Answer
There are two arc length parametrizations:
${{\bf{r}}_1}\left( s \right) = \left( {\frac{1}{9}\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right),\frac{1}{{27}}{{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)}^{3/2}}} \right)$
and
${{\bf{r}}_2}\left( s \right) = \left( {\frac{1}{9}\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right), - \frac{1}{{27}}{{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)}^{3/2}}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$.
The derivative is ${\bf{r}}'\left( t \right) = \left( {2t,3{t^2}} \right)$. So,
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {2t,3{t^2}} \right)\cdot\left( {2t,3{t^2}} \right) = 4{t^2} + 9{t^4}$
$||{\bf{r}}'\left( t \right)|| = t\sqrt {4 + 9{t^2}} $
Evaluate the arc length function:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t u\sqrt {4 + 9{u^2}} {\rm{d}}u$
Write $v = 4 + 9{u^2}$. So, $dv = 18udu$. The integral becomes
$s\left( t \right) = \frac{1}{{18}}\mathop \smallint \limits_4^{4 + 9{t^2}} {v^{1/2}}{\rm{d}}v$
$s\left( t \right) = \frac{1}{{18}}\cdot\frac{2}{3}{v^{3/2}}|_4^{4 + 9{t^2}}$
$s\left( t \right) = \frac{1}{{27}}\left( {{{\left( {4 + 9{t^2}} \right)}^{3/2}} - 8} \right)$
It follows that
$27s + 8 = {\left( {4 + 9{t^2}} \right)^{3/2}}$
${\left( {27s + 8} \right)^{2/3}} = 4 + 9{t^2}$
$t = \pm \frac{1}{3}{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)^{1/2}}$
Thus, there are two inverses of $s\left( t \right)$ corresponding to the two signs, that is,
${t_1} = \frac{1}{3}{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)^{1/2}}$
and
${t_2} = - \frac{1}{3}{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)^{1/2}}$
Substituting ${t_1}$ in ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$ gives the first arc length parametrization:
${{\bf{r}}_1}\left( s \right) = \left( {\frac{1}{9}\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right),\frac{1}{{27}}{{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)}^{3/2}}} \right)$
Substituting ${t_2}$ in ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$ gives the second arc length parametrization:
${{\bf{r}}_2}\left( s \right) = \left( {\frac{1}{9}\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right), - \frac{1}{{27}}{{\left( {{{\left( {27s + 8} \right)}^{2/3}} - 4} \right)}^{3/2}}} \right)$
