Answer
$$L = 5\pi $$
Work Step by Step
$$\eqalign{
& x = 3\cos t,\,\,\,\,y = 3\sin t,\,\,\,\,z = 4t,\,\,\,\,\,\,\,\,0 \leqslant t \leqslant \pi \cr
& \cr
& {\text{Calculate the derivatives with respect to }}t{\text{ for }}x\left( t \right),\,\,\,y\left( t \right){\text{ and z}}\left( t \right) \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {3\cos t} \right] = - 3\sin t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {3\sin t} \right] = 3\cos t \cr
& \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {4t} \right] = 4 \cr
& \cr
& {\text{Find the arc length using }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} } dt \cr
& {\text{for the interval }}0 \leqslant t \leqslant \pi ,\,\,{\text{ }}a = 0{\text{ and }}b = \pi .{\text{ Then}}{\text{,}} \cr
& L = \int_0^\pi {\sqrt {{{\left( { - 3\sin t} \right)}^2} + {{\left( {3\cos t} \right)}^2} + {{\left( 4 \right)}^2}} } dt \cr
& L = \int_0^\pi {\sqrt {9{{\sin }^2}t + 9{{\cos }^2}t + 16} } dt \cr
& {\text{factoring}} \cr
& L = \int_0^\pi {\sqrt {9\left( {{{\sin }^2}t + {{\cos }^2}t} \right) + 16} } dt \cr
& L = \int_0^\pi {\sqrt {25} } dt \cr
& L = \int_0^\pi 5 dt \cr
& \cr
& {\text{Integrate and evaluate}} \cr
& L = 5\left[ t \right]_0^\pi \cr
& L = 5\left( {\pi - 0} \right) \cr
& L = 5\pi \cr} $$
