Bizga birinchi navbatda vektor funksiyaning hosilasining kattaligi kerak. Bu,

\[\vec r'\left( t \right) = \left\langle {{t^2},4,2\sqrt 2 t} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{t^4} + 16 + 8{t^2}} = \sqrt {{t^4} + 8{t^2} + 16} = \sqrt {{{\left( {{t^2} + 4} \right)}^2}} = {t^2} + 4\]

Bunday muammolar uchun kattalikni iloji boricha soddalashtirishga ishonch hosil qiling. Bu haqiqatan ham oddiy muammo va nihoyatda qiyin muammo o'rtasidagi farqni anglatishi mumkin.

2-qadam

Keyin egri chiziq uzunligi,

\[L = \int_{0}^{2}{{{t^2} + 4\,dt}} = \left. {\left( {\frac{1}{3}{t^3} + 4t} \right)} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}}}\]

E'tibor bering, agar biz kattalikni soddalashtirmaganimizda, buni hisoblash juda qiyin integral bo'lar edi!