Page:Annales de mathématiques pures et appliquées, 1831-1832, Tome 22.djvu/39

$\mathrm {DC} n'=\operatorname {d} \varphi +\operatorname {d} \omega ,$

Ainsi, $R$ désignant le rayon de courbure, on a

$R={\frac {nn'}{\mathrm {DC} n'}}={\frac {\operatorname {d} s}{\operatorname {d} \varphi +\operatorname {d} \omega }}.\qquad$(6)

Menons présentement $nq,$ perpendiculaire sur $\mathrm {O} n',$ et posons $On=y,\ nq=\operatorname {d} x\,;$ on aura $\operatorname {d} xs=y\operatorname {d} \omega ,\ qn'=\operatorname {d} y\,;$ au moyen de quoi le rayon de courbure deviendra

$R={\frac {y\operatorname {d} s}{\operatorname {d} x+y\operatorname {d} \varphi }}\,;\qquad$(5)

mais le triangle $nqn'$ fournit les six expressions (2), et par suite les six expressions (3). Substituant donc tour à tour ces six dernières pour $d\varphi$ dans (6), on aura

${\begin{alignedat}{2}R&={\frac {y\operatorname {d} s\operatorname {d} y}{\operatorname {d} x\operatorname {d} y+y\operatorname {d} s.\operatorname {d} \left({\frac {\operatorname {d} x}{\operatorname {d} s}}\right)}},&\qquad R&={\frac {y\operatorname {d} s\operatorname {d} x}{\operatorname {d} x^{2}-y\operatorname {d} s.\operatorname {d} \left({\frac {\operatorname {d} y}{\operatorname {d} s}}\right)}},\\\\R&={\frac {y\operatorname {d} s^{3}}{\operatorname {d} s^{2}\operatorname {d} x+y\operatorname {d} y^{2}.\operatorname {d} \left({\frac {\operatorname {d} x}{\operatorname {d} y}}\right)}},&R&={\frac {y\operatorname {d} s^{3}}{\operatorname {d} s^{2}\operatorname {d} x-y\operatorname {d} x^{2}.\operatorname {d} \left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)}},\\\\R&={\frac {y\operatorname {d} s^{2}\operatorname {d} x}{\operatorname {d} s\operatorname {d} x^{2}+y\operatorname {d} y^{2}.\operatorname {d} \left({\frac {\operatorname {d} s}{\operatorname {d} y}}\right)}},&R&={\frac {y\operatorname {d} s^{2}\operatorname {d} y}{\operatorname {d} s\operatorname {d} x\operatorname {d} y-y\operatorname {d} x^{2}.\operatorname {d} \left({\frac {\operatorname {d} s}{\operatorname {d} x}}\right)}}.\end{alignedat}}$

Si, pour se conformer aux notations usitées, on remplace $y$ par $r,$ il faudra remplacer $\operatorname {d} x$ par $r\operatorname {d} \omega$ et $\operatorname {d} y$ par $\operatorname {d} r\,;$ et alors les formules ci-dessus deviendront

${\begin{alignedat}{2}R&={\frac {\operatorname {d} s\operatorname {d} r}{\operatorname {d} r\operatorname {d} \omega +\operatorname {d} s.\operatorname {d} \left({\frac {r\operatorname {d} \omega }{\operatorname {d} s}}\right)}},&\quad R&={\frac {r\operatorname {d} \omega \operatorname {d} s}{r\operatorname {d} \omega ^{2}-\operatorname {d} s.\operatorname {d} \left({\frac {\operatorname {d} r}{\operatorname {d} s}}\right)}},\\\\\end{alignedat}}$