# Given the hyperbolic metric d s 2 </msup> = d

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

I found that $d{s}^{2}=\frac{d{\theta }^{2}}{{\mathrm{cos}}^{2}\theta }$ but when I try to plug in $\pi /3,-\pi /3$, which should give me the arc length of $2\pi /3$,
I get $4\pi /3=\sqrt{\frac{{\left(\pi /3-\left(-\pi /3\right)\right)}^{2}}{co{s}^{2}\left(\pi /3\right)}}$
I feel like I'm making a simple mistake but I cant place it
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thatuglygirlyu
The circle ${x}^{2}+{y}^{2}=1$ can be parametrised by $\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)$. If $x\left(\theta \right)=\mathrm{cos}\theta$ and $y\left(\theta \right)=\mathrm{sin}\theta$ then
$d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}}{{x}^{2}}=\frac{\left({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \right)\phantom{\rule{thinmathspace}{0ex}}d{\theta }^{2}}{{\mathrm{cos}}^{2}\theta }={\mathrm{sec}}^{2}\theta \phantom{\rule{thinmathspace}{0ex}}d{\theta }^{2}.$
The arc-length that you are interested in is given by:
$s=\int \sqrt{d{s}^{2}}={\int }_{\alpha }^{\beta }|\mathrm{sec}\theta |\phantom{\rule{thinmathspace}{0ex}}d\theta \phantom{\rule{thinmathspace}{0ex}}.$