Calculating Uncertainty for ln x: A Short Guide

[fluppocinonys]

Homework Statement

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)

Homework Equations

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

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

 

[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?

 

[fluppocinonys]

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?

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

 

[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 [tex]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
$$\Delta f^2 = \left( \frac{\partial f(x_0, y_0, z_0, \cdots)}{\partial x} \Delta x \right)^2 + \left( \frac{\partial f(x_0, y_0, z_0, \cdots)}{\partial y} \Delta y \right)^2 + \left( \frac{\partial f(x_0, y_0, z_0, \cdots)}{\partial z} \Delta z \right)^2 + \cdots$$
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.

 

[fluppocinonys]

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

 

[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).