Question

Gravel is being dumped from a conveyor belt at a rate of 50 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 14 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3πr^2 h.

Answer 1

so hmm check the picture below....notice that r = h/2

thus
`\bf V=\cfrac{1}{3}\pi r^2 h\qquad \begin{cases} d=h\\ r=(d)/(2)=(h)/(2) \end{cases}\implies V=\cfrac{1}{3}\pi \left( \cfrac{h}{2} \right)^2h \\\\\\ V=\cfrac{\pi }{3}\cdot \cfrac{h^2}{2^2}\cdot h\implies V=\cfrac{\pi }{12}h^3\\\\ -----------------------------\\\\ \cfrac{dV}{dt}=\cfrac{\pi }{12}\cdot 3h^2\cfrac{dh}{dt}\implies \cfrac{dV}{dt}=\cfrac{\pi }{4}h^2\cfrac{dh}{dt}\implies \cfrac{4(dV)/(dt)}{\pi h^2}=\cfrac{dh}{dt}`

and keeping in mind that
`\bf \cfrac{dV}{dt}=50`, surely you already know what
`\bf \left. \cfrac{dh}{dt}\ \right|_(h=14)` is