Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/204

Substituting the values of $x-\alpha$ and $y-\beta$ from (D) in (A), and solving for $R$, we get

$R=\pm {\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}},$

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. Here we shall make use of the definition of circle of curvature given in §104. Draw a figure showing the tangent line, circle of curvature, radius of curvature, and center of curvature $(\alpha ,\beta )$ corresponding to the point $P(x,y)$ on the curve. Then

$\alpha \ =OA=OD-AD=OD-BP=x-BP,$

$\beta \ =AC=AB+BC=DP+BC=y+BC.$

But $BP=R\sin \tau ,BC=R\cos \tau$. Hence

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

From (29), §90, and (42), §103,

$\sin \tau ={\frac {\frac {dy}{dx}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {1}{2}}}},\ \cos \tau ={\frac {1}{\left[1+\left({\frac {dy}{dx}}\right)^{\frac {1}{2}}\right]^{\frac {1}{2}}}},\ R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{\frac {3}{2}}}{\frac {d^{2}y}{dx^{2}}}}$

Substituting these back in (A), we get

(50)	$\alpha =x-{\frac {{\frac {dy}{dx}}\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]}{\frac {d^{2}y}{dx^{2}}}};\ \beta =y+{\frac {1+\left({\frac {dy}{dx}}\right)^{2}}{\frac {d^{2}y}{dx^{2}}}}.$

From (23), §85, we know that at a point of inflection (as Q in the next figure)

${\frac {d^{2}y}{dx^{2}}}=0,$