Answer
$$L = 10\pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {3\cos t,4\cos t,5\sin t} \right\rangle {\text{, for }}0 \leqslant t \leqslant 2\pi \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {3\cos t,4\cos t,5\sin t} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle { - 3\sin t, - 4\sin t,5\cos t} \right\rangle \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^{2\pi } {\left| {\left\langle { - 3\sin t, - 4\sin t,5\cos t} \right\rangle } \right|} dt \cr
& L = \int_0^{2\pi } {\sqrt {{{\left( { - 3\sin t} \right)}^2} + {{\left( { - 4\sin t} \right)}^2} + \left( {5\cos t} \right)} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^{2\pi } {\sqrt {9{{\sin }^2}t + 16{{\sin }^2}t + 25{{\cos }^2}t} } dt \cr
& L = \int_0^{2\pi } {\sqrt {25\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} } dt \cr
& L = 5\int_0^{2\pi } {dt} \cr
& {\text{integrating}} \cr
& L = 5\left( {2\pi - 0} \right) \cr
& L = 10\pi \cr} $$
