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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.

louie mcconnell
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  • Intersection points with what? You've given only one curve here. Are you looking for the points where it intersects itself? In that case, you want to start with, given $\theta$ for what values of $\varphi$ is $\cos \theta = \cos \varphi$, and similarly for sine. Translate those values to $t$, and see if any of the values of $t$ for which the cosine condition holds is also a place where the sine condition holds. – Paul Sinclair Aug 31 '15 at 23:42
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    Yes, I'm looking for the values where the curve intersects itself. However, I'm not following your answer. – louie mcconnell Aug 31 '15 at 23:49
  • I just end up with the obvious solution that a = b, and I don't know how to manipulate the function to get any other value. – louie mcconnell Sep 01 '15 at 01:07

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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.

Paul Sinclair
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