## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises - Page 974: 7

#### Answer

$z=3x+7y-5$

#### Work Step by Step

$z=x^{2}+xy+3y^{2}$, $(1,1,5)$ Consider $f(x,y)= x^{2}+xy+3y^{2}$ $f_{x}(x,y)=2x+y$ $f_{y}(x,y)=x+6y$ At $(1,1,5)$ $f_{x}(1,1)=2.1+1=3$ $f_{y}(1,1)=1+6.1=7y$ The equation of the tangent plane to the given surface at the specified point $(1,1,5)$ 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 $z-5=3(x-1)+7(y-1)$ $z=3x-3+7y-7+5$ Hence,$z=3x+7y-5$

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