Calculus - Chapter 9 Parametric Equations and Polar Coordinates Chapter 9 Section 9.1 5 3.1 9 3.1 4.7 4.7 4.7 4.7 3.1 3.1 x = 3cos t y = 3sin t 2 2 x =

# Calculus - Chapter 9 Parametric Equations and Polar...

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Chapter 9 Parametric Equations and Polar Coordinates 302 Chapter 9 Section 9.1 5. 4.7 3.1 4.7 3.1 2 2 2 2 2 2 2 2 2 2 3cos 3sin 9cos 9sin 9(sin cos ) 9 x t y t x t y t x y t t x y = = = = + = + + = 7. 4.7 5.1 4.7 1.1 2 2 2 2 2 2 2 2 2 2 1 2cos 2 2sin 1 2cos 2 2sin ( 1) 4cos ( 2) 4sin ( 1) ( 2) 4(sin cos ) ( 1) ( 2) 4 x t y t x t y t x t y t x y t t x y = + = − + = + = = + = + + = + + + = 9. 4.7 3.1 4.7 3.1 1 2 3 3 1 2 3 2 1 3 3 3 2 2 x t y t y t y x y x y x = − + = = = − + = + = + 11. 4.7 0 4.7 6.2 2 2 2 1 2 1 ( 1) 2 2 3 x t y t t x y x y x x = + = + = = + = + 13. 4.7 3.1 4.7 3.1 2 2 2 1 2 2 1 1 1 2 4 x t y t y t y x y = = = = =
Chapter 9 Parametric Equations and Polar Coordinates 303 15. 2.7 0 6.7 6.2 2 2 2 2 1 1 1 ( 1) 2, 1 x t y t y x t t y x x = = + = + = + ≥ − 17. 4.7 3.1 4.7 3.1 19. 9.4 3.1 9.4 9.3 21. 2.35 1.55 2.35 1.55 23. 2.35 1.55 2.35 1.55 25. 15 10 15 10 27. 12 8 12 8 29. 6 4 6 4 31. 2.35 1.55 2.35 1.55 33. 2.7 1.1 6.7 5.1 35. C. Since 2 ( 1) x y + = , with x ±±²³±WKLV±LV±SDUW±RI an upward-opening parabola. 37. B. x is bounded below by –1. y is bounded below by –1 and above by 1. The graph will have x - intercepts at t = n ±RU± 2 ( ) 1. x n = π 39. A. x and y both oscillate between –1 and 1, but with different periods.
Chapter 9 Parametric Equations and Polar Coordinates 304 41. Use 0 ± t ±²³±ZLWK± t = 0 corresponding to (0, 1); then x = 0 + bt, y = 1 + dt. With t = 1, x = 3 = 0 + b , y = 4 = 1 + d . So b = 3, d = 3, the equations are x = 3 t , y = 1 + 3 t . 43. Use 0 ± t ±²³±ZLWK± t = 0 corresponding to (–2, 4); then x = –2 + bt , y = 4 + dt . With t = 1, x = 6 = –2 + b , y = 1 = 4 + d . So b = 8, d = –3, the equations are x = –2 + 8 t , y = 4 – 3 t . 45. 2 , 1 x t y t = = + for 1 ± t ±´µ 47. Since x decreases from 2 to 0, let 2 2 2 , 2 (2 ) 4 2 x t y t t t = = = − + for 0 ± t ±´µ 49. Use cos , sin x a b t y c d t = + = + , with 0 ± t ±´ µ±7KH±FHQWHU±LV±DW±¶´³±²·³±WKHUHIRUH a = 2 and c = 1. The radius is 3, therefore b = d = 3. So the equations are x = 2 + 3 cos t , y = 1 + 3 sin t . 51. Solve 2 1 1 4 t s t s = + = s = t – 1, so 2 2 1 4 ( 1) 6 0 3 t t t t t = + = = − or t = 2. So either t = –3 and s = –4, in which case x = –3 and y = 8, or t = 2 and s = 1, in which case x = 2 and y = 3. So the intersection points are (–3, 8) and (2, 3). 53. Solve 2 3 1 2 t s t s + = + = s = t + 2, so 2 2 2 ( 2) 0 0 t t t t t = + + = = or t = –1. So either t = 0 and s = 2, in which case x = 3 and y = 0, or t = –1 and s = 1, in which case x = 2 and y = 1. So the intersection points are (3, 0) and (2, 1). 55. Solve 100 t = 500 – 500( t – 2); 100 t = 1500 – 15 500 6 t t = . So at 5 2 t = , the missiles have the same x -coordinate. For the target missile, the y -coordinate is 2 15 15 80 16 100 6 6 = , and for the interceptor missile, the y- coordinate is 2 15 15 208 2 16 2 100 6 6 = . Since the y -coordinates are equal when x -coordinates are equal, the interceptor missile hits its target.
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