Answer
$${\bf{v}}\left( \pi \right) = \left\langle {4,0, - 3} \right\rangle $$$${\bf{a}}\left( \pi \right) = \left\langle {0,3,0} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {4t,3\cos t,3\sin t} \right\rangle ,{\text{ }}t = \pi \cr
& \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {4t,3\cos t,3\sin t} \right\rangle } \right] \cr
& {\bf{v}}\left( t \right) = \left\langle {4, - 3\sin t,3\cos t} \right\rangle \cr
& {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\left\langle {4, - 3\sin t,3\cos t} \right\rangle } \right\| \cr
& {\text{speed}} = \sqrt {4 + 9{{\sin }^2}t + 9{{\cos }^2}t} \cr
& {\text{speed}} = \sqrt {13} \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {4, - 3\sin t,3\cos t} \right\rangle } \right] \cr
& {\bf{a}}\left( t \right) = \left\langle {0, - 3\cos t, - 3\sin t} \right\rangle \cr
& \cr
& \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = \pi \cr
& {\bf{v}}\left( \pi \right) = \left\langle {4, - 3\sin \pi ,3\cos \pi } \right\rangle \cr
& {\bf{v}}\left( \pi \right) = \left\langle {4,0, - 3} \right\rangle \cr
& {\bf{a}}\left( \pi \right) = \left\langle {0, - 3\cos \pi , - 3\sin \pi } \right\rangle \cr
& {\bf{a}}\left( \pi \right) = \left\langle {0,3,0} \right\rangle \cr} $$
