1. **Problem statement:** A spherical balloon's volume increases at a rate of $\frac{3}{20}$ cm$^3$/s. (i) Find how fast the radius changes when the volume is 200 cm$^3$. (ii) Find the rate of change of the radius when the radius is 5 cm. 2. **Relevant formula:** The volume $V$ of a sphere with radius $r$ is given by: $$V = \frac{4}{3} \pi r^3$$ 3. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$$ This relates the rate of change of volume $\frac{dV}{dt}$ to the rate of change of radius $\frac{dr}{dt}$. 4. **Given:** $$\frac{dV}{dt} = \frac{3}{20}$$ cm$^3$/s 5. **(i) Find $\frac{dr}{dt}$ when $V=200$ cm$^3$:** First, find $r$ from volume: $$200 = \frac{4}{3} \pi r^3 \implies r^3 = \frac{200 \times 3}{4 \pi} = \frac{600}{4 \pi} = \frac{150}{\pi}$$ So, $$r = \sqrt[3]{\frac{150}{\pi}}$$ 6. Substitute into the differentiated formula: $$\frac{3}{20} = 4 \pi r^2 \frac{dr}{dt}$$ Solve for $\frac{dr}{dt}$: $$\frac{dr}{dt} = \frac{\frac{3}{20}}{4 \pi r^2} = \frac{3}{20} \times \frac{1}{4 \pi r^2} = \frac{3}{80 \pi r^2}$$ 7. Substitute $r = \sqrt[3]{\frac{150}{\pi}}$: $$r^2 = \left(\sqrt[3]{\frac{150}{\pi}}\right)^2 = \left(\frac{150}{\pi}\right)^{\frac{2}{3}}$$ So, $$\frac{dr}{dt} = \frac{3}{80 \pi \left(\frac{150}{\pi}\right)^{\frac{2}{3}}} = \frac{3}{80 \pi} \times \left(\frac{\pi}{150}\right)^{\frac{2}{3}}$$ 8. **(ii) Find $\frac{dr}{dt}$ when $r=5$ cm:** Use the formula: $$\frac{dr}{dt} = \frac{\frac{3}{20}}{4 \pi (5)^2} = \frac{3/20}{4 \pi \times 25} = \frac{3/20}{100 \pi} = \frac{3}{20} \times \frac{1}{100 \pi} = \frac{3}{2000 \pi}$$ **Final answers:** (i) $$\frac{dr}{dt} = \frac{3}{80 \pi} \times \left(\frac{\pi}{150}\right)^{\frac{2}{3}}$$ cm/s (ii) $$\frac{dr}{dt} = \frac{3}{2000 \pi}$$ cm/s