📂Geometry

Reparameterization

Definition ¹

Let $k \in \mathbb{N}$ and a curve $\alpha : (a,b) \to \mathbb{R}^{3}$ be given. If a bijection $g: (c,d) \to (a,b)$ satisfies $g , g^{-1} \in C^{k}$, then $g$ is referred to as the Reparameterization of $\alpha$.

---

• $C^{k}$ is $k$ times differentiable and its derivative is a set of continuous functions.

Explanation

Pronouncing it is hard, so even in English, it’s read as [Reparameterization]; it’s just that the word itself is long.

In fact, the concept may sound grandiose, but thinking of curve $\alpha$ represented by parameters like $\alpha (t)$ as if $\beta (t) = \alpha \left( g (t) \right)$ is essentially reparameterization. The clear part mathematically is not just leaving it as a thought or concept but defining it as a function and objectifying it.

Conservation of Regularity

The purpose of reparameterization can be easily imagined from the fact that $g$ is a bijection. It’s about changing a curve that is hard to handle into one that is easy to write and manage, using various tricks, and $g^{-1}$ to revert it back. In other words, it’s a variable substitution.

According to the chain rule of differentiation, one obtains the following for parameters $g(r) = t \in (a,b)$, $r \in (c,d)$. $$ {{ d \beta } \over { d r }} = \left( {{ d \alpha } \over { d t }} \right) \left( {{ d g } \over { d r }} \right) $$ Similarly, by applying the chain rule and differentiating both sides of $g \left( g^{-1} (t) \right) = t$ by $t$, $$ \left( {{ d g } \over { d r }} \right) \left( {{ d g^{-1} } \over { d t }} \right) = 1 $$ it is understood that $\dfrac{dg}{dr} \ne 0$. Here, if $\alpha$ is a regular curve, then since $\dfrac{d\alpha}{dt} \ne \mathbf{0}$, it is guaranteed that $\dfrac{d\beta}{dr} \ne \mathbf{0}$ and thus, reparameterization conserves the curve’s Regularity.

Lemma

Let’s say for reparameterization $g: (c,d) \to (a,b)$ that $\beta = \alpha \circ g$. If $t_{0} = g \left( r_{0} \right)$, in $t_{0}$, the tangent vector field $T$ of $\alpha$ and in $r_{0}$, the tangent vector field $S$ of $\beta$ satisfy the following. $$ S = \pm T $$ Especially if $g$ is an increasing function, then $S=T$; if it’s a decreasing function, then $S = -T$.

---

• $T = {{ d \alpha / d t } \over { \left| d \alpha / d t \right| }}$ is referred to as the tangent vector field $T$ of $\alpha$.

Proof

$$ \begin{align*} S =& {{ {{ d \beta } \over { dr }} } \over { \left| {{ d \beta } \over { dr }} \right| }} \\ =& {{ {{ d \alpha } \over { dt }} } \over { \left| {{ d \alpha } \over { dt }} \right| }} {{ {{ d g } \over { dr }} } \over { \left| {{ d g } \over { dr }} \right| }} \\ =& T \cdot ( \pm 1) \\ =& \pm T \end{align*} $$

■