## Calculus 8th Edition

$1.99$
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: