Asosiy tarkibga o'tish

12.10-bo'lim : Egrilik

1. \(\vec r\left( t \right) = \left\langle {\cos \left( {2t} \right), - \sin \left( {2t} \right),4t} \right\rangle \) ning egri chizig‘ini toping.

Yechim

Egrilikni hisoblash uchun bizda ikkita formuladan foydalanishimiz mumkin. Ulardan biri birlik tangens vektorining hosilasini olishimizni talab qiladi, ikkinchisi esa o'zaro ko'paytmani talab qiladi.

Ikkalasi ham bir xil natija beradi. Haqiqiy savol - qaysi biri foydalanish osonroq bo'ladi. O'zaro mahsulotlarni hisoblash qiyin bo'lishi mumkin, ammo keyin birlik tangens vektorlarining hosilasini olish juda tartibsiz bo'lishi mumkin. Shunday qilib, asosan, biz foydalanadigan narsa, ehtimol, foydalanish uchun eng oson bo'ladi.

Bunday holda, birlik tangens vektori bilan ishlash unchalik yomon bo'lmaganga o'xshaydi, shuning uchun biz ushbu formulaga o'tamiz. Mana birlik tangens vektor ishi.

\[\begin{array}{c}\vec r'\left( t \right) = \left\langle { - 2\sin \left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle \hspace{0.5in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{{\sin }^2}\left( {2t} \right) + 4{{\cos }^2}\left( {2t} \right) + 16} = \sqrt {20} = 2\sqrt 5 \\ \vec T\left( t \right) = \frac{1}{{2\sqrt 5 }}\left\langle { - 2\sin \left( {2t} \right), - 2\cos \left( {2t} \right),4} \right\rangle = \left\langle { - \frac{{\sin \left( {2t} \right)}}{{\sqrt 5 }}, - \frac{{\cos \left( {2t} \right)}}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right\rangle \end{array}\]
2-qadam

Endi bizga kerak bo'lgan narsa bu birlik tangens vektorining hosilasining kattaligi, shuning uchun bu ish,

\[\vec T'\left( t \right) = \left\langle { - \frac{2}{{\sqrt 5 }}\cos \left( {2t} \right),\frac{2}{{\sqrt 5 }}\sin \left( {2t} \right),0} \right\rangle \hspace{0.25in}\hspace{0.25in}\left\| {\vec T'\left( t \right)} \right\| = \sqrt {\frac{4}{5}{{\cos }^2}\left( {2t} \right) + \frac{4}{5}{{\sin }^2}\left( {2t} \right)} = \frac{2}{{\sqrt 5 }}\]
3-qadam

Keyin egrilik,

\[\kappa = \frac{{\left\| {\vec T'\left( t \right)} \right\|}}{{\left\| {\vec r'\left( t \right)} \right\|}} = \frac{{{}^{2}/{}_{{\sqrt 5 }}}}{{2\sqrt 5 }} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{5}}}\]

Shunday qilib, bu holda, egrilik \(t\) dan mustaqil bo'ladi. Bu har doim ham shunday bo'lmaydi, shuning uchun har doim shunday bo'lishini kutmang.