Thomas' Calculus 13th Edition

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

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 20

Answer

$4\sqrt 3$

Work Step by Step

Here,we have $L=\int_{m}^{n}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt$ This gives: $L=\int_{-\sqrt 3}^{\sqrt 3}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}dt=\int_{-\sqrt 3}^{\sqrt 3} \sqrt{(t^2+1)^2} dt$ This implies that $L=[\dfrac{t^3}{3}+t]_{-\sqrt 3}^{\sqrt 3}=4\sqrt 3$
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.