Answer
$$L \approx 2.73$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{e^t},2{e^{ - t}},t} \right\rangle {\text{, for }}0 \leqslant t \leqslant \ln 3 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {{e^t},2{e^{ - t}},t} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {{e^t}, - 2{e^{ - t}},1} \right\rangle \cr
& {\text{The speed is given by}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \left| {\left\langle {{e^t}, - 2{e^{ - t}},1} \right\rangle } \right| \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( {{e^t}} \right)}^2} + {{\left( { - 2{e^{ - t}}} \right)}^2} + {{\left( 1 \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{e^{2t}} + 4{e^{ - 2t}} + 1} \cr
& {\text{Find the length of the trajectory}} \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^{\ln 3} {\sqrt {{e^{2t}} + 4{e^{ - 2t}} + 1} } dt \cr
& {\text{Calculate using a computer or calculator}} \cr
& L \approx 2.73 \cr} $$
