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a sphere is expanding in such a way that the area of any circular cross section through the sphere's center is increasing at a constant rate of 2 cm^2/sec. at the instant when the radius of the sphere is 4 centimeters, what is the rate of change of the sphere's volume?

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Final answer:

The rate of change of the sphere's volume is 128π cm³/sec when the radius of the sphere is 4 centimeters.

Step-by-step explanation:

To find the rate of change of the sphere's volume, we can use the formula for the volume of a sphere. The volume of a sphere is given by V = (4/3)πr³, where V is the volume and r is the radius.

In this case, we are given that the area of any circular cross section through the sphere's center is increasing at a constant rate of 2 cm²/sec.

This means that the rate of change of the area with respect to time, dA/dt, is 2 cm²/sec.

The area of a circular cross section is given by A = πr².

To find the rate of change of the volume with respect to time, dV/dt, we need to differentiate the formula for the volume with respect to time.

Taking the derivative of V = (4/3)πr³ with respect to time, we get dV/dt = 4πr²(dr/dt).

Since we are given that dA/dt = 2 cm²/sec, we can substitute this value into the equation for the derivative of the volume to solve for dV/dt.

So, dV/dt = 4πr²(dr/dt)

= 4π(4)²(2)

= 128π cm³/sec.

Therefore, the rate of change of the sphere's volume is 128π cm³/sec when the radius of the sphere is 4 centimeters.

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