Suppose a spherical droplet of liquid evaporates at a rate proportional to its surface area.
$${dV\over dt} = -kA$$
where $V=volume (mm^3)$, $t=time(min)$, $k=$ evaporation rate $ (mm/min)$, and $A=$ surface area $(mm^2)$. Use Euler's method to compute the volume of the droplet from t=0 to 10 min using a step size of 0.25 min. Assume k = 0.1, and that the initial droplet has a radius of 3mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.
I think I have a handle of the Euler's method aspect, so now I'm trying to determine the radius of the liquid mathematically and compare the two.
I tried converting the equation into terms of radius and time and integrating:
$${d {4\over 3} \pi r^3 \over dt} = -k4\pi r^2 $$
Simplifying…
$$ {d {1\over 3} r \over dt} = -k $$
$$ \int d r = \int -3k dt$$
$$ r = -3kt + C$$
Using r(t=0) to solve for C…
$$3 = -3k(0) + C$$
$$r = -3kt + 3$$
but the result isn't right. I don't think I can convert the volume derivative to radius and integrate like that, but I don't know how else to go about it.
I would appreciate, in addition to help on the question, if anyone could point me to similar examples. Thanks!
Best Answer
What you did to solve the ODE by substituting the formula for $V$ and $A$ was correct.
The error you made is when you dropped the $r^2$ term while $r^3$ was still in differntial form. So you should have differentiated $r^3$ first then you can factorise. Anyway here's the correct approach:
$$ \frac{dV}{dt}=-kA\\ \frac{d\frac{4\pi}{3}r^3}{dt}=-k4\pi r^2\\ \frac{4\pi}{3}\frac{d}{dt}r^3=-4\pi k r^2\\ \frac{1}{3}\times 3r^2\frac{dr}{dt}=-kr^2\\ r^2\frac{dr}{dt}=-kr^2\\ \frac{dr}{dt}=-k\\ r(t)=-kt+C $$
And since $r(t=0)=3$ then:
$$ C=3 $$
So:
$$ r(t)=-kt+3 $$