Answer
$$L = \sqrt {14} $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left( {4 + 3t} \right){\bf{i}} + \left( {2 - 2t} \right){\bf{j}} + \left( {5 + t} \right){\bf{k}};\,\,\,\,\,\,\,\,\,\,\,3 \leqslant t \leqslant 4 \cr
& \cr
& {\text{Calculate }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left( {4 + 3t} \right){\bf{i}} + \left( {2 - 2t} \right){\bf{j}} + \left( {5 + t} \right){\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {4 + 3t} \right]{\bf{i}} + \left[ {2 - 2t} \right]{\bf{j}} + \left[ {5 + t} \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = 3{\bf{i}} - 2{\bf{j}} + {\bf{k}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \left\| {3{\bf{i}} - 2{\bf{j}} + {\bf{k}}} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9 + 4 + 1} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {14} \cr
& \cr
& {\text{Find the arc length using }}L = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& {\text{for the interval }}\,3 \leqslant t \leqslant 4,\,\,{\text{ }}a = 3{\text{ and }}b = 4. \cr
& L = \int_3^4 {\sqrt {14} } dt \cr
& \cr
& {\text{Integrate and evaluate}} \cr
& L = \sqrt {14} \left[ t \right]_3^4 \cr
& L = \sqrt {14} \left( {4 - 3} \right) \cr
& L = \sqrt {14} \cr} $$
