{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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”. ...
View Full Document

{[ snackBarMessage ]}