Measurement uncertainty

1. Feb 22, 2009

fluppocinonys

1. The problem statement, all variables and given/known data
If x = (7.2$$\pm$$0.6) m, determine the value of ln x with its associated uncertainty.

(Ans is 1.97$$\pm$$0.08)

2. Relevant equations
$$A = k{B^m}{\rm{ }} \Rightarrow {\rm{ }}\frac{{\Delta A}}{A} = m\frac{{\Delta B}}{B}$$
(perhaps?)

3. The attempt at a solution
i tried to ln7.2 and got 1.97, however ln0.6 = negative value. How to get 0.08?
Thanks

2. Feb 22, 2009

CompuChip

If you have any (differentiable) function f(x), and the measurement is given as $x = x_0 + \Delta x$, there is a general formula for giving the central value of f (which is just $f(x_0)$ as you would expect) and the uncertainty. Do you know what formula I'm hinting at?

3. Feb 22, 2009

fluppocinonys

No... :uhh:
I don't think I've learned that far

4. Feb 22, 2009

CompuChip

Well, maybe it's time you learn it.
It's very useful and easy to remember:

$$\Delta f = f'(x_0) \cdot \Delta x$$ (*)
so the uncertainty in f is the derivative of f w.r.t. x (evaluated at the central value) times the uncertainty in x.
The justification is of course, that very close to $$x_0$$, we can approximate the function f by a straight line with slope $f'(x_0)$. So if you vary x by an amount $\Delta x$, then you can approximate the change in f by the variation $$f'(x_0) \Delta x[/itex] of the line. If you have multiple variables, like f(x, y, z, ...) then you can simply extend this to [tex]\Delta f^2 = \left( \frac{\partial f(x_0, y_0, z_0, \cdots)}{\partial x} \Delta x \right)^2 + \left( \frac{\partial f(x_0, y_0, z_0, \cdots)}{\partial y} \Delta y \right)^2 + \left( \frac{\partial f(x_0, y_0, z_0, \cdots)}{\partial z} \Delta z \right)^2 + \cdots$$
which looks like a combination of that identity and the Pythagorean theorem.

[If you haven't learned about partial derivatives, forget about that last paragraph, you should remember formula (*) though].

When you apply (*) to the special case $f(x) = k x^m$ you will get that
$$\frac{\Delta f}{f} = m \frac{\Delta x}{x}$$
as you said. When you apply it to f(x) = ln(x) you will get the requested answer.

5. Feb 22, 2009

fluppocinonys

All right, that was rather overwhelming...
Nonetheless, thanks for helping me! you're very helpful :D

6. Feb 22, 2009

CompuChip

Hmm, looks impressive doesn't it.
Just play around with it and you'll see that it looks harder than it is (if you know how to differentiate and multiply, that is).