Answer
a) $\pi$ and b) $\pi$
Work Step by Step
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) $\pi$
