Answer
$$L = 4\sqrt {65} \pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {t,8\sin t,8\cos t} \right\rangle ,\,\,\,\,\,{\text{for 0}} \leqslant t \leqslant 4\pi \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {t,8\sin t,8\cos t} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {1,8\cos t, - 8\sin t} \right\rangle \cr
& {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& \,{\text{for 0}} \leqslant t \leqslant 4\pi \to a = 0{\text{ and }}b = 4\pi .{\text{ then}} \cr
& L = \int_0^{4\pi } {\left| {\left\langle {1,8\cos t, - 8\sin t} \right\rangle } \right|} dt \cr
& L = \int_0^{4\pi } {\sqrt {{{\left( 1 \right)}^2} + {{\left( {8\cos t} \right)}^2} + {{\left( { - 8\sin t} \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^{4\pi } {\sqrt {1 + 64{{\cos }^2}t + 64{{\sin }^2}t} } dt \cr
& L = \int_0^{4\pi } {\sqrt {1 + 64\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} } dt \cr
& L = \int_0^{4\pi } {\sqrt {1 + 64} } dt \cr
& L = \int_0^{4\pi } {\sqrt {65} } dt \cr
& {\text{integrating}} \cr
& L = \sqrt {65} \left( t \right)_0^{4\pi } \cr
& L = \sqrt {65} \left( {4\pi - 0} \right) \cr
& L = 4\sqrt {65} \pi \cr} $$
