Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 18

$${\bf{v}}\left( 2 \right) = \left\langle {\frac{1}{2}, - \frac{1}{4},32} \right\rangle $$$${\bf{a}}\left( 2 \right) = \left\langle { - \frac{1}{4},\frac{1}{4},48} \right\rangle $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {\ln t,\frac{1}{t},{t^4}} \right\rangle ,{\text{ }}t = 2 \cr & \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {\ln t,\frac{1}{t},{t^4}} \right\rangle } \right] \cr & {\bf{v}}\left( t \right) = \left\langle {\frac{1}{t}, - \frac{1}{{{t^2}}},4{t^3}} \right\rangle \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\left\langle {\frac{1}{t}, - \frac{1}{{{t^2}}},4{t^3}} \right\rangle } \right\| \cr & {\text{speed}} = \sqrt {\frac{1}{{{t^2}}} + \frac{1}{{{t^4}}} + 16{t^4}} \cr & {\text{speed}} = \sqrt {\frac{{{t^2} + 1 + 16{t^8}}}{{{t^4}}}} \cr & {\text{speed}} = \frac{{\sqrt {{t^2} + 1 + 16{t^8}} }}{{{t^2}}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {\frac{1}{t}, - \frac{1}{{{t^2}}},4{t^3}} \right\rangle } \right] \cr & {\bf{a}}\left( t \right) = \left\langle { - \frac{1}{{{t^2}}},\frac{2}{{{t^3}}},12{t^2}} \right\rangle \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ at }}t = 2 \cr & {\bf{v}}\left( 2 \right) = \left\langle {\frac{1}{2}, - \frac{1}{{{2^2}}},4{{\left( 2 \right)}^3}} \right\rangle \cr & {\bf{v}}\left( 2 \right) = \left\langle {\frac{1}{2}, - \frac{1}{4},32} \right\rangle \cr & {\bf{a}}\left( 2 \right) = \left\langle { - \frac{1}{{{2^2}}},\frac{2}{{{2^3}}},12{{\left( 2 \right)}^2}} \right\rangle \cr & {\bf{a}}\left( 2 \right) = \left\langle { - \frac{1}{4},\frac{1}{4},48} \right\rangle \cr} $$