Parametric functions

A parametric function pairs two ordinary functions of the same input: $f(t) = (x(t), y(t))$, with t drawn from a stated interval. For each t you compute both coordinates and plot the point. The variable t is the parameter; it does not appear on either axis. Often t is time, but it can be an angle, a distance, or a bare index; "t is always time" is a habit, not a rule.

To sketch, build a table of t values and their points, then connect them in order of t. For $x = t - 1$, $y = t^2$ on $-2 \le t \le 2$: at t = −2 the point is (−3, 4), at t = 0 it is (−1, 0), at t = 2 it is (1, 4). The curve is a piece of a parabola swept from the upper left down through its vertex and back up to the upper right.

Because the points arrive in t order, a parametric curve is a journey, not just a set. Three questions always make sense that never arise for y = f(x):

1. Where does it start and end? Evaluate at the interval's endpoints.
2. Which way does it travel? Check two nearby t values and see which point comes second.
3. How much of the shape is actually drawn? The traced piece is only what the t interval reaches.

The circle $x = \cos t$, $y = \sin t$ makes all three vivid: on $0 \le t \le 2\pi$ it starts at (1, 0) and sweeps the whole circle counterclockwise; on $0 \le t \le \pi$ it draws only the top half; swapping to $x = \sin t$, $y = \cos t$ starts at (0, 1) and runs clockwise instead. Same equation-shapes, three different journeys.

Solving one equation for t and substituting into the other produces a familiar x-y equation. From $x = t - 1$, $y = t^2$: t = x + 1, so $y = (x+1)^2$. That is genuinely useful; it names the shape the path lives on.

But the substitution only says the traced points satisfy the x-y equation. It does not say they cover all of it. With $-2 \le t \le 2$, x = t − 1 only reaches $-3 \le x \le 1$: the journey is a clipped arc of the parabola, not the whole thing. And the x-y equation has no memory of direction or starting point at all. The honest workflow: eliminate to identify the shape, then return to the t interval to mark off which piece is drawn, from where to where, in which direction.

Watch for parametrizations that fold over themselves: $x = t^2$, $y = t^4$ on $-1 \le t \le 1$ lives on the parabola $y = x^2$, but x never leaves [0, 1] and the point traces that little arc twice, in from (1,1), down to (0,0), and back out. Elimination alone would never tell you.

A path does not own a unique parametrization. The segment from (0, 0) to (4, 8) is traced by $x = t, y = 2t$ on [0, 4], by $x = 2t, y = 4t$ on [0, 2] (same trip, twice the speed), and by $x = 4 - t, y = 8 - 2t$ on [0, 4] (same segment, opposite direction). All three are correct descriptions of that segment; they differ in pacing and orientation, which is often the point.

So "find a parametrization" questions have many right answers, and re-parametrizing is a skill: to reverse a journey swap in $-t$ (or run the interval backward); to speed it up scale t; to start elsewhere shift t. What stays fixed is the underlying set of points; what you control is how it is traveled.