Page:Annales de mathématiques pures et appliquées, 1826-1827, Tome 17.djvu/172

ou, en vertu des formules (2),

$s={\frac {v}{\operatorname {d} v}}{\sqrt {\operatorname {d} t^{2}+\operatorname {d} u^{2}+\operatorname {d} v^{2}}},\qquad$(4)

d’où

$\operatorname {d} s={\sqrt {\operatorname {d} t^{2}+\operatorname {d} u^{2}+\operatorname {d} v^{2}}}-{\frac {v}{\operatorname {d} v}}\operatorname {d} v.{\sqrt {\operatorname {d} t^{2}+\operatorname {d} u^{2}+\operatorname {d} v^{2}}},\qquad$(5)

mais on doit avoir d’ailleurs

$\operatorname {d} s={\sqrt {\operatorname {d} t^{2}+\operatorname {d} u^{2}+\operatorname {d} v^{2}}}\,;\qquad$(6)

donc

$d.{\sqrt {\operatorname {d} t^{2}+\operatorname {d} u^{2}+\operatorname {d} v^{2}}}=0,\quad$d’où$\quad {\sqrt {\operatorname {d} t^{2}+\operatorname {d} u^{2}+\operatorname {d} v^{2}}}=A\operatorname {d} v,$

$v$ étant la variable indépendante. Cela donne en quarrant et transformant la constante

$\operatorname {d} t^{2}+\operatorname {d} u^{2}=B\operatorname {d} v^{2}.\qquad$(7)

mais, en vertu de l’équation (1)

$v={\frac {\sqrt {t^{2}+u^{2}}}{\operatorname {Tang} .\alpha }},\qquad$(8)

d’où

$\operatorname {d} v={\frac {t\operatorname {d} t+u\operatorname {d} u}{\operatorname {Tang} .\alpha .{\sqrt {t^{2}+u^{2}}}}},\quad$et$\quad \operatorname {d} v^{2}={\frac {(t\operatorname {d} t+u\operatorname {d} u)^{2}}{\left(t^{2}+u^{2}\right)\operatorname {Tang} .^{2}\alpha .}}\,;\qquad$(9)

donc, en substituant dans (7) et transformant ancore la constante