- #1
opticaltempest
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Assume we have the function [tex]z = x\sin y[/tex]
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/dp/093570275X/?tag=pfamazon01-20
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?
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/dp/093570275X/?tag=pfamazon01-20
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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