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The radius of a sphere is decreasing at a constant rate of 3 feet per minute. At the instant when the volume of the sphere is 108 cubic feet, what is the rate of change of the volume? The volume of a sphere can be found with the equation $V=\frac{4}{3} \pi r^{3}$. Round your answer to three decimal places.
Answer Attempt 1 out of 2
$\mathrm{ft}^{3}$
$\frac{\mathrm{t}^{3}}{\min }$
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Step 1 ：The problem is asking for the rate of change of the volume of a sphere when the volume is 108 cubic feet and the radius is decreasing at a rate of 3 feet per minute. This is a problem of related rates in calculus.

Step 2 ：We know that the volume of a sphere is given by \(V=\frac{4}{3} \pi r^{3}\) and we are given \(\frac{dr}{dt} = -3\) (since the radius is decreasing). We need to find \(\frac{dV}{dt}\) when \(V = 108\) cubic feet.

Step 3 ：First, we need to find the radius when the volume is 108 cubic feet. Then, we can differentiate the volume equation with respect to time to get an equation for \(\frac{dV}{dt}\), and substitute the known values to find the rate of change of the volume.

Step 4 ：Let's start by finding the radius when the volume is 108 cubic feet. Using the volume equation, we find that \(r = 2.9542350655280893\).

Step 5 ：Now that we have the radius when the volume is 108 cubic feet, we can differentiate the volume equation with respect to time to get an equation for \(\frac{dV}{dt}\), and substitute the known values to find the rate of change of the volume.

Step 6 ：Substituting \(V = 108\), \(r = 2.9542350655280893\), and \(\frac{dr}{dt} = -3\) into the equation, we find that \(\frac{dV}{dt} = -329.01918041049595\).

Step 7 ：Final Answer: The rate of change of the volume of the sphere when the volume is 108 cubic feet and the radius is decreasing at a rate of 3 feet per minute is approximately \(\boxed{-329.019}\) cubic feet per minute.