When is a ruled parametrisation an actual parametrisation? I'm trying to get an answer to the following question:

When is a ruled parametrization $ r(s, t) = c(s) + tv(s) : \mathbb{R}^2 \to \mathbb{R}^3 $ a parametrization? What constraints must $ c, v $ satisfy?

This altogether is a vague question, so to clarify, I will assume the following definition"

A smooth map $ r : \mathbb{R}^2 \to \mathbb{R}^3 $ is called a parametrization if its derivative has rank $ 2 $ at every point

This leads to the obvious observation that, in case of ruled surfaces, the following vectors:
$$ \dot{c}(s) + t\dot{v}(s) \text{ and } v(s)$$
need be linearly independent.
However, I am wondering if there is any deeper sense to it? How do I interpret this? I would love some feedback on it. Is this definition of parametrization the standard one?
 A: The meaning of many mathematical terms varies with the context - and in particular, with the author. The author will adapt the general concept to meet the particular needs of what (s)he is doing. That is the case here. 
In general, a parametrization is a map from some connected subset of $\Bbb R^n$ into some set $X$ which is locally 1-1. Beyond that, if there is additional structure on $X$, then one may want to add additional conditions on the parametrization to be compatible with it. For example, if $X$ has a topology, you would want the parametrization to be continuous. If $X$ is a manifold, you would want the parametrization to be differentiable. If the author is only discussing such a situation, then it is natural simply to add these conditions to the definition of a parametrization, instead of giving a more general definition that will not be needed.
Though you don't say directly, obviously your parametrizations are of subsets of some vector space, quite likely $\Bbb R^m$ for some $m$. This provides the set $X$ being parametrized with a natural manifold structure, allowing differentiation. Therefore your book requires differentiability.
By the Inverse Function Theorem, a differentiable map $\phi : U \to \Bbb R^m, U\subseteq \Bbb R^n$ the differentiable parametrization $r(s,t)$ is locally one-to-one around a point if and only if its derivative is non-singular. So a differentiable map qualifies as a parametrization when the derivative is non-singular.
The inverse function theorem works because if the derivative of the map is non-singular at some point $p$, and if two points near $p$ lie in different directions, then because traveling from $p$ towards them must also travel in different directions under the map, hence they cannot be mapped to the same point. And if the two points are in the same direction, but at different distances, then because the derivative in that direction is not zero, the image of the points under the map must also be at different distances, and again cannot be the same.
So the key point of a parametrization is that locally, it makes the image of the parametrization look like the domain. If the derivative were singular, then either there is some direction that moving in makes no change under the map, or else two different directions are mapped into the same direction, making domain and image look different.
