1. ## another related rates

A ferris wheel with a radius of 5 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when the rider is 9 m above ground level?

Sorry I keep posting these, I just really have trouble with finding the proper equation to start it off with. Thanks in advance.

2. We are assuming the wheel is tangent to the ground. Unrealistic, but let's go with it.

The y coordinate is given by $y=5-5cos({\theta})$...[1]

The wheel rotates $\frac{\pi}{60}$ rad/sec.

When the wheel is 9 feet off the ground, then it has swept out an angle of

$sin^{-1}(4/5)+\frac{\pi}{2}\approx{2.49809} \;\ radians$

From [1], differentiating, we get

$y'=5sin({\theta})\frac{d{\theta}}{dt}$

We know the angular velocity is $\frac{d{\theta}}{dt}=\frac{\pi}{60} \;\ rad/sec$ and ${\theta}=2.49809$

Entering in the values, we get $\frac{dy}{dt}=\frac{\pi}{20}=0.157 \;\ feet/sec$

Another way is:

$y=5-5cos(\frac{\pi}{60}t)$

$\frac{dy}{dt}=\frac{{\pi}sin(\frac{\pi}{60}t)}{12}$

Entering in the 47.71 seconds to get to 9 feet, we get:

0.157 ft/sec. as before.