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: 18



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