Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/767

Or, en regardant $\Omega$ comme fonction de $b$ et de $\varphi ,$ ou de $\beta$ et $\gamma ,$ on a l’équation identique

${\begin{aligned}{\frac {d\Omega }{db}}db+{\frac {d\Omega }{d\varphi }}d\varphi &={\frac {d\Omega }{d\beta }}d\beta +{\frac {d\Omega }{d\gamma }}d\gamma \\&=\left({\frac {d\Omega }{d\beta }}\sin \varphi +{\frac {d\Omega }{d\gamma }}\cos \varphi \right)db\end{aligned}}$

$+b\left({\frac {d\Omega }{d\beta }}\cos \varphi -{\frac {d\Omega }{d\gamma }}\sin \varphi \right)d\varphi .$

Donc

${\begin{aligned}{\frac {d\Omega }{db}}=&{\frac {d\Omega }{d\beta }}\sin \varphi +{\frac {d\Omega }{d\gamma }}\cos \varphi ,\\{\frac {d\Omega }{d\varphi }}=&b\left({\frac {d\Omega }{d\beta }}\cos \varphi -{\frac {d\Omega }{d\gamma }}\sin \varphi \right).\end{aligned}}$

De même, en regardant $\Omega$ comme fonction de $g,h,$ ou de $\varepsilon ,\lambda ,$ on a

${\begin{aligned}{\frac {d\Omega }{dg}}dg+{\frac {d\Omega }{dh}}dh&={\frac {d\Omega }{d\varepsilon }}d\varepsilon +{\frac {d\Omega }{d\lambda }}d\lambda \\&=\left({\frac {d\Omega }{d\varepsilon }}\cos h\sin g+{\frac {d\Omega }{d\lambda }}\cos h\cos g\right)dh\end{aligned}}$

$+\left({\frac {d\Omega }{d\varepsilon }}\sin h\cos g-{\frac {d\Omega }{d\lambda }}\sin h\sin g\right)dg\,;$

donc

${\begin{aligned}{\frac {d\Omega }{dg}}=&\left({\frac {d\Omega }{d\varepsilon }}\cos g-{\frac {d\Omega }{d\lambda }}\sin g\right)\sin h,\\{\frac {d\Omega }{dh}}=&\left({\frac {d\Omega }{d\varepsilon }}\sin g+{\frac {d\Omega }{d\lambda }}\cos g\right)\cos h.\end{aligned}}$

Substituant ces valeurs dans les formules précédentes, elles deviendront

${\begin{aligned}{\frac {d\beta }{dt}}=&\quad \ {\frac {\sqrt {1-b^{2}}}{na^{2}}}{\frac {d\Omega }{d\gamma }}+{\frac {1-b^{2}}{na^{2}b^{2}}}\beta {\frac {d\Omega }{dc}},\\{\frac {d\gamma }{dt}}=&-{\frac {\sqrt {1-b^{2}}}{na^{2}}}{\frac {d\Omega }{d\beta }}+{\frac {1-b^{2}}{na^{2}b^{2}}}\gamma {\frac {d\Omega }{dc}},\\{\frac {d\varepsilon }{dt}}=&\quad \ {\frac {\cos h}{na^{2}{\sqrt {1-b^{2}}}}}{\frac {d\Omega }{d\lambda }},\\{\frac {d\lambda }{dt}}=&-{\frac {\cos h}{na^{2}{\sqrt {1-b^{2}}}}}{\frac {d\Omega }{d\varepsilon }},\end{aligned}}$