Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 22

$$\eqalign{ & {\bf{v}}\left( t \right) = 3{\bf{i}} - 2{\bf{j}} - \left( {32t - 1} \right){\bf{k}} \cr & {\bf{r}}\left( t \right) = 3t{\bf{i}} - \left( {2t - 5} \right){\bf{j}} - \left( {16{t^2} - t - 2} \right){\bf{k}} \cr & {\bf{r}}\left( 2 \right) = 6{\bf{i}} + {\bf{j}} - 60{\bf{k}} \cr} $$

$$\eqalign{ & {\bf{a}}\left( t \right) = - 32{\bf{k}},{\text{ }}{\bf{v}}\left( 0 \right) = 3{\bf{i}} - 2{\bf{j}} + {\bf{k}},{\text{ }}{\bf{r}}\left( 0 \right) = 5{\bf{j}} + 2{\bf{k}} \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( { - 32{\bf{k}}} \right)} dt \cr & {\bf{v}}\left( t \right) = - 32t{\bf{k}} + {\bf{C}} \cr & {\text{Applying the initial condition }}{\bf{v}}\left( 0 \right) = 3{\bf{i}} - 2{\bf{j}} + {\bf{k}} \cr & {\bf{v}}\left( 0 \right) = - 32\left( 0 \right){\bf{k}} + {\bf{C}} \cr & {\bf{v}}\left( 0 \right) = {\bf{C}} \cr & 3{\bf{i}} - 2{\bf{j}} + {\bf{k}} = {\bf{C}} \cr & {\text{Therefore,}} \cr & {\bf{v}}\left( t \right) = - 32t{\bf{k}} + 3{\bf{i}} - 2{\bf{j}} + {\bf{k}} \cr & {\bf{v}}\left( t \right) = 3{\bf{i}} - 2{\bf{j}} - \left( {32t - 1} \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[ {3{\bf{i}} - 2{\bf{j}} - \left( {32t - 1} \right){\bf{k}}} \right]} dt \cr & {\bf{r}}\left( t \right) = 3t{\bf{i}} - 2t{\bf{j}} - \left( {16{t^2} - t} \right){\bf{k}} + {\bf{C}} \cr & {\text{Applying the initial condition }}{\bf{r}}\left( 0 \right) = 5{\bf{j}} + 2{\bf{k}} \cr & {\bf{r}}\left( 0 \right) = 3\left( 0 \right){\bf{i}} - 2\left( 0 \right){\bf{j}} - \left( {16{{\left( 0 \right)}^2} - 0} \right){\bf{k}} + {\bf{C}} \cr & {\bf{r}}\left( 0 \right) = {\bf{C}} \cr & 5{\bf{j}} + 2{\bf{k}} = {\bf{C}} \cr & {\bf{C}} = 5{\bf{j}} + 2{\bf{k}} \cr & {\text{Therefore, the position vector is}} \cr & {\bf{r}}\left( t \right) = 3t{\bf{i}} - 2t{\bf{j}} - \left( {16{t^2} - t} \right){\bf{k}} + 5{\bf{j}} + 2{\bf{k}} \cr & {\bf{r}}\left( t \right) = 3t{\bf{i}} - \left( {2t - 5} \right){\bf{j}} - \left( {16{t^2} - t - 2} \right){\bf{k}} \cr & \cr & {\text{Find the position at time }}t = 2 \cr & {\bf{r}}\left( 2 \right) = 3\left( 2 \right){\bf{i}} - \left( {2\left( 2 \right) - 5} \right){\bf{j}} - \left( {16{{\left( 2 \right)}^2} - \left( 2 \right) - 2} \right){\bf{k}} \cr & {\bf{r}}\left( 2 \right) = 6{\bf{i}} + {\bf{j}} - 60{\bf{k}} \cr} $$