Application of Derivatives — Class 12 Maths Solution

We have, rate of decrease of the volume of spherical ball of salt at any instant is proportional to surface.

Let us assume that the radius of the spherical ball of the salt be $r$.

Therefore, volume of the ball $(V) = \frac{4}{3}\pi {r^3}$
and surface area $(S) = 4\pi {r^2}$

$\frac{{dV}}{{dt}} \propto S \Rightarrow \frac{d}{{dt}}\left( {\frac{4}{3}\pi {r^3}} \right) \propto 4\pi {r^2}$

$\Rightarrow$ $\frac{4}{3}\pi \cdot 3{r^2} \cdot \frac{{dr}}{{dt}} \propto 4\pi {r^2} \Rightarrow \frac{{dr}}{{dt}} \propto \frac{{4\pi {r^2}}}{{4\pi {r^2}}}$

$\Rightarrow$ $\frac{{dr}}{{dt}} = k \cdot 1\quad$ [where, $k$ is the constant of proportionality ]

$\Rightarrow$ $\frac{{dr}}{{dt}} = k$

Hence we can say that the radius of ball is decreasing at a constant rate.