Register to reply

Propagation of uncertainties with calculus

by Darkmisc
Tags: uncertainties
Share this thread:
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


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?
Phys.Org News Partner Physics news on
Physical constant is constant even in strong gravitational fields
Physicists provide new insights into the world of quantum materials
Nuclear spins control current in plastic LED: Step toward quantum computing, spintronic memory, better displays
Apr11-12, 03:24 PM
Sci Advisor
P: 6,112
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
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]

Register to reply

Related Discussions
Errror Propagation/Uncertainties Introductory Physics Homework 0
Propagation of uncertainties Introductory Physics Homework 16
Propagation Method - Calculating Absoute/Relative/Percentage Uncertainties. Introductory Physics Homework 0
Propagation of Uncertainties using Partial Differentials and w/ and w/o Probability General Physics 2