Question

Assume that a shperical randrop evaporates at a rate proportional to its a surface area. if it's radius is originally 3 mm, and 1 minute later has been reduced to 2 mm . Find an expression for the radius of the raindrop at any time.

Answer 1

Let $V$ be the volume of the raindrop in ${mm}^3$ at time $t$ in minuts. Let $r$ be the radius of the drop at time $t$.

$\frac{dV}{dt}=-k\left(4\pi r^2\right).....\left(1\right)$ where $0<k\in R$

Since the drop is spherical, thus

$V=\frac{4}{3}\pi r^3$

Differentiate above equation w.r.t. time $t$, we have

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.....\left(2\right)$

Froom ${eq}^n\left(1\right)\&\left(2\right)$, we have

$4\pi r^2\frac{dr}{dt}=-k\left(4\pi r^2\right)$

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

Integrating the above equation, we have

$r\left(t\right)=-kt+C$

Given $r\left(0\right)=3$

$\therefore C=3$

Now,

$r\left(t\right)=-kt+3$

Again,

$r\left(1\right)=2$

$\therefore -k+3=2$

$\Rightarrow k=1$

Hence, the radius at any time $0\ge t$ is given as-

$r\left(t\right)=-2t+3$