If you've ever tried to root green cuttings, you probably know that ...
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Chapter five.
DISAPPEARANCE OF FIGURES. SECTION I
In this and the next chapters, we will follow the development of many wonderful geometric paradoxes. They all start with cutting the shape into pieces and end with the compilation of these pieces into a new shape. At the same time, it seems that part of the original figure (this may be part of the area of \u200b\u200bthe figure or one of several drawings depicted on it) has disappeared without a trace. When the pieces return to their original places, the disappeared part of the square or the drawing mysteriously reappears.
The geometric nature of these curious disappearances and reappearances justifies classifying these paradoxes as mathematical conundrums.
The line paradox
All the many paradoxes that we are going to consider here are based on the same principle, which we will call the "principle of latent redistribution." Here is one very old and very elementary paradox, which immediately explains the essence of this principle.
Draw ten vertical lines of equal length on a rectangular sheet of paper and draw a diagonal with a dotted line, as shown in Fig. fifty.
Let's look at the segments of these lines above and below the diagonal; it is easy to see that the length of the former decreases, and the latter increases accordingly.
Cut the rectangle along the dotted line and move the lower part to the left and down, as shown in Fig. 51.
If you count the number of vertical lines, you will find that there are now nine of them. Which line has disappeared and where? Move the left side to its original position, and the disappeared line will reappear.
But which line fell into place and where did it come from?
At first, these questions seem mysterious, but after a little reflection, it becomes clear that no single line disappears or appears. What happens is the following: these eight increments are exactly equal to the length of each of the original lines.
Perhaps the essence of the paradox will come out even more clearly if it is illustrated on pebbles.
Let's take five heaps of pebbles, four pebbles in a heap. Move one pebble from the second pile to the first, two pebbles from the third to the second, three from the fourth to the third, and finally all four pebbles from the fifth to the fourth. Figure: 52 explains our actions.
After such a shift, it turns out that there were only four heaps. It is impossible to answer the question which pile disappeared, since the pebbles were redistributed in such a way that each of the four piles added a pebble. Exactly the same thing happens in the line paradox. When parts of the sheet are shifted diagonally, the cut line segments are redistributed and each resulting line becomes slightly longer than the original.
Disappearance of the face
Let's move on to describing the ways in which the line paradox can be made more interesting and entertaining. This can be achieved, for example, by replacing the disappearance and appearance of lines with the same disappearance and appearance of flat figures. Images of pencils, cigarettes, bricks, high-crowned hats, glasses of water and other vertically extended objects, the nature of the image of which remains the same before and after the shift, are especially suitable here. With some artistic ingenuity, you can take more complex objects. Consider, for example, the disappearing face in fig. 53.
When you slide the bottom stripe on the top of the picture to the left, all hats remain unaffected, but one face disappears completely! (see the bottom of the figure). It makes no sense to ask which face exactly, as the shift divides the four faces into two. These parts are then redistributed, with each face gaining several additional features: one, for example, a longer nose, the other a more elongated chin, etc. However, these small redistributions are cleverly hidden, and the disappearance of the entire face, of course, is much more striking than the disappearance of a piece of the line.
"The Disappearing Warrior"
In this puzzle, the line paradox is given a circular shape and straight line segments are replaced by figures of 13 warriors (Fig. 54).
At the same time, the large arrow points to the northeast S.V. If the drawing is cut in a circle, and then the inner part begins to turn counterclockwise, then the figures will first be divided into parts, then they will be connected again, but in a different way, and when the large the arrow will point to the northwest northwest; there will be 12 warriors in the drawing (Fig. 55).
When the circle is rotated in the opposite direction to the position when the large arrow stands again on the NE, the disappeared warrior will appear again.
If fig. 54 take a closer look, you can see that the two warriors in the lower left part of the picture are located in a special way: they are opposite each other, while all the rest are placed in a chain. These two figures correspond to the extreme lines in the line segment paradox. Based on the requirements of the drawing, each of these figures should have a part of the leg missing, and so that in the turned position of the wheel this deficiency was less noticeable, it would be better to depict them side by side.
Note also that the warriors are depicted in the figure with much more ingenuity than it might seem at first glance. So, for example, in order for the figures to remain in an upright position in all places of the globe, it is necessary in one case to have the right leg instead of the left leg, and in the other, on the contrary, the left leg instead of the right leg.
The Lost Rabbit
The paradox of vertical lines can obviously be shown on more complex objects, for example, human faces, animal figures, etc. In fig. 56 shows one option.
When rectangles A and B are swapped after cutting along the thick line, one rabbit disappears, leaving behind an Easter egg. If instead of rearranging rectangles A and B, cut the right half of the drawing along the dotted line and swap the right parts, the number of rabbits will increase to 12, but one rabbit will lose ears and other funny details will appear.
Chapter six.
DISAPPEARANCE OF FIGURES. SECTION II
The Chessboard Paradox
In close connection with the paradoxes discussed in the previous chapter, there is another class of paradoxes, in which the “principle of hidden redistribution” explains the mysterious disappearance or appearance of areas. One of the oldest and simplest examples of this kind of paradox is shown in Fig. 57.
The chessboard is cut obliquely, as shown in the left half of the figure, and then part B is shifted to the left and down, as shown in the right half of the figure. If the triangle protruding in the upper right corner is cut off with scissors and placed in an empty space that looks like a triangle in the lower left corner of the figure, then you get a rectangle of 7x9 square units.
The original area was 64 square units, but now it is 63. Where has one missing square unit disappeared?
The answer is that our diagonal line runs slightly below the lower left corner of the square in the upper right corner of the board.
Due to this, the cut triangle has a height that is not 1, but 1 1/7. And thus, the height is not 9, but 9 1/7 units. A 1 / 7th unit increase in height is almost imperceptible, but when taken into account, it results in a required rectangle area of \u200b\u200b64 square units.
The paradox becomes even more striking if, instead of a checkerboard, we take just a square sheet of paper without cells, since in our case, upon close examination, an inaccurate closing of cells along the cut line is revealed.
The connection between our paradox and the vertical line paradox discussed in the previous chapter becomes clear if we trace the cells at the cut line. When moving upward along the cut line, it is found that above the line, parts of the cut cells (in the figure they are darkened) gradually decrease, and under the line they gradually increase. There were fifteen darkened cells on the chessboard, and on the rectangle obtained after rearranging the parts, there were only fourteen of them. The apparent disappearance of one darkened cell is simply another form of the paradox discussed above. When we cut and then shuffle the little triangle, we are effectively cutting part A of the chessboard into two pieces, which are then swapped along the diagonal.
For the puzzle, only the cells adjacent to the cut line are important, while the rest do not matter, playing the role of decoration. However, their presence changes the nature of the paradox. Instead of the disappearance of one of several small cells (or a slightly more complex figure, say, a playing card, a human face, etc., which could be drawn inside each cell), we are faced here with a change in the area of \u200b\u200ba large geometric figure.
The square paradox
Here's another paradox with the area. By changing the position of parts A and C, as shown in fig. 58, it is possible to transform a 30 square unit rectangle into two smaller rectangles with a total area of \u200b\u200b32 square units, thus obtaining a "payoff" of two square units. As in the previous paradox, only cells adjacent to the cut line play a role here. The rest are needed only as a decoration.
In this paradox, there are two fundamentally different ways of cutting a figure into pieces.
You can start with a large 3x10 unit rectangle (top of Figure 58) by carefully drawing a diagonal in it, then the two smaller rectangles (bottom of Figure 58) will be 1/5 units shorter than their apparent size.
But you can also start with a shape made up of two neatly drawn smaller rectangles of 2x6 and 4x5 units; then the segments connecting point X with point Y and point Y with point Z will not form a straight line. And only because the obtuse angle formed by them with the apex at point Y is very close to the unfolded one, the broken line XYZ seems to be a straight line. Therefore, a shape made up of parts of small rectangles will not really be a rectangle, since these parts will overlap slightly along the diagonal. The chessboard paradox, as well as most of the other paradoxes that we are going to consider in this chapter, can also be presented in two ways. In one of them, the paradox is obtained due to a slight decrease or increase in the height (or width) of the figures, in the other, due to the increase or loss of area along the diagonal, caused either by the overlapping of the figures, as in the case just considered, or by the appearance of empty spaces, with which we will meet soon.
By changing the size of the figures and the slope of the diagonal, this paradox can be given the most varied design. You can achieve a loss or gain in area of \u200b\u200b1 square unit or 2, 3, 4, 5 units, etc.
Square variant
In one nifty variation, the original 3x8 and 5x8 rectangles are juxtaposed to form a regular 8x8 checkerboard. These rectangles are cut into parts, which, after redistribution, form a new large rectangle with an apparent increase in area of \u200b\u200bone square unit (Fig. 59).
The essence of the paradox is as follows. With a careful construction of a square drawing, a strict diagonal of a large rectangle does not work. Instead, a diamond-shaped figure appears, so elongated that its sides seem to almost merge. On the other hand, if you carefully draw the diagonal of a large rectangle; the top of the two rectangles that make up the square will be slightly higher than it should be, and the bottom rectangle will be slightly wider. Note that inaccurate closing of parts of the figure in the second cutting method is more striking than inaccuracies along the diagonal in the first; therefore the first method is preferable. As in the previously encountered examples, circles, physiognomies or some figures can be drawn inside the cells cut by a diagonal; when rearranging the constituent parts of the rectangles, these figures will become one more or less.
Fibonacci numbers
It turns out that the lengths of the sides of the four parts that make up the figures (Figs. 59 and 60) are members of the Fibonacci series, that is, a series of numbers starting with two units: 1, 1, each of which, starting from the third, is the sum of two preceding. Our row looks like 1, 1, 2, 3, 5, 8, 13, 21, 34 ...
The arrangement of the parts that were cut into a square in the form of a rectangle illustrates one of the properties of the Fibonacci series, namely the following: when any member of this series is squared, the product of two adjacent members of the series plus or minus one is obtained. In our example, the side of the square is 8, and the area is 64. The eight in the Fibonacci row is located between 5 and 13. Since the numbers 5 and 13 become the lengths of the sides of the rectangle, its area should be equal to 65, which gives an increase in area of \u200b\u200bone unit.
Thanks to this property of the series, you can build a square, the side of which is any Fibonacci number greater than one, and then cut it in accordance with the two previous numbers of this series.
If, for example, we take a square of 13x13 units, then its three sides should be divided into segments of 5 and 8 units in length, and then cut, as shown in Fig. 60. The area of \u200b\u200bthis square is 169 square units. The sides of the rectangle formed by the parts of the squares will be 21 and 8, which gives an area of \u200b\u200b168 square units. Here, due to the overlapping of parts along the diagonal, one square unit is not added, but lost.
If you take a square with a side of 5, then there will also be a loss of one square unit. A general rule can also be formulated: taking as the side of the square any number from the "first" subsequence of Fibonacci numbers (3, 8 ...) located one after another and composing a rectangle from the parts of this square, we get a clearance along its diagonal and, as a consequence, an apparent increase in area by one unit. Taking as the side of the square any number from the "second" subsequence (2, 5, 13 ...), we get overlapping areas along the diagonal of the rectangle and the loss of one square unit of area.
It is possible to construct a paradox even on a square with a side of two units. But then there is such an obvious overlap in the 3x1 rectangle that the paradox effect is completely lost.
Using other Fibonacci series for the paradox, you can get countless options. So, for example, squares based on the series 2, 4, 6, 10, 16, 26, etc., lead to losses or gains in area of \u200b\u200b4 square units. The magnitude of these losses or gains can be found by calculating for a given series the difference between the square of any of its members and the product of its two neighboring terms on the left and right. Rows 3, 4, 7, 11, 18, 29, etc., give an increase or loss of five square units. T. de Moulidar gave a drawing of a square based on the series 1, 4, 5, 9, 14, etc. The side of this square is taken equal to 9, and after converting it to a rectangle, 11 square units are lost. Rows 2, 5, 7, 12, 19 ... also give a loss or gain of 11 square units. In both cases, the overlaps (or gaps) along the diagonal are so large that they can be seen immediately.
Denoting any three consecutive Fibonacci numbers through A, B and C, and through X - the loss or gain in area, we get the following two formulas:
A + B \u003d C
B 2 \u003d AC ± X
If you substitute for X the desired gain or loss, and instead of B, the number that is taken as the length of the side of the square, then you can construct a quadratic equation from which there are two other Fibonacci numbers, although these, of course, will not necessarily be rational numbers. It turns out, for example, that by dividing a square into figures with rational side lengths, one cannot get an gain or loss of two or three square units. With the help of irrational numbers, this can of course be achieved. So, the Fibonacci series 2 1/2, 2 2 1/2, 3 2 1/2, 5 2 1/2 gives an increase or loss of two square units, and the series 3 1/2, 2 3 1 / 2, 3 · 3 1/2, 5 · 3 1/2 results in an increase or loss of three square units.
Option with a rectangle
There are many ways in which a rectangle can be cut into a small number of pieces and then folded into another rectangle of larger or smaller area. In fig. 61 depicts a paradox also based on the Fibonacci series.
Similar to the just considered case with a square, the choice of some Fibonacci number from the "second" subsequence as the width of the first rectangle (in this case 13) leads to an increase in the area of \u200b\u200bthe second rectangle by one square unit.
If, however, some Fibonacci number from the "additional" subsequence is taken as the width of the first rectangle, then the area in the second rectangle will decrease by one unit. Area losses and gains are due to small overlaps or gaps along the diagonal section of the second rectangle. Another version of such a rectangle, shown in Fig. 62, when constructing the second rectangle, increases the area by two square units.
If you place the shaded part of the area of \u200b\u200bthe second rectangle over the unshaded part, the two diagonal cuts will merge into one large diagonal. Now rearranging parts A and B (as in Fig. 61), we get a second rectangle with a larger area.
Another version of the paradox
When summing the areas of the parts, the permutation of triangles B and C in the upper part of Fig. 63 results in an apparent loss of one square unit.
As the reader will notice, this is due to the area of \u200b\u200bthe shaded parts: there are 15 shaded squares on the top of the figure, 16 shaded squares on the bottom. By replacing the shaded parts with two special shapes covering them, we arrive at a new, startling form of paradox. Now we have a rectangle in front of us, which can be cut into 5 parts, and then, swapping them, make a new rectangle, and, despite the fact that its linear dimensions remain the same, a hole with an area of \u200b\u200bone square unit appears inside (Fig. 64).
The ability to transform one shape into another, with the same external dimensions, but with a hole inside the perimeter, is based on the following. If you take point X exactly three units from the base and five units from the side of the rectangle, then the diagonal will not pass through it. However, the polyline connecting point X to the opposite vertices of the rectangle will deviate so little from the diagonal that it will be almost invisible.
After rearranging triangles B and C in the lower half of the figure, the parts of the figure will slightly overlap along the diagonal.
On the other hand, if the line connecting the opposite vertices of the rectangle is viewed as a precisely drawn diagonal at the top of the figure, the XW line will be slightly longer than three units. And as a consequence of this, the second rectangle will be slightly higher than it seems. In the first case, the missing unit of area can be considered distributed from angle to angle and forming an overlap along the diagonals. In the second case, the missing square is distributed over the width of the rectangle. As we already know from the previous one, all paradoxes of this kind can be attributed to one of these two construction options. In both cases, the inaccuracies of the figures are so insignificant that they turn out to be completely invisible.
The most elegant form of this paradox is squares, which remain squares after redistributing parts and forming a hole.
Such squares are known in countless variations and with holes in any number of square units. Some of the most interesting of them are shown in Fig. 65 and 66.
You can point to a simple formula that relates the size of the hole to the proportions of the large triangle. The three sizes that will be discussed, we will designate through A, B to C (Fig. 67).
The area of \u200b\u200bthe hole in square units is equal to the difference between the product of A and C and the closest multiple of size B. So, in the last example, the product of A and C is 25. The closest multiple of size B to 25 is 24, so the hole is one square unit. This rule applies regardless of whether the real diagonal or the point X in Fig. 67 is drawn neatly at the intersection of the square grid lines.
If the diagonal, as it should be, is drawn as a strictly straight line or if the point X is taken exactly at one of the vertices of the square grid, then no paradox is obtained. In these cases, the formula gives a hole of zero square units, meaning, of course, that there is no hole at all.
Triangle option
Let's go back to the first example of a paradox (see Fig. 64). Note that the large triangle A does not change its position, while the rest of the parts move. Since this triangle does not play a significant role in the paradox, it can be discarded altogether, leaving only the right triangle cut into four parts. These parts can then be redistributed, thus obtaining a right-angled triangle with a hole (Fig. 68), as if equal to the original.
Composing two such right-angled triangles with legs, you can build many variants of isosceles triangles, similar to that shown in Fig. 69.
As in the previously considered paradoxes, these triangles can be constructed in two ways: either draw their sides strictly rectilinear, then the X point will not fall on the intersection of the lines of the square grid, or place the X point exactly in the intersection, then the sides will be slightly convex, or concave. The latter method seems to better mask the inaccuracies of the drawing. The paradox will seem even more surprising if the lines of a square grid are drawn on the parts that make up the triangle, thereby emphasizing that the parts were made with the necessary accuracy.
By giving our isosceles triangles different sizes, we can gain or lose any even number of square units.
Several typical examples are given in Fig. 70, 71 and 72.
Composing two isosceles triangles of any of these types with the bases, you can build a variety of options for a rhombic type; however, they do not add anything fundamentally new to our paradox.
Four-piece squares
All the types of paradoxes with a change in area that we have considered so far are closely related in terms of the method of construction. However, there are paradoxes obtained by completely different methods. You can, for example, cut a square into four parts of the same shape and size (fig. 73), and then compose them in a new way, as shown in fig. 74. This results in a square, the dimensions of which do not seem to have changed and at the same time with a hole in the middle.
Similarly, you can cut a rectangle with any aspect ratio. It is curious that the point A, at which two intersect, seem to be unchanged and at the same time with a hole in the middle.
Similarly, you can cut a rectangle with any aspect ratio. It is curious that point A, at which two mutually perpendicular cut lines intersect, can be located anywhere inside the rectangle. In each case, when the parts are redistributed, a hole appears, and its size depends on the value of the angle formed by the cut lines with the sides of the rectangle.
This paradox is comparatively simple, but it loses a lot due to the fact that even a superficial study shows that the sides of the second rectangle should be slightly larger than the sides of the first.
A more complex way of cutting a square into four parts, which produces an inner hole, is shown in Fig. 75.
It is based on the chessboard paradox that opens this chapter. Note that when redistributing parts, two of them must be turned upside down. Note also that by discarding part A, we get a right-angled triangle composed of three parts, inside which a hole can be formed.
Three-piece squares
Is there a way to cut a square into three pieces that can be rebuilt to make a square with a hole in it? The answer is yes. One neat solution is based on the application of the paradox discussed in the previous chapter.
Instead of placing the pictures in a special way, and making the cut in a straight line (horizontally), the pictures are placed on one straight line, and the cut is made in steps. The result is amazing: not only does the picture disappear, but a hole appears at the point where it disappears.
Two-piece squares
Can you do the same with two parts?
I do not think that in this case it is possible by any method to obtain an inner hole in the square by imperceptibly increasing its height or width. However, it has been shown that the paradox with a hole in a square cut in two can be built on the principle that applies to the paradox of the disappearing warrior. In this case, instead of placing the figures in a spiral or a step, they are placed strictly along the circumference, while the cut is made spiral or stepped; in the latter case, it has the form of a cogwheel with teeth of various sizes. When this wheel rotates, one figure disappears and a hole appears in its place.
The stationary and rotating parts fit neatly together only when the hole appears. In the initial position, small gaps are visible at each tooth, if the incision was stepped, or one continuous circular lumen with a spiral cut.
If the original rectangle is not a square, you can cut it in two and then get a hole inside with very little noticeable change in its outer dimensions. In fig. 76 shows one option.
Both parts are identical both in shape and size. The easiest way to demonstrate this paradox is as follows: cut out pieces of cardboard, fold them into a rectangle without a hole, put them on a piece of paper and trace around the perimeter with a pencil. Now folding the parts in a different way, you can see that they still do not go beyond the drawn line, although a hole has formed in the middle of the rectangle.
To our two parts, you can, of course, add a third, made in the form of a strip, which, when applied to one of the sides of the rectangle, turns it into a square; thus we get another way of cutting the square into three parts, giving an inner hole.
Curved and 3D options
The examples we have given clearly show that the area of \u200b\u200bparadoxes with a change in area is just beginning to be developed. Are there any curved shapes, such as circles or ellipses, that can be cut into pieces and then re-arranged so that the inner holes are obtained without noticeable distortion of the shape?
Are there three-dimensional figures that are specific to three dimensions, that is, are not a trivial consequence of two-dimensional figures? After all, it is clear that to any flat figure with which we met in this chapter, you can "add a dimension" by simply cutting it out of a fairly thick cardboard, the height of which is equal to the "length of the third dimension").
Is it possible to cut a cube or, say, a pyramid in a not very complicated way into parts so that, by composing them in a new way, you get noticeable voids inside?
The answer will be this: if you do not limit the number of parts, then such spatial figures are not at all difficult to indicate. This is clear enough in the case of a cube.
Here the inner emptiness can be obtained, but the question of the smallest number of parts with which this can be achieved is more difficult. It can certainly be made from six parts; it is possible that this can be achieved with a smaller number.
Such a cube can be effectively demonstrated as follows: take it out of a box made exactly according to the cube, disassemble it into parts, while finding a ball inside, put the parts back into a solid cube and show that it (without the ball) still fills the box tightly. We will make an assumption that there should be many such figures, both flat and spatial, which are also distinguished by simplicity and grace of form. Future explorers of this curious area will have the pleasure of discovering them.
"Application of a derivative to problem solving"
(Grade 10)
The methodological system of the teacher's activity in this lesson involves the formation of the ability of students to independently plan and carry out research work in stages. The student has the right to consult with the teacher, discuss, receive advice or tips from the teacher in order to help the child understand the variety of solutions and determine the right one.
The lesson discusses theoretical material, the class is divided into groups to ensure a variety of suggested ways of reasoning, followed by the selection of the most acceptable of them.
Along with independent activity, it is advisable to use differentiated tasks of different levels in the lesson and evaluate them accordingly.
Analysis of the results of the performance of these tasks by students, in addition to information about their assimilation, gives the teacher a picture of the main difficulties of students, their main gaps, which helps to outline the main ways of solving problems.
The purpose of the lesson: mastering skills independently in a complex to apply knowledge, skills and abilities, to carry out their transfer to new conditions, using the research method.
Tasks:
Educational and cognitive: consolidation, systematization and generalization of knowledge and skills associated with mastering the concept of "the greatest and least value of a function"; practical application of the skills and abilities being formed.
Developing: development of skills to work independently, clearly express ideas, conduct self-assessment of educational activities in the classroom.
Communicative: ability to participate in discussions, listen and hear.
During the classes
Organizing time
1. Every person from time to time finds himself in a situation when it is necessary to find the best way to solve a problem, and mathematics becomes a means of solving problems of organizing production, searching for optimal solutions. An important condition for increasing production efficiency and improving product quality is the widespread introduction of mathematical methods into technology.
Reiteration
Among the problems of mathematics, an important role is assigned to problems for extremums, i.e. the tasks of finding the highest and lowest value, the best, the most profitable, the most economical. Representatives of various specialties have to deal with such tasks: process engineers try to organize production in such a way that they get as much production as possible, designers want to plan the device on a spacecraft in such a way that the mass of the device is as small as possible, economists try to plan the attachment of factories to sources of raw materials in such a way so that transportation costs are minimal. It can be said that the tasks of finding the least and greatest value are of great practical application. Today in the lesson we will deal with such problems.
Consolidation of the studied material
2. Two "strong" pupils are called to the board to solve tasks (10 min.).
1st student: Given a tank without a lid in the form of a rectangular parallelepiped, at the base of which is a square and the volume of which is 108 cm 3. At what size of the tank will the least amount of material be used for its manufacture?
Decision: Let's denote the side of the base through x cm, express the height of the parallelepiped. Let us find the sign of the derivative on the intervals. The derivative changes its sign from "-" to "+". Hence, x \u003d 6 is the minimum point, therefore, S (6) \u003d 108 cm 2 is the smallest value. This means that the side of the base is 6 cm, the height is 12 cm.
2nd student: A rectangle of the largest area is inscribed in a circle with a radius of 30 cm. Find its dimensions.
Decision: Let's denote one side of the rectangle through x cm, then we express the area of \u200b\u200bthe rectangle. Let us find the sign of the derivative on the interval (0; 30) and on the interval (30; 60). The derivative changes its sign from "+" to "-". Hence x \u003d 30 is the maximum point. Therefore, one side of the rectangle is 30, the other is 30.
3. At this time, I dida mutual check is made on the topic "Derivative Application" (1 point is given for each correct answer). Each student answers and for verification passes his answer to a neighbor on the desk.
The questions are written on a portable board, only the answer is given:
A function is called increasing over a given interval if ...
A function is called decreasing on a given interval if ...
A point x 0 is called a minimum point if ...
A point x 0 is called a maximum point if ...
Stationary points of a function are called points ...
Write the general form of the tangent equation
The physical meaning of the derivative
Drawing conclusions
4. The class is divided into groups. Groups perform tasks to find the minimum and maximum of a function.
5. The word is given to "strong" students. Class members check their solutions (10 minutes).
6. Issues of optional tasks for each group (10 min.).
1 group.
To mark "3"
For the function f (x) \u003d x 2 * (6-x) find the smallest value on the segment.
Solution: f (x) \u003d x 2 * (6-x) \u003d 6x 2 + x 3; f / (x) \u003d 12x-3x 2; f / (x) \u003d 0; 12x-3x 2 \u003d 0; x 1 \u003d 0; x 2 \u003d 4;
f (0) \u003d 0; f (6) \u003d 0; f (4) \u003d 32-max.
To the "4"
A rectangle of the largest area should be made of a wire 20 cm long. Find its dimensions.
Solution: Let's designate one side of the rectangle through x cm, then the second will be (10-x) cm, area S (x) \u003d (10-x) * x \u003d 10x-x 2; S / (x) \u003d 10-2x; S / (x) \u003d 0; x \u003d 5. By the condition of the problem x (0; 10). Find the sign of the derivative on the interval (0; 5) and on the interval (5; 10). The derivative changes its sign from "+" to "-". Hence: x \u003d 5 is the maximum point, S (5) \u003d 25 cm 2 is the largest value. Therefore, one side of the rectangle is 5 cm, the second is 10-x \u003d 10-5 \u003d 5 cm.
To the "5"
The plot with an area of \u200b\u200b2400 m 2 must be divided into two rectangular sections so that the length of the fence is the smallest. Find the sizes of the parcels.
Solution: Let's designate one side of the site through x m, write down the length of the fence and find the derivative P / (x) \u003d 0; 3x 2 \u003d 4800; x 2 \u003d 1600; x \u003d 40. We take only a positive value according to the condition of the problem.
Let us find the sign of the derivative on the interval (0; 40) and on the interval (40 ;?). The derivative changes its sign from "-" to "+". Hence, x \u003d 40 is the minimum point, therefore, P (40) \u003d 240 is the smallest value, which means that one side is 40 m, the other is 60 m.
Group 2.
To mark "3"
For the function f (x) \u003d x 2 + (16-x) 2 find the smallest value on the segment.
Solution: f / (x) \u003d 2x-2 (16-x) x \u003d 4x-32; f / (x) \u003d 0; 4x-32 \u003d 0; x \u003d 8; f (0) \u003d 256; f (16) \u003d 256; f (8) \u003d 128-min.
To the "4"
The plot is rectangular with one side adjacent to the building. With the given dimensions of the perimeter in m, it is necessary to enclose the site so that the area is largest.
To the "5"
From a rectangular sheet of cardboard with sides 80 cm and 50 cm, you need to make a rectangular box by cutting out squares along the edges and bending the resulting edges. How high should the box be for the largest volume?
Let's denote the height of the box (this is the side of the cut-out square) through x m, then one side of the base will be (80-2x) cm, the second - (50-2x) cm, volume V (x) \u003d x (80-2x) (50-2x ) \u003d 4x 3, 260x 2 + 4000x; V / (x) \u003d 12x 2 -520x + 4000; V / (x) \u003d 0; 12x 2 -520x + 4000 \u003d 0.
By the condition of the problem x (0; 25); x 1 (0; 25), x 2 (0; 25).
Let us find the sign of the derivative on the interval (0; 10) and on the interval (10; 25). The derivative changes its sign from "+" to "-". Hence x \u003d 10 is the maximum point. Therefore, the height of the box \u003d 10 cm.
Group 3.
To mark "3"
For the function f (x) \u003d x * (60s) find the largest value on the segment.
Solution: f (x) \u003d x * (60's) \u003d 60x-x 2; f / (x) \u003d 60-2x; f / (x) \u003d 0; 60-2x \u003d 0; x \u003d 30; f (0) \u003d 0; f (60) \u003d 0; f (30) \u003d 900-max.
To the "4"
The plot is rectangular with one side adjacent to the building. With a given perimeter size of 20 m, it is necessary to fence off the site so that the area is as large as possible.
Let's designate one side of the rectangle through x m, then the second will be (20-2x) m, the area S (x) \u003d (20-2x) x \u003d 20x-2x 2; S / (x) \u003d 20-4x; S / (x) \u003d 0; 20-4x \u003d 0; x \u003d 5. By the condition of the problem x € (0; 10). Find the sign of the derivative on the interval (0; 5) and on the interval (5; 10). The derivative changes its sign from "+" to "-". Hence x \u003d 5 is the maximum point. Therefore, one side of the site \u003d 5 m, the second - 20-2 * 5 \u003d 10 m.
To the "5"
To reduce the friction of the fluid against the walls and the bottom of the channel, the area wetted by it should be made as small as possible. It is required to find the dimensions of an open rectangular channel with a cross-sectional area of \u200b\u200b4.5 m2, at which the wetted area will be the smallest.
Let's designate the depth of the ditch through x m, P / (x) \u003d 0; 2x 2 \u003d 4.5; x \u003d 1.5. We take only a positive value according to the condition of the problem. Let us find the sign of the derivative on the interval (0; 1.5) and on the interval (1.5 ;?). The derivative changes its sign from "-" to "+". Hence, x \u003d 1.5 is the minimum point, therefore, P (1.5) \u003d 6 m is the smallest value, which means that one side of the ditch is 1.5 m, the other is 3 m.
4 group.
To mark "3"
For the function f (x) \u003d x 2 (18-x) find the largest value on the segment.
f (x) \u003d x 2 (18-x) \u003d 18x 2 -x 3; f / (x) \u003d (18x 2 -x 3) /; f / (x) \u003d 0; 36x-3x 2 \u003d 0; x 1 \u003d 0; x 2 \u003d 12 f (0) \u003d 0; f (18) \u003d 0; f (12) \u003d 864-max.
To the "4" mark.
The plot is rectangular with one side adjacent to the building. With a given perimeter size of 200 m, it is necessary to fence off the site so that the area is as large as possible.
Let's designate one side of the rectangular section through x m, then the second will be (200-2x) m, the area S (x) \u003d (200-2x) x \u003d 200x-2x 2; S / (x) \u003d 200-4x; S / (x) \u003d 0; 200-4x \u003d 0; x \u003d 200/4 \u003d 50. By the condition of the problem x (0; 100). Find the sign of the derivative on the interval (0; 50) and on the interval (50; 100). The derivative changes its sign from "+" to "-". Hence x \u003d 50 is the maximum point. Therefore, one side of the site \u003d 50 m, the second - 200-2x \u003d 100 m.
To the "5"
It is required to make an open box in the form of a rectangular parallelepiped with a square base, with the smallest volume, if 300 cm 2 can be spent on its manufacture.
Let's designate one side of the base through x cm and express the volume, then V / (x) \u003d 0 300-3x 2 \u003d 0; x 2 \u003d 100; x \u003d 10. We take only a positive value according to the condition of the problem.
Let us find the sign of the derivative on the interval (0; 10) and on the interval (10; 0). The derivative changes its sign from "-" to "+". Hence, x \u003d 10 is the minimum point, therefore, V (10) \u003d 500 cm 3 is the smallest value, which means that the side of the base is 10 cm, the height is 50 cm.
Questions for the class
7. Delegates from the groups explain the solution of the selected problems (10 min.).
8. Taking into account the points in warm-up and work in groups, marks are given for the lesson.
Lesson summary
Homework
The solution to the problem is one point higher; students who complete the task on "5" are exempted from homework.
Analysis of the results of the performance of these tasks by students, in addition to information about their assimilation, gives the teacher a picture of the main difficulties of students, their main gaps, which helps to outline the main ways to eliminate them.
| FOMKINA TATIANA FEDOROVNA |
BUSINESS CARD |
|
Position | Russian language and literature teacher |
Place of work | Municipal educational institution "Secondary school No. 9" of the city of Orenburg |
Work experience in the position | |
Competitive score | |
The topic of teaching experience | Formation of linguistic competence of students on the basis of the activity-system approach in teaching the Russian language according to S.I. Lvov |
The essence of the teacher's methodological system, reflecting the leading ideas of experience | The essence of the teacher's methodological system is in the organization of educational activities as a movement from a linguistic question (allowing students to draw attention of students to the meaningful linguistic essence of a particular spelling) to a method of action (based on a rule, referring to a dictionary), and then to a result (free operating rules in the course of writing or using a spelling dictionary). |
Work on the dissemination of personal experience, presentation of the methodological system at various levels (forms, intellectual products) | Fomkina T.F. summarized in 2009 at the level of the MOU "Secondary School No. 9" and approved by the methodological council. In 2009 and 2010. represented among the teachers of the city of Orenburg at the municipal level. Tatyana Fedorovna spoke at the district methodological associations on the following issues: "The use of ICT in the lessons of the Russian language and literature as a means of forming linguistic competence", "An action-oriented approach to building educational standards." |
The effectiveness of the implementation of the methodological system | Formation of sustainable positive motivation and increasing student interest in the subject; Positive dynamics in the attitude of students to the teacher, the lessons of the Russian language and literature, the development of students' ability to predictive activity and the activation of cognitive processes; A significant increase in the quality of creative works, essays, which is confirmed by the results of the final exams: in 2007, according to the results of the SIA, academic performance was 100%, the number of those who coped with tasks for “4” and “5” was 87%; in 2008, according to the results of the Unified State Exam, academic performance was 100%, the number of those who completed the tasks for “4” and “5” was 92%, the highest score was 87; in 2009, according to the USE results, academic performance was 100%, the number of those who coped with tasks for “4” and “5” was 58%, the highest score was 96; Increasing the number of students participating in scientific and practical conferences, competitions, olympiads: X district scientific and practical conference of students "You are an Orenburzhets" (III place), XV city conference of students "Intellectuals of the XXI century" (diploma for "Comprehensive study of the family"), All-Russian correspondence competition "Knowledge and Creativity", 2010 (III place, laureate), regional intramural-extramural competition "Fatherland", 2009 (III place), VI International Olympiad in Fundamentals of Sciences, 2010 (diplomas of I and II degrees), International game-competition "Russian Bear", 2010 (15th place in the region). Monitoring of educational activities shows a high level of student learning by Tatyana Fedorovna Fomkina: Russian language - 69% (2009), literature - 77% (2009). |
EXPERIENCE MATERIALS
Lesson in assimilation of new knowledge
with multilevel differentiation of learning
"NOT with nouns"
(Grade 5)
The presented summary of the lesson is compiled in accordance with the "Program in the Russian language for grades 5-6" by S.I. Lvova (M .; "Mnemosyne", 2008). The lesson is aimed at the formation of linguistic, language and speech competence of students. The material included in the lesson is teaching, developing, educative.
Lesson Objectives:
1) develop communication skills: formulate a question and give an answer to a grammatical topic; carry out speech interaction in a mobile group; create your own texts on a given topic;
2) form linguistic and linguistic competence: know the spelling rule NOT with a noun ; be able to use the algorithm to apply this rule in practice; repeat spelling « NOT with a verb " , noun rule;
3) foster a respectful attitude towards the word as a spiritual value of the people.
Equipment:multimedia equipment, video presentation, support cards, test, files with a research task.
During the classes
Organizing time
Dear colleagues! Yes, colleagues. I, guys, did not call you that by chance. Today we will deal with a common cause: solve linguistic problems, discover the secrets of writing words. Indeed, according to Leo Nikolaevich Tolstoy, "The word is a great deed ... The word can serve love, but the word can serve enmity and hatred" (epigraph to the lesson).
Linguistic warm-up "Yes - no"
Here is a word skill and will help you cope with the linguistic workout called "Yes - no". The rules of this warm-up are as follows: I made a rule, and you will try to guess it by asking leading questions, which should be formulated in such a way that I can answer with the words "yes" or "no." I will evaluate your answers today with tokens. Ask me questions.
Pupils ask the teacher questions. For instance:
1. We learned this rule in the 5th grade? (Yes)
2. Is this a rule about spelling words? (Not)
3. Is this a rule about parts of speech? (Yes)
4. Is this a rule about a noun? (Yes)
- Well done! You guessed it!
Knowledge update
Now let's remember what a noun is. But let's talk about it in a chain, passing the baton to each other, like athletes in a competition. Anyone who wants can take advantage of the answer help cards... I will evaluate your answers with tokens ( student responses).
We did a great job! We need to know the rule about the noun in order to be able to distinguish nouns from other parts of speech.
We will test this skill by performing oral distribution dictation.
Read the words carefully (The image fades when you click on the projector screen).
But what is it? What happened to the image? Guys, there is a mistake!
Catch her! ("Catch the mistake" technique)
“Resent” must be written in one piece.Why?
This is a verb that is not used without NOT.
(Mouse click)
The task: Divide the words into two groups by parts of speech. (Students complete the assignment)
1. What parts of speech do you come across? (Nouns and verbs)
2. Name the nouns.
3. Name the verbs.
4. How to spell NOT with a verb?
Goal setting
So, knowing the rule about noun and spelling NOT with a verb will help us cope with a new topic that sounds like this: "NOT with nouns".Write it down in your notebook.
I wrote down the train of our thoughts in "Thinkingsheet", which consists of three columns: “I know”, “I want to know”, “I learned (a)”.
In the graph "I know" given the rule on which we will rely today. This is the rule about writing NOT with a verb .
In the graph "I want to know" the question of the day was formulated: "Find out when NOT with a noun is spelled together, and when - separately."
In the graph "I found out" we will write down the answer to this question.
But first let's do vocabulary work.
Guys, who are they? ignorantand ignorant?What kind of people do we call that? (Answers from students)
Write down these words and their lexical meanings in a notebook. Now make up phrases or sentences with them (optional).
Learning new material
Why do you guys think the words "ignorant" and "ignorant" are spelled together? (Because they are not used without NOT)Report
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Sections: Maths
The purpose of the lesson:
- Generalization and systematization of the knowledge gained.
- Expansion of students' ideas about solving problems to find the highest and lowest values.
During the classes
Stage 1 of the lesson
Teacher introduction: every person from time to time finds himself in a situation when it is necessary to find the best way to solve a problem.
For example: process engineers are trying to organize production in such a way as to get as much production as possible, designers want to plan instruments on a spacecraft in such a way that the mass of the instrument is as small as possible, and so on.
We can say that the tasks of finding the greatest and least important have practical applications.
To prove my words, I want to quote from the story of L.N. Tolstoy "Does a man need much land" about the peasant Pakhom, who bought land from the Bashkirians.
- And what will be the price? - says Pakhom.
- We have one price: 1000 rubles. per day.
Pakhom did not understand.
- What is this measure - a day? How many tithes will there be?
- We do not know how to count this, - he says. And we sell in a day; how much you get around in a day, so is yours, and the price is 1000 rubles.
I was surprised by Groin.
- But this, - he says, - will go around the land a lot in a day.
The foreman laughed.
“All yours,” he says. - Only one agreement: if you don't come back a day to the place from which you start, your money is gone.
- And how, - says Pakhom, - to mark where I will pass?
- And we will stand at the place where you love; we will stand, and you go, make a circle, and take a scraper with you and, where necessary, notice, at the corners of the hole, a swarm of turf; then we will go from hole to hole with a plow. Take whatever circle you want, just before sunset come to the place from which you took it. What you get around is all yours.
The figure that Pakhom got is shown in the figure. What is this figure? (Rectangular trapezoid)
Question: What do you think, is the largest area received by Pakhom. (taking into account that the plots are usually in the shape of a quadrangle)? Today in the lesson we will find out.
To solve this problem, we need to remember what stages are contained in solving extreme problems?
- The task is translated into the language of the function.
- The analysis tools are used to find the largest or smallest value.
- Find out what practical meaning the obtained result has.
Problem number 1 (We'll solve it with the whole class)
The perimeter of the rectangle is 120 cm. How long should the sides of the rectangle have in order for the area to be the largest.
We return to the problem with which the lesson began. Did Pakhom get the most area (considering that the plots are usually in the shape of a quadrangle)? We discuss with the students what the largest area Pakhom could get.
2 Stage of the lesson
In advance, the written tasks are explained on the board (there are two of them).
Problem number 1
Find under what conditions the consumption of tin for the manufacture of cans of cylindrical shape of a given capacity will be the least.
I draw the attention of the guys that hundreds of millions of cans are produced in our country and the saved consumption of tin by at least 1% will allow us to additionally produce millions of cans.
Problem number 2
The boats are located at a distance of 3 km from the nearest point A of the coast. At point B, located at a distance of 5 km from A, there is a fire. The boatman wants to come to the rescue, so he needs to get there in the shortest possible time. The boat moves at a speed of 4 km / h, and the passenger is 5 km / h. What point on the shore should the boatman dock?
Stage 3 of the lesson
Work in groups with the subsequent protection of tasks.
Problem number 1
One of the faces of a rectangular parallelepiped is a square. The sum of the lengths of the edges outgoing from one vertex of the parallelepiped is 12. Find its largest possible volume.
Problem number 2
To mount the equipment, a stand with a volume of 240 dm 3 in the form of a rectangular parallelepiped is required. The base of the stand, which will be installed in the floor, is a rectangle. The length of the rectangle is three times its width. The rear, longer wall of the stand will be built into the wall of the workshop. When installing the stand, its walls, which are not installed in the floor or in the wall, are connected to each other by welding. Determine the dimensions of the stand that will give you the shortest overall weld length.
Problem number 3
A beam with a rectangular section of the largest area is cut out of a round log. Find the dimensions of the section of the beam if the radius of the section of the log is 30 cm.
Problem number 4
From a rectangular sheet of cardboard with sides 80 cm and 50 cm, you need to make a rectangular box by cutting out squares along the edges and bending the resulting edges. How high should the box be in order for its volume to be greatest. Find this volume.
Stage 4 of the lesson
Solving problems for assessment by choice.
Problem number 1
From a wire 80 cm long, you need to make a rectangle of the largest area. Find its dimensions.
Problem number 2
The sum of the lengths of the edges of a regular triangular prism is 18√3. Find the largest possible volume of such a prism.
Problem number 3
The diagonal of a rectangular parallelepiped, one of the side faces of which is a square, is equal to 2√3. Find the largest possible volume of such a parallelepiped.
5 stage of the lesson
Example 1 ... A rectangle of the largest area should be made of a wire 20 cm long. Find its dimensions.
Decision:Let's denote one side of the rectangle through x cm, then the second will be (10-x) cm, area S (x) \u003d (10-x) * x \u003d 10x-x 2;
S / (x) \u003d 10-2x; S / (x) \u003d 0; x \u003d 5;
By the condition of the problem x (0; 10)
Find the sign of the derivative on the interval (0; 5) and on the interval (5; 10). The derivative changes the sign from “+” to “-”. Hence: x \u003d 5 maximum point, S (5) \u003d 25cm 2 - the highest value. Therefore, one side of the rectangle is 5cm, the second is 10-x \u003d 10-5 \u003d 5cm;
Example 2. The plot, with an area of \u200b\u200b2400m 2, must be divided into two rectangular sections so that the length of the fence is the smallest. Find the sizes of the parcels.
Decision:Let's designate one side of the site through x m, then the second will be m, the length of the fence P (x) \u003d 3x +;
P / (x) \u003d 3-; P / (x) \u003d 0; 3x 2 \u003d 4800; x 2 \u003d 1600; x \u003d 40. We take only a positive value according to the condition of the problem.
By the condition of the problem x (0;)
Let us find the sign of the derivative on the interval (0; 40) and on the interval (40;?). The derivative changes the sign from “-” to “+”. Hence, x \u003d 40 is the minimum point, therefore, P (40) \u003d 240m is the smallest value, which means that one side is 40m, the other \u003d 60m.
Example 3. The plot is rectangular with one side adjacent to the building. With a given perimeter size of 1 m, it is necessary to fence off the site so that the area is as large as possible.
Decision:
Let's designate one side of the rectangular section through x m, then the second will be (-2x) m, the area S (x) \u003d (-2x) x \u003d x -2x 2;
S / (x) \u003d -4x; S / (x) \u003d 0; -4x; x \u003d;
By the condition of the problem x (0;)
Let us find the sign of the derivative on the interval (0;) and on the interval (;). The derivative changes the sign from “+” to “-”. Hence x \u003d maximum point. Therefore, one side of the plot \u003d m, the second -2x \u003d m;
Example 4. From a rectangular sheet of cardboard with sides 80cm and 50cm, you need to make a rectangular box by cutting out squares along the edges and bending the resulting edges. How high should the box be for the largest volume?
Decision:We denote the height of the box (this is the side of the cut out square) through x m, then one side of the base will be (80-2x) cm, the second (50-2x) cm, volume V (x) \u003d x (80-2x) (50-2x) \u003d 4x \u200b\u200b3 -260x 2 + 4000x;
V / (x) \u003d 12x 2 -520x + 4000; V / (x) \u003d 0; 12x 2 -520x + 4000 \u003d 0; x 1 \u003d 10; x 2 \u003d
By the condition of the problem x (0; 25); x 1 (0; 25), x 2 (0; 25)
Let us find the sign of the derivative on the interval (0; 10) and on the interval (10; 25). The derivative changes the sign from “+” to “-”. Hence x \u003d 10 maximum point. Therefore, the height of the box \u003d 10cm.
Example 5. The plot is rectangular with one side adjacent to the building. With a given perimeter size of 20 m, it is necessary to fence off the site so that the area is as large as possible.
Decision:
Let's designate one side of the rectangle through x m, then the second will be (20 -2x) m, the area S (x) \u003d (20-2x) x \u003d 20x -2x 2;
S / (x) \u003d 20 -4x; S / (x) \u003d 0; 20 -4x \u003d 0; x \u003d \u003d 5;
By the condition of the problem x (0; 10)
Find the sign of the derivative on the interval (0; 5) and on the interval (5; 10). The derivative changes the sign from “+” to “-”. Hence, x \u003d 5 is the maximum point. Therefore, one side of the site \u003d 5m, the second 20 -2x \u003d 10m;
Example 6 . To reduce the friction of the fluid against the walls and the bottom of the channel, the area wetted by it should be made as small as possible. It is required to find the dimensions of an open rectangular channel with a cross-sectional area of \u200b\u200b4.5 m2, at which the wetted area will be the smallest.
Decision:
Let's designate the depth of the ditch through x m, then the width will be m, P (x) \u003d 2x +;
P / (x) \u003d 2-; P / (x) \u003d 0; 2x 2 \u003d 4.5; x \u003d 1.5. We take only a positive value according to the condition of the problem.
By the condition of the problem x (0;)
Let us find the sign of the derivative on the interval (0; 1.5) and on the interval (1.5;?). The derivative changes the sign from “-” to “+”. Hence, x \u003d 1.5 is the minimum point, therefore, P (1.5) \u003d 6m is the smallest value, which means that one side of the ditch is 1.5m, the other \u003d 3m.
Example 7. The plot is rectangular with one side adjacent to the building. With a given perimeter size of 200m, it is necessary to enclose the site so that the area is as large as possible.
Khrestina Nadezhda Mikhailovna, teacher for developing work with children, NOU DOD "DRTs" Wonderland ", Ryazan [email protected]
Application of TRIZ elements in mathematics lessons
Annotation. The article discusses the use of elements of the structure of a creative lesson in the innovative pedagogical system NFTMTRIZ in mathematics lessons. The author proposes a methodological development of a mathematics lesson in grade 5, where it is demonstrated how you can develop the creative abilities of students within the framework of the school curriculum. Key words: universal educational actions, creative thinking, system-activity approach, creative lesson, reflection.
Mathematics is a science that is vital for everyone. From a very young age, a child is surrounded by a world of numbers, shapes, etc. And at the same time, this world is very complex and multifaceted. Many children, faced with difficulties in studying the material, lose interest in the subject and "ignorance" accumulates like a snowball. Therefore, the teacher faces a problem: not only to teach, but also to instill interest, and therefore, to give the child the tools for independent mastering of new knowledge (universal learning actions). The teacher's task is to make the lesson interesting, exciting, using a variety of teaching methods, to systematically develop creative skills in the child. thinking, the ability to work with a problem and solve it, draw conclusions, look for new original approaches, see the beauty of the results obtained. The premise for this is the Federal State Educational Standard (FSES) of basic general education of December 17, 2010. It is based on a system-based activity approach, with the value of free and responsible personality of the student. The standard dictates that we move away from the classroom system of Jan Amos Comenius, in which the teacher is the “storyteller” and the students are the “retelling”. New types of lesson, such as: “brainstorming”, debate, project activities, will help the child in a constantly changing world. What results should a teacher get as a result of his work? The teacher needs to cultivate in the student patriotism, love for the motherland, history, language and culture of his people; form a responsible attitude to learning, be capable of self-development and self-education based on motivation for learning and knowledge, conscious choice of profession; to form communicative competence; the ability to set goals, look for ways to achieve them, master the basics of self-control, etc. Also, the student must have sufficient knowledge and competencies, be able to take responsibility for his actions and their consequences, respect the law, be a free and responsible, tolerant citizen. The forward movement of science and technology leads to an increase in the number of inventions and new professions, the student must be prepared for the constantly changing demands of the labor market. The foregoing allows us to conclude that in order to achieve all these results, a teacher must not only transfer knowledge, he must "teach to learn." to understand that objective results are no longer the only main ones, he also needs to form personal and metasubject ones. The very formulation of the results has changed, since the child must now master the methods of action, i.e. universal educational activities, which are metasubject results. Only a set of universal actions will make it possible to form the student's ability to learn as a system. One of the assistants for the teacher in planning the FGOS lesson is the flow chart of the lesson. It makes it possible to visually trace how and at what stage certain universal educational actions are formed. To achieve the goals, the teacher can be helped by the use of elements of the creative pedagogical system for the continuous formation of creative thinking (NFTM), in which there are tools of the theory of inventive problem solving (TRIZ). This allows students to develop creative imagination and imagination, systemic and dialectical thinking. , allows you to make the lesson brighter, less stressful for the child, to keep the child in concentration throughout the activity, and most importantly, not to provide him with ready-made knowledge, but to give him the opportunity to get it on his own. Also, an important issue is a partial transition from closed-type problems to open-type problems. type, affecting the everyday experience of students, make students think when reading the condition, since it is insufficient, "vague", may contain an excess of information. A variety of methods of solution leads to the destruction of psychological inertia - the habit of standard actions in a familiar situation or the desire to think and act in accordance with the accumulated experience. A set of possible answers helps to teach the child reflection and self-esteem. It is impossible to talk about a complete rejection of closed tasks. They are good in small quantities, when you just need to get your hands on a specific formula or property. But the explanation of the new material cannot be without a problem. After all, the first question after reading the topic in the lesson in the head of children: "Why do I need this?" or "Where will it be useful to me?" All of the above is given to us by the NFTM system - the continuous formation of creative thinking and the development of children's creative abilities. I present a 5th grade mathematics lesson, with elements of the structure of a creative lesson in the innovative pedagogical system NFTMTRIZ. Technological map of a mathematics lesson in 5 class on the topic “Area of \u200b\u200ba rectangle. Units of area "Lesson type: Lesson in studying new material. Lesson objectives: 1. Subject: to form students' idea of \u200b\u200bthe area of \u200b\u200ba figure, to establish connections between the units of measurement of area, to acquaint students with the formulas for the area of \u200b\u200ba rectangle and a square. 2. Personal: to form the ability to determine methods of action within the framework of the proposed conditions and requirements, to adjust their actions in accordance with the changing situation. Metasubject: to form the ability to see a mathematical problem in the context of a problem situation, in the surrounding life. Planned results:
students will get an idea of \u200b\u200bthe area of \u200b\u200bfigures and its properties, learn how to establish connections between units of measurement of area, apply the formulas for the area of \u200b\u200ba rectangle and a square; gain the ability to analyze, compare, generalize, draw conclusions; students will develop cognitive interest through the game moments of the "little miracle"; acquire communication skills work in a group and in pairs. Textbook: A.G. Merzlyak, V.B. Polonsky, M.S. Yakir. Grade 5 Mathematics. Textbook for students of educational institutions. 2014.
Lesson stages Stage tasks Teacher activities Student activities UUD1 Motivation Create a favorable psychological mood for work, motivate students to work Greetings, check preparedness for the class, organize children's attention Focus with game bones: first in a transparent case 1 large bone, after hitting the cover of the case in it 8 small ones appear. –How did it happen? –What did we do in the last lesson? –Today we will continue to work with rectangles. They are included in the business rhythm of the lesson.
The guys are trying to solve the trick. They activate the knowledge of the previous lesson.
Personal: self-determination. Regulatory: self-organization. Communicative: planning educational cooperation with the teacher and peers. Cognitive: skills of research activity. 2. Substantive part. Providing perception, comprehension and primary memorization of the studied topic by children: the area of \u200b\u200ba rectangle. Pictures are displayed on a multimedia projector. Problem. The neighbors have discord. The owner of the blue section, in order to get to his garden, must go through the red section of the neighbor. What to do? Entrance to the sites
Fig.1 We know from experience that equal plots of land have equal areas. - What conclusion can we draw? Problem: The man decided to paint the floor in his country house. But the floor has an unusual shape. But he does not know how much paint is needed, it says 100g per 1m2 on the paint can. The area of \u200b\u200bthe smaller figure is 12m2, the area of \u200b\u200bthe larger one is –20m 2. What to do?
They put forward versions of the settlement of the dispute. Together with the teacher, they choose the right one: the blue must take a piece of red land, and in return give him an equal size.
They conclude: equal figures have equal areas. The guys put forward versions, together we choose the correct one: we need to add the areas of the two figures and find the paint consumption. The students themselves deduce the second property: The area of \u200b\u200bthe figure is equal to the sum of the areas of the figures from which it consists. Personal: self-determination. Regulatory: development of regulation of educational activity. Communicative: the ability to work in a team, hear and respect the opinions of others, the ability to defend one's position. Cognitive: skills of research. Development of creative thinking.
Fig. 2 Heuristic conversation with elements of the method of trial and error. There is a ruler, compasses, protractor on the teacher's desk. We talked about the area, but how can we measure it? Let's measure the area of \u200b\u200bour board. –What do we have to measure segments? –– What is to measure angles? ”We conclude: for the unit of measurement of area, we choose a square, the side of which is equal to a unit segment. How do we call such a square? To measure an area, you need to calculate how many unit squares fit in it?
The guys go through all the possible tools, come to the conclusion that they are not enough.
–Rule, unit segment –Protractor, unit angle. –Individual. One of the students goes to the blackboard, counts the area of \u200b\u200bthe board using a previously prepared unit square with a side of 1 m. Two squares fit a unit square, which means the area of \u200b\u200bthe board is 2 m 2. In a notebook we write down the topic of the lesson: "Area of \u200b\u200ba rectangle". 3. Psychological relief. To give students the opportunity to change the type of activity. Problems for the development of creativity. Orientation in space. 1. A pair of horses ran 20 km. How many kilometers did each horse run? (20 km) 2. There were 4 rabbits in the cage. Four guys each bought one of these rabbits and one rabbit remained in the cage. How could this have happened? (We bought one rabbit together with the cage) 3. In two wallets there are two coins, and in one wallet there are twice as many coins as in the other. How can this be? (One wallet lies inside the other) The class is divided into groups of 6 people, in the groups the teacher selects the captain, who, after discussing the problem, chooses the correct answer. 1 minute is given for discussion.
Personal: self-determination; Regulatory: development of the regulation of educational activity; Communicative: interaction with partners in joint activities; Cognitive: skills of research activity. Development of creative thinking.
4.Two sons and two fathers ate 3 eggs. How many eggs did each one eat? (One egg each). Play: "Touch the right ear of the neighbor on the left with the elbow of your left hand" 4.Puzzle.
Introduce a system of increasingly complex puzzles, embodied in real objects. Self-solution of tasks. 1. How many centimeters in: 1 dm, 5m 3dm, 12dm 5cm; 2. How many meters in: 1 km, 4 km 16 m, 800 cm 3. The boat passed in 5 hours. 40 km. How many hours will it take at the same speed of 24 km? 4. What number should be put instead of the asterisks 1 * + 3 * + 5 * \u003d 111 to get the correct equality? 5. Fill in the magic square10
Right answers.
Fig. 3 In a notebook, only the answers are recorded, then they exchange notebooks with a neighbor on the desk and check with each other. At the end, the correct answers appear on the screen. Personal: making sense. Regulatory: self-regulation of emotional and functional states, self-organization. Communicative: ability to work in pairs. Cognitive: skills of finding solutions to problems. Development of creative thinking.
5. Intellectual warm-up. Develop logical thinking and creativity. 1. The side of a rectangular sheet of paper has an integer length (in centimeters), and the area of \u200b\u200bthe sheet is 12 cm2. How many squares with an area of \u200b\u200b4 cm2 can be cut out of this rectangle? 2. The following figure is displayed on the board through the projector Fig. 4 A rectangular hole was cut inside the rectangle ABCD. How to divide the resulting figure into two figures with equal areas in one straight-line cut. One student is at the blackboard, the rest are working from their place. Personal: meaning formation, the ability to complete the work. Regulatory: self-organization. Communicative: skills of cooperation with the teacher and peers. Cognitive: skills. research activities. 6. Substantive part.
Contains the curriculum material and provides the formation of systems thinking and the development of creativity. Was it difficult for us to calculate the area using a square? If we need to calculate the area of \u200b\u200bthe stadium, let's go try it? Then let's return to the problem with the board. If one side of the board is 2 m and the other side is 1 m, the board is rectangular, then it can be divided into 2 × 1 unit squares. Therefore, what is the area of \u200b\u200bthe board? If a and b are the adjacent sides of the rectangle, expressed in the same units. How do you find the area of \u200b\u200bsuch a rectangle?
Problem. — How to find the area of \u200b\u200ba regular quadrilateral in which all sides and angles are equal?
New units of area measurement are introduced: ar (weaving), hectare. 1 a \u003d 10 m * 10 m \u003d 100 m2
1ha \u003d 100 m * 100 m \u003d 10000 m2
What measurements are needed for such large units of area?
S \u003d a b The formula is written in a notebook. Students discuss the problem in groups that were previously formed in a psychological warm-up, the only one group becomes experts (after listening to the put forward versions, they are engaged in their processing and offer one that is correct in their opinion). There is a discussion of the solution to the problem. Then in notebooks we write down the resulting formula for the area of \u200b\u200ba square S \u003d a 2
–To measure the area of \u200b\u200bland plots, villages, stadiums, etc. Personal: self-determination. Regulatory: development of the regulation of educational activities. Communicative: the ability to work in a team, hear and respect the opinions of others, the ability to defend one's position. Cognitive: skills in research activities . Development of creative thinking.
7. Computer intellectual warm-up. Provide motivation and development of thinking. Establishment of the correctness and awareness of the study of the topic.
Test on a computer. The teacher controls the number of errors. Figure 5 (Figure is below the table)
Students work on a computer in pairs, pass the test. Personal: self-determination. Regulatory: the development of the regulation of educational activity. Communicative: the ability to work in pairs, hear and respect the opinions of others, the ability to defend their position. Cognitive: search for a solution to the problem. 8. Summary. Homework. Lesson summary. Provide feedback during the lesson. The teacher invites you to clap your hands to those who liked the lesson and stomp if they find the lesson boring. –What new things did you learn in the lesson?
Homework. Given a square with a side of 8 cm. Find its area. Using colored pieces, explain and then refute my hypothesis: 8 * 8 \u003d 65 Figure 6 Students evaluate the lesson, their actions in the lesson, the actions of their peers.
–The formula for the area of \u200b\u200ba rectangle, square, units of measurement of area. At home, students conduct an experiment with parts of a square. Control solution. Skv \u003d 8 * 8 \u003d 64 cm2. Let's make a rectangle from the pieces. Fig. 7Spr \u003d (8 + 5) * 5 \u003d 65 cm2.
Such calculations are obtained because a gap is formed between the parts during the assembly of the rectangle. Personal: self-development of moral consciousness and orientation of students in the sphere of ethical relations. Regulatory: development of regulation of educational activity. Communicative: the ability to express one's thoughts with sufficient completeness and accuracy. Cognitive: reflection.
References to sources 1. Federal state educational standard of basic general education. Federal Law of the Russian Federation of December 17, 2010 No. No. 1897FZ.2.M.Zinovkina. NFTMTRIZ: creative education of the XXI century. Moscow, 2007. –313s.
