a water tank is created by revolving the graph y=1/x about the y-axis, with the bottom of the tank at y=1. the volume of the tank is given by [v(h) = the integral from 1 to h of (pi/y^2)dy] where h is the height of the water in the tank. Initially, the tank is empty, but water begins to flow into the tank at a rate of 1.5 cubic feet per minute. Determine how fast the level of the water is rising when the water is 2 feet deep.

asked by Slaya

4 years ago

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1 answer

Since v(h) = ∫[1,h] π/y^2 dy
dv/dt = π/h^2 dh/dt
1.5 = π/2 dh/dt
3/π = dh/dt

answered by oobleck

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Question

a water tank is created by revolving the graph y=1/x about the y-axis, with the bottom of the tank at y=1. the volume of the tank is given by [v(h) = the integral from 1 to h of (pi/y^2)dy] where h is the height of the water in the tank. Initially, the tank is empty, but water begins to flow into the tank at a rate of 1.5 cubic feet per minute. Determine how fast the level of the water is rising when the water is 2 feet deep.

1 answer

Since v(h) = ∫[1,h] π/y^2 dy
dv/dt = π/h^2 dh/dt
1.5 = π/2 dh/dt
3/π = dh/dt

Ask a New Question or answer this question.

oobleck · Answer

Since v(h) = ∫[1,h] π/y^2 dy dv/dt = π/h^2 dh/dt 1.5 = π/2 dh/dt 3/π = dh/dt