Why is the Error Larger Than the Area in Calculations?

[athrun200]

Homework Statement

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Homework Equations

The Attempt at a Solution

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I don't know why the error is lager than the area.
Is it possible?

 

[Liquidxlax]

For each type of function there is a different error calculation

for let's say x = tsin(sy) the error would be

Δx = (Δy)tscos(sy) where t and s are some arbitrary constants

and something like

x = tzy were t is an arbitrary constant

then

(Δx/x)² = (Δz/z)² + (Δy/y)² + 2(Δzy)²/zy

where 2(Δzy)²/zy is the covariance factor which i doubt you need to include.

so Δx = x√(all error added together and squared individually)

so

error calculations are a pain in the but in upper level physics studies but they are a necessity.

I'll give you an different example in case it doesn't make a lot of sense

Suppose that the area of a rectangle A=LW is to be determined from the following measurements of lengths of two sides:

L = 22.1 ± 0.1cm W= 7.3 ± 0.1cm

The relative contribution of ΔA_L to the error in L will be

ΔA_L/A = ΔL/L = 0.1/22.1 = 0.005

and the corresponding contribution of ΔA_W will be

ΔA_W/A = ΔW/W = 0.1/7.3 = 0.014

Thus ΔA will equal

ΔA = A√( 0.014² + 0.005²)

ΔA = 0.015A