Suppose that we have a sphere of naphthalene that is sublimating in a tax that is proportional to its surface area. Find the radius of the sphere in a time $t$, knowing that, for a sphere, $V = \frac{4}{3} \pi R^3$ (volume) and $S = 4 \pi R^2$ (surface area).
I did the "obvious" thing here: Since the sphere is decreasing in size: $$\frac{\textrm{d}V}{\textrm{d}t} = -kS$$ And so: $$V' = - k \times 4 \pi R^2$$ Since $ V = \frac{4}{3} \pi R^3$, then $V' = 4 \pi R^2 R'$ (by the chain rule, and here $f'$ meaning derivative of $f$ with respect to $t$) And so: $$4 \pi R^2 R' = -k \times 4 \pi R^2$$ And then i get: $$R' = -k$$ So $$R(t) = -kt + c_1 \quad, c_1 \in \mathbb{R}$$ Is this correct? For me, a linear decreasing here doesn't make sense at all.
