# Thread: Related Rates and Ferris Wheel

1. ## Related Rates and Ferris Wheel

I'm not sure if this is the correct sub-forum for this question, so a mod can feel free to move it.

I've been stuck on this problem for over a week now, and even though it's not an assignment for a grade, it's eating away at me that I don't understand it.

An observer stands 20m from the bottom of a 10 meter tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of pi rad/min and the observer's line of sight with a specific seat on the ferris wheel makes an angle θ with the ground. Forty seconds after the seat leaves the lowest point on the wheel, what is the rate of change of θ? (Assume the observer's eyes are level with the bottom of the wheel.)
So, the question is asking for dθ/dt when t=40sec, i.e. when t=2pi/3 radians

What I am having trouble with is finding an equation for h(t).

The subsequent calculations should be easy:
tan(θ) = h/20
20tan(θ) = h

Differentiate both sides with respect to time:
20sec2(θ)*dθ/dt = dh/dt

But I need to find the equation for h(t) before I can find h'(t).

Can anyone shed some light on this?

A friend tried to help me with this problem and came up with h(t) = -5sin2(t-3pi/4) + 5, however I have no idea if that is correct.

2. ## Re: Related Rates and Ferris Wheel

as the seat moves from 0 to 10' and back to 0, it's height can be modeled by a cosine curve ...

$\displaystyle h(t) = 5 - 5\cos(\pi t)$

where $\displaystyle t$ is in minutes.

... remember $\displaystyle t = 40 \, seconds = \frac{2}{3} \, minute$

3. ## Re: Related Rates and Ferris Wheel

Thank you! I believe I figured it out:

At t=2pi/3: dθ/dt = 5sqrt(3)/160 = (sqrt(3))/32 radians/min

I'll talk to my T.A. in class tomorrow and hopefully he has the solution!