Answer
$$L \approx 32.85$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {2\cos t,4\sin t,6\cos t} \right\rangle {\text{, for }}0 \leqslant t \leqslant 2\pi \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {2\cos t,4\sin t,6\cos t} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle { - 2\sin t,4\cos t, - 6\sin t} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle { - 2\sin t,4\cos t, - 6\sin t} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( { - 2\sin t} \right)}^2} + {{\left( {4\cos t} \right)}^2} + \left( { - 6\sin t} \right)} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {4{{\sin }^2}t + 16{{\cos }^2}t + 36{{\sin }^2}t} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {40{{\sin }^2}t + 16{{\cos }^2}t} \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^{2\pi } {\sqrt {40{{\sin }^2}t + 16{{\cos }^2}t} } dt \cr
& {\text{Calculate using a computer or calculator}} \cr
& L \approx 32.85 \cr} $$
