Using implicit differentiation, it is found that the rate of change of the surface area of the sphere -2201.293 square centimeters per second.
[tex]V = \frac{4\pi r^3}{3}[/tex]
[tex]S = 4\pi r^2[/tex]
Applying implicit differentiation, the rate of change of both the volume and the surface area can be found, as follows:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]\frac{dS}{dt} = 8\pi r\frac{dr}{dt}[/tex]
Volume of 3131 cubic centimeters, hence:
[tex]V = \frac{4\pi r^3}{3}[/tex]
[tex]\frac{4\pi r^3}{3} = 3131[/tex]
[tex]4\pi r^3 = 3(3131)[/tex]
[tex]r = \sqrt[3]{\frac{3(3131)}{4\pi}}[/tex]
[tex]r = 9.0754[/tex]
The volume of a sphere is decreasing at a constant rate of 9988 cubic centimeters per minute, hence:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]-9988 = 4\pi (9.0754)^2\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt} = -\frac{9988}{4\pi (9.0754)^2}[/tex]
[tex]\frac{dr}{dt} = -9.651[/tex]
Then, the rate of change of the surface area of the sphere is given by:
[tex]\frac{dS}{dt} = 8\pi (9.0754)(-9.651) = -2201.293[/tex]
The rate of change of the surface area of the sphere -2201.293 square centimeters per second.
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