Solucionario de Cálculo de Varias Variables 4ta Edición, Ejercicios de Cálculo para Ingenierios

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Multivariable Calculus

Complete Solutions Manual

Brian Fulton Melanie Fulton

Fourth Edition

Chapter 10

Conics and Polar Coordinates

10.1 Conic Sections

y

x

1. vertex: (0, 0) focus: (1, 0) directrix: x = − 1 axis: y = 0

y

x

2. vertex: (0, 0) focus: (7/ 8 , 0) directrix: x = − (^78) axis: y = 0

y

x

3. vertex: (0, 0) focus: (0, −4) directrix: y = 4 axis: x = 0

668

670 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

8. vertex: (2, 0) focus:

directrix: y = (^14) axis: x = 2

y x

9. (y + 6)^2 = 4(x + 5) vertex: (− 5 , −6) focus: (− 4 , −6) directrix: x = − 6 axis: y = − 6

y x

10. (x + 3)^2 = −(y + 2) vertex: (− 3 , −2) focus: (− 3 , − 9 /4) directrix: y = − 7 / 4 axis: x = − 3

y

x

x + (^52)

= 14 (y − 1) vertex: (− 5 / 2 , −1) focus: (− 5 / 2 , − 15 /16) directrix: y = − 17 / 16 axis: x = − − 5 / 2

10.1. CONIC SECTIONS 671

y

x

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 + (^14)

vertex: (− 1 / 4 , 2) focus: (3/ 4 , 2) directrix: x = − 5 / 4 axis: y = 2

15. x^2 = 28
16. y^2 = − 16 x
17. y^2 = 10x
18. x^2 = − 40 y
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.

10.1. CONIC SECTIONS 673

y

x

x^2 2

y^2 4 = 1 center: (0, 0) foci: (0, ±

vertices: (0, ±2) endpoints of the minor axis: (±

eccentricity:

y

x

29. center: (1, 3) foci: (1 ±

vertices: (− 6 , 3), (8, 3) endpoints of the minor axis: (1, −3), (1, 9)

eccentricity:

y

x

30. center: (− 1 , 2) foci: (− 1 , 2 ±

vertices: (− 1 , −4), (− 1 , 8) endpoints of the minor axis: (− 6 , 2), (4, 2)

eccentricity:

674 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

31. center: (− 5 , −2) foci: (− 5 , − 2 ±

vertices: (− 5 , −6), (− 5 , 2) endpoints of the minor axis: (− 6 , −2), (− 4 , −2)

eccentricity:

y

x

32. center: (3, −4) foci: (3, − 4 ±

vertices: (3, −13), (3, 5) endpoints of the minor axis: (− 5 , −4), (11, −4)

eccentricity:

y

x

33. x^2 +

y + (^12)

center: (0, − 1 /2) foci: (0, − 1 / 2 ±

vertices: (0, − 5 /2), (0, 3 /2) endpoints of the minor axis: (− 1 , − 1 /2), (1, − 1 /2)

eccentricity:

676 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

x^2 9

(y + 3)^2 3

center: (0, −3) foci: (±

vertices: (± 3 , −3) endpoints of the minor axis: (0, − 3 ±

eccentricity:

y

x

38. (x − 1)^2 +

(y − 1 /2)^2 3

center: (1, 1 /2) foci: (1, 1 / 2 ±

vertices: (1, 1 / 2 ±

endpoints of the minor axis: (0, 1 /2), (2, 1 /2)

eccentricity:

39. The center is (0, 0) with the x-axis as the major axis. a = 5 and c = 3, so b = 4. Thus the

equation is

x^2 25

y^2 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

equation is

x^2 81

y^2 77

41. The center is (1, −3) with the x-axis as the major axis. a = 4 and b = 2. Thus the equation is

(x − 1)^2 16

(y + 3)^2 4

42. The center is (1, −2) with the y-axis as the major axis. a = 4 and b = 3. Thus the equation

is

(x − 1)^2 9

(y + 2)^2 16

43. The center is (0, 0) with the x-axis as the major axis. c =

2 and b = 3, so a =

11. Thus

the equation is

x^2 11

y^2 9

10.1. CONIC SECTIONS 677

44. The center is (0, 0) with the y-axis as the major axis. c =

5 and a = 8, so b =

59. Thus the equation is x^2 59

y^2 64

45. The center is (0, 0) with the y-axis as the major axis. c = 3 thus 9 = a^2 − b^2 and a =

9 + b^2.

Thus the equation is of the form

x^2 b^2

y^2 9 + b^2 = 1. The ellipse passes through the point

(− 1 , 2

2), thus

(−1)^2

b^2

2)^2

9 + b^2

= 1. Solving this for b, we obtain b =

3. Thus a^2 = 12 and

the equation is x^2 3

y^2 12

46. The center is (0, 0) with the x-axis as the major axis and a = 5. The equation is of the form x^2 25

y^2 b^2

= 1. The ellipse passes through the point (

5 , 4) so

b^2

= 1. Solving for b^2 , we

obtain b^2 = 20. Thus the equation of the ellipse is

x^2 25

y^2 20

47. The y-axis as the major axis with c = 3 and a = 4. Thus b =

7 and the equation of the ellipse is

(x − 1)^2 7

(y − 3)^2 16

48. The center is (15/ 2 , 4) with the x-axis as the major axis. a = 1/2 and c = 7/ 2 , thus b =

Thus the equation is (x − 15 /2)^2 (11/2)^2

(y − 4)^2 18

y

x

49. center: (0, 0) foci: (±

vertices: (± 4 , 0) asymptotes: y = ± 54 x

eccentricity:

y

x

50. center: (0, 0) foci: (±

vertices: (± 2 , 0) asymptotes: y = ±x eccentricity:

10.1. CONIC SECTIONS 679

y

x

54. center: (− 2 , −4) foci: (− 2 ±

vertices: (− 2 ±

asymptotes: y = − 4 ± 5 x + 10 √ 10 eccentricity:

y

x

55. center: (0, 4) foci: (0, 4 ±

vertices: (0, −2), (0, 10) asymptotes: y = 4 ± 6 x

eccentricity:

y

x

56. center: (− 3 , 1 /4) foci: (− 3 , 1 / 4 ±

vertices: (− 3 , 9 /4), (− 3 , − 7 /4) asymptotes: y =

± 23 (x + 3)

eccentricity:

680 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

(x − 3)^2 5

(y − 1)^2 25

= 1 center: (3, 1) foci: (3 ±

vertices: (3 ±

asymptotes: y = 1 ±

5(x − 3) eccentricity:

y

x

(x + 1)^2 10

(y − 1 /2)^2 50

= 1 center: (− 1 , 1 /2) foci: (− 1 ±

vertices: (− 1 ±

asymptotes: y = 1/ 2 ±

5(x + 1) eccentricity:

y

x

(x − 2)^2 6

(y − 1)^2 5

= 1 center: (2, 1) foci: (2 ±

vertices: (2 ±

asymptotes: y = 1 ±

(x − 2)

eccentricity:

682 CHAPTER 10. CONICS AND POLAR COORDINATES

The equation is

y^2 9 / 4

x^2 27 / 4

65. The center is (1, −3) with the y-axis as the transverse axis. c = 3 and a = 2, thus b =

The equation is

(y + 3)^2 4

(x − 1)^2 5

66. The center is (2, 2) with the y-axis as the transverse axis. c = 5 and a = 32, thus b = 4. The

equation is

(y − 2)^2 9

(x − 2)^2 16

67. The center is (− 1 , 3) with the y-axis as the transverse axis. a = 1 and the equation is of the

form

(y − 3)^2 1

(x + 1)^2 b^2 = 1. The hyperbola passes through the point (− 5 , 3 +

5. thus

(3 +

5 − 3)^2 −

(−5 + 1)^2

b^2

= 1. Thus b^2 = 4 and the equation is (y − 3)^2 −

(x + 1)^2 4

68. The center is (3, −5) with the y-axis as the transverse axis. a = 3 and the equation is of the form

(y − 3)^2 9

(x + 5)^2 b^2

= 1. The hyperbola passes through the point (1, −1) thus (−4)^2 9

(6)^2

b^2

= 1. Thus b^2 =

and the equation is

(y − 3)^2 9

(x + 5)^2 324 / 7

69. The center is (2, 4) with the y-axis as the transverse axis and a = 1. After solving the asymptote given in the problem for y, we obtain y = x + 6 2

x 2

• 3. The equation of the

hyperbola is of the form (y − 4)^2 −

(x − 2)^2 b^2

= 1. The asymptote equations for this hyperbola

are y − 4 = x − 2 b

and y − 4 = −x + 2 b

(these are also equivalent to y = x b

b

and

y = − x b

b

). Letting b equal 2 or -2 will yield one asymptote with the equation

y =

x 2

• 3. In either case, the equation of the hyperbola is (y − 4)^2 −

(x − 2)^2 4

70. The y-axis is the conjuate axis. The center is (− 5 , 7) with b = 3 and

c a

10. Thus c =

10 a.

Since c^2 = b^2 +a^2 then 10a^2 = 9+a^2 and a^2 = 1. Thus the equation is (x+5)^2 − (y + 7)^2 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 x^2 = 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 x^2 = 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 x^2 = 4py and contains the point (20, 1). Thus the equation of the parabola is x^2 = 400y. The towers are located at x = 175 and x = − 175. Hence the height of the towers

10.1. CONIC SECTIONS 683

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 x^2 = 4py and contains the point (125, 75). Thus the equation of the parabola is x^2 = 62512. 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 75^2 = 6253 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 x^2 = 4py and the point (4, −2) lies on the parabola. Therefore the equation is x^2 = − 8 y. The water hits the ground at y = − 20. The point on the parabola with y-value -20 is found by solving x^2 = −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 , 0). The parabola is of the form x^2 = 4p(y−5) and contains the point (

Therefore the equation of the parabola is x^2 = − 200 y. 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 10^2 = − 200 y which yields y = − 1 / 2 f 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.

y

x

3.6x10^7 a

b

c a-c

3.52x10^7

77. Taking the center of the ellipse to be at the origin, we have a = 3. 6 × 107 and b = 3. 52 × 107. Since c^2 = a^2 − b^2 , c^2 = 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.
78. Using a = 3. 6 × 107 and c = 0. 75 × 107 , we compute the eccentricity e =

0. 75 × 107

3. 6 × 107

79. From a = 1. 67 × 109 and 4. 25 × 108 we obtain

c^2 = a^2 − b^2 = 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

c a

1. 615 × 109

1. 67 × 109

10.2. PARAMETRIC EQUATIONS 685

87. Since a = 4 and b =

20, we have c^2 = a^2 + b^2 = 16 + 20 = 36 and hence c = 6. Thus the foci occur at F 1 = (− 6 , 0) and F 2 = (6, 0). The line joining (− 6 , −5) and F 2 is given by y =

x −

. The ray of light travels southwest along this line.

88. (a) The distance from (0, b) to (a, 0) is

a^2 + b^2. Thus R =

a^2 + b^2 = r. (b) From A = a + r and B = R − b =

a^2 + b^2 + r − b =

a^2 + b^2 + (A − a) − b = A − (a + b) +

a^2 + b^2 we see that

A − B = a + b −

a^2 + b^2 =

(a + b)^2 −

a^2 + b^2 =

a^2 + 2ab + b^2 −

a^2 + b^2 > 0.

Thus, A > B.

10.2 Parametric Equations

t -3 -2 -1 0 1 2 3 x -5 -3 -1 1 3 5 7 y 6 2 0 0 2 6 12

t 0 π 6 π 4 π 3 π 2 56 π^74 π x 1

√ 3 2

√ 2 2

1 2 0 -^

√ 3 2

√ 2 2 y 0 1/4 1/2 3/4 1 1/4 1/

y

x

y

x

y

x

y

x

y

x

y

x

686 CHAPTER 10. CONICS AND POLAR COORDINATES

y

x

y

x

11. y = (t^2 )^2 + 3t^2 − 1 = x^2 + 3x − 1; y = x^2 + 3x − 1 , x ≥ 0
12. − 12 y = t^3 + t; x = − 12 y + 4; 2 x + y = 4
13. x = cos 2t = cos^2 t − sin^2 t = 1 − 2 sin^2 t = 1 − 2 y^2 ; y = 1 − 2 y^2 , − 1 ≤ y ≤ 1
14. ln x = t; y = ln(ln x), x > 1. Alternatively, ey^ = t; x = ee y , x > 1
15. t = x^1 /^3 ; y = 3 ln x^1 /^3 ; y = ln x, x > 0
16. x^2 tan^2 x, y^2 = sec^2 t; x^2 + 1 = tan^2 t + 1 = sec^2 t = y^2 ; y^2 − x^2 = 1. y ≥ 1

y

x

y

x

y = x x = sin t y = sin t