Simple differential eqn's question

[DWill]

Homework Statement

A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time.

Homework Equations

The Attempt at a Solution

The answer is dV/dt = -kV^(2/3), for some k > 0.

I don't really understand this answer. The question states the volume changes at a rate proportional to its surface area (A = 4*pi*r^2), but it seems from the answer it depends on volume V. Can someone help me understand how to get to this answer? And how do I know there is a constant k in there too? The problem never states it.. am I just supposed to know this somehow?

Thanks!

 

[gabbagabbahey]

Homework Statement

A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time.

The Attempt at a Solution

The answer is dV/dt = -kV^(2/3), for some k > 0.

I don't really understand this answer. The question states the volume changes at a rate proportional to its surface area (A = 4*pi*r^2), but it seems from the answer it depends on volume V. Can someone help me understand how to get to this answer? And how do I know there is a constant k in there too? The problem never states it.. am I just supposed to know this somehow?

Thanks!

I've bolded the key phrases in the question.

To see where 'k' comes from, you need to understand what 'proportional to' means. For example, if I tell you 'y' is proportional to 'x', it could mean that $y=2x$ or $y=-7489327x$ or $y=\sqrt{\pi}x$; you don't know what the constant of proportionality is (unless otherwise told), so you usually just call it $k$ (or some other letter) so that $y=kx$

The next key phrase tells you that you should try to express the surface area of the sphere (and the entire differential equation!) in terms of the volume, not the radius.

So, if $A=4\pi r^2$ and $V=\frac{4}{3}\pi r^3$, then $A$ in terms of $V$ is____?

 

[DWill]

Thanks for the reply
Hm, then V = (1/3)*r*A, right? or A = 3V/r

where do you go from there?

 

[gabbagabbahey]

If the volume changes with time, then so does the radius. So to make a DE that involves only the Volume (and constants) you should eliminate 'r'.

If $V=\frac{4}{3}\pi r^3$, then r=___? So A=___?

 

[DWill]

r = (3V / 4pi)^(1/3), and A = 4*pi*(3V / 4pi)^(2/3) ?

 

[gabbagabbahey]

Yup, and since the volume decreases proportional to the surface area, you also know that

$\frac{dV}{dt}=-kA$ for some unknown, positive k (the negative sign and the restriction on k guarantees that the drop loses volume instead of gaining it.

So...dV/dt in terms of volume is?

 

[DWill]

OK so I see where the V^(2/3) comes from now, does that mean the rest of the numbers (constants) before it are included in the proportionality constant k?

 

[gabbagabbahey]

Yup, exactly