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

Explanation

Let at any instance (t), the radius of moth ball be r and v be its volume. Then
`v=4/3pir^(3)`
or `(dv)/(dt) = 4pir^(2)(dr)/(dt)`
Thus, as per the information
`4pir^(2)(dr)/(dt) = -k4pir^(2)`, where `k in R^(+)`
or `(dr)/(dt) = -k`
or `r=-kt+c`
at t=0, r=2cm, t=3 month, r=1 cm
`rArr c=2, k=1/3`
`rArr r=-1/3t+2`
now `r to 0`, `t to6`
Hence, it will take six months until the ball is practically gone.