Chapter 11 - Vectors and Vector-Valued Functions - Review Exercises - Page 856: 44

$$\eqalign{ & \left. a \right)\left\langle {1,0,0} \right\rangle ,\,\left\langle {0,0,\frac{\pi }{2}} \right\rangle ,\,\,\,\, \cr & \left. b \right)\left\langle { - 2{e^{ - 2t}}, - t{e^{ - t}} + {e^{ - t}},\frac{1}{{1 + {t^2}}}} \right\rangle ,\,\,\,\left\langle { - 2,1,1} \right\rangle \, \cr & \left. c \right){\bf{r}}''\left( t \right) = \left\langle {4{e^{ - 2t}},t{e^{ - t}} - 2{e^{ - t}}, - \frac{{2t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right\rangle \cr & \left. d \right)\left\langle { - \frac{1}{2}{e^{ - 2t}}, - {e^{ - t}}\left( {t + 1} \right),t{{\tan }^{ - 1}}t - \frac{1}{2}\ln \left( {1 + {t^2}} \right)} \right\rangle + {\bf{C}} \cr} $$

$$\eqalign{ & {\text{Let }}{\bf{r}}\left( t \right) = \left\langle {{e^{ - 2t}},t{e^{ - t}},{{\tan }^{ - 1}}t} \right\rangle \cr & \left. a \right){\text{ Find }}\mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right){\text{ and }}\mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \mathop {\lim }\limits_{t \to 0} \left\langle {{e^{ - 2t}},t{e^{ - t}},{{\tan }^{ - 1}}t} \right\rangle \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \left\langle {{e^{ - 2\left( 0 \right)}},0{e^{ - 0}},{{\tan }^{ - 1}}0} \right\rangle \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \left\langle {1,0,0} \right\rangle \cr & and \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \mathop {\lim }\limits_{t \to \infty } \left\langle {{e^{ - 2t}},t{e^{ - t}},{{\tan }^{ - 1}}t} \right\rangle \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \left\langle {{e^{ - 2\left( \infty \right)}},\left( \infty \right){e^{ - \infty }},{{\tan }^{ - 1}}\infty } \right\rangle \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \left\langle {0,0,\frac{\pi }{2}} \right\rangle \cr & \cr & \left. b \right){\text{ Find }}{\bf{r}}'\left( t \right){\text{ and evaluate }}{\bf{r}}'\left( 0 \right) \cr & {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {{e^{ - 2t}},t{e^{ - t}},{{\tan }^{ - 1}}t} \right\rangle \cr & {\bf{r}}'\left( t \right) = \left\langle { - 2{e^{ - 2t}},t\left( { - {e^{ - t}}} \right) + {e^{ - t}},\frac{1}{{1 + {t^2}}}} \right\rangle \cr & {\bf{r}}'\left( t \right) = \left\langle { - 2{e^{ - 2t}}, - t{e^{ - t}} + {e^{ - t}},\frac{1}{{1 + {t^2}}}} \right\rangle \cr & and \cr & {\bf{r}}'\left( 0 \right) = \left\langle { - 2{e^{ - 2\left( 0 \right)}}, - 0{e^{ - \left( 0 \right)}} + {e^{ - \left( 0 \right)}},\frac{1}{{1 + {{\left( 0 \right)}^2}}}} \right\rangle \cr & {\bf{r}}'\left( 0 \right) = \left\langle { - 2,1,1} \right\rangle \cr & \cr & \left. c \right){\text{ Find }}{\bf{r}}''\left( t \right) \cr & {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left\langle {{e^{ - 2t}},t{e^{ - t}},{{\tan }^{ - 1}}t} \right\rangle \cr & {\bf{r}}''\left( t \right) = \left\langle {4{e^{ - 2t}}, - t\left( { - {e^{ - t}}} \right) - {e^{ - t}} - {e^{ - t}}, - \frac{{2t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right\rangle \cr & {\bf{r}}''\left( t \right) = \left\langle {4{e^{ - 2t}},t{e^{ - t}} - 2{e^{ - t}}, - \frac{{2t}}{{{{\left( {1 + {t^2}} \right)}^2}}}} \right\rangle \cr & \cr & \left. d \right){\text{ Evaluate }}\int {{\bf{r}}\left( t \right)dt} \cr & \int {{\bf{r}}\left( t \right)dt} = \int {\left\langle {{e^{ - 2t}},t{e^{ - t}},{{\tan }^{ - 1}}t} \right\rangle dt} \cr & \int {{\bf{r}}\left( t \right)dt} = \left\langle { - \frac{1}{2}{e^{ - 2t}}, - {e^{ - t}}\left( {t + 1} \right),t{{\tan }^{ - 1}}t - \frac{1}{2}\ln \left( {1 + {t^2}} \right)} \right\rangle + {\bf{C}} \cr} $$