# Significant figure help?

1. Feb 22, 2015

### Sang Jun Lee

1. The problem statement, all variables and given/known data
Find the value of the function f as well as the relative and absolute error"
f(r,h,m)=

the variables were measured as follows: r=1.78cm r=0.006cm; h=2.34cm h=0.005cm; m=13.4g m=0.06g

2. Relevant equations

3. The attempt at a solution
I multiplied the whole equation with Ln, then derived it.
I got
then plugged in numbers and got 0.0044=0.004 or 0.4%
Answer says delta f is equal to 0.002
And I don't see why the answer is 0.002.

It would be helpful if you guys can tell me the correct way to do this problem.

Thanks!

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2. Feb 22, 2015

### BvU

Hello San Jung, welcome to PF

A few remarks to introduce you: in PF you post the problem, the equations you think you need to solve it, and your own attempt to sort things out and find a solution. That way we can provide adequate assistance, not just say yes or no or hand over the answer.

In this case I can' t do much, because I have no idea what you did exactly. Please indicate how you would determine the error in a product or a quotient (or a square). And show your calculation steps.

I think what you did is take logarithms (i.e. not 'multiply by Ln'), but to me that seems to complicate things.

3. Feb 22, 2015

### Sang Jun Lee

Hello.
Thank you for your reply.
I wrote down my equation using Mathtype and for some reason it isnt showing on my post.
What I did was multiply both sides by Ln
Thus, Ln(f(x))=2(deltaR/R) + (deltaH/H) - (deltaM/M) then derived the equation
deltaf/f = 2(0.006/1.78) + (0.005/2.34) - (0.06/13.4)

Therefore,
f = 1.74
delta f = 0.007564

But the answer says f = 1.74, and delta f = 0.02

4. Feb 22, 2015

### BvU

Again, you don't call that "multiply both sides by Ln". It is called: "take the logarithm on both sides".

And if you do that in a correct way, you can write: if $\displaystyle f (r,h,m) = {\pi r^2h\over m}$ then $\ln(f) = \ln\pi + 2\ln(r) + \ln(h) - \ln(m)$

However, Ln(f(x))=2(deltaR/R) + (deltaH/H) - (deltaM/M) is definitely not true.

And I really wonder how you can derive your ${\Delta f\over f} = .\;.\;.\ \$ from that.

But then, taking derivatives is indeed the right thing to do. Just not from the logarithm, but from the function itself.

Perhaps you have seen something come by in the lectures or in the textbook like $$\Delta f \approx {\partial f\over \partial r} \Delta r + {\partial f\over \partial h} \Delta h + {\partial f\over \partial m} \Delta m\ \ \ ?$$And since e.g. ${\partial f\over \partial r} = 2{f\over r}$ etc. you can divide by f on the left and right to get something like your expression.

Now comes the snag: all these contributions have to be added up. That is to say, their absolute values have to be added up. You can't have subtraction there: if a and b both have 2 %
uncertainty, a/b has 4 % uncertainty, not 0%.

You could look http://www.rit.edu/cos/uphysics/uncertainties/Uncertaintiespart2.html [Broken] to study the subject at hand. Look under (b) for multiplication and division. In your case, it looks as if the expressions under what Lindberg calls "Using simpler average errors" are to be used.

Last edited by a moderator: May 7, 2017
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