1. Muammo: Berilgan parametrik tenglamalar $x = 6 \cos^3 t$, $y = 6 \sin^3 t$ va $0 \leq t \leq \frac{\pi}{3}$ oraliqdagi egri chiziqning yoyi uzunligini topish. 2. Egri chiziqning yoyi uzunligini topish uchun formuladan foydalanamiz: $$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$$ Bu yerda $\frac{dx}{dt}$ va $\frac{dy}{dt}$ - $x$ va $y$ ning $t$ ga nisbatan hosilalari. 3. Hosilalarni hisoblaymiz: $$x = 6 \cos^3 t = 6 (\cos t)^3$$ $$\frac{dx}{dt} = 6 \cdot 3 (\cos t)^2 (-\sin t) = -18 \cos^2 t \sin t$$ $$y = 6 \sin^3 t = 6 (\sin t)^3$$ $$\frac{dy}{dt} = 6 \cdot 3 (\sin t)^2 \cos t = 18 \sin^2 t \cos t$$ 4. Formulaga qo'yamiz: $$L = \int_0^{\frac{\pi}{3}} \sqrt{(-18 \cos^2 t \sin t)^2 + (18 \sin^2 t \cos t)^2} \, dt$$ 5. Ichidagi ifodani soddalashtiramiz: $$= \int_0^{\frac{\pi}{3}} \sqrt{324 \cos^4 t \sin^2 t + 324 \sin^4 t \cos^2 t} \, dt$$ $$= \int_0^{\frac{\pi}{3}} \sqrt{324 \cos^2 t \sin^2 t (\cos^2 t + \sin^2 t)} \, dt$$ 6. Trigonometrik identifikatsiya bo'yicha $\cos^2 t + \sin^2 t = 1$, shuning uchun: $$L = \int_0^{\frac{\pi}{3}} \sqrt{324 \cos^2 t \sin^2 t} \, dt = \int_0^{\frac{\pi}{3}} 18 |\cos t \sin t| \, dt$$ 7. $0 \leq t \leq \frac{\pi}{3}$ oraliqda $\cos t$ va $\sin t$ musbat, shuning uchun modulni olib tashlash mumkin: $$L = 18 \int_0^{\frac{\pi}{3}} \cos t \sin t \, dt$$ 8. Integralni hisoblaymiz: $$\int \cos t \sin t \, dt = \frac{1}{2} \sin^2 t + C$$ 9. Shunday qilib: $$L = 18 \cdot \frac{1}{2} [\sin^2 t]_0^{\frac{\pi}{3}} = 9 (\sin^2 \frac{\pi}{3} - \sin^2 0)$$ 10. Qiymatlarni qo'yamiz: $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \Rightarrow \sin^2 \frac{\pi}{3} = \frac{3}{4}$$ 11. Yakuniy natija: $$L = 9 \left(\frac{3}{4} - 0\right) = \frac{27}{4} = 6.75$$ Demak, egri chiziqning yoyi uzunligi $\frac{27}{4}$ ga teng.