Arc Length Parametrization

Core Definition

A space curve $\vec{r}(t)$ is said to be parametrized by arc length (or given an intrinsic parametrization) when the governing parameter is the distance $s$ traversed along the curve rather than an arbitrary parameter such as time $t$ . Under this parametrization, denoted $\vec{r}(s)$ , the magnitude of the velocity vector is identically unity:

\|\vec{r}'(s)\| = \left\|\frac{d\vec{r}}{ds}\right\| = 1

A particle moving along

\vec{r}(s)

traverses the trajectory at a constant unit speed (

1\text{ unit of distance per unit of } s

Concept Explanation

Why Parametrize by Arc Length?

Standard vector-valued functions $\vec{r}(t)$ describe where an object is at a particular time $t$ . However, many physical and geometric properties of a curve—such as its intrinsic sharpness (curvature $\kappa$ ), total length, or normal vectors—depend only on the curve's spatial geometry, not on how fast an object traverses it.

When $\vec{r}$ is expressed as a function of arc length $s$ :

Unit Tangent Vector is Direct Derivative: $\hat{T}(s) = \frac{d\vec{r}}{ds} = \vec{r}'(s)$ since $\|\vec{r}'(s)\| = 1$ .
Curvature is Second Derivative Magnitude: $\kappa(s) = \left\|\frac{d^2\vec{r}}{ds^2}\right\| = \|\vec{r}''(s)\|$ .
Frenet-Serret Simplification: The principal normal vector simplifies to $\hat{N}(s) = \frac{\vec{r}''(s)}{\|\vec{r}''(s)\|}$ .

3-Step Algorithm to Find $\vec{r}(s)$

To reparametrize a given smooth curve $\vec{r}(t)$ in terms of arc length $s$ :

   r(t)  ---[ Step 1: Integrate Speed ]--->  s(t) = ∫ ||r'(u)|| du
                                                   |
                                                   v
   r(s)  <---[ Step 3: Substitute t(s) ]---  t(s) = s^(-1)(s) [Step 2: Invert]

Calculate the Arc Length Function: $s(t) = \int_{a}^{t} \|\vec{r}'(u)\|\,du$
Invert $s(t)$ to Solve for $t$ : Algebraically solve the relation $s = s(t)$ to express $t$ in terms of $s$ , obtaining $t = t(s)$ .
Substitute $t(s)$ into $\vec{r}(t)$ : Replace every occurrence of parameter $t$ in $\vec{r}(t)$ with $t(s)$ : $\vec{r}(s) = \vec{r}(t(s))$

Domain Constraints Under Parametrization

If the original curve $\vec{r}(t)$ is restricted to a closed interval $a \le t \le b$ , the arc length parameter $s$ is strictly constrained to the interval $0 \le s \le s(b)$ , where $s(b) = \int_a^b \|\vec{r}'(u)\|\,du$ .

Engineering Applications

Robotic CNC Milling & Additive Manufacturing: Programming motion controllers in terms of spatial distance $s$ to ensure constant laser/cutter feed rate across sharp contours, preventing uneven thermal melt pools or material gouging.
Autonomous Vehicle Path Tracking: Expressing highway lane centerlines as $\vec{r}(s)$ so navigation systems can index vehicle station points and curvature profiles directly from odometer logs.
Pipeline and Infrastructure Inspection: Tagging ultrasonic thickness measurements to linear distance $s$ from an access hatch rather than variable pig transit times.

Example Problems

Example 1: Reparametrizing a Line Segment

Given $\vec{r}(t) = (3t^2 - 7)\hat{i} + (-4t^2 - 10)\hat{j}$ for $0 \le t \le 4$ , find the arc length parametrization $\vec{r}(s)$ .

Solution

Compute Velocity and Speed: $\vec{r}'(t) = 6t\hat{i} - 8t\hat{j}$ $\|\vec{r}'(t)\| = \sqrt{(6t)^2 + (-8t)^2} = \sqrt{36t^2 + 64t^2} = \sqrt{100t^2} = 10t \quad (t \ge 0)$
Find Arc Length Function $s(t)$ : $s(t) = \int_{0}^{t} 10u\,du = [5u^2]_{0}^{t} = 5t^2$
Invert to Solve for $t(s)$ : $s = 5t^2 \implies t^2 = \frac{s}{5} \implies t(s) = \sqrt{\frac{s}{5}}$
Substitute $t(s)$ into $\vec{r}(t)$ : $\vec{r}(s) = \left(3\left(\sqrt{\frac{s}{5}}\right)^2 - 7\right)\hat{i} + \left(-4\left(\sqrt{\frac{s}{5}}\right)^2 - 10\right)\hat{j} = \left(\frac{3}{5}s - 7\right)\hat{i} + \left(-\frac{4}{5}s - 10\right)\hat{j}$ Since $t \in [0, 4]$ , $s \in [0, 5(4)^2] = [0, 80]$ .
Unit Speed Check: $\vec{r}'(s) = \frac{3}{5}\hat{i} - \frac{4}{5}\hat{j} \implies \|\vec{r}'(s)\| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2} = \sqrt{\frac{9+16}{25}} = 1$

Example 2: Reparametrizing a Circular Helix

Find the arc length parametrization of $\vec{r}(t) = \langle a\cos(t), a\sin(t), ct \rangle$ starting at $t = 0$ .