Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 18

$\approx 8.617$

Here, we have $L=\int_{0}^{\pi/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$ This gives: $L=\int_{0}^{1} (\dfrac{3}{\sqrt 2})(t) \sqrt{16+t^2} dt$ Plug $16+t^2=k$ or, $(2t) dt= dk$ Then, we have $L=(\dfrac{3\sqrt 2}{2})\int_{16}^{17} (\sqrt{k}) dk$ Thus, $L=(\dfrac{3\sqrt 2}{2}) [(\dfrac{2}{3})k^{3/2}]_{16}^{17} \approx 8.617$