From Chords to Curve Length
A curved path has no single straight segment whose length gives its total length. A natural way to estimate the length is to mark points along the path, join consecutive points with straight chords, and add the chord lengths. With more marked points, the polygon follows the curve more closely. The central question is whether these polygonal lengths approach a definite value, and what that value is.
For a parametric curve, the displacement across a short parameter interval is approximately its derivative vector multiplied by the interval's width. Taking the magnitude gives an approximation to the chord length. Adding these approximations leads to an integral.
Polygonal Chord Sums
Let a plane curve be given by \(\mathbf{r}(t)=(x(t),y(t))\) for \(a\le t\le b\), where \(x\) and \(y\) are continuously differentiable. A partition of the parameter interval is a finite list \(a=t_0<t_1<\cdots<t_n=b\). The curve points at consecutive partition values are joined by straight segments. The length of the chord from \(\mathbf{r}(t_{i-1})\) to \(\mathbf{r}(t_i)\) is the magnitude of their displacement vector.
The polygonal chord sum for the partition is the sum of these lengths. The arc length is defined as the supremum of all such sums: the least upper bound of the polygonal lengths obtained from partitions of the interval. This definition measures the path travelled, not merely the separation of its endpoints. If a parametrization retraces part of a curve, that travel contributes again to the length.
Why a Short Chord Is Governed by the Derivative
Write \(\mathbf{r}'(t)=(x'(t),y'(t))\). On a short interval from \(t_i\) to \(t_{i+1}\), the displacement is the change in the position vector. By the Fundamental Theorem of Calculus, that change is the integral of the derivative over the interval. If the derivative varies little there, this integral is close to the derivative at the beginning of the interval multiplied by the interval width.
The length of the chord is the magnitude of the displacement vector. Since multiplying a vector by a positive number multiplies its magnitude by that number, the chord length is therefore approximately \(\|\mathbf{r}'(t_i)\|(t_{i+1}-t_i)\). Adding these expressions gives a Riemann sum for the integral of \(\|\mathbf{r}'(t)\|\).
The approximation becomes exact in the limit as the largest partition interval tends to zero. The next result justifies that claim and identifies the limit with the arc length, rather than merely suggesting a plausible formula.
Derivation of the Arc-Length Formula
We prove both directions needed for the result. First, no polygonal chord sum can exceed the integral. Second, sufficiently fine chord sums approach the integral. These facts show that the supremum of the chord sums is exactly the stated integral.
For any partition interval, the displacement is the integral of \(\mathbf{r}'\). The triangle inequality for integrals gives \(\left\|\int_{t_i}^{t_{i+1}}\mathbf{r}'(u)\,du\right\|\le\int_{t_i}^{t_{i+1}}\|\mathbf{r}'(u)\|\,du\). Summing over the partition shows that every chord sum is at most \(\int_a^b\|\mathbf{r}'(u)\|\,du\). Thus the arc length is at most this integral.
Consider partitions whose largest interval width tends to zero. On each interval, the chord vector divided by its width is the average value of \(\mathbf{r}'\) there. Because \(\mathbf{r}'\) is continuous on the closed interval, it is uniformly continuous. Hence these averages approach the derivative at the left endpoint uniformly as the largest width shrinks.
The inequality \(\bigl|\|\mathbf{u}\|-\|\mathbf{v}\|\bigr|\le\|\mathbf{u}-\mathbf{v}\|\) shows that the magnitude of each average derivative approaches the magnitude at the left endpoint as well. Therefore the chord sum differs by an amount tending to zero from the corresponding Riemann sum for \(\|\mathbf{r}'(t)\|\). Those Riemann sums converge to the integral, since \(\|\mathbf{r}'(t)\|\) is continuous.
Fine-partition chord sums approach the integral, so the supremum of all chord sums is at least the integral. Step 1 showed that every chord sum is at most the integral. The two bounds give equality.
In more detail, on an interval of width \(\Delta t_i=t_{i+1}-t_i\), the average derivative is \(\frac{1}{\Delta t_i}\int_{t_i}^{t_{i+1}}\mathbf{r}'(u)\,du\). Its difference from \(\mathbf{r}'(t_i)\) is at most the largest variation of \(\mathbf{r}'\) over parameter distances no greater than the partition's largest interval width. Uniform continuity says this bound tends to zero. Multiplication by each \(\Delta t_i\) and summation keeps the total error bounded by that variation times \(b-a\), which also tends to zero. This completes the limit argument.
Worked Example: Chord Sums for a Parabola
Worked Example: Chord Sums for a Parabola
Find the arc length of \(\mathbf{r}(t)=(t,t^2)\) for \(0\le t\le1\), first by forming equal-interval chord sums and then by using the derived integral.
Divide \([0,1]\) into \(n\) equal intervals. Then \(t_i=i/n\) for \(i=0,1,\ldots,n\), and each interval has width \(1/n\). The change in the first coordinate on one interval is \(1/n\). The change in the second coordinate is \[ \left(\frac{i+1}{n}\right)^2-\left(\frac{i}{n}\right)^2 =\frac{(i+1)^2-i^2}{n^2} =\frac{2i+1}{n^2}. \] Thus the \(i\)-th chord has length \[ \sqrt{\left(\frac1n\right)^2+\left(\frac{2i+1}{n^2}\right)^2} =\frac1n\sqrt{1+\left(\frac{2i+1}{n}\right)^2}. \] The full chord sum is therefore \[ S_n=\frac1n\sum_{i=0}^{n-1}\sqrt{1+\left(\frac{2i+1}{n}\right)^2}. \] Since \((2i+1)/n=2(i+\tfrac12)/n\), these are midpoint Riemann sums for \(\int_0^1\sqrt{1+4t^2}\,dt\). Their limit is the curve's length.
Directly from the theorem, \(x'(t)=1\) and \(y'(t)=2t\), so the length is \[ L=\int_0^1\sqrt{1^2+(2t)^2}\,dt =\int_0^1\sqrt{1+4t^2}\,dt. \] An antiderivative is \(\frac{t}{2}\sqrt{1+4t^2}+\frac14\ln(2t+\sqrt{1+4t^2})\). Evaluating at the endpoints gives \[ L=\left[\frac{t}{2}\sqrt{1+4t^2}+\frac14\ln(2t+\sqrt{1+4t^2})\right]_0^1 =\frac{\sqrt5}{2}+\frac14\ln(2+\sqrt5). \] At \(t=0\), both terms in the antiderivative are zero. To check the antiderivative, differentiate its first term to obtain \(\frac12\sqrt{1+4t^2}+\frac{2t^2}{\sqrt{1+4t^2}}\), and differentiate its logarithmic term to obtain \(\frac{1}{2\sqrt{1+4t^2}}\). Their sum is \(\sqrt{1+4t^2}\), as required. The chord-sum limit and the integral therefore describe the same length.
Worked Example: Chords on a Quarter Circle
Worked Example: Chords on a Quarter Circle
Consider \(\mathbf{r}(t)=(\cos t,\sin t)\) for \(0\le t\le\pi/2\). Divide the interval into \(n\) equal angle increments, each of width \(\Delta t=\pi/(2n)\). Each chord is a side of an isosceles triangle with radius \(1\) and central angle \(\Delta t\). Bisecting the triangle gives half the chord as \(\sin(\Delta t/2)\), so the chord length is \(2\sin(\Delta t/2)=2\sin(\pi/(4n))\).
There are \(n\) identical chords, so their total length is \(2n\sin(\pi/(4n))\). Rewrite it as \[ \frac{\pi}{2}\frac{\sin(\pi/(4n))}{\pi/(4n)}. \] As \(n\) tends to infinity, \(\sin z/z\) tends to \(1\) as \(z\) tends to \(0\). The chord sums therefore tend to \(\pi/2\).
For a check using the integral, \(\mathbf{r}'(t)=(-\sin t,\cos t)\), whose magnitude is \(\sqrt{\sin^2t+\cos^2t}=1\). Hence \[ L=\int_0^{\pi/2}1\,dt=\frac{\pi}{2}. \] The chord-sum limit and the integral agree.
Worked Example: A Straight Path with a Simple Parametrization
Worked Example: A Straight Path with a Simple Parametrization
Let \(\mathbf{r}(t)=(3t,4t)\) for \(0\le t\le2\). With \(n\) equal parameter intervals, each has width \(2/n\). On each interval, the coordinate changes are \(3(2/n)=6/n\) and \(4(2/n)=8/n\). The chord length is \[ \sqrt{\left(\frac6n\right)^2+\left(\frac8n\right)^2} =\frac1n\sqrt{36+64} =\frac{10}{n}. \] Adding all \(n\) equal chords gives a polygonal length of \(n(10/n)=10\), for every \(n\).
The derivative is \(\mathbf{r}'(t)=(3,4)\), so the integral gives \[ L=\int_0^2\sqrt{3^2+4^2}\,dt =\int_0^2 5\,dt =10. \] This checks the chord calculation. Here every chord lies along the straight path, so even a coarse partition gives the exact length.
A Common Mistake: Confusing Path Length with Endpoint Distance
A single chord joins only the two endpoints; it does not usually follow the intervening curve. For the unit quarter circle, the endpoints are \((1,0)\) and \((0,1)\). Their straight-line distance is \(\sqrt{(0-1)^2+(1-0)^2}=\sqrt2\), which is less than the arc length \(\pi/2\).
A related error is to calculate the magnitude of the position vector \(\mathbf{r}(t)\) instead of the magnitude of the derivative \(\mathbf{r}'(t)\). For \(\mathbf{r}(t)=(t,t^2)\), this mistake would give \(\int_0^1\sqrt{t^2+t^4}\,dt\). But the short displacement is controlled by the change in position, so its first-order approximation uses \(\mathbf{r}'(t)=(1,2t)\). The correct integrand is \(\sqrt{1+4t^2}\), not \(\sqrt{t^2+t^4}\). In particular, at \(t=0\) the incorrect integrand is zero even though the curve is moving in the direction \((1,0)\).
Check Your Understanding
Use the definitions and derivation above to answer these questions.
- How is the length of a parametric curve defined using partitions?
- Why does the displacement over a short interval resemble the derivative vector multiplied by the interval width?
- What condition on \(\mathbf{r}\) is used in the arc-length theorem?
- For \(\mathbf{r}(t)=(t,3t)\) on \(0\le t\le1\), what is the length, and how does a chord sum verify it?
- Why can the straight-line distance between the endpoints be less than the arc length?