Mathematics: Gravel is being unloaded from a truck and falls into a pile shaped like a cone at a rate of 14 ft3/min. The radius of the cone base is three times the height of the cone. Find the rate (in ft/min) at which the height of the gravel changes when the pile has a height of 4 ft

Gravel is being unloaded from a truck and falls into

helplessstudent3

19.01.2022

0.031 ft/min

Step-by-step explanation:

Rate of unloading, $\frac{dV}{dt} = 14\ ft ^{3}/min$

height, h = 4ft

The formula for volume of a cone is;

$V = \frac{1}{3}\pi r^{2}h$

Given the radius of the cone base to be three times the height of the cone;

r = 3h

Substituting for r = 3h in the volume of the cone

$V = \frac{1}{3}\pi (3h)^{2}h$

$V = \frac{1}{3}\pi (9h^{2})h$

$V = 3\pi h^{3}$

Differentiating

$\frac{dV}{dt} = 9\pi h^{2}\frac{dh}{dt}$

Substituting the given values,

$14 ft^{3}/min = 9\pi (4 ft)^{2}\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{ 14 ft^{3}/min}{9\pi 16 ft^{2}}$

$\frac{dh}{dt} = 0.03094679 ft/min$

≈ 0.031 ft/min

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