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


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
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