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?

Accepted Solution

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]