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: 8


$ 2.0454$

Work Step by Step

As we are given that $r(t)=\lt t, e^{-t},te^{-t}\gt$; $1 \leq t \leq 3$ Length of the curve can be obtained by using formula, such as $L=\int_a^b |r'(t)| dt$ Now, $r'(t)=\lt 1,-e^{-t},e^{-t}-te^{-t}\gt$ ; $|r'(t)|=\sqrt {( 1)^2+(e^{-t})^2+(e^{-t}-te^{-t})^2}dt$ or,$=\sqrt{ 1+e^{-2t}+(e^{-t}-te^{-t})^2}$ or, $=\sqrt{ 1+e^{-2t}+e^{-2t}(1-t)^2}$ Since,$L=\int_{1}^3(\sqrt{ 1+e^{-2t}+e^{-2t}(1-t)^2}) dt= 2.0454$ This is the result calculated by using a calculator.
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