## University Calculus: Early Transcendentals (3rd Edition)

a) $\pi$ and b) $\pi$
a) Since, $L=\int_{0}^{\pi/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $L=\int_{0}^{\pi/2} \sqrt{(-2 \sin 2t)^2+(2 \cos 2t)^2}dt=\int_{0}^{\pi/2} \sqrt{4}dt$ or, $L=[2t]_{0}^{\pi/2}=\pi$ b) a) Since, $L=\int_{-1/2}^{1/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $L=\int_{-1/2}^{1/2}\sqrt{(\pi \cos \pi t)^2+(-\pi \sin \pi t)^2}dt=\int_{-1/2}^{1/2} \pi dt$ or, $L=[\pi t]_{-1/2}^{1/2}=\pi$ Hence, a) $\pi$ and b) $\pi$