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.

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