Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/209

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the parametric equations of the evolute. Assuming values of the parameter $t$, we calculate $x,y;\alpha ,\beta$ from (B) and (C); and tabulate the results as follows:

Now plot the curve and its evolute.

t	x	y	&aplha;	β
3	${\tfrac {5}{2}}$	${\tfrac {9}{2}}$
2	${\tfrac {5}{4}}$	${\tfrac {4}{3}}$	$-{\tfrac {35}{4}}$	${\tfrac {19}{3}}$
${\tfrac {3}{2}}$	${\tfrac {13}{16}}$	${\tfrac {9}{16}}$	$-{\tfrac {91}{32}}$	$3$
1	${\tfrac {1}{2}}$	${\tfrac {1}{6}}$	$-{\tfrac {1}{2}}$	${\tfrac {7}{6}}$
0	${\tfrac {1}{4}}$	0	${\tfrac {1}{4}}$	0
-1	${\tfrac {1}{2}}$	$-{\tfrac {1}{6}}$	$-{\tfrac {1}{2}}$	$-{\tfrac {7}{6}}$
$-{\tfrac {3}{2}}$	${\tfrac {13}{16}}$	$-{\tfrac {9}{16}}$	$-{\tfrac {91}{32}}$	$-3$
-2	${\tfrac {5}{4}}$	$-{\tfrac {4}{3}}$	$-{\tfrac {35}{4}}$	$-{\tfrac {19}{3}}$
-3	${\tfrac {5}{2}}$	$-{\tfrac {9}{2}}$

The point $({\tfrac {1}{4}},0)$ is common to the given curve and its evolute. The given curve (semi cubical parabola) lies entirely to the right and the evolute entirely to the left of $x={\tfrac {1}{4}}$.

The circle of curvature at $A({\tfrac {1}{2}},{\tfrac {1}{6}})$, where $t=1$, will have its center at $A'(-{\tfrac {1}{2}},-{\tfrac {7}{6}})$ on the evolute and radius = $AA'$. To verify our work find radius of curvature at A. From (42), §103, we get

$R={\frac {t(1+t^{2})^{\frac {3}{2}}}{2}}={\sqrt {2}},{\text{when}}t1.$

This should equal the distance

$AA'={\sqrt {({\frac {1}{2}}+{\frac {1}{2}})^{2}+({\frac {1}{6}}-{\frac {7}{6}})^{2}}}={\sqrt {2}}.$

Evolute of the curve.

Illustrative Example 4. Find the parametric equations of the evolute of the cycloid,

(C)	${\begin{cases}x=a(t-\sin t)\\y=a(1-\cos t).\end{cases}}$

Solution. As in Illustrative Example 2, §103, we get
	${\frac {dy}{dx}}={\frac {\sin t}{1-\cos t}},{\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{a(1-\cos t)^{2}}}.$
Substituting these results in formulas (50), §117, we get
(D)	${\begin{cases}\alpha =a(t+\sin t),\\\beta =-a(1-\cos t).\end{cases}}$ Ans.