University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 6 - Practice Exercises - Page 391: 20



Work Step by Step

We integrate in order to find the arc length: $$L= \int_{p}^{q} \sqrt {1+(y')^2}$$ $$ \\=\int_{1}^{2} \sqrt {1+\dfrac{y^4}{16}-\dfrac{1}{2} +\dfrac {1}{y^4}} \space dy \\=[\dfrac{y^3}{12}-\dfrac{1}{y}]_1^2 \\=[\dfrac{2^3}{12}-\dfrac{1^3}{12})-(\dfrac{1}{2}-\dfrac{1}{1})] \\= \dfrac{13}{12}$$
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.