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Find the rate of change of the radius of a sphere at the point in time when the radius is 6 feet if the volume is increasing at the rate of 8π cubic feet per second?

User Lucky
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Final answer:

The rate of change of the radius of a sphere when the volume is increasing at a rate of 8π cubic feet per second and the radius is 6 feet is 1/18 feet per second.

Step-by-step explanation:

To find the rate of change of the radius of a sphere when the radius is 6 feet and the volume is increasing at a rate of 8π cubic feet per second, we can use the formula for the volume of a sphere, which is V = (4/3)πr³. By taking the derivative with respect to time, we get dV/dt = 4πr²(dr/dt). We are given that dV/dt = 8π and we want to solve for dr/dt when r = 6 feet. Plugging in the known values, we get 8π = 4π(6²)(dr/dt), which simplifies to dr/dt = 8π / (4π· 36). Therefore, the rate of change of the radius at that instant is dr/dt = 1/18 feet per second.

User Chen Pang
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