# evolute of cycloid

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\begin{document}

We shall determine the evolute of the cycloid
\begin{align}
x \;=\; a(u-\sin{u}), \qquad y \;=\; a(1-\cos{u}),
\end{align}
where the parametre $u$ is the rolling angle of the circle with radius $a$ forming the cycloid.\\

The parametric equations of the evolute of a curve\, $x = x(u),\; y = y(u)$\, are
\begin{align}
\xi \;=\; x-\varrho\sin{\alpha}, \qquad \eta \;=\; y+\varrho\cos{\alpha}
\end{align}
with $\alpha$ the slope angle of the tangent and $\varrho$ the radius of curvature of the given curve in the point \,$(x,\,y)$\,:
$$\varrho \;=\; \frac{(x'^{\,2}+y'^{\,2})^{3/2}}{x'y''-x''y'}$$

In the case of the cycloid (1), we have
$$x' \;=\; a(1-\cos{u}), \qquad y' \;=\; a\sin{u}, \qquad x'' \;=\; a\sin{u}, \qquad y'' \;=\; a\cos{u}.$$
Now we get
$$x'^{\,2}+y'^{\,2} \;=\; 2a^2(1-\cos{u}) \;=\; 4a^2\sin^2\frac{u}{2},$$
$$x'y''-x''y' \;=\; a^2(\cos{u}-1) \;=\; -2a^2\sin^2\frac{u}{2},$$
and thus the radius of curvature (red in the diagram) is
\begin{align}
\varrho \;=\; -4a\sin\frac{u}{2}.
\end{align}
\begin{align}
1-\cos{u} \;=\; 2\sin^2\frac{u}{2}
\end{align}
(see the half angle formula of sine in the goniometric formulas).\, It is easy to show that the point, where the circle touches the $x$-axis, bisects the radius of curvature (which lies on the normal line at the point\, $(x,\,y)$\, of the cycloid).

Using then the derivative for parametric form, we obtain
$$\tan\alpha \;=\; \frac{dy}{dx} \;=\; \frac{y'}{x'} \;=\; \frac{\sin{u}}{1-\cos{u}}.$$
which implies
\begin{align}
\sin\alpha \;=\; \cos\frac{u}{2}, \quad \cos\alpha \;=\; \sin\frac{u}{2},
\end{align}
Substituting all needed expressions (1), (3), (5) into (2) and simplifying, we arrive at the result
\begin{align}
\xi \;=\; a(u+\sin{u}), \qquad \eta \;=\; -a(1-\cos{u}).
\end{align}
The equations (6) \PMlinkescapetext{represent} \textbf{the evolute of the given cycloid.\, But it is also a cycloid}, congruent to the original one, which has been \PMlinkname{translated}{Translate} the distance $\pi a$ to the left and the distance $2a$ downwards; this is seen when one performs in (6) the substitution \,$u = v-\pi$;\, then (6) reads
$$\xi \;=\; a(v+\sin{v})-\pi a, \qquad \eta \;=\; a(1-\cos{v})-2a.$$\\

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