# Error propogation in trig functions

## Homework Statement

I have to use this in my calculation $$\theta=\tan^{-1}\left(\frac{19 \pm 1}{47 \pm 1}\right)$$ where both are in mm. How would I get this into $\theta\pm \text{error}$?

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## The Attempt at a Solution

looked through my lab manual, it wasn't there.

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ok I found this online

We can express the uncertainty in R for general functions of one or more observables. If R is a function of X and Y, written as R(X,Y), then the uncertainty in R is obtained by taking the partial derivatives of R with respect to each variable, multiplication with the uncertainty in that variable, and addition of these individual terms in quadrature.

$$R=R(X,Y,....)$$
$$\delta R=\sqrt{\left(\frac{\partial R}{\partial X} \delta X\right)^2+\left(\frac{\partial R}{\partial Y}\delta Y\right)^2+...}$$

so if I say that the numerator there is x and the denominator is L, then I wind up with
$$\delta\theta=\frac{L\delta x + x\delta L}{L^2+x^2}$$

would someone be able to let me know if this is correct?

oops I made a mistake in the algebra. the formula I am getting is

$$\sqrt{ \left(\frac{L}{x^2+L^2}\delta x\right)^2+\left(-\frac{x}{x^2+L^2}\delta L\right)^2}$$

I'm getting a very small error though. less than the number of significant digits. this seems a bit off...

TSny
Homework Helper
Gold Member
oops I made a mistake in the algebra. the formula I am getting is

$$\sqrt{ \left(\frac{L}{x^2+L^2}\delta x\right)^2+\left(-\frac{x}{x^2+L^2}\delta L\right)^2}$$

I'm getting a very small error though. less than the number of significant digits. this seems a bit off...
That looks good to me. Are you expressing θ in radians?

nope, it needs to be expressed in degrees. That simplfies to $$\frac{\sqrt{(L\delta x)^2 +(x\delta L)^2}}{L^2+x^2}$$ but it's giving me a value lower than the number of sig figs I have. what do I do now? haha

is this my relative error? it comes out unitless...

bumping this. I'm really confused as to what this error means physically.

haruspex
Homework Helper
Gold Member
First, a 'soapbox' about the quadrature method. It's perfectly fine and appropriate if the input uncertainties represent some number of standard deviations (the same number in each case) of roughly normally and independently distributed measurements. The answer then represents the same number of s.d. of the output.
But it's not unusual in engineering and in lab work that the input uncertainties would be better modelled as hard limits of uniform distributions. If there are many such inputs, the best approach is probably to turn each of those into a specific number of standard deviations, use quadrature, and understand that whatever number of s.d. you chose applies to the answer. But with only two inputs, you might be better served simply considering the four extreme values generated by plugging in the extremes of the two input ranges. The resulting range won't represent a uniform distribution, more triangular perhaps, but it will give you hard limits, and in engineering that may be more relevant.
Wrt your working so far, the formula you used for derivative of arctan assumes the angle is in radians. If it's in degrees you'll need to make a correction for that. It should be dimensionless - because angles are dimensionless.

Chestermiller
Mentor
Here's another possibility. θ is maximum with 20/46, and minimum with 18/48. Determine the two angles which go along with these, average them, and then find the (equal) deviations from the average. These are the Δθs.

TSny
Homework Helper
Gold Member
the formula I am getting is

$$\sqrt{ \left(\frac{L}{x^2+L^2}\delta x\right)^2+\left(-\frac{x}{x^2+L^2}\delta L\right)^2}$$

I'm getting a very small error though. less than the number of significant digits. this seems a bit off...
The expression you derived is only valid if θ is measured in radians. When using the standard calculus formulas for derivatives of trig functions, the angles are assumed to be in radians. For example d(tanθ)/dθ = sec2θ assumes θ is in radians. So, your expression gives the error in θ measured in radians. You can then convert to degrees if you wish.

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