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: 18

Answer

$2$

Work Step by Step

$y = \sqrt{x-x^{2}} + \sin^{-1} \sqrt{x}$ then $dy/dx = \frac{1-2x}{2\sqrt{x-x^{2}}} + \frac{1}{2\sqrt{x} \sqrt{1-x^{2}}} = \frac{2-2x}{2\sqrt{x}\sqrt{1-x}} = \sqrt{\frac{1-x}{x}}$ and $1+(dy/dx)^{2} = 1+ \frac{1-x}{x} = \frac{1}{x}$ The curve has endpoint (0, 0) and (1, $\frac{\pi}{2}$) So $L = \int^{1}_{0} \sqrt{1/x} dx = \lim_{t->0+} \int ^{1}_{t} \sqrt{1/x} dx= \lim_{t->0+} [2\sqrt{x}]^{1}_{t} = lim_{t->0+} [2\sqrt{1} - 2\sqrt{t}] = 2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.