Answer
The arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (\cos \left( {{{\left( {\frac{{3s}}{2} + 1} \right)}^{2/3}} - 1} \right),\sin \left( {{{\left( {\frac{{3s}}{2} + 1} \right)}^{2/3}} - 1} \right),$
${\ \ \ \ \ \ \ \ \ }$ $\frac{2}{3}{\left( {{{\left( {\frac{{3s}}{2} + 1} \right)}^{2/3}} - 1} \right)^{3/2}})$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,\frac{2}{3}{t^{3/2}}} \right)$. So, the start point $\left( {1,0,0} \right)$ corresponds to $t=0$.
The derivative is
${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t,{t^{1/2}}} \right)$
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {{{\sin }^2}u + {{\cos }^2}u + u} {\rm{d}}u = \mathop \smallint \limits_0^t \sqrt {1 + u} {\rm{d}}u$
Write $v=1+u$. So, $dv=du$. The integral becomes
$s\left( t \right) = \mathop \smallint \limits_1^{1 + t} {v^{1/2}}{\rm{d}}v = \frac{2}{3}{v^{3/2}}|_1^{1 + t}$
$s\left( t \right) = \frac{2}{3}\left( {{{\left( {1 + t} \right)}^{3/2}} - 1} \right)$
It follows that
$\frac{{3s}}{2} + 1 = {\left( {1 + t} \right)^{3/2}}$
${\left( {\frac{{3s}}{2} + 1} \right)^{2/3}} = 1 + t$
$t = {\left( {\frac{{3s}}{2} + 1} \right)^{2/3}} - 1$
The inverse of $s\left( t \right)$ is $t = {\left( {\frac{{3s}}{2} + 1} \right)^{2/3}} - 1$. Since the start point $\left( {1,0,0} \right)$ corresponds to $t=0$, so at the start point we have $s=0$.
Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,\frac{2}{3}{t^{3/2}}} \right)$ gives the arc length parametrization:
${{\bf{r}}_1}\left( s \right) = (\cos \left( {{{\left( {\frac{{3s}}{2} + 1} \right)}^{2/3}} - 1} \right),\sin \left( {{{\left( {\frac{{3s}}{2} + 1} \right)}^{2/3}} - 1} \right),$
${\ \ \ \ \ \ }$ $\frac{2}{3}{\left( {{{\left( {\frac{{3s}}{2} + 1} \right)}^{2/3}} - 1} \right)^{3/2}})$
