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3 years ago

14

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. h

ow fast is the water draining out of the tank in cubic feet / minute if the water level is dropping at 6 inches/min?

1 answer:

Iteru [2.4K]3 years ago

6 0

<span> </span>For any prism-shaped geometry, the volume (V) is assumed by the product of cross-sectional area (A) and height (h).

<span> V = Ah </span>

<span>
Distinguishing with respect to time gives the relationship between the rates.
dV/dt = A*dh/dt</span>

<span> in the meantime the area is not altering </span>

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

<span>
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.</span>

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1)

Answer:

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Part 2)

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