The slope is straight. How to find the slope of an equation
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;
- - 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 slope the tangent will be f "(x0). Thus, it becomes clear geometric sense derivative - 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) will be obtuse, and the third (α3) zero, since the tangent line is parallel to the x-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.
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 value of the angle from the known value of the cosine - 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 angle 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.
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 answer. 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 correct solution 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 figure shows the angle of inclination of the straight line and the value of the slope coefficient for various options for the location of the straight line relative to the rectangular coordinate system.
Finding the slope of a straight line at a known angle of inclination to the Ox axis does not present any difficulties. To do this, it is enough to recall the definition of the slope coefficient and calculate the tangent of the slope angle.
Example.
Find the slope of the line if the angle of its inclination to the x-axis is equal to .
Decision.
By condition . Then, by definition of the slope of the straight line, we calculate 
.
Answer:
The task of finding the angle of inclination of a straight line to the x-axis with a known slope is a little more difficult. Here it is necessary to take into account the sign of the slope coefficient. When the angle of inclination of the straight line is acute and is found as . When the angle of inclination of a straight line is obtuse and can be determined by the formula 
.
Example.
Determine the angle of inclination of a straight line to the x-axis if its slope is 3.
Decision.
Since, by condition, the slope is positive, the angle of inclination of the straight line to the Ox axis is sharp. We calculate it according to the formula.
Answer:
Example.
The slope of the straight line is . Determine the angle of inclination of the straight line to the axis Ox.
Decision.
Denote k is the slope of the straight line, is the angle of inclination of this straight line to the positive direction of the Ox axis. As 
, then we use the formula for finding the angle of inclination of a straight line of the following form . We substitute the data from the condition into it: .
Answer:
Equation of a straight line with a slope.
Line Equation with Slope has the form , where k is the slope of the straight line, b is some real number. The equation of a straight line with a slope can specify any straight line that is not parallel to the Oy axis (for a straight line parallel to the y-axis, the slope is not defined).
Let's look at the meaning of the phrase: "a line on a plane in a fixed coordinate system is given by an equation with a slope of the form". This means that the equation is satisfied by the coordinates of any point on the line and not by the coordinates of any other point on the plane. Thus, if the correct equality is obtained when substituting the coordinates of a point, then the line passes through this point. Otherwise, the point does not lie on a line.
Example.
The straight line is given by an equation with slope . Do the points also belong to this line?
Decision.
Substitute the coordinates of the point into the original equation of a straight line with a slope: 
. We have obtained the correct equality, therefore, the point M 1 lies on a straight line.
When substituting the coordinates of the point, we get the wrong equality: 
. Thus, the point M 2 does not lie on a straight line.
Answer:
Dot M 1 belongs to the line, M 2 does not.
It should be noted that the straight line, defined by the equation of a straight line with a slope , passes through the point, since when substituting its coordinates into the equation, we get the correct equality: .
Thus, the equation of a straight line with a slope determines a straight line on a plane passing through a point and forming an angle with the positive direction of the abscissa axis, and .
As an example, let's draw a straight line defined by the equation of a straight line with a slope of the form . This line passes through the point and has a slope 
radians (60 degrees) to the positive direction of the Ox axis. Its slope is .
The equation of a straight line with a slope passing through a given point.
Now we will solve a very important problem: we will obtain the equation of a straight line with a given slope k and passing through the point .
Since the line passes through the point , then the equality 
. The number b is unknown to us. To get rid of it, we subtract from the left and right parts of the equation of a straight line with a slope, respectively, the left and right parts of the last equality. In doing so, we get 
. This equality is equation of a straight line with a given slope k that passes through a given point.
Consider an example.
Example.
Write the equation of a straight line passing through the point, the slope of this straight line is -2.
Decision.
From the condition we have 
. Then the equation of a straight line with a slope will take the form .
Answer:
Example.
Write the equation of a straight line if it is known that it passes through a point and the angle of inclination to the positive direction of the Ox axis is .
Decision.
First, we calculate the slope of the straight line whose equation we are looking for (we solved such a problem in the previous paragraph of this article). A-priory 
. Now we have all the data to write the equation of a straight line with a slope:
Answer:
Example.
Write the equation of a line with a slope that passes through a point parallel to the line.
Decision.
It is obvious that the angles of inclination of parallel lines to the axis Ox coincide (if necessary, see the article parallel lines), therefore, the slope coefficients of parallel lines are equal. Then the slope of the straight line, the equation of which we need to obtain, is equal to 2, since the slope of the straight line is 2. Now we can compose the required equation of a straight line with a slope:
Answer:
The transition from the equation of a straight line with a slope coefficient to other types of the equation of a straight line and vice versa.
With all the familiarity, the equation of a straight line with a slope is far from always convenient to use when solving problems. In some cases, problems are easier to solve when the equation of a straight line is presented in a different form. For example, the equation of a straight line with a slope does not allow you to immediately write down the coordinates of the directing vector of the straight line or the coordinates of the normal vector of the straight line. Therefore, one should learn to move from the equation of a straight line with a slope to other types of the equation of this straight line.
From the equation of a straight line with a slope, it is easy to obtain the canonical equation of a straight line on a plane of the form 
. To do this, we transfer the term b from the right side of the equation to the left side with the opposite sign, then divide both parts of the resulting equality by the slope k:. These actions lead us from the equation of a straight line with a slope to canonical equation straight.
Example.
Give the equation of a straight line with a slope 
to the canonical form.
Decision.
Let's perform the necessary transformations: .
Answer:
Example.
The straight line is given by the equation of a straight line with slope . Is the vector a normal vector of this line?
Decision.
To solve this problem, let's move from the equation of a straight line with a slope to the general equation of this straight line: 
. We know that the coefficients in front of the variables x and y in the general equation of a straight line are the corresponding coordinates of the normal vector of this straight line, that is, the normal vector of the straight line 
. Obviously, the vector is collinear to the vector , since the relation is true (if necessary, see the article). Thus, the original vector is also a normal vector of the line , and, therefore, is a normal vector and the original line .
Answer:
Yes it is.
And now we will solve the inverse problem - the problem of bringing the equation of a straight line on a plane to the equation of a straight line with a slope.
From the general straight line equation 
, where , it is very easy to pass to the slope equation. For this you need general equation direct resolve with respect to y . At the same time, we get . The resulting equality is the equation of a straight line with a slope equal to .
The slope coefficient is straight. In this article, we will consider tasks related to the coordinate plane included in the exam in mathematics. These are assignments for:
- determination of the slope of a straight line, when two points through which it passes are known;
- determination of the abscissa or ordinate of the point of intersection of two lines on the plane.
What is the abscissa and ordinate of a point was described in this section. In it, we have already considered several problems related to the coordinate plane. What needs to be understood for the type of tasks under consideration? A bit of theory.
The equation of a straight line on the coordinate plane has the form:
where k – this is the slope of the straight line.
Next moment! Slope of a straight line equal to tangent angle of inclination of a straight line. This is the angle between the given line and the axisoh.
It lies between 0 and 180 degrees.
That is, if we reduce the equation of a straight line to the form y = kx + b, then further we can always determine the coefficient k (slope coefficient).
Also, if we can determine the tangent of the slope of the straight line based on the condition, then we will thereby find its slope.
The next theoretical moment!Equation of a straight line passing through two given points.The formula looks like:
Consider problems (similar to those from open bank assignments):
Find the slope of the straight line passing through the points with coordinates (–6; 0) and (0; 6).
In this problem, the most rational way to solve this is to find the tangent of the angle between the x-axis and the given straight line. It is known that it is equal to the angular coefficient. Consider a right triangle formed by a straight line and the x and y axes:
The tangent of an angle in right triangle is the ratio of the opposite leg to the adjacent:
* Both legs are equal to six (these are their lengths).
Certainly, this task can be solved using the formula for finding the equation of a straight line passing through two given points. But it will be a longer solution path.
Answer: 1
Find the slope of the straight line passing through the points with coordinates (5;0) and (0;5).
Our points have coordinates (5;0) and (0;5). Means,
Let's bring the formula to the form y = kx + b
We got that the angular coefficient k = – 1.
Answer: -1
Straight a passes through points with coordinates (0;6) and (8;0). Straight b passes through the point with coordinates (0;10) and is parallel to the line a b with axle ox.
In this problem, you can find the equation of a straight line a, determine the slope for it. Straight line b the slope will be the same since they are parallel. Next, you can find the equation of a straight line b. And then, substituting the value y = 0 into it, find the abscissa. BUT!
In this case, it is easier to use the triangle similarity property.
The right triangles formed by the given (parallel) lines of coordinates are similar, which means that the ratios of their respective sides are equal.
The desired abscissa is 40/3.
Answer: 40/3
Straight a passes through points with coordinates (0;8) and (–12;0). Straight b passes through the point with coordinates (0; -12) and is parallel to the line a. Find the abscissa of the point of intersection of the line b with axle ox.
For this problem, the most rational way to solve it is to use the similarity property of triangles. But we will solve it in a different way.
We know the points through which the line passes a. We can write the equation of a straight line. The formula for the equation of a straight line passing through two given points is:
By condition, the points have coordinates (0;8) and (–12;0). Means,
Let's bring to mind y = kx + b:
Got that corner k = 2/3.
*The angular coefficient could be found through the tangent of the angle in a right triangle with legs 8 and 12.
We know that parallel lines have equal slopes. So the equation of a straight line passing through the point (0;-12) has the form:
Find value b we can substitute the abscissa and ordinate into the equation:
So the line looks like:
Now, to find the desired abscissa of the point of intersection of the line with the x-axis, you need to substitute y \u003d 0:
Answer: 18
Find the ordinate of the point of intersection of the axis oy and a straight line passing through point B(10;12) and a parallel line passing through the origin and point A(10;24).
Let's find the equation of a straight line passing through the points with coordinates (0;0) and (10;24).
The formula for the equation of a straight line passing through two given points is:
Our points have coordinates (0;0) and (10;24). Means,
Let's bring to mind y = kx + b
The slopes of the parallel lines are equal. Hence, the equation of a straight line passing through the point B (10; 12) has the form:
Meaning b we find by substituting the coordinates of the point B (10; 12) into this equation:
We got the equation of a straight line:
To find the ordinate of the point of intersection of this line with the axis OU must be substituted into the found equation X= 0:
*Easiest solution. With the help of parallel translation, we shift this line down along the axis OU to the point (10;12). The shift occurs by 12 units, that is, point A(10;24) "passed" to point B(10;12), and point O(0;0) "passed" to point (0;–12). So the resulting line will intersect the axis OU at the point (0;–12).
The desired ordinate is -12.
Answer: -12
Find the ordinate of the point of intersection of the line given by the equation
3x + 2y = 6, with axis Oy.
Coordinate of the point of intersection of the given line with the axis OU has the form (0; at). Substitute the abscissa into the equation X= 0, and find the ordinate:
Ordinate of point of intersection of a line with an axis OU equals 3.
* The system is being solved:
Answer: 3
Find the ordinate of the point of intersection of the lines given by the equations
3x + 2y = 6 and y = - x.
When two lines are given, and the question is about finding the coordinates of the point of intersection of these lines, the system of these equations is solved:
In the first equation, we substitute - X instead of at:
The ordinate is minus six.
Answer: – 6
Find the slope of the straight line passing through the points with coordinates (–2; 0) and (0; 2).
Find the slope of the straight line passing through the points with coordinates (2;0) and (0;2).
The line a passes through the points with coordinates (0;4) and (6;0). Line b passes through the point with coordinates (0;8) and is parallel to line a. Find the abscissa of the point of intersection of line b with the x-axis.
Find the ordinate of the point of intersection of the y-axis and the line passing through point B (6;4) and the parallel line passing through the origin and point A (6;8).
1. It is necessary to clearly understand that the slope of the straight line is equal to the tangent of the slope of the straight line. This will help you in solving many problems of this type.
2. The formula for finding a straight line passing through two given points must be understood. With its help, you can always find the equation of a straight line if the coordinates of two of its points are given.
3. Remember that the slopes of parallel lines are equal.
4. As you understand, in some problems it is convenient to use the sign of similarity of triangles. Problems are solved practically orally.
5. Tasks in which two lines are given and it is required to find the abscissa or ordinate of their intersection point can be solved graphically. That is, build them on the coordinate plane (on a sheet in a cell) and determine the intersection point visually. *But this method is not always applicable.
6. And the last. If a straight line and the coordinates of the points of its intersection with the coordinate axes are given, then in such problems it is convenient to find the slope through finding the tangent of the angle in the formed right triangle. How to "see" this triangle for various arrangements of lines on the plane is schematically shown below:
>> Line inclination angle from 0 to 90 degrees<<
>> Straight line angle from 90 to 180 degrees<<
That's all. Good luck to you!
Sincerely, Alexander.
P.S: I would be grateful if you tell about the site in social networks.
In the previous chapter it was shown that, by choosing a certain coordinate system on the plane, we can analytically express the geometric properties characterizing the points of the line under consideration by the equation between the current coordinates. Thus, we get the equation of the line. In this chapter, the equations of straight lines will be considered.
To formulate the equation of a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.
First, we introduce the concept of the slope of a straight line, which is one of the quantities characterizing the position of a straight line on a plane.
Let's call the angle of inclination of the line to the Ox axis the angle by which the Ox axis must be rotated so that it coincides with the given line (or turns out to be 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 by an angle of 180 ° will again combine it with the straight line, the angle of inclination of the straight line to the axis can be chosen ambiguously (up to a multiple of ).
The tangent of this angle is uniquely determined (since changing the angle to does not change its tangent).
The tangent of the angle of inclination of a straight line to the x-axis is called the slope of the straight line.
The slope characterizes the direction of the straight line (here we do not distinguish between two mutually opposite directions of the straight line). If the slope of the line is zero, then the line is parallel to the x-axis. With a positive slope, the angle of inclination of the straight line to the Ox axis will be sharp (we are considering here the smallest positive value of the angle of inclination) (Fig. 39); in this case, the larger the slope, the greater the angle of its inclination to the Ox axis. If the slope is negative, then the angle of inclination of the straight line to the x-axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the x-axis does not have a slope (the tangent of an angle does not exist).