ODDREV09 - 214 Chapter 9 Conics Parametric Equations and...

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Unformatted text preview: 214 Chapter 9 Conics, Parametric Equations, and Polar Coordinates ed sin 63. r1 1 ed and r2 sin 1 Points of intersection: ed, 0 , ed, dy r1: dx ed sin ed 1 sin 1 dy dx cos sin 1. At ed, cos sin 1. At ed, ed cos sin ed cos 1 sin 1 , dy dx 2 sin cos 2 At ed, 0 , 1. sin cos dy r2: dx ed sin ed 1 sin 1 dy dx ed cos 1 sin ed cos 1 sin , dy dx 2 2 At ed, 0 , 1. 11 1, and at ed, we have m1m2 1 1 1. The curves Therefore, at ed, 0 we have m1m2 intersect at right angles. Review Exercises for Chapter 9 1. Matches (d) - ellipse 5. 16x 2 x2 x 16y 2 1 4 x Circle Center: 1 , 2 3 4 16x y2 1 2 2 3. Matches (a) - parabola 24y 3 y 2 y 9 16 3 4 2 3 0 3 16 1 2 y 1 4 9 16 1 1 x 1 2 1 , 2 3 4 Radius: 1 7. 3 x2 3x 2 8x 2y 2 16 x 2 Hyperbola Center: Vertices: 4, 3 4± 2, 3 3± y 24x 2 y2 4 2 12y 6y y 3 24 9 3 2 0 24 1 48 18 9. 3 x2 3x 2 4x 2y 2 4 x 12x 2 y2 2 13 2 12y 6y y 29 9 2 0 29 1 12 18 3 12 Ellipse Center: 2, Vertices: 3 x 2 4 1 1 6 4 2 x y 3 2, 3± 2 2 Asymptotes: y x 1 2 3 2 3 4 (2, − 3) 6 4 2 Review Exercises for Chapter 9 11. Vertex: 0, 2 Directrix: x p y y2 3 2 2 215 13. Vertices: 3 3, 0 , 7, 0 15. Vertices: ± 4, 0 Foci: 0, 0 , 4, 0 Horizontal major axis Center: 2, 0 Foci: ± 6, 0 Center: 0, 0 Horizontal transverse axis 21 1 a x2 16 4, c y2 20 6, b 1 36 16 25 Parabola opens to the right 43 x 12x 4 0 0 a x 5, c 2 25 2 2, b y2 21 4y 17. x2 9 y2 4 1, a 3, b 2, c 5, e 5 3 19. y x y 2 has a slope of 1. The perpendicular slope is x2 2x 2 2x 2 1 when x y 4x 4y 1 and y 2 5 4 7 0 1x 5 . 4 1 2 1. By Example 5 of Section 9.1, 2 C 12 0 1 5 sin2 d 9 15.87. dy dx Perpendicular line: 21. (a) V (b) F a b Length 3 12 y 9 4 3 y2 3 2 16 9 192 ft 3 y2 dy 9 arcsin y 3 8 62.4 3 3 1 9 3 3 3 2 62.4 3 3 9 3 3 y2 dy 3 y9 y 2 dy 3 8 62.4 y 3 2 8 39 62.4 3 22 3 y2 32 3 9 2 8 27 62.4 3 2 7057.274 y (c) You want 4 of the total area of 12 covered. Find h so that h 2 0 4 3 h x= 9 9 y2 dy 3 9 8 9 8 9 . 4 4 3 9− y2 4 2 Area of filled tank above x-axis is 3π. h y2 dy h 0 −2 −2 −4 x 2 0 1 y 2 9 h9 y2 h2 y 9 arcsin 3 9 arcsin h 3 Area of filled tank below x-axis is 6π. By Newton’s Method, h (d) Area of ends Area of sides 2 12 2 1.212. Therefore, the total height of the water is 1.212 24 3 4.212 ft. Perimeter Length 16 0 1 2 12 4 1 7 sin 2 16 d 16 4 from Example 5 of Section 9.1 1 7 sin 2 16 8 1 2 1 7 sin 2 16 4 256 7 sin 2 0 16 1 7 3 sin2 16 8 7 sin2 16 2 353.65 Total area 24 353.65 429.05 216 23. x t Chapter 9 1 x 4 4t, y 1 ⇒y y 4y Line 3x 11 2 Conics, Parametric Equations, and Polar Coordinates 3t 2 3 x 4 0 3 x 4 11 4 1 4 y 2 1 −1 x −1 −2 1 2 3 5 25. x x 6 x2 6 cos , y 2 6 sin 1 4 2 y 27. x x x x 2 4 2 2 2 2 2 sec , y sec2 y 3 3 1 2 tan tan2 1 8 y 6 y2 2 1 y y 3 2 36 −4 −2 −2 −4 Hyperbola Circle 4 2 −4 x −2 −4 2 4 8 29. x y 3 2 3 2 6t 2t 2 3 4t 5t 31. x 3 16 x 2 y 9 2 4 2 1 y 9 4 4 2 (other answers possible) Let 3 16 cos 2 and 3 sin 2 . 3 sin . < Then x 33. x y cos 3 sin 3 5 4 cos and y 5 cos 5 sin 35. (a) x 2 cot , y 4 4 sin cos , 0 < −12 −7 8 12 −4 −5 (b) 4 x2 y 4 4 cot2 4 sin cos sin cos 16 csc2 16 cos sin 8 2 cot 8x 37. x y (a) 1 2 dy dx 4t 3t 3 4 (b) t y x 4 2 3 x 4 1 3x 4 2 1 x 1 11 (c) 5 4 y No horizontal tangents 1 2 3 5 Review Exercises for Chapter 9 1 t 2t dy dx 3 2 1 t2 2t 2 (b) t y 1 x 2 x 3 4 2 x 217 39. x y (a) (c) 6 y No horizontal tangents t0 4 2 2 4 41. x y 1 2t 1 t 2 1 2t t2 2 2t 2 2 2t 1 2 2t t t2 1 2t 1 t 22 1 3, 2 43. x y (a) 0 when t 1. (b) 1 (c) 3 2 dy dx 2 cos 5 sin 5 cos 2 sin 2.5 cot 0 when 3 3 ,. 22 dy (a) dx Points of horizontal tangency: 3, 7 , 3, x 4 y 3 2 y 25 2 2 1 Point of horizontal tangency: (b) 2t y 1 1 ⇒t x 1 11 x 2 x 1 (c) 3 2 y 11 2x 8 1 4 11 x 2 x 4x 2 4x 1 x 2 4x 2 1x x 4 4 8 x 2 5x 1 −2 −1 −1 −2 x 2 3 45. x y (a) cos 3 4 sin 3 dy dx But, 12 sin 2 cos 3 cos 2 sin dy dt y 4 y 4 sin cos 4 tan 0 when 0, . dx dt 23 0 at 1 0, . Hence no points of horizontal tangency. (b) x 2 3 (c) 4 x 4 2 2 4 4 218 47. x y Chapter 9 cot sin 2 Conics, Parametric Equations, and Polar Coordinates 49. x y dx d 3 r cos r sin r cos r sin r 0 2 sin cos 2 sin cos 2 (a), (c) −3 dy d s −2 cos 2 r 2 2 0 2 sin 2 d 1 2 2 (b) At dx , 6d dy 4, d dy 1, and dx 1 4 r 0 d r 51. x, y r 4, 42 7 4 4 4 2 y 42 1 x −1 −2 −3 1 2 3 4 5 r, 7 4 2, , 4 3 4 2, 4 −4 −5 (4, − 4) 53. r r2 x2 x2 y2 r2 r4 x2 y2 2 3 cos 3r cos 3x 0 cos 2 r2 x2 cos 2 y2 cos 2 r 2 sin 2 sin 2 55. r r2 x2 x2 y2 r y2 2x 2 21 2r 1 2± 4 x2 cos cos x2 y2 y2 2x y2 3x 57. 59. 4 cos 2 sec 4 2 cos 2 1 4 y2 4y 2 4 4 x x y x 2 1 cos r cos x x3 xy 2 y2 8 cos 2 8 x2 x2 4 4x 2 x2 61. x2 y2 2 ax 2y a r 2 cos 2 a cos 2 sin r sin 63. x 2 y2 r2 a 2 arctan a2 2 r4 r 65. r 4 π 2 67. r r cos 0 2 6 sec 1 cos 1, x 1 π 2 Circle of radius 4 Centered at the pole Symmetric to polar axis, 2, and pole Vertical line 1 0 R eview Exercises for Chapter 9 69. r 21 cos 71. r 4 3 cos 219 Cardioid Symmetric to polar axis π 2 Limaçon Symmetric to polar axis π 2 0 1 2 4 0 0 r 4 3 3 2 2 2 3 1 0 r 0 1 3 5 2 π 2 2 4 2 3 11 2 7 73. r 3 cos 2 Rose curve with four petals Symmetric to polar axis, Relative extrema: Tangents at the pole: 2 , and pole 0 4 3, 0 , 3, 2 , 3, 3 , 3, 2 3 , 44 π 2 75. r 2 r 4 sin 2 2 ± 2 sin 2 Rose curve with four petals Symmetric to the polar axis, Relative extrema: , and pole 2 3 ± 2, , ± 2, 4 4 0, 2 0 2 Tangents at the pole: 77. r 3 cos 4 3 sec rotated through an angle of 4 79. r 4 cos 2 sec Strophoid Symmetric to the polar axis r⇒ r⇒ as as 4 Graph of r 5 ⇒ ⇒ 2 2 −1 −1 8 −6 6 −4 220 81. r Chapter 9 1 2 cos Conics, Parametric Equations, and Polar Coordinates (a) The graph has polar symmetry and the tangents at the pole are 3 (b) dy dx , 3 . 1 1 2 cos cos 2 cos sin 4 cos 2 33 8 33 4 3 4 3 4 3 4 33 33 33 , arccos , ,r 1 33 8 1 8 33 8 1 8 33 33 cos 2 1 8 0.686, 0.568 0.686, 2.186, 2.206 2.186, 2.206 . 0.568 2 33 0, cos 3 4 1± 33 1 8 32 1± 8 33 2 sin 2 2 sin cos Horizontal tangents: When cos 3 1± , 1 arccos 1 , arccos , arccos Vertical tangents: sin 4 cos 0, , 1 1 , ± arccos 2 4 (c) −5 2.5 1 0, sin 1 , 4 0, cos 1, 0 , 3, 1 , 4 ± arccos 0.5, ± 1.318 1 − 2.5 83. Circle: r dy dx 3 sin 3 cos sin 3 cos cos 4 5 sin 4 4 5 sin 5 sin cos sin at dy , 6 dx 3 9 3 sin cos 3 sin sin sin 2 cos 2 sin 2 tan 2 at dy , 6 dx 3 Limaçon: r dy dx Let 5 cos sin 5 cos cos be the angle between the curves: tan 3 1 39 13 23 3 23 . 3 49.1 . Therefore, arctan Review Exercises for Chapter 9 85. r 1 cos , r 1 cos 2 are the two points of intersection (other than the pole). The slope of the graph of sin 2 sin cos cos 1 cos sin 1 cos 11 221 The points 1, 2 and 1, 3 r 1 cos is m1 At 1, m2 dy dx 2 , m1 dy dx r sin r cos 1 sin 2 sin cos r cos r sin 1 . 1. The slope of the graph of r 1 cos is 1 and at 1, 3 cos 1 cos . sin 1 cos 2 , m1 At 1, 2 , m2 1 1 1 and at 1, 3 2 , m 2 the graphs are orthogonal at 1, 2 and 1, 3 2 . 87. r A 2 2 1 2 3 11 1. In both cases, m 1 1 m 2 and we conclude that cos 2 0 89. r cos 2 sin 2 1 2 cos 2 2 d 14.14 9 2 A sin cos 2 0 2 d 0.10 −3 6 32 0.5 −3 − 0.5 − 0.1 0.5 91. r 2 A 4 sin 2 2 1 2 2 93. r 4 sin 2 d 4 A 4 cos , r 2 1 2 3 2 4d 1 2 2 3 0 4 cos 3 2 d 4.91 0 2 −3 3 −3 6 −2 −3 95. s 2 0 a2 1 1 0 cos 2 a 2 sin 2 d 2 2a 0 2 2a cos d sin 1 cos d 4 2a 1 cos 12 0 8a 97. r 1 2 ,e sin 1 99. r 3 6 2 cos 1 2 ,e 2 3 cos 2 3 Parabola π 2 Ellipse π 2 0 2 0 2 46 8 222 Chapter 9 4 3 sin Conics, Parametric Equations, and Polar Coordinates 2 ,e 3 2 sin 3 2 101. r 2 1 103. Circle 0, 5 in rectangular coordinates 2 Solution point: 0, 0 Center: x2 y y2 5 5 Hyperbola π 2 5, 25 0 0 0 2 3 4 x2 r2 10y 10r sin r 10 sin 105. Parabola Vertex: 2, Focus: 0, 0 e r 1, d 1 4 4 cos 107. Ellipse Vertices: 5, 0 , 1, Focus: 0, 0 a 3, c 2 3 1 2, e 5 2 2 cos 3 3 2 ,d 3 5 2 5 2 cos r Problem Solving for Chapter 9 1. (a) 10 8 6 4 y (4, 4) x (− 1, 1 ) 4 −6 −4 −2 2 2 4 6 −2 (b) x2 2x y y y 4y 4y 1 x 2 4 1 4 2x 1 x 2 4 ⇒y 1 ⇒y 2x 1 x 2 4 1 4 Tangent line at 4, 4 Tangent line at 1, 1 4 Tangent lines have slopes of 2 and (c) Intersection: 2x 8x 4 16 10x x 1 x 2 2x 15 3 ⇒ 2 3 , 2 1 1 4 1 1 2 ⇒ perpendicular. Point of intersection, 3 2, 1 , is on directrix y 1. Problem Solving for Chapter 9 3. Consider x2 4py with focus 0, p . F P(a, b) x y 223 B A Let p a, b be point on parabola. zx y 4py ⇒ y b a x 2p 0, y 0, b b. a x 2p Tangent line a 2p a b a2 2p b 4pb 2p b. Q For x Thus, Q FQP is isosceles because FQ FP p a b 0 2 b p 2 a2 4pb b b p. b2 b2 p 2 2bp 2bp p2 p2 Thus, FQP BPA FPQ. 2a ⇒ OB OB OA ⇒ OA 2a 5. (a) In In r OCB, cos OAC, cos OP AB 2a 2a sec . cos . cos cos (c) r cos r 3 cos x2 r 2a tan sin 2a sin2 2a r 2 sin2 2ay2 x3 2a x OB OA 2a sec 2a 2a 2a 1 cos sin2 cos y2 x y2 tan sin 2a sin2 2a tan sin2 , 2 < < 1 + t2 t (b) x y Let t r cos r sin tan , 2a tan sin 2a tan sin <t< cos sin . 2a 2 θ 1 Then sin2 t2 1 t2 and x t2 1 t2 ,y 2a t3 1 t2 . 7. y a1 arccos cos a a sin a a a a a y ⇒ cos a a y x a a 2ay − y 2 a arccos a arccos x a arccos y y y a sin arccos 2ay a 2ay y2 a a y θ a−y y2, 0 ≤ y ≤ 2a 224 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 12 rd 2 sec2 0 9. For t y 357 , , , ,. . . 2222 2 , 3 22 2 ,, ,. . . 57 2 3 1 1 3 1 4 2 5 1 5 1 6 2 7 1 7 1 8 11. (a) Area 0 x=1 r = sec θ 1 2 ... ... (b) tan d 1 1 tan 2 sec2 0 α 1 Hence, the curve has length greater that S 2 2 h ⇒ Area 1 ⇒ tan d 21 > 2 . (c) Differentiating, ... d tan d sec2 . 13. If a dog is located at r, x, y r cos , r sin , then its neighbor is at r, and x, y r sin , r cos 2 . : The slope joining these points is r cos r sin dr sin d dr cos d r sin r cos r cos r sin ⇒ dr d dr r ln r r r r 4 d ⇒r 2 d e 2 4 sin sin sin sin cos cos cos cos slope of tangent line at r, . r d C1 e Ce Ce . 4 C1 d ⇒C 2 d e 2 4 Finally, r 15. (a) The first plane makes an angle of 70 with the positive x-axis, and is 150 miles from P: x1 y1 cos 70 150 sin 70 150 375t 375t (c) 280 0 0 1 Similarly for the second plane, The minimum distance is 7.59 miles when t x2 cos 135 190 cos 45 y2 190 450t 450t 450t 450t y2 y1 2 0.4145. sin 135 190 sin 45 190 (b) d x2 cos 45 x1 2 190 450t cos 70 150 375t 2 sin 45 190 450t sin 70 150 375t 212 Problem Solving for Chapter 9 17. −6 225 4 4 4 4 6 −6 6 −6 6 −6 6 −4 −4 −4 −4 4 4 4 4 −6 6 −6 6 −6 6 −6 6 −4 −4 −4 −4 4 4 4 −6 6 −6 6 −6 6 −4 −4 −4 n 1, 2, 3, 4, 5 produce “bells”; n 1, 2, 3, 4, 5 produce “hearts”. ...
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