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

• Notes
• 47

This preview shows page 1 out of 47 pages.

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 ...
View Full Document