Final answer:
The question involves calculating the rate of change of the sphere's surface area while its volume increases at a constant rate. This requires using calculus, specifically the chain rule, to relate the rates of change of volume and surface area through the radius of the sphere.
Step-by-step explanation:
The student is asking about the rate of change of the surface area of a sphere given that the volume of the sphere is increasing at a constant rate. We know that the volume V of a sphere with radius r is given by V = 4/3 π r3 and the surface area A is given by A = 4 π r2. We are also given the volume at a particular instant to be 31353135 cubic meters and the rate of change of volume as 7857 cubic meters per minute.
First, we calculate the radius from the given volume by rearranging the volume formula: r = (3V/4π)1/3. Then, we use the chain rule from calculus to find the rate of change of the surface area with respect to time (dA/dt) by finding dr/dt from the given dV/dt and then dA/dt as dA/dt = 8 π r (dr/dt). After finding dr/dt, we substitute that and the value of the radius into the latter equation to find the rate of change of the surface area.