Identifying self-intersection points in one parametric graph.

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You are looking for various values $t, t'$ such that $\cos(\frac{\pi}{2}t^2) = \cos(\frac{\pi}{2}t'^2)$, and at the same time $\sin(4t^\frac{2}{3}) = \sin(4t'^\frac{2}{3})$. First, find solutions to these two equations separately. Any common solutions will be a point of intersection.

For the cosine equation, we let $\theta = \frac{\pi}{2}t^2$ and $\varphi = \frac{\pi}{2}t'^2$, so now the equation becomes just $\cos \theta = \cos \varphi$. A knowledge of the behavior of the cosine should make this fairly easy to solve: $\theta = \pm \varphi + 2k\pi$ for integer $k$, so we get the relation $t'^2 = 4k \pm t^2$.
A similar calculation for the sine equation gives $4t'^\frac{2}{3} = n\pi + (-1)^n4t^\frac{2}{3}$.

Now the question becomes: for various integer values of $n$ and $k$, are there any pairs of values $t, t'$ that satisfy both equations? If I replace $x^3 = t^2$ and $y^3 = t'^2$ and substitute for one equation into the other, I get $x = n\frac{\pi}{4} + (-1)^n(4k \pm x^3)^\frac{1}{3}$, or $(x - n\frac{\pi}{4})^3 = (-1)^n(4k \pm x^3)$, which for each of the choices of $n, k$ and sign provides a polynomial that can be solved for $x$, providing values for $y, t,$ and $t'$. Though the values should be checked against the introduction of false solutions when I cubed the original equation.

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Updated on May 30, 2020

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  • louie mcconnell
    louie mcconnell over 3 years

    My question for you is how to identify self-intersection points in a parametric curve of the form x = f(t), y = g(t). The specific problem asks for the t values of the intersection where $x = \sin(4t^{\frac{2}{3}})$ and $y = \cos(0.5\pi t^2)$ over $0\leq t \leq 2$, but I'm just looking for some hints to push me in the right direction, so any help or hints, generally or specifically to this problem, would be appreciated.

    Thank you so much for your help!

    Edit: I had an idea about adding in a dummy variable to create a system of equations that would let me solve the problem, and tried plugging it into wolfram alpha, to no avail.