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: 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, $x+y+z=0$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.