Ushbu bo'limdagi eslatmalardan bilamizki, birlik tangens vektorini olish uchun bizga faqat vektor funksiyasining hosilasi va uning kattaligi kerak bo'ladi. Mana bu miqdorlar.

\[\vec r'\left( t \right) = \left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)\,\vec i - 2t\,\vec j - 2{{\bf{e}}^{2t}}\vec k\] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)}^2} + {{\left( { - 2t} \right)}^2} + {{\left( { - 2{{\bf{e}}^{2t}}} \right)}^2}} = \sqrt {{{\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)}^2} + 4{t^2} + 4{{\bf{e}}^{4t}}} \]

2-qadam

Bu vektor funksiyasi uchun birlik tangens vektori quyidagicha:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec T\left( t \right) = \frac{1}{{\sqrt {{{\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)}^2} + 4{t^2} + 4{{\bf{e}}^{4t}}} }}\left( {\left( {{{\bf{e}}^{2t}} + 2t{{\bf{e}}^{2t}}} \right)\,\vec i - 2t\,\vec j - 2{{\bf{e}}^{2t}}\vec k} \right)}}\]

E'tibor bering, bu bilan bog'liq "chalkashlik" tufayli biz har bir atama uchun kattalikni taqsimlamadik, chunki biz odatdagidek. Bu muammo ular qanchalik "tartibsiz" bo'lishi mumkinligiga yaxshi misoldir.