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 the value of the gradient is nonzero, then the gradient is perpendicular to the level line passing through the given point and is directed in the direction of increasing function, while

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.

concept directional derivative considered for functions of two and three variables. To understand the meaning of a directional derivative, we need to compare derivatives by definition.

Consequently,

Now we can find the derivative in the direction of this function by its formula:

And now - homework. It gives a function of not three, but only two variables, but the direction vector is given in a slightly different way. So you have to repeat vector algebra .

Example 2 Find the derivative of a function at a point M0 (1; 2) in the direction of the vector , where M1 - point with coordinates (3; 0) .

The vector that specifies the direction of the derivative can also be given in such a form as in the following example - in the form expansions in unit vectors of coordinate axes, but this is a familiar topic from the very beginning of vector algebra.

Example 3 Find the derivative of a function at the point M0 (1; 1; 1) in the direction of the vector .

Solution. Find the direction cosines of the vector

Let's find partial derivatives of functions at a point M0 :

Therefore, we can find the derivative in the direction of this function by its formula:

.

function gradient

Gradient function of several variables at a point M0 characterizes the direction of maximum growth of this function at the point M0 and the magnitude of this maximum growth.

How to find the gradient?

Need to define vector whose projections on the coordinate axes are the values partial derivatives, , of this function at the corresponding point:

.

That is, it should be representation of a vector by the unit vectors of the coordinate axes, in which the partial derivative corresponding to its axis is multiplied by each unit vector.

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. A gradient is a vector whose direction indicates the direction of the most rapid increase in the function F. For 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 -yz / (x? (1 + (z/x)?)) = -1;F'_y = 1 arctg(z/x) = arctg 1 = ?/4;F'_z = 0 arctg(z/x) + y (arctg(z/x))'_z = y 1/(1 + (z/x)?) 1/x = y/(x (1 + (z/x)?)) = 1.grad = (- 1, ?/4, 1).

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

Instruction

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

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

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

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

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

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

You will need

• - boom;
• - pen.

Instruction

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

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

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

4. The definition of the derivative in direction, considering the point M as a variable, can be rewritten as a dot product: (du/ds)=((df/dx, df/dy,df/dz), (cos(alpha), cos(beta), cos (gamma)))=(grad u, s^o). This expression will be objective for a scalar field. If we consider an easy function, then gradf is a vector 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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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 a given function, we define a vector whose 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 at a given point is found by the 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 $\frac(\partial w)(\partial s) =|\overrightarrow(gradw) |\cdot \cos \varphi =0$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Extrema of a function of several variables

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

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

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

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

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

The points at which these equalities hold are called stationary.

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

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

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

Example 1. Find extrema of a function
.

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

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

Similarly
.

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

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

Let's find partial derivatives of the second order:

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

Similarly
;
.

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

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

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

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

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