How do I find where the graph of parametric equations crosses itself?

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!

• x does not equal y where this parametric equation crosses itself! @probably_someone Apr 2 '18 at 23:10
• Wow, yeah, that was bad. Sorry. Apr 2 '18 at 23:13

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 $(3,-2)$.
It crosses itself where, for two different values of t, you get equal x and y values. So letting $t_1 = u$ and $t_2=v$: $$u^3-u+3 = v^3-v+3$$ $$u^2-3 = v^2-3$$ These simplify to: $$u(u^2-1) = v(v^2-1)$$ $$u^2 = v^2$$ We know we want u and v to be different. From the second equation, we can deduce that $u = -v$. Plug this into the first equation: $$-v^3+v+3 = v^3-v+3$$ or $$2v^3-2v=0$$ $$v(v^2-1)=0$$ $v=0$ would make $u=0$ which disqualifies it so: $$v^2-1 = 0$$ $v\pm 1$ and $u\mp 1$ then are the two values of t where the parametric equation crosses itself.
Plug $t=1$ and $t=-1$ into your equations for x and y and you will see this is true.