Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.3 Arc Length and Curvature - 13.3 Exercises - Page 908: 9



Work Step by Step

As we are given that Given: $r(t)=\lt \cos \pi t, 2t,\sin 2 \pi t$; $0 \leq t \leq 2$ Length of the curve can be obtained by using formula, such as $L=\int_a^b |r'(t)| dt$ Now,$r'(t)=\lt -\pi \sin \pi t, 2,2 \pi \cos 2 \pi t$ ; $|r'(t)|=\sqrt {( -\pi \sin \pi t)^2+(2)^2+(2 \pi \cos 2 \pi t)^2}dt$ Since, $L=\int_{0}^2(\sqrt {( -\pi \sin \pi t)^2+(2)^2+(2 \pi \cos 2 \pi t)^2}dt) dt=10.3311$ This is the result calculated by using a calculator.
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