Related rates with a cone I am trying to figure out the rate the water level increases in a conical tank that is 3 m height, 2 m radius at top and water flows in at $2\text{m} ^3 / \text{minute}$
I know that
$$(1/3) \pi r^2 h = V$$
$$4 \pi = V$$
or at 2 seconds the volume of the water is $8.37758$
so now I have 
$$(1/3) \pi r^2 h * dh/dt = V * dV/dt$$
which gives an incorrect answer, I am not sure if I use t = 2 volume and height or what.
 A: By similar triangles, observe that:
$$
\dfrac{h}{3}=\dfrac{r}{2} \iff r=\dfrac{2h}{3}
$$
Hence, substituting into the formula for the volume of a cone will help us to avoid product rule:
$$
V=\dfrac{1}{3}\pi \left(\dfrac{2h}{3}\right)^2h = \dfrac{4\pi}{27}h^3
$$
Differentiating each side with respect to $t$ yields:
$$
\dfrac{dV}{dt} = \dfrac{4\pi}{27}(3h^2)\dfrac{dh}{dt} = \dfrac{4\pi}{9}h^2\dfrac{dh}{dt} \iff \boxed{\dfrac{dh}{dt} = \dfrac{9}{4\pi h^2}\dfrac{dV}{dt}}
$$
Since we know that $dV/dt = 2$, we simply need to plug in the height at the specific instant in time that the question is asking for into the final equation, and we are done. If the radius is given instead, simply multiply it by $3/2$ to convert to a height.

EDIT: Alright, here's my MS Paint skills. Hopefully this diagram explains where the similar triangles came from:

A: The rate of increase depends on the depth.  As you say, $V=\frac 13 \pi r^2 h$.  The calculation of the tank volume is not useful.  Then $\frac {dV}{dt}=2=\frac 13\pi(2rh\frac {dr}{dt}+r^2\frac {dh}{dt})$ using the product rule.  Now the dimensions of the cone give you $\frac {dr}{dh}$.  Can you use that to combine the two terms in the last parentheses?
