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.