Assume we have the function [tex]z = x\sin y[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Our best guest for our measurement is x=1.0 and y=2.0. The uncertainty in x is 0.05. The uncertainty in y is 0.10.

We want to calculate the final uncertainty as the initial uncertainties propagate through the function.

***** Method 1 *****

In Calculus III we find the propagation of uncertainties in multivariable functions using the following method:

[tex]

dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy

[/tex]

So the uncertainty would be

[tex]

\begin{array}{l}

dz = \sin \left( y \right)dx + x\cos \left( y \right)dy \\

dz = \sin \left( {2.0} \right)\left( {0.05} \right) + \left( {1.0} \right)\cos \left( {2.0} \right)\left( {0.10} \right) \\

dz = 0.0039 \\

\end{array}

[/tex]

***** Method 2 *****

According to

https://www.amazon.com/Introduction-Error-Analysis-Uncertainties-Measurements/dp/093570275X

It says we should use this formula to calculate the propagated uncertainty:

[tex]

\delta z = \sqrt {\left( {\frac{{\partial z}}{{\partial x}}dx} \right)^2 + \left( {\frac{{\partial z}}{{\partial y}}dy} \right)^2 }

[/tex]

Using this method the uncertainty is

[tex]

\begin{array}{l}

\delta z = \sqrt {\left[ {\sin \left( {2.0} \right)\left( {0.05} \right)} \right]^2 + \left[ {\left( {1.0} \right)\cos \left( {2.0} \right)\left( {0.10} \right)} \right]^2 } \\

\delta z = 0.062 \\

\end{array}

[/tex]

The uncertainty in method 2 is nearly 16 times larger than the uncertainty in method 1.

I am assuming method 2 represents the uncertainty better than method 1.

My question is: What is method 2 taking into account that method 1 isnt? Why does method 2 represent the uncertainty better than method 1?

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# Propagation of Uncertainties using Partial Differentials and w/ and w/o Probability

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