14.1 Functions of Several Variables
Functions of Two Variables
ļ»f ( x, y ) | ( x, y ) ļ Dļ½.
We often write z = f(x, y) to make explicit the value taken on by f at the general point (x, y).
The variables x and y are independent variables and z is the dependent variable.
Example
Find the domain of š š„, š¦ =
9 − š„ 2 − š¦2.
14.1 Functions of Several Variables
Graph
z = ax + by + c
so it is a plane.
or
ax + by − z + c = 0
14.1 Functions of Several Variables
Level Curves and Contour Maps
Definition The level curves of a function f of two variables are the curves
with equations f (x, y) = k, where k is a constant (in the range of f ). A collection of level
curves is called a contour map.
Example
Estimate the values of š(1, 3) and š(4,5).
14.1 Functions of Several Variables
Functions of Three or More Variables
Example
Find the level surfaces of š š„, š¦, š§ = š„ 2 + š§ 2 .
14.2 Limits and Continuity
Limits of Functions of Two Variables
Definition Let f be a function of two variables whose domain D includes points arbitrarily
close to (a, b). Then we say that the limit of f (x, y) as (x, y) approaches (a, b) is L and
we write
lim
( x ,y )→( a,b )
f ( x, y ) = L
if ( x, y ) ļ D and 0 ļ¼ ( x − a )2 + ( y − b )2 ļ¼ ļ¤
Review
then
f ( x, y ) − L ļ¼ ļ„
Example
Prove
lim
(š„,š¦)→(š,š)
š„=š
Example
sin(š„2 +š¦2 )
2 2
(š„,š¦)→(0,0) š„ +š¦
lim
š„2 − š¦2
2 2
(š„,š¦)→(0,0) š„ +š¦
lim
14.2 Limits and Continuity
Showing That a Limit Does Not Exist
lim
( x ,y )→( a,b )
f ( x, y )
Examples
(1)
š„š¦
lim
(š„,š¦)→(0,0) š„ 2 +š¦ 2
š„š¦ 2
(2)
lim
(š„,š¦)→(0,0) š„ 2 +š¦ 4
š„ 2 −š¦ 2
(3)
lim
(š„,š¦)→(0,0) š„ 2 +š¦ 2
14.2 Limits and Continuity
Properties of Limits
Examples
(1)
š„š¦ 2
lim
(š„,š¦)→(0,0) š„ 2 +š¦ 2
(2)
lim
(š„,š¦)→(0,0)
š š„š¦ cos š„š¦
Example
Use polar coordinates to find the limit. [ If (š, š) are polar coordinates of the point (š„, š¦)
with š > 0, note that š → 0 + as š„, š¦ → 0,0 . ]
lim
(š„,š¦)→(0,0)
š„ 2 + š¦ 2 ln(š„ 2 + š¦ 2 )
14.2 Limits and Continuity
Continuity
Definition A function f of two variables is called continuous at (a, b) if
lim
( x ,y )→( a,b )
f ( x, y ) = f (a, b )
If f is a continuous function of two variables and g is a continuous function of a single vari
able that is defined on the range of f, then
the composite function h = g f
Example
Find ā š„, š¦ = š(š š„, š¦ ) and the set of points at which ā is continuous.
1−š„š¦
š š” = š” 2 , š š„, š¦ = 1+š„ 2
14.3 Partial Derivatives
Partial Derivatives of Functions of Two Variables
Definition If f is a function of two variables, its partial derivatives are the functions fx
and fy defined by
f ( x + h, y ) − f ( x, y )
h →0
h
fx ( x, y ) = lim
f ( x, y + h ) − f ( x, y )
h →0
h
fy ( x, y ) = lim
Notations for Partial Derivatives If z = f (x, y), we write
ļ¶f
ļ¶
ļ¶z
ļ¶f
ļ¶
ļ¶z
f
(
x
,
y
)
=
f
=
=
f
(
x
,
y
)
=
= f2 = D2f = Dy f
y
fx ( x, y ) = fx =
=
f ( x, y ) =
= f1 = D1f = Dxyf
ļ¶y ļ¶y
ļ¶y
ļ¶x ļ¶x
ļ¶x
Example
G(70 + h ) − G(70)
f (96, 70 + h ) − f (96, 70)
= lim
h →0
h →0
h
h
Gļ¢(70) = lim
Examples
1 š š„, š¦ = š„ š¦ 2
(2)
š š„, š¦ = š„ š¦
š¦
(3) š š„, š¦ = ā«š„׬ā¬
š” 3 + 1 šš”
14.3 Partial Derivatives
Interpretations of Partial Derivatives
14.3 Partial Derivatives
Implicit Differentiation
Example
Find
šš§
šš§
and
šš„
šš¦
if š§ is defined implicitly as a function of š„ and š¦ by the equation
š„ 3 + š¦ 3 + š§ 3 + 6š„š¦š§ = −4
and evaluate these partial derivatives at (1, −1, 2).
14.3 Partial Derivatives
Higher Derivatives
ļ¶ ļ¦ ļ¶f ļ¶ ļ¶ 2f ļ¶ 2 z
(fx )x = fxx = f11 =
=
=
ļ¶x ļ§ļØ ļ¶x ļ·ļø ļ¶x 2 ļ¶x 2
ļ¶ ļ¦ ļ¶f ļ¶
ļ¶ 2f
ļ¶2z
(fy )x = fyx = f21 =
=
=
ļ¶x ļ§ļØ ļ¶y ļ·ļø ļ¶xļ¶y ļ¶xļ¶y
ļ¶ ļ¦ ļ¶f ļ¶
ļ¶ 2f
ļ¶ 2z
(fx )y = fxy = f12 =
=
=
ļ¶y ļ§ļØ ļ¶x ļ·ļø ļ¶y ļ¶x ļ¶y ļ¶x
ļ¶ ļ¦ ļ¶f ļ¶ ļ¶ 2f
ļ¶ 2z
(fy )y = fyy = f22 =
=
=
ļ¶y ļ§ļØ ļ¶x ļ·ļø ļ¶y 2 ļ¶y 2
Clairaut's Theorem Suppose f is defined on a disk D that contains the point (a, b). If the
functions fxy and fyx are both continuous on D, then
fxy (a, b ) = fyx (a, b )
Example
š š„, š¦
š„ 3 š¦−š„š¦ 3
, (š„,š¦)≠(0,0)
= į š„ 2+š¦2
0, š„, š¦ = (0,0)
(1) Find šš„ (š„, š¦) and šš¦ (š„, š¦) when (š„, š¦) ≠ (0,0).
(2) Find šš„ (0,0) and šš¦ (0,0).
(3) Show that šš„š¦ 0,0 = −1 and šš¦š„ 0,0 = 1.
(4) Does the result of part(3) contradict Clairaut’s Theorem?
14.4 Tangent Planes and Linear Approximations
Tangent planes
š§ = šš„ (š„0 , š¦0 )(š„ − š„0 )+(š„0 , š¦0 )(š„ − š„0 )+š(š„0 , š¦0 )
14.4 Tangent Planes and Linear Approximations
Tangent planes
If the partial derivatives šš„ and šš¦ exist near (š, š) and are continuous
at š, š , then š is differentiable at š, š .
Example
Example
14.4 Tangent Planes and Linear Approximations
Linear Approximations
Example
14.4 Tangent Planes and Linear Approximations
Differentials
Example
14.5 The Chain Rule
Case1 and Case2
Example
(1)
(2)
(3)
(4)
14.5 The Chain Rule
Implicit Function Theorem
14.5 The Chain Rule
Implicit Function Theorem
Example
(1)
(2)
14.6 Directional Derivatives and the Gradient Vector
Directional Derivatives and Gradient
The directional derivative of š at (š„0 , š¦0 ) in the direction of a unit vector
u =< š, š > is
š š„0 +āš,š¦0 +āš −š(š„0 ,š¦0 )
ā
ā→0
š·š¢ š š„0 , š¦0 = lim
if this limit exists.
14.6 Directional Derivatives and the Gradient Vector
Directional Derivatives and Gradient
If š is a function of two variables š„ and š¦, then the gradient of š is the
vector function ∇š defined by
∇š =< šš„ , šš¦ >
True or False
If š·š® š(š„0 , š¦0 ) exists for every š®, then is š differentiable at š„0 , š¦0 .
Example
(1)
(2)
14.6 Directional Derivatives and the Gradient Vector
Maximizing the Directional Derivative
Example
(1)
(2)
14.6 Directional Derivatives and the Gradient Vector
Tangent Planes and Level Surfaces
Example
14.7 Maximum and Minimum Values
Local Maximum and Minimum Values
Examples Find the all critical points.
(1)
(2)
14.7 Maximum and Minimum Values
Local Maximum and Minimum Values
Example
14.8 Lagrange Multipliers
Lagrange Multipliers: One Constraint
14.8 Lagrange Multipliers
Lagrange Multipliers: One Constraint
Example
Example