Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 10

$$\eqalign{ & {\bf{v}} = {\bf{i}} + \sqrt 2 t{\bf{j}} + {t^2}{\bf{k}} \cr & {\bf{a}} = \sqrt 2 {\bf{j}} + 2t{\bf{k}} \cr & {\text{speed}}:2 \cr & {\text{direction}}:\frac{1}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} + \frac{1}{2}{\bf{k}} \cr & {\bf{v}}\left( 1 \right) = 2\left( {\frac{1}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} + \frac{1}{2}{\bf{k}}} \right) \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {1 + t} \right){\bf{i}} + \frac{{{t^2}}}{{\sqrt 2 }}{\bf{j}} + \frac{{{t^3}}}{3}{\bf{k}},\,\,\,\,\,t = 1 \cr & {\text{Find }}{\bf{v}}\left( t \right){\text{ by differentiating }}{\bf{r}}\left( t \right) \cr & {\bf{v}}\left( t \right) = \frac{d}{{dt}}\left[ {\left( {1 + t} \right){\bf{i}} + \frac{{{t^2}}}{{\sqrt 2 }}{\bf{j}} + \frac{{{t^3}}}{3}{\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = {\bf{i}} + \frac{{2t}}{{\sqrt 2 }}{\bf{j}} + \frac{{3{t^2}}}{3}{\bf{k}} \cr & {\bf{v}}\left( t \right) = {\bf{i}} + \sqrt 2 t{\bf{j}} + {t^2}{\bf{k}} \cr & {\text{calculate }}{\bf{v}}\left( 1 \right) \cr & {\bf{v}}\left( 1 \right) = {\bf{i}} + \sqrt 2 \left( 1 \right){\bf{j}} + {\left( 1 \right)^2}{\bf{k}} \cr & {\bf{v}}\left( 1 \right) = {\bf{i}} + \sqrt 2 {\bf{j}} + {\bf{k}} \cr & \cr & {\text{Find }}{\bf{a}}\left( t \right){\text{ by differentiating }}{\bf{v}}\left( t \right) \cr & {\bf{a}}\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{i}} + \sqrt 2 t{\bf{j}} + {t^2}{\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = \sqrt 2 {\bf{j}} + 2t{\bf{k}} \cr & \cr & {\text{The speed of the particle is }}\left| {{\bf{v}}\left( t \right)} \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \left| {{\bf{i}} + \sqrt 2 t{\bf{j}} + {t^2}{\bf{k}}} \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{1^2} + {{\left( {\sqrt 2 t} \right)}^2} + {{\left( {{t^2}} \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {1 + 2{t^2} + {t^4}} \cr & {\text{find the particle's speed at }}t = 1 \cr & {\text{speed}}:\left| {{\bf{v}}\left( 1 \right)} \right| = \sqrt {1 + 2{{\left( 1 \right)}^2} + {{\left( 1 \right)}^4}} \cr & {\text{speed}}:\sqrt {1 + 2 + 1} \cr & {\text{speed}}:2 \cr & \cr & {\text{Calculate the direction of motion at the given value of }}t \cr & {\text{direction}}:\frac{{{\bf{v}}\left( 1 \right)}}{{\left| {{\bf{v}}\left( 1 \right)} \right|}} \cr & {\text{direction}}:\frac{{{\bf{i}} + \sqrt 2 {\bf{j}} + {\bf{k}}}}{2} \cr & {\text{direction}}:\frac{1}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} + \frac{1}{2}{\bf{k}} \cr & \cr & {\text{Write the particle's velocity at that time as the product of its }} \cr & {\text{speed and direction}} \cr & {\bf{v}}\left( 1 \right) = 2\left( {\frac{1}{2}{\bf{i}} + \frac{{\sqrt 2 }}{2}{\bf{j}} + \frac{1}{2}{\bf{k}}} \right) \cr} $$