Changing the Parameter Without Changing the Curve
A space curve can have more than one parametrization. Even when two formulas describe the same trace, they may use different parameter intervals, start at different points, or travel around the trace in opposite directions.
An affine change of parameter has the form \(t=a+bu\), where \(a\) and \(b\) are constants. It lets us express an existing parametrization using a new parameter \(u\). The essential bookkeeping is to transform the interval as well as the formula: using the wrong interval can produce a different part of the curve.
The condition \(b\ne 0\) matters. If \(b=0\), then \(t=a\) for every \(u\), so the new formula stays at one point instead of traversing the original curve.
Transforming a Closed Parameter Interval
Suppose the original parameter interval is \([T_0,T_1]\), where \(T_0<T_1\). Solving \(t=a+bu\) for \(u\) gives \(u=(t-a)/b\). The original endpoints therefore correspond to the two values \((T_0-a)/b\) and \((T_1-a)/b\). If \(b\) is negative, their order is reversed, so the new interval must be written in increasing order.
Equivalently, if \(b>0\), the new interval is \([ (T_0-a)/b,\,(T_1-a)/b]\). If \(b<0\), it is \([ (T_1-a)/b,\,(T_0-a)/b]\). These formulas keep the interval endpoints in increasing order; they do not determine which endpoint is encountered first as \(u\) increases.
Why the Substitution Works
The claim that the transformed parametrization traces the same points follows from the one-to-one correspondence between the two parameters. The proof also explains why the sign of \(b\) determines the direction.
For every allowed \(u\), the substitution assigns the original parameter \(t=a+bu\). Since \(b\ne0\), every \(t\) in the original interval has exactly one corresponding value \(u=(t-a)/b\).
At corresponding parameter values, \(\mathbf{q}(u)=\mathbf{r}(a+bu)=\mathbf{r}(t)\). Therefore every point in one parametrization is a point in the other, and the one-to-one correspondence shows that neither parametrization adds or loses points.
If \(u_1<u_2\), then \(a+bu_1<a+bu_2\) when \(b>0\), so increasing \(u\) follows increasing \(t\). When \(b<0\), \(a+bu_1>a+bu_2\), so increasing \(u\) follows decreasing \(t\).
This proves that the trace is unchanged on the corresponding intervals and that the direction is preserved or reversed according to the sign of \(b\). For a differentiable parametrization, the same fact is expressed by the chain rule: \(\mathbf{q}'(u)=b\,\mathbf{r}'(a+bu)\). The factor \(b\) records the effect of the change on the parameter’s direction.
Worked Examples with Lines
Worked Example: A Positive Change of Parameter on a Line
Consider \(\mathbf{r}(t)=\langle 1+2t,\,-3+t,\,4-2t\rangle\) for \(-1\le t\le2\). Use \(t=1+3u\). First transform the endpoints. When \(t=-1\), \(-1=1+3u\), so \(u=-2/3\). When \(t=2\), \(2=1+3u\), so \(u=1/3\). Because \(b=3>0\), the new interval is \([-2/3,1/3]\), and the direction is preserved.
Substitute \(1+3u\) for \(t\) in each component:
At the new starting value \(u=-2/3\), the corresponding old parameter is \(t=1+3(-2/3)=-1\). Both formulas give the starting point \(\mathbf{r}(-1)=\langle -1,-4,6\rangle\). At \(u=1/3\), \(t=2\), and both give the endpoint \(\mathbf{r}(2)=\langle 5,-1,0\rangle\). The transformed formula and interval therefore describe the same line segment, with the same direction of travel.
Worked Example: A Negative Change of Parameter on a Line
Let \(\mathbf{r}(t)=\langle t,\,2t+1,\,-t+4\rangle\) for \(-1\le t\le3\), and use \(t=1-2u\). The endpoint \(t=3\) corresponds to \(3=1-2u\), hence \(u=-1\). The endpoint \(t=-1\) corresponds to \(-1=1-2u\), hence \(u=1\). Thus \(u\in[-1,1]\). Since \(b=-2<0\), increasing \(u\) reverses the direction of travel.
Substitution gives
At \(u=-1\), the old parameter is \(t=1-2(-1)=3\), so the new traversal starts at \(\langle 3,7,1\rangle\). Substituting \(u=-1\) into \(\mathbf{q}\) gives \(\langle 3,7,1\rangle\) as well. At \(u=1\), \(t=1-2=-1\), and both formulas give \(\langle -1,-1,5\rangle\). The endpoints occur in the opposite order from the original parametrization, as required.
Phase Changes on a Circle
A standard parametrization of a circle in a coordinate plane uses cosine and sine. For example, \(\mathbf{r}(t)=\langle 2+3\cos t,\,-1+3\sin t,\,5\rangle\) traces a circle centered at \(\langle 2,-1,5\rangle\) with radius \(3\). The periodicity of sine and cosine means that shifting the parameter by \(2\pi\) does not change the point. A shift by a different amount can change the starting point while preserving the full circular trace, provided the interval still covers a full period.
Worked Example: Shifting the Starting Point on a Full Circle
Start with the circle parametrization \(\mathbf{r}(t)=\langle 2+3\cos t,\,-1+3\sin t,\,5\rangle\), \(0\le t\le2\pi\). Set \(t=u+\pi/2\). The direct image of the original interval is found from \(u=t-\pi/2\): its endpoints are \(-\pi/2\) and \(3\pi/2\). On this interval, the new formula traces the same full circle and starts at the old point for \(t=0\).
Because the circle is periodic, we can instead use \(0\le u\le2\pi\) with this same new formula. This interval also covers one full circle, but now it starts at the point corresponding to \(t=\pi/2\). Using the angle addition identities gives
At \(u=0\), the new formula gives \(\langle 2,2,5\rangle\), which is the original point \(\mathbf{r}(\pi/2)\). At \(u=2\pi\), it gives \(\langle 2,2,5\rangle\) again. As a further check, at \(u=\pi/2\), it gives \(\langle -1,-1,5\rangle\), equal to \(\mathbf{r}(\pi)\). The coefficient of \(u\) in \(t=u+\pi/2\) is positive, so the direction is preserved.
The different starting points result from shifting the interval by \(\pi/2\) while retaining the same length \(2\pi\), not from shifting by a full period. A shift by \(2\pi\) would leave the starting point unchanged because both sine and cosine have period \(2\pi\).
A Common Interval Error
A frequent mistake is to change the formula but keep the old parameter interval. For instance, take \(t=1-2u\) and an original interval \(0\le t\le4\). Keeping \(0\le u\le4\) is incorrect: as \(u\) runs over that interval, \(t=1-2u\) runs from \(1\) down to \(-7\), not from \(0\) to \(4\). The new parametrization would therefore include parameter values outside the original interval and omit part of it.
To transform the interval correctly, solve at both endpoints. The equation \(0=1-2u\) gives \(u=1/2\), and \(4=1-2u\) gives \(u=-3/2\). The new interval is \([-3/2,1/2]\). Its lower endpoint corresponds to \(t=4\), so increasing \(u\) starts at the old terminal point and travels back toward the old starting point.
Summary
An affine change of parameter modifies the formula by replacing \(t\) with \(a+bu\), and modifies the interval by solving the same relation at the original endpoints. If \(b>0\), increasing the new parameter follows the original direction; if \(b<0\), it reverses that direction. On a periodic curve such as a circle, the interval also matters when identifying the starting point: a phase shift can change that point, while a full-period shift leaves it unchanged.
Check Your Understanding
For each question, pay attention to the transformed interval as well as the new formula.
- If \(t=4+2u\) and \(t\in[0,6]\), what is the corresponding \(u\)-interval, and is the direction preserved or reversed?
- If \(t=3-5u\) and \(t\in[-2,3]\), which original endpoint corresponds to the lower endpoint of the transformed interval?
- For a full-period circle parametrization, why can shifting the phase by \(\pi/2\) change the starting point without changing the trace?
- What goes wrong if \(b=0\) in the substitution \(t=a+bu\)?
- How can you check that a transformed interval has the correct endpoints?