# Related rates and rain drops

1. Jan 6, 2005

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

2. Jan 6, 2005

### quasar987

First thing to do is realise 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?

3. Jan 6, 2005

### 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.

Last edited: Jan 6, 2005