Parametric equation

A parametric equation gives both coordinates as functions of one parameter, $x = f(t),\ y = g(t)$, for t in some range. The parameter can be an angle or a length, but here t is a clock: at each moment the pen is at $(f(t), g(t))$, and the curve is the trail it leaves. Read this way, a parametrization says more than which points lie on the curve. It also says when the pen reaches each one.

The curve and the two graphs share one unit of length. The graph of y(t), to the right of the curve, uses the curve's y axis, with t running to the right. The graph of x(t), below it, uses the curve's x axis, with t running down, so it shows x against t, not against y. Going across and down from the pen reaches the same moment on both graphs, at the red time lines. The slopes of the graphs there, $x'(t)$ and $y'(t)$, are the two components of the velocity. The t axes are scaled to fit, so how steep a graph looks is not the true size of its slope.

One curve has many parametrizations, and the clock tells them apart. On the unit circle, $(\cos t, \sin t)$ goes round once, $(\cos 2t, \sin 2t)$ twice as fast, and $(\cos t, -\sin t)$ the other way. The rational pen $\bigl(\tfrac{1 - t^2}{1 + t^2}, \tfrac{2t}{1 + t^2}\bigr)$ stays on the circle but never reaches (−1, 0); the closer it gets to that point, the slower it moves. On $y = x^2$, $(t^3, t^6)$ pauses at the vertex and $(\sin t, \sin^2 t)$ goes back and forth. Eliminating t leaves only a relation between x and y, the trace in the readout. What it loses is when the pen is where, which way it goes, how fast, and how many times it passes. The figure eight, for example, passes through the origin twice, heading a different way each time.

Play runs t at one unit a second and leaves a red tick at equal steps of time. The ticks are far apart where the pen is fast and bunched where it is slow. The ellipse $(2\cos t, \sin t)$ is fastest at the ends of its short axis. The same ellipse can also be drawn by arc length, at speed 1. Play on the first pen and then choose the arc-length one: the same moments are redrawn, now evenly spaced along the curve.

A cycloid is traced by a point on the rim of a wheel of radius 1 rolling without slipping along a line. After turning through t the wheel has rolled t forward, so $x = t - \sin t,\ y = 1 - \cos t$. At each moment the wheel turns about the point where it touches the ground. So the pen moves at right angles to the line from that point to the pen, at a speed equal to the line's length, $2\lvert\sin(t/2)\rvert$. When the pen touches the ground that length is 0, the pen stops, and the curve has a cusp. Zero velocity alone does not make a cusp: $(\sin t, \sin^2 t)$ stops and turns back at (1, 1), and $(t^3, t^6)$ pauses at the vertex, yet $y = x^2$ is smooth at both.