Propagation of errors: two different results for same question

In summary, if we write the equation ##f=\frac{uv}{u+v}## and take differentials on both sides, we get ##\frac{df}{f}=\frac{du}{u}+\frac{dv}{v}+\frac{du+dv}{u+v}##, resulting in a fractional error of 0.03. However, if we write ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## and take differentials, we get ##\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}##, giving a fractional error of 0.01.
  • #1
Terry Bing
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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 .
Relevant 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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  • #2
Terry Bing said:
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.

m
Please post your attempt, per the guidelines, and state the two results.
 
  • #3
haruspex said:
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.
 
  • #4
Terry Bing said:
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.
 
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  • #5
haruspex said:
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?
 
  • #6
Terry Bing said:
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.
 
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  • #7
haruspex said:
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.
 

Related to Propagation of errors: two different results for same question

1. What is meant by "Propagation of errors"?

Propagation of errors is the process of determining how uncertainties or errors in measured quantities affect the uncertainty of a final calculated result.

2. How is propagation of errors typically calculated?

Propagation of errors is typically calculated using the derivative method, where the uncertainties of each measured quantity are multiplied by the partial derivative of the final calculated result with respect to that quantity.

3. What can cause two different results when calculating for the same question using propagation of errors?

Two different results can occur when calculating for the same question using propagation of errors if there are discrepancies in the measured quantities or if the calculation process is done incorrectly.

4. How can one minimize the impact of errors in a calculation?

One can minimize the impact of errors in a calculation by reducing the uncertainties in the measured quantities, using more precise measuring tools, and performing multiple measurements to obtain an average value.

5. Is it possible to completely eliminate errors in a calculation?

No, it is not possible to completely eliminate errors in a calculation. However, by following proper measurement and calculation techniques, one can minimize the impact of errors and increase the accuracy of the final result.

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