Related Rates: Calculus I & II (AB & BC)

A potter places a cylindrical lump of clay on her wheel. Before she begins to shape it, it has radius 10 inches and height 4 inches. Assuming that no clay is lost in the shaping process (and that the cylindrical lump stays cylindrical throughout the process) how fast is the height of the lump changing when the radius is 7.5 inches, given the radius is shrinking by 6/5 inches/minute?

Difficulty:
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Solution:

First and foremost, you need to have the formula for the volume of a cylinder memorized. For grins, we can calculate the volume of the cylinder indicated by this problem. (It turns out to be important later ... )

As this is a related rates problem, it is now important to find the derivative with respect to time.

We are told that the volume of the lump doesn't change, so dV/dt = 0. We are given r = 7.5, dr/dt = -6/5 and dh/dt is our unknown. (Remember that dr/dt must be negative since r is decreasing in size. Anytime a value decreases, its rate of change with respect to time must be negative.) However, the problem also requires us to plug in a value for h. How do you find that? Well, if the volume doesn't change, it is still 400, and since we know the radius at this moment, we can also find the height at this moment (it will not be the same as the height at the start of the problem since the radius is not the same as the start of the problem).

Now we have all the values we need, so go back to the derivative, plug everything in, and solve for dh/dt. Don't forget to include the proper units to receive full credit. Since height is a length, it will be measured in inches to match the radius, which is also a length.