A spherical ball of salt is dissolving in water A spherical ball of salt is dissolving in water in such a way that the rate of decrease in volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate. My Approach: $$\dfrac {dV}{dt}\propto Surface (S)$$ $$\dfrac {dV}{dt}=k.S$$ where $k$ is a proportionality constant. $$\dfrac {dV}{dt}=k.4\pi r^2$$ How do I proceed?

You have $V=\frac{4}{3} \pi r^3$ so: $$\frac{d V}{dt}=\frac{d}{dt}\left(\frac{4}{3} \pi r^3 \right)=\frac{4}{3} \pi \times 3 \frac{dr}{dt} r^2 $$ so the equation is: $$4 \pi \frac{dr}{dt} r^2 = k 4 \pi r^2$$ i.e (as long as $r\
eq 0$): $$\frac{dr}{dt}=k$$.