Given the parameterization of a curve in $\mathbb{R}^n$,
$$\boldsymbol \gamma (t) = (x_1(t), x_2(t), \ldots, x_n(t))$$
I can not find a univocal definition of smoothness.
This answer requires existing and non-zero first derivatives of the components of $\boldsymbol \gamma(t)$.
This post does not provide a definitive choice.
In this ProofWiki page (linked from the smooth curve page) instead both versions are provided:
Version 1) a function is smooth if it belongs to class $C^{\infty}$;
Version 2) a function is smooth if it has continuous first derivatives everywhere in its domain.
Therefore:
Is there an unambiguous way to define smoothness, or is it an arbitrary concept, which may depend on the textbook, the author and/or his/her needs?
Assuming that all the components $x_i(t)$ (with $i = 1, 2, \ldots, n$) of $\boldsymbol \gamma (t)$ must belong to class $C^{\infty}$, does this imply that their $k$-th derivative must also be non-zero? In other words: consider the simple example parameterization of a straight line in $\mathbb{R}^1$ $\boldsymbol \gamma (t) = t$. It has existing but zero derivatives starting from the order $k = 2$. Can this curve be considered smooth?
(a) A parameterization $\gamma$ (defined on a finite or infinite interval $I$ which also can be closed or open or half-open) is onto the target curve $\Gamma$.
(b) The map $\gamma$ is at least continuous. (Most sources will assume some degree of differentiability.)
(c) The map $\gamma$ is at least locally injective, i.e. every point $t_0\in I$ has a neighborhood $U$ in $I$ such that the restriction of $\gamma$ to $U$ is 1-1. (You will likely learn (or already learned) the definition of a neighborhood in a real analysis or a topology class.) For instance, "space-filling curves" will not be parameterizations. Local injectivity is implied by the regularity of a curve, which means that $\gamma$ is $C^1$ and has nonvanishing 1st derivative.