Combining uncertainties (error propagation)

[Cheesycheese213]

Hi!
I was wondering how would I calculate the uncertainty of a value that is calculated using both multiplication and division?

For example, with something like:

Q = mc(T₂ - T₁)

I'm not sure what to do with the uncertainties for T₂ - T₁

uncertainty = Δm/m + Δc/c + (ΔT₁ + ΔT₂)/(T₁ + T₂)

Or, would I just leave one of the uncertainties (since the uncertainty for temperature is the same?) so it would be like:

uncertainty = Δm/m + Δc/c + ΔT_{1 or 2}/(T₂ - T₁)

Thanks!

 

[gneill]

For what level of course is this? Have you taken any Calculus yet (differentiation in particular)?

 

[Cheesycheese213]

For what level of course is this? Have you taken any Calculus yet (differentiation in particular)?

I have taken calculus, but I'm not sure if we're supposed to use it because we haven't been taught it before for uncertainty? (I'm taking grade 11 physics)

 

[gneill]

I have taken calculus, but I'm not sure if we're supposed to use it because we haven't been taught it before for uncertainty? (I'm taking grade 11 physics)

Ah. Okay.

So my recommendation to you would be to divide the process into several steps.

No doubt you've been given formulas for finding the error for the addition/subtraction of two values, the multiplication of two values, and the division of two values.

So perhaps begin by finding the error in the addition of the temperatures. You end up with a value for the temperature difference and its error.

Then take that temperature difference value and associated error and multiply it by m with its associated error, using the appropriate error propagation for multiplication. So now you have a new quantity with associated error.

Rinse and repeat until you've gone through all the terms of the original expression.

The Calculus-based method (which is generally preferred) takes the partial differentials of the original function with respect to each of its variables, then forms a square root of the sum of squares of each partial differential multiplied by the square of the errors. So for example, suppose you had some function of several variables $f(x_1,x_2,x_3...x_n)$ and each of the variables had associated errors $\Delta x_1, \Delta x_2, \Delta x_3...\Delta x_n$. Then the propagated error in evaluating $f$ would be given by:

$\Delta f = \sqrt{\left(\frac{\partial f}{\partial x_1}\right) ^2 \Delta x_1^2 + \left(\frac{\partial f}{\partial x_2}\right) ^2 \Delta x_2^2 + ... \left(\frac{\partial f}{\partial x_n}\right) ^2 \Delta x_n^2 }$

This works with any expression for $f$, even it contains square roots, trig functions, or anything else you can find the differentials for.

 

[Cheesycheese213]

Ah. Okay.

So my recommendation to you would be to divide the process into several steps.

No doubt you've been given formulas for finding the error for the addition/subtraction of two values, the multiplication of two values, and the division of two values.

So perhaps begin by finding the error in the addition of the temperatures. You end up with a value for the temperature difference and its error.

Then take that temperature difference value and associated error and multiply it by m with its associated error, using the appropriate error propagation for multiplication. So now you have a new quantity with associated error.

Rinse and repeat until you've gone through all the terms of the original expression.

The Calculus-based method (which is generally preferred) takes the partial differentials of the original function with respect to each of its variables, then forms a square root of the sum of squares of each partial differential multiplied by the square of the errors. So for example, suppose you had some function of several variables $f(x_1,x_2,x_3...x_n)$ and each of the variables had associated errors $\Delta x_1, \Delta x_2, \Delta x_3...\Delta x_n$. Then the propagated error in evaluating $f$ would be given by:

$\Delta f = \sqrt{\left(\frac{\partial f}{\partial x_1}\right) ^2 \Delta x_1^2 + \left(\frac{\partial f}{\partial x_2}\right) ^2 \Delta x_2^2 + ... \left(\frac{\partial f}{\partial x_n}\right) ^2 \Delta x_n^2 }$

This works with any expression for $f$, even it contains square roots, trig functions, or anything else you can find the differentials for.

Ohhhh ok thank you so much! I think I'll try both of them just in case :D Thanks again!

 

[gneill]

Very happy to be of help.

 

[jtbell]

Here's another method that's similar to gneill's method (actually based on it), but doesn't use any calculus in the calculation. I'll illustrate it with an example.

Suppose x = a / (b + c), where a = 15.4 ± 0.3, b = 5.2 ± 0.2, and c = 4.3 ± 0.3.

First calculate x using the "unvaried" values of a, b, c:
$$x_0 = 15.4 / (5.2 + 4.3) = 1.62105$$
(leaving a few extra figures for now, to prevent roundoff errors)

Then vary (only) a by its uncertainty, recalculate x, and find the difference from the "unvaried" value of x:
$$x_a = 15.7 / (5.2 + 4.3) = 1.65263 \\
\Delta x_a = x_a - x_0 = 0.03158$$
(This is what the uncertainty in x would be, if a were the only uncertain quantity in the calculation.)

Repeat, varying (only) b by its uncertainty:
$$x_b = 15.4 / (5.4 + 4.3) = 1.58763 \\
\Delta x_b = x_b - x_0 = -0.03342$$
Repeat, varying (only) c by its uncertainty:
$$x_c = 15.4 / (5.2 + 4.6) = 1.57143 \\
\Delta x_c = x_c - x_0 = -0.04962$$
Finally, add the differences "in quadrature": square them, add them, and find the square root:
$$\Delta x = \sqrt {{\Delta x_a}^2 + {\Delta x_b}^2 + {\Delta x_c}^2} \\ = \sqrt{0.03158^2 + (-0.03342)^2 + (-0.04962)^2} \\ = 0.06765$$
The result is 1.62105 ± 0.06765. If this is to be reported as a "final" answer, it should be rounded off so the uncertainty has one (maybe two) significant figures: 1.61 ± 0.07.

I hope I haven't made any errors in arithmetic, or in transcription from my calculator.

Added note: if a quantity appears more than once in the formula, vary it in all locations at once. Don't treat them as separate "variations."