Determining Rates

• Feb 18th 2009, 05:00 PM
calc_student09
Determining Rates
A conical reservoir has a depth of 24 feet and a circular top of radius 12 feet. It is being filled so that the depth of water i increasing at a constant rate of 4 feet per hour. Determine the rate in cubic feet per hour at which water is entering the reservoir when the depth is 5 feet.

I know that A= 1/3(pi)hr^2 but i havent done rates for a while and I don't remember how to do this. Its calc ab
• Feb 18th 2009, 05:23 PM
skeeter
$V = \frac{\pi}{3}r^2 h
$

$\frac{r}{h} = \frac{12}{24}$ ... $r = \frac{h}{2}$

sub in h/2 for r, and get the volume formula strictly in terms of h ...

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

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

now take the time derivative and determine $\frac{dV}{dt}$
• Feb 19th 2009, 06:19 PM
OnMyWayToBeAMathProffesor
so then $\frac{dv}{dt}=\frac{3\pi}{12}h^2=\frac{\pi}{4}h^2$. I get that, I would think you would just plug in 5 to get $\frac{dv}{dt}$ but then where does the 4 feet per hour come in? Thanks for everyones help.
• Feb 19th 2009, 06:39 PM
skeeter
Quote:

Originally Posted by OnMyWayToBeAMathProffesor
so then $\frac{dv}{dt}=\frac{3\pi}{12}h^2=\frac{\pi}{4}h^2$. I get that, I would think you would just plug in 5 to get $\frac{dv}{dt}$ but then where does the 4 feet per hour come in? Thanks for everyones help.

chain rule ...

$\frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}$
• Feb 19th 2009, 06:49 PM
OnMyWayToBeAMathProffesor
aaaa, i see. Sorry, I totally forgot about that. But is my $\frac{dv}{dt}$ formula correct? My peers got different derivative equations. If it is, this would be the subsequent work.

$\frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}$

$\frac{dV}{dt}=\frac{\pi}{4} (5)^2 \cdot (4)$

$\frac{dV}{dt}=25\pi$

correct?
• Feb 19th 2009, 06:57 PM
skeeter
units for $\frac{dV}{dt}$ ?
• Feb 19th 2009, 07:05 PM
OnMyWayToBeAMathProffesor
Thanks for catching that, so $\frac{dV}{dt}=25\pi$ feet per hour

correct?