One Curve, Different Clocks
A parametrization describes how a moving point traces a curve: the parameter plays the role of a clock. The same geometric path can be traced quickly in one part and slowly in another, or in the opposite direction. These changes affect the velocity at each parameter value, but a smooth monotone change of clock does not change the total distance travelled along the path.
This distinction matters whenever length is calculated from a parametrization. The formula involves speed, which depends on the clock, so it is important to see precisely how the change in speed is balanced by the change in parameter. The chain rule and the change-of-variables formula give that balance.
Curve Length and Reparameterization
Let \(\mathbf{r}:[a,b]\to\mathbb{R}^n\) be a continuously differentiable curve, written \(\mathbf{r}(t)=(x_1(t),\ldots,x_n(t))\). Its velocity is \(\mathbf{r}'(t)\), and its speed is the magnitude of that velocity. The curve is called regular if \(\mathbf{r}'(t)\ne\mathbf{0}\) for every \(t\in[a,b]\).
A change of parameter replaces \(t\) by a function of a new parameter \(s\). Specifically, let \(\phi:[u,v]\to[a,b]\) be continuously differentiable and monotone, with its endpoint values equal to the two endpoints \(a,b\), in either order. The new parametrization is \(\mathbf{R}(s)=\mathbf{r}(\phi(s))\). The condition on the endpoints ensures that the new parameter covers the same parameter interval of the original curve. A decreasing \(\phi\) traverses that interval in reverse.
Why the Length Stays the Same
The chain rule shows how the new velocity depends on the original velocity and the speed of the parameter change. Taking its magnitude introduces an absolute value: motion in reverse has negative parameter velocity, but speed, like distance, cannot be negative.
For an increasing change of parameter, \(\phi'(s)\ge 0\), and the ordinary change-of-variables formula converts the new length integral back to the original one. For a decreasing change, \(\phi'(s)\le 0\); the absolute value reverses the sign needed to keep length positive, while the substitution also reverses the integration limits.
For \(\mathbf{R}(s)=\mathbf{r}(\phi(s))\), differentiate to obtain \(\mathbf{R}'(s)=\mathbf{r}'(\phi(s))\phi'(s)\).
Use \(\|\mathbf{R}'(s)\|=\|\mathbf{r}'(\phi(s))\|\,|\phi'(s)|\), since multiplying a vector by a scalar multiplies its magnitude by the scalar's absolute value.
Integrate the new speed over \([u,v]\). Monotonicity lets the substitution \(t=\phi(s)\) recover the original length integral, accounting for reversed limits if \(\phi\) is decreasing.
For completeness, write \(q(t)=\|\mathbf{r}'(t)\|\), a continuous function. If \(\phi\) is increasing, then \(\phi(u)=a\), \(\phi(v)=b\), and \(|\phi'(s)|=\phi'(s)\). Thus
If \(\phi\) is decreasing, then \(\phi(u)=b\), \(\phi(v)=a\), and \(|\phi'(s)|=-\phi'(s)\). Substitution gives
This proves the result in either orientation. The key point is that changing the clock scales the speed by \(|\phi'(s)|\), and the change-of-variables formula accounts for that same factor in the parameter interval.
Worked Examples
Worked Example: A Straight Line Traced with a Changing Speed
Consider \(\mathbf{r}(t)=(3t,4t)\) for \(0\le t\le1\), and change parameter using \(\phi(s)=s^2\) for \(0\le s\le1\). First compute the original speed:
Therefore the original length is
The new parametrization is
Differentiate and find its speed:
where \(s\ge0\). Hence
The chain rule also predicts this speed: the original speed is \(5\), and \(\phi'(s)=2s\), so the new speed is \(5|2s|=10s\). The two length calculations agree, as the theorem requires.
Worked Example: A Quarter-Circle with a Nonlinear Parameter
Let \(\mathbf{r}(t)=(\cos t,\sin t)\) for \(0\le t\le\pi/2\). Its velocity and speed are
Thus its length is
Now set \(\phi(u)=\frac{\pi}{2}u^2\) on \(0\le u\le1\). This is an increasing change of parameter from \(0\) to \(\pi/2\), with \(\phi'(u)=\pi u\). By the chain rule, the reparameterized speed is
Its length is therefore
Although the new clock changes how quickly the quarter-circle is traced, it describes the same path once and gives the same length.
Worked Example: Reversing the Direction Along a Parabola
Take \(\mathbf{r}(t)=(t,t^2)\) for \(0\le t\le1\). Its velocity is \((1,2t)\), so its speed is \(\sqrt{1+4t^2}\). Reverse the parameter with \(\phi(s)=1-s\), also on \([0,1]\). The new parametrization and its derivative are
Thus the new speed is \(\sqrt{1+4(1-s)^2}\), and its length is
where the last equality uses \(t=1-s\), so \(ds=-dt\), and changes the limits from \(s=0,1\) to \(t=1,0\).
To evaluate and check this common integral, an antiderivative is
Indeed, differentiating the first term gives \(\frac12\sqrt{1+4t^2}+\frac{2t^2}{\sqrt{1+4t^2}}\), and differentiating the second gives \(\frac{1}{2\sqrt{1+4t^2}}\). Their sum is \(\sqrt{1+4t^2}\). Evaluating at the endpoints, using \(\operatorname{asinh}(2)=\ln(2+\sqrt5)\), gives
The reversed parametrization changes the direction of travel, not the distance along the parabola.
A Common Error: Forgetting the Absolute Value
For a decreasing parameter, it is incorrect to use \(\phi'(s)\) where the speed formula requires \(|\phi'(s)|\). For example, use \(\mathbf{r}(t)=(3t,4t)\) and the reversal \(\phi(s)=1-s\) on \([0,1]\). The original speed is \(5\), while \(\phi'(s)=-1\). A mistaken calculation would write
A length cannot be negative. The error is treating signed parameter velocity as speed. The chain rule gives \(\|\mathbf{r}'(\phi(s))\phi'(s)\|=\|\mathbf{r}'(\phi(s))\|\,|\phi'(s)|\), so the correct calculation is
Why Monotonicity Matters
A monotone change of parameter traverses the original interval in one direction. If a parameter instead reverses direction partway through, the path can be retraced, increasing the distance travelled. For example, let \(\mathbf{r}(t)=(t,0)\) on \([0,1]\), and let \(\phi(s)=\frac{1-\cos(2\pi s)}{2}\) on \([0,1]\). The original curve has length \(1\). The new parameter goes from \(0\) to \(1\) and back to \(0\), so its length is
This is not a contradiction: \(\phi\) is not monotone and does not describe a single traversal of the original interval. Monotonicity is a substantive condition in the invariance result, not a technical detail to omit.
When the Result Is Useful
Reparameterization invariance separates geometric length from choices made for convenience. One parametrization may produce a simple speed; another may make the endpoints or direction easier to handle. Provided the change is smooth and monotone and covers the same interval, either can be used to calculate the same length.
The distinction also clarifies what the theorem does not say. The velocity and speed as functions of the parameter can change, and a reversed parametrization changes the direction of travel. What remains fixed is the total length of the path traversed once. The proof depends on two ingredients: the chain rule for the new velocity and change of variables for the length integral.
Check Your Understanding
Use the speed formula and the conditions in the theorem when answering these questions.
- State the chain-rule expression for the speed of \(\mathbf{R}(s)=\mathbf{r}(\phi(s))\). Why does it contain an absolute value?
- For an increasing change of parameter, what are the endpoint values of \(\phi\), and how does substitution transform the length integral?
- Why does a decreasing change of parameter still give a positive length?
- For \(\mathbf{r}(t)=(2t,0)\) on \([0,1]\) and \(\phi(s)=s^3\) on \([0,1]\), compute the reparameterized speed and verify the length.
- Give a reason a nonmonotone change of parameter can produce a different length from the original parametrization.
