EVNREV09 - R eview Exercises for Chapter 9 5 4 3 2 461 50....

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Unformatted text preview: R eview Exercises for Chapter 9 5 4 3 2 461 50. a r2 4, c 5, b 3, e 52. a r2 2, b 1, c 1 3 4 cos 2 3, e 1 9 25 16 cos 2 1 54. A 2 1 2 2 2 3 2 2 sin 2 2 d 4 2 3 1 2 sin 2 d 3.37 56. (a) r 1 ed e cos 0, r c a ea a a1 e. (b) The perihelion distance is a When ,r 1 1 e2 e c a c a a1 a a1 ea e. ea e. a1 e. When Therefore, The aphelion distance is a a1 e e e2 e2 a . e cos 60. a r 1.422792505 109 1 0.0543 cos e e 1.3495139 1.5044861 b cos br cos bx 109 km 109 km 36.0 1 1 10 6 mi, e e2 a e cos ed 1 ed ed. e When 0, r 1 1 e2 a e a1 e. a1 e1 a1 Thus, r 1 1 58. a e r 1.427 0.0543 1 1 109 km 0.206 e2 a e cos 34.472 10 6 1 0.206 cos e e 28.582 43.416 10 6 mi 10 6 mi Perihelion distance: a 1 Aphelion distance: a 1 Perihelion distance: a 1 Aphelion distance: a 1 62. r r2 x2 x2 y2 bx y2 ay a sin a r sin ay 0 represents a circle. Review Exercises for Chapter 9 2. Matches (b) - hyperbola 4. Matches (c) - hyperbola 6. y 2 12y y2 8x 12y y 20 36 6 2 0 y 8x 20 2 36 16 12 42 x Parabola Vertex: 2, 6 −4 x 8 12 462 8. 4x 2 Chapter 9 y2 4x x Ellipse Center: 2, 0 Vertices: 2, ± 1 16x 4 2 Conics, Parametric Equations, and Polar Coordinates 15 y2 y2 1 1 −1 0 15 16 1 y 4 x2 (2, 0) x 1 2 3 2 14 −2 10. 4 x2 4x 2 x x Hyperbola Center: Vertices: 1 ,1 2 4y 2 1 4 4x 4 y2 8y 2y y 2 11 1 1 2 0 4 y 11 1 1 4 3 12 2 2 1 −2 −1 −2 x 1 3 4 1 ± 2 2, 1 1± x 1 2 Asymptotes: y 12. Vertex: 4, 2 Focus: 4, 0 Parabola opens downward p x x2 4 2 2 4 8y 2y 0 2 8x 14. Center: 0, 0 Solution points: 1, 2 , 2, 0 Substituting the values of the coordinates of the given points into x2 b2 y2 a2 1, 16. Foci: 0, ± 8 ± 4x Asymptotes: y Center: 0, 0 Vertical transverse axis c y 8 a x b c2 4x asymptote → a a2 64 4b 2 we obtain the system 1 b2 4 a2 1, 4 b 2 1. 4b ⇒ 17b2 64 b2 Solving the system, we have a2 16 and b 2 3 4, x2 4 3y 2 16 1. ⇒ b2 y2 1024 17 64 ⇒ a2 17 x2 64 17 1 1024 17 18. x2 4 y2 25 1, a 5, b 2, c 21, e 21 5 By Example 5 of Section 9.1, 2 C 20 0 1 21 2 sin d 25 23.01. R eview Exercises for Chapter 9 12 x 200 200y 4 50 y y 1 y 2 463 20. y (a) x 2 x2 (b) y 12 x 200 1 x 100 1 100 Focus: 0, 50 x2 10,000 x 1 x2 dx 10,000 38,294.49 S 2 0 a 22. (a) A 4 0 b a 2 a2 b x2 dx a2 2 b b2 b2 y2 4b 1 a2 y2 dy x a2 2 a2 b2 x2 b a 2 arcsin x a a ab 0 (b) Disk: V b2 0 y2 dy 0 b 2 a2 2 by b2 13 y 3 b 0 42 ab 3 S 4 0 a b b b4 a2 b2 y2 dy b b2 y2 2a cy b 4 b 2c b b2 b 4a b2 b4 0 c 2y 2 d y c 2y 2 b 4 ln cy b4 c 2y 2 0 2a 2 b c b2 b 2c 2 a2 a c2 b 4 ln cb 2 c2 b 4 ln b 2 e e 2 b2 2 ax a2 13 x 3 a 0 ab2 ca ln c e b2 2 a a2 b a a4 0 2 a2 2 b2 a2 a b2 1 ln e 1 a2 0 (c) Disk: V 2 0 a x2 dx x2 dx 4 ab2 3 S 22 0 a2 x2 a4 a2 b2 x2 a a2 x2 2b cx a 4 a 2c c a 3 x 3 x 9 Ellipse y 7 6 5 4 3 2 1 −2 −1 −2 −3 dx cx a2 a 0 4b a2 a c 2x 2 d x c 2x 2 a 4 arcsin ab 2 a c a2 a 2c 24. x t t x 4, y t2 x 4 2 c2 a 4 arcsin 2 b2 2 ab arcsin e e 2 2 26. x 3 cos , y 2 5 sin 1 28. x x 5 5 sin3 , y 23 5 cos3 1 4⇒y 3 3 y 5 y 2 2 25 Parabola y 7 6 5 4 3 2 1 −1 x 1 2 3 4 5 6 7 y 5 3 y 6 4 2 23 2 2 1 x2 y2 3 52 3 −6 −4 −4 −6 x 2 4 6 x 12345678 464 Chapter 9 2 2 Conics, Parametric Equations, and Polar Coordinates 2 2 30. x x h 5 y y k 3 r2 2 2 32. a 4 Let 4, c y2 16 5, b2 sec 2 and c2 x2 9 a2 9, y2 16 x2 9 1 tan 2 . 4 sec . Then x 34. x y (a) a x y a a b cos t b sin t 2, b cos t sin t y=0 −2 ≤ x ≤ 2 −3 3 −6 6 3 tan and y b cos b sin a b a b b b t t (b) a 3, b 2 cos t 2 sin t 4 1 cos t sin t 2 1 cos 2t sin 2t (c) a x y 4, b 3 cos t 3 sin t 1 cos 3t sin 3t 4 2 cos t 0 x y −6 6 −2 −4 −4 (d) a x y 10, b 9 cos t 9 sin t 1 cos 9t sin 9t 10 (e) a x y 15 3, b cos t sin t 2 2 cos 2 sin 4 (f) a t 2 x y 4, b cos t sin t 3 3 cos 3 sin 4 t 3 t 2 t 3 − 15 − 10 −6 6 −6 6 −4 −4 36. x t u r cos sin r sin cos r sin 38. x y t t2 dy dx 4 r cos y v w r cos (a) r sin y 2t 1 2t 0 when t 0. Point of horizontal tangency: 4, 0 (b) t y θ w v u rθ ( x, y ) x x x y 6 5 4 3 2 1 4 4 2 (c) r θ t x 1 2 3 4 5 6 R eview Exercises for Chapter 9 1 t t2 dy dx 1 x 1 x2 y 4 3 465 40. x y (a) 42. x y 2t 1 t2 2t 3 t 0 2t 1 t2 1 2t t2 1 t2 t x 2 1 x y dy (a) dx 2t 2 t 2 2 2 2t 2 No horizontal tangents (b) t y (c) 0 when t 1. 1 Point of horizontal tangency: 1, (b) t y (c) 2 1 4 3x 12 2 2x 12 x 1 2 1 −2 −2 −1 x 1 2 x 2 4 44. x y (a) 6 cos 6 sin dy dx 6 cos 6 sin cot 0 when 3 . 22 , 6 46. x y (a) et e dy dx t e et ln x e y t 1 e 2t 1 x2 Points of horizontal tangency: 0, 6 , 0, (b) (c) 4 2 −4 −2 −2 −4 x 2 4 No horizontal tangents (b) t y 1 ,x > 0 x x 6 2 y 6 2 1 y ln x e ln 1 x (c) 3 2 1 x 1 2 3 48. x y 2 2 sin cos 8 50. x y dx d 6 cos 6 sin 6 sin 6 cos 36 sin 2 0 (a), (c) −8 8 dy d s −4 36 cos 2 d 6 0 6 (b) At dy dt dx , 6d 1.134, 2 dy dx 3 , 2 (one-half circumference of circle) 0.5, and 0.441 466 Chapter 9 1, 3 1 arctan r, 2 Conics, Parametric Equations, and Polar Coordinates y 52. x, y r 32 3 10 1.89 108.43 10, 5.03 −3 (−1, 3) 2 1 −2 −1 −1 −2 −3 x 1 2 3 10, 1.89 , 54. r r2 x2 y 2 10 100 100 56. 2r 2± x2 r cos y2 4 x2 3x 2 4 3 cos 1 2 cos 3 4 3 2 sin 4y 2 3 4 tan y x y 1 1 x 2x r 2 1 1 cos x y2 1 1 x 0 1 2 58. r 4 sec 60. r cos x 3 sin 3y 8 8 62. x 2 r2 y2 4x 0 0 64. x 2 y 2 arctan y x 2 a2 2 4r cos r r2 a2 4 cos π 2 66. Line 12 68. r 3 csc , r sin 3, y 3 π 2 Horizontal line 0 1 2 0 1 2 3 4 70. r 3 4 cos Limaçon Symmetric to polar axis π 2 2 0 0 r 1 3 1 2 3 2 3 5 7 R eview Exercises for Chapter 9 72. r 2 2 467 Spiral Symmetric to π 2 0 2 4 8 0 r 0 4 5 2 3 4 3 2 2 5 4 5 2 3 2 3 74. r cos 5 π 2 Rose curve with five petals Symmetric to polar axis Relative extrema: 1, 0 , Tangents at the pole: 2 , 1, , 5 5 3 79 , ,, , 10 10 2 10 10 1, 1, 3 4 , 1, 5 5 0 1 76. r 2 cos 2 Lemniscate Symmetric to the polar axis Relative extrema: ± 1, 0 Tangents at the pole: 3 , 44 r 0 ±1 ± π 2 6 2 2 4 0 0 1 2 78. r 2 sin cos 2 2 −1 0.75 80. r 4 sec cos −1 3 Bifolium Symmetric to 1 − 0.25 Semicubical parabola Symmetric to the polar axis r⇒ r⇒ as as ⇒ ⇒ 2 2 5 −3 82. r 2 4 sin 2 dr d dr d 8 cos 2 4 cos 2 r 0, 2 (b) dy dx r cos r sin cos 2 cos 2 sin cos 4 cos 2 sin r 4 cos 2 cos r sin 2 cos sin 2 sin (a) 2r Tangents at the pole: (c) −3 2 Horizontal tangents: dy dx tan 0 when cos 2 tan 2 , sin sin 2 cos 0, 2 3, 3 0, , 0, 0 , ± 3 3 0: Vertical tangents when cos 2 cos −2 sin 2 sin 2 3, 6 tan 2 tan 1, 0, , 0, 0 , ± 6 468 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 1, 0 84. False. There are an infinite number of polar coordinate representations of a point. For example, the point x, y 1, 0 , 1, 2 , 1, , etc. has polar representations r, 86. r a sin , r a cos 2, 4 and 0, 0 . For r 2 sin cos . cos 2 0. For r cos 2 . 2 sin cos 2, a cos , a sin , The points of intersection are a m1 At a m2 At a 88. r A 51 1 2 2 dy dx 2, dy dx 2, a cos sin a cos 2 a sin cos a sin 2 4 , m1 is undefined and at 0, 0 , m1 a sin 2 a sin cos 4 , m2 sin 3 2 4 a cos 2 a cos sin 0 and at 0, 0 , m2 is undefined. Therefore, the graphs are orthogonal at a 90. r 4 sin 3 3 1 2 3 4 and 0, 0 . 2 51 sin 2 d 117.81 75 A 2 4 sin 3 0 2 d 12.57 4 8 4 −8 −6 − 12 −4 6 92. r 9 3, r 2 r2 18 sin 2 18 sin 2 −6 4 6 sin 2 1 2 −4 12 A 2 1 2 12 18 sin 2 d 0 1 2 5 12 9d 12 1 2 2 18 sin 2 d 5 12 1.2058 9.4248 1.2058 11.84 dr d 94. r A e ,0 ≤ 1 2 e 0 2 ≤ d 10 96. r 133.62 s a cos 2 , 4 2a sin 2 4a 2 sin 2 2 d 4) 2 8 0 4 a 2 cos 2 2 1 0 8a − 25 5 3 sin 2 2 d (Simpson’s Rule: n 2 1.5811 4 1.8870 a 1 6 9.69a 4 1.1997 −5 P roblem Solving for Chapter 9 2 ,e cos 4 3 sin 45 ,e 3 5 sin 3 5 469 98. r 1 1 100. r 5 1 Parabola π 2 Ellipse π 2 0 2 0 1 2 102. r 2 8 5 cos 1 4 ,e 5 2 cos 5 2 104. Line Slope: 3 3 x, r sin 3, 3 3 r cos , Solution point: 0, 0 Hyperbola π 2 y tan 0 1 2 106. Parabola Vertex: 2, 2 108. Hyperbola Vertices: 1, 0 , 7, 0 Focus: 0, 0 a 3, c 4 3 1 4, e 7 4 4 cos 3 3 4 ,d 3 7 4 Focus: 0, 0 e 1, d 4 4 r 1 sin r 7 4 cos Problem Solving for Chapter 9 2. Assume p > 0. Let y mx p be the equation of the focal chord. x2 = 4py (0, p) y First find x-coordinates of focal chord endpoints: x x2 x2 x x2 4py 4pmx 4pm ± 4p mx 4p2 p 0 16p2 2pm ± 2p m2 x . 2p 1 y = −p 16p2m2 2 4py, 2x 4py ⇒ y (a) The slopes of the tangent lines at the endpoints are perpendicular because 1 2pm 2p 2p m2 1 1 2pm 2p 2p m2 1 1 4p2m2 4p2 4p2 m2 1 1 4p2 4p2 1 —CONTINUED— 470 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 2. —CONTINUED— (b) Finally, we show that the tangent lines intersect at a point on the directrix y Let b b2 c2 b2 4p c2 4p 2 pm 8 p2m2 8 p2m2 2 pm2 2 pm2 2p 4p2 4p2 p p m2 1 and c m2 1 1 2 pm 2p m2 1. p. 8 p2m 8 p2m m2 2 pm m2 2 pm m2 b2 4p c2 4p 1 1 b x 2p c x 2p bx 2p Tangent line at x Tangent line at x b: y c: y b ⇒y c ⇒y b2 4p b2 c 1 x cx 2p 2 cx b2 bx 2p cx 2p c2 4p c2 c2 b2 4p c2 4p Intersection of tangent lines: 2 bx 2x b 2 x 4p m2 16 p2m m2 2 pm 1 Finally, the corresponding y-value is y x2 a2 y y2 b2 b2x a2y y yy0 b2 x0 x a2 x0 x a2 At x 0, y QF2 MQ QF1 x02 b2 ⇒Q y0 c2 y0 y0 y b2 y0 y b2 0, b4 y02 b2 y0 2 p, which shows that the intersection point lies on the directrix. 4. 1, a2 b2 c2, MF2 M f1 2a y β F2(− c, 0) α β M (x 0 , y 0 ) x Tangent line at M x0, y0 : y0 y02 b2x0 x a2y0 x0 x a2 x02 a2 1 b2 . y0 d f y02 b2 x02 x0 F1(c, 0) Q By the Law of Cosines, F2Q d F1Q d cos MF2 2 2 MF2 MF2 MF1 MF1 2 2 MQ f 2 2 2 MF2 MQ cos 2 f MF2 cos 2 f MF1 cos 2 f MF1 cos . MF1 2 f 2 2 f MF1 d2 2 2 2 2 f2 f 2 F2 2 f 2 d 2 , cos 2 f MF2 MF1 y0 x0 c 2a. Let z ; QF1: MF1. Slopes: MF1: —CONTINUED— b2 b2 ; QF2: y0c y0c P roblem Solving for Chapter 9 4. —CONTINUED— To show MF2 ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ Thus, a x0 c 2 2 471 , consider f2 z d2 2 f MF1 2a 2 MF1 d2 z 2 f2 f2 d2 y0 d 2 2 f MF2 d2 z 2a f2 z2 y02 z2 f2 x02 0 x0c a2b2 1. 2az 2az a2 y0 2 b2 y0 2 c2 b4 y02 az x0 x0c c 2 a2 x02b2 x0 a2 2 a2y02 y0 b2 2 and the reflective property is verified. 6. (a) y2 t2 1 1 x x 1 1 t2 2 t2 2 , x2 1 1 1 1 t2 t2 t2 t2 x . x 1 1 t2 t2 2t2 2 2 2 (b) sin2 t2 r cos r cos 1 sin2 sin2 r 2 sin2 r cos sin2 cos2 r cos r 3 ,. 44 r 2 cos2 cos2 cos2 cos2 cos 2 cos 2 1 1 1 r cos r cos r cos 1 1 r cos3 sin2 Thus, y2 x2 1 1 sec (c) π 2 (d) r 0 for Thus, y x and y the origin. 0 1 2 x are tangent lines to curve at (e) y t t4 4t2 1 t2 1 1 1 3t2 t 22 t t3 2t 1 4t2 t4 1 t2 2 1 1 3 1 0 2± 2± 5 5 5 5 3 5 1± 5 51 2 0 ⇒ t2 2± 5⇒x 51 ,± 2 51 2 2 5 472 Chapter 9 Conics, Parametric Equations, and Polar Coordinates ab a sin b cos b cos ay y b bx x a 8. (a) 10. r ,0 ≤ ab ab 1 ≤ 2 r a sin Generated by Mathematica Line segment u2 sin du is on 2 0 t (b) x, y u2 cos du, 2 0 t Area 1 ab 2 the curve whenever x, y is on the curve. (c) x t Thus, s 0 cos t2 ,y t 2 a sin t2 ,x t 2 2 yt 2 1 dt , ,s a. 2. On 12. Let r, r 2 be on the graph. 1 2r cos r2 r4 2r 2 r2 r 2 r2 1 2 1 2r cos 4r 2 cos2 4r 2 cos2 1 1 1 0 4 cos2 2 2 cos2 2 cos 2 2 1 1 4 cos 2 2 r 2 r2 r2 14. (a) r 2 −6 4 (c) r 6 2 2 cos −6 4 Circle radius 2 Cardioid 6 −4 −4 (b) r 2 cos −6 4 (d) r 6 2 3 cos −6 4 Convex limaçon Limaçon with inner loop 6 −4 −4 16. The curve is produced over the interval 0≤ ≤9. ...
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This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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