What is the method for calculating W with uncorrelated uncertainties?

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Homework Statement

For uncorrelated uncertainties [tex]\delta[/tex] x and [tex]\delta[/tex] y give the values and uncertainties for W in standard format.
A.) W=x+y with x=4.2(0.2) and y=0.7(0.1)

Homework Equations

3. The Attempt at a Solution
well i think i am supposed to take the partial derivatives of W with respect to x and y (dW/dx and dW/dy) which both =1. Then is the total derivative DW/DXorY = 1+1 dx/dy? Or do I only use the derivatives for the uncertainties and just plug x and y, 4.2 and 0.7, into W?

 

[anathema]

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it is important to approach problems with a clear and systematic method. In this case, we are given the values and uncertainties for x and y, and we are asked to find the values and uncertainties for W.

First, we can start by writing out the equation for W: W=x+y.

Next, we can calculate the partial derivatives of W with respect to x and y, which are both equal to 1.

Then, we can use the formula for the uncertainty of a sum of two variables:

δW = √(δx² + δy²)

Substituting in the given uncertainties for x and y, we get:

δW = √(0.2² + 0.1²) = √(0.04 + 0.01) = √0.05 = 0.2236

Therefore, the value and uncertainty for W in standard format is:

W = 4.2 ± 0.2 + 0.7 ± 0.1 = 4.9 ± 0.2236

It is important to note that the uncertainties for x and y are not added together, as they are uncorrelated. Instead, we use the formula for the uncertainty of a sum.

In conclusion, the value and uncertainty for W is 4.9 ± 0.2236.

 

[thomate1]

It seems like you are on the right track. In this case, since the uncertainties are uncorrelated, we can simply add the uncertainties in quadrature to get the total uncertainty for W. So, for W=x+y, the uncertainty would be:

W = 4.2 + 0.7 = 4.9(0.2+0.1) = 4.9(0.3)

So the value and uncertainty for W would be 4.9(0.3).