----- > [!proposition] Proposition. ([[smooth regular parameterized curves admit (re)parameterizations by arc length]]) > If $(\alpha, \alpha(I))$ is a [[regular curve|regular]], [[smooth]], [[parameterized curve]], then there exists a [[Euclidean diffeomorphism|diffeomorphism]] $\Phi$ such that $\tilde{\alpha} := \alpha \circ \Phi$ > is a [[parameterization by arc length]] having the same trace (geometric image) as $\alpha$. > \ > In particular, one choice is $\Phi:=s ^{-1}$, where $s$ is the [[arc length of a path|arc length]] function of $\alpha$. ^129813 > [!proof]- Proof. ([[smooth regular parameterized curves admit (re)parameterizations by arc length]]) > We first need to verify that $s ^{-1}$ exists and is [[smooth]]. > > We'll first show $s$ is smooth. Since $s'(t)=\|\alpha'(t)\|$ exists by the fundamental theorem of calculus, it suffices to show the [[derivative]] $s'$ is smooth. We note that $s'(t) = \sqrt{ \sum_{i} [\alpha'_{i}(t)]^{2} } = f \circ g$ where $g(t)=\sum_{i}[\alpha_{i}'(t)]^{2}$ and $f=\sqrt{ \cdot }$. These are each smooth functions on $(0, \infty)$, and since $\alpha$ is [[regular curve|regular]] the image of $g$ is entirely contained in this interval. Thus $s'(t)$ is smooth everywhere. Hence $s(t)$ is smooth everywhere. > > Since $s(t)$ is [[smooth]] and [[monotonically decreasing|monotonically increasing]], it follows that $s ^{-1}$ exists and is smooth. > > Now it suffices to show that $\tilde{\alpha}:=\alpha \circ \Phi$ has unit-speed, i.e., that $\|\tilde{\alpha}'(t)\|_{2}=1$. Using the [[chain rule]] and its corollary [[derivative of inverse function is inverse matrix|derivative of inverse is inverse of derivative]], we obtain $\begin{align} > \|\tilde{\alpha}'(t)\|_{} = & \|D[\alpha \circ \Phi] |_{t}\| \\ > = & \|D\alpha |_{\Phi(t)} D\Phi |_{t}\| \\ > = & \|\frac{\alpha'\big( \Phi(t) \big)}{s'(\Phi(t))}\| \\ > = & \|\frac{\alpha'\big(\Phi (t)\big)}{\|\alpha'\big( \Phi(t) \big)\|}\| \\ > = & 1, > \end{align}$where in the penultimate step we used that $s'(t) = \|\alpha'(t)\|$ (true because $s'(t) = \frac{d}{dt} \int _{t_{0}}^{t} \|\alpha'(\tau)\| \, d\tau=\|\alpha'(t)\|$ ). > > ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```