Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 12

Answer

$\frac{33}{16}$

Work Step by Step

$x = \frac{y^{4}}{8} + \frac{1}{4y^{2}}$ then $\frac{dx}{dy} = \frac{1}{2} y^{3} - \frac{1}{2} y^{-3}$ $1+(dx/dy)^{2} = 1+\frac{1}{4} y^{6} - \frac{1}{2} + \frac{1}{4} y^{-6} = \frac{1}{4} y^{6} + \frac{1}{2} + \frac{1}{4}y^{-6} = (\frac{1}{2} y^{3} + \frac{1}{2} y^{-3})^{2}$ So $L = \int^{2}_{1} \sqrt{(\frac{1}{2}y^{3} + \frac{1}{2}y^{-3})^{2}}dy = \int^{2}_{1} (\frac{1}{2}y^{3} + \frac{1}{2} y^{-3}) dy = [\frac{1}{8}y^{4} - \frac{1}{4}y^{-2}]^{2}_{1} = \frac{33}{16}$
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