Why Arc Length Requires a Limit
A parametrized curve describes a moving point by its coordinates \(x(t)\) and \(y(t)\). The point’s displacement from one time to another is easy to calculate, but displacement measures only the straight-line change between the endpoints. The length of the path can be greater, because the point may follow a curved route.
To measure that route, imagine marking many parameter values and joining the corresponding points with straight segments. The resulting polygon approximates the curve. As the parameter values become more closely spaced, the polygon follows the curve more closely, and its total length approaches the curve’s arc length.
The key question is how to express the lengths of those increasingly short segments. The coordinate changes over a short parameter interval are approximately \(x'(t)\,\Delta t\) and \(y'(t)\,\Delta t\). The distance formula then gives a short-segment length approximately equal to \(\sqrt{(x'(t))^2+(y'(t))^2}\,\Delta t\). Adding these approximations leads to an integral.
The polygonal path uses the curve’s order of traversal. If a parametrization retraces part of a curve, that portion is counted each time it is traversed. Thus arc length measures the distance travelled along the parametrized path, not merely the distance between its endpoints.
The Formula for a Smooth Parametric Curve
Suppose \(x\) and \(y\) have continuous derivatives on the closed interval \([a,b]\). This condition ensures that the velocity components vary continuously, so a short interval can be approximated using the derivative at one of its endpoints. The resulting arc length formula is:
The quantity under the integral is the magnitude of the velocity vector. In the earlier material on velocity components and speed, speed is defined as the magnitude of velocity; here, it becomes the rate at which arc length accumulates as the parameter changes.
Derivation from Short Line Segments
Divide \([a,b]\) into parameter intervals with endpoints \(a=t_0<t_1<\cdots<t_n=b\). Write \(\Delta t_i=t_i-t_{i-1}\), and let \(\delta\) be the largest of these widths. The polygonal approximation joins \(\mathbf{r}(t_{i-1})\) to \(\mathbf{r}(t_i)\) for each interval. Its length is the sum of the chord lengths.
For one interval, the coordinate changes are integrals of the derivatives, by the Fundamental Theorem of Calculus. In vector form, this means that the chord vector is the integral of the velocity vector over that interval. Because the velocity is continuous, it changes by only a small amount over a sufficiently short interval. The chord therefore has nearly the same length as the velocity at the interval’s starting point multiplied by the interval width.
For \(\mathbf{r}'(t)=(x'(t),y'(t))\), the coordinatewise Fundamental Theorem of Calculus gives \(\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})=\int_{t_{i-1}}^{t_i}\mathbf{r}'(u)\,du\).
For vectors, the difference between their magnitudes is no greater than the magnitude of their difference. Consequently, the difference between the chord length and \(\|\mathbf{r}'(t_{i-1})\|\Delta t_i\) is at most \(\int_{t_{i-1}}^{t_i}\|\mathbf{r}'(u)-\mathbf{r}'(t_{i-1})\|\,du\).
The continuous function \(\mathbf{r}'\) is uniformly continuous on \([a,b]\). Let \(\omega(\delta)\) be a bound on the change in \(\mathbf{r}'\) between parameter values at most \(\delta\) apart. Then the total difference between \(P\) and \(\sum_{i=1}^{n}\|\mathbf{r}'(t_{i-1})\|\Delta t_i\) is at most \(\omega(\delta)(b-a)\), which approaches zero as \(\delta\) approaches zero.
The sums \(\sum_{i=1}^{n}\|\mathbf{r}'(t_{i-1})\|\Delta t_i\) are Riemann sums for the continuous function \(\|\mathbf{r}'(t)\|\). As the largest interval width approaches zero, they converge to \(\int_a^b\|\mathbf{r}'(t)\|\,dt\). The polygonal lengths have the same limit.
Finally, the velocity vector has components \(x'(t)\) and \(y'(t)\). The distance formula for the magnitude of a vector gives \(\|\mathbf{r}'(t)\|=\sqrt{(x'(t))^2+(y'(t))^2}\). Substituting this magnitude into the integral proves the stated formula. The square root is not an extra rule: it comes directly from measuring the length of each short displacement vector.
Worked Examples
Worked Example: A Short Chord and Its Local Approximation
Consider \(\mathbf{r}(t)=(t^2,3t)\), and compare the points at \(t=0\) and \(t=h\), where \(h>0\).
The coordinate changes are \(x(h)-x(0)=h^2-0=h^2\) and \(y(h)-y(0)=3h-0=3h\). Therefore, the chord length is
The derivatives are \(x'(t)=2t\) and \(y'(t)=3\), so the speed at the starting parameter is \(\sqrt{(2\cdot0)^2+3^2}=3\). The local straight-segment approximation is therefore \(3h\).
To check the approximation, divide the exact chord length by \(h\): \(\frac{h\sqrt{h^2+9}}{h}=\sqrt{h^2+9}\), which approaches \(3\) as \(h\) approaches zero. Thus the chord length divided by its parameter width approaches the speed at the starting point, exactly as the derivation predicts.
Worked Example: A Constant-Velocity Parametrization
Let \(\mathbf{r}(t)=(1+3t,2-4t)\) for \(0\leq t\leq2\). Its derivatives are \(x'(t)=3\) and \(y'(t)=-4\). The speed is constant:
The formula gives \(L=\int_0^2 5\,dt=5(2-0)=10\). Check this against the endpoint coordinates. The initial point is \((1,2)\), and the final point is \((1+3\cdot2,2-4\cdot2)=(7,-6)\). Their straight-line distance is \(\sqrt{(7-1)^2+(-6-2)^2}=\sqrt{36+64}=10\), agreeing with the integral. Since the velocity is constant, every segment of the path points in the same direction, so the polygonal approximations also have total length \(10\).
Worked Example: Polygonal Approximations to a Parabola
Consider \(\mathbf{r}(t)=(t,t^2)\) on \(0\leq t\leq1\). The derivatives are \(x'(t)=1\) and \(y'(t)=2t\), so the formula predicts
To see the polygonal approximation, first use just the endpoints \((0,0)\) and \((1,1)\). Their chord length is \(\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}\).
Now include the midpoint parameter \(t=\frac12\), which gives the point \((\frac12,\frac14)\). The first chord has length \(\sqrt{(\frac12)^2+(\frac14)^2}=\frac{\sqrt{5}}{4}\). The second has length \(\sqrt{(\frac12)^2+(\frac34)^2}=\frac{\sqrt{13}}{4}\). The refined polygon therefore has length \(\frac{\sqrt{5}+\sqrt{13}}{4}\), approximately \(1.460\), while the one-chord approximation is \(\sqrt2\), approximately \(1.414\). The refined polygon is longer because it follows the bend more closely.
As the partition is refined further, the polygonal lengths approach the integral of \(\sqrt{1+4t^2}\). The derivative calculation checks the integrand, and the chord calculations show how the polygonal definition approximates the curve. No finite polygon is expected to equal the limiting arc length exactly.
Why the Conditions and the Square Root Matter
Continuity of the derivatives is what makes the short-segment approximation reliable across the whole interval. If the velocity changed abruptly, the comparison between a chord and the velocity at its starting point would require additional care. For a curve made from finitely many continuously differentiable pieces, the formula can be applied on each piece and the resulting lengths added.
The square root combines the two perpendicular coordinate changes using the distance formula. Adding the coordinate derivatives directly would not measure distance. For example, for the constant-velocity curve in the previous example, the incorrect expression \(3+(-4)=-1\) would even give a negative value for a supposed rate of distance. Taking absolute values separately would also be wrong: \(|3|+|-4|=7\), which would predict a length of \(14\) over an interval of width \(2\), rather than the verified length \(10\).
Check Your Understanding
Use the derivation and examples to answer these questions.
- What does a polygonal approximation join, and what happens to its largest parameter-interval width as it approaches arc length?
- For \(\mathbf{r}(t)=(x(t),y(t))\), why is the short-segment length approximately \(\sqrt{(x'(t))^2+(y'(t))^2}\,\Delta t\)?
- Which continuity condition is used in the derivation, and how does it control the approximation error?
- For \(\mathbf{r}(t)=(2t,2t)\) on \(0\leq t\leq1\), write the arc length integral and evaluate it.
- Why can the path length exceed the straight-line distance between the endpoints?
