Answer
$${\bf{v}}\left( t \right) = \left\langle { - 3\sin t,4\cos t} \right\rangle ,{\text{ speed}}:{\text{ }}\sqrt {16 - 7{{\sin }^2}t} {\text{ and }}{\bf{a}}\left( t \right) = \left\langle { - 3\cos t, - 4\sin t} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {3\cos t,4\sin t} \right\rangle ,\,\,\,\,\,{\text{for }}0 \leqslant t \leqslant 2\pi \cr
& \left( a \right){\text{find the velocity and speed }} \cr
& {\text{the velocity vector is given by }}\,\,\left( {{\text{see page 817}}} \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {3\cos t,4\sin t} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{v}}\left( t \right) = \left\langle { - 3\sin t,4\cos t} \right\rangle \cr
& {\text{The speed of the object is the scalar function }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ then}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \left| {\left\langle { - 3\sin t,4\cos t} \right\rangle } \right| \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( { - 3\sin t} \right)}^2} + {{\left( {4\cos t} \right)}^2}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{{\sin }^2}t + 16{{\cos }^2}t} \cr
& {\text{simplifying by the identity si}}{{\text{n}}^2}t + {\cos ^2}t = 1 \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{{\sin }^2}t + 16\left( {1 - {{\sin }^2}t} \right)} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{{\sin }^2}t + 16 - 16{{\sin }^2}t} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {16 - 7{{\sin }^2}t} \cr
& \cr
& \left( b \right){\text{find the acceleration of the object}} \cr
& {\text{the acceleration vector is given by }}\,\,{\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right)\left( {{\text{see page 817}}} \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - 3\sin t,4\cos t} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{a}}\left( t \right) = \left\langle { - 3\cos t, - 4\sin t} \right\rangle \cr} $$
