University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.6 - The Chain Rule - Exercises: 28

Answer

$\frac{dr}{d \theta} = 9 (sec \theta - tan \theta)^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$ $\frac{dr}{d \theta} = -9 sec \theta (sec \theta - tan \theta)^{\frac{3}{2}}$

Work Step by Step

$r =6(u)^{\frac{3}{2}}$ where, $u= (sec \theta - tan \theta)$ Now, $\frac{dr}{du} = 9 u^{\frac{1}{2}}$ and, $\frac{du}{d\theta} = sec \theta \times tan \theta - sec^{2} \theta = sec \theta (tan \theta - sec \theta )$ So, $\frac{dr}{d\ theta} = \frac{dr}{du} \times \frac{du}{d \theta}$ $\frac{dr}{d \theta} = 9 u^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$ $\frac{dr}{d \theta} = 9 (sec \theta - tan \theta)^{\frac{1}{2}} (sec \theta (tan \theta - sec \theta ))$ $\frac{dr}{d \theta} = -9 sec \theta (sec \theta - tan \theta)^{\frac{3}{2}}$
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.