parametrized curve কী?

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

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

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.