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Find the differential d z of the function f ( x , y ) = 4 y 2 + x 2 y 2 x y and use it to approximate Δ z at point ( 1 , −1 ) . Use Δ x = 0.03 and Δ y = −0.02 . What is the exact value of Δ z ?

d z = 0.18 Δ z = f ( 1.03 , −1.02 ) f ( 1 , −1 ) = 0.180682

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Differentiability of a function of three variables

All of the preceding results for differentiability of functions of two variables can be generalized to functions of three variables. First, the definition:

Definition

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

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

where the error term E satisfies

lim ( x , y , z ) ( x 0 , y 0 , z 0 ) E ( x , y , z ) ( x x 0 ) 2 + ( y y 0 ) 2 + ( z z 0 ) 2 = 0 .

If a function of three variables is differentiable at a point ( x 0 , y 0 , z 0 ) , then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.

Key concepts

  • The analog of a tangent line to a curve is a tangent plane to a surface for functions of two variables.
  • Tangent planes can be used to approximate values of functions near known values.
  • A function is differentiable at a point if it is ”smooth” at that point (i.e., no corners or discontinuities exist at that point).
  • The total differential can be used to approximate the change in a function z = f ( x 0 , y 0 ) at the point ( x 0 , y 0 ) for given values of Δ x and Δ y .

Key equations

  • Tangent plane
    z = f ( x 0 , y 0 ) + f x ( x 0 , y 0 ) ( x x 0 ) + f y ( x 0 , y 0 ) ( y y 0 )
  • Linear approximation
    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 )
  • Total differential
    d z = f x ( x 0 , y 0 ) d x + f y ( x 0 , y 0 ) d y .
  • Differentiability (two variables)
    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 .
  • Differentiability (three variables)
    f ( x , y ) = f ( x 0 , y 0 , z 0 ) + f x ( x 0 , y 0 , z 0 ) ( x x 0 ) + f y ( x 0 , y 0 , z 0 ) ( y y 0 ) + f z ( x 0 , y 0 , z 0 ) ( z z 0 ) + E ( x , y , z ) ,
    where the error term E satisfies
    lim ( x , y , z ) ( x 0 , y 0 , z 0 ) E ( x , y , z ) ( x x 0 ) 2 + ( y y 0 ) 2 + ( z z 0 ) 2 = 0 .

For the following exercises, find a unit normal vector to the surface at the indicated point.

f ( x , y ) = x 3 , ( 2 , −1 , 8 )

( 145 145 ) ( 12 i k )

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ln ( x y z ) = 0 when x = y = 1

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For the following exercises, as a useful review for techniques used in this section, find a normal vector and a tangent vector at point P .

x 2 + x y + y 2 = 3 , P ( −1 , −1 )

Normal vector: i + j , tangent vector: i j

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( x 2 + y 2 ) 2 = 9 ( x 2 y 2 ) , P ( 2 , 1 )

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x y 2 2 x 2 + y + 5 x = 6 , P ( 4 , 2 )

Normal vector: 7 i 17 j , tangent vector: 17 i + 7 j

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2 x 3 x 2 y 2 = 3 x y 7 , P ( 1 , −2 )

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z e x 2 y 2 3 = 0 , P ( 2 , 2 , 3 )

−1.094 i 0.18238 j

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For the following exercises, find the equation for the tangent plane to the surface at the indicated point. ( Hint: Solve for z in terms of x and y . )

−8 x 3 y 7 z = −19 , P ( 1 , −1 , 2 )

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z = −9 x 2 3 y 2 , P ( 2 , 1 , −39 )

−36 x 6 y z = −39

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x 2 + 10 x y z + y 2 + 8 z 2 = 0 , P ( −1 , −1 , −1 )

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z = ln ( 10 x 2 + 2 y 2 + 1 ) , P ( 0 , 0 , 0 )

z = 0

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z = e 7 x 2 + 4 y 2 , P ( 0 , 0 , 1 )

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x y + y z + z x = 11 , P ( 1 , 2 , 3 )

5 x + 4 y + 3 z 22 = 0

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x 2 + 4 y 2 = z 2 , P ( 3 , 2 , 5 )

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x 3 + y 3 = 3 x y z , P ( 1 , 2 , 3 2 )

4 x 5 y + 4 z = 0

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z = a x y , P ( 1 , 1 a , 1 )

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z = sin x + sin y + sin ( x + y ) , P ( 0 , 0 , 0 )

2 x + 2 y z = 0

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h ( x , y ) = ln x 2 + y 2 , P ( 3 , 4 )

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z = x 2 2 x y + y 2 , P ( 1 , 2 , 1 )

−2 ( x 1 ) + 2 ( y 2 ) ( z 1 ) = 0

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For the following exercises, find parametric equations for the normal line to the surface at the indicated point. (Recall that to find the equation of a line in space, you need a point on the line, P 0 ( x 0 , y 0 , z 0 ) , and a vector n = a , b , c that is parallel to the line. Then the equation of the line is x x 0 = a t , y y 0 = b t , z z 0 = c t . )

Questions & Answers

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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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