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The radius of a cone is increasing at a rate of 2 inches per minute. The height of the cone is always 4 times the radius. Find the rate of change of the volume when the radius is 5 inches.

User Dancer
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Final answer:

The rate of change of the volume when the radius is 5 inches is 200π cubic inches per minute.

Step-by-step explanation:

To find the rate of change of the volume when the radius is 5 inches, we need to first express the volume of the cone as a function of the radius. The formula for the volume of a cone is V = (1/3)πr^2h, where r is the radius and h is the height of the cone.

Given that the height of the cone is always 4 times the radius, we have h = 4r. Substituting this expression for h into the volume formula, we get V = (1/3)πr^2(4r) = (4/3)πr^3.

Now we can find the rate of change of the volume with respect to time, denoted as dV/dt, by differentiating the volume function with respect to r. Taking the derivative of V = (4/3)πr^3 with respect to r, we get dV/dt = 4πr^2(dr/dt).

Given that the radius is increasing at a rate of 2 inches per minute, dr/dt = 2. Substituting this value into the derivative expression, we have dV/dt = 4πr^2(2).

Finally, when the radius is 5 inches, we plug in this value for r in the rate of change equation: dV/dt = 4π(5)^2(2) = 200π cubic inches per minute.

User Ilene
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