Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/397

${\frac {\operatorname {d} x}{\operatorname {d} s}}=\operatorname {Cos} .v,\qquad {\frac {\operatorname {d} y}{\operatorname {d} s}}=\operatorname {Sin} .v,$

nous aurons

${\frac {\operatorname {d} x}{\operatorname {d} s}}\operatorname {d} .{\frac {\operatorname {d} y}{\operatorname {d} s}}-{\frac {\operatorname {d} y}{\operatorname {d} s}}\operatorname {d} .{\frac {\operatorname {d} x}{\operatorname {d} s}}=\operatorname {d} v\,;$

mais, on trouve

$\operatorname {Tang} .(v-u)={\frac {\operatorname {Tang} .v-\operatorname {Tang} .u}{1+\operatorname {Tang} .v\operatorname {Tang} .u}}={\frac {{\frac {\operatorname {d} y}{\operatorname {d} x}}-{\frac {y}{x}}}{1+{\frac {y\operatorname {d} y}{x\operatorname {d} x}}}}={\frac {x\operatorname {d} y-y\operatorname {d} x}{x\operatorname {d} x+y\operatorname {d} y}}={\frac {r\operatorname {d} u}{\operatorname {d} r}}\,;$

posant donc

${\frac {r\operatorname {d} u}{\operatorname {d} r}}=z,\quad$d’où$\quad \operatorname {d} u={\frac {z\operatorname {d} r}{r}},\qquad$(6)

on aura

$v-u=\operatorname {Arc} .(\operatorname {Tang} .=z),\qquad \operatorname {d} v=\operatorname {d} u+{\frac {\operatorname {d} z}{1+z^{2}}}.$

Par la substitution de ces valeurs, l’équation (5) devient

$(T-Rr)\operatorname {d} u+{\frac {T\operatorname {d} z}{1+z^{2}}}=0\,;$

puis, en mettant pour $\operatorname {d} u$ sa valeur (6) et décomposant,

${\frac {\operatorname {d} r}{r}}-{\frac {R\operatorname {d} r}{T}}+{\frac {\operatorname {d} z}{r}}-{\frac {z\operatorname {d} z}{1+z^{2}}}=0.$

À cause de $\operatorname {d} T=-R\operatorname {d} r,$ l’intégration par logarithmes donnera

${\frac {Trz}{\sqrt {1+z^{2}}}}=h\,;\qquad$(7)

$h$ étant la constante arbitraire.