[b]Arc length[size=150][/size][/b][br][br]Consider a differentiable vector-valued function [math]\vec{r}(t)=\langle f(t), g(t)\rangle[/math] for [math]a\leq t \leq b[/math]. It parametrizes a curve in [math]\mathbb{R}^2[/math]. How can we calculate the arc length [math]L[/math] of this curve?[br][br]First, we divide the interval [math][a,b][/math] into "subdivisions" as follows:[br][br][math]a=t_0<t_1<\cdots < t_{n-1} < t_n = b[/math] and [math]\Delta t_k=t_k-t_{k-1}[/math] for [math]k=1,2,\ldots n[/math].[br][br]Furthermore, we assume that all [math]\Delta t_k[/math] are small enough such that [br][br][math]f'(t_{k-1})\approx \frac{f(t_k)-f(t_{k-1})}{\Delta t_k}[/math][br][math]g'(t_{k-1})\approx \frac{g(t_k)-g(t_{k-1})}{\Delta t_k}[/math][br][br]Let [math]\Delta x_k = f(t_k)-f(t_{k-1})[/math] and [math]\Delta y_k = g(t_k)-g(t_{k-1})[/math] for [math]k=1,2,\ldots, n[/math]. We approximate the arc length of the curve by taking the sum of the length of the line segment joining the points on the curve corresponding to the subdivision, as shown in the applet below.[br][br]For each line segment, the length is [math]\sqrt{(\Delta x_k)^2+(\Delta y_k)^2}[/math], which can be approximated by [math]\sqrt{(f'(t_{k-1})^2+(g'(t_{k-1}))^2}\Delta t_k[/math]. Therefore, we have[br][br][math]L=\lim_{n\to\infty}\sum_{k=1}^n \sqrt{(\Delta x_k)^2+(\Delta y_k)^2} = \lim_{n\to\infty}\sum_{k=1}^n \sqrt{(f'(t_{k-1})^2+(g'(t_{k-1}))^2}\Delta t_k = \int_a^b \sqrt{(f'(t))^2+(g'(t))^2} \ dt = \int_a^b |\vec{r'}(t)| \ dt[/math][br][br](The last equality is the definition of the definite integral using Riemann sum.)[br][br]As for parametric curves in [math]\mathbb{R}^3[/math], we have the similar formula for its arc length. Suppose the curve is parametrized by [math]\vec{r}(t)=\langle f(t),g(t),h(t)\rangle[/math] for [math]a \leq t \leq b[/math]. Then the arc length [math]L[/math] of the parametric curve is[br][br][math]L=\int_a^b \sqrt{(f'(t))^2+(g'(t))^2+(h'(t))^2} \ dt = \int_a^b |\vec{r'}(t)| \ dt[/math][br][br][br]

[b][size=150]Reparametrization[/size][/b][br][br]Let [math]\phi[/math] be a smooth strictly increasing function i.e. [math]\phi[/math] is infinitely differentiable and [math]x_1 < x_2 \implies \phi(x_1) < \phi(x_2)[/math]. Given [math]\vec{r}(t)=\langle f(t),g(t),h(t)\rangle[/math] with [math]a\leq t \leq b[/math], which parametrizes a curve in [math]\mathbb{R}^3[/math]. Then we let [math]t=\phi(s)[/math] and define the vector-valued function of [math]s[/math] as follows:[br][br][math]\vec{r}(s)=\vec{r}(\phi(s))=\langle f(\phi(s)),g(\phi(s)),h(\phi(s))\rangle[/math] with [math]\phi^{-1}(a)\leq s \leq \phi^{-1}(b)[/math][br][br](Note: [math]\phi^{-1}[/math] is the inverse function of [math]\phi[/math].)[br][br]Actually, [math]\vec{r}(s)[/math] parametrizes the same curve in [math]\mathbb{R}^3[/math]. This is called a [b]reparametrization[/b] of the parametric curve.[br][br][br][u]Examples[/u]:[br][br]Suppose [math]\vec{r}(t)=\langle \cos t, \sin t, t\rangle[/math] with [math]0\leq t \leq \pi[/math].[br][br]For [math]\phi(s)=2s[/math], [math]\vec{r}(s)=\langle \cos(2s),\sin(2s), 2s\rangle[/math] with [math] 0 \leq s \leq \frac{\pi}2[/math].[br][br]For [math]\phi(s)=s^3[/math], [math]\vec{r}(s)=\langle \cos(s^3),\sin(s^3), s^3\rangle[/math] with [math] 0 \leq s \leq \sqrt[3]{\pi}[/math].[br][br][br]We would expect the arc length of a parametric curve remains unchanged after a reparametrization. The following theorem confirms this:

[u]Theorem[/u]: The arc length of the curve parametrized by [math]\vec{r}(t)[/math] from [math]a[/math] to [math]b[/math] equals the arc length of the curve parametrized by [math]\vec{r}(s)[/math], where [math]t=\phi(s)[/math], from [math]\phi^{-1}(a)[/math] to [math]\phi^{-1}(b)[/math] i.e.[br][br][math]\int_a^b |\vec{r'}(t)| \ dt = \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} |\vec{r'}(s)| \ ds[/math][br][br][u]Proof[/u]:[br][br][math] \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} |\vec{r'}(s)| \ ds = \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} |\phi'(s)\vec{r'}(\phi(s))| \ ds = \int_{\phi^{-1}(a)}^{\phi^{-1}(b)} |\vec{r'}(\phi(s))| \phi'(s)ds[/math] (since [math]\phi'(s)\geq 0[/math])[br][math]=\int_a^b |\vec{r'}(t)| \ dt [/math] (using substitution [math]t=\phi(s)[/math], we have [math]dt=\phi'(s)ds[/math]).[br][br][br]In the applet below, the right panel shows the reparametrization of the curve when [math]t=\phi(s)[/math]. You can see that the parametric curves for both parametrizations are exactly the same. [br][br][br]

[u]Reparametrization by arc length[/u][br][br]Given a vector-valued function [math]\vec{r}(t)[/math] with [math]t \geq a[/math], which parametrizes a curve in [math]\mathbb{R}^3[/math]. We want to find a reparametrization [math]t=\phi(s)[/math] such that [math]s = [/math] the arc length from [math]\vec{r}(a)[/math] to [math]\vec{r}(t)[/math] for any [math]t \geq a[/math]. Therefore, we have[br][br][math]s = \phi^{-1}(t)=\int_a^t |\vec{r'}(u)| \ du[/math][br][br]Take the helix [math]\vec{r}(t)=\langle \cos t,\sin t ,t\rangle[/math] with [math]t\geq 0[/math] as an example.[br][br][math]s=\int_0^t |\vec{r'}(u)| \ du = \int_0^t |\langle -\sin u,\cos u ,1\rangle| \ du=\int_0^t\sqrt{2} \ du=\sqrt{2}t[/math][br][br]Therefore, [math]s=\phi^{-1}(t)=\sqrt{2}t[/math], which implies that [math]t=\phi(s)=\frac{s}{\sqrt{2}}[/math]. The desired reparametrization is as follows:[br][br][math]\vec{r}(s)=\left\langle\cos\left(\frac {s}{\sqrt{2}}\right),\sin\left(\frac{s}{\sqrt{2}}\right),\frac{s}{\sqrt{2}}\right\rangle[/math], [math]s\geq 0[/math][br][br]This is called the [b]reparametrization by arc length[/b].[br][br]Suppose we want to find the point [math]\vec{r}(s)[/math] on the helix such that the arc length from [math]\vec{r}(0)[/math] to [math]\vec{r}(s)[/math] is [math]10[/math]. We just need to set [math]s=10[/math] i.e. [br][br][math]\vec{r}(10)=\left\langle\cos\left(\frac {10}{\sqrt{2}}\right),\sin\left(\frac{10}{\sqrt{2}}\right),\frac{10}{\sqrt{2}}\right\rangle[/math] is the required point on the helix.[br][br][br][u]Remark[/u]: The reparametrization of a curve in [math]\mathbb{R}^2[/math] is defined in a similar way.[br][br]