Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 14 - Section 14.5 - The Chain Rule. - 14.5 Exercise - Page 993: 47

Answer

$\dfrac{-1}{(12\sqrt 3)} rad/s$

Work Step by Step

We need to use the chain rule. $\dfrac{dA}{dt}=\dfrac{\partial A}{\partial x}\dfrac{dx}{ dt}+\dfrac{\partial A}{\partial y}\dfrac{dy}{ dt}+\dfrac{\partial A}{\partial \theta}\dfrac{d \theta}{ dt}$ Re-write as: $\dfrac{d \theta}{ dt}=-\dfrac{(\dfrac{\partial A}{\partial x}) \times (\dfrac{dx}{ dt})+(\dfrac{\partial A}{\partial y}) \times (\dfrac{dy}{ dt})}{(\dfrac{\partial A}{\partial \theta})}$ $=-\dfrac{y \sin \theta \times (\dfrac{dx}{dt})+(x \sin \theta) \times (\dfrac{dy}{ dt})}{(xy \cos \theta)}$ $=-\dfrac{(30) \times (0.5) \times (3)+(20) \times (0.5)(-2)}{(3) \times (20)(\sqrt 3/2)} $ Hence, we get $\dfrac{d \theta}{ dt}=\dfrac{-1}{(12\sqrt 3)} rad/s$
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