Answer
$\dfrac{2}{27} (13\sqrt {13}-8)$
Work Step by Step
Given: $r(t)=\lt 2t^{3/2}, \cos 2t, \sin 2t \gt$; $0 \leq t \leq 1$
The length of the curve $r(t)=\lt 2t^{3/2}, \cos 2t, \sin 2t \gt$; $0 \leq t \leq 1$ is given as:
$l=\int_m^n|r'(t)| dt$
Take derivative of the $r(t)=\lt 2t^{3/2}, \cos 2t, \sin 2t \gt$; $0 \leq t \leq 1$ with respect to $t$.
$r'(t)=\lt
\dfrac{(2)(3)}{2}t^{3/2-1}, -2\sin 2t, 2\cos 2t \gt$
This gives:
$r'(t)=\lt 3 \sqrt t, -2\sin 2t, 2\cos 2t \gt$
and $|r'(t)|=\sqrt{( 3 \sqrt t)^2+(-2\sin 2t)^2+(2\cos 2t)^2}$
or, $|r'(t)|=\sqrt{9t+4\sin^2 2t+4\cos^2 2t}=\sqrt{9t+4}$
Now, $l=\int_m^n|r'(t)| dt=\int_0^1\sqrt{9t+4} dt$
Let us consider $9t+4 =p \implies dt=\dfrac{dp}{9}$
The limits also change. When $t= 0 \implies p=4$ and when $t=1 \implies p=9(1)+4=13$
Thus, $l=(1/9) \int_{4}^{13} p^{1/2} dp$
$l=\dfrac{1}{9}[\dfrac{2}{3} p^{3/2}]_{4}^{13} =\dfrac{2}{27}[(13)^{3/2}-(4)^{3/2}]=\dfrac{2}{27} (13\sqrt {13}-8)$
