Use DE to show error percentage

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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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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]
 
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% ?
 
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.