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I'm working on a set of equations that would tell a hypothetical robot soccer player whether or not to pass a ball to a teammate. After a lot of algebra, I arrived at these equations for the partial boundary of a region:

$$x=v_1t_{\alpha}\cos\left(90-\cos^{-1}\left(-\frac{v_2^2t_{\alpha}^2-v_1^2t_{\alpha}^2-d_{\Sigma}}{2v_1t_{\alpha}d_{\Sigma}}\right)\right)$$

$$y=v_1t_{\alpha}\sin\left(90-\cos^{-1}\left(-\frac{v_2^2t_{\alpha}^2-v_1^2t_{\alpha}^2-d_{\Sigma}}{2v_1t_{\alpha}d_{\Sigma}}\right)\right)$$

They are equations for $x$ and $y$, respectively, parameterized by time ($t_{\alpha}$).

How do I go about writing $x$ as a function of $y$? I'm looking for hints, not full solutions.

HDE 226868
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1 Answers1

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Note: I am using your "90" (measuring angles in degrees) instead of the more usual "$\pi/2$" (radian measure).

Since the args of the $\sin$ and $\cos$ look the same, you have $x = A \cos(u), y=A \sin(u)$ for appropriate $A$ and $u$. Therefore $x^2+y^2 =A^2 $.

Also, since $\sin(v) =\cos(90-v) $, $90-\cos^{-1}(x) =\sin^{-1}(x) $, so the second term is $\sin(\sin^{-1}(v)) = v $, so this gives $y$ as a function without trig functions.

Similarly, you can use $\cos(\sin^{-1}(x)) =\sqrt{1-x^2} $ to eliminate the trig functions from the first equation.

marty cohen
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