Question

1. the graph below shows the height, h (in m), above the ground, of a rider on a ferris wheel after some time, t (in s). t/i - 4

create an appropriate trigonometric model of your choice (based on classroom instruction) and use your model to determine time(s) during the first revolution when the rider is 10 m above the ground.

Explanation:

Step1: Determine the general form of the trigonometric function

The general form of a sinusoidal function is \(h(t)=A\sin(B(t - C))+D\) or \(h(t)=A\cos(B(t - C))+D\).
For a cosine - based function (since the maximum occurs at \(t = 30\) s), the amplitude \(A=\frac{33 - 3}{2}=15\), the vertical shift \(D=\frac{33 + 3}{2}=18\), and the period \(T = 60\) s.
Using the formula \(B=\frac{2\pi}{T}\), we get \(B=\frac{2\pi}{60}=\frac{\pi}{30}\).
The function is \(h(t)=- 15\cos(\frac{\pi}{30}t)+18\) (the negative sign is because the cosine function is reflected to have a maximum at \(t = 30\) s).

Step2: Set up the equation to find \(t\) when \(h(t)=10\)

Set \(h(t)=10\), so \(10=-15\cos(\frac{\pi}{30}t)+18\).
First, subtract 18 from both sides: \(10 - 18=-15\cos(\frac{\pi}{30}t)\).
We get \(- 8=-15\cos(\frac{\pi}{30}t)\), then \(\cos(\frac{\pi}{30}t)=\frac{8}{15}\).

Step3: Solve for \(t\)

Using the inverse - cosine function, \(\frac{\pi}{30}t=\cos^{-1}(\frac{8}{15})\) or \(\frac{\pi}{30}t = 2\pi-\cos^{-1}(\frac{8}{15})\) (for the first revolution, \(0\leq t\leq60\)).
\(t=\frac{30}{\pi}\cos^{-1}(\frac{8}{15})\approx\frac{30}{\pi}(0.980)\approx9.3\) s.
\(t=\frac{30}{\pi}(2\pi - 0.980)=60-\frac{30}{\pi}(0.980)\approx60 - 9.3 = 50.7\) s.

Answer:

The trigonometric model is \(h(t)=-15\cos(\frac{\pi}{30}t)+18\). The times when the rider is \(10\) m above the ground during the first revolution are approximately \(t = 9.3\) s and \(t = 50.7\) s.