Related rates problem

driver327 · #1 $Old$ November 6th, 2009, 05:16 PM

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?

Scott H · #2 $Old$ November 6th, 2009, 05:43 PM

You would be correct if $h$ were a constant.

However, $h$ is a function dependent on $r$ . The correct derivative is therefore

$\frac{d}{dt}\left(\frac{1}{3}\pi r^2 h\right)=\frac{d}{dt}\left(\frac{2}{3}\pi r^3\right)=2\pi r^2\frac{dr}{dt}.$

Hope this helps.

driver327 · #3 $Old$ November 6th, 2009, 10:52 PM

That being said, then it would be:
(2/3)pi(5^2)

= (50/3)pi

which turns out to be 52.3599 ft/sec?

Scott H · #4 $Old$ November 7th, 2009, 05:22 AM

Not quite. We are given that

$\frac{dV}{dt}=2\pi r^2\frac{dr}{dt}=10\pi.$

When $r=5$ , this becomes

$2\pi\cdot5^2\cdot\frac{dr}{dt}=10\pi.$

driver327 · #5 $Old$ November 7th, 2009, 09:16 AM

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?

Scott H · #6 $Old$ November 7th, 2009, 11:59 AM

That is correct.