let the input rate be x cm^3/min
effective rate of change of volume = x-9400 cm^3/min
or
dV/dt = x-9400
At any time of t min, let the height of water be h cm and the radius of the water level be r cm
As suggested, by similar triangles,
h/r = 14/2 = 7
h = 7r or r = h/7
V = (1/3)π r^2 h = (1/3)π(h^2/49)(h)
V = (2/147) π h^3
dV/dt = (2/49)π h^2 dh/dt
given: when h = 300, dh/dt = 28
x - 9400 = (2/49)π(300^2)(28)
solve for x, it will be in cm^3/min
check my arithmetic, should have written it out on paper first
Water is leaking out of an inverted conical tank at a rate of 9400.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 meters and the diameter at the top is 4.0 meters. If the water level is rising at a rate of 28.0 centimeters per minute when the height of the water is 3.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. [answer]
Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, dVdt=R−9400.0. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by 13πr2h.
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