# Related rates problem

• Nov 6th 2009, 03:16 PM
driver327
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?
• Nov 6th 2009, 03:43 PM
Scott H
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. :)
• Nov 6th 2009, 08:52 PM
driver327
That being said, then it would be:
(2/3)pi(5^2)

= (50/3)pi

which turns out to be 52.3599 ft/sec?
• Nov 7th 2009, 03:22 AM
Scott H
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

$\displaystyle 2\pi\cdot5^2\cdot\frac{dr}{dt}=10\pi.$
• Nov 7th 2009, 07:16 AM
driver327
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?
• Nov 7th 2009, 09:59 AM
Scott H
That is correct.