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: 1

Answer

$10\sqrt{10}$

Work Step by Step

Given: $r(t)=\lt t, 3 \cos t, 3 \sin t \gt$ ; $-5 \leq t \leq 5$ To calculate the length of the curve we will have to use the formula: $L=\int_a^b |r'(t)| dt$ Thus, $r'(t)=\lt 1, -3 \sin t, 3 \cos t \gt$ and $|r'(t)|=\sqrt {1^2+(-3 \sin t)^2+(3 \cos t)^2}dt=\sqrt {1+9 \sin^2 t+9 \cos^2 t}$ $=\sqrt{10}$ $L=\int_{-5}^5 \sqrt{10}dt= \sqrt{10}t|_{-5}^5$ $=5 \sqrt{10}-(-5\sqrt{10})$ Hence, $L=10\sqrt{10}$
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