Answer:
0.031 ft/min
Step-by-step explanation:
Rate of unloading, [tex]\frac{dV}{dt} = 14\ ft ^{3}/min[/tex]
height, h = 4ft
The formula for volume of a cone is;
[tex]V = \frac{1}{3}\pi r^{2}h[/tex]
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
[tex]V = \frac{1}{3}\pi (3h)^{2}h[/tex]
[tex]V = \frac{1}{3}\pi (9h^{2})h[/tex]
[tex]V = 3\pi h^{3}[/tex]
Differentiating
[tex]\frac{dV}{dt} = 9\pi h^{2}\frac{dh}{dt}[/tex]
Substituting the given values,
[tex]14 ft^{3}/min = 9\pi (4 ft)^{2}\frac{dh}{dt}[/tex]
[tex]\frac{dh}{dt} = \frac{ 14 ft^{3}/min}{9\pi 16 ft^{2}}[/tex]
[tex]\frac{dh}{dt} = 0.03094679 ft/min[/tex]
≈ 0.031 ft/min
