A spherical mothball evaporates uniformly at a rate proportional to its surface area. Hence deduce a differential equation that links its radius with time. Given that the radius halves in one month, how long will the mothball last?
Thanks
A spherical mothball evaporates uniformly at a rate proportional to its surface area. Hence deduce a differential equation that links its radius with time. Given that the radius halves in one month, how long will the mothball last?
Thanks
A spherical mothball evaporates uniformly at a rate proportional to its surface area.
$\displaystyle \frac{dV}{dt} = -kS$
since $\displaystyle V = \frac{4}{3}\pi r^3$
$\displaystyle \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} = S \cdot \frac{dr}{dt}$
equating ...
$\displaystyle S \cdot \frac{dr}{dt} = -kS$
$\displaystyle \frac{dr}{dt} = -k$
so, the radius decreases at a constant rate ... can you take it from here?
$\displaystyle \frac{dr}{dt} = -k \Rightarrow r = -kt + C$.
When t = 0, $\displaystyle r = r_0$. Therefore $\displaystyle r = r_0 - kt$.
When t = 1, $\displaystyle r = \frac{r_0}{2} \Rightarrow k = \frac{r_0}{2}$.
Therefore $\displaystyle r = r_0 \left(1 - \frac{t}{2} \right)$.