Discussion
The surface area of a solid hemisphere is increasing at the rate of \(8\ cm^{2}\) / sec (retaining its shape). Then the rate of change of its volume (in \(cm^{3}\) / sec ), when the radius is \(5cm\), is
Answer: C
The volume increases at \(\frac{40}{3}\) cm3/s.
A solid hemisphere has total surface area \(S = 2\pi r^{2} + \pi r^{2} = 3\pi r^{2}\) (curved surface plus flat base) and volume \(V = \frac{2}{3}\pi r^{3}\).
\[ \begin{aligned} \frac{dS}{dt} &= 6\pi r\frac{dr}{dt} = 8 \Rightarrow \frac{dr}{dt} = \frac{8}{30\pi} \\ \frac{dV}{dt} &= 2\pi r^{2}\frac{dr}{dt} = 2\pi(25)\frac{8}{30\pi} = \frac{40}{3} \end{aligned} \]
A shortcut: \(\frac{dV}{dS} = \frac{2\pi r^{2}}{6\pi r} = \frac{r}{3}\), so \(\frac{dV}{dt} = \frac{5}{3} \times 8\). Using only the curved surface \(2\pi r^{2}\) gives \(\frac{r}{2} \times 8 = 20\), not offered, and \(\frac{20}{3}\) or \(\frac{80}{3}\) halve or double the answer.
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