Chapter 12 - Vector-Valued Functions - 12.3 Exercises - Page 838: 4

$${\bf{v}}\left( 2 \right) = 3{\bf{i}} + {\bf{j}}$$$${\bf{a}}\left( 2 \right) = 3{\bf{i}}$$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {\frac{1}{4}{t^3} + 1} \right){\bf{i}} + t{\bf{j}},{\text{ }}\left( {3,2} \right) \cr & {\text{Let }}t = 2 \cr & {\bf{r}}\left( 1 \right) = \left( {\frac{1}{4}{{\left( 2 \right)}^3} + 1} \right){\bf{i}} + 2{\bf{j}} \cr & {\bf{r}}\left( 1 \right) = 3{\bf{i}} + 2{\bf{j}} \cr & {\text{Then, at }}\left( {3,2} \right){\text{ }}t = 2 \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[ {\left( {\frac{1}{4}{t^3} + 1} \right){\bf{i}} + t{\bf{j}}} \right] \cr & {\bf{v}}\left( t \right) = \frac{3}{4}{t^2}{\bf{i}} + {\bf{j}} \cr & {\text{speed}} = \left\| {{\bf{v}}\left( t \right)} \right\| = \left\| {\frac{3}{4}{t^2}{\bf{i}} + {\bf{j}}} \right\| \cr & {\text{speed}} = \sqrt {{{\left( {\frac{3}{4}{t^2}} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {\frac{9}{{16}}{t^4} + 1} \cr & {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {\frac{3}{4}{t^2}{\bf{i}} + {\bf{j}}} \right] \cr & {\bf{a}}\left( t \right) = \frac{3}{2}t{\bf{i}} \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( {3,2} \right){\text{ }}t = 2 \cr & {\bf{v}}\left( 2 \right) = \frac{3}{4}{\left( 2 \right)^2}{\bf{i}} + {\bf{j}} \cr & {\bf{v}}\left( 2 \right) = 3{\bf{i}} + {\bf{j}} \cr & {\text{speed}} = \sqrt {\frac{9}{{16}}{{\left( 2 \right)}^4} + 1} = \sqrt {10} \cr & {\bf{a}}\left( 2 \right) = \frac{3}{2}\left( 2 \right){\bf{i}} \cr & {\bf{a}}\left( 2 \right) = 3{\bf{i}} \cr & \cr & \left( {\bf{c}} \right){\text{ Sketching}} \cr & {\bf{r}}\left( t \right) = \left( {\frac{1}{4}{t^3} + 1} \right){\bf{i}} + t{\bf{j}} \cr & x = \frac{1}{4}{t^3} + 1,{\text{ }}y = t \cr & x = \frac{1}{4}{y^3} + 1 \cr & y = \root 3 \of {4x - 1} \cr & {\text{Graph}} \cr} $$