# Related rates problem

1. Aug 29, 2005

### jordanfc

This is the problem as it appears in the text. "As a spherical raindrop evaporates, its volume changes at a rate proportional to its surface area A. If the constant of proportionality is 3, find the rate of change of the radius r when r=2." My first question is does the constant of proportionality refer to $$\frac{dV}{dA}$$? Secondly, am I being asked to find the rate of change of the radius with respect to time $$\frac{dr}{dt}$$ or another rate?

2. Aug 29, 2005

### quasar987

Hi, I don't think the problem makes sense. But I will try to answer your questions.

No, if you are told that A is proportional to B, it means there is a constant k such that A = kB. Here, it means that dV/dt = kA, for a certain constant k. Next, you are told that k = 3. But this is nonsense in my opinion because the raindrop is evaporating... and a positive dV/dt means V is augmenting. So I'd try doing the problem with k = -3 instead; it'd make more sense.

Yes, the rate of change of radius wrt time is dr/dt.

3. Aug 29, 2005

### jordanfc

thanks quasar, with that information ill try the problem.
$$\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}$$

Since $$\frac{dV}{dt}=-3A$$

Then $$-3(4{\pi}r^2)=\frac{dV}{dr}\frac{dr}{dt}$$

But now it seems if i differentiate the volume with respect to the radius ill get the equation for surface area and it will cancel out and leave
$$\frac{dr}{dt}=-3$$
And that cant be right, where have i erred?

4. Aug 30, 2005

### quasar987

Why can't it be right? I'd say it can't be wrong.

5. Aug 31, 2005

### jordanfc

well then, thanks again quasar. so was it irrelevant to ask to find the rate of change when the radius was 2, meaning its a constant rate of change when the volume and surface area decrease proportionally?