# Thread: Ferris Wheel Problem using derivative of sine

1. ## Ferris Wheel Problem using derivative of sine

The ferris wheel has a diameter of 40ft, axle 25ft above ground. Three seconds after starting, your seat is at its highest point. The wheel makes 3 rev/min.
I got the function, which is y(t)=25+20cos pi/10(t-3)
and its derivative y'(t)=-2sin pi/10(t-3)
what I don't understand is how do you calculate how fast/slow y(t) is increasing when t=15? And how do you determine the fastest y(t) changes, and the seat where it changes the fastest?

2. ## Re: Ferris Wheel Problem using derivative of sine

I think you forgot something in your calculation of the derivative.

The derivative y'(t) is the rate of change of y(t). In other words, how fast/slow y(t) is increasing. You just need to substitute 15 for t in your equation for y'(t) to get the answer to the first question.

For the second question, you want the maximum possible value of y'(t). You can probably just figure out the answer directly, but you also probably know how to find the maximum value for a function. In this case you find where the derivative is zero, so the derivative of y'(t) is y''(t).

If you think about the problem physically, you should be able to verify your answer. Where does a ferris wheel go fastest in the up-down direction?

Let us know if you run into any trouble.

- Hollywood

3. ## Re: Ferris Wheel Problem using derivative of sine

I did substitute 15 for y'(t), but it's not 3.7ft/s (which is what the book answered), it's a negative, which means y(t) is decreasing. I think I'm doing something wrong with the substitution, you can't plug in the numbers on the calculator, can you?

4. ## Re: Ferris Wheel Problem using derivative of sine

You should have $\displaystyle y'(t)=-2\pi\sin{\frac{\pi}{10}(t-3)}=-2\pi\sin{\frac{\pi}{10}(15-3)}$
$\displaystyle =-2\pi\sin{\frac{6\pi}{5}}=-2\pi(-0.588)=3.69$

So $\displaystyle y$ is increasing at the rate of 3.69 ft/sec.

Did you remember to put your calculator into radian mode? It's a common error.

- Hollywood

5. ## Re: Ferris Wheel Problem using derivative of sine

So to solve for the fastest y(t) change you find the third derivative which is y''(t)? Is it -1/5pi^2cospi/10(t-3)?

6. ## Re: Ferris Wheel Problem using derivative of sine

Yes, that's correct, except y''(t) is the second derivative, not the third. So where is y''(t) zero?

7. ## Re: Ferris Wheel Problem using derivative of sine

When you solve y''(t) you get 2pi, which is the fastest change, and when you plug 2pi in the original equation, it gets you the highest point? 25 ft, right?

8. ## Re: Ferris Wheel Problem using derivative of sine

I get t=8 for one of the values of t where y''(t)=0. This will be the fastest change - I'm not sure how you got $\displaystyle 2\pi$.

Plugging in to the original equation, we get $\displaystyle y=25+20\cos\frac{2\pi}{10}(8-3)=25+20\cos{\pi}=25$, as you said. And it makes sense physically - the speed will be greatest when you are at the same level as the center of rotation.

- Hollywood

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### ferris wheel mathematical derivative

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