# Differential Equation Problem

• Nov 4th 2008, 02:58 PM
eugenius
Differential Equation Problem
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 :)
• Nov 4th 2008, 04:22 PM
skeeter
Quote:

Originally Posted by eugenius
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?
• Nov 4th 2008, 04:26 PM
eugenius
thanks, i'll see how i get on
• Nov 4th 2008, 11:50 PM
eugenius
i still cant do the last bit, help please :S
• Nov 5th 2008, 01:18 AM
mr fantastic
Quote:

Originally Posted by eugenius
i still cant do the last bit, help please :S

$\displaystyle \frac{dr}{dt} = -k \Rightarrow r = -kt + C$.

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 $\displaystyle r = r_0 \left(1 - \frac{t}{2} \right)$.