Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Section 6.3 - Arc Length - Exercises 6.3 - Page 335: 6



Work Step by Step

The formula to calculate the arc length is as follows: $L=\int_m^n \sqrt {1+[f'(x)]^2} dx$ Re-write the equation as follows: $[f'(x)]^2=\dfrac{(y-1)^2}{4y}$ This implies that $L=\int_{2}^{3} [1+\dfrac{(y^4-1)^2}{4y^4}] dy $ Separate the terms and integrate as follows: $L=\int_{2}^{3} (y^2/2)+\dfrac{y^{-2}}{2} dy \\=[ \dfrac{y^3}{6}-\dfrac{1}{2y} ]_2^{3}=\dfrac{13}{4}$
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.