Question

Gravel is being dumped from a conveyor belt at a rate of 30 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 15 feet high?

Answer 1

Answer:

dh/dt = 0.0566 feet / min

Explanation:

Volume per second, dV/dt = 30 cubic feet per minute

Diameter of cone = height of cone = h

at h = 15 feet, dh/dt = ?

Volume of cone is

V = πr²h

where, r be the radius of cone

According to question

radius of cone = half of diameter of cone = h /2

So,

$V=\pi \times \left ( \frac{h^{2}}{4} \right )\times h$

$V=\pi \times \frac{h^{3}}{4}$

Differentiate both sides with respect to h

$dV/dh = \frac{\pi }{4}\times 3h^{2}\times \frac{dh}{dt}$

by substituting the values

$30=\frac{3\times3.14}{4}\times 15\times 15\times \frac{dh}{dt}$

dh/dt = 0.0566 feet / min