precalc question

Question

Shirley is on a ferris wheel which spins at the rate of 3.2 revolutions per minute. The wheel has a radius of 45 feet, and the center of the wheel is 59 feet above the ground. After the wheel starts moving, Shirley takes 16 seconds to reach the top of the wheel. How high above the ground is she when the wheel has been moving for 9 minutes?

Answer is 101.497, but please explain how you got it.

Answer 1

Having the equation of height as a function of time will help us easily solve this problem.

The generic form of such an equation is:

h(t) = Acos(B[t - C)] + D where A is the amplitude, B is 2π/Period, C is the Phase Shift, and D is the vertical shift. We'll use time in seconds and height will be in feet.

In this case:

A = 45 (the height of the Ferris wheel)

B requires the period in seconds. in 60 seconds (1 minute), there are 3.2 revolutions so the period is 60/3.2 = 18.75 s. So B = 2π/18.75 = 0.3351

C is the shift from the standard cosine function which normally has it's peak at the top when t = 0. Since the position Shirley is at requires 16 s to get to the top and there are 18.75 s in a total trip, then she is 2.75 s down from the top. So C = -2.75 making "t - C" into "t + 2.75"

D is the vertical shift, in this case 59 feet (the distance from the ground to the center of the wheel).

So the equation for the height is:

h(t) = 45cos(0.3351[t + 2.75)] + 59

Now, we can just plug in 9 minutes (actually 9•60 or 540 seconds)

h(540) = 45cos(0.3351[540 + 2.75)] + 59 = 101.4969 ft (I'd probably round to 101.5 feet) off the ground