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\begin{center}
{\Large \vspace{1pt}The semicubic parabola}
\vspace{1pt}by Arne Petersen
10/2/97
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\end{center}
\vspace{1pt}
\vspace{1pt}
\vspace{1pt}
\vspace{1pt}The semicubic \ parabola has no real usefulness. It was the
result of a problem introduced by Gilfried Wilhelm Leibniq. Leibniz for some
reason wanted the path of a particle Descending due to gravity,falling equal
vertical distance in equal time intervals with initial velocity not equal to
zero.This problem caught the interest of a (Dutch) scientist Christian
Huygens who gave the solution as a semicubic parabola. \vspace{1pt}
\vspace{1pt}
\textbf{The semicubical parabola}(the predecessor of the semicubic parabola)
\vspace{1pt}This curve is the first curve to be rectified absolutely.John
Wallis was given the credit for naming it and according to him William
Neil(1637-1670) was the first to accomplish rectifying it .Although P.Fermat
of France and Van Heuraet of Holland were also noted to be working on this
at the same time and did accomplish the rectification independently.
\vspace{1pt}
\[
ay^{2}=x^{2}
\]
\vspace{1pt}The general equation has a cusp at the origin and two
coincidental tangents with the equation $y^{2}$ $=0$. An intersecting
property of the semicubical parabola is that it can not be streched.That
is,the curve $y=bt^{2},x=at^{3}$ is equal to the curve $%
y=(b^{3}/a^{2})t^{2},x=(b^{3}/a^{2})t^{2}$
\textbf{Example. }Graph $y=bt^{2},x=at^{3}$ when $a=2$ and $b=3.$
\textbf{Solution. }Substituting $a=2$ and $b=3$ into $y=bt^{2}$ and $%
x=at^{3} $ yields
\begin{eqnarray*}
y &=&3t^{2} \\
x &=&2t^{3}
\end{eqnarray*}
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Solution:Substituting $a=3$ and $b=2$ yields
\begin{eqnarray*}
y &=&(27/4)t^{2} \\
x &=&(27/4)t^{3}
\end{eqnarray*}
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\begin{center}
\vspace{1pt}
\end{center}
If these axis were similar they would appear \ to be the same.The semicubic
parabola has the same traits as the semicubical parabola.
\begin{center}
\vspace{1pt}
\textbf{The semicubic parabola}
\vspace{1pt}System of parametric equations
\vspace{1pt}$x=t^{3}$
\vspace{1pt}$y=t^{2}$
\vspace{1pt}-$\infty