Question

Gravel is being dumped from a conveyor belt at a rate of 40 ft3/min. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 13 ft high?

Answer 1

Answer:

$\frac{dh}{dt}=\frac{160}{169\pi } ft/min$

Step-by-step explanation:

This is a classic related rates problem.  Gotta love calculus!

Start out with the formula for the volume of a cone, which is

$V=\frac{1}{3}\pi r^2h$

and with what we know, which is $\frac{dV}{dt}=40$

and the fact that the diameter = height (we will come back to that in a bit).

We need to find $\frac{dh}{dt}$ when h = 13

The thing we need to notice now is that there is no information given to us that involves the radius.  It does, however, give us a height.  We need to replace the r with something in terms of h.  Let's work on that first.

We know that d = h.  Because d = 2r, we can say that 2r = h, and solving for r gives us that $r=\frac{h}{2}$.

Now we can rewrite the formula with that replacement:

$V=\frac{1}{3}\pi (\frac{h}{2})^2h$

Simplify that all the way down to

$V=\frac{1}{12}\pi h^3$

The derivative of that function with respect to time is

$\frac{dV}{dt}=\frac{1}{12}\pi(3h^2)\frac{dh}{dt}$

Filling in what we have gives us this:

$40=\frac{1}{12}\pi (3)(13)^2\frac{dh}{dt}$

Solve that for the rate of change of the height:

$\frac{dh}{dt}=\frac{160}{169\pi } \frac{ft}{min}$

or in decimal form:

$\frac{dh}{dt}=.95\pi \frac{ft}{min}$