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 - Page 159: 89



Work Step by Step

We have $s(\theta)=\cos\theta$ and $d\theta/dt=5$ According to the Chain Rule: $$\frac{ds}{dt}=\frac{ds}{d\theta}\frac{d\theta}{dt}=5\frac{ds}{d\theta}$$ $ds/d\theta$ refers to the derivative of $s(\theta)$, which is $s'(\theta)=-\sin\theta$. That means, $$\frac{ds}{dt}=-5\sin\theta$$ For $\theta=3\pi/2$: $$\frac{ds}{dt}=-5\sin\frac{3\pi}{2}=-5(-1)=5$$
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