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 746: 15


$\theta =\dfrac{\pi}{2}$

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 $r=\lt 3t+1, \sqrt 3t ,t^2 \gt$ Now, $v(t)=r'(t)=\lt 3, \sqrt 3 ,2t \gt \\ \implies v(0)=\lt 3, \sqrt 3 ,0\gt = 3i +\sqrt 3 j$ and $a(t)=v'(t)=\lt 0,0,2 \gt \\ \implies a(0)=\lt 0,0,2 \gt= 2k$ We have found that the velocity $v(0)$ lies in xy plane and acceleration $a(0)$ lies in the direction of z-axis. This implies that $\theta =\dfrac{\pi}{2}$
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.