Answer
$$L = \sqrt 3 $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{e^t}\sin t,{e^t}\cos t,{e^t}} \right\rangle {\text{, for }}0 \leqslant t \leqslant \ln 2 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {{e^t}\sin t + {e^t}\cos t,{e^t}\cos t - {e^t}\sin t,{e^t}} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle {{e^t}\sin t + {e^t}\cos t,{e^t}\cos t - {e^t}\sin t,{e^t}} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{e^{2t}}{{\left( {\sin t + \cos t} \right)}^2} + {e^{2t}}{{\left( {\cos t - \sin t} \right)}^2} + {e^{2t}}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{e^{2t}}\left( {1 + 2\sin t\cos t} \right) + {e^{2t}}\left( {1 - 2\sin t\cos t} \right) + {e^{2t}}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{e^{2t}} + 2\sin t\cos t + {e^{2t}} - 2\sin t\cos t + {e^{2t}}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{e^{2t}} + {e^{2t}} + {e^{2t}}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt 3 {e^t} \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \sqrt 3 \int_0^{\ln 2} {{e^t}} dt \cr
& {\text{simplifying}} \cr
& L = \sqrt 3 \left( {{e^t}} \right)_0^{\ln 2} \cr
& L = \sqrt 3 \left( {{e^{\ln 2}} - {e^0}} \right) \cr
& L = \sqrt 3 \cr} $$
