Can somebody show me a proof for this? Equation of the Volume of a Cone:

$$\mathbf{V}=\frac{1}{3}\pi r^2 h$$

Taking the Derivative in perspective to time because the radii, and height are either decreasing or increasing:

$$\frac{dV}{dt}=\frac{\pi}{3}[2rh\frac{dr}{dt}+r^2\frac{dh}{dt}]$$

I understand how to take the derivative of it and all, moreover, I am lost on how to prove this a different way because I have seen it done one way substituting $r$ and $h$ for $r(t)$ and $h(t)$ but do not know any other to prove it.
Is there another way to do it, and can somebody please show me?
 A: There is actually nothing to prove here, it is simply an application of derivatives. However there are some assumptions so let's look carefully at what is happening here.
We have the equation for the volume,
$$
V =\frac{1}{3} \pi r^2 h,
$$
and we are told that both $r$ and $h$ are changing in time. Therefore writing $r = r(t)$ and $h=h(t)$ is only making it clearer what is said in the question (you should read "r changes in time" as saying "r is a function of time"). Then differentiating follows by simply applying the chain rule.
But importantly this formula applies when we know that $r$ and $h$ change in time. If they didn't then we would not formulate it this way necessarily since if they don't depend on time the derivatives would be zero.
A: You have a function in two variable, essentially.
$$V(r, h) = \frac{1}{3} \pi r^2 h$$
So when differentiation with respect to time, you have to use the chain rule:
$$\frac{\text{d}{V}}{\text{d}t} = \frac{\text{d}{V}}{\text{d}r}\frac{\text{d}r}{\text{d}t} + \frac{\text{d}{V}}{\text{d}h}\frac{\text{d}h}{\text{d}t}$$   
$$\frac{\text{d}{V}}{\text{d}r} = \frac{2}{3}\pi r h$$
$$\frac{\text{d}{V}}{\text{d}h} = \frac{1}{3}\pi r^2$$  
Hence
$$\frac{\text{d}{V}}{\text{d}t} = \frac{\text{d}{V}}{\text{d}r}\frac{\text{d}r}{\text{d}t} + \frac{\text{d}{V}}{\text{d}h}\frac{\text{d}h}{\text{d}t} = $$ $$ = 2\pi r h \frac{\text{d}r}{\text{d}t} + \pi r^2 \frac{\text{d}h}{\text{d}t} = $$
$$ = \frac{\pi}{3}\left[\text{what you have written above}\right]$$
P.s.
It's obvious that derivatives are zero if neither $r$ nor $h$ depend on time. That is as obvious as useless to remark but who knows.
