Logarithmic differentiation

Derivatives of elementary functions

Basic rules of differentiation

Function differential

Principal linear part of the function increment A D x in determining the differentiability of a function

D f=f(x)- f(x 0)=A(x - x 0)+o(x – x 0), x®x 0

called the differential of the function f(x) at the point x 0 and is denoted

df(x 0)=f¢(x 0)D x=A D x.

The differential depends on the point x 0 and from increment D x. On D x at the same time they look at it as an independent variable, so at each point the differential is a linear function of the increment D x.

If we consider as a function f(x)=x, then we get dx= D x,dy=Adx. This is consistent with Leibniz's notation

Geometric interpretation of the differential as an increment of the ordinate of a tangent.

Rice. 4.3

1) f= const , f¢= 0,df= 0D x= 0.

2) f=u+v, f¢=u¢+v¢, df = du+dv.

3) f=uv, f¢=u¢v+v¢u, df = u dv + v du.

Consequence. (cf(x))¢=cf¢(x), (c 1 f 1 (x)+…+c n f n(x))¢=c 1 f¢ 1 (x)+…+ c n f¢ n(x)

4) f=u/v, v(x 0)¹0 and the derivative exists, then f¢=(u¢v-v¢ u)/v 2 .

For brevity we will denote u=u(x), u 0 =u(x 0), then

Passing to the limit at D x® 0 we obtain the required equality.

5) Derivative of a complex function.

Theorem. If there are f¢(x 0), g¢(x 0)and x 0 =g(t 0), then in some neighborhood t 0 complex function f is defined(g(t)), it is differentiable at point t 0 And

Proof.

f(x)- f(x 0)=f¢(x 0)(x-x 0)+ a( x)(x-x 0), xÎ U(x 0).

f(g(t))- f(g(t 0))= f¢(x 0)(g(t)- g(t 0))+ a( g(t))(g(t)- g(t 0)).

Let us divide both sides of this equality by ( t - t 0) and let's go to the limit at t®t 0 .

6) Calculation of the derivative of the inverse function.

Theorem. Let f be continuous and strictly monotone on[a,b]. Let at point x 0 Î( a,b)there is f¢(x 0)¹ 0 , then the inverse function x=f -1 (y)has at point y 0 derivative equal to

Proof. We count f strictly monotonically increasing, then f -1 (y) is continuous, increases monotonically by [ f(a),f(b)]. Let's put y 0 =f(x 0), y=f(x), x - x 0 =D x,

y - y 0 =D y. Due to the continuity of the inverse function D y®0 Þ D x®0, we have

Passing to the limit, we obtain the required equality.

7) The derivative of an even function is odd, the derivative of an odd function is even.

Indeed, if x® - x 0 , That - x® x 0 , That's why

For even function for odd function

1) f= const, f¢(x)=0.

2) f(x)=x,f¢(x)=1.

3) f(x)=e x, f¢(x)= e x ,

4) f(x)=a x ,(a x)¢ = ax ln a.

5) ln a.

6) f(x)=ln x,

Consequence. (the derivative of an even function is odd)

7) (x m )¢= m x m -1 , x>0, x m =e m ln x .

8) (sin x)¢= cos x,

9) (cos x)¢=- sin x,(cos x)¢= (sin( x+ p/2)) ¢= cos( x+ p/2)=-sin x.

10) (tg x)¢= 1/cos 2 x.

11) (ctg x)¢= -1/sin 2 x.

16) sh x, ch x.

f(x),, from which it follows that f¢(x)=f(x)(ln f(x))¢ .

The same formula can be obtained differently f(x)=e ln f(x) , f¢=e ln f(x) (ln f(x))¢.

Example. Calculate the derivative of a function f=x x .

=x x = x x = x x = x x(ln x+ 1).

Geometric location of points on a plane

we will call it a graph of a function, given parametrically. They also talk about parametric specification of a function.

Note 1. If x, y continuous for [a,b] And x(t) strictly monotonic on the segment (for example, strictly monotonically increases), then on [ a,b], a=x(a) , b=x(b) function defined f(x)=y(t(x)), where t(x) – function inverse to x(t). The graph of this function coincides with the graph of the function

If the domain of definition a parametrically given function can be divided into a finite number of segments ,k= 1,2,...,n, on each of which there is a function x(t) is strictly monotonic, then the parametrically defined function decomposes into a finite number of ordinary functions fk(x)=y(t -1 (x)) with domains [ x(a k), x(b k)] for increasing sections x(t) and with domains [ x(b k), x(a k)] for areas of decreasing function x(t). The functions obtained in this way are called single-valued branches of a parametrically defined function.

The figure shows a graph of a parametrically defined function

With the selected parameterization, the definition area is divided into five sections of strict monotonicity of the function sin(2 t), exactly: tÎ tÎ ,tÎ ,tÎ , and, accordingly, the graph will split into five unambiguous branches corresponding to these sections.

Rice. 4.4

Rice. 4.5

You can choose a different parameterization of the same geometric location of points

In this case there will be only four such branches. They will correspond to areas of strict monotony tÎ ,tÎ ,tÎ ,tÎ functions sin(2 t).

Rice. 4.6

Four sections of monotonicity of the function sin(2 t) on a long segment.

Rice. 4.7

The depiction of both graphs in one figure allows you to approximately depict the graph of a parametrically specified function, using the monotonicity areas of both functions.

As an example, consider the first branch corresponding to the segment tÎ . At the ends of this section the function x= sin(2 t) takes values ​​-1 and 1 , so this branch will be defined at [-1,1] . After this, you need to look at the areas of monotony of the second function y= cos( t), she has on two sections of monotony . This allows us to say that the first branch has two sections of monotonicity. Having found the end points of the graph, you can connect them with straight lines in order to indicate the nature of the monotony of the graph. Having done this with each branch, we obtain areas of monotonicity of unambiguous branches of the graph (they are highlighted in red in the figure)

Rice. 4.8

First single-valued branch f 1 (x)=y(t(x)) , corresponding to the site will be determined for xО[-1,1] . First single-valued branch tÎ , xО[-1,1].

All other three branches will also have a domain of definition [-1,1] .

Rice. 4.9

Second branch tÎ xО[-1,1].

Rice. 4.10

Third branch tÎ xО[-1,1]

Rice. 4.11

Fourth branch tÎ xО[-1,1]

Rice. 4.12

Comment 2. The same function can have different parametric settings. Differences may concern both the functions themselves x(t), y(t) , and the domain of definition these functions.

Example of different parametric assignments for the same function

And tО[-1, 1] .

Note 3. If x,y are continuous on , x(t)- strictly monotonic on the segment and there are derivatives y¢(t 0),x¢(t 0)¹0, then there is f¢(x 0)= .

Really, .

The last statement also applies to single-valued branches of a parametrically defined function.

4.2 Derivatives and differentials of higher orders

Higher derivatives and differentials. Differentiation of functions specified parametrically. Leibniz's formula.

Derivative of a function specified implicitly.
Derivative of a parametrically defined function

In this article we will look at two more typical tasks that are often found in tests in higher mathematics. In order to successfully master the material, you must be able to find derivatives at least at an intermediate level. You can learn to find derivatives practically from scratch in two basic lessons and Derivative of a complex function. If your differentiation skills are okay, then let's go.

Derivative of a function specified implicitly

Or, in short, the derivative of an implicit function. What is an implicit function? Let's first remember the very definition of a function of one variable:

Single variable function is a rule according to which each value of the independent variable corresponds to one and only one value of the function.

The variable is called independent variable or argument.
The variable is called dependent variable or function .

So far we have looked at functions defined in explicit form. What does it mean? Let's conduct a debriefing using specific examples.

Consider the function

We see that on the left we have a lone “player”, and on the right - only "X's". That is, the function explicitly expressed through the independent variable.

Let's look at another function:

This is where the variables are mixed up. Moreover impossible by any means express “Y” only through “X”. What are these methods? Transferring terms from part to part with a change of sign, moving them out of brackets, throwing factors according to the rule of proportion, etc. Rewrite the equality and try to express the “y” explicitly: . You can twist and turn the equation for hours, but you won’t succeed.

Let me introduce you: – example implicit function.

In the course of mathematical analysis it was proven that the implicit function exists(however, not always), it has a graph (just like a “normal” function). The implicit function is exactly the same exists first derivative, second derivative, etc. As they say, all rights of sexual minorities are respected.

And in this lesson we will learn how to find the derivative of a function specified implicitly. It's not that difficult! All differentiation rules and the table of derivatives of elementary functions remain in force. The difference is in one peculiar moment, which we will look at right now.

Yes, and I’ll tell you the good news - the tasks discussed below are performed according to a fairly strict and clear algorithm without a stone in front of three tracks.

Example 1

1) At the first stage, we attach strokes to both parts:

2) We use the rules of linearity of the derivative (the first two rules of the lesson How to find the derivative? Examples of solutions):

3) Direct differentiation.
How to differentiate is completely clear. What to do where there are “games” under the strokes?

- just to the point of disgrace, the derivative of a function is equal to its derivative: .

How to differentiate
Here we have complex function. Why? It seems that under the sine there is only one letter “Y”. But the fact is that there is only one letter “y” - IS ITSELF A FUNCTION(see definition at the beginning of the lesson). Thus, sine is an external function and is an internal function. We use the rule for differentiating a complex function :

We differentiate the product according to the usual rule :

Please note that – is also a complex function, any “game with bells and whistles” is a complex function:

The solution itself should look something like this:


If there are brackets, then expand them:

4) On the left side we collect the terms that contain a “Y” with a prime. Move everything else to the right side:

5) On the left side we take the derivative out of brackets:

6) And according to the rule of proportion, we drop these brackets into the denominator of the right side:

The derivative has been found. Ready.

It is interesting to note that any function can be rewritten implicitly. For example, the function can be rewritten like this: . And differentiate it using the algorithm just discussed. In fact, the phrases “implicit function” and “implicit function” differ in one semantic nuance. The phrase “implicitly specified function” is more general and correct, – this function is specified implicitly, but here you can express the “game” and present the function explicitly. The phrase “implicit function” refers to the “classical” implicit function when the “y” cannot be expressed.

Second solution

Attention! You can familiarize yourself with the second method only if you know how to confidently find partial derivatives. Calculus beginners and dummies, please don't read and skip this point, otherwise your head will be a complete mess.

Let's find the derivative of the implicit function using the second method.

We move all the terms to the left side:

And consider a function of two variables:

Then our derivative can be found using the formula
Let's find the partial derivatives:

Thus:

The second solution allows you to perform a check. But it is not advisable for them to write out the final version of the assignment, since partial derivatives are mastered later, and a student studying the topic “Derivative of a function of one variable” should not yet know partial derivatives.

Let's look at a few more examples.

Example 2

Find the derivative of a function given implicitly

Add strokes to both parts:

We use linearity rules:

Finding derivatives:

Opening all the brackets:

We move all the terms with to the left side, the rest to the right side:

Final answer:

Example 3

Find the derivative of a function given implicitly

Full solution and sample design at the end of the lesson.

It is not uncommon for fractions to arise after differentiation. In such cases, you need to get rid of fractions. Let's look at two more examples.

Example 4

Find the derivative of a function given implicitly

We enclose both parts under strokes and use the linearity rule:

Differentiate using the rule for differentiating a complex function and the rule of differentiation of quotients :

Expanding the brackets:

Now we need to get rid of the fraction. This can be done later, but it is more rational to do it right away. The denominator of the fraction contains . Multiply on . In detail, it will look like this:

Sometimes after differentiation 2-3 fractions appear. If we had another fraction, for example, then the operation would need to be repeated - multiply each term of each part on

On the left side we put it out of brackets:

Final answer:

Example 5

Find the derivative of a function given implicitly

This is an example for you to solve on your own. The only thing is that before you get rid of the fraction, you will first need to get rid of the three-story structure of the fraction itself. Full solution and answer at the end of the lesson.

Derivative of a parametrically defined function

Let’s not stress, everything in this paragraph is also quite simple. You can write down the general formula for a parametrically defined function, but to make it clear, I will immediately write down a specific example. In parametric form, the function is given by two equations: . Often equations are written not under curly brackets, but sequentially: , .

The variable is called a parameter and can take values ​​from “minus infinity” to “plus infinity”. Consider, for example, the value and substitute it into both equations: . Or in human terms: “if x is equal to four, then y is equal to one.” You can mark a point on the coordinate plane, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the parameter “te”. As for a “regular” function, for the American Indians of a parametrically defined function, all rights are also respected: you can build a graph, find derivatives, etc. By the way, if you need to plot a graph of a parametrically defined function, you can use my program.

In the simplest cases, it is possible to represent the function explicitly. Let us express the parameter from the first equation: – and substitute it into the second equation: . The result is an ordinary cubic function.

In more “severe” cases, this trick does not work. But it doesn’t matter, because there is a formula for finding the derivative of a parametric function:

We find the derivative of the “game with respect to the variable te”:

All differentiation rules and the table of derivatives are valid, naturally, for the letter , thus, there is no novelty in the process of finding derivatives. Just mentally replace all the “X’s” in the table with the letter “Te”.

We find the derivative of “x with respect to the variable te”:

Now all that remains is to substitute the found derivatives into our formula:

Ready. The derivative, like the function itself, also depends on the parameter.

As for the notation, instead of writing it in the formula, one could simply write it without a subscript, since this is a “regular” derivative “with respect to X”. But in the literature there is always an option, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

Thus:

A special feature of finding the derivative of a parametric function is the fact that at each step it is beneficial to simplify the result as much as possible. So, in the example considered, when I found it, I opened the parentheses under the root (although I might not have done this). There is a good chance that when substituting into the formula, many things will be reduced well. Although, of course, there are examples with clumsy answers.

Example 7

Find the derivative of a function specified parametrically

This is an example for you to solve on your own.

In the article The simplest typical problems with derivatives we looked at examples in which we needed to find the second derivative of a function. For a parametrically defined function, you can also find the second derivative, and it is found using the following formula: . It is quite obvious that in order to find the second derivative, you must first find the first derivative.

Example 8

Find the first and second derivatives of a function given parametrically

First, let's find the first derivative.
We use the formula

In this case:

We substitute the found derivatives into the formula. For simplification purposes, we use the trigonometric formula:

Let’s not stress, everything in this paragraph is also quite simple. You can write down the general formula for a parametrically defined function, but to make it clear, I will immediately write down a specific example. In parametric form, the function is given by two equations: . Often equations are written not under curly brackets, but sequentially: , .

The variable is called a parameter and can take values ​​from “minus infinity” to “plus infinity”. Consider, for example, the value and substitute it into both equations: . Or in human terms: “if x is equal to four, then y is equal to one.” You can mark a point on the coordinate plane, and this point will correspond to the value of the parameter. Similarly, you can find a point for any value of the parameter “te”. As for a “regular” function, for the American Indians of a parametrically defined function, all rights are also respected: you can build a graph, find derivatives, etc. By the way, if you need to plot a graph of a parametrically specified function, download my geometric program on the page Mathematical formulas and tables.

In the simplest cases, it is possible to represent the function explicitly. Let us express the parameter from the first equation: – and substitute it into the second equation: . The result is an ordinary cubic function.

In more “severe” cases, this trick does not work. But it doesn’t matter, because there is a formula for finding the derivative of a parametric function:

We find the derivative of the “game with respect to the variable te”:

All differentiation rules and the table of derivatives are valid, naturally, for the letter , thus, there is no novelty in the process of finding derivatives. Just mentally replace all the “X’s” in the table with the letter “Te”.

We find the derivative of “x with respect to the variable te”:

Now all that remains is to substitute the found derivatives into our formula:

Ready. The derivative, like the function itself, also depends on the parameter.

As for the notation, instead of writing it in the formula, one could simply write it without a subscript, since this is a “regular” derivative “with respect to X”. But in the literature there is always an option, so I will not deviate from the standard.

Example 6

We use the formula

In this case:

Thus:

A special feature of finding the derivative of a parametric function is the fact that at each step it is beneficial to simplify the result as much as possible. So, in the example considered, when I found it, I opened the parentheses under the root (although I might not have done this). There is a good chance that when substituting into the formula, many things will be reduced well. Although, of course, there are examples with clumsy answers.

Example 7

Find the derivative of a function specified parametrically

This is an example for you to solve on your own.

In the article The simplest typical problems with derivatives we looked at examples in which we needed to find the second derivative of a function. For a parametrically defined function, you can also find the second derivative, and it is found using the following formula: . It is quite obvious that in order to find the second derivative, you must first find the first derivative.

Example 8

Find the first and second derivatives of a function given parametrically

First, let's find the first derivative.
We use the formula

In this case:

Substitutes the found derivatives into the formula. For simplification purposes, we use the trigonometric formula:

I noticed that in the problem of finding the derivative of a parametric function, quite often for the purpose of simplification it is necessary to use trigonometric formulas . Remember them or keep them handy, and don't miss the opportunity to simplify each intermediate result and answers. For what? Now we have to take the derivative of , and this is clearly better than finding the derivative of .

Let's find the second derivative.
We use the formula: .

Let's look at our formula. The denominator has already been found in the previous step. It remains to find the numerator - the derivative of the first derivative with respect to the variable “te”:

It remains to use the formula:

To reinforce the material, I offer a couple more examples for you to solve on your own.

Example 9

Example 10

Find and for a function specified parametrically

I wish you success!

I hope this lesson was useful, and you can now easily find derivatives of functions specified implicitly and from parametric functions

Solutions and answers:

Example 3: Solution:






Thus:

Consider defining a line on a plane in which the variables x, y are functions of a third variable t (called a parameter):

For each value t from a certain interval certain values ​​correspond x And y, a, therefore, a certain point M (x, y) of the plane. When t runs through all values ​​from a given interval, then the point M (x, y) describes some line L. Equations (2.2) are called parametric line equations L.

If the function x = φ(t) has an inverse t = Ф(x), then substituting this expression into the equation y = g(t), we obtain y = g(Ф(x)), which specifies y as a function of x. In this case, we say that equations (2.2) define the function y parametrically.

Example 1. Let M(x,y)– arbitrary point on a circle of radius R and centered at the origin. Let t– angle between axis Ox and radius OM(see Fig. 2.3). Then x, y are expressed through t:

Equations (2.3) are parametric equations of a circle. Let us exclude the parameter t from equations (2.3). To do this, we square each equation and add it, we get: x 2 + y 2 = R 2 (cos 2 t + sin 2 t) or x 2 + y 2 = R 2 – the equation of a circle in the Cartesian coordinate system. It defines two functions: Each of these functions is given by parametric equations (2.3), but for the first function , and for the second .

Example 2. Parametric equations

define an ellipse with semi-axes a, b(Fig. 2.4). Excluding the parameter from the equations t, we obtain the canonical equation of the ellipse:

Example 3. A cycloid is a line described by a point lying on a circle if this circle rolls without sliding in a straight line (Fig. 2.5). Let us introduce the parametric equations of the cycloid. Let the radius of the rolling circle be a, dot M, describing the cycloid, at the beginning of the movement coincided with the origin of coordinates.

Let's determine the coordinates x, y points M after the circle has rotated through an angle t
(Fig. 2.5), t = ÐMCB. Arc length M.B. equal to the length of the segment O.B. since the circle rolls without slipping, therefore

OB = at, AB = MD = asint, CD = acost, x = OB – AB = at – asint = a(t – sint),

y = AM = CB – CD = a – acost = a(1 – cost).

So, the parametric equations of the cycloid are obtained:

When changing a parameter t from 0 to 2π the circle rotates one revolution, and the point M describes one arc of a cycloid. Equations (2.5) give y as a function of x. Although the function x = a(t – sint) has an inverse function, but it is not expressed in terms of elementary functions, so the function y = f(x) is not expressed through elementary functions.

Let us consider the differentiation of a function defined parametrically by equations (2.2). The function x = φ(t) on a certain interval of change t has an inverse function t = Ф(x), Then y = g(Ф(x)). Let x = φ(t), y = g(t) have derivatives, and x"t≠0. According to the rule of differentiation of complex functions y"x=y"t×t"x. Based on the rule for differentiating the inverse function, therefore:

The resulting formula (2.6) allows one to find the derivative for a function specified parametrically.

Example 4. Let the function y, depending on x, is specified parametrically:

Solution. .
Example 5. Find the slope k tangent to the cycloid at the point M 0 corresponding to the value of the parameter.
Solution. From the cycloid equations: y" t = asint, x" t = a(1 – cost), That's why

Tangent slope at a point M0 equal to the value at t 0 = π/4:

DIFFERENTIAL FUNCTION

Let the function at the point x 0 has a derivative. A-priory:
therefore, according to the properties of the limit (Section 1.8), where a– infinitesimal at Δx → 0. From here

Δy = f "(x0)Δx + α×Δx. (2.7)

As Δx → 0, the second term in equality (2.7) is an infinitesimal of higher order, compared to , therefore Δy and f " (x 0)×Δx are equivalent, infinitesimal (for f "(x 0) ≠ 0).

Thus, the increment of the function Δy consists of two terms, of which the first f "(x 0)×Δx is main part increment Δy, linear with respect to Δx (for f "(x 0)≠ 0).

Differential function f(x) at point x 0 is called the main part of the increment of the function and is denoted: dy or df(x0). Hence,

df (x0) =f "(x0)×Δx. (2.8)

Example 1. Find the differential of a function dy and the increment of the function Δy for the function y = x 2 at:
1) arbitrary x and Δ x; 2) x 0 = 20, Δx = 0.1.

Solution

1) Δy = (x + Δx) 2 – x 2 = x 2 + 2xΔx + (Δx) 2 – x 2 = 2xΔx + (Δx) 2, dy = 2xΔx.

2) If x 0 = 20, Δx = 0.1, then Δy = 40×0.1 + (0.1) 2 = 4.01; dy = 40×0.1= 4.

Let us write equality (2.7) in the form:

Δy = dy + a×Δx. (2.9)

Increment Δy is different from differential dy to an infinitesimal of higher order, compared to Δx, therefore, in approximate calculations, the approximate equality Δy ≈ dy is used if Δx is small enough.

Considering that Δy = f(x 0 + Δx) – f(x 0), we obtain an approximate formula:

f(x 0 + Δx) ≈ f(x 0) + dy. (2.10)

Example 2. Calculate approximately.

Solution. Consider:

Using formula (2.10), we obtain:

So, ≈ 2.025.

Let us consider the geometric meaning of the differential df(x 0)(Fig. 2.6).

Let us draw a tangent to the graph of the function y = f(x) at the point M 0 (x0, f(x 0)), let φ be the angle between the tangent KM0 and the Ox axis, then f"(x 0) = tanφ. From ΔM0NP:
PN = tgφ×Δx = f "(x 0)×Δx = df(x 0). But PN is the increment of the tangent ordinate as x changes from x 0 to x 0 + Δx.

Consequently, the differential of the function f(x) at the point x 0 is equal to the increment of the ordinate of the tangent.

Let's find the differential of the function
y = x. Since (x)" = 1, then dx = 1×Δx = Δx. We will assume that the differential of the independent variable x is equal to its increment, i.e. dx = Δx.

If x is an arbitrary number, then from equality (2.8) we obtain df(x) = f "(x)dx, whence .
Thus, the derivative for a function y = f(x) is equal to the ratio of its differential to the differential of the argument.

Let's consider the properties of the differential of a function.

If u(x), v(x) are differentiable functions, then the following formulas are valid:

To prove these formulas, derivative formulas for the sum, product and quotient of a function are used. Let us prove, for example, formula (2.12):

d(u×v) = (u×v)"Δx = (u×v" + u"×v)Δx = u×v"Δx + u"Δx×v = u×dv + v×du.

Let's consider the differential of a complex function: y = f(x), x = φ(t), i.e. y = f(φ(t)).

Then dy = y" t dt, but y" t = y" x ×x" t, so dy =y" x x" t dt. Considering,

that x" t = dx, we get dy = y" x dx =f "(x)dx.

Thus, the differential of a complex function y = f(x), where x =φ(t), has the form dy = f "(x)dx, the same as in the case when x is an independent variable. This property is called invariance of the form of the differential A.

Until now, we have considered equations of lines on a plane that directly connect the current coordinates of the points of these lines. However, another method of defining a line is often used, in which the current coordinates are considered as functions of a third variable.

Let two functions of a variable be given

considered for the same values ​​of t. Then any of these values ​​of t corresponds to a certain value and a certain value of y, and therefore to a certain point. When the variable t runs through all the values ​​from the domain of definition of functions (73), the point describes a certain line C in the plane. Equations (73) are called parametric equations of this line, and the variable is called a parameter.

Let us assume that the function has an inverse function. Substituting this function into the second of equations (73), we obtain the equation

expressing y as a function

Let us agree to say that this function is given parametrically by equations (73). The transition from these equations to equation (74) is called parameter elimination. When considering functions defined parametrically, excluding the parameter is not only not necessary, but also not always practically possible.

In many cases, it is much more convenient, given different values ​​of the parameter, to then calculate, using formulas (73), the corresponding values ​​of the argument and function y.

Let's look at examples.

Example 1. Let be an arbitrary point on a circle with a center at the origin and radius R. The Cartesian coordinates x and y of this point are expressed through its polar radius and polar angle, which we denote here by t, as follows (see Chapter I, § 3, paragraph 3):

Equations (75) are called parametric equations of a circle. The parameter in them is the polar angle, which varies from 0 to .

If equations (75) are squared term by term and added, then by virtue of the identity the parameter is eliminated and the equation of a circle in the Cartesian coordinate system is obtained, which defines two elementary functions:

Each of these functions is specified parametrically by equations (75), but the parameter ranges for these functions are different. For the first of them; The graph of this function is the upper semicircle. For the second function, its graph is the lower semicircle.

Example 2. Consider simultaneously an ellipse

and a circle with a center at the origin and radius a (Fig. 138).

To each point M of the ellipse we associate a point N of the circle, which has the same abscissa as the point M and is located with it on the same side of the Ox axis. The position of point N, and therefore point M, is completely determined by the polar angle t of the point. In this case, for their common abscissa we obtain the following expression: x = a. We find the ordinate at point M from the equation of the ellipse:

The sign was chosen because the ordinate of point M and the ordinate of point N must have the same signs.

Thus, the following parametric equations are obtained for the ellipse:

Here the parameter t varies from 0 to .

Example 3. Consider a circle with center at point a) and radius a, which obviously touches the x-axis at the origin (Fig. 139). Let's assume that this circle rolls without slipping along the x-axis. Then the point M of the circle, which at the initial moment coincided with the origin of coordinates, describes a line called a cycloid.

Let us derive the parametric equations of the cycloid, taking as the parameter t the angle MSV of rotation of the circle when moving its fixed point from position O to position M. Then for the coordinates and y of point M we obtain the following expressions:

Due to the fact that the circle rolls along the axis without slipping, the length of the segment OB is equal to the length of the arc BM. Since the length of the arc BM is equal to the product of the radius a and the central angle t, then . That's why . But Therefore,

These equations are the parametric equations of the cycloid. When the parameter t changes from 0 to the circle will make one full revolution. Point M will describe one arc of the cycloid.

Excluding the parameter t here leads to cumbersome expressions and is practically impractical.

Parametric definition of lines is especially often used in mechanics, and the role of the parameter is played by time.

Example 4. Let us determine the trajectory of a projectile fired from a gun with an initial speed at an angle a to the horizontal. We neglect air resistance and the dimensions of the projectile, considering it a material point.

Let's choose a coordinate system. Let us take the point of departure of the projectile from the muzzle as the origin of coordinates. Let's direct the Ox axis horizontally, and the Oy axis vertically, placing them in the same plane with the gun muzzle. If there were no force of gravity, then the projectile would move in a straight line, making an angle a with the Ox axis, and by time t it would have traveled the distance. The coordinates of the projectile at time t would be respectively equal to: . Due to gravity, the projectile must by this moment vertically descend by an amount. Therefore, in reality, at time t, the coordinates of the projectile are determined by the formulas:

These equations contain constant quantities. When t changes, the coordinates at the projectile trajectory point will also change. The equations are parametric equations of the projectile trajectory, in which the parameter is time

Expressing from the first equation and substituting it into

the second equation, we obtain the equation of the projectile trajectory in the form This is the equation of a parabola.