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Nephroid Involute

\begin{figure}\begin{center}\BoxedEPSF{NephroidInvolute.epsf scaled 700}\end{center}\end{figure}

The Involute of the Nephroid given by

$\displaystyle x$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}[3\cos t-\cos(3t)]$  
$\displaystyle y$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}[3\sin t-\sin(3t)]$  

beginning at the point where the nephroid cuts the y-Axis is given by
$\displaystyle x$ $\textstyle =$ $\displaystyle 4\cos^3 t$  
$\displaystyle y$ $\textstyle =$ $\displaystyle 3\sin t+\sin(3t),$  

another Nephroid. If the Involute is begun instead at the Cusp, the result is Cayley's Sextic.

© 1996-9 Eric W. Weisstein