# Introduction to Functions of Several Variables ```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
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
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
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
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
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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