I am studying Calculus from the book Calculus with Concepts in Calculus 6th Edition by Denny Gulick and Robert Ellis. Within the book, in section 12.4, the concept of parametrization and the different categories of parametrization is introduced. According to it:
A curve $C$ is closed if it has a parametrization whose domain is a closed interval $[a, b]$ such that $r(a) = r(b)$, but otherwise $r(t_1) \neq r(t_2)$ for $t_1 \neq t_2$, with at most finitely many exceptions.
A vector-valued function $r$ defined on an interval $I$ is smooth if $r$ has a continuous derivative on $I$ and $r(t) \neq 0$ for each interior point $t$. A curve $C$ is smooth if it has a smooth parametrization.
Now consider this example: the parametrization of a unit circle centered at the origin given by $r(t) = cos(t)i + sin(t)j$ for $0 \leq t \leq 2\pi$ where $i$ and $j$ are unit vectors along the x and y directions. This satisfies the conditions for a closed parametrization but it also satisfies the conditions of a smooth parametrization. So is it both?
Let $$x(t)=\begin{cases}\sin^3 t& -{\pi\over 2}\le t\le {\pi \over 2}\\ 1 & {\pi\over 2} \le t\le {3\pi\over 2}\\ -\sin^3t & {3\pi\over 2}\le t\le {5\pi\over 2} \\ -1 & {5\pi \over 2}\le t\le {7\pi\over 2}\end{cases}$$ $$y(t)=\begin{cases}-1& -{\pi\over 2}\le t\le {\pi \over 2}\\ -\sin^3t & {\pi\over 2} \le t\le {3\pi\over 2}\\ 1 & {3\pi\over 2}\le t\le {5\pi\over 2} \\ \sin^3t & {5\pi \over 2}\le t\le {7\pi\over 2}\end{cases}$$ Then $r(t)=(x(t),y(t))$ parametrize the boundary of the square $|x|,|y|\le 1.$ Moreover $r\in C^{(3)}.$
Remark I guess there is a misprint in OP: the smooth parametrization requires that $r'(t)\neq 0,$ i.e. the speed does not vanish. The parametrization above does not satisfy that, as $r'(t)$ vanishes at $4$ points.
Concerning the circle $r'(t)=(-\sin t,\cos t )$ thus $\|r'(t)\|=1.$