1. **State the problem:** We are given that the volume $V$ of a sphere is increasing at a rate of $\frac{dV}{dt} = 7$ cm³/s. We need to find the rate of change of its surface area $S$ when the volume is $\frac{256\pi}{3}$ cm³. 2. **Recall formulas:** - Volume of a sphere: $$V = \frac{4}{3}\pi r^3$$ - Surface area of a sphere: $$S = 4\pi r^2$$ 3. **Find the radius when $V = \frac{256\pi}{3}$:** $$\frac{4}{3}\pi r^3 = \frac{256\pi}{3}$$ Divide both sides by $\frac{\pi}{3}$: $$4 r^3 = 256$$ Divide both sides by 4: $$r^3 = 64$$ Take cube root: $$r = 4$$ cm 4. **Differentiate volume with respect to time $t$ to relate $\frac{dr}{dt}$ and $\frac{dV}{dt}$:** $$V = \frac{4}{3}\pi r^3$$ Differentiate: $$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$$ 5. **Solve for $\frac{dr}{dt}$:** $$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{4\pi r^2} = \frac{7}{4\pi (4)^2} = \frac{7}{4\pi \times 16} = \frac{7}{64\pi}$$ cm/s 6. **Differentiate surface area $S$ with respect to time $t$ to find $\frac{dS}{dt}$:** $$S = 4\pi r^2$$ $$\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$$ 7. **Substitute $r=4$ and $\frac{dr}{dt} = \frac{7}{64\pi}$:** $$\frac{dS}{dt} = 8\pi \times 4 \times \frac{7}{64\pi} = \frac{224\pi}{64\pi} = \frac{224}{64} = \frac{7}{2}$$ cm²/s **Final answer:** The rate of change of the surface area when the volume is $\frac{256\pi}{3}$ cm³ is $$\frac{7}{2}$$ cm²/s.