Математика. Материалы курса

 

Before we start studying the concept of a function of several variables, at first, of course, we should understand what the domain of definition of such a function is, what its graph looks like, and today’s practical lesson is devoted to these issues.

So, the first task is to find the domain of the function definition of two variables.

What do we need?

We should remember how we found the domain of the function definition of a single variable. In fact, the beginning is the same: we write for which values of the variables the values of functions exist.

It is clear that the root expression must be non-negative.

At the second stage, we solved the inequality using the interval method. Here, we apply the generalized interval method. How do we implement it? We use the same steps as in the interval method. There we built points on the hour line where the numerator and denominator turn to zero. In this case, we do exactly the same – we write the zeros of the numerator and denominator.

So, for the numerator, x + y-2 =0, for the denominator, x-y = 0.

For a straight line (1), if one of the coordinates is 2, the second coordinate is zero. You don’t have to use x to express y. We mark point 2 on the numeric y-axis and point 2 on the x-axis. Let’s see, these are the zeros of the numerator, so a straight line can be drawn through these points. Let’s highlight this line in color. We construct it as a solid line since it is the zero of the numerator, and the equality is allowed here.

Let’s turn to the denominator. This line is y=x. It is clear that this is the bisector of the 1st and 3rd coordinate angles. In addition, it is a denominator, and the denominator cannot be zero, so we will build a dotted line. The line y=x will be drawn as a dotted line.

What do we have? The plane was divided into four parts. I will show these 4 parts so that you can see them (see the figure). We don’t pay attention to the coordinate axes.

We take an arbitrary point. For example, the point with coordinates (3; 0) from the

first part. The numerator is positive, the denominator is positive in this part, so the point (3; 0) satisfies this inequality, and all the points lying in this part also satisfy this inequality.

Let’s look at the point (0; 3) from the second part. The numerator is positive, the denominator is negative. The inequality does not hold, so we will not fill in the second part.

In the third part, we take the point with coordinates (-1; 0). The numerator is negative, and so is the denominator. As a result (minus by minus), the fraction turns out to be positive. The inequality is met, we fill it in.

In the fourth part, let’s take the point (0; -1). It is easy to check that the inequality does not hold at this point, and we will not fill in part 4. What is the answer to this problem? This image is the answer. We construct the domain of definition in the xOy plane.

Let’s consider the following problem.

We are given the function z=arcsin(xy). We need to find its domain of definition. Remember that an arcsine is defined if its argument satisfies the double inequality -1≤xy≤1. We start by building the boundaries of the definition domain. The boundaries are defined by the equalities xy=1 and xy= - 1. It is easy to see that these are graphs of functions of a single variable (x= 1/y and x= -1/y). These functions are well known to us, and we will build hyperbolas.

Note that x and y can be zero, all hyperbolas pass through the point (1; 1) and (1; -1), that is, when the product is 1 or -1. Let’s build it. Please note that equality is allowed, so we will draw all borders as solid lines.

What did we get? The entire plane was divided into 5 parts. We take a point in each of these parts and substitute it into the inequality that we are solving. For example, if we substitute the origin point (0; 0), it is clear that the inequality is satisfied. The entire interior is crosshatched. As for parts 1, 2, 3, and 4, if we take the point, we will get beyond this inequality at these points. The inequality will not be satisfied, so we will not fill in the remaining 4 parts. The solution to the problem looks like this.

Let’s learn how to build planes. Let me remind you that a plane in three-dimensional space is defined by the equation Ax+By+Cz=0, where the coefficients A, B, and C are not equal to zero at the same time. If C is not zero, z can be expressed, and we get a function of two variables. In each of these cases, the graph is a plane. Now we are supposed to understand how to build these planes quickly and without problems.

Where do we start? It is logical to see what intersections will be on the coordinate axes. On the Ox axis, for example, the Ox axis is set by the conditions y=0 and z=0. Substituting zero for z, and 0 for y, we get 2x=4, i.e. x=2. Mark it on the axis. On the Oy axis, the Oy axis is set by the conditions x = 0 and z=0. So, substituting it in the original equation, we get y=4. Mark it on the axis. The Oz axis is set by the conditions x=0 and y=0, then z = 4. On the z axis, we mark a point. Well, we know that if any two points lie in the plane, the line connecting these points also lies in this plane. We just need to connect the points. And we get the answer: the image of the plane.

How is a plane built when the variable y is missing from this equation? In fact, it is +0y. What do we do in this case? Let’s see, this equation is a straight line in the xOz plane, in the xOz plane, it is a straight line. If x=0, z = 4, if z =0, x = 0. Let’s build a straight line passing through these points. What should we do next? Let’s go back to the entry.

The missing y coordinate indicates that this plane is parallel to the Oy axis. If a point belongs to a straight line, this plane also belongs to a straight line passing through this point, parallel to the Oy axis. We also build through a point on the Ox axis, and build straight lines parallel to the Oy axis. We just need to crosshatch it, so that we can see this plane more clearly, and so that this edge does not stick out, you can connect it with a line. We have the image of the plane.

Here is almost exactly the same situation. This is the equation of a straight line in the yOz plane. We build this line. Each time I take the value of one of the coordinates 0 so that we could mark points on the coordinate axes.

If z=0, y = 4, and if y =0, z =4. Let’s build it. We construct a straight line given in the yOz plane, and note that the missing x coordinate indicates that the plane is parallel to the Ox axis. So, using the marked points on the coordinate axes, we build straight lines parallel to the Ox axis. To make the image look better, we draw the edge and crosshatch it.

Finally, let’s consider the case when there are no x and y coordinates at all. You know, if we didn’t mention that we are building graphs of functions of two variables, the task “to build z=4” would sound incorrect. Why? And we don’t know where we are. If we were only on the z axis, we would build point 4.If we were in a plane (with two axes), we would build a straight line parallel to the 2nd coordinate axis. Note, the Oz axis is actually located in two planes xOz and yOz, so in these planes, this equation no longer defines a point, but defines a plane, defines a straight line.

So, in these coordinate planes, this equation defines straight lines.

In the coordinate planes, in the yOz plane, we construct a straight line parallel to the Oy axis, and in the xOz plane we construct a straight line parallel to the Ox axis. Look, we have almost already received the image of the plane, we just need to connect the edges and crosshatch it.

Look again. This equation, firstly, sets a point on the z axis, we build it, and, secondly, these are straight lines parallel to the coordinate axes Oy and Ox, we build them. All we need is to complete the construction and see that we have obtained the plane specified by this equation.

Let’s move on to another task. We need to draw the graph of the function z=4-x².

A graph of a function of two variables is a surface in three-dimensional space. Let’s look closely at this equation, there is no variable y here, in fact, it is here, but it’s plus 0y, so we don’t see it. And in this form, this is the parabola equation. We start by building a parabola. We will not build a table, we just build a parabola.

The missing y coordinate indicates that the surface is parallel to the Oy axis, with any value of y. y=0, we have built it, it’s xОz plane, if y=1, i.e. at a distance 1 from the xОz plane, the equation connecting x and z will be the same. If we take any other value, we get that x and z are still connected by the same equation. See the video for building a surface.

Look at the surface which appeared in the three-dimensional space. This is the graph of this function.