Tangent equation and normal equation to the graph of a function. Normal equation to the graph of a function Normal equation online

Derivative applications.

5.1.Geometric meaning of the derivative:

Consider the graph of the function y= f (x).

From Figure 1 it is clear that for any two points A And B graph of the function: , where α is the angle of inclination of the secant AB.

Thus, the difference ratio is equal to the slope of the secant. If you fix a point A and move the point towards it B, then decreases without limit and approaches 0, and the secant AB approaching a tangent AC.

Consequently, the limit of the difference ratio is equal to the slope of the tangent at point A, i.e. . This implies: The derivative of the function at point x 0 is equal to the slope of the tangent to the graph of the function y = f(x) at this point, i.e. .

1. The tangent to the graph of a function at the point (x 0; f (x 0)) is called the limiting position of the secant (AC).

Tangent equation : y – f(x 0) =

2. The straight line perpendicular to the tangent (AC) at the point (x 0; f(x 0) is called the normal to the graph of the function.

Normal equation: y – f(x 0) =

Task: Compose equations for the tangent and normal drawn to the graph of the function y = 10x-x at the point with the abscissa equal to x 0 = 2.

Solution:

1. Find the ordinate of the tangent point: f(x 0)= f(2)=10∙2–2 2 =16,

2. Find the angular coefficient of the tangent: f "(x)= (10x-x)" =10-2x, = f"(2)=10–2∙2=6

3. We compose the tangent equation: y–16 = 6∙ (x-2), y–16 = 6x–12, y–6x–4 = 0 – tangent equation,

4. We compose the normal equation: y –16 = , 6y –96 = –x+2, 6y+x–98=0 – normal equation.

5.2. Physical meaning of the derivative:

Definition. The speed of movement of a body is equal to the first derivative of the path with respect to time:

5.3. Mechanical meaning of derivative:

Definition. The acceleration of a body is equal to the first derivative of the speed with respect to time or the second derivative of the path with respect to time:

Task: Determine the speed and acceleration of a point moving according to the law at the moment t=4c.

Solution:

1. Find the speed law: v= S"=

2. Find the speed at the moment t = 4c: v(t)= v(4)=2∙4 2 +8∙4=64 units/sec

3. Find the law of acceleration: а=v′=

4. Find the acceleration at the moment t = 4c: A(t)= A( 4)=4∙4+8=24units/sec 2

SECTION 1.3. Differential of a function and its application in approximate calculations. Concept of differential function

Function differential y=ƒ(x) at the point x is called the main part of its increment, equal to the product of the derivative of the function and the increment of the argument, and is denoted dу (or dƒ(x)): dy=ƒ"(x)∙∆х(1).

The dу differential is also called first order differential. Let's find the differential of the independent variable x, i.e. the differential of the function y=x.

Since y"=x"=1, then, according to formula (1), we have dy=dx=∆x, i.e. the differential of the independent variable is equal to the increment of this variable: dx=∆x.

Therefore, formula (1) can be written as follows: dy=ƒ"(x) ∙ dх(2) in other words, The differential of a function is equal to the product of the derivative of this function and the differential of the independent variable.

From formula (2) follows the equality dy/dx=ƒ"(x).

Example1: Find the differential of the function ƒ(x)=3x 2 -sin(l+2x).

Solution: Using the formula dy=ƒ"(x) dx we find dy=(3x 2 -sin(l+2x))"dx=(6x-2cos(l+2x))dx.

Example2: Find the second order differential of the function: y = x 3 –7x.

Solution:

SECTION 1.4. Antiderivative. Indefinite integral. Methods for calculating the indefinite integral.

Definition1. The function F(x) is called antiderivative for the function f(x) on some interval, the differential of which is equal to the expression f(x)dx. Example: f(x) = 3x 2 3x 2 dx F(x) = x 3.

However, the differential of a function corresponds not to a single antiderivative, but to many of them. Let's look at the example: F 1 (x) = x 3, F 2 (x) = x 3 + 4, F 3 (x) = x 3 - 2, in general F (x) + C, where C is an arbitrary constant . This means that for the function f(x) = 3x 2 there are many antiderivatives that differ from each other by a constant term.

Definition2. The set of all antiderivative functions f(x) on a certain interval is called the indefinite integral of the functions f(x) on this interval and is denoted by the symbol ∫f(x)dx.

This symbol reads like this: “integral of f(x) over dx”, thus by definition:

∫ (x)dx = F(x)+C.

Symbol ∫ is called the sign of the integral, f(x) is the integrand, f(x)dx is the integrand, x is the variable of integration, F(x) is some antiderivative,

C - constant.

Basic properties of the indefinite integral:

1. The differential of an indefinite integral is equal to the integrand, i.e.

d ∫ f(x)dx = f(x)dx.

2. The indefinite integral of the differential of a function is equal to this function added to an arbitrary constant: ∫ d F(x) = F(x) + C

3. The constant factor can be taken out of the integral sign: ∫ kf(x)dx = k ∫ f(x)dx , k-const.

4. The indefinite integral of an algebraic sum of functions is equal to the sum of the integrals of each of them: ∫ (f 1 (x)+f 2 (x)-f 3 (x))dx = ∫ f 1 (x)dx + ∫ f 2 (x)dx – ∫f 3 (x)dx .

Consider a curve whose equation is

The equation of the tangent to a given curve at a point has the form:

The normal to a curve at a given point is a line passing through a given point and perpendicular to the tangent at this point.

The equation of the normal to a given curve at a point has the form:

(35)

The length of the tangent segment enclosed between the point of tangency and the x-axis is called tangent length, the projection of this segment onto the x-axis is called subscript .

The length of the normal segment enclosed between the point of tangency and the x-axis is called normal length,the projection of this segment onto the x-axis is called subnormal.

Example 17

Write the equations for the tangent and normal to the curve at the point whose abscissa is equal to.

Solution:

Let's find the value of the function at the point:

Let's find the derivative of the given function at the point

Answer: Tangent equation:

Normal equation: .

Example 18

Write equations for tangent and normal, length of tangent and subtangent, length of normal and subnormal for an ellipse

at the point for which.

Solution:

Let us find as the derivative of a function specified parametrically by formula (10):

Let's find the coordinates of the point of tangency: and the value of the derivative at the point of tangency:

We find the tangent equation using formula (34):

Let's find the coordinates of the point of intersection of the tangent with the axis:

The length of the tangent is equal to the length of the segment:

By definition, the subtangent is equal to

Where angle is the angle between the tangent and the axis. Therefore, - the angular coefficient of the tangent, equal to

So the subtangent is equal to

We find the normal equation using formula (35):

Let's find the coordinates of the point of intersection of the normal with the axis:

The length of the normal is equal to the length of the segment:

According to the definition, the subnormal is equal to

Where angle is the angle between the normal and the axis. Therefore, - the angular coefficient of the normal is equal to

Therefore, the subnormal is equal to:

Answer: Tangent equation:

Normal equation:

Tangent length ; subtangent;

Normal length ; subnormal

Tasks 7. Write the tangent and normal equations:

1. To a parabola at a point whose abscissa

2. To the circle at the points of intersection with the abscissa axis

3. To the cycloid at the point for which

4. At what points of the curve tangent is parallel:

a) Ox axis; b) straight

.

10. Intervals of monotonicity of a function. Extrema of a function.

Condition for a function to be monotonic:

In order for a function differentiable by not to increase, it is necessary and sufficient that at all points belonging to it the derivative is non-positive.

In order for a function differentiable on not to decrease, it is necessary and sufficient that at all points belonging to it the derivative is non-negative.

Intervals over which the derivative of a function retains a certain sign are called intervals monotony functions

Example 19

Find the intervals of monotonicity of the function.

Solution:

Let's find the derivative of the function .

Let us find the intervals of constant sign of the resulting derivative. For this

Let's factorize the resulting quadratic trinomial:

Let us examine the sign of the resulting expression using the interval method.

Thus, according to (36), (37), we obtain that the given function increases by and decreases by.

Answer: The given function increases and decreases by.

Definition The function has at the point local maximum (minimum), if there is a neighborhood of the point such that the condition is satisfied for all

The local minimum or maximum of a function is called local extremum.

A necessary condition for the existence of an extremum.

Let the function be defined in some neighborhood of a point. If a function has an extremum at a point, then the derivative at the point is either zero or does not exist.

The point is called critical point function if the derivative at a point is either zero or does not exist.

Sufficient conditions for the presence of an extremum at a critical point.

Let the point be critical.

The first sufficient condition for an extremum:

Let the function be continuous in some neighborhood of a point and differentiable at each point.

A point is a local maximum if, when passing through

the derivative of the function changes sign from plus to minus.

A point is a local minimum if, when passing through

the derivative of the function changes sign from minus to plus.

Example 20

Find the extrema of the function.

Solution:

Let's find the derivative of the given function

Equating the numerator and denominator in the resulting derivative to zero, we find the critical points:

Let's study the sign of the derivative using the interval method.

The figure shows that when passing through a point, the derivative changes sign from plus to minus. Therefore, at the point there is a local maximum.

When passing through a point, the derivative changes sign from minus to plus.

Therefore, at the point there is a local minimum.

When passing through a point, the derivative does not change sign. Consequently, the critical point is not an extremum of the given function.

Answer:- local maximum, - local minimum.

The second sufficient condition for an extremum:

If the first derivatives of a function at a point are equal to zero, and the th derivative of the function at a point is nonzero, then the point is an extremum of the function, and,

then is a local minimum

then is a local maximum.

Example 21

Find the extrema of the function using the second derivative.

Solution:

Let's find the first derivative of the given function

Let's find the critical points of the function:

We do not consider the point, since the function is defined only in the left neighborhood.

Let's find the second derivative

We find

Thus, based on (39), we conclude that at is a local maximum.

Answer:- local maximum.

Tasks 8.

Examine for increasing and decreasing functions:

Examine for extrema of the function:

A tangent is a straight line , which touches the graph of the function at one point and all points of which are at the shortest distance from the graph of the function. Therefore, the tangent passes tangent to the graph of the function at a certain angle, and several tangents at different angles cannot pass through the point of tangency. Tangent equations and normal equations to the graph of a function are constructed using the derivative.

The tangent equation is derived from the line equation .

Let us derive the equation of the tangent, and then the equation of the normal to the graph of the function.

y = kx + b .

In him k- angular coefficient.

From here we get the following entry:

y - y 0 = k(x - x 0 ) .

Derivative value f "(x 0 ) functions y = f(x) at the point x0 equal to the slope k= tg φ tangent to the graph of a function drawn through a point M0 (x 0 , y 0 ) , Where y0 = f(x 0 ) . This is geometric meaning of derivative .

Thus, we can replace k on f "(x 0 ) and get the following equation of the tangent to the graph of a function :

y - y 0 = f "(x 0 )(x - x 0 ) .

In problems involving composing the equation of a tangent to the graph of a function (and we will move on to them soon), it is required to reduce the equation obtained from the above formula to equation of a straight line in general form. To do this, you need to move all the letters and numbers to the left side of the equation, and leave zero on the right side.

Now about the normal equation. Normal - this is a straight line passing through the point of tangency to the graph of the function perpendicular to the tangent. Normal equation :

(x - x 0 ) + f "(x 0 )(y - y 0 ) = 0

To warm up, you are asked to solve the first example yourself, and then look at the solution. There is every reason to hope that this task will not be a “cold shower” for our readers.

Example 0. Create a tangent equation and a normal equation for the graph of a function at a point M (1, 1) .

Example 1. Write a tangent equation and a normal equation for the graph of a function , if the abscissa is tangent .

Let's find the derivative of the function:

Now we have everything that needs to be substituted into the entry given in the theoretical help to get the tangent equation. We get

In this example, we were lucky: the slope turned out to be zero, so there was no need to separately reduce the equation to its general form. Now we can create the normal equation:

In the figure below: the graph of the function is burgundy, the tangent is green, the normal is orange.

The next example is also not complicated: the function, as in the previous one, is also a polynomial, but the slope will not be equal to zero, so one more step will be added - bringing the equation to a general form.

Example 2.

Solution. Let's find the ordinate of the tangent point:

Let's find the derivative of the function:

.

Let's find the value of the derivative at the point of tangency, that is, the slope of the tangent:

We substitute all the obtained data into the “blank formula” and get the tangent equation:

We bring the equation to its general form (we collect all letters and numbers other than zero on the left side, and leave zero on the right):

We compose the normal equation:

Example 3. Write the equation of the tangent and the equation of the normal to the graph of the function if the abscissa is the point of tangency.

Solution. Let's find the ordinate of the tangent point:

Let's find the derivative of the function:

.

Let's find the value of the derivative at the point of tangency, that is, the slope of the tangent:

.

We find the tangent equation:

Before bringing the equation to its general form, you need to “comb it” a little: multiply term by term by 4. We do this and bring the equation to its general form:

We compose the normal equation:

Example 4. Write the equation of the tangent and the equation of the normal to the graph of the function if the abscissa is the point of tangency.

Solution. Let's find the ordinate of the tangent point:

.

Let's find the derivative of the function:

Let's find the value of the derivative at the point of tangency, that is, the slope of the tangent:

.

We get the tangent equation:

We bring the equation to its general form:

We compose the normal equation:

A common mistake when writing tangent and normal equations is not to notice that the function given in the example is complex and to calculate its derivative as the derivative of a simple function. The following examples are already from complex functions(the corresponding lesson will open in a new window).

Example 5. Write the equation of the tangent and the equation of the normal to the graph of the function if the abscissa is the point of tangency.

Solution. Let's find the ordinate of the tangent point:

Attention! This function is complex, since the tangent argument (2 x) is itself a function. Therefore, we find the derivative of a function as the derivative of a complex function.

Definition. Normal is a straight line perpendicular to the tangent and passing through the point of tangency.

If exists final And non-zero derivative f"(x 0) then the equation of the normal to the graph of the function y=f(x) at the point x 0 expressed by the following equation:

Example 1. Write the equation of the normal to the curve y=3x-x 2 at the point x 0 =2.

Solution.

1. Find the derivative y"=3-2x

x 0 =2: f"(x 0)=f"(2)=3-2*2=-1

3. Find the value of the function at the point x 0 =2: f(x 0)=f(2)=3*2-2 2 =2

4. Substitute the found values ​​into the normal equation:

5. We obtain the normal equation: y=x

Normal equation calculator

You can find the normal equation online using this calculator.

Example 2. (Consider the special case when f"(x 0) equals zero)

Write the equation of the normal to the curve y=cos24x at the point x 0 =π/2

Solution.

1. Find the derivative y"=2cos4x*(-sin4x*4)=-4sin2x

2. Find the value of the derivative at the point x 0 =π/2:

f"(x 0)=f"(π/2)=-4sin(2*π/2)=0 , therefore the normal equation cannot be applied in this case.

Let's use the definition of a normal, first find , then find the equation of a perpendicular line passing through a given point.

Definition: the normal to the curve y = ¦(x) at the point M 0 is a straight line passing through the point M 0 and perpendicular to the tangent at the point M 0 to this curve.

Let's write the equation of the tangent and normal, knowing the equation of the curve and the coordinates of the point M 0. The tangent has an angular coefficient k = t g = ¦, (x 0). From analytical geometry it is known that a straight line has the equation y-y 0 = k(x – x 0).

Therefore, the tangent equation is: y - y 0 = ¦, (x 0)(x – x 0); (1)

The angular coefficient of the normal is Kn = (since they are perpendicular), but then the equation of the normal is:

y-y 0 =(-1/ ¦, (x 0)(x – x 0); (2)

If a derivative does not exist at a point, then a tangent does not exist at this point.

For example, the function ¦(x)=|x| at the point x=0 has no derivative.

lim D x ®0 (D y/ D x)= lim D x ®0 (| D x|/ D x)=

One-sided limits exist, but lim D x ®0 (D y/ D x) does not exist

Tangent too.

This point is called the corner point of the graph.

§4. Relationship between continuity and differentiability of a function.

The following theorem about a differentiable function is valid.

Theorem: if a function y = ¦(x) has a finite derivative at the point x 0, then the function is continuous at this point.

Proof:

Because at the point x 0 there is a derivative ¦, (x 0), i.e. there is a limit

lim D x ®0 (D y/ D x)= ¦, (x 0), then D y/ D x= ¦, (x 0)+, where

B.m.v., depending on D x. When D x®0, ®0, because = (D y/ D x) - ¦, (x 0) ®0 at D x®0

Hence we have: D y= ¦, (x 0) D x + D x.

But then

An infinitesimal increment in the argument corresponds to an infinitesimal increment in the function, therefore ¦(x) is continuous at the point x 0 .

It is important to understand that the converse theorem is not true!

Not every continuous function is differentiable.

So, ¦(x) =|x| is continuous at the point x 0 =0, the graph is a solid line, but ¦, (0) does not exist.

§5. Derivatives of constant, sine, cosine and power functions.

1. y= ¦(x) =c; y, = (c) , = 0; (1)

Proof:

a) at any point x ¦(x) = c

b) give x the increment D x, x + D x, the value of the function ¦ (x + D x) = c;

c) ¦ (x + D x)- ¦ (x)= с- с= 0;

d) D y/ D x = 0/ D x = 0

e) lim D x ®0 (D y/ D x)= lim D x ®0 0 = 0

2. y= sin x; y, = (sin x), = cos x; (2)

Proof:

a) at any point x ¦(x) = sin x;

b) give x the increment of D x, x + D x, the value of the function