Linearized uncertainty

[Dennisl]

Homework Statement

I'm doing a lab and I'm having difficulty calculating error for a linearized set of data. I need to find the error for a set of data that has been curved to a square root. for example my x= 5.00 +- .02 becomes sqrt(5) but what is the error?

Homework Equations

The Attempt at a Solution

I am debating whether to convert the original uncertainty to percent uncertainty and just multiply that by the linearized data or to convert the original uncertainty to percent uncertainty, square root the percentage and then multiply that by the linearized data.

 

[RoyalCat]

Let us denote the new quantity, $$y$$, which is a function of whatever variables you may have, $$x_1, x_2, x_3...$$, which have their respective uncertainties $$\Delta x_1, \Delta x_2, \Delta x_3...$$

Then the uncertainty in $$y$$ is given by the formula:

$$\Delta y = \sqrt{(\frac{\partial y}{\partial x_1}\cdot \Delta x_1)^2+(\frac{\partial y}{\partial x_2}\cdot \Delta x_2)^2+(\frac{\partial y}{\partial x_3}\cdot \Delta x_3)^2+...}$$

 

[Dennisl]

so for my example point 5.00 +- .02 then y=$$\sqrt{\left(\frac{\sqrt{5}}{5}\times\frac{.02}{5.00}\right)^{2}$$ = 2.2361 $$\pm$$ .0018 (ignoring sig figs)?

 

[RoyalCat]

No no, $$\frac{\partial y}{\partial x}$$ means to take the partial derivative of y with respect to x (That means taking the derivative with respect to x, while treating every other parameter as constant)

So in your case, $$y=\sqrt{x}=x^{1/2}$$
$$\frac{\partial y}{\partial x}=\frac{1}{2\sqrt{x}}$$

Now that you know what the notation means, try and find the error.

My result is 2.2360±0.0044, if you want to compare.

 

[Dennisl]

Oh I didn't realize those were d's. so it is $$\sqrt{\left(\frac{1}{2\sqrt{5}}\times.02\right)^{2}}$$
I get 2.2360±0.0044 as well. I see how it works. Thank you for your help!

 

[RoyalCat]

Sure thing. :) Just try and keep the general formula in mind, since eventually you'll have to deal with errors in quantities like:

$$z=\sqrt{x+y}$$
$$\Delta x=5.00\pm 0.02$$
$$\Delta y=6.00\pm 0.03$$

See if you can find $$\Delta z$$
My answer is: ±0.0054