Use DE to show error percentage

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In summary, the conversation discusses using differentials to show the relationship between an error in the measurement of the side of a square and the resulting error in the calculation of its area. The correct formula for area is A = s^2 and the differential of the area can be approximated as dA = (dA/ds) * ds. The attempt at a solution involves using Δx=0.02 and finding the area as s^2/2500, which results in a 0.04% error instead of the expected 4%. The conversation ends with one participant acknowledging their mistake and asking for help.
  • #1
adomad123
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Homework Statement


Use Differentials to show that an error of 2% in the measurement of the side of a square results in an error of approximately 4% in the calculation of the area


Homework Equations



I'm using Area =(sΔx)^2

The Attempt at a Solution


Using Δx=0.02,
Area= [itex]s^{2}[/itex][itex]0.02^{2}[/itex] =[itex]s^{2}[/itex]/2500=0.04%[itex]\times[/itex][itex]s^{2}[/itex]
 
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  • #2
adomad123 said:

Homework Statement


Use Differentials to show that an error of 2% in the measurement of the side of a square results in an error of approximately 4% in the calculation of the area


Homework Equations



I'm using Area =(sΔx)^2
This is incorrect. The formula for area is A = s2. The differential of the area would be dA = (dA/ds) * ds ≈ dA/ds * Δs.

Don't confuse yourself by using two variables for the length of a side of the square. You shouldn't have both s and Δx in your work.
adomad123 said:

The Attempt at a Solution


Using Δx=0.02,
Area= [itex]s^{2}[/itex][itex]0.02^{2}[/itex] =[itex]s^{2}[/itex]/2500=0.04%[itex]\times[/itex][itex]s^{2}[/itex]
 
  • #3
adomad123 said:

Homework Statement


Use Differentials to show that an error of 2% in the measurement of the side of a square results in an error of approximately 4% in the calculation of the area


Homework Equations



I'm using Area =(sΔx)^2

The Attempt at a Solution


Using Δx=0.02,
Area= [itex]s^{2}[/itex][itex]0.02^{2}[/itex] =[itex]s^{2}[/itex]/2500=0.04%[itex]\times[/itex][itex]s^{2}[/itex]

Do you really not know the difference between 4% and .04% ?
 
  • #4
Ray Vickson said:
Do you really not know the difference between 4% and .04% ?

i know... i got it wrong... hence I'm asking for help.
 

Related to Use DE to show error percentage

1. What is DE and how is it used to show error percentage?

DE stands for "difference or error". It is a mathematical calculation used to compare two values and determine the difference or error between them. To show error percentage, you would use the formula: (|Actual Value - Expected Value| / Expected Value) * 100.

2. Why is DE useful in scientific experiments?

DE allows scientists to quantify the difference between their experimental results and the expected results. This can help identify areas for improvement or potential sources of error in the experiment.

3. How do you interpret the error percentage obtained from DE?

A higher error percentage indicates a larger difference between the actual and expected values, while a lower error percentage suggests a smaller difference. Ideally, the error percentage should be as close to 0% as possible, indicating a minimal difference between the two values.

4. Are there any limitations to using DE to show error percentage?

Yes, DE does not take into account the direction of the error. In other words, a positive or negative error will have the same effect on the error percentage, even though they may have different implications for the experiment. Additionally, DE assumes that the expected value is the correct value, which may not always be the case.

5. Are there any alternative methods for showing error percentage in scientific experiments?

Yes, there are other methods such as percentage error, mean absolute percentage error, or root mean squared error. Each method has its own advantages and limitations, and the choice of which to use will depend on the specific context and purpose of the experiment.

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