Substituting the values of
x
−
α
{\displaystyle x-\alpha }
y
−
β
{\displaystyle y-\beta }
(D ) in (A ) , and solving for
R
{\displaystyle R}
R
=
±
[
1
+
(
d
y
d
x
)
2
]
3
2
d
2
y
d
x
2
,
{\displaystyle R=\pm {\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}},}
which is identical with (42), §103 . Hence
Theorem. The radius of the circle of curvature equals the radius of curvature.
117. Second method for finding center of curvature. §104 . Draw a figure showing the tangent line, circle of curvature, radius of curvature, and center of curvature
(
α
,
β
)
{\displaystyle (\alpha ,\beta )}
P
(
x
,
y
)
{\displaystyle P(x,y)}
α
=
O
A
=
O
D
−
A
D
=
O
D
−
B
P
=
x
−
B
P
,
{\displaystyle \alpha \ =OA=OD-AD=OD-BP=x-BP,}
β
=
A
C
=
A
B
+
B
C
=
D
P
+
B
C
=
y
+
B
C
.
{\displaystyle \beta \ =AC=AB+BC=DP+BC=y+BC.}
But
B
P
=
R
sin
τ
,
B
C
=
R
cos
τ
{\displaystyle BP=R\sin \tau ,BC=R\cos \tau }
(A )
α
=
x
−
R
sin
τ
,
β
=
y
+
R
cos
τ
.
{\displaystyle \alpha =x-R\sin \tau ,\ \beta =y+R\cos \tau .}
From (29 ), §90 , and (42 ), §103 ,
sin
τ
=
d
y
d
x
[
1
+
(
d
y
d
x
)
2
]
1
2
,
cos
τ
=
1
[
1
+
(
d
y
d
x
)
1
2
]
1
2
,
R
=
[
1
+
(
d
y
d
x
)
2
]
3
2
d
2
y
d
x
2
{\displaystyle \sin \tau ={\frac {\frac {dy}{dx}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}}},\ \cos \tau ={\frac {1}{\left[1+\left({\frac {dy}{dx}}\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}}},\ R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}}}
Substituting these back in (A ) , we get
(50)
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}}}}.}
From (23), §85 , we know that at a point of inflection (as Q in the next figure)
d
2
y
d
x
2
=
0
,
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=0,}
![{\displaystyle R=\pm {\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da872844e7b5d5e5540e505f0ffafc8167d79eda)
![{\displaystyle \sin \tau ={\frac {\frac {dy}{dx}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}}},\ \cos \tau ={\frac {1}{\left[1+\left({\frac {dy}{dx}}\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}}},\ R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e604690f112e70559503f60bfde8ad09c32e0e93)
![{\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/510aee465b335dc8413db651810020db3d0c5398)
