Question

a ferris wheel at the fair rotates counterclockwise. the ferris wheel measures 12m in diameter and sits 5m off the ground. sketch the height of a rider who travels around the ferris wheel in relation to the angle of rotation assuming they get on at the lowest point.

Explanation:

Identify key parameters

We extract the parameters:

• Diameter \(D = 12\text{ m}\), so radius \(R = 6\text{ m}\).
• Bottom height \(H_{\min} = 5\text{ m}\).
• Midline height \(y_{\text{mid}} = H_{\min} + R = 11\text{ m}\).
• Maximum height \(H_{\max} = H_{\min} + D = 17\text{ m}\).

Determine height function

The rider starts at the lowest point at \(\theta = 0^\circ\):

• At \(\theta = 0^\circ\), height is \(5\text{ m}\).
• At \(\theta = 90^\circ\), height is \(11\text{ m}\).
• At \(\theta = 180^\circ\), height is \(17\text{ m}\).
• At \(\theta = 270^\circ\), height is \(11\text{ m}\).
• At \(\theta = 360^\circ\), height is \(5\text{ m}\).

The model is:

$$h(\theta) = 11 - 6\cos(\theta)$$

Complete the table

We calculate the heights:

• For \(0^\circ\): \(h(0^\circ) = 11 - 6(1) = 5\text{ m}\).
• For \(90^\circ\): \(h(90^\circ) = 11 - 6(0) = 11\text{ m}\).
• For \(180^\circ\) (noting the typo "18" in the table is meant to be \(180^\circ\)): \(h(180^\circ) = 11 - 6(-1) = 17\text{ m}\).
• For \(270^\circ\): \(h(270^\circ) = 11 - 6(0) = 11\text{ m}\).
• For \(360^\circ\): \(h(360^\circ) = 11 - 6(1) = 5\text{ m}\).

Answer:

Angle (\(^\circ\))	Height (m)
90	11
180	17
270	11
360	5