Answer
$\frac{20\sqrt{10}-2}{3}$
Work Step by Step
Recall: The integral which represents the length of the parametric curve for $a\leq t\leq b$ is given by $L=\int_a^b\sqrt{(dx/dt)^2+(dy/dt)^2}dt$.
Given: $x=\frac{2}{3}t^3$ and $y=t^2-2$
Find $dx/dt$ and $dy/dt$:
$dx/dt=2t^2$ and $dy/dt=2t$
Find the length of the given curve:
$L=\int_0^3\sqrt{(2t^2)^2+(2t)^2}dt$
$L=\int_0^3\sqrt{4t^4+4t^2}$dt
$L=\int_0^32t\sqrt{t^2+1}dt$
$L=[\frac{2(t^2+1)^{3/2}}{3}]_0^3$
$L=\frac{2(3^2+1)^{3/2}}{3}-\frac{2(0^2+1)^{3/2}}{3}$
$L=\frac{20\sqrt{10}}{3}-\frac{2}{3}$
$L=\frac{20\sqrt{10}-2}{3}$
Thus, the length of the curve is $\frac{20\sqrt{10}-2}{3}$.
