Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 16

$\dfrac{13}{12} $

Here, we have $L=\int_{1}^{2}\sqrt{1+(\dfrac{dy}{dx})^2}dx$ Thus, $L=\int_{1}^{2} \sqrt{1+(1/16)(y^{4}-\dfrac{1}{2}+\dfrac{1}{y^4})} dy$ or, $L=[(1/2)y^3-y^{-1}]_1^{2}$ Therefore, $L =[\dfrac{1}{2}y^3-y^{-1}]_1^{2}=\dfrac{13}{12} $