Showing the right half of the unit hyperbola is a complete metric space.

Question

Let $f : \mathbb{R} \rightarrow \mathbb{R}^2$ be given as follows.

$$f(\theta) = (\cosh \theta, \sinh \theta)$$

I want to argue that $\mathrm{im}(f)$ is a complete metric space with respect to the Riemannian metric. What theorems are available to do this? I'm looking for something along the lines of:

1. Since $\mathbb{R}$ is complete, and
2. since $f$ is "sufficiently nice", hence

by the theorem I'm looking for, $\mathrm{im}(f)$ is a complete metric space with respect to the Riemannian metric.

Answer 1

In terms of $\theta$ the Riemannian metric on the half-hyperbola $H$ is given by

$$ds=|f'(\theta)|\>d\theta=\sqrt{\sinh^2\theta +\cosh^2\theta}\ \ d\theta\ .$$

Now the variable $\theta$, where $-\infty<\theta<\infty$, is a global coordinate on $H$. Therefore the map $$\sigma:\quad H\to{\mathbb R},\qquad f(\theta)\mapsto\int_0^\theta \sqrt{1+2\sinh^2\theta}\ \ d\theta\ ,\tag{1}$$ which sends the point $f(\theta)\in H$ to the signed length of the arc from $(1,0)\in H$ to $f(\theta)$, is an isometry from the Riemannian manifold $H$ to ${\mathbb R}$ with its standard metric. Since ${\mathbb R}$ is known to be complete it follows that $H$ is complete as well.

Unfortunately the integral on the right in $(1)$ is elliptic, so that $\sigma$ cannot be expressed in terms of elementary functions.