Gradient function and its properties. Vector analysis scalar field of surface and level line directional derivative derivative of scalar field gradient basic properties of gradient invariant definition of gradient gradient calculation rules

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; - four).

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 a
, 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 negative numbers
> 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 maximum or minimum (maximum or minimum value of the function) is somewhat more complicated than 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.

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:

Consequently,

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:

Consequently,

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 FROM, 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 equal heights on topographic map and lines of the same barometric pressure on the 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 FROM different real values, we get concentric circles centered at the origin.

Construction of level lines. Their analysis finds wide application in economic problems of 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 a C=0, then we get X2 + Z2 - Y2 =0 - cone; if C<0 , then X2 + Z2 - Y2 =C - A family of two-sheeted hyperboloids.

Some concepts and terms are used strictly within narrow limits. Other definitions are found in areas that are sharply opposed. So, for example, the concept of "gradient" is used by a physicist, and a mathematician, and a specialist in manicure or "Photoshop". What is a gradient as a concept? Let's figure it out.

What do dictionaries say?

What is a "gradient" special thematic dictionaries interpret in relation to their specifics. Translated from Latin, this word means - "the one that goes, grows." And "Wikipedia" defines this concept as "a vector indicating the direction of increasing magnitude." In explanatory dictionaries, we see the meaning of this word as "a change in any value by one value." The concept can carry both quantitative and qualitative meaning.

In short, it is a smooth gradual transition of any value by one value, a progressive and continuous change in quantity or direction. The vector is calculated by mathematicians, meteorologists. This concept is used in astronomy, medicine, art, computer graphics. Under the similar term completely different types of activities are defined.

Math functions

What is the gradient of a function in mathematics? This is which indicates the direction of growth of a function in a scalar field from one value to another. The magnitude of the gradient is calculated using the definition of partial derivatives. To find out the fastest direction of growth of the function on the graph, two points are selected. They define the start and end of the vector. The rate at which a value grows from one point to another is the magnitude of the gradient. Mathematical functions based on the calculations of this indicator are used in vector computer graphics, the objects of which are graphic images of mathematical objects.

What is a gradient in physics?

The concept of a gradient is common in many branches of physics: the gradient of optics, temperature, velocity, pressure, etc. In this industry, the concept denotes a measure of the increase or decrease in a value per unit. It is calculated as the difference between the two indicators. Let's consider some of the quantities in more detail.

What is a potential gradient? In working with an electrostatic field, two characteristics are determined: tension (power) and potential (energy). These different quantities are related to the environment. And although they define different characteristics, they still have a connection with each other.

To determine the strength of the force field, the potential gradient is used - a value that determines the rate of change in the potential in the direction of the field line. How to calculate? The potential difference of two points of the electric field is calculated from the known voltage using the intensity vector, which is equal to the potential gradient.

Terms of meteorologists and geographers

For the first time, the concept of a gradient was used by meteorologists to determine the change in the magnitude and direction of various meteorological indicators: temperature, pressure, wind speed and strength. It is a measure of the quantitative change of various quantities. Maxwell introduced the term into mathematics much later. In the definition of weather conditions, there are concepts of vertical and horizontal gradients. Let's consider them in more detail.

What is a vertical temperature gradient? This is a value that shows the change in performance, calculated at a height of 100 m. It can be either positive or negative, in contrast to the horizontal, which is always positive.

The gradient shows the magnitude or angle of the slope on the ground. It is calculated as the ratio of the height to the length of the path projection on a certain section. Expressed as a percentage.

Medical indicators

The definition of "temperature gradient" can also be found among medical terms. It shows the difference in the corresponding indicators of the internal organs and the surface of the body. In biology, the physiological gradient fixes a change in the physiology of any organ or organism as a whole at any stage of its development. In medicine, a metabolic indicator is the intensity of metabolism.

Not only physicists, but also physicians use this term in their work. What is pressure gradient in cardiology? This concept defines the difference in blood pressure in any interconnected sections of the cardiovascular system.

A decreasing gradient of automaticity is an indicator of a decrease in the frequency of excitations of the heart in the direction from its base to the top, which occur automatically. In addition, cardiologists identify the site of arterial damage and its degree by controlling the difference in the amplitudes of systolic waves. In other words, using the amplitude gradient of the pulse.

What is a velocity gradient?

When one speaks of the rate of change of a certain quantity, one means by this the rate of change in time and space. In other words, the velocity gradient determines the change in spatial coordinates in relation to temporal indicators. This indicator is calculated by meteorologists, astronomers, chemists. The shear rate gradient of fluid layers is determined in the oil and gas industry to calculate the rate at which a fluid rises through a pipe. Such an indicator of tectonic movements is the area of \u200b\u200bcalculations by seismologists.

Economic functions

To substantiate important theoretical conclusions, the concept of a gradient is widely used by economists. When solving consumer problems, a utility function is used, which helps to represent preferences from a set of alternatives. "Budget constraint function" is a term used to refer to a set of consumer bundles. The gradients in this area are used to calculate the optimal consumptions.

color gradient

The term "gradient" is familiar to creative people. Although they are far from the exact sciences. What is a gradient for a designer? Since in the exact sciences it is a gradual increase in value by one, so in color this indicator denotes a smooth, stretched transition of shades of the same color from lighter to darker, or vice versa. Artists call this process “stretching.” It is also possible to switch to different accompanying colors in the same range.

Gradient stretching of shades in the coloring of rooms has taken a strong position among design techniques. The newfangled ombre style - a smooth flow of shade from light to dark, from bright to pale - effectively transforms any room in the house and office.

Opticians use special lenses in their sunglasses. What is a gradient in glasses? This is the manufacture of a lens in a special way, when from top to bottom the color changes from a darker to a lighter shade. Products made using this technology protect the eyes from solar radiation and allow you to view objects even in very bright light.

Color in web design

Those who are engaged in web design and computer graphics are well aware of the universal tool "gradient", with which a lot of various effects are created. Color transitions are transformed into highlights, a fancy background, three-dimensionality. Hue manipulation, light and shadow creation adds volume to vector objects. For this purpose, several types of gradients are used:

Linear.
Radial.
conical.
Mirror.
Rhomboid.
noise gradient.

gradient beauty

For visitors to beauty salons, the question of what a gradient is will not come as a surprise. True, in this case, knowledge of mathematical laws and the foundations of physics is not necessary. It's all about color transitions. Hair and nails become the object of the gradient. The ombre technique, which means “tone” in French, came into fashion from sports lovers of surfing and other beach activities. Naturally burnt and regrown hair has become a hit. Women of fashion began to specially dye their hair with a barely noticeable transition of shades.

The ombre technique did not pass by nail salons. The gradient on the nails creates a coloration with a gradual lightening of the plate from the root to the edge. Masters offer horizontal, vertical, with a transition and other varieties.

Needlework

The concept of "gradient" is familiar to needlewomen from another side. A technique of this kind is used in the creation of handmade items in the decoupage style. In this way, new antique things are created, or old ones are restored: chests of drawers, chairs, chests, and so on. Decoupage involves applying a pattern using a stencil, which is based on a color gradient as a background.

Fabric artists have adopted dyeing in this way for new models. Dresses with gradient colors conquered the catwalks. Fashion was picked up by needlewomen - knitters. Knitwear with a smooth color transition is a success.

Summing up the definition of "gradient", we can say about a very extensive area of human activity in which this term has a place. The replacement by the synonym "vector" is not always appropriate, since the vector is, after all, a functional, spatial concept. What determines the generality of the concept is a gradual change in a certain quantity, substance, physical parameter per unit over a certain period. In color, this is a smooth transition of tone.

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;
- a 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 having 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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1 0 The gradient is directed along the normal to the level surface (or to the level line if the field is flat).

2 0 The gradient is directed in the direction of increasing field function.

3 0 The gradient module is equal to the largest derivative in the direction at a given point of the field:

These properties give an invariant characteristic of the gradient. They say that the gradU vector indicates the direction and magnitude of the greatest change in the scalar field at a given point.

Remark 2.1. If the function U(x,y) is a function of two variables, then the vector

(2.3)

lies in the oxy plane.

Let U=U(x,y,z) and V=V(x,y,z) functions differentiable at the point М 0 (x,y,z). Then the following equalities hold:

a) grad()= ; b) grad(UV)=VgradU+UgradV;

c) grad(U V)=gradU gradV; d) d) grad = , V ;

e) gradU( = gradU, where , U=U() has a derivative with respect to .

Example 2.1. The function U=x 2 +y 2 +z 2 is given. Determine the gradient of the function at the point M(-2;3;4).

Solution. According to formula (2.2), we have

The level surfaces of this scalar field are the family of spheres x 2 +y 2 +z 2 , the vector gradU=(-4;6;8) is the normal vector of planes.

Example 2.2. Find the gradient of the scalar field U=x-2y+3z.

Solution. According to formula (2.2), we have

The level surfaces of a given scalar field are the planes

x-2y+3z=C; the vector gradU=(1;-2;3) is the normal vector of planes of this family.

Example 2.3. Find the steepest slope of the surface U=x y at the point M(2;2;4).

Solution. We have:

Example 2.4. Find the unit normal vector to the level surface of the scalar field U=x 2 +y 2 +z 2 .

Solution. Level surfaces of a given scalar Field-sphere x 2 +y 2 +z 2 =C (C>0).

The gradient is directed along the normal to the level surface, so that

Defines the normal vector to the level surface at the point M(x,y,z). For a unit normal vector, we obtain the expression

, where

Example 2.5. Find the field gradient U= , where and are constant vectors, r is the radius vector of the point.

Solution. Let

Then:
. By the rule of differentiation of the determinant, we get

Consequently,

Example 2.6. Find the distance gradient , where P(x,y,z) is the point of the field under study, P 0 (x 0 ,y 0 ,z 0) is some fixed point.

Solution. We have - unit direction vector .

Example 2.7. Find the angle between the gradients of the functions at the point M 0 (1,1).

Solution. We find the gradients of these functions at the point M 0 (1,1), we have

; The angle between gradU and gradV at the point M 0 is determined from the equality

Hence =0.

Example 2.8. Find the derivative with respect to the direction, the radius vector is equal to

(2.4)

Solution. Finding the gradient of this function:

Substituting (2.5) into (2.4), we obtain

Example 2.9. Find at the point M 0 (1;1;1) the direction of the greatest change in the scalar field U=xy+yz+xz and the magnitude of this greatest change at this point.

Solution. The direction of the greatest change in the field is indicated by the vector grad U(M). We find it:

And, therefore, . This vector determines the direction of the greatest increase of this field at the point M 0 (1;1;1). The value of the largest change in the field at this point is equal to

Example 3.1. Find vector lines of vector field where is a constant vector.

Solution. We have so

(3.3)

Multiply the numerator and denominator of the first fraction by x, the second by y, the third by z and add it term by term. Using the proportion property, we get

Hence xdx+ydy+zdz=0, which means

x 2 +y 2 +z 2 =A 1 , A 1 -const>0. Now multiplying the numerator and denominator of the first fraction (3.3) by c 1, the second by c 2, the third by c 3 and summing it term by term, we get

From where c 1 dx+c 2 dy+c 3 dz=0

And, therefore, with 1 x+c 2 y+c 3 z=A 2 . A 2-const.

Required equations of vector lines

These equations show that vector lines are obtained as a result of the intersection of spheres having a common center at the origin with planes perpendicular to the vector . It follows that the vector lines are circles whose centers are on a straight line passing through the origin in the direction of the vector c. The planes of the circles are perpendicular to the specified line.

Example 3.2. Find vector field line passing through the point (1,0,0).

Solution. Differential equations of vector lines

hence we have . Solving the first equation. Or if we introduce the parameter t, then we will have In this case, the equation takes the form or dz=bdt, whence z=bt+c 2 .