Part2: Linearization and Tangent Planes

Definition 5.1 can be restated using the fact that Df = f( x) -f( p) and Dx = x-p. Specifically, we obtain the following:   

Definition 5.1a: A function f( x,y) is differentiable at a point ( p,q) if there exists a vector Ñf( p) for which

lim x® p     | f( x)-f( p) -Ñf( p) ·( x-p) | || x-p || = 0

(4)

When it exists, Ñf( p) is the total derivative of f( x) at p.

The definition of the limit then implies that e > 0, there is a neighborhood of p for which

| f( x) -[ f( p)+Ñf( p) ·( x-p) ] | < e || x-p ||

(5)

for all x in that neighborhood. Consequently, if we define the linearization of f( x) at p to be

Lp( x) = f( p) +Ñf( p) ·( x-p)

then | f( x) -L_p( x)| < e| | x-p|| , which means that L_p( x) is practically the same as f( x) on some neighborhood of p.

Since Ñf( p) = [ f_x( p,q),f_y( p,q) ] and x-p = [ x-p,y-q]^t, we have

Ñf( p) ·( x-p) = á fx( p,q) ,fy( p,q) ñ· á x-p,y-q ñ = fx( p,q) ( x-p) +fy( p,q) (y-q)

Thus, the linearization in non-vector form is given by

Lp( x,y) = f( p,q) +fx( p,q)( x-p) +fy( p,q) ( y-q)

The graph of L_p( x,y) is a plane that is practically the same as the surface z = f( x,y) near the point ( p,q,f( p,q) ) .
We say that z = L_p( x,y) is the tangent plane to z = f( x,y) at ( p,q) .     

EXAMPLE 2    Find the tangent plane to f( x,y) = 9-x²-y² when ( x,y) = ( 1,1) .       

Solution: To begin with, f_x( x,y) = -2x and f_y( x,y) = -2y, so that f_x( 1,1) = f_y(1,1) = -2 and

L( x,y) = f( 1,1) +fx( 1,1) (x-1) +fy( 1,1) ( y-1) = 7-2( x-1) -2( y-1)

which simplifies to L( x,y) = 11-2x-2y. Thus, the equation of the tangent plane to f( x,y) at the point ( 1,1,7) is

z = 7-2( x-1) -2( y-1)

which is shown below.

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The tangent plane is the plane that best approximates a surface in the neighborhood of a point. However, a tangent plane may intersect a surface in an infinite number of points. Thus, intuitive working definitions of tangency may not be tenable for functions of 2 variables.

EXAMPLE 3    Find the linearization of f( x,y) = x³-xy² at the point ( 1,2) .       

Solution: Since f_x = 3x²-y² and f_y = -2xy, we have

fx( 1,2) = 3-4 = -1,        fy( 1,2) = -2·1·2 = -4

Since f( 1,2) = -3, the linearization of f( x,y) =x³-xy² at ( 1,2)  is

L( x,y) = -3-1( x-1) -4( y-2)

which simplifies to L( x,y) = 6-x-4y. The graph of L(x,y) is shown versus the graph of f( x,y) below:

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