Gravel is being dumped from a conveyor belt at a rate of 40 ft3/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 13 ft high?

Answer:[tex]\frac{dh}{dt}=\frac{160}{169\pi }  ft/min[/tex]Step-by-step explanation:This is a classic related rates problem.  Gotta love calculus!Start out with the formula for the volume of a cone, which is[tex]V=\frac{1}{3}\pi r^2h[/tex]and with what we know, which is [tex]\frac{dV}{dt}=40[/tex]and the fact that the diameter = height (we will come back to that in a bit).We need to find [tex]\frac{dh}{dt}[/tex] when h = 13The thing we need to notice now is that there is no information given to us that involves the radius.  It does, however, give us a height.  We need to replace the r with something in terms of h.  Let's work on that first.We know that d = h.  Because d = 2r, we can say that 2r = h, and solving for r gives us that [tex]r=\frac{h}{2}[/tex].Now we can rewrite the formula with that replacement:[tex]V=\frac{1}{3}\pi  (\frac{h}{2})^2h[/tex]Simplify that all the way down to[tex]V=\frac{1}{12}\pi  h^3[/tex]The derivative of that function with respect to time is[tex]\frac{dV}{dt}=\frac{1}{12}\pi(3h^2)\frac{dh}{dt}[/tex]Filling in what we have gives us this:[tex]40=\frac{1}{12}\pi (3)(13)^2\frac{dh}{dt}[/tex]Solve that for the rate of change of the height:[tex]\frac{dh}{dt}=\frac{160}{169\pi } \frac{ft}{min}[/tex]or in decimal form:[tex]\frac{dh}{dt}=.95\pi  \frac{ft}{min}[/tex]