Propogation of error when taking sine inverse

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To calculate the angle of inclination of an air track using the sine inverse function, the ratio of the opposite side to the hypotenuse is first determined, yielding (0.057 +/- 0.002). The challenge arises in calculating the error when applying the sine inverse to this ratio. The error can be computed using the formula for the propagation of error for functions of two variables, specifically by finding the derivative of the arcsine function. This process may seem daunting initially, but with practice, it becomes more manageable. Understanding error propagation is crucial for accurate measurements in experimental physics.
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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:
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.
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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