Water is leaking out of an inverted conical tank at a rate of 9700.0 cubic centi

Question

Water is leaking out of an inverted conical tank at a rate of 9700.0 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 3.5 meters. If the water level is rising at a rate of 25.0 centimeters per minute when the height of the water is 4.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, rac{dV}{dt}=R-9700.0. 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 rac{1}{3}pi r^2 h

Explanation / Answer

Given,
Height=H = 14 m = 1400 cm
Diameter=D = 3.5 m = 350 cm
h = 4.5m = 450 cm
Now, we know the volume of cone:
V = (1/3) * pi * r^2 * h
Now, we need to find r when h = 450 cm
r = 0 when h = 0
r =D/2= 175 cm, when H = 1400 cm
so, by similar triangle
(1400 - 0) / (175 - 0) = 1400 /175 = 8=h/r
h = 8 * r
r = h/8
Now, V = (1/3) * pi * r^2 * h
V = (1/3) * pi * (1/8)^2 * h^2 * h
V = (1/3) * pi * (1/64) * h^3
dV/dt = 3 * (1/3) * pi * (1/64) * h^2 * dh/dt
dV/dt = ( pi /64) * h^2 * dh/dt
Now, we know that 9700 cm^3 of water is draining out every minute, so we need to modify our volume derivative:
dV/dt = (pi / 64) * h^2 * dh/dt - 9700
h = 450
dh/dt = 25
dV/dt = ?
therefor,
dV/dt = (pi /64) * 450^2 * 25 - 9700
dV/dt = (250000 * 27 * 49 * pi - 484 * 6400) / 484
dV/dt = 238804.88 cm^3 is being pumped in every minute

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