Page:Joseph Louis de Lagrange - Œuvres, Tome 6.djvu/445

${\begin{aligned}\delta f=&{\frac {y}{r^{3}}}(x\delta y-y\delta x)-\left(f+{\frac {x}{r}}\right){\frac {dy\delta x-y\,d\,\delta x}{dt{\sqrt {2h(1+m)}}}}\\&-{\frac {y}{r}}{\frac {dy\delta y-y\,d\,\delta y}{dt{\sqrt {2h(1+m)}}}}+{\frac {2dy\delta h}{dt{\sqrt {2h(1+m)}}}},\\\\\delta g=&-{\frac {x}{r^{3}}}(x\delta y-y\delta x)+\left(f+{\frac {x}{r}}\right){\frac {dx\delta x-x\,d\,\delta x}{dt{\sqrt {2h(1+m)}}}}\\&+{\frac {y}{r}}{\frac {dx\delta y-x\,d\,\delta y}{dt{\sqrt {2h(1+m)}}}}-{\frac {2dx\delta h}{dt{\sqrt {2h(1+m)}}}},\\\\\delta b=&{\frac {dy\delta z-y\,d\,\delta z}{dt{\sqrt {2h(1+m)}}}}\\\\\delta c=&{\frac {-dx\delta z+x\,d\,\delta z}{dt{\sqrt {2h(1+m)}}}}\\\\\delta i=&3t{\sqrt {\frac {2(1+m)}{a^{3}}}}{\frac {\delta a}{a}}+{\frac {2(p\delta q-q\delta p)}{a^{2}(1-4n){\sqrt {n}}}}-{\frac {q(p-4an)\delta n}{a^{2}(1-4n)n^{\frac {3}{2}}}}\,;\end{aligned}}$

mais, à cause de $b=0,\ c=0,\ g=0,$ on aura

$p=fx,\quad q=fy,\quad \delta p=f\delta x+x\delta f+y\delta g,\quad \delta q=f\delta y+y\delta f-x\delta g\,;$

donc

$p\delta q-q\delta p=f^{2}(x\delta y-y\delta x)-\left(x^{2}+y^{2}\right)f\delta g\,;$

de plus

$n={\frac {h}{a}}={\frac {1-f^{2}}{4}},\quad \delta n=-{\frac {f\delta f}{2}}\,;$

donc la dernière formule deviendra

$\delta i=3{\sqrt {\frac {2(1+m)}{a^{3}}}}{\frac {\delta a}{a}}+{\frac {2\left(x\delta y-y\delta x-{\cfrac {r^{2}\delta g}{f}}\right)}{\sqrt {a^{3}h}}}+{\frac {y(fx-4h)\delta f}{2{\sqrt {ah^{3}}}}}.$

Et l’on remarquera que, à cause de $n={\frac {h}{a}}$ et de $\delta n=-{\frac {f\delta f}{2}},$ on aura encore cette équation entre $\delta a,\delta h$ et $\delta f,$ savoir

$a\delta h-h\delta a+{\frac {a^{2}f\delta f}{2}}=0,$