Reparametrizing a Curve

Why Change a Parameter?

A curve can be described by different parameterizations. One parameter may make its formulas simpler, or a different interval may be more convenient. To compare two descriptions, track how the new parameter selects values of the original parameter.

For example, a point on a curve may be written as \(\mathbf{r}(t)=(x(t),y(t))\), with \(t\) in an interval \(I\). If a new parameter \(s\) determines an old parameter value \(t=\phi(s)\), then the resulting curve is \(\mathbf{q}(s)=\mathbf{r}(\phi(s))\). The formulas alone do not tell the whole story: the interval for \(s\), the range of \(\phi\), and whether \(\phi\) increases or decreases all matter.

Definition and the Basic Comparison

Definition: Let \(\mathbf{r}:I\to\mathbb{R}^2\) be a parametrized curve and let \(\phi:J\to I\) be a function. The curve \(\mathbf{q}:J\to\mathbb{R}^2\) defined by \(\mathbf{q}(s)=\mathbf{r}(\phi(s))\) is obtained from \(\mathbf{r}\) by a change of parameter. Its trace is the set of points \(\{\mathbf{q}(s):s\in J\}\). The original trace is \(\{\mathbf{r}(t):t\in I\}\).

The range \(\phi(J)\) tells us which original parameter values the new parameter actually uses. If \(\phi(J)=I\), then every original parameter value is reached, so the new curve covers the original trace. Conversely, if \(\phi(J)\) omits part of \(I\), it may omit points from the trace. For a curve that crosses itself, however, it is possible to omit parameter values without omitting any points: another parameter value may describe the same point.

Thus the exact test for equal traces is equality of the point sets. The condition \(\phi(J)=I\) is a useful guarantee that the traces agree, but for a curve that repeats points it is not necessary. If the original curve never takes the same value at two distinct parameter values, then \(\phi(J)=I\) is also necessary for equal traces.

Key comparison: If \(\phi:J\to I\) is onto, then \(\mathbf{r}\) and \(\mathbf{q}=\mathbf{r}\circ\phi\) have the same trace. If, in addition, \(\phi\) is strictly increasing, the new parameter follows the original parameter order. If \(\phi\) is strictly decreasing, it follows the opposite order.

Why Monotonicity Determines the Order

A function is strictly increasing if \(s_1<s_2\) implies \(\phi(s_1)<\phi(s_2)\). It is strictly decreasing if \(s_1<s_2\) implies \(\phi(s_1)>\phi(s_2)\). Suppose \(\phi\) is a one-to-one correspondence from \(J\) onto \(I\). When \(s\) increases, an increasing \(\phi\) visits the original parameter values in increasing order; a decreasing \(\phi\) visits them in decreasing order. This explains the order comparison directly.

For continuous changes of parameter between intervals, a strictly monotone map that is onto gives a one-to-one correspondence between the intervals. The endpoints, when included, must correspond as well. For an increasing change, the lower end maps to the lower end and the upper end to the upper end. For a decreasing change, those roles are reversed. This endpoint check is a practical way to catch a mistaken parameter interval.

Write the parameter map.
Express the original parameter as \(t=\phi(s)\).

Check its range.
Determine which original parameter values occur as \(s\) runs through its interval \(J\).

Check the order.
Decide whether \(\phi\) is increasing, decreasing, or nonmonotone on \(J\).

Compose and verify.
Substitute \(t=\phi(s)\) into both coordinate functions, then match parameter values and points.

Worked Examples

Worked Example: An Increasing Affine Change

Let \(\mathbf{r}(t)=(t,t^2)\) for \(-1\leq t\leq2\). Use \(t=3s-1\) with \(0\leq s\leq1\). Find the new parametrization and compare the traversal.

First check the parameter map at both endpoints:

\phi(0)=3(0)-1=-1,\qquad \phi(1)=3(1)-1=2.

The map is linear with positive coefficient \(3\), so it is strictly increasing. Its endpoint values show that its range is the full interval \([-1,2]\). Substitute \(t=3s-1\) in each coordinate:

\mathbf{q}(s)=\mathbf{r}(3s-1)=\bigl(3s-1,(3s-1)^2\bigr),\qquad 0\leq s\leq1.

At \(s=0\), \(\mathbf{q}(0)=(-1,1)=\mathbf{r}(-1)\). At \(s=\tfrac12\), \(t=3(\tfrac12)-1=\tfrac12\), so \(\mathbf{q}(\tfrac12)=(\tfrac12,\tfrac14)=\mathbf{r}(\tfrac12)\). At \(s=1\), \(\mathbf{q}(1)=(2,4)=\mathbf{r}(2)\). As \(s\) increases from \(0\) to \(1\), \(t\) increases from \(-1\) to \(2\). The trace and order are preserved.

To check the correspondence in the other direction, solve \(t=3s-1\) for \(s\): \(s=(t+1)/3\). For every \(t\in[-1,2]\), this gives \(s\in[0,1]\), confirming that no original parameter values are missed.

Worked Example: A Decreasing Change Reverses the Order

Let \(\mathbf{r}(t)=(\cos t,\sin t)\) for \(0\leq t\leq2\pi\). Use \(t=2\pi-s\) for \(0\leq s\leq2\pi\).

The map has endpoint values \(\phi(0)=2\pi\) and \(\phi(2\pi)=0\), and it is strictly decreasing. Substitution gives

\mathbf{q}(s)=\bigl(\cos(2\pi-s),\sin(2\pi-s)\bigr)=(\cos s,-\sin s).

At \(s=0\), the new curve is at \((1,0)=\mathbf{r}(2\pi)\). At \(s=\tfrac{\pi}{2}\), it is at \((0,-1)=\mathbf{r}(\tfrac{3\pi}{2})\). At \(s=\pi\), it is at \((-1,0)=\mathbf{r}(\pi)\). At \(s=2\pi\), it is back at \((1,0)=\mathbf{r}(0)\). The range of \(2\pi-s\) is the whole original interval, so the trace is unchanged. But the original parameter values are visited in decreasing order: this is the opposite traversal.

As a check, the original parameter \(t\) and the new parameter \(s\) satisfy \(s=2\pi-t\). This correspondence reverses the interval endpoints, just as the endpoint calculations show.

Worked Example: A Nonlinear Change Preserves the Order

Let \(\mathbf{r}(t)=(t,2t+1)\) for \(0\leq t\leq4\). Set \(t=u^2\) with \(0\leq u\leq2\). Then

\mathbf{q}(u)=\mathbf{r}(u^2)=(u^2,2u^2+1).

The map \(\phi(u)=u^2\) is strictly increasing on \([0,2]\): if \(0\leq u_1<u_2\), then \(u_1^2<u_2^2\). It maps the endpoints to \(\phi(0)=0\) and \(\phi(2)=4\), so its range is \([0,4]\). For instance, at \(u=1\), \(t=1\) and both descriptions give \((1,3)\). At \(u=2\), \(t=4\) and both give \((4,9)\). Every original parameter \(t\in[0,4]\) corresponds to \(u=\sqrt{t}\in[0,2]\), confirming that the full trace is covered in the same order.

This example shows that a change of parameter need not be linear to preserve the traversal. The relevant facts are its interval, range, and monotonicity.

When a Change of Parameter Retraces the Curve

A parameter map can cover the full original interval without following it in one consistent direction. If the map is not one-to-one, it may visit the same original parameter value more than once. In that case the point set may be unchanged, but the new parametrization can pause in its progress through the original parameter values, reverse direction, or retrace part of the curve.

Worked Example: Full Trace with a Repeated Visit

Let \(\mathbf{r}(t)=(t,t^2)\) for \(0\leq t\leq4\), and define \(\phi(s)=(s-1)^2\) for \(-1\leq s\leq3\). The map reaches \(\phi(-1)=4\), \(\phi(1)=0\), and \(\phi(3)=4\), so its range is the full interval \([0,4]\). The new curve is

\mathbf{q}(s)=\bigl((s-1)^2,(s-1)^4\bigr).

At the three listed parameter values, \(\mathbf{q}(-1)=(4,16)\), \(\mathbf{q}(1)=(0,0)\), and \(\mathbf{q}(3)=(4,16)\). Thus the parameter runs from the point at \(t=4\) back to the point at \(t=0\), then returns to the point at \(t=4\). Also, \(\phi(0)=1\) and \(\phi(2)=1\), so \(\mathbf{q}(0)=\mathbf{q}(2)=(1,1)\): the same point is visited twice. The trace is the same as the original trace, but the traversal is not simply in the same order or the opposite order.

Here the interval \([-1,3]\) matters. Restricting \(s\) to \([1,3]\), for example, would make \(\phi\) increasing and onto \([0,4]\), removing the initial return trip.

A Common Mistake: Ignoring the New Interval

Suppose we use the same curve \(\mathbf{r}(t)=(t,t^2)\), \(0\leq t\leq4\), and propose \(t=s^2\). A tempting claim is that taking \(-2\leq s\leq2\) gives the same traversal because \(s^2\) ranges from \(0\) to \(4\). The range does give the same trace, but not the same traversal order: \(s=-2\) gives \(t=4\), \(s=0\) gives \(t=0\), and \(s=2\) gives \(t=4\) again.

Indeed, the curve visits \((1,1)\) twice, since \(s=-1\) and \(s=1\) both give \(t=1\). The correction depends on what is intended. To preserve the same order without repeating points, use \(0\leq s\leq2\), where \(s^2\) increases from \(0\) to \(4\). To reverse the order, use \(t=(2-s)^2\) on \(0\leq s\leq2\), which decreases from \(4\) to \(0\). Always check both the parameter map and its domain.

Why Reparametrization Matters

Reparametrization separates the geometric trace from the choice of parameter used to describe it. Two formulas may describe the same set of points while moving through those points differently. For a useful comparison, state the parameter intervals, find the range of the parameter map, and determine its order of change. These checks clarify whether a new description covers only part of a trace, covers the full trace in the same or opposite order, or retraces points.

Takeaway: A one-to-one increasing change of parameter onto the original interval preserves the traversal order; a one-to-one decreasing change onto that interval reverses it. A nonmonotone change may still cover the full trace, but it can repeat or retrace points.

Check Your Understanding

For each question, identify the parameter interval and compare the range and order of the parameter map.

For \(\mathbf{r}(t)=(t,t^2)\), \(0\leq t\leq3\), use \(t=3-s\) on \(0\leq s\leq3\). Does this preserve or reverse the traversal?
For the same curve, use \(t=s^2\) on \(0\leq s\leq\sqrt{3}\). Does the new parametrization cover the full trace, and in what order?
What can go wrong if a parameter map is onto but not one-to-one?
State a condition on \(\phi:J\to I\) that guarantees \(\mathbf{r}\) and \(\mathbf{r}\circ\phi\) have the same trace.
If a continuous change of parameter is strictly decreasing and onto, what happens to the order in which the original parameter values are visited?

What You'll Learn