Arc Length as a Limit of Chord Sums

From a Curve to a Polygon

A curved path is not made of straight segments, but straight segments can approximate it. Choose points along the path in order, then join consecutive points by chords. The resulting polygonal path is an approximation to the curve. Taking more points, placed closer together, generally makes that approximation more faithful.

This idea gives a precise way to define curve length. It also clarifies an important distinction: the length of a path records the route traveled, while the distance between its endpoints measures only the shortest straight-line connection between them.

Partitions and Chord Sums

Represent a curve by a continuous function $\mathbf{r}:[a,b]\to\mathbb{R}^2$, where $\mathbf{r}(t)$ gives the point reached at parameter value $t$. A partition of $[a,b]$ is a finite ordered list of parameter values

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

The curve points $\mathbf{r}(t_0),\mathbf{r}(t_1),\ldots,\mathbf{r}(t_n)$ are joined by straight segments. The length of the resulting polygonal path is its chord sum:

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

Here $P$ denotes the partition, and $\lVert\cdot\rVert$ is the usual Euclidean distance. The mesh of the partition is the width of its largest parameter interval:

\lVert P\rVert=\max_{1\le i\le n}(t_i-t_{i-1}).

A partition $Q$ is a refinement of $P$ if it contains all the partition points of $P$, possibly with additional points in between.

Definition: The length of a continuous curve $\mathbf{r}:[a,b]\to\mathbb{R}^2$ is the supremum of all its chord sums: $$ L=\sup_P S(P). $$ If this supremum is finite, the curve is called rectifiable. For a rectifiable curve, chord sums along partitions whose mesh tends to zero converge to $L$. Thus the curve length is also the limit of increasingly fine polygonal approximations.

The supremum is the least upper bound of the chord sums over all finite partitions. It allows for the possibility that no single partition produces the exact curve length: rather, polygonal approximations can come arbitrarily close. If the chord sums have no finite upper bound, the curve has infinite length.

Why Refining a Partition Helps

Adding a point to one of the straight segments replaces that segment by two segments meeting at the new point. The triangle inequality says that the sum of the two new lengths is at least the length of the original segment. For points $A,B,C$, it states that

\lVert A-B\rVert\le \lVert A-C\rVert+\lVert C-B\rVert.

Consequently, refinement cannot decrease a chord sum. It may leave it unchanged when the added point lies on the original segment in the correct order; otherwise, it increases the sum. This is why increasingly detailed polygons can reveal more of a curve’s length.

Theorem: If $\mathbf{r}$ is continuous and rectifiable, then for every $\varepsilon>0$, there is a $\delta>0$ such that every partition $Q$ with mesh $\lVert Q\rVert<\delta$ satisfies $$ L-\varepsilon<S(Q)\le L. $$ In particular, the chord sums converge to the curve length as the mesh tends to zero.

Why Fine Chord Sums Converge

The upper bound $S(Q)\le L$ follows directly from the definition of $L$ as the supremum of all chord sums. The important point is that a sufficiently fine partition cannot miss a substantial amount of length.

Here is a proof. Fix $\varepsilon>0$. Since $L$ is the supremum of the chord sums, choose a partition $P_0$ such that

S(P_0)>L-\frac{\varepsilon}{2}.

Write $m$ for the number of interior points of $P_0$. Continuity on the closed interval $[a,b]$ implies uniform continuity: points whose parameter values are sufficiently close have curve points as close as desired. Choose $\delta>0$ so that whenever $|s-t|<\delta$,

\lVert\mathbf{r}(s)-\mathbf{r}(t)\rVert< \frac{\varepsilon}{4m}

when $m>0$. (If $m=0$, the selected partition already consists only of the endpoints.) Now take any partition $Q$ with mesh less than $\delta$, and add the interior points of $P_0$ to it to form a common refinement $R$.

Each inserted point lies in an interval of $Q$ with parameter width less than $\delta$. Inserting that point can increase the chord sum by at most the sum of its distances from the two neighboring curve points, which is less than $2\varepsilon/(4m)$. There are at most $m$ insertions, so

S(R)-S(Q)<m\left(\frac{\varepsilon}{2m}\right) =\frac{\varepsilon}{2}.

Because $R$ refines $P_0$, refinement cannot decrease its chord sum. Also, $S(R)\le L$. Therefore

S(Q)>S(R)-\frac{\varepsilon}{2} \ge S(P_0)-\frac{\varepsilon}{2} >L-\varepsilon.

Together with $S(Q)\le L$, this proves the theorem. The argument explains why both ideas matter: rectifiability supplies a finite supremum to approach, and continuity ensures that a fine partition can be adjusted to include a good approximation without changing its chord sum by much.

Worked Examples

Worked Example: A Straight-Line Path

Consider $\mathbf{r}(t)=(3t,4t)$ for $0\le t\le 1$. Take any partition $0=t_0<t_1<\cdots<t_n=1$. For each interval, the change in position is

\mathbf{r}(t_i)-\mathbf{r}(t_{i-1}) =\bigl(3(t_i-t_{i-1}),\,4(t_i-t_{i-1})\bigr).

Since $t_i-t_{i-1}>0$, the chord length is

\begin{aligned} \left\lVert\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})\right\rVert &=\sqrt{9(t_i-t_{i-1})^2+16(t_i-t_{i-1})^2}\\ &=5(t_i-t_{i-1}). \end{aligned}

Adding all the pieces gives

S(P)=5\sum_{i=1}^{n}(t_i-t_{i-1}) =5(1-0)=5.

Every partition gives the same sum, so the supremum and the limit of fine chord sums are both $5$. Checking with the endpoints, $\mathbf{r}(0)=(0,0)$ and $\mathbf{r}(1)=(3,4)$, their distance is $\sqrt{3^2+4^2}=5$, as expected for a straight path.

Worked Example: Chords Along a Semicircle

Consider the upper semicircle of radius $2$, traced once by $\mathbf{r}(t)=(2\cos t,2\sin t)$ for $0\le t\le\pi$. Divide the parameter interval into $n$ equal parts, so each angle increment is $\pi/n$. Each chord subtends a central angle $\pi/n$. Its length is $4\sin(\pi/(2n))$: the two radii and the chord form an isosceles triangle, and bisecting it gives a right triangle with hypotenuse $2$ and half-chord opposite angle $\pi/(2n)$.

There are $n$ equal chords, so this partition’s sum is

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

As $n\to\infty$, the angle $\pi/(2n)$ tends to zero, and the standard limit $\sin u/u\to1$ gives

\lim_{n\to\infty}S_n=2\pi.

For a check, with $n=2$, each chord has length $2\sqrt{2}$, so $S_2=4\sqrt{2}$. The limiting value $2\pi$ is greater than this particular polygonal approximation. The endpoints are $(2,0)$ and $(-2,0)$, whose distance is $4$, not $2\pi$: the semicircular route is longer than the straight segment joining those endpoints.

Worked Example: Chord Sums on a Parabola

Take the portion of $y=x^2$ from $x=0$ to $x=1$. The corresponding curve points are $(x,x^2)$. First use the partition $0,\frac12,1$, giving the points $(0,0)$, $(\frac12,\frac14)$, and $(1,1)$.

The first chord has length

\sqrt{\left(\frac12\right)^2+\left(\frac14\right)^2} =\frac{\sqrt{5}}{4}.

The second has length

\sqrt{\left(\frac12\right)^2+\left(\frac34\right)^2} =\frac{\sqrt{13}}{4}.

Thus this partition gives $S(P)=(\sqrt{5}+\sqrt{13})/4$, approximately $1.461$. Refine it using the points at $x=\frac14$ and $x=\frac34$. The successive changes in $x$ are all $1/4$, while the changes in $y=x^2$ are $1/16,3/16,5/16,7/16$. The four chord lengths are therefore $\sqrt{17}/16$, $5/16$, $\sqrt{65}/16$, and $\sqrt{145}/16$. The refined sum is

S(Q)=\frac{\sqrt{17}+5+\sqrt{65}+\sqrt{145}}{16} \approx 1.827.

The refined sum is greater than the original one, as the refinement principle predicts. The endpoint distance is $\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}\approx1.414$. These calculations give polygonal approximations and the endpoint distance; the exact curve length is a separate question.

Arc Length Is Not Endpoint Distance

For any two endpoints $A$ and $B$, the direct distance $\lVert B-A\rVert$ is no greater than the length of a path joining them. Indeed, for every partition, repeated use of the triangle inequality gives

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

Taking the supremum over partitions shows that endpoint distance is at most curve length. Equality holds for a straight segment traversed without reversing direction, as in the first example. A curved route generally has greater length, as the semicircle demonstrates. A path can even return to its starting point: its endpoint distance is then zero, although its traveled length can be positive.

Common pitfall: Do not calculate the length of a curved path by measuring only its endpoints. For the radius-$2$ semicircle, that mistake gives $4$, the endpoint distance. The chord-sum limit gives $2\pi$, the length of the path along the semicircle.

When This Definition Is Useful

Polygonal chord sums turn an informal idea—“how far the curve goes”—into a definition that works for many paths. They explain how a length can be approached by measurable straight pieces, and why partition choice matters when estimating it. They also provide a foundation for later formulas that compute the same length using calculus. At this stage, the essential idea is the limit itself: refine the partition, add the successive chord lengths, and distinguish the resulting path length from the direct distance between endpoints.

Check Your Understanding

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

What is a partition of a parameter interval, and how is its chord sum defined?
Why can refining a partition not decrease its chord sum?
What does it mean for a curve to be rectifiable?
For a rectifiable continuous curve, what happens to chord sums as the partition mesh tends to zero?
Can a curve have positive length when its endpoints coincide? Explain briefly.