# Quick Differentiation Qu.

1. Nov 25, 2007

### Hemmer

1. The problem statement, all variables and given/known data

I have nearly finished a problem. I am at the stage where I have:

$$B(s)\propto\frac{1}{s^{3}}$$

$$v=\frac{ds}{dt}$$

I want to find the rate of change of B(s), but expressed in terms of a velocity, rather than a displacement.

2. Relevant equations

So for displacement:

$$B'(s) \propto\frac{-3}{s^{2}}$$

3. The attempt at a solution
How can I express this in terms of v instead? My guess would be that:

$$B'(s) \propto\frac{1}{v^{3}}$$

but I'm not sure thats correct. This does seem like quite a simple problem but its really got me stumped.

2. Nov 25, 2007

### azatkgz

Is the velocity constant?
Do you mean by rate of change

$$\frac{dB(s)}{dt}$$?????
And I think it would be better if you post whole question.

Last edited: Nov 25, 2007
3. Nov 25, 2007

### Hemmer

Yes the velocity is constant. The rest of the question isn't relevant to my problem, but I can post if you want. And yes by rate of change I mean:

$$\frac{dB(s)}{dt}$$

Last edited: Nov 26, 2007
4. Nov 26, 2007

### azatkgz

Can you post?

5. Nov 28, 2007

### Hemmer

Never mind I managed to solve it (i think!).

$$B \propto \frac{-3}{s^{3}} \frac{dv}{dt}$$

using substitution of s for v.

Last edited: Nov 28, 2007