I am trying to find the value of $t$ where the graph of the following parametric equations crosses itself:
\begin{align}
x =&\; t^3 − t + 3\\
y =&\; t^2 − 3
\end{align}
I know that the next step toward solving this problem involves creating two parallel equations with a different kind of variable.
The result will look like:
\begin{align}
t^3 – t + 3 = s^3 – s + 3 \tag1\\
t^2 – 3 = s^2 – 3 \tag2
\end{align}
After this point, I am totally stumped. I know that I will have to solve for $t$, and it looks like doing so will involve some skill I have not yet acquired. Multiplying (2) by $-t$ and adding (1) to (2) helps me to eliminate the $t^3$, but I am left with
$$s^3 – s(t^3) – (s^2)t + t = 0 $$
an equation for which I have no idea how to solve for $t$.
Could someone please show me how to find $t$-values at the point(s) where the graph from these equations crosses itself? Thanks in advance!
Best Answer
Solving for $t$ from $y=t^2-3$ we get $t=\pm \sqrt{y+3}$. Substituting this into $x = t^3 − t + 3$ we get $$x = (y+3)^{(3/2)}-\sqrt{y+3}+3\quad\text{and}\quad x = -(y+3)^{(3/2)}+\sqrt{y+3}+3.$$ These two functions (which are only defined for $y\ge-3$) intersect at $$ y=-3\quad\text{and}\quad y=-2.$$ Substituting these back into the equations for $x$ we get $$x=3.$$ Since the functions are undefined for $y<-3$, $(3,-3)$ is a connection point, not a crossing. The only crossing is, therefore at $(3,-2)$.