University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 12 - Section 12.1 - Curves in Space and Their Tangents - Exercises - Page 648: 4


$9x^2+y^2=9$ or, $y=3 \sqrt {1-x^2}$; $v(0)=6j$ and $a(0)=-4i$

Work Step by Step

Since, we have $x=\cos t; y=3 \sin 2t$ This can be written as: $9x^2+y^2=9$ or, $y=3 \sqrt {1-x^2}$ Now, velocity $v(t)=r'(t)=-2 \sin 2t+6 \cos 2t$ and $v(0)=6j$ As we know acceleration $a(t)=v'(t)=-4 \cos 2t-12 \sin 2t$ and $a(0)=-4i$ Hence, we have $9x^2+y^2=9$ or, $y=3 \sqrt {1-x^2}$; $v(0)=6j$ and $a(0)=-4i$
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.