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: 17



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, $v(t)=\lt 2t(t^2+1)^{-1},(t^2+1)^{-1},t(t^2+1)^{-1/2} \gt \\ \implies v(0)=\lt 0,1,0 \gt$ and $a(t)=\lt 2t(t^2+1)^{-1}-4t^2(t^2+1)^{-2},-2t(t^2+1)^{-2},(t^2+1)^{-1/2} -t^2(t^2+1)^{-3/2}\gt \\ \implies a(0)= \lt 2,0,1 \gt$ As we know that $\theta =\cos ^{-1}\dfrac{v(0) \cdot a(0)}{|v(0)||a(0)|}$ $\theta =\cos ^{-1}(\dfrac{0}{(\sqrt{1})(\sqrt{5}}))=\dfrac{\pi}{2}$
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