Question 8 1 pts The following questions are about a spherical balloon that is being filled with air such that its radius is increasing at a constant rate of $2 \mathrm{~cm} / \mathrm{sec}$. Part A: How fast is the surface area increasing when the radius of the sphere is $10 \mathrm{~cm}$ ? Round to the nearest thousandths. Do not include units in your answer. Note, however, that on the AP exam you are required to include units. Part B: How fast is the volume increasing when the radius is $10 \mathrm{~cm}$ ? Round to the nearest thousandths. Do not include units in your answer. Note, however, that on the AP exam are required to include units.

Step 1 ：We are given a problem of related rates in calculus. The radius of a sphere is increasing at a constant rate of 2 cm/sec. We need to find how fast the surface area and the volume are increasing when the radius is 10 cm.

Step 2 ：The formula for the surface area of a sphere is \(A = 4\pi r^2\) and the formula for the volume of a sphere is \(V = \frac{4}{3}\pi r^3\). We can differentiate these with respect to time to find the rates of change of the surface area and volume with respect to time.

Step 3 ：Given that \(r = r\) and \(\frac{dr}{dt} = 2\), we can calculate \(\frac{dA}{dt} = 16\pi r\) and \(\frac{dV}{dt} = 8.0\pi r^2\).

Step 4 ：Substituting \(r = 10\) into the above equations, we find that \(\frac{dA}{dt} = 160\pi\) and \(\frac{dV}{dt} = 800.0\pi\).

Step 5 ：Final Answer: The rate of change of the surface area when the radius is 10 cm is \(\boxed{160\pi}\) cm²/sec and the rate of change of the volume when the radius is 10 cm is \(\boxed{800\pi}\) cm³/sec.