What is parametrized curve? — Mathematics

Definition 24.1 University Mathematics — Year 1 · Chapter 24 — Plane Curves

A parametrized curve is a map $f \colon I \to \R^2$ , $t \mapsto M(t) = (x(t), y(t))$ , with $x, y$ of class $C^1$ (at least) on the interval $I$ . The velocity vector is $f'(t) = (x'(t), y'(t))$ ; the point $M(t_0)$ is regular when $f'(t_0) \neq (0,0)$ , and the tangent there is the line through $M(t_0)$ directed by $f'(t_0)$ .

The Lissajous curve ( t, 2t): a double point at the origin, where two regular branches cross with velocities (1, 2) and (-1, 2), and horizontal tangents at the four summits. Symmetry reduced all the work to a quarter-period. — The Lissajous curve $(\sin t, \sin 2t)$ : a double point at the origin, where two regular branches cross with velocities $(1, 2)$ and $(-1, 2)$ , and horizontal tangents at the four summits. Symmetry reduced all the work to a quarter-period.

The astroid ( 3 t, 3 t): four arcs meeting at four cusps. It is the curve traced by a point of a circle of radius 1/4 rolling inside the unit circle. — The astroid $(\cos^3 t, \sin^3 t)$ : four arcs meeting at four cusps. It is the curve traced by a point of a circle of radius $\frac14$ rolling inside the unit circle.

Examples

Example 24.3 (Asymptotic branches, worked)

$x(t) = t$ , $y(t) = t + \dfrac1t$ on $\intoo{0}{+\infty}$ . As $t \to 0^{+}$ : $x \to 0$ while $y \to +\infty$ — the curve climbs along the vertical asymptote $x = 0$ (finite limit for one coordinate, infinite for the other). As $t \to +\infty$ : both coordinates blow up, so test a line: the difference

y(t) - x(t) = \frac1t \longrightarrow 0^{+}

exhibits the oblique asymptote $y = x$ , approached from above. In between, $y' = 1 - \frac1{t^2}$ vanishes at $t = 1$ : the point $(1, 2)$ is the low point of the branch, and $x' = 1 > 0$ throughout, so the curve always advances rightward. One arc, two asymptotes, one minimum: a complete picture from three computations — with the general recipe visible underneath: when $x, y \to \infty$ together, examine $y/x$ for a candidate slope (here $\to 1$ ), then $y - (\text{slope})\,x$ for the intercept and the side of approach.

Example 24.4 (The astroid)

$x(t) = \cos^3 t$ , $y(t) = \sin^3 t$ . Symmetries: $M(t + 2\pi) = M(t)$ (study a period); $M(-t)$ is the reflection of $M(t)$ in the $x$ -axis; $M(\pi - t)$ in the $y$ -axis; $M(\frac\pi2 - t)$ in the diagonal $y = x$ : it suffices to study $t \in \intcc{0}{\frac\pi4}$ and unfold.

Velocity: $f'(t) = 3\sin t\cos t\,(-\cos t, \sin t)$ . On $\intoo{0}{\frac\pi2}$ all points are regular with tangent directed by $(-\cos t, \sin t)$ ; at $t = 0$ (the point $(1,0)$ ) the velocity vanishes: a cusp, where the curve reverses along the tangent direction $(-1, 0)\cdot(\pm)$ — by symmetry the four cusps sit at $(\pm1, 0), (0, \pm1)$ .

Example 24.5 (A figure-eight, studied in full)

$x(t) = \sin t$ , $y(t) = \sin 2t$ (a Lissajous curve). Symmetries: $M(t + \pi) = (-x(t), y(t))$ (reflection in the $y$ -axis), $M(-t) = (-x(t), -y(t))$ (central symmetry), $M(\pi - t) = (x(t), -y(t))$ (reflection in the $x$ -axis): it suffices to study $t \in \intcc{0}{\frac\pi2}$ and unfold. Variations: $x' = \cos t \geq 0$ throughout, while $y' = 2\cos 2t$ is positive before $t = \frac\pi4$ and negative after: the arc climbs rightward to the summit $\bigl(\frac{\sqrt2}2,\ 1\bigr)$ at $t = \frac\pi4$ (horizontal tangent), then descends rightward to $(1, 0)$ at $t = \frac\pi2$ , where $x' = 0 \neq y'$ : vertical tangent. Double point: $M(0) = M(\pi) = (0,0)$ , with two different velocities

f'(0) = (1,\ 2), \qquad f'(\pi) = (-1,\ 2) :

two regular branches crossing at the origin at distinct angles — a double point, not a singular point: each passage is perfectly smooth, the two passages merely share their location. The whole curve is the figure-eight below.

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Examples

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