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Let $r$ be the radius of the hemisphere. The volume $V$ of a hemisphere is given by $V = \frac{2}{3}\pi r^3$. The surface area $S$ (including the base) is $S = 3\pi r^2$. We are given that the volume increases at a uniform rate, so $\frac{dV}{dt} = k$, where $k$ is a constant.
Differentiating $V = \frac{2}{3}\pi r^3$ with respect to $t$: $$ \frac{dV}{dt} = \frac{2}{3}\pi (3r^2) \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt} $$ Since $\frac{dV}{dt} = k$, we have $k = 2\pi r^2 \frac{dr}{dt}$, which implies $\frac{dr}{dt} = \frac{k}{2\pi r^2}$.
Differentiating $S = 3\pi r^2$ with respect to $t$: $$ \frac{dS}{dt} = 3\pi (2r) \frac{dr}{dt} = 6\pi r \frac{dr}{dt} $$
Substitute the expression for $\frac{dr}{dt}$ from Step 2 into the equation for $\frac{dS}{dt}$: $$ \frac{dS}{dt} = 6\pi r \left( \frac{k}{2\pi r^2} \right) = \frac{3k}{r} $$ Since $3$ and $k$ are constants, $\frac{dS}{dt} \propto \frac{1}{r}$. This proves that the rate of change of the surface area varies inversely as the radius.
Final Answer: Proved: \frac{dS}{dt} \propto \frac{1}{r}