Trigonometry/Worked Example: Ferris Wheel Problem

Exam Question
''"Jacob and Emily ride a Ferris wheel at a carnival in Vienna. The wheel has a $$\mathit{16}\,$$ meter diameter, and turns at three revolutions per minute, with its lowest point one meter above the ground. Assume that Jacob and Emily's height $$h$$ above the ground is a sinusoidal function of time $$t$$, where $$\mathit{t=0\,}$$ represents the lowest point on the wheel and $$t$$ is measured in seconds."

"Write the equation for $$h$$ in terms of $$t$$."''

[For those interested, the picture is actually of a Ferris wheel in Vienna.]

-Lang Gang 2016

Video Links
The Khan Academy has video material that walks through this problem, which you may find easier to follow:
 * Ferris Wheel Trig Problem
 * Ferris Wheel Trig Problem (part 2)

Solution
A $$16 \text{ m}$$ diameter circle has a radius of $$8 \text{ m}$$.

A wheel turning at three revolutions per minute is turning
 * $$\displaystyle \frac{3 \times 360^\circ}{60}$$

per second. Simplifying that's
 * $$\displaystyle 18^\circ$$

per second.

At $$t=0$$ our height $$h$$ is $$1$$. At $$t=10$$, we will have turned through $$180^\circ=10 \times 18^\circ$$, i.e. half a circle, and will be at the top most point of height $$16 + 1 = 17$$ (because the diameter of the circle is $$16$$ meters).

A cosine function, i.e. $$\displaystyle \cos \theta$$, is $$1$$ at $$\displaystyle \theta=0^\circ$$ and $$-1$$ at $$\displaystyle \theta=180^\circ$$. That's almost exactly opposite to what we want as we want the most negative value at $$0$$ and the most positive at $$180$$. Ergo, let's use the negative cosine to start our function.

At $$t=10$$ we want $$\theta=180^\circ$$, so we will multiply $$t$$ by $$18$$ so that we get $$\displaystyle -\cos( 18 t )$$. The formula we made is $$-1$$ at $$t=0$$ and $$1$$ at $$t=10$$. Multiply by $$8$$ and we get:


 * $$\displaystyle -8\cos( 18 t )$$, which is $$-8$$ at $$t=0$$ and $$8$$ at $$t=10$$

To get make sure reality is not messed up (we can't have negative height $$h$$), add $$9$$ and we get


 * $$\displaystyle 9-8\cos( 18 t )$$, which is $$1$$ at $$t=0$$ and $$17$$ at $$t=10$$

Our required formula is


 * $$\displaystyle h = 9 - 8\cos( 18 t )$$.

with the understanding that cosine is of an angle in degrees (not radians).