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: 16


$\cos ^{-1} (\dfrac{1}{\sqrt 2})$ or, $\dfrac{3\pi}{4}$

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: $a(t)=v'(t)$ Now, $v(t)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2}-32(t),0 \gt \\ \implies v(0)=\lt \dfrac{\sqrt 2}{2}, \dfrac{\sqrt 2}{2},0 \gt=\dfrac{1}{\sqrt 2}\lt 1,1,0 \gt$ and $a(t)=v'(t)=\lt 0,-32,0 \gt \\ \implies a(0)=32 \lt 0,-1,0 \gt$ As we know that $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}$ Thus, $\theta =\cos ^{-1}(\dfrac{(0-1+0)}{\sqrt{1^2+1^2+0^2}\sqrt{0^2+(-1)^2+0^2}})=\cos ^{-1} (\dfrac{1}{\sqrt 2})=\dfrac{3\pi}{4}$
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