# Trigonometry/Worked Example: Ferris Wheel Problem

## The Problem

### Exam Question

"Jacob and Emily ride a Ferris wheel at a carnival in Vienna. The wheel has a ${\displaystyle {\mathit {16}}\,}$  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 ${\displaystyle h}$  above the ground is a sinusoidal function of time ${\displaystyle t}$ , where ${\displaystyle {\mathit {t=0\,}}}$  represents the lowest point on the wheel and ${\displaystyle t}$  is measured in seconds."

"Write the equation for ${\displaystyle h}$  in terms of ${\displaystyle t}$ ."

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

-Lang Gang 2016

 Diameter to Radius A ${\displaystyle 16{\text{ m}}}$  diameter circle has a radius of ${\displaystyle 8{\text{ m}}}$ .
 Revolutions per Minute to Degrees per Second A wheel turning at three revolutions per minute is turning ${\displaystyle \displaystyle {\frac {3\times 360^{\circ }}{60}}}$ per second. Simplifying that's ${\displaystyle \displaystyle 18^{\circ }}$ per second.
 Formula for height At ${\displaystyle t=0}$  our height ${\displaystyle h}$  is ${\displaystyle 1}$ . At ${\displaystyle t=10}$ , we will have turned through ${\displaystyle 180^{\circ }=10\times 18^{\circ }}$ , i.e. half a circle, and will be at the top most point of height ${\displaystyle 16+1=17}$  (because the diameter of the circle is ${\displaystyle 16}$  meters). A cosine function, i.e. ${\displaystyle \displaystyle \cos \theta }$ , is ${\displaystyle 1}$  at ${\displaystyle \displaystyle \theta =0^{\circ }}$  and ${\displaystyle -1}$  at ${\displaystyle \displaystyle \theta =180^{\circ }}$ . That's almost exactly opposite to what we want as we want the most negative value at ${\displaystyle 0}$  and the most positive at ${\displaystyle 180}$ . Ergo, let's use the negative cosine to start our function. At ${\displaystyle t=10}$  we want ${\displaystyle \theta =180^{\circ }}$ , so we will multiply ${\displaystyle t}$  by ${\displaystyle 18}$  so that we get ${\displaystyle \displaystyle -\cos(18t)}$ . The formula we made is ${\displaystyle -1}$  at ${\displaystyle t=0}$  and ${\displaystyle 1}$  at ${\displaystyle t=10}$ . Multiply by ${\displaystyle 8}$  and we get: ${\displaystyle \displaystyle -8\cos(18t)}$ , which is ${\displaystyle -8}$  at ${\displaystyle t=0}$  and ${\displaystyle 8}$  at ${\displaystyle t=10}$ To get make sure reality is not messed up (we can't have negative height ${\displaystyle h}$ ), add ${\displaystyle 9}$  and we get ${\displaystyle \displaystyle 9-8\cos(18t)}$ , which is ${\displaystyle 1}$  at ${\displaystyle t=0}$  and ${\displaystyle 17}$  at ${\displaystyle t=10}$ Our required formula is ${\displaystyle \displaystyle h=9-8\cos(18t)}$ .with the understanding that cosine is of an angle in degrees (not radians).