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The volume of a sphere is decreasing at a constant rate of 6031 cubic centimeters per second. At the instant when the volume of the sphere is 3292 cubic centimeters, what is the rate of change of the radius? 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
$\frac{\mathrm{cm}}{\mathrm{sec}}$
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Step 1 ：Given that the volume of the sphere is decreasing at a rate of \(-6031\) cubic centimeters per second, and the volume of the sphere at a certain time \(t_0\) is \(3292\) cubic centimeters.

Step 2 ：The volume \(V(t)\) of a sphere with radius \(r(t)\) is given by the formula \(V(t)=\frac{4}{3}\pi[r(t)]^3\).

Step 3 ：Differentiating both sides with respect to \(t\), we get \(V'(t)=4\pi[r(t)]^2r'(t)\).

Step 4 ：We can find the radius of the sphere at time \(t_0\) by substituting \(V(t_0)=3292\) into the volume formula and solving for \(r(t_0)\):

Step 5 ：\[3292=\frac{4}{3}\pi[r(t_0)]^3\Rightarrow r(t_0)=\sqrt[3]{\frac{3292 \times 3}{4\pi}}\approx 8.57\] centimeters.

Step 6 ：Substitute \(V'(t_0)=-6031\), \(r(t_0)\approx 8.57\) into the expression for \(V'(t_0)\) and solve for \(r'(t_0)\):

Step 7 ：\[-6031=4\pi(8.57)^2r'(t_0)\Rightarrow r'(t_0)=\frac{-6031}{4\pi(8.57)^2}\approx -0.082\] centimeters per second.

Step 8 ：Therefore, the rate of change of the radius of the sphere at that instant is \(\boxed{-0.082}\) centimeters per second.