How to find the tangent of a tangent angle. How to find the slope

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Continuation of the topic, the equation of a line on a plane is based on the study of a straight line from algebra lessons. This article provides general information on the topic of equation of a straight line with a slope. Let's consider the definitions, get the equation itself, and identify the connection with other types of equations. Everything will be discussed using examples of problem solving.

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

Definition 1

The angle of inclination of the straight line to the O x axis, 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 the line is parallel to O x or coincides 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

Direct slope is the tangent of the angle of inclination of a given straight line.

Standard designation is k. From the definition we find that k = t g α . When a line is parallel to Oh, they say that slope does not exist, since it turns to infinity.

The slope is positive when the graph of the function increases and vice versa. The figure shows various variations of 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 angular coefficient and calculate the tangent of the angle of inclination in the plane.

Solution

From the condition we have that α = 120°. By definition, the slope must be calculated. Let's find it from the formula k = t g α = 120 = - 3.

Answer: k = - 3 .

If the angular coefficient is known, and it is necessary to find the angle of inclination to the abscissa 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 an angular coefficient of 3.

Solution

From the condition we have that the angular coefficient is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made using 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 angular coefficient, then α is the angle of inclination to a 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 π 6 .

An equation of the form y = k x + b, where k is the slope and b is some real number, is called the equation of a straight line with an angle coefficient. 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 specified by an equation with an angular coefficient that has the form y = k x + b. In this case, it means that the equation corresponds to the coordinates of any point on the line. If we substitute the coordinates of point M, M 1 (x 1, y 1) into the equation y = 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

A straight line with slope y = 1 3 x - 1 is given. 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, which means 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 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 line is defined by the equation y = k · x + b, passing through M 1 (0, b), upon substitution we obtained an equality of the form b = k · 0 + b ⇔ b = b. From this we can conclude that the equation of a straight line with an angular coefficient 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.

Let us consider, as an example, a straight line defined using an angular coefficient specified in the form y = 3 x - 1. We obtain 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 in the positive direction of the O x axis. This shows that the coefficient is 3.

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 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

Write an equation for a straight line passing through point M 1 with coordinates (4, - 1), with an angular coefficient equal to - 2.

Solution

By condition we have that x 1 = 4, y 1 = - 1, k = - 2. From here the equation of the line will be written as follows: 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 an angular coefficient that passes through the point M 1 with coordinates (3, 5), parallel to the straight line y = 2 x - 2.

Solution

By condition, we have that parallel lines have identical angles of inclination, which means that the angular coefficients are equal. To find the slope from given equation, you need to remember its basic formula y = 2 x - 2, it follows that k = 2. We create an equation with the 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 .

Transition from a straight line equation with a slope to other types of straight line equations and back

This equation is not always applicable for solving problems, since it is not very conveniently written. To do this, you need to present it in a different form. For example, an equation of the form y = k · x + b does not allow us to write down the coordinates of the direction vector of a straight line or the coordinates of a normal vector. To do this, you need to learn to represent with equations of a different type.

We can obtain the canonical equation of a line on a plane using the equation of a line with an angle coefficient. 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 resulting 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 line with a slope has become the canonical equation of this line.

Example 7

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

Solution

Let us calculate and present it 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 for this it is necessary to make transformations: y = k · x + b ⇔ k · x - y + b = 0. A transition is made from general equation straight line to equations of another type.

Example 8

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

Solution

To solve it is necessary to move 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 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 the 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 it is considered a normal vector for the line y = 1 7 x - 2.

Answer: Is

Let's solve the inverse problem of this one.

Need to move from general view equations 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

A straight line equation of the form 2 3 x - 4 y + 1 = 0 is given. Obtain the equation of a given line with an angular coefficient.

Solution

Based on the condition, it is necessary to solve for y, then we obtain 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 .

An equation of the form x a + y b = 1 is solved in a similar way, which is called the equation of a straight line in segments, or canonical type x - x 1 a x = y - y 1 a y . We need to solve it for y, only then we get an equation with the 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 an angular coefficient. 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. Reduce to the form of an equation with an angular coefficient.

Solution.

Based on the condition, it is necessary to transform, then we obtain an equation of the form _formula_. Both sides of the equation must be multiplied by - 3 to obtain 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

Reduce the straight line equation of the form x - 2 2 = y + 1 5 to a form with an angular coefficient.

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 completely enable it, to do 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 problems, parametric equations of the line of the form x = x 1 + a x · λ y = y 1 + a y · λ should be reduced to the canonical equation of the line, only after this can one proceed to the equation with the slope coefficient.

Example 12

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

Solution

It is necessary to transition from the parametric view to the 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 an angular coefficient. To do this, let's write it this way:

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

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

Answer: k = 2.

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The topic “The angular coefficient of a tangent as the tangent of the angle of inclination” is given several tasks in the certification exam. Depending on their condition, the graduate may be required to provide either a full answer or a short answer. In preparation for passing the Unified State Exam In mathematics, the student should definitely repeat problems in which it is necessary to calculate the angular coefficient of a tangent.

It will help you do this educational portal"Shkolkovo". Our specialists prepared and presented theoretical and practical material in the most accessible way possible. Having become familiar with it, graduates with any level of training will be able to successfully solve problems related to derivatives in which it is necessary to find the tangent of the tangent angle.

Basic moments

To find the correct and rational solution to such tasks in the Unified State Exam, it is necessary to remember the basic definition: the derivative represents the rate of change of a function; it is equal to the tangent of the tangent angle 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 correct solution Unified State Exam problems on the derivative, in which it is necessary to calculate the tangent of the tangent angle. For clarity, it is best to plot the graph on the OXY plane.

If you have already familiarized yourself with the basic material on the topic of derivatives and are ready to begin solving problems on calculating the tangent of the tangent angle, such as Unified State Exam assignments, you can do this online. For each task, for example, problems on the topic “Relationship of a derivative with the speed and acceleration of a body,” we wrote down the correct answer and solution algorithm. At the same time, students can practice performing tasks of varying levels of complexity. If necessary, the exercise can be saved in the “Favorites” section so that you can discuss the solution with the teacher later.

In the previous chapter it was shown that, by choosing a certain coordinate system on the plane, we can express the geometric properties characterizing the points of the line under consideration analytically by an equation between the current coordinates. Thus we get the equation of the line. This chapter will look at straight line equations.

To create an equation for a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.

First, we will introduce the concept of the angular coefficient of a line, which is one of the quantities characterizing the position of a line on a plane.

Let's call the angle of inclination of the straight line to the Ox axis the angle by which the Ox axis needs to be rotated so that it coincides with the given line (or is parallel to it). As usual, we will consider the angle taking into account the sign (the sign is determined by the direction of rotation: counterclockwise or clockwise). Since an additional rotation of the Ox axis through an angle of 180° will again align it with the straight line, the angle of inclination of the straight line to the axis can not be chosen unambiguously (to within a term, a multiple of ).

The tangent of this angle is determined uniquely (since changing the angle does not change its tangent).

The tangent of the angle of inclination of the straight line to the Ox axis is called the angular coefficient of the straight line.

The angular coefficient characterizes the direction of the straight line (we do not distinguish here between two mutually opposite directions of the straight line). If the slope of a line is zero, then the line is parallel to the x-axis. With a positive angular coefficient, the angle of inclination of the straight line to the Ox axis will be acute (we are considering here the smallest positive value tilt angle) (Fig. 39); Moreover, the greater the angular coefficient, the greater the angle of its inclination to the Ox axis. If the angular coefficient is negative, then the angle of inclination of the straight line to the Ox axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the Ox axis does not have an angular coefficient (the tangent of the angle does not exist).

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

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

Let's remember the definition:

The derivative is the rate of change of a function.

The figure shows graphs of three functions. Which one do you think is growing faster?

The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.

Here's another example.

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

The graph shows everything at once, isn’t it? Kostya’s income more than doubled in six months. And Grisha’s income also increased, but just a little. And Matvey’s income decreased to zero. The starting conditions are the same, but the rate of change of the function, that is derivative, - different. As for Matvey, his income derivative is generally negative.

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

What we're really looking at is how steeply the graph of a function goes up (or down). In other words, how quickly does y change as x changes? 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 .

We'll show you how to find it using a graph.

A graph of some function has been drawn. Let's take a point with an abscissa on it. Let us draw a tangent to the graph of the function at this point. We want to estimate how steeply the graph of a function goes up. A convenient value for this is tangent of the tangent angle.

The derivative of a function at a point is equal to the tangent of the tangent angle drawn to the graph of the function at this point.

Please note that 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 a tangent to the graph of a function is. This is a straight line that has a single common point with the graph in this section, and as shown in our figure. It looks like a tangent to a circle.

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

We found the derivative using a graph without even knowing the formula of the function. Such problems are often found in the Unified State Examination in mathematics under the number.

There is another important relationship. 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 tangent angle.

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 and decrease in others, and at different rates. And let this function have maximum and minimum points.

At a point the function increases. The tangent to the graph drawn at the point forms sharp corner; with positive axis direction. This means that the derivative at the point is positive.

At the point our function decreases. 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.

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

Point - maximum point. At this point, the increase in 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 zero, but its sign changes from “minus” to “plus”.

Conclusion: using the derivative we can find out everything that interests us about the behavior of a function.

If the derivative is positive, then the function increases.

If the derivative is negative, then the function decreases.

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”.

Let's write these conclusions 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.

It is possible that 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 is the 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 remains 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.

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