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: 5

Answer

$x+y+z=0$

Work Step by Step

$z=x+sin(x+y)$, $(-1,1,0)$ Consider $f(x,y)= x+sin(x+y)$ $f_{x}(x,y)=sin(x+y)+xcos(x+y)$ $f_{y}(x,y)=xcos(x+y)$ At $(-1,1,0)$ $f_{x}(-1,1)=sin(-1+1)+1.cos(-1+1)=-1$ $f_{y}(x,y)=-1cos(-1+1)=-1$ The equation of the tangent plane to the given surface at the specified point $(-1,1,0)$ 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 $z-0=-1(x+1)-1(y-1)$ $z=-x-1-y+1$ Hence, $x+y+z=0$
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