Sphere particle dissolution, volume loss vs time

1. Jun 13, 2013

hammal

Hi all,
it might be a silly question, but my math is "a bit" rusty.

I want to find the equation of the volume lost during the dissolution of a sphere particle as function of time:

A=area of sphere
V=volume of sphere
t=time
k=dissolution rate as volume dissolved over area time

A=(4/pi) r^2
V=(4/3)pi r^3
Volume change from dissolution Vloss=kAt

what I know is:
d_V/d_r=4 pi r^2
d_A/d_r=(8/pi) r
d_V/d_t=kA

but from here I don't know how to find V(t), so data I can plot the volume against time or integrate it to get the amount of volume lost against time.

Please any help is much appreciated.

2. Jun 13, 2013

HallsofIvy

Staff Emeritus
I presume that you are talking about a sphere, such as a moth ball, that is "evaporates" at a rate proportional to the surface area of the sphere. That is your dV/dt= kA.
Now, as you say, $A= 4\pi r^2$ and $V=n (4/3)\pi r^3$.
Differentiating with respect to t,
$$\frac{dA}{dt}= 8pi r\frac{dr}{dt}$$ and
$$\frac{dV}{dt}= 4\pi r^2\frac{dr}{dt}$$

From the first, $\frac{dr}{dt}= \frac{1}{8\pi r}\frac{dA}{dt}$
so that $\frac{dV}{dt}= 4\pi r^2\frac{1}{8\pi r}\frac{dA}{dt}= \frac{r}{4}\frac{dA}{dt}$

3. Jun 13, 2013

hammal

Yes, that is exactly what I meant. Thanks for your answer, but it is not clear how to find a volume function of time:
I need something like d_V/d_t= whatever function of t

I'm not able to do it from your last part.

Cheers

4. Jun 13, 2013

Staff: Mentor

Hi Hammel. Welcome to Physics Forums.

From HallsofIvy's response,
$$\frac{dV}{dt}= 4\pi r^2\frac{dr}{dt}=-k (4\pi r^2)$$
Canceling:
$$\frac{dr}{dt}=-k$$
or, $r=r_0-kt$

5. Jun 13, 2013

hammal

Yes, got it, thanks both of you