How Does the Radius of a Raindrop Change as It Accumulates Moisture?

[courtrigrad]

if a raindrop is a perfect sphere,and it accumulates moisture at a rate proportional to its surface area. Show that the radius increases at a constant rate.

I know surface area of sphere = 4 * pi * r^2

dV/ dr = 4 * pi * r^2

I am not sure where to go from here

Any help is appreciated

Thanks

 

[quasar987]

First thing to do is realize that the amount of moisture is nothing but the volume of the waterdrop. The rate at which it accumulate moisture is then dV/dt. We can calculate this derivative. You have done that but made a little mistake by forgetting the chain rule!

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=A\frac{dr}{dt}.

Next, we are given the fact that it accumulates moisture at a rate proportional to its surface area. The keyword in this sentence is "proportionnal" because if means that dV/dt is of the form

\frac{dV}{dt}=(a \ constant)A

Compare the two equations of dV/dt. What do you conclude?

 

[dextercioby]

You didn't hat the idea of "rate".It usually implies 'variance in time'.The problem gives that the rate the volume is increasing is proportional to its area.
Therefore
\frac{dV(r(t))}{dt}=kS
,where 'k' is a constant of proportionality.
\frac{dV}{dr}\frac{dr}{dt}=kS
But for an infinitesimal increase of the radius 'dr',the volume increases with
dV=Sdr
Therefore
S\frac{dr}{dt}=kS
The surface is nonzero.U get:
\frac{dr}{dt}=k
,which,by integration leads to the desired result.

Daniel.