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: 13

Answer

$ 2x+y-1$

Work Step by Step

Given that f is a differentiable function with $f(x,y)=x^{2}e^{y}$ at $(1,0)$ The linearization $L(x,y)$ of the function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=x^{2}e^{y}$ $f_{x}(a,b)=2xe^{y}$ $f_{y}(a,b)=x^{2}e^{y}$ At $(1,0)$ $f(1,0)=1^{2}e^{0}=1$ $f_{x}(1,0)=2(1)e^{0}=2$ $f_{y}(1,0)=1^{2}e^{0}=1$ The linearization $L(x,y)$ of function at $(1,0)$ is $L(x,y)= f(1,0)+f_{x}(1,0)(x-1)+f_{y}(1,0)(y-0)$ $=1+2(x-1)+1(y-0)$ $=1+2x-2+y$ $=2x+y-1$ Hence, the linearization $L(x,y)$ of the function at $(1,0)$ is $ 2x+y-1$
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