Let "R" be the unknown rate at which water is being pumped in.

Question

Water is leaking out of an inverted conical tank at a rate of 12500.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8.0 meters and the diameter at the top is 3.0 meters. If the water level is rising at a rate of  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. 2.9·106Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if  is volume of water, dV/dt = R -12500.0 . 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 .1/3pir^2*h

Answer 1

At what rate is the water level rising when the height of the water is 4 meters?

call this value K

V = (1/3) pi R^2 * H

Per similar triangles:

1.5 in proportion to 8

equals

radius in proportion to height

1.5/8 = r/h

1.5 h / 8 = r

15h/80 = r

3h/16 = r

V = (1/3) pi (3h/16)^2 * h

= (1/3) pi (9h^2)/256 * h

= (1/3) pi (9/256) h^3

= (3/256) pi h^3

dV/dt = (9/256) pi h^2 dH/dt

R-12500 = (9/256) pi * 4^2 * K

R = (9/256) pi * 4^2 * K + 12500