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Given the hyperbolic metric $ds^2=\frac{dx^2+dy^2}{x^2}$ on the half plane $x > 0$, find the length of the arc of the circle $x^2+y^2=1$ from $(\cos\alpha,\sin\alpha)$ to $(\cos \beta, \sin\beta)$

I found that $ds^2=\displaystyle\frac{d\theta^2}{\cos^2\theta}$ but when I try to plug in $\pi/3, -\pi/3$, which should give me the arc length of $2\pi/3$,

I get $4\pi/3=\sqrt{\displaystyle\frac{{(\pi/3-(-\pi/3))}^2}{cos^2{(\pi/3})}}$

I feel like I'm making a simple mistake but I cant place it

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What do you mean: "...plug in $\pi/3$, $-\pi/3$...? Do you mean that you want to integrate over $-\pi \le 3\theta \le \pi$? –  Fly by Night Sep 28 '12 at 19:09

1 Answer 1

up vote 3 down vote accepted

The circle $x^2 + y^2 = 1$ can be parametrised by $(\cos \theta, \sin \theta)$. If $x(\theta) = \cos \theta$ and $y(\theta) = \sin \theta$ then

$$ds^2 = \frac{dx^2+dy^2}{x^2} = \frac{(\sin^2\theta+\cos^2\theta) \, d\theta^2}{\cos^2\theta} = \sec^2\theta \, d\theta^2.$$

The arc-length that you are interested in is given by:

$$s = \int \sqrt{ds^2} = \int_{\alpha}^{\beta} |\sec \theta| \, d\theta \, . $$

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