Answer
$$14$$
Work Step by Step
$v(t)=\dfrac{dr}{dt}$
$\implies v(t)=-3 \sin t i +3 \cos t j +3\sqrt t k$
Now, $|v(t)|=\sqrt {(-3 \sin t)^2 +(3 \cos t)^2 +(3\sqrt t)^2}=3 \sqrt {1+t}$
Arc Length is:
$L =\int_a^b |v| \ dt$
or, $=\int_0^3 3 \sqrt {1+t} \ dt $
or, $=[2(1+t)^{3/2}]_0^{3}$
or, $=2(1+3)^{3/2} -2 $
or, $=14$
