Propogation of error when taking sine inverse

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Homework Help Overview

The original poster is working on calculating the angle of inclination of an air track using the sine inverse function, given measurements for the hypotenuse and the opposite side with associated errors. The challenge lies in determining how to propagate the error through the sine inverse calculation.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the ratio of the opposite side to the hypotenuse and questions how to handle the error when applying the sine inverse function. Another participant introduces a general formula for error propagation involving partial derivatives, suggesting that the derivative of the arcsine function should be considered.

Discussion Status

Participants are exploring the implications of error propagation in the context of the sine inverse function. Some guidance has been offered regarding the use of partial derivatives for error calculation, but there is no explicit consensus on the best approach to take.

Contextual Notes

The original poster's problem involves specific measurements with associated uncertainties, which are critical to the discussion but not fully resolved. The complexity of error propagation in trigonometric functions is acknowledged, indicating a need for careful consideration of the underlying mathematics.

nietzsche
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Homework Statement



I need to calculate the angle of inclination of an air track. The hypotenuse is 229.8 +/- 0.05 cm and the opposite side (the height that one side of the track is raised to) is 1.3 +/- 0.05 cm. I need to calculate the error in the angle of inclination.

Homework Equations





The Attempt at a Solution



I do the division of opposite over hypotenuse and I get (0.057 +/- 0.002) as my ratio. But what happens when I take the sine inverse of this? The angle is small, so the angle is roughly equal to the ratio (0.057 rad) but what happens to the error?

Many thanks in advance.
 
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Error calculation for a function of two variables (You can easily see how to incorporate more into this formula):

Given:
[tex]x, \Delta x[/tex]

[tex]y, \Delta y[/tex]

[tex]f(x,y)[/tex]

And what we're looking for is [tex]\Delta f(x,y)[/tex]

[tex]\Delta f(x,y)=\sqrt{(\frac{\partial f}{\partial x}\cdot \Delta x)^2+(\frac{\partial f}{\partial y}\cdot \Delta y)^2}[/tex]

The notation [tex]\frac{\partial y}{\partial x}[/tex] means the partial derivative of y, with respect to x.
What that means is that you take the function of one or more variables, and only take the derivative with respect to x, treating everything else as though it were a constant.

All you need to do now, is find the derivative of the [tex]\arcsin{x}[/tex] function and you're good to go, since it's a one-variable function.
 
Thank you very much.
 
nietzsche said:
Thank you very much.

Happy to oblige. :)

This is fairly tricky stuff. More time consuming than it is tricky, to be honest. Especially if you have a lab with a fair number of results.

It was a bit intimidating when we first learned it, but practice makes perfect, so it's just something you need to get used to. Just like how the error in a sum is the square root of the squares of the errors in each of the quantities summed up.
Haha, just writing that out still wigs me out.
 

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