Chapter 10 - Review - Exercises - Page 690: 37

$2(5\sqrt 5-1)$

$x=3 t^{2}$, $y=2t^{3}$ on $0\leq t\leq 2$ $x'=6t$, $y'=6t^{2}$ on $0\leq t\leq 2$ Length of the curve is given as: $L=\int_{0}^{2}\sqrt {(6t)^2+(6t^{2})^2} dt$ $=\int_{0}^{2}\sqrt {36t^2(1+t^{2})} dt$ Let $u=1+t^2$ then $du=2tdt$ Also, when $t=0$ then $u=1$ and $t=2$ then $u=5$ Thus, $L=3\int_{1}^{5}u^{1/2} du$ or, $L=2(5\sqrt 5-1)$