17.2-bo'lim : Parametrik yuzalar
6. \(\left( {8,14,2} \right)\) nuqtada quyidagi parametrik tenglama bilan berilgan sirtga teginish tekisligi.
\[\vec r\left( {u,v} \right) = \left( {{u^2} + 2u} \right)\vec i + \left( {3v - 2u} \right)\vec j + \left( {6v - 10} \right)\vec k\]Yechim
Tekislik tenglamasini yozish uchun bizda mavjud bo'lgan \(\left( {8,14,2} \right)\) nuqta va bizda hali mavjud bo'lmagan normal vektor kerak.
Biroq, \({\vec r_u} \times {\vec r_v}\) sirt uchun normal bo'lishini eslang. Shunday qilib, keling, buni hisoblaylik.
\[{\vec r_u} = \left( {2u + 2} \right)\vec i - 2\vec j\hspace{0.5in}{\vec r_v} = 3\vec j + 6\vec k\] \[{\vec r_u} \times {\vec r_v} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{2u + 2}&{ - 2}&0\\0&3&6\end{array}} \right| = - 12\vec i - 6\left( {2u + 2} \right)\vec j + 3\left( {2u + 2} \right)\vec k\]2-qadam
Endi \({\vec r_u} \times {\vec r_v}\)ga ega bo‘lish yaxshi va yaxshi, lekin agar biz \(\left( {8,14,2} \right)\) da bo‘lgan \(\left( {u,v} \right)\) nuqtasini bilsakgina foydali bo‘ladi, shuning uchun biz \(x\), \(y\) va \(z\) koordinatalarini \(y\) va MATHBLOCK17 ga teng qilib belgilashimiz kerak. \(u\) va \(v\) qiymatini aniqlash uchun parametrik tenglamamiz komponentlari bizga kerak.
Agar shunday qilsak, biz tenglamalarni olamiz.
\[\begin{array}{*{20}{c}}{8 = {u^2} + 2u}\\{14 = 3v - 2u}\\{2 = 6v - 10}\end{array}\hspace{0.5in} \Rightarrow \hspace{0.5in}\begin{array}{l}{0 = {u^2} + 2u - 8 = \left( {u + 4} \right)\left( {u - 2} \right)}\\{14 = 3v - 2u}\\{12 = 6v}\end{array}\]3-qadam
Yuqoridagi uchinchi tenglamadan biz \(v = 2\)ga ega bo'lishimiz kerakligini va birinchi tenglamadan bizda \(u = - 4\) yoki \(u = 2\) bo'lishi kerakligini ko'rishimiz mumkin.
\(v\) uchun yagona tanlovimizni va \(u\) uchun ikkala variantni ikkinchi tenglamaga ulab, bizda \(u = - 4\) bo'lishi kerakligini ko'ramiz.
4-qadam
\(u = - 4\) va \(v = 2\) ni \({\vec r_u} \times {\vec r_v}\) tenglamasiga ulab, biz \(\left( {8,14,2} \right)\) yuzasida quyidagi normal vektorga erishamiz.
\[\vec n = {\left( {{{\vec r}_u} \times {{\vec r}_v}} \right)_{u = - 4,v = 2}} = - 12\vec i + 36\vec j - 18\vec k\]E'tibor bering, bu holda normal vektor aslida \(v\) qiymatiga bog'liq emas edi. Bu umuman sodir bo'lmaydi, lekin biz bu erda ko'rganimizdek, ba'zida bunday narsa sodir bo'lishi mumkin, shuning uchun bu sodir bo'lganda hayajonlanmang.
\(\vec n = - 12\vec i + 36\vec j - 18\vec k\) normal vektori bilan \(\left( {8,14,2} \right)\) da sirtga teginish tekisligi tenglamasi:
\[ - 12\left( {x - 8} \right) + 36\left( {y - 14} \right) - 18\left( {z - 2} \right) = 0\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\, - 12x + 36y - 18z = 372\]5-qadam
Tangens tekislik uchun parametrik tenglamalar to'plamini olish uchun biz faqat \(z\) tenglamasini echishimiz kerak,
\[z = - \frac{{62}}{3} - \frac{2}{3}x + 2y\]Tangens tekislik uchun quyidagi parametrik tenglamalar to'plamini olish uchun biz buni \(\left\langle {x,y,z} \right\rangle \) vektoriga ulashimiz mumkin.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( {x,y} \right) = \left\langle {x,y, - \frac{{62}}{3} - \frac{2}{3}x + 2y} \right\rangle }}\]E'tibor bering, \(x\) va \(y\)da hech qanday limitlar bo'lmaydi, chunki biz to'liq tangens tekislikni xohladik.