Chapter 11 - Vectors and Vector-Valued Functions - Review Exercises - Page 856: 43

$$\eqalign{ & \left. a \right)\left\langle {1,0} \right\rangle ,\,\left\langle {1,1} \right\rangle ,\,\,\,\, \cr & \left. b \right)\left\langle { - \frac{2}{{{{\left( {2t + 1} \right)}^2}}},\frac{1}{{{{\left( {t + 1} \right)}^2}}}} \right\rangle ,\left\langle { - 2,1} \right\rangle , \cr & \,\,\left. c \right){\bf{r}}''\left( t \right) = \left\langle {\frac{8}{{{{\left( {2t + 1} \right)}^3}}}, - \frac{2}{{{{\left( {t + 1} \right)}^3}}}} \right\rangle ,\,\,\, \cr & \left. d \right)\left\langle {\frac{1}{2}\ln \left| {2t + 1} \right|,t - \ln \left| {t + 1} \right|} \right\rangle + {\bf{C}} \cr} $$

$$\eqalign{ & {\text{Let }}{\bf{r}}\left( t \right) = \left\langle {\frac{1}{{2t + 1}},\frac{t}{{t + 1}}} \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 {\frac{1}{{2t + 1}},\frac{t}{{t + 1}}} \right\rangle \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \left\langle {\frac{1}{{2\left( 0 \right) + 1}},\frac{0}{{0 + 1}}} \right\rangle \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \left\langle {1,0} \right\rangle \cr & and \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \mathop {\lim }\limits_{t \to \infty } \left\langle {\frac{1}{{2t + 1}},\frac{t}{{t + 1}}} \right\rangle \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \left\langle {\frac{1}{{2\left( \infty \right) + 1}},\frac{1}{{1 + 1/\infty }}} \right\rangle \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \left\langle {1,1} \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 {\frac{1}{{2t + 1}},\frac{t}{{t + 1}}} \right\rangle \cr & {\bf{r}}'\left( t \right) = \left\langle { - \frac{2}{{{{\left( {2t + 1} \right)}^2}}},\frac{{t + 1 - t}}{{{{\left( {t + 1} \right)}^2}}}} \right\rangle \cr & {\bf{r}}'\left( t \right) = \left\langle { - \frac{2}{{{{\left( {2t + 1} \right)}^2}}},\frac{1}{{{{\left( {t + 1} \right)}^2}}}} \right\rangle \cr & and \cr & {\bf{r}}'\left( 0 \right) = \left\langle { - \frac{2}{{{{\left( {2\left( 0 \right) + 1} \right)}^2}}},\frac{1}{{{{\left( {0 + 1} \right)}^2}}}} \right\rangle \cr & {\bf{r}}'\left( 0 \right) = \left\langle { - 2,1} \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 { - \frac{2}{{{{\left( {2t + 1} \right)}^2}}},\frac{1}{{{{\left( {t + 1} \right)}^2}}}} \right\rangle \cr & {\bf{r}}''\left( t \right) = \left\langle {4{{\left( {2t + 1} \right)}^{ - 3}}\left( 2 \right), - 2{{\left( {t + 1} \right)}^{ - 3}}} \right\rangle \cr & {\bf{r}}''\left( t \right) = \left\langle {\frac{8}{{{{\left( {2t + 1} \right)}^3}}}, - \frac{2}{{{{\left( {t + 1} \right)}^3}}}} \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 {\frac{1}{{2t + 1}},\frac{t}{{t + 1}}} \right\rangle dt} \cr & \int {{\bf{r}}\left( t \right)dt} = \left\langle {\frac{1}{2}\ln \left| {2t + 1} \right|,t - \ln \left| {t + 1} \right|} \right\rangle + {\bf{C}} \cr} $$