Question

Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

 

Answer 1

Let the surface area of the raindrop be \[A\]
Thus, the rate of evaporation will be given by \[\frac{dV}{dt}\]
As per the given condition, 
\[\frac{dV}{dt} \propto A\]
\[ \Rightarrow \frac{dV}{dt} = - kA\]
Here, k  is a constant. Also, the negative sign appears when V decreases and t  increases.
Now, \[V = \frac{4}{3}\pi r^3\]
Here, `r` is the radius of the spherical drop.

\[\therefore \frac{d}{dt}\left( \frac{4}{3} \pi r^3 \right) = - k \times 4 \pi r^2 \]

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

\[ \Rightarrow \frac{dr}{dt} = - k \]

It is therequired differential equation.