Function Gradient at a point is called a vector whose coordinates are equal to the corresponding partial derivatives and is denoted.

If we consider the unit vector e=(), then according to formula (3), the derivative in direction is the scalar product of the gradient and the unit vector that specifies the direction. It is known that the scalar product of two vectors is maximal if they have the same direction. Therefore, the gradient of the function at a given point characterizes the direction and magnitude of the maximum growth of the function at this point.

Theorem . If the function is differentiable and at the point M 0 If the gradient value is non-zero, then the gradient is perpendicular to the level line passing through given point and is directed in the direction of increasing function at the same time

CONCLUSION: 1) The derivative of a function at a point along the direction determined by the gradient of that function at the specified point has a maximum value compared to the derivative at that point along any other direction.

• 2) The value of the derivative of the function in the direction, which determines the gradient of this function at a given point, is equal to.
• 3) Knowing the gradient of the function at each point, it is possible to build level lines with some error. Let's start from the point M 0 . Let's build a gradient at this point. Set the direction perpendicular to the gradient. Let's build a small part of the level line. Consider a close point M 1 , build a gradient at it, and so on.

From school course Mathematicians 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 also be extended to an n-dimensional space (instead of two coordinates there will be n coordinates).

Gradient grad z of the function z = f(х 1 , х 2 , …х n) is the vector of partial derivatives of the function at the 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 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 be always the same in 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 straight line 1 / (x 1 x 2) = 2 is indicated by a dotted line, and the straight line
1 / (x 1 x 2) \u003d 10 - 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 depict the vector (-4; -2) in Figure 5.9, we 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).

Definition 1

If for each pair $(x,y)$ of values ​​of two independent variables from some domain a certain value of $z$ is assigned, then $z$ is said to be a function of two variables $(x,y)$. Notation: $z=f(x,y)$.

Consider the function $z=f(x,y)$, which is defined in some domain in the space $Oxy$.

Consequently,

Definition 3

If for each triple $(x,y,z)$ of values ​​of three independent variables from some domain a certain value $w$ is assigned, then $w$ is said to be a function of three variables $(x,y,z)$ in this area.

Designation:$w=f(x,y,z)$.

Consider the function $w=f(x,y,z)$, which is defined in some domain in the space $Oxyz$.

For given function define a vector for which the projections on the coordinate axes are the values ​​of the partial derivatives of the given function at some point $\frac(\partial z)(\partial x) ;\frac(\partial z)(\partial y) $.

Definition 4

The gradient of a given function $w=f(x,y,z)$ is a vector $\overrightarrow(gradw)$ of the following form:

Theorem 3

Let a gradient field be defined in some scalar field $w=f(x,y,z)$

\[\overrightarrow(gradw) =\frac(\partial w)(\partial x) \cdot \overrightarrow(i) +\frac(\partial w)(\partial y) \cdot \overrightarrow(j) +\frac (\partial w)(\partial z) \cdot \overrightarrow(k) .\]

The derivative $\frac(\partial w)(\partial s) $ in the direction of the given vector $\overrightarrow(s) $ is equal to the projection of the gradient vector $\overrightarrow(gradw) $ onto the given vector $\overrightarrow(s) $.

Example 4

Solution:

The expression for the gradient is found by the formula

\[\overrightarrow(gradw) =\frac(\partial w)(\partial x) \cdot \overrightarrow(i) +\frac(\partial w)(\partial y) \cdot \overrightarrow(j) +\frac (\partial w)(\partial z) \cdot \overrightarrow(k) .\]

\[\frac(\partial w)(\partial x) =2x;\frac(\partial w)(\partial y) =4y;\frac(\partial w)(\partial z) =2.\]

Consequently,

\[\overrightarrow(gradw) =2x\cdot \overrightarrow(i) +4y\cdot \overrightarrow(j) +2\cdot \overrightarrow(k) .\]

Example 5

Determine the gradient of a given function

at the point $M(1;2;1)$. Calculate $\left(|\overrightarrow(gradz) |\right)_(M) $.

Solution:

The expression for the gradient in given point find by formula

\[\left(\overrightarrow(gradw) \right)_(M) =\left(\frac(\partial w)(\partial x) \right)_(M) \cdot \overrightarrow(i) +\left (\frac(\partial w)(\partial y) \right)_(M) \cdot \overrightarrow(j) +\left(\frac(\partial w)(\partial z) \right)_(M) \cdot \overrightarrow(k) .\]

The partial derivatives have the form:

\[\frac(\partial w)(\partial x) =2x;\frac(\partial w)(\partial y) =4y;\frac(\partial w)(\partial z) =6z^(2) .\]

Derivatives at the point $M(1;2)$:

\[\frac(\partial w)(\partial x) =2\cdot 1=2;\frac(\partial w)(\partial y) =4\cdot 2=8;\frac(\partial w)( \partial z) =6\cdot 1^(2) =6.\]

Consequently,

\[\left(\overrightarrow(gradw) \right)_(M) =2\cdot \overrightarrow(i) +8\cdot \overrightarrow(j) +6\cdot \overrightarrow(k) \]

\[\left(|\overrightarrow(gradw) |\right)_(M) =\sqrt(2^(2) +8^(2) +6^(2) ) =\sqrt(4+64+36 ) =\sqrt(104) .\]

Let's list some gradient properties:

The derivative of a given function at a given point along the direction of some vector $\overrightarrow(s)$ has the greatest value if the direction of the given vector $\overrightarrow(s)$ coincides with the direction of the gradient. In this case, this largest value of the derivative coincides with the length of the gradient vector, i.e. $|\overrightarrow(gradw) |$.

The derivative of the given function with respect to the direction of the vector that is perpendicular to the gradient vector, i.e. $\overrightarrow(gradw) $ is equal to 0. Since $\varphi =\frac(\pi )(2) $, then $\cos \varphi =0$; hence $\frac(\partial w)(\partial s) =|\overrightarrow(gradw) |\cdot \cos \varphi =0$.

GRADIENT FUNCTION u = f(x, y, z) specified in some region. space (X Y Z), there is vector with projections denoted by symbols: grad where i, j, k- coordinate vectors. G. f. - there is a point function (x, y, z), i.e., it forms a vector field. Derivative in the direction of G. f. at this point reaches the greatest value and is equal to: The direction of the gradient is the direction of the fastest increase of the function. G. f. at a given point is perpendicular to the level surface passing through this point. Efficiency of use G. f. in lithological studies was shown in the study of eolian ex. Central Karakum.

Geological dictionary: in 2 volumes. - M.: Nedra. Edited by K. N. Paffengolts et al.. 1978 .

See what the "GRADIENT FUNCTION" is in other dictionaries:

This article is about the mathematical characteristic; about the fill method, see: Gradient (computer graphics) ... Wikipedia

- (lat.). The difference in barometric and thermometric readings in different areas. Dictionary foreign words included in the Russian language. Chudinov A.N., 1910. GRADIENT difference in the readings of a barometer and a thermometer at the same moment ... ... Dictionary of foreign words of the Russian language

gradient- Changing the value of some quantity per unit distance in a given direction. A topographic gradient is the change in elevation over a measured horizontal distance. Relay protection EN gradient of the differential protection tripping characteristic … Technical Translator's Handbook

Gradient- a vector directed towards the fastest increase of the function and equal in magnitude to its derivative in this direction: where the symbols ei denote the unit vectors of the coordinate axes (orths) ... Economic and Mathematical Dictionary

One of the basic concepts of vector analysis and the theory of nonlinear mappings. The gradient of the scalar function of the vector argument from the Euclidean space E n called. derivative of the function f (t). with respect to the vector argument t, that is, an n-dimensional vector with ... ... Mathematical Encyclopedia

physiological gradient- - a value reflecting a change in k or an indicator of a function depending on another value; e.g. gradient partial pressure the difference in partial pressures, which determines the diffusion of gases from the alveoli (accinus) into the blood and from the blood into ... ... Glossary of terms for the physiology of farm animals

I Gradient (from Latin gradiens, genus gradientis walking) A vector showing the direction of the fastest change of some quantity, the value of which changes from one point in space to another (see Field theory). If the value ... ... Great Soviet Encyclopedia

Gradient- (from lat. gradiens walking, walking) (in mathematics) a vector showing the direction of the fastest increase of some function; (in physics) a measure of increase or decrease in space or on a plane of some physical quantity per unit ... ... Beginnings of modern natural science

Books

• Methods for solving some problems of selected sections of higher mathematics. Practicum, Klimenko Konstantin Grigorievich, Levitskaya Galina Vasilievna, Kozlovsky Evgeny Alexandrovich. This workshop discusses methods for solving some types of problems from such sections of the generally accepted course of mathematical analysis as the limit and extremum of a function, gradient and derivative ...

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

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

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

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 vector field where is a constant vector.

Solution. We have so

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

Solution. Differential Equations 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 .