Asosiy tarkibga o'tish

12.13-bo'lim : Sferik koordinatalar

1. \(\left( {3, - 4,1} \right)\) uchun Dekart koordinatalarini Sferik koordinatalarga aylantiring.

Yechim

Bizga berilgan nuqtadan boshlab,

\[x = 3\hspace{0.5in}y = - 4\hspace{0.5in}z = 1\]
2-qadam

Avval \(\rho \) ni aniqlaymiz.

\[\rho = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {26} \]
3-qadam

Endi biz \(\varphi \) ni aniqlashimiz mumkin.

\[\cos \varphi = \frac{z}{\rho } = \frac{1}{{\sqrt {26} }}\hspace{0.5in}\varphi = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {26} }}} \right) = 1.3734\]
4-qadam

\(\theta \) ni aniqlash uchun \(x\) konvertatsiya formulasidan foydalanamiz.

\[\cos \theta = \frac{x}{{\rho \sin \varphi }} = \frac{3}{{\sqrt {26} \sin \left( {1.3734} \right)}} = 0.6\hspace{0.25in} \to \hspace{0.25in}{\theta _1} = {\cos ^{ - 1}}\left( {0.6} \right) = 0.9273\]

Bu burchak birinchi kvadrantda va agar biz tez birlik doirasini chizsak, to'rtinchi kvadrantdagi ikkinchi burchak \({\theta _2} = 2\pi - 0.9273 = 5.3559\) ekanligini ko'ramiz.

Agar biz uch o'lchovli koordinatalar tizimiga yuqoridan qarasak, bizning \(x\) va \(y\) koordinatalarimizdan nuqta to'rtinchi kvadrantda ekanligini ko'rishimiz mumkin. Bu, o'z navbatida, bizning fikrimiz uchun \({\theta _2}\) dan foydalanishimiz kerakligini anglatadi.

Sferik koordinatalar u holda,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\sqrt {26} ,5.3559,1.3734} \right)}}\]