12.8-bo'lim : Tangent, normal va binormal vektorlar
1. : \(\vec r\left( t \right) = \left\langle {{t^2} + 1,3 - t,{t^3}} \right\rangle \) vektor funksiyasi uchun birlik tangens vektorini toping
Yechim
Ushbu bo'limdagi eslatmalardan bilamizki, birlik tangens vektorini olish uchun bizga faqat vektor funksiyasining hosilasi va uning kattaligi kerak bo'ladi. Mana bu miqdorlar.
\[\vec r'\left( t \right) = \left\langle {2t, - 1,3{t^2}} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} = \sqrt {1 + 4{t^2} + 9{t^4}} \]2-qadam
Bu vektor funksiyasi uchun birlik tangens vektori quyidagicha:
\[\vec T\left( t \right) = \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}\left\langle {2t, - 1,3{t^2}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {\frac{{2t}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}, - \frac{1}{{\sqrt {1 + 4{t^2} + 9{t^4}} }},\frac{{3{t^2}}}{{\sqrt {1 + 4{t^2} + 9{t^4}} }}} \right\rangle }}\]