Page:Annales de mathématiques pures et appliquées, 1831-1832, Tome 22.djvu/37

en désignant par $s$ l’arc $\mathrm {A} n$ et par $R$ le rayon de courbure $n\mathrm {C} .$ On aura donc

$R={\frac {\operatorname {d} s}{\operatorname {d} \varphi }}.\qquad$(1)

Mais le triangle $nn'q,$ dans lequel l’angle $nn'g=\varphi ,\ nn'=\operatorname {d} s,$$nq=\operatorname {d} x$ et $n'q=\operatorname {d} y,$ fournit les égalités suivantes,

$\left.{\begin{alignedat}{2}\operatorname {Sin} .\varphi &={\frac {\operatorname {d} x}{\operatorname {d} s}},&\qquad \operatorname {Cos} .\varphi &={\frac {\operatorname {d} y}{\operatorname {d} s}},\\\\\operatorname {Tang} .\varphi &={\frac {\operatorname {d} x}{\operatorname {d} y}},&\operatorname {Cot} .\varphi &={\frac {\operatorname {d} y}{\operatorname {d} x}},\\\\\operatorname {S{\acute {e}}c} .\varphi &={\frac {\operatorname {d} s}{\operatorname {d} y}},&\operatorname {Cos{\acute {e}}c} .\varphi &={\frac {\operatorname {d} s}{\operatorname {d} x}}.\end{alignedat}}\right\}\quad$(2)

En différentiant les six expressions (2), on a

$\left.{\begin{alignedat}{2}\operatorname {d} \varphi &={\frac {\operatorname {d} s.\operatorname {d} \left({\frac {\operatorname {d} x}{\operatorname {d} s}}\right)}{\operatorname {d} y}},&\qquad \operatorname {d} \varphi &={\frac {\operatorname {d} s.\operatorname {d} \left({\frac {\operatorname {d} y}{\operatorname {d} s}}\right)}{\operatorname {d} x}},\\\\\operatorname {d} \varphi &={\frac {\operatorname {d} y^{2}.\operatorname {d} \left({\frac {\operatorname {d} x}{\operatorname {d} y}}\right)}{\operatorname {d} s^{2}}},&\operatorname {d} \varphi &=-{\frac {\operatorname {d} x^{2}.\operatorname {d} \left({\frac {\operatorname {d} y}{\operatorname {d} x}}\right)}{\operatorname {d} s^{2}}},\\\\\operatorname {d} \varphi &={\frac {\operatorname {d} y^{2}.\operatorname {d} \left({\frac {\operatorname {d} s}{\operatorname {d} y}}\right)}{\operatorname {d} s\operatorname {d} x}},&\operatorname {d} \varphi &=-{\frac {\operatorname {d} x^{2}.\operatorname {d} \left({\frac {\operatorname {d} s}{\operatorname {d} x}}\right)}{\operatorname {d} s\operatorname {d} y}}.\end{alignedat}}\right\}\quad$(3)

Si l’on substitue successivement chacune de ces expressions de la différentielle de l’arc $\varphi$ dans l’expression (1) ${\frac {\operatorname {d} s}{\operatorname {d} \varphi }}$ du rayon de courbure, on aura