Why does the arclength functional take same form in rotated coordinate systems? I'm going through the textbook "Emmy Noether's Wonderful Theorem" by Dwight Neuenschwander. Therein, the author defines the coordinate transformation (infinitesmal rotation of orthogonal axes) given by
$$x'=x\cos\varepsilon+y\sin\varepsilon,\quad y'=-x\sin\varepsilon+y\cos\varepsilon,\quad z'=z.$$
He goes on to say that the distance functional
$$s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$
becomes, under the infinitesmal rotation of axes,
$$s'=\int_{a'}^{b'}\sqrt{1+\left(\frac{dy'}{dx'}\right)^2}\,dx'.$$
Can someone please explain in detail how the distance functional takes the same form under this transformation? I feel like this should be easy, but I cannot see how. I tried to express $dy'/dx'$ in terms of $x$ and $y$:
$$\frac{dy'}{dx'}=\frac{dx}{dx'}\frac{dy'}{dx}=\frac{dy}{dx}-\tan\varepsilon,$$
and use this to compute the integrand, but it turns into a mess. I've tried other things in a similar vein. Any suggestions?
 A: Simple answer: Because rotations are an isometry of the euclidean metric $g_{uv} = \delta_{uv}$. The integral:
$$L = \int_a^b \sqrt{g_{uv} \frac{dx^u}{dt} \frac{dx^v}{dt} } dt =\int_a^b \sqrt{(\frac{dx^1}{dt})^2 + (\frac{dx^2}{dt})^2 }$$
Will be the same even in rotated coordinates. More detailed: After rotation, our metric is still $g_{u'v'}= \delta_{u'v'}$ , our components are:
$$L= \int_{a'}^{b'} \sqrt{ g_{u'v'} \frac{dx^{u'} }{dt} \frac{dx^{v'} }{dt}} = \int_a^b \sqrt{ (\frac{dx^{1'} }{dt})^2 + (\frac{dx^{2'} }{dt})^2} dt$$
Where $x^{1'}$ and $x^{2'}$ are $x^1$ and $x^2$ after rotation.

Previous answer on the invariance of arc length integral
The issue is you misunderstood how we are taking the functions here. We make $x,y$ depend on $\epsilon$, I'll show it for the $x$ coordinate:
$$x_{rotated} = \cos \epsilon x + \sin(\epsilon)y $$
For small values of $\epsilon$, the above becomes:
$$ x_{rotated} \approx x +\epsilon y$$
Rearranging,
$$ \delta x = x_{rot}-x = \epsilon y +\mathbb{O}(\epsilon^2)$$
Essentially, $\delta x$ here is the difference between $x$ before and after rotation. Not the 'differential step' induced in the rotated coordinate by a step in the input plane.
Similarly, we have:
$$ \delta y= -x \epsilon +O(\epsilon^2)$$
We can write:
$$ \frac{dy'}{dx'} = \frac{d( y + \delta y)}{d(x+ \delta x)} = \frac{\frac{dy}{dx} + \frac{d (\delta y)}{dx}}{1+ \frac{d (\delta x)}{dx}}= \frac{\frac{dy}{dx} -\epsilon}{1 +\epsilon \frac{dy}{dx}}$$
The last equality, I got through plugging back previous expression for $\delta x $ and $\delta y$. Now the rest is done on page-66 of the book which involves using a lot of series expansion.
