Answer
$${\bf{v}}\left( t \right) = \left\langle {3\cos t, - 5\sin t,4\cos t} \right\rangle ,{\text{ speed}}:{\text{ }}5{\text{ and }}{\bf{a}}\left( t \right) = \left\langle { - 3\sin t, - 5\cos t, - 4\sin t} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {3\sin t,5\cos t,4\sin t} \right\rangle ,\,\,\,\,\,{\text{for 0}} \leqslant t \leqslant 2\pi \cr
& \left( a \right){\text{find the velocity and speed }} \cr
& {\text{the velocity vector is given by }}\,\,\left( {{\text{see page 817}}} \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {3\sin t,5\cos t,4\sin t} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{v}}\left( t \right) = \left\langle {\frac{d}{{dt}}\left[ {3\sin t} \right],\frac{d}{{dt}}\left[ {5\cos t} \right],\frac{d}{{dt}}\left[ {4\sin t} \right]} \right\rangle \cr
& {\bf{v}}\left( t \right) = \left\langle {3\cos t, - 5\sin t,4\cos t} \right\rangle \cr
& {\text{The speed of the object is the scalar function }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ then}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \left| {\left\langle {3\cos t, - 5\sin t,4\cos t} \right\rangle } \right| \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( {3\cos t} \right)}^2} + {{\left( { - 5\sin t} \right)}^2} + {{\left( {4\cos t} \right)}^2}} \cr
& {\text{simplifying}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{{\cos }^2}t + 25{{\sin }^2}t + 16{{\cos }^2}t} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {25{{\cos }^2}t + 25{{\sin }^2}t} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {25\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = 5 \cr
& \cr
& \left( b \right){\text{find the acceleration of the object}} \cr
& {\text{the acceleration vector is given by }}\,\,{\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right)\left( {{\text{see page 817}}} \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {3\cos t, - 5\sin t,4\cos t} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{a}}\left( t \right) = \left\langle { - 3\sin t, - 5\cos t, - 4\sin t} \right\rangle \cr} $$
