# Water is leaking out of an inverted conical tank at a rate of 9500.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6.0 meters and the diameter at the top is 6.5 meters. If the water level is rising at a rate of 24.0 centimeters per minute when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Volume for a cone is:

V = pi * r^2 * h/3

We don't have enough information for the radius here.  However, we can find the radius for the height in question using a proportion of similar triangles.  We know the ratio of the height of the tank to the...

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Volume for a cone is:

V = pi * r^2 * h/3

We don't have enough information for the radius here.  However, we can find the radius for the height in question using a proportion of similar triangles.  We know the ratio of the height of the tank to the radius (diameter = 650 cm, so the radius is 325 cm).  So:

325/600 = r/h

r = 325h/600 = 13h/24

So, plugging this in:

V = (pi/3) * (13/24)^2 * h^3

Taking the derivative of each side:

dV/dt = (pi/3)(13/24)^2 * (3h^2) * dh/dt

Plugging in the numbers:

dV/dt = (pi/3)(13/24)^2 * (3*350^2) * 24

dV/dt = 2,709,950.913 cubic cm / min

Now, dV/dt = V(in) - V(out).  So

V(in) - 9500 = 2,709,950.913

V(in) = 2,719,450.913 cubic cm / min

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