Inverting Arc Length to Build a New Parameter

Why Invert Arc Length?

A curve is often given as a vector-valued function \(\mathbf{r}(t)\), where \(t\) is a parameter that describes position. The parameter might represent time, but it need not measure distance traveled. If the curve is traversed at varying speed, equal changes in \(t\) generally correspond to unequal distances along the curve.

To use distance traveled as the parameter, first calculate the cumulative arc length from a chosen starting parameter. If this cumulative length increases strictly, each distance value corresponds to exactly one original parameter value. We can then invert the cumulative-length function and substitute that inverse into \(\mathbf{r}(t)\).

Core construction: Find the cumulative arc-length function \(s(t)\), solve \(s=s(t)\) for \(t\) as a function of \(s\), and substitute that function into the original curve.

Cumulative Arc Length and Its Inverse

Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be continuously differentiable. Its speed at parameter \(t\) is \(\|\mathbf{r}'(t)\|\). The cumulative arc length measured from \(t=a\) is the function \(s:[a,b]\to\mathbb{R}\) defined by integrating this speed.

s(t)=\int_a^t \|\mathbf{r}'(u)\|\,du.

The variable \(u\) is a dummy variable of integration; the resulting value \(s(t)\) gives the length of the curve from parameter \(a\) to parameter \(t\). In particular, \(s(a)=0\), and the total length on the interval is \(L=s(b)\).

Definition: If \(s(t)\) is strictly increasing on \([a,b]\), its inverse function \(t=t(s)\) is defined for \(s\in[0,L]\) by the requirement \(s(t(s))=s\). The curve expressed in this new parameter is \(\mathbf{R}(s)=\mathbf{r}(t(s))\).

The inverse exists because a strictly increasing function takes each value in its range exactly once. For a continuously differentiable curve, a convenient sufficient condition is \(\|\mathbf{r}'(t)\|>0\) throughout \([a,b]\). More generally, strict increase holds whenever the integral of the speed over every nontrivial parameter subinterval is positive. An isolated parameter value where speed is zero does not by itself prevent strict increase.

Why the Inverse Exists

Suppose \(t_1<t_2\). The difference between the cumulative lengths at those parameters is

s(t_2)-s(t_1) = \int_{t_1}^{t_2}\|\mathbf{r}'(u)\|\,du.

Since speed is nonnegative, this difference cannot be negative. If speed is positive throughout the interval from \(t_1\) to \(t_2\), the integral is positive, so \(s(t_2)>s(t_1)\). Thus \(s(t)\) is strictly increasing. A continuous strictly increasing function on a closed interval has an inverse on its range, so each cumulative length from \(0\) to \(L\) identifies a unique original parameter value.

There is also a useful derivative formula when speed is positive. By the Fundamental Theorem of Calculus, \(ds/dt=\|\mathbf{r}'(t)\|\). Differentiating the identity \(s(t(s))=s\) with respect to \(s\) gives

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

This formula applies where the inverse is differentiable, in particular when the original speed is nonzero. It describes how the original parameter changes with cumulative distance. To construct the reparametrized curve itself, use \(\mathbf{R}(s)=\mathbf{r}(t(s))\).

Worked Examples

Worked Example: Reparametrizing a Circle Traversed at Constant Speed

Let \(\mathbf{r}(t)=(3\cos t,3\sin t)\) for \(0\leq t\leq\pi/2\). Find the curve in terms of cumulative arc length from \(t=0\).

Differentiate each component:

\mathbf{r}'(t)=(-3\sin t,3\cos t).

The speed is therefore

\|\mathbf{r}'(t)\| = \sqrt{9\sin^2t+9\cos^2t} = \sqrt{9(\sin^2t+\cos^2t)} = 3.

Integrating from \(0\) to \(t\) gives

s(t)=\int_0^t 3\,du=3t.

Solving \(s=3t\) for \(t\) gives \(t(s)=s/3\). At the endpoint, \(L=3(\pi/2)=3\pi/2\), so \(0\leq s\leq3\pi/2\). Substituting the inverse into both components of \(\mathbf{r}\) gives

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

Check the inverse by substitution: \(s(t(s))=3(s/3)=s\). At \(s=0\), the new formula gives \((3,0)=\mathbf{r}(0)\); at \(s=3\pi/2\), it gives \((0,3)=\mathbf{r}(\pi/2)\). Thus it covers the original curve segment with the correct endpoints.

Worked Example: Inverting a Quadratic Cumulative-Length Function

Let \(\mathbf{r}(t)=(t^2+t,0)\) for \(0\leq t\leq2\). Find its parametrization by cumulative arc length from \(t=0\).

Differentiate and compute the speed:

\mathbf{r}'(t)=(2t+1,0), \qquad \|\mathbf{r}'(t)\|=\sqrt{(2t+1)^2}=2t+1,

where \(2t+1>0\) on the stated interval. Hence

s(t)=\int_0^t(2u+1)\,du = \left[u^2+u\right]_0^t = t^2+t.

To invert this relation, solve \(t^2+t-s=0\) using the quadratic formula:

t=\frac{-1\pm\sqrt{1+4s}}{2}.

Since the original parameter satisfies \(t\geq0\), choose the positive-root expression, which gives \(t(0)=0\):

t(s)=\frac{-1+\sqrt{1+4s}}{2}.

The total length is \(L=s(2)=2^2+2=6\). Substitute the inverse into the curve:

\mathbf{R}(s) = \left(t(s)^2+t(s),0\right) = (s,0), \qquad 0\leq s\leq6.

For a direct check, set \(t=(-1+\sqrt{1+4s})/2\). Then \(2t+1=\sqrt{1+4s}\), and

t^2+t = \frac{(2t+1)^2-1}{4} = \frac{(1+4s)-1}{4} = s.

Thus the inverse satisfies the cumulative-length equation, and substitution produces the stated curve.

Worked Example: Undoing a Nonlinear Angle Parameter

Consider \(\mathbf{r}(t)=(\cos(t^2),\sin(t^2))\) for \(1\leq t\leq2\). Find the parametrization by cumulative arc length measured from \(t=1\).

Differentiating gives

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

Because \(t>0\) on this interval, the speed is

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

Integrate from the starting parameter \(1\):

s(t)=\int_1^t2u\,du = [t^2]_1^t = t^2-1.

Solving \(s=t^2-1\) gives \(t^2=s+1\). Since \(t\geq1\), take the positive square root, so \(t(s)=\sqrt{s+1}\). The total length is \(s(2)=4-1=3\), and therefore \(0\leq s\leq3\). Substitution yields

\mathbf{R}(s) = \left(\cos\bigl((\sqrt{s+1})^2\bigr), \sin\bigl((\sqrt{s+1})^2\bigr)\right) = (\cos(s+1),\sin(s+1)).

Check by substitution: \(s(t(s))=(\sqrt{s+1})^2-1=s\). The endpoints also agree: \(s=0\) gives \((\cos1,\sin1)=\mathbf{r}(1)\), and \(s=3\) gives \((\cos4,\sin4)=\mathbf{r}(2)\).

When an Explicit Inverse Is Not Available

The cumulative-length function may be strictly increasing even when its inverse cannot be written using familiar elementary formulas. In that situation, the inverse is still well-defined: for each \(s\) in the length range, \(t(s)\) is the unique solution of \(s(t)=s\). The new parametrization remains \(\mathbf{R}(s)=\mathbf{r}(t(s))\), even if evaluating it requires an approximate solution.

Before attempting to invert, check the parameter interval and the starting point used in the integral. The lower limit determines where cumulative length is zero. Changing that starting point changes the length coordinate, and therefore changes the inverse formula and the interval for \(s\).

Common Mistake: Inverting the Wrong Function

For the circle in the first example, the speed is \(3\), so the cumulative length is \(s(t)=3t\), not \(s(t)=t\). If we incorrectly use \(t=s\), then at \(s=\pi/2\) we obtain \(\mathbf{r}(\pi/2)=(0,3)\). But \(s=\pi/2\) corresponds to \(t=(\pi/2)/3=\pi/6\), whose point is \((3\sqrt{3}/2,3/2)\). The mistaken substitution gives a different point.

The correction is to use the inverse of the actual cumulative-length function, not the original parameter itself unless the cumulative length really equals that parameter. Always compute the speed, integrate it with the chosen lower limit, solve for \(t\), and then substitute \(t(s)\) into every component of \(\mathbf{r}\).

Compute speed.
Find \(\|\mathbf{r}'(t)\|\) on the specified parameter interval.

Build cumulative length.
Integrate speed from the chosen starting parameter to \(t\).

Check monotonicity and range.
Confirm that each length value corresponds to one original parameter value.

Invert and substitute.
Solve for \(t(s)\), then form \(\mathbf{R}(s)=\mathbf{r}(t(s))\).

Verify.
Substitute \(t(s)\) into \(s(t)\) and check the endpoints of the new parameter interval.

Check Your Understanding

For each question, identify the cumulative-length or inverse step that is needed before rewriting the curve.

What integral defines cumulative arc length from a starting parameter \(a\)?
Why does positive speed throughout an interval guarantee that cumulative arc length is strictly increasing?
If \(s(t)=4t\) for \(0\leq t\leq2\), what are \(t(s)\) and the range of \(s\)?
For \(\mathbf{r}(t)=(t,t^2)\) on an interval where the speed is positive, what equation must be solved to construct \(t(s)\)?
What can go wrong if the inverse is substituted into only one component of a vector-valued curve?