Behaviour of Parametric Equations for Folium of Descartes according to Parameter

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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.

$\Box$


Behaviour for $t < -1$

\(\ds \lim_{t \mathop \to -\infty} \dfrac {3 a t} {1 + t^3}\) \(<\) \(\ds \lim_{t \mathop \to -\infty} \dfrac {3 a t} {t^3}\)
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to -\infty} \dfrac {3 a} {t^2}\)
\(\ds \) \(=\) \(\ds \to 0^+\)


\(\ds \lim_{t \mathop \to -\infty} \dfrac {3 a t^2} {1 + t^3}\) \(<\) \(\ds \lim_{t \mathop \to -\infty} \dfrac {3 a t^2} {t^3}\)
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to -\infty} \dfrac {3 a} t\)
\(\ds \) \(=\) \(\ds \to 0^-\)

So:

$\ds \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^-}$


Then:

\(\ds \lim_{t \mathop \to -1^-} \dfrac {1 + t^3} {3 a t}\) \(\to\) \(\ds \dfrac {0^-} {-3 a}\) because $t^3 < -1$
\(\ds \) \(=\) \(\ds 0^+\)
\(\ds \lim_{t \mathop \to -1^-} \dfrac {1 + t^3} {3 a t^2}\) \(\to\) \(\ds \dfrac {0^-} {3 a}\) because $t^3 < -1$
\(\ds \) \(=\) \(\ds 0^-\)


Thus:

\(\ds \lim_{t \mathop \to -1^-} \dfrac {3 a t} {1 + t^3}\) \(\to\) \(\ds +\infty\)
\(\ds \lim_{t \mathop \to -1^-} \dfrac {3 a t^2} {1 + t^3}\) \(\to\) \(\ds -\infty\)


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

$\Box$


Behaviour for $t > 0$

\(\ds \lim_{t \mathop \to +\infty} \dfrac {3 a t} {1 + t^3}\) \(<\) \(\ds \lim_{t \mathop \to +\infty} \dfrac {3 a t} {t^3}\)
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to +\infty} \dfrac {3 a} {t^2}\)
\(\ds \) \(=\) \(\ds \to 0^+\)


\(\ds \lim_{t \mathop \to +\infty} \dfrac {3 a t^2} {1 + t^3}\) \(<\) \(\ds \lim_{t \mathop \to +\infty} \dfrac {3 a t^2} {t^3}\)
\(\ds \) \(=\) \(\ds \lim_{t \mathop \to +\infty} \dfrac {3 a} t\)
\(\ds \) \(=\) \(\ds \to 0^+\)


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 $t = +\infty$.

$\Box$


Behaviour for $-1 < t < 0$

We have that when $t = 0$, $\tuple {x, y} = \tuple {0, 0}$.

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

\(\ds x\) \(=\) \(\ds \dfrac {3 a t} {1 + t^3}\)
\(\ds \) \(<\) \(\ds 0\)
\(\ds y\) \(=\) \(\ds \dfrac {3 a t^2} {1 + t^3}\)
\(\ds \) \(>\) \(\ds 0\)


Then:

\(\ds \lim_{t \mathop \to -1^+} \dfrac {1 + t^3} {3 a t}\) \(\to\) \(\ds \dfrac {0^+} {-3 a}\) because $t^3 > -1$
\(\ds \) \(=\) \(\ds 0^-\)
\(\ds \lim_{t \mathop \to -1^-} \dfrac {1 + t^3} {3 a t^2}\) \(\to\) \(\ds \dfrac {0^+} {3 a}\) because $t^3 > -1$
\(\ds \) \(=\) \(\ds 0^+\)


Thus:

\(\ds \lim_{t \mathop \to -1^-} \dfrac {3 a t} {1 + t^3}\) \(\to\) \(\ds -\infty\)
\(\ds \lim_{t \mathop \to -1^-} \dfrac {3 a t^2} {1 + t^3}\) \(\to\) \(\ds +\infty\)


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

$\Box$


The result follows.

$\blacksquare$

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