176
PROBLÈMES
rapports différentiels partiels
. Faisant, pour abréger
![{\displaystyle P=1-\operatorname {Sin} .\lambda \operatorname {Cos} .\phi ,\qquad Q=1-\operatorname {Sin} .\mu \operatorname {Cos} .\psi ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae10070957362ebda02402165db7bcfc40b05818)
ce qui donne
![{\displaystyle \mathrm {\operatorname {d} P} =-\operatorname {d} \lambda \operatorname {Cos} .\lambda \operatorname {Cos} .\phi +\operatorname {d} \phi \operatorname {Sin} .\lambda \operatorname {Sin} .\phi ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbc7a78ac7690f3e59676b2ad586421ad9ea47b5)
![{\displaystyle \mathrm {\operatorname {d} Q} =-\operatorname {d} \mu \operatorname {Cos} .\psi \operatorname {Cos} .\phi +\operatorname {d} \psi \operatorname {Sin} .\mu \operatorname {Sin} .\psi ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/027488dddd132e0412f4171cfc3a78fb2e6b592d)
on parviendra ainsi à donner une forme un peu plus abrégée aux deux rapports
lesques deviendront
![{\displaystyle 2\varpi \left({\frac {\operatorname {d} t}{\operatorname {d} \lambda }}\right)=+{\frac {pq\operatorname {Cos} .^{2}\lambda \operatorname {Cos} .^{3}\mu \operatorname {Sin} .\phi (1+P)}{qP^{2}\operatorname {Cos} .^{3}\mu -pQ^{2}\operatorname {Cos} .^{3}\lambda }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fcdf954758016b41d98f7428dea62c5bf72e002)
![{\displaystyle 2\varpi \left({\frac {\operatorname {d} t}{\operatorname {d} \mu }}\right)=-{\frac {pq\operatorname {Cos} .^{2}\mu \operatorname {Cos} .^{3}\lambda \operatorname {Sin} .\psi (1+Q)}{qP^{2}\operatorname {Cos} .^{3}\mu -pQ^{2}\operatorname {Cos} .^{3}\lambda }}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9eb983169910fd03e245a4bc415f43e87423a2ed)
Mais il est convenable d’abréger encore. Désignons par
le dénominateur commun et les numérateurs de ces deux valeurs, de manière qu’on ait
![{\displaystyle 2\varpi \left({\frac {\operatorname {d} t}{\operatorname {d} \lambda }}\right)={\frac {M}{F}},\qquad 2\varpi \left({\frac {\operatorname {d} t}{\operatorname {d} \mu }}\right)=-{\frac {N}{F}}\,;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71538bd0b0d4fbb8dcbf25635e91e895d7677b87)
les différentiations partielles nous apprendront que
![{\displaystyle \left({\frac {\operatorname {d} F}{\operatorname {d} \lambda }}\right)=-2qP\operatorname {Cos} .^{3}\mu \operatorname {Cos} .\phi +3pQ^{2}\operatorname {Sin} .\lambda \operatorname {Cos} .^{2}\lambda }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0170196773ea21c7588775ea4f00d493500b029)
![{\displaystyle +\left({\frac {\operatorname {d} \phi }{\operatorname {d} \lambda }}\right)\left(2qP\operatorname {Cos} .^{3}\mu \operatorname {Sin} .\lambda \operatorname {Sin} .\phi -2pQ\operatorname {Cos} .^{3}\lambda \operatorname {Sin} .\mu \operatorname {Sin} .\psi \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4dc6d428bbfbae1a3cebefe8ff9d3f1c5941d6f7)
,
![{\displaystyle \left({\frac {\operatorname {d} F}{\operatorname {d} \mu }}\right)=+2pQ\operatorname {Cos} .\mu \operatorname {Cos} .^{3}\lambda \operatorname {Cos} .\psi -3qP^{2}\operatorname {Sin} .\mu \operatorname {Cos} .^{2}\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/798070f34c28970ec3fc6c135f0c517b289a5c1f)
![{\displaystyle +\left({\frac {\operatorname {d} \psi }{\operatorname {d} \mu }}\right)\left(2qP\operatorname {Cos} .^{3}\mu \operatorname {Sin} .\lambda \operatorname {Sin} .\phi -2pQ\operatorname {Cos} .^{3}\lambda \operatorname {Sin} .\mu \operatorname {Sin} .\psi \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af4535953ec67b0ef460b2ab3d1dc72df6b731f5)
,
![{\displaystyle \left({\frac {\operatorname {d} M}{\operatorname {d} \lambda }}\right)=-pq\operatorname {Cos} .\lambda \operatorname {Cos} .^{3}\mu \operatorname {Sin} .\phi \left(4\operatorname {Sin} .\lambda +\operatorname {Cos} ^{2}\lambda .\operatorname {Cos} .\phi -2\operatorname {Sin} .^{2}\lambda \operatorname {Cos} .\phi \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27a249438a5340dd3c69c6a5db95271e6df56c1c)