Question

montana boards a ferris wheel at the 3 - oclock position and rides the ferris wheel for multiple revolutions. the ferris wheel rotates at a constant angular speed. let h represent montanas height above the center of the ferris wheel (in feet) and let t represent the number of minutes since the ride started.
a. suppose the radius of the ferris wheel is 15 feet and the ferris wheel rotates at 1 radian per minute. plot the relationship between h and t for this scenario.
b. now, suppose the radius of the ferris wheel is 15 feet and the ferris wheel rotates at 2 radians per minute. plot the relationship between h and t for this scenario.

Explanation:

Step1: Determine the general form of the height function

The height \(h\) of a point on a Ferris - wheel can be modeled by a sine or cosine function. Since Montana starts at the 3 - o'clock position (where \(h = 0\) in a standard circular - motion model, but if we consider the vertical displacement from the center, and using the formula \(h = A\sin(\omega t+\varphi)\) or \(h = A\cos(\omega t+\varphi)\). Starting at the 3 - o'clock position (equivalent to \(t = 0\), \(h = 0\) for a sine function with no phase shift). The amplitude \(A\) is the radius of the Ferris - wheel. Given \(A = 15\) feet.
The angular speed is \(\omega\). For part (a), \(\omega=1\) radian per minute. The formula for the height is \(h = 15\sin(t)\) (because when \(t = 0\), \(h = 0\) which is consistent with the 3 - o'clock position).
For part (b), \(\omega = 2\) radians per minute. The formula for the height is \(h=15\sin(2t)\)

Step2: Analyze key features for plotting

For \(h = 15\sin(t)\)

• Amplitude: \(|A|=15\). The maximum value of \(y = \sin(t)\) is \(1\) and the minimum is \(- 1\). So, the maximum value of \(h\) is \(15\times1 = 15\) and the minimum is \(15\times(-1)=-15\)
• Period: The period of \(y=\sin(\omega t)\) is \(T=\frac{2\pi}{\omega}\). For \(\omega = 1\), \(T = 2\pi\)
• Key points:
• When \(t = 0\), \(h=15\sin(0)=0\)
• When \(t=\frac{\pi}{2}\), \(h = 15\sin(\frac{\pi}{2})=15\)
• When \(t=\pi\), \(h=15\sin(\pi)=0\)
• When \(t=\frac{3\pi}{2}\), \(h=15\sin(\frac{3\pi}{2})=-15\)
• When \(t = 2\pi\), \(h=15\sin(2\pi)=0\)

For \(h = 15\sin(2t)\)

• Amplitude: \(|A| = 15\)
• Period: The period of \(y=\sin(\omega t)\) is \(T=\frac{2\pi}{\omega}\). For \(\omega = 2\), \(T=\frac{2\pi}{2}=\pi\)
• Key points:
• When \(t = 0\), \(h=15\sin(0)=0\)
• When \(t=\frac{\pi}{4}\), \(h = 15\sin(\frac{\pi}{2})=15\)
• When \(t=\frac{\pi}{2}\), \(h=15\sin(\pi)=0\)
• When \(t=\frac{3\pi}{4}\), \(h=15\sin(\frac{3\pi}{2})=-15\)
• When \(t=\pi\), \(h=15\sin(2\pi)=0\)

Answer:

For part (a), the function is \(h = 15\sin(t)\) with amplitude \(15\) and period \(2\pi\). For part (b), the function is \(h=15\sin(2t)\) with amplitude \(15\) and period \(\pi\). When plotting:

• For \(h = 15\sin(t)\): Mark the points \((0,0)\), \((\frac{\pi}{2},15)\), \((\pi,0)\), \((\frac{3\pi}{2},-15)\), \((2\pi,0)\) and connect them with a smooth sine - wave.
• For \(h = 15\sin(2t)\): Mark the points \((0,0)\), \((\frac{\pi}{4},15)\), \((\frac{\pi}{2},0)\), \((\frac{3\pi}{4},-15)\), \((\pi,0)\) and connect them with a smooth sine - wave (which oscillates twice as fast as the function in part (a)).