For student 4 (functions of several variables)

Section 11.1: Functions of Several Variables

Functions of More Than One Variable

So far most of our experience has been working with functions of one variables.

Some examples are: f(x) x2, g(x)lnx, h(x)ex. In this section, we want to examine functions and

variables of multivariate equations and functions, like z f(x,y) x2 4xy or

2 ) , , (

z xe z y x g

y

 .

Example 1: Given f(x,y) x2 4xy, find f(1,2).

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Example 2: Given

2 ) , , (

z xe z y x g

y

Section 11.2: Limits and Continuity

Limits of Functions of Two Variables

A limit of a function of two variables z = f (x, y) as (x, y) approaches a specific ordered pair.

Notation: We write

L y x f b a y x



 ( , )

lim

) , ( ) , (

For a limit to exist, the function of 2 variables z = f (x, y) must approach the same z value as (x, y)

approaches (a, b) along all paths on the x-y coordinate plane. If a function of two variables is defined at a point, we can immediately substitute to find the limit.

Example 1: Evaluate lim 5 3 2 1 )

0 , 1 ( ) ,

(xy  x xy y  , if it exists, or show that the limit does not exist.

Continuity

A function of two variables f (x, y) is continuous at (a, b) if

) , ( ) , ( lim

) , ( ) ,

(x yab f x y  f a b

Note: To find where a function of 2 variables is continuous, often it suffices to find where the function is defined.

Example 5: Determine the set of points where the function

y x

x y

x f

  ) ,

Section 11.3: Partial Derivatives

Partial Derivatives

Given a function of two variables z = f (x, y). Then

h y x f y h x f y x f

x x h

) , ( ) , ( lim ) , ( respect to with Derivative Partial 0      h y x f h y x f y x f

y y h

) , ( ) , ( lim ) , ( respect to with Derivative Partial 0     

Notations For Partial Derivatives

Given z = f (x, y).

. respect to with derivative partial ) , ( ) , ( x x z x f y x f y x f

x x  

        . respect to with derivative partial ) , ( ) , ( y y z y f y x f y x f

y y  

        ) , ( point at the evaluated s derivative partial ) , ( ) , ( ) , ( ) , ( b a b a f dy z b a f dx z y b a x b a            

*Note: In partial differentiation, we treat every variable as a constant except for the one we are differentiating with respect to.

Example 2: Find the first partial derivatives of the function f(x,y)4x3y2 5x3 x2ey.

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Example 3: Find the first partial derivatives of the function f(x,y)ln(x2  y2)ex2y2.

Example 4: Find f_x(1,1) and f_y(1,1) for

2 2 ) , (

y x

xy y

x f

Note: We can also differentiate functions of more than 2 variables.

Example 5: Find the first partial derivatives of the function

) sin(

) , ,

(x y z x2 y2 z2 x2 y2

f      .

Second Order Partial Derivatives

Just as we can find second order derivatives for functions of one variable, we can do the same for functions of two variables.

Notation for Second Order Derivatives

again respect to then with , respect to with partial Take : ) , ( 2 2 2 2 x x z z x f x f dx y x f_xx                 . respect to then with , respect to with partial Take : ) , ( 2 2 y x x y z x y f x f dy y x f_yx                   . respect to then with , respect to with partial Take : ) , ( 2 2 x y y x z y x f y f dx y x f_xy                   again respect to then with , respect to with partial Take : ) , ( 2 2 2 2 y y y z y f y f dy y x f_yy                

Note: In general,

) , ( )

,

(x y f x y

The Directional Derivative

Recall that direction in the )) , ( , , ( point at the surface the to line tangent the of Slope ) , ( ) ,

( a b f a b x

x z b a f b a

x _ 

  direction in the )) , ( , , ( point at the surface the to line tangent the of Slope ) , ( ) ,

( a b f a b y

y z b a f b a

y _ 

 

Instead of restricting ourselves to the x and y axis, suppose we want to find a method for finding the slope of the surface in any desired direction.

Let u = < a, b > be the unit vector (a vector of length one) on the x-y plane which indicates the direction we are moving. Then we define the following:

Definition of the Directional Derivative

The directional derivative of a function z = f (x, y) in the direction of the unit vector

u = < a, b >, denoted by D_uf(x,y), is defined the be the following:

b y x f a y x f y x f

D_u ( , ) _x( , )  _y( , )

x

y

) 0 , , (x₀ y₀

z

0 x 0 y ) , (x y f z

) , , (x₀ y₀ z₀

dir in Slope )

,

(x₀ y₀ x

f_x 

dir in Slope )

,

(x₀ y₀ y

f_y 

dir in Slope )

,

(x₀ y₀ u

D_u 

, 

a b u

Notes

1. Geometrically, the directional derivative is used to calculate the slope of the surface

z = f (x, y). That is, to calculate the slope of the surface at the point (x₀,y₀,z₀), where z₀  f(x₀,y₀), we compute the following:

b y x f a y x f y x f D b a vector unit of direction in z y x po at Surface of Slope y

x( , ) ( , ) ) , ( . ) , , ( int 0 0 0 0 0 0 0 0

0 _{  }



 u

u

2. The vector u = < a, b > must be a unit vector. If we want to compute the directional derivative of a function in the direction of the vector vand v is not a unit vector, we compute v v | v | v u | | 1   .

3. The direction of the unit vector u can be expressed in terms of the angle  between the vector u and the x-axis. In this case, ucos,sin  (note, u is a unit vector since |u| cos2 sin2  11) and the directional derivative can be expressed as

  ( , )sin cos ) , ( ) ,

(x y f x y f x y f

D_u  _x  _y .

Example 2: Find the directional derivative of the function f(x,y)3y4xy6x at the point (1, 2) in the

direction of the unit vector that makes an angle of 3 

  radians with the x-axis.

Example 2: Find the directional derivative of the function f(x,y)3y4xy6x at the point (-3, -4) in the direction of the vector v2i3j.

Gradient of a Function

Given a function of two variables z = f (x, y), the gradient vector, denoted by f(x,y), is a vector in the x-y

plane denoted by

j

i ( , )

) , ( ) ,

(x y f x y f x y

f  _x  _y



FactsaboutGradients

1. The directional derivative of the function z = f (x, y) in the direction of the unit vector

u = < a, b > can be expressed in terms of gradient using the dot product. That is,

b y x f a y x f b a y x f y x f y x f y x f D y x y x ) , ( ) , ( , ) , ( ), , ( ) , ( ) , (           u u

2. The gradient vector f(x,y) gives the direction of maximum increase of the surface

z = f (x, y). The length of the gradient vector is the maximum value of the directional derivative (the maximum rate of change of f). That is,

| ) , ( | ) ,

(x y f x y f D Derivative l Directiona the of Value

3. The negation of the gradient vector f(x,y) gives the direction of maximum decrease.

of the surface z = f (x, y). The negation of the length of the gradient vector is the minimum value of the directional derivative. That is,

| ) , ( | ) ,

(x y f x y f

D Derivative l

Directiona the

of Value

Example 3: Given the function f(x,y) ycos(x y). a. Find the gradient of f

b. Evaluate the gradient at the point P ,0) 3 ( .

c. Use the gradient to find a formula for the directional derivative of f in the direction of the vector 

 

5 4 , 5 3

u . Use the result to result to find the rate of change of f at P in the direction of the vector u.

Directional Derivative and Gradient for Functions of 3 variables

The directional derivative of a function f (x, y, z) of 3 variables in the direction of the unit vector u = < a, b, c >, denoted by D_uf(x,y,z), is defined to be the following:

c z y x f b z y x f a z y x f z y x f

D_u ( , , ) _x( , , )  _y( , , )  _z( , , )

The gradient vector, denoted by f(x,y,z), is a vector denoted by

k j

i ( , , ) ( , , )

) , , ( ) , ,

(x y z f x y z f x y z f x y z

f  _x  _y  _z



Example 4: Find the gradient and directional derivative of f(x,y,z)5x23xyxyzat P(1, 2, 4) in the direction of the point Q(-3, 1, 2).

Example 5: Find the maximum rate of change of f(x,y,z)5x23xy xyzat the point (1, 2, 4) and the direction in which it occurs.

Recall that z = f (x, y) gives a 3D surface in space. We want to form the following functions of 3 variables

z y x f z y x

F( , , ) ( , )

Note that the function F is obtained by moving all terms to one side of an equation and setting them equal to zero. We use the following basic fact.

Fact: Given a point (x₀,y₀,z₀) on a surface, the gradient of F at this point

k j

i ( , , ) ( , , )

) , , ( ) , ,

(x₀ y₀ z₀ F x₀ y₀ z₀ F x₀ y₀ z₀ F x₀ y₀ z₀

F  _x  _y  _z



is a vector orthogonal (normal) to the surface z f(x,y).

Example 6: Find a unit normal vector to the surface x2 y2 z2 9 at the point (2, 1, 2)

Tangent Plane and Normal Line Equations to a Surface

Given a surface z = f (x, y) in 3D, form the function F(x,y,z) f(x,y)zof three variables. Then the equation of the tangent plane to the surface z = f (x, y) at the point (x₀,y₀,z₀) is given by

0 ) )(

, , ( ) )(

, , ( ) )(

, ,

(x₀ y₀ z₀ xx₀ F x₀ y₀ z₀ yy₀ F x₀ y₀ z₀ zz₀ 

F_{x y z} .

The parametric equations of the normal line through the point (x₀,y₀,z₀) are given by

t z y x F x

x ₀  _x( ₀, ₀, ₀) , y  y₀ F_y(x₀,y₀,z₀)t, z z₀ F_z(x₀,y₀,z₀)t

Note: Recall that to find the symmetric equations of a line, take the parametric equations, solve for t, and set the results equal.

Example 7: Find the equation of the tangent plane and the parametric and symmetric equations for the normal line to the surface x2  y2z2 9 at the point (2, 1, 2).

Solution:

Section 11.7: Maximum and Minimum Values

Practice HW from Larson Textbook (not to hand in) p. 717 # 7-13 odd, 23-27 odd, 45, 47

Recall that for functions of one variable of the form y = f (x), informally a local maximum represents a point on the graph where the graph changes from increasing to decreasing while a local minimum is a point where the graph changes from decreasing to increasing. We want to extend this concept to functions of two variables.

Local Maximum and Minimum Points for Functions of Two Variables

Definition: A function z = f (x, y) of two variables has a 1. local maximum (relative maximum) at a point (a, b) if

) , ( ) ,

(x y f a b

f  for x near a and y near b.

2. local minimum (relative minimum) at a point (a, b) if

) , ( ) ,

(x y f a b

f  for x near a and y near b.

x

y

) , (a b

)) , ( , ,

(a b f a b

a

b

) , (x y f z

z

Local Minimum _fy(a,b)₀

) 0 ) ,

(a b 

At both the relative maximum and relative minimum points, the slope in the x direction at (a, b) is zero. That is,

Slope of tangent of f at (a,b) = f_x(a,b)0

Similarly, in the y direction,

Slope of tangent of f at (a,b) = f_y(a,b)0

This gives the following.

First Derivative Test for a Function of Two Variables

Is z = f (x, y) has either a local (relative) maximum or a local (relative) minimum at (x, y) = (a, b) , then

0 ) , (a b 

f_x and f_y(a,b)0

or f_x(a,b) and f_y(a,b) does not exist.

x

y

) , (a b

)) , ( , ,

(a b f a b

a

b

z

) , (x y f z 

Local Maximum _fy(a,b)₀₎ ) 0 ) ,

(a b 

Notes

1. Local maximum and/or local minimum points may or may not give absolute maximum and/or minimum points.

2. The point (a, b) where f_x(a,b)0 and f_y(a,b)0 or f_x(a,b) and f_y(a,b) does not exist is called a critical point. Critical points only give candidates for local maximum and minimum points.

For functions of two variables, there is an analog for the second derivative test using concavity for one variable functions.

Second Derivative Test for Functions of Two Variables (D Test)

Suppose f (x, y) is a function and (a, b) is a point where f_x(a,b)0 and f_y(a,b)0

Let

2

)] , ( [ ) , ( ) , ( )

,

(x y f a b f a b f a b

D  _xx  _yy  _xy

1. If D(a,b)0and f_xx(a,b)0, then f has a local minimum at (a, b). 2. If D(a,b)0and f_xx(a,b)0, then f has a local maximum at (a, b).

3. If D(a,b)0, then f has neither a local maximum or local minimum. The function f has a saddle point at (a, b).

4. If D(a,b)0, the test fails.

Example 1: Determine the local maximum, local minimum, and/or saddle points for the function 2

2

) ,

(x y x y

f   .

Solution:

Example 2: Determine the local maximum, local minimum, and/or saddle points for the function

13 10 8 5 )

,

(x y  x2 y2 x y

f .

Solution:

Example 3: Determine the local maximum, local minimum, and/or saddle points for the function 2

3

9 18

) ,

(x y xy x y

f    .