The radius of the water's surface at a depth of h is
r = 11/48 h
So, the volume v is
v = 1/3 πr^2 h
= π/3 (11/48)^2 h^3
so,
dv/dt = 121π/2304 h^2 dh/dt
So, when h=400,
dv/dt = (121π/2304)(400)^2(16) = 422370 cm^3/min
That is the rate at which v is increasing, even while leaking. SO, the real rate of water pumpage is
435870 cm^3/min
Water is leaking out of an inverted conical tank at a rate of 13500.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 12.0 meters and the diameter at the top is 5.5 meters. If the water level is rising at a rate of 16.0 centimeters per minute when the height of the water is 4.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
2 answers
thank you very much! It has taken me almost 2 hours to do this