Arc Length and Reparametrization, multivariable and vector calculus lesson

A cyclist’s odometer counts meters of road, not minutes of riding. Ride the same path slowly or quickly, forward or backward, and the odometer adds up the same distance. This lesson turns that into mathematics. The length of a curve comes from integrating speed, it survives any regular change of parameter, and it gives every regular curve a natural parameter of its own.

Key terms

Speed

Definition Let $\mathbf{r}(t)$ be a vector function on an interval $I$ that is differentiable at $t$ . The velocity at $t$ is $\mathbf{r}'(t)$ and the speed at $t$ is its length

\lvert \mathbf{r}'(t) \rvert = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2},

where $\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle$ . Speed is a nonnegative scalar.

In plain words Velocity is an arrow that says where the point is heading. Speed is just the length of that arrow, how many units of distance per unit of parameter.

Example The helix $\mathbf{r}(t) = \langle 3\cos t, 3\sin t, 4t \rangle$ has velocity $\mathbf{r}'(t) = \langle -3\sin t, 3\cos t, 4 \rangle$ and speed

\begin{aligned} \lvert \mathbf{r}'(t) \rvert &= \sqrt{9\sin^2 t + 9\cos^2 t + 16} \\[0.4em] &= \sqrt{25} = 5 \end{aligned}

at every $t$ . The direction keeps changing, but the speed never does.

Regularity

Definition A parametrization $\mathbf{r}$ on an interval $I$ is of class $C^1$ if $\mathbf{r}'$ exists and is continuous on $I$ (one-sided at endpoints that belong to $I$ ). It is regular, or smooth, if, in addition,

\mathbf{r}'(t) \ne \mathbf{0} \quad \text{for every } t \in I.

It is piecewise regular if $I$ splits into finitely many subintervals on each of which it is regular.

In plain words A regular curve is traced with a velocity that varies continuously and never stops. That is what lets it have a well defined direction of travel at every point.

Example The curve $\mathbf{r}(t) = \langle t^2, t^3 \rangle$ is $C^1$ , but $\mathbf{r}'(t) = \langle 2t, 3t^2 \rangle$ is the zero vector at $t = 0$ , so it is not regular there. The picture agrees. At the origin the curve has a cusp, a sharp point where the direction of travel flips. The helix above is regular, since its speed is always $5$ .

Length of a curve

Definition Let $\mathbf{r}$ be continuous on $[a, b]$ . For a partition $a = t_0 < t_1 < \dots < t_n = b$ , the inscribed polygon joins the points $\mathbf{r}(t_0), \mathbf{r}(t_1), \dots, \mathbf{r}(t_n)$ in order, and its length is

\sum_{i=1}^{n} \lvert \mathbf{r}(t_i) - \mathbf{r}(t_{i-1}) \rvert.

The length $L$ of the parametrized curve is the supremum of these polygon lengths over all partitions. If $\mathbf{r}$ is $C^1$ on $[a, b]$ , this supremum is finite and

L = \int_a^b \lvert \mathbf{r}'(t) \rvert \, dt.

For a piecewise $C^1$ parametrization, add the integrals over the pieces.

In plain words Lay a chain of straight segments along the curve and refine it. Each tiny segment is about speed times the parameter step, $\lvert \mathbf{r}'(t_i) \rvert \Delta t$ , so the sum becomes an integral of speed.

Example For the quarter circle $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ on $[0, \pi/2]$ , split the interval into $n$ equal pieces. Each chord has length $2\sin\left(\frac{\pi}{4n}\right)$ , so the polygon has length $2n\sin\left(\frac{\pi}{4n}\right)$ .

$n$	Polygon length
$1$	$1.41421$
$2$	$1.53073$
$4$	$1.56072$
$8$	$1.56827$
$100$	$1.57078$

The integral gives the limit exactly. The speed is $1$ , so $L = \int_0^{\pi/2} 1 \, dt = \frac{\pi}{2} \approx 1.57080$ . One full turn of the helix $\langle 3\cos t, 3\sin t, 4t \rangle$ on $[0, 2\pi]$ has length $5 \cdot 2\pi = 10\pi$ .

Arc length function

Definition Let $\mathbf{r}$ be $C^1$ on an interval $I$ and fix $a \in I$ . The arc length function measured from $a$ is

s(t) = \int_a^t \lvert \mathbf{r}'(u) \rvert \, du, \qquad t \in I.

For $t > a$ it is the length of the curve from $\mathbf{r}(a)$ to $\mathbf{r}(t)$ , and for $t < a$ it is minus that length. Because $\lvert \mathbf{r}' \rvert$ is continuous, the fundamental theorem of calculus gives

\frac{ds}{dt} = \lvert \mathbf{r}'(t) \rvert.

In plain words The function $s$ is an odometer that reads zero at $t = a$ . Its rate of change is the speed, exactly as you would expect.

Example Take the conical spiral $\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, e^t \rangle$ for all real $t$ . Then

\begin{aligned} \mathbf{r}'(t) &= e^t \langle \cos t - \sin t, \\ &\qquad\ \sin t + \cos t, \ 1 \rangle, \end{aligned}

and since $(\cos t - \sin t)^2 + (\sin t + \cos t)^2 = 2$ , the speed is $\sqrt{3}\, e^t$ . Measured from $a = 0$ ,

s(t) = \int_0^t \sqrt{3}\, e^u \, du = \sqrt{3}\,(e^t - 1).

Arc length parameter

Definition Let $\mathbf{r}$ be regular on an interval $I$ and let $s$ be its arc length function from some $a \in I$ . Since $ds/dt = \lvert \mathbf{r}'(t) \rvert > 0$ , the function $s$ is strictly increasing and $C^1$ from $I$ onto an interval $J$ , and its inverse $t = t(s)$ is $C^1$ on $J$ with

\frac{dt}{ds} = \frac{1}{\lvert \mathbf{r}'(t(s)) \rvert}.

The curve parametrized by arc length is $\mathbf{R}(s) = \mathbf{r}(t(s))$ for $s \in J$ , and by the chain rule $\lvert \mathbf{R}'(s) \rvert = 1$ for every $s$ .

In plain words Use distance traveled as the clock. Then one unit of parameter is one unit of length, and the curve is traced at speed exactly $1$ .

Example For the helix $\langle 3\cos t, 3\sin t, 4t \rangle$ from $a = 0$ , $s = 5t$ , so $t = s/5$ and

\begin{gathered} \mathbf{R}(s) = \left\langle 3\cos\frac{s}{5}, \ 3\sin\frac{s}{5}, \ \frac{4s}{5} \right\rangle, \\[0.4em] s \in \mathbb{R}. \end{gathered}

For the conical spiral, $s = \sqrt{3}\,(e^t - 1)$ takes every value in $J = (-\sqrt{3}, \infty)$ , not every real number, because $e^t > 0$ . Solving gives $\theta(s) = \ln\left(1 + s/\sqrt{3}\right)$ as the value of $t$ , which is defined exactly on $J$ . Since $e^{\theta} = 1 + s/\sqrt{3}$ , the unit speed parametrization is

\begin{gathered} \mathbf{R}(s) = \bigl(1 + s/\sqrt{3}\bigr) \langle \cos\theta, \sin\theta, 1 \rangle, \\[0.4em] -\sqrt{3} < s < \infty. \end{gathered}

Reparametrization

Definition Let $\mathbf{r}$ be defined on an interval $I$ and let $\varphi$ be a $C^1$ function from an interval $K$ onto $I$ with $\varphi'(u) \ne 0$ for every $u \in K$ . Then $\varphi$ is one-to-one with a $C^1$ inverse, and

\boldsymbol{\rho}(u) = \mathbf{r}(\varphi(u)), \qquad u \in K,

is a reparametrization of $\mathbf{r}$ . If $\mathbf{r}$ is $C^1$ , the chain rule gives $\boldsymbol{\rho}'(u) = \varphi'(u)\, \mathbf{r}'(\varphi(u))$ , so a regular curve stays regular.

In plain words A reparametrization traces the same points in the same or reverse order, only with a different clock. The velocity is rescaled by the factor $\varphi'(u)$ , but never squeezed to zero.

Example Take the half circle $\mathbf{r}(t) = \langle \cos t, \sin t \rangle$ on $[0, \pi]$ . With $\varphi(u) = 2u$ on $[0, \pi/2]$ , the curve $\langle \cos 2u, \sin 2u \rangle$ covers the same half circle at speed $2$ . With $\varphi(u) = \pi - u$ on $[0, \pi]$ , the curve $\langle -\cos u, \sin u \rangle$ starts at $(-1, 0)$ and runs back to $(1, 0)$ .

Orientation

Definition Since $\varphi'$ is continuous and never zero on the interval $K$ , the intermediate value theorem forces it to keep one sign. The reparametrization is orientation preserving if $\varphi' > 0$ on $K$ and orientation reversing if $\varphi' < 0$ on $K$ . An orientation of a regular curve is a choice of direction of travel, and it fixes the unit tangent

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\lvert \mathbf{r}'(t) \rvert}.

Reversing the orientation replaces $\mathbf{T}$ by $-\mathbf{T}$ at every point.

In plain words Orientation is which way you walk along the curve. Changing speed does not change it, turning around does.

Example The segment $\mathbf{r}(t) = \langle 1 + 3t, \ 4t, \ 2 \rangle$ for $t \in [0, 1]$ runs from $(1, 0, 2)$ to $(4, 4, 2)$ with unit tangent $\langle 3/5, 4/5, 0 \rangle$ . The reparametrization $\boldsymbol{\rho}(u) = \mathbf{r}(1 - u)$ uses $\varphi(u) = 1 - u$ with $\varphi' = -1 < 0$ . It runs from $(4, 4, 2)$ to $(1, 0, 2)$ , and its unit tangent is $\langle -3/5, -4/5, 0 \rangle$ .

Invariance of geometric length

Definition Let $\mathbf{r}$ be $C^1$ on $[a, b]$ and let $\varphi$ be a $C^1$ function from $[c, d]$ onto $[a, b]$ with $\varphi' \ne 0$ on $[c, d]$ . Then $\boldsymbol{\rho} = \mathbf{r} \circ \varphi$ has the same length,

\int_c^d \lvert \boldsymbol{\rho}'(u) \rvert \, du = \int_a^b \lvert \mathbf{r}'(t) \rvert \, dt.

Proof. By the chain rule, $\lvert \boldsymbol{\rho}'(u) \rvert = \lvert \varphi'(u) \rvert \, \lvert \mathbf{r}'(\varphi(u)) \rvert$ . If $\varphi' > 0$ , then $\varphi(c) = a$ and $\varphi(d) = b$ , and the substitution $t = \varphi(u)$ turns the left side into the right side. If $\varphi' < 0$ , then $\lvert \varphi' \rvert = -\varphi'$ , $\varphi(c) = b$ and $\varphi(d) = a$ , so the same substitution gives

-\int_b^a \lvert \mathbf{r}'(t) \rvert \, dt = \int_a^b \lvert \mathbf{r}'(t) \rvert \, dt.

This length counts multiplicity. It equals the length of the traced set of points when $\mathbf{r}$ is one-to-one on $[a, b)$ , and a curve traced twice has twice the length.

In plain words Speed up, slow down or walk backward, the distance covered is the same. What does change the answer is going around the same path more than once.

Example The circle $\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$ on $[0, 2\pi]$ has speed $2$ and length $4\pi$ . Reparametrize with $\varphi(u) = \pi(u^2 + u)$ on $[0, 1]$ , which has $\varphi'(u) = \pi(2u + 1) > 0$ and maps $[0, 1]$ onto $[0, 2\pi]$ . The new speed is $2\pi(2u + 1)$ , and

\int_0^1 2\pi(2u + 1) \, du = 2\pi \cdot 2 = 4\pi,

the same length. Running the original formula on $[0, 4\pi]$ instead gives $8\pi$ , because that parametrization goes around the circle twice.

Check yourself

Find the length of $\mathbf{r}(t) = \langle 2t, \ t^2, \ \ln t \rangle$ for $1 \le t \le e$ .

Show the answer

The velocity is $\langle 2, 2t, 1/t \rangle$ , and $4 + 4t^2 + 1/t^2 = (2t + 1/t)^2$ , so the speed is $2t + 1/t$ for $t > 0$ . The length is

\begin{aligned} \int_1^e \left(2t + \frac{1}{t}\right) dt &= \Big[t^2 + \ln t\Big]_1^e \\[0.4em] &= e^2. \end{aligned}

Reparametrize the helix $\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle$ by arc length measured from $t = 0$ , keeping its orientation. Then give the arc length parametrization with the opposite orientation.

Show the answer

The speed is $\sqrt{2}$ , so $s = \sqrt{2}\, t$ and $t = s/\sqrt{2}$ . This gives

\mathbf{R}(s) = \left\langle \cos\frac{s}{\sqrt{2}}, \ \sin\frac{s}{\sqrt{2}}, \ \frac{s}{\sqrt{2}} \right\rangle.

For the opposite orientation use $t = -s/\sqrt{2}$ , which gives $\langle \cos(s/\sqrt{2}), -\sin(s/\sqrt{2}), -s/\sqrt{2} \rangle$ , again with speed $1$ .

The curve $\mathbf{r}(t) = \langle \cos 2t, \sin 2t \rangle$ on $[0, 2\pi]$ has length $4\pi$ by the integral formula, yet it traces the unit circle, whose length is $2\pi$ . Why does this not contradict the invariance of length?

Show the answer

The parametrization goes around the circle twice, so it is not one-to-one on $[0, 2\pi)$ and the integral counts every point twice. It is also not a reparametrization of a single turn $\langle \cos t, \sin t \rangle$ on $[0, 2\pi]$ , since $\varphi(t) = 2t$ maps $[0, 2\pi]$ onto $[0, 4\pi]$ , not onto $[0, 2\pi]$ .

Arc Length and Reparametrization

Key terms

Speed

Regularity

Length of a curve

Arc length function

Arc length parameter

Reparametrization

Orientation

Invariance of geometric length

Check yourself

How this lesson connects