Geometric and mechanical meaning of the first derivative. Mechanical meaning of the derivative Physical or mechanical meaning of the second derivative

Instruction card No. 20

Takyryby/Subject: « The second derivative and its physical meaning».

Maksaty/ Purpose:

Be able to find the equation of the tangent, as well as the tangent of the angle of inclination of the tangent to the OX axis. Be able to find the rate of change of a function, as well as acceleration.

Create conditions for the formation of skills to compare and classify studied facts and concepts.

Fostering a responsible attitude towards educational work, will and perseverance to achieve final results in finding the tangent equation, as well as in finding the rate of change of a function and acceleration.

Theoretical material:

(Geometric meaning derived)

The tangent equation to the graph of a function is:

Example 1: Let's find the equation of the tangent to the graph of the function at the point with obscenity 2.

Answer: y = 4x-7

The angular coefficient k of the tangent to the graph of the function at the point with the abscissa x o is equal to f / (x o) (k= f / (x o)). The angle of inclination of the tangent to the graph of the function at a given point is equal to

arctg k = arctg f / (x o), i.e. k= f / (x o)= tg

Example 2: At what angle is the sine wave intersects the x-axis at the origin?

The angle at which the graph of a given function intersects the x-axis is equal to the slope a of the tangent drawn to the graph of the function f(x) at this point. Let's find the derivative: Taking into account the geometric meaning of the derivative, we have: and a = 60°. Answer: =60 0 .

If a function has a derivative at every point in its domain of definition, then its derivative is a function of . The function, in turn, can have a derivative, which is called second order derivative functions (or second derivative) and are designated by the symbol .

Example 3: Find the second derivative of the function: f(x)=x 3 -4x 2 +2x-7.

First, let’s find the first derivative of this function f"(x)=(x 3 -4x 2 +2x-7)’=3x 2 -8x+2,

Then, we find the second derivative of the obtained first derivative

f""x)=(3x 2 -8x+2)’’=6x-8. Answer: f""x) = 6x-8.

(Mechanical meaning of the second derivative)

If a point moves rectilinearly and the law of its motion is given, then the acceleration of the point is equal to the second derivative of the path with respect to time:

The speed of a material body is equal to the first derivative of the path, that is:

The acceleration of a material body is equal to the first derivative of the speed, that is:

Example 4: The body moves rectilinearly according to the law s (t) = 3 + 2t + t 2 (m). Determine its speed and acceleration at time t = 3 s. (Distance is measured in meters, time in seconds).
Solution
v (t) = s΄ (t) =(3+2t+t 2)’= 2 + 2t
a (t) = v΄ (t) =(2+2t)’= 2 (m/s 2)
v(3) = 2 + 2∙3 = 8 (m/s). Answer: 8 m/s; 2 m/s 2 .

Practical part:

1 option	Option 2	Option 3	Option 4	Option 5
Find the tangent of the angle of inclination to the x-axis of the tangent passing through the given point M graph of function f.
f(x)=x 2 , M(-3;9)	f(x)=x 3 , M(-1;-1)
Write the equation of the tangent to the graph of the function f at the point with the abscissa x 0.
f(x)=x 3 -1, x 0 =2	f(x)=x 2 +1, x 0 =1	f(x)= 2x-x 2, x 0 = -1	f(x)=3sinx, x 0 =	f(x)= x 0 = -1
Find the slope of the tangent to the function f at the point with the abscissa x 0.
Find the second derivative of the function:
			f(x)= 2cosx-x 2	f(x)= -2sinx+x 3
The body moves rectilinearly according to the law x (t). Determine its speed and acceleration at the moment time t. (Displacement is measured in meters, time in seconds).
x(t)=t 2 -3t, t=4	x(t)=t 3 +2t, t=1	x(t)=2t 3 -t 2 , t=3	x(t)=t 3 -2t 2 +1,t=2	x(t)=t 4 -0.5t 2 =2, t=0.5

Control questions:

What do you consider the physical meaning of the derivative - is it instantaneous speed or average speed?

What is the connection between a tangent drawn to the graph of a function through any point and the concept of derivative?

What is the definition of a tangent to the graph of a function at the point M(x 0 ;f(x 0))?

What is the mechanical meaning of the second derivative?

Derivative(functions at a point) - the basic concept of differential calculus, characterizing the rate of change of a function (at a given point). It is defined as the limit of the ratio of the increment of a function to the increment of its argument as the increment of the argument tends to zero, if such a limit exists. A function that has a finite derivative (at some point) is called differentiable (at that point).

Derivative. Let's consider some function y = f (x ) at two points x 0 and x 0 + : f (x 0) and f (x 0 + ). Here, through denotes some small change in the argument, called argument increment; accordingly, the difference between two function values: f (x 0 + )  f (x 0 ) is called function increment.Derivative functions y = f (x ) at point x 0 called limit:

If this limit exists, then the function f (x ) is called differentiable at the point x 0 . Derivative of a function f (x ) is denoted as follows:

Geometric meaning of derivative. Consider the graph of the function y = f (x ):

From Fig. 1 it is clear that for any two points A and B of the graph of the function:

where is the angle of inclination of the secant AB.

Thus, the difference ratio is equal to the slope of the secant. If you fix point A and move point B towards it, then it decreases without limit and approaches 0, and the secant AB approaches the tangent AC. Therefore, the limit of the difference ratio is equal to the slope of the tangent at point A. It follows: The derivative of a function at a point is the slope of the tangent to the graph of this function at that point. This is what geometric meaning derivative.

Tangent equation. Let us derive the equation of the tangent to the graph of the function at point A ( x 0 , f (x 0 )). In general, the equation of a straight line with slope coefficient f ’(x 0 ) has the form:

y = f ’(x 0 ) · x + b .

To find b, Let's take advantage of the fact that the tangent passes through point A:

f (x 0 ) = f ’(x 0 ) · x 0 +b ,

from here, b = f (x 0 ) – f ’(x 0 ) · x 0 , and substituting this expression instead b, we will get tangent equation:

y =f (x 0 ) + f ’(x 0 ) · ( x – x 0 ) .

Mechanical meaning of derivative. Let's consider the simplest case: the movement of a material point along the coordinate axis, and the law of motion is given: coordinate x moving point - known function x (t) time t. During the time interval from t 0 to t 0 + the point moves a distance: x (t 0 + )  x (t 0) = , and her average speed is equal to: v a =  . At 0, the average speed tends to a certain value, which is called instantaneous speed v ( t 0 ) material point at time t 0 . But by the definition of a derivative we have:

from here, v (t 0 ) = x' (t 0 ) , i.e. speed is the derivative of the coordinate By time. This is what mechanical sense derivative . Likewise, acceleration is the derivative of speed with respect to time: a = v' (t).

8. Table of derivatives and differentiation rules

We talked about what a derivative is in the article “The geometric meaning of a derivative.” If a function is given by a graph, its derivative at each point is equal to the tangent of the tangent to the graph of the function. And if the function is given by a formula, the table of derivatives and the rules of differentiation will help you, that is, the rules for finding the derivative.

Let a material point on the plane be given. The law of its movement along the coordinate axis is described by the law $ x(t) $, where $ t $ specifies the time. Then in the time from $ t_0 $ to $ t_0 + \Delta t $ the point passes the path $ \Delta x = x(t_0+\Delta t) - x(t_0) $. It turns out that average speed such a point is found by the formula: $$ v_(cp) = \frac(\Delta x)(\Delta t) $$

If $ \Delta t $ tends to zero, then the value of the average speed will tend to a value called instantaneous speed at point $t_0$:

$$ \lim_(\Delta t \to 0) \frac(\Delta x)(\Delta t) = v(t_0) $$

By defining the derivative through the limit, we obtain a connection between the speed and the law of motion of the path of a material point:

$$ v(t_0) = \lim_(\Delta \to 0) \frac(\Delta x)(\Delta t) = x"(t_0) $$

Examples of solutions

Example 1

Calculate the instantaneous speed of a material point at time $ t_0 = 1 $, moving according to the law $ x(t) = t^2+3t-1 $

Solution

By defining the mechanical meaning of the derivative, we obtain the law of velocity of a material point: $$ v(t) = x"(t) = (t^2+3t-1)" = 2t + 3 $$ Knowing the moment of time $ t_0 = 1 $ from the problem conditions, we find the speed at this moment of time: $$ v(t_0) = 2\cdot 1 + 3 = 2 + 3 = 5 $$ We found that the instantaneous speed of the point at the moment $ t_0 = 1 $ is equal to $ v = 5 $ If you cannot solve your problem, then send it to us. We will provide detailed solution. You will be able to view the progress of the calculation and gain information. This will help you get your grade from your teacher in a timely manner!

Answer

$$ v(t_0) = 5 $$

Example 2

The motion of a material point is given by the law $ x(t)=t^2-t+3 $. Find at what point in time $ t_0 $ the speed of this point will be zero.

Solution

Since speed is a derivative of the law of motion path:

Mechanical meaning of derivative

The mechanical interpretation of the derivative was first given by I. Newton. It is as follows: the speed of movement of a material point at a given moment in time is equal to the derivative of the path with respect to time, i.e. Thus, if the law of motion of a material point is given by an equation, then to find the instantaneous speed of the point at any specific moment in time, you need to find the derivative and substitute the corresponding value t into it.

Second order derivative and its mechanical meaning

We get (the equation from what was done in the textbook Lisichkin V.T. Soloveichik I.L. “mathematics” p. 240):

Thus, the acceleration of the rectilinear motion of a body at a given moment is equal to the second derivative of the path with respect to time, calculated for a given moment. This is the mechanical meaning of the second derivative.

Definition and geometric meaning of differential

Definition 4. The main part of the increment of a function, linear with respect to the increment of the function, linear with respect to the increment of the independent variable, is called differential function and is denoted by d, i.e. .

The differential of a function is geometrically represented by the increment of the ordinate of the tangent drawn at the point M (x; y) for given values ​​of x and?x.

Calculation differential - .

Application of differential in approximate calculations - , the approximate value of the function increment coincides with its differential.

Theorem 1.If the differentiable function increases (decreases) in a given interval, then the derivative of this function is not negative (not positive) in this interval.

Theorem 2.If the derivative function is positive (negative) in a certain interval, then the function in this interval monotonically increases (monotonically decreases).

Let us now formulate the rule for finding intervals of monotonicity of the function

1. Calculate the derivative of this function.

2. Find the points at which it is zero or does not exist. These points are called critical for function

3. Using the found points, the domain of definition of the function is divided into intervals, at each of which the derivative retains its sign. These intervals are intervals of monotonicity.

4. Examine the sign on each of the found intervals. If on the interval under consideration, then on this interval it increases; if, then it decreases on such an interval.

Depending on the conditions of the problem, the rule for finding monotonicity intervals can be simplified.

Definition 5. A point is called a maximum (minimum) point of a function if the inequality holds for any x in some neighborhood of the point.

If is the maximum (minimum) point of the function, then they say that (minimum) at the point. The maximum and minimum functions combine the name extremum functions, and the points of maximum and minimum are called extremum points (extreme points).

Theorem 3.(a necessary sign of an extremum). If is an extremum point of a function and the derivative exists at this point, then it is equal to zero: .

Theorem 4.(a sufficient sign of an extremum). If the derivative changes sign when x passes through a, then a is the extremum point of the function.

Key points in derivative research:

1. Find the derivative.

2. Find all critical points from the domain of definition of the function.

3. Set the signs of the derivative of the function when passing through the critical points and write down the extremum points.

4. Calculate the function values ​​at each extreme point.

Let the material point M moves in a straight line according to the law S = f(t). As is already known, the derivative S t ’ equal to the speed of the point at a given time: S t ’= V.

Let at a moment in time t the speed of the point is equal to V, and at the moment t +Dt – speed is V+DV, i.e. over a period of time Dt speed changed by amount D.V..

The ratio expresses the average acceleration of a point's movement over time Dt. The limit of this ratio at Dt®0 is called the acceleration of the point M At the moment t and is designated by the letter A: So, the second derivative of the path with respect to time is the magnitude of the acceleration of the rectilinear motion of the point, i.e. .

Higher order differentials

Let y=f(x) differentiable function, and its argument X– independent variable. Then its first differential is also a function X, you can find the differential of this function.

The differential of the differential of a function is called its second differential (or second-order differential) and is denoted by: .

The second-order differential of a given function is equal to the second-order product of this function by the square of the differential of the independent variable: .

Application of differential calculus

The function is called increasing (decreasing)) on the interval ( a; b), if for any two pointsx 1 Andx 2 from the specified interval satisfying the inequality, the inequality is satisfied ().

Necessary condition for increasing (decreasing): If the function to be differentiated on the interval ( a, b) increases (decreases), then the derivative of this function is non-negative (non-positive) in this interval() .

Sufficient condition for increasing (decreasing):If the derivative of a differentiable function is positive (negative) within a certain interval, then the function increases (decreases) over this interval.

Function f(x) at the point x 1 It has maximum, if for any X f(x 1)>f(x), at x ¹x 1 .

Function f(x) at the point x 1 It has minimum, if for any X from some neighborhood of the point the following inequality holds: f(x 1) , at x ¹x 1 .

The extremum of a function is called a local extremum, since the concept of extremum is associated only with a sufficiently small neighborhood of the point x 1. So on one interval a function can have several extrema, and it may happen that the minimum at one point is greater than the maximum at another. The presence of a maximum or minimum at a particular point in the interval does not mean that at this point the function f(x) takes the largest or smallest value on this interval.

Necessary condition for an extremum: At the extremum point of a differentiable function, its derivative is equal to zero.

Sufficient condition for an extremum: If the derivative of a differentiable function at some point x 0 is equal to zero and changes its sign when passing through this value, then the number f (x 0) is an extremum of the function, and if the sign changes from plus to minus, then the maximum if from minus to plus, then the minimum.

The points at which the derivative of a continuous function is equal to zero or does not exist are called critical.

To examine a function for an extremum means to find all its extrema. Rule for studying a function for an extremum:

1). Find critical points of a function y = f(x) and select from them only those that are internal points of the domain of definition of the function;

2). Investigate the sign of the derivative f"(x) to the left and right of each of the selected critical points;

3). Based on the sufficient condition for an extremum, write down the extremum points (if any) and calculate the values ​​of the function at them.

In order to find highest and lowest value function on a segment it is necessary to perform several stages:

1). Find the critical currents of the function by solving the equation f’(x)=0.

2). If the critical points fall on a segment, then it is necessary to find the values ​​at the critical points and at the boundaries of the interval. If the critical points do not fall on the segment (or they do not exist), then the function values ​​are found only at the boundaries of the segment.

3). From the obtained function values, select the largest and smallest and write the answer, for example, in the form: ; .

Problem solving

Example 2.1. Find the differential of the function: .

Solution. Based on Property 2 of the differential of a function and the definition of a differential, we have:

Example 2.2. Find the differential of the function:

Solution. The function can be written as: , . Then we have:

Example 2.3. Find the second derivative of the function:

Solution. Let's transform the function.

Let's find the first derivative:

let's find the second derivative:

.

Example 2.4. Find the second order differential of the function .

Solution. Let's find the second order differential based on the expression for calculating:

Let's first find the first derivative:

; let's find the second derivative: .

Example 2.5. Find the angular coefficient of the tangent to the curve drawn at the point with the abscissa x=2 .

Solution. Based on the geometric meaning of the derivative, we have that the slope is equal to the derivative of the function at the point whose abscissa is equal to X . We'll find .

Let's calculate the angular coefficient of the tangent to the graph of the function.

Example 2.6. Population of bacteria at a point in time t (t measured in hours) totals individuals. Find the growth rate of bacteria. Find the growth rate of bacteria at a given time t=5 hours.

Solution. The growth rate of a bacterial population is the first derivative with respect to time t: .

If t=5 hours, then . Therefore, the growth rate of bacteria will be 1000 individuals per hour.

Example 2.7. The body's reaction to the administered drug may be expressed in an increase in blood pressure, a decrease in body temperature, a change in heart rate or other physiological indicators. The degree of reaction depends on the prescribed dose of medication. If X indicates the dose of the prescribed medication and the degree of reaction at described by the function . At what value X Is the reaction maximum?

Solution. Let's find the derivative .

Let's find critical points: ⇒ . ⇒ Consequently, we have two critical points: . The value does not satisfy the task conditions.

Let's find the second derivative . Let's calculate the value of the second derivative at . . This means - the dose level that gives the maximum response.

Examples for self-solution

Find the differential of the function:

1. .

2. .

3. .

4.

Find the second derivatives of the following functions:

6. .

Find second-order derivatives and write second-order differentials for the following functions:

9. .

11. Investigate the function for extremum.

12. Find the largest and smallest values ​​of a function on the segment.

13. Find the intervals of increase and decrease of the function, maximum and minimum points and points of intersection with the axes:

14. The law of motion of a point has the form . Determine the law of speed and acceleration of this point.

15. The equation of motion of a point has the form (m). Find 1) the position of the point at times s and s; 2) the average speed for the time elapsed between these points in time; 3) instantaneous speeds at specified times; 4) average acceleration over a specified period of time; 5) instantaneous accelerations at specified times.

Homework assignment.

Practice:

Find the differential of the function:

1. ;

2. ;

Find second order derivatives of the function:

4.

5.

Find second order differentials

6. .

7. The point moves rectilinearly according to the law. Calculate the speed and acceleration at times and .

Find the intervals of increasing and decreasing functions:

9. .

10. When glucose is infused, its content in human blood, expressed in appropriate units, after t hours will be . Find the rate of change in blood glucose at a) t =1 h; b) t =2 h.

Theory.

1. Lecture on the topic “Derivatives and differentials of functions of several arguments. Application of the differential function of several arguments."

2. Lesson 3 of this manual.

3. Pavlushkov I.V. and others pp. 101-113, 118-121.

Lesson 3. Derivatives and differentials of a function of several arguments

Relevance of the topic: this section of mathematics is widely used in solving a number of applied problems, since many physical, biological, and chemical phenomena are characterized by dependence not on one, but on several variables (factors).

Purpose of the lesson: learn to find partial derivatives and differentials of functions of several variables.

Target tasks:

know: the concept of a function of two variables; the concept of partial derivatives of a function of two variables; the concept of complete and partial differentials of a function of several variables;

be able to: find derivatives and differentials of functions of several variables.

Brief information from the theoretical course

Basic Concepts

A variable z is called a function of two arguments x and y if some pairs of values ​​are assigned a certain value z according to some rule or law. A function of two arguments is denoted by .

The function is specified as a surface in a rectangular coordinate system in space. The graph of a function of two variables is a set of points in three-dimensional space x

The work is called partial differential function z=f(x,y)by X and are designated .

Full differential function

The differential of a function is the sum of the products of the partial derivatives of this function and the increment of the corresponding independent variables, i.e. . Because And then we can write: or .