89
SECONDE PARTIE. — SECTION VII.
![{\displaystyle {\begin{aligned}\delta X'=&-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\delta \theta }{1-e}}={\frac {\mathrm {g} }{a^{2}}}{\frac {dc}{(1-e)^{2}}},\\\delta Y'=&{\sqrt {\frac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}{\frac {de}{(1-e)^{2}}}+{\frac {d\left({\sqrt {\cfrac {\mathrm {g} }{a}}}{\sqrt {1-e^{2}}}\right)}{1-e}}\\=&d\left({\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{1-e}}\right)={\frac {\sqrt {1-e^{2}}}{1-e}}d{\sqrt {\frac {\mathrm {g} }{a}}}+{\sqrt {\frac {\mathrm {g} }{a}}}{\frac {de}{(1-e){\sqrt {1-e^{2}}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/448c60d7173ee543c256014cb934be816f6406d8)
69. Ici nous avons conservé la quantité
qui est l’excentricité ; mais si, à sa place, on veut employer le paramètre
on aura, par la différentiation,
![{\displaystyle de={\frac {\left(1-e^{2}\right)da-db}{2ae}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4189ea0031f29d874ede2867bc251582caf7024e)
et les expressions de
et de
qui contiennent
deviendront
![{\displaystyle {\begin{aligned}\delta X=&{\frac {-\left(1-e^{2}\right)da+db}{2e}},\\\delta Y'=&{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{2ae}}da-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {db}{2ae(1-e){\sqrt {1-e^{2}}}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/403bb9d390fee343a01f8c412969d4367074e993)
De là nous aurons les différences partielles
![{\displaystyle {\begin{alignedat}{2}{\frac {\partial X}{\partial a}}\ =&-{\frac {\left(1-e^{2}\right)}{2e}},&{\frac {\partial X}{\partial b}}\ =&{\frac {1}{2e}},\quad {\frac {\partial X}{\partial c}}\ =0,\\{\frac {\partial Y}{\partial a}}\ =&0,&{\frac {\partial Y}{\partial b}}\ =&0,\qquad {\frac {\partial Y}{\partial c}}\ =-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{1-e}},\\{\frac {\partial X'}{\partial a}}=&0,&{\frac {\partial X'}{\partial b}}=&0,\qquad {\frac {\partial X'}{\partial c}}={\frac {\mathrm {g} }{a^{2}}}\ {\frac {1}{(1-e)^{2}}},\\{\frac {\partial Y'}{\partial a}}=&{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {\sqrt {1-e^{2}}}{2ae}},\quad &{\frac {\partial Y'}{\partial b}}=&-{\sqrt {\frac {\mathrm {g} }{a}}}\ {\frac {1}{2ae(1-e){\sqrt {1-e^{2}}}}},&\quad {\frac {\partial Y'}{\partial c}}=0,\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8113ae136920587d2847c1a77cdd264a19ec80f8)
et l’on trouvera, par la substitution de ces valeurs dans les expressions des symboles
de l’article 67,
![{\displaystyle {\begin{aligned}\,[a,b]=&0,\\\,[a,c]=&-{\frac {g}{2a^{2}e}}+{\frac {g\left(1-e^{2}\right)}{2a^{2}e(1-e)}}={\frac {g}{2a^{2}}},\\\,[b,c]=&{\frac {g}{2a^{2}e}}\left[{\frac {1}{\left(1-e^{2}\right)}}-{\frac {\sqrt {1-e^{2}}}{(1-e)^{2}{\sqrt {1-e^{2}}}}}\right]=0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f53c50665c6a2fdbd25b53efb1da0806a1d1dab)