Arc length parameter s

Question

Consider the metric $$ds^2 = \frac{dx^2+dy^2}{y^2}.$$ Assume $R>0, a\in\mathbb{R}$. Consider the curve $$\gamma(\theta)=(a+R\sin\theta,R\cos\theta)$$ for $-\frac{\pi}{2}\leq\theta\leq \frac{\pi}{2}$. Find the arclength parameter s. How would I go about doing this? Thank you in advance!

Answer 1

Since for $y>0$,

$$ds^2=\frac{dx^2+dy^2}{y^2} \iff ds=\frac{\sqrt{(dx/d\phi)^2+(dy/d\phi)^2}}{y}d\phi,$$

the arclength parameter is

$$s(\theta)=\int_0^\theta ds=\int_0^\theta \frac{\sqrt{\left(\frac{d}{d\phi}(a+R\sin\phi)\right)^2+\left(\frac{d}{d\phi}(R\cos\phi)\right)^2}}{R\cos\phi}d\phi.$$

This boils down to finding the antiderivative of $\sec\phi$.