desertiev5

2022-07-10

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 $(\mathrm{cos}\alpha ,\mathrm{sin}\alpha )$ to $(\mathrm{cos}\beta ,\mathrm{sin}\beta )$

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

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

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

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

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

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

thatuglygirlyu

Beginner2022-07-11Added 14 answers

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

$d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}}{{x}^{2}}=\frac{({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta )\phantom{\rule{thinmathspace}{0ex}}d{\theta}^{2}}{{\mathrm{cos}}^{2}\theta}={\mathrm{sec}}^{2}\theta \phantom{\rule{thinmathspace}{0ex}}d{\theta}^{2}.$

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

$s=\int \sqrt{d{s}^{2}}={\int}_{\alpha}^{\beta}|\mathrm{sec}\theta |\phantom{\rule{thinmathspace}{0ex}}d\theta \phantom{\rule{thinmathspace}{0ex}}.$

$d{s}^{2}=\frac{d{x}^{2}+d{y}^{2}}{{x}^{2}}=\frac{({\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta )\phantom{\rule{thinmathspace}{0ex}}d{\theta}^{2}}{{\mathrm{cos}}^{2}\theta}={\mathrm{sec}}^{2}\theta \phantom{\rule{thinmathspace}{0ex}}d{\theta}^{2}.$

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

$s=\int \sqrt{d{s}^{2}}={\int}_{\alpha}^{\beta}|\mathrm{sec}\theta |\phantom{\rule{thinmathspace}{0ex}}d\theta \phantom{\rule{thinmathspace}{0ex}}.$

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