Answer
$\dfrac{2}{27} (13\sqrt {13}-8)$
Work Step by Step
As we are given that $r(t)=\lt 2t^{3/2}, \cos 2t, \sin 2t \gt$; $0 \leq t \leq 1$
The length $L$ of the curve $r(t)=\lt 2t^{3/2}, \cos 2t, \sin 2t \gt$; $0 \leq t \leq 1$ is given as:
$l=\int_p^q|r'(t)| dt$
Here, we have $r'(t)=\lt
\dfrac{(2)(3)}{2}t^{3/2-1}, -2\sin 2t, 2\cos 2t \gt=\lt 3 \sqrt t, -2\sin 2t, 2\cos 2t \gt$
and $|r'(t)|=\sqrt{( 3 \sqrt t)^2+(-2\sin 2t)^2+(2\cos 2t)^2}=\sqrt{9t+4}$
Now, $L=\int_p^q|r'(t)| dt=\int_0^1\sqrt{9t+4} dt$ as $0 \leq t \leq 1$
Plug $k=9t+4 \implies dt=\dfrac{dk}{9}$
Thus, $L=(\dfrac{1}{9}) \int_{4}^{13} k^{1/2} dk$
This implies that
$L=(\dfrac{1}{9})(\dfrac{2}{3} )[k^{3/2}]_{4}^{13}$
or, $=\dfrac{2}{27}[(13)^{3/2}-(4)^{3/2}]$
or, $=\dfrac{2}{27} (13\sqrt {13}-8)$
