Behaviour of Parametric Equations for Folium of Descartes according to Parameter

Theorem
Consider the folium of Descartes $F$, given in parametric form as:
 * $\begin {cases} x = \dfrac {3 a t} {1 + t^3} \\ y = \dfrac {3 a t^2} {1 + t^3} \end {cases}$

Then:


 * $F$ has a discontinuity at $t = -1$.


 * For $t < -1$, the section in the $4$th quadrant is generated
 * For $-1 < t \le 0$, the section in the $2$nd quadrant is generated
 * For $0 \le t$, the section in the $1$st quadrant is generated.

Proof

 * FoliumOfDescartes.png

Discontinuity at $t = -1$
When $t = -1$, we have that:
 * $1 + t^3 = 0$

and so both $x$ and $y$ are undefined.

Behaviour for $t < -1$
So:
 * $\ds \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^-}$

Then:

Thus:

Thus, as $t$ goes from $-\infty$ to $-1$, $F$ goes from $\tuple {0, 0}$ in the $4$th quadrant to $\tuple {+\infty, -\infty}$.

Behaviour for $t > 0$
So:
 * $\ds \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^+}$

It is observed that for $t > 0$, we have that:
 * $x > 0$
 * $y > 0$

and so for $t > 0$, $F$ is in the $1$st quadrant.

Thus we have that the loop is traversed from $t = 0$ to $t = +\infty$.

Behaviour for $-1 < t < 0$
We have that when $t = 0$, $\tuple {x, y} = \tuple {0, 0}$.

When $-1 < t < 0$, we have that:

Then:

Thus:

Thus, as $t$ goes from $0$ to $-1$, $F$ goes from $\tuple {0, 0}$ in the $2$nd quadrant to $\tuple {-\infty, +\infty}$.

The result follows.