mathproject >> 8.6. Calculating the Length of a Path

Definition: Let

M \subset ℝ^{k}

be a subset of

ℝ^{k}

. If

[a, b]

is a closed interval in

ℝ

we call any function

w = (w_{1}, \dots w_{k}) : [a, b] \to M

[8.6.1]

a path (in M). $w (a)$ is the initial point and $w (b)$ the final point of w. The path w is closed if its initial and final points coincide: $w (a) = w (b)$ . The range ${w (t) | t \in [a, b]}$ 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: $C^{0}$ -path).
differentiable, w is a differentiable path ( $D^{1}$ -path) and the function

w^{'} ≔ ({w_{1}}^{'}, \dots, {w_{k}}^{'}) : [a, b] \to ℝ^{k}

[8.6.2]

is called the derivative of w. w is a regular path if $w^{'} (t) \neq 0$ for all $t \in [a, b]$ .

Repeatedly differentiable paths and higher derivatives are introduced analogously.

continously differentiable, w is called a smooth path ( $C^{1}$ -path).
integrable, then w is an integrable path and for any $r, s \in [a, b]$ the vector

\int_{r}^{s} w ≔ (\int_{r}^{s} w_{1}, \dots, \int_{r}^{s} w_{k})

[8.6.3]

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

Example:

With $a, b > 0$ the ellipse

$t \mapsto (a \cdot \cos t, b \cdot \sin t), t \in [0,2 π]$

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

$t \mapsto (t^{2} - 1, t^{3} - t), t \in [- \sqrt{2}, \sqrt{2}]$

is not closed and as $1 \mapsto (0,0)$ and $- 1 \mapsto (0,0)$ the curve runs through $(0,0)$ twice.
Lissajous-curves are generated by the following $C^{\infty}$ -paths:

$t \mapsto (a \cdot \sin (n t + c), b \cdot \sin t), t \in [0,2 π]$

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

The spiral
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spiral
$t \mapsto (t \cdot \cos t, t \cdot \sin t, t), t \in [0,20]$

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and the closed curve of Viviani
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curve of Viviani
$t \mapsto (1 + \cos t, \sin t,2 \cdot \sin (\frac{t}{2})), t \in [0,4 π]$

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are examples for $C^{\infty}$ -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:

For any differential paths $v, w : [a, b] \to ℝ^{k}$ each linear combination $α v + β w : [a, b] \to ℝ^{k}$ is differentiable and

(α v + β w)^{'} = α v^{'} + β w^{'}

[8.6.4]

If $w : [a, b] \to ℝ^{k}$ is a differentiable path then for any two different points $r, s \in [a, b]$ there are numbers ${\tilde{t}}_{1}, \dots, {\tilde{t}}_{k}$ between
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i.e. ${\tilde{t}}_{i} \in] r, s [$ , if $r < s$

and ${\tilde{t}}_{i} \in] s, r [$ , if $s < r$ .

r and s such that

w (s) = w (r) + (s - r) \cdot ({w^{'}}_{1} ({\tilde{t}}_{1}), \dots, {w^{'}}_{k} ({\tilde{t}}_{k}))

[8.6.5]

If $w : [a, b] \to ℝ^{k}$ is a differentiable path the derivative $w^{'}$ is integrable and

\int_{r}^{s} w^{'} = w (s) - w (r)

[8.6.6]

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

Proof:

1. ► According to the sum and factor rule all coordinate functions $α v_{i} + β w_{i}$ are differentiable and

(α v_{i} + β w_{i})^{'} = α {v_{i}}^{'} + β {w_{i}}^{'}

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

w_{i} (s) = w_{i} (r) + (s - r) \cdot {w_{i}}^{'} ({\tilde{t}}_{i}) = w_{i} (r) + (s - r) \cdot {w^{'}}_{i} ({\tilde{t}}_{i})

3. ► Every coordinate function $w_{i}$ is certainly a primitive for ${w_{i}}^{'} = {w^{'}}_{i}$ .

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

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

$\int_{r}^{r} w = 0, \int_{r}^{s} w = - \int_{s}^{r} w, \int_{r}^{s} w = \int_{r}^{t} w + \int_{t}^{s} w$

[8.6.7]

$\int_{r}^{s} α v + β w = α \int_{r}^{s} v + β \int_{r}^{s} w$

[8.6.8]

If r and s are different there are numbers ${\tilde{t}}_{1}, \dots, {\tilde{t}}_{k}$ between r and s such that

\int_{r}^{s} w = (s - r) \cdot (w_{1} ({\tilde{t}}_{1}), \dots, w_{k} ({\tilde{t}}_{k}))

[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
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The dot product

$x \cdot y ≔ \sum_{i = 1}^{k} x_{i} \cdot y_{i}, x, y \in ℝ^{k}$

in particular provides the length $| x | = \sqrt{x \cdot x} = \sqrt{\sum_{i = 1}^{k} x_{i}^{2}}$ of a vector $x \in ℝ^{k}$ . Our proof uses the simple identity

$| x |^{2} = x \cdot x$ ,

and the Cauchy-Schwarz inequality

$| x \cdot y | \leq | x | \cdot | y |$

Its proof and further properties of the dot product, as e.g. the triangle inequality

$| x + y | \leq | x | + | y |$

are to be found in part 9.13.

in $ℝ^{k}$ .

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

| \int_{a}^{b} w | \leq \int_{a}^{b} | w |

[8.6.10]

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

c ≔ \int_{a}^{b} w

, thus

c_{i} = \int_{a}^{b} 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 $| c | \neq 0$ . According to the Cauchy-Schwarz inequality and due to the integral monotony [8.2.10] we may estimate as follows:

\begin{array}{l} | c | \cdot | \int_{a}^{b} w | = | c |^{2} = c \cdot c = \sum_{i = 1}^{k} c_{i} \cdot \int_{a}^{b} w_{i} & = \int_{a}^{b} \sum_{i = 1}^{k} c_{i} \cdot w_{i} \\ = \int_{a}^{b} c \cdot w \leq \int_{a}^{b} | c \cdot w | \leq \int_{a}^{b} | c | \cdot | w | = | c | \int_{a}^{b} | w | \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 $Z = (t_{0}, \dots, t_{n})$ in $[a, b]$ , $n \geq 1$ , such that

a = t_{0} < t_{1} < \dots < t_{n - 1} < t_{n} = b

is called a partition of the interval $[a, b]$ with the number $\max {t_{1} - t_{0}, \dots, t_{n} - t_{n - 1}}$ being its fineness.

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

p_{Z} (t) = w (t_{i - 1}) + \frac{t - t_{i - 1}}{t_{i} - t_{i - 1}} (w (t_{i}) - w (t_{i - 1}))

, if

t \in [t_{i - 1}, t_{i}]

[8.6.11]

Each traverse $p_{Z}$ is a path with initial point $w (a) = w (t_{0})$ and final point $w (b) = w (t_{n})$ . Traverses approximate a continuous path w in the following way:

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

| w (t) - p_{Z} (t) | < ε

for all

t \in [a, b]

[8.6.12]

for any partition Z with a fineness less than δ.

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

s, t \in [a, b] \land | s - t | < δ_{j} \Rightarrow | w_{j} (s) - w_{j} (t) |^{2} < \frac{ε^{2}}{4 k}

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

| w (s) - w (t) | = \sqrt{\sum_{j = 1}^{k} | w_{j} (s) - w_{j} (t) |^{2}} < \sqrt{k \frac{ε^{2}}{4 k}} = \frac{ε}{2}

[1]

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

| t - t_{i - 1} | < δ

and

| t_{i} - t_{i - 1} | < δ

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

\begin{array}{l} | w (t) - p_{Z} (t) | & \leq | w (t) - w (t_{i - 1}) | + | w (t_{i - 1}) - p_{Z} (t) | \\ = | w (t) - w (t_{i - 1}) | + \underset{\leq 1}{\underset{︸}{| \frac{t - t_{i - 1}}{t_{i} - t_{i - 1}} |}} \cdot | w (t_{i}) - w (t_{i - 1}) | \\ < \frac{ε}{2} + \frac{ε}{2} = ε \end{array}

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

L (p_{Z}) = \sum_{i = 1}^{n} | w (t_{i}) - w (t_{i - 1}) |

[8.6.13]

As the line segment represents the shortest distance between $w (t_{i - 1})$ and $w (t_{i})$ we expect the length of w to be an upper bound for all $L (p_{Z})$ . Considering [8.6.12], a sound candidate for the prospective path length would be the supremum
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We remember the completeness axiom :
In $ℝ$ each non-empty bounded subset $M \subset ℝ$ has a least upper bound, its supremum (termed $\sup M$ ).
An upper bound s of M equals $\sup M$ if and only if $s - ε$ is no upper bound of M for each $ε > 0$ .

of ${L (p_{Z}) | Z partition of [a, b]}$ .

The depicted example on the right shows a part of one of René Descartes' Folium of Descartes, that is the path $t \mapsto (\frac{12 t}{1 + t^{3}}, \frac{12 t^{2}}{1 + t^{3}}), t \in [- 0.4,10]$ . For a three dimensional example we replenish the spiral
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spiral
$t \mapsto (t \cdot \cos t, t \cdot \sin t, t), t \in [0,20]$

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from a former example.

Definition: A continuous path $w : [a, b] \to ℝ^{k}$ is a path of finite length (or a rectifiable path) if the set ${L (p_{Z}) | Z partition of [a, b]}$ is bounded. In this case the number

L (w) ≔ \sup {L (p_{Z}) | Z partition of [a, b]}

[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 $w^{'}$ will provide an upper bound for all $L (p_{Z})$ . Smoothness however is only a sufficient and not a necessary condition for a finite length, as the non-smooth path $w : t \mapsto (t, | t |), t \in [- 1, - 1]$
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If $Z = (t_{0}, \dots, t_{n})$ is a partition of $[- 1,1]$ we find a suitable k such that $t_{k} \leq 0 < t_{k + 1}$ . That means:

${(| t_{i} | - | t_{i - 1} |)}^{2} = {\begin{array}{l} {(t_{i} - t_{i - 1})}^{2}, if i > k + 1 \\ {(- t_{i} + t_{i - 1})}^{2}, if i \leq k \end{array}$

and so we have ${(| t_{i} | - | t_{i - 1} |)}^{2} = {(t_{i} - t_{i - 1})}^{2}$ and consequently

$| w (t_{i}) - w (t_{i - 1}) | = \sqrt{{(t_{i} - t_{i - 1})}^{2} + {(| t_{i} | - | t_{i - 1} |)}^{2}} = \sqrt{2} (t_{i} - t_{i - 1})$

for all $i \neq k + 1$ . Taking into account that

$\begin{array}{l} | w (t_{k + 1}) - w (0) | = \sqrt{t_{k + 1}^{2} + | t_{k + 1} |^{2}} = \sqrt{2} | t_{k + 1} | = \sqrt{2} t_{k + 1} \\ | w (0) - w (t_{k}) | = \sqrt{t_{k}^{2} + | t_{k} |^{2}} = \sqrt{2} | t_{k} | = - \sqrt{2} t_{k} \end{array}$

we now may estimate $L (p_{Z})$ as follows:

$\begin{array}{l} L (p_{Z}) & = \sum_{i = 1}^{n} | w (t_{i}) - w (t_{i - 1}) | \\ \leq \sum_{i = 1}^{k} | w (t_{i}) - w (t_{i - 1}) | \\ + | w (t_{k + 1}) - w (0) | + | w (0) - w (t_{k}) | \\ + \sum_{i = k + 2}^{n} | w (t_{i}) - w (t_{i - 1}) | \\ = \sqrt{2} (\sum_{i = 1}^{k} t_{i} - t_{i - 1} + t_{k + 1} - t_{k} + \sum_{i = k + 2}^{n} t_{i} - t_{i - 1}) \\ = \sqrt{2} (t_{n} - t_{0}) = 2 \sqrt{2} \end{array}$

proves.

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

L (p_{Z}) \leq \int_{a}^{b} | w^{'} |

[8.6.15]

holds and w is thus rectifiable.

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

L (p_{Z}) = \sum_{i = 1}^{n} | w (t_{i}) - w (t_{i - 1}) | = \sum_{i = 1}^{n} | \int_{t_{i - 1}}^{t_{i}} w^{'} | \leq \sum_{i = 1}^{n} \int_{t_{i - 1}}^{t_{i}} | w^{'} | = \int_{a}^{b} | w^{'} |

Surprisingly, [8.6.15] could be replaced by a stronger result: The integral of $| w^{'} |$ is more than a common upper bound of ${L (p_{Z}) | Z partition of [a, b]}$ , it is the least one and thus the length of w.

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

L (w) = \int_{a}^{b} | w^{'} |

[8.6.16]

Proof: Due to [8.6.15] the integral $\int_{a}^{b} | w^{'} |$ is an upper bound of ${L (p_{Z}) | Z partition of [a, b]}$ . It is thus sufficient to find a partition Z for each $ε > 0$ such that

\int_{a}^{b} | w^{'} | - ε \leq L (p_{Z})

As the coordinate functions ${w_{j}}^{'}$ are continuously differentiable, each ${({w_{j}}^{'})}^{2}$ is continuous and on the closed interval $[a, b]$ even uniformly continuous. Thus there is a $δ_{j} > 0$ for $\bar{ε} ≔ \frac{ε^{2}}{k {(b - a)}^{2}} > 0$ such that

\begin{array}{l} | t - s | < δ_{j} & \Rightarrow {({w_{j}}^{'} (t))}^{2} - {({w_{j}}^{'} (s))}^{2} \leq | {({w_{j}}^{'} (t))}^{2} - {({w_{j}}^{'} (s))}^{2} | < \bar{ε} \\ \Rightarrow {({w_{j}}^{'} (t))}^{2} < {({w_{j}}^{'} (s))}^{2} + \bar{ε} \end{array}

[2]

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

${\tilde{y}}_{i} \in] t_{i - 1}, t_{i} [$ such that $\int_{a}^{b} | w^{'} | = \sum_{i = 1}^{n} \int_{t_{i - 1}}^{t_{i}} | w^{'} | = \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \cdot | w^{'} ({\tilde{y}}_{i}) |$ [3]
${\tilde{x}}_{1, i}, \dots, {\tilde{x}}_{k, i} \in] t_{i - 1}, t_{i} [$ such that

$\begin{array}{l} L (p_{Z}) = \sum_{i = 1}^{n} | w (t_{i}) - w (t_{i - 1}) | & = \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \cdot | {w_{1}}^{'} ({\tilde{x}}_{1, i}), \dots, {w_{k}}^{'} ({\tilde{x}}_{k, i}) | \\ = \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \sqrt{\sum_{j = 1}^{k} {({w_{j}}^{'} ({\tilde{x}}_{j, i}))}^{2}} \end{array}$ [4]

As ${\tilde{y}}_{i}, {\tilde{x}}_{1, i}, \dots, {\tilde{x}}_{k, i} \in] t_{i - 1}, t_{i} [$ we have for all i:

| {\tilde{y}}_{i} - {\tilde{x}}_{j, i} | < δ \leq δ_{j}

With [2] we thus know that ${({w_{j}}^{'} ({\tilde{y}}_{i}))}^{2} < {({w_{j}}^{'} ({\tilde{x}}_{j, i}))}^{2} + \bar{ε}$ and according to [3] and [4] we may estimate the integral of $| w^{'} |$ as follows (note that $\sqrt{x + y} \leq \sqrt{x} + \sqrt{y}$ for any positive x, y):

\begin{array}{l} \int_{a}^{b} | w^{'} | = \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \sqrt{\sum_{j = 1}^{k} {({w_{j}}^{'} ({\tilde{y}}_{i}))}^{2}} & < \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \sqrt{\sum_{j = 1}^{k} ({({w_{j}}^{'} ({\tilde{x}}_{j, i}))}^{2} + \bar{ε})} \\ \leq \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \sqrt{\sum_{j = 1}^{k} {({w_{j}}^{'} ({\tilde{x}}_{j, i}))}^{2}} + \sum_{i = 1}^{n} (t_{i} - t_{i - 1}) \sqrt{k \bar{ε}} \\ = L (p_{Z}) + (b - a) \sqrt{k \bar{ε}} \\ = L (p_{Z}) + ε \end{array}

This proves the required estimate $\int_{a}^{b} | w^{'} | - ε \leq L (p_{Z})$ . Therefor $\int_{a}^{b} | w^{'} |$ is the least upper bound of ${L (p_{Z}) | Z partition of [a, b]}$ , 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 $| w^{'} |$ .

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

$w : t \mapsto a + t (b - a) = (a_{1} + t (b_{1} - a_{1}), \dots, a_{k} + t (b_{k} - a_{k})), t \in [0,1]$

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

$\begin{array}{l} L (w) & = \int_{0}^{1} | b - a | \\ = | b - a | \int_{0}^{1} 1 \\ = | b - a | \end{array}$

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

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

w : t \mapsto (t \cdot \cos t, t \cdot \sin t, t), t \in [0, a]

thus measuring the spiral from our initial example:

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

\begin{array}{l} | w^{'} | & = \sqrt{{(\cos - X \cdot \sin)}^{2} + {(\sin + X \cdot \cos)}^{2} + 1^{2}} \\ = \sqrt{\cos^{2} - 2 X \cdot \cos \cdot \sin + X^{2} \cdot \sin^{2} + \sin^{2} + 2 X \cdot \sin \cdot \cos + X^{2} \cdot \cos^{2} + 1} \\ = \sqrt{X^{2} + 2} \end{array}

To calculate the integral $L (w) = \int_{0}^{a} \sqrt{X^{2} + 2}$ we need the hyperbolic functions sinh und cosh

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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

$\begin{array}{l} \sinh x ≔ \frac{\exp (x) - \exp (- x)}{2} \\ \cosh x ≔ \frac{\exp (x) + \exp (- x)}{2} \end{array}$ [0]

sinh and cosh are arbitrary often differentiable and, as $\exp^{'} = \exp$ (see [7.5.8]), their derivatives are easily calculated to

$\sinh^{'} = \cosh$ and $\cosh^{'} = \sinh$ .

The following properties are also straight forward from the definition [0]:

$\cosh + \sinh = \exp$

$\cosh^{2} - \sinh^{2} = 1$

$\cosh (x + y) = \cosh x \cdot \cosh y + \sinh x \cdot \sinh y$

$\sinh (x + y) = \sinh x \cdot \cosh y + \cosh x \cdot \sinh y$

$\cosh (- x) = \cosh x$

$\sinh (- x) = - \sinh x$

sinh is injective (according to [7.9.6] as $\sinh^{'} (x) = \cosh x > 0$ for all x) and surjective as well (a consequence of the intermediate value theorem [6.6.2], because sinh is continuous and $\lim_{x \to \pm \infty} \sinh (x) = \pm \infty$ ). Thus sinh is bijective and its inverse function
$arcsinh ≔ \sinh^{- 1} : ℝ \to ℝ$

is called arcussinus hyberbolicus. In a similar way arcuscosinus hyperbolicus is introduced, the inverse function of $\cosh | ℝ^{\geq 0}$ :

$arccosh ≔ (\cosh | ℝ^{\geq 0})^{- 1} : ℝ^{\geq 1} \to ℝ^{\geq 0}$

. As $\frac{1}{2} (\sinh \cdot \cosh + X)$ is a primitive of $\cosh^{2}$
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We use the fundamental theorem [8.2.13] and employ integration by parts (see [8.3.1]) to compute for any x the integral

$\begin{array}{l} \int_{0}^{x} \cosh^{2} & = \sinh \cdot \cosh |_{0}^{x} - \int_{0}^{x} \sinh^{2} \\ = \sinh (x) \cdot \cosh (x) - \int_{0}^{x} \cosh^{2} + \int_{0}^{x} 1 \\ \Rightarrow & 2 \int_{0}^{x} \cosh^{2} & = \sinh (x) \cdot \cosh (x) + x \end{array}$

$\frac{1}{2} (\sinh \cdot \cosh + X)$ is thus a primitive of $\cosh^{2}$ .

the integral now could be solved with the substitution formula (see [8.3.5]). We substitute $g = \sqrt{2} \sinh$ and thus get

\begin{array}{l} L (w) & = \int_{arcsinh 0}^{arcsinh \frac{a}{\sqrt{2}}} \sqrt{2 \sinh^{2} + 2} \cdot \sqrt{2} \cosh \\ = 2 \int_{0}^{arcsinh \frac{a}{\sqrt{2}}} \sqrt{\sinh^{2} + 1} \cdot \cosh \\ = 2 \int_{0}^{arcsinh \frac{a}{\sqrt{2}}} \sqrt{\cosh^{2}} \cdot \cosh \\ = 2 \int_{0}^{arcsinh \frac{a}{\sqrt{2}}} \cosh^{2} \\ = \sinh \cdot \cosh + X |_{0}^{arcsinh \frac{a}{\sqrt{2}}} \\ = \frac{a}{\sqrt{2}} \cdot \cosh (arcsinh \frac{a}{\sqrt{2}}) + arcsinh \frac{a}{\sqrt{2}} \\ = \frac{a}{\sqrt{2}} \cdot \sqrt{1 + \frac{a^{2}}{2}} + arcsinh \frac{a}{\sqrt{2}} = \frac{a}{2} \sqrt{2 + a^{2}} + arcsinh \frac{a}{\sqrt{2}} \end{array}

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

$w : t \mapsto (a \cdot \cos t, b \cdot \sin t), t \in [0,2 π]$

The integral

$\begin{array}{l} L (w) = \int_{0}^{2 π} | w^{'} | & = \int_{0}^{2 π} | (- a \cdot \sin, b \cdot \cos) | \\ = \int_{0}^{2 π} \sqrt{a^{2} \cdot \sin^{2} + b^{2} \cdot \cos^{2}} \\ = a \int_{0}^{2 π} \sqrt{1 - \cos^{2} + \frac{b^{2}}{a^{2}} \cdot \cos^{2}} \\ = a \int_{0}^{2 π} \sqrt{1 - c \cdot \cos^{2}}, wobei c ≔ \frac{a^{2} - b^{2}}{a^{2}} \end{array}$

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

$L (w) = r \int_{0}^{2 π} 1 = 2 π r$ .

Exercise:

For $w : t \mapsto (\frac{3}{2} t, \sqrt{t^{3}}), t \in [0,1]$ we have $| w^{'} | = ? | (\frac{3}{2}, \frac{3}{2} \sqrt{X}) | = \sqrt{\frac{9}{4} + \frac{9}{4} X}$ and thus

$L (w) = ? \frac{3}{2} \int_{0}^{1} \sqrt{1 + X} \underset{\begin{array}{c} substituting \\ g = X - 1 \end{array}}{=} \frac{3}{2} \int_{1}^{2} \sqrt{X} = \sqrt{X^{3}} |_{1}^{2} = \sqrt{8} - 1$
For the parabola segment $w : t \mapsto (t, t^{2}), t \in [0, 1]$ we have $| w^{'} | = ? | (1, 2 X) | = \sqrt{1 + 4 X^{2}}$ . Thus

$L (w) = ? \int_{0}^{1} \sqrt{1 + 4 X^{2}} \underset{\begin{array}{c} substituting \\ g = \frac{1}{2} \sinh \end{array}}{=} \frac{1}{2} \int_{0}^{arcsinh (2)} \underset{= \cosh}{\underset{︸}{\sqrt{1 + \sinh^{2}}}} \cdot \cosh = \frac{1}{2} \int_{0}^{arcsinh (2)} \cosh^{2}$

and as $? we know that \frac{1}{2} (\sinh \cdot \cosh + X)$ is a primitive of $\cosh^{2}$ we eventually get

$L (w) = ? \frac{1}{4} (\sinh \cdot \sqrt{1 + \sinh^{2}} + X) |_{0}^{arcsinh (2)} = \frac{1}{4} (2 \cdot \sqrt{5} + arcsinh (2))$

The length of smooth path $w : t \mapsto w (t), t \in [a, b]$ would be easy to calculate if the derivative was a vector of constant length. For $| w^{'} | = 1$ we even have

L (w | [a, x]) = \int_{a}^{x} 1 = x - a

for all $x \in [a, b]$ , 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 $w : [a, b] \to ℝ^{k}$ is a regular path then there is a $D^{1}$ -bijection $ϕ : [0, L (w)] \to [a, b]$ such that

$w \circ ϕ$ is parameterized by the arc length.
$w \circ ϕ$ and w generate the same curve.
$L (w \circ ϕ) = L (w)$ .

[8.6.17]

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

s (x) ≔ \int_{a}^{x} | w^{'} |

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

ϕ^{'} (x) = \frac{1}{s^{'} (ϕ (x))} = \frac{1}{| w^{'} | (ϕ (x))} = \frac{1}{| w^{'} |} \circ ϕ (x) \neq 0

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

\begin{array}{l} (w \circ ϕ)^{'} & = (({w_{1}}^{'} \circ ϕ) \cdot ϕ^{'}, \dots, ({w_{k}}^{'} \circ ϕ) \cdot ϕ^{'}) \\ = (({w_{1}}^{'} \circ ϕ) \cdot \frac{1}{| w^{'} |} \circ ϕ, \dots, ({w_{k}}^{'} \circ ϕ) \cdot \frac{1}{| w^{'} |} \circ ϕ) \\ = \frac{w^{'}}{| w^{'} |} \circ ϕ \end{array}

Finally we show:

1. ► $| (w \circ ϕ)^{'} | = | \frac{w^{'}}{| w^{'} |} \circ ϕ | = \frac{| w^{'} |}{| w^{'} |} \circ ϕ = 1$

2. ► As $ϕ$ is bijective we see that:

{w \circ ϕ (t) | t \in [0, L (w)]} = {w (ϕ (t)) | t \in [0, L (w)]} = {w (t) | t \in [a, b]}

3. ► $L (w \circ ϕ) = \int_{0}^{L (w)} 1 = L (w)$

8.5. 8.7.