**Toolbox:**

- If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
- $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1

Volume of the sphere = $ \large\frac{4}{3} \pi r^3$

Differentiating w.r.t $t$ we get

$ \large\frac{dv}{dt}=-\large\frac{4}{3} \pi\: \: 3r^2.\large\frac{dr}{dt}$

( -ve sign ) indicate that the volume is decreasing.

$ \Rightarrow \large\frac{dv}{dt}=-4\pi r^2\: \large\frac{dr}{dt}$

But the surface area of the sphere is $ 4\pi r^2$

i.e., $A=4 \pi r^2$

$ \therefore \large\frac{dv}{dt}=-A \large\frac{dr}{dt}$

Step 2

It is given that the rate of decrease of volume is proportional to its surface area.

i.e., $ \large\frac{dv}{dt} \: \alpha\: A$

$ \Rightarrow \large\frac{dv}{dt}=-kA$ where $ k $ is the constant of proportionality.

$ \therefore kA=-A \: \large\frac{dr}{dt}$

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

Hence the radius is decreasing at a constant rate.