## A Line Walks Into A Bar

The class notes yesterday should have included some discussion of the “self-intersecting curve” problem. Here it is a day late. Suppose x = x(t) and y = y(t) are given; a point of self-intersection is said to occur when, for two different values of the parameter—$t_1$ and $t_2$, say—one has the same point of ${\Bbb R}^2$; in other words $x(t_1) = x(t_2)$ and $y(t_1) = y(t_2)$.

In the example at hand, we have $x(t) = t^2$ and $y(t) = t^3 -6t$. Assume that the parameter values $t_1$ and $t_2$ give a point of self-intersection for this curve. From the equation for x we see that then $t_1^2 = t_2^2$; since we are assuming thqt $t_1 \not= t_2$, this implies that $t_2 = - t_1$. Substituing in the equation for y gives
$(-t_1)^3 - 6(-t_1) = t_1^3 - 6t_1$; factoring the LHS gives $-[t_1^3 - 6t_1]$, the opposite of the RHS. Since a number equal to its own opposite is zero, we have RHS = 0; factoring gives $t_1(t_1^2 - 6) = 0$; since $t_1 \not=0$ (because $t_2 = -t_1 \not= t_1$), this implies that $t_1 = \pm \sqrt 6$.

So our parametric curve intersects itself for the t values $t = \sqrt6$ and $t = -\sqrt6$. One easily obtains the desired values of ${{dy}\over{dx}}$ to complete the exercise.