Bizga birinchi navbatda vektor funksiyaning hosilasining kattaligi kerak. Bu,

\[\vec r'\left( t \right) = \left\langle {2t,6{t^2}, - 3{t^2}} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{t^2} + 36{t^4} + 9{t^4}} = \sqrt {{t^2}\left( {4 + 45{t^2}} \right)} = \sqrt {{t^2}} \sqrt {4 + 45{t^2}} = \left| t \right|\sqrt {4 + 45{t^2}} = t\sqrt {4 + 45{t^2}} \]

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

Shuni ham yodda tutingki, biz bunday muammoni \(t = 0\) dan boshlaymiz deb taxmin qilganimiz sababli, \(t \ge 0\) va shunga o'xshash \(\sqrt {{t^2}} = \left| t \right| = t\) deb taxmin qilish mumkin.

2-qadam

Yoy uzunligi funktsiyasi u holda,

\[s\left( t \right) = \int_{0}^{t}{{u\sqrt {4 + 45{u^2}} \,du}} = \frac{1}{{135}}\left. {{{\left( {4 + 45{u^2}} \right)}^{\frac{3}{2}}}} \right|_0^t = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{135}}\left[ {{{\left( {4 + 45{t^2}} \right)}^{\frac{3}{2}}} - 8} \right]}}\]