It is known from a school mathematics course that a vector on a plane is a directed segment. Its beginning and end have two coordinates. The vector coordinates are calculated by subtracting the start coordinates from the end coordinates.

The concept of a vector can also be extended to an n-dimensional space (instead of two coordinates there will be n coordinates).

Gradient gradz function z=f(x 1 , x 2 , ... x n) is the vector of partial derivatives of the function at a point, i.e. vector with coordinates.

It can be proved that the gradient of a function characterizes the direction of the fastest growth of the level of the function at a point.

For example, for the function z \u003d 2x 1 + x 2 (see Figure 5.8), the gradient at any point will have coordinates (2; 1). It can be built on a plane in various ways, taking any point as the beginning of the vector. For example, you can connect point (0; 0) to point (2; 1), or point (1; 0) to point (3; 1), or point (0; 3) to point (2; 4), or t .P. (see figure 5.8). All vectors constructed in this way will have coordinates (2 - 0; 1 - 0) = = (3 - 1; 1 - 0) = (2 - 0; 4 - 3) = (2; 1).

Figure 5.8 clearly shows that the level of the function grows in the direction of the gradient, since the constructed level lines correspond to the level values 4 > 3 > 2.

Figure 5.8 - Gradient of the function z \u003d 2x 1 + x 2

Consider another example - the function z= 1/(x 1 x 2). The gradient of this function will no longer always be the same at different points, since its coordinates are determined by the formulas (-1 / (x 1 2 x 2); -1 / (x 1 x 2 2)).

Figure 5.9 shows the level lines of the function z= 1/(x 1 x 2) for levels 2 and 10 (the line 1/(x 1 x 2) = 2 is indicated by a dotted line, and the line 1/(x 1 x 2) = 10 is solid line).

Figure 5.9 - Gradients of the function z \u003d 1 / (x 1 x 2) at various points

Take, for example, the point (0.5; 1) and calculate the gradient at this point: (-1 / (0.5 2 * 1); -1 / (0.5 * 1 2)) \u003d (-4; - 2). Note that the point (0.5; 1) lies on the level line 1 / (x 1 x 2) \u003d 2, because z \u003d f (0.5; 1) \u003d 1 / (0.5 * 1) \u003d 2. To draw the vector (-4; -2) in Figure 5.9, connect the point (0.5; 1) with the point (-3.5; -1), because (-3.5 - 0.5; -1 - 1) = (-4; -2).

Let's take another point on the same level line, for example, point (1; 0.5) (z=f(1; 0.5) = 1/(0.5*1) = 2). Calculate the gradient at this point (-1/(1 2 *0.5); -1/(1*0.5 2)) = (-2; -4). To depict it in Figure 5.9, we connect the point (1; 0.5) with the point (-1; -3.5), because (-1 - 1; -3.5 - 0.5) = (-2; - 4).

Let's take one more point on the same level line, but only now in a non-positive coordinate quarter. For example, point (-0.5; -1) (z=f(-0.5; -1) = 1/((-1)*(-0.5)) = 2). The gradient at this point will be (-1/((-0.5) 2 *(-1)); -1/((-0.5)*(-1) 2)) = (4; 2). Let's depict it in Figure 5.9 by connecting the point (-0.5; -1) with the point (3.5; 1), because (3.5 - (-0.5); 1 - (-1)) = (4 ; 2).

It should be noted that in all three cases considered, the gradient shows the direction of growth of the level of the function (toward the level line 1/(x 1 x 2) = 10 > 2).

It can be proved that the gradient is always perpendicular to the level line (level surface) passing through the given point.

Extrema of a function of several variables

Let's define the concept extremum for a function of many variables.

The function of many variables f(X) has at the point X (0) maximum (minimum), if there is such a neighborhood of this point that for all points X from this neighborhood the inequalities f(X)f(X (0)) () hold.

If these inequalities are satisfied as strict, then the extremum is called strong, and if not, then weak.

Note that the extremum defined in this way is local character, since these inequalities hold only for some neighborhood of the extremum point.

A necessary condition for a local extremum of a differentiable function z=f(x 1, . . ., x n) at a point is the equality to zero of all first-order partial derivatives at this point:
.

The points at which these equalities hold are called stationary.

In another way, the necessary condition for an extremum can be formulated as follows: at the extremum point, the gradient is equal to zero. It is also possible to prove a more general statement - at the extremum point, the derivatives of the function in all directions vanish.

Stationary points should be subjected to additional studies - whether sufficient conditions for the existence of a local extremum are satisfied. To do this, determine the sign of the second-order differential. If for any that are not simultaneously equal to zero, it is always negative (positive), then the function has a maximum (minimum). If it can vanish not only at zero increments, then the question of the extremum remains open. If it can take both positive and negative values, then there is no extremum at the stationary point.

In the general case, determining the sign of the differential is a rather complicated problem, which we will not consider here. For a function of two variables, one can prove that if at a stationary point
, then there is an extremum. In this case, the sign of the second differential coincides with the sign
, i.e. If
, then this is the maximum, and if
, then this is the minimum. If
, then there is no extremum at this point, and if
, then the question of the extremum remains open.

Example 1. Find extrema of a function
.

Let's find partial derivatives by the method of logarithmic differentiation.

ln z = ln 2 + ln (x + y) + ln (1 + xy) – ln (1 + x 2) – ln (1 + y 2)

Similarly
.

Let's find stationary points from the system of equations:

Thus, four stationary points (1; 1), (1; -1), (-1; 1) and (-1; -1) are found.

Let's find partial derivatives of the second order:

ln (z x `) = ln 2 + ln (1 - x 2) -2ln (1 + x 2)

Similarly
;
.

Because
, expression sign
depends only on
. Note that in both of these derivatives the denominator is always positive, so you can only consider the sign of the numerator, or even the sign of the expressions x (x 2 - 3) and y (y 2 - 3). Let us determine it at each critical point and check the fulfillment of the sufficient extremum condition.

For point (1; 1) we get 1*(1 2 - 3) = -2< 0. Т.к. произведение двух отрицательных чисел
> 0, and
< 0, в точке (1; 1) можно найти максимум. Он равен
= 2*(1 + 1)*(1 +1*1)/((1 +1 2)*(1 +1 2)) = = 8/4 = 2.

For point (1; -1) we get 1*(1 2 - 3) = -2< 0 и (-1)*((-1) 2 – 3) = 2 >0. Because the product of these numbers
< 0, в этой точке экстремума нет. Аналогично можно показать, что нет экстремума в точке (-1; 1).

For the point (-1; -1) we get (-1)*((-1) 2 - 3) = 2 > 0. product of two positive numbers
> 0, and
> 0, at the point (-1; -1) you can find a minimum. It is equal to 2*((-1) + (-1))*(1 +(-1)*(-1))/((1 +(-1) 2)*(1 +(-1) 2) ) = -8/4 = = -2.

Find global the maximum or minimum (the largest or smallest value of the function) is somewhat more complicated than the local extremum, since these values can be achieved not only at stationary points, but also at the boundary of the domain of definition. It is not always easy to study the behavior of a function on the boundary of this region.

Find the maximum rate of increase of the function. How to find the gradient of a function

Gradient functions is a vector quantity, the finding of which is associated with the definition of partial derivatives of the function. The direction of the gradient indicates the path of the fastest growth of the function from one point of the scalar field to another.

Instruction

1. To solve the problem on the gradient of a function, methods of differential calculus are used, namely, finding partial derivatives of the first order in three variables. It is assumed that the function itself and all its partial derivatives have the property of continuity in the domain of the function.

2. The gradient is a vector, the direction of which indicates the direction of the most rapid increase in the function F. To do this, two points M0 and M1 are selected on the graph, which are the ends of the vector. The value of the gradient is equal to the rate of increase of the function from point M0 to point M1.

3. The function is differentiable at all points of this vector, therefore, the projections of the vector on the coordinate axes are all its partial derivatives. Then the gradient formula looks like this: grad = (?F/?x) i + (?F/?y) j + (?F/?z) k, where i, j, k are the unit vector coordinates. In other words, the gradient of a function is a vector whose coordinates are its partial derivatives grad F = (?F/?х, ?F/?y, ?F/?z).

4. Example 1. Let the function F = sin (x z?) / y be given. It is required to find its gradient at the point (?/6, 1/4, 1).

5. Solution. Determine the partial derivatives with respect to any variable: F'_x \u003d 1 / y cos (x z?) z?; F'_y \u003d sin (x z?) (-1) 1 / (y?); F'_z \u003d 1/y cos(x z?) 2 x z.

6. Substitute the famous point coordinates: F'_x = 4 cos(?/6) = 2 ?3; F'_y = sin(?/6) (-1) 16 = -8; F'_z \u003d 4 cos (? / 6) 2? / 6 \u003d 2? /? 3.

7. Apply the function gradient formula: grad F = 2 ?3 i – 8 j + 2 ?/?3 k.

8. Example 2. Find the coordinates of the gradient of the function F = y arсtg (z / x) at the point (1, 2, 1).

9. Solution. F'_x \u003d 0 arctg (z / x) + y (arctg (z / x)) '_x \u003d y 1 / (1 + (z / x)?) (-z / x?) \u003d -y z / (x? (1 + (z/x)?)) = -1;F'_y = 1 arctg(z/x) = arctg 1 = ?/4;F'_z = 0 arctg(z/x) + y (arctg(z/x))'_z = y 1/(1 + (z/x)?) 1/x = y/(x (1 + (z/x)?)) = 1.grad = (- 1, ?/4, 1).

The scalar field gradient is a vector quantity. Thus, to find it, it is required to determine all the components of the corresponding vector, based on knowledge about the division of the scalar field.

Instruction

1. Read in a textbook on higher mathematics what the gradient of a scalar field is. As you know, this vector quantity has a direction characterized by the maximum decay rate of the scalar function. Such a sense of a given vector quantity is justified by an expression for determining its components.

2. Remember that every vector is defined by the values of its components. Vector components are actually projections of this vector onto one or another coordinate axis. Thus, if three-dimensional space is considered, then the vector must have three components.

3. Write down how the components of a vector that is the gradient of some field are determined. All of the coordinates of such a vector is equal to the derivative of the scalar potential with respect to the variable whose coordinate is being calculated. That is, if you need to calculate the “x” component of the field gradient vector, then you need to differentiate the scalar function with respect to the variable “x”. Note that the derivative must be quotient. This means that when differentiating, the remaining variables that do not participate in it must be considered constants.

4. Write an expression for the scalar field. As you know, this term means each only a scalar function of several variables, which are also scalar quantities. The number of variables of a scalar function is limited by the dimension of the space.

5. Differentiate separately the scalar function with respect to each variable. As a result, you will have three new functions. Write any function in the expression for the gradient vector of the scalar field. Any of the obtained functions is really an indicator for a unit vector of a given coordinate. Thus, the final gradient vector should look like a polynomial with exponents as derivatives of a function.

When considering issues involving the representation of a gradient, it is more common to think of each as a scalar field. Therefore, we need to introduce the appropriate notation.

You will need

- boom;
- pen.

Instruction

1. Let the function be given by three arguments u=f(x, y, z). The partial derivative of a function, for example with respect to x, is defined as the derivative with respect to this argument, obtained by fixing the remaining arguments. The rest of the arguments are similar. The partial derivative notation is written as: df / dx \u003d u’x ...

2. The total differential will be equal to du \u003d (df / dx) dx + (df / dy) dy + (df / dz) dz. Partial derivatives can be understood as derivatives in the directions of the coordinate axes. Consequently, the question arises of finding the derivative with respect to the direction of a given vector s at the point M(x, y, z) (do not forget that the direction s specifies a unit vector-ort s^o). In this case, the differential vector of arguments is (dx, dy, dz)=(dscos(alpha), dscos(beta), dscos(gamma)).

3. Considering the form of the total differential du, it is possible to conclude that the derivative with respect to the direction s at the point M is: (du/ds)|M=((df/dx)|M)cos(alpha)+ ((df/dy) |M) cos(beta) +((df/dz)|M) cos(gamma). If s= s(sx,sy,sz), then direction cosines (cos(alpha), cos(beta), cos( gamma)) are calculated (see Fig. 1a).

4. The definition of the derivative in direction, considering the point M as a variable, can be rewritten as a dot product: (du/ds)=((df/dx, df/dy,df/dz), (cos(alpha), cos(beta), cos (gamma)))=(grad u, s^o). This expression will be objective for a scalar field. If we consider an easy function, then gradf is a vector with coordinates coinciding with the partial derivatives f(x, y, z).gradf(x, y, z)=((df/dx, df/dy, df/ dz)=)=(df/dx)i+(df/dy)j +(df/dz)k. Here (i, j, k) are the unit vectors of the coordinate axes in a rectangular Cartesian coordinate system.

5. If we use the Hamilton Nabla differential vector operator, then gradf can be written as the multiplication of this operator vector by the scalar f (see Fig. 1b). From the point of view of the connection of gradf with the directional derivative, the equality (gradf, s^o)=0 is admissible if these vectors are orthogonal. Consequently, gradf is often defined as the direction of the fastest metamorphosis of a scalar field. And from the point of view of differential operations (gradf is one of them), the properties of gradf exactly repeat the properties of differentiation of functions. In particular, if f=uv, then gradf=(vgradu+ugradv).

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Gradient this is a tool that in graphic editors fills the silhouette with a smooth transition of one color to another. Gradient can give a silhouette the result of volume, simulate lighting, reflections of light on the surface of an object, or the result of a sunset in the background of a photograph. This tool has a wide use, therefore, for processing photographs or creating illustrations, it is very important to learn how to use it.

You will need

Computer, graphics editor Adobe Photoshop, Corel Draw, Paint.Net or other.

Instruction

1. Open the image in the program or make a new one. Make a silhouette or select the desired area on the image.

2. Turn on the Gradient tool on the toolbar of the graphics editor. Place the mouse cursor on a point inside the selected area or silhouette, where the 1st color of the gradient will start. Click and hold the left mouse button. Move the cursor to the point where the gradient should transition to the final color. Release the left mouse button. The selected silhouette will be filled with a gradient fill.

3. Gradient y it is possible to set transparency, colors and their ratio at a certain fill point. To do this, open the Gradient Edit window. To open the editing window in Photoshop, click on the gradient example in the Options panel.

4. In the window that opens, the available gradient fill options are displayed as examples. To edit one of the options, select it with a mouse click.

5. An example of a gradient is displayed at the bottom of the window in the form of a wide scale with sliders. The sliders indicate the points at which the gradient should have the specified collations, and in the interval between the sliders, the color evenly transitions from the one specified at the first point to the color of the 2nd point.

6. The sliders located at the top of the scale set the transparency of the gradient. To change the transparency, click on the desired slider. A field will appear below the scale, in which enter the required degree of transparency in percent.

7. The sliders at the bottom of the scale set the colors of the gradient. By clicking on one of them, you will be able to prefer the desired color.

8. Gradient can have multiple transition colors. To set another color, click on an empty space at the bottom of the scale. Another slider will appear on it. Set the desired color for it. The scale will display an example of a gradient with one more point. You can move the sliders by holding them with the support of the left mouse button in order to achieve the desired combination.

9. Gradient There are several types that can give shape to flat silhouettes. Let's say, in order to give a circle the shape of a ball, a radial gradient is applied, and in order to give the shape of a cone, a conical gradient is applied. A specular gradient can be used to give the surface the illusion of bulge, and a diamond gradient can be used to create highlights.

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If at each point in space or part of space the value of a certain quantity is defined, then it is said that the field of this quantity is given. The field is called scalar if the considered value is scalar, i.e. well characterized by its numerical value. For example, the temperature field. The scalar field is given by the scalar function of the point u = /(M). If a Cartesian coordinate system is introduced in space, then there is a function of three variables x, yt z - the coordinates of the point M: Definition. The level surface of a scalar field is the set of points at which the function f(M) takes the same value. Level Surface Equation Example 1. Find Level Surfaces of a Scalar Field VECTOR ANALYSIS Scalar Field Level Surfaces and Level Lines Directional Derivative Derivative Gradient of a Scalar Field Basic Gradient Properties Invariant Definition of a Gradient Rules for Calculating a Gradient -4 By definition, a level surface equation will be. This is the equation of a sphere (with Ф 0) centered at the origin. A scalar field is called flat if the field is the same in all planes parallel to some plane. If the specified plane is taken as the xOy plane, then the field function will not depend on the z coordinate, i.e., it will be a function of only the arguments x and y. and also meaning. Level line equation - Example 2. Find level lines of a scalar field Level lines are given by equations At c = 0 we get a pair of lines, we get a family of hyperbolas (Fig. 1). 1.1. Directional derivative Let there be a scalar field defined by a scalar function u = /(Af). Let's take the point Afo and choose the direction determined by the vector I. Let's take another point M so that the vector M0M is parallel to the vector 1 (Fig. 2). Let us denote the length of the MoM vector by A/, and the increment of the function /(Af) - /(Afo), corresponding to the displacement D1, by Di. The ratio determines the average rate of change of the scalar field per unit length to the given direction. Let now tend to zero so that the vector М0М remains parallel to the vector I all the time. Definition. If for D/O there exists a finite limit of the relation (5), then it is called the derivative of the function at a given point Afo to the given direction I and is denoted by the symbol zr!^. So, by definition, This definition is not related to the choice of coordinate system, that is, it has a **variant character. Let us find an expression for the derivative with respect to the direction in the Cartesian coordinate system. Let the function / be differentiable at a point. Consider the value /(Af) at a point. Then the total increment of the function can be written in the following form: where and the symbols mean that the partial derivatives are calculated at the point Afo. Hence Here the quantities jfi, ^ are the direction cosines of the vector. Since the vectors MoM and I are co-directed, their direction cosines are the same: derivatives, are derivatives of the function and along the directions of the coordinate axes with the external nno- Example 3. Find the derivative of the function towards the point The vector has a length. Its direction cosines: By formula (9) we will have The fact that, means that the scalar field at a point in a given direction of age- For a flat field, the derivative in the direction I at a point is calculated by the formula where a is the angle formed by the vector I with the axis Oh. Zmmchmm 2. Formula (9) for calculating the derivative along the direction I at a given point Afo remains in force even when the point M tends to the point Mo along a curve for which the vector I is tangent at the point PrISchr 4. Calculate the derivative of the scalar field at the point Afo(l, 1). belonging to a parabola in the direction of this curve (in the direction of increasing abscissa). The direction ] of a parabola at a point is the direction of the tangent to the parabola at this point (Fig. 3). Let the tangent to the parabola at the point Afo form an angle o with the Ox axis. Then whence directing cosines of a tangent Let's calculate values and in a point. We have Now by formula (10) we obtain. Find the derivative of the scalar field at a point in the direction of the circle The vector equation of the circle has the form. We find the unit vector m of the tangent to the circle. The point corresponds to the value of the parameter. Scalar Field Gradient Let a scalar field be defined by a scalar function that is assumed to be differentiable. Definition. The gradient of a scalar field » at a given point M is a vector denoted by the symbol grad and defined by the equality It is clear that this vector depends both on the function / and on the point M at which its derivative is calculated. Let 1 be a unit vector in the direction Then the formula for the directional derivative can be written as follows: . thus, the derivative of the function u along the direction 1 is equal to the scalar product of the gradient of the function u(M) and the unit vector 1° of the direction I. 2.1. Basic properties of the gradient Theorem 1. The scalar field gradient is perpendicular to the level surface (or to the level line if the field is flat). (2) Let us draw a level surface u = const through an arbitrary point M and choose a smooth curve L on this surface passing through the point M (Fig. 4). Let I be a vector tangent to the curve L at the point M. Since on the level surface u(M) = u(M|) for any point Mj ∈ L, then On the other hand, = (gradu, 1°). That's why. This means that the vectors grad and and 1° are orthogonal. Thus, the vector grad and is orthogonal to any tangent to the level surface at the point M. Thus, it is orthogonal to the level surface itself at the point M. Theorem 2. The gradient is directed in the direction of increasing field function . Earlier we proved that the gradient of the scalar field is directed along the normal to the level surface, which can be oriented either towards the increase of the function u(M) or towards its decrease. Denote by n the normal of the level surface oriented in the direction of increasing function ti(M), and find the derivative of the function u in the direction of this normal (Fig. 5). We have Since according to the condition of Fig. 5 and therefore VECTOR ANALYSIS Scalar field Surfaces and level lines Directional derivative Derivative Scalar field gradient Basic properties of the gradient Invariant definition of the gradient Rules for calculating the gradient It follows that grad and is directed in the same direction as the one we have chosen the normal n, i.e., in the direction of increasing function u(M). Theorem 3. The length of the gradient is equal to the largest derivative with respect to the direction at a given point of the field, (here, max $ is taken in all possible directions at a given point M to the point). We have where is the angle between the vectors 1 and grad n. Since the largest value is Example 1. Find the direction of the largest imonion of the scalar field at the point and also the magnitude of this largest change at the specified point. The direction of the greatest change in the scalar field is indicated by a vector. We have so This vector determines the direction of the greatest increase in the field to a point. The value of the largest change in the field at this point is 2.2. Invariant definition of the gradient The quantities that characterize the properties of the object under study and do not depend on the choice of the coordinate system are called the invariants of the given object. For example, the length of a curve is an invariant of this curve, but the angle of the tangent to the curve with the x-axis is not an invariant. Based on the above three properties of the scalar field gradient, we can give the following invariant definition of the gradient. Definition. The scalar field gradient is a vector directed along the normal to the level surface in the direction of increasing field function and having a length equal to the largest derivative in direction (at a given point). Let be a unit normal vector directed in the direction of increasing field. Then Example 2. Find the distance gradient - some fixed point, and M(x,y,z) - the current one. 4 We have where is the unit direction vector. Rules for calculating the gradient where c is a constant number. The above formulas are obtained directly from the definition of the gradient and the properties of the derivatives. By the rule of differentiation of the product The proof is similar to the proof of the property Let F(u) be a differentiable scalar function. Then 4 By the definition of the gradient, we have Apply to all terms on the right side the rule of differentiation of a complex function. In particular, Formula (6) follows from the formula plane to two fixed points of this plane. Consider an arbitrary ellipse with foci Fj and F] and prove that any light ray that emerges from one focus of the ellipse, after reflection from the ellipse, enters its other focus. The level lines of the function (7) are VECTOR ANALYSIS Scalar field Surfaces and level lines Directional derivative Derivative Scalar field gradient Basic properties of the gradient Invariant definition of the gradient Gradient calculation rules Equations (8) describe a family of ellipses with foci at points F) and Fj. According to the result of Example 2, we have and radius vectors. drawn to the point P(x, y) from the foci F| and Fj, and hence lies on the bisector of the angle between these radius vectors (Fig. 6). According to Tooromo 1, the gradient PQ is perpendicular to the ellipse (8) at the point. Therefore, Fig.6. the normal to the ellipse (8) at any th point bisects the angle between the radius vectors drawn to this point. From here and from the fact that the angle of incidence is equal to the angle of reflection, we obtain: a ray of light coming out of one focus of the ellipse, reflected from it, will certainly fall into the other focus of this ellipse.

Let Z= F(M) is a function defined in some neighborhood of the point M(y; x);L={ Cos; Cos} – unit vector (in Fig. 33 1=  , 2= ); L is a straight line passing through a point M; M1(x1; y1), where x1=x+x and y1=y+y- a point on a line L;  L- the size of the segment MM1;  Z= F(x+x, y+y)-F(X, Y) – function increment F(M) at the point M(x; y).

Definition. The limit of the relation, if it exists, is called Derivative function Z = F ( M ) at the point M ( X ; Y ) in the direction of the vector L .

Designation.

If the function F(M) differentiable at a point M(x; y), then at the point M(x; y) there is a derivative in any direction L coming from M; it is calculated according to the following formula:

(8)

Where Cos AND Cos- direction cosines of the vector L.

Example 46. Calculate the derivative of a function Z= X2 + Y2 X at the point M(1; 2) in the direction of the vector MM1, Where M1- point with coordinates (3; 0).

. Let's find the unit vector L, having this direction:

Where Cos= ; Cos=- .

We calculate the partial derivatives of the function at the point M(1; 2):

By formula (8) we obtain

Example 47. Find the derivative of a function U = xy2 Z3 at the point M(3; 2; 1) In vector direction MN, Where N(5; 4; 2) .

. Let's find the vector and its direction cosines:

Calculate the values of partial derivatives at the point M:

Hence,

Definition. Gradient FunctionsZ= F(M) at the point M(x; y) is a vector whose coordinates are equal to the corresponding partial derivatives u taken at the point M(x; y).

Designation.

Example 48. Find the gradient of a function Z= X2 +2 Y2 -5 at the point M(2; -1).

Solution. We find partial derivatives: and their values at the point M(2; -1):

Example 49. Find the magnitude and direction of the gradient of a function at a point

Solution. Let's find the partial derivatives and calculate their values at the point M:

Hence,

The directional derivative for a function of three variables is defined similarly U= F(X, Y, Z) , formulas are derived

The concept of a gradient is introduced

We emphasize that Basic properties of the gradient function more important for the analysis of economic optimization: in the direction of the gradient, the function increases. In economic problems, the following properties of the gradient are used:

1) Let a function be given Z= F(X, Y) , which has partial derivatives in the domain of definition. Consider some point M0(x0, y0) from the domain of definition. Let the value of the function at this point be F(X0 , Y0 ) . Consider the function graph. Through the dot (X0 , Y0 , F(X0 , Y0 )) three-dimensional space, we draw a plane tangent to the surface of the graph of the function. Then the gradient of the function calculated at the point (x0, y0), considered geometrically as a vector attached to a point (X0 , Y0 , F(X0 , Y0 )) , will be perpendicular to the tangent plane. The geometric illustration is shown in fig. 34.

2) Gradient function F(X, Y) at the point M0(x0, y0) indicates the direction of the fastest increase of the function at the point М0. In addition, any direction that makes an acute angle with the gradient is the direction of growth of the function at the point М0. In other words, a small movement from a point (x0, y0) in the direction of the gradient of the function at this point leads to an increase in the function, and to the greatest extent.

Consider a vector opposite to the gradient. It is called anti-gradient . The coordinates of this vector are:

Function anti-gradient F(X, Y) at the point M0(x0, y0) indicates the direction of the fastest decrease of the function at the point М0. Any direction that forms an acute angle with the antigradient is the direction in which the function is decreasing at that point.

3) When studying a function, it often becomes necessary to find such pairs (x, y) from the scope of the function, for which the function takes the same values. Consider the set of points (X, Y) out of function scope F(X, Y) , such that F(X, Y)= Const, where is the entry “ Const” means that the value of the function is fixed and equal to some number from the range of the function.

Definition. Function level line U = F ( X , Y ) called the lineF(X, Y)=С on the planeXOy, at the points of which the function remains constantU= C.

Level lines are geometrically depicted on the plane of change of independent variables in the form of curved lines. Obtaining level lines can be imagined as follows. Consider the set WITH, which consists of points in three-dimensional space with coordinates (X, Y, F(X, Y)= Const), which, on the one hand, belong to the graph of the function Z= F(X, Y), on the other hand, they lie in a plane parallel to the coordinate plane HOW, and separated from it by a value equal to a given constant. Then, to construct a level line, it is enough to intersect the surface of the graph of the function with a plane Z= Const and project the line of intersection onto a plane HOW. The above reasoning is the justification for the possibility of directly constructing level lines on a plane HOW.

Definition. The set of level lines is called Level line map.

Well-known examples of level lines are levels of the same height on a topographic map and lines of the same barometric pressure on a weather map.

Definition. The direction along which the rate of increase of the function is maximum is called "preferred" direction, or Direction of the fastest growth.

The "preferred" direction is given by the gradient vector of the function. On fig. 35 shows the maximum, minimum and saddle point in the problem of optimizing a function of two variables in the absence of restrictions. The lower part of the figure shows the level lines and directions of the fastest growth.

Example 50. Find feature level lines U= X2 + Y2 .

Solution. The equation of the family of level lines has the form X2 + Y2 = C (C>0) . Giving WITH different real values, we get concentric circles centered at the origin.

Construction of level lines. Their analysis is widely used in economic problems at the micro- and macrolevels, the theory of equilibrium and effective solutions. Isocosts, isoquants, indifference curves - these are all level lines built for different economic functions.

Example 51. Consider the following economic situation. Let the production of products be described Cobb-Douglas function F(X, Y)=10x1/3y2/3, Where X- amount of labor At- amount of capital. 30 USD was allocated for the acquisition of resources. units, the price of labor is 5 c.u. units, capital - 10 c.u. units Let us ask ourselves the question: what is the largest output that can be obtained under these conditions? Here, “given conditions” refers to given technologies, resource prices, and the type of production function. As already noted, the function Cobb-Douglas is monotonically increasing in each variable, i.e., an increase in each type of resource leads to an increase in output. Under these conditions, it is clear that it is possible to increase the acquisition of resources as long as there is enough money. Resource packs that cost 30 c.u. units, satisfy the condition:

5x + 10y = 30,

That is, they define the function level line:

G(X, Y) = 5x + 10y.

On the other hand, with the help of level lines Cobb-Douglas functions (Fig. 36) it is possible to show the increase of the function: at any point of the level line, the direction of the gradient is the direction of the greatest increase, and to build a gradient at a point, it is enough to draw a tangent to the level line at this point, draw a perpendicular to the tangent and indicate the direction of the gradient. From fig. 36 it can be seen that the movement of the level line of the Cobb-Douglas function along the gradient should be carried out until it becomes tangent to the level line 5x + 10y = 30. Thus, using the concepts of level line, gradient, gradient properties, it is possible to develop approaches to the best use of resources in terms of increasing the volume of output.

Definition. Function level surface U = F ( X , Y , Z ) called the surfaceF(X, Y, Z)=С, at the points of which the function remains constantU= C.

Example 52. Find feature level surfaces U= X2 + Z2 - Y2 .

Solution. The equation of the family of level surfaces has the form X2 + Z2 - Y2 =C. If C=0, then we get X2 + Z2 - Y2 =0 - cone; If C<0 , That X2 + Z2 - Y2 =C - A family of two-sheeted hyperboloids.