# Propogation of error when taking sine inverse

1. Sep 29, 2009

### nietzsche

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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?

2. Sep 29, 2009

### RoyalCat

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

Given:
$$x, \Delta x$$

$$y, \Delta y$$

$$f(x,y)$$

And what we're looking for is $$\Delta f(x,y)$$

$$\Delta f(x,y)=\sqrt{(\frac{\partial f}{\partial x}\cdot \Delta x)^2+(\frac{\partial f}{\partial y}\cdot \Delta y)^2}$$

The notation $$\frac{\partial y}{\partial x}$$ 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 $$\arcsin{x}$$ function and you're good to go, since it's a one-variable function.

3. Sep 29, 2009

### nietzsche

Thank you very much.

4. Sep 29, 2009

### RoyalCat

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.