12.13-bo'lim : Sferik koordinatalar
2. \(\left( { - 2, - 1, - 7} \right)\) uchun Dekart koordinatalarini Sferik koordinatalarga aylantiring.
Yechim
Bizga berilgan nuqtadan boshlab,
\[x = - 2\hspace{0.5in}y = - 1\hspace{0.5in}z = - 7\]2-qadam
Avval \(\rho \) ni aniqlaymiz.
\[\rho = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 7} \right)}^2}} = \sqrt {54} \]3-qadam
Endi biz \(\varphi \) ni aniqlashimiz mumkin.
\[\cos \varphi = \frac{z}{\rho } = \frac{{ - 7}}{{\sqrt {54} }}\hspace{0.5in}\varphi = {\cos ^{ - 1}}\left( {\frac{{ - 7}}{{\sqrt {54} }}} \right) = 2.8324\]4-qadam
\(\theta \) ni aniqlash uchun \(y\) konvertatsiya formulasidan foydalanamiz.
\[\sin \theta = \frac{{ - 1}}{{\rho \sin \varphi }} = \frac{{ - 1}}{{\sqrt {54} \sin \left( {2.8324} \right)}} = - 0.4472\hspace{0.25in} \to \hspace{0.25in}{\theta _1} = {\sin ^{ - 1}}\left( { - 0.4472} \right) = - 0.4636\]Bu burchak to'rtinchi kvadrantda va agar biz tez birlik doirasini chizsak, uchinchi kvadrantdagi ikkinchi burchak \({\theta _2} = \pi + 0.4636 = 3.6052\) ekanligini ko'ramiz.
Agar biz uch o'lchovli koordinatalar tizimiga yuqoridan qarasak, biz \(x\) va \(y\) koordinatalarimizdan nuqta uchinchi kvadrantda ekanligini ko'rishimiz mumkin. Bu, o'z navbatida, bizning fikrimiz uchun \({\theta _2}\) dan foydalanishimiz kerakligini anglatadi.
Sferik koordinatalar u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\left( {\sqrt {54} ,3.6052,2.8324} \right)}}\]