A sphere is partially filled with air. If the volume of the sphere is increasing at a rate of 548 cubic feet per second, what is the rate, in feet per second, at which the radius of the sphere is changing when the radius is 6 feet? Submit an exact answer. Remember that the volume of a sphere is V = (4/3) π*r^3.

Question

Thomas Waller

Mathematics

28 July, 06:39

A sphere is partially filled with air. If the volume of the sphere is increasing at a rate of 548 cubic feet per second, what is the rate, in feet per second, at which the radius of the sphere is changing when the radius is 6 feet? Submit an exact answer. Remember that the volume of a sphere is V = (4/3) π*r^3.

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Answers (1)

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Kaleigh Wood · Answer 1

137 / (36π) ft/s

Step-by-step explanation:

The rate of change of volume will be the product of the rate of change of radius and the area of the sphere. The area of the sphere is ...

A = 4πr² = 4π (6 ft) ² = 144π ft²

Then the relationship above is ...

dV/dt = A·dr/dt

548 ft³/s = (144π ft²) ·dr/dt

dr/dt = (548 ft³/s) / (144π ft²) = 137 / (36π) ft/s ≈ 1.2113 ft/s