Question

Suppose that a mothball loses volume by evaporation at a rate proportional to its instantaneous area. If the diameter of the ball decreases from $2$cm to $1$cm in $3$ months, how long(in months) will it take so the ball has practically gone

Answer 1

Let at any instance $\left(t\right),$ radius of moth ball be $r$ and $v$ be its volume.
$\Rightarrow v=\frac{4}{3}\pi r^3$
$\Rightarrow \frac{dv}{dt}=4\pi r^2\frac{dr}{dt}$
Thus, as per the information
$4\pi r^2\frac{dr}{dt}=-k4\pi r^2,$ where $k\in R^{+}$
$\Rightarrow dr=-kdt\Rightarrow r=-kt+c$
At $t=0,r=1\text{cm};t=3$ month, $r=0.5\text{cm}$
On solving,
$\Rightarrow c=1,k=\frac{1}{6}$
$\Rightarrow r=-\frac{1}{6}t+1$
Now for $r\to 0\Rightarrow t\to 6$
Hence, it will take six months until the ball is practically gone.