Speed \u200b\u200bis the first derivative of the coordinate. Derivative in physics. The geometric and physical meaning of the derivative
The physical meaning of the derivative. The structure of the exam in mathematics includes a group of tasks for the solution of which knowledge and understanding of the physical meaning of the derivative is necessary. In particular, there are tasks where the law of motion of a certain point (object) is given, expressed by the equation and it is required to find its speed at a certain moment in time of movement, or the time after which the object will acquire a certain given speed.The tasks are very simple, they are solved in one action. So:
Let the law of motion of the material point x (t) be given along the coordinate axis, where x is the coordinate of the moving point, t is time.
Speed \u200b\u200bat a particular point in time is a derivative of the time coordinate. This is the mechanical meaning of the derivative.
Similarly, acceleration is a derivative of speed with respect to time:
Thus, the physical meaning of the derivative is speed. This can be the speed of movement, the rate of change of any process (for example, the growth of bacteria), the speed of completion of work (and so on, there are many applied problems).
In addition, you need to know the table of derivatives (you need to know it as well as the multiplication table) and the rules of differentiation. Specifically, to solve the specified problems, knowledge of the first six derivatives is required (see table):
Consider the tasks:
x (t) \u003d t 2 - 7t - 20
where x t is the time in seconds, measured from the beginning of the movement. Find its speed (in meters per second) at time t \u003d 5 s.
The physical meaning of the derivative is speed (speed of movement, rate of change of the process, speed of work, etc.)
Find the law of change of velocity: v (t) \u003d x ′ (t) \u003d 2t - 7 m / s.
At t \u003d 5 we have:
Answer: 3
Decide on your own:
The material point moves rectilinearly according to the law x (t) \u003d 6t 2 - 48t + 17, where x - distance from the reference point in meters, t - time in seconds, measured from the beginning of the movement. Find its speed (in meters per second) at time t \u003d 9 s.
The material point moves rectilinearly according to the law x (t) \u003d 0.5t 3 - 3t 2 + 2t, where xt - time in seconds, measured from the beginning of the movement. Find its speed (in meters per second) at time t \u003d 6 s.
The material point moves rectilinearly according to the law
x (t) \u003d –t 4 + 6t 3 + 5t + 23
where x - distance from the reference point in meters,t - time in seconds, measured from the beginning of the movement. Find its speed (in meters per second) at time t \u003d 3 s.
The material point moves rectilinearly according to the law
x (t) \u003d (1/6) t 2 + 5t + 28
where x is the distance from the reference point in meters, t is the time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 6 m / s?
Find the law of change of speed:
In order to find at what point in timet the speed was 3 m / s, it is necessary to solve the equation:
Answer: 3
Decide for yourself:
The material point moves rectilinearly according to the law x (t) \u003d t 2 - 13t + 23, where x - distance from the reference point in meters, t - time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 3 m / s?
The material point moves rectilinearly according to the law
x (t) \u003d (1/3) t 3 - 3t 2 - 5t + 3
where x - distance from the reference point in meters, t - time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 2 m / s?
I note that focusing only on this type of tasks on the exam is not worth it. Can quite unexpectedly introduce the opposite tasks presented. When the law of change of speed is given and the question will be about finding the law of motion.
Hint: in this case it is necessary to find the integral of the velocity function (this is also a one-task task). If you need to find the distance traveled for a certain point in time, then you need to substitute the time in the resulting equation and calculate the distance. However, we will also analyze such tasks, do not miss!I wish you success!
Sincerely, Alexander Krutitsky.
P.S: I would be grateful if you talk about the site on social networks.
Algebra is generous. Often she gives more than she is asked.
J. Dalamber
Interdisciplinary communication is a didactic condition and a means of deep and comprehensive assimilation of the foundations of sciences in school.
In addition, they help to increase the scientific level of students' knowledge, the development of logical thinking and their creative abilities. The implementation of intersubject communications eliminates duplication in the study of material, saves time and creates favorable conditions for the formation of general educational skills of students.
The establishment of intersubject communications in the course of physics increases the effectiveness of the polytechnic and practical orientation of education.
In teaching mathematics, the motivational side is very important. Students perceive the mathematical problem better if it arises as if in front of their eyes, is formulated after considering some physical phenomena or technical problems.
No matter how much the teacher talks about the role of practice in the progress of mathematics and the importance of mathematics in the study of physics, the development of technology, but if he does not show how physics affects the development of mathematics and how mathematics helps practice in solving its problems, then the development of a materialistic worldview will be applied serious damage. But in order to show how mathematics helps in solving its problems, we need problems that are not invented for methodological purposes, but that actually arise in various fields of practical human activity
Historical information
Differential calculus was created by Newton and Leibniz at the end of the 17th century on the basis of two tasks:
- finding a tangent to an arbitrary line;
- about finding speed with an arbitrary law of motion.
Earlier, the notion of a derivative was encountered in the works of the Italian mathematician Nicolo Tartaglia (about 1500 - 1557) - a tangent appeared here during the study of the issue of the angle of inclination of the gun, which ensures the longest range of the projectile.
In the 17th century, on the basis of G. Galileo’s teachings on movement, the kinematic concept of the derivative actively developed.
The whole treatise on the role of the derivative in mathematics is dedicated to the famous scientist Galileo Galilei. Various expositions began to be found in the works of Descartes, the French mathematician Roberval, and the English scientist L. Gregory. A great contribution to the study of differential calculus was made by Lopital, Bernoulli, Lagrange, Euler, Gauss.
Some applications of the derivative in physics
Derivative- the basic concept of differential calculus, characterizing function change rate.
Determined by as the limit of the ratio of the increment of a function to the increment of its argument when the argument tends to increment the argument to zero, if such a limit exists.
In this way, 
Hence, to calculate the derivative of the function f (x) at the point x 0 by definition, you need:
Consider several physical problems in solving which this scheme is applied.
The problem of instant speed. The mechanical meaning of the derivative
Recall how the speed of movement was determined. The material point moves along the coordinate line. The x coordinate of this point is a known function x (t) time t. For a period of time from t 0 before t 0 + point movement equals x (t 0 +)
– x (t 0) - and its average speed is as follows: 
.
Usually, the nature of the movement is such that, at small, the average speed practically does not change, i.e. movement with a high degree of accuracy can be considered uniform. In other words, the value of the average velocity at tends to some well-defined value, which is called the instantaneous velocity v (t 0) material point at time t 0.
So, 
But by definition 
Therefore, it is believed that the instantaneous velocity at a time t 0
Arguing in a similar manner, we find that the derivative of speed with respect to time is acceleration, i.e.
The heat capacity problem of the body
So that the temperature of a body weighing 1 g rises from 0 degrees to t degrees, the body needs to report a certain amount of heat Q. Means Qthere is a temperature function t, to which the body heats up: Q \u003d Q (t). Let the body temperature rise with t 0 before t.The amount of heat expended for this heating is equal. The ratio is the amount of heat that is required on average to heat the body by 1 degree when the temperature changes
degrees. This ratio is called the average heat capacity of a given body and is denoted by from wed.
Because Since the average heat capacity does not give an idea of \u200b\u200bthe heat capacity for any temperature T, the concept of heat capacity at a given temperature is introduced t 0 (at this point t 0).
Heat capacity at temperature t 0 (at a given point) is called the limit
The linear density problem of the rod
Consider an inhomogeneous rod.
For such a rod, the question arises of the rate of change of mass depending on its length.
Average linear density 
the mass of the rod is a function of its length x.
Thus, the linear density of an inhomogeneous rod at a given point is determined as follows:
By considering such problems, one can obtain similar conclusions on many physical processes. Some of them are given in the table.
Function |
Formula |
Conclusion |
| m (t) is the dependence of the mass of spent fuel on time. |  |
Derivative mass over time there is speed fuel consumption. |
| T (t) is the dependence of the temperature of the heated body on time. |  |
Derivative temperature over time there is speed body heating. |
| m (t) is the time dependence of the mass during the decay of a radioactive substance. |  |
Derivative mass of radioactive substance over timethere is speed radioactive decay. |
| q (t) is the time dependence of the amount of electricity flowing through the conductor |  |
Derivative amount of electricity over time there is current strength. |
| A (t) - dependence of work on time |  |
Derivative work on time there is power. |
Practical exercises:
The projectile flying out of the gun moves according to the law x (t) \u003d - 4t 2 + 13t (m). Find the projectile speed at the end of 3 seconds.
The amount of electricity flowing through the conductor, starting from time t \u003d 0 c, is given by the formula q (t) \u003d 2t 2 + 3t + 1 (Cool) Find the current strength at the end of the fifth second.
The amount of heat Q (J) required to heat 1 kg of water from 0 o to t o C is determined by the formula Q (t) \u003d t + 0.00002t 2 + 0.0000003t 3. Calculate the heat capacity of water if t \u003d 100 o.
The body moves rectilinearly according to the law x (t) \u003d 3 + 2t + t 2 (m). Determine its speed and acceleration at times 1 s and 3 s.
Find the magnitude of the force F acting on a point of mass m moving according to the law x (t) \u003d t 2 - 4t 4 (m), for t \u003d 3 s.
A body whose mass is m \u003d 0.5 kg moves rectilinearly according to the law x (t) \u003d 2t 2 + t - 3 (m). Find the kinetic energy of the body 7 seconds after the start of movement.
Conclusion
You can specify many more problems from technology, for the solution of which it is also necessary to find the rate of change of the corresponding function.
For example, finding the angular velocity of a rotating body, the linear coefficient of expansion of the bodies when heated, the speed of the chemical reaction at a given time.
In view of the abundance of problems leading to the calculation of the rate of change of a function or, otherwise, to the calculation of the limit of the ratio of the increment of a function to the increment of an argument when the latter tends to zero, it turned out to be necessary to single out such a limit for an arbitrary function and study its main properties. This limit was called derivative function.
So, on a number of examples, we have shown how various physical processes are described using mathematical problems, how the analysis of solutions allows us to draw conclusions and predictions about the course of processes.
Of course, the number of examples of this kind is huge, and quite a large part of them is quite accessible to interested students.
“Music can elevate or pacify the soul,
Painting - pleasing to the eye,
Poetry - to arouse feelings
Philosophy is to satisfy the needs of the mind,
Engineering is to improve the material side of people's lives,
And mathematics can achieve all these goals. ”
So said an American mathematician Maurice Kline.
Bibliography :
- Abramov A.N., Vilenkin N.Ya. et al. Selected questions of mathematics. Grade 10. - M: Enlightenment, 1980.
- Vilenkin N.Ya., Shibasov A.P.Behind the pages of a math textbook. - M: Education, 1996.
- Dobrokhotova M.A., Safonov A.N.. Function, its limit and derivative. - M: Enlightenment, 1969.
- Kolmogorov A.N., Abramov A.M. Algebra and the beginning of mathematical analysis. - M: Education, 2010.
- Kolosov A.A. A book for extracurricular reading in mathematics. - M: Uchpedgiz, 1963.
- Fichtenholtz G.M. Fundamentals of Mathematical Analysis, Part 1 - M: Nauka, 1955.
- Yakovlev G.N. Mathematics for technical schools. Algebra and the beginning of analysis, part 1 - M: Nauka, 1987.
Solving physical problems or examples in mathematics is completely impossible without knowledge of the derivative and methods for its calculation. Derivative is one of the most important concepts of mathematical analysis. We decided to devote this article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?
The geometric and physical meaning of the derivative
Let there be a function f (x) defined in some interval (a, b) . Points x and x0 belong to this interval. When x changes, the function itself changes. Change of argument - difference of its values xx0 . This difference is written as delta x and is called an argument increment. A change or increment of a function is called the difference of the values \u200b\u200bof the function at two points. Derivative definition:
The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.
Otherwise, it can be written like this:
What is the point of finding such a limit? And here is what:
the derivative of the function at the point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at this point.
The physical meaning of the derivative: the time derivative of the path is equal to the speed of the rectilinear motion.
Indeed, since school times, everyone knows that speed is a particular way x \u003d f (t) and time t . Average speed over a period of time:
To find out the speed of a moment in time t0 need to calculate the limit:
Rule one: we issue a constant
The constant can be taken out for the sign of the derivative. Moreover - it needs to be done. When solving math examples, make it a rule - if you can simplify the expression, be sure to simplify .
Example. We calculate the derivative:
Rule two: the derivative of the sum of functions
The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.
We will not give a proof of this theorem, but rather consider a practical example.
Find the derivative of the function:
Rule Three: Derivative Product of Functions
The derivative of the product of two differentiable functions is calculated by the formula:
Example: find the derivative of a function:
Decision: 
It is important to say here about the calculation of derivatives of complex functions. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable.
In the above example, we meet the expression:
In this case, the intermediate argument is 8x to the fifth degree. In order to calculate the derivative of such an expression, we first consider the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate function itself with respect to the independent variable.
Rule Four: Derivative of the quotient of two functions
The formula for determining the derivative of the quotient of two functions:
We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so we warn you: traps are often found in examples, so be careful when calculating derivatives.
With any question on this and other topics, you can contact the student service. In a short time, we will help you to solve the most difficult control and deal with tasks, even if you have never before been involved in the calculation of derivatives.
Until now, we have associated the concept of derivative with the geometric representation of the graph of a function. However, it would be a gross mistake to limit the role of the concept of derivative to the mere task of determining the slope of the tangent to a given curve. An even more important scientific task is to calculate the rate of change of any quantity f (t)changing over time t. It was from this perspective that Newton approached differential calculus. In particular, Newton sought to analyze the phenomenon of speed, considering the time and position of a moving particle as variables (according to Newton’s expression, “fluents”). When a particle moves along the x axis, its motion is well defined, once the function x \u003d f (t)indicating the position of the particle x at any time t. "Uniform motion" with a constant speed b along the x axis is determined by a linear function x \u003d a + bt, where a is the position of the particle at the initial moment (for t \u003d 0).
The motion of a particle on a plane is already described by two functions
x \u003d f (t), y \u003d g (t),
which determine its coordinates as a function of time. In particular * two linear functions correspond to uniform motion
x \u003d a + bt, y \u003d c + dt,
where b and d are two "components" of constant velocity, and a and c are the coordinates of the initial position of the particle (for t \u003d 0); the trajectory of a particle is a straight line whose equation
(x - a) d - (y - s) b \u003d 0
obtained by eliminating t from the two relations above.
If a particle moves in the vertical plane x, y under the action of gravity alone, then its motion (this is proved in elementary physics) is determined by two equations
where a, b, c, d are constant values \u200b\u200bdepending on the state of the particle at the initial moment, and g is the acceleration of gravity of approximately 9.81 if time is measured in seconds and distance is in meters. The trajectory of motion obtained by eliminating t from two given equations is a parabola
if only b ≠ 0; otherwise, the path is a segment of the vertical axis.
If a particle is forced to move along a given curve (similar to how a train moves along rails), then its motion can be determined by a function s (t) (a function of time t) equal to the length of the arc s calculated along this curve from some initial point P 0 to the position of the particle at point P at time t. For example, if we are talking about a unit circle x 2 + y 2 \u003d 1then the function s \u003d ct determines on this circle a uniform rotational motion with speed with.
* Exercise. Draw the trajectories of plane movements defined by the equations: 1) x \u003d sin t, y \u003d cos t; 2) x \u003d sin 2t, y \u003d cos 3t; 3) x \u003d sin 2t, y \u003d 2 sin 3t; 4) in the parabolic motion described above, assume the initial position of the particle (at t \u003d 0) at the origin and assume b\u003e 0, d\u003e 0. Find the coordinates of the highest point on the path. Find the time t and x value corresponding to the secondary intersection of the trajectory with the x axis.
The first goal that Newton set himself was to find the speed of a particle moving unevenly. For simplicity, we consider the motion of a particle along a straight line defined by the function x \u003d f (t). If the motion were uniform, i.e., it was carried out at a constant speed, then this speed could be found by taking two points in time t and t 1 and the corresponding particle positions f (t) and f (t 1) and making up the relation
For example, if t is measured in hours, and x in kilometers, then at t 1 - t \u003d 1 difference x 1 - x will be the number of kilometers covered in 1 hour, and v - speed (in kilometers per hour). Saying that speed is a constant value, they mean only that the difference relation
does not change at any values \u200b\u200bof t and t 1. But if the motion is non-uniform (which, for example, takes place during the free fall of the body, the speed of which increases as it falls), then relation (3) does not give the velocity value at time t, but represents what is usually called the average speed in the time interval from t to t 1. To get the speed at time tneed to calculate the limit average speed when t 1 t Thus, following Newton, we define the speed like this:
In other words, speed is the derivative of the distance traveled (the coordinates of the particle on a straight line) with respect to time, or the "instantaneous rate of change" of the path with respect to time - in contrast middle rate of change, determined by the formula (3).
Rate of change of speed itself called acceleration. Acceleration is simply a derivative of a derivative; it is usually denoted by f "(t) and is called second derivative of the function f (t).
Sometimes, in problem B9 from the USE in mathematics, instead of everyone’s favorite graphs of a function or derivative, simply the equation of the distance from the point to the origin is given. What to do in this case? How to find speed or acceleration by distance.
In fact, everything is simple. Speed \u200b\u200bis a derivative of distance, and acceleration is a derivative of speed (or, equivalently, the second derivative of distance). In this short video you will be convinced that such tasks are solved no more complicated than the "classic" B9.
Today we will analyze two problems on the physical meaning of derivatives from the exam in mathematics. These tasks are found in Part B and are significantly different from those that most students are used to seeing on probes and exams. The thing is that they require understanding the physical meaning of the derivative function. In these problems, we will focus on functions expressing distances.
If $ S \u003d x \\ left (t \\ right) $, then we can calculate $ v $ as follows:
These three formulas are all you need to solve such examples on the physical meaning of the derivative. Just remember that $ v $ is a derivative of distance, and acceleration is a derivative of speed.
Let's see how it works in solving real problems.
Example No. 1
where $ x $ is the distance from the reference point in meters, $ t $ is the time in seconds elapsed from the beginning of the movement. Find the speed of the point (in m / s) at time $ t \u003d 2c $.
This means that we have a function that sets the distance, and we need to calculate the speed at time $ t \u003d 2c $. In other words, we need to find $ v $, i.e.
That's all we needed to find out from the condition: firstly, what the function looks like, and secondly, what is required of us to find.
Let's decide. First of all, we calculate the derivative:
\\ [(x) "\\ left (t \\ right) \u003d - \\ frac (1) (5) \\ cdot 5 ((t) ^ (4)) + 4 ((t) ^ (3)) - 3 (( t) ^ (2)) + 5 \\]
\\ [(x) "\\ left (t \\ right) \u003d - ((t) ^ (4)) + 4 ((t) ^ (3)) - 3 ((t) ^ (2)) + 5 \\]
We need to find the derivative at point 2. Let's substitute:
\\ [(x) "\\ left (2 \\ right) \u003d - ((2) ^ (4)) + 4 \\ cdot ((2) ^ (3)) - 3 \\ cdot ((2) ^ (2)) + 5 \u003d \\]
\[=-16+32-12+5=9\]
That's all, we found the final answer. Total, the speed of our material point at time $ t \u003d 2c $ will be 9 m / s.
Example No. 2
The material point moves according to the law:
where $ x $ is the distance from the reference point in meters, $ t $ is the time in seconds, measured from the beginning of the movement. At what point in time was its speed equal to 3 m / s?
Take a look, the last time we were required to find $ v $ at time 2 s, and this time we were required to find the very moment when this speed would be 3 m / s. We can say that we know the final value, and from this final value we need to find the original.
First of all, we are again looking for a derivative:
\\ [(x) "\\ left (t \\ right) \u003d \\ frac (1) (3) \\ cdot 3 ((t) ^ (2)) - 4 \\ cdot 2t + 19 \\]
\\ [(x) "\\ left (t \\ right) \u003d ((t) ^ (2)) - 8t + 19 \\]
We are asked to find at what point in time the speed will be 3 m / s. We compose and solve the equation to find the physical meaning of the derivative:
\\ [(((t) ^ (2)) - 8t + 19 \u003d 3 \\]
\\ [(((t) ^ (2)) - 8t + 16 \u003d 0 \\]
\\ [(((\\ left (t-4 \\ right)) ^ (2)) \u003d 0 \\]
The resulting number means that at time 4 with $ v $ a material point moving according to the above described law will be exactly 3 m / s.
Key points
In conclusion, let us once again go over the most important point of today's task, namely, according to the rule of converting distance to speed and acceleration. So, if we directly describe the law in the problem that directly indicates the distance from the material point to the reference point, then through this formula we can find any instantaneous speed (this is just a derivative). Moreover, we can also find acceleration. Acceleration, in turn, is equal to the derivative of speed, i.e. second derivative of distance. Such tasks are quite rare, so today we did not disassemble them. But if you see the word “acceleration” in the condition, let it not scare you, just find another derivative.
I hope this lesson will help you prepare for the exam in mathematics.
