Page:Laplace - Œuvres complètes, Gauthier-Villars, 1878, tome 2.djvu/149

que $\varphi _{\text{ı}},{\frac {d^{2}\varphi _{\text{ı}}}{ds^{2}}}$ et ${\frac {d\theta _{\text{ı}}}{ds}}$ sont des quantités de l’ordre $\alpha ,$

${\frac {d^{2}x_{\text{ı}}}{ds^{2}}}=\alpha {\frac {d^{2}u'_{\text{ı}}}{ds^{2}}}\sin \theta _{\text{ı}}+r_{\text{ı}}{\frac {d^{2}\theta _{\text{ı}}}{ds^{2}}}\cos \theta _{\text{ı}}-r_{\text{ı}}\sin \theta _{\text{ı}}{\frac {d^{2}\varphi _{\text{ı}}}{ds^{2}}}.$

On a ensuite

$\alpha {\frac {d^{2}u'_{\text{ı}}}{ds^{2}}}=\alpha {\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}{\frac {d\varphi _{\text{ı}}^{2}}{ds^{2}}}-\alpha {\frac {\partial u'_{\text{ı}}}{\partial \psi }}{\frac {d^{2}\theta _{\text{ı}}}{ds^{2}}}={\frac {\alpha {\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}}{\cos ^{2}\psi _{\text{ı}}}}-\alpha {\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} \psi _{\text{ı}}\,;$

de plus, $ds=r_{\text{ı}}\sin \theta _{\text{ı}}d\varphi _{\text{ı}}\,;$ on aura donc, en substituant pour $r_{\text{ı}},\theta _{\text{ı}},{\frac {d\varphi _{\text{ı}}}{ds}}$ et ${\frac {d^{2}\theta _{\text{ı}}}{ds^{2}}}$ leurs valeurs précédentes,

${\begin{aligned}{\frac {d^{2}x_{\text{ı}}}{ds^{2}}}=&(1-\alpha u'_{\text{ı}}){\frac {\sin ^{2}\psi _{\text{ı}}}{\cos \psi _{\text{ı}}}}+\alpha {\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} ^{2}\psi _{\text{ı}}\sin \psi _{\text{ı}}\\\\&-{\frac {1}{\cos \psi _{\text{ı}}}}\left(1-\alpha u'_{\text{ı}}+\alpha {\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} \psi _{\text{ı}}\right)+{\frac {\alpha {\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}}{\cos \psi _{\text{ı}}}}.\end{aligned}}$

On a, comme on vient de le voir, en négligeant les puissances supérieures de $s,$

${\rm {V-V_{\text{ı}}}}={\frac {s}{\cos \psi _{\text{ı}}}}\left(1-\alpha u'_{\text{ı}}+\alpha {\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} \psi _{\text{ı}}-{\frac {\alpha {\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}}{\cos ^{2}\psi _{\text{ı}}}}\right),$ et ${\frac {dy}{ds}}=1\,;$

on a donc

${\frac {dx'}{ds}}=s(1-\alpha u'_{\text{ı}}){\frac {\sin ^{2}\psi _{\text{ı}}}{\cos \psi _{\text{ı}}}}+\alpha s{\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} ^{2}\psi _{\text{ı}}\sin \psi _{\text{ı}}-\alpha s{\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}{\frac {\sin ^{2}\psi _{\text{ı}}}{\cos ^{3}\psi _{\text{ı}}}}\,;$

on trouvera semblablement

${\frac {dz}{ds}}=s(1-\alpha u'_{\text{ı}})\sin \psi _{\text{ı}}-\alpha s{\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} ^{2}\psi _{\text{ı}}\cos \psi _{\text{ı}}+\alpha s{\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}{\frac {\sin \psi _{\text{ı}}}{\cos ^{2}\psi _{\text{ı}}}}\,;$

le cosinus de l’angle azimutal à l’extrémité de l’arc $s$ sera ainsi

$s\operatorname {tang} \psi _{\text{ı}}\left(1-\alpha u'_{\text{ı}}+\alpha {\frac {\partial u'_{\text{ı}}}{\partial \psi }}\operatorname {tang} \psi _{\text{ı}}-{\frac {\alpha {\frac {\partial ^{2}u'_{\text{ı}}}{\partial \varphi ^{2}}}}{\cos ^{2}\psi _{\text{ı}}}}\right).$

Ce cosinus étant fort petit, il peut être pris pour le complément de