The modulus of the gradient of the function at a point is equal to. Vector analysis scalar field of a surface and level lines directional derivative derivative gradient of a scalar field basic properties of a gradient invariant definition of a gradient gradient calculation rules

Some concepts and terms are used within a purely narrow framework. Other definitions are found in areas that are sharply opposed. For example, the concept of “gradient” is used by a physicist, a mathematician, and a manicurist or Photoshop specialist. What is gradient as a concept? Let's figure it out.

What do dictionaries say?

Special thematic dictionaries interpret what a “gradient” is in relation to their specifics. Translated from Latin, this word means “the one who goes, grows.” And Wikipedia defines this concept as “a vector indicating the direction of increase in a quantity.” In explanatory dictionaries we see the meaning of this word as “a change in any value by one value.” A concept can have 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 and meteorologists. This concept is used in astronomy, medicine, art, and computer graphics. A similar term defines completely different types of activities.

Mathematical functions

What is the gradient of a function in mathematics? This 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 partial derivatives. To determine the fastest direction of growth of a function, two points are selected on the graph. They define the beginning 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 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 gradient is common in many branches of physics: gradient of optics, temperature, speed, pressure, etc. In this branch, the concept denotes a measure of increase or decrease of a value by one. It is calculated by calculations as the difference between two indicators. Let's look at some of the values ​​in more detail.

What is a potential gradient? When working with an electrostatic field, two characteristics are determined: tension (force) 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 of potential in the direction of the force line. How to calculate? The potential difference between two points of the electric field is calculated from a 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 gradient was used by meteorologists to determine changes in the magnitude and direction of various meteorological indicators: temperature, pressure, wind speed and strength. It is a measure of quantitative changes in various quantities. Maxwell introduced the term into mathematics much later. In determining weather conditions, there are concepts of vertical and horizontal gradients. Let's take a closer look at them.

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

The gradient shows the magnitude or angle of slope on the ground. It is calculated as the ratio of the height to the length of the projection of the path in 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 internal organs and body surface. In biology, a physiological gradient records changes in the physiology of any organ or organism as a whole at any stage of its development. In medicine, the metabolic indicator is the intensity of metabolism.

Not only physicists, but also doctors use this term in their work. What is a pressure gradient in cardiology? This concept defines the difference in blood pressure in any interconnected parts 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, occurring automatically. In addition, cardiologists identify the location of arterial damage and its degree by monitoring the difference in the amplitudes of systolic waves. In other words, using the amplitude gradient of the pulse.

What is a velocity gradient?

When they talk about the rate of change of a certain quantity, they mean by this the speed of change in time and space. In other words, the speed gradient determines the change in spatial coordinates in relation to time indicators. This indicator is calculated by meteorologists, astronomers, and chemists. The shear rate gradient of liquid layers is determined in the oil and gas industry to calculate the rate of rise of liquid through a pipe. This indicator of tectonic movements is the area of ​​calculations by seismologists.

Economic functions

Economists widely use the concept of gradient to substantiate important theoretical conclusions. When solving consumer problems, a utility function is used to help represent preferences from a set of alternatives. "Budget Constraint Function" is a term used to refer to a set of consumption bundles. Gradients in this area are used to calculate optimal consumption.

Color gradient

The term "gradient" is familiar to creative people. Although they are far from 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, extended 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 stretches of shades in painting rooms have taken a strong position among design techniques. The new-fashioned ombre style - a smooth flow of shade from light to dark, from bright to pale - effectively transforms any room in the home or office.

Opticians use special lenses in sunglasses. What is a gradient in glasses? This is the making 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 involved in web design and computer graphics are well aware of the universal “gradient” tool, which can be used to create a wide variety of effects. Color transitions are transformed into highlights, a bizarre background, and three-dimensionality. Manipulating shades and creating light and shadow gives volume to vector objects. For these purposes, several types of gradients are used:

• Linear.
• Radial.
• Cone-shaped.
• Mirror.
• Diamond-shaped.
• Noise gradient.

Gradient beauty

For visitors to beauty salons, the question of what a gradient is will not come as a surprise. True, even in this case, knowledge of mathematical laws and fundamentals of physics is not necessary. We are still talking about color transitions. The objects of the gradient are hair and nails. The ombre technique, which means “tone” in French, came into fashion from sports lovers of surfing and other beach activities. Naturally bleached and regrown hair has become a hit. Fashionistas began to specially dye their hair with a barely noticeable transition of shades.

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

Needlework

Needlewomen are familiar with the concept of “gradient” from one more side. A similar technique is used to create handmade items in the decoupage style. In this way, new antique things are created, or old ones are restored: chests of drawers, chairs, chests, etc. Decoupage involves applying a pattern using a stencil, the basis for which is a color gradient as a background.

Fabric artists have adopted this method of dyeing for new models. Dresses with gradient colors have conquered the catwalks. The fashion was picked up by needlewomen - knitters. Knitted items with a smooth color transition are popular.

To summarize the definition of “gradient,” we can say about a very broad area of ​​​​human activity in which this term has a place. Replacement with the synonym “vector” is not always suitable, since a vector is still a functional, spatial concept. What defines the generality of the concept is a gradual change in a certain quantity, substance, physical parameter by one over a certain period. In color it is a smooth transition of tone.

From the school mathematics course we know 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 be extended to n-dimensional space (instead of two coordinates there will be n coordinates).

Gradient gradzfunctionz=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 proven that the gradient of a function characterizes the direction of the fastest growth of the level of a function at a point.

For example, for the function z = 2x 1 + x 2 (see Figure 5.8), the gradient at any point will have coordinates (2; 1). You can construct it 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 so on. .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).

From Figure 5.8 it is clearly seen that the level of the function increases in the direction of the gradient, since the constructed level lines correspond to the level values ​​4 > 3 > 2.

Figure 5.8 - Gradient of function z= 2x 1 + x 2

Let's 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 function level lines z = 1/(x 1 x 2) for levels 2 and 10 (the straight line 1/(x 1 x 2) = 2 is indicated by a dotted line, and the straight line 1/(x 1 x 2) = 10 is solid line).

Figure 5.9 - Gradients of the function z= 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)) = (-4; - 2). Note that the point (0.5; 1) lies on the level line 1/(x 1 x 2) = 2, because z=f(0.5; 1) = 1/(0.5*1) = 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). Let's 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 another 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 equal to (-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 function level (towards the level line 1/(x 1 x 2) = 10 > 2).

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

Extrema of a function of several variables

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

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

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 are satisfied only for a certain 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 zero. A more general statement can also be proven: at the extremum point, the derivatives of the function in all directions vanish.

Stationary points should be subjected to additional research to determine whether sufficient conditions for the existence of a local extremum are met. To do this, determine the sign of the second order differential. If for any , not simultaneously equal to zero, it is always negative (positive), then the function has a maximum (minimum). If it can go to zero not only with 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 a stationary point.

In the general case, determining the sign of the differential is a rather complex problem, which we will not consider here. For a function of two variables, it can be proven that if at a stationary point
, then the extremum is present. 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 the extrema of the function
.

Let's find partial derivatives using the logarithmic differentiation method.

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

Likewise
.

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

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

Let's find the second order partial derivatives:

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

Likewise
;
.

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 define it at each critical point and check that the sufficient condition for the extremum is satisfied.

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 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. Because product of two positive numbers
> 0, and
> 0, at the point (-1; -1) the minimum can be found. It is equal to 2*((-1) + (-1))*(1 +(-1)*(-1))/((1 +(-1) 2)*(1 +(-1) 2) ) = -8/4 = = -2.

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

Gradient functions– a vector quantity, the determination of which is associated with the determination of the 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.

Instructions

1. To solve the problem of the gradient of a function, methods of differential calculus are used, namely, finding first-order partial derivatives with respect to three variables. It is assumed that the function itself and all its partial derivatives have the property of continuity in the domain of definition 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 magnitude 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 coordinates of the unit vector. 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 detect its gradient at the point (?/6, 1/4, 1).

5. Solution. Determine the partial derivatives with respect to each variable: F'_х = 1/y сos(х z?) z?; F'_y = sin(х z?) (-1) 1/(y?); F'_z = 1/y cos(x z?) 2 x z.

6. Substitute the famous coordinate values ​​of the point: F’_x = 4 сos(?/6) = 2 ?3; F’_y = sin(?/6) (-1) 16 = -8; F’_z = 4 cos(?/6) 2 ?/6 = 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 point (1, 2, 1).

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

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

Instructions

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 rate of decay of the scalar function. This interpretation of this vector quantity is justified by the expression for determining its components.

2. Remember that any vector is determined by the magnitudes of its components. The components of a vector 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 a certain field are determined. All of the coordinates of such a vector are 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 “x” variable. Please note that the derivative must be partial. This means that during differentiation, the remaining variables that are not involved in it must be considered constants.

4. Write an expression for the scalar field. As is well known, this term implies 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 the scalar function separately with respect to each variable. As a result, you will get three new functions. Write any function into the expression for the scalar field gradient vector. Each of the obtained functions is actually an indicator for a unit vector of a given coordinate. Thus, the final gradient vector should look like a polynomial with exponents in the form of derivatives of the function.

When considering issues involving gradient representation, it is common to think of functions as scalar fields. Therefore, it is necessary to introduce the appropriate notation.

You will need

• – boom;
• - pen.

Instructions

1. Let the function be specified 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. Similar for other arguments. The notation for the partial derivative is written in the form: df/dx = u’x ...

2. The total differential will be equal to du=(дf/дх)dx+ (дf/дy)dy+(дf/дz)dz. Partial derivatives can be understood as derivatives along 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 is determined by the unit vector s^o). In this case, the vector-differential of the arguments (dx, dy, dz) = (дscos(alpha), dscos(beta), dscos(gamma)).

3. Considering the form of the total differential du, we can conclude that the derivative in the direction s at point M is equal to: (дu/дs)|M=((дf/дх)|M)сos(alpha)+ ((дf/дy) |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 directional derivative, considering the point M as a variable, can be rewritten in the form of a scalar product: (дu/дs) = ((дf/дх, дf/дy,дf/дz), (cos(alpha), cos(beta), cos (gamma)))=(grad u, s^o). This expression will be objective for a scalar field. If a function is considered easily, 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 between gradf and the directional derivative, the equality (gradf, s^o)=0 is acceptable if these vectors are orthogonal. Consequently, gradf is often defined as the direction of the fastest metamorphosis of the 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 differentiating functions. In particular, if f=uv, then gradf=(vgradu+u gradv).

Video on the topic

Gradient This is a tool that, in graphic editors, fills a silhouette with a smooth transition from one color to another. Gradient can give a silhouette the result of volume, imitate lighting, glare of light on the surface of an object, or the result of a sunset in the background of a photograph. This tool is widely used, so 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 another.

Instructions

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

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

3. Gradient You can set transparency, colors and their ratio at a certain point of the fill. To do this, open the gradient editing window. To open the editing window in Photoshop, click on the gradient example in the Options panel.

4. The window that opens displays the available gradient fill options in the form of examples. To edit one of the options, select it with a mouse click.

5. At the bottom of the window an example of a gradient is displayed in the form of a wide scale on which sliders are located. The sliders indicate the points at which the gradient should have specified collations, and in the interval between the sliders the color evenly transitions from the color 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 required slider. A field will appear under the scale in which you enter the required degree of transparency as a percentage.

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 select the desired color.

8. Gradient may have several transition colors. To set another color, click on the free space at the bottom of the scale. Another slider will appear on it. Give it the required color. The scale will display an example of the gradient with one more point. You can move the sliders by holding them with the left mouse button to achieve the desired combination.

9. Gradient They come in several types that can give shape to flat silhouettes. For example, in order to give a circle the shape of a ball, a radial gradient is used, and in order to give the shape of a cone, a cone-shaped gradient is used. To give the surface the illusion of convexity, you can use a mirror gradient, and a diamond-shaped gradient can be used to create highlights.

Video on the topic

Video on the topic

If at each point in space or part of space the value of a certain quantity is determined, then they say that the field of this quantity is specified. A field is called scalar if the quantity under consideration is scalar, i.e. fully characterized by its numerical value. For example, the temperature field. The scalar field is given by the scalar point function 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 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 lines Directional derivative Derivative Scalar field gradient Basic properties of a gradient Invariant definition of a gradient Rules for calculating a gradient -4 According to the definition, the equation of a level surface will be. This is the equation of a sphere (with Ф 0) with its center at the origin. A scalar field is called flat if the field is the same in all planes parallel to a certain plane. If the indicated plane is taken to be 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. A plane field can be characterized using level lines - a set of points on the plane at which the function /(x, y) has one and also the meaning. Equation of a level line - Example 2. Find level lines of a scalar field Level lines are given by equations. When c = 0 we get a pair of straight lines, we get a family of hyperbolas (Fig. 1). 1.1. Directional derivative Let there be a scalar field defined by the scalar function u = /(Af). Let's take point Afo and choose the direction determined by vector I. Let's take another point M so that vector M0M is parallel to 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 movement of D1, by Di. The ratio determines the average rate of change of the scalar field per unit length in the given direction. Let now tend to zero so that the vector M0M remains parallel to the vector I all the time. Definition. If at D/O there is 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 3!^. So, by definition, This definition is not related to the choice of coordinate system, i.e., it is of a **variant nature. Let's find an expression for the directional derivative in the Cartesian coordinate system. Let the function / be differentiable at a point. Let's consider the value of /(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 codirectional, their direction cosines are the same: Since M Afo, being always on a straight line parallel to vector 1, the angles are constant therefore Finally, from equalities (7) and (8) we obtain Eamuan is 1. Particulars derivatives are derivatives of the function and along the directions of the coordinate axes, so-Example 3. Find the derivative of the function in the direction to the point The vector has a length. Its direction cosines: According to 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 with respect to 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 with respect to the direction I at a given point Afo remains in force when point M tends to point Mo along a curve for which vector I is tangent at point PrIShr 4. Calculate the derivative of the scalar field at 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 considered to be the direction of the tangent to the parabola at this point (Fig. 3). Let the tangent to the parabola at point Afo form an angle o with the Ox axis. Then where do the direction cosines of the tangent come from? Let us calculate the values ​​of and at the point. We have Now using formula (10) we get. Find the derivative of the scalar field at a point along the direction of the circle. The vector equation of a 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. The value of r at the point Afo will be equal. From here we obtain the direction cosines of the tangent to the circle at the point. Let us calculate the values ​​of the partial derivatives of the given scalar field at the point. This means the desired derivative. Scalar field gradient Let the scalar field be defined by a scalar function which is assumed to be differentiable. Definition. The gradient of the scalar field "at a given point M is a vector denoted by the symbol grad and 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 in the following form: . Thus, the derivative of the function u in direction 1 is equal to the scalar product of the gradient of the function u(M) and the unit vector 1° of direction I. 2.1. Basic properties of the gradient Theorem 1. The gradient of the scalar field 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 select on this surface a smooth curve L passing through the point M (Fig. 4). Let I be a vecgor tangent to the curve L at point M. Since on the level surface u(M) = u(M|) for any point Mj e 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 point M. Thus, it is orthogonal to the level surface itself at point M. Theorem 2. The gradient is directed towards increasing the field function . Previously, we proved that the gradient of the scalar field is directed along the normal to the level surface, which can be oriented either in the direction of increasing function u(M) or in the direction of its decreasing. Let us 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 Derivative in direction Derivative Gradient of the scalar field Basic properties of the gradient Invariant definition of the gradient Rules for calculating the gradient It follows that grad is directed in the same direction as the one we chose 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 direction at a given point of the field (here the check is taken along all possible directions at a given point M). We have where is the angle between the vectors 1 and grad n. Since the greatest value is Example 1. Find the direction of the greatest change of the scalar field at a point and also the magnitude of this greatest change at the specified point. The direction of greatest change in the scalar field is indicated by a vector. We have so that This vector determines the direction of the greatest increase in the field at a point. The magnitude of the largest field change at this point is 2.2. Invariant definition of the gradient Quantities that characterize the properties of the object under study and do not depend on the choice of coordinate system are called invariants of the given object. For example, the length of a curve is an invariant of this curve, but the tangent angle to the curve with the Ox axis is not an invariant. Based on the three properties of the scalar field gradient proven above, we can give the following invariant definition of the gradient. Definition. The scalar field gradient is a vector directed normal to the level surface in the direction of increasing the field function and having a length equal to the largest directional derivative (at a given point). Let be a unit normal vector directed in the direction of increasing field. Then Example 2. Find the gradient of the distance - 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 given formulas are obtained directly from the definition of the gradient and the properties of derivatives. According to the rule of product differentiation, the proof is similar to the proof of the property Let F(u) be a differentiable scalar function. Then 4 By definition of the fadient we have Apply the rule of differentiation of a complex function to all terms on the right side. We obtain In particular, Formula (6) follows from the formula Example 3. Find the derivative with respect to the direction of the radius vector r from the function Using formula (3) and using the formula As a result, we obtain that Example 4. Let a plane scalar field be given - distances from some point plane to two fixed points of this plane. Let us consider an arbitrary ellipse with foci Fj and F] and prove that every ray of light emerging from one focus of the ellipse, after reflection from the ellipse, ends up in its other focus. The level lines of 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 Rules for calculating the gradient Equations (8) describe a family of ellipses with foci at points F) and Fj. According to the result of Example 2, we have Thus, the gradient of a given field is equal to the vector PQ of the diagonal of the rhombus constructed on the unit vectors r? and radius vectors. drawn to the point P(x, y) from the foci F| and Fj, and therefore 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 point bisects the angle between the radius vectors drawn to this point. From this and from the fact that the angle of incidence is equal to the angle of reflection, we obtain: a ray of light emerging from one focus of the ellipse, reflected from it, will certainly fall into another focus of this ellipse.