Direct and inverse proportionality. Direct and inverse proportional relationships

>>Mathematics: Direct proportionality and its graph

Direct proportionality and its graph

Among the linear functions y = kx + m, the case when m = 0 is especially distinguished; in this case it takes the form y = kx and is called direct proportionality. This name is explained by the fact that two quantities y and x are called directly proportional if their ratio is equal to a specific
a number other than zero. Here, this number k is called the proportionality coefficient.

Many real-life situations are modeled using direct proportionality.

For example, the path s and time t at a constant speed of 20 km/h are related by the dependence s = 20t; this is direct proportionality, with k = 20.

Another example:

cost y and number x of loaves of bread at a price of 5 rubles. for the loaf are connected by the dependence y = 5x; this is direct proportionality, where k = 5.

Proof. We will implement it in two stages.
1. y = kx is a special case of a linear function, and the graph of a linear function is a straight line; let's denote it by I.
2. The pair x = 0, y = 0 satisfies the equation y - kx, and therefore the point (0; 0) belongs to the graph of the equation y = kx, i.e., straight line I.

Consequently, straight line I passes through the origin. The theorem has been proven.

You must be able to move not only from the analytical model y = kx to the geometric one (graph of direct proportionality), but also from the geometric one models to analytical. Consider, for example, a straight line on the xOy coordinate plane shown in Figure 50. It is a graph of direct proportionality; you just need to find the value of the coefficient k. Since y, then it is enough to take any point on the line and find the ratio of the ordinate of this point to its abscissa. The straight line passes through the point P(3; 6), and for this point we have: This means k = 2, and therefore the given straight line serves as a graph of direct proportionality y = 2x.

As a result, the coefficient k in the notation of the linear function y = kx + m is also called the slope coefficient. If k>0, then the straight line y = kx + m forms an acute angle with the positive direction of the x axis (Fig. 49, a), and if k< О, - тупой угол (рис. 49, б).

Calendar-thematic planning in mathematics, video in mathematics online, Mathematics at school download

A. V. Pogorelov, Geometry for grades 7-11, Textbook for educational institutions

Trikhleb Daniil, 7th grade student

acquaintance with direct proportionality and the coefficient of direct proportionality (introduction of the concept of angular coefficient”);

constructing a direct proportionality graph;

consideration of the relative position of graphs of direct proportionality and linear functions with identical angular coefficients.

Download:

Preview:

To use presentation previews, create a Google account and log in to it: https://accounts.google.com

Slide captions:

Direct proportionality and its graph

What is the argument and value of a function? Which variable is called independent or dependent? What is a function? REVIEW What is the domain of a function?

Methods for specifying a function. Analytical (using a formula) Graphical (using a graph) Tabular (using a table)

The graph of a function is the set of all points of the coordinate plane, the abscissas of which are equal to the values ​​of the argument, and the ordinates are the corresponding values ​​of the function. FUNCTION SCHEDULE

1) 2) 3) 4) 5) 6) 7) 8) 9)

COMPLETE THE TASK Construct a graph of the function y = 2 x +1, where 0 ≤ x ≤ 4. Make a table. Using the graph, find the value of the function at x=2.5. At what value of the argument does the function value equal 8?

Definition Direct proportionality is a function that can be specified by a formula of the form y = k x, where x is an independent variable, k is a non-zero number. (k-coefficient of direct proportionality) Direct proportionality

8 Graph of direct proportionality - a straight line passing through the origin of coordinates (point O(0,0)) To construct a graph of the function y= kx, two points are enough, one of which is O (0,0) For k > 0, the graph is located at I and III coordinate quarters. At k

Graphs of functions of direct proportionality y x k>0 k>0 k

Task Determine which of the graphs shows the function of direct proportionality.

Task Determine which function graph is shown in the figure. Choose a formula from the three offered.

Oral work. Can the graph of a function given by the formula y = k x, where k

Determine which of the points A(6,-2), B(-2,-10), C(1,-1), E(0,0) belong to the graph of direct proportionality given by the formula y = 5x 1) A( 6;-2) -2 = 5  6 - 2 = 30 - incorrect. Point A does not belong to the graph of the function y=5x. 2) B(-2;-10) -10 = 5  (-2) -10 = -10 - correct. Point B belongs to the graph of the function y=5x. 3) C(1;-1) -1 = 5  1 -1 = 5 - incorrect Point C does not belong to the graph of the function y=5x. 4) E (0;0) 0 = 5  0 0 = 0 - true. Point E belongs to the graph of the function y=5x

TEST 1 option 2 option No. 1. Which of the functions given by the formula are directly proportional? A. y = 5x B. y = x 2 /8 C. y = 7x(x-1) D . y = x+1 A. y = 3x 2 +5 B. y = 8/x C. y = 7(x + 9) D. y = 10x

No. 2. Write down the numbers of lines y = kx, where k > 0 1 option k

No. 3. Determine which of the points belong to the graph of direct proportionality, given by the formula Y = -1 /3 X A (6 -2), B (-2 -10) 1 option C (1, -1), E (0.0 ) Option 2

y =5x y =10x III A VI and IV E 1 2 3 1 2 3 No. Correct answer Correct answer No.

Complete the task: Show schematically how the graph of the function given by the formula is located: y =1.7 x y =-3,1 x y=0.9 x y=-2.3 x

TASK From the following graphs, select only direct proportionality graphs.

1) 2) 3) 4) 5) 6) 7) 8) 9)

Functions y = 2x + 3 2. y = 6/ x 3. y = 2x 4. y = - 1.5x 5. y = - 5/ x 6. y = 5x 7. y = 2x – 5 8. y = - 0.3x 9. y = 3/ x 10. y = - x /3 + 1 Select functions of the form y = k x (direct proportionality) and write them down

Functions of direct proportionality Y = 2x Y = -1.5x Y = 5x Y = -0.3x y x

y Linear functions that are not functions of direct proportionality 1) y = 2x + 3 2) y = 2x – 5 x -6 -4 -2 0 2 4 6 6 3 -3 -6 y = 2x + 3 y = 2x - 5

Homework: paragraph 15 pp. 65-67, No. 307; No. 308.

Let's repeat it again. What new things have you learned? What have you learned? What did you find particularly difficult?

I liked the lesson and the topic is understood: I liked the lesson, but I still don’t understand everything: I didn’t like the lesson and the topic is not clear.

I. Directly proportional quantities.

Let the value y depends on the size X. If when increasing X several times the size at increases by the same amount, then such values X And at are called directly proportional.

Examples.

1 . The quantity of goods purchased and the purchase price (with a fixed price for one unit of goods - 1 piece or 1 kg, etc.) How many times more goods were bought, the more times more they paid.

2 . The distance traveled and the time spent on it (at constant speed). How many times longer is the path, how many times more time will it take to complete it.

3 . The volume of a body and its mass. ( If one watermelon is 2 times larger than another, then its mass will be 2 times larger)

II. Property of direct proportionality of quantities.

If two quantities are directly proportional, then the ratio of two arbitrarily taken values ​​of the first quantity is equal to the ratio of two corresponding values ​​of the second quantity.

Task 1. For raspberry jam we took 12 kg raspberries and 8 kg Sahara. How much sugar will you need if you took it? 9 kg raspberries?

Solution.

We reason like this: let it be necessary x kg sugar for 9 kg raspberries The mass of raspberries and the mass of sugar are directly proportional quantities: how many times less raspberries are, the same number of times less sugar is needed. Therefore, the ratio of raspberries taken (by weight) ( 12:9 ) will be equal to the ratio of sugar taken ( 8:x). We get the proportion:

12: 9=8: X;

x=9 · 8: 12;

x=6. Answer: on 9 kg raspberries need to be taken 6 kg Sahara.

The solution of the problem It could be done like this:

Let on 9 kg raspberries need to be taken x kg Sahara.

(The arrows in the figure are directed in one direction, and up or down does not matter. Meaning: how many times the number 12 more number 9 , the same number of times 8 more number X, i.e. there is a direct relationship here).

Answer: on 9 kg I need to take some raspberries 6 kg Sahara.

Task 2. Car for 3 hours traveled the distance 264 km. How long will it take him to travel? 440 km, if he drives at the same speed?

Solution.

Let for x hours the car will cover the distance 440 km.

Answer: the car will pass 440 km in 5 hours.

Task 3. Water flows from the pipe into the pool. Behind 2 hours she fills 1/5 swimming pool What part of the pool is filled with water in 5 o'clock?

Solution.

We answer the question of the task: for 5 o'clock will be filled 1/x part of the pool. (The entire pool is taken as one whole).

ADMINISTRATION OF THE MUNICIPAL FORMATION "CITY OF SARATOV"

MUNICIPAL EDUCATIONAL INSTITUTION

"SECONDARY EDUCATIONAL SCHOOL No. 95 WITH IN-DEPTH

STUDYING INDIVIDUAL SUBJECTS"

Methodological development

algebra lesson in 7th grade

on this topic:

"Direct proportionality

and her schedule."

Mathematic teacher

1 qualification category

Goryunova E.V.

2014 – 2015 academic year

Explanatory note

for a lesson on the topic:

“Direct proportionality and its graph.”

Mathematics teacher Elena Viktorovna Goryunova.

We present to your attention a lesson in 7th grade. The teacher works according to a program compiled on the basis of Model programs of basic general education and the author’s program for general education institutions Yu.N. Makarychev. Algebra.7-9 grades //Collection of programs for algebra grades 7-9. M. Prosveshchenie, 2009 compiled by T.A. Burmistrova. The program corresponds to the algebra textbook by Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov., S.B. Suvorova., edited by S.A. Telyakovsky “Algebra 7th grade” (Prosveshcheniye publishing house, 2009).

14 hours are allotted for studying the topic “Functions”, of which 6 hours for the section “Functions and their graphs”, 3 hours for the section “Direct proportionality and its graph”, 4 hours for the section “Linear function and its graph” and 1 hour K/R.

GOALS:

Educational:

Educational:

3. Encourage students to self-control and mutual control.

Educational:

To instill a sense of respect for classmates, attention to words, to promote independence, responsibility, and accuracy when constructing drawings

Achieving these goals is accomplished through a series of tasks:

1. Formation of the ability to combine knowledge and skills that ensure the successful implementation of activities;

   Work on the development of students’ connected speech, the ability to pose and solve problems.

Lesson equipment:

The lesson used individual cards with tasks and a multimedia projector, all the facts about R. Descartes were taken by the teacher on the Internet from official media sites and revised specifically for this lesson, taking into account the topic of the lesson, textbook.

Lesson type and structure:

This lesson is a lesson in mastering new knowledge and skills (types of lessons according to V.A. Onishchuk), so it was rational to apply elements of research activity.

Implementation of training principles:

The following principles were implemented in the lesson:

Science of learning.

The principle of systematic and consistent teaching was implemented with constant reliance on previously studied material.

Consciousness, activity and independence of students were achieved in the form of stimulation of cognitive activity with the help of effective techniques and visual aids (such as showing slides, providing historical facts and information from the life of the mathematician and philosopher R. Descartes, individual printed sheets for students.

The principle of comfort was implemented in the lesson.

Forms and methods of teaching:

During the lesson, various forms of training were used - individual and frontal work, mutual testing. Such forms are more rational for this type of lesson, as they allow the child to develop independent thinking, criticality of thought, the ability to defend his point of view, the ability to compare and draw conclusions.

The main method of this lesson is the partial search method, which is characterized by the work of students in solving problematic cognitive problems.

Phys. the minute was both physical exercise and consolidation of the material just learned.

At the end of the lesson, it is advisable to summarize the work done in the lesson.

General results of the lesson:

I believe that the objectives set for the lesson were achieved, the children applied their knowledge in a new situation, everyone could express their point of view. Using visual aids in the form of presentations and individual printed sheets for students allows you to motivate students at every stage of the lesson and avoid overloading and overtiring students.

Lesson topic:

Didactic task: familiarity with direct proportionality and the construction of its graph.

Goals:

Educational:

1. Organize students’ activities on understanding the topic “Direct proportionality and its graph” and primary consolidation: defining direct proportionality and constructing its graph, developing skills in competently constructing graphs

2. Create conditions for creating a system of basic knowledge and skills in students’ memory, stimulate search activity

Educational:

1. Develop analytical-synthesizing thinking (promote the development of observation, the ability to analyze, the development of the ability to classify facts, draw generalizing conclusions).

2. Develop abstract thinking (developing the ability to identify general and essential features, distinguish unimportant features and be distracted from them).

3. Encourage students to self-control and mutual control

Educational:

To instill a sense of respect for classmates, attention to words, to promote independence, responsibility, and accuracy when constructing drawings.

Equipment: computer, presentation, printed cards with tasks for each student.

Lesson plan:

1. Organizational moment.

2.Lesson motivation.

3. Updating knowledge.

4.Learning new material.

5. Fixing the material.

6. Lesson summary.

During the classes.

1. Organizational moment.

Good morning, guys! I would like to start the lesson with the following words. (Slide 1)

French scientist René Descartes once remarked: “I think, therefore I am.”

The guys prepared a report about the French scientist R. Descartes.

René Descartes is better known as a great philosopher than a mathematician. But it was he who was the pioneer of modern mathematics, and his achievements in this field are so great that he is rightfully included among the great mathematicians of our time.

Student message:(Slide 2)

Descartes was born in France, in the small town of Lae. His father was a lawyer, his mother died when Rene was 1 year old. After graduating from a college for the sons of aristocratic families, he, following the example of his brother, began to study jurisprudence. At the age of 22, he left France and served as a volunteer officer in the forces of various military leaders who participated in the 13-year war. Descartes, in his philosophical teaching, developed the idea of ​​the omnipotence of the human mind, and therefore was persecuted by the Catholic Church. Wanting to find refuge for quiet work on philosophy and mathematics, in which he was interested since childhood, Descartes settled in Holland in 1629, where he lived almost until the end of his life. All of Descartes' major works on philosophy, mathematics, physics, cosmology and physiology were written by him in Holland.

Descartes' mathematical works are collected in his book "Geometry" (1637). In "Geometry" Descartes gave the foundations of analytical geometry and algebra. Descartes was the first to introduce the concept of a variable function into mathematics. He drew attention to the fact that a curve on a plane is characterized by an equation that has the property that the coordinates of any point lying on this line satisfy this equation. He divided the curves given by an algebraic equation into classes depending on the largest power of the unknown quantity in the equation. Descartes introduced into mathematics the plus and minus signs to denote positive and negative quantities, the notation of degree, and the sign to denote an infinitely large quantity. For variables and unknown quantities, Descartes adopted the notations x, y, z, and for known and constant quantities -a .b .c, as is known, these notations are used in mathematics to this day. Despite the fact that Descartes did not advance very far in the field of analytical geometry, his works had a decisive influence on the further development of mathematics. For 150 years, mathematics developed along the paths outlined by Descartes.

Let's follow the scientist's advice. We will be active, attentive, we will reason, think and learn new things, because knowledge will be useful to you in later life. And I would like to propose these words (Slide 3) of R. Descartes as the motto of our lesson: “Respect for others gives a reason to respect oneself.”

2.Motivation.

Let's check in what mood you came to class. Draw a smiley face in the margins.

Take the cards. The words of R. Descartes are also written here: “ In order to improve your mind, you need to reason more than memorize.” These words will guide us in our work.

Task No. 1 with mathematical terms that we will use in class. Correct any mistakes made in the spelling of these terms. (Slide 4)

Exchange leaves and check if all errors are corrected. (Slide 5) -What did you notice? Which word has no mistakes? (function, schedule)

3. Updating knowledge.

a) We became familiar with the concept of “function” in previous lessons. Let's remember the basic concepts and definitions on this topic.

We also worked with function graphs. Which of the dictation words did we use when working on the topic “Graphs of Functions”? What do they mean?

On this slide, determine which line will be the graph of the function? (Slide 6)

Who can tell us what we will talk about in this lesson? What goals will we set for the lesson? (Slide7)

Write down the number on student sheets and write the topic of the lesson: “Direct proportionality and its graph”

Let's remember the material from previous lessons

Create formulas to solve the following problems. (Slide 9,10)

Which variables are dependent and independent? What depends on what? What addiction? (Slide)

Which formula is different from the others? (Slide)

c) How can you write the formulas in general form? (Slide)

y =kx, y - dependent variable

x – independent variable

k – constant number (coefficient)

We wrote down the formula, and this is one of the ways to define a function. Direct proportional dependence is a function.

4.Learning new material.

Definition. Direct proportionality is a function that can be specified by the formula y=kx, where x is an independent variable, and k is a certain number that is not equal to zero, a coefficient of direct proportionality (a constant ratio of proportional quantities)

Let's read the rule in the textbook on page 65

What is the scope of this function? (The set of all numbers)

Fixing the material.

Complete the task in sheets No. 4 (Slide) Distribute the formulas into 2 groups in accordance with the topic of the lesson: (read the rule in the textbook on p. 65)

y=2x, y=3x-7, y=-0.2x, y=x, y=x², y=x, y=-5.8+3x, y=-x, y=50x,

Group 1:______________________________________________________________

Group 2:______________________________________________________________

Underline the coefficient of direct proportionality.

We carry out No. 298 on page 68 (orally), I dictate, you determine the formula of proportionality by ear and squint your eyes, if not by proportionality, then rotate your eyes from left to right.

Come up with and write 4 formulas for the function of direct proportionality:

1) y=_________2) y=__________3) y=_________4) y=__________

Learning new material

What is the graph of this function? Do you want to know?

We have already constructed a graph of a function in task No. 2, can we call this function pr. proportionality? This means that we have already built a graph of proportionality. The rule is in the textbook on page 67.

Let's see how we build a graph of this function (Slide)

Fixing the material.

Let's build graph No. 7 on student sheets (Slide)

What point will we have in any graph of proportionality?

We work according to ready-made drawings. (Slide)

Conclusion: the graph is a straight line passing through the origin.

T.K. The graph is a straight line, then how many points are needed to construct it? There is already one (0;0)

We carry out No. 300

Lesson summary. Let's summarize the work in today's lesson (Slide). Everything was done. What have you planned?

Reflection. (Slide)

Check the mood of the students at the end of the lesson. (smiley) (Slide)

§ 129. Preliminary clarifications.

A person constantly deals with a wide variety of quantities. An employee and a worker are trying to get to work by a certain time, a pedestrian is in a hurry to get to a certain place by the shortest route, a steam heating stoker is worried that the temperature in the boiler is slowly rising, a business executive is making plans to reduce the cost of production, etc.

One could give any number of such examples. Time, distance, temperature, cost - all these are various quantities. In the first and second parts of this book, we became acquainted with some particularly common quantities: area, volume, weight. We encounter many quantities when studying physics and other sciences.

Imagine that you are traveling on a train. Every now and then you look at your watch and notice how long you've been on the road. You say, for example, that 2, 3, 5, 10, 15 hours have passed since your train departed, etc. These numbers represent different periods of time; they are called the values ​​of this quantity (time). Or you look out the window and follow the road posts to see the distance your train travels. The numbers 110, 111, 112, 113, 114 km flash in front of you. These numbers represent the different distances the train has traveled from its departure point. They are also called values, this time of a different magnitude (path or distance between two points). Thus, one quantity, for example time, distance, temperature, can take on as many different meanings.

Please note that a person almost never considers only one quantity, but always connects it with some other quantities. He has to simultaneously deal with two, three or more quantities. Imagine that you need to get to school by 9 o'clock. You look at your watch and see that you have 20 minutes. Then you quickly figure out whether you should take the tram or whether you can walk to school. After thinking, you decide to walk. Notice that while you were thinking, you were solving some problem. This task has become simple and familiar, since you solve such problems every day. In it you quickly compared several quantities. It was you who looked at the clock, which means you took into account the time, then you mentally imagined the distance from your home to the school; Finally, you compared two values: the speed of your step and the speed of the tram, and concluded that in a given time (20 minutes) you will have time to walk. From this simple example you can see that in our practice some quantities are interconnected, that is, they depend on each other

Chapter twelve talked about the relationship of homogeneous quantities. For example, if one segment is 12 m and the other is 4 m, then the ratio of these segments will be 12: 4.

We said that this is the ratio of two homogeneous quantities. Another way to say this is that it is the ratio of two numbers one name.

Now that we are more familiar with quantities and have introduced the concept of the value of a quantity, we can express the definition of a ratio in a new way. In fact, when we considered two segments of 12 m and 4 m, we were talking about one value - length, and 12 m and 4 m were only two different values ​​of this value.

Therefore, in the future, when we start talking about ratios, we will consider two values ​​of one quantity, and the ratio of one value of a quantity to another value of the same quantity will be called the quotient of dividing the first value by the second.

§ 130. Values ​​are directly proportional.

Let's consider a problem whose condition includes two quantities: distance and time.

Task 1. A body moving rectilinearly and uniformly travels 12 cm every second. Determine the distance traveled by the body in 2, 3, 4, ..., 10 seconds.

Let's create a table that can be used to track changes in time and distance.

The table gives us the opportunity to compare these two series of values. We see from it that when the values ​​of the first quantity (time) gradually increase by 2, 3,..., 10 times, then the values ​​of the second quantity (distance) also increase by 2, 3,..., 10 times. Thus, when the values ​​of one quantity increase several times, the values ​​of another quantity increase by the same amount, and when the values ​​of one quantity decrease several times, the values ​​of another quantity decrease by the same number.

Let us now consider a problem that involves two such quantities: the amount of matter and its cost.

Task 2. 15 m of fabric costs 120 rubles. Calculate the cost of this fabric for several other quantities of meters indicated in the table.

Using this table, we can trace how the cost of a product gradually increases depending on the increase in its quantity. Despite the fact that this problem involves completely different quantities (in the first problem - time and distance, and here - the quantity of goods and its value), nevertheless, great similarities can be found in the behavior of these quantities.

In fact, in the top line of the table there are numbers indicating the number of meters of fabric; under each of them there is a number expressing the cost of the corresponding quantity of goods. Even a quick glance at this table shows that the numbers in both the top and bottom rows are increasing; upon closer examination of the table and when comparing individual columns, it is discovered that in all cases the values ​​of the second quantity increase by the same number of times as the values ​​of the first increase, i.e. if the value of the first quantity increases, say, 10 times, then the value of the second quantity also increased 10 times.

If we look through the table from right to left, we will find that the indicated values ​​​​of quantities will decrease by the same number of times. In this sense, there is an unconditional similarity between the first task and the second.

The pairs of quantities that we encountered in the first and second problems are called directly proportional.

Thus, if two quantities are related to each other in such a way that as the value of one of them increases (decreases) several times, the value of the other increases (decreases) by the same amount, then such quantities are called directly proportional.

Such quantities are also said to be related to each other by a directly proportional relationship.

There are many similar quantities found in nature and in the life around us. Here are some examples:

1. Time work (day, two days, three days, etc.) and earnings, received during this time with daily wages.

2. Volume any object made of a homogeneous material, and weight this item.

§ 131. Property of directly proportional quantities.

Let's take a problem that includes the following two quantities: working time and earnings. If daily earnings are 20 rubles, then earnings for 2 days will be 40 rubles, etc. It is most convenient to create a table in which a certain number of days will correspond to a certain earnings.

Looking at this table, we see that both quantities took 10 different values. Each value of the first value corresponds to a certain value of the second value, for example, 2 days correspond to 40 rubles; 5 days correspond to 100 rubles. In the table these numbers are written one below the other.

We already know that if two quantities are directly proportional, then each of them, in the process of its change, increases as many times as the other increases. It immediately follows from this: if we take the ratio of any two values ​​of the first quantity, then it will be equal to the ratio of the two corresponding values ​​of the second quantity. Indeed:

Why is this happening? But because these values ​​are directly proportional, i.e. when one of them (time) increased by 3 times, then the other (earnings) increased by 3 times.

We have therefore come to the following conclusion: if we take two values ​​of the first quantity and divide them by one another, and then divide by one the corresponding values ​​of the second quantity, then in both cases we will get the same number, i.e. i.e. the same relationship. This means that the two relations that we wrote above can be connected with an equal sign, i.e.

There is no doubt that if we took not these relations, but others, and not in that order, but in the opposite order, we would also obtain equality of relations. In fact, we will consider the values ​​of our quantities from left to right and take the third and ninth values:

60:180 = 1 / 3 .

So we can write:

This leads to the following conclusion: if two quantities are directly proportional, then the ratio of two arbitrarily taken values ​​of the first quantity is equal to the ratio of the two corresponding values ​​of the second quantity.

§ 132. Formula of direct proportionality.

Let's make a table of the cost of various quantities of sweets, if 1 kg of them costs 10.4 rubles.

Now let's do it this way. Take any number in the second line and divide it by the corresponding number in the first line. For example:

You see that in the quotient the same number is obtained all the time. Consequently, for a given pair of directly proportional quantities, the quotient of dividing any value of one quantity by the corresponding value of another quantity is a constant number (i.e., not changing). In our example, this quotient is 10.4. This constant number is called the proportionality factor. In this case, it expresses the price of a unit of measurement, i.e. one kilogram of goods.

How to find or calculate the proportionality coefficient? To do this, you need to take any value of one quantity and divide it by the corresponding value of the other.

Let us denote this arbitrary value of one quantity by the letter at , and the corresponding value of another quantity - the letter X , then the proportionality coefficient (we denote it TO) we find by division:

In this equality at - divisible, X - divisor and TO- quotient, and since, by the property of division, the dividend is equal to the divisor multiplied by the quotient, we can write:

y = K x

The resulting equality is called formula of direct proportionality. Using this formula, we can calculate any number of values ​​of one of the directly proportional quantities if we know the corresponding values ​​of the other quantity and the coefficient of proportionality.

Example. From physics we know that weight R of any body is equal to its specific gravity d , multiplied by the volume of this body V, i.e. R = d V.

Let's take five iron bars of different volumes; Knowing the specific gravity of iron (7.8), we can calculate the weights of these ingots using the formula:

R = 7,8 V.

Comparing this formula with the formula at = TO X , we see that y = R, x = V, and the proportionality coefficient TO= 7.8. The formula is the same, only the letters are different.

Using this formula, let's make a table: let the volume of the 1st blank be equal to 8 cubic meters. cm, then its weight is 7.8 8 = 62.4 (g). The volume of the 2nd blank is 27 cubic meters. cm. Its weight is 7.8 27 = 210.6 (g). The table will look like this:

Calculate the numbers missing in this table using the formula R= d V.

§ 133. Other methods of solving problems with directly proportional quantities.

In the previous paragraph, we solved a problem whose condition included directly proportional quantities. For this purpose, we first derived the direct proportionality formula and then applied this formula. Now we will show two other ways to solve similar problems.

Let's create a problem using the numerical data given in the table in the previous paragraph.

Task. Blank with a volume of 8 cubic meters. cm weighs 62.4 g. How much will a blank with a volume of 64 cubic meters weigh? cm?

Solution. The weight of iron, as is known, is proportional to its volume. If 8 cu. cm weigh 62.4 g, then 1 cu. cm will weigh 8 times less, i.e.

62.4:8 = 7.8 (g).

Blank with a volume of 64 cubic meters. cm will weigh 64 times more than a 1 cubic meter blank. cm, i.e.

7.8 64 = 499.2(g).

We solved our problem by reducing to unity. The meaning of this name is justified by the fact that to solve it we had to find the weight of a unit of volume in the first question.

2. Method of proportion. Let's solve the same problem using the proportion method.

Since the weight of iron and its volume are directly proportional quantities, the ratio of two values ​​of one quantity (volume) is equal to the ratio of two corresponding values ​​of another quantity (weight), i.e.

(letter R we designated the unknown weight of the blank). From here:

(G).

The problem was solved using the method of proportions. This means that to solve it, a proportion was compiled from the numbers included in the condition.

§ 134. Values ​​are inversely proportional.

Consider the following problem: “Five masons can lay the brick walls of a house in 168 days. Determine in how many days 10, 8, 6, etc. masons could complete the same work.”

If 5 masons laid the walls of a house in 168 days, then (with the same labor productivity) 10 masons could do it in half the time, since on average 10 people do twice as much work as 5 people.

Let's draw up a table by which we could monitor changes in the number of workers and working hours.

For example, to find out how many days it takes 6 workers, you must first calculate how many days it takes one worker (168 5 = 840), and then how many days it takes six workers (840: 6 = 140). Looking at this table, we see that both quantities took on six different values. Each value of the first quantity corresponds to a specific one; the value of the second value, for example, 10 corresponds to 84, the number 8 corresponds to the number 105, etc.

If we consider the values ​​of both quantities from left to right, we will see that the values ​​of the upper quantity increase, and the values ​​of the lower quantity decrease. The increase and decrease are subject to the following law: the values ​​of the number of workers increase by the same times as the values ​​of the spent working time decrease. This idea can be expressed even more simply as follows: the more workers are engaged in any task, the less time they need to complete a certain job. The two quantities we encountered in this problem are called inversely proportional.

Thus, if two quantities are related to each other in such a way that as the value of one of them increases (decreases) several times, the value of the other decreases (increases) by the same amount, then such quantities are called inversely proportional.

There are many similar quantities in life. Let's give examples.

1. If for 150 rubles. If you need to buy several kilograms of sweets, the number of sweets will depend on the price of one kilogram. The higher the price, the less goods you can buy with this money; this can be seen from the table:

As the price of candy increases several times, the number of kilograms of candy that can be bought for 150 rubles decreases by the same amount. In this case, two quantities (the weight of the product and its price) are inversely proportional.

2. If the distance between two cities is 1,200 km, then it can be covered in different times depending on the speed of movement. There are different ways to travel: on foot, on horseback, by bicycle, by boat, in a car, by train, by plane. The lower the speed, the more time it takes to move. This can be seen from the table:

With an increase in speed several times, the travel time decreases by the same amount. This means that under these conditions, speed and time are inversely proportional quantities.

§ 135. Property of inversely proportional quantities.

Let's take the second example, which we looked at in the previous paragraph. There we dealt with two quantities - speed and time. If we look at the table of values ​​of these quantities from left to right, we will see that the values ​​of the first quantity (speed) increase, and the values ​​of the second (time) decrease, and the speed increases by the same amount as the time decreases. It is not difficult to understand that if you write the ratio of some values ​​of one quantity, then it will not be equal to the ratio of the corresponding values ​​of another quantity. In fact, if we take the ratio of the fourth value of the upper value to the seventh value (40: 80), then it will not be equal to the ratio of the fourth and seventh values ​​of the lower value (30: 15). It can be written like this:

40:80 is not equal to 30:15, or 40:80 =/=30:15.

But if instead of one of these relations we take the opposite, then we get equality, i.e., from these relations it will be possible to create a proportion. For example:

80: 40 = 30: 15,

40: 80 = 15: 30."

Based on the foregoing, we can draw the following conclusion: if two quantities are inversely proportional, then the ratio of two arbitrarily taken values ​​of one quantity is equal to the inverse ratio of the corresponding values ​​of another quantity.

§ 136. Inverse proportionality formula.

Consider the problem: “There are 6 pieces of silk fabric of different sizes and different grades. All pieces cost the same. One piece contains 100 m of fabric, priced at 20 rubles. per meter How many meters are in each of the other five pieces, if a meter of fabric in these pieces costs 25, 40, 50, 80, 100 rubles, respectively?” To solve this problem, let's create a table:

We need to fill in the empty cells in the top row of this table. Let's first try to determine how many meters there are in the second piece. This can be done as follows. From the conditions of the problem it is known that the cost of all pieces is the same. The cost of the first piece is easy to determine: it contains 100 meters and each meter costs 20 rubles, which means that the first piece of silk is worth 2,000 rubles. Since the second piece of silk contains the same amount of rubles, then, dividing 2,000 rubles. for the price of one meter, i.e. 25, we find the size of the second piece: 2,000: 25 = 80 (m). In the same way we will find the size of all other pieces. The table will look like:

It is easy to see that there is an inversely proportional relationship between the number of meters and the price.

If you do the necessary calculations yourself, you will notice that each time you have to divide the number 2,000 by the price of 1 m. On the contrary, if you now start multiplying the size of the piece in meters by the price of 1 m, you will always get the number 2,000. This and it was necessary to wait, since each piece costs 2,000 rubles.

From here we can draw the following conclusion: for a given pair of inversely proportional quantities, the product of any value of one quantity by the corresponding value of another quantity is a constant number (i.e., not changing).

In our problem, this product is equal to 2,000. Check that in the previous problem, which talked about the speed of movement and the time required to move from one city to another, there was also a constant number for that problem (1,200).

Taking everything into account, it is easy to derive the inverse proportionality formula. Let us denote a certain value of one quantity by the letter X , and the corresponding value of another quantity is represented by the letter at . Then, based on the above, the work X on at must be equal to some constant value, which we denote by the letter TO, i.e.

x y = TO.

In this equality X - multiplicand at - multiplier and K- work. According to the property of multiplication, the multiplier is equal to the product divided by the multiplicand. Means,

This is the inverse proportionality formula. Using it, we can calculate any number of values ​​of one of the inversely proportional quantities, knowing the values ​​of the other and the constant number TO.

Let's consider another problem: “The author of one essay calculated that if his book is in a regular format, then it will have 96 pages, but if it is a pocket format, then it will have 300 pages. He tried different options, started with 96 pages, and then he ended up with 2,500 letters per page. Then he took the page numbers shown in the table below and again calculated how many letters there would be on the page.”

Let's try to calculate how many letters there will be on a page if the book has 100 pages.

There are 240,000 letters in the entire book, since 2,500 96 = 240,000.

Taking this into account, we use the inverse proportionality formula ( at - number of letters on the page, X - number of pages):

In our example TO= 240,000 therefore

So there are 2,400 letters on the page.

Similarly, we learn that if a book has 120 pages, then the number of letters on the page will be:

Our table will look like:

Fill in the remaining cells yourself.

§ 137. Other methods of solving problems with inversely proportional quantities.

In the previous paragraph, we solved problems whose conditions included inversely proportional quantities. We first derived the inverse proportionality formula and then applied this formula. We will now show two other solutions for such problems.

1. Method of reduction to unity.

Task. 5 turners can do some work in 16 days. In how many days can 8 turners complete this work?

Solution. There is an inverse relationship between the number of turners and working hours. If 5 turners do the job in 16 days, then one person will need 5 times more time for this, i.e.

5 turners complete the job in 16 days,

1 turner will complete it in 16 5 = 80 days.

The problem asks how many days it will take 8 turners to complete the job. Obviously, they will cope with the work 8 times faster than 1 turner, i.e. in

80: 8 = 10 (days).

This is the solution to the problem by reducing it to unity. Here it was necessary first of all to determine the time required to complete the work by one worker.

2. Method of proportion. Let's solve the same problem in the second way.

Since there is an inversely proportional relationship between the number of workers and working time, we can write: duration of work of 5 turners new number of turners (8) duration of work of 8 turners previous number of turners (5) Let us denote the required duration of work by the letter X and substitute the necessary numbers into the proportion expressed in words:

The same problem is solved by the method of proportions. To solve it, we had to create a proportion from the numbers included in the problem statement.

Note. In the previous paragraphs we examined the issue of direct and inverse proportionality. Nature and life give us many examples of direct and inverse proportional dependence of quantities. However, it should be noted that these two types of dependence are only the simplest. Along with them, there are other, more complex dependencies between quantities. In addition, one should not think that if any two quantities increase simultaneously, then there is necessarily a direct proportionality between them. This is far from true. For example, railway fares increase depending on the distance: the further we travel, the more we pay, but this does not mean that the fare is proportional to the distance.