Introduction to Functions of Several Variables

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Functions of Several Variables
Introduction to Functions of
Several Variables
Written by Karen Overman
Instructor of Mathematics
Tidewater Community College, Virginia Beach Campus
Virginia Beach, VA
With Assistance from a VCCS LearningWare Grant
Updated by Judith McKaig
Assistant Professor of Mathematics
Tidewater Community College
Norfolk, Virginia
In this lesson we will discuss:
o Notation of functions of several variables
o Domain of functions of several variables
o Graphs of functions of several variables
o Level curves for functions of several variables
Notation for Functions of Several Variables
Previously we have studied functions of one variable, y = f(x) in
which x was the independent variable and y was the dependent
variable. We are going to expand the idea of functions to include
functions with more than one independent variable. For example,
consider the functions below:
f x , y   2x 2  y 2
or
g x , y , z   2xe yz
or
h x1 , x 2 , x3, x 4   2x1  x2  4x3  x 4
Hopefully you can see the notation for functions of several
variables is similar to the notation you’ve used with single variable
functions.
z  f x , y   2x 2  y 2
or
w  g x , y , z   2xe yz
or
h x1 , x 2 , x3, x 4   2x1  x2  4x3  x 4
The function z = f(x, y) is a function of two variables. It has
independent variables x and y, and the dependent variable z.
Likewise, the function w = g(x, y, z) is a function of three
variables. The variables x, y and z are independent variables and
w is the dependent variable.
The function h is similar except there are four independent
variables.
When finding values of the several variable functions instead of
just substituting in an x-value, we will substitute in values for each
of the independent variables:
For example, using the function f on the previous slide, we will
evaluate the function f(x, y) for (2, 3) , (4, -3) and (5, y).
f x , y   2x 2  y 2
f 2,3  2  22  32  2  4  9  17
f 4,3  2  42   3  2  16  9  41
2
f 5, y   2  52  y 2  2  25  y 2  50  y 2
A Function of Two Variables: A function f of two variables x and y is
a rule that assigns to each ordered pair (x, y) in a given set D, called
the domain, a unique value of f.
Functions of more variables can be defined similarly.
The operations we performed with one-variable functions can also
be performed with functions of several variables.
For example, for the two-variable functions f and g:
f
 g x , y   f x , y   g x , y 
f  g x , y   f x , y   g x , y 
f 
f x , y 
 x , y  
, Provided g x , y   0
g


g
x
,
y
 
In general we will not
consider the composition of
two multi-variable functions.
Domains of Functions of Several Variables:
Unless the domain is given, assume the domain is the set of all
points for which the equation is defined.
For example, consider the functions
f x , y   3x 2  y 2
and
g x , y  
1
xy
The domain of f(x,y) is the entire xy-plane. Every ordered pair in
the xy-plane will produce a real value for f.
The domain of g(x, y) is the set of all points (x, y) in the xy-plane
such that the product xy is greater than 0. This would be all the
points in the first quadrant and the third quadrant.
Example 1: Find the domain of the function: f x , y   25  x 2  y 2
Solution: The domain of f(x, y) is the set of all points that satisfy the
inequality:
25  x  y  0
2
2
25  x 2  y 2
or
You may recognize that this is similar to the equation of a circle and
the inequality implies that any ordered pair on the circle or inside
2
2
the circle x  y  25 is in the domain.
y
The highlighted area is the
domain to f.
x
Example 2: Find the domain of the function:
g x , y , z   x 2  y 2  z 2  16
Solution:
Note that g is a function of three variables, so the domain is NOT an
area in the xy-plane. The domain of g is a solid in the 3-dimensional
coordinate system.
The expression under the radical must be nonnegative, resulting in
the inequality:
x 2  y 2  z 2  16  0 or x 2  y 2  z 2  16
This implies that any ordered triple outside of the sphere centered
at the origin with radius 4 is in the domain.
Example 3: Find the domain of the function: h x , y   lnxy 
Solution:
We know the argument of the natural log must be greater than zero.
So, x  y  0
This occurs in quadrant I and quadrant III. The domain is
highlighted below. Note the x-axis and the y-axis are NOT in the
domain.
y
x
Graphs of Functions of Several Variables
As you learned in 2-dimensional space the graph of a function can be
helpful to your understanding of the function. The graph gives an
illustration or visual representation of all the solutions to the
equation. We also want to use this tool with functions of two
variables.
The graph of a function of two variables, z = f(x, y), is the set of
ordered triples, (x, y, z) for which the ordered pair, (x, y) is in the
domain.
*The graph of z = f(x, y) is a surface in 3-dimensional space.
The graph of a function of three variables, w = f(x, y, z) is the set of
all points (x, y, z, w) for which the ordered triple, (x, y, z) is in the
domain.
*The graph of w = f(x, y, z) is in 4 dimensions.
We can’t draw this graph or the graphs of any functions with 3 or
more independent variables.
Example 4: Find the domain and range of the function and then
sketch the graph.
z  f x , y   25  x 2  y 2
Solution: From Example 1 we know the domain is all ordered pairs
(x, y) on or inside the circle centered at the origin with radius 5.
2
2
All ordered pairs satisfying the inequality: x  y  25
The range is going to consist of all possible outcomes for z.
The range must be nonnegative since z equals a principle square
2
2
root and furthermore, with the domain restriction: x  y  25 ,
the value of the radicand will only vary between 0 and 25.
Thus, the range is 0  z  5 .
Solution to Example 4 Continued:
Now let’s consider the sketch of the function:
z  25  x 2  y 2
2
2
2
z

25

x

y
Squaring both sides and simplifying:
x 2  y 2  z 2  25
You may recognize this equation from Chapter 7 - A sphere with
radius 5. This is helpful to sketching the function, but we must be
careful !!
2
2
The function z  25  x  y
2
2
2
x

y

z
 25
and the equation
are not exactly the same. The equation does NOT represent z as a
function of x and y – meaning there is not a unique value for z for
each (x, y). Keep in mind that the function had a range of 0  z  5 ,
which means the function is only the top half of the sphere.
As you have done before when sketching a surface in 3-dimensions
it may be helpful for you to use the traces in each coordinate plane.
1. The trace in the xy-plane, z = 0, is the equation:
0  25  x 2  y 2 or x 2  y 2  25
The circle centered at the
origin with radius 5 in the xyplane.
2. The trace in the yz-plane, x = 0, is the equation:
z  25  y 2 or y 2  z 2  25
The circle centered at the
origin with radius 5 in the yzplane.
3. The trace in the xz-plane, y = 0, is the equation:
z  25  x
2
or x  z  25
2
2
The circle centered at the
origin with radius 5 in the xzplane.
Along with sketching the traces in each coordinate plane, it may
be helpful to sketch traces in planes parallel to the coordinate
planes.
4. Let z = 3: 3  25  x 2  y 2 or x 2  y 2  16
So on the plane z = 3, parallel to the xy-plane, the trace is a circle
centered at (0,0,3) with radius 4.
5. Let z = 4: 4  25  x 2  y 2 or x 2  y 2  9
So on the plane z = 4, parallel to the xy-plane, the trace is a circle
centered at (0,0,4) with radius 3.
Here is a SKETCH with the three traces in the coordinate planes
and the additional two traces in planes parallel to the xy-plane.
Keep in mind that this is just a sketch. It is giving you a rough
idea of what the function looks like. It may also be helpful to use
a 3-dimensional graphing utility to get a better picture.
z=4
z=3
Here is a graph of the function using the 3-dimensional
graphing utility DPGraph.
z
y
x
Example 5: Sketch the surface: z  9  x 2  y 2
Solution: The domain is the entire xy-plane and the range is z  9 .
1. The trace in the xy-plane, z = 0, is the equation:
x2 y2  9
Circle
2. The trace in the yz-plane, x = 0, is the equation:
z  y 2  9
Parabola
3. The trace in the xz-plane, y = 0, is the equation:
z  x 2  9
Parabola
Solution to Example 5 Continued:
Traces parallel to the xy-plane include the following two.
4. The trace in the plane, z = 5, is the equation:
5  9x2 y2
or
x2 y2  4
Circle centered at (0, 0, 5) with
radius 2.
5. The trace in the plane, z = -7, is the equation:
7  9x2 y2
or
x  y  16
2
2
Circle centered at (0, 0, -7) with
radius 4.
z
Here is a sketch of the traces in
each coordinate plane.
This paraboloid extends below the
xy-plane.
z  y 2  9
y
z  x 2  9
x2 y2  9
x
Here’s a graph of the surface using DPGraph.
Level Curves
In the previous two examples, traces in the coordinate
planes and traces parallel to the xy-plane were used to
sketch the function of two variables as a surface in the
3-dimensional coordinate system.
When the traces parallel to the xy-plane or in other
words, the traces found when z or f(x,y) is set equal to a
constant, are drawn in the xy-plane, the traces are called
level curves. When several level curves, also called
contour lines, are drawn together in the xy-plane the
image is called a contour map.
Example 6: Sketch a contour map of the function in Example 4,
z  f x , y   25  x 2  y 2 using the level curves at c = 5,4,3,2,1 and 0.
Solution: c = a means the curve when z has a value of a.
c 5: 5 
25  x 2  y 2  x2  y 2  0  Point at 0,0 
c  4: 4 
25  x 2  y 2  x2  y 2  9  Circle with r  3
c  3: 3 
25  x 2  y 2  x2  y2  16  Circle with r  4
c 2: 2 
25  x 2  y 2  x2  y 2  21  Circle with r 
21
c 1: 1 
25  x 2  y 2  x2  y 2  24  Circle with r 
24
c 0: 0 
25  x 2  y 2  x2  y2  25  Circle with r  5
Solution to Example 6 Continued: Contour map with point (0, 0) for c
= 5 and then circles expanding out from the center for the remaining
values of c.
Note: The values of c were
uniformly spaced, but the level
curves are not.
When the level curves are
spaced far apart (in the center),
there is a gradual change in the
function values.
When the level curves are close
together (near c = 5), there is a
steep change in the function
values.
2
2
Example 7: Sketch a contour map of the function,z  f x , y   x  2y
using the level curves at c = 0, 2, 4, 6 and 8.
Solution: Set the function equal to each constant.
c  0 : 0  x 2  2y 2  Point at 0,0 
c 2: 2  x
2
 2y
2
c  4: 4 x
2
 2y
2
c  6: 6  x
2
 2y
2
c  8: 8  x
2
 2y
2
x2 y2
 Ellipse:

1
2
1
x2 y2
 Ellipse:

1
4
2
x2 y2
 Ellipse:

1
6
3
x2 y2
 Ellipse:

1
8
4
Solution to Example 7 Continued: Contour map with point (0, 0) for c
= 0 and then ellipses expanding out from the center for the
remaining values of c.
Example 7 Continued:
2
2
Here is a graph of the surface z  f x , y   x  2y .
Note:
1.
Sketching functions of two variables in 3-dimensions is
challenging and will take quite a bit of practice. Using the
traces and level curves can be extremely beneficial.
In the beginning, using 3-dimensional graphing utility will help
you visualize the surfaces and see how the traces and level
curves relate.
2. The idea of a level curve can be extended to functions of
three variables. If w = g(x, y, z) is a function of three
variables and k is a constant, then g(x, y, z) = k is considered a
level surface of the function g. Though we may be able to
draw the level surface, we still cannot draw the function g in 4
dimensions.
For comments on this presentation you may email the author
Professor Judy Gill at
[email protected] or the publisher of the VML, Dr. Julia Arnold at
[email protected]
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