Ushbu muammo uchun bizga \(ds\) kerak bo'ladi.

\[ds = \sqrt {{{\left[ 3 \right]}^2} + {{\left[ { - 2t} \right]}^2}} \,dt = \sqrt {9 + 4{t^2}} \,dt\]

U holda yoy uzunligi uchun integrali,

\[L = \int_{{}}^{{}}{{ds}} = \int_{{ - 2}}^{0}{{\sqrt {9 + 4{t^2}} \,dt}}\]

Ushbu integral trigni almashtirishni talab qiladi (bir nechta yoy uzunligi integrallari kabi!).

Mana bu integral uchun trigni almashtirish kerak.

\[t = \frac{3}{2}\tan \theta \hspace{0.25in}dt = \frac{3}{2}{\sec ^2}\theta \,d\theta \,\] \[\sqrt {9 + 4{t^2}} = \sqrt {9 + 9{{\tan }^2}\theta } = 3\sqrt {1 + {{\tan }^2}\theta } = 3\sqrt {{{\sec }^2}\theta } = 3\left| {\sec \theta } \right|\]

Sekantdagi mutlaq qiymatdan xalos bo'lish uchun chegaralarni \(\theta \) chegaralariga aylantirish kerak bo'ladi.

\[\begin{align*}t & = - 2: & \hspace{0.25in} - 2 & = \frac{3}{2}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = - \frac{4}{3}\hspace{0.25in} \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( { - \frac{4}{3}} \right) = - 0.9273\\ t & = 0:& \hspace{0.25in}0 & = \frac{3}{2}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = 0\hspace{0.25in}\, \to \hspace{0.25in}\theta = 0\end{align*}\]

Xo'sh, bu muammo uchun \(\theta \) ning mos diapazoni \( - 0.9273 \le \theta \le 0\) (to'rtinchi kvadrant) va bu diapazonda sekant ijobiy ekanligini bilamiz. Shunday qilib, ildiz bo'ladi,

\[\sqrt {9 + 4{t^2}} = 3\sec \theta \]

U holda integral,

\[\begin{align*}L & = \int_{{ - 2}}^{0}{{\sqrt {9 + 4{t^2}} \,dt}} = \int_{{ - 0.9273}}^{0}{{\left( {3\sec \theta } \right)\left( {\frac{3}{2}{{\sec }^2}\theta } \right)\,d\theta }}\\ & = \int_{{ - 0.9273}}^{0}{{\frac{9}{2}{{\sec }^3}\theta \,d\theta }} = \left. {\frac{9}{4}\left[ {\sec \theta \tan \theta + ln\left| {\sec \theta + \tan \theta } \right|} \right]} \right|_{ - 0.9273}^0 = \require{bbox} \bbox[2pt,border:1px solid black]{{7.4719}}\end{align*}\]