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Propagation of uncertainties with calculus

by Darkmisc
Tags: uncertainties
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Darkmisc
#1
Apr10-12, 09:15 PM
P: 56
Suppose F = x/y

dF= [itex]\frac{\partialF}{\partialx}[/itex][itex]\delta[/itex]x+[itex]\frac{\partialF}{\partialy}[/itex][itex]\delta[/itex]y

This gives

dF=[itex]\frac{\deltax}{Y}[/itex]-[itex]\frac{x}{y^2}[/itex][itex]\delta[/itex]y


That is, the partial derivative of y comes out negative. Should i leave it as a negative?

I see no reason to take the absolute value of the partial of y, but what happens when adding the two partials gives zero uncertainty? Would the uncertainty for that particular measurement just be zero?
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mathman
#2
Apr11-12, 03:24 PM
Sci Advisor
P: 6,040
Your expression for dF is correct. However the interpretation as far as uncertainty is concerned is flawed. To get uncertainty for 2 independent variables you need square root of sum of squares. This is what you need unless there is some relationship between x and y.
Bob S
#3
Apr11-12, 03:39 PM
P: 4,663
Statistically uncorrelated errors add in quadrature:

[tex] dF^2=\left( \frac{\partial F}{\partial x}\delta x \right)^2 +\left( \frac{\partial F}{\partial y}\delta y \right)^2 [/tex]


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