xy xy
sin xy cos xy x
=
20.
y cos2 xydx = y
+
+
2y
2 1/2
2
2 1/2
1/2
x
x
sin 2 cos 2
sin x cos x − sin x2 cos x2
sin x cos x x
x
x
=
+
−
+
=
+
2
2
2
4
2
4
Z
2
Z
x2
Z
2
(8x − 10y + 2) dy dx =
21.
1
−x
x2
2
(8xy − 5y + 2y)
1
Z 2
=
dx
−x
[(8x3 − 5x4 + 2x2 ) − (−8x2 − 5x2 − 2x)] dx
1
Z
=
2
(8x3 − 5x4 + 15x2 + 2x) dx = (2x4 − x5 + 5x3 + x2 )
1
= 44 − 7 = 37
2
1
14.2. ITERATED INTEGRALS
Z
1
y
Z
−1
0
√
Z
1
Z
y
Z
(x + y)2 dx
Z √2−y2
2
23.
−
0
√
√
Z
√
2
(2x − y) dx dy =
(x − xy)
2−y 2
0
−
√
2
0
√
Z
0
cos x
Z
√
=0
−1
2−y 2
p
2
(−2 2 − y 2 ) dy = (2 − y 2 )3−2
3
cos x
π/4
Z
2
24.
1
2−y 2
√
2
=
2
(1 + 4y tan x) dy dx =
0
y
1
h
i
p
p
(2 − y 2 − y 2 − y 2 ) − (2 − y 2 + y 2 − y 2 ) dy
=
π/4
Z
dy =
2
Z
Z
1
dy
(x + y)3
3
−1
0
−1
0
Z
Z
1 1
1 1 3
1 7 4
3
3
=
[(y + y) − (0 + y) ] dy =
7y dy =
y
3 −1
3 −1
3 4
(x + y)2 dx dy =
22.
143
Z
2
=
(y + 2y tan x)
0
0
2
=
0
2
2
4√
(0) − 23/2 = −
2
3
3
3
π/4
(cos x + 2 cos2 x tan2 x) dx
0
0
π/4
1
(cos x + 2 sin2 x) dx = sin x + x − sin 2x
2
0
√
√
2 2+π−2
2 π 1
=
+ − =
2
4
2
4
Z
π/4
=
Z
π
3y
Z
25.
0
Z
y
2
√
Z
Z
3y
1 π
1
dy
sin(2x + y)
cos(2x + y) dx dy
(sin 7y − sin 3y) dy
2 0
0 2
y
π
1
1
1
− cos 7y + cos 3y
=
2
7
3
0
1
1
1
1 1
4
− (−1) + (−1) − − +
=−
=
2
7
3
7 3
21
π
Z
x
2
26.
2y sin πx
1
27.
x
6e
x+2y
√
Z
2
dy
Z
2
=−
1
x
e
1
2
x+2y
x
Z
dx =
2
x sin πx2 dx
1
0
1
1
1
(cos 4π − cos π) = − (1 − (−1)) = −
2π
2π
π
ln 3
Z
(3e3x − 3ex ) dx = (e3x − 3ex )
dx =
1
0
3
2
y sin πx
1
1
cos πx2
2π
ln 3
2
dx =
0
dy dx =
0
√
#
x
2y sin πx
1
Z
1
"Z
dy dx =
0
ln 3
2
Z
=−
Z
0
ln 3
1
= (27 − 9) − (e − 3e) = 18 − e3 + 3e
28.
R 1 R 2y
0
0
e−y
2
dx dy =
R1
0
xe−y
2
2y
dy =
0
R1
0
2ye−y
2
dy = −e−y
2
1
0
= −e−1 − (−1) = 1 − e−1
144
CHAPTER 14. MULTIPLE INTEGRALS
Z
3
2x+1
Z
29.
x+1
0
3
Z
1
√
dy dx =
y−x
Z
1
2
2 3/2
x+1
x(y − x )
9
0
x
Z
0
Z
1/2
y
Z
32.
0
0
9
Z
1
dy dx =
x+ y 2
31.
1
1
dx dy =
0
0
Z
2
(x + 1)3/2
3
Z
1
x
0
9
1
0
y
−1
sin
0
y
16
2
10
−3 −
=
3
3
3
1
=−
5
0
0
1/2
=
0
e
y
dx dy =
x
33.
1
Z
1
4
√
Z
Z
4
Z
x
2 −x
dy dx =
y e
1
1
Z
Z √25−y2 /2
35.
0
0
1
1
p
dx dy =
(25 − y 2 ) − x2
sin−1 p
0
6
sin−1
0
Z √20−y2
36.
y2
0
Z
π
Z
√
Integration by parts
x
2
Z
Z
x
! √25−y2 /2
25 − y 2 0
Z 6
π
1
dy =
dy = π
2
0 6
2
p
0
π
e sin y dx dy =
cos y
20−y 2
Integration by parts
dy
1
1 4
2 3/2
y dx dy =
xy
dy =
(y 20 − − y ) dy = − (20 − y ) − y
3
4
0
0
y2
√
√
1
1
40 5 − 76
= − (64) − 4 − − (40 5) − 0 =
3
3
3
Z
0
37.
π/2
π
ln 9
4
= −4e−4 + e−1
6
Z
=
2
0
1
30
√
√
1 π
π + 6 3 − 12
3
−1=
+
2 6
2
12
(xe−x − e−x ) dx
1
4
e−x ) 1
Z
Z
=
1
=
4
dx =
= (−xe−x − e−x +
6
9
1
y ln x
√
−x
2ye
Z
0
1 6
y
30
y
e
x
34.
1
(−y 5 ) dy = −
Z e
dy =
y ln y dy Integration by parts
1
1
1
e
1 2 1 2
1 2
1 2
1
1
= e − e − −
=
y ln y − y
= (e2 + 1)
2
4
2
4
4
4
1
y
Z
1
sin−1 y dy
0
p
= y sin−1 y + 1 − y 2
Z
Z
1/2
Z
dy =
x
√
( x + 1 − 1) dx
π
π
dx = ln |x|
4x
4
=
1/2
Z
=2
Z
3
0
3
1
− (y 2 − x2 )5/2
5
1
y
tan−1
x
x
1
√
dx dy =
1 − x2
Z
dx = 2
y
Z
2x+1
0
=2
30.
√
2 y−x
x
e sin y
π/2
= (− cos y +
Z
2
3
π
dy =
cos y
cos y π
e
)|π/2
y2
(sin y − ecos y sin y) d
π/2
= (1 + e−1 ) − (0 + 1) = e−1
0
www.elsolucionario.org
14.2. ITERATED INTEGRALS
Z
1
145
y 1/3
Z
38.
3
6x ln(y + 1) dx dy =
0
y 1/3
1
Z
2
0
0
1
Z
dy = 2
2x ln(y + 1)
y ln(y + 1) dy
0
0
Integration by parts
1
1 2
2
= y ln(y + 1) − y + y − ln(y + 1)
2
0
1
1
= (ln 2 + 1 − ln 2) − (0 − 0 + 0 − ln 1) =
2
2
Z
2π
x
Z
Z
(cos x − sin y) dy dx =
39.
Z
2π
(x cos x + cos x − 1) dx
dx =
(y cos x + cos y)
π
0
π
x
2π
π
0
Integration by parts
2π
= (cos x + x sin x + sin x − x)|π = (1 − 2π) − (−1 − π) = 2 − π
Z
3
1/x
Z
40.
1
0
1
dy dx =
x+1
3
Z
1
y
x+1
1/x
Z
dx =
1
0
3
1
dx =
x(x + 1)
Z
1
3
1
1
−
x x+1
dx
3
= [ln x − ln(x + 1)]|1 = (ln 3 − ln 4) − (0 − ln 2) = ln 3/2
Z
5π/12
√
Z
√
2 sin 2θ
41.
π/12
Z
π/3
1
Z
2 sin 2θ
Z 5π/12 1 2
1
1
r dr dθ =
r
dθ =
sin 2θ −
dθ = − (cos 2θ + θ)
2
2
2
π/12
π/12
1
"
!
!# √
√
√
1
3 5π
3 5π
3 π
=−
−
+
−
+
=
−
2
2
12
2
12
2
6
Z
1+cos θ
42.
Z
r dr dθ =
0
3 cos θ
0
5π/12
π/3
1 2
r
2
1+cos θ
dθ
3 cos θ
Z
1 π/3
1
(1 + 2 cos θ − cos2 θ) dθ = (θ + 2 sin θ − 4θ − 2 sin 2θ)
2 0
2
√
√
1
π
= (−π + 3 − 3) = −
2
2
π/3
=
0
5π/12
π/12
146
CHAPTER 14. MULTIPLE INTEGRALS
43.
44.
y=2x+1
y
y
x=My
x=-My
x
x
45.
46.
y
y=x2+1
y
x=M16-y2
x
x
y=-x2
√
Z 4
x
x2
1 2
1 2
47.
x ydydx =
x y
dx =
x x−
dx
4
0
x/2
0 2
0 2
x/2
4
Z 4
1 4
1
1 2 1 4
x − x dx =
x − x5
=
2
8
8
40
0
0
128
32
= 32 −
=
5
5
Z 2 Z 2y
Z 2
Z 2
2y
1
1
x2 ydxdy =
x3 y dy =
y(8y 3 − y 6 )dy
0
y2
0 3
0 3
y2
2
Z 2
8 4 1 7
8 5
1 8
=
y − y dy =
y − y
3
3
15
24
0
0
256 32
32
−
=
=
3 Z 5Z
Z Z √ 15
Z
4
√
Z
x
Z
2
4
4
x
2
2y
x2 ydydx =
Therefore
0
x/2
x2 ydxdy
0
y2
y
y=Mx
y=1/2x
x
y
x=y2
x=2y
x
14.2. ITERATED INTEGRALS
Z
1
√
Z
48.
√
1−x2
1
Z
2xdydx =
√
2xy
− 1−x2
0
147
Z
0
1
=
1−x2
y
dx
√
− 1−x2
2x
p
1 − x2 + 2
y=M1-x2
p
1 − x2 dx
0
1
Z
=
4x
p
0
Z
1
4
=
3
Z √1−y2
Z
2xdxdy =
−1
√
1
1−y 2
1
y=-M1-x2
y
(1 − y 2 )dy
dy =
x
0
1
Z
x2
−1
0
4
1 − x2 dx = − (1 − x2 )3/2
3
−1
0
1
1
= y − y3
3
−1
1
4
1
− −1 +
=
= 1−
3
3
3
Z Z √ 2
2
Therefore,
Z
1
√
Z
1−x
2xdydx =
2
3
Z
x
−1
3
Z
Z
2
Z
dx dy =
−1
2
0
3
50.
−1
2
3
Z
−2
3
0
8 1
+
3 3
4
2
2
(x + 4xy)
−2
Z 2
2
dy =
(2x + 4y) dx dy =
Z
0
Z
4
Z
2
Z
−2
2
2
Z 4
=
[(16 + 16y) − (4 + 8y)] dy
2
3
Z
51.
1
0
π
4
2
2
−2
= (24 + 16) − (−24 + 16) = 48
4
[(4x + 8) − (−4 + 8)] dx
dx =
−2
2
Z
Z
(2xy + 2y 2 )
8x dx = 4x2
3
3 dy = 3y|0 = 9
−2
2
4
(2x + 4y) dy dx =
= 8 − (−1) = 9
2
dy =
−2
Z
3
dy =
(12 + 8y) dy = (12y + 4y 2 )
=
2
−1
3x2 dx = x3
−1
0
−1
Z
2
Z
dx =
x y
1
x
3
4
Z
2
dy dx =
0
0
3
2
Z
2
Z
x
0
2
x
2xdxdy
−1
49.
Z
x=M1-y2
1−y
1
√
− 1−x2
0
x=0
2
= 64 − 16 = 48
π
Z 3 2
3π 2
3 2 2
x y − 4 cos y
dx =
x − 4 − (4) dx
(3x y − 4 sin y) dy dx =
2
2
1
1
0
2
3
Z 3 2
3π 2
π 3
=
x − 8 dx =
x − 8x
2
2
1
1
27π 2
π2
=
− 24 −
− 8 = 13π 2 − 16
2
2
2
Z
3
www.elsolucionario.org
148
CHAPTER 14. MULTIPLE INTEGRALS
Z
π
3
Z
3
(x3 y − 4x sin y) 1 dy =
0
1
0
π
Z
(3x2 y − 4 sin y) dy dx =
π
Z
[(27y − 12 sin y) − (y − 4 sin y)]dy
0
π
Z
π
0
(26y − 8 sin y)dy = (13y 2 + 8 cos y)
=
0
= (13π 2 − 8) − (8) = 13π 2 − 16
Z
1
2
Z
52.
0
0
8y
2x
− 2
x+1 y +1
1
Z
dxdy =
0
x2
8y ln |x + 1| − 2
y +1
2
Z
1
dy =
8y ln 3 −
0
0
4
2
y +1
1
= (4y 2 ln 3 − 4 tan−1 y) 0 = 4 ln 3 − π
1 Z 2
Z 2Z 1
Z 2
8y
4
4y 2
2x
π
=
− 2
dydx =
− 2x tan−1 y
−
dx
x+1 y +1
x+1
x+1
2
0
0
0
0
0
π 2
= 4 ln 3 − π
= 4 ln |x + 1| − x2
4
0
Rβ
Rβ
53. We use the fact that α kF (t)dt = k α F (t)dt. Then
"Z
#
"Z
# "Z
#
Z Z
Z
d
b
d
f (x)g(y)dxdy =
c
Z
∞
Z
54.
∞
a
−(2x2 +3y 2 )
Z
b
f (x)dx dy =
c
xye
0
b
g(y)
a
∞
∞
Z
d
f (x)dx
g(y)dy
a
c
2
2
xe−2x
ye−3y dxdy
0
0
Z
Z ∞
∞
−2x2
−3y 2
=
xe
dx ·
ye
dy
dxdy =
0
0
0
=
a
Z
lim
a→∞
2
xe−2x dx ·
0
Z
lim
b→∞
!
b
−3y 2
ye
dy
0
!
!
2
2 a
e−3y
e−2x
· lim −
= lim −
a→∞
b→∞
4
6
0
#!
"
#!
"
2
2
1
−e−3b
1
−e−2a
+
· lim
+
= lim
a→∞
b→∞
4
4
6
6
1
1
1
=
·
=
4
6
24
14.3
Evaluation of Double Integrals
Z Z
x3 y 2 dA =
1.
R
Z
0
=
1
Z
x
x3 y 2 dydx =
0
1 7
x
21
Z
0
1
=
0
1
21
1
1 3 3
x y
3
x
dx =
0
1
3
Z
1
x6 dx
y
0
x
1
x
dy
14.3. EVALUATION OF DOUBLE INTEGRALS
Z Z
2
Z
2.
Z
4−x
149
R
0
y
dx
(xy + y)
x
0
4−x
2
Z
(x + 1)dydx =
(x + 1)dA =
x
2
Z
[(4x − x2 + 4 − x) − (x2 + x)]dx
=
4-x
0
2
Z
(2x − 2x2 + 4)dx
=
0
=
x
2
x − x3 + 4x
3
2
Z Z
Z
3.
1
0
20
=
3
x
x2
Z
(2x + 4y + 1)dA =
R
2
(2x + 4y + 1)dydx
Z
y
x3
0
x2
1
2
=
0
Z 1
=
x2
dx
(2xy + 2y + y)
x3
x3
x
[(2x3 + 2x4 + x2 ) − (2x4 + 2x6 + x3 )]dx
0
Z
1
3
2
6
(x + x − 2x )dx =
=
0
1 1 2
25
= + − =
4 3 7
84
Z Z
Z 1Z x
Z 1
4.
xey dA =
xey dydx =
xey
R
0
0
0
1 4 1 3 2 7
x + x − x
4
3
7
1
0
x
y
dx
0
1
Z
(xex − x)dx
=
Integration by parts
x
0
1 2
x
x
= xe − e − x
2
Z Z
Z
5.
2
Z
8
2xydA =
R
1
0
1
1
= e−e−
− (−1) =
2
2
Z
2
2xydydx =
x3
0
Z
=
0
2
(64x − x7 )dx =
8
xy
0
2
1
x
y
dx
x3
1
32x2 − x8
8
2
= 96
0
x3
x
150
CHAPTER 14. MULTIPLE INTEGRALS
Z Z
6.
R
1
Z
x
√ dA =
y
3−x2
Z
xy
1
Z
3−x2
√
dydx =
y
2x y
x2 +1
−1
Z
−1/2
dx
−1
x2 +1
x2+1
3
1
(x
=2
p
3 − x2 − x
p
x2 + 1)dx
−1
3-x2
1
1
1
= 2[− (3 − x2 )3/2 − (x2 + 1)3/2 ]
3
3
−1
2 3/2
3/2
3/2
3/2
= − [(2 + 2 ) − (2 + 2 )] = 0
3
Z Z
7.
R
1
Z
y
dA =
1 + xy
1
Z
0
Z0
y
dxdy =
1 + xy
Z
x
1
1
y
ln(1 + xy) dy
0
0
1
1
1 ln(1 + y)dy = [(1 + y) ln(1 + y) − (1 + y)]|0
=
0
= (2 ln 2 − 2) − (−1) = 2 ln 2 − 1
Z Z
8.
sin
R
y
Z 2
πx
y
πx
dxdy =
− cos
dy
y
x
y
1
0
1
0
Z 2
y
y
=
− cos πy +
dy Integration by parts
π
π
1
2
y
1
y2
= − 2 sin πy − 3 cos πy +
π
π
2π 1
1
2
1
1
3π 2 − 4
= − 3+
−
+
=
3
π
π
π
2π
2π 3
πx
dA =
y
2
Z
Z
9.
x2
y2
Z
3
Z
x
+ 1dA =
R
√
Z
3
=
(x
0
√
Z
p
2
x + 1dydx =
−x
0
√
Z
=
p
π/4
10.
x2 + 1 + x
p
p
x2 + 1dx =
Z
1
Z
xdxdy =
0
x
x2
dx
+1
y
y=x
−x
x2 + 1)dx
2 2
(x + 1)3/2
3
tan y
0
π/4
1 2
x
2
1
x
3
y=-x
0
tan y
1
dy =
2
1
1 π4
(2 − sec2 y)dy = (2y − tan y)
2 0
2
1 π
π 1
=
−1 = −
2 2
4
2
Z
y2
2 3/2
14
(4 − 13/2 ) =
3
3
xdA =
R
y
p
√
3
2x
=
Z
y
x
3
0
0
Z Z
x
sin
√
Z Z p
1
2
Z
0
y
π4
(1 − tan2 y)dy
π/4
x=tany
π/4
=
0
1
x
www.elsolucionario.org
14.3. EVALUATION OF DOUBLE INTEGRALS
Z Z
Z
11.
4
Z
2
4
Z
(x + y)dxdy
(x + y)dxdy +
(x + y)dA =
Z
4
=
0
Z 4
=
2
0
0
0
R
2
Z
151
1 2
x + xy
2
2
Z
2
dy +
0
0
Z
4
dy
2
2
[(8 + 4y) − (2 + 2y)]dy = (2y + y 2 )
(2 + 2y)dy +
0
1 2
x + xy
2
0
4
0
+ (6y + y 2 )
2
0
= 24 + 16 = 40
Z Z
Z
12.
4
Z
R
4
0
0
Z
4
=
0
Z 4
1 2
x + xy
2
= (8y + 2y )
2x−x2
13. A =
1
−x
4
0
3
1 2
x + xy
dy
2
1
1
1
+ 3y −
+ y dy
2
Z
dy −
0
3
9
2
− (4y + y 2 )
3
3
1
= 64 − (21 − 5) = 48
3
Z
y
(2x − x2 + x)dx
dydx =
0
(x + y)dxdy
4
1
2
Z
Z
0
3
3
Z
1
(8 + 4y)dy −
=
Z
3
Z
(x + y)dxdy −
(x + y)dA =
0
=
3 2 1 3
x − x
2
3
3
0
2x-x2
9
=
2
x
-x
14. Using symmetry,
Z
1
Z
y
2−y 2
A=2
Z
dxdy = 2
y2
0
0
1
2
(2−y 2 −y 2 )dy = 2 2y − y 3
3
1
0
y2
8
= .
3
2-y2
x
Z
4
Z
ex
15. A =
Z
dydx =
1
4
ln x
1
4
4
y
(ex − ln x)dx = (ex − x ln x + x)|1
ex
= (e − 4 ln 4 + 4) − (e + 1) = e4 − e − 4 ln 4 + 3
lnx
10
4
x
152
CHAPTER 14. MULTIPLE INTEGRALS
4
Z
√
(2− x)2
0
4
Z
4−x
Z
16. A =
=
4
Z
[4 − x − (2 −
dydx =
√
y
x)2 ]dx
0
√
(4 x − 2x)dx =
0
8 3/2
x − x2
3
4
=
0
16
3
4-x
(2-Mx)2
x
Z
1
Z
−2x+3
x3
−2
=
Z
1
1
−x + 3x − x4
4
2
y
(−2x + 3 − x3 )dx
dydx =
17. A =
-2x+3
−2
1
=
−2
7
63
− (−14) =
4
4
x3
x
18. Expressing y = −x2 + 3x and y = −2x + 4 as functions of
3 1√
1
y, we have x = −
9 − 4y and x = 2 − y.
2
2
2 Z 2 Z 2−y/2
Z 2 y
3 1p
−
A=
dxdy
=
2
−
−
9
−
4y
dy
√
2
2 2
0
3/2− 9−4y/2
0
2
1
1
1
27
13
1
=− − −
y − y 2 − (9 − 4y)3/2
=
=
2
4
12
12
12
6
0
19. The correct integral is ( c).
Z 2 Z √4−y2
Z
V =2
(4 − y)dxdy = 2
−2
0
√
2
(4 − y)x
−2
p
y 1
= 2 2y 4 − y 2 + 8 sin−1 + (4 − y 2 )3/2
2 3
20. The correct integral is (b).
4−y 2
Z
3/2-1/2 M9-4y
p
(4 − y) 4 − y 2 dy
−2
= 2(4π − (−4π)] = 16π
−2
2-y/2
2
dy = 2
0
2
y
x
14.3. EVALUATION OF DOUBLE INTEGRALS
Z
r
Z √r2 −y2
V =8
0
√
(r2 − y 2 )1/2 dxdy = 8
0
r
2
Z
r
(r2 − y 2 )1/2 x
0
=8
(r − y )dy = 8 ry −
0
3
r3
2r
3
=8 r −
=8
=
3
3
Z
153
2
y3
3
r 2 −y 2
dy
0
r
0
16 3
r
3
21. Setting z = 0 we have y = 6 − 2x.
Z
Z 3 Z 6−2x
(6 − 2x − y)dydx =
V =
6−2x
1 2
6y − 2xy − y
dx
2
0
0
0
0
Z 3
Z 3
1
6(6 − 2x) − 2x(6 − 2x) − (6 − 2x)2 dx =
=
(18 − 12x + 2x2 )dx
2
0
0
3
2
= 18
= 18x − 6x2 + x3
3
0
22. Setting z = 0 we have y ± 2.
R3R2
V =
(4 − y 2 )dydx =
0 0
R 3 16
dx = 16
0 3
3
y
6-2x
x
y
R3
0
1
4y − y 3
3
2
dx
=
0
x
1
1
23. Solving for z, we have x = 2 − x + y. Setting z = 0, we
2
2
see that this surface (plane) intersects the xy-plane in the
line y = x − 4. Since z(0, 0) = 2 > 0, the surface lies above
the xy-plane
over the quarter-circular region.
Z 2 Z √4−x2 1
1
V =
2 − x + y dydx
2
2
0
0
√4−x2
Z 2
1
1
=
2y − xy + y 2
dx
2
4
0
0
Z 2 p
1 p
1
=
2 4 − x2 − x 4 − x2 + 1 − x2 dx
2
4
0
p
2
1
1 3
−1 x
2 3/2
2
= x 4 − x + 4 sin
+ (4 − x ) + x − x
2 6
12
0
2
4
= 2π + 2 −
− = 2π
3
3
y
M4-x2
2
2
x
www.elsolucionario.org
154
CHAPTER 14. MULTIPLE INTEGRALS
24. Setting z = 0 we have y = 3. Using symmetry,
3
Z √3
Z √3 Z 3
1 2
(3 − y)dydx = 2
(3y − y ) dx
V =2
2
x2
0
0
x2
√
Z 3
9
1
9
1
=2
( − 3x2 + x4 )dx = 2
x − x3 + x5
2
2
2
10
0
√
√
√
√
9
24 3
9
3−3 3+
3 =
=2
.
2
10
5
y
3
√
3
x2
0
25. Note that z = 1 + x2 + y 2 is always positive. Then
Z 1 Z 3−3x
Z 1
1 3
2
2
2
(1 + x + y )dydx =
V =
y+x y+ y
3
0
0
0
Z 1
[(3 − 3x) + x2 (3 − 3x) + 9(1 − x)3 ]dx
=
x
y
3−3x
dx
0
3-3x
0
Z
=
1
(12 − 30x + 30x2 − 12x3 )dx
0
= (12x − 15x2 + 10x3 − 3x4 )
1
0
x
= 4.
26. In the first octant, z = x + y is nonnegative. Then
√9−x2
Z 3 Z √9−x2
Z 3
1
V =
(x + y)dydx =
xy + y 2
dx
2
0
0
0
0
Z 3 p
9 1 2
2
=
x 9 − x + − x dx
2 2
0
3
9
1
1
= − (9 − x2 )3/2 + x − x3
3
2
6
0
27 9
=
−
− (−9) = 18.
2
2
y
3
M9-x2
3
x
y
27. In the first octant z = 6/y is positive. Then
R6R5 6
R 6 6x 5
R 6 dy
6
V = 1 0 dxdy = 1
dy = 30 1
= 30 ln y|1 = 30 ln 6. 6
y
y 0
y
1
5
x
14.3. EVALUATION OF DOUBLE INTEGRALS
155
28. Setting z = 0, we have x2 /4 + y 2 /16 = 1. Using symmetry,
Z
2
Z
V =4
0
Z
y
√
0
2 4−x2
1
(4 − x2 − y 2 )dydx
4
2
4
√
2 4−x2
1 3
y )
dx
12
0
0
Z 2 p
p
2
=4
[8 4 − x2 − 2x2 4 − x2 − (4 − x2 )3/2 ]dx
3
0
2M4-x2
Trig substitution
p
p
x 1
x
= 4 4x 4 − x2 + 16 sin−1 − x(2x2 − 4) 4 − x2 − 4 sin−1
2 4
2
i2
p
1
x
+ x(2x2 − 20) 4 − x2 −4 sin
2 0
12
16π 4π 4π
=4
−
−
− (0) = 16π.
2
2
2
2
=4
(4y − x2 y −
29. Note that z = 4−y 2 is positive for |y| ≤ 1. Using symmetry,
√
2x−x2
Z 2 Z √2x−x2
Z 2
1
3
2
dx
V =2
(4 − y )dydx = 2
(4y − y )
3
0
0
0
0
p
Z
p
1
2
2
2
2
= 2 0 4 2x − x − (2x − x ) 2x − x dx
3
Z 2 p
p
1
=2
(4 1 − (x − 1)2 − [1 − (x − 1)2 ] 1 − (x − 1)2 )dx
3
0
x
y
1
M2x-x2
x
u = x − 1, du = dx
Z 1 p
Z 1 p
p
1
1 p
11
=2
1 − u2 + u2 1 − u2 du
[4 1 − u2 − (1 − u2 ) 1 − u2 ]du = 2
3
3
3
−1
−1
Trig substitution
p
11 p
11
1
1
−1
2
2
2
=2
u 1−u +
sin u + x(2x − 1) 1 − u +
sin u
6
6
24
24
1 π
11 π
1 π
15
11 π
=2
+
− −
−
=
.
6 2
24 2
6 2
24 2
4
1
−1
156
CHAPTER 14. MULTIPLE INTEGRALS
30. From z = 1 − x2 and z = 1 − y 2 we have 1 − x2 = 1 − y 2 or
y = x (in the first octant). Thus, the surfaces intersect in
the plane y = x. Using symmetry,
1
Z 1
Z 1Z 1
1 3
2
y− y
1 − y dydx = 2
dx
V =2
3
0
x
0
x
Z 1
2
1
=2
− x + x3 dx
3
3
0
1
1
1 2
1 4
2
= .
x− x + x
=2
3
2
12
2
0
z
1
1
y
y=x
1
x
31. From z = 4−x−2y and z = x+y, we have 4−x−2y = x+y
3
or x = 2 − y.
2
Z
Z
4/3
z
4
2−3y/2
[4 − x − 2y) − (x + y)]dxdy
V =
0
Z
0
2−3y/2
4/3
2
4x − x − 3xy
=
0
dy
2
0
2
#
4/3
3
3
3
=
4(2 − y) − 2 − y − 3 2 − y y dy
2
2
2
0
Z 4/3 9
=
4 − 6y + y 2 dy
4
0
4/3
16
3
=
= 4y − 3y 2 + y 3
4
9
0
Z
y
x=2-3y/2
"
x
32. Using symmetry,
√
Z 3 Z √9−x2
Z 3
9−x2
1 3
2
2
2
V =4
(9 − x − y )dydx = 4
[(9 − x )y − y ]
3
0
0
Z0 0
8
3
2 3/2
0 (9 − x ) dx Trig substitution
=
3
3
p
8 x
243
8 243 π
81π
x
= [− (2x2 − 45) 9 − x2 +
sin−1 ] = (
)=
.
3 8
8
3 0
3 8 2
2
z
9
y
x
33. From z = x2 and z = −x + 2 we have x2 = −x + 2 or x = 1
(in the first octant). Then
Z 5Z 1
Z 5
1
1 2
1 3
2
V =
(−x + 2 − x )dxdy =
(− x + 2x − x ) dy
2
3
0
0
0
0
Z 5
7
35
=
dy =
.
6
0 6
y=M
9-x2
z
y
x
www.elsolucionario.org
14.3. EVALUATION OF DOUBLE INTEGRALS
157
34. From 2z = 4 − x2 − y 2 and z = 2 − y we have
z
4 − x2 − y 2 = 4 − 2y or x2 + (y − 1)2 = 1. We find
the volume in√the first octant and use symmetry.
Z 2 Z 1−(y−1)2 1 2 1 2
2 − x − y − (2 − y) dxdy
V =2
2
2
0
0
√
1−(y−1)2
Z 2
1
1
dy
=2
− x3 − xy 2 + xy
y
6
2
0
0
x
Z 2
p
p
3/2 1 2
1
− 1 − (y − 1)2
=2
− y 1 − (y − 1)2 + y 1 − (y − 1)2 dy
6
2
0
Z 2
p
1
1
2 3/2
2
2
− [1 − (y − 1) ] + (2y − y ) 1 − (y − 1) dy
=2
6
2
0
Z 2
1
1
=2
− [1 − (y − 1)2 ]3/2 + [1 − (y − 1)2 ]3/2 dy
6
2
0
Z 2
2
[1 − (y − 1)2 ]3/2 dy Trig substitution
=
3 0
2
p
2 3 π 3 π
2
3
π
y−1
−1
2
2
=
=
[2(y − 1) − 5] 1 − (y − 1) + sin (y − 1)
−
−
=
−
3
8
8
3 82
8
2
4
0
√
35. Solving x = y 2 for y, we obtain y = x. Thus,
y
2Z
Z
y
2
4Z
Z
f (x, y)dxdy =
0
0
2
√
0
f (x, y)dydx.
x
x=y2
36. Solving x =
Thus,
Z Z √
p
25−y 2
5
5
Z
f (x, y)dxdy =
−5
x
√
25 − y 2 for y, we obtain y = ± 25 − x2 .
0
0
√
Z
y
x=M25-y2
25−x2
√
− 25−x2
f (x, y)dydx.
x
x
37. Solving y = ex for x, we obtain x = ln y. Thus,
Z
3
Z
ex
Z
e3
Z
3
f (x, y)dydx =
0
1
f (x, y)dxdy.
1
y
y=ex
ln y
3
x
158
CHAPTER 14. MULTIPLE INTEGRALS
38. Solving x = 3 − y and x = y/2 for y, we obtain y = 3 − x
and y = 2x. Thus,
Z 3 Z 3−x
Z 1 Z 2x
Z 2 Z 3−y
f (x, y)dydx.
f (x, y)dydx+
f (x, y)dxdy =
y/2
0
0
0
y
x=y/2
0
1
x=3-y
x
√
39. Solving y = 3 x and y = 2 − x for x, we obtain x = y 3 and
x = 2 − y. Thus,
1
Z
0
√
3
Z
x
2
Z
f (x, y)dydx+
0
2−x
Z
Z
1
Z
3
y=M
x
2−y
f (x, y)dxdy.
f (x, y)dydx =
0
1
y
y3
0
y=2-x
x
√
√
40. Solving x = y and x = 2 − y for y, we obtain y = x2
and y = 2 − x2 . Thus,
1
Z
√
Z
y
Z
2
√
Z
2−y
f (x, y)dxdy+
1
Z
2−x2
f (x, y)dxdy =
0
0
Z
0
1
y
f (x, y)dydx.
0
x2
x=My
1
x=M2-y
Z
1
Z
1
41.
x
0
x
2
p
Z
1
Z
y
Z
p
x 1 + y 4 dxdy =
1
1 3p
x 1 + y4
1 + y 4 dydx =
0
0 3
0
1
Z
1 1
1 1 3p
(1 + y 4 )3/2
=
y 1 + y 4 dy =
3 0
3 6
0
1 √
=
(2 2 − 1)
18
x
y
2
dy
y
0
1
y=x
x
Z
1
Z
2
42.
0
e−y/x dxdy =
2y
Z
2
0
Z
=
Z
x/2
e−y/x dydx =
0
−xe−y/x
0
2
(−xe−1/2 + x)dx =
0
1
= (1 − e−1/2 )x2
2
x/2
2
Z
Z
y
0
2
x=2y
(1 − e−1/2 )xdx
0
x
2
= 2(1 − e−1/2 )
0
14.3. EVALUATION OF DOUBLE INTEGRALS
Z
2
4
Z
cos x3/2 dxdy =
43.
y2
0
4
Z
=
cos x3/2 dydx =
√
x cos x3/2 dx =
√
1
Z
44.
1−x2
√
− 1−x2
−1
Z
p
x 1 − x2 − y 2 dydx =
1
y
dx
x=y2
0
2
sin x3/2
3
4
=
0
2
sin 8
3
x
Z √1−y2
−1
Z
x
y cos x3/2
0
0
Z
4
Z
0
4
Z
√
√
x
Z
0
159
y
p
2
2
√ 2 x 1 − x − y dxdy
− 1−y
√ 2
y=M1-x2
1−y
1
1
=
[− (1 − x2 − y 2 )3/2 ] √
dy
3
−1
− 1−y 2
Z
1 1
(0 − 0)dy = 0
=−
3 −1
Z
1
Z
1
1
dydx =
1 + y4
45.
0
x
Z
1
Z
0
Z
y
1
dxdy =
1 + y4
0
1
Z
1
0
0
y
x
1 + y4
y
1
dy = tan−1 y 2
4
1+y
2
=
x
y=-M1-x2
1
=
0
y
dy
y=x
0
π
8
x
Z
4
Z
46.
0
2
√
p
x3 + 1dxdy =
y
Z
2
0
Z
Z
x2
Z
p
x3 + 1dydx =
x2
2
y
p
x3 + 1
0
0
2
p
2
=
x2 x3 + 1dx = (x3 + 1)3/2
9
0
2 3/2
52
= (9 − 13/2 ) =
9
9
47. fave
1
=
A
Z
1
A
Z
=
d
c
c
b
Z
a
d
2
1
xydxdy =
A
d
Z
2
(b − a )y
1
dy =
2
A
c
x2 y
2
2
y
dx
0
2
0
b
dy
x
a
(b − a2 )y 2
4
d
c
1 (b2 − a2 )(d2 − c2 )
=
A
4
But A = (b − a)(d − c), so
fave =
x=My
(b2 − a2 )(d2 − c2 )
(b + a)(d + c)
=
4(b − a)(d − c)
4
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160
CHAPTER 14. MULTIPLE INTEGRALS
√
48. fave
1
=
A
Z
1
=
A
Z
3
√
− 3
√
3
Z √9−3y2
−
√
9−3y 2
9 − x2 − 3y 2 dxdy
√
3
9−3y 2
x
− 3y 2 x
√ 9x − −
√
3
− 3
−
√
dy
9−3y 2
3
p
p
(9 − 3y 2 )3/2
2
2
2
9 9 − 3y −
− 3y 9 − 3y
√
3
− 3
p
p
(9 − 3y 2 )3/2
− −9 9 − 3y 2 +
+ 3y 2 9 − 3y 2 dy
3
Z √3 p
1
9 − 3y 2
9 − 3y 2
2
2
2
=
9 − 3y 9 −
− 3y + 9 −
− 3y dy
A −√3
3
3
Z √3 p
1
9 − 3y 2 (12 − 4y 2 )dy
=
A −√3
"
√ !
√ !#
√
√
p
1 12
9
4
81
3y p
3y
3y
3y
√
=
9 − 3y 2 + sin−1
− √
(6y 2 − 9) 9 − 3y 2 +
sin−1
A
2
2
3
3
8
9
3
3 3
√
27π 3
=
2A
49. Let S be the solid with base R and height described by the function f (x, y). The volume of
S is equal to the volume of the solid with base R and constant height fave .
Z Z
Z dZ b
Z d
1
50. (a)
cos 2π(x + y)dA =
[sin 2π(b + y) − sin 2π(a + b)]dy
cos 2π(x + y)dxdy =
2π c
R
c
a
Z d
1
[(sin 2πb cos 2πy + cos 2πb sin 2πy) − (sin 2πa cos 2πy + cos 2πa sin 2πy)] dy
=
2π c
Z d
1
=
(S1 cos 2πy + C1 sin 2πy)dy
2π c
1
1
[S1 (sin 2πd − sin 2πc) − C1 (cos 2πd − cos 2πc)] =
(S1 S2 − C1 C2 )
=
2
2
4π
4π
Z Z
Z dZ b
Z d
1
sin 2π(x + y)dA =
sin 2π(x + y)dxdy −
[cos 2π(b + y) − cos 2π(a + y)]dy
2π
R
c
a
c
Z d
1
=−
[(cos 2πb cos 2πy − sin 2πb sin 2πy) − (cos 2πa cos 2πy − sin 2πa sin 2πy)]dy
2π c
Z d
1
(C1 cos 2πy − S1 sin 2πy)dy
=−
2π c
1
1
= − 2 [C1 (sin 2πd − sin 2πc) + S1 (cos 2πd − cos 2πc)] = − 2 (C1 S2 + S1 C2 )
4π
4π
(b) If b − a = n is an integer, then b = a + n and
1
=
A
Z
sin 2πb = sin 2π(a + n) = sin 2πa cos 2πn + cos 2πa sin 2πn = sin 2πa
cos 2πb = cos 2π(a + n) = cos 2πa cos 2πn − sin 2πa sin 2πn = cos 2πa.
√
3
√
− 3
14.4. CENTER OF MASS AND MOMENTS
161
RR
R
In this case, S1 = 0 and C1 = 0, so
cos 2π(x+y)dA = 0 and sinR sin 2π(x+y)dA =
R
0. Similarly, is d − c is an integer, the double integrals are zero.
(c) If both integrals are 0, then
0 = (S1 S2 − C1 C2 )2 + (C1 S2 + S1 C2 )2 = S12 S22 + C12 C22 + C12 S22 + S12 C22
= (S12 + C12 )(S22 + C22 ).
Thus, either S12 + C12 = 0, in which case S1 = C1 = 0, or S22 + C22 = 0, in which case
S2 = C2 = 0. Suppose S1 = C1 = 0, and b − a = k or b = a + k. We want to show that
k is an integer. Consider
S1 = sin 2πb − sin 2πa = sin 2π(a + k) − sin 2πa
= sin 2πa cos 2πk + cos 2πa sin 2πk − sin 2πa
C1 = cos 2πb − cos 2πa = cos 2π(a + k) − cos 2πa
= cos 2πa cos 2πk − sin 2πa sin 2πk − cos 2πa
S1 − C1 = (sin 2πa − cos 2πa) cos 2πk + (sin 2πa + cos 2πa) sin 2πk − (sin 2πa − cos 2πa)
= (sin 2πa − cos 2πa)(cos 2πk − 1) + (sin 2πa + cos 2πa) sin 2πk.
Since a is arbitrary we must have cos 2πk − 1 = 0 and sin 2πk = 0, which implies k is an
integer. Similarly, if S2 = C2 = 0, d − c must be an integer.
RR
RR
51. By Problem 50 (a) we have
cos 2π(x + y)dA =
sin 2π(x + y)dA = 0 for k =
Rk
Rk
1, 2, · · · , n. Then
Z Z
Z Z
Z Z
cos 2π(x + y)dA =
cos 2π(x + y)dA + · · · +
cos 2π(x + y)dA = 0 + · · · + 0 = 0
R
R1
and
Z Z
Rn
Z Z
Z Z
sin 2π(x + y)dA + · · · +
sin 2π(x + y)dA =
R
R1
sin 2π(x + y)dA = 0 + · · · + 0 = 0.
Rn
Therefore by Problem 45 (c), at least one of the two sides of R must have integer length.
14.4
Center of Mass and Moments
3
Z
Z
4
1. m =
4
3
1 2
x y dy =
2
0
xydxdy =
0
=
Z
0
2 3
4y 0 = 36
Z 3Z 4
2
My =
0
Z
=
0
0
3
64
32 2
ydy =
y
3
3
3
y
8ydy
0
3
x=4
3
Z
x ydxdy =
0
Z
0
3
= 96
0
4
1 3
x y dy
3
0
x
162
CHAPTER 14. MULTIPLE INTEGRALS
3
Z
Z
4
3
Z
2
xy dxdy =
Mx =
0
0
0
0
3
3
Z
4
1 2 2
x y
2
8 3
y
= 72
3 0
0
x = My /m = 96/36 = 8/3; y = Mx /m = 72/36 = 2. The center of mass is (8/3, 2).
Z 2 Z 4−2x
Z 2
Z 2
2
2 4−2x
y
x dydx =
2. m =
x y0
dx =
x2 (4 − 2x)dx
8y 2 dy =
=
0
0
0
2
Z
(4 − 22 − 2x3 )dx =
=
0
4 3 1 4
x − x
3
2
0
4−2x
2Z
Z
x3 dydx =
My =
2
Z
(4x3 − 2x4 )dx =
=
0
0
1
=
2
Z
y=4-2x
0
2
1 2 2
x y
2
0
4−2x
0
(16x − 16x + 4x )dx = 2
0
32
32
32
− 16 +
=
=2
15
15
15
Z
1
dx =
2
2
x2 (4 − 2x)2 dx
0
3 4
(4x − 4x x )dx = 2
0
y = Mx /m =
x
2
2
Z
4
x = My /m = 16/5 = 6/5;
x3 (4 − 2x)dx
2
2
0
3
4
2
dx =
2
x4 − x5
5
0
2
2
0
0
64
16
= 16 −
=
5
5
Z 2 Z 4−2x
Z
Mx =
x2 ydydx =
32
8
−8=
3
3
Z
x3 y
0
0
0
2
=
4−2x
3
Z
1
4 3
x − x4 + x5
3
5
2
0
32/15
= 4/5.
8/3
The center of mass is (6/5, 4/5).
3. Since both the region and ρ are symmetric with respect to
the line x = 3, x = 3.
Z 3 Z 6−y
Z 3
6−y
m=
2ydxdy =
2xy
0
Z
y
0
3
Z
0
y
3
0
4
3
0
x=y
6−y
3
2xy 2
Z
dxdy =
y
x=6-y
3
(12y − 4y 2 )dy
x
0
3
4
= 18
= 6y 2 − y 3
3
0
Z
Z 3 Z 6−y
Mx =
2y 2 dxdy =
0
3
y
2y(6 − y − y)dy =
=
y
3
2y 2 (6 − y − y)dy =
0
= (4y − y ) = 27
y = Mx /m = 27/18 = 3/2. The center of mass is (3, 3/2).
Z
0
3
(12y 2 − 4y 3 )dy
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14.4. CENTER OF MASS AND MOMENTS
163
y
4. Since both the region and ρ are symmetric with respect to
the y-axis, x = 0. Using symmetry,
3
Z
Z
y
2
2
3
Z
(x + y )dxdy =
m=
0
0
3
Z
Z
0
3
1 3
x + xy 2
3
3
y
x=y
dy
x
0
3
1
1 3
4
y 3 dy = y 4 = 27
y + y 3 dy =
3
3
3 0
0
0
y
Z 3
Z 3
Z
Z 3Z y
1 3
1 4
4 3 4
(x2 y + y 3 )dxdy =
dy =
y dy
x y + xy 3
y + y 4 dy =
Mx =
3
3
3 0
0
0
0
0
0
=
4 5
y
15
3
324
5
0
324/5
= 12/5. The center of mass is (0, 12/5).
y = Mx /m =
27
=
=
x2
y
1 2
5. m =
(x + y)dydx =
xy + y
dx
2
0
0
0
0
1
Z 1
1
7
1
1
1 4
=
=
x3 + x4 dx =
x + x5
2
4
10
20
0
0
x2
Z 1 Z x2
Z 1
1 2
2
2
My =
dx
(x + xy)dydx =
x y + xy
2
0
0
0
0
1
Z 1
1 5
17
1 5
1 6
4
=
=
x + x dx =
x + x
2
5
12
60
0
0
x2 Z 1 Z 1 Z x2
Z 1
1 5 1 6
1 2 1 3
2
Mx =
xy + y
=
x + x dx
(xy + y )dydx =
2
3
2
3
0
0
0
0
0
1
11
1
1 6
=
x + x7
=
12
21
84
0
17/60
11/84
x = My = m =
= 17/21; y + Mx /m =
= 55/147.
7/20
7/20
The center of mass is (17/21, 55/147).
Z
Z
1
4
Z
x2
√
Z
6. m =
1
Z
√
x
Z
(y + 5)dydx =
0
0
0
4
1
( y 2 + 5y)
2
4
1
x
y
x
dx
2
y=Mx
0
4
√
1
1 2 10 3/2
92
=
x + 5 x dx =
x + x
=
2
4
3
3
0
0
√x
Z 4 Z √x
Z 4
1 2
My =
(xy + 5x)dydx =
xy + 5xy
dx
2
0
0
0
0
4
1 3
224
5/2
=
x + 2x
=
6
3
0
Z
y=x2
4
x
164
CHAPTER 14. MULTIPLE INTEGRALS
Z
4
√
Z
√
x
4
Z
2
(y + 5y)dydx =
Mx =
0
0
0
1 3 5 2
y + y
3
2
x
Z
4
dx =
0
0
1 3/2 5
x + x dx
3
2
4
2 5/2 5 2
364
=
x + x
15
4
15
0
224/3
364/15
= 56/23; y = Mx /m =
= 91/115.
x = My /m =
92/3
92/3
The center of mass is (56/23, 91/115).
=
y
7. The density is ρ = ky. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. Using symmetry,
Z 1 Z 1−x2
Z 1
Z
1−x2
1 2
m=2
kydydx = 2k
dx = k 01 (1 − x2 )2 dx
y
2
0
0
0
0
Z 1
1
2
1
(1 − 2x2 + x4 )dx = k(x − x3 + x5 )
=k
3
5
0
0
2 1
8
=k 1− +
=
k
3 5
15
Z 1 Z 1−x2
Z 1
Z 1
1−x2
1 3
2
2
Mx = 2
ky dydx = 2k
y
(1 − x2 )3 dx
dx = k
3 0
0
0
0 3
0
1
Z 1
2
3
1
2
(1 − 3x2 + 3x4 − x6 )dx = k x − x3 + x5 − x7
= k
3 0
3
5
7
0
2
3 1
32
= k 1−1+ −
=
k
3
5 7
105
32k/105
y = Mx /m =
= 4/7. The center of mass is (0,4/7).
8k/15
8. The density is ρ = kx.
Z π Z sin x
Z
m=
kxdydx =
0
0
1
y=1-x2
1
y
sin x
π
y=sin x
π
dx =
kxy
0
Z
1
kx sin xdx
0
0
π
Integration by parts
π
= k(sin x − x cos x)|0 = kπ
Z
Z π Z sin x
2
My =
kx dydx =
0
0
2
x
0
sin x
π
2
kx y
Z
dx =
π
kx2 sin xdx
Integration by parts
0
0
π
= k(−x cos x + 2 cos x + 2x sin x) 0 = k[(π 2 − 2) − 2] = k(π 2 − 4)
Z π Z sin x
Z π
sin x
1
Mx =
kxydydx =
kxy 2
dx
0
0
0 2
0
Z π
Z π
1
1
=
kx sin2 xdx =
kx(1 − cos 2x)dx
4
0 2
Z π
Z π 0
1
= k
xdx −
x cos 2xdx
Integration by parts
4
0
0
x
14.4. CENTER OF MASS AND MOMENTS
π
165
π
1
1 1
1
(cos 2x + 2x sin 2x) ] = k( π 2 ) = kπ 2
4
4 2
8
0
0
k(π 2 − 4)
kπ 2 /8
x = My /m =
= π − 4/π; y = Mx /m =
= π/8.
kπ
kπ
The center of mass is (π − 4/π, π/8).
=
1 1 2
k[ x
4 2
1
Z
−
ex
Z
Z
3
9. m =
1
1 4
y
4
y dydx =
0
0
0
1 4x
e
16
Z 1Z
My =
=
0
ex
Z
1
dx =
0
0
1 4x
e dx
4
1
1 4
(e − 1)
16
0
x
Z
e
3
xy dydx =
=
0
0
1
1 4
xy
4
1
ex
dx
0
1
1
Z
y=ex
y
x
1 4x
=
xe dx Integration by parts
4
0
1
1 3 4
1 4x
1
1
1 1 4x
=
xe − e
e +
=
(3e4 + 1)
=
4 4
16
4
16
16
64
0
x
1
R1 1
R 1 R ex 4
R1 1 5 e
1 5
1 5x
dx = 0 e5x dx =
=
Mx = 0 0 y dydx = 0 y
e
(e − 1)
5 0
5
25
25
0
(3e4 + 1)/64
(e5 − 1)/25
3e4 + 1
16(e5 − 1)
x = My /m =
=
; y = Mx /m = 4
=
4
4
(e − 1)/16
4(e − 1)
(e − 1)/16
25(e4 − 1)
5
4
3e + 1 16(e − 1)
,
≈ (0.77, 1.76).
The center of mass is ( 4
4(e − 1) 25(e4 − 1)
10. Since both the region and ρ are symmetric with
respect to the y-axis, x = 0. √Using symmetry,
Z 3
Z 3 Z √9−x2
9−x2
x2 dydx = 2
m=2
x2 y
dx
0
0
Z
=2
3
x2
0
p
9 − x2 dx
y=M9-x2
Trig substitution
3
3
81 π
x
x
81π
81
=
(2x2 − 9) 9 − x2 +
sin−1
· =
.
8
8
3 0√ 4 2
8
Z 3 Z √9−x2
Z 3
Z 3
9−x2
1 2 2
2
dydx =
x2 (9 − x2 )dx
Mx = 2
x ydydx = 2
x y
0
0
0 2
0
0
=2
3
0
0
y
p
3
1
162
= (3x2 − x5 =
5 0
5
162/5
y = Mx /m =
= 16/5π. The center of mass is (0, 16/5π).
81π/8
x
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166
CHAPTER 14. MULTIPLE INTEGRALS
Z
1
y−y 2
Z
11. Ix =
0
0
0
4
5
6
(y − 2y + y )dy =
=
0
y
(y − y 2 )2 y 2 dy
dy =
x y
1
1
Z
2 2
2xy dxdy =
Z
y−y 2
1
Z
2
0
0
1 5 1 6 1 7
y − y + y
5
3
7
1
0
1
1
=
105
x=y-y2
x
Z
1
√
Z
√
x
x2
0
1
Z
x2 y 2 dydx =
12. Ix =
0
1 2
1
= ( x9/2 − x9 )
3 9
9
1
0
1 2 3
x y
3
2
= k
3
π/2
x2
1
3
1
Z
√
y=x2
0
2Z
0
1 3
y
3
cos x
dx =
0
2
k
3
1
2
cos x(1 − sin x)dx = k(sin x − sin3 x)
3
3
√
4−x2
y 3 dydx =
14. Ix =
Z
4
Z
√
Z
Z
2
15. Iy =
4
x ydxdy =
0
0
1
=
3
2
√
y
0
2 7/2
y
7
4
0
1 3
x y
3
x
y
π/2
Z
π/2
4−x2
1
cos3 xdx
0
π/2
4
= k.
9
Z
y
dy =
0
1
3
Z
4
y 3/2 ydy =
0
y=cos x
0
π/2
1 2
1 4
dx =
y
(4 − x2 )2 dx
4 0
0
0 4
0
0
2
Z
8
1
1 2
1
16x − x3 + x5
=
(16 − 8x2 + x4 )dx =
4 0
4
3
5
0
1
64 32
2 1
64
=
32 −
+
=8 1− +
=
4
3
5
3 5
15
Z
y=Mx
1
2
0
y
(x7/2 − x8 )dx
1
0
Z
dx =
1
=
27
13. Using symmetry,
Z π/2 Z cos x
Z
2
Ix = 2
ky dydx = 2k
0
x
1
3
Z
0
y
2
y=M4-x2
2
4
Z
1
Z
16. Iy =
0
x
y
y 5/2 dy
4
2 7/2
256
=
(4 ) =
21
21
x=My
2
√
x
x
x4 dydx =
x2
2
1
= ( x11/2 − x7 )
11
7
Z
√
x
1
x4 y
0
1
0
3
=
77
Z
dx =
x2
1
x
y
(x9/2 − x6 )dx
y=x2
0
y=Mx
1
1
x
14.4. CENTER OF MASS AND MOMENTS
1
Z
Z
17. Iy =
3
y
0
y
3
1
Z
(4x3 + 3x2 y)dxdy =
167
(x4 + x3 y) dy
0
x=y
1
y
1
Z
3
(81 + 27y − 2y 4 )dy
=
x
0
=
81y +
27 2 2 5
y − y
2
5
1
=
0
941
10
y
18. The density is ρ = ky. Using symmetry,
Z 1 Z 1−x2
Z 1
Z 1
1−x2
1 2 2
2
Iy = 2
kx ydydx = 2
dx = k
x2 (1 − x2 )2 dx
kx y
0
0
0
0 2
0
Z 1
1
8k
1
2
1
=k
(x2 − 2x4 + x6 )dx = k( x3 − x5 + x7 ) =
.
3
5
7
105
0
0
19. Using symmetry,
Z a Z √a2 −y2
Z
m=2
xdxdy = 2
0
0
√
a
0
1 2
x
2
1
y=1-x2
x
1
y
a2 −y 2
a
Z
(a2 − y 2 )dy
dy =
0
0
y=Ma2-x2
a
a
2
1
= (a2 y − y 3 ) = a3 .
3
3
0
Z Z √ 2 2
a y
a
x3 dxdy = 2
Iy = 2
0
0
√
Z
0
a
1 4
x
4
dy =
0
1
2
Z
a
(a2 − y 2 )2 dy
0
Z
1 a 4
2
1
1
=
(a − 2a2 y 2 + y 4 )dy = (a4 y − a2 y 3 + y 5 )
2 0
2
3
5
s
r
r
Iy
4a5 /15
2
Rg =
=
=
a
3
m
2a /3
5
a
=
0
4 5
a
15
Ra
Ra
a−x
kdydx = 0 ky 0 dx = k 0 (a − x)dx =
a
1
1
k(ax − x2 ) = ka2
2
2
0
Z a Z a−x
Z a
Z a
a−x
1 3
1
2
ky dydx =
ky
(a − x)3 dx
Ix =
dx = k
3 0
0
0
0 3
0
Z a
1
1
3
1
= k
(a3 − 3a2 x − x3 )dx = k(a3 x − a2 x2 + ax3 − x4 )
3 0 s
3
2
4
r
r
4
Ix
ka /12
1
Rg =
=
=
a
m
ka2 /2
6
20. m =
R a R a−x
0
y
0
21. (a) Using symmetry,
x
a
a2 −y 2
a
y=a-x
a
a
0
1 4
=
ka
12
x
168
CHAPTER 14. MULTIPLE INTEGRALS
√
b a2 −x2 /a
Z
4b3 a 2
(a − x2 )3/2 dx x = a sin θ, dx = a cos θdθ
Ix = 4 0
y dydx = 3
3a 0
0
Z π/2
Z π/2
4
1
4
= ab3
cos4 θdθ = ab3
(1 + cos 2θ)2 dθ
3
3
4
0
0
Z
π/2
1 3 π/2
1 1
1
3
1
1
= ab
(1 + cos 2θ + + cos 4θ)dθ = ab3 ( θ + sin 2θ + sin 4θ)
3
2 2
3
2
2
8
0
0
Z
a
Z
2
ab3 π
.
4
(b) Using symmetry,
Z a Z b√a2 −x2 /a
Z
4b a 2 p 2
x2 dydx =
x a − x2 dx x = a sin θ, dx = a cos θdθ
Iy = 4
a 0
0
0
Z π/2
Z π/2
1
sin2 θ cos2 θdθ = 4a3 b
= 4a3 b
(1 − cos2 2θ)dθ
4
0
0
Z π/2
π/2
a3 bπ
1
1
1
1
=
= a3 b
.
(1 − − cos 4θ)dθ = a3 b( θ − sin 4θ)
2 2
2
8
4
0
0
p
1p 3
1
(c) Using m = πab, Rg = Ix /m =
ab π/πab = b.
2
2
p
1p 3
1
(d) Rg = Iy /m =
a bπ/πab = a
2
2
=
22. The equation of the ellipse is 9x2 /a2 + 4y 2 /b2 = 1 and the equation of the parabola is
y = ±(9bx2 /8a2 − b/2). Letting Ie and Ip represent the moments of inertia of the ellipse and
parabola, respectively,
about the x-axis, we have
Z 0 Z b√a2 −9x2 2a
Z 0
b3
a
a
2
Ie = 2
y dydx =
(a2 − 9x2 )3/2 dx x = sin θ, dx = cos θdθ
3
12a
3
3
−a/3 0
−a/3
Z
0
b3 a4
b3 a 3π
ab3 π
=
cos4 θdθ =
=
3
12a 3 −π/3
36 16
192
and Z
3
Z
2a/3 Z b/2−9bx2 /8a2
9b
2 2a/3 b
− 2 x2 dx
Ip = 2
y 2 dydx =
3 0
2 8a
0
0
3
3 Z 2a/3
3 Z 2a/3
2b
b
243 4
729 6
9 2
27 2
=
dx =
x −
x dx
1 − 2x
1 − 2x +
3 8 0
4a
12 0
4a
16a4
64a6
2a/3
b3
9 3
243 5
729 7
8ab3
b3 32a
=
x − 2x +
x
−
x
=
=
.
12
4a
80a4
64a6
12 105
315
0
ab3 π 8ab3
Then Ix = Ie + Ip =
+
.
192
315
www.elsolucionario.org
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES
1
1 4
23. From Problem 20, m = ka2 and Ix =
ka .
2 Z
12
Z
Z a Z a−x
a−x
a
dx = k
kx2 y
kx2 dydx =
Iy =
0
0
0
0
169
y
a
a
x2 (a − x)dx
y=a-x
0
4
1 3 1 4
1 4
=
ax − x
ka
3
4
12
0
1 4
1
1
I0 = Ix + Iy =
ka + ka4 = ka4
12
12
6
=k
a
x
1
3
1
3
24. From Problem 12, Ix =
, and from Problem 16, Iy =
. Thus, I0 = Ix + Iy =
+
=
27
77
27 77
158
.
2079
25. The density is ρ = k/(x2 + y 2 ). Using symmetry,
Z √2 Z 6−y2
Z √2
k
2
2
I0 = 2
(x + y ) 2
dxdy = 2
kx
x + y2
y 2 +2
0
0
y
6−y 2
2
y 2 +2
√2
2
= 2k
(6 − y 2 − y 2 − 2)dy = 2k 4y − y 3
3
0
0
√
8√
16 2
= 2k
k.
2 =
3
3
4
Z 3Z 4
Z 3
1 3
dy
26. I0 =
k(x2 + y 2 )dxdy = k
x + xy 2
3
0
y
0
y
Z 3
1
64
=k
+ 4y 2 − y 3 − y 3 dy
3
3
0
3
64
4 3 1 4
y+ y − y
= 73k
=k
3
3
3
0
Z
x=6-y2
dy
6
x=y2+2
√
2
x
y
3
x=y
4
x
1
1
27. From Problem 20, m = ka2 , and from Problem 21, I0 = ka4 .
2
6
s
r
4
p
ka /6
1
Then Rg = I0 /m =
=
a.
ka2 /2
3
28. Since the plate is homogeneous, the density is ρ = m/lw. Using symmetry,
w/2
Z l/2 Z w/2
Z
m 2
4m l/2
1
I0 = 4
(x + y 2 )dydx =
x2 y + y 3
dx
lw
lw 0
3
0
0
0
l/2
Z
4m l/2 w 2 w3
4m w 3 w3
4m wl3
lw3
l2 + w 2
x +
dx =
x +
x
+
=m
.
=
=
lw 0
2
24
lw 6
24
lw
48
48
12
0
14.5
Double Integrals in Polar Coordinates
170
CHAPTER 14. MULTIPLE INTEGRALS
1. Using symmetry,
Z
Z π/2 Z 3+3 sin θ
rdrdθ = 2
A=2
−π/2
Z
−π/2
0
π/2
9(1 + sin θ)2 dθ = 9
=
6
π/2
Z
−π/2
1 2
r
2
3+3 sin θ
dθ
0
π/x
(1 + 2 sin θ + sin2 θ)dθ
1
= 9 θ − 2 cos θ + θ −
2
3π 3
π
=9
−
=
−
22
2
2
1
sin 2θ
4
0
0
π
π/6
1 2
r
2
2 sin θ
Z
0
π/2
dθ + 2
5. Using symmetry,
Z π/6 Z 5 cos 3θ
Z
V =2
4rdrdθ = 4
0
= 100
1
1
θ+
sin 6θ
2
12
π/6
=
0
5 cos 3θ
π/6
r2
0
2
2
1
Z π/6
Z π/2
1 2
r dθ =
4 sin2 θdθ +
dθ
π/6 2
0
0
π/6
0
0
√
π π π √3 π
4π − 3 3
π/6
−
= −
+ =
= (2θ − sin 2θ)|0 +
2
6
3
2
3
6
8
sin
4θ
Z
Z π/4 Z 8 sin 4θ
Z π/4
1 π/4
1 2
dθ =
r
4. A =
64 sin2 4θdθ
rdrdθ =
2
2 0
0
0
0
0
π/4
1
1
= 4π
= 32
θ−
sin 8θ
2
16
0
π/6
polar
axis
−π/2
3. Solving r = 2 sin θ and r = 1, we obtain sin θ = 1/2 or
θ = π/6. Using symmetry,
Z π/6 Z 2 sin θ
Z π/2 Z 1
A=2
rdrdθ + 2
rdrdθ
0
3
π/2
Z π
2+cos θ
1 2
dθ =
(2 + cos θ)2 dθ
r
0
0
0 2
0
0
π
Z π
1
1
2
=
(4 + 4 cos θ + cos θ)dθ = 4θ + 4 sin θ + θ + cos 2θ
2
4
0
0
π 1
1
9π
= 4π + +
−
=
.
2
4
4
2
=2
polar
axis
27π
2
2. Using symmetry,
Z
Z π Z 2+cos θ
rdrdθ = 2
A=2
Z
3
−π/2
25π
3
Z
dθ = 4
0
0
1 polar
axis
polar
axis
π/6
25 cos2 3θdθ
polar
axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES
2π
Z
2
Z
0
2π
0
1
=−
3
Z
0
2π
Z
3
Z
=−
1
3
Z
2
8. V =
1
2π
Z
5
π
√
r2 rdrdθ
0
2π
=
0
Z
=
0
2π
Z
9. V =
3
1
− (16 − r2 )3/2 dθ
3
1
√
√
1
2π(15 15 − 7 7)
(153/2 − 73/2 )2π =
3
3
5
1+cos θ
polar
axis
π/2
Z
0
0
1 3
r sin θ
3
π/2
1+cos θ
dθ
Z
1
0
1
1
4
− (1 + cos θ)
(1 + cos θ) sin θdθ =
3
4
0
1
5
= − (1 − 24 ) =
12
4
1
=
3
polar
axis
1 3
r dθ
3 0
(r sin θ)rdrdθ =
0
polar
axis
125
250π
dθ =
3
3
π/2
Z
(73/2 − 153/2 )dθ =
Z
2π
0
0
0
Z
0
2
1
− (9 − r2 )3/2 dθ
3
0
√
1
2π(27 − 5 5)
3/2
3/2
(5 − 27)dθ = (27 − 5 )2π =
3
3
Z
p
16 − r2 rdrdθ =
7. V =
0
2π
Z
p
9 − r2 rdrdθ =
6. V =
171
π/2
3
10. Using symmetry,
Z π/2 Z cos θ
Z
2
V =2
(2 + r )rdrdθ =
2
0
cos θ
1 4
r + r
dθ
4
0
0
0
0
Z π/2 Z π/2 1
1 1 + cos 2θ 2
2
4
2
=2
cos θ + cos θ dθ = 2
) dθ
cos θ + (
4
4
2
0
0
Z π/2 1 1
1
=
2 cos2 θ + + cos 2θ + cos2 2θ dθ
8 4
8
0
π/2
1
1
1
1
1
19π
= θ + sin 2θ + θ + sin 2θ + θ +
sin 4θ
=
.
2
8
8
16
64
32
0
π/2
polar
axis
2
1
polar
axis
www.elsolucionario.org
172
CHAPTER 14. MULTIPLE INTEGRALS
R π/2 1 2 3
1 R π/2
r dθ = k 0 8dθ = 2kπ
0
1
0
2 1
2
3
Z π/2 Z 3
Z π/2
Z π/2 Z 3
1 3
r2 cos θdrdθ = k
kxrdrdθ = k
My =
r cos θ dθ
3 polar
3
1
1
0
0
0
1
axis
Z π/2
π/2
1
26
26
= k
=
26 cos θdθ =
k sin θ
k
3 0
3
3
0
26k/3
13
x = My /m =
=
.
2kπ
3π
Since the region and density function are symmetric about the ray θ = π/4, y = x = 13/3π and
the center of mass is (13/3π, 13/3π).
11. m =
R π/2 R 3
krdrdθ = k
12. The interior of the upper-half circle is traced from θ = 0
to π/2. The density is kr. Since both the region and the
density are symmetric about the polar axis, y = 0.
Z
π/2
Z
cos θ
2
m=
Z
kr drdθ = k
0
0
0
π/2
1 3
r
3
cos θ
0
k
dθ =
3
π/2
2k
2 1
2
=
+ cos θ sin θ
3 3
9
0
Z π/2 Z cos θ
Z
My = k
(r cos θ)(r)(rdrdθ) = k
k
=
3
Z
1
1
π/2
polar
axis
3
cos θdθ
0
0
0
0
π/2
Z
cos θ
r3 cos θdrdθ = k
0
Z
0
π/2
1 4
r cos θ
4
cos θ
dθ
0
π/2
2k
2
1
k
3
5
5
=
sin θ − sin θ + sin θ
cos θdθ =
4
3
5
15
0
0
2k/15
Thus, x =
= 3/5 and the center of mass is (3/5, 0).
2k/9
k
=
4
Z
π/2
13. In polar coordinates the line x = 3 becomes r cos θ √
= 3 or
r = 3 sec θ. The angle of inclination of the line y = 3x is
π/3.
3 sec θ
Z π/3 Z
Z π/3
1 4
m=
03 sec θ r2 rdrdθ =
r
dθ
4
0
0
0
Z
Z
81 π/3
81 π/3 4
sec θdθ =
(1 + tan2 θ) sec2 θdθ
=
4 0
4 0
π/3
√
81
1
81 √
81 √
=
(tan θ + tan3 θ)
=
( 3 + 3) =
3
4
3
4
2
0
3
polar
axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES
π/3
Z
3 sec θ
Z
xr2 rdrdθ =
My =
π/3
=
0
π/3
1 5
r cos θ
5
3 sec θ
dθ =
0
3 sec θ
Z
r4 cos θdrdθ
0
0
0
0
Z
Z
173
243
5
Z
π/3
sec5 θ cos θdθ
0
Z
243 π/3 4
243 √
486 √
=
sec θdθ =
3
(2 3) =
5 0
5
5
Z
Z π/3 Z 3 sec θ
Z π/3 Z 3 sec θ
4
2
r sin θdθ =
yr rdrdθ =
Mx =
3 sec θ
π/3
1 5
r sin θ
5
0
0
0
0
0
0
Z
Z
Z π/3
243 π/3 5
243
243
=
sec θ sin θdθ =
0π/3 tan θ sec4 θdθ =
tan θ(1 + tan2 θ) sec2 θdθ
5 0
5
5 0
π/3
Z
243 3 9
243 1
243 π/3
1
729
=
(tan θ + tan3 θ) sec2 θdθ =
=
tan2 θ + tan4 θ
( + )=
5 0
5
2
4
5
2
4
4
0
√
486 3/5
729/4
√
x = My /m =
= 12/5; y = Mx /m = √
=
81 3/2
81 3/2
√
√
3 3/2. The center of mass is (12/5, 3 3/2).
14. Since both the region and the density are symmetric about
the x-axis, y = 0. Using symmetry,
4 cos 2θ
Z π/4 Z 4 cos 2θ
Z π/4
1 2
m=2
r
dθ
krdrdθ = 2k
2
0
0
0
0
Z π/4
π/4
1
1
= 16k
= 2kπ
cos2 2θdθ = 16k( θ + sin 4θ)
2
8
0
0
Z π/4 Z 4 cos 2θ
Z π/4 Z 4 cos 2θ
Z
My = 2
kxrdrdθ = 2k
r2 cos θdrdθ = 2k
0
0
128
k
3
Z
128
=
k
3
Z
=
0
π/4
cos3 2θ cos θdθ =
0
128
k
3
0
Z
0
π/4
(1 − 2 sin2 θ)3 cos θdθ
0
π/4
(1 − 6 sin2 θ + 12 sin4 θ − 8 sin6 θ) cos θdθ
0
π/4
128
12
8
k(sin θ − 2 sin3 θ +
sin5 θ − sin7 θ)
3
5
7
0
√
√
√
√ !
√
128
2
2 3 2
2
1024
=
k
−
+
−
=
2k
3
2
2
10
14
105
√
√
1024 2/105
512 2
=
.
x = My /m =
2kπ
√ 105π
The center of mass is (512 2/105π, 0) or approximately (2.20, 0).
=
polar
axis
π/4
1 3
r cos θ
3
4 cos 2θ
dθ
0
174
CHAPTER 14. MULTIPLE INTEGRALS
2
15. The density is ρ = k/r.
Z π/2 Z 2+2 cos θ
Z π/2 Z 2+2 cos θ
4 polar
k
axis
drdθ
rdrdθ = k
m=
r
2
0
2
0
Z π/2
π/2
2 cos θdθ = 2k(sin θ)|0 = 2k
=k
Z 0π/2 Z 2+2 cos θ
Z
Z π/2 Z 2+2 cos θ
2+2 cos θ
k
1
x rdrdθ = k
My =
r cos θdrdθ = k 0π/2 r2
cos θdθ
r
2 2
2
0
2
0
Z π/2
Z π/2
1
2
(8 cos θ + 4 cos θ) cos θdθ = 2k
(2 cos2 θ + cos θ − sin2 θ cos θ)dθ
= k
2 0
0
π/2
π 2
1
3π + 4
1
= 2k( + ) =
sin 2θ + sin θ − sin3 θ)
k
2
3
2
3
3
0
Z π/2 Z 2+2 cos θ
Z
Z 2+2 cos θ
Z π/2
k
1 2
Mx =
y rdrdθ = k
r sin θdrdθ = k
r
r
2
0
2
π/2 2
0
π/2
Z π/2
1
1
4
2
2
3
= k
(8 cos θ + 4 cos θ) sin θdθ = k −4 cos θ − cos θ
2 0
2
3
0
1
4
8
= k − −4 −
= k
2
3
3
3π + 4
4
(3π + 4)k/3
8k/3
=
; y = Mx /m =
=
x = My /m =
2k
6
2k
3
The center of mass is ((3π + 4)/6, 4/3).
= 2k(θ +
Z
π
Z
16. m =
2+2 cos θ
Z
π
krdrdθ = k
0
Z
= 2k
0
π
0
1 2
r
2
sin θdθ
2
2
2+2 cos θ
Z
dθ = 2k
0
2+2 cos θ
π
2
(1 + cos θ) dθ
0
4 polar
axis
(1 + 2 cos θ + cos2 θ)dθ
0
π
1
1
= 3πk
= 2k θ + 2 sin θ + θ + sin 2θ
2
4
0
Z π Z 2+2 cos θ
Z π Z 2+2 cos θ
Z π
2+2 cos θ
1 3
My =
kxrdrdθ = k
r2 cos θdrdθ = k
r
cos θdθ
0
0
0
0
0 3
0
Z π
Z π
8
8
= k
(1 + cos θ)3 cos θdθ = k
(cos θ + 3 cos2 θ + 3 cos3 θ + cos4 θ)dθ
3 0
3 0
π
8
3
3
3
1
1
= k sin θ +
θ + sin 2θ + (3 sin θ − sin3 θ) +
θ + sin 2θ +
sin 4θ
3
2
4
8
4
32
0
8
15
= k
π = 5πk
3
8
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14.5. DOUBLE INTEGRALS IN POLAR COORDINATES
π
Z
2+2 cos θ
Z
π
Z
2+2 cos θ
Z
r2 sin θdrdθ = k
kyrdrdθ = k
Mx =
0
π
0
175
0
0
π
Z
0
1 3
r
3
2+2 cos θ
sin θdθ
0
Z
Z π
8
8
(1 + cos θ)3 sin θdθ = k
(1 + 3 cos θ + 3 cos2 θ + cos3 θ) sin θdθ
k
3 0
3 0
π
8 1
8
1
15
32
= k
= k − cos θ − 32 cos2 θ − cos3 θ − cos4 θ
− −
=
k
3
4
3
4
4
3
0
5πk
32k/3
= 5/3; y = Mx /m =
= 32/9π. The center of mass is (5/3, 32/9π).
x = My /m =
3πk
3πk
=
2π
Z
a
Z
y 2 krdrdθ = k
17. Ix =
0
0
ka4
=
4
Z
0
Z
2π
=
0
Z
2π
a
ka4
sin2 θdθ =
4
a
r3 sin2 θdrdθ = k
0
2π
0
1
1
θ − sin 2θ
2
4
2π
Z
2π
0
1 4 2
r sin θ
4
1
y
rdrdθ =
1 + r4
2
Z
Z
a
dθ
0
a polar
axis
kπa4
=
4
a
r3
sin2 θdrdθ
4
0
0
0 1+r
a
1
1
1
1
ln(1 + r4 ) sin2 θdθ = ln(1 + a4 )
θ − sin 2θ
4
4
2
4
0
Z
18. Ix =
2π
0
0
2π
Z
Z
a polar
axis
2π
0
π
= ln(1 + a4 )
4
19. Solving a = 2a cos θ, cos θ = 1/2 or θ = π/3. The density
is k/r3 . Using symmetry,
Z π/3 Z 2a cos θ
Z π/3 Z 2a cos θ
2 k
cos2 θdrdθ
Iy = 2
x 3 rdrdθ = 2k
r
0
a
0
a
Z π/3
= 2k
(2a cos3 θ − a cos2 θ)dθ
a
2a
polar
axis
0
π/3
2
1
1
3
= 2ak 2 sin θ − sin θ − θ − sin 2θ
3
2
4
0
√
√ !
√
√
3 π
3
5ak 3 akπ
= 2ak
3−
− −
=
−
4
6
8
4
3
20. Solving 1 = 2 sin 2θ, we obtain sin 2θ = 1/2 or θ = π/12
and θ = 5π/12.
1
polar
axis
176
CHAPTER 14. MULTIPLE INTEGRALS
Z
5π/12
Z
2 sin 2θ
x2 sec2 θrdrdθ =
Iy =
=
π/12
5π/12
5π/12 1 4
4
2 sin 2θ
r
Z
2 sin 2θ
Z
π/12
1
π/12
Z
Z
r3 drdθ
1
5π12
sin4 2θdθ
dθ = 4
π/12
1
5π/12
3
1
1
θ − sin 4θ +
sin 8θ
4
4
32
π/12
"
√
√
√ !
√
√ !#
5π
3
3
π
3
3
8π + 7 3
=2
−
=
+
−
−
+
16
8
64
16
8
64
16
=2
21. From Problem 17, Ix = kπa4 /4. By symmetry, Iy = Ix .
Thus I0 = kπa4 /2.
a
polar
axis
π
θ
θ
π
1 5
r dθ
r (kr)rdrdθ = k
5
0
0
0
0
π
Z π
6
1
kπ
1
1
= k
=
θ6
θ5 dθ = k
5 0
5
6
30
0
Z
Z
22. The density is ρ = kr. I0 =
Z 3 Z 1/r
k
r2 rdθdr = k
f 2 dθdr
r
1
0
1
0
3
Z 3 1 2
1
= 4k
=k
dr = k
r
r2
r
2
1
1
Z
3
Z
2
1
polar
axis
1/r
Z
23. The density is ρ = k/r. I0 =
Z
π
Z
2a cos θ
Z
2a cos θ
π
Z
1 4
dθ = 4ka4
cos4 θdθ
r
4
0
0
0
0
π0
4
3
3π
1
1
3kπa
= 4ka4
= 4ka4
θ + sin 2θ +
sin 4θ
=
8
4
32
8
2
0
r2 krdrdθ = k
24. I0 =
Z
3
√
Z
25.
−3
9−x2
p
x2 + y 2 dydx =
π
Z
Z
03 |r|rdrdθ
3
polar
axis
π
2a
polar
axis
r=3
0
0
Z
=
0
π
3
1 3
r dθ = 9
3 0
Z
π
dθ = 9π
0
3 polar
axis
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES
√
Z
Z √1−y2
2/2
26.
0
y2
p
y
Z
x2 + y 2
π/4
Z
π/4
Z
1
dxdy =
0
0
Z
π/4
Z
=
0
27.
e
0
x2 +y 2
r2 sin2 θdrdθ
√
28.
√
x
Z
=
0
e rdrdθ =
1
2
0
π/2
Z
(e − 1)dθ =
0
√
π
Z
2
Z
0
Z
π−2
24
1 r2
e
2
1
dθ
r=1
0
π(e − 1)
4
1
(sin r2 )rdrdθ
0
0
sin2 θdθ
0
√
π
r=Mπ
x
1
=
− cos r2
dθ
2
0
0
Z
1 π
=−
(−1 − 1)dθ = π
2 0
Z
29.
0
1
√
Z
√
4−x2
1−x2
x2
dydx =
2
x + y2
Z
1
Z
2
Z
√
4−x2
0
π/2 Z 2
=
0
Z
1
π/2
Z
=
0
x2
2
2
r cos θ
rdrdθ
r2
2
r cos2 θdrdθ
1
=
π/2
Mπ
x2
dydx
+ y2
2
Z
1 2
3 π/2
r cos2 θdθ =
cos2 θdθ
2
2
0
0
1
π/2
3 1
1
3π
=
θ + sin 2θ
=
2 2
4
8
0
Z
polar
axis
x
sin(x + y )dydx =
√
− x
π/4
π/2
Z
r2
0
π−x2
2
1
1 3 2
r sin θ dθ =
3
3
polar
axis
π/4
Z
0
1
Z
0
=
1
1
1 1
1
( θ − sin 2θ)
3 2
4
dxdy =
0
Z
π/2
Z
r=1
0
0
Z 1 Z √1−y2
r2 sin2 θ
rdrdθ
|r|
1
=
=
177
polar
axis
r=2
1
2 polar
axis
www.elsolucionario.org
178
CHAPTER 14. MULTIPLE INTEGRALS
Z
1
Z √2y−y2
r=csc θ
2
2
(1 − x − y )dxdy
30.
0
r=2 sin θ
0
π/4
Z
Z
2 sin θ
Z
(1 − r2 )rdrdθ +
=
0
0
π/4
Z
=
0
π/2
2 sin θ
Z
(1 − r2 )rdrdθ
π/2
dθ +
π/4
0
π/4
polar
axis
1
0
π/4
1 2 1 4
r − r
2
4
csc θ
Z
π/2
1 2 1 4
r − r
2
4
csc θ
dθ
0
1
1
csc2 θ − csc4 θ dθ
2
4
0
π/4
1
3
1
1
1
1
= θ − sin 2θ −
θ − sin 2θ + sin 4θ
+ − cot θ − (− cot θ − cot3 θ)
2
2
8
2
4
3
π 1
1
1
16 − 3π
= − +
+ 0− − +
=
8
2
4 12
24
Z
(2 sin2 θ − 4 sin4 θ)dθ +
=
Z
5
√
Z
25−x2
31.
π
Z
(4r cos θ + 3r sin θ)rdrdθ
0
0
5
Z
0
0
π
=
=
0
Z √1−y2
32.
0
0
33. I 2 =
Z
∞
0
Z
=
e−(x
2
+y 2 )
5
dθ
0
500
500
cos θ + 125 sin θ dθ =
sin θ − 125 cos θ
3
3
Z
π/2
∞
Z
0
t→∞
π/2
2
1
1
− e−t +
2
2
0
Z
dθ =
0
2
e−r rdrdθ =
dxdy =
0
lim
0
4 3
r cos θ + r3 sin θ
3
π/2
Z
0
π/2
2
1
lim − e−r
t→∞
2
√
1
π
π
dθ = ; I =
2
4
2
π
= 250
0
1
Z π/2 Z 1
1
1
p
rdrdθ
dxdy =
1
+
r
1 + x2 + y 2
0
0
1
Z π/2 Z 1
Z π/2
1
=
(1 −
)drdθ =
[r − ln(1 + r)] dθ
1+r
0
0
0
0
Z π/2
π
(1 − ln 2)dθ = (1 − ln 2)
=
2
0
∞
Z
polar
axis
0
π
Z
1
5
(4r2 cos θ + 3r2 sin θ)drdθ
=
Z
r=5
0
π
Z
Z
5
5
Z
(4x + 3y)dydx =
−5
Z
r=1
1
t
dθ
0
polar
axis
π/2
π/4
14.5. DOUBLE INTEGRALS IN POLAR COORDINATES
Z Z
π/2
Z
34.
Z
2
π/2
Z
Z
2
r2 (cos θ + sin θ)drdθ
(r cos θ + r sin θ)rdrdθ =
(x + y)dA =
Z
=
0
π/2
2 sin θ
0
2 sin θ
0
R
179
1 3
r (cos θ + sin θ)
3
2
dθ =
2 sin θ
8
3
Z
π/2
(cos θ + sin θ − sin3 θ cos θ − sin4 θ)dθ
0
8
1
1
3
3
= ] sin θ − cos θ − sin4 θ + sin3 θ cos θ − θ +
sin 2θ
3
4
4
8
16
8
1 3π
28 − 3π
=
1− −
− (−1) =
3
4
16
6
π/2
0
35. The volume of the cylindrical portion of the tank is Vc = π(4.2)2 19.3 ≈ 1069.56m3 . We take
the equation of the ellipsoid to be
x2
5.15 p
x2
+
= 1 or z = ±
(4.2)2 − x2 − y 2 .
2
2
(4.2)
(5.15)
4.2
The volume of the ellipsoid is
Z Z
Z
Z
p
5.15
10.3 2π 4.2
2
2
2
Ve = 2
[(4.2)2 − r2 ]1/2 rdrdθ
(4.2) − x − y dxdy =
4.2
4.2 0
R
0
#
"
Z
Z
4.2
10.3 1 2π
10.3 2π
1 2
2
2 3/2
dθ =
=
[(4.2) − r ]
−
(4.2)3 dθ
4.2 0
2 3
4.2
3
0
0
2π 10.3
(4.2)3 ≈ 380.53.
3 4.2
The volume of the tank is approximately 1069.56 + 380.53 = 1450.09m3 .
=
36. (a) With b > 2 we have
Z Z
Z
Z
1 2π R
r
IdA =
dr u = r + c, du = dr
b
2 0
C
0 (r + c)
2
Z R+c
Z R+c
u−c
r −b
r1−b
1−b
−b
= πa
du = πa
−c
(u
− cu )du = πa
ub
2−b
1−b
c
c
2−b
2−b
2
1−b
c
(R + c)
c −b
c(R + c)
= πa
−
− πa
−
b−2
b−1
b−2
b−1
πa
1
c
=
− πa
−
.
(b − 1)(b − 2)cb−2
(b − 2)(R + c)b−2
(b − 1)(R + c)b−1
Z Z
πa
I(r)dA =
(b) lim
R→∞
(b
−
1)(b
− 2)cb−2
C
R+c
c
(c) Identifying a = 68.585, b = 2.351, and c = 0.248 in part b we find that the total number
of infections in the plane is approximately 741.25.
Z Z
Z
37. (a) P =
2π
Z
D(r)dA =
C
R
D0 e
0
0
= 2πD0 (−dre−r/d − d2 er/d )
−r/d
Z
rdrdθ = 2πD0
R
re−r/d dr
0
R
0
= 2πdD0 [d − (R + d)e−R/d ]
180
CHAPTER 14. MULTIPLE INTEGRALS
(b) Using
Z Z
Z
2π
Z
rD(r)dA =
C
R
rD0 e−r/d rdrdθ = 2πD0
0
0
Z
R
r2 e−r/d dr
0
= 2πD0 (−2d3 e−r/d − 2d2 re−r/d − dr2 e−r/d )
i
h
= 2πdD0 2d2 − (R2 + 2dR + 2d2 )e−R/d
R
0
we have
RR
rD(r)dA
2d2 − (R2 + 2dR + 2d2 )e−R/d
R RC
=
D(r)dA
d − (R + d)e−R/d
C
(c) Letting R −→ ∞ in the result of parts (a) and (b) we find that the total population is
2πd2 D0 and the average commute for the total population is 2d2 /d = 2d.
38. In the first case, let the circle centered at (D/2, 0) be described by the equation r = D cos θ
for −π/2 ≤ θ ≤ π/2 and assume that the snow is plowed to the origin. Then
Z Z
Z
π/2
D cos θ
Z
r2 drdθ =
rdA =
−π/2
R
0
3
2D
1
sin θ − sin3 θ
=
3
3
D3
3
Z
(1 − sin2 θ) cos θdθ
0
π/2
=
0
π/2
4D3
.
9
In the second case, let the circle centered at the origin be described by the equation r = D/2
for 0 ≤ θ ≤ 2π, and assume the snow is plowed to the origin. Then
Z Z
Z
2π
Z
rdA =
R
0
0
D/2
r2 drdθ =
2π 3
r
3
D/2
=
0
πD3
.
12
3
4D /9
16
=
≈ 1.698, which means that plowing snow to one
πD3 /12
3π
point on the perimeter is approximately 69.8% more costly than plowing to the center.
The ratio of these integrals is
14.6
Surface Area
1. Letting z = 0, we have 2x + 3y = 12. Using f (x, y) = z =
1
3
1
3
29
3− x− y we have fx = − , fy = − , 1+fx2 +fy2 =
.
2
4
2
4
16
Then
√ Z 6
Z 6 Z 4−2x/3 p
29
2
A=
29/16dydx =
(4 − x)dx
4
3
0
0
0
√
√
6
√
29
1
29
=
(4x − x2 ) =
(24 − 12) = 3 29.
4
3
4
0
y
4
y=4-2x/3
6 x
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14.6. SURFACE AREA
181
2. We see from the graph in Problem 1 that the plane is
entirely above the region bounded by r = sin 2θ in the
3
1
first octant. Using f (x, y) = z = 3 − x − y we have
2
4
1
3
29
2
2
fx = − , fy = − , 1 + fx + fy =
. Then
2
4
16
√
sin 2θ
Z
Z π/2 Z sin 2θ p
29 π/2 1 2
dθ
A=
29/16rdrdθ =
r
4
2
0
0
0
√
29
8
=
√
π/2
Z
2
sin 2θdθ =
0
29 1
1
( θ − sin 4θ)
8 2
8
0
π/2
r=sin 2 θ
1
polar
axis
√
29π
.
32
=
0
√
3. Using f (x, y) = z = 16 − x2 we see that for 0 ≤ x ≤ 2
and 0 ≤ y ≤ 5, z.0. Thus, the surface is entirely above the
x
, fy = 0, 1 + fx2 + fy2 =
region. Now fx = − √
16 − x2
x2
16
1+
=
and
2
16 − x
16 − x2
Z 5Z 2
Z 5
Z 5
2
4
π
10π
−1 x
√
A=
dxdy = 4
sin
dy = 4
dy =
.
2
4
6
3
16 − x
0
0
0
0
0
y
5
x=2
2
4. The region in the xy-plane beneath the surface is bounded
by the graph of x2 + y 2 = 2. Using f (x, y) = z = x2 + y 2
we have fx = 2x, fy = 2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ).
Then,
√
2π
Z
Z
√
2
p
A=
0
=
1
12
1+
4r2 rdrdθ
0
2π
Z
2π
Z
=
0
(27 − 1)dθ =
0
1
(1 + 4r2 )3/2
12
2π
Z
2
A=
0
=
1
12
Z
0
p
1+
r=M2
M2 polar
axis
2
dθ
0
13π
.
3
5. Letting z = 0 we have x2 + y 2 = 4. Using
f (x, y) = z = 4 − (x2 + y 2 ) we have fx = −2x, fy =
−2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ). Then
Z
x
4r2 rdrdθ
Z
=
0
0
2π
(173/2 − 1)dθ =
2π
2
1
(1 + 4r2 )3/2 dθ
3
0
π
(173/2 − 1).
6
r=2
2 polar
axis
182
CHAPTER 14. MULTIPLE INTEGRALS
6. The surfaces x2 + y 2 + z 2 = 2 and z 2 = x2 + y 2 intersect
on the cylinder 2x2 + 2y 2 = 2 or x2 + y 2 = 1. There are
portions of the sphere within the cone
p both above and below the xy-plane. Using f (x, y) = 2 − x2 − y 2 we have
x
y
fx = − p
, fy = − p
, 1 + fx2 + fy2 =
2
2
2
2
2−x −y
2−x −y
2
. Then
2 − x2 − y 2
"Z
#
√
1
2π Z 1
√ Z 2π p
2
√
A=2
rdrdθ = 2 2
− 2 − r2 dθ
2 − r2
0
0
0
0
Z
2π
√
√
√ √
=2 2
( 2 − 1)dθ = 4π 2( 2 − 1).
r=1
1 polar
axis
0
p
7. Using f (x, y) = z = 25 − x2 − y 2 we have
x
y
fx = − p
, fy = − p
,
2
2
25 − x − y
25 − x2 − y 2
25
. Then
1 + fx2 + fy2 =
25 − x2 − y 2
Z 5 Z √25−y2 /2
5
p
A=
dxdy
25
−
x2 − y 2
0
0
√
25−y 2 /2
Z 5
Z 5
π
25π
x
−1
=5
dy = 5
dy =
.
sin p
2
6
6
25 − y 0
0
0
8. In the first octant, the graph of z = x2 −y 2 intersects the xyplane in the line y = x. The surface is in the first octant for
x > y. Using f (x, y) = z = x2 − y 2 we have fx = 2x, fy =
−2y, 1 + fx2 + fy2 = 1 + 4x2 + 4y 2 . Then
Z
π/4
Z
2
A=
0
1
=
12
0
Z
0
π/4
Z
p
2
1 + 4r rdrdθ =
0
π/4
2
1
(1 + 4r2 )3/2 dθ
12
y
5
x=M25-y2/2
3
x
r=2
2 polar
axis
0
π
(173/2 − 1)dθ =
(173/2 − 1).
48
9. There are portions of the sphere within the cylinder both
above and below the xy-plane. Using f (x, y) = z =
p
x
a2 − x2 − y 2 we have fx = − p
, fy =
2
1 − x2 − y 2
a2
y
, 1 + fx2 + fy2 = 2
−p
. Then, using
a − x2 − y 2
a2 − x2 − y 2
symmetry,
r=a sin θ
a
polar
axis
14.6. SURFACE AREA
" Z
A=2 2
183
#
Z π/2 p
a
√
− a2 − r2
rdrdθ = 4a
a2 − r2
0
0
0
Z π/2
Z π/2
p
2
2
= 4a
(a − a 1 − sin θ)dθ = 4a
(1 − cos θ)dθ
π/2
a sin θ
Z
0
a sin θ
dθ
0
0
2
= 4a (θ − sin θ)
π/2
0
π
= 4a ( − 1) = 2a2 (π − 2).
2
2
10. There are portions of the cone within the cylinder both above and
1p 2
below the xy-plane. Using f (x, y) =
x + y 2 , we have fx =
2
y
5
x
p
, fy = p
, 1 + fx2 + fy2 = . Then, using
2
2
2
2
4
2 x +y
2 x +y
symmetry,
" Z
#
r
2 cos θ
π/2 Z 2 cos θ
√ Z π/2 1 2
5
dθ
A=2 2
rdrdθ = 2 5
r
4
2
0
0
0
0
π/2
√ Z π/2
√
√
1
1
2
=4 5
= 5π.
cos θdθ = 4 5
θ + sin 2θ
2
4
0
0
11. There are portions of the surface in each octant with areas equal
p to the
area of the portion in the first octant. Using f (x, y) = z = a2 − y 2
y
a2
. Then
we have fx = 0, fy = p
, 1 + fx2 + fy2 = 2
a − y2
a2 − y 2
Z a Z √a2 −y2
A=8
0
a2 − y 2
√ 2 2
0
Z
= 8a
0
a
a
p
p
a2 − y 2
polar
axis
y
x=Ma2-y2
a
x
a
Z
dy = 8a2 .
dy = 8a
0
2
dxdy
a −y
x
r=2 cos θ
0
12. √
From Example 1, the area of the portion of the hemisphere √with x2 + y 2 = b2 is 2πa(a −
a2 − b2 ). Thus, the area of the sphere is A = 2 lim 2πa(a − a2 − b2 ) = 2(2πa2 ) = 4πa.
b→a
13. The projection of the surface onto √
the xy-plane is shown
in the graph. Using f (x, z) = y = a2 − x2 − z 2 we have
x
z
fx = − √
, fz = − √
, 1 + fx2 +
2
2
2
2
a −x −z
a − x2 − z 2
a2
fz2 = 2
. Then
a − x2 − z 2
√ 2 2
Z 2π Z √a2 −c21
Z 2π p
a −C1
a
2
2
A=
dθ
rdrdθ = a
− a −r √
√ 2 2 √ 2
a − r2
a −c2
0
0
a2 −c22
Z 2π
=a
(c2 − c1 )dθ = 2πa(c2 − c1 ).
0
z
r=Ma2-c12
r=Ma2-c22
x
www.elsolucionario.org
184
CHAPTER 14. MULTIPLE INTEGRALS
14. The surface area of the cylinder x2 + z 2 = a2 from y = c1 to y = c2 is the area of a cylinder
of radius a and height c2 − c1 . This is 2πa(c2 − c1 ).
15. The equations of the spheres are x2 + y 2 + z 2 = a2 and x2 + y 2 + (z + a)2 = 1. Subtracting
these equations, we obtain(z − a)2 − z 2 = 1 − a2 or −2az + a2 = 1 − a2 . Thus, the spheres
intersect on the plane z = a − 1/2a. The region of integration is x2 + y 2 + (a − 1/2a)2 = a2
or r2 = 1 − 1/4a2 . The area is
Z
2π
Z √1−1/4a2
√
2 −1/r
2
2
(a − r )
2 1/2
1−1/4a2
rdrdθ = 2πa[−(a − r ) ]
0
0
0


"
1/2 !
2 #1/2
1
1
 = π.
= 2πa a −
a−
= 2πa a − a2 − 1 − 2
4a
2a
A=a
16.
(a) Both states span 7 degrees of longitude and 4 degrees of latitude, but Colorado is larger
because it lies to the south of Wyoming. Lines of longitude converge as they go north, so the
east-west dimensions of Wyoming are shorter than those of Colorado.
p
(b) We use the function f (x, y) = R2 − x2 − y 2 to describe the northern hemisphere, where R ≈ 3960 miles
is the radius of the Earth. We need to compute the
surface area over a polar rectangle P of the form
θ1 ≤ θ ≤ θ2 , R cos φ2 ≤ r ≤ R cos φ1 . We have
−y
−x
fx = p
and fy = p
2
2
2
2
2
2
R −x −y
s R −x −y
q
x2 + y 2
so that
1 + fx2 + fy2 =
1+ 2
=
R − x2 − y 2
R
√
.
2
R − r2
Thus
A=
Z Z q
1+
fx2
+
fy2 dA
P
= (θ2 − θ1 )R
p
R2 − r 2
Z
θ2
Z
R cos φ1
=
θ1
R cos φ2
R cos φ1
R cos φ2
√
φ2
φ1
R
R
θ2
θ1
R
rdrdθ
− r2
R2
= (θ2 − θ1 )R2 (sin φ2 − sin φ1 ).
The ratio of Wyoming to Colorado is then
sin 45◦ − sin 41◦
≈ 0.941. Thus Wyoming is about 6%
sin 41◦ − sin 37◦
smaller than Colorado.
(c) 97,914/104,247 ≈ 0.939, which is close to the theoretical value of 0.941. (Our formula for the
area says that the area of Colorado is approximately 103,924 square miles, while the area of
Wyoming is approximately 97,801 square miles.
14.7. THE TRIPLE INTEGRAL
14.7
1.
185
The Triple Integral
1
R4R2 1 2
dydz
(
x
+
xy
+
xz)
2 −2
2 −2 2
−1
2
R4
R4R2
R4
4
dz = 2 8zdz = 4z 2 2 = 48
= 2 −2 (2y + 2z)dydz = 2 (y 2 + 2yz)
R4R2 R1
(x + y + z)dxdydz =
−1
−2
Z
3
x
Z
xy
Z
3
Z
xy
x
Z
1
1
2
1
1
1
x
(8x2 y 3 − 24xy 2 )
1
=
Z
6
6−x
Z
6−x−z
Z
3.
0
1
Z
√
4.
0
0
y
1
Z
Z
2 3
0
1
Z
0
1
=
3
y2
y
Z
5.
0
0
0
6−x
6
1 2 3
x y
3
1
Z
Z
dydx =
x z
0
Z
y
2 4
4x z dzdydx =
0
π/2
1−x
Z
=
Z
1
14
1552
=
3
3
1
dx
(6z − xz − z 2 )
2
0
0
0
0
Z 6
Z 6
1
1
2
=
6(6 − x) − x(6 − x) − (6 − x) dx =
(18 − 6x + x2 )dx
2
2
0
0
6
1
= 36
= 18x − 3x2 + x3
6
0
0
1−x
Z
= 522 −
(6 − x − z)dzdx =
√
Z
6−x
Z
dydzdx =
0
(8x5 − 24x3 − 8x2 + 24x)dx
1
3
1
6
3
Z
dx =
4 6
8
x − 6x4 − x3 + 12x2
3
3
Z
(24x2 y 2 − 48xy)dydx
1
2
3
Z
=
x
Z
3
dydx =
24xyz
24xydzdydx =
2.
Z
0
0
1−x
dx =
0
1
3
Z
1−x
x2 y 2 dydx
0
1
x2 (1 − x)3 dx =
0
1 3 3 4 3 5 1 6
x − x + x − x
3
4
5
6
1
0
1
3
Z
1
(x2 − 3x3 + 3x4 − x5 )dx
0
1
=
180
Z π/2 Z y2
Z π/2
x
x
x
cos dzdxdy =
y cos dxdy =
y 2 sin
y
y
y
0
0
0
Z π/2
=
y 2 sin ydy Integration by parts
y2
dy
0
0
= (−y 2 cos y + 2 cos y + 2y sin y)
Z
6.
0
√
2
Z
2
√
√
2
Z
ex
Z
2
Z
xdzdxdy =
y
0
√
0
√
=
1
(ye4 − ey )
2
2
2
=π−2
√
x2
Z
xe dxdy =
y
=
0
π/2
0
0
2
2
1 x2
e
2
(e4 − ey )dy
√
y
√
√
√
1 4√
1
[(e 2 − e 2 ) − (−1)] = (1 + e4 2 − e 2 )
2
2
186
CHAPTER 14. MULTIPLE INTEGRALS
Z
1
Z
1
Z
2−x2 −y 2
1
Z
z
z
xye
xye dzdxdy =
7.
0
0
2−x2 y 2
1
Z
0
0
0
Z
1
1
Z
−y 2
− xy)dxdy
0
0
0
2
(xye2−x
dxdy =
1
Z 1
1 2−x2 −y2 1 2
1 1−y2 1
1 2−y2
=
− ye
dy =
− ye
− x y
− y + ye
dy
2
2
2
2
2
0
0
0
1 1 1 1
1
1 2
1 1−y2 1 2 1 2−y2
1
1
e
=
− − e) − ( e − e = e2 − e
=
− y − e
4
4
4
4
4
4
4
4
4
2
0
Z
1
R 4 R 1/2 R x2
1
sin−1
y
x
0
1
x2
R 4 R 1/2
dxdz = 0 0 sin−1 xdxdz
0
−
!
√
1/2
√
√
R
R4
1π
3
π
4
−1
2
dz = 0
= 0 x sin x + 1 − x
+
− 1 dz = + 2 3 − 4
26
2
3
0
Z 5Z 3
Z Z Z
Z 5 Z 3 Z y+2
2xdydz
zdxdydz =
9.
zdV =
8.
0
0
0
p
x2
y2
1
0
D
Z
=
5
R 4 R 1/2
dydxdz =
y
3
0
0
5
Z
4zdz = 2z 2
2yz dz =
0
0
1
Integration by parts
z
5
5
0
= 50
3
y
x=y
3
x=y+2
x
10. Using
Z Z Zsymmetry,
Z
(x2 + y 2 )dV = 2
D
2
Z
4
z
4−y
Z
2
4
2
(x + y )dzdydx
x2
0
Z
2
Z
0
4−y
4
2
=2
2
dydx
(x + y )z
0
Z 2
x2
Z 4
0
2
2
2
(4x − x y + 4y − y )dydx
=2
=2
0
y
y=x2
x2
0
Z
4
2
x
3
2
4
1
1
(4x y − x2 y 2 + y 3 − y 4 )
2
3
4
4
2
dx
x2
8
64
4
5
1
= 2( x3 + x − x5 − x7 + x9 )
3
3
5
42
36
2
=
0
11. The other five integrals are
R 4 R 2−x/2 R 4
R 4 R z R (z−x)/2
f (x, y, z)dydxdz,
f (x, y, z)dzdydx, 0 0 0
0 0
x+2y
R 4 R z/2 R z−2y
R 4 R 4 R (z−x)/2
f (x, y, z)dydzdx, 0 0
f (x, y, z)dxdydz,
0
R02 Rx4 R0 z−2y
f (x, y, z)dxdzdy.
0 2y 0
23, 552
.
315
z
4
z=2y
z=x
2
x+2y=4
4
x
y
www.elsolucionario.org
14.7. THE TRIPLE INTEGRAL
187
12. The √
other five integrals are
R 3 R 36−4y2 /3 R 3
f (x, y, z)dzdxdy,
R03 R02 R √36−9x2 /21
f (x, y, z)dydxdz,
1 0 0
R 3 R 3 R √36−4y2 /3
f (x, y, z)dxdydz,
1 0 0
R 3 R 3 R √36−4y2 /3
f (x, y, z)dxdzdy,
R02 R13 R0√36−9x2 /2
f (x, y, z)dydzdx.
0 1 0
z
3
y=M36-9x2 /2
3
2
R2R8 R4
R 8 R 4 R y1/3
13. (a) V = 0 x3 0 dzdydx (b) V = 0 0 0
dxdz
R4R2R8
(c) V = 0 0 x2 dydxdz
√
14. Solving z = x and x + z = 2, we obtain x = 1, z = 1.
R 3 R 1 R 2−z
R 1 R 2−z R 3
(a) V = 0 0 z2 dxdzdy (b) V = 0 z2
dydxdz
0
R 3 R 1 R √x
R 3 R 2 R 2−x
(c)V = 0 0 0 dzdxdy + 0 1 0 dzdxdy
15.
x=M
x
16.
z
y
36-4y2 /3
z
5
3
x=2-2z/3
4
2
y
3
x
y
x=M9-y2
3
x
The region in the first octant is
shown.
17.
18.
z
z
6
4
y=-M1-x2
2
y=M1-x2
2
x
y
2
2
y=M
4-x2
x
y
188
CHAPTER 14. MULTIPLE INTEGRALS
19.
20.
z
z
1
2
3
y
3
2
1
y
x
x
√
21. Solving x = y 2 and 4 − x = y 2 , we obtain x = 2, y = ± 2.
Using symmetry,
√
3
Z
Z
2
4−y 2
Z
V =2
√
3
Z
Z
2
dxdydz = 2
0
y2
0
3
Z
√
2 3
4y − y
3
=2
0
3
Z
dz = 2
0
0
(4 − 2y 2 )dydz
0
0
2
z
5
√
√
8 2
dz = 16 2.
3
4
x=y2
y
x=4-y2
5
x
2
Z
√
Z
4−x2
Z
x+y
Z
2
Z
√
4−x2
dzdydx =
22. V =
0
0
Z
2Z
0
√
0
4−x2
Z
x+y
0
0
2
z
dydx
z
√
2
4−x2
1
dx
xy + y 2
2
0
0
0
0
Z 2 p
1
1
1
=
x 4 − x2 + (4 − x2 ) dx = − (4 − x2 )3/2 + 2x − x3
2
3
6
0
4
8
16
= 4−
− −
=
3
3
3
=
(x + y)dydx =
23. Adding the two equations, we obtain 2y = 8. Thus, the
paraboloids intersect in the plane y = 4. Their intersection
is a circle of √
radius 2. Using symmetry,
Z 2 Z 4−x2 Z 8−x2 −z2
V =4
dydzdx
Z
2
√
Z
=4
0
Z
0
2
4−x2
(8 − 2x2 − 2x2 )dzdx = 4
0
2
x
0
2
2
2(4 − x2 )z − z 3
3
4
(4 − x2 )3/2 dx Trig substitution
0 3
2
p
16 h x
xi
=
− (2x2 − 20) 4 − x2 + 6 sin−1
= 16π.
3
8
2 0
=4
y
y=M4-x2
z=M4-x2
8 y
2
x
√
Z
2
z
x2 +z 2
0
0
2
2
4−x2
dx
0
14.7. THE TRIPLE INTEGRAL
189
24. Solving x = 2, y = x, and z = x2 + y 2 , we obtain the point
(2,2,8).
Z 2Z x
Z 2 Z x Z x2 +y2
(x2 + y 2 )dydx
dzdydx =
V =
0
0
2
Z
0
1
(x y + y 3 )
3
2
=
0
2
Z
dx =
0
0
(2,2,8)
0
0
x
z
4 3
1
x dx = x4
3
3
2
=
0
16
.
3
25. We are given ρ(x, y, z) = kz.
8
Z
4
Z
Z
y 1/3
8
Z
0
0
0
0
0
Z
8
y 1/3 zdzdy
8
Z 8
4
1 1/3 2
3 4/3
=k
y 1/3 dy = 8k
= 96k
y z dy = 8k
y
4
0 2
0
0
0
Z 8Z 4
Z
Z 8 Z 4 Z y1/3
y 1/3
2
2
Mxy =
kz dxdzdy = k
xz
dzdy = k
8
Z
0
0
0
0
0
0
0
8
Z
4
y 1/3 z 2 dzdy
0
8
Z 8
4
3 4/3
64
64
1 1/3 3
1/3
= 256k
y z dy =
k
k
y
y dy =
=k
3
3
4
0
0 3
0
0
1/3
Z 8Z 4
Z 8Z 4
Z 8 Z 4 Z y1/3
y
=
kyzdxdzdy = k
xyz
dzdy = i
y 4/3 zdzdy
Z
Mxz
y=x
0
0
0
y
4
Z
dzdy = k
xz
kzdxdzdy = k
m=
y 1/3
4
Z
2
2
x
8
0
0
Z
0
0
4
8
Z
0
0
0
8
0
8
3072
3 7/3
1 4/3 2
=
y z dy = 8k
y
k
y 4/3 dy = 8k
2
7
7
0
0
0
0
Z 8 Z 4 Z y1/3
Z 8Z 4
Z 8Z 4
y 1/3
1
1 2
Myz =
x z
dzdy = k
kxzdxdzdy = k
y 2/3 zdzdy
2
0
0
0
0
0
0
0 2
0
8
Z 8
Z 8
4
384
1 2/3 2
3 5/3
1
2/3
=
y z dy = 4k
y
k
y dy = 4k
= k
2 0 2
5
5
0
0
0
384k/5
3072k/7
256k
= 4/5; y = Mxz /m =
= 32/7; z = Mxy /m =
= 8/3
x = Myz /m =
96k
96k
96k
The center of mass is (4/5, 32/7, 8/3).
=k
26. We use the form of the integral in Problem 14(b) of this section. Without loss of generality,
we take
1. Z
Z 1ρZ=2−z
Z 1 Z 2−z
Z 1
3
m=
dydxdz =
3dxdz = 3
(2 − z − z 2 )dz
z2
0
0
1
= 3(2z − z 2 −
2
Z 1 Z 2−z Z
Mxy =
z2
0
Z
=3
1
1 3
z )
3
=
0
Z
0
1
Z
0
1
Z
1
Z
2−z
yz dxdz =
0
Z
3
2−z
zdydxdz =
dz = 3
z2
z2
7
2
3
0
2−z
xz
0
0
1
z2
3zdxdz
0
0
z2
1
1
(2z − z − z )dz = 3 z − z 3 − z 4
3
4
2
3
2
1
=
0
5
4
www.elsolucionario.org
190
CHAPTER 14. MULTIPLE INTEGRALS
1
Z
2−z
Z
3
Z
1
Z
2−z
Z
ydydxdz =
Mxz =
z2
1
0
0
z2
0
3
1 2
9
y dxdz =
2 0
2
1
Z
Z
2−z
dxdz
z2
0
Z
1
9
9
1
21
1
(2 − z − z 2 )dz = (2z − z 2 − z 3 ) =
2 0
2
2
3
4
0
Z 1Z
Z 1 Z 2−z
Z 1 Z 2−z Z 3
3
xy dxdz =
xdydxdz =
=
=
Myz
z2
0
0
2−z
z2
0
3xdxdz
z2
0
0
2−z
Z
1
1 2
3 1
3
16
1
1
dz =
(4 − 4z + z 2 − z 4 )dz = (4z − 2z 2 + z 3 − z 5 ) =
x
2
2
2
3
5
5
2
0
0
z
0
16/5
21/4
5/4
x = Myz /m =
= 32/35, y = Mxz /m =
= 3/2, z = Mxy /m =
= 5/14.
7/2
7/2
7/2
The centroid is (32/35, 3/2, 5/14).
1
Z
=3
z
27. The density is ρ(x, y, z) = ky. Since both the region and
the density function are symmetric with respect to the xyand yz-planes, x = z = 0. Using symmetry,
√
Z 3 Z 2 Z √4−x2
Z 3Z 2
4−k2
m=4
kydzdxdy = 4k
dxdy
yz
0
0
0
3
Z
2
Z
= 4k
0
0
Z
p
2
y 4 − x dxdy = 4k
0
0
3
Z
πydy = 4πk
= 4k
0
3
Z
2
Z
3
√
Z
1 2
y
2
0
= 4k
y
2
xp
= 18πk
√
3
Z
2
2
Z
4−
0
+ 2 sin
28. The density is ρ(x, y, z) = kz.
Z 1 Z x Z y+2
Z
m=
kzdzdydx = k
0
=
1
k
2
1
= k
2
=
1
k
6
x2 0
Z 1Z x
1
0
Z 1
0
2
Z
y2
Z
3
2
πy dy = 4πk
0
0
1 3
y
3
p
4 − x2 dxdy
3
= 36πk.
0
z
1
Z
0
x
x2
y+2
1 2
z
2
dydx
2
0
(y + 2)2 dydx
x2
0
Z
3
dxdy = 4k
2
0
36πk
y = Mxz /m =
= 2. The center of mass is (0,2,0).
18πk
0
Z
0
x
dy = 4k
2 0
−1
4−x2
2
y z
0
2
x2
y
x
0
0
0
3
Z
3
2
2
xp
−1 x
2
dy
4 − x + 2 sin
y
2
2 0
ky dzdxdy = 4k
0
2
3
4−x2
Mxz = 4
0
z=M4-x2
1
(y + 2)3
3
x
y=x
dx
1
x2
[(x + 2)3 − (x2 + 2)3 ]dx =
1
k
6
Z
x
1
[(x + 2)3 − (x6 + 6x4 + 12x2 + 8)]dx
0
1 1
1
6
= k (x + 2)4 − x7 − x5 − 4x3 − 8x
6 4
7
5
y=x2
1
=
0
407
k
840
y
14.7. THE TRIPLE INTEGRAL
1
Z
x
Z
Z
y+2
1
Z
2
Z
x
kz dzdydx = k
Mxy =
0
=
191
1
k
3
x2
Z 1
1
k
=
12
0
Z
0
x2
0
1
(y + 2)4
4
x
dx =
x2
1
k
12
Z
1 3
z
3
y+2
0
1
dydx = k
3
Z
0
1
Z
x
(y + 2)3 dydx
x2
1
[(x + 2)4 − (x2 + 2)4 ]dx
0
1
[(x + 2)4 − (x8 + 8x6 + 24x4 + 32x2 + 16)]dx
0
1
1
1493
1
1 9 8 7 24 32 3
5
=
=
k (x + 2) − x − x −
− x − 16x
k
12 5
9
7
5
3
1890
0
Z 1Z x
Z 1 Z x Z y+2
Z 1Z x
y+2
1 2
1
kyzdzdydx = k
dydx = k
y(y + 2)2 dydx
Mxz =
yz
2
2
x2 2
x2 0
0
x
0
0
0
x
Z 1Z x
Z 1
1
1
1 4 4 3
3
2
2
dx
(y + 4y + 4y)dydx = k
= k
y + y + 2y
2 0 x2
2 0
4
3
x2
Z 1
1
1
4
4
= k
− x8 − x6 − 74x4 + x3 + 2x2 dx
2 0
4
3
3
1
1
68
1
4
7
1
2
= k − x9 − x7 − x5 + x4 + x3
k
=
2
36
21
20
3
3
315
0
Z 1 Z x Z y+2
Z 1Z x
Z 1Z x
y+2
1
1 2
dydx = k
Myz =
xz
x(y + 2)2 dydx
kxzdzdydx = k
2 0 x2
0
x2 0
0
x2 2
0
Z 1
Z 1
x
1
1
1
x(y + 2)3 dx = k
[x(x + 2)3 − x(x2 + 2)3 ]dx
= k
2 0 3
6
2
0
x
Z 1
1
= k
[x4 + 6x3 + 12x2 + 8x − x(x2 + 2)3 ]dx
6 0
1
21
1 2
1 1 5 3 4
3
2
=
k
= k x + x + 4x + 4x − (x + 2)
6 5
2
8
80
0
21k/80
68k/315
x = Myz /m =
= 441/814, y = Mxz /m =
= 544/1221,
407k/840
407k/840
1493k/1890
z = Mxy /m =
= 5972/3663. The center of mass is (441/814,544/1221,5972/3663).
407k/840
29. m =
Z
1
−1
√
Z
1−x2
√
− 1−x2
Z
8−y
z
(x + y + 4)dzdydx
2+2y
8
y=-M1-x2
2
x
1
y=M1-x2
y
192
CHAPTER 14. MULTIPLE INTEGRALS
30. Both the region and the density function are symmetric with respect
to the xy- and√ yz-planes.
Thus,
Z 2 Z 1+z2 Z √1+z2 −y2
z 2 dxdydz.
m=4
−1
0
z
2
y=M1+x2
2
0
y
2
x
31. We are given ρ(x, y, z) = kz.
Z
Z 8 Z 4 Z y1/3
2
2
kz(x + z )dxdzdy = k
Iy =
8
32. We are given ρ(x, y, z) = k.
Z 1 Z 2−z Z 3
Z
Ix =
k(y 2 + z 2 )dydxdz = k
Z
Z
y 1/3
4
1
( x3 z + xz 3 )
dzdy
0 3
0
0
0
0
0
4
Z 8Z 4
Z 8
1
1 2 1 1/3 4
1/3 3
dy
=k
yz + y z dzdy = k
yz + y z
3
6
4
0
0
0
0
8
Z 8
8
2560
4 2
=k
y + 64y 1/3 dy = k
y + 48y 4/3
=
k
3
3
3
0
0r
√
p
4 5
2560k/3
=
.
From Problem 25, m = 96k. Thus, Rg = Iy /m =
96k
3
z2
0
0
0
2−z
1
Z
(9x + 3xz 2 )
=k
0
Z
dz = k
1
2−z
z2
1 3
y + yz 2
3
3
Z
1
Z
2−z
(9 + 3z 2 )dxdz
dxdz = k
z2
0
0
1
(18 − 9z − 3z 2 − 3z 3 − 3z 4 )dz
0
z2
1
223
9 2
3 4 3 5
3
=
= k 18z − z − z − z − z
k
2
4
5
20
0
Z 1 Z 2−z Z 3
Z 1 Z 2−z
Z 1
m=
kdydxdz = k
3dxdz = 3k
(2 − z − z 2 )dz
z2
0
0
0
z2
0
1
7
1
1
= k
= 3k 2z − z 2 − z 3
3
2
0
s2
r
r
Ix
223k/20
223
Rg =
=
=
m
7k/2
70
Z 1 Z 1−x Z 1−x−y
33. Iz = k
(x2 + y 2 )dzdydx
0
Z
0
1
Z
z
0
1
1−x
2
2
(x + y )(1 − x − y)dydx
=k
0
Z
0
1Z
=k
0
1
1−x
(x2 − X 3 − x2 y + y 2 − xy 2 − y 3 )dydx
1
0
x
1−x
1
1
1
dx
=k
(x2 − x3 )y − x2 y 2 + (1 − x)y 3 − y 4
2
3
4
0
0
Z 1
1
1
1
1
1
1
1
=k
[ x2 − x3 + x4 + (1 − x)4 ]dx = k x6 − x4 + x5 − (1 − x)5
2
12
6
4
10
60
0 2
Z
y
y=1-x
1
1
=
0
k
30
www.elsolucionario.org
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS
34. We are given ρ(x, y, z) = kx.
Z
Z 1 Z 2 Z 4−z
2
2
kx(x + z )dydxdz = k
Iy =
z
0
0
Z
1
0
193
z
1
Z
4−z
2
(x3 + xz 2 )y
0
y=z
dxdz
z
2
Z
3
1
y=4-z
4
2
y
x
2
(x + xz )(4 − 2z)dxdz
=k
0
0
Z 1
2
1
1
=k
( x4 + x2 z 2 )(4 − 2z) dz
2
0 4
0
Z 1
Z 1
2
=k
(4 + 2z )(4 − 2z)dz = 4k
(4 − 2z + 2z 2 − z 3 )dz
0
0
1
2 3 1 4
41
2
= 4k 4z − z + z − z
=
k
3
4
3
0
p
35. We are given ρ(x, y, z) = k x2 + y 2 + z 2 . Both the region
and the integrand are symmetric with respect to the yzand xz-planes.
Z 5 Z √25−x2 Z 5
p
Iz = 4
k(x2 + y 2 ) x2 + y 2 + z 2 dzdydx
√
0
z
5
x2 +y 2
0
5 y
5
x
36. We are given ρ(x, y, z) = kz. Both the region and the integrand are symmetric
with respect to the xz- and xy-planes.
Z 1 Z √1−x2 Z √1−z2
Iy = 4
kx(x2 + z 2 )dydzdx
−1
0
0
y=M25-x2
z
1
z=M1-x2
1
x
14.8
Triple Integrals in Other Coordinate Systems
√
√ √
√
1. x = 10 cos 3π/4 = −5 2; y = 10 sin 3π/4 = 5 2; (−5 2, 5 2, 5)
√
√
2. x = 2 cos 5π/6 = − 3; y = 2 sin 5π/6 = 1; (− 3, 1, −3)
√
√
√
√
3. x = 3 cos π/3 = 3/2; y = 3 sin π/3 = 3/2; ( 3/2, 3/2, −4)
√
√
√
√
4. x = 4 cos 7π/4 = 2 2; y = 4 sin 7π/4 = −2 2; (2 2, −2 2, 0)
5. x = 5 cos π/2 = 0; y = 5 sin π/2 = 5; (0, 5, 1)
√
√
6. x = 10 cos 5π/3 = 5; y = 10 sin 5π/3 = −5 3; (5, −5 3, 2)
y=M1-x2
1
y
194
CHAPTER 14. MULTIPLE INTEGRALS
√
7. With x = 1 and y = −1 we have r2 = 2 and tan θ = −1. The point is ( 2, −π/4, −9)
√
√
8. With x = 2 3 and y = 2 we have r2 = 16 and tan θ = 1/ 3. The point is (4, π/6, 17).
√
√
√
√
9. With x = − 2 and y = 6 we have r2 = 8 and tan θ = − 3. The point is (2 2, 2π/3, 2).
√
10. With x = 1 and y = 2 we have r2 = 5 and tan θ = 2. The point is ( 5, tan−1 2, 7).
11. With x = 0 and y = −4 we have r2 = 16 and tan θ undefined. The point is (4, −π/2, 0).
√
√
√
12. With x = 7 and y = − 7 we have r2 = 14 and tan θ = −1. The point is ( 14, −π/4, 3).
13. r2 + z 2 = 25
14. r cos θ + r sin θ − z = 1
15. r2 − z 2 = 1
16. r2 cos2 θ + z 2 = 16
17. z = x2 + y 2
18. z = 2y
19. r cos θ = 5, z = 5
√
√
√
20. tan θ = 1/ 3, y/z = 1/ 3, z = 3y, x > 0
21. The equations are
r2 = 4, r+ z 2 = 16, and z = 0.
Z 2π Z 2 Z √16−r2
Z 2π Z 2 p
V =
rdzdrdθ =
r 16 − r2 drθ
0
Z
0
2π
=
0
0
0
2
1
− (16 − r2 )3/2 dθ =
3
0
z=M16-r2
4
0
2π
Z
z
0
√
√
2π
(64 − 24 3)dθ =
(64 − 24 3)
3
2
2
y
r=2
x
22. The equation is z = 10 − r2 .
Z 2π Z 3 Z 10−r2
Z
V =
rdzdrdθ =
0
Z
=
0
0
2π
1
9 2 1 4
r − r
2
4
z
2π
0
3
Z
dθ =
0
0
2π
Z
10
3
r(9 − r2 )drdθ
z=10-r2
0
81
81π
dθ =
.
4
2
3
3
r=3
x
y
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS
23. The equations are z = r2 , r = 5, and z = 0.
Z
Z 2π Z 5
Z 2π Z 5 Z r2
r3 drdθ =
rdzdrdθ =
V =
0
0
2π
Z
0
0
0
0
195
z
2π
5
1 4
r dθ
4 0
z=r2
625
625π
dθ =
4
2
=
0
5
y
r=5
5
x
z
24. Substituting the first equation into the second, we see
y=r2
2
that the surfaces intersect in the plane y = 4. Using polar
coordinates in the xz-plane, the equations of the surfaces
become y = r2 and y = 21 r2 + 2.
Z 2π Z 2 Z r2 /2+2
Z 2π Z 2 2
2
r
rdydrdθ =
r
+ 2 − r2 drdθ x
V =
y=r2/2+2
2
0
0
r2
0
0
2
Z 2π
Z 2π Z 2π Z 2 1
1
dθ =
2dθ = 4π
r2 − r4
=
2r − r3 drdθ =
2
8
0
0
0
0
0
√
25. The equation is √
z = a2 − r2 . By symmetry, x = y = 0.
Z 2π Z a Z a2 −r2
Z 2π Z a p
m=
rdzdrdθ =
r a2 − r2 drdθ
0
0
0
0
2π
2π
a
2π
4
Z
Z
5
Z
2π
Z
kxrdxdrdθ = k
0
k
=
2
0
Z
2π
Z
Z
0
2π
0
25k
=
2
0
4
0
25k
25rdrdθ =
2
8dθ = 200kπ
0
Z
0
2π
0
4
4
y
z
4
r=4
5
1 2
rz drdθ
2
0
1 2
r dθ
2 0
a
a
x
26. We use polar coordinates in the yz-plane. The density is
ρ(x, y, z) = kz. By symmetry, y = z = 0.
Z
z=Ma2-r2
0
=
m=
y
z
Z
a
1 3
2
1
a dθ = πa3
− (a2 − r2 )3/2 dθ =
3
3
3
0
0
0
√
Z 2π Z a Z √a2 −r2
Z 2π Z a
a2 −r 2
1 2
drdθ
Mxy =
zrdzdrdθ =
rz
0
0
0
0
0 2
0
Z
Z
1 2π a
=
r(a2 − r2 )drdθ
2 0
0
a
Z
Z
1 2π 1 4
1
1 2π 1 2 2 1 4
a r − r
dθ =
a dθ = πa4
=
2 0
2
4
2
4
4
0
0
πa4 /4
z = Mxy /m =
= 3a/8. The centroid is (0, 0, 3a/8).
2πa3 /3
Z
4
4
5
x
y
www.elsolucionario.org
196
CHAPTER 14. MULTIPLE INTEGRALS
2π
Z
4
Z
5
Z
Z
2
2π
Z
kx rdxdrdθ = k
Myz =
0
0
0
0
0
4
5
1 3
1
rx drdθ = k
3
3
0
Z
2π
4
Z
125rdrdθ
0
0
Z 2π
4
125 2
1
2000
1000dθ =
r dθ = k
kπ
2
3
3
0
0
0
2000kπ/3
x = Myz /m =
= 10/3, The center of mass of the given solid is (10/3, 0, 0).
200kπ
1
= k
3
2π
Z
√
27. The equation is z√= 9 − r2 and the density is ρ = k/r2 .
When x = 2,√r = √ 5.
Z 2π Z 5 Z 9−r2
Iz =
r2 (k/r2 )rdzdrdθ(k/r2 )
0
0
2
√
2π
Z
Z
√
5
0
Z √
2π
Z
3
drdθ
2
5
0
0
2π
1
2 3/2
2
− (9 − r ) − r
3
=k
0
Z 2π
x
√
5
dθ
0
4
8
dθ = πk
3
3
=k
0
28. The equation
Z 2π Z 1 is
Z z1 = r and the density is ρ = kr.
Ix =
(y 2 + z 2 )(kr)rdzdrdθ
0
0
Z
2π
0
1
0
Z
1
1
Z
z=r
(r4 sin2 θ + r2 z 2 )dzdrdθ
r
1
1 2 3
2
4
drdθ
=k
(r sin θ)z + r z
3
0
0
r
Z 2π Z 1 1
1
=k
r4 sin2 θ + r2 − r5 sin2 θ − r4 drdθ
3
3
1
Z 2π0 0
1 5 2
1 3 1 6 2
1 6
=k
dθ
r sin θ + r − r sin θ − r
5
9
6
18
0
0
Z 2π 1
1
=k
sin2 θ +
dθ
30
18
0
2π
1
1
1
13
=k
θ−
sin 2θ + θ
=
πk
60
120
18
90
0
2π
z
r
Z
=k
Z
y
3
p
r 9 − r2 − 2r drdθ
=k
Z
z=M9-r2
3
9−r 2
rz
=k
0
z
1
1
1
y
x
√
29. (a) x = (2/3) sin(π/2) sin(π/2) cos(π/6) = 3/3; y = (2/3) sin(π/2) sin(π/6) = 1/3;
√
√
(b) With x = 3/3 and y = 1/3 we have r2 = 4/9 and tan θ = 3/3. The point is
(2/3, π/6, 0).
14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS
197
√
√
30. (a) x = 5 sin(5π/4) cos(2π/3)
2/4; y = 5√sin(5ı/4)√sin(2π/3) = −5 6/4;
√ =5 √
z = 5 cos(5π/4) = −5 2/2; (5 2/4, −5 6/4, −5 2/2)
√
√
√
(b) With x = 5 2/4√and y = −5√ 6/4 we have r2 = 25/2 and tan θ = − 3.
The point is (5/ 2, 2π/3, −5 2/2).
√
31. (a) x = 8 sin(π/4)
cos(3π/4) = −4; y = 8 sin(π/4) sin(3π/4) = 4; z = 8 cos(π/4) = 4 2;
√
(−4, 4, 4 2)
√
√
(b) With x = −4 and y = 4 we have r2 = 32 and tan θ = −1. The point is (4 2, 3π/4, 4 2).
√
32. (a) x = (1/3) sin(5π/3) cos(π/6) = −1/4;√y = (1/3) sin(5π/3) sin(π/6) = − 3/12;
z = (1/3) cos(5π/3) = 1/6; (−1/4, − 3/12, 1/6)
√
√
2
(b) With x = −1/4 and
√ y = − 3/12 we have r = 1/12 and tan θ = 3/3.
The point is (1/2 3, π/6, 1/6).
√
√
33. (a) x =
√ 4 sin(3π/4)
√ cos 0 = 2 2; y = 4 sin(3π/4) sin 0 = 0; z = 4 cos(3π/4) = −2 2;
(2 2, 0, −2 2)
√
√
√
(b) With x = 2 2 and y = 0 we have r2 = 8 and tan θ = 0. The point is (2 2, 0, −2 2).
√
34. (a) x = 1 sin(11π/6)
cos π = 1/2; y = 1 sin(11π/6) sin π = 0; z = 1 cos(11π/6) = 3/2;
√
(1/2, 0, ( 3/2)
√
(b) With x = 1/2 and y = 0 we have r2 = 1/4 and tan θ = 0. The point is (1/2, 0 3/2).
2
35. With
√ x = −5, y = −5, and z = 0, we have ρ = 50, tan θ = 1, and cos φ = 0. The point is
(5 2, π/2, 5π/4).
√
√
√
36. With z = 1, y√= − 3, and
z = 1, we have ρ2 = 5, tan θ = − 3, and cos φ = 1/ 5.
√
The point is ( 5, cos−1 1/ 5, −π/3).
√
√
√
2
37. With x = 3/2,
√ y = 1/2, and z = 1, we have ρ = 2, tan θ = 1/ 3, and cos φ = 1/ 2.
The point is ( 2, π/4, π/6).
√
38. With x = − 3/2, y = 0, and z = −1/2, we have ρ2 = 1, tan θ = 0, and cos φ = −1/2.
The point is (1, 2π/3, 0).
√
√
39. With x = 3, y = −3, and z = 3 2, we have ρ2 = 36, tan θ = −1, and cos φ = − 2/2.
The point is (6, π/4, −π/4)
√
√
40. With x = 1, y √
= 1, and z = − 6, we have ρ2 = 8, tan θ = 1, and cos φ = − 3/2
The point is (2 2, 5π/6, π/4).
41. ρ = 8
42. ρ2 = 4ρ cos φ; ρ = 4 cos φ
√
43. 4z 2 = 3x2 + 3y 2 + 3z 2 ; 4ρ2 cos2 φ = 3ρ2 ; cos φ = ± 3/2; φ = π/6, 5π/6
44. −x2 − y − z 2 = 1 − 2z 2 ; −ρ2 = 1 − 2ρ2 cos2 φ; ρ2 (2 cos2 φ − 1) = 1
45. x2 + y 2 + z 2 = 100
198
CHAPTER 14. MULTIPLE INTEGRALS
46. cos φ = 1/2; ρ2 cos2 φ = ρ2 /4; 4z 2 = x2 + y 2 + z 2 ; z 2 + y 2 = 3z 2
47. ρ cos φ = 2; z = 2
48. ρ(1 − cos2 φ) = cos φ; ρ2 − ρ2 cos2 φ = ρ cos φ; x2 + y 2 + z 2 − z 2 = z; z = x2 + y 2
z
49. The
equations are φ
=
π/4 and ρ
=
3.
3
Z 2π Z π/4
Z 2π Z π/4 Z 3
1 3
2
ρ sin φdrhodφdθ =
ρ sin φ dφdθ
V =
3
0
0
0
0
0
0
Z 2π
Z 2π Z π/4
π/4
dθ
−9 cos φ
9 sin φdφdθ =
=
0
0
0
0
!
Z 2π √
√
2
= −9
− 1 dθ = 9π(2 − 2)
2
0
50. The equations are ρ = 2, θ = π/4, and θ = π/3.
2
Z π/3 Z π/2 Z 2
Z π/3 Z π/2
1 3
ρ2 sin φdρdφdθ =
ρ sin φ dφdθ
3
π/4
0
0
π/4
0
0
Z π/3 Z π/2
8
sin φdφdθ
=
3
π/4
0
π/2
Z
8 π/3
dθ
− cos φ
=
3 π/4
0
Z
8 π/3
2π
=
(0 + 1)dθ =
3 π/4
9
0
0
π/2
=
0
=
=
0
π/2
8
3
Z
8
3
Z
0
0
Z
1 2
ρ sin φ
3
z
2
ρ=2
2
y
2
x
z
ρ=2secφ
2
2 sec φ
dφdθ
0
π/6
sec3 φ sin φdφdθ =
0
π/2
y
x
0
π/6
Z
2
2
51. Using Problem 43, the equations are φ = π/6, θ = π/2,
and ρ cos φ = 2.
Z π/2 Z π/6 Z 2 sec φ
V =
ρ2 sin φdρdφdθ
Z
ρ=3
3
1
tan2 φ
2
π/6
dθ =
0
4
3
Z
0
8
3
π/2
Z
0
π/2
Z
π/6
0
1
2
dθ = π
3
9
sec2 φ tan φdφdθ
1
x
1
y
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14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS
199
z
52. The equations are ρ = 1 and φ = π/4. We find the volume
above the xy-plane and double.
2π
Z
π/2
Z
1
Z
2π
Z
1 3
ρ sin φ dφdθ
π/4 3
0
0
π/4
0
0
Z 2π Z π/2
Z 2π
Z 2π √
π/2
2
2
2
2
sin φdφdθ =
=
− cos φ
=
dθ
3 0
3
3
2
π/4
0
0
π/4
√
2π 2
=
3
ρ2 sin φdρdφdθ = 2
V =2
ρ=1
1
π/2
Z
1
1
1
x
z
53. By symmetry, x = y = 0. The equations are φ = π/4 and
2
ρ = 2 cos φ.
2 cos φ
Z 2π Z π/4 Z 2 cos φ
Z 2π Z π/4
1 3
dφdθ
m=
ρ2 sin φdρdφdθ =
ρ sin φ
3
0
0
0
0
0
0
Z
Z
Z
π/4
8 2π π/4
8 2π 1
3
4
dθ
=
sin φ cos φdφdθ =
− cos φ
1
3 0
3 0
4
0
0
Z 2π x
2
1
=−
− 1 dθ = π
3 0
4
Z 2π Z π/4 Z 2 cos φ
Z 2π Z π/4 Z 2 cos φ
2
Mxy =
zρ sin φdρdφdθ
ρ3 sin φ cos φdρdφdθ
0
0
2π
Z
0
π/4
Z
1 4
ρ sin φ cos φ
4
=
0
0
0
0
2 cos φ
ρ=2cosφ
1
y
0
Z
2π
Z
π/4
cos5 φ sin φdφdθ
dφdθ = 4
0
0
y
0
π/4
Z
2 2π 1
7
1
6
=−
− 1 dθ = π
=4
− cos φ
6
3
8
6
0
0
0
7π/6
z = Mxy /m =
= 7/6. The centroid is (0, 0, 7/6).
π
2π
Z
z
54. We are given density= kz. By symmetry, x = y = 0. The
equation is ρ = 1.
2π
Z
π/2
Z
Z
1
kzρ2 sin φdρdφdθ = k
m=
0
0
2π
Z
0
π/2
Z
=k
0
=
=
0
2π
1
k
4
Z
1
k
4
Z
Z
2π
0
Z
0
1
π/2
0
0
2π
1
sin2 φ
2
π/2
dθ =
0
1
k
8
Z
2π
dθ =
0
ρ=1
1
ρ3 sin φ cos φdρdφdθ
0
x
sin φ cos φdφdθ
0
Z
1
1 4
ρ sin φ cos φ dφdθ
4
0
Z
π/2
1
kπ
4
1
y
200
CHAPTER 14. MULTIPLE INTEGRALS
2π
Z
π/2
Z
1
Z
kz 2 ρ2 sin φdρdφdθ = k
Mxy =
2π
Z
Z
2π
Z
1
ρ3 cos2 φ sin φdρdφdθ
0
0
1
π/2
Z
π/2
Z
0
0
0
0
2π
Z
Z
π/2
1 5
1
cos2 φ sin φdφdθ
ρ cos2 φ sin φ dφdθ = k
5
5
0
0
0
0
0
Z 2π
Z 2π
π/2
1
1
2
1
− cos3 φ
=− k
(0 − 1)dθ =
= k
kπ
5 0
3
15
15
0
0
2kπ/15
z = Mxy /m =
= 8/15. The center of mass is (0, 0, 8/15).
kπ/4
=k
55. We are given density= k/ρ.
Z 2π Z cos−1 4/5 Z 5
k 2
m=
ρ sin φdρdφdθ
4 sec φ ρ
0
0
5
Z 2π Z cos−1 4/5
1 2
dφdθ
ρ sin φ
=k
2
0
0
z
5
ρ=4sec φ
4 sec φ
Z
2π
Z
3
cos−1 4/5
1
= k
(25 sin φ − 16 tan φ sec φ)dφdθ
2 0
0
Z 2π
cos−1 4/5
1
k
dθ
(−25 cos φ − 16 sec φ)
2 0
0
Z 2π
1
= k
[−25(4/5) − 16(5/4) − (−25 − 16)]dθ
2 0
Z 2π
1
dθ = kπ
= k
2 0
Z 2π Z π Z a
56. Iz =
(x2 + y 2 )(kρ)ρ2 sin φdρdφdθ
0
0
Z
2π
z
a
2
=k
2
2
2
2
ρ=a
3
sin π cos θ + ρ sin φ sin θ)ρ sin φdρdφdθ
0
Z
0
2π
Z
0
π
Z
a
ρ5 sin3 φdρdφdθ = k
=k
0
14.9
x
a
Z
y
3
0
π
Z
ρ=5
1
= ka6
6
Z
1
= ka3
6
Z
0
2π
0
0
2π
Z
0
Z
π
Z
2π
0
2π
Z
Z
0
π
π
a
1 6 3
ρ sin φ dφdθ
6
0
a
y
a
1
sin φdφdθ = ka3
(1 − cos2 φ) sin φdφdθ x
6
0
0
0
π
Z 2π
1
1
4
4π 6
3
− cos φ + cos φ
dθ = ka3
dθ =
ka
3
6
3
9
0
0
3
Change of Variables in Multiple Integrals
1. T : (0, 0) −→ (0, 0); (0, 2) −→ (−2, 8); (4, 0) −→ (16, 20); (4, 2) −→ (14, 28)
2. Writing x2 = v − u and y = v + u and solving for u and v, we obtain u = (y − √
x2 )/2 and
2
−1
v = (x +y)/2. Then the images under T are (1, 1) −→ (0, 1); (1, 3) −→ (1, 2); ( 2, 2) −→
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
201
(0, 2).
3. The uv-corner points (0, 0), (2, 0), (2, 2) correspond to
xy-points (0, 0), (4, 2), (6, −4).
v = 0 : x = 2u, y = u =⇒ y = x/2
y
u = 2 : x = 4 + v, y = 2 − 3v =⇒
y = 2 − 3(x − 4) = −3 + 14
v = u : x = 3u, y = −2u =⇒ y = −2x/3
v=u
y
y=x/2
y=14-3x
x
u=2
y=-2x/3
S
v=0
4. Solving for x and y we see that the
transformation is x = 2u/3 + v/3, y =
−u/3 + v/3 > The uv-corner points
(−1, 1), (4, 1), (4, 5), (−1, 5) correspond u=-1
to the xy-points (−1/3, 2/3), (3, −1),
(13/3, 1/3), (1, 2). v = 1 : x + 2y = 1;
v = 5 : x + 2y = 5;
u = −1 : x − y = −1;
u=4:x−y =4
y
y
v=5
x+2y=5
x-y=-1
u=4
S
v=1
x
x-y=4
x+2y=1
x
5. The uv-corner points
(0.0),
(1, 0),
(1, 2),
(0, 2) correspond to the
u=0
xy-points (0, 0), (1, 0),
(−3, 2), (−4, 0).
v = 0 : x = u2 , y = 0 =⇒ y = 0 and 0 ≤ x ≤ 1
u = 1 : x = 1 − v 2 , y = v =⇒ x = 1 − y 2
v = 2 : x = u2 − 4, y = 2u =⇒ x = y 2 /4 − 4
u = 0 : x = −v 2 , y = 0 =⇒ y = 0 and −4 ≤ x ≤ 0
6. The uv-corner points (1, 1), (2, 1), (2, 2),
(1, 2) correspond to the xy-points (1, 1),
(2, 1), (4, 4), (2, 4).
v = 1 : x = u, y = 1 =⇒ y = 1, 1 ≤
x≤2
u = 2 : x = 2v, y = v 2 =⇒ y = x2 /4
v = 2 : x = 2u, y = 4 =⇒ y = 4, 2 ≤
x≤4
u = 1 : x = v, y = v 2 =⇒ y = x2
x
y
y
v=2
S u=1
v=0
x=y2/4-4
x=1-y2
x
y=0
x
y
y
y=4
y=x2
y=x2/4
v=2
u=1
S u=2
v=1
y=1
x
x
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202
CHAPTER 14. MULTIPLE INTEGRALS
7.
∂(x, y)
=
∂(u, v)
−ve−u
veu
e−u
eu
8.
∂(x, y)
=
∂(u, v)
3e3u sin v
3e3u cos v
9.
∂(u, v)
=
∂(x, y)
−2y/x3
−y 2 /x2
= −2v
e3u cos v
−e3u sin v
1/x2
2y/x
=−
= −3e6u
3y 2
y
∂(x, y)
1
1
= −3( 2 )2 = −3u2 ;
=− 2
=
x4
x
∂(u, v)
−3u2
3u
2(y 2 − x2 )
−4xy
2
2
2
2
4
x +y )
(x + y 2 )2 =
2
4xy
2(y 2 − x2 )
(x + y 2 )2
2
2
2
2
2
2
(x + y )
(x + y )
From u = 2x/(x2 + y 2 ) and v = −2y(x2 + y 2 ) we obtain u2 = v 2 = 4/(x2 + y 2 ). Then
x2 + y 2 = 4/(u2 + v 2 ) and ∂(x, y)/∂(u, v) = (x2 + y 2 )2 /4 = 4/(u2 + v 2 )2 .
∂(u, v)
10.
=
∂(x, y)
11.
(a) The uv-corner points (0, 0), (1, 0), (1, 1), (0, 1)
correspond to the xy-points (0, 0), (1, 0), (0, 1),
(0, 0).
v = 0 : x = u, y = 0 =⇒ y = 0, 0 ≤ x ≤ 1
u=0
u = 1 : x = 1 − v, y = v =⇒ y = 1 − x
v = 1 : x = 0, y = u =⇒ x = 0, 0 ≤ y ≤ 1
u = 0 : x = 0, y = 0
y
y
v=1
y=1-x
S
u=1
v=0
x=0
y=0
x
x
(b) Since the segment u = 0, 0 ≤ v ≤ 1 in the uv-plane maps to the origin in the xy-plane, the
transformation is not one-to-one.
12.
∂(x, y)
=
∂(u, v)
1−v
−u
v
u
= u. The transformation is 0 when u is 0, for 0 ≤ v ≤ 1.
y
13. R1 : x + y = −1 =⇒ v = −1
R4
R2 : x − 2y = 6 =⇒ u = 6
R3
R
R3 : x + y = 3 =⇒ v = 3
R4 : x − 2y = −6 =⇒ u = −6
R1
∂(u, v)
∂(x, y)
1
1 −2
R2
=
= 3 =⇒
=
1 1
∂(x, y)
∂(u, v)
3
RR
RR
1
1 R3 R6
vdudv =
(x + y)dA =
v( )dA0 =
R
S
3
3 −1 −6
3
R3
1
1
(12) −1 vdv = 4( )v 2
= 16
3
2
−1
y
S
x
x
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
14. R1 : y = −3x + 3 =⇒ v = 3
y
R2 : y = x − π =⇒ u = π
R3 : y = −3x + 6 =⇒ v = 6
R4
R4 : y = x =⇒ u = 0
R3
∂(x, y)
∂(u, v) 1 −1
1
= 4 =⇒
=
∂(x, y) 3 1
∂(u, v)
4
R1 R
Z Z cos 1 (x − y)
Z Z
cos u/2 1
R2
2
dA =
( )dA0
3x
+
y
v
4
R
S
Z Z
Z
1 6 π cos u/2
1 6 2 sin u/2
=
dudv =
4 3 0
v
4 3
v
Z 6
6
1
dv
1
1
=
= ln v = ln 2
2 3 v
2
2
3
203
y
S
x
x
π
dv
0
15. R1 : y = x2 =⇒ u = 1
y
R2 : x = y 2 =⇒ v = 1
1
R4
R3 : y = x2 =⇒ u = 2
2
R R3
R1
1
R4 : x = y 2 =⇒ v = 2
R2
2
∂(u, v)
2x/y
−x2 /y 2
=3
=
2
2
x
−y /x
2y/x
∂(x, y)
∂(x, y)
1
=⇒
=
∂(u, v)
3
2
RR
R R y2
1
1 R2R2
1
1 2
1 R2
0
dA
=
v(
)dA
=
v
=
vdv
=
vdudv
=
S
R x
1
1
1
3
3
3
6 1
2
y
S
x
x2 + y 2 = 2y =⇒ v = 1
y
x2 + y 2 = 2x =⇒ u = 1
y
x2 + y 2 = 6y =⇒ v = 1/3
x2 + y 2 = 4x =⇒ u = 1/2
R4
2(y 2 − x2 )
−4xy
S
R
2
2 2
∂(u, v)
(x2 + y 2 )2
R1
= (x + y )
R3
2
2
−4xy
2(x
−
y
)
∂(x, y)
R2
x
(x2 + y 2 )2 (x2 + y 2 )2
x
−4
= 2
(x + y 2 )2
2
Using u + v 2 = 4/(x2 + y 2 ) we see that ∂(x, y)/∂(u, v) = −4/(u2 + v 2 )2 .
RR
RR
4
−4
1 R1 R1
115
(u2 + v 2 )dudv =
(x2 + y 2 )−3 dA =
(
)−3 | 2
|dA0 =
R
S u2 + v 2
(u + v 2 )2
16 1/3 1/2
5184
16. R1 :
R2 :
R3 :
R4 :
204
CHAPTER 14. MULTIPLE INTEGRALS
y
y
17. R1 : 2xy = c =⇒ v = c
R2 : x2 − y 2 = b =⇒ u = b
d
d
S
R3 : 2xy = d =⇒ v = d
R4 : x2 − y 2 = a =⇒ u = a
c
R3
c
R
R4
∂(u, v)
2x −2y
R2
2
2
= 4(x + y )
=
R1
2y 2x
∂(x, y)
a
b
b
a
x
∂(x, y)
1
=⇒
=
∂(u, v)
4(x2 + y 2 )
RR
R
R
1
1 RdRb
1
(x2 + y 2 )dA =
(x2 + y 2 )
dA0 =
dudv = (b − a)(d − c)
R
S
4(x2 + y 2 )
4 c a
4
x
y
18. R1 : xy = −2 =⇒ v = −2
R2 : x2 − y 2 = 9 =⇒ u = 9
R3 : xy = 2 =⇒ v = 2
R4 : x2 − y 2 = 1 =⇒ u = 1
∂(u, v)
2x −2y
= 2(x2 + y 2 )
=
y
x
∂(x, y)
∂x, y)
1
=⇒
=
2
∂(u, v)
2(x + y 2 )
Z Z
Z Z
(x2 + y 2 ) sin xydA =
(x2 + y 2 ) sin v(
R
1
=
2
Z
S
2
R3
R4
R
R2
4
2
y
x
S
R1
x
1
1
)dA0 =
2(x2 + y 2 )
2
Z
2
Z
9
sin vdudv
−2
1
8 sin vdv = 0
−2
19. R1 : y = x2 =⇒ v + u = v − u =⇒ u = 0
R2 : y = 4 − x2 =⇒ v + u = 4 − (v − u)
=⇒ v + u = 4 − v + u =⇒ v = 2
R3 : x = 1 =⇒ v − u = 1 =⇒ v = 1 + u
1
1
√
∂(x, y)
1
− √
=
2 v − u 2 v − u = −√
∂(u, v)
v−u
1
1
y
y
R2
R3
R
2
R1
x
S
x
www.elsolucionario.org
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
Z Z
R
205
Z Z √
v−u
1
dA0
−√
2v
v−u
S
Z Z
Z
1 1 2 1
1 1
=
[ln 2 − ln(1 + u)]du
dvdu =
2 0 1+u v
2 0
x
dA =
y + x2
1
1
ln 2 − [(1 + u) ln(1 + u) − (1 + u)]
2
2
1 1
= − ln 2
2 2
1
=
=
0
1
1
ln 2 − [2 ln 2 − 2 − (0 − 1)]
2
2
20. Solving x = 2u − 4v, y = 3u + v for
y
2
1
x + y,
u and v we obtain u =
14
7
3
1
y
v = − x + y. The xy-corner points (−4, 1),
14
7
(0, 0), (2, 3) correspond to the uv-points
R3
S
R2
R
(0, 1), (0, 0), (1, 0).
R1
∂(x, y)
2 −4
= 14
=
x
3 1
∂(u, v)
Z Z
Z Z
Z 1 Z 1−u
Z 1
1
ydA =
(3u + v)(14)dA0 = 14
(3u + v)dvdu = 14
(3uv + v 2 )
2
R
S
0
0
0
Z 1
1
28
1
5
35
= 14
( + 2u − u2 )du = (7u + 14u2 − u3 ) =
2
2
3
3
0
0
21. R1 :
R2 :
R3 :
R4 :
y
y
y
y
= 1/x =⇒ u = 1
= x =⇒ v = 1
= 4/x =⇒ u = 4
= 4x =⇒ v = 4
y
4
=
1
du
0
y
R3
R1
21 2
v
4
1−u
R4
2y
∂(u, v)
y
x
=
=
=⇒
2
−y/x 1/x
∂(x, y)
x
Z Z
Z Z
1
4
y dA =
u2 v 2 ( )dudv =
2v
R
S
=
x
S
R2
∂(x, y)
x
=
x
∂(u, v)
2y
Z 4
Z 4
Z
4
1
1
1 3
1 4
2
u vdudv =
u v dv =
63vdv
2 1
2 1 3
6 1
1
x
315
4
22. Under the transformation u = y + z, v = −y + z, w = x − y the parallelepiped D is mapped
206
CHAPTER 14. MULTIPLE INTEGRALS
to the parallelepiped E : 1 ≤ u ≤ 3, −1 ≤ v ≤ 1, 0 ≤ w ≤ 3.
0 1 1
∂(x, y, z)
∂(u, v, w)
1
= 0 −1 1 = 2 =⇒
=
∂(x, y, z)
∂(u, v, w)
2
1 −1 0
Z Z
Z Z
Z Z Z
1 3 1 3
1 0
(2u + 2v + 2w)dudvdw
(4z + 2x − 2y)dV =
(2u + 2v + 2w) dV =
2
2 0 −1 1
D
E
Z Z
Z
3
1 3 1 2
1 3
=
(u + 2uv + 2uw) dvdw =
int1−1 (8 + 4v + 4w)dvdw
2 0 −1
2
0
1
Z 3
Z 3
1
3
(8 + 4w)dw = (8w + 2w2 ) 0 = 42
dw =
(4v + v 2 + 2vw)
=
0
0
−1
23. We let u = y − x and V=y+x.
R1 : y = 0 =⇒ u = −x, v = x =⇒ v = −u
R2 : x + y = 1 =⇒ v = 1
R3 : x = 0 =⇒ u = y, v = y, =⇒ v = u
∂(u, v)
−1 1
= −2
=
1 1
∂(x, y)
∂(x, y)
1
=⇒
=−
∂(u,
v)
2
Z Z
Z Z
1
(y−x)/(y+x)
e
dA =
eu/v − dA0
2
R
S
Z Z
1 1 v u/v
=
e dudv =
2 0 −v
Z
1 1
v(e − e−1 )dv =
=
2 0
y
S
2
−2xy+x2
1
2
Z
x
x
v
1
veu/v
0
dv
−v
1
1
(e − e−1 ) v 2
2
2
1
=
0
1
(e − e−1 )
4
y
y
R2
R1
2
R
R
R1
eu |−1| dA0
dA =
R2
R3
24. We let u = y − x and v = y.
R1 : y = 0 =⇒ v = 0, u = −x =⇒ v = 0, 0 ≤ u ≤ 2
R2 : x = 0 =⇒ v = u
R3
R3 : y = x + 2 =⇒ u = 2
∂(x,
y)
∂(u, v)
−1 1
R
= −1 =⇒
=
= −1
0
1
∂(x,
y)
∂(u,
v)
Z Z
Z Z
ey
y
1
S
x
x
S
Z
2
Z
=
0
0
u
2
eu dvdu =
Z
0
2
2
ueu du =
1 u2
e
2
2
=
0
1 4
(e − 1)
2
14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
25. Noting that R2, R3, and R4 have equations y + 2x =
8, y − 2x = 0, and y + 2x = 2, we let u = y/x and
v = y + 2x.
R1 : y = 0 =⇒ u = 0, v = 2x =⇒ u = 0, 2 ≤ v ≤ 8
y
R2 : y + 2x = 8 =⇒ v = 8
R3 : y − 2x = 0 =⇒ u = 2
R3
R4 : y + 2x = 2 =⇒ v = 2
y + 2x
∂(u, v)
−y/x2 1/x
=
=
R
2
1
∂(x, y)
x2
R4
∂(x, y)
x2
=⇒
=
∂(u, v)
y + 2x
RR
RR
x2
−
(6x
+
3y)dA
=
3
(y
+
2x)
dA0 =
R
S
y
+
2x
RR
3 S x2 dA0
From y = ux we see that v = ux + 2x and x = v/(u + 2).
Then
8
RR
R2R8
R2
v3
3 S x2 dA0 = 3 0 2 v 2 (u+2)2 dvdu = 0
du =
(u + 2)2 2
2
R2
du
504
504 0
=
−
= 126.
(u + 2)2
u+2 0
207
y
S
R2
R1
x
x
y
y
26. We let u = x + y and v = x − y.
R1 : x + y = 1 =⇒ u = 1
R4
R3
R2 : x − y = 1 =⇒ v = 1
R
S
R3 : x + y = 3 =⇒ u = 3
R2
R1
R4 : x − y = −1 =⇒ v = −1
x
∂(x, y)
∂(u, v)
1
1 1
= −2 =⇒
=
=−
1 −1
∂(x,
2
Z Z y)
Z Z ∂(u, v)
1
(x + y)4 ex−y dA =
u4 ev − dA0
2
R
S
Z 3Z 1
Z
1
1
1 3 4 v
4 v
=
u e dvdu =
u e
du
2 1 −1
2 1
−1
Z
3
e − e−1 5
242(e − e−1
121
e − e−1 3 4
u du =
u
=
=
(e − e−1 )
=
2
10
10
5
1
1
27. The image of the ellipse is the unit circle x2 + y 2 = 1. From
∂(x, y)
=
∂(u, v)
5
0
0
3
x
= 15 we
www.elsolucionario.org
208
CHAPTER 14. MULTIPLE INTEGRALS
obtain
Z Z
y2
x2
( + )dA =
9
R 25
Z Z
sin ω cos θ
sin ω sin θ
cos ω
∂(x, y, z)
=
∂(ρ, ω, θ)
2π
Z
0
2
Z
(u + v )15dA = 15
15
4
0
0
s
=
28.
2
2π
Z
15
r rdrdθ =
4
2
Z
1
2π
r4 dθ
0
0
15π
.
2
dθ =
0
ρ cos ω cos θ
ρ cos ω sin θ
−ρ sin ω
1
−ρ sin ω sin θ
ρ sin ω cos θ
0
= cos ω(ρ2 sin ω cos ω cos2 θ + ρ2 sin ω cos ω sin2 θ) + ρ sin ω(ρ sin2 ω cos2 θ + ρ sin2 ω sin2 θ)
= ρ2 sin ω cos2 ω(cos2 θ + sin2 θ) + ρ2 sin3 ω(cos2 θ + sin2 θ) = ρ2 sin ω(cos2 ω + sin2 ω)
= ρ2 sin ω
29. The image of the ellipsoid x2 /a2 + y 2 /b2 + z 2 /c2 = 1 under the transformation u = x/a, v =
4
y, w = z/c, is the unit sphere u2 + v 2 + w2 = 1. The volume of this sphere is π. Now
3
¯
a 0 0
∂(x, y, z)
= 0 b 0 = abc
∂(u, v, w)
0 0 c
and
Z Z Z
Z Z Z
dV =
EabcdV 0 = abc
Z Z Z
D
dV 0 = abc
E
4
π
3
=
4
πabc.
3
30. Let u = xy and v = xy 1.4 . Then xy 1.4 = c =⇒ v = c; xy = b =⇒ u = b; xy 1.4 = d =⇒ v =
d; xy = a =⇒ u = a.
∂(x, y)
∂(u, v)
5
y
x
= 0.4xy 1.4 = 0.4v =⇒
=
=
y 1.4 1.4xy 0.4
∂(x, y)
∂(u, v)
2v
Z Z
Z Z
Z dZ b
Z d
5
5
5
dv
5
dA =
dA0 =
dudv = (b − a)
= (b − a)(ln d − ln c)
2v
2v
2
v
2
R
S
c
c
a
Chapter 14 in Review
A. True/False
1. True; use ex
2
−y
2
= ex e−y and Problem 53 in Section 14.2
2. True
3. True
4. False; consider f (x, y) = x.
5. False; both the density function and the lamina must be symmetric about an axis.
6. True; the equation of the plane is θ =
π
4,
θ=
5π
4
CHAPTER 14 IN REVIEW
209
B. Fill in the Blanks
Z
5
1.
y 2 +1
5y
8y −
x
3
dx = (8xy 3 − 5y ln x)
5
y 2 +1
= −8y 5 + 32y 3 + 5y ln
2. 16
6.
Z
a
Z b√1−x2 /a2
7.
y2 + 1
y
3. square
Z √ 2 2
√
4. II
5. f (x, 4) − f (x, 2)
1−x /a −y 2 /b2
c
√
−b 1−x2 /a2 −c
R 4 R √x
f (x, y)dydx
0 x/2
−a
= 40y 3 − 5y ln 5 − [8(y 2 + 1)y 3 − 5y ln(y 2 + 1)]
ρ(x, y, z)dzdydx
1−x2 /a2 −y 2 /b2
y
y=Mx
y=x/2
1
1
4
x
√
8. x = 6 sin(5π/3) cos(5π/6)
=
9/2;
y
=
6
sin(5π/3)
sin(5π/6)
=
−3
3/2;
z
=
6
cos(5π/3)
=
3
√
The point is (9/2, −3 3/2, 3).
√
√
√
√
9. r = 2 sin(π/4) = 2; θ = 2π/3; z = 2 cos(π/4) = 2; ( 2, 2π/3, 2)
Z 4 Z √4−y
y
10.
f
(x,
y)dxdy
4
√
y=4-x
2
− 4−y
0
1
11. z = r2 ; ρ = cot φ csc φ
12. circle
C. Exercises
Z
1. Holding x fixed,
12x2 e−4xy
= −5xy + y + c1 (x)
−4x
= −3xe−4xy − 5xy + y + c1 (x)
(12x2 e−4xy − 5x + 1)dy =
2. Holding y fixed,
R
1
ln |3xy + 4|
dx =
+ c2 (y)
4 + 3xy
3y
Z y
y
y 2 sin xydx = −y cos xy|y3 = y(cos y 4 − cos y 2 )
3.
y3
Z
ex
4.
1/x
x
x
dy = −
2
y
y
ex
= x2 − x/ex
1/x
x
210
CHAPTER 14. MULTIPLE INTEGRALS
Z
2
2x
Z
y−x
ye
5.
y−x
dydx =
0
0
2x
2
Z
−e
dx
Integration by parts
(2xex − ex + e−x )dx
Integration by parts
(ye
0
Z 2
=
y−x
)
0
0
= (2xex − 2ex − ex − e−x )
Z
4
Z
x
1
√
Z
x
7.
x
0
Z
1
dydx =
16 + x2
6.
0
= e2 − e−2 + 4
Z 4
4
4
y
x
1
−1 x
2
=
−
dx
=
tan
−
ln(16
+
x
)
2
16 + x2
16 + x2
4 2
0
0 16 + x x
π 1
1
π 1
π 1 1
=
− ln 32 − 0 − ln 16 = + (ln 16 − ln 32) = + ln
4
2
2
4
2
4
2 2
4
Z
2
0
sin y
dydx =
y
1
Z
4
Z
y
y2
0
Z
sin y
dxdy =
y
1
y
sin y
x
y
0
y
dy
y=x
1
y2
y=Mx
1
Z
(sin y − y sin y)dy
=
Integration by parts
1 x
0
1
= (− cos y − s ∈ y + y cos y)|0 = (− cos 1 − sin 1 + cos 1) − (−1) = 1 − sin 1
8.
Z
e2
1/x
Z
Z
ln xdydx =
e
Z
0
5
Z
5
Z
2
e
0
r
0
0
0
5Z
Z
3
5
Z
0
e2
=
e
1
2
(2x2 − 12 ) =
2
3
π/2
Z
cos3 θdθdz
dθdz =
0
0
π/2
(1 − sin2 θ) cos θdθdz =
=
0
1
1
ln xdx = (ln x)2
x
2
cos θ
π/2
Z
3r drdθdz =
0
e2
dx =
0
cos θ
9.
Z
y ln x
e
π/2
Z
1/x
e2
5
Z
sin θ −
0
0
1
sin3 θ
3
5
Z
=
2/3dz = 10/3
0
Z
π/2
Z
sin x
Z
10.
ln x
Z
y
π/2
e dydxdz =
π/4
0
ln x
sin x
Z
e
0
π/4
π/2
0
Z
π/2
Z
π/4
0
1 2
x −x
2
sin z
sin x
(x − 1)dxdz
dxdz =
π/2
Z
=
0
1
sin2 z − sin z dz
2
π/4
π/4
0
√ !
π/2
1
1
π
π
1
2
=
z − sin 2z + cos z
= −
− +
4
8
8
16
8
2
π/4
√
π+2−8 2
=
16
Z
=
y
π/2
dz
0
4
0
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CHAPTER 14 IN REVIEW
211
11. Using polar coorindates,
r=8
Z Z
Z
2π
8
Z
Z
5rdrdθ = 5
5dA =
R
0
0
0
Z Z
12. Using symmetry,
Z Z
8
1 2
r dθ = 5
2 0
1
Z
y 2 +1
(2x + y)dA =
32dθ = 320π.
8 polar
axis
0
0
Z
=
y 2 +1
1
Z
2
(2x + y)dxdy =
R
2π
Z
Z π
Z π Z 1+cos θ
1+cos θ
1 2
rdrdθ = 2
dθ
dA = 2
r
0
0 2
R
0
0
Z π
(1 + 2 cos θ + cos2 θ)dθ
=
0
π
1
1
= θ + 2 sin θ + θ + sin 2θ
= 3π/2.
2
4
0
Z
13.
2π
2y
0
2 polar
axis
x=2y
y
dy
(x + xy)
r=1+cos θ
x=y2+1
2y
1
[(y 2 + 1)2 + (y 2 + 1)y − (4y 2 + 2y 2 )]dy
x
0
= (y 4 + y 3 − 4y 2 + y + 1)dy =
1 5 1 4 4 3 1 2
y + y − y + y +y
5
4
3
2
14. Substracting z = 6 − x − y from z = x + y, we obtain
x + y = 3.
Z Z Z
Z 3 Z 3−x Z 6−x−y
Z 3 Z 3−x
6−x−y
xdV =
xdzdydx =
dydx
xz
R
0
Z
0
3
Z
=
0
Z
x+y
0
z
6
z=6-x-y
x+y
3−x
dx
3 y
0
[6x(3 − x) − 2x2 (3 − x) − x(3 − x)2 ]dx
0
3
(9x − 6x2 + x3 )dx
0
0
37
60
(6x − 2x2 − 2xy)dydx
(6xy − 2x2 y − xy 2 )
=
=
=
3−x
3
0
Z 3
Z
1
0
=
=
0
9 2
1
x − 2x3 + x4
2
4
3
=
0
27
4
2
x
212
CHAPTER 14. MULTIPLE INTEGRALS
√
15. The circle x2 + y 2 = 1 intersects y = x at x√= 1/ 2. The
circle x2 + y 2 = 9 intersects
y = x at x = 3/ 2.
Z Z
Z 1/√2 Z √9−x2
1
1
dA =
dydx
√
2
2
2
2
R x +y
0
1−x2 x + y
Z 3/√2 Z √9−x2
1
+
dydx
√
2
x + y2
1/ 2
x
y
y=M9-x2
y=x
x
y
16. The circles are r = 1 and r = 3; the line is θ = π/4.
3
Z π/2 Z 3
Z π/2
Z Z
1
1
dθ
ln
r
dA
=
rdrdθ
=
2
2
2
1 r
π/4
π/4
R x +y
1
Z π/2
π
=
ln 3dθ = ln 3
4
π/4
r=3
r=1
x
17.
y
y=x2
4
2
x
y=-x2
z
18. The region is symmetric with respect to the xz- and yzplanes and is shown in the first octant.
z=x2+y2
1
1
y
1
x
Z
1
Z
19.
√
3
y
Z
2
1
Z
x
cos x dxdy =
0
y
2
Z
cos x dydx =
x3
0
Z
=
0
0
1
y
x
y cos x2 x3
1
(x cos x2 − x3 cos x2 )dx
y=x
dx
1
y=x3
1
x
CHAPTER 14 IN REVIEW
1
1
sin x2
2
=
Z
=
213
1
x2 (x cos x2 )dx
0
0
Integration by parts
1 2
1
1
2
2
x sin x + cos x
= sin 1 −
2
2
2
1
1
1
1
= sin 1 −
sin 1 + cos 1 −
2
2
2
2
1 − cos 1
=
2
1
0
z
20. The
six Z
Z 2 Z 4−2x
forms
of
the
integral
are:
0
4
0
2−y/2
Z
y+z=8
8−2x−y
F (x, y, z)dzdydx;
Z
8
2x+z=8
4
8−2x−y
Z
F (x, y, z)dzdxdy;
Z
0
8
Z
0
8−z
2x+y=4
4
4−y/2−z/2
Z
F (x, y, z)dxdydz;
4
0
4
Z
8−y
Z
y
0
4−y/2−z/2
Z
x
F (x, y, z)dxdzdy;
0
4
8
Z
0
4−z/2
Z
Z
8−2x−z
F (x, y, z)dydxdz;
4
0
Z
0Z
0
8−2x Z
8−2x−z
F (x, y, z)dydzdx.
0
Z
2
4
Z
1
0
√
Z
x−x2
2
Z
Z
1
(4z − 1)
(4z + 1)dydxdz =
21.
0
1/2
0
0
Z
2
Z
=
(4z + 1)
0
1/2
2
x−2 dxdz
1/2
s
1
p
2
1
1
− x−
dxdz
4
2
Trig substitution
x − 1/2 p
1
x − 1/2
=
(4z + 1)
x − x2 + sin−1
2
8
1/2
0
Z 2
π
2
π
5π
=
(4z + 1)
− 0 dz =
(2z 2 + z) =
16
16
8
0
0
Z
1
dz
1/2
22. The region is the portion of the sphere of radius 1 centered at the origin in the first octant
and the octant below that. Using spherical coordinates, we have
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214
CHAPTER 14. MULTIPLE INTEGRALS
Z
1
√
Z
0
Z √1−x2 −y2
1−x2
−
0
√
(x2 + y 2 + z 2 )4 dzdydx =
π/2
Z
1−x2 −y 2
π/2
π
Z
1 1
ρ 1 sin φ
11
=
0
0
=
1
11
− cos φ
0
ρ8 ρ2 sin φdρdφdθ
0
1
Z
π/2
Z
π
dφdθ =
0
0
π
π/2
Z
1
Z
0
0
Z
π
Z
dθ =
0
1
11
Z
π/2
2dθ =
0
0
1
sin φdφdθ
11
π
11
23. fx = y; fy = z, 1 + fx2 + x2y = 11 + x2 + y 2 . Using cylindrical coordinates,
2π
Z
1
Z
A=
0
Z
p
2
1 + r rdrdθ =
0
1
2π
1
1
(1 + r2 )3/2 dθ =
3
3
0
0
2π
Z
(23/2 − 1)dθ =
0
2π √
(2 2 − 1).
3
3
Z √3 Z √3 2 3
2 6
2 2
2
dx =
6y − y
(18 − 6) − 6x − x
dx
24. V =
6 − y dydx =
3
9
9
0
0
0
x2
x2
√3
Z √3 √
√
2 6
2 7
6√
48 √
2
3
=
12x − 6x + x dx = 12x − 2x + x
= 12 3 − 6 3 +
3=
3
9
63
7
7
0
0
√
Z
3
Z
3
Z
1
Z
2x
25. (a) V =
0
1
= − (1 − x2 )3/2
3
1
Z
Z
y
0
y
p
1−
x2
0
x
(b) V =
2x
1
Z
p
2
1 − x dydx =
p
1
=
0
Z
dx =
1
p
x 1 − x2 dx
0
x
1
3
2
Z
1 − x2 dxdy +
y/2
1
Z
1
p
1 − x2 dxdy
y/2
y
26. We are given ρ = k(x2 + y 2 ).
Z 1Z x
Z
m=
k(x2 + y 2 )dydx = k
x3
0
Z
=k
0
=k
1
1
0
1
x y + y3
3
2
1
x2
dx
x3
y=x2
y=x3
1
1
x4 + x− x5 − x9 dx
3
3
1 5
1
1
1
x + x7 − x6 − x10
5
21
6
30
1
1
0
k
=
21
x
CHAPTER 14 IN REVIEW
1
Z
Z
215
x
3
Z
2
1
k(x + xy )dydx = k
My =
x3
0
0
1
x y + xy 3
3
3
x2
Z
1
dx = k
0
x3
1
1
x5 + x7 − x6 − x10 dx
3
3
1
1 6
65k
1
1
1
=
x + x8 − x7 − x11
6
24
7
33
1848
0
x2
Z 1
Z 1Z x
1 2 2 1 4
2
3
k(x y + y )dydx = k
dx
Mx =
x y + y
2
4
0
x3
0
x3
Z 1
1 6 1 8 1 8 1 12
=k
x + x − x − x
dx
2
4
2
4
0
1
20k
1 9
1 13
1 7
x − x − x
=
=k
14
36
52
819
0
65k/1848
20k/819
x = My /m =
= 65/88; y = Mx /m =
= 20/39
k/21
k/21
The center of mass is (65/88, 20/39).
=k
x2
1 2 3
dx
27. Iy =
k(x + x y )dydx = k
x y+ x y
3
0
x3
0
x3
Z 1
1 7
1
1
1
1
1
=k
x + x9 − x8 − x12
x6 + x8 − x7 − x11 dx = k
3
3
7
27
8
36
0
Z
1
Z
x2
4
Z
2 2
1
4
28. (a) Using symmetry,
Z a Z √a2 −x2 Z √a2 −x2 −y2
Z
V =8
dzdydx = 8
0
0
0
a
Z
0
0
Z
0
√
a2 −r 2
Z
Z
a
2
1
− (a2 − r2 )3/2 dθ =
3
3
0
0
Z 2π Z π Z a
Z
2
(c) V =
ρ sin φdρdφdθ =
a
r
0
2π
2π
rdzdrdθ = 2
Z
0
2π
0
Z
0
41
k
1512
p
a2 − x2 − y 2 dydx
0
√
a2 −x2
Z
a
dx = 8
0
0
a2 − r2 drdθ
0
a3 dθ =
=2
2π
p
=
√
a2 −x2
Trig substitution
Z a p
a2 − x2
y
y
sin−1 √
=8
a2 − x2 − y 2 +
2
2
a2 − x2
0
a
4
1
= πa3
= 2π a2 x − x3
3
3
0
(b) Using symmetry,
Z 2π Z a Z
V =2
1
π
4 3
πa
3
a
1 3
ρ sin φ dφdθ
0
0
0
0
0 3
0
Z 2π Z π
Z 2π
Z
π
1
1 2π 3
4
1
=
a3 sin φdφdθ =
−a3 cos φ dθ =
2a dθ = πa3
3 0
3 0
3
3
0
0
0
29. We use spherical coordinates.
π a2 − x2
dx
2
2
216
CHAPTER 14. MULTIPLE INTEGRALS
2π
Z
π/4
Z
3 sec φ
Z
ρ2 sin φdρdφdθ
V =
tan−1
0
2π
Z
Z
1/2
π/4
=
tan−1 1/2
0
=
1
3
2π
Z
2π
=9
0
Z
2π
Z
1 3
ρ sin φ
3
3 sec φ
dφdθ
0
π/4
27 sec3 φ sin φdφdθ = 9
tan−1
0
Z
Z
0
π/6
Z
2π
Z
0
1/2
1
tan2 φ
2
Z
π/4
dθ =
tan−1 1/2
2
9
2
Z
0
Z
2π
π/4
tan φ sec2 φdφdθ
tan−1
1/2
1
1−
dθ = 8π
9
2π
Z
2
π/6
1 2
ρ sin φdρdφdθ =
30. V =
ρ sin φ dφdθ
3
0
0
1
0
0
1
Z 2π Z π/6 Z 2π Z π/6
Z
1
7
8
7 2π
sin φ − sin φ dφdθ =
− cos φ
sin φdφdθ =
=
3
3
3 0
3 0
0
0
0
" √
#
√ !
Z
√
7 2π
7
7π
3
3
=
−
− (−1) dθ =
1−
2π =
(2 − 3)
3 0
2
3
2
3
2
dθ
0
y
y
2
31. x = 0 =⇒ u = 0, v = −y =⇒ u = 0, −1 ≤ v ≤ 0
x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4
x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4
y = 0 =⇒ u = 0, v = x2 =⇒ u = 0, 0 ≤ v ≤ 1
y = 1 =⇒ u = 2x, v = x2 − 1 = u2 /4 − 1
∂(u, v)
2y 2x
= −4(x2 + y 2 )
=
2x −2y
∂(x, y)
π/6
1
1
R
S
1
1
x
x
-1
∂(x, y)
1
=−
∂(u, v)
4(x2 + y 2 )
Z Z
Z Z
p
√
(x2 + y 2 ) 3 x2 + y 2 dA =
(x2 + y 2 ) 3 v −
=⇒
R
2
Z Z
1
1 2 1−u /4 1/3
0
dA
=
v dvdu
4(x2 + y 2 )
4 0 u2 /4−1
S
Z
Z 2h
1−u2 /4
i
3
1 2 3 4/3
v
du =
(1 − u2 /4)4/3 − (u2 /4 − 1)4/3 du
=
4 0 4
16 0
u2 /4−1
Z 2h
i
3
(1 − u2 /4)4/3 − (1 − u2 /4)4/3 du = 0
=
16 0
y
32. y = x =⇒ u + uv = v + uv2 =⇒ v = u
2
x = 2 =⇒ u + uv = 2 =⇒ v = (2 − u)/u
y = 0 =⇒ v = 0 or u = −1
∂(x, y)
1−w
u
we take v = 0
=
=1+u+v
v
1+u
∂(u, v)
v
2
R
2
x
2
u
www.elsolucionario.org
CHAPTER 14 IN REVIEW
217
Using x = u + uv and y = v + uv we find
(x − y)2 = (u + uv − v − uv)2 = (u − v)2 = u2 − 2uv + v 2
x + y = u + uv + v + uv = u + v + 2uv
2
(x + y) + 2(x + y) + 1 = u2 + 2uv + v 2 + 2(u + v) + 1 = (u + v)2 + 2(u + v) + 1 = (u + v + 1)2 .
Then
Z Z
R
Z 1 Z 2/(1+v)
1
0
p
dA =
dudv
(u + v + 1)dA =
(x − y)2 + 2(x + y) + 1
S u+v+1
v
0
1
Z 1
2
1 2
1
=
− v dv = 2 ln(1 + v) − v
= 2 ln 2 − .
1
+
v
2
2
0
0
1
Z Z
www.elsolucionario.org
Chapter 15
Vector Integral Calculus
15.1
Line Integrals
Z
1.
Z
π/4
π/4
2(5 cos t)(5 sin t)(−5 sin t) dt = −250
2xy dx =
C
Z
0
sin2 t cos t dt
0
√
π/4
125 2
1
=−
sin3 t
= −250
3
6
0
Z
Z π/4
Z π/4
1
2xy dy =
2(5 cos t)(5 sin t)(5 cos t) dt = 250
cos2 t sin t dt = 250 − cos3 t
3
C
0
0
√ !
√
250
2
125
=
1−
=
(4 − 2)
3
4
6
Z
Z π/4
Z π/4
p
2
2
2xy ds =
2(5 cos t)(5 sin t) 25 sin t + 25 cos t dt = 250
sin t cos t dt
C
0
0
= 250
Z
2.
1
sin2 t
2
(x3 + 2xy 2 + 2x) dx =
π/4
=
0
1
Z
C
125
2
[8t3 + 2(2t)(t4 ) + 2(2t)]2 dt = 2
0
1
Z
(8t3 + 4t5 + 4t) dt
0
1
Z
(8t3 + 4t5 + 4t) dt
=2
0
1
2 6
28
4
2
= 2 2t + t + 2t
=
3
3
0
Z
Z 1
Z 1
(x3 + 2xy 2 + 2x) dy =
[8t3 + 2(2t)(t4 ) + 2(2t)]2t dt = 2
(8t4 + 4t6 + 4t2 ) dt
C
0
=2
0
8 5 4 7 4 3
t + t + t
5
7
3
218
1
=
0
736
105
π/4
0
15.1. LINE INTEGRALS
Z
219
(x3 + 2xy 2 + 2x) ds =
Z
1
Z
p
[8t3 + 2(2t)(t4 ) + 2(2t)] 4 + 4t2 dt = 8
0
C
1
t(1 + t2 )5/2 dt
0
1
8 7/2
(2 − 1)
7
0
Z
Z 0
Z 0
3.
(3x2 + 6y 2 ) dx =
[3x2 + 6(2x + 1)2 ] dx =
(27x2 + 24x + 6) dx = (9x3 + 12x2 + 6x)
=
1
(1 + t2 )7/2
7
=
−1
C
−1
= −(−9 + 12 − 6) = 3
Z
Z 0
[3x2 + 6(2x + 1)2 ]2 dx = 6
(3x2 + 6y 2 ) dy =
−1
C
Z
Z 0
√
√
2
2
(3x + 6y ) ds =
[3x2 + 6(2x + 1)2 ] 1 + 4 dx = 3 5
−1
C
Z 8
8
x2
x2
8
56
dx
=
dx
=
dx =
3
2 /8
y
27x
27
27
1
ZC 2
Z 18
Z 8
8
x
8
4
x2
4 2/3
−1/3
−1/3
dy
=
x
dx
=
x
=
x
dx
=
3
2
y
27x /8
27 1
9
3
1
ZC 2
Z 18
Z 8
8
p
x
x2 p
8
−1/3
2/3 3/2
−2/3 dx =
2/3 dx =
ds
=
1
+
x
x
1
+
x
(1
+
x
)
3
2
27
C y
1 27x /8
1
1
8 3/2
=
(5 − 23/2 )
27
Z
Z 2π
Z 2π
p
2
2
2
2
t
2
5.
(x + y )ds =
(25 cos −25 sin t) 25 sin t + 25 cos tdt = 125
(cos2 t − sin2 t)dt
Z
Z
4.
C
0
0
Z
= 125
0
Z
6.
Z
125
sin 2t
cos 2tdt =
2
2π
=0
0
π
(2x + 3y) d =
C
2π
(6 sin 2t + 6 cos 2t)(−4 sin 2t) dt
Z0 π
−24 sin2 2t − 24 sin 2t cos 2t dt
Z0 π
1
=
−24 (1 − cos 2t) − 24 sin 2t cos 2t dt
2
0
=
= −12t + 6 sin2 2t − 12 sin2 2t
π
0
= −12π
Z
7.
Z
z dx =
C
π/2
t(− sin t) dt
Integration by parts
0
π/2
Z
C
= (t cos t − sin t)|0 = −1
Z π/2
z dy =
t cos t dt Integration by parts
0
π/2
= (t sin t + cos t)|0
=
π
−1
2
0
−1
220
CHAPTER 15. VECTOR INTEGRAL CALCULUS
Z
Z
π/2
z dz =
ZC
=
π2
8
0
Z0 π/2 p
√ Z
2
z dx =
t sin t + cos2 t + 1 dt = 2
C
8.
π/2
1 2
t
2
t dt =
π/2
t dt =
0
0
√
π2 2
8
1
8 9
8
1 3
8 1 8
t dt =
t (t2 )(2t)t2 dt =
t
=
3
3
27
27
0
0
ZC
Z 01 Z
1
1
1 3
2
2
16
4xyz dy =
4
t7 dt = t8 =
t (t2 )(2t)2t dt =
3
3
3
3
0
Z0
ZC
Z0 1 1
16
16
1 3
16 1 6
2
t dt =
=
4
t (t )(2t)2 dt =
4xyz dz =
3
21
21
21
0
0
Z 1
ZC
Z 01 p
8 1 9 2 7
8
1 3
t6 (t2 + 2) dt =
t (2t) t4 + 4t2 + 4 dt =
t + t
4xyz ds =
4
3
3 0
3 9
7
C
0
Z
Z
4xyz dx =
1
Z
4
1
=
0
200
189
9. Using x as the parameter, dy = dx and
Z
Z
2
Z
2
= (2x + y) dx + xy dy =
−1
C
=
1 3
x + 3x2 + 3x
3
−1
=
(x2 + 6x + 3) dx
−1
2
= 21.
−1
10. Using
x as the parameter, dy
Z
Z 2= 2x dx and
Z
2
(2x + y) dx + xy dy =
(2x + x + 1) dx +
C
2
(2x + x + 3 + x + 3x) dx =
2
Z
2
−1
2 5
x + x3 + x2 + x
5
2
(2x4 + 3x2 + 2x + 1) dx
x(x + 1)2x dx =
2
=
−1
−1
141
.
5
11. From (−1, 2) to (2, 2) we use x as a parameter with y = 2 and dy = 0. From (2, 2) to (2, 5)
we use y as a parameter with x = 2 and dx = 0.
Z
Z
2
Z
(2x + y) d + xy dy =
(2x + 2) dx +
−1
C
5
2y dy = (x2 + 2x)
2
2
−1
+ y2
2
−1
= 9 + 21 = 30
12. From (−1, 2) to (−1, 0) we use y as a parameter with x = −1 and dx = 0. From (−1, 0) to
(2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (2, 5) we use y as a parameter
with x = 2 and dx = 0.
Z
Z
(2x + y) d + xy dy =
C
0
Z
2
(−1)y dy +
2
= 2 + 3 + 25 = 30
Z
2x dx +
−1
0
5
1
2y dy = − y 2
2
0
+ x2
2
2
−1
+ y2
2
0
www.elsolucionario.org
15.1. LINE INTEGRALS
221
13. Using x as a the parameter, dy = 2xdx.
Z
1
Z
x2 dx +
y dx + x dy =
1
Z
3x2 dx = x3
x(2x) dx =
0
0
0
C
1
Z
1
0
=1
14. Using x as a the parameter, dydx.
Z
1
Z
y dx + x dy =
1
Z
x dx +
C
1
Z
2x dx = x2
x dx =
0
0
0
1
0
=1
15. From (0, 0) to (0, 1) we use y as a parameter with x = dx = 0. From (0, 1) to (1, 1) we use x
as a parameter with y = 1 and dy = 0.
Z
1
Z
y dx + x dy = 0 +
1 dx = 1
C
0
16. From (0, 0) to (1, 0) we use x as a parameter with y = dy = 0. From (1, 0) to (1, 1) we use y
as a parameter with x = 1 and dx = 0.
Z
Z
y dx + x dy = 0 +
C
Z
17.
1 dy = 1
0
Z
2
1
9
(6x + 2y + 2) dx + 4xy dy =
C
4
= (2t
3/2
+ 2t
3/2
9
Z
1
(6t + 2t ) t−1/2 dt +
2
2
Z 9
√
4 tt dt +
(3t1/2 + 5t3/2 ) dt
4
4
9
)
= 460
4
18.
R
(−y 2 ) dx + xy dy =
C
Z
19.
R2
(−t6 )2 dt +
0
2x3 y d + (3x + y) dy =
Z
=
Z
Z
1
2(y 6 )y2y dy +
Z
4 9
1
y + y3 + y2
9
2
Z
2
2
4(y 3 + 1)3y 2 dy +
4x dx + 2y dy =
C
(2t)(t3 )3t2 dt =
0
−1
C
20.
R2
−1
6
= 2y + 4y +
2
y 2 −1
0
4t6 dt =
1
(3y 2 + y) dy =
−1
1
−1
Z
4 7
t
7
2
=
0
512
7
1
(4y 8 + 3y 2 + y) dy
−1
=
26
9
Z
2
2y dy =
−1
3
R2
(12y 5 + 12y 2 + 2y) dy
−1
= 165
21. From (−2, 0) to (2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (−2, 0) we
parameterize the semicircle as x = 2 cos θ and y = 2 sin θ for 0 ≤ θ ≤ π.
222
CHAPTER 15. VECTOR INTEGRAL CALCULUS
Z
2
Z
2
2
(x + y ) dx − 2xy dy =
x
2
1 3
x
3
=
2
4(−2 sin θ dθ) −
8 cos θ sin θ(2 cos θ dθ)
0
0
π
Z
(sin θ + 2 cos2 θ sin θ) dθ
−8
0
−2
16
=
−8
3
π
Z
dx +
−2
C
π
Z
2
3
− cos θ − cos θ
3
π
16 80
64
−
=−
3
3
3
=
0
22. We start at (0, 0) and use x as a parameter.
Z
2
Z
1
Z
1
Z
0
(x2 + x) dx
xx (2x dx) +
(x + x ) dx − 2
1
0
0
Z 0
√
1 −1/2
x x
−2
x
dx
2
1
Z 1
Z 0
Z 1
3
=
(x2 − 3x4 ) dx +
x2 dx =
(−3x4 ) dx = − x5
5
0
1
0
2
2
(x + y ) dx − 2xy dy =
C
4
2
1
=−
0
3
5
23. From (1, 1) to (−1, 1) and (−1, −1) to (1. − 1) we use x as a parameter with y = 1 and
y = −1, respectively, and dy = 0. From (−1, 1) to (−1, −1) and (1, −1) to (1, 1) we use y as
a parameter with x = −1 and z = 1, respectively, and dx = 0.
Z
x2 y 3 dx − xy 2 dy =
C
Z
−1
x2 (1) dx +
1
=
−1
Z
−(−1)y 2 dy +
Z
−1
+
1
1 3
y
3
x2 (−1)3 dx +
Z
−1
1
1 3
x
3
1
1
1 3
x
3
−
−1
1
−
−1
1 3
y
3
1
=−
−1
1
−(1)y 2 dy
−1
8
3
24. From (2, 4) to (0, 4) we use x as a parameter with y = 4 and dy = 0. From (0, 4) to (0, 0) we
use y as a parameter with x = dx = 0. From (0, 0) to (2, 4) we use y = 2x and dy = 2dx.
Z
x2 y 3 dx − xy 2 dy =
C
Z
0
x2 (64) dx −
Z
Z
4
0
4
+ x6
3
2
2
−
0
π
Z
y dx − x dy =
25.
3 sin t(−2 sin t) dt −
0
Z π
= −6
dt = −6π
C
Z
0
y dx − x dy = 6π.
Thus,
−C
Z
0 dy +
2
64 3
=
x
3
0
Z
2
x2 (8x3 ) dx −
0
2
2x4 0
2
x(4x2 )2 dx
0
512 256
352
=−
+
− 32 = −
3
3
3
π
Z
2 cos t(3 cos t) dt = −6
0
Z
0
π
(sin2 t + cos2 ) dt
15.1. LINE INTEGRALS
Z
26.
x2 y 3 + x3 y 2 dy =
223
Z
1
x
=
−1
Z 1
=
1
x3 (x8 )(4x3 ) dx
−1
−1
Z 1
C
Z
x2 (x12 ) dx +
14
1
Z
4x14 dx
dx +
−1
1
5 15
x
15
5x14 dx =
−1
−1
5
5
2
=
+
=
15 15
3
27. We parameterize the line segment from (0, 0, 0) to (2, 3, 4) by x = 2t, y = 3y, z = 4t for
0 ≤ t ≤ 1. We parameterize the line segment from (2, 3, 4) to (6, 8, 5) by x = 2 + 2t, y =
3 + 5t, z = 4 + t, 0 ≤ t ≤ 1.
Z
1
Z
0
Z 1
C
1
Z
Z
3t(2 dt) +
y dx + z dy + x dz =
0
(3 + 5t)(4 dt)
0
(2 + 4t) dt
0
1
(55t + 34) dt =
0
Z
y dx + z dy + x dz =
C
2
Z
t3 (3 dt) +
2
55 2
t + 34t
2
1
=
0
123
2
Z 2
5 2
5
t (3t2 dt) +
(3t)
t dt
4
2
0
0
0
2
Z 2
15 4 15 2
3 4 3 5 5 3
3
=
3t + t + t
= 56
dt =
t + t + t
4
2
4
4
2
0
0
Z
28.
1
1
Z
0
=
Z
2t(4 dt) +
0
(4 + t)(5 dt) +
Z
1
4t(3 dt) +
29. From (0, 0, 0) to (6, 0, 0) we use x as a parameter with y = dy = 0 and z = dz = 0. From
(6, 0, 0) to (6, 0, 5) we use z as a parameter with x = 6 and dx = 0 and y = dy = 0. From
(6, 0, 5) to (6, 8, 5) we use y as a parameter with x = 6 and dz = 0 and z = 5 and dz = 0.
Z
Z 6
Z 5
Z 8
y dx + z dy + z dz =
0 dx +
6 dz +
5 dy = 70
C
0
0
0
30. We parameterize the line segment from (0, 0, 0) to (6, 8, 0) by x = 6t, y = 8t, z = 0 for
0 ≤ t ≤ 1. From (6, 8, 0) to (6, 8, 5) we use z as a parameter with x = 6, dx = 0, and
y = 8, dy = 0.
Z
Z
y dx + z dy + z dz =
Z
C
Z
0
10x dx − 2xy 2 dy + 6xz dz =
Z
6 dz = 24t2
0
1
Z
10(t) dt −
0
= 5t2
5
8t(6 dt) +
C
31.
1
1
2(t)(t2 )2 (2t) dt +
0
1
0
− 4t6
1
0
+ 18t6
= 5 − 4 + 18 = 19
1
0
+ 30 = 54
Z
0
1
0
1
6(t)(t3 )(3t2 ) dt
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224
CHAPTER 15. VECTOR INTEGRAL CALCULUS
32. Parametrize the line segments as follows:
C1 : r1 (t) = ti + tj, 0 ≤ t ≤ 1
C2 : r2 (t) = i + j + tk, 0 ≤ t ≤ 1
C3 : r3 (t) = (1 − t)i + (1 − t)j + (1 − t)k, 0 ≤ t ≤ 1
We
Z then have
Z
Z
1
3x dx − y 2 dy + z 2 dz =
C1
1
t2 dt
3t dt −
0
0
7
3 1
− =
2 3
6
Z 1
Z 1
Z
3(1)(0) dt −
(1)2 (0) dt +
3x dx − y 2 dy + z 2 dz =
=
Z
0
C2
0
1
t2 dt
0
1
=
3
Z 1
Z 1
Z 1
Z
2
2
2
(1 − t)2 (−1) dt
(1 − t) (−1) dt +
3(1 − t)(−1) dt −
3x dx − y dy + z dz =
0
0
0
C3
3
1
1
3
= −
− −
+ −
=−
2
3
3
2
Z
33.
Z
2
y dx + xydy =
C1
1
2
Z
(4t + 2) 2dt +
0
1
Z
(2t + 1)(4t + 2)4dt =
0
1
(64t2 + 64t + 16)dt
0
1
64
208
64 3
=
t + 32t2 + 16t
+ 32 + 16 =
3
3
3
0
√
Z
Z √3
Z √3
Z √3
3
8
8
208
y 2 dx + xydy =
4y 4 (2t)dt +
2t4 (4t)dt =
16t5 dt = t6
= 72 − =
3
3
3
C2
1
1
1
1
Z
Z e3
Z e3
Z e3
e3
1
8
8
2
y 2 dx + xydy =
4(ln t)2 dt +
2(ln t)2 dt =
(ln t)2 dt = (ln t)3
t
t
t
3
C3
e
e
e
e
8
208
= (27 − 1) =
3
3
√
2
√ R2 2
√
R
R2
√
16 5
1 3
=
34. C1 xyds = 0 t(2t) 1 + 4dt = 2 5 0 t dt = 2 5
t
3
3
0
Z
Z 2
Z 2 p
p
1
xyds =
t(t2 ) 1 + 4t2 dt =
t3 1 + 4t2 dt u = 1 + 4t2 , du = 8tdt; t2 = (u − 1)
4
C2
0
0
Z 17
Z 17
1
1
1
=
(u − 1)u1/2 du =
(u3/2 − u1/2 )du
4
8
32
1
1
17
1
2 5/2 2 3/2
=
u − u
32 5
3
1
√
391 17 + 1
=
√
3
Z
Z 3 120
√ Z 3
√ 1
√
16 5
xyds =
(2t − 4)(4t − 8) 4 + 16dt = 16 5
(t − 2)2 dt 16 5 (t − 2)3
=
3
3
C3
2
2
2
C1 and C3 are different parameterization of the same curve, while C1 and C2 are different
=
15.2. LINE INTEGRALS OF VECTOR FIELDS
225
curves.
35. We are given ρ = kx. Then
Z
m=
Z
π
C
Z
kx ds = k
ρ dx =
π
Z π
p
(1 + cos t) sin2 t + cos2 t dt = k
(1 + cos t) dt
0
0
π
0
= k (t + sin t)|0 = kπ.
36. From Problem
35,Zm = kπ and Zds = dt.
π
Z
π
1
(1 + cos t) sin tdtk − cos t + sin2 t
Mx =
yρds =
kxyds = k
= 2k
2
0Z
C
C
0
Z
π
π
My = intC xρds = intC kx2 ds = k
(1 + cos t)2 dt = k
(1 + 2 cos t + cos2 t)dt
0
0
π
1
3
1
= kπ
= k t + 2 sin t + t + sin 2t
2
4
2
0
3
2
3kπ/2
2k
= ; y = Mx /m =
= . The center of mass is (3/2, 2/π).
x = My /m =
kπ
2
kπ
π
15.2
1.
Line Integrals of Vector Fields
2.
y
x
y
x
226
3.
CHAPTER 15. VECTOR INTEGRAL CALCULUS
4.
y
y
x
5.
x
6.
y
y
x
7. Since each vector points in a northeasterly direction, the vector field must have positive i and
j components. Therefore, the answer is (b).
8. Since each vector points in a northwesterly direction, the vector field must have negative i
and positive j components. Therefore, the answer is (a).
x
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15.2. LINE INTEGRALS OF VECTOR FIELDS
227
9. Since each vector points in a southwesterly direction, the vector field must have negative i
and j components. Therefore, the answer is (d).
10. Since each vector points in a southeasterly direction, the vector field must have positive i and
negative j components. Therefore, the answer is (c).
11. Note that the k component of each vector is always positive. Therefore, the answer is (d).
12. Note that the i component of each vector is always positive. Therefore, the answer is (c).
13. Note that each vector points directly away from the origin. Therefore, the answer is (a).
14. Note that the i and j components of each vector are zero. Therefore, the answer is (b).
F = e3t i − (e−4t )et j = e3t i − e−3t j; dr = (−2e−2t i + et j)dt; F · dr = (−2et − e−2t )dt;
Z
Z ln 2
ln 2
15.
31
3
19
1
= − − (− ) = −
(−2et − e−2t )dt = (−2et + e−2t )
F · dr =
2
8
2
8
0
c
0
16. F = 2(t)(t2 )i + t2 j = 2t3 i + t2 j; dr = (i + 2tj)dt;
F · dr = 4t3 dt;
R
R2
2
F · dr = 0 4t2 dt = t4 0 = 16
C
17. F = 2(2t − 1)i − 2(6t + 1)j = (4t − 2)i + (12t + 2)j; dr = (2i + 6j)dt;
F · dr = −64t − 16;
R
R1
1
F · dr = −1 (−64t − 16) dt = −32t2 − 16t −1 = −32
C
18. F = cos2 ti + sin tj; dr = (− sin ti + cos tj);
F · dr = (− cos2 t sin t + sin t cos t)dt;
Z
Z π/6
F · dr =
(− cos2 t sin t + sin t cos t) dt
C
0
π/6
cos3 t sin2 t
+
=
3
2 0
√ !3
2 1 1
1
1
3
=
+
−
+0
3
2
2 2
3
√
√
5
3 1 1
3
=
+ − =
−
8
8 3
8
24
19. F = −3 sin ti + 2 cos tj + 6tk; dr = (−2 sin ti + 3 cos tj + 3k)dt;
2
2
F
Z · dr = (−6Zsin t + 6 cos t + 18t)dt;
π
(−6 sin2 t + 6 cos2 t + 18t) dt
Z π
1
1
=
−6 (1 − cos 2t) + 6 (1 + cos 2t) + 18t dt = 3 sin 2t + 9t2
2
2
0
F · dr =
C
0
= 9π 2
π
0
228
CHAPTER 15. VECTOR INTEGRAL CALCULUS
3
6
20. F = et i + tet j + t3 et k; dr = (i + 2tj + 3t2 k)dt;
Z
Z 1
3
6
2 3 1 6
F · dr =
(et + 2t2 et + 3t5 et )dt = (et + et + et )
3
2
C
0
1
=
0
13
(e − 1)
6
21. Using x as a parameter, r(x) = xi + ln xj. Then F = ln xi + xj, dr = (i +
Z
Z
F · dr =
W =
1
c
e
1
j)dx, and
x
e
(ln x + 1)dx = (x ln x)|1 = e.
22. Let r1 = (−2+2t)i+(2−2t)j and r2 = 2ti+3tj for 0 ≤ t ≤ 1.
Then
dr1 = 2i − 2j, dr2 = 2i + 3j,
y
F1 = 2(−2 + 2t)(2 − 2t)i + 4(2 − 2t)2 j = (−8t2 + 16t − 8)i +
(16t2 − 32t + 16)j,
c1
F2 = 2(2t)(3t)i + 4(3t)2 j = 12t2 i + 36t2 j,
and Z
W =
Z
F1 · dr1 +
C1
Z 1
F2 · dr2
x
cC2
2
1
Z
2
(24t2 + 108t2 )dt
(−16t + 32t − 16 − 32t + 64t − 32)dt +
=
0
Z
c2
0
1
(84t2 + 96t − 48)dt = (28t3 + 48t2 − 48t)
=
0
1
0
= 28.
23. Let r1 = (1 + 2t)i + j,
r2 = 3i + (1 + t)j, and
r3 = (3 − 2t)i + (2 − t)j for 0 ≤ t ≤ 1. Then
dr1 = 2i,
dr2 = j,
dr3 = −2i − j,
y
c3
F1 = (1 + 2t + 2)i + (6 − 2 − 4t)j = (3 + 2t)i + (4 − 4t)j,
c1
F2 = (3 + 2 + 2t)i + (6 + 6t − 6)j = (5 + 2t)i + 6tj,
F3 = (3−2t+4−2t)i+(12−6t−6+4t)j = (7−4t)i+(6−2t)j,
and
Z
Z
Z
F1 · dr1 +
W =
C1
Z 1
=
F2 · dr2 +
Z
(6 + 4t)dt +
0
Z
=
0
F3 · dr3
c2
c3
1
Z
0
1
(−14 + 8t − 6 + 2t)dt
6tdt +
0
1
(−14 + 20t)dt = (−14t + 10t2 )
1
0
= −4.
24. F = t3 i + t4 j + t5 k; dr = 3t2 i + 2tj + k;
R
R3
R3
W = C F · dr = 1 (3t5 + 2t5 = t5 )dt = 1 6t5 dt = t6
3
1
c2
= 728
x
15.2. LINE INTEGRALS OF VECTOR FIELDS
229
25. r = 3 Rcos ti + 3 sinRtj, 0 ≤ t ≤ 2π; dr = −3 sin ti + 3 cos tj; F = ai + bj;
2π
2π
W = C F · dr = 0 (−3a sin t + 3b cos t)dt = (3a cos t + 3b sin t)|0 = 0
26. Let r = ti + tj + tk for 1 ≤ t ≤ 3. Then dr = i + j + k, and
ct
c
(ti + tj + tk) = √
(i + j + k) =
3
|r|
( 3t2 )3
Z 3
Z
Z 3
c
1
c
√ (1 + 1 + 1)dt = √
dt =
W =
F · dr =
2
t
3 1 2
C
1 3 3t
c
1
2c
= √ (− + 1) = √ .
3 3
3 3
F=
c
√ (i + j + k),
3 3t2
3
1
c
√
−
t 1
3
27. F = 10 cos ti − 10 sin tj; dr = (−5 sin ti + 5 cos tj)dt;
F · dr = (−100 cos t sin t)dt;
R
R 2π
2π
F · dr = 0 (−100 cos t sin t) dt = 50 cos2 t 0 = 0
C
28. F = 10 cos ti + 2 sin tj; dr = (2 cos ti − 10 sin tj)dt;
F · dr = (10 cos2 t − 20 sin2 t)dt;
Z
Z 2π
F · dr =
(10 cos2 t − 20 sin2 t) dt
C
0
Z 2π 1
1
15
=
10 (1 + cos 2t) − 20 (1 − cos 2t) dt = −5t +
sin 2t
2
2
2
0
π
0
= −10π
29. On C1 , T = i and F · TcompT F ≈ 1. On C2 , T = −j and F · T = compT F ≈ 2. OnC3 , T = −i
and F · T = compT F ≈ 1.5. Using the fact that the lengths of C1 , C2 , and C3 are 4, 5, and
5, respectively,
we have
R
R
R
R
W = C F · Tds = C1 F · Tds + C2 F · Tds + C3 F · Tds ≈ 1(4)+2(5)+1.5(5)=21.5 ft-lb.
Z
Z
F · dr =
30. W =
C
b
b
(ma · r0 (t)) dt
a
b
dv
m(a · v) dt =
m
=
· v dt
dt
a
a
Z b
Z b
m dv
dv
m d
=
·v+v·
dt =
(v · v) dt
dt
dt
a 2
a 2 dt
Z b
m d 2
m 2b
=
(v ) dt =
v
2
a
a 2 dt
1
1
= m[v(b)]2 − m[v(a)]2
2
2
= K(B) − K(A)
Z
Z
1
1
(3x − 6y)3i + (3x − 6y)(−6)j
3
3
= (3x − 6y)i + (−6x + 12y)j
31. ∇f (x, y) =
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230
CHAPTER 15. VECTOR INTEGRAL CALCULUS
32. ∇f (x, y) = (1 + 2 cos 5xy − 10xy sin 5xy)i + (−1 − 10x2 sin 5xy)j
33. ∇f (x, y, z) = tan−1 zyi +
xz
xy
j+ 2 2
k
+1
y z +1
y2 z2
34. ∇f (x, y, z) = (1 − 2xyz 4 )i − x2 z 4 j − 4x2 yz 3 k
2
2
35. ∇f (x, y, z) = e−y i + 1 + 2xye−y j + k
36. ∇f (x, y, z) =
8y 3
18z 5
2x
i
+
j
+
k
x2 + 2y 4 + 3z 6
x2 + 2y 4 + 3z 6
x2 + 2y 4 + 3z 6
37. ∇ x2 + 21 y 2 = 2xi + yj = F(x, y). Therefore, the answer is (b).
38. ∇
1 2
2x
+ y 2 − 4 = xi + 2yj = F(x, y). Therefore, the answer is (c).
39. ∇ 2x + 12 y 2 + 1 = 2i + yj = F(x, y). Therefore, the answer is (d).
40. ∇
1 2
2x
+ 13 y 3 − 5 = xi + y 2 j = F(x, y). Therefore, the answer is (a).
41. φ(x, y) = sin x + y + cos y
42. φ(x, y) = xe−y
43. φ(x, y) = x + y 2 − 4z 3
44. φ(x, y) = xy 2 z 3
15.2. LINE INTEGRALS OF VECTOR FIELDS
45.
231
46.
y
y
x
47.
x
48.
y
x
y
x
232
CHAPTER 15. VECTOR INTEGRAL CALCULUS
49.
50.
y
y
x
x
51. Let φ(x, y, z) = −c(x2 + y 2 + z 2 )−1/2 . Then
cx
cy
cz
i+ 2
j+ 2
k
(x2 + y 2 + z 2 )3/2
(x + y 2 + z 2 )3/2
(x + y 2 + z 2 )3/2
c(xi + yj + zk)
= 2
(x + y 2 + z 2 )3/2
cr
= 3 =F
|r|
∇φ(x, y, z) =
52. Yes; if f and g differ by a constant, they will have the same gradient field.
15.3
Independence of the Path
1 3
x + g(y), φy =
3
(2,2)
R (2,2)
1
1
1
16
1
g 0 (y) = y 2 , g(y) = y 3 , φ = x3 + y 3 , (0,0) x2 dx + y 2 dy = (x3 + y 3 )
=
3
3
3
3
3
(0,0)
1. (a) Py = 0 = Qx and the integral is independent of path. φx = x2 , φ =
(b) Use y = x for 0 ≤ x ≤ 2.
R (2,2)
(0,0)
x2 dx + y 2 dy =
R2
(x2 + x2 )dx =
0
2 3
x
3
2
=
0
16
3
2. (a) Py = 2x = Qx and the integral is independent of path. φx = 2xy, φ = x2 y +g(y), φy =
R (2,4)
(2,4)
x2 + g 0 (y) = x2 , g(y) = 0, φ = x2 y, (1,1) 2xydx + x2 dy = x2 y (1,1) = 16 − 1 = 15
(b) Use y = 3x − 2 for 1 ≤ x ≤ 2.
R (2,4)
R2
R2
2xydx + x2 dy = 1 [2x(3x − 2) + x2 (3)]dx = 1 (9x2 − 4x)dx = (3x3 − 2x2 )
(1,1)
2
1
= 15
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15.3. INDEPENDENCE OF THE PATH
233
1
3. (a) Py = 2 = Qx and the integral is independent of path. φx = x + 2y, φ = x2 + 2xy +
2
1 2
1 2
1 2
0
g(y), φy = 2x + g (y) = 2x − y, g(y) = − y , φ = x + 2xy − y ,
2
2
2(3,2)
R (3,2)
1 2
1 2
(x + 2y)dx + (2x − y)dy =
= 14
x + 2xy − y
(1,0)
2
2
(1,0)
(b) Use y = x − 1 for 1 ≤ x ≤ 3.
Z
(3,2)
3
Z
(x + 2y)dx + (2x − y)dy =
[x + 2(x − 1) + 2x − (x − 1)]dx
(1,0)
1
3
Z
(4x − 1)dx = (2x2 − x)
=
1
3
1
= 14
4. (a) Py = − cos x sin y = Qx and the integral is independent of path. φx = cos x cos y, φ =
sin x cos y + g(y), φy = − sin x sin y + g 0 (y) = 1 − sin x sin y, g(y) = y, φ =
R (π/2,0)
(π/2,0)
sin x cos y + y, (0,0) cos x cos ydx + (1 − sin x sin y)dy = (sin x cos y + y)|(0,0) = 1
(b) Use y = 0 for 0 ≤ x ≤ π/2.
Z
(π/2,0)
Z
cos x cos ydx + (1 − sin x sin y)dy =
(0,0)
0
π/2
π/2
cos xdx = sin x|0
=1
x
1
5. (a) Py = 1/y 2 = Qx and the integral is independent of path. φx = − , φ = − +
y
y
(4,4)
R
x
x
x
x
x
1
(4,4)
=3
g(y), φy = 2 + g 0 (x) = 2 , g(y) = 0, φ = − , (4,1) − dx + 2 dy = (− )
y
y
y
y
y
y (4,1)
(b) Use x = 4 for 1 ≤ y ≤ 4.
Z
(4,4)
(4,1)
1
x
− dx + 2 dy =
y
y
Z
1
4
4
4
dy = −
y2
y
4
=3
1
x
6. (a) Py = −xy(x2 +y 2 )−3/2 = Qx and the integral is independent of path. φx = p
, φ=
x2 + y 2
p
p
y
y
x2 + y 2 + g(y), φy = p
+ g 0 (y) = p
, g(y) = 0, φ = x2 + y 2 ,
2
2
2
2
x +y
x +y
(3,4)
Z (3,4)
xdx + ydy p 2
p
= x + y2
=4
x2 + y 2
(1,0)
(1,0)
(b) Use y = 2x − 2 for 1 ≤ x ≤ 3.
Z 3
Z 3
Z (3,4)
x + (2x − 2)2
5x − 4
xdx + ydy
p
p
√
=
dx =
5x2 − 8x + 4
x2 + (2x − 2)2
x2 + y 2
1
1
(1,0)
p
3
= 5x2 − 8x + 4 = 4
1
234
CHAPTER 15. VECTOR INTEGRAL CALCULUS
7. (a) Py = 4xy = Qx and the integral is independent of path. φx = 2y 2 x − 3, φ =
x2 y 2 − 3x + g(y), φy = 2x2 y + g 0 (y) = 2x2 y + 4, g(y) = 4y, φ = x2 y 2 − 3x +
R (3,6)
(3,6)
4y, (1,2) (2y 2 x − 3)dx + (2yx2 + 4)dy = x2 y 2 − 3x + 4y) (1,2) = 330
(b) Use y = 2x for 1 ≤ x ≤ 3.
Z
(3,6)
(2y 2 x − 3)dx + (2yx2 + 4)dy =
(1,2)
3
Z
([2(2x)2 x − 3] + [2(2x)x2 + 4]2)dx
1
3
Z
(16x3 + 5)dx = (4x4 + 5x)
=
1
3
1
= 330
5
8. (a) Py = 4 = Qx and the integral is independent of path. φx = 5x + 4y, φ = x2 + 4xy +
2
5
g(y), φy = 4x + g 0 (y) = 4x − 8y 3 , g(y) = −2y 4 , φ = x2 + 4xy − 2y 4 ,
2
(0,0)
R (0,0)
7
5
=
x2 + 4xy − 2y 4
(5x + 4y)dx + (4x − 8y 3 )dy =
(−1,1)
2
2
(−1,1)
(b) Use y = −x for −1 ≤ x ≤ 0.
Z
(−1, 1)
(0,0)
Z
3
0
[(5x − 4x) + (4x + 8x3 )(−1)]dx
(5x + 4y)dx + (4x − 8y )dy =
−1
Z 0
3
(−3x − 8x3 )dx = (− x2 − 2x4 )
2
−1
=
0
=
−1
7
2
9. (a) Py = 3y 2 + 3x2 = Qx and the integral is independent of path. φx = y 3 + 3x2 y,
; φ = xy 3 + x3 y + g(y), φy = 3xy 2 + x3 + g 0 (y) = x3 + 3y 2 x + 1, g(y) = y, φ =
(2,8)
R (2,8)
xy 3 + x3 y + y, (0,0) (y 3 + 3x2 y)dx + (x3 + 3y 2 x + 1)dy = .(xy 3 + x3 y + y)
= 1096
(0,0)
(b) Use y = 4x for 0 ≤ x ≤ 2.
Z (2,8)
Z 2
3
2
3
2
(y + 3x y)dx + (x + 3y x + 1)dy =
[(64x3 + 12x3 ) + (x3 + 48x3 + 1)(4)]dx
(0,0)
0
Z
=
0
2
(272x3 + 4)dx = (68x4 + 4x)
2
0
= 1096
10.
11. Py = 12x3 y 2 = Qx throughout the plane and the vector field is a conservative field. φx =
4x3 y 3 +3, φ = x4 y 3 +3x+g(y), φy = 3x4 y 2 +g 0 (y) = 3x4 y 2 +1, g(y) = y, φ = x4 y 3 +3x+y
12. Py = 6xy 2 = Qx throughout the plane and the vector field is a conservative field. φx =
2xy 3 , φ = x2 y 3 + g(y), φy = 3x2 y 2 + g 0 (y) = 3x2 y 2 + 3y 2 , g(y) = y 3 , φ = x2 y 3 + y 3
13. Py = −2xy 3 sin xy 2 + 2y cos xy 2 , Qx = −2xy 3 cos xy 2 − 2y sin xy 2 throughout the plane and
the vector is not a conservative field.
15.3. INDEPENDENCE OF THE PATH
235
14. Py = −4xy(x2 + y 2 + 1)−3 = Qx throughout the plane and the vector field is a conservative
1
field. φx = x(x2 + y 2 + 1)−2 , φ = − (x2 + y 2 + 1)−1 + g(y), φy = y(x2 + y 2 + 1)−2 + g 0 (y) =
2
1
y(x2 + y 2 + 1)−2 , g(y) = 0, φ = − (x2 + y 2 + 1)−1
2
15. Py = 1 = Qx throughout the plane and the vector field is a conservative field. φx = x3 +
1
1
1
1
y, φ = x4 + xy + g(y), φy = x + g 0 (y) = x + y 3 , g(y) = − y 4 , φ = x4 + xy + y 4
4
4
4
4
16. Py = 4e2y , Qx = e2y throughout the plane and the vector field is not a conservative field.
17. Py = 0 = Qx , Px = 0 = Rx , Qz = −1 = Ry throughout 3-space and the vector field is a
conservative field.
∂g
φx = 2x, φ = x2 + g(y, z)φy =
= 3y 2 − x,
∂y
g(y, z) = y 3 − yz + h(z), φ = x2 + y 3 − yz + h(z),
φz = −y + h0 (z) = −y, h(z) = 0, φ = x2 + y 3 − yz
18. Py = 2x = Qx , Pz = 0 = Rx , Qz = −e−y = Ry throughout 3-space and the vector field is a
conservative field.
∂g
φx = 2xy, φ = x2 y + g(y, z), φy = x2 +
= x2 − ze−y ,
∂y
g = ze−y + h(z), φ = x2 y + ze−y + h(z),
φz = e−y + h0 (z) = ey − 1, h(z) = −z, φ = x2 y + ze−y − z
R
19. Since Py = −e−y = Qx , F is conservative and C F · dr is independent of the path. Thus,
instead of the given curve we may use the simpler curve C1 : y = x, 0 ≤ x ≤ 1. Then
Z
(2x + e−y )dx + (4y − xe−y )dy
W =
C1
1
Z
=
(2x + e−x )dx +
0
Z
1
(4x − e−x )dx
Integration by parts
0
= (x2 − e−x )
1
0
1
0
−1
+ (2x2 + xe−x )
= [(1 − e−1 ) − (−1)] + [(2 + e
+ e−1 ) − (1)] = 3 + e−1 .
R
20. Since Py = −e−y = Qx , F is conservative and C F · dr is independent of the path. Thus,
instead of the given curve we may use the simpler curve C1 : y = 0 − 2 ≤ −x ≤ 2. Then
dy = 0 and
R
R −2
−2
W = C1 (2x+e−y )dx+(4y −xe−y )dy = 2 (2x+1)dx = (x2 + x) 2 = (4−2)−(4+2) = −4.
21. Py = z = Qx , Qz = x = Ry , Rx = y = Pz , and the integral is independent of path.
Parameterize the line segment between the points by x = 1 + t, y = 1 + 3t, z = 1 + 7t, for
0 ≤ t ≤ 1. Then dx = dt, dy = 3dt, dz = 7dt and
Z (2,4,8)
Z 1
yzdx + xzdy + xydz =
[(1 + 3t)(1 + 7t) + (1 + t)(1 + 7t)(3) + (1 + t)(1 + 3t)(7)]dt
(1,1,1)
0
Z
=
0
1
(11 + 62t + 63t2 )dt = (11t + 31t2 + 21t3 )
1
0
= 63.
www.elsolucionario.org
236
CHAPTER 15. VECTOR INTEGRAL CALCULUS
22. Py = 0 = Qx , Qz = 0 = Ry , Rx = 0 = Pz and the integral is independent of path.
Parameterize the line segment between the points by x = t, y = t, z = t, for 0 ≤ t ≤ 1.
Then dx = dy = dz = dt and
R (1,1,1)
R1
1
2xdx + 3y 2 dy + 4z 3 dz = 0 (2t + 3t2 + 4t3 )dt = (t2 + t3 + t4 ) 0 = 3.
(0,0,0)
23. Py = 2x cos y = Qx , Qz = 0 = Ry , Rx = 3e3z = Pz , and the integral is independent
of path. Integrating φx = 2x sin y + e3z we find φ = x2 sin y + xe3z + g(y, z). Then φy =
x2 cos y + gy = Q = x2 cos y, so gy = 0, g(y, z) = h(z), and φ = x2 sin y + xe3z + h(z). Now
φz = 3xe3z + h0 (z) = R = 3xe3z + 5, so h0 (z) = 5 and h(z) = 5z. Thus φ = x2 sin y + xe3z + 5z
and
Z (2,π/2,1)
(2x sin y + e3z )dx + x2 cos ydy( 3xe3z + 5)dz
(1,0,0)
= (x2 sin y + xe3z + 5z)
(2,π/2,1)
(1,0,0)
= [4(1) + 2e3 + 5] − [0 + 1 + 0] = 8 + 2e3 .
24. Py = 0 = Qx , Qz = 0 = Ry , Rx = 0 = Pz , and the integral is independent of path.
Parameterize the line segment between the points by x = 1 + 2t, y = 2 + 2t, z = 1, for
0 ≤ t ≤ 1. Then dx = 2dt, dz = 0 and
(3,4,1)
Z
(1,2,1)
Z 1
1
[(2 + 4t + 1)2 + 3(2 + 2t)2 2]dt
(2x + 1)dx + 3y 2 dy + dz =
z
0
Z 1
1
=
(24t2 + 56t + 30)dt = (8t3 + 28t2 + 30t) 0 = 66.
0
25. Py = 0 = Qx ; Qz = 0 = Ry , Rx = 2e2z = Pz and the integral is independent of path.
Parameterize the line segment between the points by x = 1 + t, y = 1 + t, z = ln 3, for
0 ≤ t ≤ 1. Then dx = dy = dt, dz = 0 and
Z
(2,2,ln 3)
2z
2
Z
2z
e dx + 3y dy + 2xe dz =
(1,1,ln 3)
1
[e2 ln 3 + 3(1 + t)2 ]dt = [9t + (1 + t)3 ]
0
1
0
= 16
26. Py = 0 = Qx , Qz = 2y = Ry , Rx = 2x = Pz and the integral is independent of path.
Parameterize the line segment between the points by x = −2(1 − t), y = 3(1 − t), z = 1 − t,
for 0 ≤ t ≤ 1. Then dx = 2dt, dy = −3dt, dz = −dt, and
Z
(0,0,0)
2
2
1
Z
[−4(1 − t)2 (2) + 6(1 − t)2 (−3)
2xzdx + 2yzdy + (x + y )dz =
(−2,3,1)
0
+ 4(1 − t)2 (−1) + 9(1 − t)2 (−1)]dt
Z 1
1
=
−39(1 − t)2 dt = 13(1 − t)3 = −13.
0
0
27. Py = 1 − z sin x = Qx , Qz = cos x = Ry , Rx = −y sin x = Pz and the integral is independent of path. Integrating θx = y − yz sin x we find θ = xy + yz cos x + g(y, z).
Then θy = x + z cos x + gy (y, z) = Q = x + z cos x, so gy = 0, g(y, z) = h(z), and
15.3. INDEPENDENCE OF THE PATH
237
θ = xy + yz cos x + h(z). Now θz = y cos x + h(z) = R = y cos x, so h(z) = 0 and
θ = xy + yz cos x. Since r(0) = 4j and r(π/2) = πi + j + 4k,
Z
π,1,4)
F · dr = (xy + yz cos x)|(0,4,0) = (π − 4) − (0 + 0) = π − 4.
C
28. P − y = 0 = Qx , Qz = 0 = Ry , Rz = −ez = Pz and the integral is independent
of path. Integrating φx = 2 − ez we find φ = 2x − xez + g(y, z). Then φ − Y = gy =
2y − 1, so g(y, z) = y 2 − y + h(z) and φ = 2x − xez + y 2 − y + h(z). Now φz = −xez +
hZ0 (z) = R = 2 − xez , so h0 (z) = 2, h(z) = 2z, and φ = 2x − xez + y 2 − y + 2z. Thus
F · dr = (2x − xez + y 2 − y + 2z)
C
(2,4,8)
(−1,1,−1)
= (4 − 2e8 + 16 − 4 + 16) − (−2 + e−1 + 1 − 1 − 2) = 36 − 2e8 − e−1
29. Since Py = Gm1 m2 (2xy/|r|5 ) = Qx , Qz = Gm1 m2 (2yz/|r|5 ) = Ry , and Rx = Gm1 m2 (2xz/|r|5 ) =
Pz , the force field is conservative.
x
, θ = Gm1 m2 (x2 + y 2 + z 2 )−1/2 + g(y, z),
θx = −Gm1 m2 2
2
(x + y + z 2 )3/2
y
y
θy = −Gm1 m2 2
+ gy (y, z) = −Gm1 m2 2
, g(y, z) = h(z),
(x + y 2 + z 2 )3/2
(x + y 2 + z 2 )3/2
θ = Gm1 m2 (x2 + y 2 + z 2 )−1/2 + h(z),
z
z
θz = −Gm1 m2 2
+ h0 (z) = −Gm1 m2 2
,
2
2
3/2
2
(x + y + z )
(x + y + z 2 )3/2
Gm1 m2
Gm1 m2
h(z) = 0, θ = p
=
2
2
2
|r|
x +y +z
30. Since Py = 24xy 2 z = Qx , Qz = 12x2 y 2 = Ry , and Rx = 8xy 3 = Pz , F is conservative.
Thus, the work done between two points is independent of the path. From θx = 8xy 3 z we
obtain θ = 4x2 y 3 z which is a potential function for F. Then
Z
√
(1, 3,π/3)
F · dr = 4x2 y 3 z
W =
(2,0,0)
√
(1, 3,π/3)
(2,0,0)
√
= 4 3π
R (0,2,π/2)
F · dr = 0.
R
R
31. Since F is conservative, C1 F · dr = −C2 F · dr. Then, since the simply
closed curve C is composed of C1 and C2 ,
Z
Z
Z
Z
Z
F · dr =
F · dr +
F · dr =
F · dr −
F · dr = 0.
and W =
(2,0,0)
C
C1
C2
c2
c1
−C2
C1
32. From F = (x2 + y 2 )n/2 (xi + yj) we obtain Py = nxy(x2 + y 2 )n/2−1 = Qx , so that F is conservative. From θx = x(x2 + y 2 )n/2 we obtain the potential function θ + (x2 + y 2 )(n+2)/2 /(n + 2).
Then
Z
(x2 ,y2 )
W =
F·dr = (
(x1 ,y1 )
(x2 + y 2 )(n+2)/2
)
n+2
(x2 ,y2 )
=
(x1 ,y1 )
i
1 h 2
(x2 + y22 )(n+2)/2 − (x21 + y12 )(n+2)/2 .
n+2
238
CHAPTER 15. VECTOR INTEGRAL CALCULUS
33. P − y = −2x sin y = Qx throughout the plane and the vector field F is a conservative field.
The path starts at point (1, 0) and ends at point (2, 1). Since F is conservative, the integral
is path independent so we can use any path C starting at (1, 0) and ending at (2, 1). Use the
path y = x − 1, 1 ≤ x ≤ 2. Then
Z
Z
F · dr =
C
2
(2x cos(x − 1) − x2 sin(x − 1))dx
1
= x2 cos(x − 1)
2
1
= 4 cos 1 − 1
34. Py = cos y + Qx , P − z = 0 = Rx , Qz = 0 = Ry throughout the plane and the vector field
F is a conservative field. The path starts at the point (0, 0, 0) and ends at the point (1, 1, 1).
Since F is conservative, the integral is path independent so we can use any path C starting
at (0, 0, 0) and ending at (1, 1, 1). Use the path y = z = x, 0 ≤ x ≤ 1. Then
Z
Z
F · dr =
1
(sin x + x cos x + x2 )dx
0
C
= x sin x +
x3
3
1
= sin 1 +
0
1
3
35. F cannot be a conservative field in the region.
36. (a) Py =
y 2 − x2
= Qx . Using the hint, we have
(y 2 + x2 )2
Z
Z 2π
Z
F · dr =
(sin2 t + cos2 t)dt =
C
0
2π
dt = 2π.
0
Since C is a closed path, the integral would be zero if F were conservative.
(b) Any simply connected region containing the path C would have to contain the origin.
But F and the partials Py and Qx are not defined at the origin. Therefore, the theorem
does not apply.
dv dr
dv
1 d 2
dp
∂p dx
37. From Problem 45 in Exercises 15.2,
·
=
·v =
v . Then, using
=
+
dt dt
dt
2 dt
dt
∂x dt
dr
∂p dy
= ∇p · , we have
∂y dt
dt
Z
Z
Z
dv
dr
m
· drdtdt + ∇p ·
= 0dt
dt
dt
Z
Z
1
d 2
dp
m
v dt +
dt = constant
2
dt
dt
1
mv 2 + p = constant.
2
38. By Problem 37, the sum of kinetic and potential energies in a conservative force field is
constant. That is, it is independent of points A and B, so p(B) + K(B) = p(A) + K(A).
www.elsolucionario.org
15.4. GREEN’S THEOREM
15.4
239
Green’s Theorem
1. The sides of the triangle are C1 : y = 0, 0 ≤ x ≤ 1; C2 :
x = 1, 0 ≤ y ≤ 3;
C
Z3 : y = 3x, 0 ≤ −x ≤Z 1.1
Z 3
Z 0
Z
(x − y)dx + xydy =
xdx +
ydy +
(x − 3x)dx +
C
0
0
1
1
1 2
1 2
x
+
y
2
2
0
1 9
= + +1−3=3
2 2
Z Z
Z
Z 1 Z 3x
(y + 1)dydx =
(y + 1)dA =
R
0
0
=
0
3 3 3 2
x + x
2
2
0
x(3x)dx
x=1
1
3
+ (−x2 )
=
y
0
1
0
+ (3x2 )
0
1
1
1
1 2
y +y
2
3x
Z
1
dx =
0
0
x
9 2
x + 3x dx
2
1
=3
0
2. The sides of the rectangle are C1 : y = 0, −1 ≤ x ≤ 1; C2 : x = 1, 0 ≤ y ≤ 1; C3 :
yZ = 1, 1 ≥ x ≥ −1; C4 : Z
x = −1, 1Z≥ y ≥ 0.
Z 1
Z 1
1
1
2
2
3x ydx+(x − 5y)dy =
0dx +
(1 − 5y)dy =
0dx +
(1 − 5y)dy
−1
C
−1
0
0
1
5
−1
=
3x2 dx +
(1 − 5y)dy = y − y 2
+ x3 1 + y −
2
1
1
0
1
RR
R1R1
R1 2
R1
2
2
3
(2x
−
3x
)dA
=
(2x
−
3x
)dxdy
=
(x
−
x
)
dy = 0 (−2)dy
R
0 −1
0
Z
−1
Z
0
−1
Z
2
2
Z
2π
−y dx + x dy =
3.
C
Z
2
1
= −2
0
= −2
2π
(−9 sin t)(−3 sin t)dt +
0
5 2
y
2
9 cos2 t(3 cos t)dt
0
2π
Z
[(1 − cos2 t) sin t + (1 − sin2 t) cos t]dt
= 27
0
2π
1
1
= 27(0) = 0
= 27 − cos t + cos3 t + sin t − sin3 t
3
3
0
Z Z
Z 2π Z 3
Z 2π Z 3
(2x + 2y)dA = 2
(r cos θ + r sin θ)rdrdθ = 2
r2 (cos θ + sin θ)drdθ
R
0
Z
=2
0
0
2π
0
0
3
Z 2π
1 3
r (cos θ + sin θ) dθ = 18
(cos θ + sin θ)dθ
3
0
0
2π
= 18(sin θ − cos θ)|0 = 18(0) = 0
240
CHAPTER 15. VECTOR INTEGRAL CALCULUS
4. The sides of the region are C1 : y √
= 0, 0 ≤ x ≤ 2; C2 :
y = −x + 2, 2 ≥ x ≥ 1; C3 : y = x, 1 ≥ x ≥ 0.
Z 1
Z
Z 2
−2(−x + 2)2 dx
0dx +
−2y 2 dx + 4xydy =
C
y
2
2
0
1
Z
4x(−x + 2)(−dx)
+
R
2
0
Z
Z
−2xdx +
+
0
√
4x x
1
1
1
√
2 x
dx
2
2 8
10
+ +1−1=
3 3
3
RR
R 1 R 2−y
R1
8 3
2
2
4
8ydA
=
8ydxdy
=
8y(2
−
y
−
y
)dy
=
8y
−
y
−
2y
R
0 y2
0
3
x
=0+
1
=
0
5. PZ = 2y, Py = 2, Q
Z =Z 5x, Qx = 5
2ydx + 5xdy =
(5 − 2)dA
C
Z RZ
=3
dA = 3(25π) = 75π
10
3
y
x
R
R
6. PZ = x + y 2 , Py = 2y, Q = 2x2Z−Zy, Qx = 4x
(x + y 2 )dx + (2x2 − y)dy =
(4x − 2y)dA
C
y
4
R
Z
2
4
Z
R
(4x − 2y)dydx
=
x2
−2
4
2
Z
2
(4xy − y )
=
−2
Z 2
dx
2
x
x2
(16x − 16 − 4x3 + x4 )dx
=
−2
=
1
8x − 16x − x + x5
5
2
4
2
=−
−2
96
5
y
7. P = x4 − 2y 3 , Py = −6y 2 , Q = 2x3 − y 4 , Qx = 6x2 .
Using
polar coordinates,
Z
Z Z
4
3
3
4
(x − 2y )dx + (2x − y )dy =
(6x2 + 6y 2 )dA
C
R
2π Z 2
Z
=
2
R
2
6r rdrdθ
0
Z
=
0
0
2π
3 4
r
2
2
Z
dθ =
0
2π
24dθ = 48π.
0
r=2
2
x
15.4. GREEN’S THEOREM
241
y
8. RP = x − 3y, Py = −3, Q = 4x
R R+ y, Qx = 4
(x
−
3y)dx
+
4(x
+
y)dy
=
(4 + 3)dA = 7(10) = 70
C
R
3
R
3
y
9. P = 2xy, Py = 2x, Q = 3xy 2 , Qx = 3y 2
Z
Z Z
Z
2xydx + 3xy 2 dy =
(3y 2 − 2x)dA =
C
R
Z
2
(y 3 − 2xy)
=
1
=
2
Z
2x
(3y 2 − 2x)dydx
4
2
1
2x
y=2x
2
Z
(8x3 − 4x2 − 8 + 4x)dx
dx =
R
1
2
4
2x4 − x3 − 8x + 2x2
3
2
=
1
40
16
56
− −
=
3
3
3
2x
2x
10. RP = e2x sin 2y, Py = 2e2x cos 2y, Q
R R= e cos 2y, Qx = 2e cos 2y
2x
2x
= e sin 2ydx + e cos 2ydy =
0dA = 0
C
R
11. P = xy, Py = x, Q = x2 , Qx = 2x. Using polar
coordinates,
Z
Z Z
Z π/2 Z 1
2
xydx + x dy =
(2x − x)dA =
r cos θrdrdθ
C
−π/2
R
Z
π/2
=
−π/2
=
1
sin θ
3
1 3
r cos θ
3
π/2
=
−π/2
x
1
x
y
1
r=1
0
Z
π/2
dθ =
0
2
−π/2
1
cos θdθ
3
R
1
x
2
3
2
2
y
12. P = ex , Py = 0, Q = 2 tan−1 x, Qx =
2
1
+
Z
Z Z
Zx 0 Z 1
2
2
2
1
ex dx + 2 tan−1 xdy =
dA =
dydx
2
2
1
+
x
1
+
x
R
C
R
−1 −x
y=-x
1
Z 0
2y
=
dx
1
+
x2 −x
−1
-1
Z 0
2x
2
+
dx
=
1 + x2
1 + x2
−1
π
π
0
= [2 tan−1 x + ln(1 + x2 )] −1 = 0 − − + ln 2 = − ln 2
2
2
x
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242
CHAPTER 15. VECTOR INTEGRAL CALCULUS
1
13. P = y 3 , Py = y 2 , Q = xy + xy 2 , Qx = y + y 2
3
Z
Z Z
Z 1/√2 Z 1−y2
1 3
2
ydxdy
y dx + (xy + xy )dy =
ydA =
C 3
0
y2
R
√
=
(xy)
1 2 1 4
y − y
2
2
√
1/ 2
Z
=−
R
x2
1
0
0
=
0
Z
Z
x2
2xydydx
x3
y=x2
(x3 − x4 )dx
R
0
x3
1 4 1 5
x − x
4
5
y
1
1
dx = −
(xy)
x
1 1
1
− =
4 8
8
=
14. P = xy 2 , Py = 2xy, Q = 3 cos y, Qx = 0
Z
Z Z
Z
xy 2 dx + 3 cos ydy =
(−2xy)dA = −
C
1
(y − y 2 − y 3 )dy
0
=
x=1-y2
y2
√
1/ 2
=
R
dy
0
Z
x=y2
1
1−y 2
1/ 2
Z
y
1
=−
0
1
20
y=x3
x
15. P
R = ay, Py = a, RQR= bx, Qx = b.
aydx + bxdy =
(b − a)dA = (b − a) × (area bounded by C)
C
R
16. P = P (x), Py = 0, Q = Q(y), Qx = 0.
R
C
P (x)dx + Q(y)dy =
RR
R
0dA = 0
R
RR
17. For the first integral: P = 0, Py = 0, Q = x, Qx = 1; C Rxdy = − RR R1dA =area of R.
For the second
integral:
R
R P = y, Py = 1, Q = 0, Q = 0; − C ydx = − R −dA =area of
R. Thus, C xdy = − C ydx.
18. P = −y, Py = −1, Q + x, Qx = 1.
Z Z
Z
19. A =
dA =
R
= 3a2
= ab
a cos3 t(3a sin2 t cos tdt) = 3a2
0
Z
Z
xdy =
C
1
1
t + sin 2t
2
4
2π
=
2π
0
3 2
πa
8
2π
Z
a cos t(b cos tdt) = ab
0
2π
= πab
0
Z
0
1
1
1
t−
sin 4t +
sin3 2t
16
64
48
dA =
R
2π
xdy =
C
Z Z
20. A =
Z
RR
1RR
1R
−ydx + xdy =
2dA =
dA =area of R
C
R
R
2
2
0
2π
cos2 tdt
sin2 t cos4 tdt
15.4. GREEN’S THEOREM
243
21. (a) Parameterize C by x = x1 + (x2 − x1 )t and y = y1 + (y2 − y1 )t for 0 ≤ t ≤ 1. Then
Z
Z 1
Z 1
−ydx + xdy =
−[y1 + (y2 − y1 )t](x2 − x1 )dt +
[x1 + (x2 − x1 )t](y2 − y1 )dt
C
0
0
1
1
1
= −(x2 − x1 )[y1 t + (y2 − y1 )t2 ] + (y2 − y1 )[x1 t + (x2 − x1 )t2 ]
2
2
0
1
1
= −(x2 − x1 ) y1 + (y2 − y1 ) + (y2 − y1 ) x1 + (x2 − x1 )
2
2
1
0
= x1 y2 − x2 y1 .
(b) Let Ci be the line segment from (xi , yi ) to (xi+1 , yi+1 ) for i = 1, 2, · · · , n − 1, and C2
the lineZsegment from (xn , yn )to (x1 , y1 ). Then
1
A=
−ydx + xdy Problem 18
2 C
"Z
#
Z
Z
Z
1
=
−ydx + xdy +
−ydx + xdy + · · · +
−ydx + xdy +
−ydx + xdy
2 C1
C2
Cn−1
Cn
=
1
1
1
1
(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + (xn−1 yn − xn yn−1 ) + (xn y1 − x1 yn ).
2
2
2
2
22. From part
(b) of Problem 21
1
1
1
1
A=
(−1)(1) − (1)(3)] + [(1)(2) − (4)(1) +
(4)(5) − (3)(2)] + [(3)(3) − (−1)(5)
2
2
2
2
1
= (−4 − 2 + 14 + 14) = 11.
2
23. P = 4x2 − y 3 , Py = −3y 2 ; Q = x3 + y 2 , Qx = 3x2 .
Z
Z Z
Z
(4x2 − y 3 )dx + (x3 + y 2 )dy =
(3x2 + 3y 2 )dA =
C
Z
=
0
R
2π
2π
Z
0
2
2
Z
3r (rdrdθ) =
1
0
2π
3 4
r
4
2
dθ
1
45
45π
dθ =
4
2
p
24. PI = cos2 x − y, Py = −1; Q = Zy 2Z+ 1, Qx = 0 Z Z
p
(cos2 x − y)dx + y 2 + 1dy =
(0 + 1)dA =
dA
C
R
R
√
= (6 2)2 − π(2)(4) = 72 − 8π
1
25. We first observe that Py + (y 4 − 3x2 y 2 )/(x2 + y 2 )3 = Qx . Letting C 0 be the circle x2 + y 2 =
4
we have
244
CHAPTER 15. VECTOR INTEGRAL CALCULUS
Z
C
−y 3 dx + xy 2 dy
=
(x2 + y 2 )2
−y 3 dx + xy 2 dy
(x2 + y 2 )2
Z
C0
1
1
1
1
cos t, dx = − sin tdt, y = sin t, dy = cos tdt
4
4
4
4
1
1
1
1
1
3
2
Z 2π − sin t − sin tdt + cos t
sin t
cos tdt
64
4
4
16
4
=
1/256
0
Z 2π
Z 2π
=
(sin4 t + sin2 t cos2 t)dt =
(sin4 t + (sin2 t − sin4 t)dt
x=
0
2π
Z
sin2 tdt =
=
0
1
1
t − sin 2t
2
4
0
2π
=π
0
26. We first observe that Py = [4y 2 − (x + 1)2 ]/[(x + 1)2 + 4y 2 ]2 = Qx . Letting C 0 be the ellipse
2
2
(x
Z + 1) + 4y = 4 we have
Z
−y
x+1
−y
x+1
dx+
dy =
dx +
dy
2
2
2
2
2
2
(x + 1) + 4y
(x + 1)2 + 4y 2
C (x + 1) + 4y
C 0 (x + 1) + 4y
x + 1 = 2 cos t, dx = −2 sin tdt, y = sin t, dy = cos tdt
Z 2π Z
2 cos t
1 2π
− sin t
=
(−2 sin t) +
cos t dt =
(sin2 t + cos2 t)dt = π.
4
4
2 0
0
RR 2
RR
27. Writing
x dA =
(Qx − Py )dA we identify Q = 0 and P = −x2 y. Then, with
R
R
C : x = 3 cos t, y = 2 sin t, 0 ≤ t ≤ 2π, we have
Z Z
x2 dA =
R
Z
Z
P dx + Qdy =
C
=
54
4
27
=
4
−x2 ydx = −
C
4 sin2 t cos2 tdt =
0
1
t − sin 4t
4
2π
9 cos2 t(2 sin t)(−3 sin t)dt
0
2π
Z
Z
2π
=
0
27
2
Z
2π
sin2 2tdt =
0
27
4
Z
2π
(1 − cos 4t)dt
0
27π
.
2
RR
RR
28. Writing
[1 − 2(y − 1)]dA =
(Qx − Py )dA we identify Q = x and P = (y − 1)2 . Then,
R
R
with
Z Z C1 : x = cos t, y −Z1 = sin t, −π/2 ≤
Z t ≤ π/2, and C2Z: x = 0, 2 ≥ y ≥ 0, Z
[1 − 2(y − 1)]dA =
R
P dx + Qdy +
C1
Z π/2
=
[sin2 t(− sin t) + cos t cos t]dt =
−π/2
π/2
P dx + Qdy =
C2
(y − 1)2 dx + xdy +
C1
Z π/2
[cos2 t − (1 − cos2 t) sin t]dt
−π/2
1
(1 + cos 2t) − sin t + cos2 t sin t dt
=
−π/2 2
π/2
1
1
1
π π π
=
t + sin 2t + cos t − cos3 t
= − −
= .
2
4
3
4
4
2
−π/2
Z
0dy
C2
www.elsolucionario.org
15.5. PARAMETRIC SURFACES AND AREA
245
29. P = x − y, Py = −1, Q = x + y, Qx = 1;
R
RR
3π
3
W = C F · dr =
2dA = 2 × area = 2( ) = π
R
4
2
30. P = −xy 2 , Py = −2xy, Q = x2 y, Qx = 2xy. Using polar coordinates,
2
Z
Z Z
Z π/2
Z π/2 Z 2
4
W =
F · dr =
4(r cos θ)(r sin θ)rdrdθ =
(r cos θ sin θ) dθ
4xydA =
C
R
Z
= 15
1
0
π/2
sin θ cos θdθ =
0
15
sin2 θ
2
0
π/2
1
15
.
2
=
0
2
31. Let P = 0 and W = x . Then Qx − Py = 2x and
RR
I
Z Z
xdA
1
1
2
R
x dy =
2xdA =
= x.
2A C
2A
A
R
Let P = y 2 and Q = 0. Then Qx − Py = −2y and
RR
I
I Z Z
ydA
1
1
2
R
y dx = −
−2ydA =
= y.
−
2A C
2A C
A
R
32. Using Green’s Theorem,
Z
Z
Z Z
Z
W =
F · dr =
−ydx + xdy =
2dA = 2
C
C
2π
Z
=2
0
1 2
r
2
R
1+cos θ
Z
2π
0
Z
1+cos θ
rdrdθ
0
2π
dθ =
(1 + 2 cos θ + cos2 θ)dθ
0
0
1
1
= θ + 2 sin θ + θ + sin 2θ
2
4
2π
= 3π.
0
RB
33. Since A P dx + Qdy is independent of path, Py = Qx by Theorem 17.3. Then, by Green’s
Theorem
Z
Z Z
Z Z
P dx + Qdy =
(Qx − Py )dA =
0dA = 0.
C
R
R
34.
15.5
Parametric Surfaces and Area
1. x = u, y = v, z = 4u + 3v − 2, −∞ < u < ∞, −∞ < v < ∞
2. x = u, y = 1 − 2u, z = v, −∞ < u < ∞, −∞ < v < ∞
√
3. x = u, y = − 1 + u2 + v 2 , z = v, −∞ < u < ∞, −∞ < v < ∞
4. x = u, y = v, z = 5 − u2 − v 2 , −∞ < u < ∞, −∞ < v < ∞
5. r(u, v) = ui + vj + (1 − v 2 )k, −2 ≤ u ≤ 2, −3 ≤ v ≤ 3
246
CHAPTER 15. VECTOR INTEGRAL CALCULUS
6. r(u, v) = 2 cos ui + 3 sin uj + vk, 0 ≤ u ≤ 2π, −∞ < v < ∞
7. x2 + y 2 = cos2 u + sin2 u = 1, circular cylinder
8. z = x2 + y 2 , paraboloid
9. x = sin u, y = sin u cos v, z = sin u sin v
y 2 + z 2 = sin2 u cos2 v + sin2 u sin2 v = sin2 (cos2 v + sin2 v) = sin2 u = z 2 ,
so x2 = y 2 + z 2 , portion of a circular cone
10. x = 2 sin φ cos θ, y = 3 sin φ sin θ, z = 4 cos φ,
y2
z2
x2
+
+
= sin2 φ cos2 θ + sin2 φ sin2 θ + cos2 φ
4
9
16
= sin2 (cos2 θ + sin2 θ) + cos2 φ
= sin2 φ + cos2 φ = 1,
ellipsoid
11. Surface is parameterzied by x = u, y = sin v, z = cos v so R is defined by 0 ≤ u ≤ 4, 0 ≤
v ≤ π2
12. Surface is parameterzied by x = u,
2, − π2 ≤ v ≤ π2
y = sin v,
z = cos v so R is defined by −2 ≤ u ≤
13. Surface is parameterzied by x = sin φ cos θ, y = sin φ sin θ, z = cos φ so R is defined by
0 ≤ θ ≤ 2π, π2 ≤ φ ≤ π
14. Surface is parameterzied by x = sin φ cos θ, y = sin φ sin θ, z = cos φ so R is defined by
0 ≤ θ ≤ π, 0 ≤ φ ≤ π2
√
15. At u = π/6, v = 2, we have x = 5, y = 5 3, z = 2.
√
∂x π ∂z π ∂y π , 2 = 5 3,
, 2 = −5,
,2 = 0
∂u 6
∂u 6
∂u 6
∂y π ∂z π ∂x π , 2 = 0,
, 2 = 0,
, 2 = 1.
∂v 6
∂v 6
∂v 6
i
j k
√
√
A normal vector is given by n = 5 3 −5 0 = −5i − 5 3j.
0
0 1
√
√
√
The tangent plane is −5(x − 5) − 5 3(y − 5 3) = 0 or x + 3y = 20.
16. At u = 1, v = 0, we have x = 1, y = 0, z = 1.
∂y
∂z
∂x
(1, 0) = 1,
(1, 0) = 0,
(1, 0) = 2
∂u
∂u
∂u
∂x
∂y
∂z
(1, 0) = 0,
(1, 0) = 1,
(1, 0) = 0.
∂v
∂v
∂v
i j k
A normal vector is given by n = 1 0 2 = −2i + k.
0 1 0
The tangent plane is −2(x − 1) + (z − 1) = 0 or −2x + z = −1.
15.5. PARAMETRIC SURFACES AND AREA
247
17. At u = 1, v = 2, we have x = 3, y = 3, z = −3.
∂x
∂y
∂z
(1, 2) = 2,
(1, 2) = 1,
(1, 2) = 2
∂u
∂u
∂u
∂x
∂y
∂z
(1, 2) = 1,
(1, 2) = 1,
(1, 2) = −4.
∂v
∂v
∂v
i j k
A normal vector is given by n = 2 1 2 = −6i + 10j + k.
1 1 −4
The tangent plane is −6(x − 3) + 10(y − 3) + (z + 3) = 0 or −6x + 10y + z = 9.
√
we have x = −4, y = 32 , z = 3 2 3 .
√
∂y
∂z
−1, π3 = −3,
−1, π3 = −3 3
∂u
∂u
√
−3 3 ∂z
3
∂y
π
,
−1, 3 =
−1, π3 = .
∂v
2
∂v
2
i
j
k√
√
4
−3
−3 3
√
= −18i − 6j − 6 3k.
A normal vector is given by n =
3 3
3
0 −
2
√ √
√2 The tangent plane is −18(x + 4) − 6(y − 32 ) − 6 3 z − 3 2 3 = 0 or 3x + y + 3z = −6.
18. At u = −1, v =
∂x
−1, π3 = 4,
∂u
∂x
−1, π3 = 0,
∂v
π
3,
19. At u = 3, v = 3, we have x = 3, y = 3, z = 9.
∂x
∂y
∂z
(3, 3) = 1,
(3, 3) = 0,
(3, 3) = 3
∂u
∂u
∂u
∂x
∂y
∂z
(3, 3) = 0,
(3, 3) = 1,
(3, 3) = 3.
∂v
∂v
∂v
i j k
A normal vector is given by n = 1 0 3 = −3i − 3j + k.
0 1 3
The tangent plane is −3(x − 3) − 3(y − 3) + (z − 9) = 0 or 3x + 3y − z = 9.
√
√
20. At u = 1, v = π/4, we have x = 22 , y = 22 , z = 1.
√
√
2 ∂y
2 ∂z
∂x
1, π4 =
,
1, π4 =
,
1, π4 = 1
∂u
∂u
2√ ∂u
√2
2 ∂y
2 ∂z
∂x
π
π
1, 4 =
,
1, 4 = −
,
1, π4 = 0.
∂v
2
∂v
2
∂v
√
√
i
j
k
√
√
2
2
2
2
1 =
A normal vector is given by n = √2
i+
j − k.
2√
2
2
2
2
−
0
√
√ !2 √ 2
√ !
√
2
2
2
2
The tangent plane is
x−
+
y−
− (z − 1) = 0 or x + y − 2z = 0.
2
2
2
2
21. At u = −2, v = 1, we have x = −1, y = 3, z = −2.
∂x
∂y
∂z
(−2, 1) = 1,
(−2, 1) = −1,
(−2, 1) = 1
∂u
∂u
∂u
∂x
∂y
∂z
(−2, 1) = 1,
(−2, 1) = 1,
(−2, 1) = −2.
∂v
∂v
∂v
www.elsolucionario.org
248
CHAPTER 15. VECTOR INTEGRAL CALCULUS
i
j
k
1 −1 1 = i + 3j + 2k.
1 1 −2
The tangent plane is (x + 1) + 3(y − 3) + 2(z + 2) = 0 or x + 3y + 2z = 4.
A normal vector is given by n =
22. At u = 0, v = ln 3, we have x = 0, y = ln 3 + 1, z = 3.
∂x
∂y
∂z
(0, ln 3) = ln 3,
(0, ln 3) = 1,
(0, ln 3) = 1
∂u
∂u
∂u
∂x
∂y
∂z
(0, ln 3) = 0,
(0, ln 3) = 1,
(0, ln 3) = 3.
∂v
∂v
∂v
i
j k
A normal vector is given by n = ln 3 1 1 = 2i − 3 ln 3j + ln 3k.
0 1 3
The tangent plane is 2(x − 0) − 3 ln 3(y − ln 3 − 1) + ln 3(z − 3) = 0 or 2x − 3(ln 3)y + (ln 3)z =
−3(ln 3)2 .
23. At (1, 7, 5), we have u = 2, v = 1.
∂x
∂y
∂z
(2, 1) = 1,
(2, 1) = 2,
(2, 1) = 4
∂u
∂u
∂u
∂x
∂y
∂z
(2, 1) = −1,
(2, 1) = 3,
(2, 1) = 2.
∂v
∂v
∂v
i
j k
1 2 4 = −8i − 6 ln 3j + 5k.
A normal vector is given by n =
−1 3 2
The tangent plane is −8(x − 1) − 6(y − 7) + 5(z − 35) = 0 or 8x + 6y − 5z = 25.
24. At (1, 3, 16), we have u = 4, v = 1.
∂y
∂z
∂x
(4, 1) = 0,
(4, 1) = 1,
(4, 1) = 8
∂u
∂u
∂u
∂x
∂y
∂z
(4, 1) = 2,
(4, 1) = −1,
(4, 1) = 0.
∂v
∂v
∂v
i
j k
A normal vector is given by n = 0 1 8 = 8i − 16j − 2k.
2 −1 0
The tangent plane is 8(x − 1) − 16(y − 3) − 2(z − 16) = 0 or 4x − 8y − z = −36.
25.
∂r
= h2, 1, i,
∂u
∂r
∂r
×
=
∂u ∂v
∂r
= h−1, 1, 0i
∂v
i
j k
2 1 1 = h−1, −1, 3i
−1 1 0
√
√
∂r
∂r
×
= 1 + 1 + 9 = 11
∂u ∂v
√
R2R1 √
A = 0 −1 11 dv du = 4 11
26. Let x = u, y = √
v, z = 1 − u − v.
√
R 1 R 1−u2 p
Then A = −1 −√1−u2 1 + (−1)2 + (−1)2 dv du = 3π
27.
∂r
∂r
= h1, 0, 2ui,
= h0, 1, 2vi
∂u
∂v
15.5. PARAMETRIC SURFACES AND AREA
249
i j k
1 0 2u = h−2u, −2v, 1i
0 1 2v
√
∂r
∂r
= 4u2 + 4v 2 + 1
×
∂u ∂v
Since 0 ≤ z √≤ 4, we have 0 ≤ u2 + v 2 ≤ 4. So
Z 2 Z 4−u2 p
A=
4u2 + 4v 2 + 1 dv du
√
∂r
∂r
×
=
∂u ∂v
−2
− 4−u2
√
2
4−u
p
vp 2
(4u2 + 1)
4u + 4v 2 + 1 +
=
ln |2v + 4u2 + 4v 2 + 1| √
4
−2 2
− 4−u2
i
h
p
p
√
1
=
2(4u2 + 1) ln |2 4 − u2 + 17| − (4u2 + 1) ln(4u2 + 1) + 4 −17(u2 − 4)
4
Z 2π Z 2 p
=
4r2 + 1r drdθ polar transformation
Z
2
0
Z
0
2π
=
0
√
=
28.
(4r2 + 1)3/2
12
17 17 − 1
θ
12
2π
0
2
Z
dθ =
0
0
2π
√
17 17 − 1
dθ
12
√
(17 17 − 1)π
=
6
∂r
∂r
= hcos θ, sin θ, 1i,
= h−r sin θ, r cos θ, 0i
∂r
∂θ
i
j
k
∂r ∂r
sin θ 1 = h−r cos θ, −r sin θ, ri
×
= cos θ
∂r ∂θ
r sin θ r cos θ 0
p
√
√
∂r ∂r
×
= r2 cos2 θ + r2 sin2 θ + r2 = r2 + r2 = r 2.
∂r ∂θ
√
R 2 R 2π √
A = 0 0 r 2 dθ dr = 4π 2
29. r = (r cos θ)i + (r sin θ)j + θk
∂r
∂r
= hcos θ, sin θ, 0i,
= h−r sin θ, r cos θ, f i
∂r
∂θ
i
j
k
∂r ∂r
cos θ
sin θ 0 = hsin θ, − cos θ, ri
×
=
∂r ∂θ
−r sin θ r cos θ 1
p
√
∂r ∂r
×
= sin2 θ + cos2 θ + r2 = 1 + r2 .
∂r ∂θ
√
√
R 2π R 2 √
A = 0 0 1 + r2 dr dθ = 2 5π + π ln(2 + 5)
30. r = (a sin φ cos θ)i + (a sin φ sin θ)j + (a cos φ)k
∂r
∂r
= h−a sin φ sin θ, a sin φ cos θ, 0i,
= ha cos φ cos θ, a cos φ sin θ, −a sin φi
∂θ
∂φ
250
CHAPTER 15. VECTOR INTEGRAL CALCULUS
∂r
∂r
×
=
∂θ ∂φ
i
− sin φ sin θ
a cos φ cos θ
j
a sin φ cos θ
a cos φ sin θ
k
0
−a sin φ
= h−a2 sin2 φ cos θ, −a sin2 φ sin θ, −a2 sin φ cos φi
q
∂r
∂r
= a4 sin4 φ cos2 θ + a4 sin4 φ sin2 θ + a4 sin2 φ cos2 θ
×
∂θ ∂φ
q
q
4
2
4
4
2
= a sin φ + a sin φ cos φ = a4 sin2 φ = a2 sin φ
R π R 2π 2
A = 0 0 a sin φdθdφ = 4a2 π
31. We have a = 2, so x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ,
R π R 2π
A = π/3 0 4 sin φ dθ dφ = 12π
32. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ,
R π R 2π
A = π/3 0 4 sin φ dθ dφ = 4π
π
3
π
2,
≤φ≤
33. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, 0 ≤ φ ≤
√
R π/4 R 2π
A= 0
4 sin φ dθ dφ = 4π(2 − 2)
0
π
4,
π
3
≤ φ ≤ π, 0 ≤ θ ≤ 2π,
0 ≤ θ ≤ 2π,
0 ≤ θ ≤ 2π,
34. x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ; the spehre intersects the cylinder when z 2 = 2,
so the region outside the cylinder is described by π4 ≤ φ 3π
4 , 0 ≤ θ ≤ 2π
√
R 3π/4 R 2π
A = π/4 0 4 sin φdθdφ = 8π 2
35.
(b)
(a)
(c)
z
z
1
1
1
1
y
1
y
y
x
(b)
(c)
z
z
y
x
1
1
x
(a)
37. (f )
1
1
x
36.
z
z
y
x
y
x
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15.5. PARAMETRIC SURFACES AND AREA
251
38. (e)
39. (d)
40. (a)
41. (c)
42. (b)
43.
z
y
x
44.
∂r
= h− cos φ cos θ, − cos φ sin θ, − sin φi,
∂φ
∂r
= h(sin φ − R) sin θ, (R − s ∈ φ) cos θ, 0i
∂θ
i
j
∂r
∂r
− cos φ cos θ
− cos φ sin θ
×
=
∂φ ∂θ
(sin φ − R) sin θ (R − sin φ) cos θ
k
− sin φ
0
= h(r − sin φ) sin φ cos θ, (R − sin φ) sin φ sin θ, −(R − sin φ) cos φi
q
∂r
∂r
×
= (R − sin φ)2 sin2 φ cos2 θ + (R − sin φ)2 sin2 φ sin2 θ + (R − sin φ)2 cos2 φ
∂φ ∂θ
q
= (R − sin φ)2 sin2 φ + (R − sin φ)2 cos2 φ
p
= (R − sin φ)2 = R − sin φ
R 2π R 2π
A = 0 0 R − sin φdθdφ = 4π 2 R
45. x = 2u, y = 2v, z = 8u + 6v − 2, −∞ < u < ∞, −∞ < v < ∞
46. The surface area of a circular cylinder with height h and radius r is A = 2πrh. The surface
pictures is one quarter of a circular cylinder with height 4 and radius 1. Therefore, A = 2π.
47. We have r = ui + f (u) cos vj + f (u) sin vk.
∂r
∂r
= h1, f 0 (u) cos v, f 0 (u) sin vi,
= h0, −f (u) sin v, f (u) cos vi,
∂u
∂v
i
j
k
∂r
∂r
×
= 1 f 0 (u) cos v f 0 (u) sin v
∂u ∂v
0 −f (u) sin v f (u) cos v
= hf (u)f 0 (u), −f (u) cos v
−f (u) sin vi
252
CHAPTER 15. VECTOR INTEGRAL CALCULUS
∂r
∂r
=
×
∂u ∂v
q
[f (u)f 0 (u)]2 + [f (u)]2 cos2 v + [f (u)]2 sin2 v
p
[f (u)]2 [f 0 (u)]2 + [f (u)]2
p
= f (u) 1 + [f 0 (u)]2
p
p
R b R 2π
Rb
By (11), A = a 0 f (u) 1 + [f 0 (u)]2 dvdu = 2π a f (u) 1 + [f 0 (u)]2 du
=
48. (a) x = u, y = sin u cos v, z = sin u sin v, −2π ≤ u ≤ 2π 0 ≤ v ≤ 2π
(b)
z
y
x
(c) Let S1 be the surface corresponding to the parameter domain 0 ≤ u ≤ pi, 0 ≤ v ≤ 2π.
Then
Z π
p
A(S1 ) = 2π
sin(x) 1 + cos2 xdx
0
√
√
√ = ln( 2 + 1) − ln( 2 − 1) = 2 2 π
√
√ √
Finding the area of the entire surface S, we have A(S) = 4A(S1 ) = ln( 2 + 1) − ln( 2 − 1) = 2 2 4π
49. The surface is a plane passing through the point (x0 , y0 , z0 ) with a normal vector n = v1 × v2 .
50. x = 5 sin φ cos θ + 2, y = 5 sin φ sin θ + 3, z = 5 cos φ + 4
15.6
Surface Integrals
√
z
4x2 dA
1. zx = −2x, zy = 0; dS = 1 +
Z Z
Z 4 Z √2 p
Z
xdS =
x 1 + 4x2 dxdy =
S
0
Z
=
0
0
4
13
26
dy =
6
3
0
√
4
1
(1 + 4x2 )3/2
12
2
2
z=2-x2
dy
0
R
2
x
4
y
15.6. SURFACE INTEGRALS
253
2. See Problem 1.
Z Z
Z Z
Z
xy(9 − 4z)dS =
xy(1 + 4x2 )dS =
S
S
4
Z
y
(1 + 4x2 )5/2
20
=
0
3. zx = p
x
x2
y2
, zy = p
y
x2
y2
dy =
; dS =
√
0
4
242
121
ydy =
20
10
Z
4
0
2dA.
121
ydy =
10
2π
=
√ Z
2
2π
2π
0
2π
0
x
y2
0
(r cos θ)r3/2 rdrdθ
1
R
1
Z
r7/2 cos θdrdθ
y
1
x
1
2 9/2
r cos θ dθ
9
0
√
2
2 2
cos θdθ =
sin θ
9
9
y
2π
= 0.
0
√
z
2dA.
+
+
Using
Z Z polar coordinates,
Z Z
p
√
(x + y + z)dS =
R(x + y + x2 + y 2 ) 2dA
x2
=
0
0
√ Z
= 2
√ Z
= 2
4
z=Mx2+y2
0
0
1
1
Z
1 2
y
2
z
R
√ Z
= 2
4. zx = p
xy(1 + 4x2 )3/2 dxdy
0
Z
0
√
2
Z
0
√
2
+
+
Using
Z Z polar coordinates,
Z Z
√
xz 3 dS =
x(x2 + y 2 )3/2 2dA
S
4
, zy = p
x2
y2
; dS =
z=Mx2+y2
4
S
=
√ Z
2
0
√ Z
= 2
0
2π
Z
4
(r cos θ + r sin θ + r)rdrdθ
1
2π Z
4
R
4
r2 (1 + cos θ + sin θ)drdθ
y
4
x
1
4
√ Z 2π 1 3
2
r (1 + cos θ + sin θ) dθ
3
0
1
√ Z 2π
√
63 2
=
(1 + cos θ + sin θ)dθ = 21 2(θ + sin θ − cos θ)
3
0
=
2π
0
√
= 42 2π.
484
5
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254
CHAPTER 15. VECTOR INTEGRAL CALCULUS
x
36 − x2 − y 2 , zx = − p
,
36 − x2 − y 2
y
;
zy = −
2 − y2
36
−
x
s
y2
x2
dS = 1 +
+
dA
36 − x2 − y 2
36 − x2 − y 2
6
=p
dA.
36 − x2 − y 2
Using polar coordinates,
5. z =
p
z
6
z=M36-x2-y2
6
R
y
6
x
Z Z
2
Z Z
2
(x + y )zdS =
S
p
(x + y ) 36 − x2 − y 2 p
2
R
Z
2π
=6
0
2
6
1 4
r dθ = 6
4 0
Z
−1
√ Z
= 2
=
√
2
y(x + 1)2
z=x+1
(1 + 2x − 2x3 − x4 )dx
1
1
1
√
8 2
=
5
y
R
x
z
2
0
1
2
dx
0
2
0
Z
z
1−x2
1
−1
1
1
2 x + x2 − x4 − x5
2
5
−1
p
7. zx = −x, zy = −y; dS = 1 + x2 + y 2 dA
Z Z
Z 1Z 1 p
xydS =
xy 1 + x2 + y 2 dxdy
S
0
0
(1 − x )(x + 1) dx
−1
Z 1
√
r2 rdrdθ
dA = 6
324dθ = 972π.
−1
=
6
Z
0
0
1
2
36 − x2 = y 2
2π
2π
√
6. zx = 1, zy = 0; dS = 2dA
Z Z
Z 1 Z 1−x2
√ Z
√
z 2 dS =
(x + 1)2 2dydx = 2
S
Z
6
1
1
y(1 + x2 + y 2 )3/2 dy
3
0
0
Z 1
1
1
=
y(2 + y 2 )3/2 − y(1 + y 2 )3/2 dy
3
3
0
1
1
1
2 5/2
2 5/2
=
(2 + y ) − (1 + y )
15
15
0
1 5/2
7/2
=
(3 − 2 + 1)
15
z=2-x2/2-y2/2
=
1
1
x
y
15.6. SURFACE INTEGRALS
255
z
1
1
1
8. z =
+ x2 + y 2 ,
zx = x,
zy = y;
2
2
2
p
1 + x2 + y 2 dA.
Using
Z Z polar coordinates,
Z Z
p
2zdS =
(1 + x2 + y 2 ) 1 + x2 + y 2 dA
S
Z
R
π/2 Z 1
=
1
z=1/2+x2/2+y2/2
1
p
(1 + r ) 1 + r2 rdrdθ
2
1
R
y
x
0
π/3
Z
dS =
π/2
Z
=
1
(1 + r2 )3/2 rdrdθ
0
π/3
1
Z
1 π/2 5/2
1
(2 − 1)dθ
(1 + r2 )5/2 dθ =
5 π/3
π/3 5
0
√
√
4 2 − 1 π π (4 2 − 1)π
=
−
=
.
5
2
3
30
Z
π/2
=
z
√
3
9. yZx =
= 1 + 4x2 dA
Z 2x, yz = 0; ZdS3 Z
2
p
√
24 yzdS =
24xz 1 + 4x2 dxdz
S
0
Z
=
y=x2
R
0
2
3
2 3/2
2z(1 + 4x )
0
4
y
2
dz
x
0
= 2(173/2 − 1)
3
Z
zdz = 2(173/2 − 1)
0
1 2
z
2
3
0
= 9(173/2 − 1)
z
p
10. xy = −2y, xz = −2z; dS = 1 + 4y 2 + 4z 2 dA
Using polar coordinates,
Z Z
Z π/2 Z 2
(1 + 4y 2 + 4z 2 )1/2 dS =
(1 + 4r2 )rdrdθ
S
0
Z
π/2
0
1
16
R
1
=
=
2
2
1
(1 + 4r2 )2 dθ
16
2
1
Z
π/2
12dθ =
0
3π
.
8
1
(6 − x − 3z).
2
√
p
1
3
14
yz = − , yz = − ; dS = 1 + 1/4 + 9/4 =
.
2
2
2
11. Write the equation of the surface as y =
3
x
z=4-y2-z2
y
256
CHAPTER 15. VECTOR INTEGRAL CALCULUS
Z Z
Z
2
2
Z
6−3z
(3z + 4yz)dS =
S
0
0
√
14
1
2
3z + 4z (6 − x − 3z)
dxdz
2
2
√
6−3z
Z
14 2 2
dz
3z x − z(6 − x − 3z)2
2
0
0
√ Z 2
14
[3z 2 (6 − 3z) − 0] − [0 − z(6 − 3z)2 ]dz
=
2
0
√
√
√ Z 2
2
√
14
14
14
2
2
3
(36z − 18z )dz =
(18z − 6z ) =
(72 − 48) = 12 14
=
2
2
2
0
=
0
12. Write
the equation
of the surface as x = 6 − 2y − 3z. Then xy = −2, xz = −3; dS =
√
√
1 + 4 + 9 = 14.
Z Z
Z 2 Z 3−3z/2
3−3z/2
√ Z 2
√
(3z 2 + 4yz)dS =
dz
(3yz + 2y 2 z)
(3z 2 + 4yz) 14dydz = 14
S
0
0
0
0
√ Z 2
z 2
z
45 2 9 3
+ 18z 1 −
dz = 14
9z 1 −
27z − z + z dz
= 14
2
2
2
2
0
0
2
√
√
√
27 2 15 3 9 4
= 14(54 − 60 + 18) = 2 14
z − z + z
= 14
2
2
8
0
√
Z
2
z
1
z=1-x-y
2
13. The density is ρ = kx . The
√ surface is z = 1 − x − y. Then
zx = −1, zy = −1; dS = 3dA.
Z Z
Z 1 Z 1−x √
√ Z 1 1 3 1−x
2
2
m=
kx dS = k
x 3dydx = 3k
x
dx
S
0
0
0 3
1
0
√ Z 1
√ √
1
x
3
3
1
3
3
4
=
k
(1 − x) dx =
k − (1 − x)
=
k
3
3
4
12
0
1
y
0
z
2
z=M4-x2-y2
R
2
x
2
y
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15.6. SURFACE INTEGRALS
257
y
x
, zy = − p
; dS =
14. zx = − p
2
2
4−x −y
4 − x2 − y 2
s
y2
2
x2
+
dA = p
dA.
1+
4 − x2 − y 2
4 − x2 − y 2
4 − x2 − y 2
Using symmetry and polar coordinates,
Z Z
Z π/2 Z 2
2
m=4
(r2 cos θ sin θ) √
rdrdθ
|xy|dS = 4
4
− r2
0
S
0
Z π/2 Z 2
r2 (4 − r2 )−1/2 sin 2θ(rdr)dθ
=4
0
0
u = 4 − r2 , du = −2rdr, r2 = 4 − u
Z π/2 Z 0
Z π/2 Z 0
1
−1/2
(4 − u)u
sin 2θ − du dθ = −2
(4u−1/2 − u1/2 ) sin 2θdudθ
=4
2
4
4
0
0
0
Z π/2 Z π/2 2 3/2
32
1/2
= −2
8u − u
sin 2θdθ = −2
− sin 2θ dθ
3
3
0
0
4
π/2
64
64
1
=
=
− cos 2θ
.
3
2
3
0
15. The surface is g(x, y, z) = y 2 + z 2 − 4 = 0. ∇g = 2yj + 2zk,
p
yi + zk
|∇g| = 2 y 2 + z 2 ; ∇ p
;
y2 + z2
yz
3yz
2yz
+ p
= p
;; z =
F·∇ = p
2
2
2
2
y +z
y +z
y2 + z2
p
4 − y 2 , zx = 0,
s
y
y2
2
zy = − p
dA = p
; dS = 1 +
dA
2
2
4
−
y
4 − y2
Z 4Z− y
Z Z
3yz
2
p
p
Flux =
F · ndS =
dA
2 + z2
y
4
− y2
S
R
p
Z Z
3y 4 − y 2
2
p
p
=
dA
y2 + 4 − y2 4 − y2
R
Z 3Z 2
Z 3
Z 3
2
3 2
=
y dx =
6dx = 18
3ydydx =
0
0
0
0 2
0
z
2
z=M4-y2
R
3
x
2
y
258
CHAPTER 15. VECTOR INTEGRAL CALCULUS
16. The surface is g(x, y, z) = x2 +py 2 + z − 5 = 0. ∇g =
2xi + 2yj + k,
|∇g| =
1 + 4x2 + 4y 2 ;
n =
z
2xi + 2yj + k
p
;
F · n = p
;
zx =
1 + 4x2 + 4y 2
1 + 4x2 + 4y 2
p
−2x, zy = −2y, dS = 1 + 4x2 + 4y 2 dA. Using polar
coordinates,
Z Z
Z Z
p
z
p
Flux =
1 + 4x2 + 4y 2 dA
F · ndS =
1 + 4x2 + 4y 2
S
R
Z Z
=
(5 − x2 − y 2 )dA
R
2π Z 2
Z
5
z=5-x2-y2
2
2
2
x
(5 − r )rdrdθ
=
z
y
R
0
0
2π
Z
5 2 1 4
r − r
2
4
=
0
2
Z
2π
6dθ = 12π.
dθ =
0
0
2xi + 2yj + k
2x2 + 2y 2 + z
17. From Problem 16, n = p
. Then F · n = p
. Also, from Problem
1 + 4x2 + 4y 2
1 + 4x2 + 4y 2
p
16, dS = 1 + 4x2 + 4y 2 dA. Using polar coordinates,
Z Z
Z Z
Z Z
2x2 + 2y 2 + z p
2
2
p
Flux =
F · ndS =
1 + 4x + 4y dA =
(2x2 + 2y 2 + 5 − x2 − y 2 )dA
1 + 4x2 + 4y 2
S
R
R
2
Z 2π Z 2
Z 2π Z 2π
1 4 5 2
2
=
(r + 5)rdrdθ =
dθ =
14dθ = 28π.
r + r
4
2
0
0
0
0
0
18. The surface is g(x, y, z) = z − x − 3 = 0.
∇g =
√
−i + k
−i + k, |∇g| = 2; n = √ ;
2
√
1
1
F · n = √ x3 y + √ xy 3 zx = 1, zy = 0, dS = 2dA.
2
2
Using polar
Z Z coordinates,Z Z
Z Z
√
1 3
3
√ (x y + xy ) 2dA =
Flux =
F · ndS =
xy(x2 + y 2 )dA
2
S
R
R
Z π/2 Z 2 cos θ
=
(r2 cos θ sin θ)r2 rdrdθ
0
5
z=x+3
1 y
0
π/2
Z
z
Z
=
5
x R
r cos θ sin θdrdθ
0
Z
=
0
r=2cos θ
2
2 cos θ
0
π/2
1 6
r cos θ sin θ
6
2 cos θ
0
1
dθ =
6
Z
0
π/2
32
64 cos θ sin θdθ =
3
7
1
8
− cos θ
8
π/2
=
0
4
.
3
15.6. SURFACE INTEGRALS
259
z
19. The surfacep
is g(x, y, z) = x2 + y 2 + z − 4. ∇g = 2xi + 2yj +
k, |∇g| = 4x2 + 4y 2 + 1;
2xi + 2yj + k
x3 + y 3 + z
n = p
; F·n = p
; zx =
4x2 + 4y 2 + 1
4x2 + 4y 2 + 1
−2x, p
zy = −2y,
1 +Z4x2 + 4y 2 dA.Z Using
polar coordinates,
dS = Z
Z
3
Flux =
F · ndS =
(x + y 3 + z)dA
R
Z ZS
2
2
=
(4 − x − y + x3 + y 3 )dA
Z
R
2π Z 2
=
4
z=4-x2-y2
2
2
y
R
x
(4 − r2 + r3 cos3 θ + r3 sin3 θ)rdrdθ
0
0
2
1 5 3
1 4 1 5
3
dθ
2r − r + r cos θ + r sin θ
=
4
5
5
0
0
Z 2π 32
32
2π
cos3 θ +
sin3 θ dθ = 4θ|0 + 0 + 0 = 8π.
=
4+
5
5
0
Z
2π
2
z
20. The surface is g(x,√
y, z) = x + y + z −√6.
∇g =
i + j + k, |∇g| √
= 3; n = (i + j + k)/ 3; F · n =
y
x
(e
zx = −1,
zy = −1,
dS =
√ + e + 18y)/√ 3;
1 + 1 + 1dA = 3dA.
Z Z
Z Z
Flux =
F · ndS =
(ey + ex + 18y)dA
S
Z
6
Z
=
0
Z
z=6-x-y
R
6
y
6
x
r
6−x
(ey + ex + 18y)dydx
0
6−x
6
=
0
Z 6
=
6
(ey + yex + 9y 2 )
dx
0
[e6−x + (6 − x)ex + 9(6 − x)2 − 1]dx
0
= [−e6−x + 6ex − xex + ex − 3(6 − x)3 − x]
6
0
= (−1 + 6e6 − 6e6 + e6 − 6) − (−e6 + 6 + 1 − 648) = 2e6 + 634 ≈ 1440.86
p
21. For S1 : g(x, y, z) = x2 + y 2 − z, ∇g = 2xi + 2yj − k, |∇g| = 4x2 + 4y 2 + 1; n1 =
p
2xy 2 + 2x2 y − 5z
2xi + 2yj − k
p
; F · n1 = p
; zx = 2x, zy = 2y, dS1 = 1 + 4x2 + 4y 2 dA.
2
2
2
2
4x + 4y + 1
4x + 4y + 1
For S2 : g(x, y, z) = z − 1, ∇g = k; |∇g| = 1; n2 = k; F · n2 = 5z; zx = 0, zy = 0, dS2 =
dA. Using polar coordinates and R : x2 + y 2 ≤ 1 we have
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260
CHAPTER 15. VECTOR INTEGRAL CALCULUS
Z Z
Z Z
Z Z
2
Z Z
2
F · n1 dS1 +
F · n2 dS2 =
(2xy + 2x y − 5z)dA +
S2
R
Z ZS1
=
[2xy 2 + 2x2 y − 5(x2 + y 2 ) + 5(1)]dA
Flux =
R
2π Z 1
Z
=
5zdA
R
(2r3 cos θ sin2 θ + 2r3 cos2 θ sin θ − 5r2 + 5)rdrdθ
0
0
1
2 5
2
5
5
r cos θ sin2 θ + r5 cos2 θ sin θ − r4 + r2
dθ
5
5
4
2
0
0
2π
Z 2π 2π
2
2 1
1
5
5
2
3
2
3
=
(cos θ sin θ + cos θ sin θ) +
dθ =
sin θ − cos θ
+ θ
5
4
5 3
3
4 0
0
0
2
1
1
5
5
=
− − −
+ π = π.
5
3
3
2
2
p
22. For S1 : g(x, y, z) = x2 + y 2 + z − 4, ∇g = 2xi + 2yj + k, |∇g| = 4x2 + 4y 2 + 1; n1 =
p
p
2xi + 2yj + k
p
; F·n1 = 6z 2 / 4x2 + 4y 2 + 1; zx = −2x, zy = −2y, dS1 = 1 + 4x2 + 4y 2 dA.
4x2 + 4y 2 + 1
p
For S2 : g(x, y, z) = x2 + y 2 − z, ∇g = 2xi + 2yj − k, |∇g| = 4x2 + 4y 2 + 1; n2 =
p
p
2xi + 2yj − k
p
; F·n2 = −6z 2 / 4x2 + 4y 2 + 1; zx = 2x, zy = 2y, dS2 = 1 + 4x2 + 4y 2 dA.
4x2 + y 2 + 1
2
2
Using polar
Z Z
Z R
Z : x + y ≤ 2 ZweZhave
Z Z coordinates and
F · n2 dS2 =
6z 2 dA +
−6z 2 dA
F · n1 dS1 +
Flux =
Z
2π
=
R
S1
S1
Z Z
2
2 2
2
Z
2 2
2π
√
Z
2
[(4 − r2 )2 − r4 ]rdrdθ
[6(4 − x − y ) − 6(x + y ) ]dA = 6
=
R
Z
=6
0
√
2π
1
1 6
2 3
− (4 − r ) − r
6
6
0
2
Z
dθ = −
0
0
2π
3
3
√
6
Z
2π
[(2 − 4 ) + ( 2) ]dθ =
0
48dθ = 96π.
0
23. The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. ∇g = 2xi + 2yj + 2zk, |∇g| =
p
xi + yj + zk
xi + yj + zk
2 x2 + y 2 + z 2 ; n = p
; F · n = −(2xi + 2yj + 2zk) · p
=
2
2
2
x +y +z
x2 + y 2 + z 2
p
2x2 + 2y 2 + 2z 2
−p
= −2 x2 + y 2 + z 2 = −2a.
2
2
x2 +
R Ry + z
Flux =
−2adS
= −2a × area = −2a(4πa2 ) = −8πa3
S
24. n1 = k, n2 = −i, n3 = j, n4 = −k, n5 = i, n6 = −j; F · n1 = z = 1, F · n2 = −x =
0, F · nR3 R= y = 1, RF R· n4 = −z R=R0, F · n5 = x = 1, F · n6 = −y = 0;
Flux =
1dS +
1dS +
1dS = 3
S1
S3
S5
a
xi + yj + zk
and dS = p
dA.
25. Refering to the solution to Problem 23, we find n = p
x2 + y 2 + z 2
a2 − x2 − y 2
Now
F · n = kq
r
r
kq
kq
kq
kq
·
= 4 |r|2 = 2 = 2
= 2
|r|3 |r|
|r|
|r|
x + y2 + z2
a
15.6. SURFACE INTEGRALS
261
Z Z
and
Z Z
F · ndS =
Flux =
S
S
26. We are given σ = kz. Now zx − p
y
kq
kq
kq
dS = 2 × area = 2 (4πa2 ) = 4πkq.
a2
a
a
x
16 − x2 − y 2
,
z
zy = − p
;
2
2
s 16 − x − y
x2
y2
+
dA
dS = 1 +
16 − x2 − y 2
16 − x2 − y 2
4
=p
dA
16 − x2 − y 2
UsingZpolar
Z coordinates,
Z Z p
4
Q=
16 − x2 − y 2 p
dA
kzdS = k
16 − x2 − y 2
S
R
Z 2π Z 3
= 4k
rdrdθ
0
Z
4
z=M16-x2-y2
3
R
y
3
x
0
2π
= 4k
0
3
1 2
r dθ = 4k
2 0
Z
2π
0
9
dθ = 36πk.
2
27. The surface√ is z = 6 − √
2x − 3y. Then zx = −2, zy =
−3, dS = 1 + 4 + 9 = 14dA. The area of the surface is
Z Z
Z
A(s) =
3
Z
dS =
S
0
0
2−2x/3
√
14dydx =
√
Z
14
0
3
z
6
2
2 − x dx
3
z=6-2x-3y
3
√
1
= 14(2x − x2 ) = 3 14.
3
0
Z Z
Z 3 Z 2−2x/3 √
1
1
2 y
x= √
14xdydx
xdS = √
3
R
3 14
3 14 0 0
s
Z
Z 2−2x/3
x
1 3
1 3
2 2
=
dx =
xy
2x − x dx
3 0
3 0
3
0
3
2
1
=1
x2 − x3
=
3
9
0
Z Z
Z 3 Z 2−2x/3 √
Z
2−2x/3
1
1
1 31 2
√
√
ydS =
y=
14ydydx =
y
dx
3 0 2 0
3 14
3 14 0 0
S
2
3
Z 1 3
2
1
1
2
2
=
2 − x dx =
− (2 − x)3
=
6 0
3
6
2
3
3
0
Z Z
Z 3 Z 2−2x/3
√
1
1
z= √
zdS = √
(6 − 2x − 3y) 14dydx
3 14
3 14 0 0
S
2−2x/3
Z 3
Z 1
3 2
1 3
2 2
1
2 3
2
=
6y − 2xy − y
dx =
6 − 4x + x dx =
6x − 2x + x
3 0
2
3 0
3
3
9
0
The centroid is (1, 2/3, 2).
√
3
=2
0
262
CHAPTER 15. VECTOR INTEGRAL CALCULUS
28. The area of the hemisphere is A(s) = 2πa2 . By symmetry,
s x = y = 0.
y
x2
x
y2
, zy = − p
; dS = 1 + 2
zx = − p
+ 2
dA =
2
2
a −x −z
a − x2 − y 2
a2 − x2 − y 2
a 2 − x2 − y 2
a
p
dA
a2 − x2 − y 2
UsingZ polar
coordinates,Z Z
Z
Z 2π Z a
p
1
a
1
zdS
2
2
2
rdrdθ
z=
=
a −x −y p
dA =
2
2πa2
2πa 0
a2 − x2 − y 2
0
S 2πa
R
Z 2π
Z 2π
a
1
1 2
1
1 2
a
=
r dθ =
s dθ = .
2πa 0 2 0
2πa 0 2
2
The centroid is (0, 0, a/2).
p
2
2
2
2
29. (a) The region
pin the xy-plane is x + y ≤ 16. From zx = −x/ x + y and
2
2
zy = −y/ x + y we see that
dS =
p
√
1 + x2 /(x2 + y 2 ) + y 2 /(x2 + y 2 )dA = 2da
Z Z
and
A(S) =
dS =
S
Then
x=
Z Z
1
√
1
dS = √
16 2π
S
Z Z √
2dA =
√
√
2π42 = 16 2π.
R
Z Z √
1
2xdA =
16π
Z
2π
Z
4
r cos θrdrdθ
16 2π
R
0
0
Z 2π
Z 2π
4
4
1 3
1
r cos θ dθ =
cos θdθ = 0
=
16π 0 3
3π 0
0
Z Z
Z Z √
Z 2π Z 4
1
1
1
ydS = √
y= √
2ydA =
r sin θrdrdθ
16π 0
16 2π
16 2π
S
R
0
Z 2π
Z 2π
4
4
1
1 3
=
r cos θ dθ =
sin θdθ = 0
16π 0 3
3π 0
0
Z 2π Z 4
Z Z
Z Z √
p
1
1
1
z= √
zdS = √
2(4 − x2 + y 2 )dA =
(4 − r)rdrdθ
16π 0
16 2π
16 2π
S
R
0
4
Z 2π Z 2π
2
1 3
4
1
2
dθ =
2r − r
dθ = .
=
16π 0
3
3π
3
0
0
The centroid is (0,0,4/3).
√ Z Z
√ Z
2
2
(x + y )kdS = k 2
(x + y )dA = k 2
Z Z
2
(b) Iz =
2
S
R
√ Z
4
√ Z 2π
√
k 2 2π 4
=
r dθ = 64k 2
dθ = 128kπ 2
4
0
0
0
2π
Z
4
r2 rdrdθ
0
0
30. The surface is g(x, y, z) = z − f (x, y) = 0. ∇g = −fx i − fy j + k, |∇g| =
q
fx2 + fy2 + 1; n =
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15.7. CURL AND DIVERGENCE
263
−fx i − fy j + k
−P fx − Qfy + R
q
; F·n= q
;
1 + fx2 + fy2
1 + fx2 + fy2
q
dS = 1 + fx2 + fy2 dA
RR
R R −P fx − Qfy + R q
RR
q
F · ndS =
1 + fx2 + fy2 dA =
(−P fx − Qfy + R)dA
s
R
R
1 + fx2 + fy2
15.7
Curl and Divergence
1. curlF = (x − y)i + (x − y)j; divF = 2z
2. curlF = −2x2 i + (10y − 18x2 )j + (4xz − 10z)k; divF = 0
3. curlF = 0; divF = 4y + 8z
4. curlF = (xe2y + ye−yz + 2xye2y )i − ye2y j + 3(x − y)2 k; divF = 3(x − y)2 − ze−yz
5. curlF = (4y 3 − 6xz 2 )i + (2x3 − 3x2 )k; divF = 6xy
3
6. curlF = −x3 zi + (3x2 yz − z)j + ( x2 y 2 − y − 15y 2 )k; divF = (x3 y − x) − (x3 y − x) = 0
2
7. curlF = (3e−z − 8yz)i − xe−z j; divF = e−z + 4z 2 − 3ye−z
8. curlF = (2xyz 3 + 3y)i + (y ln x − y 2 z 3 )j + (2 − z ln x)k; divF =
yz
− 3z + 3xy 2 z 2
x
9. curlF = (xy 2 ey + 2xyey + x3 yez + x3 yzez )i − y 2 ey j + (−3x2 yzez − xex )k;
divF = xyex + yex − x3 zez
10. curlF = (5xye5xy + e5xy + 3xz 3 sin xz 3 − cos xz 3 )i + (x2 y cos yz − 5y 2 e5xy )j
+ (−z 4 sin xz 3 − x2 z cos yz)k; divF = 2x sin yz
11. div r = 1 + 1 + 1 = 3
12. curl r =
i
∂/∂x
x
j
∂/∂y
y
k
∂/∂z
z
13. a×∇ =
i
a1
∂/∂x
j
a2
∂/∂y
k
a3
∂/∂z
(a × ∇) × r =
i
∂
∂
a2
− a3
∂z
∂y
x
= 0i − 0j + 0k = 0
=
∂
∂
a2
− a3
∂z
∂y
j
∂
∂
a3
− a1
∂x
∂z
y
∂
∂
i+ a3
− a1
∂x
∂z
∂
∂
j+ a1
− a2
k
∂y
∂x
k
∂
∂
a1
− a2
∂y
∂x
z
= (−a1 − a1 )i − (a2 + a2 )j + (−a3 − a3 )k = −2a
∂
∂
∂
+ a2
+ a3
r
14. ∇ × (a × r) = (∇ · r)a − (∇ · a)r = (1 + 1 + 1)a − a1
∂x
∂y
∂z
= 3i − (a1 i + a2 j + a3 k) = 2a
264
CHAPTER 15. VECTOR INTEGRAL CALCULUS
15. ∇ · (a × r) =
∂/∂x
a1
x
16. ∇ × r =
i
∂/∂x
x
17. r × a =
i
x
a1
j
y
a2
∇ × [(r · r)a] =
∂/∂y
a2
y
j
∂/∂y
y
k
z
a3
∂/∂z
a3
z
k
∂/∂z
z
=
∂
∂
∂
(a2 z − a3 y) −
(a1 z − a3 x) +
(a1 y − a2 x) = 0
∂x
∂y
∂z
= 0; a × (∇ × r) = a × 0 = 0
= (a3 y − a2 z)i − (a3 x − a1 z)j + (a2 x − a1 y)k; r · r = x2 + y 2 + z 2
i
∂/∂x
(r · r)a1
j
∂/∂y
(r · r)a2
k
∂/∂z
(r · r)a3
= (2ya3 − 2za2 )i − (2xa3 − 2za1 )j + (2xa3 − 2ya1 )k = 2(r × a)
18. r · a = a1 x + a2 y + a3 z; r · r = x2 + y 2 + z 2 ; ∇ · [(r × r)a] = 2xa1 + 2ya2 + 2za3 = 2(r · a)
19. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k.
∇ · (F + G) = ∇ · [(P + S)i + (Q + T )j + (R + U )k] = Px + Sx + Qy + Ty + Rx + Uz
= (Px + Qy + Rz ) + (Sx + Ty + Uz ) = ∇ · F + ∇ · G
20. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k.
j
j
k
∂/∂z
∇ × (F + G) = ∂/∂x ∂/∂y
P +S Q+T R+U
= (Ry + Uy − Qz − Tz )i − (Rx + Ux − Pz − Sz )j + (Qx + Tx − Py − Sy )k
= (Ry − Qz )i − (Rx − Rz )j + (Qx − Py )k + (Uy − Tz )i − (Ux − Sz )j + (Tx − Sy )k
=∇×F+∇×G
21. ∇ · (f F) = ∇ · (f P i + f Qj + f Rk) = f Px + P fx + f Qy + Qfy + f Rz + Rfz
= f (Px + Qy + Rz ) + (P fx + Qfy + Rfz ) = f (∇ · F) + F · (∇f )
22. ∇ × (f F) =
j
∂/∂x
fP
j
∂/∂y
fQ
k
∂/∂z
fR
= (f Ry + Rfy − f Qz − Qfz )i − (f Rx + Rfx − f Pz − P fz )j
(f Qx + Qfx − f Py − P fy )k
= (f Ry − f Qz )i − (f Rx − f Pz )j + (f Qx − f Py )k + (Rfy − Qfz )i
− (Rfx − P fz )j + (Qfx − P fy )k
f [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k] +
i
fx
P
j
fy
Q
k
fz
R
= f (∇ × F) + (∇f ) × F
15.7. CURL AND DIVERGENCE
265
23. Assuming continuous second partial derivatives,
curl(gradf ) = ∇ × (fx i + fy j + fz k) =
j
∂/∂x
fx
j
∂/∂y
fy
k
∂/∂z
fz
= (fzy − fyz )i − (fzx − fxz )j + (fyx − fxy )k = 0.
24. Assuming continuous second partial derivatives,
div(curlF) = ∇ · [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k]
= (Ryx − Qzx − (Rxy − Pzy ) + (Qxz − Pyz ) = 0.
25. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + Y (x, y, z)j + U (x, y, z)k.
i j k
F × G = P Q R = (QU − RT )i − (P U − RS)j + (P T − QS)k
S T U
div (F × G) = (QUx + Qx U − RTx − Rx T ) − (P Uy + Py U − RSy − Ry S)
+ (P Tz + Pz T − QSz − Qz S)
= S(Ry − Qz ) + T (Pz − Rx ) + U (Qx − Py ) − P (U − y − Tz ) − Q(Sz − Ux )
− R(Rx − Sy )
= G · (curlF) − F · (curlG)
26. Using Problems 20 and 23,
curl(curlF + gradf ) = ∇ × (curlF + gradf ) = ∇ × (curlF) + ∇ × (gradf )
= curl(curlF) + curl(gradf ) = curl(curlF) + 0 = curl(curlF).
27. curl F = −8yzi − 2zj − xk; curl (curl F) = 2i − (8y − 1)j + 8zk
28. For F = P i + Qj + Rk,
curl (curl F) = (Qxy − Pyy − Pzz + Rxz )i + (Ryz − Qzz − Qxx + Pyx )j
+ (Pzx − Rxx − Ryy + Qzy )k
and
−∇2 F + grad(divF) = −(Pxx + Pyy + Pzz )i − (Qxx + Qyy + Qzz )j − (Rxx + Ryy + Rzz )k
+ grad(Px + Qy + Rz )
= −Pxx i − Qyy j − Rzz k + (−Pyy − Pzz )i + (−Qxx − Qzz )j
+ (−Rxx − Ryy )k + (Pxx + Qyx + Rzx )i + (Pxy + Qyy + Rzy )i
+ (Pxz + Qyz + Rzz )k
= (−P − P + Q + R)i + (−Qxx − Qzz + Pxy + Rzy )j
+ (−Rxx − Ryy + Pxz + Qyz )k.
Thus, curl(curlF) = −∇2 F + grad(divF).
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266
CHAPTER 15. VECTOR INTEGRAL CALCULUS
29. fz = 6x + 4y − 9z; fxx = 6; fy = 10y + 4z; fyy = 10; fz = −9x − 16z; fzz = −16;
∇2 f = fxx + fyy + fzz = 6 + 10 − 16 = 0
30. Using Problem 21, ∇ · (f ∇f ) = f (∇ · ∇f ) + ∇f · ∇f = f (∇2 f ) + |∇f |2 .
31. fx = 6x + 4y − 9z; fxx = 6; fy = 10y + 4x; fyy = 10; fz = −9x − 16z; fzz = −16;
∇2 f + fx x + fy y + zz z = 6 + 10 − 16 = 0
(a − x)A
∂f
=
3/2
2
∂x
[(x − a) + (y − b)2 + (z − c)2 ]
2
2
2
A 2(x − a) − (y − b) − (z − c)2
∂ f
=
5/2
∂x2
[(x − a)2 + (y − b)2 + (z − c)2 ]
∂f
(b − y)A
=
3/2
2
∂y
[(x − a) + (y − b)2 + (z − c)2 ]
2
2
2
A 2(y − b) − (x − a) − (z − c)2
∂ f
=
5/2
∂y 2
[(x − a)2 + (y − b)2 + (z − c)2 ]
(c − z)A
∂f
=
3/2
2
∂z
[(x − a) + (y − b)2 + (z − c)2 ]
A 2(z − c)2 − (x − a)2 − (y − b)2
∂2f
=
5/2
∂z 2
[(x − a)2 + (y − b)2 + (z − c)2 ]
2
2
2
∂ f
∂ f
∂ f
Now
+ 2 + 2 = 0. Hence f is harmonic.
∂x2
∂y
∂z
1
4xy
4xy
33. fx =
− 2
=− 2
2
2
2
2
(x + y − 1)
(x + y − 1)2 + 4y 2
4y
1+ 2
2
2
(x + y − 1)
12x4 y − 4y 5 + 8x2 y 3 − 8x2 y − 8y 3 − 4y
[(x2 + y 2 − 1)2 + 4y 2 ]4y − 4xy[4x(x2 + y 2 − 1)]
fxx = −
=
[(x2+ y 2 − 1) + 4y 2 ]2
[(x2 + y 2 − 1)2 + 4y 2 ]2
2(x2 + y 2 − 1) − 4y 2
1
2(x2 + y 2 − 1)2
fy =
= 2
(x2 + y 2 − 1)2
(x + y 2 − 1)2 + 4y 2
4y 2
1+ 2
2
2
(x + y − 1)
[(x2 + y 2 − 1)2 + 4y 2 ](−4y) − 2(x2 + y 2 − 1)2 [4y(x2 + y 2 − 1)2 + 8y]
fyy =
[(x2 + y 2 − 1)2 + 4y 2 ]2
4
5
2 3
−12x y + 4y − 8x y + 8x2 y + 8y 3 + 4y
=
[(x2 + y 2 − 1)2 + 4y 2 ]2
∇2 f = fxx + fyy = 0
32.
34.
∂f
∂2f
= 4x3 − 12xy 2 ,
= 12x2 − 12y 2 ,
∂x
∂x2
∂f
∂2f
= −12x2 y + 4y 3 ,
= −12x2 + 12y 2 ,
∂y
∂y 2
∂2f
∂2f
Now
+ 2 = 0. Hence f is harmonic.
2
∂x
∂y
35. Using Problems 25 and 23,
˙
∇ · F = div (∇f × ∇g) = ∇g (curl
∇f ) − ∇f · (curl g) = ∇g · 0 − ∇f · 0 = 0.
15.7. CURL AND DIVERGENCE
267
36. Recall that a · (a × b) = 0. Then, using Problmems 25, 23, and 22,
∇ · F = (∇f × f ∇g) = f ∇g · (curl ∇f ) − ∇f · (curl f ∇g) = f ∇g · 0 − ∇f · (∇ × f ∇g)
= −∇f · [f (∇ × ∇g) + (∇f × ∇g)] = −∇f · [f curl ∇g + (∇f × ∇g)]
= −∇f · [f 0 + (∇f × ∇g)] = −∇f · (∇f × ∇g) = 0.
2
37. The surface is g(x, y) = x2 + y 2 +
z
p4z − 4 = 0.
2
∇g = 2xi + 2yj + 8zk, |∇g| = 2 x + y 2 + 16z 2 ;
1
z=M1-x2/4-y2/4
2xi + 2yj + 8zk
xi + yj + 4zk
n= p
=p
;
2 x2 + y 2 + 16z 2
x2 + y 2 + 16z 2
2
y
12z(x2 + y 2 )
2
2
R
∇ × F = (3x − 3y )k, (∇ × F) · n = p
x2 + y 2 + 16z 2
2
r=2
Writing
the
equation
of
the
surface
as z
=
p
2
2
1 − x /4 − y /4, we have
x
x
zx
=
− p
,
zy
=
4 1 − x2 /4 − y 2p
/4
y
16 − 3x2 − 3y 2
− p
, and dS = p
dA.
2
2
4 1 − x /4 − y /4
2 4 − x2 − y 2
Then, using polar coordinates,
p
Z Z
Z Z
16 − 3x2 − 3y 2
12z(x2 − y 2 )
p
p
dA
Flux =
(∇ × F) · ∇dS =
x2 + y 2 + 16z 2 2 4 − x2 − y 2
S
R
p
Z Z p
6 1 − x2 /4 − y 2 /4(x2 − y 2 ) 16 − 3x2 − 3y 2 dA
p
p
=
x2 + y 2 + 16 − 4x2 − 4y 2 4 − x2 − y 2
R
Z π/4 Z 2 p
Z π/4 Z 2
p
=
1 − r2 /4(r2 cos2 θ − r2 sin2 θ)rdrdθ 4 − r2 =
3r2 cos 2θdrdθ
0
Z
0
π/4
=
0
38.
0
2
3 4
r cos 2θ dθ =
4
0
1
1
1
curl v = curl (ω × r) =
2
2
2
=
Z
π/4
0
π/4
12 cos 2θdθ = 6 sin 2θ|0
i
∂/∂x
ω2 z − ω3 y
j
∂/∂y
ω3 x − ω1 z
0
= 6.
k
∂/∂z
ω1 y − ω2 x
1
[(ω1 + ω1 )i − (ω2 − ω2 )j + (ω3 + ω3 )k] = ω1 i + ω2 j + ω3 k = ω
2
39. curl F = −Gm1 m2
i
∂/∂x
x/|r|3
j
∂/∂y
y/|r|3
k
∂/∂z
z/|r|3
= −Gm1 m2 [(−3yz/|r|5 + 3yz/|r|5 )i − (−3xz/|r|5 + 3xz/|r|5 )j + (−3xy/|r|5 + 3xy/|r|5 )k]
=0
div F = −Gm1 m2
−2x2 + y 2 + z 2
x2 − 2y 2 + z 2
x2 + y 2 − 2z 2
+
+
=0
|r|5/2
|r|5/2
|r|5/2
268
CHAPTER 15. VECTOR INTEGRAL CALCULUS
40. (a) Expressing the vertical component of V in polar coordinates, we have
2xy
2r2 sin θ cos θ
sin 2θ
=
=
2
2
2
4
(x + y )
r
r2
Similarly,
x2 − y 2
r2 (cos2 θ − sin2 θ)
cos 2θ
=
=
.
(x2 + y 2 )2
r4
r2
Since lim (sin 2θ)/r2 = lim (cos 2θ)/r2 , V ≈ Ai for r large or (x, y) far from the origin.
r→∞
r→∞
x2 − y 2
2Axy
(b) Identify P (x, y) = A 1 − 2
, Q(x, y) = − 2
, and R(x, y) = 0, we
(x − y 2 )2
(x − y 2 )2
have
Py =
2Ay(3x2 − y 2 )
2Ay(3x2 − y 2 )
,
Q
=
,
x
(x2 + y 2 )3
(x2 + y 2 )3
and Pz = Qz = Rx = Ry = 0.
Thus, curlV = (Ry − Qz )i + (Pz − Rx )j + (Qx − Py )k = 0 and V is irrotational.
2Ax(3y 2 − x2 )
2Ax(x2 − 3y 2 )
, Qy =
, and Rz = 0, ∇·F = Px +Qy +Rz = 0
2
2
3
(x + y )
(x2 + y 2 )3
and V is incompressible.
(c) Since Px =
41. We first note that curl (∂H/∂t) = ∂(curl H)/∂t and curl (∂E/∂t) = ∂(curl E)/∂t. Then,
from Problem 30,
−∇2 E = −∇2 E + 0 = −∇2 E + grad 0 = −∇2 E + grad (div E) = curl (curl E)
1 ∂H
1 ∂
1 ∂ 1 ∂E
1 ∂2E
= curl −
=−
curl H = −
=− 2
c ∂t
c ∂t
c ∂t c ∂t
c ∂t
and ∇2 E =
2
1 2
2
c2 ∂ E/∂t .
Similarly,
2
−∇ H = −∇ H + grad (div H) = curl (curl H) = curl
1 ∂
1 ∂H
1 ∂2H
=
−
=− 2
c ∂t
c ∂t
c ∂t2
and ∇2 H =
1 ∂E
c ∂t
=
1 ∂
curl E
c ∂t
1 2
∂ H/∂t2 .
c2
42. We note that div F = 2xyz − 2xyz + 1 = 1 6= 0. If F = curl G, then div (curl G) = div F = 1.
But, by problem 24, for any vector field G, div (curl G) = 0. Thus, F cannot be the curl of
G.
15.8
Stokes’ Theorem
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15.8. STOKES’ THEOREM
269
1. Surface Integral: curlF = −10k. Letting g(x, y, z) = −1,
we have ∇g = k and n = k. Then
Z Z
Z Z
(curlF)·ndS =
(−10)dS = −10×(area of S) = −10(4π) = −40π.
S
z
3
z=1
C
3 y
R
3
S
x
Line Integral: Parameterize the curve C by x = 2 cos t, y = 2 sin t, z = 1, for 0 ≤ t ≤ 2π. Then
I
I
Z
F · dr =
2π
[10 sin t(−2 sin t) − 10 cos t(2 cos t)]dt
5ydx − 5xdy + 3dz =
0
2π
Z
(−20 sin2 t − 20 cos2 t)dt =
=
Z
2π
−20dt = −40π.
0
0
2. Surface Integral: curlF = 4i−2j−3k. Letting g(x, y, z) =
x2 + y 2 +p
z − 16, ∇g = 2xi + 2yj + k, and n = (2xi +
2yj + k)/ 4x2 + 4y 2 + 1. Thus,
Z Z
Z Z
8x − 4y − 3
p
dS.
(curlF) · ndS =
4x2 + 4y 2 + 1
S
S
z
16
z=16-x2-y2
Letting the surface be z =p16 − x2 − y 2 , we have zx =
−2x, zy = −2y, and dS = 1 + 4x2 + 4y 2 dA. Then, using
polar
Z Z coordinates,
Z Z
(curlF) · ndS =
(8x − 4y − 3)dA
S
Z
2π
Z
0
y
C
4
4
(8r cos θ − 4r sin θ − 3)rdrdθ
=
4
R
R
x
r=4
0
4
4 3
3 2
8 3
r cos θ − r sin θ − r
dθ =
=
3
3
2
0
0
Z 2π 512
256
cos θ −
sin θ − 24 dθ
3
3
0
2π
512
256
= −48π.
=
sin θ +
cos θ − 24θ
3
3
0
Line Integral: Parameterize the curve C by x = 4 cos t, y = 4 sin t, z = 0, for 0 ≤ t ≤ 2π.
Then,
I
I
Z
Z
2π
2π
F · dr =
C
2zdx − 3xdy + 4ydz =
c
Z
=
0
[−12 cos t(4 cos t)]dt
0
2π
2π
−48 cos2 tdt = (−24t − 12 sin 2t)|0 = −48π.
270
CHAPTER 15. VECTOR INTEGRAL CALCULUS
3. Surface Integral: curlF
=
i + j + k. Letting
z=3-y/2-x
g(x, y, z) = 2x + y + 2z − 6, we have ∇g = 2i + 2j + 2k and
z
RR
RR 5
n = (2i + j + 2k)/3. Then
(curlF) · ndS =
dS.
C3
S
S 3
1
Letting the surface be z = 3 − y − x we have zx = C1
2
r
1
3
1
R
y
−1, zy = − , and dS = 1 + (−1)2 + (− )2 dA = dA.
2
2
2
C2
Then
x
Z Z
Z Z
5 3
5
5
45
(curlF)·ndS =
dA = ×(area of R) = (9) =
.
2
2
2
2
S
R 3
Line Integral: C1 : z = 3 − x, 0 ≤ x ≤ 3, y = 0; C2 : y = 6 − 2x, 3 ≥ x ≥ 0, z = 0; C3 : z =
3I− y/2, 6 ≥ y ≥ 0, x Z= Z0.
Z
Z
zdx + xdy + ydz =
zdx +
C
C1
Z
xdy +
C2
3
0
Z
(3 − x)dx +
=
0
ydz
C3
Z
x(−2dx) +
3
0
y(−dy/2)
6
3
0
1
9
1
45
1
0
− x2 3 − y 2 = − (0 − 9) − (0 − 36) =
3x − x2
2
4
2
4
2
0
6
RR
4. Surface Integral: curlF = 0 and
(curlF) · ndS = 0.
S
Line Integral: The curve is x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π.
=
I
Z
xdx + ydy + zdz =
C
2π
[cos t(− sin t) + sin t(cos t)]dt = 0.
0
5. curlF = 2i + j. A
√ unit vector normal to the plane is
n = (i + j + k)/ 3. Taking the equation of the plane
to be √z = 1 − x − y,
√ we have zx = zy = −1. Thus,
dS = 1 + 1 + 1dA = 3dA and
I
Z Z
Z Z √
√ Z Z √
F · dr =
S(curlF) · ndS =
3dS = 3
3dA
C
S
= 3 × (area of R) = 3(1/2) = 3/2.
z
1
C
z=1-x-y
R
R
1
y
1
x
√
2. From z = 1 − y,
6. curlF = −2xzi + x2 k. A unit vector normal
to
the
plane
is
n
=
(j
+
k)/
√
√
we have zx = 0 and zy = −1. Thus, dS = 1 + 1dA = 2dA and
Z Z
I
Z Z
Z Z
1 2√
√ z 2dA =
(1 − y)2 dA
F · dr =
(curlF) · ndS =
2
R
C
S
R
Z 2Z 1
Z 2
Z 2
1
1
1
2
=
(1 − y)2 dydx =
− (1 − y)3 dx =
dx = .
3
3
3
0
0
0
0
0
15.8. STOKES’ THEOREM
271
√
7. curlF = −2yi − zj − xk. A unit vector normal
√ is n = (j + k)/ 2. From z = 1 − y
√ to the plane
we have zx = 0 and zy = −1. Then dS = 1 + 1dA = 2dA and
Z Z
Z Z √
1
2dA =
(y − x − 1)dA
− √ (z + x)
2
R
S
R
1
Z 2
Z 2
Z 2Z 1
1
1 2
dx =
−x −
(y − x − 1)dydx =
y − xy − y
dx
=
2
2
0
0
0
0
0
2
1
1
= − x2 − x
= −3.
2
2
0
I
Z Z
F · dr =
c
(curlF) · ndS =
8. curlF = 2i + 2j + 3k. Letting g(x, y, z) = x + 2y√+ z − 4,
we have ∇g = i + 2j + k and n = (i + 2j + k)/ 6. From
z = 4√− x − 2y we have zx = −1 and zy = −2. Then
dS = 6dA and
z
4
z=4-x-2y
C
√
RR
RR 1
√ (9) 6dA =
F · dr =
(curlF) · ndS =
C
S
R
6
RR
9dA
=
9x
(area
of
R)=
9(4)
=
36.
R
H
2 y
R
4
9. curlF = (−3x2√
−3y 2 )k. A unit vector normal to the plane is
3. From z = 1−x−y, we have zx = zy = −1
n = (i+j+k)/
√
and dS = 3dA. Then, using polar coordinates,
I
Z Z
Z Z
√
√
√
F · dr =
(curlF) · ndS =
(− 3x2 − 3y 2 ) 3dA
C
S
z
3
C
z=1-x-y
R
Z Z
2
Z
2
2π
R
2π
=3
0
0
1
1
− r4 dθ = 3
4 0
Z
0
2π
R
1
Z
(−r2 )rdrdθ
(−x − y )dA = 3
=3
Z
x
1 y
0
1
3π
− dθ =
.
4
2
x
2yj + k
10. curlF = 2xyzi − y 2 zj + (1 − x2 )k. A unit vector normal to the surface is n = p
. From
4y 2 + 1
p
z = 9 − y 2 we have zx = 0, zy = −2y and dS = 1 + 4y 2 dA. Then
I
Z Z
Z Z
Z 3 Z y/2
F · dr =
(curlF) · ndS =
(−2y 3 z + 1 − x2 )dA =
[−2y 3 (9 − y 2 ) + 1 − x2 ]dxdy
C
S
Z
3
R
0
y/2
1
=
−18y x + 2y x + x − x3
dy =
3
0
0
3
9 5 1 7 1 2
1 4
= − y + y + y − y
≈ 123.57.
5
7
4
96
0
3
5
Z
0
3
0
1
1 3
−9y + y + y − y dy
2
24
4
6
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272
CHAPTER 15. VECTOR INTEGRAL CALCULUS
11. curlF = 3x2 y 2 k. A unit vector normal to the surface is
8xi + 2yj + 2zk
4xi + yj + zk
n= p
=p
.
2
2
2
64x + 4y + 4z
16x2 + y 2 + z 2
From zx = − p
4x
4 − 4x2 − y 2
, zy = − p
s
y
4 − 4x2 − y 2
we obtain dS = 2
1 + 3x2
dA.
4 − 4x2 − y 2
Then
I
Z Z
F · dr =
(curlF) · ndS =
C
S
Z Z
R
3x2 y 2 dA
=
s
3x2 y 2 z
Z Z
p
(2
16x2 + y 2 + z 2
Using symmetry
R
Z
1
√
2 1−x2
Z
x2 y 2 dydx = 12
= 12
0
Z
= 32
1 + 3x2
)dA
4 − 4x2 − y 2
0
Z
1
0
1 2 3
x y
3
√
2 1−x2
dx
0
1
x2 (1 − x2 )3/2 dx
x = sin t, dx = cos tdt
0
Z
= 32
π/2
sin2 t cos4 tdt = π.
0
12. curlF = i + j + k. A unit vector normal to the surface is
2xi + 2yj + 2zk
xi + yj + zk
n= p
=p
2
2
2
4x + 4y + 4z
x2 + y 2 + z 2
= xi + yj + zk.
p
y
x
, zy = − p
and dS =
From z = 1 − x2 − y 2 , we have zx = − p
2
2
1−x −y
1 − x2 − y 2
1
p
dA. Then
1 − x2 − y 2
!
I
Z Z
Z Z
1
dA
F · dr =
(curl F) · ndS =
(x + y + z) p
1 − x2 − y 2
C
S
R
p
Z Z
Z Z
Z Z
x + y + 1 − x2 − y 2
x+y
p
p
=
dA =
1dA +
dA
2
2
1−x −y
1 − x2 − y 2
R
R
R
Z Z
=
1dA + 0 Using Symmetry
R
π
=
2
since R is the disk x2 + y 2 ≤
1
2
with radius
√1 .
2
13. Parameterize C by x = 4 cos t, y = 2 sin t, z = 4, for 0 ≤ t ≤ 2π. Then
15.8. STOKES’ THEOREM
273
Z Z
I
I
(curlF) · ndS =
F · dr =
S
2
6yzdx + 5xdy + yzex dz
C
2π
Z
=
[6(2 sin t)(4)(−4 sin t) + 5(4 cos t)(2 cos t) + 0]dt
0
2π
Z
(−24 sin2 t + 5 cos2 t)dt = 8
=8
0
Z
2π
(5 − 29 sin2 t)dt = −152π.
0
14. Parameterize C by x = 5 cos t, y = 5 sin t, z = 4, for 0 ≤ t ≤ 2π. Then,
Z Z
I
I
(curlF) · ndS =
F·r=
ydx + (y − x)dy + z 2 dz
S
C
2π
C
Z
[(5 sin t)(−5 sin t) + (5 sin t − 5 cos t)(5 cos t)]dt
=
0
2π
Z
(25 sin t cos t − 25)dt =
=
0
25
sin2 t − 25t
2
15. Parameterize C by C1 : x = 0, z = 0, 2 ≥ y ≥ 0; C2 :
z = x, y = 0, 0 ≤ x ≤ 2; C3 : x = 2, z = 2, 0 ≤ y ≤
2;
Z ZC4 : z = x, y = 2,
I 2 ≥ x ≥I0. Then
(curlF) · ndS =
F·r=
3x2 dx + 8x3 ydy + 3x2 ydz
S
C
ZC
Z
=
0dx + 0dy + 0dz +
3x2 dx
C1
C2
Z
Z
+
64dy +
3x2 dx + 6x2 dx
Z
C3
2
=
C4
2
3x2 dx +
Z
0
= x3
Z
64dy +
0
2
0
2
+ 64y|0 + 3x3
2π
= −50π.
0
z
2
C3
C4
C2
C1
2
y
2
0
9x2 dx
2
0
2
x
= 112.
16. Parameterize C by x = cos t, y = sin t, z = sin t, 0 ≤ t ≤ 2π. Then
Z Z
I
I
(curlF) · ndS =
F·r=
2xy 2 zdx + 2x2 yzdy + (x2 y 2 − 6x)dz
S
C
2π
Z
C
[2 cos t sin2 t sin t(− sin t) + 2 cos2 t sin t sin t cos t
=
0
+ (cos2 t sin2 t − 6 cos t) cos t]dt
Z
=
2π
(−2 cos t sin4 t + 3 cos3 t sin2 t − 6 cos2 t)dt = −6π.
0
17. We take the surface to be z = 0. Then n = k and dS = dA. Since curlF =
1
2
i + 2zex j +
2
1+y
274
CHAPTER 15. VECTOR INTEGRAL CALCULUS
y 2 k,
I
2 x2
z e dx + xydy + tan
Z Z
−1
Z Z
(curlF) · ndS =
ydz =
C
S
Z
2π
Z
3
0
Z
0
2π
0
r2 sin2 θrdrdθ =
y 2 dA
y dS =
S
=
81
=
4
Z Z
2
R
2π
Z
0
3
1 4 2
r sin θ dθ
4
0
81π
sin2 θdθ =
.
4
2xi + 2yj + k
and
18. (a) curlF = xzi − yzj. A unit vector normal to the surface is n = p
4x2 + 4y 2 + 1
p
dS = 1 + 4x2 + 4y 2 dA. Then, using x = cos t, y = sin t, 0 ≤ t ≤ 2π, we have
Z Z
Z Z
Z Z
(curlF) · ndS =
(2x2 z − 2y 2 z)dA =
(2x2 − 2y 2 )(1 − x2 − y 2 )dA
S
R
R
Z Z
=
(2x2 − 2y 2 − 2x4 + 2y 4 )dA
R
2π Z 1
Z
=
0
(2r2 cos2 θ − 2r2 sin2 θ − 2r4 cos4 θ + 2r4 cos4 θ)rdrdθ
0
2π
Z
Z
1
[r3 cos 2θ − r5 (cos2 θ − sin2 θ)(cos2 θ + sin2 θ)]drdθ
=2
0
0
2π
Z
Z
0
1
6
3
Z
5
(r cos 2θ − r cos 2θ)drdθ = 2
=2
=
1
Z
0
2π
cos 2θ
0
1
dθ
0
2π
cos 2θdθ = 0.
0
(b) RWe
R take the surface to be z = 0. Then n = k,
curlF · ndS = 0.
S
curlF · n = curlF · k = 0 and
(c) By Stoke’s Theorem, using z = 0, we have
Z Z
I
I
I
curlF · ndS =
F · dr =
xyzdz = xy(0)dz = 0.
S
15.9
1 4 1 6
r − r
4
6
Divergence Theorem
C
C
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15.9. DIVERGENCE THEOREM
275
z
1
1. divF = y + x + z
The
Integral:
Z Z Triple
Z
Z 1Z
divFdV =
D
1
1
Z
0
0
1
Z
1
=
Z
0
1
=
0
0
1
Z
=
Z
0
1
S4
1
(x + y + z)dxdydz
0
0
Z
Z
S6
1
1
( x2 + xy + xz) dydz
2
0
1
y
1
1
( + y + z)dydz
2
S1
x
1
1
1
( y + y 2 + yz) dz
2
2
0
1
1
3
1
(1 + z 2 ) = 2 − =
2
2
2
0
0
The Surface Integral: Let the surfaces be S1 in z = 0, S2 in z = 1, S3 in y = 0, S4 in
y = 1, S5 in x = 0, and S6 in x = 1. The unit outward normal vectors are −k, k, −j, j, −i
and i, respectively. Then
Z Z
Z Z
Z Z
Z Z
Z Z
F · ndS =
F · (−k)dS1 +
F · kdS2 +
F · (−j)dS3 +
F · jdS4
S
S
S
S3
S4
Z Z1
Z Z2
+
F · (−i)dS5 +
F · idS6
S6
Z Z S5
Z Z
Z Z
Z Z
=
(−xz)dS1 +
xzdS2 +
(−yz)dS3 +
yzdS4
S3
S4
Z SZ1
Z SZ2
+
(−xy)dS 5 +
xydS6
S5
S6
Z Z
Z Z
Z Z
=
xdS2 +
zdS4 +
ydS6
(1 + z)dz =
=
S2
1
Z
1
Z
=
Z
S4
1Z 1
xdxdy +
0
0
1
Z
=
0
Z
1
dy +
2
0
1
S6
Z
1
Z
zdzdx +
0
0
1
dx +
2
Z
ydydz
0
0
1
1
0
3
1
dz = .
2
2
z
1
2. divF = 6y + 4z
S3
The
Integral:
Z Z Triple
Z
Z 1Z
divFdV =
D
0
Z
1−x
Z
S4
1−x−y
(6y + 4z)dzdydx
0
1Z
=
0
S1
0
1
0
1−x−y
1−x
(6yz + 2z 2 )
1
dydx
0
x
S2
y
276
CHAPTER 15. VECTOR INTEGRAL CALCULUS
Z
1
Z
=
1−x
(−4y 2 + 2y − 2xy + 2x2 − 4x + 2)dydx
0
0
1−x
4 3
2
2
2
− y + y − xy + 2x y − 4xy + 2y
=
dx
3
0
0
1
Z 1
5
5
5
5
5
5
5
=
− x3 + 5x2 − 5x +
=
dx = − x4 + x3 − x2 + x
3
3
12
3
2
3
12
0
0
The Surface Integral: Let the surfaces be S1 in the plane x+y+z = 1, √S2 in z = 0, S3 in x = 0,
and S4 in y = 0. The unit outward normal vectors
√ are n1 = (i + j + k)/ 3, n2 = −k, n3 = −i,
and n4 = −j, respectively. Now on S1 , dS1 = 3dA1 , on S3 , x = 0, and on S4 , y = 0, so
Z Z
Z Z
Z Z
Z Z
Z Z
F · ndS =
F · n1 dS1 +
F · (−k)dS2 +
F · (−j)dS3 +
F · (−i)dS4
Z
1
S
S1
Z
=
1
s2
S3
1−x
Z
(6xy + 4y(1 − x − y) + xe−y )dydx +
Z Z
Z Z
+
(−6xy)dS3 +
(−4yz)dS4
0
0
S3
Z
S4
Z
1
=
0
Z 1
1
Z
1−x
(−xe−y )dydx
0
0
S4
4
xy + 2y − y 3 − xe−y
3
2
2
1−x
Z
1−x
1
dx +
xe
−y
0
0
dx + 0 + 0
0
Z 1
4
x(1 − x)2 + 2(1 − x)2 − (1 − x)3 − xex−1 + x dx +
(xex−1 − x)dx
3
0
0
1
5
1 2 2 3 1 4 2
1
3
4
=
x − x + x − (1 − x) + (1 − x)
.
=
2
3
4
3
3
12
0
=
3. divF = 3x2 + 3y 2 + 3z 2 . Using spherical coordinates,
Z Z
Z Z Z
Z
F · ndS =
3(x2 + y 2 + z 2 )dV =
S
Z
D
2π Z π
2π
Z
π
Z
a
3ρ2 ρ2 sin φdρdφdθ
0
0
0
5 Z 2π Z π
a
3a
3 5
ρ sin φ dφdθ =
sin φdφdθ
5
5 0
0
0
0
0
Z
Z
π
6a5 2π
3a5 2π
12πa5
.
=
− cos φ dθ =
dθ =
5 0
5 0
5
0
=
4. divF = 4 + 1 + 4 = 9. Using the formula for the volume of a sphere,
Z Z
Z Z Z
4 3
F · ndS =
9dV = 9
π2 = 96π.
3
S
D
5. divF = 2(z − 1). Using cylindrical coordinates,
Z Z
Z Z Z
Z 2π Z 4 Z
F · ndS =
2(z − 1)V =
S
D
Z
2π
Z
=
0
4
Z
16rdrdθ =
0
0
0
Z
2
Z
Z
2π
dθ = 256π.
0
5
4
(z − 1)
0
dθ = 128
0
2π
2(z − 1)dzrdrdθ =
1
4
2π
8r
0
5
0
2
rdrdθ
1
15.9. DIVERGENCE THEOREM
277
2
6. divF
Z Z = 2x + 2z +Z 12z
Z Z.
Z
F · ndS =
divFdV =
S
2
Z
Z
(x2 + 2xz + 12xz 2 ) dydz
=
0
Z 3
0
Z 2
0
0
=
0
(1 + 2z + 12z 2 )dydz
3
Z
2(1 + 2z + 12z 2 )dz = (2z + 2z 2 + 8z 3 )
=
3
(2x + 2z + 12z 2 )dxdydz
1
2
Z
z
1
0
0
0
D
3
Z
3
0
3
0
Z Z Z
F · ndS =
2π
√
Z
D
√
3
Z
=
0
Z 2π
=
0
0
2π
Z
√
1
− (4 − r2 )5/2
5
r=M4-r2
2
4−r 2
3z 2 rdzdrdθ
0
2
r(4 − r2 )3/2 drdθ
0
0
0
√
√
3
Z
drdθ =
0
2π
0
4−r 2
rz 3
=
0
3Z
divFdV =
S
Z
z
√
2π Z
Z
y
x
7. divF = 3z 2 . Using cylindrical coordinates,
Z Z
2
1
= 240
y
r=M3
2
x
3
Z
2π
dθ =
0
0
1
− (1 − 32)dθ
5
31
62π
dθ =
.
5
5
8.
z
divF = 2x.
Z Z
Z Z Z
F · ndS =
divFdV
S
Z
3
Z
9
z=9-y
D
9 Z 9−y
=
2xdzdydx
x2
0
Z
3
Z
0
9
2x(9 − y)dydx =
=
x2
0
Z
9
3
Z
−x(9 − y)
0
3
2
dx
x2
y=x2
x
x(9 − x)2 dx
=
9
4
0
Z
=
3
(x3 − 18x2 + 81x)dx =
0
891
=
4
81
1 4
x − 6x3 + x2
4
2
3
0
y
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278
CHAPTER 15. VECTOR INTEGRAL CALCULUS
9. divF =
x2
Z Z
1
. Using spherical coordinates,
+ y2 + z2
Z Z Z
2π
Z
Z
π
Z
y
b
1 2
ρ sin φdρdφdθ
2
a ρ
0
D
0
Z 2π
Z 2π Z π
π
− cos φ dθ
(b − a) sin φdφdθ = (b − a)
=
F · ndS =
S
a
divFdV =
0
0
0
b
x
0
2π
Z
= (b − a)
2dθ = 4π(b − a).
0
Z Z
Z Z Z
F · ndS =
10. Since divF = 0,
0dV = 0.
S
D
11. divF = 2z + 10y − 2z = 10y.
Z Z
Z Z Z
D
2
Z
2−x2 /2
Z
0
5y 2
0
0
0
Z
0
2
0
Z
2−x
Z
3
3
z
x+y
2−x
2
(90xy − 30x2 y − 30xy 2 )dydx
y
2
2−x
2
2 2
3
(45xy − 15x y − 10xy )
x
dx
0
(−5x4 + 45x3 − 120x2 + 100x)dx =
−x5 +
0
13. divF = 6xy 2 + 1 − 6xy 2 = 1. Using cylindrical coordinates,
Z Z
Z Z Z
Z π Z 2 sin θ Z 2r sin θ
Z
F · ndS =
dV =
dzrdrdθ =
S
D
π
0
0
r2
2 sin θ
0
45 4
x − 40x3 + 50x2
4
π
Z
2
= 28
0
2 sin θ
(2r sin θ − r2 )rdrdθ
0
Z π
2 3
1
16
r sin θ − r4
dθ =
sin4 θ − 4 sin4 θ dθ
3
4
3
0
0
0
π
Z π
4
4 3
1
1
π
4
=
sin θdθ =
θ − sin 2θ +
sin 4θ
=
3 0
3 8
4
32
2
0
Z
= 128
0
0
Z 2
=
2
0
30xydzdydx
0
2
=
=
(80 − 5x4 )dx = (80x − x5 )
x+y
=
Z
2
dydx
30xyz
2
Z
3
2−x
Z
2−x2 /2
0
dx =
12. divF
Z Z = 30xy. Z Z Z
Z
F · ndS =
30xydV =
=
Z
(80 − 40z)dzdx
0
0
D
2
Z
(80z − 20z 2 )
S
z
z
2−x2 /2
0
0
0
dzdx =
2
Z
=
Z
4−z
Z
10ydydzdx
0
4−z
=
2
2−x2 /2
Z
10ydV =
S
Z
2
Z
F · ndS =
15.9. DIVERGENCE THEOREM
279
14. divF = y 2 + x2 . Using spherical coordinates, we have x2 + y 2 = ρ2 sin2 ω and z = ρ cos ω or
ρ = z sec ω. Then
Z Z
Z Z Z
(x2 + y 2 )dS =
F · ndS =
S
D
2π
Z
π/4
Z
=
0
0
2π
Z
Z
π/4
=
0
=
992
5
0
Z 2π
0
1 5 3
ρ sin φ
5
Z
2π
0
4 sec φ
Z
π/4
Z
4 sec φ
ρ2 sin2 φρ2 sin φdρdφdθ
2 sec φ
0
Z
2π
π/4
Z
dφdθ =
0
0
2 sec φ
992
992
tan3 φ sec2 φdφdθ =
5
5
2π
Z
0
992
sec5 φ sin3 φdφdθ
5
1
tan4 φ
4
π/4
dθ
0
1
496π
dθ =
.
4
5
15. Since div a = 0, by the divergence Theorem
Z Z
Z Z Z
Z Z Z
(a · n)dS =
div adV =
0dV = 0.
S
D
D
16. By the Divergence Theorem and Problem 24 in Section 15.7,
Z Z
Z Z Z
Z Z Z
(curlF · n)dS =
div(curlF)dV =
0dV = 0.
S
D
D
x2 − 2y 2 + z 2
x2 + y 2 − 2z 2
−2x2 + y 2 + z 2
+
+ 2
=0
17. (a) divE = q
(x2 + y 2 + Zz 2 )Z5/2Z (x2 + y 2 +Zz 2Z)5/2
(x + y 2 + z 2 )5/2
Z Z
Z
(E · n)dS =
divEdV =
SU Sa
D
0dV = 0
D
RR
RR
RR
RR
(E · n)dS = − Sa (E · n)dS. on
(b) From (a),
(E · n)dS + Sa (E · n)dS = 0 and
S
S
Sa , |r| = a, n = −(xi + yj + zk)/a = −r/a and E · n = (qr/a3 ) · (−r/a) = −qa2 /a4 =
−qa2 . Thus
RR
RR
q
q RR
q
q
(E · n)dS = − Sa (− 2 )dS = 2
dS = 2 × (area of Sa ) = 2 (4πa2 ) = 4πq.
S
S
a
a
a
a
a
RR
RRR
RR
18. (a) By Gauss’
(E · n)dS R=R R D 4πρdV, Rand
(E ·
S
R RLaw
R
R R by the Divergence
R R RTheorem
n)dS =
divEdV.
Thus
4πρdV
=
divEdV
and
(4πρ−divE)dV
=
D
D
D
D
0. Since this holds for all regions D, 4πρ − divE = 0 and divE = 4πρ.
(b) Since E is irrotational, E = ∇φ and ∇2 φ = ∇ · ∇φ = ∇E = divE = 4πρ.
19. By the Divergence Theorem and Problem 21 in Section 15.7,
Z Z
Z Z Z
Z Z Z
Z Z Z
(f ∇g) · ndS =
div(f ∇g)dV =
∇ · (f ∇g)dV =
[f (∇ · ∇g) + ∇g · ∇f ]dV
S
D
D
Z Z ZD
=
(f ∇2 g + ∇g · ∇f )dV.
D
280
CHAPTER 15. VECTOR INTEGRAL CALCULUS
20. By the Divergence Theorem and Problem 19 and 21 in Section 15.7,
Z Z
Z Z Z
Z Z Z
(f ∇g − g∇f ) · ndS =
div(f ∇g − g∇f )dV =
∇ · (f ∇g − g∇f )dV
S
D
Z Z ZD
=
[f (∇ · ∇g) + ∇g · ∇f − g(∇ · ∇f ) − ∇f · ∇g]dV
Z Z ZD
=
(f ∇2 g − g∇2 f )dV.
D
21. If G(x, y, z) is a vector valued function then we define surface integrals and triple integrals of
G component-wise. In this case, if a is a constant vector it is easily shown that
Z Z
Z Z
Z Z Z
Z Z Z
a · GdS = a ·
GdSand
a · GdV = a ·
GdV.
S
S
Now let F = f a. Then
Z Z
D
D
Z Z
Z Z
F · ndS =
S
(f a) · ndS =
a · (f n)dS
S
S
and, using Problem 21 in Section 15.7 and the fact that ∇ · a = 0, we have
Z Z Z
Z Z Z
Z Z Z
Z Z Z
divFdV =
∇ · (f a)dV =
[f (∇ · a) + a · ∇f ]dV =
a · ∇f dV.
D
D
D
D
By the Divergence Theorem,
Z Z
Z Z
Z Z Z
Z Z Z
a · (f n)dS =
F · ndS =
divFdV =
a · ∇f dV
S
S
D
D
and
Z Z
a·
f ndS
Z Z Z
=a·
∇f dV
S
Z Z
Z Z Z
or a ·
f ndS −
D
∇f dV
S
D
Since a is arbitrary,
Z Z
Z Z Z
Z Z
Z Z Z
f ndS −
∇f dV = 0 and
f ndS =
∇f dV.
S
D
Z Z
S
Z Z Z
pndS + mg = mg −
∇pdV = mg −
D
Z Z Z
= mg −
ρdV g = mg − mg = 0
D
Z Z Z
22. B + W = −
S
D
Chapter 15 in Review
A. True/False
1. True; the value is 4/3.
ρgdV
D
= 0.
www.elsolucionario.org
CHAPTER 15 IN REVIEW
281
2. True; since 2xydx − x2 dy is not exact.
R
3. False; C xdx + x2 dy = 0 from (−1, 0) to (1, 0) along the x-axis and along the semicircle
√
y = 1 − x2 , but since xdx + x2 dy is not exact, the integral is not independent of path.
4. True
5. True; assuming that the first partial derivatives are continuous.
6. True
7. True
8. True; since curlF = 0 when F is a conservative vector field.
9. True
10. True
11. True
12. True
B. Fill in the Blanks
1. F = ∇φ = −x(x2 + y 2 )−3/2 i − y(x2 + y 2 )−3/2 j
2. curlF =
i
∂/∂x
f (x)
j
∂/∂y
g(y)
k
∂/∂z
h(z)
=0
3. 2xy + 2xy + 2xy = 6xy
4.
5.
i
∂/∂x
x2 y
j
∂/∂y
xy 2
k
∂/∂z
2xyz
= 2xzi − 2yzj + (y 2 − x2 )k
∂
∂ 2
∂
(2xz) −
(2yz) +
(y − x2 ) = 0
∂x
∂y
∂z
6. ∇(6xy) = 6yi + 6xj
3
7. 0; since (y − 7ex )dx + (x + ln
√
y)dy is exact.
8. Irrotational
9. At u = 1, v = 4, we have r = h1, 4, 4i.
∂r
∂r
(1, 4) = h1, 0, 2i,
(1, 4) = h0, 1, 1/2i
∂u
∂v
i j
k
A normal vector is given by n = 0 1 2
= h−2, −1/2, 1i.
0 1 1/2
The tangent plane is −2(x − 1) − 12 (y − 4) + (z − 4) = 0 or 4x + y − 2z = 0.
10. r(2, v) = (8 + v)i + (2 + 2v)j + (2 + v)k
So the parametric equations are x = 8 + v, y = 2 + 2v, z = 2 + v
282
CHAPTER 15. VECTOR INTEGRAL CALCULUS
C. Exercises
Z
1.
C
2π
p
4t2
4 sin2 2t + 4 cos2 2t + 4dt =
cos2 2t + sin 2t
π
√
√
2π
56 2π 3
8 2 3
=
=
t
3
3
z2
ds =
x2 + y 2
Z
Z
2π
√
8 2t2 dt
π
π
√
√
[x(2 − 2x) + 4x] 1 + 4dx = 5
0
√
0
√
7 5
2 3
2
= 5 3x − x
=−
3
3
1
Z
2.
Z
1
1
Z
(6x − 2x2 )dx
(xy + 4x)ds =
C
0
3. Since P − y = 6x2 y = Qx , the integral is independent of path.
φx = 3x2 y 2 , φ = x3 y 2 + g(y), φy = 2x3 y + g 0 (y) = 2x3 y − 3y 2 ;
g(y) = −y 3 ; φ = x3 y 2 − y 3 ;
R (−1,2) 2 2
(−1,2)
3x y dx + (2x3 y − 3y 2 )dy = (x3 y 2 − y 3 ) (0,0) = −12
(0,0)
4. By Green’s Theorem,
I
(x2 + y 2 )dx+(x2 − y 2 )dy =
C
=2
0
5.
Z
2π
2π
3
Z
(2x − 2y)dA = 2
R
Z
Z
Z Z
(r cos θ − r sin θ)rdrdθ
0
1
1 3
r cos θ − r3 sin θ
3
3
0
3
Z
2π
27
(cos θ − sin θ)dθ = 0.
3
Z
1
dθ = 2
0
0
y sin πzdx + x2 ey dy + 3xyzdz
C
Z
1
=
2
[t2 sin πt3 + t2 et (2t) + 3tt2 t3 (3t2 )]dt =
0
0
1
cos πt3 + t9
= −
3π
1
Z
+2
1
2
t3 et dt
Integration by parts
0
0
2
2
2
+ 1 + (t2 et − et )
3π
=
2
(t2 sin πt3 + 2t3 et + 9t8 )dt
1
=
0
2
+2
3π
6. Parameterize C by x = cos t, y = sin t; 0 ≤ t ≤ 2π. Then
I
Z
2π
F · dr =
C
Z
2π
[4 sin t(− sin tdt) + 6 cos t(cos t)dt] =
0
Z
0
2π
2π
(6 cos2 t − 4 sin2 t)dt
5
sin 2t − 4t)
= 2π.
2
0
0
H
RR
Using Green’s Theorem, Qz − Py = 6 − 4 = 2 and C F · dr =
2dA = 2(π · 12 ) = 2π.
R
=
(10 cos2 t − 4)dt = (5t +
CHAPTER 15 IN REVIEW
283
π
π
7. Let r1 = ti and r2 = i + πtj for 0 ≤ t ≤ 1. Then
2
2
π
dr1 = i, dr2 = πj, F1 = 0,
2
F2 =
y
π
C2
π
π
π
sin πti + πt sin j = sin πti + πtj,
2
2
2
C1
and
π/2
Z
Z
Z
F1 ·dr1 +
W =
C1
F2 ·dr2 =
C2
0
1
1
1
π 2 tdt = π 2 t2
2
x
π2
=
.
2
0
8. Parameterize the line segment from (-1/2,1/2) TO (-1,1) using y = −x as x goes from -1/2
to -1. Parameterize the line segment from (-1,1)
√ to (1,1) using y = 1 as x goes from
√ -1 to 1.
Parameterize the line segment from (1,1) to (1, 3) using x = 1 as y goes from 1 to 3. Then
Z
Z
−1
F · dr =
W =
Z
F · (dxi − dxj) +
−1
3
F · (dyj)
F · (dxi) +
−1/2
C
√
1
Z
1
Z 1
Z √3
1
2
1
2
− 2
)dx +
dx +
dy
=
( 2
2
2
2+1
x
+
(−x)
x
+
(−x)
x
1
+
y2
−1
−1/2
1
Z −1
Z 1
Z √3
1
2
1
=
dx +
dx +
dy
2
2
2x
1
+
x
1
+
y2
−1/2
1
−1
Z
=−
−1
1
2x
−1
+ 2 tan−1 x
−1/2
1
−1
√
+ tan−1 y
3
1
1
π
π
13π − 6
= − + 2( ) +
=
.
2
2
12
12
9. Py = 2x = Qx , Qz = 2y = Ry , Rx = 0 = Pz and the integral is independent of path.
Parameterize the line segment between the points by x = 1, y = 1, z = t, 0 ≤ t ≤ π. Then
dx = dy = 0, dz = dt, and
Z
(1,1,π)
2xydx + (x2 + 2yz)dy + (y 2 + 4)dz =
(1,1,0)
Z
π
[2(0) + (1 + 2t)(0) + (1 + 4)]dt = 5π.
0
10. Py = 0 = Qx , Qz = 0 = Ry , Rx = 2e2x = Pz and the integral is independent of path. From
ω = x2 + y 2 − y + ze2x we obtain
Z
(3,2,0)
(2x + 2ze2x )dx + (2y − 1)dy + e2x dz = (x2 + y 2 − y + ze2x )
(0,0,1)
(3,2,0)
(0,0,1)
= 11 − 1 = 10.
11. Using Green’s Theorem,
I
Z Z
Z Z
−4ydx + 8xdy =
[8 − (−4)]dA = 12
dA = 12 × (area of R)
C
R
= 12(16π − π) = 180π.
R
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284
CHAPTER 15. VECTOR INTEGRAL CALCULUS
12. Py = [(x − 1)2 − (y − 1)2 ]/[(x − 1)2 + (y − 1)2 ]2 = Qx . When (1,1) is outside C, Green’s
Theorem applies and
I
Z Z
Z Z
P dx + Qdy =
(Qx − Py )dA =
0dA = 0.
C
R
R
For (1,1) inside C, let Ca be a circle of radius a centered at (1,1) and lying entirely inside C.
Using x − 1 = a cos θ and y − 1 = a sin θ for 0 ≤ θ ≤ 2π we obtain
I
I
I
1
(y − 1)dx + (1 − x)dy
P dx + Qdy =
P dx + Qdy = 2
a Ca
C
Ca
Z 2π
1
= 2
[a sin θ(−a sin θ) − a cos θ(a cos θ)]dθ
a 0
Z 2π
=−
(sin2 θ + cos2 θ)dθ = −2π.
0
√
13. zx = 2x, zy = 0; dS = 1 + 4x2 dA
2
Z 3Z 2 2p
Z Z
Z 3 z
x
1 1
2 3/2
2
dy
dS =
(1 + 4x )
1 + 4x dxdy =
S xy
1
1 xy
1 y 12
1
√
√
3
Z 3 3/2
1
17 − 53/2
17 17 − 5 5
=
dy =
ln y
12 1
y
12
1
√
√
2
17 17 − 5 5
x
=
ln 3
12
z
4
z=x2
3
14. n = k, R F
R · n = 3;
RR
flux =
F·ndS = 3 S dS = 3×(area of S) = 3(1) = 3
S
y
z
2
S
1
1
y
x
√
−x
−x
−x
−2x + 1,
15. The surface
√is g(x, y, z) = y +e −2 = 0. Then ∇g = −e i+j, n = (−e i+j)/ e
and dS = 1 + e−2x dA.
Z Z
Z Z
Z 2Z 3
Z 2
3
flux =
(F · n)dS =
(−4e−x + 2 − y)dA =
(−3e−x )dxdz =
3e−x dz
S
Z
=
R
0
0
0
0
2
(3e−3 − 3)dz = 6e−3 − 6.
0
16. Solving y = 2−e−x for x, we obtain x = − ln(2−y). The surface is g(x, y, z) = x+ln(2−y) = 0.
p
1
Then ∇g = i −
j,
and |∇g| =
1 + 1/(2 − y)2 . Due to the orientation of S we
2−y
want the j component of the unit normal vector to be positive. Since y < 2 we shall take
CHAPTER 15 IN REVIEW
285
p
p
n = [−i + (1/2 − y)j]/ 1 + 1/(2 − y)2 . Now dS = 1 + 1/(2 − y)2 dA and the region R in
the yz -plane is 0 ≤ z ≤ 2 and 1 ≤ y ≤ 2 − e−3 . Then
Z Z
Z Z
flux =
(F · n)dS =
(−4 + 1)dA = −3 × (area of R)
S
R
= −3 2 (2 − e−3 ) − 1 = −6 + 6e−3 .
17. The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. ∇g = 2(xi + yj + zk) = 2r n = r/|r|, F =
−xi − yj − zk
= cr/|r|3
c∇(1/|r|) + c∇(x2 + y 2 + z 2 )−1/2 = c 2
(x + y 2 + z 2 )3/2
r
r
|r|2
c
c
r·r
F·n=− 3 ·
= −c 4 = −c 4 = − 2 = − 2
|r| |r|
|r|
|r|
|r|
a
RR
c
c
c RR
dS = − 2 × (area of S) = − 2 (4πa2 ) = −4πc
flux =
F · ndS = − 2
S
S
a
a
a
18. In Problem 17, F is not continuous at (0, 0, 0) which is in any acceptable region containing
the sphere.
19. Since F = c∇(1/r), divF = ∇ · (c∇(1/r)) = c∇2 (1/r) = c∇2 [(x2 + y 2 + z 2 )−1/2 ] = 0 by
Problem 31 in Section 17.5. Then, by the Divergence Theorem,
Z Z
Z Z Z
Z Z Z
fluxF =
F · ndS =
divFdV =
0dV = 0.
S
D
D
20. Parameterize C by x = 2 cos t, y = 2 sin t, z = 5, for
0Z ≤Zt ≤ 2π. Then
I
I
(curlF · n)dS =
F · dr =
6xdx + 7zdy + 8ydz
S
C
2π
z
9
C
Z
=
[12 cos t(−2 sin t) + 35(2 cos t)]dt
0
Z
2π
C
(70 cos t − 24 sin t cos t)dt
=
0
= (70 sin t − 12 sin2 t)
2π
0
2
= 0.
y
2
21. Identify F = −2yi + 3xj + 10zk. Then curlF = 5k. The
curve C lies in the plane z = 3, so n = k and dS = dA.
Thus,
I
Z Z
F · dr =
(curlF) · ndS
C
Z ZS
=
5dA = 5 × (area of R) = 5(25π) = 125π.
R
22. Since curlF = 0,
H
F · dr =
RR
S
(curlF · n)dS =
RR
S
0dS = 0.
z
x
C
6
R
6
x
10
y
286
CHAPTER 15. VECTOR INTEGRAL CALCULUS
z
1
23. divF
Z Z = 1 + 1 + 1Z=Z3; Z
F · ndS =
divFdV
S
Z Z ZD
=
3dV = 3 × (volume of D) = 3π
1
D
y
1
x
24. divF = x2 + y 2 + z 2 . Using cylindrical coordinates,
Z Z
Z Z Z
Z Z Z
F · ndS =
(x2 + y 2 + z 2 )dV =
divFdV =
S
D
2π
Z
D
Z
0
1
Z
0
1
(r2 + z 2 )rdzdrdθ
0
1
Z 2π Z 1 1
1
drdθ =
r3 + r drdθ
=
r3 z + rz 3
3
3
0
0
0
0
0
1
Z 2π
Z 2π 5
1 4 1 2
5π
dθ =
=
r + r
dθ =
.
4
6
12
6
0
0
0
Z
2π
Z
1
z
25. divF
Z Z = 2x + 2(x Z+ y)
Z −
Z 2y = 4x
Z Z Z
F · ndS =
divFdV =
4xdV
S
Z
1
Z
D
1−x2
z=1-x2
D
Z
=
2−z
4xdydzdx
0
Z
0
1
Z
y=2-z
1
2
1
0
y
x
1−x2
4x(2 − z)dzdx
=
0
Z
0
1
Z
1−x2
2
(8x − 4xz)dzdx =
=
0
Z
=
1−x2
1
Z
(8xz − 2xz )
0
0
dx
0
1
[8x(1 − x2 ) − 2x(1 − x2 )2 ]dx
0
1
= −2(1 − x2 )2 + (1 − x2 )3
3
1
=
0
5
3
S3
z
p
26. For S1 , n = (xi + yj)/ x2 + y 2 ; for S2 , n2 = −k and
z = 0; and for S3 , n3 = k and z = c. Then
c
S1
a
x
S2
y
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CHAPTER 15 IN REVIEW
287
Z Z
Z Z
Z Z
F · ndS =
Z Z
F · n1 dS1 +
S
F · n2 dS2 +
S1
S2
F · n3 dS3
S3
Z Z
Z Z
x2 + y 2
2
p
dS1 +
(−z − 1)dS2 +
(z 2 + 1)dS3
=
x2 + y 2
s2
S3
S1
Z Z
Z Z
Z Z p
2
2
x + y dS1 +
(−1)dS2 +
(c2 + 1)dS3
=
S2
S3
S1
Z Z
Z Z
Z Z
2
=a
dS1 −
dS2 + (c + 1)
dS3
Z Z
S1
S2
S3
= a(2πac) − πa2 + (c2 + 1)πa2 = 2πa2 c + πa2 c2 .
27. x2 − y 2 = u2 (cosh v)2 − u2 (sinh v)2 hyperbolic paraboloid
= u2 (cosh v)2 − (sinh v)2
u2 = z;
28. z = x2 + y 2 ; paraboloid
29. y = x2 ; parabolic cylinder
30. x2 + y 2 − z 2 = (cos u cosh v)2 + (sin u cosh v)2 − (sinh v)2
= (cosh v)2 − (sinh v)2 = 1; z 2 = x2 + y 2 − 1
frustum of a cone
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Chapter 16
Higher-Order Differential
Equations
16.1
Exact First-Order Equations
1. Since Py = 0 = Qx , the equation is exact.
3
fx = 2x + 4, f = x2 + 4x + g(y), fy = g 0 (y) = 3y − 1, g(y) = y 2 − y
2
3
The solution is x2 + 4x + y 2 − y = C.
2
2. Since Py = 1 and Qx = −1, the equation is not exact.
3. Since Py = 4 = Qx , the equation is exact.
5
fx = 5x + 4y, f = x2 + 4xy + g(y), fy = 4x + g 0 (y) = 4x − 8y 3 , g(y) = −2y 4
2
5 2
The solution is x + 4xy − 2y 4 = C.
2
4. Since Py = cos y − sin x = Qx , the equation is exact.
fx = sin y−y sin x, f = x sin y+y cos x+g(y), fy = x cos y+cos x+g 0 (y) = cos x+x cos y−y,
1
g(y) = − y 2
2
1
The solution is x sin y + y cos x − y 2 = C.
2
5. Since Py = 4xy = Qx , the equation is exact.
fx = 2y 2 x − 3, f = y 2 x2 − 3x + g(y), fy = 2yx2 + g 0 (y) = 2yx2 + 4, g(y) = 4y
The solution is y 2 x2 − 3x + 4y = C.
y
1
1
3
6.
− 4x + 3y sin 3x dx + 2y − + cos 3x dy = 0. Since Py = 2 + 3 sin 3x and Qx =
2
x
x
x
1
− 3 sin 3x, the equation is not exact.
x2
7. (x2 − y 2 )dx + (x2 − 2xy)dy = 0. Since Py = −2y and Qx = 2x − 2y, the equation is not exact.
288
16.1. EXACT FIRST-ORDER EQUATIONS
8.
289
y
1
dx + (ln x − 1)dy = 0. Since Py =
= Qx , the equation is exact. fy =
x
x
y
y
0
ln x − 1, f = y ln x − y + g(x), fx = + g (x) = 1 + ln x + , g 0 (x) = 1 + ln x, g(x) = x ln x
x
x
The solution is y ln x − y + x ln x = C.
1 + ln x +
9. (y 3 − y 2 sin x − x)dx + (3xy 2 + 2y cos x)dy = 0. Since Py = 3y 2 − 2y sin x = Qx , the equation
is exact.
fx = y 3 − y 2 sin x − x, f = xy 3 + y 2 cos x − 21 x2 + g(y), fy = 3xy 2 + 2y cos x + g 0 (y) =
3xy 2 + 2y cos, g(y) = 0
The solution is xy 3 + y 2 cos x − 21 x2 = C
10. Since Py = 3y 2 = Qx , the equation is exact. fx = x3 + y 3 , f =
3xy 2 + g 0 (y) = 3xy 2 , g(y) = 0
1
The solution is x4 + xy 3 = C.
4
1 4
x + xy 3 + g(y), fy =
4
11. Since Py = 1 + ln y + xe−xy and Qx = ln y, the equation is not exact.
12. Since Py = 3x2 + ey = Qx , the equation is exact. fx = 3x2 y + ey , f = x3 y + xey + g(y), fy =
x3 + xey + g 0 (y) = x3 + xey − 2y, g(y) = −y 2
The solution is x3 y + xey − y 2 = C.
13. (2xex − y + 6x2 )dx − xdy = 0. Since Py = −1 = Qx , the equation is exact. fx = 2xex − y +
6x2 , f = 2xex − 2ex − yx + 2x3 + g(y), fy = −x + g 0 (y) = −x, g(y) = 0
The solution is 2xex − 2ex − yx + 2x3 = C.
3
3
14. 1 − + y dx + 1 − + x dy = 0. Since Py = 1 = Qx , the equation is exact.
x
y
3
3
3
fx = 1 − + y, f = x − 3 ln |x| + xy + g(y), fy = x + g 0 (y) = 1 − + x, g 0 (y) = 1 − ,
x
y
y
g(y) = y − 3 ln |y|
The solution is x − 3 ln |xy| + xy + y = C.
15. Since Py = 3x2 y 2 = Qx , the equation is exact.
1
1
1
fy = x3 y 2 , f = x3 y 3 + g(x), fx = x2 y 3 + g 0 (x) = x2 y 3 −
, g 0 (x) = −
=
2
3
1 = 9x
1 + 9x2
1
1 1
x
1
1
, g(x) = −
tan−1
= − tan1 3x
−
9 1/9 + x2
9 1/3
1/3
3
1
1
The solution is x3 y 3 − tan−1 3x = C or x3 y 3 = tan −13x = C1 .
3
3
16. 2ydx − (5y − 2x)dy = 0. Since Py = 2 = Qx , the equation is exact.
5
fx = 2y, f = 2xy + g(y), fy = 2x + g 0 (y) = −5y + 2x, g(y)= − y 2
2
5
The solution is 2xy − y 2 = C.
2
17. Since Py = sin x cos y = Qx , the equation is exact.
fy = cos x cos y, f = cos x sin y +g(x), fx = − sin x sin y +g 0 (x) = tan x−sin x sin y, g 0 (x) =
tan x, g(x) = ln | sec x|
The solution is cos x sin y + ln | sec x| = C or cos x sin y − ln | cos x| + C.
290
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
2
2
18. (2y sin x cos x − y + 2y 2 exy )dx + (sin2 x + 4xyexy − x)dy = 0. Since Py = 2 sin x cos x −
2
2
2
1 + 4xy 3 exy + 4yexy = Qz , the equation is exact. fx = 2y sin x cos x − y + 2y 2 exy =
1
2
2
y sin 2x − y + 2y 2 exy , f = − y cos 2x − xy + 2exy + g(y),
2
2
2
1
1
fy = − cos 2x − x + 4xyexy + g 0 (y) = − (1 − 2 sin2 x) − x + 4xyexy + g 0 (y)
2
2
2
2
1
= − + sin2 x − x + 4xyexy + g 0 (y) = sin2 x + 4xyexy − x
2
1
1
g 0 (y) = , g(y) = y
2
2
1
1
2
The solution is − y cos 2x − xy + 2exy + y = C.
2
2
19. Since Py = 4t3 − 1 = Qt , the equation is exact.
ft = 4t3 y − 15t2 − y, f = t4 y − 5t3 − yt + g(y),
fy = t4 − t + g 0 (y) = t4 + 3y 2 − t, g 0 (y) − 3y 2 , g(y) = y 3 .
The solution is t4 y − 5t3 − yt + y 3 = C.
y 2 − t2
(t2 + y 2 ) + y(2y)
=
(t2 + y 2 )2
(t2 + y 2 )2
−2t
, the equation is not exact.
and Qy = 2
(t + y 2 )2
20. Since Py = −
21. Since Py = 2(x + y) = Qx , the equation is exact.
1
fx = (x+y)2 = x2 +2xy+y 2 , f = x3 +x2 y+xy 2 +g(y), fy = x2 +2xy+g 0 (y) = 2xy+x2 −1
3
1
g 0 (y) = −1, g(y) = −y A family of solutions is x3 + x2 y + xy 2 − y = C. Substituting x = 1
3
1
4
and y = 1 we obtain + 1 + 1 − 1 = = C. The solution subject to the given condition is
3
3
1 3
4
2
2
x + x y + xy − y = .
3
3
22. Since Py = 1 = Qx , the equation is exact.
fx = ex + y, f = ex + xy + g(y), fy = x + g 0 (y) = 2 + x + yey , g 0 (y) = 2 + yey Using
integration by parts, g(y) = 2y + yey − y. A family of solutions is ex + xy + 2y + yey − ey = C.
Substituting x = 0 and y = 1 we obtain 1 + 2 + e − e = 3 = C. The solution subject to the
given condition is ex + xy + 2y + yey − ey = 3.
23. Since Py = 4 = Qt , the equation is exact.
ft = 4y + 2t − 5, f = 4ty + t2 − 5t + g(y), fy = 4t + g 0 (y) = 6y + 4t − 1, g 0 (y) = 6y − 1, g(y) =
3y 2 − y A family of solutions is 4ty + t2 − 5t + 3y 2 − y = C. Substituting t = −1 and y = 2
we obtain −8 + 1 + 5 + 12 − 2 = 8 = C. The solution subject to the given condition is
4ty + t2 − 5t + 3y 2 − y = 8.
24. Since Py = 2y cos x − 3x2 = Qx , the equation is exact.
fx = y 2 cos x − 3x2 y − 2x, f = y 2 sin x − x3 y − x2 + g(y), fy = 2y sin x − x3 + g 0 (y) =
2y sin x − x3 + ln y,
g 0 (y) = ln y, g(y) = y ln y − y A family of solutions is y 2 sin x − x3 y − x2 + y ln y − y = C.
Substituting x = 0 and y = e we obtain e − e = 0 = C. The solution subject to the given
condition is y 2 sin x − x3 y − x2 + y ln y − y = 0.
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16.2. HOMOGENEOUS LINEAR EQUATIONS
291
25. We want Py = Qx or 3y 2 + 4kxy 3 = 3y 2 + 40xy 3 . Thus, 4k = 40 and k = 10.
26. We want Py = Qx or 18xy 2 − sin y = 4kxy 2 − sin y. Thus 4k = 18 and k = 92 .
27. We need Py = Qx , so we must have
1 xy
xe
+
∂M
= exy + xyexy + 2y −
∂y
1
x2 .
This gives M (x, y) =
(yx − 1)exy
y
+ y 2 − 2 + g(x) for some function g.
x
x
28. We need Py = Qx , so we must have
x1/2 y −1/2 +
∂N
x
. This gives N (x, y) =
= 12 x−1/2 y −1/2 − 2
∂x
(x + y 2 )2
1
+ g(y) for some function g.
2(x2 + y)
∂
∂ 4
[µ(x, y)M (x, y)] =
xy = 4xy 3
∂y
∂y
∂
∂ 2 3
[µ(x, y)N (x, y)] =
2x y + 3y 5 − 20y 3 = 4xy 3
∂z
∂x
Therefore, µ(x, y)M (x, y)dx + µ(x, y)N(x, y) = 0 is exact, and µ(x, y) is an integrating factor.
Now, if y 3 xydx + (2x2 + 3y 2 − 20)dy = 0, then xydx + (2x2 + 3y 2 − 20)dy = 0, provided
y 6= 0. Therefore, to solve the original DE, we solve xy 4 dx + 2x2 y 3 + 3y 5 − 20y 3 dy = 0.
fx = xy 4 , f = 12 x2 y 4 + g(y), fy = 2x2 y + g 0 (y) = 2x2 y 3 + 3y 5 − 20y 3 ,
g 0 (y) = 3y 5 − 20y , g(y) = 21 y 6 − 5y 4 , f = 21 x2 y 4 + 12 y 6 − 5y 4 .
The solution is therefore 21 x2 y 4 + 12 y 6 − 5y 4 = C.
29. Let µ(x, y = y 3 . Then
1
dy − g(x)dx = 0. Since g is a function of x
h(y)
only and h is a function of y only, we have Py = Qx = 0.
30. True; a separable equation can be written as
16.2
Homogeneous Linear Equations
1. 3m2 − m = 0 =⇒ m(3m − 1) = 0 =⇒ m = 0, 1/3; y + C1 + C2 ex/3
2. 2m2 + 5m = 0 =⇒ m(2m + 5) = 0 =⇒ m = 0, −5/2; y = C1 + C2 e−5x/2
3. m2 − 16 = 0 =⇒ m2 = 16 =⇒ m = −4, 4; y = C1 e−4x + C2 e4x
√
√
√
√
4. m2 − 8 = 0 =⇒ m2 = 8 =⇒ m = −2 2, 2 2; y = C1 e−2 2x + C2 e2 2x
5. m2 + 9 = 0 =⇒ m2 = −9 =⇒ m = −3i, 3i; y = C1 cos 3x + C2 sin 3x
1
1
6. 4m2 + 1 = 0 =⇒ m2 = −1/4 =⇒ m = −i/2, i/2; y = C1 cos x + C2 sin x
2
2
7. m2 − 3m + 2 = 0 =⇒ (m − 1)(m − 2) = 0 =⇒ m = 1, 2; y = C1 ex + C2 e2x
8. m2 − m − 6 = 0 =⇒ (m + 2)(m − 3) = 0 =⇒ m = −2, 3; y = C1 e−2x + C2 33x
9. m2 + 8m + 16 = 0 =⇒ (m + 4)2 = 0 =⇒ m = −4, −4; y = C1 e−4x + C2 xe−4x
10. m2 − 10m + 25 = 0 =⇒ (m − 5)2 = 0 =⇒ m = 5, 5; y = C1 e5x + C2 xe5x
292
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
11. m2 + 3m − 5 = 0 =⇒ m = −3/2 ±
12. m2 + 4m − 1 = 0 =⇒ m = −2 ±
√
√
29/2; y = C1 e(−3/2−
5; y = C1 e(−2−
√
5)x
√
29/2)x
√
+ C2 e(−3/2+
√
+ C2 e(−2+
29/2)x
5)x
13. 12m2 − 5m − 2 = 0 =⇒ (3m − 2)(4m + 1) = 0 =⇒ m = −1/4, 2/3; y = C1 e−x/4 + C2 e2x/3
14. 8m2 + 2m − 1 = 0 =⇒ (4m − 1)(2m + 1) = 0 =⇒ m = −1/2, 1/4; y = C1 e−x/2 + C2 ex/4
15. m2 − 4m + 5 = 0 =⇒ m = 2 ± i; y = e2x (C1 cos x + C2 sin x)
√
16. 2m − 3m + 4 = 0 =⇒ m = 3/4 ± ( 23/4)i; y = e3x/4
2
√
!
√
23
23
C1 cos
x + C2 sin
x
4
4
√
√
17. 3m2 + 2m + 1 = 0 =⇒ m = −1/3 ± ( 2/3)i; y = e−x/3
2
18. 2m + 2m + 1 = 0 =⇒ m = −1/2 ± (1/2)i; y = e
−x/2
√ !
2
2
C1 cos
x + C2 sin
x
3
3
1
1
C1 cos x + C2 sin x
2
2
19. 9m2 + 6m + 1 = 0 =⇒ (3m + 1)2 = 0 =⇒ m = −1/3, −1/3; y = C1 e−x/3 + C2 xe−x/3
20. 15m2 − 16m − 7 = 0 =⇒ (3m + 1)(5m − 7) = 0 =⇒ m = −1/3, 7/5; y = C1 e−x/3 + C2 e7x/5
21. m2 + 16 = 0 =⇒ m2 = −16 =⇒ m = ±4i; y = C1 cos 4x + C2 sin 4x; y 0 = −4C1 sin 4x +
C2 cos 4x
Using y(0) = 2 we obtain 2 = C1 . Using y 0 (0) = −2 we obtain −2 = 4C2 or C2 = −1/2. The
1
solution is y = 2 cos 4x − sin 4x.
2
22. m2 − 1 = 0 =⇒ m2 = 1 =⇒ m = ±1; y = C1 ex + C2 e−x ; y 0 = C1 ex − C2 e−x . Using
y(0) = Y 0 (0) = 1 we obtain the system C1 + C2 = 1, C1 − C2 = 1. Thus, C1 = 1 and C2 = 0.
The solution is y = ex .
23. m2 + 6m + 5 = 0 =⇒ (m + 1)(m + 5) = 0 =⇒ m = −5, −1; y = C1 e−5x + C2 e−x ;
y 0 = −5C1 e−5x − C2 e−x . Using y(0) = 0 and y 0 (0) = 3 we obtain the system C1 + C2 = 0,
3
3
− 5C1 − C2 = 3. Thus, C1 = −3/4 and c2 = 3/4. The solution is y = − e−5x + e−x .
4
4
24. m2 − 8m + 17 = 0 =⇒ m = 4 ± i; y = e4x (C1 cos x + C2 sin x);
y 0 = e4x [(4C1 + C2 ) cos x + (−C1 + 4C2 ) sin x] .
Using y(0) = 4 and y 0 (0) = −1 we obtain the system C1 = 4, 4C1 + C2 = −1. Thus, C1 = 4
and C2 = −17. The solution is y = e4x (4 cos x − 17 sin x).
1
1
1
25. 2m2 − 2m + 1 = 0 =⇒ m = 1/2 ± (1/2)i; y = ex/2 (C1 cos x + C2 sin x); y 0 = ex/2 [ (C1 +
2
2
2
1
1
1
C2 ) cos x− (C1 −C2 ) sin x]. Using y(0) = −1 and y 0 (0) = 0 we obtain the system C1 = −1,
2
2
2
1
1
1
1
x/2
C1 + C2 = 0. Thus, C1 = −1 and C2 = 1. The solution is y = e
sin x − cos x .
2
2
2
2
16.2. HOMOGENEOUS LINEAR EQUATIONS
293
26. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; y = C1 ex + C2 xex ; y 0 = (C1 + C2 )ex +
C2 xex .
Using y(0) = 5 and y 0 (0) = 10 we obtain the system C1 = 5, C1 + C2 = 10. Thus,
C1 = C2 = 5. The solution is y = 5ex + 5xex .
√
√
√
7
7
−x/2
2
(C1 cos
x + C2 sin
x);
27. m + m + 2 = 0 =⇒ m = −1/2 ± ( 7/2)i; y = e
2
2
"
#
√
√
√
√
1
7
7
7
7
1
y 0 = e−x/2 (− C1 +
C2 ) cos
x + (−
C1 − C2 ) sin
x .
2
2
2
2
2
2
√
1
7
C2 = 0. Thus, C1 = C2 = 0.
Using y(0) = y 0 (0) = 0 we obtain the system C1 = 0, − C1 +
2
2
The solution is y = 0.
28. 4m2 − 4 − 3 = 0 =⇒ (2m − 3)(2m + 1) = 0 =⇒ m = −1/2, 3/2; y = C1 e−x/2 + C2 e3x/2 ;
1
3
y 0 = − C1 e−x/2 + C2 e3x/2 . Using y(0) = 1 and y 0 (0) = 5 we obtain the system C1 +C2 = 1,
2
2
3
7
11
1
− C1 + C2 = 5. Thus, c1 = −7/4 and C2 = 11/4. The solution is y = − e−x/2 + e3x/2 .
2
2
4
4
29. m2 −3m+2 = 0 =⇒ (m−1)(m−2) = 0 =⇒ m = 1, 2; y = C1 ex +C2 e2x ; y 0 = C1 ex +2C2 e2x .
Using y(1) = 0 and y 0 (1) = 1 we obtain the system eC1 + e2 C2 = 0, eC1 + 2e2 C2 = 1. Thus,
C1 = −e−1 and C2 = e−2. The solution is y = −ex−1 + e2x−2 .
0
30. m2 +1 = 0 =⇒ m2 = −1 =⇒ m = ±i; y = C1 cos x+C2 sin x;
x+C2 cos x. Using
√ y = C1 sin√
1
3
3
1
y(π/3) = 0 and y 0 (π/3) = 2 we obtain the system C1 +
C2 = 0, −
C1 + C2 = 2.
2
2
2
2
√
√
Thus, C1 = − 3 and C2 = 1. The solution is y = − 3 cos x + sin x.
31. The auxiliary equation is (m − 4)(m + 5) = m2 + m − 20 = 0. The differential equation is
y 00 + y 0 − 20y = 0.
32. The auxiliary equation is [(m − 3) − i] [(m − 3) + i] = (m − 3)2 − i2 = m2 − 6m + 10 = 0. The
differential equation is y 00 = 6y 0 + 10y = 0.
33. The auxiliary equation is m2 +1 = 0, so m = ±i. The general solution is y = C1 cos x+C2 sin x.
The boundary conditions yield y(0) = C1 = 0, y(π) = −C1 = 0, so y = C2 sin x.
34. The general solution is y = C1 cos x + C2 sin x. The boundary conditions yield y(0) = C1 =
0, y(π) = −C1 = 1, which is a contradiction. No solution.
35. The general
solution is y = C1 cos x + C2 sin x. The boundary conditions yield y 0 (0) = C2 =
0 1
0, y 2 = −C1 = 2, so y = −2 cos x.
36. The auxiliary equation is m2 − 1 = 0, so m = ±1. The general solution is y = C1 ex + C2 e−x .
The boundary conditions yield y(0) = C1 + C2 = 1, y(1) = C1e + C2 e−1 = −1, or C1 =
−1
−1
−1 − e
e+1
−1 − e
e+1
and C2 =
, so y =
ex +
e−x.
e − e−1
e − e−1
e − e−1
e − e−1
37. The auxiliary equation is m2 − 2m + 2 = 0, so m = 1 ± i. The general solution is y =
ex (C1 cos x + C2 sin x) . The boundary conditions yield y(0) = C1 = 1 and y(π) = −eπ C1 =
−1, which is a contradiction. No solution.
www.elsolucionario.org
294
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
38. The general solution is y = ex (C1 cos x + C2 sin x) . The boundary
conditions yield y(0) =
C1 = 1 and y (π/2) = C2 eπ/2 = 1, so y = ex cos x + e−π/2 sin x
39. The auxiliary equation is m2 − 4m + 4 = 0, so m = 2 is a repeated root. The general solution
is y = C1 e2x + C2 xe2x . The boundary conditions yield y(0) = C1 = 0 and y(1) = C2 e2 = 1,
so y = xe−2 e2x = xe2(x−1) .
40. The general solution is y = C1 e2x +C2 xe2x . The boundary conditions yield y 0 (0) = 2C1 +C2 =
1 and y(1) = (C1 + C2 ) e2 = 2, or C1 = 1 − 2e−2 and C2 = −1 + 4e−2 , so y = (1 − 2e−2 )e2x +
(−1 + 4e−2 )xe2x .
41. Assuming a solution of the form y = emx we obtain the auxiliary equation m3 − 9m2 + 25m −
17 = 0. Since y1 = ex is a solution we know that m1 = 1 is a root of the auxiliary equation.
The equation can then be written as (m−1)(m2 −8m+17) = 0. The roots of this equation are 1
and 4±i. The general solution of the differential equation is y = C1 ex +e4x (C2 cos x+C3 sin x).
42. Assuming a solution of the form y = emx we obtain the auxiliary equation m3 +6m2 +m−34 =
0. Since y1 = e−4x cos x is a solution, we know that m1 = −4 + i is a root of the auxiliary
equation. Using the fact that complex roots of real polynomial equations occur in conjugate
pairs we have that m2 = −4−i is also a root. Thus [m−(−4+i)][m−(−4−i)] = m2 +8m+17 is
a factor of the auxiliary equation and we can write it as m3 +6m2 +m−34 = (m2 +8m+17)(m−
2) = 0. The general solution of the differential equation is y = C1 e2x +e−4x (C1 cos x+C2 sin x).
43. y 0 = memx , y 00 = m2 emx , y 000 = m3 emx ; m3 emx − 4m2 emx − 5memx = 0 =⇒ (m3 − 4m2 −
5m)emx = 0 =⇒ m3 − 4m2 − 5m = 0 =⇒ m(m − 5)(m + 1) = 0 =⇒ m = 0, −1, 5; y =
C1 + C2 e−x + C3 e5x
44. y 0 = memx , y 00 = m2 emx , y 000 = m3 emx ; m3 emx + 3m2 emx − 4memx − 12emx = 0 =⇒
(m3 + 3m2 − 4m − 12)emx = 0 =⇒ m3 + 3m2 − 4m − 12 = 0 =⇒ m2 (m + 3) − 4(m + 3) =
0 =⇒ (m2 − 4)(m + 3) = 0 =⇒ m = −3, −2, 2; y = C1 e−3x + C2 e−2x + C3 e2x
45. Case 1: λ = −α2 < 0
Auxiliary equation is m2 − α2 = 0, so m = ±α and general solution is y = C1 eαx + C2 e−αx .
Boundary conditions yield y(0) = C1 +C2 = 0 and y(1) = C1 eα +C2 e−α = 0, or C1 = C2 = 0.
So Case 1 yields no nonzero solutions.
Case 2: λ = 0
Auxiliary equation is m2 = 0, so m = 0 is a repeated root and general solution is y = C1 +C2 x.
Boundary conditions yield y(0) = C1 = 0 and y(1) = C2 = 0. So Case 2 yields no nonzero
solutions.
Case 3: λ = α2 > 0
Auxiliary equation is m2 + α2 = 0, so m = ±αi and the general solution is y = C1 cos αx +
C2 sin x. Boundary conditions yield y(0) = C1 = 0 and y(1) = C2 sin α = 0. Hence, nonzero
solutions exist only when sin α = 0, which implies α = ±nπ so that λ = n2 π 2 for n = 1, 2, 3, . . .
(n = 0 is excluded since that would give λ = 0).
16.3. NONHOMOGENEOUS LINEAR EQUATIONS
295
4
4
46. (a) If the earth has density ρ then M = ρ πR3 and Mr = ρ πr3 , so that M/Mr = R3 /r3
3
3
and Mr = r3 M/R3 . Then
F = −k
r3 M m/R3
mM
Mr m
=
−k
= −k 3 r.
r2
r2
R
d2 r
mM
d2 r
kM
d2 r
=
−k
r
=⇒
+
r
=
0
=⇒
+ ω2 r =
dt2
R3
dt2
R3
dt2
0whereω 2 = kM/R3 . Since kmM/R2 = mg we have ω 2 = kM/R3 = g/R.
(b) Since a = d2 r/dt2 ,F = ma = m
(c) The general solution of the differential equation in part (b) is r(t) = c1 cos ωt + c2 sin ωt.
The initial conditions r(0) = R and r0 (0) = 0 imply c1 = R and c2 = 0. Then r(t) =
R cos ωt. The mass oscillates back and forth from one side of the earth to the other with
a period of T = 2π/ω. If we use R=3960 mi and g=32 ft/s2 , then T ≈ 5079 s or 1.41 h.
47.
48.
16.3
Nonhomogeneous Linear Equations
1. m2 − 9 = 0 =⇒ m = −3, 3; yc = C1 e−3x + C2 e3x ; yp = A, yp0 = yp00 = 0; −9A = 54 =⇒
A = −6; yp = −6; y = C1 e−3x + C2 e3x − 6
2. 2m2 − 7m + 5 = 0 =⇒ (2m − 5)(m − 1) = 0 =⇒ m = 1, 5/2; yc = C1 ex + C2 e5x/2 ; yp =
29
A, yp0 = yp00 = 0; 5A = −29 =⇒ A = −29/5; y = C1 ex + C2 e5x/2 −
5
3. m2 + 4m + 4 = 0 =⇒ (m + 2)2 = 0 =⇒ m = −2, −2; yc = C1 e−2x + C2 xe−2x ; yp =
Ax + B, yp0 = A, yp00 = 0; 4A + 4(Ax + B) = 2x + 6 =⇒ 4Ax + 4(A + B) = 2x + 6
Solving 4A = 2, 4A + 4B = 6, we obtain A = 1/2 and B = 1. Thus, y = C1 e−2x + C2 xe−2x +
1
x + 1.
2
4. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; yc = C1 ex + C2 xex ; yp = Ax3 + Bx2 +
Cx + d, yp0 = 3Ax2 + 2Bx + C, yp00 = 6Ax + 2B
(6Ax + 2B) − 2(3Ax2 + 2Bx + c) + (Ax3 + Bx2 + Cx + D) = x3 + 4x
=⇒ Ax3 + (−6A + B)x2 + (6A − 4B + C)x + (2B − 2C + D) = x3 + 4x
Solving A = 1, −6A + B = 0, 6A − 4B + c = 4, 2B − 2C + D = 0, we obtain A = 1, B =
6, C = 22, and D = 32. Thus, y = C1 ex + C2 xex + x3 + 6x2 + 22x + 32.
5. m2 + 25 = 0 =⇒ m = ±5i; yc = C1 cos 5x + C2 sin 5x; yp = A sin x + B cos x, yp0 = A cos x −
B sin x, yp00 = −A sin x − B cos x; −A sin x − B cos x + 25(A sin x + B cos x) = 6 sin x =⇒
1
24A sin x + 24B cos x = 6 sin x; A = 1/4, B = 0; y = C1 cos 5x + C2 sin 5x + sin x
4
6. m2 − 4 = 0 =⇒ m = −2,
2;
yc = C1 e−2x + C2 e2x ;
16Ae4x − 4Ae4x = 7e4x =⇒ 12Ae4x = 7e4x
yp0 = 4Ae4x , yp00 =
7
=⇒ A = 7/12; y = C1 e−2x + C2 e2x + e4x
12
yp = Ae4x ,
296
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
7. m2 − 2m − 3 = 0 =⇒ (m − 3)(m + 1) = 0 =⇒ m = −1, 3; yc = C1 e−x + C2 e3x
yp = Ae2x + Bx3 + Cx2 + Dx + E, yp0 = 2Ae2x + 3Bx2 + 2Cx + D, yp00 = 4Ae2x + 6Bx + 2C
(4Ae2x + 6Bx + 2C) − 2(2Ae2x + 3Bx2 + 2Cx + D) − 3(Ae2x + Bx3 + Cx2 + Dx + E) =
4e2x +2x3 =⇒ −3Ae2x 3Bx3 +(−6B −3C)x2 +(6B −4C −3D)x+(2C −2D −3E) = 4e2x +2x3
Solving −3A = 4, −3B = 2, −6B − 3C = 0, 6B − 4C − 3D = 0, 2C − 2D − 3E = 0, we
obtain A = −4/3, B = −2/3, C = 4/3, D = −28/9, and E = 80/27. Thus,
2
4
28
80
4
y = C1 e−x + C2 e3x − e2x − x3 + x2 − x + .
3
3
3
9
27
√
√
√
1
3
8. m2 + m + 1 = 0 =⇒ m = − ±
i; yc = e−x/2 (C1 cos 3x/2 + C2 sin 3x/2)
2
2
yp = Ax2 ex + Bxex + Cex + D, yp0 = Ax2 ex + (2A + B)xex + (B + C)ex
yp00 = Ax2 ex + (4A + B)xex + (2A + 2B + C)ex
2 x
Ax
(4A + B)xex + (2A+ 2B + C)ex + Ax2 ex + (2A + B)xex + (B + C)ex
e2 +
+ Ax ex + Bxex + Cex + D = x2 ex +3 =⇒ 3Ax2 ex +(6A+3B)xex (2A+3B +3C)ex +D =
x2 ex + 3
Solving 3A = 1, 6A + 3B = 0, 2A + 3B + 3C = 0, D = 3, we obtain A = 1/3, B =
−2/3, C = 4/9, and D = 3. Thus,
y = e−x/2 (C1 cos
√
√
1
2
4
3x/2 + C2 sin 3x/2) + x2 ex − xex + ex + 3.
3
3
9
9. m2 − 8m + 25 = 0 =⇒ m = 4 ± 3i; yc = e4x (C1 cos 3x + C2 sin 3x); yp = Ae3x + B sin 2x +
C cos 2x, yp0 = 3Ae3x + 2B cos 2x − 2C sin 2x, yp00 = 9Ae3x − 4B sin 2x − 4C cos 2x
(9Ae3x − 4B sin 2x − 4C cos 2x) − 8(3Ae3x + 2B cos 2x − 2C sin 2x) + 25(Ae3x + B sin 2x +
C cos 2x) = e3x − 6 cos 2x =⇒ 10Ae3x + (21B + 16C) sin 2x + (−16B + 21C) cos 2x = e3x −
6 cos 2x
Solving 10A = 1, 21B + 16C = 0, −16B + 21C = −6, we obtain A = 1/10, B = 96/697,
and C = −126/697. Thus,
y = e4x (C1 cos 3x + C2 sin 3x) +
1 3x
96
126
e +
sin 2x −
cos 2x.
10
697
697
10. m2 − 5m + 4 = 0 =⇒ (m − 1)(m − 4) = 0 =⇒ m = 1, 4; yc = C1 ex + C2 e4x
yp = A sinh 3x + B cosh 3x, yp0 = 3A cosh 3x + 3B sinh 3x, yp00 = 9A sinh 3x + 9B cosh 3x
(9A sinh 3x+9B cosh 3x)−5(3A cosh 3x+3B sinh 3x)+4(A sinh 3x+B cosh 3x) = 2 sinh 3x =⇒
(13A − 15B) sinh 3x + (−15A + 13B) cosh 3x = 2 sinh 3x
Solving 13A − 15B = 2, −15A + 13B = 0, we obtain A = −13/28 and B = −15/28. Thus,
y + C1 ex + C2 e4x −
13
15
sinh 3x −
cosh 3x.
28
28
11. m2 − 64 = 0 =⇒ m = −8, 8; yc = C1 e−8x + C2 e8x ; yp = A, yp0 = yp00 = 0
1
− 64A = 16 =⇒ A = −1/4; y = C1 e−8x + C2 e8x − , y 0 = −8C1 e−8x + 8C2 e8x .
4
1
Using y(0) = 1 and y 0 (0) = 0 we obtain C1 + C2 − = 1, −8C1 + 8C2 = 0, or C1 = C2 = 5/8.
4
5
5
1
Thus, y = e−8x + e8x − .
8
8
4
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16.3. NONHOMOGENEOUS LINEAR EQUATIONS
297
12. m2 + 5m − 6 = 0 =⇒ (m + 6)(m − 1) = 0 =⇒ m = −6, 1; yc = C1 e−6x + C2 ex ; yp =
Ae2x , yp0 = 2Ae2x , yp00 = 4Ae2x ; Ae2x + 5(2Ae2x ) − 6(Ae2x ) = 10e2x =⇒ 8Ae2x = 10e2x =⇒
5
5
A = 5/4; y = C1 e−6x + C2 ex + e2x , y 0 = −6C1 e−6x + C2 ex + e2x .
4
2
5
5
9
Using y(0) = 1 and y 0 (0) = 0 we obtain C1 + C2 + = 1, −6C1 + C2 + = 0, or C1 =
4
2
28
4
9 −6x 4 x 5 2x
and C2 = − . Thus, y =
e
− e + e .
7
28
7
4
cos x sin x
=1
− sin x cos x
u01 = − sin x sec x = − tan x, u1 = ln | cos x|; u02 = cos x sec x = 1, u2 = x
yp = cos x ln | cos x| + x sin x; y = C1 cos x + C2 sin x + cos x ln | cos x| + x sin x
13. m2 + 1 = 0 =⇒ m = −i; i; yc = C1 cos x + C2 sin x; W =
14. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W =
cos x sin x
− sin x cos x
=1
sin2 x
1 − cos2 x
=−
= − sec x + cos x, u1 = − ln | sec x + tan x| + sin x
cos x
cos x
u02 = cos x tan x = sin x; u2 = − cos x
yp = − cos x ln | sec x + tan x| + sin x cos x − sin x cos x = − cos x ln | sec x + tan x|
y = C1 cos x + C2 sin x − cos x ln | sec x + tan x|
u01 = − sin x tan x = −
cos x sin x
=1
− sin x cos x
1
1
1
u01 = − sin2 x, u1 = − x + sin x cos x; u02 = sin x cos x, u2 = sin2 x
2
2
2
1
1
1
yp = − x cos x + sin x cos2 x + sin3 x
2
2
2
1
1
1
y = C1 cos x + C2 sin x − x cos x + sin x cos2 x + sin3 x
2
2
2
1
1
1
2
= C1 cos x + C2 sin x − x cos x + sin x(cos x + sin2 x) = C1 cos x + C3 sin x − x cos x
2
2
2
15. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W =
cos x sin x
=1
− sin x cos x
2
0
2
0
u1 = − sin x sec x tan x = − tan x = 1−sec x, u1 = x−tan x; u2 = cos x sec x tan x = tan x
u2 = − ln | cos x|; yp = cos x(tan x − x) = x cos x − sin x − sin x ln | cos x|
y = C1 cos x + C2 sin x + x cos x − sin x − sin x ln | cos x| = C1 cos x + C3 sin x + x cos x − sin x ln | cos x|
16. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W =
cos x sin x
=1
− sin x cos x
u01 = − sin x cos2 x u1 = 31 cos3 x; u02 = cos x cos2 x = cos x−cos x sin2 x, u2 = sin x− 13 sin3 x
yp = 31 cos4 x + sin2 x − 13 sin4 x = sin2 x + 13 (cos2 x − sin2 x) = sin2 x + 13 cos 2x
1
1 1
1
y = C1 cos x + C2 sin x + sin2 x + cos 2x = C1 cos x + C2 sin x + − cos 2x + cos 2x
3
2 2
3
1 1
= C1 cos x + C2 sin x + − cos 2x
2 6
17. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W =
18. m2 + 1 = 0 =⇒ m = −i, i; yc = C1 cos x + C2 sin x; W =
cos x sin x
− sin x cos x
=1
298
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
u01 = − sin x sec2 x = − tan x sec x, u1 = − sec x; u02 = cos x sec2 x = sec x;
u2 = ln | sec x + tan x|
yp = − cos x sec x + sin x ln | sec x + tan x| = −1 + sin x ln | sec x + tan x|
y = C1 cos x + C2 sin x − 1 + sin x ln | sec x + tan x|
19. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1 e−x + C2 ex ; W =
e−x
−e−x
ex
ex
=2
1 x
1
1
1
e cosh x = − (e2x + 1), u1 = − e2x − x;
2
4
8
4
1
1
1
1
u02 = e−x cosh x = (1 + e−2x ) u2 = x − e−2x ,
2
4
4
8
1
1
1
1
1
1
1
1
yp = e−x (− e2x − x) + ex ( x − e−2x = − ex − xe−x + xex − d−x
8
4
4
8
8
4
4
8
1 x 1 −x 1
= − e − e + x sinh x
8
8
2
1
1
1
1
y = C1 e−x + C2 ex − ex − e−x + x sinh x = C3 e−x + C4 ex + x sinh x
8
8
2
2
u01 =
20. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1 e−x + C2 ex ; W =
e−x
−e−x
ex
ex
=2
1
1
1
1
u01 = − ex sinh 2x = − (e3x − e−x ), u1 = − e3x − e−x
2
4
12
4
1
1
1
1
u02 = d−x sinh 2x = (ex − e−3x ), u2 = ex + e−3x
2
4
4
12
1
1
1
1
1
1
1
1
yp = e−x (− e3x − e−x ) + ex ( ex + e−3x ) = − e2x − e−2x + e2x + e−2x
12
4
4
12
12
4
4
12
1
1
1
= e2x − e−2x = sinh 2x
6
6
3
1
−x
x
y = C1 e + C2 e + sinh 2x
3
e−2x
e2x
21. m2 − 4 = 0 =⇒ m = −2, 2; yc = C1 e−2x + C2 e2x ; W =
=4
−2x
−2e
2e2x
2x
Z x 4t
2x
4x
1
e
1e
1
e
1
e
1
1
u01 = − e2x (
)=−
, u1 = −
dt; u02 = e−2x
=
, u2 = ln |x|
4
x
4 x
4 x0 t
4
x
4x
4
Z
Z x 4t
1 2x
1
e
1
1 −2x x e4t
dt + e ln |x|; y = C1 e−2x + C2 e2x − e2x
dt + e2x ln |x|
yp = − e
4
4
4
4
x0 t
x0 t
e−3x
−3e−3x
1
9x
3
3
9x
3
1
u01 = − e3x ( 3x = − x, u1 = − x2 ; u02 = e−3x ( 3x ) = xe−6x
6
e
2
4
6
e
2
22. m2 − 9 = 0 =⇒ m = −3, 3; yc = C1 e−3x + C2 e3x ; W =
Z
e3x
3e3x
=6
3 −6x
xe
dx Integration by parts
2
1
1
= − xe−6x − e−6x
4
24
3
1
1
3
1
yp = − x2 e−3x − xe−3x − e−3x ; y = C3 e−3x + C2 e3x − x2 e−3x − xe−3x
4
4
24
4
4
u2 =
16.3. NONHOMOGENEOUS LINEAR EQUATIONS
299
23. m2 + 3m + 2 = 0 =⇒ (m + 2)(m + 1) = 0 =⇒ m = −2, −1; yc = C1 e−2x + C2 e−x
1
e−x
e2x
e−2x
e−x
= e−3x ; u01 = − −3x
W =
=−
−2x
−x
x
−2e
−e
e
1+e
1 + ex
Z
2x
e
dx v = 1 + ex , dv = ex dx, ex = v − 1
u1 = −
1 + ex
Z
v−1
=−
dv = −v + ln |v| = −1 − ex + ln(1 + ex )
v
yp = e−2x [−1 − ex + ln(1 + ex )] + e−x ln(1 + ex ) = −e−2x − e−x + e−2x ln(1 + ex )
+ e−x ln(1 + ex )
y = C1 e−2x + C2 e−x − e−2x − e−x + e−2x ln(1 + ex ) + e−x ln(1 + ex )
= C3 e−2x + C4 e−x + e−2x ln(1 + ex ) + e−x ln(1 + ex )
24. m2 − 3m + 2 = 0 =⇒ (m − 1)(m − 2) = 0 =⇒ m = 1, 2; yc = C1 ex + C2 e2x ;
e3x
e2x
1
ex e2x
= e3x ; u01 = − 3x e2x
=−
W =
x
2x
x
e
2e
e
1+e
1 + ex
Z
2x
e
u1 = −
dx v = 1 + ex , dv = ex dx, ex = v − 1
1 + ex
Z
v−1
dv = −v + ln |v| = −1 − ex + ln(1 + ex )
=−
v
1
e3x
ex
u02 = 3x ex
=
, u2 = ln(1 + ex )
x
e
1+e
1 + ex
yp = ex [−1 − ex + ln(1 + ex )] + e2x ln(1 + ex ) = −ex − e2x + ex ln(1 + ex ) + e2x ln(1 + ex )
y = C1 ex + C2 e2x − ex − e2x + ex ln(1 + ex ) + e2x ln(1 + ex )
= C3 ex + C4 e2x + ex ln(1 + ex ) + e2x ln(1 + ex )
25. m2 + 3m + 2 = 0 =⇒ (m + 2)(m + 1) = 0 =⇒ m = −2, −1; yc = C1 e−2x + C2 e−x
1
e−2x
e−x
= e−3x ; u01 = − −3x e−x sin ex = −e2x sin ex
W =
−2x
−x
−2e
−e
e
Z
u1 = −
e2x sin ex dx
Integration by parts
= ex cos ex − sin ex
1
u02 = −3x e−2x sin ex = ex sin ex , u2 = − cos ex
e
yp = e−2x (ex cos ex −sin ex )+e−x (− cos ex ) = −e−2x sin ex ; y = C1 e−2x +C2 e−x −e−2x sin ex
26. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; yc = C1 ex + C2 xex ;
1
ex
xex
W =
= e2x ; u01 = − 2x xex ex tan−1 x = −x tan−1 x
x
x
e
xe + ex
e
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300
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
Z
u1 = −
x tan−1 xdx
Integration by parts
1
1
1
= − x2 tan−1 x − tan−1 x + x
2
2
2
1
1 x x
−1
−1
0
u2 = 2x e e tan x = tan x, u2 = x tan−1 x − ln(1 + x2 )
e 2
1
1
1
1 2
−1
−1
x
x
yp = e − x tan x − tan x + x + xe x tan−1 x − ln(1 + x2 )
2
2
2
2
1 2 x
1
1
1
= x e tan−1 x − ex tan−1 x + xex − xex ln(1 + x2 )
2
2
2
2
1 x
1
1 2 x
−1
x
x
y = C1 e + C3 xe + x e tan x − e tan−1 x − xex ln(1 + x2 )
2
2
2
27. m2 − 2m + 1 = 0 =⇒ (m − 1)2 = 0 =⇒ m = 1, 1; yc = C1 ex + C2 xex ;
1
ex
x
1
ex
xex
= e2x ; u01 = − 2x xex
=−
, u1 = − ln(1 + x2 );
W =
x
x
x
2
2
e
xe + e
e
1+x
1+x
2
1 x ex
1
−1
0
u2 = 2x e
=
, u2 = tan x
e
1 + x2
1 + x2
1
1
yp = − ex ln(1 + x2 ) + xex tan−1 x; y = C1 ex + C2 xex − ex ln(1 + x2 ) + xex tan−1 x
2
2
28. m2 − 2m + 2 = 0 =⇒ m = 1 ± i; yc = ex (C1 cos x + C2 sin x);
ex cos x
ex sin x
= e2x
W =
x
x
x
−e sin x + e cos x e cos x + ex sin x
1
1
u01 = − 2x ex sin x ex sec x = − tan x, u1 = ln | cos x|; u02 = 2x ex sec x = 1, u2 = x
e
e
yp = ex cos x ln | cos x| + xex sin x; y = ex (C1 cos x + C2 sin x) + ex cos x ln | cos x| + xex sin x
29. m2 + 2m + 1 = 0 =⇒ (m + 1)2 = 0 =⇒ m = −1, −1; yc = C1 e−x + C2 xe−x
1
e−x
xe−x
= e−2x ; u01 = − −2x xe−x e−x ln x = −x ln x
W =
−x
−x
−x
−e
−xe
+
e
e
Z
u1 = −
x ln xdx
Integration by parts
1 2 1 2
x − x ln x
4
2
1
u02 = −2x e−x e−x ln x = ln x, u2 = x ln x − x
e 1 2 1 2
1
3
−x
yp = e
x − x ln x + xe−x (x ln x − x) = x2 e−x ln x − x2 e−x
4
2
2
4
1
3
y = C1 e−x + C2 xe−x + x2 e−x ln x − x2 e−x
2
4
=
30. m2 + 10m + 25 = 0 =⇒ (m + 5)2 = 0 =⇒ m = −5, −5; yc = C1 e−5x + C2 xe−5x
1
e−10x
e−5x
xe−5x
e−5x
W =
= e−10x ; u01 = −10x xe−5x 2 = −
−5x
−5x
−5x
−5e
−5xe
+e
e
x
x
Z −5x
Z x −5t
−10x
−5x
e
1
e
e
e
u1 = −
dx = −
dt; u02 = −10x e−5x 2 = 2 ,
x
t
e
x
x
x
0
Z x −5t
Z −5x
Z x −5t
Z x −5x
e
e
e
e
−5x
−5x
u2 =
dx =
dt; yp = −e
dt + xe
dt
2
2
x2
t
t
x0
x0
x0 t
16.3. NONHOMOGENEOUS LINEAR EQUATIONS
y = C1 e−5x + C2 xe−5x − e−5x
Z
x
x0
e−5t
dt + xe−5x
t
301
Z
x
x0
e−5t
dt
t2
31. 4m2 − 4m + 1 = 0 =⇒ (2m − 1)2 = 0 =⇒ m = 1/2, 1/2; yc = C1 ex/2 + C2 xex/2
ex/2
xex/2
1
1
1
W = 1 x/2 1 x/2
= ex ; u01 = − x xex/2 (2e−x + x) = −2xe−3x/2 − x2 e−x/2
x/2
e
4
4
e
xe
+e
2Z
2
Z
1
u1 = −2 xe−3x/2 dx −
x2 e−x/2 dx Integration by parts
4
4
8
1
= xe−3x/2 + e−3x/2 + x2 e−x/2 + 2xe−x/2 + 4e−x/2
3
9
2
1
1
1
u02 = x ex/2 (2e−x + x) = 2e−3x/2 − xe−x/2
eZ
4 Z
4
1
−3x/2
−x/2
u2 = 2 e
dx +
xe
dx Integration by parts
4
1
4
= − e−3x/2 − xe−x/2 − e−x/2
3
2
8
1
x/2 4
−3x/2
yp = e ( xe
+ e−3x/2 + x2 e−x/2 + 2xe−x/2 + 4e−x/2 )
3
9
2
4 −3x/2 1 −x/2
8
x/2
+ xe (− e
− xe
− e−x/2 ) = e−x + x + 4
3
2
9
8 −x
x/2
x/2
y = C1 e
+ C2 xe
+ e +x+4
9
32. 4m2 − 4m + 1 = 0 =⇒ (2m − 1)2 = 0 =⇒ m = 1/2, 1/2; yc = C1 ex/2 + C2 xex/2
√
ex/2
xex/2
1 x/2 ex/2 √
x 1 − x2 0
x
0
2
= e ; u1 = − x xe
1−x = −
u1 =
W = 1 x/2 1 x/2
e
4
4
e
xe
+ ex/2
2
2
√
1
1
ex/2 √
1 − x2
(1 − x2 )3/2 ; u02 = x ex/2
1 − x2 =
12
e
4
4
Z
1 p
2
u2 =
1 − x dx Trig substitution
4
1 p
1
= sin−1 x + x 1 − x2
8
8
√
1
1
1 x/2
e (1 − x2 )3/2 + xex/2 sin−1 x + x2 ex/2 1 − x2
yp =
12
8
8
√
1
1
1 x/2
2 3/2
x/2
x/2
y = C1 e
+ C2 xe
+ e (1 − x ) + xex/2 sin−1 x + x2 ex/2 1 − x2
12
8
8
33. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1 e−x + C2 ex ; W =
e−x
−e−x
ex
ex
=2
1 x x
1
e xe = − xe2x
2 Z
2
1
2x
u1 = −
xe dx Integration by parts
2
1
1
= e2x − xe2x
8
4
1
1
1
1
1
1
1
1
1
u02 = e−x xex = x, u2 = x2 ; yp = e−x ( e2x − xe2x ) + ex ( x2 ) = ex − xex + x2 ex
2
2
4
8
4
4
8
4
4
u01 =
302
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
1
1
1
1
1
y = C1 e−x + C3 ex − xex + x2 ex ; y 0 = −C1 e−x + C3 ex − ex + xex + x2 ex
4
4
4
4
4
1
0
Using y(0) = 1 and y (0) = 0 we have C1 + C3 = 1, −C1 + C3 − = 0, or C1 = 3/8 and
4
3
5
1
1
C3 = 5/8. Thus, y = e−x + ex − xex + x2 ex .
8
8
4
4
34. 2m2 + m − 1 = 0 =⇒ (2m − 1)(m + 1) = 0 =⇒ m = −1, 1/2; yc = C1 e−x + C2 ex/2
e−x
ex/2
3
(x + 1)
1
1
2
= − (xex + ex ), u1 = − xex
W =
1 x/2 = e−x/2 ; u01 = − −x/2 ex/2
2
2
3
3
−e−x
e
3e
2
2
1
−x (x + 1)
0
u2 = −x/2 e
= e−x/2 (x + 1)
2
3
3eZ
1
e−x/2 (x + 1)dx Integration by parts
u2 =
3
2
= − xe−x/2 − 2e−x/2
3
2
1
−x
yp = e (− xex ) + ex/2 (− xe−x/2 − 2e−x/2 ) = −x − 2
3
3
1
y = C1 e−x + C2 ex/2 − x − 2; y 0 = −C1 e−x + C2 ex/2 − 1
2
1
Using y(0) = 1 and y 0 (0) = 0 we obtain C1 + C2 − 2 = 1, −C1 + C2 − 1 = 0, or C1 = 1/3
2
1
8
and C2 = 8/3. Thus, y = e−x + ex/2 − x − 2.
3
3
1 0
1
4
x x ln x
=x
y + 2 y = ln x; yc = C1 x + C2 x ln x; W =
1 1 + ln x
x
x
x
1
4
4
4
4
4
1
u01 = − (x ln x)( ln x) = − (ln x)2 , u1 = − (ln x)3 ; u02 = (x)( ln x) = ln x,
x
x
x
3
x
x
x
4
2
2
u2 = 2(ln x)2 ; yp = − x(ln x)3 + 2x(ln x)3 = x(ln x)3 ; y + C1 x + C2 x ln x + x(ln x)3
3
3
3
35. y 00 −
4 0
6
1
x2 x3
= x4 ;
y + 2 y = 3 ; yc = C1 x2 + C2 x3 ; W =
2x 3x2
x
x x
1
1
1
1
1
1
1
1
u01 = − 4 (x3 )
= − 4 ; u1 = 3 ; u02 = 4 (x2 )
= 5 , u2 = − 4 ;
3
3
x
3x
x
x
x
4x
x x 1
1
1
2
3
yp = x
+x − 4 =
3x3
4x
12x
1
2
3
y = C1 x + C2 x +
12x
36. y 00 −
37. Writing the differential equation in the form d2 C/dx2 − (1/λ2 )C = −C(∞)/λ2 we see that
the auxiliary equation is m2 − 1/λ2 = 0. Thus, Cc = c1 ex/λ + c2 e−x/λ . Using undetermined
coefficients with Cp = A we find that A = C(∞). Then C(x) = c1 ex/λ + c2 e−x/λ + C(∞).
Since C(0) = c1 +c2 +C(∞) = 0 and lim C(x) = C(∞) we see that c1 = 0 and c2 = −C(∞).
x→∞
Thus, C(x) = C(∞)(1 − e−x/λ ).
38. If yc is the complementary function and yp is a particular solution, we have
ayc00 + byc0 + cyc = 0
and ayp00 + byp0 + cyp = g(x).
www.elsolucionario.org
16.4. MATHEMATICAL MODELS
303
Therefore, letting y = yc + yp , we have
ay 00 = by 0 + cy 0 = a(yc + yp )00 + b(yc + yp )0 + c(yc + yp )
= ayc00 + ayp00 + byc0 + byp0 + cyc + cyp
= [ayc00 + byc0 + cyc ] + [ayp00 + byp0 + cyp ]
= ayp00 + byp0 + cyp = g(x)
39. (a) Substituting Aex in for y in the DE, we have Aex + 2Aex − 3Aex = 10ex or 0 = 10ex ,
which is a contradiction for any value of A.
(b) Substituting Axex for y, we have
A(x + 2)ex + A(2x + 2)ex − 3Axex = 10ex .
Equating coefficients of xex and coefficients of ex , we get
A + 2A − 3A = 0
and
2A + 2A = 10
5
5
. Therefore, yp = xex .
2
2
(c) The auxiliary equation is m2 + 2m − 3 = 0, so m = −3 or m = 1. This gives yc =
C1 e−3x + C2 ex . Therefore, the general solution is
which gives A =
5
y = yc + yp = C1 e−3x + C2 ex + xex
2
40. The auxiliary equation is m2 − 1 = 0, so m = ±1. This gives yc = C1 e−x + C2 ex . We look for a
particular solution of the form yp = Axex + B(x − 2)e−x − Axex − Bxex = e−x − ex . Equating
coefficients of xex , ex , xe−x , and e−x , we get A − A = 0, 2A = −1, B − B = 0, −2B = 1,
1
1
1
1
which gives A = − , B = − . Therefore, yp = − xex − xe−x and the general solution is
2
2
2
2
1
1
y = yc + yp = C1 e−x + C2 ex − xex − xe−x
2
2
16.4
Mathematical Models
1. A weight of 4 pounds is pushed up 3 feet above the equilibrium position. At t = 0 it is given
an initial speed upward of 2 feet per second.
2. A mass of 2 pounds is pulled down 0.7 feet below the equilibrium position and held. At t = 0
it is released from rest.
1
1
3. Using m = W/g = 8/32 = 1/4, the initial value problem is x00 + x = 0; x(0) = , x0 (0) =
4
2
3
1
. The auxiliary equation is m2 + 1 = 0, so m = ±2i and x = C1 cos 2t + C2 sin 2t,
2
4
3
x0 = −2C1 sin 2t + 2C2 cos 2t. Using the initial condition, we obtain C1 = 1/2 and C2 = .
4
1
3
The equation of motion is x(t) = cos 2t + sin 2t.
2
4
304
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
4. From Hooke’s law we have 24 = k(1/3), so k = 72. Using m = W/g = 24/32 = 3/4, the initial
3
3
value problem is x00 +72x = 0; x(0) = −3, x0 (0) = 0. The auxiliary equation is m2 +72 =
√
√
√
√
√ 4
√
√ 4
0, so m = ±4 6i and x = C1 cos 4 6t+C2 sin 4 6t, x0 = −4 6C1 sin 4 6+4 6C2 cos 4 6t.
√
1
Using the initial conditions, we obtain C1 = −1/4 and C2 = 0. Thus, x(t) = − cos 4 6t.
4
5. From Hooke’s law we have 400 = k(2), so k = 200. The initial value problem is 50x00 + 200x =
0; x(0) = 0, x0 (0) = −1 = . The auxiliary equation is 50m2 + 200 = 0, so m = ±2i and
x = C1 cos 2x + C2 sin 2x, x0 = −2C1 sin 2x + 2C2 cos 2x. Using the initial conditions, we
obtain C1 = 0 and C2 = −5. Thus, x(1) = −5 sin 2x.
1 00
x + 4x = 0; x(0) =
6. Using m = W/g = 2/32 = 1/16, the initial value problem is
16
2
4
, x0 (0) = − . The auxiliary equation is m2 /16 + 4 = 0, so m = ±8i and x = C1 cos 8x +
3
3
C2 sin 8x, x0 = −8C1 sin 8x + 8C2 cos 8x. Using the initial conditions, we obtain C1 = 2/3
1
2
and C2 = −1/6. Thus, x(t) = cos 8t − sin 8t.
3
6
7. A 2 pound weight is released from the equilibrium position with an upward speed of 1.5 ft/s.
A damping force numerically equal to twice the instantaneous velocity acts on the system.
8. A 16 pound weight is released from 2 feet above the equilibrium position with a downward
speed of 1 ft/s. A damping force numerically equal to the instantaneous velocity acts on the
system.
1 00
x + x0 + 2x = 0; x(0) =
8
−1, x0 (0) = 8. The auxiliary equation is m2 /8 + m + 2 = 0 or (m + 4)2 = 0, so m = −4, −4
and x = C1 e−4t + C2 te−4t , x0 = (C2 − 4C1 )e−4t − 4C2 te−4t . Using the initial conditions, we
obtain C1 = −1 and C2 = 4. Thus, x(t) = −e−4t +4te−4t . Solving x(t) = −e4t +4te−4t = 0, we
see that the weight passes through the equilibrium position at t = 1/4s. To find the maximum
displacement we solve x0 (t) = 8e−4t − 16te−4t = 0. This gives t = 1/2. Since x(1/2) = e−2 ≈
0.14, the maximum displacement is approximately 0.14 feet below the equilibrium position at
t = 1/2s.
9. Using m = W/g = 4/32 = 1/8, the initial value problem is
10. From Hooke’s law we have 40(980) = k(10), so k = 3920. The initial value problem is 40x00 +
560x0 + 3920x = 0; x(0) = 0, x0 (0) − 2. The auxiliary equation is 40m2 + 560m + 3920 = 0
or m2 + 14m + 98 = 0, so m = −7 ± 7i and x = e−7t (C1 cos 7t + C2 sin 7t),
x0 = −7(C1 + C2 )e−7t sin 7t − 7(C1 − C2 )e−7t cos 7t. Using the initial conditions, we obtain
2
C1 = 0 and C2 = 2/7. Thus, x(t) = e−7t sin 7t.
7
11. From Hooke’s law we have 10 = k(7 − 5), so k = 5. Using m = W/g = 8/32 = 1/4, the
1
1
initial value problem is x00 + x0 + 5x = 0; x(0) = ; x0 (0) = 1. The auxiliary equation is
4
2
m2 /4+m+5 = 0 or m2 +4m+20 = 0, so m = −2±4i. Thus, x = e−2t (C1 cos 4t+C2 sin 4t) and
x0 = −2(C1 − 2C2 )e−2t cos 4t − 2(2C1 + C2 )e−2t sin 4t. Using the initial conditions, we obtain
1
1
1
= C1 and 1 = −2( − 2C2 ), so C1 = 1/2 and C2 = 1/2. Therefore x(t) = e−2t (cos 4t +
2
2
2
sin 4t).
16.4. MATHEMATICAL MODELS
305
12. From Hooke’s law we have 24 = k(4), so k = 6. Using m = W/g = 24/32 = 3/4, the initial
3
value problem is x00 + βx0 + 6x = 0; x(0) = 0, x0 (0) = −2. The auxiliary equation
4
p
3
is m2 + βm + 6 = 0. Using the quadratic formula, m = (−β ± β 2 − 18/(3/2). When
4
√
√
2
2p 2
2
2p 2
β > 18 = 3 2, we have m1 = − β +
β − 18 and m2 = − β −
β − 18. Thus,
3
3
3
3
√ 2
√ 2
x(t) = C1 e−2βt/3+2t β −18/3 + C2 e−2βt/3−2t β −18/3
p
p
2
2
−2βt/3
2
2
=e
C3 cosh
β − 18 t + C4 sinh
β − 18 t
3
3
see Example 5 in Section 16.2.
2p 2
β − 18t). The velocity is
3
2p 2
2p 2
2p 2
2β
x0 (t) =
C4 e−2βt/3 sinh(
β − 18C4 e−2βt/3 cosh(
β − 18t) −
β − 18t).
3
3
3
3
p
2p 2
From x0 (0) = −2 we obtain −2 =
β − 18C4 or C4 = −3/ β 2 − 18. Therefore,
3
2p 2
−3
e−2βt/3 sinh(
β − 18t).
x(t) = p
3
β 2 − 18
From x(0) = 0 we obtain C3 = 0 so that x(t) = C4 e−2βt/3 sinh(
13. From Hooke’s law we have 10 = k(2), so k = 5. Using m = W/g = 10/32 = 5/16, the
5 00
5 2
differential equation is
x + βx0 + 5 = 0. The auxiliary is
m + βm + 5 = 0 Using the
16
16
p
2
quadratic formula, m = (−β ± β − 25/4)/(5/8). For β > 0 the motion is
(a) overdamped when β 2 − 25/4 > 0 or β > 5/2
(b) critically damped when β 2 − 25/4 = 0 or β = 5/2
(c) underdamped when β 2 − 25/4 < 0 or β < 5/2.
14. Since W = mg = 1(32) = 32, we have from Hooke’s law 32 = k(2), so k = 16. The initial
value problem is x00 + 8x0 + 16x = 8 sin 4t; x(0) = x0 (0) = 0. The auxiliary equation is
m2 + 8m + 16 = (m + 4)2 = 0 so m = −4, −4, and xc = C1 e−4t + C2 te−4t . Using
xp = A sin 4t + B cos 4t we find A = 0 and B = −1/4. Thus,
x(t) = C1 e−4t + C2 te−4t −
1
cos 4t and x0 (t) = −4C1 e−4t − 4C2 te−4t + C2 e−4t + sin 4t.
4
1
Using the initial conditions, we obtain 0 = C1 − and 0 = −4C1 + C2 . Thus, C1 = 1/4 and
4
1
1
C2 = 4C1 = 1. Therefore x(t) = e−4t + te−4t − cos 4t.
4
4
www.elsolucionario.org
306
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
15. The initial value problem is x00 + 8x0 + 16x = e−t sin 4t; x(0) = x0 (0) = 0. Using xp =
Ae−t sin 4t + Be−t cos 4t we find A = −7/625 and B = −24/625. Thus,
x(t) = C1 e−4t + C2 te−4t −
x0 (t) = −4C1 e−4t − 4C2 te−4t + C2 e−4t −
7 −t
24 −t
e sin 4t −
e cos 4t,
625
625
28 −t
7 −t
96 −t
e cos 4t +
e sin 4t +
e sin 4t
625
625
625
24 −t
e cos 4t.
625
Using the initial conditions, we obtain C1 = 24/625 and C2 = 100/625. Thus,
+
x(t) =
24 −4t 100 −4t
7 −t
24 −t
e
te
e sin 4t −
e cos 4t.
+
−
625
625
625
626
As t −→ ∞, e−t −→ 0 and x(t) −→ 0.
16. A 32 pound weight is pulled 2 feet below the equilibrium position and held. At time t =
0 an external force equal to 5 sin 3t is applied to the system. The auxiliary equation is
m2 + 9 = 0, so m = ±3i and xc = C1 cos 3t + C2 sin 3t. Using variation of parameters
5
5
sin 3t, so
xp = − t cos 3t +
6
18
5
x(t) = C1 cos 3t + C3 sin 3t − t cos 3t
6
5
5
x0 (t) = −3C1 sin 3t + 3C2 cos 3t + t sin 3t − cos 3t.
2
6
Using the initial conditions, we obtain C1 = 2 and C2 = 5/18. Thus, x(t) = 2 cos 3t +
5
5
sin 3t − t cos 3t. The spring-mass system is in pure resonance.
18
6
17. The DE describing charge is .05q 00 + 2q 0 + 100q = 0. The auxiliary equation is 0.5m2 + 2m +
100 = 0, so m = −20 ± 40i. The general solution is q = e−20t (C1 cos 40t + C2 sin 40t). The
0
initial conditions yield q(0) = C1 = 5 and i(0)
= q (0) = −20C1+ 40C2 = 0, which gives
5
C1 = 5 and C2 = 25 . Therefore q(t) = e−20t 5 cos 40t + sin 40t , and q(0.01) = 4.568C.
2
q(t) = 0 when 5 cos 40t + 52 sin 40t = 0 which first occurs at t = 0.0509 s.
18. The DE describing charge is 41 q 00 +20q 0 +300q = 0. The auxiliary equation is 41 m2 +20m+300 =
0, so m = −20 or m = −60. The general solution is q(t) = C1 e−20t + C2 e−60t . The initial
conditions yield q(0) = C1 + C2 = 4 and i(0) = q 0 (0) = −20C1 − 60C2 = 0, which give C1 = 6
and C2 = −2. Therefore, q(t) = 6e−20t − 2e−60t . The charge is never equal to zero.
5 00
q + 10q 0 + 30q = 300. The auxiliary equation is 53 m2 + 10m + 30 = 0. so
3
m = −3 ± 3i. This gives qc = e−3t (C1 cos 3t + C2 sin 3t). Assume a particular solution of the
form qp = A. Substituting into the DE, we have 30A = 300 so that A = 10 and therefore
qp = 10. Thus, the general solution is q = qc + qp = e−3t (C1 cos 3t + C2 sin 3t) + 10. The initial
conditions yield q(0) = C1 +10 = 0 and i(0) = q 0 (0) = −3(C1 −C2 ) = 0, which gives C1 = −10
and C2 = −10. Therefore, q(t) = e−3t (−10 cos 2t − 10 sin 3t) + 10 − 10 − 10e−3t (cos 3t + sin 3t),
i(t) = q 0 (t) = 60e−3t sin 3t. The charge q(t) attains a maximum of 10.432 C at t = π3 .
19. The DE is
16.5. POWER SERIES SOLUTIONS
307
20. The DE is q 00 +100q 0 +2500q = 30. The auxiliary equation is m2 +100m+250000, so m = −50
is a repeated root. This gives qc = C1 e−50t + C2 te−50t . Assume a particular solution of the
3
form qp = A. Substituting into the DE, we have 2500A = 30 so that A = 250
and therfore
3
3
−50t
−50t
+ C2 te
+ 250 . The general solution
qp = 250 . The general solution is q = qc + qp = C1 e
3
is q = qc + qp = C1 e−50t + C2 te−50t + 250
. The initial conditions yield q(0) = C1 = 0 and
3
0
i(0) = q (0) = −50C! + C2 = 2 which give C1 = 0 and C2 = 2. Therefore, q(t) = 2te−50t + 250
1
0
−50t
and i(t) = q (t) = (2 − 100t)e
. The charge q(t) attains a maximum of 0.0267 C at t = 50
s.
21.
16.5
1.
∞
X
Power Series Solutions
n(n − 1)cn xn−2 +
cn xn =
n=0
n=2
|
∞
X
{z
k=n−2
∞
X
(k + 2)(k + 1)ck+2 xk +
k=0
∞
X
ck xk
k=0
}
=
∞
X
[(k + 2)(k + 1)ck+2 + ck ]xk = 0
k=0
ck
, k = 0, 1, 2, . . .
(k + 2)(k + 1)ck+2 + ck = 0; ck+2 = −
(k + 2)(k + 1)
c0
c0
c1
c1
c2
c0
c0
c2 = − = − , c 3 = −
= − , c4 = −
=
= ,
2
2!
3·2
3!
4 cos 3
4 · 3 · 2!
4!
c1
c1
c4
c0
c0
c3
=
= , c6 = −
=−
=− ,
c5 = −
5·4
5 · 4 · 3!
5!
6·5
6 · 5 · 4!
6!
c5
c1
c1
c7 = −
=−
=−
7 · 6 · 5!
7!
7 · 6
1 4
1 6
1 5
1 7
1 2
1 3
y = c0 1 − x + x − x + · · · + c1 x − x + x − x + · · ·
2!
4!
6!
3!
5!
7!
∞
∞
X
X
1
1
x2n + c1
(−1)n
x2n+1
= c0
(−1)n
(2n)!
(2n
+
1)!
n=0
n=0
2.
∞
X
n(n − 1)cn xn−2 −
n=2
|
∞
X
cn xn =
n=0
{z
k=n−2
∞
X
(k + 2)(k + 1)ck+2 xk −
k=0
∞
X
ck xk
k=0
}
=
∞
X
[(k + 2)(k + 1)ck+2 − ck ]xk = 0
k=0
c0
ck
(k + 2)(k + 1)ck+2 − ck = 0; ck+2 =
, k = 0, 1, 2, . . . ; c2 = ,
(k + 2)(k + 1)
2!
c1
c1
c2
c0
c3
c1
c4
c0
c5
c1
c3 =
= , c4 =
= , c5 =
= , c6 =
= , c7 =
=
3 ·2
3!
4·3
4!
6·5
6!
7!
5 · 4 5!
7 · 6
1
1
1
1
1
1
y = c0 1 + x2 + x4 + x6 + · · · + c1 x + x3 + x5 + x7 + · · ·
2!
4!
6!
3!
5!
7!
∞
∞
X 1
X
1
= c0
x2n + c1
x2n+1
(2n)!
(2n
+
1)!
n=0
n=0
308
3.
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
∞
X
n(n − 1)cn xn−2 −
ncn xn−1 =
n=1
n=2
|
∞
X
{z
k=n−2
|
}
∞
X
(k + 2)(k + 1)ck+2 xk −
k=0
{z
∞
X
(k + 1)ck+1 xk
k=0
}
k=n−1
=
∞
X
[(k + 2)(k + 1)ck+2 − (k + 1)ck+1 ]xk = 0
k=0
c1
c1
ck+1
, k = 0, 1, 2, . . . ; c2 =
= ,
(k + 2)(k + 1)ck+2 − (k + 1)ck+1 = 0; ck+2 =
(k + 2)
2
2!
c2
c1
c3
c1
c3 =
= , c4 =
= ,
3
3!
4
4!
P∞ 1
1 2
1 3
y = c0 + c1 x + x + x + · · · = c0 + c1 n=1 xn
2!
3!
n!
4.
∞
X
n(n − 1)cn x
n−2
ncn x
n−1
∞
X
=2
n=1
n=2
|
−
∞
X
{z
k=n−2
}
|
k
(k + 2)(k + 1)ck+2 x +
k=0
{z
∞
X
(k + 1)ck+1 xk
k=0
}
k=n−1
=
∞
X
[2(k + 2)(k + 1)ck+2 − (k + 1)ck+1 ]xk = 0
k=0
ck+1
2(k + 2)(k + 1)ck+2 + (k + 1)ck+1 = 0; ck+2 = −
, k = 0, 1, 2, . . . ;
2(k + 2)
c1
c1
c2
c1
c3
c1
c2 = −
=−
, c3 = −
= 2
, c4 = −
=− 3
,
2·2
2 · 2!
2·3
2 · 3!
2 · 4!
2 · 4
1
1
1 2
x + 2
x3 − 3
x4 + · · ·
y = c0 + c1 x −
2 · 2!
2 · 3!
2 · 4!
5.
∞
X
n(n − 1)cn xn−2 − x
n=2
|
∞
X
cn xn =
n=0
{z
k=n−2
}
|
{z
k=n+1
∞
X
(k + 2)(k + 1)ck+2 xk −
∞
X
ck−1 xk
k=1
k=1
}
= 2c2 +
∞
X
[(k + 2)(k + 1)ck+2 − ck−1 ]xk = 0
k=0
ck−1
c2 = 0; (k + 2)(k + 1)ck+2 − ck−1 = 0; ck+2 =
, k = 1, 2, 3, . . . ;
(k + 2)(k + 1)
c0
c2
c3
c0
c4
c1
c3 =
, c5 =
= 0, c6 =
=
, c7 =
=
3·2
5·4
6·5
6·5·3·2
7·6
7·6·4·3
c5
c6
c0
c7
c1
c9 =
= 0, c9 =
= c9 =
, c10 =
=
8 ·7
9·8
9·8·6·5·3·2
10 ·9
10 · 9 · 7 · 6 · 4 · 3
1 3
1
1
y = c0 1 +
x +
x6 +
x9 + · · ·
3·2
6·5·3·2
9·8·6·5·3·2
1 4
1
1
7
10
+ c1 x +
x +
x +
x + ···
4·3
7·6·4·3
10 · 9 · 7 · 6 · 4 · 3
www.elsolucionario.org
16.5. POWER SERIES SOLUTIONS
6.
∞
X
n(n − 1)cn xn−2 +x2
∞
X
cn xn =
n=0
n=2
{z
|
k=n−2
309
∞
X
(k + 2)(k + 1)ck+2 xk +
k=0
∞
X
ck−2 xk
k=2
| {z }
}
k=n+2
= 2c2 + +c3 x +
∞
X
[(k + 2)(k + 1)ck+2 + ck−2 ]xk = 0
k=2
ck−2
, k = 2, 3, 4, . . . ;
c2 = c3 = 0; (k + 2)(k + 1)ck+2 + ck−2 = 0; ck+2 = −
(k + 2)(k + 1)
c0
c1
c2
c3
c4
c0
c4 = −
, c5 = −
, c6 = −
= 0, c7 = −
, c8 = −
=
4·3
5·4
7·6
7·6
8·7
8·7·4·3
c5
c1
c6
c7
c9 = −
=
, c10 = −
= 0, c11 −
= 0,
9·8
9·8·5·4
10 · 9
11 · 10
c8
c0
c9
c1
c12 = −
=−
, c13 = −
=−
,
12
·
11
12
·
11
·
8
·
7
·
4
·
3
13
·
12
13
·
12
·
9 · 8 · 5 · 4
1
1
1
1 4
x +
x8 − · · · c1 x −
x5 +
x9 − · · ·
y = c0 1 −
4·3
8·7·4·3
5·4
9·8·5·4
7.
∞
X
n(n − 1)cn xn−2 −2x
n=2
∞
X
cn xn−1 +
n=1
|
{z
k=n−2
∞
X
cn x n
n=0
}
=
∞
X
k
(k + 2)(k + 1)ck+2 x −
k=0
∞
X
k=1
= c0 + 2c2 +
∞
X
k
kck x +
∞
X
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (2k − 1)ck ]xk = 0
k=1
c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 − (2k − 1)ck = 0; c2 = −
c0
2
c1
c1
3c2
3c0
(2k − 1)ck
, k = 1, 2, 3, . . . ; c3 =
= , c4 =
=−
,
(k + 2)(k + 1)
3·2
3!
4·3
4!
5c3
5c1
7c4
7 · 3c0
9c5
9 · 5c1
c5 =
=
, c6 =
=
, c7 =
=
5 · 4
5!
6·5
6!
7
·
6
7!
1 2
3 4 7·3 6
5
9·5 7
1 3
y = c0 1 − x − x −
x − · · · + c1 x + x + x 5 +
x + ···
2!
4!
6!
3!
5!
7!
ck+2 =
8.
∞
X
n(n − 1)cn xn−2 −x
n=2
|
∞
X
ncn xn−1 + 2
n=1
{z
k=n−2
∞
X
cn xn
n=0
}
=
∞
X
(k + 2)(k + 1)ck+2 xk −
k=0
= 2c0 + 2c2 +
∞
X
k=1
∞
X
kck xk + 2
∞
X
ck x k
k=0
[(k + 2)(k + 1)ck+2 − (k − 2)ck ]xk = 0
k=1
2c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 − (k − 2)ck = 0; c2 = −c0
(k − 2)ck
c1
c1
c3
c1
ck+2 =
, k = 1, 2, 3, . . . ; c3 = −
= − , c4 = 0, c5 =
=− ,
(k + 2)(k + 1)
3·2
3!
5·4
5!
310
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
3c5
3c1
5c7
5 · 3c1
c6 = c8 = c10 = 0, c7 =
=−
, c9 =
=−
7
·
6
7!
9
·
8
9! 1 5
3 7 5·3 9
1 3
2
x + ···
y = c0 (1 − x ) + c1 1 − x − x − x −
3!
5!
7!
9!
9.
∞
X
n(n − 1)cn xn−2 + x2
ncn xn−1 + x
n=1
n=2
|
∞
X
{z
k=n−2
}
|
=
∞
X
cn x n
n=0
{z
k=n+1
∞
X
}
|
{z
k=n+1
}
(k + 2)(k + 1)ck+2 xk +
k=0
∞
X
k=2
= 2c2 + (6c3 + c0 )x +
∞
X
∞
X
(k − 1)ck−1 xk +
ck−1 xk
k=1
[(k + 2)(k + 1)ck+2 + kck−1 ]xk = 0
k=2
c0
2c2 = 0, 6c3 + c0 = 0 (k + 2)(k + 1)ck+2 + kck−1 = 0; c2 = 0, c3 = −
3·2
kck−1
2c1
4c3
4c0
ck+2 = −
, k = 2, 3, 4, . . . ; c4 = −
, c5 = 0, c6 = −
=
(k + 2)(k + 1)
4·3
6·5
6·5·3·2
5c4
5 · 2c1
7c6
7 · 4c0
c7 = −
=
, c8 = c11 = c14 = · · · = 0, c9 = −
=−
7·6
7·6·4·3
9·8
9·8·6·5·3·2
8c7
8 · 5 · 2c1
c10 = −
=−
10 · 9 · 7 · 6 · 4 · 3
10 · 9
22 4 52 · 22 7 82 · 52 · 22 10
1 3 42 6 72 · 42 9
x · · · + c1 x − x +
x −
x + ···
y = c0 1 − x + x −
3!
6!
9!
4
7!
10!
10.
∞
X
n(n − 1)cn xn−2 +2x
n=2
|
∞
X
ncn xn−1 + 2
n=1
{z
k=n−2
∞
X
cn x n
n=0
}
=
∞
X
(k + 2)(k + 1)ck+2 xk + 2
k=0
= 2c2 + 2c0 +
∞
X
k=1
∞
X
kck xk + 2
∞
X
ck xk
k=0
[(k + 2)(k + 1)ck+2 + 2(k + 1)ck ]xk = 0
k=1
2c0 + 2c2 = 0; (k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0; c2 = −c0
2c1
2c2
2c0
2c3
22 c1
ck
, k = 1, 2, 3, . . . ; c3 = −
, c4 = −
=
, c5 = −
=
,
ck+2 = −
k+2
3
4
4
5
5·3
2
3
3
2c4
2 c0
2c5
2 c1
2c6
2 c0
c6 = −
=−
, c7 = −
=−
, c8 = −
=
6·4
7
7·5·3
8·6·4
6
8
2 4
22 6
23
2 3
22 5
23
2
8
7
y = c0 1 − x + x −
x +
x + · · · +c1 x − x +
x −
x + ···
4
6·4
8·6·4
3
5·3
7·5·3
16.5. POWER SERIES SOLUTIONS
11. (x − 1)
∞
X
n(n − 1)cn xn−2 +
n=2
∞
X
311
ncn xn−1
n=1
=
∞
X
n(n − 1)cn xn−1 −
∞
X
n(n − 1)cn xn−2 +
n=2
n=2
{z
|
=
∞
X
}
k=n−1
(k + 1)kck+1 xk −
k=1
ncn xn−1
n=2
|
{z
k=n−2
∞
X
∞
X
}
|
{z
k=n−1
∞
X
(k + 2)(k + 1)ck+2 xk +
(k + 1)ck+1 xk
k=0
= c1 − 2c2 +
∞
X
}
k=0
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 ]xk = 0
k=1
c1 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 = 0; c2 =
(k + 1)ck+1
2c2
c1
3c3
c1
, k = 1, 2, 3, . . . ; c3 =
= , c4 =
=
k+ 2
3
4
4
3
∞
X
1 n
1 2 1 3 1 4
x
y = c0 + c1 x + x + x + x + · · · = c0 + c1
2
3
4
n
n=1
c1
2
ck+2 =
12. (x + 2)
∞
X
n(n − 1)cn xn−2 + x
∞
X
ncn xn−1 −
=
∞
X
n(n − 1)cn xn−1 +2
=
∞
X
n(n − 1)cn xn−2 +
n=2
n=2
|
cn xn
n=0
∞
X
n=1
n=2
∞
X
{z
}
k=n−1
(k + 1)kck+1 xk + 2
k=1
= 4c2 − c0 +
|
∞
X
∞
X
n=1
{z
k=n−2
cn xn
n=0
∞
X
(k + 2)(k + 1)ck+2 xk +
kck xk −
k=1
∞
X
ck xk
k=0
[(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck ]xk = 0
k=1
4c2 − c0 = 0; (k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck = 0; c2 =
c0
4
(k + 1)kck+1 + (k − 1)ck
kck+1
(k − 1)ck
=−
−
, k = 1, 2, 3, . . .
2(k + 2)(k + 1)
2(k + 2) 2(k + 2)(k + 1)
c2
c0
2c3
c2
c0
c0
c3 = −
=−
, c4 = −
−
=
−
=0
2
2·3
2·3·4
2·4 2·4·3
2·3·4
2 · 3 · 42
c9
4c5
c0
2c3
=
, c6 = −
−0=−
c5 = 0 −
5 · 42 · 3 · 2
2·6
2·5·4
6 · 5 · 4 · 3 · 22
1
1
1
y = c0 1 + x2 −
x3 +
x5 − · · · + c1 x
2
4
4·3·2
5·4 ·3·2
ck+2 = −
∞
X
}
k=0
∞
X
ncn xn −
www.elsolucionario.org
312
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
13. (x2 − 1)
∞
X
n(n − 1)cn xn−2 + 4x
n=2
=
∞
X
∞
X
ncn xn−1 + 2
n=1
∞
X
n(n − 1)cn xn −
n=2
∞
X
cn xn
n=0
n(n − 1)cn xn−2 +4
k(k − 1)ck xk −
k=2
∞
X
∞
X
ncn xn + 2
n=1
n=2
{z
|
=
∞
X
cn xn
n=0
}
k=n−2
(k + 2)(k + 1)ck+2 xk + 4
k=0
∞
X
kck xk + 2
k=1
= (2c0 − 2c2 ) + (2c1 + 4c1 − 6c3 )x +
∞
X
∞
X
∞
X
ck xk
k=0
[k(k − 1)ck − (k + 2)(k + 1)ck+2 + 4kck + 2ck ]xk
k=2
=0
2c0 − 2c2 = 0; 6c1 − 6c3 = 0; (k + 2)(k + 1)ck − (k + 2)(k + 1)ck+2 = 0; c2 = c0 , c3 = c1 ;
ck+2 =ck , k = 2, 3, 4, . . .; c4 = c2 = c0 , c5 = c3 = c1 ,Pc6 = c4 = c0 , Pc7 = c5 = c1
∞
∞
y = c0 1 + x2 + x4 + · · · + c1 [x + x3 + x5 + · · · ] = c0 n=0 x2n + c1 n=0 x2n+1
14. (x2 + 1)
∞
X
n(n − 1)cn xn−2 − 6
∞
X
cn xn
n=0
n=2
=
∞
X
n(n − 1)cn xn +
n=2
∞
X
=
k=2
k
k(k − 1)ck x +
∞
X
∞
X
ncn xn
n=0
n=2
{z
|
∞
X
n(n − 1)cn xn−2 −6
k=n−2
}
k
(k + 2)(k + 1)ck+2 x − 6
k=0
= (2c2 − 6c0 ) + (6c3 − 6c1 )x +
∞
X
ck x k
k=0
∞
X
[k(k − 1)ck + (k + 2)(k + 1)ck+2 − 6ck ]xk = 0
k=2
2c2 − 6c0 = 0; 6c3 − 6c1 = 0; (k − 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0; c2 = 3c0 , c3 = c1 ;
−c2
c4
c0
(k − 3)ck
, k = 2, 3, 4, . . . ; c4 = −
= c0 , c5 = 0, c6 = − = −
ck+2 = −
k+1
3
5
5
3c6
3c0
5c8
5 · 3c0
c7 = c9 = c11 = · · · = 0, c8 = −
=
, c10 = −
=−
7 7·5
9
9·7·5
3 6
2
4
3
y = c0 1 + 3x + x −
x + · · · + c1 (x + x )
5·3
16.5. POWER SERIES SOLUTIONS
15. (x2 + 2)
∞
X
n(n − 1)cn xn−2 + 3x
n=2
313
∞
X
ncn xn−1 −
n=1
=
∞
X
n(n − 1)cn xn + 2
n=2
∞
X
cn x n
n=0
∞
X
n(n − 1)cn xn−2 +3
k(k − 1)ck xk + 2
{z
k=2
ncn xn −
∞
X
cn xn
n=0
}
k=n−2
∞
X
∞
X
n=1
n=2
|
=
∞
X
(k + 2)(k + 1)ck+2 xk + 3
k=0
∞
X
kck xk −
k=1
= (4c2 − c0 ) + (12c3 + 3c1 − c1 )x +
∞
X
∞
X
ck xk
k=0
[k(k − 1)ck + 2(k + 2)(k + 1)ck+2 + 3kck − ck ]xk
k=2
=0
c1
c0
, c3 = − ;
4
6
(k 2 + 2k − 1)ck
7c2
7
14c3
=−
, k = 2, 3, 4, . . . ; c4 = −
=−
c0 , c5 = −
=
2(k + 2)(k + 1)
2·4·3
4 · 4!
2·5·4
4c2 − c0 = 0; 12c3 + 2c1 = 0; 2(k + 2)(k + 1)ck+2 + (k 2 + 2k − 1)ck = 0; c2 =
ck+2
14
c1
2 · 5!
23c4
23 · 7
34c5
34 · 14
7 4 23 · 7 6
1
c6 = −
= 3
c0 , c7 = −
=−
c1 y = c0 1 + x2 −
x +
x − ··· +
2·6·5
2 · 6!
2 · 7 ·6
4 · 7!
4
4 · 4!
8 · 6!
14 5 34 · 14 7
1 3
c1 x − x +
x −
x + ···
6
2 · 5!
4 · 7!
16. (x2 − 1)
∞
X
n(n − 1)cn xn−2 + x
∞
X
ncn xn−1 −
=
∞
X
n(n − 1)cn xn −
∞
X
|
=
k=2
n(n − 1)cn xn−2 +
n=2
n=2
∞
X
cn xn
n=0
n=0
n=2
∞
X
k(k − 1)ck xk −
∞
X
∞
X
ncn xn −
}
k=n−2
(k + 2)(k + 1)ck+2 xk +
k=0
= −(2c2 + c0 ) + (c1 − 6c3 − c1 )x +
cn x n
n=0
n=1
{z
∞
X
∞
X
k=1
∞
X
kck xk −
∞
X
ck x k
k=0
[k(k − 1)ck − (k + 2)(k + 1)ck+2 + kck − ck ]xk
k=2
=0
c0
, c3 = 0;
2
(k − 1)ck
c2
c0
3c4
=
, k = 2, 3, 4, . . . ; c4 =
=−
, c5 = c7 = c9 = · · · = 0, c6 =
=
k+2
4
4·2
6
2c2 + c0 = 0; −6c3 = 0; (k + 1)(k − 1)ck − (k + 2)(k + 1)ck+2 = 0; c2 = −
ck+2
c0
−
4 · 22
y = c0
1 2 1 4
1 6
1 − x − x − x − · · · + c1 x
2
8
16
314
17.
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
∞
X
n(n − 1)cn xn−2 − (x + 1)
n=2
∞
X
ncn xn−1 −
n=1
=
∞
X
n(n − 1)cn xn−2 −
∞
X
ncn xn −
n=1
{z
|
∞
X
cn xn
n=0
n=2
=
∞
X
ncn xn−1 −
{z
|
(k + 2)(k + 1)ck+2 xk −
k=0
∞
X
= 2c2 − c1 − c0 +
cn x n
}
k=n−1
∞
X
kck xk −
k=1
∞
X
∞
X
n=0
n=1
}
k=n−2
∞
X
(k + 1)ck+1 xk −
k=0
∞
X
ck x k
k=0
[(k + 2)(k + 1)ck+2 − kck − (k + 1)ck+1 − ck ]xk = 0
k=1
c0 + c1
2
ck + ck+1
c1 + c2
c1 + c0 /2 + c1 /2
c0 + 3c1
ck+2 =
, k = 1, 2, 3, . . . ; c3 =
=
=
k+2
3
3
6
c2 + c3
c0 /2 + c1 /2 + c0 /6 + c1 /2
2c0 + 3c1
c4 =
=
=
4
4
12
c3 + c4
c0 /6 + c1 /2 + c0 /6 + c1 /4
4c0 + 9c1
c5 =
=
=
5
60
5
1 2 1 3 1 4
1 2 1 3 1 4
y = c0 1 + x + x + x + · · · + c1 x + x + x + x + · · ·
2
6
6
2
2
4
2c2 − c1 − c0 = 0; (k + 2)(k + 1)ck+2 − (k + 1)ck+1 − (k + 1)ck = 0; c2 =
18.
∞
X
n(n − 1)cn xn−2 − x
n=2
∞
X
ncn xn−1 − (x + 2)
n=1
=
∞
X
=
∞
X
n(n − 1)cn x
n−2
−
∞
X
n
ncn x −
n=1
{z
(k + 2)(k + 1)ck+2 xk −
k=0
|
∞
X
k=1
∞
X
∞
X
cn x
n+1
n=0
}
k=n−2
= 2c2 − 2c0 +
cn xn
n=0
n=2
|
∞
X
−2
∞
X
cn xn
n=0
{z
k=n+1
∞
X
kck xk −
}
ck−1 xk − 2
k=1
∞
X
ck x k
k=0
[(k + 2)(k + 1)ck+2 − kck − ck−1 − 2ck ]xk = 0
k=1
2c2 − 2c0 = 0; (k + 2)(k + 1)ck+2 − (k + 2)ck − ck−1 = 0; c2 = 0
ck−1
c1
c0
c0
c1
ck
++
, k = 1, 2, 3, . . . ; c3 =
+
=
+
ck+2 =
k+1
(k + 2)(k + 1)
2
3·2
3!
2
c2
c1
2c0
2c1
c3
c2
c1
c0
c0
11c0
3c1
c4 =
+
=
+
, c5 =
+
=
+
+
=
+
3
4·3
3!
4!
4
5·4
4·2
4!
5·4
5!
4!
c4
c3
2c0
2c1
c1
c0
52c0
4c1
c6 =
+
=
+
+
+
=
+
5 6 · 5
5 · 3!
5!
6 · 5 · 2 6 · 5 · 3!
6!
5!
1 3
2 4 11 5
1 3
2 4
3 5
2
y = c0 1 + x + x + x + x + · · · + c1 x + x + x + x + · · ·
3!
3!
5!
2
4!
4!
www.elsolucionario.org
16.5. POWER SERIES SOLUTIONS
19. (x − 1)
∞
X
n(n − 1)cn xn−2 − x
n=2
315
∞
X
ncn xn−1 +
n=1
=
∞
X
=
n(n − 1)cn xn−1 −
n(n − 1)cn xn−2 −
n=2
{z
}
k=n−1
∞
X
cn xn
n=0
∞
X
n=2
|
∞
X
k=1
{z
k=n−2
∞
X
= c0 − 2c2 +
∞
X
cn xn
n=0
}
(k + 2)(k + 1)ck+2 xk −
k=0
∞
X
ncn xn +
n=1
|
(k + 1)kck+1 xk −
∞
X
∞
X
kck xk +
k=1
∞
X
ck xk
k=0
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 − kck + ck ]xk = 0
k=1
c0 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 − (k − 1)ck = 0; c2 =
1
c0
2
2c2
c0
(k + 1)kck+1 − (k − 1)ck
, k = 1, 2, 3, . . . ; c3 =
=
(k + 2)(k + 1)
3·2
3·2
3 · 2c3 − c2
c0 − c0 /2
c0
c4 =
=
=
4
·
3
4
·
3
4
·
3·2
1 3
1
1
1 2
x +
x4 + · · · + c1 x; y 0 = c0 x + x2 + · · · + c1
y = c0 1 + x +
2
3·2
4·3·2
2
Using the initial conditions, we obtain −2 = y(0) = c0 and 6 = y 0 (0) = c1 . The solution is
ck+2 =
1 3
1
1
1 4
1
x +
x4 + · · · + 6x = −2 + 6x − x2 − x3 −
x + ···
y = −2 1 + x2 +
2
3·2
4·3·2
3
4·3
20. (x − 1)
∞
X
n(n − 1)cn x
n−2
− 2x
n=2
∞
X
ncn x
n−1
+8
n=1
=
∞
X
=
∞
X
cn x n
n=0
n(n − 1)cn xn−2 −2
n=2
|
∞
X
∞
X
ncn xn + 8
n=1
{z
(k + 2)(k + 1)ck+2 xk − 2
k=0
= 2c2 + 8c0 +
cn xn
n=0
}
k=n−2
∞
X
k=1
∞
X
∞
X
kck xk + 8
∞
X
ck xk
k=0
[(k + 2)(k + 1)ck+2 − 2kck + 8ck ]xk = 0
k=1
2c2 + 8c0 = 0; (k + 2)(k + 1)ck+2 − 2(k − 4)ck = 0; c2 = −4c0
2(k − 4)ck
−2 · 3c1
−2 · 2c2
4
ck+2 =
, k = 1, 2, 3, . . . ; c3 =
= −c1 , c4 =
= c0
(k + 2)(k + 1)
3·2
4·3
3
−2 · 1c3
1
2 · 0c4
c5 =
=
c1 , c 6 =
= 0, c8 = c10 = c12 = · · · = 0
5·2 5 · 4
6·5
4 4
1 5
2
3
y = c0 1 − 4x + x + c1 x − x + x + · · ·
3 10
16
1
0
3
2
4
y = c0 −8x + x + c1 1 − 3x + x + · · ·
3
2
316
CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
Using the initial conditions, we obtain 3 = y(0) = c0 and 0 = y 0 (0) = c1 . The solution is
4 4
2
y = 3 1 − 4x + x = 3 − 12x2 + 4x4
3
Chapter 16 in Review
A. True/False
1. True
2. True. We know a general solution is y = Aex + Be−x . Now
x
x
e + e−x
e − e−x
C1 cosh x + C2 sinh x = C1
+ C2
2
2
C1
C
C2
C
1
2
x
=
+
e +
−
e−x .
2
2
2
2
By varying C1 and C2 , we see that the two equations are different forms of the same general
solution.
3. False. y2 is a constant multiple of y1 . Specifically, y2 = 0 · y1 .
4. False. Plugging yp = A into the DE gives 0 = 10, a contradiction.
5. True. Any constant function solves the DE.
6. False. Py = 2x while Qx = −2x.
7. True
8. True
B. Fill in the Blanks
1. By inspection, the constant function y = 0 solves the DE.
2. The auxiliary equation is m2 − m = 0, so m = 0 or m = 1. The general solution is y =
C1 + C2 ex . Boundary conditions yield y(0) = C1 + C2 = 1 and y(1)
= C1 + C2 e = 0, which
e
−1
e
1
and C2 =
. Therefore, y =
−
ex .
give C1 =
e−1
e−1
e−1
e−1
3. 10 = k(2.5) =⇒ k = 4 lb/ft;
32 = 4x =⇒ x = 8 ft
4. We have a repeated root m = −7. Therefore, y = C1 e−7x + C2 xe−7x .
5. yp = Ax2 + Bx + C + Dxe2x + Ee2x
CHAPTER 16 IN REVIEW
317
C. Exercises
1. Py = −6xy 2 sin y 3 = Qx , and the equation is exact.
fx = 2x cos y 3 , f = x2 cos y 3 + g(y), fy = −3x2 y 2 sin y 3 + g 0 (y) = −1 − 3x2 y 2 sin y 3 ,
g 0 (y) = −1, g(y) = −y, f = x2 cos3 y − y.
Therefore, the solution is x2 cos y 3 − y = C.
2. Py = 6y 2 = Qx , and the equation is exact. fx = 3x2 + 2y 3 , f = x3 + 2xy 3 + g(y), fy =
6xy 2 + g 0 (y) = 6xy 2 + y 2 ,
y3
y3
g 0 (y) = y 2 , g(y) = , f = x3 + 2xy 3 + .
3
3
3
y
Therefore, the solution is x3 + 2xy 3 +
= C.
3
3. Py = −2xy −5 = Qx , and the equation is exact.
fx = 21 xy −4 , f = 14 x2 y −4 + g(y), fy = −x2 y −5 + g 0 (y) = 3y −3 − x2 y −5
g 0 (y) = 3y −3 , g(y) = − 32 y −2 , f = 14 x2 y −4 − 32 y −2 .
Therefore, the general solution is 14 x2 y −4 − 32 y −2 = C. Since y(1) = 1, we have 41 (1)(1)− 32 (1) =
C or C = − 54 . Thus, the solution is 41 x2 y −4 − 32 y −2 = − 54 .
4. Py = 2x + sin x = Qx and the equation is exact.
1
fx = y 2 + y sin x, f = xy 2 − y cos x + g(y), fy = 2xy − cos x + g 0 (y) = 2xy − cos x − 1+y
2,
1
, g(y) = tan−1 (y), f = xy 2 − y cos x + tan−1 (y). Therefore, the general
g 0 (y) =
1 + y2
solution is xy 2 − y cos x + tan−1 (y) = C. Since y(0) = 1, we have −1 + π4 = C. Thus, the
π
solution is xy 2 − y cos x + tan−1 (y) = − 1
4
√
√
√
5. m2 − 2m − 2 = 0 =⇒ m = 1 ± 3; y = C1 e(1− 3)x + C2 e(1+ 3)x
√
√
√
6. m2 − 8 = 0 =⇒ m = ±2 2; y = C1 e−2 2x + C2 e2 2x
7. m2 − 3m − 10 = 0 =⇒ (m − 5)(m + 2) = 0 =⇒ m = −2, 5; y = C1 e−2x + C2 e5x
8. 4m2 + 20m + 25 = 0 =⇒ (2m + 5)2 = 0 =⇒ m = −5/2, −5/2; y = C1 e−5x/2 + C2 xe−5x/2
1
x
x
9. 9m2 + 1 = 0 =⇒ m = ± i; y = C1 cos + C2 sin
3
3
3
10. 2m2 − 5m = 0 =⇒ m(2m − 5) = 0 =⇒ m = 0, 5/2; y = C1 + C2 e5x/2
11. Letting y = ux we have
dx
− eu du = 0
x
=⇒ ln |x| − eu = C1 =⇒ ln |x| − ey/x = C1 .
(x + uxeu )dx − xeu (udx + xdu) = 0 =⇒ dx−xeu du = 0 =⇒
Using y(1) = 0 we find C1 = −1. The solution of the initial-value problem is ln |x| = ey/x − 1.
12. The auxiliary equation is m = m2 + 4m + 4 = 0, so m = −2 is a repeated root. The
general solution is y = C1 e−2x + C2 xe−2x . Initial conditions yield y(0) = C1 = −2 and
y 0 (0) = −2C1 +C2 = 0 which give C1 = −2 and C2 = −4. The solution is y = −2e−2x −4xe−2x .
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CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
13. m2 − m − 12 = 0 =⇒ (m − 4)(m + 3) = 0 =⇒ m = −3, 4; yc = C1 e−3x + C2 e4x
yp = Axe2x + Be2x , yp0 = 2Axe2x + (A + 2B)e2x ; yp00 = 4Axe2x + 4(A + B)e2x
[4Axe2x +4(A + B)e2x ] − [2Axe2x + (A + 2B)e2x ] − 12[Axe2x + Be2x ]
= −10Axe2x + (3A − 10B)e2x = xe2x + ex
Solving −10A = 1, 3A − 10B = 1 we obtain A = −1/10 and B = −13/100. Thus,
y = C1 e−3x + C2 e4x −
1 2x
13 2x
xe −
e .
10
100
14. The auxiliary equation is m2 + 4 = 0, so m = ±2i. Therefore, yc = C1 cos 2x + C2 sin 2x.
Assume a particular solution of the form yp = Ax2 + Bx + C. Substituting into the DE, we
have
2A + Ax2 + Bx + C = 16x2 .
Equating coefficients, we get 2A+C = 0, B = 0, and A = 16. This gives C = −32. Therefore,
yp = 16x2 − 32. The general solution is y = yc + yp = C1 cos 2x + C2 sin 2x + 16x2 − 32.
15. m2 − 2m + 2 = 0 =⇒ m = 1 ± i; yc = ex (C1 cos x + C2 sin x)
ex cos x
ex sin x
= e2x
W =
x
x
x
−e sin x + e cos x e cos x + ex sin x
sin2 x
cos2 x − 1
1
=
= cos x−sec x, u = sin x−ln | sec x+tan x|
u0 = − 2x ex sin x ex tan x = −
e
cos x
cos x
1
v 0 = 2x ex cos x ex tan x = sin x, v = − cos x
e
yp = ex cos x(sin x − ln | sec x + tan x|) − ex sin x cos x = −ex cos x ln | sec x + tan x|
y = ex (C1 cos x + C2 sin x) − ex cos x ln | sec x + tan x|
16. m2 − 1 = 0 =⇒ m = −1, 1; yc = C1 e−x + C2 ex ; W =
e−x
−e−x
ex
ex
=2
1
2ex
e2x
e3x
u0 = − ex x
=− x
= − 2x
−x
−x
2Z e + e
e +e
e +1
e3x
x
x
u=−
dx t = e , dt = e dx
e2x + 1
Z Z
1
t2
dt
=
−
1
−
=−
dt = tan−1 t − t = tan−1 ex − ex
t2 + 1
t2 + 1
1
2ex
1
ex
v 0 = e−x x
=
=
e + e−x
ex + e−x
e2x + 1
Z2
ex
v=
dx t = ex , dt = ex dx
e2x + 1
Z
dt
= tan−1 t = tan−1 ex
=
t2 + 1
yp = e−x (tan−1 ex − ex ) + ex tan−1 ex = (ex + e−x ) tan−1 ex − 1
y = C1 e−x + C2 ex + (ex + e−x ) tan−1 ex − 1
17. m2 + 1 = 0 =⇒ m = ±i; yc = C1 cos x + C2 sin x; W =
cos x
− sin x
sin x
cos x
=1
1
u0 = − sin x sec3 x = − tan x sec2 x, u = − sec2 x; v 0 = cos x sec3 x = sec2 x, v = tan x
2
CHAPTER 16 IN REVIEW
319
1
1
sin2 x
1
yp = − cos x sec2 x + sin x tan x = sin x tan x − sec x =
−
2
2
cos x
2 cos x
2 sin2 x − 1
sin2 x − cos2 x
1
1
=
=
= sin x tan x − cos x
2 cos x
2 cos x
2
2
1
1
0
y = C3 cos x + C2 sin x + sin x tan x, y = −C3 sin x + C2 cos x + sin x sec2 x + sin x
2
2
Using the initial conditions, we obtain C3 = 1 and C2 = 1/2. Thus,
1
2 cos2 x
sin2 x
1
1
sin x + sin x tan x =
+
+ sin x
2
2
2 cos x
2 cos x 2
cos2 x + 1 1
1
=
+ sin x = (sin x + cos x + sec x).
2 cos x
2
2
y = cos x +
18. The auxiliary equation is m2 +2m+2 = 0, so m = −1±i. Therefore, yc = e−x (C1 cos x + C2 sin x) .
Assume a particular solution of the form yp = A. Substituting this into the DE, we have
2A = 1, or A = 21 . Therefore, the general solution is y = yc +yp = e−x (C1 cos x + C2 sin x)+ 21 .
The initial conditions yield y(0) = C1 + 12 = 0 and y 0 (0) = −C1 + C2 = 1 which give C1 = − 21
and C2 = 12 . Thus, the solution is y = e−x − 21 cos x + 12 sin x + 12 .
19.
∞
X
n(n − 1)cn xn−2 + x
cn xn =
n=0
n=2
|
∞
X
{z
k=n−2
}
|
{z
k=n+1
∞
X
(k + 2)(k + 1)ck+2 xk +
k=0
∞
X
ck−1 xk
k=1
}
= 2c2 +
∞
X
[(k + 2)(k + 1)ck+2 + ck−1 ]xk = 0
k=1
ck−1
, k = 1, 2, 3, . . .
c2 = 0; (k + 2)(k + 1)ck+2 + ck−1 = 0; ck+2 = −
(k + 2)(k + 1)
c0
c1
c3
c0
c3 = −
, c4 = −
, c5 = 0, c6 = −
=
,
3·2
4·3
6·5
6·5·3·2
c4
c1
c7 = −
=
7·6
7·6·4·3
c6
c0
c7
c1
c8 = 0. c9 = −
=−
, c10 = −
=−
9·8
9·8·6·5·3·2
10 · 9
10
·9·7·6·4·3
1 3
1
1
y = c0 1 −
x +
x6 −
x9 + · · ·
3·2
6·5·3·2
9·8·6·5·3·2
1 4
1
1
7
10
+ c1 x −
x +
x −
x + ···
4·3
7·6·4·3
10 · 9 · 7 · 6 · 4 · 3
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CHAPTER 16. HIGHER-ORDER DIFFERENTIAL EQUATIONS
20. (x − 1)
∞
X
n(n − 1)cn xn−2 + 3
n=2
∞
X
cn xn
n=1
=
∞
X
(n)(n − 1)cn xn−1 −
=
∞
X
n(n − 1)cn xn−2 +3
n=2
n=2
|
∞
X
{z
}
k=n−1
(k + 1)kck+1 xk −
k=1
= 3c0 − 2c2 +
|
∞
X
cn xn
n=0
{z
k=n−2
}
(k + 2)(k + 1)ck+2 xk + 3
k=0
∞
X
∞
X
∞
X
ck x k
k=0
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck ]xk = 0
k=1
3c0
;
3c0 − 2c2 = 0; (k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck = 0; c2 =
2
kck+1
3ck
c2 3c1
c0 c1
2c3 3c2
ck+2 =
+
, k = 1, 2, 3, . . . ; c3 = +
= + , c4 =
+
=
k + 2 (k + 2)(k + 1)
3 3·2
2 2
4
4·3
c0
c1
3c0
5c0
c1
+
+
=
+ ,
4
4
8
8
4
3c4
3c3
3c0
3c1
3c0
3c1
9c0
9c1
c5 =
+
=
+
+
+
=
+
5
5·4
8
20
40 40 20
40
1 3 1 4
3 3 1 3 5 4
y = c0 1 + x + x + x + · · · + c1 x + x + x + · · ·
2
2
8
2
4
21. The differential equation is mx00 + 4x0 + 2x = 0. The solutions of the auxiliary equation are
√
√
1
1
(−4 ± 16 − 8m) = (−2 ± 4 − 2m).
2m
m
The motion will be non-oscillatory when 4 − 2m ≥ 0 or 0 < m ≤ 2.
22. Substituting xp = αA into the differential equation we obtain ω 2 αA = A, so α = 1/ω 2 and
xp = A/ω 2 .
1 00
x + x0 + 3x = e−t ; x(0) = 2,
8
x0 (0) =√0. The auxiliary equation is√m2 /8 + m + √
3 = 0. Using the quadratic formula, m =
−4 ± 2 2i. Thus, xc = e−4t (C1 cos 2 2t + C2 sin 2 2t). Using xp = Ae−t , we find A = 8/17.
Thus,
√
√
8
x(t) = e−4t (C1 cos 2 2t + C2 sin 2 2t) + e−t
17
23. Using m = W/g = 4/32 = 1/8, the inital value problem is
√
√
√
√
8
and x0 (t) = e−4t [(2 2C2 − 4C1 ) cos 2 2t − (2 2C1 − 4C2 ) sin 2 2t] − e−t .
17
√
Using the initial conditions,
√ we obtain 2 = C1 + 8/17 and 0 = 2 2C2 − 4C1 − 8/17. Then
C1 = 26/17 and C2 = 28 2/17 and
√
√
28 √
8
26
cos 2 2t +
2 sin 2 2t + e−t .
x(t) = e−4t
17
17
17
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CHAPTER 16 IN REVIEW
321
24. (a) From k1 = 2W and k2 = 4W we find 1/k = 1/2W + 1/4W = 3/4W. Then k = 4W/3 =
4mg/3. The differential equation
mx00 + kxp= 0 then becomes x00 + (4g/3)x = 0. The
p
sin 2 g/3t. The initial conditions x(0) = 1 and
solution is x(t) = C1 cos 2 g/3t + C2 √
x0 (0) = 2/3 imply C1 = 1 and C2 = 1/ 3g.
(b) To find the maximum speed of the weight we compute
r
r
r
r
g 4
g
g 2
g
2p
0
3g + 1.
x (t) = 2
sin 2
+ cos 2
t and |x (t)| = 4 + =
3
3 3
3
3 9
3
0
25. The auxiliary equation is m2 /4 + m + 1 = 0 or (m + 2)2 = 0, so m = −2, −2 and x(t) =
C1 e−2t + C2 te−2t and x0 (t) = −2C1 e−2t − 2C2 te−2t + C2 e−2t . Using the initial conditions, we
obtain 4 = C1 and 2 = −2C1 +C2 . Thus, C1 = 4 and C2 = 10. Therefore x(t) = 4e−2t +10te−2t
and x0 (t) = 2e−2t − 20te−2t . Setting x0 (t) = 0 we obtain the critical point t = 1/10. The
maximum vertical displacement is x(1/10) = 5e−0.2 ≈ 4.0937.