I need to prove that a pseudosphere $S$ is a regular surface. I have found for $S$ the parametrization $X(t,\theta) = (\mathbb{e}^{t} cos{(\theta)}, \mathbb{e}^{t} sin{(\theta)}, \pm (arctan{(\sqrt{1-\mathbb{e}^{2t}})}-\sqrt{1-\mathbb{e}^{2t}}))$. How do I prove its regularity?
I was kind of thinking that I would find the vectors $t$ and $\theta$ in the tangent plane (at point $p \in S$) $T_{p}(S)$ and try to wedge (^) them together, showing that the result is nowhere zero. But I am not sure that this is correct, and I do not know how I would do it if it was.
The surface $X(t,\theta)$ is regular if $X_t,X_\theta$ (the partial derivatives wrt each variable) are linearly independent for all $(t,\theta)\in domain$.
Compute $X_t\times X_\theta$. Showing that it is nowhere zero would imply regularity.
Let $$f(t)=\pm (arctan{(\sqrt{1-\mathbb{e}^{2t}})}-\sqrt{1-\mathbb{e}^{2t}}).$$ Then $$f'(t)=\pm\frac{e^{2 t} \sqrt{(1-e^{2 t})}}{(2-e^{2 t})}$$ and $X_t=(e^t\cos\theta,e^t\sin\theta,f'(t))$, $\;$$X_\theta=(-e^t\sin(\theta),e^t\cos(\theta),0)$.
Now
$X_t\times X_\theta=\left(\array{-f'(t)e^t\cos(\theta)\\-f'(t)e^t\sin(\theta)\\e^{2t}}\right)$
Whether it is the zero vector depends on your domain. Though, it looks like it is the zero vector only when $t=-\infty$.
Disclaimer: I did this quickly. There may be a mistake. :)