1. ## 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

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

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

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

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

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

now take the time derivative and determine $\displaystyle \frac{dV}{dt}$

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

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

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

5. aaaa, i see. Sorry, I totally forgot about that. But is my $\displaystyle \frac{dv}{dt}$ formula correct? My peers got different derivative equations. If it is, this would be the subsequent work.

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

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

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

correct?

6. units for $\displaystyle \frac{dV}{dt}$ ?

7. Thanks for catching that, so $\displaystyle \frac{dV}{dt}=25\pi$ feet per hour

correct?