Calculating Uncertainty in z with Method 1 & Method 2

[opticaltempest]

Assume we have the function z = x\sin y
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:

<br /> dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy<br />

So the uncertainty would be

<br /> \begin{array}{l}<br /> dz = \sin \left( y \right)dx + x\cos \left( y \right)dy \\ <br /> dz = \sin \left( {2.0} \right)\left( {0.05} \right) + \left( {1.0} \right)\cos \left( {2.0} \right)\left( {0.10} \right) \\ <br /> dz = 0.0039 \\ <br /> \end{array}<br />***** 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:

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

Using this method the uncertainty is

<br /> \begin{array}{l}<br /> \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 } \\ <br /> \delta z = 0.062 \\ <br /> \end{array}<br />

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?

 

[mathman]

To reconcile the 2 approaches, I suggest you modify method 1 to use absolute value for both terms and then add. This would bring them closer.

Method 2 is the usual statistical approach, since errors can be negative or positive.

 

[Chronos]

I believe the law of averages would rapidly assert itself in this scenario. The Chi squared probability is the most reliable method, IMO.