A conical tank( with vertex down) is 10 feet across the top and 18 feet deep. As the water flows into the tank, the change is the radius of the water at a rate of 2 feet per minute, find the rate of change of the volume of the water when the radius of the water is 2 feet.

Answer 1

Let y be the water level height above the vertex. The volume of water is
V = (pi/3)r^2 y

From the dimensions you have provided, r = (5/18) y
y = (18/5) r

V = (pi/3)(18/5)^2 r^3

Calculate dV/dt = (dV/dr)*(dr/dt) and evaluate it when r = 2 ft.

In your case, dr/dt = 2 ft/min