Answer
The length of the path:
$s = \mathop \smallint \limits_1^2 \sqrt {\frac{1}{{{t^2}}} + 1 + {{\rm{e}}^{2t}}} {\rm{d}}t$
$s \simeq 4.85$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\ln t,t,{{\rm{e}}^t}} \right)$ for $1 \le t \le 2$. So,
${\bf{r}}'\left( t \right) = \left( {\frac{1}{t},1,{{\rm{e}}^t}} \right)$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {\frac{1}{t},1,{{\rm{e}}^t}} \right)\cdot\left( {\frac{1}{t},1,{{\rm{e}}^t}} \right)} $
$ = \sqrt {\frac{1}{{{t^2}}} + 1 + {{\rm{e}}^{2t}}} $
By Theorem 1 of Section 14.3, the length of the path is given by
$s = \mathop \smallint \limits_1^2 ||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$s = \mathop \smallint \limits_1^2 \sqrt {\frac{1}{{{t^2}}} + 1 + {{\rm{e}}^{2t}}} {\rm{d}}t$
Using a computer algebra system we compute the definite integral and obtain $s \simeq 4.847$.
