Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/271

$\psi \quad$ l’angle de la latitude de $\mathrm {B}$ par rapport à ce même plan,

$\varphi '\quad$la longitude de $\mathrm {C} ,$

$\psi '\quad$sa latitude,

et qu’on fasse, pour abréger,

${\begin{aligned}\mathrm {P} \ \,=&pp'+pp''+p'p''=r^{2}r'^{2}-p''^{2}\\=&{\frac {1}{4}}\left(2r^{2}r'^{2}+2r^{2}r''^{2}+2r'^{2}r''^{2}-r^{4}-r'^{4}-r''^{4}\right)\\=&-{\frac {1}{4}}(r+r'+r'')(r+r'-r'')(r-r'+r'')(r-r'-r''),\\\\\Sigma \ \,=&p\ \,\left({\frac {d\left(r^{2}\right)}{dt}}\ \,\right)^{2}+p'\left({\frac {dp''}{dt}}\right)^{2}+p''\left({\frac {dp'}{dt}}\right)^{2}\\&\ \qquad \qquad \qquad -2\left(p''{\frac {dp'}{dt}}-p'{\frac {dp''}{dt}}\right){\frac {d\rho }{dt}}+r^{2}\left({\frac {d\rho }{dt}}\right)^{2},\\\Sigma '\,=&p'\,\left({\frac {d\left(r'^{2}\right)}{dt}}\,\right)^{2}+p\,\left({\frac {dp''}{dt}}\right)^{2}+p''\left({\frac {dp}{dt}}\,\right)^{2}\\&\ \qquad \qquad \qquad +2\left(p''{\frac {dp}{dt}}\,-p\,{\frac {dp''}{dt}}\right){\frac {d\rho }{dt}}+r'^{2}\left({\frac {d\rho }{dt}}\right)^{2},\\\Sigma ''=&p''\left({\frac {d\left(r''^{2}\right)}{dt}}\right)^{2}+p'\left({\frac {dp}{dt}}\ \right)^{2}+p\ \,\left({\frac {dp'}{dt}}\right)^{2}\\&\ \qquad \qquad \qquad +2\left(p\ \,{\frac {dp'}{dt}}-p'{\frac {dp}{dt}}\ \ \right){\frac {d\rho }{dt}}+r''^{2}\left({\frac {d\rho }{dt}}\right)^{2},\\\\\Pi \ \,=&\ \,r^{2}u^{2}\ -\left({\frac {d\left(r^{2}\right)}{2dt}}\right)^{2},\\\Pi '\,=&\,r'^{2}u'^{2}\,-\left({\frac {d\left(r'^{2}\right)}{2dt}}\right)^{2},\\\Pi ''=&r''^{2}u''^{2}-\left({\frac {d\left(r''^{2}\right)}{2dt}}\right)^{2},\\\\\Psi \ \,=&\ \,pv\ \,-\left({\frac {dp}{2dt}}\,\right)^{2}+\left({\frac {d\rho }{2dt}}\right)^{2},\\\Psi '\,=&\,p'v'\,-\left({\frac {dp'}{2dt}}\,\right)^{2}+\left({\frac {d\rho }{2dt}}\right)^{2},\\\Psi ''=&p''v''-\left({\frac {dp''}{2dt}}\right)^{2}+\left({\frac {d\rho }{2dt}}\right)^{2},\end{aligned}}$

en supposant

$v={\frac {u'^{2}+u''^{2}-u^{2}}{2}},\quad v'={\frac {u^{2}+u''^{2}-u'^{2}}{2}},\quad v''={\frac {u^{2}+u'^{2}-u''^{2}}{2}},$

on aura sur-le-champ

$\sin \psi ={\frac {{\cfrac {1}{\mathrm {B} }}{\sqrt {4\mathrm {P} u'^{2}-\Sigma '}}+{\cfrac {1}{\mathrm {A} }}{\sqrt {4\mathrm {P} u''^{2}-\Sigma ''}}}{2hr}},\quad {\frac {d\varphi }{dt}}={\frac {\mathrm {{\cfrac {\Pi }{C}}+{\cfrac {\Psi '}{A}}+{\cfrac {\Psi ''}{B}}} }{hr^{2}\cos ^{2}\psi }},$