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 22 2 2 3cos 3sin 9cos 9sin 9(sin cos ) 9 xt yt xy t t = = = = += + 7. 4.7 5.1 4.7 1.1 2 2 12 c o s s i n c o s s i n (1 ) 4 c o s (2 )4 s i n ( 1) ( 2) 4(sin cos ) )( 2 t t =+ =− + −= −+ + = + + = 9. 4.7 3.1 4.7 3.1 3 3 3 2 1 3 33 y t y x yx = =    11. 4.7 0 4.7 6.2 2 2 2 1 2 1 )2 23 tx yx x =− =−+ 13. 4.7 3.1 4.7 3.1 2 2 2 1 2 2 1 11 24 y t y = = =
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Chapter 9 Parametric Equations and Polar Coordinates 303 15. 2.7 0 6.7 6.2 2 2 22 1 1 1( 1 ) 2, 1 xt yt yxt t yx 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 ) xy += , with x ±²³´µ²WKLV²LV²SDUW²RI 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 ± ²RU² 2 () 1 . xn =π− 39. A. x and y both oscillate between –1 and 1, but with different periods.
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Chapter 9 Parametric Equations and Polar Coordinates 304 41. Use 0 ±² t ±²³´²ZLWK² 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 ±² t ±²³´²ZLWK² 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 xtyt == + for 1 ±² t ±²µ¶ 47. Since x decreases from 2 to 0, let 22 2, 2( 2) 42 xt y tt t =− =− − = −+ − for 0 ±² t ±²µ¶ 49. Use cos , sin xab tycd t =+ , with 0 ±² t ±²µ ± ¶²7KH²FHQWHU²LV²DW²·µ´²³¸´²WKHUHIRUH 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 14 ts −= − s = t – 1, so 14( 1 ) 60 3 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 31 2 +=+ s = t + 2, so 2( 2 ) 0 0 ttt 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 ⇒= . 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 66  −=   , and for the interceptor missile, the y- coordinate is 2 15 15 208 2 16 2 100 −− = . Since the y -coordinates are equal when x -coordinates are equal, the interceptor missile hits its target. 57. For the interceptor missile, y = 0 at t = time delay, and here y = 0 at t = 2. To find a time delay d that leads to missile collision, solve 100 500 200( ) 80 16 80( ) 16( ) .
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Calculus - Chapter 9 Parametric Equations and Polar...

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