1. **State the problem:** A spherical balloon's surface area is increasing at a rate of 8 cm²/s. We need to find how fast the volume is changing when the radius is 5 cm. 2. **Relevant formulas:** - Surface area of a sphere: $$S = 4\pi r^2$$ - Volume of a sphere: $$V = \frac{4}{3}\pi r^3$$ 3. **Differentiate both formulas with respect to time $t$:** - $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$$ - $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ 4. **Given:** $$\frac{dS}{dt} = 8$$ cm²/s and $$r = 5$$ cm. 5. **Find $$\frac{dr}{dt}$$ using the surface area rate:** $$8 = 8\pi (5) \frac{dr}{dt}$$ $$8 = 40\pi \frac{dr}{dt}$$ $$\frac{dr}{dt} = \frac{8}{40\pi} = \frac{1}{5\pi}$$ cm/s 6. **Find $$\frac{dV}{dt}$$ using $$\frac{dr}{dt}$$:** $$\frac{dV}{dt} = 4\pi (5)^2 \times \frac{1}{5\pi}$$ $$\frac{dV}{dt} = 4\pi \times 25 \times \frac{1}{5\pi}$$ $$\frac{dV}{dt} = \cancel{4\pi} \times 25 \times \frac{1}{\cancel{5\pi}} = 4 \times 5 = 20$$ cm³/s **Final answer:** The volume is increasing at a rate of 20 cm³/s when the radius is 5 cm.