Let $\mathbf{x}=\mathbf{x}(s)$, with $s\in I_s \subset\mathbb R $, be a natural representation of a regular curve $C$. According to Lipschutz's Differential Geometry book, the unit tangent vector to $C$ at the point $\mathbf{x}(s)$ is defined to be
$$\mathbf{t}=\mathbf{t}(s)=\frac{d\mathbf{x}(s)}{ds}$$
and the curvature vector on $C$ at the point $\mathbf{x}(s)$ is defined to be
$$\mathbf{k}=\mathbf{k}(s)=\frac{d\mathbf{t}(s)}{ds}=\frac{d^2\mathbf{x}(s)}{ds^2}$$
However, in case that $C$ is represented in terms of a non-natural parameter $t$, with $\mathbf{x}=\mathbf{x}(t)$ and $t\in I_t \subset\mathbb R $, the book only gives a formula for the curvature, but not for the curvature vector:
$$ |\kappa|= \frac{\left|\mathbf{x}^{\prime} \times \mathbf{x}^{\prime \prime}\right| }{\left|\mathbf{x}^{\prime}\right|^{3}} $$
So, what would the expression of the curvature vector $\mathbf{k}=\mathbf{k}(t)$ be in the case of an arbitrary parametrization? Since $|\kappa|=| \mathbf k|$, would it just be
$$\mathbf k (t)= \frac{\mathbf{x}^{\prime} \times \mathbf{x}^{\prime \prime}}{\left|\mathbf{x}^{\prime}\right|^{3}}?$$