Parabola Involute

$\begin{figure}\begin{center}\BoxedEPSF{parabola_involute.epsf}\end{center}\end{figure}$

$\begin{displaymath} {d{\bf r}\over dt} = \left[{\matrix{1\cr 2t\cr}}\right] \end{displaymath}$

(1)

$\begin{displaymath} \hat{\bf T} = {1\over\sqrt{1+4t^2}} \left[{\matrix{1\cr 2t\cr}}\right] \end{displaymath}$

(2)

$\begin{displaymath} ds^2=\vert d{\bf r}\vert^2 = (1+4t^2)\,dt^2 \end{displaymath}$

(3)

$\begin{displaymath} ds=\sqrt{1+4t^2}\,dt \end{displaymath}$

(4)

$\begin{displaymath} s=\int \sqrt{1+4t^2}\,dt = {\textstyle{1\over 2}}t\sqrt{1+4t^2}+{\textstyle{1\over 4}}\sinh^{-1}(2t). \end{displaymath}$

(5)

So the equation of the Involute is

$\displaystyle {\bf r}_i$	$\textstyle =$	$\displaystyle {\bf r}-s\hat{\bf T} = \left[\begin{array}{c}t\\ t^2\end{array}\... ...t)\over \sqrt{1+4t^2}} \left[\begin{array}{c}1\\ 2t\end{array}\right]\nonumber$
	$\textstyle =$	$\displaystyle {1\over 2\sqrt{1+4t^2}} \left[\begin{array}{c}t-{\textstyle{1\over 2}}\sinh^{-1}(2t)\\ -\sinh^{-1}(2t)\end{array}\right].$	(6)