Calculus: Early Transcendentals 9th Edition

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

Chapter 16 - Section 16.5 - Curl and Divergence - 16.5 Exercise - Page 1169: 25

Answer

$div (F+G)=div F+div G$

Work Step by Step

Suppose $F=ai+b j$ $div F=\dfrac{\partial a}{\partial x}+\dfrac{\partial b}{\partial y}$ Suppose $F=a_1i+b_1j; G=a_2i+b_2j$ Now, $div (F+G)=\dfrac{\partial a_1}{\partial x}+\dfrac{\partial b_1}{\partial y}+\dfrac{\partial a_2}{\partial x}+\dfrac{\partial b_2}{\partial y}$ Hence, we get $div (F+G)=div F+div G$
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