Functions of Several Variables: Learn It 1

Giselle Miller

Functions of Several Variables: Learn It 1

Recognize a function of two variables and identify its domain and range
Sketch a graph of a function of two variables
Analyze functions by looking at their level curves and traces
Identify functions of three or more variables and understand their level surfaces

Functions of Two Variables

Our first step is to explain what a function of more than one variable is, starting with functions of two independent variables. This includes identifying the domain and range of such functions and learning how to graph them. We also examine ways to relate graphs of functions in three dimensions to graphs of more familiar planar functions.

The definition of a function of two variables is very similar to the definition for a function of one variable. The main difference is that instead of mapping values of one variable to values of another variable, we map ordered pairs of variables to another variable.

function of two variables

A function of two variables [latex]z=f(x, y)[/latex] maps each ordered pair [latex](x, y)[/latex] in a subset [latex]{\bf{D}}[/latex] of the real plane [latex]\mathbb{R}^{2}[/latex] to a unique real number [latex]z[/latex].

The set [latex]{\bf{D}}[/latex] is called the domain of the function
The range of [latex]f[/latex] is the set of all real numbers [latex]z[/latex] that has at least one ordered pair [latex](x, y)\in{\bf{D}}[/latex] such that [latex]f(x, y)=z[/latex]

A bulbous shape is marked domain and it contains the point (x, y). From this point, there is an arrow marked f that points to a point z on a straight line marked range. — Figure 1. The domain of a function of two variables consists of ordered pairs [latex](x,y)[/latex].

Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist.

Look for restrictions similar to those in single-variable calculus:

Square roots require non-negative expressions under the radical
Denominators cannot equal zero
Logarithms require positive arguments

The difference is that these restrictions now define regions in the [latex]xy[/latex]-plane rather than intervals on the number line.

Find the domain and range of each of the following functions:

[latex]f\,(x,\ y)=3x+5y+2[/latex]
[latex]g\,(x,\ y)=\sqrt{9-x^{2}-y^{2}}[/latex]

This is an example of a linear function in two variables. There are no values or combinations of [latex]x[/latex] and [latex]y[/latex] that cause [latex]f\,(x,\ y)[/latex] to be undefined, so the domain of [latex]f[/latex] is [latex]\mathbb{R}^{2}[/latex]. To determine the range, first pick a value for [latex]z[/latex]. We need to find a solution to the equation [latex]f\,(x,\ y)=z[/latex], or [latex]3x-5y+2=z[/latex]. One such solution can be obtained by first setting [latex]y=0[/latex], which yields the equation [latex]3x+2=z[/latex]. The solution to this equation is [latex]x=\frac{z-2}{3}[/latex], which gives the ordered pair [latex](\frac{z-2}{3},\ 0)[/latex] as a solution to the equation [latex]f\,(x,\ y)=z[/latex] for any value of [latex]z[/latex]. Therefore, the range of the function is all real numbers, or [latex]\mathbb{R}[/latex].
For the function [latex]g\,(x,\ y)[/latex] to have a real value, the quantity under the square root must be nonnegative:
[latex]9-x^{2}-y^{2}\,\geq\,0.[/latex]

This inequality can be written in the form

[latex]x^{2}+y^{2}\,\leq\,9.[/latex]

Therefore, the domain of [latex]g\,(x,\ y)[/latex] is [latex]{\{(x,\ y)\in\mathbb{R}^{2}\,|\,x^{2}+y^{2}\,\leq\,9\}}[/latex]. The graph of this set of points can be described as a disk of radius [latex]3[/latex] centered at the origin. The domain includes the boundary circle as shown in the following graph.

Figure 2. The domain of the function [latex]\small{g(x,y)=\sqrt{9-x^{2}-y^{2}}}[/latex] is a closed disk of radius [latex]\small{3}[/latex].

To determine the range of [latex]g\,(x,\ y)=\sqrt{9-x^{2}-y^{2}}[/latex] we start with a point [latex](x_0,\ y_0)[/latex] on the boundary of the domain, which is defined by the relation [latex]x^{2}+y^{2}=9[/latex]. It follows that [latex]x^2_0+y^2_0=9[/latex] and

[latex]g\,(x_0,\ y_0)=\sqrt{9-x^2_0-y^2_0}=\sqrt{9-(x^2_0+y^2_0)}=\sqrt{9-9}=0[/latex].

If [latex]x^2_0+y^2_0=0[/latex] (in other words, [latex]x_0=y_0=0[/latex]), then

[latex]g\,(x_0,\ y_0)=\sqrt{9-x^2_0-y^2_0}=\sqrt{9-(x^2_0+y^2_0)}=\sqrt{9-0}=3[/latex]

This is the maximum value of the function. Given any value [latex]c[/latex] between 0 and 3, we can find an entire set of points inside the domain of [latex]g[/latex] such that [latex]g\,(x,\ y)=c[/latex]:

[latex]\begin{array}{ccc} \hfill {\sqrt{9-x^{2}-y^{2}}} & =\hfill & {c} \hfill \\ \hfill {9-x^{2}-y^{2}} & =\hfill & {c^{2}}\hfill \\ \hfill {x^{2}+y^{2}} & =\hfill & {9-c^{2}.}\hfill \\ \hfill \end{array}[/latex]

Since [latex]9-c^{2}\,>\,0[/latex], this describes a circle of radius [latex]\sqrt{9-c^{2}}[/latex] centered at the origin. Any point on this circle satisfies the equation [latex]g\,(x,\ y)=c[/latex]. Therefore, the range of this function can be written in interval notation as [latex][0,\ 3][/latex].