# Self intersection

• Jul 31st 2010, 08:44 PM
ulysses123
Self intersection
Show that the following curve is not closed and that it has exactly one self-
intersection

γ(t)=(t²-1.t³-t)
• Jul 31st 2010, 10:36 PM
TheEmptySet
Let $\displaystyle a,b$ be the times where the curve intersects with $\displaystyle a \ne b$then
$\displaystyle x=a^2-1$ and $\displaystyle y=a^3-a$ and
$\displaystyle x=b^2-1$ and $\displaystyle y=b^3-b$

Subtracting the x equations gives
$\displaystyle 0=a^2-b^2 \implies a=\pm b$
Since $\displaystyle a\ne b \implies a=-b$

Plugging into the y equations and subtracting gives
$\displaystyle -2b^3+2b=0 \implies b=0,\pm 1$
So the curve intersects itself when t=-1 and t=1 at the point (0,0)
• Jul 31st 2010, 11:36 PM
ulysses123
thanks thats nice and easy to understand, my teacher had confused me with his explanation.
• Aug 1st 2010, 04:15 AM
HallsofIvy
Also, note that as t goes to infinity, both $\displaystyle t^2-1$ and $\displaystyle t^3- 1$ go to infinity but as t goes to negative infinity, [tex]t^2- 1[tex] goes to infinity while $\displaystyle t^3- 1$ goes to negative infinity. That shows that the path is not closed.