Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 8 cubic feet per minute. If the pool has radius 7 feet and height 6 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 3 feet?
We are given that the pool has radius 7 feet and height 6 feet. The volume of the cylinder is $\pi\cdot7^2\cdot6=294\pi$. We are told that water is entering at a rate of 8 cubic feet per minute, so the rate of change of the volume of water in the pool is 8. When the depth of the water in the pool is 3 feet, we have a similar situation: the bottom base of the cylinder is still a circle of radius 7 feet, but the height of the water is now 3 feet. Let $x$ be the length of time in minutes that the water has been entering the pool. When the depth of the water is 3 feet, the volume of water in the pool is $\pi\cdot7^2\cdot3=147\pi$ cubic feet. Since the water has been entering the pool at a rate of 8 cubic feet per minute for $x$ minutes, there is now $8x$ cubic feet of water in the pool. We solve \begin{align*}
147\pi+8x&=294\pi
\\\Rightarrow\qquad 8x&=147\pi,
\end{align*}so $x=\frac{147\pi}{8}$.
[asy]
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[/asy]
The height of the water is a function of time, $h(x)$. We are asked for the rate of change of the height of the water with respect to time at the moment when the depth of the water is 3 feet, so we are looking for $h'(x)$ at $x=\frac{147\pi}{8}$. We have that \[
294\pi=\pi\cdot7^2\cdot h(x)=147\pi\cdot h(x).
\]So we see that when the depth of the water is 3 feet (i.e., at time $x=\frac{147\pi}{8}$), $h(x)=2$. Therefore, we have that when the water depth is 3 feet, the height of the water is changing at a rate of $\boxed{0}$ feet per minute. (Notice that this is consistent with what we would expect: once the water depth reaches 3 feet, the height of the water will remain constant, and so the rate of change will be 0.)