Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.4 - Tangent Planes and Linear Approximation - 14.4 Exercise - Page 934: 6


$z=x-2y-1$ OR $x-2y-z-1=0$

Work Step by Step

$z=ln(x-2y)$, $(3,1,0)$ Consider $f(x,y)= ln(x-2y)$ $f_{x}(x,y)=\frac{1}{x-2y}$ $f_{y}(x,y)=\frac{-2}{x-2y}$ At $(3,1,0)$ $f_{x}(3,1)=\frac{1}{3-2.1}=1$ $f_{y}(3,1)=\frac{-2}{3-2.1}=-2$ The equation of the tangent plane to the given surface at the specified point $(3,1,0)$ is given by $z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$ On substituting the values, we get $z-0=1(x-3)+(-2)(y-1)$ $z=x-3-2y+2$ Hence, $z=x-2y-1$ OR $x-2y-z-1=0$
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