info prev up next book cdrom email home

Parabola Evolute

Given a Parabola

\end{displaymath} (1)

the parametric equation and its derivatives are
\matrix{x=t\cr y=t^2\cr}\qquad
\matrix{x'=t\cr x''=0\cr}\qquad
\matrix{y'=2t\cr y''=2.\cr}
\end{displaymath} (2)

The Radius of Curvature is
R={(x'^2+y'^2)^{3/2}\over x'y''-x''y'} = {(1+4t^2)^{3/2}\over 2}.
\end{displaymath} (3)

The Tangent Vector is
\hat {\bf T} = {1\over\sqrt{1+4t^2}} \left[{\matrix{1\cr 2t\cr}}\right],
\end{displaymath} (4)

so the parametric equations of the evolute are
$\displaystyle \xi$ $\textstyle =$ $\displaystyle -4t^3$ (5)
$\displaystyle \eta$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}+3t^2,$ (6)

$\displaystyle -{\textstyle{1\over 4}}\xi$ $\textstyle =$ $\displaystyle t^3$ (7)
$\displaystyle {\textstyle{1\over 3}} (\eta-{\textstyle{1\over 2}})$ $\textstyle =$ $\displaystyle t^2$ (8)

{\textstyle{1\over 3}}(\eta-{\textstyle{1\over 2}})=(-{\textstyle{1\over 4}}\xi)^{2/3}
\end{displaymath} (9)

{\textstyle{1\over 3}}(\eta-h) = \left({-{2\xi\over 8}}\right)^{2/3}= {\textstyle{1\over 4}}(2\xi)^{2/3}.
\end{displaymath} (10)

The Evolute is therefore
\eta={\textstyle{3\over 4}} (2\xi)^{2/3}+{\textstyle{1\over 2}}.
\end{displaymath} (11)

This is known as Neile's Parabola and is a Semicubical Parabola. From a point above the evolute three normals can be drawn to the Parabola, while only one normal can be drawn to the Parabola from a point below the Evolute.

See also Neile's Parabola, Parabola, Semicubical Parabola

© 1996-9 Eric W. Weisstein