Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.4 - The Chain Rule - Exercises 14.4 - Page 817: 26



Work Step by Step

Consider $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ Plug the derivatives in equation $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ This implies that $\dfrac{(y-3)}{(x+2y)} (-1)=\dfrac{3-y}{(x+2y)} $ For point $(-1,1)$, we have $\dfrac{3-y}{x+2y} =\dfrac{3-1}{-1+2(1)}=2$
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