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: 14

Answer

$-x+\frac{y}{2}+\frac{3}{2}$

Work Step by Step

Given that f is a differentiable function with $f(x,y)=\frac{1+y}{1+x}$ at $(1,3)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=\frac{1+y}{1+x}$ $f_{x}(a,b)=\frac{-1-y}{(1+x)^{2}}$ $f_{y}(a,b)=\frac{1}{1+x}$ At $(1,3)$ $f(1,3)=\frac{1+3}{1+1}=2$ $f_{x}(1,3)=\frac{-1-3}{(1+1)^{2}}=-1$ $f_{y}(1,3)=\frac{1}{1+1}=\frac{1}{2}$ The linearization $L(x,y)$ of function at $(1,3)$ is $L(x,y)= f(1,3)+f_{x}(1,3)(x-1)+f_{y}(1,3)(y-3)$ $=2+(-1)(x-1)+\frac{1}{2}(y-3)$ $=2-x+1+\frac{y}{2}-\frac{3}{2}$ $=-x+\frac{y}{2}+\frac{3}{2}$ Hence, the linearization $L(x,y)$ of the function at $(1,3)$ is $-x+\frac{y}{2}+\frac{3}{2}$
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