   Chapter 14.4, Problem 20E

Chapter
Section
Textbook Problem

Find the linear approximation of the function f(x, y) = 1 − xy cos πy at (1, 1) and use it to approximate f(1.02, 0.97). Illustrate by graphing f and the tangent plane.

To determine

To find: The linearization of the function f(x,y)=1xycosπy at the point (1,1) and use it approximate the value of f(1.02,0.97) ;illustrate using the graph and the tangent plane.

Explanation

Given:

The function is, f(x,y)=1xycosπy .

The point is P(a,b)=(1,1) .

Theorem used:

“If the partial derivatives fx and fy exist near (a,b) and are continuous at (a,b) , then f is differentiable at (a,b) .”

Result used:

“Linear approximation of the function f(x,y) at (a,b) is f(x,y)f(a,b)+fx(a,b)(xa)+fy(a,b)(yb)

“Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z=f(x,y) at the point P(x0,y0,z0) is zz0=fx(x0,y0)(xx0)+fy(x0,y0)(yy0)

Calculation:

Take partial derivativewith respect to x in the function f(x,y)=1xycosπy ,

fx(x,y)=0(1)ycosπy=ycosπy

Take partial derivative with respect to y in the function f(x,y)=1xycosπy ,

fy(x,y)=0x[y(sinπy)(π)+(cosπy)(1)][y(uv)=u'v+uv']=πxysinπyxcosπy

Obatin the partial derivative of x at the point (1, 1),

fx(a,b)=fx(1,1)=1(cosπ.1)=1(1)[cos(π)=1]=1

Obtain the partial derivative of y at the point (1,1) ,

fy(a,b)=fy(1,1)=π(1)(1)sin(π.1)1cos(π

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