Constructing An Arc-Length Parameter

Why Use Arc Length as a Parameter?

A parameter \(t\) describes where a curve is, but it need not measure distance along the curve. Equal changes in \(t\) can correspond to unequal distances traveled. When the speed is positive, accumulated arc length provides another parameter: the distance traveled from the starting point. The resulting parameter increases at a constant rate of one unit of distance per unit of parameter.

The construction has two parts. First, accumulate the speed to obtain a function \(s(t)\). Then invert that function to express the original parameter \(t\) in terms of \(s\). Replacing \(t\) by this inverse gives a new position function. The chain rule explains why its speed is exactly one.

Accumulated Arc Length

Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be a continuously differentiable parametrized curve. Its speed at parameter value \(t\) is the magnitude of its derivative, \(\|\mathbf{r}'(t)\|\), where \(\|\cdot\|\) is the Euclidean norm. The accumulated arc length from \(a\) to \(t\) is the integral of that speed.

Definition: The arc-length function measured from \(a\) is \(s(t)=\int_a^t \|\mathbf{r}'(u)\|\,du\). Its value \(s(t)\) is the distance traveled along the curve from parameter \(a\) to parameter \(t\).

By the Fundamental Theorem of Calculus, \(s'(t)=\|\mathbf{r}'(t)\|\) when the speed is continuous. If the speed is positive throughout the interval, then \(s'(t)>0\), so \(s(t)\) is strictly increasing. It therefore has an inverse: each accumulated length corresponds to exactly one original parameter value.

Condition for the construction: Suppose \(\mathbf{r}\) is continuously differentiable on \([a,b]\) and \(\|\mathbf{r}'(t)\|>0\) for every \(t\in[a,b]\). Then \(s(t)\) is strictly increasing from \(s(a)=0\) to \(s(b)=L\), where \(L\) is the curve’s length on the interval. Thus \(s\) has an inverse \(t=t(s)\) on \([0,L]\).

Since \(s'(t)=\|\mathbf{r}'(t)\|\) is positive, the inverse function theorem gives, for interior values of \(s\),

\frac{dt}{ds} = \frac{1}{s'(t)} = \frac{1}{\|\mathbf{r}'(t)\|}.

At the endpoints, the same relationship is interpreted using one-sided derivatives. The new position function is obtained by substituting the inverse parameter into the original curve.

Definition: The arc-length reparameterization of \(\mathbf{r}\) from \(a\) is \(\mathbf{R}:[0,L]\to\mathbb{R}^n\), defined by \(\mathbf{R}(s)=\mathbf{r}(t(s))\), where \(t(s)\) is the inverse of the accumulated arc-length function.

Why the Reparameterized Curve Has Unit Speed

The key fact is a direct application of the chain rule. Differentiating \(\mathbf{R}(s)=\mathbf{r}(t(s))\) gives the velocity of the new position function. Its magnitude combines the original speed with the rate at which the original parameter changes with \(s\).

Theorem: If \(\mathbf{r}\) is continuously differentiable and has positive speed on \([a,b]\), then its arc-length reparameterization satisfies \(\|\mathbf{R}'(s)\|=1\) for \(0<s<L\), with the corresponding one-sided result at the endpoints.

Differentiate the accumulated length.
By the Fundamental Theorem of Calculus, \(ds/dt=\|\mathbf{r}'(t)\|\).

Differentiate the inverse.
Since the speed is positive, \(dt/ds=1/\|\mathbf{r}'(t)\|\).

Apply the chain rule.
For \(\mathbf{R}(s)=\mathbf{r}(t(s))\), the derivative is \(\mathbf{R}'(s)=\mathbf{r}'(t(s))\,dt/ds\).

Taking magnitudes in the last equation proves the theorem:

\|\mathbf{R}'(s)\| = \|\mathbf{r}'(t(s))\| \left|\frac{dt}{ds}\right| = \|\mathbf{r}'(t(s))\| \frac{1}{\|\mathbf{r}'(t(s))\|} = 1.

The absolute value is appropriate because speed is a magnitude. Here \(dt/ds\) is positive, since both \(s(t)\) and its inverse are increasing. The calculation shows precisely why the inverse parameter matters: it cancels the original speed.

Worked Examples

Worked Example: A Line Traversed at Constant Speed

Let \(\mathbf{r}(t)=(3t,4t)\) for \(0\leq t\leq2\). Find its arc-length reparameterization and check its speed.

Differentiate the position function and calculate its speed:

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

The speed is positive. The accumulated arc length from \(0\) to \(t\) is

s(t)=\int_0^t 5\,du=5t.

At \(t=2\), the total length is \(L=5(2)=10\). Inverting \(s=5t\) gives \(t(s)=s/5\), for \(0\leq s\leq10\). Substitute into the position function:

\mathbf{R}(s)=\mathbf{r}(s/5) = \left(\frac{3s}{5},\frac{4s}{5}\right).

Differentiate and check:

\mathbf{R}'(s)=\left(\frac35,\frac45\right), \qquad \|\mathbf{R}'(s)\| = \sqrt{\left(\frac35\right)^2+\left(\frac45\right)^2} = \sqrt{\frac{9+16}{25}} = 1.

The parameter \(s\) runs from \(0\) to \(10\), and the reparameterized curve has unit speed throughout.

Worked Example: A Circle Traversed at Variable Speed

Let \(\mathbf{r}(t)=(\cos(t^2),\sin(t^2))\) for \(1\leq t\leq2\). Find the arc-length reparameterization.

Differentiate both coordinates:

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

Its speed is

\|\mathbf{r}'(t)\| = \sqrt{4t^2\sin^2(t^2)+4t^2\cos^2(t^2)} = \sqrt{4t^2} = 2t.

Because \(t\geq1\), this speed is positive. Accumulating length from \(t=1\) gives

s(t)=\int_1^t 2u\,du = \left[u^2\right]_1^t = t^2-1.

Thus \(L=s(2)=4-1=3\), and \(t(s)=\sqrt{1+s}\) for \(0\leq s\leq3\). Since \(t(s)^2=1+s\), substitution yields

\mathbf{R}(s) = \bigl(\cos(1+s),\sin(1+s)\bigr).

Differentiate and verify the speed:

\mathbf{R}'(s) = \bigl(-\sin(1+s),\cos(1+s)\bigr), \qquad \|\mathbf{R}'(s)\| = \sqrt{\sin^2(1+s)+\cos^2(1+s)} = 1.

The original parameter changes the rate at which the circle is traversed, but the arc-length parameter makes the speed one.

Worked Example: A Nonlinear Parameter on a Straight Path

Let \(\mathbf{r}(t)=(t^2,t^2)\) for \(1\leq t\leq3\). Construct its arc-length reparameterization.

First compute the derivative and speed:

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

where the last equality uses \(t>0\) on the interval. The speed is positive. Accumulating length from \(t=1\) gives

s(t) = \int_1^t 2\sqrt{2}\,u\,du = \sqrt{2}(t^2-1).

At \(t=3\), the length is \(L=\sqrt{2}(9-1)=8\sqrt{2}\). Solve \(s=\sqrt{2}(t^2-1)\) for \(t\):

\frac{s}{\sqrt{2}}=t^2-1, \qquad t(s)=\sqrt{1+\frac{s}{\sqrt{2}}}.

Substitute this into \(\mathbf{r}\). Since \(t(s)^2=1+s/\sqrt{2}\),

\mathbf{R}(s) = \left(1+\frac{s}{\sqrt{2}},1+\frac{s}{\sqrt{2}}\right), \qquad 0\leq s\leq8\sqrt{2}.

Now differentiate and check:

\mathbf{R}'(s)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right), \qquad \|\mathbf{R}'(s)\| = \sqrt{\frac12+\frac12} = 1.

Although the original parameter does not advance at a constant rate along the line, the new parameter measures distance traveled.

A Common Mistake: Substituting Without Differentiating the Inverse

A frequent error is to take the original velocity and merely replace \(t\) by \(s\). But the reparameterized function is a composition, \(\mathbf{r}(t(s))\), so its derivative must include the factor \(dt/ds\). The circle example makes the error visible. There \(t(s)=\sqrt{1+s}\). At \(s=1\), the corresponding original parameter is \(t=\sqrt{2}\), and the original speed is \(2\sqrt{2}\), not one.

If one incorrectly treats the new derivative as just the old derivative evaluated at \(t(s)\), its magnitude at \(s=1\) would be \(2\sqrt{2}\). The omitted factor is

\frac{dt}{ds} = \frac{1}{2\sqrt{1+s}}, \qquad \left.\frac{dt}{ds}\right|_{s=1} = \frac{1}{2\sqrt{2}}.

Including it gives the correct speed:

\|\mathbf{R}'(1)\| = (2\sqrt{2})\left(\frac{1}{2\sqrt{2}}\right) = 1.

Always differentiate the composition by the chain rule. Simply substituting the new parameter into the old speed does not account for how quickly the original parameter changes.

When This Construction Applies

The positive-speed condition has a specific role. It ensures that \(s'(t)>0\), so the accumulated length is strictly increasing and can be inverted over the whole interval. If the speed is zero somewhere, this argument does not guarantee such an inverse. Therefore, before constructing an arc-length parameter by this method, check the speed on the entire interval—not just at its endpoints.

Once the inverse exists, the new parameter starts at \(s=0\), ends at \(s=L\), and describes the original curve in the same direction. Its derivative has magnitude one because each unit increase in \(s\) corresponds to one unit of distance along the curve. This makes arc length a useful parameter whenever the rate of travel in the original parameter is uneven.

Check Your Understanding

Use the definitions and chain-rule argument above to answer these questions.

For a continuously differentiable curve, what integral defines accumulated arc length from a starting parameter \(a\)?
Why does positive speed ensure that the accumulated arc-length function has an inverse?
If \(s'(t)=\|\mathbf{r}'(t)\|>0\), what is \(dt/ds\) in terms of the original speed?
For \(\mathbf{r}(t)=(t,2t)\) on \(0\leq t\leq1\), find \(s(t)\), the total length, and \(t(s)\).
Which factor is missing if one differentiates \(\mathbf{r}(t(s))\) as though its derivative were just \(\mathbf{r}'(t(s))\)?