Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 974: 3



Work Step by Step

Given: $z=e^{x-y}$, $(2,2,1)$ Consider $f(x,y)= e^{x-y}$ $f_{x}(x,y)=e^{x-y}$ $f_{y}(x,y)=-e^{x-y}$ At $(2,2,1)$ $f_{x}(2,2)=e^{2-2}=1$ $f_{y}(2,2)=-e^{2-2}=-1$ The equation of the tangent plane to the given surface at the specified point $(2,2,1)$ 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 $z-1=1(x-2)-1(y-2)$ $z=x-2-y+2+1$ Hence, $z=x-y+1$
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