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I have a curve parameterized by $$\left\{\begin{array}{ll}x(t)=\frac{k\sin{t}}{t}\\y(t)=\frac{k(1-\cos{t})}{t}.\\\end{array}\right.$$ In polar form, this is $$r^2=\frac{2k^2(1-\cos{\theta})}{\theta^2}.$$ I was wondering if this curve has a name. I found it to be the trajectory of the center of mass of a uniform sector as the central angle of the sector varies, where $k=\frac{2}{3}R$ and $R$ is the radius of the sector.

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    $\begingroup$ Note that this is equivalent to $$r=\pm\frac{k\sin{(\theta/2)}}{\theta/2}$$which is the sinc function. $\endgroup$ – Peter Foreman Jul 16 at 23:04

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