Definition:Derivative of Smooth Path/Real Cartesian Space

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Let $\R^n$ be a real cartesian space of $n$ dimensions.

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $\rho: \left[{a \,.\,.\, b}\right] \to \R^n$ be a smooth path in $\R^n$.

For each $k \in \left\{ {1, 2, \ldots, n}\right\}$, define the real function $\rho_k: \left[{a \,.\,.\, b}\right] \to \R$ by:

$\forall t \in \left[{a \,.\,.\, b}\right]: \rho_k \left({t}\right) = \pr_k \left({\rho \left({t}\right)}\right)$

where $\pr_k$ denotes the $k$th projection from the image $\operatorname{Im} \left({\rho}\right)$ of $\rho$ to $\R$.

It follows from the definition of a smooth path that $\rho_k$ is continuously differentiable for all $k$.

Let $\rho_k' \left({t}\right)$ denote the derivative of $\rho_k$ with respect to $t$.

The derivative of $\rho$ is the continuous vector-valued function $\rho': \left[{a \,.\,.\, b}\right] \to \R^n$ defined by:

$\forall t \in \left[{a \,.\,.\, b}\right]: \rho' \left({t}\right) = \displaystyle \sum_{k \mathop = 1}^n \rho_k' \left({t}\right) \mathbf e_k$

where $\left({\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}\right)$ denotes the standard ordered basis of $\R^n$.