University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.3 - Arc Length - Exercises - Page 370: 4

Answer

$\dfrac{32}{3}$

Work Step by Step

Formula to calculate the arc length is: $l=\int_p^q \sqrt {1+[f'(x)]^2} dy$ The given equation can be differentiated and re-written as: $[f'(x)]^2=\dfrac{(y-1)^2}{4y}$ Now, $l=\int_1^9 [1+\dfrac{(y-1)^2}{4y}] dx =\dfrac{1}{2}[\int_1^9\dfrac{y}{y^{1/2}}+\int_1^9\dfrac{1}{y^{1/2}}]$ Since, we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ or,$ \dfrac{1}{2}[\dfrac{y^{3/2}}{3/2}+\dfrac{y^{1/2}}{1/2}]_1^9=\dfrac{32}{3}$
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