Unit-Speed Reparametrizations

Why Reparametrize by Arc Length?

A curve can describe a geometric path while its parameter moves along that path at a changing rate. The derivative of a parametrized curve records that rate: its magnitude is the speed. Reparametrizing by distance traveled gives the parameter a direct geometric meaning and makes the speed exactly one.

The construction uses two functions. First, measure the arc length accumulated from a chosen starting point. Then use the inverse of that increasing function to recover the original parameter from the distance traveled. Composing the original curve with this inverse changes the parameter without changing the path.

Cumulative Arc Length and Its Inverse

Let \(\mathbf r:I\to\mathbb R^n\) be a continuously differentiable curve on an interval \(I\), and suppose it is regular on \(I\), meaning \(\mathbf r'(t)\ne\mathbf 0\) for every \(t\in I\). Its speed is \(\|\mathbf r'(t)\|\), which is therefore positive. Choose a starting parameter \(t_0\in I\). The cumulative arc-length function measures the distance traveled from \(t_0\) to \(t\).

Definition: The cumulative arc-length function based at \(t_0\) is \(s(t)=\int_{t_0}^{t}\|\mathbf r'(u)\|\,du\). Its values form an interval \(J=s(I)\). A unit-speed reparametrization is obtained by composing \(\mathbf r\) with the inverse function \(t=\tau(s)=s^{-1}(s)\), so that \(\mathbf R(s)=\mathbf r(\tau(s))\), for \(s\in J\).

Here, \(s\) denotes the new parameter, whose value is distance traveled from the chosen starting point. To avoid confusing the function \(s(t)\) with its input, it can help to write the inverse as \(\tau:J\to I\): it satisfies \(s(\tau(s))=s\). Since the derivative of the cumulative function is \(\|\mathbf r'(t)\|>0\), the function is strictly increasing and has an inverse on its image.

The inverse-function theorem also gives the derivative of \(\tau\). At interior points of the relevant intervals, differentiating \(s(\tau(s))=s\) gives \(s'(\tau(s))\tau'(s)=1\). Because \(s'(t)=\|\mathbf r'(t)\|\), it follows that \(\tau'(s)=1/\|\mathbf r'(\tau(s))\|\). This factor is what compensates for the original curve’s speed.

Theorem: If \(\mathbf r\) is continuously differentiable and regular on an interval \(I\), then the arc-length reparametrization \(\mathbf R(s)=\mathbf r(\tau(s))\), where \(\tau\) is the inverse of the cumulative arc-length function, has velocity of magnitude one at every interior point of its parameter interval.

Why the Construction Has Unit Speed

The result follows from the chain rule and the derivative of the inverse. The derivative of the composed curve is the original velocity multiplied by the rate at which the original parameter changes with respect to arc length.

Differentiate the cumulative length.
By the fundamental theorem of calculus, \(s'(t)=\|\mathbf r'(t)\|\). Regularity ensures this derivative is positive.

Differentiate the inverse relation.
From \(s(\tau(s))=s\), the chain rule gives \(\|\mathbf r'(\tau(s))\|\tau'(s)=1\), so \(\tau'(s)=1/\|\mathbf r'(\tau(s))\|\).

Differentiate the reparametrized curve.
Applying the chain rule to \(\mathbf R(s)=\mathbf r(\tau(s))\) gives \(\mathbf R'(s)=\mathbf r'(\tau(s))\tau'(s)\).

Take the magnitude.
Substituting the inverse derivative yields \(\|\mathbf R'(s)\|=\|\mathbf r'(\tau(s))\|/\|\mathbf r'(\tau(s))\|=1\).

This proves the theorem wherever the derivatives are defined. The positivity assumption matters: if the original velocity vanishes, the cumulative length need not have a differentiable inverse there, and the displayed reciprocal formula cannot be applied at that point.

Worked Examples

Worked Example: A Circle Traversed at Variable Speed

Consider \(\mathbf r(t)=(\cos(t^2),\sin(t^2))\) for \(1\le t\le2\). Find its arc-length reparametrization based at \(t_0=1\), and verify its speed.

First differentiate each coordinate:

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

The speed is

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

where \(t>0\) on this interval. Integrating from the starting value \(1\) gives \(s(t)=\int_1^t2u\,du=t^2-1\). Thus \(0\le s\le3\), and solving \(s=t^2-1\) for \(t\) gives \(\tau(s)=\sqrt{s+1}\); the positive root is required because \(t\ge1\). The new curve is

\mathbf R(s)=\mathbf r(\sqrt{s+1}) =(\cos(s+1),\,\sin(s+1)).

Differentiate and check its speed:

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

As a further check, substituting \(s=t^2-1\) into the new curve gives \(\mathbf R(s(t))=(\cos(t^2),\sin(t^2))=\mathbf r(t)\), so the same path is traced.

Worked Example: A Circle with Constant Original Speed

Let \(\mathbf r(t)=(2\cos(3t),2\sin(3t),0)\) for \(0\le t\le\pi/3\), based at \(t_0=0\). Its derivative and speed are

\mathbf r'(t)=(-6\sin(3t),\,6\cos(3t),\,0), \qquad \|\mathbf r'(t)\|=\sqrt{36\sin^2(3t)+36\cos^2(3t)}=6.

Therefore \(s(t)=\int_0^t6\,du=6t\). The new parameter ranges from \(0\) to \(2\pi\), and its inverse is \(\tau(s)=s/6\). Composing gives

\mathbf R(s)=\mathbf r(s/6) =(2\cos(s/2),\,2\sin(s/2),\,0).

Its derivative has components \((-\sin(s/2),\cos(s/2),0)\), so

\mathbf R'(s)=(-\sin(s/2),\,\cos(s/2),\,0), \qquad \|\mathbf R'(s)\|=\sqrt{\sin^2(s/2)+\cos^2(s/2)}=1.

Finally, \(\mathbf R(6t)=(2\cos(3t),2\sin(3t),0)=\mathbf r(t)\), confirming that the reparametrization preserves the curve.

Worked Example: A Curve Whose Inverse Need Not Be Explicit

Consider the plane curve \(\mathbf r(t)=(t,t^2)\) on \(0\le t\le1\), based at \(t_0=0\). We can construct its unit-speed parametrization without solving explicitly for the inverse arc-length function.

Differentiating gives \(\mathbf r'(t)=(1,2t)\), so the speed is \(\sqrt{1+4t^2}\), which is positive throughout the interval. Define

s(t)=\int_0^t\sqrt{1+4u^2}\,du.

This function is strictly increasing, so it has an inverse \(\tau(s)\). The new parametrization is \(\mathbf R(s)=(\tau(s),\tau(s)^2)\), where \(0\le s\le\int_0^1\sqrt{1+4u^2}\,du\). The inverse-function derivative is

\tau'(s)=\frac{1}{\sqrt{1+4\tau(s)^2}}.

Applying the chain rule to both coordinates gives

\mathbf R'(s) =\bigl(\tau'(s),\,2\tau(s)\tau'(s)\bigr) =\frac{(1,\,2\tau(s))}{\sqrt{1+4\tau(s)^2}}.

Hence the speed is

\|\mathbf R'(s)\| =\frac{\sqrt{1+4\tau(s)^2}}{\sqrt{1+4\tau(s)^2}} =1.

To check that this is the original curve with a new parameter, use \(\tau(s(t))=t\). Then \(\mathbf R(s(t))=(t,t^2)=\mathbf r(t)\). An explicit formula for \(\tau\) is not necessary: the inverse relation and its derivative are enough to define and verify the reparametrization.

A Common Chain-Rule Error

A frequent mistake is to compose with the inverse arc-length function but forget to differentiate that inverse. In the variable-speed circle example, the original derivative has magnitude \(2t\). The incorrect calculation would treat the new velocity as though it were just \(\mathbf r'(\tau(s))\), and would therefore claim that its magnitude is \(2\tau(s)=2\sqrt{s+1}\), not one.

The error is that \(\mathbf R(s)=\mathbf r(\tau(s))\) is a composition. Its derivative must include the factor \(\tau'(s)\). In this example, \(\tau'(s)=1/(2\sqrt{s+1})\), so the correct magnitude is \(2\sqrt{s+1}\cdot1/(2\sqrt{s+1})=1\). The inverse derivative cancels the original speed.

What the Construction Preserves

Arc-length reparametrization changes how quickly the curve is traced, not which points it traces or their order. In the examples, substituting \(s=s(t)\) into \(\mathbf R(s)\) recovers \(\mathbf r(t)\). The new parameter starts at zero because the cumulative length was based at \(t_0\); choosing a different additive constant for the length would shift the parameter values but would not change the unit-speed calculation.

For practical work, it is useful to separate the construction into three tasks: compute the original speed, integrate it to obtain cumulative length, and invert that function (explicitly if possible, implicitly otherwise). Then verify the result by differentiating the composition and taking the magnitude. The regularity condition is not merely technical: it ensures the length function is strictly increasing, so distance can serve as a parameter along the curve.

Check Your Understanding

Use the definition and chain-rule calculation to answer these questions.

For a regular curve \(\mathbf r(t)\) based at \(t_0\), what integral defines its cumulative arc length?
Why does regularity ensure that the cumulative arc-length function is strictly increasing?
If \(s(t)=\int_{t_0}^t\|\mathbf r'(u)\|\,du\) and \(\tau\) is its inverse, what is \(\tau'(s)\)?
For \(\mathbf r(t)=(\cos(t^2),\sin(t^2))\) with \(1\le t\le2\), what is the cumulative length based at \(1\), and what is its inverse?
Why is an explicit formula for the inverse arc-length function unnecessary when verifying unit speed?

What You'll Learn