# Error propogation in trig functions

1. Feb 17, 2013

### richyw

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

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 $\theta\pm \text{error}$?

2. Relevant equations

shown above

3. The attempt at a solution

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

2. Feb 17, 2013

### richyw

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.

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

3. Feb 17, 2013

### richyw

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?

4. Feb 17, 2013

### richyw

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...

5. Feb 17, 2013

### TSny

That looks good to me. Are you expressing θ in radians?

6. Feb 17, 2013

### richyw

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

7. Feb 17, 2013

### richyw

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

8. Feb 19, 2013

### richyw

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

9. Feb 19, 2013

### haruspex

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.

10. Feb 19, 2013

### Staff: 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.

11. Feb 19, 2013

### TSny

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.

Last edited: Feb 19, 2013