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6 November, 06:13

A water heater that has the shape of a right cylindrical tank with a radius of 1 foot and a height of 4 feet is being drained. how fast is the water draining out of the tank in cubic feet / minute if the water level is dropping at 6 inches/min?

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  1. 6 November, 07:56
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    For any prism-shaped geometry, the volume (V) is assumed by the product of cross-sectional area (A) and height (h).

    V = Ah

    Distinguishing with respect to time gives the relationship between the rates.

    dV/dt = A*dh/dt

    in the meantime the area is not altering

    dV/dt = π * (1 ft) ^2 * (-0.5 ft/min)

    dV/dt = - π/2 ft^3/min ≈ - 1.571 ft^3/min

    Water is draining from the tank at the rate of π/2 ft^3/min.
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