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: 20

Answer

$1.99$

Work Step by Step

Given that the function is a differentiable function with $f(x,y)=1-xycos(\pi y)$ at $1,1)$ The linearization $L(x,y)$ of function at $(a,b)$ is given by $L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ $f(x,y)=1-xycos(\pi y)$ $f_{x}(a,b)=-ycos(\pi y)$ $f_{y}(a,b)=\pi xysin(\pi y)-xcos(\pi y)$ At $(1,1)$ $f(1,1)=1-(1)(1)cos(\pi (1))=1-(-1)=2$ $f_{x}(1,1)=-1cos(\pi (1))=1$ $f_{y}(1,1)=\pi (1)(1)sin(\pi (1))-1cos(\pi (1))=\pi(0)-1(-1)=1$ The linearization $L(x,y)$ of function at $(1,1)$ is $L(x,y)= f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$ $=2+1(x-1)+1(y-1)$ $=2+x-1+y-1$ $L(x,y)=x+y$ Therefore, the linearization $L(x,y)$ of the function $f(x,y)=1-xycos(\pi y)$at $(1.02,0.97)$ is $f(1.02,0.97)=1.02+0.97$ $=1.99$ The graph of function and tangent plane at (1, 1) is depicted below:
Small 1521552218
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.