Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 658: 41

(a) $\pi$ (b) $\pi$

(a) Here, we have $L=\int_{0}^{\pi/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt=\int_{0}^{\pi/2} \sqrt{(-2 \sin 2t)^2+(2 \cos 2t)^2}dt$ $\implies L=\int_{0}^{\pi/2} \sqrt{4}dt=[2t]_{0}^{(\pi/2)}=\pi$ (b) Here, we have $L=\int_{-1/2}^{1/2}\sqrt{(\pi \cos \pi t)^2+(-\pi \sin \pi t)^2}dt$ This implies that $L=(\pi) \int_{-1/2}^{1/2} dt$ Therefore, $L=[\pi t]_{-1/2}^{1/2}=\pi$