Answer
(a) $s\left( t \right) = \sqrt {29} t$
(b) the inverse of $s\left( t \right)$ is $t = \frac{s}{{\sqrt {29} }}$.
(c) ${{\bf{r}}_1}\left( s \right) = \left( {\frac{{3s}}{{\sqrt {29} }} + 1,\frac{{4s}}{{\sqrt {29} }} - 5,\frac{{2s}}{{\sqrt {29} }}} \right)$ is an arc length parametrization.
Work Step by Step
(a) We have ${\bf{r}}\left( t \right) = \left( {3t + 1,4t - 5,2t} \right)$. So, ${\bf{r}}'\left( t \right) = \left( {3,4,2} \right)$.
Evaluate the arc length integral:
$s\left( t \right) = \mathop \smallint \limits_0^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_0^t \sqrt {\left( {3,4,2} \right)\cdot\left( {3,4,2} \right)} {\rm{d}}u$
$s\left( t \right) = \sqrt {29} t$
(b) From part (a) we get $s\left( t \right) = \sqrt {29} t$. The inverse of $s\left( t \right)$ is $t = \frac{s}{{\sqrt {29} }}$.
(c) From part (b) we get $t = \frac{s}{{\sqrt {29} }}$. Substituting $t$ in ${\bf{r}}\left( t \right) = \left( {3t + 1,4t - 5,2t} \right)$ gives
${{\bf{r}}_1}\left( s \right) = \left( {\frac{{3s}}{{\sqrt {29} }} + 1,\frac{{4s}}{{\sqrt {29} }} - 5,\frac{{2s}}{{\sqrt {29} }}} \right)$
Taking the derivative of ${{\bf{r}}_1}\left( s \right)$ gives
${{\bf{r}}_1}'\left( s \right) = \left( {\frac{3}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }},\frac{2}{{\sqrt {29} }}} \right)$
So,
$||{{\bf{r}}_1}'\left( s \right)|| = \sqrt {\left( {\frac{3}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }},\frac{2}{{\sqrt {29} }}} \right)\cdot\left( {\frac{3}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }},\frac{2}{{\sqrt {29} }}} \right)} $
$||{{\bf{r}}_1}'\left( s \right)|| = \sqrt {\frac{{29}}{{29}}} = 1$
Thus, it is verified that ${{\bf{r}}_1}'\left( s \right)$ has unit speed. Hence, ${{\bf{r}}_1}\left( s \right)$ is an arc length parametrization.
