Cone differentiation question help! A right circular cone has a constant volume. The height h and the base radius r can both vary. Find the rate at which h is changing with respect to r at the instant when r and h are equal.
For this question the answer I got was 1, but its actually -2? I got my answer by finding dh/dt using chain rule so dh/dv * dv/dr. I derived the volume after eliminating r for one and h for the other and so got 1. Does anyone know where I went wrong? And how to approach this question?
Thank you 
 A: There is no explicit $t$ here, so using it is not necessary, and will generally just complicate things. As you don't show more details of what you did, I don't know what you specifically did incorrectly. However, here is how I would approach the problem.
First, the equation for the volume, $V$, of a right circular code of height $h$ and base radius $r$ is given by
$$V = \cfrac{\pi r^2 h}{3} \tag{1}\label{eq1}$$
You are given that $V$ is a constant, and want to find how $h$ is changing wrt $r$. As such, consider $h$ to be a function of $r$ so \eqref{eq1} becomes
$$V = \cfrac{\pi r^2 h\left(r\right)}{3} \tag{2}\label{eq2}$$
We could use the product rule when differentiating wrt $r$, but as the LHS is a constant and we can easily get $h\left(r\right)$ by itself by moving everything else to that side, I will do that as it simplifies the calculations somewhat.
$$h\left(r\right) = \cfrac{3V}{\pi}r^{-2} \tag{3}\label{eq3}$$
Differentiating wrt $r$ now gives that
$$\cfrac{dh}{dr} = \cfrac{3V}{\pi}\left(-2\right)r^{-3} = \cfrac{-6V}{\pi}r^{-3} \tag{4}\label{eq4}$$
Note we just left $V$ alone as it's given to be constant. For some $r = r_0$ where $h\left(r_0\right) = r_0$, use \eqref{eq2} to get $V$ and then substitute this into \eqref{eq4} to get your stated correct answer. I will leave it to you to finish these final details.
One final note is, in case you weren't aware, that you should be able to tell immediately that $1$ is not the right answer. This is because from \eqref{eq1}, when $r$ is increasing, then $h$ must be decreasing for $V$ to not be changing. Thus, the rate at which $h$ is changing wrt to $r$ must be negative, so you can tell something is wrong even without knowing the answer. This sort of sanity type check won't catch all mistakes, but it's often useful to consider whether or not your answer makes reasonable sense.
