Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 21


$$ \ln (5+\sqrt{26})$$

Work Step by Step

Since \begin{align*} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{-a}^{a} \sqrt{1+(\sinh x)^{2}} d x\\ &=\int_{-a}^{a} \sqrt{1+\sinh ^{2} x} d x\\ &=\int_{-a}^{a} \sqrt{\cosh ^{2} x} d x\\ &=\int_{-a}^{a} \cosh x d x\\ &=[\sinh x]_{-a}^{a}\\ &=2 \sinh a \end{align*} Setting this expression equal to 10 and solving for $a$ yields $$a=\sinh ^{-1}(5)=\ln (5+\sqrt{26})$$
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