Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 13

$\dfrac{10}{3}$

Here, we have $L=\int_{1}^{4}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt=\int_{1}^{4} \sqrt{1+(1/4)(1/x-2+x)} dx$ This gives: $L= (\dfrac{1}{2}) \int_{1}^{4}\sqrt{x^{-1/2}+x^{1/2}}dx=(\dfrac{1}{2})[2x^{(\frac{1}{2})}+(\dfrac{2}{3})x^{(3/2)}]_{1}^{4}$ Therefore, $L=(\dfrac{1}{2})[2+\dfrac{14}{3}]=\dfrac{10}{3}$