12.13-bo'lim : Sferik koordinatalar
6. Sferik koordinatalarda yozilgan tenglamani Dekart koordinatalaridagi tenglamaga aylantiring.
\[\csc \varphi = 2\cos \theta + 4\sin \theta \]Yechim
Bu erda, albatta, ko'p narsa qilish mumkin emas. Biz qilishimiz kerak bo'lgan yagona narsa, iloji bo'lsa (va agar) tenglamada quyidagi konversiya formulalaridan foydalanishdir
\[\begin{array}{c}x = \rho \sin \varphi \cos \theta \hspace{0.5in}y = \rho \sin \varphi \sin \theta \hspace{0.5in}z = \rho \cos \varphi \\ {\rho ^2} = {x^2} + {y^2} + {z^2}\end{array}\]2-qadam
Bu muammoni biroz osonlashtirish uchun, avvalo, tenglamani qayta yozamiz.
Birinchidan, keling, kosekant bilan shug'ullanamiz.
\[\frac{1}{{\sin \varphi }} = 2\cos \theta + 4\sin \theta \hspace{0.25in}\,\,\, \to \hspace{0.5in}1 = 2\sin \varphi \cos \theta + 4\sin \varphi \sin \theta \]Keyin, keling, hamma narsani \(\rho \) ga ko'paytiramiz,
\[\rho = 2\rho \sin \varphi \cos \theta + 4\rho \sin \varphi \sin \theta \]Buni qilish o'ngdagi shartlarni tanib olishni biroz osonlashtiradi.
3-qadam
1-bosqichdagi tegishli konvertatsiya formulalaridan foydalanish,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt {{x^2} + {y^2} + {z^2}} = 2x + 4y}}\]