Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.4 Motion in Space: Velocity and Acceleration - 13.4 Exercises - Page 918: 16


$r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$

Work Step by Step

As we know $v(t)=\int a(t)$ and $r(t)=\int v(t)$ $v(t)=\int (\sin t i+2 \cos tj+6tk) dt$ and $v(t)=(1-\cos t) i+2 \sin tj+(3t^2-1)k$ Solve $r(t)=\int [(1-\cos t) i+2 \sin tj+(3t^2-1)k] dt=(t-\sin t) i-2 \cos tj+(t^3-t)k+c$ Also, $r(0)=j-4k $ and $c= j-4k$ Therefore, $r(t)=(t-\sin t) i-2 \cos tj+(t^3-t)k+j-4k$ Hence, the result. $r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.