Answer
$$s = \frac{{3\sqrt {17} \pi }}{2}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {4t, - \cos t,\sin t} \right\rangle ,{\text{ }}\left[ {0,\frac{{3\pi }}{2}} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {4t, - \cos t,\sin t} \right\rangle } \right] \cr
& {\bf{r}}'\left( t \right) = \left\langle {4,\sin t,\cos t} \right\rangle \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( 4 \right)}^2} + {{\left( {\sin t} \right)}^2} + {{\left( {\cos t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {16 + {{\sin }^2}t + {{\cos }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {17} \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^{3\pi /2} {\sqrt {17} } dt \cr
& {\text{Integrate }} \cr
& s = \left[ {\sqrt {17} t} \right]_0^{3\pi /2} \cr
& s = \frac{{3\sqrt {17} \pi }}{2} \cr
& \cr
& {\text{Graph}} \cr} $$
