Deriving the Arc-Length Integral from Chord Sums

From Chords to Curve Length

A smooth curve is not made of straight segments, yet its length can be approached by joining nearby points on it with chords. As the points get closer together, those straight segments follow the curve more closely. The key question is how to turn that geometric idea into a precise formula.

For a parametrized curve, the velocity vector describes how position changes with the parameter, and its magnitude is the speed. Over a short parameter interval, the displacement is approximately the velocity at one point multiplied by the interval’s width. Taking magnitudes suggests that a short chord has length approximately equal to speed times parameter width. Adding those estimates leads to the integral of speed.

Partitions and Polygonal Length

Let $\mathbf{r}:[a,b]\to\mathbb{R}^m$ be a curve, where $m$ is the number of coordinates. A partition of $[a,b]$ is a finite increasing list of parameter values

a=t_0<t_1<\cdots<t_n=b.

The curve points $\mathbf{r}(t_0),\mathbf{r}(t_1),\ldots,\mathbf{r}(t_n)$, joined in order by straight line segments, form a polygonal approximation. Its length is the sum of the chord lengths:

L(P)=\sum_{i=1}^{n}\left\lVert \mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\right\rVert.

The width of the $i$th parameter interval is $\Delta t_i=t_i-t_{i-1}$. The mesh of the partition is the width of its largest interval, $\lVert P\rVert=\max_i\Delta t_i$. A sequence of partitions becomes finer when its mesh tends to zero.

Definition: For a continuously differentiable curve on $[a,b]$, its arc length is the supremum of the polygonal lengths $L(P)$ over all partitions $P$ of $[a,b]$. The theorem below shows that this length equals the integral of speed.

Why a Short Chord Measures Speed

Assume that $\mathbf{r}$ is continuously differentiable on the closed interval $[a,b]$. Its velocity is $\mathbf{r}'(t)$, and its speed at $t$ is $\lVert\mathbf{r}'(t)\rVert$. By the fundamental theorem of calculus, the displacement over an interval $[t_{i-1},t_i]$ is the integral of the velocity over that interval:

\mathbf{r}(t_i)-\mathbf{r}(t_{i-1}) = \int_{t_{i-1}}^{t_i}\mathbf{r}'(t)\,dt.

If the velocity changes very little across this short interval, it is nearly constant there. If its approximate value is $\mathbf{r}'(\xi_i)$, then the displacement is approximately $\mathbf{r}'(\xi_i)\Delta t_i$. Taking magnitudes gives the local estimate

\left\lVert\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\right\rVert \approx \left\lVert\mathbf{r}'(\xi_i)\right\rVert\Delta t_i.

The factor $\Delta t_i$ matters: speed is distance per unit parameter, not distance by itself. Summing the local estimates gives a Riemann sum for the integral of speed. The following proof makes the approximation precise and shows that its total error tends to zero.

The Chord-Sum Theorem

Theorem: If $\mathbf{r}:[a,b]\to\mathbb{R}^m$ is continuously differentiable, then for any sequence of partitions whose mesh tends to zero, the polygonal lengths converge to $$ \int_a^b\left\lVert\mathbf{r}'(t)\right\rVert\,dt. $$ Consequently, the supremum of all polygonal lengths is this integral.

Proof. Since $\mathbf{r}'$ is continuous on the closed, bounded interval $[a,b]$, it is uniformly continuous. Define $\omega(h)$ to be the greatest possible change in $\mathbf{r}'$ between two parameter values at distance at most $h$. Uniform continuity means that $\omega(h)\to0$ as $h\to0$.

For each partition interval, choose any $\xi_i\in[t_{i-1},t_i]$, and write $\Delta t_i=t_i-t_{i-1}$. Every $t$ in this interval is at most $\lVert P\rVert$ from $\xi_i$, so $\lVert\mathbf{r}'(t)-\mathbf{r}'(\xi_i)\rVert\leq\omega(\lVert P\rVert)$. Using the integral expression for displacement and the inequality $\lvert\lVert\mathbf{u}\rVert-\lVert\mathbf{v}\rVert\rvert\leq\lVert\mathbf{u}-\mathbf{v}\rVert$, we obtain

\begin{aligned} \left| \left\lVert\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\right\rVert -\left\lVert\mathbf{r}'(\xi_i)\right\rVert\Delta t_i \right| &\leq \int_{t_{i-1}}^{t_i} \left\lVert\mathbf{r}'(t)-\mathbf{r}'(\xi_i)\right\rVert\,dt\\ &\leq \omega(\lVert P\rVert)\Delta t_i. \end{aligned}

The same continuity estimate compares the integral of speed on this interval with the constant-speed approximation:

\begin{aligned} \left| \int_{t_{i-1}}^{t_i}\left\lVert\mathbf{r}'(t)\right\rVert\,dt -\left\lVert\mathbf{r}'(\xi_i)\right\rVert\Delta t_i \right| &\leq \int_{t_{i-1}}^{t_i} \left| \left\lVert\mathbf{r}'(t)\right\rVert-\left\lVert\mathbf{r}'(\xi_i)\right\rVert \right|\,dt\\ &\leq \omega(\lVert P\rVert)\Delta t_i. \end{aligned}

Combining these two bounds shows that the difference between the chord length and the integral of speed on the interval is at most $2\omega(\lVert P\rVert)\Delta t_i$. Summing across the partition and using $\sum_i\Delta t_i=b-a$ gives

\left| L(P)-\int_a^b\left\lVert\mathbf{r}'(t)\right\rVert\,dt \right| \leq 2(b-a)\omega(\lVert P\rVert).

As the mesh tends to zero, the right-hand side tends to zero. Thus the chord sums converge to the stated integral. Moreover, the triangle inequality gives, for every partition interval,

\left\lVert\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\right\rVert \leq \int_{t_{i-1}}^{t_i}\left\lVert\mathbf{r}'(t)\right\rVert\,dt.

Therefore every polygonal length is at most the integral, so their supremum is at most the integral. Since chord sums along partitions with mesh tending to zero converge to the integral, their supremum is also at least the integral. The supremum and integral are equal. This completes the proof.

Worked Examples

Worked Example: A Straight Line

Consider $\mathbf{r}(t)=(3t,4t)$ for $0\leq t\leq2$. Divide the interval into $n$ equal parts, so each parameter width is $\Delta t=2/n$. Between consecutive points, the coordinate changes are $3\Delta t=6/n$ and $4\Delta t=8/n$. Thus each chord has length

\sqrt{\left(\frac{6}{n}\right)^2+\left(\frac{8}{n}\right)^2} =\frac{\sqrt{36+64}}{n} =\frac{10}{n}.

The polygon has $n$ such chords, so its length is $n(10/n)=10$, for every $n$. Independently, $\mathbf{r}'(t)=(3,4)$, and its speed is $\sqrt{3^2+4^2}=5$. The speed integral is

\int_0^2 5\,dt=5(2)=10.

The chord sum and the integral both give $10$. The result is also the distance between the endpoints: $\mathbf{r}(0)=(0,0)$, $\mathbf{r}(2)=(6,8)$, and $\sqrt{6^2+8^2}=10$.

Worked Example: A Quarter-Circle

Let $\mathbf{r}(t)=(\cos t,\sin t)$ for $0\leq t\leq\pi/2$. Divide the quarter-circle into $n$ equal parameter intervals. Each interval has angle $\Delta t=\pi/(2n)$. The points at its ends lie on a unit circle, so their chord length is $2\sin(\Delta t/2)=2\sin(\pi/(4n))$. The polygonal length is therefore

L_n=2n\sin\left(\frac{\pi}{4n}\right) =\frac{\pi}{2}\, \frac{\sin(\pi/(4n))}{\pi/(4n)}.

As $n\to\infty$, the standard limit $\sin x/x\to1$ gives $L_n\to\pi/2$. To check this against speed, differentiate the coordinates: $\mathbf{r}'(t)=(-\sin t,\cos t)$, so $\lVert\mathbf{r}'(t)\rVert=\sqrt{\sin^2t+\cos^2t}=1$. Hence

\int_0^{\pi/2}1\,dt=\frac{\pi}{2}.

The limiting chord sum agrees with the speed integral.

Worked Example: A Helix

Consider the helix $\mathbf{r}(t)=(\cos t,\sin t,t)$ for $0\leq t\leq2\pi$. Divide the interval into $n$ equal parts, each of width $2\pi/n$. The horizontal coordinates complete a circle. Over one step, their chord length is $2\sin(\pi/n)$; the vertical coordinate changes by $2\pi/n$. By the three-dimensional distance formula, each chord has length

\sqrt{4\sin^2\left(\frac{\pi}{n}\right)+\left(\frac{2\pi}{n}\right)^2}.

Thus the polygonal length is

\begin{aligned} L_n &=n\sqrt{4\sin^2\left(\frac{\pi}{n}\right)+\left(\frac{2\pi}{n}\right)^2}\\ &=\sqrt{4n^2\sin^2\left(\frac{\pi}{n}\right)+4\pi^2}. \end{aligned}

Since $n\sin(\pi/n)\to\pi$, this tends to $\sqrt{4\pi^2+4\pi^2}=2\pi\sqrt{2}$. For the integral check, $\mathbf{r}'(t)=(-\sin t,\cos t,1)$, so its speed is $\sqrt{\sin^2t+\cos^2t+1}=\sqrt{2}$. Therefore

\int_0^{2\pi}\sqrt{2}\,dt=2\pi\sqrt{2},

which confirms the chord-sum limit.

A Common Mistake: Forgetting the Parameter Width

Speed alone is not the length of a short segment. For the straight-line example, the speed is $5$, but on a partition with $n$ equal intervals each chord has length $5\Delta t=5(2/n)=10/n$, not $5$. Incorrectly using $5$ as every chord length would give a total of $5n$, which grows without bound as the partition is refined.

The correction is to include the interval width: a short chord is approximately $\lVert\mathbf{r}'(\xi_i)\rVert\Delta t_i$. The approximation becomes accurate when the velocity changes little across each interval. The proof establishes this for continuously differentiable curves by bounding the total error and showing it vanishes as the mesh tends to zero.

Why the Derivation Matters

The integral formula is not merely a convenient rule for calculating length. It explains how local motion accumulates into a global distance. Each small contribution is speed multiplied by a small parameter change; the integral is the limit of adding those contributions. Polygonal approximations provide the geometric link: they replace curved pieces by chords, and continuity of the velocity ensures that the accumulated difference disappears as the chords become short.

Check Your Understanding

Use the definitions and reasoning in this tutorial to answer the following questions.

What information about a partition is measured by its mesh?
Why does a short chord involve both speed and the parameter interval width?
Which continuity property of the velocity makes the total approximation error tend to zero?
Why is every polygonal length at most the integral of speed?
For the helix example, what is the length of each chord when the interval is divided into $n$ equal parts?