I got this question and it seemed to be tricky!
You are standing at a base of a Ferris Wheel which is 4 m above ground while your friend is riding. The Ferris Wheel is 8m in diameter. Describe how the shape of the sine curve models the distance your friend is to the platform you are on. Identify the function that will model this situation as well as a function that will model the if we measure his distance to the ground. In your explanation use the following terms:
Sine=
Function=
Radius=
- Repeat=
- Rotate=
- Amplitude=
- Period=
- Intercept=
- Maximum=
- Minimum=
- Axis of the curve=
I think that we can model it by using y = a sin (x - d) + c
c is zero here because there is no time
Radius= diameter/2 8/2 = 2 a = 8 which is the amplitude.
Repeat= undefined
Rotate= undefined
*I don't know if it is right or not, just confused. I need some ideas or some keys to solve this situation.
Thanks
It helps to draw a diagram.
Method 1: We define $\theta$ to be the angle formed with the radius that connects the center of the ferris wheel with the point directly beneath it on the platform.
Then the vertical displacement above the platform is $h = r - r\cos\theta$. If the angular velocity of the ferris wheel is $$\omega = \frac{\theta}{t}$$ then $$h(t) = r - r\cos(\omega t)$$
Since a complete revolution is $2\pi$ radians, the period $T$ of the ferris wheel must satisfy the equation $$2\pi = \omega T$$ from which we obtain $$T = \frac{2\pi}{\omega}$$
Method 2: We define $\theta$ to be the angle formed with the positive $x$-axis, where the origin is the center of the ferris wheel.
Then the vertical displacement above the platform is $h = r + r\sin\theta$. However, we can't simply write $$h(t) = r + r\sin(\omega t)$$ unless we assume that the friend starts from the horizontal. Initially, the friend is at the bottom of the ferris wheel, so $\theta = -\dfrac{\pi}{2}$. Hence, $$h(t) = r + r\sin\left(\omega t - \frac{\pi}{2}\right)$$ where $\omega$ and $T$ are defined as above.
From here, you should be able to use the given information to complete the problem.