Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/30

26

PARALLÉLISME DES LIGNES

${\frac {\operatorname {d} \Sigma }{\operatorname {d} t}}={\frac {\operatorname {d} \Sigma }{\operatorname {d} x}}.{\frac {\operatorname {d} x}{\operatorname {d} u}}+{\frac {\operatorname {d} \Sigma }{\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} u}},$

d’où

${\frac {\operatorname {d} .{\frac {\operatorname {d} \Sigma }{\operatorname {d} t}}}{\operatorname {d} x}}={\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} x^{2}}}.{\frac {\operatorname {d} x}{\operatorname {d} u}}+{\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} x\operatorname {d} y}}.{\frac {\operatorname {d} y}{\operatorname {d} u}},$

${\frac {\operatorname {d} .{\frac {\operatorname {d} \Sigma }{\operatorname {d} t}}}{\operatorname {d} y}}={\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} x\operatorname {d} y}}.{\frac {\operatorname {d} x}{\operatorname {d} u}}+{\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} y^{2}}}.{\frac {\operatorname {d} y}{\operatorname {d} u}},$

ce qui donnera, en substituant

${\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} t\operatorname {d} u}}={\frac {\operatorname {d} ^{2}S}{\operatorname {d} x\operatorname {d} y}}={\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} x^{2}}}\left({\frac {\operatorname {d} x}{\operatorname {d} u}}\right)^{2}+{\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} y^{2}}}\left({\frac {\operatorname {d} y}{\operatorname {d} u}}\right)^{2}+2{\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} x\operatorname {d} y}}.{\frac {\operatorname {d} x}{\operatorname {d} u}}.{\frac {\operatorname {d} y}{\operatorname {d} u}}.$

Cette équation différentielle contient implicitement l’expression cherchée de ${\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} x\operatorname {d} y}}$ en $S,x,y.$ On pourra la résoudre dans des cas particuliers.

13.

Ayant trouvé les expressions des élémens

${\frac {\operatorname {d} ^{2}S}{\operatorname {d} x\operatorname {d} y}},\qquad {\frac {\operatorname {d} ^{2}\Sigma }{\operatorname {d} t\operatorname {d} u}}$

en $x,y,$ on aura aussi celle de l’élément du volume compris entre les deux surfaces parallèles ; car cet élément peut être considéré comme une pyramide tronquée, dont nos deux surfaces élémentaires sont les deux bases parallèles et dont la hauteur est $k.$ Il serait intéressant d’appliquer ces formules aux surfaces développables.