Trig chapter 5 ferris wheel question

Question

A Ferris Wheel has a maximum height of 44, and a minimum height of 4. It takes 3 minutes to go around twice. If we were going to model the height of a rider over time with a cos function, we could write h(t)= a cos(bx) +d, where is h(t) the height of the rider at time t, in minutes. Assume the rider must be at the minimum height at time 0.

I keep trying to do these questions but I never get the right answers

Answer 1

The diameter of the Ferris Wheel is 40 (presumably the units are feet even though you don't say) and since it sit 4 off the ground the center is at 24. The amplitude is the radius which is 20. The period is 1.5 minutes so B = 2pi/1.5 or B = 4pi/3. Since we take time 0 to be the low then we write as a negative cosine.

H(t) = - 20Cos[(4pit)/3] + 24

Answer 2

Make a sketch:

The red dot is the place the rider gets on at time zero (t = 0). Since the max height is 44 and the min height is 4, the Ferris Wheel must have a diameter of 40. That makes the radius 20. So the distance from the ground to the midline is 24.

For h(t) = Acos(Bt) + D, A is the amplitude which is the max - midline = 44 - 24 = 20. B is calculated from the period. The period is the time it takes to go around ONCE so in this case is 1.5 minutes. To find B, we start with 2π and we divide the period. B = 2π/1.5 = 2π/(3/2) = 2π•(2/3) = 4π/3. D is the midline location so D = 24.

The only issue is that a cosine function starts at a maximum instead of a minimum. So to fix that we need to use a negative cosine function. So:

h(t) = -20cos((4π/3)t) + 24