2.2 Parametric Playground

Parametric curves earn their keep in the plane long before they need a third dimension. The pair $x = f(t)$ , $y = g(t)$ describes a motion, and curves that are awkward or impossible to write as $y = F(x)$ often have short, transparent parametric stories. This section is a playground with two classic inhabitants, each carrying a slider so you can interrogate the formula by moving it.

First, the cycloid: the path traced by a pebble stuck to the rim of a wheel of radius $a$ rolling along a line. Rolling means the wheel's rotation and its forward travel are locked together — after turning through angle $t$ , the axle has advanced exactly $at$ . Subtracting the pebble's position on the rim from the axle's motion gives

$x = a(t - \sin t), \qquad y = a(1 - \cos t).$

A red curve of three identical arches resting on a gray horizontal baseline, like the path traced by a pebble stuck to a rolling wheel: each arch rises smoothly from a sharp downward point (a cusp) on the line, crests, and returns to the next cusp. An amber horizontal line runs at twice the wheel's radius and touches the crest of every arch, and a short sky-blue vertical line stands where the first arch ends, one wheel circumference from the start. A slider named a, starting at 1, sets the rolling wheel's radius: dragging it up makes the arches and both reference lines taller and wider in the same proportion, and dragging it down shrinks them. The curve always has three arches, but at larger radii, or on narrow screens, the later arches run past the right edge of the view. — Three arches of the cycloid $\big(a(t - \sin t),\ a(1 - \cos t)\big)$ over a gray baseline, with two reference lines that move with the slider: an amber line at height $2a$ and a sky-blue vertical line at $x = 2\pi a$ . Drag the slider for $a$ — the wheel's radius — and watch the arches scale. The default view faces the curve head-on.

Explore the cycloid

Each arch ends in a sharp point — a cusp — where the curve touches the baseline. Physically, the pebble is at the bottom of the wheel there. At which parameter values $t$ does that happen, and why does the motion momentarily stall?
Drag $a$ from 0.5 to 2. Height and width scale together. The amber line sits at height $2a$ and the sky-blue line at $x = 2\pi a$ : does the crest of every arch stay on the amber line, and does the first arch always end at the foot of the sky-blue line? What do the two lines tell you about the arch height and the arch width as multiples of $a$ ?
The curve never dips below the baseline and never doubles back in $x$ . Which parts of the two formulas guarantee each fact?

Second, the polar rose. A polar curve $r = g(\theta)$ is secretly parametric: since $x = r\cos\theta$ and $y = r\sin\theta$ , the graph is the parametric curve

$x = g(t)\cos t, \qquad y = g(t)\sin t,$

with the polar angle as parameter. Take $g(t) = \cos(bt)$ for a whole number $b$ and you get a rose.

A violet rose-shaped curve of identical petals radiating from the origin, drawn flat in the plane and inscribed in a gray unit circle, shown face-on; at the starting value b equals 3 it has three petals. A slider named b sets the frequency in r equals the cosine of b theta: odd values of b draw exactly b petals, while even values draw twice as many, 2b, which is the parity surprise the scene is built to expose. A second slider named s, starting at 1, traces the rose: an amber curve retraces it from angle 0 up to angle s, and a short sky-blue segment from the origin points in the direction of angle s. Wherever the cosine of b times s is negative, as it is at the starting values, the end of the amber curve lies on the opposite side of the origin from the sky-blue segment. — The rose $r = \cos(b\theta)$ , drawn in violet inside a gray unit circle. The slider $b$ sets the frequency from 1 to 6 in whole-number steps. The slider $s$ traces the rose: the amber curve is the part drawn for $0 \le \theta \le s$ , and the sky-blue segment points in the direction $\theta = s$ . Count petals at each value of $b$ before you trust any rule you have heard.

Explore the rose

Set $b = 1$ and identify the curve (it is a familiar shape in disguise — a circle of radius one half through the origin). Then count petals at $b = 2, 3, 4, 5, 6$ and record the counts.
The pattern splits by parity: odd $b$ gives $b$ petals, even $b$ gives $2b$ . Set $b = 3$ and drag $s$ slowly from 0 to 6.28, watching the end of the amber curve near values where $\cos(bs)$ is negative: a negative radius plots the point on the opposite side of the origin from the sky-blue segment. For odd $b$ , where do those negative-radius petals land?
Every petal tip touches the gray circle. Why must the whole rose live inside it?

Why parity decides the petal count

A petal is traced each time $\cos(bt)$ runs from 0 up to $\pm 1$ and back to 0, which happens $2b$ times as $t$ crosses $[0, 2\pi]$ . When $b$ is even, those $2b$ passes land in $2b$ distinct directions: $2b$ petals. When $b$ is odd, the negative-radius passes retrace petals already drawn on the opposite side — each petal is drawn exactly twice — leaving only $b$ visible. The slider makes this a two-minute experiment; the algebra behind it is the single identity that $(r, \theta)$ and $(-r, \theta + \pi)$ name the same point.

Example 1 Dimensions of a cycloid arch

A wheel of radius $a = 1.5$ rolls along a line. Find the height and width of one arch of the cycloid its rim point traces, and the coordinates of the first arch's crest.

Show solution

Height: $y = a(1 - \cos t)$ is largest when $\cos t = -1$ , at $t = \pi$ , giving $y = 2a = 3$ . Width: one arch spans one full revolution, $t \in [0, 2\pi]$ , and $x$ advances by $a \cdot 2\pi = 3\pi \approx 9.42$ . The crest sits at $t = \pi$ : $x = a(\pi - \sin \pi) = 1.5\pi \approx 4.71$ , $y = 3$ — halfway across the arch, at twice the wheel's radius. Dragging the slider to $a = 1.5$ in the cycloid figure above lets you check all three numbers against the reference lines: the crest touches the amber line at height 3, halfway between the first cusp and the sky-blue line at $x = 3\pi$ , where the arch ends.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.