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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
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.
672
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 xintercept set y = 0. Solving 42 = 4(x + 1) gives x = 3. The xintercept is (3, 0). To find the yintercept set x = 0. Solving (y + 4)2 = 4 gives y = −4 ± 2. The yintercepts are (0, −2) and (0, −6).
√ 2 24. To find √set y = 0. Solving (x − 1) = 2 gives x = 1 ± 2. The xintercepts are √ the xintercept (1 + 2, 0) and (1 − 2, 0). To find the yintercept set x = 0. Solving 1 = −2(y − 1) gives y = 12 . The yintercept 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
674
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
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 xaxis 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 xaxis 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 xaxis 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 yaxis 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 xaxis 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 yaxis 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 yaxis 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 xaxis 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 yaxis 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 xaxis 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
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
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 yaxis 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 yaxis as the transverse axis. c = 3 and a = 3/2, thus b =
x
√
12.
√ 3 3 . 2
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 yaxis 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 yaxis 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 yaxis 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 yaxis 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 yaxis 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 yaxis 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 yvalue of the point on the parabola when we are 50 ft from the tower or when x = 75f t. Hence this yvalue 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 yvalue 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 xaxis along the ground and the yaxis 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 yvalue of the point on the parabola corresponding to the xvalue 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
ac
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
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 xdirection. 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
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
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
690
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 .
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 ).
694
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
75 tan 33◦ ≈ 3.044, and therefore x ≈ 228.3 ft, y ≈ −148.25 ft. 16
Calculus and Parametric Equations
dx 1. = 3t2 − 2t; dt 2.
dx 4 = − 2; dt t
3.
dx t =√ ; 2 dt t +1
4.
5.
6.
7.
16 y = t. x 15
dy = 2t + 5; dt
dy = 6t2 − 1; dt dy = 4t3 ; dt
dy 3 = dx t=−1 5 dy 6t2 − 1 dy 6t4 − t2 92 = ; = −23 = − =− dx −4/t2 4 dx t=2 4 p √ dy 4t3 dy 2 2 = √ = 4(3) 4 = 24 = 4t t + 1; √ 2 dx dx t/ t + 1 t= 3 dy 2t + 5 = 2 ; dx 3t − 2t
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
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
3
t
0
p
p
25t4
t2 + 1dt
+
144t4 dt
= 13
Z
0
2
2 13 3 104 t dt = t = 3 3 0 2
u = t2 + 1, du = 2tdt
4 1 1/2 1 1 3/2 7 u du = u = (8 − 1) = 2 3 3 3 1
4
1
27. x0 (t)Z= et cos t + et sin t; π
s=
0
=
√
2
Z
π
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
√
π √ 2et = 2(eπ − 1) 0
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 = 3b 0
= 3b
Z
0
0
π/2
π/2
sin θ cos θ
π/2 1 3 3 3 sin 2θdθ = − b cos 2θ = − b(−1 − 1) = b 2 4 4 2 0
30. x0 (θ) = −4a sin θ + 4a sin 4θ;
y 0 (θ) = 4a cos θ − 4a cos 4θ
p
cos2 θ + sin2 θdθ
698
CHAPTER 10. CONICS AND POLAR COORDINATES s=
Z
2π/3
0
= 4a
Z
[(−4a sin θ + 4a sin 4θ)2 + (4a cos θ − 4a cos 4θ)2 ]1/2 dθ 2π/3
0
= 4a
Z
2π/3
0
Z √ = 4 2a
(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 2a
[1 − (cos 4θ cos θ + sin 4θ sin θ)]1/2 dθ
2π/3
0
Z √ = 4 2a
2π/3
0
p q
Z √ 1 − cos(4θ − θ)dθ = 4 2a 2 sin2 3θ/2dθ = 8a
Z
2π/3
√
0
2π/3
sin
0
1 − cos 3θdθ
3θ dθ 2
2π/3 3θ 16 32 2 = − a(cos π − cos 0) = a = 8a − cos 3 2 0 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 π 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 0 3 = 2a2 π = 3(πa2 ). 2
Thus, the area under an arch of the cycloid is three times the area of the circle.
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
700
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 3140, 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
702
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 4152, 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 yintercept 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 yintercept is thus , in polar coordinates. The xintercept 4 4 2
704
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 xintercept 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
706
CHAPTER 10. CONICS AND POLAR COORDINATES 23.
22.
24.
polar axis
polar axis
polar axis
rose curve
circle
circle with center on xaxis
25.
26.
27.
circle with center on yaxis 28.
polar axis
polar axis
polar axis
circle with center on yaxis
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
708
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)
710
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 xaxis. 57.
58. (r,θ)
θ
(−r, −θ)
(r,θ) θ
(−r,θ+2π)
Symmetric with respect to the origin.
Symmetric with respect to the yaxis .
10.6. CALCULUS IN POLAR COORDINATES
711
59. Symmetric with respect to the xaxis. 60. The graph is symmetric with respect to the yaxis. 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
712
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
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 17. A = 4 sin2 θdθ = θ − sin 2θ = π 2 0 2 0
3π 2
3π 2 .
1
18. A =
1 2
1 19. A = 2
Z
π
100 cos2 θdθ =
0
Z
0
2π
25θ +
(4 + 4 cos θ) dθ = 8 2
Z
π 25 sin 2θ = 25π 2 0
2π
polar axis
polar axis
(1 + 2 cos θ + cos2 θ)dθ
0 2π
= (8θ + 16 sin θ + 4θ + 2 sin 2θ)0 = 24π polar axis
716
CHAPTER 10. CONICS AND POLAR COORDINATES
20. A = =
21. A = =
22. A = =
23. A =
1 2
Z
1 2
Z
1 2
1 2
9 = 4
24. A = =
1 2 1 6
1 = 6
2π
0
(1 − sin θ)2 dθ =
1 2
Z
2π
0
(1 − 2 sin θ + sin2 θ)dθ
2π 1 3 1 1 θ + cos θ + θ − sin 2θ = π 2 4 8 2 0 2π
(3 + 2 sin θ)2 dθ =
0
1 2
Z
2π
(9 + 12 sin θ + 4 sin2 θ)dθ
0
2π 9 1 θ − 6 cos θ + θ − sin 2θ = 11π 2 2 0
Z
2π
(2 + cos θ)2 dθ =
0
1 2
Z
2π
2π
0
Z
4π
sin udu = 2
0
Z
π
cos2 3θdθ
polar axis
u = 2θ du = 2dθ
9 sin2 2θdθ
0
polar axis
(4 + 4 cos θ + cos2 θ)dθ
2π 9 1 1 2θ + 2 sin θ + θ + sin 2θ = π 4 8 2 0 Z
polar axis
4π 9 9 9 u− sin 2u = π 8 16 2 0
polar axis
u = 3θ, du = 3dθ
0
Z
3π
cos2 udu
0
3π 1 1 π u + sin 2u = 2 4 4 0
polar axis
10.6. CALCULUS IN POLAR COORDINATES 1 25. A = 2
Z
3π/2
717
3π/2 29 3 2 3 4θ dθ = θ = π 3 0 4 2
0
polar axis
26. A =
1 2
Z
π
π/2
π θ
dθ = −
π π π π 2 =− −π = 2θ π/2 2 2
polar axis
27. A =
1 2
1 28. A = 2
Z
π
e2θ dθ =
0
Z
2
π 1 1 2θ 1 e = e2θ − 4 4 4 0
100e−2θ dθ
1
2 = −25e−2θ 1 = −25(e−4 − e−2 ) = 25e−2 − 25e−4
29. A = =
30. A =
1 2
Z
1 2
Z
=−
polar axis
π/4
0
tan2 θdθ =
1 2
Z
0
π/4
(sec2 θ − 1)dθ
π/4 1 π 1 1 4−π = tan θ − θ − −0= 2 2 2 8 8 0
π/6
25 2
π/3 25 cot θ 2 π/6 ! √ √ 3 √ 25 3 − 3 = 3 3
π/3
polar axis
polar axis
25 csc2 θdθ = −
polar axis
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 4π/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 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
=
√
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/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
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
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=
√
4 √ √ 5 R 4 θ/2 e dθ = 5eθ/2 0 = 4(e2 − 1) 0 2
2 dr dr −θ 42. = −2e ; + r2 = 4e−2θ + 4e−2θ = 8e−2θ ; dθ √ dθ π √ √ √ Rπ 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
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
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 halfpetal =
1 2
1 = 4
Z
1
π/8
sin2 2θdθ
polar axis
u = 2θ, du = 2dθ
0
Z
π/4
sin2 udu
0
π/4 1 1 1 u − sin 2u 4 2 4 0 π−2 = 32 π−2 π−2 Total area = 16 · = 32 2 =
48. Each of the areas will equal 1 + cos θ.
polar axis
3π 2
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
10.7. CONIC SECTIONS IN POLAR COORDINATES
polar axis
723
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 π 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 3 π 2π 3 = − + + − 6 4 12 3 4
polar axis
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
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
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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 xaxis. 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 =
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
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 xaxis. 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 xaxis.
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
731
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 yintercepts 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 yaxis 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.
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 xintercepts are 7 and 5. √ 13. Setting x = 0 and solving, we have y 2 − 4 = 1, y 2 = 5, or y = ± 5. The yintercepts 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 yintercepts 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 xaxis. 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 xaxis. 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
=
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 2π 19π 9 1 1 = θ + 3 sin θ + θ + sin 2θ = . 2 4 8 2 0
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 π/2 9 1 9 1 = + θ − 2 cos θ + θ − sin 2θ θ − sin 2θ 2 4 2 4 0 π/6 √ ! " √ !# √ 3π 9 3 3π π 5π 3 = + = − − 3− . 4 8 4 4 8 4 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 θ=π/4
and
polar axis
√ ! √ ! 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 yaxis. 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
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 xaxis, 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
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’Hˆ opital’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
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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 yvalue 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
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