Question

Suppose that a moth- ball loses volumes by evaporation at a rate proportional to its instantaneous surface area . If radius of the ball decreases from 2cm to 1 cm in 3 months, how long will it take until the ball has practically gone?

Answer 1

$\begin{array}{c}u=\frac{4\pi }{3}{r}^3\\ \Rightarrow \frac{dv}{dt}=4\pi r^2.\frac{dr}{dt}\\ 4\pi r^2.\frac{dr}{dt}--u\left(4\pi r^2\right)\\ \frac{dr}{dt}=-k\\ \int dr=\int -kdt\\ r=-Kt+c\\ r=1cmt=0r=\frac{1}{2}cm,t=3\\ \Rightarrow 1=-k\left(0\right)+c\therefore c=1\\ \Rightarrow \frac{1}{2}=-k\left(3\right)+1\\ \Rightarrow -3k=-\frac{1}{2}\\ r=6r=\frac{-t}{6}+1\Rightarrow 0=\frac{-t}{6}+1\\ \Rightarrow \frac{t}{6}=1\\ t=6months\end{array}$