Answer
The arc length parametrization is
${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {17} }},\frac{{4s}}{{\sqrt {17} }} + 9} \right)$
Work Step by Step
Setting $t=x$, the line $y=4x+9$ can be parametrized by ${\bf{r}}\left( t \right) = \left( {t,4t + 9} \right)$.
The derivative is ${\bf{r}}'\left( t \right) = \left( {1,4} \right)$.
Evaluate the arc length function
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {\left( {1,4} \right)\cdot\left( {1,4} \right)} {\rm{d}}u$
$s\left( t \right) = \sqrt {17} t$
The inverse of $s\left( t \right)$ is $t = \frac{s}{{\sqrt {17} }}$. Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {t,4t + 9} \right)$ gives
${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {17} }},\frac{{4s}}{{\sqrt {17} }} + 9} \right)$
Thus, the arc length parametrization is ${{\bf{r}}_1}\left( s \right) = \left( {\frac{s}{{\sqrt {17} }},\frac{{4s}}{{\sqrt {17} }} + 9} \right)$
