Water is leaking out of an inverted conical tank at a rate of 10600.0 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 m and the the diameter at the top is 6.0 m. If the water level is rising at a rate of 27.0 cm/min when the height of the water is 1.5 m, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

draw a cross-section. When the water has depth y, the radius of the surface is r = 3/14 y

So, the volume of the water at depth y is

v = 1/3 πr^2y
= π/3 (3/14 y)^2 y
= 3π/196 y^3
dv/dt = 9π/196 y^2 dy/dt
If the water is being pumped in at a rate of k cm^3/min, then

dv/dt = k-10600

dv/dt = dv/dy dy/dt, so

k-10600 = 9π/196 y^2 dy/dt
Plugging in the given values, we have
k-10600 = 9π/196 * 150^2 * 27
k-10600 = 87636
k = 98436 cm^3/min