A mechanic has a bicycle wheel with a two-foot diameter in a clamp and is looking at a blemish in the tire. The center of the wheel is at countertop level, and the blemish is level with the center of the wheel (at the three oclock position). Suppose that she then spins the wheel in a clockwise direction.

A) How far above the countertop will the blemish be after the wheel turns radians? To indicate that the blemish is below the countertop, give a negative answer to this question.

B) How far above the countertop will the blemish be after the wheel turns 4 radians? Give your answer to the nearest hundredth of a foot. To indicate that the blemish is above the countertop, give a positive answer to this question.

C) Write a formula for the blemishs height, y, above the countertop after the wheel has turned through an angle of radians. Again, negative heights would indicate that the blemish is below the countertop.

D) Now consider the blemishs horizontal distance in other words, the distance between the blemish and an imaginary vertical line running through the center of the wheel. Lets call this distance positive if the blemish is to the right of the imaginary vertical line and negative if the blemish is to the left of the imaginary vertical line. What is this distance before the wheel is spun?

E) What is this distance after the wheel turns radians? 4 radians? Round your answers to the nearest hundredth of a foot.

F) Write a formula for the blemishs horizontal distance, x, after the wheel has turned through an angle of radians.

G) Suppose that the wheel is spinning so that it makes one complete turn every 12 seconds. Rewrite your formulas from (c) and (f) so that they give the vertical and horizontal distance of the blemish when the wheel has spun for t seconds. (Hint: Think about how many radians per second the wheel turns.)