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

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Chapter 9 Parametric Equations and Polar Coordinates302Chapter 9Section 9.15.4.73.14.73.122222222223cos3sin9cos9sin9(sincos)9xtytxtytxyttxy====+=++=7.4.75.14.71.1222222222212cos22sin12cos22sin(1)4cos(2)4sin(1)(2)4(sincos)(1)(2)4xtytxtytxtytxyttxy=+= −+=+==+=++=+++=9.4.73.14.73.112331232133322xtytytyxyxyx= − +=== − +=+=+11.4.704.76.2222121(1)223xtyttxyxyxx=+=+==+=+13.4.73.14.73.122212211124xtytytyxy=====
Chapter 9 Parametric Equations and Polar Coordinates30315.2.706.76.22222111(1)2,1xtytyxttyxx==+=+=+≥ −17.4.73.14.73.119.9.43.19.49.321.2.351.552.351.5523.2.351.552.351.5525.1510151027.12812829.646431.2.351.552.351.5533.2.71.16.75.135.C. Since 2(1)xy+=, with x±±²³±WKLV±LV±SDUW±RIan upward-opening parabola.37.B.xis bounded below by –1. yis bounded belowby –1 and above by 1. The graph will have x-intercepts at t = n ±RU±2()1.xn=π39.A.xand yboth oscillate between –1 and 1, butwith different periods.
Chapter 9 Parametric Equations and Polar Coordinates30441.Use 0 ±t±²³±ZLWK±t= 0 corresponding to(0, 1); then x= 0 + bt, y = 1 + dt.Witht = 1,x= 3 = 0 + b,y= 4 = 1 + d. So b= 3, d= 3, theequations are x= 3t,y= 1 + 3t.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, theequations are x= –2 + 8t,y= 4 – 3t.45.2,1xtyt==+for 1 ±t±´µ47.Sincexdecreases from 2 to 0, let222,2(2)42xtyttt=== −+for0±t±´µ49.Use cos ,sinxabtycdt=+=+, with0±t±´µ±7KH±FHQWHU±LV±DW±¶´³±²·³±WKHUHIRUHa= 2 and c= 1. The radius is 3, thereforeb= d= 3. So the equations are x= 2 + 3 cos t,y= 1 + 3 sin t.51.Solve2114tsts=+=s= t– 1, so2214(1)603ttttt=+== −ort= 2. So either t= –3 and s= –4, in which casex= –3 and y= 8, or t= 2 and s= 1, in which casex= 2 and y= 3. So the intersection points are (–3,8) and (2, 3).53.Solve2312tsts+=+=s= t+ 2, so 222(2)00ttttt=++==ort= –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 100t= 500 – 500(t– 2);100t= 1500 –155006tt=. So at 52t=, themissiles have the same x-coordinate.For the target missile, the y-coordinate is21515801610066=, and for the interceptormissile, the y-coordinate is21515208216210066=. Since they-coordinates are equal when x-coordinates areequal, the interceptor missile hits its target.

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