The discussion revolves around using differentials to analyze the impact of measurement errors on the area of a square. The original poster presents a problem involving a 2% error in the side length and its effect on the area calculation.
Some participants have pointed out potential misunderstandings in the original poster's approach, particularly regarding the use of variables and the calculation of percentage errors. There is an ongoing exploration of the correct application of differentials in this context.
There appears to be confusion regarding the correct interpretation of percentage errors, as well as the proper formulation of the area in relation to the differential. The original poster acknowledges a mistake in their calculations.
This is incorrect. The formula for area is A = s2. The differential of the area would be dA = (dA/ds) * ds ≈ dA/ds * Δs.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
adomad123 said:The Attempt at a Solution
Using Δx=0.02,
Area= s^{2}0.02^{2} =s^{2}/2500=0.04%\timess^{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
The Attempt at a Solution
Using Δx=0.02,
Area= s^{2}0.02^{2} =s^{2}/2500=0.04%\timess^{2}
Ray Vickson said:Do you really not know the difference between 4% and .04% ?