The radius of a spherical balloon is increasing at a rate of 3 centimeters per minute. How fast is the surface area changing when the radius is 8 centimeters? Hint: The surface area is $S=4 \pi r^{2}$. Rate of change of surface area $=$

Step 1 ：We are given that the radius of a spherical balloon is increasing at a rate of 3 centimeters per minute. We are asked to find how fast the surface area is changing when the radius is 8 centimeters.

Step 2 ：The surface area of a sphere is given by the formula \(S=4 \pi r^{2}\).

Step 3 ：To find the rate of change of the surface area with respect to time, we differentiate the surface area with respect to time. This gives us \(\frac{dS}{dt} = 8\pi r \frac{dr}{dt}\).

Step 4 ：We are given that \(\frac{dr}{dt} = 3\) cm/min.

Step 5 ：Substituting \(r = 8\) cm and \(\frac{dr}{dt} = 3\) cm/min into the equation, we get \(\frac{dS}{dt} = 24\pi r\).

Step 6 ：Substituting \(r = 8\) cm into the equation, we get \(\frac{dS}{dt} = 192\pi\).

Step 7 ：Final Answer: The rate of change of the surface area when the radius is 8 cm is \(\boxed{192\pi}\) cm²/min.