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The volume of a sphere is increasing at a constant rate of 5617 cubic centimeters per minute. At the instant when the volume of the sphere is 1461 cubic centimeters, what is the rate of change of the surface area of the sphere? The volume of a sphere can be found with the equation $V=\frac{4}{3} \pi r^{3}$ and the surface area can be found with $S=4 \pi r^{2}$. Round your answer to three decimal places (if necessary).
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$\mathrm{cm}^{2}$
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Step 1 ：First, we need to calculate the radius of the sphere when the volume is 1461 cubic centimeters. We can use the formula for the volume of a sphere, \(V=\frac{4}{3} \pi r^{3}\), and solve for \(r\).

Step 2 ：Then, we can calculate the rate of change of the radius with respect to time, \(\frac{dr}{dt}\), using the formula for the rate of change of the volume with respect to time, \(\frac{dV}{dt}=4 \pi r^{2} \frac{dr}{dt}\), and the given rate of change of the volume, \(\frac{dV}{dt}=5617\).

Step 3 ：Finally, we can calculate the rate of change of the surface area with respect to time, \(\frac{dS}{dt}\), using the formula for the rate of change of the surface area with respect to time, \(\frac{dS}{dt}=8 \pi r \frac{dr}{dt}\), and the calculated values of \(r\) and \(\frac{dr}{dt}\).

Step 4 ：The rate of change of the surface area of the sphere at the instant when the volume of the sphere is 1461 cubic centimeters is approximately \(\boxed{1595.930}\) cubic centimeters per minute.