Water is leaking out of an inverted conical tank at a rate of 2300.0 cubic centi

Question

Water is leaking out of an inverted conical tank at a rate of 2300.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 11.0 meters and the diameter at the top is 3.0 meters. If the water level is rising at a rate of 17.0 centimeters per minute when the height of the water is 3.5 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-2300.0. 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

First, we need everything in common terms (centimeters) H = 11 m = 1100 cm D = 3 m = 300 cm h = 3.5m = 350 cm Now, we need to set up a formula for volume: V = (1/3) * pi * r^2 * h Now, we need to find r when h = 350 r = 0 when h = 0 r = 150 when h = 1100 (1100 - 0) / (150 - 0) = 110 / 15 = 22 / 3 h = (22/3) * r r = (3/22) * h V = (1/3) * pi * r^2 * h V = (1/3) * pi * (3/22)^2 * h^2 * h V = (1/3) * pi * (9/484) * h^3 dV/dt = 3 * (1/3) * pi * (9/484) * h^2 * dh/dt dV/dt = (9 * pi / 484) * h^2 * dh/dt Now, we know that 2300 cm^3 of water is draining out every minute, so we need to modify our volume derivative: dV/dt = (9 * pi / 484) * h^2 * dh/dt - 2300 h = 350 dh/dt = 17 dV/dt = ? dV/dt = (9 * pi / 484) * 350^2 * 17 - 2300 dV/dt = (122500 * 17 * 9 * pi - 484 * 2300) / 484 dV/dt = 119355.5792 So, 119355.5792 cm^3 is being pumped in every minute

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