Page:Annales de mathématiques pures et appliquées, 1830-1831, Tome 21.djvu/249

${\frac {\operatorname {d} S}{\operatorname {d} x}}t+{\frac {\operatorname {d} S}{\operatorname {d} x}}u+{\frac {\operatorname {d} S}{\operatorname {d} x}}v=0\,;\qquad$(58)

${\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}(t-t')+{\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}(u-u')+{\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}(v-v')=0.\qquad$(54)

En combinant la première avec l’équation (69) et la seconde avec l’équation (72), on obtiendra, pour les équations des tangentes à la section, en ces deux points,

${\frac {t}{B{\frac {\operatorname {d} S}{\operatorname {d} z}}-C{\frac {\operatorname {d} S}{\operatorname {d} y}}}}={\frac {u}{C{\frac {\operatorname {d} S}{\operatorname {d} x}}-A{\frac {\operatorname {d} S}{\operatorname {d} z}}}}={\frac {v}{A{\frac {\operatorname {d} S}{\operatorname {d} y}}-B{\frac {\operatorname {d} S}{\operatorname {d} x}}}},\qquad$(73)

${\frac {t-t'}{B{\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}-C{\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}}}={\frac {u-u'}{C{\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}-A{\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}}}={\frac {v-v'}{A{\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}-B{\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}}},\qquad$(74)

En conséquence, les plans conduits par ces mêmes points, perpendiculairement à ces mêmes tangentes, auront respectivement pour équations

$\left(B{\frac {\operatorname {d} S}{\operatorname {d} z}}-C{\frac {\operatorname {d} S}{\operatorname {d} y}}\right)t+\left(C{\frac {\operatorname {d} S}{\operatorname {d} x}}-A{\frac {\operatorname {d} S}{\operatorname {d} z}}\right)u$

$+\left(A{\frac {\operatorname {d} S}{\operatorname {d} y}}-B{\frac {\operatorname {d} S}{\operatorname {d} x}}\right)v=0,\quad$(75)

$\left(B{\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}-C{\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}\right)(t-t')+\left(C{\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}-A{\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}\right)(u-u')$

$+\left(A{\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}-B{\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}\right)(v-v')=0,\quad$(76)

et couperont le plan (69) suivant les normales menées à la section par l’origine et par le point $(t',u',v')$ ; on aura doncle point de concours de ces deux normales, en combinant entre elles les trois équations (69), (75), (76). En traitant donc ces trois équations comme celles d’un même problème déterminé, opérant toutes les simplifications que pourra permettre la relation (70), et posant, pour abréger,

$M=\left\{{\begin{aligned}&(Bv'-Cu'){\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}\\\\+&(Ct'-Av'){\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}\\\\+&(Au'-Bt'){\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}\end{aligned}}\right\},\mathrm {(77)} \ N=\left\{{\begin{aligned}&\left(B{\frac {\operatorname {d} S}{\operatorname {d} z}}-C{\frac {\operatorname {d} S}{\operatorname {d} y}}\right){\frac {\operatorname {d} \Omega '}{\operatorname {d} t'}}\\\\+&\left(C{\frac {\operatorname {d} S}{\operatorname {d} x}}-A{\frac {\operatorname {d} S}{\operatorname {d} z}}\right){\frac {\operatorname {d} \Omega '}{\operatorname {d} u'}}\\\\+&\left(A{\frac {\operatorname {d} S}{\operatorname {d} y}}-B{\frac {\operatorname {d} S}{\operatorname {d} x}}\right){\frac {\operatorname {d} \Omega '}{\operatorname {d} v'}}\end{aligned}}\right\}\,;\ \mathrm {(78)}$