# Propagation of errors: two different results for same question

#### Terry Bing

Homework Statement
The focal length of a spherical mirror is given by $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ . If $u=30\pm 0.3$ cm and $v=60\pm 0.6$ cm. Find the percentage error in the measurement of focal length of the mirror .
Homework Equations
We estimate errors by using formulae for differentials.
If I write $f=\frac{uv}{u+v}$ and then take differentials on both sides, I get $\frac{df}{f}=\frac{du}{u}+\frac{dv}{v}+\frac{du+dv}{u+v}$, I get the fractional error as 0.03. (I have replaced the negative signs that come as a result of quotient rule with positive signs, since we are asked to estimate the maximum error. This leads to the formula that when measurements are multiplied or divided together, their relative errors just add up.)
If instead I write $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ and then take differentials on both sides, then I get $\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}$. Using this, I get the fractional error to be 0.01. Which of these is a better estimate of the error? Also, why are there two different answers based on how I write the function. Basically I am taking first order approximations in both cases, right?

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#### haruspex

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Problem Statement: The focal length of a spherical mirror is given by $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ . If $u=30\pm 0.3$ cm and $v=60\pm 0.6$ cm. The maximum percentage error in the measurement of focal length of the mirror is $n%$. Find the value of n.
Relevant Equations: We estimate errors by using formulae for differentials.

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Please post your attempt, per the guidelines, and state the two results.

#### Terry Bing

Please post your attempt, per the guidelines, and state the two results.
I am sorry, the question got posted before I could type it out completely. I have updated it.

#### haruspex

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I have replaced the negative signs that come as a result of quotient rule with positive signs, since we are asked to estimate the maximum error
But the terms are not independent. The max you get by treating them as independent cannot be achieved.

#### Terry Bing

But the terms are not independent. The max you get by treating them as independent cannot be achieved.
I see. So if I assumed du and dv to be positive in the first two terms, then they must remain positive in the last term. So the negative sign remains negative. In general, if the numbers were different, and if I want to do it the first way, to find the maximum error, should I try different combinations of +ve and -ve du and dv and see which one gives me maximum error?

#### haruspex

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Gold Member
2018 Award
I see. So if I assumed du and dv to be positive in the first two terms, then they must remain positive in the last term. So the negative sign remains negative. In general, if the numbers were different, and if I want to do it the first way, to find the maximum error, should I try different combinations of +ve and -ve du and dv and see which one gives me maximum error?
You can rewrite the equation, with signs correct, into one du term and one dv term. Since u and v are both positive (yes?) it should be evident what the max df/f can be.

#### Terry Bing

You can rewrite the equation, with signs correct, into one du term and one dv term. Since u and v are both positive (yes?) it should be evident what the max df/f can be.
Thanks a lot. It is clear to me now.

"Propagation of errors: two different results for same question"

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