1. **State the problem:** A balloon is being filled with helium at a rate of 4 ft³/min. We want to find the rate at which the surface area is increasing when the volume is $$\frac{32\pi}{3}$$ ft³. 2. **Relevant formulas:** - Volume of a sphere: $$V = \frac{4}{3}\pi r^3$$ - Surface area of a sphere: $$S = 4\pi r^2$$ 3. **Given:** - $$\frac{dV}{dt} = 4$$ ft³/min - Volume at the instant: $$V = \frac{32\pi}{3}$$ ft³ 4. **Find:** $$\frac{dS}{dt}$$ when $$V = \frac{32\pi}{3}$$. 5. **Step 1: Find radius $$r$$ at given volume.** $$\frac{4}{3}\pi r^3 = \frac{32\pi}{3}$$ Divide both sides by $$\pi$$: $$\frac{4}{3} r^3 = \frac{32}{3}$$ Multiply both sides by 3: $$4 r^3 = 32$$ Divide both sides by 4: $$r^3 = 8$$ Take cube root: $$r = 2$$ ft 6. **Step 2: Differentiate volume formula with respect to time $$t$$:** $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ 7. **Step 3: Solve for $$\frac{dr}{dt}$$:** $$4 = 4\pi (2)^2 \frac{dr}{dt}$$ $$4 = 4\pi \times 4 \frac{dr}{dt} = 16\pi \frac{dr}{dt}$$ Divide both sides by $$16\pi$$: $$\frac{dr}{dt} = \frac{4}{16\pi} = \frac{1}{4\pi}$$ ft/min 8. **Step 4: Differentiate surface area formula with respect to time:** $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$$ 9. **Step 5: Substitute $$r=2$$ and $$\frac{dr}{dt} = \frac{1}{4\pi}$$:** $$\frac{dS}{dt} = 8\pi \times 2 \times \frac{1}{4\pi} = \frac{16\pi}{4\pi} = 4$$ ft²/min **Final answer:** The surface area is increasing at a rate of 4 ft²/min when the volume is $$\frac{32\pi}{3}$$ ft³.