1. **Problem Statement:** We have a spherical balloon with radius $r$ increasing at a constant rate of $\frac{dr}{dt} = 2$ cm/sec. Part A: Find the rate of change of the surface area $S$ when $r=10$ cm. Part B: Find the rate of change of the volume $V$ when $r=10$ cm. 2. **Formulas and Rules:** - Surface area of a sphere: $$S = 4\pi r^2$$ - Volume of a sphere: $$V = \frac{4}{3}\pi r^3$$ - To find how fast these quantities change with respect to time, use the chain rule: $$\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt}$$ $$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$$ 3. **Part A: Rate of change of surface area** - Differentiate surface area with respect to $r$: $$\frac{dS}{dr} = 8\pi r$$ - Substitute $r=10$ and $\frac{dr}{dt} = 2$: $$\frac{dS}{dt} = 8\pi (10) \times 2 = 160\pi$$ - Calculate numerical value: $$160\pi \approx 160 \times 3.1416 = 502.655$$ - Rounded to nearest thousandths: $$502.655$$ 4. **Part B: Rate of change of volume** - Differentiate volume with respect to $r$: $$\frac{dV}{dr} = 4\pi r^2$$ - Substitute $r=10$ and $\frac{dr}{dt} = 2$: $$\frac{dV}{dt} = 4\pi (10)^2 \times 2 = 800\pi$$ - Calculate numerical value: $$800\pi \approx 800 \times 3.1416 = 2513.274$$ - Rounded to nearest thousandths: $$2513.274$$ **Final answers:** - Part A: $502.655$ - Part B: $2513.274$