
Related rates problem
First, I want to say I am terrible with word problems... here is the problem:
Sand falling from a hopper at a rate of 10pi ft^3/sec forms a conical pile whose radius is always equal to half its height. How fast is the radius of the pile increasing when the radius is 5ft.
I believe the equation is V = (1/3)*pi*r^2*h
here is what I know:
the derivative of the equation is: (2/3)pi*rh
r = (1/2)h = 5
h= 2r = 10
To plug all this in (2/3)pi (5)(10)
= (100/3)pi
=104.7198 ft^2/ sec
Is this right?

You would be correct if $\displaystyle h$ were a constant.
However, $\displaystyle h$ is a function dependent on $\displaystyle r$. The correct derivative is therefore
$\displaystyle \frac{d}{dt}\left(\frac{1}{3}\pi r^2 h\right)=\frac{d}{dt}\left(\frac{2}{3}\pi r^3\right)=2\pi r^2\frac{dr}{dt}.$
Hope this helps. :)

That being said, then it would be:
(2/3)pi(5^2)
= (50/3)pi
which turns out to be 52.3599 ft/sec?

Not quite. We are given that
$\displaystyle \frac{dV}{dt}=2\pi r^2\frac{dr}{dt}=10\pi.$
When $\displaystyle r=5$, this becomes
$\displaystyle 2\pi\cdot5^2\cdot\frac{dr}{dt}=10\pi.$

I think I see what you are trying to say now. Get dr/dt on one side so it will be:
10pi/50pi = .2 ft/sec?
