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

Question

jamesj · Answer

Nice question.  This means change in mass, delta m, is proportional to r^2.delta t.  i.e.,
$$ dm = kr^2 \ dt $$
for some constant k.  Thus
$$ \frac{dm}{dt} = kr^2 $$
Now you want to show that $ dr/dt = constant $.

anonymous · Answer

does the d in "kr2 dt" mean delta? 

and where would I go from there? would i find the derivative of the surface area?

jamesj · Answer

the d is short hand for delta, yes.

Now, as you know the volume of a sphere of radius r is $ (4/3)\pi r^3 $, hence is rho is the density of water, the mass of the sphere is
$$ m = \frac{4\pi\rho}{3}r^3 $$
and it follows that
$$ \frac{dm}{dr} = (4\pi\rho)r^2 $$

We now have an expression for $ dm/dr $ and for $ dm/dt $.  Using the chain rule and/or whatever other techniques you like, find an expression for $ dr/dt $ and show it is constant.