Gravel off of a conveyor belt problem. Gravel is being dumped from a conveyor belt at a rate of $30\, \mathrm{ft^3/min}$ . It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is $23\, \mathrm{ft}$ high? I'm solving for the $dh/dt$. How do I eliminate the $dr/dt$ term? Could you explain it step by step?

You eliminate the $\frac {dr}{dt}$ term because you have $2r=h$, so your volume is proportional to $h^3$. You have $V=\frac 13 \pi r^2h=\frac 13 \pi \frac {h^3}4=\frac {\pi h^3}{12}$ Now you can use what you know about $\frac {dh}{dt}$