Why Arc Length Is Invariant Under Reparametrization

Why the Parameter Should Not Change the Length

A parametrized curve can be described using different parameters. One parameter may move steadily along the curve, while another may move faster in some places, slower in others, or traverse the same path in the opposite direction. These changes can alter the velocity vector at each instant, but they do not necessarily alter the distance traveled.

The essential condition is that the parameter change is monotone: it keeps the traversal in the same order, or reverses that order throughout. The chain rule changes the velocity by a factor equal to the rate of change of parameter. Speed uses the magnitude of that factor, which is why the arc-length integral remains unchanged even when the direction is reversed.

Key idea: A monotone change of parameter may change how quickly a curve is traversed, but it does not change the total distance along that traversal.

Length and Reparametrization

Let $\mathbf{r}:[a,b]\to\mathbb{R}^n$ be a continuously differentiable curve. Its arc length on $[a,b]$ is defined by integrating its speed:

L=\int_a^b \|\mathbf{r}'(t)\|\,dt.

Here $\mathbf{r}'(t)$ is the velocity vector, and its norm $\|\mathbf{r}'(t)\|$ is the speed. This definition applies whether or not the curve is regular; at a point where the velocity is zero, the speed is simply zero.

Now let $\phi:[c,d]\to[a,b]$ be a continuously differentiable, strictly monotone function that maps the endpoints of its interval onto the endpoints of the original interval. The new parametrization is the composition

\widetilde{\mathbf{r}}(u)=\mathbf{r}(\phi(u)).

If $\phi$ is increasing, then $\phi(c)=a$ and $\phi(d)=b$. If it is decreasing, then $\phi(c)=b$ and $\phi(d)=a$. In either case, the new parameter covers the same parameter interval in one direction or the other.

By the chain rule, the velocity of the new parametrization is

\widetilde{\mathbf{r}}'(u) = \mathbf{r}'(\phi(u))\phi'(u).

Taking the norm gives its speed:

\|\widetilde{\mathbf{r}}'(u)\| = \|\mathbf{r}'(\phi(u))\phi'(u)\| = \|\mathbf{r}'(\phi(u))\|\,|\phi'(u)|.

The absolute value is crucial. A decreasing change of parameter has $\phi'(u)\leq 0$, but speed cannot be negative.

Definition: A monotone reparametrization of a curve is a composition $\widetilde{\mathbf{r}}=\mathbf{r}\circ\phi$, where $\phi$ is a continuously differentiable monotone map that covers the original parameter interval. The reparametrization preserves length if the arc-length integrals for the two parametrizations are equal.

The Invariance Result

Theorem: Let $\mathbf{r}:[a,b]\to\mathbb{R}^n$ be continuously differentiable. Suppose $\phi:[c,d]\to[a,b]$ is continuously differentiable, strictly monotone, and maps the endpoints of $[c,d]$ onto the endpoints of $[a,b]$. Then $\widetilde{\mathbf{r}}(u)=\mathbf{r}(\phi(u))$ has the same arc length as $\mathbf{r}$: $$ \int_c^d \|\widetilde{\mathbf{r}}'(u)\|\,du = \int_a^b \|\mathbf{r}'(t)\|\,dt. $$

The proof follows from the chain rule and one-variable substitution. It is useful to treat the increasing and decreasing cases separately, because the direction of the parameter change affects the integration limits.

Write the new speed.
With $g(t)=\|\mathbf{r}'(t)\|$, the chain rule gives $\|\widetilde{\mathbf{r}}'(u)\|=g(\phi(u))|\phi'(u)|$.

Use monotonicity.
If $\phi$ is increasing, $|\phi'|=\phi'$; if it is decreasing, $|\phi'|=-\phi'$.

Substitute the original parameter.
The one-variable substitution $t=\phi(u)$, with the corresponding endpoint order, makes the transformed length equal to $\int_a^b g(t)\,dt$.

For completeness, define $G(x)=\int_a^x g(t)\,dt$. Since $g$ is continuous, the Fundamental Theorem of Calculus gives $G'(x)=g(x)$. If $\phi$ is increasing, then $\phi'\geq 0$, and the chain rule gives

\int_c^d g(\phi(u))|\phi'(u)|\,du = \int_c^d g(\phi(u))\phi'(u)\,du = G(\phi(d))-G(\phi(c)) = G(b)-G(a) = \int_a^b g(t)\,dt.

If $\phi$ is decreasing, then $\phi'\leq 0$, so $|\phi'|=-\phi'$. Therefore,

\begin{aligned} \int_c^d g(\phi(u))|\phi'(u)|\,du &= -\int_c^d g(\phi(u))\phi'(u)\,du\\ &= -\bigl(G(\phi(d))-G(\phi(c))\bigr)\\ &= -\bigl(G(a)-G(b)\bigr)\\ &= G(b)-G(a)\\ &= \int_a^b g(t)\,dt. \end{aligned}

Thus the length is unchanged in either direction. The proof does not require $\phi'(u)$ to be nonzero at every point: isolated points or intervals where the parametrization momentarily stops do not add distance. The stated hypotheses ensure that the parameter still covers the original interval monotonically.

Worked Examples

Worked Example: A Nonlinear Change of Speed on a Line Segment

Consider $\mathbf{r}(t)=(t^2,0)$ for $0\leq t\leq 1$. Reparametrize it using $t=\phi(u)=u^3$, with $0\leq u\leq 1$. First compute the original velocity and speed:

\mathbf{r}'(t)=(2t,0), \qquad \|\mathbf{r}'(t)\|=\sqrt{(2t)^2+0^2}=2t \quad (0\leq t\leq1).

The original length is

L=\int_0^1 2t\,dt=[t^2]_0^1=1^2-0^2=1.

The reparametrized curve is $\widetilde{\mathbf{r}}(u)=((u^3)^2,0)=(u^6,0)$. Its velocity and speed are

\widetilde{\mathbf{r}}'(u)=(6u^5,0), \qquad \|\widetilde{\mathbf{r}}'(u)\|=\sqrt{(6u^5)^2}=6u^5 \quad (0\leq u\leq1).

Thus its length is

\widetilde L=\int_0^1 6u^5\,du=[u^6]_0^1=1^6-0^6=1.

As a direct check, $t=u^3$ gives $dt=3u^2\,du$, so the original integrand $2t\,dt$ becomes $2u^3(3u^2\,du)=6u^5\,du$, exactly the transformed speed times $du$. Both calculations give length $1$.

Worked Example: Reversing a Quarter-Circle

Let $\mathbf{r}(t)=(\cos t,\sin t)$ for $0\leq t\leq\frac{\pi}{2}$. The velocity and speed are

\mathbf{r}'(t)=(-\sin t,\cos t), \qquad \|\mathbf{r}'(t)\| = \sqrt{\sin^2t+\cos^2t} =1.

The original length is $\int_0^{\pi/2}1\,dt=\frac{\pi}{2}$. Now take the decreasing change of parameter $\phi(u)=\frac{\pi}{2}-u$, for $0\leq u\leq\frac{\pi}{2}$. It maps the new interval from $t=\frac{\pi}{2}$ back to $t=0$. The new curve is

\widetilde{\mathbf{r}}(u) = \left(\cos\left(\frac{\pi}{2}-u\right), \sin\left(\frac{\pi}{2}-u\right)\right) = (\sin u,\cos u).

Differentiate and calculate its speed:

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

Consequently, $\widetilde L=\int_0^{\pi/2}1\,du=\frac{\pi}{2}$, equal to the original length. The curve is traversed in reverse, but its distance is positive and unchanged.

Worked Example: A Nonlinear Reparametrization of a Helix

Consider the helix $\mathbf{r}(t)=(3\cos t,3\sin t,4t)$ for $0\leq t\leq2\pi$. Its velocity is $\mathbf{r}'(t)=(-3\sin t,3\cos t,4)$, so its speed is

\|\mathbf{r}'(t)\| = \sqrt{9\sin^2t+9\cos^2t+16} = \sqrt{9+16} =5.

Its length is therefore $L=\int_0^{2\pi}5\,dt=10\pi$. Now set $t=\phi(u)=u^2$ for $0\leq u\leq\sqrt{2\pi}$. This is increasing and maps the new interval onto $[0,2\pi]$. The reparametrized curve is $(3\cos(u^2),3\sin(u^2),4u^2)$, with velocity

\widetilde{\mathbf{r}}'(u) = (-6u\sin(u^2),6u\cos(u^2),8u).

Its speed is

\begin{aligned} \|\widetilde{\mathbf{r}}'(u)\| &=\sqrt{36u^2\sin^2(u^2)+36u^2\cos^2(u^2)+64u^2}\\ &=\sqrt{100u^2} =10u, \end{aligned}

because $u\geq0$. Thus $\widetilde L=\int_0^{\sqrt{2\pi}}10u\,du=[5u^2]_0^{\sqrt{2\pi}}=5(2\pi)-0=10\pi$, matching the original length. Substitution also confirms this: $dt=2u\,du$, so the original speed $5$ becomes $5(2u)=10u$.

Why Monotonicity Matters

A monotone change covers the original parameter interval in order, either forward or backward. If a change of parameter reverses direction partway through, the new parametrization can retrace part of the curve. The arc-length integral counts every traversal, so its value can then be larger than the length of the original single traversal.

For a concrete example, let $\mathbf{r}(t)=(t,0)$ on $0\leq t\leq1$, whose length is $\int_0^1 1\,dt=1$. Use the non-monotone parameter change $\phi(u)=u^2$ on $-1\leq u\leq1$. The new curve is $\widetilde{\mathbf{r}}(u)=(u^2,0)$, which moves from $t=1$ to $t=0$ and then back to $t=1$. Its speed is $|2u|$, and its length is

\int_{-1}^{1}|2u|\,du = \int_{-1}^{0}(-2u)\,du+\int_0^1 2u\,du = 1+1=2.

The segment is traversed twice, giving length $2$, not $1$. This does not contradict the invariance theorem: $\phi$ is not monotone on the whole interval.

A Common Sign Error

When reversing a curve, it is incorrect to use the signed parameter derivative as though it were the speed. In the quarter-circle example, $\phi'(u)=-1$. Omitting the absolute value would incorrectly produce

\int_0^{\pi/2} 1\cdot(-1)\,du=-\frac{\pi}{2}.

That is a signed change in parameter, not a distance. The speed of the reparametrized curve is $\|\mathbf{r}'(\phi(u))\|\,|\phi'(u)|=1\cdot1=1$, so the correct length is $\int_0^{\pi/2}1\,du=\frac{\pi}{2}$. Always take the norm of the transformed velocity, or equivalently use the absolute value in the speed formula.

What the Invariance Does—and Does Not—Say

Reparametrization invariance means the speed profile may change while the total length of a monotone traversal stays fixed. A parameter that advances quickly contributes more distance per unit parameter, while one that advances slowly contributes less; the chain-rule factor and substitution balance each other exactly.

The result concerns the length of the traversal, not merely the set of points in its image. If a curve intersects itself, the length integral counts each portion as often as it is traversed. A monotone change preserves that order and multiplicity; a change that retraces part of the path may increase the length.

Takeaway: For a monotone parameter change, the transformed speed is $\|\mathbf{r}'(\phi(u))\|\,|\phi'(u)|$. Substitution converts its integral to the original arc-length integral, whether the change preserves or reverses direction.

Check Your Understanding

Use the chain rule, the speed formula, and parameter substitution to answer the following.

If $\widetilde{\mathbf{r}}(u)=\mathbf{r}(\phi(u))$, what is the formula for $\|\widetilde{\mathbf{r}}'(u)\|$, and why does it include an absolute value?
For an increasing change of parameter, how do the endpoints of the new parameter interval correspond to those of the original interval?
Why does a decreasing change of parameter preserve length even though its derivative is nonpositive?
For $\mathbf{r}(t)=(t,0)$, $0\leq t\leq1$, explain why $\phi(u)=u^2$ on $[-1,1]$ does not satisfy the invariance theorem’s monotonicity condition.
Can the parameter derivative be zero at some points without invalidating the length calculation? Explain using the transformed-speed formula.