Page:Mémoires de l’Académie des sciences, Tome 7.djvu/445

${\begin{aligned}(e,c)&=-{\frac {\mathrm {(C-A)(C-B)} e}{n\mathrm {C} }},\\(e,g)&=-{\frac {\mathrm {(C-A)(C-B)} e\cos .\lambda }{n\mathrm {C} }},\\(e,l)&={\frac {\mathrm {(C-A)(C-B)} e}{n\mathrm {C} }},\\(\lambda ,g)&=-Cn\sin .\lambda .\end{aligned}}$

(7) Au moyen de ces résultats, l’équation (6) donne :

${\begin{aligned}{\frac {d\Omega }{dn}}dt&=\mathrm {C} \cos .\lambda dg-\mathrm {C} dl,\\{\frac {d\Omega }{de}}dt&={\frac {\mathrm {(C-A)(C-B)} }{n\mathrm {C} }}(edl-edc-e\cos .\lambda dg),\\{\frac {d\Omega }{dc}}dt&={\frac {\mathrm {(C-A)(C-B)} }{n\mathrm {C} }}ede,\\{\frac {d\Omega }{d\lambda }}dt&=-\mathrm {C} n\sin .\lambda dg,\\{\frac {d\Omega }{dg}}dt&=-\mathrm {C} \cos .\lambda dn+{\frac {\mathrm {(C-A)(C-B)\cos .\lambda } }{n\mathrm {C} }}ede+\mathrm {C} n\sin .\lambda d\lambda ,\\{\frac {d\Omega }{dl}}dt&=\mathrm {C} dn-{\frac {\mathrm {(C-A)(C-B)\cos .\lambda } }{n\mathrm {C} }}edc\,;\end{aligned}}$

d’où l’on tire réciproquement

$\left.{\begin{aligned}&dn=\left({\frac {d\Omega }{dc}}+{\frac {d\Omega }{dl}}\right){\frac {dt}{\mathrm {C} }},\\&ede={\frac {d\Omega }{dc}}{\frac {\mathrm {C} ndt}{\mathrm {(C-A)(C-B)} }},\\&edc=-{\frac {d\Omega }{de}}{\frac {\mathrm {C} ndt}{\mathrm {(C-A)(C-B)} }}-{\frac {d\Omega }{dn}}{\frac {edt}{\mathrm {C} }},\\&d\lambda ={\frac {d\Omega }{dg}}{\frac {dt}{\mathrm {C} n\sin .\lambda }}+{\frac {d\Omega }{dl}}{\frac {\cos .\lambda dt}{\mathrm {C} n\sin .\lambda }},\\&dg=-{\frac {d\Omega }{d\lambda }}{\frac {dt}{\mathrm {C} n\sin .\lambda }},\\&dl=-{\frac {d\Omega }{d\lambda }}{\frac {\mathrm {C} n\sin .\lambda }{\cos .\lambda dt}}-{\frac {d\Omega }{dn}}{\frac {dt}{\mathrm {C} }}.\end{aligned}}\right\}$

(7)