en désignant par
l’arc
et par
le rayon de courbure
On aura donc
![{\displaystyle R={\frac {\operatorname {d} s}{\operatorname {d} \varphi }}.\qquad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d526a04cb2bc1b81c1b67b19a89a93627c74357)
(1)
Mais le triangle
dans lequel l’angle ![{\displaystyle nn'g=\varphi ,\ nn'=\operatorname {d} s,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a97926804d115aa878da171a18920b4c56d761a7)
et
fournit les égalités suivantes,
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/042dbcfad1f2656988643b2635c8af5251a5dcb2)
(2)
En différentiant les six expressions (2), on a
![{\displaystyle \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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9a2308d9997efa159119b89b2b488024b3c097e)
(3)
Si l’on substitue successivement chacune de ces expressions de la différentielle de l’arc
dans l’expression (1)
du rayon de courbure, on aura