Answer
$\dfrac{2}{27} (13\sqrt {13}-8)$
Work Step by Step
Here, we have $r'(t)=\lt
3t^{3/2-1}, -2\sin (2t), 2\cos (2t) \gt$
$\implies r'(t)=\lt 3 \sqrt t, -2\sin 2t, 2\cos 2t \gt$
and $|r'(t)|=\sqrt{( 3 \sqrt t)^2+(-2\sin 2t)^2+(2\cos 2t)^2}$
$\implies |r'(t)|=\sqrt{9t+4\sin^2 (2t)+4\cos^2 (2t)}=\sqrt{9t+4}$
The length of the arc is $l=\int_0^1\sqrt{9t+4} dt$
Suppose $u=9t+4\implies dt=\dfrac{1}{9}du$
$l=(\dfrac{1}{9}) \int_{4}^{13} u^{1/2} du=\dfrac{2}{27}[u^{3/2}]_{4}^{13} $
Therefore, $l=[(13)^{(3/2)}-(4)^{(3/2)}]=\dfrac{2}{27} (13\sqrt {13}-8)$
