Answer
The arc length function: $s\left( t \right) = 2t + \ln t - 2$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {4{t^{1/2}},\ln t,2t} \right)$ and $a=1$.
The derivative is ${\bf{r}}'\left( t \right) = \left( {2{t^{ - 1/2}},\frac{1}{t},2} \right)$. So,
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {2{t^{ - 1/2}},\frac{1}{t},2} \right)\cdot\left( {2{t^{ - 1/2}},\frac{1}{t},2} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = \frac{4}{t} + \frac{1}{{{t^2}}} + 4 = \frac{{4t + 1 + 4{t^2}}}{{{t^2}}}$
$||{\bf{r}}'\left( t \right)|{|^2} = {\left( {\frac{{2t + 1}}{t}} \right)^2}$
The arc length function is
$s\left( t \right) = \mathop \smallint \limits_1^t ||{\bf{r}}'\left( u \right)||{\rm{d}}u$
$s\left( t \right) = \mathop \smallint \limits_1^t \frac{{2u + 1}}{u}{\rm{d}}u = \mathop \smallint \limits_1^t \left( {2 + \frac{1}{u}} \right){\rm{d}}u$
$s\left( t \right) = \left( {2u + \ln u} \right)|_1^t = 2t - 2 + \ln t$
$s\left( t \right) = 2t + \ln t - 2$
