Workshop: Building and Comparing Space-Curve Parametrizations

Building a Parametrization from Geometric Conditions

A geometric description tells us what a curve must satisfy, but it does not always specify a unique formula. A line segment can be traversed at different rates; a circular arc can be described using different angle variables; and a spatial curve can use a parameter tied to its height rather than its angle. The practical task is to choose a parameter, encode the conditions, and then check that the formula produces exactly the intended curve.

Those checks have several parts. We verify the stated geometric constraints, confirm the parameter interval produces the intended endpoints or extent, and inspect the order in which points are traversed. These parts should not be collapsed into one another: having the right points in the trace does not by itself establish that the parametrization traverses them in the intended direction.

Definition: A parametrization of a curve is a vector-valued function $\mathbf r:I\to\mathbb R^3$, where $I$ is an interval of real parameter values. Its trace is the set of points it reaches: $$ \{\mathbf r(t):t\in I\}. $$ The parameter interval determines which parameter values are used, while the formula and interval together determine the trace and the order of traversal.

When a curve lies in a plane, we still write its points as vectors in $\mathbb R^3$; one coordinate or a plane equation can express the planar constraint. A parametrization is successful only relative to the geometric description and the specified parameter interval. A formula may satisfy a surface equation but still trace too much, too little, or the same portion in the wrong direction.

Start with the Conditions

Before choosing a formula, list what the description requires. For a segment, the endpoints are usually the most useful data. For a planar arc, identify its center, radius, plane, and angle range. For a spatial curve, identify equations or relationships that every point must satisfy, together with any required start, end, or extent conditions.

Then choose a parameter that makes those requirements simple. Substitution is the main verification tool: if the curve lies on a cylinder, substitute the coordinate formulas into the cylinder equation; if the endpoints are prescribed, evaluate the vector at the parameter endpoints. A check should address the interval as well as the formula, since the same formula on a larger interval can describe more of a curve.

Worked Example: Parametrizing and Reversing a Line Segment

Find a parametrization of the segment from $A=(1,-2,0)$ to $B=(5,1,2)$, and compare it with a parametrization that traverses the segment in reverse.

The displacement from $A$ to $B$ is $\mathbf B-\mathbf A=(5-1,1-(-2),2-0)=(4,3,2)$. Starting at $A$ and taking the full displacement as the parameter runs from $0$ to $1$ gives

\mathbf r(t)=(1,-2,0)+t(4,3,2) =(1+4t,-2+3t,2t),\qquad 0\le t\le1.

At the first endpoint, $\mathbf r(0)=(1,-2,0)=A$. At the other endpoint, $\mathbf r(1)=(1+4,-2+3,2)=(5,1,2)=B$. For any $t$ in the interval, the point is $A+t(B-A)$, a point on the line through $A$ and $B$. Because $0\le t\le1$, it lies between the endpoints, including both. Thus the trace is the required segment.

For the reverse traversal, start at $B$ and use the displacement $A-B=(-4,-3,-2)$:

\mathbf q(s)=(5,1,2)+s(-4,-3,-2) =(5-4s,1-3s,2-2s),\qquad 0\le s\le1.

Here $\mathbf q(0)=B$, and $\mathbf q(1)=(1,-2,0)=A$. Moreover, $\mathbf q(s)=\mathbf r(1-s)$, since $\mathbf r(1-s)=(1+4(1-s),-2+3(1-s),2(1-s))=(5-4s,1-3s,2-2s)$. As $s$ ranges from $0$ to $1$, $1-s$ ranges over the whole interval from $1$ down to $0$. Therefore the two parametrizations have the same trace but traverse it in opposite directions.

Encoding a Planar Arc

For a circular arc in a coordinate-parallel plane, the familiar sine and cosine coordinates provide a direct construction. A circle of radius $R$, centered at $(a,b,c)$ in a horizontal plane, has coordinates $x=a+R\cos\theta$, $y=b+R\sin\theta$, and $z=c$. The permitted angle interval selects the arc. Restricting the angle is essential: allowing a full revolution or several revolutions would not give just a shorter specified arc.

Worked Example: A Quarter-Circle Arc in a Horizontal Plane

Parametrize the counterclockwise quarter-circle of radius $3$ centered at $(1,-1,2)$, beginning at angle $0$ and ending at angle $\pi/2$.

Use the angle itself as parameter. The center fixes the coordinate offsets, the radius multiplies the trigonometric coordinates, and the horizontal plane keeps the third coordinate equal to $2$:

\mathbf r(t)=(1+3\cos t,-1+3\sin t,2), \qquad 0\le t\le\frac{\pi}{2}.

To check the circle condition, subtract the center coordinates: $x-1=3\cos t$ and $y+1=3\sin t$. Hence $(x-1)^2+(y+1)^2=9\cos^2t+9\sin^2t=9$, and $z=2$. At $t=0$, the point is $(1+3,-1+0,2)=(4,-1,2)$. At $t=\pi/2$, it is $(1+0,-1+3,2)=(1,2,2)$.

For an additional check of the traversal, at the intermediate value $t=\pi/4$ the point is $(1+3\sqrt2/2,-1+3\sqrt2/2,2)$, which lies halfway around the quarter-circle in angle. The angle increases from $0$ to $\pi/2$, so the arc is traversed counterclockwise when viewed from above.

An alternative formula uses $u\in[0,1]$ and sets $t=\pi u/2$:

\mathbf q(u)= \left(1+3\cos\frac{\pi u}{2}, -1+3\sin\frac{\pi u}{2},2\right).

As $u$ runs from $0$ to $1$, the angle $\pi u/2$ runs through exactly the same values as $t$. Thus the trace and traversal direction agree, although the parameter intervals and the formulas as written differ. In particular, both formulas start at $(4,-1,2)$ and end at $(1,2,2)$.

Building a Spatial Curve from Relationships

A spatial curve can be specified by simultaneous conditions. For example, the equation $x^2+y^2=4$ restricts points to a cylinder of radius $2$ around the $z$-axis. A further relationship involving $z$ and the angular position can make the curve rise as it goes around that cylinder. To construct a parametrization, choose an angle parameter and then express all coordinates in terms of it.

Worked Example: A Helix on a Cylinder

Construct the portion of the curve on the cylinder $x^2+y^2=4$ for which the angle increases from $0$ to $\pi$ while the height satisfies $z=1+2t$, where $t$ is the angle.

The cylinder equation suggests $x=2\cos t$ and $y=2\sin t$. Using the given height relationship gives

\mathbf r(t)=(2\cos t,2\sin t,1+2t), \qquad 0\le t\le\pi.

For every parameter value, $x^2+y^2=(2\cos t)^2+(2\sin t)^2=4(\cos^2t+\sin^2t)=4$, so the curve lies on the specified cylinder. Its height is exactly the required $z=1+2t$. At $t=0$, the point is $(2,0,1)$. At $t=\pi$, it is $(-2,0,1+2\pi)$. The parameter interval gives a half-turn, not a complete revolution.

We can use a unit interval instead by setting $t=\pi s$:

\mathbf q(s)=(2\cos(\pi s),2\sin(\pi s),1+2\pi s), \qquad 0\le s\le1.

Substituting $t=\pi s$ into $\mathbf r(t)$ gives exactly $\mathbf q(s)$. Since $s\mapsto\pi s$ takes $[0,1]$ onto $[0,\pi]$ in increasing order, these are two descriptions of the same trace with the same traversal direction. Specifically, $\mathbf q(0)=(2,0,1)$, $\mathbf q(1)=(-2,0,1+2\pi)$, and both satisfy the cylinder equation throughout their intervals.

Comparing Trace, Interval, and Orientation

Two parametrizations can be compared at three distinct levels. Their traces are the sets of points reached. Their parameter intervals specify the allowed parameter values. Their traversals describe the order in which those points are reached. Different intervals do not prevent two parametrizations from having the same trace, as the examples showed. Conversely, a curve equation alone may describe a full curve even when the intended interval selects only an arc.

A useful comparison method is to look for a change of parameter. Suppose $\mathbf q(s)=\mathbf r(\phi(s))$, where $\phi$ maps the entire $s$-interval onto the entire $t$-interval. Then every point reached by $\mathbf q$ is reached by $\mathbf r$, and surjectivity of $\phi$ ensures that every point reached by $\mathbf r$ is also reached by $\mathbf q$. Therefore their traces agree. If both parametrizations are injective and $\phi$ is increasing, the points are encountered in the same order; if $\phi$ is decreasing, the order is reversed.

Theorem: Let $I$ and $J$ be intervals, let $\phi:J\to I$ be a bijection, and define $\mathbf q(s)=\mathbf r(\phi(s))$. Then $\mathbf q$ and $\mathbf r$ have the same trace. If $\phi$ is increasing, the change of parameter preserves the traversal order; if $\phi$ is decreasing, it reverses that order for an injective parametrization of a simple arc.

To justify the trace claim, take any $s\in J$. Since $\phi(s)\in I$, the point $\mathbf q(s)=\mathbf r(\phi(s))$ belongs to the trace of $\mathbf r$. This proves one trace is contained in the other. Conversely, for any $t\in I$, bijectivity gives an $s\in J$ with $\phi(s)=t$; thus $\mathbf r(t)=\mathbf q(s)$, proving the reverse containment. For traversal order, an increasing $\phi$ takes successive parameter values in $J$ to values in the same order in $I$; a decreasing $\phi$ takes them in reverse order. Injectivity on a simple arc ensures that distinct points on the arc correspond to distinct parameter values, so this comparison of parameter order is also a comparison of point order.

For an injective parametrization of a simple arc with distinct endpoints, the trace and the ordered endpoints identify its direction along that arc. Do not extend this endpoint test to arbitrary curves. For a closed curve, for example, $\mathbf r(t)=(\cos 2\pi t,\sin 2\pi t,0)$ and $\mathbf q(t)=(\cos 2\pi t,-\sin 2\pi t,0)$, both on $0\le t\le1$, have the same trace and both start and end at $(1,0,0)$. Yet at $t=1/4$, the first is $(0,1,0)$ and the second is $(0,-1,0)$; they traverse the circle in opposite directions. In general, compare how points are traversed, not endpoints alone.

A Common Parametrization Error

A frequent mistake when building a segment formula is to use the endpoint vector itself as the displacement. For the segment example, this incorrect expression would be

\mathbf w(t)=(1,-2,0)+t(5,1,2), \qquad 0\le t\le1.

At $t=0$, it does start at $(1,-2,0)$, but at $t=1$ it gives $(1+5,-2+1,0+2)=(6,-1,2)$, not the required endpoint $(5,1,2)$. The mistake is adding the endpoint coordinates to the starting point. The needed displacement is endpoint minus starting point: $(5,1,2)-(1,-2,0)=(4,3,2)$. Using that difference gives $(1,-2,0)+t(4,3,2)$, which passes the endpoint check.

For any proposed parametrization, substitute the parameter endpoints and test the defining geometric conditions. Then check that the interval selects only the intended portion and compare the traversal order if direction matters. These checks are useful precisely because a plausible-looking formula can still have a wrong endpoint, an overly broad interval, or an unintended direction.

Check Your Understanding

For each question, distinguish the formula, interval, trace, and traversal where relevant.

Write a parametrization of the segment from $(0,1,-1)$ to $(2,4,3)$ on $0\le t\le1$, and verify both endpoints.
For a circle of radius $2$ centered at $(0,0,3)$ in the plane $z=3$, what angle interval would select the upper semicircle?
Explain why $\mathbf q(s)=\mathbf r(1-s)$ reverses the traversal of a simple arc when both parameters range over $[0,1]$.
Two parametrizations have the same trace but use different parameter intervals. What additional information would help determine whether their traversal directions agree?
Why do the start and end points alone fail to determine traversal direction for a closed curve?