35
INDÉTERMINÉES.
![{\displaystyle \int \operatorname {d} z{\sqrt {1+x'^{2}+y'^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9edef9ea2853ccffcfc1d82c71deae9b6e947e34)
minimum, entre les limites
et ![{\displaystyle c_{1}\,:}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ede39d63f4ed010f0ab6c5b44bb667d9653cbb75)
Nous aurons donc ici
![{\displaystyle V={\sqrt {1+x'^{2}+y'^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c8fe04f3d815ba88c28f5bc0a3fcd0a72002635)
d’où (21)
![{\displaystyle {\begin{alignedat}{3}\left({\frac {\operatorname {d} V}{\operatorname {d} x}}\right)=&0,\qquad &\left({\frac {\operatorname {d} V}{\operatorname {d} x'}}\right)=&{\frac {x'}{\sqrt {1+x'^{2}+y'^{2}}}},\qquad &\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)=&0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} y}}\right)=&0,&\left({\frac {\operatorname {d} V}{\operatorname {d} y'}}\right)=&{\frac {y'}{\sqrt {1+x'^{2}+y'^{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)=&0,\ldots \end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3b31ca1ca17e5a08cc714994881085f4b183221)
et de là
![{\displaystyle {\begin{alignedat}{2}\left({\frac {\operatorname {d} V}{\operatorname {d} x'}}\right)'=&{\frac {\left(1+y'^{2}\right)x''-x'y'y''}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}},\qquad &\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)'=&0,\ldots \\\\\left({\frac {\operatorname {d} V}{\operatorname {d} y'}}\right)'=&{\frac {\left(1+x'^{2}\right)y''-x'y'x''}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}},&\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)'=&0,\ldots \\\\&&\left({\frac {\operatorname {d} V}{\operatorname {d} x''}}\right)''=&0,\ldots \\\\&&\left({\frac {\operatorname {d} V}{\operatorname {d} y''}}\right)''=&0,\ldots \end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c444654abfc00ac39ef9c0a799b5ef1f88fbe2e4)
au moyen de quoi l’équation (XIV) deviendra
![{\displaystyle {\frac {\left(1+y'^{2}\right)x''-x'y'y''}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}X+{\frac {\left(1+x'^{2}\right)y''-x'y'x''}{(1+x'^{2}+y'^{2})^{\frac {3}{2}}}}Y=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbacce27872b420628dbd9ec91a16630e9e7bd9a)
Si la courbe n’est assujettie à aucune autre condition qu’à celle d’être minimum entre les deux plans donnés,
et
devront demeurer indépendans, et conséquemment cette équation se partagera en ces deux-ci :