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A simple, closed, planar curve can be given by the following parametric function: $$ \gamma(t)=\left(\cos t,\sin t+\frac{\sin^2t}{2}\right) $$ This function on $t=0$ to $t=2\pi$ gives the following image: enter image description here

Is there a name for this particular curve? It looks a lot like the bean curve, but I'm not sure how I could show that the implicit equation for the bean curve is equivalent to the parametric equation I've given above.

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  • $\begingroup$ Have you tried factoring $\sin t$ in the second term and then writing $x=\cos t$ and $y= \sin t$? I get then $(x, y(1+y/2))$ but am not sure how to proceed from there. $\endgroup$ – Chill2Macht Apr 28 '16 at 2:09
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From the parametric equation $x = \cos t$, so $x^2 = \cos^2 t$. Recall that $\sin^2 x = 1 - \cos^2 x$. We can use this to re-write $y$ \begin{align} y &= \sqrt{1-\cos^2 t} + \frac{1 - \cos^2 t}{2}\\ y &= \sqrt{1 - x^2} + \frac{1 - x^2}{2}\\ 2y + x^2 - 1 &= 2\sqrt{1 - x^2}\\ 4y^2 + 4y(x^2 - 1) + x^4 - 2x^2 + 1 &= 4 - 4x^2 \end{align} This means the implicit equation of this curve is \begin{equation} 4y^2 - 4y +4x^2y + 2x^2 + x^4 = 3 \end{equation} Considering that the bean curve does not contain a constant, this may not be a bean curve, BUT maybe with some clever factorization it can be put into the same form. Unfortunately I have not been able to figure out a transformation that (i.e let $x' = x - 2$) that gets rid of the constant.

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