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
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.
$\blacksquare$