Deriving Arc Length for Parametric Curves

From Chords to Speed

A parametrized curve is described by giving the coordinates of a moving point as functions of a parameter $t$. Its path may bend, and the parameter need not represent distance along that path. To calculate the length traversed, we need to account for the point’s motion in both coordinate directions.

The basic idea is to approximate the curve by short straight chords. On a sufficiently small parameter interval, the point moves approximately in the direction of its velocity, and the chord length is approximately the speed multiplied by the interval width. Adding these local approximations leads to the parametric arc-length formula.

The earlier material on arc length as a limit of chord sums gives the geometric starting point: refine a partition of the parameter interval and add the lengths of the successive chords. Here we justify why, for a sufficiently smooth parametrization, those sums approach the integral of the speed.

Parametrizations and Chord Sums

Definition: A parametrized plane curve on $[a,b]$ is a map $\mathbf{r}:[a,b]\to\mathbb{R}^2$ of the form $\mathbf{r}(t)=(x(t),y(t))$. A partition is a finite sequence $a=t_0<t_1<\cdots<t_n=b$. Its chord sum is $\sum_{i=1}^{n}|\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})|$, where $|\cdot|$ denotes Euclidean length.

The chord joining $\mathbf{r}(t_{i-1})$ to $\mathbf{r}(t_i)$ replaces the portion of the curve between those parameter values by a straight segment. The arc length of the parametrized curve is the supremum of the chord sums over all partitions. In the smooth setting considered here, increasingly fine partitions have chord sums converging to that length.

We assume throughout the derivation that $x$ and $y$ have continuous first derivatives on the closed interval $[a,b]$. Equivalently, $\mathbf{r}$ is continuously differentiable there. This condition ensures that the velocity varies continuously and that the fundamental theorem of calculus applies to each coordinate.

Regularity condition: The formula below is valid when $\mathbf{r}$ is continuously differentiable on $[a,b]$. A regular curve has the additional property that $\mathbf{r}'(t)\ne\mathbf{0}$ throughout the interval. That nonzero-speed condition is convenient for describing a smooth moving point, but it is stronger than necessary for the length formula: continuous differentiability alone suffices.

The velocity is $\mathbf{r}'(t)=(x'(t),y'(t))$. Its magnitude, called the speed, is $|\mathbf{r}'(t)|=\sqrt{[x'(t)]^2+[y'(t)]^2}$. Speed is nonnegative and measures how quickly the point moves per unit of parameter.

Deriving the Parametric Arc-Length Formula

Consider one subinterval $[t_{i-1},t_i]$ of a partition, with width $\Delta t_i=t_i-t_{i-1}$. By applying the fundamental theorem of calculus to each coordinate, the displacement vector across this interval is the integral of the velocity:

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

If the velocity were exactly constant on this interval, its chord length would be $|\mathbf{r}'(t)|\Delta t_i$. For a varying velocity, the difference between these quantities can be bounded by how much the velocity changes over the interval. The following argument makes that approximation precise.

Theorem: If $\mathbf{r}(t)=(x(t),y(t))$ is continuously differentiable on $[a,b]$, then its arc length is $$ L=\int_a^b|\mathbf{r}'(t)|\,dt =\int_a^b\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt. $$

Proof. Since $\mathbf{r}'$ is continuous on the closed interval, it is uniformly continuous. Thus, for any $\varepsilon>0$, there is a $\delta>0$ such that $|\mathbf{r}'(s)-\mathbf{r}'(t)|<\varepsilon$ whenever $s,t\in[a,b]$ and $|s-t|<\delta$.

For any partition whose largest subinterval width is less than $\delta$, choose a point $\tau_i\in[t_{i-1},t_i]$ for each $i$. Using the reverse triangle inequality and the displacement integral gives:

\begin{aligned} \left| |\mathbf{r}(t_i)-\mathbf{r}(t_{i-1})| -|\mathbf{r}'(\tau_i)|\Delta t_i \right| &\le \left| \mathbf{r}(t_i)-\mathbf{r}(t_{i-1}) -\mathbf{r}'(\tau_i)\Delta t_i \right|\\ &= \left| \int_{t_{i-1}}^{t_i} \bigl(\mathbf{r}'(t)-\mathbf{r}'(\tau_i)\bigr)\,dt \right|\\ &\le \int_{t_{i-1}}^{t_i} |\mathbf{r}'(t)-\mathbf{r}'(\tau_i)|\,dt\\ &\le \varepsilon\Delta t_i. \end{aligned}

Adding these inequalities over all subintervals shows that the difference between the chord sum and the Riemann sum $\sum_{i=1}^{n}|\mathbf{r}'(\tau_i)|\Delta t_i$ is at most $\varepsilon(b-a)$. Because $|\mathbf{r}'|$ is continuous, these Riemann sums converge to $\int_a^b|\mathbf{r}'(t)|\,dt$ as the largest subinterval width tends to zero. The chord sums therefore converge to the same integral.

It remains to connect this limit with the definition of arc length as a supremum. For every subinterval, the triangle inequality for integrals gives:

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

Summing shows that every chord sum is at most $\int_a^b|\mathbf{r}'(t)|\,dt$, so the supremum is at most that integral. On the other hand, chord sums from partitions with widths tending to zero converge to the integral, so their supremum is at least the integral. The supremum equals the integral, proving the formula. Finally, the coordinate expression follows from $\mathbf{r}'(t)=(x'(t),y'(t))$ and the definition of Euclidean length. $\square$

Worked Examples

Worked Example: A Parametrized Line Segment

Find the length of $\mathbf{r}(t)=(3t,4t)$ for $0\le t\le2$.

Differentiate each coordinate:

x'(t)=3,\qquad y'(t)=4.

The speed is constant:

|\mathbf{r}'(t)|=\sqrt{3^2+4^2}=\sqrt{9+16}=5.

Therefore the length is:

L=\int_0^2 5\,dt=5[t]_0^2=5(2-0)=10.

As a check, the endpoints are $\mathbf{r}(0)=(0,0)$ and $\mathbf{r}(2)=(6,8)$. Their distance is $\sqrt{(6-0)^2+(8-0)^2}=\sqrt{36+64}=10$, matching the calculated length of this straight segment.

Worked Example: A Quarter-Circle Parametrization

Find the length of $\mathbf{r}(t)=(2\cos t,2\sin t)$ for $0\le t\le\frac{\pi}{2}$.

Differentiate the coordinates:

x'(t)=-2\sin t,\qquad y'(t)=2\cos t.

Then the speed is:

|\mathbf{r}'(t)| =\sqrt{(-2\sin t)^2+(2\cos t)^2} =2\sqrt{\sin^2 t+\cos^2 t} =2.

Integrating over the parameter interval gives:

L=\int_0^{\pi/2}2\,dt =2\left[t\right]_0^{\pi/2} =2\left(\frac{\pi}{2}-0\right) =\pi.

For a check of the integrand, an antiderivative of the constant speed $2$ is $2t$, whose derivative is $2$. The endpoints are $(2,0)$ and $(0,2)$; their chord length is $2\sqrt{2}$, which is less than the computed arc length $\pi$, as expected for a curved path.

Worked Example: A Curve with a Simplifiable Speed

Find the length of $\mathbf{r}(t)=\left(t-\frac{t^3}{3},t^2\right)$ for $0\le t\le1$.

Differentiate both coordinates:

x'(t)=1-t^2,\qquad y'(t)=2t.

Substitute these derivatives into the speed formula and simplify:

\begin{aligned} |\mathbf{r}'(t)| &=\sqrt{(1-t^2)^2+(2t)^2}\\ &=\sqrt{1-2t^2+t^4+4t^2}\\ &=\sqrt{1+2t^2+t^4}\\ &=\sqrt{(1+t^2)^2} =1+t^2. \end{aligned}

The last equality holds because $1+t^2>0$ on the interval. Thus:

L=\int_0^1(1+t^2)\,dt =\left[t+\frac{t^3}{3}\right]_0^1 =1+\frac13 =\frac43.

To check the integration, the derivative of $t+\frac{t^3}{3}$ is $1+t^2$, the speed. Also, the endpoint displacement is $\left(\frac23,1\right)$, with length $\frac{\sqrt{13}}{3}<\frac43$. The chord is shorter than the curved path, consistent with the result.

A Common Mistake: Mixing Parameter Derivatives with Coordinate Changes

The parametric formula uses derivatives with respect to the parameter $t$, and it integrates with respect to $t$. Replacing $x'(t)$ with a coordinate differential, or adding an extra factor of $dx/dt$, changes the calculation and generally gives the wrong length.

For example, for $\mathbf{r}(t)=(3t,4t)$ on $0\le t\le2$, one might incorrectly calculate $\int_0^2 5\,dx$. Since $dx/dt=3$, this expression would give $\int_0^2 5(3)\,dt=30$, not the correct length $10$. The error is using $dx$ as though the parameter interval were an $x$-interval. The correct calculation is $\int_0^2\sqrt{3^2+4^2}\,dt=\int_0^2 5\,dt=10$.

Another common error is to omit the square root and integrate $[x'(t)]^2+[y'(t)]^2$. That quantity is the square of the speed, not the speed. For the same line it is $3^2+4^2=25$, which would produce $25(2)=50$. The square root is required to obtain Euclidean distance per unit of parameter.

Why the Formula Matters

The derivation explains the geometric meaning of the formula: each short piece contributes approximately speed times parameter width, and the integral adds these contributions continuously. The parameter interval records how the curve is traversed, while the speed converts parameter change into distance traveled.

The assumptions also matter. Continuous differentiability supplies a continuous velocity and makes the chord approximation argument work. If a parametrization is not continuously differentiable on the whole interval, this particular theorem does not directly apply as stated; additional care is needed to establish a length formula. Under the stated assumptions, however, the speed integral gives the curve’s length, even if the speed happens to be zero at some parameter value.

Check Your Understanding

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

What does a chord sum measure for a partition of a parametrized curve?
For $\mathbf{r}(t)=(x(t),y(t))$, how is speed expressed in terms of the coordinate derivatives?
Why does continuous differentiability allow the chord sums to approach the integral of speed?
Is nonzero speed required for the length formula proved here? State the condition that is sufficient.
For $\mathbf{r}(t)=(t,2t)$ on $0\le t\le3$, set up the length integral and evaluate it.