University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 189: 36

Answer

1 rad/sec

Work Step by Step

We have: $y=x^2$ $\tan\theta=\frac{y}{x}$ Using the Pythagorean Theorem: $\tan\theta =\frac{x^2}{x}=x$ and given $dx=10\frac{m}{sec}$ on differentiating the above: $\sec^2\theta d\theta=dx$ $d\theta=\frac{dx}{ \sec^2\theta}=\frac{10}{(\frac{\sqrt(90)}{3})^2}=1 rad/sec$ Thus, the final answer is: 1 rad/sec
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.