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Reversing a Curve's Orientation

Reverse a curve’s direction by reversing its parameter interval, and track how this changes its starting point and direction of travel.

Beginner 9 min read

What You'll Learn

  • Reverse a parametrization on an interval from a to b using the parameter value a + b - t.
  • Identify the reversed curve’s starting and ending points from the original parametrization.
  • Reverse the direction of a parametrized line segment without changing its trace.
  • Reverse the direction of a circular traversal and distinguish a full circle from an arc.
  • Check a reversed parametrization by comparing its points and direction of travel with the original.
  • Recognize why negating a position vector does not generally reverse a curve.

Reversing the Direction of Travel

A parametrized curve has a direction of travel: as the parameter increases through its interval, the point on the curve moves along a particular route. Reversing that direction means traversing the same route in the opposite order. For a segment, this swaps the endpoints. For a circle completed in one full turn, it changes the direction around the circle, even though the starting point can remain the same.

The key is to make the new parametrization follow the original one’s parameter values in reverse order. This changes how the curve is traversed without changing which points the curve visits.

Definition: Suppose a curve is parametrized by \(\mathbf{r}(t)\) for \(a\leq t\leq b\). Its reversed parametrization on the same interval is
$$ \mathbf{r}_{\mathrm{rev}}(t)=\mathbf{r}(a+b-t), \qquad a\leq t\leq b. $$
As \(t\) increases from \(a\) to \(b\), the input \(a+b-t\) decreases from \(b\) to \(a\). Thus the new parametrization follows the original curve backwards.

The expression \(a+b-t\) reverses the endpoints of the parameter interval. When \(t=a\), its value is \(b\); when \(t=b\), its value is \(a\). It also takes every value between \(a\) and \(b\), so the reversed parametrization visits exactly the points visited by the original one.

What Happens to the Starting and Ending Points?

Evaluating the reversed parametrization at the endpoints gives the most direct way to track where it starts and ends:

$$ \mathbf{r}_{\mathrm{rev}}(a) =\mathbf{r}(a+b-a) =\mathbf{r}(b), \qquad \mathbf{r}_{\mathrm{rev}}(b) =\mathbf{r}(a+b-b) =\mathbf{r}(a). $$

So the reversed curve starts at the original curve’s ending point and finishes at its original starting point. This endpoint swap is especially visible for a line segment or a circular arc. For a closed curve, the two original endpoints may be the same point, so reversing direction need not change the starting location.

Key takeaway: Reversing a curve changes the order in which its points are visited, not its trace. On an interval \([a,b]\), the reversed parametrization starts at \(\mathbf{r}(b)\) and ends at \(\mathbf{r}(a)\).

Worked Example: Reversing a Parametrized Line Segment

Worked Example: Reversing a Parametrized Line Segment

Consider the parametrization

$$ \mathbf{r}(t)=(1+2t,\;3-t,\;2+t), \qquad 0\leq t\leq 1. $$

First find its endpoints. At \(t=0\),

$$ \mathbf{r}(0)=(1+2(0),\;3-0,\;2+0)=(1,3,2). $$

At \(t=1\),

$$ \mathbf{r}(1)=(1+2(1),\;3-1,\;2+1)=(3,2,3). $$

The interval is \([0,1]\), so the reversed parameter input is \(0+1-t=1-t\). Substitute \(1-t\) wherever the original formula has \(t\):

$$ \begin{aligned} \mathbf{r}_{\mathrm{rev}}(t) &=\mathbf{r}(1-t)\\ &=(1+2(1-t),\;3-(1-t),\;2+(1-t))\\ &=(3-2t,\;2+t,\;3-t), \qquad 0\leq t\leq 1. \end{aligned} $$

Check the endpoints. At \(t=0\), the reversed curve is \((3,2,3)\), the original endpoint. At \(t=1\), it is \((1,3,2)\), the original starting point. More generally, \(\mathbf{r}_{\mathrm{rev}}(t)=\mathbf{r}(1-t)\), where \(1-t\) stays in \([0,1]\). Therefore the reversed parametrization visits the same points as the original and visits them in the opposite order.

Why the Reversal Formula Works

The formula rests on two simple facts: its new input runs through the original interval backwards, and the chain rule reverses the tangent direction. The first fact proves that the trace is unchanged. The second verifies the change in direction wherever the original curve is differentiable.

Derivation: For \(a\leq t\leq b\), the value \(a+b-t\) lies in \([a,b]\), since \(a\leq a+b-t\leq b\). As \(t\) runs from \(a\) to \(b\), this value runs from \(b\) down to \(a\). Thus \(\mathbf{r}(a+b-t)\) is always a point on the original curve and visits its parameter values in reverse order. If \(\mathbf{r}\) is differentiable, the chain rule gives
$$ \mathbf{r}_{\mathrm{rev}}'(t) =\mathbf{r}'(a+b-t)\frac{d}{dt}(a+b-t) =-\mathbf{r}'(a+b-t). $$
Where the original velocity is nonzero, the reversed velocity points in the opposite direction at the corresponding point.

The derivative statement concerns direction at corresponding points; it does not say that the reversed velocity must be the negative of the original velocity evaluated at the same parameter value. In general, it is the original velocity at \(a+b-t\) that appears. The negative sign comes from the fact that the original parameter input decreases as the new parameter increases.

Worked Example: Reversing a Full Circle

Worked Example: Reversing a Full Circle

Take a circle of radius \(3\), centered at \((2,-1,4)\) in a plane parallel to the \(xy\)-plane, with the parametrization

$$ \mathbf{r}(t)=(2+3\cos t,\;-1+3\sin t,\;4), \qquad 0\leq t\leq 2\pi. $$

The interval-reversal formula uses \(2\pi-t\). Substituting this into the original parametrization gives

$$ \begin{aligned} \mathbf{r}_{\mathrm{rev}}(t) &=\mathbf{r}(2\pi-t)\\ &=(2+3\cos(2\pi-t),\;-1+3\sin(2\pi-t),\;4)\\ &=(2+3\cos t,\;-1-3\sin t,\;4). \end{aligned} $$

Here we used \(\cos(2\pi-t)=\cos t\) and \(\sin(2\pi-t)=-\sin t\). The reversed parametrization starts at

$$ \mathbf{r}_{\mathrm{rev}}(0)=(5,-1,4)=\mathbf{r}(0), $$

and finishes at the same point, since a full circle has matching start and end points. To check that it remains on the same circle, subtract the center coordinates and calculate:

$$ \begin{aligned} (x-2)^2+(y+1)^2 &=(3\cos t)^2+(-3\sin t)^2\\ &=9\cos^2t+9\sin^2t\\ &=9. \end{aligned} $$

Thus the radius remains \(3\). The direction is reversed: the original sine coordinate increases initially from \(-1\), whereas the reversed sine coordinate decreases initially from \(-1\). The same circle is traversed in the opposite direction, but its starting point does not change because the full-turn interval begins and ends at the same point.

A Circular Arc Has Different Endpoints

Reversing an interval does not always leave the starting point unchanged. A circle arc is not necessarily a full closed traversal: its parameter interval may stop before it returns to its starting point.

Worked Example: Reversing a Quarter-Circle Arc

Consider the unit-circle arc

$$ \mathbf{r}(t)=(\cos t,\;\sin t,\;0), \qquad 0\leq t\leq \frac{\pi}{2}. $$

The original arc starts at \(\mathbf{r}(0)=(1,0,0)\) and ends at \(\mathbf{r}(\pi/2)=(0,1,0)\). The interval-reversal formula is \(\pi/2-t\), so

$$ \begin{aligned} \mathbf{r}_{\mathrm{rev}}(t) &=\mathbf{r}\left(\frac{\pi}{2}-t\right)\\ &=\left(\cos\left(\frac{\pi}{2}-t\right),\; \sin\left(\frac{\pi}{2}-t\right),\;0\right)\\ &=(\sin t,\;\cos t,\;0). \end{aligned} $$

At \(t=0\), this gives \((0,1,0)\), the original endpoint. At \(t=\pi/2\), it gives \((1,0,0)\), the original starting point. Also, \((\sin t)^2+(\cos t)^2=1\), so the points remain on the unit circle. This arc’s starting point changes under reversal because its endpoints are distinct.

A Common Mistake: Negating the Position Vector

Reversing a curve does not generally mean multiplying its position vector by \(-1\). That operation reflects every point through the origin, which usually creates a different trace rather than reversing the order of points.

Worked Example: Correcting a False Reversal

For the segment in the first example, \(\mathbf{r}(0)=(1,3,2)\) and \(\mathbf{r}(1)=(3,2,3)\). A mistaken attempt at reversal would be

$$ -\mathbf{r}(t)=(-1-2t,\;-3+t,\;-2-t). $$

Its starting point is \(-\mathbf{r}(0)=(-1,-3,-2)\), not the original ending point \((3,2,3)\). It has reflected the curve through the origin. The correct reversal is instead

$$ \mathbf{r}(1-t)=(3-2t,\;2+t,\;3-t), $$

which starts at \((3,2,3)\) and ends at \((1,3,2)\). The correction is to reverse the parameter input, not to negate the position vector.

When Reversal Is Useful

Orientation matters whenever a parametrization is meant to describe motion or an ordered traversal. Reversing a segment can make a path start at the opposite endpoint; reversing an arc can make it proceed around the same portion of a circle in the other direction. The same interval formula applies to any parametrized curve for which the original parameter values are defined on the whole interval.

Before using a reversed parametrization, check three things:

  • The interval: use the original interval endpoints \(a\) and \(b\) to form \(a+b-t\).
  • The endpoints: confirm the new starting point is \(\mathbf{r}(b)\) and the new ending point is \(\mathbf{r}(a)\).
  • The trace and direction: confirm the new parameter input stays in \([a,b]\) and decreases as \(t\) increases.

These checks distinguish reversal from changing the curve itself. A reversed parametrization should preserve the set of points while switching their order of traversal. For a closed curve, matching endpoints can conceal the endpoint swap, so checking how the intermediate points are visited is important.

Check Your Understanding

Use the interval-reversal formula to answer each question.

  1. If \(\mathbf{r}(t)\) is defined on \([2,5]\), what parameter input gives the reversed parametrization on that same interval?
  2. For a curve on \([a,b]\), what are the starting and ending points of its reversed parametrization?
  3. Why does reversing one full turn around a circle leave its starting point unchanged?
  4. What is the reversed parametrization of \(\mathbf{r}(t)=(\cos t,\sin t,0)\) on \([0,\pi]\)? State its starting point.
  5. Why does \(-\mathbf{r}(t)\) generally fail to reverse a curve’s orientation?

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