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Definition

Given a function z = f ( x , y ) with continuous partial derivatives that exist at the point ( x 0 , y 0 ) , the linear approximation    of f at the point ( x 0 , y 0 ) is given by the equation

L ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x x 0 ) + f y ( x 0 , y 0 ) ( y y 0 ) .

Notice that this equation also represents the tangent plane to the surface defined by z = f ( x , y ) at the point ( x 0 , y 0 ) . The idea behind using a linear approximation is that, if there is a point ( x 0 , y 0 ) at which the precise value of f ( x , y ) is known, then for values of ( x , y ) reasonably close to ( x 0 , y 0 ) , the linear approximation (i.e., tangent plane) yields a value that is also reasonably close to the exact value of f ( x , y ) ( [link] ). Furthermore the plane that is used to find the linear approximation is also the tangent plane to the surface at the point ( x 0 , y 0 ) .

A paraboloid with surface z = f(x, y). There is a point given on the paraboloid P (x0, y0) with a tangent plane at that point. There is a point on the plane which is marked as the linear approximation L(x, y) to f(x0, y0), which is close to the corresponding point on the paraboloid.
Using a tangent plane for linear approximation at a point.

Using a tangent plane approximation

Given the function f ( x , y ) = 41 4 x 2 y 2 , approximate f ( 2.1 , 2.9 ) using point ( 2 , 3 ) for ( x 0 , y 0 ) . What is the approximate value of f ( 2.1 , 2.9 ) to four decimal places?

To apply [link] , we first must calculate f ( x 0 , y 0 ) , f x ( x 0 , y 0 ) , and f y ( x 0 , y 0 ) using x 0 = 2 and y 0 = 3 :

f ( x 0 , y 0 ) = f ( 2 , 3 ) = 41 4 ( 2 ) 2 ( 3 ) 2 = 41 16 9 = 16 = 4 f x ( x , y ) = 4 x 41 4 x 2 y 2 so f x ( x 0 , y 0 ) = 4 ( 2 ) 41 4 ( 2 ) 2 ( 3 ) 2 = −2 f y ( x , y ) = y 41 4 x 2 y 2 so f y ( x 0 , y 0 ) = 3 41 4 ( 2 ) 2 ( 3 ) 2 = 3 4 .

Now we substitute these values into [link] :

L ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x x 0 ) + f y ( x 0 , y 0 ) ( y y 0 ) = 4 2 ( x 2 ) 3 4 ( y 3 ) = 41 4 2 x 3 4 y .

Last, we substitute x = 2.1 and y = 2.9 into L ( x , y ) :

L ( 2.1 , 2.9 ) = 41 4 2 ( 2.1 ) 3 4 ( 2.9 ) = 10.25 4.2 2.175 = 3.875 .

The approximate value of f ( 2.1 , 2.9 ) to four decimal places is

f ( 2.1 , 2.9 ) = 41 4 ( 2.1 ) 2 ( 2.9 ) 2 = 14.95 3.8665 ,

which corresponds to a 0.2 % error in approximation.

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Given the function f ( x , y ) = e 5 2 x + 3 y , approximate f ( 4.1 , 0.9 ) using point ( 4 , 1 ) for ( x 0 , y 0 ) . What is the approximate value of f ( 4.1 , 0.9 ) to four decimal places?

L ( x , y ) = 6 2 x + 3 y , so L ( 4.1 , 0.9 ) = 6 2 ( 4.1 ) + 3 ( 0.9 ) = 0.5 f ( 4.1 , 0.9 ) = e 5 2 ( 4.1 ) + 3 ( 0.9 ) = e −0.5 0.6065 .

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Differentiability

When working with a function y = f ( x ) of one variable, the function is said to be differentiable at a point x = a if f ( a ) exists. Furthermore, if a function of one variable is differentiable at a point, the graph is “smooth” at that point (i.e., no corners exist) and a tangent line is well-defined at that point.

The idea behind differentiability of a function of two variables is connected to the idea of smoothness at that point. In this case, a surface is considered to be smooth at point P if a tangent plane to the surface exists at that point. If a function is differentiable at a point, then a tangent plane to the surface exists at that point. Recall the formula for a tangent plane at a point ( x 0 , y 0 ) is given by

z = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x x 0 ) + f y ( x 0 , y 0 ) ( y y 0 ) ,

For a tangent plane to exist at the point ( x 0 , y 0 ) , the partial derivatives must therefore exist at that point. However, this is not a sufficient condition for smoothness, as was illustrated in [link] . In that case, the partial derivatives existed at the origin, but the function also had a corner on the graph at the origin.

Definition

A function f ( x , y ) is differentiable    at a point P ( x 0 , y 0 ) if, for all points ( x , y ) in a δ disk around P , we can write

f ( x , y ) = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x x 0 ) + f y ( x 0 , y 0 ) ( y y 0 ) + E ( x , y ) ,

where the error term E satisfies

lim ( x , y ) ( x 0 , y 0 ) E ( x , y ) ( x x 0 ) 2 + ( y y 0 ) 2 = 0 .
Practice Key Terms 4

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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