Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. 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 11 feet high?

Question

margotlakus · Answer

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

holsteremission · Answer

You're looking for $\dfrac{\mathrm dh}{\mathrm dt}$ given that $\dfrac{\mathrm dv}{\mathrm dt}=40$, and you're considering the pile at the point when $h=11$.

Differentiating with respect to time, you have
$$\frac{\mathrm dv}{\mathrm dt}=\frac{\pi}{3}\left(2rh\frac{\mathrm dr}{\mathrm dt}+r^2\frac{\mathrm dh}{\mathrm dt}\right)$$You have everything you need *except* the requisite information about the conical pile's radius. But since you know the pile's base diameter (i.e. twice its radius) is the same as the height, you have $h=2r$, or $r=\dfrac{h}{2}$. Differentiating this, you know that $\dfrac{\mathrm dr}{\mathrm dt}=\dfrac{1}{2}\dfrac{\mathrm dh}{\mathrm dt}$. Substituting, you get
$$\frac{\mathrm dv}{\mathrm dt}=\frac{\pi}{3}\left(\frac{h^2}{2}\frac{\mathrm dh}{\mathrm dt}+\frac{h^2}{4}\frac{\mathrm dh}{\mathrm dt}\right)$$Plug in everything you do know, then solve for $\dfrac{\mathrm dh}{\mathrm dt}$. (Don't forget your units)