Answer
Arc length integrals for the two parametrizations:
${s_1}\left( t \right) = \mathop \smallint \limits_0^t \sqrt {1 + 9{u^4}} {\rm{d}}u$
${s_2}\left( t \right) = 3\mathop \smallint \limits_0^t {u^2}\sqrt {1 + 9{u^{12}}} {\rm{d}}u$
They yield the same result.
Work Step by Step
We have two parametrizations of the same curve $y = {x^3}$ for $0 \le x \le 8$, that is,
${{\bf{r}}_1}\left( t \right) = \left( {t,{t^3}} \right)$ ${\ \ }$ and ${\ \ }$ ${{\bf{r}}_2}\left( t \right) = \left( {{t^3},{t^9}} \right)$
The derivatives of the two parametrizations are
${{\bf{r}}_1}'\left( t \right) = \left( {1,3{t^2}} \right)$ ${\ \ }$ and ${\ \ }$ ${{\bf{r}}_2}'\left( t \right) = \left( {3{t^2},9{t^8}} \right)$
The speeds of the two parametrizations are
$||{{\bf{r}}_1}'\left( t \right)|| = \sqrt {1 + 9{t^4}} $
and
$||{{\bf{r}}_2}'\left( t \right)|| = \sqrt {\left( {3{t^2},9{t^8}} \right)\cdot\left( {3{t^2},9{t^8}} \right)} $
$||{{\bf{r}}_2}'\left( t \right)|| = \sqrt {9{t^4} + 81{t^{16}}} = \sqrt {9{t^4}\left( {1 + 9{t^{12}}} \right)} $
$||{{\bf{r}}_2}'\left( t \right)|| = 3{t^2}\sqrt {1 + 9{t^{12}}} $
Evaluate the arc length functions:
${s_1}\left( t \right) = \mathop \smallint \limits_0^t ||{{\bf{r}}_1}'\left( u \right)||{\rm{d}}u$
(1) ${\ \ \ }$ ${s_1}\left( t \right) = \mathop \smallint \limits_0^t \sqrt {1 + 9{u^4}} {\rm{d}}u$
and
${s_2}\left( t \right) = \mathop \smallint \limits_0^t ||{{\bf{r}}_2}'\left( u \right)||{\rm{d}}u$
${s_2}\left( t \right) = 3\mathop \smallint \limits_0^t {u^2}\sqrt {1 + 9{u^{12}}} {\rm{d}}u$
Write $v = {u^3}$, so $dv = 3{u^2}du$. The integral becomes
(2) ${\ \ \ }$ ${s_2}\left( t \right) = \mathop \smallint \limits_0^t \sqrt {1 + 9{v^4}} {\rm{d}}v$
Since $u$ and $v$ are dummy variables, the two integrals in equation (1) and equation (2) yield the same result.
