Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 2

$${\bf{v}}\left( 1 \right) = {\bf{i}} - 2{\bf{j}}$$$${\bf{a}}\left( 1 \right) = - 2{\bf{j}}$$

$$\eqalign{ & {\bf{r}}\left( t \right) = t{\bf{i}} + \left( { - {t^2} + 4} \right){\bf{j}},{\text{ }}\left( {1,3} \right) \cr & {\text{Let }}t = 1 \cr & {\bf{r}}\left( 1 \right) = \left( 1 \right){\bf{i}} + \left( { - {{\left( 1 \right)}^2} + 4} \right){\bf{j}} \cr & {\bf{r}}\left( 1 \right) = {\bf{i}} + 3{\bf{j}} \cr & {\text{Then, at }}\left( {1,3} \right){\text{ }}t = 1 \cr & \left( {\bf{a}} \right){\text{Find the vectors: }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed}}{\text{.}} \cr & {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {t{\bf{i}} + \left( { - {t^2} + 4} \right){\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = {\bf{i}} - 2t{\bf{j}} \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {{\bf{i}} - 2t{\bf{j}}} \right\| \cr & {\text{speed}} = \sqrt {\left( 1 \right) + {{\left( { - 2t} \right)}^2}} = \sqrt {1 + 4{t^2}} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{i}} - 2t{\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = - 2{\bf{j}} \cr & \cr & \left( {\bf{b}} \right){\text{Evaluating }}{\bf{v}}\left( t \right),{\text{ }}{\bf{a}}\left( t \right){\text{ and speed at the given point}}{\text{.}} \cr & {\text{At }}\left( {1,3} \right){\text{ }}t = 1 \cr & {\bf{v}}\left( 1 \right) = {\bf{i}} - 2\left( 1 \right){\bf{j}} \cr & {\bf{v}}\left( 1 \right) = {\bf{i}} - 2{\bf{j}} \cr & {\text{speed}} = \sqrt {1 + 4{{\left( 1 \right)}^2}} = \sqrt 5 \cr & {\bf{a}}\left( 1 \right) = - 2{\bf{j}} \cr & \cr & \left( {\bf{c}} \right){\text{ Sketching}} \cr & {\bf{r}}\left( t \right) = t{\bf{i}} + \left( { - {t^2} + 4} \right){\bf{j}} \cr & x = t,{\text{ }}y = - {t^2} + 4 \cr & y = - {x^2} + 4,{\text{ parabola}} \cr & {\text{Graph}} \cr} $$