Topic: physical meaning of derivative. Physical meaning of the derivative

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Sometimes in problem B9 from the Unified State Examination in mathematics, instead of everyone’s favorite graphs of a function or derivative, simply the equation of the distance from a point to the origin is given. What to do in this case? How to find speed or acceleration from distance.

It's actually simple. Velocity is the derivative of distance, and acceleration is the derivative of velocity (or, equivalently, the second derivative of distance). In this short video you will see that such problems are solved no more difficult than the “classic” B9.

Today we will analyze two problems on the physical meaning of derivatives from the Unified State Examination in mathematics. These tasks are found in part B and are significantly different from those that most students are used to seeing on samples and exams. The thing is that they require understanding the physical meaning of the derivative of a function. In these problems we will talk about functions expressing distances.

If $S=x\left(t \right)$, 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 the derivative of distance and acceleration is the derivative of speed.

Let's see how this works in solving real problems.

Example #1

where $x$ is the distance from the reference point in meters, $t$ is the time in seconds that has passed since the beginning of the movement. Find the speed of the point (in m/s) at time $t=2c$.

This means that we have a function that specifies the distance, but we need to calculate the speed at time $t=2c$. In other words, we need to find $v$, i.e.

That's all we needed to figure out from the condition: firstly, what the function looks like, and secondly, what we are required to find.

Let's decide. First of all, let's calculate the derivative:

\[(x)"\left(t \right)=-\frac(1)(5)\cdot 5((t)^(4))+4((t)^(3))-3(( t)^(2))+5\]

\[(x)"\left(t \right)=-((t)^(4))+4((t)^(3))-3((t)^(2))+5\]

We need to find the derivative at point 2. Let's substitute:

\[=-16+32-12+5=9\]

\[(x)"\left(2 \right)=-((2)^(4))+4\cdot ((2)^(3))-3\cdot ((2)^(2)) +5=\] That's it, we have found the final answer. In total, our speed material point

at time $t=2c$ will be 9 m/s.

Example No. 2

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 at last time We were required to find $v$ at a time of 2 s, and this time we are required to find the very moment when this speed is equal to 3 m/s. We can say that we know the final value, and from this final value we need to find the initial one.

First of all, we look for the derivative again:

\[(x)"\left(t \right)=\frac(1)(3)\cdot 3((t)^(2))-4\cdot 2t+19\]

\[(x)"\left(t \right)=((t)^(2))-8t+19\]

We are asked to find at what point in time the speed will be 3 m/s. We compose and solve an equation to find the physical meaning of the derivative:

\[((t)^(2))-8t+19=3\]

\[((t)^(2))-8t+16=0\]

\[((\left(t-4 \right))^(2))=0\]

The resulting number means that at time 4 s $v$ of a material point moving according to the law described above will be exactly 3 m/s.

Key points

In conclusion, let's once again go over the most important point of today's task, namely, the rule for converting distance into speed and acceleration. So, if the problem directly describes to us a law that directly indicates the distance from a material point to a reference point, then through this formula we can find any instantaneous speed (this is just a derivative). And what's more, we can also find acceleration. Acceleration, in turn, is equal to the derivative of speed, i.e. second derivative of distance. Such problems are quite rare, so we didn’t look at them today. But if you see the word “acceleration” in the condition, don’t let it scare you, just find another derivative.

I hope this lesson will help you prepare for the Unified State Exam in mathematics.

IN coordinate plane xOy consider the graph of the function y=f(x). Let's fix the point M(x 0 ; f (x 0)). Let's add an abscissa x 0 increment Δх. We will get a new abscissa x 0 +Δx. This is the abscissa of the point N, and the ordinate will be equal f (x 0 +Δx). The change in the abscissa entailed a change in the ordinate. This change is called the increment of the function and is denoted Δy.

Δy=f (x 0 +Δx) - f (x 0). Through dots M And N let's draw a secant MN, which forms an angle φ with positive axis direction Oh. Let's determine the tangent of the angle φ from right triangle MPN.

Let Δх tends to zero. Then the secant MN will tend to take a tangent position MT, and the angle φ will become an angle α . So, the tangent of the angle α is the limiting value of the tangent of the angle φ :

The limit of the ratio of the increment of a function to the increment of the argument, when the latter tends to zero, is called the derivative of the function at a given point:

Geometric meaning derivative lies in the fact that the numerical derivative of the function at a given point is equal to the tangent of the angle formed by the tangent drawn through this point to the given curve and the positive direction of the axis Oh:

Examples.

1. Find the increment of the argument and the increment of the function y= x 2, if the initial value of the argument was equal to 4 , and new - 4,01 .

Solution.

New argument value x=x 0 +Δx. Let's substitute the data: 4.01=4+Δx, hence the increment of the argument Δх=4.01-4=0.01. The increment of a function, by definition, is equal to the difference between the new and previous values ​​of the function, i.e. Δy=f (x 0 +Δx) - f (x 0). Since we have a function y=x2, That Δу=(x 0 +Δx) 2 - (x 0) 2 =(x 0) 2 +2x 0 · Δx+(Δx) 2 - (x 0) 2 =2x 0 · Δx+(Δx) 2 =

2 · 4 · 0,01+(0,01) 2 =0,08+0,0001=0,0801.

Answer: argument increment Δх=0.01; function increment Δу=0,0801.

The function increment could be found differently: Δy=y (x 0 +Δx) -y (x 0)=y(4.01) -y(4)=4.01 2 -4 2 =16.0801-16=0.0801.

2. Find the angle of inclination of the tangent to the graph of the function y=f(x) at the point x 0, If f "(x 0) = 1.

Solution.

The value of the derivative at the point of tangency x 0 and is the value of the tangent of the tangent angle (the geometric meaning of the derivative). We have: f "(x 0) = tanα = 1 → α = 45°, because tg45°=1.

Answer: the tangent to the graph of this function forms an angle with the positive direction of the Ox axis equal to 45°.

3. Derive the formula for the derivative of the function y=x n.

Differentiation is the action of finding the derivative of a function.

When finding derivatives, use formulas that were derived based on the definition of a derivative, in the same way as we derived the formula for the derivative degree: (x n)" = nx n-1.

These are the formulas.

Table of derivatives It will be easier to memorize by pronouncing verbal formulations:

1. Derivative constant value equal to zero.

2. X prime is equal to one.

3. The constant factor can be taken out of the sign of the derivative.

4. The derivative of a degree is equal to the product of the exponent of this degree by a degree with the same base, but the exponent is one less.

5. The derivative of a root is equal to one divided by two equal roots.

6. The derivative of one divided by x is equal to minus one divided by x squared.

7. The derivative of the sine is equal to the cosine.

8. The derivative of the cosine is equal to minus sine.

9. The derivative of the tangent is equal to one divided by the square of the cosine.

10. The derivative of the cotangent is equal to minus one divided by the square of the sine.

We teach differentiation rules.

1. The derivative of an algebraic sum is equal to the algebraic sum of the derivatives of the terms.

2. The derivative of a product is equal to the product of the derivative of the first factor and the second plus the product of the first factor and the derivative of the second.

3. The derivative of “y” divided by “ve” is equal to a fraction in which the numerator is “y prime multiplied by “ve” minus “y multiplied by ve prime”, and the denominator is “ve squared”.

4. A special case of the formula 3.

The derivative of the function f (x) at the point x0 is the limit (if it exists) of the ratio of the increment of the function at the point x0 to the increment of the argument Δx, if the increment of the argument tends to zero and is denoted by f '(x0). The act of finding the derivative of a function is called differentiation.
The derivative of a function has the following physical meaning: the derivative of a function in given point- rate of change of the function at a given point.

Geometric meaning of derivative. The derivative at point x0 is equal to slope tangent to the graph of the function y=f(x) at this point.

Physical meaning of derivative. If a point moves along the x axis and its coordinate changes according to the law x(t), then the instantaneous speed of the point is:

The concept of differential, its properties. Rules of differentiation. Examples.

Definition. The differential of a function at some point x is the main, linear part of the increment of the function. The differential of the function y = f(x) is equal to the product of its derivative and the increment of the independent variable x (argument).

It is written like this:

or

Or


Differential properties
The differential has properties similar to those of the derivative:





TO basic rules of differentiation include:
1) placing a constant factor outside the sign of the derivative
2) derivative of a sum, derivative of a difference
3) derivative of the product of functions
4) derivative of the quotient of two functions (derivative of a fraction)

Examples.
Let us prove the formula: By definition of derivative we have:

An arbitrary factor can be taken beyond the sign of passage to the limit (this is known from the properties of the limit), therefore

For example: Find the derivative of a function
Solution: Let's use the rule of placing the multiplier outside the sign of the derivative :

Quite often it is necessary to first simplify the form of the differentiable function in order to use the table of derivatives and the rules for finding derivatives. The following examples clearly confirm this.

Differentiation formulas. Application of differential in approximate calculations. Examples.

Using a differential in approximate calculations allows you to use a differential to approximate the values ​​of a function.
Examples.
Using the differential, calculate approximately
To calculate given value let's apply the formula from theory
Let us introduce a function into consideration and represent the given value in the form
then let's calculate

Substituting everything into the formula, we finally get
Answer:

16. L'Hopital's rule for disclosing uncertainties of the form 0/0 Or ∞/∞. Examples.
Limit of the ratio of two infinitesimals or two infinitesimals large quantities is equal to the limit of the ratio of their derivatives.

1)

17. Increasing and decreasing function. Extremum of the function. Algorithm for studying a function for monotonicity and extremum. Examples.

Function increases on an interval if for any two points of this interval connected by the relation , the inequality is true. That is, higher value the argument corresponds to a larger value of the function, and its graph goes “from bottom to top”. The demonstration function increases over the interval

Likewise, the function decreases on an interval if for any two points of a given interval such that , the inequality is true. That is, a larger value of the argument corresponds to a smaller value of the function, and its graph goes “from top to bottom”. Ours decreases at intervals decreases at intervals .

Extremes A point is called the maximum point of the function y=f(x) if the inequality is true for all x in its vicinity. The value of the function at the maximum point is called maximum of the function and denote .
A point is called the minimum point of the function y=f(x) if the inequality is true for all x in its vicinity. The value of the function at the minimum point is called minimum function and denote .
The neighborhood of a point is understood as the interval , where is a sufficiently small positive number.
The minimum and maximum points are called extremum points, and the function values ​​corresponding to the extremum points are called extrema of the function.

To explore the function to monotony, use the following scheme:
- Find the domain of definition of the function;
- Find the derivative of the function and the domain of definition of the derivative;
- Find the zeros of the derivative, i.e. the value of the argument at which the derivative is equal to zero;
- On numerical rays mark common part the domain of definition of a function and the domain of definition of its derivative, and on it - the zeros of the derivative;
- Determine the signs of the derivative on each of the resulting intervals;
- Using the signs of the derivative, determine on which intervals the function increases and on which it decreases;
- Write the appropriate intervals separated by semicolons.

Research algorithm continuous function y = f(x) for monotonicity and extrema:
1) Find the derivative f ′(x).
2) Find stationary (f ′(x) = 0) and critical (f ′(x) does not exist) points of the function y = f(x).
3) Mark stationary and critical points on the number line and determine the signs of the derivative on the resulting intervals.
4) Draw conclusions about the monotonicity of the function and its extremum points.

18. Convexity of function. Inflection points. Algorithm for studying a function for convexity (concavity) Examples.

convex down on the X interval if its graph is located not lower than the tangent to it at any point of the X interval.

The function to be differentiated is called convex up on the X interval if its graph is located no higher than the tangent to it at any point in the X interval.

The point formula is called inflection point of the graph function y=f(x), if at a given point there is a tangent to the graph of the function (it can be parallel to the Oy axis) and there is such a neighborhood of the point formula, within which to the left and right of the point M the graph of the function has different directions of convexity.

Finding intervals for convexity:

If the function y=f(x) has a finite second derivative on the interval X and if the inequality holds (), then the graph of the function has a convexity directed downwards (upwards) at X.
This theorem allows you to find the intervals of concavity and convexity of a function; you only need to solve the inequalities and, respectively, on the domain of definition of the original function.

Example: Find out the intervals on which the graph of the function Find out the intervals on which the graph of the function has a convexity directed upward and a convexity directed downward. has a convexity directed upward and a convexity directed downward.
Solution: The domain of definition of this function is the entire set real numbers.
Let's find the second derivative.

The domain of definition of the second derivative coincides with the domain of definition of the original function, therefore, to find out the intervals of concavity and convexity, it is enough to solve and accordingly. Therefore, the function is convex downward on the interval formula and convex upward on the interval formula.

19) Asymptotes of the function. Examples.

The straight line is called vertical asymptote graph of the function if at least one of the limit values ​​is either equal to or .

Comment. A straight line cannot be a vertical asymptote if the function is continuous at the point. Therefore, vertical asymptotes should be sought at the discontinuity points of the function.

The straight line is called horizontal asymptote graph of the function if at least one of the limit values ​​or is equal to .

Comment. The graph of a function can have only a right horizontal asymptote or only a left one.

The straight line is called oblique asymptote function graph if

EXAMPLE:

Exercise. Find asymptotes of the graph of a function

Solution. Function definition scope:

a) vertical asymptotes: straight line - vertical asymptote, since

b) horizontal asymptotes: we find the limit of the function at infinity:

that is, there are no horizontal asymptotes.

c) oblique asymptotes:

Thus, the oblique asymptote is: .

Answer. The vertical asymptote is straight.

The oblique asymptote is straight.

20) General scheme for studying a function and plotting a graph. Example.

a.
Find the ODZ and discontinuity points of the function.

b. Find the points of intersection of the graph of the function with the coordinate axes.

2. Conduct a study of the function using the first derivative, that is, find the extremum points of the function and the intervals of increase and decrease.

3. Investigate the function using the second-order derivative, that is, find the inflection points of the function graph and the intervals of its convexity and concavity.

4. Find the asymptotes of the graph of the function: a) vertical, b) oblique.

5. Based on the research, construct a graph of the function.

Note that before plotting a graph, it is useful to determine whether a given function is odd or even.

Recall that a function is called even if changing the sign of the argument does not change the value of the function: f(-x) = f(x) and a function is called odd if f(-x) = -f(x).

In this case, it is enough to examine the function and plot its graph at positive values arguments belonging to ODZ. At negative values argument, the graph is completed on the basis that for even function it is symmetrical about the axis Oy, and for odd relative to the origin.

Examples. Explore functions and build their graphs.

Function Domain D(y)= (–∞; +∞). There are no breaking points.

Intersection with axis Ox: x = 0,y= 0.

The function is odd, therefore, it can only be studied on the interval )