Using a Free Coordinate as the Parameter

Choosing a Coordinate as the Parameter

When a curve is described by equations relating \(x\), \(y\), and \(z\), it can be useful to let one coordinate vary freely and express the other two in terms of it. That coordinate then serves as the parameter. This approach can turn a system of curve constraints into a direct vector-valued parametrization.

The word “freely” needs care. A chosen coordinate is free only over values for which the constraints can actually be solved with real coordinates. A square root, for example, restricts which parameter values are allowed. Also, solving an equation may produce multiple roots, so one formula might describe only one branch of the curve.

Definition: A coordinate is used as a parameter when it is set equal to the parameter, such as \(z=t\), and the curve constraints are solved for the other coordinates as functions of \(t\). The parameter interval or domain must include exactly the values used for the intended branch or portion of the curve.

The Basic Procedure

Suppose a curve is specified by two constraints. To use \(z\) as the parameter, set \(z=t\), substitute this into both constraints, and solve for \(x\) and \(y\). If the solutions are \(x=f(t)\) and \(y=g(t)\), the resulting parametrization is

\mathbf{r}(t)=\langle f(t),g(t),t\rangle.

The allowed parameter values are those for which the formulas are defined and real, and for which the original constraints hold. If solving for the coordinates gives more than one choice, each choice may define a separate branch. The same reasoning applies if \(x\) or \(y\), rather than \(z\), is chosen as the parameter.

Choose the coordinate.
Set that coordinate equal to \(t\), and substitute it into the curve constraints.

Solve for the remaining coordinates.
Keep track of every solution, including distinct signs when a square root appears.

Find the permitted parameter values.
Require the coordinate formulas to be defined and real, and check any other restrictions imposed by the constraints.

Write and check the parametrization.
Substitute its coordinates back into the original constraints and verify which branch or branches it describes.

Why This Produces the Curve

The key is to preserve both directions of the relationship between the equations and the parametrization. A parametrization must produce points that satisfy the constraints, but it must also include every point on the intended branch.

Result: Suppose a curve is defined by constraints on \((x,y,z)\), and choose \(z\) as the parameter. If, for every permitted \(t\), the constraints have the solution \(x=f(t)\), \(y=g(t)\), then \(\mathbf{r}(t)=\langle f(t),g(t),t\rangle\) lies on the curve. If every point on the intended branch has \(z\)-coordinate \(t\) and its other coordinates equal \(f(t)\) and \(g(t)\), the parametrization covers that branch.

Indeed, for any permitted \(t\), the coordinates of \(\mathbf{r}(t)\) are precisely the solution of the constraints obtained after setting \(z=t\). Thus the point satisfies the constraints. Conversely, take a point on the intended branch. Its \(z\)-coordinate supplies a parameter value \(t\). If the chosen solution formulas give that point’s \(x\)- and \(y\)-coordinates, the point is \(\mathbf{r}(t)\). This second check matters: satisfying the equations proves that the parametrization stays on the curve, but does not by itself prove that it covers all of the desired branch.

Worked Examples

Worked Example: Solving Two Linear Constraints with \(z\) as Parameter

Find a parametrization of the curve defined by \(x+y+z=6\) and \(2x-y+z=3\), using \(z\) as the parameter.

Set \(z=t\). The two equations for \(x\) and \(y\) become

x+y=6-t,\qquad 2x-y=3-t.

Add the equations to eliminate \(y\):

3x=9-2t,\qquad x=3-\frac{2t}{3}.

Substitute this value into \(x+y=6-t\):

y=6-t-\left(3-\frac{2t}{3}\right) =3-\frac{t}{3}.

There are no square roots or other restrictions, so \(t\) can be any real number. The parametrization is

\mathbf{r}(t)=\left\langle 3-\frac{2t}{3},\,3-\frac{t}{3},\,t\right\rangle,\qquad t\in\mathbb{R}.

Check the first constraint:

\left(3-\frac{2t}{3}\right)+\left(3-\frac{t}{3}\right)+t =6-\frac{3t}{3}+t=6.

Check the second:

2\left(3-\frac{2t}{3}\right)-\left(3-\frac{t}{3}\right)+t =6-\frac{4t}{3}-3+\frac{t}{3}+t =3.

Both constraints hold for every real \(t\). Since the equations determine \(x\) and \(y\) uniquely once \(z=t\) is fixed, this parametrization gives the entire curve.

Worked Example: A Plane-Sphere Intersection and Its Two Branches

Parametrize the curve of intersection of the sphere \(x^2+y^2+z^2=25\) and the plane \(y=0\), using \(z\) as the parameter. Describe the two branches.

Set \(z=t\) and \(y=0\). The sphere equation reduces to

x^2+t^2=25,\qquad x^2=25-t^2.

Solving for \(x\) gives two possibilities:

x=\sqrt{25-t^2} \qquad\text{or}\qquad x=-\sqrt{25-t^2}.

For these coordinates to be real, \(25-t^2\) must be nonnegative. Equivalently, \(t^2\leq 25\), so \(-5\leq t\leq 5\). The two parametrizations are

\mathbf{r}_{+}(t)=\left\langle \sqrt{25-t^2},\,0,\,t\right\rangle,\qquad -5\leq t\leq 5,

\mathbf{r}_{-}(t)=\left\langle -\sqrt{25-t^2},\,0,\,t\right\rangle,\qquad -5\leq t\leq 5.

To check either formula, let \(x=\pm\sqrt{25-t^2}\), \(y=0\), and \(z=t\). Then

x^2+y^2+z^2=(25-t^2)+0+t^2=25,\qquad y=0.

Thus both parametrizations lie on the intersection. For any point on that intersection, its \(z\)-coordinate is some \(t\) in \([-5,5]\), and its \(x\)-coordinate is one of the two displayed roots. The two formulas therefore cover the full intersection, with the branches meeting at \(t=-5\) and \(t=5\), where \(x=0\).

Worked Example: Choosing \(y\) as the Parameter

A curve is defined by \(y=x^2\) and \(z=x+1\). Use \(y\) as the parameter and give parametrizations for the branches.

Set \(y=t\). The first constraint becomes \(t=x^2\), so \(t\geq 0\) and

x=\sqrt{t}\qquad\text{or}\qquad x=-\sqrt{t}.

Using \(z=x+1\), the corresponding \(z\)-coordinates are \(1+\sqrt{t}\) and \(1-\sqrt{t}\). Hence the two branch formulas are

\mathbf{r}_{+}(t)=\langle \sqrt{t},\,t,\,1+\sqrt{t}\rangle,\qquad t\geq 0,

\mathbf{r}_{-}(t)=\langle -\sqrt{t},\,t,\,1-\sqrt{t}\rangle,\qquad t\geq 0.

For either sign, \(x^2=t=y\), so the first constraint is satisfied. Also, \(z=x+1\) by construction, which verifies the second constraint. Conversely, any point on the curve has \(y=x^2\geq 0\), so \(t=y\) is an allowed parameter value. Its \(x\)-coordinate is either \(\sqrt{t}\) or \(-\sqrt{t}\), and its \(z\)-coordinate must then be \(x+1\). Therefore the two formulas cover the whole curve. At \(t=0\), both give the shared point \(\langle 0,0,1\rangle\).

Parameter Intervals and Branches

A parameter domain is part of the parametrization, not an afterthought. It records which coordinate values actually produce points on the curve through the chosen solution formulas. For a square root \(\sqrt{q(t)}\), the real-valued formula requires \(q(t)\geq 0\). Other constraints may impose additional restrictions.

A branch is often identified by a consistent choice among multiple solutions, such as the positive or negative square root. The parameter interval then determines how much of that branch is traced. Endpoints should be included when they are allowed by the original constraints and are part of the requested curve.

Common pitfall: In the sphere-plane example, writing \(x=\sqrt{25-t^2}\) with \(t\in\mathbb{R}\) is invalid: for \(t=6\), the radicand is \(25-36=-11\), so the coordinate is not real. Moreover, using only the positive root misses points with \(x<0\). The correction is to restrict \(t\) to \([-5,5]\) and use both signs when the full intersection is required.

A useful final check is to ask two separate questions. First, does every point produced by the formula satisfy every original constraint? Second, can every point on the intended branch be obtained from some allowed parameter value? These checks catch different errors: the first detects formulas that leave the curve, while the second detects omitted branches or an interval that is too small.

Check Your Understanding

For each question, identify both the coordinate formulas and the permitted parameter values.

If \(x+y=5\) and \(y-z=2\), set \(z=t\) and solve for \(x\) and \(y\). What is the parameter domain?
For \(x^2+z^2=16\) and \(y=2\), use \(z=t\). What interval is allowed, and how many root choices are there for \(x\)?
For \(x=y^2\) and \(z=3y\), use \(x=t\). What restriction does \(x=y^2\) place on \(t\), and what are the possible values of \(y\)?
Why does verifying that a parametrization satisfies the constraints not, by itself, prove that it covers the entire curve?
For the sphere-plane intersection in the worked example, which points are omitted if only the positive square root is used?