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 649: 15


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

Work Step by Step

Since, we have $r=\lt 3t+1, \sqrt 3t ,t^2 \gt$ Now, velocity $v(t)=r'(t)=\lt 3, \sqrt 3 ,2t \gt$ and $v(0)=\lt 3, \sqrt 3 ,0\gt = 3i +\sqrt 3 j$ As we know acceleration $a(t)=v'(t)=\lt 0,0,2 \gt$ and $a(0)=\lt 0,0,2 \gt= 2k$ Hence, we can see that $v(0)$ is in the xy plane and $a(0)$ is in the direction of k . Thus, we have $\theta =\dfrac{\pi}{2}$
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