Two Equations, One Point at a Time
A pair of parametric equations assigns a point in the plane to each allowed value of a parameter. The essential step is to use the same parameter value in both equations: its output is the ordered pair \((x,y)\). The equations are not two unrelated lists of possible \(x\)- and \(y\)-coordinates.
This viewpoint is useful whenever a curve is described parametrically rather than by a single equation involving \(x\) and \(y\). A value table makes the pairing explicit. The parameter interval then tells you which parameter values are allowed and, consequently, which points the equations produce.
Making a Value Table
A useful table has three columns: the parameter value, the corresponding \(x\)-coordinate, and the corresponding \(y\)-coordinate. Choose parameter values that include interval endpoints when they are allowed, as well as convenient interior values. For trigonometric functions, familiar angles often give exact coordinates.
| Parameter \(t\) | \(x=f(t)\) | \(y=g(t)\) | Point \((x,y)\) |
|---|---|---|---|
| \(t_1\) | \(f(t_1)\) | \(g(t_1)\) | \((f(t_1),g(t_1))\) |
| \(t_2\) | \(f(t_2)\) | \(g(t_2)\) | \((f(t_2),g(t_2))\) |
The last column is the important one when plotting: it records the coordinates as a pair. Plotting every \(x\)-value against every \(y\)-value would create points that the equations may never produce.
Worked Example: A Line Segment from Paired Coordinates
Interpret \(x=2t-1\) and \(y=t+2\) for \(-1\leq t\leq 2\). First make a table using the two endpoints and convenient interior values.
| \(t\) | \(x=2t-1\) | \(y=t+2\) | \((x,y)\) |
|---|---|---|---|
| \(-1\) | \(2(-1)-1=-3\) | \(-1+2=1\) | \((-3,1)\) |
| \(0\) | \(2(0)-1=-1\) | \(0+2=2\) | \((-1,2)\) |
| \(1\) | \(2(1)-1=1\) | \(1+2=3\) | \((1,3)\) |
| \(2\) | \(2(2)-1=3\) | \(2+2=4\) | \((3,4)\) |
To identify the curve, solve \(y=t+2\) for the parameter: \(t=y-2\). Substitute this into the \(x\)-equation:
The points lie on a line. Since \(t\) ranges from \(-1\) to \(2\), the endpoint values produce \((-3,1)\) and \((3,4)\), so the represented part is the line segment between those endpoints, including both. The calculation checks: at \((-3,1)\), \((x+5)/2=(-3+5)/2=1=y\); at \((3,4)\), \((x+5)/2=(3+5)/2=4=y\). The intermediate table points also satisfy the equation: \((-1+5)/2=2\) and \((1+5)/2=3\).
The Parameter Interval Selects the Portion
Eliminating the parameter, when possible, can reveal an equation that the points satisfy. But that equation alone may describe more points than the parametrization produces. The parameter interval is part of the description of the curve, not a detail to discard.
For instance, \(x=t^2,\ y=t\) satisfies \(x=y^2\). If \(t\) ranges over all real numbers, the parametrization produces the full right-opening parabola \(x=y^2\). If \(t\) is restricted, it produces only the points corresponding to values in that restricted interval.
Worked Example: A Restricted Portion of a Parabola
Interpret \(x=t^2\) and \(y=t\) for \(-2\leq t\leq 1\). Evaluate each coordinate at the same parameter value.
| \(t\) | \(x=t^2\) | \(y=t\) | \((x,y)\) |
|---|---|---|---|
| \(-2\) | \((-2)^2=4\) | \(-2\) | \((4,-2)\) |
| \(-1\) | \((-1)^2=1\) | \(-1\) | \((1,-1)\) |
| \(0\) | \(0^2=0\) | \(0\) | \((0,0)\) |
| \(1\) | \(1^2=1\) | \(1\) | \((1,1)\) |
Because \(y=t\), the parameter interval becomes \(-2\leq y\leq 1\). Also \(x=t^2=y^2\). Thus the represented curve is the part of \(x=y^2\) whose \(y\)-coordinates range from \(-2\) to \(1\), including the points \((4,-2)\) and \((1,1)\). Check the table against the equation: for \((4,-2)\), \(y^2=4=x\); for \((1,-1)\), \(y^2=1=x\); for \((0,0)\), \(y^2=0=x\); and for \((1,1)\), \(y^2=1=x\).
The equation \(x=y^2\) by itself does not specify this restricted portion: it also includes points with \(y<-2\) or \(y>1\). The interval restriction is what excludes them.
Example: Reading a Semicircle from a Table
Worked Example: A Trigonometric Parametrization
Interpret \(x=3\cos t\) and \(y=3\sin t\) for \(0\leq t\leq\pi\). Use familiar angles to create sample points.
| \(t\) | \(x=3\cos t\) | \(y=3\sin t\) | \((x,y)\) |
|---|---|---|---|
| \(0\) | \(3(1)=3\) | \(3(0)=0\) | \((3,0)\) |
| \(\pi/6\) | \(3(\sqrt{3}/2)=3\sqrt{3}/2\) | \(3(1/2)=3/2\) | \((3\sqrt{3}/2,3/2)\) |
| \(\pi/2\) | \(3(0)=0\) | \(3(1)=3\) | \((0,3)\) |
| \(5\pi/6\) | \(3(-\sqrt{3}/2)=-3\sqrt{3}/2\) | \(3(1/2)=3/2\) | \((-3\sqrt{3}/2,3/2)\) |
| \(\pi\) | \(3(-1)=-3\) | \(3(0)=0\) | \((-3,0)\) |
The points lie on the circle \(x^2+y^2=9\), since
Furthermore, \(\sin t\geq 0\) for \(0\leq t\leq\pi\), so \(y\geq 0\). These facts show that every produced point lies on the upper semicircle.
To check that the parametrization produces every point of that semicircle, take any point \((x,y)\) with \(x^2+y^2=9\) and \(y\geq 0\). Then \(x/3\) lies between \(-1\) and \(1\). Choose \(t=\arccos(x/3)\), so \(0\leq t\leq\pi\) and \(3\cos t=x\). Also, \(\sin t\geq0\), and
Both \(3\sin t\) and \(y\) are nonnegative, so \(3\sin t=y\). Thus this parameter value produces the chosen point. The represented curve is exactly the upper semicircle, including both endpoints \((3,0)\) and \((-3,0)\). Direct substitution checks them: \(t=0\) gives \((3,0)\), and \(t=\pi\) gives \((-3,0)\).
Why the Paired-Coordinate Method Works
The definition of the curve is a set of outputs: for every allowed parameter value \(t\), the two functions give one ordered pair \((f(t),g(t))\). A value table simply evaluates that definition at selected inputs. Plotting its ordered pairs displays samples of the set.
When \(f\) and \(g\) are continuous on an interval, the point \((f(t),g(t))\) changes continuously as \(t\) changes. This is why the graph is read as a continuously varying curve rather than as unrelated table entries. However, a finite table alone does not prove the exact shape between its rows. To identify the entire curve, use the formulas and interval as well as the sample points. For example, the circle example used the identity \(\cos^2t+\sin^2t=1\) and the interval’s sine values to justify the full semicircle, not just the five plotted samples.
Endpoint Parameters and Endpoint Points
An interval specifies which parameter values are allowed. A closed endpoint, such as \(t=b\) in \([a,b]\), is allowed; an open endpoint, such as \(t=b\) in \([a,b)\), is not. But excluding a parameter value does not automatically mean that the point it would produce is absent. The same point might be produced by another allowed parameter value.
For a simple case where the omitted endpoint point is absent, let \(x=t,\ y=t\) with \(0\leq t<1\). The parameter \(t=1\) is not allowed, and no other allowed \(t\) gives \((1,1)\), so that point is absent. In contrast, for \(x=\cos t,\ y=\sin t\) with \(0\leq t<2\pi\), the parameter \(2\pi\) is excluded, but its point \((1,0)\) is still included because the allowed value \(t=0\) also produces \((1,0)\).
Common Mistake: Mixing Coordinates from Different Rows
Suppose \(x=t^2\) and \(y=t\), with \(t\) taking the values \(0\) and \(1\). The table gives \((0,0)\) at \(t=0\), and \((1,1)\) at \(t=1\). A mistaken plot might combine \(x=0\) from the first row with \(y=1\) from the second row, claiming that \((0,1)\) is on the curve.
That is incorrect: no single parameter value produced that pair. At \(t=0\), the coordinates are \((0^2,0)=(0,0)\); at \(t=1\), they are \((1^2,1)=(1,1)\). The point \((0,1)\) also fails the relation \(x=y^2\), since \(0\neq1^2\). Always keep coordinates from the same row together.
A Practical Reading Process
Write down the interval and note whether each endpoint is included.
Include allowed endpoints and convenient interior values; use exact function values when available.
For each chosen \(t\), calculate \(x=f(t)\) and \(y=g(t)\), then record \((x,y)\).
Use the formulas to determine what curve contains the points and the interval to determine which portion is represented.
An omitted parameter’s point is absent only when no other allowed parameter yields it.
This process helps prevent two errors: treating the coordinate equations independently and mistaking an equation containing the curve for a complete description of the parametrized portion. A correct interpretation accounts for both the paired outputs and the parameter domain.
Check Your Understanding
For each question, keep the two coordinate equations paired and pay attention to the parameter interval.
- For \(x=t+1,\ y=2t\) on \(-1\leq t\leq 1\), make a table at \(t=-1,0,1\). What are the endpoint points?
- For \(x=t^2,\ y=t\) on \(0\leq t\leq 3\), what relation do the points satisfy, and what range of \(y\)-values is represented?
- For \(x=2\cos t,\ y=2\sin t\) on \(0\leq t\leq\pi\), what circle equation do the points satisfy, and which half of the circle is included?
- Why is a table of four parameter values not, by itself, a proof that the entire curve is a line?
- If an endpoint parameter is excluded, what must be checked before deciding that its corresponding point is absent?
