Page:Annales de mathématiques pures et appliquées, 1822-1823, Tome 13.djvu/95

91

INDÉTERMINÉES.

les coefficiens de $X_{1},Y_{1},Z_{1},$ on élimine $\lambda$ et $\mu$ entre les équations résultantes, il viendra

$\left.{\begin{aligned}&x_{1}'\left[\left({\frac {\operatorname {d} R}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)_{1}\right]\\\\+&y_{1}'\left[\left({\frac {\operatorname {d} R}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} x}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}\right]\\\\+&z_{1}'\left[\left({\frac {\operatorname {d} R}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}\right]\end{aligned}}\right\}=0$

mais, si l’on différentie les deux équations $R=0,\ S=0,$ en y considérant $x$ et $y$ comme des fonctions de $z,$ il viendra, en passant à la limite,

${\begin{aligned}&\left({\frac {\operatorname {d} R}{\operatorname {d} x}}\right)_{1}{\frac {\operatorname {d} x_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} R}{\operatorname {d} y}}\right)_{1}{\frac {\operatorname {d} y_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} R}{\operatorname {d} z}}\right)_{1}=0,\\\\&\left({\frac {\operatorname {d} S}{\operatorname {d} x}}\right)_{1}{\frac {\operatorname {d} x_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)_{1}{\frac {\operatorname {d} y_{1}}{\operatorname {d} z_{1}}}+\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}=0\,;\end{aligned}}$

équations d’où l’on tirera

${\begin{aligned}&\left[\left({\frac {\operatorname {d} R}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)_{1}\right]\\\\=&\left[\left({\frac {\operatorname {d} R}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}\right]{\frac {\operatorname {d} x_{1}}{\operatorname {d} z_{1}}},\\\\&\left[\left({\frac {\operatorname {d} R}{\operatorname {d} z}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} x}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}\right]\\\\=&\left[\left({\frac {\operatorname {d} R}{\operatorname {d} x}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} y}}\right)_{1}-\left({\frac {\operatorname {d} R}{\operatorname {d} y}}\right)_{1}\left({\frac {\operatorname {d} S}{\operatorname {d} z}}\right)_{1}\right]{\frac {\operatorname {d} y_{1}}{\operatorname {d} z_{1}}}\,;\end{aligned}}$

mettant les valeurs données par ces deux dernières dana l’équation