One Curve, Different Parameter Speeds
A parameter describes how a curve is traced, but it is not the curve’s length. A change of parameter can make the same path be traversed faster in some places and slower in others. It can also reverse the direction of traversal. The total length, however, is unchanged when the new parameter traces the same path exactly once.
The reason is that the chain rule adjusts the velocity by the rate at which the parameter changes. In the length integral, this new speed is paired with a matching change of variables. The resulting factors cancel in the integral, with an absolute value accounting for reversed direction.
Length and Reparameterization
Let \(\mathbf r:[a,b]\to\mathbb R^n\) be a continuously differentiable curve. Its speed at parameter value \(t\) is \(\|\mathbf r'(t)\|\), the magnitude of its derivative. Its arc length on \([a,b]\) is the integral of that speed:
Now let \(\phi:[c,d]\to[a,b]\) be a continuously differentiable, strictly monotone function whose range is all of \([a,b]\). It may be increasing, with \(\phi(c)=a\) and \(\phi(d)=b\), or decreasing, with \(\phi(c)=b\) and \(\phi(d)=a\). Define a new curve by composing the original curve with \(\phi\):
The chain rule gives \(\mathbf q'(u)=\mathbf r'(\phi(u))\phi'(u)\). Taking magnitudes produces the speed of the new parametrization. Since \(\phi'(u)\) is a scalar, its magnitude factors out as an absolute value:
The absolute value is essential: speed is nonnegative, whether the parameter is increasing or decreasing. The length of \(\mathbf q\) is therefore
The Invariance Theorem
Here is a complete justification. The chain rule provides the speed formula. If \(\phi\) is increasing, then \(\phi'(u)\geq 0\), so \(|\phi'(u)|=\phi'(u)\). The ordinary change-of-variables formula, with \(t=\phi(u)\), turns the new length integral into the original one. If \(\phi\) is decreasing, then \(\phi'(u)\leq 0\), so \(|\phi'(u)|=-\phi'(u)\). The change of variables reverses the integration limits; reversing those limits introduces the second minus sign needed for a positive length.
By the chain rule, \(\mathbf q'(u)=\mathbf r'(\phi(u))\phi'(u)\).
Thus \(\|\mathbf q'(u)\|=\|\mathbf r'(\phi(u))\|\,|\phi'(u)|\).
Then \(|\phi'|=\phi'\), and \(t=\phi(u)\) changes the limits from \(c,d\) to \(a,b\).
Then \(|\phi'|=-\phi'\); the reversed limits contribute the compensating sign, again giving the integral from \(a\) to \(b\).
In equations, the two cases are:
The argument does not require \(\phi'(u)\) to be nonzero at every point. What matters here is that the change of parameter is continuously differentiable, monotone, and covers the original parameter interval. A zero derivative at an individual parameter value does not invalidate the speed formula or the integral equality.
Worked Examples
Worked Example: A Nonuniformly Traversed Line Segment
Let \(\mathbf r(t)=(t,0)\), for \(0\leq t\leq 2\). Reparameterize it using \(t=\phi(u)=2u^2\), for \(0\leq u\leq1\). First, the original derivative and speed are
Therefore its length is
The new curve is \(\mathbf q(u)=\mathbf r(2u^2)=(2u^2,0)\). Differentiating gives \(\mathbf q'(u)=(4u,0)\), so its speed on \(0\leq u\leq1\) is \(4u\). Its length is
The check agrees with the theorem. The endpoints also match the original segment: \(\mathbf q(0)=(0,0)\) and \(\mathbf q(1)=(2,0)\). The speed is not constant, but the total length is still \(2\).
Worked Example: A Circular Arc With a Nonlinear Parameter
Consider the radius-\(3\) quarter-circle \(\mathbf r(t)=(3\cos t,3\sin t)\) for \(0\leq t\leq\pi/2\). Its derivative and speed are
Thus the original arc length is
Now use \(t=\phi(u)=\frac{\pi}{2}u^2\) on \(0\leq u\leq1\). The reparameterized curve is \(\mathbf q(u)=\mathbf r(\phi(u))\). Since \(\phi'(u)=\pi u\), the chain rule gives its speed as
Integrating over the new parameter interval gives
This matches the original length. The nonlinear parameter changes the angular rate but not the length of the quarter-circle.
Worked Example: A Reparameterization That Reverses Direction
Let \(\mathbf r(t)=(t,\frac{2}{3}t^{3/2})\) for \(0\leq t\leq1\). Its derivative is \(\mathbf r'(t)=(1,\sqrt t)\), so its speed is \(\sqrt{1+t}\). The original length is
Use the decreasing parameter change \(t=\phi(u)=1-u\), with \(0\leq u\leq1\). Then \(\phi'(u)=-1\), and the new curve is \(\mathbf q(u)=(1-u,\frac{2}{3}(1-u)^{3/2})\). Its speed, using the chain rule and the absolute value, is
Consequently,
The lengths agree. The endpoints confirm that the traversal is reversed: \(\mathbf q(0)=\mathbf r(1)\) and \(\mathbf q(1)=\mathbf r(0)\). Reversing the direction changes neither the positive accumulated length nor the portion of the curve traced.
Why the Same Path Matters
The theorem compares two parametrizations that trace the same curve, with the same traversal apart from a possible reversal. The condition that \(\phi\) is onto ensures that every parameter value in \([a,b]\) is included. Its monotonicity ensures the traversal does not move back and forth through the original parameter interval.
This distinction matters for curves that cross themselves. A curve’s geometric image is the set of points it reaches, but its length as a parametrized traversal accounts for how the curve is traced. Visiting parts of a path more than once can add length, even if the set of points visited does not change. The theorem applies to a monotone reparameterization, not to an arbitrary new traversal of the same set.
Arc-length invariance is useful because it separates geometric quantities from choices of parameter. Different formulas may be convenient for different calculations, but the length remains a property of the traced curve. It is not a consequence of a particular clock, variable, or rate of motion.
A Common Sign Error
When the parameter decreases, \(\phi'(u)\) is negative. A common mistake is to use \(\phi'(u)\) directly as a factor in the speed, rather than using \(|\phi'(u)|\). For example, take the line \(\mathbf r(t)=(t,0)\), \(0\leq t\leq2\), with \(t=2(1-u)\), \(0\leq u\leq1\). Its reparameterized derivative is \(\mathbf q'(u)=(-2,0)\).
This is a signed change in position, not a length. Speed is the magnitude of the derivative, so \(\|\mathbf q'(u)\|=2\). The corrected calculation is
This equals the original length, \(\int_0^2 1\,dt=2\). The absolute value in the speed formula prevents a direction reversal from producing a negative length.
Check Your Understanding
Use the speed integral and the conditions on the parameter change to answer these questions.
- For \(\mathbf q(u)=\mathbf r(\phi(u))\), what does the chain rule give for \(\mathbf q'(u)\)?
- Why does the speed formula contain \(|\phi'(u)|\) rather than \(\phi'(u)\)?
- What happens to the integration limits when \(\phi\) is decreasing?
- For \(\mathbf r(t)=(t,0)\), \(0\leq t\leq1\), find the speed of \(\mathbf q(u)=\mathbf r(u^3)\) on \(0\leq u\leq1\), and verify its length.
- Why does the invariance theorem not imply that every different traversal of the same geometric image has the same length?
