Are Differential Distances Along Curved Surfaces Larger Than Fixed Distances?

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I recently did a problem with some electron constraint to move on a hoop. It kind of surprised me that you just could take the old Schrödinger-equation with and let your
dx ->dβ, where β is the distance along the hoop.
Saying it in a less mathematical way, isn't a differential distance along something curved larger than a differential distance in a fixed direction? I do realize that a rigorous mathematician would shoot me for saying something like this, so how would he say it?
 
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In a problem like this the best suited thing to do is to pass to polar coordinates, so you can describe a loop more simply. Then in these coordinates with origin at the center of the loop which is at a fixed radius r, you have dx^2+dy^2=r^2d\theta^2=d\beta^2. So it's simply the old good polar coordinates.
 
Remember this? ## \displaystyle \lim_{x \rightarrow 0 } \frac {\sin x} {x} = 1 ##.

It ensures that the length of a chord and the corresponding arc are about the same when they are small, let alone differential. It might be useful for you to follow the proof of the statement.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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