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

$$\eqalign{ & {\bf{v}}\left( t \right) = \frac{2}{{t + 1}}{\bf{i}} + 2t{\bf{j}} + t{\bf{k}} \cr & {\bf{a}}\left( t \right) = - \frac{2}{{{{\left( {t + 1} \right)}^2}}}{\bf{i}} + 2{\bf{j}} + {\bf{k}} \cr & {\text{speed}}:\sqrt 6 \cr & {\text{direction}}:\frac{1}{{\sqrt 6 }}{\bf{i}} + \frac{2}{{\sqrt 6 }}{\bf{j}} + \frac{1}{{\sqrt 6 }}{\bf{k}} \cr & {\bf{v}}\left( 1 \right) = \sqrt 6 \left( {\frac{1}{{\sqrt 6 }}{\bf{i}} + \frac{2}{{\sqrt 6 }}{\bf{j}} + \frac{1}{{\sqrt 6 }}{\bf{k}}} \right) \cr} $$

$$\eqalign{ & {\bf{r}}\left( t \right) = \left( {2\ln \left( {t + 1} \right)} \right){\bf{i}} + {t^2}{\bf{j}} + \frac{{{t^2}}}{2}{\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( {2\ln \left( {t + 1} \right)} \right){\bf{i}} + {t^2}{\bf{j}} + \frac{{{t^2}}}{2}{\bf{k}}} \right] \cr & {\bf{v}}\left( t \right) = \frac{2}{{t + 1}}{\bf{i}} + 2t{\bf{j}} + t{\bf{k}} \cr & \cr & {\text{calculate }}{\bf{v}}\left( 1 \right) \cr & {\bf{v}}\left( 1 \right) = \frac{2}{{1 + 1}}{\bf{i}} + 2\left( 1 \right){\bf{j}} + \left( 1 \right){\bf{k}} \cr & {\bf{v}}\left( 1 \right) = {\bf{i}} + 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[ {\frac{2}{{t + 1}}{\bf{i}} + 2t{\bf{j}} + t{\bf{k}}} \right] \cr & {\bf{a}}\left( t \right) = - \frac{2}{{{{\left( {t + 1} \right)}^2}}}{\bf{i}} + 2{\bf{j}} + {\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| {\frac{2}{{t + 1}}{\bf{i}} + 2t{\bf{j}} + t{\bf{k}}} \right| \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{{\left( {\frac{2}{{t + 1}}} \right)}^2} + {{\left( {2t} \right)}^2} + {{\left( t \right)}^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {\frac{4}{{{{\left( {t + 1} \right)}^2}}} + 4{t^2} + {t^2}} \cr & \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {\frac{4}{{{{\left( {t + 1} \right)}^2}}} + 5{t^2}} \cr & {\text{find the particle's speed at }}t = 1 \cr & {\text{speed}}:\left| {{\bf{v}}\left( 1 \right)} \right| = \sqrt {\frac{4}{{{{\left( {1 + 1} \right)}^2}}} + 5{{\left( 1 \right)}^2}} \cr & {\text{speed}}:\sqrt 6 \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}} + 2{\bf{j}} + {\bf{k}}}}{{\sqrt 6 }} \cr & {\text{direction}}:\frac{1}{{\sqrt 6 }}{\bf{i}} + \frac{2}{{\sqrt 6 }}{\bf{j}} + \frac{1}{{\sqrt 6 }}{\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( {{\bf{i}} + \frac{2}{3}{\bf{j}} + \frac{2}{3}{\bf{k}}} \right) \cr} $$