Answer
$${\bf{v}}\left( t \right) = \left\langle {4{e^{2t}},2{e^{2t}},4{e^{2t}}} \right\rangle ,{\text{ speed}}:{\text{ }}6{e^{2t}}{\text{ and }}{\bf{a}}\left( t \right) = \left\langle {8{e^{2t}},4{e^{2t}},8{e^{2t}}} \right\rangle $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {2{e^{2t}} + 1,{e^{2t}} - 1,2{e^{2t}} - 10} \right\rangle ,\,\,\,\,\,{\text{for }}t \geqslant 0 \cr
& \left( a \right){\text{find the velocity and speed }} \cr
& {\text{the velocity vector is given by }}\,\,\left( {{\text{see page 817}}} \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {2{e^{2t}} + 1,{e^{2t}} - 1,2{e^{2t}} - 10} \right\rangle } \right] \cr
& {\text{differentiating each component}}{\text{, use }}\frac{d}{{dt}}\left[ {{e^{at}}} \right] = a{e^{at}} \cr
& {\bf{v}}\left( t \right) = \left\langle {2{e^{2t}} + 1,{e^{2t}} - 1,2{e^{2t}} - 10} \right\rangle \cr
& {\bf{v}}\left( t \right) = \left\langle {4{e^{2t}},2{e^{2t}},4{e^{2t}}} \right\rangle \cr
& {\text{The speed of the object is the scalar function }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ then}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \left| {\left\langle {4{e^{2t}},2{e^{2t}},4{e^{2t}}} \right\rangle } \right| \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( {4{e^{2t}}} \right)}^2} + {{\left( {2{e^{2t}}} \right)}^2} + {{\left( {4{e^{2t}}} \right)}^2}} \cr
& {\text{simplifying}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {16{e^{4t}} + 4{e^{4t}} + 16{e^{4t}}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {36{e^{4t}}} \cr
& \left| {{\bf{v}}\left( t \right)} \right| = 6{e^{2t}} \cr
& \cr
& \left( b \right){\text{find the acceleration of the object}} \cr
& {\text{the acceleration vector is given by }}\,\,{\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right)\left( {{\text{see page 817}}} \right) \cr
& {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {4{e^{2t}},2{e^{2t}},4{e^{2t}}} \right\rangle } \right] \cr
& {\text{differentiating each component}} \cr
& {\bf{a}}\left( t \right) = \left\langle {8{e^{2t}},4{e^{2t}},8{e^{2t}}} \right\rangle \cr} $$
