Chapter 11 - Vectors and Vector-Valued Functions - Review Exercises - Page 856: 42

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

$$\eqalign{ & {\text{Let }}{\bf{r}}\left( t \right) = \left\langle {t + 1,{t^2} - 3} \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 {t + 1,{t^2} - 3} \right\rangle \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \left\langle {0 + 1,{0^2} - 3} \right\rangle \cr & \mathop {\lim }\limits_{t \to 0} {\bf{r}}\left( t \right) = \left\langle {1, - 3} \right\rangle \cr & and \cr & \mathop {\lim }\limits_{t \to \infty } {\bf{r}}\left( t \right) = \mathop {\lim }\limits_{t \to 0} \left\langle {\infty + 1,{\infty ^2} - 3} \right\rangle \cr & {\text{The limit does not exist}} \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 {t + 1,{t^2} - 3} \right\rangle \cr & {\bf{r}}'\left( t \right) = \left\langle {1,2t} \right\rangle \cr & and \cr & {\bf{r}}'\left( 0 \right) = \left\langle {1,2\left( 0 \right)} \right\rangle \cr & {\bf{r}}'\left( 0 \right) = \left\langle {1,0} \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 {1,2t} \right\rangle \cr & {\bf{r}}''\left( t \right) = \left\langle {0,2} \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 {t + 1,{t^2} - 3} \right\rangle dt} \cr & \int {{\bf{r}}\left( t \right)dt} = \left\langle {\frac{{{t^2}}}{2} + t,\frac{{{t^3}}}{3} - 3t} \right\rangle + {\bf{C}} \cr & \cr & \left. a \right)\left\langle {1, - 3} \right\rangle ,\,\,\,\, \cr & \left. b \right)\left\langle {1,2t} \right\rangle ,\,\,\left\langle {1,0} \right\rangle ,\,\, \cr & \,\left. c \right){\bf{r}}''\left( t \right) = \left\langle {0,2} \right\rangle ,\,\,\,\, \cr & \left. d \right)\left\langle {\frac{{{t^2}}}{2} + t,\frac{{{t^3}}}{3} - 3t} \right\rangle + {\bf{C}} \cr} $$