Precalculus Trigonometry Question

Question

The Giant Sky Wheel is a Ferris wheel in Tokyo with a diameter of 100 meters that completes one full revolution every 16 minutes.¹ A rider boards at the bottom of the Ferris wheel which is 15 meters above the ground and rides for 32 minutes.



(a) Graph of the height, h, of the rider as a function of the time, t, after the rider boards the Ferris wheel. Use your graph to estimate the times during the first half hour of the ride at which the rider is 100 meters above the ground.

First time ≈  _minutes.

Second time ≈ _minutes.

Third time ≈ _minutes.

Fourth time ≈ _minutes.

(b) Find, to the nearest second, the times during the first half hour of the ride at which the rider is 100 meters above the ground.

First time = _seconds and _minutes

Second time = _seconds and _minutes

Third time =_seconds and _minutes

Fourth time = _seconds and _minutes

Answer 1

We can model the rider's height as a function of time using either sine, cosine, - sine, or - cosine. Since the rider boards at the bottom, - cosine is most convenient (the - cosine graph is at its lowest y-value when x = 0).

The center of the wheel is 65 m above ground (15m + 50m since the radius = 50m). The period = 16 mins.

Putting all of these numbers into an equation of the form h(t) = - Acos(Bt) + D we get the following:

h(t) = - 50cos(pi/8⋅t) + 65

Graph this on a calculator or in Desmos -- be sure it is in radians mode. You will see the graph of the rider's height as a function of time. You can manually set a window 0 < x < 32 and 0 < y < 125 to see the whole graph

You can enter y₂ = 100 to graph the horizontal line and calculate the pts of intersection for part a. You will get

x ~ 6 mins, 10 mins, 22 mins, and 26 mins. (Note: the 1st 2 times are symmetrical across x = 8 mins, the time of max. height. The 3rd and 4th times are 16 mins after the 1st and 2nd times, one full period later.)

Solving for these times algebraically (analytically) requires inverse trig and looks like this:

h(t) = - 50cos(pi/8⋅t) + 65 = 100 - 50cos(pi/8⋅t) = 35 cos(pi/8⋅t) = -.7 pi/8⋅t = cos^-1(-.7) ~ 2.346 and t ~ 6