ODD09 - CHAPTER 9 Conics Parametric Equations and Polar...

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Unformatted text preview: CHAPTER 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1 Section 9.2 Section 9.3 Section 9.4 Section 9.5 Section 9.6 Conics and Calculus . . . . . . . . . . . . . . . . . . . . 177 Plane Curves and Parametric Equations . . . . . . . . . . 188 Parametric Equations and Calculus . . . . . . . . . . . . 192 Polar Coordinates and Polar Graphs . . . . . . . . . . . . 198 Area and Arc Length in Polar Coordinates . . . . . . . . 205 Polar Equations of Conics and Kepler’s Laws . . . . . . . 210 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 CHAPTER 9 Conics, Parametric Equations, and Polar Coordinates Section 9.1 Conics and Calculus Solutions to Odd-Numbered Exercises 1. y 2 p 4x 1>0 3. x p 3 2 2y 3, 2 <0 2 Vertex: 0, 0 Vertex: 1 2 Opens to the right Matches graph (h). Opens downward Matches graph (e). 5. x2 9 y2 4 1 7. y2 16 x2 1 1 Center: 0, 0 Ellipse Matches (f) Hyperbola Center: 0, 0 Vertical transverse axis. Matches (c) 3 2 9. y 2 6x 3 2, 4 x y 11. x 8 3 y y 2 2 2 2 0 4 1 4 Vertex: 0, 0 Focus: 0 3 2 12 8 4 4 8 x 3 y 4 Vertex: 4 3, 2 3.25, 2 2.75 −8 Directrix: x Focus: (0, 0) x Directrix: x (− 3, 2) −6 −4 −2 2 x −2 −4 13. y2 y2 4y 4y y 4x 4 2 2 0 4x 4 1 y 15. x2 4x x2 4y 4x x 2 4 4 2 0 4y 4 4 4 2 y 4 41 x 1y Vertex: 1, 2 6 Vertex: Focus: (− 1, 2) 2, 2 2, 1 3 −6 −4 Focus: 0, 2 Directrix: x 2 4 Directrix: y (− 2, 2) x 2 −2 −4 −2 x 2 2 2 4 6 177 178 17. y 2 y 2 Chapter 9 x y y Vertex: y 1 4 12 2 1 4, 1 2 1 2 1 2 Conics, Parametric Equations, and Polar Coordinates 19. y 2 x 1 4 1 4 0 4x 4 y 2 0 4x 4 1 4 4 x 1 4 2 41 x Vertex: 2 1, 0 2 −6 Focus: 0, Directrix: x Focus: 0, 0 −5 Directrix: x 6 −3 −4 21. y2 4y y 8x 2 2 4 0 2x 3 23. x h 2 4p y 46 y 0 k 4 20 x2 x2 24y 96 25. x2 y y 4 4 0 x2 27. Since the axis of the parabola is vertical, the form of the equation is y ax2 bx c. Now, substituting the values of the given coordinates into this equation, we obtain 3 c, 4 9a 3b c, 11 16a 5 3, b 29. x2 x2 4 a2 4y 2 y2 1 4, b2 4 2 y 1 1, c2 (0, 0) 4b 14 3, c. c 3. 0. 3 −1 x 1 Solving this system, we have a Therefore, y 52 3x 14 3x Center: 0, 0 Foci: ± 3, 0 −2 3 or 5x 2 14x 3y 9 Vertices: ± 2, 0 e 3 2 31. x 9 a2 1 2 y 25 5 9, c2 2 y 1 16 12 33. (1, 5) 9x2 9 x2 4x 4y 2 4 36x 4 y2 24y 6y 36 9 0 36 36 36 36 25, b2 Center: 1, 5 Foci: 1, 9 , 1, 1 Vertices: 1, 10 , 1, 0 e 4 5 8 4 4 x 4 8 x 4 a2 9, b2 4, c2 2, 3 2, 3 ± 2, 6 , 5 3 2 2 y 9 3 2 1 5 y Center: Foci: Vertices: e 5 2, 0 (− 2, 3) 6 2 x 6 4 2 2 S ection 9.1 35. 12x2 12 x2 x 20y 2 1 4 12x 20 y 2 40y 2y 37 1 0 37 60 x a2 5, b2 1 , 2 1 ± 2 12 5 2 2 Conics and Calculus 4y 2y y 2 0.25 1 1 2 179 37. 3 20 x2 x2 3x x a2 4, b2 2y 2 9 4 3x 2 y2 32 4 2 0 1 4 1 9 4 2 4 y 3 1 2 1 2, c 2 1 2, 1 , 2 1 1, 2 3, c2 1 Center: Foci: 3 , 2 3 ± 2 Center: Foci: 2, 1 Vertices: 5, 1 Vertices: 1 ± 2 7 , 2 1 1 1 2 Solve for y: 2 y 2 2y y x2 17 24 7 3x 3x 1 4 x2 2 Solve for y: 20 y 2 2y y 1 1 2 12x2 57 12 x 12x 20 57 37 12x2 20 y 12x 20 12x2 1± 12x 8 4x2 y 1± (Graph each of these separately.) 1 −2 4 (Graph each of these separately.) 1 −3 3 −3 −3 39. Center: 0, 0 Focus: 2, 0 Vertex: 3, 0 Horizontal major axis a x2 9 3, c y2 5 2⇒b 1 5 41. Vertices: 3, 1 , 3, 9 Minor axis length: 6 Vertical major axis Center: 3, 5 a x 9 4, b 3 2 3 y 16 5 2 1 43. Center: 0, 0 Horizontal major axis Points on ellipse: 3, 1 , 4, 0 Since the major axis is horizontal, x2 a2 y2 b2 1. Substituting the values of the coordinates of the given points into this equation, we have 9 a2 1 b2 1, and 16 a2 1. 16, b2 16 7. The solution to this system is a2 Therefore, x2 16 y2 16 7 1, x2 16 7y 2 16 1. 180 y2 1 a Chapter 9 x2 4 1, b Conics, Parametric Equations, and Polar Coordinates x 4 a 2, b 1, c 2 1, 5, 2 , 3, 2 2± 1 x 2 1 2 1 2 45. 1 2, c 5 47. y 1 2 2 1 5 Center: 0, 0 Vertices: 0, ± 1 Foci: 0, ± 5 ±x Center: 1, Vertices: Foci: 1 ± 1 2 Asymptotes: y y Asymptotes: y y 4 2 x 1 x 1 2 1 2 3 4 2 2 2 4 4 4 5 49. 9 x2 9x2 4x y2 4 x 1 36x y2 2 2 6y 6y y 9 18 9 3 2 0 18 1 36 9 51. x2 x2 2x 9y 2 1 x 2x 9 y2 1 2 54y 6y 9y y 80 9 3 2 0 80 0 ± 1 81 0 a 1, b 3, c 3 3 , 3, 10, 3 10 3 1 x 3 1 Center: 2, Vertices: 1, Foci: 2 ± y 3 2 Degenerate hyperbola is two lines intersecting at 1, 3 . y Asymptotes: y 3±3x x 4 2 2 2 x 2 2 4 6 2 4 6 4 6 53. 9 y2 9y 2 6y x2 9 y 2 2x x2 3 2 54y 2x x 18 25 62 1 1 2 0 62 1 1 −5 7 55. 1 81 18 3 x2 3x2 2x 2y 2 1 x 4 6x 2 y2 1 2 12y 6y y 6 10 27 9 3 2 0 27 1 1 3 18 12 a 2, b 3 2, c 3 3± 2 3±2 5 a 2, b 6, c 3 1, 10, 3 , 3, 3 Center: 1, Vertices: 1, Foci: 1, Solve for y: 9 y2 Center: 1, Vertices: Foci: 1 ± −5 7 3 −7 −7 Solve for y: 2 y2 6y y 9 3 2 6y y 9 3 2 x2 x2 2x 2x 9 3± 1 3 62 19 81 3x2 3x2 6x 6x 2 27 9 3 x2 18 y x2 2x 19 y 3± 2x 2 3 (Graph each curve separately.) (Graph each curve separately.) Section 9.1 57. Vertices: ± 1, 0 Asymptotes: y ± 3x Horizontal transverse axis Center: 0, 0 a 1, ± b a x2 1 ± Conics and Calculus 181 59. Vertices: 2, ± 3 Point on graph: 0, 5 Vertical transverse axis Center: 2, 0 a 3 3 y2 9 2 b 1 y2 9 ±3 ⇒ b Therefore, the equation is of the form x b2 2 1. Therefore, 1. Substituting the coordinates of the point 0, 5 , we have 25 9 4 b2 1 or b2 y2 9 9 . 4 x 2 94 2 Therefore, the equation is 1. 61. Center: 0, 0 Vertex: 0, 2 Focus: 0, 4 Vertical transverse axis a 2, c 4, b2 y2 4 x2 12 c2 1. a2 12 63. Vertices: 0, 2 , 6, 2 2 x, y 4 3 Horizontal transverse axis Asymptotes: y Center: 3, 2 a 3 b a ± 2 x 3 Therefore, Slopes of asymptotes: ± Thus, b x 9 3 2. Therefore, 2 2 3 y 4 2 2 1. 65. (a) x2 9 At x y2 1, 6: y 2x 9 ± 2yy 3, y 3 0, x 9y y ±2 (b) From part (a) we know that the slopes of the normal lines must be 9 2 3 . 3 At 6, 3:y or 9x 3 9 9 x 23 2 3y 3 9 23 60 x 6 0 6 0 ±6 93 At 6, 3:y or 2x 23 x 9 3 3y 3 3 3y 3 6 0 6 At 6, 3 :y or 9x At 6, 3:y or 2x 23 x 9 3 0 2 3y 60 67. x2 A AC 4y 2 1, C 6x 4 16y 21 0 69. y2 A 4y 0, C 4x 1 0 71. 4x2 A C Circle 4y 2 4 16y 15 0 4>0 Parabola Ellipse 73. 9x2 A C Circle 9y 2 9 36x 6y 34 0 75. 3x2 A 2y2 3, C 3x2 6x 6x 4y 3 5 6 0 2y2 4y 2 2, AC < 0 Hyperbola 182 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 79. (a) A hyperbola is the set of all points x, y for which the absolute value of the difference between the distances from two distance fixed points (foci) is constant. (b) x a2 k± b x a h 2 77. (a) A parabola is the set of all points x, y that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line. (b) x h 2 4p y k or y k 2 4p x h y b2 k 2 1 or y a2 a x b k 2 x b2 h h 2 1 (c) See Theorem 9.2. (c) y h or y k± 81. Assume that the vertex is at the origin. x2 3 2 83. y y ax2 2ax 4py 4p 1 p The equation of the tangent line is y Let y ax02 2ax0 x x0 or y 2ax 0 x ax 02. 9 4 0. Then: ax02 ax02 2ax0x 2ax0x x0 2 x is the x-intercept. 2ax02 The pipe is located 9 meters from the vertex. 4 y 3 Focus 2 Therefore, (3, 1) y (− 3, 1) 1 −3 −2 −1 x 1 2 3 y = ax 2 0 (x0, ax02 ) ( x2 , 0) x 85. (a) Consider the parabola x2 4py. Let m0 be the slope of the one tangent line at x1, y1 and therefore, the second at x2, y2 . From the derivative given in Exercise 32 we have: m0 1 m0 1 x or x1 2p 1 1 x or x2 2p 2 2pm0 2p m0 1 m0 is the slope of Substituting these values of x into the equation x2 4py, we have the coordinates of the points of tangency 2pm0, pm02 and 2p m0, p m02 and the equations of the tangent lines are y pm02 m0 x 2pm0 and y p m02 1 x m0 2p . m0 The point of intersection of these lines is p m02 1 , m0 p and is on the directrix, y p. y x 2 = 4py ( 2p p −,2 m0 m0 ) 2 (2pm0, pm0) y = −p 0 x ( p(mm− 1) , − p) 0 —CONTINUED— Section 9.1 85. —CONTINUED— (b) x2 4x 4y x 2x 4 4 2 8 2 Conics and Calculus 183 0 4y 0 1 x 2 1 1 2. 1 . Vertex 2, 1 dy dx dy dx At 2, 5 , dy dx 2. At 3, 5 , dy dx 4 2, 5 : y 5 4 Tangent line at Tangent line at 3, Since m1m2 5 5 4 1 2 2x x 2 ⇒ 2x 4y y 1 1 0. 0. :y 1 2 3 ⇒ 2x 2 1, the lines are perpendicular. 2x 1 5 x 2 x y 1 2 0 1 2, Point of intersection: 1 x 2 5 4 1 4 Directrix: y 87. y dy dx x 1 x2 2x 0 and the point of intersection 0 lies on this line. At x1, y1 on the mountain, m y1 x1 x1 x1 x1 2 2 1 2x1. Also, m y1 x1 1 . 1 1 1 1 1 2 x1 1 1 2x1 2x1 x1 2x1 2 x12 x1 2x1 1 1 x1 0 2± 22 4 1 21 2 1 2±2 3 2 3. y 2 1± 3 Choosing the positive value for x1, we have x1 m m Thus, 1 0 x0 1 x0 3 3 1 1 23 23 3 2 1 1 3 x0 1 1 x0 3 1 1 3 23 (− 1, 1) −2 −1 1 ( x1 , y1 ) (x0, 0) 1 x 23 1 x0 23 3 x0. 1 0.155. −1 −2 The closest the receiver can be to the hill is 2 3 3 184 Chapter 9 Conics, Parametric Equations, and Polar Coordinates Circle Center: 0, k Radius: 8 y 89. Parabola Vertex: 0, 4 x2 42 p x2 y 4 4p y 4p 0 1 4y x2 4 4 −6 −4 −2 x −2 −4 −6 −8 2 4 6 4 4 x2 42 y 0 k k 2 2 64 64 48 43 64 4 3± 64 x2 (Center is on the negative y-axis.) k2 k x2 y 43 2 8 y Since the y-value is positive when x 4 0, we have y 64 x2 x2 dx 43 64 x2. A 2 0 4 x3 12 64 12 x2 4 4 3x 16 3 2 43 2 4x 2 16 16 4 1 x 64 2 2 48 64 arcsin 1 2 x 8 4 0 32 arcsin 33 3 ax2. 2 15.536 square feet 91. (a) Assume that y 20 (b) f x S a 60 y (− 60, 20) (60, 20) 20 15 10 5 ⇒a 1 x 90 2 360 1 ⇒y 180 12 x 180 12 x,f x 180 60 2 0 1 1 x 90 x2 2 dx 2 90 60 902 0 x2 x2 dx 60 −60 − 45 −30 −15 x 15 30 45 60 21 x 902 90 2 1 60 11,700 90 1 1800 13 90 20 13 10 2 13 90 ln 902 ln x 902 ln 60 902 11,700 30 13 (formula 26) 0 902 ln 90 902 ln 90 902 ln 60 60 2 3 30 13 90 13 128.4 m 9 ln 93. x2 4py, p 11 3 , , 1, , 2 42 2 95. 6 y 15 16 17 As p increases, the graph becomes wider. 1 2 3 4 7 8 9 10 11 12 13 14 5 y p= 1 4 p=1 p=2 p= p= 1 2 3 2 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 x 24 −16 −8 x 8 16 Section 9.1 5 ,b 2 5 2 2 Conics and Calculus 185 97. a 2, c 2 2 3 2 5 feet long. The tacks should be placed 1.5 feet from the center. The string should be 2a c a 2a c 99. A e P a c e y A 2 a c a P P y2 52 2yy 52 y P A P (a, 0) x P A A A 2 P P A A A 2 P P P P2 P2 101. e A A x2 102 35.34au 35.34au 0.59au 0.59au 0.9672 103. 1 0 52x 102y 8 12 x 4y 2 3 2 3 2 3 105. 16x2 32x 9y 2 96x 36y 96 36 36y 36 y 0 0 2x 102 18yy y 18y 32x 96 32x 96 18y 36 2. At 8, 3 : y y At x x 3 8 . It 25 3. 0 when x 3, y 3. y is undefined when y 2 or 6. 3, 2 , 3, 6 The equation of the tangent line is y 3 will cross the y-axis when x 0 and y Endpoints of major axis: At y 2, x 0 or 6. 8 Endpoints of minor axis: 0, Note: Equation of ellipse is x 2, 3 9 2 6, 2 y 16 2 2 1 2 107. (a) A 4 0 1 2 4 x2 dx V y y 2 x4 2 x2 4 arcsin 1 2 x 2 4x 2 2 0 or, A 2 0 ab 21 2 (b) Disk: 0 1 4 4 x2 x x2 dx 13 x 3 8 3 1 2 4 24 1 x2 x2 16 23 4x2 3x 16 16 4y 3x2 16 arcsin 3x 4 2 0 1 2 y 2 3x2 2 9 9 S 22 0 y 16 3x2 dx 4y 43 21.48 —CONTINUED— 186 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 107. —CONTINUED— 2 2 (c) Shell: V x x 1 1 2 0 x4 y2 y2 1 x2 dx 0 2x 4 x2 12 dx 2 3 2 4 x2 32 0 16 3 21 2y 1 x 2 4y 2 1 y2 3y2 dy y2 ln 3y 8 1 1 1 3y 2 y2 1 3y 2 dy 1 S 22 0 21 3y 1 y2 3y 2 1 1 0 8 23 109. From Example 5, 2 1 3y 2 0 4 6 3 3 ln 2 3 34.69 111. Area circle Area ellipse 1 e2 sin2 d 2 100 r2 ab 100 a 10 20 40. C For a C x2 25 4a 0 10 a ⇒ a y2 49 5, c 2 1, we have 49 1 25 2 6, e d c a 26 . 7 Hence, the length of the major axis is 2a 7, b 47 0 24 2 sin 49 28 1.3558 37.9614 115. 2a c 10 ⇒ a 6⇒b 5 11 113. The transverse axis is horizontal since 2, 2 and 10, 2 are the foci (see definition of hyperbola). Center: 6, 2 c 4, 2a 6, b2 c2 a2 7 Therefore, the equation is x 9 6 2 y 7 2 2 1. 117. Time for sound of bullet hitting target to reach x, y : Time for sound of rifle to reach x, y : 2c Since the times are the same, we have: vm 4c2 vm2 4c vmvs x c 2 2c vm y2 x c vs 2 y2 y ( x, y ) x c vs 2 x x c vs 2 y2 y2 x x c vs c2 vs2 2 y2 y2 (− c, 0) rifle x (c, 0) target y2 c2 vs2 c 2 x 1 x2 c2vs2 vm2 c2 vm2 y2 y2 vm2x vs2c vsvm vs2 vm2 1 1 c2 vm2 2 x vs2 y2 vs2 vm2 Section 9.1 119. The point x, y lies on the line between 0, 10 and 10, 0 . Thus, y x2 36 y2 64 1. Using substitution, we have: x2 36 16x2 7x 2 Conics and Calculus 187 10 x. The point also lies on the hyperbola 10 64 9 10 180x x x 2 1 2 576 0 180 ± 1802 4 7 27 1476 180 ± 192 2 14 90 ± 96 2 7 1476 x Choosing the positive value for x we have: x 90 7 x2 a2 x2 a2 1 x2 a2 x2 2y2 b2 1 y2 b2 a2 2y 2 b2 2y 2 b2 x2 b2 2a2 a2 2a2 96 2 6.538 and y 160 96 2 7 3.462 121. 1⇒ 1⇒ 2y 2 b2 2y 2 b2 x2 ± 1 x2 a2 1 a2 x2 2 ,c a2 b2 1 a2 b2 1 a2 b2 1⇒2 b2 ⇒x b2 ⇒ 2y 2 b2 2a a2 b2 2a2 b2 b2 2a2 b2 b2 ,± b2 2 2a2 2ac 2a2 b2 ± 2ac 2a2 b2 1 2a2c 2 a2 2a2 b2 b4 ⇒y 2 2a2 b2 ± There are four points of intersection: x2 a2 x2 a2 At b2 2ac 2a2 2y2 b2 2y 2 b2 , b2 b2 y e b2 2 2a2 b2x 2a2y b 2x 2c 2y b2 , 2ac 2a2 b2 ,± b2 2 2a2 b2 1⇒ 1⇒ b2 2 2a2 2x a2 2x c2 4yy b2 4yy b2 0 ⇒ ye 0⇒y h b2 , the slopes of the tangent lines are: 2ac 2a2 b2 b2 2 2a2 b2 2a2 2ac 2a2 b2 b2 2 2a2 b2 c a b2 and y h 2c 2 a c Since the slopes are negative reciprocals, the tangent lines are perpendicular. Similarly, the curves are perpendicular at the other three points of intersection. 123. False. See the definition of a parabola. 127. False. y2 lines. x2 125. True 129. True 2x 2y 0 yields two intersecting 188 Chapter 9 Conics, Parametric Equations, and Polar Coordinates Section 9.2 1. x (a) t x y (c) −1 Plane Curves and Parametric Equations 1 t (b) y t, y 0 0 1 2 1 1 0 2 2 1 3 3 2 4 1 2 −1 1 −1 −2 −3 2 3 x 3 3 (d) x2 −4 t 1 x2, x ≥ 0 y 3. x y y 2x 3t 2t 2 x 1 1 1 3 5 0 1 5. x y y t t2 x y 1 7. x y t3 12 2t 1 2 x y t 3 implies t 1 23 2x y x1 3 3y 4 y 1 4 x 3 2 2 x x 2 1 1 2 3 2 4 4 2 2 9. x y y y 4 3 2 1 t, t ≥ 0 t x2 2 2, x ≥ 0 11. x y y t t t x x 1 1 1 y 13. x y y 2t t x 2 y 2 2 x 2 4 8 4 4 x −1 −2 2 2 3 4 5 6 2 4 x x 4 8 12 2 15. x y y et, x > 0 e3t x3 1 1, x > 0 5 4 3 2 1 −2 −1 y x −1 1 2 3 4 Section 9.2 17. x y 0≤ xy y 1 1 x 4 2 2 4 4 2 x Plane Curves and Parametric Equations 21. x y x 16 y2 4 x2 16 y2 4 2 189 sec cos < , < 22 ≤ 19. x 3 cos , y 3 sin 4 sin 2 2 cos 2 sin2 2 cos2 2 1 4 Squaring both equations and adding, we have x2 y2 y 9. x ≥ 1, y ≤ 1 y 2 4 3 2 1 x −6 6 1 2 3 2 3 −4 23. x y x 4 y 1 x 4 2 4 1 cos2 2 cos sin 25. x y x 4 y 16 x 4 3 4 1 cos2 2 cos 4 sin 4 1 1 1 2 4 1 1 16 2 2 2 sin2 2 sin2 2 4 2 y 1 4 2 y 1 −1 8 −2 10 −4 −5 27. x y x2 16 y2 9 x2 16 y2 9 4 sec 3 tan sec2 tan2 29. x y y t3 3 ln t 3 ln 3 x 2 31. x y e e 3t t ln x et et 5 1 x 3 y −1 3 y y 1 x 1 x3 1 −2 6 x>0 y>0 9 3 −9 −6 −1 −1 5 190 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 2x 1. They differ from each other in orientation and in restricted (b) x cos 1≤x≤1 dx d dy d y x 1 2 3 2 1 x 1 −1 2 33. By eliminating the parameters in (a) – (d), we get y domains. These curves are all smooth except for (b). (a) x t, y y 3 2 1 2t 1 y 2 cos 1 1≤y≤3 0, ± , ± 2 , . . . . 0 when −2 −1 −1 −2 −1 (c) x e t y 2e t 1 (d) x et y 2et 1 x>0 y 4 3 2 1 y>1 x>0 y 4 3 2 1 x y>1 −1 1 2 3 −1 x 1 2 3 35. The curves are identical on 0 < 37. (a) −6 4 < . They are both smooth. Represent y 4 21 x2 (b) The orientation of the second curve is reversed. (c) The orientation will be reversed. 6 6 −6 (d) Many answers possible. For example, x y 1 2t, and x 1 t, x 1 2t. −4 −4 1 t, 39. x y x x2 x1 x1 y y y y1 y1 x1 y1 t y1 y2 x2 mx t x2 t y2 x1 y1 41. x y x a h k b x a 2 h k cos sin 1 a cos b sin x x2 y1 x x1 x1 x1 y x1 2 x1 y1 h 2 y y b 2 k 2 43. From Exercise 39 we have x y 5t 2t. 45. From Exercise 40 we have x y 2 1 4 cos 4 sin . 47. From Exercise 41 we have a 5, c x y 4⇒b 5 cos 3 sin . 3 Solution not unique Solution not unique Center: 0, 0 Solution not unique Section 9.2 49. From Exercise 42 we have a 4, c x y 5⇒b 4 sec 3 tan . 3 51. y 3x 2 Plane Curves and Parametric Equations x3 191 53. y Example x x t, t 3, y y 3t 3t 2 11 Example x x x t, 3 y t, y y t3 t tan3 t Center: 0, 0 Solution not unique 55. x y 2 21 5 tan t, sin cos 57. x y 1 5 3 2 3 2 sin cos 59. x y 3 cos3 3 sin3 4 −6 −2 −1 16 −2 −1 7 6 −4 Not smooth at 2n Not smooth at x, y 1 0, ± 3 , or 2n . 63. See definition on page 665. ± 3, 0 and 61. x y 2 cot 2 sin 2 −6 4 6 −4 Smooth everywhere 65. A plane curve C, represented by x f t , y g t , is smooth if f and g are continuous and not simultaneously 0. See page 670. 67. x y 4 cos 2 sin 2 69. x y cos sin sin cos Matches (d) 71. When the circle has rolled sin cos Therefore, x 73. False x x t2 ⇒ x ≥ 0 t2 ⇒ y ≥ 0 sin 180 cos 180 a b sin Matches (b) radians, we know that the center is at a , a . C b AP b and y a BD b or or AP b cos . y BD b sin P b b cos A C θa D x B The graph of the parametric equations is only a portion of the line y x. 192 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 100 5280 3600 t t t 440 cos 3 16t 2 16t 2 440 ft sec 3 t 75. (a) 100 mi hr x y v0 cos h 3 (b) 30 (d) We need to find the angle x y 3 440 cos 3 t (and time t) such that 400 t 16t 2 10. v0 sin 440 sin 3 440 sin 3 From the first equation t into the second equation, 10 7 3 440 sin 3 16 16 1200 440 cos . Substituting 1200 440 cos 120 44 120 44 2 16 1200 440 cos 2 400 tan 400 tan sec2 2 0 0 400 It is not a home run—when x (c) 60 400, y ≤ 20. tan2 1. We now solve the quadratic for tan : 16 0 0 400 120 44 2 tan2 400 tan 19.4 7 16 120 44 2 0 tan 400, y > 10. 0.35185 ⇒ Yes, it’s a home run when x Section 9.3 1. dy dx dy dt dx dt Parametric Equations and Calculus 4 2t 2 t 3. dy dx dy dt dx dt y 2 cos t sin t 2 sin t cos t 1⇒y 1 1 x and dy dt 1 Note: x 5. x 2t, y dy dx d 2y dx2 3t d y dt dx dt 0 Line 1 3 2 7. x t dy dx d 2y dx2 1, y 2t 1 t2 3 3t 1 when t 1. 2 concave upwards 9. x 2 cos , y dy dx d 2y dx2 2 cos 2 sin 2 sin cot csc3 2 1 when 2 when 4 . . 11. x 2 dy dx sec , y 2 sec2 sec tan 2 sec tan 1 2 tan csc2 2 sin 4 2 csc 4 when 6 . concave downward d 2y dx2 2 csc cot sec tan 2 cot3 6 3 when 6 . concave downward Section 9.3 13. x cos3 , y dy dx sin3 15. x Parametric Equations and Calculus 2 sin2 2 sin3 2 dy , and 3 dx y 3 3x 8y dy dx 0 , and y 3x 8y t3 dy dx 1 2 10 3t, t 0 1 3 . 8 3 x 8 23 3 2 18 0. cos 33 . 8 33 x 8 0 2 3 193 2 cot , y dy dx 3 sin2 cos 3 cos2 sin tan 1 when 4 . At 4 sin cos 2 csc2 23 ,, 32 d 2y dx2 sec2 3 cos 2 sin sec4 3 csc 1 3 cos4 sin 42 when 3 4 . Tangent line: concave upward At 0, 2 , Tangent line: y At 2 3, 1 , 2 2 , and 2 6 Tangent line: 17. x (a) 2t, y t2 10 1, t 2 19. x (a) t2 5 t 2, y −6 −4 6 −1 8 −3 (b) At t dx dt (c) dy dx 2, x, y 2, dy dt 4, 4, 3 , and dy dx 2 3 y 2x 2x 4 5 (b) At t dx dt (c) dy dx 5 1, x, y 3, dy dt 0, 4, 2 , and dy dx 2 y 0 0x 2 4 2. At 4, 3 , y 0. At 4, 2 , y (d) 10 (d) (4, 3) (4, 2) 5 −1 8 −5 −4 −3 21. x 2 sin 2t, y 3 sin t crosses itself at the origin, x, y 0 or t . 0, 0 . At this point, t dy dx At t At t 0: , 3 cos t 4 cos 2t dy dx dy dx 3 and y 4 3 and y 4 3 x. Tangent Line 4 3 x Tangent Line 4 194 23. x Chapter 9 cos Conics, Parametric Equations, and Polar Coordinates sin , y dy d 1 sin sin , 1, 2n 1, cos 2n 2 ,1 , 1 , 3 , 2 , 1, cos 0 when 0, , 2 , 3 ,. . .. Horizontal tangents: Points: 1, 2n where n is an integer. 2 2 , 35 , ,. . .. 22 Points shown: 1, 0 , Vertical tangents: Points: 1 dx d n 1 0 when 1 n 1 Points shown: 2 1, 5 ,1 2 27. x 1 t, y t3 3t 3 0 when t ± 1. 25. x 1 t, y t2 dy dt dx dt 2t 0 when t 0. Horizontal tangents: Point: 1, 0 Vertical tangents: 3 Horizontal tangents: Points: 0, dy 3t 2 dt 2 , 2, 2 dx dt 1 1 0; none Vertical tangents: 3 0; none (2, 2) −4 −2 −1 4 5 (1, 0) −3 (0, − 2) 29. x 3 cos , y 3 sin dy d 3 dx d 3, 0 3 sin 0 when 0, . 3 cos 0 when 3 ,. 22 31. x 4 2 cos , y dy d 1 sin cos 0 when 3 ,. 22 Horizontal tangents: Points: 0, 3 , 0, Vertical tangents: Points: 3, 0 , 4 Horizontal tangents: Points: 4, 0 , 4, Vertical tangents: Points: 6, 2 2 dx d 2 sin 1 0 when x 0, . 1 , 2, (0, 3) (4, 0) 0 −6 9 6 (− 3, 0) (3, 0) (0, − 3) −4 (2, − 1) (4, − 2) −4 (6, − 1) 33. x sec , y tan dy d dx d 1, 0 sec2 sec tan 0; none 0 when x 0, . 35. x dx dt t 2, y 2t, s dy dt 2t, 0 ≤ t ≤ 2 2, 2 Horizontal tangents: Vertical tangents: Points: 1, 0 , 4 dx dt 2 dy dt 2 4t 2 4 4 t2 1 2 0 t2 t t2 1 1 dt 2 ln t 5 t2 1 0 −6 (− 1, 0) (1, 0) 6 25 ln 2 5.916 −4 S ection 9.3 Parametric Equations and Calculus dx dt 9 dt u2 du 6 195 37. x dx dt e t cos t, y e t e t sin t, 0 ≤ t ≤ dy dt dy dt e 2 t 2 sin t 39. x S t, y 1 3t 1 4t 6 1, 1 dy , 2 t dt 1 2 1 0 3 36t t dt sin t 2 cos t , dx dt 2 cos t 1 0 s 0 2 dt 2 1 6 e t 1 0 2e 0 2t dt 2 2 0 1 dt 1.12 u 1 ln 12 1 ln 12 6 t, du 1 37 3 t dt u2 6 u u1 u2 0 2e t 0 21 e 2 6 37 3.249 41. x dy d a cos3 , y 3a sin2 2 a sin3 , dx d 3a cos2 sin , 43. x dx d a a1 S 2 sin cos ,y , dy d a1 cos , cos 9a2 cos4 sin2 cos2 9a2 sin4 cos2 d a sin cos 2 S 4 0 a2 1 0 a2 sin2 d 2 12a 0 2 sin cos sin 2 d sin2 d 2 2 2a 0 1 sin 0 cos d d 8a 0 6a 0 3a cos 2 0 6a 2 2a 1 cos cos 4 2a 1 45. x (a) 90 cos 30 t, y 35 90 sin 30 t 16t 2 (d) y x 0 ⇒ 90 sin 90 cos t t 16t2 ⇒ t 90 sin 16 902 sin 2 32 45 45 4 902 cos sin 16 0⇒ 0 0 240 x 902 2 cos 2 32 (b) Range: 219.2 ft (c) dx dt y s 0 By the First Derivative Test, maximizes the range. dy dt dx dt dy dt 2 90 cos 30 , 0 for t 45 16 90 sin 30 32t. 90 cos , 90 sin 90 16 sin 45 . 16 90 cos 30 90 sin 30 32t 2 dt 32 90 sin 90 cos 2 32 90 sin 16 90 sin 2 90 sin dt s 0 90 16 sin 230.8 ft 0 90 16 sin 90 dt 90 sin 16 ds d 902 cos 16 0⇒ 2 90t 0 2 90 maximizes the By the First Derivative Test, arc length. 196 47. (a) Chapter 9 x y t 1 Conics, Parametric Equations, and Polar Coordinates sin t cos t x y 2t 1 sin 2t cos 2t (b) The average speed of the particle on the second path is twice the average speed of a particle on the first path. (c) x y 1 2t 0≤t≤2 3 0≤t≤ 3 sin 1 t 2 1 cos 2 t 1 − 3 − 3 The time required for the particle to traverse the same path is t 4 . −1 −1 49. x t, y 2 2t, 4 dx dt 2t 1 1, dy dt 4 dt 32 2 4 51. x t dt 0 4 cos , y 2 4 sin , 4 cos dx d 4 sin 4 sin , 2 dy d 4 cos 2 4 cos 2 (a) S 45 5 4 S 2 0 d 0 4 2 2 5 t2 0 4 32 0 cos d 32 sin 0 32 (b) S 2 0 t1 4 4 dt 16 25 0 t dt 5 t2 0 5 53. x a cos3 , y 2 a sin3 , a sin3 dx d 3a cos2 sin , dy d 3a sin2 d cos 2 S 4 0 9a2 cos4 sin2 9a2 sin4 cos2 12a2 0 sin4 cos d 12 a2 sin5 5 b 2 0 12 2 a 5 2 55. dy dx dy dt dx dt 57. One possible answer is the graph given by x t, y y 4 2 x 2 −2 −4 4 59. s a dx dt 2 dy dt dt See Theorem 9.7, page 675. t. See Theorem 9.8, page 678. −4 −2 61. x S r cos , y 2 0 r sin y r sin sin 0 r 2 sin2 d r 2 cos2 d θ 2 r2 x 2 r 2 cos 0 2 r2 1 cos Section 9.3 63. x A 0 Parametric Equations and Calculus 197 t, y 4 4 4 t 4 0 4 t, 0 ≤ t ≤ 4 1 2t t t 2 dt t 1 1 2 4 4t 0 12 t1 4 2 dt t dt 8t1 2 1 8t 2 3 4t 32 t3 2 2 tt 3 t2 2 4 0 4 0 16 3 3 4 16 tt 3 22 tt 5 4 0 x y x, y 3 16 3 32 4 2t 1 dt dt 3 64 3 32 4 4 0 12 4 0 2t 16t 0 dt 3 32 t 64 8 5 38 , 45 dx d 3 sin d cos2 cos3 3 sin 0 65. x 3 cos , y 0 3 sin , 3 sin 2 0 2 3 sin d V 2 54 sin3 2 0 54 2 1 cos d 36 2 54 67. x y dx d A 2 sin2 2 sin2 tan π 0≤θ< 2 −2 −1 −1 −2 2 1 y 4 sin cos 2 2 x 1 2 sin2 tan 0 4 sin cos 3 sin cos 8 d 3 8 8 0 2 0 sin4 3 2 d 8 sin3 4 cos 69. ab is area of ellipse (d). t2 ,y t2 71. 6 a2 is area of cardioid (f). 73. 8 ab is area of hourglass (a). 3 2 75. (a) x 1 1 2t 1 t2 , 20 ≤ t ≤ 20 y2 t2 t2 2 The graph is the circle x 2 Verify: x2 y2 1 1 1, except the point 2t 1 t 2 2 1, 0 . t 4 4t 2 t2 2 1 1 t2 t2 2 2 −3 3 1 2t 2 1 1 −2 (b) As t increases from 77. False d 2y dx2 dgt dt f t ft 20 to 0, the speed increases, and as t increases from 0 to 20, the speed decreases. f tg t ft g tf t 3 198 Chapter 9 Conics, Parametric Equations, and Polar Coordinates Section 9.4 1. x y x, y 4, Polar Coordinates and Polar Graphs 3. 0 4 x y x, y 4, 5. x 2 23 2, 2.36 2 cos 2.36 2 sin 2.36 1.004 0.996 2 4 cos 4 sin 2 2 0, 4 π 2 3 4 cos 4 sin 3 3 2, 2 3 π 2 y x, y 1.004, 0.996 π 2 ( 2, 2.36 ) (4, 36π ) (− 4, − π ) 3 0 3 1 0 1 1 2 0 7. r, x, y 5, 3 4 9. r, x, y y y 1 3.5, 2.5 2.804, 2.095 11. x, y r tan ± 1, 1 2 1 5 ,, 44 2, 4 , 2, 5 4 3.5355, 3.5355 4 (− 3.54, 3.54) 3 2 1 −1 −1 −2 x 1 −3 x 1 2 3 y 2 (2.804, − 2.095) (1, 1) 1 −4 −3 −2 −1 −1 x 1 2 13. x, y r tan ± 3, 4 9 4 3 y 16 ±5 (− 3, 4) 5 4 3 2.214, 5.356, 5, 2.214 , 5, 5.356 2 1 −4 −3 −2 −1 x 1 15. x, y r, 19. (a) x, y 3, 2 0.588 y 4 3 2 17. x, y r, (b) r, (4, 3.5) 54 2, 3 3.606, 2.833, 0.490 4, 3.5 π 2 4, 3.5 0 1 (4, 3.5) 1 x 1 2 3 4 S ection 9.4 21. x2 y2 r a2 a 0 a Polar Coordinates and Polar Graphs y 4 4 r 4 csc 0 2 4 199 π 2 23. r sin π 2 25. 3r cos 3x r sin y 2 2 0 0 2 27. r sin 2 y2 2 9x 9r cos r 3 cos sin r r 2 sin π 2 9 cos sin2 9 csc2 cos 3 cos r π 2 0 0 1 2 1 2 3 4 5 6 7 29. r r2 x2 y2 y 3 9 9 31. r r2 x2 x2 y y2 y sin r sin y 2 33. r tan r tan x2 y2 y2 y tan y x arctan y x y2 1 2 y 1 4 0 2 1 x x2 x2 1 2 2 1 1 2 2π π 1 2 π −π x 2π x 1 2 1 2 − 2π 35. r cos r 3 sec 3 3 y 37. r 0≤ 3 4 cos <2 − 12 6 x x 3 3 0 6 2 −6 1 x 1 2 39. r 0≤ 2 sin <2 −4 4 41. r 1 2 cos − 10 5 5 Traced out once on << 5 −5 −2 200 Chapter 9 3 2 Conics, Parametric Equations, and Polar Coordinates 43. r 0≤ 2 cos 2 45. r2 3 4 sin 2 < 2 −3 2 <4 −3 0≤ 3 −2 −2 47. r r2 r2 x2 x x 2 2 2 h cos 2r h cos 2 h r cos 2 hx 0 0 h2 h2 k2 k2 ky k sin k sin k r sin Radius: h2 k2 Center: h, k y2 2ky k k 2 2 y 2 2 2 2hx 2ky y 2hx h 2 y x h 49. 4, 2 , 2, 3 6 d 42 20 22 2 2 4 2 cos 3 2 25 4.5 6 51. 2, 0.5 , 7, 1.2 d 22 53 72 2 2 7 cos 0.5 0.7 5.6 1.2 28 cos 16 cos 53. r dy dx 2 3 sin cos sin 2 2 3 sin 3 sin 2 cos 6 cos2 3 sin 2 sin 1 3 55. (a), (b) r 31 cos 4 3 cos sin 3 cos cos 2 cos 3 sin 1 3 cos 2 2 sin dy 0. 2 dx dy 2 At 2, , . dx 3 3 dy 1, , 0. At 2 dx At 5, , −8 4 −4 r, 3, 2 ⇒ x, y 3 y 1x x 0, 3 0 3 Tangent line: y dy , 2 dx (c) At 1.0. 57. (a), (b) r 3 sin 5 −4 −1 5 r, 33 , 23 9 4 y ⇒ x, y 3x 3x 3 3 39 , 44 33 4 9 2 Tangent line: y (c) At dy , 3 dx 1.732. S ection 9.4 59. r dy d 1 1 cos cos sin sin 1 cos 2 sin 1 ⇒ 2 2, cos sin 0 3 5 , ,, 2266 61. r dy d Polar Coordinates and Polar Graphs 2 csc 2 csc 3 cos 3 , 22 3 3 cos 0 2 csc cot sin 201 0, sin Horizontal tangents: dx d 1 sin 2 sin2 2 sin sin 1, sin sin 3 1 15 ,,,, 2 26 26 cos cos sin2 1 1 0 1 Horizontal: 5, 2 , 1, 3 2 sin sin2 sin 1 sin 1 ⇒ 2 7 11 ,, 26 6 Vertical tangents: 37 3 11 , ,, 26 26 65. r 2 csc 10 63. r 4 sin cos2 2 5 −3 3 −12 12 −2 −6 Horizontal tangents: 0, 0 , 1.4142, 0.7854 , 1.4142, 2.3562 67. r r2 x2 x 2 Horizontal tangents: 7, 2 , 3, 3 2 π 2 3 sin 3r sin 3y 2 π 2 69. r 21 sin Cardioid 0 1 2 3 y2 3 2 3 2 3 2 Symmetric to y-axis, 0 1 2 3 y 9 4 2 Circle r Center: 0, Tangent at the pole: 71. r 2 cos 3 0 Rose curve with three petals Symmetric to the polar axis Relative extrema: 2, 0 , 0 r 2 2, 3 2 0 , 2, 2 3 π 2 0 2 6 0 4 2 3 2 2 3 2 5 6 0 2 Tangents at the pole: 5 ,, 62 6 202 73. r Chapter 9 3 sin 2 Conics, Parametric Equations, and Polar Coordinates Rose curve with four petals Symmetric to the polar axis, Relative extrema: ± 3, Tangents at the pole: ,3 75. r 5 , ± 3, 0, 2 5 4 , and pole π 2 4 0 3 2 2 give the same tangents. π 2 77. r 41 cos π 2 Circle radius: 5 x2 y2 25 0 2 4 6 Cardioid 0 2 4 6 10 79. r 3 2 cos π 2 81. r sin r 3 csc 3 π 2 Limaçon Symmetric to polar axis 0 r 1 2 3 4 5 0 2 y 3 0 1 2 Horizontal line 3 2 2 3 83. r 2 π 2 Spiral of Archimedes Symmetric to 2 3 4 3 2 0 2 5 4 5 2 3 2 3 0 1 2 3 0 r 0 4 2 2 Tangent at the pole: 85. r2 4 cos 2 Lemniscate Symmetric to the polar axis, Relative extrema: ± 2, 0 2 , and pole π 2 0 1 0 r ±2 ± 6 2 4 0 3 , 44 Tangents at the pole: Section 9.4 2 Polar Coordinates and Polar Graphs 203 87. Since r 2 sec 2 1 , cos 89. r Hyperbolic spiral r⇒ r y −6 6 the graph has polar axis symmetry and the lengths at the pole are 3 , 3 . x = −1 4 as 2 ⇒ ⇒0 2 r 2 sin r sin 2 sin y 2 sin 2 sin 3 Furthermore, r⇒ r⇒ Also, r rx r as 2 2x 2x 1 x as ⇒ ⇒ 1 cos r . 1. 2 2 . 2 lim −4 →0 lim →0 2 cos 1 2 y=2 r r cos 2 r x −3 −1 3 Thus, r ⇒ ± as x ⇒ 91. The rectangular coordinate system consists of all points of the form x, y where x is the directed distance from the y-axis to the point, and y is the directed distance from the x-axis to the point. Every point has a unique representation. The polar coordinate system uses r, to designate the location of a point. is the angle the point makes with the positive x-axis, r is the directed distance to the origin and measured clockwise. Point do not have a unique polar representation. 93. r a circle b line 95. r 2 sin circle 97. r 31 cos Matches (c) Cardioid Matches (a) 99. r 4 sin ≤ (b) 2 π 2 (a) 0 ≤ 2 ≤ ≤ π 2 (c) 2 ≤ ≤ π 2 2 0 1 2 0 1 2 1 2 0 204 Chapter 9 Conics, Parametric Equations, and Polar Coordinates be rotated by to form the curve r is on r g . That is, f , or f 1 1 1 101. Let the curve r f r f , then r1, g 1 1 g . If r1, 1 is a point on π 2 1 r1 . , we see that f . φ θ (r, θ +φ ) Letting g g (r, θ ) 1 0 103. r 2 sin 2 sin 4 (a) r 4 2 2 sin 2 cos (b) r 2 cos 4 2 cos −6 6 −6 6 −4 −4 (c) r 2 sin 4 2 sin (d) r 2 cos 4 −6 6 −6 6 −4 −4 105. (a) r 1 sin π 2 (b) r 1 sin 4 π 2 Rotate the graph of 0 1 2 r 1 sin 1 2 0 through the angle 4. 107. tan At r dr d , tan 3 2 1 cos 2 sin is undefined ⇒ 2 . 109. tan At r dr d 6 , tan 2 2 cos 3 6 sin 3 0⇒ 0. −6 3 −3 3 −3 −2 Section 9.5 6 cos 1 dr d 6 sin cos Area and Arc Length in Polar Coordinates 205 111. r 1 r dr d 61 cos 1 ⇒ 1 2 7 ψ θ tan 6 cos 6 sin 1 cos 1 1 2 cos sin −8 −3 7 At 2 , tan 3 1 2 3 2 3. 3 113. True , 60 115. True Section 9.5 1. (a) r 8 sin π 2 Area and Arc Length in Polar Coordinates (b) A 2 64 0 2 1 2 2 2 8 sin 0 2 d sin2 d 1 0 32 32 cos 2 2 d 16 0 2 4 sin 2 2 0 A 1 2 4 2 16 2d 6 3. A 2 2 cos 3 0 2 1 sin 6 6 6 0 3 5. A 2 1 2 1 2 4 cos 2 0 2 d 1 sin 4 4 1 2 3 4 0 8 7. A 2 1 2 3 2 2 1 2 sin 2 d 2 2 9. A 3 2 2 1 2 3 2 cos sin 2 2 d 2 2 3 2 cos 1 sin 2 4 4 sin 33 2 2 −1 4 −2 11. The area inside the outer loop is 2 1 2 2 0 3 2 1 2 cos 2 2 3 d 3 4 sin sin 2 0 4 2 33 . −1 4 From the result of Exercise 9, the area between the loops is −2 A 4 2 33 2 2 33 3 3. 206 Chapter 9 Conics, Parametric Equations, and Polar Coordinates 13. r r 1 1 cos cos 15. r r 1 1 cos sin Solving simultaneously, 1 cos 2 cos 1 0 3 ,. 22 Replacing r by r and by in the first equation 1 cos , cos 1, 0. and solving, 1 cos Both curves pass through the pole, 0, , and 0, 0 , respectively. Points of intersection: 1, 2 , 1, 3 , 0, 0 2 cos Solving simultaneously, 1 cos cos tan 1 sin sin 1 37 ,. 44 Replacing r by r and by in the first equation 1 sin , sin cos 2, and solving, 1 cos which has no solution. Both curves pass through the pole, 0, , and 0, 2 , respectively. Points of intersection: 2 2 23 2 27 , , , , 0, 0 4 2 4 17. r r 4 3 sin 5 sin 19. r r 3 sin 1 2 5 ,. 66 2 2 π 2 Solving simultaneously, 4 5 sin sin Solving simultaneously, we have 0 2 2, 4. 1 Points of intersection: 2, 4 , 2, 4 Both curves pass through the pole, 0, arcsin 4 5 , and 0, 0 , respectively. Points of intersection: 3 35 ,,, , 0, 0 26 26 π 2 21. r r r 4 sin 2 2 4 sin 2 is the equation of a rose curve with four petals and is symmetric to the polar axis, 2, and the pole. Also, r 2 is the equation of a circle of radius 2 centered at the pole. Solving simultaneously, 4 sin 2 2 2 5 , 66 5 ,. 12 12 Therefore, the points of intersection for one petal are 2, 12 and 2, 5 12 . By symmetry, the other points of intersection are 2, 7 12 , 2, 11 12 , 2, 13 12 , 2, 17 12 , 2, 19 12 , and 2, 23 12 . 0 1 3 Section 9.5 23. r r 2 sec 2 −4 8 Area and Arc Length in Polar Coordinates cos 2 3 sin 207 3 cos 4 r = sec θ 2 25. r r Points of intersection: 0, 0 , 0.935, 0.363 , 0.535, 1.006 −4 r = 2 + 3 cos θ The graphs reach the pole at different times ( values). r = cos θ 5 The graph of r 2 3 cos is a limaçon with an inner loop b > a and is symmetric to the polar axis. The graph of r sec 2 is the vertical line x 1 2. Therefore, there are four points of intersection. Solving simultaneously, 2 6 cos2 3 cos 1 cos sec 2 0 2± 6 arccos arccos Points of intersection: 10 2 6 2 6 10 10 1 −4 −5 r = 2 − 3 sin θ 4 cos 1.376 2.6068. 0.581, ± 2.607 , 2.581, ± 1.376 12 and 2, 5 2 27. From Exercise 21, the points of intersection for one petal are 2, 1 2 16 0 12 12 . The area within one petal is 4 4 sin 2 0 12 2 d 2 1 2 5 5 12 2d 12 12 2 1 2 2 4 sin 2 5 12 d −6 sin2 2 1 sin 4 4 4 4 3 d 12 d (by symmetry of the petal) 12 5 12 12 6 8 Total area 2 0 4 3 4 4 3 3. 33 −4 3 16 3 43 29. A 4 1 2 2 3 0 2 sin 2 6 d 2 −9 9 2 11 12 cos sin 2 0 11 24 −6 31. A 2 1 2 1 2 6 4 sin 0 2 d 6 1 2 4 2 2d 6 2 6 2 33. A 2 a2 1 2 a1 0 cos 2 d a2 4 a2 4 16 8 3 1 sin 2 4 23 5 0 3 2 2 sin a2 4 5a 2 4 sin 2 4 0 2 4 3 33 3a 2 2 −6 6 −3 208 Chapter 9 a2 8 a2 8 a2 8 a2 8 1 2 a2 2 a2 3 22 a2 2 a cos2 ar2 cos2 y2 32 Conics, Parametric Equations, and Polar Coordinates 35. A a1 2 cos 2 cos 2 d cos 2 2 d π 2 2 3 2 0 2 sin 3 4 2 sin 2 4 a2 2 a 2a 2 3 2 2 37. (a) r r3 x2 (b) a=4 −6 4 a=6 6 ax 2 −4 (c) A 4 1 2 2 2 2 6 cos2 0 2 2 4 cos2 1 2 d 40 0 cos4 d 3 2 10 0 1 2 0 cos 2 15 2 2 d 10 0 1 2 cos 2 cos 4 2 d 10 sin 2 1 sin 4 8 39. r a cos n 1: a cos a 2 π 2 For n r A For n r 2 2: a cos 2 8 π 2 a2 4 A 1 2 4 a cos 2 0 2 d a2 2 a 0 a 0 For n r A 3: a cos 3 6 π 2 For n r 6 4: a cos 4 16 π 2 1 2 a cos 3 0 2 d a2 4 A 1 2 8 a cos 4 0 2 d a2 2 a 0 a 0 In general, the area of the region enclosed by r a 2 2 if n is even. a cos n for n 1, 2, 3, . . . is a 2 4 if n is odd and is Section 9.5 41. r r s 0 Area and Arc Length in Polar Coordinates 1 cos 3 2 209 a 0 2 43. r r a2 02 d 2 sin a 0 2a s 2 2 3 2 1 1 2 3 2 2 sin 2 cos 2 d (circumference of circle of radius a) 22 22 42 42 1 1 2 sin d d 2 2 cos 1 sin 3 sin 0 8 45. r 2 ,0 ≤ 4 ≤ 2 47. r , ≤ 0.5 ≤2 49. r sin 3 cos 1 ,0 ≤ ≤ − 0.5 −1 −1 2 − 0.5 −1 0.5 2 −1 Length 4.16 Length 0.71 Length 4.39 51. r r S 6 cos 6 sin 2 53. r r 6 cos sin 36 cos2 36 sin2 d S ea aea 2 2 0 2 0 ea cos 2 ea 2 aea 2 d 2 72 0 sin cos d 2 2 2 2 1 1 a2 0 e2a cos d e2a 2a cos 4a2 1 e a 2 36 sin2 0 a2 a2 sin 0 36 1 4a2 1 2a 55. r r S 4 cos 2 8 sin 2 4 2 0 4 cos 2 sin 4 16 cos2 2 cos2 2 64 sin2 2 d 21.87 32 0 cos 2 sin 4 sin2 2 d 57. Area Arc length 1 2 f f 2 2 d f 1 2 2 r2 d d r2 dr d 2 59. (a) is correct: s d 33.124. 210 Chapter 9 Conics, Parametric Equations, and Polar Coordinates π 2 2a a 61. Revolve r f f S a about the line r a 0 2 b sec where b > a > 0.` 2 0 b a cos 2 a2 02 d a b 0 2ab 2 a2 b a sin 0 4 2ab 63. False. f 1 and g 1 have the same graphs. 65. In parametric form, b s a dx dt 2 dy dt 2 dt. r cos and dy d f f cos sin and y f r sin cos . f sin . Thus, Using dx d instead of t, we have x f cos f sin It follows that dx d 2 dy d 2 f f 2 2 f f 2 2. Therefore, s d Section 9.6 1. r 2e e cos 1, r 0.5, r 1.5, r 1 Polar Equations of Conics and Kepler’s Laws 3. r 2 , parabola cos 1 1 1 0.5 cos 3 1.5 cos 2 2 2 , ellipse cos 6 , hyperbola 3 cos 2e e sin 1, r 0.5, r 1.5, r e = 1.0 e = 1.5 8 1 1 (a) e (b) e (c) e (a) e (b) e (c) e 1 1 1 4 2 , parabola sin 1 0.5 sin 3 1.5 sin 2 2 2 , ellipse sin 6 , hyperbola 3 sin 4 e = 0.5 e = 0.5 −4 −9 9 e = 1.5 e = 1.0 −4 −8 S ection 9.6 4 e sin 5 − 30 Polar Equations of Conics and Kepler’s Laws 211 5. r (a) 1 e = 0.1 30 (b) − 30 5 30 e = 0.25 e = 0.5 e = 0.75 e = 0.9 − 40 − 40 The conic is an ellipse. As e → 1 , the ellipse becomes more elliptical, and as e → 0 , it becomes more circular. (c) 80 The conic is a parabola. e = 1.1 e=1 e=2 − 90 90 − 40 The conic is a hyperbola. As e → 1 , the hyperbolas opens more slowly, and as e → 7. Parabola; Matches (c) 1 sin 1 1 9. Hyperbola; Matches (a) 6 cos 3 1 2 cos 1 <1 2 , they open more rapidly. 11. Ellipse; Matches (b) 13. r 1 15. r 2 17. r 2 r 2 1 sin 4 sin 4 Parabola since e Vertex: 13 , 22 π 2 Ellipse since e 2 1 2 sin 1 <1 2 Vertices: 2, 0 , 6, 0 1 2 Ellipse since e Vertices: π 2 3 4 , , 4, 32 2 π 2 0 1 3 0 1 3 4 19. r 1 5 2 cos 1 5 2 cos 21. r 2 3 6 sin 1 32 3 sin 3>1 Hyperbola since e Vertices: 5, 0 , π 2 2>1 5 , 3 Hyperbola since e Vertices: 3 ,, 82 π 2 33 , 42 0 4 6 8 1 0 212 23. −2 Chapter 9 Conics, Parametric Equations, and Polar Coordinates Ellipse 2 −2 1 1 25. 2 Parabola −2 −2 27. r 1 sin 1 4 29. r 2 6 cos 6 Rotate the graph of r 1 1 sin Rotate the graph of r 2 6 cos counterclockwise through the angle . 4 2 −2 10 clockwise through the angle . 6 5 −8 4 −6 −3 31. Change to 4 :r 5 5 3 cos 4 . 33. Parabola e r 1, x 1 1, d ed e cos 1 1 1 cos 35. Ellipse e r 1 ,y 2 1 1 2 41. Ellipse Vertices: 2, 0 , 8, e r 3 ,d 5 1 1 5 16 3 ed e cos 16 5 3 5 cos 16 3 cos 1, d ed e sin 12 1 2 sin 1 sin 1 37. Hyperbola e r 2, x 1 1, d ed e cos 1 1 2 2 cos 39. Parabola Vertex: 1, e 1, d 2 2, r 1 2 sin 43. Hyperbola Vertices: 1, e r 5 ,d 4 1 1 4 3 3 , 9, 2 2 45. Ellipse if 0 < e < 1, parabola if e 1, hyperbola if e > 1. 9 5 ed e sin 94 5 4 sin 9 5 sin S ection 9.6 47. (a) Hyperbola e (b) Ellipse e (c) Parabola e 2>1 1 <1 2 1 3 5 3 Polar Equations of Conics and Kepler ’s Laws 49. a r2 5, c 4, e 4 ,b 5 3 213 1 9 16 25 cos 2 (d) Rotated hyperbola e 51. a r2 3, b 4, c 5, e 53. A 2 9 1 2 0 2 3 cos 1 cos 2 2 d d 10.88 1 16 25 9 cos 2 0 2 55. Vertices: 126,000, 0 , 4119, a r 126,000 2 1 ed e cos 60 , r 4119 65,059.5, c 65,059.5 4119 60,940.5, e c a 40,627 ,d 43,373 4119 84,000 40,627 1 4119 84,000 43,373 40,627 43,373 cos 15,004.49. 345,996,000 43,373 40,627 cos When 345,996,000 23,059.5 Distance between the surface of the earth and the satellite is r 57. a r 92.957 1 1 106 mi, e 0.0167 4000 59. a r 11,004.49 miles. 5.900 1 1 109 km, e 1 0.2481 e2 a e cos 92,931,075.2223 1 0.0167 cos e e 91,404,618 mi 94,509,382 mi e2 a e cos 5.537 10 9 0.2481 cos e e 4.436 7.364 10 9 km 10 9 km Perihelion distance: a 1 Aphelion distance: a 1 5.537 109 1 0.2481 cos 1 2 9 0 9 0 2 0 Perihelion distance: a 1 Aphelion distance: a 1 61. r (a) A 5.537 109 1 0.2481 cos 5.537 109 0.2481 cos 5.537 109 1 0.2481 cos 1 2 2 d 2 9.341 1018 km2 1 2 248 1 2 (b) 1 2 d 2 21.867 yr d 5.537 109 1 0.2481 cos 0.8995 rad d 9.341 1018 In part (a) the ray swept through a smaller angle to generate the same area since the length of the ray is longer than in part (b). (c) r s 0 5.537 10 9 0.2481 sin 1 0.2481 cos 2 9 5.537 109 1 0.2481 cos 1.17 2 1.3737297 109 sin 1 0.2481 cos 2 2 d 2.559 109 km 2.559 109 km 21.867 yr 0.899 108 km yr 2 s 4.119 109 km 21.867 yr 1 5.537 109 0.2481 cos 108 km yr 1.3737297 109 sin 1 0.2481 cos 2 2 d 4.119 109 km 1.88 214 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 ...
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