Did you know that many mathematical models involve functions of several variables?
It’s true.
For example…
- The ability to represent points on the surface of the earth.
- Finding the work done by a force acting on an object.
- Calculating the profit made from selling various items.
- Or even finding the volume of a rectangular solid (i.e., V=lwh) all involve functions of two or more variables.
Domain And Range Of Multivariable Functions
So, a function \(f\) of two variables is a rule that assigns to each ordered pair of real numbers \(\left( {x,y} \right)\) in a set \(D\) a unique real number denoted \(f\left( {x,y} \right)\). The set \(D\) is the domain of \(f\) and its range is the set of values that \(f\) takes on, such that \(\left\{ {f\left( {x,y} \right):\left( {x,y} \right) \in D} \right\}\).
And for convenience we say \(z = f\left( {x,y} \right)\), just like we let \(y = f\left( x \right)\) for functions of single variable. Thus, creating an explicit equation where \(x\) and \(y\) are independent variables and \(z\) is the dependent variable.
But don’t worry.
Functions of several variables are just like functions of single variables.
Example #1 – Evaluate
For instance, suppose \(f\left( {x,y} \right) = x + {y^2}\), and we wish to evaluate at \(f\left( {5,2} \right)\) and \(f\left( { – 5, – 2} \right)\).
Well, all we have to do to determine the value of \(f\left( {5,2} \right)\) and \(f\left( { – 5, – 2} \right)\) is to substitute the respective \(x\) and \(y\) coordinates into the given function and simplify.
When \(f(5,2)\), then \(f\left( {5,2} \right) = \left( 5 \right) + {\left( 2 \right)^2} = 9\).
And when \(f\left( { – 5, – 2} \right)\), then \(f\left( { – 5, – 2} \right) = \left( { – 5} \right) + {\left( { – 2} \right)^2} = – 1\)
Therefore, we have found two ordered triples.
- If \(x = 5,\) \(y = 2\) and \(z = f\left( {5,2} \right) = 9\), then \(\left( {5,2,9} \right)\)
- If \(x = – 5,\) \(y = – 2\) and \(z = f\left( { – 5, – 2} \right) = – 1\), then \(\left( { – 5, – 2, – 1} \right)\)
But this now leads us to ask about “all” points \(\left( {x,y} \right)\).
In other words, if \(z = f\left( {x,y} \right)\) is the value of the function at \(\left( {x,y} \right)\), then the domain consists of all points \(\left( {x,y} \right)\) for which the formula makes sense and yields a valid or truthful result. And the range of the function is the set of its values for all \(\left( {x,y} \right)\) in its domain.
So, to determine the domain and range of a function of two variables or more, we simply need to determine where the function is defined.
Example #2 – Find The Domain & Range
For example, let’s find the domain and range of the following functions:
- \(f(x, y)=e^{x^{2}-y}\)
- \(f(x, y, z)=x^{2} \ln (x-y+z)\)
Problem #1: \(f\left( {x,y} \right) = {e^{{x^2} – y}}\)
Since \(f\left( {x,y} \right) = {e^{{x^2} – y}}\) is an exponential function, it is defined everywhere. Therefore, no matter what values we substitute for \(x\) and \(y\), we will have a defined \(z = f\left( {x,y} \right)\) value. So, the domain of \(f\left( {x,y} \right) = {e^{{x^2} – y}}\) is \(\left\{f(x, y): \mathbb{R}^{2}\right\}\).
Moreover, because we know that \(y = {e^x}\) has a horizontal asymptote at \(y = 0\), and is only defined for when \(y>0\) the range of \(f\left( {x,y} \right) = {e^{{x^2} – y}}\) is \(\left\{ {z = f\left( {x,y} \right):z>{0}} \right\}\)
Problem #2: \(f\left( {x,y,z} \right) = {x^2}\ln \left( {x – y + z} \right)\)
Okay, so \(f\left( {x,y,z} \right) = {x^2}\ln \left( {x – y + z} \right)\) is a logarithmic function. And we know that \(y = \ln x\) has a vertical asymptote at \(x = 0\) and is defined when \(x>0\). Therefore,\(f\left( {x,y,z} \right) = {x^2}\ln \left( {x – y + z} \right)\) is defined when \(x – y + z>0\). So, the domain of \(f\left( {x,y,z} \right) = {x^2}\ln \left( {x – y + z} \right)\) is \(\left\{ {f\left( {x,y,z} \right):x – y + z>0} \right\}\).
And because \(f\left( {x,y,z} \right) = {x^2}\ln \left( {x – y + z} \right)\) is a logarithmic function, the range is any real number. Thus, the range is \(\left\{ {f\left( {x,y,z} \right):\mathbb{R}} \right\}\).
Knowing how to describe functions of several variables by their domain and range is an important skill we will practice throughout our video lesson.
Level Curves
But sometimes, it is easier to understand a function’s behavior if we can visualize it.
So, now it’s time to consider the graph of a function of several variables.
If \(f\) is a function of two variables with domain \(D\), then the graph of \(f\) is the set of all points \(\left( {x,y,z} \right)\) in \({\mathbb{R}^3}\) such that \(z = f\left( {x,y} \right)\)and \(\left( {x,y} \right)\) is in \(D\).
It may be helpful to think of it this way.
Imagine the graph of the surface, \(S\) of \(f\), as lying directly above its domain \(D\) in the xy-plane.
And the easiest way to create such a graph is by utilizing the method of level curves, sometimes called contour curves.
The level curves of a function \(f\) of two variables are the curves with equations \(f\left( {x,y} \right) = k\), where \(k\) is a constant and represents the set of all points in the domain of \(f\) at which \(f\) takes on a given value of \(k\).
But here’s what is important to note!
The level curves \(f\left( {x,y} \right) = k\) are nothing more than the traces of the graph of \(f\) in the horizontal plane \(z=k\) projected down onto the xy-plane.
Example #1 – Sketch The Graph
For instance, suppose we are asked to sketch the graph \(f\left( {x,y} \right) = \sqrt {25 – {x^2} – {y^2}} \).
First, we let \(z = f\left( {x,y} \right)\) such that \(z = \sqrt {25 – {x^2} – {y^2}} \).
Next, we will simplify our equation by squaring both sides to obtain a more recognizable function.
\begin{equation}
\begin{aligned}
&z^{2}=25-x^{2}-y^{2} \\
&x^{2}+y^{2}+z^{2}=25
\end{aligned}
\end{equation}
Now, some of you may instantly recognize the function is the top half of a sphere (i.e., hemisphere) where \(z \ge 0\) due to the square root, and with radius 5.
But for the rest of us, we still might need a bit more help.
Example #2 – Project Our Surface Onto The XY-Plane
Not to fear!
We simply create level curves to help us make sense of our surface.
Let \(z = k\), where \(k\) is a constant, so we can project our surface onto the xy-plane.
\begin{equation}
\begin{aligned}
&x^{2}+y^{2}+k^{2}=25 \\
&x^{2}+y^{2}=\left(25-k^{2}\right)
\end{aligned}
\end{equation}
Now, we have a circle, and if we plug in values for \(k\), such as \(k = 0,1,2,3,4,5\) and graph becomes:
See, all we have to do is create a discrete set of height values by letting \(z = k\) and then sketching the resulting curve in the xy-plane, and the shape is instantly more easily identifiable.
And lastly, we simply must visualize these level curves “lifted up” to form the surface in three-dimensional space.
Together we will study the functions of several variables and explore how independent and dependent variables impact a function’s domain and range. And we will learn how graph level curves and surfaces for functions of two variables or more.
Let’s jump right in.
Video Tutorial w/ Full Lesson & Detailed Examples (Video)
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