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 - Page 974: 12



Work Step by Step

Given that f is a differentiable function with $f(x,y)=\sqrt {xy}$ at $(1,4)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=\sqrt {xy}$ $f_{x}(a,b)=\frac{\sqrt y}{2\sqrt x}$ $f_{y}(a,b)=\frac{\sqrt x}{2\sqrt y}$ At $(1,4)$ $f(1,4)=\sqrt {1\times 4}=2$ $f_{x}(1,4)=\frac{\sqrt 4}{2\sqrt 1}=1$ $f_{y}(1,4)=\frac{\sqrt 1}{2\sqrt 4}=\frac{1}{4}$ The linearization $L(x,y)$ of function at $(1,4)$ is $L(x,y)= f(1,4)+f_{x}(1,4)(x-1)+f_{y}(1,4)(y-4)$ $=2+1(x-1)+\frac{1}{4}(y-4)$ $=2+x-1+\frac{y}{4}-1$ $=x+\frac{y}{4}$ Hence, the linearization $L(x,y)$ of the function at $(1,4)$ is $ x+\frac{y}{4}$
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