Question

Gravel is being dumped from a conveyor belt at a rate of 30 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 25 ft high? The height is increasing at ft/min.

Answer 1

Answer:

1.528 ft/min

Explanation:

$\frac{dv}{dt}$= Rate of change of volume = 30 ft³/min

d = Diameter of cone = Height of cone = h = 25 ft

Volume of cone

$\frac{1}{3}\pi \frac{d^2}{4}h=v\\\Rightarrow \frac{1}{3}\pi \frac{h^3}{4}=v$

Differentiating with respect to time

$\frac{1}{3}\pi \frac{3h^2}{4}\frac{dh}{dt}=\frac{dv}{dt}\\\Rightarrow \frac{dh}{dt}=\frac{dv}{dt}\frac{4}{\pi h^2}\\\Rightarrow \frac{dh}{dt}=30\times \frac{4}{\pi 25^2}\\\Rightarrow \frac{dh}{dt}=1.528\ ft/min$

∴ The height is increasing at 1.528 ft/min