Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Section 13.3 - Arc Length and Curvature - 13.3 Exercise - Page 868: 6

Answer

$42$

Work Step by Step

Given: $r(t)=t^2i+9tj+4t^{3/2}k$ ; $1 \leq t \leq 4$ To calculate the length of the curve we will have to use the formula: $L=\int_a^b |r'(t)| dt$ Thus, $r'(t)=\lt 2t,9,6t^{1/2}\gt$ and $|r'(t)|=\sqrt {( 2t)^2+(9)^2+(6t^{1/2})^2}dt$ $=\sqrt {4t^2+81+36t}dt$ $=2t+9$ $L=\int_{1}^4(2t+9) dt$ $\implies L=(t^2+9t)|_{1}^4$ $=((4)^2-(1)^2)+(9(4)-9(1))$ $=42$ Hence, $L=42$
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