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: 16

Answer

$\dfrac{3\pi}{4}$

Work Step by Step

Since we know, velocity $v(t)=r'(t)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2}-32t,0 \gt$ and $v(0)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2},0 \gt=\dfrac{1}{\sqrt 2}\lt 1,1,0 \gt$ As we know acceleration $a(t)=v'(t)=\lt 0,-32,0 \gt$ and $a(0)=32 \lt 0,-1,0 \gt$ Hence, $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}=\cos ^{-1}(\dfrac{0-1+0}{\sqrt{1^2+1^2+0^2}\sqrt{0^2+(-1)^2+0^2}})=\cos ^{-1} \dfrac{1}{\sqrt 2}$ or, $\theta=\dfrac{3\pi}{4}$
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