Measurement uncertainty

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

(Ans is 1.97[tex] \pm [/tex]0.08)


2. Relevant equations
[tex]A = k{B^m}{\rm{ }} \Rightarrow {\rm{ }}\frac{{\Delta A}}{A} = m\frac{{\Delta B}}{B}[/tex]
(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
 

CompuChip

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If you have any (differentiable) function f(x), and the measurement is given as [itex]x = x_0 + \Delta x[/itex], there is a general formula for giving the central value of f (which is just [itex]f(x_0)[/itex] as you would expect) and the uncertainty. Do you know what formula I'm hinting at?
 
If you have any (differentiable) function f(x), and the measurement is given as [itex]x = x_0 + \Delta x[/itex], there is a general formula for giving the central value of f (which is just [itex]f(x_0)[/itex] as you would expect) and the uncertainty. Do you know what formula I'm hinting at?
No... :uhh:
I don't think I've learned that far
 

CompuChip

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

[tex]\Delta f = f'(x_0) \cdot \Delta x[/tex] (*)
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 [tex]x_0[/tex], we can approximate the function f by a straight line with slope [itex]f'(x_0)[/itex]. So if you vary x by an amount [itex]\Delta x[/itex], 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
[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[/tex]
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 [itex]f(x) = k x^m[/itex] you will get that
[tex]\frac{\Delta f}{f} = m \frac{\Delta x}{x}[/tex]
as you said. When you apply it to f(x) = ln(x) you will get the requested answer.
 
All right, that was rather overwhelming...
Nonetheless, thanks for helping me! you're very helpful :D
 

CompuChip

Science Advisor
Homework Helper
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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).
 

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