# 8.6. Calculating the Length of a Path

We now consider paths and their lengths. This will finish the geometry section in our integral course.

Definition:  Let $M\subset {ℝ}^{k}$ be a subset of ${ℝ}^{k}$. If $\left[a,b\right]$ is a closed interval in $ℝ$ we call any function

 $w=\left({w}_{1},\dots {w}_{k}\right):\left[a,b\right]\to M$ [8.6.1]

a path (in M). $w\left(a\right)$ is the initial point and $w\left(b\right)$ the final point of w. The path w is closed if its initial and final points coincide: $w\left(a\right)=w\left(b\right)$. The range $\left\{w\left(t\right)|t\in \left[a,b\right]\right\}$ is called the curve belonging to w and the path [8.6.1] is then regarded as a parametrization of its curve.

If all the coordinate functions ${w}_{1},\dots ,{w}_{k}$ are

• continuous then w is a continuous path (occasionally: ${\mathcal{C}}^{0}$-path).

• differentiable, w is a differentiable path (${\mathcal{D}}^{1}$-path) and the function

 ${w}^{\prime }≔\left({{w}_{1}}^{\prime },\dots ,{{w}_{k}}^{\prime }\right):\left[a,b\right]\to {ℝ}^{k}$ [8.6.2]

is called the derivative of w. w is a regular path if ${w}^{\prime }\left(t\right)\ne \mathbf{0}$ for all $t\in \left[a,b\right]$.

Repeatedly differentiable paths and higher derivatives are introduced analogously.

• continously differentiable, w is called a smooth path (${\mathcal{C}}^{1}$-path).

• integrable, then w is an integrable path and for any $r,s\in \left[a,b\right]$ the vector

 $\underset{r}{\overset{s}{\int }}w≔\left(\underset{r}{\overset{s}{\int }}{w}_{1},\dots ,\underset{r}{\overset{s}{\int }}{w}_{k}\right)$ [8.6.3]

is read as the integral of w from r to s.

Consider:

• If $f:\left[a,b\right]\to ℝ$ is a common real-valued function, its graph is the curve belonging to the path

$w:t↦\left(t,f\left(t\right)\right),\text{ }t\in \left[a,b\right]$.

As X is arbitrary often differentiable (thus continuous and integrable as well), w has the same qualities as f has.

• The path concept is, among others, designed to describe movement. In this case it is the movement of a particle along its way, with the interval $\left[a,b\right]$ being the section of time in which the movement is monitored. The notation

$t↦\left({w}_{1}\left(t\right),\dots ,{w}_{k}\left(t\right)\right),\text{ }t\in \left[a,b\right]$

(instead of $x↦\left({w}_{1}\left(x\right),\dots ,{w}_{k}\left(x\right)\right),\text{ }x\in \left[a,b\right]$ ) supports this perception. If w is ${\mathcal{D}}^{1}$ we interpret the derivative ${w}^{\prime }\left(t\right)$ as the speed of the particle at time t.

• There is a clear difference between a path and the associated curve. A single curve may belong to several different paths. For an example consider the paths

$\begin{array}{l}w:t↦\left(t,0\right),\text{ }t\in \left[0,1\right]\hfill \\ v:t↦\left(1-t,0\right),\text{ }t\in \left[0,1\right]\hfill \end{array}$

Both of them produce the line segment between $\left(0,0\right)$ and $\left(1,0\right)$, but they differ in their running direction: Whereas w generates the line segment from left to right, v does it the reverse way.

Also, running through a curve repeatedly will change the path but not the curve. If we substitute in the following example the interval $\left[0,2\pi \right]$ by $\left[0,2k\pi \right]$ the ellipse will be cycled k-times. But this of course won't alter its shape.

Finally we note that cutting constant segments (i.e. ${w}^{\prime }\left(t\right)=\mathbf{0}$ for all $t\in \left[r,s\right]\subset \left[a,b\right]$) will leave the curve untouched, but not the path.

Example:

• With $a,b>0$ the ellipse

$t↦\left(a\cdot \mathrm{cos}t,b\cdot \mathrm{sin}t\right),\text{ }t\in \left[0,2\pi \right]$

is a closed ${\mathcal{C}}^{\infty }$-path in ${ℝ}^{2}$. The sketch shows the associated curve for $a=2$ and $b=1$.

• The ${\mathcal{C}}^{\infty }$-path

$t↦\left({t}^{2}-1,{t}^{3}-t\right),\text{ }t\in \left[-\sqrt{2},\sqrt{2}\right]$

is not closed and as $1↦\left(0,0\right)$ and $-1↦\left(0,0\right)$ the curve runs through $\left(0,0\right)$ twice.

• Lissajous-curves are generated by the following ${\mathcal{C}}^{\infty }$-paths:

$t↦\left(a\cdot \mathrm{sin}\left(nt+c\right),b\cdot \mathrm{sin}t\right),\text{ }t\in \left[0,2\pi \right]$

The sketch shows the Lissajous-curve for $a=b=c=1$ and $n=3$. Lissajou-curves are not always closed.

• The spiral  i left mouse: rotate right mouse: context menu spiral$t↦\left(t\cdot \mathrm{cos}t,t\cdot \mathrm{sin}t,t\right),\text{ }t\in \left[0,20\right]$ Display by JavaView
and the closed curve of Viviani  i left mouse: rotate right mouse: context menu curve of Viviani$t↦\left(1+\mathrm{cos}t,\mathrm{sin}t,2\cdot \mathrm{sin}\left(\frac{t}{2}\right)\right),\text{ }t\in \left[0,4\pi \right]$ Display by JavaView
are examples for ${\mathcal{C}}^{\infty }$-curves in ${ℝ}^{3}$.

• The Famous Curves Index is a large collection of curves.

Some basic derivation rules and and a version of the mean value thoerem are valid for differentiable paths.

Proposition:

1. For any differential paths $v,w:\left[a,b\right]\to {ℝ}^{k}$ each linear combination $\alpha v+\beta w:\left[a,b\right]\to {ℝ}^{k}$ is differentiable and

 $\left(\alpha v+\beta w{\right)}^{\prime }=\alpha {v}^{\prime }+\beta {w}^{\prime }$ [8.6.4]
1. If $w:\left[a,b\right]\to {ℝ}^{k}$ is a differentiable path then for any two different points $r,s\in \left[a,b\right]$ there are numbers ${\stackrel{˜}{t}}_{1},\dots ,{\stackrel{˜}{t}}_{k}$  between
 i i.e.  ${\stackrel{˜}{t}}_{i}\in \right]r,s\left[$, if $r and  ${\stackrel{˜}{t}}_{i}\in \right]s,r\left[$, if $s.
r and s such that

 $w\left(s\right)=w\left(r\right)+\left(s-r\right)\cdot \left({{w}^{\prime }}_{1}\left({\stackrel{˜}{t}}_{1}\right),\dots ,{{w}^{\prime }}_{k}\left({\stackrel{˜}{t}}_{k}\right)\right)$ [8.6.5]
1. If $w:\left[a,b\right]\to {ℝ}^{k}$ is a differentiable path the derivative ${w}^{\prime }$ is integrable and

 $\underset{r}{\overset{s}{\int }}{w}^{\prime }=w\left(s\right)-w\left(r\right)$ [8.6.6]

for all $r,s\in \left[a,b\right]$.

Proof:

1.   According to the sum and factor rule all coordinate functions $\alpha {v}_{i}+\beta {w}_{i}$ are differentiable and

$\left(\alpha {v}_{i}+\beta {w}_{i}{\right)}^{\prime }=\alpha {{v}_{i}}^{\text{'}}+\beta {{w}_{i}}^{\text{'}}$

2.   Let ${w}_{i}$ be an arbitrary coordinate function. The mean value theorem [7.9.5] provides a ${\stackrel{˜}{t}}_{i}$ between r and s such that

${w}_{i}\left(s\right)={w}_{i}\left(r\right)+\left(s-r\right)\cdot {{w}_{i}}^{\text{'}}\left({\stackrel{˜}{t}}_{i}\right)={w}_{i}\left(r\right)+\left(s-r\right)\cdot {{w}^{\prime }}_{i}\left({\stackrel{˜}{t}}_{i}\right)$

3.   Every coordinate function ${w}_{i}$ is certainly a primitive for ${{w}_{i}}^{\text{'}}={{w}^{\prime }}_{i}$.

The integral of a path satisfies the same calaculation rules and has the same properties just as a usual integral.

Proposition:  If $v,w:\left[a,b\right]\to {ℝ}^{k}$ are integrable paths we have for all $r,s,t\in \left[a,b\right]$ :

 $\underset{r}{\overset{r}{\int }}w=\mathbf{0},\text{ }\underset{r}{\overset{s}{\int }}w=-\underset{s}{\overset{r}{\int }}w,\text{ }\underset{r}{\overset{s}{\int }}w=\underset{r}{\overset{t}{\int }}w+\underset{t}{\overset{s}{\int }}w$ [8.6.7]
 $\underset{r}{\overset{s}{\int }}\alpha v+\beta w=\alpha \underset{r}{\overset{s}{\int }}v+\beta \underset{r}{\overset{s}{\int }}w$ [8.6.8]
1. If r and s are different there are numbers ${\stackrel{˜}{t}}_{1},\dots ,{\stackrel{˜}{t}}_{k}$ between r and s such that

 $\underset{r}{\overset{s}{\int }}w=\left(s-r\right)\cdot \left({w}_{1}\left({\stackrel{˜}{t}}_{1}\right),\dots ,{w}_{k}\left({\stackrel{˜}{t}}_{k}\right)\right)$ [8.6.9]

Proof:  We prove the respective identity coordinatewise and succeed in

1.    with [8.2.2] - [8.2.4].

2.    with [8.2.5]/[8.2.7].

3.    with [8.2.8].

The estimate [8.2.11] is transferable to continuous paths, an essential step in our theory. The notation and the proof as well use the common dot product
 i The dot product $\mathbit{x}·\mathbit{y}≔\sum _{i=1}^{k}{x}_{i}\cdot {y}_{i}\text{\hspace{0.17em}},\text{ }\mathbit{x},\mathbit{y}\in {ℝ}^{k}$ in particular provides the length  $|\mathbit{x}|=\sqrt{\mathbit{x}·\mathbit{x}}=\sqrt{\sum _{i=1}^{k}{x}_{i}^{2}}\text{\hspace{0.17em}}$ of a vector $\mathbit{x}\in {ℝ}^{k}$. Our proof uses the simple identity $|\mathbit{x}{|}^{2}=\mathbit{x}·\mathbit{x}$, and the Cauchy-Schwarz inequality $|\mathbit{x}·\mathbit{y}|\le |\mathbit{x}|\cdot |\mathbit{y}|$ Its proof and further properties of the dot product, as e.g. the triangle inequality $|\mathbit{x}+\mathbit{y}|\le |\mathbit{x}|+|\mathbit{y}|$ are to be found in part 9.13.
in ${ℝ}^{k}$.

Proposition:  For any continuous path $w:\left[a,b\right]\to {ℝ}^{k}$ the following inequality holds:

 $|\underset{a}{\overset{b}{\int }}w|\le \underset{a}{\overset{b}{\int }}|w|$ [8.6.10]

Proof:  First we note that the paths w and $|w|$ are continuous and thus integrable. With the abbreviation

$c≔\underset{a}{\overset{b}{\int }}w$,  thus  ${c}_{i}=\underset{a}{\overset{b}{\int }}{w}_{i}$

the following calculation is easier to follow. For $|c|=0$ there is nothing to prove because the right side of [8.6.10] is always positive. So we may assume $|c|\ne 0$. According to the Cauchy-Schwarz inequality and due to the integral monotony [8.2.10] we may estimate as follows:

$\begin{array}{ll}\hfill |c|\cdot |\underset{a}{\overset{b}{\int }}w|=|c{|}^{2}=c·c=\sum _{i=1}^{k}{c}_{i}\cdot \underset{a}{\overset{b}{\int }}{w}_{i}& =\underset{a}{\overset{b}{\int }}\sum _{i=1}^{k}{c}_{i}\cdot {w}_{i}\hfill \\ \hfill & =\underset{a}{\overset{b}{\int }}c·w\le \underset{a}{\overset{b}{\int }}|c·w|\le \underset{a}{\overset{b}{\int }}|c|\cdot |w|=|c|\underset{a}{\overset{b}{\int }}|w|\hfill \end{array}$

We are now going to calculate the length of a continuous path w. The actual key concept in calculating areas then was to approximate the region in question by a sequence of elementary regions (union of rectangles) with a known area. Analogously we will now try to approximate the path w by elementary paths, namely the union of line segments.

We start with some technical preparations: A finite sequence $Z=\left({t}_{0},\dots ,{t}_{n}\right)$ in $\left[$, $n\ge 1$, such that

$a={t}_{0}<{t}_{1}<\dots <{t}_{n-1}<{t}_{n}=b$

is called a partition of the interval $\left[$ with the number $\mathrm{max}\left\{{t}_{1}-{t}_{0},\dots ,{t}_{n}-{t}_{n-1}\right\}$ being its fineness.

For a given path w any partition Z features  $n+1$ points $w\left({t}_{0}\right),\dots ,w\left({t}_{n}\right)$ of w and thus n consecutive line segments. We use them to create the traverse   ${p}_{Z}$:

 ${p}_{Z}\left(t\right)=w\left({t}_{i-1}\right)+\frac{t-{t}_{i-1}}{{t}_{i}-{t}_{i-1}}\left(w\left({t}_{i}\right)-w\left({t}_{i-1}\right)\right)$,  if $t\in \left[{t}_{i-1},{t}_{i}\right]$ [8.6.11]

Each traverse ${p}_{Z}$ is a path with initial point $w\left(a\right)=w\left({t}_{0}\right)$ and final point $w\left(b\right)=w\left({t}_{n}\right)$. Traverses approximate a continuous path w in the following way:

Proposition:  Let $w:\left[a,b\right]\to {ℝ}^{k}$ be a continuous path. Then there is $\delta >0$ for each $\epsilon >0$ such that

 $|w\left(t\right)-{p}_{Z}\left(t\right)|<\epsilon$  for all $t\in \left[a,b\right]$. [8.6.12]

for any partition Z with a fineness less than δ.

Proof:  Take an arbitrary $\epsilon >0$. As each coordinate function ${w}_{j}$ is continuous function on the closed interval $\left[a,b\right]$ it is actually uniformly continuous (see [6.5.5]). Thus there is a ${\delta }_{j}>0$ for $\frac{\epsilon }{2\sqrt{k}}>0$ such that

$s,t\in \left[a,b\right]\text{ }\wedge \text{ }|s-t|<{\delta }_{j}\text{ }⇒\text{ }|{w}_{j}\left(s\right)-{w}_{j}\left(t\right){|}^{2}<\frac{{\epsilon }^{2}}{4k}$

Setting $\delta ≔\mathrm{min}\left\{{\delta }_{1},\dots ,{\delta }_{k}\right\}$ we therefor have for any $s,t\in \left[a,b\right]$ with $|s-t|<\delta$:

$|w\left(s\right)-w\left(t\right)|=\sqrt{\sum _{j=1}^{k}|{w}_{j}\left(s\right)-{w}_{j}\left(t\right){|}^{2}}<\sqrt{k\frac{{\epsilon }^{2}}{4k}}=\frac{\epsilon }{2}$[1]

Now let $Z=\left({t}_{0},\dots ,{t}_{n}\right)$ be an arbitrary partition with a fineness not exceeding δ. If $t\in \left[a,b\right]$, say $t\in \left[{t}_{i-1},{t}_{i}\right]$, we thus know:

$|t-{t}_{i-1}|<\delta$  and  $|{t}_{i}-{t}_{i-1}|<\delta$.

Using [8.6.11], the triangle inequality and the estimate [1] we now see that

$\begin{array}{ll}|w\left(t\right)-{p}_{Z}\left(t\right)|\hfill & \le |w\left(t\right)-w\left({t}_{i-1}\right)|+|w\left({t}_{i-1}\right)-{p}_{Z}\left(t\right)|\hfill \\ \hfill & =|w\left(t\right)-w\left({t}_{i-1}\right)|+\underset{\le 1}{\underbrace{|\frac{t-{t}_{i-1}}{{t}_{i}-{t}_{i-1}}|}}\cdot |w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|\hfill \\ \hfill & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \hfill \end{array}$

The length $L\left({p}_{Z}\right)$ of a traverse is easily calculated by adding up the lengths of the line segments involved. Thus we set

 $L\left({p}_{Z}\right)=\sum _{i=1}^{n}|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|$ [8.6.13]

As the line segment represents the shortest distance between $w\left({t}_{i-1}\right)$ and $w\left({t}_{i}\right)$ we expect the length of w to be an upper bound for all $L\left({p}_{Z}\right)$. Considering [8.6.12], a sound candidate for the prospective path length would be the supremum
 i We remember the completeness axiom :In $ℝ$ each non-empty bounded subset $M\subset ℝ$ has a least upper bound, its supremum (termed $\mathrm{sup}M$).An upper bound s of M equals $\mathrm{sup}M$ if and only if $s-\epsilon$ is no upper bound of M for each $\epsilon >0$.
of $\left\{L\left({p}_{Z}\right)|Z\phantom{\rule{0.3em}{0ex}}\text{partition of}\phantom{\rule{0.3em}{0ex}}\left[a,b\right]\right\}$.

The depicted example on the right shows a part of one of René Descartes'  Folium of Descartes, that is the path  $t↦\left(\frac{12t}{1+{t}^{3}},\frac{12{t}^{2}}{1+{t}^{3}}\right)\text{\hspace{0.17em}},\text{\hspace{0.17em}\hspace{0.17em}}t\in \left[-0.4,10\right]$. For a three dimensional example we replenish the spiral
 i left mouse: rotate right mouse: context menu spiral$t↦\left(t\cdot \mathrm{cos}t,t\cdot \mathrm{sin}t,t\right),\text{ }t\in \left[0,20\right]$ Display by JavaView
from a former example.

Definition: A continuous path $w:\left[a,b\right]\to {ℝ}^{k}$ is a path of finite length (or a rectifiable path) if the set $\left\{L\left({p}_{Z}\right)|Z\phantom{\rule{0.3em}{0ex}}\text{partition of}\phantom{\rule{0.3em}{0ex}}\left[a,b\right]\right\}$ is bounded. In this case the number

 $L\left(w\right)≔\mathrm{sup}\left\{L\left({p}_{Z}\right)|Z\phantom{\rule{0.3em}{0ex}}\text{partition of}\phantom{\rule{0.3em}{0ex}}\left[a,b\right]\right\}$ [8.6.14]

is called the length of w.

We will find out that smooth paths are rectifiable and, in addition, integrating the length of their derivative ${w}^{\prime }$ will provide an upper bound for all $L\left({p}_{Z}\right)$. Smoothness however is only a sufficient and not a necessary condition for a finite length, as the non-smooth path  $w:t↦\left(t,|t|\right),\text{\hspace{0.28em}}t\in \left[-1,-1\right]$
 i If $Z=\left({t}_{0},\dots ,{t}_{n}\right)$ is a partition of $\left[-1,1\right]$ we find a suitable k such that ${t}_{k}\le 0<{t}_{k+1}$. That means: and so we have ${\left(|{t}_{i}|-|{t}_{i-1}|\right)}^{2}={\left({t}_{i}-{t}_{i-1}\right)}^{2}$ and consequently $|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|=\sqrt{{\left({t}_{i}-{t}_{i-1}\right)}^{2}+{\left(|{t}_{i}|-|{t}_{i-1}|\right)}^{2}}=\sqrt{2}\left({t}_{i}-{t}_{i-1}\right)$ for all $i\ne k+1$. Taking into account that $\begin{array}{l}|w\left({t}_{k+1}\right)-w\left(0\right)|=\sqrt{{t}_{k+1}^{\phantom{\rule{1.5em}{0ex}}2}+|{t}_{k+1}{|}^{2}}=\sqrt{2}|{t}_{k+1}|=\sqrt{2}\text{\hspace{0.17em}}{t}_{k+1}\hfill \\ |w\left(0\right)-w\left({t}_{k}\right)|=\sqrt{{t}_{k}^{\phantom{\rule{0.4em}{0ex}}2}+|{t}_{k}{|}^{2}}=\sqrt{2}|{t}_{k}|=-\sqrt{2}\text{\hspace{0.17em}}{t}_{k}\hfill \end{array}$ we now may estimate $L\left({p}_{Z}\right)$ as follows: $\begin{array}{ll}L\left({p}_{Z}\right)\hfill & =\sum _{i=1}^{n}|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|\hfill \\ \hfill & \le \sum _{i=1}^{k}|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|\hfill \\ \hfill & +|w\left({t}_{k+1}\right)-w\left(0\right)|+|w\left(0\right)-w\left({t}_{k}\right)|\hfill \\ \hfill & +\sum _{i=k+2}^{n}|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|\hfill \\ \hfill & =\sqrt{2}\left(\sum _{i=1}^{k}{t}_{i}-{t}_{i-1}+{t}_{k+1}-{t}_{k}+\sum _{i=k+2}^{n}{t}_{i}-{t}_{i-1}\right)\hfill \\ \hfill & =\sqrt{2}\left({t}_{n}-{t}_{0}\right)=2\sqrt{2}\hfill \end{array}$
proves.

Proposition:  If $w:\left[a,b\right]\to {ℝ}^{k}$ is a smooth path then for any partition Z of $\left[a,b\right]$ the estimate

 $L\left({p}_{Z}\right)\le \underset{a}{\overset{b}{\int }}|{w}^{\prime }|$ [8.6.15]

holds and w is thus rectifiable.

Proof:  If $Z=\left({t}_{0},\dots ,{t}_{n}\right)$ is an arbitrary partition we see from the properties [8.6.6], [8.6.7] and the estimate [8.6.10] that

$L\left({p}_{Z}\right)=\sum _{i=1}^{n}|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|=\sum _{i=1}^{n}|\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}{w}^{\prime }|\le \sum _{i=1}^{n}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}|{w}^{\prime }|=\underset{a}{\overset{b}{\int }}|{w}^{\prime }|$.

Surprisingly, [8.6.15] could be replaced by a stronger result: The integral of $|{w}^{\prime }|$ is more than a common upper bound of $\left\{L\left({p}_{Z}\right)|Z\phantom{\rule{0.3em}{0ex}}\text{partition of}\phantom{\rule{0.3em}{0ex}}\left[a,b\right]\right\}$, it is the least one and thus the length of w.

Proposition:  For the length of a smooth path $w:\left[a,b\right]\to {ℝ}^{k}$ we have:

 $L\left(w\right)=\underset{a}{\overset{b}{\int }}|{w}^{\prime }|$ [8.6.16]

Proof:  Due to [8.6.15] the integral $\underset{a}{\overset{b}{\int }}|{w}^{\prime }|$ is an upper bound of . It is thus sufficient to find a partition Z for each $\epsilon >0$ such that

$\underset{a}{\overset{b}{\int }}|{w}^{\prime }|-\epsilon \le L\left({p}_{Z}\right)$.

As the coordinate functions ${{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }$ are continuously differentiable, each ${\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\right)}^{2}$ is continuous and on the closed interval $\left[a,b\right]$ even uniformly continuous. Thus there is a ${\delta }_{j}>0$ for $\overline{\epsilon }≔\frac{{\epsilon }^{2}}{k{\left(b-a\right)}^{2}}>0$  such that

$\begin{array}{ll}|t-s|<{\delta }_{j}\hfill & \text{ }⇒\text{ }{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left(t\right)\right)}^{2}-{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left(s\right)\right)}^{2}\le |{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left(t\right)\right)}^{2}-{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left(s\right)\right)}^{2}|<\overline{\epsilon }\hfill \\ \hfill & \text{ }⇒\text{ }{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left(t\right)\right)}^{2}<{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left(s\right)\right)}^{2}+\overline{\epsilon }\hfill \end{array}$[2]

for all $s,t\in \left[a,b\right]$. Now take a partition $Z=\left({t}_{0},\dots ,{t}_{n}\right)$ with a fineness less than $\delta ≔\mathrm{min}\left\{{\delta }_{1},\dots ,{\delta }_{k}\right\}$. According to [8.2.8] and [8.6.5] resp. we find numbers

• ${\stackrel{˜}{y}}_{i}\in \right]{t}_{i-1},{t}_{i}\left[$  such that  $\underset{a}{\overset{b}{\int }}|{w}^{\prime }|=\sum _{i=1}^{n}\underset{{t}_{i-1}}{\overset{{t}_{i}}{\int }}|{w}^{\prime }|=\sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\cdot |{w}^{\prime }\left({\stackrel{˜}{y}}_{i}\right)|$[3]

• ${\stackrel{˜}{x}}_{1,i},\dots ,{\stackrel{˜}{x}}_{k,i}\in \right]{t}_{i-1},{t}_{i}\left[$  such that

$\begin{array}{ll}L\left({p}_{Z}\right)=\sum _{i=1}^{n}|w\left({t}_{i}\right)-w\left({t}_{i-1}\right)|\hfill & =\sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\cdot |{{w}_{\phantom{\rule{-0.5em}{0ex}}1}}^{\prime }\left({\stackrel{˜}{x}}_{1,i}\right),\dots ,{{w}_{\phantom{\rule{-0.5em}{0ex}}k}}^{\prime }\left({\stackrel{˜}{x}}_{k,i}\right)|\hfill \\ \hfill & =\sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\sqrt{\sum _{j=1}^{k}{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left({\stackrel{˜}{x}}_{j,i}\right)\right)}^{2}}\hfill \end{array}$[4]

As  ${\stackrel{˜}{y}}_{i},{\stackrel{˜}{x}}_{1,i},\dots ,{\stackrel{˜}{x}}_{k,i}\in \right]{t}_{i-1},{t}_{i}\left[$ we have for all i:

$|{\stackrel{˜}{y}}_{i}-{\stackrel{˜}{x}}_{j,i}|<\delta \le {\delta }_{j}$.

With [2] we thus know that ${\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left({\stackrel{˜}{y}}_{i}\right)\right)}^{2}<{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left({\stackrel{˜}{x}}_{j,i}\right)\right)}^{2}+\overline{\epsilon }$ and according to [3] and [4] we may estimate the integral of $|{w}^{\prime }|$ as follows (note that $\sqrt{x+y}\le \sqrt{x}+\sqrt{y}$ for any positive x, y):

$\begin{array}{ll}\underset{a}{\overset{b}{\int }}|{w}^{\prime }|=\sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\sqrt{\sum _{j=1}^{k}{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left({\stackrel{˜}{y}}_{i}\right)\right)}^{2}}\hfill & <\sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\sqrt{\sum _{j=1}^{k}\left({\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left({\stackrel{˜}{x}}_{j,i}\right)\right)}^{2}+\overline{\epsilon }\right)}\hfill \\ \hfill & \le \sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\sqrt{\sum _{j=1}^{k}{\left({{w}_{\phantom{\rule{-0.5em}{0ex}}j}}^{\prime }\left({\stackrel{˜}{x}}_{j,i}\right)\right)}^{2}}+\sum _{i=1}^{n}\left({t}_{i}-{t}_{i-1}\right)\sqrt{k\overline{\epsilon }}\hfill \\ \hfill & =L\left({p}_{Z}\right)+\left(b-a\right)\sqrt{k\overline{\epsilon }}\hfill \\ \hfill & =L\left({p}_{Z}\right)+\epsilon \hfill \end{array}$

This proves the required estimate $\underset{a}{\overset{b}{\int }}|{w}^{\prime }|-\epsilon \le L\left({p}_{Z}\right)$. Therefor $\underset{a}{\overset{b}{\int }}|{w}^{\prime }|$ is the least upper bound of , i.e. its supremum.

In a former comment paths were considered as means of describing the movement of a particle. With this perception [8.6.16] just says that the length of the path traced by the particle is the integral of its magnitude of speed $|{w}^{\prime }|$.

Although [8.6.16] could replace computing a supremum (usually a difficult task) with calculating an integral, the length of a path is rarely simple to get as the integrand is always the length of a vector, which actually is the root of a sum. The following examples will demonstrate this.

Example:

• The curve belonging to the smooth path

$w:t↦a+t\left(b-a\right)=\left({a}_{1}+t\left({b}_{1}-{a}_{1}\right),\dots ,{a}_{k}+t\left({b}_{k}-{a}_{k}\right)\right),\text{ }t\in \left[0,1\right]$

is the line segment joining the points $a=\left({a}_{1},\dots {a}_{k}\right)$ and $b=\left({b}_{1},\dots {b}_{k}\right)$ in ${ℝ}^{k}$. With the constant derivative ${w}^{\prime }=\left({b}_{1}-{a}_{1},\dots ,{b}_{k}-{a}_{k}\right)=b-a$ the length of w straight away calculates to

$\begin{array}{ll}L\left(w\right)\hfill & =\underset{0}{\overset{1}{\int }}|b-a|\hfill \\ \hfill & =|b-a|\underset{0}{\overset{1}{\int }}1\hfill \\ \hfill & =|b-a|\hfill \end{array}$

which, as expected, is the distance between a and b.

• For an arbitrary $a>0$ we compute the length of

$w:t↦\left(t\cdot \mathrm{cos}t,t\cdot \mathrm{sin}t,t\right),\text{ }t\in \left[0,a\right]$

thus measuring the spiral from our initial example:

From ${w}^{\prime }=\left(\mathrm{cos}-\mathrm{X}\cdot \mathrm{sin},\mathrm{sin}+\mathrm{X}\cdot \mathrm{cos},1\right)$ we get (note the Pythagorean theorem : ${\mathrm{sin}}^{2}+{\mathrm{cos}}^{2}=1$)

$\begin{array}{ll}|{w}^{\prime }|\hfill & =\sqrt{{\left(\mathrm{cos}-\mathrm{X}\cdot \mathrm{sin}\right)}^{2}+{\left(\mathrm{sin}+\mathrm{X}\cdot \mathrm{cos}\right)}^{2}+{1}^{2}}\hfill \\ \hfill & =\sqrt{{\mathrm{cos}}^{2}-2\mathrm{X}\cdot \mathrm{cos}\cdot \mathrm{sin}+{\mathrm{X}}^{2}\cdot {\mathrm{sin}}^{2}+{\mathrm{sin}}^{2}+2\mathrm{X}\cdot \mathrm{sin}\cdot \mathrm{cos}+{\mathrm{X}}^{2}\cdot {\mathrm{cos}}^{2}+1}\hfill \\ \hfill & =\sqrt{{\mathrm{X}}^{2}+2}\hfill \end{array}$

To calculate the integral $L\left(w\right)=\underset{0}{\overset{a}{\int }}\sqrt{{\mathrm{X}}^{2}+2}$ we need the hyperbolic functions sinh und cosh  i Hyperbolic sine (sinus hyperbolicus) and hyperbolic cosine (cosinus hyperbolicus) are two functions based on the exponential function exp (see [5.9.18]). For $x\in ℝ$ we set  $\begin{array}{l}\mathrm{sinh}x≔\frac{\mathrm{exp}\left(x\right)-\mathrm{exp}\left(-x\right)}{2}\hfill \\ \mathrm{cosh}x≔\frac{\mathrm{exp}\left(x\right)+\mathrm{exp}\left(-x\right)}{2}\hfill \end{array}$[0]  sinh and cosh are arbitrary often differentiable and, as ${\mathrm{exp}}^{\prime }=\mathrm{exp}$ (see [7.5.8]), their derivatives are easily calculated to  ${\mathrm{sinh}}^{\prime }=\mathrm{cosh}$  and  ${\mathrm{cosh}}^{\prime }=\mathrm{sinh}$.  The following properties are also straight forward from the definition [0]: $\mathrm{cosh}+\mathrm{sinh}=\mathrm{exp}$ ${\mathrm{cosh}}^{2}-{\mathrm{sinh}}^{2}=1$ $\mathrm{cosh}\left(x+y\right)=\mathrm{cosh}x\cdot \mathrm{cosh}y+\mathrm{sinh}x\cdot \mathrm{sinh}y$ $\mathrm{sinh}\left(x+y\right)=\mathrm{sinh}x\cdot \mathrm{cosh}y+\mathrm{cosh}x\cdot \mathrm{sinh}y$ $\mathrm{cosh}\left(-x\right)=\mathrm{cosh}x$ $\mathrm{sinh}\left(-x\right)=-\mathrm{sinh}x$ sinh is injective (according to [7.9.6] as ${\mathrm{sinh}}^{\prime }\left(x\right)=\mathrm{cosh}x>0$ for all x) and surjective as well (a consequence of the intermediate value theorem [6.6.2], because sinh is continuous and $\underset{x\to ±\infty }{\mathrm{lim}}\mathrm{sinh}\left(x\right)=±\infty$ ). Thus sinh is bijective and its inverse function $\mathrm{arcsinh}≔{\mathrm{sinh}}^{-1}:ℝ\to ℝ$ is called arcussinus hyberbolicus. In a similar way arcuscosinus hyperbolicus is introduced, the inverse function of $\mathrm{cosh}|{ℝ}^{\ge 0}$:  $\mathrm{arccosh}≔\left(\mathrm{cosh}|{ℝ}^{\ge 0}{\right)}^{-1}:{ℝ}^{\ge 1}\to {ℝ}^{\ge 0}$
. As $\frac{1}{2}\left(\mathrm{sinh}\cdot \mathrm{cosh}+\mathrm{X}\right)$ is a primitive of ${\mathrm{cosh}}^{2}$  i We use the fundamental theorem [8.2.13] and employ integration by parts (see [8.3.1]) to compute for any x the integral $\begin{array}{lll}\hfill & \phantom{\rule{0.6em}{0ex}}\underset{0}{\overset{x}{\int }}{\mathrm{cosh}}^{2}\hfill & =\mathrm{sinh}\cdot \mathrm{cosh}\phantom{\phantom{\rule{0pt}{12pt}}}{|}_{0}^{x}-\underset{0}{\overset{x}{\int }}{\mathrm{sinh}}^{2}\hfill \\ \hfill & \hfill & =\mathrm{sinh}\left(x\right)\cdot \mathrm{cosh}\left(x\right)-\underset{0}{\overset{x}{\int }}{\mathrm{cosh}}^{2}+\underset{0}{\overset{x}{\int }}1\hfill \\ ⇒\text{ }\hfill & 2\underset{0}{\overset{x}{\int }}{\mathrm{cosh}}^{2}\hfill & =\mathrm{sinh}\left(x\right)\cdot \mathrm{cosh}\left(x\right)+x\hfill \end{array}$ $\frac{1}{2}\left(\mathrm{sinh}\cdot \mathrm{cosh}+\mathrm{X}\right)$ is thus a primitive of ${\mathrm{cosh}}^{2}$.
the integral now could be solved with the substitution formula (see [8.3.5]). We substitute $g=\sqrt{2}\mathrm{sinh}$ and thus get

$\begin{array}{ll}L\left(w\right)\hfill & =\underset{\mathrm{arcsinh}0}{\overset{\mathrm{arcsinh}a}{\sqrt{2}}}{\int }}\sqrt{2{\mathrm{sinh}}^{2}+2}\cdot \sqrt{2}\mathrm{cosh}\hfill \\ \hfill & =2\underset{0}{\overset{\mathrm{arcsinh}a}{\sqrt{2}}}{\int }}\sqrt{{\mathrm{sinh}}^{2}+1}\cdot \mathrm{cosh}\hfill \\ \hfill & =2\underset{0}{\overset{\mathrm{arcsinh}a}{\sqrt{2}}}{\int }}\sqrt{{\mathrm{cosh}}^{2}}\cdot \mathrm{cosh}\hfill \\ \hfill & =2\underset{0}{\overset{\mathrm{arcsinh}a}{\sqrt{2}}}{\int }}{\mathrm{cosh}}^{2}\hfill \\ \hfill & =\mathrm{sinh}\cdot \mathrm{cosh}+\mathrm{X}\phantom{\phantom{\rule{0pt}{12pt}}}{|}_{0}^{\mathrm{arcsinh}a}{\sqrt{2}}}\hfill \\ \hfill & =\frac{a}{\sqrt{2}}\cdot \mathrm{cosh}\left(\mathrm{arcsinh}\frac{a}{\sqrt{2}}\right)+\mathrm{arcsinh}\frac{a}{\sqrt{2}}\hfill \\ \hfill & =\frac{a}{\sqrt{2}}\cdot \sqrt{1+\frac{{a}^{2}}{2}}+\mathrm{arcsinh}\frac{a}{\sqrt{2}}=\frac{a}{2}\sqrt{2+{a}^{2}}+\mathrm{arcsinh}\frac{a}{\sqrt{2}}\hfill \end{array}$

• Surprisingly we can't find a single term that calculates the length of the ellipse given by

$w:t↦\left(a\cdot \mathrm{cos}t,b\cdot \mathrm{sin}t\right),\text{ }t\in \left[0,2\pi \right]$

The integral

is a so called elliptical integral that could only be solved in the trivial cases $c=1$ and $c=0$. In the latter however, i.e. $a=b$, the ellipse is circle with radius $r≔a=b$. Its circumference thus is

$L\left(w\right)=r\underset{0}{\overset{2\pi }{\int }}1=2\pi r$.

Exercise:

• For $w:t↦\left(\frac{3}{2}t,\sqrt{{t}^{3}}\right),\text{\hspace{0.28em}}t\in \left[0,1\right]$  we have  $\phantom{|}|{w}^{\prime }|={\text{?}}|\left(\frac{3}{2},\frac{3}{2}\sqrt{\mathrm{X}}\right)|=\sqrt{\frac{9}{4}+\frac{9}{4}\mathrm{X}}$  and thus

$\phantom{|}L\left(w\right)={\text{?}}\frac{3}{2}\underset{0}{\overset{1}{\int }}\sqrt{1+\mathrm{X}}\underset{\begin{array}{c}\mathit{\text{substituting\hspace{0.28em}}}\\ \text{\hspace{0.28em}}g=\mathrm{X}-1\end{array}}{=}\frac{3}{2}\underset{1}{\overset{2}{\int }}\sqrt{\mathrm{X}}=\sqrt{{\mathrm{X}}^{3}}\phantom{\phantom{\rule{0pt}{12pt}}}{|}_{1}^{2}=\sqrt{8}-1$

• For the parabola segment  $w:t↦\left(t,{t}^{2}\right),\text{\hspace{0.28em}}t\in \left[0,1\right]$  we have $\phantom{|}|{w}^{\prime }|={\text{?}}|\left(1,2\mathrm{X}\right)|=\sqrt{1+4{\mathrm{X}}^{2}}$. Thus

$\phantom{|}L\left(w\right)={\text{?}}\underset{0}{\overset{1}{\int }}\sqrt{1+4{X}^{2}}\underset{\begin{array}{c}\mathit{\text{substituting}}\\ g=1}{2}\mathrm{sinh}\end{array}}{=}\frac{1}{2}\underset{0}{\overset{\mathrm{arcsinh}\left(2\right)}{\int }}\underset{=\mathrm{cosh}}{\underbrace{\sqrt{1+{\mathrm{sinh}}^{2}}}}\cdot \mathrm{cosh}=\frac{1}{2}\underset{0}{\overset{\mathrm{arcsinh}\left(2\right)}{\int }}{\mathrm{cosh}}^{2}$

and as $\phantom{|}{\text{?}}{\text{we know}}\phantom{\rule{0.3em}{0ex}}\text{that}\phantom{\rule{0.3em}{0ex}}\frac{1}{2}\left(\mathrm{sinh}\cdot \mathrm{cosh}+\mathrm{X}\right)$ is a primitive of ${\mathrm{cosh}}^{2}$ we eventually get

$\phantom{|}L\left(w\right)={\text{?}}\frac{1}{4}\left(\mathrm{sinh}\cdot \sqrt{1+{\mathrm{sinh}}^{2}}+\mathrm{X}\right)\phantom{\phantom{\rule{0pt}{12pt}}}{|}_{0}^{\mathrm{arcsinh}\left(2\right)}=\frac{1}{4}\left(2\cdot \sqrt{5}+\mathrm{arcsinh}\left(2\right)\right)$

The length of smooth path $w:t↦w\left(t\right),\text{\hspace{0.28em}}t\in \left[a,b\right]$ would be easy to calculate if the derivative was a vector of constant length. For $|{w}^{\prime }|=1$ we even have

$L\left(w|\left[a,x\right]\right)=\underset{a}{\overset{x}{\int }}1=x-a$

for all $x\in \left[a,b\right]$, i.e. the section of the path is as long as the section of the respective interval. In that case we say that w is parameterized by the arc length. Regular paths are always parameterizable in this way:

Proposition:  If $w:\left[a,b\right]\to {ℝ}^{k}$ is a regular path then there is a ${\mathcal{D}}^{1}$-bijection  $\varphi :\left[0,L\left(w\right)\right]\to \left[a,b\right]$ such that

 $w\circ \varphi$  is parameterized by the arc length. $w\circ \varphi$  and w generate the same curve. $L\left(w\circ \varphi \right)=L\left(w\right)$. [8.6.17]

Proof:  Due to the fundamental theorem [8.2.13] the function  $s:\left[a,b\right]\to \left[0,L\left(w\right)\right]$ given by

$s\left(x\right)≔\underset{a}{\overset{x}{\int }}|{w}^{\prime }|$

is a prinitive of $|{w}^{\prime }|$. As w is regular we have: $s\left(x\right)>0$ for all $x\in \right]a,b\right]$. From [7.9.6] we thus know that s is injective and that $\varphi ≔{s}^{-1}$ is differentiable (see [7.5.4]) at each $x\in s\left(\left[a,b\right]\right)$ with

${\varphi }^{\prime }\left(x\right)=\frac{1}{{s}^{\prime }\left(\varphi \left(x\right)\right)}=\frac{1}{|{w}^{\prime }|\left(\varphi \left(x\right)\right)}=\frac{1}{|{w}^{\prime }|}\circ \varphi \left(x\right)\ne 0$

Further, the intermediate value theorem [6.6.2] guarantees that $s\left(\left[a,b\right]\right)=\left[0,L\left(w\right)\right]$ because $s\left(a\right)=0$ and $s\left(b\right)=L\left(w\right)$. Now due to the chain rule [7.7.8] the path $w\circ \varphi :\left[0,L\left(w\right)\right]\to {ℝ}^{k}$ is regular and

$\begin{array}{ll}\left(w\circ \varphi {\right)}^{\prime }\hfill & =\left(\left({{w}_{1}}^{\prime }\circ \varphi \right)\cdot {\varphi }^{\prime },\dots ,\left({{w}_{k}}^{\prime }\circ \varphi \right)\cdot {\varphi }^{\prime }\right)\hfill \\ \hfill & =\left(\left({{w}_{1}}^{\prime }\circ \varphi \right)\cdot \frac{1}{|{w}^{\prime }|}\circ \varphi ,\dots ,\left({{w}_{k}}^{\prime }\circ \varphi \right)\cdot \frac{1}{|{w}^{\prime }|}\circ \varphi \right)\hfill \\ \hfill & =\frac{{w}^{\prime }}{|{w}^{\prime }|}\circ \varphi \hfill \end{array}$

Finally we show:

1.   $|\left(w\circ \varphi {\right)}^{\prime }|=|\frac{{w}^{\prime }}{|{w}^{\prime }|}\circ \varphi |=\frac{|{w}^{\prime }|}{|{w}^{\prime }|}\circ \varphi =1$

2.   As $\varphi$ is bijective we see that:

$\left\{w\circ \varphi \left(t\right)|t\in \left[0,L\left(w\right)\right]\right\}=\left\{w\left(\varphi \left(t\right)\right)|t\in \left[0,L\left(w\right)\right]\right\}=\left\{w\left(t\right)|t\in \left[a,b\right]\right\}$

3.   $L\left(w\circ \varphi \right)=\underset{0}{\overset{L\left(w\right)}{\int }}1=L\left(w\right)$

 8.5. 8.7.