:500px === Discontinuity at $t = -1$ === When $t = -1$, we have that: :$1 + t^3 = 0$ and so both $x$ and $y$ are undefined. {{qed|lemma}} === Behaviour for $t < -1$ === {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -\infty} \dfrac {3 a t} {1 + t^3}       | o = <       | r = \lim_{t \mathop \to -\infty} \dfrac {3 a t} {t^3}       | c =  }} {{eqn | r = \lim_{t \mathop \to -\infty} \dfrac {3 a} {t^2}       | c =  }} {{eqn | r = \to 0^+       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -\infty} \dfrac {3 a t^2} {1 + t^3}       | o = <       | r = \lim_{t \mathop \to -\infty} \dfrac {3 a t^2} {t^3}       | c =  }} {{eqn | r = \lim_{t \mathop \to -\infty} \dfrac {3 a} t       | c =  }} {{eqn | r = \to 0^-       | c =  }} {{end-eqn}} So: :$\displaystyle \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^-}$ Then: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {1 + t^3} {3 a t}       | o = \to       | r = \dfrac {0^-} {-3 a}       | c = because $t^3 < -1$ }} {{eqn | r = 0^+       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {1 + t^3} {3 a t^2}       | o = \to       | r = \dfrac {0^-} {3 a}       | c = because $t^3 < -1$ }} {{eqn | r = 0^-       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {3 a t} {1 + t^3}       | o = \to       | r = +\infty       | c =  }} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {3 a t^2} {1 + t^3}       | o = \to       | r = -\infty       | c =  }} {{end-eqn}} Thus, as $t$ goes from $-\infty$ to $-1$, $F$ goes from $\tuple {0, 0}$ in the $4$th quadrant to $\tuple {+\infty, -\infty}$. {{qed|lemma}} === Behaviour for $t > 0$ === {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to +\infty} \dfrac {3 a t} {1 + t^3}       | o = <       | r = \lim_{t \mathop \to +\infty} \dfrac {3 a t} {t^3}       | c =  }} {{eqn | r = \lim_{t \mathop \to +\infty} \dfrac {3 a} {t^2}       | c =  }} {{eqn | r = \to 0^+       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to +\infty} \dfrac {3 a t^2} {1 + t^3}       | o = <       | r = \lim_{t \mathop \to +\infty} \dfrac {3 a t^2} {t^3}       | c =  }} {{eqn | r = \lim_{t \mathop \to +\infty} \dfrac {3 a} t       | c =  }} {{eqn | r = \to 0^+       | c =  }} {{end-eqn}} So: :$\displaystyle \lim_{t \mathop \to -\infty} \tuple {x, y} = \tuple {0^+, 0^+}$ 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 $t = +\infty$. {{Qed|lemma}} === Behaviour for $-1 < t < 0$ === We have that when $t = 0$, $\tuple {x, y} = \tuple {0, 0}$. When $-1 < t < 0$, we have that: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {3 a t} {1 + t^3}       | c =  }} {{eqn | o = <       | r = 0       | c =  }} {{eqn | l = y       | r = \dfrac {3 a t^2} {1 + t^3}       | c =  }} {{eqn | o = >       | r = 0       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -1^+} \dfrac {1 + t^3} {3 a t}       | o = \to       | r = \dfrac {0^+} {-3 a}       | c = because $t^3 > -1$ }} {{eqn | r = 0^-       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {1 + t^3} {3 a t^2}       | o = \to       | r = \dfrac {0^+} {3 a}       | c = because $t^3 > -1$ }} {{eqn | r = 0^+       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {3 a t} {1 + t^3}       | o = \to       | r = -\infty       | c =  }} {{eqn | l = \lim_{t \mathop \to -1^-} \dfrac {3 a t^2} {1 + t^3}       | o = \to       | r = +\infty       | c =  }} {{end-eqn}} Thus, as $t$ goes from $0$ to $-1$, $F$ goes from $\tuple {0, 0}$ in the $2$nd quadrant to $\tuple {-\infty, +\infty}$. {{qed|lemma}} The result follows. {{qed}} Category:Folium of Descartes h80kc9adekvjjv9reoytdl18b938o7q