Calculus: Early Transcendentals 8th Edition

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

Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 549: 24

Answer

$L=\int_0^{\frac{\pi}{2}}\sqrt{1+(1-\sin(x))^2}dx=1.7294$

Work Step by Step

More than 1.67, line length $\sqrt{\Big(\frac{\pi}{2}\Big)^2+\Big(\frac{\pi}{2}+\cos\Big(\frac{\pi}{2}\Big)-1\Big)^2}=1.67$ $L=\int_a^b\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}dx$ $y=x+\cos(x)$ $\frac{dy}{dx}=1-\sin(x)$ $a=0$, $b={\frac{\pi}{2}}$ $L=\int_0^{\frac{\pi}{2}}\sqrt{1+(1-\sin(x))^2}dx=1.7294$
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