# Chapter 10 - Practice Exercises - Page 593: 18

$\approx 8.617$

#### Work Step by Step

Since, $L=\int_{0}^{\pi/2}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$ Thus, $L=\int_{0}^{1} (3\sqrt 2)t \sqrt{16+t^2} dt$ Plug $16+t^2=u \implies du=2tdt$ Then, we have $L=(\dfrac{3\sqrt 2}{2})\int_{16}^{17} \sqrt{p} dp=(\dfrac{3\sqrt 2}{2}) [(2/3)u^{3/2}]_{16}^{17}$ Thus, $L \approx 8.617$

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