# Variation of a quantity

1. Jul 4, 2014

### ChrisVer

Suppose I have such an equation:
$A= f(B)$
so $A$ is a function of $B$.

Can I really use the fact that a variation of $A$ is like taking the differential of it?
$\delta A= dA$

so that:

$\delta A = \frac{d f(B)}{dB} \delta B$
?

2. Jul 4, 2014

### mathman

Superficially it looks OK. You need to look at the definition involved for variation.

3. Jul 4, 2014

### ChrisVer

the reason I ask is because I have a confusion with one more way to see variation.
Suppose I vary $B$, so that $A$ is going to be varied too:

$A+ \delta A = f(B+\delta B)$
And suppose $f(x) = \frac{1}{\sqrt{x}}$

Then:

$A( 1+ \frac{\delta A}{A}) = \frac{1}{\sqrt{B+\delta B}}= \frac{1}{\sqrt{B}} \frac{1}{\sqrt{1+\frac{\delta B}{B}}}≈ A -A \frac{1}{2} \frac{\delta B}{B}$

$\frac{\delta A}{A}= - \frac{1}{2} \frac{\delta B}{B}$

This result would imply that:
$\frac{d f(B)}{dB} = -\frac{1}{2}\frac{A}{B}$

in my OP...

with a simple check, we have:

$\frac{d (1/ \sqrt{B})}{dB} = -\frac{1}{2} \frac{1}{B^{3/2}}$

This method works fine for such an easy function as $f(x)=\frac{1}{\sqrt{x}}$
$f(x)= \frac{1}{\sqrt{x}} + c$

with $c$ a constant. The same method won't work because I will end up with something like:

$\frac{\delta A}{A} = - \frac{1}{A} \frac{1}{2} \frac{1}{\sqrt{B}} \frac{\delta B}{B}$

unfortunately now $\frac{1}{A}$ can't be cancelled with the $\frac{1}{\sqrt{x}}$ ... Only maybe, up to some error, if I consider $c$ very small... The derivative however will kill $c$ and give some other result (the same as for without $c$ which I think is more physical- isn't it? a constant shouldn't destroy the variations)..

Last edited: Jul 4, 2014
4. Jul 4, 2014

### ChrisVer

I think I found my misconception... the variation is still independent of $c$... it's 1/A which gives me that dependence. So there's no way to escape from c rather than neglecting it :) that's fine...i guess

5. Jul 4, 2014

### Matterwave

You can basically use Taylor expansion to show your equality.

If $A=f(B)$ and we add a small quantity $\delta B$ to $B$ then by Taylor's theorem we will have:

$$A'\equiv A+\delta A=f(B+\delta B)=f(B)+\frac{df}{dB}\delta B+\mathcal{O}(\delta B^2)$$

This gives as the limit $\delta B\rightarrow 0$:

$$\delta A=\frac{df}{dB}\delta B$$

Of course, this is usually much more interesting when we consider functionals rather than simple functions.