$A$ spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to its surface area. Prove that the radius is decreasing at a constant rate.

(N/A) Let the radius of the spherical ball of salt at any time $t$ be $r$.
The volume of the ball is $V = \frac{4}{3} \pi r^{3}$ and the surface area is $S = 4 \pi r^{2}$.
According to the problem,the rate of decrease of volume is proportional to the surface area:
$-\frac{dV}{dt} \propto S$
This implies $-\frac{dV}{dt} = kS$,where $k$ is a positive constant of proportionality.
Substituting the expressions for $V$ and $S$:
$-\frac{d}{dt} \left( \frac{4}{3} \pi r^{3} \right) = k(4 \pi r^{2})$
Using the chain rule:
$-\frac{4}{3} \pi \cdot 3r^{2} \cdot \frac{dr}{dt} = k(4 \pi r^{2})$
$-4 \pi r^{2} \cdot \frac{dr}{dt} = k(4 \pi r^{2})$
Dividing both sides by $4 \pi r^{2}$ (assuming $r \neq 0$):
$-\frac{dr}{dt} = k$
$\frac{dr}{dt} = -k$
Since $k$ is a constant,the rate of change of the radius $\frac{dr}{dt}$ is a constant. Thus,the radius is decreasing at a constant rate.