Water is leaking out of an inverted conical tank at a rate of 4100.0 cubic centi

Question

Water is leaking out of an inverted conical tank at a rate of 4100.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 10.0 meters and the diameter at the top is 3.0 meters. If the water level is rising at a rate of 19.0 centimeters per minute when the height of the water is 1.0 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 is volume of water, dV/dt= R - 4100.0 . 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 1/3*pi*r^2*h

Explanation / Answer

Let's start out with the fact that Volume,henceforth v, = (? r^2 h / 3) Now in this case the change in volume is (? r^2 h) - 4100 and dh/dt = 19. Since the height is 10 when the radius is 1.5, h/r=10/1.5 h= 20r/3 So letting v = the changing volume, we have v = (20/9?r^3) - 4100t dv/dt = 20/3?r^2 dr/dt - 4100 = 20/3?r^2dr/dt. But we know that dh/dt = 20/3*dr/dt and that dh/dt = 19, so dr/dt = (3/20)*19 = 57/20 and r = 3/20 m = 300/20 cm = 15cm. So now we have dv/dt = ((20/3) * (15)^2 * (57/20) * pi -4100 dv/dt =9330.31 So the water is being pumped in at 9330.31 cubic cm/min

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