# Gasoline is pouring into a cylindrical tank of radius 3ft. When the depth of the gasoline is 7ft the depth is increasing at 0.5 ft/sec.?

How fast is the volume of gasoline changing at that instant?

Round answer to three decimal places.

Relevance

V = pi * r^2 * h

r = 3

V = pi * 9 * h

dV/dt = pi * 9 * dh/dt

The depth doesn't matter

dh/dt = 0.5

dV/dt = 9 * pi * 0.5 = 4.5 * pi

Your calculator can do the rest.

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• We would have to know the orientation of the tank, and that information is not given.

The simplest case would be a cylinder with a vertical axis. That would given it a constant surface area, regardless of depth.

A horizontal axis would be more typical of a cylindrical fuel tank. That would make the solution a little more difficult, and in that case we would also need to know the length of the cylinder.

An axis with an oblique angle of elevation would be unusual, but certainly possible. Things could get very involved in that case, and again we would still need more information on its dimensions.

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