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!


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:

$$ C=3 $$


$$ r(t)=-kt+3 $$

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  • $\begingroup$ Thanks a lot! I completely forgot that was something you could do. $\endgroup$ – SBoots Sep 7 '15 at 21:28
  • $\begingroup$ @SBoots Glad I helped. Always remember to differentiate first until you get $f(r)dr$ then you can deal with that $f(r)$. $\endgroup$ – Oussama Boussif Sep 7 '15 at 21:30

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