Hence, by Rolle's Theorem (§105 ),
F
′
(
x
)
{\displaystyle F'(x)}
x
{\displaystyle x}
x
0
{\displaystyle x_{0}}
x
1
{\displaystyle x_{1}}
x
′
{\displaystyle x'}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
″
{\displaystyle x''}
F
′
(
x
′
)
=
0
,
F
′
(
x
″
)
=
0.
{\displaystyle F'(x')=0,\ F'(x'')=0.}
Again, for the same reason,
F
″
(
x
)
{\displaystyle F''(x)}
x
{\displaystyle x}
x
′
{\displaystyle x'}
x
″
{\displaystyle x''}
x
3
{\displaystyle x_{3}}
F
″
(
x
3
)
=
0.
{\displaystyle F''(x_{3})\ =\ 0.}
Therefore the elements
α
′
,
β
′
,
R
′
{\displaystyle \alpha ',\beta ',R'}
P
0
,
P
1
,
P
2
{\displaystyle P_{0},P_{1},P_{2}}
F
(
x
0
)
=
0
,
F
′
(
x
′
)
=
0
,
F
″
(
x
3
)
=
0.
{\displaystyle F(x_{0})=0,\ F'(x')=0,\ F''(x_{3})=0.}
Now let the points
P
1
{\displaystyle P_{1}}
P
2
{\displaystyle P_{2}}
P
0
{\displaystyle P_{0}}
x
1
,
x
2
,
x
′
,
x
″
,
x
3
{\displaystyle x_{1},x_{2},x',x'',x_{3}}
x
0
{\displaystyle x_{0}}
α
,
β
,
R
{\displaystyle \alpha ,\beta ,R}
F
(
x
0
)
=
0
,
F
′
(
x
0
)
=
0
,
F
″
(
x
0
)
=
0
;
{\displaystyle F(x_{0})=0,\ F'(x_{0})=0,\ F''(x_{0})=0;}
or, dropping the subscripts, which is the same thing,
(A)
(
x
−
α
)
2
+
(
y
−
β
)
2
=
R
2
,
{\displaystyle (x-\alpha )^{2}+(y-\beta )^{2}\ =\ R^{2},}
(B)
(
x
−
α
)
+
(
y
−
β
)
d
y
d
x
=
0
,
{\displaystyle \left(x-\alpha \right)+\left(y-\beta \right){\frac {dy}{dx}}=0,}
differentiating (A ) .
(C)
1
+
(
d
y
d
x
)
2
+
(
y
−
β
)
d
2
y
d
x
2
=
0
,
{\displaystyle 1+\left({\frac {dy}{dx}}\right)^{2}+\left(y-\beta \right){\frac {d^{2}y}{dx^{2}}}=0,}
differentiating (B ) .
Solving (B ) and (C ) for
x
−
α
{\displaystyle x-\alpha }
y
−
β
{\displaystyle y-\beta }
(
d
2
y
d
x
2
≠
0
)
{\displaystyle \left({\tfrac {d^{2}y}{dx^{2}}}\neq 0\right)}
(D)
{
x
−
α
=
d
y
d
x
[
1
+
(
d
y
d
x
)
2
]
d
2
y
d
x
2
y
−
β
=
−
1
+
(
d
y
d
x
)
2
d
2
y
d
x
2
;
{\displaystyle {\begin{cases}x-\alpha ={\frac {{\frac {dy}{dx}}\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]}{\frac {d^{2}y}{dx^{2}}}}\\y-\beta =-{\frac {1+\left({\frac {dy}{dx}}\right)^{2}}{\frac {d^{2}y}{dx^{2}}}};\end{cases}}}
Radius and center of curvature.
hence the coördinates of the center of curvature are
(E)
α
=
x
−
d
y
d
x
[
1
+
(
d
y
d
x
)
2
]
d
2
y
d
x
2
;
β
=
y
+
1
+
(
d
y
d
x
)
2
d
2
y
d
x
2
.
{\displaystyle \alpha =x-{\frac {{\frac {dy}{dx}}\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]}{\frac {d^{2}y}{dx^{2}}}};\beta =y+{\frac {1+\left({\frac {dy}{dx}}\right)^{2}}{\frac {d^{2}y}{dx^{2}}}}.}
(
d
2
y
d
x
2
≠
0
)
{\displaystyle \left({\frac {d^{2}y}{dx^{2}}}\neq 0\right)}
![{\displaystyle {\begin{cases}x-\alpha ={\frac {{\frac {dy}{dx}}\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]}{\frac {d^{2}y}{dx^{2}}}}\\y-\beta =-{\frac {1+\left({\frac {dy}{dx}}\right)^{2}}{\frac {d^{2}y}{dx^{2}}}};\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcb254a3edd437c68d24ebbe1e18925b7f052cd5)
![{\displaystyle \alpha =x-{\frac {{\frac {dy}{dx}}\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]}{\frac {d^{2}y}{dx^{2}}}};\beta =y+{\frac {1+\left({\frac {dy}{dx}}\right)^{2}}{\frac {d^{2}y}{dx^{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f5f77eead5d051f5e9196899dddfcf98f86ed8d)
