Page:Annales de mathématiques pures et appliquées, 1814-1815, Tome 5.djvu/61

57

RÉSOLUES.

$\operatorname {d} t{\sqrt {g}}={\frac {2\operatorname {d} \phi \operatorname {Cos} .\phi }{\operatorname {Sin} .\phi }}+{\frac {\operatorname {d} \phi \operatorname {Sin} .\phi }{\operatorname {Cos} .\phi }}={\frac {\operatorname {d} .\operatorname {Sin} .\phi }{\operatorname {Sin} .\phi }}-{\frac {\operatorname {d} .\operatorname {Cos} .\phi }{\operatorname {Cos} .\phi }},$

d’où en intégrant et faisant la constante nulle, attendu que $t$ et $tang.4\phi$ doivent être nuls en même temps,

$t{\sqrt {g}}=Log.\operatorname {Sin} .\phi -Log.\operatorname {Cos} .\phi =Log.{\frac {\operatorname {Sin} .\phi }{\operatorname {Cos} .\phi }}=Log.\operatorname {Tang} .\phi ,$

et par conséquent

$\operatorname {Tang} .\phi =e^{t{\sqrt {g}}}.$

de là on tire

$\operatorname {Tang} .4\phi ={\frac {4\operatorname {Tang} .\phi -4\operatorname {Tang} .^{3}\phi }{1-6\operatorname {Tang} .^{2}\phi +\operatorname {Tang} .^{4}\phi }}={\frac {4e^{t{\sqrt {g}}}-4e^{3t{\sqrt {g}}}}{1-6e^{2t{\sqrt {g}}}+e^{4t{\sqrt {g}}}}}$

et par suite

$\operatorname {Sin} .4\phi ={\frac {\operatorname {Tang} .4\phi }{\sqrt {1+\operatorname {Tang} .^{2}4\phi }}}={\frac {4e^{t{\sqrt {g}}}\left(1-4e^{2t{\sqrt {g}}}\right)}{\left(1+e^{2t{\sqrt {g}}}\right)^{2}}},$

$\operatorname {Cos} .4\phi ={\frac {1}{\sqrt {1+\operatorname {Tang} .^{2}4\phi }}}={\frac {1-6e^{2t{\sqrt {g}}}+e^{4t{\sqrt {g}}}}{\left(1+e^{2t{\sqrt {g}}}\right)^{2}}}.$

Donc

$x=bt+4e^{t{\sqrt {g}}}{\frac {1-e^{2t{\sqrt {g}}}}{\left(1+e^{2t{\sqrt {g}}}\right)^{2}}},\qquad y={\frac {1-6e^{2t{\sqrt {g}}}+e^{4t{\sqrt {g}}}}{\left(1+e^{2t{\sqrt {g}}}\right)^{2}}}.$