#### 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 the surface. Prove that the radius is decreasing at a constant rate.

Given: When a spherical ball salt is dissolved, the rate of decrease of the volume at any instant is proportional to the surface

To prove: The rate of decrease of radius is constant at any given time

Explanation: Take the radius of the spherical ball at any time t be ‘r’

Assume S as the surface area of the spherical ball

Then, $S = 4\pi r^2$……….(i)

Take the volume of the spherical ball be V

Then, $\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$
According to the given criteria,

$-\frac{\mathrm{dV}}{\mathrm{dt}} \propto \mathrm{S}$
The rate of decrease of volume is indicated by the negative sign It can also be written as

$-\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\mathrm{kS}$
Here K is the proportional constant

After substitution of values from equation (i) and (ii), we get

$\Rightarrow-\frac{\mathrm{d}\left(\frac{4}{3} \pi r^{3}\right)}{\mathrm{dt}}=\mathrm{k} 4 \pi \mathrm{r}^{2}$
When the constant term is taken outside the LHS, we get

$\Rightarrow-\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dt}}=\mathrm{k} 4 \pi \mathrm{r}^{2}$

After the derivatives are applied with respect to t, we get

$\Rightarrow-\frac{4}{3} \pi \times 3 \mathrm{r}^{2} \times \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k} 4 \pi \mathrm{r}^{2}$

After cancelling of the like terms, we get

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

Hence the rate of decrease of the radius of the spherical ball is constant.

Hence Proved