12.8-bo'lim : Tangent, normal va binormal vektorlar
4. \(t = 2\) da \(\displaystyle \vec r\left( t \right) = \left\langle {7{{\bf{e}}^{2 - t}},\frac{{16}}{{{t^3}}},5 - t} \right\rangle \) ga teguvchi chiziqni toping.
Yechim
Birinchidan, vektor funktsiyasiga teginish vektorini olishimiz kerak. Tangens vektori,
\[\vec r'\left( t \right) = \left\langle { - 7{{\bf{e}}^{2 - t}}, - \frac{{48}}{{{t^4}}}, - 1} \right\rangle \]E'tibor bering, agar biz xohlasak, bu erda birlik tangens vektoridan foydalanishimiz mumkin, ammo ular qanchalik chalkash bo'lishini hisobga olsak, biz buni amalga oshiramiz.
2-qadam
Endi bizga "to'liq" tangens vektori emas, balki muammo bayonotida berilgan \(t\) qiymatidagi tangens vektori kerak. Bizga muammo bayonidagi \(t\) qiymatidagi vektor funksiyasidagi nuqta ham kerak bo'ladi. Bular,
\[\begin{align*}\vec r\left( 2 \right) & = \left\langle {7,2,3} \right\rangle \\ \vec r'\left( 2 \right) & = \left\langle { - 7, - 3, - 1} \right\rangle \end{align*}\]3-qadam
Tangens chiziq tenglamasini yozish uchun bizga chiziq ustidagi nuqta va chiziqqa parallel vektor kerak. Albatta, bu faqat oldingi bosqichda topilgan miqdorlar.
Tangens chiziq u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {7,2,3} \right\rangle + t\left\langle { - 7, - 3, - 1} \right\rangle = \left\langle {7 - 7t,2 - 3t,3 - t} \right\rangle }}\]