Answer
The arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (\left( {\frac{s}{{\sqrt 3 }} + 1} \right)\sin \left( {\ln \left( {\frac{s}{{\sqrt 3 }} + 1} \right)} \right),$
${\ \ \ \ }$ $\left( {\frac{s}{{\sqrt 3 }} + 1} \right)\cos \left( {\ln \left( {\frac{s}{{\sqrt 3 }} + 1} \right)} \right),\frac{s}{{\sqrt 3 }} + 1)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t}\sin t,{{\rm{e}}^t}\cos t,{{\rm{e}}^t}} \right)$.
The derivative is
${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}\sin t + {{\rm{e}}^t}\cos t,{{\rm{e}}^t}\cos t - {{\rm{e}}^t}\sin t,{{\rm{e}}^t}} \right)$
${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}\left( {\sin t + \cos t} \right),{{\rm{e}}^t}\left( {\cos t - \sin t} \right),{{\rm{e}}^t}} \right)$
So,
$||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}{\left( {\sin t + \cos t} \right)^2} + {{\rm{e}}^{2t}}{\left( {\cos t - \sin t} \right)^2} + {{\rm{e}}^{2t}}$
$||{\bf{r}}'\left( t \right)|{|^2} = {{\rm{e}}^{2t}}\left( {1 + 2\sin t\cos t} \right)$
${\ \ \ \ \ }$ $ + {{\rm{e}}^{2t}}\left( {1 - 2\cos t\sin t} \right) + {{\rm{e}}^{2t}}$
$||{\bf{r}}'\left( t \right)|{|^2} = 3{{\rm{e}}^{2t}}$, ${\ \ \ }$ $||{\bf{r}}'\left( t \right)|| = \sqrt 3 {{\rm{e}}^t}$
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \sqrt 3 \mathop \smallint \limits_0^t {{\rm{e}}^u}{\rm{d}}u = \sqrt 3 \left( {{{\rm{e}}^t} - 1} \right)$
It follows that $\frac{s}{{\sqrt 3 }} + 1 = {{\rm{e}}^t}$.
The inverse of $s\left( t \right)$ is $t = \ln \left( {\frac{s}{{\sqrt 3 }} + 1} \right)$.
Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t}\sin t,{{\rm{e}}^t}\cos t,{{\rm{e}}^t}} \right)$ gives the arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (\left( {\frac{s}{{\sqrt 3 }} + 1} \right)\sin \left( {\ln \left( {\frac{s}{{\sqrt 3 }} + 1} \right)} \right),$
${\ \ \ \ }$ $\left( {\frac{s}{{\sqrt 3 }} + 1} \right)\cos \left( {\ln \left( {\frac{s}{{\sqrt 3 }} + 1} \right)} \right),\frac{s}{{\sqrt 3 }} + 1)$
