Evolute of a cycloid. NOTE. If we eliminate
t
{\displaystyle t}
(D ) , there results the rectangular equation of the evolute
O
O
′
Q
v
{\displaystyle OO'Q^{v}}
O
′
α
{\displaystyle O'\alpha }
O
′
β
{\displaystyle O'\beta }
O with respect to these axes are
(
−
π
a
,
−
2
a
)
{\displaystyle (-\pi a,-2a)}
(D ) to the new set of axes OX and OY . Then
α
=
x
−
π
a
,
β
=
y
−
2
a
,
t
=
t
′
−
π
.
{\displaystyle \alpha =x-\pi a,\ \beta =y-2a,\ t=t'-\pi .}
Substituting in (D ) and reducing, the equations of the evolute become
(E )
{
x
=
a
(
t
′
−
sin
t
′
)
,
y
=
a
(
1
−
cos
t
′
)
.
{\displaystyle {\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. (A ), §117 ,
(A )
α
=
x
−
R
sin
τ
,
β
=
y
+
R
cos
τ
.
{\displaystyle \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
,
α
,
β
{\displaystyle x,y,R,T,\alpha ,\beta }
s . Differentiating (A ) with respect to
s
{\displaystyle s}
(B )
d
α
d
s
=
d
x
d
s
−
R
cos
τ
d
τ
d
s
−
sin
τ
d
R
d
s
,
{\displaystyle {\frac {d\alpha }{ds}}={\frac {dx}{ds}}-R\cos \tau {\frac {d\tau }{ds}}-\sin \tau {\frac {dR}{ds}},}
d
β
d
s
=
d
y
d
s
−
R
sin
τ
d
τ
d
s
+
cos
τ
d
R
d
s
.
{\displaystyle {\frac {d\beta }{ds}}={\frac {dy}{ds}}-R\sin \tau {\frac {d\tau }{ds}}+\cos \tau {\frac {dR}{ds}}.}
But
d
x
d
s
=
cos
τ
,
d
y
d
s
=
sin
τ
{\displaystyle {\tfrac {dx}{ds}}=\cos \tau ,{\tfrac {dy}{ds}}=\sin \tau }
(26 ), §90 ; and
d
τ
d
s
=
1
R
{\displaystyle {\tfrac {d\tau }{ds}}={\tfrac {1}{R}}}
(38 ) in § 100 and (39 ) in §101 .
Substituting in (B ) and (C ) , we obtain
(D )
d
α
d
s
=
cos
τ
−
R
cos
τ
⋅
1
R
−
sin
τ
d
R
d
s
=
−
sin
τ
d
R
d
s
,
{\displaystyle {\frac {d\alpha }{ds}}=\cos \tau -R\cos \tau \cdot {\frac {1}{R}}-\sin \tau {\frac {dR}{ds}}=-\sin \tau {\frac {dR}{ds}},}
(E )
d
β
d
s
=
sin
τ
−
R
sin
τ
⋅
1
R
+
cos
τ
d
R
d
s
=
cos
τ
d
R
d
s
.
{\displaystyle {\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 )
d
β
d
α
=
−
cot
τ
=
−
1
tan
τ
=
−
1
d
y
d
x
.
{\displaystyle {\frac {d\beta }{d\alpha }}=-\cot \tau =-{\frac {1}{\tan \tau }}=-{\frac {1}{\frac {dy}{dx}}}.}
