Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.1 - Curves in Space and Their Tangents - Exercises 13.1 - Page 745: 4


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

Work Step by Step

We know that the velocity and acceleration is defined as: velocity is given as: $v(t)=r'(t)$ and acceleration is given as: $a(t)=v'(t)$ In the given problem, we have $x=\cos t; y=3 \sin 2t$ Re-write as follows: $9x^2+y^2=9 \\ \implies y=3 \sqrt {1-x^2}$ $v(t)=r'(t)=-2 \sin 2t+6 \cos 2t \\ \implies v(0)=6j$ and $a(t)=v'(t)=-4 \cos 2t-12 \sin 2t\\ \implies a(0)=-4i$
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