Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.3 Arc Length and Curvature - 13.3 Exercises - Page 908: 23


$\dfrac{\sqrt 6}{2(3t^2+1)^2}$

Work Step by Step

Solve $r'(t)=2\sqrt6ti+2j+6t^2k$ ; $r''(t)=2\sqrt6i+12tk$ This yields, $|r'(t)|=\sqrt {(2\sqrt6t)^2+(2)^2+(6t^2)^2}$ or, $|r'(t)|=6t^2+2$ Write Theorem 10. $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (2\sqrt6ti+2j+6t^2k)\times(2\sqrt6i+12tk|}{|6t^2+2|^3}$ or, $=\dfrac{|24t-12\sqrt 6 t^2j-4\sqrt6 k|}{|6t^2+2|^3}$ or, $=\dfrac{\sqrt{(24)^2+(-12\sqrt 6 t^2)^2+(-4\sqrt6)^2}}{|6t^2+2|^3}$ or, $=\dfrac{\sqrt 6}{(6t^2+2)^3}$ or, $=\dfrac{\sqrt 6}{2(3t^2+1)^2}$
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.