Explanation
Given
From the question, we can see that the bottom of the ferris wheel is 30 feet about the ground and also while rotating, it can move to a height of 550 feet off the ground. In essence the actual height the ferris wheel can attain is
[tex]h=550-30=520[/tex]
The amplitude then becomes half of the height which is
[tex]A=\frac{h}{2}=\frac{520}{2}=260[/tex]
The vertical shift the Ferris wheel undergoes becomes the sum of the amplitude and its distance above the ground.
[tex]D=260+30=290[/tex]
Since it takes the Ferris wheel 15 minutes to move from bottom to top, it will take it twice that to complete one revolution which will be its period.
[tex]T=2\times15mins=30[/tex]
Therefore, the frequency B, becomes;
[tex]B=\frac{2\pi}{T}=\frac{2\pi}{30}=\frac{\pi}{15}[/tex]
We can then place in the above parameters to form the equation.
[tex]y=Acos(B(t+C))+D\Rightarrow260cos(\frac{\pi}{15}(t+C)+290[/tex]
The last missing parameter is the phase shift C. At time t =0, the function (y) has a position at 30. Therefore,
[tex]\begin{gathered} 30=260cos(\frac{\pi}{15}C)+290 \\ 260cos(\frac{\pi}{15}C)=-290+30 \\ 260cos(\frac{\pi}{15}C)=-260 \\ divide\text{ both sides by 260} \\ \frac{\begin{equation*}260cos(\frac{\pi}{15}C)\end{equation*}}{260}=\frac{-260}{260} \\ cos(\frac{\pi}{15}C)=-1 \\ \frac{\pi}{15}C=cos^{-1}(-1) \\ \frac{\pi}{15}C=\pi \\ C=\frac{15\pi}{\pi} \\ C=15 \end{gathered}[/tex]
Therefore, the function y becomes
Answer:
[tex]y=260cos(\frac{\pi}{15}(t+15)+290[/tex]
