Trace, Parametrization, and Cartesian Equation

From a Formula to a Set of Points

A vector-valued function describes a curve by assigning a point in space to each parameter value. But the function and the geometric curve it produces are not the same kind of object: the function is a rule with a domain, while the curve’s trace is a set of points.

Eliminating the parameter can reveal a Cartesian equation for that set. The important qualification is that equations alone may not record the parameter interval. A complete description may also need inequalities or other restrictions.

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 all points attained by the function: $$ \operatorname{Trace}(\mathbf{r}) = \{\mathbf{r}(t):t\in I\}. $$

If $\mathbf{r}(t)=(x(t),y(t),z(t))$, then the trace consists of the points $(x(t),y(t),z(t))$ as $t$ ranges over $I$. A Cartesian description gives conditions directly on $x,y,z$, without using the parameter $t$.

To obtain that description, solve one or more coordinate equations for $t$, substitute into the other coordinates, and keep track of which parameter values are allowed. The final check is to confirm both directions: every parametrized point satisfies the Cartesian conditions, and every point satisfying those conditions is attained for an allowed parameter value.

A Line: Eliminating a Parameter

Worked Example: A Line in Space

Consider the parametrization $\mathbf{r}(t)=(1+2t,-1+t,3)$, where $t\in\mathbb{R}$. Write its trace as Cartesian equations.

The component equations are $x=1+2t$, $y=-1+t$, and $z=3$. Solve the second equation for the parameter:

$$ y=-1+t \quad\Longrightarrow\quad t=y+1. $$

Substitute $t=y+1$ into the expression for $x$:

$$ x=1+2(y+1)=1+2y+2=3+2y. $$

Thus $x-2y=3$, and the third coordinate gives $z=3$. The Cartesian equations are

$$ x-2y=3, \qquad z=3. $$

To check, substitute the parametrized coordinates into these equations:

$$ x-2y =(1+2t)-2(-1+t) =1+2t+2-2t =3, \qquad z=3. $$

Conversely, suppose a point $(x,y,z)$ satisfies $x-2y=3$ and $z=3$. Set $t=y+1$. Then the parametrization gives the second coordinate $y=-1+t$, the third coordinate $z=3$, and the first coordinate

$$ 1+2t =1+2(y+1) =3+2y =x, $$

where the final equality follows from $x-2y=3$. So every point satisfying the Cartesian equations is attained. Since $t$ ranges over all real numbers, there are no additional bounds to include.

In three dimensions, two independent Cartesian equations can describe a line: here one equation places the point in a plane, and the other restricts it to the intersection of two planes. The converse check matters because algebraic manipulation can show that trace points satisfy equations without, by itself, proving that the equations describe no extra points.

A Circle: Using a Trigonometric Identity

Worked Example: A Circle in a Horizontal Plane

Let $\mathbf{r}(t)=(1+3\cos t,-2+3\sin t,2)$, with $0\leq t\leq 2\pi$. Find a Cartesian description of its trace.

The coordinate equations are $x=1+3\cos t$, $y=-2+3\sin t$, and $z=2$. Rearrange the first two:

$$ x-1=3\cos t, \qquad y+2=3\sin t. $$

Square each equation and add them. Using $\cos^2 t+\sin^2 t=1$, we obtain

$$ (x-1)^2+(y+2)^2 =9\cos^2 t+9\sin^2 t =9(\cos^2 t+\sin^2 t) =9. $$

Every point on the trace therefore satisfies

$$ (x-1)^2+(y+2)^2=9, \qquad z=2. $$

For the converse, suppose a point $(x,y,z)$ satisfies these equations. Define $a=(x-1)/3$ and $b=(y+2)/3$. The circle equation gives $a^2+b^2=1$. Thus the pair $(a,b)$ is a point on the unit circle, so there is a value $t\in[0,2\pi]$ for which $\cos t=a$ and $\sin t=b$. For that $t$,

$$ 1+3\cos t =1+3a =1+(x-1) =x, $$

and

$$ -2+3\sin t =-2+3b =-2+(y+2) =y. $$

The parametrization’s third coordinate is $2=z$. Hence every point satisfying the Cartesian equations occurs in the trace. The equation describes a circle of radius $3$, centered at $(1,-2,2)$ in the plane $z=2$.

The identity proves that the equations are necessary. The converse uses the fact that each point on the unit circle has a cosine-sine representation for some angle in the stated interval. Checking both directions prevents a common gap in parameter elimination.

A Parabola: Keep the Interval Restriction

Worked Example: A Restricted Part of a Parabola

Consider $\mathbf{r}(t)=(2t+1,t^2,0)$, where $0\leq t\leq 2$. Find Cartesian conditions for its trace.

The coordinate equations are $x=2t+1$, $y=t^2$, and $z=0$. Solve the first equation for $t$:

$$ x=2t+1 \quad\Longrightarrow\quad x-1=2t \quad\Longrightarrow\quad t=\frac{x-1}{2}. $$

Substituting this expression into $y=t^2$ gives

$$ y=\left(\frac{x-1}{2}\right)^2 =\frac{(x-1)^2}{4}. $$

Also, $z=0$. But the parameter restriction $0\leq t\leq 2$ gives an additional condition on $x$: multiplying by $2$ and adding $1$ yields $1\leq x\leq 5$. A complete Cartesian description is therefore

$$ y=\frac{(x-1)^2}{4}, \qquad z=0, \qquad 1\leq x\leq 5. $$

To check that the parametrization satisfies these conditions, use $x=2t+1$: then $x-1=2t$, so $(x-1)^2/4=4t^2/4=t^2=y$. The third coordinate is $0$, and $0\leq t\leq2$ implies $1\leq2t+1\leq5$.

For the reverse direction, take any point satisfying the displayed Cartesian conditions and set $t=(x-1)/2$. The bounds $1\leq x\leq5$ imply $0\leq t\leq2$. The first coordinate of $\mathbf{r}(t)$ is $2t+1=x$; its second is

$$ t^2 =\left(\frac{x-1}{2}\right)^2 =\frac{(x-1)^2}{4} =y; $$

and its third is $0=z$. Thus the conditions describe exactly the trace, not a larger curve.

Why Elimination Works

Elimination is substitution combined with a check on the allowed parameters. When one coordinate determines the parameter uniquely, the argument has a particularly direct form.

Proposition: Suppose $x(t)$ is one-to-one on the parameter interval $I$, and its values determine $t$ through $t=g(x)$. If the other coordinates are $y(t)$ and $z(t)$, then a point belongs to the trace precisely when $$ x\in x(I), \qquad y=y(g(x)), \qquad z=z(g(x)). $$ The condition $x\in x(I)$ must be retained if the range of $x(t)$ is not all real numbers.

Here is the justification. If a point belongs to the trace, it equals $(x(t),y(t),z(t))$ for some $t\in I$. Therefore $x=x(t)\in x(I)$, and the unique parameter producing this $x$ is $g(x)=t$. Substitution gives $y=y(g(x))$ and $z=z(g(x))$. This proves necessity.

Conversely, suppose the stated conditions hold. Since $x\in x(I)$, the parameter $t=g(x)$ is in $I$. The point produced at this parameter is $(x(t),y(t),z(t))$. By the conditions, its first coordinate is $x$, and its other coordinates are $y(g(x))=y$ and $z(g(x))=z$. It is therefore exactly the proposed point, proving sufficiency.

The circle example uses a different version of the same logic: the squared equation follows from the trigonometric identity, while the converse establishes that every point satisfying the equation has an allowed parameter value. In every case, a Cartesian description is complete only when it captures exactly the trace.

A Common Pitfall: Dropping a Condition

Suppose we eliminate $t$ from the line example and write only $x-2y=3$. That calculation is not false: every point on the trace does satisfy this equation. But it is an incomplete description of the trace in space. The equation alone describes a plane, because $z$ can take any value. For example, $(3,0,10)$ satisfies $x-2y=3$, but it cannot be on the parametrized line, whose third coordinate is always $3$.

The correction is to retain both equations, $x-2y=3$ and $z=3$. The same issue appears with a restricted parameter interval. In the parabola example, the equation $y=(x-1)^2/4$ and the condition $z=0$ describe the full planar parabola. They do not describe only the trace for $0\leq t\leq2$: the point $(7,9,0)$ satisfies those equations, but its corresponding parameter would be $t=(7-1)/2=3$, which is outside the allowed interval. Adding $1\leq x\leq5$ corrects the description.

Key takeaway: Eliminating a parameter gives equations that points on the trace must satisfy. To describe the trace exactly, also check the converse and retain any restrictions imposed by the parameter interval.

Check Your Understanding

For each question, distinguish the equations a point must satisfy from any restrictions needed to describe the full trace.

For $\mathbf{r}(t)=(t+4,2t-1,6)$, with $t\in\mathbb{R}$, eliminate $t$ and give Cartesian equations for the trace.
For $\mathbf{r}(t)=(2\cos t,2\sin t,1)$, with $0\leq t\leq2\pi$, what Cartesian equations follow from the trigonometric identity?
For $\mathbf{r}(t)=(t,3t^2,0)$, with $1\leq t\leq3$, eliminate $t$ and state the necessary bound on $x$.
Why is it not enough to show that every point produced by a parametrization satisfies a proposed Cartesian equation?
In the restricted-parabola example, which parameter value corresponds to $x=5$, and is it allowed?