Too small error on a physics lab

Question

I have this function: $$\lambda=d \sin\left(\arctan\left(\frac{x}{z}\right)\right)$$ and I want to find its absolute error. d is a constant ($10^{-6}$), x is $(0.716 \pm 0.001) m $ , and z is $(1.000 \pm 0.001) m $. For the error of $\lambda$, $$\Delta\lambda=\sqrt{\left(\frac{\partial\lambda}{\partial x}\Delta x\right)^2 + \left(\frac{\partial\lambda}{\partial z}\Delta z\right)^2}=d\sqrt{\left(\frac{z\Delta x}{(z^2+x^2)(\sqrt{\frac{x^2}{z^2}+1})}\right)^2 + \left(\frac{x\Delta z}{(z^2+x^2)(\sqrt{\frac{x^2}{z^2}+1})}\right)^2}$$

And I have obtained $6.6 \cdot 10^{-10} m$. I expect a much larger error. What's wrong? The formula?

Thanks a lot

Answer 1

A suggestion: when dealing with error propagation, try always to reduce the problem as much as possible. It's terribly easy to make a mistake when dealing with those little numbers. Now, note that for positive $\cos t$ you have: $$\sin t= \tan t \cos t=\frac{\tan t}{\sqrt{1+\tan ^2t}}.$$ Substituting $t=\arctan \xi$, you get:$$ \sin (\arctan \xi)=\dfrac{\xi}{\sqrt{1+\xi ^2}}=\dfrac{1}{\sqrt{\xi ^{-2}+1}}.$$ Now:

a) Compute the uncertainty for $\xi ^{-1}=z/x.$

b) $\Delta({\sqrt{1+t}})\approx\dfrac{d{\sqrt{1+t}}}{dt}\cdot \Delta t=\dfrac{\Delta t}{2\sqrt{1+t}}.$

c) The relative uncertainty for $1/p$ is the same of $p$ (why?).