# Thread: Trouble with related rates.

1. ## Trouble with related rates.

The problem is as follows:

A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 $\displaystyle m^3/min$, find the rate at which water level is rising when the water is 3 m deep.

So the problem tells me that $\displaystyle dv/dt=2$, h=4, and r=2. After taking my derivative in respect to time, I have to solve $\displaystyle dr/dt$..but then what do I plug in for $\displaystyle dh/dt$? Is that where the "3 m deep" part comes into play?

Thank you.

2. isn't dr/dt =2m^3??

so find dh/dt

3. I thought that because it's cubed, that's the volume.
Plus, the question asks for the rate, not the height. :/

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

$\displaystyle \frac{r}{h} = \frac{2}{4}$ , so $\displaystyle r = \frac{h}{2}$ ...

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

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

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

plug in your values and determine the value of $\displaystyle \frac{dh}{dt}$

5. Originally Posted by skeeter
$\displaystyle V = \frac{\pi}{3} r^2 h$

$\displaystyle \frac{r}{h} = \frac{2}{4}$ , so $\displaystyle r = \frac{h}{2}$ ...

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

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

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

plug in your values and determine the value of $\displaystyle \frac{dh}{dt}$
Oh, okay..thank you..