## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 746: 9

#### Answer

\eqalign{ & {\bf{v}} = {\bf{i}} + 2t{\bf{j}} + 2{\bf{k}} \cr & {\bf{a}} = 2{\bf{j}} \cr & {\text{speed}}:3 \cr & {\text{direction}}:\frac{1}{3}{\bf{i}} + \frac{2}{3}{\bf{j}} + \frac{2}{3}{\bf{k}} \cr & {\bf{v}}\left( 1 \right) = 3\left( {\frac{1}{3}{\bf{i}} + \frac{2}{3}{\bf{j}} + \frac{2}{3}{\bf{k}}} \right) \cr}

#### Work Step by Step

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

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