12.8-bo'lim : Tangent, normal va binormal vektorlar
5. Quyidagi vektor funksiya uchun normal va binormal vektor birligini toping.
\[\vec r\left( t \right) = \left\langle {\cos \left( {2t} \right),\sin \left( {2t} \right),3} \right\rangle \]Yechim
Bizga birinchi navbatda birlik tangens vektori kerak, shuning uchun uni olamiz.
\[\begin{array}{c}\vec r'\left( t \right) = \left\langle { - 2\sin \left( {2t} \right),2cos\left( {2t} \right),0} \right\rangle \hspace{0.5in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{{\sin }^2}\left( {2t} \right) + 4{{\cos }^2}\left( {2t} \right)} = 2\\ \vec T\left( t \right) = \frac{1}{2}\left\langle { - 2\sin \left( {2t} \right),2cos\left( {2t} \right),0} \right\rangle = \left\langle { - \sin \left( {2t} \right),cos\left( {2t} \right),0} \right\rangle \end{array}\]2-qadam
Birlik normal vektor u holda,
\[\begin{array}{c}\vec T'\left( t \right) = \left\langle { - 2\cos \left( {2t} \right), - 2\sin \left( {2t} \right),0} \right\rangle \hspace{0.5in}\left\| {\vec T'\left( t \right)} \right\| = \sqrt {4{{\cos }^2}\left( {2t} \right) + 4{{\sin }^2}\left( {2t} \right)} = 2\\ \require{bbox} \bbox[2pt,border:1px solid black]{{\vec N\left( t \right) = \frac{1}{2}\left\langle { - 2\cos \left( {2t} \right), - 2\sin \left( {2t} \right),0} \right\rangle = \left\langle { - \cos \left( {2t} \right), - \sin \left( {2t} \right),0} \right\rangle }}\end{array}\]3-qadam
Nihoyat, binormal vektor,
\[\begin{align*}\vec B\left( t \right) & = \vec T\left( t \right) \times \vec N\left( t \right) = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - \sin \left( {2t} \right)}&{\cos \left( {2t} \right)}&0\\{ - \cos \left( {2t} \right)}&{ - \sin \left( {2t} \right)}&0\end{array}} \right|\\ & = {\sin ^2}\left( {2t} \right)\vec k - \left( { - {{\cos }^2}\left( {2t} \right)\vec k} \right) = \left( {{{\sin }^2}\left( {2t} \right) + {{\cos }^2}\left( {2t} \right)} \right)\vec k = \require{bbox} \bbox[2pt,border:1px solid black]{{\vec k = \left\langle {0,0,1} \right\rangle = \vec B\left( t \right)}}\end{align*}\]