Related Rates: Cone-shaped pile

Related Rate

Find the rate of change of the height of a cone, knowing the rate of change of the volume.

 

A cone-shaped pile of dirt is growing at the rate of 10 cubic feet/minute. The diameter of its base is 3 times its height.

How fast is the height of the pile growing when it's 15feet high?

We'll ask some questions.

What do we want? - We want the rate of change of the height of the cone. We'll call the height h, and its rate of change dh/dt.

What do we know already? -We know the rate of change of the volume of the cone. We'll call the volume V, and its rate of change dV/dt. We also know that dV/dt = 10 cubic feet/minute.

What geometric formulas can help us? We know the formula for the volume of the cone. V = (1/3)(pi)(r^2)h

What substitutions can we make? (or,  How can we get rid of that r? ) – We have the relationship 2r = 3h, or r = 3h/2.

(The first time I did this problem,  I momentarily forgot that the diameter is twice the radius.)

Now we can set up our geometric equation.

V = (pi/3)(r^2)h

Substitute 3h/2 for r.

V = (3pi/4)(h^3)

We take the derivative of each side, over time.

dV/dt =(9pi/4)(h^2)(dh/dt)

We know the values of dV/dt and h, so we substitute those values into our equation.

10 cubic feet/minute = (9*15^2 feet squared *pi) /4* dh/dt

Solve for dh/dt.

8/(405pi) feet/minute = dh/dh

 

Related rate problems are just algebra equations. If I'd known that when I first encountered them back in the day, I'd have liked them a lot better a lot sooner. I also should have brushed up on my solid geometry, or at least learned some of it first.

 

Source:
 Larsen, Hostetler and Edwards,  Calculus, Fifth Edition)