Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

Question

anonymous · Answer

You know that the moisture increases at a rate proportional to the surface area.  I am interpreting this as meaning the volume increases at a rate proportional to the surface area.  You can express this statement mathematically as $$dv/dt=a4 \Pi r(t)^{2}$$
where a is an arbitrary constant and r(t) is the rate at which the radius grows.  I got this simply by using the equation for the surface area of a sphere.

anonymous · Answer

That's perfect. I was on the right track. Thank you!

anonymous · Answer

next we can use the equation for the volume of a sphere$$v=(4/3) \pi r(t)^{3}$$so $$dv/dt=(4/3) \pi d/dt(r(t)^{3)}$$

anonymous · Answer

substitute that into the first equation and simplify and you have a differential equation in terms of just r(t)

anonymous · Answer

Dont you have to differentiate r(t)^3 as 3r(t)^2

anonymous · Answer

$$r \prime(t)^{3}=3ar(t)^{2}$$

anonymous · Answer

no you cannot differentiate it like that because r is a function not a variable.  you must factor out all the constants and leave it as a differential.  Then you can obtain the differential equation I have written above.

anonymous · Answer

So I set the two equations equal pretty much?

anonymous · Answer

You probably think I'm an idiot. This question is confusing me so bad for some reason.

anonymous · Answer

ya, and the solution to the differential equation is r(t)=bt+c where b and c are arbitrary constants.  This means that r(t) is increasing at a constant rate.

anonymous · Answer

ok... thank you!

anonymous · Answer

If you could help me with this last question that I just posted that would be great!