Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/210

Evolute of a cycloid.

NOTE. If we eliminate $t$ between equations (D), there results the rectangular equation of the evolute $OO'Q^{v}$ referred to the axes $O'\alpha$ and $O'\beta$. The coördinates of O with respect to these axes are $(-\pi a,-2a)$. Let us transform equations (D) to the new set of axes OX and OY. Then

$\alpha =x-\pi a,\ \beta =y-2a,\ t=t'-\pi .$

Substituting in (D) and reducing, the equations of the evolute become

(E)	${\begin{cases}x=a(t'-\sin t'),\ \\y=a(1-\cos t').\end{cases}}$

Since (E) and (C) are identical in form, we have:

The evolute of a cycloid is itself a cycloid whose generating circle equals that of the given cycloid.

120. Properties of the evolute. From (A), §117,

(A)	$\alpha =x-R\sin \tau ,\ \beta =y+R\cos \tau .$

Let us choose as independent variable the lengths of the arc on the given curve; then $x,y,R,T,\alpha ,\beta$ are functions of s. Differentiating (A) with respect to $s$ gives

(B)	${\frac {d\alpha }{ds}}={\frac {dx}{ds}}-R\cos \tau {\frac {d\tau }{ds}}-\sin \tau {\frac {dR}{ds}},$
	${\frac {d\beta }{ds}}={\frac {dy}{ds}}-R\sin \tau {\frac {d\tau }{ds}}+\cos \tau {\frac {dR}{ds}}.$

But ${\tfrac {dx}{ds}}=\cos \tau ,{\tfrac {dy}{ds}}=\sin \tau$, from (26), §90; and ${\tfrac {d\tau }{ds}}={\tfrac {1}{R}}$, from (38) in § 100 and (39) in §101.

Substituting in (B) and (C), we obtain

(D)	${\frac {d\alpha }{ds}}=\cos \tau -R\cos \tau \cdot {\frac {1}{R}}-\sin \tau {\frac {dR}{ds}}=-\sin \tau {\frac {dR}{ds}},$
(E)	${\frac {d\beta }{ds}}=\sin \tau -R\sin \tau \cdot {\frac {1}{R}}+\cos \tau {\frac {dR}{ds}}=\cos \tau {\frac {dR}{ds}}.$

Dividing (E) by (D) gives

(F)	${\frac {d\beta }{d\alpha }}=-\cot \tau =-{\frac {1}{\tan \tau }}=-{\frac {1}{\frac {dy}{dx}}}.$