Arc Length | Sahithyan's Notes

Let $\boldsymbol{r}(t)$ be defined on $[a, b]$ , and let $P = \{t_0, t_1, \dots, t_n\}$ , $a = t_0 < t_1 < \dots < t_n = b$ , be a partition of $[a, b]$ . The polygonal length corresponding to $P$ is

s(\boldsymbol{r}, P) = \sum_{k=1}^{n} \lVert \boldsymbol{r}(t_k) - \boldsymbol{r}(t_{k-1}) \rVert

The curve is rectifiable iff these polygonal lengths have a finite least upper bound, which is the arc length of the curve.

Examples:

$\boldsymbol{r}(t) = a\cos t\,\boldsymbol{i} + a\sin t\,\boldsymbol{j}$ , $0 \le t \le 2\pi$ , is rectifiable: $\boldsymbol{r}'(t)$ is continuous, so the arc length integral applies directly, giving $L = 2\pi a$ .
$f(x) = x\sin(1/x)$ for $0 < x \le 1$ , $f(0) = 0$ , is continuous but not rectifiable: near $x = 0$ the graph oscillates infinitely often, and the amplitude does not decay fast enough to keep the polygonal lengths bounded.

If $x = x(t)$ , $y = y(t)$ , $z = z(t)$ have continuous derivatives on $[a, b]$ , the arc length from $t = a$ to $t = b$ is

L = \int_a^b \sqrt{\left( \frac{\text{d}x}{\text{d}t} \right)^2 + \left( \frac{\text{d}y}{\text{d}t} \right)^2 + \left( \frac{\text{d}z}{\text{d}t} \right)^2} \, \text{d}t

Equivalently, for $C$ the curve traced by continuously differentiable $\boldsymbol{r}(t)$ ,

L = \int_a^b \left\lVert \frac{\text{d}\boldsymbol{r}}{\text{d}t} \right\rVert \text{d}t

Change of Parameter

A change of parameter in $\boldsymbol{r}(t)$ is a substitution $t = g(\tau)$ producing $\boldsymbol{r}(g(\tau))$ , tracing the same curve as $\boldsymbol{r}(t)$ , possibly in a different direction as $\tau$ increases.

If $\boldsymbol{r}(t)$ is differentiable with respect to $t$ , and $t = g(\tau)$ is a change of parameter with $g$ differentiable with respect to $\tau$ , then $\boldsymbol{r}(g(\tau))$ is differentiable with respect to $\tau$ , and

\frac{\text{d}}{\text{d}\tau}[\boldsymbol{r}(g(\tau))] = \boldsymbol{r}'(g(\tau))g'(\tau)

Equivalently,

\frac{\text{d}\boldsymbol{r}}{\text{d}\tau} = \frac{\text{d}\boldsymbol{r}}{\text{d}t} \cdot \frac{\text{d}t}{\text{d}\tau}

Arc-Length Parametrization

Let $C$ be the curve traced by a continuously differentiable $\boldsymbol{r}(t)$ with $\boldsymbol{r}'(t) \ne \vec{0}$ , and let $\boldsymbol{r}(t_0)$ be a reference point on $C$ . The signed arc-length parameter $s$ measured from $\boldsymbol{r}(t_0)$ is

s = s(t) = \int_{t_0}^{t} \left\lVert \frac{\text{d}\boldsymbol{r}}{\text{d}u} \right\rVert \text{d}u

This is an odometer reading, integrating speed over time gives distance travelled.

Since $\text{d}s/\text{d}t = \lVert \boldsymbol{r}'(t) \rVert > 0$ , $s = s(t)$ is strictly increasing, hence invertible into $\boldsymbol{r}(s)$ . A parametrization expressed in terms of $s$ is an arc-length parametrization.

Properties

$C$ is the curve traced by a continuously differentiable $\boldsymbol{r}(t)$ with $\boldsymbol{r}'(t) \ne \vec{0}$ , and $s$ is the signed arc-length parameter defined above.

$s = s(t)$ is the arc-length parameter for $C$ iff

\frac{\text{d}s}{\text{d}t} = \left\lVert \frac{\text{d}\boldsymbol{r}}{\text{d}t} \right\rVert

This is the defining integral restated as a derivative: $s$ advances at exactly the rate $C$ is traced, not some rate tied to an arbitrary parameter $t$ .

If $\boldsymbol{r}$ is expressed as a function of $s$ instead of $t$ , then

\left\lVert \frac{\text{d}\boldsymbol{r}}{\text{d}s} \right\rVert = 1

everywhere, almost by definition: since $s$ counts distance travelled, advancing it by one unit moves $\boldsymbol{r}(s)$ exactly one unit along $C$ . This is why an arc-length parametrization is also called a unit-speed parametrization.

Conversely, if $\lVert \text{d}\boldsymbol{r}/\text{d}t \rVert = 1$ for all $t$ , then for any $t_0$ in the domain of $\boldsymbol{r}$ ,

s = t - t_0

is already an arc-length parameter with reference point $\boldsymbol{r}(t_0)$ : a unit-speed parametrization needs no reparametrization through the integral formula, only a shift so that $s = 0$ at the reference point.

Examples:

For $x^2 + y^2 = a^2$ , $a > 0$ , with counterclockwise orientation and reference point $(a, 0)$ : $\boldsymbol{r}(t) = a\cos t\,\boldsymbol{i} + a\sin t\,\boldsymbol{j}$ , with $t_0 = 0$ giving $\boldsymbol{r}(t_0) = (a, 0)$ . Since $\lVert \boldsymbol{r}'(t) \rVert = a$ is constant, $s = at$ , so the arc-length parametrization is $\boldsymbol{r}(s) = a\cos(s/a)\,\boldsymbol{i} + a\sin(s/a)\,\boldsymbol{j}$ .