Received via Twitter:
A trough is 10 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 12f^3/min, how fast is the water level rising when the water is 6 inches deep?
NOTE:
Before we get start, let me just say that this is outside the scope of what I envision for this blog. My primary intent is to help middle and high school students. Having said that, I haven't received any other questions yet, so I decided to work this one out and post it.
THE SOLUTION:
First of all, it's important to understand which values are changing over time and which are fixed. The volume (V) and the dimensions of the triangle (b and h) are changing as the trough fills with water. The length of the trough, however, remains constant, so

.
I always find it helpful to draw a picture so I can see what information I have and what I'm trying to find:
The trough is a triangular prism, meaning the volume can be represented by the equation

. Since we're interested in how fast the water is rising, we'll look for

as your professor suggested. To do this, we start by rewriting the volume equation so h is on one side of the equation:

.
Next, we differentiate the whole thing with respect to t:

From there, we can simplify a little since

. We need to calculate the values for volume (V) and width (w) when h = 1.
Alternate Solution:
It was proposed to me that using the Quotient Rule was overly cumbersome (in fact, I myself forgot to apply the chain rule when I first posted the solution). Therefore, here's an alternate, slightly more elegant solution based on the product rule.