Regularity at Points and Arc-Length Reparametrization

Why Regularity Matters for Arc Length

A curve can stop momentarily under a particular parametrization even when its geometric path is smooth. Conversely, a parametrization can fail at a point because the curve itself has a sharp feature that cannot be smoothed away by changing parameters. Distinguishing these situations is important when using arc length as a new parameter.

The key quantity is velocity: the derivative of the position vector with respect to the parameter. If velocity is nonzero, the curve is regular at that parameter value. Regularity ensures that the cumulative arc-length function has a nonzero derivative, so it has a differentiable local inverse. When velocity vanishes, that standard argument no longer applies. But this failure alone does not settle whether another parametrization of the same path can be regular.

Definition: Let \(\mathbf{r}:I\to\mathbb{R}^n\) be a differentiable curve on an interval \(I\). The curve is regular at \(t_0\in I\) if \(\mathbf{r}'(t_0)\neq\mathbf{0}\), and it is regular on \(I\) if \(\mathbf{r}'(t)\neq\mathbf{0}\) for every \(t\in I\). A parameter value where \(\mathbf{r}'(t_0)=\mathbf{0}\) is called a singular parameter value of that parametrization.

The speed at \(t\) is \(\lVert\mathbf{r}'(t)\rVert\). Thus, regularity at \(t_0\) is equivalent to positive speed there. This condition concerns the parametrization: it says that the position is changing to first order as the parameter changes. It does not, by itself, describe whether the geometric trace has a corner, cusp, or other feature.

Arc Length as a New Parameter

Choose a reference parameter \(t_0\), and define the signed cumulative arc-length function by integrating the speed:

s(t)=\int_{t_0}^{t}\lVert\mathbf{r}'(u)\rVert\,du.

For increasing \(t\), this function records distance traveled from \(t_0\); for \(t<t_0\), the integral is signed. The fundamental theorem of calculus gives \(s'(t)=\lVert\mathbf{r}'(t)\rVert\) when the speed is continuous. If the curve is regular near \(t_0\), then \(s'(t_0)>0\). The inverse function theorem therefore gives a differentiable local inverse \(t=t(s)\). Substituting that inverse into the original curve gives an arc-length reparametrization.

Theorem: If \(\mathbf{r}\) is continuously differentiable and regular near \(t_0\), then its cumulative arc-length function has a differentiable local inverse near \(s(t_0)\). The reparametrized curve \(\mathbf{q}(s)=\mathbf{r}(t(s))\) has unit speed there.

Here is the justification for the unit-speed conclusion. Since \(s(t(s))=s\), differentiating gives \(s'(t(s))t'(s)=1\). The chain rule then yields:

\mathbf{q}'(s) =\mathbf{r}'(t(s))t'(s) =\frac{\mathbf{r}'(t(s))}{\lVert\mathbf{r}'(t(s))\rVert}, \qquad \lVert\mathbf{q}'(s)\rVert=1.

The denominator is nonzero precisely where regularity holds. This argument explains why regularity is needed for the standard local inverse construction: it makes the arc-length function’s derivative positive and permits differentiation of the inverse. If the original velocity vanishes, the construction cannot be justified by this argument at that point. Whether the curve can nevertheless be reparametrized must be checked separately.

Finding the Singular Parameter Values

To test regularity, differentiate each component of \(\mathbf{r}(t)\). A parameter value is singular only when all components of \(\mathbf{r}'(t)\) are zero at once. Finding a zero in just one component is not enough. The following example shows the test and its consequence for arc length.

Worked Example: Locate the Singular Value

Consider \(\mathbf{r}(t)=(t^3,2t^2)\). Differentiate each component:

\mathbf{r}'(t)=(3t^2,4t).

For the velocity to be zero, both \(3t^2=0\) and \(4t=0\) must hold. Each equation gives \(t=0\), so the only singular parameter value is \(t=0\). The speed is:

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

At \(t=0\), the speed is \(0\). At \(t=1\), the velocity is \((3,4)\), whose norm is \(5\); at \(t=-1\), it is \((3,-4)\), also of norm \(5\). Thus the curve is regular at these two values but not at \(0\). For any reference value, the derivative of its cumulative arc-length function at \(0\) is the speed there, namely \(s'(0)=0\). The usual inverse-function argument therefore does not give a differentiable local inverse at that value.

This calculation identifies where the standard method loses its guarantee. It does not yet prove whether some other parametrization can pass regularly through the corresponding point.

A Singular Parameter Can Be Repaired

A zero velocity may reflect a poor choice of parameter rather than a defect in the path. If the same path can be described with a parameter that moves through it at nonzero speed, then the singularity belongs to the original description, not to the geometry. Arc length can sometimes reveal that replacement directly.

Worked Example: Reparametrize a Stopped Straight-Line Motion

Let \(\mathbf{r}(t)=(t^3,0)\), using \(t_0=0\). Its velocity and speed are:

\mathbf{r}'(t)=(3t^2,0), \qquad \lVert\mathbf{r}'(t)\rVert=3t^2.

The velocity vanishes at \(t=0\). Compute signed cumulative arc length from \(0\):

s(t)=\int_0^t 3u^2\,du =\left[u^3\right]_0^t =t^3.

This function is strictly increasing, so \(t=\sqrt[3]{s}\). Substitute this expression into the original curve:

\mathbf{q}(s) =\mathbf{r}\!\left(\sqrt[3]{s}\right) =\left((\sqrt[3]{s})^3,0\right) =(s,0).

Check the replacement: \(\mathbf{q}(s(t))=(t^3,0)=\mathbf{r}(t)\), so it describes the same motion in the same direction. Also, \(\mathbf{q}'(s)=(1,0)\), and \(\lVert\mathbf{q}'(s)\rVert=1\) for every \(s\), including \(s=0\). Thus this parametrization passes regularly through the point where the original velocity vanished. The inverse \(t=\sqrt[3]{s}\) is not differentiable at \(s=0\), but the composed curve \(\mathbf{q}\) is differentiable there. The composition, not the inverse considered in isolation, is what matters for the replacement.

This example also shows why it would be incorrect to conclude that a singular parameter makes any arc-length description impossible. The standard inverse function theorem cannot be used at the singular value, but a replacement may still exist and may be perfectly regular.

When the Geometric Path Is the Obstruction

Not every singularity can be removed. A cusp gives a contrasting case. Its two branches approach the singular point in the same direction, whereas a differentiable regular parametrization through an interior point must approach from opposite directions on the two sides of its parameter.

Worked Example: A Cusp That Has No Regular Passage Through Its Point

Consider the curve \(\mathbf{r}(t)=(t^2,t^3)\) near \(t=0\). Its velocity and speed are:

\mathbf{r}'(t)=(2t,3t^2), \qquad \lVert\mathbf{r}'(t)\rVert =\sqrt{4t^2+9t^4} =|t|\sqrt{4+9t^2}.

Both velocity components vanish at \(t=0\), and the speed is zero there. To test whether a regular reparametrization can pass through the origin, suppose there were a differentiable parametrization \(\mathbf{q}(u)\) of both nearby branches with \(\mathbf{q}(0)=(0,0)\) and \(\mathbf{q}'(0)=\mathbf{v}\neq\mathbf{0}\).

Differentiability implies \(\mathbf{q}(u)=u\mathbf{v}+o(|u|)\) as \(u\to0\). Consequently, the directions of \(\mathbf{q}(u)\) from the origin approach \(\mathbf{v}/\lVert\mathbf{v}\rVert\) for \(u>0\), and \(-\mathbf{v}/\lVert\mathbf{v}\rVert\) for \(u<0\). These are opposite directions.

But every nonzero point on this cusp has first coordinate \(x=t^2>0\), and its direction from the origin approaches the positive horizontal direction on both branches. Indeed, \(\mathbf{r}(t)/\lVert\mathbf{r}(t)\rVert\to(1,0)\) as \(t\to0\) from either side. The two directions cannot simultaneously be the opposite limits required by a regular differentiable parametrization. This contradiction proves that no such regular parametrization can pass through the cusp while covering both branches.

The obstruction is geometric: the path arrives at and leaves the origin along the same ray. A change of parameter cannot turn those two same-side approaches into the opposite-side approaches required for a regular passage through an interior parameter value.

What the Arc-Length Function Can and Cannot Tell You

The speed is nonnegative, so the cumulative arc-length function is nondecreasing. More precisely, if the speed is continuous and is not identically zero on any subinterval, then \(s(t)\) is strictly increasing: for \(a<b\), continuity and nonnegativity imply \(\int_a^b\lVert\mathbf{r}'(u)\rVert\,du>0\). Strict increase gives a continuous inverse on the range, but it does not guarantee a differentiable inverse at points where the speed is zero.

If the velocity vanishes throughout a whole interval, the arc-length function is constant there. It cannot distinguish different parameter values in that interval, so it has no inverse there. An isolated zero is different: the function may remain strictly increasing, as it did for \(s(t)=t^3\), even though its derivative vanishes at that point. Strict increase alone therefore does not establish regularity of the inverse or of the resulting parametrization.

Differentiate the curve.
Find all parameter values at which every component of \(\mathbf{r}'(t)\) vanishes.

Check the arc-length function.
Positive speed permits the usual differentiable local inverse argument. At zero speed, that argument no longer applies.

Test for a replacement.
Ask whether the path admits a differentiable regular parametrization through the point, rather than inferring the answer from the original velocity alone.

Common Mistake: Treating a Failed Test as a Proof of Impossibility

A tempting but invalid conclusion is: “The speed is zero at \(t_0\), so the arc-length function has no usable inverse and the curve cannot have a regular parametrization.” The first part overstates what the derivative test shows; the second does not follow from it. For \(\mathbf{r}(t)=(t^3,0)\), for example, the calculation \(s'(0)=0\) is correct, but the conclusion that no regular replacement exists is false: \(\mathbf{q}(s)=(s,0)\) is a regular unit-speed parametrization of the same path.

The correction is to separate two questions. First, does the original parametrization have positive speed? If yes, the standard local arc-length inversion works. Second, if it does not, can the path itself be described regularly through that point? The straight-line example answers yes; the cusp example answers no for a passage covering both branches. In either case, the reasoning must concern the path and the required differentiability—not just the fact that one derivative vanishes.

Why This Distinction Is Useful

Arc-length parameters are useful because they measure distance traveled directly and give the resulting curve unit speed wherever the reparametrization is differentiable and regular. Checking regularity first tells you when the standard construction is justified. Checking singular points separately prevents two opposite errors: rejecting a smooth path because its original parameter stalls, or assuming that a change of parameter can remove a genuine cusp.

In practical work, start by solving \(\mathbf{r}'(t)=\mathbf{0}\) component by component. At each solution, compute or inspect the speed and the local shape of the path. If proposing a replacement, verify that it describes the intended portion of the same path, is differentiable through the point, and has nonzero velocity there. These checks distinguish a valid regularization from a formula that only works away from the singular value.

Check Your Understanding

Use the definitions and examples above to answer these questions.

For \(\mathbf{r}(t)=(t^4,t^3)\), find every parameter value at which the velocity vanishes.
Why does positive speed allow the cumulative arc-length function to have a differentiable local inverse?
For \(\mathbf{r}(t)=(t^3,0)\), verify that the arc-length reparametrization has unit speed at the singular point of the original parametrization.
Why does strict increase of the arc-length function not, by itself, guarantee a differentiable inverse?
What feature of the cusp \(\mathbf{r}(t)=(t^2,t^3)\) prevents a regular parametrization from passing through the origin while covering both branches?