Why the Choice of Parameter Does Not Change Length
A parameter tells us how to describe a curve as it is traced. The same geometric path can be traversed quickly or slowly, and it can be traversed in either direction. Those changes affect the velocity, but they do not alter the distance along the path.
Arc length is calculated by integrating speed. When a curve is reparameterized, the chain rule changes its velocity by a factor involving the rate of change of the new parameter. Substitution accounts for precisely that factor. If the new parameter runs backward, speed uses the absolute value of the rate, while the integration bounds reverse as well.
Set-Up and Definitions
Let \(\mathbf r:[a,b]\to\mathbb R^n\) be a continuously differentiable curve. Its arc length on \([a,b]\) is
Here, \(\|\mathbf r'(u)\|\) is the speed with respect to the parameter \(u\). Now let \(\phi:[c,d]\to[a,b]\) be a continuously differentiable, one-to-one map onto \([a,b]\). Since it is continuous and one-to-one on an interval, it is strictly increasing or strictly decreasing. Define the reparameterized curve by
The map \(\phi\) changes which parameter value on \(\mathbf r\) is used at each \(t\). It does not change the points of the path that are covered.
Why the Formula Works
The chain rule gives the velocity of the reparameterized curve. Differentiating \(\mathbf q(t)=\mathbf r(\phi(t))\) yields
For a scalar \(k\) and vector \(\mathbf v\), \(\|k\mathbf v\|=|k|\|\mathbf v\|\). Therefore,
This is the key step: the new speed includes the factor \(|\phi'(t)|\), which is exactly the factor needed when changing variables from \(t\) back to \(u=\phi(t)\). The two possible directions can be checked separately.
Then \(\phi(c)=a\), \(\phi(d)=b\), and \(\phi'(t)\geq 0\). Thus \(|\phi'(t)|=\phi'(t)\), and substitution \(u=\phi(t)\) gives the original integral from \(a\) to \(b\).
Then \(\phi(c)=b\), \(\phi(d)=a\), and \(\phi'(t)\leq 0\). Thus \(|\phi'(t)|=-\phi'(t)\). Substitution first produces an integral from \(b\) to \(a\); the negative factor reverses those bounds and makes the length positive.
In the decreasing case, writing \(g(u)=\|\mathbf r'(u)\|\), the calculation is
This completes the justification for both directions. The absolute value in the speed formula and the reversed limits work together; neither should be omitted.
Worked Example: Increasing the Parameter Rate on a Line
Worked Example: Increasing the Parameter Rate on a Line
Let \(\mathbf r(u)=(3u,4u)\), for \(0\leq u\leq 1\). Reparameterize it using \(u=\phi(t)=t^2\), for \(0\leq t\leq 1\), and compare the arc lengths.
First compute the original velocity and speed:
Therefore, the original length is
The reparameterized curve is
Differentiate each coordinate and calculate the speed:
because \(t\geq 0\) on this interval. Hence
The substitution check is \(u=t^2\), so \(du=2t\,dt\). The new speed \(10t\) is \(5(2t)\), and therefore
Both parameterizations give length \(5\). The transformed curve moves at a different rate, but covers the same line segment.
Worked Example: Reversing Direction on a Circle
Worked Example: Reversing Direction on a Circle
Consider the quarter-circle \(\mathbf r(u)=(\cos u,\sin u)\) for \(0\leq u\leq \frac{\pi}{2}\). Reparameterize it with \(\phi(t)=\frac{\pi}{2}(1-t^2)\), for \(0\leq t\leq 1\). This map decreases from \(\frac{\pi}{2}\) to \(0\), so the curve is traversed in reverse.
For the original parameter,
Thus the original length is
For the reparameterization, differentiate \(\phi\):
The chain rule gives
Its speed is
Since \(t\geq 0\), this is nonnegative. The reparameterized length is
For a direct substitution check, \(u=\frac{\pi}{2}(1-t^2)\) gives \(du=-\pi t\,dt\). As \(t\) goes from \(0\) to \(1\), \(u\) goes from \(\frac{\pi}{2}\) to \(0\). Therefore,
The reversed traversal has the same length. The changed order of the bounds is accounted for by the negative sign in \(du\).
Worked Example: A Nonlinear Change on a Parabola
Worked Example: A Nonlinear Change on a Parabola
Let \(\mathbf r(u)=(u,u^2)\), for \(0\leq u\leq 1\), and set \(\phi(t)=\frac{t+t^2}{2}\), for \(0\leq t\leq 1\). Here \(\phi(0)=0\), \(\phi(1)=1\), and \(\phi'(t)=\frac{1+2t}{2}>0\), so this is an increasing one-to-one change of parameter.
The original velocity and speed are
Thus the original length is
For example, an antiderivative is
Differentiating its first term gives \(\frac{1}{2}\sqrt{1+4u^2}+\frac{2u^2}{\sqrt{1+4u^2}}\), and differentiating its logarithmic term gives \(\frac{1}{2\sqrt{1+4u^2}}\). Adding these terms gives \(\sqrt{1+4u^2}\), as required. Evaluating \(F(1)-F(0)\) gives the displayed length.
Now write \(\mathbf q(t)=\mathbf r(\phi(t))\). By the chain rule,
where the factor \(\phi'(t)\) is positive on \([0,1]\). Substituting \(u=\phi(t)\) and \(du=\frac{1+2t}{2}\,dt\) gives
The transformed integral agrees with the original one exactly. The substitution accounts for the changing rate at which the parabola is traced.
A Common Bounds Mistake
A frequent error is to use a signed coordinate derivative as though it were speed, or to reverse the substitution bounds without also changing the sign. For example, let \(\mathbf r(u)=(u,0)\) on \(0\leq u\leq 1\), and reverse it using \(u=1-t\). Then \(\mathbf q(t)=(1-t,0)\), so \(x'(t)=-1\).
The incorrect calculation \(\int_0^1 -1\,dt=-1\) is not a length: it has used signed velocity instead of speed. The correct speed is \(\|\mathbf q'(t)\|=|-1|=1\), and the length is
The same correction appears in substitution. The original speed is \(1\), and \(u=1-t\) takes the bounds \(t=0,1\) to \(u=1,0\). Hence
When This Result Is Useful
The invariance of arc length lets you choose a parameter that makes a calculation or description more convenient. One parameterization may simplify the coordinates, while another may give simpler bounds or make the direction of travel clearer. The resulting speed functions can look different; the substitution theorem explains why their integrals still agree.
The conditions matter. The parameter map must cover the corresponding interval one-to-one, so it traces the same parameter values exactly once. If a map instead goes back and forth over the interval, it is not a one-to-one change of parameter; its arc-length integral can count portions of the path more than once. For a valid one-to-one change, check the endpoint correspondence, apply the chain rule, use the absolute value in speed, and match the bounds to the direction.
Check Your Understanding
Use the chain rule and substitution, keeping track of whether the parameter map increases or decreases.
- If \(\mathbf q(t)=\mathbf r(\phi(t))\), what is \(\mathbf q'(t)\), and why does its speed contain \(|\phi'(t)|\)?
- A parameter map \(\phi:[c,d]\to[a,b]\) is decreasing. What are \(\phi(c)\) and \(\phi(d)\), and how do the bounds change under substitution?
- For \(\mathbf r(u)=(u,0)\) on \([0,1]\), reparameterize by \(u=t^2\) on \([0,1]\). Compute the new speed and its arc length.
- For \(\mathbf r(u)=(\cos u,\sin u)\) on \([0,\frac{\pi}{2}]\), explain why \(u=\frac{\pi}{2}(1-t)\) gives the same length despite reversing direction.
- Why can a parameter map that runs back and forth over an interval fail to preserve the length of a single traversal?
