Answer
$L=\dfrac{\pi}{4}\sqrt {1+\dfrac{\pi^2}{16}}+\ln |\dfrac{\pi}{4}+\sqrt {1+\dfrac{\pi^2}{16}}|$
Work Step by Step
$v(t)=\dfrac{dr}{dt}$
$\implies v(t)=-2 \sin t i +2 \cos t j +2 t k$
$\implies |v(t)|=\sqrt {(-2 \sin t)^2 +(2 \cos t)^2 +(2 t)^2}=2 \sqrt {1+t^2}$
Arc Length is:
$L=\int_a^b 2 \sqrt {1+t^2} \ dt =[t \sqrt {1+t^2}+\ln |t + \sqrt {1+t^2}|]_0^{\pi/4} \\=\dfrac{\pi}{4}\sqrt {1+\dfrac{\pi^2}{16}}+\ln |\dfrac{\pi}{4}+\sqrt {1+\dfrac{\pi^2}{16}}|$
