Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 8

$${\bf{v}}\left( 0 \right) = \left\langle { - 1,1} \right\rangle $$$${\bf{a}}\left( 0 \right) = \left\langle {1,1} \right\rangle $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {{e^{ - t}},{e^t}} \right\rangle ,{\text{ }}\left( {1,1} \right) \cr & {\text{Let }}t = 0 \cr & {\bf{r}}\left( 0 \right) = \left\langle {{e^0},{e^0}} \right\rangle \cr & {\bf{r}}\left( 0 \right) = \left\langle {1,1} \right\rangle \cr & {\text{Then, at }}\left( {1,1} \right){\text{ }}t = 0 \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 {{e^{ - t}},{e^t}} \right\rangle } \right] \cr & {\bf{v}}\left( t \right) = \left\langle { - {e^{ - t}},{e^t}} \right\rangle \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\left\langle { - {e^{ - t}},{e^t}} \right\rangle } \right\| \cr & {\text{speed}} = \sqrt {{{\left( { - {e^{ - t}}} \right)}^2} + {{\left( {{e^t}} \right)}^2}} = \sqrt {{e^{ - 2t}} + {e^{2t}}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle { - {e^{ - t}},{e^t}} \right\rangle } \right] \cr & {\bf{a}}\left( t \right) = \left\langle {{e^{ - t}},{e^t}} \right\rangle \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed at the given point}}{\text{.}} \cr & {\text{At }}\left( {1,1} \right){\text{ }}t = 0 \cr & {\bf{v}}\left( 0 \right) = \left\langle { - {e^0},{e^0}} \right\rangle \cr & {\bf{v}}\left( 0 \right) = \left\langle { - 1,1} \right\rangle \cr & {\text{speed}} = \sqrt {{e^0} + {e^0}} \cr & {\text{speed}} = \sqrt 2 \cr & {\bf{a}}\left( 0 \right) = \left\langle {{e^0},{e^0}} \right\rangle \cr & {\bf{a}}\left( 0 \right) = \left\langle {1,1} \right\rangle \cr & \cr & \left( {\bf{c}} \right){\text{ Sketching}} \cr & {\bf{r}}\left( t \right) = \left\langle {{e^{ - t}},{e^t}} \right\rangle \cr & x = {e^{ - t}},{\text{ }}y = {e^t} \cr & {\text{Graph}} \cr} $$