Deriving the Arc-Length Element for Curve Integrals

Why Arc Length Needs a Speed Factor

A curve is described by a parameter, but the parameter is not itself a measure of distance along the curve. A small change in the parameter can produce a small or large movement, depending on how the curve is parametrized. To integrate a scalar quantity over a curve, we need to weight each value by the actual distance traveled along the curve.

The conversion from parameter change to distance is supplied by the curve’s speed. We will derive that conversion from the lengths of short curve segments and then use it to understand the arc-length element $ds$.

Short Segments and Their Lengths

Let $\mathbf r:[a,b]\to\mathbb R^n$ be a continuously differentiable parametrization. At a parameter value $t$, the point on the curve is $\mathbf r(t)$. For a small positive increment $\Delta t$, the displacement to the next point is

\mathbf r(t+\Delta t)-\mathbf r(t).

The straight-line distance between these two points, called the chord length, is

\left\|\mathbf r(t+\Delta t)-\mathbf r(t)\right\|.

Differentiability tells us that, for a sufficiently small increment, the displacement is approximately the derivative times the parameter change:

\mathbf r(t+\Delta t)-\mathbf r(t) = \mathbf r'(t)\Delta t+\mathbf e(\Delta t),

where the error vector $\mathbf e(\Delta t)$ satisfies $\|\mathbf e(\Delta t)\|/\Delta t\to 0$ as $\Delta t\to 0^+$. Taking lengths and dividing by $\Delta t$ gives the local distance-per-parameter-change relationship:

\lim_{\Delta t\to 0^+} \frac{\left\|\mathbf r(t+\Delta t)-\mathbf r(t)\right\|}{\Delta t} = \left\|\mathbf r'(t)\right\|.

The quantity $\|\mathbf r'(t)\|$ is the speed of the parametrized curve. It tells us how much distance is covered per unit change in the parameter, in the limiting sense of increasingly short segments.

Definition: For a differentiable curve $\mathbf r(t)$, its speed at $t$ is $\|\mathbf r'(t)\|$. The arc-length element is $$ ds=\left\|\mathbf r'(t)\right\|\,dt $$ when the parameter increases. It represents the infinitesimal distance along the curve corresponding to a parameter increment $dt$.

The notation $ds$ is not an independent parameter increment: it is the distance increment obtained after accounting for how quickly the curve moves as $t$ changes. Since speed is nonnegative, the arc-length increment is nonnegative when $dt$ is positive.

From Chords to the Length of the Curve

For a partition $a=t_0<t_1<\cdots<t_m=b$, connect successive curve points with straight segments. The resulting polygonal length is

\sum_{i=0}^{m-1} \left\|\mathbf r(t_{i+1})-\mathbf r(t_i)\right\|.

The length of a continuously differentiable curve is the limit of these polygonal lengths as the partition is refined; equivalently, it is the supremum of the polygonal lengths over all partitions. For such a curve, this length is the integral of its speed.

Theorem: If $\mathbf r:[a,b]\to\mathbb R^n$ is continuously differentiable, then its arc length $L$ is $$ L=\int_a^b \left\|\mathbf r'(t)\right\|\,dt. $$ The formula also applies piece by piece to a piecewise continuously differentiable curve, with the lengths of its pieces added.

Here is a justification of the formula. On any subinterval $[t_i,t_{i+1}]$, the fundamental theorem of calculus, applied to each component, gives the displacement as the integral of the derivative. The triangle inequality for integrals then bounds the chord by the integral of the speed:

\left\|\mathbf r(t_{i+1})-\mathbf r(t_i)\right\| = \left\|\int_{t_i}^{t_{i+1}}\mathbf r'(u)\,du\right\| \leq \int_{t_i}^{t_{i+1}}\left\|\mathbf r'(u)\right\|\,du.

Adding these inequalities over a partition shows that every polygonal length is at most $\int_a^b\|\mathbf r'(u)\|\,du$. Thus the curve length cannot exceed that integral.

For the reverse inequality, consider partitions whose largest subinterval length tends to zero. Because $\mathbf r'$ is continuous on the closed interval, it is uniformly continuous. Let $\omega(\delta)$ be an upper bound on how much $\mathbf r'$ can change between parameter values at most $\delta$ apart; then $\omega(\delta)\to0$ as $\delta\to0$. On a subinterval of width $\Delta t_i=t_{i+1}-t_i$, compare the displacement with $\mathbf r'(t_i)\Delta t_i$:

\left\| \mathbf r(t_{i+1})-\mathbf r(t_i)-\mathbf r'(t_i)\Delta t_i \right\| \leq \Delta t_i\,\omega(\Delta t_i).

Consequently, the chord length differs from $\|\mathbf r'(t_i)\|\Delta t_i$ by at most $\Delta t_i\omega(\Delta t_i)$. Summing over the partition, the total error is at most $(b-a)\omega(\delta)$, where $\delta$ is the partition’s largest subinterval width. This error tends to zero as the partition is refined. The sums $\sum_i\|\mathbf r'(t_i)\|\Delta t_i$ converge to the Riemann integral of the continuous speed. Therefore the polygonal lengths converge to $\int_a^b\|\mathbf r'(t)\|\,dt$, proving the formula.

Why Scalar Values Are Weighted by Distance

Suppose $f$ is a continuous scalar field defined at points of the curve. Over a short segment, the field value is approximately $f(\mathbf r(t_i))$, while the segment’s length is approximately $\|\mathbf r'(t_i)\|\Delta t_i$. A distance-weighted sum therefore has terms of the form

f(\mathbf r(t_i))\left\|\mathbf r'(t_i)\right\|\Delta t_i.

As the segments become shorter, these sums approach the scalar line integral. In parameter form, it is

\int_C f\,ds = \int_a^b f(\mathbf r(t))\left\|\mathbf r'(t)\right\|\,dt.

This expression requires a parametrized curve that is continuously differentiable, or piecewise continuously differentiable with the integral evaluated piece by piece, and a scalar field continuous along the curve. The factor $\|\mathbf r'(t)\|$ is essential: it makes the contribution depend on the distance represented by each parameter interval, not just on the interval’s width.

If $t$ represents time and the coordinates of $\mathbf r(t)$ are measured in metres, then $\|\mathbf r'(t)\|$ has units of metres per second. Multiplying by $dt$, in seconds, gives $ds$ in metres. If the parameter is not physical time, speed instead has units of distance per unit of that parameter, and the product still has units of distance.

Worked Examples

Worked Example: A Scalar Field Along a Straight Curve

Let $\mathbf r(t)=(3t,4t)$ for $0\leq t\leq2$, and let $f(x,y)=x$. First find the velocity vector and speed:

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

Thus $ds=5\,dt$. The field restricted to the curve is $f(\mathbf r(t))=3t$, so the scalar line integral is

\int_C f\,ds = \int_0^2 (3t)(5)\,dt = \int_0^2 15t\,dt = \left[\frac{15}{2}t^2\right]_0^2 = \frac{15}{2}(4)-0 = 30.

As a check, differentiating the antiderivative gives $\frac{d}{dt}(\frac{15}{2}t^2)=15t$, the integrand. The curve’s total length is also $5(2-0)=10$, consistent with its displacement $(6,8)$, whose length is $\sqrt{6^2+8^2}=10$.

Worked Example: A Curve With Changing Speed

Let $\mathbf r(t)=(t,t^2)$ for $0\leq t\leq1$, and take $f(x,y)=\sqrt{1+4x^2}$. Differentiate the curve and compute its speed:

\mathbf r'(t)=(1,2t), \qquad \left\|\mathbf r'(t)\right\| = \sqrt{1^2+(2t)^2} = \sqrt{1+4t^2}.

Along the curve, $x=t$, so $f(\mathbf r(t))=\sqrt{1+4t^2}$. The distance element is $ds=\sqrt{1+4t^2}\,dt$, giving

\int_C f\,ds = \int_0^1 \sqrt{1+4t^2}\sqrt{1+4t^2}\,dt = \int_0^1(1+4t^2)\,dt = \left[t+\frac{4}{3}t^3\right]_0^1 = 1+\frac{4}{3} = \frac{7}{3}.

For a check, the derivative of $t+\frac{4}{3}t^3$ is $1+4t^2$, which is the simplified integrand, and evaluating the antiderivative at the endpoints gives $\frac{7}{3}$.

Worked Example: A Scalar Field Along a Space Curve

Let $\mathbf r(t)=(t,2t,2t)$ for $0\leq t\leq1$, and let $f(x,y,z)=x+y-z$. The derivative and speed are

\mathbf r'(t)=(1,2,2), \qquad \left\|\mathbf r'(t)\right\| = \sqrt{1^2+2^2+2^2} = \sqrt{1+4+4} = 3.

Substitution into the scalar field gives $f(\mathbf r(t))=t+2t-2t=t$. Therefore

\int_C f\,ds = \int_0^1 t(3)\,dt = \int_0^1 3t\,dt = \left[\frac{3}{2}t^2\right]_0^1 = \frac{3}{2}-0 = \frac{3}{2}.

The check is $\frac{d}{dt}(\frac{3}{2}t^2)=3t$, exactly the parameter integrand, and the endpoint evaluation yields $\frac{3}{2}$.

A Common Mistake: Omitting the Speed

A frequent error is to write $\int_a^b f(\mathbf r(t))\,dt$ when the quantity being integrated is meant to accumulate per unit of curve length. That expression weights by parameter change, not by distance.

For the first example, $f(\mathbf r(t))=3t$. Omitting the speed would give the wrong calculation

\int_0^2 3t\,dt = \left[\frac{3}{2}t^2\right]_0^2 = 6.

But the curve moves at speed $5$, so each parameter increment $dt$ corresponds to $5\,dt$ units of distance. The distance-weighted calculation is $\int_0^2 3t\cdot5\,dt=30$, not $6$. The correction is to include $\|\mathbf r'(t)\|$ whenever converting a scalar line integral into a parameter integral.

The Geometric Meaning

The formula $ds=\|\mathbf r'(t)\|\,dt$ is a local conversion rule. It says that the length covered over a tiny parameter increment is approximately the speed at that point multiplied by the increment. Summing these local distances produces the total arc length.

For a scalar line integral, each local field value is multiplied by that same distance increment. The speed factor therefore does not change the field; it changes the measure over which the field is accumulated. This is why the parameterized formula represents accumulation over physical distance along the curve.

Check Your Understanding

Use the definitions and derivation above to answer these questions.

What limit of short chord lengths gives the speed of a parametrized curve?
For $\mathbf r(t)=(t,2t)$, what is $\|\mathbf r'(t)\|$, and what is $ds$?
State the condition on $\mathbf r$ in the arc-length theorem.
Why does a scalar line integral include $\|\mathbf r'(t)\|$ in its parameter formula?
If coordinates are measured in metres and $t$ in seconds, what are the units of $\|\mathbf r'(t)\|$ and $ds$?