Reversed and Rescaled Curve Parameters

Why Change a Curve’s Parameter?

A curve can be described by a vector-valued function $\mathbf{r}(t)$, where $t$ ranges over an interval. The parameter is a way of tracing the curve: as $t$ increases, the point $\mathbf{r}(t)$ moves along the path. The same path can be traced with a different parameter, perhaps in the opposite direction or at a different rate.

This tutorial studies a particular, useful family of parameter changes: affine changes of the form $t=au+b$. The value of $a$ controls both the rate and the direction of traversal, while $b$ shifts which original parameter value corresponds to $u=0$. The interval for $u$ must be chosen so that $au+b$ remains in the original domain of $\mathbf{r}$.

Definition: Let $\mathbf{r}$ be a differentiable curve defined on an interval containing the values $au+b$, and let $a\ne 0$. The affine reparametrization is $\mathbf{q}(u)=\mathbf{r}(au+b)$. The original parameter value is $t=au+b$.

When $a>0$, increasing $u$ makes $t$ increase, so the curve is traversed in the same direction over the part of the path selected by the interval. When $a<0$, increasing $u$ makes $t$ decrease, so that part of the path is traversed in reverse. The magnitude $|a|$ determines how quickly the original parameter changes with $u$.

Velocity and Speed Under an Affine Change

The velocity of a parametrized curve is its derivative with respect to its parameter. Applying the chain rule to $\mathbf{q}(u)=\mathbf{r}(au+b)$ gives

\mathbf{q}'(u)=a\,\mathbf{r}'(au+b).

Speed is the magnitude of velocity. Since the magnitude of a scalar multiple of a vector satisfies $\|a\mathbf{v}\|=|a|\|\mathbf{v}\|$, the new speed is

\|\mathbf{q}'(u)\|=|a|\,\|\mathbf{r}'(au+b)\|.

The absolute value is essential: speed cannot be negative. A negative $a$ changes the direction of the velocity vector, but not its magnitude apart from the factor $|a|$. If $|a|>1$, the new parameter runs through the original parameter values faster per unit $u$; if $0<|a|<1$, it runs through them more slowly per unit $u$.

Length and the Parameter Interval

For a continuously differentiable curve $\mathbf{r}$ on $[\alpha,\beta]$, its arc length is the integral of its speed over that interval. Suppose $\mathbf{q}(u)=\mathbf{r}(au+b)$ is defined for $u\in[c,d]$. Its length over this $u$-interval is

L_{\mathbf{q}} = \int_c^d \|\mathbf{q}'(u)\|\,du = \int_c^d |a|\,\|\mathbf{r}'(au+b)\|\,du.

The original parameter values covered by this interval are the values between $ac+b$ and $ad+b$. Consequently, the length of $\mathbf{q}$ is the length of $\mathbf{r}$ over the interval with those endpoints, taken in increasing order. The order matters when describing traversal direction; the length integral always accumulates a nonnegative speed.

Theorem: If $\mathbf{r}$ is continuously differentiable on an interval containing all values $au+b$ for $c\le u\le d$, and $a\ne0$, then $$ \int_c^d \|\mathbf{q}'(u)\|\,du = \int_{\min(ac+b,\,ad+b)}^{\max(ac+b,\,ad+b)} \|\mathbf{r}'(t)\|\,dt, \qquad \mathbf{q}(u)=\mathbf{r}(au+b). $$

This result concerns affine changes of parameter. It does not require $\mathbf{r}'(t)$ to be nonzero at every point: the speed and the length formula still make sense if the curve has a point where its velocity is zero. The condition $a\ne0$, however, excludes a constant parameter map, which would select just one point rather than traverse an interval of the curve.

Why the Length Formula Works

Here is a direct justification. The chain rule gives $\|\mathbf{q}'(u)\|=|a|\|\mathbf{r}'(au+b)\|$. If $a>0$, use the substitution $t=au+b$, so $dt=a\,du=|a|\,du$. The endpoints $u=c,d$ map in increasing order to $ac+b,ad+b$, giving the theorem’s integral. If $a<0$, then $dt=a\,du$ has the opposite sign and the endpoints map in decreasing order. Reversing the limits contributes another negative sign, so the resulting length integral has increasing endpoints.

Differentiate the changed curve.
For $\mathbf{q}(u)=\mathbf{r}(au+b)$, the chain rule gives $\mathbf{q}'(u)=a\mathbf{r}'(au+b)$.

Take the magnitude.
The speed is $|a|\|\mathbf{r}'(au+b)\|$, not $a\|\mathbf{r}'(au+b)\|$ when $a$ is negative.

Transform the length integral.
Substitute $t=au+b$ and account for the endpoint order. The integral covers precisely the original parameter interval reached by $u\in[c,d]$.

Worked Examples

Worked Example: Reversing a Helix

Let $\mathbf{r}(t)=(\cos t,\sin t,t/2)$ for $0\le t\le2\pi$. Reverse its traversal using $\mathbf{q}(u)=\mathbf{r}(2\pi-u)$ for $0\le u\le2\pi$.

Here $a=-1$ and $b=2\pi$. Differentiating the original curve gives

\mathbf{r}'(t)=(-\sin t,\cos t,1/2), \qquad \|\mathbf{r}'(t)\| =\sqrt{\sin^2t+\cos^2t+\frac14} =\frac{\sqrt5}{2}.

By the chain rule,

\mathbf{q}'(u) =-\mathbf{r}'(2\pi-u) =(\sin(2\pi-u),-\cos(2\pi-u),-1/2).

Therefore $\|\mathbf{q}'(u)\|=\sqrt5/2$. The endpoints show the reversal explicitly:

\mathbf{q}(0)=\mathbf{r}(2\pi)=(1,0,\pi), \qquad \mathbf{q}(2\pi)=\mathbf{r}(0)=(1,0,0).

Both lengths are $(\sqrt5/2)(2\pi)=\pi\sqrt5$. Thus the direction is reversed, the speed is unchanged in magnitude, and the length is the same. As a check, the new velocity has third component $-1/2$, confirming that the helix moves downward as $u$ increases.

Worked Example: Reversing a Curve with Nonconstant Speed

Let $\mathbf{r}(t)=(t,t^2)$ on $0\le t\le2$, and set $\mathbf{q}(u)=\mathbf{r}(2-u)$ on $0\le u\le2$. The original velocity and speed are

\mathbf{r}'(t)=(1,2t), \qquad \|\mathbf{r}'(t)\|=\sqrt{1+4t^2}.

The reversed curve is $\mathbf{q}(u)=(2-u,(2-u)^2)$, so

\mathbf{q}'(u)=(-1,-2(2-u))=(-1,-4+2u), \qquad \|\mathbf{q}'(u)\|=\sqrt{1+4(2-u)^2}.

The speed varies with $u$, and it is the original speed evaluated at $2-u$. Its length is

L_{\mathbf{q}} =\int_0^2\sqrt{1+4(2-u)^2}\,du =\int_0^2\sqrt{1+4t^2}\,dt,

where the last equality uses $t=2-u$, with the limits reversed by the substitution. An antiderivative is $F(t)=\frac{t}{2}\sqrt{1+4t^2}+\frac14\operatorname{asinh}(2t)$. Indeed, differentiating gives $F'(t)=\sqrt{1+4t^2}$. Hence

L_{\mathbf{q}}=F(2)-F(0) =\sqrt{17}+\frac14\operatorname{asinh}(4) =L_{\mathbf{r}}.

The endpoints also check the direction: $\mathbf{q}(0)=(2,4)=\mathbf{r}(2)$ and $\mathbf{q}(2)=(0,0)=\mathbf{r}(0)$. The curve is traversed backward, but its total length is unchanged.

Worked Example: Rescaling a Nonconstant-Speed Curve

Let $\mathbf{r}(t)=(t,t^2)$ for $0\le t\le3$, and define $\mathbf{q}(u)=\mathbf{r}(1+2u)$ for $0\le u\le1$. The mapped original parameter is $t=1+2u$, which runs from $1$ to $3$; the new curve covers only that portion of the original curve.

The derivative of the original curve is $\mathbf{r}'(t)=(1,2t)$. Applying the chain rule yields

\mathbf{q}'(u)=2\mathbf{r}'(1+2u) =(2,4+8u),

and therefore

\|\mathbf{q}'(u)\| =\sqrt{4+(4+8u)^2} =2\sqrt{1+4(1+2u)^2}.

The factor $2$ reflects that the original parameter changes twice as fast per unit $u$. To compute the length, substitute $t=1+2u$, so $dt=2\,du$:

L_{\mathbf{q}} =\int_0^1 2\sqrt{1+4(1+2u)^2}\,du =\int_1^3\sqrt{1+4t^2}\,dt.

Using the antiderivative $F(t)=\frac{t}{2}\sqrt{1+4t^2}+\frac14\operatorname{asinh}(2t)$, the length is

L_{\mathbf{q}} =F(3)-F(1) =\frac32\sqrt{37}-\frac12\sqrt5 +\frac14\bigl(\operatorname{asinh}(6)-\operatorname{asinh}(2)\bigr).

Finally, $\mathbf{q}(0)=\mathbf{r}(1)=(1,1)$ and $\mathbf{q}(1)=\mathbf{r}(3)=(3,9)$, confirming that the parameter interval maps to $1\le t\le3$, not to the entire original interval $0\le t\le3$.

A Common Sign and Interval Mistake

A frequent error is to treat the scalar factor in the velocity as though it were also a factor in speed without taking its absolute value. For instance, let $\mathbf{r}(t)=(t,t^2)$ and $\mathbf{q}(u)=\mathbf{r}(3-2u)$ for $0\le u\le1$. The chain rule gives $\mathbf{q}'(u)=-2\mathbf{r}'(3-2u)$. The incorrect calculation $\|\mathbf{q}'(u)\|=-2\|\mathbf{r}'(3-2u)\|$ would give a negative speed. The correction is

\|\mathbf{q}'(u)\| =\|-2\mathbf{r}'(3-2u)\| =2\|\mathbf{r}'(3-2u)\|.

There is a second related error: using a $u$-interval that maps outside the domain of $\mathbf{r}$. Before differentiating or integrating, compute the endpoint images $ac+b$ and $ad+b$. For example, if $\mathbf{r}$ is defined only on $0\le t\le3$, then $\mathbf{r}(1+3u)$ is not defined for all $0\le u\le1$, because $1+3u$ reaches $4$. The interval $0\le u\le1$ is valid for $\mathbf{r}(1+2u)$, whose original parameter values range from $1$ to $3$.

What to Track in Practice

When comparing a curve with an affine reparametrization, keep the three effects separate:

Interval: Map both new parameter endpoints through $t=au+b$ to identify the original portion of the curve.
Direction: If $a>0$, parameter order is preserved; if $a<0$, it is reversed.
Speed and length: Speed is multiplied by $|a|$ at the corresponding original parameter value. The length integral covers the mapped original interval, with its endpoints written in increasing order.

These facts describe affine parameter changes without requiring a curve to have constant speed. The examples show why the chain rule and the interval map must be considered together: the velocity factor alone does not tell you which portion of the path is being traversed.

Check Your Understanding

For each question, track the parameter interval as well as the derivative.

If $\mathbf{q}(u)=\mathbf{r}(5-3u)$ for $0\le u\le1$, what original parameter interval is covered, and is its traversal direction preserved or reversed?
If $\mathbf{q}(u)=\mathbf{r}(2u+1)$, how does its speed compare with the speed of $\mathbf{r}$ at the corresponding parameter value?
Why must the speed factor for $\mathbf{q}(u)=\mathbf{r}(b-4u)$ use $4$, rather than $-4$?
For $\mathbf{r}(t)=(t,t^2)$ on $0\le t\le3$, is $\mathbf{r}(1+3u)$ defined throughout $0\le u\le1$? Explain using the mapped endpoints.
State the length of $\mathbf{q}(u)=\mathbf{r}(2-u)$, $0\le u\le2$, in terms of the speed of $\mathbf{r}$.