![{\displaystyle z+A={\frac {ASin.\omega +{\sqrt {x^{2}+y^{2}}}Cos.\omega }{Sin.\omega }}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/027562ea6ed8c1fe2ebd0924124b0a9cd14ec3cd)
il viendra donc, en substituant,
![{\displaystyle \left(x^{2}+y^{2}\right)\left\{A\operatorname {Sin} .\omega +{\sqrt {x^{2}+y^{2}}}\operatorname {Cos} .\omega \right\}^{2}\times }](https://wikimedia.org/api/rest_v1/media/math/render/svg/579c6d6e0708bd8c9f46fbfd6e55444b8e539e98)
![{\displaystyle \left\{\left(x^{2}+y^{2}\right)\left(\operatorname {d} x^{2}+\operatorname {d} y^{2}\right)-(x\operatorname {d} x+y\operatorname {d} y)^{2}\right\}\operatorname {Cos} .^{2}\omega }](https://wikimedia.org/api/rest_v1/media/math/render/svg/efce2203f7ed090967632c8111429c807d2a6fc0)
![{\displaystyle =a^{2}c^{2}\left\{\left(x^{2}+y^{2}\right)\left(\operatorname {d} x^{2}+\operatorname {d} y^{2}\right)\operatorname {Sin} .^{2}\omega +(x\operatorname {d} x+y\operatorname {d} y)^{2}\operatorname {Cos} .^{2}\omega \right\}\operatorname {Sin} .^{2}\omega \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e627ac245dd487290ff281b3c4dc1ca212434aab)
et telle est l’équation differentielle de la projection de la chaînette sur le plan des ![{\displaystyle xy.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0add5fb23e378ec970ad47ea154f8a6431843a8f)
Si, pour passer aux coordonnées polaires nous posons
![{\displaystyle x=r\operatorname {Sin} .\theta ,\qquad y=r\operatorname {Cos} .\theta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c35a22afae7f8541bf94871a5c023313507ca70a)
il en résultera
![{\displaystyle \operatorname {d} x=\operatorname {d} r\operatorname {Sin} .\theta +r\operatorname {d} \theta \operatorname {Cos} .\theta ,\qquad \operatorname {d} y=\operatorname {d} r\operatorname {Cos} .\theta -r\operatorname {d} \theta \operatorname {Sin} .\theta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0633724f7d1ff38affc13002bb0b297e593fde46)
![{\displaystyle x^{2}+y^{2}=r^{2},\qquad \operatorname {d} x^{2}+\operatorname {d} y^{2}=\operatorname {d} r^{2}+r^{2}\operatorname {d} \theta ^{2},\qquad x\operatorname {d} x+y\operatorname {d} y=r\operatorname {d} r\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f65a0be1cc68287d65028f3c0286fdd35ef3af64)
ce qui donnera, en substituant,
![{\displaystyle r^{4}(A\operatorname {Sin} .\omega +r\operatorname {Cos} .\omega )^{2}\operatorname {d} \theta ^{2}\operatorname {Cos} .^{2}\omega =a^{2}c^{2}\left(\operatorname {d} r^{2}+r^{2}\operatorname {d} \theta ^{2}\operatorname {Sin} .^{2}\omega \right)\operatorname {Sin} .^{2}\omega \,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1bcad352a970824c16ef571c3727b9cd5e2879a)
équation d’où on tirera
![{\displaystyle \operatorname {d} \theta ={\frac {ac\operatorname {d} r\operatorname {Sin} .\omega }{r{\sqrt {(A\operatorname {Sin} .\omega +r\operatorname {Cos} .\omega )^{2}r^{2}\operatorname {Cos} .^{2}\omega -a^{2}c^{2}\operatorname {Sin} .^{4}\omega }}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ba6badb31df330a2feaa3797ce3dd2d6df3d884)
Il ne paraît pas que cette valeur soit généralement intégrable sous forme finie.
En conséquence nous nous bornerons à considérer le cas où
et
il vient alors
![{\displaystyle \operatorname {d} \theta ={\frac {a^{2}\operatorname {d} r\operatorname {Sin} .\omega }{r{\sqrt {r^{4}\operatorname {Cos} .^{4}\omega -a^{4}\operatorname {Sin} .^{4}\omega }}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c874cf47741996a279a65d3804893b075186788b)
ce qui donne en intégrant,