Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 21

$$\eqalign{ & {\bf{v}}\left( t \right) = \left( {\frac{{{t^2}}}{2} + \frac{9}{2}} \right){\bf{j}} + \left( {\frac{{{t^2}}}{2} - \frac{1}{2}} \right){\bf{k}} \cr & {\bf{r}}\left( t \right) = \left( {\frac{{{t^3}}}{6} + \frac{9}{2}t} \right){\bf{j}} + \left( {\frac{{{t^3}}}{6} - \frac{1}{2}t} \right){\bf{k}} + {\bf{C}} \cr & {\bf{r}}\left( 2 \right) = \frac{{17}}{3}{\bf{j}} + \frac{2}{3}{\bf{k}} \cr} $$

$$\eqalign{ & {\bf{a}}\left( t \right) = t{\bf{j}} + t{\bf{k}},{\text{ }}{\bf{v}}\left( 1 \right) = 5{\bf{j}},{\text{ }}{\bf{r}}\left( 1 \right) = {\bf{0}} \cr & {\text{The velocity vector is}} \cr & {\bf{v}}\left( t \right) = \int {{\bf{a}}\left( t \right)} dt \cr & {\bf{v}}\left( t \right) = \int {\left( {t{\bf{j}} + t{\bf{k}}} \right)} dt \cr & {\bf{v}}\left( t \right) = \frac{1}{2}{t^2}{\bf{j}} + \frac{1}{2}{t^2}{\bf{k}} + {\bf{C}} \cr & {\text{Applying the initial condition }}{\bf{v}}\left( 1 \right) = 5{\bf{j}} \cr & {\bf{v}}\left( 1 \right) = \frac{1}{2}{\left( 1 \right)^2}{\bf{j}} + \frac{1}{2}{\left( 1 \right)^2}{\bf{k}} + {\bf{C}} \cr & {\bf{v}}\left( 1 \right) = \frac{1}{2}{\bf{j}} + \frac{1}{2}{\bf{k}} + {\bf{C}} \cr & 5{\bf{j}} = \frac{1}{2}{\bf{j}} + \frac{1}{2}{\bf{k}} + {\bf{C}} \cr & {\bf{C}} = 5{\bf{j}} - \frac{1}{2}{\bf{j}} - \frac{1}{2}{\bf{k}} \cr & {\bf{C}} = \frac{9}{2}{\bf{j}} - \frac{1}{2}{\bf{k}} \cr & {\text{Therefore,}} \cr & {\bf{v}}\left( t \right) = \frac{1}{2}{t^2}{\bf{j}} + \frac{1}{2}{t^2}{\bf{k}} + \frac{9}{2}{\bf{j}} - \frac{1}{2}{\bf{k}} \cr & {\bf{v}}\left( t \right) = \left( {\frac{{{t^2}}}{2} + \frac{9}{2}} \right){\bf{j}} + \left( {\frac{{{t^2}}}{2} - \frac{1}{2}} \right){\bf{k}} \cr & \cr & {\text{The position vector is}} \cr & {\bf{r}}\left( t \right) = \int {{\bf{v}}\left( t \right)} dt \cr & {\bf{r}}\left( t \right) = \int {\left[ {\left( {\frac{{{t^2}}}{2} + \frac{9}{2}} \right){\bf{j}} + \left( {\frac{1}{2}{t^2} - \frac{1}{2}} \right){\bf{k}}} \right]} dt \cr & {\bf{r}}\left( t \right) = \left( {\frac{{{t^3}}}{6} + \frac{9}{2}t} \right){\bf{j}} + \left( {\frac{{{t^3}}}{6} - \frac{1}{2}t} \right){\bf{k}} + {\bf{C}} \cr & {\text{Applying the initial condition }}{\bf{r}}\left( 1 \right) = {\bf{0}} \cr & {\bf{r}}\left( 1 \right) = \left( {\frac{{{{\left( 1 \right)}^3}}}{6} + \frac{9}{2}\left( 1 \right)} \right){\bf{j}} + \left( {\frac{{{{\left( 1 \right)}^3}}}{6} - \frac{1}{2}\left( 1 \right)} \right){\bf{k}} + {\bf{C}} \cr & {\bf{r}}\left( 1 \right) = \frac{{14}}{3}{\bf{j}} - \frac{1}{3}{\bf{k}} + {\bf{C}} \cr & {\bf{0}} = \frac{{14}}{3}{\bf{j}} - \frac{1}{3}{\bf{k}} + {\bf{C}} \cr & {\bf{C}} = - \frac{{14}}{3}{\bf{j}} + \frac{1}{3}{\bf{k}} \cr & {\text{Therefore, the position vector is}} \cr & {\bf{r}}\left( t \right) = \left( {\frac{{{t^3}}}{6} + \frac{9}{2}t} \right){\bf{j}} + \left( {\frac{{{t^3}}}{6} - \frac{1}{2}t} \right){\bf{k}} - \frac{{14}}{3}{\bf{j}} + \frac{1}{3}{\bf{k}} \cr & {\bf{r}}\left( t \right) = \left( {\frac{{{t^3}}}{6} + \frac{9}{2}t - \frac{{14}}{3}} \right){\bf{j}} + \left( {\frac{{{t^3}}}{6} - \frac{1}{2}t + \frac{1}{3}} \right){\bf{k}} \cr & \cr & {\text{Find the position at time }}t = 2 \cr & {\bf{r}}\left( 2 \right) = \left( {\frac{{{{\left( 2 \right)}^3}}}{6} + \frac{9}{2}\left( 2 \right) - \frac{{14}}{3}} \right){\bf{j}} + \left( {\frac{{{{\left( 2 \right)}^3}}}{6} - \frac{1}{2}\left( 2 \right) + \frac{1}{3}} \right){\bf{k}} \cr & {\bf{r}}\left( 2 \right) = \frac{{17}}{3}{\bf{j}} + \frac{2}{3}{\bf{k}} \cr} $$