Calculus 8th Edition

Published by Cengage

Chapter 14 - Partial Derivatives - 14.4 Tangent Planes and Linear Approximations - 14.4 Exercises: 16

Answer

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

Work Step by Step

Given that f is a differentiable function with $f(x,y)=y+sin(\frac{x}{y})$ at $(0,3)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=y+sin(\frac{x}{y})$ $f_{x}(a,b)=\frac{1}{y}+cos(\frac{x}{y})$ $f_{y}(a,b)=1-\frac{x}{y^{2}}cos(\frac{x}{y})$ At $(0,3)$ $f(0,3)=3+sin(\frac{0}{3})=3$ $f_{x}(0,3)=\frac{1}{3}+cos(\frac{0}{3})=\frac{1}{3}$ $f_{y}(a,b)=1-\frac{0}{3^{2}}cos(\frac{0}{3})=1$ The linearization $L(x,y)$ of function at $(0,3)$ is $L(x,y)= f(0,3)+f_{x}(0,3)(x-0)+f_{y}(0,3)(y-3)$ $=3+\frac{1}{3}(x-0)+1(y-3)$ $=3+\frac{x}{3}+y-3$ $=\frac{x}{3}+y$ Hence, the linearization $L(x,y)$ of the function at $(1,3)$ is $\frac{x}{3}+y$

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