I'm trying to understand the concept of ruled surfaces but I'm still a bit confused about how they are parametrized. Let $M$ be: $$M = \{(x,y,z \in \mathbb{R}^3: x^2 + y^2 - z^2 = 1\}$$
A parametrization of $M$ is: $$\Psi:]-\pi, \pi[\times\mathbb{R} \to \mathbb{R}^3$$ $$(u,v)\mapsto(\cos(u)-v \sin (u), \sin(u) + v\cos(u), v)$$ I have two questions:
- Show that the parametrization in one-one in the given domain;
- Show that the angle between the unit circunference and the lines that generate $M$ is $\frac{\pi}{4}$.
In the first question my doubt is how to arrive at a conclusion that $\psi(u_1,v_1)=\psi(u_2,v_2)$ because I can't simplify the expression enough to arrive at that.
Regarding the second question: this is an Hyperboloid of One Sheet so we know it is a ruled surface. What is the equation of the lines that generate this surface? I believe if I have the vector of these lines I would be able to apply the formula: $$\cos(\alpha)=\frac{uv}{|u||v|}$$
For the parametrization to be one-to-one, you need to show that it's both injective and surjective (and of course that its maps to the original surface, which in this case is pretty trivial).
Injectivity is quite easy: suppose $\Psi(u_1, v_1) = \Psi(u_2, v_2) = (x, y, z)$, we have immediately that $v_1 = v_2 = z$. To prove that $u_1 = u_2$, note that $\Psi_v(u)=\Pi_{xy}\circ\Psi(u, v)$ (projecting to the first two coordinates) is the composition of three injective functions:
\begin{bmatrix} \frac{1}{1+v^2}&\frac{v}{1+v^2}\\\frac{-v}{1+v^2}&\frac{1}{1+v^2} \end{bmatrix}
So it's indeed injective.
Surjectivity fails because the point $(-1, 0, 0)$ (and a whole line passing through it) is missing. You can fix it by adding either $\pi$ or $-\pi$ to the first coordinate of the domain of $\Psi$. In that case it's easy because for any point of $M$ you can use again the above composition, inverting each step, and get your preimage.
Regarding the second question, don't overwork. The whole parametrization is rotation-invariant, all the lines are preserved by rotation around the $z$ axis. You only have to prove this, and moreover that the line corresponding to $u=0$ (which intersects the unit circle at $(1,0,0)$) has a specific angle with the line $y \to (1, y, 0)$ which is the tangent of the unit circle.
This line is $(0, v) \to (1, v, v)$ which is exactly at $\frac{\pi}{2}$ angle to the tangent to the circle.