The tangent of the angle of inclination is called. How to find the slope

The derivative of a function is one of difficult topics in the school curriculum. Not every graduate will answer the question of what a derivative is.

This article simply and clearly explains what a derivative is and why it is needed.. We will not now strive for mathematical rigor of presentation. The most important thing is to understand the meaning.

Let's remember the definition:

The derivative is the rate of change of the function.

The figure shows graphs of three functions. Which one do you think grows the fastest?

The answer is obvious - the third. She has the most high speed changes, that is, the largest derivative.

Here is another example.

Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:

You can see everything on the chart right away, right? Kostya's income has more than doubled in six months. And Grisha's income also increased, but just a little bit. And Matthew's income decreased to zero. The starting conditions are the same, but the rate of change of the function, i.e. derivative, - different. As for Matvey, the derivative of his income is generally negative.

Intuitively, we can easily estimate the rate of change of a function. But how do we do it?

What we are really looking at is how steeply the graph of the function goes up (or down). In other words, how fast y changes with x. Obviously, the same function at different points can have different meaning derivative - that is, it can change faster or slower.

The derivative of a function is denoted by .

Let's show how to find using the graph.

A graph of some function is drawn. Take a point on it with an abscissa. Draw a tangent to the graph of the function at this point. We want to evaluate how steeply the graph of the function goes up. A handy value for this is tangent of the slope of the tangent.

The derivative of a function at a point is equal to the tangent of the slope of the tangent drawn to the graph of the function at that point.

Please note - as the angle of inclination of the tangent, we take the angle between the tangent and the positive direction of the axis.

Sometimes students ask what is the tangent to the graph of a function. This is a straight line that has the only common point with the graph in this section, moreover, as shown in our figure. It looks like a tangent to a circle.

Let's find . We remember that the tangent of an acute angle in a right triangle is equal to the ratio of the opposite leg to the adjacent one. From triangle:

We found the derivative using the graph without even knowing the formula of the function. Such tasks are often found in the exam in mathematics under the number.

There is another important correlation. Recall that the straight line is given by the equation

The quantity in this equation is called slope of a straight line. It is equal to the tangent of the angle of inclination of the straight line to the axis.

.

We get that

Let's remember this formula. She expresses geometric meaning derivative.

The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.

In other words, the derivative is equal to the tangent of the slope of the tangent.

We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.

Let's draw a graph of some function. Let this function increase in some areas, decrease in others, and with different speed. And let this function have maximum and minimum points.

At a point, the function is increasing. The tangent to the graph, drawn at the point, forms an acute angle; with positive axis direction. So the derivative is positive at the point.

At the point, our function is decreasing. The tangent at this point forms an obtuse angle; with positive axis direction. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.

Here's what happens:

If a function is increasing, its derivative is positive.

If it decreases, its derivative is negative.

And what will happen at the maximum and minimum points? We see that at (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the slope of the tangent at these points zero, and the derivative is also zero.

The point is the maximum point. At this point, the increase of the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from "plus" to "minus".

At the point - the minimum point - the derivative is also equal to zero, but its sign changes from "minus" to "plus".

Conclusion: with the help of the derivative, you can find out everything that interests us about the behavior of the function.

If the derivative is positive, then the function is increasing.

If the derivative is negative, then the function is decreasing.

At the maximum point, the derivative is zero and changes sign from plus to minus.

At the minimum point, the derivative is also zero and changes sign from minus to plus.

We write these findings in the form of a table:

Let's make two small clarifications. You will need one of them when solving the problem. Another - in the first year, with a more serious study of functions and derivatives.

A case is possible when the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This so-called :

At a point, the tangent to the graph is horizontal and the derivative is zero. However, before the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - it has remained positive as it was.

It also happens that at the point of maximum or minimum, the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.

But how to find the derivative if the function is given not by a graph, but by a formula? In this case, it applies

The continuation of the topic of the equation of a straight line on a plane is based on the study of a straight line from algebra lessons. This article gives generalized information on the topic of the equation of a straight line with a slope. Consider the definitions, get the equation itself, reveal the relationship with other types of equations. Everything will be discussed on examples of problem solving.

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Before writing such an equation, it is necessary to define the angle of inclination of a straight line to the O x axis with their slope. Let us assume that a Cartesian coordinate system O x is given on the plane.

Definition 1

The angle of inclination of the straight line to the axis O x, located in the Cartesian coordinate system O x y on the plane, this is the angle that is measured from the positive direction O x to the straight line counterclockwise.

When a line is parallel to Ox or coincidence occurs in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is defined on the interval [ 0 , π) .

Definition 2

Slope of a straight line is the tangent of the slope of the given line.

The standard notation is k. From the definition we get that k = t g α . When a line is parallel Oh, they say that slope does not exist, since it goes to infinity.

The slope is positive when the graph of the function is increasing and vice versa. The figure shows various variations of the location right angle relative to the coordinate system with the coefficient value.

To find this angle, it is necessary to apply the definition of the slope coefficient and calculate the tangent of the inclination angle in the plane.

Solution

From the condition we have that α = 120 °. By definition, you need to calculate the slope. Let's find it from the formula k = t g α = 120 = - 3 .

Answer: k = - 3 .

If the angular coefficient is known, but it is necessary to find the angle of inclination to the x-axis, then the value of the angular coefficient should be taken into account. If k > 0, then the right angle is acute and is found by the formula α = a r c t g k . If k< 0 , тогда угол тупой, что дает право определить его по формуле α = π - a r c t g k .

Example 2

Determine the angle of inclination of the given straight line to O x with a slope equal to 3.

Solution

From the condition we have that the slope is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made according to the formula α = a r c t g k = a r c t g 3 .

Answer: α = a r c t g 3 .

Example 3

Find the angle of inclination of the straight line to the O x axis, if the slope = - 1 3 .

Solution

If we take the letter k as the designation of the slope, then α is the angle of inclination to the given straight line in the positive direction O x. Hence k = - 1 3< 0 , тогда необходимо применить формулу α = π - a r c t g k При подстановке получим выражение:

α = π - a r c t g - 1 3 = π - a r c t g 1 3 = π - π 6 = 5 π 6 .

Answer: 5 pi 6 .

An equation of the form y \u003d k x + b, where k is a slope, and b is some real number, is called the equation of a straight line with a slope. The equation is typical for any straight line that is not parallel to the O y axis.

If we consider in detail a straight line on a plane in a fixed coordinate system, which is given by an equation with a slope that looks like y \u003d k x + b. In this case, it means that the coordinates of any point on the line correspond to the equation. If we substitute the coordinates of the point M, M 1 (x 1, y 1), into the equation y \u003d k x + b, then in this case the line will pass through this point, otherwise the point does not belong to the line.

Example 4

Given a straight line with slope y = 1 3 x - 1 . Calculate whether the points M 1 (3 , 0) and M 2 (2 , - 2) belong to the given line.

Solution

It is necessary to substitute the coordinates of the point M 1 (3, 0) into the given equation, then we get 0 = 1 3 3 - 1 ⇔ 0 = 0 . The equality is true, so the point belongs to the line.

If we substitute the coordinates of the point M 2 (2, - 2), then we get an incorrect equality of the form - 2 = 1 3 · 2 - 1 ⇔ - 2 = - 1 3 . We can conclude that the point M 2 does not belong to the line.

Answer: M 1 belongs to the line, but M 2 does not.

It is known that the straight line is defined by the equation y = k · x + b passing through M 1 (0 , b) , substitution yielded an equality of the form b = k · 0 + b ⇔ b = b . From this we can conclude that the equation of a straight line with a slope y = k · x + b on the plane defines a straight line that passes through the point 0, b. It forms an angle α with the positive direction of the O x axis, where k = t g α .

Consider, for example, a straight line defined using a slope given by the form y = 3 · x - 1 . We get that the straight line will pass through the point with coordinate 0, - 1 with a slope of α = a r c t g 3 = π 3 radians along the positive direction of the O x axis. From this it can be seen that the coefficient is 3.

The equation of a straight line with a slope passing through a given point

It is necessary to solve a problem where it is necessary to obtain the equation of a straight line with a given slope passing through the point M 1 (x 1, y 1) .

The equality y 1 = k · x + b can be considered valid, since the line passes through the point M 1 (x 1 , y 1) . To remove the number b, it is necessary to subtract the equation with the slope coefficient from the left and right sides. It follows from this that y - y 1 = k · (x - x 1) . This equality is called the equation of a straight line with a given slope k, passing through the coordinates of the point M 1 (x 1, y 1) .

Example 5

Compose the equation of a straight line passing through the point M 1 with coordinates (4, - 1), with a slope equal to - 2.

Solution

By condition, we have that x 1 \u003d 4, y 1 \u003d - 1, k \u003d - 2. From here, the equation of the straight line will be written in this way y - y 1 = k · (x - x 1) ⇔ y - (- 1) = - 2 · (x - 4) ⇔ y = - 2 x + 7.

Answer: y = - 2 x + 7 .

Example 6

Write the equation of a straight line with a slope that passes through the point M 1 with coordinates (3, 5) parallel to the straight line y \u003d 2 x - 2.

Solution

By condition, we have that parallel lines have coinciding angles of inclination, hence the slope coefficients are equal. To find the slope from given equation, it is necessary to recall its basic formula y = 2 x - 2, hence it follows that k = 2 . We compose an equation with a slope coefficient and get:

y - y 1 = k (x - x 1) ⇔ y - 5 = 2 (x - 3) ⇔ y = 2 x - 1

Answer: y = 2 x - 1 .

The transition from the equation of a straight line with a slope to other types of equations of a straight line and vice versa

Such an equation is not always applicable for solving problems, since it has a not very convenient notation. To do this, it must be presented in a different form. For example, an equation of the form y = k · x + b does not allow you to write down the coordinates of the direction vector of the straight line or the coordinates of the normal vector. To do this, you need to learn how to represent equations of a different kind.

We can get canonical equation a straight line in a plane using the equation of a straight line with a slope. We get x - x 1 a x = y - y 1 a y . It is necessary to move the term b to the left side and divide by the expression of the obtained inequality. Then we get an equation of the form y = k x + b ⇔ y - b = k x ⇔ k x k = y - b k ⇔ x 1 = y - b k .

The equation of a straight line with a slope has become the canonical equation of a given straight line.

Example 7

Bring the equation of a straight line with slope y = - 3 x + 12 to canonical form.

Solution

We calculate and represent in the form of a canonical equation of a straight line. We get an equation of the form:

y = - 3 x + 12 ⇔ - 3 x = y - 12 ⇔ - 3 x - 3 = y - 12 - 3 ⇔ x 1 = y - 12 - 3

Answer: x 1 = y - 12 - 3.

The general equation of a straight line is easiest to obtain from y = k x + b, but this requires transformations: y = k x + b ⇔ k x - y + b = 0. The transition is made from general equation direct to equations of another kind.

Example 8

An equation of a straight line of the form y = 1 7 x - 2 is given. Find out if the vector with coordinates a → = (- 1 , 7) is a normal straight line vector?

Solution

To solve it, it is necessary to switch to another form of this equation, for this we write:

y = 1 7 x - 2 ⇔ 1 7 x - y - 2 = 0

The coefficients in front of the variables are the coordinates of the normal vector of the straight line. Let's write it like this n → = 1 7 , - 1 , hence 1 7 x - y - 2 = 0 . It is clear that the vector a → = (- 1 , 7) is collinear to the vector n → = 1 7 , - 1 , since we have a fair relation a → = - 7 · n → . It follows that the original vector a → = - 1 , 7 is a normal vector of the line 1 7 x - y - 2 = 0 , which means that it is considered a normal vector for the line y = 1 7 x - 2 .

Answer: Is

Let's solve the problem inverse to this one.

It is necessary to move from the general form of the equation A x + B y + C = 0 , where B ≠ 0 , to an equation with a slope. To do this, we solve the equation for y. We get A x + B y + C = 0 ⇔ - A B · x - C B .

The result is an equation with a slope equal to - A B .

Example 9

An equation of a straight line of the form 2 3 x - 4 y + 1 = 0 is given. Get the equation of a given line with a slope.

Solution

Based on the condition, it is necessary to solve for y, then we get an equation of the form:

2 3 x - 4 y + 1 = 0 ⇔ 4 y = 2 3 x + 1 ⇔ y = 1 4 2 3 x + 1 ⇔ y = 1 6 x + 1 4 .

Answer: y = 1 6 x + 1 4 .

In a similar way, an equation of the form x a + y b \u003d 1 is solved, which is called the equation of a straight line in segments, or the canonical form x - x 1 a x \u003d y - y 1 a y. It is necessary to solve it with respect to y, only then we get an equation with a slope:

x a + y b = 1 ⇔ y b = 1 - x a ⇔ y = - b a x + b .

The canonical equation can be reduced to a form with a slope. For this:

x - x 1 a x = y - y 1 a y ⇔ a y (x - x 1) = a x (y - y 1) ⇔ ⇔ a x y = a y x - a y x 1 + a x y 1 ⇔ y = a y a x x - a y a x x 1 + y 1

Example 10

There is a straight line given by the equation x 2 + y - 3 = 1 . Bring to the form of an equation with a slope.

Solution.

Based on the condition, it is necessary to transform, then we get an equation of the form _formula_. Both sides of the equation should be multiplied by -3 to get the required slope equation. Transforming, we get:

y - 3 = 1 - x 2 ⇔ - 3 y - 3 = - 3 1 - x 2 ⇔ y = 3 2 x - 3 .

Answer: y = 3 2 x - 3 .

Example 11

The straight line equation of the form x - 2 2 \u003d y + 1 5 is brought to the form with a slope.

Solution

It is necessary to calculate the expression x - 2 2 = y + 1 5 as a proportion. We get that 5 · (x - 2) = 2 · (y + 1) . Now you need to fully enable it, for this:

5 (x - 2) = 2 (y + 1) ⇔ 5 x - 10 = 2 y + 2 ⇔ 2 y = 5 x - 12 ⇔ y = 5 2 x

Answer: y = 5 2 x - 6 .

To solve such tasks, parametric equations of the straight line of the form x \u003d x 1 + a x λ y \u003d y 1 + a y λ should be reduced to the canonical equation of the straight line, only after that you can proceed to the equation with the slope.

Example 12

Find the slope of the straight line if it is given by parametric equations x = λ y = - 1 + 2 · λ .

Solution

You need to transition from parametric view to slope. To do this, we find the canonical equation from the given parametric one:

x = λ y = - 1 + 2 λ ⇔ λ = x λ = y + 1 2 ⇔ x 1 = y + 1 2 .

Now it is necessary to resolve this equality with respect to y in order to obtain the equation of a straight line with a slope. To do this, we write this way:

x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1

It follows that the slope of the straight line is equal to 2. This is written as k = 2 .

Answer: k = 2 .

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The topic "The angular coefficient of the tangent as the tangent of the angle of inclination" in the certification exam is given several tasks at once. Depending on their condition, the graduate may be required to provide both a full answer and a short one. In preparation for passing the exam in mathematics, the student should definitely repeat the tasks in which it is required to calculate the slope of the tangent.

Doing this will help you educational portal"Shkolkovo". Our experts have prepared and presented theoretical and practical material as accessible as possible. Having become acquainted with it, graduates with any level of training will be able to successfully solve problems related to derivatives, in which it is required to find the tangent of the slope of the tangent.

Basic moments

To find the correct and rational solution to such tasks in the exam, you need to remember basic definition: the derivative is the rate of change of the function; it is equal to the tangent of the slope of the tangent drawn to the graph of the function at a certain point. It is equally important to complete the drawing. It will allow you to find the right decision USE problems on the derivative, in which it is required to calculate the tangent of the slope of the tangent. For clarity, it is best to plot a graph on the OXY plane.

If you have already familiarized yourself with the basic material on the topic of the derivative and are ready to start solving problems for calculating the tangent of the angle of inclination of a tangent, similar to USE assignments you can do it online. For each task, for example, tasks on the topic "Relationship of the derivative with the speed and acceleration of the body", we wrote down the correct answer and the solution algorithm. In this case, students can practice completing tasks. different levels difficulties. If necessary, the exercise can be saved in the "Favorites" section, so that later you can discuss the decision with the teacher.

The line y \u003d f (x) will be tangent to the graph shown in the figure at the point x0 if it passes through the point with coordinates (x0; f (x0)) and has a slope f "(x0). Find such a coefficient, knowing the features of the tangent, it is not difficult.

You will need

• - mathematical reference book;
• - a simple pencil;
• - notebook;
• - protractor;
• - compass;
• - a pen.

Instruction

If the value f‘(x0) does not exist, then either there is no tangent, or it passes vertically. In view of this, the presence of the derivative of the function at the point x0 is due to the existence of a non-vertical tangent that is in contact with the graph of the function at the point (x0, f(x0)). In this case, the slope of the tangent will be equal to f "(x0). Thus, the geometric meaning of the derivative becomes clear - the calculation of the slope of the tangent.

Draw on additional tangents that would be in contact with the graph of the function at points x1, x2 and x3, and also mark the angles formed by these tangents with the abscissa axis (such an angle is counted in the positive direction from the axis to the tangent line). For example, the angle, that is, α1, will be acute, the second (α2) is obtuse, and the third (α3) is zero, since the tangent line is parallel to the OX axis. In this case, the tangent of an obtuse angle is negative, the tangent of an acute angle is positive, and for tg0 the result is zero.

note

Correctly determine the angle formed by the tangent. To do this, use a protractor.

Useful advice

Two oblique lines will be parallel if their slopes are equal to each other; perpendicular if the product of the slopes of these tangents is -1.

Sources:

• Tangent to function graph

Cosine, like sine, is referred to as "direct" trigonometric functions. The tangent (together with the cotangent) is added to another pair called "derivatives". There are several definitions of these functions that make it possible to find the tangent given by known value cosine of the same value.

Instruction

Subtract the quotient from unity by the cosine of the given angle raised to the value, and extract the square root from the result - this will be the value of the tangent from the angle, expressed by its cosine: tg (α) \u003d √ (1-1 / (cos (α)) ²) . At the same time, pay attention to the fact that in the formula the cosine is in the denominator of the fraction. The impossibility of dividing by zero excludes the use of this expression for angles equal to 90°, as well as differing from this value by multiples of 180° (270°, 450°, -90°, etc.).

There is also alternative way calculating the tangent from the known value of the cosine. It can be used if there is no restriction on the use of other . To implement this method, first determine the angle value from a known cosine value - this can be done using the arccosine function. Then simply calculate the tangent for the angle of the resulting value. AT general view this algorithm can be written as follows: tg(α)=tg(arccos(cos(α))).

There is also an exotic option using the definition of cosine and tangent through sharp corners right triangle. The cosine in this definition corresponds to the ratio of the length of the leg adjacent to the considered angle to the length of the hypotenuse. Knowing the value of the cosine, you can choose the lengths of these two sides corresponding to it. For example, if cos(α)=0.5, then the adjacent can be taken equal to 10 cm, and the hypotenuse - 20 cm. Specific numbers do not matter here - you will get the same and correct with any values ​​\u200b\u200bthat have the same. Then, using the Pythagorean theorem, determine the length of the missing side - the opposite leg. She will be equal square root from the difference between the lengths of the squared hypotenuse and the known leg: √(20²-10²)=√300. By definition, the tangent corresponds to the ratio of the lengths of the opposite and adjacent legs (√300/10) - calculate it and get the tangent value found using the classical definition of cosine.

Sources:

• cosine through tangent formula

One of trigonometric functions, most often denoted by the letters tg, although the designations tan are also found. The easiest way is to represent the tangent as the ratio of the sine corner to its cosine. This is an odd periodic and not continuous function, each cycle of which is equal to the number Pi, and the break point corresponds to half that number.